qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,072,561 | <p>can you help me with this quest?</p>
<p>About composition $f$ and vector space $\mathbf{V}=\mathbb{Z^4_2}$ we know the following:</p>
<p>$f \circ f = id_V$,$~~f $
$
\left(\begin{array}{ccc}
1\\
0\\
1\\
0\\
\end{array}\right)=
\left(\begin{array}{ccc}
1\\
1\\
0\\
0\\
\end{array}\right)
$,
$~~f $
$
\left(\begin{array}{ccc}
0\\
1\\
0\\
1\\
\end{array}\right)=
\left(\begin{array}{ccc}
1\\
1\\
0\\
1\\
\end{array}\right)
$</p>
<p>$id_V$ is identity. </p>
<p>Find $f~((x_1,x_2,x_3,x_4)^T).$</p>
<p>I would be gratefull for any kind of advice.</p>
<p>Thanks</p>
| Ilmari Karonen | 9,602 | <p>Let $f(\mathbf v_E) = \mathbf v_B$. Can you show that the map $f$ satisfies the <a href="//en.wikipedia.org/wiki/Linear_map#Definition_and_first_consequences" rel="nofollow">definition of a linear map</a>, i.e. that $f(\mathbf u + \mathbf v) = f(\mathbf u) + f(\mathbf v)$ and that $f(\alpha\,\mathbf u) = \alpha\, f(\mathbf u)$? (Hint: You should be able to do this just by looking at the definition of $\mathbf v_B$.)</p>
<hr>
<p>OK, let me unpack that a little:</p>
<p>Given a basis $B = \{\mathbf b_1, \mathbf b_2, \dotsc, \mathbf b_n\}$, let $\mathbf u = c_1 \mathbf b_1 + \dotsb + c_n \mathbf b_n$ and $\mathbf v = d_1 \mathbf b_1 + \dotsb + d_n \mathbf b_n$, so that, by definition, $\mathbf u_B = (c_1, \dotsc, c_n)$ and $\mathbf v_B = (d_1, \dotsc, d_n)$. From this, it directly follows that $\mathbf u + \mathbf v = (c_1 + d_1) \mathbf b_1 + \dotsb + (c_n + d_n) \mathbf b_n$, and thus that $(\mathbf u + \mathbf v)_B = \mathbf u_B + \mathbf v_B$.</p>
<p>By a similar argument, you can show that $(\alpha\, \mathbf u)_B = \alpha\, (\mathbf u_B)$ for $\alpha \in \mathbb R$. Together, these results prove that the map $\mathbf v \mapsto \mathbf v_B$ is linear, as the question asks.</p>
<p>(In the question you quoted, the vector $\mathbf v$ is identified with its standard representation $\mathbf v_E = (a_1, \dotsc, a_n)$ given by $\mathbf v = a_1 \mathbf e_1 + \dotsb + a_n \mathbf e_n$, where $E = \{\mathbf e_1, \dotsc, \mathbf e_n\}$ is the standard basis of $\mathbb R^n$. If you want to distinguish between $\mathbf v$ and $\mathbf v_E$, and specifically show that the map $\mathbf v_E \mapsto \mathbf v_B$ is linear, it's enough to show that $\mathbf v_E \mapsto \mathbf v$ is also (trivially) linear, and then apply the lemma that the composition of two linear maps is linear.)</p>
|
446,959 | <p>I am searching two simple/efficient/generic algorithms to generate a uniform distribution of random points:</p>
<ul>
<li>in the volume of a n-dimensional hypersphere</li>
<li>on the surface of a n-dimensional hypersphere</li>
</ul>
<p>knowing the dimension $n$, the center of the hypersphere $\vec{x}$ and its radius $r$.</p>
<p>How to do that ?</p>
| sheß | 86,591 | <p>For the volume I'd say the simlest algorithm would be a standard accept-reject algorithm.</p>
<pre><code>(1) draw a uniform random point within an n-dimensional hypercube
(2) repeat (1) if the distance exceeds $r$, otherwise done
</code></pre>
<p>if im not mistaken though, accept-rates decrease as $n$ increases, maybe there is something more efficient</p>
<p><strong>edit</strong>
you can also find some ideas here: <a href="https://math.stackexchange.com/questions/87230/picking-random-points-in-the-volume-of-sphere-with-uniform-probability?rq=1">Picking random points in the volume of sphere with uniform probability</a></p>
<p><strong>edit2</strong> thanks to @awkward for pointing out that my suggestion to draw uniformly distributed angles is not a valid method to obtain points on the surface</p>
|
1,680,210 | <p>Three cards are drawn at random from a full deck. what is the probability of getting a three , a seven, and an ace?</p>
<p>Can someone help me with this question please, thanks!</p>
| Ojas | 154,392 | <p>Substituting <span class="math-container">$x = 1 \implies a \mid 92$</span></p>
<p>Substituting <span class="math-container">$x = -1 \implies (a+2)\mid 88$</span></p>
<p>Divisors of <span class="math-container">$92 = 1,2,4,23,46,92$</span></p>
<p>Divisors of <span class="math-container">$88 = 1,2,4,8,11,22,44,88$</span></p>
<p>Thus, <span class="math-container">$a = 2$</span>.</p>
<p>Indeed, checks out with Wolfram : <a href="https://i.stack.imgur.com/A03iH.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A03iH.gif" alt="enter image description here"></a></p>
|
1,680,210 | <p>Three cards are drawn at random from a full deck. what is the probability of getting a three , a seven, and an ace?</p>
<p>Can someone help me with this question please, thanks!</p>
| Jean Marie | 305,862 | <p>A remark and an alternative proof.</p>
<p><strong>Remark</strong>: this problem as already been posed in the 1963 Putnam competition</p>
<p><a href="https://mks.mff.cuni.cz/kalva/putnam/psoln/psol637.html" rel="nofollow noreferrer">https://mks.mff.cuni.cz/kalva/putnam/psoln/psol637.html</a></p>
<p><strong>An alternative proof</strong>: </p>
<p>The arithmetical proof of @Ojas is perfect, but another one is possible.</p>
<p>I was in fact puzzled by the fact that when I solved (with the aid of Mathematica) the long division by $x^2-x+a$ ($a$ unknown), I obtained the same solution as you, @Ojas,<em>without assuming integer (or rational) values for the coefficients</em>; an afterthought said me that it is perfectly normal because long division with a single unknown coefficient gives (at most) a solution.</p>
<p>Then I thought that there should exist a non-arithmetical proof. Here is one.</p>
<p>Let $P(x)=x^{13}+x+90$; having an odd degree and being strictly increasing (its derivative is everywhere $>0$), $P$ has a unique real root. Thus its factorisation into irreducible factors in $\mathbb{R}[x]$ is a product of a first degree polynomial and 6 second degree polynomials, the latter having complex conjugate roots of $P$. </p>
<p>As a second degree factor $F(x)=x^2-x+a$ is imposed in the question, and because there is only one real root, this real root is a root of $Q(x)$. As the sum of the roots of $F$ is $1$, we have to look for conjugate roots of the form $\frac12 \pm bi$. There are such roots: $r_1, r_2=\frac12(1 \pm i\sqrt{7})$, whence $a=r_1r_2=\frac84=2$, proving the result.</p>
<p>Frankly speaking, I admit that $\frac12(1 \pm i\sqrt{7})$ is the rabbit pulled out of the hat... </p>
<p>By mere curiosity, I have represented on the same figure the roots $z_k$ of $P$ by blue dots and the roots of the "perturbated" polynomial $z^{13}+90$ i.e., $\zeta_k=re^{i(2k+1)\pi/13} \ (k \in \mathbb{Z})$ by black stars, with $r=\sqrt[13]{90}$. The two families are rather close. Roots $z_3$ and $z_{11}$ correspond to $r_1$ and $r_2$ and $z_7$ is the real root.</p>
<p><a href="https://i.stack.imgur.com/IOO1V.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IOO1V.jpg" alt="enter image description here"></a></p>
|
4,070,996 | <p>I'm a bit lost in this integral: <span class="math-container">$$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$</span>
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.</p>
<p>Do you have any ideas? :)</p>
<p>EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution <span class="math-container">$t=\tan(x)$</span>, I got to</p>
<p><span class="math-container">$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$</span></p>
<p>By expanding with 1:
<span class="math-container">$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$</span>
<span class="math-container">$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$</span>
<span class="math-container">$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$</span></p>
<p>And using the substitution: <span class="math-container">$t=\tan\left(x\right)$</span></p>
<p><span class="math-container">$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$</span></p>
<p><span class="math-container">$$t^2=\tan^2\left(x\right)$$</span></p>
<p><span class="math-container">$$t^2=\frac{\sin^2x}{\cos^2x}$$</span></p>
<p><span class="math-container">$$t^2=\frac{1-\cos^2x}{\cos^2x}$$</span></p>
<p><span class="math-container">$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$</span>
<span class="math-container">$$t^2+1=\frac{1}{\cos^2x}$$</span></p>
<p>Using it:
<span class="math-container">$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$</span></p>
<p>I don't think I got to the expected result but I can't seem to be able to find why…</p>
| Claude Leibovici | 82,404 | <p>For <span class="math-container">$n=4$</span>, you can make it shorter since
<span class="math-container">$$A=\frac 1{1+\sin^4(x)}=\frac 1{(\sin^2(x)+i)(\sin^2(x)-i)}$$</span></p>
<p>Using partial fractions and double angle formula
<span class="math-container">$$A=\frac{i}{\cos (2 x)-(1-2 i)}-\frac{i}{\cos (2 x)-(1+2 i)}$$</span> Using now the tangent half-angle
<span class="math-container">$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan
^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$</span></p>
|
918,049 | <p>Let $f: X \to Y$ be a map of sets. We are given that $X$ is a topological space. We are to show that there is a topology on $Y$ making $f$ continuous, and moreover, determine if this topology is unique. </p>
<p>Should I read it as "$X$ is just a set on which there is given a topology, i.e sets that are open in $X$ are predetermined" or "$X$ is a collection of subsets that form a topological space"?</p>
<p>I am not asking for assistance on the exercise itself.</p>
| almagest | 172,006 | <p>The first thing is to establish some notation so that you can apply induction.</p>
<p>Put $a_1=\sqrt{\alpha},a_n=\sqrt{\alpha+a_n}$. Check that that is the series you want. </p>
<p>Now you go about proving it by induction in the usual way. Start by showing that $a_1<(1+\sqrt{4\alpha+1})/2$. Then show that if $a_n<(1+\sqrt{4\alpha+1})/2$, so is $a_{n+1}$.</p>
|
137,813 | <p>Let $k$ be an algebraically closed field of characteristic $0$, let $C_{/k}$ be a nice (smooth, projective, geometrically integral curve), let $K = k(C)$, and let $\overline{K}$ be an algebraic closure of $K$. Let $E_{/K}$ be an elliptic curve with $j(E) \notin k$. Let $P \in E(\overline{K}) \setminus E(K)$ be a point of infinite order. Let $K(P)$ be the field of definition of $P$ (equivalently, the field obtained by adjoining to $K$ the coordinates of $P$ in a Weierstrass equation for $E$). Then </p>
<p>$\langle P, E(K) \rangle \subset E(K(P))$. </p>
<p>Must we have equality? </p>
<p>Comments:</p>
<p>1) For my application, I may assume that $E_{/K}$ is semistable, so please feel free to address the question under that additional hypothesis if it helps. (But I don't see how it does...)</p>
<p>2) I am not able to assume anything about $C$.</p>
<p>3) For my application, I have already dealt with the case in which $P$ has finite order, and in that case I could assume that $C = X_m(n)$ -- the elliptic modular curve parameterizing $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ torsion structures -- and $E$ is the universal elliptic curve over $C$ (the four pairs $(m,n) \in \{(1,1), (1,2), (1,3), (2,2)\}$ in which there is no universal elliptic curve are excluded). In this case the result follows from work of Shioda and Cox-Parry. </p>
<p>4) If this is true, it seems to be closely related to Shioda's landmark 1972 paper on elliptic modular surfaces. I confess that I am asking this question before I have fully absorbed this important paper: I have several collaborators who would be happy if my plate were cleaner. </p>
<p><b>Added</b>: If I may, I'll try one variant of the question: what if $[n]P \in E(K)$ for some $n \in \mathbb{Z}^+$? </p>
| Remke Kloosterman | 8,621 | <p>I will sketch a counterexample for the modified question. The idea behind the construction is similar to the counterexample for the original question. Only the geometric details of this construction are trickier and does not integrate so well in my previous answer. For this reason I will post this a new answer.</p>
<p>For this example we take $k=\mathbb{C}$ and $C=\mathbb{P}^1$.
Take two polynomials $a(t), b(t)$ of degree 2 and 4 and consider the elliptic curve $E_{a,b}: y^2=x(x^2+a(t)x+b(t))$. This is a rational elliptic surface, with Mordell-Weil group $\mathbb{Z}^4 \times \mathbb{Z}/2\mathbb{Z}$, provided that $a$ and $b$ are sufficiently general, e.g., $a$ and $b$ have no common zero and both $b$ and $a^2-4b$ are squarefree.</p>
<p>Take a point $Q\in E_{a,b}(K(t))$ of infinite order. Let $T$ be a point of order 2, different from $(0,0)$ and take for $P$ the point $Q+T$. Then $2P\in E_{a,b}(\mathbb{C}(t))$ and if $a^2-4b$ is not a square then $P$ is not in $E_{a,b}(\mathbb{C}(t))$.</p>
<p>As in my previous answer, it suffices to find a $(a,b)$ such that the twist of $E_{a,b}$ over $K(P)=K(\sqrt{a^2-4b})$ has positive rank. I will sketch two arguments why such an $(a,b)$ exists.</p>
<p>Let $U \subset \mathbb{C}[t]_2\times \mathbb{C}[t]_4$ be the locus of polynomials such that $E_{a,b}$ defines a rational elliptic surface with positive Mordell-Weil rank and precisely one point of order $2$. (The complement of $U$ is the union of the locus where $a$ is a multiple of $b^2$, together with the locus where $a^2-4b$ is a square, and finitely many codimension 4 components parametrizing the Mordell-Weil rank $0$ surfaces.)</p>
<p>For any integer $m$ consider the locus $L_m$ of $(a,b)$ such that the twisted elliptic curve $(a^2-4b)y^2=x(x^2+ax+b)$ has a section of infinite order which intersects the zero section in $m$ points.
Using Hodge theory, Lefschetz $(1,1)$ and the fact that the corresponding elliptic surface has $h^{2,0}=2$ it follows that this locus is locally given by two equations. This implies that $L_m$ is either contained in the complement of $U$ or $L_m$ contains a component of codimenion at most two in $U$.
Now the complement of $U$ has only one component of codimension two. Since for most $m,m'$ we have that $L_m\neq L_{m'}$ we find at least one component of some $L_m$ which is not contained in $U$. Actually, there is a standard argument in Noether-Lefschetz theory to check whether the union of $L_m$ is dense in the analytic topology on $U$ and I expect that this argument applies also here.</p>
<p>A second way to construct an explicit example. It is rather easy to find equations for the loci $L_m$. For $L_0$ you have $19$ equations in $25$ variables. Since the equations are of small degree I expect that with some help from a computer you can find an explicit example. Moreover, modern computer algebra software should be able to check whether $L_0$ is in the complement of $U$ or not.</p>
<p>Edit: If you want that $E_{a,b}$ is semistable then you need to do a bit more work. In this case you need to exclude all $(a,b)$ with a common zero from $U$. This defines a codimension one locus. Hence, depending on the strategy you need to show that $\cup L_m$ is dense or that $L_0$ is not contained in the complement of $U$.</p>
|
2,338,508 | <p>Given this matrix that stretches to infinity to the right and up:
$$
\begin{matrix}
...&...&...\\
\frac{1}{4}& \frac{1}{8}& \frac{1}{16}&... \\
\frac{1}{2} & \frac{1}{4}& \frac{1}{8}&... \\
1 & \frac{1}{2}& \frac{1}{4}&... \\
\end{matrix}
$$</p>
<p>I was trying to find the total sum of this matrix. I know the answer should be $4$. I came up with a different solution and a different answer. What is wrong with that solution? Here it is:</p>
<p>The first row sums to $2$. The second row to $2-1$. The third row to $2-1-\frac{1}{2}$ etc... So we get:</p>
<p>$$
\begin{matrix}
2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}&-\frac{1}{16}\\
2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}\\
2&-1&-\frac{1}{2}&-\frac{1}{4}\\
2&-1&-\frac{1}{2}\\
2&-1 \\
2 \\
\end{matrix}
$$</p>
<p>Now for each "$2$" there is a diagonal that gives the sequence $2-1-\frac{1}{2}-\frac{1}{4}...=0$ (since the matrix goes on forever) Therefore, the sum of the matrix must be $0$!</p>
<p>Apparently that's wrong; but why? Thanks!</p>
<p>EDIT: I am looking for an answer to the question what is fundamentally <strong>wrong</strong> with my method plus an explanation for why that is wrong. I am not looking for an explanation of the correct method.</p>
| Bob Jones | 370,252 | <p>Notice that the second matrix you come up with is not absolutely convergent: you have an infinite number of $2$'s whose sum is $\infty$, and same with an infinite number of $-1$'s. So you can't just rearrange terms like that, and if you do, you get a wrong answer. What you need to do is not write $1$ as $2-1$, but just as $1$. And then $\frac{1}{2}$ is not $2-1-\frac{1}{2}$, but just $\frac{1}{2}$. If you do this, the sequence should be familiar and you should be able to easily find the sum.</p>
|
3,520,538 | <blockquote>
<p>Let <span class="math-container">$X$</span> be a Hausdorff topological space.
