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 |
|---|---|---|---|---|
160,818 | <p>Could someone help me with an simple example of a profinite group that is not the p-adics integers or a finite group? It's my first course on groups and the examples that I've found of profinite groups are very complex and to understand them requires advanced theory on groups, rings, field and Galois Theory. Know a simple example?</p>
<p>Last, how to prove that that $\mathbb{Z}$ not is a profinite group?</p>
| Pete L. Clark | 299 | <p>Yes. Let $G = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ be the direct product of (countably) infinitely many copies of the cyclic group of order $2$. This profinite group, sometimes called (well, by me at least) the <em>Bernoulli group</em>, occurs naturally in probability theory. As a topological space it is homeomorphic to the standard Cantor set.</p>
<p>(For that matter, any Cartesian product of finite groups is a profinite group, and this is an important example, because any profinite group is a closed subgroup of such a product.)</p>
<p>As for your second question, a profinite group is in particular a compact (Hausdorff) topological group and thus carries a Haar measure, i.e., a translation-invariant probability measure. Thus it cannot be countably infinite, and in particular $\mathbb{Z}$ is not (or more precisely, cannot be endowed with the structure of) a profinite group.</p>
|
4,056,073 | <p>I need help with this task, if anyone had a similar problem it would help me !</p>
<p>The task is: Determine the type of interruption at the point x = 0 for the function</p>
<p><span class="math-container">$$f(x)=2^{-\frac{1}{x^{2}}}$$</span></p>
<p>I did:</p>
<p><span class="math-container">$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$L=R=x=0$$</span></p>
<p>And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?</p>
<p>Thanks in advance !</p>
| jjagmath | 571,433 | <p>What is happening here is just a consequence that an infinite set and a proper subset can be in bijective correspondence. That's an well known fact about infinite sets. And it is a paradox in the sense that it is anti-intuitive, but not in the sense that it leads to a contradiction.</p>
|
4,056,073 | <p>I need help with this task, if anyone had a similar problem it would help me !</p>
<p>The task is: Determine the type of interruption at the point x = 0 for the function</p>
<p><span class="math-container">$$f(x)=2^{-\frac{1}{x^{2}}}$$</span></p>
<p>I did:</p>
<p><span class="math-container">$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$L=R=x=0$$</span></p>
<p>And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?</p>
<p>Thanks in advance !</p>
| Paul Frost | 349,785 | <p>The core of your paradox seems to be that you claim that in some sense there are "more" (ordered) tuples <span class="math-container">$(c_{n-1},\dots,c_0)$</span> than (unordered) sets <span class="math-container">$\{r_{n-1},\dots,r_0\}$</span>. This is not true for infinite sets like <span class="math-container">$\mathbb C$</span>, only for finite sets, and even then it would not be significant for your problem as will be seen below.</p>
<p>In the sequel we only consider polynomials having leading coefficient <span class="math-container">$1$</span> to avoid a "trival non-uniqueness".</p>
<ol>
<li><p>It is in general not true that the set <span class="math-container">$R(p)$</span> of roots of a polynomial <span class="math-container">$p = p(x)$</span> allows to reconstruct <span class="math-container">$p$</span>. Note that <span class="math-container">$R(p)$</span> is nothing else than a <strong>set</strong> and does not contain information about the multiplicity of the roots. The number <span class="math-container">$m(p)$</span> of elements of <span class="math-container">$R(p)$</span> may be anything between <span class="math-container">$1$</span> and <span class="math-container">$n = \deg(p)$</span>, i.e. we have <span class="math-container">$1 \le m(p) \le n$</span>. [Note that for <span class="math-container">$\deg(p) = 0$</span> this says that the set <span class="math-container">$R(p)$</span> is empty.] The notation <span class="math-container">$R(p) = \{r_{n-1},\dots,r_0\}$</span> is misleading since it suggests that the set <span class="math-container">$R(p)$</span> has <span class="math-container">$n$</span>-elements. Instead we have <span class="math-container">$R(p) = \{ \rho_1, \dots, \rho_{m(p)}\}$</span> with <span class="math-container">$\rho_i \ne \rho_j$</span> for <span class="math-container">$i \ne j$</span>.<br />
Therefore the problem is this: To each ordered tuple <span class="math-container">$\tau = (c_{n-1},\dots,c_0)$</span> of finite length we can associate a unique polynomial <span class="math-container">$p_\tau$</span> having the <span class="math-container">$c_i$</span> as coefficients, but to a finite set <span class="math-container">$R$</span> we can associate infinitely many polynomials having <span class="math-container">$R$</span> as a root set. There is exactly one such polynomial with minimal degree (in which each <span class="math-container">$\rho \in R$</span> occurs with multiplicity <span class="math-container">$1$</span>, i.e.the degree equals the number of elements of <span class="math-container">$R$</span>), but if we associate it to <span class="math-container">$R$</span>, we do not get all polynomials having <span class="math-container">$R(p) = R$</span>. The only exception is <span class="math-container">$R =\emptyset$</span>; in that case <span class="math-container">$p(x) = 1$</span>.</p>
</li>
<li><p>What we really need to consider is the set of pairs <span class="math-container">$\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$</span> where <span class="math-container">$k_i \ge 1$</span> is the multiplicity of the root <span class="math-container">$\rho_i$</span>. Then we get a unique polynomial <span class="math-container">$p$</span> of degree <span class="math-container">$n = \sum k_i$</span> such that <span class="math-container">$R(p) = \{ \rho_1, \dots, \rho_{m(p)}\}$</span>.<br />
Alternatively we can consider <span class="math-container">$\mathbb C^n/\mathfrak S_n$</span>, the quotient of <span class="math-container">$\mathbb C^n$</span> by the operation of the symmetric group of <span class="math-container">$n$</span> elements. That is, tupels are identified if they are related by permuting their components. One might be tempted to think of the elements of <span class="math-container">$\mathbb C^n/\mathfrak S_n$</span> as simple (unordered) sets, but it is not true. In fact, <span class="math-container">$\mathbb C^n/\mathfrak S_n$</span> can be identified with the collection of pairs <span class="math-container">$\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$</span> with <span class="math-container">$k_i \ge 1$</span> and <span class="math-container">$n = \sum k_i$</span>.</p>
</li>
</ol>
<p>See also <a href="https://math.stackexchange.com/q/3907735">How to show symmetric product of complex space is homeomorphic to the complex space</a> .</p>
|
698,743 | <blockquote>
<p>Let the real coefficient polynomials
$$f(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$$
$$g(x)=b_{m}x^m+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}$$
where $a_{n}b_{m}\neq 0,n\ge 1,m\ge 1$, and let
$$g_{t}(x)=b_{m}x^m+(b_{m-1}+t)x^{m-1}+\cdots+(b_{1}+t^{m-1})x+(b_{0}+t^m).$$
Show that</p>
<p><strong>there exist positive $\delta$, such for any $t$ such that $0<|t|<\delta$, and such
$f(x)$ and $g_{t}(x)$ be relatively prime.</strong></p>
</blockquote>
<p>I fell this result is very well, because although two polynomial are not relatively prime, we can do it to one of the polynomial tiny perturbation makes relatively prime.</p>
<p>But I can't prove my problem.</p>
<p>Thank you very much</p>
| Calvin Lin | 54,563 | <p>Here is a series of steps, which seem mostly true to me.</p>
<p><strong>Fact:</strong> $f(x)$ has at most $n$ distinct roots.</p>
<p><strong>Claim:</strong> There exists a map $G: [0,1] \rightarrow (\alpha_1, \alpha_2, \ldots, \alpha_m)$ which is differentiable in each coordinate, and $ \alpha_i$ are roots of $ g_t(x)$ (with multiplicity).</p>
<p>Possible Proof: Use Inverse/Implicit function theorem till 2 roots meet. Then be very careful?</p>
<p><strong>Claim:</strong> On any open interval, $\frac{dG}{d\alpha_i}$ is not identically 0.</p>
<p>Proof: If it does, then there is a number $ \alpha$ such that $ g_t(\alpha) = 0 $ on that open interval. But this implies that $ \sum t^i \alpha^{m-i} = 0 $ infinitely often, which is a contradiction (since there is at most $m$ values of $t$ that can satisfy the polynomial).</p>
<p>Hence, the result follows, where we take a small enough $t$, such that the roots of $g_t$ all avoid the roots of $f$. The final claim allows us to move off of common roots of $f$ and $g_0$ .</p>
|
48,629 | <p>Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$.</p>
<p>On my desk you could find the following books: <strong>Algebraic Geometry</strong> by <em>Hartshorne</em>, <strong>Ideals, Varieties, and Algorithms</strong> by <em>Cox & Little & O'Shea</em>, <strong>Algorithms in Real Algebraic Geometry</strong> by <em>Basu & Pollack & Roy</em> and <strong>A SINGULAR Introduction to Commutative Algebra</strong> by <em>Greuel & Pfister</em>. Unfortunately, neither of them introduced notions and ideas I'm looking for.</p>
<p>If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: <em>is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space?</em></p>
<p>Further questions I have are, for example:</p>
<ol>
<li>Can I define <em>geodesics</em> (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities?</li>
<li>How about <em>curvature</em>? Is it defined for these objects?</li>
<li>Can we talk about <em>convexity</em> of subsets of the algebraic surface?</li>
<li>What other tools and term can be imported from differential/Riemannian geometry?</li>
</ol>
<p>I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help.</p>
| Daniel Loughran | 5,101 | <p>Any non-singular complex variety $V$ of dimension $n$ (in either affine space or projective space) can be endowed with the structure of a complex manifold of dimension $n$. Moreover as a submanifold of a Kahler manifold, it will also be Kahler.</p>
<p>The passage to a real manifold can be slightly subtle, as there are two approaches you can take. Firstly, you can simply consider the complex manifold as a real manifold of dimension $2n$, with a complex structure.</p>
<p>Alternatively, if the variety is defined over $\mathbb{R}$ (that is, it has equations with real coefficients), then you can look at the set of real points $V(\mathbb{R})$, which you can also endow with the struture of a real manifold. As a submanifold of a Riemannian manifold it is also a Riemannian manifold.</p>
<p>I should note that some funny things can happen though, for example even if $V$ is connected then $V(\mathbb{R})$ may not be.</p>
<p>If your variety has singularities, these always occur on a closed subset so you can just remove them to get a non-singular variety. Alternatively you can resolve them as Ariyan suggests.</p>
|
88,199 | <p>Is there a function that would satisfy the following conditions?:</p>
<p>$\forall x \in X, x = f(f(x))$ and $x \not= f(x)$,</p>
<p>where the set $X$ is the set of all triplets $(x_1,x_2,x_3)$ with $x_i \in \{0,1,\ldots,255\}$.</p>
<p>I would like to find a function that will have as an input RGB color values (triplets) and return the original color after two applications of the function.</p>
| Dilip Sarwate | 15,941 | <p>If you are doing these calculations on a computer and can get away from
insisting on interpreting $x_1$, $x_2$, $x_3$ as integers in the range
$[0,255]$, consider thinking of $x_1$, $x_2$, $x_3$ as eight-bit <em>bytes</em>
or vectors of length $8$ over $\mathbb F_2$ to be a bit more formal about it.
Then, for any three bytes $a$, $b$, $c$ with at least one being nonzero,
$$f(x_1, x_2, x_3) = (x_1\oplus a, x_2\oplus b, x_3\oplus c)$$
has the desired properties that
$$\begin{align*}
f(f(x_1, x_2, x_3)) &= f(x_1\oplus a, x_2\oplus b, x_3\oplus c)\\
&= (x_1\oplus a\oplus a, x_2\oplus b\oplus b, x_3\oplus c \oplus c)\\
&= (x_1, x_2, x_3),
\end{align*}$$
and $(x_1, x_2, x_3) \neq f(x_1, x_2, x_3)$.
As an added bonus (not necessarily important to
math.SE readers), the XOR operation on bytes is a
machine-language instruction on most computers
and in some cases can be faster than the ADD instruction
which is often defined for full (multiple-byte) words only. </p>
<p>Note that $f(x_1,x_2,x_3)=(255−x_1,255−x_2,255−x_3)$ as suggested
by @ofer is just
$f(x_1,x_2,x_3)
= (x_1 \oplus \mathbf 1, x_2\oplus \mathbf 1, x_3\oplus \mathbf 1)$
where $\mathbf 1 = (1,1,1,1,1,1,1,1)$ is the eight-bit all-ones byte.</p>
|
1,643,201 | <p>The spectrum-functor
$$
\operatorname{Spec}: \mathbf{cRng}^{op}\to \mathbf{Set}
$$
sends a (commutative unital) ring $R$ to the set $\operatorname{Spec}(R)=\{\mathfrak{p}\mid \mathfrak{p} \mbox{ is a prime ideal of R}\}$ and a morpshim $f:S\to R$ to the map $\operatorname{Spec}(R)\to \operatorname{Spec}(S)$ with $\mathfrak{p}\mapsto f^{-1}(\mathfrak{p})$. Does this functor send pullback squares
\begin{eqnarray}
S\times_R T&\to& T\\
\downarrow && \downarrow\\
S&\to& R
\end{eqnarray}
of (commutative unital) rings to pushout squares
\begin{eqnarray}
\operatorname{Spec}(R)&\to& \operatorname{Spec}(T)\\
\downarrow && \downarrow\\
\operatorname{Spec}(S)&\to& \operatorname{Spec}(S\times_R T)
\end{eqnarray}
of sets? Put in other words, does the functor $\operatorname{Spec}$ from above preserve pushouts?</p>
| Travis Willse | 155,629 | <p>Yes, except that:</p>
<ol>
<li>I would replace $\frac{100}{7}$ with $\frac{99}{7}$ (to see why this is important consider the analogous question asking for how many multiples of $7$ there are in $\{700, \ldots, 999\}$), and</li>
<li>the quotients aren't quite right as written, but we can repair them with floor notation, e.g., $\lfloor\frac{999}{7}\rfloor = 142$.</li>
</ol>
|
1,386,682 | <p>How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?</p>
<p>I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>And that I could not get out, can anyone help me?</p>
| Empty | 174,970 | <p><strong>From your calculation :</strong></p>
<p>$$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>$$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$</p>
<p>From here, show that both the limits are <strong>zero</strong> by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$.</p>
<p><strong>For the first limit,</strong></p>
<p>$$\left|\frac{r^3\cos^3\theta-3r^3\cos \theta\sin^2\theta}{r^2}\right|\le 4r<\epsilon$$whenever, $r^2<\epsilon^2/16$ i,e, whenever $|x|<\epsilon/\sqrt 8=\delta(say)$ , and $|y|<\epsilon/\sqrt 8=\delta(say)$.</p>
<p>Similarly the second limit will be zero and hence the given limit will be $0$.</p>
|
2,113,777 | <p>I have the following IVP (Initial value problem, Cauchy-Problem), and I do not know how to solve this.</p>
<p>$$y'=e^{-x}-\frac{y}{x} \qquad \qquad y(1)=2$$</p>
<p>I hope you can help me, cause I really do not know how to start.</p>
<p>Thank you! :)</p>
| Mercy King | 23,304 | <p>First, one need to solve the homogeneous differential equation associated to the original equation, that is:
$$
y'=-\dfrac{y}{x}
$$
We have
\begin{eqnarray}
\dfrac{y'}{y}&=&-\dfrac1x\\
\int\dfrac{dy}{y}&=&-\int\dfrac{dx}{x}\\
\ln|y|&=&\ln|C|-\ln|x|\\
y&=&\dfrac{A}{x}.
\end{eqnarray}
Now, we use the variation of constants:
$$
y'=\dfrac{A'}{x}-\dfrac{A}{x^2}=e^{-x}-\dfrac{A}{x^2},
$$
therefore
$$
A'=xe^{-x},
$$
and using integration by parts we get
$$
A=\int xe^{-x}\,dx=-xe^{-x}+\int e^{-x}\,dx=-xe^{-x}-e^{-x}+B=-(x+1)e^{-x}+B
$$
We deduce that
$$
y=-\left(1+\dfrac{1}{x}\right)e^{-x}+\dfrac{B}{x}.
$$
using the initial condition $y(1)=2$, we deduce that
$$
B=2+2e^{-1},
$$
thus
$$
y=-\left(1+\dfrac{1}{x}\right)e^{-x}+\dfrac{2+2e^{-1}}{x}.
$$</p>
|
3,916,092 | <blockquote>
<p>A ball rotates at a rate <strong><span class="math-container">$r$</span></strong> rotations per second and simultaneously revolves around a stationary point <strong><span class="math-container">$O$</span></strong> at a rate <strong><span class="math-container">$R$</span></strong> revolutions per second <strong><span class="math-container">$(R<r)$</span></strong>. The rotation and revolution are in the same sense. A certain point on the ball is in the line of the centre of the ball and point <strong><span class="math-container">$O$</span></strong> at a certain time. This configuration repeats after a time</p>
</blockquote>
<blockquote>
<p><span class="math-container">$(1)\ $</span> <span class="math-container">$\frac{1}{r-R}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(2)\ $</span> <span class="math-container">$\frac{1}{R}-\frac{1}{r}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(3)\ $</span> <span class="math-container">$\frac{1}{r+R}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(4)\ $</span> <span class="math-container">$\frac{1}{R}+\frac{1}{r}$</span></p>
</blockquote>
<p>How do I solve this aptitude question? Thanks for your time.</p>
<p><strong>Source</strong> <span class="math-container">$:$</span> CSIR NET JUNE <span class="math-container">$2019.$</span></p>
| Prakasan S.P | 1,142,365 | <p>Given:
Frequency of rotation = 'r' times/sec
Frequency of revolution = 'R' times/sec.<br />
(the rotation and the revolution are in the same sense), hence
Relative frequency of rotation = (r-R) times/sec.<br />
As per this relative frequency of rotation, the centre of the ball and the point of its surface come in line with the external fixed point O, at each full rotation.</p>
<p>Time taken for one full rotation = 1/f = 1/(r-R) in sec</p>
|
3,357,841 | <p>In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.</p>
<p><a href="https://i.stack.imgur.com/DK8sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DK8sn.png" alt="enter image description here"></a></p>
| Quanto | 686,284 | <p><a href="https://i.stack.imgur.com/RZvDU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RZvDU.png" alt="enter image description here"></a></p>
<p>It can be deduced that the area <span class="math-container">$[FED] = 7\cdot 4/14=2$</span>. </p>
<p>Furthermore, examine the ratios below,</p>
<p><span class="math-container">$$\frac{[FDC]}{[BDC]}=\frac{9}{18}=\frac{\frac{a}{a+b} \frac{d}{c+d}[ABC] }{ \frac{b}{a+b}[ABC]} =\frac ab \frac{1}{\frac cd +1} \tag{1}$$</span></p>
<p><span class="math-container">$$\frac{[FDB]}{[FCB]}=\frac{6}{21}=\frac{\frac{c}{c+d} \frac{b}{a+b} [ABC] }{ \frac{d}{c+d} [ABC]} =\frac cd \frac{1}{\frac ab +1} \tag{2}$$</span></p>
<p>From (1) and (2), </p>
<p><span class="math-container">$$ \frac cd = \frac 12$$</span></p>
<p>Then, the area of the quadrilateral is</p>
<p><span class="math-container">$$[AFED]= 2+ [AFD] = 2+\frac 12\cdot 9 = \frac {13}{2}$$</span></p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| NaOH | 398,864 | <p>There is still some hope in taking the first minus the second in this case:
$$6-2\sqrt{3} - (3\sqrt{2}-2) = 8 - (2\sqrt{3} + 3\sqrt{2})$$</p>
<p>So now the question boils down to if the expression with the square root exceeds $8$. We know that $8^{2} = 64$ and:
$$(2\sqrt{3}+3\sqrt{2})^{2}=4*3+2(2\sqrt{3})(3\sqrt{2})+9*2 =30+12\sqrt{3}\sqrt{2}$$</p>
<p>Conversely:
$$8^{2} = 64 = 28+36=28+6*6 = 28+6*\sqrt{6}\sqrt{6}$$</p>
<p>Subtracting them yields:
$$ 28+6*\sqrt{6}\sqrt{6}-30+12\sqrt{3}\sqrt{2} = -2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2}$$</p>
<p>Thus this is only positive if the square root terms are positive and exceed 2. Comparing the square root terms:
$$6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = 6*\sqrt{2}\sqrt{3}(\frac{\sqrt{2}}{\sqrt{2}})\sqrt{6}-12\sqrt{6}=12\frac{\sqrt{3}}{\sqrt{2}}\sqrt{6}-12\sqrt{6}=12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1)$$</p>
<p>We conclude that the square root term is positive since $\sqrt{3}>\sqrt{2}$ . But is it greater than $-2$? Or rather, how much do we need to multiply to the expression $\frac{\sqrt{3}}{\sqrt{2}}-1$ for it to be greater than $2$?
