qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,112,043 | <p>The first part of the problem is:</p>
<p>Prove that for all integers <span class="math-container">$n \ge 1$</span> and real numbers <span class="math-container">$t>1$</span>,
<span class="math-container">$$ (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1)$$</span></p>
<p>I have done the first part by induction on <span class="math-container">$n$</span> for any real <span class="math-container">$t>1$</span>.</p>
<p>However, I don't know how to do the second part, which is:</p>
<p>Use this to prove that if <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are positive integers,</p>
<p><span class="math-container">$$ \frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<\Big(1+\frac{1}{m}\Big)^{n+1}\frac{m^{n+1}}{n+1} $$</span></p>
<p>I factorized <span class="math-container">$t^{n+1}-1$</span> into <span class="math-container">$(t-1)(1+t+t^2+\dots+t^n)$</span> and cancelled <span class="math-container">$t-1$</span> to obtain <span class="math-container">$(n+1)t^n>1+t+t^2+\dots+t^n>n+1$</span>. And have no idea what to do next. This is the only way I could think of in order to get a "sum", but couldn't see any relation between the two sum (if there are any...).</p>
<p>The last part of the question is:</p>
<p>Find
<span class="math-container">$$ \lim_{m\rightarrow \infty} \frac{1^n+2^n+\dots+m^n}{m^{n+1}}$$</span></p>
<p>I think I know how to do the last part. It should be divide the inequality by <span class="math-container">$m^{n+1}$</span>, then apply Squeeze Theorem. I believe the answer is <span class="math-container">$\frac{1}{n+1}$</span>.</p>
| Hypergeometricx | 168,053 | <p>Let <span class="math-container">$S=1^n+2^n+3^n+\cdots+m^n$</span>. </p>
<p>From a quick graph sketch it is clear that:
<span class="math-container">$$\begin{align}
\int_0^m x^n dx \;\;&<\qquad \qquad \quad S &&< \int_0^m (x+1)^n dx\\
\left[\frac {x^{n+1}}{n+1}\right]_0^m\;\;&<\qquad \qquad \quad S &&<\;\;\left[\frac {(x+1)^{n+1}}{n+1}\right]_0^m\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\frac {(m+1)^{n+1}}{n+1}-\frac 1{n+1}\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \quad S &&<\;\;\left[\left(1+\frac 1m\right)^{m+1}-1\right]\frac {m^{n+1}}{n+1}<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad1^n+2^n+3^n+\cdots+m^n &&<\left(1+\frac 1m\right)^{m+1}\frac {m^{n+1}}{n+1}\\
\end{align}$$</span></p>
|
3,466,870 | <p>Suppose </p>
<p><span class="math-container">$$a^2 = \sum_{i=1}^k b_i^2$$</span> </p>
<p>where <span class="math-container">$a, b_i \in \mathbb{Z}$</span>, <span class="math-container">$a>0, b_i > 0$</span> (and <span class="math-container">$b_i$</span> are not necessarily distinct).</p>
<p>Can any positive integer be the value of <span class="math-container">$k$</span>?</p>
<hr>
<p>The reason I am interested in this: in a irreptile tiling where the smallest piece has area <span class="math-container">$A$</span>, we have <span class="math-container">$a^2A = \sum_{i=1}^k b_i^2A$</span>, where we have <span class="math-container">$k$</span> pieces scaled by <span class="math-container">$b_i$</span> to tile the big figure, which is scaled by <span class="math-container">$a$</span>. I am wondering what constraints there are on the number of pieces.</p>
<p>Here is an example tiling that realizes <span class="math-container">$4^2 = 3^2 + 7 \cdot 1^2$</span>, so <span class="math-container">$k = 8$</span>.</p>
<p><a href="https://i.stack.imgur.com/LdwQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LdwQy.png" alt="enter image description here"></a></p>
| Herman Tulleken | 67,933 | <p>Here is a geometric solution (for <span class="math-container">$k > 5$</span>).</p>
<p>The following are solutions for 6, 7, and 8 squares. </p>
<p><a href="https://i.stack.imgur.com/JGaFA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/JGaFA.png" alt="enter image description here"></a></p>
<p>We can replace a square in each of these with four equal size squares to find a tiling with 3 more squares, so we can get 9, 10, and 11 squares. Repeating this we can get any number of squares larger than 5.</p>
<p>I show one iteration below:</p>
<p><a href="https://i.stack.imgur.com/9BuOC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9BuOC.png" alt="enter image description here"></a></p>
|
3,466,870 | <p>Suppose </p>
<p><span class="math-container">$$a^2 = \sum_{i=1}^k b_i^2$$</span> </p>
<p>where <span class="math-container">$a, b_i \in \mathbb{Z}$</span>, <span class="math-container">$a>0, b_i > 0$</span> (and <span class="math-container">$b_i$</span> are not necessarily distinct).</p>
<p>Can any positive integer be the value of <span class="math-container">$k$</span>?</p>
<hr>
<p>The reason I am interested in this: in a irreptile tiling where the smallest piece has area <span class="math-container">$A$</span>, we have <span class="math-container">$a^2A = \sum_{i=1}^k b_i^2A$</span>, where we have <span class="math-container">$k$</span> pieces scaled by <span class="math-container">$b_i$</span> to tile the big figure, which is scaled by <span class="math-container">$a$</span>. I am wondering what constraints there are on the number of pieces.</p>
<p>Here is an example tiling that realizes <span class="math-container">$4^2 = 3^2 + 7 \cdot 1^2$</span>, so <span class="math-container">$k = 8$</span>.</p>
<p><a href="https://i.stack.imgur.com/LdwQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LdwQy.png" alt="enter image description here"></a></p>
| Bill Dubuque | 242 | <p>This holds far more generally. OP is the special case <span class="math-container">$S$</span> = integer squares, which is closed under multiplication <span class="math-container">$\,a^2 b^2 = (ab)^2,\,$</span> and has an element that is a sum of <span class="math-container">$\,2\,$</span> others, e.g. <span class="math-container">$\,5^2 = 4^2+3^2 $</span>.</p>
<p><strong>Theorem</strong> <span class="math-container">$ $</span> If <span class="math-container">$\,S\,$</span> is a set of integers <span class="math-container">$\rm\color{#0a0}{closed}$</span> under multiplication then<br />
<span class="math-container">$\qquad\qquad \begin{align}\phantom{|^{|^|}}\forall\,n\ge 2\!:\text{ there is a }\,t_n\in S\,&\ \text{that is a sum of $\,n\,$ elements of $\,S$ }\\[.1em]
\iff\! \text{ there is a }\,t_2\in S\,&\ \text{that is a sum of $\,2\,$ elements of $\,S\!$}\\
\end{align}$</span></p>
<p><strong>Proof</strong> <span class="math-container">$\ \ (\Rightarrow)\ $</span> Clear. <span class="math-container">$\ (\Leftarrow)\ $</span> We induct on <span class="math-container">$n$</span>. The base case <span class="math-container">$\,n = 2\,$</span> is true by hypothesis, i.e. we are given that <span class="math-container">$\,\color{#c00}{a = b + c}\,$</span> for some <span class="math-container">$\,a,b,c\in S.\,$</span> If the statement is true for <span class="math-container">$\,n\,$</span> elements then</p>
<p><span class="math-container">$$\begin{align}
s_0 &\,=\, s_1\ \ +\ s_2\ + \cdots +s_n,\ \ \ \ \,{\rm all}\ \ s_i\,\in\, S\\
\Rightarrow\ s_0a &\,=\, s_1 a + s_2 a + \cdots +s_n\color{#c00} a,\ \ \color{#0a0}{\rm all}\ \ s_ia\in S \\[.1em]
&\,=\, s_1 a + s_2 a + \cdots + s_n \color{#c00}b + s_n \color{#c00}c \end{align}\qquad\qquad$$</span></p>
<p>so <span class="math-container">$\,s_0 a\in S\,$</span> is a sum of <span class="math-container">$\,n\!+\!1\,$</span> elements of <span class="math-container">$S$</span>, completing the induction.</p>
<p><strong>Remark</strong> <span class="math-container">$ $</span> A comment asks for further examples. Let's consider some "minimal" examples. <span class="math-container">$S$</span> contains <span class="math-container">$\,a,b,c\,$</span> wth <span class="math-container">$\,a = b + c\:$</span> so - being closed under multiplication - <span class="math-container">$\,S\,$</span> contains all products <span class="math-container">$\,a^j b^j c^k\ne 1$</span>. But these products are already closed under multiplication so we can take <span class="math-container">$S$</span> to be the set of all such products. Let's examine how the above inductive proof works in this set.</p>
<p><span class="math-container">$$\begin{align} \color{#c00}a &= b + c\\
\smash{\overset{\times\ a}\Longrightarrow}\qquad\qquad\, a^2 = ab+\color{#c00}ac\, &= b(a+c)+c^2\ \ \ {\rm by\ substituting}\,\ \color{#c00}a = b+c\\
\smash{\overset{\times\ a}\Longrightarrow}\ \ a^3 = b(a^2+ac)+ \color{#c00}ac^2 &= b(a^2+ac+c^2)+c^3 \\[.4em]
{a^{n}} &\ \smash{\overset{\vdots_{\phantom{|^|}\!\!}}= \color{#0a0}b (a^{n-1} + \cdots + c^{n-1}) + c^{n}\ \ \text{ [sum of $\,n\!+\!1\,$ terms]}}\\[.2em]
{\rm by}\ \ \ a^{n}-c^{n} &= (\color{#0a0}{a\!-\!c}) (a^{n-1} + \cdots + c^{n-1})\ \ \ {\rm by}\ \ \color{#0a0}{b = a\!-\!c} \end{align}\quad\ \ \ $$</span></p>
<p>So the proof's inductive construction of an element that is a sum of <span class="math-container">$n+1$</span> terms boils down here to writing <span class="math-container">$\,a^n\,$</span> that way using the above well known factorization of <span class="math-container">$\,a^n-c^n\,$</span> via the <a href="https://math.stackexchange.com/a/738715/242">Factor Theorem.</a></p>
<p>By specializing <span class="math-container">$\,a,b,c\,$</span> one obtains many examples, e.g. using <span class="math-container">$\,5^2,4^2,3^2$</span> as in the OP then <span class="math-container">$S$</span> is set of squares composed only of those factors, and the <span class="math-container">$\,n\!+\!1\,$</span> element sum constructed is</p>
<p><span class="math-container">$$ 25^n =\, 9^n + 16(25^{n-1}+ \cdots + 9^{n-1})$$</span></p>
|
649,502 | <p>What do we mean when we talk about a topological <em>space</em> or a metric <em>space</em>? I see some people calling metric topologies metric spaces and I wonder if there is some synonymity between a topology and a space? What is it that the word means, and if there are multiple meanings how can one distinguish them?</p>
| Emi Matro | 88,965 | <p>A metric space is a space in which the notion of distance is defined: there is a distance function that exists in that space, and it is true for all points in that space. For example, the distance formula between two points in Euclidean space. </p>
<p>A topological space is a space in which a notion of "closeness" or "nearness" is understood as existing between some points.</p>
|
649,502 | <p>What do we mean when we talk about a topological <em>space</em> or a metric <em>space</em>? I see some people calling metric topologies metric spaces and I wonder if there is some synonymity between a topology and a space? What is it that the word means, and if there are multiple meanings how can one distinguish them?</p>
| monroej | 121,862 | <p>In mathematics, you usually call a set (a collection of objects) with some additional structures a space. So for example, a set with a certain distance function is called a metric space, and a set with certain subsets defined to be open is called a topological space. (of course in these two examples the distance function and open sets have to satisfy certain axioms.) There are tons of other spaces too, vector spaces etc. </p>
<p>The reason some people refer to a topological space as as having a metric topology, is because a metric space is a specific example of a topological space. hope that helps.</p>
|
649,502 | <p>What do we mean when we talk about a topological <em>space</em> or a metric <em>space</em>? I see some people calling metric topologies metric spaces and I wonder if there is some synonymity between a topology and a space? What is it that the word means, and if there are multiple meanings how can one distinguish them?</p>
| Community | -1 | <p>Probably the best way to answer your question is to describe an observation: </p>
<p>Let $X:=(-1,1)$. Define two metrics on $X$ as follows
$$
d_1(x,y):=|x-y|, \forall x,y\in X,
$$
and
$$
d_2(x,y):=|arctan(x)-arctan(y)|, \forall x,y\in X.
$$</p>
<p>Then you can see that metrics $d_1$ and $d_2$ define the same topology on $X$. However they define different metric structures on $X$. The reason is that $(X,d_1)$ is not a complete metric space (every sequence approaching to 1 is a Cauchy sequence in $(X,d_1)$ but it is not convergent to any point in $X$.), while $(X,d_2)$ is a complete metric space (because it is basically the same as $\mathbb{R}$ with the usual metric which has been shrunk in $X$).</p>
|
2,238,614 | <p>For $n\ge3$ a given integer, find a Pythagorean Triple having n as one of its members.<br>
Hint: For n an odd integer, consider the triple $$\left(n, \frac 12\left(n^2-1\right), \frac 12(n^2+1)\right);$$ For n even, consider the triple $$\left(n, \left(\frac{n^2}{4}\right)-1, \left(\frac{n^2}{4}\right)+1 \right)$$</p>
<p>I have been trying to solve this problem by letting $n=x-y$ or $n=2xy$ but have been unable to use the hints properly. Do I just add the squares of the first 2 and see if it equals the square of the last? I am just missing the concept here. Any suggestions would be appreciated.</p>
<p>$$n^2+\left(\frac{n^2-1}{2}\right)^2=n^2+\frac{n^4-2n^2+1}{4}=\frac{n^4+2n^2+1}{4}$$</p>
<p>$$\left( \frac{n^2+1}{2}\right)^2=\frac{n^4+2n+1}{4}$$</p>
<p>Is this all I have to do for odd?</p>
| poetasis | 546,655 | <p><span class="math-container">$\\ \textbf{Matching sides using Euclid's F(m,n)}: $</span> Solving for <span class="math-container">$n$</span>, any values of <span class="math-container">$m$</span> that yield integers provide the <span class="math-container">$F(m,n)$</span> to identify a triple. Examples follow the solutions</p>
<p><span class="math-container">$$\mathbf{A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil}$$</span>
The lower limit ensures <span class="math-container">$m^2>A$</span> and the upper limit ensures <span class="math-container">$m-n\ge 1$</span>.</p>
<p><span class="math-container">$$\text{Example:}\qquad A=21\Rightarrow \lceil\sqrt{21+1}=5 \le m \le \biggl\lceil\frac{21}{2}\biggr\rceil =11\quad and \quad m\in\{5,11\}\Rightarrow n\in\{2,10\} $$</span></p>
<p><span class="math-container">$F(5,2)=(21,20,29)\qquad\qquad F(11,10)=(21,220,221)\\ $</span></p>
<p><span class="math-container">$$\mathbf{B=2mn\Rightarrow n=\frac{B}{2m}\qquad\qquad \bigl\lceil\sqrt{B}\bigr\rceil\le m \le \frac{B}{2}}$$</span>
The lower limit ensures <span class="math-container">$m>n$</span> and the upper limit ensures <span class="math-container">$m-n\ge 1$</span>.</p>
<p><span class="math-container">$$\text{Example:}\qquad B=44\Rightarrow \lceil\sqrt{88}\rceil =10 \le m \le \frac{44}{2}=22\quad and \quad m\in\{11,22\}\Rightarrow n\in\{2,1\}$$</span></p>
<p><span class="math-container">$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)\\$</span></p>
<p><span class="math-container">$$\mathbf{C=m^2+n^2\Rightarrow n=\sqrt{C-m^2}\qquad\qquad \biggl\lceil\sqrt{\frac{C}{2}}\biggr\rceil \le m < \sqrt{C}}$$</span></p>
<p><span class="math-container">$$\text{Example:}\qquad C=1105\Rightarrow \biggl\lfloor\sqrt{\frac{1105}{2}}\biggr\rfloor=23 \le m < \lfloor\sqrt{1105}=33\quad and \quad m\in\{24,31,32,33\}\Rightarrow n\in\{23,12,9,4\}\\$$</span></p>
<p><span class="math-container">$F(24,23)=(47,1104,1105)\quad F(31,12)=(817,744,1105)\\ $</span>
<span class="math-container">$F(32,9)=(943,576,1105)\quad F(33,4)=(1073,264,1105)\\$</span></p>
|
2,421,145 | <p>I am practicing problems around NFA and DFA.</p>
<p>I have seen many questions on how to convert NFA to DFA and DFA to Regular expression etc.</p>
<p>But I have seen very different question and I am stuck on how to proceed with the following question? </p>
<p>Given DFA. Convert this DFA to NFA with 5 states.
<a href="https://i.stack.imgur.com/GyDyp.png" rel="nofollow noreferrer">DFA IMAGE ATTACHED</a></p>
<p>I plan is to find a language and regular expression first and then try to convert regular expression to NFA with 5 states. But I had hard time just to find the language it accepts.</p>
<p>How to approach these problems. And to how to find language or regular expressions for large DFA's? Are there any algorithms or rules?</p>
<p>Edit: </p>
<p>Regular exp for the language and NFA with 5 states are added in the diagram.</p>
<p><a href="https://i.stack.imgur.com/t2z3x.jpg" rel="nofollow noreferrer">reg exp and NFA</a></p>
| Aflah | 478,577 | <p>You can refer to the following links
<a href="https://www.tutorialspoint.com/automata_theory/dfa_minimization.htm" rel="nofollow noreferrer">https://www.tutorialspoint.com/automata_theory/dfa_minimization.htm</a></p>
<p><a href="https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://web.cs.ucdavis.edu/~rogaway/classes/120/winter12/min.pdf&ved=0ahUKEwjp6MHp9pTWAhVEM48KHdu3BxEQFghnMAw&usg=AFQjCNEMrxLBaLEEZD6aR9w28vUe_6w4kw" rel="nofollow noreferrer">https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://web.cs.ucdavis.edu/~rogaway/classes/120/winter12/min.pdf&ved=0ahUKEwjp6MHp9pTWAhVEM48KHdu3BxEQFghnMAw&usg=AFQjCNEMrxLBaLEEZD6aR9w28vUe_6w4kw</a></p>
|
2,005,649 | <p>I am struggling to find a parameterization for the following set : </p>
<p>$$F=\left\{(x,y,z)\in\mathbb R^3\middle| \left(\sqrt{x^2+y^2}-R\right)^2 + z^2 = r^2\right\}
\quad\text{with }R>r$$</p>
<p>I also have to calculate the area. </p>
<p>I know its a circle so we express it in terms of the angle but my problem is with the $x$ and $y$ . They are not defined uniquely by the angle.<br>
Please explain with details because it is more important for me to understand than the answer itself</p>
| Pieter21 | 170,149 | <p>I think it is a bit easier than what you did if you handle $D$ more freely.</p>
<ol>
<li>Take all combinations: $(A+B+C+D)^8$</li>
<li>Exclude combinations that do not have (at least) one of $A, B, C$ = $ (B+C+D)^8 + (A+C+D)^8 + (A+B+D)^8$</li>
<li>Include combinations that do not have (at least) two of $A, B, C$ = $ (C+D)^8 + (B+D)^8 + (A+D)^8$</li>
<li>Exclude combinations that do not have three of $A, B, C$ = $ D^8$</li>
</ol>
<p>The inclusion/exclusion principle lies in the fact that if you exclude and item because it does not have an $A$, but it also does not have a $B$, you have to include it again.</p>
|
1,548,130 | <p>Let's say that $F$ is a nice well-behaved function. How would I compute the following derivative?</p>
<p>$\frac{\partial}{\partial t} \left\{ \int_{0}^{t} \int_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi d\eta \right\}$</p>
<p>I'm guessing I need the fundamental theorem of calculus, but the double integral is REALLY throwing me off - especially that the $t$ is contained in the limits of both integrals. Can someone help me out?</p>
<p>EDIT: Given the problem that this came up in, I have hunch that the above derivative is zero.</p>
| Ekaveera Gouribhatla | 31,458 | <p>since $gof$ is Surjective for every $y \in A$ $\exists$ $x \in A$ such that</p>
<p>$$gof(x)=y$$ $\implies$</p>
<p>$$g(f(x))=y \tag{1}$$</p>
<p>But since we know that $g$ is surjective, for every $y \in A$ $\exists$ $z \in B$ such that</p>
<p>$$g(z)=y \tag{2}$$ Using $(2)$ in $(1)$ we get</p>
<p>$$g(f(x))=g(z)$$ But since $g$ is Injective</p>
<p>$$f(x)=z$$ which proves $f$ is Surjective.</p>
|
350,747 | <p>Base case: $n=1$. Picking $2n+1$ random numbers 5,6,7 we get $5+6+7=18$. So, $2(1)+1=3$ which indeed does divide 18. The base case holds.
Let $n=k>=1$ and let $2k+1$ be true. We want to show $2(k+1)+1$ is true. So, $2(k+1)+1=(2k+2) +1$....</p>
<p>Now I'm stuck. Any ideas?</p>
| Taladris | 70,123 | <p>A variation on Marvis answer: let $a$ be the middle number. So the numbers are </p>
<p>$$
a-n,\dots,a-2,a-1,a,a+1,a+2,\dots,a+n
$$</p>
<p>So their sum is clearly $(2n+1)a$ which is divisible by $2n+1$.</p>
|
1,955,591 | <p>I have to prove that ' (p ⊃ q) ∨ ( q ⊃ p) ' is a tautology.I have to start by giving assumptions like a1 ⇒ p ⊃ q and then proceed by eliminating my assumptions and at the end i should have something like ⇒(p ⊃ q) ∨ ( q ⊃ p) but could not figure out how to start.</p>
| Doug Spoonwood | 11,300 | <p>Start off by showing the law of the excluded middle (p V $\lnot$ p).</p>
<p>Put the proof of law of excluded middle here such that it ends as follows:</p>
<p>(p V $\lnot$ p)</p>
<pre><code>Hypothesize p
Hypothesize r
.
.
.
p
(r -> p)
((p -> q) V (r -> p))
</code></pre>
<p>(p -> ((p -> q) V (r -> p)))</p>
<pre><code> Hypothesize ~p
hypothesize p
.
.