Prove that if <span class="math-container">$\{ C_i | i \in I \}$</span> is an infinite collection of compact subsets of <span class="math-container">$X$</span> such that <span class="math-container">$\cap_{i \in I} C_i = \emptyset$</span>, then some finite sub collection of <span class="math-container">$\{ C_i | i \in I \}$</span> also has an empty intersection. </p>
</blockquote>
<p>Well my idea is that I want to show that <span class="math-container">$X$</span> is compact. Since each <span class="math-container">$C_i$</span> is a compact subset of <span class="math-container">$X$</span>, this means each <span class="math-container">$C_i$</span> is closed, which means <span class="math-container">$(X \setminus C_i)$</span> is an open set. I want to say that <span class="math-container">$\cup_{i \in I} (X \setminus C_i)$</span> is an open cover of X but I don't know if this proves <span class="math-container">$X$</span> is a compact set. Do you think I am on the right track? Thank you very much. </p>
| Kavi Rama Murthy | 142,385 | <p>How about <span class="math-container">$u_n=\frac 1 n$</span> for <span class="math-container">$n$</span> even and <span class="math-container">$\frac 1 {n^{2}}$</span> ( or <span class="math-container">$\frac 1 {2^{n}})$</span> for <span class="math-container">$n$</span> odd?</p>
|
1,927,599 | <p>In first-order logic, there is normally a formal distinction between <em>constants</em> and <em>variables</em>. Namely:</p>
<ul>
<li><p>A <em>constant symbol</em> is a $0$-ary function symbol in a language $\mathcal{L}$.</p></li>
<li><p>A <em>variable</em> is one of countably many special symbols used for first-order reasoning, and can be quantified over.</p></li>
</ul>
<p>Instead, suppose we insist that there are no variables, but only constants. And specifically:</p>
<ul>
<li><p>We may quantify over constant symbols, but not over general function or relation symbols.</p></li>
<li><p>There is an infinite supply of unused constant symbols available for use.</p></li>
</ul>
<p><strong>Question: can we develop first-order-logic this way?</strong> Has it been done, and if not, what goes wrong?</p>
<hr>
<h1>Why would you want to do that?</h1>
<p>Both the syntax and the semantics of constants and variables suggest that they are more naturally understood as the same thing. Consider:</p>
<ul>
<li><p><strong>Logical implication for formulas</strong> Let $\varphi, \psi$ be any formulas. We say that $\varphi \vDash \psi$ if, under any model( interpretation of the function symbols and relation symbols), <em>and</em> any assignment of the variables to elements of our model, if $\varphi$ is true then $\psi$ is true.</p>
<p>If variables are just constant symbols, then this simply says that under any interpretation (which must automatically include interpreting the constant symbols), if $\varphi$ is true then $\psi$ is true.</p></li>
<li><p><strong>Axioms with existential statements vs. axioms with constant symbols</strong> Consider the group axioms. There are two different formal presentations of the identity axiom, both commonly used: option (i) is to have a constant symbol $e$ in the language, and assert the axiom that $\forall x (x e = ex = x$). Option (ii) is to assert that $\exists y \forall x (xy = yx = x)$. If you do the first axiom, then your language of groups is $\mathcal{L} = \{*, e\}$; if you do the second axiom, your language is just $\{*\}$. (The unary function for inverse is also sometimes included, $^{-1}$.)</p>
<p>The two formulations give two different theories over two different languages, but they amount to the same thing. However, the equivalence is not immediate because while we can show the theory with the constant symbol $e$ in it can prove the other theory's axioms, going the other way requires "defining" a new symbol $e$ and adding it to the language, which has to be done on a meta level.</p></li>
<li><p><strong>$\exists$ elimination</strong> The $\exists$ elimination rule tells us that from $\exists x \; \varphi(x)$, we may deduce that $\varphi(x)$, where $x$ is a new variable that does not occur free in the current scope. But informally, what is this saying? <em>If we know that there exists an object satisfying a property, we may give it a name</em>. And I think of the <em>name</em> as a constant symbol even though it's formally a variable.</p>
<p>Consider defining $i$ in the complex numbers given the axiom $\exists x \; : \; x^2 = -1$. We employ $\exists$ elimination to obtain $x^2 = -1$ for some variable $x$. But it would be helpful for this variable to then become part of our language, so that $i$ can be considered a constant instead of a variable.</p></li>
<li><p><strong>$\forall$ introduction</strong> If $a$ is a constant symbol, it should be valid to conclude $\forall x \; \varphi(x)$ from $\varphi(a)$, if there are no formulas involving $a$ in scope. But, it isn't allowed. Because we insist on only generalizing from a <em>variable</em> which doesn't occur free in any other statement, we can't generalize from a <em>constant</em> even though it seems we should be able to.</p></li>
</ul>
<h1>Possible issues</h1>
<ul>
<li><p><strong>How do we distinguish between sentences and formulas?</strong> For any language $\mathcal{L}$, we can define a <em>sentence</em> to be a formula where all the constant symbols are either bound, or elements of $\mathcal{L}$.</p></li>
<li><p><strong>What about quantifying over an important constant?</strong> If we are in the langauge of fields, it does seem a bit odd to allow a sentences such as $\forall 0 \; \forall 1 \; (0 + 1 = 1 + 0)$. But this is more bad style than anything else, and doesn't strike me as a problem with the formal system. Bound variables are just dummy variables whose names don't matter; quantifying over constants that are mentioned in the axioms would be discouraged, but not disallowed formally. We would have to say that within the scope of the quantification the axioms about the constant will not apply.</p></li>
</ul>
| Eric Wofsey | 86,856 | <p>The obvious reason to distinguish constants from variables is that sometimes we really do want constants as a part of our structure, and not constants that can be uniquely determined from the other structure like the identity of a group. For instance, a perfectly natural sort of mathematical structure to study is a commutative ring with a distinguished element called $x$ (these are also known as commutative $\mathbb{Z}[x]$-algebras). If you don't distinguish between constants and variables, you can't distinguish between this structure and the structure of a commutative ring. The key reason that this distinguished constant $x$ is different from a variable is that to define a model of this theory, you need to define an interpretation for $x$, but you don't need to define an interpretation for all the infinitely many variables you have (or "unused constant symbols", as you call them).</p>
<p><s>It gets even worse if you try to add axioms. For instance, you might also study commutative rings with a distinguished element $x$ such that $x^2=0$ (these are also known as commutative $\mathbb{Z}[x]/(x^2)$-algebras). If you add an axiom $x^2=0$ to express this, then by your $\forall$ introduction rule, your structure would need to satisfy $(\forall x)x^2=0$, which actually together with the other ring axioms implies that your ring has only one element.</s> (This was based on a misinterpretation of what you meant)</p>
<p>Your argument about "Axioms with existential statements vs. axioms with constant symbols" strikes me as kind of silly. Yes, sometimes the presence of constant symbols that aren't really necessary can make it slightly more complicated to compare two theories. But it is extremely common throughout logic for theories to have multiple different "equivalent" formulations which live over different languages, possibly in much more complicated ways than just inclusion of constant symbols. For instance, you can axiomatize groups using just a binary operation which sends $(x,y)$ to $xy^{-1}$. Eliminating the distinction between constants and variables is hardly going to eliminate the need to have a robust way of defining what this sort of "equivalence" really is.</p>
<p>It's also unclear to me what you're trying to do with $\exists$ elimination. Yes, you can go from $\exists x:x^2=-1$ to $x^2=-1$. But stating the axioms $\exists x:x^2=-1$ does <em>not</em> mean that $x^2=-1$ is true and you can refer to $x$ in other axioms as though this were true. It is only true in the context of an argument in which you have derived $x^2=-1$ as an instantiation of $\exists x:x^2=-1$, in which case I don't see what you have to gain by thinking of $x$ as a "constant" instead of a "variable". Note that including an axiom $x^2=-1$ where $x$ is a constant symbol actually gives you a genuinely different mathematical structure than including an axiom $\exists x:x^2=-1$ does: in the first case, complex conjugation is not an automorphism of $\mathbb{C}$ (since it doesn't fix the distinguished constant $x$), but in the second case it is. If you're trying to say that whenever you have an axiom $\exists x:x^2=-1$ you should also be able to get $x^2=-1$ for free, then it sounds like you're saying you don't want to allow $\mathbb{C}$ as a structure in which conjugation is an automorphism.</p>
<p>That said, the idea of treating certain free variables as though they are new constant symbols is a very natural one that is used all the time in model theory. For instance, it is convenient to think of a formula with $n$ free variables as really being a sentence over an enlarged language which has $n$ new constant symbols (this is basically the idea of "types" in model theory).</p>
|
4,333,566 | <p>Say that <span class="math-container">$\{a_n\}^\infty_\mathrm{n=1}$</span> is a sequence of real numbers so that for all odd <span class="math-container">$n$</span>, <span class="math-container">$a_n \in (0, 1)$</span> and for all even <span class="math-container">$n$</span>, <span class="math-container">$a_n \not\in (0, 1)$</span>.</p>
<p>Say we also know that for some <span class="math-container">$L \in \mathbb{R}$</span>, <span class="math-container">$lim_\mathrm{n\rightarrow\infty}(a_n) = L$</span>.</p>
<p>I'm trying to figure out what I can conclude from this and why. (Assuming undergraduate knowledge of real analysis)</p>
<p>It looks like we know that a limit for this sequence exists in the real numbers, hence the sequence is convergent.</p>
<p>Convergent sequences are bounded by definition?</p>
<p>Any thoughts for what I should be seeing from this?</p>
<p>EDIT: As a commenter pointed out, if considering subsequences, then we know that the subsequential limits of a convergent sequence are all the same. (Or stated: Subsequences of a convergent sequence are convergent)</p>
| Henry | 6,460 | <p>You do not need any property for <span class="math-container">$a$</span>, since its square is a square.</p>
<p>If <span class="math-container">$b=v^2+w^2+z^2$</span> (copying Simon Fox's description) then</p>
<p><span class="math-container">$$a^2b =a^2v^2+a^2w^2+a^2z^2 = (av)^2+(aw)^2+(az)^2$$</span></p>
|
3,818,445 | <p>I'm trying to compute <span class="math-container">$\operatorname{Ext}_{\mathbb{Z}}^1(\mathbb{Z}[1/p],\mathbb{Z})\cong \mathbb{Z}_p/\mathbb{Z}$</span>.</p>
<p>Now I have the projective resolution
<span class="math-container">$$0\rightarrow \bigoplus_{i>1}\mathbb{Z}\xrightarrow{\alpha} \mathbb{Z}\oplus \bigoplus_{i>1}\mathbb{Z}\xrightarrow{\beta} \mathbb{Z}[1/p]\rightarrow 0 .$$</span> The map <span class="math-container">$\alpha$</span> is given by <span class="math-container">$(a_i)_{i>0}\mapsto (-\Sigma a_i, a_ip^i)$</span> and <span class="math-container">$\beta$</span> is given by <span class="math-container">$(b_i)_{i\geq 0}\mapsto \Sigma_{i\geq 0} b_i/p^i$</span>.</p>
<p>Now apply the <span class="math-container">$\operatorname{Hom}(-,\mathbb{Z})$</span>, I want to calculate the kernel of the dualised map <span class="math-container">$\prod_{i>1}\mathbb{Z}\xleftarrow{\alpha^*} \mathbb{Z}\prod (\prod_{i>1} \mathbb{Z})$</span> which is given by <span class="math-container">$(f_0,0,\dots)\mapsto f_0'$</span> and <span class="math-container">$(0,\dots,f_i,\dots )\mapsto (0,\dots,p^if_i,\dots )$</span>, where <span class="math-container">$f_0': \prod_{i>1}\mathbb{Z}\to \mathbb{Z}$</span>, <span class="math-container">$f_0'((a_i))=f_0(\Sigma a_i)$</span>. Is there any way to see what is this kernel and how the quotient of <span class="math-container">$\prod_{i>1} \mathbb{Z}$</span> by this kernel is <span class="math-container">$\mathbb{Z}_p/\mathbb{Z}$</span>?</p>
| Angina Seng | 436,618 | <p>Let's do this by injective resolutions instead. I'll write <span class="math-container">$A$</span> for <span class="math-container">$\Bbb Z[1/p]$</span>.
Then
<span class="math-container">$$0\to\Bbb Z\to\Bbb Q\to\Bbb Q/\Bbb Z\to0$$</span>
gives an injective resolution of <span class="math-container">$\Bbb Z$</span>. Therefore <span class="math-container">$\text{Ext}^1(A,\Bbb Z)$</span> is the
cokernel of
<span class="math-container">$$\text{Hom}(A,\Bbb Q)\to\text{Hom}(A,\Bbb Q/\Bbb Z).$$</span>
It's easy to see that <span class="math-container">$\text{Hom}(A,\Bbb Q)\cong\Bbb Q$</span> via <span class="math-container">$f\mapsto f(1)$</span>.
What is an element of <span class="math-container">$\text{Hom}(A,\Bbb Q/\Bbb Z)$</span>? It is described completely
by <span class="math-container">$f(1/p^k)=a_k+\Bbb Z$</span> where <span class="math-container">$a_k\in\Bbb Q$</span> and <span class="math-container">$pa_{k+1}-a_k\in\Bbb Z$</span>.
The image of <span class="math-container">$\text{Hom}(A,\Bbb Q)$</span> consists of those <span class="math-container">$f$</span> where <span class="math-container">$f(1/p^k)=a/p^k$</span>
for some <span class="math-container">$a\in\Bbb Q$</span>. We can subtract one of these from general <span class="math-container">$f$</span> and
assume that <span class="math-container">$a_0=0+\Bbb Z$</span> and then still <span class="math-container">$pa_{k+1}-a_k\in\Bbb Z$</span>. Then <span class="math-container">$a_k=b_k/p^k$</span>
where <span class="math-container">$b_k\in\Bbb Z$</span> and <span class="math-container">$b_k$</span> is defined modulo <span class="math-container">$p^k$</span>; also <span class="math-container">$b_{k+1}\equiv b_k
\pmod{p^k}$</span>. Thus the <span class="math-container">$(b_k)$</span> represents an element <span class="math-container">$b$</span> of the <span class="math-container">$p$</span>-adic integers <span class="math-container">$\Bbb Z_p$</span>.</p>
<p>We still have some freedom in choosing <span class="math-container">$a$</span>; we need <span class="math-container">$a+\Bbb Z=f(0)+\Bbb Z$</span>, so we
can still change <span class="math-container">$a$</span> by an integer, which changes <span class="math-container">$b$</span> by an integer. So the cokernel
is isomorphic to <span class="math-container">$\Bbb Z_p/\Bbb Z$</span>.</p>
<p>I'm sure all of this can be done by direct and inverse limits....</p>
|
3,414,072 | <blockquote>
<p>Let <span class="math-container">$U \subset \mathbb{R}^N$</span> be an open set, let <span class="math-container">$f : U \times [a, b] \to \mathbb{R}$</span> be a continuous function. Consider the function <span class="math-container">$$g(x):= \int_a^b f(x,y) \,dy$$</span>
with <span class="math-container">$x \in U$</span>.</p>
<p>i) Prove that <span class="math-container">$g$</span> is continuous in <span class="math-container">$U$</span>.</p>
<p>ii) consider the function <span class="math-container">$f : [−1, 1] \times [−1, 1] \to \mathbb{R}$</span> defined by
<span class="math-container">$$f (x, y) :=\begin{cases} \frac {|y|−|x|} {y^2}&\text{ if $|x|<|y|$}\\
0 &\text{ if $|x|\geq |y|$}
\end{cases}$$</span>
Let <span class="math-container">$g(y):= \int_{-1}^1 f(x,y) \,dx$</span> for <span class="math-container">$y \in [-1,1]$</span>
Study the continuity of <span class="math-container">$g$</span>.</p>
</blockquote>
<p>So I am a bit stuck on how to prove the continuity from the basics: i know how to prove that if <span class="math-container">$f$</span> is continuous on <span class="math-container">$[a,b]$</span>, then <span class="math-container">$g=\int f$</span> is continuous on <span class="math-container">$[a,b]$</span> but i assume because of the different notation and dimension here, i have to prove a different way? In addition, how would i study the continuity?</p>
| Marios Gretsas | 359,315 | <p>Let <span class="math-container">$x \in U$</span> such that <span class="math-container">$x_n \to x$</span></p>
<p>Then <span class="math-container">$x_n$</span> is bounded,thus exists a closed ball <span class="math-container">$B:=\overline{B(0,M)}$</span> such that <span class="math-container">$x_n \in B,\forall n \in \Bbb{N}$</span></p>
<p>Take <span class="math-container">$f_n(y)=f(x_n,y)$</span> we have that <span class="math-container">$f(x,y)$</span> on <span class="math-container">$[a,b]$</span> by continuity of <span class="math-container">$f$</span> on <span class="math-container">$B \times [a,b]$</span></p>
<p>In this cartesian product we have the metric <span class="math-container">$d((x_1,y_1),(x_2,y_2))=\sqrt{||x_1-x_2||_2^2+|y_1-y_2|^2}$</span> where <span class="math-container">$||x_1-x_2||_2$</span> is the usual metric on <span class="math-container">$\Bbb{R}^N$</span></p>
<p>Since <span class="math-container">$B\times [a,b]$</span> is compact then <span class="math-container">$f$</span> is uniformly continuous with respect to this metric on <span class="math-container">$B \times [a,b]$</span></p>
<p>Let <span class="math-container">$\epsilon>0$</span></p>
<p><span class="math-container">$\delta>0$</span> and <span class="math-container">$n_0 \in \Bbb{N}$</span> such that <span class="math-container">$||x_n-x||_2<\delta,\forall n \geq n_0$</span></p>
<p>So <span class="math-container">$d((x_n,y),(x,y))=||x_n-x||_2<\delta$</span> for every <span class="math-container">$y \in [a,b]$</span> and <span class="math-container">$\forall n \geq n_0$</span></p>
<p>and <span class="math-container">$|f(x_n,y)-g(x,y)| <\epsilon,\forall y \in [a,b]$</span> by uniform continuity.</p>
<p>Thus <span class="math-container">$\sup_{y \in [a,b]}|f(x_n,y)-f(x,y)| \leq \epsilon, \forall n \geq n_0$</span></p>
<p>Thus <span class="math-container">$f_n(y) \to f(x,y)$</span> uniformply on <span class="math-container">$[a,b]$</span></p>
<p>So by using this and the interchange of limit and riemman integral under uniform convergence,you have the desired conclusion.</p>
<p>For the second part,just study and calculate the integral use the first part.</p>
|
3,969,679 | <p><strong>The letters in the word GUMTREE and KOALA are rearranged to form a 12-letter word where KOALA appears precisely in order but not necessarily together. How many ways can this happen?</strong></p>
<p>So I attmepted it via this method:</p>
<p>Firstly arrange like this since KOALA must be in order but not necessarily together: (let the fullstops (.) be spots for the letters in GUMTREE. There are <span class="math-container">$7$</span> letters in GUMTREE and <span class="math-container">$6$</span> full stops so <span class="math-container">$6^7$</span>.</p>
<p>.K.O.A.L.A.</p>
<p>Butthere are two E's so <span class="math-container">$\frac{6^7}{2!}$</span>. The letters in KOALA are fixed so they have <span class="math-container">$1$</span> way each, except for A (there are two so the first A has two choices and the second A has one choice)</p>
<p>Therefore, <span class="math-container">$$2\cdot\frac{6^7}{2!}=279936$$</span></p>
<p>But the answer is 1995840 arrangements</p>
<p>I belive my method is very close but I am forgetting to multiply by something. Can someone point out my logical flaw? Otherwise the worked solutions propose <span class="math-container">$\frac{12!}{5!2!}$</span>, but I don't get why you divide by 5! for the KOALA, since they are not identical letters... regardless it would be great to understand both ways!</p>
<p>Thanks</p>
| heropup | 118,193 | <p>You are assuming that the letters for the two words are interleaved, when they do not need to be. For example, "KOALAGUMTREE" is a permissible arrangement, as is "GUMTREEKOALA" or "GUMKOALATREE." The problem only stipulates that the letters in "KOALA" appear in sequence.</p>
<p>So, count the number of ways to order the letters in "GUMTREE." There are <span class="math-container">$7$</span> letters, two of which are identical; therefore, there are <span class="math-container">$7!/2!$</span> such arrangements of the letters. For "KOALA," there is only one permissible arrangement. So now all that is left is to figure out how many combined arrangements exist; to do this, note there are a total of <span class="math-container">$7 + 5 = 12$</span> letters in total, hence there are <span class="math-container">$\binom{12}{5}$</span> ways to select the positions of the letters in "KOALA" among the <span class="math-container">$12$</span> total letters.</p>
<p>Hence the correct number of arrangements is <span class="math-container">$$\frac{7!}{2} \binom{12}{5} = 1995840,$$</span> as claimed.</p>
|
758,158 | <p>I am trying to use <span class="math-container">$f(x)=x^3$</span> as a counterexample to the following statement. </p>
<p>If <span class="math-container">$f(x)$</span> is strictly increasing over <span class="math-container">$[a,b]$</span> then for any <span class="math-container">$x\in (a,b), f'(x)>0$</span>. </p>
<p>But how can I show that <span class="math-container">$f(x)=x^3$</span> is strictly increasing?</p>
| evil999man | 102,285 | <p>$$\frac{d x^3}{dx}=3x^2\geqslant0, \forall x \in[1,2]$$</p>
<p>Function is strictly increasing in $[1,2]$ and...</p>
|
758,158 | <p>I am trying to use <span class="math-container">$f(x)=x^3$</span> as a counterexample to the following statement. </p>
<p>If <span class="math-container">$f(x)$</span> is strictly increasing over <span class="math-container">$[a,b]$</span> then for any <span class="math-container">$x\in (a,b), f'(x)>0$</span>. </p>
<p>But how can I show that <span class="math-container">$f(x)=x^3$</span> is strictly increasing?</p>
| André Nicolas | 6,312 | <p>Consider $a^3-b^3$, where $a\gt b$. We want to show this is <strong>positive</strong>.
We have
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
Note that $a^2+ab+b^2$ is positive unless $a=b=0$. There are many ways to do this. For example,
$$a^2+ab+b^2=\frac{1}{2}\left(a^2+b^2+(a+b)^2\right).$$
More conventionally, complete the square. We have $4(a^2+ab+b^2)=(2a+b)^2+3b^2$. </p>
<p><strong>Remark:</strong> If we are in a calculus mood, note that
$$\frac{a^3-b^3}{a-b} =3c^2$$
for some $c$ between $a$ and $b$. This argument breaks down if $c=0$. But that can only happen when $0$ is in the interval $(b,a)$. That means $b$ is negative and $a$ is positive, making the inequality $a^3\gt b^3$ obvious. </p>
|
4,694 | <p>While teaching the concept of vector spaces, my professor mentioned that addition and multiplication aren't necessarily what we <em>normally</em> call addition and multiplication, but any other function that complies with the eight axioms needed by the definition of a vector space (for instance, associativity, commutativity of addition, etc.). Is there any widely used vector space in which alternative functions are used as addition/multiplication?</p>
| whuber | 1,489 | <p>Many of the operations commonly encountered in vector spaces, such as modular arithmetic, functional convolution, or even p-adic arithmetic, ultimately are related to the addition and multiplication with which you are familiar. One interesting exception is Conway's Field of <em>nimbers</em> (which is effectively a one-dimensional vector space). I wouldn't say it's "widely used," but it's well enough known to qualify as a genuine example.</p>
<p>References: <a href="https://mathoverflow.net/questions/6455/nimber-multiplication">https://mathoverflow.net/questions/6455/nimber-multiplication</a></p>
<p><a href="http://en.wikipedia.org/wiki/Surreal_number" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Surreal_number</a> (nimbers are related to the surreal numbers).</p>
|
4,694 | <p>While teaching the concept of vector spaces, my professor mentioned that addition and multiplication aren't necessarily what we <em>normally</em> call addition and multiplication, but any other function that complies with the eight axioms needed by the definition of a vector space (for instance, associativity, commutativity of addition, etc.). Is there any widely used vector space in which alternative functions are used as addition/multiplication?</p>
| Qiaochu Yuan | 232 | <p>My favorite example is the set of subsets of a set under the operation of <a href="http://en.wikipedia.org/wiki/Symmetric_difference">symmetric difference</a> (otherwise known as bitwise XOR). This forms a vector space over the <a href="http://en.wikipedia.org/wiki/Finite_field">finite field</a> $\mathbb{F}_2$. This example is important in computer science, coding theory, combinatorics, ...</p>
|
148,853 | <p>Here is my code;</p>
<pre><code>f =
Exp[-2*x^2]*(1 - Exp[-1/y])*Sqrt[2/π]*
Sum[
Exp[-n/y]*Exp[4*n*χ*x*Sin[θ]]*Exp[-n^2*χ^2*x*(Sin[θ])^2],
{n, 0, 10}];`
g = D[f, y];
f1 = g^2/f;
f2 = Integrate[f1, {x, -∞, ∞}]
</code></pre>
<p>When I evaluate the expression for <code>f2</code>, Mathematica just keeps running without producing andy result. Where may I be wrong?</p>
| Bill | 18,890 | <p>Let's see if we can demonstrate a little progress while it is working. We can use <code>Simplify</code> to sometimes make integrals a little easier. Then because you have a fraction we can use <code>Apart</code> to see if we can break this into a sum of simpler terms. Then we use <code>Expand</code> to try to really force it into separate simpler terms. Finally we use a trick to watch it in the process of doing your problem one piece at a time.</p>
<p>Note: You had <code>Sqrt[2/pi]</code> in your equation. Did you mean that to be <code>Sqrt[2/Pi]</code>? I assume you did.</p>
<pre><code>f = Exp[-2*x^2]*(1 - Exp[-1/y])*Sqrt[2/Pi]*
Sum[Exp[-n/y]*Exp[4*n*χ*x*Sin[θ]]*Exp[-n^2*χ^2*x*(Sin[θ])^2], {n,0,10}];
g = D[f, y];
f1 = g^2/f;
Map[(Print[{"int",#}];zed=Integrate[#,{x,-Infinity,Infinity}]; Print[{"=",zed}]; zed)&,
Expand[Apart[Simplify[f1]]]]
</code></pre>
<p>The <code>Map</code> is going to take the huge sum of terms making up your integral and one term at a time show you what it is about to integrate and then the result of the integration while it is working on the next one.</p>
<p>Thus far mine gets through about 60 terms while I have been formatting this. It may finish. Or it may show you a really ugly integrand and take forever working on it. But at least you can see some indication of progress while it is working on your problem so that you don't have to just wonder if it is doing anything. If it does finish then the result will be a list of the result of the individual integrations and you can use <code>Total</code> to add them all up. <code>Length[Expand[Apart[Simplify[f1]]]]</code> tells me there are going to be 470 separate integrations that it will have to get through.</p>
|
148,853 | <p>Here is my code;</p>
<pre><code>f =
Exp[-2*x^2]*(1 - Exp[-1/y])*Sqrt[2/π]*
Sum[
Exp[-n/y]*Exp[4*n*χ*x*Sin[θ]]*Exp[-n^2*χ^2*x*(Sin[θ])^2],
{n, 0, 10}];`
g = D[f, y];
f1 = g^2/f;
f2 = Integrate[f1, {x, -∞, ∞}]
</code></pre>
<p>When I evaluate the expression for <code>f2</code>, Mathematica just keeps running without producing andy result. Where may I be wrong?</p>
| m_goldberg | 3,066 | <p>I don't think you have a chance of finding a symbolic result for your integral. I think where you went astray was in not exploring the behavior of the integrand before you tried to evaluate the integral. If you had done that, I think you would have gone for a numerical analysis of <code>f2</code>'s behavior, which is quite feasible.</p>
<h3>Basics</h3>
<pre><code>Clear[f, f1, f2, g]
f[x_, y_] :=
Exp[-2*x^2]*(1 - Exp[-1/y])*Sqrt[2/π]*
Sum[
Exp[-n/y]*Exp[4*n*χ*x*Sin[θ]]*Exp[-n^2*χ^2*x*(Sin[θ])^2],
{n, 0, 10}];
g[x_, y_] = D[f[x, y], y];
f1[x_, y_] = g[x, y]^2/f[x, y];
</code></pre>
<h3>Behavior of f1</h3>
<p>To do numerical explorations, we need to pick value the parameters; I will use χ = 1 and θ = 30 ° (I tried several value for each parameter; the values I chose give fairly typical results).</p>
<pre><code>Block[{χ = 1, θ = 30 °},
Plot3D[f1[x, y], {x, -3, 4}, {y, 0, 5},
AxesLabel -> {x, y, z},
PlotRange -> All,
ImageSize -> 500]]
</code></pre>
<p><a href="https://i.stack.imgur.com/ORdXN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ORdXN.png" alt="f1Plot"></a></p>
<p>Experimentation with domain of the plot will show the rectangle given by -3 <= x <= 4 and 0 <= y, <= 5 is where all the action is. So when we build an approximation function for <code>f2</code>, there is no need to integrate over the whole real line.</p>
<h3>Approximating f2</h3>
<pre><code>f2[1, 30 °] =
Block[{χ = 1, θ = 30 °},
Interpolation[Table[{y, NIntegrate[f1[x, y], {x, -3, 4}]}, {y, .5, 4, .01}]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/NTz6T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NTz6T.png" alt="f2Interp"></a></p>
<pre><code>Plot[f2[1, 30 °][y], {y, .5, 4}]
</code></pre>
<p><a href="https://i.stack.imgur.com/U8jHc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U8jHc.png" alt="f2plot"></a></p>
|
725,241 | <p>Where s=circumcenter, H= orthocenter, and A'= midpoint of one side of triangle.