$$n*(\frac{\sqrt{3}}{\sqrt{2}}-1)>2$$
$$n>\frac{2}{\frac{\sqrt{3}}{\sqrt{2}}-1}*\frac{\frac{\sqrt{3}}{\sqrt{2}}+1}{\frac{\sqrt{3}}{\sqrt{2}}+1}\approx\frac{2*(1.5+1)}{0.5} = 10$$
So the factor in front of the term $\frac{\sqrt{3}}{\sqrt{2}}-1$ should be at least 10. But since $12>10$ and $\sqrt{6}>1$, we conclude that $12\sqrt{6}>10$. Thus, substituting back to the original equation:
$$-2+6*\sqrt{6}\sqrt{6}-12\sqrt{3}\sqrt{2} = -2+12\sqrt{6}(\frac{\sqrt{3}}{\sqrt{2}}-1) > 0$$
And so we conclude that:
$$64>(2\sqrt{3}+3\sqrt{2})^{2}$$
and so, for positive root,
$$8>(2\sqrt{3}+3\sqrt{2})$$
and
$$ 8 - (2\sqrt{3} + 3\sqrt{2}) > 0 $$</p>
<p>You may use other approaches. Notice that I split $8^{2} = 28+36$. Equally viable is to split $8^{2} = 30+34$, and you might have to use a similar (Squaring than rooting) trick to show that $34 > 12\sqrt{6}$, and then finally conclude that $8>(2\sqrt{3} + 3\sqrt{2})$. This alternative approach should be able to give you a smaller threshold compared to the approximation I made midway.</p>
|
192,125 | <p>Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<</p>
| Amzoti | 38,771 | <p>Questions:</p>
<ol>
<li><p>Is the problem written correctly.</p></li>
<li><p>Are there restrictions on x?</p></li>
</ol>
<p>Something does not seem right in the problem as posed.</p>
<p>Hint: Plot the left hand side and then plot the right hand side and see what it looks like.</p>
|
583,030 | <p>I have to show that the following series convergences:</p>
<p>$$\sum_{n=0}^{\infty}(-1)^n \frac{2+(-1)^n}{n+1}$$</p>
<p>I have tried the following:</p>
<ul>
<li>The alternating series test cannot be applied, since $\frac{2+(-1)^n}{n+1}$ is not monotonically decreasing.</li>
<li>I tried splitting up the series in to series $\sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n \frac{2}{n+1}$ and $\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}(-1)^n \frac{(-1)^n}{n+1}$. I proofed the convergence of the first series using the alternating series test, but then i realized that the second series is divergent.</li>
<li>I also tried using the ratio test: for even $n$ the sequence converges to $\frac{1}{3}$, but for odd $n$ the sequence converges to $3$. Therefore the ratio is also not successful.</li>
</ul>
<p>I ran out of ideas to show the convergence of the series.</p>
<p>Thanks in advance for any help!</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}%
\newcommand{\yy}{\Longleftrightarrow}$
\begin{align}
&\sum_{n=0}^{\infty}\pars{-1}^{n}\,{2+ \pars{-1}^{n} \over n + 1}
=
\sum_{n=0}^{\infty}\bracks{{3 \over 2n + 1} - {1 \over 2n + 2}}
=
\sum_{n=0}^{\infty}{4n + 5 \over \pars{2n + 1}\pars{2n + 2}}
\end{align}
The 'serie general term' is $\quad\sim 1/n\quad$ when $\quad n \gg 1.\quad$The series $\color{#ff0000}{\large diverges}$ as the harmonic one.</p>
|
111,183 | <p><img src="https://i.stack.imgur.com/1MOuo.jpg" alt="Problem">
<img src="https://i.stack.imgur.com/bdRXi.png" alt="New Solution"></p>
<p>I believe I have gotten all of the ways now - thanks for the hints below Yun, Andre Nicolas, and Gerry Myerson. If anyone could confirm my answer (I feel there should be more possibilities, but for the numbers to be increasing left-right, I have found them all I believe). Sorry I don't have the headings on this chart, but from left to right it is: Numbers 1-6 (one for each column) with the last column being "Sum," if it was not already obvious. </p>
| Lin | 348,610 | <p>2 + 4 + 4 + 4 + 4 + 8 also equals 26. I didn't see that on the list.</p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| Christian Blatter | 1,303 | <p>The function
$$f(x):=\cos(\sin x)-\sin(\cos x)$$
is even and $2\pi$-periodic; therefore it suffices to consider $x\in[0,\pi]$. When $x=0$ or $x\in\bigl[{\pi\over2},\pi\bigr]$ then obviously $f(x)>0$. Finally, when $0<x<{\pi\over2}$ then $\cos x$ and $\sin x$ both lie in the interval $\ ]0,1[\ \subset\ ]0,{\pi\over2}[\ $. Therefore we also have
$$\sin(\cos x)<\cos x<\cos(\sin x)\qquad\bigl(0<x<{\pi\over2}\bigr).$$</p>
|
670,781 | <p>Given $y=x\sqrt{a+bx^2}$. the tangent to $y$ at point $x=\sqrt5$ is also passing at point </p>
<p>$(3\sqrt5,\sqrt5).$ the area between $y=x\sqrt{a+bx^2}$ and the $x$-axis is equal to $18$.</p>
<p>Need to find $a,b$.</p>
<p>I have tried to differentiate and eliminate but failed...</p>
| 5xum | 112,884 | <p>Hint: For a curve $y=f(x)$, the slope of the tangent at $x_0$ is given by $f'(x_0)$. This means that you have a tangent with the formula $y=f'(x_0)\cdot x + n$ for some real $n$. Plugging in the value $x=x_0$ and $y=f(x_0)$ can then help you calculate $n$ in terms of $a$ and $b$, giving you one of the equations for $a$ and $b$.</p>
<p>For the second part, it's clear that if $b>0$, you need some limit where $x$ may be, otherwise the limit of $a+bx^2$ is $\infty$.</p>
|
3,077,629 | <p>Assume <span class="math-container">$(f_i)_{i\in I}$</span> is an orthonormal/orthogonal system in an (complex) inner product space. Does <span class="math-container">$$\sum_{i\in I}\langle f_i,f\rangle f_i$$</span> always converges for any <span class="math-container">$f$</span> (may not to <span class="math-container">$f$</span>)? Especially, when <span class="math-container">$I$</span> is uncountable.</p>
<p>And we define convergency of a sum <span class="math-container">$\sum_I x_i$</span> on an arbitrary set by: if there exists <span class="math-container">$g$</span> and for any <span class="math-container">$\epsilon>0$</span>, there exists a finite set <span class="math-container">$F$</span> and for all other finite set <span class="math-container">$H\supseteq F,|g-\sum_H x_i|<\epsilon$</span> as an analogue to the limit of series in analysis.</p>
| boojum | 882,145 | <p>The Lagrange equations are not all consistent, as we obtain
<span class="math-container">$$ \frac{-2x}{(x^2+y^2+z^2)^2} \ = \ 2x· \lambda \ \ \Rightarrow \ \ -2x \ · \left[ \ \lambda \ + \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ , $$</span>
<span class="math-container">$$ \frac{-2y}{(x^2+y^2+z^2)^2} \ = \ -2y· \lambda \ \ \Rightarrow \ \ 2y \ · \left[ \ \lambda \ - \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ , $$</span>
<span class="math-container">$$ \frac{-2z}{(x^2+y^2+z^2)^2} \ = \ -2z· \lambda \ \ \Rightarrow \ \ 2z \ · \left[ \ \lambda \ - \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ . $$</span></p>
<p>[The denominator in the derivatives given in the post is inaccurate, but does not materially affect the conclusion.]</p>
<p>The expression in brackets in the latter two equations doesn't tell us much. We can extract that <span class="math-container">$ \ x \ , \ y \ , \ \text{or} \ \ z = 0 \ \ , $</span> but we will see that this is not giving us a complete description of the solution.</p>
<p>We can arrive at a geometrical interpretation of the situation by treating the function <span class="math-container">$ \ f(x,y,z) \ = \ \frac{1}{x^2 + y^2 + z^2} \ $</span> as the "reciprocal distance-squared" for a point <span class="math-container">$ \ (x,y,z) \ $</span> measured from the origin. Expressed in "standard form" for a quadric surface, the region <span class="math-container">$ \ R \ $</span> is <span class="math-container">$ \ -x^2+y^2+z^2 \ \ge \ 16 \ $</span> with the boundary <span class="math-container">$ \ \partial R \ $</span> being <span class="math-container">$ \ -x^2+y^2+z^2 \ = \ 16 \ \ . $</span> We see then that this boundary is an "inner surface" which is a hyperboloid of one sheet with its symmetry axis being the <span class="math-container">$ \ x-$</span>axis [depicted in the graph below]. The region <span class="math-container">$ \ R \ $</span> comprises all points "outside" of this surface, extending infinitely.</p>
<p>Writing the hyperboloid equation as <span class="math-container">$ \ y^2+z^2 \ = \ 16 + x^2 \ \ , $</span> we observe that the "narrowest" cross-section occurs at <span class="math-container">$ \ x = 0 \ \ , $</span> giving us a circle in the <span class="math-container">$ \ yz-$</span>plane with a radius of <span class="math-container">$ \ 4 \ $</span> centered on the <span class="math-container">$ \ x-$</span> axis. So the critical points <span class="math-container">$ \ ( 0 \ , \ \pm4 \ , \ 0) \ $</span> and <span class="math-container">$ \ ( 0 \ , \ 0 \ , \ \pm4) \ $</span> are "degenerate" and belong to the minimal-distance "ring" <span class="math-container">$ \ y^2+z^2 \ = \ 16 \ \ , $</span> as <strong>lonza leggiera</strong> also describes. [This also indicates that the specified exclusion of the origin from <span class="math-container">$ \ R \ $</span> and <span class="math-container">$ \ \partial R \ $</span> in the problem statement is unnecessary, as the given inequality already prevents it from being included.]</p>
<p>Since no point in <span class="math-container">$ \ R \ $</span> is closer to the origin than <span class="math-container">$ \ 4 \ $</span> units, the <strong>global maximum</strong> for <span class="math-container">$ \ f(x,y,z) \ $</span> is <span class="math-container">$ \ \frac{1}{16} \ \ . $</span> There is no upper limit to distances from the origin for points in the region, so our function is bounded below by zero, but <em>does not</em> possess a global minimum.</p>
<p><a href="https://i.stack.imgur.com/OBn9u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OBn9u.png" alt="enter image description here" /></a></p>
|
2,336,535 | <p>I have a limit:</p>
<p>$$\lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^4+y^2}$$</p>
<p>I need to show that it doesn't equal 0.</p>
<p>Since the power of $x$ is 3 and 4 down it seems like that part could go to $0$ but the power of $y$ is 3 and 2 down so that seems like it's going to $\infty$.</p>
<p>I wonder if that even makes sense.</p>
<p>So how can I solve this limit, with substitution or changing it into a polar representation?</p>
| Rom | 446,025 | <p><em>Graphically</em> <strong>f</strong> looks like this : <a href="https://i.stack.imgur.com/23eVc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/23eVc.png" alt="pic1"></a>
You can see that the value limit $(x,y) \rightarrow (0,0)$ <strong>depends</strong> on the path we take. </p>
<p>$\\
\\ $</p>
<p>If we take the path $\lim_{(x,0) \rightarrow (0_{+},0)} f$, <strong>graphically</strong> it is this path : <a href="https://i.stack.imgur.com/2aIPE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2aIPE.png" alt="1st path"></a> </p>
<p>We find $+ \infty$. And if we plug this limit in <strong>f</strong> we find : $\lim_{(x,0) \rightarrow (0_{+},0)} \frac{x^3}{x^4} = \lim_{(x,0) \rightarrow (0_{+},0)} \frac{1}{x} = +\infty$</p>
<p>$\\ $ </p>
<p>And the other limit we will consider is $\lim_{(x,0) \rightarrow (0_{-},0)} f$, <strong>graphically</strong> it is this path : <a href="https://i.stack.imgur.com/vJBL0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vJBL0.png" alt="2nd path"></a> </p>
<p>We find $- \infty$. And if we plug this limit in <strong>f</strong> we find : $\lim_{(x,0) \rightarrow (0_{-},0)} \frac{x^3}{x^4} = \lim_{(x,0) \rightarrow (0_{-},0)} \frac{1}{x} = -\infty$ </p>
<p><strong>To conclude</strong>, we have : $$\lim_{(x,0) \rightarrow (0_{-},0)} f \neq \lim_{(x,0) \rightarrow (0_{+},0)} f$$ </p>
<p>So the limit $\lim_{(x,y) \rightarrow (0,0)} f$ doesn't exist !</p>
|
3,561,664 | <p>I did part of this question but am stuck and don't know how to continue</p>
<p>I let <span class="math-container">$x= 2k +1$</span></p>
<p>Also noticed that <span class="math-container">$x^3+x = x(x^2+1)$</span></p>
<p>therefore
<span class="math-container">$4m+2 = 2k+1((2k+1)^2+1)$</span></p>
<p>I simplified this and ended up with</p>
<p><span class="math-container">$4m+2 = 8k^3+12k^2+8k+2$</span></p>
<p>I don't know how to continue from and prove that <span class="math-container">$x^3+x$</span> has remainder 2 when divided by 4</p>
| fleablood | 280,126 | <p>You didn't do your math right Let <span class="math-container">$x= 2k + 1$</span> so <span class="math-container">$x(x^2 + 1) = (2k+1)((2k+1)^2 + 1)=$</span></p>
<p><span class="math-container">$(2k+1)((2k+1)^2 + 1) = 8k^3 + 12k^2 + 8k +2$</span> so you had a typo.</p>
<p>But it's easier to do</p>
<p><span class="math-container">$(2k+1)^3 + (2k + 1)=$</span></p>
<p><span class="math-container">$(2k)^3 + 3(2k)^2 + 3(2k)+ 1 + (2k + 1)=$</span></p>
<p><span class="math-container">$8k^3 + 12k^2 + 6k + 1 + 2k + 1=$</span></p>
<p><span class="math-container">$8k^3 + 12k^2 + 8k + 2$</span>.</p>
<p>So <span class="math-container">$4|8, 12, 8$</span> the <span class="math-container">$4|8k^3 + 12k^2 + 8k$</span> and the remainder of <span class="math-container">$8k^3 + 12k^2 + 8k + 2$</span> is <span class="math-container">$2$</span>.</p>
<p>....</p>
<p>Also ... this may be overly abstract: If we let <span class="math-container">$x = 2k + 1 = m + 1$</span> where <span class="math-container">$m=2k$</span> is an even number, then <span class="math-container">$x^3 +x = (m+1)^3 + (m+1) = a_3m^3 + a_2m^2 + a_1m + a_0$</span>, a polynomial. For <span class="math-container">$j\ge 2; m^j = (2k)^j = 2^j*k^j=4*2^{j-2}{k^j}$</span> we have <span class="math-container">$4|a_jm^j$</span>. </p>
<p>So the remainder when divided by <span class="math-container">$4$</span> will be the same as the remainder of <span class="math-container">$a_1m + a_0$</span>. </p>
<p>By binomial theorem: <span class="math-container">$(m+1)^3 = \sum_{i=0}^3 {3\choose i}m^i$</span> so for <span class="math-container">$(m+1)^3 + (m+ 1)$</span> we have <span class="math-container">$a_1 = {3\choose 1} + 1$</span> and <span class="math-container">$a_0 = {3\choose 0} + 1$</span>.</p>
<p><span class="math-container">${3\choose 1} = 3$</span> and <span class="math-container">$3+1=4$</span> so <span class="math-container">$4|a_jm$</span> and the remainder when divided by <span class="math-container">$4$</span> will be the same as the remainder of <span class="math-container">$a_0 = {3\choose 0} + 1$</span> when divided by <span class="math-container">$4$</span>.</p>
<p><span class="math-container">${3\choose 0} = 1$</span> and <a href="https://math.stackexchange.com/questions/278974/prove-that-11-2">1+1 = 2</a></p>
|
246,071 | <p>How do I solve the following equation?</p>
<p>$$x^2 + 10 = 15$$</p>
<p>Here's how I think this should be solved.
\begin{align*}
x^2 + 10 - 10 & = 15 - 10 \\
x^2 & = 15 - 10 \\
x^2 & = 5 \\
x & = \sqrt{5}
\end{align*}
I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.</p>
<p>I've also seen another equation like this:
\begin{align*}
x^2 & = 4 \\
x^2 + 4 & = 0 \\
(x - 2)(x + 2) & = 0 \\
x & = 2 \text{ or } -2
\end{align*}
So I guess I could near the end of my equation do the following:</p>
<p>$$x^2 + 5 = 0$$</p>
<p>and then go from there?</p>
<p>Is my first attempt at solving correct?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ \ $ If $\rm\ \ f(n) =\, \dfrac{5}6 - \dfrac{1}{n\!+\!1} - \dfrac{1}{n\!+\!2} - \,\cdots\, - \dfrac{1}{2n\!+\!1}\ $ then $$\rm f(n\!+\!1)-f(n) = \dfrac{1}{n\!+\!1}-\dfrac{1}{2n\!+\!2}-\dfrac{1}{2n\!+\!3} = \dfrac{1}{2(n\!+\!1)(2n\!+\!3)} > 0$$</p>
<p>thus $\rm\:f(n)\:$ is increasing, so $\rm\:f(n) \ge \cdots \ge f(1) \ge f(0) = \frac{5}6\!-\!\frac{1}2\!-\!\frac{1}3 = 0\:$ by induction.</p>
<p><strong>Remark</strong> $\ $ In the same way, one can often reduce inductive proofs to <em>trivial</em> inductions, such as the above induction that an increasing function stays $\ge$ its initial value. For further discussion and examples see my <a href="https://math.stackexchange.com/search?q=user%3A242+telescopy">many prior posts on <strong>telescopy</strong>.</a> There, you'll learn that, by summing, the above proof can be viewed as a trivial induction that a sum of nonnegative integers stays nonnegative. The sooner one learns how to restructure inductive proofs to make the induction "obvious", the sooner one will be able to tackle much more complicated inductions that arise in the wild.</p>
|
3,703,981 | <p>If we consider an equation <span class="math-container">$x=2x^2,$</span> we find that the values of <span class="math-container">$x$</span> that solve this equation are <span class="math-container">$0$</span> and <span class="math-container">$1/2$</span>. Now, if we differentiate this equation on both sides with respect to <span class="math-container">$x,$</span> we get <span class="math-container">$1=4x.$</span> Now, I know that it is wrong to say that the value of <span class="math-container">$x=1/4,$</span> but then, what does <span class="math-container">$x=1/4$</span> signify? Is it related to maxima or minima? Please help me with this.</p>
| Batominovski | 72,152 | <p>Alternatively, if <span class="math-container">$p$</span> is a prime natural number that divides <span class="math-container">$x^2+xy+y^2$</span>, then <span class="math-container">$$(2x+y)^2+3y^2=4(x^2+xy+y^2)\equiv 0\pmod{p}\,.$$</span> Thus, either <span class="math-container">$p$</span> divides both <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, or <span class="math-container">$\left(\dfrac{-3}{p}\right)=1$</span>. Now, by quadratic reciprocity,
<span class="math-container">$$1=\left(\dfrac{-3}{p}\right)=\left(\dfrac{p}{-3}\right)=\left(\dfrac{p}{3}\right)\,,$$</span>
whence <span class="math-container">$p\equiv 1\pmod{3}$</span>. Because <span class="math-container">$$(x-y)(x^2+xy+y^2)=x^3-y^3=2020=2^2\cdot 5\cdot 101$$</span> with <span class="math-container">$5\not\equiv 1\pmod{3}$</span> and <span class="math-container">$101\not\equiv 1\pmod{3}$</span>, we conclude that <span class="math-container">$5$</span> and <span class="math-container">$101$</span> cannot divide <span class="math-container">$x^2+xy+y^2$</span>. Thus, the only possible prime divisor of <span class="math-container">$x^2+xy+y^2$</span> is <span class="math-container">$2$</span>, and if <span class="math-container">$2\not\equiv 1\pmod{3}$</span> is a factor of <span class="math-container">$x^2+xy+y^2$</span>, we must have <span class="math-container">$2\mid x$</span> and <span class="math-container">$2\mid y$</span>. Since <span class="math-container">$x^2+xy+y^2\geq 0$</span>, this implies
<span class="math-container">$$x^2+xy+y^2=1\text{ or }x^2+xy+y^2=4\,.$$</span>
The only solutions <span class="math-container">$(x,y)\in\mathbb{Z}\times\mathbb{Z}$</span> to <span class="math-container">$x^2+xy+y^2=1$</span> are <span class="math-container">$$(x,y)=\pm (1,0),\pm(0,1),\pm(1,-1)\,.$$</span>
The only solutions <span class="math-container">$(x,y)\in\mathbb{Z}\times\mathbb{Z}$</span> to <span class="math-container">$x^2+xy+y^2=4$</span> are <span class="math-container">$$(x,y)=\pm (2,0),\pm(0,2),\pm(2,-2)\,.$$</span>
None of these solutions satisfies <span class="math-container">$x^3-y^3=2020$</span>.</p>
|
3,274,172 | <p>Let <span class="math-container">$X$</span> a compact set. Prove that if every connected component is open then the number of components is finite.</p>
<p>Ok, <span class="math-container">$X = \bigcup C(x)$</span> where <span class="math-container">$C(x)$</span> is the connected component of <span class="math-container">$x \in X.$</span> I know that <span class="math-container">$X \subset UA_\lambda$</span>, where <span class="math-container">$A_\lambda$</span> is a finite family of opens set but how can i conclude that the components are finite??</p>
| Paulo Mourão | 673,659 | <p><span class="math-container">$\textbf{Hint:}$</span> Argue by contradiction: assume <span class="math-container">$X$</span> has infinite <span class="math-container">$\textit{open}$</span> connected components and prove this implies that <span class="math-container">$X$</span> is not compact.</p>
<p>Recall that <span class="math-container">$X$</span> being non-compact is equivalent to the existence of an open cover of <span class="math-container">$X$</span> that has no finite subcover. Can you think of such a cover in this case?</p>
|
3,757,213 | <blockquote>
<p>Prove that the maximum area of a rectangle inscribed in an ellipse <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span> is <span class="math-container">$2ab$</span>.</p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>Equation of ellipse: <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span>.</p>
<p>Assume that <span class="math-container">$a>b$</span> & let <span class="math-container">$l$</span> & <span class="math-container">$w$</span> are length and width of rectangle then area will be <span class="math-container">$$A=l\times w \tag{1}$$</span></p>
<p>Now, I substituted the point <span class="math-container">$\left(\dfrac {l}{2}, \dfrac {w}{2}\right)$</span> in the equation of ellipse
<span class="math-container">$$\frac{(l/2)^2}{a^2}+\frac{(w/2)^2}{b^2}=1,$$</span>
<span class="math-container">$$b^2l^2+a^2w^2=4a^2b^2.\tag{2}$$</span></p>
<p>I am not sure how to proceed from here.</p>
| heropup | 118,193 | <p>Consider the coordinate transformation <span class="math-container">$$(u,v) = (x/a, y/b),$$</span> which maps the ellipse <span class="math-container">$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$</span> to the unit circle <span class="math-container">$$u^2 + v^2 = 1.$$</span> Such a transformation preserves the ratios of areas; in particular, if a quadrilateral inscribed in the ellipse has maximal area among all such quadrilaterals, then the transformed quadrilateral under this mapping will also have maximal area; and vice versa. This is not difficult to show.</p>
<p>Therefore, it suffices to consider all quadrilaterals inscribed in the unit circle in the transformed coordinate system. Since we know that the quadrilateral of maximal area in a circle is a square (proof below), it follows that the family of quadrilaterals of maximal area inscribed in the ellipse are those parallelograms that arise from the inverse transformation. These squares all have diagonal with length <span class="math-container">$2$</span>, thus area of <span class="math-container">$2$</span>, and upon the inverse transformation, the corresponding parallelograms all have area <span class="math-container">$2ab$</span>. Of these, there is only one that is a rectangle, corresponding to the configuration where the sides are parallel to the coordinate axes; although merely observing existence is sufficient (uniqueness is irrelevant).</p>
<hr />
<p>For the proof that a quadrilateral of maximal area inscribed in a circle is a square, let <span class="math-container">$ABCD$</span> be any quadrilateral inscribed in the circle. Consider a diagonal, say <span class="math-container">$AC$</span>. Then <span class="math-container">$\triangle ABC$</span> is an inscribed triangle whose area is maximized when the height <span class="math-container">$h_B$</span> perpendicular to <span class="math-container">$AC$</span> is maximized, since <span class="math-container">$AC$</span> is fixed. This corresponds to choosing the midpoint of <span class="math-container">$AC$</span> as the foot of the perpendicular. Similarly, the area of <span class="math-container">$\triangle ADC$</span> is maximized in the same manner. But this means <span class="math-container">$BD$</span> must be a diameter, since the feet of the perpendiculars coincide, both being the midpoint of <span class="math-container">$AC$</span>, and the perpendicular to any chord through its midpoint is a diameter. Thus among all quadrilaterals with fixed diagonal <span class="math-container">$AC$</span>, the one that maximizes the area is one where the other diagonal <span class="math-container">$BD$</span> is a diameter. Then among all such quadrilaterals with <span class="math-container">$BD$</span> as diameter, the one that maximizes the area is the one where <span class="math-container">$AC$</span> is also a diameter. Since <span class="math-container">$AC \perp BD$</span>, it follows <span class="math-container">$ABCD$</span> is a square.</p>
|
1,040,932 | <p>I have a system of congruence equations</p>
<p>$$
\begin{cases}
x \equiv 17 \pmod{15} \\
x \equiv 14 \pmod{33}
\end{cases}
$$</p>
<p>I need to investigate the system and see if they've got any solutions.</p>
<p>I know that I should use the Chinese remainder theorem "in a reverse order" so I think I should split each congruence equation in two new systems of two congruence equations.</p>
<p>From the CRT two congruence equations can be joined in a single congruence equation by</p>
<p>$$
x \equiv b_1 + c n_1 (b_2 - b_1) \pmod{n_1 n_2}
$$</p>
<p>From the first congruence equation I can get these two
$$
b_1 + c n_1 (b_2 - b_1) = 17 \\
n_1 n_2 = 15
$$</p>
<p>and from the second I can get
$$
b_1 + cn_1 (b_2 - b_1) = 14 \\
n_1 n_2 = 33
$$</p>
<p>but the unknown variables are not combined so I cannot just solve the system of four equations.</p>
<p>I need a hint :-)</p>
| Landon Carter | 136,523 | <p>$x\equiv2(\mod15)\implies x=15k+2$ for some integer $k$.
