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 |
|---|---|---|---|---|
562,707 | <p>This is a famous rudimentary problem : how to use mathematical operations (not any other temporary variable or storage) to swap two integers A and B. The most well-known way is the following:</p>
<pre><code>A = A + B
B = A - B
A = A - B
</code></pre>
<p>What are some of the alternative set of operations to achieve this?</p>
| Steven Stadnicki | 785 | <p>Any <em>group operator</em> — that is, any operator that is associative and has an identity and inverse — provides a solution to this problem. Suppose that the operator is $\bigtriangleup$, and that every element $X$ has an inverse $X^{-1}$ such that $X\bigtriangleup X^{-1} = X^{-1}\bigtriangleup X = e_{\triangle}$ for all $X$, where $e_\triangle$ is the identity for the $\bigtriangleup$ operator: $Y\bigtriangleup e_\triangle = e_\triangle\bigtriangleup Y = Y$ for all $Y$. Then starting with $A=A_0$, $B=B_0$ the sequence of operations
$$\begin{align}A&=A\bigtriangleup B\\
B&=A\bigtriangleup B^{-1}\\
A&=B^{-1}\bigtriangleup A\end{align}$$
successively sets $A=A_0\bigtriangleup B_0$, then $B=(A_0\bigtriangleup B_0)\bigtriangleup B^{-1}_0$ $=A_0\bigtriangleup (B_0\bigtriangleup B^{-1}_0)$ $= A_0$, and finally $A=A_0^{-1}\bigtriangleup (A_0\bigtriangleup B_0)$ $= (A_0\bigtriangleup A_0^{-1})\bigtriangleup B_0$ $= B_0$. Note that associativity is used in both the second and third operations. This approach also requires an 'easy' way of writing the inverse operation.</p>
<p>All of the 'classic' examples (the addition/subtraction version that you give, as well as the XOR version) can be seen as examples of this general scheme; in the case of addition and subtraction the $B=A\bigtriangleup B^{-1}$ step can be written as $B=A+(-B) = A-B$, and using the commutativity of addition the $A=B^{-1}\bigtriangleup A$ step can be written as $A=(-B)+A = A+(-B) = A-B$. In the case of XOR, $A$ and $B$ are their own inverses, and the operation is again commutative, so we get the $A=A\bigoplus B,\ B=A\bigoplus B,\ A=A\bigoplus B$ solution.</p>
|
533,399 | <p>Starting with the classical propositional logic, is there a rather canonical way to prove that $$p\wedge q=q\wedge p$$ for the commutativity of the conjunction and analogously for the other properties and connectives, other than using truth tables, visualizing with Venn diagrams akin <a href="http://en.wikipedia.org/wiki/Logical_conjunction" rel="nofollow">Wikipedia's approach</a>, or verbal philosophical reasoning?</p>
<p>Put it other way, can we well-define the connectives from a deeper foundation than that?</p>
<p>For example, in set theory, we define a union of the two sets $A$, $B$ as $$A\cap B:=\{x\,|\,x\in A\wedge x\in B\}$$
to then move on and prove that $\cap$ is commutative. By doing so we simply delegate the proof to the very propositional (or whichever) logic we defined the operator with $$A\cap B\overset{\mathrm{def}}{=}\{x\,|\,x\in A\wedge x\in B\}\overset{\mathrm{com}}{=}\{x\,|\,x\in B\wedge x\in A\}\overset{\mathrm{def}}{=}B\cap A.\square$$</p>
| hmakholm left over Monica | 14,366 | <p>Once you select a particular proof system, you should be able to write down a <em>formal proof</em> of $(p\land q)\leftrightarrow (q \land p)$. How such a proof will look will vary wildly between different proof systems, though.</p>
<p>For example, in (classical or intuitionistic) sequent calulus, the formal proof might look like this:
$$
\begin{array}{rll}
1) & q \vdash q & \text{axiom} \\
2) & p,q \vdash q & 1, \text{weakening} \\
3) & p \vdash p & \text{axiom} \\
4) & p,q \vdash p & 3, \text{weakening} \\
5) & p,q \vdash q\land p & 2,4,{\land}\mathrm I \\
6) & p\land q \vdash q \land p & 5, {\land}\mathrm E \\
7) & \vdash (p\land q) \to (q\land p) & 6, {\to}\mathrm I \\
8-14) & \vdash (q\land p) \to (p\land q) & \text{(repeat the above with $p$ and $q$ interchanged)} \\
15 & \vdash (p\land q) \leftrightarrow (q\land p) & 7, 14, {\leftrightarrow}I
\end{array}
$$</p>
<p>However, such a proof doesn't really offer any deep insight into how conjunction works. It says more about the proof system than about conjunction, namely that the proof system works "in a sane way" with respect to conjunction.</p>
<p>If you're not interested in proof systems, but in semantics, then you don't get around the fact that truth tables <em>is the most fundamental way</em> to define the semantics of the propositional connectives -- much more suitable as a definition than verbal descriptions. Therefore a truth-table based proof <em>is the most fundamental proof</em> you can possibly get about semantics here, and everything else will just be more or less convincingly paraphrased guises for the fundamental truth-table based proof.</p>
|
1,237,595 | <p>Random variable $X$ has probability density function $g(x)=\frac{3}{7}x^2\mathbf{1}_{[1,2]}$. Is there a function $F: \mathbb{R}\to\mathbb{R}$ for which $F(X)$ has an exponential distribution with parameter 1? Thanks for help.</p>
| mathifold.org | 231,554 | <p>The function $F$ does exist, because you only have to redistribute the values going to $[1,2]$ to values going to $(0,\infty)$. But how to find $F$ explicitely? It must satisfy $F(1)=0$, $F(2)=\infty$ (so strictly speaking the function <strong>will not</strong> be $\mathbb{R}\longrightarrow \mathbb{R}$, or at least not continuously). What we need is</p>
<p>$\int_1^x g(t)dt=\int_0^{F(x)}e^{-t}dt=1-e^{-F(x)}$, and you may calculate $F$ (just calculate the first integral as a function of $x$ and rearrange).</p>
|
1,237,595 | <p>Random variable $X$ has probability density function $g(x)=\frac{3}{7}x^2\mathbf{1}_{[1,2]}$. Is there a function $F: \mathbb{R}\to\mathbb{R}$ for which $F(X)$ has an exponential distribution with parameter 1? Thanks for help.</p>
| zoli | 203,663 | <p>The corresponding cdf is:</p>
<p>$$G_X(x)=\begin{cases}0, & \text{ if }x<1\\
\frac{1}{7}x^3& \text{ if }1\le x<2\\
0, & \text{ otherwise. }\end{cases}.$$
Let us calculate the distribution of the random variable $G(X)$. Let it be denoted by $H$:
$$H(x)=P(G_X(X)<x)=P(X<G_X^{-1}(x))=G_X(G_X^{-1}(x))=x$$
over the interval $[1,2]$. That is, the distribution of $Y=G_X(X)-1$ is uniform over $[0,1]$. </p>
<p>Now, the question is the following: Can one generate an exponential distribution if a random variable of uniform distribution over $[0,1]$ is given?</p>
<p>I hope that you already know the answer.</p>
|
90,480 | <p>Given two simplicial topological spaces $X_{\bullet}$ and $Y_{\bullet}$ (i.e. a simplicial object in Top) and a continuous map between their geometric realizations $f \colon \lvert X_{\bullet} \rvert \to \lvert Y_{\bullet} \rvert$. Is $f$ homotopic to $\lvert \varphi_{\bullet} \rvert$ for a map $\varphi_{\bullet}$ of simplicial spaces?</p>
| Thomas Nikolaus | 11,002 | <p>The answer is no. For an arbitrary simplicial space $X_\bullet$ we can consider $|X_\bullet|$ as a constant simplicial space, lets call this $Y_\bullet$. Then there is clearly the identity $|X_\bullet| \to |Y_\bullet|$, but there is in general no nontrivial map $X_\bullet \to Y_\bullet$ (take e.g. BG for a top. group $G$).</p>
|
36,568 | <p>To do Algebraic K-theory, we need a technical condition that a ring $R$ satisfies $R^m=R^n$ if and only if $m=n$. I know some counterexamples for a ring $R$ satisfies $R=R^2$. </p>
<p>Are there any some example that $R\neq R^3$ but $R^2 = R^4$ or something like that?</p>
<p>(c.f. if $R^2=R^4$, then we need that $R^3=R^5=\ldots =R^{2n+1}$ for any $n>1$)</p>
| Seamus | 6,701 | <p>Rings that satisfy the condition
$R^n \cong R^m \iff n=m$ are said to have <em>invariant basis number</em> or the <em>invariant basis property</em>. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$. </p>
<p>Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that </p>
<p>$$
R^m\cong R^{m'} \iff m=m'\ \mbox{or}\ m,m'\ge h \ \mbox{and} \ m\equiv m'\ (\mod k)
$$</p>
<p>A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$. </p>
<p>There are earlier constructions of rings of type $(h,k)$ by Leavitt [W. G. Leavitt:
Modules without invariant basis number, Proc. Am. Math. Sot. 8 (1957), 322-328], [W. G. Leavitt:
The module type of a ring, Trans. Am. Math. Sot. 103 (1962), 113-130], but they are far more complicated. </p>
<p>Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K_0(R)\cong {\bf Z} \oplus \tilde{K}_0(R)$. </p>
|
2,909,022 | <p>I don't understand how to get from the first to the second step here and get $1/3$ in front.</p>
<p>In the second step $g(x)$ substitutes $x^3 + 1$.</p>
<p>\begin{align*}
\int_0^2 \frac{x^2}{x^3 + 1} \,\mathrm{d}x
&= \frac{1}{3} \int_{0}^{2} \frac{1}{g(x)} g'(x) \,\mathrm{d}x
= \frac{1}{3} \int_{1}^{9} \frac{1}{u} \,\mathrm{d}u \\
&= \left. \frac{1}{3} \ln(u) \,\right|_{1}^{9}
= \frac{1}{3} ( \ln 9 - \ln 1 )
= \frac{\ln 9}{3}
\end{align*}</p>
| MPW | 113,214 | <p>Looking at the denominator, you define $g(x)=x^3+1$. This means that $g'(x)=3x^2$.</p>
<p>Your goal is to manipulate the integrand so that it is in the form $g'(x)/g(x)$, that is, $3x^2/(x^3+1)$. As it stands, the factor $3$ you need is missing. You can't just throw it in, because that changes the integrand. But you can throw it in and pull it out at the same time, since that amounts to multiplying by $1$, which doesn't change anything:
$$\frac{x^2}{x^3-1}=\frac{\overbrace{(\frac13\cdot 3)}^1 x^2}{x^3-1}$$
$$=\frac{\frac13\cdot (3 x^2)}{x^3-1}$$
$$=\tfrac13\cdot \frac{3x^2}{x^3-1}$$
So you proceed from here.</p>
<p>Generally, if you need a missing constant factor of $c$ in the integrand, you can put it in if you also compensate by putting its reciprocal $\frac1c$ in as well. And remember, constant factors can be pulled to the outside of the integral. The rule is</p>
<p>$$\int f(x)\; dx = \frac1c\int cf(x)\; dx$$</p>
|
440,082 | <blockquote>
<p><span class="math-container">$$ \int_{1}^{\infty} \frac{\sin^2 (\mu \sqrt{x^2 -1})}{(x+1)^{\frac{9}{2}} (x-1)^{\frac{3}{2}}} \,dx $$</span>
Note: <span class="math-container">$\mu$</span> here is an extremely small constant.</p>
</blockquote>
<p>I have tried:</p>
<ol>
<li>Estimating the integral by Taylor expansion of <span class="math-container">$\sin^2(\mu \sqrt(x^2 - 1)$</span> but the it diverges after few terms.</li>
<li>I have also tried to split the integral into two regions, solving the two integrals differently and then adding them in the hope that the arbitrary parameter <span class="math-container">$\alpha$</span> will vanish (asymptotic matching/splitting). While I have been able to solve the two integrals but <span class="math-container">$\alpha$</span> does not vanish: <span class="math-container">$$\int_{1}^{\alpha} f(x)\,dx + \int_{\alpha}^{\infty} f(x)\,dx$$</span>
Approximations in for the two regions:<br />
<span class="math-container">$1$</span> to <span class="math-container">$\alpha$</span> region: <span class="math-container">$x \tilde = 1$</span> and hence <span class="math-container">$(x+1) \tilde= 2$</span> <br />
<span class="math-container">$\alpha$</span> to <span class="math-container">$\infty$</span> region: <span class="math-container">$(x^2 -1) \tilde= x^2$</span></li>
</ol>
| AccidentalFourierTransform | 106,114 | <p>It is more or less straightforward to write down an asymptotic expansion around <span class="math-container">$\mu\to0$</span>,
<span class="math-container">$$
I(\mu)\sim \frac{2 \mu ^2}{15}-\frac{\mu ^4}{9}+\frac{\pi \mu ^5}{15}+\frac{1}{450} \mu ^6 (60 \log (\mu )+60 \gamma -67)+\cdots
$$</span>
where <span class="math-container">$\gamma$</span> is Euler-Mascheroni. Given this result it should be rather clear why the naive Taylor expansion diverges after the first two terms: the next terms have odd powers of <span class="math-container">$\mu$</span> and even non-analytic factors of <span class="math-container">$\log(\mu)$</span>, while your function <span class="math-container">$f$</span> only has even powers of <span class="math-container">$\mu$</span>.</p>
<p>Here is a log-plot of the error:</p>
<p><a href="https://i.stack.imgur.com/C7ra7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C7ra7.png" alt="enter image description here" /></a></p>
<p>(Blue means one-term approximation, Red is two-term, Green is three-term, and Orange is four-term.)</p>
<p>A simple Mathematica code to reproduce the formula above (and easily generalized to yield more terms):</p>
<pre><code>Sin[µ Sqrt[x^2 - 1]]^2/((x + 1)^(9/2) (x - 1)^(3/2))
Series[%, {µ, 0, 10}]
small = Integrate[Normal[%], {x, 1, Λ}, Assumptions -> Λ > 1];
Sin[µ x]^2 Series[1/((x + 1)^(9/2) (x - 1)^(3/2)), {x, ∞, 8}] // Normal
large = Integrate[%, {x, Λ, ∞}, Assumptions -> Λ > 1 && 0 < µ < 1];
Assuming[Λ > 1 && 0 < µ < 1, Series[small + large, {µ, 0, 10}] // Normal] // Short
Assuming[Λ > 1 && 0 < µ < 1, Series[%, {Λ, ∞, 1}] // Normal]
</code></pre>
|
937,064 | <p>The title pretty much says it all:</p>
<p>If supposing that a statement is false gives rise to a paradox, does this prove that the statement is true?</p>
<p><em>Edit:</em> Let me attempt to be a little more precise:</p>
<p>Suppose you have a proposition. Furthermore, suppose that assuming the proposition is false leads to a paradox. Does this imply the proposition is true?
In other words, can I replace the "contradiction" in "proof by contradiction" with "paradox." </p>
<p>This question might still be somewhat ambiguous; I'm reluctant to attempt to precisely define "paradox" here.
As a (somewhat loose) example however, consider some proposition whose negation leads to, for example, Russell's paradox. Would this prove that the proposition is true? </p>
| Mark Bennet | 2,906 | <p>It depends on the statement. Some statements e.g.</p>
<blockquote>
<p>This statement is false.</p>
</blockquote>
<p>lead to a contradiction whether you assume them true or false, so don't have an assignable truth value.</p>
<p>You also need to know or prove that your statement has a truth value (i.e. is either true or false) before you can conclude your argument.</p>
|
2,989,494 | <p>I am trying to derive properties of natural log and exponential just from the derivative properties.</p>
<p>Let <span class="math-container">$f : (0,\infty) \to \mathbb{R}$</span> and <span class="math-container">$g : \mathbb{R} \to \mathbb{R}$</span>.
Without knowing or stating that <span class="math-container">$f = \ln(x)$</span> and <span class="math-container">$g = e^x$</span>, but only knowing that <span class="math-container">$f'(x) = \frac{1}{x}$</span> and <span class="math-container">$g'(x) = g(x)$</span> how do I show from just the derivative that:</p>
<ul>
<li><span class="math-container">$f(xy) = f(x)+f(y)$</span> and <span class="math-container">$g(x+y) = g(x)g(y)$</span></li>
<li><span class="math-container">$f(x^y) = yf(x)$</span></li>
<li><span class="math-container">$\lim_{x \to \infty}f(x) = \infty$</span> and <span class="math-container">$\lim_{x \to 0}f(x) = -\infty$</span></li>
<li><span class="math-container">$\lim_{x\to \infty}g(x) = \infty$</span> and <span class="math-container">$\lim_{x \to -\infty}g(x) = 0$</span></li>
<li><span class="math-container">$g \circ f$</span> is an identity of <span class="math-container">$(0,\infty)$</span> and <span class="math-container">$f \circ g$</span> is an identity on <span class="math-container">$\mathbb{R}$</span>.</li>
</ul>
<p>I know how to do that in general, but using <span class="math-container">$\textbf{only}$</span> the derivative, I am a bit stuck, and would appreciate help. The text that inspired me to do this problem (I expanded it a bit), also expects me not to use integration.</p>
| Derek Elkins left SE | 305,738 | <p>Do a case analysis on either <span class="math-container">$x\geq y$</span> or <span class="math-container">$y > x$</span>. In the <span class="math-container">$x \geq y$</span> case, <span class="math-container">$x+x \geq x+y$</span> therefore <span class="math-container">$2x\geq 100$</span> so <span class="math-container">$x \geq 50$</span> and thus, trivially, <span class="math-container">$x\geq 50\lor y\geq 50$</span>. The <span class="math-container">$y > x$</span> case proceeds similarly.</p>
|
221,053 | <p>Given two uncorrelated random variables $X,Y$ with the same variance $\sigma^2 $ I need to compute $\rho= \frac{COV(X,Y)}{\sigma(X)\sigma(Y)}$ between $X+Y$ and $2X+2Y$. I know it should be a number between $-1$ and $1$ and I don't understand how come I get $4$. </p>
<p>Here's what I did:</p>
<p>$COV(X+Y,2X+2Y)=COV(X+Y,2X)+COV(X+Y,2Y)=COV(2X,X)+COV(2X,Y)+COV(2Y,Y)+COV(2Y,X)=2COV(X,X)+2COV(Y,Y)+4COV(X,Y)=2\sigma^2+2\sigma^2=4\sigma^2$ so final result is $\rho=4$ since $\sigma(X)=\sqrt{Var(x)}$.</p>
<p>What's wrong with what I did?</p>
| André Nicolas | 6,312 | <p>The reason things went wrong is probably due to an unfortunate choice of notation, the use of $X$ and $Y$ with two different meanings. </p>
<p>We want the correlation coefficient $\rho(U,V)$, where $U=X+Y$ and $V=2(X+Y)$. So we need to divide $\text{Cov}(U,V)$ by the product of the standard deviations of $U$ and of $V$ (<strong>not</strong> of $X$ and of $Y$).</p>
<p>You did divide, but by the wrong thing. For the denominator, calculate and use $\sigma_U\sigma_V$.</p>
|
1,076,292 | <p>I wish to use two points say $(x_1$,$y_1)$ and $(x_2$,$y_2)$ and obtain the coefficients of the line in the following form: $$ Ax + By + C = 0$$</p>
<p>Is there any direct formula to compute.</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Set the values of $x,y$ to form two linear simultaneous equation for the unknowns $A,B$</p>
<p>Solve for $A,B$ in terms of $C, x_1,y_1,x_2,y_2$</p>
|
3,738,508 | <p>If <span class="math-container">$G$</span> is order <span class="math-container">$p^2q$</span>, where <span class="math-container">$p$</span>, <span class="math-container">$q$</span> are primes, prove that either a Sylow <span class="math-container">$p$</span>-subgroup or a Sylow <span class="math-container">$q$</span>-subgroup must be normal in <span class="math-container">$G$</span>.</p>
| Angina Seng | 436,618 | <p>If we defined the Frobenius map by <span class="math-container">$F(a)=a^5$</span>, then <span class="math-container">$a$</span> lies in <span class="math-container">$\Bbb F_{5^k}$</span>
iff <span class="math-container">$F^k(a)=a$</span>.</p>
<p>Here let <span class="math-container">$a$</span> be a root of <span class="math-container">$x^4+2=0$</span>. Then <span class="math-container">$a^4=-2$</span> and <span class="math-container">$F(a)=a^5=-2a\ne a$</span>.
So <span class="math-container">$a\notin\Bbb F_5$</span>.</p>
<p>Then <span class="math-container">$F^2(a)=F(-2a)=(-2a)^5=4a=-a\ne a$</span>. Therefore <span class="math-container">$a\notin\Bbb F_{25}$</span>.</p>
<p>Then <span class="math-container">$F^3(a)=F(-a)=(-a)^5=2a\ne a$</span>. Therefore <span class="math-container">$a\notin\Bbb F_{125}$</span>.</p>
<p>Of course, <span class="math-container">$F^4(a)=F(2a)=-4a=a$</span>, so <span class="math-container">$a\in\Bbb F_{625}$</span>.</p>
<p>Therefore <span class="math-container">$a$</span> generates the field extension of degree <span class="math-container">$4$</span> over <span class="math-container">$\Bbb 5$</span>, and
so <span class="math-container">$X^4+2$</span> is ireeducible.</p>
|
2,339,974 | <p>I know that there is a theorem that states that If $(G, *)$ and $(H, •)$ are groups, $e_G$ (identity of $G$) and $e_H$ (identity of $H$). Let $f: G\to H$ be a homomorphism. Then </p>
<ol>
<li>$f(e_G) = e_H$.</li>
</ol>
<p>I don't know how to use this, or begin my proof or should I use kernel for this problem?</p>
| Lukas Heger | 348,926 | <p>Let $f:G \to H$ be a homomorphism of groups, where $|G|=20$ and $|L|=17$.