<img src="https://i.stack.imgur.com/Pg6rv.png" alt="enter image description here"></p>
<p>How can can I determine the location of the three vertices of the triangle? </p>
| robjohn | 13,854 | <p>Compare the coefficients of $x^{j-1}$ from
$$
\begin{align}
(1+x)^{k-1}(1+x)^n
&=\sum_{i=1}^k\binom{k-1}{i-1}x^{i-1}\sum_{j=0}^n\binom{n}{j}x^j\\
&=\sum_{i=1}^k\sum_{j=i}^{n+i}\binom{k-1}{i-1}\binom{n}{j-i}x^{j-1}\\
&=\sum_{j=1}^{n+k}\sum_{i=1}^j\binom{k-1}{i-1}\binom{n}{j-i}x^{j-1}
\end{align}
$$
and
$$
(1+x)^{n+k-1}=\sum_{j=1}^{n+k}\binom{n+k-1}{j-1}x^{j-1}
$$
to get that
$$
\sum_{i=1}^j\binom{k-1}{i-1}\binom{n}{j-i}=\binom{n+k-1}{j-1}
$$
Plug in $j=n$, to get
$$
\sum_{i=1}^n\binom{k-1}{i-1}\binom{n}{n-i}=\binom{n+k-1}{n-1}
$$
Since $\binom{n}{n-i}=\binom{n}{i}$, we get
$$
\sum_{i=1}^n\binom{k-1}{i-1}\binom{n}{i}=\binom{n+k-1}{n-1}
$$</p>
|
1,912,660 | <p>I am trying to really understand why the gradient of a function gives the direction of steepest ascent intuitively.</p>
<p>Assuming that the function is differentiable at the point in question,<br>
a) I had a look at a few resources online and also looked at this
<a href="https://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent">Why is gradient the direction of steepest ascent?</a> , a popular question on this stackexchange site.<br>
The accepted answer basically says that we multiply the gradient with an arbitrary vector and then say that the product is maximum when the vector points in the same direction as the gradient? This to me really does not answer the question, but it has 31 upvotes so can someone please point out what I am obviously missing?</p>
<p>b) Does the gradient of a function tell us a way to reach the maxima or minima? if yes, then how and which one - maxima or minima or both?<br>
Edit: I read the gradient descent algorithm and that answers this part of my question.</p>
<p>c) Since gradient is a feature of the function at some particular point - am I right in assuming that it can only point to the local maxima or minima?</p>
| jnyan | 365,230 | <p>This answer is not as detailed as other answers. Start from single dimension. In y=f(x), gradient gives derivative. Now derivative gives direction if steepest ascent.
In two dimensions, the partial derivatives are taken to get the direction in both directions individually. So the movement should be in such a direction in which function increases in both dimensions. So you make the vector addition of partial derivatives. Same logic applies to higher dimensions. If you understand that derivative gives direction of steepest ascent then gradient will make sense.
Hope this helps!!!</p>
<p>Because derivative gives the direction of tangent at a point. By the definition of derivative, tangent direction is obtained. Now, a tangent is the instantaneous movement direction of a curve. If the movement increases the value of function, keep moving in that direction. Derivative doesn't give Maxima or minima direction. It just gives instantaneous movement direction. Now, if the instantaneous movement gives lesser value of the function, meaning the function is decreasing, go in opposite direction. Meaning negative derivative. In one dimension, it means go in left on x-axis</p>
|
1,912,660 | <p>I am trying to really understand why the gradient of a function gives the direction of steepest ascent intuitively.</p>
<p>Assuming that the function is differentiable at the point in question,<br>
a) I had a look at a few resources online and also looked at this
<a href="https://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent">Why is gradient the direction of steepest ascent?</a> , a popular question on this stackexchange site.<br>
The accepted answer basically says that we multiply the gradient with an arbitrary vector and then say that the product is maximum when the vector points in the same direction as the gradient? This to me really does not answer the question, but it has 31 upvotes so can someone please point out what I am obviously missing?</p>
<p>b) Does the gradient of a function tell us a way to reach the maxima or minima? if yes, then how and which one - maxima or minima or both?<br>
Edit: I read the gradient descent algorithm and that answers this part of my question.</p>
<p>c) Since gradient is a feature of the function at some particular point - am I right in assuming that it can only point to the local maxima or minima?</p>
| amd | 265,466 | <p>Let’s try coming at it from a different direction, so to speak. </p>
<p>Consider the plane in $\mathbb R^3$ given by $ax+by=z$. The vector $\mathbf n=\langle a,b,-1\rangle$ is normal to this plane. A bit of thought should convince you that the projection of $\mathbf n$ onto the $xy$ plane, $\langle a,b\rangle$, points in the direction in which this plane is steepest. It’s fairly straightforward to prove this analytically, but you can also see this by visualizing cutting a cylinder centered on the $z$-axis with this plane and imagining what happens to the high point of the cut as you tilt the plane in various directions. Displacing the plane from the origin doesn’t change its inclination, so $\langle a,b\rangle$ also gives the steepest direction for any other plane with the same normal, i.e., for $ax+by-z=c$. </p>
<p>Moving now to a curved surface, by analogy with functions of one dimension, we define instantaneous rates of change in terms of tangents to the surface. We’re assuming that the function which defines our surface is suitably well-behaved, so all of these tangents lie in a well-defined <em>tangent plane</em> to the surface. Looking at it another way, this tangent plane captures the rates of change of the function in all directions. As above, then, a “downward” normal to this plane will give us the direction of fastest increase. All we need to do now is find such a normal vector. </p>
<p>Let a surface in $\mathbb R^3$ be given by $F(x,y,z)=c$. Consider a curve $\gamma: t\mapsto(x(t),y(t),z(t))$ on this surface that passes through the point $P_0 = \gamma(0)$, so that we have $(F\circ\gamma)(t)=c$. (Again, we’re assuming that these functions are suitably well-behaved so that this parametrization exists.) Differentiating both sides with respect to $t$ and applying the chain rule gives $$F_x(P_0)x'(0)+F_y(P_0)y'(0)+F_z(P_0)z'(0)=\nabla F(P_0)\cdot\gamma'(0)=0.$$ Now, $\gamma'(0)$ is tangent to $\gamma$ at $P_0$ and thus lies in the tangent plane. Since $\gamma$ was arbitrary, we can conclude that $\nabla F$ is orthogonal to every tangent vector to the surface at $P_0$, i.e., that it is normal to the tangent plane. </p>
<p>For a surface given by $z=f(x,y)$ this normal vector is $\langle f_x,f_y,-1\rangle$, and its projection $\nabla f$ thus points in the direction of steepest ascent along the surface, i.e., the direction in which $f$ increases fastest. </p>
<p><em>Afterthought:</em> Going back to the original plane example at the top, we can see why this result is plausible. A plane in $\mathbb R^3$ is completely specified by its $x$-slope/rate-of-change $a$, its $y$-slope $b$ and a point on the plane. For the tangent plane to the surface $z=f(x,y)$, these rates of change in the directions of the coordinate axes are given by the partial derivatives of $f$, which are encoded in its gradient.</p>
|
3,144,398 | <p>Q) Suppose the rank of the matrix</p>
<p><span class="math-container">$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$</span></p>
<p>is <span class="math-container">$2$</span> for some real numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. Then <span class="math-container">$b$</span> equals</p>
<p><span class="math-container">$(A)$</span> <span class="math-container">$1\;\;\;$</span> <span class="math-container">$(B)$</span> <span class="math-container">$3\;\;\;$</span> <span class="math-container">$(C)$</span> <span class="math-container">$1/2\;\;\;$</span> <span class="math-container">$(D)$</span> <span class="math-container">$1/3\;\;\;$</span></p>
<p><strong>My Approach :-</strong> Since rank is <span class="math-container">$2$</span> , So, the determinant of all the submatrices of order <span class="math-container">$3 \times 3$</span> must be zero.</p>
<p>So, <span class="math-container">$\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$</span> = <span class="math-container">$0$</span>. After solving it, I am getting <span class="math-container">$a=b$</span> </p>
<p>and <span class="math-container">$\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$</span> = <span class="math-container">$0$</span>. After Solving it , <span class="math-container">$b=\frac{1}{3}$</span> </p>
<p>Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then</p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$</span> </p>
<p>Applying <span class="math-container">$R_{2} \leftarrow R_{2}-R_{1}$</span> </p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$</span></p>
<p>Now, Applying <span class="math-container">$R_{3}\leftarrow R_{3}-aR_{1}$</span> </p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$</span> </p>
<p>Now, to make rank of this matrix = <span class="math-container">$2$</span> means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.</p>
<p>So, <span class="math-container">$(b-a) =0$</span> and <span class="math-container">$(b-2a) =0$</span> and <span class="math-container">$(1-2a) =0$</span> </p>
<p>So, <span class="math-container">$b=a$</span> and <span class="math-container">$b=2a$</span> and <span class="math-container">$a=\frac{1}{2}$</span> </p>
<p>Now, My doubt is how <span class="math-container">$b=a$</span> and <span class="math-container">$b=2a$</span> is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.</p>
| Michael Albanese | 39,599 | <p>Let <span class="math-container">$M$</span> be a manifold with boundary. Here's a trick you can use if <span class="math-container">$\partial M$</span> is connected.</p>
<p>The double of <span class="math-container">$M$</span> is defined as <span class="math-container">$D(M) = M\cup_{\partial M} M$</span>; it is a closed manifold of the same dimension. If <span class="math-container">$X$</span> is an open manifold which embeds in <span class="math-container">$M$</span>, then it embeds in <span class="math-container">$D(M)$</span>. Using the fact that <span class="math-container">$\partial M$</span> is connected, it follows from the Seifert-van Kampen Theorem that if <span class="math-container">$M$</span> is simply connected, then so is <span class="math-container">$D(M)$</span>.</p>
<p>Note, this trick doesn't work if <span class="math-container">$\partial M$</span> is disconnected. For example, the open disc <span class="math-container">$D^n$</span> embeds in <span class="math-container">$M = S^{n-1}\times [0,1]$</span>, so it embeds in the double <span class="math-container">$D(M) = S^{n-1}\times S^1$</span>, but this is no longer simply connected.</p>
|
3,144,398 | <p>Q) Suppose the rank of the matrix</p>
<p><span class="math-container">$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$</span></p>
<p>is <span class="math-container">$2$</span> for some real numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. Then <span class="math-container">$b$</span> equals</p>
<p><span class="math-container">$(A)$</span> <span class="math-container">$1\;\;\;$</span> <span class="math-container">$(B)$</span> <span class="math-container">$3\;\;\;$</span> <span class="math-container">$(C)$</span> <span class="math-container">$1/2\;\;\;$</span> <span class="math-container">$(D)$</span> <span class="math-container">$1/3\;\;\;$</span></p>
<p><strong>My Approach :-</strong> Since rank is <span class="math-container">$2$</span> , So, the determinant of all the submatrices of order <span class="math-container">$3 \times 3$</span> must be zero.</p>
<p>So, <span class="math-container">$\begin{vmatrix} 1&1 &2 \\ 1&1 &1 \\ a&b &b \end{vmatrix}$</span> = <span class="math-container">$0$</span>. After solving it, I am getting <span class="math-container">$a=b$</span> </p>
<p>and <span class="math-container">$\begin{vmatrix} 1&2 &2 \\ 1&1 &3 \\ b&b &1 \end{vmatrix}$</span> = <span class="math-container">$0$</span>. After Solving it , <span class="math-container">$b=\frac{1}{3}$</span> </p>
<p>Now, Rank of a matrix is also defined as no. of non-zero rows in row echelon form of that matrix. So, If I convert it into Row Echelon form then</p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{bmatrix}$</span> </p>
<p>Applying <span class="math-container">$R_{2} \leftarrow R_{2}-R_{1}$</span> </p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ a& b &b &1 \end{bmatrix}$</span></p>
<p>Now, Applying <span class="math-container">$R_{3}\leftarrow R_{3}-aR_{1}$</span> </p>
<p><span class="math-container">$\begin{bmatrix} 1 &1 &2 &2 \\ 0&0 &-1 &1 \\ 0&(b-a) &(b-2a) &(1-2a) \end{bmatrix}$</span> </p>
<p>Now, to make rank of this matrix = <span class="math-container">$2$</span> means I should have 2 non-zero rows or all the elements of the last row must be zero simultaneously.</p>
<p>So, <span class="math-container">$(b-a) =0$</span> and <span class="math-container">$(b-2a) =0$</span> and <span class="math-container">$(1-2a) =0$</span> </p>
<p>So, <span class="math-container">$b=a$</span> and <span class="math-container">$b=2a$</span> and <span class="math-container">$a=\frac{1}{2}$</span> </p>
<p>Now, My doubt is how <span class="math-container">$b=a$</span> and <span class="math-container">$b=2a$</span> is possible here simultaneously and why this method is giving wrong result. I must be doing some mistake here but I am not getting what mistake I am doing. Please help.</p>
| Moishe Kohan | 84,907 | <p>Let me answer the second part of your question, namely, about the uniqueness of the boundary. More precisely, the setup for Siebenmann's compactification is that you have an open manifold <span class="math-container">$M$</span> (every component is noncompact). Then Siebemnann gives you some conditions under which there exists a compact manifold <strong>necessarily with nonempty boundary</strong> <span class="math-container">$N$</span> such that <span class="math-container">$M$</span> is homeomorphic to the interior of <span class="math-container">$N$</span>, i.e. to <span class="math-container">$N- \partial N$</span>. In this situation one can ask if <span class="math-container">$\partial N$</span> is uniquely determined by <span class="math-container">$M$</span>. The answer to this question is negative, see Danny Ruberman's example <a href="https://mathoverflow.net/questions/307704/is-the-boundary-of-a-manifold-topologically-unique?noredirect=1&lq=1">here</a>. </p>
|
1,506,805 | <p>$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$</p>
| Olivier Oloa | 118,798 | <p><strong>Hint.</strong> You may observe that
$$
\cos\left(x+\frac{\pi }{2}\right)=-\sin x
$$ then your limit is equal to
$$
-\lim _{x\to 0}\left(\frac{\sin x}{x}\right)=?
$$ which is easier to obtain.</p>
|
2,849,303 | <p>Given two distinct factorizations of a positive integer with the same number of factors (not necessarily prime or all distinct), must the sums of the respective sets of factors also be distinct? This question arises frequently in puzzles of the KenKen or Killer Sudoku type. I have found no obvious counter examples searching by hand. For the purpose at hand, the numbers being factored may be limited to less than 1000, say.</p>
| dxiv | 291,201 | <p>The sums can match, for example $\,144 = 8 \cdot 6 \cdot 3 = 4 \cdot 4 \cdot 9\,$ with $\,8+6+3=4+4+9\,$.</p>
<p><hr>
[ <em>EDIT</em> ] Also, $144 = 2\cdot8\cdot9 = 3 \cdot 4 \cdot 12$ with $\,2+8+9=3+4+12\,$, so multiple such factorizations may exist for the same number.</p>
<p>Morevover, there exist such with the same sum e.g. $\,1680 = 4 \cdot 20 \cdot 21 = 5 \cdot 12 \cdot 28 = 7 \cdot 8 \cdot 30\,$ with $\,4+20+21=5+12+28=7+8+30\,$.</p>
<p><hr>
[ <em>EDIT #2</em> ] The $\scriptsize\color{silver}{\text{(quick-and-dirty)}}$ Python code used to lookup the triplets of factors:</p>
<pre><code>n = 2000 # upper bound of range to check
k = 2 # minimum number of matching triples that get listed
m = 2 # change to 1 to allow unit factors
o = 0 # change to 1 to disallow identical factors in a triple
px = [{} for i in range(n)]
for a in range(m, n):
for b in range(a + o, n // a):
for c in range(b + o, n // ( a * b)):
p = a * b * c; s = a + b + c
px[p][s] = px[p].get(s, []) + [(a, b, c)]
for i in range((o+1)**3, n):
for j in sorted(px[i].keys()):
if len(px[i][j]) >= k:
print(str(i) + "\t+" + str(j) + "\t" + str(px[i][j])[1:-1])
</code></pre>
<p>Some more:</p>
<ul>
<li>smallest number that has $3$ sets of $3$ triples each that sum to different values:</li>
</ul>
<p>$$
\begin{matrix}
5400 &= 5 \cdot 30 \cdot 36 &= 6 \cdot 20 \cdot 45 &= 9 \cdot 12 \cdot 50 &\quad\quad \style{font-family:inherit}{\text{sum}} &= 71\\
&= 5 \cdot 24 \cdot 45 &= 6 \cdot 18 \cdot 50 &= 10 \cdot 10 \cdot 54 & & = 74\\
&= 4 \cdot 30 \cdot 45 &= 5 \cdot 20 \cdot 54 &= 9 \cdot 10 \cdot 60 & &= 79\\
\end{matrix}
$$</p>
<ul>
<li>smallest number that has $4$ sets of $4$ triples each that sum to different values:</li>
</ul>
<p>$$
\small\begin{matrix}
166320 &= 20 \cdot 77 \cdot 108 &= 22 \cdot 63 \cdot 120 &= 24 \cdot 55 \cdot 126 &= 28 \cdot 45 \cdot 132 &\quad \style{font-family:inherit}{\text{sum}} &= 205\\
&= 16 \cdot 99 \cdot 105 &= 18 \cdot 70 \cdot 132 &= 21 \cdot 55 \cdot 144 &= 30 \cdot 36 \cdot 154 & & = 220\\
&= 11 \cdot 105 \cdot 144 &= 14 \cdot 66 \cdot 180 &= 16 \cdot 55 \cdot 189 &= 20 \cdot 42 \cdot 198 & & = 260 \\
&= 5 \cdot 154 \cdot 216 &= 6 \cdot 105 \cdot 264 &= 8 \cdot 70 \cdot 297 &= 21 \cdot 24 \cdot 330 & & = 375
\end{matrix} \\
$$</p>
|
3,557,398 | <p>Here are two second-order differential equations.</p>
<p><span class="math-container">$$ y''+9y=\sin(2t) \tag 1 $$</span></p>
<p><span class="math-container">$$ y'' +4y =\sin(2t) \tag 2 $$</span></p>
<p>I am told to use undetermine coefficients method to solve.</p>
<p>For 1), I use <span class="math-container">$y_p=A \cos(2t)+B \sin(2t)$</span> to get <span class="math-container">$A=0$</span> and B=<span class="math-container">$\frac{1}{5}$</span> and get <span class="math-container">$y_p=\frac{1}{5} \sin(2t)$</span></p>
<p>For 2), I realize that that method doesn't work and told to do <span class="math-container">$y_p=t(A \cos(2t)+B \sin(2t)$</span> Why does it work then?</p>
| Surajit | 528,430 | <p>The relation is symmetric in <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Now divide it in three cases, and deduce the inequality. </p>
<p><strong>Case I:</strong> <span class="math-container">$x+y<z$</span>.</p>
<p><strong>Case II:</strong> <span class="math-container">$z\leq \min(x,y)$</span>.</p>
<p><strong>Case III:</strong> <span class="math-container">$x+y\geq z$</span> and <span class="math-container">$x\leq z \leq y$</span>.</p>
|
1,614,899 | <p>Construct Context-free Grammar for integers. Integer can begin with + or - and after that we have non-empty string of digits. Integer must not contain unnecessary leading zeros and zero should not be preceded by + or -.
For example: 0; 123; -15; +9999 are correct, but +0; 01; +-3; +09; + are incorrect.</p>
<p>I have something like this:</p>
<p>(number) ::= (unsigned number) | (sign)(unsigned number)</p>
<p>(sign) ::= + | – </p>
<p>(unsigned number) ::= (digit) | (unsigned number)(digit) </p>
<p>(digit) ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |</p>
<p>| - or</p>
<p>Is it okay? ;)</p>
| Elle Najt | 54,092 | <p>Hint: (0,1), or any bounded open set.</p>
|
1,832,515 | <p>I know that you almost always set domains for a summation function $\left( \sum \right)$, but can you also set an interval for that domain? Say the domain was 1 to 10, could I set an increment of 0.5 instead of the standard 1? I know that there are ways around this such as setting another variable for the previous number, subtracting 0.5 from the current loop variable, and so on, but is there a way to simply set an interval so that I don't have to do any of that? </p>
| lab bhattacharjee | 33,337 | <p>As $8-3=x^2-2x+8-(x^2-2x+3)$
$=(\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3})(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})$</p>
<p>$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125\iff(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})=\dfrac{8-3}{125}$$</p>
<p>Adding we get $2\sqrt{x^2-2x+8}=125+\dfrac1{25}=?$</p>
<p>Now square both sides. </p>
|
2,653,713 | <p>Let $\varphi:\mathbb{R}\rightarrow \mathbb{R}$ be a convex function. If $y<\varphi(x)$ why does there exist a line through $(x,y)$ which lies strictly below the graph of $\varphi$? I ask because this is a step in the proof of Jensen's inequality for conditional expectation.</p>
| BoolHool | 532,226 | <p>Okay so, for your first question, all the points in the form of the line $y=x$ is in the form $(a,a)$ (So this is our best alternative, keep in mind however that the $x$'s in $y=x$ is not the same as the $x$'s in $y=e^x$, as discussed in the comments). As you mentioned, you are using the Euclidean Metric. So we need to find the minimum distance between $(a,a)$ and $(x,e^x)$. We can start at:
$$g(x) = \sqrt{(a-x)^2+(a-e^x)^2}$$
So we would like to find $x$ such that this distance is minimal for a given $a$.
We can take the derivative of $g$, treating $a$ as a constant to find that:
$$g'(x) = \frac{-2e^x(a-e^x)-2(a-x)}{2\sqrt{(a-e^x)^2+(a-x)^2}}$$
So we set $g'(x) = 0$ and then solving for $a$ would give us the closest point of the form $(a,a)$ for a given point $(x,e^x)$. We do the math and we find that:
$$a = \frac{e^{2x}+2x}{e^x+2}$$
Since the $a$ minimizes the distance $g$ we just plug it back in. Therefore, the function that represents the minimum distance for each point in the form $(a,a)$ to the curve $y=e^x$, as you said, "for every $x$" is:
$$g(x) = \sqrt{(\frac{e^{2x}+2x}{e^x+2}-x)^2+(\frac{e^{2x}+2x}{e^x+2}-e^x)^2}$$</p>
<p>Your second question is too unclear, and I'll have to make too many assumptions to answer it</p>
|
2,283,367 | <p>I know product of nth roots of unity is 1 or -1 depending whether n is odd or even.
