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 |
|---|---|---|---|---|
47,890 | <p>hey so i'm programming something that finds an angle of a line between 0 and 180 degrees based on two points.... </p>
<p>the equation to find the answer is <code>Angle = sin-1((1/Hypotenuse)*B)</code>where B is the vertical side of the triangle formed and the hypotenuse is the distance between point 1 and 2. </p>
<p>However the inverse sin function in my program only takes and outputs radians so instead the equation to get degrees becomes </p>
<p><code>(Angle = sin-1(((1/Hypotenuse)*B *3.14) /180) *180) /3.14</code></p>
<p>This does not however seem to be right for some reason, as when putting in the parameters of <code>Hypotenuse=150</code>, <code>B=149.6</code> i get the answer of 85.8 (right) for the original equation and then .9973 degrees for the new equation?? </p>
<p><strong>Please help me fix this!</strong></p>
| Nick Dixon | 12,462 | <p>You would calculate the answer in radians, and then convert to degrees. Inside of the inverse sin should just be (1/hypotenuse*B) since its a ratio of side lengths.</p>
<p>You're overthinking it, I'm guessing.</p>
|
3,691,692 | <p>Find all real values of a such that <span class="math-container">$x^2+(a+i)x-5i=0$</span> has at least one real solution. </p>
<p><span class="math-container">$$x^2+(a+i)x-5i=0$$</span></p>
<p>I have tried two ways of solving this and cannot seem to find a real solution.</p>
<p>First if I just solve for <span class="math-container">$a$</span>, I get <span class="math-container">$$a=-x+i\frac{5-x}{x}$$</span>
Which is a complex solution, not a real solution...</p>
<p>Then I tried using the fact that <span class="math-container">$x^2+(a+i)x-5i=0$</span> is in quadratic form of <span class="math-container">$x^2+px+q=0$</span> with <span class="math-container">$p=(a+i)$</span> and <span class="math-container">$q=5i$</span></p>
<p>So I transform <span class="math-container">$$x^2+(a+i)x-5i=0$$</span> to <span class="math-container">$$(x+\frac{a+i}{2})^2=(\frac{a+i}{2})^2+5i$$</span></p>
<p>Now it is in the form that one side is the square of the other but I don't know how to find the roots since I'm not sure if I'm supposed to convert <span class="math-container">$(\frac{a+i}{2})^2+5i$</span> to polar form since I can't take the modulus of <span class="math-container">$(\frac{a+i}{2})^2+5i$</span> (or at least I don't know how).</p>
<p>At thins point I feel like I'm just using the wrong method if anyone could guide me in the right direction I would very much appreciate it. Thank you. </p>
| PythonSage | 596,065 | <p>Use AM-GM. <span class="math-container">$\frac{ad + bc}{2} \ge \sqrt{abcd}$</span>. Squaring both sides, you get the answer. A tiny tip: If everything is positive, and you have an inequality, think about AM-GM once at least. </p>
|
3,631,903 | <p>A Calculus A level trigonometry problem:</p>
<blockquote>
<p>Solve <span class="math-container">$\tan x = \dfrac{p}{q}$</span> where <span class="math-container">$p,q\in\mathbb{Z}$</span> such that <span class="math-container">$$3\cos x\ - 4\sin x = -5$$</span></p>
</blockquote>
<p>I tried moving terms to one side, but that doesn't help much.</p>
<p>Any ideas?</p>
| Zarrax | 3,035 | <p>For <span class="math-container">$x < -1$</span>, you have <span class="math-container">$|x + 1| = x + 1$</span> and for <span class="math-container">$x \geq -1$</span> you have <span class="math-container">$|x + 1| = -x - 1$</span>. So if you split into <span class="math-container">$x < -1$</span> and <span class="math-container">$x \geq -1$</span> cases you can get rid of the absolute value sign for that term. You can do a similar thing for the other absolute value symbols that show up, so you end out having to separately consider the following cases:</p>
<p><span class="math-container">$$ x < -1$$</span>
<span class="math-container">$$-1 \leq x < 0$$</span>
<span class="math-container">$$0 \leq x < 1$$</span>
<span class="math-container">$$1 \leq x < 2$$</span>
<span class="math-container">$$x \geq 2$$</span>
Note that you can put the endpoints in either of the adjacent intervals, but you should specify one or the other to make sure you get all solutions and don't have duplicates.</p>