Let $g \in G$ be any element, then the order of $g$ divides $20$, thus $g^{20}=e_G$, so after applying $f$ we get $f(g)^{20}=f(e_G)=e_L$, so the order of $f(g)$ divides $20$ as well. Now as $f(g)$ is an element of $L$, the order of $f(g)$ also divides $17$. As $17$ and $20$ are coprime, the order of $f(g)$ must be one, i.e. $f(g)=e_L$.</p>
|
4,185,100 | <p>In parallelogram <span class="math-container">$ABCD$</span> <span class="math-container">$CE=ED$</span>. <span class="math-container">$O$</span> is the intersection of <span class="math-container">$AE$</span> and the bisector of <span class="math-container">$\angle ABC$</span>. Given <span class="math-container">$AB=b$</span>, <span class="math-container">$BC=a$</span> and <span class="math-container">$\angle ABO = \alpha$</span>, find the are of <span class="math-container">$OBCE$</span>. I have given my calculations below and would like to know if it is correct.</p>
| Overdrowsed | 849,029 | <p>I have extended <span class="math-container">$AE$</span> and <span class="math-container">$BC$</span>, with their intersection point being <span class="math-container">$K$</span>, as shown below</p>
<p><a href="https://i.stack.imgur.com/9Nqk3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Nqk3.jpg" alt="enter image description here" /></a></p>
<p>As <span class="math-container">$AB \parallel CE$</span> it is the median of triangle <span class="math-container">$ABK$</span>, hence, due to triangle similarity, <span class="math-container">$BC=CK=a$</span>.</p>
<p>Area of <span class="math-container">$ABK = absin(2\alpha)$</span>, area of <span class="math-container">$ECK$</span> will be <span class="math-container">$\frac14$</span> of <span class="math-container">$ABK$</span> due to similarity, or <span class="math-container">$\frac14absin(2\alpha)$</span>.</p>
<p>As <span class="math-container">$BO$</span> is a bisector, area <span class="math-container">$A(ABO)/absin(2\alpha) = b/(b+2a) => A(ABO)=ab^2sin(2\alpha)/(2a+b)$</span></p>
<p>Area <span class="math-container">$A(OBCE) = A(ABK)-A(ECK)-A(ABO) = \frac34absin(2\alpha)-ab^2sin(2\alpha)/(2a+b)$</span></p>
|
4,185,100 | <p>In parallelogram <span class="math-container">$ABCD$</span> <span class="math-container">$CE=ED$</span>. <span class="math-container">$O$</span> is the intersection of <span class="math-container">$AE$</span> and the bisector of <span class="math-container">$\angle ABC$</span>. Given <span class="math-container">$AB=b$</span>, <span class="math-container">$BC=a$</span> and <span class="math-container">$\angle ABO = \alpha$</span>, find the are of <span class="math-container">$OBCE$</span>. I have given my calculations below and would like to know if it is correct.</p>
| Mick | 42,351 | <p>Note: This is essentially what you have done. I found that out after I have the solution written up.</p>
<p>Let [ABO] = X, [the required] = Y, [CEK] = Z.</p>
<p><a href="https://i.stack.imgur.com/rc7qU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rc7qU.png" alt="enter image description here" /></a></p>
<p>By area formula, <span class="math-container">$X + Y + Z = 0.5(b)(2a) \sin 2 \theta = ab \sin 2 \theta $</span>.</p>
<p>By midpoint theorem, 4Z = X + Y + Z.</p>
<p>Therefore, <span class="math-container">$Z = \dfrac {ab \sin 2 \theta}{4}$</span>.</p>
<p>By angle bisector theorem, <span class="math-container">$\dfrac {X}{b} = \dfrac {Y + Z}{2a}$</span>. That is, <span class="math-container">$2aX = b(Y + Z)$</span>.</p>
<p>Then, <span class="math-container">$2aX + bX = b(Y + Z) + bX = b(X + Y + Z)$</span>.</p>
<p>That is, <span class="math-container">$X = \dfrac {b(ab \sin 2 \theta )}{2a + b}$</span></p>
<p>Y can therefore be found.</p>
|
351,226 | <p>I am trying to brush up on my regular grammar knowledge to prepare for an interview, and I just am not able to solve this problem at all. This is NOT for homework, it is merely me trying to solve this.</p>
<p>I want to give a regular grammar for the language of the finite automaton whose screen shot is below, please help me, and if you can, a step by step answer would be of great assistance. Thank you!</p>
<p><img src="https://i.stack.imgur.com/U02gk.jpg" alt="enter image description here"></p>
| MJD | 25,554 | <p>To convert an automaton to a regular grammar is easy. You have one symbol for each state of the automaton, in this case $A, B,$ and $C$. Each symbol has one production for each transition that the automaton has out of the corresponding state. For example, when you have a transition $A\stackrel{0}{\to} B$ as you do here, the corresponding production is $A\to 0B$. </p>
<p>Then you let the start symbol be $S$, and add a single production $S\to A$, where $A$ is the start state. And for each accepting state, say $C$, you add a production $C\to\epsilon$.</p>
<p>So the solution here is:</p>
<p>$$\begin{align}
S & \to A\\
A & \to 1A \mid 0B\\
B & \to 1A \mid 0C\\
C & \to 1C \mid 0C \mid \epsilon
\end{align}
$$</p>
<p>The idea here is that the grammar symbol $B$ can produce all the strings that the automata would accept as input from state $B$. Since $B$ has a transition on <code>0</code> to state $C$, the $B$ symbol should be able to produce any string that has a <code>0</code> followed by some string that would be accepted from state $C$, so we add a production $B\to 0C$ so that the $B$ symbol can produce such strings.</p>
<p>Then in accepting states, the automaton will accept $\epsilon$, so we add those productions too.</p>
|
1,552 | <p>Closely related: what is the smallest known composite which has not been factored? If these numbers cannot be specified, knowing their approximate size would be interesting. E.g. can current methods factor an arbitrary 200 digit number in a few hours (days? months? or what?).
Can current methods certify that an arbitrary 1000 digit number is prime, or composite in a few hours (days? months? not at all?).</p>
<p>Any broad-brush comments on the current status of primality proving, and how active this field is would be appreciated as well. Same for factoring.</p>
<hr>
<p>Edit: perhaps my header question was something of a troll. I am not interested in lists. But if anyone could shed light on the answers to the portion of my question starting with "E.g.". it would be appreciated. (I could answer it in 1990, but what is the status today?)</p>
| Ilya Nikokoshev | 65 | <p>What you're asking is (or will be after a slight change of question) essentially a form of "what's the smallest number that cannot be written in words" paradox.</p>
|
1,552 | <p>Closely related: what is the smallest known composite which has not been factored? If these numbers cannot be specified, knowing their approximate size would be interesting. E.g. can current methods factor an arbitrary 200 digit number in a few hours (days? months? or what?).
Can current methods certify that an arbitrary 1000 digit number is prime, or composite in a few hours (days? months? not at all?).</p>
<p>Any broad-brush comments on the current status of primality proving, and how active this field is would be appreciated as well. Same for factoring.</p>
<hr>
<p>Edit: perhaps my header question was something of a troll. I am not interested in lists. But if anyone could shed light on the answers to the portion of my question starting with "E.g.". it would be appreciated. (I could answer it in 1990, but what is the status today?)</p>
| Harrison Brown | 382 | <p>Kenny's point in the comments is a good one as well, and it's also worth keeping in mind that the number of atoms in the observable universe is at most a few orders of magnitude more than 10^80, so writing down all the numbers with 100 or fewer digits is a hopeless task. (Certainly we haven't checked them all for primality!) That said, I'll try to give a rather rough answer to your questions.</p>
<p>1 terabyte, which is a good estimate for the size of a good commercially-available hard drive, is about 2^40 bits. Testing a number that small for primality is very easy (in fact, it's not terribly slow even with naive methods), so you could quite easily and cheaply list all the primes between about 1 and 2^35. Pushing higher than that, things start to get problematic -- there's just not enough space to hold all these numbers! If you gave a 1 TB hard drive to everyone on earth, there wouldn't be enough space to list all the primes between 1 and 2^70, which has about 23 digits. If you pick a random 25-digit number, odds are that it's never been tested for primality before. But it's easy to test such numbers for primality quickly, so this isn't a problem about primes so much as it is about the fact that the exponential function is <em>big</em>. So what about prime testing? How fast can we do that?</p>
<p>In practice, the best general method for primality testing seems to be <a href="http://en.wikipedia.org/wiki/Elliptic_curve_primality_proving" rel="noreferrer">elliptic curve primality proving</a>, probably combined with trial division by small integers. To give you some idea of how efficient it is, the biggest number that's been proved prime by ECPP methods has around 20,000 digits, and the proof took 9 months via a distributed computing project (equivalent to several years on a top-of-the-line general-purpose processor.) A rough back-of-the-envelope calculation suggests that if you want to test for primality in at most a couple of weeks on a single machine, 10000 digits is probably too much, although 1000 seems very reasonable.</p>
<p>Factoring seems much harder. 200 digits is reasonable with a wide distributed computing project, but 2000 is likely to be way out of reach.</p>
|
4,214,474 | <p>Consider <span class="math-container">$\mathbb{R}^\omega$</span> (countably infinite product of <span class="math-container">$\mathbb{R}$</span>) with the uniform metric.</p>
<p>Let <span class="math-container">$A$</span> be the set of infinite bounded sequences of <span class="math-container">$\mathbb{R}$</span>, i.e. <span class="math-container">$$A = \{x = (x_1, x_2, \dots ) \mid |x_k| \leq M \ \text{for all $k$ , for some positive number $M$}\}$$</span></p>
<p>Consider the metric <span class="math-container">$d(x,y) = \mathop{\text{sup }}_k \{\text{min}\{|x_k-y_k|, 1\}\}$</span>.</p>
<p>I want to prove <span class="math-container">$A$</span> is closed in the topology generated by <span class="math-container">$d$</span> (uniform topology).</p>
<p>My attempt:</p>
<p>We want to prove <span class="math-container">$A^c$</span> is open. Let <span class="math-container">$x = (x_k) \in A^c$</span>, so <span class="math-container">$x = (x_k)$</span> is a unbounded sequence. Suppose there exist a <span class="math-container">$M$</span> such that <span class="math-container">$|x_k| > M$</span> for infinitely many <span class="math-container">$k$</span>. I want to show there exist some <span class="math-container">$r>0$</span> such that <span class="math-container">$B_d(x,r) \subset A^c$</span>. Take <span class="math-container">$r = \frac{M}{2}$</span>, and <span class="math-container">$B_d(x, \frac{M}{2}) \subset A^c$</span>. Let <span class="math-container">$y = (y_k) \in B_d(x, \frac{M}{2})$</span>.</p>
<p>But I have no idea how to proceed further?</p>
<p>Actually, I want to prove <span class="math-container">$y = \{y_k\}$</span> is unbounded.</p>
<p>Please help me.</p>
| Lukas Betz | 238,388 | <p>For your unbounded <span class="math-container">$x$</span> you are only using the fact that there is an <span class="math-container">$M$</span> such that <span class="math-container">$|x_n| > M$</span> for infinitely many <span class="math-container">$n$</span>. This is to weak since it is true for any <span class="math-container">$x$</span> which doesn't converge to <span class="math-container">$0$</span> hence also for many <span class="math-container">$x\in A$</span>. You would need to use that there is a subsequence <span class="math-container">$(x_{n_k})_k$</span> with <span class="math-container">$|x_{n_k}| > k$</span> for all <span class="math-container">$k\in\mathbb N$</span> or something similar. Then look at <span class="math-container">$B := B_d(x, \frac 12)$</span>. It is important that the distance is less than <span class="math-container">$1$</span>, otherwise your ball is the whole space. Now for each <span class="math-container">$y\in B$</span> it holds <span class="math-container">$$k < |x_{n_k}| \leq |x_{n_k}-y_{n_k}|+|y_{n_k}|\leq \frac 12 + |y_{n_k}|.$$</span></p>
<p>From here it is easy to see that <span class="math-container">$y$</span> is unbounded.</p>
|
1,470,760 | <p>Okay so $A=0.2, B=0.5$ and the probability that both $A$ and $B$ occur is equal to $0.12$.</p>
<p>What is $P((A \cap B) \cup A^c)$?</p>
<p>What I basically did was $0.12 \times 0.5+0.5+0.2-0.12 = 1.2$. </p>
<p>Am I doing it right?</p>
| barak manos | 131,263 | <p>HINT:</p>
<p><a href="https://i.stack.imgur.com/od1UI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/od1UI.png" alt="enter image description here"></a></p>
<p>You need the yellow part plus the blue part.</p>
<p>BTW, probability is a value between $0$ and $1$ <strong>by definition</strong>, so that rules out your answer.</p>
|
3,439,223 | <blockquote>
<p>Given the metric space <span class="math-container">$(X,d)$</span>:</p>
<p>If <span class="math-container">$M\subset N \subset X$</span>, <span class="math-container">$M\neq 0$</span>, we have <span class="math-container">$\text{diam}(M)\leq
\text{diam}(N)$</span></p>
</blockquote>
<p>Negating the statement, we assume that - given the assumptions - we deduce <span class="math-container">$\text{diam}(M):=\sup\limits_{x,y\in M}d(x,y)>\sup\limits_{u,v\in N}d(x,y)=:\text{diam}(N)$</span>. This can't be true, since <span class="math-container">$x,y\in M \implies x,y\in N$</span> but <span class="math-container">$u,v\in N$</span> does not necessarily imply that <span class="math-container">$u,v\in M$</span>.</p>
<p>I don't really know how to put it formally though.</p>
| José Carlos Santos | 446,262 | <p>The number <span class="math-container">$\operatorname{diam}M$</span> is the supremum of a set <span class="math-container">$S_M$</span> of numbers and the number <span class="math-container">$\operatorname{diam}N$</span> is the supremum of a set <span class="math-container">$S_N$</span> of numbers. Since <span class="math-container">$S_M\subset S_N$</span>, <span class="math-container">$\sup S_M\leqslant\sup S_N$</span>.</p>
|
1,620,686 | <p>Prove that this works for all $x$ and and only some $y$
$$\sqrt{(x-1)^2-(y+2)^2}=0.$$</p>
<p>This is as far as I got so far</p>
<p>Difference of squares:</p>
<p>$\sqrt{(x-1-y-2)(x-1+y+2)}=0$<br>
$\sqrt{x-y-3}\sqrt{x+y+1}=0$</p>
<p>Therefore $x-y-3=0 \implies y=x-3$ </p>
<p>$x+y+1=0$ and $y=-1-x$</p>
<p>I just don't know where to go from here?</p>
| BrianO | 277,043 | <p>What you need to show is the following:</p>
<blockquote>
<p>For every real number $x$, there exists a real number $y$ such that
$$\sqrt{(x-1)^2-(y+2)^2}=0,\tag{Eqn}$$</p>
</blockquote>
<p>An equivalent way of saying "there exists a real $y$" is to say "for some real $y$". Saying "for some <em>of</em> $y$" brings to mind other things that aren't relevant. For each $x$, you only need to show that there is (at least) one $y$ satisfying (Eqn).</p>
<p>We write "for all x" and "there is y" using the <em>quantifiers</em> $\forall x$ and $\exists y$. In what follows, both variables $x$ and $y$ are assumed to be real numbers.</p>
<p>Let's rewrite what you have to show using quantifiers:
$$
(\forall x)(\exists y)\, \sqrt{(x-1)^2-(y+2)^2}=0,\tag{*}
$$
Clearly (yes?), the following statement implies (*):
$$
(\forall x)(\exists y)\, [(x-1)^2 = (y+2)^2].\tag{**}
$$</p>
<p>So, given $x$, simply take $y=x-3$. Then $x-1 = y+2$, so $(x-1)^2 = (y+2)^2$. This proves (**), which implies (*).</p>
<p><hr>
Your question (but not your title) says "only some", so it seems you also want to show the following:</p>
<blockquote>
<p>For every real number $x$, there exists a real number $y$ such that
$$\sqrt{(x-1)^2-(y+2)^2}\ne0,\tag{not-Eqn}$$</p>
</blockquote>
<p>This is also straightforward. Given $x$, here's how to find a $y$ such that (not-Eqn) is true:</p>
<ul>
<li>if $x \ne 1$, let $y=-2$, and</li>
<li>if $x = 1$, let $y = 0$.</li>
</ul>
<p>So the second highlighted statement is also true.</p>
|
2,483,794 | <p>I'm trying to figure out the equality $$\frac{1}{y(1-y)}=\frac{1}{y-1}-\frac{1}{y}$$</p>
<p>I have tried but keep ending up with RHS $\frac{1}{y(y-1)}$.</p>
<p>Any help would be appreciated.</p>
| Michael Rozenberg | 190,319 | <p>$$\frac{1}{y(y-1)}=\frac{y-(y-1)}{y(y-1)}=\frac{y}{y(y-1)}-\frac{y-1}{y(y-1)}=\frac{1}{y-1}-\frac{1}{y}$$</p>
|
2,118,761 | <p>How can I show that there are an infinite number of primes by using the Fundamental Theorem of Arithmetic?</p>
| marty cohen | 13,079 | <p>The standard way
is by assuming that
there are only a finite number of primes
and deducing that
if all terms of the form
$\prod_{p \in P} p^{a(p)}$
are counted,
there are not enough of them.</p>
<p>I don't remember the details,
but it might go
something like this:</p>
<p>The number of integers
of the form
$\prod_{p \in P} p^{a(p)}$
which are less that $x$
are of the form
$\prod_{p \in P} p^{a(p)}
\le x$
or
$\sum_{p \in P} a(p)\ln p
\le \ln x$.</p>
<p>We certainly have
$a(p) \le \ln x/\ln p$.</p>
<p>The total number of possibilities
is certainly less than,
where $N$ is the number of primes,
$\prod_{p \in P}(\ln x)/(\ln p)
=(\ln x)^{N}\prod_{p \in P}1/(\ln p)
$.</p>
<p>However,
for every finite $N$,
$(\ln x)^{N}$ will be less than $x$
for large enough $x$.</p>
<p>So $N$ must be infinite.</p>
<p>I just did this off
the top of my head,
so I'm not completely sure that
it is correct,
but it seems OK to me.</p>
<p>Anyway,
there it is.</p>
|
342,306 | <p>An elementary embedding is an injection $f:M\rightarrow N$ between two models $M,N$ of a theory $T$ such that for any formula $\phi$ of the theory, we have $M\vDash \phi(a) \ \iff N\vDash \phi(f(a))$ where $a$ is a list of elements of $M$.</p>
<p>A critical point of such an embedding is the least ordinal $\alpha$ such that $f(\alpha)\neq\alpha$. </p>
<p>A large cardinal is a cardinal number that cannot be proven to exist within ZFC.
They often appear to be critical points of an elementary embedding of models of ZFC where $M$ is the von Neumann hierarchy, and $N$ is some transitive model. Is this in fact true for all large cardinal axioms?</p>
| Paul McKenney | 53,995 | <p>No. If $\kappa$ is the critical point of a full elementary embedding $j : V\to N$, with $N$ transitive, then $\kappa$ is <a href="http://en.wikipedia.org/wiki/Measurable_cardinal" rel="nofollow">measurable</a>. Yet not all large cardinals are measurable; see, for instance, <a href="http://en.wikipedia.org/wiki/Weakly_compact_cardinal" rel="nofollow">weakly compact cardinals</a>. Even large cardinals with higher consistency strength (like <a href="http://en.wikipedia.org/wiki/Woodin_cardinal" rel="nofollow">Woodin cardinals</a>) are not always measurable.</p>
|
342,306 | <p>An elementary embedding is an injection $f:M\rightarrow N$ between two models $M,N$ of a theory $T$ such that for any formula $\phi$ of the theory, we have $M\vDash \phi(a) \ \iff N\vDash \phi(f(a))$ where $a$ is a list of elements of $M$.</p>
<p>A critical point of such an embedding is the least ordinal $\alpha$ such that $f(\alpha)\neq\alpha$. </p>
<p>A large cardinal is a cardinal number that cannot be proven to exist within ZFC.
They often appear to be critical points of an elementary embedding of models of ZFC where $M$ is the von Neumann hierarchy, and $N$ is some transitive model. Is this in fact true for all large cardinal axioms?</p>
| zyx | 14,120 | <p>There is an online PDF of lecture slides by Woodin on the "Omega Conjecture" in which he axiomatizes the type of formulas that are large cardinals. I do not know how exhaustive his formulation is. See the references under</p>
<p><a href="http://en.wikipedia.org/wiki/Omega_conjecture" rel="nofollow">http://en.wikipedia.org/wiki/Omega_conjecture</a></p>
|
140,358 | <p>Let $X$ and $Y$ be two topological spaces with $C(X) \cong C(Y)$ (where $C(X)$ is the ring of all continuous real valued functions on $X$). I know that we can not conclude that $X$ and $Y$ are homeomorphic. But I wonder how independent $X$ and $Y$ could be ? For example is there any forced relation between their cardinality ?</p>
| Community | -1 | <p>Let $X,Y$ be arbitrary sets (of arbitrary cardinals) armed with topologies $\tau_1 = \lbrace \emptyset, X\rbrace$ and $\tau_2 = \lbrace \emptyset, Y\rbrace$. Then it is clear that $C(X) \cong \Bbb{R} \cong C(Y)$. So there is no forced relation between the cardinal numbers. </p>
|
2,309,721 | <p>The problem is: Prove that $7|x^2+y^2$ only if $7|x$ and $7|y$ for $x,y∈Z$.</p>
<p>I found a theorem in my book that allows to do the following transformation:
if $a|b$ and $a|c$ -> $a|(b+c)$</p>
<p>So, can I prove it like this: $7|x^2+y^2 =>7|x^2, 7|y^2 => 7|x*x, 7|y*y => 7|x, 7|y$ ?</p>
<p>I am not really sure because I have this simple example in my head that even if $6|18$ => $6|14+4$, $6 ∤ 14$ and $6 ∤ 4$.</p>
<p>Any help will be appreciated, thank you!</p>
| knm | 447,358 | <p>There is a theorem that if a prime $p$ of the form $p \equiv 3\pmod4$, and $p \mid x^2+y^2$, then $p \mid x$ and $p \mid y$.</p>
<p>At your problem take $p=7$, and with the above theorem you are done.</p>
|
119,722 | <p>For a hyperplane arrangement $\mathcal{A}$ over a vector space $V$, we define its intersection poset, $L(\mathcal{A})$, as the set of all nonempty intersections of hyperplanes in $\mathcal{A}$ ordered by reverse inclusion. The empty intersection, $V$ itself, is the unique minimal element of $L(\mathcal{A})$.</p>
<p>It is known that $L(\mathcal{A})$ is a ranked meet-semilattice, and moreover, any interval $[x,y]$ in $L(\mathcal{A})$ is a geometric lattice. But these properties alone are not sufficient for some poset $P$ to be the intersection poset of a hyperplane arrangement. Consider the following poset:<a href="http://www.freeimagehosting.net/newuploads/b4quk.png" rel="nofollow noreferrer">poset http://www.freeimagehosting.net/newuploads/b4quk.png</a></p>
<p>If this were the intersection poset of some arrangement, then $a$ would be parallel to $d$ and to $c$, $b$ would be parallel to $d$, and thus $b$ and $c$ would be parallel. But $b$ and $c$ have nonempty intersection, so this is nonsense.</p>
<p>Is there a known characterization of hyperplane arrangement intersection posets?</p>
| Michael Falk | 15,365 | <p>The intersection poset of a (not necessarily central) hyperplane arrangement is a geometric semi-lattice, as defined by Bjorner and Wachs, who show that every such poset is isomorphic to the subposet of $x \not \geq a$ of a geometric lattice and an atom a. This corresponds geometrically to putting the arrangement into projective space, and adding the hyperplane at infinity, which corresponds to $a$. I haven't looked at the definition, but I suspect your example above is not a geometric semi-lattice; if it were it should be "realizable" by an arrangement of pseudo-hyperplanes, but your argument shows it cannot. </p>
<p>Given a geometric semi-lattice, it is realizable by an arrangement of affine hyperplanes if and only if the associated geometric lattice is realizable by linear hyperplanes, which is equivalent to the matroid-realizability question as Richard points out above. For this you need to specify a field - some lattices are realizable over $\mathbb C$ or over a finite field, but not over $\mathbb R$. Oriented matroids and arrangements of pseudo-spheres are relevant only to realizability over $\mathbb R$. </p>
|
354,642 | <p>Show that each of the following initial-value problems has a unique solution ($0 ≤ t ≤ 1 , y(0) = 1$).</p>
<p>$$y' = \exp(t-y)$$</p>
<p><strong>Theorem 1</strong>: Suppose that $D=\{(t,y)|a≤t≤b, −∞< y<∞\}$ and that $f(t,y)$ is continuous on $D$. If $f$ satisfies a Lipschitz condition on $D$ in the variable $y$, then the initial-value problem $y′(t)=f(t,y)$, $a≤t≤b$, $y(a)=α$, has a unique solution $y(t) for a ≤ t$ ≤ b.</p>
<p>Can theorem 1 be applied?</p>
| Community | -1 | <p>Consider the sequences $$x_n = \dfrac1{2n \pi} \,\,\,\, \text{ and } x_n = \dfrac1{2 n \pi + \dfrac{\pi}2}$$ Both tend to zero. What happens to $\cos(1/x)$ along these sequences? (Recall that if a limit exists it has to be unique.)</p>
|
468,784 | <p>Two disjoint sets $A$ and $B$, neither empty, are said to be <strong>mutually separated</strong> if neither contains a boundary point of the other. A set is disconnected if it is the union of separated subsets, and is called <strong>connected</strong> if it is not disconnected.</p>
<p>With the above definition of connected set, how to prove that the set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected.</p>
<p>Exercise should try using the definition of previous related and for this I guess I should do the proof by contradiction but do not know how to proceed.</p>
| Betty Mock | 89,003 | <p>If |x| < |y| then those x's are all in the disk of radius y about the origin. That is surely connected</p>
|
468,784 | <p>Two disjoint sets $A$ and $B$, neither empty, are said to be <strong>mutually separated</strong> if neither contains a boundary point of the other. A set is disconnected if it is the union of separated subsets, and is called <strong>connected</strong> if it is not disconnected.</p>
<p>With the above definition of connected set, how to prove that the set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected.</p>
<p>Exercise should try using the definition of previous related and for this I guess I should do the proof by contradiction but do not know how to proceed.</p>
| Balbichi | 24,690 | <p>Do you know straight lines are connected set in $\mathbb{R}^2$? do you know union of connected set is connected if they have one point in common?</p>
<p>Now can you see your set with above stated properties?</p>
|
2,881,020 | <p>We're trying to figure out something, and things aren't adding up. The senario is that you make $\$ 50,000$ a year. Every year you get a $15\%$ bonus of that income, which then gets added to your next year's income. So, the first year, you get $\$ 7,500$ which then makes your base $\$ 57,500$ the next year. When we try to figure out what the bonus will be past $6$ or $7$ years the numbers just sort of plateau at around $\$ 8,823$. All we were doing is adding the amount of the bonus to $\$ 50,000$ and then finding $15\%$ of the resulting number, and repeating. Is this correct? Why are the numbers plateauing? What would the bonus actually be after $6$ or $7$ years? I'm sorry if this is a simple question for this forum, it was recommended to me. Thanks so much in advance.</p>
| mzp | 287,326 | <p>I am not sure this is what you want, but here is the calculation for the bonuses in the first $3$ years which should get you going:</p>
<p><strong>1st Year:</strong></p>
<p>$$ \text{Base} = 50,000 \quad \Rightarrow \quad \text{Bonus}=0.15*50,000 = 7,500. $$</p>
<p><strong>2nd Year:</strong></p>
<p>$$ \text{Base} = 50,000+7,500=57,500 \quad \Rightarrow \quad \text{Bonus}=0.15*57,500 = 8,625. $$</p>
<p><strong>3rd Year:</strong></p>
<p>$$ \text{Base} = 57,500+8,625=66,125 \quad \Rightarrow \quad \text{Bonus}=0.15*57,500 \approx 9,919. $$</p>
|
2,881,020 | <p>We're trying to figure out something, and things aren't adding up. The senario is that you make $\$ 50,000$ a year. Every year you get a $15\%$ bonus of that income, which then gets added to your next year's income. So, the first year, you get $\$ 7,500$ which then makes your base $\$ 57,500$ the next year. When we try to figure out what the bonus will be past $6$ or $7$ years the numbers just sort of plateau at around $\$ 8,823$. All we were doing is adding the amount of the bonus to $\$ 50,000$ and then finding $15\%$ of the resulting number, and repeating. Is this correct? Why are the numbers plateauing? What would the bonus actually be after $6$ or $7$ years? I'm sorry if this is a simple question for this forum, it was recommended to me. Thanks so much in advance.</p>
| Ross Millikan | 1,827 | <p>The general rule is that in year $n$ the base is $50,000\cdot 1.15^n$ and the bonus is $7,500\cdot 1.15^n$ This increases without bound.