.
q
(p -> q)
((p -> q) V (r -> p))
</code></pre>
<p>(~p -> ((p -> q) V (r -> p)))</p>
<p>By disjunction elimination, </p>
<p>((p -> q) V (r -> p)).</p>
<p>Therefore,</p>
<p>((p -> q) V (q -> p)).</p>
<p>End of proof.</p>
<p>One might see that:</p>
<p>(q -> ((p -> q) V (r -> p)))</p>
<p>is a theorem also.</p>
<p>In Polish notation one can write:</p>
<p>CqACpqCrp. This is a special case of a more general tautology:</p>
<p>CpACqrCsq. If you did all of calculations involving CpACqrCsq and CpACpqCrp you would find that the calculations involving CpACqrCsq has all the instances of CqACpqCrp. Additionally, say we start with </p>
<ol>
<li>CpACqrCsq. </li>
</ol>
<p>Now, we'll put 'a' for every instance of 'p', 'b' for every instance of 'q', 'c' for every instance of 'r', and 'd' for every instance of 's' in CpACqrCsq. Doing that, we obtain:</p>
<ol start="2">
<li>CaACbcCdb. </li>
</ol>
<p>Now we put 'q' for 'a', 'p' for every instance of 'b', 'q' for every instance of 'c', and 'r' for 'd' obtaining:</p>
<ol start="3">
<li>CqACpqCrp.</li>
</ol>
<p>Thus, CqACpqCrp can get obtained from CpACqrCsq and using just a rule of substitution. But, we cannot obtain CpACqrCsq from CqACpqCrp, since CpACqrCsq has more distinct letters in it and they have the same length and the same pattern of big letters.</p>
<p>CpACqrCsp is also a tautology.</p>
<p>CpACpqCrp is also a tautology. </p>
<p>CNpACpqCrs is also a tautology.</p>
<p>CNpACqrCsq is also a tautology. </p>
|
96,468 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/22537/how-many-fixed-points-in-a-permutation">How many fixed points in a permutation</a> </p>
</blockquote>
<p>Suppose we have a collection of n objects, numbered from 1 to n. These objects are placed in a random order.</p>
<p>What is the probability that p of the objects are in the position in the order corresponding to their order. </p>
<p>For instance, </p>
<p>For n=3, 1-2-3 has all objects in the correct position, p = 3, and has probability P(p=3) = 1/3! = 1/6.</p>
<p>However P(p=2) = 0</p>
<p>P(p=1) = 3/3! = 1/2. (1-3-2, 3-2-1, 2-1-3)</p>
<p>P(p=0) = 2/3!. (2-3-1, 3-1-2)</p>
| yoyo | 6,925 | <p>if a permutation $\pi\in S_n$ has exactly $k$ fixed points, you can consider it a permutation in $S_{n-k}$ with no fixed points (a <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">derangement</a>). so you choose your fixed points ${n \choose k}$ and multiply by the number of derangements on $n-k$ elements, $(n-k)!\sum_{i=0}^{n-k}(-1)^i/i!$, then divide by $n!$ for the probability
$$
\frac{1}{k!}\sum_{i=0}^{n-k}\frac{(-1)^i}{i!}
$$
so for $n=3$ as you gave you get
$$
k=0:1/3, k=1: 1/2, k=2: 0, k=3: 1/6
$$
as desired (as a check i guess)</p>
|
281,450 | <p>How can I find the area bound by $\;x=0,\, x=1,\;$ the $\;x$-axis ($y = 0$) and $\;y=x^2+2x\;$ using Riemann sums? </p>
<p>I want to use the right-hand sum. Haven't really found any good resources online to explain the estimation of areas bounded by curves, hoping anyone here can help?</p>
<p>By the way, I would like there to be 100 intervals.</p>
| Adi Dani | 12,848 | <p>$$(1 + x + x^{10})^{20}=((1+(x+x^{10}))^{20}=\sum_{k=0}^{20}\binom{20}{k}(x+x^{10})^k=$$
$$=\sum_{k=0}^{20}\binom{20}{k}x^k(1+x^{9})^k=\sum_{k=0}^{20}\binom{20}{k}x^k\sum_{j=0}^{k}\binom{k}{j}x^{9j}=\sum_{k=0}^{20}\sum_{j=0}^{k}\binom{20}{k}\binom{k}{j}x^{k+9j}$$ let $0\leq k+9j=t\leq200\Rightarrow j=\frac{t-k}{9},k=t-9j$
$$\sum_{k=0}^{20}\sum_{j=0}^{k}\binom{20}{k}\binom{k}{j}x^{k+9j}=\sum_{k=0}^{20}\sum_{t=0}^{200}\binom{20}{k}\binom{k}{\frac{t-k}{9}}x^{t}=\sum_{t=0}^{200}\sum_{j=0}^{20}\binom{20}{t-9j}\binom{t-9j}{j}x^{t}$$
$\binom{k}{\frac{t-k}{9}}=0$ if $\frac{t-k}{9}$ is not integer</p>
|
121,546 | <p>Consider the 2-Wasserstein distance between probability measures $\mu$ and $\nu$ (on $\mathbb{R}^d$), defined as
$$
d_{W_2}(\mu,\nu) = \inf_{\gamma} \Big[\int \|x-y\|^2 d\gamma(x,y)\Big]^{1/2}
$$
where the $\inf$ is over all couplings $\gamma$ of $\mu$ and $\nu$. Can we define a norm (or something norm-like) on the space of signed measures (or a linear subspace of it containing the cone of probability measures) which gives rise to $W_2$ for probability measures. (I suppose not, but why?)</p>
<p>If not, can we approximate $d_{W_2}$ by a norm?</p>
| mcuturi | 9,804 | <p>This paper has several links to relevant literature by Kantorovich & Rubinstein who define an OT inspired norm for signed measures.</p>
<p><a href="https://hal.inria.fr/inria-00072186/en" rel="nofollow">https://hal.inria.fr/inria-00072186/en</a></p>
|
121,546 | <p>Consider the 2-Wasserstein distance between probability measures $\mu$ and $\nu$ (on $\mathbb{R}^d$), defined as
$$
d_{W_2}(\mu,\nu) = \inf_{\gamma} \Big[\int \|x-y\|^2 d\gamma(x,y)\Big]^{1/2}
$$
where the $\inf$ is over all couplings $\gamma$ of $\mu$ and $\nu$. Can we define a norm (or something norm-like) on the space of signed measures (or a linear subspace of it containing the cone of probability measures) which gives rise to $W_2$ for probability measures. (I suppose not, but why?)</p>
<p>If not, can we approximate $d_{W_2}$ by a norm?</p>
| leo monsaingeon | 33,741 | <p>For the Earth-mover <span class="math-container">$W_1$</span> distance (based upon the cost function <span class="math-container">$c(x,y)=\|x-y\|$</span>) this is exactly the purpose of <a href="https://arxiv.org/abs/1910.05105" rel="nofollow noreferrer">this paper</a>. Note however that their construction does <strong>not</strong> work for the quadratic cost <span class="math-container">$c(x,y)=\|x-y\|^2$</span> as you seem to be hoping for.</p>
|
1,297,863 | <p>Is it possible to write the following function:
$$
f(x) = \begin{cases}
\frac{x-\sin x}{1- \cos x}& x\neq 0\\
0 & x=0
\end{cases}
$$
as a composition of elementary functions (including $\mathrm{sinc} (x) = (\sin x) / x)$ so that I get not large numerical errors for $x$ close to zero?</p>
<p>This is the complete list of functions I can use: <a href="http://docs.scipy.org/doc/numpy/reference/routines.math.html" rel="nofollow">http://docs.scipy.org/doc/numpy/reference/routines.math.html</a></p>
<p>This formula is used to compute the area of a circular segment with fixed chord length and given angle.</p>
<p><strong>addendum</strong></p>
<p>I found I can write:
$$
f(x) = \frac{\frac{x}{\sin x} - 1}{x} \frac{x}{\sin x}.
$$
But this is not resolutive. Seems to me that the derivative of the $\mathrm{sinc}$ function cannot be explicitly written in terms of the extended elementary function listed in the link above.</p>
| Claude Leibovici | 82,404 | <p>In the same spirit as Yves Daoust, for more accuracy than using Taylor series, you could use Pade approximants. The simplest one would be $$\frac{x-\sin x}{1- \cos x}\approx \frac{x \left(420-x^2\right)}{45 \left(28-x^2\right)}$$ The error is extremely small : $\approx 10^{-15}$ over the interval $(-0.01,0.01)$. </p>
<p>Another could be $$\frac{x-\sin x}{1- \cos x}\approx \frac{x \left(x^4+12600\right)}{1260 \left(30-x^2\right)}$$</p>
|
1,297,863 | <p>Is it possible to write the following function:
$$
f(x) = \begin{cases}
\frac{x-\sin x}{1- \cos x}& x\neq 0\\
0 & x=0
\end{cases}
$$
as a composition of elementary functions (including $\mathrm{sinc} (x) = (\sin x) / x)$ so that I get not large numerical errors for $x$ close to zero?</p>
<p>This is the complete list of functions I can use: <a href="http://docs.scipy.org/doc/numpy/reference/routines.math.html" rel="nofollow">http://docs.scipy.org/doc/numpy/reference/routines.math.html</a></p>
<p>This formula is used to compute the area of a circular segment with fixed chord length and given angle.</p>
<p><strong>addendum</strong></p>
<p>I found I can write:
$$
f(x) = \frac{\frac{x}{\sin x} - 1}{x} \frac{x}{\sin x}.
$$
But this is not resolutive. Seems to me that the derivative of the $\mathrm{sinc}$ function cannot be explicitly written in terms of the extended elementary function listed in the link above.</p>
| Doggyshakespeare | 243,086 | <p>You can use
$$\frac{x}{\sin ^2x}+\frac{x}{\tan \left(x\right)\sin \left(x\right)}-\frac{1}{\sin \left(x\right)}-\frac{1}{\tan \left(x\right)}$$
which is equal to the original function, though it is still undefined at 0. It is also equal to
$$\frac{x+x\cos \left(x\right)-\sin \left(x\right)-\sin \left(x\right)\cos \left(x\right)}{\sin ^2\left(x\right)}$$</p>
|
3,075,979 | <p>Prove that <span class="math-container">$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$</span> is an integer using mathematical induction.</p>
<p>I tried using mathematical induction but using binomial formula also it becomes little bit complicated.</p>
<p>Please show me your proof.</p>
<p>Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.</p>
| pwerth | 148,379 | <p>@I like Serena has a great answer but since the OP asked for a proof by induction, I'll show what that would look like. Define
<span class="math-container">$$f(k)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{15k^7 + 21k^5+70k^3-k}{105}$$</span></p>
<p>For our base case, let <span class="math-container">$k=1$</span>. Then we have
<span class="math-container">$$f(1)=\frac{15+21+70-1}{105}=1$$</span>
which is an integer. Now suppose <span class="math-container">$f(k)$</span> is an integer for some <span class="math-container">$k\geq 1$</span>. We want to prove that <span class="math-container">$f(k+1)$</span> is also an integer. To that end, observe that
<span class="math-container">\begin{align}
f(k+1)&=\frac{15(k+1)^7 + 21(k+1)^5+70(k+1)^3-(k+1)}{105}\\
&=\frac{15k^7 + 105k^6+336k^5+630k^4 + 805k^3+735k^2+419k+105}{105}
\end{align}</span>
Therefore
<span class="math-container">\begin{align}
f(k+1)-f(k)&=\frac{105k^6+315k^5+630k^4+735k^3+735k^2+420k+105}{105}\\
&=\frac{105(k^6+3k^5+6k^4+7k^3+7k^2+4k+1)}{105}\\
&= k^6+3k^5+6k^4+7k^3+7k^2+4k+1
\end{align}</span>
Which is an integer, say <span class="math-container">$N$</span>. Rearranging this gives <span class="math-container">$f(k+1)=f(k)+N$</span> and since <span class="math-container">$f(k)$</span> is assumed to be an integer from the induction hypothesis, <span class="math-container">$f(k+1)$</span> is the sum of two integers, hence an integer.</p>
|
7,110 | <p>In 1974, Aharoni proved that every separable metric space (X, d) is Lipschitz isomorphic to a subset of the Banach space c_0.
Thus, for some constant L, there is a map K: X --> c_0 that satisfies the inequality d(u,v) <= || Ku - Kv || <= Ld(u,v) for all u and v in X.
Now, suppose X = l_1 (in this case, L = 2 is best possible). I have the following</p>
<p><strong>Conjecture</strong>: Let K: l_1 --> c_0 be a Lipschitz embedding. Then K cannot be <em>monotone</em> w.r.t. the natural duality pairing (.,.) between l_1 and c_0,
i.e., there are some u and v in l_1 such that (u - v, Ku - Kv) < 0.</p>
| Bill Johnson | 2,554 | <p>Ady, this could be a hard problem. Why are you interested in the answer?</p>
|
2,461,962 | <p>I'm looking to reproduce
\begin{align}
\partial_{j_m,j_n}\bigg|_{\bf{j}=0}\exp\left(\frac12\bf{j}^\top\bf{B}\bf{j}\right) = B_{mn}
\end{align}
where $B_{mn}=B_{nm}$ is a real, symmetric, positive-definite $N\times N$ matrix. I have tried the following, and I know this is incorrect due to the surplus of indices.
\begin{align}
\partial_{j_m,j_n}\bigg|_{\bf{j}=0}\exp\left(\frac12j_mB_{mn}j_n\right)
&= \left[\partial_{j_m} \frac12B_{mn}j_m\exp\left(\frac12j_mB_{mn}j_n\right)\right]_{\bf{j}=0}
\\
&= \left[\frac12B_{mn}\exp\left(\frac12j_mB_{mn}j_n\right) + \frac14B_{mn}j_mB_{mn}j_n\exp\left(\frac12j_mB_{mn}j_n\right)\right]_{\bf{j}=0}
\\
&= \frac12B_{mn}
\end{align}</p>
<p>I'm naively working with the indices it seems. Can someone clarify my mistakes?</p>
<p>(PS. Please edit the title to something more descriptive if you can think of something.)</p>
<p>With Jiaqi Li's answer, I think I've understood how to go about this:
\begin{align}
\partial_{j_m,j_n}\exp\left(\frac12j_rB_{rs}j_s\right)
&= \frac12\partial_{j_m}\left[\left(B_{ns}j_s+j_rB_{rn}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \right]
\\
&= \frac12\left[\left(B_{nm}+B_{mn}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \\
\qquad\qquad\qquad\qquad+ \frac12\left(B_{ns}j_s+j_rB_{rn}\right)\left(B_{ms}j_s+j_rB_{rm}\right)\exp\left(\frac12j_rB_{rs}j_s\right) \right]
\\
\end{align}</p>
<p>This appears to evaluate to the desired result for $\bf{B}$ symemtric. The two major mistakes were the initial usage of the same indices inside the exponential and forgetting that $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$.</p>
| Jiaqi Li | 480,797 | <p>The way you work with indices are incorrect. It should be
$$\partial_{j_m,j_n}\bigg|_{\mathbf{j=0}}\exp\left(\frac12\bf{j}^\top\bf{B}\bf{j}\right)=\partial_{j_m,j_n}\bigg|_{\mathbf{j=0}}\exp\left(\frac12j_rB_{rs}j_s\right)$$
since when a index appears twice, it means summation over all possible values. See the wikipedia page <a href="https://en.wikipedia.org/wiki/Einstein_notation" rel="nofollow noreferrer">Einstein notation</a>. So in the $\exp()$ function, you use indices to imply summation. But when taking the partial derivative, you don't want summation over all possible values for $m,n$. Therefore, you need to use different indices for the summation and the partial derivative.</p>
|
170,830 | <p>I have two circles with the same radius and I want to calculate the points of tangency. </p>
<p>For example, in the picture below, I want to calculate $(x_3, y_3)$ and $(x_4,y_4)$. I have the radius and the distance between the two circles as shown below:</p>
<p><img src="https://i.stack.imgur.com/tQ2qu.png" alt=""></p>
| user29999 | 29,999 | <p>\begin{eqnarray}
y_3 = y_1 + R \cos \theta \\
x_3= x_1 + R \sin \theta \\
y_4= y_2 +R \cos \theta \\
x_4 = x_2 +R \sin \theta \\
\end{eqnarray}</p>
|
458,779 | <p>I need to find the volume of an object restricted with the $x^{2}+z^{2} < 8$ and $0 < y < 2$ planes. It would be easy if the cylinder were "parallel" to the XY plane, because then:</p>
<p>$$0 < r < 2\sqrt{2}$$
$$0 < \phi < 2\pi$$
$$0 < z < 2$$</p>
<p>But well, how should I handle this here?</p>
| amWhy | 9,003 | <p>Think of this as a cylinder whose base lies in the $x$-$z$ plane, encompassed by the region $x^{2}+z^{2} < r^2 = 8$ and $0 < y < 2$, so that:</p>
<p>$$0 < r < 2\sqrt{2}$$
$$0 < \phi < 2\pi$$
$$0 < y < 2$$</p>
|
2,554,448 | <p>Beside using l'Hospital 10 times to get
$$\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} = 3$$ and lots of headaches, what are some elegant ways to calculate the limit?</p>
<p>I've tried to write the functions as powers of $e$ or as power series, but I don't see anything which could lead me to the right result.</p>
| Kenny Lau | 328,173 | <p>$$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\
=& 3
\end{array}$$</p>
|
1,457,623 | <p>A floor is paved with rectangular marble blocks,each of length $a$ and breadth $b$.A circular block of diameter $c(c<a,b)$ is thrown on the floor at random.Show that the chance that it falls entirely on one rectangular block is $\frac{(a-c)(b-c)}{ab}$<br></p>
<p>I thought over this problem,i found total number of cases as area of rectangular marble block,but i cannot find the favorable number of cases.Favorable number of cases cannot be area of rectangular marble block$-$ area of circular block,as the answer suggests.What should be the correct logic,please help.</p>
| marty cohen | 13,079 | <p>The center of the disc
must be at least
c/2 from any of the edges.</p>
|
299,405 | <p>(a) In how many ways can the students answer a 10-question true false examination? </p>
<p>(b) In how many ways can the student answer the test in part (a) if it is possible to leave a question unanswered in order to avoid an extra penalty for a wrong answer</p>
<hr>
<p>For part (a) I've got the answer, it is $2^{10}$.</p>
<p>For part (b) I think the answer is $ 10 \times 2^9 $ because the number of ways to choose the question to answer is 10 and in each selection the number of ways to answer the question is $2^9$ but the answer provided in the book is $3^{10}$.</p>
<p>Can someone explain to me?</p>
| Peter Phipps | 15,984 | <p>In part b), for simplicity, let's reduce the problem to just two questions. </p>
<p>There are $2^2$ ways in which both questions may be answered true/false. </p>
<p>If a student does not answer question 1, there are still $2^1$ ways in which they can answer question 2, and vice versa. Thus there are $2\times 2^1$ ways in which only one question is answered. </p>
<p>Finally, there is just one way in which neither question is answered. </p>
<p>Putting this together, there are $2^2 + 2\times 2^1 + 1 = (2+1)^2 = 3^2$ ways of answering the questions.</p>
<p>Extending this to ten questions, there are $2^{10}$ ways to answer all ten questions, $10\times 2^9$ for answering all but one question, $45 \times 2^8$ ways of answering all but two questions, etc, giving a total of $(2+1)^{10} = 3^{10}$.</p>
|
3,609,191 | <p>Actually I am not very comfortable with using blocks, I understand the definition that it is a maximal <span class="math-container">$2$</span>-connected graph, though.</p>
<p><strong>My attempt</strong> Suppose not. Then there exists a maximal graph <span class="math-container">$G$</span> which cannot be written as an edge disjoint union of its blocks.</p>
<p>I cannot go further. I am not even sure why maximal exists, because the number of vertices of <span class="math-container">$G$</span> are not predetermined here. I am not even sure how can we write complete graph in such a fashion.</p>
<p>Please help</p>
| Jonas Linssen | 598,157 | <p>I think we need maximality of the blocks rather than maximality of the graph here.</p>
<p>Assume there is a graph, which cannot be written as edge disjoint union of blocks, that is to say there are two <em>distinct</em> blocks <span class="math-container">$A,B$</span>, which share a common edge <span class="math-container">$xy$</span>. Without loss of generality(1) we can assume that <span class="math-container">$x$</span> has a neighbor <span class="math-container">$a$</span> in <span class="math-container">$A\setminus B$</span>. As <span class="math-container">$y$</span> and <span class="math-container">$a$</span> lie in <span class="math-container">$A$</span> we can find an <span class="math-container">$ay$</span>-path <span class="math-container">$p$</span> in <span class="math-container">$A$</span> avoiding <span class="math-container">$x$</span> and hence <span class="math-container">$xy$</span>. But this means that <span class="math-container">$B \cup p \cup xa$</span> is a 2-connected subgraph strictly containing <span class="math-container">$B$</span>, a contradiction.</p>
<p>(1) Take an edge <span class="math-container">$uv$</span> in the intersection and a vertex <span class="math-container">$w$</span> in <span class="math-container">$A\setminus B$</span>. By connectivity we have a <span class="math-container">$vw$</span>-path. Traveling along this path we find the first edge that leaves the intersection (it may go into <span class="math-container">$B\setminus A$</span> but that is fine: just swap the meaning of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>).</p>
|
84,605 | <p>Let $w$ be a word in letters $x_1,...,x_n$. A value of $w$ is any word of the form $w(u_1,...,u_n)$ where $u_1,...,u_n$ are words. For example, $abaaba$ is a value of $x^2$. A word $u$ is called unavoidable if every infinite word in a finite alphabet contains a value of $u$ as a subword. There is a nice characterization of unavoidable words due to Zimin. A word $u$ in $n$ letters is unavoidable if and only if a value of $u$ is a subword of the $n$th Zimin word $Z_n$ defined by induction: $Z_1=x_1$,...,$Z_n=Z_{n-1}x_nZ_{n-1}$, that is $Z_1=x_1, Z_2=x_1x_2x_1, Z_3=x_1x_2x_1x_3x_1x_2x_1,...$. Zimin words appear very often in algebra. For example, if one lists binary expressions of all numbers $1,2,3,...$ and records the numbers of 0s at the end of the numbers plus 1, one will get $12131214121...$ which is the infinite Zimin word. Values of Zimin words also appeared as $m$-sequences in Levitzki's description of Baer radical (see Jacobson's book "Structure of rings") and in the work of Schutzenberger. The Zimin words have obvious fractal structure, so these words could have appeared in other areas of mathematics as well. </p>
<p><b> Question. </b> Do Zimin words appear in your area of mathematics?</p>
<p>This might be a "big list" question. But I do not know how big the list is, it may be empty. If it turn out to be big, I will make the question "community wiki". </p>
<p><b> Update 1. </b> Googling 121312141213121 ($=Z_4$) returns 439 results including a discussion at reddit. </p>
<p><b> Update 2. </b> The most curious among these links is <a href="http://webcache.googleusercontent.com/search?q=cache:wJzPCA5sw0cJ:www.patents.com/us-4389538.html+&cd=1&hl=en&ct=clnk&gl=us">this link</a> to a US patent. It looks like Zimin words up to $Z_5$ at least have been patented before Zimin introduced them. </p>
<p><b> Update 3. </b> This is needed for the book with the current title "Words and their meaning" which we are writing together with Mikhail Volkov. There we already have four different applications of Zimin words to different areas of algebra and would like to mention applications outside algebra as well. </p>
