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 |
|---|---|---|---|---|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| J Marcos | 65,869 | <p>Proofs by contradiction prove that your assumptions are <em>jointly incompatible</em>. In this case, from $P\to Q$ and $R\to\neg Q$ you may conclude that $P\uparrow R$, where $\uparrow$ denotes <a href="http://en.wikipedia.org/wiki/Sheffer_stroke" rel="nofollow">NAND</a>. Note that $P\uparrow R$ is true iff $P$ and $R$ cannot be simultaneously true. Negation is a particular case of NAND, when a sentence is 'self-incompatible' (i.e., $S\equiv(S\uparrow S)$). When reasoning by contradiction, you have in general no reason to conclude that a single one of your assumptions is false (in other words, to conclude that its negation is true), unless you have reasons to maintain the truth of all the other assumptions (which appears not to be the case in the statement of your <strong>Theorem</strong>, where no reason is given for one to prefer $P$ over $R$).</p>
<p>Of course, $P\uparrow R$ <em>is</em> equivalent to $P\to\neg R$ and to $R\to\neg P$ (and to $\neg P\lor\neg R$), but I feel this facile observation does not tell the full story. Because many people seem not to be comfortable with a <em>disjunctive conclusion</em> such as $\neg P\lor\neg R$, they seem to prefer restating this as an implication.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| Tanner Swett | 13,524 | <p>Great question. The answer is that, <em>at that point in the proof</em>, you're still assuming that $P$ is true. With the assumptions stated more prominently, the proof goes like this:</p>
<blockquote>
<p>We are given that $P \to Q$ and that $R \to \neg Q$. <strong>Now we begin imagining that $P$ is true.</strong> Since $P$ is true and $P \to Q$, we know that $Q$ is true. <strong>Now we begin imagining that $R$ is true.</strong> Since $R$ is true and $R \to \neg Q$, we know that $\neg Q$ is true as well. So, $Q$ and $\neg Q$ are both true, which is a contradiction. <strong>Now we stop imagining that $R$ is true.</strong> Since assuming that $R$ is true leads to a contradiction, $R$ must be false. <strong>Now we stop imagining that $P$ is true.</strong> Since assuming that $P$ is true leads to the conclusion that $R$ is false, $P \to \neg R$.</p>
</blockquote>
<p>Sure, either $P$ or $R$ could be false, and the other one true. Why do we say that $R$ is false instead of allowing the possibility that $P$ is false? Because we haven't yet stopped imagining that $P$ is true.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| Taemyr | 114,582 | <p>We are trying to prove that given $P \rightarrow Q$ and $R \rightarrow \neg Q$, we get $P \rightarrow \neg R$</p>
<p>The proof you describe derives a contradiction by assuming $P$ and $R$, so at least one of the assumptions have to be false.</p>
<p>You wonder why we have to select $R$ as the false assumption. We do not have to do so, but it does not matter. If $\neg P$ then $P\rightarrow \neg R$ is vacously true.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| Dan Christensen | 3,515 | <blockquote>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
</blockquote>
<p>If you look at a proof as an ordered (numbered) sequence of statements, the assumption that is negated by a contradiction is always the last assumption previously introduced that has yet to be discharged.</p>
|
313,522 | <p>The problem I am currently working on is:</p>
<blockquote>
<p>Consider the following information: where <ul><li>$A$={Visa Card}</li> <li>$B$={MasterCard}</li></ul>
$P(A)=0.5$, $P(B)=0.4$, and $P(A \cap B) = 0.25$</p>
</blockquote>
<p>The part I am having difficulty with is part (e):</p>
<blockquote>
<p>Given that an individual is selected at random and that he or she has at least one card, what is the probability that he or she has a Visa Card?</p>
</blockquote>
<p>For some reason it is just eluding me. Could someone help me?</p>
| gelichor | 62,894 | <p>The probability of having at least one card is: $$P(A)+P(B)-P(A\cap B) = 0.65 $$</p>
<p>Denote $C$={ <em>at least one card</em> }.</p>
<p>The probability you need (definition of conditional probability):</p>
<p>$$ P(A\;|\;C) = \frac {P(A\cap C)}{P(C)}$$</p>
<p>If you have a Visa card, you have at least one card, so $P(A\cap C)=P(A)$</p>
<p>$$ P(A\;|\;C) = \frac {P(A)}{P(C)} = \frac{0.5}{0.65}$$</p>
|
313,522 | <p>The problem I am currently working on is:</p>
<blockquote>
<p>Consider the following information: where <ul><li>$A$={Visa Card}</li> <li>$B$={MasterCard}</li></ul>
$P(A)=0.5$, $P(B)=0.4$, and $P(A \cap B) = 0.25$</p>
</blockquote>
<p>The part I am having difficulty with is part (e):</p>
<blockquote>
<p>Given that an individual is selected at random and that he or she has at least one card, what is the probability that he or she has a Visa Card?</p>
</blockquote>
<p>For some reason it is just eluding me. Could someone help me?</p>
| Gerry Myerson | 8,269 | <p>Suppose there are $100$ people. How many have a Visa card? how many a Mastercard? how many both? of those who have at least one, how many have a Visa card?</p>
|
4,039,424 | <p>I would like to calculate the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$</span>.
I know that the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2})$</span> have the form of <span class="math-container">${a+b\sqrt[3]{2}+c\sqrt[3]{4}}$</span>, where a,b,c <span class="math-container">$\in \mathbb{Q}$</span> and the elements of <span class="math-container">$\mathbb{Q}(\sqrt{3})$</span> have the form of <span class="math-container">$a+b\sqrt{3}$</span>, where a,b <span class="math-container">$\in \mathbb{Q}$</span>.
I also calculated the minimal polynomial of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span> over <span class="math-container">$\mathbb{Q}$</span> , which is: <span class="math-container">$x^6−9x^4−4x^3+27x^2−36x−23$</span>.</p>
<p>Can you help me to calculate the form of the elements of <span class="math-container">$\mathbb{Q}(\sqrt[3]{2}+\sqrt{3})$</span>? I have to find the elements of the linear combinations which form the higher powers of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span> which are also powers of <span class="math-container">$(\sqrt[3]{2}+\sqrt{3})$</span>?
Could you give me a proper method to find the solution? Also, could you write down your calculation in your answer? Thank you for helping me!</p>
| TravorLZH | 748,964 | <p>Similar to how Gronwall deduced</p>
<p><span class="math-container">$$
\limsup_{n\to\infty}{\sigma(n)\over e^\gamma n\log\log n}=1
$$</span></p>
<p>We consider defining <span class="math-container">$a_n$</span> such that</p>
<p><span class="math-container">$$
\prod_{p\le p_{a_n}}p\le n\le\prod_{p\le p_{a_n+1}}p
$$</span></p>
<p>Taking logarithms gives</p>
<p><span class="math-container">$$
{\vartheta(p_{a_n})\over p_{a_n}}\le{\log n\over p_{a_n}}\le{p_{a_n+1}\over p_m}\cdot{\vartheta(p_{a_n+1})\over p_{a_n+1}}
$$</span></p>
<p>Now, due to prime number theorem we have <span class="math-container">$p_n\sim p_{n+1}$</span> and <span class="math-container">$\vartheta(x)\sim x$</span>, so that</p>
<p><span class="math-container">$$
\log n\sim p_{a_n}\tag1
$$</span></p>
<p>With these priliminaries, we can begin working on the problem</p>
<p><span class="math-container">$$
\sum_{p|n}\frac1p\le\sum_{p\le p_{a_n}}\frac1p=\log\log p_{a_n}+\mathcal O(1)
$$</span></p>
<p>By (1), we also see that</p>
<p><span class="math-container">$$
\sum_{p\le p_{a_n}}\frac1p\sim\log\log\log n
$$</span></p>
<p>To see how tight this bound is, we set</p>
<p><span class="math-container">$$
n_k=(p_1p_2\cdots p_k)^{\lfloor\log p_k\rfloor}
$$</span></p>
<p>so that</p>
<p><span class="math-container">$$
\log n_k=\lfloor\log p_k\rfloor\vartheta(p_k)\sim p_k\log p_k
$$</span></p>
<p>and</p>
<p><span class="math-container">$$
\log\log n_k\sim\log p_k
$$</span></p>
<p>Plugging these into the original formula, we get</p>
<p><span class="math-container">$$
\sum_{p|n}\frac1p=\sum_{p\le p_k}\frac1p\sim\log\log p_k\sim\log\log\log n
$$</span></p>
<p>Therefore, <span class="math-container">$\log\log\log n$</span> is not only an upper bound for <span class="math-container">$\sum_{p|n}\frac1p$</span>, but also a maximal order for <span class="math-container">$\sum_{p|n}\frac1p$</span>:</p>
<p><span class="math-container">$$
\limsup_{n\to\infty}{1\over\log\log\log n}\sum_{p|n}\frac1p=1
$$</span></p>
|
3,613,180 | <p>Let <span class="math-container">$f:[0, \pi] \to \mathbb{R}$</span> with the <span class="math-container">$L^2$</span> inner product</p>
<p><span class="math-container">$$
\langle f,g \rangle = \int_0^{\pi} f(x)g(x) \mathrm{d}x
$$</span></p>
<p>I want to find a projection of <span class="math-container">$f(x)=1$</span> onto <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\sin(3x)$</span> and use that to find a function of the form <span class="math-container">$1+k_1\sin(x)+k_3\sin(3x)$</span> that is orthogonal to <span class="math-container">$\sin(x), \sin(2x), \sin(3x), \sin(4x)$</span>.</p>
<p><strong>My work:</strong></p>
<p>I calculate the projection (is this correct?):</p>
<p><span class="math-container">$$
Pr_{\sin(x),\sin(3x)}(1)=\frac{\langle 1, \sin(x) \rangle}{\mid \mid \sin(x) \mid \mid^2}\sin(x)+\frac{\langle 1, \sin(3x) \rangle}{\mid \mid \sin(3x) \mid \mid^2}\sin(3x) \\
= \int_0^{\pi} 1\cdot\sin(x)dx\cdot\frac{1}{(\int_0^{\pi}\sin(x)\sin(x)dx)^2}\sin(x) + \int_0^{\pi} 1\cdot\sin(3x)dx\cdot\frac{1}{(\int_0^{\pi}\sin(3x)\sin(3x)dx)^2}\sin(3x) \\
=\frac{8}{\pi^2}\sin(x)+\frac{8}{3\pi^2}\sin(3x)
$$</span></p>
<p>Now I have the function <span class="math-container">$\frac{8}{\pi^2}\sin(x)+\frac{8}{3\pi^2}\sin(3x)$</span>, but how do I use that to find a function of the form <span class="math-container">$1+k_1\sin(x)+k_3\sin(3x)$</span> that is orthogonal to <span class="math-container">$\sin(x), \sin(2x), \sin(3x), \sin(4x)$</span>?</p>
<p>I am unsure of how to proceed.</p>
| Robert Lewis | 67,071 | <p>Having inspected the cited problem and it's surroundings, I note that part (b) of problem (2) asks us to show that
<span class="math-container">$(y')^2 + y^2$</span> is constant; this is seen to be an easy consequence of</p>
<p><span class="math-container">$y'' + y = 0 \tag 1$</span></p>
<p>as follows: we have</p>
<p><span class="math-container">$((y')^2 + y^2)' = 2y'y'' + 2yy' = -2yy' + 2yy' = 0, \tag 2$</span></p>
<p>using (1). Now with</p>
<p><span class="math-container">$y(0) = y'(0) = 0, \tag 3$</span></p>
<p>we have</p>
<p><span class="math-container">$(y')^2 + y^2 = (y'(0))^2 + y^2(0) = 0 \tag 4$</span></p>
<p>everywhere; thus</p>
<p><span class="math-container">$y' = y = 0, \tag 5$</span></p>
<p>that is, <span class="math-container">$y$</span> is the trivial solution to (1). Then if <span class="math-container">$y$</span> is not trivial, (3) cannot bind and at least one of <span class="math-container">$y(0)$</span>, <span class="math-container">$y'(0)$</span> does not vanish. Note that we may in fact have</p>
<p><span class="math-container">$y(0) \ne 0 \ne y'(0). \tag 6$</span></p>
|
3,613,180 | <p>Let <span class="math-container">$f:[0, \pi] \to \mathbb{R}$</span> with the <span class="math-container">$L^2$</span> inner product</p>
<p><span class="math-container">$$
\langle f,g \rangle = \int_0^{\pi} f(x)g(x) \mathrm{d}x
$$</span></p>
<p>I want to find a projection of <span class="math-container">$f(x)=1$</span> onto <span class="math-container">$\sin(x)$</span> and <span class="math-container">$\sin(3x)$</span> and use that to find a function of the form <span class="math-container">$1+k_1\sin(x)+k_3\sin(3x)$</span> that is orthogonal to <span class="math-container">$\sin(x), \sin(2x), \sin(3x), \sin(4x)$</span>.</p>
<p><strong>My work:</strong></p>
<p>I calculate the projection (is this correct?):</p>
<p><span class="math-container">$$
Pr_{\sin(x),\sin(3x)}(1)=\frac{\langle 1, \sin(x) \rangle}{\mid \mid \sin(x) \mid \mid^2}\sin(x)+\frac{\langle 1, \sin(3x) \rangle}{\mid \mid \sin(3x) \mid \mid^2}\sin(3x) \\
= \int_0^{\pi} 1\cdot\sin(x)dx\cdot\frac{1}{(\int_0^{\pi}\sin(x)\sin(x)dx)^2}\sin(x) + \int_0^{\pi} 1\cdot\sin(3x)dx\cdot\frac{1}{(\int_0^{\pi}\sin(3x)\sin(3x)dx)^2}\sin(3x) \\
=\frac{8}{\pi^2}\sin(x)+\frac{8}{3\pi^2}\sin(3x)
$$</span></p>
<p>Now I have the function <span class="math-container">$\frac{8}{\pi^2}\sin(x)+\frac{8}{3\pi^2}\sin(3x)$</span>, but how do I use that to find a function of the form <span class="math-container">$1+k_1\sin(x)+k_3\sin(3x)$</span> that is orthogonal to <span class="math-container">$\sin(x), \sin(2x), \sin(3x), \sin(4x)$</span>?</p>
<p>I am unsure of how to proceed.</p>
| marty cohen | 13,079 | <p>If <span class="math-container">$y+y''=0$</span>, then <span class="math-container">$yy'+y'y''=0$</span>
so <span class="math-container">$(y^2)'+((y')^2)' =0$</span>
so <span class="math-container">$y^2+(y')^2=c$</span>
for some real <span class="math-container">$c$</span>.</p>
<p>Since <span class="math-container">$y$</span> is non-trivial, <span class="math-container">$y(x) \ne 0$</span> for some <span class="math-container">$x$</span>,
so <span class="math-container">$c \ne 0$</span>.</p>
<p>Therefore if
<span class="math-container">$y(x)=0$</span>
then <span class="math-container">$(y'(x))^2 = c \ne 0$</span>
and if
<span class="math-container">$y'(x)=0$</span>
then <span class="math-container">$(y(x))^2 = c \ne 0$</span>.</p>
|
86,536 | <p>Considering we have a an association:</p>
<pre><code>assc = <|"A" -> <|"a" -> 1, "aa" -> 2|>, "B" -> 0, "C" -> 5,"D" -> <|"d" -> 2, "dd" -> 12|>|>
</code></pre>
<p>Let's also consider we have 2 known lists and one list for nested keys
(**
- <em>how to create this list?</em>
**):</p>
<pre><code>bigKeys = {"A", "D"}
lst1 = {"a", "dd", "B"}
lst2 = {"aa", "d", "C"}
</code></pre>
<p>I would like to loop over <code>assc</code> and assign the values for the keys that are from <code>lst1</code> and are <strong>not</strong> <code>bigKeys</code> - <strong>0</strong>, and to the values for the keys that are from <code>lst2</code> and again are <strong>not</strong> <code>bigKeys</code> - <strong>1</strong>. </p>
<p>As a result to have: </p>
<pre><code>asscNew = <|"A" -> <|"a" -> 0, "aa" -> 1|>, "B" -> 0, "C" -> 1, "D" -> <|"d" -> 1, "dd" -> 0|>|>
</code></pre>
<p>Thanks!</p>
| WReach | 142 | <p><code>MapIndexed</code> can be used to map a function to the inner-most values in nested associations, for example:</p>
<pre><code>MapIndexed[f, <| "A" -> <|"a" -> 1|>|>, {-1}]
(* <|"A" -> <|"a" -> f[1, {Key["A"], Key["a"]}]|>|> *)
</code></pre>
<p>The level specification <code>{-1}</code> selects only the deepest level in the expression. Note how <code>f</code> receives not only the value (<code>1</code>), but also the keys used to locate that particular value (<code>{Key["A"], Key["a"]}</code>).</p>
<p>Knowing this, we can define a helper function <code>val</code> which applies the rules specified in the question:</p>
<pre><code>val[_, {___, Key[k_]}] /; MemberQ[lst1, k] := 0
val[_, {___, Key[k_]}] /; MemberQ[lst2, k] := 1
val[v_, _] := v
</code></pre>
<p>Now, <code>MapIndexed</code> can provide the desired result:</p>
<pre><code>MapIndexed[val, assc, {-1}]
(* <|"A"-> <|"a"->0, "aa"->1|>, "B"-> 0, "C"-> 1, "D"-> <|"d"->1, "dd"->0|>|> *)
% === asscNew
(* True *)
</code></pre>
<p>Note that this solution automatically ignores keys that have nested associations -- there is no need to use <code>bigKeys</code>. If it is important to take <code>bigKeys</code> into consideration for some other reason, then we need only add one more rule (which must be the <em>first</em> rule):</p>
<pre><code>ClearAll[val]
(* new rule *)
val[v_, {Key[k_], ___}] /; MemberQ[bigKeys, k] := v
(* same rules as before *)
val[_, {___, Key[k_]}] /; MemberQ[lst1, k] := 0
val[_, {___, Key[k_]}] /; MemberQ[lst2, k] := 1
val[v_, _] := v
</code></pre>
<p>This technique generalizes nicely to complex nested structures, allowing each value to be transformed by an arbitrary function of that value and its location ("path") within the structure. It also works with nested mixtures of lists and associations. By relaxing the level specification in <code>MapIndexed</code>, any or all levels of a structure can be transformed.</p>
|
3,300,469 | <p>I have a problem counting all the possible ways of "pairing" two datasets of size n and m, including partial pairing. </p>
<p>Example:
Assume we have two sets <span class="math-container">$\{A,B\}$</span> and <span class="math-container">$\{1,2,3\}$</span>. My aim is to find all ways of pairing letters with numbers, including the consideration of situations, where only a fraction of possible pairs is actually present, including a case of "no pairing at all".</p>
<p>In this case I would get the following ways of pairing:<br>
<span class="math-container">$\{A,1\},\{B,2\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,1\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B,3\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,3\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,2\},\{B\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,3\},\{B\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,1\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,2\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,3\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B\},\{1\},\{2\},\{3\}$</span> (zero pairs out of two coexisting pairs possible) </p>
<p>How can I generalize it to bigger sets of size n and m?</p>
| JMoravitz | 179,297 | <p>Assume our two sets are <span class="math-container">$A$</span> and <span class="math-container">$B$</span> where <span class="math-container">$A\cap B = \emptyset$</span>
Break into cases based on the number of pairs used. With <span class="math-container">$k$</span> pairs used, choose which <span class="math-container">$k$</span> elements are involved in the pairs from the first set and which <span class="math-container">$k$</span> from the second set.</p>
<p>Then, choose how the selected elements are paired together.</p>
<p>This gives a total of:</p>
<p><span class="math-container">$$\sum\limits_{k=0}^{\min(|A|,|B|)}\binom{|A|}{k}\binom{|B|}{k}k!$$</span></p>
|
812,345 | <p><a href="https://math.stackexchange.com/a/87705/53259">This</a> proves: Similar matrices have the same characteristic polynomial. (Lay P277 Theorem 4)</p>
<p>I prefer <a href="https://math.stackexchange.com/a/8407/53259">https://math.stackexchange.com/a/8407/53259</a>, but this proves that they have the same eigenvalues. </p>
<p>Are they equivalent? What about in general, even for matrices which are NOT similar? </p>
| Omran Kouba | 140,450 | <ol>
<li>If $A$ and $B$ have the same characteristic polynomial, then clearly the have the same eigenvalues, these are the zeros of the characteristic polynomial.</li>
<li>The converse is generally not true: for example
$$
A=\left[\matrix{1&0&0\cr
0&0&1\cr 0&0&0}\right],\quad
B=\left[\matrix{1&1&0\cr
0&1&0\cr 0&0&0}\right]
$$
we have $\sigma(A)=\sigma(B)=\{0,1\}$, but $\chi_A(X)=X^2(X-1)$, $\chi_B(X)=X(X-1)^2$.</li>
</ol>
|
1,248,068 | <p>Let $S$ be a set of cardinality $\aleph_1$. Consider the directed family $\mathcal{C}$ (here <em>directed</em> means <em>directed with respect to the inclusion</em>) of all countably infinite subsets of $S$. Suppose that</p>
<p>$$\mathcal{C} = \bigcup_{n=1}^\infty \mathcal{C}_n$$</p>
<p>for some families $\mathcal{C}_n$. Does it follow that for some $n_0$ the family $\mathcal{C}_{n_0}$ contains an uncountable, directed subfamily?</p>
<p>Of course, at least one $\mathcal{C}_n$ is uncountable, so let us take this one. Must it contain an uncountable directed subfamily?</p>
| hot_queen | 72,316 | <p>Yes. Letting $S = \omega_1$, one of the $C_n$'s must contain uncountably many ordinals and hence an uncountable linearly ordered subfamily. </p>
<p>Also not every uncountable family contains one. For example, an almost disjoint family.</p>
|
1,254,189 | <p>I know that I have to study the order of every element in $\mathbb{Q/Z}$.