<a href="https://i.stack.imgur.com/0VDsp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0VDsp.png" alt="enter image description here"></a></p>
|
162,655 | <p>Does there exist a Ricci flat Riemannian or Lorentzian manifold which is geodesic complete but not flat? And is there any theorm about Ricci-flat but not flat? </p>
<p>I am especially interset in the case of Lorentzian Manifold whose sign signature is (- ,+ ,+ , + ). Of course, the example is not constricted in Lorentzian case.</p>
<p>I know there are many Ricci flat case in General Relativiy which is the vacuum solution to Einstein's equation. But what I know, such as Kerr solution, are all geodesic incomplete. So I want a geodesic complete example and reference. Thanks!</p>
| Yiyan | 22,880 | <p>Eguchi - Hanson metric over $T^*S^2$ is complete Ricci flat but not flat, which can be written down explicitly. In fact, $T^*S^n$ admits Calabi-Yau structure for each $n$. </p>
<p>Ref: Stenzel, Ricci flat metrics on the complexification of a compact rank one symmetric space.</p>
|
1,779,088 | <blockquote>
<p>Prove
$$\sum_{i=1}^n i^{k+1}=(n+1)\sum_{i=1}^n i^k-\sum_{p=1}^n\sum_{i=1}^p i^k \tag1$$
for every integer $k\ge0$. </p>
</blockquote>
<p>By principle of induction,</p>
<p>$$\sum_{i=1}^n i = n(n+1)- \sum_{p=1}^n p$$
$$2\sum_{i=1}^n i = n(n+1)$$
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
$\implies$$(1)$ is true for k equals to zero.</p>
<p>Assume $(1)$ is true.
$$\sum_{i=1}^n i^{k+2}=(n+1)\sum_{i=1}^n i^{k+1}-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\tag2$$
We prove $(2)$ is true.</p>
<p>From $(2)$,</p>
<p>$$\begin{align}
RHS & = (n+1)\left[(n+1)\sum_{i=1}^n i^k-\sum_{p=1}^n\sum_{i=1}^p i^k \right]-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\\
& = (n+1)^2\sum_{p=1}^n p^k-(n+1)\sum_{p=1}^n\sum_{i=1}^p i^k-\sum_{p=1}^n\sum_{i=1}^p i^{k+1}\\
& = \sum_{p=1}^n\left[(n+1)^2p^k-(n+1)\sum_{i=1}^p i^k-\sum_{i=1}^p i^{k+1}\right]\\
& = \sum_{p=1}^n\left[(n+1)^2p^k-\sum_{i=1}^p (n+1+i)i^k\right]\\
\end{align}$$</p>
<p>By examining $(2)$,</p>
<p>$$\sum_{p=1}^n\left[(n+1)^2p^k-\sum_{i=1}^p (n+1+i)i^k\right]=\sum_{p=1}^n p^{k+2}=LHS\tag3$$</p>
<p>We should be able to get $(3)$.</p>
<p>Anyone knows how to prove $(3)$?</p>
| Mark Viola | 218,419 | <p>Rather than proceed using induction, I thought it might be instructive to present a straightforward approach. </p>
<p>To that end we proceed by using the <a href="https://en.wikipedia.org/wiki/Summation_by_parts#Newton_series" rel="nofollow">Newton Series</a> for summation by parts with $f_i=i$ and $g_i=i^k$.</p>
<p>Then, we can write</p>
<p>$$\begin{align}
\sum_{i=1}^n i^{k+1}&=\sum_{i=1}^n (i)(i)^k\\\\
&=n\sum_{i=1}^n i^k - \sum_{j=1}^{n-1}\sum_{i=1}^j i^k\\\\
&=n\sum_{i=1}^n i^k - \sum_{j=1}^{n}\sum_{i=1}^j i^k +\sum_{i=1}^n i^k\\\\
&=(n+1)\sum_{i=1}^n i^k - \sum_{j=1}^{n}\sum_{i=1}^j i^k
\end{align}$$</p>
<p>as was to be shown!</p>
|
2,314,327 | <p>I have a quick question here.</p>
<p>For an exercise, I was asked to factor:</p>
<p>$$11x^2 + 14x - 2685 = 0$$</p>
<p>How do I figure this out quickly without staring at it forever? Is there a quicker mathematical way than guessing number combinations, or do I have to guess until I find the right combination of numbers?</p>
<p>The answer is:</p>
<p>$$(11x + 179)(x - 15) = 0 $$</p>
| Ross Millikan | 1,827 | <p>You want to know the <a href="https://en.wikipedia.org/wiki/Divisibility_rule" rel="noreferrer">divisibility rules</a> for small numbers. We know $2685$ is divisible by $5$ because of the last digit and by $3$ because of the sum of the digits. Once you find those factors, divide them out, getting $179$. The rules show it is not divisible by $2,3,5,11$ (or $7$ if you know that one, but it is less common. I like the double the ones digit and subtract, which just gives a yes/no answer). Since you only need to check primes up to the square root of the number, you would just have to trial divide by $13$ and maybe $7$ to find that $179$ is prime. In exercises there will always be small factors. In RSA encryption, not so much.</p>
|
341,648 | <p>I'm trying to understand what a tableaux ring is (it's not clear to me reading Young Tableaux by Fulton).</p>
<p>I studied what a monoid ring is on Serge Lang's Algebra, and then I read about modules, modules homomorphism. I'm trying to prove what is stated at page 121 (S. Lang, Algebra) while talking about algebras: "we note that the group ring $A[G]$ (or monoid ring when $G$ is a monoid) is an $A$-algebra, also called group (or monoid) algebra."</p>
<p>Correct me if I'm wrong, I believe that i should start proving that the monoid ring $A[G]$ (here $G$ is a monoid) is an $A$-module. Well I can't figure out how it could be! This is what i've tried:</p>
<p>notation: $A$ ring, $G$ monoid, $a \in A, x \in G$.
$a \cdot x$ is the map $G \rightarrow A$ such that $a \cdot x (x) = a$ and $a \cdot x (y) = 0$ for every $y \neq x$</p>
<p>$(a+b) \cdot x = a \cdot x + b\cdot x$ for every $a,b \in A$ and every $x \in G$ follows from definition of $a \cdot x$. I don't know how to show $a \cdot (x+y) = a \cdot x + a \cdot y$, honestly I'm about to think it is not true.</p>
<p>Thanks in advance,</p>
<p>sciamp</p>
| Did | 6,179 | <p>The integral $I$ to be computed is
$$
I=2\int_0^{1/2}\log(\sin\pi x)\mathrm dx\stackrel{x\to 1/2-x}{=}2\int_0^{1/2}\log(\cos\pi x)\mathrm dx.
$$
Summing up yields
$$
2I=2\int_0^{1/2}\log(\cos\pi x\sin\pi x)\mathrm dx=2\int_0^{1/2}\log(\sin2\pi x)\mathrm dx-2\int_0^{1/2}\log(2)\mathrm dx.
$$
The first integral on the RHS is
$$
\int_0^{1/2}\log(\sin2\pi x)\mathrm dx\stackrel{x\to2x}{=}\int_0^{1}\log(\sin\pi x)\frac{\mathrm dx}2=\frac{I}2,
$$
and the second integral on the RHS is easy, hence $2I=I-\log2$ and, finally,
$$
I=-\log2.
$$</p>
|
977,956 | <p>Can you help me solve this problem?</p>
<blockquote>
<p>Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.</p>
</blockquote>
| Community | -1 | <p>Take the terms in opposite pairs, and note the change of sign,</p>
<p>$$\sin \dfrac{2k\pi}{n}+\sin \dfrac{2\pi(n-k)}{n}=\sin \dfrac{2k\pi}{n} +\sin(2\pi-\dfrac{2k\pi}{n})=\sin \dfrac{2k\pi}{n} -\sin\dfrac{2k\pi}{n}=0.$$
In case that $n$ is even, the central term remains, but $$\sin \dfrac{2n\pi}{2n}=0.$$</p>
|
4,216,105 | <p>In the <a href="https://www.feynmanlectures.caltech.edu/I_22.html#Ch22-S5" rel="nofollow noreferrer">Algebra chapter</a> of the Feynman Lectures on Physics, Feynman introduces complex powers:</p>
<blockquote>
<p>Thus <span class="math-container">$$10^{(r+is)}=10^r10^{is}\tag{22.5}$$</span>
But <span class="math-container">$10^r$</span> we already know how to compute, and we can always multiply anything by anything else; therefore the problem is to compute only <span class="math-container">$10^{is}$</span>. Let us call it some complex number, <span class="math-container">$x+iy$</span>. Problem: given <span class="math-container">$s$</span>, find <span class="math-container">$x$</span>, find <span class="math-container">$y$</span>. Now if
<span class="math-container">$$10^{is}=x+iy$$</span> then the complex conjugate of this equation must also be true, so that
<span class="math-container">$$10^{−is}=x−iy$$</span></p>
</blockquote>
<p>I don't think it's all that easy to guess/infer intuitively, this fact (about complex conjugates). Especially when you are a beginner (that's who the author addresses this chapter to, building steadily from arithmetic through algebra, logarithms, etc... guided by the intellectual beacons of Abstraction and Generalisation), it isn't at all convincing why if <span class="math-container">$10^{is}$</span> equals some <span class="math-container">$x + iy$</span>, then <span class="math-container">$10^{-is}$</span> must be <span class="math-container">$x-iy$</span>.</p>
<p>The only definition of <span class="math-container">$i$</span> is that its square is <span class="math-container">$-1$</span> (Feynman's reason for this to be true). What am I missing here?</p>
| Brick | 263,389 | <p>Feynman in that whole section is using results that he knows to be true and trying to provide some intuition, but you've picked one of several things in that chapter that are not mathematically rigorous. Even just with the equations you've shown, a mathematician would want to prove that <span class="math-container">$10^{r+is} = 10^r 10^{is}$</span>. It's true, but you cannot <em>assume</em> it's true just from the definition <span class="math-container">$i^2 = -1$</span>. (For other structures, like matrix exponentiation, it's not generally true, just for example.)</p>
<p>Similarly, he's assuming that <span class="math-container">$(10^{is})^* = 10^{-is}$</span>. Again, that's definitely true but it also does not immediately follow from the definitions. It should be proved if you want to be mathematically rigorous.</p>
<p>I don't know that this is a "cop out" because I also don't think he's claiming rigor here at all. He's trying to give some quick intuition and, like other teachers, he's approaching it with foresight to the answers in a way that he thinks is useful to the student. I think there's no more or less to it than that.</p>
|
43,513 | <p>Please help me with the following question.</p>
<p>Let $F:\mathbb{R}^{k}\to\mathbb{R}^{k}$ be a continuously differentiable mapping;</p>
<p>$F_{n}(x)$ be $n$-th iteration of $F(x)$, i.e. $F_{1}(x)=F(x)$, $F_{n}(x)=F(F_{n-1}(x))$;</p>
<p>$J_{n}(x)=(F_{n}(x))'$ be Jacobian matrix of $F_{n}(x)$;</p>
<p>$\lambda_{n}^{(1)}(x), \lambda_{n}^{(2)}(x), \ldots, \lambda_{n}^{(k)}(x)$ be eigenvalues of $J_{n}(x)$;</p>
<p>$\mu_{n}^{(1)}(x), \mu_{n}^{(2)}(x), \ldots, \mu_{n}^{(k)}(x)$ be eigenvalues of $(J_{n}(x))^{T}J_{n}(x)$.</p>
<p>How to prove that for all $i=\overline{1,k}$ there exists $j=\overline{1,k}$ such that $\lim\limits_{n\to\infty}\big|\lambda_{n}^{(i)}(x)\big|^{\frac{1}{n}}=\lim\limits_{n\to\infty}\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2n}}$? (if this statement is true)</p>
<p>As I see, $\big|\lambda_{n}^{(i)}(x)\big|\neq\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2}}$ in common case...</p>
| rpotrie | 5,753 | <p>In general, this may not be true. </p>
<p>EDIT (Atending OP´s objection) The important thing is that as stated, the problem is reduced to a linear algebra problem since it is possible to construct a diffeomorphism of $\mathbb{R}^n$ such that matrix $J_n(0)$ for the orbit of $0$ is the product of any sequence of invertible matrices. </p>
<p>To construct this, consider a translation of $\mathbb{R}^n$ (say $F(x)= x+b$) and in a neighborhood of $nb$ modify the diffeomorphism so that the derivative in that point is the desired matrix $A_n$. </p>
<p>In dimension $2$, a way of getting the desired counterexample is to consider the two times two upper triangular matrices $A_n$ with both eigenvalues $1$ and $K^n$ in the upper right corner (I am not being able to write matrices). </p>
<p>We get that the eigenvalues of $J_n$ will be always $1$, but the norm of $J_n$ grows exponentially, so, it is not true that the limits coincide. </p>
<p>I haven't thought on how to make a counterexample where the norms of $A_n$ are bounded but it should be not very difficult. </p>
<p>However, when some recurrence is added into the game, some results in the direction of what you are looking for are available. A key word for searching is <a href="http://www.scholarpedia.org/article/Oseledets_theorem" rel="nofollow">Oseledets Theorem</a> (or Multiplicative ergodic theorem, notice that in some places it is named Oseledec, or with some variations, othe key word for searching is: Lyapunov exponents). In particular, given an invariant probability measure, what you look for is satisfied for almost every point. </p>
<p>Playing with the proofs of this results, other more ``natural'' counterexamples can be constructed. </p>
|
43,513 | <p>Please help me with the following question.</p>
<p>Let $F:\mathbb{R}^{k}\to\mathbb{R}^{k}$ be a continuously differentiable mapping;</p>
<p>$F_{n}(x)$ be $n$-th iteration of $F(x)$, i.e. $F_{1}(x)=F(x)$, $F_{n}(x)=F(F_{n-1}(x))$;</p>
<p>$J_{n}(x)=(F_{n}(x))'$ be Jacobian matrix of $F_{n}(x)$;</p>
<p>$\lambda_{n}^{(1)}(x), \lambda_{n}^{(2)}(x), \ldots, \lambda_{n}^{(k)}(x)$ be eigenvalues of $J_{n}(x)$;</p>
<p>$\mu_{n}^{(1)}(x), \mu_{n}^{(2)}(x), \ldots, \mu_{n}^{(k)}(x)$ be eigenvalues of $(J_{n}(x))^{T}J_{n}(x)$.</p>
<p>How to prove that for all $i=\overline{1,k}$ there exists $j=\overline{1,k}$ such that $\lim\limits_{n\to\infty}\big|\lambda_{n}^{(i)}(x)\big|^{\frac{1}{n}}=\lim\limits_{n\to\infty}\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2n}}$? (if this statement is true)</p>
<p>As I see, $\big|\lambda_{n}^{(i)}(x)\big|\neq\big(\mu_{n}^{(j)}(x)\big)^{\frac{1}{2}}$ in common case...</p>
| Alexandra Korobeynikova | 10,343 | <p>But this is not a counterexample. You proposed $F(x)=Ax+b$ if $\|x\|\leq\delta$, $F(x)=x+b$ if $\|x\|>\delta$ where $\|b\|$ is sufficiently large.</p>
<p>Here $A$ is a matrix with eigenvalues $\lambda^{(1)}, \lambda^{(2)}, \ldots\lambda^{(k)}$;</p>
<p>$\mu^{(1)}, \mu^{(2)}, \ldots\mu^{(k)}$ are eigenvalues of $A^{T}A$;</p>
<p>$A$ is such that $\big|\lambda^{(i)}\big|\neq\big(\mu^{(i)}\big)^{\frac{1}{2}}$.</p>
<p>Then, as you write, $J_{n}(0)=A$ and we have</p>
<p>$\lim\limits_{n\to\infty}\big|\lambda_{n}^{(i)}(0)\big|^{\frac{1}{n}}=\lim\limits_{n\to\infty}\big(\mu_{n}^{(i)}(0)\big)^{\frac{1}{2n}}=1$</p>
<p>since $\lambda_{n}^{(i)}(0)=\lambda^{(i)}$ do not depend on $n$.</p>
|
332,170 | <p>How can I solve <span class="math-container">$$T(n) = aT(n-1) + bT(n-2)+ cn $$</span>; where <span class="math-container">$a,b,c$</span> are constants. I could not figure it out :(</p>
<p>There are T(0) = d and T(1) = e,</p>
<p>Thanks in advance.</p>
| Brian M. Scott | 12,042 | <p>Your general problem is significantly harder than the specific problem that gave rise to it. I would not use the characteristic equation at all for the specific problem. For the specific recurrence $T(n)=T(n-2)+4n$ with initial conditions $T(0)=2$ and $T(1)=3$, I’d separate it into two sequences, one corresponding to even $n$ and the other to odd $n$.</p>
<p>Let $E(n)=T(2n)$; then $E(n)=E(n-1)+8n$, and $E(0)=2$. It follows readily that</p>
<p>$$E(n)=E(n-1)+8n=E(n-2)+8n+8(n-1)=\ldots E(0)+8\sum_{k=1}^nk=2+4n(n+1)$$</p>
<p>for all $n$. (If this is insufficiently rigorous, you can prove by induction that $E(n)=2+4n(n+1)$ for all $n$.) We can now infer that $T(2n)=E(n)=2+4n(n+1)$ for all $n$.</p>
<p>Let $O(n)=T(2n+1)$; then $O(n)=O(n-1)+4(2n+1)$, and $O(0)=3$, and we can make the same sort of calculation:</p>
<p>$$\begin{align*}
O(n)&=O(n-1)+4(2n+1)\\
&=O(n-2)+4(2n-1)+4(2n+1)\\
&\;\vdots\\
&=O(0)+4\sum_{k=1}^n(2k+1)\\
&=3+8\sum_{k=1}^nk+4n\\
&=3+4n(n+1)+4n\\
&=3+4n(n+2)\;.
\end{align*}$$</p>
<p>Again, the final result, $O(n)=3+4n(n+2)$, can be proved by induction on $n$. Combining results, we find that</p>
<p>$$T(n)=\begin{cases}
2+4\lfloor n/2\rfloor(\lfloor n/2\rfloor+1),&\text{if }n\text{ is even}\\\\
3+4\lfloor n/2\rfloor(\lfloor n/2\rfloor+2),&\text{if }n\text{ is odd}\;.
\end{cases}$$</p>
<p><strong>Added:</strong> Suppose that you have the recurrence $T(n)=aT(n-1)+bT(n-2)+cn$, with initial values $T(0)=d$ and $T(1)=e$. The characteristic polynomial is $x^2-ax-b$. If $1$ is not a zero of the characteristic equation, there is a particular solution of the form</p>
<p>$$P(n)=p_0+p_1n\;;$$</p>
<p>if $1$ is a zero of multiplicity $m$ of the characteristic polynomial, there is a particular solution of the form</p>
<p>$$P(n)=n^m(p_0+p_1n)\;.$$</p>
<p>In your specific problem the characteristic polynomial is $x^2-1=(x-1)(x+1)$, so there is a particular solution of the form</p>
<p>$$P(n)=n(p_0+p_1n)\;.$$</p>
<p>The general solution to the homogeneous recurrence has the form $A+B(-1)^n$, so for the given initial values we must have $A+B=2$ and $A-B=3$, $A=\frac52$, and $B=-\frac12$. Thus,</p>
<p>$$T(n)=\frac52-\frac12(-1)^n+n(p_0+p_1n)$$</p>
<p>for some constants $p_0$ and $p_1$. Setting $n=1$, we see that we must have $p_0+p_1=0$, and setting $n=2$ yields the equation $10=T(2)=2+2(p_0+2p_1)$, or $4=p_0+2p_1$. Thus, $p_1=4$, $p_0=-4$, and the general solution is</p>
<p>$$T(n)=\frac52-\frac12(-1)^n+4n(n-1)\;.$$</p>
<p>More generally, if instead of $cn$ you have $k^np(n)$ for some constant $k$ and polynomial $p$ of degree $d$, there is a particular solution of the form</p>
<p>$$P(n)=k^n\left(p_0+p_1n+\ldots+p_dn^d\right)$$</p>
<p>if $k$ is not a zero of the characteristic polynomial, and there is one of the form</p>
<p>$$P(n)=k^nn^m\left(p_0+p_1n+\ldots+p_dn^d\right)$$</p>
<p>if $k$ is a zero of multiplicity $m$ of the characteristic polynomial.</p>
|
332,170 | <p>How can I solve <span class="math-container">$$T(n) = aT(n-1) + bT(n-2)+ cn $$</span>; where <span class="math-container">$a,b,c$</span> are constants. I could not figure it out :(</p>
<p>There are T(0) = d and T(1) = e,</p>
<p>Thanks in advance.</p>
| André Nicolas | 6,312 | <p>We use your specific equation. Look for a <strong>particular</strong> solution of the shape $an^2+bn+c$, actually in this case $an^2+bn$ is good enough.</p>
<p>Substituting in the equation, we get $an^2+bn=a(n-2)^2+b(n-2)+4n$, Comparing coefficients, we find that $-4a+4=0$ and $4a-2b=0$. Thus $a=1$, $b=2$. So we have found the particular solution $n^2+2n$. </p>
<p>Now let $T(n)=S(n)+n^2+2n$. Then we find that $S(n)$ satisfies the <em>homogeneous</em> equation $S(n)=S(n-2)$. The method of characteristic equations shows that the general solution of the homogeneous equation is $A(1^n)+B(-1)^n$. Use our initial conditions to determine $A$ and $B$. </p>
<p>Similar ideas will work for, say, $T(n)=pT(n-1)+qT(n-2)+P(n)$, where $p$ and $q$ are constants, and $P(n)$ is a polynomial. </p>
|
28,568 | <p>Recently, I answered to this problem:</p>
<blockquote>
<p>Given <span class="math-container">$a<b\in \mathbb{R}$</span>, find explicitly a bijection <span class="math-container">$f(x)$</span> from
<span class="math-container">$]a,b[$</span> to <span class="math-container">$[a,b]$</span>.</p>
</blockquote>
<p>using an "iterative construction" (see below the rule).</p>
<p>My question is: is it possible to solve the problem finding a less <em>exotic</em> function?</p>
<p>I mean: I know such a bijection cannot be monotone, nor globally continuous; but my <span class="math-container">$f(x)$</span> has <em>a lot of</em> jumps... Hence, can one do without so many discontinuities?</p>
<hr>
<p>W.l.o.g. assume <span class="math-container">$a=-1$</span> and <span class="math-container">$b=1$</span> (the general case can be handled by translation and rescaling).
