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 |
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4,544,787 | <p>These sums showed up in a probability problem I was working on. They're not quite the Stirling numbers of the first kind since it's possible to have e.g. <span class="math-container">$i_1 = i_2$</span>. Denoting the sum by <span class="math-container">$(k\mid n)$</span> we have the recurrence relation</p>
<p><span class="math-container">$(k\mid n) = n(k-1\mid n) + (k\mid n-1)$</span></p>
<p>I thought maybe the average term of the sum would be similar to the average product over a random list of <span class="math-container">$k$</span> numbers from <span class="math-container">$1$</span> to <span class="math-container">$n$</span>. But it seems to always be a few times greater.</p>
<p>I'm beginning to despair of there being a closed form expression for <span class="math-container">$(k\mid n)$</span> in terms of factorials and powers. Does anyone know how to analyze the sum further and perhaps find a good approximation?</p>
| Parcly Taxel | 357,390 | <p>Here I will show by induction that the Stirling numbers of the second kind give the general answer for the multisubset sum: with <span class="math-container">$Q_{n,k}$</span> being the set of all <span class="math-container">$k$</span>-multisubsets of <span class="math-container">$[n]=\{1,\dots,n\}$</span>,
<span class="math-container">$$\sum_{s\in Q_{n,k}}\prod_{i=1}^ks_i=S(n+k,n)$$</span>
For <span class="math-container">$n=1$</span> the equation holds since the only multisubset in <span class="math-container">$Q_{1,k}$</span> is <span class="math-container">$\{1,1,\dots,1\}$</span>, leading to a sum of <span class="math-container">$1$</span>, while <span class="math-container">$S(1+k,1)=1$</span>.</p>
<p>For <span class="math-container">$k=1$</span> the equation holds because the multisubset sum reduces to a sum of the numbers in <span class="math-container">$[n]$</span>:
<span class="math-container">$$\sum_{i=1}^ni=\binom{n+1}2=S(n+1,n)$$</span>
Now for <span class="math-container">$n,k>1$</span> the multisubsets in <span class="math-container">$Q_{n,k}$</span> can be separated into those with the largest possible element <span class="math-container">$n$</span> and those without it. The first subset of multisubsets is <span class="math-container">$Q_{n,k-1}$</span> with <span class="math-container">$n$</span> added to each multisubset; the second subset is <span class="math-container">$Q_{n-1,k}$</span>. Thus
<span class="math-container">$$\sum_{s\in Q_{n,k}}\prod_{i=1}^ks_i=n\sum_{s\in Q_{n,k-1}}\prod_{i=1}^ks_i+\sum_{s\in Q_{n-1,k}}\prod_{i=1}^ks_i$$</span>
Assuming the relation is true for <span class="math-container">$(n,k-1)$</span> and <span class="math-container">$(n-1,k)$</span> – inducting on <span class="math-container">$n+k$</span>:
<span class="math-container">$$\sum_{s\in Q_{n,k}}\prod_{i=1}^ks_i=nS(n+k-1,n)+S(n+k-1,n-1)$$</span>
But <span class="math-container">$S(n,k)=kS(n-1,k)+S(n-1,k-1)$</span>, so the left-hand side is <span class="math-container">$S(n+k,n)$</span> by substituting <span class="math-container">$(n,k)\leftarrow(n+k,n)$</span>. This proves the induction, so the multinomial sum is <span class="math-container">$S(n+k,n)$</span> for all <span class="math-container">$n,k\ge1$</span>.</p>
|
4,544,787 | <p>These sums showed up in a probability problem I was working on. They're not quite the Stirling numbers of the first kind since it's possible to have e.g. <span class="math-container">$i_1 = i_2$</span>. Denoting the sum by <span class="math-container">$(k\mid n)$</span> we have the recurrence relation</p>
<p><span class="math-container">$(k\mid n) = n(k-1\mid n) + (k\mid n-1)$</span></p>
<p>I thought maybe the average term of the sum would be similar to the average product over a random list of <span class="math-container">$k$</span> numbers from <span class="math-container">$1$</span> to <span class="math-container">$n$</span>. But it seems to always be a few times greater.</p>
<p>I'm beginning to despair of there being a closed form expression for <span class="math-container">$(k\mid n)$</span> in terms of factorials and powers. Does anyone know how to analyze the sum further and perhaps find a good approximation?</p>
| Mike Earnest | 177,399 | <p>Parcly has given a proof that
<span class="math-container">$$
\sum_{1\le i_1\le \dots \le i_k\le n}i_1\cdots i_k={n+k \brace n}\tag{$\star$}
$$</span>
Surely, such a beautiful equation must have a combinatorial proof? Indeed it does!</p>
<p>Given a partition of <span class="math-container">$\{1,2,\dots,n+k\}$</span> into <span class="math-container">$n$</span> unlabeled parts, define the "leader" of each part to be its smallest element. Every other element of <span class="math-container">$\{1,\dots,n+k\}$</span> is a "follower." Let <span class="math-container">$F_1,\dots,F_k$</span> be the list of followers, sorted in increasing order, so <span class="math-container">$F_1<\dots<F_k$</span>.</p>
<p>Then, define the "signature" of a partition to be the sequence <span class="math-container">$(i_1,\dots,i_k)$</span>, where for each <span class="math-container">$r\in \{1,\dots,k\}$</span>, <span class="math-container">$i_r$</span> is the number of leaders which are less than <span class="math-container">$F_r$</span>. Here is an example:
<span class="math-container">$$
\color{red}{1},2,5,6 \mid \color{red}{3}, 9 \mid \color{red}{4}\mid \color{red}7,8
$$</span>
The leaders are in red. The signature of this partition is <span class="math-container">$(1,3,3,4,4)$</span>, because the followers are <span class="math-container">$2,5,6,8,9$</span>, and there is <span class="math-container">$1$</span> leader which is numerically less than <span class="math-container">$2$</span> (only the leader <span class="math-container">$1$</span> is less than <span class="math-container">$2$</span>), there are <span class="math-container">$3$</span> leaders less than <span class="math-container">$5$</span> (namely, <span class="math-container">$1,3$</span> and <span class="math-container">$4$</span>), there are <span class="math-container">$3$</span> leaders less than <span class="math-container">$6$</span>, and there are four leaders less than both <span class="math-container">$8$</span> and <span class="math-container">$9$</span>.</p>
<p>Note that a signature must be a weakling increasing sequence. I claim that for each potential signature <span class="math-container">$(i_1,i_2,\dots,i_k)$</span>, there are exactly <span class="math-container">$i_1\times \dots \times i_k$</span> partitions with that signature. Indeed, imagine choosing a partition with signature <span class="math-container">$(i_1,\dots,i_k)$</span> one element at a time, starting at <span class="math-container">$1$</span> and proceeding in numerical order.</p>
<ul>
<li><p>The signature <span class="math-container">$(i_1,\dots,i_k)$</span> completely determines the set of leaders. There is only one way to choose the part for a leader, since they must be put in a new part, and all the currently empty parts are identical.</p>
</li>
<li><p>When choosing the <span class="math-container">$r^\text{th}$</span> follower, it must be placed in a currently nonempty part, with some other leader. There are currently <span class="math-container">$i_r$</span> nonempty parts, so there are <span class="math-container">$i_r$</span> ways to place this follower.</p>
</li>
</ul>
<p>Summing over all possible signatures, we have proved <span class="math-container">$(\star)$</span>.</p>
|
1,135,045 | <p>I need to compute
\begin{align}
S = \sum_{k=-\infty}^j \sum_{m=-1}^2 w_{k,m} f_{k+m-1}
\end{align}
but I only want to access the elements of $f$ once, so I would prefer something like
\begin{align}
\sum_k f_k \sum_m ...
\end{align}
Here is what I did: substitute $l=m-1+k$ to get
\begin{align}
S &= \sum_{k=-\infty}^j \sum_{m=-1}^2 w_{k,m} f_{k+m-1}
\\
&=\sum_{k=-\infty}^j \sum_{l+1-k=-1}^2 w_{k,l+1-k} f_{l}
\\
&=\sum_{k=-\infty}^j \sum_{l=k-2}^{k+1} w_{k,l+1-k} f_{l}
\end{align}
But when I try to get $f_l$ out of the inner sum I'm messing something up.
Can anyone produce the correct sum for looping over $f$ only once? Thanks in advance.</p>
<p>Since the highest index of $f$ that is accessed is $j+1$, I assume that the outer sum should be
\begin{align}
\sum_{k=-\infty}^{j+1} f_k
\end{align}</p>
| PdotWang | 212,686 | <p>Or you can continue on your method, just make the range of $l$ large enough:</p>
<p>Let $l \in (- \infty, + \infty)$, </p>
<p>$$S=\sum_{k=-\infty}^{j}\sum_{l=- \infty}^{+ \infty} \Omega_{k,l-k+1} \cdot {f_l}=\sum_{l=- \infty}^{+ \infty} \sum_{k=-\infty}^{j}\Omega_{k,l-k+1} \cdot {f_l}$$
where:
$$
\Omega=\begin{cases}
\omega_{k,l-k+1}, (-1\leq l-k+1 \leq 2) \\0, therwise
\end{cases}
$$
The fact is that you can always switch the order (under some condition ?) and you can always assume a infinite range. Please comment is I am less than 100% correct.</p>
|
1,271,942 | <p>I am a little bit confused with the definition of finitely presented modules. In Lang's <em>Algebra</em> he defines a module <span class="math-container">$M$</span> to be finitely presented if and only if there is a exact sequence <span class="math-container">$F'\to F\to M \to 0$</span> such that both <span class="math-container">$F', F$</span> are free. However the standard definition I have seen elsewhere only demands <span class="math-container">$F'$</span> be finitely generated. Are these two definitions equivalent?</p>
<p>Looking at the situation of a non-principal ideal of a ring, say <span class="math-container">$(x, y)$</span> of <span class="math-container">$\mathbb{R}[x, y]$</span>, it appears that this is finitely presented, by the usual definition, but I don't see any way of making it finitely presented by Lang's definition.</p>
| user26857 | 121,097 | <p>In Lang's <em>Algebra</em> he defines a module <span class="math-container">$M$</span> to be finitely presented if and only if there is an exact sequence <span class="math-container">$F'\to F\to M \to 0$</span> such that both <span class="math-container">$F', F$</span> are free <em>of finite rank</em>, and this is the definition of <a href="http://en.wikipedia.org/wiki/Finitely-generated_module#Finitely_presented_module" rel="noreferrer">finitely presented modules</a>. (Note that for each module <span class="math-container">$M$</span> there is an exact sequence <span class="math-container">$F'\to F\to M\to 0$</span> with <span class="math-container">$F,F'$</span> free modules.) </p>
<p>"<em>However the standard definition I have seen elsewhere only demands <span class="math-container">$F'$</span> be finitely generated</em>." This is the definition of <a href="http://en.wikipedia.org/wiki/Finitely-generated_module#Finitely_presented_module" rel="noreferrer">finitely related modules</a>. </p>
<p>"<em>Are these two definitions equivalent</em>?" In general they aren't: Let <span class="math-container">$M$</span> be a finitely presented module, and <span class="math-container">$L$</span> a free module which is not finitely generated. Then <span class="math-container">$M\oplus L$</span> is finitely related, but not finitely presented. However, if the module is finitely generated the two definitions coincide.</p>
|
2,634,791 | <blockquote>
<p>How can I show that the map $f: GL_n(\mathbb R)\to GL_n(\mathbb R)$ defined by $f(A)=A^{-1}$ is continuous?</p>
</blockquote>
<p>The space $GL_n(\mathbb R)$ is given the operator norm and so I want to show for all $\epsilon$ there exists $\delta$ such that $\|A-B\|<\delta \implies \|A^{-1}-B^{-1}\|<\epsilon$.</p>
| D_S | 28,556 | <p>Use the fact that $A^{-1} = \frac{1}{\textrm{det(A)}}\textrm{adj}(A)$ and the following general principles:</p>
<p>1 . The topology on $\textrm{GL}_n(\mathbb{R})$, with the operator norm, is the subspace topology coming from $\mathbb{R}^{n^2}$, the $n^2$-fold product of copies of $\mathbb{R}$. In general, norm topology on finite dimensional vector spaces coincides with product topology.</p>
<p>2 . If $Z$ is a subspace of a topological space $Y$, then a function $f: X \rightarrow Z$ is continuous if and only if $f$ is continuous as a map from $X$ to $Y$. This follows directly from the definition of the subspace topology. Similarly, if $S$ is a subspace of $X$, and $f: X \rightarrow Y$ is a continuous function, then the restriction of $f$ to $S$ is a continuous function from $S$ to $Y$.</p>
<p>3 . If $Y$ is a topological space, and $n$ is an integer, let $Y^n$ be the $n$-fold product of copies of $Y$. Let $\pi_i: Y^n \rightarrow Y$ be the function $(y_1, ... , y_n) \mapsto y_i$. A function $f: X \rightarrow Y^n$ is continuous if and only if each $\pi_i \circ f: X \rightarrow Y$ is continuous. This follows directly from the definition of the product topology.</p>
<p>Using these principles, you can reduce the problem to verifying that certain functions going into $\mathbb{R}$ are continuous. For example, the determinant function $\textrm{det}$ is continuous, and $\frac{1}{\textrm{det(g)}}$ is a continuous function $\textrm{GL}_n(\mathbb R) \rightarrow \mathbb R$, because it is the composition of the determinant function and the function $\frac{1}{x}$.</p>
|
3,566,603 | <p>I need a simple way to show <span class="math-container">$\mathbb R^2$</span> is not isomorphic <span class="math-container">$\mathbb{R}[x]/(x^2)$</span>. Both are not integral domains, and both are not fields, so I’m not sure how to go about it.</p>
| Marc Olschok | 19,950 | <p>One possibility (among many others) is to combine the following observations:</p>
<p>(1) <span class="math-container">$\epsilon = x + (x^2) \in \mathbb{R}[x]/(x^2)$</span> satisfies
<span class="math-container">$\epsilon^2 = 0$</span>.</p>
<p>(2) any isomorphism must map nonzero nilpotent elements to nonzero nilpotent elements</p>
<p>(3) <span class="math-container">$\mathbb{R}^2$</span> does not have any nonzero nilpotent elements.</p>
|
3,695,868 | <p>In right triangle <span class="math-container">$ABC,$</span> <span class="math-container">$\angle C = 90^\circ.$</span> Let <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> be points on <span class="math-container">$\overline{AC}$</span> so that <span class="math-container">$AP = PQ = QC.$</span> If <span class="math-container">$QB = 67$</span> and <span class="math-container">$PB = 76,$</span> find <span class="math-container">$AB.$</span></p>
<p><a href="https://i.stack.imgur.com/BIPQ0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BIPQ0.png" alt="enter image description here"></a></p>
<p>How do I use ratios and given side lengths to create a proportion to solve for <span class="math-container">$AB$</span>? Is there any other way to solve this?</p>
<p>I would think the best way to approach this is to relate <span class="math-container">$QB/CB = AB/CB$</span>, though that would make <span class="math-container">$CB$</span> for both the same. I guess the relation of <span class="math-container">$AB/AC = QB/QC$</span> can also be used.</p>
| Jan Eerland | 226,665 | <p>Well, we can use Pythagoras twice (as @Narasimham pointed out):</p>
<p><span class="math-container">$$
\begin{cases}
\text{BP}^2=\text{CP}^2+\text{BC}^2\\
\\
\text{BQ}^2=\text{CQ}^2+\text{BC}^2\\
\\
\text{CP}=\text{CQ}+\text{PQ}\\
\\
\text{CQ}=\text{PQ}\color{\white}{+}\\
\\
\text{BP}=76\\
\\
\text{BQ}=67
\end{cases}\tag1
$$</span></p>
<p>In order to solve this, I used Mathematica (it is also easy to solve by hand) with the following code:</p>
<pre><code>In[1]:=BP = 76;
BQ = 67; FullSimplify[
Solve[{BP^2 == CP^2 + BC^2, BQ^2 == CQ^2 + BC^2, CP == CQ + PQ,
CQ == PQ, CP > 0 && BC > 0 && CQ > 0 && PQ > 0}, {CP, BC, CQ, PQ}]]
Out[1]={{CP -> 2 Sqrt[429], BC -> 2 Sqrt[1015], CQ -> Sqrt[429],
PQ -> Sqrt[429]}}
</code></pre>
<p>So, we know:</p>
<p><span class="math-container">$$\text{AB}=\sqrt{\left(2\sqrt{1015}\right)^2+\left(3\cdot\sqrt{429}\right)^2}=89\tag2$$</span></p>
|
3,695,868 | <p>In right triangle <span class="math-container">$ABC,$</span> <span class="math-container">$\angle C = 90^\circ.$</span> Let <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> be points on <span class="math-container">$\overline{AC}$</span> so that <span class="math-container">$AP = PQ = QC.$</span> If <span class="math-container">$QB = 67$</span> and <span class="math-container">$PB = 76,$</span> find <span class="math-container">$AB.$</span></p>
<p><a href="https://i.stack.imgur.com/BIPQ0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BIPQ0.png" alt="enter image description here"></a></p>
<p>How do I use ratios and given side lengths to create a proportion to solve for <span class="math-container">$AB$</span>? Is there any other way to solve this?</p>
<p>I would think the best way to approach this is to relate <span class="math-container">$QB/CB = AB/CB$</span>, though that would make <span class="math-container">$CB$</span> for both the same. I guess the relation of <span class="math-container">$AB/AC = QB/QC$</span> can also be used.</p>
| Quanto | 686,284 | <p>Let AP = PQ = QC = <span class="math-container">$x$</span>. Note that PB and QB are the medians of the triangles AQB and PCB respectively. Per the Apollonius's Theorem for medians</p>
<p><span class="math-container">$$AB^2 + QB^2 = 2x^2 + 2PB^2\tag1 $$</span>
<span class="math-container">$$CB^2 +PB^2 = 2x^2 + 2QB^2\tag2 $$</span></p>
<p>Take (1) - (2) and recognize <span class="math-container">$AB^2 -CB^2=(3x)^2$</span> to get</p>
<p><span class="math-container">$$ x^2=\frac13(PB^2-QB^2)$$</span></p>
<p>Then, substitute <span class="math-container">$x^2$</span> in (1) to obtain </p>
<p><span class="math-container">$$AB= \sqrt{ \frac83 PB^2 - \frac53 QB^2}= 89
$$</span></p>
|
13,843 | <p>We have a natural number $n>1$. We want to determine whether there exist
natural numbers $a, k>1$ such that $n = a^k$. </p>
<p>Please suggest a polynomial-time algorithm.</p>
| Felipe Voloch | 2,290 | <p>For each $k \le \log n/\log 2$, compute an approximation to the positive real $k$-th root of $n$ using Newton's method to enough precision to check if it is an integer. Alternatively, use $p$-adic roots for a suitable $p$, with Newton turning into Hensel.</p>
|
3,959,178 | <p>Let <span class="math-container">$G$</span> be a group and <span class="math-container">$\mathbb k$</span> a field. Then, let <span class="math-container">$\mathfrak 1$</span> be the trivial <span class="math-container">$\mathbb k G$</span>-module.</p>
<p>According to Lorenz's A Tour of Representation Theory, he states "it turns out <span class="math-container">$\mathfrak 1$</span> has a great significance, if it is a projective as a <span class="math-container">$\mathbb kG$</span>-module, then all <span class="math-container">$\mathbb kG$</span>-module is projective" and goes on to reference Maschke's Theorem.</p>
<p>I can understand the following claim, that a ring <span class="math-container">$\mathbb kG$</span> is semisimple if and only if all its modules are projective, so Maschke's Theorem does indeed show that <span class="math-container">$\mathfrak 1$</span> is projective for a semisimple <span class="math-container">$\mathbb k G$</span>. I can also see that if <span class="math-container">$\mathfrak 1$</span> is projective, then it is direct summand of some free module.</p>
<p>However, I don't understand how the converse implication (i.e. <span class="math-container">$\mathfrak 1$</span> projective implies <span class="math-container">$\mathbb kG$</span> semisimple) suggested by Lorenz's remark is shown. It seems to suggest that we want <span class="math-container">$\mathfrak 1$</span> to be the direct summand of <span class="math-container">$\mathbb kG$</span> as a <span class="math-container">$\mathbb kG$</span>-module, but I don't quite see how this is done and even if it is, how the remark is proven.</p>
| Qiaochu Yuan | 232 | <p>I'll write <span class="math-container">$k$</span> for the trivial module. There's a natural map <span class="math-container">$\varepsilon : k[G] \to k$</span> of <span class="math-container">$k[G]$</span>-modules given by sending every <span class="math-container">$g \in G$</span> to <span class="math-container">$1$</span>, and <span class="math-container">$k$</span> is projective iff this map splits as a map of <span class="math-container">$k[G]$</span>-modules. A splitting of this map must send <span class="math-container">$1 \in k$</span> to a <span class="math-container">$G$</span>-invariant element of <span class="math-container">$k[G]$</span>, which must have the form <span class="math-container">$c \sum_{g \in G} g$</span> for some constant <span class="math-container">$c$</span>; moreover, we must have</p>
<p><span class="math-container">$$\varepsilon \left( c \sum_{g \in G} \right) = c |G| = 1$$</span></p>
<p>so a splitting exists iff <span class="math-container">$|G|$</span> is invertible in <span class="math-container">$k$</span>, and by Maschke's theorem this condition is necessary and sufficient for <span class="math-container">$k[G]$</span> to be semisimple. (Maschke's theorem isn't always stated as an if-and-only-if but it is.)</p>
|
907,893 | <p>I wanted to know about this convention :</p>
<p>By rate of growth of R, we normally mean : (change in R) / (change in Time)</p>
<p>But Rate of growth of a geometric sequence "a(1+r)^n" is r, which is strange i feel</p>
<p>I am kind of confused, can anyone clear it </p>
| Did | 6,179 | <blockquote>
<p>By rate of growth of R, we normally mean : (change in R) / (change in Time)</p>
</blockquote>
<p>Actually no, the rate of growth of some quantity $R$ over the time interval $(t_1,t_2)$ is not $$\frac{R(t_2)-R(t_1)}{t_2-t_1}$$ but $$\frac{R(t_2)-R(t_1)}{R(t_1)}$$ and, when the time is continuous, the (instantaneous) rate of growth of $R$ at time $t$ is $$\frac1{R(t)}\cdot\frac{\mathrm dR(t)}{\mathrm dt}.$$ In discrete time, if $R(n)=a\,(1+r)^n$ for every $n$ then the rate of growth over the time interval $(n,n+1)$ is $r$ for every $n$. In continuous time, if $R(t)=a\,\mathrm e^{rt}$ for every $t$ then the rate of growth at time $t$ is $r$ for every $t$.</p>
|
521,740 | <p>$x$,$y$ are real numbers satisfying $(x-1)^{2}+4y^{2}=4$<br>
find the maximum of $xy$ and justify it without calculus.<br>
Does there exist a tricky solution using elementary inequalities (AM-GM or Cauchy-Schwarz) ?</p>
<p>I tried and got it's when $x=\dfrac{3+\sqrt{33}}{4}$</p>
| Elmar Zander | 10,076 | <p>If you need to <em>justify only the solution</em> without calculus, you can do it the following way: The solution you have is
$$
x=\frac{1}{4} \left(3+\sqrt{33}\right) \qquad y=\frac{1}{4} \sqrt{\frac{1}{2} \left(15+\sqrt{33}\right)}
$$
If you go along the ellipse from that $(x,y)$ by a very small amount it will be along the direction
$$
\Delta x = - 2 a y \qquad \Delta y = a (x-1) / 2
$$
for some very small $a$ (infinitesimal, so to say). Now compute
$$
(x+\Delta x)(y + \Delta y) - x y = -\frac{1}{16} \left(\sqrt{33}-1\right) \sqrt{\frac{1}{2}
\left(15+\sqrt{33}\right)} a^2,
$$
which is obviously negative, so the $x$ and $y$ you found must be a maximum.</p>
<p>Maybe, you can also find $x$ and $y$ in a similar way, by requiring that the linear term in the equation above (i.e. $(x+\Delta x)(y + \Delta y) - x y$) vanishes, but I haven't tried that.</p>
<p>EDIT: I tried it now, and it gives you the additional equation $-4y^2 + x(x-1)=0$, which you can combine with the original equation to easily find the solution first for $x$ and then for $y$.</p>
|
1,701,176 | <p>The problem I'm having is with the logs. I go:</p>
<p>$$\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}} \cdot \frac{n-2}{n-1} \Big)$$</p>
<p>$$=\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}}\Big) \cdot \lim_{n \to \infty} \Big(\frac{n-2}{n-1} \Big)$$</p>
<p>and here I know that $$\lim_{n \to \infty} \Big(\frac{n-2}{n-1} \Big) = \lim_{n \to \infty} \Bigg(\frac{1-\frac{2}{n}}{1-\frac{1}{n}} \Bigg) = \frac{\lim_{n \to \infty} ({1-\frac{2}{n}})}{\lim_{n \to \infty} (1-\frac{1}{n})} = 1$$</p>
<p>However, I don't know how to do the equivalent for $$\lim_{n \to \infty} \Big( \frac{\log{(n+1)}}{\log{(n)}}\Big)$$</p>
<p>I know that the numerator and denominator functions converge as $n$ grows, but I don't know how to compute the limit algebraically and show that it's also $1$.</p>
| quasicoherent_drunk | 195,334 | <p>We can write $\log{(n+1)}$ as $$\log{(n(1+\frac{1}{n}))}=\log{n}+\log\left(1+\frac{1}{n}\right).$$</p>
<p>Now $\log\left(1+\frac{1}{n}\right)$ is bounded, so is insignificant compared to $\log{n}.$ So the limit of $$\frac{\log(n+1)}{\log(n)}=\frac{\log{(n)}}{\log{(n)}}+\frac{\log{\left(1+\frac{1}{n}\right)}}{\log{(n)}}$$ tends to $1+0=1$.</p>
|
4,196,109 | <p>While studying about inequalities, I came across the following definition (<span class="math-container">$\forall a > 0)$</span>:</p>
<p><span class="math-container">$$
\begin{alignat}{1}
& |x| > a \iff \{ x \mid x < -a \text{ or } x > a \} \\
& |x| < a \iff \{ x \mid -a < x < a \}
\end{alignat}
$$</span></p>
<p>Naturally, as <span class="math-container">$\{ -a < x < a \}$</span> could be rewritten as <span class="math-container">$\{ -a < x \text{ and } x < a \}$</span>, I wonder if is valid to rewrite <span class="math-container">$\{ x < -a \text{ or } x > a \}$</span> as <span class="math-container">$\{ -a > x > a \}$</span>.</p>
<p>I don't know if that would be valid because, while <span class="math-container">$\{ -a < x < a \}$</span> represents only one interval, <span class="math-container">$\{ -a > x > a \}$</span> would represent two in a single expression. Is that notation valid?</p>
| user97357329 | 630,243 | <p>A framework proposed by Cornel (<strong>answer to the main integral</strong>)</p>
<p>The skeleton of the solution may be immediately obtained by using the strategy given for the generalization in Sect. <span class="math-container">$1.24$</span>, page <span class="math-container">$14$</span>, <strong>(Almost) Impossible Integrals, Sums, and Series</strong> (more precisely, see pages <span class="math-container">$142-145$</span>).