|
3,743,743 | <p>I have this condition:</p>
<p><strong>(A is true OR B is true OR C is true) OR (A is false AND B is false AND C is false)</strong></p>
<p><em>(edit: It's been pointed out that this formula is wrong for what I want)</em></p>
<p>So as the title says, I want the condition to be true if only 1 of A, B or C is true, or if they're all false.</p>
<p>Is there a better way to write this condition?</p>
<p>edit 2: The context is a SQL Server validation check.</p>
<p>Thanks.</p>
| Robert Israel | 8,508 | <p>If the change in <span class="math-container">$\mu$</span> is small, you might use the series</p>
<p><span class="math-container">$$ (B+\lambda I)^{-1} = (B+\mu I)^{-1} - (\lambda - \mu) (B + \mu I)^{-2} + (\lambda - \mu)^2 (B + \mu I)^{-3} - \ldots $$</span></p>
|
3,743,743 | <p>I have this condition:</p>
<p><strong>(A is true OR B is true OR C is true) OR (A is false AND B is false AND C is false)</strong></p>
<p><em>(edit: It's been pointed out that this formula is wrong for what I want)</em></p>
<p>So as the title says, I want the condition to be true if only 1 of A, B or C is true, or if they're all false.</p>
<p>Is there a better way to write this condition?</p>
<p>edit 2: The context is a SQL Server validation check.</p>
<p>Thanks.</p>
| jsievers | 805,998 | <p>Do you need to have the matrix in explicit form, or are you happy for it to be factored? If the latter, then note that adding a constant to the diagonal of a matrix leaves the eigenvectors unchanged and shifts all the eigenvalues by that constant. That's easy to see since if A<em>x</em>=<span class="math-container">$\lambda$</span><em>x</em>, then (A+<span class="math-container">$\mu$</span>I)<em>x</em> = (<span class="math-container">$\lambda+\mu$</span>)x. If you've done the eigendecomposition, then the new inverse is just V <span class="math-container">$(\Lambda + \mu I)^{-1}$</span> V<span class="math-container">$^T$</span>. Since the thing being inverted is diagonal, this is very fast.</p>
|
2,079,822 | <p>I am asked to find the maximum velocity of a mass. </p>
<p>I know that the equation for maximum acceleration is </p>
<p>$$a = w^2A$$</p>
<p>However I do not know how to find the maximum velocity. Is velocity just the same as acceleration? </p>
| Anthony Storace | 403,229 | <p>acceleration depends on the applied force and the mass</p>
<p>acceleration = force/mass</p>
<p>for uniform acceleration velocity = acceleration X time</p>
<p>if acceleration varies you must integrate to find
velocity = integral of acceleration over time</p>
|
1,893,540 | <p>I've been asked to prove the following,
if $x - ε ≤ y$ for all $ε>0$ then $x ≤ y$.
I tried proof by contrapositive, but I keep having trouble choosing the right $ε$. Can you guys help me out? </p>
| Paramanand Singh | 72,031 | <p>First of all, understand that most of the problems dealing with $\epsilon, \delta$ are an exercise in inequalities and use very very elementary ideas. The difficulty of such problems arises mainly because of use Greek letters and partially because students don't really understand inequalities as much as they understand $+, -, \times, /, =$.</p>
<p>You wish to start by contrapositive which is almost similar to method of contradiction which we use in this answer. Let us then assume that $x > y$ i.e. $x - y > 0$. Now our goal is to use the hypotheses $x - y > 0$ to contradict the following statement $$\text{for all }\epsilon > 0, x - \epsilon \leq y$$ One way to contradict this statement is to find an $\epsilon > 0$ such that $x - \epsilon > y$ i.e. $\epsilon < x - y$.</p>
<p>Clearly this is possible because of the hypotheses that $x - y > 0$ and we just need to choose a positive $\epsilon$ smaller than positive number $x - y$. We don't specifically need to have an explicit expression for $\epsilon$ like $\epsilon = (x - y)/2$ (doing this always looks magical to a beginner). What is really needed is to understand that by hypotheses $A = x - y$ is positive and any positive $\epsilon$ less that this specific positive $A$ will do our job.</p>
<p>So after all this understanding given in last few paragraphs you just write the answer like this:</p>