Similarly $x=33m+14$.</p>
<p>Thus $15k+2=33m+14\implies15k-33m=12\implies5k-11m=4$.</p>
<p>$\gcd(5,11)=1$ so the equation above has solutions.</p>
|
1,462,379 | <p>I have been given the task to compute $\int_{-1}^1 \sqrt{1-x^2} dx$ by means of calculus. We got the hint to substitute $x=\sin u$, but that only seems to make things more complicated:</p>
<p>$$\int_{-1}^1 \sqrt{1-x^2}dx = \int_{\arcsin-1}^{\arcsin1} \sqrt{1-\sin^2u} \frac{d\arcsin u}{du} du = \int_{\arcsin-1}^{\arcsin1} \sqrt{\frac{(1-\sin u)(1+\sin u)}{(1-u)(1+u)}} du.$$</p>
<p>I know that the answer should be $\frac\pi2$, because this is the area under one half of the unit circle, but how to arrive there by means of calculus is completely unclear to me. Could someone point me in the right direction?</p>
| Hosein Rahnama | 267,844 | <p>This is how it works:</p>
<p>$$\left\{ \matrix{
x = \sin (u) \hfill \cr
x = 1\, \to \,\,\,\,\,\,\,u = {\pi \over 2} \hfill \cr
x = - 1 \to u = - {\pi \over 2} \hfill \cr} \right.$$</p>
<p>and also</p>
<p>$$\sqrt {1 - {x^2}} = \sqrt {1 - {{\sin }^2}(u)} = \sqrt {{{\cos }^2}(u)} = \left| {\cos (u)} \right| = \cos (u)$$</p>
<p>The last equality holds since $ - {\pi \over 2} \le u \le {\pi \over 2}$. One more thing</p>
<p>$$dx = \cos (u)du$$</p>
<p>Finally, putting all this into the integral you find</p>
<p>$$\eqalign{
& \int_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}(u)du} = 2\int_0^{{\pi \over 2}} {{{\cos }^2}(u)du} = 2\int_0^{{\pi \over 2}} {\left( {{{1 - \cos (2u)} \over 2}} \right)du} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{{\pi \over 2}} {\left( {1 - \cos (2u)} \right)du} = \left. {\left( {u - {1 \over 2}\sin (2u)} \right)} \right|_0^{{\pi \over 2}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\pi \over 2} \cr} $$</p>
|
22,839 | <p>Is it possible to have the text generated by <code>PlotLabel</code> (or any other function) aligned to the left side of the plot instead of in the center?</p>
| Mr.Wizard | 121 | <p>Not directly that I am aware of. You could of course fake it with spacing:</p>
<pre><code>Plot[Sinc[x], {x, 0, 9}, PlotLabel -> Row@{"Text", Spacer[300]}]
</code></pre>
<p>Simpler may be use <code>Column</code> but it is not part of the <code>Graphics</code> object itself:</p>
<pre><code>Column@{"Text", Plot[Sinc[x], {x, 0, 9}]}
</code></pre>
|
22,839 | <p>Is it possible to have the text generated by <code>PlotLabel</code> (or any other function) aligned to the left side of the plot instead of in the center?</p>
| DavidC | 173 | <p>You can use Epilog to insert the plot label wherever you wish within the <code>PlotRange</code>. </p>
<p><em>Edit</em>: There is a drawback to this approach. Because the title will be within the graph region, there is a possibility that the title will be appear over part of the graph of the function.</p>
<pre><code>Plot[Sinc[x], {x, 0, 9},
Epilog -> Text[Style["This is where my plot label goes", {Blue, 16}],
Offset[{30, -10}, Scaled[{0, 1}]], {-1, 0}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/Aiur1.png" alt="plot"></p>
<p>In the example, the text, the title is offset 30 printer's points (30/72 in) to the right and 10 printers points (10/72 in) under the upper left corner (<code>Scaled[{0,1}]</code>) of the Graphic. It is left-justified, <code>{-1,0}</code>.</p>
|
256,666 | <p>Let $X$ be a set. Suppose $\beta$ is a basis for the topology $\tau_\beta$ of $X$. Since each base element is open (with respect to $\tau_\beta$) we have that $$B\in \beta\Rightarrow B\in \tau_\beta.$$ Thus, $\beta\subset \tau_\beta$. </p>
<p>However, since $\beta$ is a union of base elements (I assume a set can always be written as a union of its elements) and topologies are closed under arbitrary unions, we have $\beta \in \tau_\beta.$ Is it possible for a set to be a subset an an element of another set?</p>
| Tom Oldfield | 45,760 | <p>$\beta$ is a set of sets, and is not the union of it's elements .$\beta$ is not a union of basis elements, but a union of <em>sets containing</em> basis elements. Consider for example $A = \{[0,1], [1,2]\}$. It should be clear that $A \not= [0,2]$.</p>
<p>However, in general it is possible for a set to be both a subset and an element of another. Take for example $A = \{1\}, B =\{1,\{1\}\}$. Certainly $A \subset B$ and $A \in B$.</p>
|
3,438,653 | <p>I have this thing written on my notes: let <span class="math-container">${x}, {y}\in\mathbb R^n$</span> be two distinct points, then the set <span class="math-container">$$\{ \lambda x + (1-\lambda){y}\;\lvert\; \lambda \in [0,1] \}$$</span> contains all the points on the line segment that connects <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.
I can't seem to understand why it is so, and why we need to require that <span class="math-container">$\lambda \in [0,1]$</span>.
Thanks for any clarification.</p>
| Matematleta | 138,929 | <p>We use your (correct) definition of the line <span class="math-container">$\vec x(t)= \vec x_0 + t\vec{v}$</span>, where <span class="math-container">$\vec v$</span> marks the direction of <span class="math-container">$l$</span>. Note that <span class="math-container">$\textit{any}$</span> vector that has the same direction as <span class="math-container">$\vec v$</span> will do to give the set of points lying on <span class="math-container">$l$</span>, because when you change the magnitude of <span class="math-container">$\vec v$</span> you are just reparameterizing <span class="math-container">$l.$</span> </p>
<p>With the foregoing in mind, fix <span class="math-container">$\vec x_0,\vec y_0\in \mathbb R^n$</span> and take <span class="math-container">$\vec v=\vec y_0-\vec x_0.$</span> This makes sense because the vector difference <span class="math-container">$\vec y_0-\vec x_0$</span> is a vector in the same direction as <span class="math-container">$l.$</span></p>
<p>Therefore, a parametric equation of <span class="math-container">$l$</span> is <span class="math-container">$\vec x(t)=\vec x_0+t(\vec y_0-\vec x_0),$</span> which gives, on rearranging, </p>
<p><span class="math-container">$\vec x(t)=t\vec y_0+(1-t)\vec x_0$</span>. Now, as <span class="math-container">$t$</span> ranges from <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, <span class="math-container">$\vec x(t)$</span> goes from <span class="math-container">$\vec y_0$</span> to <span class="math-container">$\vec x_0$</span>. So values ot <span class="math-container">$t$</span> between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> correspond to points on the segment <span class="math-container">$\overline{\vec y_0 \vec x_0}.$</span></p>
<p>In fact, you can check that <span class="math-container">$\vec x_0$</span> and <span class="math-container">$\vec y_0$</span> split the line into three regions: <span class="math-container">$t<0;\ 0\le t\le 1;\ t>1$</span>, which if you make a graph, are easily located on <span class="math-container">$l$</span>.</p>
|
2,386,689 | <p>I know I can find solution for These equations using subtraction</p>
<p>$3x+2y=14.....(\text{i})$</p>
<p>$4x+3y=20.....(\text{ii})$</p>
<p>My question is can I divide $(\text{i})$ by $(\text{ii})$ to get values of $x$ and $y$ ?</p>
| Bill Dubuque | 242 | <p>Yes, eliminatiion by division always works. But you also need to separately consider the case that the eliminated variable $=0.\,$ Let's compare eliminating vs. isolating $x$ then dividing for the system</p>
<p>$$\begin{align} a x + b y = c\\
d x + e y = f \end{align}$$</p>
<p>Eliminating $x$ we obtain the equation $\ (bd-ae)\, y = cd-af$</p>
<p>Isolating $x$ and dividing we obtain $\ \ a/d = (by-c)/(ey-f)$</p>
<p>Clearing denominators shows that the latter is equivalent to the former, assuming that $d\neq 0$ (which we can assume wlog). Note that $\,ey-f = dx\neq 0\,$ by $\,d,x\neq 0$.</p>
|
92,983 | <p><strong>Does every polyhedron in $\mathbb{R}^3$ with $n$ triangular facets have a <em>topological</em> triangulation with complexity $O(n)$?</strong></p>
<p>Suppose $P$ is a non-convex polyhedron in $\mathbb{R}^3$ with $n$ triangular facets, possibly with positive genus. A <em>topological</em> triangulation of $P$ is a simplicial complex whose underlying space is the closure of the interior of $P$, such that every facet of $P$ is a cell in the complex. These boundary facets are true geometric triangles, but interior simplices may be arbitrarily bent and twisted. In the more standard <em>geometric</em> triangulations, every simplex is the convex hull of its vertices.</p>
<p>Results of <a href="http://www.cs.princeton.edu/~chazelle/pubs/BoundsSizeTetrahedral.pdf" rel="nofollow">Chazelle and Shouraboura</a> imply that every polyhedron has a geometric triangulation with complexity $O(n^2)$. Moreover, a classical construction of <a href="http://www.cs.princeton.edu/~chazelle/pubs/ConvexPartitionPolyhedra.pdf" rel="nofollow">Chazelle</a> implies that the $O(n^2)$ bound is is optimal in the worst case, even when the genus is zero.</p>
<p>But we can get tighter bounds for topological triangulations, at least for genus-zero polyhedra. If $P$ has genus zero, <a href="http://en.wikipedia.org/wiki/Steinitz%27s_theorem" rel="nofollow">Steinitz's theorem</a> implies that there is a <em>convex</em> polyhedron $Q$ that is combinatorially equivalent to $P$. <a href="http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085499/" rel="nofollow">Alexander's extension of the Schönflies theorem</a> implies that the interiors of $P$ and $Q$ are both homeomorphic to open balls. Thus, applying a suitable homeomorphism to a minimal <em>geometric</em> triangulation of $Q$ gives us a <em>topological</em> triangulation of $P$ with complexity $O(n)$. (Alternatively, we can triangulate $P$ by joining an arbitrary interior point to every facet.)</p>
<p>What makes the question tricky for higher-genus polyhedra is the possibility of knottedness; the topology of the interior of $P$ is not determined by its genus. Intuitively, the question is how knotted the interior of a polyhedron can be, as a function of the number of facets.</p>
<p>The following question may be equivalent: Let $K$ be a closed polygonal chain (or "stick knot") in $S^3$ with $n$ edges. Is there a <em>topological</em> triangulation of $S^3$ with complexity $O(n)$ that includes $K$ in its 1-skeleton? Again, if we insist on <em>geometric</em> triangulations, $\Theta(n^2)$ tetrahedra are always sufficient and sometimes necessary, even if $K$ is unknotted.</p>
<p><strong>Added for bounty (Apr 13):</strong> Partial results, subquadratic upper bounds, or references that imply this problem is open (or the crossing-number problem in my comment on the first answer) would be welcome.</p>
| Misha | 21,684 | <p>Correction: My "answer" below has a fatal mistake, but the idea still could be useful, although, seems to be hard to implement. One could try to use the fact that complexity of a topological triangulation is bounded from below by hyperbolic volume and that for alternating knots/links hyperbolic volume is $O(t)$ where $t$ is the twist number of the alternating diagram (in "most cases" it is just the crossing number). The problem is how to find alternating knot/link diagrams where the crossing number grows quadratically (or, at least, superlinearly) with respect to the number of edges. It is harder than I originally thought and I do not know how to do so. </p>
<p>The original "answer": The answer is negative. I will consider a similar problem: Let $K$ be a polygonal knot in ${\mathbb R}^3$; assume that $K$ has $n$ edges. We will estimate the minimal number of simplices needed to (topologically) triangulate ${\mathbb R}^3$ so that $K$ is contained in the 1-skeleton of the triangulation. By topological triangulation I mean one where simplices are not required to be linear with respect to the standard affine structure of ${\mathbb R}^3$. (Instead of a knot $K$ you would be considering a torus which is the boundary of a tubular neighborhood of $K$, but the size of the linear triangulation of the torus is $O(n)$ and, thus, my setup is equivalent to a special case of yours for knotted tori in ${\mathbb R}^3$.) Assuming that the exterior $ext(K)$ of $K$ is hyperbolic, the number of simplices in this triangulation is bounded from below by the hyperbolic volume of $ext(K)$. Now, suppose that $K$ is an alternating knot and, moreover, projection of $K$ to a generic plane is an alternating knot diagram $D$. (Thus, $ext(K)$ is almost always hyperbolic with the, very rare, exception of an alternating torus knot diagram, due to a result of Menasco.) If you take $K$ so that the vertices $K$ project to points on the unit circle and all edges project to segments of roughly length 2, then the crossing number $c=c(D)$ in $D$ is $O(n^2)$ as projections of all edges of $K$ intersect. Marc Lakenby proved in his <a href="http://arxiv.org/abs/math.GT/0012185" rel="nofollow">"The volume of hyperbolic alternating link complements"</a> paper that the volume of $ext(K)$ (for alternating hyperbolic knot $K$) is $O(t)$, where $t=t(D)$ is the number of "twist regions" in $D$. If $D$ contains no bigons (as in the construction I described with vertices on the unit circle) then $t=c$. Thus,
$vol(ext(K))=O(n^2)$ for such knots. </p>
<p>On the other hand, if you assume that your polyhedral surface is bounds, say, a handlebody, then complexity of minimal (topological) triangulation of the handlebody is $O(n)$. </p>
|
155,455 | <p>I want to find the maximum of a function (f) over a variable (t). The function is huge and it's not possible to maximize f(t) directly. So I want to create f inside a Table and then find the highest value over a small range of t. How can I add the steps to construct f into a Table? It seems "/." is not working. </p>
<p>A simplified version of my problem is:</p>
<pre><code> v = Table[f = TR1 + TR2
/. TR1 = Solve[TR1 - t = 0, {TR1}]
/. TR2 = t + 1
, {t, 1, 3, 1}];
tstar = Max[v]
</code></pre>
<p>f(t) has many components like TR1 and TR2 which makes it so huge.</p>
| bbgodfrey | 1,063 | <p>The code contains numerous syntax errors. It could, for instance, be written as</p>
<pre><code>v = Table[(TR1 + TR2) /. Flatten@Solve[TR1 - t == 0, TR1] /. TR2 -> t + 1, {t, 1, 3, 1}]
tstar = Max[v]
(* {3, 5, 7} *)
(* 7 *)
</code></pre>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| The Mathemagician | 3,546 | <p>Grubb's recent <em>Distributions And Operators</em> is supposed to be quite good. </p>
<p>There's also the recommended reference work, Strichartz, R. (1994), <em>A Guide to Distribution Theory and Fourier Transforms</em> </p>
<p>The comprehensive treatise on the subject-although quite old now-is Gel'fand, I.M.; Shilov, G.E. (1966–1968), <em>Generalized functions, 1–5,</em>. </p>
<p>A very good,though quite advanced,source that's now available in Dover is Trèves, François (1967), <em>Topological Vector Spaces, Distributions and Kernels</em> That book is one of the classic texts on functional analysis and if you're an analyst or aspire to be,there's no reason not to have it now. But as I said,it's quite challenging. </p>
<p>That should be enough to get you started.And of course,if you read French,you really should go back and read Schwartz's original treatise. </p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| abcd | 4,582 | <p>If you want a comparatively elementary approach to distribustion theory with applications to integral equations and difference equation no books come close to <strong>Distribution Theory and Transform Analysis: An Introduction to Generalized Functions, with Applications</strong> by A H Zemanian. another plus is it is Dover paperback, so cheap. Check this out.
<a href="http://rads.stackoverflow.com/amzn/click/0486654796" rel="nofollow">http://www.amazon.com/Distribution-Theory-Transform-Analysis-Introduction/dp/0486654796/ref=cm_cr_pr_product_top</a>.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| wildildildlife | 10,729 | <p>I'd like to point out a recent (Birkhäuser Cornerstones) <a href="http://www.springer.com/birkhauser/mathematics/book/978-0-8176-4672-1?cm_mmc=Google-_-Book%2520Search-_-Springer-_-0" rel="noreferrer">textbook</a> on Distribution Theory by Duistermaat and Kolk. </p>
<blockquote>
<p>The present text has evolved from a set of notes for courses taught at Utrecht University over the last twenty years, mainly to bachelor-degree students in their third
year of theoretical physics and/or mathematics. </p>
</blockquote>
<p>(I have followed this course, which was quite fun.)</p>
<p>For a more advanced exposition, Knapp's <a href="http://www.springer.com/birkhauser/mathematics/book/978-0-8176-4382-9" rel="noreferrer">Advanced Real Analysis</a> is great.</p>
<p>Very complete and advanced (and dry) is Hörmander's <a href="http://www.springer.com/mathematics/analysis/book/978-3-540-00662-6" rel="noreferrer">The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis</a>, which has already been mentioned.</p>
|
4,291,864 | <p>I have the following equation:</p>
<p><span class="math-container">$y=\frac{3x}{x^{2}+1}$</span></p>
<p>and I want to obtain x in terms of y, so far what I have done is the following:</p>
<p><span class="math-container">$3x=y(x^{2}+1)$</span></p>
<p><span class="math-container">$3x=x^{2}y+y$</span></p>
<p><span class="math-container">$3x-x^{2}y=y$</span></p>
<p>and at this point I got stucked, because that cuadratic term would not allow me to just get one x in one side of the equation. Any advice? Thanks</p>
| Mr.Gandalf Sauron | 683,801 | <p><span class="math-container">$(a,b)$</span> itself is infinite.</p>
<p>Any open set is of the form <span class="math-container">$\mathbb{R}\setminus F$</span> where <span class="math-container">$F$</span> is some finite set.</p>
<p>So you can see that any open set only excludes atmost finitely many elements of <span class="math-container">$(a,b)$</span>. So any point is a limit point of the set <span class="math-container">$(a,b)$</span>.</p>
<p>So <span class="math-container">$\overline{(a,b)}=\mathbb{R}$</span>.</p>
<p>Also for any infinite subset of a infinite space , the set of limit points will be the entire space under cofinite topology</p>
<p>Also I would like to reword your definition:-</p>
<p>A collection of sets <span class="math-container">$\delta$</span> is called a topology over a set <span class="math-container">$X$</span> if :-</p>
<ol>
<li><p><span class="math-container">$X\in \delta$</span></p>
</li>
<li><p>Arbitrary union of open sets is always open . (The members of <span class="math-container">$\delta$</span> are called open sets).</p>
</li>
<li><p>Intersection of finite number of open sets is open .( Or equivalently , intersection of two open sets is open. Hence you can proceed inductively and conclude that intersections of finite number of open sets is open).</p>
</li>
</ol>
<p>Now as to your question why <span class="math-container">$X$</span> and <span class="math-container">$\phi$</span> are closed.</p>
<p>By definition of topology <span class="math-container">$X$</span> is open hence <span class="math-container">$X\setminus X = \phi$</span> is closed.
And <span class="math-container">$\phi$</span> is open , so <span class="math-container">$X\setminus \phi=X$</span> is closed.</p>
|
842,266 | <p>I have a tiny little doubt related to one proof given in Ahlfors' textbook. I'll copy the statement and the first part of the proof, which is the part where my doubt lies on.</p>
<p><strong>Statement</strong>
The stereographic projection transforms every straight line in the $z$-plane into a circle on $S$ which passes through the pole $(0,0,1)$ and the converse is also true. More generally, any circle on the sphere corresponds to a circle or straight line in the $z$-plane.</p>
<p><strong>Proof</strong></p>
<p>To prove this we observe that a circle on the sphere lies in a plane $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=\alpha_0$, where we can assume ${\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2=1$ and $0\leq \alpha_0 <1$</p>
<p>I don't understand why it is always the case that the condition $0\leq \alpha_0 <1$ can be satisfied. I mean, a plane can be described as:</p>
<p>$$\Pi: \space n.(v-v_0)=0 \tag{1}$$ where $v$ and $v_0$ are two vectors with endpoints lying on $\Pi$. I know that $n$ is a perpendicular vector to the plane, and I understand that if $n=(\alpha_1,\alpha_2,\alpha_3)$ doesn't satisfy $||n||=1$, then the vector $n'=\dfrac{n}{||n||}$ is a unit vector which also satisfies equation (1).</p>
<p>Equation (1) is the same as $$\space n.v=n.v_0 \tag{2}$$</p>
<p>In this problem, $\alpha_0=n.v_0$, I don't understand why we can always choose $n$ and $v_0$ such that all the conditions said in my previous lines are satisfied.</p>
<p>I put the title "complex-analysis" but I am not sure if it is the proper tag, if anyone can think of a better tag for this post, tell me and I'll change it.</p>
| Joshua P. Swanson | 86,777 | <p>Here's another way to say it. Suppose $n \cdot v = n \cdot v_0$ is the equation of a plane (so $v = (x, y, z)$ are the variables, $n$ is a non-zero vector, and $v_0$ is an arbitrary point on the plane). It turns out $n \cdot v_0$ is $|n|$ times the minimum distance from the origin to the plane (times $\pm 1$).</p>
<p>For proof, let $v_1$ be the point on the plane closest to the origin, or equivalently it's the point you hit on the plane if you travel from the origin in a direction parallel to $n$. My previous comment says $v_1$ (as a vector starting at the origin) is parallel to $n$. From the usual formula for the magnitude of a dot product, $n \cdot v_1 = \pm |n||v_1|$, which I've claimed is $n \cdot v_0$. But this is true since $n \cdot v_1 = n \cdot v_0$.</p>
<p>In particular, if $n$ has length $1$, then $|n \cdot v_0|$ is the distance from the plane to the origin. For Ahlfors, $n \cdot v_0 = \alpha_0$. Since the plane in question intersects the unit sphere, $|n \cdot v_0| < 1$. The rest follows.</p>
|
3,369,069 | <p>Let <span class="math-container">$l_1$</span> and <span class="math-container">$l_2$</span> be two distributions in disjoint variables <span class="math-container">$x_1, ..., x_n$</span> and <span class="math-container">$y_1, ..., y_m$</span>. Then it is said to be possible to define a product distribution.</p>
<p>However, I am fundamentally confused. Distributions are in fact linear functionals on the space of smooth and compactly supported functions. Then, how does the 'product' of linear functionals again become a linear functional?</p>
<p>Especially, what can be a definition of <span class="math-container">$\delta(x_1)\delta(x_2)$</span> such that it is equal to <span class="math-container">$\delta(x_1, x_2)$</span>? I am just stuck......</p>
| reuns | 276,986 | <p>Concretely for <span class="math-container">$\phi \in C^\infty_c(\Bbb{R}^2)$</span> take <span class="math-container">$\varphi_n,\psi_n \in C^\infty_c(\Bbb{R})$</span> such that <span class="math-container">$$\sum_{n=1}^N \varphi_n(x)\psi_n(y)\to \phi$$</span> in test function topology.</p>
<p>Then for <span class="math-container">$T,S\in D'(\Bbb{R})$</span> <span class="math-container">$$\langle T(x)S(y),\phi\rangle =\sum_{n=1}^\infty \langle T(x)S(y),\varphi_n(x)\psi_n(y)\rangle=\sum_{n=1}^\infty \langle T,\varphi_n\rangle \langle S,\psi_n\rangle$$</span></p>
<p>To find <span class="math-container">$\varphi_n,\psi_n$</span>, for <span class="math-container">$\phi$</span> supported on <span class="math-container">$(-T,T)\times (-T,T)$</span>, take <span class="math-container">$\rho \in C^\infty_c(\Bbb{R})$</span>, <span class="math-container">$\rho = 1$</span> on <span class="math-container">$(-T,T)$</span> and look at the Fourier series <span class="math-container">$$\phi(x,y) = \sum_{m,k}c_{m,k}e^{2i\pi mx/T}\rho(x)e^{2i \pi ky/T}\rho(y)$$</span></p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Hagen von Eitzen | 39,174 | <p>Clearly, <span class="math-container">$x=y$</span> is impossible as <span class="math-container">$x\mid 2x^n=x^n+y^n=z^n$</span> leads to <span class="math-container">$x=z$</span>, which is absurd.
So wlog. <span class="math-container">$x<y<z$</span>.
