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 |
|---|---|---|---|---|
185,549 | <blockquote>
<p>Suppose $F$ is a nondecreasing and right continuous function, and the sequence $\{x_n\}_{n\geq1}$ converges to $x$. Then $\liminf\limits_{n\to\infty}F(x_n)\leq F(x)$. </p>
</blockquote>
<p>How can I prove this?</p>
| Brian M. Scott | 12,042 | <p>HINT: Let $\sigma_R=\langle x_{n_k}:k\in\Bbb N\rangle$ be the subsequence of terms greater than or equal to $x$, and let $\sigma_L=\langle x_{m_k}:k\in\Bbb N\rangle$ be the subsequence of terms less than $x$.</p>
<ol>
<li><p>If either subsequence is finite, it can be ignored.</p></li>
<li><p>If $\sigma_R$ is infinite, $\lim\limits_{k\to\infty}F(x_{n_k})=F(x)$.</p></li>
<li><p>For each $k\in\Bbb N$, $F(x_{m_k})\le F(x)$.</p></li>
<li><p>If both subsequences are infinite, $\liminf_k x_k=\min\{\liminf\sigma_L,\liminf\sigma_R\}$.</p></li>
</ol>
|
2,502,255 | <p>We say that if $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ then $(f\circ g)(x)$ is continuous at $a$. </p>
<p>The converse: If $(f\circ g)(x)$ is continuous at $a$ then $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$ is not necessarily true.</p>
<p>For example consider $g(x) = x+1 -2 H(x)$ where $H(x)$ is the Heaviside step function. $g(x)$ has a jump discontinuity at $x=0$. In fact,</p>
<p>$g(x) = \left\{\begin{array}{l}x+1\qquad x<0\\x-1\qquad x\geq0 \end{array}\right.$</p>
<p>However, if we choose$f(x)=x^2$ then </p>
<p>$(f\circ g)(x) = \left\{\begin{array}{l}(x+1)^2\qquad x<0\\(x-1)^2\qquad x\geq0 \end{array}\right.$</p>
<p>which is continuous for all real $x$. </p>
<p>Its seems plausible that if $g(x)$ has a jump discontinuity at $x=a$ then we will always be able to find some non-constant $f(x)$ that makes it continuous through composition. Can anyone state a counter example for this or maybe have seen a proof? </p>
<p>Extensions of this would be to a finite collection of discontinuities and then countable collections. </p>
| eyeballfrog | 395,748 | <p>If $g$ has a jump discontinuity at $a$, then </p>
<p>$$f(x) = \left[x - \frac{g(a^+) + g(a^-)}{2}\right]^2$$</p>
<p>will have the property that $f(g(x))$ is continuous.</p>
|
3,855,736 | <p>Mathematics is not my primary discipline, but I know enough about both it and academics in general to know that many to most mathematical researchers do what they do because they enjoy doing it. This would seem to make "recreational mathematics" a rhetorical tautology, yet the term is used as if it were a subdiscipline in its own right. Universities offer courses on recreational mathematics. There are academic journals on recreational mathematics. There's a tag right here on SE Mathematics that reads <code>recreational-mathematics</code>. It's defined as "Mathematics done just for fun, often disjoint from typical school mathematics curriculum." Yet it does seem to be applied to a certain specific, albeit rather eclectic category of mathematical problems.</p>
<p>A lot of math that is called recreational falls quite neatly under some other category of math, often logic or combinatorics. In some cases, a kind of recreational math seems to be characterized as such only because some other formulation of the same ideas "got there first": it's relaxing to draw shapes with a compass and straightedge in much the same way that folding paper is relaxing, and the mathematics behind origami apprehend most if not all of the same mathematical truths that Euclidean geometry does, but mathematical origami is considered recreational while Euclidean geometry generally isn't.</p>
<p>With that in mind, what <em>is</em> recreational math? "Math that's done for fun" doesn't seem to make sense, because, again, that applies in some capacity to all math. Is it math that's done casually, with less concern for rigorous proof? Is it math that has not (yet) found an application in engineering or the empirical sciences? Is it the mathematical counterpart to popular science? What <em>is</em> it?</p>
<p>It's good math, but I don't understand this label attached to it. What makes a particular bit of math recreational?</p>
| Ross Millikan | 1,827 | <p>I don't think there is a definitive answer. Some characteristics I have noted that set recreational problems apart are</p>
<ul>
<li>Particular instead of general problems</li>
<li>Depend on the digit representation of a number</li>
<li>Involve finding a trick of logic to find the answer</li>
<li>Worded in a way to make finding the right approach difficult</li>
</ul>
|
1,550,923 | <p>I know of two applications of vector spaces over $\mathbb Q$ to problems posed by people not specifically interested in vector spaces over $\mathbb Q$:</p>
<ul>
<li>Hilbert's third problem; and</li>
<li>The Buckingham pi theorem.</li>
</ul>
<p>What others are there?</p>
| Brian M. Scott | 12,042 | <p>Assuming $\mathsf{CH}$, one can show that $\Bbb R$ is the union of countably many metrically rigid subsets by viewing it as a vector space over $\Bbb Q$. (A set $D$ in a metric space $\langle X,d\rangle$ is <em>metrically rigid</em> if no two distinct two-point subsets of $D$ are isometric.) This is mentioned in Brian M. Scott and Ralph Jones, <em>Metric rigidity in</em> $E^n$, Proceedings of the American Mathematical Society, Vol. $53$, $1975$, $219{-}222$, though the paper itself contains a different proof of a more general result.</p>
|
4,380,274 | <p>Indefinite integral is pretty easy to solve, I did it by substitution and I'm pretty sure it can be done relatively easy via integration by parts.
The problem are boundaries.</p>
<p>After substitution <span class="math-container">$arcsin x=t$</span> we get</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{(\sin t)^3}dt$$</span>
so we have</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{\cos t}{(\sin t)^2} dt+ \int_0^\frac{\pi}{2} \frac{t \cos t}{(\sin t)^3}dt$$</span></p>
<p>Now for example first integral is easy to compute and I get <span class="math-container">$\frac{-1}{\sin t}$</span> from 0 to <span class="math-container">$\frac{\pi}{2}$</span>.But in 0 the value is <span class="math-container">$\infty$</span>.
The second integral can also be solved using partial integration with <span class="math-container">$t$</span> and <span class="math-container">$\frac{cost}{(sint)^3}$</span>.</p>
<p>The only method to bypass this that I know of is linear substitution (I'm not sure how you call it) but that doesn't give anything useful.</p>
<p>Any hints?</p>
| user170231 | 170,231 | <p><span class="math-container">$$\begin{align*}
I &= \int_0^1 \frac{x - \arcsin(x)}{x^3} \, dx \\[1ex]
&= -\frac12\left(1 - \frac\pi2\right) + \frac12 \int_0^1 \left(1-\frac1{\sqrt{1-x^2}}\right) \, \frac{dx}{x^2} \tag{1} \\[1ex]
&= \frac\pi4-\frac12 + \frac12 \int_0^1 \frac{\sqrt{1-x^2}-1}{x^2 \sqrt{1-x^2}} \, dx \\[1ex]
&= \frac\pi4 - \frac12 + \frac12 \int_0^{\frac\pi2} (\cot(x)\csc(x) - \csc^2(x)) \, dx \tag{2} \\[1ex]
&= \boxed{\frac\pi4-1}
\end{align*}$$</span></p>
<hr />
<ul>
<li><span class="math-container">$(1)$</span> : integrate by parts</li>
<li><span class="math-container">$(2)$</span> : substitute <span class="math-container">$x\mapsto\sin(x)$</span></li>
</ul>
|
656,423 | <p>This is a really simple problem but I am unsure if I have proved it properly.</p>
<p>By contradiction:</p>
<p>Suppose that $x \geq 1$ and $x< \sqrt{x}$. Then $x\cdot x \geq x \cdot 1$ and $x^2 < x$ (squaring both sides), which is a contradiction.</p>
| Asinomás | 33,907 | <p>Here is a direct proof (without contradiction)</p>
<p>$x=1+r$ with $r\geq0$ . Then $x^2=(1+r)^2=1+2r+r^2\geq1+r=x\rightarrow x^2\geq x\rightarrow x\geq\sqrt x$</p>
|
1,784,634 | <p>Here is what I am tasked with..</p>
<p>Find a solution to the recurrence relation: </p>
<p>$F(0) = 2$</p>
<p>$F(n+1) = F(n) + 2n^2 - 1$</p>
<p>as a formula involving the summation operator</p>
<p>$$\sum_{i=1}^n$$</p>
<p>Sorry for the wonky formatting. Anyway, I am mostly looking for pointers/hints as to where or how to begin as I know it is heavily encouraged to try to work it out and not just ask for the answer. I have worked a little with finding closed-form solutions to factorial, fibonacci etc. But this problem states to write as a summation, not a closed form solution. Thanks!</p>
| Roby5 | 243,045 | <p><strong>Hint:</strong>(Use telescoping)</p>
<p>$F(n+1) -F(n) =(2n^2 -1)$</p>
<p>...</p>
<p>$F(1) -F(0)=(2\cdot 0^2 -1)$</p>
<p>Now add these up.</p>
<p>Can you take it from here?</p>
|
1,784,634 | <p>Here is what I am tasked with..</p>
<p>Find a solution to the recurrence relation: </p>
<p>$F(0) = 2$</p>
<p>$F(n+1) = F(n) + 2n^2 - 1$</p>
<p>as a formula involving the summation operator</p>
<p>$$\sum_{i=1}^n$$</p>
<p>Sorry for the wonky formatting. Anyway, I am mostly looking for pointers/hints as to where or how to begin as I know it is heavily encouraged to try to work it out and not just ask for the answer. I have worked a little with finding closed-form solutions to factorial, fibonacci etc. But this problem states to write as a summation, not a closed form solution. Thanks!</p>
| sinbadh | 277,566 | <p>Note that $F(n+1)-F(n)=2n^2-1$. Then $\sum_{n=0}^k 2n^2-1=\sum_{n=0}^kF(n+1)-F(n)$. But in the RHS:</p>
<p>$$\sum_{n=0}^kF(n+1)-F(n)=\sum_{n=0}^kF(n+1)-\sum_{n=0}^kF(n)\\
=F(k+1)+\sum_{n=0}^{k-1}F(n+1)-F(0)-\sum_{n=1}^kF(n)\\
=F(k+1)-F(0)+\sum_{n=0}^{k-1}F(n+1)-\sum_{n=0}^{k-1}F(n+1)\\
=F(k+1)-F(0)\\
=F(k+1)-2$$.</p>
<p>So, $F(k+1)=\sum_{n=0}^k(2n^2-1)+2$</p>
|
816,249 | <p>I just need some references which studies examples of skew adjoint differential operators generating unitary strongly continuous groups of operators, and its applications to partial differential equations.</p>
<p>The example I know is the differential operator defined on the hilbert space $H=L^2(\mathbb{R})$ by
$$Af=f'$$,
which has as domain $$D(A)=\{f \in L^2(\mathbb{R}), absolutely \ continuous, \ with \ f'\in L^2(\mathbb{R}) \}.$$</p>
<p>This operator generates a strongly continuous unitary group: $$(U(t)f)=f(t+s).$$
By unitary I mean $U(t)^{-1}=U(t)^*$. By a <a href="http://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups" rel="nofollow">Stone's Theorem</a>, this implies that $A$ must be skew adjoint.</p>
| Community | -1 | <p>You can take any self-adjoint operator and multiply it by $i$. Example: $i\Delta$ generates the <a href="http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Time-dependent_equation" rel="nofollow">Schrödinger equation</a> for a free particle (the potential $V$ is identically zero). The wave equation $u_{tt}=c^2u_{xx}$ can also be interpreted in this way, by considering it as evolution of $(u,cu_x)$ in phase space: the generating operator is $\begin{pmatrix} 0 & c\frac{d}{dx} \\ c\frac{d}{dx} & 0\end{pmatrix}$, which is skew-adjoint.</p>
<p>Reference: <a href="http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/" rel="nofollow">Mathematical Methods in Quantum Mechanics</a> by Gerald Teschl: very readable and free to download. Or pretty much any PDE book with "Hamiltonian" or "quantum mechanics" in it.</p>
|
4,146,858 | <blockquote>
<p>Q) For every twice differentiable function <span class="math-container">$f:\mathbb{R}\longrightarrow [-2,2] $</span> with <span class="math-container">$[f(0)]^2+[f'(0)]^2=85$</span> , which of the following statement(s) is(are) TRUE?</p>
</blockquote>
<blockquote>
<p>(A) There exists <span class="math-container">$r,s \in\mathbb{R}$</span> , where <span class="math-container">$r<s$</span> , such that f is one-one on the open interval <span class="math-container">$(r,s)$</span></p>
</blockquote>
<blockquote>
<p>(B) There exists <span class="math-container">$x_0 \in (-4,0)$</span> such that <span class="math-container">$|f'(x_0)|\leq 1$</span></p>
</blockquote>
<blockquote>
<p>(C) <span class="math-container">$lim_{x\to \infty}f(x)=1$</span></p>
</blockquote>
<blockquote>
<p>(D) There exists <span class="math-container">$\alpha \in (-4,4)$</span> such that <span class="math-container">$f(\alpha)+f"(\alpha)=0 and f'(\alpha)=0$</span></p>
</blockquote>
<p>I have problem with option B. Here's the given solution-
<a href="https://i.stack.imgur.com/DV1ND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DV1ND.jpg" alt="enter image description here" /></a></p>
<p>But I think this is wrong because <span class="math-container">$f(0)$</span> cannot be equal to <span class="math-container">$2$</span> as that gives us the value of <span class="math-container">$f'(0)=9$</span> (from the condition given in the question). If the slope is positive at <span class="math-container">$x=0$</span> and the function is achieving its highest value there then for the points in the right neighbourhood of <span class="math-container">$x=0$</span> the slope would have to abruptly change to <span class="math-container">$0$</span> otherwise the function would obtain values greater than <span class="math-container">$2$</span> which are not in its co-domain. But it cannot abruptly change either coz it's given to be a twice differentiable function.</p>
<p>And I took <span class="math-container">$f(x)=9$</span> only but of course similar argument can be made with the negative value and a similar argument can be made if they took <span class="math-container">$f(0)=-2$</span> and <span class="math-container">$f(-4)=2$</span> in the solution.</p>
<p>My question-</p>
<ol>
<li>Is the given solution wrong because of what I just said? Or did I go wrong somewhere? If not then can this be put into some more concrete words or if there's a theorem related to this?</li>
<li>Can an alternate solution be proposed for option B?</li>
</ol>
<p>I'm under confident about this because this is a JEE Advanced 2018 question and no objections were made against this that I'm aware of.</p>
| user21820 | 21,820 | <p>Personally, I have never used the quotient rule as I have always found it ugly and inconvenient. For example try computing the derivative of <span class="math-container">$\large\frac{(x+1)^3}{(2x-4)^5}$</span> with respect to <span class="math-container">$x$</span>. Since the quotient rule can be derived from the product rule and chain rule, it is indeed redundant if you have the other two as well as the derivative of <span class="math-container">$x^{-1}$</span> with respect to <span class="math-container">$x$</span>.</p>
<p>One might point out that we could just as well say that all the rules are redundant if you have the definition of derivative. However, there is something called the <a href="https://en.wikipedia.org/wiki/Formal_derivative" rel="nofollow noreferrer">formal derivative</a> that is defined algebraically and not via a limit-based definition (because it makes no sense in that context). So there is some benefit (beyond just for ease of computation) in finding purely algebraic rules for differentiation.</p>
|
1,585,772 | <p>I am finding this problem confusing :</p>
<blockquote>
<p>If,for all $x$,$f(x)=f(2a)^x$ and $f(x+2)=27f(x)$,then find $a$.</p>
</blockquote>
<p>When $x=1$ I have that $f(1)=f(2a)$ using the first identity.</p>
<p>Then when $x=2a$ I have by the second identity that $f(2a+2)=27f(2a)$,after that I simple stare at the problem without having a clue of how to proceed.</p>
<p>What's the trick the problem is calling for ?</p>
<p>I've thought of finding the inverse of the function $f(x)$ but It's not really clear to me how to apply this idea as I don't have linear functions .</p>
<p>Can you guys give me a hint ?</p>
| J.Gudal | 225,386 | <p>Note that:</p>
<p>$f(2a)=f(2a)^{2a} \Rightarrow 1=f(2a)^{2a-1}$</p>
<p>Hence either $f(2a)=1$ or $2a-1=0$.</p>
<p>But if $f(2a)=1$, then,</p>
<p>$f(x)=1^{x}$ for all $x$, but $f(2)=27$ and so this is false.</p>
<p>Consequently $a=\frac{1}{2}$.</p>
|
3,310,193 | <p>I want to show:
<span class="math-container">$$\mu(E)=0 \rightarrow \int_E f d\mu=0$$</span> </p>
<p><span class="math-container">$f:X \rightarrow [0, \infty] $</span> is measurable and <span class="math-container">$E \in \mathcal{A} $</span></p>
<p>Consider a step function <span class="math-container">$s=\sum_i a_i \chi_{A_i}$</span></p>
<p>Then I get: <span class="math-container">$\int_E s d \mu = \sum_i a_i \mu(A_i\cap E) $</span></p>
<p>Can I conclude that <span class="math-container">$ \mu(A_i \cap E) \leq \mu(E)=0 $</span>?</p>
| Aphelli | 556,825 | <p>Consider the smooth surjective exponential map, which is Lipschitz continuous <span class="math-container">$e: T_S S^n = \mathbb{R}^n \rightarrow S^n$</span> (<span class="math-container">$S$</span> being the south pole). </p>
<p>Consider a large <span class="math-container">$M>0$</span> and a small <span class="math-container">$r > 0$</span> and define <span class="math-container">$F(x_1, \ldots, x_{n+1})=r \cdot e(Mx_1, \ldots, Mx_n) \in S^n \subset \mathbb{R}^{n+1}$</span>. </p>
<p>If <span class="math-container">$M$</span> is large enough, then <span class="math-container">$F(B_1(0))=rS^n$</span> for any <span class="math-container">$r$</span>. If you choose then <span class="math-container">$r$</span> small enough, <span class="math-container">$F$</span> is a contraction, and the image of <span class="math-container">$B_1(0)$</span> is homeomorphic to <span class="math-container">$S^n$</span>, which is not contractible.</p>
|
327,291 | <p>I am struggling to find the English translation of Malcev's paper "On a class of homogenous spaces" providing foundational material for nil-manifolds. To be precise this paper: <i>Malcev, A. I. On a class of homogeneous spaces. Amer. Math. Soc. Translation 1951, (1951). no. 39, 33 pp.</i> (<a href="https://mathscinet.ams.org/mathscinet-getitem?mr=39734" rel="noreferrer">mathscinet link</a>) . It would be really important, for a project I am doing, to find this paper and I did not succeed neither on the website of the AMS nor by standard googling, which gives tons of papers referring to it. </p>
<p>Can anyone provide a reference to a place where to download the paper? I am at an institution with free access virtually everywhere, I just need a place with the actual paper in English (yeah in Russian I could find it). </p>
| R W | 8,588 | <p>This place is the interlibrary loan of your institution</p>
|
1,548,159 | <p>This is a question asked in India's CAT exam: <a href="http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html" rel="nofollow noreferrer">http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html</a> </p>
<blockquote>
<p>How many numbers with distinct digits are possible product of whose digits is 28?</p>
<p>A. 6</p>
<p>B. 4</p>
<p>C. 8</p>
<p>D. 12 </p>
</blockquote>
<p>Firstly, I couldn't even understand the question because the English seems grammatically incorrect. </p>
<p>Secondly, I couldn't understand how the answer was arrived at either.</p>
<pre><code>Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
</code></pre>
<p>If you can't have three digits, then how can four digits even be considered? And how on earth did they eventually reach 8 numbers? What does this even mean? </p>
<p>ps: I considered asking on <a href="https://puzzling.stackexchange.com/">puzzling.stackexchange</a>, but felt that it'd be more appropriate in a math forum.</p>
| Graham Kemp | 135,106 | <blockquote>
<blockquote>
<p>How many numbers with distinct digits are possible product of whose digits is 28?</p>
</blockquote>
</blockquote>
<p>Best interpretation of that is "How many numbers with distinct digits are possible, when the product of those digits must be 28?"</p>
<p>Well, the prime factorisation of $28$ is $2^2\cdot 7^1$. That is $28=2\cdot 14 = 4\cdot 7$. Also, of course, anything multiplied by $1$ is itself. (Well there's also $2\cdot 2\cdot 7$ but $[2,2,7]$ is <em>not</em> a list of <em>distinct</em> digits. )</p>
<p>So the only lists of <em>distinct</em> <em>digits</em> we can use are $[4, 7]$ and $[1,4,7]$. There are $2!$ permutations of the former and $3!$ permutations of the later. Thus we have counted $8$ possibilities.</p>
|
1,503,958 | <p>In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?</p>
<p>I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!</p>
| Piquito | 219,998 | <p>$$x+3-4\sqrt{x-1}=x-1+4-4\sqrt{x-1}=(x-1)-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$ Similarly $$x+8-6\sqrt{x-1}=(\sqrt{x-1}-3)^2$$ from which the answer.</p>
<p>It is not an equation but an identity in its domain of definition.This is the reason there are infinitely many solutions and not the expected finite number of them of a true equation with coefficients in a field. For example the "equation"</p>
<p>$$\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}=x$$ has all the reals as solutions.</p>
|
3,847,209 | <p>I'm currently trying to solve the question below.</p>
<h2>Abed is sitting in front of a large screen tv. He thinks he gets the best view when the screen takes up the maximum angle in his field of view.
What is the optimal point where he sits(to get the largest angle in his view)?</h2>
<p>I think it has something to do with a geometric inequality, but I'm stuck at the first step, I can't find any way to progress. I've drawn a few constructions, but I still can't find the optimal point.</p>
<p>EDIT:</p>
<p>I forgot to clarify. We treat the TV as a line, and Abed is looking at it. He cannot leave the 'floor' so he can only move from left to right, not up and down.</p>
| vvg | 820,543 | <p>If you consider the screen to be the base of a triangle and the person sitting at the apex of the triangle at a distance of <span class="math-container">$a$</span> measured from the center of the base, what describes the locus of the apex for angle <span class="math-container">$\alpha \in (0^{\circ}, +180^{\circ})$</span> measured relative to the base? <em><strong>Hint:</strong></em> a semi-circle.</p>
<p>So, you describe the equation for the angle <span class="math-container">$\theta$</span> subtended by a symmetric line segment of the diameter, centered on the circle's center by point on a semi circle and maximize it.</p>
<p><a href="https://i.stack.imgur.com/qHRoC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qHRoC.png" alt="Visual of the problem statement" /></a></p>
<p>As you vary the position of the viewing point on the circle, you get different angles of view. You also get different positions by varying the radius.</p>
<p>What radius of semi-circle maximizes the angle of view?</p>
<p>Theoretically, if the person places his eyes on the center of the screen, they get maximum angle of view <span class="math-container">$(180^{\circ})$</span>. But, it is not practical. So, a radius is chosen depending on the person's viewing ability (at what distance is the individual able to focus).</p>
<p>Now, we just need to determine the maximum angle for a given <span class="math-container">$r$</span>.</p>
|
1,179,195 | <p>Good day everyone. </p>
<p>I need to know automata theory. Can you advice me the best way to study math?
What themes will I need to know to understand automata theory. What a sequence of study? What level will I need to study intermediate themes? Maybe can you say something yet, what can help me quickly learn automata theory?</p>
| Pedro | 23,350 | <p>If $a'$ and $a$ are fixed points, then $d(f(a'),f(a))=d(a',a)<\lambda d(a',a)$ with $0<\lambda <1$, which implies that $d(a,a')=0$ and hence $a=a'$.</p>
|
2,065,639 | <p>$\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$</p>
<p>$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ </p>
<p>Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?</p>
| Alex Macedo | 400,433 | <p>If you would like a geometric argument (that can be better visualized for your first integral), note that </p>
<p>$\int_a^b (x - a)(x - b)\, dx = -\int_a^b\int_{x - b}^0\int_0^{x - a} 1 \, dzdydx$</p>
<p>Which is negative the volume of a pyramid with base area $(b - a)^2/2$ and height $b -a$. Therefore the value of your integral is $-(b - a)^3/6$.</p>
|
2,475,617 | <p>EDIT: I first designated $x$, $y$ as irrational numbers. I mean rational.</p>
<p>I have this, In the question it says: For every $x$, $y$ being rational,there exists $z$ being rational so that: $x<z$ or $z<y$ Now, I have this: $\forall(x,y) \in\Bbb Q^2, \exists z\in\Bbb Q/(x<z)∨(z<y)$ Does this the signify the same as $\forall(x,y)\in\Bbb Q^2 \Rightarrow\exists z\in\Bbb Q/(x<z)\lor(z<y)$</p>
| amWhy | 9,003 | <p>Your first version says "For all rational ordered pairs (x, y), there exists a rational number z such that $x\lt z \lor z\lt y$."</p>
<p>Here is another way of stating what you seem to want to express, in the form of an implication: </p>
<p>$$\forall x, \forall y \Big((x\in \mathbb Q \land y \in \mathbb Q)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y)\big)\Big)\tag 1$$</p>
<p>$$\forall (x, y) \Big(\big (x, y) \in \mathbb Q^2)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y))\big)\Big)\tag 2$$</p>
<p>You've done most of the work, but in, example, $(1)$ above, we have (in loglish): For all $x, y$, (if $x$ and $y$ are rational, then there exists a rational number $z$ such that $(x\lt z \lor z\lt y))$. </p>
<hr>
<p>But you pretty much said the same thing, in your first proposal. My versions show the explicit implication operator; but your expression is another way to say the same thing. In math, you'll often see statements written in your fashion. It a more strict logic class, we use set membership as a predicate.</p>
|
185,867 | <p>I hear that the axiom of choice (AC) derives from
The generalized continuum hypothesis(GCH).
