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 |
|---|---|---|---|---|
3,516,776 | <p>I've been trying to solve this for limit comparison test with <span class="math-container">$a_n=a^\frac{1}{n}+a^{-\frac{1}{n}}-2 , b_n= \frac{1}{n}$</span>,
but
<span class="math-container">$\frac{a_n}{b_n}\rightarrow\ln{a}(a^{\frac{1}{x}}-a^{-\frac{1}{x}})\rightarrow 0$</span>.
Any help appreciated.</p>
| Pythagoras | 701,578 | <p>Assume <span class="math-container">$a>0$</span>. Let <span class="math-container">$x_n=\frac 1{2n}.$</span> Then <span class="math-container">$$\sum_{n=1}^\infty \left(a^{1/n}+a^{-1/n}-2\right)=\sum_{n=1}^\infty(a^{x_n}-a^{-x_n})^2$$</span> <span class="math-container">$$=\sum_{n=1}^\infty a^{2x_n}(1-a^{-2x_n})^2=\sum_{n=1}^\infty a^{2x_n}(1-e^{-2x_n\ln a})^2,$$</span> which by mean value theorem equals
<span class="math-container">$$\sum_{n=1}^\infty a^{2x_n}(e^{-2y_n\ln a}\cdot 2x_n\ln a)^2,$$</span> where <span class="math-container">$0<y_n<x_n$</span>. Clearly the series is bounded by <span class="math-container">$$C\cdot \sum_{n=1}^\infty (2x_n)^2=C\sum_{n=1}^\infty\frac 1{n^2}$$</span> where <span class="math-container">$C$</span> is a bounded constant. It follows that the original series is convergent.</p>
<p><strong>Note</strong>: <span class="math-container">$1-e^x=e^0-e^x=e^{\xi}(0-x),$</span> where <span class="math-container">$\xi$</span> is between <span class="math-container">$0$</span> and <span class="math-container">$x$</span>.</p>
|
2,519,620 | <blockquote>
<p><strong>Question :</strong> Three balls are to be randomly selected without replacement from an urn containing $20$ balls numbered $1$ through $20$. If we bet that at least one of the balls that are drawn hasa number as large as or larger than $17$, what is the probability that we win the bet?</p>
</blockquote>
<p>I am sorry that I neither have any approach nor any solution. I am not getting question at all. I am completely new to random variable concept, thus will need some help initially.
Please help me out.</p>
<p>P.S - This is not a homework question.</p>
| drhab | 75,923 | <p>For $i=1,2,3$ let $E_i$ denote the event that the number of the $i$-th drawn ball is smaller than $17$. </p>
<p>Then you are looking for: $$1-P(E_1\cap E_2\cap E_3)=1-P(E_1)P(E_2\mid E_1)P(E_3\mid E_1\cap E_2)=1-\frac{16}{20}\frac{15}{19}\frac{14}{18}$$</p>
|
2,519,620 | <blockquote>
<p><strong>Question :</strong> Three balls are to be randomly selected without replacement from an urn containing $20$ balls numbered $1$ through $20$. If we bet that at least one of the balls that are drawn hasa number as large as or larger than $17$, what is the probability that we win the bet?</p>
</blockquote>
<p>I am sorry that I neither have any approach nor any solution. I am not getting question at all. I am completely new to random variable concept, thus will need some help initially.
Please help me out.</p>
<p>P.S - This is not a homework question.</p>
| Kyky | 423,726 | <p>We need to find the probability of not getting a ball above $17$ first. In the urn, there are $4$ balls that are equal or larger than $17$ ($17,18,19,20$). Since there are $20$ balls, there are $20-4=16$ number of balls that are below $17$ in the urn. That means there is a $\frac{16}{20}$ chance that the first ball is less than $17$, $\frac{16-1}{20-1}$ chance that the second ball is less than $17$ provided the first ball is below $17$, and $\frac{16-2}{20-2}$ chance that the third ball is below $17$ provided the first two are below $17$. This gives us:
$$\frac{16}{20}\cdot\frac{16-1}{20-1}\cdot\frac{16-2}{20-2}$$
$$=\frac{16}{20}\cdot\frac{15}{19}\cdot\frac{14}{18}$$
$$=\frac{3360}{6840}$$
$$=\frac{28}{57}$$
Now, $1-\frac{28}{57}$=$\frac{29}{57}$ which is roughly $0.50877$.</p>
|
527,576 | <blockquote>
<p>Three men (out of 7) and three women (out of 6) will be chosen to
serve on a 7 member committee. In how many ways can the committee be
formed?</p>
</blockquote>
<p>I did 7C3 to get 35 men.</p>
<p>Then i did 6C3 to get 20 women.</p>
<p>Then i decide to add up 20 + 35 and get 55 but it is suggested i have to multiply 35 and 20 instead. I want to know why is it that we are multiplying 35 and 20 instead of adding them up?</p>
| Bill Cook | 16,423 | <p>The choices of men and women are made independently. Independent $\Longleftrightarrow$ Multiply.</p>
<p>Why? A concrete example might help.</p>
<p>I want to choose 1 fruit from $\{apple,orange,banana\}$ and 1 drink from $\{water,tea\}$. Then I have $3 \times 2 = 6$ choices: $(apple,water)$, $(orange,water)$, $(banana,water)$, $(apple,tea)$, $(orange,tea)$, and $(banana,tea)$.</p>
<p>For each fixed choice of the first item I can choose any of the second item. So if I have $n$ choices for item #1 and $m$ choices for item #2: I get $\underbrace{m+m+\cdots+m}_{n-\mbox{times}} = n \cdot m$ choices.</p>
|
129,788 | <blockquote>
<p>Let be A and B two events from the same sample set. If $\space P(A)+P(B)=1$, can one say that they are opposite events?</p>
</blockquote>
<p>In my thought:</p>
<p>$\space P(A)+P(B)=1$</p>
<p>$\space P(A)=1-P(B)$</p>
<p>So they are opposite events. But my book says no! It says that is not necessary true.</p>
<p>Can you explain me, why not?</p>
<p>Thanks</p>
| Yang | 28,639 | <p>If B is the complementary event of A, then they satisfy the property:</p>
<p>$P(A)=1-P(B)$</p>
<p>But the reverse (your claim) is not true. The key thing is to realize that events are not necessarily mutually exclusive (<em>disjoint</em>, $P(A\cap B)=\emptyset$). </p>
<p>If they happen to be disjoint, then your claim would indeed be true.</p>
|
4,644,186 | <p>Let m be a positive integer.Find the values of <span class="math-container">$$\sum_{k=0}^n \frac{{n\choose k }}{k+1}$$</span>. Leave your answer in terms of n where appropriate.</p>
<p>Remark. There is an alternative method for computing the sums described here: make use of integration.</p>
<p>I can only list out the terms
<span class="math-container">$$\sum_{k=0}^n \frac{{n\choose k }}{k+1}=1+\frac{\binom{n}{1}}{2}+\frac{\binom{n}{2}}{3}+...+\frac{1}{m+1}$$</span>
I can't think of how to simplify them and get the answer.</p>
<p>Also, the question said I can use integration to solve it, but I have no idea how to start.I would greatly appreciate it if someone could show how to solve this.</p>
| Z Ahmed | 671,540 | <p><span class="math-container">$$\sum_{k=0}^{n}{n \choose k} x^k= (1+x)^n$$</span>
<span class="math-container">$$\implies \int_{0}^{1}\sum_{k=0}^{n} {n \choose k} x^k dx=\int_{0}^{1} (1+x)^n dx.$$</span>
<span class="math-container">$$\implies \sum_{k=0}^{n} \frac{{n \choose k}}{k+1}=\frac{2^{n+1}-1}{n+1}.$$</span></p>
|
3,059,571 | <p><span class="math-container">$$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$</span></p>
<p>I only know of L'hopital method but that is very long. Is there a shorter method to solve this?</p>
| lab bhattacharjee | 33,337 | <p>Set <span class="math-container">$\pi-2x=4y$</span> to find <span class="math-container">$$\lim_{y\to0}\dfrac{\tan y(1-\cos2y)}{(2y)^3}=\lim{...}\left(\dfrac{\sin y}y\right)^3\dfrac1{4\cos y}=?$$</span></p>
|
2,699,942 | <p>I am confused about one thing during the lecture. </p>
<p>Let $x_n = n$ and $A_n = \{x_k | k \ge n\} = \{n, n+1, n+2, ...\}$.</p>
<p>Then, $\inf A_n = n $, and $\sup A_n = \infty$. </p>
<p>My lecturer also said that $\lim\inf x_n = \lim\inf A _n=\lim n$. </p>
<p>My thinking is that $\{x_n\}_{n=1}^{\infty}=\{1, 2, 3, .....\}$. Shouldn't $\inf x_n = 1$?? Then, $\lim \inf x_n =1$, which is not equal to $\lim n$.</p>
<p>Could you tell me if I am wrong? </p>
| user284331 | 284,331 | <p>Because we define $\liminf\limits_{n\rightarrow\infty}x_{n}=\lim\limits_{n\rightarrow\infty}\left(\inf\limits_{k\geq n}x_{k}\right)=\lim\limits_{n\rightarrow\infty}\left(\inf A_n\right)$.</p>
<p>It is not defined as $\lim\limits_{n\rightarrow\infty}\inf\{x_{k}:k=1,2,...\}$. Note that $\inf\{x_{k}: k=1,2,...\}$ is an extended real number independent of $n$, hence $\lim\limits_{n\rightarrow\infty}\inf\{x_{k}: k=1,2,...\}$ is simply $\inf\{x_{k}: k=1,2,...\}$.</p>
<p>@Did has noted a good comment.</p>
|
21,156 | <p>The title says it all, is there a way to get in contact which users who consistently post answers without using <span class="math-container">$\LaTeX$</span>? I've come across a user who does that and (as I had some free time) edited about 10-15 of his posts, some of his answers were barely readable; on each post I left a comment including a link to the MathJax tutorial. He still keeps posting answers without using <span class="math-container">$\LaTeX$</span>, so is there anything else besides editing and commenting one can do?</p>
| nbubis | 28,743 | <p>I know of some experienced users on Math.SE who regularly post excellent answers, but due to physical disabilities or technological inexpertise have a lot of difficulty typing in $\LaTeX$.</p>
<p>I'm not saying all users have an actual problem, but for the sake of those who do, it's worth giving them the benefit of the doubt. </p>
<p>To @NajibIdrissi's comment - People who have sight problems may not use regular monitors or browsers, and can't always see the outputted $\LaTeX$, so I can understand that they’d rather not type it in the first place. As for technical expertise, Learning such skills above a certain age may not be trivial, and there's no reason why we'd want to lose a professional (though older) Mathematician just because it means some of us younger members will have to do some editing work.</p>
|
21,156 | <p>The title says it all, is there a way to get in contact which users who consistently post answers without using <span class="math-container">$\LaTeX$</span>? I've come across a user who does that and (as I had some free time) edited about 10-15 of his posts, some of his answers were barely readable; on each post I left a comment including a link to the MathJax tutorial. He still keeps posting answers without using <span class="math-container">$\LaTeX$</span>, so is there anything else besides editing and commenting one can do?</p>
| ASCII Advocate | 260,903 | <p>ASCII is more universal than LaTeX. Where's the problem?</p>
|
1,720,053 | <p>The PDF describes the probability of a random variable to take on a given value:</p>
<p>$f(x)=P(X=x)$</p>
<p>My question is whether this value can become greater than $1$?</p>
<p>Quote from wikipedia:</p>
<p>"Unlike a probability, a probability density function can take on values greater than one; for example, the uniform distribution on the interval $[0, \frac12]$ has probability density $f(x) = 2$ for $0 \leq x \leq \frac12$ and $f(x) = 0$ elsewhere."</p>
<p>This wasn't clear to me, unfortunately. The question has been asked/answered here before, yet used the same example. Would anyone be able to explain it in a simple manner (using a real-life example, etc)?</p>
<p>Original question:</p>
<p>"$X$ is a continuous random variable with probability density function $f$. Answer with either True or False.</p>
<ul>
<li>$f(x)$ can never exceed $1$."</li>
</ul>
<p>Thank you!</p>
<p>EDIT: Resolved.</p>
| drhab | 75,923 | <p>Your conception of <a href="https://en.wikipedia.org/wiki/Probability_density_function" rel="noreferrer">probability density function</a> is wrong.</p>
<p>You are mixing it up with <a href="https://en.wikipedia.org/wiki/Probability_mass_function" rel="noreferrer">probability mass function</a>.</p>
<p>If <span class="math-container">$f$</span> is a PDF then <span class="math-container">$f(x)$</span> is not a probability and does not have the restriction that it cannot exceed <span class="math-container">$1$</span>.</p>
|
1,720,053 | <p>The PDF describes the probability of a random variable to take on a given value:</p>
<p>$f(x)=P(X=x)$</p>
<p>My question is whether this value can become greater than $1$?</p>
<p>Quote from wikipedia:</p>
<p>"Unlike a probability, a probability density function can take on values greater than one; for example, the uniform distribution on the interval $[0, \frac12]$ has probability density $f(x) = 2$ for $0 \leq x \leq \frac12$ and $f(x) = 0$ elsewhere."</p>
<p>This wasn't clear to me, unfortunately. The question has been asked/answered here before, yet used the same example. Would anyone be able to explain it in a simple manner (using a real-life example, etc)?</p>
<p>Original question:</p>
<p>"$X$ is a continuous random variable with probability density function $f$. Answer with either True or False.</p>
<ul>
<li>$f(x)$ can never exceed $1$."</li>
</ul>
<p>Thank you!</p>
<p>EDIT: Resolved.</p>
| Shikhar Srivastava | 1,131,158 | <p>Here's an intuition:</p>
<p>Probability Density exists in the continuous space. Probability Mass exists in the discrete space.</p>
<p>The PDF <span class="math-container">$f(x)$</span> is the derivative of the CDF <span class="math-container">$F(x)$</span>:
<span class="math-container">$$ f(x) = \frac{d(F(X))}{d(x)} $$</span></p>
<p>Thus, for a given range of <span class="math-container">$x \in (x_1, x_2]$</span>, we can say that the pdf is the <em>unit change in cumulative probability</em> when moving from <span class="math-container">$x_1$</span> to <span class="math-container">$x_2$</span>, i.e. <span class="math-container">$f(x)_{\{x_1,x_2\}} = \frac{F(x_2) - F(x_1)}{x_2 - x_1}$</span>.</p>
<p>Or, "How much will my probability of <span class="math-container">$X \in \{0,x_0\}$</span> increase if I include <span class="math-container">$\{x_1, x_2\}$</span> in my range, normalised by the size of the range <span class="math-container">$|\{x_1, x_2\}|$</span>"?.</p>
<p>You can now imagine, if there is a highly <em>dense</em> range <span class="math-container">$(0, \frac{1}{2})$</span> with probability of 1 within it's range (i.e. <span class="math-container">$F(1/2) - F(0) = 1$</span>), then it's density would therefore be 2.</p>
<p><span class="math-container">$$ f(x)_{(0, \frac{1}{2}]} = \frac{F(1/2) - F(0)}{1/2 - 0} = 2$$</span></p>
|
2,258,557 | <p>Why the equation of an arbitrary straight line in complex plane is $zz_o + \bar z \bar z_0 = D$ where D $\in R$</p>
<p>I understand that a vertical straight line can be defined by the equation $z+\bar z= D$ because suppose $z =x+yi$ then $\bar z = x-yi$ Thus, $z+\bar z = x+yi+x-yi=2x$ which is an arbitrary vertical straight line in w-plane.</p>
<p>But why $zz_o + \bar z \bar z_0 = D$ is an arbitrary straight line in complex plane?</p>
| Community | -1 | <p>Hack the equation.</p>
<p>Substitute:</p>
<p>$$
\begin{cases}
z = x + iy \\
z_0 = x_0 + iy_0
\end{cases}
$$</p>
<p>Do some algebraic manipulations and you'll obtain the equation of a line.</p>
<p>Maybe what confuses you is that neither $z_0$ nor $D$ have a clear geometric interpretation in terms of intercepts, distance to the origin, etc...</p>
<p>Hope this helps!</p>
|
59,495 | <p>Suppose $K$ is an $n$-dimensional $C^2$ convex body in $\mathbb{R}^{n+1}$. We choose two distinct directions $z_0, z_1\in\mathbb{S^{n}}$. If $P_1$ and $P_2$ are the corresponding hyperplanes($z_0\perp P_1$ and $z_1\perp P_2$) and $K'$ is the projection of $K$ on $P_1\cap P_2$, what is the $Vol(K')$?
We know the support function, and for simplicity let's suppose the body is symmetric and centered at the origin.