Let:</p>
<p>(<strong>1</strong>) <span class="math-container">$X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$</span>, and</p>
<p>(<strong>2</strong>) <span class="math-container">$f_0(x):=\begin{cases}
-x-\frac{3}{2} &\text{, if } -1<x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &\text{,
if } \frac{1}{2}\leq
x<1\\ 0 &\text{, otherwise} \end{cases}$</span>,</p>
<p>so that the graph of <span class="math-container">$f_0(x)$</span> is made of two segments (parallel to the line <span class="math-container">$y=x$</span>) and one segment laying on the <span class="math-container">$x$</span> axis; then define by induction:</p>
<p>(<strong>3</strong>) <span class="math-container">$X_{n+1}:=\frac{1}{2} X_n$</span>, and</p>
<p>(<strong>4</strong>) <span class="math-container">$f_{n+1}(x):= \frac{1}{2} f_n(2 x)$</span></p>
<p>for <span class="math-container">$n\in \mathbb{N}$</span> (hence <span class="math-container">$X_n=\frac{1}{2^n} X_0$</span> and <span class="math-container">$f_n=\frac{1}{2^n} f_0(2^n x)$</span>).</p>
<p>Then the function <span class="math-container">$f:]-1,1[\to \mathbb{R}$</span>:</p>
<p>(<strong>5</strong>) <span class="math-container">$f(x):=\sum_{n=0}^{+\infty} f_n(x)$</span></p>
<p>is a bijection from <span class="math-container">$]-1,1[$</span> to <span class="math-container">$[-1,1]$</span>.</p>
<p><em>Proof</em>: <strong>i</strong>. First of all, note that <span class="math-container">$\{ X_n\}_{n\in \mathbb{N}}$</span> is a pairwise disjoint covering of <span class="math-container">$]-1,1[\setminus \{ 0\}$</span>. Moreover the range of each <span class="math-container">$f_n(x)$</span> is <span class="math-container">$f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$</span>.</p>
<p><strong>ii</strong>. Let <span class="math-container">$x\in ]-1,1[$</span>. If <span class="math-container">$x=0$</span>, then <span class="math-container">$f(x)=0$</span> by (<strong>5</strong>). If <span class="math-container">$x\neq 0$</span>, then there exists only one <span class="math-container">$\nu\in \mathbb{N}$</span> s.t. <span class="math-container">$x\in X_\nu$</span>, hence <span class="math-container">$f(x)=f_\nu (x)$</span>. Therefore <span class="math-container">$f(x)$</span> is <em>well defined</em>.</p>
<p><strong>iii</strong>. By <strong>i</strong> and <strong>ii</strong>, <span class="math-container">$f(x)\lesseqgtr 0$</span> for <span class="math-container">$x\lesseqgtr 0$</span> and the range of <span class="math-container">$f(x)$</span> is:</p>
<p><span class="math-container">$f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$</span>,</p>
<p>therefore <span class="math-container">$f(x)$</span> is surjective.</p>
<p><strong>iv</strong>. On the other hand, if <span class="math-container">$x\neq y \in ]-1,1[$</span>, then: if there exists <span class="math-container">$\nu \in \mathbb{N}$</span> s.t. <span class="math-container">$x,y\in X_\nu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$</span> (for <span class="math-container">$f_\nu (x)$</span> restrited to <span class="math-container">$X_\nu$</span> is injective); if <span class="math-container">$x\in X_\nu$</span> and <span class="math-container">$y\in X_\mu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$</span> (for the restriction of <span class="math-container">$f_\nu (x)$</span> to <span class="math-container">$X_\nu$</span> and of <span class="math-container">$f_\mu(x)$</span> to <span class="math-container">$X_\mu$</span> have disjoint ranges); finally if <span class="math-container">$x=0\neq y$</span>, then <span class="math-container">$f(x)=0\neq f(y)$</span> (because of <strong>ii</strong>).
Therefore <span class="math-container">$f(x)$</span> is injective, hence a bijection between <span class="math-container">$]-1,1[$</span> and <span class="math-container">$[-1,1]$</span>. <span class="math-container">$\square$</span></p>
| mjqxxxx | 5,546 | <p>Define a bijection <span class="math-container">$f:(-1,1)\rightarrow[-1,1]$</span> as follows: <span class="math-container">$f(x)=2x$</span> if <span class="math-container">$|x|=2^{-k}$</span> for some <span class="math-container">$k\in\mathbb{N}$</span>; otherwise <span class="math-container">$f(x)=x$</span>.</p>
|
28,568 | <p>Recently, I answered to this problem:</p>
<blockquote>
<p>Given <span class="math-container">$a<b\in \mathbb{R}$</span>, find explicitly a bijection <span class="math-container">$f(x)$</span> from
<span class="math-container">$]a,b[$</span> to <span class="math-container">$[a,b]$</span>.</p>
</blockquote>
<p>using an "iterative construction" (see below the rule).</p>
<p>My question is: is it possible to solve the problem finding a less <em>exotic</em> function?</p>
<p>I mean: I know such a bijection cannot be monotone, nor globally continuous; but my <span class="math-container">$f(x)$</span> has <em>a lot of</em> jumps... Hence, can one do without so many discontinuities?</p>
<hr>
<p>W.l.o.g. assume <span class="math-container">$a=-1$</span> and <span class="math-container">$b=1$</span> (the general case can be handled by translation and rescaling).
Let:</p>
<p>(<strong>1</strong>) <span class="math-container">$X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$</span>, and</p>
<p>(<strong>2</strong>) <span class="math-container">$f_0(x):=\begin{cases}
-x-\frac{3}{2} &\text{, if } -1<x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &\text{,
if } \frac{1}{2}\leq
x<1\\ 0 &\text{, otherwise} \end{cases}$</span>,</p>
<p>so that the graph of <span class="math-container">$f_0(x)$</span> is made of two segments (parallel to the line <span class="math-container">$y=x$</span>) and one segment laying on the <span class="math-container">$x$</span> axis; then define by induction:</p>
<p>(<strong>3</strong>) <span class="math-container">$X_{n+1}:=\frac{1}{2} X_n$</span>, and</p>
<p>(<strong>4</strong>) <span class="math-container">$f_{n+1}(x):= \frac{1}{2} f_n(2 x)$</span></p>
<p>for <span class="math-container">$n\in \mathbb{N}$</span> (hence <span class="math-container">$X_n=\frac{1}{2^n} X_0$</span> and <span class="math-container">$f_n=\frac{1}{2^n} f_0(2^n x)$</span>).</p>
<p>Then the function <span class="math-container">$f:]-1,1[\to \mathbb{R}$</span>:</p>
<p>(<strong>5</strong>) <span class="math-container">$f(x):=\sum_{n=0}^{+\infty} f_n(x)$</span></p>
<p>is a bijection from <span class="math-container">$]-1,1[$</span> to <span class="math-container">$[-1,1]$</span>.</p>
<p><em>Proof</em>: <strong>i</strong>. First of all, note that <span class="math-container">$\{ X_n\}_{n\in \mathbb{N}}$</span> is a pairwise disjoint covering of <span class="math-container">$]-1,1[\setminus \{ 0\}$</span>. Moreover the range of each <span class="math-container">$f_n(x)$</span> is <span class="math-container">$f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$</span>.</p>
<p><strong>ii</strong>. Let <span class="math-container">$x\in ]-1,1[$</span>. If <span class="math-container">$x=0$</span>, then <span class="math-container">$f(x)=0$</span> by (<strong>5</strong>). If <span class="math-container">$x\neq 0$</span>, then there exists only one <span class="math-container">$\nu\in \mathbb{N}$</span> s.t. <span class="math-container">$x\in X_\nu$</span>, hence <span class="math-container">$f(x)=f_\nu (x)$</span>. Therefore <span class="math-container">$f(x)$</span> is <em>well defined</em>.</p>
<p><strong>iii</strong>. By <strong>i</strong> and <strong>ii</strong>, <span class="math-container">$f(x)\lesseqgtr 0$</span> for <span class="math-container">$x\lesseqgtr 0$</span> and the range of <span class="math-container">$f(x)$</span> is:</p>
<p><span class="math-container">$f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$</span>,</p>
<p>therefore <span class="math-container">$f(x)$</span> is surjective.</p>
<p><strong>iv</strong>. On the other hand, if <span class="math-container">$x\neq y \in ]-1,1[$</span>, then: if there exists <span class="math-container">$\nu \in \mathbb{N}$</span> s.t. <span class="math-container">$x,y\in X_\nu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$</span> (for <span class="math-container">$f_\nu (x)$</span> restrited to <span class="math-container">$X_\nu$</span> is injective); if <span class="math-container">$x\in X_\nu$</span> and <span class="math-container">$y\in X_\mu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$</span> (for the restriction of <span class="math-container">$f_\nu (x)$</span> to <span class="math-container">$X_\nu$</span> and of <span class="math-container">$f_\mu(x)$</span> to <span class="math-container">$X_\mu$</span> have disjoint ranges); finally if <span class="math-container">$x=0\neq y$</span>, then <span class="math-container">$f(x)=0\neq f(y)$</span> (because of <strong>ii</strong>).
Therefore <span class="math-container">$f(x)$</span> is injective, hence a bijection between <span class="math-container">$]-1,1[$</span> and <span class="math-container">$[-1,1]$</span>. <span class="math-container">$\square$</span></p>
| CopyPasteIt | 432,081 | <p>For background to this 'turn-the-crank' technique see this <a href="https://math.stackexchange.com/a/3156178/432081">answer</a>.</p>
<p>Let <span class="math-container">$A = (0,1)$</span> and <span class="math-container">$B = [0,1]$</span>. Let <span class="math-container">$f: A \to B$</span> be the inclusion mapping <span class="math-container">$f(x) = x$</span> and <span class="math-container">$g: B \to A$</span> be given by</p>
<p><span class="math-container">$$ g(x) = \frac{x}{2} + \frac{1}{4}$$</span></p>
<p>We can build our bijective mapping using these two injective functions (thanks to Julius König of Schröder–Bernstein-Cantor theorem fame).</p>
<p>The number <span class="math-container">$0 \in B = [0,1]$</span> is a <em>B-stopper</em>; let</p>
<p><span class="math-container">$$ D_0 = \{\, \frac{1}{2} - \frac{1}{2^n} \, | \, n \text{ is a positive integer}\,\}$$</span></p>
<p>The number <span class="math-container">$1 \in B = [0,1]$</span> is a <em>B-stopper</em>; let</p>
<p><span class="math-container">$$ D_1 = \{\, \frac{1}{2} + \frac{1}{2^n} \, | \, n \text{ is a positive integer}\,\}$$</span></p>
<p>Define <span class="math-container">$h: A \to B$</span> as follows:</p>
<p><span class="math-container">$$h(x) = \left\{
\begin{array}{1 1}
\frac{1}{2} - \frac{1}{2^{n-1}} & \mbox{if } x = \frac{1}{2} - \frac{1}{2^{n}} \, \land \, n \gt 1\\
\frac{1}{2} + \frac{1}{2^{n-1}} & \mbox{if } x = \frac{1}{2} + \frac{1}{2^{n}} \, \land \, n \gt 1\\
x & \mbox{otherwise}
\end{array}
\right.$$</span></p>
<p>The function <span class="math-container">$h: A \to B$</span> is a bijection with <span class="math-container">$h(\frac{1}{2}) = \frac{1}{2}$</span>.</p>
|
4,132,402 | <p>Can this be solved without trigonometry?</p>
<blockquote>
<p><span class="math-container">$AB$</span> is the base of an isosceles <span class="math-container">$\triangle ABC$</span>. Vertex angle <span class="math-container">$C$</span> is <span class="math-container">$50^\circ$</span>. Find the angle between the altitude and the median drawn from vertex <span class="math-container">$A$</span> to the opposite side.</p>
<p><a href="https://i.stack.imgur.com/YwI7Dm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YwI7Dm.png" alt="enter image description here" /></a></p>
</blockquote>
<p>I think I know how to do this using the Law of Cosines.</p>
<p>Call the side of the isosceles triangle <span class="math-container">$x$</span>, <span class="math-container">$CA=CB=x$</span>, and let <span class="math-container">$E$</span> be the midpoint of <span class="math-container">$BC$</span>. Then in the triangle formed by the median, <span class="math-container">$\triangle CAE$</span>, using the Law of Cosines:</p>
<p><span class="math-container">$$AE = \sqrt{x^2 + \frac{x^2}4-2\frac{x^2}{2}\cos50^\circ}$$</span></p>
<p>From this you can find <span class="math-container">$AE$</span> in terms of <span class="math-container">$x$</span>.</p>
<p>Then apply the Law of Cosines again to find <span class="math-container">$\angle CAE$</span> using
<span class="math-container">$$CE=\frac{x}{2}= \sqrt{x^2 + AE^2-2xAE\cos\angle CAE}$$</span>
and from <span class="math-container">$\cos\angle CAE$</span> you find <span class="math-container">$\angle CAE$</span>, and the angle we want is <span class="math-container">$40^\circ-\angle CAE$</span>.</p>
<p>But is it possible w/o trig?</p>
| quasi | 400,434 | <p>Using trigonometry, the angle in question is approximately equal, in degrees, to
<span class="math-container">$$
10.558536057412143196227467316938626443256567512439
$$</span>
which, given the lack of an apparent repeating block, is almost certainly not a rational number.</p>
<p>
Hence you should not expect a solution vis synthetic geometry.
|
947,618 | <p>For T: V2->V2</p>
<p>T maps each point with polar coordinate (r.theta) to each point with polar coordinate (r,2theta) and T maps 0 onto itself.</p>
<p>Hi,</p>
<p>I was trying to do this by letting r= square root of x^2 + y^2 and theta=arctan(y/x) </p>
<p>but I failed.</p>
<p>can anybody please explain it? </p>
| Mohamed | 33,307 | <p>$T$ is not linear since $$T(1+i)=T\left({\sqrt 2} e^{i\frac{\pi}4} \right)={\sqrt 2} e^{i\frac{\pi}2} =i {\sqrt 2}$$
and: $$T(1)+T(i)=1+(-1)=0$$</p>
|
3,601,552 | <p>A school has <span class="math-container">$500$</span> girls and <span class="math-container">$500$</span> boys. A simple random sample is obtained by selecting names from a box (with replacement) to a get a sample of <span class="math-container">$10$</span>. </p>
<p>Find the probability of someone being picked more than once.</p>
<p>My working is:
<span class="math-container">\begin{align}
P(\text{being picked more than once}) &= 1 - P(\text{not being picked}) - P(\text{being picked exactly once})\\
&=1 - \left(\frac{999}{1000}\right)^{10} - 10\left(\frac{1}{1000}\right)\left(\frac{999}{1000}\right)^9 = 0.00004476
\end{align}</span>
However, this is not correct. Any ideas?</p>
| Rezha Adrian Tanuharja | 751,970 | <p>Probability of anybody being picked more than once:</p>
<p><span class="math-container">$$
\frac{\binom{10^{3}}{10}\times 10!}{\left(10^{3}\right)^{10}}
$$</span></p>
<p>Probability of a particular person being picked more than once:</p>
<p><span class="math-container">$$
1-\frac{\left(10^{3}-1\right)^{10}+10\times\left(10^{3}-1\right)^{9}}{\left(10^{3}-1\right)^{10}}
$$</span></p>
|
666,461 | <p>The function $f(x)=x+\log x$ has only one root on $(0,\infty)$ which is in $(0,1)$.</p>
<p>Using the Intermediate value theorem: $f$ is continuous on $(0,\infty)$ and $f(0)=0+\log(0)=-\infty<0$ and $f(1)=1+\log(1)=1>0$. So there exists an $x$ such $f(x)=0$.</p>
<p>But how to show that this $x$ is the only root? </p>
| zhw. | 228,045 | <p>Note $f(1/e) = 1/e-1<0,$ $f(1) = 1>0.$ By the IVT, $f$ has a root in $(1/e,1).$ Also note $f'(x) = 1 + 1/x > 0.$ Thus $f$ is strictly increasing, and therefore the root just found is the unique root.</p>
|
4,373,055 | <p>We define a bound vector to be a quantity with a defined starting point, magnitude and direction. A free vector has no defined starting point, just magnitude and direction.</p>
<p>So what is a position vector (of a point)? It is defined relative to something else (Origin), so it has a starting point, size and direction like a bound vector. So it makes sense to call position vector a bound vector. However if the point is moving around (say some physical object), then position vector direction and magnitude is changing - so is it still bound vector?</p>
<p>And if it is neither a free vector nor bound, does this mean position vectors is a "3rd type" of a vector?</p>
| Karl | 279,914 | <p>The "free"/"bound" terminology doesn't have a precise mathematical meaning, it's just an informal description of how we're thinking about a vector quantity or what we're using it to represent. The vector itself never contains information about a starting point; it's always just a direction with a magnitude.</p>
|
323,783 | <p>How do I evaluate this definite integral?
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$
I know this is a trig. function. </p>
| Community | -1 | <p>We have
$$\sin^4x\cos^4x=\frac{1}{16}\sin^4(2x)=\frac{1}{16}(\frac{e^{2ix}-e^{-2ix}}{2i})^4=\frac{1}{16^2}(2\cos(8x)-8\cos(4x)+6), $$
so we integrate and we find
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}=\frac{1}{512}π-\frac{7}{2048}\sqrt{3}.$$</p>
|
945,651 | <p>Use mathematical induction to prove the following statement:</p>
<p>For all $b\in\mathbb R$, and for all $n\in\mathbb N$, $$b>-1\implies (1+b)^n \geq 1+nb$$</p>
<p>When $n=1$, the inequality still holds
$1+b \geq 1+b$.</p>
<p>For n+1$:
$$(1+b)^{n+1} \geq 1+(n+1)b$$
Here I'm not sure the best way to simplify...
$$(1+b)^n(1+b)\geq 1+bn+b$$</p>
| Kim Jong Un | 136,641 | <p>For the induction step, if $(1+b)^n\geq 1+bn$, then
$$
(1+b)^{n+1}-(1+b(n+1))\geq (1+b)(1+bn)-(1+b(n+1))=b^2n\geq 0.
$$
Note that the first inequality above uses both the induction hypothesis and $b>-1$.</p>
|
945,651 | <p>Use mathematical induction to prove the following statement:</p>
<p>For all $b\in\mathbb R$, and for all $n\in\mathbb N$, $$b>-1\implies (1+b)^n \geq 1+nb$$</p>
<p>When $n=1$, the inequality still holds
$1+b \geq 1+b$.</p>
<p>For n+1$:
$$(1+b)^{n+1} \geq 1+(n+1)b$$
Here I'm not sure the best way to simplify...
$$(1+b)^n(1+b)\geq 1+bn+b$$</p>
| Community | -1 | <p>$$(1+b)^{n+1} =(1+b)^n (1+b) \geq (1+nb)(1+b) =1+(n+1)b +nb^2 \geq 1+ (n+1)b.$$</p>
|
288,051 | <p>In enumerative combinatorics, a <i>bijective proof</i> that $|A_n| = |B_n|$ (where $A_n$ and $B_n$ are finite sets of combinatorial objects of size $n$) is a proof that constructs an explicit bijection between $A_n$ and $B_n$.
Bijective proofs are often prized because of their beauty and because of the insight that they often provide. Even if a combinatorial identity has already been proved (e.g., using generating functions), there is often interest in finding a bijective proof.</p>
<p>In spite of the importance of bijective proofs, the process of discovering or constructing a bijective proof seems to be an area that has been relatively untouched by computers. Of course, computers are often enlisted to generate all small examples of $A_n$ and $B_n$, but then the process of searching for a bijection between $A_n$ and $B_n$ is usually done the "old-fashioned" way, by playing around with pencil and paper and using human insight.</p>
<p>It seems to me that the time may be ripe for computers to search directly for bijections. To clarify, I do not (yet) envisage computers autonomously producing full-fledged bijective proofs. What I want computers to do is to search empirically for a <i>combinatorial rule</i>—that says something like, take an element of $A_n$ and do $X$, $Y$, and $Z$ to produce an element of $B_n$—that appears to yield a bijection for small values of $n$.</p>
<p>One reason that such a project has not already been carried out may be that the sheer diversity of combinatorial objects and combinatorial rules may seem daunting. How do we even describe the search space to the computer?</p>
<p>It occurs to me that, now that proof assistants have "come of age," people may have already had to face, and solve (at least partially), the problem of systematically encoding combinatorial objects and rules. This brings me to my question:</p>
<blockquote>
<p>Does there exist a robust framework for encoding combinatorial objects and combinatorial rules in a way that would allow a computer to empirically search for bijections? If not, is there something at least close, that could be adapted to this end with a modest amount of effort?</p>
</blockquote>
<p>In my opinion, Catalan numbers furnish a good test case. There are many different types of combinatorial objects that are enumerated by the Catalan numbers. As a first "challenge problem," a computer program should be able to discover bijections between different kinds of "Catalan objects" on its own. If this can be done, then there is no shortage of more difficult problems to sink one's teeth into.</p>
| Per Alexandersson | 1,056 | <p>It is perhaps hard still to automate bijection finding, but if you have a database of statistics on A and B, you can automatically check if there is (empirically) a bijection which sends some statistic on A, to some other statistic on B. That is, you refine the bijection.</p>
<p>I have used this approach successfully in a number of projects, in one case it even required the bijection to be canonical (i.e. there was a unique one that preserved the statistics).</p>
|
4,138,292 | <p>Can anybody help me understand why these terms are always reciprocals? (theta <= 45°)</p>
<p><span class="math-container">$$ x = \frac{1}{\cos \theta} + \tan{\theta} $$</span>
<span class="math-container">$$ \frac{1}{x} = \frac{1}{\cos \theta} - \tan{\theta} $$</span></p>
<p>I understand that if we multiply them, they equal <span class="math-container">$1$</span> (because of the equation for a circle).</p>
<p><span class="math-container">$$\begin{align}
1 &= (\frac{1}{\cos \theta} + \tan{θ})(\frac{1}{\cos \theta} - \tan{θ}) \\[4pt]
1 &= \frac{1}{(\cos{\theta})^{2}} - \frac{\tan{\theta}}{\cos{\theta}} + \frac{\tan{\theta}}{\cos{\theta}} - (\tan{\theta})^2 \\[4pt]
1 &= \frac{1}{(\cos{\theta})^2} - (\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\cos{\theta})^2(\tan{\theta})^2 \\[4pt]
(\cos{\theta})^2 &= 1 - (\sin{\theta})^2
\end{align}$$</span></p>
<p>But I am looking for a deeper understanding? Regards</p>
| David K | 139,123 | <p>There is a theorem (or set of theorems) of geometry called the
<a href="https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem" rel="noreferrer">Power of a Point</a>.