But what do I do?
I've been struggling of what to do for this question</p>
| Dietrich Burde | 83,966 | <p>The abelian group $(\mathbb{Q},+)$ is torsion-free, but the abelian group
$(\mathbb{Q/Z},+)$ is not torsion-free. Hence these two groups cannot be isomorphic. Also, the $\mathbb{Z}$-module $\mathbb{Q}$ is flat, whereas $\mathbb{Q/Z}$ is not flat.</p>
|
59,846 | <p>In "The New Book of Prime Number Records", Ribenboim reviews the known results on the degree and number of variables of prime-representing polynomials (those are polynomials such that the set of positive values they obtain for nonnegative integral values of the variables coincides with the set of primes). For example, it is known that there is such a polynomial with 42 variables and degree 5, as well as one with 10 variables and astronomical degree.</p>
<p>Ribenboim mentions that it's an open problem to determine the least number of variables possible for such a polynomial, and remarks "it cannot be 2". It's a fairly simple exercise to show that it cannot be 1, but why can't it be 2?</p>
<p>EDIT: here's the relevant excerpt from Ribenboim's book. Given that nobody seems to be familiar with such a proof, I'm inclined to assume that this is a typo and he just meant "it cannot be 1". </p>
<p><img src="https://i.stack.imgur.com/R1Hcw.png" alt="Excerpt from "The New Book of Prime Number Records""></p>
| Alon Amit | 308 | <p>Given all the evidence so far, I'm inclined to declare Ribenboim's parenthetic remark a typo. He probably meant either "it cannot be 1" or "it must be at least 2". It would be confusing and unusual for him to mention this off-hand with no reference as if it's a simple observation. <em>That</em> it certainly isn't. </p>
|
409,689 | <p>I have $(x_1, y_1), (x_2, y_2)$.</p>
<p>How do I find the point that's $d$ distance away from $(x_1, y_1)$ on a straight line to $(x_2, y_2)$?</p>
<p>I know I can get the length of the line with Pythagoras. I know if I drew a circle I could use the radius as distance and the point would be where the line and the circle intersect.</p>
<p>Could someone briefly explain each step to me please?</p>
<p>I don't understand <a href="https://math.stackexchange.com/questions/175896/finding-a-point-along-a-line-a-certain-distance-away-from-another-point">Finding a point along a line a certain distance away from another point!</a></p>
| George V. Williams | 54,806 | <p>Consider a circle with radius $d$ and center $(x_1, y_1)$. This is the equation:</p>
<p>$$ (x - x_1)^2 + (y - y_1)^2 = d^2 $$</p>
<p>Let $m$ be the slope of the line from $(x_1, y_1)$ to $(x_2, y_2)$. ($m = \dfrac{y_2 - y_1}{x_2 - x_1}$). Our line must satisfy the equation:</p>
<p>$$ y - y_1 = m(x - x_1) $$
$$ y = m(x - x_1) + y_1 $$</p>
<p>We want to find when these two will intersect, so substituting:</p>
<p>$$ (x - x_1)^2 + (m(x - x_1) + y_1 - y_1)^2 = d^2 $$
$$ (x - x_1)^2 + m^2(x - x_1)^2 = d^2 $$
$$ (1 + m^2)(x - x_1)^2 = d^2 $$
$$ (x-x_1)^2 = \frac{d^2}{1+m^2} $$
$$ x - x_1 = \pm \sqrt\frac{d^2}{1+m^2} $$
$$ x = x_1 \pm \sqrt\frac{d^2}{1+m^2} $$</p>
|
3,865,388 | <p>Let <span class="math-container">$A$</span> be the set of all <span class="math-container">$2\times2$</span> boolean matrices and <span class="math-container">$R$</span> be a relation defined on <span class="math-container">$A$</span> as <span class="math-container">$M \mathrel{R} N$</span> if and only if <span class="math-container">$m_{ij} \leqslant n_{ij}$</span>, where <span class="math-container">$1 \leqslant i, j \leqslant 2$</span>. Is <span class="math-container">$(A,R)$</span> a lattice? Justify.</p>
<p>I am unable to solve the above question. I already know the following: There are 16 elements in set <span class="math-container">$A$</span>. Now when i try to calculate relation <span class="math-container">$R$</span> according to the given conditions i am getting 81 pairs in relation <span class="math-container">$R$</span>. Drawing a hasse diagram and calculating <strong>least upper bound</strong> and <strong>greatest lower bound</strong> for each point in the hasse diagram will go too lengthy. Is there a better method to do this?? Please help.</p>
| J.-E. Pin | 89,374 | <p>It looks like your structure is simply <span class="math-container">${\Bbb B}^4$</span>, the product of four copies of the Boolean lattice <span class="math-container">${\Bbb B} = \{0, 1\}$</span>, ordered by <span class="math-container">$0 \leqslant 1$</span>. Thus, yes, <span class="math-container">$(A, R)$</span> is a lattice.</p>
|
3,865,388 | <p>Let <span class="math-container">$A$</span> be the set of all <span class="math-container">$2\times2$</span> boolean matrices and <span class="math-container">$R$</span> be a relation defined on <span class="math-container">$A$</span> as <span class="math-container">$M \mathrel{R} N$</span> if and only if <span class="math-container">$m_{ij} \leqslant n_{ij}$</span>, where <span class="math-container">$1 \leqslant i, j \leqslant 2$</span>. Is <span class="math-container">$(A,R)$</span> a lattice? Justify.</p>
<p>I am unable to solve the above question. I already know the following: There are 16 elements in set <span class="math-container">$A$</span>. Now when i try to calculate relation <span class="math-container">$R$</span> according to the given conditions i am getting 81 pairs in relation <span class="math-container">$R$</span>. Drawing a hasse diagram and calculating <strong>least upper bound</strong> and <strong>greatest lower bound</strong> for each point in the hasse diagram will go too lengthy. Is there a better method to do this?? Please help.</p>
| Hans | 602,799 | <p>In general, if <span class="math-container">$\{ L_\alpha \}$</span> is a family of lattices, then the Cartesian product <span class="math-container">$\prod_{\alpha} L_{\alpha}$</span> also has the structure of a lattice given by <span class="math-container">$\mathbf{x} \wedge \mathbf{y} = (x_\alpha \wedge y_\alpha)_{\alpha}$</span>, <span class="math-container">$\mathbf{x} \vee \mathbf{y} = (x_\alpha \vee y_\alpha)_{\alpha}$</span> for <span class="math-container">$\mathbf{x}, \mathbf{y}$</span> in the product. This fact, although obvious, is actually very convenient as it allows you to point out complicated lattices as the product of less-complicated ones, or as sublattices of the product of less-complicated lattices. (If you are interested, Google "Dilworth's Embedding Theorem.")</p>
<p>Sometimes is is useful to indentify matrices with elements in <em>blank</em> as a product of <em>blank</em> the number of entries in such matrices times. You see this in many areas of mathematics. Lie group theory comes to mind. If you want to think about <span class="math-container">$GL_n(\mathbb{R})$</span> as a space, then you can think about how it sits inside <span class="math-container">$\mathbb{R}^{n \times n}$</span>.</p>
|
1,398,956 | <p>I saw from literature that the expected value of a random variable $f(X)$ is either $E f(X)$, $E(f(X))$ or $E[f(X)]$. Is there a standard which one notation should one use? Is the expected value a function $f(X)\to\mathbb R$?</p>
| Karl | 203,893 | <p>I've seen $\langle X \rangle$ used as well for expected value. I quite like this as it makes moment generating functions look nice $\langle e^{tX} \rangle$ only has one 'e' compared with $\mathbb{E}\left[e^{tX}\right]$
I guess that's your answer, people choose notation to balance form and function. I suspect there is no standard notation just some are more common than others.</p>
|
3,100,957 | <p>A fair coin is tossed until one of the patterns show up: TTH or THT.
Let A be the event that TTH shows up before THT.</p>
<p>What is P(A)?</p>
<p>Here is my solution but I am not sure if it is correct or there is a better solution.</p>
<p>Let <span class="math-container">$p=P(A)$</span>. Define </p>
<p><span class="math-container">$A_1=$</span> the event that the first toss is H</p>
<p><span class="math-container">$A_2=$</span> the event that the first two tosses are TT</p>
<p><span class="math-container">$A_3=$</span> the event that the first three tosses are THT</p>
<p><span class="math-container">$A_4=$</span> the event that the first three tosses are THH</p>
<p>Then this is a partition for the sample space. </p>
<p><span class="math-container">$p=P(A|A_1)P(A_1)+P(A|A_2)P(A_2)+P(A|A_3)P(A_3)+P(A|A_4)P(A_4)$</span>.</p>
<p>Then</p>
<p><span class="math-container">$p=p\frac{1}{2}+1\frac{1}{4}+0\frac{1}{8}+p\frac{1}{8}$</span>
which implies that <span class="math-container">$p=\frac{2}{3}.$</span></p>
| DanielWainfleet | 254,665 | <p>If <span class="math-container">$(X,d)$</span> is a metric space and <span class="math-container">$Y\subset X$</span> then for <span class="math-container">$Y$</span> to be compact it is necessary that <span class="math-container">$Y$</span> is bounded. That is, it is necessary that <span class="math-container">$B_d(x,r)\supset Y$</span> for some <span class="math-container">$x\in X$</span> and some <span class="math-container">$r>0.$</span> Otherwise, take some <span class="math-container">$x=y_1\in Y$</span> and for <span class="math-container">$n\in \Bbb Z^+$</span> take <span class="math-container">$y_{n+1}\in Y$</span> with <span class="math-container">$d(y_1,y_{n+1})>1+\max \{d(y_1,y_j): 1\le j\le n)\}.$</span> By the triangle inequality, <span class="math-container">$d(y_i,y_j)>1$</span> whenever <span class="math-container">$i\ne j.$</span> So no subsequence of <span class="math-container">$(y_n)_{n\in \Bbb Z^+} $</span> is even a Cauchy sequence.</p>
|
2,995,327 | <p>Suppose a,b ∈ Z. If 4 | <span class="math-container">$(a^2 + b^2)$</span> then a and b are not both odd.</p>
<p>So, assuming that 4 | <span class="math-container">$(a^2 + b^2)$</span> and <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are odd</p>
<p>this gives <span class="math-container">$4k=(2l+1)^2+(2u+1)^2$</span> for some <span class="math-container">$k,l,u\in z$</span></p>
<p>eventually leading to <span class="math-container">$4k=4(l^2+l+u)+2(u^2+1)$</span></p>
<p>The RHS is not a multiple of 4 when <span class="math-container">$u=2$</span> contradiction.</p>
<p>Is this valid, thanks.</p>
| Martund | 609,343 | <p>This is invalid. You have done a mistake in calculation.
RHS will be <span class="math-container">$4(l^2+u^2+l+u)+2$</span>, which is not divisible by 4.
Hope it helps:)</p>
|
1,199,304 | <p>Let $M\neq \{0\}$ be a semi-simple left $R$ module .Prove that it contains a simple sub-module.</p>
<p>An $R-$ module $M$ is said to be semi-simple if every submodule of $M$ is a direct summand of M
<strong>My solution</strong></p>
<p>Since $M\neq \{0\}$; $\exists m\in M$ such that $m\neq 0$.Then I can consider the left $R-$ module $Rm$ ;By hypothesis it is a direct summand of $M$.Thus $M=N+Rm$ where $N$ is a sub module of $M$ </p>
<p>How to proceed next?</p>
| Math137 | 60,099 | <p>Your definition is equivalent to "An $R$-module $M$ is semisimple if it can be written as a direct sum of family of simple modules" for example <a href="http://nptel.ac.in/courses/111102009/module2/lec4.pdf" rel="nofollow">see here page 2</a> for the proof, in which your method is subsidised by the proof of the theorem. </p>
<p>Also what you want is a direct consequence of the above definition. </p>
|
3,965,164 | <p>I know the standard and expanded forms of the equation of the circle in the simple 2d space,</p>
<p><span class="math-container">${(x-a)}^2+{(y-b)}^2=r^2$</span></p>
<p><span class="math-container">$x^2-2ax+y^2-2by=c$</span></p>
<p>So in 3d space what are the equations for a circle laying in an arbitrary plane,
and what is the 3d version of the polar form of the circle equation</p>
<p><span class="math-container">$$(a+rcos(t),b+rsin(t))$$</span></p>
<p>How the two equations (that I mentioned up) look in 3d space and are there more abstractive formulae for the circle equation in more than 3 dimensions?</p>
| Brian M. Scott | 12,042 | <p>In order to make the intersection equal to <span class="math-container">$\Bbb R\setminus\Bbb Q$</span>, you need to make sure that each <span class="math-container">$A_n$</span> contains every irrational number, and that each rational number is excluded from at least one <span class="math-container">$A_n$</span>. The set of rational numbers is countably infinite, and you have countably infinitely many sets <span class="math-container">$A_n$</span> at your disposal, so the easiest way to do this is to use each <span class="math-container">$A_n$</span> to exclude a different rational number.</p>
<p>Let <span class="math-container">$\Bbb Q=\{q_n:n\in\Bbb Z^+\}$</span> be an enumeration of the rationals, and for each <span class="math-container">$n\in\Bbb N$</span> let <span class="math-container">$A_n=\Bbb R\setminus\{q_n\}$</span>. That takes care of the intersection, but unfortunately, these sets aren’t nested. However, that’s easily fixed: instead of removing <strong>one</strong> rational to get <span class="math-container">$A_n$</span>, remove the first <span class="math-container">$n$</span> rationals in the list: let <span class="math-container">$A_n=\Bbb R\setminus\{q_k:k=1,\ldots,n\}$</span>. The intersection is still <span class="math-container">$\Bbb R\setminus\Bbb Q$</span>, and now <span class="math-container">$A_{n+1}\subseteq A_n$</span> for each <span class="math-container">$n\in\Bbb Z^+$</span>.</p>
|
2,545,516 | <p>So I have to assess the convergence of $$\displaystyle\sum_{n=1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right).$$</p>
<p>I'm told that it diverges, but can't really see why.</p>
<p>The divergence test doesn't really help, because
$\lim\limits_{x\to\infty}\displaystyle\frac{1}{\sqrt{n}}=0$, so</p>
<p>$\lim\limits_{x\to\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right)=0$, which doesn't conclude its divergence.</p>
<p>I doubt the ratio test would be much of use in this situation.</p>
<p>I can't imagine using the integral comparison test, as I wouldn't know where to start with $\displaystyle\int_{1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right) \mathrm dx$.</p>
| Angel Moreno | 327,493 | <p>The function $\sin(x) \geq\dfrac{2}{\pi}x$ for $x\in[0,\pi/2]$</p>
<p>$\sum_{n} 1/n^{1/2}$ diverge</p>
<p>$\forall n: \sin(1/n^{1/2}) \geq \dfrac{2}{\pi n^{1/2}}$</p>
|
1,796,792 | <p>Is $\log_27$ a rational number?</p>
| DanielWainfleet | 254,665 | <p>If $a,b$ are positive integers and $2^{a/b}=7$ then $2^a=7^b$ which make an even number equal to an odd number.</p>
|
52,079 | <p>I'm in doubt about the topology of maps between fibres of vector bundles.</p>
<p>Consider $E$ and $F$ vector bundles and the set of all linear maps from a fibre of $E$ to a fibre of $F$, ie, the set of all linear maps $T:E_x \rightarrow F_y$, where $E_x$ is the fiber over $x$ and $F_y$ is the fiber over $y$.</p>
<p>I want to know how to define the topology of this set.</p>
<p>I need this topology for this question: Consider $f: E \rightarrow F$, a map that preserves each fiber and its restriction to each fibre, $f_x : E_x \rightarrow F_y$, is differentiable. The differential of $f_x$ calculated in a vector $v \in E_x$ is the linear maps $df_x(v):E_x \rightarrow F_y$. I want to say that $f$ is a $C^1$ map if the function $v \in E \rightarrow df_{\pi(v)} (v)$ is continuous. And for this I need a topology for the set defined above.</p>
<p>Does anybody know how to define the topology? What does it mean that two of these maps are close to each other?</p>
<p>Note that $df_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.</p>
| HenrikRüping | 3,969 | <p>If both bundles were trivial, say $X\times \mathbb{R}^n\rightarrow X$ and $Y\times \mathbb{R}^n\rightarrow Y$ one could just take $X\times Y\times M(m,n,\mathbb{R})$.
Different choices of trivializations should give homeomorphic spaces, so this topology seems to be right.</p>
<p>In general the bundles are just locally trivial. However this is enough to write down a basis for the topology. </p>
<p>I don't see, what nice properties this topology might have but it seems to be the canonical choice.</p>
<p>EDIT: I think one can reduce this to classical vector bundle constructions in the following way. Given two vector bundles $p_i:E_i\rightarrow X_i$ (for $i=1,2$). Let $pr_i:X_1\times X_2\rightarrow X_i$ denote the projections. Then the desired bundle is just Hom$(pr_1^*(p_2,E_2,X_2),pr_2^*(p_1,E_1,X_1))$.</p>
|
52,079 | <p>I'm in doubt about the topology of maps between fibres of vector bundles.</p>
<p>Consider $E$ and $F$ vector bundles and the set of all linear maps from a fibre of $E$ to a fibre of $F$, ie, the set of all linear maps $T:E_x \rightarrow F_y$, where $E_x$ is the fiber over $x$ and $F_y$ is the fiber over $y$.</p>
<p>I want to know how to define the topology of this set.</p>
<p>I need this topology for this question: Consider $f: E \rightarrow F$, a map that preserves each fiber and its restriction to each fibre, $f_x : E_x \rightarrow F_y$, is differentiable. The differential of $f_x$ calculated in a vector $v \in E_x$ is the linear maps $df_x(v):E_x \rightarrow F_y$. I want to say that $f$ is a $C^1$ map if the function $v \in E \rightarrow df_{\pi(v)} (v)$ is continuous. And for this I need a topology for the set defined above.</p>
<p>Does anybody know how to define the topology? What does it mean that two of these maps are close to each other?</p>
<p>Note that $df_x (v)$ is not necessarily a homomorphism, because this is only defined over the fiber that contains $v$.</p>
| Patrick I-Z | 11,885 | <p>I elaborate a little bit on what your question inspires me. I will treat a more general question, but you can reduce it to finite linear dimensional fiber bundles over manifolds. Let $\pi : E \to X$ and $\pi' : E' \to X$ be two projections (smooth in some sense, and you can assume that you deal with manifolds). You consider the set
$$
M = \{ f \in C^\infty(E_x,E'_y) \mid \mbox{for some } x, y \in X \}
$$
You have two natural projections ${\it src} : f \mapsto x$ and ${\it trg} : f \mapsto y$, with
$ \in C^\infty(E_x,E'_y)$. Let $r \mapsto f_r$ be a parametrization, that means a map defined on some open subset $U$ of some vector space ${\bf R}^n$, for some $n \in {\bf N}$. We will say that $r \mapsto f_r$ is smooth if </p>
<p>1) The map $r \mapsto x_r = {\it src}(f_r)$ is smooth.</p>
<p>2) The evaluation map $(r,e) \mapsto f_r(e)$ defined
$$
\mbox{from } \ \{ (r,e) \in U \times E \mid e \in E_{x_r} \} \ \mbox{ to }\ E'
$$
is smooth. The source of the evaluation map has a nice induced smooth structure.</p>
<p>These parametrizations define on $M$ what is called a diffeology. Now, every time you have a diffeology you get a topology by considering the finest topology such that the smooth parametrizations are continuous. You may change everywhere $C^\infty$ by $C^0$ I don't think it will change much, if you prefer continuous maps. So, a subset $\Omega \subset M$ is open iff its preimage by any smooth parametrization is open. </p>
<p>Writing this stuff down, I realize that it is maybe overkilling for the question you ask. Since the bundle are locally trivial, by using a trivialization and using the fact that you consider just linear maps (changing $f \in C^\infty(E_x,E'_y)$ to $f \in {\rm L}(E_x,E'_y)$, you may simplify this construction. But it will get you a natural topology. </p>
<p>I don't know if it may help you but it can inspire you to build what you are looking for.</p>
<hr>
<p>Actually, in your case, this construction gives you a fiber bundle, as notices Johannes Ebert in next answer. So, for finite dimension fiber bundle, it is really overkill. Sorry.</p>
|
2,365,933 | <p>I'm aware of how we can simplify functions which have $Arc$ as an argument . For example $\sin(\cos^{-1}(x)) = \sqrt{1-x^2}$ but what about cases which $Arc$ is out of the parentheses ? For instance consider this : $\sin^{-1}(\tan x)$ . Is there any way for simplification ? </p>
| Vassilis Markos | 460,287 | <p>Well, let $f(x)=\arcsin(\tan(x))$, $x\in(-\frac{\pi}{4},\frac{\pi}{4})$. Now, since:
$$\arcsin(x)=\int_0^{x}\frac{1}{\sqrt{1-t^2}}dt$$
we have, that:
$$f(x)=\int_0^{\tan(x)}\frac{1}{\sqrt{1-t^2}}dt$$
So:
$$\begin{align*}f'(x)=&\frac{1}{\sqrt{1-\tan^2(x)}}\frac{1}{\cos^2(x)}=\frac{1}{\cos^2(x)\sqrt{1-\frac{\sin^2(x)}{\cos^2(x)}}}=\\=&\frac{1}{\cos^2(x)\sqrt{\frac{cos^2(x)-\sin^2(x)}{\cos^2(x)}}}=\frac{1}{\cos(x)\sqrt{\cos(2x)}}\end{align*}$$
So:
$$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}dt+c$$
and since $f(0)=\arcsin(\tan(0))=0$, we have $c=0$, so:
$$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}d,\ x\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$$</p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Gabe K | 125,275 | <p>Mohammed Ghomi's <a href="https://mathoverflow.net/a/376809/2383">answer</a> reminds me of a related picture that Cedric Villani drew to depict Ricci curvature ([1] Chapter 14). Similar to the <span class="math-container">$\operatorname{CAT}(\kappa)$</span> inequality, this idea can be used to derive notions of Ricci curvature for more general metric measure spaces.