And also hear that both AC and GCH are independent of
Zermelo–Fraenkel set theory(ZF).</p>
<p>So, I'm just curious why don't expert mathematicians use ZF+GCH
instead of ZF+AC(ZFC).</p>
| Timothy Chow | 3,106 | <p>There are two distinct questions that you might be asking.</p>
<blockquote>
<ol>
<li><p>Why has the mathematical community adopted ZFC as a standard foundation and not ZF+GCH?</p>
</li>
<li><p>What mathematical and philosophical arguments can be advanced for and against adopting AC and/or GCH as a fundamental axiom?</p>
</li>
</ol>
</blockquote>
<p>Much as we might like to believe that the reason for the sociological acceptance of ZFC is that there are strong rational arguments for it, and that mathematicians accept those strong rational arguments because mathematicians are strongly rational, I do not believe that this is true. I believe that ZFC has been adopted as the standard largely for historical reasons. At some point, the mathematical community recognized the value of having <i>some</i> standard, because it would demonstrate that all of mathematics could in principle be formalized in a single system that avoided all the known paradoxes, and ZFC just happened to show up at the right place at the right time. I suspect that various other candidates could have "landed the job," including ZF + V=L, or even Z + C, and that ZFC was just a bit lucky.</p>
<p>After a choice was made, most of the mathematical community lost interest in foundations and so had no real interest in tinkering with this or that axiom to get "better" foundations. I don't think that ZF+GCH was ever a serious contender, because GCH was still considered an open problem by the time ZFC had already secured its status as "the" foundation. If a statement X is considered an "open problem" then people generally do not also consider X to be a candidate for a basic axiom. By the time the independence of GCH was established, it was too late to apply for the job.</p>
<p>Having said all that, I should add that if by "expert mathematicians" you mean experts in set theory, then the question is a bit different, because set theorists are more interested in these sorts of questions than the mathematical community as a whole is. Some of them will tell you that they reject GCH as an axiom simply because they don't believe that GCH is <i>true</i>. Still, even among set theorists, a sizable proportion take what you might call a "pragmatic" approach. They care mostly about whether the standard base theory is a technically convenient one for the investigations that they are interested in. Then the considerations that Asaf Karagila mentions come into play. GCH just isn't the most natural or convenient axiom to use for most things that set theorists currently care about. If it does happen to be useful in a certain context then they won't hesitate to assume it, but such occasions don't come up that often. (By the way, often <a href="http://en.wikipedia.org/wiki/Martin%27s_axiom" rel="nofollow noreferrer">Martin's axiom</a> turns out to be what you really need when you might think you need CH.)</p>
|
55,965 | <p>I'm a games programmer with an interest in the following areas:</p>
<ul>
<li>Calculus</li>
<li>Matrices</li>
<li>Graph theory</li>
<li>Probability theory</li>
<li>Combinatorics</li>
<li>Statistics</li>
<li>More linguistic related fields of logic such as natural language processing, generative grammars</li>
</ul>
<p>Here are some examples of topics I've come across in the last 18 months in my design/development work, that have been of interest in solving certain problems. I grasp the outlines of these topics enough to know how they would help me to solve certain problems in my designs, but I don't even scratch the surface in understanding how to apply the math involved.</p>
<ul>
<li>Matrix math for spatial transformations</li>
<li>Minkowski sums for spatial expansion</li>
<li><a href="http://www.cs.brown.edu/~rt/gdhandbook/chapters/planarity.pdf" rel="nofollow">Planarity testing and embedding to convert logical non-planar to planar graphs</a></li>
<li>Generative grammars and natural language for narrative generation (linguistics / logic)</li>
</ul>
<p>My maths ability is sorely lacking. I know enough to get by for the relatively simple games I write. My logical and analytical skills are generally good, being a programmer. I enjoyed math in high school, but college was a different story -- my lecturer was terrible, and I didn't get any individual tutoring as I did before that. Anything that was in my head has long since departed. I would need to relearn what I learnt, which in mostly centred around "the calculus".</p>
<p>Bearing in mind that I need to balance my time between improving as a game designer, developer and mathematician/logician, what is the best way for me to tackle these gaping holes in my knowledge, enough to work in-depth mathematical descriptions into working algorithms?</p>
| Chris Eagle | 5,203 | <p>By Cauchy's theorem, $G$ contains an element of order 2. How does this act in the regular representation?</p>
|
3,019,506 | <p>I am stuck on this problem during my review for my stats test. </p>
<p>I know I have to use the convolution formula, and I understand that:</p>
<p><span class="math-container">$f_{U_1}(U_1) = 1$</span> for <span class="math-container">$0≤U_1≤1$</span> </p>
<p><span class="math-container">$f_{U_2}(U_2) = 1$</span> for <span class="math-container">$0≤U_2≤1$</span></p>
<p>but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks</p>
| Oscar Lanzi | 248,217 | <p>Take a cube and orient it so you're looking down a body diagonal. The edges that appear to be at the boundary of this view actually form a nonplanar object with six congruent edges, right angles wherever two adjacent edges meet, and <span class="math-container">$D_{3d}$</span> point group symmetry.</p>
|
1,998,391 | <p>To define a <strong>real</strong> exponential function
$$f(x)=a^x=e^{x \,\mathrm{lg} a}$$</p>
<p>It is strictly necessary that $a>0$.</p>
<p>But is the same true if the exponent is a <strong>natural</strong> number?</p>
<hr>
<p>In function series and moreover in power series I see things like
$$\sum_{n\geq0} f(n) x^n, \,\,\,\, x\in \mathbb{R}$$
So there is not the resctriction $x>0$.</p>
<p>It makes sense because natural number in the exponent do not create problems like even roots of negative numbers, so I think I'm ok with this.</p>
<p>Nevertheless if I rewrite $x^n$ using the definition of logaritm (for example this passage is useful while evaluating limits of such expressions), I get</p>
<p>$$\sum_{n\geq0} f(n) \,\, e^{n \, \mathrm{lg}x}$$</p>
<p>So suddenly I get a $\mathrm{lg}x$ and I need the condition $x>0$. </p>
<p>On the one hand this passage should be valid because it is just the definition of logaritm, but on the other hand it looks like I cannot perform it without meet a restriction on $x$ that was not present in the form $x^n$.</p>
<hr>
<p>So what is the point here? Is the use of the definition of logaritm not allowed in these cases, or is it wrong to avoid setting the condition $x>0$ in $x^n$?</p>
| qman | 366,057 | <p>The key is to realize that there are two distinct exponentiation functions, where the one becomes a motivating model for (but does not necessarily embed in) the other. These are unfortunately often conflated, and the identical notation does not help.</p>
<p>The first is the case of natural number exponents ($\Bbb N_+$ is the positive integers, and $X$ is any power-associative magma, which you can take as $\Bbb R$. If the identity element $1$ is defined, it can be extended to $n=0$, and if $x$ has a multiplicative inverse, to all $n$.):</p>
<p>$$X \times \Bbb{N}_+ \to X : (x,n) \mapsto \prod_{i=1}^n x$$</p>
<p>The second is the case of a base that is in $\Bbb{R}_+$, the positive reals, but the exponent can be from any real algebra $A$:</p>
<p>$$\Bbb{R}_+ \times A \to A : (x, y) \mapsto \exp(y \ln x)$$</p>
<p>In both cases, you can derive single-parameter functions by fixing one parameter, but this does not change the conclusion.
You will see that the added restriction arises when you try to use the second in place of the first, but the substitution is not valid where the domain of the second mapping does not contain that of the first.</p>
<p>So the answer to your first question is no, it need not be true of integer exponents, depending on which definition you choose to use for when the exponent is restricted to the natural numbers. The answer to your second question is that inc general, in expressions such as power series, the first case is usually the intended definition, not the second.</p>
|
2,262,661 | <p>The question is in how many ways can we select 20 different items from the empty set?</p>
<h3>ans:</h3>
<p>Obviously in 0 ways since the empty set has no items. I mean, this seem obvious, but maybe there is a trick to this question.</p>
| szw1710 | 130,298 | <p>One second ago there was a very interesting solution here: $$\binom{x}{k}=\frac{x(x-1)\cdot\ldots\cdot(x-k+1)}{k!}$$ for $x\in\Bbb R$ and $k\in\Bbb N$. The number of choices of $k$ distinct elements from $n$-element set is $\binom{n}{k}$ and $\binom{0}{20}=0$. I like this solution, but I am not the author. If it will return here, I will delete my answer.</p>
|
478,713 | <p>I have this logic statement:</p>
<pre><code> (A and x) or (B and y) or (not (A and B) and z)
</code></pre>
<p>The problem is that accessing A and B are rather expensive. Therefore I'd like to access them only once each. I can do this with an if-then-else construct:</p>
<pre><code>if A then
if x then
true
endif
else
if B then
if y then
true
endif
else
if z then
true
endif
endif
endif
</code></pre>
<p>Is there a way to express this as a boolean expression? I have "and", "or" (both short-circuit) but no "xor".</p>
<p>I thought this would work:</p>
<pre><code>X and (A or (Y and (B xor Z)))
</code></pre>
<p>But my test program (<a href="http://pastebin.com/EjURvpM4" rel="nofollow">http://pastebin.com/EjURvpM4</a>) shows it doesn't.</p>
| G Tony Jacobs | 92,129 | <p>The x-intercept is the value that you can plug in for x and have f(x) come out to 0. Try evaluating f(3), and see what happens.</p>
<p>Since f(3)=0, that means the point (3,0) is on the graph, so it is an x-intercept.</p>
<p>Seeing the minus sign inside the parentheses can be confusing, because it makes you think of a negative number, but remember to ask, "what value of x would make this equal 0?"</p>
|
511,150 | <p>I want to check if I understand proof by induction, so I want to proof the following:</p>
<p>$a^n<b^n$ for $a,b \in \mathbb{R}$, $0<a<b$, $n \in \mathbb{N}$ and $n>0$</p>
<p>Here's my attempt:</p>
<blockquote>
<p><strong>Base Case</strong></p>
<p>If $a=1$ and $b=2$, then $1^n < 2^n$ for any $n$</p>
<p><strong>Induction Step</strong></p>
<p>Now I need to show that if $a^n < b^n$ is true, then $a^{n+1} < b^{n+1}$ is true too.</p>
<p>So, I have:</p>
<p>$a<b$</p>
<p>and</p>
<p>$a^n < b^n$</p>
<p>So, assume $a^n < b^n$, then
$ a^n . a = a^{n+1} < b^n .a \tag{1}$
$a^n . b < b^n . b = b^{n+1} \tag{2}$</p>
</blockquote>
<p>I don't know how to bring (1) and (2) together.</p>
<p>Assuming my workings follows the steps for a proof by induction, how do I complete the proof? If this workings is not consistent with a proof by induction, what is the proper way to proof the assertion by a proof of induction?</p>
| Neal | 20,569 | <p>Since $a < b$, for any number $m>0$, $ma < mb$. In your case, $b^n\cdot a < b^n\cdot b$.</p>
<hr>
<p>I didn't read your question carefully enough. I assumed you were proving this:</p>
<blockquote>
<p>If $a,b\in\mathbb{R}$ and $a<b$, then for every $n\in\mathbb{N}$, $a^n<b^n$.</p>
</blockquote>
<p>But the first step in your proof is wrong. You're doing induction on $n$, not on $a$ and $b$, so you should start with $n=1$, not by setting values for $a$ and $b$.</p>
|
511,150 | <p>I want to check if I understand proof by induction, so I want to proof the following:</p>
<p>$a^n<b^n$ for $a,b \in \mathbb{R}$, $0<a<b$, $n \in \mathbb{N}$ and $n>0$</p>
<p>Here's my attempt:</p>
<blockquote>
<p><strong>Base Case</strong></p>
<p>If $a=1$ and $b=2$, then $1^n < 2^n$ for any $n$</p>
<p><strong>Induction Step</strong></p>
<p>Now I need to show that if $a^n < b^n$ is true, then $a^{n+1} < b^{n+1}$ is true too.</p>
<p>So, I have:</p>
<p>$a<b$</p>
<p>and</p>
<p>$a^n < b^n$</p>
<p>So, assume $a^n < b^n$, then
$ a^n . a = a^{n+1} < b^n .a \tag{1}$
$a^n . b < b^n . b = b^{n+1} \tag{2}$</p>
</blockquote>
<p>I don't know how to bring (1) and (2) together.</p>
<p>Assuming my workings follows the steps for a proof by induction, how do I complete the proof? If this workings is not consistent with a proof by induction, what is the proper way to proof the assertion by a proof of induction?</p>
| Siddhant Trivedi | 28,392 | <p>Now, suppose the result is true for $n=k$ then</p>
<p>$$a^k<b^k$$</p>
<p>we have to show that the result is true for $k+1$</p>
<p>$$a^k<b^k$$</p>
<p>$$a^k.a<\overbrace{b^k.a}^{3^k.2}$$</p>
<p>$$a^{k+1}<\underbrace{b^k.b}_{3^k.3}$$</p>
<p>because $a<b$ so,"a" is replaced by b on rhs</p>
<p>$$a^{k+1}<b^{k+1}$$</p>
<p>Therefore the result is true for $n=k+1$</p>
<p>$\textbf{For example}$:take $a=2$ and $b=3$ then see what happen </p>
|
1,528,501 | <p>Let $P(n)$ be a property for all $n \geq 1$. For the phrase "there is some $N \geq 1$ such that $P(n)$ holds for all $n \geq N$" there are some suggestive, convenient abbreviations such as "$P(n)$ holds for large $n$" or "$P(n)$ holds eventually" and so on.</p>
<p>I wonder if there is in literature a like abbreviation for "$P(n)$ holds for infinitely many $n$"? I am aware that in probability theory some authors would write "$P(n)$ holds infinitely often"; but, in my humble opinion, this abbreviation would be not that useful in contexts other than probability theory.</p>
| bof | 111,012 | <p>From <a href="https://en.wikipedia.org/wiki/John_L._Kelley" rel="nofollow">John L. Kelley</a>'s <em>General Topology</em> (available at the <a href="https://archive.org/details/GeneralTopology" rel="nofollow">Internet Archive</a>), p. 65:</p>
<blockquote>
<p>A <strong>directed set</strong> is a pair $(D,\ge)$ such that $\ge$ directs $D.$ [. . . .] A net $\{S_n,n\in D,\ge\}$ is <strong>in</strong> a set $A$ iff $S_n\in A$ for all $n$; it is <strong>eventually in</strong> $A$ iff there is an element $m$ of $D$ such that, if $n\in D$ and $n\ge m,$ then $S_n\in A.$ The net is <strong>frequently in</strong> $A$ iff for each $m$ in $D$ there is $n$ in $D$ such that $n\ge m$ and $S_n\in A.$</p>
</blockquote>
<p>Similarly, I guess you could say "$P(n)$ holds <strong>eventually</strong>" if $P(n)$ holds for all sufficiently large $n,$ and "$P(n)$ holds <strong>frequently</strong>" if $P(n)$ holds for infinitely many $n.$</p>
|
3,934,974 | <p>I'm wondering how to prove the associativity and identity to prove that the Möbius transformations forms a group.</p>
<p>A Möbius transformation is a complex function of the form <span class="math-container">$M(z)=\dfrac{az+b}{cz+d}$</span>.</p>
<p>Thank you in advance :)</p>
| Mo145 | 857,217 | <p>Since you are explicitly asking for the direct proof:</p>
<ol>
<li>concerning the associativity, this is just the same proof as for the associativity of any group of functions. So consider three moebius transforms <span class="math-container">$M_1,M_2,M_3$</span> then for any complex number <span class="math-container">$z$</span> we have <span class="math-container">$$M_1\circ( M_2\circ M_3)(z)=M_1((M_2\circ M_3)(z))=M_1(M_2(M_3(z)))=(M_1\circ M_2)(M_3(z))=(M_1\circ M_2)\circ M_3(z)$$</span> Note that we did not use any of the properties of the moebius transforms since the associativity just follows from the definition of the concatenation of functions.</li>
<li>For the identity consider the element <span class="math-container">$M(z)=id(z)=z$</span>, i.e. choose <span class="math-container">$a=d=1,b=c=0$</span> then clearly <span class="math-container">$ad-bc\neq 0$</span> so this is a moebius transform and you can easily verify that this is the identity for your group.</li>
<li>I guess you already figured out the formula for the inverse transformation since you were only asking for the other two properties, but still for completeness: it is easy to check that for any moebius transform <span class="math-container">$M(z)=\frac{az+b}{cz+d}$</span> we have an inverse given by <span class="math-container">$M^{-1}(z)=\frac{dz-b}{-cz+a}$</span> and this also is a moebius transform</li>
<li>We actually also have to show that the concatenation of two moebius transforms is again a moebius transform: but <span class="math-container">$$M_1\circ M_2(z)=M_1(\frac{a_2z+b_2}{c_2z+d_2})=\frac{a_1(\frac{a_2z+b_2}{c_2z+d_2})+b_2}{c_1(\frac{a_2z+b_2}{c_2z+d_2})+d_2}=\frac{(a_1a_2+b_1c_1)z+(a_1b_2+b_1d_2)}{(a_2c_1+c_2d_1)z+(b_2c_1+d_1d_2)}$$</span> Now denoting the coefficients of the concatenation by <span class="math-container">$a',b',c',d'$</span> we have <span class="math-container">$a'd'-c'b'\neq 0$</span> which can either be verified by direct calculation or the fact that <span class="math-container">$0\neq det(\left(\begin{array}{l}a_1&b_1\\c_1&d_1\end{array}\right)\left(\begin{array}{l}a_2&b_2\\c_2&d_2\end{array}\right))=\det(\left(\begin{array}{l}a'&b'\\c'&d'\end{array}\right))$</span>so this claim also holds.</li>
</ol>
<p>So the set of moebius transforms forms a group under concatenation.<br />
Note that it actually suffices to prove 3 and 4 as this shows that the moebius transforms form a subgroup of the group of all bijections of the extended complex numbers to themselves. But since you were explicitly asking for the first two steps i provided them too :)<br />
Lg Mo</p>
|
1,176,098 | <p>Here are some of my ideas:</p>
<p><strong>1. Addition Formula:</strong> <span class="math-container">$\sin{x}$</span> and <span class="math-container">$\cos{x}$</span> are the unique functions satisfying:</p>
<ul>
<li><p><span class="math-container">$\sin(x + y) = \sin x \cos y + \cos x \sin y $</span></p>
</li>
<li><p><span class="math-container">$\cos(x + y) = \cos x \cos y - \sin x \sin y$</span></p>
</li>
<li><p><span class="math-container">$\sin 0 = 0\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{\sin x }{x} = 1}$</span></p>
</li>
<li><p><span class="math-container">$\cos 0 = 1\quad$</span> and <span class="math-container">$\quad\displaystyle{\lim_{x \rightarrow 0} \frac{1-\cos x}{x} = 0}$</span></p>
</li>
</ul>
<p><strong>2. Taylor Series:</strong></p>
<ul>
<li><p><span class="math-container">$\displaystyle{\sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}\;x^{2n+1}}$</span></p>
</li>
<li><p><span class="math-container">$\displaystyle{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\;x^{2n}}$</span></p>
</li>
</ul>
<p><strong>3. Differential Equations:</strong> <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\cos(x)$</span> are the unique solutions to <span class="math-container">$y'' = -y$</span>, where <span class="math-container">$\sin(0) = \cos^\prime(0) = 0$</span> and <span class="math-container">$\sin^\prime(0) = \cos(0) = 1$</span>.</p>
<p><strong>4. Inverse Formula:</strong> We have:</p>
<p><span class="math-container">$$\begin{align}
\arcsin x &= \phantom{\frac{\pi}{2} + } \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt \\[6pt]
\arccos x &= \frac{\pi}{2} - \int_0^x \frac{1}{\sqrt{1 - t^2}}\, dt
\end{align}$$</span></p>
<p>Then <span class="math-container">$\sin x$</span> is the inverse of <span class="math-container">$\arcsin x$</span>, extended appropriately to the real line, and <span class="math-container">$\cos x$</span> is similar.</p>
<p><strong>Question:</strong> Are there any others that you like? In particular, are there any good rigorous ones coming from the original geometric definition?</p>
| md2perpe | 168,433 | <p>Here's one way:</p>
<ol>
<li>Define some basic operations like integration and analytical continuation.</li>
<li>Set <span class="math-container">$\ln x=\int_1^x \frac{dt}{t}.$</span></li>
<li>Set <span class="math-container">$\exp=\ln^{-1}$</span> (the inverse function).</li>
<li>Extend <span class="math-container">$\exp$</span> from <span class="math-container">$\mathbb R$</span> to <span class="math-container">$\mathbb C$</span> by analytical continuation.</li>
<li>Finally set
<span class="math-container">$$
\sin(z)=\frac{\exp(iz)-\exp(-iz)}{2i},\quad
\cos(z)=\frac{\exp(iz)+\exp(iz)}{2}.
$$</span></li>
</ol>
|
4,116,134 | <p>Find angle between <span class="math-container">$y=\sin x$</span> and <span class="math-container">$y=\cos x$</span> at their intersection point.</p>
<p>Intersection points are <span class="math-container">$\frac{\pi}{4}+\pi k$</span> and to find angle between them we need to compute derivatives at intersection points but then I can't combine them to get an answer which is <span class="math-container">$\arctan2\sqrt2$</span>. Will be thankful for your help.</p>
| Gary Moon | 477,460 | <p>If you think about these as parametrized curves <span class="math-container">$(t,\sin t)$</span> and <span class="math-container">$(t,\cos t)$</span>, then the angle between them at a point <span class="math-container">$t_0$</span> is the angle between <span class="math-container">$(1,\cos t_0)$</span> and <span class="math-container">$(1,-\sin t_0)$</span>. If <span class="math-container">$t_0 = \frac{\pi}{4} + \pi k$</span>, then we get
<span class="math-container">$$\cos\theta = \frac{(1,\pm\frac{1}{\sqrt{2}})\cdot(1,\mp\frac{1}{\sqrt{2}})}{\frac{3}{2}} = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$</span>
This, of course, yields an angle of <span class="math-container">$\arccos\frac{1}{3}$</span>, which is the same as <span class="math-container">$\arctan 2\sqrt{2}$</span>.</p>
|
4,116,134 | <p>Find angle between <span class="math-container">$y=\sin x$</span> and <span class="math-container">$y=\cos x$</span> at their intersection point.</p>
<p>Intersection points are <span class="math-container">$\frac{\pi}{4}+\pi k$</span> and to find angle between them we need to compute derivatives at intersection points but then I can't combine them to get an answer which is <span class="math-container">$\arctan2\sqrt2$</span>. Will be thankful for your help.</p>
| Bread Zeppelin | 919,679 | <p>f(x): y=sin(x)
g(x): y=cos(x)
let m1 and m2 be the slopes of f(x) and g(x)
m1 = d(sin(x))/dx = cos(x)
m2 = d(cos(x))/dx = -sin(x)</p>
<p>We know that if k is the angle between two lines with slope ma and mb</p>
<p>Tan(k)=|(ma-mb)/(1+ma*mb)|</p>
<p>At the point of intersection, the angle between the curves is the same as angle between tangents of the curves at the point of intersection (say (x0,y0)) and these tangents have slope f'(x0) and g'(x0).</p>
<pre><code> [f'(x)= d(f(x))/dx and g'(x)=d(g(x))/d(x)]
</code></pre>
<p>we know x0=pi/4
Therefore at x=pi/4 :</p>
<p>m1=cos(pi/4)=1/sqrt(2)</p>
<p>m2=-sin(pi/4)=-1/sqrt(2)</p>
<p>If @(I don't have theta on my keyboard, sorry) is the angle between f(x) and g(x) at x=pi/4</p>
<p>tan(@)=|(m1-m2)/(1-m1*m2)|</p>
<p>=|{(1/sqrt(2))-[-(1/sqrt(2))]}/{1-(1/sqrt(2)^2)}|</p>
<p>=|{2/sqrt(2)}/{1-(1/2)}|</p>
<p>=|sqrt(2)/(1/2)|</p>
<p>=|2*sqrt(2)|</p>
<p>=2*sqrt(2)</p>
<p>Therefore @=Tan^-1(2<em>sqrt(2)) or arctan(2</em>sqrt(2))</p>
|
1,461,254 | <p>I am about to take an undergraduate course in Mathematical logic any textbooks to recommend.Want it to be rigorous and not missing things. I am an Math undergraduate. Got still 2 years for my degree. I have taken courses mostly in Algebra (ring theory etc.). Also taken Real analysis courses. So my mathematical maturity is mediocre I would say. (I have also done calculus courses etc on my first year.)</p>
| Peter Smith | 35,151 | <p>Take a look at my <a href="http://www.logicmatters.net/resources/pdfs/TeachYourselfLogic2015.pdf">Teach Yourself Logic Study Guide</a>, which gives a lot of detailed advice about logic books at different levels, suitable to different backgrounds. </p>
<p>Check out the proposed syllabus of the course you are about to start to see what it covers and hence what's relevant in the Guide. </p>
<hr>
|
3,155,229 | <p>Stumbled across this weird phenomenon using the equation <span class="math-container">$y = \frac{1}{x} $</span>.</p>
<p><strong>Surface Area:</strong>
When you calculate the surface area under the curve from 1 to <span class="math-container">$\infty$</span></p>
<p><span class="math-container">$$\int_1^\infty \frac{1}{x}dx = \lim_{a \to \infty} \int_1^a \frac{1}{x}dx = \lim_{a \to \infty} \left[\ln\left|x\right|\right]^a_1 = \lim_{a \to \infty} (\ln\left|a\right|-ln\left|1\right|) = \infty$$</span></p>
<p><strong>Volume of revolution
:</strong>
When you calculate the volume of the revolution from 1 to <span class="math-container">$\infty$</span></p>
<p><span class="math-container">$$\pi\int_1^\infty \left(\frac{1}{x}\right)^2dx = \pi\lim_{a \to \infty} \int_1^a \frac{1}{x^2}dx = \pi\lim_{a \to \infty}\left[-\frac{1}{x}\right]^a_1 = \pi *(1-0) = \pi $$</span></p>
<p>How can it be that an object with an infinite surface area under his curve has a finite volume when you rotate it around the axis?</p>
<p>I get the math behind it and I'm assuming there is nothing wrong with the math. But it seems very contra-intuitive because if you rotate an infinite surface area just a little fraction it should have an infinite volume, that's what my intuition tells me?. So can someone explain to me why this isn't like that, that an infinite surface area rotated around the axis can have a finite volume?</p>
| Cye Waldman | 424,641 | <p>A simple way to visualize this is in terms of Pappus's <span class="math-container">$2^{nd}$</span> Centroid Theorem.</p>
<p>Pappus's <span class="math-container">$2^{nd}$</span> Centroid Theorem says the volume of a planar area of revolution is the product of the area <span class="math-container">$A$</span> and the length of the path traced by its centroid <span class="math-container">$R$</span>, i.e., <span class="math-container">$2πR$</span>. The bottom line is that the volume is given simply by <span class="math-container">$V=2πRA$</span>. The centroid of a volume is given by</p>
<p><span class="math-container">$$\mathbf{R}=\frac{\int_A \mathbf{r}dA}{\int_A dA}=\frac{1}{A} \int_A \mathbf{r}dA$$</span></p>
<p>Now you can see that the product <span class="math-container">$RA$</span> essentially eliminates any problems with the area and you are left with a <em>proper</em> intergral.</p>
|
4,277,924 | <p><strong>1. p ∧ ¬q = T</strong>
<br><br>
<strong>2. (q ∧ p) → r = T</strong>
<br><br>
<strong>3.¬p → ¬r = T</strong>
<br><br>
<strong>4.(¬q ∧ p) → r = T</strong>
<br><br>
From <strong>Eq 1</strong>, we got <strong>p = T</strong> and <strong>q = F</strong>
<br>
Now Apply value of <strong>P</strong> in <strong>Eq 3</strong>, we get:
<br>
<span class="math-container">$$\begin{array}{cc}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$</span></p>
<p><strong>Now there are two possibilities when ¬p→r is T, and ¬p is F but the r has two separate values.</strong></p>
<p><strong>Is this System consistent or inconsistent?</strong></p>
| James | 917,059 | <p>For a system to be consistent it must have <span class="math-container">$\textbf{one}$</span> outcome that is true, <span class="math-container">$T$</span>. For a system to be inconsistent it would no true values in the outcome, in other words is all false values, <span class="math-container">$F$</span>.</p>
<p>In your example we see we have <span class="math-container">$T$</span>'s in the right most column meaning this system <span class="math-container">$\textbf{is}$</span> consistent.</p>
<p><span class="math-container">$$\begin{array}{c|c}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$</span></p>
<p>For a system to be in consistent, here is an example:
<span class="math-container">$$\begin{array}{c|c} p & (p\land \lnot p) \\ \hline T & F \\ F &F
\end{array}$$</span>
The right most column is always <span class="math-container">$F$</span> so this system is inconsistent.</p>
<p>Note: If the right most column is always <span class="math-container">$T$</span>, that is called a tautology.</p>
<p>Also, for these problems I would do a truth table for each one. You won't have that many rows and it would be easy to see if the system is consistent or not.</p>
|
1,794,855 | <p>I need to prove that for every three integers $(a,b,c)$, the $\gcd(a-b,b-c) = \gcd(a-b,a-c)$. Assuming that a $a \ne b$.</p>
<p>Having:</p>
<p>$d_1 = \gcd(a-b,b-c)$</p>
<p>$d_2 = \gcd(a-b,a-c)$</p>
<p>How do i prove $d_1 = d_2$?</p>
| Michael Hardy | 11,667 | <p>We have this sum:
$$
\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1
$$
First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then
$$
\frac k {k!} = \frac 1 {(k-1)!}
$$
so that
$$
k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}.
$$
The two expressions inside the parentheses in the denominator now add up to $n-1$ rather $\text{than } n.$ So we can write it like this:
$$
\frac{n!}{(k-1)!(n-k)!} = n\cdot \frac{(n-1)!}{(k-1)!(n-k)!} = n\cdot \binom{n-1}{k-1}.