So, following the suggested strategy, we have that</p>
<p><span class="math-container">$$\int_0^1 \frac{\arctan^2(x)\log(x)}{1+x}\textrm{d}x$$</span>
<span class="math-container">$$=\frac{1}{2}\int_0^1 \frac{\arctan^2(x)\log(x)}{x}\textrm{d}x-\frac{1}{2}\int_0^{\infty} \frac{\arctan^2(x)\log(x)}{x(1+x)}\textrm{d}x+\frac{\pi}{2}\int_0^1 \frac{\arctan(x)\log(x)}{1+x}\textrm{d}x$$</span>
<span class="math-container">$$-\frac{\pi^2}{8}\int_0^1\frac{\log(x)}{1+x}\textrm{d}x.$$</span></p>
<p>The resulting integrals are either known or manageable.</p>
<ul>
<li><p>Behind the first resulting integral lies a very difficult harmonic series which is calculated by real methods here <a href="https://math.stackexchange.com/q/3803762">https://math.stackexchange.com/q/3803762</a>.</p>
</li>
<li><p>The integral <span class="math-container">$\displaystyle \int_0^{\infty} \frac{\arctan^2(x)\log(x)}{x(1+x)}\textrm{d}x$</span> is reducible to the following integrals, <span class="math-container">$\displaystyle \int_0^1 \frac{x\log(1-x)}{1+x^2}\textrm{d}x$</span>, <span class="math-container">$\displaystyle \int_0^1 \frac{x\log(1+x)}{1+x^2}\textrm{d}x$</span>, <span class="math-container">$\displaystyle \int_0^1 \frac{x\log^2(x)\log(1-x)}{1+x^2}\textrm{d}x$</span>, <span class="math-container">$\displaystyle \int_0^1 \frac{x\log^2(x)\log(1+x)}{1+x^2}\textrm{d}x$</span>,
<span class="math-container">$\displaystyle \int_0^1 \frac{\log(x)\log(1-x)}{1+x^2}\textrm{d}x$</span>,<span class="math-container">$\displaystyle \int_0^1 \frac{\log(x)\log(1+x)}{1+x^2}\textrm{d}x$</span>, where most of them are known or easily reducible to known harmonic series like the ones given at the previous link (see the third resulting integral). Here one also needs to calculate <span class="math-container">$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}^{(3)}}{n}$</span> which is derived in a generalized form in the second theorem of the paper <a href="https://www.researchgate.net/publication/333339284_A_simple_strategy_of_calculating_two_alternating_harmonic_series_generalizations" rel="nofollow noreferrer">https://www.researchgate.net/publication/333339284_A_simple_strategy_of_calculating_two_alternating_harmonic_series_generalizations</a>.</p>
</li>
</ul>
<p><strong>End of story</strong></p>
|
2,867,207 | <p>This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.</p>
<p>Define the cone $\Sigma(\theta)$ for $\theta\in(0,\pi/2]$,
$$\Sigma(\theta) = \left\{ z = x+iy : x>0, |y|<(\tan\theta)x\right\}, $$
and define the norm $\|f\|_\theta$ for functions $f$ analytic on $\Sigma(\theta)$ by</p>
<p>$$\|f\|_\theta = \sup_{z\in \Sigma(\theta)}|f|$$
Let $Y_\theta$ be the space of functions analytic on $\Sigma(\theta)$ with $\|f\|_\theta < \infty$. Also let $\chi$ be a distinguished analytic function in $Y_{\pi/2}$ with $\chi(0) = 0$. It would seem then, that the following inequality is true: Let $f\in Y_{\theta}$. Then for any $\theta'<\theta$,
$$ \|\chi f'\|_{\theta'} \leq C \frac{\|f\|_{\theta}}{\theta-\theta'} $$
where the constant $C$ can depend on $\chi$. How is this proven, and how does $\chi$ help?</p>
<hr>
<p>Remarks</p>
<ul>
<li>(basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $\sup_{z\in B(r)} |f(z)|$. Lets say $f\in X_r$ if $f:B(r)\to \mathbb C$ is analytic on its domain, with $|f|_r<\infty$. From the usual Cauchy formula $f'(z) = \frac1{2π i}\int_{\partial D} \frac{f(w) dw}{(w-z)^2}$ it is not hard to prove that for any $f\in X_r$, with any $r'<r$,
$$ |f'|_{r'} \leq C \frac{|f|_{r}}{r-r'}.$$</li>
</ul>
| Ben West | 37,097 | <p>Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $k\in\mathbb{Z}$. Then
$$
(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1.
$$</p>
<p>Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.</p>
|
1,320,874 | <p>I am trying to answer the following: Does the congruence $x^2 \equiv -1$ (mod $p$) have any solutions if $p \equiv 3$ (mod $4$)? If so, how many incongruent solutions does it have? If not, why not?</p>
<p>I know from the previous part of the question that if $p$ is a prime and $p \equiv 1$ (mod $4$), then the congruence $x^2 \equiv -1$ (mod $p$) has two incongruent solutions, namely $x \equiv \pm (\dfrac{p-1}{2})!$ (mod $p$).</p>
<p>I am completely unsure how to even approach solving this problem. Any hints would be appreciated.</p>
| Zev Chonoles | 264 | <p>André Nicolas has an approach that uses Wilson's theorem, as requested by the OP, but I guess I'll leave my answer up.</p>
<hr>
<p>No, it has no solutions. The first step is to observe that, for an odd prime $p$,
$$x^2\equiv -1\bmod p\iff x\text{ has order 4 in }(\mathbb{Z}/p\mathbb{Z})^\times$$
But the group $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic, so
$$\text{there exists an element of order $d$ in }(\mathbb{Z}/p\mathbb{Z})^\times\iff d\text{ divides }\lvert(\mathbb{Z}/p\mathbb{Z})^\times\rvert=p-1$$
Thus, for an odd prime $p$, there exist solutions to $x^2\equiv -1\bmod p$ if and only if $4$ divides $p-1$, i.e., if and only if $p\equiv 1\bmod 4$.</p>
<p>Thus, there are no solutions when $p\equiv 3\bmod 4$.</p>
|
1,265,801 | <p>Let $p$ be an odd prime and $a, b \in \Bbb Z$ with $p$ doesn't divide $a$ and $a$ doesn't divide $b$. Prove that among the congruence's $x^2 \equiv a \mod p$, $\ x^2 \equiv b \mod p$, and $x^2 \equiv ab \mod p$, either all three are solvable or exactly one.</p>
<p>Please help I'm trying to study for final in number theory and I can't figure out this proof.</p>
| lhf | 589 | <p>You can <a href="https://en.wikipedia.org/wiki/Euler%27s_criterion" rel="nofollow">Euler's criterion</a>.
Let
$$
A=a^{\frac{p-1}{2}} \bmod p,
\quad
B=b^{\frac{p-1}{2}} \bmod p,
\quad
C=(ab)^{\frac{p-1}{2}} \bmod p
$$
Then $C=AB$. This implies that if two of the equations are solvable, then so is the third.</p>
|
73,238 | <p>How can I calculate the solid angle that a sphere of radius R subtends at a point P? I would expect the result to be a function of the radius and the distance (which I'll call d) between the center of the sphere and P. I would also expect this angle to be 4π when d < R, and 2π when d = R, and less than 2π when d > R.</p>
<p>I think what I really need is some pointers on how to solve the integral (taken from <a href="http://en.wikipedia.org/wiki/Solid_angle" rel="nofollow">wikipedia</a>) $\Omega = \iint_S \frac { \vec{r} \cdot \hat{n} \,dS }{r^3}$ given a parameterization of a sphere. I don't know how to start to set this up so any and all help is appreciated!</p>
<p>Ideally I would like to derive the answer from this surface integral, not geometrically, because there are other parametric surfaces I would like to know the solid angle for, which might be difficult if not impossible to solve without integration.</p>
<p>*I reposted this from mathoverflow because this isn't a research-level question.</p>
| Zarrax | 3,035 | <p>Let $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ be two unit vectors perpendicular to the direction of the axis and each other, and let $(c_1,c_2,c_3)$ be any point on the axis. (If ${\bf v} = (v_1,v_2,v_3)$ is a unit vector in the direction of the axis, you can choose ${\bf a} = (a_1,a_2,a_3)$ by solving ${\bf a} \cdot {\bf v} = 0$, scaling ${\bf a}$ to make $\|{\bf a}\| = 1$, then letting ${\bf b} = {\bf a} \times {\bf v}$.)</p>
<p>Then for any $r$ and $\theta$, the point $(c_1,c_2,c_3) + r\cos(\theta)(a_1,a_2,a_3) + r\sin(\theta)(b_1,b_2,b_3)$ will be at distance $r$ from $(c_1,c_2,c_3)$, and as $\theta$ goes from $0$ to $2\pi$, the points of distance $r$ from $(c_1,c_2,c_3)$ on the plane containing $(c_1,c_2,c_3)$ perpendicular to the axis will be traced out. </p>
<p>So the parameterization of the circle of radius $r$ around the axis, centered at $(c_1,c_2,c_3)$, is given by
$$x(\theta) = c_1 + r\cos(\theta)a_1 + r\sin(\theta)b_1$$
$$y(\theta) = c_2 + r\cos(\theta)a_2 + r\sin(\theta)b_2$$
$$z(\theta) = c_3 + r\cos(\theta)a_3 + r\sin(\theta)b_3$$</p>
|
2,059,604 | <p>Let's say I have a two periodic functions f(x) and g(x) each with the same period of p. Is it always the case that the sum of these two functions will also have the period of p? Is there any counter example?</p>
| Antoine | 73,561 | <p>If you define period of a function $h$ as the number $p$, such that $h(x + p) = h(x)$, for all $x$, then yes (try it).</p>
<p>If you define period as the smallest positive number, such that $h(x + p) = h(x)$, then no, for example: $g = \sin$ and $f = -\sin$ will give you $h(x) = \sin(x) - \sin (x) = 0$.</p>
|
19,842 | <p>Seeing <a href="https://stackoverflow.com/help/self-answer">this</a> I though this thing was promoted, and for avoiding for the question becoming boring, I didn't answer it suddenly and waited and I did mentioned that I knew the answer, maybe it's just misunderstanding that I don't know the answer. Anyways, what's the state/condition regarding such things?</p>
<p>See <a href="https://math.stackexchange.com/questions/1176644/unspecific-function-integration">this</a> and <a href="https://math.stackexchange.com/questions/1176622/differential-equation-sin2x-left-frac-rm-dy-rm-dx-sqrt-tan-x-ri">this</a>. I edited one to include answer in question, I just didn't edited other just because.</p>
| apnorton | 23,353 | <p>Self-answering a question means that you actually post an answer to your question, not that you had already solved the question in the past. Thus, your second link isn't a "self-answered question" by any measure. </p>
<p>Instead of editing the question to include an answer, it is best to <em>post an answer</em> to the question, <strong>and</strong> place a note in the question saying "this is a self-answered question." The note in the question reduces the likelihood that someone closes or downvotes by mistake.</p>
|
490,641 | <p>In Niels Lauritzen, <em>Concrete Abstract Algebra</em>, I'm having trouble showing the following:</p>
<p>The problem starts out like this:</p>
<p>$f(X)=a_nX^n+\cdots+a_1X+a_0, a_i \in \mathbb Z, n \in \mathbb N$ </p>
<p>Part (i) which I think I've done right:</p>
<p>i) Show $X-a \mid X^n-a^n$:
$X^n - a^n = (X-a)(X^{n-1} + X^{n-2}a + \cdots + Xa^{n-2} + a^{n-1}), n\ge2 \Rightarrow X-a \mid X^n-a^n$.</p>
<p>Part (ii) which I am having trouble solving:</p>
<blockquote>
<p>ii) Show that if: $a, N \in \mathbb Z$, $f$ has degree $n$ modulo $N$, $f(a) \equiv 0$ (mod $N$) then $f(X) \equiv (X-a)g(X)$ (mod $N$) and $g$ has degree $n-1$ modulo $N$.</p>
</blockquote>
<p>Hint: use (i) and $f(X) \equiv f(X) - f(a)$ (mod $N$).</p>
<p>$f$ has degree $n$ modulo $N$ means $N$ does not divide $a_n$.</p>
<p>Thanks for your time.</p>
| Martin Brandenburg | 1,650 | <p>Somehow the statement is unnatural und unnecessarily complicated.</p>
<p>If $R$ is an arbitrary commutative ring, and $r \in R$ is a root of a polynomial $f \in R[X]$, then $X-r$ divides $f$. Proof: Using the transformation $X \mapsto X-r$ it suffices to check the case $r=0$. But then $f$ has no constant term and the claim follows.</p>
<p>Now apply this to $R=\mathbb{Z}/N$.</p>
|
2,466,949 | <p>Room coordinates are following my walls, to use the guidance system I build the position from various other sensors & built a GPS position from it.</p>
<p>As I also need the a "fake" compass I'm trying to interface a moving robot with a sensor I made.</p>
<p>Robot expect compass to send him the values of a 3-axis magnometer.
As my sensor gives me the orientation pitch & roll I have this formula:</p>
<p>$\text{Orientation}=\text{atan2}( (-\text{ymag}*\cos(\text{Roll}) + \text{zmag}*\sin(\text{Roll}) ) , (\text{xmag}*\cos(\text{Pitch}) + \text{ymag}*\sin(\text{Pitch})*\sin(\text{Roll})+ \text{zmag}*\sin(\text{Pitch})*\cos(\text{Roll})))$</p>
<p>as I've 3 unknown variables & one equation I need more equations.
But I'm stuck, there should be a way based on Orientation values to get constraints (i.e in $\text{atan2}(y,x) = \arctan(y/x)$ if $x > 0$, etc.) but I can translate those relations to equations.</p>
<p>Am I missing something or is it impossible?</p>
<p>What Im trying to do:</p>
<p>-get Xmag,Ymag and Zmag, those are the expected output of the fake compass. </p>
<p>-Known variables are: Orientation (Yaw) Pitch & Roll, on the robot system (X: right of robot, Y: front of robot, Z: going up) Yaw is the rotation on Z in reference to a "North" arbitrary selected, Pitch the rotation on X and Roll the rotation on Y.</p>
| Allawonder | 145,126 | <p>I think what you notice is that the process reminds you of the number of permutations (not selections, as you wrote) of <span class="math-container">$4$</span> out of <span class="math-container">$7$</span> objects. This is indeed the case.</p>
<p>Now, first note that to count the number of arrangements of <span class="math-container">$n$</span> <em>distinct</em> objects, you may proceed as follows:</p>
<p>Partition the set in two, one containing <span class="math-container">$r$</span> of these <span class="math-container">$n$</span> objects. Then the number of arrangements is (by the multiplication principle) <span class="math-container">$$nPr\times (n-r)P(n-r)=nPr\times (n-r)!=n!.$$</span> This means that for each arrangement of the <span class="math-container">$r$</span> objects in the first part, we have <span class="math-container">$(n-r)!$</span> arrangements of the <span class="math-container">$n$</span> objects. <em>Now, if these <span class="math-container">$n-r$</span> objects are indistinguishable, then the number of permutations of the <span class="math-container">$n$</span> (previously distinct) objects should be divided by <span class="math-container">$(n-r)!.$</span></em> This is what you noticed; and, as I have explained, hopefully you can see why it is true that <strong>the number of arrangements of <span class="math-container">$n$</span> objects, <span class="math-container">$n-r$</span> of which are indistinguishable, is the same as the number of arrangements of <span class="math-container">$r$</span> out of <span class="math-container">$n$</span> <em>distinct</em> objects.</strong></p>
|
858,494 | <p>Where does the definition of the $L_\infty$ norm come from?</p>
<p>$$\|x\|_\infty=\max \{|x_1|,\dots,|x_k|\}$$</p>
| Cameron Williams | 22,551 | <p>Here's the way I like to think about it (which is not too rigorous but can be made rigorous). We have, more generally, the $L^p$ norms ($1\le p < \infty$):</p>
<p>$$(\|x\|_p)^p:=\sum_{i=1}^n |x_i|^p.$$</p>
<p>The Euclidean norm is a special case of this (take $p = 2$); the taxicab norm is also a special case (take $p=1$). Suppose $|x_i|\ge |x_j|$ for all $1\le j\le n$. What happens if $p$ gets really large? Well we would see that the $|x_i|$ term would dominate the sum and asymptotically, all of the others would be inconsequential (due to the function being very convex). So what we could say is that for large $p$,</p>
<p>$$(\|x\|_p)^p \approx |x_i|^p.$$</p>
<p>Taking a $p$th root of both sides gives us</p>
<p>$$\|x\|_p \approx |x_i|.$$</p>
<p>However note that we said that $|x_i|$ was the largest of the components of our vector. Thus, informally, we would say that $\|x\|_{\infty} = |x_i| = \max \{|x_1|,\ldots,|x_n|\}$.</p>
|
3,026,097 | <p>I was studying an article where I encountered <span class="math-container">$\mathbb{R}^E_{\gt 0}$</span>. I couldn't find out what does this notation mean exactly.