If we just consider one hyperplane say, $P_0$, and want to compute the area of projection of $K$ on $P_0$ then the answer is $\frac{1}{2}\int_{\mathbb{S}^{n-1}}\frac{|\langle z,z_0\rangle|}{G}d\mu$ where $G$ is the Guass curvature of the boundary of $K$. I am looking for a solution of this type, possibly involving other symmetric functions of principle curvatures.</p>
| Dick Palais | 7,311 | <p>There is a paper here:</p>
<p><a href="http://www.math.poly.edu/~alvarez/pdfs/crofton.pdf">http://www.math.poly.edu/~alvarez/pdfs/crofton.pdf</a></p>
<p>that develops a theory of "Gelfand Transforms" which in a sense made precise there is a generalization of both the Radon Transform and the Cauchy-Crofton formula.</p>
|
4,450,169 | <p>Here is a (seemingly) simple problem in group theory. Given a non-elementary finite nilpotent group <span class="math-container">$N$</span>, show there exist <span class="math-container">$p \neq q$</span> primes such that <span class="math-container">$N$</span> has a quotient <span class="math-container">$\Bbb Z_{pq}^{2}$</span>.</p>
<p>Here, an elementary group is defined to be a direct product of a <span class="math-container">$p$</span> group and a cyclic group of order coprime to p. That is, <span class="math-container">$E$</span> elementary <span class="math-container">$\iff \exists P, C : E = P \times C$</span> where <span class="math-container">$|P| = p^{k}$</span>, and <span class="math-container">$C$</span> is cyclic such that <span class="math-container">$(|C|, p) = 1$</span>.</p>
<p>A nilpotent group is defined in the standard way, a group <span class="math-container">$N$</span> is nilpotent <span class="math-container">$\iff$</span> the central series of <span class="math-container">$N$</span> is finite.</p>
<p>I'm not sure how to approach this one-- does anyone have any pointers?</p>
| ancient mathematician | 414,424 | <p>(i) A finite nilpotent group is the direct product of its Sylow-subgroups.</p>
<p>(ii) The quotient of a finite <span class="math-container">$p$</span>-group by its Frattini subgroup is elementary abelian.</p>
<p>(iii) If the quotient of a finite group by its Frattini subgroup (the set of non-generators) is cyclic then the group itself is cyclic.</p>
<p>(iv) The direct product of two finite cyclic groups of coprime orders is itself cyclic.</p>
<p>From these standard results it is clear that a finite nilpotent group which is not "elementary" must have at least two Sylow subgroups which are non-cyclic, for primes <span class="math-container">$p,q$</span> say. Hence we get a quotient <span class="math-container">$\mathbb{Z}_p\times\mathbb{Z}_p\times\mathbb{Z}_q\times\mathbb{Z}_q\simeq\mathbb{Z}_{pq}\times\mathbb{Z}_{pq}$</span>.</p>
|
2,255,617 | <p>I am trying to learn how to do proofs by contradiction. The proof is,</p>
<p>"Prove by Contradiction that there are no positive real roots of $x^6 + 2x^3 +4x + 5$"</p>
<p>I understand that now I am attempting to prove that there is a positive real root of this equation, so I am able to contradict myself within the proof. I just don't even know where to start.</p>
| Community | -1 | <p>No. $\mathbb{Z} \to \mathbb{F}_2$ and $\mathbb{Z} \to \mathbb{Q}$ are counterexamples.</p>
|
21,372 | <blockquote>
<p>Let $ y = \min \{ (x + 6), (4 – x) \}$, then find $y$.</p>
</blockquote>
<p>How to solve this problem?</p>
| Bill Dubuque | 242 | <p><strong>HINT</strong> $\ $ For any continuous functions $\rm\:f,g,\:$ the intermediate value theorem implies that $\rm\ f-g\ $ will have constant sign between its roots. So you need only partition $\mathbb R$ by these roots and then evaluate the function at any test point of each interval to determine the sign on the whole interval.</p>
<p>In fact this decomposition technique generalizes to higher dimensions. It is known as the <a href="https://math.stackexchange.com/questions/13654/13681#13681">cylindrical decomposition algorithm</a> - an effective interpretation of Tarski's decision procedure for the first-order theory of the reals.</p>
|
3,433,492 | <p>I know that a function can admitted multiple series representation (according to Eugene Catalan), but I wonder if there is a proof for the fact that each analytic function has only one unique Taylor series representation. I know that Taylor series are defined by derivatives of increasing order. A function has one and only one unique derivative. So can this fact be employed to prove that each function only has one Taylor series representation?</p>
| Φίλ λιπ | 494,571 | <p>I think this simple proof is sufficient. I'm going to do it in two cases, but really the first case is a special case of the second.</p>
<p>Suppose a function <span class="math-container">$f(x)$</span> has two taylor series representations.</p>
<p><span class="math-container">$$f(x)=\sum a_n x^n$$</span></p>
<p><span class="math-container">$$f(x) = \sum b_n x^n$$</span></p>
<p>we know that <span class="math-container">$f(x) - f(x) = 0$</span>, so just plug in each of the representations</p>
<p><span class="math-container">$$f(x) - f(x) = \sum b_n x^n - \sum a_n x^n = 0$$</span></p>
<p><span class="math-container">$$\sum (b_n-a_n) x^n = 0$$</span></p>
<p>The only way we can get 0 is if the coefficients are separately equal, since there is no cancellation, in general for all x, for monomials of different degree.</p>
<p><span class="math-container">$$b_n-a_n = 0 $$</span>
<span class="math-container">$$b_n =a_n $$</span></p>
<p>Now suppose we center the series at different points for each representation, i.e.</p>
<p><span class="math-container">$$f(x)=\sum a_n (x-a)^n$$</span></p>
<p><span class="math-container">$$f(x) = \sum b_n (x-b)^n$$</span></p>
<p>The binomial theorem is helpful here</p>
<p><span class="math-container">$$f(x)=\sum a_n (x-a)^n = \sum a_n\sum\binom{n}{k}a^{n-k}x^k =\sum a'_kx^k $$</span></p>
<p>so <span class="math-container">$a'_k$</span> is just a new constant. The same will happen with the other representation, just set <span class="math-container">$a$</span> to <span class="math-container">$b$</span>, and you will get again that</p>
<p><span class="math-container">$$b'_k =a'_k$$</span></p>
<p>So the Taylor series representation is unique.</p>
|
396,794 | <p>Let's say that a (right) module <span class="math-container">$M$</span> is <em>well complemented</em> if every non-zero submodule of <span class="math-container">$M$</span> has an indecomposable direct summand (by the way, is there a better or more standard name for this property?). For instance, every module of finite <a href="https://en.wikipedia.org/wiki/Uniform_module#Uniform_dimension_of_a_module" rel="nofollow noreferrer">uniform dimension</a> is well complemented.</p>
<blockquote>
<p><strong>Question.</strong> Is the regular right module <span class="math-container">$R_R$</span> of a von Neumann regular ring <span class="math-container">$R$</span> well complemented?</p>
</blockquote>
<p>As a recall, a ring <span class="math-container">$R$</span> is <em>von Neumann regular</em> if, for every <span class="math-container">$x \in R$</span>, there exists <span class="math-container">$y \in R$</span> such that <span class="math-container">$x = xyx$</span>.</p>
| Luc Guyot | 84,349 | <p><strong>No</strong>, a <a href="https://en.wikipedia.org/wiki/Free_Boolean_algebra" rel="nofollow noreferrer">free Boolean algebra</a> <span class="math-container">$R$</span> on an infinite cardinal <span class="math-container">$\kappa$</span> (e.g., if <span class="math-container">$\kappa = \aleph_0$</span>, <span class="math-container">$R$</span> is the Cantor algebra), is a commutative von Neumann regular ring which is not well complemented as an <span class="math-container">$R$</span>-module.</p>
|
945,395 | <p>Let $a_1$ be real, and define $$a_{n+1}=\frac{2a_n^3}{1+a_n^4}$$ How can I prove that this $\{a_n\}$ to have limit. </p>
<p>I find it is hard to track. What I can do is just when $a_1=1$ then $a_n=1$; when $a_1=-1$, then $a_n=-1$; when $|a_1|<1$, $a_n\to 0$. When $|a_1|>1$, I have not find any idea.</p>
| Christian Blatter | 1,303 | <p>We are given the recursion
$$a_0:=a>0,\qquad a_{n+1}:=f(a_n)\quad(n\geq0)$$
with
$$f(x):={2x^3\over 1+x^4}\ .$$
The following figure shows the graph of $f$ for $x\geq0$:</p>
<p><img src="https://i.stack.imgur.com/8lDnT.jpg" alt="enter image description here"></p>
<p>Since
$$f(x)\geq0,\qquad x-f(x)={x(1-x^2)^2\over 1+x^4}\geq0$$
(with strict inequality unless $x=0$ or $x=1$), we deduce that the the sequence $(a_n)_{n\geq0}$ is monotonically decreasing towards a limit $\alpha\geq0$. This limit necessarily satisfies $f(x)=x$, whence $\alpha\in\{0,1\}$, and may depend on the starting value $a$.</p>
<p>It follows that for $0<a<1$ we have $\lim_{n\to\infty} a_n=0$.</p>
<p>When $a=1$ one has $a_n=1$ for all $n\geq0$.</p>
<p>In order to analyze the case $a>1$ we write $f$ in the form
$$f(x)=1+{x-1\over1+x^4}(1+x+x^2-x^3)\ .\tag{1}$$
Here the right hand side is $>1$ for $1<x<\xi\doteq1.83929$, where $\xi$ is the real zero of the polynomial on the right of $(1)$, and is $<1$ when $x>\xi$; see also the figure. </p>
<p>It follows that for $1<a<\xi$ we have $\lim_{n\to\infty} a_n=1$. When $a=\xi$ we obtain $a_1=1$ and then $a_n=1$ for all $n\geq1$. When $a>\xi$ then $a_1<1$ and as a consequence $\lim_{n\to\infty} a_n=0$.</p>
|
120,687 | <p>Consider the following code</p>
<pre><code>styles = {Red, Blue, {Red, Dashed}, {Blue, Dashed}}
pt1 = Plot[{x^2, 2 x^2, 1/x^2, 2/x^2}, {x, 0, 3}, Frame -> True,
PlotStyle -> styles, PlotLegends -> {"1", "2", "1", "2"}]
</code></pre>
<p>I would like the two red lines to carry the same label "1" and the two blue lines the same label "2". That is, in the legend I would like a red line and a red-dashed line below each other and then one label right of it. Similarly for the blue lines. Does anybody know how to do this?</p>
| Michael E2 | 4,999 | <p>First, <code>NIntegrate[f1[x], {x, xmin, xmax}]</code> usually proceeds by constructing an <code>Experimental`NumericalFunction</code> from the expression for <code>f1[x]</code>. This will circumvent an attempt to memoize <code>f1</code> in the OP's manner, <code>f1[x_] := f1[x] =...</code>. One can prevent this by memoizing the function with <code>?NumericQ</code> checks via <code>f2[x_?NumericQ] := f2[x] = ...</code>. One thing to consider is that the <code>NumericalFunction</code> constructed in each case is different, and if one of them is more efficient, I would bet it's the first one; see below for evidence of this. Anton Antonov <a href="https://mathematica.stackexchange.com/questions/120678/how-to-avoid-repetitive-calculation-when-doing-numerical-integral#comment327037_120722">alluded to this in a comment</a>.</p>
<p>Second, as <a href="https://mathematica.stackexchange.com/questions/120678/how-to-avoid-repetitive-calculation-when-doing-numerical-integral#comment326894_120678">Mr.Wizard has pointed out</a>, <code>NIntegrate</code> on a list of integrands just call <code>NIntegrate</code> on each integrand. To expand further, the reason there is no speed-up is that the first integral samples only <code>116721</code> points, whereas the second samples <code>1063425</code>. The time it takes to do the second integral dwarfs the first. Memoizing the first integral isn't going to help much, even if all the saved values are reused. (The reason for the difference is that the <code>Sqrt</code> creates some singularities, as well as complex values, which makes the second integral more difficult to compute.)</p>
<p><em>Code for testing the sampling:</em></p>
<pre><code>ClearAll[g];
g[x_, y_, z_] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
{pts1, pts2} =
Last@Last@Reap@
NIntegrate[#, {x, 0, 10}, {y, 0, 10}, {z, 0, 10},
Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0},
PrecisionGoal -> 7,
EvaluationMonitor :> Sow[{x, y, z}]] & /@
{g[x, y, z], Sqrt[g[x, y, z]] + x}; // AbsoluteTiming
(* {10.162, Null} *)
ClearAll[g];
g[x_, y_, z_] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
ptsall = Last@Last@Reap@
NIntegrate[{Sqrt[g[x, y, z]] + x, g[x, y, z]},
{x, 0, 10}, {y, 0, 10}, {z, 0, 10},
Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0},
PrecisionGoal -> 7,
EvaluationMonitor :> Sow[{x, y, z}]]; // AbsoluteTiming
(* {10.1304, Null} *)
Length /@ {pts1, pts2}
Total@%
(*
{116721, 1063425}
1180146
*)
Length@ptsall (* same as the preceeding Total *)
(* 1180146 *)
</code></pre>
<p><em>Code for estimating how much time could be saved by memoizing ~120K sample values, if it worked as expected (about half a second):</em></p>
<pre><code>ClearAll[g];
g[x_, y_, z_] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
Table[g[x, y, z], {x, 0., 10, 1./8}, {y, 0., 10, 1./4}, {z, 0., 10, 1./4}]
// Flatten // Length // AbsoluteTiming
Table[Sqrt[g[x, y, z]] + x, {x, 0., 10, 1./8}, {y, 0., 10, 1./4}, {z, 0., 10, 1./4}]
// Flatten // Length // AbsoluteTiming
(*
{0.81421, 136161}
{0.243886, 136161}
*)
</code></pre>
<p><em>Code to compare the <code>Experimental`NumericalFunction</code> of the symbolic integrand <code>g</code> and the <code>?NumericQ</code> version:</em></p>
<pre><code>ClearAll[g];
g[x_, y_, z_] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
NIntegrate[g[x, y, z], {x, 0, 10}, {y, 0, 10}, {z, 0, 10},
Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0},
PrecisionGoal -> 7,
IntegrationMonitor :> ((numfnSymbolic = First[#]["NumericalFunction"]) &)];
ClearAll[g];
g[x_?NumericQ, y_?NumericQ, z_?NumericQ] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
NIntegrate[g[x, y, z], {x, 0, 10}, {y, 0, 10}, {z, 0, 10},
Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0},
PrecisionGoal -> 7,
IntegrationMonitor :> ((numfnNumeric = First[#]["NumericalFunction"]) &)];
Table[numfnSymbolic[x, y, z], {x, 0., 10, 1./8}, {y, 0., 10, 1./8}, {z, 0., 10, 1./8}]
// Flatten // Length // AbsoluteTiming
Table[numfnNumeric[x, y, z], {x, 0., 10, 1./8}, {y, 0., 10, 1./8}, {z, 0., 10, 1./8}]
// Flatten // Length // AbsoluteTiming
(*
{0.747187, 531441}
{3.70936, 531441}
*)
</code></pre>
<p>Note this suggests that the difference between the OP's <code>g</code> and a truly memoized <code>g</code> with <code>?NumericQ</code> should be around 6 sec. Well, it is:</p>
<pre><code>ClearAll[g];
g[x_?NumericQ, y_?NumericQ, z_?NumericQ] := g[x, y, z] = Exp[Sin[x]] + Cos[y + z];
NIntegrate[{g[x, y, z], Sqrt[g[x, y, z]] + x}, {x, 0, 10}, {y, 0, 10}, {z, 0, 10},
Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0},
PrecisionGoal -> 7]; // AbsoluteTiming
(* {16.2901, Null} *)
</code></pre>
<p>One caveat: In the <code>Table[]</code> comparison, I tested just one numerical function (from one integration subregion). I cannot say authoritatively that all numerical functions are the same. (Help, anyone?)</p>
|
12,927 | <p>The problem:</p>
<p><strong><em>Three poles standing at the points $A$, $B$ and $C$ subtend angles $\alpha$, $\beta$ and $\gamma$ respectively, at the circumcenter of $\Delta ABC$.If the heights of these poles are in arithmetic progression; then show that $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression.</em></strong></p>
<p>Now, what I could not understand is subtending of the angle part,precisely <strong>how a point subtends angle at another point?</strong> So, what I am looking for a proper explanation of the problem statement with a figure, since it's troubling me from sometime.</p>
<p>PS: I am not looking for the solution (as of now) or any hint regarding the solution, just a clear explanation will be appreciated.</p>
<hr>
<p>My solution using <a href="https://math.stackexchange.com/users/1102/moron">Moron</a>'s <a href="https://math.stackexchange.com/questions/12927/help-to-understanding-the-problem/12933#12933">interpretation</a>,</p>
<p>Let $a$ $b$ and $c$ are the length of three sides of the poles and $O$ be the circumcenter then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$</p>
<p>$$ \cot \gamma = \frac{OC}{c}$$</p>
<p>As $O$ is the circumcenter,$OA = OB = OC = k $(say) </p>
<p>Again, $a$ $b$ and $c$ are in arithmetic progression, hence</p>
<p>$$2 \cdot \frac{k}{\cot \alpha} = \frac{k}{\cot \beta} + \frac{k}{\cot \gamma}$$</p>
<p>Canceling $k$ from both sides,</p>
<p>$$2 \cdot \frac{1}{\cot \alpha} = \frac{1}{\cot \beta} + \frac{1}{\cot \gamma}$$</p>
<p>Hence, $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression. (<strong>QED</strong>)</p>
| Ross Millikan | 1,827 | <p>I would read it that the poles have a diameter>0. The pole at A has diameter so the angle seen from the circumcenter is $\alpha$. See if that works.</p>
|
2,884,785 | <p>Find all integral pairs (x,y) such that - $$( xy - 1)^2 = (x +1)^2 + ( y+1)^2$$</p>
<hr>
<p><strong>My Approach :</strong></p>
<p>I just expanded this equation and wrote it in another form - $$\frac{(xy+1)(xy-1)}{(x+y)}-2=x+y$$ and from this we can say that $(x+y)|(xy+1) \ \mathrm{or}\ (x+y)|(xy-1) $ . But i don't know how to solve it further. Please help me with this. </p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>$$x^2(y^2-1)-2x(1+y)-(y+1)^2=0$$</p>
<p>What if $y+1=0?$</p>
<p>Else $$x^2(y-1)-2x-(y+1)=0$$</p>
<p>Method$\#1:$</p>
<p>$$x=\dfrac{2\pm\sqrt{4+4(y^2-1)}}{2(y-1)}=\dfrac{1\pm y}{1+y}$$</p>
<p>Now $\dfrac{1-y}{1+y}=\dfrac2{1+y}-1$</p>
<p>$\implies1+y$ must divide $2$</p>
<p>Method $\#2:$</p>
<p>By symmetry, $x+1$ must be a factor of $$x^2(y-1)-2x-(y+1)$$</p>
<p>What is the quotient?</p>
|
1,141,074 | <p>I need help with this integral: $$\int\frac{\sqrt{\tan x}}{\cos^2x}dx$$ I tried substitution and other methods, but all have lead me to this expression: $$2\int\sqrt{\tan x}(1+\tan^2 x)dx$$ where I can't calculate anything... Any suggestions? Thanks!</p>
| Emilio Novati | 187,568 | <p>$$
\int \dfrac{\sqrt{\tan x}}{\cos^2 x} dx =
\int \sqrt{\tan x} \; d(\tan x) = \dfrac{2}{3}\sqrt{(\tan x)^3} +C
$$</p>
|
62,790 | <p>Among several possible definitions of ordered pairs - see below - I find Kuratowski's the least compelling: its membership graph (2) has one node more than necessary (compared to (1)), it is not as "symmetric" as possible (compared to (3) and (4)), and it is not as "intuitive" as (4) - which captures the intuition, that an ordered pair has a first and a second element.</p>
<p><img src="https://i.stack.imgur.com/RVRjq.png" alt="alt text"><a href="http://epublius.de/mathoverflow/orderedpairs.png" rel="noreferrer">(source)</a></p>
<p><sup><em>Membership graphs for possible definitions of ordered pairs (≙ top node, arrow heads omitted)</em></sup></p>
<pre><code>1: (x,y) := { x , { x , y } }
2: (x,y) := { { x } , { x , y } } (Kuratowski's definition)
3: (x,y) := { { x } , { { x } , y } }
4: (x,y) := { { x , 0 } , { 1 , y } } (Hausdorff's definition)
</code></pre>
<p>So my question is: </p>
<blockquote>
<p>Are there good reasons to choose
Kuratowski's definition (or did
Kuratowski himself give any) instead of
one of the more "elegant" - sparing,
symmetric, or intuitive -
alternatives?</p>
</blockquote>
| Andreas Blass | 6,794 | <p>Kuratowski's definition arose naturally out of Kuratowski's idea for representing any linear order of a set $S$ in terms of just sets, not ordered pairs. The idea was that a linear ordering of $S$ can be represented by the set of initial segments of $S$. Here "initial segment" means a nonempty subset of $S$ closed under predecessors in the ordering. When applied to the special case of two-element sets $S$, this gives the Kuratowski ordered pair.</p>
|
1,144,141 | <p>I have a question for my exam and I find it hard to understand.</p>
<p>I have to prove that the following formula is logically valid:</p>
<p><img src="https://i.stack.imgur.com/kdELq.jpg" alt="Example"></p>
<p>The professor told me to "push" all the symbols inside the brackets, and use the deduction theorem.</p>
<p>But I don't know how to do it, because I can't find the identities to push the "exist" symbol inside the brackets.</p>
<p>Your help is appriciated, thank you.</p>
<p>Alan</p>
| Hagen von Eitzen | 39,174 | <p>Assume $\forall yp(y)$. Then $p(x)\to\forall yp(y)$ is true for any $x$.