Note that this theorem is easily proved <a href="https://www.cut-the-knot.org/pythagoras/PPower.shtml" rel="noreferrer">without using any trigonometry</a>.</p>
<p><a href="https://i.stack.imgur.com/nXqDQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/nXqDQ.png" alt="enter image description here" /></a></p>
<p>A particular case of the power of a point says that if you have a line through the point <span class="math-container">$A$</span> that is tangent to a circle at <span class="math-container">$B,$</span>
and another line through <span class="math-container">$A$</span> that intersects the same circle at <span class="math-container">$C$</span> and <span class="math-container">$D,$</span>
as shown in the figure above, then</p>
<p><span class="math-container">$$ AB^2 = AC \cdot AD. $$</span></p>
<p>Now lets add some more specific properties to the figure. First, suppose we take the distance <span class="math-container">$AB$</span> as our unit of length, so <span class="math-container">$AB = 1.$</span>
Next, suppose <span class="math-container">$\angle BAC = \theta.$</span>
Finally, suppose the line <span class="math-container">$AD$</span> passes through the center of the circle, <span class="math-container">$O$</span>.</p>
<p><a href="https://i.stack.imgur.com/3q8St.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3q8St.png" alt="enter image description here" /></a></p>
<p>Now <span class="math-container">$BO = \tan \theta$</span> and <span class="math-container">$AO = \frac{1}{\cos\theta} = \sec \theta$</span>.
Observe that <span class="math-container">$AC = \sec\theta - \tan\theta$</span> and that <span class="math-container">$AD = \sec\theta + \tan\theta$</span>.
Recalling the formula for the power of a point in a case like this,
<span class="math-container">$AB^2 = AC \cdot AD,$</span>
and putting the particular lengths of the segments in this example into that formula,
we find that</p>
<p><span class="math-container">$$ 1^2 = (\sec\theta - \tan\theta)(\sec\theta + \tan\theta), $$</span></p>
<p>and therefore</p>
<p><span class="math-container">$$ \sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta}. $$</span></p>
<p>So your trigonometric identity is simply a special case of the power of a point.
Note that it is not restricted only to <span class="math-container">$0 \leq \theta \leq 45^\circ.$</span>
The geometric theorem shows that the identity is true for any acute angle.</p>
<p>If you use the usual definition of trigonometric functions for angles outside the range from zero to a right angle, the identity is good for all angles for which the cosine is not zero. But that takes a bit more interpretation if you want to make something geometric out of it. (In particular, you have to deal with the fact that the cosine and tangent are sometimes negative.)</p>
|
256,612 | <p>I've found assertions that recognising the unknot is NP (but not explicitly NP hard or NP complete). I've found hints that people are looking for untangling algorithms that run in polynomial time (which implies they may exist). I've found suggestions that recognition and untangling require exponential time. (Untangling is a form of recognising.) </p>
<p>I suppose I'm asking whether there exists
(1) a "diagram" of a knot,
(2) a "cost" measure of the diagram,
(3) a "move" which can be applied to the diagram,
(4) the "move" always reduces the "cost",
(5) the "move" can be selected and applied in polynomial time,
(6) the "cost" can be calculated in polynomial time.</p>
<p>For instance, Reidemeister moves, fail on number (4) if the "cost" is number of crossings.</p>
<p>So what is the current status of the problem?</p>
<p>Thanks</p>
<p>Peter</p>
| Ian Agol | 1,345 | <p>Recently, Marc Lackenby discussed a new algorithm for unknot recognition in a <a href="https://www.newton.ac.uk/seminar/20170130113012301" rel="noreferrer">talk at the Newton Institute</a> (see time around 1:03). He conjectures (but indicates at the time that it is work-in-progress) that his algorithm runs in quasi-polynomial time ($c^{\log c}$, where $c$ is the crossing number). In any case, his talk discusses the state of the art of the subject. More recently, he is <a href="http://www.ub.edu/focm2017/content/viewAbstract.php?code=1726" rel="noreferrer">announcing this as a conjecture</a>, so presumably it is still a work in progress after 6 months. </p>
|
95,314 | <p>To evaluate this type of limits, how can I do, considering $f$ differentiable, and $ f (x_0)> 0 $</p>
<p>$$\lim_{x\to x_0} \biggl(\frac{f(x)}{f(x_0)}\biggr)^{\frac{1}{\ln x -\ln x_0 }},\quad\quad x_0>0,$$</p>
<p>$$\lim_{x\to x_0} \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0},\quad\quad n\in\mathbb{N}.$$</p>
| N. S. | 9,176 | <p>For the second limit you can also observe that if $x_0 \neq 0$ then</p>
<p>$$\lim_{x\to x_0} \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0} = \lim_{x \to x_0}x^nx_0^n \frac{ \frac{f(x)}{x^n}- \frac{f(x_0)}{x_0^n}}{x-x_0}= (x_0)^{2n} (\frac{f(x)}{x^n})'(x_0) \,.$$</p>
<p>The case $x_0=0$ is trivial.</p>
|
4,236,148 | <p><span class="math-container">$R=\{(x,y):x^2=y^2\}$</span> and I have to determine whether its an equivalence relation.</p>
<p>I found that it's reflexive but for the symmetry part I got confused as <span class="math-container">$x=y$</span> is sometimes said to be symmetric others not so I don't know what to take it as.</p>
| Ivo Terek | 118,056 | <p>You can even replace taking squares with an arbitrary function. Namely, if <span class="math-container">$X$</span> is a set, <span class="math-container">$Y$</span> is a second set, and <span class="math-container">$f\colon X\to Y$</span> is a function, then define <span class="math-container">$x\sim x'$</span> iff <span class="math-container">$f(x)=f(x')$</span>. In other words, <span class="math-container">$R = \{ (x,x') \in X\times X \mid f(x)=f(x')\}$</span>. This is an equivalence relation:</p>
<ul>
<li><span class="math-container">$x\sim x$</span> because <span class="math-container">$f(x)=f(x)$</span>.</li>
<li><span class="math-container">$x\sim x'$</span> implies that <span class="math-container">$f(x)=f(x')$</span>, so <span class="math-container">$f(x')=f(x)$</span>, thus <span class="math-container">$x'\sim x$</span>.</li>
<li><span class="math-container">$x\sim x'$</span> and <span class="math-container">$x'\sim x''$</span> imply that <span class="math-container">$f(x)=f(x')$</span> and <span class="math-container">$f(x')=f(x'')$</span>, so <span class="math-container">$f(x)=f(x'')$</span> and <span class="math-container">$x\sim x''$</span>.</li>
</ul>
<p>The corresponding partition is the partition of <span class="math-container">$X$</span> by level sets of <span class="math-container">$f$</span>. It doesn't get any more geometric than that.</p>
|
222,093 | <p>For what value of m does equation <span class="math-container">$y^2 = x^3 + m$</span> has no integral solutions?</p>
| LieX | 46,134 | <p>LHS being a perfect square must have "digital root" $1$,$4$,$7$ or $9$. Cubes have digital root $1$,$8$ or $9$. Hence RHS doesn't have same digital roots as LHS so can't be equal.</p>
|
688,742 | <p>Given $P\colon\mathbb{R} \to \mathbb{R}$ , $P$ is injective (one to one) polynomial function i need to formally prove that $P$ is onto $\mathbb{R}$</p>
<p>my strategy so far .......
polynomial function is continuous and since it one-to-one function it must be strictly monotonic and now i have no idea what to do .... </p>
<p>there is a theorem saying all continuous and monotonic functions have an inverse function and another theorem saying function have an inverse function if and only if its one-to-one and onto ... </p>
<p>but for formal proof of the on-to function property i think i need to show here that every element in the co-domain/target-set have a "source" in the domain .</p>
<p>i don't think its useful for this proof but only a polynomials of odd degrees have the one-to-one property . </p>
| Eric Astor | 16,599 | <p>Option (A) is almost right!</p>
<p>Specifically, it is a property of limits that if $c$ is a constant,
\begin{equation*}
\lim_{x\to a}{c}=c.
\end{equation*}
You can prove this using the definition of a limit (that is, using a $\delta$-$\epsilon$ proof).</p>
<p>So $\lim_{x\to\infty}{\frac{11}{7}}=\frac{11}{7}$, having no limit on the right-hand side, just like $\lim_{x\to\infty}{e^{-x}}=0$, having no limit on the right-hand side.</p>
|
228,036 | <p>I quote from the <a href="http://en.wikipedia.org/wiki/Von_Neumann_cardinal_assignment" rel="nofollow">Wikipedia article</a>:</p>
<p>"So (assuming the axiom of choice) we identify $\omega_\alpha$ with $\aleph_\alpha$, except that the notation $\aleph_\alpha$ is used for writing cardinals, and $\omega_\alpha$ for writing ordinals. "</p>
<p>Let's remind ourselves of the definitions: </p>
<p><strong>(Def.1)</strong> The $\aleph_\alpha$ numbers are defined recursively, with $\aleph_0 = |\mathbb N| = \omega = \omega_0$ and $\aleph_{\alpha + 1} =$ the least ordinal such that it has strictly greater cardinality than $\aleph_\alpha$ for $\alpha \in \mathbf{ON}$.</p>
<p><strong>(Def.2)</strong> The initial ordinal of a cardinal $\kappa$ is defined to be the least ordinal of cardinality $\kappa$. Then $\omega_\alpha$ are the <a href="http://mathworld.wolfram.com/InitialOrdinal.html" rel="nofollow">transfinite initial ordinals</a>.</p>
<p>Then by definition, every $\aleph$ is an ordinal and also by definition, $\aleph_\alpha = \omega_\alpha$, with or without AC. What am I missing? Thanks for clarification. </p>
| Cameron Buie | 28,900 | <p>Perhaps they're alluding to the fact that without the axiom of choice, not all infinite cardinals are alephs; with it, the finite ordinals and the $\aleph_\alpha$ take care of all the cardinals.</p>
|
1,946,438 | <p>I solved the equation $e^{e^z}=1$ and it seemed to easy so I suspect I must be missing something.</p>
<blockquote>
<p>Would someone please check my answer?</p>
</blockquote>
<p>My original answer:</p>
<p>$e^{e^z}=1$ if and only if $e^z = 2\pi i k$ for $k\in \mathbb Z$ if and only if $z=\ln(2\pi i k)$ for $k\in \mathbb Z$.</p>
<p><strong>Edit</strong></p>
<p>After reading the comments and answers I tried to do it again. Unfortunately, I still do not get the same result as in the answers.</p>
<p><strong>My second attempt:</strong></p>
<p>We have </p>
<p>$$ e^x = 1 \iff x = 2 \pi i k$$</p>
<p>hence </p>
<p>$$ e^z = 2 \pi i k$$</p>
<p>for some $k$ in $\mathbb Z$. </p>
<p>Letting $e^z = e^x (\cos y + i \sin y)$ we get </p>
<p>$$ e^x \cos y + i e^x \sin y = 2 \pi k i$$</p>
<p>which implies that $\cos y = 0$ which happens if and only if $y_j = {\pi \over 2} + \pi j$ where $j\in \mathbb Z$. At $y_j$ we have
$\sin y = \pm 1$ hence if $j$ is even</p>
<p>$$ e^x = 2 \pi i k$$</p>
<p>and if $j$ is odd </p>
<p>$$ e^x = -2 \pi i k$$</p>
<p>Hence if $j$ is even,</p>
<p>$$ x = {\pi \over 2} + \ln(2 \pi k)$$</p>
<p>and if $j$ is odd,</p>
<p>$$ x = {3\pi \over 2} + \ln(2 \pi k)$$</p>
<p>So we see that the solutions are</p>
<p>$$
z_{t,k}=\begin{cases}
{\pi \over 2} + \ln(2 \pi k) + i ({\pi \over 2} + 2t \pi )\\
{3\pi \over 2} + \ln(2 \pi k) + i({\pi \over 2} + (2 +1)t \pi )
\end{cases}
$$
for $k,t \in \mathbb Z$.</p>
<blockquote>
<p>What am I doing wrong?</p>
</blockquote>
| MPW | 113,214 | <p>The problem is that your expression "$\ln 2\pi ik$" is multivalued (worse, it is undefined if $k=0$).</p>
<p>One may write $$e^{e^z}=1 \iff e^z = 2\pi i k \stackrel{k\neq 0}{\iff} z =
\begin{cases}
\ln 2\pi k + \frac{\pi i}{2}(4n+1), & k>0\\
\ln -2\pi k + \frac{\pi i}{2}(4n-1), & k<0
\end{cases}$$</p>
<p>So the solutions are $$z_{n,k} = \ln2\pi |k| + \frac{\pi i}{2}\left(4n+\frac{k}{|k|}\right)$$ for integral $n$ and nonzero integral $k$, and "$\ln$" is the real-valued function of a positive real variable (note that since $k$ is nonzero, one has that $\frac{k}{|k|} = \pm 1$ according to the sign of $k$).</p>
|
69,590 | <p>Consider the following code.</p>
<pre><code>f[a_,b_]:=x
x=a+b;
f[1,2]
(* a + b *)
</code></pre>
<p>From a certain viewpoint, one might expect it to return <code>3</code> instead of <code>a + b</code>: the symbols <code>a</code> and <code>b</code> are defined during the evaluation of <code>f</code> and <code>a+b</code> should evaluate to their sum.</p>
<p>Why is this viewpoint wrong? What's the right way to make it behave the way I want it to? (Something more clever than <code>f[p_,q_]:=x/.{a->p,b->q};</code>.) </p>
| Carl Woll | 45,431 | <p>I gave the following answer to essentially the same <a href="https://mathematica.stackexchange.com/q/184624/45431">question</a>:</p>
<pre><code>TagSetDelayed[x, lhs_, rhs_] ^:= SetDelayed @@ (Hold[lhs, rhs] /. OwnValues[x])
</code></pre>
<p>For your example:</p>
<pre><code>x = a + b;
x /: f[a_, b_] := x
</code></pre>
<p>Then:</p>
<pre><code>f[1, 2]
</code></pre>
<blockquote>
<p>3</p>
</blockquote>
<p>The main difference between this answer and @Mr.Wizard's answer is the time at which <code>x</code> is evaluated. In my answer, the <a href="http://reference.wolfram.com/language/ref/OwnValues" rel="nofollow noreferrer"><code>OwnValues</code></a> of <code>x</code> at the time of function <em>definition</em> is used. In @Mr.Wizard's answer, the <a href="http://reference.wolfram.com/language/ref/OwnValues" rel="nofollow noreferrer"><code>OwnValues</code></a> of <code>x</code> at the time of function <em>application</em> is used. Let's see an example where this difference manifests:</p>
<pre><code>x = a + b;
x /: f[a_, b_] := x
</code></pre>
<p>The <a href="http://reference.wolfram.com/language/ref/DownValues" rel="nofollow noreferrer"><code>DownValues</code></a> of <code>f</code> are:</p>
<pre><code>DownValues[f]
</code></pre>
<blockquote>
<p>{HoldPattern[f[a_, b_]] :> a + b}</p>
</blockquote>
<p>Notice that no trace of <code>x</code> remains. If I change <code>x</code> to something else, and then evaluate <code>f</code>:</p>
<pre><code>x = 1;
f[1, 2]
</code></pre>
<blockquote>
<p>3</p>
</blockquote>
<p>The function <code>f</code> does not change when <code>x</code> is changed. Now, for @Mr.Wizard's answer:</p>
<pre><code>x = a + b;
blockSet[f[a_, b_] := x]
</code></pre>
<p>Notice that the <a href="http://reference.wolfram.com/language/ref/OwnValues" rel="nofollow noreferrer"><code>OwnValues</code></a> of <code>x</code> have not been used in the definition of <code>f</code>:</p>
<pre><code>DownValues[f]
</code></pre>
<blockquote>
<p>{HoldPattern[f[a<span class="math-container">$29662_, b$</span>29663_]] :> Block[{a = a<span class="math-container">$29662, b = b$</span>29663}, x]}</p>
</blockquote>
<p>If I change <code>x</code>:</p>
<pre><code>x = 1;
f[1, 2]
x = a-b;
f[1, 2]
x = 22;
f[1, 2]
</code></pre>
<blockquote>
<p>1</p>
<p>-1</p>
<p>22</p>
</blockquote>
<p>Notice how the definition of <code>f</code> is dependent on the value of <code>x</code>. This difference in behavior should be considered when choosing which approach to take.</p>
|
4,151 | <p>Since it's currently summer break, and I've a bit more time than normal, I've been organizing my old notes. I seem to have an almost unwieldy amount of old handouts and tests from classes previously taught. I'm hesitant to get rid of these, as they may provide useful for some future course. Because I adjunct at a few different colleges, with slightly different course content, I find that a lot of these materials are useful for multiple classes. I've begun organizing them according to specific content, but I'm curious, does anyone have an organizational style that they have found useful?</p>
| user1984 | 1,984 | <p>I find that the most efficient way to store things is to scan them in. You will hopefully find that the photocopier at your school can batch-scan and email the copies to you, or put them on a USB.</p>
<p>After that I give them very descriptive names according to a naming scheme, so that they easy to search for. By storing them in my dropbox or google-drive folders, they easy to access on my iPad etc. as well as my laptop.</p>
|
3,014,438 | <p>Find Number of Non negative integer solutions of <span class="math-container">$x+2y+5z=100$</span></p>
<p>My attempt: </p>
<p>we have <span class="math-container">$x+2y=100-5z$</span> </p>
<p>Considering the polynomial <span class="math-container">$$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$</span></p>
<p><span class="math-container">$\implies$</span></p>
<p><span class="math-container">$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$</span> </p>
<p>we need to collect coefficient of <span class="math-container">$100-5z$</span> in the above given by</p>
<p><span class="math-container">$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$</span></p>
<p>Total number of solutions is</p>
<p><span class="math-container">$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$</span></p>
<p><span class="math-container">$$S(z)=540.5$$</span></p>
<p>what went wrong in my analysis?</p>
| Key Flex | 568,718 | <p>An alternative way.</p>
<p>Given <span class="math-container">$x+2y+5z=100$</span> and it is clear that <span class="math-container">$0\le z\le20$</span>.</p>
<p>For any possible values of <span class="math-container">$z$</span>, <span class="math-container">$x+2y=100-5z$</span>.</p>
<p>Let us take <span class="math-container">$p=100-5z\ge0$</span>. Solving the equation <span class="math-container">$x+2y=p$</span>,
<span class="math-container">$(-p,p)$</span> is a solution. The general solution of <span class="math-container">$(x,y)$</span> is <span class="math-container">$$x=-p+2q,\ \ y=p-q,\ \ q\in\mathbb{Z}$$</span></p>
<p>If <span class="math-container">$p=2k$</span>, then <span class="math-container">$k=\dfrac p2\le q\le p=2k$</span>.</p>
<p>So, there are <span class="math-container">$k+1=\dfrac p2+1$</span> solutions for <span class="math-container">$(x,y)$</span></p>
<p>So, we have the following numbers as follows
<span class="math-container">$$p=100,95,90,85,80,75,......,15,10,5,0$$</span>and<span class="math-container">$$k+1=51,48,46,43,41,38,......,8,6,3,1$$</span></p>
<p>The total number of solutions are <span class="math-container">$$4(10+20+30+40)+5(8+6+3+1)+51=541$$</span></p>
|
4,389,441 | <p>I'm dealing with a sample problem where I want to work out the probability of a fair coin toss landing <em>heads</em> and a fair die roll landing <em>6</em>. We are then told that <strong>at least</strong> one of those events has happened.</p>
<p>Why is the probability of this not as simple as <em>P(C)P(D) = 0.083</em>? How would one work out the correct probability using the Bayes theorem?</p>
<p>Thanks</p>
| Siong Thye Goh | 306,553 | <p>Because now you have the additional information that at least one of those events has happened.</p>
<p>Hence the corresponding probability is <span class="math-container">$$\frac{P(CD)}{P(CD^c) + P(C^cD) + P(CD)}=\frac{P(CD)}{1-P(C^cD^c)}$$</span></p>
<p>That is I have excluded the possibility of <span class="math-container">$C^cD^c$</span> from happening.</p>
<p>Try to compute this quantity.</p>
|
1,512,528 | <p>As the title says, I'm looking to find all solutions to $$x^2 \equiv 4 \pmod{91}$$ and I am not exactly sure how to proceed.</p>
<p>The hint was that since 91 is not prime, the Chinese Remainder Theorem might be useful.</p>
<p>So I've started by separating into two separate congruences:
$$x^2 \equiv 4 \pmod{7}$$ $$x^2 \equiv 4 \pmod{13}$$</p>
<p>but now I'm confused about how to apply the CRT so I'm a bit stuck, and I'd appreciate any help or hints!</p>
| MrMazgari | 284,607 | <p>We begin with your system of equations: $$\begin{cases} x^2 \equiv 4 \pmod{7} \\ x^2 \equiv 4 \pmod{13} \end{cases}$$</p>
<p>Then, solving each congruence, we obtain the system: $$\begin{cases} x \equiv \pm 2 \pmod{7} \\ x \equiv \pm 2 \pmod{13} \end{cases}$$</p>
<p>We therefore have four systems of linear congruences, each with a unique solution by the Chinese Remainder Theorem:</p>
<p>$$\begin{cases} x \equiv 2 \pmod{7} \\ x \equiv 2 \pmod{13} \end{cases} \tag{1}$$</p>
<p>$$\begin{cases} x \equiv 2 \pmod{7} \\ x \equiv - 2 \pmod{13} \end{cases} \tag{2}$$</p>
<p>$$\begin{cases} x \equiv - 2 \pmod{7} \\ x \equiv 2 \pmod{13} \end{cases} \tag{3}$$</p>
<p>$$\begin{cases} x \equiv - 2 \pmod{7} \\ x \equiv -2 \pmod{13} \end{cases} \tag{4}$$</p>
<p>These four solutions will be distinct modulo $91$, it remains to find them!</p>
|
1,862,232 | <p>I'm studying basic Ring Theory. And in my textbook, the author states the definition of Euclidean domain:<br>
An integral domain $R$ is called to be a <em>Euclidean domain</em> precisely when there is a function $f: R\setminus\{0\}\rightarrow\Bbb N_0$, called degree function of $R$, such that:<br>
(i) If $a,b \in R\setminus\{0\}$ and there exists $c\in R$ such that $ac=b$ then $f(a)\le f(b)$.<br>
(ii) $a,b\in R$ with $b\neq 0$, then there exist $q,r\in R$ such that
$a=bq + r$ with $r=0$ or $r\neq 0$ and $f(r)\lt f(b)$. </p>
<p>I know the fact: all units in $R$ have smallest degree, and this question pops into my head: </p>
<blockquote>
<p>I want to prove that all units have degree $0$. </p>
</blockquote>
<p>Unfortunately I have no idea for it. Can anyone has an answer for my question or give me a readable explanation about it? I really appreciate ! </p>
| egreg | 62,967 | <p>You can't prove that units have degree $0$: if $f\colon R\setminus\{0\}\to\mathbb{N}_0$ is a degree function, also
$$
f_k\colon R\setminus\{0\}\to\mathbb{N}_0,
\qquad f_k(a)=f(a)+k
$$
is a degree function as well, for every integer $k\ge -f(1)$ (because $f(1)$ is the minimum value attained by $f$).</p>
<p>You can <em>normalize</em> your degree function, by defining
$$
\hat{f}(a)=f(a)-f(1)
$$
and now units will have degree $0$ with respect to $\hat{f}$.</p>
|
1,487,966 | <p>I have been looking at stereographic projections in books, online but they all seem...I don't know how else to put this, but very pedantic yet skipping the details of calculations.</p>
<p>Say, I have a problem here which asks;</p>
<blockquote>
<p>Let <span class="math-container">$n \geq 1$</span> and put <span class="math-container">$S^n=\{(x_0,x_1...,x_n) \in \mathbb{R}^{n+1}|{x_0}^2+...+{x_n}^2=1\}$</span> (So I understand this is a unit sphere in <span class="math-container">$n+1$</span> dimensions). Let <span class="math-container">$P=\{(1,0,...,0)\}$</span> and consider <span class="math-container">$S^n$</span>\ <span class="math-container">$P$</span> and <span class="math-container">$Y=\{(y_0...y_n)\in \mathbb{R}^{n+1}|y_0=0\}$</span> both with Euclidean metric. Thus <span class="math-container">$X$</span> is an <span class="math-container">$n$</span>-sphere with a point removed and <span class="math-container">$Y \cong \mathbb{R}^n$</span>.</p>
</blockquote>
<p>That is the set up. The problem I don't know how to approach is,</p>
<blockquote>
<p><span class="math-container">$i$</span>) For <span class="math-container">$x=(x_0,...,x_n) \in X$</span> and let <span class="math-container">$f(x)$</span> be a unique point of <span class="math-container">$Y$</span> such that <span class="math-container">$P=(1,0...0)$</span> and <span class="math-container">$x$</span>, <span class="math-container">$f(x)$</span> are collinear. So, find <span class="math-container">$ \lambda(x)$</span> such that <span class="math-container">$f(x)=\lambda(x)P+(1-\lambda(x))x$</span> fo some <span class="math-container">$\lambda(x) \in \mathbb{R}$</span>.</p>
<p><span class="math-container">$ii$</span>) For <span class="math-container">$y=(y_0,...y_n) \in Y$</span> let <span class="math-container">$g(y)$</span> be a unique point of <span class="math-container">$X$</span> such that <span class="math-container">$(1,0,...,0)$</span>, <span class="math-container">$y$</span> and <span class="math-container">$g(y)$</span> are collinear, by similar method to <span class="math-container">$i$</span>), find a formula for <span class="math-container">$g(y)$</span>.</p>
<p><span class="math-container">$iii$</span>) Prove that <span class="math-container">$f : X \rightarrow Y$</span> and <span class="math-container">$g : Y \rightarrow X$</span> are inverse to each other and deduce that <span class="math-container">$X$</span> is homeomorphic to <span class="math-container">$Y$</span></p>
</blockquote>
<p>Leaving aside part iii, I don't get how to do i.