Note that <span class="math-container">$y^n=z^n-x^n$</span> a multiple of <span class="math-container">$z-x$</span>, which must therefore be a power of <span class="math-container">$y$</span> (and <span class="math-container">$>y^0$</span>). Thus <span class="math-container">$z\ge x+y$</span>. But <span class="math-container">$(x+y)^n>x^n+y^n$</span>.</p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Barry Cipra | 86,747 | <p>The three primes cannot all be odd, so one of them must be <span class="math-container">$2$</span>. It cannot be <span class="math-container">$z$</span>, so let's let it be <span class="math-container">$x$</span>, in which case <span class="math-container">$y$</span> and <span class="math-container">$z$</span> are odd and we have</p>
<p><span class="math-container">$$2^n=z^n-y^n=(z-y)(z^{n-1}+\cdots+y^{n-1})$$</span></p>
<p>which implies <span class="math-container">$z-y=2^k$</span> for some <span class="math-container">$1\le k$</span> (ruling out <span class="math-container">$k=0$</span> since <span class="math-container">$y$</span> and <span class="math-container">$z$</span> are odd). Writing <span class="math-container">$z=2^k+y$</span>, we see that, since <span class="math-container">$n\ge2$</span>, we have</p>
<p><span class="math-container">$$2^n\ge2^k(2^k+y)^{n-1}=2^k(2^{k(n-1)}+\cdots+y^{n-1})\gt2^{kn}\ge2^n$$</span></p>
<p>which is a contradiction because of the strict inequality.</p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Jyrki Lahtonen | 11,619 | <p>Reposting my answer from the deleted thread here by Martin's request.</p>
<p>Modulo two consideration shows that one of <span class="math-container">$a,b,c$</span> needs to be even. Given that there is a single even prime, we can conclude that <span class="math-container">$a$</span> or <span class="math-container">$b$</span> must be two (leaving the case <span class="math-container">$c=2$</span> as an exercise). W.l.o.g. <span class="math-container">$a=2$</span> and <span class="math-container">$b,c $</span> are both odd. But then <span class="math-container">$c\ge b+2$</span>, and proving the inequality
<span class="math-container">$$
2^n+b^n<(b+2)^n
$$</span>
is another very simple exercise.</p>
<p>Observe that the arguments works even when <span class="math-container">$n=2$</span>.</p>
|
2,647,000 | <p>Consider a function $ϕ$ such that $$\lim_{h→0} ϕ(h) = L$$ and $$L − ϕ(h) ≈ ce^{−1/h}$$ for some constant $c$. By combining $ϕ(h)$, $ϕ(h/2)$, and $ϕ(h/3)$, find an accurate estimate of $L$.</p>
<p>Isn't $ϕ(h)=-ce^{−1/h}+L$? I think I am over-simplfying this...</p>
| Ri-Li | 152,715 | <p>Normally in my opinion in research there is nothing called number theory. We start with a subject called number theory and we develop tools to solve some problems. </p>
<p>Yes, it is true that to work in number theory we need to know a bit of stuffs but not a lot always. It depends on problems and possible projects of the supervisor.I would like to second @videlity that it depends on the present institutions and faculties.</p>
<p>E.g Representation theory and finite dimensional algebras are used in number theory, at the same time there is something called probabilistic number theory. Again some professors like Prof Alexandru Zaharescu works on Ramanujan Number theory, Prof G. Andrews works on Partition theory.</p>
<p>I hope it helps.</p>
|
3,807,550 | <p>I am stuck in a true/false question. It is</p>
<p>In a finite commutative ring, every prime ideal is maximal.</p>
<p>The answer says it's false.</p>
<p>Well what I can say is (Supposing the answer is right)</p>
<p><span class="math-container">$(1)$</span> The ring can't be Integral domain since finite integral domain is a field.</p>
<p><span class="math-container">$(2)$</span> There can't be unity in the ring since in that case the result would be true.(By the Theorem that if <span class="math-container">$R$</span> is a commutative ring with unity then an ideal <span class="math-container">$I$</span> is prime iff <span class="math-container">$R/I$</span> is Integral Domain)</p>
<p><span class="math-container">$(3)$</span> All the elements are zero divisors since if there is at least one non- zero divisor, there will be a unity and so <span class="math-container">$(2)$</span> would follow.</p>
<p>So at the end, I am in search of a finite commutative with all elements as zero -divisors, having no unity and obviously a prime ideal in it which is not maximal.</p>
<p>What kind of strange looking ring is this (if possible) ? Any hints??</p>
| markvs | 454,915 | <p>The answer is false. <span class="math-container">$I$</span> is prime means <span class="math-container">$R/I$</span> is a domain. Which implies <span class="math-container">$R/I$</span> is a field which implies that <span class="math-container">$I$</span> is maximal.</p>
|
2,035,186 | <p>This is a probability question where I am asked to integrate a region that represents the probability of a scenario. X, Y, and U are random variables, where U = X-Y. I need to find the probability </p>
<p>$P(U \leq u) = P(X-Y \leq u)$, where the density function I'm integrating over is defined by f(x,y) = 1, for 0 $\leq x \leq 2 \quad 0 \leq y \leq 1 \quad 2 y \leq x$. </p>
<p>Why can't I capture this region with a single double integral going from $\int_{0}^{1}$$\int_{2y}^{u+y}$dxdy ? </p>
| Community | -1 | <p>$\newcommand{\cm}{\mathrm{cm}}$Call $AOB$ the triangle, and label the tangency points $P,\,Q,\,R$ going from left to right.</p>
<p>You know $AO=\frac{120\cm^2}{15\cm}=8\cm$ and $AB=\sqrt{AO^2+BO^2}=\sqrt{289\cm^2}=17\cm$.</p>
<p>Moreover, you know that $PA=AQ$, that $QB=BR$, that $PA+AO=BR+BO$ and that $QB+PA=AB$. Putting the last two together: $$\begin{cases}PA+8\cm=BR+15\cm\\ PA+BR=17\cm\end{cases}$$</p>
<p>And this can be solved for $PA$ and $BR$. Since $RO\perp PO$, you have that $RO=BO+BR$ has the same length as the radius.</p>
|
1,724,554 | <p>Say, $A$ is an $ n\times n $ matrix over $\Bbb R$, with</p>
<p>$$ A_{ij} = \begin{cases} a \qquad \text{if } i=j\\
b \qquad \text{otherwise.}
\end{cases} $$</p>
<p>How do we compute the determinant of this symmetrix matrix $A$?</p>
| lhf | 589 | <p>We can write $A=B-(b-a)I$, where $B$ is the matrix with all entries equal to $b$.</p>
<p>Therefore, $\det A=(-1)^n\chi(b-a)$, where $\chi$ is the characteristic polynomial of $B$.</p>
<p>Since $B$ has rank $1$, we have $\chi(x)=x^n-tr(B)x^{n-1}=x^n-nbx^{n-1}$ (see <a href="https://math.stackexchange.com/a/458125">this</a> for instance, or use induction in this simpler case).</p>
<p>Finally,</p>
<p>$\det A = (-1)^n\chi(b-a)=$</p>
<p>$\qquad=(-1)^n((b-a)^n-nb(b-a)^{n-1})$</p>
<p>$\qquad=(a-b)^n+nb(a-b)^{n-1}$</p>
<p>$\qquad =(a-b)^{n-1}(a+(n-1)b)$</p>
|
161,024 | <p>I was recently having a discussion with someone, and we found that we could not agree on what an exponential function is, and thus we could not agree on what exponential growth is. </p>
<p>Wikipedia claims it is $e^x$, whereas I thought it was $k^x$, where k could be any unchanging number. For example, when I'm doing Computer Science classes, I would do everything using base 2. Is $2^x$ not an exponential function? The classical example of exponential growth is something that doubles every increment, which is perfectly fulfilled by $2^x$. I'd also thought $10^x$ was a common case of exponential growth, that is, increasing by an order of magnitude each time. Or am I wrong in this, and only things that follow the natural exponential are exponential equations, and thus examples of exponential growth?</p>
| Qiaochu Yuan | 232 | <p>If $x$ has units (e.g. time), then there's no way to distinguish between these possibilities; they're all equivalent up to change of units. </p>
|
3,491,816 | <p>Find the min and max values of the function <span class="math-container">$$f(x,y)=10y^2-4x^2$$</span> with the constraint <span class="math-container">$$g(x,y)=x^4+y^4=1$$</span>
I have done the following working;
<span class="math-container">$$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$</span><span class="math-container">$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$</span><span class="math-container">$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$</span><span class="math-container">$$\lambda x^3=2\\\lambda y^3=5$$</span> My question is how can I find the value of lambda for the constraint to hold.</p>
| Siddhartha | 257,185 | <p>A different approach without using Lagrange multipliers:</p>
<p>Let <span class="math-container">$x^2=\cos\theta$</span> and <span class="math-container">$y^2=\sin\theta$</span> where <span class="math-container">$\displaystyle 0\le\theta\le\frac{\pi}{2}$</span>.</p>
<p><span class="math-container">$f(\theta)=10\sin\theta-4\cos\theta$</span></p>
<p><span class="math-container">$f'(\theta)=10\cos\theta+4\sin\theta$</span></p>
<p><span class="math-container">$f'(\theta)=0$</span></p>
<p><span class="math-container">$\implies 10\cos\theta+4\sin\theta=0$</span></p>
<p>We note that this is impossible for any value of <span class="math-container">$\theta$</span> in our domain. So, the maximum and minimum occur at the end-points.</p>
<p><span class="math-container">$f(0)=-4$</span></p>
<p><span class="math-container">$\displaystyle f\left(\frac{\pi}{2}\right)=10$</span></p>
<p>So, the minimum value is <span class="math-container">$-4$</span> and the maximum value is <span class="math-container">$10$</span>.</p>
|
3,676,284 | <p><a href="https://i.stack.imgur.com/9xfxz.png" rel="nofollow noreferrer">This is link to question</a>
[Here is my attempt, but the answer key is convergent. I dont think I count it wrong.<a href="https://i.stack.imgur.com/nAcEs.jpg" rel="nofollow noreferrer">][1]</a></p>
| J.G. | 56,861 | <p>The usual proof that well-ordering implies a choice function <span class="math-container">$f$</span> exists on <span class="math-container">$X\not\owns\emptyset$</span> notes that if <span class="math-container">$\le$</span> well-orders <span class="math-container">$\bigcup X$</span>, any <span class="math-container">$x\in X$</span> has a <span class="math-container">$\le$</span>-minimal element, say <span class="math-container">$y$</span>, because <span class="math-container">$x\subseteq\bigcup X$</span>; we can thus define <span class="math-container">$f(x):=y$</span>.</p>
<p>This argument requires <span class="math-container">$\bigcup X$</span> to exist for each such <span class="math-container">$X$</span>. In set theory, that's no problem; it's the <a href="https://en.wikipedia.org/wiki/Axiom_of_union" rel="nofollow noreferrer">axiom of union</a>. For a theory of sorted lists to work the same way, we need the union to admit a specific ordering as well.</p>
<p>So if your universe of objects is well-ordered, every list can just be an ordered sample therein, in which case choice is trivial. But then you're basically studying, in a non-standard formalism, the subsets of some universe nowhere near as general as we like to consider in foundational mathematics.</p>
<p>For example, our ordered lists could be of ordinals, so <span class="math-container">$<=\in$</span>. It's true we can construct an easy choice function on any set of nonempty sets of ordinals. But one cannot, for example, easily identify the real numbers (or which are less naturally ordered, the complex ones) with ordinals in a way that makes this useful.</p>
|
2,969,203 | <p>Let <span class="math-container">$f$</span> be a <span class="math-container">$C''$</span> function on <span class="math-container">$(a, b)$</span> and suppose there is a point <span class="math-container">$c$</span> in (a, b) with <span class="math-container">$$f(c)= f'(c)=f''(c) = 0$$</span> Show that there is a continuous function <span class="math-container">$h$</span> on <span class="math-container">$(a, b)$</span> with <span class="math-container">$$f(x) =(x-c)^2h(x)$$</span> for all <span class="math-container">$x$</span> in <span class="math-container">$(a, b)$</span>.</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>For <span class="math-container">$x$</span> in <span class="math-container">$(a,b)$</span>,</p>
<p><span class="math-container">$$f(x)=f(c)+(x-c)f'(c)+\frac{(x-c)^2}{2}f''(c)+(x-c)^2\epsilon(x)$$</span></p>
|
4,086,485 | <blockquote>
<p>We can regard <span class="math-container">$\pi_1(X,x_0)$</span> as the set of basepoint-preserving homotopy classes of maps <span class="math-container">$(S^1,s_0)\rightarrow(X,x_0$</span>). Let <span class="math-container">$[S^1,X]$</span> be the set of homotopy classes of maps <span class="math-container">$S^1\rightarrow X$</span>, with no conditions on basepoints. Thus there is a natural map <span class="math-container">$\Phi:\pi_1 (X,x_0)\rightarrow[S^1,X]$</span> obtained by ignoring basepoints. Show that <span class="math-container">$\Phi$</span> is onto if <span class="math-container">$X$</span> is path-connected, and that <span class="math-container">$\Phi([f])=\Phi([g])$</span> iff <span class="math-container">$[f]$</span> and <span class="math-container">$[g]$</span> are conjugate in <span class="math-container">$\pi_1(X,x_0)$</span>. Hence <span class="math-container">$\Phi$</span> induces a one-to-one correspondence between <span class="math-container">$[S^1,X]$</span> and the set of conjugacy classes in <span class="math-container">$\pi_1(X)$</span>, when <span class="math-container">$X$</span> is path-connected.</p>
</blockquote>
<p>We think of <span class="math-container">$\pi(X, X_0)$</span> as homotopy classes of basepoint preserving maps <span class="math-container">$(S^1, s_0) \rightarrow (X, x_0)$</span>. Recall that a map is basepoint preserving iff <span class="math-container">$f(s_0) = x_0$</span>.</p>
<p>Define <span class="math-container">$[S^1, X]$</span> to be homotopy classes of maps <span class="math-container">$S^1 \rightarrow X$</span> with no condition on basepoints.
The exercise asks us to show that the map <span class="math-container">$\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$</span> is an onto map if <span class="math-container">$X$</span> is path-connected.</p>
<p>I don't think is true. Consider a wedge of circles. This is path-connected. Now consider all loops based at <span class="math-container">$x_0$</span>, a point on the left circle that is not on the right circle. In the figure, this is in blue. These loops are in <span class="math-container">$\pi(X, x_0)$</span>. Now, on the other hand, consider a loop on the circle on the right hand side. In the figure, this loop is in pink. It seems impossible that any blue loop can be homotped into a pink loop.</p>
<p>Indeed, by reading ahead, we know that the fundamental group of the wedge of circles is <span class="math-container">$W \equiv \mathbb Z * \mathbb Z \simeq \langle a, b\rangle$</span>. All loops on the left circle will be of the form <span class="math-container">$a^n$</span>, while a loop on the right circle will be of the form <span class="math-container">$b^m$</span>. The map <span class="math-container">$\Phi$</span> that produces all the <span class="math-container">$a^n$</span> cannot produce a <span class="math-container">$b^m$</span>, and thus the map <span class="math-container">$\Phi$</span> cannot be surjective.</p>
<p>What am I missing?</p>
<p><a href="https://i.stack.imgur.com/4WUVa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4WUVa.png" alt="enter image description here" /></a></p>
| feynhat | 359,886 | <p>One can write an explicit expression for the homotopy but I am quite convinced by this picture, <img src="https://i.imgur.com/IU7h3A5.jpg" alt="this picture" /></p>
<p>where <span class="math-container">$x_1$</span> is a point on the pink loop <span class="math-container">$b$</span> (more generally suppose <span class="math-container">$[b]$</span> is any member of <span class="math-container">$[S^1, X]$</span>) and <span class="math-container">$h$</span> is a path joining <span class="math-container">$x_0$</span> and <span class="math-container">$x_1$</span>. The top edge of the square is <span class="math-container">$\beta_h(b) = h \cdot b \cdot \overline{h}$</span>.</p>
<p>The only thing that might need some explanation is the green portion.</p>
<p>On the left green portion, we can define the homotopy to be <span class="math-container">$h$</span> along the line segment emanating from <span class="math-container">$x_0$</span> and meeting the left slant edge. And do something similar on the right green portion.</p>
|
508,791 | <p>I have an integer list that is <code>n</code> long and each value can be ranging from <code>1 .. n</code>.</p>
<p>I need a formula that tells me how many of all possible lists for a given n, that have one or more consecutive sequences of a length of exactly 2 of the same number and no other consecutive sequences that are longer than 2.</p>
<p>For example for n=5:</p>
<p>These two should count: </p>
<pre><code>{ 1, 1, 5, 3, 3 }
{ 2, 3, 2, 5, 5 }
</code></pre>
<p>Where as these should not:</p>
<pre><code>{ 1, 1, 1, 2, 2 }
{ 1, 3, 2, 5, 4 }
</code></pre>
<p>I've been attempting to do this by looking at the possible sequences using the following formula where x = n-1:</p>
<pre><code>(n) x n n = x * n^3
x (n) x n = x^2 * n^2
n x (n) x = x^2 * n^2
n n x (n) = x * n^3
</code></pre>
<p>And sum these four up.</p>
<p>However, these also needs to take overlaps between the four into account. This is where I could use a bit of help..? What would the formulas be for excluding the overlaps?</p>
<p>An alternative approach would also be welcome.</p>
<p>Going trough all sequences counting manually is not an option - I need this to work for for n larger than what makes that approach computationally feasible.</p>
<p>If it helps anyone, I've written a little program that runs trough all the sequences and counts have the following results:</p>
<pre><code>n = 2
L[2]: 2
L[1]: 2
n = 3
L[3]: 3
L[2]: 12
L[1]: 12
n = 4
L[4]: 4
L[3]: 24
L[2]: 120
L[1]: 108
n = 5
L[5]: 5
L[4]: 40
L[3]: 280
L[2]: 1520
L[1]: 1280
</code></pre>
<p>Where <code>n = 5</code>, <code>L[2]: 1520</code> is the result I've asked for a formula to in the above question.</p>
| Marko Riedel | 44,883 | <p>Here is a different solution that may interest you. Introduce three sequences
$a_{n,k}$, $b_{n,k}$ and $c_{n,k}$ that count the number of strings over $\Sigma^k$ where $|\Sigma|=n,$ that end in a digit that is not repeated, a digit that is repeated twice and a digit that is repeated at least three times. In fact we take these sequences to be generating functions in two variables, where the variable $v$ counts occurrences of subsequences of length exactly two and $w$ counts occurrences of subsequences of length at least three.</p>
<p>This gives the following set of recurrences:
$$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v b_{n,k-1} + (n-1) c_{n,k-1}, $$
$$b_{n,k} = a_{n,k-1} \quad\text{and}\quad
c_{n,k} = w b_{n,k-1} + c_{n,k-1}.$$</p>
<p>With these settings the generating function of all elements of $\Sigma^k$ classified according to the number of length 2 and length more than two subsequences is given by
$$a_{n,k} + v b_{n,k} + w c_{n,k}.$$</p>
<p>Observe that $a_{n,1} = 1$ and $b_{n,1} = c_{n,1} = 0.$ Furthermore the last recurrence implies that $$c_{n,k} = w \sum_{q=1}^{k-1} b_{n,q} = w \sum_{q=1}^{k-2} a_{n,q}.$$</p>
<p>Taken together this yields a recurrence for $a_{n,k}:$
$$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v a_{n,k-2} + (n-1) w \sum_{q=1}^{k-3} a_{n,q}. $$</p>
<p>Introduce the trivariate generating function $$G_n(z) = \sum_{k\ge 1} a_{n,k} z^k.$$</p>
<p>Multiply the recurrence by $z^k$ and sum for $k\ge 4:$
$$\sum_{k\ge 4} a_{n,k} z^k =
(n-1) z \sum_{k\ge 4} a_{n,k-1} z^{k-1} +
(n-1) vz^2 \sum_{k\ge 4} a_{n,k-2} z^{k-2} \\+
\sum_{k\ge 4} (n-1) w z^k [z^{k-3}] \frac{1}{1-z} G_n(z). $$
Now $a_{n,1} = n, \; a_{n,2} = n(n-1)$ and
$$a_{n,3} = n(n-1)^2 + n(n-1)v.$$
The equation derived from the recurrence now becomes
$$ G_n(z) - (n(n-1)^2 + n(n-1)v)z^3 - n(n-1)z^2 -nz \\=
(n-1)z (G_n(z) - n(n-1)z^2 -nz)
+ (n-1)vz^2 (G_n(z) - nz) \\
+ (n-1)w z^3 \sum_{k\ge 4} z^{k-3} [z^{k-3}] \frac{1}{1-z} G_n(z).$$
The sum term simplifies to $$(n-1)w z^3 \frac{1}{1-z} G_n(z).$$
We may now solve for $G_n(z),$ getting
$$G_n(z) = -{\frac {nz \left( -1+z \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{
z}^{2}-v{z}^{3}-w{z}^{3}n+w{z}^{3}+1}}.$$
Now the generating function for $b_{n,k}$ is $$z G_n(z)
\quad\text{and the one for}\; c_{n,k} \; \text{is}\quad
\frac{wz^2}{1-z} G_n(z).$$
It follows that the generating function $H_n(z)$ of $a_{n,k} + v b_{n,k} + w c_{n,k}$
is $$\left(1 + vz + \frac{w z^2}{1-z}\right) G_n(z)$$
or equivalently
$$H_n(z) = -{\frac {zn \left( -1+z-vz+v{z}^{2}-w{z}^{2} \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{z}^{2}-v{z}^{3}-w{
z}^{3}n+w{z}^{3}+1}}.$$
Now we do not permit sequences of length at least three, so we take
$$[w^0] H_n(z) =
-{\frac { \left( vz+1 \right) nz}{v{z}^{2}n-v{z}^{2}+nz-z-1}}.$$
In fact there must be at least one two-sequence, so we first calculate
$$[w^0] H_n(z) - [v^0] [w^0] H_n(z) =
{\frac {v{z}^{2}n}{ \left( nz-z-1 \right) \left( v{z}^{2}n-v{z}^{2}+nz-z-1
\right) }}$$ and put $v=1$, finally obtaining the generating function
$$M_n(z) = {\frac {n{z}^{2}}{ \left( nz-z-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1
\right) }}.$$
The partial fraction decomposition of $M_n(z)$ is given by
$$-{\frac {n}{ \left( n-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1 \right) }}+{
\frac {n}{ \left( n-1 \right) \left( nz-z-1 \right) }}.$$</p>
<p>Now the singularities are
at $$\rho_0 = \frac{1}{n-1} \quad\text{and}\quad
\rho_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{n^2+2n-3}}{2(n-1)}.$$
Expanding $M_n(z)$ into series we obtain the following closed form result for the number of admissible strings of length $k$ and an alphabet of $n$ symbols:
$$ [z^k] M_n(z) =
n\frac{\rho_0^2 \rho_1 \rho_2}{(\rho_0-\rho_1)(\rho_0-\rho_2)} \rho_0^{-k}\\
+ n\frac{\rho_0\rho_1^2 \rho_2}{(\rho_1-\rho_0)(\rho_1-\rho_2)} \rho_1^{-k}
+ n\frac{\rho_0\rho_1 \rho_2^2}{(\rho_2-\rho_0)(\rho_2-\rho_1)} \rho_2^{-k}.$$
The reader is asked to verify that indeed
$$ [z^n] M_n(z) =
\sum_{k=1}^{\lfloor n/2 \rfloor} {n-k\choose k} n (n - 1)^{n-k-1}.$$
The above formula for $[z^k] M_n(z)$ is quite powerful. It gives the exact value of the number of admissible strings with any alphabet of size $n$ and of length $k$. For example, when there are $n=2$ symbols, the sequence starting at length $k=1$ is:
$$0, 2, 4, 8, 14, 24, 40, 66, 108, 176.$$ With $n=7$ we get the following sequence:
$$0, 7, 84, 798, 6804, 54684, 423360, 3194856, 23668848, 172939536.$$</p>
<p>The generating function $H_n(z)$ encapsulates the complete distribution of the $n^k$ strings classified according to the number of two-sequences (counted by $v$) and sequences of length at least three (counted by $w$).</p>
<p>Here is an example:
$$[z^4] H_3(z) = 6\,{v}^{2}+36\,v+15\,w+24$$
for a total of $3^4=81$ terms (strings of length $4$ over $3$ symbols).
Another example is:
$$[z^6] H_5(z) = 80\,{v}^{3}+1920\,{v}^{2}+520\,vw+20\,{w}^{2}+6400\,v+1565\,w+5120$$
for a total of $5^6=15625$ terms (strings of length $6$ over $5$ symbols).