If on th eother hand $\neg \forall yp(y)$, then $\exists y\neg p(y)$. Let $x$ be such an $y$ then again $p(x)\to\forall yp(y)$ is true, this time because the antecedent is false.</p>
|
3,772,923 | <p>My child's teacher raised a quesion in class for students who are interested to prove. The teacher says that the volume of a cube is the greatest among rectangular-faced shapes of the same perimeter and asks his students to prove this proposition.</p>
<p>I considered the relationship between the length of the sides of a cube and the lengths of the sides of rectangular-faced shapes in different situation. But when the calculations came down to polynomials, I couldn't proceed due to the uncertainty of the variables in the polynomials.</p>
<p>Can anyone please find a good way to prove the above proposition? Or is there already a proof? Thank you for your help!</p>
| mjw | 655,367 | <p>If you mean by "perimeter" the sum of the edges, then yes, the cube is the maximal rectangular parallelepiped among those with the same "perimeter".</p>
<p>Let the edges have lengths <span class="math-container">$(a,b,c)$</span>.</p>
<p>Then the volume is <span class="math-container">$V=abc$</span> and the "perimeter" is <span class="math-container">$p=4(a+b+c).$</span></p>
<p>We can maximize volume while constraining the sum of the edges using Lagrange multipliers:</p>
<p><span class="math-container">$$\begin{aligned}
L &= abc-\lambda \left(a+b +c-\frac{p}{4}\right)\\
0&=\frac{\partial L}{\partial a} = bc - \lambda\\
0&=\frac{\partial L}{\partial b} = ac - \lambda\\
0&=\frac{\partial L}{\partial c} = ab - \lambda\\
\end{aligned}$$</span>
so that
<span class="math-container">$$bc=ac=ab$$</span> and
<span class="math-container">$$a=b=c.$$</span></p>
|
295,545 | <p>The following figure depicts the paths from home to work. SAM never travels through the park when going to work.</p>
<p><img src="https://i.stack.imgur.com/IANqM.png" alt="enter image description here"></p>
| Mårten W | 58,780 | <p>What the author meant was that you want two vectors satisfying the equation of the plane (that is, they should lie in the plane). Since the equation of the plane is $x-y-z=0$, one can choose two of the coordinates, and then solve for the third.</p>
<p>In your case $a$ is obtained by choosing $x=0$ and $y=1$, and plugging that into the equation of the plane in order to find out that $z=-1$. In the same way, $b$ is obtained by choosing two coordinates and solving for the third.</p>
<p>As a matter of fact, a better approach would be to choose orthogonal vectors in the plane, because that would facilitate an orthonormal change of basis (you can just transpose the matrix instead of inverting it the painful way).</p>
|
7,981 | <p>I've read so much about it but none of it makes a lot of sense. Also, what's so unsolvable about it?</p>
| vonjd | 346 | <p>In very layman's terms it states that there is some order in the distribution of the primes (which seem to occur totally chaotic at first sight). Or to say it like Shakespeare: "Though this be madness, yet there is method in 't."</p>
<p>If you want to know more there is a new trilogy about that topic where the first volume has just arrived:
<a href="http://www.secretsofcreation.com/volume1.html">http://www.secretsofcreation.com/volume1.html</a></p>
<p>It is a marvelous and easy to understand book from a number theorist who knows his stuff! </p>
|
7,981 | <p>I've read so much about it but none of it makes a lot of sense. Also, what's so unsolvable about it?</p>
| Community | -1 | <p>Here is a very simply description of the Riemann Hypothesis that requires nothing more than a 3rd grade education to understand:</p>
<p><a href="http://www.jstor.org/pss/2323497" rel="nofollow">http://www.jstor.org/pss/2323497</a></p>
<p>There is also a beautiful proof linking the Farey sequence of fractions to the Riemann hypothesis by Jerome Franel. It's only three pages long and should be able to be understood by any undergraduate mathematics major.</p>
|
2,107,854 | <p>What is the limit when $n \to \infty$?</p>
<p>$$\lim_{n \to \infty} \frac{1}{n^4} \sum_{J=0}^{2n-1} J^3=?$$</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{n \to \infty}\bracks{{1 \over n^{4}}\sum_{J = 0}^{2n - 1}J^{3}} &\ =\
\overbrace{\lim_{n \to \infty}\bracks{{1 \over \pars{n + 1}^{4} - n^{4}}
\pars{\sum_{J = 0}^{2n + 1}J^{3} - \sum_{J = 0}^{2n - 1}J^{3}}}}
^{\ds{\mbox{Stolz-Ces$\mrm{\grave{a}}$ro Theorem}}}
\\[5mm] & =
\lim_{n \to \infty}\,\,
{\pars{2n + 1}^{3} + \pars{2n}^{3} \over 4n^{3} + 6n^{2} + 4n + 1} =
\lim_{n \to \infty}\,\,
{16n^{3} + 12n^{2} + 6n + 1\over 4n^{3} + 6n^{2} + 4n + 1} = \bbx{\ds{4}}
\end{align}</p>
|
9,437 | <p>I love the way that Mathematica allows me to type in of formulas. It is really easy to type complicated expressions with shortcuts on the keyboard. It would be great if I could use Mathematica completely to publish my articles. The biggest reason I don't already do this is:</p>
<p>I can't find the proper tutorial for styling notebooks for PDF export. How is it possible to deal with page numbering, headers, footers, page breaks, or placing graphs in specific positions, instead of being limited to line by line text? Is this possible? I believe it is -- I'm fascinated with Mathematica's capabilities, but I don't yet have the skills to take advantage of all the features it offers. </p>
<p>If someone would write tutorial on styling notebooks, formatting, and exporting to PDF, I believe that it would be appreciated by many others. I have seen some texts where it's emphasized that they are written totally in Mathematica (and they are really styled well). </p>
<p>Thank you for any tips, ways of accomplishing these tasks and for sharing your experience. </p>
| NCSNY | 19,431 | <p>It is easier to create a notebook object and set options. For example, if I want a PDF with some plots and an equation and save it as a PDF in Landscape form (so its not cut off):</p>
<pre><code>newbook = CreateDocument[{plot1, plot2, plot3, equations1}]
SetOptions[newbook,
PrintingOptions -> {"PaperOrientation" -> "Landscape"}]
Export[NotebookDirectory[] ~~ "myfile.pdf", newbook]
</code></pre>
<p>(The export command puts the new PDF in the same directory as the notebook with all theses commands.) You can get any number of notebook options by going to the "Format->option inspector" choose your notebook, find the option you want, change it, then view as text or do a </p>
<pre><code>Options[newbook]
</code></pre>
<p>and see what option was added (and do a copy and paste)</p>
<p>Once you determine what options you want, you can programmatically set all aspects of the new notebook (Headers, Footers, etc.) and then save it as a PDF. I think this is the easiest approach to doing this.</p>
|
2,508,011 | <blockquote>
<p>find the <span class="math-container">$x$</span> :</p>
<p><span class="math-container">$$x^2(x-1)^2+x^2=8(x-1)^2$$</span></p>
</blockquote>
<hr />
<p>My Try :</p>
<p><span class="math-container">$$x^2(x-1)^2+x^2=8(x-1)^2\\ x^2(x^2-2x+1)+x^2=8(x^2-2x+1)\\x^4-2x^3+x^2+x^2=8x^2-16x+8\\x^4-2x^3-6x^2+16x-8=0$$</span></p>
<p>Now What ?</p>
| C. Dubussy | 310,801 | <p>Hint : Try the divisors of $8$.</p>
|
2,508,011 | <blockquote>
<p>find the <span class="math-container">$x$</span> :</p>
<p><span class="math-container">$$x^2(x-1)^2+x^2=8(x-1)^2$$</span></p>
</blockquote>
<hr />
<p>My Try :</p>
<p><span class="math-container">$$x^2(x-1)^2+x^2=8(x-1)^2\\ x^2(x^2-2x+1)+x^2=8(x^2-2x+1)\\x^4-2x^3+x^2+x^2=8x^2-16x+8\\x^4-2x^3-6x^2+16x-8=0$$</span></p>
<p>Now What ?</p>
| thesmallprint | 438,651 | <p>So it can be checked that your polynomial $$x^4-2x^3-6x^2+16x-8=0$$ has a factor of $(x-2)$. We check that this is so by the <a href="https://en.wikipedia.org/wiki/Factor_theorem" rel="nofollow noreferrer">factor theorem</a>. Then (and it should be checked by you too) that by carrying out the appropriate algebraic division that</p>
<p>$$x^4-2x^3-6x^2+16x-8=(x-2)(x^3-6x+4).$$</p>
<p>Then by a similar argument, we may divide this through by $(x-2)$ again to give</p>
<p>$$(x^3-6x+4)=(x-2)(x^2+2x-2).$$</p>
<p>Then apply the quadratic formula to yield</p>
<p>$$x^2+2x-2=0 \Rightarrow -1\pm\sqrt{3},$$</p>
<p>so altogether $$x=2, -1\pm\sqrt{3}.$$</p>
|
3,615,117 | <p>I want to find the intersection of the sphere <span class="math-container">$x^2+y^2+z^2 = 1$</span> and the plane <span class="math-container">$x+y+z=0$</span>. </p>
<p><span class="math-container">$z=-(x+y)$</span> that gives <span class="math-container">$x^2+y^2+xy= \frac 12$</span></p>
<p>How do I represent this in the standard form of ellipse? Any help is appreciated to proceed further. Thanks in advance.</p>
| Ninad Munshi | 698,724 | <p>Let <span class="math-container">$\sqrt{2}x = u+v$</span> and <span class="math-container">$\sqrt{2}y=u-v$</span>. Then the resulting expression is</p>
<p><span class="math-container">$$3u^2 + v^2 = 1$$</span></p>
<p>which is the standard form of an ellipse and since the transformation is a pure rotation, the shapes of the objects haven't been distorted at all.</p>
|
2,274,736 | <p>I am finding particular subgroups of $Q_{12}$ and had a couple of questions about it.</p>
<p>$Q_{12}=\langle a,b:a^6=1,b^2=a^3,ba=a^{-1}b\rangle$</p>
<p>Firstly here is part of a solution I came across: </p>
<p>The first step is to establish the orders of the elements. So $1$ has order 1, $a^3$ has order 2, $a^2$ and $a^4$ have order 3, $a^ib$ has order 4 for all $i$, and $a$ and $a^{−1}$ have order 6. Therefore there is a unique subgroup of order 1, namely {1}, a unique subgroup of order 2, namely $\langle a^3 \rangle$ and
a unique subgroup of order 3, namely $\langle a^2\rangle$</p>
<p>How are the subgroups $\langle a^3 \rangle$ and $\langle a^2 \rangle$ identified straight away as unique? I understand that the fact that they are unique means they are normal and the fact they are normal means they are unique. But I wanted to know why they are identified as unique first hand in this example which then goes on to say because they are unique they are normal (I'm not interested in the other way round). Is it because they are cyclic?</p>
<p>Also the subgroups of order 4 are $\langle b \rangle$, $\langle ab \rangle$ and $\langle a^2b \rangle$. I'm struggling to see how because take $\langle b \rangle$, are the elements $\{1,b,a^3b\}$? but I thought a subgroup of order 4 had to have 4 elements?</p>
| caverac | 384,830 | <p>Recall that for a multinomial distribution the PDF is</p>
<p>$$
f(x_1, \ldots, x_k; n, \pi_1, \ldots, \pi_k) = \frac{n!}{x_1!\cdots x_k!}\pi^{x_1} \cdots \pi^{x_k} = {{n} \choose {x_1,\ldots,x_k}}\pi^{x_1} \cdots \pi^{x_k}
$$</p>
<p>with</p>
<p>$$
\sum_{i= 1}^k x_i = n
$$</p>
<p>In your case $k = 3$. The likelihood evaluated at the point ${\bf p} = (1 - 2p, p, p)$ is then</p>
<p>$$
L({\bf p}) = {{n} \choose {x_1,x_2,x_3}}(1- 2p)^{x_1} p^{x_2} p^{x_3} = {{n} \choose {x_1,x_2,x_3}}(1- 2p)^{x_1} p^{x_2+ x_3} = L(p)
$$</p>
<p>Now, to find the value of $p$ for which $L(p)$ we need to solve $dL/dp = 0$ </p>
<p>\begin{eqnarray}
\frac{dL}{dp} &=& (1-2 p)^{x_1} (x_2 + x_3) p^{x_2 + x_3-1}-2
x_1 (1-2 p)^{x_1-1} p^{x_2 + x_3} \\
&=& (1 - 2p)^{x_1}p^{x_2 + x_3} \left(\frac{x_2 + x_3}{p} - 2\frac{x_1}{1 - 2p} \right)
\end{eqnarray}</p>
<p>Therefore the value of $\hat{p}$ for which $dL(\hat{p})/dp = 0$ is</p>
<p>$$
\frac{x_2 + x_3}{\hat{p}} - 2\frac{x_1}{1 - 2\hat{p}} = 0 ~~~\Rightarrow ~~~ \hat{p} = \frac{x_2 + x_3}{2(x_1 + x_2 + x_3)}
$$</p>
<p>or equivalently</p>
<p>$$
\hat{p} = \frac{x_2 + x_3}{2n}
$$</p>
|
853,659 | <p>Evaluate the integral:</p>
<p>$$\int \frac{x^6}{x^4-1} \, \mathrm{d}x$$</p>
<p>After a lot of help I have reached this point:</p>
<p>$x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D$</p>
<p>But now I don't really know how to solve for $A, B, C$, and $D$. Please help!</p>
| M. Strochyk | 40,362 | <p>Integrand can be transformed without long divison
$$\begin{gathered}\frac{x^6}{x^4 - 1}=\frac{x^6-1+1}{x^4 - 1}=\frac{(x^2-1)(x^4+x^2+1)}{x^4 - 1}+\frac{1}{x^4 - 1}=\\
=\frac{x^4+x^2+1}{x^2 + 1} + \frac{1}{x^4 - 1}=\\
=x^2+\frac{1}{x^2 + 1}+\frac{1}{2}\left(\frac{1}{x^2-1} - \frac{1}{x^2+1} \right)=\\
=x^2 + \frac{1}{2}\left(\frac{1}{x^2-1} + \frac{1}{x^2+1} \right).
\end{gathered}$$</p>
|
1,284,938 | <p>I was revising for one of my end of year maths exams, then I came across this example on how to find lines of tangents to ellipses outside the curve. Personally, I'd use differentiation and slopes to find such lines, but the lecturer does something simpler and more elegant.</p>
<p>The question is: "Find the equations of the lines through (1, 4) which are tangents to the ellipse $x^2 + 2y^2 = 6$"</p>
<p>And then we put the lines into the standard form, which comes out as $ y = mx − (m − 4)$ where $m$ is the slope of the line.</p>
<p>Then, the lecturer substitutes the equation we got into the original equation of the curve, which we get $x^2 + 2[mx − (m − 4)]^2 = 6$.</p>
<p>Now, the lecturer goes from the equation above to something I can't understand how to derive. With the explanation "We now look for repeated roots in the equation, as each tangent meets the line exactly once, we get":</p>
<p>$[4m(m − 4)]^2 − 4(1 + 2m^2
)(2m^2 − 16m + 26) = 0$</p>
<p>Can you guys please help me understand how to get to the equation above? I tried using the quadratic formula, where x has repeated roots (in other words, the rational bit is zero), but I still got something entirely different.</p>
<p>Thanks.</p>
| mickep | 97,236 | <p>If I understand your question correct, rewrite it as
$$
y'=2e^x-y,
$$
which is first order and linear,
$$
y'+y=2e^x,\quad\text{so}\quad D(e^x y)=2e^{2x},\quad\text{and hence}\quad e^xy=e^{2x}+C.
$$</p>
|
4,312,323 | <p>I get why <span class="math-container">$\sqrt{9} = \pm 3$</span>. But (at least I think) the ± is there because there's a certain ambiguity as to which number was squared to obtain <span class="math-container">$9$</span>.</p>
<p>Does that mean that if we remove the ambiguity <span class="math-container">$\sqrt{3^2} = 3$</span> ?</p>
<p>One argument could be that since <span class="math-container">$\sqrt{3^2} = \sqrt{9} = \pm 3$</span>.
Then again we could argue that we know for a fact that <span class="math-container">$9$</span> is the result of squaring the number <span class="math-container">$3$</span> and should therefore be <span class="math-container">$\sqrt{3^2} = 3$</span>.</p>
<p>I apologize as I'm only a beginner and this may perhaps seem too basic.</p>
| 欲しい未来 | 772,122 | <p>When we have <span class="math-container">$\sqrt{x}$</span> it is usually assumed to be the <em><strong>principal square root operator</strong></em>, which means it returns only the positive root. We have
<span class="math-container">$$\sqrt{\cdot}: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}\\
x \mapsto \sqrt{x}$$</span></p>
<p>In your case, by definition</p>
<p><span class="math-container">$$\sqrt{x^2} =\left|x\right| = \begin{cases} x & \mbox{ if } x > 0 \\
-x & \mbox{ if } x < 0 \\
0 & \mbox{ if } x = 0\end{cases}$$</span></p>
|
23,911 | <p>I am teaching a course on Riemann Surfaces next term, and would <strong>like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties</strong> (over, say, $\mathbb{C}$). I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.</p>
<p>Here are some examples that I thought of:</p>
<p><strong>1.</strong> Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$. By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$. Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.</p>
<p><strong>2.</strong> Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$. Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that
agrees with $f$ on $V$ and is identically zero outside of $U$.</p>
<p>By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function. It may be impossible find a global regular function $g$ that agrees with $f$ on $V$. When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.</p>
<p><strong>3.</strong> If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian. The local ring of a variety at a point is always noetherian. </p>
<p><em><strong>What are some more examples?</em></strong></p>
<p>Answers illustrating the difference between real manifolds and complex manifolds are also welcome.</p>
| Henri | 5,659 | <p>I think there's some big difference concerning the metric approach too.</p>
<p>In fact, the Gram-Schmidt process (which is real analytic) enables us -in real differential geometry- to find some local orthonormal frames (for any hermitian bundle, and in particular for the tangent bundle), whereas in the holomorphic case, very subtle differences may occur there.</p>
<p>For example, in the Kähler case, we can find "orthonormal" frames for the tangent bundle at order 2, which is the key for the Kähler identities, leading to fundamental results like the equality of all Laplacians and thus the Hodge decomposition theorem in the compact case.</p>
|
23,911 | <p>I am teaching a course on Riemann Surfaces next term, and would <strong>like a list of facts illustrating the difference between the theory of real (differentiable) manifolds and the theory non-singular varieties</strong> (over, say, $\mathbb{C}$). I am looking for examples that would be meaningful to 2nd year US graduate students who has taken 1 year of topology and 1 semester of complex analysis.</p>
<p>Here are some examples that I thought of:</p>
<p><strong>1.</strong> Every $n$-dimensional real manifold embeds in $\mathbb{R}^{2n}$. By contrast, a projective variety does not embed in $\mathbb{A}^n$ for any $n$. Every $n$-dimensional non-singular, projective variety embeds in $\mathbb{P}^{2n+1}$, but there are non-singular, proper varieties that do not embed in any projective space.</p>
<p><strong>2.</strong> Suppose that $X$ is a real manifold and $f$ is a smooth function on an open subset $U$. Given $V \subset U$ compactly contained in $U$, there exists a global function $\tilde{g}$ that
agrees with $f$ on $V$ and is identically zero outside of $U$.</p>
<p>By contrast, consider the same set-up when $X$ is a non-singular variety and $f$ is a regular function. It may be impossible find a global regular function $g$ that agrees with $f$ on $V$. When $g$ exists, it is unique and (when $f$ is non-zero) is not identically zero on outside of $U$.</p>
<p><strong>3.</strong> If $X$ is a real manifold and $p \in X$ is a point, then the ring of germs at $p$ is non-noetherian. The local ring of a variety at a point is always noetherian. </p>
<p><em><strong>What are some more examples?</em></strong></p>
<p>Answers illustrating the difference between real manifolds and complex manifolds are also welcome.</p>
| Heinrich Hartmann | 5,714 | <p>For a closed analytic subset Z ⊂ S of a (say compact) complex manifold
with complement U=S-Z one has additivity of the (topological) Euler characteristic:</p>
<p>Χ(S)=Χ(Z)+Χ(U).</p>
<p>This is wrong for if S and Z are topological spaces or smooth manifolds.
Indeed, take for Z a point on a circle S.