<a href="https://i.stack.imgur.com/rdgDr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/rdgDr.png" alt="Because of positive curvature effects, the observer overestimates the surface of the light source" /></a></p>
<p>[1] <em>Villani, Cédric</em>, <a href="http://dx.doi.org/10.1007/978-3-540-71050-9" rel="noreferrer"><strong>Optimal transport. Old and new</strong></a>, Grundlehren der Mathematischen Wissenschaften 338. Berlin: Springer (ISBN 978-3-540-71049-3/hbk). xxii, 973 p. (2009). <a href="https://zbmath.org/?q=an:1156.53003" rel="noreferrer">ZBL1156.53003</a>.</p>
|
4,338,382 | <blockquote>
<p>Let <span class="math-container">$a,b>0$</span>. Prove that: <span class="math-container">$$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$</span></p>
</blockquote>
<p>Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:</p>
<p>Case 1: <span class="math-container">$b-a+1>0$</span> then squaring both side, we get equivalent inequality: <span class="math-container">$$\frac{1}{a^4}+b^4+2\frac{b^2}{a^2}\ge2\left(\frac{1}{a^2}+a^2\right)(b^2+a^2+1-2ab-2a+2b)$$</span> Or: <span class="math-container">$$\frac{1}{a^4}+b^4\ge\frac{2}{a^2}(a^2+1-2ab-2a+2b)+2a^2(b^2+a^2+1-2ab-2a+2b)$$</span>
The rest is so complicated. Is there nice idea etc: AM-GM, C-S to prove this inequality.</p>
<p>Case 2: <span class="math-container">$b-a+1<0$</span> which is obviously true.</p>
<p>I hope we can find a better approach for the inequality. Thank you very much!</p>
| Rushabh Mehta | 537,349 | <p>Well, let's run through the possibilities.</p>
<p><strong>Case 1</strong>: A knight</p>
<p>In this case, B is a spy and C is a knave.</p>
<p><strong>Case 2</strong>: A knave</p>
<p>In this case, B is not a spy, and thus a knight.</p>
<p><strong>Case 3</strong>: A spy</p>
<p>In this case, C is a knave, and B is a knight.</p>
<p>Thus, the knight is either A or B.</p>
<p>Now let's think about B's statement. Case 1 is a valid possibility, and so is Case 2 (ignoring B's statement). Thus, we can't determine the knight without B's statement, and thus, B is not telling the truth.</p>
<p>Thus, A is the knight, B is the spy, and C the knave.</p>
|
9,918 | <p>I recently flagged as "rude or offensive" the comment </p>
<blockquote>
<p>As the tone should suggest, he’s a crank. It’s a hysterical screed with a few nuggets of fact surrounded by a great deal of nonsense. E.g., he may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it. – <a href="https://math.stackexchange.com/questions/356264#comment765805_356264">Brian M. Scott</a></p>
</blockquote>
<p>to the question</p>
<p><a href="https://math.stackexchange.com/questions/356264">Infinite sets don't exist!?</a></p>
<p>My complaint is very narrow: I am not arguing that the subject of the comment is right in his critique of infinite sets, only that the word "crank" is inappropriate in this context. In particular, none of the answers to the question provide evidence of crankdom, but rather disagreement over philosophy. If the comment had simply been </p>
<p>"He may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it."</p>
<p>I would have no objection.</p>
<p>I could also have flagged it as "not constructive/off topic" or as "too chatty". But to my mind it is the unwarranted rudeness that is the real problem here. More generally, I am quite upset that a productive mathematician is being called a crank. I don't know the subject of the comment personally, but I am familiar with his work and he is most definitely <em>not</em> a crank. </p>
<p>Comments like this reflect quite poorly on math.stackexchange and I hope not to see any more of them. What is the best way to encourage commenters to refrain from baseless accusations of crankdom? Is flagging the appropriate course of action?</p>
| Stephen | 146,439 | <p>It seems that flagging a comment in which (you believe) a baseless accusation of crankdom has been made is the only recourse. If the moderators doesn't agree with you, it's probably best to shrug and move on. </p>
<p>At the moment, this question has 3 votes to close, so I am answering it even though it doesn't seem all that helpful, just in order to have a future record. I'm making this post community wiki in case anyone with a better idea would like to chime in.</p>
|
275,539 | <p>Kind of leading on from my other question, how would I solve for $i$? Or how would I check that it is possible to have such an $i$?</p>
<p>First I had to check for all $2^i$ and clearly this doesn't happen as all $2^i$ are even and so I will just get even $x's$ such that $2^i \equiv x \mod 28$. So the next one I go onto is $3$.</p>
<p>Now how do I go about doing this? </p>
| Calvin Lin | 54,563 | <p>To solve $a^i \equiv 1 \pmod{n}$, first note that we must have $\gcd(a,n) = 1$ in order to get a positive (non-zero) integer solution. Like you realized, since $\gcd(2, 28)=2$, thus $2^i$ will always be a multiple of $2$, and hence cannot be of the from $28k+1$.</p>
<p>Given that condition, such an $i$ always exists, by <a href="http://en.wikipedia.org/wiki/Euler%27s_theorem">Euler's theorem</a>, which states that $ a^{\phi(n)} \equiv 1 \pmod{n}$. The solution, is known as the order. I.e. the smallest positive integer such that $3^k \equiv 1 \pmod{28}$ is called the order of 3 modulo 28.</p>
<p>In this case, we calculate that $ \phi(28) = 28 \times \frac {1}{2} \times \frac {6}{7} = 12$, and so we know that $3^{12} \equiv 1 \pmod{28}$. From here, we only need to check the factors of 12, which are 1, 2, 3, 4, 6, 12.</p>
<p>$3^1 \equiv 3, 3^2 \equiv 9, 3^3\equiv 27 \equiv -1, 3^4 \equiv -3, 3^6 \equiv (-1)^2 \equiv 1 \pmod{28}$. Hence, the order of 3 modulo 28 is 6.</p>
|
200,903 | <p>My teacher was explaining quadratics in my class and it was a little bit unclear to me. The problem was <br> <br>
Suppose $at^2 + 5t + 4 > 0$, show that $a > 25/16$ . <br> <br></p>
<p>My teacher said that there are no solutions for this function when it is greater than $0$ and used $b^2-4ac \lt 0$, and this is the part that confused me. I understand why he used $b^2-4ac \lt 0$ but I cannot understand why there are no solutions by just looking at the function. Could someone explain this to me?</p>
| Thomas | 26,188 | <p>Edit: I will try to say something general first. Hopefully that will make my answer a bit easier to understand. But as a first thing, it might be helpful to take a look at for example the Wikipedia article on <a href="http://en.wikipedia.org/wiki/Quadratic_equation" rel="nofollow">quadratic equations</a>.</p>
<p>If you have a function $f(t)$, then $t$ is a variable (called the independent variable). When we change the value of $t$, then the value of $f(t)$ changes. For example if $f(t) = t +2$, then for $t = 1, f(t) = f(1) = 3$. For $t = 2, f(t) = t(2) = 4$. </p>
<p>Now given a function $f(t)$, you can ask the question: does there exist a value or $t$ such that for that value $f(t)$ is zero. You are asking if there are any zeros. Anything that makes that function equal to zero.</p>
<p>A special type of function are the quadratic polynomials. Here $f(t)$ looks like this:</p>
<p>$$
f(t) = at^2 + bt + c.
$$
($a,b,c$ are fixed constants.) For example $f(t) = t^2 + 3t + 2$. Now we ask the question: does there exist a number such that the value of $f(t)$ is $0$? The answer is yes because $$f(-1) = (-1)^2 + 3\cdot(-1) + 2 = 0.$$</p>
<p>Now if you know about a function that $f(t)>0$ for all $t$, then no matter what $t$ is $f(t)$ is never going to be equal to zero. It is always positive. That is exactly what it means to say that $f(t) > 0$ for all $t$.</p>
<p>Another way to answer the question about whether a function is ever equal to zero is to draw the graph of the function. If the function intersects the $t$-axis (think: $x$-axis) then there is some number for which the value of the function is zero. That is exactly what it means to intersect the $t$-axis.</p>
<p>So to echo what other people have already mentioned: you are considering a polynomial:
$$
P(t) = at^2 +5t + 4.
$$
If you sketch the graph of this polynomial you get a <a href="http://en.wikipedia.org/wiki/Parabola" rel="nofollow">parabola</a>. The question can be asked whether the graph of the polynomial intersects the $t$-axis. Asking this question is the same as asking for roots (or zeros) of the polynomial. I.e. you want to know whether of not there is a number $t_0$ such that $P(t_0) = 0$.</p>
<p>You are assuming (supposing) that $P(t) > 0$ for all $t$. I.e. that means that there are no roots of the polynomial, i.e. no solution to the equation $P(t) =0$. First you might notice that if $P(t) >0$ for all $t$, then $a$ has to be a positive number. If $a$ was negative, then for $t = -100000$, $P(t)$ would be negative, right?</p>
<p>So what more can we say? We have the discriminant that tells you something about how many roots the polynomial has. For a general <a href="http://en.wikipedia.org/wiki/Quadratic_equation" rel="nofollow">quadratic equation</a>: $at^2 + bt + c= 0$ ($a,b,c$ being constants here and $t$ the variable), the discriminant is defined as
$$
d = b^2 - 4ac.
$$</p>
<p>It is a fact that if
$$
\begin{align}
d &> 0\quad \text{ means that there are exactly two distinct roots}\\
d &= 0\quad \text{ means that there are exactly one root}\\
d &< 0\quad \text{ means that there are no roots.}
\end{align}
$$</p>
<p>For your polynomial, you would then need $d = 5^2 - 4a\cdot 4 < 0$. So
$$
\begin{align}
5^2 - 4a\cdot 4 &< 0 &\Rightarrow\\
- 4a\cdot 4 &< -25 &\Rightarrow\\
a &> \frac{25}{16}.
\end{align}
$$
And there you go.</p>
|
3,826,994 | <p>I would like to find <span class="math-container">$z$</span> which minimizes the below, when <span class="math-container">$x$</span> is held at a specific value.</p>
<p><span class="math-container">$f(x,z) =\sqrt{\sqrt{x^2 + z^2} - 0.25}$</span></p>
<p>For example; I would like to find the value of <span class="math-container">$z$</span> which minimizes the function when <span class="math-container">$x = 0.5$</span></p>
| Toby Mak | 285,313 | <p>You have not made a mistake. Observe that <span class="math-container">$x$</span> is a multiple of <span class="math-container">$y$</span>:</p>
<p><span class="math-container">$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$</span></p>
<p>which is what the answer states.</p>
|
2,512,294 | <p>So I've been given the following problem:</p>
<p>How many positive integers are there that can not be written as a sum of 5's and 7's? For example, 4 is one of those integers, but 19 is not because 19 = 5 + 7 + 7. How to solve this? </p>
| paw88789 | 147,810 | <p>Hints:</p>
<p>(1) Start at the beginning: Can you get $1$, $2$,...</p>
<p>(2) If you can write a number $n$ as a sum of $5$s and $7$s, you can write $n+5$ as a sum of $5$s and $7$s.</p>
<p>(3) If you ever achieve five consecutive numbers that you can write as a sum of $5$s and $7$s, what will hint (2) allow you to conclude?</p>
|
38,659 | <p>I know how to use Matrix Exponentiation to solve problems having linear Recurrence relations (for example Fibonacci sequence). I would like to know, can we use it for linear recurrence in more than one variable too? For example can we use matrix exponentiation for calculating ${}_n C_r$ which follows the recurrence C(n,k) = C(n-1,k) + C(n-1,k-1). Also how do we get the required matrix for a general recurrence relation in more than one variable?</p>
| Shahab | 10,575 | <p>The product rule is that if we have x ways of doing something and y ways of doing another thing, then there are xy ways of performing both actions. Here's how you can think of the product rule. The things in question are choosing a first chapter and choosing a second chapter.</p>
<p>First consider each of the ways of choosing a first chapter. There are 10 such ways, namely chapter 1 to 10. Put all these ways in a set A.
$A=\{way_1,way_2,...,way_{10}\}$</p>
<p>Now each of the ways of choosing a second chapter can be considered. There are nine such ways determined after leaving out an unknown chapter. (Whichever chapter was selected earlier, the number of ways will always be 1 less) Put all these ways in the set B.
$B=\{Way_1,Way_2,...,Way_9\}$</p>
<p>By the product rule, total number of ways is the number of elements in A x B i.e. 90.</p>
<p>In general if you have to choose k chapters out of n, where order is important, we can extend the same logic to see that the number of ways is $n(n-1)...(n-(k-1))$. If we are not bothered about the order then note that any such way appears exactly k! times in the above (since k chapters may be permuted in k! ways). So if ways without regard to order are considered, and we suppose there are X such ways then,
$(k! + k! + .... + k!)$(X times) $= n(n-1)...(n-(k-1))$ or</p>
<p>$X=\frac{n(n-1)...(n-k+1)}{k!}=\frac{n!}{k!(n-k)!}$</p>
<p>As you are aware the usual way to denote X is by the symbol $\tbinom{n}{k}$</p>
<p>Hope this helps. </p>
|
3,490,329 | <blockquote>
<p>Show that a 2-dimensional subspace of the space of <span class="math-container">$2\times2$</span> matrices contains a non-zero symmetric matrix. </p>
</blockquote>
<p>I don't know if it should be written like the addition of two symmetric and skew-symmetric matrix or there is another way to show it. </p>
| almagest | 172,006 | <p>Take any two linearly independent matrices in the subspace:</p>
<p><span class="math-container">$\begin{pmatrix}a_1 & b_1\\ c_1 & d_1\end{pmatrix}$</span> and <span class="math-container">$\begin{pmatrix}a_2 & b_2\\ c_2 & d_2\end{pmatrix}$</span></p>
<p>If <span class="math-container">$b_1=c_1$</span> or <span class="math-container">$b_2=c_2$</span>, then we are done. So assume <span class="math-container">$b_1\ne c_1$</span> and <span class="math-container">$b_2\ne c_2$</span>. Now take the following linear combination (which must be in the subspace):
<span class="math-container">$(c_2-b_2)\begin{pmatrix}a_1 & b_1\\ c_1 & d_1\end{pmatrix}+(b_1-c_1)\begin{pmatrix}a_2 & b_2\\ c_2 & d_2\end{pmatrix}=\begin{pmatrix}x & b_1c_2-b_2c_1\\ b_1c_2-b_2c_1 & y\end{pmatrix}$</span> where <span class="math-container">$x=a_1(c_2-b_2)+a_2(b_1-c_1),y=d_1(c_2-b_2)+d_2(b_1-c_1)$</span>.</p>
<p>It is clearly symmetric, so we are done provided at least one of <span class="math-container">$x,y,b_1c_2-b_2c_1$</span> is non-zero. If <span class="math-container">$b_1c_2-b_2c_1=0$</span> then <span class="math-container">$b_1:b_2=c_1:c_2$</span>. wlog we can assume <span class="math-container">$b_1=\lambda b_2,c_1=\lambda c_2$</span>. If also <span class="math-container">$x=0$</span> then <span class="math-container">$a_1=\lambda a_2$</span>. If also <span class="math-container">$y=0$</span>, then <span class="math-container">$d_1=\lambda d_2$</span> and so the first matrix is <span class="math-container">$\lambda$</span> times the second, contradicting independence. So <span class="math-container">$x,y,b_1c_2-b_2c_1$</span> are not all zero and we have found a nonzero symmetric matrix in the subspace.</p>
|
2,996,920 | <p>Please recommend me a good book to study interpolation techniques such as polynomial interpolation, cubic, spline interpolations, if possible tell me the branch of mathematics that deals with this subject. I want to go in depth with this topic.</p>
| cqfd | 588,038 | <p>The topics mentioned above are usually dealt under numerical analysis. Some of the textbooks are:</p>
<ol>
<li>S.D.Conte and C.deBoor, <em>Elementary Numerical Analysis-an algorithmic approach</em>,
3rd Edn., McGraw Hill, 1980.</li>
<li><p>K. E. Atkinson, <em>An Introduction to Numerical Analysis</em>, 2nd Edn., John
Wiley, 1989.</p></li>
<li><p>C.deBoor, <em>A practical guide to splines</em>.</p></li>
</ol>
|
1,270,042 | <p>$$(a+5)(b-1)=ab-a+5b-5=20-5=15.$$</p>
<p>So, both $a + 5$ and $b-1$ divide $15$. </p>
<p>Then, $a + 5$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $a$ is one of $10, -20, -2, -8, 0, -10, -4, -6$ and $b – 1$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $b = 14, -14, 4, -2, 6, -4, 2, 0$.</p>
<p>Could all possibilities for $a, b$ found by considering $(a+5)(b-1)$ be just random(not in a probability sense) and not connected to $ab = a - 5b + 20$ at all? In other words, could it be that if some of the possible $a, b$ found this way happen to satisfy $ab = a - 5b + 20$, then it's just a coincidence? </p>
| mathlove | 78,967 | <p>Having $$(a+5)(b-1)=15$$
gives you $$(a+5,b-1)=(1,15),(3,5),(5,3),(15,1),(-1,-15),(-3,-5),(-5,-3),(-15,-1),$$
i.e.
$$(a,b)=(-4,16),(-2,6),(0,4),(10,2),(-6,-14),(-8,-4),(-10,-2),(-20,0).$$
Since $a,b$ are positive integers, $(a,b)=(10,2)$ is the only solution.</p>
|
2,512,363 | <h2>Defining the barycentre and finding its variance</h2>
<p>I have a set of $N$ points at the locations $x_i$ which has weights $W_i$, $i=1,\ldots, N$ and want to find the barycenter (or center of gravity) </p>
<p>$$ B = {\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i}.$$</p>
<p>I want to find the variance of $B$.
I assume that $W_i$ are i.i.d. and positive with expected value $Ew_i=\mu$ and variance $Var[W_i]=E[w_i-\mu]^2=\sigma^2 $.
I also assume that $N$ is large, so $\sum_{i=1}^N w_i$ is well approximated by $N\mu$ (justified below).</p>
<p>When I do the calculations (see one example below) I end up with
$$ Var [B] = {\sigma^2 \over N^2 \mu^2} \sum_{i=1}^N x_i^2. $$</p>
<h2>The contradiction</h2>
<p>As explained below, I think that the barycentre variance should be independent of a shift of all the $x_i$, that is $Var[B]$ should not change if all $x_i$ is shifted to $x_i+c$. But $\sum_{i=1}^N x_i^2$, and therefore my estimated $Var[B]$, will change a shift of the $x_i$. </p>
<p>Where is the error: In my calculations, or the expectation that the variance should be independent of shift in $x$.</p>
<hr>
<h2>One derivation of the variance</h2>
<p>$$Var [B ]
= Var \left[{\sum_{i=1}^N W_i x_i \over \sum_{i=1}^N w_i}\right]
= {1\over N^2 \mu^2} \sum_{i=1}^N x_i^2 Var [W_i]
={\sigma^2\over N^2 \mu^2} \sum_{i=1}^N x_i^2$$</p>
<p>Here I used that $Var [\sum_i X_i] = \sum_i Var[ X_i]$ when the $X_i$ are independent, that $Var [cX]=c^2 Var[X]$, and that $Var[ w_i] =\sigma^2$.</p>
<p>I have also done the same in a more direct fashion, with the same result. It is kind of messy, so I have omitted it.</p>
<h2>Why do I expect $Var [B]$ to be invariant to a shift of $x_i$?</h2>
<p>The expected value of $B$ is the mean of the $x_i$:</p>
<p>$$E[B]=E\left[{\sum_{i=1}^N E[W_i] x_i\over \sum_{i=1}^N W_i}\right]
={\sum_{i=1}^N E[W_i] x_i\over N\mu}
={\mu\sum_{i=1}^N x_i\over N\mu}={\sum_{i=1}^N x_i\over N}=\bar x_i.$$ </p>
<p>Moving all $x_i$ by $c$ will move every barycentre estimate by $c$:
$$ \tilde B = {\sum_{i=1}^N W_i (x_i+c)\over \sum_{i=1}^N W_i}
= {\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i} +{\sum_{i=1}^N W_i c\over \sum_{i=1}^N W_i}=
{\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i} +c=B+c.$$</p>
<p>The variance of $B$ is $Var[B]=E[(B-E[B])^2]$.