I'm sorry if my question is basic, I searched this community but I didn't find the answer to my question.</p>
<p>Here is a part of that article:</p>
<blockquote>
<p>Given an undirected graph <span class="math-container">$G=(V, E)$</span> with positive edge lengths <span class="math-container">$l\in\mathbb{R}^E_{\gt 0}$</span> on which one desires to compute the shortest path from a vertex <span class="math-container">$s$</span> to <span class="math-container">$t$</span>, ...</p>
</blockquote>
| YiFan | 496,634 | <p>Hint: break the integral up into
<span class="math-container">$$\int_0^\infty=\int_0^1+\int_1^2+\int_2^3+\cdots.$$</span>
On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.</p>
|
3,757,763 | <p>Let <span class="math-container">$T: V\rightarrow V$</span> be a linear operator of the vector space <span class="math-container">$V$</span>.</p>
<p>We write <span class="math-container">$V=U\oplus W$</span>, for subspaces <span class="math-container">$U,W$</span> of <span class="math-container">$V$</span>, if <span class="math-container">$U\cap W=\{0\}$</span> and <span class="math-container">$V=U+W$</span>.</p>
<p>If we assume <span class="math-container">$\dim V<\infty$</span>, then by the <a href="https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" rel="nofollow noreferrer">rank-nullity</a> theorem, <span class="math-container">$\ker T\cap {\rm Im}\,T=\{0\}$</span> implies <span class="math-container">$V=\ker T\oplus {\rm Im}\,T$</span>.</p>
<blockquote>
<p>However, my question is about the case <span class="math-container">$\dim V$</span> is infinite. Is it still true? What if <span class="math-container">$T$</span> has a minimal polynomial?</p>
</blockquote>
<p>Thanks.</p>
| Tsemo Aristide | 280,301 | <p>Consider the shift operator <span class="math-container">$s$</span>, defined on <span class="math-container">$\text{Vect}(e_i, i\in\mathbb{N})$</span>, where <span class="math-container">$s(e_n)=e_{n+1}$</span> for <span class="math-container">$n\in\mathbb{N}$</span>. Note that <span class="math-container">$\ker(s)=0$</span> but <span class="math-container">$s$</span> is not surjective.</p>
|
917,302 | <p>If $p(x)$ is a polynomial of degree 4 such that $p(2)=p(-2)=p(-3)=-1$ and $p(1)=p(-1)=1$, then find $p(0)$.</p>
| Community | -1 | <p>This is a case of Lagrangian interpolation.</p>
<p>$$P(x)=\sum_i\left(y_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}\right).$$</p>
<p>It is most efficiently computed using <a href="http://en.wikipedia.org/wiki/Neville%27s_algorithm" rel="nofollow noreferrer">Neville's scheme</a>.</p>
<p>Using the abscissas in the order $-3, -2, -1, 1, 2$, the triangle of interpolated values is:
$$-1,\ -1,\ 1,\ 1,\ -1$$
$$-1,\ 3,\ 1,\ 3$$
$$5,\ \frac53,\ \frac53$$
$$\frac52,\ \frac53$$
$$\color{blue}2$$</p>
<p><img src="https://i.stack.imgur.com/yjseX.png" alt="enter image description here"></p>
|
589,309 | <p>Finding all sets of primes $p$ and $q$ such that $p$ divides $q^2 -4$ and $q$ divides $p^2-1$.</p>
| Robert Israel | 8,508 | <p>Hint: $q^2-4$ and $p^2-1$ can be factored.</p>
|
589,309 | <p>Finding all sets of primes $p$ and $q$ such that $p$ divides $q^2 -4$ and $q$ divides $p^2-1$.</p>
| Superguy | 510,499 | <p>Four cases would be formed and only two would be left .</p>
<p>$p+1 \equiv 0 \pmod q$</p>
<p>$q+2 \equiv 0 \pmod p$</p>
<p>After some estimation it would be found that only $p=5$ and $q=3$ can be the solution.</p>
|
844,355 | <p>Let $T\colon V \rightarrow W$ a linear transformation between the real vector spaces $V$ and $W$ both with finite dimension.</p>
<p>How can i prove that $\dim(V) = \dim T(V) + \dim T^{-1}(0)$.</p>
<p>I can't understand this problem and how to solve it , if you can help me please.</p>
| mookid | 131,738 | <p>Consider a space $F$ such as $V = F\oplus T^{-1}(0)$. Consider the restriction $T'$ of $T$ from
$F$ to $T(V)$.</p>
<ol>
<li>Let $x,y\in F.$</li>
</ol>
<p>$$
T'(x) = T'(y)\implies T(x-y) = T(x)-T(y) = 0\implies
x-y\in T^{-1}(0)
$$but as $x,y\in F$:
$$
x-y\in T^{-1}(0)\cap F= \{0\}\implies x=y.
$$2. Let $y\in T(V)$. $\exists x\in V\,\,y=T(x)$. Let us write $x = x_0 + x_1$, $x_0\in T^{-1}(0)
$ and $x_1\in F$.</p>
<p>$$
y = T(x) = T(x_0+x_1) = T(x_0)+T(x_1) = 0+T(x_1) = T(x_1) = T'(x_1)
$$hence $T'$ is onto.</p>
<p>Conclusion: this proves the second equality in
$$
\dim V - \dim T^{-1}(0) = \dim F = \dim T'(F) = \dim T(V)
$$</p>
|
2,531,714 | <p>Given is: </p>
<p>$(1-x^2)\dfrac{d^2y(x)}{dx^2} + 2x\dfrac{dy(x)}{dx} - 2y(x) = 0 $</p>
<p>The Solution is: </p>
<p>$y(x) =C_1x + C_2(x^2+1)$ </p>
<p>How do I factor the $x$ out in order to get it into a normal "linear" form that contains only coefficients to show that the solution is valid? </p>
<p>Edit: The equation should be both Linear and Homogeneous</p>
| Jack D'Aurizio | 44,121 | <p>We have that
$$ f(x)=\sum_{n\geq 0}a_n x^n\quad\Longleftrightarrow\quad \frac{f(x)}{1-x}=\sum_{n\geq 0}A_n x^n $$
where $A_n = a_0+a_1+\ldots+a_n$. In order to find the OGF of $H_{n}^2$ it is enough to find the OGF of
$$ H_{n+1}^2-H_{n}^2 = \left(H_{n+1}-H_n\right)\left(H_{n+1}+H_n\right)=\frac{2H_n}{n+1}+\frac{1}{(n+1)^2}, $$
and while the OGF of $\frac{1}{(n+1)^2}$ is clearly related with $\text{Li}_2(x)=\sum_{n\geq 1}\frac{x^n}{n^2}$, the OGF of $\frac{H_n}{n+1}$ can be deduced by applying termwise integration to the OGF of $H_n$. It follows that
$$ \sum_{n\geq 1}H_n^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}\tag{A}$$
and the (more involved) OGF of $H_n^3$ can be computed through the same trick.
It is useful to consider that the Taylor series of $\log(1-x)^k$ is related to <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind" rel="noreferrer">Stirling numbers of the first kind</a>.</p>
|
3,884,581 | <p>Please don't just throw an answer at me, please explain how you arrived at it cause I've been fiddling with this for the past 30min...</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$a^2+b^2\pm2ab=(a\pm b)^2.$$</span></p>
<hr />
<p>The standard way is to try and eliminate one of the unknonws, for instance getting <span class="math-container">$b$</span> from the second equation and plugging in the first. Then solve the equation in a single unknown.</p>
<p>But the above hint is more direct, IMO.</p>
|
3,884,581 | <p>Please don't just throw an answer at me, please explain how you arrived at it cause I've been fiddling with this for the past 30min...</p>
| Spectre | 799,646 | <p>A larger hint :</p>
<p><span class="math-container">$a^2 + b^2 = 6, ab = 4$</span></p>
<p><span class="math-container">$(a+b)^2 = a^2 + b^2 + 2ab = 6 + 2 \times 4 = 14 \implies a + b = \pm \sqrt{14}\longrightarrow(1)$</span>
<span class="math-container">$(a-b)^2 = a^2+b^2-2ab = 6 - 2\times 4 = -2 \implies a - b = \sqrt{2}i\longrightarrow(2)$</span></p>
|
3,103,991 | <p>Suppose <span class="math-container">$X$</span> is a topological space, <span class="math-container">$C$</span> is a closed subset of <span class="math-container">$X$</span>, <span class="math-container">$U$</span> is an open subset of <span class="math-container">$X$</span> and <span class="math-container">$U$</span> is dense in <span class="math-container">$X$</span>, is <span class="math-container">$U\cap C$</span> dense in <span class="math-container">$C$</span>?</p>
| tomasz | 30,222 | <p>This is true if and only if <span class="math-container">$C$</span> is the closure of its interior. Such sets are called <em>regular closed sets</em>.</p>
<p>Indeed, if <span class="math-container">$C=\overline{V}$</span> for some open <span class="math-container">$V$</span>, then <span class="math-container">$U\cap V$</span> is dense in <span class="math-container">$V$</span>, and hence in <span class="math-container">$C$</span>, so <span class="math-container">$U\cap C\supseteq U\cap V$</span> is dense in <span class="math-container">$C$</span>. (Note that we do not really use the assumption that <span class="math-container">$U$</span> is open --- only density matters.)</p>
<p>Otherwise, if <span class="math-container">$C$</span> is not the closure of its interior, <span class="math-container">$U=(X\setminus C)\cup \operatorname{int}(C)$</span> is open and dense, and <span class="math-container">$U\cap C=\operatorname{int}(C)$</span>, so <span class="math-container">$U\cap C$</span> is not dense in <span class="math-container">$C$</span>.</p>
|
2,429,769 | <p>I watched a <a href="https://www.youtube.com/watch?v=lXkRj6MKbZs" rel="nofollow noreferrer">great youtube video</a> about how to prove a limit of a multivariable function exists. It explained that one method is by substitution. For example, we can solve $$lim_{(x, y) \to 0,0} \frac{xy}{\sqrt{x^2 + y^2}}$$</p>
<p>By substituting with polar coordinates - namely letting $x = rcos\theta$ and $y=rsin\theta$ </p>
<p>The youtube video mentioned that polar coordinates are not the only accepted form of substitution. </p>
<p><strong>What are other forms of substitution one can use? When do you know which form to substitute with?</strong></p>
| Raphael Rafatpanah | 61,853 | <p>The formula to sum an arithmetic sequence is:</p>
<p><a href="https://i.stack.imgur.com/NQt1T.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NQt1T.gif" alt="enter image description here"></a></p>
<p><code>n</code> = <code>iterations</code> = <code>initial velocity</code> / <code>drag</code></p>
<p><code>a</code> = first term = <code>initial velocity</code></p>
<p><code>d</code> = common difference = <code>drag</code></p>
<p><a href="https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html" rel="nofollow noreferrer">https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html</a></p>
<p>A comparison of the imperative and formula methods in JavaScript show the same result for both: <a href="https://jsfiddle.net/persianturtle/we1bkk9d/1/" rel="nofollow noreferrer">https://jsfiddle.net/persianturtle/we1bkk9d/1/</a></p>
<pre><code>var initial_velocity = 1.8509277593973181
var drag = 0.0175
var iterations = Math.floor(initial_velocity / drag)
// Imperative
var sum = 0
for (var i = 0; i < iterations; i++) {
sum += initial_velocity - drag * i
}
console.log('imperative', sum)
// Formula
sum = (iterations / 2) * (2 * initial_velocity - (iterations - 1) * drag)
console.log('formula', sum)
// Estimate
sum = (initial_velocity * initial_velocity) / (2 * drag);
console.log('estimate', sum)
</code></pre>
|
2,231,092 | <p>I am reading <a href="http://people.ucalgary.ca/~rzach/static/open-logic/open-logic-complete.pdf" rel="nofollow noreferrer">Open Logic TextBook</a>. In which there is a proposition about Extensionality of first order sentences (6.12) It goes like this, </p>
<p>Let $\phi$ be a sentence, and $M$ and $M'$
be structures.
If $c_M = c_{M'}$
, $R_M = R_{M'}$
, and $f_M = f_{M'}$
for every constant symbol $c$,
relation symbol $R$, and function symbol $f$ occurring in $\phi$, then $ M \models \phi$ iff $M' \models \phi$</p>
<p>Does this statement implicitly imply that the Domain is exactly the same set, since $f_M = f_{M'}$ I am confused at this statement, does it mean, $f_M = f_{M'}$ only on the domain of constant values (or other covered terms?)</p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>You can write the equation in cartesian coordinate as $y=mx+q$, than substitute
$y=\rho \sin \theta$ and $x=\rho \sin \theta$. From this it is easy to obtain $\rho$ as a function of $\theta$.</p>
|
622,090 | <p>We are asked to solve the following linear system</p>
<p>$$x_1-3x_2+x_3=1$$
$$2x_1-x_2-2x_3=2$$
$$x_1+2x_2-3x_3=-1$$</p>
<p>by using gauss-jordan elimination method. The augmented matrix of the linear system is $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\2 & -1 & -2 & 2 \\1 & 2 & -3 & -1\end{array}\right).$$ By a series of elementary row operations, we have $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right).$$ My question is, although the question asked us to solve the linear system using gauss-jordan elimination method, can we stop immediately and conclude that the linear system is inconsistent without further apply any elementary row operation to the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ until the matrix $$\left(\begin{array}{ccc|c}1 & -3 & 1 & 1 \\0 & 5 & -4 & 0 \\0 & 0 & 0 & -2\end{array}\right)$$ is transformed into reduced-row echelon form?</p>
| Zhoe | 99,231 | <p>Yes, you can stop there and conclude that the system is inconsistent as $0\ne-2$. If you were to continue to reduce the matrix to reduced-row echelon form, row $3$'s inconsistency would remain unaffected.
$$\left(\begin{array}{ccc|c}1 & 0 & -\frac{7}{5} & 0 \\0 & 1 & -\frac{4}{5} & 0 \\0 & 0 & 0 & 1\end{array}\right)$$
$R_3\to-\frac{1}{2}R_3$ was performed to get the new row $3$ and notice that the completely reduced-row echelon form above also has $0x_1+0x_2+0x_3=1 \implies 0=1$ which is not possible, and hence the system still maintains its inconsistency. </p>
|
4,564,882 | <p>Suppose there are two types of weathers. Sunny and Rainy. <br />
The probability that a sunny day is followed by a sunny day is 70% and followed by a rainy day is 30%. <br />
The probability that a rainy day is followed by a rainy day is 60% and followed by a sunny day is 40%. <br />
In a year (365 days), how many days do we expect to be sunny?</p>
<p>Based on the question above, I only get the transition matrix
<span class="math-container">$
\begin{bmatrix}
0.7 & 0.3\\
0.4 & 0.6
\end{bmatrix}
$</span>
May I ask how do I calculate the expected number of sunny days in a year?
Thanks in advance.</p>
| true blue anil | 22,388 | <p>Taking that the steady state probability exists, let these probabilitie be <strong>s</strong> for sunny and <strong>r</strong> for rainy, then one more iteration won't change these probabiliies, hence</p>
<p><span class="math-container">$s*0.7 + r*0.4 = s$</span></p>
<p><span class="math-container">$s*0.3 + r*0.6 = r$</span></p>
<p><span class="math-container">$s+r=1$</span></p>
<p>which yields <span class="math-container">$s = \frac47, r = \frac37$</span></p>
<p>The required answer <span class="math-container">$=\frac47\times 365$</span></p>
|
959,525 | <p>Could someone tell me what i've done wrong?</p>
<p>I tried to find out the derivative of $3^(2x)-2x+1$ but I got it wrong.