This (surprising) difference was recently pointed out to me by Manfred Lehn.</p>
<p>Of course there is also no additivity of Poincare-Polynomials or other
"motivic" invariants of complex varieties.</p>
|
1,424,273 | <p>Let $(a_n)$ be a convergent sequence of positive real numbers. Why is the limit nonnegative?</p>
<p>My try: For all $\epsilon >0$ there is a $N\in \mathbb{N}$ such that $|a_n-L|<\epsilon$ for all $n\ge N$. And we know $0< a_n$ for all $n\in \mathbb{N}$, particularly $0<a_n$ for all $n\ge N$. Maybe by contradiction: suppose that $L<0$, then $L<0<a_n$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. Then $0<-L<a_n-L$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. It follows: for all $\epsilon >0$, there is a $N\in \mathbb{N}$ such that $0<|-L|=-L<|a_n-L|<\epsilon$ for all $n\ge N$, which can't be true.</p>
<p>Is my proof ok?</p>
| Math1000 | 38,584 | <p>If $a_n>0$ for all $n$, then $\liminf_{n\to\infty} a_n \geqslant 0$, for if not, there would exist an $n$ such that $\inf_{k\geqslant n}a_k<0$. But the $\inf$ of a set of positive numbers cannot be negative, so $\liminf_{n\to\infty} a_n\geqslant0$, and because $a_n$ is convergent, $$\lim_{n\to\infty}a_n=\liminf_{n\to\infty} a_n \geqslant 0.$$</p>
|
1,424,273 | <p>Let $(a_n)$ be a convergent sequence of positive real numbers. Why is the limit nonnegative?</p>
<p>My try: For all $\epsilon >0$ there is a $N\in \mathbb{N}$ such that $|a_n-L|<\epsilon$ for all $n\ge N$. And we know $0< a_n$ for all $n\in \mathbb{N}$, particularly $0<a_n$ for all $n\ge N$. Maybe by contradiction: suppose that $L<0$, then $L<0<a_n$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. Then $0<-L<a_n-L$ for all $n\in \mathbb{N}$, particularly for all $n\ge N$. It follows: for all $\epsilon >0$, there is a $N\in \mathbb{N}$ such that $0<|-L|=-L<|a_n-L|<\epsilon$ for all $n\ge N$, which can't be true.</p>
<p>Is my proof ok?</p>
| John Dawkins | 189,130 | <p>Using your notation, if $n\ge N$ then $|a_n-L|<\epsilon$. This absolute-value inequality if equivalent to the inequalities $a_n-\epsilon< L < a_n+\epsilon$. In particular (use only the left-hand inequality and take $n=N$),
$$
-\epsilon<a_N-\epsilon<L,
$$
Summarizing: $-\epsilon <L$ for each $\epsilon>0$. This implies that $0\le L$.</p>
|
4,080,776 | <p>I am doing an individual study of an abstract algebra for number theory course online. I just started, so I hope my question just note come off as too trivial. The lecture notes state that the ring of <span class="math-container">$p$</span>-adic integers does not have a ring endomorphism.</p>
<h3>Questions:</h3>
<p><strong>1.</strong> Does not the identity mapping work as a counterexample?</p>
<p>Then, assuming they meant: "no endomorphism except the trivial case", so the entire thing is not just a mistake:</p>
<p><strong>2.</strong> I still cannot convince myself that there is no other ring endomorphism of <span class="math-container">$p$</span>-adic integers. Could you please give me a hint how to prove it or point me to literature where such a proof is shown?</p>
| Torsten Schoeneberg | 96,384 | <p>Re Question 1) Yes, the identity <span class="math-container">$id: \mathbb Z_p \rightarrow \mathbb Z_p$</span> is of course a ring endomorphism.</p>
<p>Re Question 2) To show that there is no other, assume <span class="math-container">$f: \mathbb Z_p \rightarrow \mathbb Z_p$</span> is any ring endomorphism. We have <span class="math-container">$f(1)=1$</span> hence <span class="math-container">$f(1+1)=1+1$</span> etc. as well as <span class="math-container">$f(-1)=-1$</span> etc., so that</p>
<p><span class="math-container">$$(*) \qquad \qquad f(x) =x \text{ at least for all } x \in \mathbb Z.$$</span></p>
<p>We now will show it for all <span class="math-container">$x \in \mathbb Z_p$</span>. Generally one could say that <span class="math-container">$\mathbb Z$</span> is dense in <span class="math-container">$\mathbb Z_p$</span>, so it suffices to show continuity of <span class="math-container">$f$</span>. But we'll be a bit more down to earth.</p>
<p><span class="math-container">$(*)$</span> in particular means <span class="math-container">$f(p^n)=p^n$</span> and hence, by <span class="math-container">$f$</span> being and endomorphism (edit_corrected, thanks @KCd), <span class="math-container">$f(p^n \mathbb Z_p) \subseteq p^n\mathbb Z_p$</span> for all <span class="math-container">$n \in \mathbb N$</span>. By the definition of the <span class="math-container">$p$</span>-adic value, this implies that</p>
<p><span class="math-container">$$\lvert f(x) \rvert_p \le \lvert x \rvert_p$$</span> for all <span class="math-container">$x \in \mathbb Z_p$</span>, hence by <span class="math-container">$f$</span> being a ring endomorphism also</p>
<p><span class="math-container">$$\lvert f(x)-f(y) \rvert_p \le \lvert x-y\rvert_p$$</span></p>
<p>for any <span class="math-container">$x,y \in \mathbb Z_p$</span>, i.e. <span class="math-container">$f$</span> is necessarily (uniformly) continuous.</p>
<p>Now let <span class="math-container">$x \in \mathbb Z_p$</span>. There exists some sequence of integers (!) <span class="math-container">$x_n \in \mathbb Z$</span> such that <span class="math-container">$\lim_{n \to \infty} x_n =x$</span>. (This is the "density" argument in a nutshell.) Then</p>
<p><span class="math-container">$$f(x) \stackrel{cont.}= \lim_{n \to \infty}f(x_n) \stackrel{(*)}=\lim_{n\to \infty} x_n \stackrel{def.}=x.$$</span></p>
<p>Since <span class="math-container">$x \in \mathbb Z_p$</span> was arbitrary, we just showed <span class="math-container">$f=id$</span>.</p>
|
2,355,852 | <p>Given are $m$ bins with equal probability of choosing one of them. Unknown number of balls $n$ is placed into the bins, and, at the end of placement, we observe number of empty bins $m_e$ and non-empty bins $m_{n}$.</p>
<p>Given $m$, $m_e$, $m_n$, what is the most likely number of balls $n$, which have been placed into bins?</p>
<p>(UPD) possible additional information: number of bins with <em>exactly</em> one ball can be also known.</p>
| Henry | 6,460 | <p>Presumably $m=m_e+m_n$. </p>
<p>The likelihood of seeing $m_n$ out of $m$ occupied is $\dfrac{S_2(n,m_n)\, m!}{m^n \;(m-m_n)!}$ where $S_2(x,y)$ is a <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling number of the second kind</a>. I do not know an easy way of finding the $n$ which maximises this in general but it is easy enough to calculate for given reasonably small $m$. </p>
<p>One possible alternative estimator is $\dfrac{\log(m)-\log(m-m_n)}{\log(m)-\log(m-1)}$, which follows from saying $E[m_n \mid m, n] = m\left(1 -\left(\frac{m-1}{m}\right)^n\right)$. This is neither a maximum likelihood estimator nor an unbiased estimator for $n$, and it is usually not an integer, but it comes fairly close to the maximum likelihood estimator; it is slightly biased upwards when $1 \lt n \lt m$. The table below compares the values when $m=10$ for different $m_n$</p>
<pre><code> m m_n n_MLE n_alt.est
10 0 0 0
10 1 1 1
10 2 2 2.12
10 3 3 3.39
10 4 4 or 5 4.85
10 5 6 6.58
10 6 8 8.70
10 7 11 11.43
10 8 15 15.25
10 9 22 21.85
10 10 infinite n/a
</code></pre>
|
2,853,989 | <p>I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.</p>
<p>What am I missing? Also, is there a method to demonstrate that without induction?</p>
| mechanodroid | 144,766 | <p>\begin{align}
\left(1+\frac1n\right)^n &= \sum_{k=0}^n {n \choose k} \frac1{n^k} \\
&= \sum_{k=0}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k} \\
&= \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)
\end{align}</p>
<p>so</p>
<p>\begin{align}
\left(1+\frac1{n+1}\right)^{n+1} &= \sum_{k=0}^{n+1} \frac1{k!}\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\cdots \left(1-\frac{k-1}{n+1}\right)\\
&> \sum_{k=0}^{n} \frac1{k!}\left(1-\frac1{n}\right)\left(1-\frac2{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\\
&= \left(1+\frac1n\right)^n
\end{align}</p>
<p>On the other hand, for $n = 1$ we have</p>
<p>$$\left(1+\frac11\right)^1 = 2$$</p>
<p>so $\left(1+\frac1n\right)^n > 2, \forall n \ge 2$.</p>
|
2,853,989 | <p>I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.</p>
<p>What am I missing? Also, is there a method to demonstrate that without induction?</p>
| Momo | 384,029 | <p>Another way is to prove first that your sequence is monotonically increasing like has been done here:</p>
<p><a href="https://math.stackexchange.com/questions/167843/i-have-to-show-1-frac1nn-is-monotonically-increasing-sequence">I have to show $(1+\frac1n)^n$ is monotonically increasing sequence</a></p>
<p>... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.</p>
|
2,853,989 | <p>I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.</p>
<p>What am I missing? Also, is there a method to demonstrate that without induction?</p>
| copper.hat | 27,978 | <p>Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x \ge 0$ and so
$f(x) \ge f(0)+f'(0) x = 1+xn$. Hence $f({1 \over n}) \ge 2$.</p>
|
12,204 | <p>A tag named <a href="https://math.stackexchange.com/questions/tagged/tricks" class="post-tag" title="show questions tagged 'tricks'" rel="tag">tricks</a> has recently been created in <a href="https://math.stackexchange.com/questions/616672/2-tricks-to-prove-every-group-with-an-identity-and-xx-identity-is-abelian-f">this question</a>. So far there is <a href="https://math.stackexchange.com/tags/tricks/info">no tag wiki</a> to indicate intended usage.</p>
<p>Do we really need such a tag? It veers slightly towards <a href="https://math.meta.stackexchange.com/questions/2498/the-meta-tags">meta tags</a>, which are generally discouraged.</p>
<hr>
<p>To me it seems somewhat similar to <a href="https://math.stackexchange.com/questions/tagged/proof-strategy" class="post-tag" title="show questions tagged 'proof-strategy'" rel="tag">proof-strategy</a>, see the <a href="https://math.stackexchange.com/tags/proof-strategy/info">tag-wiki</a>. (I would characterize this tag as questions that are similar to <a href="http://www.tricki.org/" rel="nofollow noreferrer">tricki</a> articles, see <a href="http://chat.stackexchange.com/rooms/3740/conversation/proof-strategy-tag">this conversation in chat</a>.) However, <a href="https://math.stackexchange.com/questions/tagged/proof-strategy" class="post-tag" title="show questions tagged 'proof-strategy'" rel="tag">proof-strategy</a> tag is often used incorrectly.</p>
| Martin Sleziak | 8,297 | <p>Just offering another possibility to vote for/against: </p>
<p>Keep the <a href="https://math.stackexchange.com/questions/tagged/tricks" class="post-tag" title="show questions tagged 'tricks'" rel="tag">tricks</a> tag, but make it a synonym of <a href="https://math.stackexchange.com/questions/tagged/proof-strategy" class="post-tag" title="show questions tagged 'proof-strategy'" rel="tag">proof-strategy</a>.</p>
|
1,693,045 | <p>I know if $x=e^{\frac{2\pi i}{17}}$ then $x^{17}=1$ and $\Re(x)=\cos\left(\frac{2\pi}{17}\right)$.</p>
<p>But how do I form a polynomial which has root $\cos\left(\frac{2\pi}{17}\right)$.</p>
<p>I know you can consider de Moivre's theorem and expand the LHS using binomial theorem but that will take a long time.</p>
| David C. Ullrich | 248,223 | <p>Probably the best way is to just show that the sum of two algebraic numbers is algebraic. This is not obvious, but if you look at it just right it's much easier than it seems at first.</p>
<p>Regard $\Bbb C$ as a vector space over $\Bbb Q$. Any linear-algebra concepts below refer to $\Bbb Q$-linear subspaces of $\Bbb C$.</p>
<p>Lemma 0. The number $a\in \Bbb C$ is algebraic if and only if the span of $1,a,a^2\dots$ is finite-dimensional.</p>
<p>Proof: Easy exercise. QED.</p>
<p>Lemma 1. Suppose $A,B\subset\Bbb C$ are subspaces, and let $C$ be the span of the $xy$ with $x\in A$, $y\in B$. If $A$ is spanned by $a_1,\dots,a_n$ and $B$ is spanned by $b_1,\dots,b_m$ then $C$ is spanned by $a_jb_k$, $1\le j\le n$, $1\le k\le m$ (so in particular $C$ is finite dimensional).</p>
<p>Proof: Easy exercise. QED.</p>
<p>Theorem. If $a,b\in\Bbb C$ are algebraic then $a+b$ is algebraic.</p>
<p>Proof. Let $A$ be the span of the powers of $a$ and $B$ the span of the powers of $b$. Let $C$ be as in Lemma 1. Now there exist $n$ and $m$ such that $A$ is spanned by $1,a,\dots, a^n$ and $B$ is spanned by $1,b,\dots,b^m$. So Lemma 1 shows that $C$ is finite dimensional.</p>
<p>But every power of $a+b$ lies in $C$. So the span of the powers of $a+b$ is finite dimensional, and hence Lemma 0 shows that $a+b$ is algebraic. QED.</p>
<p>This shows that your number is algebraic since $\cos(t)=(e^{it}+e^{-it})/2$.</p>
<p><strong>Edit.</strong> One could use the argument above to find $P$ with $P(a+b)=0$, given $p(a)=0$ and $q(b)=0$. Any power of $A$ can be written explicitly as a linear combination of $1,a,\dots a^n$, and similarly for $b$. So any $a^jb^k$ can be written explicitly as a linear combination of $a^jb^k$ with $0\le j\le n$ and $0\le k\le m$. Hence the same is true of any power of $a+b$. So write down the powers of $a+b$ as such linear combinations, one by one, and check the vectors of coefficients for linear dependence. Eventually a dependence relation appears, and that gives you $P$ with $P(a+b)=0$.</p>
|
1,693,045 | <p>I know if $x=e^{\frac{2\pi i}{17}}$ then $x^{17}=1$ and $\Re(x)=\cos\left(\frac{2\pi}{17}\right)$.</p>
<p>But how do I form a polynomial which has root $\cos\left(\frac{2\pi}{17}\right)$.</p>
<p>I know you can consider de Moivre's theorem and expand the LHS using binomial theorem but that will take a long time.</p>
| Ron Gordon | 53,268 | <p>Consider that</p>
<p>$$\sin{\left ( \frac{9 \pi}{17} \right )} = \sin{\left ( \frac{8 \pi}{17} \right )} $$</p>
<p>or, letting $y = \frac{\pi}{17}$,</p>
<p>$$3 \sin{3 y} - 4 \sin^3{3 y} = 2 \sin{4 y} \cos{4 y} $$</p>
<p>or</p>
<p>$$9 \sin{y} - 120 \sin^3{y} + 432 \sin^5{y} - 576 \sin^7{y} + 256 \sin^9{y} \\= 8 \sin{y} \cos{y} (2 \cos^2{y}-1) [2 (2 \cos^2{y}-1)^2-1] $$</p>
<p>Note that $\sin{y}$ cancels on both sides. Using $\sin^2{y}=1-\cos^2{y}$, we get an $8$th degree polynomial in $\cos{y}$, from which $\cos{2 y} = 2 \cos^2{y}-1$ may be determined.</p>
|
4,178,548 | <p>I'm starting a Linear Algebra course and I'm a bit confused.</p>
<p>Say we have a vector
<span class="math-container">$x = \begin{pmatrix}
x_1\\
x_2\\
\end{pmatrix}$</span>, and another vector <span class="math-container">$y = \begin{pmatrix}
y_1\\
y_2\\
\end{pmatrix}$</span></p>
<p>When we have a matrix composed of the two vectors, instead of saying <span class="math-container">$\begin{pmatrix}
x_1&y_1\\
x_2&y_2\\
\end{pmatrix}$</span></p>
<p>we say that the first column corresponds to <span class="math-container">$x_1$</span>, and the second corresponds to <span class="math-container">$x_2$</span> so
<span class="math-container">$\begin{pmatrix}
x_1&x_2\\
y_1&y_2\\
\end{pmatrix}$</span></p>
<p>Maybe I misunderstood something, but I feel like this is a bit confusing to me, may someone explain to me how this works?</p>
<p>Thanks in advance!</p>
| Acccumulation | 476,070 | <p>There are contexts where matrices can be considered to be a vector of vectors. There are two types of vectors: row vectors and column vectors. A matrix can be viewed as either column vector where all the elements are row vectors, or a row vectors where all the elements are column vectors.</p>
<p>Thus <span class="math-container">$\left[\begin{bmatrix} x_1 \cr x_2 \end{bmatrix} , \begin{bmatrix} y_1 \cr y_2 \end{bmatrix} \right]$</span> can be considered to represent <span class="math-container">$\begin{bmatrix} x_1& y_1 \cr x_2& y_2 \end{bmatrix}$</span>. If we're representing the matrix <span class="math-container">$\begin{bmatrix} x_1 & x_2 \cr y_1& y_2 \end{bmatrix}$</span>, then the we should write <span class="math-container">$x$</span> and <span class="math-container">$y$</span> as row vectors: <span class="math-container">$x = [x_1, x_2]$</span>, <span class="math-container">$y=[y_1,y_2]$</span>, so that <span class="math-container">$\begin{bmatrix} x\cr y \end{bmatrix}=\begin{bmatrix} x_1& x_2 \cr y_1& y_2 \end{bmatrix}$</span>. It looks to me that your textbook and/or instructor is failing to distinguish between row and column vectors and treating <span class="math-container">$[x_1, x_2]$</span> and <span class="math-container">$\begin{bmatrix} x_1\cr x_2 \end{bmatrix}$</span> as interchangeable, either out of ignorance, laziness, or a belief that doing so would overly confuse students (and in doing so, confusing at least one student anyway).</p>
|
3,053,975 | <p><span class="math-container">$3^6-3^3 +1$</span> factors?, 37 and 19, but how to do it using factoring, <span class="math-container">$3^3(3^3-1)+1$</span>, can't somehow put the 1 inside </p>
| Oscar Lanzi | 248,217 | <p>Numbers having the form <span class="math-container">$n^2-n+1$</span> can never have prime factors congruent to <span class="math-container">$2$</span> modulo <span class="math-container">$3$</span>. Since <span class="math-container">$3$</span> is obviously out and any factorization must include a prime factor less than or equal to the square root if the number is composite, only factors <span class="math-container">$7,13,19$</span> need further investigation. The factorization <span class="math-container">$19×37$</span> is tgen fairly easy to find.</p>
|
275,430 | <p>I'm trying to give an $\epsilon$-$\delta$ proof that the following function $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$. </p>
<p>Let $f:\mathbb{A\subset R\to R}, \mathbb{A=\{x\in R| x>0\}}$ be given by:
$$
f(x) = \begin{cases}
1/n,&x=m/n\in\mathbb Q
\\
0,&x\notin\mathbb Q
\end{cases}
$$</p>
<p>where $m/n$ is in the lowest terms.</p>
<p>Can anyone help me with this proof (I'd prefer an answer with an $\epsilon$-$\delta$ proof).</p>
<p>Thank you very much!</p>
| Brian M. Scott | 12,042 | <p>HINTS: To show that $f$ is continuous at an irrational $x$, show for any $n\in\Bbb Z^+$ you can choose $\delta>0$ small enough so that the interval $(x-\delta,x+\delta)$ contains no fraction with a denominator $\le n$ in lowest term. Use the fact that between two rationals with denominator $m$ there is a gap of at least $\frac1m$.</p>
<p>To show that $x$ is discontinuous at a rational $\frac{m}n$, let $\epsilon=\frac{m}n$ (or any smaller positive real), and just observe that every non-empty open interval in $\Bbb R$ contains an irrational number.</p>
|
275,430 | <p>I'm trying to give an $\epsilon$-$\delta$ proof that the following function $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$. </p>
<p>Let $f:\mathbb{A\subset R\to R}, \mathbb{A=\{x\in R| x>0\}}$ be given by:
$$
f(x) = \begin{cases}
1/n,&x=m/n\in\mathbb Q
\\
0,&x\notin\mathbb Q
\end{cases}
$$</p>
<p>where $m/n$ is in the lowest terms.</p>
<p>Can anyone help me with this proof (I'd prefer an answer with an $\epsilon$-$\delta$ proof).</p>
<p>Thank you very much!</p>
| Hagen von Eitzen | 39,174 | <p>For $x=\frac mn\in\mathbb Q$ select $\epsilon=\frac1{2n}>0$. Assume there is $\delta>0$ such that $|x-y|<\delta$ implies $|f(y)-f(x)|<\epsilon$.
Then especially $f(y)>\frac{1}{2n}$ for such $y$, which means that alls such $y$ are rational and have denominator $<2n$. Even without knowing that the irrationals are dense, we can see that $y=\frac ab$ with $b<2n$ and $y\ne x$ implies
$$|y-x|=\frac{|an-bm|}{bn}\ge \frac1{bn}>\frac1{2n^2}$$
because the numerator must be $\ge 1$. But of course there <em>are</em> (rational) numbers $y$ with $|y-x|\le \frac1{2n^2}$ (for example $y=x+\frac1{2n^2+1}$), hence no such $\delta$ exists, $f$ is not continuous.</p>
<p>Let $x\notin\mathbb Q$ and $\epsilon>0$ be given.