But in this way I am getting 1. Where am I wrong?</p>
<p>$ \text{Let }\alpha = \cos \frac{2 \pi}{n} + \iota \sin \frac{2 \pi}{n} \text{ be a root of }x^n=1 \\
\text{Then product of nth roots will be } 1\cdot \alpha \cdot \alpha^2 ... \alpha^{n-1} = \alpha^{\frac{(n)(n-1)}{2}}\\
=\left( \alpha^n \right)^{\frac{n-1}{2}}\\
=1^{\frac{n-1}{2}} \text{......By the definition of alpha ??}\\
=1 $</p>
<p>I can see this doesn't even work for n=2.</p>
| florence | 343,842 | <p>Consider the following:
$$(-1)^1 = (-1)^{2\cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}=1^{\frac{1}{2}} = 1$$
You did something like that. $n(n-1)/2$ is always an integer when $n$ is a natural number, but when you take out the $n$ you are left with $(n-1)/2$ which is not always an integer, which messes things up when $n$ is even. In short, the rule $a^{bc} = (a^b)^c$ does not always hold for complex numbers, in particular when $b$ or $c$ are not integers. </p>
|
1,255,215 | <p>How to prove that:</p>
<p>$$\cos{n\theta}=\cos^n{\theta}-
\binom {n} {2}\cos^{n-2} \theta \cdot \sin^2 \theta+
\binom {n} {4}\cos^{n-4} \theta \cdot \sin^{4} \theta -\cdots$$</p>
<p>$$\sin n\theta =
\binom {n} {1}\cos^{n-1} \theta \cdot \sin \theta -
\binom {n} {3}\cos^{n-3} \theta \cdot \sin^3\theta +\cdots$$</p>
<p>I don't see how to start.</p>
| Pauly B | 166,413 | <p>$$\begin{align}
\frac{dy}{dx}&=\frac{\sin t-\cos t}{\cos t+\sin t}\\
&=\frac{\sqrt2(\sin t \cos\frac\pi4-\cos t\sin\frac\pi4)}{\sqrt2(\cos t \cos\frac\pi4+\sin t\sin\frac\pi4)}\\
&=\frac{\sin(t-\frac\pi4)}{\cos(t-\frac\pi4)}\\
&=\tan(t-\frac\pi4)
\end{align}$$</p>
|
101,256 | <p>Can you please help me finding an exact description of the set:</p>
<p>$$ E_{R}=\{\cos{z} | z \in \mathbb{C}, |z|>R\} $$</p>
<p>For any $0<R \in \mathbb{R}$.</p>
<p>My feeling is the $E_R = \mathbb{C}$, for any $R$, but I don't know how to show it, if it's true.</p>
| Christian Blatter | 1,303 | <p>The function $\cos$ is periodic with period $2\pi$; therefore any vertical strip of width $2\pi$ in the $z$-plane, $z=x+iy$, will give rise to the full set of values of $\cos$. By excluding the $z$'s with $|z|\leq R$ there is an infinity of such strips left, so there are no values excluded. It follows that we may as well look at the values of $\cos$ in the strip $0\leq x\leq 2\pi$. Now
$$\cos z=\cosh y \ \cos x - i\ \sinh y\ \sin x\ .$$
Keeping $y\geq0$ fixed and letting $x$ go from $0$ to $2\pi$ the point $w:=\cos z$ describes an ellipse with horizontal semiaxis $\cosh y$ and vertical semiaxis $\sinh y$, and it is easily seen that the family of these ellipses covers all of ${\mathbb C}$.</p>
<p>In this very special example we needed neither the theorem of Casorati-Weierstrass nor the really deep Picard's theorem, but of course the function $\cos$ may serve as an explicit instance for these two theorems.</p>
|
118,680 | <p>The question is motivated by this one <a href="https://mathoverflow.net/questions/118626/real-symmetric-matrix-has-real-eigenvalues-elementary-proof">real symmetric matrix has real eigenvalues - elementary proof</a>: </p>
<p>Are there other fields $F$ than $\mathbb{R}$ (maybe some valued fields or real closed fields) with the property that every symmetric matrix in $M_n(F)$ is diagonalizable ? </p>
| Will Sawin | 18,060 | <p>This is a countable family of first-order statements, so it holds for every real-closed field, since it holds over $\mathbb R$. </p>
<p>From a square matrix, we immediately derive that such a field must satisfy the property that the sum of two perfect squares is a perfect square. Indeed, the matrix:</p>
<p>$ \left(\begin{array}{cc} a & b \\ b & -a \end{array}\right)$</p>
<p>has characteristic polynomial $x^2-a^2-b^2$, so it is diagonalizable as long as $a^2+b^2$ is a pefect square.</p>
<p>Moreover, $-1$ is not a perfect square, or else the matrix:</p>
<p>$ \left(\begin{array}{cc} i & 1 \\ 1 & -i \end{array}\right)$</p>
<p>would be diagonalizable, thus zero, an obvious contradiction.</p>
<p>So the semigroup generated by the perfect squares consists of just the perfect squares, which are not all the elements of the field, so the field can be ordered.</p>
<hr>
<p>However, the field need not be real-closed. Consider the field $\mathbb R((x))$. Take a matrix over that field. Without loss of generality, we can take it to be a matrix over $\mathbb R[[x]]$. Looking at it mod $x$, it is a symmetric matrix over $\mathbb R$, so we can diagonalize it using an orthogonal matrix. If its eigenvalues mod $x$ are all distinct, we are done, because we can find roots of its characteristic polynomial in $\mathbb R[[x]]$ by Hensel's lemma. If they are all the same, say $\lambda$ we can reduce: subtract $\lambda I$, divide by $x$ and diagonalize again. The only remaining case is if some are the same and some are distinct. If we can handle that case, then we can diagonalize any matrix.</p>
<p>Lemma: Let $M$ be a symmetric matrix over $\mathbb R[[x]]$ such that some eigenvalues are distinct mod $x$. There exists an orthogonal matrix $A$ such that $AMA^{-1}$ is block diagonal, with the blocks symmetric.</p>
<p>Proof: Consider the scheme of such orthogonal matrices. Each connected component of this scheme corresponds to a partition of the eigenvalues into blocks. Choose one. Since we can diagonalize the matrix with an orthogonal matrix mod $x$, there is certainly a mod $x$ point on this component. We want to lift this to a point on the whole ring. We can do this if the scheme is smooth over $\mathbb R[[x]]$.</p>
<p>Assuming the blocks have distinct eigenvalues, the variety of ways to do this looks, over an algebraically closed field, like $O(n_1) \times O(n_2) \times.. \times O(n_k)$ where $n_1,\dots,n_k$ are the sizes of the blocks. This is because the only way to keep a diagonal matrix block diagonal is to hit it with one of those. So as long as the blocks are chosen such that the eigenvalues in different blocks are distinct and remain so on reduction mod $x$, the variety is smooth over $\mathbb R((x))$ and smooth over $\mathbb R$, and has the same dimension over both, so is smooth over $\mathbb R[[x]]$. (This bit might not be entirely correct.) Thus there is a lift and the matrix can be put in this form.</p>
<p>Then we do an induction on dimension. The only way we would be unable to put a matrix in a form where two of its eigenvalues are distinct mod $x$ is if its eigenvalues are all the same, in which case,since $\mathbb R((x))$ is contained in a real closed field, it's a scalar matrix and we're done. </p>
|
1,483,343 | <p>For equation
$$
x_1+x_2+x_3 = 15
$$
Find number of positive integer solutions on conditions:
$$
x_1<6, x_2 > 6
$$
Let: $y_1 = x_1, y_2 = x_2 - 6, y_3 = x_3$
than, to solve the problem, equation $y_1+y_2 +y_3 = 9$ where $y_1 < 6,0<y_2, 0<y_3 $ has to be solved. <strong>Is this correct?</strong></p>
<p>To solve this equation by <strong>inclusion-exclusion</strong>, number of solution without restriction have to be found $C_1 (3+9-1,9)$ and this value should be subtracted by $C_2 (3+9-7-1,2)$ , (as the negation of $y_1 < 6$ is $y_1 \geq 7$).
Thus:
$$
55-6=49
$$
Is this the correct answer ? <br>
<strong>Problem must be solved using inclusion-exclusion...</strong></p>
| Klaramun | 189,687 | <p>If such a function $f\colon \mathbb{N} \rightarrow \mathbb{N}$ exists, consider the image set of it
\begin{equation*}
Im(f) = \{ f(n) \}_{n \in \mathbb{N}}
\end{equation*}
Which is a subset of $\mathbb{N}$, so there exists an element there $f(n_0)$ which is the smallest, i.e. $f(n_0) \leq f(n)$ for all $n \in \mathbb{N}$. But then
\begin{equation*}
f(n_0) > f(n_0+1)
\end{equation*}
contradicting the minimality of $f(n_0)$. So such a function cannot exist.</p>
|
125,393 | <p>As we know, Linux’s Bash shell is coming to Windows 10. Now, I want to know, how to using Run[] command or other commands to run some commands like gfortran installed in Linux of Win10 ? And, is it possible to call mathematica using Bash shell in Windows 10 ? </p>
| Emilio Pisanty | 1,000 | <p>As we <a href="http://www.howtogeek.com/249966/how-to-install-and-use-the-linux-bash-shell-on-windows-10/" rel="noreferrer">also</a> <a href="http://www.howtogeek.com/265900/everything-you-can-do-with-windows-10s-new-bash-shell/" rel="noreferrer">know</a> <a href="https://msdn.microsoft.com/en-us/commandline/wsl/about" rel="noreferrer">from</a> <a href="http://www.hanselman.com/blog/DevelopersCanRunBashShellAndUsermodeUbuntuLinuxBinariesOnWindows10.aspx" rel="noreferrer">a</a> <a href="https://blogs.windows.com/buildingapps/2016/03/30/run-bash-on-ubuntu-on-windows/" rel="noreferrer">number</a> of sources, bash-over-windows-10 is in beta, not yet feature complete, and it is not really meant to interface with Windows apps at the moment. If you can get a command to run from a <code>cmd</code> shell (like, <a href="http://www.howtogeek.com/265900/everything-you-can-do-with-windows-10s-new-bash-shell/" rel="noreferrer">say</a>, running <code>bash -c "vi"</code>, which will hopefully launch a graphical <code>vi</code> instance) then it <em>should</em> work just fine through <code>Run[]</code>. Otherwise, you're probably asking too much from asystem that is too young at the moment. Similarly, launching a Mathematica kernel from inside bash seems unlikely to be possible anytime soon.</p>
|
3,572,721 | <p>A finite sum of cosine functions weighted with different amplitude and phase, but with a fixed frequency,<span class="math-container">$$f(x) = \sum_{n=1}^{N}A_{n}cos(x+\phi_{n})$$</span>
the question is if I were to fit <span class="math-container">$f(x)$</span> with <span class="math-container">$cos(x)$</span>, what will be the amplitude and phase offset?</p>
| G Cab | 317,234 | <p>You just obtain a cosine
<span class="math-container">$$
\eqalign{
& \sum\limits_{n = 1}^N {A_{\,n} \cos \left( {x + \phi _{\,n} } \right)} = {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{n = 1}^N {A_{\,n} e^{ix + i\phi _{\,n} } } } \right) = \cr
& = {\mathop{\rm Re}\nolimits} \left( {\left( {\sum\limits_{n = 1}^N {A_{\,n} e^{i\phi _{\,n} } } } \right)e^{ix} } \right) = {\mathop{\rm Re}\nolimits} \left( {\left( {Ce^{i\theta } } \right)e^{ix} } \right) = \cr
& = C\cos \left( {x + \theta } \right) \cr}
$$</span>
whatever be the number and value of the amplitudes and phases, as electrical engineers know very well.</p>
|
240,620 | <p>I would like to train some recreational probability (Puzzles). </p>
<p>Does any of you know a good collection? Preferably with hints or answers.</p>
<p>I've been studying quite a bit of probability theory, but I don't suspect that will do me any real good.</p>
<p>Thanks in advance,</p>
| broccoli | 50,577 | <p>There is a book, <a href="http://books.google.com/books/about/Fifty_Challenging_Problems_in_Probabilit.html?id=QiuqPejnweEC" rel="nofollow">Fifty Challenging Problems in Probability With Solutions</a> by Frederick Mosteller which is quite a good read.
You can look up this <a href="http://bayesianthink.blogspot.com/" rel="nofollow">blog</a> which has a good list of puzzles worked out.</p>
|
139,818 | <p>I'm currently working with large datasets of "discrete data". Each of these is a <em>multiset</em> of real numbers of the form $kS + B$, where $k$ is an integer, and $S$ (the <em>"scale</em>") and $B$ (the <em>"bias"</em>) are real values.</p>
<p>These datasets are such that it is often economical to represent them as "tallies":</p>
<pre><code>dd = <|
"scale" -> 1.234,
"bias" -> 5.678,
"tally" -> {{-5, 2},
{-4, 251},
{-3, 5941},
{-2, 60383},
{-1, 241185},
{ 0, 383613},
{ 1, 241644},
{ 2, 61035},
{ 3, 5686},
{ 4, 259},
{ 5, 1}}
|>
</code></pre>
<p>The <code>"scale"</code> and <code>"bias"</code> entries are $S$ and $B$, as described already. The <code>"tally"</code> entry is a list of pairs $\{k, n\}$, where $n$ is the number of times that $kS + B$ appears in the data.</p>
<p>(The example above is just a toy; in practice, the <code>"tally"</code> elements are much longer, but still far more compact than the simple <code>List</code> representation of the data.)</p>
<p>One can recover the fully expanded representation from the tally representation with the function</p>
<pre><code>(#[["bias"]] + #[["scale"]] Flatten[ ConstantArray @@@ #[["tally"]] ]) &
</code></pre>
<hr>
<p>Even though the tally representation is not a sparse array, it would be nice to give it an interface similar to that of a <code>SparseArray</code> object.</p>
<p>The most important feature of such an interface is <em>the ability to behave correctly in any <strong>context</strong> where normally a list would be expected.</em></p>
<p>For example, it would be nice if <code>Length[dd]</code> automatically behaved like <code>Plus @@ (Last /@ dd[["tally"]])</code> when given a discrete data object <code>dd</code> as argument.</p>
<p>And more generally, if a list is expected, and no special method is available (like the one for <code>Length</code> above), the discrete data object would be automatically expanded to its full list form.</p>
<p>How does one do this in <em>Mathematica</em>?</p>
<hr>
<p>Python, for one, solves this problem by specifying a set of standard methods that will get called on an object when it is in one of a corresponding set of contexts. For example, if for some Python object <code>x</code> the method <code>__length__</code> exists, then the value of the expression <code>len(x)</code> will evaluate to whatever <code>x.__length__()</code> returns. In contexts where <code>x</code> is expected to be iterated over, <code>x.__iterator__</code> will be invoked. Etc.</p>
<p>Therefore, by implementing a few standard methods (like <code>__length__</code> and <code>__iterator__</code>), one can define new classes that can be used like the standard classes (lists, dicts, etc) even if they have completely different internal implementations.</p>
<p>Is there something like this in <em>Mathematica</em>?</p>
| gwr | 764 | <h2>Making discrete Data an abstract data type</h2>
<p>I would try to do something like the following:</p>
<pre><code>(* write a constructor function for the data type 'discrete data' *)
discreteData[
scale_?NumericQ ,
bias_?NumericQ ,
tally : { { _?NumericQ , _?NumericQ } .. }
] := discreteData @ Association[
"scale" -> scale,
"bias" -> bias,
"tally" -> tally
]
(* define upvalues *)
discreteData /: Length[ d_discreteData ] := With[
{ assoc = d[[1]] },
Plus @@ ( Last /@ assoc[["tally"]] )
]
discreteData /: Normal[ d_discreteData ] := With[
{ assoc = d[[1]] },
assoc[["bias"]] + assoc[[ "scale" ]] Flatten [
ConstantArray @@@ assoc[["tally"]]
]
]
(* If you prefer, you can make Dataset the printed form of this data type *)
discreteData/: Format[ d_discreteData ] := With[
{ assoc = d[[1]] },
Dataset @ assoc
]
</code></pre>
<h2>Working with <code>discreteData</code></h2>
<p>Then this will work out as follows: We can construct <code>disctreteData</code> using the constructor function.</p>
<pre><code>data = discreteData[
1.234 , (* scale *)
5.678 , (* bias *)
{ {-5, 2}, {-4, 251}, {-3, 5941}, {-2, 60383},
{-1, 241185}, {0, 383613}, {1, 241644}, {2, 61035}, {3, 5686},
{4,259}, {5, 1} } (* tally *)
]
</code></pre>
<p><a href="https://i.stack.imgur.com/V7X9S.png" rel="noreferrer"><img src="https://i.stack.imgur.com/V7X9S.png" alt="Dataset"></a></p>
<p>While <code>discreteData</code> are displayed as a Dataset, they internally are simply an association given an expressive <code>Head</code>:</p>
<pre><code>InputForm @ data
</code></pre>
<blockquote>
<p>discreteData[<|"scale" -> 1.234, "bias" -> 5.678,
"tally" -> {{-5, 2}, {-4, 251}, {-3, 5941}, {-2, 60383}, {-1, 241185}, {0, 383613},
{1, 241644}, {2, 61035}, {3, 5686}, {4, 259}, {5, 1}}|>]</p>
</blockquote>
<p>We can extract information comfortably using <em>Mathematica's</em> built-in functions (e.g. <a href="http://reference.wolfram.com/language/ref/Length.html" rel="noreferrer"><code>Length</code></a>, <a href="http://reference.wolfram.com/language/ref/Normal.html" rel="noreferrer"><code>Normal</code></a>).</p>
<pre><code>Length @ data
</code></pre>
<blockquote>
<p>1 000 000</p>
</blockquote>
<pre><code>Normal @ data
</code></pre>
<blockquote>
<p>{-0.492, -0.492, 0.742, 0.742, 0.742,...,10.614, 10.614, 10.614, 10.614, 11.848}</p>
</blockquote>
|
2,702,347 | <blockquote>
<p>Prove that $$\left|\dfrac{1}{z^4-4z^2+3}\right|\leq \dfrac{1}{3},\quad\text{ if}\quad |z|=2$$</p>
</blockquote>
<p>I know that I can complete squares on the denominator ($(z^2-2)^2-1$). I know the triangular inequality and some other few things... but I can't work so well with absolute value on $\mathbb{C}$, if anyone could help me...</p>
| Martin R | 42,969 | <p>For $|z|=2$ the triangle inequality gives
$$
|z^2 -3| \ge |z^2| - 3 = |z|^2 - 3 = 1 \, ,\\
|z^2 -1| \ge |z^2| - 1 = |z|^2 - 1 = 3 \, ,
$$
and therefore
$$
|z^4 - 4z^2 +3| = |(z^2-3)(z^2-1)| = |z^2-3 | \cdot | z^2-1 | \ge 1 \cdot 3 = 3 \, .
$$</p>
<p><em>Alternatively:</em> You already found that
$$
z^4 - 4z^2 +3 = (z^2 -2)^2 -1
$$
and for $|z| = 2$ the triangle inequality gives
$$
|z^2 -2| \ge |z|^2 - 2 = 2^2 - 2 = 2 \\
\Longrightarrow
|z^4 - 4z^2 +3| \ge | z^2 -2|^2 -1 \ge 2^2 - 1 = 3 \, .