It is a useful combinatorics excercise to verify some of these values with pen and paper. For example, a string of length four over three symbols containing a sequence of length at least three can consist of four equal symbols, giving a contribution of three, or a length three sequence at the front for a contribution of three times two or a length three sequence at the end, again for three times two, for a total of $3+6+6 = 15$ which is indeed the value from the generating function.</p>
|
2,073,410 | <p>If $$ a-(a \bmod x)<b$$ how do I prove that $$c-(c\bmod x)<b \;\forall c<a?$$ </p>
| Bart Michels | 43,288 | <p>We have to prove that
$$c-(c\bmod x)\leq a-(a\bmod x)$$
for all $a,c,x$, $c\leq a$.</p>
<p>Both are multiples of $x$, and their difference is
$$\begin{align*}a-(a\bmod x)-(c-(c\bmod x))&=(a-c)+(c\bmod x-a\bmod x)\\&\geq 0+c\bmod x-a\bmod x\\&>0+0-x\end{align*}$$
and it is a multiple of $x$, hence $\geq0$.</p>
|
22,207 | <p>How to make a defined symbol stay in symbol form?</p>
<pre><code>w = 3; g = 4;
{w, g}[[2]]
</code></pre>
<blockquote>
<p><code>3</code></p>
</blockquote>
<p>I want the output to be <strong><code>g</code></strong> and not <code>3</code>. For example, if I want to save different definitions by <code>DumpSave</code> in different files like below:</p>
<p><code>Table[DumpSave["/Users/simonlausen/Desktop/Input/ex"<>ToString[i]<>".mx",
{w,g}[[i]]],{i,1,2}]</code></p>
<p>Any suggestions?</p>
| Mr.Wizard | 121 | <p>I would do this:</p>
<pre><code>w = 3; g = 4;
HoldForm[w, g][[{2}]]
</code></pre>
<blockquote>
<pre><code> g (* wrapped in HoldForm *)
</code></pre>
</blockquote>
<p>The <code>{}</code> brackets are critical. See <a href="https://mathematica.stackexchange.com/a/2450/121">this answer</a> for an explanation.</p>
<hr>
<p>I just read Jens' answer and now realize I missed an important part of the question. I would do something similar to what he did, but I would write it a bit differently:</p>
<pre><code>Table[
DumpSave["path" <> ToString[i] <> ".mx", #] & @ {Unevaluated[w, g]}[[1, {i}]],
{i, 1, 2}
]
</code></pre>
<p>Ignore the syntax highlighting in <code>Unevaluated[w, g]</code>.</p>
|
372,198 | <blockquote>
<p>If $G$ is a group, $H$ and $K$ both subgroups of $G$, $K \subseteq H$, $\left[G:H\right]$ and $\left[H:K\right]$ both finite then $\left[G:K\right]=\left[G:H\right]\cdot\left[ H:K \right].$</p>
</blockquote>
<p>I am not sure if this is standard notation but $\left[ G : K \right]$ denotes the number of right or left cosets of $K$ in $G$.</p>
<p>I haven't tried to do the case where if $G$ is finite but I imagine the result would immediately follow by using Lagrange's Theorem. I am trying to think about the case where $G$ is infinite.</p>
<p>I at least made an example to show myself that the index $\left[ G: K \right]$ could be a finite number but $K$ could have size of infinity. Any leads?</p>
<p>Thanks very much</p>
| bfhaha | 128,942 | <p>Consider the splitting field $E$ for $f(x)$ over $F$.<br>
There are three possible factorizations of $f(x)$ in $E[x]$.<br>
(i) $f(x)=(x-r_1)(x-r_2)\cdots (x-r_p)$.<br>
(ii) $f(x)=(x-r_1)^s(x-r_2)^s\cdots (x-r_t)^s$, where $s\geq 2$, $t\geq 2$.<br>
(iii) $f(x)=(x-r_1)^p$.<br>
Since $\gcd{(f(x),f'(x))}\neq 1$,
the case (i) is impossible.<br>
Since $p$ is a prime (the characteristic of a field),
the case (ii) is impossible.<br>
The remain possible is $f(x)=(x-r_1)^p\in E[x]$.
Let $r=r_1$.</p>
<p>If $r\in F$,
then $f(x)$ is splits in $F[x]$.</p>
<p>Suppose that $r\notin F$.
Since $F[x]$ is a U.F.D.,<br>
write $f(x)$ as a product of some irreducible polynomials $c_1(x), c_2(x), ..., c_n(x)$.</p>
<p>For each $i=1,2,...,n$,
$f(x)$ has a root in $K_i=F[x]/\langle c_i(x)\rangle$.<br>
But the only one root of $f(x)$ is $r$.
Hence, $f(x)=(x-r)^p\in K_i[x]$.</p>
<p>Since $c_i(x)\mid f(x)$, we have $c_i(x)=(x-r)^{q_i}\in K_i[x]$</p>
<p>Note that $K_i=E$ for each $i=1,2,...,n$ by the uniqueness of the splitting field.<br>
Therefore, $q_i=\deg{c_i(x)}=[K_i:F]=[E:F]$.<br>
Says $q_1=q_2=\cdots=q_n=q=[E:F]$.<br>
Then $p=\deg{f(x)}=n\cdot q$ and $n=1$ and $q=p$.<br>
$f(x)=c_1(x)$ is irreducible.<br>
(If $n=p$ and $q=1$, then $\deg{c_i(x)}=1$ and $K_i=F$ and $r\in F$,
contrary to the hypothesis.)</p>
|
2,797,717 | <p>Prove that exist function $\varphi :\left( {0,\varepsilon } \right) \to \mathbb{R}$ such that
$$\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right) = 0,\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right)\ln x = - \infty .$$
I think $\varphi \left( x \right) = \frac{1}{{\ln \left( {\ln \left( {\Gamma \left( x \right)} \right)} \right)}}$ is fine. But i can't check that.
If we replace $ln$ by $f\left( x \right)$ such that $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \infty$, can we find $\varphi$?</p>
| Ceeerson | 427,680 | <p>I guess you want something that shrinks to zero, but does so slower than $\ln$ grows. So what about $$\varphi(x) = |\ln(x)|^{-1/2}$$ I think that should work.</p>
|
3,800,521 | <p>Let <span class="math-container">$x=\tan y$</span>, then
<span class="math-container">$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$</span></p>
<p>Given answer is <span class="math-container">$0$</span></p>
<p>What’s wrong here?</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p><span class="math-container">$$\sin^{-1}\dfrac{2(-10)}{1+(-10)^2}=\sin^{-1}\left(-\dfrac{20}{101}\right)$$</span></p>
<p><span class="math-container">$$\tan^{-1}\dfrac{2(-10)}{1-(-10)^2}=\tan^{-1}\dfrac{20}{99}=u(\text{say})$$</span></p>
<p><span class="math-container">$\implies\dfrac\pi2>u>0$</span></p>
<p><span class="math-container">$\sec u=+\sqrt{1+\left(\dfrac{20}{99}\right)^2}=\dfrac{101}{99}$</span></p>
<p><span class="math-container">$\implies\sin u=\dfrac{\tan u}{\sec u}=?$</span></p>
<p><span class="math-container">$u=\sin^{-1}?$</span></p>
|
1,240,212 | <blockquote>
<p>How to find the degree of an extension field ?</p>
</blockquote>
<p>Let $f:=T^3-T^2+2T+8\in\mathbb Z[T]$ and $\alpha$ be the real root of $f$. Why is then $\mathbb Q(\alpha)$ is a number field of degree $3$ ?</p>
<p>I've seen somewhere that $[\mathbb Q(r):\mathbb Q]\le n$ if $r$ is a root of an irreducible polynomial with coefficients in $\mathbb Q$ of degree $n$. What does it change in my case, if the extension field would contain also the other roots, they're also roots of the polynomial $f$, how does the degree increase ?</p>
<p>Obviously, by finding an element in $\mathbb Q(\alpha)$, which is not in $\mathbb Q$, the degree cannot be $1$, so it remains to show that, it is also not $2$. Or is there a better way, can we find $3$ field embeddings ?</p>
| egreg | 62,967 | <p>The polynomial is irreducible over the rationals, because its possible rational roots are to be found among $\pm1$, $\pm2$, $\pm4$ and $\pm8$. A direct check shows these numbers are not roots.</p>
<p>Since the polynomial has degree $3$, reducibility over $\mathbb{Q}$ coincides with having a rational root.</p>
<p>So, if $\alpha$ is a root of the polynomial, $f$ is its minimum polynomial and it's a standard result that the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ equals the degree of the minimum polynomial.</p>
|
1,178,361 | <p>The surface with equation $z = x^{3} + xy^{2} $ intersects the plane with equation $2x-2y = 1$ in a curve. What is the slope of that curve at $x=1$ and $ y = \frac{1}{2} $</p>
<p>So I put $ x^{3} + xy^{2} = 2x - 2y - 1 $</p>
<p>We have $ x^{3} + xy^{2} - 2x + 2y + 1 $</p>
<p>Do I then differentiate wrt x and y simultaneously?</p>
<p>I know how to differentiate at a point with directional derivatives. But how do I go about the above question considering the fact that direction isn't mentioned.</p>
<p>Maybe I'm going completely down the wrong route... any help is hugely appreciated !</p>
| drhab | 75,923 | <p>Let $c\in C$ and $c\notin D$. </p>
<p>Since $f$ is onto the set $f^{-1}(\{c\})\subset f^{-1}(C)$ is not empty and its elements do not belong to $f^{-1}(D)$ so that $f^{-1}(C)\neq f^{-1}(D)$.</p>
<p>Of course this also works if $d\notin D$ and $d\in D$ and proved is now $C\neq D\implies f^{-1}(C)\neq f^{-1}(D)$. So the statement is true.</p>
|
4,058,884 | <p>I have an orthonormal basis <span class="math-container">${\bf{b}}_1$</span> and <span class="math-container">${\bf{b}}_2$</span> in <span class="math-container">$\mathbb{R}^2$</span>. I want to find out the angle of rotation. I added a little picture here. I essentially want to find <span class="math-container">$\theta$</span></p>
<p><a href="https://i.stack.imgur.com/YH2Vt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YH2Vt.png" alt="enter image description here" /></a></p>
<p>I know that I can compute the angle between two vectors but then there are <span class="math-container">$4$</span> combinations here</p>
<ul>
<li><span class="math-container">$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$</span></li>
<li><span class="math-container">$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$</span></li>
</ul>
<p>How would one know which angle is correct? Importantly, here I used the labels <span class="math-container">$b_1$</span>, <span class="math-container">$b_2$</span> in the same order as <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> but that's not necessarily the same order geometically!</p>
| David K | 139,123 | <p>You have drawn the angle <span class="math-container">$\theta$</span> as an arc from <span class="math-container">${\bf e}_1$</span> to <span class="math-container">${\bf b}_1,$</span>
so apparently you already know intuitively that this is the angle you want,
implying that you want you use the angle
<span class="math-container">$\theta_{11} = \arccos( {\bf b}_1^\top {\bf e}_1).$</span>
But this is only half the story.</p>
<p>You are rotating both basis vectors in the same direction in the same plane, so you will always find that</p>
<p><span class="math-container">$$
\cos(\theta_{11}) = {\bf b}_1^\top {\bf e}_1 = {\bf b}_2^\top {\bf e}_2 = \cos(\theta_{22}).
$$</span></p>
<p>Choose either one, it's the same either way.</p>
<p>Another way to see this is to observe from the figure that the pair of vectors <span class="math-container">$({\bf e}_2, {\bf b}_2)$</span> is just the
pair of vectors <span class="math-container">$({\bf e}_1, {\bf b}_1)$</span>
rotated by a right angle counterclockwise.
The angle between each pair of vectors is the same.</p>
<p>Now for the other half of the story:</p>
<p>When you take the arc cosine, you get an angle <span class="math-container">$\theta$</span> in the range
<span class="math-container">$0 \leq \theta \leq \pi,$</span>
indicating that you can rotate <span class="math-container">${\bf e}_1$</span> and <span class="math-container">${\bf e}_2$</span>
to <span class="math-container">${\bf b}_1$</span> and <span class="math-container">${\bf b}_2$</span> by rotating them through <span class="math-container">$\theta$</span> radians
in either the counterclockwise or clockwise direction.
But this result does not tell you whether the rotation is clockwise or counterclockwise.</p>
<p>By convention, we usually want <span class="math-container">$\theta$</span> to be positive for a counterclockwise rotation, negative for a clockwise rotation.</p>
<p>Assuming the vectors <span class="math-container">${\bf e}_1$</span> and <span class="math-container">${\bf e}_2$</span> are arranged in the usual counterclockwise orientation as shown in your figure,
a counterclockwise rotation more than <span class="math-container">$0$</span> but less than <span class="math-container">$\pi$</span> radians will put
<span class="math-container">${\bf b}_1$</span> on the same side of <span class="math-container">${\bf e}_1$</span> as <span class="math-container">${\bf e}_2$</span>,
so <span class="math-container">${\bf b}_2^\top {\bf e}_1$</span> will be positive.
With a clockwise rotation, <span class="math-container">${\bf b}_2^\top {\bf e}_1$</span> will be negative.
So a rule to decide the direction of rotation is:</p>
<ul>
<li>Counterclockwise if <span class="math-container">${\bf b}_2^\top {\bf e}_1 > 0$</span>,
so let <span class="math-container">$\theta = \arccos({\bf b}_1^\top {\bf e}_1).$</span></li>
<li>Clockwise if <span class="math-container">${\bf b}_2^\top {\bf e}_1 < 0$</span>,
so let <span class="math-container">$\theta = -\arccos({\bf b}_1^\top {\bf e}_1).$</span></li>
</ul>
<p>If <span class="math-container">${\bf b}_2^\top {\bf e}_1 = 0$</span> then either direction of rotation has the same result, so you can choose arbitrarily.</p>
<p>Note that in general you will find that
<span class="math-container">${\bf b}_2^\top {\bf e}_1 = - {\bf b}_1^\top {\bf e}_2.$</span></p>
|
2,646,363 | <p>Let $A_1, A_2, \ldots , A_{63}$ be the 63 nonempty subsets of $\{ 1,2,3,4,5,6 \}$. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_i$. Then what is the value of $\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})$?</p>
<p>Here is the solution </p>
<p>For size 1: sum of the elements, which is 21
For size 2: $ 1 \cdot (2 + 3 + 4 + 5 + 6) = 20 $, $ 2 \cdot (3 + 4 + 5 + 6) = 36 $, $ 3 \cdot (4 + 5 + 6) = 45 $, $ 4 \cdot (5 + 6) = 44 $, $ 5 \cdot 6 = 30 $. Sum is 175.
For size 3: Those with least element 1: $ 6, 8, 10, 12, 12, 15, 18, 20, 24, 30 = 155 $. Those with least element 2: $ 24, 30, 36, 40, 48, 60 = 238 $. Those with least element 3: $ 60 + 72 + 90 = 222 $. Those with least element 4: only one possible subset, which is $ \{4, 5, 6\} $, the $ \pi $ of which is 120. The total sum here is 735.
For size 4: Least element 1: $ 24 + 30 + 36 + 40 + 48 + 60 + 60 + 72 + 90 + 120 = 580 $; least element 2: $ 120 + 144 + 180 + 240 + 360 = 1044 $; least element 3: only one, which is $ 3 \cdot 4 \cdot 5 \cdot 6 = 360 $. The total sum here is 1984.
For size 5: Exclude each one individually to get $ 720 + 360 + 240 + 180 + 144 + 120 = 1764 $
For size 6: $ 6! = 720 $</p>
<p>The final answer is $ 21 + 175 + 735 + 1984 + 1764 + 720 = \boxed{5399} $</p>
<p>Is there any shorter way for doing this ?</p>
<p>Thank a lot </p>
| Community | -1 | <p>I have a way of doing it, but for some reason I don't get the same result that you've got.</p>
<p>Generally, if you have a finite set $A$ of numbers, and you want $\sum_{X\subseteq A}\prod_{x\in X}x$, the result will be $\prod_{x\in A}(x+1)$.</p>
<p>In your case it will be $(1+1)(2+1)(3+1)(4+1)(5+1)(6+1)=2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7=5040$: take away the empty product for $X=\emptyset, \prod_{x\in\emptyset}x=1$ and you will get the final result $5039$.</p>
<p><strong>Proof</strong>: Induction on $n=|A|$. For $n=0$ the equality holds trivially. Assume it holds for all sets of size $n$. Let $A$ be a set of size $n+1$ and let $a\in A$ and $B=A\setminus\{a\}$. Now, we can break up all subsets of $A$ into those that contain $a$ and those that don't:</p>
<p>$$\sum_{X\subseteq A}\prod_{x\in X}x=\sum_{X\subseteq B}\prod_{x\in X}x+\sum_{X\subseteq B}a\prod_{x\in X}x=(a+1)\sum_{X\subseteq B}\prod_{x\in X}x=(a+1)\prod_{x\in B}(x+1)\text{ (inductive hypothesis) }=\prod_{x\in A}(x+1)$$</p>
|
3,101,098 | <p>From 11, 12 in the book Logic in Computer Science by M. Ryan and M. Huth:</p>
<p>**</p>
<blockquote>
<p>"What we are saying is: let’s make the assumption of ¬q. To do this,
we open a box and put ¬q at the top. Then we continue applying other
rules as normal, for example to obtain ¬p. But this still depends on
the assumption of ¬q, so it goes inside the box. Finally, we are ready
to apply →i. It allows us to conclude ¬q → ¬p, but that conclusion no
longer depends on the assumption ¬q. Compare this with saying that ‘If
you are French, then you are European.’ The truth of this sentence
does not depend on whether anybody is French or not. Therefore, we
write the conclusion ¬q → ¬p outside the box."</p>
</blockquote>
<p>**</p>
<p>My question is about the scope of assumptions in propositional logic and proving techniques. I am not sure I fully understand what this text is trying to say. </p>
<p>How can an assumption only have scope inside the box, but once you finish what you want to prove it is no more part of the assumption box and is accessible universally in the proof? WHY is this possible? Why does it not break things in the proof? This looks too convenient and random.</p>
<p>Secondly, I do not understand the French and European example connection to what is written in this text. If somebody could please connect this example to what the author is actually trying to explain through this. </p>
| Dan Christensen | 3,515 | <p>I don't find the "box" analogy to be very helpful. Just understand that, in classical logic:</p>
<p>(1) If you assume only that proposition P is true and you are subsequently able to prove that proposition Q is true, then you can conclude P implies Q. Having done so, you can no longer assume that P is true. That is the standard practice in formal proofs.</p>
<p>(2) If you assume only that proposition P is true and are subsequently able to prove that proposition Q is true <em>and</em> that it is false (an contradiction), then you can conclude P is false. </p>
|
143,324 | <p>I want to know how to simplify the following expression by using the fact that $\sum_{i=0}^\infty \frac{X^i}{i!}=e^X$. The expression to be simplified is as follows:</p>
<p>$$\sum_{i=0}^{\infty} \sum_{j=0}^i \frac{X^{i-j}}{(i-j)!} \cdot \frac{Y^j}{j!}\;,$$ where $X$ and $Y$ are square matrices (not commutative). (That is, $X\cdot Y \neq Y \cdot X$).</p>
| Community | -1 | <p>Multiply and divide the innermost term by $\displaystyle i!$ and use binomial theorem. Move the mouse over the gray area to get the answer.</p>
<blockquote class="spoiler">
<p>This gives us $$\displaystyle \sum_{i=0}^{\infty} \frac1{i!} \sum_{j=0}^{i} \frac{x^{i-j} y^j i!}{(i-j)!j!} = \sum_{i=0}^{\infty} \frac1{i!} \sum_{j=0}^{i} \binom{i}{j} x^{i-j} y^j = \sum_{i=0}^{\infty} \frac{(x+y)^i}{i!} = e^{x+y}$$ where $\displaystyle \sum_{j=0}^{i} \binom{i}{j} x^{i-j} y^j = (x+y)^i$ from binomial theorem.</p>
</blockquote>
|
1,222,064 | <p>Given is an ellipse with $x=a\cos(t),~~y=b\sin(t)$</p>
<p>I do this by using $S=|\int_c^d x(t)y'(t) dt|$, so calculating the area regarding the vertical axis.
Since $t$ runs from $0$ to $2\pi$ I figured I only had to calculate it from $c=\pi/2$ to $d=3\pi/2$ and then this times $2$. But when I integrate over those I get zero...</p>
<p>My steps:
\begin{align*}
0.5S & = |\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d ab\cos^2(t) dt|\\
S & = 2ab|\int_c^d 2\cos(t)*(-\sin(t)) dt|\\
S & = 4ab|\int_c^d -\cos(t)\sin(t) dt|
\end{align*}</p>
<p>On those bounds $\cos(t)$ is zero, so how will this work?!</p>
<p>I notice that to integrate $\cos^2(x)$, most use $\cos^2x = 1/2\cos(2x) + 1/2 $ and thus find $1/4 sin(2x) + 1/2 x$. </p>
<p>This works for those boundaries (except for a factor 2 somewhere ???), but how do I know NOT to use $\cos(x)\sin(x)$ ?!</p>
| neslrac | 1,124,426 | <p>The "easy" one is the perspective divide in matrix form, which has to follow the "awful" one.</p>
<p>The "easy" matrix shows the right column starting with <span class="math-container">$x' = x_c$</span>. x, y and z all have identity; only w is z/D. But with contradicting indices (s, c, slash) and an undefined <code>D</code> (for depth?) I don't see much sense in this matrix. The thing itself - divide by z - is important, of course.</p>
<p>The complicated (real) perspective matrix has a "1" on the bottom row. This stores the z value aka depth into the w coordinate, while the x, y and z get scaled into clip space, according to the angle of view and the near and far planes.</p>
<p>But the division by z can't be coded in a matrix. This perspective divide is done afterwards. Maybe the "easy" matrix is for this step, but it seems overkill. The perspective divide is</p>
<p><span class="math-container">$x_{ndc} = x_{clip} / w_{clip}$</span></p>
<p>and same for y and z. No need for a matrix, only the z value saved into the w coordinate by the perspective matrix. In OpenGl the variable <code>gl_FragCoord</code> holds the current x, y and z values ie. NDC scaled to window size (for x and y) or depth range (for x). The 4th coordinate contains <code>1/w</code>. The "easy" matrix would make more sense with a <span class="math-container">$w_s = 1 / z_c$</span>. Dividing <span class="math-container">$w_c$</span> by itself would be just another way to say "1".</p>
<p>So the "google" matrix does the perspective projection of a geometric frustum into a normalized cube, but minus the final division, which is only prepared.</p>
<p>It is not a question of converting, but of doing the second first and then replacing the first with a simple perspective divide.</p>
|
2,414,965 |
<p>I am following along and reading this notes: <a href="https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf" rel="nofollow noreferrer">https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf</a></p>
<p>I am having trouble understanding why we necessarily have $e_n=d_n+1$,
$d_{n+1}= d_{n+2} =···= 2$
and $e_{n+1} = e_{n+2} = ··· = 0$
when $d_n > e_n$. It would much appreciated if someone can guide me through this. </p>
| eyeballfrog | 395,748 | <p>All numbers have at least one ternary expansion. For example,
\begin{eqnarray}
\frac{1}{2} &=& 0.11111111... \\
\frac{2}{9} &=& 0.02000000... \\
\frac{\pi}{8} &=& 0.21001211... \\
\gamma &=& 0.12012021...
\end{eqnarray}
However, some numbers have two ternary expansions
\begin{eqnarray}
\frac{2}{9} &=& 0.02000000... = 0.01222222...\\
\frac{19}{27} &=& 0.20100000... = 0.20022222...\\
\frac{100}{243} &=& 0.10201000... = 0.10200222...
\end{eqnarray}
Note that whenever a number has two ternary expansions, it always has one that ends in $222222...$ and one that ends in $000000...$, with the preceding digit being one larger for the $000000...$ expansion. That's what the article is talking about.</p>
<p>You'll also notice that such numbers are always fractions with a power of 3 in the denominator. This is no coincidence, as they're exactly the numbers with terminating ternary expansions. As for why the other expansion has $222222...$, consider that $\sum_1^\infty (2/3)^n = 1$.</p>
|
2,414,965 |
<p>I am following along and reading this notes: <a href="https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf" rel="nofollow noreferrer">https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf</a></p>
<p>I am having trouble understanding why we necessarily have $e_n=d_n+1$,
$d_{n+1}= d_{n+2} =···= 2$
and $e_{n+1} = e_{n+2} = ··· = 0$
when $d_n > e_n$. It would much appreciated if someone can guide me through this. </p>
| Community | -1 | <p>As in the case of $\frac 13$ there are two ternary expressions. In this example $n $ as they have defined it $=1$. That is, the first place they don't agree is the first place after the decimal. ..</p>
<p>So, $0.e_1e_2\dots e_n=0.10\dots0$ and $0.d_1d_2\dots d_n=0.02\dots 2\dots $. Looking at this you should be able to see the $e_n's $ and $d_n's $ are as described. ..