What I did was derivate $3^a-2x+1$ where a = 2x then multiply those two.</p>
<p>$(ln3*3^a - 2)*2$ = $2ln3*3^(2x)-4$</p>
<p>Ps. x = 2 so the answer is supposed to be 176.</p>
| Arodi007 | 135,954 | <p>Find value of C <br/><br/></p>
<p>from Left-side <br/>
$C=2\sqrt{a^2-m^2}$</p>
<p>From right side
$m^2=\sqrt{b^2+a^2}/(2)$</p>
<p>You might continue the solution from these...</p>
|
2,750,783 | <p>There is a problem I am having trouble understanding. </p>
<p>We are asked to evaluate the definite integral:</p>
<p>$$\int_0^2\sin(e^x+x^2)(e^x+2x)\,dx$$ </p>
<p>If anyone would be so kind as to walk me through it, I would be extremely grateful. </p>
| giobrach | 332,594 | <p>I do not understand how you can postulate the existence of $y \notin A^c$ such that $y$ is a limit point of $A^c$ just from the fact that $A^c$ is open (i.e., just by using your definition of openness). I think you may be implicitly using the fact that $A$ is closed, and as such it contains all its border points, which are also the border points of $A^c$; i.e. you are using the thesis to prove the thesis. Furthermore, you it is not immediately true that if $y$ is a limit point for $A^c$ then it is also a limit point for $A$: what if $y$ is an isolated point in $A$? Any ball centered in $y$ will intersect $A^c$, but small enough balls will <em>not</em> contain points $\neq y$ in $A$, by definition of isolated point. So you also implicitly assumed that $A$ is <em>perfect</em>, i.e. all its points are limit points.</p>
<p>Here's a simpler way. Basically, to prove that $A$ is closed, you want to prove that a point not in $A$ cannot be a limit point of $A$. Let $x$ be such a point. Then $x \in A^c$. But $A^c$ is open, so there exists a small enough ball around $x$ that is entirely contained within $A^c$. Then this ball does not contain any points in $A$, and $x$ cannot be a limit point of $A$.</p>
|
2,750,783 | <p>There is a problem I am having trouble understanding. </p>
<p>We are asked to evaluate the definite integral:</p>
<p>$$\int_0^2\sin(e^x+x^2)(e^x+2x)\,dx$$ </p>
<p>If anyone would be so kind as to walk me through it, I would be extremely grateful. </p>
| Nuntractatuses Amável | 537,135 | <p>If $A^C$ is open, then for each $x \in A^C$ there is an $\epsilon>0$ such that $B(x; \epsilon) = \{ y \in \mathbb{R}^2 \colon |x - y| < \epsilon\} \subseteq A^C$. </p>
<p>If $x \in \bar{A}$, then, for each $\epsilon > 0$, $B(x; \epsilon) \cap A \neq \emptyset$.</p>
<p>These are your definitions, I guess. Now suppose some $x \in \bar{A} $. $x$ can't be in $A^C$, for then no $\epsilon$-neighbourhood of $x$ would be contained in $A^C$, contrary to the hypothesis that $A^C$ is open. Then, if $x \in \bar{A}$, $x \in (A^C)^C = A$. Therefore, $\bar{A} = A$ and $A$ is closed.</p>
|
2,750,783 | <p>There is a problem I am having trouble understanding. </p>
<p>We are asked to evaluate the definite integral:</p>
<p>$$\int_0^2\sin(e^x+x^2)(e^x+2x)\,dx$$ </p>
<p>If anyone would be so kind as to walk me through it, I would be extremely grateful. </p>
| aschepler | 2,236 | <p>There are a number of problems with your proof. Taking it one bit at a time,</p>
<blockquote>
<p>We want to show if $A^c$ is open, the $A$ is closed. Let $X=\mathbb{R}^2$ and $A\subseteq X$. Then, denote $A^c$ as the complement of A such that $A^c=\{x\in X|x\notin A\}$.</p>
</blockquote>
<p>So far, just restating the problem, maybe a bit more clearly. Okay, that doesn't really hurt.</p>
<blockquote>
<p>Take an arbitrary point $x$ in $A^c$. Since $x$ is in $A^c$, we know $x$ is a limit point of $A^c$...</p>
</blockquote>
<p>It's true every element of an open set in $\mathbb{R}^2$ is a limit point of that set. But that's not a direct application of your definition of open, so is this a theorem you've already seen proved, or can you see why it must be true?</p>
<blockquote>
<p>... but we also know that $\exists y \notin A^c$ such that $y$ is a limit point of $A^c$</p>
</blockquote>
<p>This is not certain. The main premise of this proof should be that $A^c$ is open. How can you say anything about the limit points of an open set, since your definition of open involves open balls surrounding points, and it's the definition of closed that mentions limit points?</p>
<p>In particular, remember that the entire set $\mathbb{R}^2$ is an open set in $\mathbb{R}^2$. And if $A^c = \mathbb{R}^2$, it's not true that there is a limit point $y$ of $\mathbb{R}^2$ which is not an element of $\mathbb{R}^2$.</p>
<blockquote>
<p>Since these arbitrary point(s) $y$ are not in $A^c$, they must be in $A$.</p>
</blockquote>
<p>Okay, except be careful about switching between singular and plural. You originally claimed there exists one such point $y$. Saying there exists an object that satisfies a predicate does mean there might be just one such object or more than one, but if you need to deduce anything else from that claim, the one object needs to be enough.</p>
<p>The word "arbitrary" doesn't have enough context here to be meaningful. It normally means there was some earlier choice, and then you showed that in some sense that choice didn't matter. But if only one point satisfied the property of $y$, and you didn't say anything otherwise, there was no choice in the first place.</p>
<blockquote>
<p>We have generated an arbitrary point $y$ that is not in $A^c$ but in $A$ that is a limit point of both sets.</p>
</blockquote>
<p>You claimed $y$ is a limit point of $A^c$, but now how do you also conclude $y$ is also a limit point of $A$? The only similar thing I see was that $x$ is a limit point of $A$, but you haven't established any relationship between $x$ and $y$.</p>
<blockquote>
<p>Thus we can conclude $A$ is closed since it contains all its limit points.</p>
</blockquote>
<p>You claimed something about one point $y$, not something about every limit point of $A^c$. So $A$ being closed doesn't follow.</p>
<hr>
<p>So what is a good approach here?</p>
<p>The statement to prove is "if $A^c$ is open, then $A$ is closed". Let's start by expanding those with the definitions:</p>
<ul>
<li><p>$A^c$ is open means that $\forall x \in A^c : \exists \epsilon >0 : B_\epsilon(x) \subseteq A^c $.</p></li>
<li><p>$A$ is closed means that every limit point $y$ of $A$ is an element of $A$.</p></li>
</ul>
<p>Probably the simplest thing from this point is a proof by contradiction: Suppose $A^c$ is open and $A$ is NOT closed.</p>
<ul>
<li>$A$ is NOT closed means that there exists some limit point $y \in X$ of $A$ such that $y \notin A$.</li>
</ul>
<p>But if $y \notin A$, then $y \in A^c$. Since the meaning above for "$A^c$ is open" shows a property for every element of $A^c$, this property must apply to $y$ in particular, so we can plug in $y=x$ to that property:</p>
<ul>
<li>$ \exists \epsilon>0 : B_\epsilon(y) \subseteq A^c $</li>
</ul>
<p>Can you take it from here and find the contradiction?</p>
|
299,170 | <p>Given the following integer programming formulation, how can I specify that the variables are unique and none of them has the same value as the other one. basically <code>x1</code>, <code>x2</code>, <code>x3</code> , and <code>x4</code> need to get only one unique value from 1, 2, 3 or 4. and same applies to <code>y1</code>, <code>y2</code>, <code>y3</code>, and <code>y4</code>.</p>
<pre><code>Minimize
obj: +2*x1 -3*y1 +3*x2 -3*y2 +1*x3 -2*y3 +4*x4 -2*y4
Constraints
+2*x1 -3*y1 +3*x2 -3*y2 +1*x3 -2*y3 +4*x4 -2*y4 >= 0
Bounds
1 <= x1 <= 4
1 <= x2 <= 4
1 <= x3 <= 4
1 <= x4 <= 4
1 <= y1 <= 4
1 <= y2 <= 4
1 <= y3 <= 4
1 <= y4 <= 4
</code></pre>
| Is7aq | 61,660 | <p>I found a workaround by adding the following constraints, to ensure that the sum of every 2 numbers, every 3 number, and every 4 numbers is at the least their minumum sum if they are to be distinct. For the above problem, the following additional constraints ensured that values are distinct.</p>
<pre><code>+x1 +x2 >= 3
+x1 +x3 >= 3
+x1 +x4 >= 3
+x2 +x3 >= 3
+x2 +x4 >= 3
+x3 +x4 >= 3
+y1 +y2 >= 3
+y1 +y3 >= 3
+y1 +y4 >= 3
+y2 +y3 >= 3
+y2 +y4 >= 3
+y3 +y4 >= 3
+x1 +x2 +x3 >= 6
+x1 +x3 +x4 >= 6
+x2 +x3 +x4 >= 6
+x1 +x2 +x3 +x4 >= 10
</code></pre>
|
943,048 | <p><strong>Question:</strong></p>
<blockquote>
<p>let $x_{i}=1$ or $-1$,$i=1,2,\cdots,1990$, show that
$$x_{1}+2x_{2}+\cdots+1990x_{1990}\neq 0$$</p>
</blockquote>
<p>this problem it seem is easy,But I think is not easy. </p>
<p>I think note
$$1+2+3+\cdots+1990\equiv \pmod { 1990}?$$</p>
| extremeaxe5 | 174,546 | <p>Consider the equation modulo 2. Regardless of whether $x_i = 1,or -1$, $x_i\equiv 1\pmod{2}$.</p>
<p>Thus $\sum_{i=1}^{1990}{ix_i}\equiv 1+0+1+\cdots +0\pmod{2}$, where there are $\dfrac{1990}{2}=995$ $1$'s in the summation. </p>
<p>We conclude that $\sum_{i=1}^{1990}{ix_i}\equiv 1\pmod{2}$, so it definitely cannot be equal to $0$.</p>
|
1,355,684 | <p>How to find the lower and upper focus? Hyperbola </p>
<p>I started with this
$$ 9x^2 + 54x - y^2 + 10y + 81 = 0 $$</p>
<p>and broke it down to</p>
<p>$$ \frac{9(x+3)^2}{25} - \frac{(y-5)^2}{25} = -1 $$</p>
<p>center = (-3,5)
Lower Vertex = (-3,0)
Upper Vertex = (-3,10)</p>
<p>How to get the foci? </p>
<p>foci / focus = (h, k +- c)</p>
<p>b = 5 but what is a?</p>
<p>$$ c^2 = a^2 + b^2 $$</p>
<p>Thank you.</p>
| Daniel Griscom | 253,093 | <p>Forming a number by "repeating a two digit number three times" is the equivalent of "multiplying a two digit number by 10101". And, as @achuille-hui said, $10101 = 3 \times 7 \times 17 \times 37$. So, any number in your form will be a multiple of 3, 7, 17 and 37.</p>
|
1,996,290 | <p>I don't know how to solve the following integral:</p>
<p>$$ \frac{2\pi}R\int_{r_1}^{r_2}r(r+R-|r-R|)dr $$</p>
<p>I have solved it when $R \le r_1$ and $R \ge r_2$ but I need the answer for $ r_1\lt R \lt r_2$. <br/>
R is a constant as well as $r_1$ and $r_2$.</p>
<p>I appreciate your help.</p>
| Alexis Olson | 11,246 | <p>If $r_1 < R < r_2$, then</p>
<p>\begin{eqnarray}
\int_{r_1}^{r_2}r(r+R-|r-R|)dr
&=& \int_{r_1}^{R}r(r+R+(r-R))dr + \int_{R}^{r_2}r(r+R-(r-R))dr\\
&=& \int_{r_1}^{R}2r^2dr + \int_{R}^{r_2}2rR\,dr
\end{eqnarray}</p>
|
4,115,069 | <p>I understand 'functionals' as functions of functions, for example:</p>
<p><span class="math-container">$$ S[y(x)]= \int_{t_1}^{t_2} \sqrt{1+(y')^2} dx$$</span></p>
<p>Which is the famous arc length integral</p>
<p>Now, in a similar way, a limit we can write as:</p>
<p><span class="math-container">$$L(a, [y(x)] ) = \lim_{x \to a} y(x) \tag{1}$$</span></p>
<p>In this way, we can think of a limit as a function of a 'function' and a 'number'. So, would it be correct to call the above object a functionals? Why/Why not?</p>
<p>Examples of (1):</p>
<p><span class="math-container">$$L(0,\frac{\sin x}{x}) = \lim_{x \to 0} \frac{\sin x}{x} = 1$$</span></p>
<p><span class="math-container">$$L(0,e^x) = 1$$</span></p>
<p>etc</p>
<hr />
<p>This doubt mainly emerged while I was answering through <a href="https://math.stackexchange.com/questions/4114333/is-l-a-function-of-a/4115070#4115070">this post</a></p>
| peek-a-boo | 568,204 | <p>A functional is just a function whose domain and target space are of a specific type. Typically the domain is assumed to be a space like <span class="math-container">$L^p, C^k$</span>, or some other Hilbert/Banach/Frechet space (or maybe you may not even want all this structure, and just assume the domain is some real/complex vector space) etc, while the target space is assumed to be a field <span class="math-container">$\Bbb{R}$</span> or <span class="math-container">$\Bbb{C}$</span>. So, tbh this whole "function vs functional" business is in my opinion a pointless terminology chasing game with no mathematical content.</p>
<p>Anyway, here's what you can do. Let
<span class="math-container">\begin{align}
S:=\{f:\Bbb{R}\to\Bbb{C}\,|\,\text{for every $a\in \Bbb{R}$, $\lim\limits_{x\to a}f(x)$ exists in $\Bbb{C}$}\}
\end{align}</span></p>
<p>Now, we can define the "limiting" operation as <span class="math-container">$L:\Bbb{R}\times S\to \Bbb{C}$</span> as
<span class="math-container">\begin{align}
L(a,f):=\lim_{x\to a}f(x).
\end{align}</span>
So, sure, the domain of <span class="math-container">$L$</span> is a vector space and it maps into a field, so you can call this a functional (though it is not a linear functional, only for fixed <span class="math-container">$a$</span>, is <span class="math-container">$L(a,\cdot)$</span> linear). If you want to talk about limits at <span class="math-container">$\infty$</span>, things can get more tricky since the domain will no longer be a vector space, so I guess you can't call them a functional in the typical sense of the word.</p>
<p>Note also, that sometimes the term "functional" means "continuous linear map from a (topological/normed) vector space into the field (<span class="math-container">$\Bbb{R}$</span> or <span class="math-container">$\Bbb{C}$</span>)". So, if you want to talk about continuity, you of course have to equip <span class="math-container">$S$</span> with a topology, you can of course do this, but as of right now I don't see a point to this. Terminology can be confusing and is not always standardized; one must always refer to contextual clues for which meaning is intended.</p>
<p>As you can see, it is a little unweidly to define <span class="math-container">$L$</span> in full generality; this is not to say it isn't useful to think of it in this way. There are some problems where viewing a "basic vanilla" object from a more abstract perspective leads to quicker solutions (sometimes it's just a simple rephrasing which allows us to invoke more power theorems, eg from functional analysis)... though I can't think of any super convincing examples off the top of my head.</p>
|
1,665,443 | <p>How do we show the ring homomorphism for </p>
<p>$\phi :\mathbb F_p(\alpha) \rightarrow\mathbb F_p(\alpha)$ which is defined as $ \phi(\alpha)=\alpha +1$.</p>
<p>This is a very basic fact but I am unable to prove it by the definition of ring homomorphism. Same thing happens for ring homomorphism over $\phi :K[x] \rightarrow K[X]$ defined as $\phi (X)=X+1$. I have studied this earlier and assumed it as trivial but never tried to see the proof. </p>
<p>Thanks in advance.</p>
| DonAntonio | 31,254 | <p>Ok, so $\;\Bbb F_p(\alpha)\;$ is a finite, and thus algebraic, extension of $\;\Bbb F_p\;$ , but also</p>
<p>$$\alpha^p-\alpha+a=0\implies (\alpha+1)^p-(\alpha+1)+a=\alpha^p+1-\alpha-a+a=0$$</p>
<p>so both $\;\alpha,\,\alpha+1\;$ are roots of the same polynomial irreducible polynomial over $\;\Bbb F_p[x]\;$ and thus there always exist a $\;\Bbb F_p$-automorphism of $\;\Bbb F_p(\alpha)\;$ mapping one of these roots to the other one, which is given by your map.</p>
<p>For example:</p>
<p>$$\phi(a_{p-1}\alpha^{p-1}+\ldots+a_1\alpha+a_0)=0\iff$$</p>
<p>$$0=a_{p-1}(\alpha+1)^{p-1}+\ldots+a_1(\alpha+1)+a_0=$$</p>
<p>$$=a_{p-1}\alpha^{p-1}+\ldots+a_1\alpha+\left(a_0+a_{p-1}+\ldots+a_1\right)\stackrel{\text{by lin. indep.}}\implies$$</p>
<p>$$\implies\begin{cases}a_{p-1}=0\\a_p=0\\...\\a_1=0\\a_0+\ldots+a_{p-1}=0\end{cases}\;\;\;\implies a_i=0\;\;\;\forall\;1=0,1,2,...,p-1$$</p>
<p>and $\;\phi\;$ is injective then. And this is enough, because $\;\phi\;$ is in particular a linear operator on a finite dimensional linear , so it is injective$\;\iff\;$ it is surjective $\;\iff\;$ it is bijective.</p>
<p>By the way, you could as well take $\;\alpha+x\;,\;\;\forall\,x\in\Bbb F_p\;$ . Any of these is a root of the above polynomial.</p>
|
1,821,248 | <p>Which of the following are true?</p>
<ol>
<li><p>$\sigma \circ \sigma(j)=j~\forall j, 1 \leq j \leq 5$.</p></li>
<li><p>$\sigma^{-1}(j)=\sigma(j)~\forall j, 1 \leq j \leq 5$.</p></li>
<li><p>The set $ \{k: \sigma(k) \neq k \}$ has even number of elements.</p></li>
<li><p>The set $\{k:\sigma(k) =k \}$ has an odd number of elements . </p></li>
</ol>
<p>Can someone tell me to solve it in $1$ or $2$ min....trick..or some concept behind it?</p>
| Brian Cheung | 248,555 | <p>$\sigma^{-1}(j)\leq\sigma(j)\forall 1\leq j \leq 5$<br>
since $\sigma$ is a permutation,<br>
$j\leq\sigma^2(j)\forall 1\leq j\leq 5$<br>
since $\sigma^2$ is also a permutation, we have $j=\sigma^2(j)\forall 1\leq j\leq 5$<br>
So the longest cycle in $\sigma$ is at most of length $2$.<br>
That is, an element is either mapped to itself, or it is 'paired up' with another element which mapped back to itself. ($\sigma(j)=j$ <strong>or</strong> $\exists i$ s.t. $\sigma(j)=i$ and $\sigma(i)=j$)<br>
So all of 1,2,3,4 are true.</p>
|
122,776 | <p>Let $\Gamma$ be a lattice in a (real or p-adic) Lie group.