$$
Then the sum $(1)$ becomes
$$
\sum_{k=1}^n n\cdot \binom{n-1}{k-1} p^k (1-p)^{n-k}.
$$
Since $n$ does not change as $k$ goes from $1$ to $n$, we can pull $n$ out, getting
$$
n \sum_{k=1}^n \binom{n-1}{k-1} p^k (1-p)^{n-k}.
$$
Now let $j=k-1$ and observe that as $k$ goes from $1$ to $n$ then $j$ goes from $0$ to $n-1$, and $k = j+1$, so we have
$$
n \sum_{j=0}^{n-1} \binom{n-1} j p^{j+1} (1-p)^{(n-1)-j}.
$$
Since $p$ does not change as $j$ goes from $0$ to $n-1$, we can pull $p$ out, getting
$$
np \sum_{j=0}^{n-1} \binom{n-1} j p^j (1-p)^{(n-1)-j}.
$$
Now let $m= n-1$, so we have
$$
np \sum_{j=0}^m \binom m j p^j (1-p)^{m-j}.
$$
This sum is $1$ since it's the sum of probabilities assigned by the $\mathrm{Binomial}(m,p)$ distribution. Hence we get
$$
np\cdot 1.
$$</p>
|
124,060 | <p>I have a small dark object in the upper left corner of an <code>image</code>.</p>
<p>How can I separate it from the noisy rest and determine its <code>IntensityCentroid</code>?</p>
<p><a href="https://i.stack.imgur.com/D7pHv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D7pHv.png" alt="enter image description here" /></a></p>
<p>The problem is that the objects intensity is not significantly lower than the intensity in the lower part of the image:</p>
<pre><code>ListPlot3D[ImageData[image] // N]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/GoLaS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GoLaS.png" alt="enter image description here" /></a></p>
</blockquote>
<p><strong>UPDATE</strong>:</p>
<p>Following the question <a href="https://mathematica.stackexchange.com/questions/109658/finding-objects-in-images-with-inhomogeneous-background">Finding objects in images with inhomogeneous background</a> I tried:</p>
<pre><code>imageN = ColorNegate[image];
background = ImageConvolve[imageN, GaussianMatrix[7]];
subImage = ImageSubtract[imageN, background];
t = FindThreshold[subImage, Method -> "Entropy"];
binImg = DeleteSmallComponents[Binarize[subImage, t], 5];
pts = ComponentMeasurements[ImageMultiply[image, binImg],
"IntensityCentroid"]
Show[image, Graphics[{Red, Point[pts[[All, 2]]]}]]
</code></pre>
<p>which gives:</p>
<blockquote>
<pre><code>{1 -> {77.2279, 397.997}}
</code></pre>
<p><a href="https://i.stack.imgur.com/8RCob.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8RCob.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Unfortunately the other solutions for the mentioned question cannot detect the object here.</p>
<p>As Rahul mentioned also <code>background = MedianFilter[imagen, 10]</code> can be used to detect the object. But this is much slower:</p>
<pre><code>background = ImageConvolve[imageN, GaussianMatrix[7]]; // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{0.005, Null}
</code></pre>
</blockquote>
<pre><code>background = MedianFilter[imageN, 20]; // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{22.8, Null}
</code></pre>
</blockquote>
| Wjx | 6,084 | <p>There is a way to simplify all these process:</p>
<pre><code>img=--your image here--;
ComponentMeasurements[
DeleteSmallComponents@
ColorNegate@
LocalAdaptiveBinarize[img, 50, {1, -2, -.01}], {"Centroid",
"EquivalentDiskRadius"}]
</code></pre>
<blockquote>
<p>{1 -> {{77.0441, 397.853}, 4.65243}}</p>
</blockquote>
<p>The first is the Centroid of your region, and the second is equivalent disk radius.</p>
<p>The basic idea is to use <code>LocalAdaptiveBinarize</code>. Firstly, I checked the image's size and have a approximate point size, about 10 pixels, so we set the local adaptive range to <code>50</code>. Then, we want to eliminate the average, so the first argument in <code>{1,-2,-.01}</code> should be 1, representing the elimination of <code>1*average</code>. then we want to neglect those noises, and from your <code>ListPlot3D</code>, we can easily see that there's almost no noise over <code>2*sigma</code>, so a proper value for the second part is <code>2</code>, then the third represents a slight shift in mean, so <code>-0.01</code> should be proper.</p>
<p>Check the result, it's beautiful!</p>
<p>Then use <code>ComponentMeasurements</code> and everything will be fine.</p>
<hr />
<h1>Edit</h1>
<p>Note that you need to know the "IntensityCentroid", I made some slight modification to let it show you exactly the "IntensityCentroid" as in previous answer, Binarize will mop out any intensity information.</p>
<pre><code>TakeLargestBy[
ComponentMeasurements[
ImageMultiply[
Dilation[
DeleteSmallComponents@
ColorNegate@LocalAdaptiveBinarize[img, 50, {1, -2, -.01}],
DiskMatrix[10]],
ImageAdjust@
ImageSubtract[Blur[img, 50], img]], {"IntensityCentroid",
"EquivalentDiskRadius"}], #[[2, 2]] &, 1]
</code></pre>
<p>The basic idea is to use <code>Blur</code> and <code>ImageSubtract</code> to get out the background information then subtract it. Then we can use the method in the previous part to narrow down the selection. Finally determin the IntensityCentroid in this way.</p>
<blockquote>
<p>{1 -> {{76.4188, 397.894}, 10.66}}</p>
</blockquote>
<p>A sight difference, but this time the first part: <code>{76.4188, 397.894}</code> tells you the intensity centriod instead of morphological centroid. also, the radius is more accurate~~~ :)</p>
|
215,898 | <p>This is a quick follow up to my other <a href="https://mathematica.stackexchange.com/q/215884/44420">question</a> which I thought was different enough to warrant a separate post.</p>
<p><strong>My Question</strong></p>
<blockquote>
<p>How do you plot a region (such as <span class="math-container">$f(x,y)>z$</span>) on a 2D surface <span class="math-container">$g(x,y)=z$</span> and change the color of the region?</p>
</blockquote>
<p><strong>A Simple Example</strong></p>
<pre><code>f[x_, y_] := Sin[x] + Sin[y];
g[x_, y_] := y x Sin[x y];
img = RegionPlot[3/2 > f[x, y] > 1, {x, 0, 2}, {y, 0, 2},
Frame -> False, PlotRangePadding -> None, PlotStyle -> Orange,
BoundaryStyle -> Black]
Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> {Texture[img]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/yQF6w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yQF6w.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/W3V5K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W3V5K.png" alt="enter image description here"></a></p>
<p>So I have figured out how to overlay the region plot on the surface, but the only color the region appears is black. This is in spite that I specified the <code>RegionPlot</code> to be <code>Orange</code>. How can I change the region plot on the surface to anything but black.</p>
<p><strong>Notes</strong></p>
<ul>
<li>If you have any questions or need clarification please ask.</li>
<li>Again this is a quick follow up on my previous <a href="https://mathematica.stackexchange.com/q/215884/44420">question</a>.</li>
</ul>
| kglr | 125 | <h3>MeshFunctions + MeshShading</h3>
<pre><code>Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2},
MeshFunctions -> {f[#, #2] &},
Mesh -> {{1, 3/2}},
MeshShading -> {White, Orange},
Lighting -> "Neutral"]
</code></pre>
<p><a href="https://i.stack.imgur.com/6JtZI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6JtZI.png" alt="enter image description here" /></a></p>
<h3>BoundaryDiscretizeGraphics</h3>
<p>Get a <em>single</em> polygon from <code>img</code> using <a href="https://reference.wolfram.com/language/ref/BoundaryDiscretizeGraphics.html" rel="nofollow noreferrer"><code>BoundaryDiscretizeGraphics</code></a> and use it as the texture:</p>
<pre><code>img2 = BoundaryDiscretizeGraphics[img, MeshCellStyle -> {2 -> Orange, 1 -> Black},
ImagePadding -> 0, PlotRangePadding -> 0];
Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> Texture[img2],
Mesh -> None, Lighting -> "Neutral"]
</code></pre>
<p><a href="https://i.stack.imgur.com/QSy8p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QSy8p.png" alt="enter image description here" /></a></p>
<h3>Plot3D + RegionFunction</h3>
<pre><code>Show[Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> None, Mesh -> None],
Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> Orange,
BoundaryStyle -> Black, Mesh -> None, RegionFunction -> (1 < f[#, #2] < 3/2 &)]]
</code></pre>
<p><a href="https://i.stack.imgur.com/EvOHw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EvOHw.png" alt="enter image description here" /></a></p>
<h3>ImageMultiply</h3>
<p>Minimal change in OP's code that gives the desired result is to use <code>Texture[ImageMultiply[img, White]]</code> instead of <code>Texture[img]</code>:</p>
<pre><code>Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2},
PlotStyle -> Texture[ImageMultiply[img, White]],
Mesh -> None, Lighting -> "Neutral"]
</code></pre>
<p><a href="https://i.stack.imgur.com/jPzNk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jPzNk.png" alt="enter image description here" /></a></p>
<h3>Post-process RegionPlot output into 3D polygons:</h3>
<p>You can use <code>PlotStyle -> None</code> in <code>Plot3D</code> and <em>lift</em> the 2D region plot surface to 3D replacing coordinate <code>{x,y}</code> with <code>{x,y, g[x,y]}</code>:</p>
<pre><code>Show[Plot3D[g[x, y], {x, 0, 2}, {y, 0, 2}, PlotStyle -> None, Mesh -> None],
Graphics3D[Cases[img, GraphicsComplex[a_, b__] :>
GraphicsComplex[{##, g[##]} & @@@ a, b]]], Lighting -> "Neutral"]
</code></pre>
<p><a href="https://i.stack.imgur.com/UznVt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UznVt.png" alt="enter image description here" /></a></p>
|
267,051 | <p>Games appear in pure mathematics, for example, <a href="https://en.wikipedia.org/wiki/Ehrenfeucht%E2%80%93Fra%C3%AFss%C3%A9_game" rel="noreferrer">Ehrenfeucht–Fraïssé game</a> (in mathematical logic) and <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_game" rel="noreferrer">Banach–Mazur game</a> (in topology).</p>
<p>But the Game Theory behind those applications is not so deep, and we don't need to know some fundamental theorems for them. Maybe except Zarmelo Theorem.</p>
<p>Are there applications of <strong>advanced</strong> (anything behind the basic definitions) game theory ideas in pure mathematics?</p>
<p>Thanks!</p>
| Alec Rhea | 92,164 | <p>I would nominate the following paper of Victoria Gitman and Joel David Hamkins for this purpose:</p>
<p><a href="https://arxiv.org/abs/1509.01099" rel="nofollow noreferrer">https://arxiv.org/abs/1509.01099</a>.</p>
<blockquote>
<p>The principle of open determinacy for class games---two-player games of perfect information with plays of length $\omega$, where the moves are chosen from a possibly proper class, such as games on the ordinals---is not provable in Zermelo-Fraenkel set theory ZFC or G\"odel-Bernays set theory GBC, if these theories are consistent, because provably in ZFC there is a definable open proper class game with no definable winning strategy. In fact, the principle of open determinacy and even merely clopen determinacy for class games implies Con(ZFC) and iterated instances Con(Con(ZFC)) and more, because it implies that there is a satisfaction class for first-order truth, and indeed a transfinite tower of truth predicates $\text{Tr}_\alpha$ for iterated truth-about-truth, relative to any class parameter. This is perhaps explained, in light of the Tarskian recursive definition of truth, by the more general fact that the principle of clopen determinacy is exactly equivalent over GBC to the principle of transfinite recursion over well-founded class relations. Meanwhile, the principle of open determinacy for class games is provable in the stronger theory GBC+$\Pi^1_1$-comprehension, a proper fragment of Kelley-Morse set theory KM.</p>
</blockquote>
|
361,171 | <p>If I am about to fabricate a bracelet and I can select $24$ pearls out of a total of $500$ pearls ($300$ white, $150$ red and $50$ green) how many possible bracelets can I create?</p>
<p>The (official) solution to this question is
$$ \frac{500!}{476!} = 3.4\cdot10^{64} $$
But how can it be that the colors of the pearls were totally neglected? I mean if there would just be red pearls then this solution would be clear to me. But since there are different colors I am confused...</p>
| Douglas S. Stones | 139 | <p>Mathematically, <a href="http://en.wikipedia.org/wiki/Necklace_%28combinatorics%29" rel="nofollow">bracelets</a> are defined inequivalent under rotations and reflections (i.e., the <a href="http://en.wikipedia.org/wiki/Dihedral_group" rel="nofollow">dihedral group</a> action). Moreover, e.g. one red bead should be indistinguishable from another red bead, otherwise we should just call them "labelled beads".</p>
<p>Under this interpretation, the fact that there are $300$, $150$ and $50$ beads of the various colours, essentially means we have an unlimited supply of each colour (since there's more than $24$ in each case). In this context, the question should be written along the lines:</p>
<blockquote>
<p>How many $24$-bead necklaces can be made with beads of $3$ colours?</p>
</blockquote>
<p>The best method for answering this question is via <a href="http://en.wikipedia.org/wiki/Burnside%27s_lemma" rel="nofollow">Burnside's Lemma</a>.</p>
<p>Let $L$ be the set of $24$-element sequences of $\{\text{white},\text{red},\text{green}\}$. We see $|L|=3^{24}$. The dihedral group $D_{48}$ <a href="http://en.wikipedia.org/wiki/Group_action" rel="nofollow">acts on</a> $L$ by permuting the indices of the elements. To use Burnside's Lemma, we need to count how many necklaces are fixed by the action of $\alpha$, for each $\alpha \in D_{48}$.</p>
<p>If $\alpha \in D_{48}$, and $\alpha N=N$ for some $N \in L$ (i.e. $\alpha$ is an automorphism of $N$), then the group generated by $\alpha$, namely $\langle \alpha \rangle$, acts on the elements of $N$. The beads in the same orbit under this action have the same colour (since $\alpha N=N$). So there are $3^{\#\text{orbits}}$ necklaces that are fixed by any $\alpha \in D_{48}$ (we pick a bead colour for each orbit). The number of orbits is the number of cycles in the cycle decomposition of $\alpha$. Hence $\alpha$ fixes exactly $3^{\#\text{cycles of } \alpha}$ necklaces.</p>
<p>We can do some bookkeeping, e.g. with <a href="http://www.gap-system.org/" rel="nofollow">GAP</a>, to obtain:</p>
<p>$$\begin{array}{c|cc}
\text{cycle structure} & \text{nr perms in } D_{48} \text{ with that cycle structure} & \text{nr cycles} \\
\hline
1^{24} & 1 & 24 \\
1^2 2^{11} & 12 & 13 \\
2^{12} & 13 & 12 \\
3^8 & 2 & 8 \\
4^6 & 2 & 6 \\
6^4 & 2 & 4 \\
8^3 & 4 & 3 \\
12^2 & 4 & 2 \\
24 & 8 & 1 \\
\end{array}$$</p>
<p>So, by Burnside's Lemma, the number of necklaces is $$\frac{1}{48} \left( 3^{24} + 12 \times 3^{13} + 13 \times 3^{12} + 2 \times 3^8 + 2 \times 3^6 + 2 \times 3^4 + 4 \times 3^3 + 4 \times 3^2 + 8 \times 3^1 \right)$$ which is $5884491500$.</p>
|
3,697,062 | <p>How many solutions are there for this equation <span class="math-container">$$x_1+x_2+x_3+x_4 = 49\,,$$</span> where <span class="math-container">$x_i,\;i= 1,2,3,4$</span> is a non negative integer such that:<span class="math-container">$$ 1\le x_1\le 8,\;3\le x_2\le 9,\;10\le x_3\le 20,\ 0\le x_4\,?$$</span></p>
| Jan Eerland | 226,665 | <p>Well, just a quick search with Mathematica gives:</p>
<pre><code>In[1]:=Length[Solve[{x1 + x2 + x3 + x4 == 49,
1 <= x1 <= 8 && 3 <= x2 <= 9 && 10 <= x3 <= 20 && 0 <= x4}, {x1,
x2, x3, x4}, Integers]]
Out[1]=616
</code></pre>
<p>So, there are <span class="math-container">$616$</span> solutions.</p>
<hr>
<p>And the solutions are:</p>
<pre><code>In[2]:=Solve[{x1 + x2 + x3 + x4 == 49,
1 <= x1 <= 8 && 3 <= x2 <= 9 && 10 <= x3 <= 20 && 0 <= x4}, {x1, x2,
x3, x4}, Integers]
Out[2]={{x1 -> 1, x2 -> 3, x3 -> 10, x4 -> 35}, {x1 -> 1, x2 -> 3, x3 -> 11,
x4 -> 34}, {x1 -> 1, x2 -> 3, x3 -> 12, x4 -> 33}, {x1 -> 1,
x2 -> 3, x3 -> 13, x4 -> 32}, {x1 -> 1, x2 -> 3, x3 -> 14,
x4 -> 31}, {x1 -> 1, x2 -> 3, x3 -> 15, x4 -> 30}, {x1 -> 1,
x2 -> 3, x3 -> 16, x4 -> 29}, {x1 -> 1, x2 -> 3, x3 -> 17,
x4 -> 28}, {x1 -> 1, x2 -> 3, x3 -> 18, x4 -> 27}, {x1 -> 1,
x2 -> 3, x3 -> 19, x4 -> 26}, {x1 -> 1, x2 -> 3, x3 -> 20,
x4 -> 25}, {x1 -> 1, x2 -> 4, x3 -> 10, x4 -> 34}, {x1 -> 1,
x2 -> 4, x3 -> 11, x4 -> 33}, {x1 -> 1, x2 -> 4, x3 -> 12,
x4 -> 32}, {x1 -> 1, x2 -> 4, x3 -> 13, x4 -> 31}, {x1 -> 1,
x2 -> 4, x3 -> 14, x4 -> 30}, {x1 -> 1, x2 -> 4, x3 -> 15,
x4 -> 29}, {x1 -> 1, x2 -> 4, x3 -> 16, x4 -> 28}, {x1 -> 1,
x2 -> 4, x3 -> 17, x4 -> 27}, {x1 -> 1, x2 -> 4, x3 -> 18,
x4 -> 26}, {x1 -> 1, x2 -> 4, x3 -> 19, x4 -> 25}, {x1 -> 1,
x2 -> 4, x3 -> 20, x4 -> 24}, {x1 -> 1, x2 -> 5, x3 -> 10,
x4 -> 33}, {x1 -> 1, x2 -> 5, x3 -> 11, x4 -> 32}, {x1 -> 1,
x2 -> 5, x3 -> 12, x4 -> 31}, {x1 -> 1, x2 -> 5, x3 -> 13,
x4 -> 30}, {x1 -> 1, x2 -> 5, x3 -> 14, x4 -> 29}, {x1 -> 1,
x2 -> 5, x3 -> 15, x4 -> 28}, {x1 -> 1, x2 -> 5, x3 -> 16,
x4 -> 27}, {x1 -> 1, x2 -> 5, x3 -> 17, x4 -> 26}, {x1 -> 1,
x2 -> 5, x3 -> 18, x4 -> 25}, {x1 -> 1, x2 -> 5, x3 -> 19,
x4 -> 24}, {x1 -> 1, x2 -> 5, x3 -> 20, x4 -> 23}, {x1 -> 1,
x2 -> 6, x3 -> 10, x4 -> 32}, {x1 -> 1, x2 -> 6, x3 -> 11,
x4 -> 31}, {x1 -> 1, x2 -> 6, x3 -> 12, x4 -> 30}, {x1 -> 1,
x2 -> 6, x3 -> 13, x4 -> 29}, {x1 -> 1, x2 -> 6, x3 -> 14,
x4 -> 28}, {x1 -> 1, x2 -> 6, x3 -> 15, x4 -> 27}, {x1 -> 1,
x2 -> 6, x3 -> 16, x4 -> 26}, {x1 -> 1, x2 -> 6, x3 -> 17,
x4 -> 25}, {x1 -> 1, x2 -> 6, x3 -> 18, x4 -> 24}, {x1 -> 1,
x2 -> 6, x3 -> 19, x4 -> 23}, {x1 -> 1, x2 -> 6, x3 -> 20,
x4 -> 22}, {x1 -> 1, x2 -> 7, x3 -> 10, x4 -> 31}, {x1 -> 1,
x2 -> 7, x3 -> 11, x4 -> 30}, {x1 -> 1, x2 -> 7, x3 -> 12,
x4 -> 29}, {x1 -> 1, x2 -> 7, x3 -> 13, x4 -> 28}, {x1 -> 1,
x2 -> 7, x3 -> 14, x4 -> 27}, {x1 -> 1, x2 -> 7, x3 -> 15,
x4 -> 26}, {x1 -> 1, x2 -> 7, x3 -> 16, x4 -> 25}, {x1 -> 1,
x2 -> 7, x3 -> 17, x4 -> 24}, {x1 -> 1, x2 -> 7, x3 -> 18,
x4 -> 23}, {x1 -> 1, x2 -> 7, x3 -> 19, x4 -> 22}, {x1 -> 1,
x2 -> 7, x3 -> 20, x4 -> 21}, {x1 -> 1, x2 -> 8, x3 -> 10,
x4 -> 30}, {x1 -> 1, x2 -> 8, x3 -> 11, x4 -> 29}, {x1 -> 1,
x2 -> 8, x3 -> 12, x4 -> 28}, {x1 -> 1, x2 -> 8, x3 -> 13,
x4 -> 27}, {x1 -> 1, x2 -> 8, x3 -> 14, x4 -> 26}, {x1 -> 1,
x2 -> 8, x3 -> 15, x4 -> 25}, {x1 -> 1, x2 -> 8, x3 -> 16,
x4 -> 24}, {x1 -> 1, x2 -> 8, x3 -> 17, x4 -> 23}, {x1 -> 1,
x2 -> 8, x3 -> 18, x4 -> 22}, {x1 -> 1, x2 -> 8, x3 -> 19,
x4 -> 21}, {x1 -> 1, x2 -> 8, x3 -> 20, x4 -> 20}, {x1 -> 1,
x2 -> 9, x3 -> 10, x4 -> 29}, {x1 -> 1, x2 -> 9, x3 -> 11,
x4 -> 28}, {x1 -> 1, x2 -> 9, x3 -> 12, x4 -> 27}, {x1 -> 1,
x2 -> 9, x3 -> 13, x4 -> 26}, {x1 -> 1, x2 -> 9, x3 -> 14,
x4 -> 25}, {x1 -> 1, x2 -> 9, x3 -> 15, x4 -> 24}, {x1 -> 1,
x2 -> 9, x3 -> 16, x4 -> 23}, {x1 -> 1, x2 -> 9, x3 -> 17,
x4 -> 22}, {x1 -> 1, x2 -> 9, x3 -> 18, x4 -> 21}, {x1 -> 1,
x2 -> 9, x3 -> 19, x4 -> 20}, {x1 -> 1, x2 -> 9, x3 -> 20,
x4 -> 19}, {x1 -> 2, x2 -> 3, x3 -> 10, x4 -> 34}, {x1 -> 2,
x2 -> 3, x3 -> 11, x4 -> 33}, {x1 -> 2, x2 -> 3, x3 -> 12,
x4 -> 32}, {x1 -> 2, x2 -> 3, x3 -> 13, x4 -> 31}, {x1 -> 2,
x2 -> 3, x3 -> 14, x4 -> 30}, {x1 -> 2, x2 -> 3, x3 -> 15,
x4 -> 29}, {x1 -> 2, x2 -> 3, x3 -> 16, x4 -> 28}, {x1 -> 2,