| Petar Markovic | 26,899 | <p>I am about to go chair a dissertation defence where the candidate uses an endomorphic image of Zimin words (of his invention) to construct various counterexamples to questions in combinatorics of finite and infinite words, mainly to do with counting palindromes in subwords and scattered subwords of finite and infinite words. The image in question are words on a three-letter alphabet of the form w_0=a, w_1=aba, w_{n+1}=w_nc^nw_n, which the candidate Bojan Basic calls highly potential words. This is obviously a related sequence to Zimin words, and the candidate managed to construct several examples and counterexamples using these, including a counterexample to an already published "theorem" by non-unknown authors.</p>
<p>I suppose it qualifies as an application of Zimin words as highly potential ones are clearly homomorphic images of Zimin words and inspired by them. Don't know if it is too late for your book. Counter-question: has anybody seen these highly potential words before?</p>
<p>Petar Markovic</p>
|
1,050,382 | <p>In $\mathbb{R}^5$ there is given vector space $V$. Its dimension is 3. In $\mathbb{R}^{6,5}$ consider the subset $X = \{A \in \mathbb{R}^{6,5} : V \subset \ker A\}$. I have to show that $X$ is a vector space in $\mathbb{R}^{6,5}$ and find its dimension. To show that $X$ is vector space consider $x_1, x_2 \in X$ and $v \in V$. We know that $x_1 v = 0$ and $x_2 v = 0$ so $(\alpha x_1 + \beta x_2) v = \alpha (x_1 v) + \beta (x_2 v) = 0$ so $V \subset \ker (\alpha x_1 + \beta x_2)$. But I don't know how to find $X$'s dimension. Any ideas?</p>
| dami | 197,371 | <p>I understand $V \subseteq \mathbb{R}^5$ is a subspace, $\dim V = 3$ </p>
<p>$X = \{A \in \mathbb{R}^{6 \times 5} : V \subseteq \ker A \}$</p>
<p>To show that $X$ is a vector space, it suffices to show it is a subspace of $\mathbb{R}^{6 \times 5}$. </p>
<ol>
<li>$0 \in X$, clearly because $V \subseteq \mathbb{R}^5 = \ker 0$</li>
<li>For $\alpha_i \in \mathbb{R}$ and $A_i \in X$, if $v \in V$, $A_i v = 0$, and so $ (\sum_i \alpha_i A_i) v = 0$. </li>
</ol>
<p>So it is a subspace.</p>
<p>Let $B = \{v_1, v_2, v_3 \}$ be a basis of $V$, and extend it to a basis $\{v_1, v_2, v_3, v_4, v_5 \}$ of $\mathbb{R}^5$. As $Av_i = 0$ for $1 \leq i \leq 3$, you only have to say where goes $A v_i$ for $i=4$ and $5$.</p>
<p>So you have 5-3=2 degrees of freedom in the domain and 6 in the codomain, that gives $2 \cdot 6 = 12$. I suspect the dimension is 12. Hope that helps.</p>
<p>I add this: you can think of $A$ as the matrix representation of some linear transformation $f: \mathbb{R}^5 \to \mathbb{R}^6$ with respect to the bases the extended version of $B$ above for $\mathbb{R}^5$ and and the standard basis for $\mathbb{R}^6$, so $f(v_i) = 0$ for $1 \leq i \leq 3$, and you can decide where goes $f(v_i)$ for $i=4,5$.</p>
|
2,184,776 | <p>So there's an almost exact question like this here: </p>
<p><a href="https://math.stackexchange.com/questions/576268/use-a-factorial-argument-to-show-that-c2n-n1c2n-n-frac12c2n2-n1#576280">Use a factorial argument to show that $C(2n,n+1)+C(2n,n)=\frac{1}{2}C(2n+2,n+1)$</a></p>
<p>However, I'm getting stuck in just figuring out the lcds for the factorials.</p>
<p>I end up with this after the <strong>CNR</strong>:</p>
<p>$$\frac{(2n)!}{(n-1)!(n+1)!} + \frac{(2n)!}{n!n!}$$</p>
<p>When I try to find the common denominator, I do:</p>
<p>$$\frac{(2n)!n}{(n-1)!(n+1)n!n} + \frac{(2n)!(n+1)}{n(n-1)!n!(n+1)}$$</p>
<p>Putting it together I get:</p>
<p>$$\frac{(2n)!(n) + (2n)!(n+1)}{ (n)(n+1)(n-1)!n!}$$</p>
<p>Which is wrong because according to the other answer, it should be:</p>
<p>$$\frac{(2n+1)!}{n!(n+1)!}$$</p>
<p>Not sure how they got there. I guess that's my question, how did they get that?</p>
<p>I've been googling for hours on how to find common denominators of factorials but can't seem to find anything. I mean, what happened to the $(n-1)!$ ?</p>
<p>Thanks.</p>
| Giuseppe | 243,672 | <p>You have it right!</p>
<p>Notice that</p>
<p>$$\frac{(2n)! n + (2n)!(n+1)}{(n+1)n(n-1)!n!} = \frac{(2n)!(2n+1)}{[(n+1)n!][n(n-1)!]}$$</p>
<p>Do you see any simplifications?</p>
|
2,792,770 | <p>I found the following question in a test paper:</p>
<blockquote>
<p>Suppose $G$ is a monoid or a semigroup. $a\in G$ and $a^2=a$. What can we say
about $a$?</p>
</blockquote>
<p>Monoids are associative and have an identity element. Semigroups are just associative. </p>
<p>I'm not sure what we can say about $a$ in this case other than that $a$ could be other things apart from the identity. Any idea if there's a definitive answer to this question?</p>
| BCLC | 140,308 | <p>Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space for a fair coin flip $(\{H,T\}, 2^{\Omega}, \mathbb{P}(\omega)=\frac12)$, and let $(\Omega',\mathcal{F}')$ be a (probable?) measurable space for a die roll $(\{1,2,3,4,5,6\}, 2^{\Omega'})$. Let $X(\omega) = 1_{H}(\omega)$, a random variable indicating payoff of flipping heads in the fair coin.T hen $X$ is $\mathcal{F}$-measurable but not $\mathcal{F'}$ measurable.</p>
|
2,450,245 | <p>A set $Q$ contains $0$, $1$ and the average of all elements of every finite non-empty subset of $Q$. Prove that $Q$ contains all rational numbers in $[0,1]$.</p>
<p>This is the exact wording, as it was given to me.
Obviously, the elements that correspond to the average, are rational, since they can be expressed as $\frac{(q_1+q_2+\dots q_k)}{k}$, where $0\leq k\leq1$. But I don't know how to proceed. </p>
| Christian Blatter | 1,303 | <p>By successively taking midpoints we can produce all numbers of the form
$${k\over2^n}\qquad(n\geq0, \ 0\leq k\leq 2^n)\ .$$
Now let two numbers $0<p<q$ be given. Write $q-p=:p'$ for simplicity. Choose a sufficiently large $n$ (see below), and put
$$k_i:=i-1\quad(0< i\leq p'),\qquad k_{q-i}:=2^n-i\quad(0\leq i<p)\ .$$
One then has
$$\sum_{i=1}^q k_i=p\cdot 2^n+{(p'-1)p'\over2}-{(p-1)p\over2}=:p\cdot 2^n+\Delta\ .$$
Note that $|\Delta|<{q^2\over2}$. If $\Delta>0$ replace the <em>first</em> "large" $k_i\,$, namely $k_{p'+1}$, by $\ k_{p'+1}^{\rm new}:=k_{p'+1}-\Delta$. If $n$ is large enough then one still has $k_{p'+1}^{\rm new}>k_{p'}$. Make a similar correction if $\Delta<0$. The numbers
$$x_i:={k_i\over 2^n}\in Q\quad(1\leq i\leq q)$$
now form an increasing sequence of $q$ <em>different</em> numbers, and one has
$${1\over q}\sum_{i=1}^q x_i={1\over q\cdot 2^n}\sum_{i=1}^q k_i={p\cdot 2^n\over q\cdot 2^n}={p\over q}\ ,$$
hence ${p\over q}\in Q$.</p>
|
665,596 | <p>Let $b_n$ be the number of lists of length $100$ from the set $\{0,1,2\}$ such that the sum of their entries is $n$. How does $b_{198}$ equal ${100\choose 2}+100$?</p>
| Karthik C | 35,357 | <p>$b_{198}$ calculation:</p>
<p>99 twos and a zero - $100$</p>
<p>98 twos and 2 ones - $^{100}C_2$</p>
|
1,929,698 | <p>Let $f(x)=\chi_{[a,b]}(x)$ be the characteristic function of the interval $[a,b]\subset [-\pi,\pi]$. </p>
<p>Show that if $a\neq -\pi$, or $b\neq \pi$ and $a\neq b$, then the Fourier series does not converge absolutely for any $x$. [Hint: It suffices to prove that for many values of $n$ one has $|\sin n\theta_0|\ge c \gt 0$ where $\theta_0=(b-a)/2.$]</p>
<p>However, prove that the Fourier series converges at every point $x$.</p>
<p>I've computed the Fourier series and got $\frac{b-a}{2\pi}+\sum_{n\neq 0}\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}.$</p>
<p>Also, $|e^{-ina}-e^{-inb}|=2|\sin n\theta_0|$, and $\theta_0\in (0,\pi)$, so I can see that for infinitely many values of $n$, we have $|\sin n\theta_0|\ge c \gt 0$. But this does not guarantee $\sum_{n\neq 0}|\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}|\ge \sum \frac{c}{n}$, and in fact we might have this inequality only for the squares of integers, in which case the right hand side converges. So how does the hint solve the problem?</p>
<p>Moreover, for the second problem, to show that the Fourier series converges at every point, I think I need to use Dirichlet's test, using $1/n$ as the decreasing sequence to $0$, but how can I show that $\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}$ has bounded partial sums?</p>
<p>I would greatly appreciate any help.</p>
| ThePhantomE | 820,746 | <p>I see that both of the previous answers to part (b) of this exercise do not fully use the hint in the statement, so I'll attempt to do so. For part (c), the first few paragraphs of <a href="https://math.stackexchange.com/a/3021029/820746">fonini's answer</a> should suffice.</p>
<p>We need to show that the series
<span class="math-container">$$
\sum_{n\ne0} \left| \frac{e^{-ina}-e^{-inb}}{2\pi in} e^{inx} \right|
$$</span>
converges. Rewriting the expression in terms of <span class="math-container">$\theta_0=(b-a)/2$</span> by factoring out <span class="math-container">$e^{-in\frac{1}{2}(a+b)}$</span> and recalling that <span class="math-container">$|e^{in\varphi}|=1$</span> for any <span class="math-container">$\varphi$</span> gives us
<span class="math-container">$$
\sum_{n\ne0} \left| \frac{\sin n\theta_0}{\pi n} \right|
= \frac{2}{\pi} \sum_{n=1}^\infty \frac{|\sin n\theta_0|}{n},
$$</span>
since the value of the argument of the sum does not change if <span class="math-container">$n\to-n$</span>.</p>
<p><strong>Now comes the part of the proof where we use the 'for many values of <span class="math-container">$n$</span> one has …' part of the hint.</strong> To prove convergence, it suffices to show that in a <em>countably infinite</em> number of terms, <span class="math-container">$|\sin n\theta_0| \ge c > 0$</span>, for if so, the sum of all these terms diverges by comparison with <span class="math-container">$\sum_{n=1}^\infty \frac{c}{n}$</span>, which, in turn, implies divergence of our expression by comparison with this smaller series (insert zeroes in between the terms of the smaller series so that the identical terms in both series have the same index, i.e. match up).</p>
<p>To do this, we show that, for all integers <span class="math-container">$k>0$</span>, we can choose another integer <span class="math-container">$n_k>0$</span> such that the desired property holds when <span class="math-container">$n=n_k$</span>, dependent upon the value of <span class="math-container">$k$</span>. First note that <span class="math-container">$0<\theta_0<\pi$</span>, so there exists <span class="math-container">$0<c<1$</span> such that <span class="math-container">$\pi-\theta_0 < 2 \arcsin c$</span>. (<em>The choice of <span class="math-container">$2\arcsin c$</span> in favour of <span class="math-container">$\arcsin c$</span> is done to facilitate the key argument below.</em>) We claim that, for each <span class="math-container">$k$</span>, there exists a positive integer <span class="math-container">$n_k$</span> such that
<span class="math-container">$$
n_k\theta_0 \in \big( (k-1)\pi + \arcsin c , k\pi - \arcsin c \big),
$$</span>
the interval on the right hand side we denote as <span class="math-container">$I_k$</span>. For if so, we have <span class="math-container">$(k-1)\pi + \arcsin c < n_k\theta_0 < k\pi - \arcsin c$</span>, <strong>which may be rearranged to give the desired condition</strong>. (Note that <span class="math-container">$|\sin(n_k\theta_0-(k-1)\pi)|=|\sin(n_k\theta_0-k\pi)|=|\sin(n_k\theta_0)|$</span>.)</p>
<p>The claim can be proved as follows. (It helps to have a sketch of <span class="math-container">$y=|\sin x|$</span> on hand and draw on it as you read this argument.) Since <span class="math-container">$\theta_0<\pi$</span>, it is obvious (by contradiction; simple exercise) that, for all <span class="math-container">$k$</span>, each interval <span class="math-container">$((k-1)\pi,k\pi)$</span> includes at least one <span class="math-container">$n\theta_0$</span>. Now there are two cases: either <span class="math-container">$n\theta_0 \in I_k$</span> or <span class="math-container">$n\theta_0 \notin I_k$</span>. In the first case, we are done, and can set <span class="math-container">$n_k=n$</span>. In the second case, either <span class="math-container">$n\theta_0 \in ((k-1)\pi, (k-1)\pi+\arcsin c)$</span> or <span class="math-container">$n\theta_0 \in (k\pi-\arcsin c, k\pi)$</span>. But since <span class="math-container">$|I_k|>n_k\theta_0$</span>, in the first subcase <span class="math-container">$(n+1)\theta_0 \in I_k$</span> (so <span class="math-container">$n_k=n+1)$</span>, whereas in the second subcase <span class="math-container">$(n-1)\theta_0 \in I_k$</span> (so <span class="math-container">$n_k=n-1$</span>), and we are also done. (<em>Now you should be able to see why we chose <span class="math-container">$2\arcsin c$</span>.</em>) Therefore, each interval <span class="math-container">$((k-1)\pi,k\pi)$</span> (the number of which is countably infinite) can be assigned to an integer <span class="math-container">$n_k$</span>, so we conclude that there are infinitely many <span class="math-container">$n_k$</span>, such that if <span class="math-container">$n$</span> does not satisfy our desired condition, then either <span class="math-container">$n-1$</span> or <span class="math-container">$n+1$</span> will, i.e. there cannot be consecutive values of <span class="math-container">$n$</span> which do not satisfy the condition. We conclude that, in the worst case scenario where the <span class="math-container">$n_k$</span> are most sparsely distributed, either the <span class="math-container">$n_k$</span> are all the odd positive integers or <span class="math-container">$n_k$</span> are all the even positive integers. If we take the reciprocal of each and sum over each collection separately, both diverge (since they are related to the harmonic series).</p>
<p>Thus there exists an infinite number of <span class="math-container">$n_k$</span> such that <span class="math-container">$|\sin n_k\theta_0| \ge c$</span>, enumerated by the positive integers <span class="math-container">$k$</span>. This implies that
<span class="math-container">$$
\sum_{n=1}^\infty \frac{|\sin n\theta_0|}{n}
\ge \sum_{k=1}^\infty \frac{|\sin n_k\theta_0|}{n_k}
\ge \sum_{n\,\text{odd}} \frac{c}{n} \to +\infty,
$$</span>
by the arguments presented in paragraph 3. We are done.</p>
<p>[Note: The book claims to assume very few prerequisites, so in all candour, I don't know why the authors assigned this as Exercise 9(b) in a collection of 20 exercises without more hints, since this argument is somewhat intricate.]</p>
|
2,268,947 | <p><a href="https://i.stack.imgur.com/6hCd2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6hCd2.png" alt="enter image description here"></a></p>
<p>($\dot I = \{0,1\}$)</p>
<p>The homotopy I've constructed is:
$$G(t_1, t_2) =
\begin{cases}
\alpha(t_1), & \text{if $(t_1,t_2) \in I \times \{0\}$} \\
\gamma * \beta* \delta^{-1}(t_1) , & \text{otherwise}
\end{cases}$$</p>
<p>But this isn't correct since the domain of this function isn't a union of two open or closed sets as per the gluing lemma.</p>
<p>Anyone have any ideas as to how to re-write the domain so that this becomes the required homotopy? </p>
| Ronnie Brown | 28,586 | <p>Let $S$ be a simply connected space and let $g: S \to X$ be a map. Let $a,b: [0,1] \to S$ be paths in $S$ with the same end points. Then $ga,gb: [0,1] \to X$ are homotopic rel end points. </p>
<p><strong>Proof</strong> Since $S$ is simply connected there is a homotopy $H_t:a \simeq b$ rel end points. Then $gH_t:ga \simeq gb$. </p>
<p>In the case we need, $S$ is a square. It is convex and we can define $H_t(s)= (1-t)a(s) + t b(t)$. Then $H_0(s) =a(s), H_1(s) = b(s)$. </p>
|
2,249,109 | <p><strong>Question:</strong> Three digit numbers in which the middle one is a perfect square are formed using the digits $1$ to $9$.Then their sum is?</p>
<p>$A. 134055$<br>
$B.270540$<br>
$C.170055$<br>
D. None Of The Above</p>
<p>Okay, It's pretty obvious that the number is like $XYZ$ where $X,Z\in[1,9] $ and $Y\in\{1,4,9\}$</p>
<p>I'm facing problems in finding a method to evaluate such a sum?</p>
<p>(obviously, I can't afford to use a calculator)</p>
| Cye Waldman | 424,641 | <p>Here is a formal derivation of your result. The sequence you have found is a generalization of the Fibonacci sequence.</p>
<p>There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at <a href="http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf" rel="nofollow noreferrer">http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf</a>.)</p>
<p>We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that</p>
<p>$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n) $$</p>
<p>where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.</p>
<p>The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$.</p>
<p>So, specializing to your case, we can say</p>
<p>$$E_n=\left(B-\frac{A}{2} \right)F_n+\frac{A}{2}L_n$$</p>
<p>where</p>
<p>$$F_n,L_n=\frac{\varphi^n \mp \psi^n}{\varphi \mp \psi}\\
\varphi,\psi=(1\pm \sqrt{5})/2
$$</p>
<p>Ergo,</p>
<p>$$
E_n=BF_n+\frac{A}{2}(L_n-F_n)=AF_{n-1}+BF_n
$$</p>
<p>since $L_n-F_n=2F_{n-1}$.</p>
<p>This proves the OP's assertion.</p>
|
2,034,523 | <p>I am sure there is a general and simplified way to solve this problem, I am just unable to figure out the generalized formula (if there is one). </p>
<p>Say we have to write a <strong>code with 4 digits</strong>, the digits can range from <strong>0</strong> to <strong>9</strong>. </p>
<p>All digits in the code <strong>must be unique</strong>.<br>
All of the digits cannot be <strong>neither increasing nor decreasing</strong>. </p>
<p>For example, "1234" is not allowed, neither is "1289" nor "9821". </p>
<p>How many code combinations are there in total?</p>
| barak manos | 131,263 | <p>The total number of combinations is $\binom{10}{4}$.</p>
<p>For each combination there are $4!$ different arrangements.</p>
<p>Exactly $1$ of these arrangements is strictly increasing.</p>
<p>Exactly $1$ of these arrangements is strictly decreasing.</p>
<p>Hence the number of valid arrangements is $\binom{10}{4}\cdot(4!-1-1)$.</p>
|
277,135 | <p><code>Graphics[{{Blue, Line[{{-5, 3}, {5, 8}}]}, {Dashed, Arrow[{{0, 0}, P}]}, Red, Arrow[{{0, 0}, v}]}, Axes -> True, AxesLabel -> {x, y}]</code>
I've tried combining Graphics and Plot to manipulate my graphic but it says that I can't.</p>
<p><a href="https://i.stack.imgur.com/8slBl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8slBl.png" alt="enter image description here" /></a></p>
| Daniel Huber | 46,318 | <p>With the answer from user29378 you may get a solution that is close enough to what you want:</p>
<pre><code>d1 = {-1, -1, -1, 1, 1, 1};
d2 = {-2, -2, 2, 2, 2, 2};
d3 = {-4, 4, 4, 4, 4, 4};
dis1 = EmpiricalDistribution[d1];
dis2 = EmpiricalDistribution[d2];
dis3 = EmpiricalDistribution[d3];
</code></pre>
<p>To test this we may use RandomVariate to generate samples from the distribution, count how many are o.k. and calculate the probability:</p>
<pre><code>n = 10^6;
Count[RandomVariate[dis1, n], x_ /; x > 0]/n // N
Count[RandomVariate[dis1, n] + RandomVariate[dis2, n], x_ /; x > 0]/n // N
Count[RandomVariate[dis1, n] + RandomVariate[dis2, n] + RandomVariate[dis3, n], x_ /; x > 0]/n // N
</code></pre>
<p><a href="https://i.stack.imgur.com/SOhrf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SOhrf.png" alt="enter image description here" /></a></p>
|
2,876,050 | <p>I have There was a cowboy and 3 donkeys (Donkey A, Donkey B, Donkey C). The cowboy wears an eye cover and shoots randomly. What is the probability of donkey A to still be there after the cowboy shot 2 bullets? There is equal chances that the cowboy hits Donkey A, Donkey B, Donkey C or a Miss.</p>
<p>I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that donkey A still alive. But I want to solve it straight forward and directly. Is it possible? If yes, how?</p>
| mdnestor | 519,413 | <p>Here is a tree diagram representing the situation. For example, $$\text{Pr}(\text{Hit A on First shot)}=1/4$$ To solve the problem, add up all the cases where $\text{Donkey A}$ survives.
$$\text{Pr}(\text{Miss A on First shot and Second shot})$$
$$=\text{Pr}(\text{Hit B or C on First shot and Miss A on Second Shot})$$
$$+\text{Pr}(\text{Miss on First shot and Miss A on Second Shot})$$
$$=\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{4}\cdot\frac{3}{4}=\frac{25}{48}\approx0.52.$$</p>
<p><a href="https://i.stack.imgur.com/FLBr8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FLBr8.png" alt="enter image description here"></a></p>
<p>We must split into cases because the probability of hitting $\text{Donkey A}$ on the second shot depends on the outcome of the first shot. If the first shot is a miss, there is a $1/4$ probability $\text{Donkey A}$ is shot second. However if $\text{Donkey B or C}$ was shot first, then there is a $1/3$ probability $\text{Donkey A}$ is shot second</p>
|
265,067 | <p>$$\lim_{n\to\infty}\frac{(2n-1)!}{3^n(n!)^2}$$</p>
<p>How can I associate limit problem with series? And how can i find limits from series?