I have seen some examples on removing the North Pole(which is NOT my case) but then the equation for <span class="math-container">$f(x)$</span> appears out of nowhere in those cases.
I don't know where and how they were obtained, no explanations and steps were given.</p>
<p>So, <span class="math-container">$\lambda(x)$</span> is a real number or, at least a scalar I understand. Given my conditions, I tried substituting the points to the given collinear form but found only that <span class="math-container">$\frac{y_0-x_0}{1-x_0} = \lambda(x)$</span> and also <span class="math-container">$\frac{x_i-y_i}{x_i} = \lambda(x)$</span> unless <span class="math-container">$i=0$</span>. Which doesn't make sense to me, as I found well, 2 different <span class="math-container">$\lambda(x)$</span>s (haven't I??)</p>
<p>I just don't get this stereographic projection thing "analytically". I have seen pictures and diagrams which visualises it and that's all very nice but algebraically/analytically, I cannot make sense of it.</p>
<p>Would anyone care helping me out at all?? Thank you so much....</p>
| David K | 139,123 | <p>The point of finding equations with $\lambda(x)$ and individual coordinates
$x_k$ and $y_k$ is that you do not already know what all the values of all
the coordinates are. You may have been given the coordinates $(x_0, \ldots, x_n)$
but not the coordinates $(y_0, \ldots, y_n)$,
and you want to find those coordinates.</p>
<p>Or in other words, given the stereographic projection as defined in the question, and the coordinates of a point $x \in X$, find the coordinates of $f(x)$.</p>
<p>So initially, you do not know $y_i$ in general, and you cannot
use $\frac{x_i - y_i}{x_i} = \lambda(x)$ to derive $\lambda(x)$.
But you <em>do</em> know from the definition of the projection that $y_0 = 0$,
so you can use $\frac{y_0 - x_0}{1 - x_0} = \lambda(x)$
to find $\lambda(x)$,
and then for $i\neq 0$ you can use $\frac{x_i - y_i}{x_i} = \lambda(x)$
with the known value of $\lambda(x)$ to derive the (previously unknown)
value of $y_i$.</p>
<p>If you <em>did</em> know the values of all the coordinates already,
that is, if you were given then coordinates $(x_0, \ldots, x_n)$ of $x$
and the coordinates $(y_0, \ldots, y_n)$ of $f(x)$,
then the $n+1$ equations of the form $\frac{y_0 - x_0}{1 - x_0} = \lambda(x)$
and $\frac{x_i - y_i}{x_i} = \lambda(x)$
must all agree on the same value of $\lambda(x)$.
If solving any of these equations for $\lambda(x)$
resulted in a value of $\lambda(x)$ different from the value that
solves any other equation, the points you were given would not be a correct
stereographic projection.</p>
|
190,948 | <p>So I have </p>
<pre><code>Emin[T_, d_] := If[T == 0, 0, 1/(-1 + E^(1/T)) - d/(-1 + E^(d/T))];
</code></pre>
<p>which I can solve wonderfully for </p>
<pre><code>Solve[Emin[T, 10] == 0.5, T, Reals]
{{T -> 0.910427}}
</code></pre>
<p>Now as you see <code>Emin[T_, d_]</code> is also dependent on d which I set to d=10 in the second line.</p>
<p>All I'd like is to now Plot this solve for different d.</p>
<p>I've got so far this:</p>
<pre><code>Tlist = Table[Solve[Emin[T, d] == 0.5, T, Reals] // Values // Flatten // N, {d, 5, 10}]
{{0.936864}, {0.919994}, {0.913886}, {0.911604}, {0.910747}, {0.910427}}
</code></pre>
<p>But as you see I cannot plot this because d is missing in those entries. </p>
<p>So can anyone either help me add the correct d value to the entries or maybe help me plot this even simpler? Thanks in advance !</p>
| bill s | 1,783 | <p>Reformatting a little:</p>
<pre><code>Emin[T_, d_] := If[T == 0, 0, 1/(-1 + E^(1/T)) - d/(-1 + E^(d/T))];
Tlist = Table[{d, First[Solve[Emin[T, d] == 0.5, T, Reals]
// Values // Flatten // N]}, {d, 5, 10}] // Quiet
ListPlot[Tlist]
</code></pre>
<p><a href="https://i.stack.imgur.com/boO5k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/boO5k.png" alt="enter image description here"></a></p>
|
8,023 | <p>I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for $n=1$, all Riemannian manifolds are modeled on $\mathbb R$. When $n=2$, I believe that it suffices for the scalar curvature to vanish everywhere (this is certainly necessary). But my intuition is poor for higher-dimensional structures.</p>
<p>Put another way: given a Riemannian structure $g$ on a smooth manifold, when can I find coordinates $x^1,\dots,x^n$ so that $g_{ij}(x) = \delta_{ij}$?</p>
| Kevin H. Lin | 83 | <p>This is just a rephrasing of Deane's answer, but let me add one general comment. To any Riemannian metric (or pseudo-Riemannian metric) $g$ on a manifold $M$, you can associate a <a href="http://en.wikipedia.org/wiki/Levi-Civita_connection" rel="nofollow">Levi-Civita connection</a> $\nabla : T_M \to T_M \otimes \Omega_M$, where $T_M$ is the cotangent sheaf and $\Omega_M$ is the cotangent sheaf. Just like how the de Rham $d : \mathcal{O}_M \to \Omega_M$ can be extended to $d : \Omega_M^i \to \Omega_M^{i+1}$, the connection $\nabla : T_M \to T_M \otimes \Omega_M$ can be extended to $\nabla : T_M \otimes \Omega_M^i \to T_m \otimes \Omega_M^{i+1}$. </p>
<p>Then being able to find local coordinates $x_i$ such that $g_{ij} = \delta_{ij}$ is equivalent to $\nabla^2 = \nabla \circ \nabla : T_M \to T_M \otimes \Omega_M^2$, which corresponds to the Riemann curvature tensor that Deane mentions, being zero. This should be reminiscent of the $d^2 = 0$ of de Rham cohomology, or homological algebra in general... ;-)</p>
<p>It is perhaps not-so-standard (or not as standard as I would like) to talk about connections in terms of sheaves, but this point of view is better because it generalizes better. What I said above works in all of the "standard" geometric categories: $C^\infty$, real analytic, complex analytic, super-, etc. (Maybe it even works in the algebraic category? But I'm not sure, as the notion of local coordinates is more complicated in algebraic geometry.)</p>
|
3,421,858 | <p><span class="math-container">$\sqrt{2}$</span> is irrational using proof by contradiction.</p>
<p>say <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span> where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. </p>
<p><span class="math-container">$b\sqrt{2}$</span> is an integer. ----[Understood]</p>
<p>Let <span class="math-container">$b$</span> denote the smallest such positive integer.----[My understanding of this is </p>
<p>that were are going to assume b is the smallest possible integer such that <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span>, ... Understood]</p>
<p>Then <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>----[I'm not sure I understand the point that's being made here,</p>
<p>from creating a variable <span class="math-container">$b^{*} = a - b$</span> ]</p>
<p>Next, <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span> is a positive integer such that <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer.----[ I get that (<span class="math-container">$a - b$</span>) has to</p>
<p>be a positive integer, why does it follow that then <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer?]</p>
<p>Lastly, <span class="math-container">$b^{*}<b$</span>, which is a contradiction.----[I can see that given <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>, we then have </p>
<p><span class="math-container">$b^{*}<b$</span>, I don't get how that creates a contradiction]</p>
<p>Any help is appreciated, thank you. </p>
| fleablood | 280,126 | <p>So you got to <span class="math-container">$b\sqrt 2$</span> is an integer.</p>
<p>Thus <span class="math-container">$b\sqrt 2- b$</span> is an integer because if you subtract two integers you get an integer.</p>
<p><span class="math-container">$b\sqrt 2-b = b(\sqrt 2 -1):=b^*$</span> is an integer.</p>
<p>Hopefully that clears the first issue.</p>
<p>Now notice that <span class="math-container">$b^*\sqrt 2 = (b\sqrt 2-b)\sqrt 2 = b\sqrt 2\sqrt 2 + b\sqrt 2 = 2b + b\sqrt 2$</span>. </p>
<p>Now <span class="math-container">$b$</span> is an integer so <span class="math-container">$2b$</span> is an integer. And <span class="math-container">$b\sqrt 2$</span> was an integer (that was our first observation). And so integers plus integers are integers.</p>
<p>So <span class="math-container">$2b + b\sqrt 2= b^*\sqrt 2$</span> is an integer.</p>
<p>Hopefully that clears the second issue.</p>
<p><span class="math-container">$b^*=b(\sqrt 2 - 1)$</span> and <span class="math-container">$0 < (\sqrt 2 - 1) < 1$</span>. (The text should have proven this. But you claim you were okay with this...[1]) <span class="math-container">$0*b < (\sqrt 2-1)*b < 1*b$</span> and So <span class="math-container">$0< b^*< b$</span>.</p>
<p>Why is this a contradiction?</p>
<p>Because you assumed <span class="math-container">$b$</span> was the <em>smallest</em> positive integer where <span class="math-container">$b\sqrt 2$</span> is an integer.</p>
<p>Consider <span class="math-container">$1,2,3,.....= \mathbb N$</span> and consider the list <span class="math-container">$1\sqrt 2, 2\sqrt 2, 3\sqrt 2,....$</span></p>
<p>Now either none of them are integers or there is a first item on the list that is an integer. If we assume <span class="math-container">$b\sqrt 2$</span> is the first on the list then <span class="math-container">$b$</span> is the smallest positive integer so that <span class="math-container">$b\sqrt 2=a$</span> is an integer.</p>
<p>But we have just discovered that <span class="math-container">$b^*\sqrt 2 = b(\sqrt 2-1)*\sqrt 2= 2b-b\sqrt 2=2b-a$</span> is a <em>smaller</em> integer. </p>
<p>That contradicts that <span class="math-container">$b$</span> is the smallest. Which means there can't be any smallest value where <span class="math-container">$b\sqrt 2$</span> is an integer.</p>
<p>That is the contradiction.</p>
<p>In my opinion, the text maybe should have <em>defined</em> <span class="math-container">$b$</span> as the smallest integer <span class="math-container">$b$</span> so that <span class="math-container">$b\sqrt 2$</span> is an integer first <em>before</em> defining <span class="math-container">$\sqrt 2 =\frac ab$</span>. There's no indication that <span class="math-container">$b$</span> had to be the smallest positive integer. But <em>something</em> has to be. And if <span class="math-container">$\beta \sqrt 2 = \alpha$</span> is such that <span class="math-container">$\beta$</span> is the <em>smallest</em> then <span class="math-container">$\sqrt 2 =\frac{\alpha}{\beta}$</span>.</p>
<p><del>Note.... the contradiction arises because we assume all rational numbers <span class="math-container">$r$</span> can be written as a fraction <span class="math-container">$\frac ab$</span> where <span class="math-container">$a$</span> is an integer and <span class="math-container">$b$</span> is a positive integer. This is a fair assumption as that is simply the <em>definition</em> of "rational number". <em>AND WE ASSUMED</em> there is always a smallest such integer <span class="math-container">$b$</span>.</del></p>
<p><del>I think it's worth questioning our assumptions. <em>Why</em> do we assume all rational numbers can be expressed in such a "lowest term"?</del></p>
<p><del>Food for thought.</del></p>
<p>=======</p>
<p>[1] Oh what the heck. If <span class="math-container">$0\le x \le 1$</span> then <span class="math-container">$0 \le x^2 \le 1$</span> and <span class="math-container">$x^2 \ne 2$</span>. And if <span class="math-container">$x\ge 2$</span> then <span class="math-container">$x^2 \ge 4$</span> and <span class="math-container">$x^2 \ne 2$</span>. So if a positive <span class="math-container">$\sqrt 2$</span> exists at all [2] then <span class="math-container">$1 < \sqrt 2 < 2$</span> so <span class="math-container">$0 < \sqrt 2 -1< 1$</span>.</p>
<p>=======</p>
<p>[2] Actually it's important to realize that this proof doesn't prove <span class="math-container">$\sqrt 2$</span> is irrational. It just proves that a rational <span class="math-container">$\sqrt 2$</span> does not exist. Proving that there are irrational numbers and that there is a <span class="math-container">$\sqrt 2$</span> is another issue altogether.</p>
<p><em>BUT</em> that's okay. In this proof <em>by contradiction</em> you are assuming that <span class="math-container">$\sqrt 2 = \frac ab \in \mathbb Q$</span> does exist.</p>
|
4,383,800 | <p>I can already see that the <span class="math-container">$\lim_\limits{n\to\infty}\frac{n^{n-1}}{n!e^n}$</span> converges by graphing it on Desmos, but I have no idea how to algebraically prove that with L’Hopital’s rule or induction. Where could I even start with something like this?</p>
<p>Edit: For context, I came across this limit while studying the series expansion for the Lambert W Function, <span class="math-container">$W(x)= \sum_{n=0}^{\infty}\frac{(-n)^{n-1}x^n}{n!}$</span> . By the ratio test, it is clear that <span class="math-container">$|x|<\frac1e$</span> in order to converge, but I needed to use the Alternating Series Test to see whether this series converges at <span class="math-container">$x= \pm\frac1e$</span>. Finding <span class="math-container">$\lim_\limits{n\to\infty}|a_n|$</span> is the first step of the test.</p>
| xpaul | 66,420 | <p>By the Stirling approximation <a href="https://en.wikipedia.org/wiki/Stirling%27s_approximation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Stirling%27s_approximation</a>,
<span class="math-container">$$ n!> \sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n e^{\frac1{12n+1}}$$</span>
one has
<span class="math-container">$$0<\frac{n^{n-1}}{n!e^n} < \frac{n^{n-1}}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac1{12n+1}}e^n} < \frac 1{n}.$$</span></p>
|
3,730,083 | <p>If <span class="math-container">$a,b>0$</span> and <span class="math-container">$Q=\{x_1, x_2, x_3,..., x_a\}$</span> a subset of the natural numbers <span class="math-container">$1, 2, 3,..., b$</span> such that, for <span class="math-container">$x_i+x_j<b+1$</span> with <span class="math-container">$1 ≤ i ≤ j ≤ a$</span>, then <span class="math-container">$x_i+x_j$</span> is also an element of Q. Prove that:<br> <br>
<span class="math-container">$ \frac{x_1+x_2+x_3+...+x_a}{a} ≥ \frac{b+1}{2}$</span> <br>so basically, you have to prove that the arithmetic mean of Q satisfying the condition is greater than equal to the arithmetic mean of the set the natural numbers <span class="math-container">$1, 2, 3,..., b$</span>.<br><br>
Any help would be appreciated. Thanks in advance!</p>
| 5xum | 112,884 | <p>Literally from the first paragraph from the <a href="https://en.wikipedia.org/wiki/Hypercube_graph" rel="nofollow noreferrer">Wikipedia article on the hypercube</a>:</p>
<blockquote>
<p><span class="math-container">$Q_n$</span> has <span class="math-container">$2^n$</span> vertices, <span class="math-container">$2^{n−1}n$</span> edges.</p>
</blockquote>
|
225,866 | <p>If I define, for example,</p>
<pre><code>f[OptionsPattern[{}]] := OptionValue[a]
</code></pre>
<p>Then the output for <code>f[a -> 1]</code> is 1.</p>
<p>However, in my code, I have a function that must be called using the syntax <code>f[some parameters][some other parameters]</code>, and I want to add options to the <strong>second</strong> set of square brackets. So I tried:</p>
<pre><code>g[][OptionsPattern[{}]] := OptionValue[a]
</code></pre>
<p>But then, the output for <code>g[][a -> 1]</code> is <code>OptionValue[a]</code> instead of 1. I'm not sure why this is not working. Shouldn't <code>OptionsPattern[{}]</code> match <strong>any</strong> set of options, no matter where they are located?</p>
<p>How can I add options that can be provided in the second set of square brackets instead of the first?</p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
vl = {6, 9, 10};
vars3d = Array[Through[{x, y, z}@vl[[#]]] &, 3]
(* {{x[6], y[6], z[6]}, {x[9], y[9], z[9]}, {x[10], y[10], z[10]}} *)
{x[#], y[#], z[#]} & /@ vl
(* {{x[6], y[6], z[6]}, {x[9], y[9], z[9]}, {x[10], y[10], z[10]}} *)
Table[{x[n], y[n], z[n]}, {n, vl}]
(* {{x[6], y[6], z[6]}, {x[9], y[9], z[9]}, {x[10], y[10], z[10]}} *)
% == %% == %%%
(* True *)
</code></pre>
|
4,549,898 | <p>I need some help with solving the following problem: Let <span class="math-container">$Q(n)$</span> be the number of partitions of <span class="math-container">$n$</span> into distinct parts. Show that <span class="math-container">$$\sum_{n=1}^\infty\frac{Q(n)}{2^n}$$</span> is convergent by estimating <span class="math-container">$Q(n)$</span> in some way. I have been trying to do this for a few days now, but without succeeding. Notice that if we only want to show that the series converges and don't care about which method we use, we can do this by considering the infinite product <span class="math-container">$\prod_{n=1}^\infty\left(1+\frac{1}{2^n}\right)$</span>.</p>
| TravorLZH | 748,964 | <p>Let <span class="math-container">$p(n)$</span> be the unrestricted partition function. Then we have <span class="math-container">$Q(n)\le p(n)$</span>. Note that for <span class="math-container">$|z|<1$</span></p>
<p><span class="math-container">$$
1+\sum_{n\ge1}p(n)z^n=\prod_{k\ge1}(1-z^k)^{-1},
$$</span></p>
<p>so we have</p>
<p><span class="math-container">$$
\sum_{n\ge1}{Q(n)\over2^n}\le1+\sum_{n\ge1}p(n)\left(\frac12\right)^n=\prod_{k\ge1}\left(1-{1\over2^k}\right)^{-1}<\infty.
$$</span></p>
|
76,600 | <p>The group of three dimensional rotations $SO(3)$ is a subgroup of the Special Euclidean Group $SE(3) = \mathbb{R}^3 \rtimes SO(3)$. The manifold of $SO(3)$ is the three dimensional real projective space $RP^3$. Does $RP^3$ cause a separation of space in the manifold of $SE(3)$? </p>
<p>(edit) Sorry about lack of clarity. My question should be worded as 'does $SO(3)$ partition any four dimensional subspace of $SE(3)$ into exactly two disjoint pieces?'</p>
<p>I am basically interested in understanding whether a generalization of the Jordan curve separation theorem works in such non Euclidean spaces. In particular, I want to know if (non) orientability of $SO(3)$ affects the generalization, especially since it is used to construct $SE(3)$ as a product space with $\mathbb{R}^3$.</p>
| Ryan Budney | 1,465 | <p>Okay, now I think I understand your question. This is the question I will answer:</p>
<ul>
<li>Question: Let $X$ be a connected $4$-dimensional subspace of $SE(3)$ that contains $SO(3)$. Is it possible for $X \setminus SO(3)$ to be connected? Disconnected? </li>
</ul>
<p>The answer to both questions is yes. So there is no Jordan separation theorem for $4$-dimensional subspaces of $SE(3)$ containing $SO(3)$. </p>
<p>Observation 1: As a space, $SE(3)$ is just the cartesian product of $SO(3)$ with $\mathbb R^3$. Explicitly, we will think of $SE(3)$ as the set $SO(3) \times \mathbb R^3$. </p>
<p>Observation 2: If $X := SO(3) \times \mathbb R$ embeds in $SE(3)$, therefore $SO(3) \times \{0\}$ disconnects it. </p>
<p>Observation 3: If $X := SO(3) \times S^1$, where $S^1 = \{ x \in \mathbb R^2 : |x|=1\}$, then the map $X \to SO(3) \times \mathbb R^3$ given by $(p,x) \longmapsto (p,x,0)$ is an embedding. In particular, $X \setminus (SO(3) \times \{1\})$ is connected. </p>
<p>So the answer to both your questions is yes. </p>
<p>I'd like to suggest looking at the proof of the generalized Jordan-Brouwer theorem in Guillemin and Pollack, or perhaps in an algebraic topology textbook like Bredon's. This will give you a very flexible set of tools that will let you know quite generally when you can expect a separation theorem, and when you can't. </p>
<p>Notice: my answer had nothing to do with the fact that $SO(3)$ has a non-trivial fundamental group, or whether or not it is orientable. The key part of the construction is that $SO(3)$ has co-dimension <em>at least</em> $2$ (And actually co-dimension $3$) in $SE(3)$. </p>
|
2,489,988 | <p>A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five consecutive positions in which the numbers are in increasing order?</p>
<p>I approached this problem in the following way, but I am wondering if there is a better way, since my approach is quite complicated.</p>
<p><strong>My Approach:</strong> There are three possibilities: a sequence have $7$ consecutive positions in which numbers increase, have $6$ consecutive positions in which numbers increase, and $5$ consecutive positions in which numbers increase.</p>
<p>There is only $1$ sequence that have $7$ consecutive positions. Namely, the sequence $(1,2,3,4,5,6,7)$.</p>
<p>There are $12$ sequences that have $6$ consecutive positions. Namely, we choose $1$ number from $(1,2,3,4,5,6,7)$, and move it to either sides. As an illustration, if we choose $3$, then we can get $(3,1,2,4,5,6,7)$ or $(1,2,4,5,6,7,3)$.</p>
<p>Now consider when there are $5$ consecutive positions in which numbers increase. We choose $2$ numbers that are <em>not</em> in the increasing subsequence. </p>
<p>If $1$ and $7$ are not chosen, we can place them in front of the subsequence, of after. For example, if we choose $(2,5)$, then we will have $(2,5,1,3,4,6,7)$,$(5,2,1,3,4,6,7)$, $(1,3,4,6,7,2,5)$ and $(1,3,4,6,7,5,2)$. This is $\binom{5}{2}\times4$.</p>
<p>Then I'm not sure how to proceed when we choose $1$ and/or $7$?</p>
| N. F. Taussig | 173,070 | <p><em>This solution does not differ in any essential way from that of N. Shales. I am posting this here so N. Shales can compare our approaches.</em></p>
<p>Since the sequence contains seven numbers, any block of five consecutive increasing numbers must start in the first, second, or third positions. Let $A_1$, $A_2$, and $A_3$ denote, respectively, the set of sequences containing five consecutive increasing numbers that begin in the first, second, and third positions.</p>
<p>By the <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">Inclusion-Exclusion Principle</a>, the number of sequences containing a block of at least five consecutive increasing numbers is
$$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$</p>
<p>$|A_1|$: There are $\binom{7}{5}$ ways to select five of the seven numbers to be in a block of five consecutive increasing numbers and one way to arrange the numbers within that block so that they are increasing. There are $2!$ ways to arrange the remaining two numbers in the remaining two positions. Hence,
$$|A_1| = \binom{7}{5}2!$$</p>
<p>By symmetry, $|A_1| = |A_2| = |A_3|$.</p>
<p>$|A_1 \cap A_2|$: If both the first five and second five numbers are increasing, then the first six numbers must form an increasing sequence. There are $\binom{7}{6}$ ways to select six of the seven numbers to be in the block of six consecutive increasing numbers and one way to arrange the numbers within the block so that they are increasing. There is one way to place the remaining number in the remaining position. Hence,
$$|A_1 \cap A_2| = \binom{7}{6}1!$$</p>
<p>By symmetry, $|A_1 \cap A_2| = |A_2 \cap A_3|$.</p>
<p>$|A_1 \cap A_3|$: If the first five and last five of the seven numbers are increasing, then all seven numbers must be increasing. There is only one way to arrange all seven numbers in an increasing sequence. Hence,
$$|A_1 \cap A_3| = \binom{7}{7}$$</p>
<p>$|A_1 \cap A_2 \cap A_3|$: There is only one way for all seven numbers to be arranged in increasing order. Hence,
$$|A_1 \cap A_2 \cap A_3| = \binom{7}{7}$$</p>
<p>Therefore,
$$|A_1 \cup A_2 \cup A_3| = 3\binom{7}{5}2! - 2\binom{7}{6}1! - \binom{7}{7} + \binom{7}{7} = 3\binom{7}{5}2! - 2\binom{7}{6}$$</p>
<p>Hence, the probability that a sequence of seven numbers contains a block of at least five consecutive increasing numbers is
$$\frac{3\dbinom{7}{5}2! - 2\dbinom{7}{6}1!}{7!}$$</p>
|
256,549 | <p>Matsumura, in his "Commutative Ring Theory" p. 14 proves that "A partially ordered set $\Gamma$ satisfies the ascending chain condition $\Leftrightarrow$ every nonempty subset of $\Gamma$ has a maximal element." </p>
<p>In proving the $\Rightarrow$ direction, he uses the following argument: Consider a nonempty subset $\Gamma'$ of $\Gamma$. Suppose that $\Gamma'$ has no maximal element. Then "by the axiom of choice" for any $\gamma \in \Gamma'$ we can select an element $\phi(\gamma')$ such that $\phi(\gamma')> \gamma'$. In that way we obtain a nonterminating ascending chain $\gamma' < \phi(\gamma')<\phi^2 (\gamma') <\cdots$, contradiction." </p>
<p>My question is: why is the axiom of choice necessary here?