In particular, $e_1=1=0+1=d_1+1$, and after that all the $e_n's $ are $0$, and all the $d_n's $ are $2$...</p>
<p>It turns out this is always the situation when ambiguity occurs (that is, there is more than one expression, in ternary,for a number in the cantor set )... </p>
|
4,289,129 | <p>Let <span class="math-container">$H$</span> be a group with identity <span class="math-container">$1_H$</span> that is generated by 2 elements <span class="math-container">$a,b$</span> that commute (<span class="math-container">$ab=ba$</span>) and where each has at most order <span class="math-container">$3$</span>. In symbols (I hope I translated correctly):</p>
<p><span class="math-container">$$H=\langle a,b\rangle \ \text{, where} \ a^3=b^3=1_H=a^{-1}b^{-1}ab$$</span></p>
<p>Assuming <span class="math-container">$H$</span> has exactly order 9 and assuming <span class="math-container">$\{a,b,1_H\}$</span> are all distinct, <strong>what is <span class="math-container">$H$</span> isomorphic to?</strong> (<a href="https://groupprops.subwiki.org/wiki/Groups_of_order_9" rel="nofollow noreferrer">order 9 possibilities</a> are <span class="math-container">$\mathbb Z_9$</span> and <span class="math-container">$\mathbb Z_3 \times \mathbb Z_3$</span>)</p>
<hr />
<p><strong>For order 9</strong>: Assuming <span class="math-container">$H$</span> is of order 9, I believe <span class="math-container">$H$</span> is isomorphic to <span class="math-container">$\mathbb Z_3 \times \mathbb Z_3$</span>.</p>
<p>Construct a map <span class="math-container">$\gamma: \mathbb Z_3^2 \to H$</span>, <span class="math-container">$\gamma(c \times d)=b^ca^d$</span>, where <span class="math-container">$c,d \in \{0,1,2\}$</span>.</p>
<p>Show <span class="math-container">$\gamma$</span> is bijective: obvious</p>
<p>Show <span class="math-container">$\gamma$</span> is a homomorphism: For each <span class="math-container">$c,d,e,f \in \{0,1,2\}$</span>, we must show that</p>
<p>(<strong>notation</strong>: instead of <span class="math-container">$(c,d) \in \mathbb Z_3^2$</span>, I'll say <span class="math-container">$c \times d$</span>)</p>
<p><span class="math-container">$$\gamma(c \times d + e \times f) = \gamma(c \times d) \gamma(e \times f)$$</span></p>
<p>I believe this is equivalent to</p>
<p><span class="math-container">$$b^{c+d}a^{e+f} = b^ca^d b^ea^f$$</span></p>
<p>Finally, because <span class="math-container">$a^db^e=b^ea^d$</span> for <span class="math-container">$d,e \in \{0,1,2\}$</span>, we have that</p>
<p><span class="math-container">$$RHS = b^ca^d b^ea^f = b^cb^e a^da^f = LHS$$</span></p>
| Asinomás | 33,907 | <p>Every element of <span class="math-container">$H$</span> is a finite product of <span class="math-container">$a,b$</span> and <span class="math-container">$a^{-1}$</span> and <span class="math-container">$b^{-1}$</span>. Fortunately <span class="math-container">$a^{-1}$</span> is <span class="math-container">$a^2$</span> and <span class="math-container">$b^{-1}$</span> is <span class="math-container">$b^2$</span>. It follows every element of <span class="math-container">$H$</span> is a finite product of <span class="math-container">$a$</span>'s and <span class="math-container">$b$</span>'s. Since <span class="math-container">$a$</span> and <span class="math-container">$b$</span> commute we can show by induction over the number of factors that every expression is of the form <span class="math-container">$a^nb^m$</span>. To do this take an expression <span class="math-container">$(x_1x_2\dots x_s)x_{s+1}$</span> and see it is equal to <span class="math-container">$(a^nb^m)x_{s+1}$</span> and do the case when <span class="math-container">$x_{s+1}$</span> is <span class="math-container">$a$</span> or <span class="math-container">$b$</span>.</p>
<p>Finally since every <span class="math-container">$a^n$</span> is equal to one of <span class="math-container">$1,a,a^2$</span> and every <span class="math-container">$b^m$</span> is equal to one of <span class="math-container">$1,b,b^2$</span> you will be done.</p>
|
1,920,994 | <p>My calculus teacher gave us this interesting problem: Calculate</p>
<p>$$ \int_{0}^{1}F(x)\,dx,\ $$ where $$F(x) = \int_{1}^{x}e^{-t^2}\,dt $$</p>
<p>The only thing I can think of is using the Taylor series for $e^{-t^2}$ and go from there, but since we've never talked about uniform convergence and term by term integration, I suppose that there is an easier way to do this.</p>
| grand_chat | 215,011 | <p>Hint: First (1) flip the limits on the inner integral (assuming those limits are correctly stated), then (2) switch the order of integration:
$$
\int_{x=0}^1\int_{t=1}^x e^{-t^2}dt\,dx
\stackrel{(1)}=-\int_{x=0}^1\int_{t=x}^1 e^{-t^2}dt\,dx
\stackrel{(2)}=-\int_{t=0}^1\int_{x=0}^t e^{-t^2}dx\,dt
=-\int_{t=0}^1t e^{-t^2}dt
$$</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| dxiv | 291,201 | <blockquote>
<p>you can also repeat the numbers</p>
</blockquote>
<p>Wonder if that means $\,11,5+13,5+5=30\,$ (where the $\,,\,$ comma works as
decimal separator).</p>
|
2,854,671 | <p>Consider a $n\times n$ Hankel Matrix</p>
<p>$$
H = \begin{bmatrix}
x_{1} & x_{2} & \dots & x_{n} \\
x_{2} & x_{3} & \dots & x_{n+1} \\
\vdots \\
x_{n} & x_{n+1} & \dots & x_{2n}
\end{bmatrix}
$$
, where all $x_i \in \mathbb{Z}_p = \{ 0,\dots,p-1 \}$, where $p$ is prime.</p>
<p>What is the most efficient way to test whether the matrix is invertible or not. More concretely:
Is there a more efficient than computing the determinant?
If not, is there a more efficient way of computing the determinant of such a matrix?</p>
| Community | -1 | <p>There is an algorithm called Levinson Recursion for Toeplitz matrices which is <span class="math-container">$\mathcal{O}(n^{2})$</span>. There is exists a similar algorithm for Hankel matrices called <a href="https://ac.els-cdn.com/S0024379510006294/1-s2.0-S0024379510006294-main.pdf?_tid=2586b0e6-a947-4789-8049-4a102e630b12&acdnat=1531854737_7fa5e3ff800b926b927a5d3cab9d0d97" rel="nofollow noreferrer">Hankel Recursion.</a> It appears to be based on the Lanczos algorithm. People don't generally compute determinants the normal way. E.g. they form a matrix decomposition since the following</p>
<p><span class="math-container">$$ A = LU \implies det(A) = det(LU) =det(L)det(U)$$</span></p>
<p>after this is done.</p>
<p><span class="math-container">$$ det(L)det(U) =\prod_{i=1}^{n} l_{ii} \prod_{i=1}^{n} u_{ii} $$</span></p>
<p>Similarly with the QR decomp</p>
<p><span class="math-container">$$ A =QR \implies det(A) = det(Q)det(R) $$</span></p>
<p>since the determinant of <span class="math-container">$ Q $</span> is 1
<span class="math-container">$$ det(A) = 1 \cdot \prod_{i=1}^{n} r_{ii} $$</span></p>
<p>However, in general, you don't want to use determinant to see if it is invertible. Just extra steps...</p>
|
1,555,697 | <p>So, I'm working out one of my assignments and I'm a little bit stuck on this problem:</p>
<blockquote>
<p>A fish store is having a sale on guppies, tiger barbs, neons,
swordtails, angelfish, and siamese fighting fish (6 kinds). How many
ways are there to choose 24 fish with at least 1 guppy, at least 2
tiger barbs, at least 3 neons, exactly 1 swordtail, at least two
angelfish, and no more than 3 siamese fighting fish?</p>
</blockquote>
<p>I feel like the way to approach this problem is to use r-combinations with repetition (because the fish store isn't running out of any of the fish) and subtract the combinations that I don't want. Here's what I've got so far:</p>
<p>We need <em>at least</em> 1 guppy, 2 tiger barbs, 3 neons, and 2 angelfish. If we set those aside (all 8 of those), we're down to $24-8$, or 16 remaining fish to choose. Add the 1 swordtail, and we only need to choose $24-9$ or 15 fish out of 5 ($(6-1)$ because there were 6 kinds, but we can only have 1 of the swordtail).</p>
<p>All this accounted for, utilizing r-combination with repetition, $\binom{n+r-1}{r}$:
$$\binom{15+5-1}{15}=3,876$$</p>
<p><strong>BUT</strong> I can't have more than 3 siamese fighting fish! How do I subtract all the groups with more than 3 siamese fighting fish?</p>
<p>Any help would be much appreciated! Also, if I've made any other errors, please point those out and I'll edit the question appropriately. We covered it in class, but it went so quick I didn't get it down in my notes. Thanks!</p>
| Community | -1 | <p>Here is an approach using generating functions. The generating function for the guppies
is $g+g^2+g^3+\cdots ={g\over 1-g}$. In the same way, the generating functions for the other
5 types of fish are ${t^2\over 1-t}$ (tiger barbs), ${n^3\over 1-n}$ (neons), $sw$ (swordtails),
${a^2\over 1-a}$ (angelfish), and $1+s+s^2+s^3$ (siamese fighting fish).
The generating function in six variables is
$$\Phi(g,t,n,sw,a,s)={g\, t^2\, n^3\, sw\, a^2\, (1+s+s^2+s^3)\over (1-g)(1-t)(1-n)(1-a)},$$
whose coefficients give the counts for all acceptable purchases of fish. </p>
<p>Since we only care that the total number of fish is 24, and not the particular makeup
of types, we may look at a simpler generating function where all six variables
are replaced by the variable "$f$":
\begin{eqnarray*}G(f)&=&\Phi(f,f,f,f,f,f)\\[5pt]
&=&{f^9(1+f+f^2+f^3)\over (1-f)^4}\\[5pt]
&=&\sum_{j\geq 0}{j+3\choose 3} (f^{9+j}+f^{10+j}+f^{11+j}+f^{12+j}).
\end{eqnarray*}</p>
<p>Extracting the coefficient of $f^{24}$ in $G(f)$ gives
${18\choose 3}+{17\choose 3}+{16\choose 3}+{15\choose 3}=2511.$</p>
<p>This isn't as easy as mathochist's solution, but you can
use the same idea for more complicated problems. </p>
|
539,027 | <p>How many ways are there to distribute $18$ different toys among $4$ children?</p>
<ol>
<li><p>without restrictions</p></li>
<li><p>if $2$ children get $7$ toys each and $2$ children get $2$ toys each.</p></li>
</ol>
<p>For $1$ since toys are different, then there are $4^{18}$ ways to distribute .</p>
<p>for $2$, Im kind of stuck. I know I have two do it in $2$ cases. Can someone help me?</p>
| Community | -1 | <ol>
<li><p>$4^18$ works.</p></li>
<li><p>Choose 2 of them. Then choose 7 toys. There are 7^2 ways to distribute them. Then choose 2 toys and give 1 to 1 child and the other to the other. So we have ${4 \choose 2}{18 \choose 7}7^2{11 \choose 2}2^2$. </p></li>
</ol>
<p>I know that Andres' answer assumes that the children each get 7 and they each get 2. Mine assumes that we don't care about the other 9 lost toys. Hope you learn from both examples.</p>
|
1,105,971 | <p>I've got two independent bernoulli distributed random variables $X$ and $Y$ with parameter $\frac{1}{2}$. Based on those I define two new random variables </p>
<p>$X' = X + Y , E(X') = 1$</p>
<p>$Y' = |X - Y|, E(Y') = \sum_{x=0}^1\sum_{y=0}^1|x-y|*P(X=x)*P(Y=y) = \frac{1}{2}$ </p>
<p><strong>How can I calculate E(X'Y')?</strong> As X' and Y' and not independent (e.g. it is impossible for Y' to assume 1 if X' is 0) I must not use the sum of all possible outcomes multiplied with their likelihood as I did for $E(Y')$ but I cannot find another formula to calculate the expected value.</p>
| tau_cetian | 206,595 | <p>Someone else can answer more authoritatively for the general case, but for a small experiment such as this one can we build up all possible values of $X' \cdot Y'$ from the four possible outcomes of $(X,Y)$?</p>
<p>$$
\begin{array}{l|l|l|l|l}
(X,Y) & X' & Y' & X' \cdot Y' & P(\ \ ) \\
\hline
(0,0) & 0 & 0 & 0 & \frac{1}{4} \\
(0,1) & 1 & 1 & 1 & \frac{1}{4} \\
(1,0) & 1 & 1 & 1 & \frac{1}{4} \\
(1,1) & 2 & 0 & 0 & \frac{1}{4}
\end{array}
$$</p>
<p>So $P(X'Y'=0) = P(X'Y'=1) = \frac{1}{2}$ and $E(X'Y') = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}$.</p>
|
1,118,259 | <p>Consider a sphere of radius $a$ with 2 cylindrical holes of radius $b<a$ drilled such that both pass through the center of the sphere and are orthogonal to one another. What is the volume of the remaining solid?</p>
<p>Can someone help me at least setting up the integral? I know that there is a similar problem but it was a sphere with one hole. </p>
| Christian Blatter | 1,303 | <p>(<strong>Edit:</strong> This answer does not cover the case of large holes; see the comments by coffeemath.)</p>
<p>A hint:</p>
<p>If $b<{a\over\sqrt{2}}$ then the intersection $B$ of the two cylinders is completely in the interior of the sphere. In this case you can proceed as follows:</p>
<p>Do the problem with $1$ hole, then compute the volume of $B$, and use inclusion-exclusion.</p>
<p>For the computation of ${\rm vol}(B)$ consider $$B_1:=\{(x,y,z)\>|\>0\leq y\leq x, \ x^2+z^2\leq b^2\}\ ,$$
whose volume is ${1\over16}$ of the volume of $B$.</p>
|
405,205 | <p>Some friends and I have a family of polynomials (in one variable) with rational coefficients and we would very much like a formula for them. Grasping at straws, we computed many examples and wrote them in the basis of binomial coefficients. Specifically, I mean the basis <span class="math-container">$\left\{\binom{x}{0},\binom{x}{1},\binom{x}{2},\ldots\right\}$</span> of the ring of rational polynomials over <span class="math-container">$x$</span>. We were surprised to find that our polynomials all expand positively and integrally in that basis.</p>
<p>That is, if one of our polynomials <span class="math-container">$p(x)$</span> of degree d is written as <span class="math-container">$\sum_{k=0}^da_k\binom{x}k$</span>, each of the <span class="math-container">$a_k$</span> is a nonnegative integer.</p>
<p>We can't make much sense of the coefficients. But we're wondering if the positivity of our polynomials in the binomial coefficients basis is a sort of "shadow" of some other stronger phenomenon. Suppose there were some other basis <span class="math-container">$\{b_0(x),b_1(x),\ldots\}$</span> such that each <span class="math-container">$b_i(x)$</span> expands nonnegatively in the basis of binomial coefficients. If our polynomials expand nonnegatively (and in some understandable way) in the basis <span class="math-container">$\{b_0(x),b_1(x),\ldots\}$</span>, that's our desired formula.</p>
<p>If you're thinking "They're still grasping at straws", you're right. But it can't hurt to ask:</p>
<p>What bases for the polynomial ring should we try? Are there some well-known bases that expand positively in the binomial coefficients basis?</p>
| Ira Gessel | 10,744 | <p>The basis <span class="math-container">$\binom{x}{0}$</span>, <span class="math-container">$\binom{x+1}{1}$</span>, <span class="math-container">$\binom{x+2}{2}$</span>, <span class="math-container">$\dots$</span> has this property. More generally, if <span class="math-container">$i$</span> is a nonnegative integer then <span class="math-container">$\binom{x+i}{j}$</span> is a nonnegative linear combination of the <span class="math-container">$\binom{x}{k}$</span>.</p>
<p>A useful basis for polynomials of degree at most <span class="math-container">$n$</span> with this property is <span class="math-container">$\binom{x}{n}$</span>, <span class="math-container">$\binom{x+1}{n}$</span>, <span class="math-container">$\dots$</span>, <span class="math-container">$\binom{x+n}{n}$</span>.</p>
|
4,460,778 | <p>I have a more general question on the importance of fixed-point theorems. In mathematics youre being introduced to so many fixed-point theorems but i still could not figure out why they are so important. Why would be a simply looking statement as <span class="math-container">$f(x)=x$</span> be so important. I would appreciate any answer. Thanks in advance. On wikipedia it says nothing about the importance, contextualisation of theorems in mathematics is sometimes not given.</p>
| Michael Greinecker | 21,674 | <p>One important reason is that the existence of solutions to systems of equations are equivalent to fixed-points of appropriate functions. Suppose you want to show <span class="math-container">$f(x)=0$</span> for some <span class="math-container">$x$</span>. This is equivalent to <span class="math-container">$f(x)+x=x$</span>, which means that the function <span class="math-container">$F$</span> defined by <span class="math-container">$F(x)=f(x)+x$</span> has a fixed-point.</p>
<p>If you want to discuss properties of solutions to equations that you might not be able to solve explicitly, it is useful to know that such solutions exist in the first place.</p>
|
4,460,778 | <p>I have a more general question on the importance of fixed-point theorems. In mathematics youre being introduced to so many fixed-point theorems but i still could not figure out why they are so important. Why would be a simply looking statement as <span class="math-container">$f(x)=x$</span> be so important. I would appreciate any answer. Thanks in advance. On wikipedia it says nothing about the importance, contextualisation of theorems in mathematics is sometimes not given.</p>
| ml0105 | 135,298 | <p>I second lisyarus' link. Nash's thesis on the existence of Nash equilibria in normal form games relied on the Brouwer Fixed Point Theorem. It was famously dismissed by von Neumann (who did work in both Game Theory and Functional Analysis) as being "just a fixed point theorem." More modern proofs rely on Kakutani's Fixed Point Theorem instead.</p>
<p>Computing Brouwer Fixed Points is also an interesting question. This problem is complete for the complexity class PPAD, as are a number of associated problems in theoretical economics (<a href="https://en.wikipedia.org/wiki/List_of_PPAD-complete_problems" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/List_of_PPAD-complete_problems</a>). Note that Nash's theorem guarantees the existence of Nash equilibria. So deciding whether a normal form game admits a Nash equilibrium is in NP, where every instance has an answer of YES. Thus, computing Nash equilibria is very unlikely to be NP-complete. However, it is not easy to find Nash equilibria. Thus, PPAD-completeness is a weaker notion of intractability than NP-completeness.</p>
|
4,545,364 | <blockquote>
<p>Solve the quartic polynomial :
<span class="math-container">$$x^4+x^3-2x+1=0$$</span>
where <span class="math-container">$x\in\Bbb C$</span>.</p>
<p>Algebraic, trigonometric and all possible methods are allowed.</p>
</blockquote>
<hr />
<p>I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.</p>
<p>I realized there is no any rational root, by the rational root theorem.</p>
<p>The harder part is, WolframAlpha says the factorisation over <span class="math-container">$\Bbb Q$</span> is impossible.</p>
<p>Another solution method can be considered as the quasi-symmetric equations approach. (divide by <span class="math-container">$x^2$</span>).</p>
<p><span class="math-container">$$x^2+\frac 1{x^2}+x-\frac 2x=0$$</span></p>
<p>But the substitution <span class="math-container">$z=x+\frac 1x$</span> doesn't make any sense.</p>
<p>I want to ask the question here to find possible smarter ways to solve the quartic.</p>
| Qiaochu Yuan | 232 | <p>We can look for a difference of squares factorization. Completing the square gives</p>
<p><span class="math-container">$$\left( x^2 + \frac{1}{2} x + c \right)^2 - \left( 2c + \frac{1}{4} \right) x^2 - (c + 2) x - (c^2 - 1)$$</span></p>
<p>and we want to find a value of <span class="math-container">$c$</span> such that the discriminant of the quadratic on the right is equal to zero. This gives</p>
<p><span class="math-container">$$\Delta = (c + 2)^2 - 4 \left( 2c + \frac{1}{4} \right) \left( c^2 - 1) \right) = - 8c^3 + 12c + 5$$</span></p>
<p>which happily has a rational root <span class="math-container">$c = - \frac{1}{2}$</span> (I guess we must be essentially using the <a href="https://en.wikipedia.org/wiki/Resolvent_cubic" rel="noreferrer">resolvent cubic</a> here). This gives us a factorization</p>
<p><span class="math-container">$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} \right)^2 + \frac{3}{4} (x - 1)^2$$</span></p>
<p>which gives a difference of squares factorization</p>
<p><span class="math-container">$$\left( x^2 + \frac{1}{2} x - \frac{1}{2} + \frac{i \sqrt{3}}{2} (x - 1) \right) \left( x^2 + \frac{1}{2} x - \frac{1}{2} - \frac{i \sqrt{3}}{2} (x - 1) \right)$$</span></p>
<p>and we can use the quadratic formula from here; if you want to know what the roots end up looking like you can <a href="https://www.wolframalpha.com/input/?i=roots+of+x%5E4+%2B+x%5E3+-+2x+%2B+1" rel="noreferrer">ask WolframAlpha</a>.</p>
|
4,545,364 | <blockquote>
<p>Solve the quartic polynomial :
<span class="math-container">$$x^4+x^3-2x+1=0$$</span>
where <span class="math-container">$x\in\Bbb C$</span>.</p>
<p>Algebraic, trigonometric and all possible methods are allowed.</p>
</blockquote>
<hr />
<p>I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.</p>
<p>I realized there is no any rational root, by the rational root theorem.</p>
<p>The harder part is, WolframAlpha says the factorisation over <span class="math-container">$\Bbb Q$</span> is impossible.</p>
<p>Another solution method can be considered as the quasi-symmetric equations approach. (divide by <span class="math-container">$x^2$</span>).</p>
<p><span class="math-container">$$x^2+\frac 1{x^2}+x-\frac 2x=0$$</span></p>
<p>But the substitution <span class="math-container">$z=x+\frac 1x$</span> doesn't make any sense.</p>
<p>I want to ask the question here to find possible smarter ways to solve the quartic.</p>
| Anders Kaseorg | 38,671 | <p>Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics:</p>
<p><span class="math-container">$$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \overline a)x^3 + (a\overline a + b + \overline b)x^2 + (a\overline b + \overline ab)x + b\overline b.$$</span></p>
<p>The <span class="math-container">$x^3$</span> coefficient is <span class="math-container">$1 = a + \overline a$</span>, so we must have</p>
<p><span class="math-container">$$a = \frac12 + si$$</span></p>
<p>for some <span class="math-container">$s ∈ ℝ$</span>. The constant term is <span class="math-container">$1 = b\overline b$</span>, so <span class="math-container">$b$</span> lies on the complex unit circle, and we must have</p>
<p><span class="math-container">$$b = \cos θ + i \sin θ = \frac{1 - t^2}{1 + t^2} + \frac{2t}{1 + t^2}i$$</span></p>
<p>where <span class="math-container">$t = \tan \frac θ2 ∈ ℝ$</span>. The <span class="math-container">$x$</span> coefficient is now</p>
<p><span class="math-container">$$1 = a\overline b + \overline ab = \frac{1 - 4st - t^2}{1 + t^2},$$</span></p>
<p>so we must have</p>
<p><span class="math-container">$$s = -\frac t2.$$</span></p>
<p>Finally, the <span class="math-container">$x^2$</span> coefficient is</p>
<p><span class="math-container">$$0 = a\overline a + b + \overline b = \frac{t^4 - 6t^2 + 9}{4t^2 + 4} = \frac{(t^2 - 3)^2}{4t^2 + 4},$$</span></p>
<p>so <span class="math-container">$t = \sqrt3$</span> (or <span class="math-container">$-\sqrt3$</span>, which would give the same solution with the quadratics swapped). Then <span class="math-container">$s = -\frac{\sqrt3}2$</span>, <span class="math-container">$a = \frac{1 - i\sqrt3}2$</span>, <span class="math-container">$b = \frac{-1 + i\sqrt3}2$</span>, and the above factorization becomes</p>
<p><span class="math-container">$$\left(x^2 + \frac{1 - i\sqrt3}2x + \frac{-1 + i\sqrt3}2\right)\left(x^2 + \frac{1 + i\sqrt3}2x + \frac{-1 - i\sqrt3}2\right).$$</span></p>
|
68,145 | <p>All the statements below are considered over local rings, so by regular, I mean a regular local ring and so on;</p>
<p>It is well-known that every regular ring is Gorenstein and every Gorenstein ring is Cohen-Macaulay. There are some examples to demonstrate that the converse of the above statements do not hold. For example, $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy, yz, xz)$ where $k$ is a field, is Gorenstein but not regular, or $k[[x^3, x^5, x^7]]$ is C.M. but not Gorenstein.