Can anyone help?</p>
| doniyor | 32,885 | <p>by ratio rule:<br>
$\dfrac{(2(n+1)-1)!}{3^{n+1}((n+1)!)^2}\cdot\dfrac{3^n(n!)^2}{(2n-1)!}= \dfrac{4n^2+2n}{3n^2+6n+3} \rightarrow \dfrac{4}{3}$</p>
<p>thus the series doesnot converge as the quotient and thus limsup is bigger than 1</p>
|
3,453,175 | <p>If <span class="math-container">$y=\dfrac {1}{x^x}$</span> then show that <span class="math-container">$y'' (1)=0$</span></p>
<p>My Attempt:</p>
<p><span class="math-container">$$y=\dfrac {1}{x^x}$$</span>
Taking <span class="math-container">$\ln$</span> on both sides,
<span class="math-container">$$\ln (y)= \ln \left(\dfrac {1}{x^x}\right)$$</span>
<span class="math-container">$$\ln (y)=-x.\ln (x)$$</span>
Differentiating both sides with respect to <span class="math-container">$x$</span>
<span class="math-container">$$\dfrac {1}{y}\cdot y'=-(1+\ln (x))$$</span></p>
| Contestosis | 462,389 | <p>As <span class="math-container">$x^x$</span> is <span class="math-container">$\exp(x \ln(x))$</span>, we have <span class="math-container">$y(x) = e ^ {- x \ln(x)}$</span>. It is easy to see that <span class="math-container">$y$</span> is indefinitely derivable over <span class="math-container">$\mathbb{R}^\star_+$</span>.</p>
<p>Thus, <span class="math-container">$y'(x) = - y(x) (1+ \ln(x))$</span> and <span class="math-container">$y''(x) = - \left( y'(x)(1+ \ln(x)) + y(x) \frac{1}{x} \right)$</span>.</p>
<p>We obtain:</p>
<p><span class="math-container">$y''(1) = - y'(1) - y(1) $</span> as <span class="math-container">$\ln(1) = 0$</span>.</p>
<p><span class="math-container">$y(1) = 1$</span> and <span class="math-container">$y'(1) = -y(1)(1+\ln(1)) = -1$</span>.</p>
<p>Finally we can say that <span class="math-container">$y''(1) = 0$</span>. </p>
|
2,807,478 | <p>If $f(x)$ and $g(x)$ are continuous at $x=c$ </p>
<p>then show that:</p>
<p>$h(x)=f(g(x))$ is also continuous at $x=c$.
(Given that $c$ belongs to the Domain of $h$)</p>
| drhab | 75,923 | <p>Let $g:\mathbb R\to\mathbb R$ be the function $x\mapsto x+1$ and let $f:\mathbb R\to\mathbb R$ be prescribed by $x\to x$ if $x<1$ and $x\mapsto x+1$ otherwise.</p>
<p>Then $g$ is continuous at every $0\in\mathbb R$ but can the same be said about $f\circ g$?</p>
|
2,807,478 | <p>If $f(x)$ and $g(x)$ are continuous at $x=c$ </p>
<p>then show that:</p>
<p>$h(x)=f(g(x))$ is also continuous at $x=c$.
(Given that $c$ belongs to the Domain of $h$)</p>
| Badr B | 350,060 | <p>Another counter example: let $f(x)=\frac{1}{x+1}$ and $g(x)=x-1$. Both are continuous at $x=0$, but $f \circ g$ is not. In order for your statement to be always true, then we need both functions to be continuous on $(-\infty, \infty)$ or for $f(c)=c$ and $g(c)=c$. </p>
|
1,703,120 | <p>So I have a vector <span class="math-container">$a =( 2 ,2 )$</span> and a vector <span class="math-container">$b =( 0, 1 )$</span>.<br />
As my teacher told me, <span class="math-container">$ab = (-2, -1 )$</span>.</p>
<p><span class="math-container">$ab = b-a = ( 0, 1 ) - ( 2, 2 ) = ( 0-2, 1-2 ) = ( -2, -1 )$</span><br />
<span class="math-container">$ab = a-b = ( 2 ,2 ) - ( 0 ,1 ) = ( 2-0,2-1 ) = ( 2 ,1 )$</span></p>
<p>Seems like its the same but the negative signs are gone.</p>
<p>Why do I have to subtract b from a to get ab? Why not a-b or a+b?</p>
| DonAntonio | 31,254 | <p>The correct thing is $\;b-a\;$ for the direction vector $\;\vec{ab}\;$. The substraction $\;a-b\;$ gives <em>the opposite</em> direction vector, namely $\;\vec
{ba}\;$</p>
|
66,009 | <p>Hi I have a very simple question but I haven't been able to find a set answer. How would I draw a bunch of polygons on one graph. The following does not work:</p>
<pre><code>Graphics[{Polygon[{{989, 1080}, {568, 1080}, {834, 711}}],
Polygon[{{1184, 1080}, {989, 1080}, {834, 711}, {958, 541}}],
Polygon[{{1379, 1080}, {1184, 1080}, {958, 541}, {1082, 370}}],
Polygon[{{1470, 1080}, {1379, 1080}, {1082, 370}, {1140, 291}}],
Polygon[{{1665, 1080}, {1470, 1080}, {1140, 291}, {1263, 120}}],
Polygon[{{1756, 1080}, {1665, 1080}, {1263, 120}, {1321, 41}}],
Polygon[{{1394, 0}, {1920, 0}, {1920, 1080}, {1846, 1080}}],
Polygon[{{1352, 0}, {1394, 0}, {1846, 1080}, {1756, 1080}, {1321,
41}}], Polygon[{{931, 0}, {1352, 0}, {1084, 367}}],
Polygon[{{736, 0}, {931, 0}, {1084, 367}, {961, 537}}],
Polygon[{{540, 0}, {736, 0}, {961, 537}, {836, 708}}],
Polygon[{{450, 0}, {540, 0}, {836, 708}, {779, 788}}],
Polygon[{{255, 0}, {450, 0}, {779, 788}, {654, 958}}],
Polygon[{{164, 0}, {255, 0}, {654, 958}, {597, 1038}}],
Polygon[{{73, 0}, {164, 0}, {597, 1038}, {568, 1080}, {524, 1080}}],
Polygon[{{0, 0}, {73, 0}, {524, 1080}, {0, 1080}}]}]
</code></pre>
<p>I apologize for how basic this question is. If anyone could steer me in the right direction it would be greatly appreciated.</p>
| Simon Woods | 862 | <p>Another possible workaround is to wrap <code>Dispatch</code> with a memoized function, so that both expressions <code>a</code> and <code>b</code> contain references to the same internal dispatch table.</p>
<p>i.e. define</p>
<pre><code>mem : disp[x_] := mem = Dispatch[x]
</code></pre>
<p>then use <code>disp</code> in place of <code>Dispatch</code> in your code.</p>
|
2,093,720 | <p>$$y~ dy+(2+x^2-y^2)dx$$</p>
<p>I try to solve this equation by putting standard form but becomes more challenge . So your answer is helpful </p>
| Yami Kanashi | 403,840 | <p>Applying AM-GM inequality
We get,
((abc)/(bca))^1/3 ≤ ((a/b) + (b/c) + (c/a))/3</p>
<p>1 ≤ ((a/b) + (b/c) + (c/a))/3</p>
<p>3≤ (a/b) + (b/c) + (c/a)</p>
<p>But since 2017/1000< 3
Therefore there does not exist any such triad.</p>
|
3,109,482 | <p><a href="https://i.stack.imgur.com/awp2x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/awp2x.png" alt="enter image description here"></a></p>
<p>I'm stumped on determining P(ABC) of Part A. My understanding is:</p>
<ul>
<li><p>Calculate the total number of patients (100)</p></li>
<li><p>Calculate individual <span class="math-container">$P(A), P(B),$</span> and <span class="math-container">$P(C)$</span> <span class="math-container">$(0.4; 0.35; 0.24 $</span> respectively)</p></li>
<li><p>Multiply <span class="math-container">$P(A)P(B)P(C) = ANS (0.0336)$</span></p></li>
</ul>
<p>This answer should be the same as <span class="math-container">$8/20$</span> or the population of serious patients, under 40, whose parents had diabetes but this fraction comes out to 0.08--what am I missing here? I appreciate your help.</p>
| John Wayland Bales | 246,513 | <p>You were close, but you chose a straight line segment. You want each point to be a unit distance from the origin. So, provided <span class="math-container">$A\ne-B$</span> you should divide each vector by its magnitude.</p>
<p><span class="math-container">$$R(x)=\frac{A+x(B-A)}{\Vert A+x(B-A)\Vert}$$</span></p>
<p>If <span class="math-container">$A=-B$</span> the straight line will pass through <span class="math-container">$(0,0,0)$</span> and you would be dividing by <span class="math-container">$0$</span>.</p>
|
3,353,826 | <p>All the vertices of quadrilateral <span class="math-container">$ABCD$</span> are at the circumference of a circle and its diagonals intersect at point <span class="math-container">$O$</span>. If <span class="math-container">$∠CAB = 40°$</span> and <span class="math-container">$∠DBC = 70°$</span>, <span class="math-container">$AB = BC$</span>, then find <span class="math-container">$∠DCO$</span>.</p>
| Arthur | 15,500 | <p>I have seen the notation <span class="math-container">$[x]$</span>. However, that is some times used as the floor function when TeX is unavailable, or the author is unfamiliar with it (I'm sure there are plenty of examples on this site, for instance).</p>
<p>The safest bet is to say something along the lines of</p>
<blockquote>
<p>Let <span class="math-container">$[x]$</span> mean the integer closest to <span class="math-container">$x$</span> (rounding up for half-integer values).</p>
</blockquote>
<p>or </p>
<blockquote>
<p>Let <span class="math-container">$[\phantom x]$</span> denote the standard rounding function.</p>
</blockquote>
<p>That is, explicitly defining the notation yourself, so that anyone who reads your text knows exactly what you're talking about. If you do this, you are of course entirely free to "invent" your own notation (within reason) for this if there is some other notation you prefer.</p>
|
3,353,826 | <p>All the vertices of quadrilateral <span class="math-container">$ABCD$</span> are at the circumference of a circle and its diagonals intersect at point <span class="math-container">$O$</span>. If <span class="math-container">$∠CAB = 40°$</span> and <span class="math-container">$∠DBC = 70°$</span>, <span class="math-container">$AB = BC$</span>, then find <span class="math-container">$∠DCO$</span>.</p>
| Especially Lime | 341,019 | <p>Whatever notation you use (punctured dusk gives some good suggestions), you should <em>always</em> define this explicitly if you are going to use it, since there is no standard way to treat half-integers. (I recently found this out the hard way when I assumed the rounding method I was always taught was standard, but python's default does something different.) </p>
|
3,353,826 | <p>All the vertices of quadrilateral <span class="math-container">$ABCD$</span> are at the circumference of a circle and its diagonals intersect at point <span class="math-container">$O$</span>. If <span class="math-container">$∠CAB = 40°$</span> and <span class="math-container">$∠DBC = 70°$</span>, <span class="math-container">$AB = BC$</span>, then find <span class="math-container">$∠DCO$</span>.</p>
| Greg Nisbet | 128,599 | <p>It might be too verbose, but something like the following is unlikely to be misinterpreted.</p>
<p><span class="math-container">$$\mathrm{RoundToEven}(5.5) = 6$$</span></p>
<p>If you need another convention such as rounding to the nearest odd number, rounding towards infinity, or rounding towards negative infinity I'd define my own function and include some examples.</p>
<p>For instance:</p>
<hr>
<p>Let <span class="math-container">$R \mathop: \mathbb{R} \to \mathbb{Z}$</span> denote the function that rounds each real number to the nearest integer, rounding ties towards negative infinity.</p>
<p><span class="math-container">$$ R(-0.5) = -1 $$</span>
<span class="math-container">$$ \left\{ R(-0.3)\;,\; R(0)\;,\; R(0.3)\;,\; R(0.5) \right\} = \left\{ 0 \right\} $$</span>
<span class="math-container">$$ R(1.5) = 1 $$</span>
<span class="math-container">$$ R(1.7) = 2 $$</span></p>
|
894,152 | <blockquote>
<p>Let $x_{i}\ge 0$ for $i\in\{1,2,\cdots,n\}$ and $x_{1}+x_{2}+\cdots+x_{n}=n$ for $n\ge 3$</p>
<p>Show that for all strictly positive integers $k\ge2$ the following inequality holds :
$$\sum_{i=1}^{n}x^k_{i}\ln{x_{i}}\ln{\dfrac{x_{i}}{n}}\le 0$$</p>
</blockquote>
<p>We consider
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
Then
$$f'(x)=kx^{k-1}\ln{x}\cdot\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+\dfrac{x^{k-1}}{n}\ln{x}$$
$$\Longrightarrow
f''(x)=k(k-1)x^{k-2}\cdot\ln{x}\cdot\ln{\dfrac{x}{n}}+kx^{k-2}\ln{\dfrac{x}{n}}+\dfrac{kx^{k-2}}{n}\ln{x}+(k-1)x^{k-2}\ln{\dfrac{x}{n}}+\dfrac{x^{k-2}}{n}+\dfrac{k-1}{n}x^{k-2}\ln{x}+\dfrac{x^{k-2}}{n}.$$
Unfortunatly I can't know the sign of $f''(x)$ because I want to use <a href="http://mathworld.wolfram.com/JensensInequality.html" rel="nofollow">Jensen's Inequality</a> to prove it.</p>
<p>So how can we prove this inequality ? </p>
| RE60K | 67,609 | <p>Firstly, you diffrentiated it wrongly:
$$f(x)=x^k\ln{x}\ln{\dfrac{x}{n}}$$
$$f'(x)=kx^{k-1}\ln{x}\ln{\dfrac{x}{n}}+x^{k-1}\ln{\dfrac{x}{n}}+x^{k-1}\ln{x}$$
$$f'(x)=x^{k-1}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-1}\left(\frac kx\ln\frac xn+\frac kx\ln x+\frac1x+\frac1x\right)$$
$$f''(x)=(k-1)x^{k-2}\left(k\ln x\ln\frac xn+\ln \frac xn+\ln x\right)+x^{k-2}\left(k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(k-1)\ln\frac xn+(k-1)\ln x+k\ln\frac xn+k\ln x+2\right)$$
$$f''(x)=x^{k-2}\left(k(k-1)\ln x\ln\frac xn+(2k-1)\ln\frac {x^2}n+2\right)$$</p>
<hr>
<p>As $k>2$, $x^{k-2}>0$ Also $x<n$.See this graph:
<img src="https://i.stack.imgur.com/YTZFV.png" alt="enter image description here"></p>
|
2,428,009 | <p>I want to solve the equation $2^n=2k$ for $n$ even with $n,k \in \Bbb{N}$. I'm not sure how to go about this, using logarithm makes me enter the reals.</p>
| ksoileau | 480,055 | <p>$2^n=2k$ implies $2^{n-1}=k,$ so the solution set (n,k) is (2 r,2^{2 r-1}) for $r \geqslant 1.$</p>
|
405,087 | <blockquote>
<p>Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$?</p>
</blockquote>
<p>So I started by saying that by the geometric series test where $a=x^2$ and $|r| = |\frac{1}{e^x}| \leq 1$, the series converges pointwise.</p>
<p>But how do I exactly prove that it converges uniformly? I am quite sure it is by weistrass test but I can not find an upper bound to compare it to! Any direction would be appreciated!</p>
| Community | -1 | <p>Let
$$f_n(x)=x^2e^{-nx}$$
then we have
$$f'_n(x)=e^{-nx}\left(2x-nx^2\right)=0\iff x=0\ \text{or}\ x=\frac{2}{n}$$
so
$$||f_n||_\infty=f_n\left(\frac{2}{n}\right)=\frac{4}{n^2}e^{-2}$$
hence the series $\displaystyle \sum_{n=1}^\infty ||f_n||_\infty$ is convergent and then the series $\displaystyle \sum_{n=1}^\infty f_n$ is uniformly convergent.</p>
|
405,087 | <blockquote>
<p>Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$?</p>
</blockquote>
<p>So I started by saying that by the geometric series test where $a=x^2$ and $|r| = |\frac{1}{e^x}| \leq 1$, the series converges pointwise.</p>
<p>But how do I exactly prove that it converges uniformly? I am quite sure it is by weistrass test but I can not find an upper bound to compare it to! Any direction would be appreciated!</p>
| Hans Engler | 9,787 | <p>You can compute the remainder term explicitly, using the formula for the geometric series:
$$ r_N(x) = \sum_{n=N}^\infty x^2 e^{-nx} = x^2e^{-Nx} \sum_{j = 0}^\infty e^{-jx} = \dots
$$
Now find the maximum of $r_N$ on $[0, \infty)$. (It's obviously a positive function.) </p>
<p>If this maximum tends to $0$ as $N \to \infty$, then the convergence is uniform. Otherwise it isn't. </p>
|
964,999 | <p>If A and B are two closed sets of $R$ is A.B closed?
By A.B I mean the set $\sum_{i=1}{^ n} a_ib_i$ where $a_i \in A,b_i\in B,n\in N$
How to view A.B geometrically?