<hr>
<p><em>Suppose on the contrary that $x > y$ so that $x - y$ is positive and hence there is a positive $\epsilon $ which is smaller than $x - y$ so that $0 < \epsilon < x - y$ and therefore $x - \epsilon > y$ which contradicts the fact that $x - \epsilon \leq y$ for all $\epsilon > 0$. Therefore our supposition is wrong and we must have $x \leq y$.</em></p>
<hr>
<p>In the above answer the fundamental idea used is the fact that <em>given any positive number, there exists a smaller positive number</em>. This is a fact of inequalities which is so obvious yet students find it difficult to use in problems such as these because of the use of Greek letters.</p>
|
167,946 | <p>I seek to replace derivatives like <code>D[u[x, y], x, x]</code> which are evaluated as $u^{(2,0)}[x,y]$ by variables with names like <code>uxx</code>. Derivatives that I work with are denoted by their "order-vector", for example <code>{2,0}</code> is the vector for this particular derivative and <code>{1,1}</code> is for <code>uxy</code>. I have a list of different derivative vectors and I want to create rule for replacement which is comprised of elements like $u^{(2,0)}[x,y]\rightarrow uxx$. But for that, I need to transform {2,0} into (2,0) which mathematica doesn't want to do:</p>
<pre><code>dsT={1,0,0,0};
ToExpression[StringReplace[ToString[dsT], {"}" -> ")", "{" -> "("}]]
</code></pre>
<p>This returns an error though without "ToExpression" the output is exaxtly what I need "(1, 0, 0, 0)". How can I do this conversion for any length of derivative vector?</p>
| J. M.'s persistent exhaustion | 50 | <p>Here is one method:</p>
<pre><code>derivativeToSymbol[expr_] /; Nest[Head, expr, 3] === Derivative := Block[{args, tmp},
args = List @@ expr; tmp = Head[expr];
Symbol[(ToString @@ tmp) <>
MapThread[ConstantArray, {ToString /@ args, List @@ Head[tmp]}]]]
</code></pre>
<p>Examples:</p>
<pre><code>derivativeToSymbol[D[u[x, y], x, x]]
uxx
derivativeToSymbol[D[u[x, y, z], {x, 3}, z]]
uxxxz
</code></pre>
|
1,902,455 | <p>$x=e^t$ $y=te^(-t)$</p>
<p>$\frac{dy}{dx}= \frac{e^(-t)(1-t)}{e^(t)}$</p>
<p>$\frac{d^2y}{dx^2}= \frac{\frac{dy}{dx}}{\frac{dx}{dt}}= \frac{e^(-t)(1-t)}{e^t}$</p>
<p>any t's without proper enclosement are meant to be to the power...I don't know why its giving me this trouble. I entered these answers into my homework and it said it could not understand my answers and could not be graded.... Also it asked the interval over which it is is concave upward and if im not mistaken from my work it would be from 0 to infinity.</p>
| Behrouz Maleki | 343,616 | <p>Set $y=\tan u$ we have
$$\int \frac{1-y^2}{(1+y^2)^2} dy=\int\frac{1-\tan^2u}{1+\tan^2 u}du=\int \cos 2u\,du=\frac{1}{2}\sin 2u+C= \frac{\tan u}{1+\tan^2 u}+C$$</p>
|
2,913,974 | <p>In an additive category, we say that an object $A$ is compact if the functor $\text{Hom}(A, -)$ respects coproducts. That is, if the canonical morphism
$$
\coprod_{i} \text{Hom} \left( A, X_{i} \right) \longrightarrow \text{Hom} \left( A, \coprod_{i} X_{i} \right)
$$
is a bijection. Suppose $A \oplus B$ is compact. Why are the summands $A$ and $B$ compact? Everywhere claims this is obvious and provides no justification, but I cannot see why this is true. </p>
| Hetong Xu | 749,743 | <p>In what follows, I shall add some detail to user14972's answer from the perspective of retract. Some more details and conventions are introduced in my question (post) <a href="https://math.stackexchange.com/questions/4155082/complete-compactness-and-retract">Complete compactness and retract</a> .</p>
<p>Let <span class="math-container">$\alpha: \mathsf{I} \rightarrow \mathsf{C}$</span> be a functor (i.e. a diagram in <span class="math-container">$\mathsf{C}$</span>), where <span class="math-container">$\mathsf{I}$</span> is <strong>filtered</strong>, viewing the retract as a coequalizer (which is a special finite colimit), we have
<span class="math-container">\begin{align}
\varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}(Y, \alpha(i)) &= \varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}\left(\varinjlim_{j \in \mathsf{J} := \{0,1\} } X, \alpha(i)\right) \\
&= \varinjlim_{i \in \mathsf{I}} \varprojlim_{j \in \mathsf{J} := \{0,1\} } \mathrm{Hom}_{\mathsf{C}}( X, \alpha(i)) \\