We want a $\delta>0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$.
Select $n>\frac1\epsilon$. Then we want that $|y-x|<\delta$ implies that $y$ is either irrational or has a denominator $\ge n$.
For each $k< n$, there are only finitely many rationals $\frac mk$ with denominator $k$ and $|\frac mk-x|<1$. Then
$$\delta =\min\left\{\left|\frac mk-x\right|\colon k<n, \left|\frac mk-x\right|<1\right\} $$
will do the trick.</p>
|
2,292,520 | <p>I know that the logical negation of $$\neg(a \rightarrow b)= a \wedge \neg b $$ I am not clear what that means in the following simple setting:</p>
<p>So its clear that $$x\geq 2 \to x^2\geq 4.$$ Now I can write the logical negation of $a\to b$ as $a \wedge \neg b$, but what does that intuitively mean? </p>
<p>Suppose I want to prove "$a \wedge \neg b$", what do i need to prove mathematically?</p>
<p>thnks</p>
| Mirko | 188,367 | <p>Say $a$ is $x\ge2$ and $b$ is $x^2\ge14$ (and formally an universal quantifier should be involved as in @JMoravitz comment, i.e. $\forall x, x\ge2\to x^2\ge14$ ). Pick $x=3$, then $a$ is true, but $b$ is false. In other words, $x\ge2$ does not imply that $x^2\ge14$. Formally, the negation here is $\exists x, (x\ge2 \land\neg x^2\ge14)$. </p>
<p>Formally, one may separate syntax from sematic (or form from meaning). Given any statements $a$ and $b$, the negation of the formula $a\to b$ is the formula $a\land\neg b$. If you want to prove that $a\land\neg b$ then you need to either use some previously proven formulas, or axioms (accepted without proof), or to interpret $a$ and $b$ in some known model (as for the reals above), giving each of $a$ and $b$ meaning and truth values. </p>
|
1,804,042 | <p><strong>Edit:</strong> Here is the original problem; it is possible that my recurrence for the stationary distribution $\pi$ is incorrect.</p>
<blockquote>
<p>Consider a single server queue where customers arrive according to a Poisson process with intensity $\lambda$ and request i.i.d. $\mathsf{Exp}(\mu)$ service times. The server is subject to failures and repairs. The lifetime of a working server is an $\mathsf{Exp}(\theta)$ random variable, while the repair time is an $\mathsf{Exp}(\alpha)$ random variable. Successive lifetimes and repair times are independent, and are independent of the number of customers in the queue. When the server fails, all the customers in the queue are forced to leave, and while the server is under repair no new customers are allowed to join.</p>
</blockquote>
<p><strong>Edit:</strong> I have revised the recurrence.</p>
<p>In a problem on queueing theory I've derived the following recurrence:
\begin{align}
\pi_1 &=\left(\frac{\lambda+\theta}\mu\right)\pi_0 - \frac{\alpha\theta}{\mu(\alpha+\theta)}\\
\pi_{n+1} &= \left(1+\frac{\lambda+\theta}\mu\right)\pi_n - \frac\lambda\mu\pi_{n-1},\ n\geqslant1.
\end{align}
where $\lambda$, $\mu$, $\theta$, and $\alpha$ are positive constants and $$\sum_{i=0}^\infty \pi_i = \frac\alpha{\alpha+\theta}. $$ </p>
<p>After a lot of tedious algebra, I found that
$$\scriptsize\pi_n = \left(\frac{\alpha \theta \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\theta +\lambda +\mu+ \sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}\right)^n}{(\alpha +\theta ) (2 \mu )^n \sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}}\right)(1+\pi_0) $$ for $n\geqslant 1$. To save space, let $$\mathcal C:=\sqrt{\theta ^2+2 \theta (\lambda +\mu )+(\lambda -\mu )^2}. $$</p>
<p>Summing over $n$ and solving for $\pi_0$, I found
$$\pi_0 =\frac{\alpha \mu \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\lambda -\mu-\theta-\mathcal C \right)}{2 \theta (\alpha +\theta ) \mathcal C}, $$</p>
<p>and so
$$
\pi_n=\left(\frac{ \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right) \left(\lambda -\mu-\theta-\mathcal C \right)+2 \theta (\alpha +\theta ) \mathcal C }{2(\alpha +\theta )^2\mathcal C^2\left(\alpha^2\mu \left(2 \theta (\lambda +\mu )+(\lambda -\mu )^2\right)\right)^{-1}} \right)\left(\frac{\lambda+\mu+\theta+\mathcal C }{2\mu}\right)^n.
$$</p>
<p>If you see any errors let me know...</p>
<p>I'm also wondering what conditions on $\lambda,\mu,\theta$, and $\alpha$ are necessary for $\sum_{i=0}^\infty \pi_i$ to converge. For context, this is a $M/M/1$ queue with arrival rate $\lambda$, service rate $\mu$, but with an added state $D$ with transitions of rate $\theta$ from each state $n$ to $D$ and a transition of rate $\alpha$ from $D$ to $0$.</p>
| Przemo | 99,778 | <p>The solution to the recurrence in question is definitely governed by the roots of the characteristic equation which reads:
<span class="math-container">\begin{equation}
\lambda z^2- \left( \lambda + \mu + \theta \right) z + \mu =0
\end{equation}</span></p>
<p>where </p>
<p><span class="math-container">\begin{eqnarray}
z_\pm = \frac{1}{2 \lambda} \left( \lambda + \mu+\theta \pm \sqrt{-4 \lambda \mu + (\lambda +\mu+\theta)^2}\right)
\end{eqnarray}</span></p>
<p>Also define:
<span class="math-container">\begin{eqnarray}
(A,B):= \left(\pi_0 \frac{\mu}{\lambda}, -\frac{(\alpha \theta + \alpha \mu \pi_0 + \mu \theta \pi_0)}{(\lambda(\alpha+\theta))} \right)
\end{eqnarray}</span></p>
<p>Then the solution to the recurrence reads:</p>
<p><span class="math-container">\begin{eqnarray}
\pi_n = \frac{1}{z_- - z_+} \left( -(A+B z_-) (\frac{1}{z_-})^{n+1} + (A+B z_+) (\frac{1}{z_+})^{n+1} \right)
\end{eqnarray}</span>
for <span class="math-container">$n=0,1,\cdots$</span>.</p>
<pre><code>In[109]:= a =.; mu =.; l =.; th =.;
{a, mu, l, th} = RandomInteger[{1, 10}, 4]/10;
{A, B} = {(mu pi[0])/
l, -((a th + a mu pi[0] + mu th pi[0])/(a l + l th))};
{zm, zp} = {(l + mu + th - Sqrt[-4 l mu + (l + mu + th)^2])/(2 l), (
l + mu + th + Sqrt[-4 l mu + (l + mu + th)^2])/(2 l)};
mpi = 1/(zm - zp)
Table[-(A + B zm) (1/zm)^(n + 1) + (A + B zp) (1/zp)^(n + 1), {n,
0, 20}];
(mpi[[2]] - ((l + th)/mu mpi[[1]] - a th/(mu (a + th)))) // Simplify
Table[mu mpi[[n + 2]] - (mu + l + th) mpi[[n + 1]] + l mpi[[n]], {n,
1, Length[mpi] - 2}] // Simplify
Out[114]= 0
Out[115]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
</code></pre>
|
2,196,037 | <p>Let $E$ be a universal set and $\{A_{\alpha}\}_{\alpha \in J},$ for some index set $J$ be a family of subsets of $E.$</p>
<p>Prove that:
(a)$E-\bigcup_{\alpha \in J}A_{\alpha} = \bigcap_{\alpha \in J}($R$-A_{\alpha}).$</p>
<p>I do not know what is $R$ or it is a mistake in the question, Could anyone help me ? </p>
<p>(b)$E-\bigcap_{\alpha \in J}A_{\alpha} = \bigcup_{\alpha \in J}($E$-A_{\alpha}).$</p>
<p>Shall I prove it by induction? but what about the index set is it countably infinite or finite or uncountable, and how the proof will differ? </p>
| Axion004 | 258,202 | <p>\begin{eqnarray*}
(\bigcup_{i\in\Lambda}A_i)^c &=& \{x|x\notin \bigcup_{i\in\Lambda}A_i\} \\
&=& \{x|\forall i\in\Lambda,~~x\notin A_i\}\\
&=& \{x|\forall i\in\Lambda,~~x\in A_i^c\} \\
&=& \bigcap_{i\in\Lambda}A_i^c
\end{eqnarray*}</p>
|
3,554,891 | <p>Let's take a look back at this familiar "Law of cosines":</p>
<blockquote>
<p>Consider the triangle <span class="math-container">$\triangle ABC$</span>. Let <span class="math-container">$a = BC, b = AC, c = AB$</span>; <span class="math-container">$\angle A, \angle B, \angle C$</span> are the angles of the triangle opposite to side <span class="math-container">$a, b, c,$</span> respectively. By the Law of Cosines:
<span class="math-container">$$a^{2} = b^{2} + c^{2} - 2bc \cdot \cos \angle A$$</span></p>
</blockquote>
<p>This formula can apply for any triangle.</p>
<p>But what about quadrilaterals? Is there a formula, which shows the relationship between sides and angles, similar to the Law of Cosines? Can we extend the Law of Cosines???</p>
<p>This is the way to approach the formula for quadrilaterals (<em>It's not (really) a proof</em>):</p>
<blockquote>
<p>Given the quadrilateral ABCD. Let <span class="math-container">$a = BC, b = CD, c = AB, d = AD$</span>. Let <span class="math-container">$E = AB \cap CD$</span> and <span class="math-container">$G = AC \cap BD$</span></p>
</blockquote>
<p>Let consider <span class="math-container">$\triangle ABC$</span> as a "special quadrilateral" (where <span class="math-container">$d=0$</span>). Then by the Law of Cosines:</p>
<p><span class="math-container">$$a^{2} = b^{2} + c^{2} - 2bc \cdot \cos \angle BEC = b^{2} + c^{2} - 2bc \cdot \cos \angle BGC$$</span></p>
<p>(because when <span class="math-container">$d=0$</span>, <span class="math-container">$E \equiv G \equiv A \Rightarrow \angle BEC = \angle BGC$</span>)</p>
<p>Notice that when <span class="math-container">$d=0$</span> then <span class="math-container">$CA = CD = CE = b$</span>; <span class="math-container">$BD = BE = BA = c$</span>. So we can guess the general formula for a quadrilateral will be one of these two formulas:</p>
<blockquote>
<p><span class="math-container">$$ a^{2} + Kd^{2} = b^{2} + c^{2} - 2 \cdot BE \cdot CE \cdot \cos \angle BEC \text{ (1)}$$</span>
<span class="math-container">$$ a^{2} + Kd^{2} = b^{2} + c^{2} - 2 \cdot BD \cdot CA \cdot \cos \angle BGC \text{ (2)}$$</span></p>
</blockquote>
<p>(where <span class="math-container">$K$</span> is a constant)</p>
<p>The reason we add <span class="math-container">$Kd^{2}$</span> is to make the formula homogeneous (since the Law of Cosines is also homogeneous), and when <span class="math-container">$d=0$</span>, the <span class="math-container">$Kd^{2}$</span> term is gone. Moreover, from our intuition, if the formula contains <span class="math-container">$\angle BEC$</span>, then two sides, which multiply to its cosines, have to be <span class="math-container">$BE$</span> and <span class="math-container">$CE$</span>. Otherwise, those two sides will be <span class="math-container">$BD$</span> and <span class="math-container">$CA$</span> multiplied by <span class="math-container">$\cos \angle BGC$</span></p>
<p>To see which one is possibly correct, we can try to apply the formula to a special quadrilateral: square. In a square, <span class="math-container">$a=b=c=d$</span>, "<span class="math-container">$BE = CE = \infty$</span>", "<span class="math-container">$\angle BEC = \infty$</span>", <span class="math-container">$\angle BGC = 90^{\circ}$</span>. Apply <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span>:</p>
<p><span class="math-container">$$(1): a^{2} + Ka^{2} = a^{2} + a^{2} - \infty$$</span>
<span class="math-container">$$(2): a^{2} + Ka^{2} = a^{2} + a^{2}$$</span></p>
<p><span class="math-container">$(1)$</span> is definitely wrong. The formula <span class="math-container">$(2)$</span> can be true if <span class="math-container">$K=1$</span>, so let re-written it:</p>
<p><span class="math-container">$$a^{2} + d^{2} = b^{2} + c^{2} - 2 \cdot BD \cdot CA \cdot \cos \angle BGC$$</span></p>
<p>To be sure that this formula is correct, let's apply this in another quadrilateral. This time is a rectangle, where <span class="math-container">$\angle BGC = 60^{\circ}$</span>. We have <span class="math-container">$a=d, b=c=a\sqrt{3}$</span>, <span class="math-container">$BD = AC = 2a$</span>. Apply the formula that we've just found, we get:</p>
<p><span class="math-container">$$a^{2} + a^{2} = 3a^{2} + 3a^{2} - 2 \cdot 4a^{2} \cdot \frac{1}{2}$$</span></p>
<p>And this is true. You can verify it with some other quadrilaterals, and it'll also true. So, our new extended "Law of Cosines" is:</p>
<blockquote>
<p><span class="math-container">$$a^{2} + d^{2} = b^{2} + c^{2} - 2 \cdot BD \cdot CA \cdot \cos \angle BGC$$</span></p>
</blockquote>
<p>So that seems fine. But</p>
<blockquote>
<p><em>Is there a proof of the formula above</em>?</p>
</blockquote>
<p>Now, my main question (and my main focus) is:</p>
<blockquote>
<p><em><strong>Can we extend the formula (find a general formula) for polygons with n sides</strong></em>?</p>
</blockquote>
<p>This question is what I'm looking for (<em>This isn't a homework question</em>). I'm really curious about this. If you have an answer (or just an idea) to approach, please provide it. </p>
<p>Thank you a lot and have a nice day :D</p>
| Intelligenti pauca | 255,730 | <p>In quadrilateral <span class="math-container">$ABCD$</span>, in addition to <span class="math-container">$a = BC$</span>, <span class="math-container">$b = CD$</span>, <span class="math-container">$c = AB$</span>, <span class="math-container">$d = AD$</span>,
also set:
<span class="math-container">$$
AG=e,\quad CG=f,\quad BG=g,\quad DG=h,\quad \angle BGC=\alpha,
$$</span>
where <span class="math-container">$G$</span> is the intersection point of diagonals <span class="math-container">$AC$</span> and <span class="math-container">$BD$</span>.
By the cosine law we get then:
<span class="math-container">$$
\begin{align}
a^2 &=f^2+g^2-2fg\cos\alpha \\
d^2 &=e^2+h^2-2eh\cos\alpha \\
b^2 &=f^2+h^2+2fh\cos\alpha \\
c^2 &=e^2+g^2+2eg\cos\alpha \\
\end{align}
$$</span>
and from that we obtain:
<span class="math-container">$$
b^2+c^2-a^2-d^2 = 2(fh+eg+fg+eh)\cos\alpha=2(e+f)(g+h)\cos\alpha,
$$</span>
which is precisely your formula.</p>
|
129 | <p>Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient. Are there any other special homotopical properties of manifolds?</p>
| John Klein | 8,032 | <p>Sean: this gives a Poincare space which is not homotopy equivalent to a closed manifold. the idea is that the Spivak fibration of the <span class="math-container">$5$</span> dimensional Poincare space doesn't lift to a stable vector bundle. One can prove this as follows: let <span class="math-container">$X^5$</span> be as in Madsen and Milgram. Then <span class="math-container">$X$</span> fibers over <span class="math-container">$S^3$</span> with fiber <span class="math-container">$S^2$</span>. Call this fibration <span class="math-container">$\xi$</span>, and let the projection <span class="math-container">$X \to S^3$</span> be <span class="math-container">$p$</span>.
Note that <span class="math-container">$p$</span> has a section, call it <span class="math-container">$s$</span>.</p>
<p>It's not hard to prove that the Spivak fibration of <span class="math-container">$X$</span> in this case is just <span class="math-container">$p^*\xi$</span>. One then needs to verify that <span class="math-container">$p^*\xi$</span> doesn't lift to a stable vector bundle. But if it did, then
so would <span class="math-container">$$\xi = (p \circ s)^*\xi = s^*p^* \xi\ .$$</span> But it's very easy to see that <span class="math-container">$\xi$</span> doesn't lift
(<span class="math-container">$\xi$</span> is given by the nontrivial element of <span class="math-container">$\pi_2(F) = \mathbb{Z}_2$</span>, where <span class="math-container">$F$</span> classifies spherical fibrations with section, whereas <span class="math-container">$\pi_2(O)$</span> is trivial).</p>
|
456,892 | <p>Find all solutions of $4\cos^2(x)-4\sin(x)-5=0$ in the interval $(6\pi, 8\pi)$.</p>
<p>I tried to work it out and got: $4y^2-4y -9 = 0$, but I can't figure out what $\cos x = $from there to finish the problem.</p>
| Robert Israel | 8,508 | <p>If $\sin(x) = y$, $\cos(x) = \pm \sqrt{1-y^2}$. But your equation for $y$ is not quite right.</p>
|
456,892 | <p>Find all solutions of $4\cos^2(x)-4\sin(x)-5=0$ in the interval $(6\pi, 8\pi)$.</p>
<p>I tried to work it out and got: $4y^2-4y -9 = 0$, but I can't figure out what $\cos x = $from there to finish the problem.</p>
| lab bhattacharjee | 33,337 | <p>As already found, $\sin x=\frac12$ which is $=\sin(-\frac\pi6)$</p>
<p>$$\implies x=n\pi+(-1)^n\left(-\frac\pi6\right)$$</p>
<p>If $n$ is even, $=2m$(say), </p>
<p>$x=2m\pi-\frac\pi6=(12m-1)\frac\pi6 $ and we need $6\pi<x<8\pi\implies 6\pi<(12m-1)\frac\pi6<8\pi \implies\frac{37}{12}<m<\frac{49}{12}\implies m=4$
$\implies x=(12\cdot4-1)\frac\pi6=\frac{47\pi}6 $</p>
<p>Similarly, if $n$ is odd $=2m+1$(say)</p>
|
1,441,624 | <p>Let <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> be elements of a group <span class="math-container">$G$</span>, how can I prove that <span class="math-container">$abc$</span> and <span class="math-container">$cba$</span> do not necessarily have the same order?</p>
<p>I know that this cannot hold for abelian groups, but unsure how to start otherwise.</p>
<p>Also, it is impossible to find a counterexample if we change the order to <span class="math-container">$cab$</span>, see here: <a href="https://math.stackexchange.com/q/2238562">Let <span class="math-container">$G$</span> be a group. Show that <span class="math-container">$\forall a, b, c \in G$</span>, the elements <span class="math-container">$abc, bca, cab$</span> have the same order.</a>.</p>
| Shaun | 104,041 | <p><strong>Hint:</strong> Try the dihedral group <span class="math-container">$D_4$</span> of eight elements.</p>
|
204,365 | <p>Consider a positive matrix <code>M</code> and a positive vector <code>b</code>, e.g.</p>
<pre><code>nn = 1000;
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
</code></pre>
<p>I would like to find a positive vector <code>X</code></p>
<pre><code>X = Array[x,nn];
</code></pre>
<p>(each <code>x[i]>0</code>) such that given</p>
<pre><code>expr = M.X-b;
</code></pre>
<p>the quantity <code>expr.expr</code> is minimized. Is it possible to do that in Mathematica efficiently (so that it finishes within a few seconds/minutes)?</p>
| Bob Hanlon | 9,362 | <p><strong>THIS IS AN EXTENDED COMMENT RATHER THAN AN ANSWER.</strong></p>
<p>It is inefficient to use <code>Table</code> to generate random numbers.</p>
<pre><code>Clear["Global`*"]
nn = 1000;
SeedRandom[10];
t1 = AbsoluteTiming[
{M1 = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}],
b1 = Table[RandomReal[{0, 100}], {i, 1, nn}]};][[1]]
(* 0.202688 *)
SeedRandom[10];
t2 = AbsoluteTiming[
{M2 = RandomReal[{0, 100}, {nn, nn}],
b2 = RandomReal[{0, 100}, nn]};][[1]]
(* 0.005653 *)
</code></pre>
<p>The values are identical</p>
<pre><code>{M1 === M2, b1 === b2, t1/t2}
(* {True, True, 35.8549} *)
</code></pre>
|
3,557,840 | <p>Find the quadratic polynomial <span class="math-container">$p(x)$</span> for given data points <span class="math-container">$$p(x_0)=y_0, p'(x_1)=y_1', p(x_2)=y_2 \text{ with } x_0 \neq x_2.$$</span></p>
<p><strong>My approach</strong></p>
<p>I tried the problem taking <span class="math-container">$p(x)=a+bx+c x^2$</span> but I am not sure about.</p>
<p>Any help is appreciated.</p>
| Alain Remillard | 278,299 | <p>Starting with a generic quadratic polynomial, you could create three linear equations.