Moving all the $x_i$ by $c$ adds $c$ to $B$ and $E[B]$, so $B-E[B]$ and the variance should stay the same.</p>
<p>To check this I did a <a href="https://gist.github.com/anonymous/30f09acb70b920b63f110bd1de32ee79" rel="nofollow noreferrer">MATLAB simulation</a>. In that simulation the estimated variances and barycentres (after i remove the displacement $c$) are identical.</p>
<h2>Why do I think $\sum_{i=1}^N w_i =N\mu$ is a good approximation</h2>
<p>Although the variance of $\sum_{i=1}^N w_i$ grow with $N$, this means that the standard deviation grow with $\sqrt N$.
And since the expected value grow with $N$, the relative spread will decrease. </p>
<p>It is the relative spread that is important for this case. Say that for one $N$ the expected value is 100 and the standard deviation is 1. Then, if $\sum_{i=1}^N w_i $ miss $N\mu$ with one standard deviation my estimate of $B$ is off by 1%.
If I increase N by a factor of a 100, the expected value will be 10 000 and the standard deviation will be 10.
So if $\sum_{i=1}^N w_i $ miss $N\mu$ with one standard deviation my estimate of $B$ is off by 0.1%.</p>
<p>Which mean that for large enough N the error will be negligible.</p>
<hr>
<h2>Context</h2>
<p>I'm trying to repeat a derivation for the variance of an estimated arrival time from <a href="http://ieeexplore.ieee.org/document/5607320/" rel="nofollow noreferrer">this paper</a>. The barycentre is used to find the time when the echo from the seafloor is received for seabed mapping echo sounder. </p>
<p>As a simplified model, the part of the time sequence used in the barycentre calculation can be modelled as a sequence of independent, identically (Rayleigh) distributed, amplitude values. These amplitude values are equally spaced in time.</p>
<p>In this question the $W_i$ represent the amplitudes, the $x_i$ it the sample times and $B$ is the estimated time of arrival.
The paper states that it can be show analytically that </p>
<p>$$Var[B]= \frac{\Delta x^2}{12(N+1)}\left(\frac{4}{\pi}-1\right)N(N+2),$$</p>
<p>where $\Delta x$ is the distance between the $x_i$ ($x_i=x_0+i \Delta x$).
I have translated the equation to the notation used in this question.
No reference is given, although I am searching in related papers.</p>
<p>The middle factor in that equation $\left(4/\pi-1\right)$ is equal to the ratio the variance and the mean value squared of a <a href="https://en.wikipedia.org/wiki/Rayleigh_distribution" rel="nofollow noreferrer">Rayleigh</a> distributed variable.</p>
| Dean | 393,411 | <p>There are a couple of mistakes in your work here.</p>
<p>In your derivation you replace the denominator, $\sum W_i$ with its expectation value. You say you do this because it is a good approximation, but you are ignoring the variance it contributes. Would you still do that if the numerator was a constant? If so, you would have found that $V[\hat{X}]=0$.</p>
<p>The more important issue is that you are assuming the $W_i$ are independent. If we make the replacement,
$$F_i = {W_i\over\sum W_i}$$
then,
$$B = \sum x_i F_i$$
Because $\sum F_i = 1$, there are only $n-1$ independent random variables $F_i,\ \ i =2...n$. The other random variable is
$$F_1 = 1 - \sum_{i=2}^{n}F_i$$
Write $B$ in terms of the remaining independent random variables:
$$B = x_1\left(1 - \sum_{i=2}^{n}F_i\right)+\sum_{i=2}^n x_i F_i=x_1+\sum_{i=2}^n (x_i-x_1) F_i$$
Then,
$$E[B] = x_1 +E[F]\sum_{i=2}^n (x_i-x_1)= x_1 + {1\over n}\sum_{i=2}^n (x_i-x_1) = {1\over n}\sum_{i=1}^n x_i = \bar{x}$$
and
$$V[B] = V[F]\sum_{i=2}^n (x_i-x_1)^2$$</p>
<p>For this example the variance is invariant under translation of $x_i$. In reducing the degrees of freedom to $n-1$, the term $x_1$ is treated differently from the rest; note that $V[F_1]=(n-1)V[F]$.</p>
|
817,386 | <p>For a function $f$ I know that: $$\int{f'(r)dr}=f(r)$$ where $f(r)$ is known. knowing the result of this integral how can i calculate $$\int{(f'(r))^2dr}$$ Is there any relation between these integrals?</p>
| Gerry Myerson | 8,269 | <p>Let $f(x)=e^{x^2}$, so $(f'(x))^2=4x^2e^{2x^2}$. But $\int4x^2e^{2x^2}\,dx$ can't be evaluated in terms of elementary functions (exponentials, trig functions, polynomials, $n$th roots, etc.). </p>
|
226,551 | <p><strong>(1)</strong> Which graph classes are extremely tough to test for graph non-isomorphic pairs from isomorphic pairs?</p>
<p><strong>(2)</strong> Is there a repository of adjacencies from such classes?</p>
| joro | 12,481 | <p>Paper: <a href="http://mivia.unisa.it/wp-content/uploads/2013/05/desanto02.pdf" rel="noreferrer">A large database of graphs and its use for benchmarking
graph isomorphism algorithms</a></p>
<p>The graph instances and graph generator software are here:</p>
<p><a href="http://mivia.unisa.it/datasets/graph-database/arg-database/" rel="noreferrer">http://mivia.unisa.it/datasets/graph-database/arg-database/</a></p>
|
671,407 | <p>I have problem with equation: $4^x-3^x=1$. </p>
<p>So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions? </p>
| imranfat | 64,546 | <p>You can make a graph of the function $y=4^x-3^x-1=0$. It has a y-intercept at (0,-1) and for $x<0$ the curve stays under the y-axis and for $X>0$ the curve is only increasing. </p>
|
129,293 | <p>I'm writing a survey that involves Levy processes and wanted to mention the different forms of the Levy-Khintchine formula found in literature.</p>
<p>The most common version seems to give the Levy symbol as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - i\langle u,y \rangle\mathbf{1}_{|y|\le1}{)}\, dK(y)$$ </p>
<p>while in other versions it seems to be given as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|^2}{)} \, dK(y)$$</p>
<p>while at <a href="http://almostsure.wordpress.com/2010/09/15/processes-with-independent-increments/" rel="nofollow">almostsure blog</a> it is given as</p>
<p>$$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d}{(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|}{)} \, dK(y).$$</p>
<p>Are all of these correct and equivalent? If the last one is, does anyone know a published source I could cite that mentions it?</p>
| Community | -1 | <p>Here is part of <strong>Exercise 3.2.40</strong> from <em>Probability Theory: An Analytic View</em> by Daniel W. Stroock. He refers to
$$\int_{\{0<|y|<1\}} |y|\,K(dy)+K((-1,1)^c)<\infty\tag{3.2.2}$$ </p>
<blockquote>
<p>The difficulty of distinguishing between the drift and small jumps when (3.2.2) fails has a purely analytic antecedent which is reflected by an inherent arbitrariness in the Lévy-Khinchine formula. Namely, there is nothing sacrosanct, or even particularly compelling, about the way in which we <em>corrected</em> $e^{i\xi y}-1$ in order to accommodate $K$'s for which (3.2.2) fails. Indeed, show that we could have equally well taken any function of the form $$e^{i\xi y}-1-i\xi \psi(y),$$ where $\psi:\mathbb{R}\to\mathbb{R}\setminus\{0\}$ is any bounded measurable function with the property that $$\sup_{y\neq 0}\left| {\psi(y)-y\over y^2}\right|<\infty.$$ </p>
</blockquote>
|
1,689,523 | <p>I need help with this Laplace question.
<span class="math-container">$$f(t) = e^{-t} \sin(t) $$</span></p>
<hr />
<p>Answer should be <span class="math-container">$\dfrac{1}{s^2 + 2s + 2}$</span></p>
<hr />
<p>What I'm currently doing is as follows:</p>
<p><span class="math-container">$u = \sin(t)\qquad$</span> <span class="math-container">$dv = e^{-(s+1)t}dt$</span></p>
<p><span class="math-container">$du = \cos(t)dt\qquad$</span> <span class="math-container">$v = \dfrac{e^{-(s+1)t}}{-(s+1)}$</span></p>
<p><span class="math-container">$\dfrac{-\sin(t) e^{-(s+1)t}}{-(s+1)} - \int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$</span></p>
<p>But even if I solved the integral, I wouldn't get this (which is what I should, see picture).</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/BhMOx.png" alt="enter image description here" /></p>
</blockquote>
| Mark Viola | 218,419 | <p>You need not integrate by parts to evaluate this integral. In fact, one would need to integrate by parts twice. <strong>See the section following the highlighted SPOILER ALERT</strong></p>
<p>So, I thought it would be instructive to present a "trick" that we can use to quickly evaluate the integral of interest and other similar integrals.</p>
<p><strong>HERE IS A HINT:</strong></p>
<p><span class="math-container">$$\sin(t)=\text{Im}\left(e^{it}\right) \tag 1$$</span></p>
<p><strong>SPOILER ALERT:</strong> Scroll over the highlighted area to reveal the solution</p>
<blockquote class="spoiler">
<p>Using <span class="math-container">$(1)$</span>, we have<span class="math-container">$$\begin{align}\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\text{Im}\left(\int_0^\infty e^{(-1-s+i)t}\,dt\right)\\\\&=\text{Im}\left(\frac{1}{-(1+s)+i)}\right)\\\\&=\text{Im}\left(\frac{-(s+1)-i}{(s+1)^2+1}\,dt\right)\\\\&=\frac{1}{s^2+2s+2}\end{align}$$</span>as was to be shown! Quick, painless, and less prone to careless errors.</p>
</blockquote>
<hr />
<p>The last result in the OP was</p>
<p><span class="math-container">$$\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt=\left.\left(\frac{-\sin(t) e^{-(s+1)t}}{-(s+1)}\right)\right|_0^\infty - \int_0^{\infty}\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$$</span></p>
<p>Continuing we have</p>
<p><span class="math-container">$$\begin{align}
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1} \int_0^{\infty} e^{-(s+1)t}\cos(t)\,dt \tag 2
\end{align}$$</span></p>
<p>We integrate by parts <span class="math-container">$(2)$</span> with <span class="math-container">$u=\cos(t)$</span> and <span class="math-container">$v=-\frac{e^{-(s+1)t}}{s+1}$</span>. Then, we obtain</p>
<p><span class="math-container">$$\begin{align}
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1}\left.\left(-\frac{\cos(t)e^{-(s+1)t}}{s+1}\right)\right|_0^\infty -\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\
&=\frac{1}{(s+1)^2}-\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\
((s+1)^2+1)\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=1\\\\
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s^2+2s+2}
\end{align}$$</span></p>
<p>as expected!</p>
|
3,031,290 | <p>Can you choose <span class="math-container">$11$</span> different numbers among them so that the numbers <span class="math-container">$|a_1-a_2|, |a_2-a_3|,\ldots,|a_{10}-a_{11}|,|a_{11}-a_{1}|$</span> are all different. The smartest thing that my dumbest mind could accomplish is that all those differences are <span class="math-container">$1,2,3,...,11$</span>. From this, there are two ways - construct an example which is quite painful - but I did it for <span class="math-container">$n=4,5$</span> and tried for <span class="math-container">$6$</span> so I couldn't find example for <span class="math-container">$3$</span>, so may be there something about <span class="math-container">$6$</span> and its multiplicators? another way - prove by something, may be algebra or properties of numbers of <span class="math-container">$1,2,3,\ldots,11$</span> (sum of squares so we could get rid of modulus), try to prove that there always will be at least two equal differences and so on. Can you give me some hint? </p>
| saulspatz | 235,128 | <p>I haven't been able to do this by any reasonable method. I worked out that the sum of the numbers that are bigger than both their neighbors (considering the numbers as arranged on a circle) minus the sum of the numbers that are smaller than both their neighbors must be <span class="math-container">$33,$</span> but I wasn't able to turn that to account.</p>
<p>I started to do it by brute force, but I was afraid of making a mistake, so I wrote a simple-minded python script to do it. I figured it would take longer to write an intelligent script than it would to write and run the dumb one.</p>
<p>I also listed the elements that were larger than both of their neighbors and the elements smaller than both their neighbors. In all cases there were <span class="math-container">$5$</span> of each. I don't know what, if anything can be made of that. </p>
<p>There are <span class="math-container">$208$</span> solutions.</p>
<p>Here's the script</p>
<pre><code>from itertools import permutations
N = 12
def uniqueDiffs(x):
y = x+(x[0],)
a = [abs(y[t]-y[(t+1)]) for t in range(N-1)]
return len(set(a))==N-1
def minors(x):
y = (x[-1],) + x + (x[0],)
return sorted([y[i] for i in range(1,N) if y[i-1] > y[i] < y[i+1]])
def majors(x):
y = (x[-1],) + x + (x[0],)
return sorted([y[i] for i in range(1,N) if y[i-1] < y[i] > y[i+1]])
good = []
S= range(1,N+1)
start = (1,N)
S = set(range(2,N))
for x in S:
T=S.copy()
T.remove(x)
for p in permutations(T):
s = start + p
if uniqueDiffs(s):
good.append(s)
print(len(good), "solutions")
for g in good:
print(g, minors(g), majors(g))
</code></pre>
<p>Here's the first solution it printed</p>
<p><code>(1, 12, 2, 7, 8, 6, 9, 5, 11, 3, 10)</code></p>
<p>I don't imagine anyone wants to see the other <span class="math-container">$207.$</span></p>
<p><strong>P.S.</strong></p>
<p>I got to wondering what values other than <span class="math-container">$12$</span> work. The problem makes no sense for <span class="math-container">$N<4.$</span> I tested for <span class="math-container">$4\le N\le13$</span> and found that there are solutions only for <span class="math-container">$N=4,5,8,9,12,13.$</span> Also, in all cases the number of elements larger than both their neighbors ("majors") is the same as the number of elements smaller than both their neighbors ("minors"), but this number is not necessarily the same in all solutions. For <span class="math-container">$N=13,$</span> there are solution with <span class="math-container">$5$</span> majors and <span class="math-container">$5$</span> minors, and also solutions with <span class="math-container">$6$</span> of each. It is of course, tempting to guess that the problem can be solved only when <span class="math-container">$N\equiv0\pmod{4}$</span> or <span class="math-container">$N\equiv1\pmod{4},$</span> but I have far too little data. I will have to write a more intelligent program before trying to test larger cases. </p>
<p><strong>P.P.S</strong></p>
<p>By considering the difference of the majors and the minrs it's easy to show that the problem is impossible when <span class="math-container">$N\equiv2\pmod{4}$</span> or <span class="math-container">$N\equiv3\pmod{4}$</span> </p>
|
4,177,639 | <p>I have an object with known coordinates in in 3D but on the ground (<code>z=0</code>). The object has a direction vector. My goal is to move this object on the ground (so <code>z</code> stays <code>0</code>) using its direction vector and via randomly-generated velocity vectors with one condition: I want to ensure that the generate velocity vector can only move the object within a "valid arc", defined in degrees with respect to the direction vector of the object. More specifically, the way I determine valid range is by ensuring that the new position is within 45 degrees of the old object's direction vector (-45 degrees to the left and +45 degrees to the right). Can someone write a pseudocode on how I can achieve this?</p>
<p>Here's my attempt to do this but this doesn't seem to be the correct way to help me achieve what I want:</p>
<pre><code>object_dir = object_position # the direction could be the same as the object's coortinates
while True:
vel_vec=[uniform(-max_vel, max_vel), uniform(-max_vel, max_vel)] # generate a random velocity vector
new_pos = object_dir + vel_vec # compute a new position (and/or object direction vector) for the object
if (compute_angle(new_pos, object_dir) < 45 or compute_angle(new_pos, object_dir) > 315):
break
</code></pre>
| Claude Leibovici | 82,404 | <p>In the most general case, there is no analytical solution for the zero of function
<span class="math-container">$$f(x)=a^x+b^x-1$$</span> and numerical iterative methods would be required.</p>
<p>If wa assume <span class="math-container">$a>1$</span> and <span class="math-container">$b>1$</span>, <span class="math-container">$f(x)$</span> is not very pleasant to look at if graphing but this is not the case of its logarithmic tansform
<span class="math-container">$$g(x)=\log(a^x+b^x)$$</span> which looks quite close to a straight line.</p>
<p>Being lazy, expand <span class="math-container">$g(x)$</span> as a Taylor series around <span class="math-container">$x=0$</span> and obtain
<span class="math-container">$$g(x)=\log(2)+\log(\sqrt{ab})x+O(x^2)\implies x_0=-\frac {2\log(2) } {\log({ab}) }$$</span> and start Newton method for generating the sequence
<span class="math-container">$$x_{n+1}=x_n-\frac{\left(a^{x_n}+b^{x_n}\right) \log \left(a^{x_n}+b^{x_n}\right)}{a^{x_n} \log (a)+b^{x_n} \log (b)}$$</span></p>
<p>For illustration, using <span class="math-container">$a=3$</span> and <span class="math-container">$b=7$</span>, the iterates will be
<span class="math-container">$$\left(
\begin{array}{cc}
n & x_n \\
0 & -0.455340 \\
1 & -0.468168 \\
2 & -0.468178
\end{array}
\right)$$</span> which is quite fast.</p>
<p>But we could have a still better approximation performing one single iteration of Halley method starting at <span class="math-container">$x=0$</span>. This would give, as an <em>approximation</em>,
<span class="math-container">$$x_0=\frac {4\log(2)\log(ab) } {(\log (2)-2) \left(\log ^2(a)+\log ^2(b)\right)-2 (2+\log (2)) \log (a) \log (b) }$$</span></p>
<p>For the worked example, this would give <span class="math-container">$x_0=-0.46790$</span>.</p>
<p>This estimate could still be improved using one single iteration of higher order methods (we still get analytical expressions. The formulae start to be too long for typing them, but for the worked example, as a function of the order of the method, the results would be
<span class="math-container">$$\left(
\begin{array}{ccc}
n & x_0^{(n)} & \text{method} \\
2 & -0.4553404974 & \text{Nexton} \\
3 & -0.4679002951 & \text{Halley} \\
4 & -0.4682565630 & \text{Householder} \\
5 & -0.4681819736 & \text{no name} \\
6 & -0.4681774686 & \text{no name} \\
7 & -0.4681781776 & \text{no name} \\
8 & -0.4681782373 & \text{no name}
\end{array}
\right)$$</span></p>
|
1,908,923 | <p>Let $X$ be a Riemannian manifold*, and $S$ a compact submanifold of $X$. </p>
<p>Assume there exists an <strong>open, dense</strong> subset $Y$ of $\,X$, such that for any element $y \in Y$, there exists a unique element in $S$ closest to $y$; i.e there is a function $\tilde s:Y \to S$ such that $$ d(y,\tilde s(y))=d_S(y)=\inf\{d(s,y)|\, \,s \in S\},$$</p>
<p>and $\tilde s(y)$ is the only element in $S$ satisfying the above equality.</p>
<p><strong>Question:</strong></p>
<p>Let $x \in X$, and let $\tilde s \in S$ be a closest element to $x$ in $S$. (such an element exists by compactness of $S$). </p>
<p>Is it true that there exists a sequence $y_n \in Y$, $y_n \to x$, such that $\tilde s(y_n) \to \tilde s$?</p>
<p>Note that I do <strong>not</strong> assume $x$ has a unique closest point on $S$. (If this were true the question would become trivial; On any subset consisting of elements which have unique closest points, the "closest point mapping" $\tilde s$ is continuous).</p>
<hr>
<p>*(Actually, one could ask the question in a more general setting when $X$
is a metric space, and $S$ is a subset of it, but I am not sure which "strange pathologies" can arise then)</p>
| HK Lee | 37,116 | <p>Yes </p>
<p>$c(t)$ is a shortest geodesic from $c(0)=s_0\in S$ to $c(1)=x$ For some $0<t_0<1$ assume that $$d(c(t_0),S)=d(c(t_0),s_1) \leq d(c(0),c(t_0)),\ s_1\in S$$</p>
<p>Then $$ d(x,S)\geq d(c(t_0),x) + d(c(t_0),s_1) \geq d(x,S) $$</p>
<p>It is a contradiction So $s_1=s_0$ If $y_n\in Y$ goes to $ c(t_0)$, then by uniqueness of $s_1$ corresponded points in $S$ wrt $y_n$ goes to $c(0)$ When $t_0$ goes to $1$, then we have $y_n$ wrt $c(t_0)$ So we can take $y_n$ converging to $x$</p>
|
1,185,108 | <p>empty set is an subset of any sets maybe any collection of sets.</p>
<p>I wonder what about the case of the empty set being a member,not subset, of any collection (family) of sets.</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Counterexample: the only element of $\{\{\emptyset\}\}$ is $\{\emptyset\}\ne\emptyset$.</p>
|
1,257,598 | <p>Suppose A is a family of subsets of R with the property that the intersection of any two sets in A is finite. Show that $|A|\leq 2^{\aleph_0}$.</p>
<p>I was told that choosing a countable $D \subset B$ for all $B \in A$ would be helpful. I'm just really not sure where to go with this. Any hints would be appreciated! </p>
| Brian M. Scott | 12,042 | <p>Here’s an argument that’s a bit closer to the hint. Suppose that $|A|>2^{\aleph_0}$. $\Bbb R$ has only $2^{\aleph_0}$ finite subsets, so without loss of generality we may assume that every member of $A$ is infinite. For each $B\in A$ let $C_B$ be a countably infinite subset of $A$. $\Bbb R$ has $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$ countably infinite subsets, so there must be distinct $B,D\in A$ such that $C_B=C_D$ and hence $B\cap D$ is not finite. In fact there must be some countably infinite $C\subseteq\Bbb R$ such that</p>
<p>$$|\{B\in A:C_B=C\}|>2^{\aleph_0}\;.$$</p>
|
1,297,690 | <p>If I have a program that creates, let's say, one billion integers, with each having a pure $50 - 50$ chance to be one or zero,</p>
<p>what is the chance of finding $x$ zeros in a row?</p>
<p>for brownie points, instead of the program creating a set billion numbers, what would the equation be with $z$ numbers?</p>
| Nick Peterson | 81,839 | <p><strong>Hint:</strong> If you never had $x$ zeros in a row, then (at least) one of the first $x$ numbers must be a one.</p>
<p>If $n_z$ is the number of arrangements, then you can partition $n_z$ based on which digit (of the first $x$) is the first to be a one. This will give you a recurrence relation in terms of numbers $n_{z_0}$, where each $z_0<z$.</p>
<p>Doing this will let you come up with the probability that such an arrangement never occurs; the event that such an arrangement DOES occur is the complement of this event.</p>
|
942,030 | <p>Given a Hilbert space $\mathcal{H}$.</p>
<p>Consider spectral measures:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$</p>
<p>Define its support:
$$\operatorname{supp}(E):=\bigg(\bigcup_{U=\mathring{U}:E(U)=0}U\bigg)^\complement=\bigcap_{C=\overline{C}:E(C)=1}C$$</p>
<p>By second countability:
$$E\bigg(\operatorname{supp}E^\complement\bigg)\varphi=E\left(\bigcup_{k=1}^\infty B_k'\right)\varphi=\sum_{k=1}^\infty E(B_k')\varphi=0$$</p>
<blockquote>
<p>But it may happen:
$$\Omega\subsetneq\operatorname{supp}E:\quad E(\Omega)=E(\operatorname{supp}E)=1$$</p>
</blockquote>
<p>What is an example?</p>
| Community | -1 | <p>$\frac{d(\dot\theta^2)}{dt}$ is not obtained from $b'(\theta)\dot\theta$, but from $b(\theta)=\dot\theta^2$, by derivation.</p>
|
4,616,559 | <p>For each day we store a snapshot of data in a database. We want to balance the storage costs with the densitiy of snapshots.