I am new to this subject.Sorry if the question sounds something wrong</p>
| Chris Culter | 87,023 | <p>First, see user5527's answer.</p>
<p>Now, if $A$ and $B$ are closed <em>and bounded</em> subsets of $\mathbb R$, so that they're compact, then the answer is yes. The Cartesian product $A\times B$ is a compact subset of $\mathbb R^2$, and the set of products $A\cdot B$ is its image under a continuous map.</p>
|
377,354 | <p>I'm referencing this page: <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html#sqrtalgsalg" rel="nofollow">An Introduction to the Continued Fraction</a>, where they explain the algebraic method of solving the square root of $14$.</p>
<p>$$\sqrt{14} = 3 + \frac1x$$</p>
<p>So, $x_0 = 3$, Solving for $x$, we get</p>
<p>$$x = \frac{\sqrt{14} + 3}5$$</p>
<p>However, in the next step, how do we get the whole number $x_1$ = 1?</p>
<p>$$\frac{\sqrt{14} + 3}5 = 1 + \frac1x$$</p>
<p>My understanding is we would substitute for $x$ in the original equation for $\sqrt{14}$ whereas $$\sqrt{14} = 3 + \frac1{\frac{\sqrt{14} + 3}5}$$</p>
<p>Then substitute the $\sqrt{14}$ again here for $x = \frac{\sqrt{14} + 3}5$ to get the $x_1$ of the continued fraction? Am I just getting the algebra at this point wrong or am I botching steps? </p>
| genepeer | 50,955 | <p>$\sqrt{14}=3+\sqrt{14}-3=3+\frac{1}{\frac{\sqrt{14}+3}{5}}\implies x_0 = 3$</p>
<p>$\frac{\sqrt{14}+3}{5}=\frac{6+\sqrt{14}-3}{5}=1+\frac{\sqrt{14}-2}{5}=1+\frac{1}{\frac{\sqrt{14}+2}{2}} \implies x_1 = 1$</p>
<p>$\frac{\sqrt{14}+2}{2}=\frac{5+\sqrt{14}-3}{2}=2+\frac{\sqrt{14}-2}{2}=2+\frac{1}{\frac{\sqrt{14}+2}{5}} \implies x_2 = 2$</p>
<p>etc.</p>
|
2,950,813 | <blockquote>
<p>Take <span class="math-container">$G$</span> to be a group of order <span class="math-container">$600$</span>. Prove that for any element <span class="math-container">$a$</span> <span class="math-container">$\in$</span> G there exist an element <span class="math-container">$b$</span> <span class="math-container">$\in$</span> G such that <span class="math-container">$a = b^7$</span>. </p>
</blockquote>
<p>My thought process: since <span class="math-container">$a = b^7$</span> <span class="math-container">$\implies$</span> <span class="math-container">$|a| = |b^7|$</span>. Consequently
<span class="math-container">$\operatorname{lcm}(1,|a|) = \dfrac{1}{7}\operatorname{lcm}(7,|b|)$</span> implies <span class="math-container">$7|a| = 7|b|$</span> so <span class="math-container">$|a| = |b|$</span>. I don't know where I would go from here or if this is even the right approach. </p>
| Bill Dubuque | 242 | <p>First I show how to reduce it to computing <span class="math-container">$\,7^{-1}\!\pmod{\!600};$</span> then I explain why this reduction works so universally due to <em>persistence</em> of GCDs having linear (Bezout) representation.</p>
<p><span class="math-container">$\langle a\rangle$</span> has order <span class="math-container">$\,n\mid 600\,$</span> so <span class="math-container">$\langle a\rangle$</span> is an image of <span class="math-container">$\,\Bbb Z/600,\,$</span> so it suffices to solve it mod <span class="math-container">$600;\,$</span> said more concretely <span class="math-container">$\,a\equiv 7b\pmod{\!600\!=\!nk}\,\Rightarrow\,a\equiv 7b\pmod{\!n}.$</span> <span class="math-container">$\,7^{-1}$</span> exists by <span class="math-container">$(7,600)=1;\,$</span> explicitly</p>
<p><span class="math-container">$\!\bmod 600\!:\ 7b\equiv a\iff b\equiv a/7\equiv\color{#c00}{343}\,a,\ $</span> by <span class="math-container">$ \underbrace{\ 7^{\large 4}\equiv 1\,\Rightarrow\, 1/7\equiv 7^{\large 3}\!\equiv \color{#c00}{343}}_{\Large\ \ 7^{\LARGE 2}\, \equiv\ \pm1\ \bmod\, 24\ \,\&\, \ 25\qquad}$</span></p>
<p><strong>Remark</strong> <span class="math-container">$\ $</span> The proof requires only the coprimality of <span class="math-container">$\,k,n = 7,600\,$</span> since then we have a Bezout equation <span class="math-container">$\, k'k - n' n = 1\,$</span> <span class="math-container">$\,\Rightarrow\,\bmod n\!:\ k'\equiv k^{-1} =: (1/k)_n\,$</span> so we can proceed as above, i.e.</p>
<p><span class="math-container">$$\begin{align} \color{#c00}{na = 0}\,\Rightarrow&\,\ k'(ka) = (1\!+\!n'n)a = a\! +\! n'(\color{#c00}{na}) = a\ \quad\rm Additive\ Form\\[.4em]
\color{#c00}{a^{\large n} = 1}\,\Rightarrow&\,\ (a^{\large k})^{\large k'} = a^{\large 1\ +\ n'n}\ \ \ =\ \ a\,(\color{#c00}{a^{\large n}})^{\large n'} =\ a\ \quad\rm Multiplicative\ Form\end{align}\quad$$</span></p>
<p>hence <span class="math-container">$\quad\ \ \bbox[5px,border:1px solid #c00]{\begin{align}
\text{if $\,\ n\,a = 0 = n\,b\ $ then }\ kb = a &\iff b = (1/k)_n\, a\\[.4em]
\text{if $\:\!\ a^n\, =\, 1\, =\, b^n \ $ then }\ b^{\large k} = a &\iff b = a^{\large (1/k)_n}
\end{align}}$</span> </p>
<p>This works so nicely because whenever a gcd has a linear representation (Bezout) it is <em>universal</em>, i.e. the <a href="https://math.stackexchange.com/a/1073618/242">gcd persists in extension rings /algebras / modules</a> - just as above</p>
<hr>
<p><strong>Note</strong> <span class="math-container">$\ $</span> Alternatively we can invert <span class="math-container">$7$</span> mod <span class="math-container">$\ 600 = 24\cdot 25\ $</span> using <a href="https://math.stackexchange.com/a/20259/242">Easy CRT</a> as below</p>
<p><span class="math-container">$\!\bmod 24\!:\ \dfrac{1}7\ \equiv\ \dfrac{49}7\ \equiv\ 7\qquad\qquad\qquad\ $</span> </p>
<p><span class="math-container">$\!\bmod 25\!:\ \dfrac{1}7\equiv \dfrac{-49}7\equiv-7\ \ {\rm so}\ \smash{\ \dfrac{1}7\equiv \overbrace{-7\!+\!25\left[\dfrac{14}{25}\bmod 24\right]}^{\text{by Easy CRT}}\! \equiv -7\!+\!25[14]\equiv \color{#c00}{343}\pmod{600}}$</span></p>
|
1,305,935 | <p>Let $f(n)$ be non-negative real valued function defined for each natural number $n$.</p>
<p>If $f$ is convex and $lim_{n\to\infty}f(n)$ exists as a finite number, then can we conclude that $f$ is non-increasing?</p>
| PSPACEhard | 140,280 | <p>Let $\textbf{x}_0$ be a point in the hyperplane $\textbf{wx} - b = -1$, i.e., $\textbf{wx}_0 - b = -1$. To measure the distance between hyperplanes $\textbf{wx}-b=-1$ and $\textbf{wx}-b=1$, we only need to compute the perpendicular distance from $\textbf{x}_0$ to plane $\textbf{wx}-b=1$, denoted as $r$.</p>
<p>Note that $\frac{\textbf{w}}{\|\textbf{w}\|}$ is a unit normal vector of the hyperplane $\textbf{wx}-b=1$. We have
$$
\textbf{w}(\textbf{x}_0 + r\frac{\textbf{w}}{\|\textbf{w}\|}) - b = 1
$$
since $\textbf{x}_0 + r\frac{\textbf{w}}{\|\textbf{w}\|}$ should be a point in hyperplane $\textbf{wx}-b = 1$ according to our definition of $r$.</p>
<p>Expanding this equation, we have
\begin{align*}
& \textbf{wx}_0 + r\frac{\textbf{w}\textbf{w}}{\|\textbf{w}\|} - b = 1 \\
\implies &\textbf{wx}_0 + r\frac{\|\textbf{w}\|^2}{\|\textbf{w}\|} - b = 1 \\
\implies &\textbf{wx}_0 + r\|\textbf{w}\| - b = 1 \\
\implies &\textbf{wx}_0 - b = 1 - r\|\textbf{w}\| \\
\implies &-1 = 1 - r\|\textbf{w}\|\\
\implies & r = \frac{2}{\|\textbf{w}\|}
\end{align*}</p>
|
1,253,475 | <p>I'm trying to prove the following statement: if E is a subspace of V, then dim E + dim $E^{\perp}$ = dim V. I know this is true because when these two subspaces are added, they are equal to V, but I'm not sure how to rigorously say this, could I get a little help?</p>
| Quality | 153,357 | <p>First understand the following</p>
<p>$\mathbf{Thereom:}$ Let $\{v_1,…,v_n\}$ be any basis of an inner product space V. Then there exists an orthonormal basis $\{u_1,…,u_n\}$ of V such that the change of basis matrix from $\{v_i\} to \{u_i\}$ is triangular i.e. for $k=1,2.., n$,</p>
<p>$u_k= a_{k1}v_1+a_{k2}v_2+..+a_{kk}v_{k}$</p>
<p>The proof comes from applying the Gram Schmidt algorithm to $\{v_i\}$ to obtain an orthogonal basis and the normalize it to obtain a orthonormal basis of V.</p>
<p>$\mathbf{Theorem:}$ Let W be a subspace of V. Then $V=W \oplus W^{\perp}$</p>
<p>$ \mathbf{Proof:}$ We know that there exists an orthogonal basis $\{u_1,…,u_r\}$ of W and we can extend it to an orthogonal basis , $\{u_1,..,u_n\}$ of V hence we have that $u_{r+1},…,u_n \in W^{\perp}$ , if $v \in V$ then</p>
<p>$v=a_1u_1+…+a_nu_n$, where $a_1u_1+…+a_ru_r \in W$ and $a_{r+1}u_{r+1}+…+a_nu_n \in W^{\perp}$</p>
<p>that is, $V= W+W^{\perp}$</p>
<p>Now suppose that if $w \in W \cap W^{\perp}$ , then $\langle w , w \rangle =0 $ and by properties of inner product this implies $w=0$ hence we have that $W \cap W^{\perp}= \{0\}$ and this completes the proof.</p>
|
176,059 | <p>I asked this question in MSE, but I did not received any answer, so I repeat it here:</p>
<p><a href="https://math.stackexchange.com/questions/858238/a-question-on-fixed-point-property">https://math.stackexchange.com/questions/858238/a-question-on-fixed-point-property</a></p>
<p>Assume that $0<k<n-1$, Note that $\mathbb{C}P^{k}$ can be considered as a closed subset of $\mathbb{C}P^{n}$, in a natural way. We collapse $\mathbb{C}P^{k}$ to a point. The resulting space is denoted by $\mathbb{C}P^{n}/\mathbb{C}P^{k}$ </p>
<p><strong>My fixed point question:</strong></p>
<blockquote>
<p>Does $\mathbb{C}P^{n}/\mathbb{C}P^{k}$ satisfies fixed point property?(At least when $n$ is even)</p>
</blockquote>
<p>This question is motivated by:</p>
<p><a href="https://math.stackexchange.com/questions/845057/show-mathbbcp2-cp1-is-not-a-retract-of-mathbbcp4-cp1#comment1754879_845057">https://math.stackexchange.com/questions/845057/show-mathbbcp2-cp1-is-not-a-retract-of-mathbbcp4-cp1#comment1754879_845057</a></p>
| Eric Wofsey | 75 | <p>Here is a partial affirmative answer using mod 2 Steenrod operations; the simplest case of this (for $n$ and $k$ even) is just a correction of the slightly incorrect answer originally posted by Włodzimierz Holsztyński. The result is that if $k+1$ and $n+1$ are both odd multiples of $2^d$ for some integer $d\geq 0$, then $\mathbb{C}P^n/\mathbb{C}P^k$ has the fixed point property. In particular, for $d=0$ we get the fixed point property whenever $n$ and $k$ are both even. All cohomology in this answer will have coefficients in $\mathbb{F}_2$.</p>
<p>Let's start by describing the action of the Steenrod squares on the cohomology $H^*(\mathbb{C}P^n)=\mathbb{F}_2[x]/(x^{n+1})$. The following formulas are easy to prove by induction using the Cartan formula (induct on $d$ and for fixed $d$ induct on $m$):</p>
<p>$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=x^{2^d(m+1)} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm}\right)=0 \text{ if $m$ is even}$$</p>
<p>From these, we deduce the following for all $0\leq \ell<2^d$:
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=x^{2^d(m+1)+\ell} \text{ if $m$ is odd}$$
$$Sq^{2^{d+1}}\left(x^{2^dm+\ell}\right)=0 \text{ if $m$ is even}$$</p>
<p>The quotient map $\mathbb{C}P^n\to\mathbb{C}P^n/\mathbb{C}P^k$ identifies $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$ with the subring of $H^*(\mathbb{C}P^n)$ which has as a basis $\{1,x^{k+1},x^{k+2},\dots, x^n\}$, and so the same relations hold in $H^*(\mathbb{C}P^n/\mathbb{C}P^k)$.</p>
<p>Suppose now that $f:\mathbb{C}P^n/\mathbb{C}P^k\to\mathbb{C}P^n/\mathbb{C}P^k$ is any map. For $k<i\leq n$, let $a_i\in \mathbb{F}_2$ be such that $f^*(x^i)=a_ix^i$. By the Lefschetz fixed point theorem, $f$ must have a fixed point if $1+\sum_{k+1}^n a_i\neq 0$ (the $1$ coming from $H^0$), or equivalently if $\sum a_i=0$.</p>
<p>Since $f^*$ must commute with Steenrod operations, we must have $a_{2^dm+\ell}=a_{2^d(m+1)+\ell}$ for $m$ odd and $0\leq \ell<2^d$, as long as $k<2^dm+\ell<2^d(m+1)+\ell\leq n$. Together, these relations imply that if $m$ is odd and $k<2^dm<2^d(m+2)-1\leq n$, then all the $a_i$ for $2^dm\leq i \leq 2^d(m+2)-1$ are equal to each other (everything below $2^d(m+1)$ can be related to $2^d(m+1)$ using the smaller Steenrod squares, and everything above $2^d(m+1)$ can be related to something below it using $Sq^{2^{d+1}}$). That is, the $a_i$ are constant in blocks of length $2^{d+1}$ starting from an odd multiple of $2^d$.</p>
<p>Now suppose that $k+1$ and $n+1$ are both odd multiples of $2^d$. The numbers from $k+1$ to $n$ can be broken into blocks of length $2^{d+1}$, each starting with an odd multiple of $2^d$. All of the $a_i$ in each block are equal to each other, and hence their sum is zero since there are an even number of them. Thus the sum of all of the $a_i$ is zero, and so $f$ must have a fixed point.</p>
<p>Let me conclude with a couple remarks on this result. First, as Włodzimierz observed, this argument works equally well for projective spaces over $\mathbb{R}$ or $\mathbb{H}$ (for $\mathbb{R}$, replace $Sq^{2^{d+1}}$ with $Sq^{2^d}$ and for $\mathbb{H}$ replace it with $Sq^{2^{d+2}}$). Second, the condition obtained here is sufficient but not necessary for $\mathbb{C}P^n/\mathbb{C}P^k$ to have the fixed point property. Indeed, in the comments I sketched an argument using cup products and integer coefficients rather than Steenrod squares and mod 2 coefficients which shows that the fixed point property holds when $n\gg k$ as long as either $n$ is even or $k$ is odd (note that in fact, using only mod 2 coefficients there is no hope of proving the fixed point property in cases where $n$ and $k$ have different parity).</p>
|
1,749,340 | <p>I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\
=\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\
=\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=...
\end{gather}
Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.</p>
| StackTD | 159,845 | <p>For simplicity, let $x = \frac{\pi}{14}$, then we want to simplify:
$$\sin 3x-\sin x -\sin 5x$$
Multiply by $\cos x$ to get:
$$\color{blue}{\sin 3x\cos x}-\color{green}{\sin x\cos x} -\color{red}{\sin 5x\cos x} \quad (*)$$
With $\sin\alpha\cos\beta = \tfrac{1}{2}\left( \sin(\alpha+\beta)+\sin(\alpha-\beta) \right)$, you have:
$$\color{blue}{\sin 3x\cos x = \tfrac{1}{2}\left( \sin 4x +\sin 2x \right)} \quad \mbox{and} \quad \color{red}{\sin 5x\cos x = \tfrac{1}{2}\left( \sin 6x +\sin 4x \right)} $$
and $\color{green}{\sin x\cos x = \tfrac{1}{2}\sin 2x}$; so:
$$\require{cancel} (*) \quad
\tfrac{1}{2}\left( \cancel{\sin 4x} +\bcancel{\sin 2x} \right) - \bcancel{\tfrac{1}{2}\sin 2x} - \tfrac{1}{2}\left( \sin 6x +\cancel{\sin 4x} \right) = - \tfrac{1}{2}\sin 6x
$$
Divide again by $\cos x$:
$$- \tfrac{1}{2}\frac{\sin 6x}{\cos x} = - \tfrac{1}{2}\frac{\sin \frac{6\pi}{14}}{\cos \frac{\pi}{14}}= - \tfrac{1}{2}\frac{\cos\left( \frac{\pi}{2}-\frac{6\pi}{14}\right)}{\cos \frac{\pi}{14}} =- \tfrac{1}{2}\frac{\cos\frac{\pi}{14}}{\cos \frac{\pi}{14}} = -\frac{1}{2}$$</p>
|
1,749,340 | <p>I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\
=\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\
=\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=...
\end{gather}
Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.</p>
| lab bhattacharjee | 33,337 | <p>Let $14x=\pi$</p>
<p>$$S=\sin3x-\sin x-\sin5x=\sin3x+\sin(-x)+\sin(-5x)$$</p>
<p>Using <a href="https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro">How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?</a> ,</p>
<p>$$\sin(-2x)\cdot S=\cos5x-\cos7x$$</p>
<p>$$-\sin2x\cdot S=\cos5x=\sin2x$$</p>
<p>As $\cos7x=0,\cos5x=\sin2x$ as $5x+2x=\dfrac\pi2$</p>
<p>Can you take it from here?</p>
|
302,179 | <p>The question I am working on is:</p>
<blockquote>
<p>"Use a direct proof to show that every odd integer is the difference of two squares."</p>
</blockquote>
<p>Proof:</p>
<p>Let n be an odd integer: $n = 2k + 1$, where $k \in Z$</p>
<p>Let the difference of two different squares be, $a^2-b^2$, where $a,b \in Z$.</p>
<p>Hence, $n=2k+1=a^2-b^2$...</p>
<p>As you can see, this a dead-end. Appealing to the answer key, I found that they let the difference of two different squares be, $(k+1)^2-k^2$. I understand their use of $k$--$k$ is one number, and $k+1$ is a different number--;however, why did they choose to add $1$? Why couldn't we have added $2$?</p>
| DonAntonio | 31,254 | <p>Directly:</p>
<p>$$2k+1=k^2+2k+1-k^2=(k+1)^2-k^2\ldots$$</p>
|
1,406,280 | <p>Why does taylor series have ample amount of importance in calculus? </p>
<p>I like to know some insights behind taylor series. </p>
| mathlove | 78,967 | <p>I assume that $a,b,c\ge 1$.</p>
<p>You can use the method you write for
$$a+c=100-2b$$
where $b=1,2,\cdots,49$. </p>
<p>Then, the answer is
$$\sum_{b=1}^{49}\binom{99-2b}{1}=99\times 49-2\cdot\frac{49\cdot 50}{2}=49(99-50)=49^2=\color{red}{2401}.$$</p>
|
4,550,991 | <p>This is question is taken from an early round of a Norwegian national math competition where you have on average 5 minutes to solve each question.</p>
<p>I tried to solve the question by writing every number with four digits and with introductory zeros where it was needed. For example 0001 and 0101 would be the numbers 1 and 101. I then divided the different numbers into four groups based on how many times the digit 1 appeared in the number. I called these group 1,2,3 and 4. 0001 would then belong to group 1 and 0101 to group 2. I first found out in how many ways I could place the digit 1 in each group, then multiplied it by the number of combinations with the other possible eight digits (0,2,3,5,6,7,8,9). This would be the number of combinations for each group and I lastly multiplied it with the number of times 1 appeared in the number. This is done for all of the groups under:</p>
<p><span class="math-container">$\binom{4}{1}\cdot8^3\cdot1$</span> times in group 1</p>
<p><span class="math-container">$\binom{4}{2}\cdot8^2\cdot2$</span> times in group 2</p>
<p><span class="math-container">$\binom{4}{3}\cdot8^1\cdot3$</span> times in group 3</p>
<p><span class="math-container">$\binom{4}{4}\cdot8^0\cdot4$</span> times in group 4</p>
<p>The sum of all these calculations will be the 2916 and the correct number of times 1 appears (I think). Is this calculation/way of thinking correct? And is there a more efficient way to do it?</p>
| Abel Wong | 1,090,313 | <p>The answer is correct. I try to do it in another way:</p>
<p>Count the <span class="math-container">$1$</span>'s in thousandth place first.</p>
<p>There is 1000 <span class="math-container">$1$</span>'s from <span class="math-container">$1$</span> to <span class="math-container">$9999$</span>. But we need to exclude those contain <span class="math-container">$4$</span>.</p>
<p>There is <span class="math-container">$3\times 100-3\times10+1 = 271$</span> numbers contain <span class="math-container">$4$</span>.</p>
<p>So there is <span class="math-container">$1000-271 = 729$</span> <span class="math-container">$1$</span>'s in thousandth place.</p>
<p>Use the same argument. There is <span class="math-container">$729$</span> <span class="math-container">$1$</span>'s in hundredth place, tenth place and unit place.
So totally , there is <span class="math-container">$729 \times 4=2916$</span> <span class="math-container">$1$</span>'s</p>
|
4,550,991 | <p>This is question is taken from an early round of a Norwegian national math competition where you have on average 5 minutes to solve each question.</p>
<p>I tried to solve the question by writing every number with four digits and with introductory zeros where it was needed. For example 0001 and 0101 would be the numbers 1 and 101. I then divided the different numbers into four groups based on how many times the digit 1 appeared in the number. I called these group 1,2,3 and 4. 0001 would then belong to group 1 and 0101 to group 2. I first found out in how many ways I could place the digit 1 in each group, then multiplied it by the number of combinations with the other possible eight digits (0,2,3,5,6,7,8,9). This would be the number of combinations for each group and I lastly multiplied it with the number of times 1 appeared in the number. This is done for all of the groups under:</p>
<p><span class="math-container">$\binom{4}{1}\cdot8^3\cdot1$</span> times in group 1</p>
<p><span class="math-container">$\binom{4}{2}\cdot8^2\cdot2$</span> times in group 2</p>
<p><span class="math-container">$\binom{4}{3}\cdot8^1\cdot3$</span> times in group 3</p>
<p><span class="math-container">$\binom{4}{4}\cdot8^0\cdot4$</span> times in group 4</p>
<p>The sum of all these calculations will be the 2916 and the correct number of times 1 appears (I think). Is this calculation/way of thinking correct? And is there a more efficient way to do it?</p>
| user2661923 | 464,411 | <p>Your <strong>direct approach</strong> looks good, and may well be the easiest approach for this particular problem. The alternative approach, which generalizes better and is used below is Inclusion-Exclusion.</p>
<p>See <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow noreferrer">this article</a> for an
introduction to Inclusion-Exclusion.