Is it true that for a given natural number $n$ there exists a finite index subgroup $\Sigma\subset\Gamma$ such that each $\sigma\in\Sigma$ is an $n$-th power of some element of $\Gamma$?</p>
<p>In other words, is it true that for given $\sigma\in\Sigma$ there exists $\gamma\in\Gamma$ such that $\sigma=\gamma^n$?</p>
<p>If not true for general lattices, are there some restrictions under which this holds (cocompactness, arithmeticity,...)?</p>
| Ian Agol | 1,345 | <p>This is false for uniform lattices in rank one semi simple Lie groups and large $n$ by a result of <a href="http://www.ams.org/journals/tran/1996-348-06/S0002-9947-96-01510-3/home.html" rel="nofollow">Ivanov and Olshanskii</a>, which implies that the normal subgroup generated by $n$th powers is infinite index for certain large $n$. </p>
|
2,421,771 | <p>I’m attempting to explain curvature in layman’s terms to my class before explaining the formula. I like to do this first to give my students an idea of what we are finding. </p>
<p>Some people explain curvature as a “measure of how fast a curve is changing direction at a given point.” But this seems misleading to me. It seems to me that when someone say “how fast” most people interpret that as a change in direction per unit time. But time has nothing to do with it correct? </p>
<p>For example the curvature is the same for any particular curve regardless of the speed of its parametrization. So time has nothing to do with it. The way I’ve been explaining curvature in laymen’s terms is that it is a “measure of how ‘hard’ a curve is changing direction at a given point”. </p>
<p>Do you think this is sufficient for a laymen’s term explanation without leading the students astray with a time component. Perhaps you have another way to explain it? Or perhaps other people’s laymen definition is perfectly fine and I’m over analyzing. </p>
<p>Advice? </p>
| James Arathoon | 448,397 | <p>The standard definition for the curvature at a particular point on a curved path is just $\frac{1}{R}$, where R is the radius of curvature back to some notional associated point (valid for that particular point on the curved path only). The definition need not involve time, but most applications involve time and motion.</p>
<p>e.g. if in a racing car the absolute centripetal acceleration, $a$ (measured using an 3 axis-accelerometer) and forward speed, $v$ are measured regularly as the car drives around a racing track, the curvature can be determined at all the points of measurement using $a=\frac{v^2}{r}$. From this data a map of the track can be drawn. The usefulness of allowing a ever changing curvature is that any arbitrary track can be mapped not just circular ones, with a fixed radius of curvature. The forward speed of the car can vary as this acceleration/deceleration is perpendicular to the centripetal acceleration. </p>
<p>Another application for the concept of curvature and radius of curvature is refraction of light by a varying density gas, such as for light from the sun or stars passing down through the earths atmosphere.</p>
|
2,421,771 | <p>I’m attempting to explain curvature in layman’s terms to my class before explaining the formula. I like to do this first to give my students an idea of what we are finding. </p>
<p>Some people explain curvature as a “measure of how fast a curve is changing direction at a given point.” But this seems misleading to me. It seems to me that when someone say “how fast” most people interpret that as a change in direction per unit time. But time has nothing to do with it correct? </p>
<p>For example the curvature is the same for any particular curve regardless of the speed of its parametrization. So time has nothing to do with it. The way I’ve been explaining curvature in laymen’s terms is that it is a “measure of how ‘hard’ a curve is changing direction at a given point”. </p>
<p>Do you think this is sufficient for a laymen’s term explanation without leading the students astray with a time component. Perhaps you have another way to explain it? Or perhaps other people’s laymen definition is perfectly fine and I’m over analyzing. </p>
<p>Advice? </p>
| ChocolateAndCheese | 148,590 | <p>I think about it as related to the radius of a circle or ball that will juuust lie tangent to the curve at that point. Any larger radius and the circle will be too large to touch at that point, and so will necessarily touch the curve in 2 other points.</p>
<p>Of course this isn't exactly rigorous, but I found it to be a helpful conceptualization.</p>
|
118,763 | <p>Hello,</p>
<p>Let $X'X$ be a positive definite matrix and let $\mathbf{1}$ denote the vector of ones. </p>
<p>I'm hoping to construct a positive, diagonal matrix $W$ such that
$$(W X'X W) \mathbf{1} = \mathbf{1}$$</p>
<p>$X$ and $W$ are all assumed to have real-valued entries, and $X'$ denotes the transpose of $X$.</p>
<p>I don't, yet, have a proof that such a matrix $W$ always exists, but strongly suspect it. Any ideas on algorithms, proofs, or counter-examples would be gratefully received.</p>
<p>The problem arises from work in statistics. </p>
<p>thanks,</p>
<p>David.</p>
| Will Sawin | 18,060 | <p>Consider the simplex of nonzero diagonal matrices W with nonnegative entries up to scaling, and the simplex of nonzero vectors V with nonnegative entries up to scaling.</p>
<p>There is a map, $V=\max(WX′XW\mathbf 1,0)$, from the first simplex to the second, with $\max(a,0)$ interpreted entrywise. This is well-defined because $WX'XW\mathbf 1$ always has some positive entry, because the sum of its entries is $1'W X' X W1$, with $W1$ a nonzero vector and $X'X$ positive-definite.</p>
<p>This map clearly sends each k-cell of the first simplex into the corresponding k-cell of the second simplex, since if some of the coordinates of $W$ are $0$ then some of the coordinates of $V$ are $0$.</p>
<p>Every such map on simplices must be surjective. This is because the map from the boundary sphere of one simplex to the boundary sphere of the other is degree one, because every such map on simplices has a boundary-preserving homotopy to the standard isomorphism between those simplices, by induction.</p>
<p>So there is some $W$ such that $\max(WX′XW\mathbf 1,0)=\mathbf 1$. So $WX'XW\mathbf 1=\mathbf 1$.</p>
|
15,237 | <p><a href="https://matheducators.stackexchange.com/questions/176/knowing-mathematics-does-not-translate-to-knowing-to-teach-mathematics-why">A question</a> has been asked about why great mathematicians are not necessarily great teachers. On the other hand, I am wondering if knowing more mathematics actually helps with one's teaching of lower level courses in mathematics. For example, I believe that a good student with bachelor's degree in mathematics should have sufficient knowledge to teach calculus. However, how does having a master's or doctorate degree help one's teaching in calculus, if any at all?</p>
<p>I am teaching calculus now and I do not understand commutative algebra; I took a course on commutative algebra long time ago; I did poorly in the course and now I could hardly recall anything from this course. If I invest substantial among of time studying this subject well now, will it help me in my calculus course in any sense?</p>
| Tommi | 2,083 | <p>There are four ways an advanced knowledge of mathematics can help, in rough order of importance: Mastery, understanding student thinking, context and content and tricks. However, general teaching skill also matters, and I would guess that it does not reliably increase along with mathematics studies, or at least that the effect is small.</p>
<h1>Mastery</h1>
<p>All other things being equal, someone with more experience with difficult mathematics is likely to have mastered lots of easy mathematics. Even when the particular subject is new to them, they have learned how to easily learn a new mathematical idea or abstraction.</p>
<p>Mastery gives confidence, speeds up preparation and makes it easier to find mistakes. It is all around helpful.</p>
<h1>Understanding student thinking</h1>
<p>You need to master several ways of thinking about something as simple as multiplication, not to speak of more advanced mathematics. It helps to have developed mathematical thinking to be able to understand these different ways of seeing (say) multiplication, and also thereby being able to understand non-standard solutions and solving methods of students. The same is true of common misunderstandings and explaining what is the problem, or maybe giving a spontaneous counterexample of exercise that illustrates the different ways of thinking.</p>
<p>In particular, you need to do this on the fly, in front of a classroom. It is not easy!</p>
<h1>Context</h1>
<p>Having learned higher mathematics gives context for the basics. This helps in making the content meaningful and in finding examples and applications, which can be made into exercises or project work.</p>
<h1>Tricks and content</h1>
<p>Sometimes, though rarely, the specific techniques or content one has worked on perfectly line up with the teaching. I doubt this is very common, and the more advanced mathematics one is teaching, the more likely it is. Applied research and courses for students of relevant subjects might also increase the incidence.</p>
|
481,313 | <p>Show that in an abelian group the product of two elements of finite order is itself an element of finite order.</p>
<p>I need some hint to start with, I am familiar with the basic</p>
| Dan Rust | 29,059 | <p>If $a$ has order $l$ and $b$ has order $m$, can you find an $n\geq 1$, in terms of $l$ and $m$, such that $a^n=b^n=e$?</p>
|
2,462,722 | <p><a href="https://i.stack.imgur.com/jtWGA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jtWGA.jpg" alt="enter image description here"></a></p>
<p>I got this from QFT Demystified in the author attempt to derive the Euler Lagrange equation. But isn't the Taylor expansion for $f(x+a)$ supposed to be:
$$f(x+a)=f(a)+x\frac{df(a)}{dx}+...(1)$$
Even if I exchange $x\rightarrow \epsilon, a \rightarrow x$, it should have been:
$$f(\epsilon+x)=f(x)+\epsilon\frac{df(x)}{d\epsilon}+...(2)$$</p>
<p>I understand that in (1), $a$ is the point where we start or "nail" the fitting process, and $x$ is the independent variable. But what are $\epsilon$ and $x$ in (2)? Feel free to be rigorous if there's no intuitive way to answer, cause I really don't understand Taylor expansion at all and I need any answer. :((</p>
<p>Thank you! :D</p>
| alexjo | 103,399 | <p>$$
\frac{\mathrm d V(x)}{\mathrm d x}=\lim_{\varepsilon\to 0}\frac{V(x+\varepsilon)-V(x)}{\varepsilon}
$$
so for $\varepsilon\ll 1$ we can write</p>
<p>$$
\frac{\mathrm d V(x)}{\mathrm d x}\approx \frac{V(x+\varepsilon)-V(x)}{\varepsilon}
$$
that is
$$
V(x+\varepsilon)\approx V(x) + \varepsilon \frac{\mathrm d V(x)}{\mathrm d x}
$$</p>
|
3,395,910 | <p>I understand how to apply the trapezoidal rule to approximate the area under a curve.</p>
<p>But I'm not sure how to apply it when approximating areas between two functions. </p>
<ul>
<li>Do you use the formula like how you normally would, except apply it to the <strong>first function - the second function</strong>?</li>
</ul>
<p>Or is there a completely different approach?</p>
<p>Thanks!</p>
| cqfd | 588,038 | <p>Suppose <span class="math-container">$y\in[a,b]\setminus T$</span>. Then <span class="math-container">$a:=|f(y)-g(y)|\gt 0$</span>. As <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are continuous, we can find a <span class="math-container">$\delta\gt0$</span> such that <span class="math-container">$$|f(x)-f(y)|\lt \frac a3,\:|g(x)-g(y)|\lt \frac a3$$</span>whenever <span class="math-container">$|x-y|\lt\delta$</span>. So if <span class="math-container">$|x-y|\lt\delta$</span>, by triangle inequaltiy, <span class="math-container">$$\begin{align*}a=|f(y)-g(y)|&\leq |f(y)-f(x)|+|f(x)-g(x)|+|g(x)-g(y)|\\&\lt\frac a3+|f(x)-g(x)|+\frac a3\end{align*}.$$</span> So <span class="math-container">$|f(x)-g(x)|\gt \frac a3\gt 0$</span> whenever <span class="math-container">$|x-y|\lt\delta$</span>. This shows that <span class="math-container">$[a,b]\setminus T$</span> is open in <span class="math-container">$[a,b]$</span>. So <span class="math-container">$T$</span> is closed. </p>
<p>Observe that if <span class="math-container">$T$</span> is dense in <span class="math-container">$[a,b]$</span>, <span class="math-container">$f\equiv g$</span> in <span class="math-container">$[a,b]$</span>.</p>
|
56,162 | <p>I'm trying to understand the Cartan decomposition of a semisimple Lie algebra, $\mathfrak g=\mathfrak k \oplus \mathfrak p$, where $[\mathfrak k,\mathfrak p] \subseteq \mathfrak p$, cf. the wikipedia article on <a href="http://en.wikipedia.org/wiki/Cartan_decomposition" rel="noreferrer">Cartan decomposition</a>.</p>
<p>I posted the following question on math.stackexchange.com, where Darij suggested to repost the question here as an answer is not completely obvious, I suppose.</p>
<p>Let $\mathfrak {so}_{n}$ denote the skew-symmetric complex $n \times n$-matrices and let $M$ denote the symmetric $n \times n$-matrices of trace 0.</p>
<p>Then $M$ is a module over the Lie algebra $\mathfrak {so}_n$ (this comes from the Cartan decomposition of $\mathfrak {sl}_n$).</p>
<blockquote>
<p>What is the decomposition of $M$ into irreducible $\mathfrak {so}_n$-modules? </p>
</blockquote>
<p>The standard representation of $\mathfrak {so}_n$ has dimension $n$, the adjoint representation has dimension $\frac 1 2 n \cdot (n-1)$ and there are two spin representations of small dimension. But I don't see a way how these, together with trivial representations, should add up to the dimension of $M$, which is $\frac 1 2 n \cdot (n+1)-1$. </p>
| Konrad Waldorf | 3,473 | <p>I think the important information is that $H^3(G,\mathbb{R}) \neq 0$. </p>
<p>By the way, every $G$-bundle over $M$ is trivializable, under the conditions you have mentioned. That's why a gauge transformation can be regarded as a (smooth) map $g: M \to G$. </p>
<p>Now you look at the behaviour of the Chern-Simons 3-form $CS(A)$ of a connection $A$ on $E$ under a gauge transformation $g$. The formula is
$$
CS(g^*A) = CS(A) + g^*H + \text{exact terms}.
$$
where $H$ is the canonical 3-form of $G$ that represents a non-trivial element of $H^3(G,\mathbb{R})$. Now you can find a gauge transformation $g$ such that $g^*H$ is not exact. In that sense you have non-trivial gauge transformations. </p>
<p>EDIT: The comment that every $G$-bundle is trivializable is only true if $G$ is additionally assumed to be simply-connected, sorry. So you either assume that (so does Witten) or you must see gauge transformations as maps $g:P \to G$, rather, and the Chern-Simons form $CS(A)$ as a form on $P$, not on $M$.</p>
|
3,756,649 | <p>If <span class="math-container">$f:\mathbb{R}\to \mathbb{R}$</span> such that <span class="math-container">$\lim\limits_{x \to \infty}xf(x)=L$</span>. Then <span class="math-container">$\lim\limits_{x \to \infty}f(x)=0$</span>.</p>
<p>My proof is as follows:</p>
<p>Let <span class="math-container">$g(x)=xf(x)$</span>. Fix an <span class="math-container">$\epsilon>0$</span>, then by definition, there exists a <span class="math-container">$\delta>0$</span> such that <span class="math-container">$$x>\delta \implies |g(x)-L|<\epsilon$$</span> Then <span class="math-container">$$|f(x)|=\frac{|g(x)|}{|x|}\le \frac{L+\epsilon}{\delta}$$</span> claiming the result.</p>
<p>Are there any flaws in my argument?</p>
| Community | -1 | <p>Unfortunately,</p>
<p><span class="math-container">$$|f(x)|\le\frac{L+\epsilon}\delta$$</span> is not conclusive. Nothing says that <span class="math-container">$\delta$</span> is even large.</p>
<hr />
<p>A simple proof is</p>
<p><span class="math-container">$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\left(xf(x)\cdot\frac1x\right)=\lim_{x\to\infty}(xf(x))\cdot\lim_{x\to\infty}\frac1x=L\cdot0.$$</span></p>
|
7,268 | <p>I'm a private tutor working with a 7th grader who is struggling with solving equations. Given a simple equation, he is able to solve it using a formulaic procedure, but it is very obvious that he has no idea what the solution really means. Hence, if he gets a problem that's slightly different from ones he's solved before, he's completely lost. </p>
<p>Working with him yesterday, I realized that he doesn't understand what a variable is, or what he's doing when he's solving an equation -- he's just following the steps his teacher told him for that specific problem. </p>
<p>I'm trying to think of a way to visually demonstrate what's happening when he's solving an equation -- something visual that he can see. Kind of an algebraic equivalent to putting four coins on the table and adding one more to show 4 + 1 = 5. </p>
<p>Any ideas? Thanks!
-Ian </p>
| Tamisha Thompson | 4,702 | <p>I would use Algebra tiles (like <a href="http://illuminations.nctm.org/Activity.aspx?id=3482" rel="nofollow">these</a>). I've found that as students see the difference between an x tile and a unit tile, for example, they tend to make fewer mistakes with combining like terms. When they have to physically remove the same amount from both sides of the "equal sign", it's harder for them to make careless mistakes, especially with negatives. It takes a little practice (and more so for students who have a traditional method sort of stuck in their heads), but moving to abstraction seems to take less time.</p>
|
4,374,307 | <p>Problem:<br />
Suppose there are <span class="math-container">$7$</span> chairs in a row. There are <span class="math-container">$6$</span> people that are going to randomly
sit in the chairs. There are <span class="math-container">$3$</span> females and <span class="math-container">$3$</span> males. What is the probability that
the first and last chairs have females sitting in them?</p>
<p>Answer:<br />
Let <span class="math-container">$p$</span> be the probability we seek. Out of <span class="math-container">$3$</span> females, only <span class="math-container">$2$</span> can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
<span class="math-container">\begin{align*}
p &= \dfrac{ {3 \choose 2 } 3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}</span>
Am I right?
Here is an updated solution.</p>
<p>Let <span class="math-container">$p$</span> be the probability we seek. Out of <span class="math-container">$3$</span> females, only <span class="math-container">$2$</span> can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
<span class="math-container">\begin{align*}
p &= \dfrac{ 3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}</span>
Now is my answer right?</p>
| Misha Lavrov | 383,078 | <p>We can think of the seating assignment as a random permutation of <span class="math-container">$7$</span> items (the three males, the three females, and the empty seat). This random permutation puts a female in the first chair with probability <span class="math-container">$\frac37$</span>. Conditional on having done that, there are <span class="math-container">$2$</span> females left, so one of them ends up in the last chair with probability <span class="math-container">$\frac26$</span>.</p>
<p>Overall, <span class="math-container">$p = \frac37 \cdot \frac26 = \frac17$</span>.</p>
<p>There's also the brute force approach: <span class="math-container">$p = \frac{20}{140}$</span> by counting.</p>
<p><a href="https://i.stack.imgur.com/MtYXY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MtYXY.png" alt="enter image description here" /></a></p>
<p>With an approach whose denominator is <span class="math-container">$7!$</span>, the numerator should be <span class="math-container">$\binom32 \cdot 2 \cdot 5!$</span>: we pick the <span class="math-container">$2$</span> females at the ends, pick the order they sit in, and then pick the permutation of the middle. This also gives <span class="math-container">$\frac{\binom 32 \cdot 2 \cdot 5!}{7!} = \frac{3\cdot 2}{7 \cdot 6} = \frac17$</span>.</p>
|
2,679,393 | <p>If $X=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\}\}$ and $Y=X\setminus\{\{\emptyset\}\}$ </p>
<p>then what element is excluded from $X$? Is it $\{\{\emptyset\}\}$, or $\{\emptyset\}$?</p>
<p>In a similar vein, if $Z=\{a, b, c\}$, does it make sense to say $Z\setminus a$?</p>
<p>Thanks</p>
| Pietro Paparella | 414,530 | <p>For your first question, if $A$ and $B$ are sets, then $A \setminus B := \{ x \mid x \in A \wedge x \not \in B\}$. Thus, $X \setminus \{\{\emptyset\}\} = \{\emptyset,\{\{\emptyset\}\}\}$.</p>
<p>For the second question, if you follow the previous definition strictly, then it doesn't make sense to write $Z \setminus a$ although it might be allowed via convention given how clunky $Z \setminus \{a\}$ looks (especially in view of the example you provided).</p>
|
2,679,393 | <p>If $X=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\}\}$ and $Y=X\setminus\{\{\emptyset\}\}$ </p>
<p>then what element is excluded from $X$? Is it $\{\{\emptyset\}\}$, or $\{\emptyset\}$?</p>
<p>In a similar vein, if $Z=\{a, b, c\}$, does it make sense to say $Z\setminus a$?</p>
<p>Thanks</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$X=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\}\}$$</p>
<p>and $$ Y=X\setminus\{\{\emptyset\}\}= \{\emptyset,\{\{\emptyset\}\}\} $$ because $\{\emptyset\}$ is removed from your $X$.</p>
<p>For your next question regarding $$ Z=\{a, b, c\}$$ $Z\setminus a$ does not make sense unless $a$ is a set.</p>
|
1,666,977 | <p><strong>Background</strong></p>
<p>This is purely a "sate my curiosity" type question.</p>
<p>I was thinking of building a piece of software for calculating missing properties of 2D geometric shapes given certain other properties, and I got to thinking of how to failsafe it in case a user wants to calculate the area of a $2$-gon, $1$-gon, $0$-gon, 'aslkfn'-gon or maybe even $-4$-gon.</p>
<p><strong>Question</strong></p>
<p>Are there any definitions for $n$-gons where $n < 0$?</p>
<p><strong>Valid assumptions</strong></p>
<p>Let's, for the sake of simplicity (if possible) say that $n \in \mathbb Z$, although I might come back later and ask what a $\pi$-gon is.</p>
| Bobson Dugnutt | 259,085 | <p>For regular $n$-gons with side-length $1$, the area is given as $$\frac{1}{4}n \cot \frac{\pi}{n}$$</p>
<p>Here are some values of the formula for negative values of $n$:</p>
<p>\begin{array}{c|c}
n & \cot\frac{\pi}{n} \\
\hline -1 & \text{complex } \infty \\
-2 & 0\\
-3 & -\frac{1}{\sqrt{3}} \\
n \leq 3 & <0
\end{array}</p>
<p>How you want to interpret that is up to you, but the function is there. </p>
<p>This kind of thing (where you extend something beyond what's intuitive) is done many places within mathematics. Take for instance the $\Gamma$-function (<a href="https://en.wikipedia.org/wiki/Factorial#Extension_of_factorial_to_non-integer_values_of_argument" rel="nofollow">see here</a>), which is an extension of the factorial function. However, I don't know if it is useful in this particular case (with the $n$-gons), but why not try? </p>
|
1,666,977 | <p><strong>Background</strong></p>
<p>This is purely a "sate my curiosity" type question.</p>
<p>I was thinking of building a piece of software for calculating missing properties of 2D geometric shapes given certain other properties, and I got to thinking of how to failsafe it in case a user wants to calculate the area of a $2$-gon, $1$-gon, $0$-gon, 'aslkfn'-gon or maybe even $-4$-gon.</p>
<p><strong>Question</strong></p>
<p>Are there any definitions for $n$-gons where $n < 0$?</p>
<p><strong>Valid assumptions</strong></p>
<p>Let's, for the sake of simplicity (if possible) say that $n \in \mathbb Z$, although I might come back later and ask what a $\pi$-gon is.</p>
| zyx | 14,120 | <p>For regular n-gons inscribed in a given oriented circle, a regular $(-n)$-gon can be defined to be regular n-gon with the opposite orientation. </p>
<p>This is not a definition I have ever seen in a publication, but it is consistent with most of the standard conventions.</p>
<p>Extending the idea a bit, one can take an orientation of the plane and consider the data of an n-gon, not necessarily regular, to always include a cyclic ordering of the vertices, and $n \to (-n)$ being orientation reversal. Whether a given convex n-gon has positive or negative $n$ can be interpreted as the question of whether its area, in the given cyclic ordering, is positive relative to the orientation of the plane.</p>
|
3,367,672 | <p>Assume <span class="math-container">$a_n$</span> is a non-negative sequence, that is, <span class="math-container">$a_n \geq 0$</span> for every <span class="math-container">$n \in \mathbb{N}$</span>. Prove that if <span class="math-container">$a_n$</span> converges then the limit is non-negative. Clue: prove it by negation.</p>
<p>I think proof by contradiction is a good method, which is <span class="math-container">$a_n$</span> converges and the limit is negative. Since <span class="math-container">$a_n$</span> converges, <span class="math-container">$\lim_{n\to\infty}a_n = L$</span>. Also, since <span class="math-container">$a_n$</span> is non-negative sequence, <span class="math-container">$L \geq 0$</span>. This is a contradiction. Is my method correct? </p>
| msm | 340,064 | <p>One way to do this is by contrapositive: if <span class="math-container">$L \in \mathbb{R}$</span> is negative, show that the sequence does not converge to <span class="math-container">$L$</span>. Take <span class="math-container">$L \in \mathbb{R}$</span> with <span class="math-container">$s<0$</span>. Then you know that for every <span class="math-container">$n$</span> you must have <span class="math-container">$|a_n - L| \geq |s|$</span> (why?). </p>
<p>Now, use the definition of the limit of a sequence to show that the sequence cannot converge to <span class="math-container">$L$</span>. </p>
|
3,995,272 | <p>I am a master student in mathematical physics. I study soliton and traveling wave solutions of the differential equations.</p>
<p>Let's consider the following ODE:
<span class="math-container">$$Q^{\prime}(\xi)=ln(A)(\alpha+\beta Q(\xi)+\sigma Q^2(\xi))$$</span>
where
<span class="math-container">$A \neq 0,1.$</span></p>
<p>In a book, the solutions of the ODE are given as follows: <strong>(But, I don't understand how to derive it.)</strong></p>
<p>There are twelve solution cases w.r.t coefficients of ODE.