The older a time frame is the fewer snapshots from this time frame we need.
For example: if we store 10 snapshots from last year then we would like to store only 1 snpashot from the time ten years ago.
The density of snapshots should be roughly 1/x where x is the age of the snapshot.
How can we determine when to delete a given snapshot?</p>
| joriki | 6,622 | <p>When you need to delete a snapshot, delete the one with minimal ratio <span class="math-container">$T/t$</span>, where <span class="math-container">$t$</span> is the time of the snapshot and <span class="math-container">$T$</span> is the length of the interval without snapshot that would result from deleting it.</p>
|
659,256 | <p>This might be a silly question to some, but I need some help in this topic. <br />
Iota, denoted as <em>'i'</em> is equal to the principal root of -1.
Therefore, </p>
<p>$\iota^2 = -1$</p>
<p>When studying Modulus, I was wondering..</p>
<p>$|\iota| = ?$</p>
<p>A Google search revealed that the value is <strong>+1</strong>. This is because of the equation:</p>
<p>$z = x + y\iota$;
$Therefore, |z| = \sqrt(x^2 + y^2)$</p>
<p>Substituting the value of $0 + 1\iota$, we get our value. But how was this equation derived, and why would it hold true, since, if $\iota$ is +1, $\iota$ must be equal to either +1 or -1, making it a Real Number. My doubt is not analytical but intuitive. As it is an abstract concept, I
am having trouble understanding it. </p>
<p>Sorry for the lack of understanding of any fundamental concepts that render this question redundant. But, sadly, I do not know them, and would appreciate any help you gave me for understanding these.</p>
| Kindeep Singh Kargil | 422,502 | <p>Well, we can define modulus as <span class="math-container">$|x|$</span>= <span class="math-container">$\sqrt{x^2}$</span>.</p>
<p>So, since i=<span class="math-container">$\sqrt{(-1)}$</span></p>
<p>|i|=<span class="math-container">$\sqrt{(-1)^2}$</span></p>
<p>=<span class="math-container">$\sqrt{1}$</span></p>
<p>=1</p>
<p>Edit: </p>
<p>Actually can someone confirm if this makes sense?</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Henry | 6,460 | <p>Imagine a vertical sea cliff. Floating in the sea some distance away
from the cliff there is a boat, which is attached to a rope which goes
diagonally up to the top edge of the cliff and then continues on the
field at the top to a tractor. (Draw a picture.) The tractor then moves away from the
cliff edge, pulling the rope, and so pulling the boat towards the
cliff.</p>
<p>For the whole class to discuss:</p>
<ul>
<li>Which moves faster, the tractor or the boat? Or do they move at
the same speed?</li>
</ul>
<p>For the clever ones</p>
<ul>
<li>If the tractor moves at a constant speed, does the boat accelerate
towards the cliff or decelerate? Or does the boat travel at a constant
speed?</li>
</ul>
|
3,779,589 | <p>Let the metric <span class="math-container">$d$</span> be defined as
<span class="math-container">$$
d(f,g) =\sup_{x\in[0,1]}|f(x)-g(x)|,
$$</span>
and let<br />
<span class="math-container">$$
H(x) = \begin{cases} 0 \text{ if } x \leq \frac{1}{2}\\ 1 \text { if } x > \frac{1}{2} \end{cases}.
$$</span>
Is <span class="math-container">$f(x) = x$</span> in <span class="math-container">$B_\frac{1}{2}(H)$</span> ?</p>
<p><strong>My answer</strong>. No, because
<span class="math-container">$$
d(H(x),f(x)) = \sup_{x\in[0,1]}|f(x)-H(x)| = \frac{1}{2}.
$$</span>
Therefore, <span class="math-container">$f(x) \not \in B_\frac{1}{2}(H)$</span></p>
<p>I am not sure if my answer is correct, and I found that it is hard to visualize this metric. Can someone helps me on this?</p>
| Àlex Rodríguez | 813,535 | <p>I think that is much easier to think in terms of the euler function. We know that when n=200, this function give us the value 80, so there are 80 numbers between 1 and 200 that do not have a 2 or 5 in its factorisation, and the problem is over.</p>
|
3,779,589 | <p>Let the metric <span class="math-container">$d$</span> be defined as
<span class="math-container">$$
d(f,g) =\sup_{x\in[0,1]}|f(x)-g(x)|,
$$</span>
and let<br />
<span class="math-container">$$
H(x) = \begin{cases} 0 \text{ if } x \leq \frac{1}{2}\\ 1 \text { if } x > \frac{1}{2} \end{cases}.
$$</span>
Is <span class="math-container">$f(x) = x$</span> in <span class="math-container">$B_\frac{1}{2}(H)$</span> ?</p>
<p><strong>My answer</strong>. No, because
<span class="math-container">$$
d(H(x),f(x)) = \sup_{x\in[0,1]}|f(x)-H(x)| = \frac{1}{2}.
$$</span>
Therefore, <span class="math-container">$f(x) \not \in B_\frac{1}{2}(H)$</span></p>
<p>I am not sure if my answer is correct, and I found that it is hard to visualize this metric. Can someone helps me on this?</p>
| Jan Eerland | 226,665 | <blockquote>
<p>Not a 'real' answer, but it was too big for a comment.</p>
</blockquote>
<p>I wrote and ran some Mathematica code:</p>
<pre><code>In[1]:=Length[ParallelTable[
If[TrueQ[If[IntegerQ[n/2] \[Or] IntegerQ[n/5], True, False]],
Nothing, n], {n, 1, 200}]]
</code></pre>
<p>Running the code gives:</p>
<pre><code>Out[1]=80
</code></pre>
<p>So, when we look at your question there are <span class="math-container">$80$</span> numbers in the range <span class="math-container">$1\le\text{n}\le200$</span> such that <span class="math-container">$\text{n}$</span> does not divide <span class="math-container">$2$</span> and <span class="math-container">$5$</span>.</p>
<hr />
<p>Using Mathematica we can look at more complicated versions of this statement:</p>
<p>In the range <span class="math-container">$1\le\text{n}\le10^9$</span> there are <span class="math-container">$400000000$</span> numbers that does not divide <span class="math-container">$2$</span> and <span class="math-container">$5$</span>:</p>
<pre><code>In[2]:=Length[ParallelTable[
If[TrueQ[If[IntegerQ[n/2] \[Or] IntegerQ[n/5], True, False]],
Nothing, n], {n, 1, 10^9}]]
Out[2]=400000000
</code></pre>
<p>In the range <span class="math-container">$1\le\text{n}\le200$</span> there are <span class="math-container">$80$</span> numbers that does not divide <span class="math-container">$2$</span> and <span class="math-container">$5$</span> and <span class="math-container">$8$</span>:</p>
<pre><code>In[3]:=Length[ParallelTable[
If[TrueQ[If[IntegerQ[n/2] \[Or] IntegerQ[n/5] \[Or] IntegerQ[n/8], True, False]],
Nothing, n], {n, 1, 200}]]
Out[3]=80
</code></pre>
<p>In the range <span class="math-container">$1\le\text{n}\le10^9$</span> there are <span class="math-container">$400000000$</span> numbers that does not divide <span class="math-container">$2$</span> and <span class="math-container">$5$</span> and <span class="math-container">$8$</span>:</p>
<pre><code>In[4]:=Length[ParallelTable[
If[TrueQ[If[IntegerQ[n/2] \[Or] IntegerQ[n/5] \[Or] IntegerQ[n/8], True, False]],
Nothing, n], {n, 1, 10^9}]]
Out[4]=400000000
</code></pre>
<p>In the range <span class="math-container">$1\le\text{n}\le10^9$</span> there are <span class="math-container">$285714286$</span> numbers that does not divide <span class="math-container">$2$</span> and <span class="math-container">$3$</span> and <span class="math-container">$4$</span> and <span class="math-container">$7$</span> and <span class="math-container">$9$</span>:</p>
<pre><code>In[5]:=Length[ParallelTable[
If[TrueQ[If[
IntegerQ[n/2] \[Or] IntegerQ[n/3] \[Or] IntegerQ[n/4] \[Or]
IntegerQ[n/7] \[Or] IntegerQ[n/9], True, False]], Nothing,
n], {n, 1, 10^9}]]
Out[5]=285714286
</code></pre>
|
2,469,798 | <p>Let $S = \left\{x \in \mathbb{Q} \mid 1 \leqslant {x}^2 \leqslant 29 \right\}$</p>
<p>What can we say about the supremum and infimum of this set? Would it be non-existent?</p>
<p>Would it be correct to say the following?</p>
<p>Suppose $ \sup S < \sqrt{29} $ then $ \exists x \in S $ such that $ x > \sup S$ </p>
<p>Therfore $ \sqrt{29} \leqslant \sup S$</p>
<p>But on the other hand $\sup S > \sqrt{29}$ then $\exists x \in \mathbb{Q}$ such that $ x < \sup S$</p>
<p>Therefore $\sup S \leqslant \sqrt{29}$</p>
<p>Therefore $\sup S $ can only be equal to $\sqrt{29}$</p>
<p>However $\sqrt{29} \notin \mathbb{Q}$ therefore $\sqrt{29} \notin S$</p>
<p>Therefore there can not be a supremum to this set.</p>
<p>(I am assuming the supremum of a set must belong to the set. Is this correct?)</p>
| orangeskid | 168,051 | <p>Let's consider the set of rational numbers
$$\{ r \in \mathbb{Q} \mid r \ge 1 \text{ and } r^2 \le 29\}$$</p>
<p>The supremum of the set equals $\sqrt{29}$. Perhaps it is more interesting to show that there does not exist a supremum of this set in $\mathbb{Q}$. That is in some way obvious. But we may still play with it and show the following:</p>
<ol>
<li><p>$S$ does not have a largest element. </p></li>
<li><p>While $S$ has majorants in $\mathbb{Q}$ ( for instance $6$), it does not have a smallest majorant in $\mathbb{Q}$. </p></li>
<li><p>Let $r = \frac{m}{n}$ in $S$. We have $\left( \frac{m}{n}\right)^2 \le 29$, so $29 n^2 - m^2 \ge 0$. Some number theory shows that we cannot have equality. So $29 n^2 - m^2 \ge 1$. Rewrite it as
$$(\sqrt{29} n - m)(\sqrt{29}n + m) \ge 1 \\
\sqrt{29}n - m \ge \frac{1}{\sqrt{29}n + m} \\
\sqrt{29} - \frac{m}{n} \ge \frac{1}{n^2( \sqrt{29}+ \frac{m}{n})}\ge \frac{1}{n^2 \cdot 2 \sqrt{29}} > \frac{1}{12 n^2}$$</p></li>
</ol>
<p>We can rephrase this as follows:</p>
<p>If $\frac{m}{n} \in S$ then $\frac{m}{n} + \frac{1}{12 n^2} \in S$. So clearly $S$ does not have a largest element. </p>
<ol start="2">
<li>From the above, the set rationals so that $t \ge s$ for all $s \in S$ is
$$T = \{ t \in \mathbb{Q} \mid t \ge 0 \text{ and } t^2 > 29 \}$$</li>
</ol>
<p>Let $ \frac{m}{n}\in T$. In a similar way as above we can show that $\frac{m}{n}- \frac{1}{2 m n} \in T$. Conclusion: the set $T$ does not have a smallest element.</p>
<p>We conclude that $S$ does not have a supremum in $\mathbb{Q}$.</p>
|
34,215 | <p>How do professional mathematicians learn new things? How do they expand their comfort zone? By talking to colleagues? </p>
| Gerry Myerson | 3,684 | <p>Talking to colleagues is good. Also, attending talks, reading papers and books, and teaching - I never knew much about differential equations until my department made me teach it (I still don't know much about differential equations, but at least I know enough to do a decent job of teaching it). </p>
|
188,087 | <p>Is there a function that can extract a list of variables in an expression?
For example, assume we have an expression</p>
<pre><code>x^2+y^3+z
</code></pre>
<p>This expression has variables x, y and z. The result should be</p>
<pre><code>{x, y, z}
</code></pre>
<p>. Is there a way to get this?</p>
| TimRias | 10,587 | <p>For polynomial expressions @Buddha_the_Scientist's suggestion <code>Variables</code> will work. For more general expressions</p>
<pre><code>expr = x^2 + y^3 + z
DeleteDuplicates@Cases[expr, _Symbol, ∞]
</code></pre>
<p>Should do the trick in most situations.</p>
|
188,087 | <p>Is there a function that can extract a list of variables in an expression?
For example, assume we have an expression</p>
<pre><code>x^2+y^3+z
</code></pre>
<p>This expression has variables x, y and z. The result should be</p>
<pre><code>{x, y, z}
</code></pre>
<p>. Is there a way to get this?</p>
| Michael E2 | 4,999 | <p>The undocumented <code>Integrate`getAllVariables</code> is a somewhat more robust version of <code>Variables</code>. It has a required second argument that specifies a variable to be excluded from the output. It just goes to show that internal functions are not always defined with the general user in mind.</p>
<pre><code>Integrate`getAllVariables[x^2 + y^3 + z, {}] (* delete {} from output: *)
Variables[x^2 + y^3 + z] (* {} can't happen anyway *)
(*
{x, y, z}
{x, y, z}
*)
</code></pre>
<p>A case <code>Variables</code> does not handle:</p>
<pre><code>Integrate`getAllVariables[{x[0]'[t] + a t == 0,
y[1] == Sin[b[t]] x[0][t]^2}, {}]
Variables[{x[0]'[t] + a t == 0, y[1] == Sin@b[t] x[0][t]^2}]
(*
{a, t, b[t], y[1], x[0][t]}
{}
*)
</code></pre>
<p>Note how <code>b[t]</code> is treated differently than <code>Sin[t]</code>, etc.:</p>
<pre><code>Integrate`getAllVariables[a + b[t] - c[t + s] + x^y, {}]
Integrate`getAllVariables[a + Sin[t] - Cos[t + s] + x^y, {}]
(*
{a, x, y, b[t], c[s + t]}
{a, s, t, x, y}
*)
</code></pre>
<p>The second argument is useful in problems in which there is a principal independent variable and you want to get all the others. Excluding more than one, however, can only be achieved by hacking:</p>
<pre><code>Integrate`getAllVariables[x^2 + y^3 + z, x]
(* {y, z} *)
xc /: {xc[a__]} := {a}; (* except the variables a.. *)
Integrate`getAllVariables[x^2 + y^3 + z, xc[x, y]]
(* {z} *)
</code></pre>
|
1,913,320 | <blockquote>
<p>Let <span class="math-container">$A=(a_{ij})_{n\times n}$</span> and <span class="math-container">$A=(a_{ij})_{n\times n}$</span> be two upper triangular matrices, i.e. <span class="math-container">$a_{ij}=b_{ij}=0$</span> whenever <span class="math-container">$i>j$</span>.</p>
<p><span class="math-container">$(a)$</span> Show that the <span class="math-container">$(i,j)$</span>-entry of <span class="math-container">$AB$</span> is <span class="math-container">$0$</span> if <span class="math-container">$i>j$</span>, i.e <span class="math-container">$AB$</span> is an upper triangular matrix.</p>
<p><span class="math-container">$(b)$</span> Find the <span class="math-container">$(i,i)$</span>-entry of <span class="math-container">$AB$</span>.</p>
</blockquote>
<p>I have already proven part (a). How do I go about finding part (b)? Any help would be greatly appreciated! Thank you so much!</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ Suppose for contradiction that $\,2\,$ and $\,7\,$ are the only primes. Then $\,15 = 2*7+1\,$ has no smaller prime factors so is prime, contradiction. But $15$ is not prime in the <em>real</em> integers. Rather, it is prime only in the <em>hypothetical</em> integers having only the primes $\,2\,$ and $\,7\,$. </p>
|
3,524,550 | <p>Lines in <span class="math-container">$\mathbb{R}^3$</span> are all congruent to one another, but circles in <span class="math-container">$\mathbb{R}^3$</span> are not all congruent to one another (because two different circles may have different radii). Visually, this is completely obvious. However, I would like a <strong>group-theoretic</strong> explanation for this.</p>
<blockquote>
<p>I am thinking of <span class="math-container">$\mathbb{R}^3$</span> as the homogeneous space <span class="math-container">$\mathbb{R}^3 = \frac{G}{G_0} = \frac{\text{SE}(3)}{\text{SO}(3)}$</span>, where <span class="math-container">$G = \text{SE}(3)$</span> is the group of (orientation-preserving) rigid motions and <span class="math-container">$G_0 = \text{SO}(3)$</span> is the stabilizer of the origin.</p>
<p>A <strong>line in <span class="math-container">$\mathbb{R}^3$</span></strong> is an orbit of a point in <span class="math-container">$\mathbb{R}^3$</span> by a subgroup <span class="math-container">$H \leq G$</span> that is conjugate to the subgroup <span class="math-container">$\{ (x_1, x_2, x_3) \mapsto (x_1 + t, x_2, x_3) \colon t \in \mathbb{R}\}$</span> of translations by the vector <span class="math-container">$(1,0,0)$</span>.</p>
<p>A <strong>circle in <span class="math-container">$\mathbb{R}^3$</span></strong> is an orbit of a point in <span class="math-container">$\mathbb{R}^3$</span> by a subgroup <span class="math-container">$K \leq G$</span> that is conjugate to the subgroup <span class="math-container">$\{ (x_1 + ix_2, x_3) \mapsto (e^{i\theta}(x_1 + ix_2), x_3) \colon e^{i\theta} \in \mathbb{S}^1\}$</span> of rotations around the <span class="math-container">$x_3$</span>-axis.</p>
<p>Two subsets <span class="math-container">$S_1, S_2$</span> of <span class="math-container">$\mathbb{R}^3$</span> are <strong>congruent</strong> if there exists <span class="math-container">$g \in \text{SE}(3)$</span> such that <span class="math-container">$S_2 = g \cdot S_1$</span>.</p>
</blockquote>
<p>Given these definitions of "line" and "circle" --- as orbits of subgroups --- how could we have known that all lines in <span class="math-container">$\text{SE}(3)/\text{SO}(3)$</span> are congruent, but not all circles in <span class="math-container">$\text{SE}(3)/\text{SO}(3)$</span> have this property?</p>
<p>In other words: What are the relevant aspects of the subgroups <span class="math-container">$H$</span>, <span class="math-container">$K$</span>, and <span class="math-container">$G_0$</span> that explain the <span class="math-container">$G$</span>-equivalence of <span class="math-container">$H$</span>-orbits in <span class="math-container">$G/G_0$</span>, as opposed to the non-<span class="math-container">$G$</span>-equivalence of all <span class="math-container">$K$</span>-orbits in <span class="math-container">$G/G_0$</span>?</p>
| Jean Marie | 305,862 | <p>This is not a direct answer to your question, but an enlargment of its scope.</p>
<p>There exists a group which is transitive on the union of lines and circles : it is the <strong>anallagmatic group</strong>, existing in any dimension. I mention it in the third paragraph of <a href="https://math.stackexchange.com/q/2626356">this question</a> and in particular i give a linear representation through generators that can be helpful for obtaining for example, given two 3D circles, the operation that will map the one onto the other.</p>
|
1,272,499 | <p>Definite integral of $$\int_0^{2\pi} \frac{1}{2+\cos x}$$ without using improper integral, I want to solve this without having to use $-\infty$ and $\infty$ on the integrals limits. Is that possible?</p>
<p>The only way I can think of solving that is by using Weierstrass. $u = \tan \frac{x}{2}$, don't you have to modify the lower and upper limits with that substitution? The only way I can think of to progress in this is to change the limits to $-\pi$ and $\pi$, but when you'll get $-\infty$ and $\infty$ upper and lower limits.</p>
| mickep | 97,236 | <p>Split the integral as $\int_0^{2\pi}=\int_0^{\pi}+\int_{\pi}^{2\pi}$. Do the substitution $u=x-\pi$ in the second one, and put the expressions under common denominator, and simplify. You should end up with
$$
4\int_0^{\pi}\frac{1}{4-\cos^2x}\,dx
$$
The integrand is now symmetric in $x=\pi/2$, so the integral equals
$$
8\int_0^{\pi/2}\frac{1}{4-\cos^2x}\,dx.