<span class="math-container">$$p(x)=a+bx+cx^2$$</span>
<span class="math-container">$$p'(x)=b+2cx$$</span>
Then
<span class="math-container">$$\begin{cases}y_0=a+bx_0+cx_0^2\\y_1'=b+2cx_1\\y_2=a+bx_2+cx_2^2\end{cases}$$</span>
Finally, solve for <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span>.</p>
|
292,948 | <p>I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?</p>
| Brian M. Scott | 12,042 | <p>It is not true.</p>
<p>John Thomas, <a href="http://www.jstor.org/stable/2317272"><em>A regular space, not completely regular</em></a>, Amer. Math. Monthly <strong>76</strong> (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:X\to\Bbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, $\{p\}$ is a closed $G_\delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.</p>
<p>A. Mysior, <a href="http://www.ams.org/journals/proc/1981-081-04/S0002-9939-1981-0601748-4/"><em>A regular space which is not completely regular</em></a>, Proc. Amer. Math. Soc. <strong>81</strong> (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:X\to\Bbb R$ and $x\in F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again $\{p\}$ is a closed $G_\delta$ that cannot be a zero-set.</p>
|
292,948 | <p>I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?</p>
| Martin | 49,437 | <p>Brian's answer covers the question fully. For fun, here's another example:</p>
<p>Bing's <a href="http://dx.doi.org/10.1090/S0002-9939-1953-0060806-9" rel="nofollow">irrational slope space</a> is a countable and connected Hausdorff space. </p>
<p>Now observe:</p>
<ol>
<li><p>If $f \colon X \to \mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $\mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f \colon X \to \mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.</p></li>
<li><p>Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_\delta$-set: $F = \bigcap_{x \in X \setminus F} X \setminus \{x\}$, in particular, there is an abundance of closed $G_\delta$ sets that are not zero sets.</p></li>
</ol>
|
292,948 | <p>I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?</p>
| GEdgar | 442 | <p>Not even in completely regular Hausdorff spaces. In general we have
$$
\text{compact $G_\delta$}\qquad\Longrightarrow\qquad
\text{zero-set}\qquad\Longrightarrow\qquad
\text{closed $G_\delta$}
$$
but none reversible.</p>
|
218,915 | <blockquote>
<p>Prove that for any integer $n$, $\gcd (3n^2+5n+7, n^2+1)=1$ or $41$.</p>
</blockquote>
<p>The following answer is convoluted because I've intentionally created excess solutions. However, I can't figure out how to eliminate them! Anyone?</p>
<p>Let $$d=\gcd (3n^2+5n+7, n^2+1).$$
Then $$d|[(3n^2+5n+7)-3(n^2+1)]$$
$$d |(5n+4)$$
And
$$d | [5(3n^2+5n+7)-3n(5n+4)]$$
$$d |(13n+35)$$
And
$$d |[5(13n+35)-13(5n+4)]$$
$$d |123$$
Therefore, $d= 1$ or $3$ or $41$ or $123$.</p>
| robjohn | 13,854 | <p>Suppose that
$$
(3n^2+5n+7,n^2+1)=(5n+4,n^2+1)\ne1\tag{1}
$$
then either
$$
(5n+4,n+i)=(4-5i,n+i)\ne1\tag{2}
$$
or
$$
(5n+4,n-i)=(4+5i,n-i)\ne1\tag{3}
$$
Since $4-5i$ is a Gaussian prime, $(2)\Rightarrow4-5i\,|\,n+i$. That is,
$$
\frac{n+i}{4-5i}=\frac{(4n-5)+(5n+4)i}{41}\in\mathbb{Z}[i]\tag{4}
$$
which is true if and only if $n\equiv32\pmod{41}$.</p>
<p>Since $4+5i$ is a Gaussian prime, $(3)\Rightarrow4+5i\,|\,n-i$. That is,
$$
\frac{n-i}{4+5i}=\frac{(4n-5)-(5n+4)i}{41}\in\mathbb{Z}[i]\tag{5}
$$
which is true if and only if $n\equiv32\pmod{41}$.</p>
<p>Thus, $(1)$ implies either
$$
(2)\Rightarrow4-5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{6}
$$
or
$$
(3)\Rightarrow4+5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{7}
$$
Therefore,
$$
(1)\Rightarrow n\equiv32\pmod{41}\tag{8}
$$
It is easy to verify that
$$
n\equiv32\pmod{41}\Rightarrow41\,|\,(3n^2+5n+7,n^2+1)\tag{9}
$$</p>
<hr>
<p>Again, $(1)$ implies either
$$
(2)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\tag{10}
$$
or
$$
(3)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\tag{11}
$$
Therefore,
$$
(1)\Rightarrow41=(4-5i)(4+5i)\,|\,(3n^2+5n+7,n^2+1)\tag{12}
$$
Finally, as Pambos points out, the Euclidean Algorithm yields
$$
(15n+13)(n^2+1)-(5n-4)(3n^2+5n+7)=41\tag{13}
$$
Therefore,
$$
(3n^2+5n+7,n^2+1)\,|\,41\tag{14}
$$</p>
|
2,042,428 | <p>If I'm correct, hidden induction is when we use something along the lines of "etc..." in a proof by induction. Are there any examples of when this would be appropriate (or when it's not appropriate but used anyway)?</p>
| Sean Keeler | 395,712 | <p>First, here is an example of when this works. Let $X$ and $Y$ be Hausdorff spaces. This implies $X\times Y$ with the product topology is Hausdorff. Therefore any finite product of Hausdorff spaces is Hausdorff. The "hidden induction" is the idea that $X\times Y$ is a Hausdorff space, which implies any Hausdorff space times $X\times Y$ is also Hausdorff. This argument can continue for any finite number of spaces.</p>
<p>An example of when you can't use this argument is the proof of $\Sigma_{i=1}^n=\frac{n(n-1)}{2}$. Here, showing the truth for $n=1$ does not make it immediately clear that the rest follows by induction.</p>
<p>It was appropriate to use this in the first problem as it was clear that it follows. But the second problem isn't as obvious. It is like a lot of other problems in mathematics; the mathematician must develop an intuition for when something is obvious enough to not require an explicit explanation (e.g. $2*1=2$).</p>
|
2,042,428 | <p>If I'm correct, hidden induction is when we use something along the lines of "etc..." in a proof by induction. Are there any examples of when this would be appropriate (or when it's not appropriate but used anyway)?</p>
| Jack M | 30,481 | <p>Hidden induction happens a lot in cases where you go <em>backwards</em> from $n$ to $1$, using some kind of reduction argument. For example, the proof that every number can be written as a product of primes:</p>
<blockquote>
<p>Let $n$ be some number. If it's prime, then we're done. Otherwise it can be written as $ab$, with $a, b < n$. Again, if both $a$ or $b$ are prime, we're done, otherwise they can be broken up in the same way. Since the factors are getting smaller and smaller, this process must stop eventually, but the only way it can stop is if one of all of the numbers involved are prime.</p>
</blockquote>
<p>Induction serves to tighten up the structure of the argument, replacing a vague "this process must stop" with an explicit invocation of an axiom:</p>
<blockquote>
<p>Suppose every number less than $n$ can be written as a product of primes. If $n$ is prime, we're done. Otherwise, $n=ab$ with $a, b < n$, so $a$ and $b$ can be written as products of primes. Therefore $n$ can be written as a product of primes.</p>
</blockquote>
|
1,309,728 | <p>I know what a 3x10 looks like, but I cannot seem to find a distinguishable pattern to extend it to a 3x14.</p>
<p>The 3x10 pattern I'm using looks like the one at the top right of figure 6 of <a href="http://faculty.olin.edu/~sadams/DM/ktpaper.pdf" rel="nofollow">this paper</a>.</p>
<p>Any help would be greatly appreciated.</p>
| user26857 | 121,097 | <p>Maybe you want to know more, namely: $$f(X,Y)\in(X-a,Y-b) \text{ iff } f(a,b)=0.$$</p>
<p>This solves instantly your question: $f(a,b)=0$ for $f(X,Y)=XY-1$ means $ab=1$.</p>
|
3,362,115 | <blockquote>
<p>Find the maximum value of <span class="math-container">$y/x$</span> if it satisfies <span class="math-container">$(x-5)^2+(y-4)^2=6$</span>.</p>
</blockquote>
<p>Geometrically, this is finding the slope of the tangent from the origin to the circle. Other than solving this equation with <span class="math-container">$x^2+y^2=35$</span>, I cannot see any synthetic geometry solution. Thanks!</p>
| Jan-Magnus Økland | 28,956 | <p>See <a href="https://en.wikipedia.org/wiki/Pole_and_polar" rel="nofollow noreferrer">Pole and polar</a>. </p>
<p>The polar line of the origin <span class="math-container">$-5x-4y+35=0$</span> intersects the circle as seen in the image below, giving the tangents from the origin to the circle.</p>
<p><a href="https://i.stack.imgur.com/5Z9NF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Z9NF.png" alt="pp"></a></p>
<p>Now you get the two points <span class="math-container">$(x_1,y_1)=(\frac{175+4\sqrt{210}}{41}, \frac{140-5\sqrt{210}}{41}),(x_2,y_2)=(\frac{175-4\sqrt{210}}{41},\frac{140+5\sqrt{210}}{41})$</span>. The maximal ratio <span class="math-container">$\frac{y}{x}$</span> is then <span class="math-container">$$\frac{y_2}{x_2}={{140+5\,\sqrt{210}}\over{175-4\,\sqrt{210}}}\approx 1.815335618220497\approx \frac{20+\sqrt{210}}{19},$$</span> the minimal given by <span class="math-container">$\frac{y_1}{x_1}.$</span></p>
|
3,362,115 | <blockquote>
<p>Find the maximum value of <span class="math-container">$y/x$</span> if it satisfies <span class="math-container">$(x-5)^2+(y-4)^2=6$</span>.</p>
</blockquote>
<p>Geometrically, this is finding the slope of the tangent from the origin to the circle. Other than solving this equation with <span class="math-container">$x^2+y^2=35$</span>, I cannot see any synthetic geometry solution. Thanks!</p>
| Cesareo | 397,348 | <p>The line <span class="math-container">$y = \lambda x$</span> and the circle <span class="math-container">$(x-5)^2+(y-4)^2= 6$</span> should intersect and <span class="math-container">$\max \frac xy =\max \lambda$</span> should be located at a tangency point hence solving for <span class="math-container">$x$</span></p>
<p><span class="math-container">$$
(x-5)^2+(\lambda x-4)^2= 6
$$</span></p>
<p>we have</p>
<p><span class="math-container">$$
x = \frac{4 \lambda +5\pm\sqrt{-19 \lambda ^2+40 \lambda -10}}{\lambda ^2+1}
$$</span></p>
<p>but at tangency</p>
<p><span class="math-container">$$
-19 \lambda ^2+40 \lambda -10 = 0
$$</span></p>
<p>with</p>
<p><span class="math-container">$$
\lambda^* = \frac {1}{19}(20+\sqrt{210})
$$</span></p>
|
3,362,115 | <blockquote>
<p>Find the maximum value of <span class="math-container">$y/x$</span> if it satisfies <span class="math-container">$(x-5)^2+(y-4)^2=6$</span>.</p>
</blockquote>
<p>Geometrically, this is finding the slope of the tangent from the origin to the circle. Other than solving this equation with <span class="math-container">$x^2+y^2=35$</span>, I cannot see any synthetic geometry solution. Thanks!</p>
| Narasimham | 95,860 | <p>Locate the circle center and draw the displaced circle with its radius including other sides using Pythagoras thm.
Add two angles at max slope tangent point around origin O as shown directly:</p>
<p><a href="https://i.stack.imgur.com/dD8na.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dD8na.png" alt="enter image description here"></a></p>
<p><span class="math-container">$$ \tan^{-1}{\dfrac{y_{ tgt}}{x_{ tgt}}}=\tan^{-1}\frac{4}{5}+\tan^{-1}\sqrt{\frac{6}{35}} $$</span></p>
<p>Now apply arctangent sum formula and complete the same.</p>
|
78,478 | <blockquote>
<p>Prove that $\frac{1}{n} \sum_{k=2}^n \frac{1}{\log k}$ converges to $0.$</p>
</blockquote>
<p>Okay, seriously, it's like this question is mocking me. I know it converges to $0$. I can feel it in my blood. I even proved it was Cauchy, but then realized that didn't tell me what the limit <em>was</em>. I've been working on this for an hour, so can one of you math geniuses help me?</p>
<p>Thanks!</p>
| hmakholm left over Monica | 14,366 | <p>In general, if $a_n\to 0$, then $\frac1n \sum_{k=0}^n a_k \to 0$ too.</p>
<p>(For any $\varepsilon>0$, find an $N$ such that $|a_n|<\varepsilon/2$ for all $n>N$, and then take enough terms beyond $N$ that they dominate whatever the terms <em>before</em> $N$ contribute to the average).</p>
<p>Even more generally (and as an easy consequence), whenever $a_n$ converges, $\frac1n \sum_{k=0}^n a_k$ converges to the same limit.</p>
|
352,849 | <p>I have to show that $\lim \limits_{n\rightarrow\infty}\frac{n!}{(2n)!}=0$ </p>
<hr>
<p>I am not sure if correct but i did it like this :
$(2n)!=(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))\cdot (n!)$ so I have $$\displaystyle \frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}$$ and $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$ is this correct ? If not why ?</p>
| Elias Costa | 19,266 | <p>If so addressing trivial rigorously I suggest using the notation produtory to fatorial use the formula $n!=\prod_{k=1}^{n}$ .
\begin{align}
0\leq \frac{n!}{(2n)!}
=
&
\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{2n}k\big)}
\\
=
&
\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=n+1}^{2n}k\big)\big(\prod_{k=1}^{n}k\big)}
\\
=
&
\frac{1}{\big(\prod_{k=n+1}^{2n}k\big)}\frac{\big(\prod_{k=1}^{n}k\big)}{\big(\prod_{k=1}^{n}k\big)}
\\
=
&
\frac{1}{\big(\prod_{k=n+1}^{k=2n}k\big)}
\\
=
&
\frac{1}{2n\big(\prod_{k=n+1}^{2n-1}k\big)}
\\
=
&
\frac{1}{2n}\frac{1}{\big(\prod_{k=n+1}^{2n-1}k\big)}
\\
\leq
&
\frac{1}{2n}
\end{align}</p>
|
2,596,213 | <p>I'm having huge troubles with problems like this. I know the following:</p>
<p>$$\frac{\sin{x}}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^7)$$</p>
<p>and </p>
<p>$$\ln{(1+t)}=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)$$</p>
<p>So</p>
<p>$$\ln{\left(1+\left(-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^7)\right)\right)}=\\\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^7)\right]-\frac{\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^7)\right]^2}{2}+\frac{\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^7)\right]^3}{3}+O(x^8).$$</p>
<p>But how on earth would one simplify this? Obviously I should not need to manually expand something of the form $(a+b+c+d+e)^n$. Seriously don't understand what is happening here.</p>
<p>Also, how should I know to what $O(x^?)$ I should expand the initial functions to? </p>
| user | 505,767 | <p>You need to consider</p>
<p>$$\frac{\sin{x}}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^8)$$</p>
<p>and</p>
<p>$$\ln{(1+t)}=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)$$</p>
<p>then substitute and expand keeping only the terms with order less than $x^8$ thus you don't need to expand all the expression but only the parts you need.</p>
<p>That is</p>
<p>$$\ln{\left(1+\left(-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^8)\right)\right)}
=\\\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^8)\right]-\frac12\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^8)\right]^2+\frac13\left[-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+O(x^8)\right]^3+O(x^8)
=\\-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}
-\frac12\frac{x^4}{(3!)^2}+\frac12\frac{2x^6}{3!5!}+\frac12\frac{x^8}{(5!)^2}-\frac13\frac{x^6}{(3!)^3}+O(x^8)$$</p>
|
51,752 | <p>Can someone give an argument, if possible using only the axioms of set theory, because I'm <strong>very</strong> weak there and have virtually no background, except the usual knowledge of the operation with sets one has to have when doing non-set theoretic non-research mathematics, why $\emptyset \in \emptyset$ or $\emptyset \subseteq \emptyset$ should or should not hold?</p>
| Asaf Karagila | 622 | <p>The empty set $\varnothing$ is the only set which satisfies $\forall x(x\notin y)$ (that means the formula is true if and only if $y=\varnothing$)</p>
<p>There are many ways to define the empty set (the set of all $x$ such that $x\neq x$ - we will use this formula later on) but by the axiom of extensionality it is unique.</p>
<p>The axiom of extensionality is, in simple words, two sets are equal if and only if they have the same elements. Or formally: $$\forall x\forall y\bigg(\forall z(z\in x\leftrightarrow z\in y)\leftrightarrow x=y\bigg)$$
So since $\varnothing=\{x\mid x\neq x\}$ we can deduce that $\varnothing\notin\varnothing$, otherwise $\varnothing$ would be such $x$ for which $x\neq x$. </p>
<p>Even worse, if $\varnothing\in\varnothing$ then we contradict another axiom of ZFC - the axiom of foundation (or regularity) which asserts that there is a $\in$-minimal element in every non-empty set. Since $\varnothing\in\varnothing$, we will have that $\varnothing$ is not empty.</p>
<p>What does that mean? It means that if I consider a set, there is someone in that set that no one else in that set is its element. A quick corollary is that $\forall x (x\notin x)$, in particular for $\varnothing$.</p>
<p>Lastly, $\varnothing\subseteq\varnothing$. Let us consider the meaning of $\subseteq$: $$\forall x\forall y\bigg(\forall z(z\in x\rightarrow z\in y)\leftrightarrow x\subseteq y\bigg)$$
Informally we have that $x$ is a subset of $y$ if and only if all the elements of $x$ are elements of $y$, with this we can reformulate the axiom of extensionality as $\forall x\forall y(x=y\leftrightarrow(x\subseteq y\land y\subseteq x))$.</p>
<p>In particular, $\varnothing=\varnothing$ so $\varnothing\subseteq\varnothing$.</p>
<p>A stronger conclusion, however, can be drawn from the definition of $\subseteq$ by the idea of <em>vacuous truth</em>, since for all $z$ we have $z\notin\varnothing$ the clause $z\in\varnothing\rightarrow z\in y$ is <strong>always</strong> true, therefore $\varnothing\subseteq y$, for every $y$.</p>
|
184,824 | <p>I have two piecewise function</p>
<pre><code>equ1 = Piecewise[{{0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322}}]
equ2 = Piecewise[{{-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085}}]
</code></pre>
<p>Now, I am trying to solve <code>equ1 = equ2</code>.</p>
<p>Firstly I tried <code>FindRoot</code>: </p>
<pre><code>FindRoot[equ1 == equ2, x]
</code></pre>
<p>But the output is <code>x = 0</code>. I can only get the correct answer by set search starting point <code>0.7</code>. How can I direct get the answer without set starting point? </p>
<p>Secondly, I tried code Reduce: </p>
<pre><code>Reduce[equ1 == equ2, x]
</code></pre>
<p>However, the error appear. The good news is <code>Reduce</code> do provide the correct answer for my equation. The error is: </p>
<pre><code>Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.