Don't we already have by hypothesis (no maximal element exists) a rule $\phi : \Gamma' \rightarrow \Gamma'$ to assign to each $\gamma'$ the element $\phi(\gamma')$? </p>
| user642796 | 8,348 | <p>The Axiom of Choice is the mechanism that allows you to construct this rule/function. Even though you know that to each $\gamma \in \Gamma^\prime$ there is some $\gamma^\prime \in \Gamma^\prime$ greater than it, this only means that for each $\gamma$ the collection $$A_\gamma = \{ \gamma^\prime \in \Gamma^\prime : \gamma^\prime > \gamma \}$$ is nonempty. It does not provide you a means for constructing a rule to say for each $\gamma$ <em>this is the element of $A_\gamma$ I want!</em> The whole point of the Axiom of Choice is to be able to choose particular elements from possibly infinitely many different nonempty sets where there is no rule to uniformly select a unique element from each.</p>
<p>Note, also, that in this instance, since you are only concerned with generating an increasing sequence of elements of $\Gamma^\prime$, you could get away with using only the <a href="http://en.wikipedia.org/wiki/Axiom_of_dependent_choice">Axiom of Dependent Choice</a>), which is quite a bit weaker.</p>
|
256,549 | <p>Matsumura, in his "Commutative Ring Theory" p. 14 proves that "A partially ordered set $\Gamma$ satisfies the ascending chain condition $\Leftrightarrow$ every nonempty subset of $\Gamma$ has a maximal element." </p>
<p>In proving the $\Rightarrow$ direction, he uses the following argument: Consider a nonempty subset $\Gamma'$ of $\Gamma$. Suppose that $\Gamma'$ has no maximal element. Then "by the axiom of choice" for any $\gamma \in \Gamma'$ we can select an element $\phi(\gamma')$ such that $\phi(\gamma')> \gamma'$. In that way we obtain a nonterminating ascending chain $\gamma' < \phi(\gamma')<\phi^2 (\gamma') <\cdots$, contradiction." </p>
<p>My question is: why is the axiom of choice necessary here?
Don't we already have by hypothesis (no maximal element exists) a rule $\phi : \Gamma' \rightarrow \Gamma'$ to assign to each $\gamma'$ the element $\phi(\gamma')$? </p>
| Rudy the Reindeer | 5,798 | <p>No, "no maximal element" doesn't give you a choice function. A choice function is a function $\phi: \mathcal P(\Gamma')\setminus \{\varnothing\} \to \Gamma'$ such that $\phi(x) \in x \subset \Gamma'$ for all subsets $x$ of $\Gamma'$. </p>
<p>If $\Gamma'$ has no maximal element then for every $\gamma \in \Gamma'$ you can find $\gamma'$ such that $\gamma < \gamma'$. Then to construct your infinite ascending chain you use that together with a choice function as follows: start with an element $\gamma$ and consider the set of all elements that are bigger than it (which is non-empty since $\Gamma'$ has no maximal element). You use the choice function to pick an element from this set. To construct the infinite chain you need to make infinitely many choices which is the reason why you need a choice function. If you only made finitely many choices there would not be a need for a choice function.</p>
<p>Whenever you need to make infinitely many choices for which you cannot write down an explicit rule you need the axiom of choice. In this case, you have infinitely many sets from which you need to choose: the set of elements that are bigger than the ones you have already chosen. </p>
|
4,224,417 | <p>I'm trying to learn a bit of Number Theory. And while I understand the definition of congruence relations modulo <span class="math-container">$n$</span> and that they are an equivalence relations, I fail to see the <em>motivation</em> for it. So what is congruence relation <span class="math-container">$\bmod n$</span> <em>intuitively</em>? (The "bold lines" below are my questions that I'm seeking answer to.)</p>
<p>Definition: For <span class="math-container">$a,b \in \mathbb{Z}$</span> and <span class="math-container">$n \in \mathbb{N}$</span>, <span class="math-container">$a\equiv b \bmod n \Leftrightarrow n|(a-b)$</span></p>
<hr />
<p>Okay, so let's start with the definition, what is really the point of "<span class="math-container">$n | (a-b)$</span>?" That <span class="math-container">$a=nq + b$</span>, for some <span class="math-container">$q \in \mathbb{Z}$</span>? So <strong>what do I do with this and why is it so important?</strong></p>
<p>Secondly, if <span class="math-container">$a$</span> and <span class="math-container">$b$</span> leave the same residue or remainder upon division by <span class="math-container">$n$</span> then <span class="math-container">$a \equiv b \bmod n$</span>, again I don't see <strong>why are we so interested in remainders?</strong></p>
<p>And lastly, I keep seeing examples of clocks, days of the week and months. That's good but is that all there's to it?</p>
<p>I have a strong feeling, I'm grossly underestimating congruence relations modulo <span class="math-container">$n$</span>, perhaps that's because I don't have the intuition for it and where should I should use it. So <strong>any intuitive explanations of it and where should one use them</strong> would be really really really nice. I'm desperately trying to figure this out. Thanks in advance.</p>
| Claudio Buffara | 58,890 | <p>The short answer is that there are many problems in number theory whose solutions depend on looking at the remainders of numbers when they are divided by a specific number m. I can think of no simpler example to illustrate this than the problem of deciding when a given natural number is divisible by 3 or by 9 or by 11. In this case, one needs to look at the remainders of the powers of 10 when they are divided by 3 or by 9 or by 11. For high powers, the theory of congruences makes the job significantly easier. A more sophisticated example is given by certain investigations of Pierre de Fermat, menioned in the previous answer. Fermat was interested, for example, in deciding, for a given prime number p, which natural n makes 2^n - 1 a multiple of p. This line of inquiry led him to his famous Little Theorem: a prime p divides n^(p-1) - 1 whatever natural n one plugs in the expression.The easiest way to prove this is by using congruences, and observing that, for any n not divisible by p, the numbers n, 2n, 3n, ... (p-1)n leave p-1 distinct non-zero remainders when divided by p. These remainders can only be 1, 2, 3, ..., p-1, so that we can say that n*(2n)<em>(3n)</em>...<em>(p-1)n == 1</em>2<em>3</em>...*(p-1) mod p. Then, dividing both sides by (p-1)! (which is allowed since (p-1)! and p are coprime), we obtain the theorem.</p>
|
1,566,111 | <p>prove $(n)$ prime ideal of $\mathbb{Z}$ iff $n$ is prime or zero</p>
<hr>
<p><strong>Defintions</strong></p>
<p>Def of prime Ideal (n)
$$ ab\in (n) \implies a\in(n) \vee b\in(n) $$
Def 1] integer n is prime if $n \neq 0,\pm 1 $ and only divisors are $\pm n,\pm 1$ </p>
<p>Def 2 of n is prime] If $n\neq0,\pm1$ only divisors of n are $\pm1,\pm n$
$$ n|ab \implies n|a \vee n|b $$</p>
<hr>
<p>$\Rightarrow $] (<em>Prime Ideal $(n)$ of $\mathbb{Z} $$\Rightarrow$ $n$ prime or zero</em>)</p>
<p>Now consider the case where $(n)\neq (0)$. that is $n\neq 0$ </p>
<p>Using the def of prime Ideals $$\begin{aligned} ab \in (n) \implies a \in(n) \vee b \in (n) \end{aligned} $$</p>
<p>Well, If an element $x\in(n) \iff x=q*n \iff n|x$ </p>
<p>$$n|ab \implies n|a \vee n|b$$
So, $n$ is prime ,nonzero. </p>
<p>In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $\mathbb{Z}$</p>
<p>$\Leftarrow$] <em>(n is prime or zero $\Rightarrow $ $(n)$ is a prime ideal of $\mathbb{Z}$)</em></p>
<p>Consider the case where $p is prime$ so $n\neq 0, \pm1$</p>
<p>$$ \begin{aligned}
n|ab &\implies n|a \vee n|b \\
ab \in (n) &\implies a \in (n) \vee b\in(n) \end{aligned}$$</p>
<p>in the case $n=0$, since $\mathbb{Z}$ has no zero divisors
$$ab=0 \implies a=0 \vee b=0 $$
So (0) is a prime ideal. </p>
<hr>
<p>Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.</p>
| qwr | 122,489 | <p>Yes, from <span class="math-container">$p \mid a \iff a \in (p)$</span> we have more generally in a commutative ring with unity, <span class="math-container">$(p)$</span> is a prime ideal iff <span class="math-container">$p$</span> is prime (in the ring theory sense: <span class="math-container">$p \mid ab \implies p \mid a \vee p \mid b$</span>)</p>
|
2,130,658 | <p>How would I go about proving this mathematically? Having looked at a proof for a similar question I think it requires proof by induction. </p>
<p>It seems obvious that it would be even by thinking about the first few cases. As for $n=0$ there will be no horizontal dominoes which is even, and for $n=1$ there can only be one vertical domino so there are $0$ horizontal dominoes, which is again even. Then for $n=2$ you can have either two horizontal or two vertical dominoes which again gives $0$ or $2$ horizontal dominoes which is again an even number. And so on for n greater than $2$.</p>
<p>I would like to prove that the number of way of dividing a 2-high-by-n-wide rectangle into dominoes so that $2j$ dominoes are horizontal is ${n-j\choose j}$ and deduce that $U_n$ (where $U_n$ is the number of ways to divide a 2-high-by-n-wide rectangle into 2-wide-by-1-high dominoes) = $$\sum_j {n-j\choose j}$$ where this sum is over all the integers $j$ with $0\le j\le \frac{n}{2}$.</p>
<p>I understand that trivially for a 2-high-by-n-wide rectangle you can divide it by exactly $2j=n$ horizontal dominoes or by $n$ vertical dominoes or some combination of vertical and horizontal dominoes, but how can I use this knowledge to construct the proof?</p>
| samerivertwice | 334,732 | <p>If it's only $2$ high then any domino laid horizontally, and offset from the one above or below, leaves a single space above and a single below, which cannot be filled without leaving another single space.</p>
<p>Therefore any horizontal domino must be one of a pair laid one directly above the other.</p>
|
2,130,658 | <p>How would I go about proving this mathematically? Having looked at a proof for a similar question I think it requires proof by induction. </p>
<p>It seems obvious that it would be even by thinking about the first few cases. As for $n=0$ there will be no horizontal dominoes which is even, and for $n=1$ there can only be one vertical domino so there are $0$ horizontal dominoes, which is again even. Then for $n=2$ you can have either two horizontal or two vertical dominoes which again gives $0$ or $2$ horizontal dominoes which is again an even number. And so on for n greater than $2$.</p>
<p>I would like to prove that the number of way of dividing a 2-high-by-n-wide rectangle into dominoes so that $2j$ dominoes are horizontal is ${n-j\choose j}$ and deduce that $U_n$ (where $U_n$ is the number of ways to divide a 2-high-by-n-wide rectangle into 2-wide-by-1-high dominoes) = $$\sum_j {n-j\choose j}$$ where this sum is over all the integers $j$ with $0\le j\le \frac{n}{2}$.</p>
<p>I understand that trivially for a 2-high-by-n-wide rectangle you can divide it by exactly $2j=n$ horizontal dominoes or by $n$ vertical dominoes or some combination of vertical and horizontal dominoes, but how can I use this knowledge to construct the proof?</p>
| Batominovski | 72,152 | <p><strong>Coloring Argument</strong></p>
<p>Let's say that the $n$-by-$2$ board is colored black on one $n$-by-$1$ row and white on the other row. Every vertical domino takes a square of each color, whilst a horizontal domino can only take two squares of the same color. Since the number of black squares is the same as the number of white squares, we conclude that the number of black horizontal dominoes is the same as the number of white horizontal dominoes, rendering the total number of horizontal dominoes even.</p>
<hr>
<p><strong>Algebraic Argument</strong></p>
<p>Suppose that, for some $A(x,y),B(x,y)\in\mathbb{Z}[x,y]$,
$$\left(1+x+x^2+\ldots +x^{n-1}\right)\,(1+y)=A(x,y)\cdot (1+x)+B(x,y)\cdot (1+y)\,.$$
(In other words, if $x^{i-1}y^{j-1}$ is a term in $A(x,y)$, then there is a horizontal domino with the left square on the $i$-th row and $j$-th column. Likewise, if $x^{i-1}y^{j-1}$ is a term in $B(x,y)$, then there is a vertical domino with the top square on the $i$-th row and $j$-th column.)
Plugging in $x=1$ and $y=-1$ to get
$$A(1,-1)=0\,.$$
However, $A(1,1)$ is the number of horizontal dominoes and
$$A(1,1)\equiv A(1,-1)=0\pmod{2}\,.$$
Thus, for any such arrangement, there are always an even number of horizontal dominoes. </p>
|
2,628,220 | <p>Let $(a_{n})_{n \in \mathbb N_{0}}$ be a sequence in $\mathbb Z$, defined as follows:
$a_{0}:=0,
a_{1}:=2,
a_{n+1}:= 4(a_{n}-a_{n-1}) \forall n \in \mathbb N$. </p>
<p>Required to prove: $a_{n}=n2^{n} \forall n \in \mathbb N_{0}$</p>
<p>I have gone about it in the following: </p>
<p>Induction start: $n=0$ (condition fulfilled)</p>
<p>Induction premise: $a_{n}=n2^{n}$ for a specific $n \in \mathbb N_{0}$</p>
<p>Induction step: </p>
<p>$a_{n+1}=4(a_{n}-a_{n-1})$, and here the first problem arises, since I can say that (given the premise)
$4(a_{n}-a_{n-1})=4(n2^{n}-a_{n-1})$, yet how do I get rid of the $a_{n-1}$? Surely stating the $a_{n-1}=(n-1)2^{n-1}$ is false, given that my premise is only based on an $n$ and not $n-1$. </p>
| Patrick Stevens | 259,262 | <p>Note that this can be done in an alternative way, using generating functions: $$\sum_{n=0}^{\infty} a_n x^n = 0 + 2x + 4x \sum_{n=1}^{\infty} (a_n - a_{n-1}) x^n = 2x + 4x f(x) - 4x^2 \sum_{n=0}^{\infty} a_{n} x^{n}$$</p>
<p>so $$f(x) = 2x+4x \left[f(x) - x f(x)\right]$$</p>
<p>Solving, $$f(x) = \frac{2x}{(2x-1)^2}$$</p>
<p>And it's easy to find the $x^n$ coefficient in this expansion.</p>
|
1,707,132 | <blockquote>
<p>Let $X$ be a contractible space (i.e., the identity map is homotopic to the constant map). Show that $X$ is simply connected.</p>
</blockquote>
<p>Let $F$ be the homotopy between $\mathrm{id}_X$ and $x_0$, that is $F:X\times [0,1]\to X$ is a continuous map such that
$$ F(x,0)=x,\quad F(x,1)=x_0$$
for all $x\in X$.
Next let $f:[0,1]\to X$ be a loop based $x_0$, I need to show that $f$ is path homotopic to the constant path $e_{x_0}$, The continuous map $G:[0,1]\times [0,1]\to X$ defined by $G(s,t)=F(f(s),t))$ is a homotopy between $f$ and $e_{x_0}$, but it may not be a path homotopy (i.e., $G(0,t)=G(1,t)=x_0$ for all $t\in [0,1]$), how to get around this?</p>
| Xiang Yu | 187,406 | <p>We can multiply $G$ by a homopoty $H$ to get around the obstacle. More precisely, let $H:[0,1]\times [0,1]\to X$ be the homopoty defined by
$H(s,t)=G(0,st)$. We note that $H(1,t)=G(0,t)=G(1,t)=H(1,t) $ for all $t\in[0,1]$, thus for every $t\in[0,1]$, the product $H_t*G_t*\overline{H_t}$ can be defined, where $H_t:[0,1]\to X, s\mapsto H(s,t)$ is the path of $H$ at time $t$. $\overline{H_t}$ is the reverse path of $H_t$ (i.e., $\overline{H_t}(s)=H_t(1-s)$), the product of paths induces a product of homotopies. We can thus define a homotopy $K:[0,1]\times [0,1]\to X$ by
$$K(s,t)=\begin{cases} H_t(4s)= H(4s,t)=G(0,4st),& s\in[0,1/4]\\
G_t(4s-1)=G(4s-1,t),&s\in[1/4,1/2]\\
\overline{H_t}(2s-1,t)=H(2-2s,t)=G(0,(2-2s)t),&s\in[1/2,1]
\end{cases}$$
Note that $K$ is continuous by the pasting lemma. </p>
<p>Now the path $K(s,0)=H_{0}*G_{0}*\overline{H_0}=e_{x_0}*f*\overline{e_{x_0}}$ is path homotopic to $f$ (note that $[e_{x_0}*f*\overline{e_{x_0}}]=[e_{x_0}]*[f]*[\overline{e_{x_0}}]=[f]$). Similarily, the path $K(s,1)=H_{1}*G_{1}*\overline{H_1}=H_t*e_{x_0}*\overline{H_t}$ is path homotopic to $e_{x_0}$. We also have $K(0,t)=K(1,t)=G(0,0)=x_0$ for all $t\in[0,1]$, thus $K$ is a path homotopy between $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$, thus $e_{x_0}*f*\overline{e_{x_0}}$ and $H_{1}*G_{1}*\overline{H_1}$ are path homotopic, by previous discussion, we see that $f$ and $e_{x_0}$ are also path homotopic.</p>
|
195,333 | <p>I'm looking to make a graph with the x-axis reversed and a frame and tick marks that are thicker than default. However, the x-axis tick marks do not maintain the specified thickness once I reverse the x-axis. I can't restore the thickness of the tick marks using TicksStyle or FrameTicksStyle. How can I get around this?</p>
<pre><code>ListLinePlot[{},
Frame -> True,
FrameStyle ->
{{Black, Thickness[0.005]},
{Black, Thickness[0.005]},
{Black, Thickness[0.005]},
{Black, Thickness[0.005]}},
ScalingFunctions -> {"Reverse", Identity}
]
</code></pre>
<p>Notice how the tick marks on the y-axis are made thick (as desired), but not along the x-axis:</p>
<p><a href="https://i.stack.imgur.com/0a3gc.png" rel="noreferrer"><img src="https://i.stack.imgur.com/0a3gc.png" alt="enter image description here"></a></p>
| m_goldberg | 3,066 | <p>I think you have found a bug. As far as I can tell, using any scaling option (other than Automatic or None, doesn't annihilate all tick styling, but it does annihilate any attempt to change tick thickness. </p>
<p>Here is supporting evidence.</p>
<p>We start with a <code>ListLinePlot</code> in which we override the frame styling in the top and bottom ticks and their labels.</p>
<pre><code>tickStyle = Directive[Red, AbsoluteThickness[5], FontSize -> 14];
ListLinePlot[Table[{x, Sin[x]}, {x, Subdivide[0, 2 Pi, 100]}],
Frame -> True,
FrameStyle -> Thick,
FrameTicks -> All,
FrameTicksStyle -> {{None, None}, {tickStyle, tickStyle}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/de7WM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/de7WM.png" alt="plot1"></a></p>
<p>Now let's add scaling.</p>
<pre><code>ListLinePlot[
Table[{x, Sin[x]}, {x, Subdivide[0, 2 Pi, 100]}],
Frame -> True,
FrameStyle -> Thick,
FrameTicks -> All,
FrameTicksStyle -> {{None, None}, {tickStyle, tickStyle}},
ScalingFunctions -> {"Reverse", None}]\
</code></pre>
<p><a href="https://i.stack.imgur.com/tOrAJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tOrAJ.png" alt="plot2"></a></p>
<p>Note that, while the <code>AbsoluteThickness</code> setting has been ignored, the other style setting are respected. The ticks are still red and the tick labels are still 14 points.</p>
|
1,093,396 | <p>I've been working on a problem from a foundation exam which seems totally straightforward but for some reason I've become stuck:</p>
<p>Let $f: \mathbb{ R } \rightarrow \mathbb{ R } ^n$ be a differentiable mapping with $f^\prime (t) \ne 0$ for all $t \in \mathbb{ R } $, and let $p \in \mathbb{ R } ^n$ be a point NOT in $f(\mathbb{ R }) $. </p>
<ol>
<li>Show that there is a point $q = f(t)$ on the curve $f(\mathbb{ R }) $ which is closest to $p$. </li>
<li>Show that the vector $r := (p - q)$ is orthogonal to the curve at $q$. </li>
</ol>
<p>Hint: Consider the function $t \mapsto |p - f(t)|$ and its derivative. </p>
<p>I can solve the second part of the problem assuming that I've found the point $q$ requested in the first part: consider the square difference function $\varphi(t) = |p - f(t)|^2 = (p - f(t)) \cdot (p - f(t))$, derive and at $q = f(t_0)$ we will have $\varphi ^\prime(t_0) = 0$. How do we prove the existence of the point $q$? The square distance function $\varphi(t)$ is a function from $\mathbb{ R } \rightarrow \mathbb{ R } $, so maybe something like Rolle's Theorem? Not too sure about this. Any help will be greatly appreciated! </p>
| Jihad | 191,049 | <p>Take some $s$ on the curve. Take circle $C$ with center in $p$ of radius $|p-s|$. $\overline C \cap f(\mathbb{R}) - $ closed and bounded set (i.e. compact in $\mathbb{R}^n$). $t \to |p-f(t)|$ is continuous function. What do you know about continuous functions on compacts?</p>
|
394,321 | <p>I'm studying the asymptotic behavior <span class="math-container">$(n \rightarrow \infty)$</span> of the following formula, where <span class="math-container">$k$</span> is a given constant.