&\cong \varprojlim_{j \in \mathsf{J} := \{0,1\} } \varinjlim_{i \in \mathsf{I}} \mathrm{Hom}_{\mathsf{C}}( X, \alpha(i)) \\
&\cong \varprojlim_{j \in \mathsf{J} := \{0,1\} } \mathrm{Hom}_{\mathsf{C}}( X, \varinjlim_{i \in \mathsf{I}} \alpha(i)) \\
&\cong \mathrm{Hom}_{\mathsf{C}}\left(\varinjlim_{j \in \mathsf{J} := \{0,1\} } X, \varinjlim_{i \in \mathsf{I}} \alpha(i)\right) \\
&\cong \mathrm{Hom}_{\mathsf{C}}(Y, \varinjlim_{i \in \mathsf{I}} \alpha(i)).
\end{align}</span>
On the third line, we interchanged the order of limits and colimits because <strong>filtered colimits commute with finite limits in the category of sets.</strong> (See post <a href="https://math.stackexchange.com/questions/3075524/about-a-specific-step-in-a-proof-of-the-fact-that-filtered-colimits-and-finite-l">About a specific step in a proof of the fact that filtered colimits and finite limits commute in $\mathbf{Set}$</a> .) The fourth line holds since <span class="math-container">$X$</span> is <strong>compact</strong>.</p>
|
2,435 | <p>I'm not sure we already have something similar, but I'm working on more code inspections for the IntelliJ plugin and it's always a good idea to ask the community. Since it doesn't really fit on main, I'm posting it here on Meta.</p>
<p>Linting is an excellent way to point the developer to probable errors that he might have overlooked. With a dynamic language like the one of Mathematica, we are a bit restricted with what we can do, since we cannot evaluate code and since most things require evaluation to be sure if they are a bug or not. Nevertheless, there are checks we can do. For instance <code>If[a=b, ..]</code> is most likely a bug and even if the developer knew what he did, it is a bad style.</p>
<p>There are trickier examples like <code>If[a<5,...]</code>. This looks okay but knowing that <code>a<5</code> stays unevaluated if the comparison cannot be done, it is a source of error because you end up with the unevaluated <code>If</code> expression in your wrong result and debugging might be complicated.</p>
<p>In both examples, wrapping <code>TrueQ</code> around the condition resolves the issue and although there might still be a bug, at least you can be sure your <code>If</code> expression is evaluated to some branch.
Other common sources of error are, e.g. <code>x_?testFunc[#]&</code> or implicit multiplication through linebreaks.</p>
<p><strong>Question:</strong> What are common bugs in your code and could they have been pointed out by a linter? If you like to share your thoughts, please provide one issue per answer, so that others can vote. I'm looking forward to your suggestions and see if I can implement some of them in IntelliJ.</p>
<hr>
<p>Example issue: With the <a href="https://mathematica.stackexchange.com/a/176489/187">alternative layout for packages</a> which was pointed out by Leonid, we can use <em>directives</em> for a static code analyzer to easily export symbols or declare them as package symbols. As Leonid pointed out, the directives need to be on their own source-line with nothing else on it. So for the directives</p>
<pre><code>PackageScope["myFunc"]
PackageExport["MyExportedFunc"]
</code></pre>
<p>I implemented the following rules</p>
<ol>
<li>They need to be on their own source line with nothing else on it</li>
<li>Their string argument must be a valid identifier</li>
</ol>
<p><a href="https://i.stack.imgur.com/3bO61.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3bO61.gif" alt="enter image description here"></a></p>
| Kuba | 5,478 | <blockquote>
<h1>Status Partially Completed</h1>
</blockquote>
<h2>1. Invalid number of arguments for some of built-in functions e.g.</h2>
<ul>
<li><p><code>Which</code> called with an odd number of arguments</p>
</li>
<li><p><code>Switch</code> called with an even number of arguments</p>
</li>
<li><p><code>Module/With/...</code> called with more than two arguments</p>
<p><br>I don't think it needs to work for every built-in function but will probably be useful for scoping and conditional expressions.</p>
</li>
</ul>
<h2>2. Backtick in <code>Begin</code> and friends</h2>
<ul>