</code></pre>
<p>Do I have other way to solve those two piecewise function? </p>
| Bill | 18,890 | <p>If you have 100 equations and you want to find the unique solution to the whole set and each of your <code>Piecewise</code> are written in exactly the same form as the two you showed and the solution must lie in the intersection of all those inequalities then perhaps</p>
<pre><code>equ1 = Piecewise[{{0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322}}];
equ2 = Piecewise[{{-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085}}];
funs = {equ1, equ2};
Join[{Equal @@ Map[#[[1, 1, 1]] &, funs]},
{Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]}]
</code></pre>
<p>will give you</p>
<pre><code>{0.524324 + 0.0376478 x == -0.506432 + 1.48068 x, 0.658914 < x < 0.77085}
</code></pre>
<p>and</p>
<pre><code>x /. Solve[
Join[{Equal @@ Map[#[[1, 1, 1]] &, funs]},
{Max[Map[#[[1, 1, 2, 1]] &, funs]] < x < Min[Map[#[[1, 1, 2, 3]] &, funs]]}]
, x][[1]]
</code></pre>
<p>will give you</p>
<pre><code>0.714299
</code></pre>
<p>You can make the list of funs longer and longer and still get the solution.</p>
<p>The <code>Max</code> part of that is extracting all your lower bounds and finding the largest one. The <code>Min</code> part of that is extracting all your upper bounds and finding the smallest one. The <code>Equal</code> part of that is extracting all your equations and setting them equal to each other. We assemble those pieces and give them to Solve.</p>
<p>You should check carefully to make certain that there will always be one unique solution to your set of equations.</p>
<p>You might click on <code>Piecewise</code> and then <code>Details</code> in the help pages and read very carefully all the information there. Because you have not given any other information about the value of your functions if x happens to be outside the bounds you have given then the value of that <code>Piecewise</code> function will be zero for x outside those bounds. That might surprise you.</p>
|
2,012,947 | <p>I'm trying to prove that if f,g are continuous functions, and if E is a dense subset of X $(\text{or } Cl(E) = X)$ and if $f(x)=g(x) \forall x \in E$ then $f(x)=g(x) \forall x \in X$. </p>
<p>I understand that if f,g are continuous, then:</p>
<blockquote>
<p>$\exists \delta_1, \delta_2$ such that $\forall X \in E$ with $d(x,p)< \delta_1$, $|f(x) - f(p)| < \epsilon$ and similarly $\forall X \in E$ with $d(x,p)< \delta_2$, $|g(x) - g(p)| < \epsilon$</p>
</blockquote>
<p>And by definition of closure, I know that:</p>
<blockquote>
<p>$Cl(E) = E \cup E'$ where E' is the set accumulation points of E, where p is an accumulation point if $\forall r>0, (E\cup N_r(p)) \backslash \{p\} \neq \emptyset $</p>
</blockquote>
<p>I have zero clue on how to approach this problem. If $f(x) = g(x)$, then I'm guessing it implies that $|f(x) - f(p)| = |g(x) - g(p)|$. And so I'm guessing that $\delta_1 = \delta_2$. </p>
<p>Help would be very much appreciated. </p>
| Learnmore | 294,365 | <p>Suppose that $f(a)\neq g(a);a\in X$.</p>
<p>Let $d(f(a),g(a))=r>0$.</p>
<p>Since $f$ is continuous at $a$ so $\exists \delta_1>0$ such that $f(B(a,\delta_1))\subset B(f(a),\frac{r}{3})$.</p>
<p>Since $g$ is continuous at $a$ so $\exists \delta_2>0$ such that $g(B(a,\delta_2))\subset B(g(a),\frac{r}{3})$.</p>
<p>Take $\delta=\min\{\delta_1,\delta_2\}$.</p>
<p>Then $f(B(a,\delta))\subset B(f(a),\frac{r}{3})$ and $g(B(a,\delta))\subset B(g(a),\frac{r}{3})$.</p>
<p>Since $E$ is dense in $X$ so $B(a,\delta)\cap E\neq \emptyset$. Let $k\in B(a,\delta)$. Then $f(k)=g(k)$</p>
<p>Then $f(k)\in B(f(a),\frac{r}{3})$ and $g(k)\in B(g(a),\frac{r}{3})$ .</p>
<p>Hence we have $d(f(a),g(a))\le d(f(a),f(k))+d(f(k),g(k))+d(g(k),g(a))<\dfrac{2r}{3}$</p>
<p>which is false as $d(f(a),g(a))=r$.</p>
|
3,391,225 | <p>What is the term for a (connected?) set <span class="math-container">$S$</span> of the plane <span class="math-container">$\mathbb{R}^2$</span> such that the intersection of <span class="math-container">$S$</span> with every horizontal line <span class="math-container">$\ell_{b}: y=b$</span> is either empty, or an interval of the line <span class="math-container">$\ell_b$</span>? </p>
| Pacciu | 8,553 | <p>From where I’m from, such a set is called <em><span class="math-container">$y$</span>-simple domain</em> or <em><span class="math-container">$y$</span>-normal domain</em>.</p>
|
4,394,247 | <p>I know how to represent the sentence “there is exactly one person that is happy”,</p>
<p>∀y∀x((Happy(x)∧Happy(y))→(x=y))</p>
<p>Edit: ∃x∀y(y=x↔Happy(y))
(NOW, I actually know how to represent it)</p>
<p>Where x and y represent a person.</p>
<p>However, my problem is that I can’t figure out how to say “there are exactly 3 people that are happy” in predicate logic.</p>
| A J | 934,287 | <p>You can extend this simply as: if 4 people are happy then at least one is equal to the other.
<span class="math-container">$\forall x,y,z,w((Happy(x)\land Happy(y) \land Happy(z) \land Happy(w))\rightarrow(x=y \lor x=z \lor x=w \lor y=z \lor y=w \lor z=w)) \land \exists x,y,z (Happy(x)\land Happy(y) \land Happy(z) \land x\neq y \land y\neq z \land x \neq z) $</span></p>
|
869,218 | <p>I am reading Stillwell's <em>Numbers and Geometry</em>. There is an exercise about Egyptian fractions which is the following:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/JiSpw.png" alt="enter image description here"></p>
</blockquote>
<p>I've tried to do it in the following way - Expressing an arbitrary fraction $\frac{n}{m}$ as the sum of two aribtrary egipcian fractions:</p>
<p>$$\frac{1}{a}+\frac{1}{b}=\frac{n}{m}$$</p>
<p>$$\frac{1}{a}+\frac{1}{b}-\frac{n}{m}=0$$</p>
<p>$$\frac{bm+am-abn}{abm}=0$$</p>
<p>Then solving for $a$ yields:</p>
<p>$$a=\frac{b m}{b n-m}$$</p>
<p>And then expressing it as a function:</p>
<p>$$f(b)= \frac{b m}{b n-m}$$</p>
<p>I have tested with $\frac{n}{m}=\frac{3}{4}$ on Mathematica and all the sums of unit fractions I've obtained equals $\frac{3}{4}$. Is this correct? I'm afraid there's something wrong and I'm not seeing it.</p>
| Deathkamp Drone | 56,720 | <p>Here's one systematic method to see if there are ways of expressing a fraction as a sum of two Egyptian fractions:</p>
<p>$$\begin{align}\frac{4}{5}=\frac{1}{m}+\frac{1}{n} &\Leftrightarrow 4mn=5m+5n \\
&\Leftrightarrow 5m-4mn+5n=0 \\
&\Leftrightarrow \left(2m-\frac{5}{2}\right)\left(\frac{5}{2}-2n\right)=-\frac{25}{4}\\
&\Leftrightarrow (4m-5)(5-4n)=-25\end{align}$$</p>
<p>Notice that $m,n$ are integers, and so $(4m-5),(5-4n)$ are integers too. Therefore, those have to be divisors of $25$ and there are finitely many divisors of $25$, allowing you to test all possible combinations.</p>
<p>You can use a similar method for three fractions (if no representation using two fractions is possible), but as it has been pointed out in the comments already, it's better to test some small fractions until you get what you want.</p>
|
869,218 | <p>I am reading Stillwell's <em>Numbers and Geometry</em>. There is an exercise about Egyptian fractions which is the following:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/JiSpw.png" alt="enter image description here"></p>
</blockquote>
<p>I've tried to do it in the following way - Expressing an arbitrary fraction $\frac{n}{m}$ as the sum of two aribtrary egipcian fractions:</p>
<p>$$\frac{1}{a}+\frac{1}{b}=\frac{n}{m}$$</p>
<p>$$\frac{1}{a}+\frac{1}{b}-\frac{n}{m}=0$$</p>
<p>$$\frac{bm+am-abn}{abm}=0$$</p>
<p>Then solving for $a$ yields:</p>
<p>$$a=\frac{b m}{b n-m}$$</p>
<p>And then expressing it as a function:</p>
<p>$$f(b)= \frac{b m}{b n-m}$$</p>
<p>I have tested with $\frac{n}{m}=\frac{3}{4}$ on Mathematica and all the sums of unit fractions I've obtained equals $\frac{3}{4}$. Is this correct? I'm afraid there's something wrong and I'm not seeing it.</p>
| individ | 128,505 | <p>As I said a decision for 3 terms there. <a href="https://math.stackexchange.com/questions/450280/erd%C3%B6s-straus-conjecture/831870#831870">Erdős-Straus conjecture</a> </p>
<p>As for the answers of two terms, the decision is not always exist.</p>
<p>For the equation: $$\frac{1}{X}+\frac{1}{Y}=\frac{b}{A}$$ </p>
<p>You can write a simple solution if the number on the decomposition factors as follows: $$A=(k-t)(k+t)$$<br>
then: $$X=\frac{2k(k+t)}{b}$$ $$Y=\frac{2k(k-t)}{b}$$ </p>
<p>or: $$X=\frac{2t(k-t)}{b}$$ $$Y=\frac{-2t(k+t)}{b}$$</p>
<p>First decide so $b=1$ and then I look at what other values have solutions.</p>
|
205,926 | <p>I'm trying to understand a proof about density of a subset $X$ in its one-point compactification $Y$.</p>
<p>We can do this proof by contradiction, suppose we don't have $\operatorname{cl}(X) = Y$.
This implies that $\operatorname{cl}(X) = X$. </p>
<p>Why? Can anyone help me?</p>
<p>Thanks</p>
| Idan | 37,998 | <p>Suppose $\operatorname{cl}(X)\not=Y$. We know $X\subseteq \operatorname{cl}(X)$ so we get $\operatorname{cl}(X)=X$ and $\infty \notin \operatorname{cl}(X).$ So by definition of closure, there exists a (wlog, open) neighborhood $U$ of $\infty$ s.t. $U \cap X=\emptyset$. The topology of the extension is defined to be all open subsets of $X$ together with all sets $V$ that contain $\infty$ and such that $X\setminus V$ is closed and compact. $\infty \in U$ so $U$ is of the second kind of open sets, meaning $X \setminus U$ is closed and compact. But remember that $U \cap X=\emptyset$ so $X\setminus U=X$. We get that $X$ is compact, in contradiction to the assumption (if $X$ was compact, we wouldn't have needed the compactification).</p>
|
979,432 | <p>i was recently watching a single variable calculus video of mit 18.01, lecture 23. in that it is said that average height of a point on semicircle with respect to arc length is 2/pi.I have a hard time to understand that point. i understand why average height of point on semi circle with respect to x is pi/4. but i dont understand with respect to arc length. plz can somebody help me. </p>
| Jim H | 473,669 | <p>The average value with respect to $x$ over interval $[a,b]$ is found by: </p>
<p>Cut the interval into $n$ equal subintervals of length $\Delta x = \frac{b-a}{n}$. Note that $n = \frac{\Delta x}{b-a}$.
On each subinterval, use a single value of $x$ (which I'll call $x_i$).
Sum the values of $f\left( x_i \right)$ and divide by $n$</p>
<p>And the limit as $n$ increases without bound is the average value of $f$ on $[a,b]$.</p>
<p>$$\lim_{n \to \infty} \sum_{i=1}^n \frac{f(x_i)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \frac{f(x_i)\Delta x}{b-a} = \frac{1}{b-a} \int_a^b f(x) dx$$</p>
<p>For the upper unit semicircle, we have an average height</p>
<p>$$\frac 1 2 \lim_{n\to \infty}\sum_{i=1}^n \sqrt{1-x_i^2} \Delta x = \frac{1}{2} \int_{-1}^1 \sqrt{1-x^2}dx$$</p>
<p>The average with respect to arc length cuts the curve of total length $T$ into $n$ equal subarcs each of length $\Delta t = \frac{T}{n}$ and again uses a representative value, then takes the limit as the number of subarcs of the curve increases without bound.</p>
<p>$$\lim_{n \to \infty} \sum_{i=1}^n \frac{f(t_i)}{n} = \lim_{n \to \infty} \sum_{i=1}^n \frac{f(t_i)\Delta t}{T} = \frac{1}{T} \int_0^T f(t) dt$$ </p>
<p>For the upper unit semicircle, we have $t \in [0,\pi]$ we cut the semicircle into $n$ equal arcs, each of length $\Delta t = \frac{\pi}{n}$. Note that $n = \frac{\pi}{\Delta t}$.
For each subarc, we use a $t_i$
The height of the point on the semicircle as a function of $t_i$ is $\sin{t_i}$. So the average value of height w.r.t. arc length is </p>
<p>$$\lim_{n \to \infty} \sum_{i=1}^n \frac {\sin{t_i}}{n} = \lim_{n \to \infty}\sum_{i=1}^n \frac {\sin{t_i} \Delta t}{T} = \frac{1}{T} \int_0^\pi \sin{t} dt$$</p>
|
4,540,637 | <blockquote>
<p>Given a line <span class="math-container">$y=kx$</span> on a Cartesian coordinate, I want to find an equation of a parabola, whose base is on that line at point <span class="math-container">$(x_1,y_1)$</span> and passes through point <span class="math-container">$(x_2,y_2)$</span>.</p>
</blockquote>
<p>Then, an equation of a red line that is perpendicular to line <span class="math-container">$y=kx$</span> at point <span class="math-container">$(x_1,y_1)$</span> is <span class="math-container">$y=-\frac{1}{k}x+y_1+\frac{1}{k}x_1$</span>. I tried to assume that <span class="math-container">$y=kx$</span> line is my new <span class="math-container">$x$</span> axis, and <span class="math-container">$y=-\frac{1}{k}x+y_1+\frac{1}{k}x_1$</span> is my new <span class="math-container">$y$</span> axis and do calculations, but the computation became messy and I couldn't finish. Is there simple way to solve this problem?</p>
<p><a href="https://i.stack.imgur.com/PuEIL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PuEIL.jpg" alt="enter image description here" /></a></p>
| Abel Wong | 1,090,313 | <p><span class="math-container">$x_1$</span> and <span class="math-container">$y_1$</span> make the problem complicate. First, we do a translation to make <span class="math-container">$x_1, y_1$</span> move to origin in <span class="math-container">$x'-y'$</span> coordinate system.</p>
<p>Now, the parabola touch <span class="math-container">$y=kx$</span> at origin and pass <span class="math-container">$(x'_0,y'_0)$</span> where <span class="math-container">$x'_0=x_2-x_1, y'_0=y_2-y_1$</span>.</p>
<p>Then we make a rotation to <span class="math-container">$x''-y''$</span> coordinate system. by a rotation of <span class="math-container">$\theta$</span>.</p>
<p><span class="math-container">$x'=x''\cos\theta + y''\sin\theta$</span></p>
<p><span class="math-container">$y'=-x''\sin\theta + y''\cos\theta$</span> (Hope I did not rotation at wrong direction).</p>
<p><span class="math-container">$(x'_0,y'_0)$</span> now rotated to <span class="math-container">$(x''_0,y''_0)$</span>.</p>
<p>Then the parabola touch x-axis at origin.</p>
<p>The parabola is <span class="math-container">$y''=ax''^2$</span>. Put <span class="math-container">$(x''_0,y''_0)$</span> into the equation, you can solve for <span class="math-container">$a$</span> and fix the parameter of parabola equation in <span class="math-container">$x''-y''$</span> coordinate. Then you can work back to a equation in <span class="math-container">$x-y$</span> coordinate. The equation could be complicated so you can write it in terms of <span class="math-container">$a$</span> since you already have the expression of <span class="math-container">$a$</span>.</p>
<p>Method II:
Start from general equation of parabola:
<span class="math-container">$\frac{(ax+by+c)^2}{a^2+b^2}=(x-x_f)^2+(y-y_f)^2$</span></p>
<p>Assume the signed distance between focus <span class="math-container">$(x_f,y_f)$</span> to <span class="math-container">$y=kx$</span> is <span class="math-container">$d$</span>. Then you can write down the equation of directrix <span class="math-container">$ax+by+c$</span> and focus <span class="math-container">$(x_f,y_f)$</span> in term of <span class="math-container">$d$</span>. Then put <span class="math-container">$(x_2,y_2)$</span> into the equation, you can find <span class="math-container">$d$</span> and then fix all parameters <span class="math-container">$a,b,c,x_f,y_f$</span>.</p>
|
691,734 | <p>Consider the sequence defined recursively by $x_1$=$\sqrt2$ and where $x_n$=$\sqrt2$ + $x_n$$_-$$_1$. </p>
<p>Find a explicit formula for the $n^t$$^h$ term.</p>
<p>I considered using the general equation to find an explicit formula for any term in an arithmetic sequence. a$_n$ = a$_1$ + $d(n-1)$, but I came to no conclusion helping my argument. </p>
<p>Am I using the correct method? </p>
| Mhenni Benghorbal | 35,472 | <p>Here is an approach.</p>
<p>$$ x_{n+1}-x_{n}=\sqrt{2} \implies \sum_{i=0}^{n-1}( x_{i+1}-x_{i}) = \sqrt{2}\sum_{i=0}^{n-1}1 $$</p>
<p>$$ \implies x_n-x_0=\sqrt{2} n .$$</p>
|
234,340 | <p>Suppose that I have two real-valued matrices $\bf{A}$ and $\bf{B}$. Both matrices are exactly the same size. I multiply both matrices together in a point-by-point fashion similar to the Matlab <code>A .* B</code> operation.</p>
<p>Under what conditions can I approximately separate $\bf{A}$ and $\bf{B}$ using Principle Components Analysis (PCA)? Would it be possible to remove some components of the product <code>A .* B</code> to get an approximation of $\bf{A}$ or $\bf{B}$?</p>
<p>What algorithm might be best suited for this operation?</p>
<p>I am not looking for an exact separation of the matrices, but a separation using some sort of (statistical or numerical?) constraints. How would I set this problem up, and is there a good example of how to do this?</p>
| Bitwise | 42,051 | <p>It seems to me that you can't separate the matrices from their pointwise multiplication (Hadamard/Schur product) without additional constraints.</p>
<p>Consider some matrix C. Any number in C is decomposable into an infinite number of products of two real numbers... which would give you an infinite number of "perfect" decompositions.</p>
<p>For example, you can always decompose C into 1 (a matrix of ones) and C. In fact, for any choice of A you can find a B such that their Schur product will result in any C that is specified (ignoring some problems when zero appears)...</p>
|
3,662,286 | <p>Can you raise the imaginary number i to a power that is an irrational number?</p>
| Lukas Rollier | 737,665 | <p>Take <span class="math-container">$c \in \mathbb{C}$</span> arbitrarily. One might define <span class="math-container">$i^c := e^{\ln(i^c)} = e^{c \ln(i)} = e^{c \cdot \frac{i\pi}{2}}$</span>, which works. There is a catch though: this makes use of the logarithm, which is not nicely defined on the entire complex plane. The set of all 'reasonable' outcomes for <span class="math-container">$i^c$</span> would be <span class="math-container">$e^{ic \cdot \left(\frac{ \pi}{2} + 2k\pi \right)}$</span> for <span class="math-container">$k \in \mathbb{Z}$</span>. Hence, it is not uniquely defined. This problem already arises when taking rational powers of <span class="math-container">$i$</span> though, so nothing's really changed.</p>