x2 -> 3, x3 -> 17, x4 -> 27}, {x1 -> 2, x2 -> 3, x3 -> 18,
x4 -> 26}, {x1 -> 2, x2 -> 3, x3 -> 19, x4 -> 25}, {x1 -> 2,
x2 -> 3, x3 -> 20, x4 -> 24}, {x1 -> 2, x2 -> 4, x3 -> 10,
x4 -> 33}, {x1 -> 2, x2 -> 4, x3 -> 11, x4 -> 32}, {x1 -> 2,
x2 -> 4, x3 -> 12, x4 -> 31}, {x1 -> 2, x2 -> 4, x3 -> 13,
x4 -> 30}, {x1 -> 2, x2 -> 4, x3 -> 14, x4 -> 29}, {x1 -> 2,
x2 -> 4, x3 -> 15, x4 -> 28}, {x1 -> 2, x2 -> 4, x3 -> 16,
x4 -> 27}, {x1 -> 2, x2 -> 4, x3 -> 17, x4 -> 26}, {x1 -> 2,
x2 -> 4, x3 -> 18, x4 -> 25}, {x1 -> 2, x2 -> 4, x3 -> 19,
x4 -> 24}, {x1 -> 2, x2 -> 4, x3 -> 20, x4 -> 23}, {x1 -> 2,
x2 -> 5, x3 -> 10, x4 -> 32}, {x1 -> 2, x2 -> 5, x3 -> 11,
x4 -> 31}, {x1 -> 2, x2 -> 5, x3 -> 12, x4 -> 30}, {x1 -> 2,
x2 -> 5, x3 -> 13, x4 -> 29}, {x1 -> 2, x2 -> 5, x3 -> 14,
x4 -> 28}, {x1 -> 2, x2 -> 5, x3 -> 15, x4 -> 27}, {x1 -> 2,
x2 -> 5, x3 -> 16, x4 -> 26}, {x1 -> 2, x2 -> 5, x3 -> 17,
x4 -> 25}, {x1 -> 2, x2 -> 5, x3 -> 18, x4 -> 24}, {x1 -> 2,
x2 -> 5, x3 -> 19, x4 -> 23}, {x1 -> 2, x2 -> 5, x3 -> 20,
x4 -> 22}, {x1 -> 2, x2 -> 6, x3 -> 10, x4 -> 31}, {x1 -> 2,
x2 -> 6, x3 -> 11, x4 -> 30}, {x1 -> 2, x2 -> 6, x3 -> 12,
x4 -> 29}, {x1 -> 2, x2 -> 6, x3 -> 13, x4 -> 28}, {x1 -> 2,
x2 -> 6, x3 -> 14, x4 -> 27}, {x1 -> 2, x2 -> 6, x3 -> 15,
x4 -> 26}, {x1 -> 2, x2 -> 6, x3 -> 16, x4 -> 25}, {x1 -> 2,
x2 -> 6, x3 -> 17, x4 -> 24}, {x1 -> 2, x2 -> 6, x3 -> 18,
x4 -> 23}, {x1 -> 2, x2 -> 6, x3 -> 19, x4 -> 22}, {x1 -> 2,
x2 -> 6, x3 -> 20, x4 -> 21}, {x1 -> 2, x2 -> 7, x3 -> 10,
x4 -> 30}, {x1 -> 2, x2 -> 7, x3 -> 11, x4 -> 29}, {x1 -> 2,
x2 -> 7, x3 -> 12, x4 -> 28}, {x1 -> 2, x2 -> 7, x3 -> 13,
x4 -> 27}, {x1 -> 2, x2 -> 7, x3 -> 14, x4 -> 26}, {x1 -> 2,
x2 -> 7, x3 -> 15, x4 -> 25}, {x1 -> 2, x2 -> 7, x3 -> 16,
x4 -> 24}, {x1 -> 2, x2 -> 7, x3 -> 17, x4 -> 23}, {x1 -> 2,
x2 -> 7, x3 -> 18, x4 -> 22}, {x1 -> 2, x2 -> 7, x3 -> 19,
x4 -> 21}, {x1 -> 2, x2 -> 7, x3 -> 20, x4 -> 20}, {x1 -> 2,
x2 -> 8, x3 -> 10, x4 -> 29}, {x1 -> 2, x2 -> 8, x3 -> 11,
x4 -> 28}, {x1 -> 2, x2 -> 8, x3 -> 12, x4 -> 27}, {x1 -> 2,
x2 -> 8, x3 -> 13, x4 -> 26}, {x1 -> 2, x2 -> 8, x3 -> 14,
x4 -> 25}, {x1 -> 2, x2 -> 8, x3 -> 15, x4 -> 24}, {x1 -> 2,
x2 -> 8, x3 -> 16, x4 -> 23}, {x1 -> 2, x2 -> 8, x3 -> 17,
x4 -> 22}, {x1 -> 2, x2 -> 8, x3 -> 18, x4 -> 21}, {x1 -> 2,
x2 -> 8, x3 -> 19, x4 -> 20}, {x1 -> 2, x2 -> 8, x3 -> 20,
x4 -> 19}, {x1 -> 2, x2 -> 9, x3 -> 10, x4 -> 28}, {x1 -> 2,
x2 -> 9, x3 -> 11, x4 -> 27}, {x1 -> 2, x2 -> 9, x3 -> 12,
x4 -> 26}, {x1 -> 2, x2 -> 9, x3 -> 13, x4 -> 25}, {x1 -> 2,
x2 -> 9, x3 -> 14, x4 -> 24}, {x1 -> 2, x2 -> 9, x3 -> 15,
x4 -> 23}, {x1 -> 2, x2 -> 9, x3 -> 16, x4 -> 22}, {x1 -> 2,
x2 -> 9, x3 -> 17, x4 -> 21}, {x1 -> 2, x2 -> 9, x3 -> 18,
x4 -> 20}, {x1 -> 2, x2 -> 9, x3 -> 19, x4 -> 19}, {x1 -> 2,
x2 -> 9, x3 -> 20, x4 -> 18}, {x1 -> 3, x2 -> 3, x3 -> 10,
x4 -> 33}, {x1 -> 3, x2 -> 3, x3 -> 11, x4 -> 32}, {x1 -> 3,
x2 -> 3, x3 -> 12, x4 -> 31}, {x1 -> 3, x2 -> 3, x3 -> 13,
x4 -> 30}, {x1 -> 3, x2 -> 3, x3 -> 14, x4 -> 29}, {x1 -> 3,
x2 -> 3, x3 -> 15, x4 -> 28}, {x1 -> 3, x2 -> 3, x3 -> 16,
x4 -> 27}, {x1 -> 3, x2 -> 3, x3 -> 17, x4 -> 26}, {x1 -> 3,
x2 -> 3, x3 -> 18, x4 -> 25}, {x1 -> 3, x2 -> 3, x3 -> 19,
x4 -> 24}, {x1 -> 3, x2 -> 3, x3 -> 20, x4 -> 23}, {x1 -> 3,
x2 -> 4, x3 -> 10, x4 -> 32}, {x1 -> 3, x2 -> 4, x3 -> 11,
x4 -> 31}, {x1 -> 3, x2 -> 4, x3 -> 12, x4 -> 30}, {x1 -> 3,
x2 -> 4, x3 -> 13, x4 -> 29}, {x1 -> 3, x2 -> 4, x3 -> 14,
x4 -> 28}, {x1 -> 3, x2 -> 4, x3 -> 15, x4 -> 27}, {x1 -> 3,
x2 -> 4, x3 -> 16, x4 -> 26}, {x1 -> 3, x2 -> 4, x3 -> 17,
x4 -> 25}, {x1 -> 3, x2 -> 4, x3 -> 18, x4 -> 24}, {x1 -> 3,
x2 -> 4, x3 -> 19, x4 -> 23}, {x1 -> 3, x2 -> 4, x3 -> 20,
x4 -> 22}, {x1 -> 3, x2 -> 5, x3 -> 10, x4 -> 31}, {x1 -> 3,
x2 -> 5, x3 -> 11, x4 -> 30}, {x1 -> 3, x2 -> 5, x3 -> 12,
x4 -> 29}, {x1 -> 3, x2 -> 5, x3 -> 13, x4 -> 28}, {x1 -> 3,
x2 -> 5, x3 -> 14, x4 -> 27}, {x1 -> 3, x2 -> 5, x3 -> 15,
x4 -> 26}, {x1 -> 3, x2 -> 5, x3 -> 16, x4 -> 25}, {x1 -> 3,
x2 -> 5, x3 -> 17, x4 -> 24}, {x1 -> 3, x2 -> 5, x3 -> 18,
x4 -> 23}, {x1 -> 3, x2 -> 5, x3 -> 19, x4 -> 22}, {x1 -> 3,
x2 -> 5, x3 -> 20, x4 -> 21}, {x1 -> 3, x2 -> 6, x3 -> 10,
x4 -> 30}, {x1 -> 3, x2 -> 6, x3 -> 11, x4 -> 29}, {x1 -> 3,
x2 -> 6, x3 -> 12, x4 -> 28}, {x1 -> 3, x2 -> 6, x3 -> 13,
x4 -> 27}, {x1 -> 3, x2 -> 6, x3 -> 14, x4 -> 26}, {x1 -> 3,
x2 -> 6, x3 -> 15, x4 -> 25}, {x1 -> 3, x2 -> 6, x3 -> 16,
x4 -> 24}, {x1 -> 3, x2 -> 6, x3 -> 17, x4 -> 23}, {x1 -> 3,
x2 -> 6, x3 -> 18, x4 -> 22}, {x1 -> 3, x2 -> 6, x3 -> 19,
x4 -> 21}, {x1 -> 3, x2 -> 6, x3 -> 20, x4 -> 20}, {x1 -> 3,
x2 -> 7, x3 -> 10, x4 -> 29}, {x1 -> 3, x2 -> 7, x3 -> 11,
x4 -> 28}, {x1 -> 3, x2 -> 7, x3 -> 12, x4 -> 27}, {x1 -> 3,
x2 -> 7, x3 -> 13, x4 -> 26}, {x1 -> 3, x2 -> 7, x3 -> 14,
x4 -> 25}, {x1 -> 3, x2 -> 7, x3 -> 15, x4 -> 24}, {x1 -> 3,
x2 -> 7, x3 -> 16, x4 -> 23}, {x1 -> 3, x2 -> 7, x3 -> 17,
x4 -> 22}, {x1 -> 3, x2 -> 7, x3 -> 18, x4 -> 21}, {x1 -> 3,
x2 -> 7, x3 -> 19, x4 -> 20}, {x1 -> 3, x2 -> 7, x3 -> 20,
x4 -> 19}, {x1 -> 3, x2 -> 8, x3 -> 10, x4 -> 28}, {x1 -> 3,
x2 -> 8, x3 -> 11, x4 -> 27}, {x1 -> 3, x2 -> 8, x3 -> 12,
x4 -> 26}, {x1 -> 3, x2 -> 8, x3 -> 13, x4 -> 25}, {x1 -> 3,
x2 -> 8, x3 -> 14, x4 -> 24}, {x1 -> 3, x2 -> 8, x3 -> 15,
x4 -> 23}, {x1 -> 3, x2 -> 8, x3 -> 16, x4 -> 22}, {x1 -> 3,
x2 -> 8, x3 -> 17, x4 -> 21}, {x1 -> 3, x2 -> 8, x3 -> 18,
x4 -> 20}, {x1 -> 3, x2 -> 8, x3 -> 19, x4 -> 19}, {x1 -> 3,
x2 -> 8, x3 -> 20, x4 -> 18}, {x1 -> 3, x2 -> 9, x3 -> 10,
x4 -> 27}, {x1 -> 3, x2 -> 9, x3 -> 11, x4 -> 26}, {x1 -> 3,
x2 -> 9, x3 -> 12, x4 -> 25}, {x1 -> 3, x2 -> 9, x3 -> 13,
x4 -> 24}, {x1 -> 3, x2 -> 9, x3 -> 14, x4 -> 23}, {x1 -> 3,
x2 -> 9, x3 -> 15, x4 -> 22}, {x1 -> 3, x2 -> 9, x3 -> 16,
x4 -> 21}, {x1 -> 3, x2 -> 9, x3 -> 17, x4 -> 20}, {x1 -> 3,
x2 -> 9, x3 -> 18, x4 -> 19}, {x1 -> 3, x2 -> 9, x3 -> 19,
x4 -> 18}, {x1 -> 3, x2 -> 9, x3 -> 20, x4 -> 17}, {x1 -> 4,
x2 -> 3, x3 -> 10, x4 -> 32}, {x1 -> 4, x2 -> 3, x3 -> 11,
x4 -> 31}, {x1 -> 4, x2 -> 3, x3 -> 12, x4 -> 30}, {x1 -> 4,
x2 -> 3, x3 -> 13, x4 -> 29}, {x1 -> 4, x2 -> 3, x3 -> 14,
x4 -> 28}, {x1 -> 4, x2 -> 3, x3 -> 15, x4 -> 27}, {x1 -> 4,
x2 -> 3, x3 -> 16, x4 -> 26}, {x1 -> 4, x2 -> 3, x3 -> 17,
x4 -> 25}, {x1 -> 4, x2 -> 3, x3 -> 18, x4 -> 24}, {x1 -> 4,
x2 -> 3, x3 -> 19, x4 -> 23}, {x1 -> 4, x2 -> 3, x3 -> 20,
x4 -> 22}, {x1 -> 4, x2 -> 4, x3 -> 10, x4 -> 31}, {x1 -> 4,
x2 -> 4, x3 -> 11, x4 -> 30}, {x1 -> 4, x2 -> 4, x3 -> 12,
x4 -> 29}, {x1 -> 4, x2 -> 4, x3 -> 13, x4 -> 28}, {x1 -> 4,
x2 -> 4, x3 -> 14, x4 -> 27}, {x1 -> 4, x2 -> 4, x3 -> 15,
x4 -> 26}, {x1 -> 4, x2 -> 4, x3 -> 16, x4 -> 25}, {x1 -> 4,
x2 -> 4, x3 -> 17, x4 -> 24}, {x1 -> 4, x2 -> 4, x3 -> 18,
x4 -> 23}, {x1 -> 4, x2 -> 4, x3 -> 19, x4 -> 22}, {x1 -> 4,
x2 -> 4, x3 -> 20, x4 -> 21}, {x1 -> 4, x2 -> 5, x3 -> 10,
x4 -> 30}, {x1 -> 4, x2 -> 5, x3 -> 11, x4 -> 29}, {x1 -> 4,
x2 -> 5, x3 -> 12, x4 -> 28}, {x1 -> 4, x2 -> 5, x3 -> 13,
x4 -> 27}, {x1 -> 4, x2 -> 5, x3 -> 14, x4 -> 26}, {x1 -> 4,
x2 -> 5, x3 -> 15, x4 -> 25}, {x1 -> 4, x2 -> 5, x3 -> 16,
x4 -> 24}, {x1 -> 4, x2 -> 5, x3 -> 17, x4 -> 23}, {x1 -> 4,
x2 -> 5, x3 -> 18, x4 -> 22}, {x1 -> 4, x2 -> 5, x3 -> 19,
x4 -> 21}, {x1 -> 4, x2 -> 5, x3 -> 20, x4 -> 20}, {x1 -> 4,
x2 -> 6, x3 -> 10, x4 -> 29}, {x1 -> 4, x2 -> 6, x3 -> 11,
x4 -> 28}, {x1 -> 4, x2 -> 6, x3 -> 12, x4 -> 27}, {x1 -> 4,
x2 -> 6, x3 -> 13, x4 -> 26}, {x1 -> 4, x2 -> 6, x3 -> 14,
x4 -> 25}, {x1 -> 4, x2 -> 6, x3 -> 15, x4 -> 24}, {x1 -> 4,
x2 -> 6, x3 -> 16, x4 -> 23}, {x1 -> 4, x2 -> 6, x3 -> 17,
x4 -> 22}, {x1 -> 4, x2 -> 6, x3 -> 18, x4 -> 21}, {x1 -> 4,
x2 -> 6, x3 -> 19, x4 -> 20}, {x1 -> 4, x2 -> 6, x3 -> 20,
x4 -> 19}, {x1 -> 4, x2 -> 7, x3 -> 10, x4 -> 28}, {x1 -> 4,
x2 -> 7, x3 -> 11, x4 -> 27}, {x1 -> 4, x2 -> 7, x3 -> 12,
x4 -> 26}, {x1 -> 4, x2 -> 7, x3 -> 13, x4 -> 25}, {x1 -> 4,
x2 -> 7, x3 -> 14, x4 -> 24}, {x1 -> 4, x2 -> 7, x3 -> 15,
x4 -> 23}, {x1 -> 4, x2 -> 7, x3 -> 16, x4 -> 22}, {x1 -> 4,
x2 -> 7, x3 -> 17, x4 -> 21}, {x1 -> 4, x2 -> 7, x3 -> 18,
x4 -> 20}, {x1 -> 4, x2 -> 7, x3 -> 19, x4 -> 19}, {x1 -> 4,
x2 -> 7, x3 -> 20, x4 -> 18}, {x1 -> 4, x2 -> 8, x3 -> 10,
x4 -> 27}, {x1 -> 4, x2 -> 8, x3 -> 11, x4 -> 26}, {x1 -> 4,
x2 -> 8, x3 -> 12, x4 -> 25}, {x1 -> 4, x2 -> 8, x3 -> 13,
x4 -> 24}, {x1 -> 4, x2 -> 8, x3 -> 14, x4 -> 23}, {x1 -> 4,
x2 -> 8, x3 -> 15, x4 -> 22}, {x1 -> 4, x2 -> 8, x3 -> 16,
x4 -> 21}, {x1 -> 4, x2 -> 8, x3 -> 17, x4 -> 20}, {x1 -> 4,
x2 -> 8, x3 -> 18, x4 -> 19}, {x1 -> 4, x2 -> 8, x3 -> 19,
x4 -> 18}, {x1 -> 4, x2 -> 8, x3 -> 20, x4 -> 17}, {x1 -> 4,
x2 -> 9, x3 -> 10, x4 -> 26}, {x1 -> 4, x2 -> 9, x3 -> 11,
x4 -> 25}, {x1 -> 4, x2 -> 9, x3 -> 12, x4 -> 24}, {x1 -> 4,
x2 -> 9, x3 -> 13, x4 -> 23}, {x1 -> 4, x2 -> 9, x3 -> 14,
x4 -> 22}, {x1 -> 4, x2 -> 9, x3 -> 15, x4 -> 21}, {x1 -> 4,
x2 -> 9, x3 -> 16, x4 -> 20}, {x1 -> 4, x2 -> 9, x3 -> 17,
x4 -> 19}, {x1 -> 4, x2 -> 9, x3 -> 18, x4 -> 18}, {x1 -> 4,
x2 -> 9, x3 -> 19, x4 -> 17}, {x1 -> 4, x2 -> 9, x3 -> 20,
x4 -> 16}, {x1 -> 5, x2 -> 3, x3 -> 10, x4 -> 31}, {x1 -> 5,
x2 -> 3, x3 -> 11, x4 -> 30}, {x1 -> 5, x2 -> 3, x3 -> 12,
x4 -> 29}, {x1 -> 5, x2 -> 3, x3 -> 13, x4 -> 28}, {x1 -> 5,
x2 -> 3, x3 -> 14, x4 -> 27}, {x1 -> 5, x2 -> 3, x3 -> 15,
x4 -> 26}, {x1 -> 5, x2 -> 3, x3 -> 16, x4 -> 25}, {x1 -> 5,
x2 -> 3, x3 -> 17, x4 -> 24}, {x1 -> 5, x2 -> 3, x3 -> 18,
x4 -> 23}, {x1 -> 5, x2 -> 3, x3 -> 19, x4 -> 22}, {x1 -> 5,
x2 -> 3, x3 -> 20, x4 -> 21}, {x1 -> 5, x2 -> 4, x3 -> 10,
x4 -> 30}, {x1 -> 5, x2 -> 4, x3 -> 11, x4 -> 29}, {x1 -> 5,
x2 -> 4, x3 -> 12, x4 -> 28}, {x1 -> 5, x2 -> 4, x3 -> 13,
x4 -> 27}, {x1 -> 5, x2 -> 4, x3 -> 14, x4 -> 26}, {x1 -> 5,
x2 -> 4, x3 -> 15, x4 -> 25}, {x1 -> 5, x2 -> 4, x3 -> 16,
x4 -> 24}, {x1 -> 5, x2 -> 4, x3 -> 17, x4 -> 23}, {x1 -> 5,
x2 -> 4, x3 -> 18, x4 -> 22}, {x1 -> 5, x2 -> 4, x3 -> 19,
x4 -> 21}, {x1 -> 5, x2 -> 4, x3 -> 20, x4 -> 20}, {x1 -> 5,
x2 -> 5, x3 -> 10, x4 -> 29}, {x1 -> 5, x2 -> 5, x3 -> 11,
x4 -> 28}, {x1 -> 5, x2 -> 5, x3 -> 12, x4 -> 27}, {x1 -> 5,
x2 -> 5, x3 -> 13, x4 -> 26}, {x1 -> 5, x2 -> 5, x3 -> 14,
x4 -> 25}, {x1 -> 5, x2 -> 5, x3 -> 15, x4 -> 24}, {x1 -> 5,
x2 -> 5, x3 -> 16, x4 -> 23}, {x1 -> 5, x2 -> 5, x3 -> 17,
x4 -> 22}, {x1 -> 5, x2 -> 5, x3 -> 18, x4 -> 21}, {x1 -> 5,
x2 -> 5, x3 -> 19, x4 -> 20}, {x1 -> 5, x2 -> 5, x3 -> 20,
x4 -> 19}, {x1 -> 5, x2 -> 6, x3 -> 10, x4 -> 28}, {x1 -> 5,
x2 -> 6, x3 -> 11, x4 -> 27}, {x1 -> 5, x2 -> 6, x3 -> 12,
x4 -> 26}, {x1 -> 5, x2 -> 6, x3 -> 13, x4 -> 25}, {x1 -> 5,
x2 -> 6, x3 -> 14, x4 -> 24}, {x1 -> 5, x2 -> 6, x3 -> 15,
x4 -> 23}, {x1 -> 5, x2 -> 6, x3 -> 16, x4 -> 22}, {x1 -> 5,
x2 -> 6, x3 -> 17, x4 -> 21}, {x1 -> 5, x2 -> 6, x3 -> 18,
x4 -> 20}, {x1 -> 5, x2 -> 6, x3 -> 19, x4 -> 19}, {x1 -> 5,
x2 -> 6, x3 -> 20, x4 -> 18}, {x1 -> 5, x2 -> 7, x3 -> 10,
x4 -> 27}, {x1 -> 5, x2 -> 7, x3 -> 11, x4 -> 26}, {x1 -> 5,
x2 -> 7, x3 -> 12, x4 -> 25}, {x1 -> 5, x2 -> 7, x3 -> 13,
x4 -> 24}, {x1 -> 5, x2 -> 7, x3 -> 14, x4 -> 23}, {x1 -> 5,
x2 -> 7, x3 -> 15, x4 -> 22}, {x1 -> 5, x2 -> 7, x3 -> 16,
x4 -> 21}, {x1 -> 5, x2 -> 7, x3 -> 17, x4 -> 20}, {x1 -> 5,
x2 -> 7, x3 -> 18, x4 -> 19}, {x1 -> 5, x2 -> 7, x3 -> 19,
x4 -> 18}, {x1 -> 5, x2 -> 7, x3 -> 20, x4 -> 17}, {x1 -> 5,
x2 -> 8, x3 -> 10, x4 -> 26}, {x1 -> 5, x2 -> 8, x3 -> 11,
x4 -> 25}, {x1 -> 5, x2 -> 8, x3 -> 12, x4 -> 24}, {x1 -> 5,
x2 -> 8, x3 -> 13, x4 -> 23}, {x1 -> 5, x2 -> 8, x3 -> 14,
x4 -> 22}, {x1 -> 5, x2 -> 8, x3 -> 15, x4 -> 21}, {x1 -> 5,
x2 -> 8, x3 -> 16, x4 -> 20}, {x1 -> 5, x2 -> 8, x3 -> 17,
x4 -> 19}, {x1 -> 5, x2 -> 8, x3 -> 18, x4 -> 18}, {x1 -> 5,
x2 -> 8, x3 -> 19, x4 -> 17}, {x1 -> 5, x2 -> 8, x3 -> 20,
x4 -> 16}, {x1 -> 5, x2 -> 9, x3 -> 10, x4 -> 25}, {x1 -> 5,
x2 -> 9, x3 -> 11, x4 -> 24}, {x1 -> 5, x2 -> 9, x3 -> 12,
x4 -> 23}, {x1 -> 5, x2 -> 9, x3 -> 13, x4 -> 22}, {x1 -> 5,
x2 -> 9, x3 -> 14, x4 -> 21}, {x1 -> 5, x2 -> 9, x3 -> 15,
x4 -> 20}, {x1 -> 5, x2 -> 9, x3 -> 16, x4 -> 19}, {x1 -> 5,
x2 -> 9, x3 -> 17, x4 -> 18}, {x1 -> 5, x2 -> 9, x3 -> 18,
x4 -> 17}, {x1 -> 5, x2 -> 9, x3 -> 19, x4 -> 16}, {x1 -> 5,
x2 -> 9, x3 -> 20, x4 -> 15}, {x1 -> 6, x2 -> 3, x3 -> 10,
x4 -> 30}, {x1 -> 6, x2 -> 3, x3 -> 11, x4 -> 29}, {x1 -> 6,
x2 -> 3, x3 -> 12, x4 -> 28}, {x1 -> 6, x2 -> 3, x3 -> 13,
x4 -> 27}, {x1 -> 6, x2 -> 3, x3 -> 14, x4 -> 26}, {x1 -> 6,
x2 -> 3, x3 -> 15, x4 -> 25}, {x1 -> 6, x2 -> 3, x3 -> 16,
x4 -> 24}, {x1 -> 6, x2 -> 3, x3 -> 17, x4 -> 23}, {x1 -> 6,
x2 -> 3, x3 -> 18, x4 -> 22}, {x1 -> 6, x2 -> 3, x3 -> 19,
x4 -> 21}, {x1 -> 6, x2 -> 3, x3 -> 20, x4 -> 20}, {x1 -> 6,
x2 -> 4, x3 -> 10, x4 -> 29}, {x1 -> 6, x2 -> 4, x3 -> 11,
x4 -> 28}, {x1 -> 6, x2 -> 4, x3 -> 12, x4 -> 27}, {x1 -> 6,
x2 -> 4, x3 -> 13, x4 -> 26}, {x1 -> 6, x2 -> 4, x3 -> 14,
x4 -> 25}, {x1 -> 6, x2 -> 4, x3 -> 15, x4 -> 24}, {x1 -> 6,
x2 -> 4, x3 -> 16, x4 -> 23}, {x1 -> 6, x2 -> 4, x3 -> 17,
x4 -> 22}, {x1 -> 6, x2 -> 4, x3 -> 18, x4 -> 21}, {x1 -> 6,
x2 -> 4, x3 -> 19, x4 -> 20}, {x1 -> 6, x2 -> 4, x3 -> 20,
x4 -> 19}, {x1 -> 6, x2 -> 5, x3 -> 10, x4 -> 28}, {x1 -> 6,
x2 -> 5, x3 -> 11, x4 -> 27}, {x1 -> 6, x2 -> 5, x3 -> 12,
x4 -> 26}, {x1 -> 6, x2 -> 5, x3 -> 13, x4 -> 25}, {x1 -> 6,
x2 -> 5, x3 -> 14, x4 -> 24}, {x1 -> 6, x2 -> 5, x3 -> 15,
x4 -> 23}, {x1 -> 6, x2 -> 5, x3 -> 16, x4 -> 22}, {x1 -> 6,
x2 -> 5, x3 -> 17, x4 -> 21}, {x1 -> 6, x2 -> 5, x3 -> 18,
x4 -> 20}, {x1 -> 6, x2 -> 5, x3 -> 19, x4 -> 19}, {x1 -> 6,
x2 -> 5, x3 -> 20, x4 -> 18}, {x1 -> 6, x2 -> 6, x3 -> 10,
x4 -> 27}, {x1 -> 6, x2 -> 6, x3 -> 11, x4 -> 26}, {x1 -> 6,
x2 -> 6, x3 -> 12, x4 -> 25}, {x1 -> 6, x2 -> 6, x3 -> 13,
x4 -> 24}, {x1 -> 6, x2 -> 6, x3 -> 14, x4 -> 23}, {x1 -> 6,
x2 -> 6, x3 -> 15, x4 -> 22}, {x1 -> 6, x2 -> 6, x3 -> 16,
x4 -> 21}, {x1 -> 6, x2 -> 6, x3 -> 17, x4 -> 20}, {x1 -> 6,
x2 -> 6, x3 -> 18, x4 -> 19}, {x1 -> 6, x2 -> 6, x3 -> 19,
x4 -> 18}, {x1 -> 6, x2 -> 6, x3 -> 20, x4 -> 17}, {x1 -> 6,
x2 -> 7, x3 -> 10, x4 -> 26}, {x1 -> 6, x2 -> 7, x3 -> 11,
x4 -> 25}, {x1 -> 6, x2 -> 7, x3 -> 12, x4 -> 24}, {x1 -> 6,
x2 -> 7, x3 -> 13, x4 -> 23}, {x1 -> 6, x2 -> 7, x3 -> 14,
x4 -> 22}, {x1 -> 6, x2 -> 7, x3 -> 15, x4 -> 21}, {x1 -> 6,
x2 -> 7, x3 -> 16, x4 -> 20}, {x1 -> 6, x2 -> 7, x3 -> 17,
x4 -> 19}, {x1 -> 6, x2 -> 7, x3 -> 18, x4 -> 18}, {x1 -> 6,
x2 -> 7, x3 -> 19, x4 -> 17}, {x1 -> 6, x2 -> 7, x3 -> 20,
x4 -> 16}, {x1 -> 6, x2 -> 8, x3 -> 10, x4 -> 25}, {x1 -> 6,
x2 -> 8, x3 -> 11, x4 -> 24}, {x1 -> 6, x2 -> 8, x3 -> 12,
x4 -> 23}, {x1 -> 6, x2 -> 8, x3 -> 13, x4 -> 22}, {x1 -> 6,
x2 -> 8, x3 -> 14, x4 -> 21}, {x1 -> 6, x2 -> 8, x3 -> 15,
x4 -> 20}, {x1 -> 6, x2 -> 8, x3 -> 16, x4 -> 19}, {x1 -> 6,
x2 -> 8, x3 -> 17, x4 -> 18}, {x1 -> 6, x2 -> 8, x3 -> 18,
x4 -> 17}, {x1 -> 6, x2 -> 8, x3 -> 19, x4 -> 16}, {x1 -> 6,
x2 -> 8, x3 -> 20, x4 -> 15}, {x1 -> 6, x2 -> 9, x3 -> 10,
x4 -> 24}, {x1 -> 6, x2 -> 9, x3 -> 11, x4 -> 23}, {x1 -> 6,
x2 -> 9, x3 -> 12, x4 -> 22}, {x1 -> 6, x2 -> 9, x3 -> 13,
x4 -> 21}, {x1 -> 6, x2 -> 9, x3 -> 14, x4 -> 20}, {x1 -> 6,
x2 -> 9, x3 -> 15, x4 -> 19}, {x1 -> 6, x2 -> 9, x3 -> 16,
x4 -> 18}, {x1 -> 6, x2 -> 9, x3 -> 17, x4 -> 17}, {x1 -> 6,
x2 -> 9, x3 -> 18, x4 -> 16}, {x1 -> 6, x2 -> 9, x3 -> 19,
x4 -> 15}, {x1 -> 6, x2 -> 9, x3 -> 20, x4 -> 14}, {x1 -> 7,
x2 -> 3, x3 -> 10, x4 -> 29}, {x1 -> 7, x2 -> 3, x3 -> 11,
x4 -> 28}, {x1 -> 7, x2 -> 3, x3 -> 12, x4 -> 27}, {x1 -> 7,
x2 -> 3, x3 -> 13, x4 -> 26}, {x1 -> 7, x2 -> 3, x3 -> 14,
x4 -> 25}, {x1 -> 7, x2 -> 3, x3 -> 15, x4 -> 24}, {x1 -> 7,
x2 -> 3, x3 -> 16, x4 -> 23}, {x1 -> 7, x2 -> 3, x3 -> 17,
x4 -> 22}, {x1 -> 7, x2 -> 3, x3 -> 18, x4 -> 21}, {x1 -> 7,
x2 -> 3, x3 -> 19, x4 -> 20}, {x1 -> 7, x2 -> 3, x3 -> 20,
x4 -> 19}, {x1 -> 7, x2 -> 4, x3 -> 10, x4 -> 28}, {x1 -> 7,
x2 -> 4, x3 -> 11, x4 -> 27}, {x1 -> 7, x2 -> 4, x3 -> 12,
x4 -> 26}, {x1 -> 7, x2 -> 4, x3 -> 13, x4 -> 25}, {x1 -> 7,
x2 -> 4, x3 -> 14, x4 -> 24}, {x1 -> 7, x2 -> 4, x3 -> 15,
x4 -> 23}, {x1 -> 7, x2 -> 4, x3 -> 16, x4 -> 22}, {x1 -> 7,
x2 -> 4, x3 -> 17, x4 -> 21}, {x1 -> 7, x2 -> 4, x3 -> 18,
x4 -> 20}, {x1 -> 7, x2 -> 4, x3 -> 19, x4 -> 19}, {x1 -> 7,
x2 -> 4, x3 -> 20, x4 -> 18}, {x1 -> 7, x2 -> 5, x3 -> 10,
x4 -> 27}, {x1 -> 7, x2 -> 5, x3 -> 11, x4 -> 26}, {x1 -> 7,
x2 -> 5, x3 -> 12, x4 -> 25}, {x1 -> 7, x2 -> 5, x3 -> 13,
x4 -> 24}, {x1 -> 7, x2 -> 5, x3 -> 14, x4 -> 23}, {x1 -> 7,
x2 -> 5, x3 -> 15, x4 -> 22}, {x1 -> 7, x2 -> 5, x3 -> 16,
x4 -> 21}, {x1 -> 7, x2 -> 5, x3 -> 17, x4 -> 20}, {x1 -> 7,
x2 -> 5, x3 -> 18, x4 -> 19}, {x1 -> 7, x2 -> 5, x3 -> 19,
x4 -> 18}, {x1 -> 7, x2 -> 5, x3 -> 20, x4 -> 17}, {x1 -> 7,
x2 -> 6, x3 -> 10, x4 -> 26}, {x1 -> 7, x2 -> 6, x3 -> 11,
x4 -> 25}, {x1 -> 7, x2 -> 6, x3 -> 12, x4 -> 24}, {x1 -> 7,
x2 -> 6, x3 -> 13, x4 -> 23}, {x1 -> 7, x2 -> 6, x3 -> 14,
x4 -> 22}, {x1 -> 7, x2 -> 6, x3 -> 15, x4 -> 21}, {x1 -> 7,
x2 -> 6, x3 -> 16, x4 -> 20}, {x1 -> 7, x2 -> 6, x3 -> 17,
x4 -> 19}, {x1 -> 7, x2 -> 6, x3 -> 18, x4 -> 18}, {x1 -> 7,
x2 -> 6, x3 -> 19, x4 -> 17}, {x1 -> 7, x2 -> 6, x3 -> 20,
x4 -> 16}, {x1 -> 7, x2 -> 7, x3 -> 10, x4 -> 25}, {x1 -> 7,
x2 -> 7, x3 -> 11, x4 -> 24}, {x1 -> 7, x2 -> 7, x3 -> 12,
x4 -> 23}, {x1 -> 7, x2 -> 7, x3 -> 13, x4 -> 22}, {x1 -> 7,
x2 -> 7, x3 -> 14, x4 -> 21}, {x1 -> 7, x2 -> 7, x3 -> 15,
x4 -> 20}, {x1 -> 7, x2 -> 7, x3 -> 16, x4 -> 19}, {x1 -> 7,
x2 -> 7, x3 -> 17, x4 -> 18}, {x1 -> 7, x2 -> 7, x3 -> 18,
x4 -> 17}, {x1 -> 7, x2 -> 7, x3 -> 19, x4 -> 16}, {x1 -> 7,
x2 -> 7, x3 -> 20, x4 -> 15}, {x1 -> 7, x2 -> 8, x3 -> 10,
x4 -> 24}, {x1 -> 7, x2 -> 8, x3 -> 11, x4 -> 23}, {x1 -> 7,
x2 -> 8, x3 -> 12, x4 -> 22}, {x1 -> 7, x2 -> 8, x3 -> 13,
x4 -> 21}, {x1 -> 7, x2 -> 8, x3 -> 14, x4 -> 20}, {x1 -> 7,
x2 -> 8, x3 -> 15, x4 -> 19}, {x1 -> 7, x2 -> 8, x3 -> 16,
x4 -> 18}, {x1 -> 7, x2 -> 8, x3 -> 17, x4 -> 17}, {x1 -> 7,
x2 -> 8, x3 -> 18, x4 -> 16}, {x1 -> 7, x2 -> 8, x3 -> 19,
x4 -> 15}, {x1 -> 7, x2 -> 8, x3 -> 20, x4 -> 14}, {x1 -> 7,
x2 -> 9, x3 -> 10, x4 -> 23}, {x1 -> 7, x2 -> 9, x3 -> 11,
x4 -> 22}, {x1 -> 7, x2 -> 9, x3 -> 12, x4 -> 21}, {x1 -> 7,
x2 -> 9, x3 -> 13, x4 -> 20}, {x1 -> 7, x2 -> 9, x3 -> 14,
x4 -> 19}, {x1 -> 7, x2 -> 9, x3 -> 15, x4 -> 18}, {x1 -> 7,
x2 -> 9, x3 -> 16, x4 -> 17}, {x1 -> 7, x2 -> 9, x3 -> 17,
x4 -> 16}, {x1 -> 7, x2 -> 9, x3 -> 18, x4 -> 15}, {x1 -> 7,
x2 -> 9, x3 -> 19, x4 -> 14}, {x1 -> 7, x2 -> 9, x3 -> 20,
x4 -> 13}, {x1 -> 8, x2 -> 3, x3 -> 10, x4 -> 28}, {x1 -> 8,
x2 -> 3, x3 -> 11, x4 -> 27}, {x1 -> 8, x2 -> 3, x3 -> 12,
x4 -> 26}, {x1 -> 8, x2 -> 3, x3 -> 13, x4 -> 25}, {x1 -> 8,
x2 -> 3, x3 -> 14, x4 -> 24}, {x1 -> 8, x2 -> 3, x3 -> 15,
x4 -> 23}, {x1 -> 8, x2 -> 3, x3 -> 16, x4 -> 22}, {x1 -> 8,
x2 -> 3, x3 -> 17, x4 -> 21}, {x1 -> 8, x2 -> 3, x3 -> 18,
x4 -> 20}, {x1 -> 8, x2 -> 3, x3 -> 19, x4 -> 19}, {x1 -> 8,
x2 -> 3, x3 -> 20, x4 -> 18}, {x1 -> 8, x2 -> 4, x3 -> 10,
x4 -> 27}, {x1 -> 8, x2 -> 4, x3 -> 11, x4 -> 26}, {x1 -> 8,
x2 -> 4, x3 -> 12, x4 -> 25}, {x1 -> 8, x2 -> 4, x3 -> 13,
x4 -> 24}, {x1 -> 8, x2 -> 4, x3 -> 14, x4 -> 23}, {x1 -> 8,
x2 -> 4, x3 -> 15, x4 -> 22}, {x1 -> 8, x2 -> 4, x3 -> 16,
x4 -> 21}, {x1 -> 8, x2 -> 4, x3 -> 17, x4 -> 20}, {x1 -> 8,
x2 -> 4, x3 -> 18, x4 -> 19}, {x1 -> 8, x2 -> 4, x3 -> 19,