<span class="math-container">$$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$</span></p>
<p>I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.</p>
<p>I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used).
<a href="http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29" rel="nofollow noreferrer">http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29</a></p>
<p>However in series expansion at <span class="math-container">$n \rightarrow \infty$</span>, the result still has <span class="math-container">$\ln n$</span>. This is actually a form I prefer, compared to the form
<span class="math-container">$$a_0 + a_1x + a_2 x^2+...$$</span>
Does anyone see how the result may be produced? Any help is much appreciated. Thanks.</p>
| Adi Dani | 12,848 | <p>HINT </p>
<p>$$\frac{y}{(x-y)(y-1)} = \frac{Ax+By+C}{x-y} + \frac{Dx+Ey+F}{y-1}$$</p>
|
4,231,509 | <p>I'm trying to prove that the group <span class="math-container">$(\mathbb{R}^*, \cdot)$</span> is not cyclic (similar to [1]). My efforts until now culminated into the following sentence:</p>
<blockquote>
<p>If <span class="math-container">$(\mathbb{R}^*,\cdot)$</span> is cyclic, then <span class="math-container">$\exists x \in \mathbb{R}^*$</span> such that <span class="math-container">$x \cdot x \neq x \in \mathbb{R}^*$</span>.</p>
</blockquote>
<p>The assumption written above is only not true for the neutral element on <span class="math-container">$\mathbb{R}^*$</span>. Are there any follow-ups that I should do to improve that sentence?</p>
<p>[1] <a href="https://math.stackexchange.com/questions/1491510/show-that-q-and-r-are-not-cyclic-groups">Show that (Q, +) and (R, +) are not cyclic groups.</a></p>
| Shaun | 104,041 | <p><strong>Hint:</strong> If <span class="math-container">$G$</span> is a cyclic group and <span class="math-container">$H\cong G$</span>, then <span class="math-container">$H$</span> is also cyclic. Now use the fact that <span class="math-container">$(\Bbb R, +)$</span> is not cyclic.</p>
<p>Also, every group satisfies <span class="math-container">$x^2\neq x$</span> for each <span class="math-container">$x\neq e$</span>. Indeed, if <span class="math-container">$x^2=x$</span>, then <span class="math-container">$x^2=ex$</span>, so, multiplying by <span class="math-container">$x^{-1}$</span> on the right, we get <span class="math-container">$x=e$</span>.</p>
|
1,722,287 | <p>So far I know that when matrices A and B are multiplied, with B on the right, the result, AB, is a linear combination of the columns of A, but I'm not sure what to do with this. </p>
| Daniel Akech Thiong | 169,316 | <p>Let $A$ be $m \times n$ and $B$ be $n \times p$ both with entries in the field $F$. Consider the maps: $L_{A}: F^{n} \rightarrow F^{m}: L_{A}(X) = AX$ and $L_{B}: F^{p} \rightarrow F^{n}: L_{B}(X) = BX$. </p>
<p>Then $L_{A} \circ L_{B}: F^{p} \rightarrow F^{m}$ satisfies $Y \in \text{Image } L_{A} \circ L_{B} \implies L_{A}(L_{B}(X)) = Y \implies Y \in \text{Image } L_{A} $. It follows that $\text{Image } L_{A} \circ L_{B} \subset \text{Image } L_{A}$. The inequality in question then follows. </p>
|
637,819 | <p>$$x\in(\cap F)\cap(\cap G)=[\forall A\in F(x\in A)]\land[\forall A\in G(x\in A)]$$</p>
<p>Since the variable $A$ is bounded by universal quantifier, it is regarded as bounded variable, according to the rules, the variable is free to change to other letters while the meaning statement remains unchanged. But,the above statements mention two different families of sets, $F$ and $G$, why is it correct to state the sets of $F$ and $G$ by using the same letter $A$, for the first $A$ in the first part of the conjunction stands for sets in $F$ while the latter stands for sets in $G$? Isn't different letters should be used to refer those sets($A$ for $F$ while $B$ for $G$)? I am extremely confused with the usage of bound variables. Please explain, thanks!</p>
| user121173 | 121,173 | <p>The reason for this is that when you take the conjunction of two statements, its truth depends on the truth of each of the two components, and you don't look any further than that.</p>
<p>For example, the statements</p>
<p>$$\forall x \geq 5 \quad 2x \geq 10$$
and<br>
$$\forall x \leq - 3 \quad x^2 \geq 9$$
are both true, taken separately. Therefore the conjunction
$$(\forall x \geq 5 \quad 2x \geq 10) \wedge (\forall x \leq - 3 \quad x^2 \geq 9)$$
must be true, regardless of the fact that $x$ represents different possible numbers in each of the two statements.</p>
<p>Edit: Nonetheless, it is true that $x$ is taken as belonging to some overarching "universe" of all objects. In my example, this is the set of all real numbers, and in your example, $A$ belongs to the universe of all sets, regardless of whether they belong to $F$ or $G$ or not. A statement such as
$$ \forall A \in F \quad x \in A$$
really means
$$\forall A \quad A \in F \Longrightarrow x \in A,$$
where the part "$\forall A$" means "for all sets $A$." Analyzed this way, you see that really $A$ represents the same type of object, namely a set.</p>
|
85,470 | <p>We decided to do secret Santa in our office. And this brought up a whole heap of problems that nobody could think of solutions for - bear with me here.. this is an important problem.</p>
<p>We have 4 people in our office - each with a partner that will be at our Christmas meal.</p>
<p>Steve,
Christine,
Mark,
Mary,
Ken,
Ann,
Paul(me),
Vicki</p>
<p>Desired outcome</p>
<blockquote>
<p>Nobody can know who is buying a present for anybody else. But we each
want to know who we are buying our present for before going to the
Christmas party. And we don't want to be buying presents for our partners.
Partners are not in the office.</p>
</blockquote>
<p>Obvious solution is to put all the names in the hat - go around the office and draw two cards.</p>
<p>And yes - sure enough I drew myself and Mark drew his partner. (we swapped)</p>
<p>With that information I could work out that Steve had a 1/3 chance of having Vicki(he didn't have himself or Christine - nor the two cards I had acquired Ann or Mary) and I knew that Mark was buying my present. Unacceptable result.</p>
<p>Ken asked the question: "What are the chances that we will pick ourselves or our partner?"</p>
<p>So I had a stab at working that out.</p>
<p>First card drawn -> 2/8
Second card drawn -> 12/56</p>
<p>Adding them together makes 28/56 i.e. 1/2.</p>
<p>i.e. This method won't ever work... half chances of drawing somebody you know means we'll be drawing all year before we get a solution that works.</p>
<p>My first thought was that we attach two cards to our backs... put on blindfolds and stumble around in the dark grabbing the first cards we came across... However this is a little unpractical and I'm pretty certain we'd end up knowing who grabbed what anyway.</p>
<p>Does anybody have a solution for distributing cards that results in our desired outcome?</p>
<hr>
<p><em><strong>I'd prefer a solution without a third party..</em></strong></p>
| Ilmari Karonen | 9,602 | <p>If each of you just draws a random name out of the hat, the probability that nobody gets their own or their partner's name is</p>
<p>$$\frac{4752}{40320} = \frac{33}{280} \approx 11.8\%.$$</p>
<p>Thus, the expected number of times you'll need to repeat the process before getting "a solution that works" is $280/33 \approx 8.5$. This could get a bit tedious, but it'll hardly take "all year", at least unless you're both really slow and really unlucky with the draw.</p>
<hr>
<p><em>Ps.</em> I went and calculated the odds for other numbers of couples too:</p>
<ul>
<li>$2$ couples: $\frac{4}{24} = \frac{1}{6} \approx 16.7\%$ chance of nobody getting their own or their partner's name</li>
<li>$3$ couples: $\frac{80}{720} = \frac{1}{9} \approx 11.1\%$</li>
<li>$4$ couples: $\frac{4752}{40320} = \frac{33}{280} \approx 11.8\%$</li>
<li>$5$ couples: $\frac{440192}{3628800} = \frac{3439}{28350} \approx 12.1\%$</li>
<li>$6$ couples: $\frac{59245120}{479001600} = \frac{16831}{136080} \approx 12.4\%$</li>
</ul>
<p>If I'm not mistaken, as the number of couples involved tends to infinity, the probability of nobody getting their or their partner's name should tend towards the limit $e^{-2} \approx 13.5\%$, and thus the expected number of retries needed should tend towards $e^2 \approx 7.4$.</p>
|
126,052 | <p>I have no doubt that the following observation is quite well known. Let $\varphi:[0,1]\to [0,1]$ be a continuous map. Assume that the iterates $\varphi^n$ converge pointwise to some continuous map $\varphi_\infty$. Then the convergence is in fact uniform. However, I was unable to locate a reference. Does anybody know where I can find it written? Thanks in advance.</p>
| Sergei Ivanov | 4,354 | <p>I don't know a reference but maybe the following proof is shorter than yours.</p>
<p>By continuity, $\varphi\circ\varphi_\infty=\varphi_\infty$. Hence $\varphi$ is identity on the set $I:=\varphi_\infty([0,1])$. Hence $I$ is the set of fixed points of $\varphi$. And it is compact and connected. Then there are two cases: either $I$ is a nontrivial interval $[a,b]$ or it is a single point $\{a\}$.</p>
<p>In the first case, we have $f(x)<x$ for all $x>b$ and $f(x)>x$ for all $x<a$, otherwise there is a fixed point outside $I$. There is an $\varepsilon>0$ such that $f(x)>a$ for all $x\in(b,b+\varepsilon)$ and $f(x)<b$ for all $x\in(a-\varepsilon,a)$. Let $U=(a-\varepsilon,b+\varepsilon)$. Then $\varphi(U)\subset U$ and for every $x\in U$ we have
$$
|\varphi^n(x)-x|<\delta(\varepsilon)
$$
where $\delta(\varepsilon)\to $ as $\varepsilon\to 0$ ($\delta(\varepsilon)$ is the maximum of the diameters of the sets $\varphi([a-\varepsilon,a])$ and $\varphi([b,b+\varepsilon])$) For every $x\in[0,1]$, the iterations $\varphi^n(x)$ eventually get into $U$, hence
$$
\bigcup_n \varphi^{-n}(U) = [0,1] .
$$
By compactness, this implies that $[0,1]$ is covered by finitely many sets $f^{-n}(U)$ and hence there exist $n$ such that $\varphi^n([0,1])\subset U$. After this $n$, all iterations stay within the distance $\delta(\varepsilon)$ from the limit. Since $\varepsilon$ can be arbitrarily small, this implies uniform convergence.</p>
<p>In the second case (when $I$ is a single point $a$), similarly observe that $f(x)<x$ for $x>a$ and $f(x)>x$ for $x<a$, and the same holds for $\varphi^2$, $\varphi^3$, etc. This implies that if e.g. $x>a$ and $f(x)<a$ then all iterations $\varphi^n(x)$ lie between $x$ and $f(x)$. This easily implies that there is an arbitrarily small neighborhood $U$ of $a$ such that $f(U)\subset U$. Then uniform convergence follows similarly to the first case.</p>
|
3,510,233 | <blockquote>
<p>If <span class="math-container">$\sin\left(\operatorname{cot^{-1}}(x + 1)\right) = \cos\left(\tan^{-1}x\right)$</span>, then find the value of <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Please solve this question by using <span class="math-container">$\cos\left(\dfrac\pi2 - \theta\right) = \sin\theta$</span> by changing <span class="math-container">$\cos\left(\tan^{-1}x\right) = \sin\left(\dfrac\pi2 - \tan^{-1}x\right)$</span> and then equate both LHS and RHS. If not then why? How does the contradiction below occur?</p>
<p><a href="https://i.stack.imgur.com/WDXoM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WDXoM.jpg" alt="enter image description here"></a></p>
| Quanto | 686,284 | <p>The growth is the fastest when the slope of <span class="math-container">$P(t)$</span> is the largest. Evaluate <span class="math-container">$
P'(t) = \frac{ke^{-t/2}}{\left(1000 e^{-t/2}+1\right)^2}$</span>, with <span class="math-container">$k=25mm$</span>. Then, cast it in the form</p>
<p><span class="math-container">$$P'(t) = \frac k{(1000e^{-t/4}-e^{t/4})^2+2000}\le \frac k{2000}$$</span></p>
<p>where the equality, or, the maximal growth, occurs at <span class="math-container">$1000e^{-t/4}=e^{t/4}$</span>, i.e. <span class="math-container">$t=2\ln 1000$</span>.</p>
|
3,459,532 | <p>I have a pretty straightforward linear programming problem here:</p>
<p><span class="math-container">$$ maximize \hskip 5mm -x_1 + 2x_2 -3x_3 $$</span></p>
<p>subject to</p>
<p><span class="math-container">$$ 5x_1 - 6x_2 - 2x_3 \leq 2 $$</span>
<span class="math-container">$$ 5x_1 - 2x_3 = 6 $$</span>
<span class="math-container">$$ x_1 - 3x_2 + 5x_3 \geq -3 $$</span>
<span class="math-container">$$ 1 \leq x_1 \leq 4 $$</span>
<span class="math-container">$$ x_3 \leq 3 $$</span></p>
<p>Convert to standard form.</p>
<p>what boggles me is how to substitute <span class="math-container">$x_1$</span> since it’s restricted from both sides and I can’t move forward in the problem until I figure it out...</p>
<p>I’m not asking for the whole standard form, just how to approach this one variable. :)</p>
| Anurag A | 68,092 | <p>Use the fact that <span class="math-container">$u^Tv=u \cdot v$</span>, thus <span class="math-container">$u^Tu=\|u\|^2$</span> and note the following: </p>
<ul>
<li><span class="math-container">$a_ia_i^T$</span> is a rank one matrix.</li>
<li><span class="math-container">$a_i^Ta_j=0$</span> for <span class="math-container">$i \neq j$</span> because we are given that <span class="math-container">$a_i \perp a_j$</span>.</li>
<li><span class="math-container">$a_i^Ta_i=1$</span> because we are given that <span class="math-container">$a_i$</span> are unit vectors.</li>
</ul>
<p><strong>Claim:</strong> <span class="math-container">$A^2=A$</span>.</p>
<p><strong>Proof:</strong>
Let <span class="math-container">$P_1=a_1a_1^T$</span> and <span class="math-container">$P_2=a_2a_2^T$</span>, then <span class="math-container">$P_1P_2=a_1a_1^Ta_2a_2^T=0$</span>, likewise <span class="math-container">$P_2P_1=0$</span> and since <span class="math-container">$P_i$</span> are projection matrices, therefore <span class="math-container">$P_i^2=P_i$</span> (this can be verified directly as well).
<span class="math-container">\begin{align*}
A^2 & = (I-P_1-P_2)^2\\
&=I-2P_1-2P_2+P_1P_2+\color{red}{P_1^2}+P_2P_1+\color{blue}{P_2^2}\\
&=I-2P_1-2P_2+\color{red}{P_1}+\color{blue}{P_2}\\
& = I-P_1-P_2\\
&=A.
\end{align*}</span>
This suggests that the eigenvalues of <span class="math-container">$A$</span> are either <span class="math-container">$0$</span> or <span class="math-container">$1$</span>. </p>
<p>Now consider
<span class="math-container">\begin{align*}
Aa_2 & =(I-a_1a_1^T-a_2a_2^T)a_2\\
&=a_2-a_1a_1^Ta_2-a_2a_2^Ta_2\\
& =a_2-0-a_2 && (\because a_1 \perp a_2 \& \|a_2\|=1)\\
& = 0.
\end{align*}</span>
Thus <span class="math-container">$0$</span> is an eigen value with <span class="math-container">$a_2$</span> as the corresponding eigenvector.</p>
<p>Since <span class="math-container">$d \geq 3$</span>, this means there is at least one non-zero vector <span class="math-container">$u$</span> such that <span class="math-container">$u \perp a_1$</span> and <span class="math-container">$u \perp a_2$</span> (same as saying <span class="math-container">$a_i^Tu=0$</span>). Now consider,
<span class="math-container">\begin{align*}
Au & =(I-a_1a_1^T-a_2a_2^T)u\\
&=u-a_1a_1^Tu-a_2a_2^Tu\\
& =u-0-0 && (\because u \perp a_i)\\
& = u.
\end{align*}</span>
Thus <span class="math-container">$1$</span> is also an eigenvalue with eigenvector <span class="math-container">$u$</span>.</p>
|
3,459,532 | <p>I have a pretty straightforward linear programming problem here:</p>
<p><span class="math-container">$$ maximize \hskip 5mm -x_1 + 2x_2 -3x_3 $$</span></p>
<p>subject to</p>
<p><span class="math-container">$$ 5x_1 - 6x_2 - 2x_3 \leq 2 $$</span>
<span class="math-container">$$ 5x_1 - 2x_3 = 6 $$</span>
<span class="math-container">$$ x_1 - 3x_2 + 5x_3 \geq -3 $$</span>
<span class="math-container">$$ 1 \leq x_1 \leq 4 $$</span>
<span class="math-container">$$ x_3 \leq 3 $$</span></p>
<p>Convert to standard form.</p>
<p>what boggles me is how to substitute <span class="math-container">$x_1$</span> since it’s restricted from both sides and I can’t move forward in the problem until I figure it out...</p>
<p>I’m not asking for the whole standard form, just how to approach this one variable. :)</p>
| user1551 | 1,551 | <p>Extend <span class="math-container">$\{a_1,a_2\}$</span> to an orthonormal basis <span class="math-container">$\{a_1,a_2,\ldots,a_d\}$</span> of <span class="math-container">$\mathbb R^d$</span>. Then
<span class="math-container">$$
Aa_i=(I-a_1a_1^T-a_2a_2^T)a_i=
\begin{cases}
0,&i=1,2,\\
a_i,&i\ge3.
\end{cases}
$$</span>
Therefore the eigenvalues of <span class="math-container">$A$</span> are <span class="math-container">$0$</span> (of multiplicity <span class="math-container">$2$</span>) and <span class="math-container">$1$</span> (of multiplicity <span class="math-container">$d-2$</span>) and the <span class="math-container">$a_i$</span>s are eigenvectors of <span class="math-container">$A$</span>.</p>
|
3,510,156 | <p>This is a duplicate question of <a href="https://math.stackexchange.com/questions/2088815/find-integers-solutions-of-x27-y5#">Find integers solutions of $x^2+7=y^5$</a>, however there was no full answer. The solutions <span class="math-container">$(\pm5, 2)$</span> and <span class="math-container">$(\pm 181, 8)$</span> have been found. </p>
<p>The usual strategy for such a question is to work inside the ring of integers of <span class="math-container">$\mathbb{Q}(\sqrt{-7})$</span>, which is <span class="math-container">$\mathcal{O} = \mathbb{Z}[ \frac{1+\sqrt{-7}}{2}]$</span>. It turns out that this is a unique factorisation domain (which one can figure out by calculating its Class group). So it is natural to factor the equation as <span class="math-container">$(x - \sqrt{-7})(x+\sqrt{-7}) = y^5$</span>. If we assume that <span class="math-container">$x-\sqrt{-7}$</span> and <span class="math-container">$x+\sqrt{-7}$</span> are coprime, we find that <span class="math-container">$x+\sqrt{-7} = \beta^5$</span> for a certain <span class="math-container">$\beta = a + b\frac{1+\sqrt{-7}}{2}\in \mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$</span>. Writing <span class="math-container">$c= 2a+b$</span> and expanding the fifth power, this gives the system of equations
<span class="math-container">$$ c^5 -70 c^3 b^2 + 245 c^4 b = 32 x, $$</span>
<span class="math-container">$$ 5 c^4 b -70 c^2 b^3 + 49 b^5 = 32. $$</span>
Now with enough patience, one can show that this system has no solutions with <span class="math-container">$b \equiv c \pmod{2}$</span>. </p>
<p>However this contradicts the solutions that we have found. And indeed there's no reason for <span class="math-container">$x \pm \sqrt{-7}$</span> to be coprime when <span class="math-container">$x$</span> is odd. </p>
<p>What is the approach to solve the remaining case of this diophantine equation?</p>
<p>One approach that I have tried is that the coprime condition holds inside the ring <span class="math-container">$\cal{O}[\frac{1}{2}]$</span>. This gives the equation <span class="math-container">$x + \sqrt{-7} = (a+b\sqrt{-7})^5$</span> with <span class="math-container">$a,b \in \mathbb{Z}[\frac{1}{2}]$</span>, which I am unable to solve.</p>
| Community | -1 | <p>Consider the general case
<span class="math-container">$$x^2+7=y^m \tag{1}$$</span>
(Integers <span class="math-container">$(x, y, m), \, m \geq 3)$</span>
Let
<span class="math-container">$$\rho = (1+\sqrt{-7})/2$$</span></p>
<p>Then as you are aware <span class="math-container">$(1, \rho)$</span> is a basis for the ring of integers of the field <span class="math-container">$\mathbb{Q}(-7)$</span>. A standard factorization argument then compels us to devise a
<span class="math-container">$$\frac{x-1}{2}+\rho = \rho^{m-5}(U+\rho V)^m$$</span></p>
<p>For the case where <span class="math-container">$m=5$</span> one needs only consider the coefficients of <span class="math-container">$(U, V)$</span> and form
<span class="math-container">$$−U^5 − 15U^4V − 10U^3V^2 + 50U^2V^3 + 35UV^4 − 3V^5 = 1$$</span>
This is a version of the <a href="https://en.wikipedia.org/wiki/Thue_equation" rel="noreferrer">Thue equation</a> which I can solve using the <a href="https://cran.r-project.org/" rel="noreferrer">R</a> package <a href="https://cran.r-project.org/web/packages/nilde/nilde.pdf" rel="noreferrer">NILDE</a>.</p>
<p>Using this, the only solution is the only solutions to the above equation are <span class="math-container">$(U, V ) = (−1, 0),(2, −1)$</span>, which then give solutions proper as <span class="math-container">$(5, x, y)=(5, ±5, 2),(5, ±181, 8)$</span></p>
<p>Now, in <a href="http://J.L.%20Lesage,%20Diff%C2%B4erence%20entre%20puissances%20et%20carr%C2%B4es%20d%E2%80%99entiers,%20Journal%20of%20Number%20Theory%2073%20(1998),%20390%E2%80%93425." rel="noreferrer">Lesage</a>, the author shows various partial results concerning the equation <span class="math-container">$(1)$</span>, including the following.</p>
<ul>
<li>There are integer solutions to equation (1) with <span class="math-container">$m = 5, 7, 13$</span>, this he proves by reducing to Thue equations, which he then solves by hand.</li>
<li>There are no solutions to equation (1) for <span class="math-container">$m = 11$</span> and for <span class="math-container">$m$</span> prime
and <span class="math-container">$17 \leq m \leq 5000$</span>. This he proves using classical algebraic number
theory, and a computational method.</li>
<li>If <span class="math-container">$(x, y, m)$</span> is a solution to (1) then <span class="math-container">$m \leq 6.6 × 10^{15}$</span>. This he proves
using lower bounds for linear forms in logarithms.</li>
</ul>
|
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