|
661,269 | <p>Check if $\mathbb{Z}_5/x^2 + 3x + 1$ is a field. Is $(x+2)$ a unit, if so calculate its inverse. </p>
<p>I would say that this quotient ring is not a field, because $<x^2 + 3x + 1>$ is not a maximal ideal, since $x^2 + 3x + 1 = (x+4)^2$ is not irreducible. </p>
<p>However, the result should still be a ring, right? How do I check if $(x+2)$ is a unit in that ring. Should I just "try" to invert it, or is there a better way.</p>
<p>Thanks </p>
| André Nicolas | 6,312 | <p>You are asking for the distribution of $Y=a_1X_1+\cdots +a_n X_n$, where more generally the $X_i$ are independent normal, means $\mu_i$, variances $\sigma_i^2$. The random variable $Y$ has normal distribution, mean $\sum_1^n a_i \mu_i$, variance $\sum_1^n a_i^2 \sigma_i^2$. </p>
|
3,646,911 | <p>Exercise 14.7.4 from Dummit and Foote</p>
<blockquote>
<p>Let <span class="math-container">$K=\mathbb{Q}(\sqrt[n]{a})$</span>, where <span class="math-container">$a\in \mathbb{Q}$</span>, <span class="math-container">$a>0$</span> and suppose <span class="math-container">$[K:\mathbb{Q}]=n$</span>(i.e., <span class="math-container">$x^n-a$</span> is irreducible). Let <span class="math-container">$E$</span> be any subfield of <span class="math-container">$K$</span> and let <span class="math-container">$[E:\mathbb{Q}]=d$</span>. Prove that <span class="math-container">$E=\mathbb{Q}(\sqrt[d]{a})$</span>. [Consider <span class="math-container">$ N_{K/E}(\sqrt[n]a)\in E$</span>]</p>
</blockquote>
<p>Here is a solution in MSE. </p>
<p><a href="https://math.stackexchange.com/q/1785837/517603">Subfield of $\mathbb{Q}(\sqrt[n]{a})$</a></p>
<p>I rewrite the solution in that answer here. </p>
<p>Let <span class="math-container">$\alpha=\sqrt [n]a\in \mathbb R_+$</span> be the real positive <span class="math-container">$n$</span>-th root of <span class="math-container">$a$</span>, so that <span class="math-container">$K=\mathbb Q(\alpha)$</span>.<br>
Consider some intermediate field <span class="math-container">$\mathbb Q\subset E\subset K$</span> (with <span class="math-container">$d:=[E:\mathbb Q]$</span>) and define <span class="math-container">$\beta=N_{K/E}(\alpha)\in E$</span>.<br>
We know that <span class="math-container">$\beta=\Pi _\sigma \sigma (\alpha) $</span> where <span class="math-container">$\sigma$</span> runs through the <span class="math-container">$E$</span>-algebra morphisms <span class="math-container">$K\to \mathbb C$</span>.<br>
Now, <span class="math-container">$\sigma (\alpha)=w_\sigma \cdot\alpha$</span> for some suitable complex roots <span class="math-container">$w_\sigma$</span> of <span class="math-container">$1$</span> so that <span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span> Remembering that <span class="math-container">$\beta\in E\subset K\subset \mathbb R$</span> is real and that the only real roots of unity are <span class="math-container">$\pm 1$</span> we obtain <span class="math-container">$\Pi _\sigma w_\sigma=\frac {\beta}{\alpha^e}=\pm 1$</span> and <span class="math-container">$\beta=\pm \alpha^e=\pm \sqrt [d]a$</span>.<br>
Thus we have <span class="math-container">$\mathbb Q(\beta)=\mathbb Q(\sqrt [d]a)\subset E$</span> with <span class="math-container">$ \sqrt [d]a$</span> of degree <span class="math-container">$d$</span> over <span class="math-container">$\mathbb Q$</span>.<br>
Since <span class="math-container">$[E:\mathbb Q]=d$</span> too we obtain <span class="math-container">$E=\mathbb Q(\sqrt [d]a)$</span>, just as claimed in the exercise.</p>
<p>Question:
In this line,
<span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span>
<span class="math-container">$K$</span> is not necessarily galois extension.<span class="math-container">$|Aut(K/E)|$</span> need not be equal to <span class="math-container">$[K:E]=\frac nd$</span>.</p>
<p>It's there in all the solution I came across. I wonder why it has to be true. Please explain why <span class="math-container">$e$</span> in above equation has to be <span class="math-container">$[K:E]$</span></p>
| nonuser | 463,553 | <p><span class="math-container">$$32=\binom50x^{5}-\binom52x^{3} + \binom52x^{1}-\binom53x^{-1} + \binom54x^{-3}-\binom55x^{-5} $$</span></p>
<p>Or
<span class="math-container">$$32x^5=\binom50x^{10}-\binom52x^{8} + \binom52x^{6}-\binom53x^{4} + \binom54x^{2}-\binom55 $$</span></p>
<p><span class="math-container">$$ 32x^5 = (x^2-1)^5\implies 2x= x^2-1 \implies x=...$$</span></p>
|
613,940 | <p>Given two parameters $a$ and $b$ (both positive integers), please estimate
the order of growth of the following function:</p>
<p>$$F(t)=\left\{\begin{array}{ll} 1, \, &t\le a \\ F(t-1) + b\cdot F(t-a),&t>a\end{array}\right.$$ </p>
<p>My guess is $\Theta\left(b^{t/a}\right)$. Any answer that might help to confirm or deny this is welcome. The same with helpful references or suggestions.</p>
| Peter | 82,961 | <p>How about ${b}^{\frac{t}{a}}$ ? Because if t increases by a, F approximately is multiplied with b ? </p>
|
19,521 | <p>I am trying to integrate a hat function for a project that I am doing and have found a method to do so but I find it sloppy. Currently I have the basis function</p>
<pre><code>\[Psi][z_] := z - Subscript[Z, i]/ \[CapitalDelta]z + 1;
</code></pre>
<p>which I am trying to integrate from $z_{i-1}$ to $z_{i+1}$. I break the basis function up into two pieces and integrate the left side from $z_{i-1}$ to $z_{i}$ and then the right side from $z_i$ to $z_{i+1}$. My first question is, is there a way to integrate piecewise functions? The second question I have is, is there a way to set global assumptions like $z_{i-1} < z_i < z_{i+1}$, $z_i - z_{i-1} = \Delta z$ , etc?</p>
<p><strong>Edit</strong>: This is the piece wise function taken directly from my code I am trying to integrate</p>
<pre><code>\[Psi][z_, c_] := Piecewise[{{(z - c)/\[CapitalDelta]z + 1,
z <= c}, {-(z - c)/\[CapitalDelta]z + 1, z > c}}];
</code></pre>
<p>where $c$ is the center of the hat function. Here is my attempt to integrate the piece wise function</p>
<pre><code> FullSimplify[
Integrate[\[Psi][z, Subscript[Z, i]], {z, Subscript[Z, i - 1],
Subscript[Z, i + 1]}],
Assumptions -> {-(Subscript[Z, i + 1] - Subscript[Z,
i ]) == -\[CapitalDelta]z,
Subscript[Z, i + 1] - Subscript[Z,
i ] == \[CapitalDelta]z, -(Subscript[Z, i] - Subscript[Z,
i - 1 ]) == -\[CapitalDelta]z,
Subscript[Z, i] - Subscript[Z, i - 1 ] == \[CapitalDelta]z}]
</code></pre>
<p>I do not get a usable answer. Am I doing something wrong (ie can one integrate a piece wise function)?</p>
| Xerxes | 5,406 | <p>Short answer: <em>Mathematica</em> has no problem integrating piecewise or hat functions.</p>
<p>Your notation seems to me to be needlessly complex. Why bother to define $Z_i$ when it's just $Z_0+i\Delta z$? Isn't your $\psi$ just <code>1-Abs[z-c]/Δz</code>? However, I'll try to adhere to the spirit of your notation. Here's some code that works for me:</p>
<pre><code>Zdefs = {Subscript[Z, i] - Subscript[Z, i - 1] == Δz,
Subscript[Z, i + 1] - Subscript[Z, i] == Δz};
Integrate[ψ[z, Subscript[Z, i]],
{z, Subscript[Z, i - 1], Subscript[Z, i + 1]} /.
Solve[Zdefs, {Subscript[Z, i + 1], Subscript[Z, i - 1]}][[1]],
Assumptions -> Δz > 0 &&
Subscript[Z, i] ∈ Reals]
</code></pre>
|
4,236,878 | <p>Given a symmetric matrix <span class="math-container">$S$</span> and positive definite matrix <span class="math-container">$B$</span>, with <span class="math-container">$S,B \in \mathbb{R}^{n \times n}$</span> can one prove that</p>
<p><span class="math-container">\begin{align*}
\text{tr}((S-B)B) \le -\mu(S) \text{tr}(B)
\end{align*}</span></p>
<p>where <span class="math-container">$\mu(S) < 0$</span> is the largest eigenvalue of <span class="math-container">$S$</span>? And does this hold if <span class="math-container">$\mu(S) > 0$</span>?</p>
| user1551 | 1,551 | <p>If <span class="math-container">$\lambda_\max(S)\le0$</span>, then <span class="math-container">$S$</span> is negative semidefinite and <span class="math-container">$S-B$</span> is negative definite. Therefore <span class="math-container">$\operatorname{tr}((S-B)B)<0$</span>. On the other hand, <span class="math-container">$-\lambda_\max(S)\operatorname{tr}(B)$</span> is nonnegative. Hence <span class="math-container">$\operatorname{tr}((S-B)B)<-\lambda_\max(S)\operatorname{tr}(B)$</span>. (Note that strict inequality holds.)</p>
<p>If <span class="math-container">$\lambda_\max(S)>0$</span>, the inequality <span class="math-container">$\operatorname{tr}((S-B)B)\le-\lambda_\max(S)\operatorname{tr}(B)$</span> doesn't hold, as shown in the counterexample where <span class="math-container">$S=B$</span>.</p>
|
1,560,050 | <p>I want to solve the homogenous part of a stretched string problem where $y=y(x)$.</p>
<p>$$y'' + y = 0$$</p>
<p>with the boundary conditions such that: $y(0)=y(\pi/2)=0$</p>
<p>The differential equation gives rise to a solution on the form:
$$y = a \cos(x) + b \sin(x)$$</p>
<p>But when applying the boundary conditions I end up with only trivial solution ($a=b=0$).</p>
<p>Have I made a mistake or does these B.C only lead to $a=b=0$?</p>
| Simon S | 21,495 | <p>Usually this problem in physics--with such boundary conditions--has ODE $$y'' + k^2y = 0$$ The problem then turns into finding values of $k$ for which the boundary conditions are met. In this case we find infinitely many discrete $k$, corresponding to the fundamental and the harmonics above:</p>
<p>$$y_k(x) = A_k\sin(kx), \quad k = 2m, \text{ where } m \text{ is a positive integer }$$</p>
<p>If the string is constrained to move according to $$y'' + y = 0$$ alone, then the problem only admits the trivial solution. In the language of physics, $k = 1$ is not a mode of the system.</p>
|
3,872,033 | <p>Currently I meet with the following interesting problem.</p>
<p>Let <span class="math-container">$x_1,\cdots,x_n$</span> be i.i.d standard Gaussian variables. How to calculate the probability distribution of the sum of their absoulte value, i.e., how to calculate
<span class="math-container">$$\mathbb{P}(|x_1|+\cdots+|x_n|\leq nt).$$</span>
Here I use <span class="math-container">$nt$</span> instead of <span class="math-container">$t$</span> for sake of possible concise formula.</p>
<p>I cannot find out the exact value. However, a practical lower bound is also good. Here practical means the ratio between the exact value and lower bound is independent of <span class="math-container">$n$</span> and <span class="math-container">$t$</span>, and is small.</p>
<p>Thanks very much!</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\mathbb{P}\pars{\verts{x_{1}} + \cdots + \verts{x_{n}} \leq nt}}
\\[5mm] \equiv &\
\int_{-\infty}^{\infty}
{\exp\pars{-x_{1}^{2}/2} \over \root{2\pi}}\cdots
\int_{-\infty}^{\infty}
{\exp\pars{-x_{n}^{2}/2} \over \root{2\pi}}\ \times
\\[2mm] &\
\underbrace{\bracks{%
nt - \verts{x_{1}} - \cdots - \verts{x_{n}} > 0}}
_{\mbox{Heaviside Theta/Step Function}}\
\dd x_{1}\ldots\dd x_{n}
\\[5mm] = &\
{1 \over \pars{2\pi}^{n/2}}\int_{-\infty}^{\infty}
\exp\pars{-x_{1}^{2}/2}\cdots
\int_{-\infty}^{\infty}
\exp\pars{-x_{n}^{2}/2}\ \times
\\[2mm] &\
\underbrace{\bracks{%
\int_{-\infty}^{\infty}{%
\expo{\ic k\pars{nt - \verts{x_{1}} - \cdots - \verts{x_{n}}}} \over k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}}}
_{\substack{\mbox{Heaviside Theta/Step Function}
\\ \mbox{Integral Representation}}}
\dd x_{1}\ldots\dd x_{n}
\\[5mm] = &\
{1 \over \pars{2\pi}^{n/2}}
\int_{-\infty}^{\infty}{\expo{\ic k n t} \over k - \ic 0^{+}}\ \times
\\[2mm] &\
\pars{\int_{-\infty}^{\infty}\exp\pars{-x^{2}/2}
\expo{-\ic k\verts{x}}\dd x}^{n}{\dd k \over 2\pi\ic}
\\[5mm] = &\
\int_{-\infty}^{\infty}{\expo{\ic k n t} \over k - \ic 0^{+}}\ \times
\\[2mm] &\
\braces{\expo{-k^{2}/2}\bracks{1 - \ic\on{erfi}\pars{k \over \root{2}}}}^{n}{\dd k \over 2\pi\ic}
\\[5mm] = &\
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic k n t} \over k}\,\expo{-nk^{2}/2}\ \times
\\[2mm] &\
\bracks{1 - \ic\on{erfi}\pars{k \over \root{2}}}^{n}{\dd k \over 2\pi\ic} + {1 \over 2}
\\[1cm] = &\
{1 \over 2} +
{1 \over \pi}\int_{0}^{\infty}
\Im\left\{\expo{-nk^{2}\,/\,2\ +\ \ic k n t}
\right.
\\[2mm] &\ \phantom{{1 \over 2} +
{1 \over \pi}}
\left.\bracks{1 - \ic\on{erfi}\pars{k \over \root{2}}}^{n}\right\}{\dd k \over k}
\end{align}</span></p>
<p><a href="https://i.stack.imgur.com/zJDFE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zJDFE.png" alt="enter image description here" /></a>
<span class="math-container">$\qquad\qquad\qquad\qquad\qquad t$</span></p>
|
134,444 | <p>I have the following code that determines when the second business day of each month is (given a start and end date). I have a few If statements I would like to replace with functional programming.</p>
<pre><code>getAccrualDates[fromDate_List,toDate_List]:=
(
today = fromDate;
projectionDate =toDate;
(*If the projected date is before a payroll, we don't want to grab any business days from that month. This will prevent the case where we try an grab the second business day of the month (i.e. payroll day), but it not existing*)
If[dateBeforePayrollQ[projectionDate],projectionDate=DatePlus[{projectionDate[[1]],projectionDate[[2]],1},{-1,"Day"}],projectionDate];
(*Determine if today is before the payroll date of the current month. If it is, include all business days of the current month. Otherwise, create a list of business days starting with next month*)
If[dateBeforePayrollQ[fromDate],allBusinessDays = Take[DateList[#],3]&/@DayRange[DateObject[firstDayOfThisMonth],DateObject[projectionDate],"BusinessDay"];,allBusinessDays = Take[DateList[#],3]&/@DayRange[DateObject[firstDayOfNextMonth],DateObject[projectionDate],"BusinessDay"];];
(*Partition data by month*)
groupedBusinessDays = Flatten[GatherBy[Flatten[GatherBy[allBusinessDays,First],1],#[[2]]&],0];
(*Find all first payrolls between today and projectionDate*)
payrollDays = #[[2]]&/@groupedBusinessDays
)
</code></pre>
<p>Helper function below:</p>
<pre><code>(*Create a helper function that returns true if the given date is before payroll, false otherwise*)
dateBeforePayrollQ[theDate_List]:=
(
firstDayOfThisMonth = {theDate[[1]],theDate[[2]],1};
firstDayOfNextMonth = DatePlus[firstDayOfThisMonth,{1,"Month"}];
businessDaysOfThisMonth = Take[DateList[#],3]&/@DayRange[DateObject[firstDayOfThisMonth],DateObject[DatePlus[firstDayOfNextMonth,{-1,"Day"}]],"BusinessDay"];
If[businessDaysOfThisMonth[[2]][[3]]<= theDate[[3]],False,True]
)
</code></pre>
<p>Does anyone have any suggestions/better ways to accomplish this?</p>
<pre><code>getAccrualDates[{2016,9,1},{2017,1,4}]
(*{{2016,9,2},{2016,10,4},{2016,11,2},{2016,12,2},{2017,1,4}}*)
</code></pre>
| Edmund | 19,542 | <p>You may use the <a href="http://reference.wolfram.com/language/guide/DateAndTime.html" rel="nofollow noreferrer">Date & Time</a> guide functions to greatly simplify your code.</p>
<pre><code>ClearAll[getAccrualDates];
getAccrualDates[start_DateObject, end_DateObject, frequency_, location_] :=
DatePlus[#, location] & /@
DateRange[DatePlus[start, {-1, frequency}], end, frequency] //
Select[Between[{start, end}]]
</code></pre>
<p>Then</p>
<pre><code>getAccrualDates[
DateObject@{2016, 9, 1}, DateObject@{2017, 1, 4}, "EndOfMonth", {2, "BusinessDay"}]
</code></pre>
<blockquote>
<pre><code>{{2016, 9, 2}, {2016, 10, 4}, {2016, 11, 2}, {2016, 12, 2}, {2017, 1, 4}}
</code></pre>
</blockquote>
<p><a href="http://reference.wolfram.com/language/ref/DateRange.html" rel="nofollow noreferrer"><code>DateRange</code></a> gives the range of dates in <code>frequency</code> increments. <code>start</code> is adjusted backwards one unit of <code>frequency</code> to ensure no dates are missed. <a href="http://reference.wolfram.com/language/ref/DatePlus.html" rel="nofollow noreferrer"><code>DatePlus</code></a> is mapped over the <code>frequency</code> dates to adjust them to the required <code>location</code>. Finally, the dates that are <a href="http://reference.wolfram.com/language/ref/Between.html" rel="nofollow noreferrer"><code>Between</code></a> <code>start</code> and <code>end</code> are <a href="http://reference.wolfram.com/language/ref/Select.html" rel="nofollow noreferrer"><code>Select</code></a>ed. <a href="http://reference.wolfram.com/language/ref/DateObject.html" rel="nofollow noreferrer"><code>DateObject</code></a> is required for <code>start</code> and <code>end</code> in order for <code>Between</code> to know what to do with them.</p>
<p>Try others like the second Tuesday of each month.</p>
<pre><code>getAccrualDates[
DateObject@{2016, 9, 1}, DateObject@{2017, 1, 4}, "EndOfMonth", {2, Tuesday}]
</code></pre>
<blockquote>
<pre><code>{{2016, 9, 13}, {2016, 10, 11}, {2016, 11, 8}, {2016, 12, 13}}
</code></pre>
</blockquote>
<p><code>DateObjects</code> actually returned but you can easily convert with <a href="http://reference.wolfram.com/language/ref/DateList.html" rel="nofollow noreferrer"><code>DateList</code></a>.</p>
<p>Hope this helps.</p>
|
2,296,544 | <p>Let $\{F_n\}, n\in \mathbb{N}$ be the sequence of Fibonacci numbers such that $F_1=1$, $F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ $\forall n\geq2$.</p>
<p>Define a new sequence $\{S_n\}$ such that $S_n=F_n+1$ $\forall n\in \mathbb{N}$.</p>
<p>Now the question is: For every prime $p$, does there exist an $N\in \mathbb{N}$, such that $p|S_N$ ?</p>
| Robert Z | 299,698 | <p>Hint. The answer is yes. Show that for any prime $p\not=5$,
$$p\;\mbox{divides}\;S_{p^2-3}=F_{p^2-3}+1.$$
See for example Jack D'Aurizio's answer here:<a href="https://math.stackexchange.com/questions/1985541/fibonacci-sequence-problem-prove-that-there-are-infinitely-many-prime-numbers-s/1985568#1985568">Fibonacci Sequence problem. Prove that there are infinitely many prime numbers such that $p$ divides $F_{p-1}$</a></p>
|
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