x4 -> 18}, {x1 -> 8, x2 -> 4, x3 -> 20, x4 -> 17}, {x1 -> 8,
x2 -> 5, x3 -> 10, x4 -> 26}, {x1 -> 8, x2 -> 5, x3 -> 11,
x4 -> 25}, {x1 -> 8, x2 -> 5, x3 -> 12, x4 -> 24}, {x1 -> 8,
x2 -> 5, x3 -> 13, x4 -> 23}, {x1 -> 8, x2 -> 5, x3 -> 14,
x4 -> 22}, {x1 -> 8, x2 -> 5, x3 -> 15, x4 -> 21}, {x1 -> 8,
x2 -> 5, x3 -> 16, x4 -> 20}, {x1 -> 8, x2 -> 5, x3 -> 17,
x4 -> 19}, {x1 -> 8, x2 -> 5, x3 -> 18, x4 -> 18}, {x1 -> 8,
x2 -> 5, x3 -> 19, x4 -> 17}, {x1 -> 8, x2 -> 5, x3 -> 20,
x4 -> 16}, {x1 -> 8, x2 -> 6, x3 -> 10, x4 -> 25}, {x1 -> 8,
x2 -> 6, x3 -> 11, x4 -> 24}, {x1 -> 8, x2 -> 6, x3 -> 12,
x4 -> 23}, {x1 -> 8, x2 -> 6, x3 -> 13, x4 -> 22}, {x1 -> 8,
x2 -> 6, x3 -> 14, x4 -> 21}, {x1 -> 8, x2 -> 6, x3 -> 15,
x4 -> 20}, {x1 -> 8, x2 -> 6, x3 -> 16, x4 -> 19}, {x1 -> 8,
x2 -> 6, x3 -> 17, x4 -> 18}, {x1 -> 8, x2 -> 6, x3 -> 18,
x4 -> 17}, {x1 -> 8, x2 -> 6, x3 -> 19, x4 -> 16}, {x1 -> 8,
x2 -> 6, x3 -> 20, x4 -> 15}, {x1 -> 8, x2 -> 7, x3 -> 10,
x4 -> 24}, {x1 -> 8, x2 -> 7, x3 -> 11, x4 -> 23}, {x1 -> 8,
x2 -> 7, x3 -> 12, x4 -> 22}, {x1 -> 8, x2 -> 7, x3 -> 13,
x4 -> 21}, {x1 -> 8, x2 -> 7, x3 -> 14, x4 -> 20}, {x1 -> 8,
x2 -> 7, x3 -> 15, x4 -> 19}, {x1 -> 8, x2 -> 7, x3 -> 16,
x4 -> 18}, {x1 -> 8, x2 -> 7, x3 -> 17, x4 -> 17}, {x1 -> 8,
x2 -> 7, x3 -> 18, x4 -> 16}, {x1 -> 8, x2 -> 7, x3 -> 19,
x4 -> 15}, {x1 -> 8, x2 -> 7, x3 -> 20, x4 -> 14}, {x1 -> 8,
x2 -> 8, x3 -> 10, x4 -> 23}, {x1 -> 8, x2 -> 8, x3 -> 11,
x4 -> 22}, {x1 -> 8, x2 -> 8, x3 -> 12, x4 -> 21}, {x1 -> 8,
x2 -> 8, x3 -> 13, x4 -> 20}, {x1 -> 8, x2 -> 8, x3 -> 14,
x4 -> 19}, {x1 -> 8, x2 -> 8, x3 -> 15, x4 -> 18}, {x1 -> 8,
x2 -> 8, x3 -> 16, x4 -> 17}, {x1 -> 8, x2 -> 8, x3 -> 17,
x4 -> 16}, {x1 -> 8, x2 -> 8, x3 -> 18, x4 -> 15}, {x1 -> 8,
x2 -> 8, x3 -> 19, x4 -> 14}, {x1 -> 8, x2 -> 8, x3 -> 20,
x4 -> 13}, {x1 -> 8, x2 -> 9, x3 -> 10, x4 -> 22}, {x1 -> 8,
x2 -> 9, x3 -> 11, x4 -> 21}, {x1 -> 8, x2 -> 9, x3 -> 12,
x4 -> 20}, {x1 -> 8, x2 -> 9, x3 -> 13, x4 -> 19}, {x1 -> 8,
x2 -> 9, x3 -> 14, x4 -> 18}, {x1 -> 8, x2 -> 9, x3 -> 15,
x4 -> 17}, {x1 -> 8, x2 -> 9, x3 -> 16, x4 -> 16}, {x1 -> 8,
x2 -> 9, x3 -> 17, x4 -> 15}, {x1 -> 8, x2 -> 9, x3 -> 18,
x4 -> 14}, {x1 -> 8, x2 -> 9, x3 -> 19, x4 -> 13}, {x1 -> 8,
x2 -> 9, x3 -> 20, x4 -> 12}}
</code></pre>
|
4,416,001 | <p>Given <span class="math-container">$f(xy) = f(x+y)$</span> and <span class="math-container">$f(11) = 11$</span>, what is <span class="math-container">$f(49)$</span>?</p>
| MangoPizza | 956,791 | <p><span class="math-container">$f(xy) = f(x + y)$</span> implies <span class="math-container">$f(x) = f(x \cdot 1) = f(x + 1)$</span>. So, <span class="math-container">$11 = f(11) = f(12) = \cdots = f(49)$</span></p>
|
4,047,784 | <p>Let <span class="math-container">$f_{avg}$</span> be the average value of a continuous function f(x) on the interval [a,b]. Assume that there's only one c <span class="math-container">$\in$</span> [a , b] for which <span class="math-container">$f(c)=f_{avg}$</span>. The question is, if this is the case, then must the function be either increasing or decreasing on [a, b]? This questions seems to jump from the assumption that <span class="math-container">$f(c)=f_{avg}$</span> for 1 value of c to the conclusion above, and I'm having trouble making sense of the question - obviously this is about the average value of a function, which is denoted by <span class="math-container">$\frac{1}{b-a}\int_{a}^{b}f(x)dx$</span>, but how does it relate to whether the function is increasing or decreasing? One way I know of finding out is by taking the derivative of f(x), but we don't know what f is in the first place. So...plz help :D</p>
| Kenny Lau | 328,173 | <p>I am not sure what you want. You have essentially specified the answer to be a root of a certain polynomial equation.</p>
<p>If you want to express the solution in terms of radicals (i.e. <span class="math-container">$n$</span><sup>th</sup> roots) like <span class="math-container">$\phi = \frac{1+\sqrt5}2$</span> then I am afraid this is impossible, and the reason has to do with Galois theory, so you can ignore the following if you are unfamiliar with Galois theory.</p>
<p>But if you are, then e.g. the case where <span class="math-container">$m=7$</span> we have the polynomial <span class="math-container">$x^7 - x^6 - 1$</span> whose Galois group is <span class="math-container">$S_7$</span> (computed with PARI/GP) which is not solvable, so its roots cannot be expressed using radicals.</p>
|
4,047,784 | <p>Let <span class="math-container">$f_{avg}$</span> be the average value of a continuous function f(x) on the interval [a,b]. Assume that there's only one c <span class="math-container">$\in$</span> [a , b] for which <span class="math-container">$f(c)=f_{avg}$</span>. The question is, if this is the case, then must the function be either increasing or decreasing on [a, b]? This questions seems to jump from the assumption that <span class="math-container">$f(c)=f_{avg}$</span> for 1 value of c to the conclusion above, and I'm having trouble making sense of the question - obviously this is about the average value of a function, which is denoted by <span class="math-container">$\frac{1}{b-a}\int_{a}^{b}f(x)dx$</span>, but how does it relate to whether the function is increasing or decreasing? One way I know of finding out is by taking the derivative of f(x), but we don't know what f is in the first place. So...plz help :D</p>
| Cye Waldman | 424,641 | <p>The sequence that you are looking at is similar to the Fibonacci-Narayana sequence, which is given by <span class="math-container">$f_n=f_{n-1}+f_{n-1-m}, m\ge 1$</span>. The generating function is given by</p>
<p><span class="math-container">$$
\frac{1}{1-x-x^{n+1}}
$$</span></p>
<p>and the limiting ratio is given by</p>
<p><span class="math-container">$$\lim_{n\to\infty}\frac{f_{n+1}}{f_{n}}=\chi\\
\chi-1=\chi^{-n}
$$</span></p>
<p>This sequence can be found in OEIS A000930 and related. <span class="math-container">$n=1, 2, 4$</span> are the golden ratio, supergolden ratio, and the plastic number, respectively.</p>
<p>Additional properties of this sequence are</p>
<p><span class="math-container">$$
\sum_{n=0}^{\infty}\frac{1}{\chi^n}=\chi^{n+1}\\
S_n=\sum_{j=1}^{n}f_j \quad \text{(cumulative sum)}\\
\lim_{n\to\infty}\frac{S_{n}}{S_{n+1}}=\chi
$$</span></p>
<p>You might also be interested in the Fibonacci-Padovan series, <span class="math-container">$f_n=f_{n-m+1}+f_{n-m}$</span> with a limiting value given by <span class="math-container">$X+1=X^m$</span>.</p>
|
3,213,854 | <p>From the definition of a inverse standpoint (<span class="math-container">$f^{-1}(f(x))=f(f^{-1}(x))=x$</span>), why does interchanging variables (<span class="math-container">$x$</span> and <span class="math-container">$y$</span>) work to find the inverse? It seems logical to me but I cannot come up with a coherent argument as to why.</p>
| 雨が好きな人 | 438,600 | <p>When we find the inverse in this way, we are not really ‘interchanging variables’. We’re just rearranging to find the input in terms of the output (when usually we have the output defined explicitly in terms of the input), and then relabelling to make it clearer. It just happens that we usually use <span class="math-container">$y$</span> to refer to the output of a function and <span class="math-container">$x$</span> to refer to the input.</p>
<p>An inverse function <span class="math-container">$f^{-1}$</span> is just a function that ‘undoes’ the operation of <span class="math-container">$f$</span>, so when we find the input in terms of the output we are simply finding what operations are required to turn the output of <span class="math-container">$f$</span> back into the input, which we often call <span class="math-container">$x$</span>.</p>
<p>As a simple example, suppose <span class="math-container">$y = 3x + 2$</span>. Then <span class="math-container">$\frac{y - 2}{3} = x$</span>.</p>
<p>We didn’t interchange any variables and we have found the inverse. However, if we want to write it in the ‘standard’ <span class="math-container">$y = f(x)$</span> way, we can relabel to find <span class="math-container">$y = \frac{x-2}{3}$</span>.</p>
<p>This is all quite loose because we are not really manipulating functions—just expressions. Really, we are taking <span class="math-container">$f(x)$</span> in terms of <span class="math-container">$x$</span> and finding <span class="math-container">$x$</span> in terms of <span class="math-container">$f(x)$</span>. But you can see hopefully more intuitively why this technique works when we have relatively simple functions defined like this.</p>
<p><span class="math-container">$$\rule{4cm}{0.4pt}$$</span></p>
<p>Since I feel like I have tried to fully address your concern about ‘interchanging variables’, let me try to give some extra reasoning about why we can ‘rearrange’ an equation like we did above to find the inverse.</p>
<p>Any function that we can write like we did above can be expressed as a composition of intermediate functions. Consider what happens to the input <span class="math-container">$x$</span> in our example above: first it gets multiplied by <span class="math-container">$3$</span> and then it gets added to <span class="math-container">$2$</span>.</p>
<p>As I said, the inverse is just an ‘undoing’ function. It takes us back from the output to the input again. When we rearrange an equation, what we are doing is repeatedly applying functions to both sides.</p>
<p>When we find the inverse using our rearranging technique...</p>
<p><span class="math-container">$$y = 3x + 2$$</span></p>
<p><span class="math-container">$$\text{(apply $f(\lambda) = \lambda - 2$ to both sides)}$$</span></p>
<p><span class="math-container">$$y - 2 = 3x$$</span></p>
<p><span class="math-container">$$\text{(apply $f(\lambda) = \frac{1}{3}\lambda$ to both sides)}$$</span></p>
<p><span class="math-container">$$\frac{y - 2}{3} = x$$</span></p>
<p>...we are just applying the inverse operations of the function, in the reverse order, which of course will undo the operation of the function.</p>
<p>This is also why functions that are not one-to-one do not have inverse functions—we can’t undo their operation as we did here because we don’t know which input lead to a specific output.</p>
|
3,146,161 | <blockquote>
<p>Find particular solution of <span class="math-container">$\dfrac{dx}{dy} +x\cot y =y\cot y$</span> given <span class="math-container">$x=0$</span> when <span class="math-container">$y= π/2$</span>.</p>
</blockquote>
<p>Please help me figure this out
I can't separate them out in order to integrate</p>
| Lucas | 444,015 | <p>The function <span class="math-container">$e^{x^{2}}$</span> is continuous increasing on <span class="math-container">$[0, \infty]$</span>. Take
<span class="math-container">$$F(t) = \int_{0}^{t}e^{x^{2}}dx.$$</span>
In each compact interval <span class="math-container">$[0,a]$</span>, <span class="math-container">$e^{x^{2}}$</span> attains a minimum <span class="math-container">$m_{a}$</span>. If <span class="math-container">$m_{a_{n}} \leq e^{x^{2}}$</span>, then <span class="math-container">$m_{a_{n}}a_{n} \leq F(a_{n})$</span> on <span class="math-container">$[0,a_{n}]$</span>. Also,
<span class="math-container">$$\infty = \lim_{n \to \infty}m_{a_{n}}a_{n} \leq \lim_{n \to \infty}F(a_{n})$$</span></p>
|
1,683,414 | <p>$$\begin{cases}
x^2 = yz + 1 \\
y^2 = xz + 2 \\
z^2 = xy + 4
\end{cases}
$$</p>
<p>How to solve above system of equations in real numbers? I have multiplied all the equations by 2 and added them, then got $(x - y)^2 + (y - z)^2 + (x - z)^2 = 14$, but it leads to nowhere.</p>
| Ross Millikan | 1,827 | <p>Now I would note that $(\pm 3)^2+(\pm 2)^2+(\pm 1)^2=14$. Since $(x-y)+(y-z)=(x-z)$ we can't assign the signs and values arbitrarily. For example, from $x-y=1, y-z=2, x-z=3$ I get $x=1,y=0,z=-2$ and from $x-y=-1, y-z=-2, x-z=-3$ I get $x=-1,y=0,z=2$ The other possibilities do not give solutions.</p>
|
1,575,107 | <p>$$L^{-1}\frac{4s}{(s-6)^{3}}$$
$$4L^{-1}\frac{s}{s^{3}}|s=s-6$$
$$4L^{-1}\frac{1}{s^{2}}|s=s-6$$
$$4L^{-1}\frac{1!}{s^{1+1}}|s=s-6$$
$$4te^{6t}$$</p>
<p>Is this correct? symbolab and Wolfram are giving me different answers...</p>
| Ron Gordon | 53,268 | <p>The ILT may be computed via the residue theorem as follows:</p>
<p>$$\frac{4}{2!} \frac{d^2}{ds^2} \left [s e^{s t} \right ]_{s=6} = 2 \frac{d}{ds} \left [(1+s t) e^{s t} \right ]_{s=6} = 2 \left [(t (1+s t) + t) e^{s t} \right ]_{s=6} = (4 t+12 t^2) e^{6 t}$$</p>
<p>You seem to be missing the $t^2$ term.</p>
|
815,739 | <p>Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?</p>
| Lucian | 93,448 | <p>Since $\mathbb R^a\times\mathbb R^b\cong\mathbb R^{a+b}$, it would follow that $\mathbb R^a\times\mathbb R^0\cong\mathbb R^a$, and therefore $\mathbb R^0\cong\{\varnothing\}$. But I have to admit I never actually saw this notation being used anywhere. As for $\mathbb R^{-n}$, we would have $\mathbb R^n\times$ $\times\mathbb R^{-n}\cong\mathbb R^0\cong\{\varnothing\}$, and likewise, $\underbrace{\mathbb R^{\frac1n}\times\mathbb R^{\frac1n}\times\ldots\times\mathbb R^{\frac1n}}_{n\text{ times}}\cong\mathbb R$. But, again, I am unaware of such abstractions appearing anywhere in literature.</p>
|
1,116,435 | <p>How do I get the value of </p>
<p>$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }?$$ </p>
<p>I need the steps without using L'hospital.</p>
| Math-fun | 195,344 | <p>Firs note that $ \displaystyle \tan x - \cot x = - 2 \frac{\cos 2x}{\sin 2x}$. Then consider a change of variable $\displaystyle u=x-\frac{\pi}{4}$ to obtain </p>
<p>$$ \lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }= \lim_{u \rightarrow 0 } [\frac{\sin 2u}{u}\times \frac{2}{\cos2u}]=\lim_{u \rightarrow 0 } [\frac{\sin 2u}{u}] \times \lim_{u \rightarrow 0 } [ \frac{2}{\cos2u}].$$
Now note that $ \displaystyle \lim_{u \rightarrow 0 } \frac{\sin 2u}{u} = 2 \lim_{u \rightarrow 0 } \frac{\sin 2u}{2 u} = 2 \lim_{t \rightarrow 0 } \frac{\sin t}{t} =2$, which has received some good bit of attention <a href="https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1">here</a>.</p>
<p>The limit will hence be $\displaystyle 2 \times 2 =4$.</p>
|
1,027,235 | <p>Let N = 12345678910111213141516171819. How can I use modular arithmetic to show that N is (or isn't) divisible by 11? In general, how can I apply modular arithmetic to determine the divisibility of an integer by a smaller integer? I am finding modular arithmetic very confusing and unintuitive. I can understand "simple" modular arithmetic like the 24-hour day etc. but when it comes down to finding the modulo of high raised powers, or checking divisibility of large integers, I am totally lost. </p>
| Mike Earnest | 177,399 | <p>First, since $10\equiv -1$ (mod $11$), notice that $10^k\equiv (-1)^k$ (mod $11$). Then, given your number, which can be written more generally as
$$
d_n10^n+d_{n-1}10^{n-1}+\dots+d_1\cdot10+d_0,
$$
consider what the above expression is (mod $11$). When the above is looked at "with mod 11 goggles on," all of the instances of $10^k$ can be replaced with $(-1)^k$, which leaves an alternating sum of digits. </p>
|
3,305,677 | <p>I am learning Asymptotic complexity of functions from CLRS. I know that exponentiation functions like <span class="math-container">$a^n$</span>,<span class="math-container">$(a>0)$</span> are faster than <span class="math-container">$n!$</span> But what about <span class="math-container">$a^{a^n}$</span> vs <span class="math-container">$n!$</span> How do they compare? A proof would really help me understand the concept. Thanks!</p>
| epi163sqrt | 132,007 | <blockquote>
<p>Applying <span class="math-container">$\ln\circ\ln$</span> we obtain
<span class="math-container">\begin{align*}
\ln\left(\ln\left(a^{a^n}\right)\right)
&=\ln\left(a^n\ln a\right)\\
&=\ln a^n+\ln\left(\ln\left(a\right)\right)\\
&=n\ln a+\ln\left(\ln\left(a\right)\right)\\
&\sim n\ln a\tag{1}
\end{align*}</span></p>
</blockquote>
<p>Recalling Stirlings formula <span class="math-container">$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$</span> we obtain</p>
<blockquote>
<p><span class="math-container">\begin{align*}
\ln\left(\ln\left(n!\right)\right)&\sim\ln\left(\ln\left(\left(\frac{n}{e}\right)^n\sqrt{2\pi n}\right)\right)\\
&=\ln\left(\ln\left(\frac{n}{e}\right)^n+\ln\sqrt{2\pi n}\right)\\
&=\ln\left(n\ln n-n+\ln\sqrt{2\pi}+\ln n\right)\\
&\sim \ln\left(n\ln n\right)\tag{2}
\end{align*}</span></p>
</blockquote>
<p>We conclude from (1) and (2)</p>
<blockquote>
<p><span class="math-container">\begin{align*}
\ln\left(n\ln n\right)=\mathcal{O}\left(n\ln a\right)
\end{align*}</span>
from which
<span class="math-container">\begin{align*}
\ln\left(\ln\left(n!\right)\right)=\mathcal{O}\left(\ln\ln a^{a^n}\right)
\end{align*}</span>
and finally
<span class="math-container">\begin{align*}
\color{blue}{n!=\mathcal{O}\left(a^{a^n}\right)}
\end{align*}</span>
follows.</p>
</blockquote>
|
1,590,817 | <p>I was hoping someone could explain to me how to prove a sequence is Cauchy. I've been given two definitions of a Cauchy sequence:</p>
<p>$\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n,m> N$ $\Rightarrow |a_n - a_m| ≤ \epsilon$</p>
<p>and equivalently $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n> N$ $\Rightarrow |a_{n+p} - a_n| ≤ \epsilon$, $\forall p \in \mathbb{N}$</p>
<p>I understand that proving a sequence is Cauchy also proves it is convergent and the usefulness of this property, however, it was never explicitly explained how to prove a sequence is Cauchy using either of these two definitions. I'd appreciate it if someone could explain me how to prove a sequence is Cauchy perhaps $a_n = \sqrt{n+1} - \sqrt{n}$ ? or another example just for me to grasp the concept.</p>
| Ian Cavey | 293,726 | <p>As an easier example of how to apply the definition of a Cauchy sequence, define the sequence $\{\frac{1}{n}\}$. Given any $\epsilon>0$, you would like to find an $N$ such that for any $n,m>N$, $\left|\frac{1}{n}-\frac{1}{m}\right|<\epsilon$. This would certainly be the case if $\frac{1}{n},\frac{1}{m}<\frac{\epsilon}{2}$ since $\left|\frac{1}{n}-\frac{1}{m}\right|\leq\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|$. Therefore you need to force $n,m>\frac{2}{\epsilon}$. Conveniently enough, you are allowed to choose whichever $N$ you like, so choosing $N>\frac{2}{\epsilon}$ will do the trick. The above constitutes the work you do beforehand, now the proof.</p>
<p>Claim: The sequence $\{\frac{1}{n}\}$ is Cauchy.</p>
<p>Proof: Let $\epsilon>0$ be given and let $N>\frac{2}{\epsilon}$. Then for any $n,m>N$, one has $0<\frac{1}{n},\frac{1}{m}<\frac{\epsilon}{2}$. Therefore, $\epsilon>\frac{1}{n}+\frac{1}{m}=\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|\geq\left|\frac{1}{n}-\frac{1}{m}\right|$. Thus, the sequence is Cauchy as was to be shown.</p>
|
2,094,243 | <p>Given the norm $||(x,y)|| = 2|x| +\frac{1}{3}|y|$.