Now, here is my question:</p>
<p>I want to know where these examples have come from, I mean, have they been created by the existence of some logical translations to the Algebraic combinatorics (like Stanley did), or even algebraic geometry, or they are as they are and they are some kind of lights that have been descended from heaven to their creators by any reason!</p>
| Hailong Dao | 2,083 | <p>I will argue that the examples you gave are "simplest" in some strong sense, so although they look unnatural, if Martians study commutative algebra they will have to come up with them at some point. </p>
<p>Let's look at the first one $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy,yz,zx)$. Suppose you want </p>
<blockquote>
<p>a $0$-dimensional Gorenstein ring which is <em>not</em> a complete intersection (complete intersections are the cheapest way to get Gorenstein but non-regular, for example $k[[x]]/(x^2)$). </p>
</blockquote>
<p>Then it would look like $A=R/I$, where $R$ is regular and $I$ is of height equals to $\dim R$. If $\dim R=1$ or $2$, then $I$ would have to be a complete intersection, no good (now, the poor Martian may not know this at the begining, but after trying for so long she will have to give up and move on to higher dimension, or prove that result for herself). Thus $\dim R=3$ at least, and we may assume $R=k[[x,y,z]]$ (let's suppose everything are complete and contains a fied). </p>
<p>Since $I$ is not a complete intersection, it must have more than $3$ minimal generators. If it has $4$ then it would be an almost complete intersection, and by an amazing result by Kunz (see this <a href="https://mathoverflow.net/questions/36396/length-of-i-i2-versus-anni-anni2-in-artinian-rings/41527#41527">answer</a>), those are <em>never</em> Gorenstein.</p>
<p>So, in summary, a simplest $0$-dimensional Gorenstein but not complete intersection would have to be $k[[x,y,z]]/I$, where $I$ is generated by at least $5$ generators. At this point our Martian would just play with the simplest non-degenerate generators: quadrics, and got lucky! </p>
<p>(You can look at this from other point of view, Macaulay inverse system or Pfaffians of alternating matrices, see Bruns-Herzog, but because of the above reasons all the simplest examples would be more or less the same, up to some linear change of variables)</p>
<p>On to your second example, $k[[t^3,t^5,t^7]]$. Now very reasonably, our Martian wants </p>
<blockquote>
<p>a one dimensional Cohen-Macaulay ring B that is not Gorenstein. </p>
</blockquote>
<p>Since $\dim B=1$, being a domain would automatically make it Cohen-Macaulay. The most natural way to make one dimensional domains is to use monomial curves, so $B=k[[t^{a_1},...,t^{a_n}]]$. But if $n=2$ or $n=3$ and $a_1=2$ you will run into complete intersections, so $a_1$ would have to be $3$ and $n=3$ at least, and on our Martian goes...</p>
<p>(Again, you can arrive at this example by looking at things like non-symmetric numerical semi-groups, but again you would end up at the same simplest thing). </p>
<p>Reference: I would recommend <a href="http://arxiv.org/abs/math/0209199" rel="nofollow noreferrer">this survey</a> for a nice reading on Gorenstein rings. </p>
|
1,537,676 | <p>$f(x,y)=\begin{cases}
|\frac{y}{x^2}|exp(-|\frac{y}{x^2}|) & , x\ne0\\
0 & , x=0
\end{cases}$</p>
<p>I need to show that $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ the limit does not exist in $(0,0)^T$. I tried to prove it with the sequence criteria but I could not find a good sequence. Polar form doesn't get me further. </p>
| MASL | 266,735 | <p>Choose two paths:</p>
<p>1) $y=e^x-1$. Then $\lim_{(x,y)\to 0^+}f(x,y)=\lim_{x\to0^+}\left|\frac{e^x-1}{x^2}\right|\exp(-\frac{e^x-1}{x^2})=\lim_{x\to 0^+}\left|\frac{1}{x}\right|\exp(-\frac{1}{x})=0$</p>
<p>This is a path for which $y$ reaches zero at the same pace as $x$. You can convince yourself that the limit is then $0$.</p>
<p>2) $y=x^2$, for the which the limit is easily seen to be $e^{-1}$.</p>
<p>As the limit depends on how we take it (the path we folow) it is not well defined.
Therefore, the limit at $(0,0)$ does not exist for this function.</p>
<p>(A third alternative path could be taking the one in 1) and considering the limit as $x\to0^-$. In this case the exponential diverges as well as the multiplicative factor and, thus, so does the limit as well. EDIT: I forgot take the absolute value of the exponent. Hence, this third path won't apply to the function in the problem statement. The other two are still valid though.)</p>
|
3,164,280 | <p>By accident, I find this summation when I pursue the particular value of <span class="math-container">$-\operatorname{Li_2}(\tfrac1{2})$</span>, which equals to integral <span class="math-container">$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$</span>.</p>
<p>Notice this observation</p>
<p><span class="math-container">$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$</span></p>
<p>And using the Taylor series of <span class="math-container">$\ln(1-x^{2})$</span>, I found this summation <span class="math-container">$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$</span>, where <span class="math-container">$H_{n}$</span> is the harmonic-numbers.</p>
<p>If the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$</span> is given, the result can be easily deduced, which is</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$</span></p>
</blockquote>
<p>For the original goal is to calculate <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>, I expect some other approaches to the summation without using the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>. I already knew some famous problem like <a href="https://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk?noredirect=1&lq=1">Euler's Sum</a>, which holds very similar form to this summation, but still in trouble finding the appropriate path.</p>
| user90369 | 332,823 | <p>With this answer I show an indirect method to the wished result of the integral <span class="math-container">$~\int\limits_0^1\frac{\ln(1-x)}{1+x}dx~$</span>, </p>
<p>and <em>indirect</em> means here: It’s used <span class="math-container">$~\text{Li}_2\left(\frac{1}{2}\right)~$</span> <em>without</em> knowing it’s value, only as a catalyst. </p>
<p><em>First a general formula.</em> It’s not difficult to find out, that formally holds:</p>
<p><span class="math-container">$$-\frac{d}{dx}(x+z)^y \sum\limits_{k=1}^\infty\frac{\left(\frac{x+z}{a+z}\right)^k}{k+y} = \frac{(x+z)^y}{x-a}$$</span> </p>
<p>With the integration to <span class="math-container">$x$</span> and using Taylor series for <span class="math-container">$~y~$</span> around <span class="math-container">$~0~$</span> we get:</p>
<blockquote>
<p><span class="math-container">$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{x-a}dx = \sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\text{Li}_{n-k+1}\left(\frac{x+z}{a+z}\right) + C$$</span></p>
</blockquote>
<p>First we use <span class="math-container">$~(n;z;a):=(1;1;1)~$</span> :</p>
<p><span class="math-container">$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = -\int\limits_{-1}^0\frac{\ln(1+x)}{1-x}dx =$</span></p>
<p><span class="math-container">$\displaystyle = -\text{Li}_2\left(\frac{x+1}{2}\right)|_{-1}^0 + \ln(x+1)\text{Li}_1\left(\frac{x+1}{2}\right)|_{-1}^0 = -\text{Li}_2\left(\frac{1}{2}\right) $</span></p>
<p>Our next step is to transform the integral by <em>partial integration</em> :</p>
<p><span class="math-container">$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = (\ln(1-x))(\ln(1+x) - \ln 2)|_0^1 + \int\limits_0^1\frac{\ln(1+x) - \ln 2}{1-x}dx = $</span></p>
<p><span class="math-container">$\displaystyle = 0 + 2\int\limits_0^{1/2}\frac{\ln(1+2x) - \ln 2}{1-2x}dx = -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx$</span></p>
<p>Now we use <span class="math-container">$~(n;z;a):=(1;\frac{1}{2};\frac{1}{2})~$</span> and <span class="math-container">$~\text{Li}_1\left(\frac{1}{2}\right)=\ln 2~$</span> :</p>
<p><span class="math-container">$\displaystyle -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx = -\text{Li}_2\left(x+\frac{1}{2}\right)|_0^{1/2} + \ln\left(x+\frac{1}{2}\right) \text{Li}_1\left(x+\frac{1}{2}\right)|_0^{1/2}$</span></p>
<p><span class="math-container">$\displaystyle = -\frac{\pi^2}{6} + \text{Li}_2\left(\frac{1}{2}\right) + (\ln 2)^2 \enspace\enspace$</span> which is, as found before, the <em>same</em> as <span class="math-container">$~\displaystyle -\text{Li}_2\left(\frac{1}{2}\right)~$</span> . </p>
<p>Comparing both results we get the wished formula.</p>
<p><em>Note:</em> <span class="math-container">$~$</span> Here we see very well that the partial integration leads to the second representation of the result and both representations have as a common base the (yellow marked) general formula.</p>
<hr>
<p><em>Hint:</em></p>
<p><span class="math-container">$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{(x-a)^{m+1}}dx =\\ =\frac{(-1)^m}{m!(a+z)^m}\sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\sum\limits_{j=0}^m\begin{bmatrix}m~\\j~\end{bmatrix}\text{Li}_{n-k+1-j}\left(\frac{x+z}{a+z}\right) + C$$</span></p>
<p>for <span class="math-container">$~m\in\mathbb{N}_0~$</span> and with </p>
<p>the <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind" rel="nofollow noreferrer">Stirling numbers of the first kind</a> <span class="math-container">$\begin{bmatrix}n~\\k~\end{bmatrix}~$</span> defined by <span class="math-container">$~\displaystyle\sum\limits_{k=0}^n\begin{bmatrix}n~\\k~\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k)~$</span> </p>
<p>A simple example: <span class="math-container">$\enspace\displaystyle\int\limits_0^1\frac{\ln(1-x)}{(1+x)^3}dx = \displaystyle -\int\limits_{-1}^0\frac{\ln(x+1)}{(x-1)^3}dx =-\frac{1+\ln 2}{8}$</span></p>
|
1,921,302 | <p>I can't believe I am asking such a silly question. So I have the function
$$\ln\tan^{-1}x$$
I am asked to find the range of this function. I know that the range of $\ln x$ is all real numbers and that the range of $\tan^{-1}(x)$ is $(-\frac\pi2$, $\frac\pi2)$. Wouldn't the range of $\ln\tan^{-1}x$ also be $(-\frac\pi2$, $\frac\pi2)$, or am I doing something really stupid?</p>
| Ege Erdil | 326,053 | <p>You are given the inclusion $ \beta I \subset I $, multiplying both sides by the fractional ideal $ I^{-1} $ gives $ (\beta) \subset \mathcal O_K $, in particular, $ \beta \in \mathcal O_K $.</p>
|
1,921,302 | <p>I can't believe I am asking such a silly question. So I have the function
$$\ln\tan^{-1}x$$
I am asked to find the range of this function. I know that the range of $\ln x$ is all real numbers and that the range of $\tan^{-1}(x)$ is $(-\frac\pi2$, $\frac\pi2)$. Wouldn't the range of $\ln\tan^{-1}x$ also be $(-\frac\pi2$, $\frac\pi2)$, or am I doing something really stupid?</p>
| tracing | 200,415 | <p>You don't need to argue with principal ideals if you don't want to. Instead, note that $I$ is finitely generated over $R$, say by $x_1,\ldots,x_n$.
Then $\alpha \cdot x_i = \sum_j r_{ij} x_j$ for some $r_{ij} \in R$
(by the assumption), and so the matrix $\bigl( \alpha\delta_{ij} - r_{ij}
\bigr)$ annihilates $I$. Thus so does its determinant; since $I \neq 0,$
we conclude that this determinant equals zero.</p>
<p>This determinant is a monic polynomial with entries in $R$, and thus $\alpha$ satisfies a monic polynomial with algebraic integer coefficients, and so is itself an algebraic integer.</p>
|
3,294,369 | <p>I am reading a paper which asserts that the value of the function <span class="math-container">$f(x) = x^{\frac{1}{x}}$</span> at <span class="math-container">$x=0$</span> is equal to <span class="math-container">$0$</span>. I can believe that this is so if I write <span class="math-container">$f(x) = e^{{\frac{1}{x}}{{\rm ln}(x)}}$</span>, but I want a rigorous proof of this fact! Is this function even defined at <span class="math-container">$x=0?$</span></p>
| hmakholm left over Monica | 14,366 | <p>The expression <span class="math-container">$x^{1/x}$</span> is not defined when <span class="math-container">$x=0$</span>, because <span class="math-container">$1/0$</span> has no value.</p>
<p>What you <em>can</em> ask is whether the function defined on <span class="math-container">$(0,\infty)$</span> can be <em>extended continuously</em> to a function defined on <span class="math-container">$[0,\infty)$</span>. (Such an extended function cannot be defined by the expression <span class="math-container">$x^{1/x}$</span> at <span class="math-container">$0$</span>, of course). This is the same as asking for
<span class="math-container">$$ \lim_{x\to 0^+} x^{1/x} $$</span>
and the way to investigate that would indeed be to look at how its logarithm behaves as <span class="math-container">$x$</span> approaches <span class="math-container">$0$</span> from above.</p>
<p>Since <span class="math-container">$$ \log(x^{1/x}) = \frac1x\log(x) = - \frac1x \log \frac1x $$</span>
which clearly goes to <span class="math-container">$-\infty$</span> when <span class="math-container">$\frac1x$</span> goes to <span class="math-container">$+\infty$</span>, it is easy to show explicitly that the limit above must be <span class="math-container">$0$</span>.</p>
|
510,130 | <p>Let $(r_i)_{i=1}^m$ be a sequence
of positive reals such that
$\sum_i r_i < 1$
and let $t$ be a positive real.
Consider the sequence $T(n)$
defined by $T(0) = t$,
$T(n) = \sum_i T(\lfloor r_i n \rfloor) $
for $n \ge 1$.</p>
<p>Show that
$T(n) = o(n)$,
that is,
$\lim_{n \to \infty} \dfrac{T(n)}{n}
= 0
$.</p>
<p>Note:
This is a variation on
<a href="https://math.stackexchange.com/questions/506489/if-tn-un-sum-i-t-lfloor-r-i-n-rfloor-show-that-tn-thetan">If $T(n) = un + \sum_i T(\lfloor r_i n \rfloor) $, show that $T(n) = \Theta(n)$</a>.
It is gotten by setting $u=0$ there.</p>
<p>I am close to a solution,
and hope to have one in a few days.
If I find one,
I will post it.</p>
<p>Note:
It is easy to prove that
$T(n) = O(n)$.
The problem is showing that
$T(n)/n \to 0$.</p>
| Marko Riedel | 44,883 | <p>Here is a proof for a special case that is not too involved yet does occur in actual settings. Suppose your $r_k$ are all inverse integer powers of some positive integer $p$, where $p\ge 2,$ so that $r_k = 1/p^{q_k}$ with $q_k\ge 1$ and the $q_k$ distinct.</p>
<p>Your recurrence now looks like this:
$$ T(n) = \sum_{k=1}^m T(\lfloor n/p^{q_k} \rfloor).$$</p>
<p>Let $D = \max_{k} q_k$ and introduce $$S(n) = \sum_{k=1}^m S(n-q_k)$$
where the initial values are $S(n) = T(p^n)$ for $0\le n < D.$ Then it is not difficult to see that the following exact formula holds: $$ T(n) = S(\lfloor\log_p n\rfloor).$$</p>
<p>Now $S(n)$ is linear and homogeneous with characteristic equation
$$\lambda^D = \sum_{k=1}^m \lambda^{D-q_k}.$$
Note that the maximum modulus of the roots of this equation is strictly less than two, because for $|\lambda|=R$ we have
$$\left| \sum_{k=1}^m \lambda^{D-q_k} \right|
\le \sum_{k=1}^m |\lambda|^{D-q_k} = \sum_{k=1}^m R^{D-q_k}
\le \sum_{d=0}^{D-1} R^d = \frac{R^D-1}{R-1}.$$
But for $R\ge 2$ we have
$$ \frac{R^D-1}{R-1} < R^D $$
since this is just $$ R^D - 1 < R^{D+1} - R^D \quad\text{or}\quad 2R^D < R^{D+1} + 1,$$ producing a contradiction to the fact that $\lambda$ was supposed to be a root (modulus of LHS $>$ RHS in the original equation).</p>
<p>Let $\rho$ be the root with maximum modulus of the characteristic equation. Then by the basic theory of recurrence relations the asymptotically dominant term of the solution to the recurrence satisfies $$S(n) \sim c n^{a-1} \rho^n$$ with $c$ a constant and $a$ the multiplicity of the root. An easy continuity argument shows that in fact $\rho$ is real and $1<\rho<2.$ (The root $\rho$ cannot correspond to a pair of complex conjugate roots, because these would generate a fluctuating trigonometric term which, its modulus being the largest, would eventually produce a pair of consecutive values of $S(n)$ with different signs or one of them being zero, contradicting the fact that $S(n)$ is easily seen to be strictly increasing.)</p>
<p>This yields the following for $T(n):$
$$ T(n) \sim c \lfloor\log_p n\rfloor^{a-1} \rho^{\lfloor\log_p n\rfloor}.$$
Hence
$$\frac{T(n)}{n} \in
\Theta\left( \lfloor\log_p n\rfloor^{a-1}
\left(\frac{\rho}{p}\right)^{\lfloor\log_p n\rfloor}\right).$$
This shows the claim that $T(n)/n\to 0$ because $\rho/p$ is strictly less than one and its positive powers vanish as $\log_p n$ goes to infinity, canceling the polynomial in $\log_p n$ along the way. (The trick was that $\rho<2$ and $p\ge 2.$)</p>
<p><strong>Addendum I.</strong> The reader may wish to verify that $\sum_{k=1}^m \frac{1}{p^{q_k}}$ is indeed less than one. <strong>II.</strong> The argument goes through even if the set of roots with maximum modulus (which is finite) were to include complex numbers, because the bound on $|\rho|$ continues to hold.</p>
|
1,811,528 | <p>The definition of the order of an element in a group is:</p>
<blockquote>
<p>The order of an element $x$ of a group $G$ is the smallest positive integer $n$ such that $x^{n}=e$.</p>
</blockquote>
<p>Doesn't this definition assume that the integers are somehow relevant to every group? </p>
<p>All of the other definitions concerning Groups do not invoke the integers in any way.</p>
<p>I would expect that Groups would contain elements that would somehow represent exponentiation.</p>
<p>This question may reflect some deep misunderstandings on my part so please feel free to offer guidance not directly relevant to the question.</p>
| Ben Grossmann | 81,360 | <p>Every group has an associative "multiplication" (binary operation), so there is no ambiguity in referring to the integer power of an element.</p>
<p>In particular, $x^n$ is just short hand for
$$
x^n = \overbrace{x x \cdots x}^n
$$
where we note that this expression is this same, no matter how the $x$s are "grouped together" within that product. This only defines exponentiation for positive integers; however, because groups have an identity and group elements have inverses, we may also define $x^0$ to be the identity element and
$$
x^{-n} = (x^{-1})^{n}
$$
for any $n \geq 1$.</p>
|
992,068 | <p>I am having a little trouble understanding this question.</p>
<p>For a DFA M = (Q, Σ, δ, q0, F), we say that a state q ∈ Q is reachable if there
exists some string w ∈ Σ∗ such that q = δ∗(q0, w).</p>
<p>Give an algorithm that, given as input a DFA expressed as a five-tuple M =
(Q, Σ, δ, q0, F), returns the set of all of M’s reachable states. </p>
<p>Can anyone help me understand and answer this question?</p>
| Brian M. Scott | 12,042 | <p>Someone hands you a DFA $M=\langle Q,\Sigma,\delta,q_0,F\rangle$, i.e., a set of states, an input alphabet, a transition function, an initial state, and a set of acceptor states. You’re to come up with an algorithm — a systematic procedure — that takes this information as input and produces as output a list of the reachable states of $M$.</p>
<p>A first easy observation is that you can ignore $F$ altogether: you don’t care which states are accepting, just which ones can actually be reached from $q_0$ with some input. You will need to use the other four items, however. A second easy observation is that $q_0$ is reachable, since $\delta^*(q_0,\epsilon)=q_0$, where $\epsilon$ is the empty word.</p>
<p>If $\Sigma=\{\sigma_1,\ldots,\sigma_n\}$, the natural thing to do is to start by running $M$ with all $n$ one-symbol inputs; that will tell you which states are reachable in one transition. Then you could systematically test inputs of length $2$: </p>
<p>$$\sigma_1\sigma_1,\sigma_1\sigma_2,\ldots,\sigma_1\sigma_n,\sigma_2\sigma_1,\sigma_2\sigma_2,\ldots,\sigma_2\sigma_n,\ldots,\sigma_n\sigma_n\;.$$</p>
<p>And you could go on to longer and longer inputs. The only real question is how to know when to stop: at what point can you be sure that you’ve found all of the reachable states?</p>
<p>HINT: If there are $m$ states, how long a word must you input in order to guarantee that the DFA will visit some state twice while reading the word?</p>
|
4,324,493 | <p>According to Rick Miranda's Algebraic curves and Riemann surfaces, a hyperelliptic curve is defined as the Riemann surface obtained by gluing two algebraic curves, <span class="math-container">$y^2=h(x)$</span> and <span class="math-container">$w^2 = k(z)$</span> (where <span class="math-container">$h$</span> has distinct roots and <span class="math-container">$k(z) := z^{2g+2} h(1/z)$</span>) through the map <span class="math-container">$(x,y) \mapsto (z,w) := (1/x, y/x^{g+1})$</span>.