$$
Now, doing $t=\tan(x/2)$ gives you (or at least me) the integral
$$
16\int_0^1\frac{(t^2+1)}{3t^4+10t^2+3}\,dt,
$$
which you can do with the methods of partial fraction. The result is already mentioned in other comments/answers.</p>
|
1,272,499 | <p>Definite integral of $$\int_0^{2\pi} \frac{1}{2+\cos x}$$ without using improper integral, I want to solve this without having to use $-\infty$ and $\infty$ on the integrals limits. Is that possible?</p>
<p>The only way I can think of solving that is by using Weierstrass. $u = \tan \frac{x}{2}$, don't you have to modify the lower and upper limits with that substitution? The only way I can think of to progress in this is to change the limits to $-\pi$ and $\pi$, but when you'll get $-\infty$ and $\infty$ upper and lower limits.</p>
| Jack Tiger Lam | 186,030 | <p>Make the substutition</p>
<p>$x \mapsto 2\theta$</p>
<p>$$\int_0^\pi \frac{2\text{d}\theta}{\sin^2{\theta}+3\cos^2{\theta}}$$</p>
<p>The polar integral for (half) the area of an ellipse is almost of this form, but some adjustments are required before we can proceed.</p>
<p>Factor out the 2 and insert a factor of 3 into the numerator of the integral, and a factor of 2 into the denominator.</p>
<p>$$\frac{4}{3}\int_0^\pi \frac{1^2 \times \sqrt{3}^2 \, \text{d}\theta}{2(\sin^2{\theta}+3\cos^2{\theta})}$$</p>
<p>The polar form of an ellipse centered on the center of the ellipse with vertical semi-major axis $a$ and lateral semi-minor axis $b$ is</p>
<p>$$r(\theta) = \frac{ab}{\sqrt{b^2\sin^2{\theta} + a^2\cos^{\theta}}}$$</p>
<p>The area of the ellipse with major and minor axes $a,b$ is $πab$.</p>
<p>In this case, $a=\sqrt{3}, b=1$</p>
<p>However, in this case, the integral is a half-revolution, which is half the area of the ellipse, giving us</p>
<p>$$\frac{4}{3} \times \frac{\pi \times \sqrt{3} \times 1}{2} = \frac{2\pi}{\sqrt{3}}$$</p>
|
2,745,918 | <p>The numbers $1,2, \ldots, n$ are written in a board, with $n \in \mathbb{N}$. In every move, we can choose two numbers of the board, find their $\rm lcm$, and replace the two numbers with it. After $k$ moves, we find the sum of the numbers in the board, and we name it $S$. Find the minimum and the maximum value of $n$, such that there is $k$ such that $S=2017$.</p>
<p>This is an exercise I show at a Mathematical Forum, posted on 2017, and I have been trying this for a long time actually, and the things I have proven aren't something important, so I would prefer full answers.</p>
<p>As far as I now, it isn't from an exam or a Math Olympiad. </p>
| didgogns | 392,996 | <p>It is possible to make $S=2017$ starting from $n=11$. Merge $4, 5, 9, 11$ in any order to make the sequence $1, 2, 3, 6, 7, 8, 10, 1980$ which achieves the sum $2017$.</p>
<p>Now what is left is to prove that it is impossible to achieve the sum of $2017$ with $n=10$. Since $\gcd(1, \cdots, 10)=2520>2017$, there should be at least $2$ numbers after the process. Let them $a, b$.</p>
<p>It is obvious that only one of them can be multiple of $7$. WLOG, let $7|a$. Then $a\le1260$, $b\le360$. Since their sum is too small ($1260+360+360<2017$), there need to be at least $4$ numbers to make sum $2017$. Let them $a, b, c, d$.</p>
<p>At most $2$ of $a, b, c, d$ is multiple of $5$ and only one of them is multiple of $7$. WLOG, $a$ is multiple of $7$. If $a$ is not multiple of $5$, WLOG $b, c$ is multiple of $5$. So $a\le504$, $b\le360$, $c\le360$ and $a+b+c\le1224$. If $a$ is multiple of $5$ either, then WLOG $a, b$ is multiple of $5$ thus $a\le1260$, $b\le360$, $c\le72$ and $a+b+c\le1692$. Note that in any cases, $a+b+c\le1692$.</p>
<p>In all cases $d$ is not multiple of $5$ nor $7$, which means $d\le72$. Since $1692+72\times4<2017$, at least $7$ numbers at the end of process is required to make $S=2017$.</p>
<p>Since $k$, the number of replacements, is at most $3$ now, there are only $2$ possible cases for replacement.</p>
<ul>
<li><p>Four numbers were merged, remaining six are not touched: In this case, replaced numbers' lcm is either $2520$ or at most $1260$ and remaining numbers are too small to sum from $1260$ to $2017$.</p></li>
<li><p>Three numbers merged into one, other two numbers merged, and remaining five not touched: The sum is at most $10\times9\times8+10\times9+10+\cdots+1=865<2017$.</p></li>
</ul>
<p>Therefore, one cannot achieve $S=2017$ from $n=10$ and $n=11$ is the smallest possible $n$.</p>
|
2,507,613 | <p>I am trying to teach myself group theory and I recently came across the topic of Isomorphisms.
I know that 2 groups are isomorphic if there is a one-on-one correspondence between their elements.
So if the groups have a different order, does that mean they are not isomorphic? Such as a group $S$ and its permutation group $S_{n}$ for $n>1$.</p>
| Community | -1 | <p>If two groups are isomorphic then their orders are the same. It follows from the fact that isomorphism between groups is a bijection. </p>
|
3,145,896 | <h1>Solve for <span class="math-container">$x$</span></h1>
<p>I have an equation that I have been working on solving; I know the solution, but I cannot get to it myself. Almost every simplification I do reverts back to a previous step. Can anyone show me how to solve for <span class="math-container">$x$</span> in this equation?</p>
<h3>Equation:</h3>
<p><span class="math-container">$$\log_6(2x-3)+\log_6(x+5)=\log_3x$$</span></p>
<h3>Solution:</h3>
<p><span class="math-container">$$x ≅ \frac{3347}{2000} ≅ 1.6735$$</span>
<br>
<strong>Note:</strong> upon further analysis of the answer, while close, it does not seem to be the <em>exact</em> solution.</p>
<hr>
<h3>What I Have Tried So Far</h3>
<p><span class="math-container">$$\log_6(2x-3) + \log_6(x + 5) = \log_3x$$</span>
<span class="math-container">$$\frac{\log(2x-3)}{\log6} + \frac{\log(x + 5)}{\log6} = \frac{\log x}{\log3}$$</span>
<span class="math-container">$$\log3 \cdot \log(2x-3) + \log3 \cdot \log(x + 5) = \log6 \cdot \log x$$</span>
<span class="math-container">$$\log3 \cdot \log \left[(2x - 3)(x + 5)\right] = \log6 \cdot \log x$$</span>
<span class="math-container">$$\frac{\log \left[(2x - 3)(x + 5)\right]}{\log_3 10} = \frac{\log6}{\log_x10}$$</span>
<span class="math-container">$$\log_x10 \cdot \log \left[(2x - 3)(x + 5)\right] = \log_3 10 \cdot \log6$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right] = \log_3 6$$</span>
<span class="math-container">$$\log_x3 \cdot \log_x \left[(2x - 3)(x + 5)\right] = \frac{\log_3 6}{\log_3 x}$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right]^{\ \log_x3} = \log_x 6$$</span>
<span class="math-container">$$\left[(2x - 3)(x + 5)\right]^{\ \log_x3} = 6$$</span>
<span class="math-container">$$(2x - 3)(x + 5) = x^{\log_3 6}$$</span></p>
<p><br>
I know these steps aren't really working towards the solution at points; I was sort of just playing around with the equation. Regardless, I really don't know how to go about moving forward from here.</p>
<hr>
| heropup | 118,193 | <p>If <span class="math-container">$x \in \mathbb R$</span>, the equation <span class="math-container">$$\log_6 (2x-3) + \log_6 (x+5) = \log_3 x$$</span> requires <span class="math-container">$x > 3/2$</span>. Under such an assumption, the LHS becomes <span class="math-container">$$\log_6 (2x-3)(x+5),$$</span> and the RHS, using the change-of-base formula, is <span class="math-container">$$\log_3 x = \frac{\log_6 x}{\log_6 3}.$$</span> Thus <span class="math-container">$$(2x-3)(x+5) = 6^{\log_6 x / \log_6 3} = x^{\log_3 6} = x^{\log_3 3 + \log_3 2} = x^{1 + \log_3 2}.$$</span></p>
<p>Let <span class="math-container">$a = \log_3 2 < 1$</span>. Then to seek a numerical root, we apply Newton's method to <span class="math-container">$$f(x) = 2x^2 - x^{a+1} + 7x - 15$$</span> by computing iterates of <span class="math-container">$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{2x_n^2 - x_n^{a+1} + 7x_n - 15}{4x_n - (a+1)x_n^a + 7}.$$</span> A suitable initial guess has already been provided, namely <span class="math-container">$x_0 = \frac{3347}{2000}$</span>, from which we iterate with a computer to obtain to 75 digits of precision
<span class="math-container">$$x_0 = \color{green}{1.6735}0000000000000000000000000000000000000000000000000000000000000000000000 \\
x_1 = \color{green}{1.6735161761}6028420105594976526962483014865547809090912454018791698117840074 \\
x_2 = \color{green}{1.673516176124260238848}57028078132556340450344946364614508901732583022202367 \\
x_3 = \color{green}{1.67351617612426023884839162222058963917030308}792976477026920554313588396492 \\
x_4 = \color{green}{1.67351617612426023884839162222058963917030308353547657451861889646281230396} \\
x_5 = \color{green}{1.67351617612426023884839162222058963917030308353547657451861889646281230396}
$$</span>
where green digits indicate correct values, demonstrating the rapid convergence of the iterates. Moreover, we are assured that this is the unique root, since <span class="math-container">$f$</span> is a monotonically increasing function on <span class="math-container">$x > 3/2$</span> (its derivative being trivially greater than <span class="math-container">$0$</span> on this interval).</p>
|
312,145 | <p>OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\frac{2I}{\pi}=\int_{0}^{\pi}\frac{1}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What should I do ahead to get the value I need? Any tips? (Thanks in advance)</p>
| Sangchul Lee | 9,340 | <p>I prefer to the following method:</p>
<p>\begin{align*}
I
:= \int_{0}^{\pi} \frac{x}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \, dx
&= \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x )^2} \, \sec^2 x \, dx.
\end{align*}</p>
<p>Now we make the substitution $b \tan x \mapsto a \tan x$. Then</p>
<p>\begin{align*}
I
&= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} \frac{b^2 + a^2\tan^2 x}{(1 + \tan^2 x )^2} \, \sec^2 x \, dx \\
&= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} ( b^2 \cos^2 x + a^2\sin^2 x ) \, dx \\
&= \frac{\pi}{(ab)^3} \cdot \frac{\pi}{4} \left( a^2 + b^2 \right)
= \frac{\pi^2(a^2 + b^2)}{4(ab)^3}.
\end{align*}</p>
|
312,145 | <p>OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\frac{2I}{\pi}=\int_{0}^{\pi}\frac{1}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What should I do ahead to get the value I need? Any tips? (Thanks in advance)</p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
\because I&=\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\& \stackrel{x\mapsto\pi-x}{=} \int_0^\pi \frac{\pi-x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} d x \\
&=\pi \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}-I \\
\therefore I&=\frac{\pi}{2} \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\
&=\pi \int_0^{\frac{\pi}{2}} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}
\end{aligned}
$$</span></p>
<p>By my <a href="https://www.quora.com/What-is-the-value-of-int-frac-%CF%80-2-0-frac-1-a-2-cos-2-x-b-2-sin-2-x-2-dx/answer/Lai-Johnny?ch=17&oid=349365811&share=74f378b8&srid=CCOmo&target_type=answer" rel="nofollow noreferrer">post</a>, <span class="math-container">$$
\boxed{I =\pi\left[\frac{\pi\left(a^2+b^2\right)}{4 a^3 b^3}\right]=\frac{\pi^2\left(a^2+b^2\right)}{4 a^3 b^3}}
$$</span></p>
|
1,505,920 | <p>(I think) My textbook says something is a linear transformation if </p>
<ul>
<li>$L(ax) = aL(x)$</li>
<li>$L(x+y) = L(x) + L(y)$</li>
<li>$L(x) = A(x)$</li>
</ul>
<p>But a lot of sites I've been on haven't proved these 3 things, so I just wanted to make sure that this is the proper way to prove it.</p>
<p>For example, if $L(x) = (x_1, x_2, x_1 + 2x_2)^T$, then is it a linear transformation from $R^2$ to $R^3$? I know that it is, but I'm not sure if what I have above is how I'm supposed to prove it.</p>
| Red | 232,153 | <p>Yes your textbook is right, basically a function is a linear transformation if and only if scalar multiplicity is reserved meaning that letting <span class="math-container">$a $</span> be a real number then</p>
<p><span class="math-container">$L(a*x)=a*L(x)$</span></p>
<p>In your example if you wanted to show this property holds you show that</p>
<p><span class="math-container">$2L(x)=2(x_1,x_2,x_1+2x_2)=(2x_1,2x_2,2x_1+4x_2)$</span></p>
<p>The second property that linear transformations must satisfy is preservation or distribution over vector addition. Let's say <span class="math-container">$v$</span> and <span class="math-container">$u$</span> are vectors then</p>
<p><span class="math-container">$L(x+v)=L(x)+L(v)$</span></p>
<p>Meaning you can add the vectors and then transform them or you can transform them individually and the sum should be the same. If in any case it isn't, then it isn't a linear transformation.</p>
<p>The third property you mentioned basically says that linear transformation are the same as matrix transformations. So every linear transformation has a matrix corresponding to it called the standard matrix. Hope this helps.</p>
|
1,505,920 | <p>(I think) My textbook says something is a linear transformation if </p>
<ul>
<li>$L(ax) = aL(x)$</li>
<li>$L(x+y) = L(x) + L(y)$</li>
<li>$L(x) = A(x)$</li>
</ul>
<p>But a lot of sites I've been on haven't proved these 3 things, so I just wanted to make sure that this is the proper way to prove it.</p>
<p>For example, if $L(x) = (x_1, x_2, x_1 + 2x_2)^T$, then is it a linear transformation from $R^2$ to $R^3$? I know that it is, but I'm not sure if what I have above is how I'm supposed to prove it.</p>
| Eric Wofsey | 86,856 | <p>This is almost right, but that third condition shouldn't be there (in fact, I don't know what it means in general--what is "$A$"?). That is, a map $L$ between two vector spaces is a linear transformation if and only if it satisfies $L(ax)=aL(x)$ and $L(x+y)=L(x)+L(y)$ (for any scalar $a$ and any elements $x,y$ in the domain of $L$).</p>
<p>You can check this directly for your example $L(x)=(x_1,x_2,x_1+2x_2)$ (for $x=(x_1,x_2)$). For any scalar $a$, we have $ax=(ax_1,ax_2)$, so $$L(ax)=L(ax_1,ax_2)=(ax_1,ax_2,ax_1+2ax_2)=a(x_1,x_2,x_1+2x_2)=aL(x).$$</p>
<p>If $x=(x_1,x_2)$ and $y=(y_1,y_2)$, then $x+y=(x_1+y_1,x_2+y_2)$, so $$\begin{align}
L(x+y)=L(x_1+y_1,x_2+y_2)={} &(x_1+y_1,x_2+y_2,x_1+y_1+2x_2+2y_2)\\={}&(x_1,x_2,x_1+2x_2)+(y_1,y_2,y_1+2y_2)\\={}&L(x)+L(y).