Any help would be appreciated.</p>
<p><strong>CASE I)</strong> When <span class="math-container">$\beta^{2}-4 \alpha \sigma<0$</span> and <span class="math-container">$\sigma \neq 0$</span>, then
<span class="math-container">$$
\begin{array}{l}
Q_{1}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{2}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma} \cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2} \xi\right) \\
Q_{3}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\tan _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \sec _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right) \xi}\right)\right) \\
Q_{4}(\xi)=-\frac{\beta}{2 \sigma}-\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{2 \sigma}\left(\cot _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right) \pm \sqrt{p q} \csc _{A}\left(\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)} \xi\right)\right) \\
Q_{5}(\xi)=-\frac{\beta}{2 \sigma}+\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4 \sigma}\left(\tan _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha \sigma\right)}}{4} \xi\right)-\cot _{A}\left(\frac{\sqrt{-\left(\beta^{2}-4 \alpha\right) \sigma}}{4} \xi\right)\right)
\end{array}
$$</span></p>
<p><strong>CASE II:</strong></p>
<p><span class="math-container">$\vdots$</span></p>
<p><strong>CASE XII:</strong>
When <span class="math-container">$\beta=\lambda, \sigma=m \lambda(m \neq 0)$</span> and <span class="math-container">$\alpha=0,$</span> then
<span class="math-container">$$
Q_{37}(\xi)=\frac{p A^{\lambda \xi}}{q-m p A^{\lambda \xi}}
$$</span>
where triangular functions are defined as
<span class="math-container">\begin{array}{l}
\sin _{A}(\xi)=\frac{p A^{i \xi}-q A^{-i \xi}}{2 i}, \quad \cos _{A}(\xi)=\frac{p A^{i \xi}+q A^{-i \xi}}{2} \\
\tan _{A}(\xi)=-i \frac{p A^{i \xi}-q A^{-i \xi}}{p A^{i \xi}+q A^{-i \xi}}, \quad \cot _{A}(\xi)=i \frac{p A^{i \xi}+q A^{-i \xi}}{p A^{i \xi}-q A^{-i \xi}} \\
\sec _{A}(\xi)=\frac{2}{p A^{i \xi}+q A^{-i \xi}}, \quad \csc _{A}(\xi)=\frac{2 i}{p A^{i \xi}-q A^{-i \xi}}
\end{array}</span>
where <span class="math-container">$\xi$</span> is an independent variable, <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are constants greater than zero and called deformation parameters.</p>
| Aatmaj | 769,348 | <p>I have a lenthier method, put sinx=<span class="math-container">$e^t$</span> As x→π/2,sinx→1, so the limit t tends to 0. (<span class="math-container">$e^0=1$</span>)</p>
<p>we get</p>
<blockquote>
<p><span class="math-container">$lim_{t\rightarrow 0} \frac{t}{1-e^{2t}}$</span>=</p>
</blockquote>
<blockquote>
<p><span class="math-container">$lim_{t\rightarrow 0} \frac{t}{\left(1-e^t\right)\left(1+e^t\right)}$</span>=</p>
</blockquote>
<blockquote>
<p><span class="math-container">$lim_{t\rightarrow 0} \frac{t}{2\left(1-e^t\right)}$</span></p>
</blockquote>
<p>Now using taylor series expansion for <span class="math-container">$e^x$</span> we easily get the limit as -1/2. Can you proceed or should I explain ahead?</p>
|
1,879,076 | <p>How to show that $(1+x/n)^n\geq (1+x/10)^{10}$? (for $n\geq 10$)</p>
<p>I see that if I consider $n\to\infty$, the LHS approaches $e^x$, but that's all I could really see.</p>
<p>Please assume that $x\in[0,\infty)$</p>
| H. H. Rugh | 355,946 | <p>$$(1+\frac{x}{n})^n = \sum_{k=0}^n \frac{1 \times (1-\frac1n) ... (1-\frac{k-1}{n})}{k!} x^k \geq \sum_{k=0}^{10} \frac{1 \times (1-\frac{1}{10}) ... (1-\frac{k-1}{10})}{k!} x^k = (1+ \frac{x}{10})^{10}$$</p>
<p>In fact, same proof shows that $(1+x/n)^n$ is increasing in $n$ (fixed $x>0$).</p>
|
2,762,230 | <blockquote>
<p>Let $I:=[a,b]$ a perfect interval and $\gamma\in C(I,\Bbb R^n)$ an injective path such that $\Gamma:=\gamma(I)$ is rectifiable. Show that $\dim_H(\Gamma)=1$.</p>
</blockquote>
<p>Here $\dim_H$ is the Hausdorff dimension. My work so far: </p>
<p>Note that the canonical projections $\pi_k$ are Lipschitz, and because $\gamma$ is continuous and it domain is compact and connected then $\Gamma$ is also compact and connected, thus $\pi_k(\Gamma)\subset\Bbb R$ is compact and connected. </p>
<p>Because $I$ is perfect and $\gamma$ injective then $\Gamma$ is not a singleton, so there is some $k\in\{1,\ldots,n\}$ such that $\pi_k(\Gamma)$ is a perfect closed interval, thus setting
$$
f:\Gamma\to\Bbb R^n,\, x\mapsto (\pi_k(x),0,\ldots,0)\tag1
$$
we can see that $f$ is also Lipschitz and we find that $\dim_H(\pi_k(\Gamma))=\dim_H(f(\Gamma))=1\le\dim_H(\Gamma)$ by some elementary identities of the Hausdorff outer measures.</p>
<p>However Im unable to find a way to show that $\dim_H(\Gamma)\le 1$. I dont have a clue about how to do it. </p>
<p>Some random ideas that I had: I tried to relate that $\gamma$ have a continuous inverse in $\Gamma$, or some uniform polynomial approximation to $\Gamma$, or the fact that $\Gamma$ is rectifiable and compact with the definition of Hausdorff outer measure, but I dont found something.</p>
<p>Some help will be appreciated, thank you.</p>
| Jonas | 382,832 | <p>Let $H^1$ be the 1-dimensional outer Hausdorff measure on $\mathbb{R}^n$. Since the curve is injective, you can show that $H^1(\Gamma) =$ "Length of $\gamma$". See for instance <a href="https://math.stackexchange.com/questions/1186574/hausdorff-measure-of-rectifiable-curve-equal-to-its-length">here</a>. </p>
<p>Since $\gamma$ is rectifiable, its length $L$ is finite and by your argument not $0$. Thus $H^1(\Gamma) = L \in (0, \infty)$. However, if that is the case, we have $\dim_H(\Gamma) = 1$, by the definition of the Hausdorff dimension.</p>
|
137,595 | <p>I was using the <code>StreamPlot</code> function to plot the direction field of a system of two first order differential equations. Is there any way I could add solution curves to my direction field with this function? Or is there another function that could do that for me? I looked around, but I couldn't find anything. </p>
<p>Edit: The system of equations is: </p>
<p>$$x' = -2x+y-11\quad \& \quad y' = -5x+4y-35$$</p>
<p>Any help would be appreciated. Thanks!</p>
| Nasser | 70 | <blockquote>
<p>Is there any way I could add solution curves to my direction field
with this function</p>
</blockquote>
<h2>First method</h2>
<p>One direct way, is to use <code>Show</code> and simply add the solution to the Stream plot. Here is a quick example (since you did not give one)</p>
<pre><code>f[x_, y_] := y - x
p1 = StreamPlot[{1, f[x, y]}, {x, -5, 6}, {y, -4, 3}, Frame -> False,
Axes -> True, AspectRatio -> 1/GoldenRatio,
AxesLabel -> {"x", "y(x)"}, BaseStyle -> 12]
</code></pre>
<p><img src="https://i.stack.imgur.com/O2xdU.png" alt="Mathematica graphics"></p>
<p>Now to add solution curve, use <code>DSolve</code> to find the solution and add it using <code>Show</code></p>
<pre><code>ic = y[1] == .5;
sol = y[x] /. First@DSolve[{y'[x] == y[x] - x, ic}, y[x], x];
p2 = Plot[sol, {x, -4, 6}, PlotStyle -> Red];
Show[p1, p2]
</code></pre>
<p><img src="https://i.stack.imgur.com/DARhM.png" alt="Mathematica graphics"></p>
<h2>Second method</h2>
<p>Use the option <code>StreamPoints</code> to select stream line, which passes through the initial conditions. This is automatically then the solution curve. This does not require one to solve the ODE and obtain the solution like the above.</p>
<pre><code>f[x_, y_] := y - x
p1 = StreamPlot[{1, f[x, y]}, {x, -5, 6}, {y, -4, 3}, Frame -> False,
Axes -> True, AspectRatio -> 1/GoldenRatio,
AxesLabel -> {"x", "y(x)"}, BaseStyle -> 12,
StreamPoints -> {{{{1, .5}, Red}, Automatic}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/iK7Yu.png" alt="Mathematica graphics"></p>
|
269,450 | <p><strong>Bug introduced in 9.0 or earlier and persisting through 13.0.1 or later</strong></p>
<p>I noticed that initially valid syntax becomes invalid within nested <code>RootSum</code> objects. Consider for example the expression</p>
<pre><code>test=RootSum1[
Function[{x},2 x^7 Log[5805]-(25 Log[5805]^6 RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,Log[#1]/(-2+#1)&]^2)/16384],
Function[{x},(E^x (48 x^2 RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,(-11 Log[#1]-4 Log[#1] #1+Log[#1] #1^2)/(-174+95 #1-6 #1^2+#1^3)&]))/(64 x^3-48 x^2 Log[5805]+100 x RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,Log[#1]/(-2+#1)&]^2)]
]
</code></pre>
<p>Which is pretty valid expression where <code>RootSum</code> is replaced by <code>RootSum1</code> for demonstration. It contains two pure functions with valid syntax.</p>
<p>Now replace <code>RootSum1</code> by <code>RootSum</code>:</p>
<pre><code>(test /. RootSum1 -> RootSum) // N
</code></pre>
<blockquote>
<p>(3.63594 + 0. I) RootSum[
Function[{x}, {-32768. x^7 + 25. 48889.1 (0.416541 + 0. I)}],
Function[{x}, {(
2.71828^x x)/((10.4135 + 0. I) - 103.998 x + 16. x^2)}]
]</p>
</blockquote>
<p>Note that syntax for <code>Function</code> became invalid: additional internal <code>List</code> appeared, which prevents outer computation. With the internal list removed computations complete:</p>
<pre><code>(3.63594 + 0. I) RootSum[
Function[{x}, -32768. x^7 + 25. 48889.1 (0.416541 + 0. I)],
Function[{x}, (2.71828^x x)/((10.4135 + 0. I) - 103.998 x + 16. x^2)]
]
(* -0.283396 - 3.9421*10^-17 I *)
</code></pre>
<p>Is this a bug? Can it be avoided?</p>
| cvgmt | 72,111 | <pre><code>styles = Directive[Black, Dotted, Thickness[0.003], Opacity[1]];
Plot[x^2, {x, 0, 5}, GridLines -> {{{3, styles}, {4, styles}}},
PlotRange -> {5, 30}, PlotStyle -> {Black, Dotted, Thickness[0.003]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/MV8P4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MV8P4.png" alt="enter image description here" /></a></p>
|
269,450 | <p><strong>Bug introduced in 9.0 or earlier and persisting through 13.0.1 or later</strong></p>
<p>I noticed that initially valid syntax becomes invalid within nested <code>RootSum</code> objects. Consider for example the expression</p>
<pre><code>test=RootSum1[
Function[{x},2 x^7 Log[5805]-(25 Log[5805]^6 RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,Log[#1]/(-2+#1)&]^2)/16384],
Function[{x},(E^x (48 x^2 RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,(-11 Log[#1]-4 Log[#1] #1+Log[#1] #1^2)/(-174+95 #1-6 #1^2+#1^3)&]))/(64 x^3-48 x^2 Log[5805]+100 x RootSum[5805-696 #1+190 #1^2-8 #1^3+#1^4&,Log[#1]/(-2+#1)&]^2)]
]
</code></pre>
<p>Which is pretty valid expression where <code>RootSum</code> is replaced by <code>RootSum1</code> for demonstration. It contains two pure functions with valid syntax.</p>
<p>Now replace <code>RootSum1</code> by <code>RootSum</code>:</p>
<pre><code>(test /. RootSum1 -> RootSum) // N
</code></pre>
<blockquote>
<p>(3.63594 + 0. I) RootSum[
Function[{x}, {-32768. x^7 + 25. 48889.1 (0.416541 + 0. I)}],
Function[{x}, {(
2.71828^x x)/((10.4135 + 0. I) - 103.998 x + 16. x^2)}]
]</p>
</blockquote>
<p>Note that syntax for <code>Function</code> became invalid: additional internal <code>List</code> appeared, which prevents outer computation. With the internal list removed computations complete:</p>
<pre><code>(3.63594 + 0. I) RootSum[
Function[{x}, -32768. x^7 + 25. 48889.1 (0.416541 + 0. I)],
Function[{x}, (2.71828^x x)/((10.4135 + 0. I) - 103.998 x + 16. x^2)]
]
(* -0.283396 - 3.9421*10^-17 I *)
</code></pre>
<p>Is this a bug? Can it be avoided?</p>
| LouisB | 22,158 | <p>Another way (without <code>GridLines</code>) is use <code>Epilog</code> or <code>Prolog</code>. An advantage of this is grid lines can be added to the plot with the simplest <code>GridLines->Automatic</code> option without affecting the the vertical lines. Here is an example that uses color, thickness and dashing to distinguish the function, the grid lines and the vertical lines.</p>
<pre><code>vlines = {Black, Dotted, Thickness[0.005],
InfiniteLine[{#, 0}, {0, 1}] & /@ {3, 4}};
Plot[x^2, {x, 0, 5}, GridLines -> Automatic,
PlotRange -> {5, 30}, Prolog -> vlines]
</code></pre>
<p><a href="https://i.stack.imgur.com/b9PWC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b9PWC.png" alt="enter image description here" /></a></p>
|
868,935 | <p>I saw a proof that
$$
\lim_{x\to 0} \ln|x|\cdot x = 0
$$
where is is argued that for $x \in (0,1)$ we have
$$
| \ln(x) x | = \left| \int_1^x x/t ~\mathrm d t \right| = \left| \int_x^1 x/t ~\mathrm d t \right| \le \left|\int_x^1 1 dt\right| = |1 - x| \le 1
$$
and therefore the result follows, but why should the fact that $|\ln(x)x|$ is bounded imply it converges to zero?</p>
| 5xum | 112,884 | <p>By itself, what you have shown does not imply that the limit is zero. After all, the value of $\frac1x\cdot x$ is also bounded by $1$ but does not have a limit of $0$.</p>
<p>If your book says that the result follows soley from the bounds you wrote, it is wrong. I suggest, however, that you reread the entire argument made in the book.</p>
|
1,639,081 | <p>I have been unable to solve the following question, </p>
<p>If $$\sin(2x) - \tan(x) = 0$$</p>
<p>Find $x$ , $-\pi\le x\le \pi$</p>
<p>So far my workings have been
Use following identity: </p>
<p>$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0\\
2\frac{\sin(x)\cos(x)}{1} - \frac{\sin(x)}{\cos(x)} = 0$$</p>
<p>Then cross multiply to give :</p>
<p>$$-\sin x+((2\cos(x)\sin(x))\cos(x))/\cos(x) = 0$$</p>
<p>$$-\sin x+(2\cos^2(x)\sin(x))/ \cos(x) = 0$$</p>
<p>However, I have been unable to get any further.</p>
<p>If someone could help me find a solution to this question it would be very much appreciated.Thank you. </p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice, $$\sin 2x-\tan x=0$$
$$\frac{2\tan x}{1+\tan^2x}-\tan x=0$$
$$\tan x\left(\frac{1-\tan^2 x}{1+\tan^2x}\right)=0$$
$$\color{blue}{\tan x\cos 2x=0}$$
Now, solving for $x$,<br>
$$\tan x=0\iff x=n\cdot 180^\circ$$
where $n$ is any integer </p>
<p>For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1 $, one should get $$\color{blue}{x=-180^\circ, 0, 180^\circ}$$</p>
<p>or $$\cos 2x=0\iff 2x=(2n-1)\cdot 90^\circ\ \ $$$$or \ \ x=(2n-1)\cdot 45^\circ$$
where $n$ is any integer</p>
<p>For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1, 2 $, one should get $$\color{blue}{x=-135^\circ, -45^\circ, 45^\circ, 135^\circ}$$</p>
<p>hence, the complete solution is
$$\color{red}{x\in\{-180^\circ, -135^\circ, -45^\circ, 0, 45^\circ, 135^\circ, 180^\circ\}}$$ </p>
|
204,218 | <p>I can't bear an expression containing radicals of imaginary numbers,
in case it can be expressed as in terms of radicals of real numbers only.</p>
<p>For example, I can't bear the expression</p>
<pre><code>Sqrt[2 + I]
</code></pre>
<p>because
it can be expressed as </p>
<pre><code>Sqrt[1/2 (2 + Sqrt[5])] + Sqrt[1/2 (-2 + Sqrt[5])] I
</code></pre>
<p>But it seems there is no easy way to do it in Mathematica. I have tried many commands (in Mathematica) but all in vain.</p>
<p>Is there a systematic way to do such job ?</p>
<p><code>Sqrt[2 + I]</code> was a very simple example. I hope the method work for much more complicated expression.</p>
<p>P.S
I know that there are many algebraic numbers that cannot be expressed as in terms of radicals, For example, <code>Root[#^5 + # - 1 &, 1]</code>.</p>
| J. M.'s persistent exhaustion | 50 | <p>Yet another possibility:</p>
<pre><code>ComplexExpand[Sqrt[2 + I], TargetFunctions -> {Re, Im}] // FunctionExpand // Simplify
(5^(1/4) ((2 + I) + Sqrt[5]))/Sqrt[10 + 4 Sqrt[5]]
</code></pre>
|
4,461,327 | <p>To show that they are equal, I need to show</p>
<p><span class="math-container">$\bigcap_{n=1}^{\infty}[0,1+1/n) \subset [0,1]$</span> and <span class="math-container">$[0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$</span></p>
<p>My attempt is: let <span class="math-container">$x \in [0,1] \Rightarrow 0 \leq x \leq 1$</span>, since <span class="math-container">$1 < 1+1/n, \ \forall n \geq1 \Rightarrow x \in \bigcap_{n=1}^{\infty}[0,1+1/n) \Rightarrow [0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$</span></p>
<p>However, I don't know how to show <span class="math-container">$\bigcap_{n=1}^{\infty}[0,1+1/n) \subset [0,1]$</span>. It seems obvious since <span class="math-container">$\lim_{n \to \infty} 1+1/n = 1$</span>, but I am having trouble to proving that. Any help or hint would be appreciated</p>
| xpaul | 66,420 | <p>Showing
<span class="math-container">$$[0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$$</span>
is equivalent to showing
<span class="math-container">$$ \overline{[0,1]} \supset \overline{\bigcap_{n=1}^{\infty}[0,1+1/n)}. \tag1$$</span>
Since
<span class="math-container">$$ \overline{[0,1]}=(-\infty, 0) \cup (1, \infty)$$</span>
and
<span class="math-container">$$\overline{\bigcap_{n=1}^{\infty}[0,1+1/n)}=\bigcup_{n=1}^{\infty}(-\infty,0)\cup[1+1/n,\infty)=(-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty), $$</span>
(1) becomes
<span class="math-container">$$ (-\infty, 0) \cup (1, \infty) \supset (-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty). \tag2$$</span>
Now we show
<span class="math-container">$$ \bigcup_{n=1}^{\infty}[1+1/n,\infty)\subset(1,\infty). \tag3 $$</span>
In fact, for <span class="math-container">$\forall x\in\bigcup_{n=1}^{\infty}[1+1/n,\infty)$</span>, then <span class="math-container">$\exists n$</span> such that <span class="math-container">$x\in[1+1/n,\infty)$</span> and so <span class="math-container">$x\in(1,\infty)$</span>; namely, (3) is true. Thus
<span class="math-container">$$ (-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty)\subset(-\infty, 0) \cup (1, \infty) $$</span>
or (2) is true.</p>
|
1,115,545 | <p>In my lecture notes we have the following:</p>
<p>The set <span class="math-container">$$\mathbb{P}^2(K)=\{[x, y, z] | (x, y, z) \in (K^3)^{\star}\}$$</span> is called projective plane over <span class="math-container">$K$</span>.</p>
<p>There are the following cases:</p>
<ul>
<li><p><span class="math-container">$z \neq 0$</span>
<span class="math-container">$$\left [x, y, z\right ]=\left [\frac{x}{z}, \frac{y}{z}, 1\right ]$$</span>
<span class="math-container">$$\mathbb{P}^2(K) \ni \left [x, y, z\right ] \to \left (\frac{x}{z}, \frac{y}{z}\right ) \in A^2(K)$$</span>
These points are the finite points of <span class="math-container">$\mathbb{P}^2(K)$</span>.</p>
</li>
<li><p><span class="math-container">$z=0$</span></p>
<p>The points <span class="math-container">$\left [x, y, 0\right ]$</span> are called points at infinity of <span class="math-container">$\mathbb{P}^2(K)$</span>.</p>
</li>
</ul>
<p>The finite points <span class="math-container">$\left [x, y, 1\right ] \in \mathbb{P}^2(K)$</span> geometrically correspond to the intersection points of the lines from <span class="math-container">$(0,0,0)$</span> with the plane <span class="math-container">$z=1$</span>.</p>
<p>The points at infinity <span class="math-container">$\left [x, y, 0\right ]$</span> geometrically correspond to the lines of the plane <span class="math-container">$x0y$</span> that pass through <span class="math-container">$(0, 0, 0)$</span>.</p>
<p>We define as line in <span class="math-container">$\mathbb{P}^2(K)$</span> each equation of the form <span class="math-container">$$ax+by+cz=0, (a,b,c) \neq (0,0,0)$$</span> that means the set <span class="math-container">$$E=\{\left [x, y, z\right ] \in \mathbb{P}^2(K) | ax+by+cz=0, (a, b, c) \neq (0,0,0) \}$$</span></p>
<p>If <span class="math-container">$z \neq 0$</span> : <span class="math-container">$\left [x, y, z\right ]=\left [x, y, 1\right ]$</span>, that means that the equation is written <span class="math-container">$$ax+by+c=0$$</span> (and if <span class="math-container">$ab \neq 0$</span> then it is the known affine line)</p>
<p><span class="math-container">$$$$</span></p>
<p>Can you explain to me what <span class="math-container">$$\mathbb{P}^2(K) \ni \left [x, y, z\right ] \to \left (\frac{x}{z}, \frac{y}{z}\right ) \in A^2(K)$$</span> means?</p>
<p>Also, at the end, why does it stand that <span class="math-container">$\left [x, y, z\right ]=\left [x, y, 1\right ]$</span> ? At the beginning didn't we have that <span class="math-container">$\left [x, y, z\right ]=\left [\frac{x}{z}, \frac{y}{z}, 1\right ]$</span> ?</p>
| Redundant Aunt | 109,899 | <p>Use Cauchy-Schwarz:
$$ \frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}=\left(\frac{a^2}{a+b}+\frac{d^2}{a+d}+\frac{b^2}{b+c}+\frac{c^2}{c+d}\right)\cdot\left((a+b)+(a+d)+(b+c)+(c+d)\right)\cdot 0.5\geq (a+b+c+d)^2\cdot 0.5= 0.5 $$</p>
|
129,439 | <p>This code in Mathematica 9 returns two graphs that are NOT complementary.</p>
<pre><code>{GraphData[{7, 172}], GraphData[{7, 172}, "ComplementGraph"]}
</code></pre>
| Community | -1 | <p>In addition to the mapping presented in <a href="https://mathematica.stackexchange.com/a/129442/31159">Feyre's answer</a>, it is possible to check that those two graphs are complementary by generating the complement of the first graph, and verifying that it is isomorphic to the second graph.</p>
<pre><code>{graph1, graph2} = {GraphData[{7, 172}], GraphData[{7, 172}, "ComplementGraph"]};