Sketch the open ball at the on the origin $(0,0)$, and radius $1$.</p>
<p>I understand that the sketch of an open ball withina set looks like the image attached, <a href="https://i.stack.imgur.com/j9HN4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j9HN4.jpg" alt="enter image description here"></a> in a general case, but have no idea how to sketch one applying the above norm to the situation. </p>
<p>i understand that in the case of $B_{r}(a)=\{x \in X | d(x,a) < r\}$ in this case, $a = (0,0)$, and $r = 1$. Could someone please help as to how to sketch it?</p>
<p>Thanks</p>
| Mercy King | 23,304 | <p>The open ball centered at the origin, with radius $1$, is the region containing the center $(0,0)$ and bounded by the line-segments
\begin{eqnarray}
2x+\dfrac13y&=&1, 0 \le x\le 0.5\\
-2x+\dfrac13y&=&1, -0.5\le x\le 0\\
-2x-\dfrac13y&=&1, -0.5\le x\le 0\\
2x-\dfrac13y&=&1, 0\le x\le 0.5
\end{eqnarray}
In other words, your open ball is the region bounded by the polygon with vertices at
$$
(0.5,0), (0,3), (-0.5,0), (0,-3).
$$
<a href="https://i.stack.imgur.com/eWJ8p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eWJ8p.png" alt="enter image description here"></a></p>
|
743,819 | <p>I'm studying for my exam and I came across the following draw without replacement problem :</p>
<blockquote>
<p><span class="math-container">$N$</span> boxes filled with red and green balls.
The box <span class="math-container">$r$</span> contains <span class="math-container">$r-1$</span> red balls and <span class="math-container">$N-r$</span> green balls.
We pick a box at random and we take <span class="math-container">$2$</span> random balls inside it, without putting them back.</p>
<p><strong>1)</strong> What is the probability that the second ball is green ?</p>
<p><strong>2)</strong> What is a probability that the second ball is green, knowing the first one is green.</p>
</blockquote>
<p>I don't know where to start, all those dependance (to <span class="math-container">$r$</span> and <span class="math-container">$N$</span>) are blowing my mind.</p>
<p>I don't know if I should concider it as a Binomial law (with Bernoulli : ball = green, <span class="math-container">$n=2, p = ?$</span>) or go with the formula <span class="math-container">$$p(X=k)=\frac{C_{m}^{k} C_{N-m}^{n-k}}{C_{N}^{n}}$$</span>
or something else...</p>
<p>Could someone advise me ?</p>
| Jared | 138,018 | <p>All of the boxes contain $N - 1$ balls. This is just a complicated conditional probability problem. Lets look at a single box with $r$ red balls and $g$ green balls. What would the probability be of getting green on the second? Well it depends on whether or not you draw a red or green first. If you draw a red first, then there are $\left.p(\text{second green } \right| \text{ first red}) = \frac{g}{r + g - 1}$. However, if you draw a green ball first then you have one less green to choose from giving: $\left.p(\text{second green } \right| \text{ first green}) = \frac{g - 1}{g + r - 1}$. So what are the chances of each condition happening? $p(\text{first red}) = \frac{r}{g + r}$ and $p(\text{first green}) = \frac{g}{r + g}$. Therefore we can finally write:</p>
<p>\begin{align}
p(\text{second green}) =& \left.p(\text{second green } \right| \text{ first red})p(\text{first red}) + \left.p(\text{second green } \right| \text{ first green})p(\text{first green})\\
=& \frac{r}{r + g}\frac{g}{r+g-1} + \frac{g}{r + g}\frac{g-1}{r+g-1} = \frac{g(r + g - 1)}{(r + g)(r + g - 1)} = \frac{g}{r + g}
\end{align}</p>
<p>Not surprising that drawing the second green has just as good of a chance of being green as the first pick.</p>
<p>Therefore for each of the $N$ boxes you need to compute $p(\text{second green})$ (which is just the probability of drawing a green on the first try). Now the condition is that we choose box $r$ which has $p(\text{second green}) = p(\text{first green}) = \frac{N - r}{N - 1}$. The probability of choosing box $r$ among $N$ boxes is just $\frac{1}{N}$ which gives:</p>
<p>$$
p(\text{second green}) = \sum_1^N \frac{1}{N}\frac{N - r}{N - 1} = \frac{1}{N(N - 1)}\sum_1^r (N - r)
$$</p>
<p>The first sum is very easy (you're just summing the same number, $N$, $N$ times) $\sum_1^N N = N\cdot N = N^2$. The second part is easy if you remember the sum of the first $n$ consecutive integers is $\sum_1^n i = \frac{n(n + 1)}{2}$. So this gives:</p>
<p>$$
p(\text{green}) = \frac{N^2 - \frac{N(N + 1)}{2}}{N(N - 1)} = \frac{2N^2 - N^2 - N}{2N(N - 1)} = \frac{N^2 - N}{2\left(N^2 - N\right)} = \frac{1}{2}
$$</p>
<p>For part $2$), we actually already computed that above: $\left.p(\text{second green }\right|\text{ first green}) = \frac{g - 1}{g + r - 1}$. But now you need to sum over the condition that it could be <em>any</em> of the $N$ boxes (<i><b>edit:</b> However, the last box, box $N$, has $0$ green balls (and thus seeing green first means it definitely wasn't this box. So we should only sum over the first $N - 1$ boxes and divide by $N - 1$, not $N$.</i>):</p>
<p>\begin{align}
\left.p(\text{second green }\right|\text{ first green}) =& \sum_1^{N - 1}
\frac{1}{N - 1}\frac{N - r - 1}{N - 2} \\
=& \frac{N(N - 1) - (N - 1) - \frac{N(N - 1)}{2}}{N(N - 2)} \\
=& \frac{2N(N - 1) - 2(N - 1) - N(N - 1)}{2(N - 1)(N - 2)}\\
=& \frac{N(N - 1) - 2(N - 1))}{2(N - 1)(N - 2)} \\
=& \frac{(N - 1)(N - 2)}{2(N - 1)(N - 2)} \\
=& \frac{1}{2}
\end{align}</p>
<p>This is only valid for $N > 2$ (since if $N = 1$ there are <i>no</i> balls in each box and if $N = 2$ there is only <i>one</i> ball in each box). This result just confirms that drawing balls are independent events.</p>
|
1,521,649 | <p>I need to show that at most finitely many terms of this sequence are greater than or equal to $c$. </p>
<p>I don't know if it is the wording of the problem but I don't know what this is asking me to do. Help on this would be amazing! And thank you in advance.</p>
| John Douma | 69,810 | <p>$b\ln a=b\int_1^a\frac{1}{x}dx=\int_1^a\frac{b}{x}dx$</p>
<p>Let $u=x^b$</p>
<p>Then $du= bx^{b-1}dx$ and $\frac{du}{u}=\frac{b}{x}dx$</p>
<p>Therefore, $\int_1^a\frac{b}{x}dx = \int_1^{a^b}\frac{du}{u}=\ln {a^b}$</p>
|
1,521,649 | <p>I need to show that at most finitely many terms of this sequence are greater than or equal to $c$. </p>
<p>I don't know if it is the wording of the problem but I don't know what this is asking me to do. Help on this would be amazing! And thank you in advance.</p>
| BLAZE | 144,533 | <p>Assuming knowledge of </p>
<p>$(a^m)^n=a^{mn}\tag{1}$</p>
<p>Now let $a^m=b$ and $a^n = c$</p>
<p>Then from $(1)$</p>
<p>we have $b^n=(a^m)^n=a^{mn}$</p>
<p>so $\log_a(b^n)=mn$ </p>
<p>therefore $n\log_a b=\log_a(b^n)$</p>
<p>Hence </p>
<p>$\color{blue}{\fbox{$\log_a (b^n) = n\log_a b$}}$</p>
<p>I did this for general logarithms, as I think it's good practice to generalize proofs, but the above proof works exactly the same for natural logarithms (which you were asking about) also. </p>
|
1,710,517 | <blockquote>
<p>Show that 13 is the largest prime which divides two consecutive terms of $n^2 + 3$.</p>
</blockquote>
<p>The integers are $39$ and $52$. First of all, I set the variable for the number as $k$. So, $k|n^2 +3$ and $k|n^2 + 2n+ 4$ which imply that $k|2n+1$. $n=6$ over here. And the fact that 13 is the largest 'prime' makes me feel it is hard to prove. That's all I have managed to get. I need a few hints to set me in the right direction. Thanks.</p>
| lulu | 252,071 | <p>Let $a_n=n^2+3$. If $p|a_n$ then $n^2=-3\pmod p$. </p>
<p>A quick calculation shows that $a_{n+1}-a_n=2n+1$ so if $p$ also divides $a_{n+1}$ we must have $2n=-1\pmod p$. Square this to see that it implies $4n^2=1\implies -12=1 \pmod p$. Thus, $p|13$ and we are done.</p>
|
453,212 | <p>Consider a number Q in a made up base system:</p>
<p>The base system is as follows:</p>
<p>It encodes a number as a sum of odd numbers:</p>
<p>1 3 5 7 9 ...</p>
<p>If the number can be expressed as a sum of unique odds. For example, the number 16 is expressed as:</p>
<p>1110 = 7 + 5 + 3 + 1</p>
<p>The system is also redundant as 16 can also be expressed as:</p>
<p>11000</p>
<p>My question:</p>
<p>Given a natural number u, how can u be expressed in this system quickly if u can be expressed in the system, quickly.</p>
| Zev Chonoles | 264 | <p>If $u$ is odd, write $u$ as the following sum of odd numbers:
$$u$$
If $u$ is even (and $\geq 2$), write $u$ as the following sum of odd numbers:
$$(u-1)+1$$
(Converting into a string of ones and zeros is straightforward.)</p>
|
1,479,822 | <p>$f\cdot g$ is Lebesgue integrable, g is Lebesgue integrable, can we deduce that f is Lebesgue integrable? </p>
<p>$f\cdot g$ is integrable, g is integrable, can we deduce that f is finite a.e.? </p>
| Silvia Ghinassi | 258,310 | <p>For a nontrivial example, take $f=\frac{1}{x}\chi_{[1,+\infty)}$ and $g=\frac{1}{x^2}\chi_{[1,+\infty)}$. Then $fg$ is integrable, $g$ is integrable, but $f$ is not.</p>
<p>As for the second question, in general you can't conclude anything, as pointed out already.</p>
|
162,611 | <p>I am working with the square-roots of square symmetric matrices. The answers are to be binary symmetric matrices.</p>
<p>If we take the matrix $$M = \begin{pmatrix}1&1&1&0&0&0&0\\1&0&0&1&1&0&0\\1&0&0&0&0&1&1\\0&1&0&0&1&0&1\\0&1&0&1&0&1&0\\0&0&1&0&1&1&0\\0&0&1&1&0&0&1 \end{pmatrix},$$ then $M^2$ will yield a $7 \times 7$ matrix with $3$ down the main diagonal and $1$ elsewhere.Once we have the matrix $M^2$ then we want to take the square root of it in order to get M.Now this is the simplest of all examples that I can give. M is the incidence matrix of a symmetric balanced incomplete block design. This design has 7 varieties, 7 blocks,each variety occurs in 3 blocks,each block contains 3 varieties, each pair of varieties occurs in exactly one block. If you are familiar with block designs then the parameters are (v,b,r,k,l)=(7,7,3,3,1) known as the fano plane.</p>
| Will Jagy | 10,400 | <p>I believe what you have found is what happens when you take the Fano plane, that is a finite projective plane of "order" 2, and write it as a bipartite graph on 7 plus 7 vertices. I think it is an artifact of small dimension that you are able to have your "binary" matrix also be symmetric. There is a projective plane of order 3, see <a href="http://en.wikipedia.org/wiki/Finite_projective_plane#Finite_projective_planes" rel="nofollow">FANO</a>.<br>
I think you can begin with a 13 by 13 symmetric matrix with 4's on the main diagonal and 1's elsewhere, and write it as $P P^T$ where $P$ is "binary" but not necessarily symmetric. I will see if I can construct such a $P$ myself. Meanwhile, it would appear to be impossible to do this with the 43 by 43 matrix with 6's on the main diagonal and 1's elsewhere.</p>
<p>The reason that it is impossible to do this with a 43 by 43 matrix with 7's rather than 6's on the main diagonal and 1's elsewhere is that the design (43,43,7,7,1) does not exist. This design is known as a symmetric balanced incomplete block design.</p>
<p>EDIT: if we take
$$ P \; = \;
\left( \begin{array}{ccccccccccccc}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0
\end{array}
\right),
$$
then $P \; P^T$ is the 13 by 13 matrix with 4 on the main diagonal and 1 elsewhere. I do not see how to do this with symmetric $P.$ Proof of impossibility would be another matter. </p>
<p>EDITEDIT: Note that $P^T \; P = P \; P^T.$</p>
|
28,456 | <p>I built a <code>Graph</code> based on the permutations of city's connections from :</p>
<pre><code>largUSCities =
Select[CityData[{All, "USA"}], CityData[#, "Population"] > 600000 &];
uScityCoords = CityData[#, "Coordinates"] & /@ largUSCities;
Graph[#[[1]] -> #[[2]] & /@ Permutations[largUSCities, {2}] ,
VertexCoordinates -> Reverse[uScityCoords, 2], VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},ImageSize -> 650]
</code></pre>
<p>It looks like this:
<img src="https://i.stack.imgur.com/1ea6c.png" alt="graph"></p>
<p>My question, is there any way to have the Graph like this?
<img src="https://i.stack.imgur.com/ajwOT.png" alt="enter image description here"></p>
| cormullion | 61 | <h2>Revised answer</h2>
<p>This uses the connectivity between states to create the graph, and uses the coordinates of the center of each state rather than the cities. I couldn't find a way to get these easily from <em>Mathematica</em> or from <em>WolframAlpha</em> (I'm no Harry Potter, and failed to discover the correct incantation for the latter). But I found a table somewhere:</p>
<pre><code>stateConnections = {{"NV", "CA", "AZ", "UT", "ID", "OR"}, {"OR", "CA",
"NV", "ID", "WA"}, {"TX", "OK", "LA", "NM", "AR"}, {"DC", "VA",
"MD"}, {"FL", "GA", "AL"}, {"RI", "MA", "CT"}, {"SC", "GA",
"NC"}, {"WA", "OR", "ID"}, {"CA", "NV", "OR", "AZ"}, {"CT", "RI",
"MA", "NY"}, {"DE", "MD", "PA", "NJ"}, {"LA", "TX", "MS",
"AR"}, {"MI", "IN", "OH", "WI"}, {"ND", "SD", "MN", "MT"}, {"NH",
"ME", "VT", "MA"}, {"NJ", "NY", "PA", "DE"}, {"VT", "NH", "MA",
"NY"}, {"AL", "GA", "MS", "TN", "FL"}, {"AZ", "CA", "NM", "UT",
"NV"}, {"IN", "OH", "MI", "IL", "KY"}, {"KS", "OK", "CO", "MO",
"NE"}, {"MD", "DE", "PA", "VA", "WV"}, {"MN", "WI", "IA", "SD",
"ND"}, {"MS", "AL", "LA", "AR", "TN"}, {"MT", "ID", "WY", "SD",
"ND"}, {"NC", "SC", "VA", "TN", "GA"}, {"NM", "TX", "AZ", "CO",
"OK"}, {"WI", "IL", "MI", "IA", "MN"}, {"GA", "FL", "SC", "NC",
"AL", "TN"}, {"IL", "IA", "WI", "IN", "KY", "MO"}, {"MA", "VT",
"NH", "NY", "RI", "CT"}, {"NV", "CA", "AZ", "UT", "ID",
"OR"}, {"NY", "NJ", "VT", "PA", "MA", "CT"}, {"OH", "IN", "WV",
"PA", "KY", "MI"}, {"UT", "CO", "WY", "ID", "NV", "AZ"}, {"VA",
"WV", "MD", "NC", "TN", "KY"}, {"WV", "VA", "OH", "PA", "MD",
"KY"}, {"AR", "TX", "LA", "OK", "MO", "TN", "MS"}, {"CO", "UT",
"WY", "NM", "NE", "KS", "OK"}, {"IA", "IL", "WI", "MN", "SD",
"NE", "MO"}, {"ID", "WA", "OR", "NV", "UT", "WY", "MT"}, {"NE",
"KS", "CO", "WY", "SD", "IA", "MO"}, {"OK", "TX", "CO", "KS",
"NM", "AR", "MO"}, {"PA", "WV", "DE", "MD", "NJ", "NY",
"OH"}, {"SD", "ND", "MT", "WY", "NE", "IA", "MN"}, {"WY", "MT",
"ID", "UT", "CO", "NE", "SD"}, {"KY", "IL", "MO", "TN", "VA",
"WV", "OH", "IN"}, {"MO", "IA", "NE", "KS", "OK", "AR", "TN",
"KY", "IL"}, {"TN", "KY", "MO", "AR", "MS", "AL", "GA", "NC",
"VA"}, {"ME", "NH"}};
stateData = {"AK,61.3850,-152.2683", "AL,32.7990,-86.8073",
"AR,34.9513,-92.3809", "AZ,33.7712,-111.3877",
"CA,36.1700,-119.7462", "CO,39.0646,-105.3272",
"CT,41.5834,-72.7622", "DC,38.8964,-77.0262",
"DE,39.3498,-75.5148", "FL,27.8333,-81.7170",
"GA,32.9866,-83.6487", "HI,21.1098,-157.5311",
"IA,42.0046,-93.2140", "ID,44.2394,-114.5103",
"IL,40.3363,-89.0022", "IN,39.8647,-86.2604",
"KS,38.5111,-96.8005", "KY,37.6690,-84.6514",
"LA,31.1801,-91.8749", "MA,42.2373,-71.5314",
"MD,39.0724,-76.7902", "ME,44.6074,-69.3977",
"MI,43.3504,-84.5603", "MN,45.7326,-93.9196",
"MO,38.4623,-92.3020", "MS,32.7673,-89.6812",
"MT,46.9048,-110.3261", "NC,35.6411,-79.8431",
"ND,47.5362,-99.7930", "NE,41.1289,-98.2883",
"NH,43.4108,-71.5653", "NJ,40.3140,-74.5089",
"NM,34.8375,-106.2371", "NV,38.4199,-117.1219",
"NY,42.1497,-74.9384", "OH,40.3736,-82.7755",
"OK,35.5376,-96.9247", "OR,44.5672,-122.1269",
"PA,40.5773,-77.2640", "RI,41.6772,-71.5101",
"SC,33.8191,-80.9066", "SD,44.2853,-99.4632",
"TN,35.7449,-86.7489", "TX,31.1060,-97.6475",
"UT,40.1135,-111.8535", "VA,37.7680,-78.2057",
"VT,44.0407,-72.7093", "WA,47.3917,-121.5708",
"WI,44.2563,-89.6385", "WV,38.4680,-80.9696",
"WY,42.7475,-107.2085"} ;
stateAbbreviations = Union[Flatten[stateConnections]];
stateToNumber =
MapThread[
Rule, {stateAbbreviations, Range[Length[stateAbbreviations]]}];
numberToState =
MapThread[
Rule, {Range[Length[stateAbbreviations]], stateAbbreviations}];
allConnections =
Flatten[Function[e, Map[UndirectedEdge[First[e], #] &, Rest[e]]] /@
stateConnections];
connections = Union[Sort /@ allConnections];
stateCenters =
First[StringSplit[#, ","]] ->
ToExpression /@ RotateLeft @ Rest[StringSplit[#, ","]] & /@
stateData;
stateCoords = (# & /@ stateAbbreviations) /. stateCenters;
temp = Graph[connections /. stateToNumber];
vertexCoordinates = stateCoords[[VertexList[temp]]];
g = Graph[connections /. stateToNumber,
VertexCoordinates -> vertexCoordinates,
VertexLabels -> numberToState,
VertexShapeFunction -> "Square",
VertexSize -> 3,
VertexLabelStyle -> Directive[Black, 12]];
Show[Graphics[{LightGray, CountryData["USA", "Polygon"]}], g,
ImageSize -> 700]
</code></pre>
<p><img src="https://i.stack.imgur.com/53ZwY.png" alt="graph of the usa" /></p>
<p>Apparently the order of the vertices is required from the graph before you can draw the vertices at the right coordinates on the graph - hence the weird use of <code>temp = Graph[connections /. stateToNumber]</code> before creating the graph again for real.</p>
|
28,456 | <p>I built a <code>Graph</code> based on the permutations of city's connections from :</p>
<pre><code>largUSCities =
Select[CityData[{All, "USA"}], CityData[#, "Population"] > 600000 &];
uScityCoords = CityData[#, "Coordinates"] & /@ largUSCities;
Graph[#[[1]] -> #[[2]] & /@ Permutations[largUSCities, {2}] ,
VertexCoordinates -> Reverse[uScityCoords, 2], VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},ImageSize -> 650]
</code></pre>
<p>It looks like this:
<img src="https://i.stack.imgur.com/1ea6c.png" alt="graph"></p>
<p>My question, is there any way to have the Graph like this?
<img src="https://i.stack.imgur.com/ajwOT.png" alt="enter image description here"></p>
| PlatoManiac | 240 | <p>Another trivial way to keep the cities connected and at the same time reduce the number of edges in the graph.</p>
<pre><code><< ComputationalGeometry`;
rule = MapThread[#1 -> #2 &, {uScityCoords, largUSCities}];
graph = (Rest@PlanarGraphPlot[uScityCoords][[1, 2]] /.
Line[{a_, b_}] -> UndirectedEdge[a, b]) /. rule;
gorg = Graph[graph, VertexCoordinates -> Reverse[uScityCoords, 2],
EdgeStyle -> Directive[Opacity[.6], Gray, Thick],
VertexSize -> 0.8,Prolog -> {LightRed, CountryData["USA", "FullPolygon"]},
ImageSize -> 450, AspectRatio -> .5];
convexhull = ConvexHull[uScityCoords];
conv =PlanarGraphPlot[Reverse[uScityCoords, 2], convexhull,LabelPoints -> False];
edge = EdgeList[g];
g = gorg;
For[i = 1, i <= 1500, i++,
ng = EdgeDelete[g, RandomChoice@edge];
g = If[ConnectedGraphQ[ng] === True, edge = EdgeList[ng]; ng, g];
];
Row@{gorg, Spacer[60], Show[g, conv, AspectRatio -> .5]}
</code></pre>
<p><img src="https://i.stack.imgur.com/JWHzw.gif" alt="enter image description here"></p>
|
4,263,631 | <p>so i have question about existence of function <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> such <span class="math-container">$f$</span> is not the pointwise limit of a sequence of continuous functions <span class="math-container">$\mathbb{R} \to \mathbb{R}$</span>.</p>
<p>i'm created a family of continuous functions <span class="math-container">$k_i:\mathbb{R} \to \mathbb{R}$</span> for any function <span class="math-container">$f:\mathbb{R} \to \mathbb{R}$</span> such that <span class="math-container">$\lim_{n \to \infty} k_n(x)=f(x)$</span> , how ever i am pretty sure my construction have a flaw in it but i could not understand why it's wrong can some one tell me what am i doing wrong , because it seems very unrealistic to be able to do it for any <span class="math-container">$f$</span>.</p>
<p>here is my construction steps:</p>
<p>1.pick any arbitrary <span class="math-container">$x,y \in \mathbb{R}$</span> such <span class="math-container">$x<y$</span> , and connect <span class="math-container">$f(x)$</span> to <span class="math-container">$f(y)$</span>.</p>
<p>2.pick any arbitrary <span class="math-container">$z \in \mathbb{R}-\{x,y\}$</span> , if <span class="math-container">$f(y)<f(z)$</span> connect <span class="math-container">$f(z)$</span> to <span class="math-container">$f(y)$</span> and if <span class="math-container">$f(z)<f(x)$</span> connect <span class="math-container">$f(z)$</span> to <span class="math-container">$f(x)$</span>, if <span class="math-container">$f(x)<f(z)<f(y)$</span> , connect <span class="math-container">$f(x)$</span> to <span class="math-container">$f(z)$</span> and <span class="math-container">$f(z)$</span> to <span class="math-container">$f(y)$</span>.</p>
<p>now we are again doing this uncountable times with other elemnts in <span class="math-container">$\mathbb{R}-\{x,y,z\}$</span> and do the step 2 for that element over all choosed elements by uncountable number of comparisions (after uncountable number of iteration we will have uncountable number of picked elements of <span class="math-container">$\mathbb{R}$</span>).</p>
<p>call set of picked elements <span class="math-container">$S$</span>, when we pick an other element from <span class="math-container">$\mathbb{R}-S$</span> , like <span class="math-container">$t$</span> we will consider <span class="math-container">$ x_1= \max \{ x ; x \in S \land x<t \}$</span> and <span class="math-container">$x_2 = \min \{ x ; x \in S \land x>t \}$</span> and we connect <span class="math-container">$f(x_1)$</span> to <span class="math-container">$f(t)$</span> and then <span class="math-container">$f(t)$</span> to <span class="math-container">$f(x_2)$</span> and call all the connected line with <span class="math-container">$S \cup {t}$</span> function <span class="math-container">$k_t$</span>. we will do this uncountable number of times.</p>
<p>i'm pretty sure there is a flaw in my argument , but can you please help me to find it.i really appreciate your kindness and support.</p>
| Lázaro Albuquerque | 85,896 | <p>Consider the cover <span class="math-container">$\cup [k, k+1] \times [0, \frac{1}{k^2}]$</span>. The sum of these rectangle's measures is <span class="math-container">$\sum \frac{1}{k^2} < \infty$</span>.</p>
|
839,431 | <p>Can someone be kind enough to show me the steps to integrate this, I'm sure it's by parts but how do I split up the sin part?
$$x\sin(1+2x)$$</p>
| Anurag A | 68,092 | <p>Let $u=x$ and $dv =\sin (1+2x) \, dx$
Then $v=-\frac{1}{2}\cos(1+2x)$. Now perform integration by parts.</p>
|
253,746 | <p>I am currently aware of the following two versions of the global Cauchy Theorem. Which one is stronger?</p>
<p>1.)If the region $U$ is simply connected, then for every closed curve contained therein, the integral of the holomorphic function $f$ defined on $U$ over the curve is zero.</p>
<p>2.) If the domain $D$ is an arbitrary open set, then for every closed curve contained therein with the property that the index is zero for points outside $D$, the integral of the holomorphic function $f$ defined on $D$ is zero.