\end{align}$$</p>
<p>As you can see, checking something like this can be a kind of lengthy computation sometimes! But it's not hard--you just plug in the definitions and rearrange the terms appropriately.</p>
|
1,501,595 | <blockquote>
<p>Let $A$ be an integral domain. Show that $\dim(A)=0 \iff A$ is a field.</p>
</blockquote>
<p>The backward implication is trivial.</p>
<p>For the forward implication, if we can show that $1 \in <a>$, where $a(\neq 0) \in A$. Then, we are done. However, I don't know how to show it. </p>
<p>Any suggestions are appreciated.</p>
| Bernard | 202,857 | <p>This means $(0)$ is the only prime ideal, since $A$ is a domain. As the set of non-invertible elements is the union of all prime/maximal ideals, it implies all non-zero elements are invertible.</p>
|
493,104 | <p>I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating </p>
<p>$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$</p>
<p>So we should be able to find the area of the ellipse by </p>
<p>$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$</p>
<p>$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$</p>
<p>$$=\pi\frac{a^2+b^2}{2}.$$</p>
<p>First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as <a href="http://www.wolframalpha.com/input/?i=1%2F2+*+integral+from+0+to+2*pi+of+%28cos%28x%29%29%5E2+%2B+2*+%28sin%28x%29%29%5E2" rel="nofollow">Wolfram Alpha tells me</a>.</p>
<p>What am I doing wrong?</p>
| André Nicolas | 6,312 | <p>Your question has been answered, so now we look at how to find the area, using your parametrization, which is a perfectly good one. </p>
<p>The area is the integral of $|y\,dx|$ (or alternately of $|x\,dy|$. over the appropriate interval.</p>
<p>We have $y=b\sin\theta$ and $dx=-a\sin\theta\,d\theta$. So the area is
$$\int_0^{2\pi} |-ab\sin^2\theta|\,d\theta.$$
Using $\sin^2\theta=\frac{1-\cos 2\theta}{2}$, we find that the area is
$$\int_0^{2\pi} ab\frac{1-\cos 2\theta}{2}\,d\theta.$$
This is indeed $\pi ab$. </p>
|
493,104 | <p>I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating </p>
<p>$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$</p>
<p>So we should be able to find the area of the ellipse by </p>
<p>$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$</p>
<p>$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$</p>
<p>$$=\pi\frac{a^2+b^2}{2}.$$</p>
<p>First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as <a href="http://www.wolframalpha.com/input/?i=1%2F2+*+integral+from+0+to+2*pi+of+%28cos%28x%29%29%5E2+%2B+2*+%28sin%28x%29%29%5E2" rel="nofollow">Wolfram Alpha tells me</a>.</p>
<p>What am I doing wrong?</p>
| Rogelio Molina | 87,320 | <p>This is another way to do it when one know the area of a circle: Consider the area of a circle with radius 1 in coordinates $(\xi, \eta)$ this is:</p>
<p>$$
\int d\xi d \eta = \pi
$$</p>
<p>now if you define new coordinates in your ellipse equation $\xi = \frac{x}{a}, \quad \eta= \frac{y}{b}$ you obtain a circle of radius one: $\xi^2 + \eta^2 =1$ </p>
<p>The area of the ellipse you want is $ \int dx dy = ab \int d\xi d\eta = \pi ab$.</p>
|
2,868,595 | <p>A Vitali set is a subset $V$ of $[0,1]$ such that for every $r\in \mathbb R$ there exists one and only one $v\in V$ for which $v-r \in \mathbb Q$. Equivalently, $V$ contains a single representative of every element of $\mathbb R / \mathbb Q$.</p>
<p>The proof I read is in this short article on Wikipedia: <a href="https://en.wikipedia.org/wiki/Vitali_set" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Vitali_set</a></p>
<p>Under "proof", the second to last inequality $1 \leq \sum \lambda (V_k) \leq 3$ is claimed to result from the previous inequality $[0,1] \subset \bigcup V_k \subset [-1,2]$ simply using sigma-additivity. There must be some missing argument to claim that the sum of the measures, although greater than the measure of the union, is still less than the measure of $[-1,2]$.</p>
<p>What is the missing argument ?</p>
| Alex R. | 22,064 | <p>The sets $V_k$ are disjoint and countable, hence the measure of the union is exactly equal to the sum of measures.</p>
|
164,152 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/33215/what-is-48293">What is 48÷2(9+3)?</a> </p>
</blockquote>
<p>Please, look at the picture?</p>
<p><img src="https://i.stack.imgur.com/3Tauh.jpg" alt="http://s16.radikal.ru/i190/1206/3b/7411739c6d9a.jpg"></p>
<p>Which calculator shows true result and how to prove achieved result?</p>
| Siminore | 29,672 | <p>Despite every personal belief, a notation like $a/b(c+d)$ <strong>is</strong> ambiguous. At school I learned that this should be
$$
\frac{a}{b}(c+d),
$$
but even in research papers somebody could read
$$
\frac{a}{b(c+d)}.
$$</p>
|
164,152 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/33215/what-is-48293">What is 48÷2(9+3)?</a> </p>
</blockquote>
<p>Please, look at the picture?</p>
<p><img src="https://i.stack.imgur.com/3Tauh.jpg" alt="http://s16.radikal.ru/i190/1206/3b/7411739c6d9a.jpg"></p>
<p>Which calculator shows true result and how to prove achieved result?</p>
| Aang | 33,989 | <p>when two operations have same precedence, then the operations are done from left to right.So here, division is done first and then multiplication.This is just a convention.Without this, this expression is ambiguous.</p>
|
2,197,790 | <h3>Question</h3>
<blockquote>
<p>A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for
each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies
that $|a_n − a_m| <\epsilon$.</p>
<p>Prove that every convergent sequence is a Cauchy sequence</p>
</blockquote>
<hr>
<h3>Attempt</h3>
<p>This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$? </p>
<p>Attempt:</p>
<p>WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a_m| < \epsilon$</p>
<p>Let $\epsilon > 0$ be arbitrary</p>
<p>Choose N such that for $n > N$ we have $|a_n - a_m| < \epsilon$</p>
<p>Suppose $n > N$, then </p>
<p>??</p>
<p>Could someone point me to the right direction? Thx.</p>
| Yes | 155,328 | <p>The sequence $(a_{n})$ is convergent by assumption; let $l := \lim_{n}a_{n}$. Let $\varepsilon > 0$. Then there is some $N$ such that $|a_{n} - l| < \varepsilon/2$ for all $n \geq N$. Note that
$$
|a_{n} - a_{m}| \leq |a_{n}-l| + |l-a_{m}| < \varepsilon/2 + \varepsilon/2 = \varepsilon
$$
for all $n,m \geq N$.</p>
<p>The major tool here is the triangle inequality.</p>
|
25,260 | <h2>TL;DR:</h2>
<hr />
<p>Tell me which topics should i study the most, based on this three tests:</p>
<p>Mathematics (A):
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2020_ga_math_a.pdf" rel="nofollow noreferrer">2020</a>
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2019_ga_math_a.pdf" rel="nofollow noreferrer">2019</a>
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2018_ga_math_a.pdf" rel="nofollow noreferrer">2018</a></p>
<p>This question may sound a bit weird, since the natural answer would be "study whats already is on the test" but i'm wanna share a bit of context for this.</p>
<h4>Context:</h4>
<hr />
<p>So i'm currently studying for the Japanese Scholarship known as MEXT in which every year the japanese embassy of each country gave the opportunity to enroll on a japanese university and go to study in that country. As an undergraduate student this is a big deal specially if there's a lack of opportunities in your current country.</p>
<p>I also want you to know that here in my country, there is a lack of proper mathematical education when it comes to high school, pretty much anyone who obtained a bachelor degree's are not well prepared to face one of this japanese test designed for undergraduate students since the test, present to you calculus, trigonometry, modular-arithmetic, number-theory, etc. problems which you don't see until you reach third semester of any university engineer-based career (at least here where i live)</p>
<p>Finally i want to tell that i've been studying calculus for a bit and i do know the basics and i don't have any problem with arithmetic or algebra.</p>
<h2>So here is the question:</h2>
<hr />
<p>Could somebody look at those test and tell me which topics should i study the most? Below (in the comment section) you will find the three test that were used in past examinations and contains all the questions, i also will leave a link to the answers in case someone want to deep onto this:</p>
<h4>Current approach and status:</h4>
<hr />
<p>In my current approach i tried solving most of the exercises presented and always fall in the same loop of no knowing, research, try solving, fail, research again, try solving again, failing, and finally making a question here. This of course may not be the best for a huge variety of topic since you have to make sure to <strong>understand</strong> what you are doing before going into any kind of examination but with so many themes its hard to pick one thing to study since you're pretty much dropping everything else in order to solve only one thing.</p>
<p><a href="https://i.stack.imgur.com/MhGqK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MhGqK.png" alt="flow chart" /></a></p>
<h3>The ideal:</h3>
<hr />
<p>When making this post i'm hoping someone could orient me through a path of things i should know and preferably master in order to be well prepared for the test.</p>
| James S. Cook | 128 | <p>It is discussed in all the introductory DEqns texts of which I've used. It's needed to complete the discussion of linear independence of solution sets. Together with Abel's formula it provides some rather general theorems for linear n-th order ODEs. The Wronskian lies at the heart of variation of parameters which is the general method to solve nonhomogeneous n-th order ODEs.</p>
<p>It's also a bit tricky since the theorems which hold for solution sets do not hold for arbitrary sets of functions. In particular, the Wronksian can vanish identically for linearly independent functions. But, that can never happen for a solution set thanks to Abel's formula.</p>
<p>Incidentally, all these things generalize to calculus over a finite dimensional commutative algebra.</p>
|
670,522 | <p>In my very young mathematical career, I have worked a lot with modular forms. Recently, I worked as a teaching assistant in a course about geometry. At the end of the course, we dealt with hyperbolic geometry. It seems as if there is some relation between hyperbolic geometry and modular forms, for example, why is it precisely the set $\mathbb{H}$ (from which modular forms map into $\mathbb{C}$) that is also a model for a "weird" geometry in which the sum over the angles in a triangle is not $\pi$ or in which some axiom about parallel lines does not hold? It seems at first sight, as if these two mathematical areas are quite distant from each other.</p>
<p>If there is such a relation, can someone solve the following equation:</p>
<p>$$ \frac{\text{modular forms}}{\text{hyperbolic geometry}} = \frac{???}{\text{euclidean geometry}}$$</p>
<p>Of course, one can reinterpret modular forms as certain sections of line bundles over ... blah blah blah, but this is not the way you would ever describe what a modular form is to someone who has never heard about them.</p>
<p>cheers,</p>
<p>FW</p>
| Brian Fitzpatrick | 56,960 | <p>I think there's a much easier way to go about this. Let $W$ be a proper subspace of $V$. Then $W$ has a basis $\{w_1,\dotsc,w_k\}$. A standard theorem of linear algebra says this basis for $W$ can be extended to a basis $\{w_1,\dotsc,w_k,v_1,\dotsc,v_n\}$ for $V$. Now, let $f:W\rightarrow U$ be a linear map. Then let $\overline{f}(w_i)=f(w_i)$ for $1\leq i\leq k$ and let $\overline{f}(v_i)=\mathbf0$ for $1\leq i\leq n$. Extending linearly gives an extension $\overline{f}:V\rightarrow U$ of $f$ to $V$.</p>
|
291,957 | <p>Does there exist a simple expression for integrals of the form,</p>
<p>$I = \int_{-\infty}^0 H_n(u) H_m(u)\, \mathrm{e}^{-u^2}\,du$,</p>
<p>where $m$ and $n$ are nonnegative integers and $H_n$ is the $n$'th (physicists') Hermite polynomial?</p>
<p>When $n+m$ is even, the symmetry of the integrand and the orthogonality of $H_n$ imply,</p>
<p>$I = \sqrt{\pi} \,2^{n-1} n! \,\delta_{n,\,m}$ (for $n+m$ even).</p>
<p>For $n+m$ odd, $I$ is nonzero and increases in magnitude with $n+m$, but I have been unable to find a general formula.</p>
| Robert Israel | 8,508 | <p>It looks to me like we have exponential generating functions</p>
<p>$$\sum_{n=0}^\infty I(n,n+2k+1) t^n/n! = \dfrac{(-1)^{k+1}(2k)!}{k! (1-2t)^{k+3/2} (1+2t)^{k+1/2}}$$</p>
<p>EDIT: Hmm, these can be combined into a bivariate exponential generating function</p>
<p>$$ \sum_{n=0}^\infty \sum_{k=0}^\infty I(n,n+2k+1) \frac{s^k t^n}{k! n!}
= \frac{1}{(-1+2t) \sqrt{1+4s-4t^2}}$$</p>
|
291,957 | <p>Does there exist a simple expression for integrals of the form,</p>
<p>$I = \int_{-\infty}^0 H_n(u) H_m(u)\, \mathrm{e}^{-u^2}\,du$,</p>
<p>where $m$ and $n$ are nonnegative integers and $H_n$ is the $n$'th (physicists') Hermite polynomial?</p>
<p>When $n+m$ is even, the symmetry of the integrand and the orthogonality of $H_n$ imply,</p>
<p>$I = \sqrt{\pi} \,2^{n-1} n! \,\delta_{n,\,m}$ (for $n+m$ even).</p>
<p>For $n+m$ odd, $I$ is nonzero and increases in magnitude with $n+m$, but I have been unable to find a general formula.</p>
| user132949 | 132,949 | <p>This question is getting a little old now, but I feel I can add something here, for my own conscience, if nothing else.</p>
<p>My take on this problem is - simply define $u = \sqrt{v}$ in the integral and take advantage of the relation between the Hermite and Laguerre polynomials. i.e.</p>
<p>$$
H_{2n}(\sqrt{v}) = (-1)^n 2^{2n}n! L^{\left(-\frac{1}{2}\right)}_n(v)\\
H_{2n+1}(\sqrt{v}) = (-1)^n 2^{2n+1}n! \sqrt{v} L^{\left(\frac{1}{2}\right)}_n(v)
$$</p>
<p>So, assuming that $m$ and $n$ are both even we get:
\begin{eqnarray}
\int^0_{-\infty}du e^{-u^2} H_n(u) H_m(u)= (-1)^{n+m}\int^{\infty}_0 dx e^{-x^2} H_n(x) H_m(x) \;\; \textrm{changing order of integration and changing variables $u \rightarrow -x$}\\
= \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} H_n(\sqrt{v}) H_m(\sqrt{v}) \;\; \textrm{changing variables $x = \sqrt{v}$}\\
= \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} (-1)^{\frac{n}{2}} 2^{n}\left(\frac{n}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)(-1)^{\frac{m}{2}} 2^{m}\left(\frac{m}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{applying the above relation between Hermite and Laguerre polys. - the even case}\\
=\frac{(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n}}{2} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)!\int^{\infty}_0 dv v^{-\frac{1}{2}} e^{-v}L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{decluttering algebra}\\
=(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n-1} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)! \frac{\Gamma\left( \frac{n+1}{2}\right)}{\left(\frac{n}{2}\right)!}\delta_{n,m} \;\; \textrm{applying Laguerre orthogonality}\\
=(-1)^{n}2^{2n-1} \left(\frac{n}{2}\right)! \Gamma\left( \frac{n+1}{2}\right)\delta_{n,m}
\end{eqnarray}</p>
<p>Since $n$ is even the expression is positive. Equivalent expressions exist for all other odd/even combinations of $m$ and $n$. I have probably screwed up the algebra in places, but I think the general argument is valid.</p>
|
2,359,700 | <p>Given the vector space, $ C(-\infty,\infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{a^x\mid a \ge 1 \}$, a subspace of the given vector space?</p>
<p>As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it). </p>
<p>My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-\infty,\infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).</p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>$$f (x,0)=0$$
$$f (x,x^2)=\frac {1}{2} $$</p>
|
3,099,815 | <p>I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone has an approach example to post I would be very grateful:</p>
<p>"Consider a relation <span class="math-container">$R$</span> defined on the set of integers. Determine for the following if the relation is reflexive, symmetric, and transitive: <span class="math-container">$R = \{(x, y)|x = 2y \}.$</span>"</p>
| Fred | 380,717 | <ol>
<li><p>is it true that for all <span class="math-container">$x \in \mathbb Z$</span> we have <span class="math-container">$x=2x$</span> ? If yes, then <span class="math-container">$R$</span> is reflexive, if no, then <span class="math-container">$R$</span> is not reflexive.</p></li>
<li><p>suppose that <span class="math-container">$x,y \in \mathbb Z$</span> and <span class="math-container">$x=2y$</span>. Does it always follow that <span class="math-container">$y=2x$</span> ? If yes, then <span class="math-container">$R$</span> is symmetric, if no, then <span class="math-container">$R$</span> is not symmetric.</p></li>
<li><p>suppose that <span class="math-container">$x,y,z \in \mathbb Z$</span> and <span class="math-container">$x=2y$</span> and <span class="math-container">$y=2z.$</span> Does it always follow that <span class="math-container">$x=2z$</span> ? If yes, then <span class="math-container">$R$</span> is transitive, if no, then <span class="math-container">$R$</span> is not transitive.</p></li>
</ol>
|
42,301 | <p>everyone! I am sorry, but I am an abcolute novice of Mathematica (to be more precise this is my first day of using it) and even after surfing the web and all documents I am not able to solve the following system: </p>
<pre><code>Solve[{y*(((y*x)/(beta*b))^(1/(beta - 1)) - v) - c*alpha ==
0, ((x/alpha))*(((y*x)/(alpha*beta*b))^(1/(beta - 1)) -
v) + (((y*x)/alpha) -
2*alpha*((yx)/(2*beta*b))^(1/(beta - 1)))*(1/(beta -
1))*(x/(alpha*beta*
b))*((y*x)/(alpha*beta*b))^((2 - beta)/(1 - beta)) == 0}, {x,
y}]
</code></pre>
<p>What I need is to solve following systems, getting x and y expressed through all these symbols. Is it even possible? Thank you in advance. </p>
| halirutan | 187 | <p>How about using <code>SortBy</code> to sort your list by the last element and then take the first entry?</p>
<pre><code>First[SortBy[mya, Last]]
(* {0, 2, 5} *)
</code></pre>
<p>A simple iterative approach to go through your list exactly once and remember the minimum element can be written as</p>
<pre><code>Block[{min = First[mya]},
Do[If[Last[min] > Last[elm], min = elm], {elm, Rest[mya]}];
min
]
</code></pre>
<p>Although my tests showed that this is a bit slower (about 2 seconds for 10^7 elements) as the first approach.</p>
<p>An faster approach then the two above is to first extract the minimum of all z-values and then go through the list until you hit the first match</p>
<pre><code>Block[{min = Min[Last[Transpose[mya]]]},
Do[If[Last[elm] === min, Return[elm]], {elm,mya}]
]
</code></pre>
|
42,301 | <p>everyone! I am sorry, but I am an abcolute novice of Mathematica (to be more precise this is my first day of using it) and even after surfing the web and all documents I am not able to solve the following system: </p>
<pre><code>Solve[{y*(((y*x)/(beta*b))^(1/(beta - 1)) - v) - c*alpha ==
0, ((x/alpha))*(((y*x)/(alpha*beta*b))^(1/(beta - 1)) -
v) + (((y*x)/alpha) -
2*alpha*((yx)/(2*beta*b))^(1/(beta - 1)))*(1/(beta -
1))*(x/(alpha*beta*
b))*((y*x)/(alpha*beta*b))^((2 - beta)/(1 - beta)) == 0}, {x,
y}]
</code></pre>
<p>What I need is to solve following systems, getting x and y expressed through all these symbols. Is it even possible? Thank you in advance. </p>
| Yi Wang | 7,253 | <p>I think @halirutan's answer is quite nice and clean. Nevertheless just give an alternative one:</p>
<pre><code>findLastMin[mat_] := Cases[mat, {__, Min@mat[[All, -1]]}]
findLastMin[mya]
</code></pre>
<blockquote>
<p>{{0, 2, 5}}</p>
</blockquote>
<p>There is additional {...} outside the desired output by the OP, because if there are multiple equal minimal values, it returns them all. </p>
|
212,949 | <p>A simple question:</p>
<p>I have this equation:</p>
<pre><code>eq1=Derivative[0, 1][T1][x, t] - Derivative[1, 0][T0][x, t]^2 -
T0[x, t]*Derivative[2, 0][T0][x, t] - Derivative[2, 0][T1][x, t] == 0;
</code></pre>
<p>I want only to select terms that contain T0 or its derivatives only, that is:</p>
<pre><code>-Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]
</code></pre>
<p>Thanks in anticipation.</p>
| NonDairyNeutrino | 46,490 | <pre><code>Block[{T1, Equal = Plus}, SetAttributes[T1, Constant]; eq1]
</code></pre>
<blockquote>
<pre><code>-Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]
</code></pre>
</blockquote>
<p>Or (thanks to Mr. Wizard <a href="https://mathematica.stackexchange.com/a/75295/46490">here</a>)</p>
<pre><code>eq1 /. {s_Symbol /; StringMatchQ[SymbolName[Unevaluated@s], "T" ~~ Except["0"]] -> 0, Equal -> Plus}
</code></pre>
<blockquote>
<pre><code>-Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]
</code></pre>
</blockquote>
|
212,949 | <p>A simple question:</p>
<p>I have this equation:</p>
<pre><code>eq1=Derivative[0, 1][T1][x, t] - Derivative[1, 0][T0][x, t]^2 -
T0[x, t]*Derivative[2, 0][T0][x, t] - Derivative[2, 0][T1][x, t] == 0;
</code></pre>
<p>I want only to select terms that contain T0 or its derivatives only, that is:</p>
<pre><code>-Derivative[1, 0][T0][x, t]^2 - T0[x, t]*Derivative[2, 0][T0][x, t]
</code></pre>
<p>Thanks in anticipation.</p>
| Michael E2 | 4,999 | <p>While the structural operation</p>
<pre><code>DeleteCases[eq1 /. Equal -> Subtract, _?(FreeQ[#,T0]&)]
</code></pre>
<p>works, I prefer using an algebraic approach on a algebraic problem.</p>
<pre><code>vars = Select[Not@*FreeQ[T0]]@Variables[eq1 /. Equal -> Subtract];
coeffs = CoefficientArrays[eq1 /. Equal -> Subtract, vars];
Fold[#2 + #1.vars &, Reverse@ReplacePart[coeffs, 1 -> 0]]
</code></pre>
<p><img src="https://i.stack.imgur.com/NSfIo.png" alt="Mathematica graphics"></p>
<p>The structural approach relies on the equation being in a particular form, a flat sum of terms, which does not always happend. The algebraic approach does not. However, <code>CoefficientArrays</code> does rely on it being a polynomial in the variables <code>vars</code>.</p>
|
18,686 | <p>Suppose you have an arbitrary triangle with vertices $A$, $B$, and $C$. <a href="http://www.cs.princeton.edu/~funk/tog02.pdf">This paper (section 4.2)</a> says that you can generate a random point, $P$, uniformly from within triangle $ABC$ by the following convex combination of the vertices:</p>
<p>$P = (1 - \sqrt{r_1}) A + (\sqrt{r_1} (1 - r_2)) B + (r_2 \sqrt{r_1}) C$</p>
<p>where $r_1, r_2 \sim U[0, 1]$.</p>
<p>How do you prove that the sampled points are uniformly distributed within triangle $ABC$?</p>
| mercio | 17,445 | <p>Pick $A,B,C = (0,0),(1,0),(1,1)$.