$$</p>
|
949,320 | <p>how can I solve this:</p>
<blockquote>
<p>How many different ways can 8 children be divided into two groups of 3
and one group of 2?</p>
</blockquote>
<p>My method was:</p>
<p>8C3*5C3*2C2+8C3*5C2*3C3+8C2*6C3*3C3=560+560+560=1680</p>
<p>But the answer is:</p>
<p>8C3*5C3*2C2=560</p>
<p>I feel close to getting the logic but I always seem to miss it... Can anyone point it out to me? I'm bad in perms and combs. Thanks.</p>
| Varun Iyer | 118,690 | <p>The logic behind here is the set of actions that take place.</p>
<p>First, we have $8$ people choosing $3$ kids for the first group. That is $8C3$</p>
<p>Next, we have $5$ kids left, who need to be divided into $3$ kids to form the second group.</p>
<p>Finally, we only have $2$ kids left and one group $2$, so that is $2C2$, or $1$ way to arrange the kids.</p>
<p>So, we get that:</p>
<p>$$8C3*5C3*2C2 = 560$$</p>
|
4,132,907 | <p>I'm trying to figure out the identity above though I'm having difficulties towards figuring it out and would kindly appreciate your support!</p>
<p><span class="math-container">$n\binom{n-1}{r-1} = r\binom{n}{r}$</span></p>
<p>What I have tried:
Given that</p>
<p><span class="math-container">$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$$</span>
Then by rearranging for <span class="math-container">$\binom{n-1}{r-1}$</span> I get</p>
<p><span class="math-container">$$\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}$$</span></p>
<p>Which simplifys to:</p>
<p><span class="math-container">$$n\left[\frac{n!}{(n-r)!r!}-\frac{(n-1)!}{(n-r!)r!}\right]=n\cdot\frac{n!-(n-1)!}{(n-r!)r!}$$</span></p>
<p>I'm stuck here on how to simply this any further to get the result I'm after.</p>
| Vishu | 751,311 | <p><span class="math-container">$$\frac nr \binom{n-1}{r-1} = \frac nr \frac{(n-1)!}{(r-1)! (n-r)!} = \frac{n!}{r!(n-r)!} = \binom nr$$</span></p>
|
41,940 | <p>For example, the square can be described with the equation $|x| + |y| = 1$. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required?</p>
<p>Using the Wolfram Alpha site, this input gave an almost-square:
<code>PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi)</code></p>
<p>This input gave an almost-octagon:
<code>PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi)</code></p>
<p>The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on).</p>
<p>It can be assumed that the centre of the regular polygon is at the origin $(0,0)$, and the radius is $1$ unit.</p>
<p>If there's no such equation, can the non-existence be proven? If there <em>are</em> equations, but only for certain polygons (for example, only for $n < 7$ or something), can those equations be provided?</p>
| jimjim | 3,936 | <p>Depends what you accept as "a general equation" and what is "a plot". For example, under the $l_1$ metric, the equation of what is perceived as a circle ( points having the same distance to a center) ends up looking like a square. Now would that count as genuine plot? Or, are we only allowed the Euclidean metric for plotting? Also, is an answer containing a limit acceptable? For example $$x^{2n}+y^{2n}=r^{2n} \lim_{n \to +\infty}$$ is a square ( or is it? ) see <a href="http://www.wolframalpha.com/input/?i=x%5E18%2by%5E18=1" rel="nofollow">this for $n=18$</a></p>
<p>Also is the equation implicit in x and y or polar and parametric plots are also allowed? </p>
|
2,440,478 | <p>So, I understand that for something to be <em>rigorous</em> it is inclined to meet strict and occasionally extreme criteria. Within the scope of mathematics, this therefore implies that anything that is truly rigorous may not defy any of the rules of mathematics, and as it should with everything in mathematics, but my question is: do the abstract rules align with the geometric rules when it comes to proving statements in question?</p>
<p>For example, I will take the proof that $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$</p>
<p>the most renowned proof of this statement entails the following geometric shape:<img src="https://i.stack.imgur.com/QtBB9.gif" alt="enter image description here"><a href="https://i.stack.imgur.com/QtBB9.gif" rel="nofollow noreferrer">1</a></p>
<p>based on this shape, the following inequality can be seen:
$$\triangle ACB\leq ACB \leq \triangle ADB$$ which can be rewritten as $$\frac{1}{2}\cdot1\cdot1\cdot\sin(x)\leq\frac{1}{2}1^2x\leq\frac{1}{2}\cdot1\cdot\tan(x)$$ $$\frac{1}{2}\sin(x)\leq\frac{1}{2}x\leq\frac{1}{2}\tan(x)$$ then, dividing through by $\frac{1}{2}$, $$\sin(x)\leq x\leq \tan(x)$$ and then by $\sin(x)$, $$1\leq\frac{x}{\sin(x)}\leq\frac{\tan(x)}{\sin(x)}$$ and since $\tan(x)=\frac{\sin(x)}{\cos(x)}, \frac{\tan(x)}{\sin(x)}=\frac{1}{\cos(x)}$ $$1\geq\frac{\sin(x)}{x}\geq\cos(x)$$ from here, the limit can be taken: $$\lim_{x\to0}1\geq\lim_{x\to0}\frac{\sin(x)}{x}\geq\lim_{x\to0}\cos(x)$$ $$1\geq\lim_{x\to0}\frac{\sin(x)}{x}\geq1$$ hence, by the squeeze theorem $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$</p>
<p>the reason for my question is this: why, despite the fact that we identify the inequality by means of observation, and that hence it can clearly be observed that the three identified areas are each one bigger than the other, do we still use the $\leq$ symbol as oppose to the $<$ symbol? Isn't it true that these areas can never be <em>exactly</em> equal to each other? Hence, does the use of the symbol <em>truly</em> act in accordance with the "extreme conditions" of abstract mathematics? <em>(Note: in this context, I am referring to abstract mathematics as anything that is <strong>not</strong> represented visually)</em></p>
<p>I understand that explaining the scenario for one geometric proof does not mean it applies to all geometric proofs, but the question is merely a curious one, and anyone with more experience in mathematics than myself is more than welcome to explain any flaws in my logic. Thank you.</p>
| Theo Bendit | 248,286 | <p>There's nothing fundamentally different about a geometric proof compared to other proofs. It refers to a style of proof, where the steps are designed to be pictured, operating on some kind of prototypical example. When people don't follow along with a picture, the proofs tend to be more dense and less intelligble. Such proofs don't always need to be accompanied by a diagram, but there's always an assumption that the reader is keeping some kind of picture in their head or on paper.</p>
<p>While it is common in geometric proofs to be less rigourous, it is not essential! The choice to be less rigourous is usually chosen because rigour is often accompanied by mind-numbing detail, which spoils the evocative nature of geometric proofs. The details can still be filled in, and there should be rigorous rules that underpin them.</p>
<p>For example, one can show both rigorously and geometrically that any equilateral triangle has angles $\pi/3$, like so:</p>
<p>Suppose $\Delta ABC$ is equilateral. Then $AB = BC = AC$. Since $AB = BC$, using a theorem about isoceles triangles, we obtain $\angle BAC = \angle BCA$. Similarly, from $BC = AC$, we obtain $\angle BCA = \angle ABC$. Using the angle sum theorem, we see that $\angle BCA + \angle ABC + \angle BAC = \pi$, and since they're equal, $\angle ABC = \pi /3$ (and similarly for the other angles).</p>
<p>Everything in the above proof could be verified by a computer, from Euclid's axioms (or maybe just inner product space axioms). It's perfectly rigorous, but it's difficult to read through without having some picture in your head. That's what makes it a geometric proof.</p>
|
3,753,205 | <p>In my general topology textbook there is the following exercise:</p>
<blockquote>
<ul>
<li>(i) - Let <span class="math-container">$\mathcal B$</span> be a basis for a topology <span class="math-container">$\tau$</span> on a non-empty set <span class="math-container">$X$</span>. If <span class="math-container">$\mathcal B_1$</span> is a collection of subsets of <span class="math-container">$X$</span> such that <span class="math-container">$\mathcal B \subseteq \mathcal B_1 \subseteq \tau$</span>, prove that <span class="math-container">$\mathcal B_1$</span> is also a basis of <span class="math-container">$\tau$</span></li>
<li>(ii) - Deduce from (i) that there exists an uncountable number of distinct basis of the euclidean topology on <span class="math-container">$\mathbb R$</span></li>
</ul>
</blockquote>
<p>I already proved statement number (i), but I'm having some trouble to think of a proof for the second statement.</p>
<p>I think that if we consider <span class="math-container">$\mathcal B= \{]a,b[:a,b\in \mathbb R\}$</span> as a basis for the euclidean topology <span class="math-container">$\tau$</span>, then we just need to show that there exists an uncountable number of sets <span class="math-container">$\mathcal B_1$</span> such that <span class="math-container">$\mathcal B \subseteq \mathcal B_1 \subseteq \tau$</span>. But how can I show that?</p>
| Henno Brandsma | 4,280 | <p>One base for <span class="math-container">$\Bbb R$</span> is the countable family <span class="math-container">$\mathcal{B}=\{(a,b): a < b, a,b \in \Bbb Q\}$</span>.</p>
<p>Note that for any irrational <span class="math-container">$w$</span>, <span class="math-container">$\mathcal{B}(w):=\mathcal{B} \cup \{(0,w)\}$</span> is a family of open sets as described in (i). So for every irratonal <span class="math-container">$w$</span>, we have a different new base <span class="math-container">$\mathcal{B}(w)$</span> and as the are uncountably many irrationals, we have at least that many bases for <span class="math-container">$\Bbb R$</span>.</p>
|
703,758 | <p>I need help with a set problem</p>
<p>Given:</p>
<p>$$A=\{(\sqrt{n}+2) \in \Bbb Z \ /\ \ 16\le n^2 \le 1296 \}$$</p>
<p>$$B=\{({3m-2}) \in A \ /\ \ 4 \le 4m+3 \le 17 \}$$</p>
<p>Calculate the value of :
$$n(A)\times n(B)$$</p>
<p>So far I've got into $$ A = \{-8;-7;-6;-5;-4;4;5;6;7;8\} $$
$$B= \{4;5;6;7;8\}$$</p>
<p>therefore
$$n(A) = 10$$
$$n(B) = 5$$
However I don't know if this is correct as the result does not match any of the options given as answer.</p>
| Community | -1 | <p>Not quite: The symbol $\sqrt{n}$ usually refers to the nonnegative square root of $n$, so $A$ should only contain positive numbers. Now</p>
<p>$$\sqrt{1296} = \sqrt{6^4} = 36$$</p>
<p>so the smallest element of $A$ corresponds to $\sqrt{4} + 2$, while the largest corresponds to $\sqrt{36} + 2$; that is, it's equivalent to write</p>
<p>$$A = \{\sqrt{n} + 2 : 4 \le n \le 36\}$$</p>
|
703,758 | <p>I need help with a set problem</p>
<p>Given:</p>
<p>$$A=\{(\sqrt{n}+2) \in \Bbb Z \ /\ \ 16\le n^2 \le 1296 \}$$</p>
<p>$$B=\{({3m-2}) \in A \ /\ \ 4 \le 4m+3 \le 17 \}$$</p>
<p>Calculate the value of :
$$n(A)\times n(B)$$</p>
<p>So far I've got into $$ A = \{-8;-7;-6;-5;-4;4;5;6;7;8\} $$
$$B= \{4;5;6;7;8\}$$</p>
<p>therefore
$$n(A) = 10$$
$$n(B) = 5$$
However I don't know if this is correct as the result does not match any of the options given as answer.</p>
| David | 119,775 | <p><strong>Hint</strong>. The symbol $\sqrt{\ }$ means the positive square root. It doesn't mean plus- or- minus unless you actually write $\pm$. </p>
|
578,248 | <p>Let $X$ be a random variable taking values in some metric space $M$. Let $\{\phi_n\}$ be a sequence of measurable functions from $M$ to another metric space $\tilde M$. Suppose that $\phi_n(X)$ converges in law to a random variable $Y$. Must it be the case that the pairs $(X, \phi_n(X))$ converge in law to a pair $(X, \phi(X))$, where $\phi$ is a measurable function such that $\phi(X)$ has the same law as $Y$?</p>
| grand_chat | 215,011 | <p>A more elementary counterexample: Let $X$ have uniform distribution over $[0,1]$ and define $\phi_n(x):=x$ when $n$ is odd, and $\phi_n(x):=1-x$ when $n$ is even. It's clear that every $\phi_n(X)$ has the same distribution (namely, that of $X$), hence we have convergence in law. However, the pair $(X,\phi_n(X))$ doesn't converge in law to anything, since its distribution keeps flipping back and forth.</p>
|
2,323,235 | <p>When I was trying to evaluate a definite integral (given below the question for those who're curious), I came across a paper of Ramanujan pertaining to the evaluation of the very same integral. Amidst the paper's proofs, Ramanujan mentions the following:</p>
<hr>
<p><em>It's easy to see that</em>:
$$\prod^{n=\alpha}_{n=1}\left\{\frac{\left(1+\frac{\alpha+2\beta}{n}\right)\left(1+\frac{\beta+2\alpha}{n}\right)}{\left(1+\frac{\alpha}{n}\right)^3\left(1+\frac{\beta}{n}\right)^3}\right\}=\frac{[\Gamma(1+\alpha)\Gamma(1+\beta)]^3}{\Gamma(1+\alpha+2\beta)\Gamma(1+\beta+2\alpha)}$$</p>
<p><strong>Right. So how do I go about proving this?</strong></p>
<hr>
<hr>
<hr>
<p>Note: The integral I was attempting to evaluate:
$$\int^{\infty}_0\frac{\tan^{-1}x^3}{e^{4\pi x}-1}dx$$</p>
| Angina Seng | 436,618 | <p>Use the infinite product formula for $\Gamma$:
$$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n}$$
which yields
$$\frac1{\Gamma(z+1)}=\frac1{z\Gamma(z)}=e^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n}$$
etc.</p>
|
2,323,235 | <p>When I was trying to evaluate a definite integral (given below the question for those who're curious), I came across a paper of Ramanujan pertaining to the evaluation of the very same integral. Amidst the paper's proofs, Ramanujan mentions the following:</p>
<hr>
<p><em>It's easy to see that</em>:
$$\prod^{n=\alpha}_{n=1}\left\{\frac{\left(1+\frac{\alpha+2\beta}{n}\right)\left(1+\frac{\beta+2\alpha}{n}\right)}{\left(1+\frac{\alpha}{n}\right)^3\left(1+\frac{\beta}{n}\right)^3}\right\}=\frac{[\Gamma(1+\alpha)\Gamma(1+\beta)]^3}{\Gamma(1+\alpha+2\beta)\Gamma(1+\beta+2\alpha)}$$</p>
<p><strong>Right. So how do I go about proving this?</strong></p>
<hr>
<hr>
<hr>
<p>Note: The integral I was attempting to evaluate:
$$\int^{\infty}_0\frac{\tan^{-1}x^3}{e^{4\pi x}-1}dx$$</p>
| robjohn | 13,854 | <p>I assume that the upper limit of the product is supposed to be $\infty$.</p>
<p><strong>Hint:</strong></p>
<p>By <a href="https://math.stackexchange.com/a/103028/13854">Gautschi's Inequality</a>, $\lim\limits_{m\to\infty}\frac{\Gamma(m+\alpha+1)}{m^\alpha\,\Gamma(m+1)}=1$. Therefore,
$$
\begin{align}
\lim_{m\to\infty}\frac1{m^\alpha}\prod_{n=1}^m\left(1+\frac{\alpha}n\right)
&=\lim_{m\to\infty}\frac1{m^\alpha}\prod_{n=1}^m\left(\frac{\color{#C00}{n+\alpha}}{\color{#090}{n}}\right)\\
&=\lim_{m\to\infty}\frac1{m^\alpha}\color{#C00}{\frac{\Gamma\left(m+\alpha+1\right)}{\Gamma(\alpha+1)}}\color{#090}{\frac{\Gamma(1)}{\Gamma\left(m+1\right)}}\\
&=\lim_{m\to\infty}\frac1{\Gamma(\alpha+1)}\frac{\Gamma\left(m+\alpha+1\right)}{m^\alpha\,\Gamma(m+1)}\\
&=\frac1{\Gamma(\alpha+1)}
\end{align}
$$</p>
|
343,993 | <p>Let $I$ be an ideal of $\mathbb{C}[x,y]$ such that its zero set in $\mathbb{C}^2$ has cardinality $n$. Is it true that $\mathbb{C}[x,y]/I$ is an $n$-dimensional $\mathbb{C}$-vector space (and why)?</p>
| Martin Brandenburg | 1,650 | <p>Here is a positive result (see Fulton, <em>Algebraic curves</em>, Corollary I.7.4 on page 23). Let $k$ be some algebraically closed field. If $I \subseteq k[x_1,\dotsc,x_n]$ is an ideal, then $V(I) \subseteq \mathbb{A}^n(k)$ is finite if and only if $k[x_1,\dotsc,x_n]/I$ is a finite-dimensional $k$-vector space. In this case, we have $|V(I)| \leq \dim_k(k[x_1,\dotsc,x_n]/I)$.</p>
<p>A more precise result is that that $k[x_1,\dotsc,x_n]/\sqrt{I} \cong \prod_{P \in V(I)} k$, in particular $\dim_k(k[x_1,\dotsc,x_n]/\sqrt{I})=|V(I)|$.</p>
|
2,870,039 | <p>Just for the sake of simplicity, assume that the centre of rotation is $(0, 0)$. I know that I can find the destination point $(x', y')$ given angle of rotation $a$ and source point $(x, y)$ using
$$\begin{cases}
x' = x \cos a - y \sin a \\
y' = x \sin a + y \cos a \\
\end{cases}$$</p>
<p>However, I want to ask how can I find the angle of rotation $a$ given the source point $(x, y)$ and only the x-coordination of the destination point $x'$? The y-coordinate of destination point $y'$ would be according to the angle of rotation $a$ that we get. I am having trouble trying to solve the above equations for the desired value because we don't have exact value of $y'$. It would be great if someone could help. Thanks.</p>
| md2perpe | 168,433 | <p>First we replace $\sin a$ with $\sqrt{1-\cos^2a}$:
$$x' = x \cos a - y \sqrt{1-\cos^2 a}$$</p>
<p>Then we eliminate the square root by isolating it and squaring:
$$y \sqrt{1-\cos^2 a} = x \cos a - x'$$
$$y^2 (1-\cos^2 a) = (x \cos a - x')^2 = x^2 \cos^2 a - 2 x x' \cos a + (x')^2$$</p>
<p>This can now be rewritten as a quadric in $\cos a$:
$$0 = (x^2+y^2) \cos^2 a - 2xx' \cos a + ((x')^2-y^2)$$
$$\cos^2 a - 2\frac{xx'}{x^2+y^2} \cos a + \frac{(x')^2-y^2}{x^2+y^2} = 0$$</p>
<p>It can be solve using the "$p$-$q$-formula":
$$
\cos a = \frac{xx'}{x^2+y^2} \pm \sqrt{\left(\frac{xx'}{x^2+y^2}\right)^2 - \frac{(x')^2-y^2}{x^2+y^2}} \\
= \frac{xx' \pm \sqrt{(x^2+y^2)y^2-(x')^2y^2}}{x^2+y^2}
$$</p>
<p>A couple of notes:</p>
<ol>
<li>The identity $\sin a = \sqrt{1-\cos^2a}$ is not always valid. For some $a$ we instead have $\sin a = -\sqrt{1-\cos^2a}.$ You could do this case yourself.</li>
<li>When we square the equation we might add fake solutions. Therefore the solutions should be tested before accepted.</li>
</ol>
|
2,110,111 | <p>For example, concerning de Rham cohomology as explained on the lower part of page 5 <a href="http://www.math.uni-hamburg.de/home/wockel/teaching/data/cohomology_of_lie_algebras_mavraj.pdf">here</a>, one considers the vector space of all forms on a manifold $M$, $A_{DR}(M)$ in the notation of the paper (?).</p>
<p>The $p$-th de Rham cohomology is then defined as the quotient of the kernel and the image of the exterior derivative $d$ acting on the space of $p$ and $p-1$ forms respectively as</p>
<p>$$
H_{DR}^p(M) = \frac{Ker(d: A_{DR}^{p}(M)\rightarrow A_{DR}^{p+1}(M))} {Im(d: A_{DR}^{p-1}(M)\rightarrow A_{DR}^{p}(M))}
$$</p>
<p>Can cohomology always be defined as the kernel divided by the image of a certain operation?</p>
<p>What does cohomology mean or "measure" in as simple as possible intuitive terms in the de Rham case but also more generally?</p>
<p>PS: I tried to read wikipedia of course, but it was (not yet?) very enlightening for me ...</p>
| Michael Albanese | 39,599 | <p>A <em>cochain complex</em> $(A^{\bullet}, d^{\bullet})$ is a sequence of abelian groups $\dots, A^{-2}, A^{-1}, A^0, A^1, A^2, \dots$ and group homomorphisms $d^k : A^k \to A^{k+1}$ such that $d^k\circ d^{k-1} = 0$. That is, we have a complex </p>
<p>$$\dots \xrightarrow{d^{-3}} A^{-2} \xrightarrow{d^{-2}} A^{-1} \xrightarrow{d^{-1}} A^0 \xrightarrow{d^0} A^1 \xrightarrow{d^1} A^2 \xrightarrow{d^2} \dots$$</p>
<p>As $d^k\circ d^{k-1} = 0$, $\operatorname{im} d^{k-1} \subseteq \ker d^k$. Furthermore, as the group $A^k$ is abelian, $\operatorname{im} d^{k-1}$ is a normal subgroup, so we can form the quotient group</p>
<p>$$H^k(A^{\bullet}, d^{\bullet}) := \frac{\ker d^k : A^k \to A^{k+1}}{\operatorname{im} d^{k-1} : A^{k-1} \to A^k}.$$</p>
<p>This is called the $k^{\text{th}}$ <em>cohomology group</em> of the cochain complex $(A^{\bullet}, d^{\bullet})$. </p>
<p>Note that $H^k(A^{\bullet}, d^{\bullet}) = 0$ if and only if the complex is <em>exact</em> at $A^k$ (i.e. $\operatorname{im} d^{k-1} = \ker d^k$). So we can view $H^k(A^{\bullet}, d^{\bullet})$ as a measure of how close the complex is to being exact at $A^k$ - that is, how close the inclusion $\operatorname{im} d^{k-1} \subseteq \ker d^k$ is to being an equality (the bigger $H^k(A^{\bullet}, d^{\bullet})$ is, the more elements of $\ker d^k$ there are which are not in $\operatorname{im} d^{k-1}$). From this point of view, it seems that considering $\ker d^k\setminus \operatorname{im} d^{k-1}$ is a much more natural thing to do. One problem with this approach is that while $A^{k-1}$, $A^k$, $\ker d^k$, and $\operatorname{im} d^{k-1}$ are all groups, $\ker d^k\setminus \operatorname{im} d^{k-1}$ is not a group whereas $H^k(A^{\bullet}, d^{\bullet})$ is. The group structure is very useful, which is why cohomology is defined as it is.</p>
<p>The example you gave, de Rham cohomology, is the cohomology of the cochain complex $(A^{\bullet}_{DR}(M), d^{\bullet})$ where $d^k : A^k_{DR}(M) \to A^{k+1}_{DR}(M)$ is the exterior derivative. Note that $\ker d^k$ is the set of closed $k$-forms and $\operatorname{im} d^{k-1}$ is set of exact $k$-forms, so $H^k_{DR}(M)$ measures how close the statement "every closed $k$-form is exact" is to being true.</p>
<p>Your first question seems to be: </p>
<blockquote>
<p>Do all cohomology groups arise from taking the cohomology of a cochain complex? </p>
</blockquote>
<p>What it means to be a cohomology group (or more precisely, a cohomology theory) can be defined axiomatically using the Eilenberg-Steenrod axioms.</p>
<p>While there are many different ways of defining a collection of cohomology groups $\dots H^{-2}, H^{-1}, H^0, H^1, H^2, \dots$, some of which are not <em>defined</em> as the cohomology of a cochain complex, there does exist a cochain complex $(A^{\bullet}, d^{\bullet})$ whose cohomology groups are isomorphic to the given ones: just take $A^k = H^k$ and $d^k = 0$, then</p>
<p>$$H^k(A^{\bullet}, d^{\bullet}) = \frac{\ker d^k : A^k \to A^{k+1}}{\operatorname{im} d^{k-1} : A^{k-1} \to A^k} = \frac{A^k}{0} \cong A^k = H^k.$$</p>
|
2,110,111 | <p>For example, concerning de Rham cohomology as explained on the lower part of page 5 <a href="http://www.math.uni-hamburg.de/home/wockel/teaching/data/cohomology_of_lie_algebras_mavraj.pdf">here</a>, one considers the vector space of all forms on a manifold $M$, $A_{DR}(M)$ in the notation of the paper (?).</p>
<p>The $p$-th de Rham cohomology is then defined as the quotient of the kernel and the image of the exterior derivative $d$ acting on the space of $p$ and $p-1$ forms respectively as</p>
<p>$$
H_{DR}^p(M) = \frac{Ker(d: A_{DR}^{p}(M)\rightarrow A_{DR}^{p+1}(M))} {Im(d: A_{DR}^{p-1}(M)\rightarrow A_{DR}^{p}(M))}
$$</p>
<p>Can cohomology always be defined as the kernel divided by the image of a certain operation?</p>
<p>What does cohomology mean or "measure" in as simple as possible intuitive terms in the de Rham case but also more generally?</p>
<p>PS: I tried to read wikipedia of course, but it was (not yet?) very enlightening for me ...</p>
| Andrew D. Hwang | 86,418 | <p>$\newcommand{\dd}{\partial}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$To give a more elementary answer intended to complement the existing answers, here are some multivariable calculus-type observations. Throughout, $M$ denotes a compact, smooth manifold of dimension $n$ (possibly disconnected), and differential forms are smooth. Briefly,</p>
<ul>
<li>$p$-dimensional de Rham coholomogy measures "how many linearly-independent $p$-dimensional real homology classes $M$ contains". (William S. Massey's Algebraic Topology article in the <em>Encyclopedia Britannica</em> gives a clear, geometric introduction to homology.)</li>
</ul>
<p>If $X$ is a <a href="https://en.wikipedia.org/wiki/Singular_homology" rel="nofollow noreferrer">smooth $k$-chain</a> and $\omega$ is a smooth $p$-form on $M$, let
$$
\Brak{X, \omega} = \begin{cases}
\int_{X} \omega & k = p, \\
0 & k \neq p,
\end{cases}
$$
denote the "integration pairing". If $X$ is a $(p + 1)$-chain, Stokes's theorem gives
$$
\Brak{\dd X, \omega} = \int_{\dd X} \omega = \int_{X} d\omega = \Brak{X, d\omega}.
$$
With respect to the integration pairing, the boundary operator on chains and the exterior derivative operator on forms are formally adjoint.</p>
<ol>
<li><p>A <em>closed</em> form $\omega$ (i.e., with $d\omega = 0$) satisfies
$$
\Brak{\dd X, \omega} = \Brak{X, d\omega} = 0.
$$
In words, the integral of $\omega$ over a $p$-chain bounding some $(p + 1)$-chain vanishes. Consequently, if $X_{1}$ and $X_{2}$ are homologous $p$-chains (i.e., whose difference is a boundary, say $X_{1} - X_{2} = \dd Y$), then
$$
\Brak{X_{1}, \omega} = \Brak{X_{2}, \omega}.