Then, see <a href="https://math.stackexchange.com/questions/4427605/inclusion-exclusion-principle-what-is-1n1/4427645#4427645">this answer</a> for an explanation of and justification for the Inclusion-Exclusion <em>formula</em>.</p>
<p>It is harmless to zero fill the numbers, since neither of the digits being interrogated, <span class="math-container">$(1)$</span>'s and <span class="math-container">$(4)$</span>'s are <span class="math-container">$(0)$</span>'s. Further, you can harmlessly include the number <span class="math-container">$(0000)$</span>, as the first number under investigation. Therefore, you are examining all <span class="math-container">$[(10)^4]$</span> 4 digit numbers.</p>
<p>Let <span class="math-container">$S$</span> denote the enumeration of all <span class="math-container">$(1)$</span>'s, among the 4 digit numbers, where the prohibition against the digit <span class="math-container">$(4)$</span> is ignored. Clearly, enumerating column my column, since each column will have exactly <span class="math-container">$(1/10)$</span> of its digits be a <span class="math-container">$(1)$</span>, the enumeration here is</p>
<p><span class="math-container">$$T_0 = \frac{4 \times [(10)^4]}{10} = 4000. \tag1 $$</span></p>
<p>For <span class="math-container">$k \in \{1,2,3,4\}$</span> let <span class="math-container">$S_k$</span> denote the portion of the <span class="math-container">$4000$</span> occurrences of the digit <span class="math-container">$(1)$</span> that occur when the <span class="math-container">$4$</span> digit number has a <span class="math-container">$(4)$</span> in the <span class="math-container">$k$</span>-th digit place, reading from left to right.</p>
<p>Then, the desired enumeration will be</p>
<p><span class="math-container">$$T_0 - ~\text{the corresponding enumeration from} ~|S_1 \cup S_2 \cup S_3 \cup S_4|.$$</span></p>
<p>For <span class="math-container">$r \in \{1,2,3,4\}$</span> let <span class="math-container">$T_r$</span> denote the enumeration of the <span class="math-container">$(1)$</span>'s in the <span class="math-container">$~\displaystyle \binom{4}{r}$</span> terms represented by the summation of</p>
<p><span class="math-container">$$\sum_{1 \leq i_1 < \cdots < i_r \leq 4} |S_{i_1} \cap \cdots \cap S_{i_r}|.$$</span></p>
<p>Then, in accordance with Inclusion-Exclusion, the desired enumeration will be</p>
<p><span class="math-container">$$\sum_{r=0}^4 (-1)^r T_r.$$</span></p>
<hr />
<p>Actually, I haven't worded the above definitions very well. However, the details below will clarify my intent. Further, there will be many considerations of symmetry which will simplify the computations.</p>
<p>Under the assumption that the leftmost digit is <span class="math-container">$(4)$</span>, with the other <span class="math-container">$(3)$</span> digits unconstrained, it is clear that there will be <span class="math-container">$1000$</span> numbers under consideration. At random, within these <span class="math-container">$(1000)$</span> numbers, the probability of a <span class="math-container">$(1)$</span> in each of the other <span class="math-container">$(3)$</span> columns is <span class="math-container">$(1/10)$</span>.</p>
<p>So, focusing only on the <span class="math-container">$(1000)$</span> numbers that have a <span class="math-container">$(4)$</span> in the leftmost digit (i.e. focusing on the set <span class="math-container">$S_1$</span>),
the number of occurrences of a <span class="math-container">$(1)$</span> in one of the other <span class="math-container">$(3)$</span> digits will be</p>
<p><span class="math-container">$$3 \times \frac{1000}{10} = 300.$$</span></p>
<p>Further, by considerations of symmetry, you have the same enumeration for <span class="math-container">$S_2, S_3,$</span> and <span class="math-container">$S_4$</span>. Therefore,</p>
<p><span class="math-container">$$T_1 = 1200.$$</span></p>
<hr />
<p>For <span class="math-container">$S_1 \cap S_2$</span>, you have <span class="math-container">$2$</span> columns unconstrained, the 3rd and the 4th column, reading left to right. Further, the same symmetry considerations prevail.</p>
<p>Examining (in effect) only <span class="math-container">$S_1 \cap S_2$</span>, you are examining <span class="math-container">$(100)$</span> numbers. In these <span class="math-container">$(100)$</span> numbers, the number of occurrences of a <span class="math-container">$(1)$</span> in the 3rd digit, combined with the number of occurrences of a <span class="math-container">$(1)$</span> in the 4th digit will be</p>
<p><span class="math-container">$$2 \times \frac{100}{10} = 20.$$</span></p>
<p>With the same considerations of symmetry in the computation of <span class="math-container">$T_2$</span> as was present in the computation of <span class="math-container">$T_1$</span>, you have that</p>
<p><span class="math-container">$$T_2 = \binom{4}{2} \times 20 = 120.$$</span></p>
<hr />
<p>Similarly,</p>
<p><span class="math-container">$$T_3 = \binom{4}{3} \times \frac{10}{10} = 4.$$</span></p>
<p>That is, <span class="math-container">$T_3$</span> is computed by consideration of the 4 numbers <span class="math-container">$4441, 4414, 4144, 1444.$</span></p>
<p>Clearly,</p>
<p><span class="math-container">$$T_4 = 0,$$</span></p>
<p>by consideration of the number <span class="math-container">$4444.$</span></p>
<hr />
<p>Therefore, the final computation is</p>
<p><span class="math-container">$$4000 - 1200 + 120 - 4 + 0 = 2916.$$</span></p>
|
1,464,143 | <p>$\lim_{n \to \infty} n\ln\left(1+\frac{1}{n}\right)$ using L'Hòpital rule show that this is $1$. Can you do this since there isn't a division and $n$ will obviously tend to infinity and $\ln\left(1+\frac{1}{n}\right)$ will tend to $0$? So there limits aren't matching?</p>
<p>So I set $u=n $</p>
<p>$du=1$</p>
<p>$v= \ln\left(1+\frac{1}{n}\right)$</p>
<p>$dv= -\frac{1}{n^2+n}$</p>
<p>Hence $\ln\left(1+\frac{1}{n}\right) - \frac{1}{n+1}$ in which both of these tend to $0$ so I am complete lost.</p>
| mfl | 148,513 | <p><strong>Hint</strong></p>
<p>$$\lim_{n\to \infty} n\ln \left(1+\frac1n\right)=\lim_{n\to \infty} \frac{\ln \left(1+\frac1n\right)}{\frac 1n}.$$</p>
|
302 | <p>I know that the Fibonacci numbers converge to a ratio of .618, and that this ratio is found all throughout nature, etc. I suppose the best way to ask my question is: where was this .618 value first found? And what is the...significance?</p>
| S. Carnahan | 121 | <p>Golden ratio came first. <a href="http://en.wikipedia.org/wiki/Golden_ratio" rel="noreferrer">Wikipedia</a> has a rather thorough article on it. It's not nearly as pervasive in nature or architecture as people like to say it is. It will show up in anything with regular pentagons, though.</p>
|
255,773 | <p>As it is known that
Integrate[A+B]= Integrate[A] + Integrate[B]</p>
<p>I am facing problem with the following integral, when I integrate</p>
<pre><code>Integrate[(-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) -
2 (EulerGamma + Log[(b x)/2])) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2) , {x,
1, \[Infinity]}]
</code></pre>
<p>I obtain an answer but when I wish to solve a part</p>
<pre><code>Integrate[-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2) , {x,
1, \[Infinity]}]
</code></pre>
<p>Mathematica 11 do not returns any value and processing goes on.</p>
| Carl Woll | 45,431 | <p>The distributive property only holds when both integrals are convergent. For example:</p>
<pre><code>Integrate[Exp[-x] + Pi/(2x) - ArcTan[x]/x, {x, 1, Infinity}]
</code></pre>
<blockquote>
<p>Catalan + 1/E</p>
</blockquote>
<p>However:</p>
<pre><code>Integrate[Exp[-x] + Pi/(2x), {x, 1, Infinity}]
Integrate[ - ArcTan[x]/x, {x, 1, Infinity}]
</code></pre>
<blockquote>
<p>Integrate::idiv: Integral of E^-x+[Pi]/(2 x) does not converge on {1,[Infinity]}.</p>
</blockquote>
<blockquote>
<p>Integrate[E^(-x) + Pi/(2*x), {x, 1, Infinity}]</p>
</blockquote>
<blockquote>
<p>Integrate::idiv: Integral of -(ArcTan[x]/x) does not converge on {1,[Infinity]}.</p>
</blockquote>
<blockquote>
<p>Integrate[-(ArcTan[x]/x), {x, 1, Infinity}]</p>
</blockquote>
<p>The same thing is happening for your example.</p>
|
1,853,846 | <p>Prove that the equation <span class="math-container">$$x^2 - x + 1 = p(x+y)$$</span> has integral solutions for infinitely many primes <span class="math-container">$p$</span>.</p>
<p>First, we prove that there is a solution for at least one prime, <span class="math-container">$p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is always odd so there is no solution for <span class="math-container">$p=2$</span>. We prove there is a solution for <span class="math-container">$p=3$</span>. If <span class="math-container">$p=3$</span>, <span class="math-container">$y = (x-2)^2/3-1$</span>. We get integral solutions whenever we get <span class="math-container">$x = 3m +2$</span>, where <span class="math-container">$m$</span> is any integer.<span class="math-container">$\\$</span>
We provide a proof by contradiction which is similar to Euclid's proof of there being infinitely many primes. \Let us assume that it is is true for only finitely many primes, and name the largest prime for which the equation is true <span class="math-container">$P$</span>.\\We set <span class="math-container">$$x = 2\cdot3\cdot5\dots P$$</span> <span class="math-container">$x$</span> is the product of all primes upto <span class="math-container">$P.\\$</span>
Then, the term <span class="math-container">$x(x-1) + 1$</span> is either prime or composite. If it is prime, then we set <span class="math-container">$p = x(x-1) + 1, y = 1 - x$</span> and get a solution. If it is composite, we write <span class="math-container">$x(x-1) + 1 = p\times q$</span>, where <span class="math-container">$p$</span> is any prime factor of <span class="math-container">$x(x-1)+1$</span>, and <span class="math-container">$q$</span> is an integer, <span class="math-container">$q = (x(x-1)+1)/p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is not divisible by any prime upto <span class="math-container">$P$</span> since it leaves a remainder of <span class="math-container">$1$</span> with all of them. So, <span class="math-container">$p > P$</span>. We set <span class="math-container">$y=q-x$</span> for a solution.<span class="math-container">$\\$</span>In either case, we get a solution for a prime <span class="math-container">$p > P$</span>, which means there's no largest prime for which this equation has solutions. This contradicts the assumption that there are solutions for only finitely many primes.</p>
<p>I feel like I'm missing some step. Is this correct ?</p>
| velut luna | 139,981 | <p>$$8\sin\theta=4+\cos\theta$$
$$8\sin\theta-\cos\theta=4$$
$$\sqrt{65}\sin\alpha\sin\theta-\sqrt{65}\cos\alpha\cos\theta=4$$
where $\alpha=\tan^{-1}8$
$$-\sqrt{65}\cos(\theta+\alpha)=4$$
Can you continue from here?</p>
|
542,148 | <p>Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)</p>
<p>\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}\\
0 & 0 & 0 & x^{12} & x^{13}\\
0 & 0 & 0 & x^{14} & x^{15}
\end{bmatrix}</p>
<p>Can someone help me with this question?</p>
| 2012ssohn | 103,274 | <p>First, note that the 5th column is a multiple of the 4th column. That is,</p>
<p>\begin{bmatrix}
x^4\\
x^9\\
x^{11}\\
x^{13}\\
x^{15}\\
\end{bmatrix}</p>
<p>is $x$ times
\begin{bmatrix}
x^3\\
x^8\\
x^{10}\\
x^{12}\\
x^{14}\\
\end{bmatrix}.</p>
<p>Because the determinant of a matrix does not change when you subtract a multiple of one column from another column, we get that that matrix has the same determinant as that of</p>
<p>\begin{bmatrix}
1 & x & x^2 & x^3 & 0\\
x^5 & x^6 & x^7 & x^8 & 0\\
0 & 0 & 0 & x^{10} & 0\\
0 & 0 & 0 & x^{12} & 0\\
0 & 0 & 0 & x^{14} & 0\\
\end{bmatrix}</p>
<p>and you can easily tell that the determinant of that matrix is $0$.</p>
|
542,148 | <p>Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)</p>
<p>\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}\\
0 & 0 & 0 & x^{12} & x^{13}\\
0 & 0 & 0 & x^{14} & x^{15}
\end{bmatrix}</p>
<p>Can someone help me with this question?</p>
| Robert Israel | 8,508 | <p>Another way to look at this: the bottom three rows can't have rank more than $2$, since they have only two nonzero columns, so the whole matrix can't have rank more than $4$, and therefore is singular.</p>
|
8,567 | <p>When highlighting text using <code>Style</code> and <code>Background</code>, as in <code>Style["Test ", White, Background -> Lighter@Blue]</code> is there a way to pad (ie, enlarge) the bounding box? </p>
<p>The bottom of the background seems coincident with the base of the text: <img src="https://i.stack.imgur.com/SLimA.jpg" alt="here"></p>
| Mr.Wizard | 121 | <p>To answer my own question and further illustrate the kind of operation I am describing, here is a method using <code>Set</code> itself:</p>
<pre><code>SetAttributes[f, HoldAllComplete]
f[args___] :=
Module[{h},
h[args] = 1;
Level[DownValues@h, {4}, HoldComplete]
]
f[own, down[1], sub[1][2], N[n], up]
</code></pre>
<blockquote>
<pre><code>HoldComplete["OwnValue", "DownValue", "SubValue", 3.14, up]
</code></pre>
</blockquote>
|
3,009,112 | <p>I am a geographer/ecologist and I want to know how to accurately calculate volume of a lake or a reservoir? I am not looking for a vague estimate which is generally calculated using surface area and mean height parameters assuming the body is of a certain shape (truncated cone/triangle or circular). Since reservoirs are completely irregular in shape I am having difficulties in using the traditional volume formulae. Any help or suggestion would be greatly appreciated. Thank you. </p>
| the_fox | 11,450 | <p>Corollary <span class="math-container">$(2I)$</span> in the paper "On groups of even order" by Brauer and Fowler says that if <span class="math-container">$G$</span> is a simple group which contains <span class="math-container">$n$</span> involutions and <span class="math-container">$t= \frac{|G|}{n}$</span> then <span class="math-container">$|G| < \lceil t(t+1)/2 \rceil !$</span>. </p>
<p>This is a simple consequence of Theorem <span class="math-container">$(2F)$</span> or Theorem <span class="math-container">$(2H)$</span> in that paper so check if something along these lines is in your book. (Which book is it by the way?)</p>
<p>Arguing by contradiction, suppose that <span class="math-container">$n \geq \frac{|G|}{3}$</span>. Then <span class="math-container">$t \leq 3$</span> so <span class="math-container">$|G|<720$</span>. There are just <span class="math-container">$5$</span> non-abelian simple groups of order less than <span class="math-container">$720$</span>, which are <span class="math-container">$A_5$</span>, <span class="math-container">$A_6$</span>, <span class="math-container">$\operatorname{PSL}_2(7)$</span>, <span class="math-container">$\operatorname{PSL}_2(8)$</span> and <span class="math-container">$\operatorname{PSL}_2(11)$</span>, and these have <span class="math-container">$15$</span>, <span class="math-container">$45$</span>, <span class="math-container">$21$</span>, <span class="math-container">$63$</span>, and <span class="math-container">$55$</span> involutions respectively. In no case does <span class="math-container">$n \geq \frac{|G|}{3}$</span> hold, which is a contradiction.</p>
<hr>
<p><strong>Added.</strong> You can use directly Theorem <span class="math-container">$6.7$</span> in Rose's "A Course on Group Theory" which says: </p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a group of even order with precisely <span class="math-container">$n$</span> involutions, and suppose that <span class="math-container">$|Z(G)|$</span> is odd. Let <span class="math-container">$a = |G|/n$</span>. Then <span class="math-container">$G$</span> has a proper subgroup <span class="math-container">$H$</span> such that either <span class="math-container">$|G:H|=2$</span> or <span class="math-container">$|G:H|<\frac{1}{2}a(a+1)$</span>.</p>
</blockquote>
<p>Now suppose that <span class="math-container">$G$</span> is a finite simple group with precisely <span class="math-container">$n$</span> involutions. Since <span class="math-container">$|Z(G)|=1$</span> the preceding theorem applies. Note that <span class="math-container">$|G:H| \neq 2$</span> since otherwise <span class="math-container">$H$</span> is normal in <span class="math-container">$G$</span>. In fact, the stronger claim is true (which Derek mentioned in his answer), that <span class="math-container">$|G:H| \geq 5$</span>.</p>
<p>Assume for a contradiction that <span class="math-container">$n \geq |G|/3$</span>. Then <span class="math-container">$a := |G|/n \leq 3$</span>, so <span class="math-container">$G$</span> has a proper subgroup <span class="math-container">$H$</span> such that <span class="math-container">$|G:H|<6$</span> by Thm. <span class="math-container">$6.7$</span>, thus <span class="math-container">$|G:H|=5$</span> by the preceding observation. But <span class="math-container">$|G:H|=5$</span> is only possible if <span class="math-container">$G \cong A_5$</span> (do you see why?) and you are given that <span class="math-container">$A_5$</span> has less than <span class="math-container">$60/3=20$</span> involutions. That is a contradiction, however, and the proof is complete.</p>
|
3,997,632 | <p>Use the Chain Rule to prove the following.<br />
(a) The derivative of an even function is an odd function.<br />
(b) The derivative of an odd function is an even function.</p>
<p><strong>My attempt:</strong></p>
<p>I can easily prove these using the definition of a derivative, but I'm having trouble showing them using the chain rule.</p>
<p>(a) <span class="math-container">$f(x)$</span> is even <span class="math-container">$ \therefore f(-x) = f(x)$</span>.<br />
We need to show that <span class="math-container">$f'(-x) = -f'(x)$</span>.</p>
<p>Let <span class="math-container">$u = -x$</span>.<br />
My reasoning for the next step is that if we want to find the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$x$</span> from <span class="math-container">$f(u)$</span> we need to use the chain rule and first find the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$u$</span> and multiply that with the derivative of <span class="math-container">$u$</span> with respect to <span class="math-container">$x$</span>.<br />
<span class="math-container">$f'(x) = f'(u) \cdot u' = - f'(u) = -f'(-x)$</span>.<br />
<span class="math-container">$f'(x) = -f'(-x)$</span>.<br />
<span class="math-container">$f'(-x) = -f'(x)$</span>.</p>
<p>The problem with this is that I never used the fact that <span class="math-container">$f$</span> is even and I feel like there is a mistake in my approach, most likely where I explained my reasoning. I just can't see it. Can you please point out the mistake and point me to the right direction?</p>
<p>Thanks!</p>
| Paul Frost | 349,785 | <p>You <em>did</em> use that <span class="math-container">$f$</span> is even. Let us look at the following more general situation that
<span class="math-container">$$ f = v \circ f \circ u .$$</span>
For an even function we have <span class="math-container">$u(x) = -x$</span> and <span class="math-container">$v(x) = x$</span>, for an odd function <span class="math-container">$u(x) = v (x) = -x$</span>. Then
<span class="math-container">$$f'(x) = v'(f(u(x)))f'(u(x))u'(x) .$$</span>
For <span class="math-container">$u(x) = -x$</span> and <span class="math-container">$v(x) = (-1)^\epsilon x$</span> we get
<span class="math-container">$$f'(x) = (-1)^{\epsilon +1}f'(-x).$$</span></p>
|
264,587 | <p><strong>NOTE</strong></p>
<p>I'm sorry, my question was not clear. I want to know all the ways to split a list with a given length simply, <strong>rather than split a cyclic substitution</strong>.
If a given list has length <span class="math-container">$N$</span> and the rule is <span class="math-container">${m, n, p, ...}$</span>, we should get a list of length <span class="math-container">${}_{N} C_{m} {}_{N-m} C_{n} {}_{N-m-n} C_{p} \dots = \frac{N!}{m! n! p! \cdots}$</span> if all elements of <span class="math-container">$N$</span> are independent.</p>
<p>Other examples:
<code>Length@partitionList[{a, b, c, d, e, f, g, h}, {2, 2, 4}]</code>
returns <span class="math-container">$420 = \frac{8!}{2! 2! 4!}$</span></p>
<p><code>Length@partitionList[{a, b, b, c, d, e, e, f}, {2, 2, 4}]</code>
returns 173</p>
<hr>
<p><strong>Question</strong></p>
<p>I want to split a given list into sets of lists, whose lengths are given.
For example, this means if we split a list <code>{a, b, c, d}</code> (length 4) to two lists with length <code>{1, 3}</code> (the sum of lengths should be 4), we obtain</p>
<pre><code>{{{a}, {b, c, d}}, {{b}, {c, d, a}}, {{c}, {d, e, a}}, {{d}, {e, a, b}}
</code></pre>
<p>(here, we don't care about ordering for elements in each sublist).</p>
<p>To achieve this, I prepared the following function:</p>
<pre class="lang-Mathematica prettyprint-override"><code>partitionList[l_List, p_List] :=
DeleteDuplicates@(Module[{$tmp, $deleteList, $lastchoose,
l2 = Range[Length@l]},
$tmp = Subsets[l2, {p[[1]]}];
$deleteList = Flatten /@ $tmp;
If[Length@p > 1,
Do[
$lastchoose =
Table[Subsets[
Delete[l2, {#} & /@ $deleteList[[$j]]], {p[[$i]]}], {$j,
Length@$deleteList}];
$tmp =
Replace[Flatten[
Tuples /@ Transpose[{{#} & /@ $tmp, $lastchoose}], 1],
x_ /; Depth@x > 2 :> Sequence @@ x, {2} ];
$deleteList = Flatten /@ $tmp;
, {$i, 2, Length@p}]
];
Map[l[[#]] &, $tmp, {2}]
]
)
</code></pre>
<p>Here, the argument <span class="math-container">$l$</span> is a list which we want to split, and <span class="math-container">$p$</span> is a list of lengths of sublists.
In the previous example, <span class="math-container">$l$</span> is <code>{a, b, c, d}</code> and <span class="math-container">$p$</span> is <code>{1, 3}</code>.</p>
<p>However, since it is based on procedural programming, I believe there are more efficient ways.