For any point $(x,y)$, we have that $(x,y)$ is in the triangle if and only if $0 < x < 1$ and $0 < y/x < 1$.</p>
<p>Now, we look for the distribution of $x$ and $y/x$. </p>
<p>Computing a few triangle areas, we can easily check that $P(0 < x < x_0) = x_0^2$.
Hence $P(0 < x^2 < a) = P(0 < x < \sqrt a) = a$, so that $x^2$ is uniformly distributed in the unit interval. </p>
<p>Again with an area computations, we can check that $P(0 < y/x < k) = k$. Hence $y/x$ is also uniformly distributed in the unit interval.</p>
<p>Finally we have to check (again computing a simple area) that $P(0 < x < x_0 \land 0 < y/x < k) = x_0^2k$ which proves that $x^2$ and $y/x$ are independant.</p>
<p>So we have found that to generate a point uniformly in the triangle is the same as picking $x^2 = r_1$ and $y/x = r_2$ uniformly in the unit interval, and then form $(x,y) = (\sqrt r_1, r_2 \sqrt r_1)$, which is the barycenter of $(A,1- \sqrt {r_1})(B,\sqrt {r_1}(1-r_2))(C,\sqrt{r_1}r_2)$. </p>
|
1,942,578 | <p>Consider the following wedge</p>
<p><a href="https://i.stack.imgur.com/xiaPX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xiaPX.png" alt=""></a> cut from a cylinder of radius r.
The plane that cuts the wedge goes through the very bottom of the cylinder leading to an ellipse as the cross section of the wedge.</p>
<p>The long axis of the ellipse is $2R$ in length. </p>
<p>How can I prove that the minor axis of the ellipse is $2r$ in length where $r$ is the radius of the cylinder?</p>
| user296113 | 296,113 | <p>First, notice that the functions $f_n$ are not defined on $0$ so there is no sense to say that the uniform convergence is actually on all $\Bbb R$. Secondly the point-wise convergence of $f_n$ is the null function on $(0,1]$ and we have</p>
<p>$$\Vert f_n\Vert_\infty=\sup_{x\in(0,1]}\vert f_n(x)\vert\ge f_n\left(\frac1n\right)=1\not\xrightarrow{n\to\infty}0$$
so the convergence is not uniform on $(0,1]$.</p>
|
137,414 | <p>I am trying to evaluate the following expression numerically
$$\frac{d^2}{dt^2}e^{-2t^2}\int_0^\infty\frac{\xi/\sqrt{2}}{\xi^{3/2}}e^{(-\xi^2/2-2\xi t))}$$</p>
<p>My code is as follows</p>
<pre><code>f[t_]:=Exp[-2*t^2]*NIntegrate[Erf[\[Xi]/Sqrt[2]]/\[Xi]^(3/2)*Exp[-(\[Xi]^2/2)-2*\[Xi]*t],{\[Xi],0,Infinity}]
Der[t_] := f''[t]
</code></pre>
<p>But when I evaluate, say by entering </p>
<pre><code>Der[5]
</code></pre>
<p>I am getting an error that the integrand has evaluated to non-numerical values for all sampling points in the region. Now I know that this is a common error, and other threads have revealed that this is happening because Mathematica is not recognizing something in my expression (likely $\xi$) as a numerical variable. However, I am not sure how I would go about fixing this. I have tried putting in ?NumericQ beside all instances of $\xi$ but that didn't seem to work.</p>
<p>How would I go about numerically evaluating this expression (and ideally, plotting it as a function of t) without the NIntegrate::inumr error?</p>
<p>Thank you in advance!</p>
| Dr. Wolfgang Hintze | 16,361 | <p>Notice: my formula contains a sign error (-t instaed of +t). But the derivation is still valid. Only the final numeric value must be taken at t = -5.</p>
<p>The integral can be written as</p>
<pre><code>f[t_] :=
NIntegrate[\[Xi]^(-3/2 )
Erf[\[Xi]/Sqrt[2]] Exp[-(1/2) (\[Xi] - 2 t)^2], {\[Xi],
0, \[Infinity]}]
</code></pre>
<p>Let us form the second derivative with respect to <code>t</code> under the integral:</p>
<pre><code>D[Exp[-1/2 (\[Xi] - 2 t)^2], {t, 2}]
(* Out[22]= -4 E^(-(1/2) (-2 t + \[Xi])^2) +
4 E^(-(1/2) (-2 t + \[Xi])^2) (-2 t + \[Xi])^2 *)
</code></pre>
<p>and call <code>f''[t] = f2[t]</code> which is</p>
<pre><code>f2[t_] :=
NIntegrate[\[Xi]^(-3/2 )
Erf[\[Xi]/Sqrt[2]] 4 E^(-(1/
2) (-2 t + \[Xi])^2) (-1 + (-2 t + \[Xi])^2), {\[Xi],
0, \[Infinity]}]
</code></pre>
<p>The value requested is then</p>
<pre><code>f2[5]
(* Out[24]= 0.0129202 *)
</code></pre>
<p>This is very different from Bob Hanlon's result which is, in fact</p>
<pre><code>f2[-5]
(* Out[30]= 3.43309*10^-20 *)
</code></pre>
|
3,966 | <p>This type of answer is what I'm looking for:</p>
<pre><code>In[58]:= ArcTan @ 1
Out[58]= π/4
</code></pre>
<p>This is what mathematica gives me:</p>
<pre><code>In[59]:= ArcTan@2
Out[59]= ArcTan[2]
</code></pre>
<p>Is it possible to express <code>ArcTan</code> in terms of $\pi$? I understand some fractions would be hairy.</p>
<p>I am using <code>Mathematica 8</code>.</p>
| Artes | 184 | <p>Maybe this </p>
<pre><code>HoldForm[Pi] (1/Pi ArcTan@2.)
</code></pre>
<p><img src="https://i.stack.imgur.com/6PKA9.gif" alt="enter image description here"></p>
<p>or if you want a nicer way </p>
<pre><code>Rationalize /@ (HoldForm[Pi] N@(1/Pi ArcTan@Range[5]))
</code></pre>
<p><img src="https://i.stack.imgur.com/xgSEx.gif" alt="enter image description here"></p>
<p><strong>Edit</strong></p>
<p>The latter method works well in cases when there is a rational fraction of $\pi$ :</p>
<pre><code>Rationalize /@ (HoldForm[Pi] N @ (1/Pi ArcTan @ { Sqrt[1 - 2/Sqrt[5]], 2 - Sqrt[3],
1/Sqrt[3], Sqrt[3], 1}))
</code></pre>
<p><img src="https://i.stack.imgur.com/NM3v6.gif" alt="enter image description here">
<img src="https://i.stack.imgur.com/lTpfg.gif" alt="enter image description here"></p>
<p>To sum up : Mathematica does what it should do, namely <code>ArcTan[2]</code> is not a rational fraction of $\pi$ and that's why it returns <code>ArcTan[2]</code> unlike in case <code>ArcTan[1]</code>. The above method is to express <code>ArcTan[x]</code> in terms of a real multiple of $\pi$. </p>
<p>If you want to get back what you have evaluated you shoud use <code>ReleaseHold</code>, e.g. </p>
<pre><code>Tan @ ReleaseHold @ %
</code></pre>
<p><img src="https://i.stack.imgur.com/DmHNI.gif" alt="enter image description here"></p>
|
2,169,845 | <p>Basically it says given that s is a root of this polynomial: <span class="math-container">$(\sqrt3-\sqrt2)x^3 + \sqrt2x -\sqrt3 + 1$</span>, find another polynomial with integer coefficients that has the same root s as well.
I'm super stuck and am unsure on how to approach this problem.
I attempted to square some stuff but it didn't really work out.</p>
| zoli | 203,663 | <p>You can calculate the appropriate conditional density directly</p>
<p>$$f_{X\mid X>Y}(x)=\frac{dF_{X\mid X>Y}(x)}{d x}=$$
$$=\lim_{\Delta x\to 0}\frac{F_{X\mid X>Y}(x+\Delta x)-F_{X\mid X>Y}(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{P(X<x+\Delta x\mid X>Y)-P(X<x\mid X>Y)}{\Delta x}=$$
$$=\lim_{\Delta x\to 0}\frac{P(x<X<x+\Delta x\mid X>Y)}{\Delta x}=$$
$$=\frac1{P(X>Y)}\lim_{\Delta x\to 0}\frac{P(x<X<x+\Delta x\cap X>Y)}{\Delta x}.$$</p>
<p>Now,</p>
<p>$$P(x<X<x+\Delta x\cap X>Y)=\int_{x}^{x+\Delta x}P(u>Y)\ f_X(u)\ du$$
$$=\int_{x}^{x+\Delta x}F_Y(u)\ f_X(u)\ du.$$</p>
<p>Then</p>
<p>$$\lim_{\Delta x\to 0}\frac{\int_{x}^{x+\Delta x}F_Y(u)\ f_X(u)\ du}{\Delta x}=$$
$$=\frac {d}{dx}\int_{-\infty}^x F_Y(u)\ f_X(u)\ du =f_X(x)F_Y(x).$$</p>
<p>That is,
$$f_{X\mid X>Y}(x)=\frac{f_X(x)F_Y(x)}{P(X>Y)}=\frac{f_X(x)F_X(x)}{P(X>Y)}$$</p>
<p>because the two random variables has a common distribution.</p>
<p>Then, the conditional expectation is</p>
<p>$$E[X\mid X>Y]=2\int_{-\infty}^{\infty}xf_X(x)F_X(x)\ dx$$</p>
<p>because $P(X>Y)=\frac12$ .</p>
<p>So,
$$E[X\mid X>Y]=2\int_{-\infty}^{\infty}xf_X(x)F_Y(x)\ dx=\frac1{\pi}\int_{-\infty}^{\infty}xe^{-\frac{x^2}2}\int_{-\infty}^xe^{-\frac{u^2}2} \ du\ dx.$$</p>
|
2,169,845 | <p>Basically it says given that s is a root of this polynomial: <span class="math-container">$(\sqrt3-\sqrt2)x^3 + \sqrt2x -\sqrt3 + 1$</span>, find another polynomial with integer coefficients that has the same root s as well.
I'm super stuck and am unsure on how to approach this problem.
I attempted to square some stuff but it didn't really work out.</p>
| Gordon | 169,372 | <p>Note that
\begin{align*}
E(X \mid X > Y) = \frac{E(X\mathbb{1}_{X>Y})}{P(X>Y)}= 2E(X\mathbb{1}_{X>Y}).
\end{align*}
Moreover,
\begin{align*}
E(X\mathbb{1}_{X>Y}) &= E\big(E(X\mathbb{1}_{X>Y} \mid X) \big)\\
&=E(X\Phi(X))\\
&=\int_{-\infty}^{\infty}x\Phi(x)\phi(x) dx,
\end{align*}
where $\Phi$ is the cumulative distribution of a standard normal and $\phi$ is the density.</p>
|
4,570,329 | <p>In the textbook that I am working through, it is left as an exercise to prove the following claim</p>
<blockquote>
<p>Consider two linear maps <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> from <span class="math-container">$\mathbb{R}^n$</span> to <span class="math-container">$\mathbb{R}^m$</span> (for <span class="math-container">$m \le n$</span>) such that <span class="math-container">$P$</span> is an injective linear map. If we assume that the first <span class="math-container">$m$</span> columns of P are linearly independent, then we need to show that <span class="math-container">$\exists \delta >0$</span> such that <span class="math-container">$$ \lvert \lvert Q-P \rvert \rvert < \delta \implies rank(Q) = m$$</span> where our choice of norm is the operator norm.</p>
</blockquote>
<p>This exercise is supposed to be equivalent to showing that the set of matrices in <span class="math-container">$\mathbb{R}^{mn}$</span> that have full rank form an open subset of <span class="math-container">$\mathbb{R}^{mn}$</span></p>
<p>As someone that doesn't have a math background, I find these types (epsilon-delta) of proofs particularly challenging, as I often find them difficult to build up an intuition for. As there are no solutions to the exercises, I would be grateful for any guidance here on how to prove the statement.</p>
| belkacem abderrahmane | 660,639 | <p>Second proof
Remark that my first proof is more general in that it shows that the set of right invertible linear continous maps(linear maps <span class="math-container">$P:X\mapsto Y$</span> such that there's Some continous <span class="math-container">$T:Y\to X$</span> such that <span class="math-container">$PT=I$</span>) between two normed spaces is open. <span class="math-container">$$ $$</span>
<span class="math-container">$P$</span> has <span class="math-container">$m$</span> linearly independent columns iff there's some minor (detrrminent of a square m×m sub-matrix) is not zero, Since the determinznt (any minor) is a continuous map(polynomial in the entries), there's a <span class="math-container">$\delta$</span> such that if <span class="math-container">$||P-Q||<\delta $</span>, <span class="math-container">$Q$</span> has a non-vanishing <span class="math-container">$m$</span>-minor, i.e <span class="math-container">$Q$</span> has <span class="math-container">$m$</span> linearly independent columns, equivelently <span class="math-container">$Rank(Q) =m$</span></p>
|
173,466 | <p>For a given matrix <code>M[n]</code> of size $ n\times n $ I want to define the following list of matrix-expressions:</p>
<pre><code>n=1
{Tr[M[1]]}
n=2
{Tr[M[2]]^2,Tr[M[2].M[2]]}
n=3
{Tr[M[3]]^3,Tr[M[3]]Tr[M[3].M[3]],Tr[M[3].M[3].M[3]]}
</code></pre>
<p>How could I generalize this relation for arbitrary $ n $? I tried <code>Nest</code> and <code>NestList</code>.
Thanks!</p>
| kglr | 125 | <pre><code>f[k_] := Table[Tr[M @ k]^(k - i) Array[M @ k &, i, 1, Tr @ Dot @ ## &], {i, k}]
f /@ Range[5] // Column // TeXForm
</code></pre>
<blockquote>
<p>$\tiny\begin{array}{l}
\{\text{Tr}[M(1)]\} \\
\left\{\text{Tr}[M(2)]^2,\text{Tr}[M(2).M(2)]\right\} \\
\left\{\text{Tr}[M(3)]^3,\text{Tr}[M(3).M(3)] \text{Tr}[M(3)],\text{Tr}[M(3).M(3).M(3)]\right\} \\
\left\{\text{Tr}[M(4)]^4,\text{Tr}[M(4).M(4)] \text{Tr}[M(4)]^2,\text{Tr}[M(4).M(4).M(4)] \text{Tr}[M(4)],\text{Tr}[M(4).M(4).M(4).M(4)]\right\}
\\
\left\{\text{Tr}[M(5)]^5,\text{Tr}[M(5).M(5)] \text{Tr}[M(5)]^3,\text{Tr}[M(5).M(5).M(5)] \text{Tr}[M(5)]^2,\text{Tr}[M(5).M(5).M(5).M(5)]
\text{Tr}[M(5)],\text{Tr}[M(5).M(5).M(5).M(5).M(5)]\right\} \\
\end{array}$</p>
</blockquote>
|
188,938 | <p>Hyperbolic "trig" functions such as $\sinh$, $\cosh$, have close analogies with regular trig functions such as $\sin$ and $\cos$. Yet the hyperbolic versions seem to be encountered relatively rarely. (My frame of reference is that of someone with college freshman/sophomore, but not advanced math.)</p>
<p>Why is that? Is it because the hyperbolic versions of these functions are less common/useful than the circular versions? </p>
<p>Can you do the "usual" applications (Taylor series, Fourier series) with hyperbolic functions as you can with trigonometric?</p>
<p>I'm not a professional mathematician. I've had three semesters of calculus and one of linear algebra/differential equations, and "barely" know about hyperbolic functions. The question is with that frame of reference. </p>
| Fly by Night | 38,495 | <p>Well, yes, I would say so. After all, sine and cosine describe everything from lengths of sides of right-angled triangle to angular momentum. Rotation is so fundamental that we're bound to see sine and cosine very often. Of course $\sin$ and $\cos$ are just the opposite side of the exponential coin from $\sinh$ and $\cosh$ since</p>
<p>$$ \cosh x = \frac{1}{2}(e^{x} + e^{-x}) \, , $$
$$ \sinh x = \frac{1}{2}(e^{x} - e^{-x}) \, , $$
$$ \cos x = \frac{1}{2}(e^{ix} + e^{-ix}) \, , $$
$$ \sin x = \frac{1}{2i}(e^{ix} - e^{-ix}) \, . $$</p>
|
188,938 | <p>Hyperbolic "trig" functions such as $\sinh$, $\cosh$, have close analogies with regular trig functions such as $\sin$ and $\cos$. Yet the hyperbolic versions seem to be encountered relatively rarely. (My frame of reference is that of someone with college freshman/sophomore, but not advanced math.)</p>
<p>Why is that? Is it because the hyperbolic versions of these functions are less common/useful than the circular versions? </p>
<p>Can you do the "usual" applications (Taylor series, Fourier series) with hyperbolic functions as you can with trigonometric?</p>
<p>I'm not a professional mathematician. I've had three semesters of calculus and one of linear algebra/differential equations, and "barely" know about hyperbolic functions. The question is with that frame of reference. </p>
| Doc | 86,414 | <p>Tom,</p>
<p>Anon's answer is a good one, with one small excepton (or perhaps one wide misconception). What people commonly refer to as the complex plane is an unfortunate misnomer. It should be called the "complex number plane". By calling it the complex plane (which, by the way, is a quite widespread phenomenon), one is led to think of it as being C^2. It is not. It is actually the real plane R^2 used as a device to represent the complex numbers C (= C^1). Thus I would maintain (in support of Anon) that the best way to understand the hyperbolic sine and cosine functions is via restriction of the analytic continuation of the real sin and cosine functions to the imaginary number line of the complex number plane. In a sense. the analytic continuation of the real exponential function does both for the price of one. </p>
<p>One more thing, regarding your (over)reaction to Edgar's comment. Being a trained mathematician is <em>vastly</em> beyond taking courses in which you would first encounter hyperbolic trig functions. If you are a major in the hard sciences or in engineering, you will be exposed to them in the near future. I think that was Edgar's intended message, nothing more. </p>
|
230,204 | <p>Let $X$ be a compact, oriented Riemann manifold. Let $\pi_{P}: P \rightarrow X$ be a principal $G$-bundle over $X$, for a compact Lie group $G$. Let $(M, \omega)$ be a symplectic manifold endowed with a symplectic action of $G$. Denote by $\mathcal{N}:=C^{\infty}(P,M)^{U(1)}$ the space of smooth $G$-equivariant maps $u:P \rightarrow M$. Then $C^{\infty}(P,M)^{G}$ is a smooth Frechet manifold. The total space of the tangent bundle $T\mathcal{N} = C^{\infty}(P,TM)^{G}$. At a point $u \in \mathcal{N}$, $T_{u}\mathcal{N} = \Gamma(P, u^{\ast}TM)^{G}$.</p>
<p>For $\xi_{1}, \xi_{2} \in T_{u}\mathcal{N}$, define $\Omega(\xi_{1}, \xi_{2}) = \displaystyle \int_{X} \omega_{u}(\xi_{1}, \xi_{2}) ~ dvol_{\scriptscriptstyle X}$, where $\omega_{u}(\cdot, \cdot)$ denotes the restric tion of $\omega$ along $u$. Will $\Omega(\cdot, \cdot)$ be a symplectic form on $\mathcal{N}$? More precisely, is $\Omega(\cdot, \cdot)$ closed?</p>
| Tobias Diez | 17,047 | <p>Yes, the form $\Omega$ is closed and defines indeed a weak symplectic structure. This can be verified by a direct but a bit messy calculation. A cleaner way would be to generalize the ideas of <a href="http://arxiv.org/abs/1111.3889" rel="nofollow">Vizman: Induced differential forms on manifolds of functions</a> to the case of sections of non-trivial bundles.</p>
<p>In case you wonder what the momentum maps are:</p>
<ul>
<li><p>For the gauge group of $P$ the momentum map $\Gamma^\infty(P \times_G M) \to \Gamma(P \times_G \mathfrak{g}^*)$ is the composition with the momentum map $J_M: M \to \mathfrak{g}^*$ of the $G$-action on $M$. </p></li>
<li><p>Assume that the symplectic form on $M$ is exact, $\omega = d \theta$. Then the action of the group of volume-preserving diffeomorphisms on $X$ has a momentum map $\Gamma^\infty(P \times_G M) \to \Omega^1(X) / d \Omega^0(X)$ which is essentially the pullback of $\alpha$ by a section, twisted by the above momentum map for the gauge group. For more details see <a href="http://intlpress.com/site/pub/files/_fulltext/journals/sdg/2003/0008/0001/SDG-2003-0008-0001-a006.pdf" rel="nofollow">Donaldson: Moment maps in differential geometry</a>.</p></li>
</ul>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.