(* first approach *)
graph1Complement1 = GraphComplement[graph1]
</code></pre>
<p><a href="https://i.stack.imgur.com/MeXpd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MeXpd.png" alt="enter image description here"></a></p>
<pre><code>(* second approach *)
graph1Complement2 = With[{adj = AdjacencyMatrix[graph1]},
AdjacencyGraph[1 - IdentityMatrix[Length[adj]] - adj]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/IrFN3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IrFN3.png" alt="enter image description here"></a></p>
<p>Both are isomorphic to <code>graph2</code>:</p>
<pre><code>IsomorphicGraphQ[graph1Complement1, graph2]
(* True *)
IsomorphicGraphQ[graph1Complement2, graph2]
(* True *)
</code></pre>
<hr>
<p>As mentioned by Feyre, the drawing of the graphs does not make it obvious that <code>graph2</code> is the complement of <code>graph1</code>. In particular, it cannot be easily seen that one of the vertex is of degree 5,</p>
<pre><code>VertexDegree[graph2]
(* {3, 3, 1, 0, 4, 4, 5} *)
</code></pre>
<p>This is however clearer in the drawing of <code>graph1Complement1</code>.</p>
|
3,946,591 | <blockquote>
<p>The sides of a triangle are on the lines <span class="math-container">$2x+3y+4=0$</span>, <span class="math-container">$ \ \ x-y+3=0$</span>, and <span class="math-container">$5x+4y-20=0$</span>. Find the equations of the altitudes of the triangle.</p>
</blockquote>
<p>Should I find the vertices first? Or is there a direct way? Actually, I tried finding the vertices, using the substitution method but I find it hard to turn it into an equation.</p>
| user2661923 | 464,411 | <p>Hint</p>
<p><span class="math-container">$\sum_{k=1}^n [f(k) \times g(k)].$</span></p>
<p><span class="math-container">$f(k)$</span> is the # of possible sequences of length <span class="math-container">$k$</span>, where the very first occurrence of the letter <span class="math-container">$b$</span> is on the <span class="math-container">$k$</span>-th letter.</p>
<p>Note that this categorization facilitates mutually exclusive groupings.</p>
<p><span class="math-container">$g(k)$</span> is the # of possible sequences of length <span class="math-container">$(n-k)$</span> where the letter <span class="math-container">$a$</span> is excluded from this sequence.</p>
<p>Two items need special handling:</p>
<p>What happens if the first <span class="math-container">$b$</span> in the sequence is on the last (i.e. <span class="math-container">$n$</span>-th) letter?</p>
<p>What happens if the sequence does not contain any <span class="math-container">$b$</span>'s.</p>
|
3,231,271 | <blockquote>
<p>Suppose <span class="math-container">$X$</span> is Banach and <span class="math-container">$T\in B(X)$</span> (i.e. <span class="math-container">$T$</span> is a linear and continuous map and <span class="math-container">$T:X \to X$</span>). Also, suppose <span class="math-container">$\exists c > 0$</span>, s.t. <span class="math-container">$\|Tx\| \ge c\|x\|, \forall x\in X$</span>. Prove <span class="math-container">$T$</span> is a compact operator if and only if <span class="math-container">$X$</span> is finite dimensional.</p>
</blockquote>
<p>"<span class="math-container">$X$</span> is finite dimensional <span class="math-container">$\implies$</span> <span class="math-container">$T$</span> is compact" is easy to show. To prove the other side, at first, I made a mistake, thinking <span class="math-container">$X$</span> is reflexive. Then this work can be easily done by the fact that any sequence of a reflexive linear space has a weakly convergent subsequence and <span class="math-container">$T$</span> is completely continuous (since <span class="math-container">$T$</span> is compact). But this is not the situation. </p>
<blockquote>
<p>So how to prove "<span class="math-container">$T$</span> is compact <span class="math-container">$\implies X$</span> is finite dimensional"?</p>
</blockquote>
| Martin Argerami | 22,857 | <p>Suppose that <span class="math-container">$X$</span> is infinite-dimensional. Then its unit ball is not compact, so there exists a sequence <span class="math-container">$\{x_n\}$</span> in the unit ball that admits no convergent subsequence; by replacing with a subsequence if necessary, we may assume that there exists <span class="math-container">$\delta>0$</span> with <span class="math-container">$\|x_n-x_m\|\geq\delta$</span> for all <span class="math-container">$n\neq m$</span>. Now
<span class="math-container">$$
\|Tx_n-Tx_m\|=\|T(x_n-x_m)\|\geq c\|x_n-x_m\|\geq c\delta>0,
$$</span>
so <span class="math-container">$\{Tx_n\}$</span> does not admit a convergent subsequence. </p>
|
3,814,199 | <p>On a Hilbert space <span class="math-container">$\mathcal{H}$</span>, it is well known that trace class operators have a finite trace. However, there are operators which are not trace class but have a finite trace, e.g. the <a href="https://math.stackexchange.com/a/3039962/138382">unilateral shift</a>. Then, consider a (bounded) operator <span class="math-container">$A$</span> such that
<span class="math-container">$$
c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle<\infty,
$$</span>
for some orthonormal basis <span class="math-container">$\{x_1,x_2,\ldots\}$</span>. I'm not sure whether this is enough for the trace of <span class="math-container">$A$</span> to be defined in a basis independent way. Namely, <span class="math-container">$\mathrm{tr}(A)=\sum_{n=1}^{\infty}\langle y_n,A y_n\rangle=c$</span> for <strong>any</strong> orthonormal basis <span class="math-container">$\{y_1,y_2,\ldots\}$</span>.</p>
<p>In the classics von Neumann's book on Mathematical Foundations of Quantum Mecanics, Sec. II.11, he commented <strong>this is true</strong> (in fact, he did not consider trace class operators) using essentially the following argument:
<span class="math-container">$$
c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\langle x_n,y_m\rangle \langle y_m, A x_n\rangle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\langle y_m, A x_n\rangle\langle x_n,y_m\rangle=\sum_{m=1}^{\infty}\langle y_m,A y_m\rangle.
$$</span>
However, it is not clear to me that the double sum can be switched. Using that <span class="math-container">$|\langle u,y_m \rangle \langle y_m, v\rangle|\leq \frac12 (|\langle u,y_m \rangle|^2+|\langle v,y_m \rangle|^2)$</span> it is obtained that
<span class="math-container">$$
\sum_{m=1}^{\infty}\langle x_n,y_n\rangle \langle y_n, A x_n\rangle
$$</span>
is absolutely convergent. Is this enough for the summation switch?</p>
<p>Thank you in advance.</p>
<p><strong>EDIT</strong>: As Robert Israel points, <span class="math-container">$\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle$</span> must be absolutely convergent. Is it enough?</p>
| Peter | 463,556 | <p>Your second approach seems right. Every assignment of lectures to teachers corresponds to a mapping from the set of lectures <span class="math-container">$S_M$</span> to the set of teachers <span class="math-container">$S_T$</span>. The number of ways to assign lectures to teachers such that every teacher has at least one lecture is then precisely, as you said, the number of surjective mappings from <span class="math-container">$S_M$</span> to <span class="math-container">$S_T$</span>.</p>
<p>To get to this number, we partition the set of lectures into <span class="math-container">$T$</span> subsets, one for each teacher. Since every teacher must get at least one lecture, each of these partitions must be nonempty. Now the number of ways to partition an <span class="math-container">$M$</span>-element set into <span class="math-container">$T$</span> nonempty subsets is given by the <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling number of the second kind</a>, <span class="math-container">$$S(M,T) = \frac{1}{T!} \sum_{k=0}^T (-1)^k \binom{T}{k} (T-k)^M.$$</span>
If we now fix one such partition, there are still <span class="math-container">$T!$</span> ways to assign the teachers to the <span class="math-container">$T$</span> sets of lectures, so the total number of such assignments is given by <span class="math-container">$T! S(M,T)$</span>. If each possible assignment of lectures to teachers has the same probability, then the probability that every teacher gets at least one lecture is given by <span class="math-container">$$p = \frac{T! S(M,T)}{T^M} = \sum_{k=0}^T (-1)^k \binom{T}{k} \left(\frac{T-k}{T}\right)^M.$$</span></p>
|
3,814,199 | <p>On a Hilbert space <span class="math-container">$\mathcal{H}$</span>, it is well known that trace class operators have a finite trace. However, there are operators which are not trace class but have a finite trace, e.g. the <a href="https://math.stackexchange.com/a/3039962/138382">unilateral shift</a>. Then, consider a (bounded) operator <span class="math-container">$A$</span> such that
<span class="math-container">$$
c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle<\infty,
$$</span>
for some orthonormal basis <span class="math-container">$\{x_1,x_2,\ldots\}$</span>. I'm not sure whether this is enough for the trace of <span class="math-container">$A$</span> to be defined in a basis independent way. Namely, <span class="math-container">$\mathrm{tr}(A)=\sum_{n=1}^{\infty}\langle y_n,A y_n\rangle=c$</span> for <strong>any</strong> orthonormal basis <span class="math-container">$\{y_1,y_2,\ldots\}$</span>.</p>
<p>In the classics von Neumann's book on Mathematical Foundations of Quantum Mecanics, Sec. II.11, he commented <strong>this is true</strong> (in fact, he did not consider trace class operators) using essentially the following argument:
<span class="math-container">$$
c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\langle x_n,y_m\rangle \langle y_m, A x_n\rangle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\langle y_m, A x_n\rangle\langle x_n,y_m\rangle=\sum_{m=1}^{\infty}\langle y_m,A y_m\rangle.
$$</span>
However, it is not clear to me that the double sum can be switched. Using that <span class="math-container">$|\langle u,y_m \rangle \langle y_m, v\rangle|\leq \frac12 (|\langle u,y_m \rangle|^2+|\langle v,y_m \rangle|^2)$</span> it is obtained that
<span class="math-container">$$
\sum_{m=1}^{\infty}\langle x_n,y_n\rangle \langle y_n, A x_n\rangle
$$</span>
is absolutely convergent. Is this enough for the summation switch?</p>
<p>Thank you in advance.</p>
<p><strong>EDIT</strong>: As Robert Israel points, <span class="math-container">$\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle$</span> must be absolutely convergent. Is it enough?</p>
| Math Lover | 801,574 | <p>While neither of your answers seem to be correct, your starting approach is fine. The problem is that you get duplicates and they need to be removed.<br />
For example, say, there are <span class="math-container">$4$</span> lectures <span class="math-container">$L1, L2, L3, L4$</span> and <span class="math-container">$3$</span> teachers <span class="math-container">$T1, T2, T3$</span>. Now the first thing you do is to assign teachers to at least one lecture. A few of them as -</p>
<p><span class="math-container">$a) (T1,L1), (T2,L2), (T3,L3)$</span><br />
<span class="math-container">$b) (T1,L2), (T2,L3), (T3,L4)$</span><br />
<span class="math-container">$c) (T1,L3), (T2,L1), (T3,L2)$</span><br />
<span class="math-container">$d) (T1,L4), (T2,L2), (T3,L3)$</span><br />
...<br />
...</p>
<p>Now when you take the first arrangement <span class="math-container">$(a)$</span>, <span class="math-container">$L4$</span> is left and it can go to any teacher. Say, an arrangement where it goes to teacher <span class="math-container">$T1$</span>. Similarly take arrangement <span class="math-container">$(d)$</span> and say lecture <span class="math-container">$L1$</span> which is left goes to teacher <span class="math-container">$T1$</span>,</p>
<p>Now these two arrangements become</p>
<p><span class="math-container">$a) (T1,L1), (T2,L2), (T3,L3), (T1, L4)$</span><br />
<span class="math-container">$d) (T1,L4), (T2,L2), (T3,L3), (T1, L1)$</span></p>
<p>If you notice, they are duplicate.</p>
<p>That is why we need to apply principle of inclusion-exclusion (PIE).</p>
<p>Now based on the question, we need to make sure each teacher teaches at least one lecture.</p>
<p>One way to do it is to find all arrangements where at least one teacher is not assigned a lecture using PIE. Then subtract it from total arrangements. That will give you arrangements where every teacher is assigned at least one lecture.</p>
<p>Let <span class="math-container">$N_k$</span> denote that teacher <span class="math-container">$k$</span> does not have a lecture. So we need to find <span class="math-container">$|\bigcup \limits_{k=1}^{T}{N_k}|$</span>. Number of ways to assign <span class="math-container">$M$</span> lectures to <span class="math-container">$(T-1)$</span> teachers leaving teacher <span class="math-container">$k$</span> is <span class="math-container">$(T-1)^M$</span>.</p>
<p>So for all teachers, it comes to <span class="math-container">$T(T-1)^M$</span>. But out of these arrangements, there are arrangements where two teachers (say, teacher <span class="math-container">$k=1$</span> and teacher <span class="math-container">$k=2$</span>) did not have any lecture. This will be repeated when you are counting for <span class="math-container">$k=1$</span> and for <span class="math-container">$k=2$</span>. So you need to find ways in which two teachers will have no lectures and remove them but you will end up removing some arrangements where <span class="math-container">$k=3$</span> is also empty so you should add those back...and this goes on. So you finally have -</p>
<p>Number of arrangements where at least one teacher has no lecture assigned <span class="math-container">$(N)$</span></p>
<p><span class="math-container">$= {{T}\choose{1}} {(T-1)}^M - {{T}\choose{2}} {(T-2)}^M + {{T}\choose{3}} {(T-3)}^M...+(-1)^{(T-2)} {{T}\choose{T-1}} 1^M + (-1)^{(T-1)} {{T} \choose{T}} 0^M$</span></p>
<p>Subtract it from <span class="math-container">$T^M$</span> to get the number of arrangements where every teacher has at least one lecture assigned</p>
<p><span class="math-container">$= {T \choose 0}T^M - {T \choose 1}(T-1)^M + {T \choose 2}(T-2)^M - {T \choose 3}(T-3)^M ... + (-1)^{(T-1)} {{T}\choose{T-1}} 1^M$</span></p>
<p><span class="math-container">$= \sum \limits_{k=0}^{T} (-1)^{k}{{T}\choose {k}} {(T-k)}^M$</span></p>
|
1,190,345 | <p>If $f$ is Riemann integrable on $[a,b]$ , is $|f|$ Riemann integrable on $[a,b]$ ?
(The metric is $\mathbb R$ usual)</p>
<p>The other is question is $f$ is Riemann integrable on $[a,b]$ , can I claim $f$ is bounded on $[a,b]$ ? (I think the answer can be either yes or no that depend on considerating generalized function or not)</p>
<p>Update: I think I made a mistake on second question that both definitions of Riemann integral and generalized Riemann integral on [a,b] request the f being bounded on [a,b]. Sorry for my mistake.</p>
| RRL | 148,510 | <p>Using the reverse triangle inequality we have</p>
<p>$$\begin{align}\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x) \\&=\sup_{x,y \in [x_{j-1},x_j]}| |f(x)|-|f(y)|| \\ &\leqslant \sup_{x,y \in [x_{j-1},x_j]}| f(x)-f(y)| \\ & = \sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x).\end{align}$$</p>
<p>For a given partition $P = (x_0,x_1, \ldots, x_n)$, the difference of upper and lower sums
is</p>
<p>$$U(P,|f|) - L(P,|f|) = \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x)\right)(x_j - x_{j-1})\\ \leqslant \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x)\right)(x_j - x_{j-1})\\= U(P,f) - L(P,f)$$</p>
<p>Since $f$ is Riemann integrable, for any $\epsilon > 0$ there exists a partition $P$ such that the difference of upper and lower Darboux sums satisfies</p>
<p>$$U(P, |f|) - L(P, |f|) \leqslant U(P,f)-L(P,f) < \epsilon.$$</p>
<p>Hence, $|f|$ is Riemann integrable.</p>
|
1,301,937 | <p>A curve $C$ is said to be <em>trigonal</em> if it admits a rational function $f: C \to \mathbb{CP}^1$ of degree $3$. Is any nonsingular plane curve of degree four trigonal? Can the map be chosen so that it has exactly $10$ ramification points?</p>
| user149792 | 149,792 | <p>The answer to both of the questions is yes.</p>
<p>Construct an $($almost everywhere$)$ $3$-sheeted covering map by choosing a point on our curve $C$ and a hyperplane not passing through it, and projecting from the point to the hyperplane. If we worked out the coordinates, this would clearly be given by a rational function, and it has degree $3$ by Bézout's Theorem. By Riemann-Hurwitz, which applies in full generality by using the $$e_\varphi(P) = v_P(\varphi^*t_{\varphi(P)})$$"vanishing of defining polynomial $P$ of $C$ along the projection line" definition, we have that$$\sum e_\varphi(P) - 1 = 4 + 3 \cdot 2 = 10,$$so we have exactly $10$ ramification points when there is no higher order ramification. Ramification points of order higher than $2$ correspond to degree $3$ or higher vanishing of $P$ along a projection line, i.e. tangency of the projection line at an inflection point. There are finitely many of these, so a generic choice of point and hyperplane will not any higher ramification, and the desired properties.</p>
|
103,540 | <p>Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$. A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.</p>
<p>What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?</p>
<p>More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?</p>
| Joseph O'Rourke | 6,094 | <p>A visualization aid, $n=10$:
<br />
<img src="https://i.stack.imgur.com/bBQwy.jpg" alt="HexChess n=10"><br /></p>
<p>And now here are Will Swain's rook placements:
<br /><img src="https://i.stack.imgur.com/mHq1b.jpg" alt="HexRooks n=6,5"><img src="https://i.stack.imgur.com/XEJo5.jpg" alt="HexRooks n=9,7"></p>
|
103,540 | <p>Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$. A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.</p>
<p>What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?</p>
<p>More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?</p>
| Noam D. Elkies | 14,830 | <p>Nice question!</p>
<p>For the maximum number of pairwise non-defending rooks,
Will Sawin proved an upper bound of $(2n/3) + 1$
in his comment to the original question. This bound is attained,
at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each,
starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$
and proceeding by steps of $(-1,-1,2)$ until reaching the
$y=0$ or $x=0$ edge of the triangle. This construction
generalizes Sawin's five-Rook placement for $n=6$.</p>
<p>On further thought, it seems we actually achieve
$\lfloor (2n/3) + 1 \rfloor$ exactly for all $n$.
Here's how it works for $n=12$ and $n=15$,
with $(2n/3)+1 = 9$ and $11$ respectively:</p>
<pre><code> .
. .
. . .
. . . . .
. . . . . . .
. . . R . . . . .
. . . . . . . . . . R
R . . . . . R . . . . . .
. . . . . R . . . . . . . R .
. R . . . . . . . R . . . . . . .
. . . . . . R . . . . . . . . . R . .
. . R . . . . . . . . . R . . . . . . . .
. . . . . . . R . . . . . . . . . . . R . . .
. . . R . . . . . . . . . . . R . . . . . . . . .
. . . . . . . . R . . . . . . . . . . . . . R . . . .
. . . . R . . . . . . . . . . . . . R . . . . . . . . . .
</code></pre>
<p>Starting from such a solution with $n=3m$, we can add an empty row
to get an optimal solution for $n=3m+1$, and remove an edge
(and the Rook it contains) to get an optimal solution for $n=3m-1$.
So this should solve the problem for all $n$.</p>
<p>Jeremy Martin also asks:</p>
<blockquote>
<p>More generally, is anything known about the graph whose vertices
are these ordered triples and whose edges are rook moves?</p>
</blockquote>
<p>I don't remember reading about this graph before.
Experimentally (for $3 \leq n \leq 16$) its adjacency matrix
has all eigenvalues integral, the smallest being $-3$ with huge multiplicity
$n-1\choose 2$; more precisely:</p>
<blockquote>
<p><strong>Conjecture.</strong> For $n \geq 3$ the eigenvalues of the adjacency matrix are:
a simple eigenvalue at the graph degree $2n$; a $n-1\choose 2$-fold
eigenvalue at $-3$; and a triple eigenvalue at each integer
$\lambda \in [-2,n-2]$, except that $\mu := \lfloor n/2 \rfloor - 2$
is omitted, and $\mu - (-1)^n$ has multiplicity only $2$.</p>
</blockquote>
<p>This is probably not too hard to show. For example, the $\lambda = -3$
eigenvectors constitute the codimension-$3n$ space of functions
whose sum over each of the $3(n+1)$ Rook lines vanishes.
<em>[Added later: in the comment Jeremy Martin reports that
he and Jennifer Wagner already made</em> <strong>and proved</strong> <em>the same conjecture.]</em></p>
<p>Given that the minimal eigenvalue is $-3$,
it follows by a standard argument in "spectral graph theory"
that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$.