Thanks in advance.</p>
| Christian Blatter | 1,303 | <p>If $\Omega$ is simply connected then all closed curves in $\Omega$ have index $0$ with respect to points outside $\Omega$. Therefore the first statement follows from the second.</p>
<p>The second statement actually is the strongest form of Cauchy's theorem in the plane.</p>
|
13,989 | <p>Suppose $E_1$ and $E_2$ are elliptic curves defined over $\mathbb{Q}$.
Now we know that both curves are isomorphic over $\mathbb{C}$ iff
they have the same $j$-invariant.</p>
<p>But $E_1$ and $E_2$ could also be isomorphic over a subfield of $\mathbb{C}$.
As is the case for $E$ and its quadratic twist $E_d$. Now the question general is.</p>
<blockquote>
<p>$E_1$ and $E_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$
the smallest subfield of $\mathbb{C}$ such that $E_1$ and $E_2$ become isomorphic over $K$.
What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be
said about the extension $K|\mathbb{Q}$.</p>
</blockquote>
<p>My second question is something goes something like in the opposite direction. I start again with
quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension
$\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$)
which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E_d$.
Again I would like to know what happens if we take a larger extension.</p>
<blockquote>
<p>Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension.
Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic
to $E$ over K.</p>
</blockquote>
<p>I have no idea what is the right context to answer such questions.</p>
| JSE | 431 | <p>Seconding the pointer to Silverman's book. Once you learn the basic definitions of Galois cohomology, you can check as an exercise that the set of isomorphism classes in your second question, when K is a Galois extension of Q, is in natural bijection with</p>
<p>H^1(Gal(K/Q), Aut(E)(K)).</p>
|
3,540,045 | <p>The definite integral, <span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\:dt$$</span></p>
<p><strong>My question</strong>: for the trigonometric integral above the answer is <span class="math-container">$\frac{3\pi}{16}$</span>. What I want to know is how can I compute these integrals easily. Is there more than one way to solve it? If so, is the key to solving these integrals, just recognizing some trig identities and using u-sub until it looks like a simpler integral?</p>
<p><strong>Here's what I tried (Why doesn't it work!):</strong></p>
<p>I rewrote the integrand as: <span class="math-container">$3(1-\cos^2t)\cos^4t\:dt$</span> then foiled it in,</p>
<p><span class="math-container">$3\cos^4t-3\cos^6t dt$</span>
, then I used the power rule and multipied through by chain rule and then did </p>
<p><span class="math-container">$F(\pi)-F(0)$</span> and got the answer: <span class="math-container">$\frac{6}{35}$</span></p>
<p>Why does this not work?!?</p>
| Community | -1 | <p>Use walli's formula given here to do it faster <a href="https://math.stackexchange.com/questions/2833731/reduction-formula-for-integral-sinm-x-cosn-x-with-limits-0-to-pi-2">Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$</a></p>
<p><span class="math-container">$I=3\displaystyle\int_{0}^{\pi} \sin^2t\cos^4t\ dt=6 \displaystyle\int_{0}^{\pi/2} \sin^2t\cos^4t\ dt=6\times \dfrac{(2-1)(4-1)(4-3)}{6(6-2)(6-4)}\times \dfrac{\pi}{2}=\dfrac{3\pi}{16}$</span></p>
<p><strong>your mistake :</strong> Use of Power rule in wrong way</p>
<p>we know,</p>
<p><span class="math-container">$\displaystyle\int x^m \ dx=\dfrac{x^{m+1}}{m+1}+C$</span> but this is not analogous to </p>
<p><span class="math-container">$3 \displaystyle\int \cos^6t\ dt\neq\dfrac{3}{7}cos^7t + C $</span> as you've written in comments.</p>
<p>but ,</p>
<p><span class="math-container">$\displaystyle\int \cos^6t\ d(cost)\ =-\displaystyle\int \cos^6t \ sint\ dt \ =\dfrac{3}{7}cos^7t + C $</span> </p>
|
2,292,096 | <p>I'm learning inequalities for the first time, and except a paragraph by Paul Zeitz in his book Art and Craft of problem solving, none actually give much motivation of why should I care about inequalities.</p>
<p>The example given by Paul Zeitz was that to prove $b^2-b+1$ is never a perfect square for integer $b$. Well- that kinda motivates a tiny bit, but the inequality used is triviality obvious; I want much deeper.</p>
<blockquote>
<p>What are some easy-to-state, moderately hard to solve number theoretic or combinatorics problem which requires applying a nontrivial inequality ?</p>
</blockquote>
| Somos | 438,089 | <p>The pigeonhole principle depends on an inequality and is nontrivial because it is so useful. Another is $x^2 \ge 0$ for $x$ real which can be used to prove some polynomials have no real roots.</p>
|
1,788,435 | <p>My math book says, a Linear equation has exactly one solution. Because $ax + b = 0$; $x =-\frac{b}{a}$. But I've solved many linear equations with multiple solutions before.
(I'm not very good in math. Need help...)</p>
| Jared | 138,018 | <p>If we assume that <em>all</em> linear equations have the form:</p>
<p>$$
ax + b= 0
$$</p>
<p>(which is completely valid and <em>should</em> be how we view linear equations)</p>
<p>then linear equations have either 1, 0, or infinite solutions. It's quite simple if $a \neq 0$ then they have <em>exactly</em> one solution: $x = -\frac{b}{a}$.</p>
<p>On the other hand, if $a = 0$ then if $b = 0$ we have infinite solutions (any value of $x$ solves $0x + 0 = 0$) and if $a = 0$ and $b \neq 0$ then there are <em>no solutions</em> (there is no value of $x$ that makes $0x + 1 = 0$, for example).</p>
|
3,878,380 | <p>Do the columns of a matrix always represent different vectors? If so, I don't understand how if I have a <span class="math-container">$3\times3$</span> matrix where the rows represent the dimensions and I multiply it by a <span class="math-container">$3\times1$</span> column vector with the same dimensions, it will give me a vector. Some sources say the result comes from the dot product of each line - is this correct? If so, <span class="math-container">$a_{12}\cdot b_{21}$</span> would give zero right?</p>
| Laars Helenius | 112,790 | <p>Suppose <span class="math-container">$\mathbb{R}^m$</span> and <span class="math-container">$\mathbb{R}^n$</span> are vector spaces and there is a linear transformation <span class="math-container">$f: \mathbb{R}^m \to \mathbb{R}^n$</span>. Then <span class="math-container">$f$</span> can be represented by a <span class="math-container">$m\times n$</span> matrix <span class="math-container">$M$</span> so that if <span class="math-container">$v\in\mathbb{R}^m$</span>, then <span class="math-container">$f(v)$</span> is computed as <span class="math-container">$M\cdot v$</span>. This is why the multiplication of a matrix with a vector is always another vector.</p>
<p>The action of the linear transformation organizes vectors vectors from both <span class="math-container">$\mathbb{R}^m$</span> and <span class="math-container">$\mathbb{R}^n$</span> in important ways.</p>
<p>The linearly independent columns of <span class="math-container">$M$</span> form a basis for a subspace of <span class="math-container">$\mathbb{R}^n$</span> called the column space of <span class="math-container">$M$</span> and is denoted as <span class="math-container">$C(M)$</span>. The column space is where all transformed vectors “live”.</p>
<p>The linearly independent rows of <span class="math-container">$M$</span> form a sub space of <span class="math-container">$\mathbb{R}^m$</span> called the row space of <span class="math-container">$M$</span> and is denoted as <span class="math-container">$R(M)$</span>, whose orthogonal complement, <span class="math-container">$R(M)^{\perp}$</span>, is the subspace of <span class="math-container">$\mathbb{R}^m$</span> such that for any <span class="math-container">$v\in R(M)^{\perp}$</span> we have <span class="math-container">$M\cdot v=0$</span>. This is also known as the null space of <span class="math-container">$M$</span> or the kernel of <span class="math-container">$M$</span>.</p>
|
3,656,978 | <p>I understand to say that a bounded linear operator <span class="math-container">$T$</span> is called "polynomially compact" if there is a nonzero polynomial <span class="math-container">$p$</span> such that <span class="math-container">$p(T)$</span> is compact. </p>
<p>Can anyone help me with examples of polynomially compact operators? </p>
| Aditya Dev | 746,162 | <p>The error is in the <span class="math-container">$2^{nd}$</span> step of your calculation. Matrix multiplication is <a href="https://en.wikipedia.org/wiki/Matrix_multiplication#Non-commutativity" rel="nofollow noreferrer">Non- Commutative</a> (i.e. <span class="math-container">$ \mathbf {A} \mathbf {B} \not = \mathbf {B}\mathbf {A}$</span> ) .The correct way to do this is as mentioned by @RedLapm.</p>
|
1,132,599 | <p>For which prime numbers p does the congruence $x^2+x+1\equiv0$ mod p have solutions? </p>
<p>I am new to the topic of quadratic reciprocity and I know how to answer this question had it been for which prime numbers p does the congruence $x^2\equiv-6$ mod p have solutions? </p>
<p>Can I perhaps split the congruence into two parts, solve them individually and then combine solutions? </p>
<p>Thanks in advance. </p>
| marwalix | 441 | <p>If $p\neq 2$ we have equivalence with $4(x^2+x+1)\equiv 0 \pmod p$. We ca now complete the square $(2x+1)^2\equiv {-3} \pmod p$ this means $-3$ is a quadratic residue.
For $p=2$ we have $\forall x, x^2+x+1\equiv 1\pmod 2$</p>
|
739,516 | <p>Is there a closed form for $$\int_{0}^{\infty}e^{ax^3+bx^2}\,\mathrm{d}x $$?</p>
| Keshav Srinivasan | 71,829 | <p>Here is what WolframAlpha gives me, just for $a=-1$:</p>
<p>$$\begin{align*}
\int^\infty_0 e^{-x^3+bx^2}\,dx=-\frac1{27}b\,\left(2\sqrt{3}\pi e^{\tfrac{2b^3}{27}}\left(I_{-\frac13}\left(-\dfrac{2b^3}{27}\right)+I_{\frac13}\left(-\dfrac{2b^3}{27}\right)\right)-9_2F_2\left(\dfrac{1}{2},1;\dfrac23,\dfrac43;\dfrac{4b^3}{27}\right)\right)\,\,\,\\\color{grey}{\text{ for }}\,\operatorname{Re}(b)<0
\end{align*}
$$</p>
<p>It's expressed using Bessel functions and hypergeometric functions, and WolframAlpha was just barely able to compute it with extra computational time. So just imagine how complicated an expression you'd have in the general case.</p>
|
2,506,182 | <p>The question speaks for itself, since the question comes from a contest environment, one where the use of calculators is obviously not allowed, can anyone perhaps supply me with an easy way to calculate the first of last digit of such situations?</p>
<p>My intuition said that I can look at the cases of $3$ with an index ending in a $1$, so I looked at $3^1=3$, $3^{11}=177,147$ and $3^{21}= 10,460,353,203$. So there is a slight pattern, but I'm not sure if it holds, and even if it does I will have shown it holds just for indices with base $3$, so I was wondering whether there is an easier way of knowing.</p>
<p>Any help is appreciated, thank you.</p>
| D.R. | 405,572 | <p>The last digits of 3 repeat every 4 powers: $3^1=3$, $3^2=9$, $3^3=27$, $3^4=81$, and because the last digit of $3^4$ is one, multiplying by three just gives us three again, and the pattern continues.</p>
<p>So taking $2011 \mod 4$, we get $3 \mod 4$, so the last digit is the same as that of $3^3=27$, which is 7.</p>
|
2,506,182 | <p>The question speaks for itself, since the question comes from a contest environment, one where the use of calculators is obviously not allowed, can anyone perhaps supply me with an easy way to calculate the first of last digit of such situations?</p>
<p>My intuition said that I can look at the cases of $3$ with an index ending in a $1$, so I looked at $3^1=3$, $3^{11}=177,147$ and $3^{21}= 10,460,353,203$. So there is a slight pattern, but I'm not sure if it holds, and even if it does I will have shown it holds just for indices with base $3$, so I was wondering whether there is an easier way of knowing.</p>
<p>Any help is appreciated, thank you.</p>
| Roman83 | 309,360 | <p>$3^1$: 3 is last digit </p>
<p>$3^2$: 9 is last digit </p>
<p>$3^3$: 7 is last digit </p>
<p>$3^4$: 1 is last digit </p>
<p>$3^5$: 3 is last digit </p>
<p>$3^6$: 9 is last digit </p>
<p>$3^7$: 7 is last digit </p>
<p>$3^8$: 1 is last digit </p>
<p>$(3,9,7,1)$ is period.</p>
<p>$$3^{2011}=3^{502\cdot4+3}$$
Then 7 is last digit</p>
|
369,589 | <p>I am not a math student, and only kind of picking up something whenever I need it. After emerged in the field of machine learning, probability, measure theory and functional analysis seem to be quite intriguing. I am considering learning stochastic calculus myself, but do not quite know what kind of prerequisites should I have. Anyone got some suggestions about the things I need to check before I challenge myself in stochastic calculus?</p>
| John Doe | 200,484 | <p>To gain a working knowledge of stochastic calculus, you don't need all that functional analysis/ measure theory. What you need is a good foundation in probability, an understanding of stochastic processes (basic ones [markov chains, queues, renewals], what they are, what they look like, applications, markov properties), calculus 2-3 (Taylor expansions are the key) and basic differential equations.</p>
<p>Some people here are trying to scare you away. It is certainly possible to apply stochastic calculus and gain an intuitive understanding of what's going on without knowing the details of a mean square limit or how to prove a function is square integrable in Lp space. </p>
<p>After all, it is a tool that first came into being for thermodynamic processes. Many of the Physicists who use it don't concern themselves with martingales and measure theory and all that in their applications. </p>
<p>Don't get me wrong, Functional analysis/ pdes/ measure theory are all interesting topics, but they are not necessary.</p>
<p>I'd recommend the book: "Paul Wilmott Introduces Quantitative Finance" </p>
|
104,592 | <p>Hello everyone,</p>
<p>Suppose that I am defining a function which embeds a surface (manifold) in $\mathbb{R}^3$.</p>
<p>Is there a standard symbol or letter that is used for this function?</p>
<p>Additionally, is there any other classical or standard notation (such as the hooked arrow for inclusion maps) of which I ought to be aware?</p>
<p>Regards,</p>
<p>Christopher</p>
| Dox | 25,356 | <p>No, there is none.</p>
|
72,478 | <p>I am trying to find the intervals on which f is increasing or decreasing, local min and max, and concavity and inflextion points for $f(x)=\sin x+\cos x$ on the interval $[0,\pi]$.</p>
<p>I know at $\pi/4$ the derivative will equal zero. So that gives me my critical numbers, positive and negative $\pi/4$ so now I need to find the intervals which is not making any sense to me, I thought they could only change at critical numbers but $\pi$ and $2\pi$ are different values. I am getting a positive for $2\pi$ and a negtive for $\pi$. How can this happen if the only critical number is $\pi/4$?</p>
| davidlowryduda | 9,754 | <p>In short, $\pi / 4$ is not the only critical value. In general, we know $\sin$ and $\cos$ are periodic, so we expect infinitely many critical values. We also know that if $\sin a = 0$, then $\sin (a + \pi) = 0$.</p>
<p>So when you take the derivative, you get $f'(x) = \cos x - \sin x$. You look for critical points - so you check when $\sin x = \cos x$, and you get when they're both positive (which you got), and when they're both negative (which you did not).</p>
<p>Note that $- \pi / 4$ is <em>not</em> a solution.</p>
|
3,784,484 | <p>Is there a bounded linear operator <span class="math-container">$T \in l^2(\mathbb{N})$</span> have for the essential spectrum the unit disk of <span class="math-container">$\mathbb{C}$</span>; i.e, such that <span class="math-container">$\sigma_{e}(T)=\textbf{D}(0, 1)$</span>; where <span class="math-container">$\textbf{D}(0, 1):=\{\lambda \in \mathbb{C} : |\lambda|\leq 1\}.$</span></p>
| Martin Argerami | 22,857 | <p>Given any compact <span class="math-container">$K\subset \mathbb C$</span>, there exists <span class="math-container">$T\in B(\ell^2(\mathbb N))$</span> with <span class="math-container">$\sigma_e(T)=K$</span>.</p>
<p>Let <span class="math-container">$\{P_n\}$</span> be a pairwise orthogonal sequence of infinite projections that add to <span class="math-container">$I$</span> (construction at the end). Take <span class="math-container">$\{q_n\}\subset K$</span> to be a countable dense set. Now define <span class="math-container">$T$</span> such that
<span class="math-container">$$
TP_n=q_nP_N.
$$</span>
That is, <span class="math-container">$T$</span> is <span class="math-container">$q_nI$</span> on each subspace <span class="math-container">$P_n\ell^2(\mathbb N)$</span>. First, note that <span class="math-container">$\sigma(T)=K$</span>. Indeed, each <span class="math-container">$q_n$</span> is an eigenvalue. If <span class="math-container">$r\in\mathbb C\setminus K$</span>, let <span class="math-container">$d=\operatorname{dist}(r,K)$</span>. Define <span class="math-container">$S$</span> to be given by
<span class="math-container">$$
SP_n=\frac1{q_n-r}\,P_n.
$$</span>
Then <span class="math-container">$S$</span> is bounded (because <span class="math-container">$|q_n-r|\geq d$</span> for all <span class="math-container">$n$</span>) and <span class="math-container">$S(T-r I)=I$</span>; this shows that <span class="math-container">$\sigma(T)\subset\overline{\{q_n\}}=K$</span>. As the multiplicity of each <span class="math-container">$q_n$</span> is infinite, <span class="math-container">$\{q_n\}\subset \sigma_e(T)$</span>. As the essential spectrum is closed and contained in <span class="math-container">$\sigma(T)$</span>, we get that <span class="math-container">$\sigma_e(T)=K$</span>.</p>
<hr />
<p><strong>Construction of the projections.</strong> Let <span class="math-container">$\{e_k\}\subset\ell^2(\mathbb N)$</span> be an orthonormal basis. Let <span class="math-container">$\{G_n\}\subset \mathbb N$</span> be a partition of <span class="math-container">$\mathbb N$</span> into infinite sets. Now let <span class="math-container">$P_n$</span> be the projection onto the span of <span class="math-container">$\{e_k:\ k\in G_n\}$</span>.</p>
|
1,725,945 | <p>I'm reading the proof of "Fundamental Theorem of Finite Abelian Groups" in Herstein Abstract Algebra, and I've found this statement in the proof that I don't see very clear.</p>
<p>Let $A$ be a normal subgroup of $G$. And suppose $b\in G$ and the order of $b$ is prime $p$, and $b$ is not in $A$. Then $A \cap (b)=(e)$.</p>
| David C. Ullrich | 248,223 | <p>If $x\in A\cap (b)$ then $x=b^n$ for some $n$. If the order of $b$ is $p$ and $x\ne e$ then we can take $0<n<p$. Since $p$ is prime there exists $m$ so that $b^{nm}=b$; hence $b\in A$.</p>
|
2,075,485 | <p>Let $[a,b]$ be a finite closed interval on $\mathbb{R}$, $f$ be a continuous differentible function on $[a,b]$. Prove that
$$\max_{x\in [a,b]} |f(x)|\le \Bigg|\frac{1}{b-a} \int_a^b f(x)dx\Bigg|+\int_a^b |f'(x)|dx$$</p>
<p>I think this is similar to Sobolev embedding theorem but have no idea about how to use it. I have also tried to transform the inequality into $$(b-a)\int_a^b (|f(t)|-|f'(x)|)dx\le \Bigg|\int_a^b f(x)dx\Bigg|,$$ where $f(t)$ is the maximum, but still don't know how to proceed. Thanks for any help.</p>
| kobe | 190,421 | <p>There is a point $c\in (a,b)$ such that </p>
<p>$$f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx$$</p>
<p>Let $x\in [a,b]$. Since </p>
<p>$$f(x) = f(c) + \int_c^x f'(t)\, dt$$</p>
<p>then </p>
<p>$$\lvert f(x)\rvert \le \lvert f(c)\rvert + \int_a^b \lvert f'(t)\rvert\, dt$$</p>
<p>Now finish the argument.</p>
|
736,749 | <p>Let us consider the Fourier transform of <span class="math-container">$\mathrm{sinc}$</span> function. As I know it is equal to a rectangular function in frequency domain and I want to get it myself, I know there is a lot of material about this, but I want to learn it by myself. We have <span class="math-container">$\mathrm{sinc}$</span> function whhich is defined as
<span class="math-container">$$
\mathrm{sinc}(\omega_0\,t) = \sin(\omega_0\,t)/(\omega_0\,t).
$$</span>
Its Fourier transform
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0\,t) \,e^{-i\,\omega\,t}/(\omega_0\,t)\,\mathrm dt
$$</span>
can be represented as
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0\,t)\,(\cos(\omega\,t) - i \,\sin(\omega\,t))/(\omega_0\,t)\,\mathrm dt.
$$</span>
Because we can distribute in brackets and consider that integral of difference is equal differences of integrals, we get
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0 \,t) \cos(\omega\,t)/(\omega_o\,t)\,\mathrm dt - \int_{\Bbb R} \sin(\omega_0\,t) \sin(\omega\,t)/(\omega_o\,t)\,\mathrm dt
$$</span></p>
<p>but first product is zero, right? Because sine and cosine are orthogonal, so how could continue? Please help me.</p>
| Alijah Ahmed | 124,032 | <p>In terms of deriving the Fourier Transform, I will make some use of techniques highlighted in <a href="http://www.claysturner.com/dsp/FTofSync.pdf" rel="noreferrer">http://www.claysturner.com/dsp/FTofSync.pdf</a></p>
<p>Let us start with your expression</p>
<p>$$\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)\cos(\omega t)}{(\omega_ot)}dt-\int_{-\infty}^{\infty}\frac{j\sin(\omega_0t)\sin(\omega t)}{(\omega_ot)}dt$$ </p>
<p>Examining the integrals, the term $\omega_0t$ in the denominator makes evaluating the integral more involved.</p>
<p>We can use some relevant trigonometric identities so that we can express
$$\sin(\omega_0t)\cos(\omega t)=\frac{1}{2}[\sin((\omega+\omega_0)t)-\sin((\omega-\omega_0)t)]$$</p>
<p>$$\sin(\omega_0t)\sin(\omega t)=\frac{1}{2}[\cos((\omega-\omega_0)t)-\cos((\omega+\omega_0)t)]$$</p>
<p>To deal with the awkward $\omega_0t$ term, we can use the following identity to convert the single integral into a double integral that is far nicer to evaluate:
$$\frac{1}{\omega_0t}=\int_0^{\infty}e^{-\omega_0ts}ds$$</p>
<p>Thus the integral to evaluate is
$$\int_0^{\infty}\int_{-\infty}^{\infty}\sin(\omega_0t)\cos(\omega t)e^{-\omega_0tx}dtdx-j\int_0^{\infty}\int_{-\infty}^{\infty}\sin(\omega_0t)\sin(\omega t)e^{-\omega_0tx}dtdx$$ </p>
<p>which expands to
$$\frac{1}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\sin((\omega+\omega_0)t)e^{-\omega_0tx}dtdx-\frac{1}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\sin((\omega-\omega_0)t)e^{-\omega_0tx}dtdx\\-\frac{j}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\cos((\omega-\omega_0)t)e^{-\omega_0tx}dtdx+\frac{j}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\cos((\omega+\omega_0)t)e^{-\omega_0tx}dtdx$$</p>
<p>We can exploit Fubini's theorem to rewrite the integral as</p>
<p>$$\frac{1}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\sin((\omega+\omega_0)t)e^{-\omega_0xt}dt\right]dx-\frac{1}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\sin((\omega-\omega_0)t)e^{-\omega_0xt}dt\right]dx\\-\frac{j}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\cos((\omega-\omega_0)t)e^{-\omega_0xt}dt\right]dx+\frac{j}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\cos((\omega+\omega_0)t)e^{-\omega_0xt}dt\right]dx$$</p>
<p>We shall use the following integral identities to calculate the above integrals :-
$$\int_0^{\infty}\sin(at)e^{-st}dt=\frac{a}{a^2+s^2}$$
$$\int_0^{\infty}\cos(at)e^{-st}dt=\frac{s}{a^2+s^2}$$</p>
<p>This results in
$$\frac{1}{2}\int_{-\infty}^{\infty}\left[\frac{(\omega+\omega_0)}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx-\frac{1}{2}\int_{-\infty}^{\infty}\left[\frac{(\omega-\omega_0)}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx\\-\frac{j}{2}\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx+\frac{j}{2}\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx\text{ (Eq. 1)}$$</p>
<p>To evaluate the real component integrals in (Eq. $1$), we use the following result:-
$$\int_{-\infty}^{\infty}\frac{a}{a^2+s^s}ds=\frac{|a|}{a}\int_{-\infty}^{\infty}\frac{1}{1+y^2}dy=sgn(a)\left[\arctan y\right]^{\infty}_{-\infty}=sgn(a)\pi$$</p>
<p>This leads to
$$\int_{-\infty}^{\infty}\left[\frac{(\omega+\omega_0)}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx=sgn(\omega+\omega_0)\pi$$
$$\int_{-\infty}^{\infty}\left[\frac{(\omega-\omega_0)}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx=sgn(\omega-\omega_0)\pi$$</p>
<p>As regards the imaginary component integrals, note that the numerator is simply a constant times the derivative of the denominator, so we have</p>
<p>$$\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx=\frac{1}{2\omega_0}\left[\ln\{(\omega-\omega_0)^2+\omega_0^2x^2\}\right]_{-\infty}^{\infty}$$ </p>
<p>$$\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx=\frac{1}{2\omega_0}\left[\ln\{(\omega+\omega_0)^2+\omega_0^2x^2\}\right]_{-\infty}^{\infty}$$ </p>
<p>Combining the imaginary component integrals as per the integral in (Eq. $1$) we wish to evaluate and noting that they have opposite signs, we have
$$-\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx+\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx\\=\lim_{x\rightarrow\infty}\frac{1}{2\omega_0}ln\left[\frac{(\omega+\omega_0)^2+\omega_0^2x^2}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]+\lim_{x\rightarrow -\infty}\frac{1}{2\omega_0}ln\left[\frac{(\omega-\omega_0)^2+\omega_0^2x^2}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]=\frac{2}{\omega_0}\ln(1)=0$$
Thus the imaginary terms in the integral cancel out, leading to the integral in (Eq. $1$) being a real result, as follows
$$\frac{\pi}{2}\left[sgn(\omega+\omega_0)-sgn(\omega-\omega_0)\right]$$</p>
<p>Putting everything together we have:
$$\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)\cos(\omega t)}{(\omega_ot)}dt-\int_{-\infty}^{\infty}\frac{j\sin(\omega_0t)\sin(\omega t)}{(\omega_ot)}dt=\frac{\pi}{2}\left[sgn(\omega+\omega_0)-sgn(\omega-\omega_0)\right]$$</p>
<p>The result is a rectangular function that starts from frequency $-\omega_0$ and ends at frequency $\omega_0$.</p>
|
736,749 | <p>Let us consider the Fourier transform of <span class="math-container">$\mathrm{sinc}$</span> function. As I know it is equal to a rectangular function in frequency domain and I want to get it myself, I know there is a lot of material about this, but I want to learn it by myself. We have <span class="math-container">$\mathrm{sinc}$</span> function whhich is defined as
<span class="math-container">$$
\mathrm{sinc}(\omega_0\,t) = \sin(\omega_0\,t)/(\omega_0\,t).