$$</p></li>
<li><p>An <em>exact</em> $p$-form $\omega$ (i.e., with $\omega = d\eta$ for some $(p - 1)$-form $\eta$) satisfies
$$
\Brak{X, \omega} = \Brak{X, d\eta} = \Brak{\dd X, \eta}.
$$
Particularly, the integral of $\omega$ over a boundaryless $p$-dimensional chain vanishes. Consequently, if $X_{1}$ and $X_{2}$ are $p$-chains having the same boundary, then
$$
\Brak{X_{1}, \omega} = \Brak{X_{2}, \omega}.
$$</p></li>
</ol>
<p>Via the integration pairing, a $p$-form determines a real-valued function on $p$-chains (with real coefficients). Item 1 says that a <em>closed</em> $p$-form determines a real-valued function on <em>real homology classes</em>. Item 2 says this function is unchanged upon addition of an exact $p$-form, i.e., is constant on de Rham cohomology classes.</p>
|
1,950,381 | <p>If $H$ is a subgroup of $S_n$ and if H is not contained in $A_n$ prove that precisely one half of the elements of H are even permutations. I know that multiplying two odd permutations gives a even, and that one even and one odd return an odd permutation but i have no idea where to go from there. </p>
| DonAntonio | 31,254 | <p>Hint:</p>
<p>Let $\;\sigma\in H\;$ be <em>an odd</em> permutation, and now define </p>
<p>$$\;f:A_N\cap H\to H\setminus(A_n\cap H)\;,\;\;f(x):=\sigma x\;$$ </p>
<p>Prove $\;f\;$ is a bijection.</p>
|
2,480,493 | <p>I have a problem with what looks like a very easy equation to solve $\left|\frac{1 + z}{1- i\overline z}\right| = 1$ . ($z$ is a complex number, $\overline z$ is a conjugate of $z$) I got stuck at the point when after replacing $\overline z = a-bi $ and $z = a +bi$ and getting rid of absolute value I end up with $a^2 +a-b =0$. I have no idea how to follow this up or wether I should take totally different approach from the begining. I'd be very gratefull if someone could guide me into the right solution.</p>
| ivanculet | 493,187 | <p>Take this
$$|1+z|=|1-i\bar{z}|$$
then square this babies
$$(1+z)(1+\bar{z})=(1-i\bar{z})(1+iz)$$
$$1+z+\bar{z}+z\bar{z}=1-i\bar{z}+iz+z\bar{z} $$
so
$$ z+\bar{z}=i(z-\bar{z})$$
$$2Re(z)=i(i2Im(z)$$
therefore $$Re(z)=-Im(z)$$</p>
|
485,340 | <p>My question is to to find the period of
$$\sin(8\pi\{x\}),$$
where $\{\cdot\}$-is the fractional part of function.</p>
<p>I know that the period of $\{\cdot\}$ is 1 and the period of $\sin(8\pi x)$ is $1/4$. But how to find the overall period of the given function? </p>
| An_Elephant | 1,020,656 | <p>Let <span class="math-container">$g(f(x))$</span> be the composition of <span class="math-container">$g$</span> and <span class="math-container">$f$</span>.</p>
<p>Further, suppose that <span class="math-container">$g(x)$</span> is an odd function, so that <span class="math-container">$g(f(x)) = -g(f(-x))$</span> .</p>
<p>Then it follows , that the period of <span class="math-container">$g(f(x))$</span> must be same as period of <span class="math-container">$f(x)$</span> because :</p>
<p><span class="math-container">$g(f(x+T)) = g(f(x))$</span></p>
<p>In your example, let <span class="math-container">$g(x) =$</span> sin<span class="math-container">$(8πx)$</span> and <span class="math-container">$f(x) =$</span> {<span class="math-container">$x$</span>}</p>
<p>As established above, period of <span class="math-container">$g(f(x))$</span> is same as that of <span class="math-container">$f(x)$</span> . Hence, period of sin<span class="math-container">$(8π$</span>{<span class="math-container">$x$</span>}) is same as that of {<span class="math-container">$x$</span>} , which is 1 .</p>
|
359,528 | <p>Let $P$ be the set of permutations all of whose cycles are of even length. Prove that the exponential generating function for $P$ is $\dfrac{1}{\sqrt{1-x^2}}$.</p>
| Marko Riedel | 44,883 | <p>This is definitely straightforward using symbolic combinatorics as formalized by Flajolet and Sedgewick. Let $\mathcal{Q}$ be the combinatorial class of permutations consisting only of even cycles.</p>
<p>We have the combinatorial specification
$$ \mathcal{Q} =
\mathfrak{P}(\mathfrak{C}_2(\mathcal{Z}) + \mathfrak{C}_4(\mathcal{Z}) + \mathfrak{C}_6(\mathcal{Z}) + \cdots)$$
where $\mathcal{Z}$ is the singleton class.</p>
<p>Translating this into generating functions we obtain for the required EGF $G(z)$ that
$$ G(z) = \exp\left(\frac{z^2}{2} + \frac{z^4}{4} + \frac{z^6}{6}+\cdots\right) =
\exp
\left(\frac{1}{2}\left(\frac{(z^2)}{1} + \frac{(z^2)^2}{2} + \frac{(z^2)^3}{3}
+\cdots\right)\right) \\=
\exp\left(\frac{1}{2} \log \frac{1}{1-z^2}\right) =
\left(\frac{1}{1-z^2}\right)^{1/2} = \frac{1}{\sqrt{1-z^2}}.$$</p>
|
621 | <p>The OP of this question is asking if <a href="https://mathematica.stackexchange.com/questions/9639">https://mathematica.stackexchange.com/questions/9639</a> could be reopened. I did not want to act unilaterally, so I want to ask the rest of you whether you think the question should be reopened. Plead your case for or against reopening that question here.</p>
<hr>
<p>The intent with the poll answers seems to have been misunderstood, so: <em>they're not supposed to be downvoted</em>. It's a poll; vote up whichever of the two you agree with. Otherwise, this confounds the vote counting.</p>
| rm -rf | 5 | <p>I can speak for myself and why <em>I</em> voted to close. First, let me note that I voted to close as "Not a real question", as did 3 other voters. It just so happens that a mod's choice of close reason trumps that of others and hence it looks like we all voted to close as "Too localized". I do not think it fits the localized description, but I do agree with:</p>
<blockquote>
<p>It is difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered <strong>in its current form</strong>.</p>
</blockquote>
<p>I must, at this point, urge the OP to not read into the above sentences personally, but objectively as these are the system's stock descriptions for close votes.</p>
<p>Coming to the question, here are some observations:</p>
<ul>
<li><p>the title asks <em>"Why is WolframAlpha more intelligent than Mathematica?"</em> while the body offers no reason as to why the choice of adding <code>==0</code> to whatever is specified is an "intelligent" option. As several commenters have observed, <code>==0</code> is a rather arbitrary choice and might make sense in a program like W|A, which needs to take into account the fact that the average user might not necessarily know how to state it mathematically or know W|A's syntax. However, in a programming language, such <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27904_9639">"interpretations of intent" will result in ambiguity and is often undesirable</a>.</p></li>
<li><p>I felt <a href="https://mathematica.stackexchange.com/a/9640/5">acl's answer</a> showed how one can, if they so desired, program rather loosely with <em>Mathematica</em> by getting W|A to interpret its input. But the OP <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27896_9640">dismissed it</a> and insisted on knowing why <em>Mathematica</em> can't interpret itself, which wasn't what the question asked.</p></li>
<li><p>The confusion started when the OP started posting a series of comments and update to the question along the lines of <em>"It is not about syntax or mathematical ability of the user"</em>, which made <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27911_9639">people wonder</a> what it really was about then. There were some more comments on using <em>Mathematica</em>'s intuition and guessing in <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27907_9639">debugging</a>... It was also <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27915_9639">mentioned</a> that the existing scheme provides an excellent way to debug by generating an error message and pointing to the docs.</p></li>
<li><p>Finally, after it was closed, the <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27926_9641">OP mentioned</a> that they thought their question was correctly worded to mean "Is there an option?". Well yes, that was in the question, but given the title and the additional ambiguous comments, it wasn't clear if this really was the crux of the question.</p></li>
</ul>
<p>It is very easy to ask <em>"Why is X so?"</em> or <em>"Why is X not Y?"</em> but it is harder to justify why the "Why..." question is important. This is the distinction between a question that arises from idle curiosity (often construed as a lazy question, placing the intellectual burden solely on the answerer) and one that has been given some thought and consideration with reasonable arguments (often construed as a good question that shares the intellectual burden, even if it is unequal). The latter is what makes a question's purpose clear and motivates an answer and makes it a good fit for this site.</p>
<p>Now in light of <a href="https://mathematica.stackexchange.com/questions/9639/why-is-wolframalpha-more-intelligent-than-mathematica#comment27935_9639">Daniel's comment</a>, it looks like the OP is trying to <a href="https://mathematica.meta.stackexchange.com/questions/621/a-plea-for-reopening-why-is-wolframalpha-more-intelligent-than-mathematica#comment1993_622">reposition their intent</a> as being focussed on the technical aspects/limitations/design considerations etc. of having such a "feature" in <em>Mathematica</em>. This is patently <strong>not</strong> the intent of the question, at least not of the one that has been asked, without having to resort to mind reading.</p>
<p>This brings us to the OP's choices (in my view):</p>
<ul>
<li><p>Rephrase the question body and title to better reflect what exactly they're after (now that they have additional info). Perhaps gear it to the specific details so that DL could provide an authoritative answer (to the extent he can) on the implementation details.</p></li>
<li><p>Let this question run its course (eventual deletion sometime in the future) and ask a fresh question that asks for technical limitations and/or design considerations.</p></li>
</ul>
<p>I would suggest the latter option.</p>
<hr>
<p>At the end of the day, I don't think any of us (I certainly don't) rejoice in closing questions. However, it is necessary at times and things are always done with the community's goals and interests in mind. So given my views above, I would not vote to reopen the question as it stands. However if the OP were to follow the suggestions and rephrase their question, I would gladly vote to reopen.</p>
|
1,869,981 | <p>I have a question in a book which says in how many ways can 6 different things be divided between 2 boys and (my understanding of) the explanation goes something along the lines of:</p>
<p>Items: 1 1 1 1 1 1</p>
<p>The first item can be distributed to either boy 1 or boy 2, i.e. 2 choices the second item could be distributed to either boy 1 or boy 2 therefore you get 2 choices again.
Therefore the number of choices are $2\times2\times2\times2\times2\times2$.</p>
<p>However if I thought of it in another way such as you have 6 items therefore boy 1 can get all 6 and boy 2 could get 0, you could organise it into pairs such that:</p>
<p>\begin{array}{c c}
boy 1 & boy 2 \\
0 & 6 \\
1 & 5 \\
2 & 4 \\
3 & 3 \\
4 & 2 \\
5 & 1 \\
6 & 0
\end{array}</p>
<p>Therefore there are 7 ways of distributing the items. Where am I wrong with my thoughts? I cannot see why my way of thinking is wrong.</p>
| Kiran | 82,744 | <p>The second method does not take care of the fact that items are distinct.</p>
<p>Assume items are $A,B,C,D,E,F$</p>
<p>In the second method, you counted $(1,5)$ as just one.</p>
<p>In fact, it can be</p>
<p>$A~~(B,C,D,E,F)\\
B~~(A,C,D,E,F)\\
C~~(A,B,D,E,F)\\
D~~(A,B,C,E,F)\\
E~~(A,B,C,D,F)\\
F~~(A,B,C,D,E)$</p>
<p>i.e., this (1,5) alone can happen in 6 different ways.</p>
<p>Each of the other cases can be considered in a similar way. Note that your count 7 is true if things are identical.</p>
|
1,057,172 | <p>How many 3digit numbers can be written with $2,4,4,6,6$ ? </p>
<p>I tried $\frac{5.4.3}{2!.2!} = 15$ but it's wrong. </p>
<p>when I solved the question "how many 3digit numbers can be written with $1,1,2$"
the solution $\frac{3.2.1}{2!}$ was correct but why this way doesn't work for above question?</p>
| Mark Joshi | 106,024 | <p>if you are allowed to wait forever: 1. </p>
<p>There are lots of ways to do this. Suppose we ask the question hit 1 or -n first and stop when this happens. </p>
<p>Then since hitting time is a finite stopping time, the value of the random walk stopped at the hitting is still a martingale and so has expectation 0. So probability of hitting $1$ is $n$ times that of hitting $-n.$ So the probability of hitting $1$, is $n/(n+1).$</p>
<p>The probability you want must be bigger than this for any $n$ so it must be $1.$ </p>
|
2,419,785 | <blockquote>
<p>Prove the statement is true using mathematical induction: $$2n-1 \leq n!$$</p>
</blockquote>
<p>My attempt: this is true for $n=1$. </p>
<p>Suppose it is true for $n$, i.e., $2n-1 \leq n!$</p>
<p>Now, $2n-1 \leq n!\implies 2n-1+2 \leq n!+2$ </p>
<p>From here, how do I proceed? </p>
| Robert Israel | 8,508 | <p>By the laws of exponents
$$x^{10n} x^5 \left(x^{-5}\right)^n = x^{10 n + 5 - 5 n} = x^{5 + 5 n}$$
so you want
$$ x^{5+5n} = x^{-10}$$
Assuming $x$ is not $0$ or $\pm 1$, $x^t$ is a one-to-one function of $t$, so this will
imply $$5 + 5 n = -10$$</p>
|
2,546,434 | <p>I have trouble calculating this limit algebraically (without L'Hospital's rule):</p>
<p>$$\lim_{x \to 1} \frac{x^{\pi} - x^e}{x-1}$$</p>
<p>Substituting 1 gives indeterminate form. If this was e.g. $(x^5 - x^2)$ in the numerator, one could easily factor out the $(x-1)$, because $(x^5 - x^2) = x^2 (x - 1) (x^2 + x + 1)$. But I don't know what to do in my problem, because the exponents are not integers.</p>
<p>EDIT: The problem is from a calculus textbook and is before derivatives are introduced so I assume it can be found without derivatives.</p>
| carmichael561 | 314,708 | <p>Hint: write
$$ \lim_{x\to 1}\frac{x^{\pi}-x^e}{x-1}=\lim_{x\to 1}\frac{x^{\pi}-1}{x-1}-\lim_{x\to 1}\frac{x^e-1}{x-1}$$
and then note that each term is a derivative.</p>
|
2,546,434 | <p>I have trouble calculating this limit algebraically (without L'Hospital's rule):</p>
<p>$$\lim_{x \to 1} \frac{x^{\pi} - x^e}{x-1}$$</p>
<p>Substituting 1 gives indeterminate form. If this was e.g. $(x^5 - x^2)$ in the numerator, one could easily factor out the $(x-1)$, because $(x^5 - x^2) = x^2 (x - 1) (x^2 + x + 1)$. But I don't know what to do in my problem, because the exponents are not integers.</p>
<p>EDIT: The problem is from a calculus textbook and is before derivatives are introduced so I assume it can be found without derivatives.</p>
| orion | 137,195 | <p>Try</p>
<p>$$\lim_{x\to 1}x^{e}\frac{x^{\pi-e}-1}{x-1}$$</p>
<p>Now it depends on what you are "allowed" to use. If you assume binomial expansion works for real exponends, then that's it (this is the Taylor expansion, or the well-known first order approximation $(1+\epsilon)^n\approx 1+n\epsilon$).</p>
|
427,663 | <p>If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $</p>
<p>i found one link that had a similar problem.
<a href="https://math.stackexchange.com/questions/343982/prove-if-z-1-and-w-1-then-1-zw-neq-0-and-z-w-over-1">Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$</a></p>
| vadim123 | 73,324 | <p>We have $$\frac{z-w}{1-zw}=\frac{(z-w)(1-\overline{zw})}{(1-zw)(1-\overline{zw})}$$
Note that $(1-zw)(1-\overline{zw})=1-(zw+\overline{zw})+|zw|^2$ is real, so we need only consider the numerator (as Berci hinted).</p>
<p>We calculate $(z-w)(1-\overline{zw})=z-w-|z|^2\overline{w}+|w|^2\overline{z}=(z+\overline{z})-(w+\overline{w})$, which is real.</p>
|
3,715,674 | <p>Let <span class="math-container">$X$</span> be a scheme and <span class="math-container">$U$</span> an open subset of <span class="math-container">$X$</span>. Let <span class="math-container">$s \in \Gamma(U, O_X)$</span> and <span class="math-container">$x \in U$</span>.<br>
I am getting confused with what the difference is between
<span class="math-container">$s_x$</span> and <span class="math-container">$s(x)$</span>... Are they the same thing? </p>
| Jean-Armand Moroni | 1,064,750 | <p>An <em>exact</em> differential has the property that its integral over a path is path-independent: it does not depend upon the taken path, but only upon the origin and end of the path.</p>
<p>In mechanics, the work of a force can be <em>path-dependent</em> (e.g. friction), or <em>path-independent</em> (e.g. gravity), in which case it is also called <em>conservative</em>. When it is path-independent, the force can be associated with a <em>potential energy</em>, and the work is only the (opposite of the) potential energy variation; i.e. the difference between potential energies at origin and end of the path.
<a href="https://en.wikipedia.org/wiki/Work_(physics)#Work_and_potential_energy" rel="nofollow noreferrer">Work and potential energy (Wikipedia)</a></p>
<p>Then why the path-independent case was called "exact"? Probably because calculating the work involves no numerical approximation: it is the difference between two values of a potential energy, which has a known expression for usual forces such as gravity.</p>
<p>On the other hand, when there is a non-conservative force such as friction, its contribution must be calculated by an integral and may not have a closed expression, hence the computing should be numerical and give an inexact result.</p>
<p>See also <a href="https://physics.stackexchange.com/questions/603147/why-does-an-exact-differential-mean-a-force-is-conservative">https://physics.stackexchange.com/questions/603147/why-does-an-exact-differential-mean-a-force-is-conservative</a>.</p>
|
2,315,236 | <blockquote>
<p>Let $I := [0, \frac{\pi}{2}]$ and let $f : I \to \Bbb{R}$ be defined by $f(x) = \sup \{x^2, \cos(x)\}$ for $x \in I$. Show there exists an absolute minimum point $x_0 \in I$ for $f$ on $I$. Show that $x_0$ is a solution to the equation $x^2 = \cos(x)$. </p>
</blockquote>
<p>I am having a little trouble with this problem. From the max/min theorem, we know that such an $x_0$ described above exists; moreover, by the intermediate value theorem it is clear that there exists a $z \in I$ such that $z^2 = \cos(z)$. My strategy is to show that $x_0 = z$. Since $f(x_0)$ is minimum, we know that $\sup \{x_0^2, \cos(x_0) \} \le \max \{z^2, \cos(z)\} = z^2$. Hence $x_0^2 \le z^2$, $\cos(x_0) \le z^2$, and $\cos (x_0) \le \cos(z)$, and therefore $(x_0 - z)(x_0 + z) \le 0$. </p>
<p>Now, if $x_0 \neq z$ were true, then $x_0 + z > 0$ and so we would need $x_0 - z \le 0$ or $x_0 \le z$. Since $\cos()$ is decreasing on $I$, $\cos (z) \le \cos (x_0)$, which contradicts the fact that $\cos (x_0) \le \cos (z)$. However, I don't yet have access to the concept of a derivative yet to show that $\cos()$ is decreasing. </p>
<p>I am not sure how to finish this problem without the assumption that $\cos()$ is decreasing on $I$. I could use some hints.</p>
| Hagen von Eitzen | 39,174 | <p>The argument works for the general case of $f(x)=\sup\{g(x),h(x)\}$ where $g,h\colon [a,b]\to \Bbb R$ are continuos, $g$ strictly increasing, $h$ strictly decreasing, $h(a)>g(a)$, and $g(b)>h(b)$:</p>
<p>A minimum point $x_0\in [a,b]$ exists because $[a,b]$ is compact and $f$ is continuous.</p>
<p>If $f(x_0)\ne g(x_0)$, then $f(x)\ne g(x)$ in a neighbourhood of $x_0$. But then $f(x)=h(x)$ in that neighbourhood. As $h$ is strictly decreasing, this is only possible if said neighbourhood does not extend to the right of $x_0$, i.e., $x_0=b$. But then $f(x_0)=g(x_0)$, contradiction. We conclude $f(x_0)=h(x_0)$.</p>
<p>The proof for $f(x_0)=g(x_0)$ is essentially the same.</p>
<p>Note that you do not need derivatives to show that $\cos x$ is strictly decreasing (or that $x^2$ is strictly increasing):
If $0\le x<y\le\pi$ then $\cos y-\cos x=-2\sin\frac{x+y}{2}\sin\frac{y-x}2<0$ because $0<\frac{y-x}{2}<\frac{x+y}2<\pi$.</p>
|
1,623,796 | <p>In the structure $\langle \mathbb{Q}, < \rangle$ which of the following axioms hold? How about when we use the weak versions of the axioms (all $\leftrightarrow$ replaced with $\rightarrow$ )? </p>
<p>Extensionality, Empty Set, Power Set, Infinity</p>
<p>Find an instance of Separation that is true in $\langle \mathbb{Q}, < \rangle$, and on that is false. </p>
<p>What I'm struggling with is:
What does the $\in$ in the axioms mean in$\langle \mathbb{Q}, < \rangle$? I assume it has to mean
<p>In which case for the strong axioms, I'd say the Extensionality, Power Set and Infinity holds and Empty set doesn't as there's no minimal element, is that right?