But that's asymptotically worse than $2n/3 + O(1)$, though it's still
good enough to prove the optimality of Will Sawin's cocliques of size
$5$ for $n=6$ and of size $7$ for $n=9$.</p>
<p>Here's some <strong>gp</strong> code to play with this graph and its spectrum:</p>
<pre><code>{
R(n)=
l = [];
for(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));
matrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1
}
</code></pre>
<p>running "R($n$)" puts a list of the vertices in "l" and returns
the adjacency matrix with the corresponding labeling. So for instance</p>
<pre><code>matkerint(R(7)-2)~
matkerint(R(8)-1)~
</code></pre>
<p>returns matrices whose rows are nice generators of the
$2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.</p>
|
2,710,200 | <p>I need to find the norm of an operator from $l^2 \to l^1$, but I'm struggling because of the different norms on $l^2$ and $l^1$. </p>
<p>The operator is defined by $T:l^2 \to l^1, x_i \mapsto 2^{-i}x_i$. </p>
<p>Using the canonical basis, I have that $||T||\geq 1/2$, but I have a feeling this is not a very good lower bound. I also cant seem to find any upper bound, because I have that
$$||Tx||_1 = \sum_{i=1}^{\infty}2^{-i}x_i$$ but I can't relate this to $||x||_2$ because $||x||_2= (\sum_i^\infty x_i^2)^{1/2}$. </p>
<p>Thanks for any help!</p>
| Arian | 172,588 | <p>By Cauchy-Schwartz
$$\Big(\sum_{i=1}^{n}\frac{|x_i|}{2^i}\Big)^2\leqslant\Big(\sum_{i=1}^{n}\frac{1}{2^{2i}}\Big)\Big(\sum_{i=1}^nx^2_i\Big)$$
Taking limit $n\to\infty$
$$\lim_n\Big(\sum_{i=1}^{n}\frac{|x_i|}{2^i}\Big)^2\leqslant\lim_n\Big(\sum_{i=1}^{n}\frac{1}{2^{2i}}\Big)\lim_n\Big(\sum_{i=1}^nx^2_i\Big)$$
which yields
$$||Tx||_1^2=\Big(\sum_{i=1}^{\infty}\frac{|x_i|}{2^i}\Big)^2\leqslant\Big(\sum_{i=1}^{\infty}\frac{1}{2^{2i}}\Big)\Big(\sum_{i=1}^{\infty}x^2_i\Big)\Rightarrow ||Tx||_1^2\leqslant C||x||^2_2$$
with $$C:=\sum_{i=1}^{\infty}\frac{1}{2^{2i}}=\frac{1}{4}\cdot\frac{1}{1-\frac{1}{4}}=\frac{1}{3}$$
Therefore
$$||T||:=\sup_{||x||_2\leqslant 1}||Tx||_1\leqslant \frac{\sqrt{3}}{3}$$
We need to show that $$||T||\geqslant\frac{\sqrt{3}}{3}$$
You can pick $x^*$ as given by @user545497 in the comments above with $x^*_k:=1/2^k$ then $x^*_k\in\mathcal{l}^2(\mathbb{N})$ and $||x^*||_2<1$. Also $||Tx^*||_1=C||x^*||_2$. Thus $||T||=\sqrt{3}/3$.</p>
|
668,291 | <p>If $h$ and $k$ are any two distinct integers, then $h^n-k^n$ is divisible by $h-k$.</p>
<p>Let's start with the basis. Let $n=1$, then
$h^1-k^1 = h-k$</p>
<p>Now for the induction, I can't use $k$ because I don't want to be confused. So let $P(r)$ for $h^n-k^n$ and that's $h^r-k^r$</p>
<p>$h^r-k^r = h-k$</p>
<p>$h^r = h-k +k^r$</p>
<p>So, for $P(r+1)$</p>
<p>$h^{r+1}-k^{r+1}$</p>
<p>$h^r * h^1 - k^r * k^1$</p>
<p>$ (h-k +k^r) * h -k^r *k $</p>
<p>This is the point where I'm not certain if I should distribute the $h $ all over the place...so here it is</p>
<p>$ (h*h-k*h +k^r*h) -k^r *k $</p>
<p>$ (h*h)+(-k*h) +(k^r*h) -k^r *k $</p>
<p>$ (h)*(h-k) + (k^r)*(h-k)$</p>
<p>$(h-k) * (h+k^r)$</p>
| Gurvan | 127,051 | <p>Notice that you have
$$
\begin{align}
&a^2-b^2 = (a-b)(a+b) \\
&a^3-b^3 = (a-b)(a^2+ab+b^2) \\
&a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3) \\
etc
\end{align}
$$
In fact for every integer $n \ge 1$ you have
$$
a^n-b^n = (a-b) \cdot \sum_{k=0}^{n-1}a^kb^{n-1-k}
$$
Of course you can prove this formula by induction, or just develop the expression and order the terms of the sum to check the equality.</p>
<p>With this formula you see immediately the divisibility.</p>
|
213,102 | <p>Is it true than an aperiodic, ergodic, minimal and equicontinuous dynamical system on a compact metric space is totally ergodic ?</p>
<p>According to some results I found in some books, a rotation on a compact metric group is equicontinous, and it is minimal <del>and totally ergodic</del> whenever it is ergodic.</p>
| V. Delecroix | 15,342 | <p>The answer is no in this generality! If you consider the classical odometer (i.e. addition by 1 on 2-adic integers) then its second power (addition by 2) is not minimal. This second power preserves the numbers starting with 0 (the "even numbers") and the numbers starting with 1 (the "odd numbers"). But of course this example is totally disconnected.</p>
|
2,960,734 | <p>So basically, I am given the following to prove:</p>
<blockquote>
<p>Let <span class="math-container">$+\gamma$</span> be a positively oriented smooth Jordan arc, and let <span class="math-container">$\omega$</span> denote the interior of <span class="math-container">$+\gamma$</span>. Recall that if <span class="math-container">$F = (F_1, F_2):D \to \mathbb{R}^2$</span> is a continuously differentiable vector field in an open set <span class="math-container">$D$</span> containing <span class="math-container">$\omega \cup(+\gamma)$</span>, then</p>
<p><span class="math-container">$$\iint_\omega \left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) dxdy = \oint_{+\gamma}F \cdot \overrightarrow{ds} $$</span>
where the right hand-side is the line-integral of <span class="math-container">$F$</span> along the path <span class="math-container">$+\gamma$</span>.</p>
<p>By suitably choosing <span class="math-container">$F$</span>, prove that <span class="math-container">$$ \DeclareMathOperator{\Area}{Area} \DeclareMathOperator{\diameter}{diameter} \DeclareMathOperator{\length}{length} 2\Area(\omega) \leq \diameter(+\gamma) \length(+\gamma)$$</span>
where</p>
<p><span class="math-container">$\diameter(+\gamma) = \sup\{|z(s)-z(t)| : s,t \in [a,b]\}$</span> and <span class="math-container">$\length(+\gamma) = \int_{a}^{b} |z'(t)| dt$</span>.</p>
</blockquote>
<p>The only thing I know so far is that I need to find an <span class="math-container">$F$</span> such that <span class="math-container">$\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} =1$</span> because then <span class="math-container">$$\iint_{\omega} dxdy = \Area(\omega).$$</span> However, I do not know where to proceed from there!</p>
| Gibbs | 498,844 | <p>Choose <span class="math-container">$F(x,y) = (-y+y(a),x-x(a))$</span> and define a <span class="math-container">$1$</span>-form <span class="math-container">$\alpha := (x-x(a))\mathrm{d}y+(y(a)-y)\mathrm{d}x = F_2(x,y)\mathrm{d}y+F_1(x,y)\mathrm{d}x$</span>. Then
<span class="math-container">$$\mathrm{d}\alpha = 2\mathrm{d}x \wedge \mathrm{d}y$$</span>
and so
<span class="math-container">$$\int_{\omega} \mathrm{d}\alpha= \int_{\omega}\left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \mathrm{d}x\mathrm{d}y= 2\int_{\omega} \mathrm{d}x \wedge \mathrm{d}y = 2\mathrm{Area}(\omega). $$</span>
By Stokes theorem (which reduces exactly to the curl theorem you are recalling in this case), you also get
<span class="math-container">\begin{align}
\int_{\omega} \mathrm{d}\alpha &= \int_{\partial \omega} \alpha = \int_{\gamma+} \alpha\\
&= \int_a^b (x(t)-x(a))y'(t)+(y(a)-y(t))x'(t)\,\mathrm{d}t = \int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t,
\end{align}</span>
where <span class="math-container">$\langle \cdot, \cdot \rangle$</span> denotes the dot product. By Cauchy-Schwarz inequality you get
<span class="math-container">\begin{align}
\int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t & \leq \int_a^b \sqrt{(x(t)-x(a))^2+(y(t)-y(a))^2}\sqrt{x'(t)^2+y'(t)^2}\, \mathrm{d}t \\ &
\leq \mathrm{diameter}(\omega)\int_a^b \lvert z'(t)\rvert \, \mathrm{d}t\\
&\leq \mathrm{diameter}(\omega)\mathrm{length}(\gamma+).
\end{align}</span></p>
|
1,710,523 | <p>Im asked to find the arc length of :</p>
<p>$$
\int_{-2}^{x}\sqrt{3w^4-1}dw
$$
where x is between -2 and -1.</p>
<p>Do I find the integral just as I would normally find it then find the arc length of that? Im a little confused on the notation I guess.</p>
| Plutoro | 108,709 | <p>The function $$f(x)=\int_{-2}^x\sqrt{3w^4-1}\,dw$$ is a curve in the plane. The formula for arc-length of a curve is $$L=\int_a^b\sqrt{(f'(x))^2+1}\,dx.$$ The fundamental theorem of calculus tells us that $$f'(x)=\sqrt{2x^4-1}.$$ Therefore, you need to solve the integral $$L=\int_{-2}^{-1}\sqrt{(\sqrt{3x^4-1})^2+1}\,dx.$$</p>
|
1,816,109 | <p>Does anyone know how to deal with the integral
$$\int_0^{\pi /2}\cot^n(x)dx$$</p>
<p>with $n\in (-1,1)$?</p>
<p>Apparently it is a well known identity: it is listed in the <a href="http://mathworld.wolfram.com/Cotangent.html" rel="nofollow">page of wolfram for the cotangent</a>, where it says that it equals $2^{-1}\pi \sec [2^{-1} (\pi n)]$</p>
<p>Thanks in advance for any solution or hint!</p>
| Mark Viola | 218,419 | <p>For $|n|<1$, we can write</p>
<p>$$\begin{align}
\int_0^{\pi/2}\cot^n(x)\,dx&=\frac12 B\left(\frac{1-n}{2},\frac{1+n}{2}\right)\\\\
&=\frac12 \Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{1+n}{2}\right)\\\\
&=\frac{\pi}{2\sin\left(\pi \frac{1+n}{2}\right)}\\\\
&=\frac{\pi}{2\cos(n\pi/2)}
\end{align}$$</p>
|
4,496,736 | <p>Question: Use the variation of parameter method to find the general solution of the following differential equation
<span class="math-container">$$(\cos x) y''+(2\sin x) y'-(\cos x) y =0\;\;\;\;,\;\;\;\;0<x<1$$</span></p>
<p><strong>My Try:</strong></p>
<p>I think the question is wrong, since the right hand side term is 0, so the particular integral will also be zero. Thus, the general solution will be equal to homogenous solution. So, I think no use of using the variation of parameter formulas since <span class="math-container">$y_p(x)=0$</span> always.
I reduced the equation as in the subject or title and then used integrating factor <span class="math-container">$$y=(\cos x )z$$</span> to eliminate the term <span class="math-container">$y'$</span> but I got another difficult DE as <span class="math-container">$z''-2\sec^2xz=0$</span></p>
<blockquote>
<p>Please help with any suggestions or do you think question is correct. Is there a way to solve it ?</p>
</blockquote>
| Doug | 1,064,129 | <p>By trial and error, you can show that one of the fundamental solutions is <span class="math-container">$y_1(x) = c\sin(x)$</span>. Now, Abel's Theorem tells us that the Wronskian of solutions is given by
<span class="math-container">$$W(x) = c\text{exp}\left(-\int 2\tan(x) dx\right) = c\cos^2(x).$$</span>
But, <span class="math-container">$W$</span> is also equal to the determinant of the Wronskian matrix:
<span class="math-container">$$W(x) = (\cos(x))y_2(x)-(\sin(x))(y_2)'(x).$$</span>
So, solve the equation
<span class="math-container">$$(\sin(x))(y_2)'(x)-(\cos(x))y_2(x) = \cos^2(x),$$</span>
using VOP (or just integrating factor) to find the other fundamental solution, <span class="math-container">$y_2(x)$</span>.
I find that <span class="math-container">$y_2(x) = c\sin(x)-x\sin(x)-\cos(x)$</span>.
Then, the general solution would be
<span class="math-container">$$y(x) = c_1\sin(x)+c_2(x\sin(x)+\cos(x)).$$</span></p>
<p>Notice that we do not need to use Abel's theorem. You could also sub in <span class="math-container">$z(x) = \sin(x)y(x)$</span> and obtain a first-order equation in the same spirit of what you were trying. This approach is know as Reduction of Order.</p>
|
15,871 | <p>I would like to state something about the existence of solutions $x_1,x_2,\dots,x_n \in \mathbb{R}$ to the set of equations</p>
<p>$\sum_{j=1}^n x_j^k = np_k$, $k=1,2,\dots,m$</p>
<p>for suitable constants $p_k$. By "suitable", I mean that there are some basic requirements that the $p_k$ clearly need to satisfy for there to be any solutions at all ($p_{2k} \ge p_k^2$, e.g.).</p>
<p>There are many ways to view this question: find the coordinates $(x_1,\dots,x_n)$ in $n$-space where all these geometric structures (hyperplane, hypersphere, etc.) intersect. Or, one can see this as determining the $x_j$ necessary to generate the truncated Vandermonde matrix $V$ (without the row of 1's) such that $V{\bf 1} = np$ where ${\bf 1} = (1,1,\dots,1)^T$ and $p = (p_1,\dots,p_m)^T$.</p>
<p>I'm not convinced one way or the other that there has to be a solution when one has $m$ degrees of freedom $x_1,\dots,x_m$ (same as number of equations). In fact, it would be interesting to even be able to prove that for finite number $m$ equations $k=1,2,\dots,m$ that one could find $x_1,\dots,x_n$ for bounded $n$ (that is, the number of data points required does not blow up).</p>
<p>A follow on question would be to ask if requiring ordered solutions, i.e. $x_1 \le x_2 \le \dots \le x_n$, makes the solution unique for the cases when there is a solution.</p>
<p>Note: $m=2$ is easy. There is at least one solution = the point(s) where a line intersects a circle given that $p_2 \ge p_1^2$. </p>
<p>Any pointers on this topic would be helpful -- especially names of problems resembling it.</p>
| Douglas S. Stones | 2,264 | <p>A specific (and rather well-known) example exists for prime numbers. There are necessary conditions (such as Fermat's Little Theorem) which is satisfied by all primes, but sometimes satisfied by composites as well. These composites are called <em><a href="http://en.wikipedia.org/wiki/Pseudoprime" rel="nofollow">pseudoprimes</a></em>. The property D could be considered to be <em>non-pseudoprimality</em>.</p>
|
3,433,249 | <p>I need to find the number of <span class="math-container">$7$</span>s if we write all the numbers from <span class="math-container">$1$</span> to <span class="math-container">$1000000$</span>(so <span class="math-container">$77$</span>, for example, counts as two <span class="math-container">$7$</span>s and not one).</p>
<p>Here's what I did:</p>
<p>I split the problem into <span class="math-container">$7$</span> sections:</p>
<p>The number of <span class="math-container">$7$</span>s in numbers with one seven: <span class="math-container">$\displaystyle \binom{7}{1}$$*$</span> <span class="math-container">$9^6$</span>(number of ways to place one <span class="math-container">$7$</span> times the number of possible numbers we could make with each displacement. Note that leading zeros wouldn't be a problem since they would result in numbers with less than <span class="math-container">$7$</span> digits which we need)</p>
<p>The number of <span class="math-container">$7$</span>s in numbers with two sevens:
<span class="math-container">$\displaystyle\binom{7}{2} * 9^5 * 2$</span>(same logic, but we multiplied it by <span class="math-container">$2$</span> since there's two sevens)</p>
<p>...</p>
<p>So my answer would be <span class="math-container">$\sum_{i=1}^7 \displaystyle \binom{7}{i}9^ii$</span> but my textbook says the right answer is <span class="math-container">$600000$</span>. I do understand its solution but I don't know why mine is wrong.</p>
<p>Thanks in advance!</p>
| kingW3 | 130,953 | <p>You're counting <span class="math-container">$7$</span> digit numbers but you only need <span class="math-container">$6$</span>.</p>
|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| Asinomás | 33,907 | <p>The probability is <span class="math-container">$\frac{1}{2}$</span> because the last flip determines it.</p>
|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| Arthur | 15,500 | <p>All the possible sequences of tosses (which are all equally likely) may be divided into the two categories "even number of heads" and "odd number of heads". You can then pair up each sequence in one category with the "flipped" version (flip every coin in the sequence around) in the other category. This shows that there are equally many sequences in each of the two categories. So the probability of landing in a specific one of them must be <span class="math-container">$\frac12$</span>.</p>
|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| Frxstrem | 1,556 | <p>There are two cases here:</p>
<ul>
<li>There's an even number of heads: 0, 2, 4, 6 or 8 heads</li>
<li>There's an odd number of heads: 1, 3, 5, 7 or 9 heads</li>
</ul>
<p>But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:</p>
<ul>
<li>There's an even number of tails: 0, 2, 4, 6 or 8 tails</li>
</ul>
<p>Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability <span class="math-container">$1/2$</span>.</p>
|
2,406,061 | <p>I am also confused about whether these are symbols or have some meaning of their own.
PS- I know that <span class="math-container">$\operatorname{d}y\over\operatorname{d}x$</span> geometrically represents the slope. But, I've come across <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> to make problems easier. What does <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> mean?</p>
| Community | -1 | <p><a href="https://i.stack.imgur.com/XWjZz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XWjZz.png" alt="enter image description here"></a></p>
<p>$dy$ represents the $y$ increment along the tangent. It is strictly proportional to $dx$.</p>
<p>$\Delta y$ represents the $y$ increment along the curve. It depends on the shape of the curve.</p>
<p>$\dfrac{dy}{dx}$ is the slope of the tangent, i.e. the derivative. $\dfrac{\Delta y}{\Delta x}$ approximates it for small $\Delta x$.</p>
|
1,458,405 | <p>(p ∨ q) → r ≡ (p → q) ∨ (p → r) could be valid or invalid</p>
<p>I need to prove it using logical equivalences (can't use truth table)</p>
<p>This is how far I've gotten by working with the right side:</p>
<p><-> p→(q v r)</p>
<p><-> ¬p v (q v r) </p>
<p>then commutative law</p>
<p><-> (q v r) v ¬p </p>
<p>then commutative law</p>
<p><-> (r v q) v ¬p </p>
<p>then associative law</p>
<p><-> r v ( q v ¬p ) </p>
<p>then commutative law</p>
<p><-> (q v ¬p) v r</p>
<p>then commutative law</p>
<p><-> (¬p v q) v r</p>
<p><-> ¬(¬p v q) → r</p>
<p>What do I do next? Or did I do this wrong? Thanks for the assistance</p>
| Brian M. Scott | 12,042 | <p>When you get to $\neg p\lor(q\lor r)$, I would apply associativity:</p>
<p>$$\begin{align*}
(p\to q)\lor(p\to r)&\equiv p\to(q\lor r)\\
&\equiv\neg p\lor(q\lor r)\\
&\equiv(\neg p\lor q)\lor r\;.
\end{align*}$$</p>
<p>Your target, after all, is $(p\lor q)\to r$, which you know is equivalent to $\neg(p\lor q)\lor r$, with $r$ hanging out there on the end, so it makes sense to go for something with the same general form. However, $\neg(p\lor q)\equiv\neg p\land\neg q$, and it should be pretty clear intuitively that $(\neg p\lor q)\lor r$ is <strong>not</strong> equivalent to $(\neg p\land\neg q)\lor r$: they’re both true if $r$ is true, but if $r$ is false and $p$ and $q$ are both true, $(\neg p\land\neg q)\lor r$ is false, but $(\neg p\lor q)\lor r$ is still true.</p>
|
2,002,201 | <p>simplify <span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$</span>.</p>
<blockquote>
<p>1.<span class="math-container">$90^{\frac{3}{2}}$</span></p>
<p>2.<span class="math-container">$106\sqrt{41}$</span></p>
<p>3.<span class="math-container">$4\sqrt{41}$</span></p>
<p>4.<span class="math-container">$504$</span></p>
<p>5.<span class="math-container">$508$</span></p>
</blockquote>
<p><strong>My attempt</strong>:I do like this but I didn't get any of those five.</p>
<p><span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$</span></p>
<p><span class="math-container">$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$</span></p>
<p>Now I do the nested radicals formula and I get <span class="math-container">$254\sqrt{41}$</span> which is none of those where did I mistaked?</p>
| Bernard | 202,857 | <p><strong>Hint:</strong>
$$45\pm4\sqrt{41}=(2\pm\sqrt{41})^2.$$</p>
|
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