$$</span>
Its Fourier transform
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0\,t) \,e^{-i\,\omega\,t}/(\omega_0\,t)\,\mathrm dt
$$</span>
can be represented as
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0\,t)\,(\cos(\omega\,t) - i \,\sin(\omega\,t))/(\omega_0\,t)\,\mathrm dt.
$$</span>
Because we can distribute in brackets and consider that integral of difference is equal differences of integrals, we get
<span class="math-container">$$
\int_{\Bbb R} \sin(\omega_0 \,t) \cos(\omega\,t)/(\omega_o\,t)\,\mathrm dt - \int_{\Bbb R} \sin(\omega_0\,t) \sin(\omega\,t)/(\omega_o\,t)\,\mathrm dt
$$</span></p>
<p>but first product is zero, right? Because sine and cosine are orthogonal, so how could continue? Please help me.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\fermi\pars{t} \equiv {\sin\pars{\omega_{0}t} \over \omega_{0}t}
=\int_{-\infty}^{\infty}\hat{\fermi}\pars{\omega}\expo{-\ic\omega t}\,
{\dd\omega \over 2\pi}\,,\qquad\qquad\hat{\fermi}\pars{\omega}:\ {\large ?}}$</p>
<blockquote>
<p>\begin{align}
\color{#c00000}{\hat{\fermi}\pars{\omega}} &= \int_{-\infty}^{\infty}
{\rm sinc}\pars{\omega_{0}t}
\expo{\ic\omega t}\,\dd t
={1 \over \verts{\omega_{0}}}\int_{-\infty}^{\infty}{\sin\pars{t} \over t}\,
\expo{\ic\omega t/\verts{\omega_{0}}}\,\dd t
\\[3mm]&={1 \over \verts{\omega_{0}}}\int_{-\infty}^{\infty}
\pars{\half\int_{-1}^{1}\expo{\ic\nu t}\,\dd\nu}
\expo{\ic\omega t/\verts{\omega_{0}}}\,\dd t
=\color{#c00000}{{1 \over 2\verts{\omega_{0}}}\int_{-1}^{1}\dd\nu\
\overbrace{\int_{-\infty}^{\infty}
\expo{\ic\pars{\nu + \omega/\verts{\omega_{0}}}t}\,\dd t}
^{\ds{2\pi\,\delta\pars{\nu + {\omega \over \verts{\omega_{0}}}}}}}
\end{align}
where $\ds{\delta\pars{x}}$ is the
<a href="http://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow noreferrer">Dirac Delta 'Function'</a>.</p>
</blockquote>
<p>Then
\begin{align}
\color{#c00000}{\hat{\fermi}\pars{\omega}}&={\pi \over \verts{\omega_{0}}}\,
\Theta\pars{1 - \verts{\omega \over \verts{\omega_{0}}}}
\end{align}
where $\ds{\Theta\pars{x}}$ is the
<a href="http://en.wikipedia.org/wiki/Unit_step_function" rel="nofollow noreferrer">Heaviside Step Function</a>.</p>
<blockquote>
<p>$$\color{#00f}{\large%
\hat{\fermi}\pars{\omega}={\pi \over \verts{\omega_{0}}}\,
\Theta\pars{\verts{\omega_{0}} - \verts{\omega}}}
$$</p>
</blockquote>
|
2,181,540 | <p>What approach we have to solve the following differential equation?</p>
<p>$$y''(x)- \frac{y'(x)^2}{y} + \frac{y'(x)}{x}=0$$</p>
<p>The known solution is $y(x) = c_2 * x^{c_1}$</p>
| Claude Leibovici | 82,404 | <p><strong>Hint</strong></p>
<p>Let $y=e^z$ to make the equation $$\frac{e^{z} \left(x z''+z'\right)}{x}=0$$ This reduces to $$x z''+z'=0$$ Reduce the order $z'=t$ and continue.</p>
<p>I am sure that you can take it from here.</p>
|
2,181,540 | <p>What approach we have to solve the following differential equation?</p>
<p>$$y''(x)- \frac{y'(x)^2}{y} + \frac{y'(x)}{x}=0$$</p>
<p>The known solution is $y(x) = c_2 * x^{c_1}$</p>
| Lutz Lehmann | 115,115 | <p>In principle the same reduction as the other answer, but the other way around:</p>
<p>Substitute $u=\frac{y'}{y}=(\log|y|)'$ to find $u'=\frac1y(y''-\frac{y'^2}y)$ and consequently
$$
u'+\frac{u}x=0\iff (ux)'=0.
$$</p>
|
2,953,757 | <p>My try to solve this question goes as follows:</p>
<p><span class="math-container">$g=gcd(n^2+1, (n+1^2)+1) = gcd(n^2+1, 2n+1) = gcd(n^2-2n, 2n+1)$</span>.</p>
<p>By long division: </p>
<p><span class="math-container">$$n^2-2n = -2n(2n+1) + 5n^2$$</span></p>
<p>Since <span class="math-container">$g$</span> divides <span class="math-container">$n^2-2n$</span> and <span class="math-container">$g$</span> divides <span class="math-container">$(2n+1)$</span>, <span class="math-container">$g$</span> divides <span class="math-container">$5n^2$</span>. However, I need to show that <span class="math-container">$g$</span> is prime in order to show that it divides <span class="math-container">$5$</span> or <span class="math-container">$n^2$</span>. I am stuck at this point.</p>
<p>Any Ideas,,</p>
| Hagen von Eitzen | 39,174 | <p>Clearly <span class="math-container">$$\gcd(n^ 2+1,(n+1)^2+1)\mid \bigl((n+1)^2+1\bigr)-\bigl(n^ 2+1\bigr)=2n+1,$$</span>
hence also
<span class="math-container">$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=n-2 $$</span>
as well as
<span class="math-container">$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=2n+1-2(n-2)=5. $$</span>
This leaves only the possibilites <span class="math-container">$1$</span> and <span class="math-container">$5$</span> (which are both possible as can be seen by checking the cases <span class="math-container">$n=1$</span> and <span class="math-container">$n=2$</span>)</p>
|
2,953,757 | <p>My try to solve this question goes as follows:</p>
<p><span class="math-container">$g=gcd(n^2+1, (n+1^2)+1) = gcd(n^2+1, 2n+1) = gcd(n^2-2n, 2n+1)$</span>.</p>
<p>By long division: </p>
<p><span class="math-container">$$n^2-2n = -2n(2n+1) + 5n^2$$</span></p>
<p>Since <span class="math-container">$g$</span> divides <span class="math-container">$n^2-2n$</span> and <span class="math-container">$g$</span> divides <span class="math-container">$(2n+1)$</span>, <span class="math-container">$g$</span> divides <span class="math-container">$5n^2$</span>. However, I need to show that <span class="math-container">$g$</span> is prime in order to show that it divides <span class="math-container">$5$</span> or <span class="math-container">$n^2$</span>. I am stuck at this point.</p>
<p>Any Ideas,,</p>
| Barry Cipra | 86,747 | <p>Since <span class="math-container">$\gcd(n,2n+1)=\gcd(n,1)=1$</span>, you can keep going:</p>
<p><span class="math-container">$$\gcd(n^2-2n,2n+1)=\gcd(n(n-2),2n+1)=\gcd(n-2,2n+1)=\gcd(n-2,5)\in\{1,5\}$$</span></p>
<p>The key reduction here is the gcd property <span class="math-container">$\gcd(a,b)=1\implies\gcd(ac,b)=\gcd(c,b)$</span>.</p>
|
1,440,106 | <p>I am currently studying how to prove the Fibonacci Identity by Simple Induction, shown <a href="http://mathforum.org/library/drmath/view/52718.html">here</a>, however I do not understand how $-(-1)^n$ becomes $(-1)^{n+1}$. Can anybody explain to me the logic behind this?</p>
| Robert Soupe | 149,436 | <p>The powers of $-1$ are $$-1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, \ldots$$ If $n$ is odd, then $(-1)^n = -1$, but if $n$ is even then $(-1)^n = 1$. So if you multiply $(-1)^n$ once by $-1$, it is the same as incrementing $n$. (Technically you can also decrement, but you can't generalize that to other negative numbers).</p>
|
3,098,245 | <p><span class="math-container">$$ \frac{d}{dx}e^x =\frac{d}{dx} \sum_{n=0}^{ \infty} \frac{x^n}{n!}$$</span>
<span class="math-container">$$ \sum_{n=0}^{ \infty} \frac{nx^{n-1}}{n!}$$</span>
<span class="math-container">$$ \sum_{n=0}^{ \infty} \frac{x^{n-1}}{(n-1)!}$$</span>
This isn't as straightforward as I thought it would be. I'm assuming there is some way to shift indices to complete the proof, but I'm not sure exactly how. </p>
| Hyperion | 636,641 | <p>Rewrite the infinite sum as
<span class="math-container">$$\sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty}\frac{x^n}{n!}$$</span>
Then
<span class="math-container">$$\frac{d}{dx} (1 + \sum_{n=1}^{\infty}\frac{x^n}{n!}) = \sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$</span></p>
|
2,230,897 | <blockquote>
<p>A line passing through $P=(\sqrt3,0)$ and making an angle of $\pi/3$ with the positive direction of x axis cuts the parabola $y^2=x+2$ at A and B, then:<br>
(a)$PA+PB=2/3$<br>
(b)$|PA-PB|=2/3$<br>
(c)$(PA)(PB)=\frac{4(2+\sqrt3)}{3}$<br>
(d)$\frac{1}{PA}+\frac{1}{PB}=\frac{2-\sqrt3}{2}$ </p>
</blockquote>
<p>Equation of line:<br>
$y=\sqrt3(x-\sqrt3)$<br>
$y=\sqrt3x-3$ </p>
<p>substituting $y=\sqrt3x-3$ in $y^2=x+2$ </p>
<p>$(\sqrt3x-3)^2=x+2$<br>
$3x^2-6\sqrt3x+9=x+2$<br>
$3x^2-(6\sqrt3+1)x+7=0$ </p>
<p>$x_A=\frac{(6\sqrt3+1)+\sqrt{(6\sqrt3+1)^2-84}}{6}$<br>
$x_B=\frac{(6\sqrt3+1)-\sqrt{(6\sqrt3+1)^2-84}}{6}$ </p>
<p>But this gives a very complicated value of $x$ and that makes me feel that there ought to be a shorter way to solve this question, just can't figure out what it is.</p>
| mathlove | 78,967 | <p>Let $C,D$ be the point on $x$-axis such that $AC\perp CP,BD\perp PD$ respectively.</p>
<p>Then, $\triangle{PAC},\triangle{PBD}$ are triangles with $30^\circ,60^\circ,90^\circ$ from which we have
$$PA=\frac{2}{\sqrt 3}AC=\frac{2}{\sqrt{3}}|A_y|,\qquad PB=\frac{2}{\sqrt 3}|B_y|$$</p>
<p>Eliminating $x$ from the system gives
$$y^2=\frac{y+3}{\sqrt 3}+2\implies \sqrt 3\ y^2-y-3-2\sqrt 3=0$$
from which we have
$$A_y+B_y=\frac{1}{\sqrt 3},\qquad A_yB_y=\frac{-3-2\sqrt 3}{\sqrt 3}$$</p>
<p>(a) is not correct since
$$PA+PB\gt \frac{2(A_y+B_y)}{\sqrt 3}=\frac{2}{\sqrt 3}\cdot \frac{1}{\sqrt 3}=\frac 23$$</p>
<p>(b) is correct since
$$|PA-PB|=\left|\frac{2(A_y+B_y)}{\sqrt 3}\right|=\frac 23$$</p>
<p>(c) is correct since
$$PA\times PB=\frac 43\left|\frac{-3-2\sqrt 3}{\sqrt 3}\right|=\frac{4(2+\sqrt 3)}{3}$$</p>
<p>(d) is not correct since
$$\frac{1}{PA}+\frac{1}{PB}=\frac{PA+PB}{PA\cdot PB}\gt \frac{2(A_y+B_y)}{\sqrt 3}\div \frac{4(2+\sqrt 3)}{3}=\frac{2-\sqrt 3}{2}$$</p>
|
3,596,514 | <p>Let <span class="math-container">$0\le x \le 1$</span>
find the maximum value of <span class="math-container">$x(9\sqrt{1+x^2}+13\sqrt{1-x^2})$</span></p>
<p>I try to use am-gm inequality to solve this because it's similar to CMIMC2020 team prob.12 but i don't know how to do next. </p>
| Dr. Mathva | 588,272 | <p>Even though I still believe that the calculus approach is the most suited for the problem, consider the somehow <em>magical</em> ansatz I found: <span class="math-container">\begin{align*}x\cdot\left(9\sqrt{1+x^2}+13\sqrt{1-x^2}\right)&=2\cdot \frac34\cdot 3x\cdot 2\cdot \sqrt{1+x^2}+2\cdot\frac{13}4\cdot x\cdot2\sqrt{1-x^2}\\&\stackrel{AM-GM}{\leqslant}\frac34\cdot\left(9x^2+4\cdot(1+x^2)\right)+\frac{13}4\cdot \left(x^2+4\cdot (1-x^2)\right)\\&=16\end{align*}</span></p>
|
259,119 | <p>I have a string like this:</p>
<p><code>string="there is a humble-bee in Hanna's garden";</code></p>
<p>Now I want to exclude those words that contain "-" and "'". My own solution would be:</p>
<p><code>StringDelete[string,Cases[StringSplit[string," "], _?(StringContainsQ[#, {"'", "-"}] &)]]</code></p>
<p>so the outcome is:</p>
<p><code>"there is a in garden"</code></p>
<p>But I was wondering whether there is a more elegant solution?</p>
| kglr | 125 | <pre><code>StringDelete[string, WordCharacter .. ~~ "'" | "-" ~~ WordCharacter ..]
</code></pre>
<blockquote>
<pre><code>"there is a in garden"
</code></pre>
</blockquote>
<pre><code>StringRiffle @ Select[StringFreeQ["'" | "-"]] @ StringSplit[string]
</code></pre>
<blockquote>
<pre><code>"there is a in garden"
</code></pre>
</blockquote>
|
182,101 | <p>With respect to assignments/definitions, when is it appropriate to use $\equiv$ as in </p>
<blockquote>
<p>$$M \equiv \max\{b_1, b_2, \dots, b_n\}$$</p>
</blockquote>
<p>which I encountered in my analysis textbook as opposed to the "colon equals" sign, where this example is taken from Terence Tao's <a href="http://terrytao.wordpress.com/">blog</a> :</p>
<blockquote>
<p>$$S(x, \alpha):= \sum_{p\le x} e(\alpha p) $$</p>
</blockquote>
<p>Is it user-background dependent, or are there certain circumstances in which one is more appropriate than the other?</p>
| paul garrett | 12,291 | <p>Upvoting the other answers and comments... and: conventions vary. The only way to know with reasonable confidence is from <em>context</em>. However, one can't know whether an author "believes in" setting context. The most important criterion may be whether or not one is tracking things well enough to reasonably infer which equalities are assignments, and which are assertions. If this seems to be an issue, likely one should back-track a bit, anyway.</p>
<p>In practice, assignment will be clear because the left-hand side is a single symbol (even if composed of several marks), and is appearing for the first time. The first-time-appearance criterion is obviously more easily applied if all first-time appearances are highlighted by a consistent convention (rather than being buring in-line, without emphasis or fanfare).</p>
|
3,247,176 | <p>I have this statement:</p>
<blockquote>
<p>It can be assured that | p | ≤ 2.4, if it is known that:</p>
<p>(1) -2.7 ≤ p <2.3</p>
<p>(2) -2.2 < p ≤ 2.6</p>
</blockquote>
<p>My development was:</p>
<p>First, <span class="math-container">$ -2.4 \leq p \leq 2.4$</span></p>
<p>With <span class="math-container">$1)$</span> by itself, that can't be insured, same argument for <span class="math-container">$2)$</span></p>
<p>Now, i will use <span class="math-container">$1)$</span> and <span class="math-container">$2)$</span> together, and the intersection between this intervals are: <span class="math-container">$(-2.2, 2.3)$</span>. So, this also does not allow me to ensure that | p | ≤ 2.4, since, there are some numbers that are outside the intersection of these two intervals, for example 2.35 is outside this interval. </p>
<p>But according to the guide, the correct answer must be <span class="math-container">$1), 2)$</span> together. And i don't know why.</p>
<p>Thanks in advance.</p>
| Ross Millikan | 1,827 | <p>The point is that if you use <span class="math-container">$1$</span> and <span class="math-container">$2$</span> together you know <span class="math-container">$-2.2 \lt p \lt 2.3$</span>. This does allow you to ensure <span class="math-container">$|p| \le 2.4$</span> because all numbers in the interval are less than <span class="math-container">$2.4$</span> in absolute value. The fact that there are numbers outside the interval that also qualify does not matter at all. If I told you that <span class="math-container">$0.1 \le p \le 0.9$</span> you could assure me that <span class="math-container">$|p| \le 2.4$</span>. You could also make a stronger statement, like <span class="math-container">$|p| \le 1$</span> or <span class="math-container">$|p| \le 0.9$</span>, but that doesn't mean the one with <span class="math-container">$2.4$</span> is incorrect.</p>
|
317,980 | <p>Suppose that $(P,\leq)$ is a countably infinite linear ordering, and $U$ is a p-point ultrafilter on $P$. Show that there is an $X\in U$ such that $X$ has order type $\omega$ or $\omega^*$. (This is <a href="http://books.google.com/books?id=WTAl997XDb4C&pg=PA86" rel="nofollow">Exercise 7.8</a> in Jech)</p>
<p>I know that if $U=\{P\}$, or if $U$ is Ramsey, then this is true; enumerate $P=\{a_n\}_{n\in\omega}$ and color pairs $\{m,n\}$ according to whether or not their order in $\omega$ agrees with the order of $\{a_m,a_n\}$. But if $U$ is a p-point, I don't see how to adapt this argument.</p>
<p>(Edit) A definition: $U$ is a p-point ultrafilter on a countably infinite set $P$ if it is non-principle and for every partition $\{A_n:n\in\omega\}$ of $P$ into $\aleph_0$ many pieces such that $A_n\notin U$ for all $n$, there is an $X\in U$ such that $X\cap A_n$ is finite for all $n$.</p>
| Camilo Arosemena-Serrato | 33,495 | <p>We shall make use of the following facts:</p>
<ol>
<li>If $U$ is an ultrafilter and $X\cup Y\in U$, then $X\in U$ or $Y\in U$.</li>
<li>(Exercise $7.7$ of Jech's <em>Set Theory</em>)If $D$ is a $p$-point on $\omega$, then for any decreasing sequence $A_0\supset A_1\supset \cdots \supset A_n\supset \cdots$ of elements of $D$, there exists $X\in D$ such that for each $n\in \mathbb N$, $X-A_n$ is finite. </li>
</ol>
<p>With no loss of generality we may assume that $P\subseteq \mathbb Q$ and $<=<^{\mathbb Q}\cap P\times P$, and furthermore we may assume $P$ is bounded in $(\mathbb Q,<)$; $P\subseteq [a,b]$ for some $a,b\in \mathbb Q$. </p>
<p>Let $D$ be a $p$-point on $P$.</p>
<p>By fact $(1)$ we have that $[a,\frac{a+b}{2}]\cap P\in D$ or $[\frac{a+b}{2},b]\cap P\in D$, let $[a_1,b_1]$ be one of these half-intervals such that $[a_1,b_1]\cap P\in D$.
Continuing this process we get a shrinking sequence of nested intervals $\{ [a_k,b_k]\}_{k\in \mathbb N}$; with $[a_o,b_0]=[a,b]$, such that $b_k-a_k=\frac{b-a}{2^k}$ and $[a_k,b_k]\cap P\in D$ for all $k\in \mathbb N$</p>
<p>By fact $(2)$ there exists $X\in D$ such that $X-([a_k,b_k]\cap P)$ is finite for all $k\in \mathbb N$. Let $r\in \mathbb R$ be such that $r\in \bigcap_{k\in \mathbb N}[a_k,b_k]$, then either $[a,r]\cap P\in D$ or $[r,b]\cap P\in D$.</p>
<p>If $[a,r]\cap P\in D$, then $[a,r]\cap P\cap X\in D$, and hence since $D$ is nonprincipal, $[a,r]\cap P\cap X$ is infinite, but $a_k\rightarrow r$ as $k\rightarrow \infty$ then $X\cap P\cap [a,r]$ is a discrete infinite linear ordering, since $X-[a_k,b_k]$ is finite for all $k$, since $X\cap P\cap [a_k,b_k]\neq \emptyset $ for all sufficiently large $k$; otherwise $[a,r]\cap P\cap X$ would be finite, and since the intersection of the nested intervals is $\{ r\}$, the order-type of $[a,r)\cap P\cap X$ must be $\omega$, but since $D$ is non-principal $[a,r)\cap P\cap X\in D$.</p>
<p>Similarly if $[r,b]\cap P \in D$, the order-type of $(r,b]\cap X$ must be $\omega^*$, and $(r,b]\cap X\in D$. Since one of this cases must hold by fact $(1)$, we are done.</p>
|
317,980 | <p>Suppose that $(P,\leq)$ is a countably infinite linear ordering, and $U$ is a p-point ultrafilter on $P$. Show that there is an $X\in U$ such that $X$ has order type $\omega$ or $\omega^*$. (This is <a href="http://books.google.com/books?id=WTAl997XDb4C&pg=PA86" rel="nofollow">Exercise 7.8</a> in Jech)</p>
<p>I know that if $U=\{P\}$, or if $U$ is Ramsey, then this is true; enumerate $P=\{a_n\}_{n\in\omega}$ and color pairs $\{m,n\}$ according to whether or not their order in $\omega$ agrees with the order of $\{a_m,a_n\}$. But if $U$ is a p-point, I don't see how to adapt this argument.</p>
<p>(Edit) A definition: $U$ is a p-point ultrafilter on a countably infinite set $P$ if it is non-principle and for every partition $\{A_n:n\in\omega\}$ of $P$ into $\aleph_0$ many pieces such that $A_n\notin U$ for all $n$, there is an $X\in U$ such that $X\cap A_n$ is finite for all $n$.</p>
| Brian M. Scott | 12,042 | <p>I’m going to call the ultrafilter $\mathscr{U}$. Let $P_0=P\in\mathscr{U}$. Given $P_\xi\in\mathscr{U}$ for some $n\in\omega$, choose a non-endpoint $p_\xi\in P_\xi$. Exactly one of the sets $\{p\in P_\xi:p<p_\xi\}$ and $\{p\in P_\xi:p_\xi\le p\}$ belongs to $\mathscr{U}$; let it be $P_{\xi+1}$. If $\eta$ is a limit ordinal, $P_\xi$ has been constructed for each $\xi<\eta$, let $P_\eta=\bigcap_{\xi<\eta}P_\xi$. If $P_\eta\in\mathscr{U}$, continue the construction; otherwise, stop.</p>
<p>Since $P$ is countable, there must be a countable limit ordinal $\eta$ such that $P_\eta\notin\mathscr{U}$. For $\xi<\eta$ let $D_\xi=P_\xi\setminus P_{\xi+1}\notin\mathscr{U}$, and note that $\bigcup_{\xi<\eta}D_\xi=P\setminus P_\eta\in\mathscr{U}$, while each $D_\xi\notin\mathscr{U}$. Moreover, the sets $D_\xi$ are order-convex and pairwise disjoint, so the linear order on $P$ induces a natural linear order on $\mathscr{D}=\{D_\xi:\xi<\eta\}$. Each $D_\xi$ is either the left or the right end of the corresponding $P_\xi$; let $L=\{\xi<\eta:D_\xi\text{ is the left end of }P_\xi\}$, and let $R=\eta\setminus L$. Then exactly one of $\bigcup_{\xi\in L}D_\xi$ and $\bigcup_{\xi\in R}D_\xi$ belongs to $\mathscr{U}$; without loss of generality assume that $\bigcup_{\xi\in L}D_\xi\in\mathscr{U}$.</p>
<p>Let $L=\{\lambda_\xi:\xi<\alpha\}$ be an increasing enumeration of $L$, and let $\beta$ be minimal such that $\bigcup_{\xi<\beta}D_{\lambda_\xi}\in\mathscr{U}$; clearly $\beta$ is a limit ordinal, so there is a strictly increasing sequence $\langle\beta_n:n\in\omega\rangle$ of ordinals such that $\beta_0=\lambda_0$ and $\beta=\sup_n\beta_n$. For $n\in\omega$ let $E_n=\bigcup\{D_{\lambda_\xi}:\beta_n\le\lambda_\xi<\beta_{n+1}\}$; then $\bigcup_nE_n\in\mathscr{U}$, but $E_n\notin\mathscr{U}$ for $n\in\omega$, so there is a $U\in\mathscr{U}$ such that $U\subseteq\bigcup_nE_n$ and $U\cap E_n$ is finite for each $n\in\omega$. The construction then ensures that the order-type of $U$ as a subset of $P$ is $\omega$. (Had $\bigcup_{\xi\in R}D_\xi$ been a member of $\mathscr{U}$, we would instead have got a member of $\mathscr{U}$ of type $\omega^*$.)</p>
|
223,008 | <p>Ok so my teacher said we can use this sentence:
<strong>If $a$ is not a multiple of $5$, then $a^2$ is not a multiple of $5$ neither.</strong></p>
<p>to prove this sentence:
<strong>If $a^2$ is a multiple of $5$, then $a$ itself is a multiple of $5$</strong></p>
<p>I don't understand the logic behind it, I mean what's the link between them, how can we conclude the 2nd sentence to be true if the 1st one is true?</p>
<p>Thanks a lot guys!</p>
| Cameron Buie | 28,900 | <p>Given statements <span class="math-container">$p,q$</span>, the following are equivalent:</p>
<blockquote>
<p><span class="math-container">$p\Rightarrow q$</span></p>
<p><span class="math-container">$\neg q\Rightarrow\neg p$</span></p>
</blockquote>
<p>To check that, you can use a truth table. These two are called <em>contrapositives</em> of each other.</p>
<p>In particular, let <span class="math-container">$p$</span> be the statement "<span class="math-container">$a^2$</span> is a multiple of <span class="math-container">$5$</span>", and <span class="math-container">$q$</span> be the statement "<span class="math-container">$a$</span> is a multiple of <span class="math-container">$5$</span>." Your teacher is saying that you can use <span class="math-container">$\neg q\Rightarrow\neg p$</span> to prove <span class="math-container">$p\Rightarrow q$</span>.</p>
|
223,008 | <p>Ok so my teacher said we can use this sentence:
<strong>If $a$ is not a multiple of $5$, then $a^2$ is not a multiple of $5$ neither.</strong></p>
<p>to prove this sentence:
<strong>If $a^2$ is a multiple of $5$, then $a$ itself is a multiple of $5$</strong></p>
<p>I don't understand the logic behind it, I mean what's the link between them, how can we conclude the 2nd sentence to be true if the 1st one is true?</p>
<p>Thanks a lot guys!</p>
| sperners lemma | 44,154 | <p>For $p$ prime, $p|ab$ implies $p|a$ or $p|b$.</p>
<p>Since 5 is prime we have $5|a^2$ implies $5|a$ or $5|a$.</p>
|
3,535,316 | <p><span class="math-container">$$\int_{0}^{\pi}e^{x}\cos^{3}(x)dx$$</span></p>
<p>I tried to solve it by parts.I took <span class="math-container">$f(x)=\cos^{3}(x)$</span> so <span class="math-container">$f'(x)=-3\cos^{2}x\sin x$</span> and <span class="math-container">$g'(x)=e^{x}$</span> and I got"</p>
<p><span class="math-container">$$\int_{0}^{\pi}e^{x}\cos^{3}(x)dx=e^{x}\cos(x)+3\int_{0}^{\pi}e^{x}\cos^{2}(x)\sin(x)dx.$$</span></p>
<p>How to approach the second integral?</p>
| aud098 | 744,646 | <p>Break each term as <span class="math-container">$\frac{n}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$</span> now just add these two series of the form <span class="math-container">$\sum \frac {1}{(n!)}=e$</span></p>
|
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