For the weak versions, just the same ones?</p>
<p>And what is meant by "An instance of Separation"?</p>
| egreg | 62,967 | <p>You have to interpret $\in$ with $<$. In order to check whether extensionality holds you have to consider the sentence</p>
<blockquote>
<p>for all $x$ and $y$, $x=y$ if and only if, for every $z$, $z<x$ if and only if $z<y$</p>
</blockquote>
<p>or, in symbolic form,
$$
\forall x\forall y
(x=y \leftrightarrow (\forall z(z<x\leftrightarrow z<y)))
$$
where the universe is, of course, $\mathbb{Q}$.</p>
<p>It's easy to see this sentence is true.</p>
<p>On the other hand, it's also true that
$$
\forall x\exists y(y<x)
$$
so the interpretation of the empty set axiom is false.</p>
<p>Can you go on?</p>
<p>You can notice that the relation “$x$ is a subset of $y$” is interpreted into $x\le y$. The power set of $x$ is a set $y$ such that, for every $z$, if $z\subseteq x$, then $z\in y$. So, given $x\in\mathbb{Q}$, you should find $y$ such that, for all $z$, if $z\le x$ then $z<y$. Note that such an element exists, but it's not unique: $x+1$ and $x+2$ both work. Can you conclude something about the power set axiom?</p>
<p>For the axiom of separation, which is an axiom scheme, you have to find a case where it holds and one where it doesn't.</p>
|
3,577,098 | <p>In metric space there is a well-known result that <span class="math-container">$X$</span> is a separable metric space <span class="math-container">$\implies$</span>Every uncountable set in <span class="math-container">$X$</span> has a limit point.My question is does the result hold if <span class="math-container">$(X,\tau)$</span> is arbitrary topological space?I think that it may not hold for Non-Hausdorff spaces where we do not have disjoint open sets around two points.</p>
| Aiden Chow | 743,034 | <p>The number of ways that Carlos getting a <span class="math-container">$6$</span> while still being higher than Eric is <span class="math-container">$5$</span>(Eric can get <span class="math-container">$1$</span> to <span class="math-container">$5$</span>). The total number of ways to pick out of the urn when Carlos's number is higher than Eric's number is <span class="math-container">$\frac{10*9}{2}=45$</span>(See saulpatz's <a href="https://math.stackexchange.com/questions/3577053/picking-out-balls-in-an-urn/3577073?noredirect=1#comment7354631_3577073">comment</a>). That makes the conditional probability:<span class="math-container">$$\frac{5}{45}=\frac{1}{9}$$</span>(I think I may have oversimplified the problem but in my head this makes perfect sense. Please tell me if I got anything wrong. I am only in the eighth grade so I am quite new to these maths)</p>
|
1,713,841 | <p>I think $\pi$ is algebraic of degree 3 over $\mathbb{Q}(\pi^3)$. To prove it, I need to show that $\pi \notin \mathbb{Q}(\pi^3)$ (which implies that $x^3-\pi^3$ is the minimum polynomial of $\pi$ over $\mathbb{Q}(\pi^3)$). I have no experience solving this kind of problem for transcendental elements like $\pi$.</p>
| may | 1,061,720 | <p><span class="math-container">$|(\mathbb{Q}(\pi^3))(\pi):\mathbb{Q}(\pi^3)|=irr(\mathbb{Q}(\pi),\mathbb{Q}(\pi^3)$</span></p>
<p><span class="math-container">$irr(\pi,\mathbb{Q}(\pi^3)=x^3-\pi^3$</span></p>
<p><span class="math-container">$\pi$</span> is the root of <span class="math-container">$x^3-\pi^3$</span></p>
<p><span class="math-container">$\pi^3$</span> is in <span class="math-container">$Q(pi^3)$</span></p>
<p>Then, <span class="math-container">$\pi$</span> is algebraic of degree 3 over <span class="math-container">$\mathbb{Q}(\pi^3)$</span></p>
|
2,755 | <p>As Akhil had great success with his <a href="https://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry">question</a>, I'm going to ask one in a similar vein. So representation theory has kind of an intimidating feel to it for an outsider. Say someone is familiar with algebraic geometry enough to care about things like G-bundles, and wants to talk about vector bundles with structure group G, and so needs to know representation theory, but wants to do it as geometrically as possible.</p>
<p>So, in addition to the algebraic geometry, lets assume some familiarity with representations of finite groups (particularly symmetric groups) going forward. What path should be taken to learn some serious representation theory?</p>
| Ben Webster | 66 | <p>I second the suggestion of Fulton and Harris. It's a funny book, and definitely you want to keep going after you finish it, but it's a good introduction to the basic ideas.</p>
<p>You specifically might be happier reading a book on algebraic groups.</p>
<p>While I third the suggestion of Ginzburg and Chriss, I wouldn't call it a "second course." Maybe if what you really wanted to do was serious, Russian-style geometric representation theory, but otherwise you might want to try something a little less focused, like Knapp's "Lie Groups Beyond an Introduction."</p>
<p>If you want Langlandsy stuff, then Ginzburg and Chriss is actually a bit of a tangent; good enrichment, but not directly what you want, since it skips over all the good stuff with D-modules. Look in the background reading for the graduate student seminar we're having in Boston this year: <a href="http://www.math.harvard.edu/~gaitsgde/grad_2009/">http://www.math.harvard.edu/~gaitsgde/grad_2009/</a></p>
|
2,755 | <p>As Akhil had great success with his <a href="https://mathoverflow.net/questions/1291/a-learning-roadmap-for-algebraic-geometry">question</a>, I'm going to ask one in a similar vein. So representation theory has kind of an intimidating feel to it for an outsider. Say someone is familiar with algebraic geometry enough to care about things like G-bundles, and wants to talk about vector bundles with structure group G, and so needs to know representation theory, but wants to do it as geometrically as possible.</p>
<p>So, in addition to the algebraic geometry, lets assume some familiarity with representations of finite groups (particularly symmetric groups) going forward. What path should be taken to learn some serious representation theory?</p>
| Chirag Lakhani | 2,565 | <p>I ran across an excellent book by Lakshmibai and Brown called <a href="http://www.ams.org/bookstore?fn=20&arg1=hinseries&ikey=HIN-40">Flag Varieties: an Interplay of Geometry, Combinatorics, and Representation Theory</a>. It seems like an excellent book for an algebraic geometer who is interested in representation theory and algebraic groups.</p>
|
2,925,320 | <p>I was wondering how did Spectral graph theory, this multidisciplinary area between Linear Algebra and Graph theory start?</p>
<p>How did it emerge? </p>
<p>Was there a certain problem (maybe graph isomorphism problem?) that evoked its formation? </p>
| Mike | 544,150 | <p>Well for one thing, certain combinatorial properties of a graph that are of interest to graph theorists and theoretical computer scientists, can be gleaned from analysing the adjacency matrix of the graph. For example, (put informally) a <span class="math-container">$k$</span>-regular graph <span class="math-container">$G$</span> (where <span class="math-container">$k \ge 3$</span>) is an expander (i.e., lots of edges between <span class="math-container">$S$</span> and <span class="math-container">$V(G) \setminus S$</span> for arbitrary nonempty subsets <span class="math-container">$S$</span> of <span class="math-container">$V(G)$</span>) iff every eigenvalue except the largest is less than <span class="math-container">$(1-\epsilon)k$</span>. </p>
<p>Surely there are other interesting combinatorial properties of a graph that can be gleaned from its adjacency matrix, but this comes to mind to me now.</p>
|
2,167,598 | <p>I've seen the answer for gcd$(a,b)$ but never for the lcm$(a,b)$?</p>
| edupppz | 217,861 | <p>Well this actually follows from the answer for the gcd. Here's a big hint, suppose x' and y' are such that gcd(a,b) = ax'+by'. Now, we know the gcd divides both a and b and that lcm is divisible by both a and b. So try working from there toward a solution.</p>
|
1,072,634 | <p>I need to solve the differential equation $$y\frac{dy}{dx} = (x+7)(y^2+6)$$</p>
<p>I know that the first step is to isolate both term each side and then integrate...</p>
<p>But I can't figure out how to isolate term on this one, I would probably be able to solve the rest of the equation, just need to know how to start.</p>
<p>Thanks.</p>
| Asaf Karagila | 622 | <p>The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.</p>
<p>This means that if $f\colon A\to B$, then for every $a\in A$, there is a unique $b\in B$ such that the pair $(a,b)\in f$.</p>
<p>So the converse holds just for it to be a function from $A$ to $B$.</p>
|
3,258,256 | <p>I know that if we have a matrix A defined for a space V with dimension n that has n distinct eigenvalues, we can write it as <span class="math-container">$X^{-1}DX$</span> where D is a diagonal matrix with the eigenvalues of A on its diagonal and X is a matrix of eigenvectors. </p>
<p>It seems that what we are doing is using the fact that we have n distinct eigenvalues to find a basis of eigenvectors of our space. In this basis, the linear transformation which corresponds to A simply rescales each eigenvector so A can be written as a diagonal matrix in this basis.</p>
<p>I understand that we need the matrix X to transform a vector defined in a basis <span class="math-container">$f_1,...,f_n$</span>that was used to define A to the eigenbasis <span class="math-container">$e_1,...e_n$</span>. However, I do not understand why we have to use X as the linear map that does this. Why can we not just define a map <span class="math-container">$L:V \to V$</span> where <span class="math-container">$L(f_i) = e_i$</span> <span class="math-container">$\forall 1\leq i \leq n $</span> and then X is just the identity matrix.</p>
| Osvaldo Paniccia | 352,453 | <p>Because in that basis, the algebra of matrices becomes really simple and immediate and you can get lot of information just by looking at the matrix itself.
Also, the representative matrix of <span class="math-container">$L$</span> is exactly <span class="math-container">$X^{-1}$</span>.</p>
<p>At the end you will have like: <span class="math-container">$A=X^{-1}ID I^{-1} X$</span></p>
|
4,013,630 | <blockquote>
<p>The population of bacteria triples each day on a petri dish. If it takes 20 days for the population of bacteria to fill the entire dish, how many days will it take bacteria to fill half of the petri dish?</p>
</blockquote>
<ul>
<li><p>My doubt 1: if we don't have a initial population can be solve this problem?
Like <span class="math-container">$A(d)$</span>= initial population times <span class="math-container">$3^{20}$</span> days but as I don't have this I am stuck.</p>
</li>
<li><p>My doubt 2: Also, can we assume a number for population at day 20 and move our way backwards?</p>
</li>
</ul>
| José Carlos Santos | 446,262 | <p>The natural way is induction. You have <span class="math-container">$a_2=-3$</span>. Now take <span class="math-container">$k\in\Bbb N$</span> and assume that<span class="math-container">$$a_k=\begin{cases}-3&\text{ if $k$ is even}\\7&\text{ if $k$ is odd.}\end{cases}$$</span>Then,<span class="math-container">\begin{align}a_{k+1}&=4-a_k\\&=\begin{cases}7&\text{ if $k$ is even}\\-3&\text{ if $k$ is odd.}\end{cases}\\&=\begin{cases}-3&\text{ if $k+1$ is even}\\7&\text{ if $k+1$ is odd.}\end{cases}\end{align}</span></p>
|
4,013,630 | <blockquote>
<p>The population of bacteria triples each day on a petri dish. If it takes 20 days for the population of bacteria to fill the entire dish, how many days will it take bacteria to fill half of the petri dish?</p>
</blockquote>
<ul>
<li><p>My doubt 1: if we don't have a initial population can be solve this problem?
Like <span class="math-container">$A(d)$</span>= initial population times <span class="math-container">$3^{20}$</span> days but as I don't have this I am stuck.</p>
</li>
<li><p>My doubt 2: Also, can we assume a number for population at day 20 and move our way backwards?</p>
</li>
</ul>
| cosmo5 | 818,799 | <p>Use <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow noreferrer">Iverson bracket</a>.</p>
<p><span class="math-container">$$a_n=-3[n \equiv 0\pmod 2]+7[n \equiv 1\pmod 2]$$</span></p>
|
325,236 | <blockquote>
<p>Is there a bijection between $\mathbb N$ and $\mathbb N^2$?</p>
</blockquote>
<p>If I can show $\mathbb N^2$ is equipotent to $\mathbb N$, I can show that $\mathbb Q$ is countable. Please help. Thanks,</p>
| Josephine Moeller | 950 | <p>Yes. Imagine starting at $(1,1)$ and then zig-zagging diagonally across the quadrant. I'll leave you to formulate it.</p>
<p>Hint: for every natural number $>1$ there's a set of elements of $\mathbb{N}^2$ that add up to that number. For $2$, there's $(1,1)$. For $3$, there's $(2,1)$ and $(1,2)$...</p>
|
4,623,427 | <p><span class="math-container">$f$</span> is a twice differentiable function from <span class="math-container">$R^n $</span> to <span class="math-container">$R$</span>. I want to show that
<span class="math-container">$$ \| \nabla^2 f(x) \| \le L \implies \| \nabla f(x) - \nabla f(y) \| \le L \| x-y \| $$</span> for all <span class="math-container">$x,y \in R^n $</span> and <span class="math-container">$L \ge 0 \in R$</span></p>
<p>I know that based on the mean value theorem that <span class="math-container">$$ f(x) - f(y) = (\nabla f (z) )^T (x-y) \implies \| f(x) - f(y) \| \le \| \nabla f (z) \| \| x-y| $$</span> for some point z on the line going thru <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. But not sure how to utilize this.</p>
<p>Thanks!</p>
| Tomer | 578,689 | <p>Based on the hint given in the comment below:</p>
<p>From the mean value theorem we know that
<span class="math-container">$$g(y) - g(x) = (\nabla g (z))^T (x-y) $$</span> for some <span class="math-container">$z$</span> on the line going thru <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>Take <span class="math-container">$g(x) = w^T \nabla f(x) $</span> for an arbitrary <span class="math-container">$w \in R^n$</span> , observe that <span class="math-container">$$\nabla g(x)= \nabla^2 f(x) w$$</span> and have:</p>
<p><span class="math-container">\begin{align*}
&w^T(\nabla f(x) - \nabla g(y)) = (\nabla^2 f(z) w )^T (x-y) \\
& \implies \| w^T(\nabla f(x) - \nabla g(y)) \| \\
&= \| (\nabla^2 f(z) w )^T (x-y) \| \le \| w \| \| \nabla^2 f(z) \| \| x-y \|\\
& \le L \| w \| x-y \| \;\;\;\ (\text{using } \| \nabla^2 f(x) \| \le L) \\
\end{align*}</span></p>
<p>Take <span class="math-container">$w = \frac { \nabla f(x) - \nabla g(y) } {\| \nabla f(x) - \nabla g(y) \| } $</span>, a unit vector and get</p>
<p><span class="math-container">$$ \| \nabla f(x) - \nabla f(y) \| \le L \| x-y \| $$</span></p>
|
3,799,421 | <p>While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
<span class="math-container">$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$$</span>
where <span class="math-container">$R$</span> and <span class="math-container">$x$</span> are positive constants.
Can this integral be solved using substitution?</p>
| SarGe | 782,505 | <p>Once even I tried to derive this result and got stuck on the same step. After searching many articles and abstracts on internet, I concluded that there is no simple analytical solution of this integral rather a lot of heavy computation is required.</p>
<p>The potential equation was like <span class="math-container">$$V=\int_0^{2\pi}k\cdot\frac{Q\ d\theta}{2\pi}\cdot\frac{1}{\sqrt{R^2+x^2-2Rx\cos\theta}}=\frac{k\ Q}{\pi\ R}\int_0^{\pi}\frac{d\theta}{\sqrt{a-b\cos\theta}}$$</span></p>
<p>where <span class="math-container">$\displaystyle a=1+\frac{x^2}{R^2}$</span> and <span class="math-container">$\displaystyle b=\frac{2x}{R}.$</span></p>
<h3>1. One possibility is to express the integrand as a power series in <span class="math-container">$\cosθ$</span>, and then integrate term by term.</h3>
<p>Integrate this series, using <span class="math-container">$$\int_0^{\pi} \cos^nθ\ dθ=\begin{cases}\frac{(n−1)!!π}{n!!} &, \text{if $n$ is even}\\ 0 &, \text{if $n$ is odd}\end{cases}$$</span></p>
<p><span class="math-container">$$V=\frac{k\ Q}{R}(1+\frac{3}{16}c^2+\frac{105}{1024}c^4+\frac{1155}{16384}c^6+\frac{25025}{4194304}c^8+...)$$</span></p>
<p>where <span class="math-container">$\displaystyle c=\frac{b}{a}=\frac{2(x/R)}{(x/R)^2+1}.$</span></p>
<h3>2. We can obtain a power series in <span class="math-container">$x/R$</span> .</h3>
<p>Consider the expression <span class="math-container">$\displaystyle \frac{1}{\sqrt{R^2+x^2−2Rx\cosθ}}=\frac{1}{R\sqrt{1+(x/R)^2−2(x/R)\cosθ}}$</span></p>
<p>Expanding this trinomial, we get a Legendre Polynomial whose co-efficients can be calculated. After integrating, we get,</p>
<p><span class="math-container">$$V=\frac{k\ Q}{R}\left(1+\frac{1}{4}(\frac{x}{R})^2+\frac{9}{64}(\frac{x}{R})^4+\frac{25}{256}(\frac{x}{R})^6+\frac{1225}{16384}(\frac{x}{R})^8...)\right)$$</span></p>
|
1,137,914 | <p>Can Someone please help me prove that the number 1729 is a Pseudoprime?</p>
<p>So a pseudoprime is a composite $n$ such that $n |(2^n − 2)$. And every prime
number also has this property.</p>
| Derek Holt | 2,820 | <p>Let $G$ be the semidirect product ${\mathbb Q}^2 \rtimes {\rm SL}(2,{\mathbb Q})$ with the natural action. Then $G$ has a chief series $1 < {\mathbb Q}^2 < {\mathbb Q}^2 \rtimes \{\pm I_2\} < G$ and, since the action on ${\mathbb Q}^2$ is irreducible and ${\rm PSL}(2,{\mathbb Q})$ is simple, these are the only normal subgroups of $G$. Since ${\mathbb Q}^2$ has no finite composition series, neither does $G$.</p>
<p>It would guess that having a finite composition series implies having a finite chief series. </p>
|
4,133,172 | <p>There are special arguments to trigonometric functions which produce closed-form results, e.g. <span class="math-container">$\sin(\frac{\pi}{4}) = \frac{1}{\sqrt 2}$</span>. I also recently learned that <span class="math-container">$\cos(\frac{\pi}{5})=\frac{\phi}{2}$</span> as shown <a href="https://www.youtube.com/watch?v=EzCBUY6htwk" rel="nofollow noreferrer">here</a>.</p>
<p>We can extend these results, for example by using the angle-sum formule and half-angle formulae and get closed-form results for e.g. <span class="math-container">$\sin({\frac{1}{2}\frac{\pi}{5}})$</span> or <span class="math-container">$\sin({\frac{\pi}{4}+\frac{\pi}{5}})$</span>.</p>
<p>Is there a completely-known set of all the closed-form values of the trig functions?</p>
| TomKern | 908,546 | <p>Wikipedia <a href="https://en.wikipedia.org/wiki/Trigonometric_constants_expressed_in_real_radicals#Scope_of_this_article" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Trigonometric_constants_expressed_in_real_radicals#Scope_of_this_article</a> claims that:</p>
<blockquote>
<p>expressions in real radicals exist for a trigonometric function of a rational multiple of π if and only if the denominator of the fully reduced rational multiple is a power of 2 by itself or the product of a power of 2 with the product of distinct Fermat primes, of which the known ones are 3, 5, 17, 257, and 65537.</p>
</blockquote>
|
4,133,172 | <p>There are special arguments to trigonometric functions which produce closed-form results, e.g. <span class="math-container">$\sin(\frac{\pi}{4}) = \frac{1}{\sqrt 2}$</span>. I also recently learned that <span class="math-container">$\cos(\frac{\pi}{5})=\frac{\phi}{2}$</span> as shown <a href="https://www.youtube.com/watch?v=EzCBUY6htwk" rel="nofollow noreferrer">here</a>.</p>
<p>We can extend these results, for example by using the angle-sum formule and half-angle formulae and get closed-form results for e.g. <span class="math-container">$\sin({\frac{1}{2}\frac{\pi}{5}})$</span> or <span class="math-container">$\sin({\frac{\pi}{4}+\frac{\pi}{5}})$</span>.</p>
<p>Is there a completely-known set of all the closed-form values of the trig functions?</p>
| Tyma Gaidash | 905,886 | <p>Technically, every value of sin can be expressed in closed form by means of using the exponential definition of sine:
<span class="math-container">$$\mathrm{sinθ=\frac{e^{iθ}-e^{-iθ}}{2i}}$$</span></p>
<p>Simply substitute the value of θ wanted and you will get a form in terms of the natural base exponential function</p>
<p>I assume you probably mean for θ=qπ,<span class="math-container">$ q\in \Bbb Q$</span>. In this case, the sine can be written in terms of a polynomial root in which case wikipedia considers closed form solutions, even though a general formula for degree 5 and above is impossible using finite numbers radicals, arithmetic operations, and integers where the imaginary unit can be expressed by the square root and the integer -1. Even this is true:
<span class="math-container">$$\mathrm{sin\frac π7=a\ root(64x^6-112x^4+56x^2-7)}=$$</span><a href="https://i.stack.imgur.com/W4nrB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W4nrB.jpg" alt="enter image description here" /></a></p>
<p>This result was gotten using Mathematica, which would be quite the pain to type out, but still shows how complex these answers can get using complex numbers...</p>
<p>As seen by from <a href="https://en.m.wikipedia.org/wiki/Closed-form_expression" rel="nofollow noreferrer">Wikipedia</a></p>
<blockquote>
<p>In mathematics, a closed-form expression is a mathematical expression expressed using a finite number of standard operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., <strong><em>nth root</em></strong>, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but usually no limit, differentiation, or integration. The set of operations and functions admitted in a closed-form expression may vary with author and context.</p>
</blockquote>
<p>If you want even more specific answers, if you do not consider the root function, for which the polynomial has no closed form in terms of finite radicals, integers, and arithmetic operations, then one should consider composite numbers, and certain prime numbers as mentioned nicely in another answer. These prime numbers are <a href="https://mathworld.wolfram.com/FermatPrime.html" rel="nofollow noreferrer">Fermat Primes</a> for which I am not sure why these are the only ones for trigonometric functions.</p>
<p>Nevertheless, if a number can be decomposed into a its prime factorization in terms of Fermat Primes and 2, then it seems that the trigonometric function does have a value. The numerator can then be evaluated, after the form of θ=<span class="math-container">$\frac{\pi}{r\in \Bbb N}$</span> using multiple angle formulas and making θ=<span class="math-container">$\mathrm{\frac{π}{r}⇔x=rθ}$</span>, through <a href="https://mathworld.wolfram.com/Multiple-AngleFormulas.html" rel="nofollow noreferrer">multiple angle formulas</a>. Also, the even and odd properties can be used to determine the answer based in the sign of the argument. Sorry for one of my answers, but this</p>
<blockquote>
<p><a href="https://math.stackexchange.com/questions/3948418/how-to-find-sin-of-any-fraction-angles-and-how-do-you-find-them-in-fraction-for/4088241#4088241">Link to the Answer is a Great Example This Process.</a>.</p>
</blockquote>
|
137,915 | <blockquote>
<p>Given two continuously differentiable functions $f,\
g:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x_0) = g(x_0)$ and
$f'(x_0) < g'(x_0)$, there exists $\epsilon > 0$ such that $f(x) <
g(x)$ for $x \in (x_0, x_0 + \epsilon)$.</p>
</blockquote>
<p>The above result is not too difficult to prove, but I was wondering if the condition that the functions be <strong>continuously</strong> differentiable is absolutely necessary. I have not taken much analysis, but I'm wondering if the fact that the derivative exists at $x_0$ would be enough to prove this result without needing the derivative to be continuous. The reason I ask this is because it seems that the existence of a derivative function already implies it must satisfy some quite stringent requirements (i.e. Darboux's Theorem), perhaps these are enough?</p>
| leslie townes | 18,076 | <p>You are right, it would be enough to assume that $f$ and $g$ are differentiable at $x_0$.</p>
<p>By considering a translate of the difference $g - f$ it suffices to prove that if $h$ is differentiable at $0$ and $h(0) = 0$ and $h'(0) > 0$, then there is $\delta > 0$ with the property that $h(x) > 0$ for all $x \in (0, \delta)$. </p>
<p>But this is clear: put the positive number $\epsilon = h'(0)/2$ into the "$\epsilon-\delta$" definition of the statement $\lim_{x \to 0} \frac{h(x) - h(0)}{x - 0} = h'(0)$ to learn that there is $\delta > 0$ with the property that for all $x \in (0, \delta)$ one has
$$
\left|\frac{h(x)}{x} - h'(0)\right| < \frac{h'(0)}{2}.