He also mentions that this is topologically a genus <span class="math-container">$g$</span> surface; and has a notion of involution <span class="math-container">$\sigma : (x,y)\mapsto(x,-y)$</span></p>
<p>I am trying to visualise how all these actually looks like, and I have a couple of questions:</p>
<ol>
<li><p>Is there any geometric way to visualise the construction? Formally, is there any embedding/immersion of this construction into <span class="math-container">$\mathbb{R}^3$</span>, that helps see, atleast topologically, atleast for specific examples, what parts of genus <span class="math-container">$g$</span> surface are being glued together?</p>
</li>
<li><p>What does the <span class="math-container">$\sigma$</span> 'looks like' as a map on genus <span class="math-container">$g$</span> surface, <span class="math-container">$\Sigma_g$</span>, imagined as embedded in <span class="math-container">$\mathbb{R}^3$</span> in the standard way? Does it look like a reflection? Or maybe a <span class="math-container">$180^\circ$</span> rotation?</p>
</li>
<li><p>I think this brings about some sort of duality between the polynomials <span class="math-container">$h$</span> and <span class="math-container">$k$</span>. Is this somehow important? Specifically, is there anything special about the Riemann surfaces that are obtained by gluing the same algebraic curve? As an example for such a curve, if <span class="math-container">$x\neq 0$</span>, for <span class="math-container">$g=3$</span>, <span class="math-container">$$y^2 = 3x^8 +10x^4 +3 \Leftrightarrow \bigg(\frac{y}{x^{3+1}}\bigg)^2 = 3\bigg(\frac{1}{x}\bigg)^8 +10\bigg(\frac{1}{x}\bigg)^4 +3 $$</span></p>
</li>
</ol>
<p>Sorry if the questions are a bit vague. I am just trying to get more intuition about the subject.</p>
| hm2020 | 858,083 | <p><strong>Question1:</strong> "Is there any geometric way to visualise the construction? Formally, is there any embedding/immersion of this construction into R3, that helps see, atleast topologically, atleast for specific examples, what parts of genus g surface are being glued together?"</p>
<p><strong>Answer:</strong> About question 1: When <span class="math-container">$E$</span> is an elliptic curve over the complex number field, there is a result in Hartshorne (Thm.IV.4.16) that realizes E topologically as a quotient <span class="math-container">$\mathbb{C}/\Gamma$</span>. Maybe</p>
<p>"Farkas; Riemann Surfaces. Graduate Texts in Mathematics. Springer-Verlag 1980."</p>
<p>for the general case.</p>
<p><strong>Note:</strong> Naively when an algebraic group <span class="math-container">$G$</span> acts on a commutative ring <span class="math-container">$A$</span> and an affine scheme <span class="math-container">$X:=Spec(A)$</span>, we may construct the "quotient" <span class="math-container">$X/G:=Spec(A^G)$</span> using the invariant ring. The above "topological quotient" <span class="math-container">$\mathbb{C}/\Gamma$</span> is projective, hence we cannot naively construct <span class="math-container">$\mathbb{C}/\Gamma$</span> using <span class="math-container">$Spec(A^G)$</span> for some <span class="math-container">$G$</span> acting on <span class="math-container">$A$</span>.</p>
<p><strong>Question2:</strong> "What does the map <span class="math-container">$\sigma$</span> 'looks like' as a map on genus g surface, Σg, imagined as embedded in R3 in the standard way? Does it look like a reflection? Or maybe a 180∘ rotation?"</p>
<p><strong>Answer question 2:</strong> Since <span class="math-container">$C$</span> is hyper-elliptic, there is a degree 2 separable map</p>
<p><span class="math-container">$$\phi_K:C \rightarrow \mathbb{P}^1_{\mathbb{C}}$$</span></p>
<p>coming from the canonical divisor. You involution <span class="math-container">$\sigma$</span> permutes the points in a fiber: If <span class="math-container">$p,q \in \phi_K^{-1}(x)$</span> it follows <span class="math-container">$\sigma(p)=q, \sigma(q)=p$</span>.
You find this discussed in the same book HH Chapter IV.</p>
<p><strong>Note:</strong> There is a classical result saying the following: Let <span class="math-container">$C \subseteq \mathbb{P}^n_k$</span> be a smooth projective curve over an algebraically closed field <span class="math-container">$k$</span>. If <span class="math-container">$g(C)=g \geq 2$</span>, let <span class="math-container">$K_C$</span> be the canonical divisor with associated morphism</p>
<p><span class="math-container">$$\phi_C: C \rightarrow \mathbb{P}^{g-1}.$$</span></p>
<p>Then either <span class="math-container">$\phi$</span> is a closed immersion or <span class="math-container">$Im(\phi_C) \cong \mathbb{P}^1$</span> and <span class="math-container">$deg(\phi_C)=2$</span>. Moreover: <span class="math-container">$C$</span> is hyperelliptic iff there is an element <span class="math-container">$\sigma \in Aut_k(C)$</span> with <span class="math-container">$\sigma^2=Id$</span> and <span class="math-container">$C/<\sigma> \cong \mathbb{P}^1_k$</span>.</p>
<p>Being a hyper elliptic curve is an "intrinsic property" (independent of any embedding into projective space), and when the base field is algebraically closed this property is detected by the canonical divisor <span class="math-container">$K_C$</span>. Hence if someone "gives you such a curve" that is not embedded into a projective space and asks: "Is it hyperelliptic?" you can check this using <span class="math-container">$K_C$</span>.</p>
|
2,193,171 | <p>Question: Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences with $a_n \Rightarrow L$ and $b_n \Rightarrow M$ as $n \Rightarrow \infty$. </p>
<p>Prove that $a_nb_n \Rightarrow LM$</p>
<p>Solution: (My Attempt). Instead of redoing it could someone just tell me what I'm doing wrong. Thx</p>
<p>WTS: </p>
<p>(1) $\exists L \in R, \forall \epsilon > 0, \exists N_1 > 0$ such that for all $n \in N_1$, if $n > N_1$, then </p>
<p>$|a_n - L| < \text{(We dont know yet)}$</p>
<p>(2) $\exists M \in R, \forall \epsilon > 0, \exists M > 0$, such that for all $m \in M$, if $m > M$, then </p>
<p>$|b_n - M| < \text{(We dont know yet)}$</p>
<p>Choose N = $\text{we dont know yet} > 0$</p>
<p>Suppose $n > N$ and $m > M$, then</p>
<p>$$|a_nb_n - LM| = |a_nb_n - a_nM + a_nM - LM| $$</p>
<p>$$= |a_n(b_n - M) + M(a_n - L)| \text{ by algebra}$$</p>
<p>$$\leq |a_n(b_n-M)| + |M(a_n - L)| \text{ triangle inequality}$$ </p>
<p>$$= |a_n||b_n - M| + |M||a_n - L|$$</p>
<p>Can we say $|a_n||b_n - M| = \epsilon/2$ same with $|M||a_n - L| = \epsilon/2$ ? Then Q.E.D? With N = $max(N_1, M)$ ? </p>
<p>I have no idea what I'm doing. </p>
| Ethan Alwaise | 221,420 | <p>Pick $\epsilon > 0$. By taking the derivative one sees that $e^{-1/t}t^{-k}$ is increasing on some interval $(0,a)$. Choose $m \in \mathbb{N}$ such that $2^{-m} < a$ and
$$mk - 2^m < \log_2\epsilon.$$
Therefore if $0 < t < 2^{-m}$ we have
$$\frac{e^{-1/t}}{t^k} < \frac{2^{-1/t}}{t^k} < \frac{2^{mk}}{2^{2^m}} = 2^{mk - 2^m} < \epsilon.$$</p>
|
2,246,629 | <p>I'm trying to evaluate the following limit but I'm stuck.</p>
<p>$$
\lim_{x\to +\infty} {x^3\cos(1/x)\over \sin x}
$$ </p>
<p>I tried the squeeze theorem but I was led to a dead-end. Any help would be appreciated.</p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>prove that for any $N>0$ we can find $x_1>N$ such that the function is positive and $x_2>N$ such that the function is negative. So the limit cannot exists.</p>
|
2,246,629 | <p>I'm trying to evaluate the following limit but I'm stuck.</p>
<p>$$
\lim_{x\to +\infty} {x^3\cos(1/x)\over \sin x}
$$ </p>
<p>I tried the squeeze theorem but I was led to a dead-end. Any help would be appreciated.</p>
| Lakshya Gupta | 439,351 | <p>We can exclude cos(1/x) as when x approaches infinity cos(1/x) would approach 1.
Secondly.
We are left with x^3/sin(x).
If we apply L hospital rule thrice. We are left with -6/cos(x) and cos(x) is not defined when x approaches infinity. Thus the limit is not defined.</p>
|
1,237,450 | <p>I couldn't follow a step while reading this <a href="https://math.stackexchange.com/a/1237316/135088">answer</a>. Since I do not have enough reputation to post this as a comment, I'm asking a question instead. The answer uses "partial integration" to write this $$ \int \frac{dv}{(v^2 + 1)^\alpha} = \frac{v}{2(\alpha-1)(v^2+ 1)^{\alpha - 1}} + \frac{2\alpha -3}{2\alpha - 2}\int \frac{dv}{(v^2 + 1)^{\alpha -1}} $$
I would like to know what this technique is, and how this equality follows from it.</p>
| N. S. | 9,176 | <p>Try integration by parts
$$f= (v^2+1)^{-\alpha} ; g' =1 \\
f'=(-\alpha) (v^2+1)^{-\alpha-1}(2v) ; g=v$$</p>
<p><strong>P.S.</strong> After integration by Parts, you write the $v^2$ in the numerator as
$$v^2=(v^2+1)-1$$
and split the integral in two.</p>
|
122,546 | <p>There is a famous proof of the Sum of integers, supposedly put forward by Gauss.</p>
<p>$$S=\sum\limits_{i=1}^{n}i=1+2+3+\cdots+(n-2)+(n-1)+n$$</p>
<p>$$2S=(1+n)+(2+(n-2))+\cdots+(n+1)$$</p>
<p>$$S=\frac{n(1+n)}{2}$$</p>
<p>I was looking for a similar proof for when $S=\sum\limits_{i=1}^{n}i^2$</p>
<p>I've tried the same approach of adding the summation to itself in reverse, and I've found this:</p>
<p>$$2S=(1^2+n^2)+(2^2+n^2+1^2-2n)+(3^2+n^2+2^2-4n)+\cdots+(n^2+n^2+(n-1)^2-2(n-1)n$$</p>
<p>From which I noted I could extract the original sum;</p>
<p>$$2S-S=(1^2+n^2)+(2^2+n^2-2n)+(3^2+n^2-4n)+\cdots+(n^2+n^2-2(n-1)n-n^2$$</p>
<p>Then if I collect all the $n$ terms;</p>
<p>$$2S-S=n\cdot (n-1)^2 +(1^2)+(2^2-2n)+(3^2-4n)+\cdots+(n^2-2(n-1)n$$</p>
<p>But then I realised I still had the original sum in there, and taking that out mean I no longer had a sum term to extract.</p>
<p>Have I made a mistake here? How can I arrive at the answer of $\dfrac{n (n + 1) (2 n + 1)}{6}$ using a method similar to the one I expound on above? <strong>I.e following Gauss' line of reasoning</strong>?</p>
| Thomas Andrews | 7,933 | <p>We use that <span class="math-container">$n^2=\sum_{k=1}^{n} (2k-1).$</span> Then:</p>
<p>Not quite Gaussian, but similar:</p>
<p><span class="math-container">$$\begin{align}S&=\sum_{n=1}^{m} n^2 \\&= \sum_{n=1}\sum_{k=1}^n (2k-1)\\
&=\sum_{k=1}^{m}\sum_{n=k}^{m}(2k-1)\\
&=\sum_{k=1}^m(2k-1)\sum_{n=k}^{m}1\\
&=\sum_{k=1}^m(2k-1)(m+1-k)\\
&=-2S -m(m+1)+ (2m+3)\sum_{k=1}^m k
\end{align}$$</span></p>
<p>So we get;</p>
<p><span class="math-container">$$3S = m(m+1)\left(\frac{2m+3}{2}-1\right)=\frac{m(m+1)(2m+1)}{2}$$</span></p>
|
1,747,696 | <p>First of all: beginner here, sorry if this is trivial.</p>
<p>We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .</p>
<p>My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ?
The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on.
In this case, this is also true but I can't seem to find a way to generalize it.</p>
| C. Dubussy | 310,801 | <p>Assume you have a sequence $$x, x+a, x+2a, x+3a, ..., x+na.$$</p>
<p>Let us note $$S = \sum_{i=0}^n (x+ia).$$ Then by the trick you mentioned, we see that $$S+S = (2x+na)+(2x+na)+...+(2x+na) = (n+1)(2x+na).$$ Hence $$S = \frac{(n+1)(2x+na)}{2}.$$</p>
|
3,554,646 | <p>Let <span class="math-container">$f:(a,b)\to \mathbb{R}$</span> be injective and continuous. Prove that</p>
<ul>
<li><span class="math-container">$f$</span> is monotonic.</li>
<li>The image of <span class="math-container">$f$</span> is <span class="math-container">$(c,d)$</span> or maybe (<span class="math-container">$-\infty,+\infty$</span>).</li>
</ul>
<p>I proved <span class="math-container">$f$</span> is monotonic but I don't know how to the second proposition begin. </p>
<p>Thankyou so much.</p>
| Martund | 609,343 | <p>Without loss of generality, we may assume that <span class="math-container">$f$</span> is monotonic increasing function (Consider <span class="math-container">$-f$</span> otherwise). Let <span class="math-container">$$c:=\lim_{x\to a+}f(x)\in\mathbb R\cup\{-\infty\}$$</span>
<span class="math-container">$$d:=\lim_{x\to b-}f(x)\in\mathbb R\cup\{\infty\}$$</span>
Note that these limits exist because of monotonicity of <span class="math-container">$f$</span>. Now, <span class="math-container">$f$</span> takes every value in <span class="math-container">$(c,d)$</span> by intermediate value theorem. Also any other value is not attained by <span class="math-container">$f$</span> by monotonicity and we are done.</p>
|
1,932,961 | <p>Prove by mathematical induction that
$$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$
holds $\forall n\in\mathbb{N}$.</p>
<hr>
<p>(1) Assume that $n=1$. Then left side is $1^2 =1$ and right side is $6/6 = 1$, so both sided are equal and expression holds for $n = 1$.</p>
<p>(2) Let $k \in \mathbb{N}$ is given. Assume that for $n = k$ expression holds. Then for $n = k+1$ we get
$$\sum_{i = 1}^{k+1} i^2 = \left(\sum_{i = 1}^{k} i^2\right) + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + k^2 + 2k + 1 = \frac{2k^3 + 9k^2 + 13k + 6}{6}.$$
Factoring the result we get that $\frac{2k^3 + 9k^2 + 13k + 6}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$ and thus expression holds for $n = k+1$.</p>
<p>Combining (1) and (2) we can conclude that the expression holds $\forall n \in \mathbb{N}$.</p>
<hr>
<p>I have a few questions:</p>
<ol>
<li>Is my proof correct?</li>
<li>If you would be a math professor, is this style of writing math proofs right and sufficient for freshman? Or is there something I miss?</li>
</ol>
| Dr. Sonnhard Graubner | 175,066 | <p>it should be $$\frac{(xy+yz+zx-3)^2}{(x-1)^2(y-1)^2(z-1)^2}\geq 0$$
Hint: set $$x=a/b,y=b/c,z=c/a$$ in the given term
after the substituion we obtain
$${\frac {{a}^{2}}{{b}^{2}} \left( {\frac {a}{b}}-1 \right) ^{-2}}+{
\frac {{b}^{2}}{{c}^{2}} \left( {\frac {b}{c}}-1 \right) ^{-2}}+{
\frac {{c}^{2}}{{a}^{2}} \left( {\frac {c}{a}}-1 \right) ^{-2}}-1
$$
factoring all we get
$${\frac { \left( b{a}^{2}-3\,bca+{c}^{2}a+{b}^{2}c \right) ^{2}}{
\left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( -c+a \right) ^
{2}}}
\geq 0$$</p>
|
185,177 | <p>Let $X$ be a smooth finite type separated connected Deligne-Mumford stack over $\mathbb C$.</p>
<p>Does there exist a finite etale morphism $Y\to X$ with $Y$ a scheme?</p>
<p>What if $X$ is an algebraic space (i.e., trivial stabilizers)?</p>
<p>Edit: I changed the old question to a different question which should be more clear. An answer to the new question would help a lot in answering the old question.</p>
| Daniel Litt | 6,950 | <p>It seems to me that the answer is NO if $X$ is a DM stack. If I'm not mistaken, it suffices to give a smooth finite type separated connected Deligne-Mumford stack over $\mathbb{C}$ which is simply connected (since such a thing has no non-trivial finite etale covers, let alone finite etale covers by a scheme). But <a href="http://arxiv.org/pdf/math/0504309.pdf">this paper</a> of Behrend and Noohi shows that the weighted projective lines $\mathcal{P}(m, n)$ (constructed by taking the stack quotient of $\mathbb{A}^2\setminus\{0\}$ by the $\mathbb{G}_m$-action $\lambda\cdot(x,y):=(\lambda^m x, \lambda^n y)$) are simply connected. The proof is easy; one just uses the long exact sequence for homotopy groups associated to the fibration $$\mathbb{G}_m\to \mathbb{A}^2\setminus\{0\}\to \mathcal{P}(m, n).$$</p>
<p><strong>Added later</strong>: The answer seems to be no for algebraic spaces as well. Example 5.7 <a href="http://projecteuclid.org/download/pdf_1/euclid.pja/1117805146">here</a> is simply connected if I'm not mistaken, and is not a scheme by Remark 3.4 in the same paper.</p>
|
2,631,230 | <p>So, I'm studying mathematics on my own and I took a book about Proofs in Abstract Mathematics with the following exercise:</p>
<p>For each $k\in\Bbb{N}$ we have that $\Bbb{N}_k$ is finite</p>
<p>Just to give some context on what theorems and definitions we can use:</p>
<ol>
<li>Definition: $\Bbb{N}_k = \{1, 2, ..., k \} $</li>
<li>Definition: A set $S$ is infinite iff there exists a one-to-one but not onto $\ f:S\to S$</li>
<li>Definition: $A\sim B$ means $A$ is equipotent to (or same cardinality of) $B$</li>
<li>Theorem: if $A$ is infinite and $A\sim B$, then $B$ is infinite</li>
<li>Theorem: if $A$ is infinite and $f:A\to B$ is one-to-one, then $B$ is infinite</li>
<li>Theorem: Let $\ f:A \to B$ be one-to-one and $C\subseteq A$ then $\ g:C \to B$, $\ g(x)=f(x)\ $ for any $\ x\in C$, is also one-to-one</li>
<li>Lemma: Let $k\in\Bbb{N}$, then $\Bbb{N}_k- \{x\} \sim \Bbb{N}_{k-1}$ for any $x\in \Bbb{N}_k$</li>
</ol>
<p>What I did was:</p>
<p>Suppose that $\Bbb{N}_K$ is not finite for every $k\in\Bbb{N}$, then by the Well-Ordering Principle, there is a smallest element $k\in\Bbb{N}$ such that $\Bbb{N}_k$ is infinite. Let $x_0\in\Bbb{N}_k\ $ be the smallest element of $\Bbb{N}_k$ and define $C=\Bbb{N}_k - \{x_0\}$. Let $f:\Bbb{N}_k \to C\ $ be $\ f(n)=n+1$.
We will prove that $f$ is one-to-one. Let $x_1,x_2\in\Bbb{N}_k$ such that $f(x_1)=f(x_2)$, then $x_1+1=x_2+1$. Hence $x_1=x_2$, what proves that $f$ is one-to-one. Thus we have that $C$ is infinite. Then $C\sim \Bbb{N}_{k-1}$ and thus we must have that $\Bbb{N}_{k-1}$ is infinite. However this contradicts our hypothesis that $k$ is the least element such that $\Bbb{N}_k$ is infinite. Thus it must be that for each $k\in \Bbb{N}$ we have $\Bbb{N}_k$ is finite.</p>
<p>My question is if the proof above, especially when creating the function $f:\Bbb{N}_k\to C$ has any flaw. The book explicitly says we should use the 6th theorem listed above, but I didn't find any explicitly use of it. Maybe is there another way to prove it?</p>
<p><strong>Edited:</strong> </p>
<p>As some of you commented, the proof above was wrong. The function I created was not defined to $k+1$. I think this one is correct:</p>
<p>If $\Bbb{N}_k$ is not finite for every $k \in \Bbb{N}$, then by the Well-Ordering principle there exists a least element $k \in \Bbb{N}$ such that $\Bbb{N}_k$ is infinite. By definition, there exists $f:\Bbb{N}_k \to \Bbb{N}_k$ such that $f$ is one-to-one but not onto. Then, because $f$ is not onto, there exists $y\in\Bbb{N}_k$ such that $y\neq f(x)$ for every $x\in \Bbb{N}_k$. Pick $x_0\neq y$ and define $A=\Bbb{N}_k-\{x_0\}$. Let $g:A\to A$ be defined as:
$$g(x)=
\begin{cases}
f(x) \ if \ f(x)\neq x_0 \\
f(x_0) \ if \ f(x) = x_0
\end{cases}$$</p>
<p>We will prove that $g$ is one-to-one but not onto. </p>
<p>First we show $g$ is one-to-one. Let $x_1,x_2 \in A$ such that $x_1\neq x_2$. Since $f$ is one-to-one, $f(x_1)\neq f(x_2)$. If $f(x_1)=x_0$, then $f(x_2)\neq x_0$. Hence $g(x_1)=f(x_0)$ and $g(x_2)=f(x_2)$. Since $x_0\neq x_2$, then $f(x_0)\neq f(x_2)$ and thus $g(x_1)\neq g(x_2)$. Without loss of generality, if $f(x_2)=x_0$, then $g(x_1)\neq g(x_2)$. If $f(x_1)\neq x_0$ and $f(x_1)\neq x_0$, then $g(x_1)=f(x_1)$ and $g(x_2)=f(x_2)$. Hence $g(x_1)\neq g(x_2)$. We have that $g$ is one-to-one.</p>
<p>We now show that $g$ is not onto. Note that, because $x_0\neq y$, such that $y\neq f(x)$ for all $x\in\Bbb{N}_k$, we have $y\in A=\Bbb{N}_k-\{x_0\}$. Let $x\in A$. If $f(x)=x_0$, then $g(x)=f(x_0)\neq y$. If $f(x)\neq x_0$, then $g(x)=f(x)\neq y$. Hence there exists $y \in A$ such that for any $x \in A$ we have $g(x)\neq y$. Thus, $g$ is not onto.</p>
<p>We have demonstrated that $g:A\to A$ is one-to-one, but not onto, hence A is infinite by definition. Giving that $A=\Bbb{N}_k-\{x_0\}$ and our lemma, we have that $\Bbb{N}_{k-1}$ is also infinite. However this contradicts our hypothesis that $k$ is the smallest element such that $\Bbb{N}_k$ is infinite. Hence it must be that for every $k\in\Bbb{N}$ we have $\Bbb{N}_k$ is finite.</p>
<p>Sorry if my proof writing is bad in anyway. If you have any stylistic suggestion, or any suggestion at all, I would gladly read it :) </p>
| Jack D'Aurizio | 44,121 | <p>An alternative approach:
$$ \begin{eqnarray*}\iint_{(0,1)^2}4xy\sqrt{x^2+y^2}\,dx\,dy&=&\iint_{(0,1)^2}\sqrt{X+Y}\,dX\,dY\\&=&2\iint_{0\leq Y\leq X\leq 1}\sqrt{X+Y}\,dX\,dY\\&=&2\iint_{(0,1)^2}X\sqrt{X}\sqrt{K+1}\,dX\,dK\\&=&2\int_{0}^{1}X\sqrt{X}\,dX\int_{0}^{1}\sqrt{K+1}\,dK\end{eqnarray*}$$
by exploting the substitution $x=\sqrt{X}, y=\sqrt{Y}$, then symmetry, then the substitution $Y=KX$, then Fubini's theorem. The last integrals are elementary and they respectively equal $\frac{2}{5}$ and $\frac{2}{3}(2\sqrt{2}-1)$.</p>
|
424,514 | <p>Suppose one has a generating function <span class="math-container">$$F(z) = \sum_{k\ge 0} f(k) z^k$$</span>
for some <span class="math-container">$f:\mathbb{Z}\rightarrow \mathbb{Z}$</span>. Is there a way to express an iteration of <span class="math-container">$f$</span> in terms of <span class="math-container">$F(z)$</span>. E.g., <span class="math-container">$$G(z) = \sum_{k\ge 0} f(f(k)) z^k$$</span>
Can <span class="math-container">$G(z)$</span> be expressed in terms of <span class="math-container">$F(z)$</span>?</p>
| Joe | 171,026 | <p>You could explore this conjecture by the following method: Suppose <span class="math-container">$f(f(k))= h(f(k))$</span> for different specified <span class="math-container">$h$</span>, then look for <span class="math-container">$G(z) = H(F(z))$</span>. So eg. <span class="math-container">$h(k) = a k + b$</span> gives <span class="math-container">$G(z) = a F(z) +\frac{b}{1-z}.$</span> Then you need to solve the functional equation <span class="math-container">$f(f(k)) = h(f(k))$</span>, and this will give you some sets of pairs <span class="math-container">$(h, H)$</span> which might inform your conjecture</p>
<p>The example that you give in the comments on the other question is <span class="math-container">$(h,H)=(1,1)$</span>, but this case is very simple, I don't know that I would take it to be the basis for a conjecture.</p>
|
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