Could you please suggest such a method?</p>
| user1066 | 106 | <pre><code>TakeDrop[#,1]&/@NestList[RotateLeft, {a,b,c,d},3]
(* {{{a}, {b, c, d}}, {{b}, {c, d, a}}, {{c}, {d, a, b}},
{{d}, {a, b, c}}} *)
</code></pre>
<p>And</p>
<pre><code>TakeDrop[#,2]&/@NestList[RotateLeft, {a,b,c,d},3]
( {{{a, b}, {c, d}}, {{b, c}, {d, a}}, {{c, d}, {a, b}},
{{d, a}, {b, c}}} *)
</code></pre>
|
18,444 | <p>I am a student, in my last year of school(17 years old)</p>
<p>When I was about 13 years old I fell into the <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">calculus trap</a> by starting off learning trigonometry on my own, when I was supposed to factor equations or solve basic probability questions. Being good at math(fast arithmetic skills, could grasp new concepts easily etc.) I came across an interschool math competition at that time which was purely based on geometry. I studied by myself for that comp, mainly the properties of circles, triangles and as I earlier mentioned, was introduced to trigonometry. Even though I didn't score well in that comp, I was mesmerized studying those new topics which led me to discover more and more, and finally started learning calculus just one year after that time.</p>
<p>I soon realized that I have fallen into a large hole with no end, though going deeper and deeper was becoming more interesting, it started affecting my performance in other subjects. I also realized that instead of starting with calculus, I could have solved much harder problems related to my school curriculum and hence could score better in Olympiads. </p>
<p>Unfortunately I am still in that hole, going deeper and deeper, studying more abstract, higher level concepts. But I will be passing out of my school next year and will get admission in a nice college aiming to study higher maths and everything will become normal. </p>
<p>My main question is that should I encourage my juniors to do the same thing which I did? Or should I guide them to study maths in a more systematic way.</p>
<p>The <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">link</a> that I had provided is from art of problem solving and hence I found it a bit harsh towards my situation, so I decided to ask here for guidance. I also read <a href="https://matheducators.stackexchange.com/questions/7718/effects-of-early-study-of-advanced-books">this</a> question but that is not exactly my case, I was thorough in my study, also took help from my teachers if I did not understand anything.</p>
| paul garrett | 63 | <p>Echoing @AndreasBlass' remark, and having experienced somewhat similar episodes, it is already precarious enough to make such choices _for_oneself_. So, to directly answer your question: I think "no, do not encourage others to (too violently) disconnect from the math curriculum at school". I don't think it's about problem-solving versus calculus, at all. And, no, I'm not a fan of typical school math curricula.</p>
<p>If nothing else, given the way many kids think about things (as I did, for sure), there is the risk of providing a convenient rationalization for disconnection and non-conformity that is not replaced by anything constructive. And then it's hard to get back in sync even if one wants to.</p>
<p>E.g., when I learned about the expressibility of trig functions in terms of exponentials, via (basic things about) complex numbers, the game/challenge of proving trig identities lost its charm. I could not make myself care very much. But the instructor for that class did not care about "better alternatives", so, after some earlier class-room debacles, I decided to play along to a sufficient degree to avoid trouble. And "physics without calculus" seems ridiculous, too, but, ... and "economics without calculus"...</p>
<p>So I'd recommend maintaining at least two threads: one following your own curiosity and "genuine" mathematics and science (rather than school curricula), but another to maintain "presentability", in effect showing that you can understand the ambient social constructs and cooperate with other people at least minimally.</p>
<p>No, I am no fan of conformity for its own sake! :) But it is certainly very convenient for <em>individuals</em> to be <em>able</em> to "code switch", to survive. Or "to survive long enough to get to a position to have less obligation to conform"? A tricky balance, for sure.</p>
|
18,444 | <p>I am a student, in my last year of school(17 years old)</p>
<p>When I was about 13 years old I fell into the <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">calculus trap</a> by starting off learning trigonometry on my own, when I was supposed to factor equations or solve basic probability questions. Being good at math(fast arithmetic skills, could grasp new concepts easily etc.) I came across an interschool math competition at that time which was purely based on geometry. I studied by myself for that comp, mainly the properties of circles, triangles and as I earlier mentioned, was introduced to trigonometry. Even though I didn't score well in that comp, I was mesmerized studying those new topics which led me to discover more and more, and finally started learning calculus just one year after that time.</p>
<p>I soon realized that I have fallen into a large hole with no end, though going deeper and deeper was becoming more interesting, it started affecting my performance in other subjects. I also realized that instead of starting with calculus, I could have solved much harder problems related to my school curriculum and hence could score better in Olympiads. </p>
<p>Unfortunately I am still in that hole, going deeper and deeper, studying more abstract, higher level concepts. But I will be passing out of my school next year and will get admission in a nice college aiming to study higher maths and everything will become normal. </p>
<p>My main question is that should I encourage my juniors to do the same thing which I did? Or should I guide them to study maths in a more systematic way.</p>
<p>The <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">link</a> that I had provided is from art of problem solving and hence I found it a bit harsh towards my situation, so I decided to ask here for guidance. I also read <a href="https://matheducators.stackexchange.com/questions/7718/effects-of-early-study-of-advanced-books">this</a> question but that is not exactly my case, I was thorough in my study, also took help from my teachers if I did not understand anything.</p>
| Community | -1 | <p>The article at artofproblemsolving seems silly to me. The author's idiosyncratic opinion seems to be that students who are ready to take calculus should refrain from taking calculus and instead do math contests. People are all different, and there is not just one appropriate path for a mathematically precocious student. Some people might want to take calculus <em>and</em> do math contests -- it's not like they're mutually exclusive.</p>
<p>One good argument for learning calculus early is that at many universities, the quality of instruction in first-semester calculus is atrocious. It might be better to learn it on your own or in an AP high school class than to be subjected to that.</p>
|
18,444 | <p>I am a student, in my last year of school(17 years old)</p>
<p>When I was about 13 years old I fell into the <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">calculus trap</a> by starting off learning trigonometry on my own, when I was supposed to factor equations or solve basic probability questions. Being good at math(fast arithmetic skills, could grasp new concepts easily etc.) I came across an interschool math competition at that time which was purely based on geometry. I studied by myself for that comp, mainly the properties of circles, triangles and as I earlier mentioned, was introduced to trigonometry. Even though I didn't score well in that comp, I was mesmerized studying those new topics which led me to discover more and more, and finally started learning calculus just one year after that time.</p>
<p>I soon realized that I have fallen into a large hole with no end, though going deeper and deeper was becoming more interesting, it started affecting my performance in other subjects. I also realized that instead of starting with calculus, I could have solved much harder problems related to my school curriculum and hence could score better in Olympiads. </p>
<p>Unfortunately I am still in that hole, going deeper and deeper, studying more abstract, higher level concepts. But I will be passing out of my school next year and will get admission in a nice college aiming to study higher maths and everything will become normal. </p>
<p>My main question is that should I encourage my juniors to do the same thing which I did? Or should I guide them to study maths in a more systematic way.</p>
<p>The <a href="https://artofproblemsolving.com/news/articles/avoid-the-calculus-trap" rel="noreferrer">link</a> that I had provided is from art of problem solving and hence I found it a bit harsh towards my situation, so I decided to ask here for guidance. I also read <a href="https://matheducators.stackexchange.com/questions/7718/effects-of-early-study-of-advanced-books">this</a> question but that is not exactly my case, I was thorough in my study, also took help from my teachers if I did not understand anything.</p>
| Daniel R. Collins | 5,563 | <p>I will also chime in and say that the argument on the linked Art of Problem Solving site is unpersuasive, and somewhat misses a broader point. </p>
<p>Ultimately, the <em>real</em> point of the mathematical discipline is to identify patterns in systems and prove their correctness (hopefully in an insightful, persuasive, explanatory style). The "trap" that I would identify is the standard calculus-track being about numerical calculations for grades K-14, and then mathematics majors needing to switch tracks to proof-based courses, which is the essence of the profession of mathematics. It's really a huge irritation that so many students are "tricked" into thinking that being good at following rules for calculations means that majoring in math is a good choice. (E.g., in reality, computers can compute any such K-14 exercise instantaneously, so humans doing calculations is not really in itself of any practical use.) I may argue that "problem solving" in terms of applications with numerical answers at the end is not any better in this regard.</p>
<p>So my top suggestion would be to supplement the time between high school geometry and the 2nd year of college with some kind of practice at reading and writing proofs (number theory being a common starting sandbox). I might recommend Richard Hammack's free, open-textbook <a href="http://www.people.vcu.edu/~rhammack/BookOfProof/" rel="nofollow noreferrer">Book of Proof</a> as a good starting point, even though it's written for a college audience -- if others have better suggestions I'd like to hear them. </p>
|
2,612,416 | <p>Can you please help me with this limit? I can´t use L'Hopital rule.</p>
<p>$$\lim_{x\to \infty} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^4}-1} $$</p>
| Atmos | 516,446 | <p>Make the ratio of the high degree terms which gives here
$$
\frac{2x}{x^{4/3}} \underset{x \rightarrow +\infty}{\rightarrow}0
$$
EDIT : </p>
<p>I will propose my way of doing things. You are studying the limit around $1$ then, makes it move to $0$. Let $x=1+h$
$$
\frac{\sqrt{4\left(1+h\right)^2+5}-3}{\sqrt[3]{1+h}-1}=\frac{\sqrt{8h+4h^2+9}-3}{\sqrt[3]{1+h}-1}
$$
First
$$
\sqrt{8h+4h^2+9}=3\sqrt{1+\frac{8}{9}h+\frac{4}{9}h^2}=3+\frac{8}{18}h+o\left(h\right)
$$
and
$$
\sqrt[3]{1+h}-1=1+\frac{h}{3}-1+o\left(h\right)=\frac{h}{3}+o\left(h\right)
$$
So</p>
<blockquote>
<p>$$
\frac{\sqrt{4\left(1+h\right)^2+5}-3}{\sqrt[3]{1+h}-1}\underset{(0)}{=}\frac{3+\frac{8}{18}h+o\left(h\right)-3}{\frac{h}{3}+o\left(h\right)}\underset{h \rightarrow 0}{\rightarrow}\frac{24}{18}=\frac{4}{3}
$$</p>
</blockquote>
<p>Then we conclude that</p>
<blockquote>
<p>$$
\frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^4}-1}\underset{x \rightarrow 1}{\rightarrow}\frac{4}{3}$$</p>
</blockquote>
|
404,574 | <p>Suppose that:</p>
<p>$Y \pmod B = 0$</p>
<p>$Y \pmod C = X$</p>
<p>I know $B$ and $C$. $Y$ is unknown, it might be an extremely large number, and it does not interest me. </p>
<p>The question is: Is it possible to find $X$, and if so, how?</p>
| André Nicolas | 6,312 | <p>In the case where $b$ aand $c$ are relatively prime, knowing $y\bmod b$ gives <strong>absolutely</strong> no information about $y\bmod c$. It could be any of $0,1,2, \dots,c-1$.</p>
<p>In the general case where $b$ and $c$ are not necessarily prime, let $d=\gcd(b,c)$. Knowing $y\bmod b$ tells us what $y\bmod d$ is. But $y\bmod c$ can take on <strong>any</strong> value compatible with the known value of $y\bmod d$. That is, if $y\bmod d=y_1$, then $y\bmod c$ can take on any of the values $y_1,y_1+d, y_1+2d, \dots,y_1+kd$, where $k=\frac{c}{d}-1$. </p>
|
90,459 | <p>I want to find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$. I observe that
$3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2$ so
$$
\mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2}+1)=\mathbb{Q}(\sqrt{2})
$$
so the degree is 2.</p>
<p>Is there a more mechanical way to show this without noticing the factorization?</p>
| Bruno Stonek | 2,614 | <p>Here's how I would do it:</p>
<p>Observe first that there is a tower of fields $\mathbb{Q}\subset \mathbb{Q}(\sqrt2)\subset \mathbb{Q}\left(\sqrt{3+2\sqrt2}\right)$.</p>
<p>Now, the first extension has degree 2. To determine the degree of the extension you're asking, by transitivity of degrees it suffices to compute the degree of the second extension.</p>
<p>This extension has degree $\leq 2$, since $\left(\sqrt{3+2\sqrt2}\right)^2=3+2\sqrt2\in \mathbb{Q}(\sqrt2)$.</p>
<p>Let us try to prove (or disprove) it is of degree 1, i.e. let us try to find $a,b\in \mathbb{Q}$ such that $\sqrt{3+2\sqrt2}=a+b\sqrt2$.</p>
<p>Squaring both sides, we get $3+2\sqrt2=a^2+2b\sqrt2+2b^2$.</p>
<p>This forces $b=1$ and then $a=1$, so actually $\sqrt{3+2\sqrt2}=1+\sqrt2$.</p>
<p>So $\left\lvert\mathbb{Q}\left(\sqrt{3+2\sqrt2}\right):\mathbb{Q}(\sqrt2)\right\rvert=1$, therefore $\left\lvert \mathbb{Q}\left(\sqrt{3+2\sqrt2}\right):\mathbb{Q}\right\rvert=2$.</p>
|
363,767 | <p>An ellipse is specified $ x^2 + 4y^2 = 4$, and a line is specified $x + y = 4$. I need to find the max/min distances from the ellipse to the line.</p>
<p>My idea is to find two points $(x_1, y_1)$ and $(x_2,y_2)$ such that the first point is on the ellipse and the second point is on the line. Furthermore, the line segment formed by these two points should be perpendicular to the line (slope = 1). This gives 3 constraints $g_i$ and an objective $f$. </p>
<p>$$ g_1: x_1 ^2 + 4y_1 ^2 - 4 = 0$$
$$ g_2: x_2 + y_2 - 4= 0 $$
$$ g_3:\frac{(y_2 - y_1)}{(x_2 - x_1)} - 1 = 0 $$
$$ f: (y_2 - y_1)^2 + (x_2- x_1)^2 $$ </p>
<p>Then I compute $\nabla g_i$ and $\nabla f$:</p>
<p>$$ \nabla g_1 = (2x_1, 8y_1, 0, 0) $$
$$ \nabla g_2 = (0, 0, 1, 1)$$
$$ \nabla g_3 = ( (y_2 - y_1)(x_2 - x_1)^{-2}, -(x_2 - x_1)^{-1}, -(y_2 - y_1)(x_2 - x_1)^{-2}, (x_2 - x_1)^{-1} )$$
$$ \nabla f = (-2(x_2-x_1), -2(y_2-y_1), 2(x_2-x_1), 2(y_2 - y_1))$$ </p>
<p>At this point I try to solve $\nabla f = \sum\lambda_i\nabla g_i$, which, together with the constraints, gives me 7 equations with 7 variables. I'm not sure how to solve this system. </p>
<p>$$\lambda_1 2 x_1 + \lambda_3(y_2-y_1)(x_2-x_1)^{-2} = -2(x_2-x_1)$$
$$\lambda_1 8 y_1 - \lambda_3 (x_2 - x_1)^{-1} = -2 (y_2 - y_1)$$
$$\lambda_2 - \lambda_3 (y_2 - y_1)(x_2 - x_1)^{-2} = 2(x_2 - x1)$$
$$\lambda_2 + \lambda_3 (x_2 - x_1)^{-1} = 2(y_2 -y_1)$$ </p>
<p>Is there an easy way to see the solutions of this system in $x_1, y_1, x_2, y_2$? If not, is there an easier formulation of this optimization problem?</p>
| DonAntonio | 31,254 | <p>The ellipse is</p>
<p>$$x^2+4y^2=4\iff \frac{x^2}{2^2}+y^2=1$$</p>
<p>and the line is $\,y=-x+4\,$ , which is then "above" the ellipse all the time and, in particular, in the first quadrant (draw an approximate sketck of the functions to see why is this relevant).</p>
<p>Thus, we can write down a point on the ellipse in the first quadrant as </p>
<p>$$\left(x\,,\,\frac{1}{2}\sqrt{4-x^2}\right)\;,\;\;x\ge 0$$</p>
<p>so using the formula for the distance between a point as above and the line $\,x+y-4=0\,$ we get</p>
<p>$$\frac{\left|\;x+\frac{1}{2}\sqrt{4-x^2}-4\;\right|}{\sqrt2}=\frac{\left|\;2x+\sqrt{4-x^2}-8\;\right|}{2\sqrt2}$$</p>
<p>But since the expression with the absolute value is always <strong>negative</strong> (in $\,0\le x\le 2\,$ , that is)(why?) , we must get the minimum of:</p>
<p>$$f(x)=\frac{-2x-\sqrt{4-x^2}+8}{2\sqrt 2}\implies f'(x)=\frac{-2+\frac{x}{\sqrt{4-x^2}}}{2\sqrt 2}=0\iff$$</p>
<p>$$-2\sqrt{4-x^2}+x=0\iff x^2=16-4x^2\iff x=\frac{4}{\sqrt 5}$$</p>
<p>Check the above indeed is a minimum point and input in the resp. equation...</p>
<p>Of course, the maximal distance is...</p>
|
2,266,573 | <p>I am working through some problems about probability and seem to be having trouble working through this one in particular. I'd love some help learning how to go about solving problems such as this.</p>
<p>A website estimates that 19% of people have a phobia regarding public speaking. If three students are assigned to a project group, what's the probability...</p>
<p>a.) That all 3 students have a fear of public speaking.</p>
<p>b.) That none have a fear of public speaking</p>
<p>c.) That at least one of the students has a fear of public speaking.</p>
| The Dead Legend | 433,379 | <p>For this question,:</p>
<p>A) If one kid has a fear then probability of picking him will be $0.19$ while picking one more reduces the probability to $(0.19)*(0.19)$ . For three people: it becomes $(0.19)^3$
This is from selection via independent events.</p>
<p>B)None have fear: $(1-0.19)^3$<br>
This is via noting that 81% population won't be fearful as such. and following the above principal again.</p>
<p>C) At least one have fear: $1-(1-0.19)^3$<br>
This one is by asking in account that for at least one to be fearful, we may take it as Anti-event when none is fearful. Thus 1-P(part B) </p>
|
2,266,573 | <p>I am working through some problems about probability and seem to be having trouble working through this one in particular. I'd love some help learning how to go about solving problems such as this.</p>
<p>A website estimates that 19% of people have a phobia regarding public speaking. If three students are assigned to a project group, what's the probability...</p>
<p>a.) That all 3 students have a fear of public speaking.</p>
<p>b.) That none have a fear of public speaking</p>
<p>c.) That at least one of the students has a fear of public speaking.</p>
| John Doe | 399,334 | <p>Use the fact that these students are chosen independently, and so $$\Bbb P(A\text{ and }B\text{ and }C)=\Bbb P(A)\Bbb P(B)\Bbb P(C)$$ For your problem, you could say $A$ is the event that the first student has a fear of public speaking, etc. You can also use this property for part $B$.</p>
<p>For questions where they say something like "at least", think of <em>the opposite</em> - the opposite of at least one student having this fear is for none of them to have this fear. This is easier to calculate, and in fact you have already worked it out in another question. Then the probability you'd want is $1-\Bbb P(\text{the opposite})$.</p>
|
1,278,329 | <p>Solve the recurrence $a_n = 4a_{n−1} − 2 a_{n−2}$</p>
<p>Not sure how to solve this recurrence as I don't know which numbers to input to recursively solve?</p>
| k1.M | 132,351 | <p>Hint: Let $r_1,r_2$ be two distinct real roots of the equation
$$
r^2-4r+2
$$
then this recurrence equation has a solution of the form
$$
C_1 r_1^n+C_2 r_2^n
$$
which the constants $C_1,C_2$ can be found by initial values condition.</p>
|
1,902,138 | <p>It's common to see a plus-minus ($\pm$), for example in describing error
$$
t=72 \pm 3
$$
or in the quadratic formula
$$
x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
$$
or identities like
$$
\sin(A \pm B) = \sin(A) \cos(B) \pm \cos(A) \sin(B)
$$</p>
<p>I've never seen an analogous version combining multiplication with division, something like $\frac{\times}{\div}$</p>
<blockquote>
<p>Does this ever come up, and if not why?</p>
</blockquote>
<p>I suspect it simply isn't as naturally useful as $\pm$. </p>
| marty cohen | 13,079 | <p>As you indicated, square root can be + or -.