$$
This implies, in particular, that for all $x \in (0, \delta)$ one has $-\frac{h'(0)}{2} < \frac{h(x)}{x} - h'(0)$ and hence that $h(x) > \frac{x}{2} h'(0) > 0$ whenever $x$ is in $(0, \delta)$.</p>
|
2,235,304 | <blockquote>
<p>Let <span class="math-container">$f(z) = \frac{z^2}{(z^2+1)^2(z^2+2z+2)}$</span>.</p>
<p>Find singularities of this function. For each singularity determine if it is removable, a pole (if a pole determine it's order,) or essential.</p>
</blockquote>
<p>I have worked out that the singularities of this function are <span class="math-container">$z=\pm i$</span>, <span class="math-container">$z=-1\pm i$</span>.</p>
<p>However, I'm struggling to determine whether these singularities are removable, a pole or essential. Is <span class="math-container">$z=\pm i$</span> a pole of order <span class="math-container">$2$</span> and <span class="math-container">$z=-1\pm i$</span> a simple pole?</p>
<p>Any help will be appreciated.</p>
| Davide F. | 208,720 | <p>Let start with the poles: as your function is rational the pole are the zero of the denominator, so being $z^2+1=0 \iff z=\pm 1$ and $z^2+2z+2=0\iff z=-1\pm i$ we have $4$ singularities: $\{i,-i,-1+i,-1-i\}$.</p>
<p>Now note that $z^2+2z+2\neq 0$ if $z=\pm i$, and the same is true for $z^2$.</p>
<p>Let be $g(z)=\frac{1}{(z^2+1)^2}$, therefore $f(z)=\frac{z^2}{z^2+2z+2}g(z)$ and to study the singularities $\pm i$ we can study directly $g(z)$.</p>
<p>If we do a series expansion we get $g(z)=\sum_{n=-\infty}^{\infty}a_n(z-i)^n$ where $a_n=0$ if $n\leq 3$ and $a_n\neq 0$ if $n\geq 2$ therefore $-i$ is a singularity of order $2$.</p>
<p>The exact same rasonement is true for $-i$ and a very similar one for the other two poles.</p>
<p>I've gave to you the general procedure and a partial example, I'll leave to you the details of the calculations and the other cases.</p>
<p>You can also prove in a more general way (with the same procedure that I used in this answer) that the order of a pole in a rational function is exactly the multiplicity of the $0$ in the polynomial.</p>
|
2,877,032 | <p>$\int y^{2}dx+x^{2}dy$ and $C\left( r\right) =\left(a \cos t,b\sin t\right)$ ,$0<t< \pi$</p>
<p>Could someone give me a hint to evaluate this integral?</p>
<p>my efforts:
$C^{'}\left( r\right)=\left(-a \sin t,b\cos t\right)$ then $\left\| C^{'}\left( r\right) \right\|$=$\sqrt {a^{2}\sin ^{2}t+b^{2}\cos ^{2}t}$ then I didn't get anywhere from here. Should I employ Green's theorem?</p>
| Fred | 380,717 | <p>Let $x(t)=a \cos t$ and $y(t)= b \sin t$.</p>
<p>Then $\int_C y^{2}dx+x^{2}dy=\int_0^{\pi}(y(t)^2 x'(t)+x(t)^2 y'(t)) dt.$</p>
|
1,826,435 | <p>I was given a problem:
(Edited) </p>
<blockquote>
<p>Show $$ S=\bigl\{S_p:p\in\mathbb{P}\bigr\}\cup\bigl\{\{1\}\bigr\} $$
where $\mathbb{P}$ is the set of prime numbers, and $S = \{n \in
\mathbb{N}: n \text{ is a multiple of } p\}$ is a subbasis on
$\mathbb{N}$</p>
</blockquote>
<p>The easiest way to do this in my opinion to show that the $\mathcal{S}$ covers $\mathbb{N}$</p>
<p>i.e. $\bigcup S= \mathbb{N}$</p>
<p>Does this actually cover $\mathbb{N}$?</p>
<p>Take $n' \in \mathbb{N}$, then we need to show that $n' \in\bigcup S$</p>
<p>Is there some result that all natural number are multiple of primes? Without knowing that I don't see how I can show that $\mathcal{S}$ covers $\mathbb{N}$</p>
| egreg | 62,967 | <p>The formulation is quite ambiguous; I believe your set $S$ consists of the sets $\{1\}$ and of the sets $S_p=\{n\in\mathbb{N}: p\mid n\}$ as $p$ varies through the prime numbers. In other terms,
$$
S=\bigl\{S_p:p\in\mathbb{P}\bigr\}\cup\bigl\{\{1\}\bigr\}
$$
where $\mathbb{P}$ is the set of prime numbers.</p>
<p>You want to show that
$$
\mathbb{N}=\bigcup S=\{1\}\cup\bigcup_{p\in\mathbb{P}}S_p
$$</p>
<p>Suppose $n\in\mathbb{N}$. If $n=1$ then $n\in\bigcup S$. Otherwise $n$ is divisible by a prime $p$, so $n\in S_p$ and $n\in\bigcup S$ as well.</p>
<hr>
<p>How do you show that $n\ne1$ is divisible by at least a prime? If $n=0$, it's obvious. Otherwise $n>1$; consider the set
$$
D'(n)=\{k\in\mathbb{N}:k>1,k\mid n\}
$$
The set is not empty because $n\mid n$ and $n>1$. Let $p$ be the smallest element in $D'(n)$; then $p$ is prime.</p>
|
951,524 | <p>How do I prove, that $(\mathbb R×\mathbb R;+,*) $ is a ring, but not a field, where the $+$ and $*$ operations are: $(a,b)+(c,d):=(a+c,b+d)$ and $(a,b)*(c,d):=(ac,bd)$?</p>
<p>For the solution: so I would have to first show, that $(\mathbb R×\mathbb R;+,*) $ is a ring, I have to prove that using the definition of a ring:</p>
<ol>
<li>$(\mathbb R×\mathbb R;+)$ has to be commutative group</li>
<li>$(\mathbb R×\mathbb R;*)$ has to be a semigroup</li>
<li>$*$ must be distributive over $+$ (from both sides)</li>
</ol>
<hr>
<p>If 1. 2. and 3. can be proven, then $(\mathbb R×\mathbb R;+,*) $ is a ring.</p>
<ol>
<li>here I have to prove that $(\mathbb R×\mathbb R;+) $ is an algebraic structure where $+$ is associative and commutative; the identity element is $0$ and that all elements have an inverse</li>
<li>$*$ has to be associative</li>
<li>? (How do I prove distributivity in this particular example?)</li>
</ol>
<hr>
<p>Now I have to prove that $(\mathbb R×\mathbb R;+,*) $ is not a field. (How do I do that?)
Also how can I show whether or not $(\mathbb R×\mathbb R;+,*) $ is a <em>commutative</em> ring?</p>
| lab bhattacharjee | 33,337 | <p>From
<a href="https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series">Sum of tangent functions where arguments are in specific arithmetic series</a>,</p>
<p>$$\tan180x=\frac{\binom{180}1t-\binom{180}3t^3+\cdots+\binom{180}{177}t^{177}-\binom{180}{179}t^{179}}{1-\binom{180}2t^2+\cdots-\binom{180}{178}t^{178}+\binom{180}{180}t^{180}}$$ where $t=\tan x$</p>
<p>If we set $\tan180x=0,180x=180^\circ r$ where $r$ is any integer</p>
<p>$\implies x=r^\circ$ where $0\le r<180$</p>
<p>So, the roots of $\binom{180}1t-\binom{180}3t^3+\cdots+\binom{180}{177}t^{177}-\binom{180}{179}t^{179}=0$
$\iff\binom{180}{179}t^{179}-\binom{180}{177}t^{177}+\cdots+\binom{180}3t^3-\binom{180}1t=0$ are $\tan r^\circ$ where $0\le r<180,r\ne90$ ($r=90$ corresponds to the denominator $=\infty$)</p>
<p>So, the roots of $\binom{180}{179}t^{178}-\binom{180}{177}t^{176}+\cdots+\binom{180}3t^2-\binom{180}1=0$ are $\tan r^\circ$ where $0<r<180,r\ne90$</p>
<p>So, the roots of $\binom{180}{179}u^{89}-\binom{180}{177}u^{88}+\cdots+\binom{180}3u-\binom{180}1=0$ are $\tan^2r^\circ$ where $0<r<90$</p>
<p>Using Vieta's formula, $\sum_{r=1}^{89}\tan^2r^\circ=\dfrac{\binom{180}{177}}{\binom{180}{179}}$</p>
|
1,552,501 | <p><a href="https://i.stack.imgur.com/zFz40.jpg" rel="nofollow noreferrer">My Question</a></p>
<p>First I let u = y' and employed the Chain Rule to obtain du/dx = du/dy * u</p>
<p>But I am not sure where to go from there. Any tips, suggestions, or solutions to the problem would be much appreciated!</p>
<p>I suspect there may also be different families of solutions to this problem.</p>
| Claude Leibovici | 82,404 | <p>You already received the explanation : close to $x=0$, both $\sin(x)$ and $\tan(x)$ are $\simeq x$.</p>
<p>Let me go a bit further, using more terms in Taylor expansion $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}+O\left(x^7\right)$$ $$\tan(x)\sin^2(x)=x^3+\frac{x^7}{15}+O\left(x^8\right)\approx x^3\left(1+\frac{x^4}{15}\right)$$ You see how small is the error. Just for fun, plot the function and its approximations for $0 \leq x \leq 1$; you will be surprized to notice how close are the two curves.</p>
|
1,178,969 | <p>$$\sum_{k=1}^{\infty}\ln\frac{k+5}{k+4}$$</p>
<p>$$
\frac{\ln\frac{k+6}{k+5}}{\ln\frac{k+5}{k+4}}
=\cdots
$$</p>
<p>(ln(k+6/k+5)/ln(k+5/k+4))</p>
<p>ln((k+6/k+5)*(k+4/k+5))</p>
<p>simplify</p>
<p>lim as k approaches infinity of ln(kˆ2 +10k +24/ kˆ2 +10k+25)= ln(1) = 0.</p>
<p>why doesn't the series converge?</p>
| Bernard | 202,857 | <p>With equivalents (we can use equivalents because we have a series with positive termes): $$\ln \Bigl(\dfrac{k+5}{k+4}\Bigr)=\ln\Bigl(1+\dfrac 1{k+4}\Bigr)\sim_{+\infty} \dfrac 1{k+4}\sim_{+\infty} \dfrac 1k .$$
The series has a general term equivalent to the general term of the harmonic series, which diverges, hence it diverges.</p>
|
2,476,267 | <p>Here's a problem presented in my proofs class that I cannot for the life of me figure out. Apparently the answer is P2 according to the answer book but I have no idea how Help is much appreciated, thanks in advance!</p>
<p>Anna wrote down the following 3 statements, denoted P1, P2, P3, on a blank piece of paper:</p>
<p>P1 : There is exactly one FALSE statement written on this piece of paper.</p>
<p>P2 : There are exactly two FALSE statements written on this piece of paper .</p>
<p>P3 : There are exactly three FALSE statements written on this piece of paper.</p>
<p>One of the above statements is TRUE. What statement is TRUE?</p>
| Brian Tung | 224,454 | <p><strong>Hint.</strong> If exactly one of the above statements is TRUE, and there are exactly three statements in all, then exactly how many are FALSE?</p>
|
1,160,726 | <p>I am trying to figure out why the equation,</p>
<p>$2x^2\:-5y^2\:=\:1$</p>
<p>has no solutions. I have tried taking the equation mod $2$,$4$, and $5$, but when I do this I am not sure what I am supposed to look for. I would really appreciate an explanation on how to solve this.</p>
<p>Thanks!</p>
| Lucian | 93,448 | <p>Modulo $5$, <em>x</em> can be only $0,\pm1,\pm2.~$ Squaring and then doubling, we have $0,2,3$, all of which are $\neq1\bmod5.~$ $($We were required to find $2x^2\equiv1\bmod5)$.</p>
|
2,212,946 | <blockquote>
<p><strong>Question:</strong>
Show that if $f(x)$ is a continuous function, then:
$$\int_{0}^{1} \int_{0}^{y} f(x) \, dx \, dy = \int_{0}^{1}(1-x)\ f(x) \ dx$$</p>
</blockquote>
<p>How can this be done? I was taught to visualise double integrals as either a summation of horizontal or vertical strips but I am unsure as to how to apply that idea to this.</p>
<p>Many thanks in advance</p>
| Rab | 417,087 | <p>The area you're looking for is the one highlighted in the image below.<a href="https://i.stack.imgur.com/wzMYV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wzMYV.png" alt="enter image description here"></a></p>
<p>Your integral integral is one that runs from x=0 to x=y, the external one runs from y=0 to y=1.</p>
<p>Using Cauchy's repeated integral formula.</p>
<p>The internal integral runs from y=0 to y=1 and the external one runs from x=0 to x=y.</p>
<p>You get $\int_{x=0}^{x=1}\int_{y=x}^{y=1} f(x)dydx$</p>
<p>Can you finish the rest?</p>
|
18,174 | <p>I am an undergraduate secondary math education major. In <span class="math-container">$2$</span> weeks I have to give a <a href="https://www.mathmammoth.com/lessons/number_talks.php" rel="nofollow noreferrer">Number Talk</a> in my math ed class on the problem "<span class="math-container">$3.9$</span> times <span class="math-container">$7.5$</span>". I need to come up with as many different solution methods as possible. </p>
<p>Here is what I have come up with so far:</p>
<ol>
<li><p>The most common way: multiply the two numbers "vertically", ignoring the decimal, to get <span class="math-container">$2925$</span>:
<span class="math-container">\begin{array}
{}\hfill {}^6{}^439\\
\hfill \times\ 75 \\\hline
\hfill {}^1 195 \\
\hfill +\ 273\phantom{0} \\\hline
\hfill 2925
\end{array}</span> Since there are two numbers that are to the right of the decimal, place the decimal after the <span class="math-container">$9$</span> to get the answer <span class="math-container">$29.25$</span>.</p></li>
<li><p>Write both numbers as improper fractions: <span class="math-container">$$3.9= \dfrac{39}{10}$$</span> and <span class="math-container">$$7.5=\dfrac{75}{10}$$</span>Then multiply <span class="math-container">$$\dfrac{39}{10}\cdot\dfrac{75}{10}$$</span> to get <span class="math-container">$\dfrac{2925}{100}$</span> which simplifies to 29.25.</p></li>
<li><p>Use lattice multiplication. This is a very uncommon method that I doubt the students will use, and I need to review it myself before I consider it.</p></li>
<li><p>Since <span class="math-container">$3.9$</span> is very close to <span class="math-container">$4$</span>, we could instead do <span class="math-container">$4\cdot7.5=30$</span> and then subtract <span class="math-container">$0.1\cdot7.5=0.75$</span> to get <span class="math-container">$30 - 0.75=29.25$</span></p></li>
<li><p>Similarly, since <span class="math-container">$7.5$</span> rounds up to <span class="math-container">$8$</span>, we can do <span class="math-container">$3.9\cdot 8=31.2$</span> and then subtract .<span class="math-container">$5\cdot 3.9=1.95$</span> to get <span class="math-container">$31.2-1.95=29.25$</span> </p></li>
</ol>
<p>Are there any other possible methods the students might use? (<strong>Note:</strong> they are junior college math ed students.) Thanks!</p>
| TheSimpliFire | 8,598 | <p>Consider the following solution. <span class="math-container">\begin{align*}7.5\times3.9&=(0.1\times75)\times(0.1\times39)\\\\&=0.01\times75\times39\\\\&=75\%\,\,\text{of}\,\,39\quad(\text{as}\,\,\%\,\,\text{means}\,\,1/100=0.01)\tag{1}\\\\&=\frac34\times39\quad\small{\left[=\frac{4-1}4\times39=\left(\frac44-\frac14\right)\times39\right]}\tag2\\\\&=\left(1-\frac14\right)\times39\quad\small{\left[=1\times39-\frac14\times39\right]}\\\\&=39-\frac{39}4\\\\&=39-\frac{36+3}4\quad\small{\left[=39-\left(\frac{36}4+\frac34\right)\right]}\\\\&=39-\left(9+\frac34\right)\\\\&=39-9-\frac34\\\\&=30-0.75\\\\&=29.25\end{align*}</span> <span class="math-container">$(1)$</span> As an aside, it may be of interest to mention that <span class="math-container">$75\%\times39=39\%\times75$</span>.</p>
<p><span class="math-container">$(2)$</span> Comparing the length of this solution to <a href="https://matheducators.stackexchange.com/a/18180/8598">Martin's</a>, it shows the importance of choosing whether to express <span class="math-container">$3=4-1$</span> (as in this answer), or <span class="math-container">$39=40-1$</span> which yields a quicker solution.</p>
|
112,575 | <p>If we set $\exp(x)=\sum x^k/k!$, then $\exp(x+y)=\exp(x)\cdot \exp(y)$. In terms of coefficients it means that $(x+y)^n=\sum \frac{n!}{k!(n-k)!} x^ky^{n-k}$, i.e. just binomial expansion.</p>
<p>Now consider logarithm. Set $L(x):=\sum_{k>0} x^k/k$, then $L(x)=-\log(1-x)$ in a sense, and hence $$L(u+v-uv)=L(u)+L(v),$$ i.e. $\sum (u+v-uv)^n/n=\sum (u^n+v^n)/n$, or, if we pass to coefficients of $u^av^b$ ($a,b\geq 1$), we get
$$
\sum_k (-1)^k\frac{(a+b-k-1)!}{(a-k)!(b-k)!k!}=0
$$</p>
<p>The question is what is combinatorial meaning of this identity. Maybe, it is some exclusion-inclusion formula, as it is usual for alternating sums?</p>
| Ira Gessel | 10,744 | <p>As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can give a combinatorial interpretation to the identity that we obtain (though doing this destroys the symmetry between $a$ and $b$).</p>
<p>If we multiply by $b$, the sum may be written
$$\sum_{k=0}^a (-1)^k \binom{a+b-k-1}{a-k}\binom{b}{k}.$$
This is a special case ($m=a+b-1$) of the more general identity
$$\sum_{k=0}^a (-1)^k\binom{m-k}{a-k}\binom{b}{k} = \binom{m-b}{a},$$
which we can prove combinatorially. (Incidentally, this identity is a form of Vandermonde's theorem.)</p>
<p>To prove this formula, we start with a set $M$ of size $m$, with a subset $B$ of size $b$. To interpret $\binom{m-k}{a-k}\binom{b}{k}$, we choose a $k$-subset $K$ of $B$ and then choose an $(a-k)$-subset $C$ of $M-K$. The right side $\binom{m-b}{a}$ counts pairs $(K,C)$ in which $K$ is empty and $C$ is an $a$-subset of $M-B$. To prove the identity, we find an involution on the set of pairs $(K,C)$ not of this form that changes the parity of $|K|$. The pairs $(K,C)$ to be canceled are those in which $K\cup C$ contains at least one element of $B$. Then the involution moves the smallest element of $(K\cup C) \cap B$ from $K$ to $C$ or from $C$ to $K$.</p>
|
1,735,653 | <p>So I need to use the fact that: $$\cos(4x) + i\sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$ to derive identities for $\cos(4x)$ and $\sin(4x)$ in terms of $\cos(x)$ and $\sin(x)$. I'm not sure how to go about this, could I please get some help.</p>
| nbubis | 28,743 | <p><strong>Hint</strong>:</p>
<p>Try replacing $x$ by $-x$, and then adding the expressions. Using the fact that $\sin(-x) = -\sin(x)$ and that $\cos(-x) = \cos(x)$:</p>
<p>$$\cos(4x) + i \sin(4x) = \left(\cos(x) + i\sin(x)\right)^4$$
$$\cos(4x) - i \sin(4x) = \left(\cos(x) - i\sin(x)\right)^4$$
So:
$$2\cos(4x) = \left(\cos(x) + i\sin(x)\right)^4 + \left(\cos(x) - i\sin(x)\right)^4 $$
Can you take it from there?</p>
|
1,843,369 | <p>Consider the congruence </p>
<p>$$2x+7y \equiv 5\pmod{12}$$</p>
<p>Here $(2,7,12)=1$. Since $(2,12)=2$, we must have
$$7y \equiv 5\pmod{2}$$</p>
<p>Which clearly gives $y \equiv 1\pmod{2}$, or $y \equiv 1,3,5,7,9,11\pmod{12}$</p>
<p>Why does the previous statement follow?. This is not a problem. It's something I can't understand from the chapter.</p>
| Pjotr5 | 157,405 | <p>The statement $2x + 7y \equiv 5 \pmod{12}$ means that $$2x + 7y = 5 + 12k$$ for some $k \in \mathbb{Z}$. We can consider this equation modulo $2$ to get $7y \equiv 5 \pmod{2}$, since both $2x$ and $12k$ are divisible by $2$. Because $7 \equiv 5 \equiv 1 \pmod{2}$ we get $y \equiv 1 \pmod{2}$, which means that we can write $$y = 1 + 2m$$ for some $m \in \mathbb{Z}$. Now you can consider this equation modulo $12$ to get $y \equiv 1 + 2m \pmod{12}$, by trying values for $m$ you can quickly see that $y$ can only be equivalent to $1,3,5,7,9$ or $11$ modulo $12$.</p>
<hr>
<p>It might give a little insight to look at the points in $\mathbb{Z}^2$ that correspond to the solutions of the original equation $2x + 7y \equiv 5 \pmod{12}$ shown below. It turns out that this this equation can be rewritten as $y \equiv 11-2x \pmod{12}$. We see a pattern of diagonal lines of dots appear. If you want to move around on a single diagonal line for every step you take in the $x$ direction you must take two steps in the $y$ direction because of the factor $2x$. This means that on a diagonal line the parity of the $y$ values can not change. But if you jump from one diagonal line up or down to a different diagonal line your $y$ value wil always change a multiple of $12$. Since twelve is even, this will also not change the parity of $y$. So we see that the $2x$ and the fact that the equation is in $\mathbb{Z}/12\mathbb{Z}$ work together to ensure that $y$ is odd for all solutions to the equation.</p>
<p><a href="https://i.stack.imgur.com/aRfkk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aRfkk.png" alt=""></a></p>
|
58,009 | <p>Let $f: X \to Y$ be a morphism of varieties such that its fibres are isomorphic to $\mathbb{A}^n$. Since the definition of a vector bundle stipulates that $f$ be locally the projection $U \times \mathbb{A}^n \to U$, it is likely that there exist morphisms that are not locally of that form, but I can't come up with an example.</p>
<p>So the question is: what is an example of a morphism with fibres $\mathbb{A}^n$ that is not locally trivial? not locally isotrivial? </p>
<p>UPDATE: what if one assumes vector space structure on the fibres?</p>
| Sasha | 4,428 | <p>I think such map is locally trivial if and only if it is smooth. The "only if" part is clear. So, assume $f$ is smooth. Take any point $y_0 \in Y$ and $n+1$ affine-linearly independent points in the fiber over $y_0$. Then choose local sections through this points (this is where you need smoothness), denote them by $x_0,\dots,x_n$. Let $U$ be the set of $y \in Y$ over which $x_i$ are affine-lnearly independent. Then we have a map $f^{-1}(U) \to A^n$, taking a point $x$ to $(t_0,t_1,\dots,t_n) \in \{ \sum t_i = 1 \} \subset A^{n+1}$ such that $x = \sum t_ix_i$. This gives a local trivialization.</p>
|
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