$\pm$ shows this ambiguity.</p>
<p>As far as I know,
there is no similar use case
where the choice is
to multiply or divide
by an expression. </p>
|
1,043,956 | <p>Find a normal vector and a tangent vector to the curve given by the equation: $x^5 + y ^5 =2x^3$ at the point $P(1, 1)$. Find the equation of the tangent line. <br/>
Edit: The notes I have:<img src="https://i.stack.imgur.com/arOee.png" alt="enter image description here"></p>
<p>Taking $f(x, y) = x^5 - 2x^3 + y^5 = 0$, I got $m_n = (-1,5), m_t = (5, 1)$ and $x = 5y-4$ for the tangent line. <br/></p>
<p>I'm very much unsure on if this formula is what I should/can use and if I've used it correctly. </p>
| ploosu2 | 111,594 | <p>Differentiate explicitly:</p>
<p>$$5x^4 + 5y^4y' = 6x^2$$
Solve for $y'$</p>
|
1,262,174 | <p>I am currently teaching Physics in an Italian junior high school. Today, while talking about the <a href="http://en.wikipedia.org/wiki/Dipole#/media/File:Dipole_Contour.svg" rel="noreferrer">electric dipole</a> generated by two equal charges in the plane, I was wondering about the following problem:</p>
<blockquote>
<p>Assume that two equal charges are placed in <span class="math-container">$(-1,0)$</span> and <span class="math-container">$(1,0)$</span>.</p>
<p>There is an equipotential curve through the origin, whose equation is
given by:
<span class="math-container">$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 $$</span> and
whose shape is very <a href="http://en.wikipedia.org/wiki/Lemniscate" rel="noreferrer">lemniscate</a>-like:</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/oojJw.png" alt="enter image description here" /></p>
<blockquote>
<p><strong>Is there a fast&tricky way to compute the area enclosed by such a curve?</strong></p>
</blockquote>
<p>Numerically, it is <span class="math-container">$\approx 3.09404630427286$</span>.</p>
| Hosein Rahnama | 267,844 | <p>At the first step, I will introduce a proper curve linear coordinates for this problem. This will help to construct the integral for area. We can write the equation of these equi-potential curves as</p>
<p>$$\frac{1}{r_1}+\frac{1}{r_2}=C \tag{1}$$</p>
<p>where $C$ is some real constant and $r_1$ and $r_2$ are defined as</p>
<p>$$r_1=\sqrt{(x-a)^2+y^2} \\
r_2=\sqrt{(x+a)^2+y^2} \tag{2}$$</p>
<p>where $2a$ is the distance between the two like charges on the $x$-axis placed at $x=a$ and $x=-a$. Equation $(2)$ is introducing a new curve-linear coordinates $(r_1,r_2)$ which is called the <strong>two-center bipolar coordinates</strong>. The geometric interpretation is easy as it just describes the coordinates of a point in $xy$ plane via the distance of that point through two other points which are called the <strong>centers</strong>. You can take look at <a href="https://en.wikipedia.org/wiki/Two-center_bipolar_coordinates" rel="nofollow noreferrer">this link in WIKI</a> or <a href="https://math.stackexchange.com/questions/346696/how-to-handle-two-center-bipolar-coordinates">this post on MSE</a>. However, they doesn't contain that much information.</p>
<p><a href="https://i.stack.imgur.com/zSVgy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zSVgy.png" alt="enter image description here"></a></p>
<p>Then we find $x$ and $y$ from equations in $(2)$ in terms of $r_1$ and $r_2$. For this purpose, subtract and add the equations in $(4)$ to get</p>
<p>$$\begin{align}
r_2^2-r_1^2&=4ax \\
r_2^2+r_1^2&=2(x^2+y^2+a^2)
\end{align}
\tag{3}$$</p>
<p>after some simplifications, $x$ and $y$ in terms of $r_1$ an $r_2$ will be</p>
<p>$$
\begin{align}
x &= \frac{1}{4a} (r_{2}^{2}-r_{1}^{2})\\
y &= \pm \frac{1}{4a} \left( \sqrt{16 a^2 r_{2}^{2}-(r_{2}^{2}-r_{1}^{2}+4a^2)^2} \right)
\end{align}
\tag{4}$$</p>
<p>For a given $(r_1,r_2)$ we will find two pairs $(x,y)$ and $(x,-y)$. It is evident from the above picture that why this happens.</p>
<p>The next step will be the construction of the integral for area. We set $C=2a=2$ and find the intersection of the $\infty$ shaped curve with the $x$-axis</p>
<p>$$y=0 \qquad \to \qquad \frac{1}{\sqrt{(x-1)^2}}+\frac{1}{\sqrt{(x+1)^2}}=2 \qquad \to \qquad x=-\phi,0,\phi \tag{5}$$ </p>
<p>where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio number. So according to $(1)$, $(4)$, and $(5)$ the parametric equations of the curve in the <strong>second quadrant</strong> of $xy$ plane will be</p>
<p>$$
\begin{align}
x &= \frac{1}{4} \left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}\right)\\
y &= \frac{1}{4} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2}
\end{align}
\qquad \qquad \phi-1 \le r_2 \le 1
\tag{6}$$</p>
<p><a href="https://i.stack.imgur.com/2J6Ds.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2J6Ds.png" alt="enter image description here"></a></p>
<p>Finally, the integral to be evaluated for the area will be</p>
<p>$$\begin{align}
\text{Area} &=4 \int_{-\phi}^{0} y dx \\
&=4 \int_{\phi-1}^{1}y \frac{dx}{dr_2}dr_2 \\
&= \int_{\phi-1}^{1} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2} \left(\frac{r_2}{2}-\frac{r_2}{2(2r_2-1)^2}+\frac{r_{2}^{2}}{(2r_2-1)^3}\right) dr_2
\end{align}$$</p>
<p>The numerical value of area up to fifty digits is</p>
<p>$$\text{Area}=3.0940463058814386237217800770286020796565427678113$$</p>
<p>as stated in the question. However, finding a tricky way to evaluate the definite integral of the area in a closed form should be investigated. I did the computations in <a href="http://pc.cd/EYcrtalK" rel="nofollow noreferrer">this MAPLE file</a> which may be useful for anyone who reads this post.</p>
|
1,262,174 | <p>I am currently teaching Physics in an Italian junior high school. Today, while talking about the <a href="http://en.wikipedia.org/wiki/Dipole#/media/File:Dipole_Contour.svg" rel="noreferrer">electric dipole</a> generated by two equal charges in the plane, I was wondering about the following problem:</p>
<blockquote>
<p>Assume that two equal charges are placed in <span class="math-container">$(-1,0)$</span> and <span class="math-container">$(1,0)$</span>.</p>
<p>There is an equipotential curve through the origin, whose equation is
given by:
<span class="math-container">$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 $$</span> and
whose shape is very <a href="http://en.wikipedia.org/wiki/Lemniscate" rel="noreferrer">lemniscate</a>-like:</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/oojJw.png" alt="enter image description here" /></p>
<blockquote>
<p><strong>Is there a fast&tricky way to compute the area enclosed by such a curve?</strong></p>
</blockquote>
<p>Numerically, it is <span class="math-container">$\approx 3.09404630427286$</span>.</p>
| Hosein Rahnama | 267,844 | <p>Here is another method based on the curve-linear coordinates introduced by <em>Achille Hui</em>. He introduced the following change of variables</p>
<p>$$\begin{align}
\sqrt{(x+1)^2+y^2} &= u+v\\
\sqrt{(x-1)^2+y^2} &= u-v
\end{align} \tag{1}$$</p>
<p>Then solving for $x$ and $y$ we shall get</p>
<p>$$\begin{align}
x &= u v\\
y &= \pm \sqrt{-(u^2-1)(v^2-1)}
\end{align} \tag{2}$$</p>
<p>required that</p>
<p>$$-(u^2-1)(v^2-1) \ge 0 \tag{3}$$</p>
<p>It does not look familiar but in fact it is! Taking into account the equations $(2)$ and $(3)$, we can consider the following as a parameterization for the <strong>first quadrant</strong> of the $xy$ plane</p>
<p>$$\boxed{
\begin{array}{}
x=uv & & 1 \le u \lt \infty \\
y=\sqrt{-(u^2-1)(v^2-1)} & & 0 \le v \le 1
\end{array}} \tag{4}$$</p>
<p>I tried to draw the coordinate curves of this curve-linear coordinates and I just noticed that it is exactly the same as the <strong>Elliptic Coordinates</strong> and nothing else! You can show this analytically by the change of variables</p>
<p>$$\begin{align}
u &= \cosh p \\
v &= \cos q
\end{align} \tag{5}$$</p>
<p>I leave the further details in this avenue to the reader.</p>
<p><a href="https://i.stack.imgur.com/RgOZC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RgOZC.png" alt="enter image description here"></a></p>
<p>Let us go back to the problem of calculating the area. The equation of the $\infty$ curve was</p>
<p>$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 \tag{6}$$</p>
<p>so combining $(1)$ and $(6)$ leads to</p>
<p>$$v=\pm \sqrt{u^2-u} \tag{7}$$ </p>
<p>and hence the <strong>parametric equation</strong> of the $\infty$ curve in the first quadrant by considering $(4)$ and $(7)$ will be</p>
<p>$$\boxed{
\begin{array}{}
x=u\sqrt{u^2-u} & & 1 \le u \lt \phi \\
y=\sqrt{-(u^2-1)(u^2-u-1)}
\end{array}}
\tag{8}$$</p>
<p>and finally the integral for the area is</p>
<p>$$\begin{align}
\text{Area} &=4 \int_{0}^{\phi} y dx \\
&=4 \int_{1}^{\phi}y \frac{dx}{du}du \\
&=2 \int_{1}^{\phi} (4u-3)\sqrt{-u(u+1)(u^2-u-1)}du \\
&\approx 3.09405
\end{align} \tag{9}$$</p>
|
102,963 | <p>Could someone please explain the difference between the group of all icosahedral symmetries and S5? I know that the former is a direct product, but don't they work the same? Say I have an icosahedron, why wouldn't S5 work as a description of its symmetries? Thank you very much.</p>
<p><strong>Added:</strong> When counting the symmetries of a platonic solid, in this case the icosahedron, Does it include reflecting along a plane cutting through the solid, in a sort of turing itself inside out reflection? I read that the symmetries counted should be "orientation-preserving". What does that mean?</p>
| Arkady | 23,522 | <p>Two groups are treated as same if there is an isomorphism between them.A simple reason why $S_5$ cannot be used to describe the symmetries of an icosahedron(whose group of symmetries we will call $I_h$) is that its structure is fundamentally different from that of $I_h$. For starters, $S_5$ cannot be expressed as a direct product of two groups unlike $I_h$, for which it is possible to do so. So $S_5$ cannot be isomorphic to $I_h$.</p>
|
126,553 | <p>In <a href="http://www.icpr2010.org/pdfs/icpr2010_MoAT5.1.pdf" rel="nofollow">this paper</a>, in the Formula at the beginning of 2.2, we have</p>
<p>$B=\{b_i(O_t)\}$</p>
<p>where </p>
<p>$i=0,1$ - the number of probability formula</p>
<p>$O_t$ - the state at moment $t$</p>
<p>$b_i(O_t)$ - two probabilities or estimations for the state $O_t$</p>
<p>The result ($B$) is named "likelihood". How can likelihood be obtained from 2 numbers? Is this a weighted average like</p>
<p>$b_i(O_t)= 0*b_0(O_t) + 1*b_1(O_t)$</p>
<p>or </p>
<p>$b_i(O_t)= -1*b_0(O_t) + 1*b_1(O_t)$</p>
<p>or something?</p>
| joriki | 6,622 | <p>This is a good example of why it makes sense to quote texts with more context, since it's often the context that supplies the clues for interpretation.</p>
<p>$B$ is referred to not just as a likelihood, but as "the likelihood [...] of [...] the frame [...] being a speech or a noise frame". This is rather badly phrased (like some other things in that paper), but it's clear that a single number cannot be simultaneously the likelihood of being a speech or a noise frame, so $B$ is the pair of likelihoods for these two cases. The curly brace notation for this pair is somewhat unfortunate in my view.</p>
|
16,105 | <p>The answer to this question should be obvious, but I can't seem to figure it out. Suppose we have a surface $F$, and a representation $\rho : \pi_1(F)\to SU(n)$. We can define the homology with local coefficients $H_*(F,\rho)$ straightforwardly as the homology of the twisted complex $$C_*(F,\rho):=C_*(\widetilde{F};\mathbf{Z})\otimes_{\mathbf{Z}[\pi_1(F)]} \mathbf{C}^n$$ where $\widetilde{F}$ is the universal cover, and $\mathbf{Z}[\pi_1(F)]$ acts on each side in the obvious way. </p>
<p>Now, this complex is actually very easy to compute explicitly: just lift a nice basis of cells in $F$ to $\widetilde{F}$, and write down the boundary maps explicitly. For example, if $F$ is a torus and we take $n=2$, say, we can choose a natural meridian-longitude basis $(x,y)$ for $H_1(F)$, and the twisted boundary map $\partial_1:C_1(F,\rho)=\mathbf{C}^4\to C_2(F,\rho)=\mathbf{C}^2$ is $$ \left( \begin{array}{ccc}
\rho(x)-Id \newline\rho(y)-Id\end{array} \right)$$</p>
<p>So, here's my question. Since $\rho$ is a unitary representation, we should get a twisted intersection form on $H_1(F)$, simply by combining the untwisted intersection form with the standard hermitian product on $\mathbf{C}^2$, right? And I would imagine this is also really easy to compute, in a similar basis, say? I can't seem to figure out how it would go. Could anyone help me, even show me how it works for the same torus example?</p>
<p>Or, if I've said anything wrong, tell me where?</p>
| Paul | 3,874 | <p>What you say is right, and makes sense on any even dimensional manifold. Computing it can be tricky: a useful approach is to use a regular cell complex and the dual complex, then on the chain level the intersection form is given by the identity matrix (see the first couple pages of Milnor's "a duality theorem for Reidemeister torsion").</p>
<p>One suggestion for calculation is to assume $\rho$ is irreducible, since if $C^n$ splits invariantly under the $\pi_1F$ action so does the cohomology. In your torus example, since $\pi_1=Z\oplus Z$ is abelian, the only irreducible reps are 1-dimensional. For Euler characteristic reasons (and Poincare duality) in this case it turns out either the rep is trivial in which case you know the answer, or else the rep is non-trivial in which case the homology vanishes and the intersection form is trivial. For higher genus surfaces you will get something non-zero, but in this dimension you get a skew-hermitian form, which is determined up to iso by its rank, if I'm thinking clearly. For dimensions divisible by 4, you can get interesting (i.e. non-zero) signature, but for a closed manifold it will just equal n times the ordinary signature by the twisted form of the Hirzebruch signature theorem. </p>
<p>But the twisted intersection form is interesting when your manifold has non-empty boundary, since it gives invariants of the boundary. Hundreds of papers are based on this observation. Even when your surface has non-empty boundary you get something interesting.</p>
|
2,352,313 | <p>If $f_n$ is the number of permutations of numbers $1$ to $n$ that no number is in it's place(I think same as $D_n$)and $g_n$ is the number of the same permutations with exactly one number in it's place Prove that $\mid f_n-g_n \mid =1$.</p>
<p>I need a proof using mosly combinatorics not mostly algebra.I think we should find sth like below:</p>
<p>$f_n-g_n=g_{n-1}-f_{n-1}$</p>
<p>But I can't do that.</p>
| Christian Blatter | 1,303 | <p>The <em><a href="https://en.wikipedia.org/wiki/Rencontres_numbers" rel="nofollow noreferrer">rencontres numbers</a></em> $D_n$ satisfy the recursion
$$D_n=(n-1)(D_{n-1}+D_{n-2})\ ,\tag{1}$$
which can be proven as follows: You obtain a derangement $\pi\in{\cal S}_n$ by picking a derangement $\pi'\in{\cal S}_{n-1}$ and writing the entry $n$ at an arbitrary place in the cycle representation of $\pi'$, or by choosing a $k\in[n-1]$ and adding the transposition $(n,k)$ to a derangement of $[n-1]\setminus\{k\}$.</p>
<p>From $(1)$ one easily deduces
$$D_n-nD_{n-1}=-\bigl(D_{n-1}-(n-1)D_{n-2}\bigr)=\ldots=(-1)^n\ .$$
This should solve your problem.</p>
|
2,386,602 | <p>This is a question from an exam I recently failed. </p>
<p>What is the radius of convergence of the following power series? $$(a) \sum_{n=1}^\infty(n!)^2x^{n^2}$$ and $$(b) \sum_{n=1}^\infty \frac {x^{n^2}}{n!}$$</p>
<p>Edit: Here's my attempt at the first one, if someone could tell me if it's any good...</p>
<p>$\sum_{n=1}^\infty(n!)^2x^{n^2}$=$\sum_{n=1}^\infty a_nx^n$ where $a_n= 0$ for any $n\notin \{k\in N| \exists t\in N: k=t^2\}$ and $a_n = (n!)^2$ else. So the radius of convergence would be the inverse of $\lim_{n\rightarrow \infty}{(n!)^{2/n}}=\lim e^{2/n\cdot log(n!) }$. The exponent with log of factorial becomes a series, $\sum_{n=1}^{\infty} \frac{logn}{n}$ which diverges by comparison test with $\frac{1}{n}$, so the radius of convergence would be equal to $0$.</p>
<p>Second edit: This is wrong. Correct answer below.</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>The ration test gives for $a) $,</p>
<p>$$\lim_\infty (n+1)^2e^{(2n+1)\ln (|x|)}$$
$$=0$$ if $|x|<1$.</p>
|
3,062,701 | <p>I want to solve this system by Least Squares method:<span class="math-container">$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$</span> This symmetric matrix is singular with one eigenvalue <span class="math-container">$\lambda1 = 0$</span>, so <span class="math-container">$\ A^t\cdot A$</span> is also singular and for this reason I cannot use the normal equation: <span class="math-container">$\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b $</span>.
So I performed Gauss-Jordan to the extended matrix to come with <span class="math-container">$$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$</span>
Finally I solved the <span class="math-container">$\ 2x2$</span> system: <span class="math-container">$$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$</span> taking into account that the best <span class="math-container">$\ \hat b\ $</span> is <span class="math-container">$\begin{pmatrix}1\\3\\0\end{pmatrix}$</span></p>
<p>The solution is then <span class="math-container">$\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$</span></p>
<p>Is this approach correct ? </p>
<p><strong>EDIT</strong></p>
<p>Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: <span class="math-container">$(A^tA)\hat x=A^t b $</span></p>
<p><span class="math-container">$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$</span>
The reduced echelon from the augmented is:
<span class="math-container">$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$</span> for the independent variable case that <span class="math-container">$z=\alpha , \alpha=0 $</span> </p>
| Damien | 621,834 | <p>The RLS solution is given by
<span class="math-container">$$ \hat x = A^+ \, b$$</span>
where <span class="math-container">$A^+$</span> is the pseudo inverse of <span class="math-container">$A$</span>.</p>
<p>As <span class="math-container">$A$</span> is not full rank, it is not possible effectively to calculate it with the simple formula you mentioned.<br>
However, it is still possible to calculate it with numerical solutions, for example based on the SVD.</p>
<p>By doing so, we get:</p>
<p><span class="math-container">$$ \hat x = \begin{pmatrix}-41/12\\-1/2\\29/12\end{pmatrix} $$</span></p>
<p>For a value <span class="math-container">$A \, \hat x$</span> equal to:</p>
<p><span class="math-container">$$ A \, \hat x = \begin{pmatrix}17/6\\4/3\\-1/6\end{pmatrix} $$</span></p>
<p>Your method does not seem to work. I can only give my interpretation of what happens:</p>
<p>The issue is your choice of the "best" <span class="math-container">$\hat b$</span> value. The vector that you considered is not directly related to the real <span class="math-container">$b$</span> vector, but to something obtained after some manipulations on the rows of the linear system matrix. </p>
<p>Difficult in this situation to rely it to a selection of a "best" <span class="math-container">$\hat b$</span> vector</p>
|
1,515,776 | <p>How can I solve something like this?</p>
<p>$$3^x+4^x=7^x$$</p>
<p>I know that $x=1$, but I don't know how to find it. Thank you!</p>
| cr001 | 254,175 | <p>For $x>1$ obviously $(3+4)^x > 3^x+4^x$ by binomial theorem.</p>
<p>For $1>x>0$, we have $(3^x+4^x)^{1\over x} > 3^{x{1\over x}}+4^{x{1\over x}}$ since ${1\over x} > 1$ and hence $3^x+4^x > (3+4)^x$</p>
<p>For $x< 0$, let $y=-x$ then $({1\over 3})^y+({1\over 4})^y>({1\over3})^y > ({1\over7})^y$</p>
|
1,515,776 | <p>How can I solve something like this?</p>
<p>$$3^x+4^x=7^x$$</p>
<p>I know that $x=1$, but I don't know how to find it. Thank you!</p>
| Narasimham | 95,860 | <p>By putting the given equation in the form:</p>
<p>$$ {\left( \dfrac {3}{3+4} \right ) } ^x + {\left( \dfrac {4}{3+4} \right ) } ^x =1 $$ </p>
<p>we find it is satisfied by $x=1$. The monotonic nature of exp function gives no other real roots.</p>
|
2,806,858 | <p>There is an equation
$$\sin2\theta=\sin\theta$$
We need to show when the right-hand side is equal to the left-hand side for $[0,2\pi]$.
<hr>
Let's rewrite it as
$$2\sin\theta\cos\theta=\sin\theta$$
Let's divide both sides by $\sin\theta$ (then $\sin\theta \neq 0 \leftrightarrow \theta \notin \{0,\pi,2\pi\}$)
$$2\cos\theta=1$$
$$cos\theta=\frac{1}{2}$$
$$\theta\in\left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
<hr>
Now, let's try something different.
$$2\sin\theta\cos\theta=\sin\theta$$
$$2\sin\theta\cos\theta-\sin\theta=0$$
$$\sin\theta(2\cos\theta-1)=0$$
We can have the solution when $\sin\theta=0$.
$$\sin\theta=0$$
$$\theta \in \left\{0,\pi,2\pi\right\}$$
And when $2\cos\theta-1=0$.
$$2\cos\theta-1=0$$
$$2\cos\theta=1$$
$$\cos\theta=\frac{1}{2}$$
$$\theta \in \left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
Therefore the whole solution set is
$$\theta \in \left\{0,\pi,2\pi,\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
<strong>This is the correct solution.</strong>
<hr>
Why is this happening? In the first approach, the extra solution given by $\sin\theta=0$ is not only non-appearing but actually banned. Both approaches look valid to me, yet the first one yields less solutions than the second one. Is the first approach invalid in some cases? This is not the only case when this happens, so I'd like to know when I need to use the second approach to solve the equation, so I don't miss any possible solutions.</p>
| Misha Lavrov | 383,078 | <p>Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by $0$ is not valid.</p>
<p>So going from $2 \sin \theta \cos \theta = \sin \theta$ to $2 \cos\theta = 1$ is only valid when $\sin\theta \ne 0$.</p>
<p>In general, all this means is that you need to check the $\sin \theta = 0$ case separately. For example, going from $2 \sin\theta \cos\theta = 1$ to $2 \cos\theta = \frac1{\sin\theta}$ is also only valid when $\sin\theta \ne 0$, but you don't lose any solutions by doing this, because values of $\theta$ for which $\sin\theta=0$ weren't solutions to begin with.</p>
<p>But in this particular case, when $\sin\theta = 0$, the equation $2\sin\theta \cos\theta = \sin\theta$ is satisfied, so it's correct to go from this to
$$
2\cos\theta = 1 \text{ or } \sin\theta = 0.
$$</p>
|
4,090,408 | <p>Show that <span class="math-container">$A$</span> is a whole number: <span class="math-container">$$A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}.$$</span>
I don't know if this is necessary, but we can compare <span class="math-container">$40\sqrt{2}$</span> and <span class="math-container">$57$</span>: <span class="math-container">$$40\sqrt{2}\Diamond57,\\1600\times2\Diamond 3249,\\3200\Diamond3249,\\3200<3249\Rightarrow 40\sqrt{2}<57.$$</span> Is this actually needed for the solution? So <span class="math-container">$$A=\sqrt{57-40\sqrt2}-\sqrt{40\sqrt2+57}.$$</span> What should I do next?</p>
| Angelo | 771,461 | <p>Actually it is not necessary to know if <span class="math-container">$\;40\sqrt2\;$</span> is greater or less than <span class="math-container">$\;57$</span>.
Moreover we do not need to square <span class="math-container">$\;A\;$</span> or to solve any system of equations.</p>
<p>The check that the original poster does about the comparison between <span class="math-container">$\;40\sqrt2\;$</span> and <span class="math-container">$\;57\;$</span> is not necessary, indeed just the same we can remove the absolute value inside the first radical as soon as we get the square of a subtraction.</p>
<p><span class="math-container">$\begin{align}
A&=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}=\\
&=\sqrt{\left|40\sqrt2-32-25\right|}-\sqrt{40\sqrt2+32+25}=\\
&=\sqrt{\left|40\sqrt2-\left(4\sqrt2\right)^2-5^2\right|}-\sqrt{40\sqrt2+\left(4\sqrt2\right)^2+5^2}=\\
&=\sqrt{\left|\left(4\sqrt2\right)^2+5^2-40\sqrt2\right|}-\sqrt{\left(4\sqrt2\right)^2+5^2+40\sqrt2}=\\
&=\sqrt{\left|\left(4\sqrt2-5\right)^2\right|}-\sqrt{\left(4\sqrt2+5\right)^2}=\\
&=\sqrt{\left(4\sqrt2-5\right)^2}-\sqrt{\left(4\sqrt2+5\right)^2}=\\
&=\left|4\sqrt2-5\right|-\left(4\sqrt2+5\right)=\\
&\underset{\color{blue}{\overbrace{\text{but }\;4\sqrt2-5=\sqrt{32}-\sqrt{25}>0}}}{=}\left(4\sqrt2-5\right)-\left(4\sqrt2+5\right)=\\
&=-10\;.
\end{align}$</span></p>
<p>Hence <span class="math-container">$A$</span> is the integer <span class="math-container">$-10$</span>.</p>
|
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