<li><p><code>Begin</code> called with <code>"Private`"</code> instead of <code>"`Private`"</code></p>
<p>etc.</p>
</li>
</ul>
<hr />
<p><strong>Implemented</strong></p>
<p><img src="https://i.imgur.com/A29kV7P.png" alt="img" /></p>
<hr />
<p>Oh, I almost forgot my personal number one problem:</p>
<h2>3. <code>ClearAll @ foo</code></h2>
<p>when it is called after options/attributes or any definition for <code>foo</code> was defined.</p>
<h2>4. Need for explicit <code>Optional</code>.</h2>
<p>This gives a syntax error, it is debatable but certainly can be unexpected:</p>
<pre><code> {_} : {1}
</code></pre>
|
2,435 | <p>I'm not sure we already have something similar, but I'm working on more code inspections for the IntelliJ plugin and it's always a good idea to ask the community. Since it doesn't really fit on main, I'm posting it here on Meta.</p>
<p>Linting is an excellent way to point the developer to probable errors that he might have overlooked. With a dynamic language like the one of Mathematica, we are a bit restricted with what we can do, since we cannot evaluate code and since most things require evaluation to be sure if they are a bug or not. Nevertheless, there are checks we can do. For instance <code>If[a=b, ..]</code> is most likely a bug and even if the developer knew what he did, it is a bad style.</p>
<p>There are trickier examples like <code>If[a<5,...]</code>. This looks okay but knowing that <code>a<5</code> stays unevaluated if the comparison cannot be done, it is a source of error because you end up with the unevaluated <code>If</code> expression in your wrong result and debugging might be complicated.</p>
<p>In both examples, wrapping <code>TrueQ</code> around the condition resolves the issue and although there might still be a bug, at least you can be sure your <code>If</code> expression is evaluated to some branch.
Other common sources of error are, e.g. <code>x_?testFunc[#]&</code> or implicit multiplication through linebreaks.</p>
<p><strong>Question:</strong> What are common bugs in your code and could they have been pointed out by a linter? If you like to share your thoughts, please provide one issue per answer, so that others can vote. I'm looking forward to your suggestions and see if I can implement some of them in IntelliJ.</p>
<hr>
<p>Example issue: With the <a href="https://mathematica.stackexchange.com/a/176489/187">alternative layout for packages</a> which was pointed out by Leonid, we can use <em>directives</em> for a static code analyzer to easily export symbols or declare them as package symbols. As Leonid pointed out, the directives need to be on their own source-line with nothing else on it. So for the directives</p>
<pre><code>PackageScope["myFunc"]
PackageExport["MyExportedFunc"]
</code></pre>
<p>I implemented the following rules</p>
<ol>
<li>They need to be on their own source line with nothing else on it</li>
<li>Their string argument must be a valid identifier</li>
</ol>
<p><a href="https://i.stack.imgur.com/3bO61.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3bO61.gif" alt="enter image description here"></a></p>
| Jason B. | 9,490 | <p>These are really common mistakes that go unnoticed because they are valid syntax:</p>
<ul>
<li><code>Which</code> statement using <code>_</code> as the final predicate</li>
<li><code>Switch</code> statement using <code>True</code> as the final pattern</li>
<li>A list with a random dot product in the middle, usually means someone typed a period instead of a comma in a long multi-line list</li>
</ul>
|
2,435 | <p>I'm not sure we already have something similar, but I'm working on more code inspections for the IntelliJ plugin and it's always a good idea to ask the community. Since it doesn't really fit on main, I'm posting it here on Meta.</p>
<p>Linting is an excellent way to point the developer to probable errors that he might have overlooked. With a dynamic language like the one of Mathematica, we are a bit restricted with what we can do, since we cannot evaluate code and since most things require evaluation to be sure if they are a bug or not. Nevertheless, there are checks we can do. For instance <code>If[a=b, ..]</code> is most likely a bug and even if the developer knew what he did, it is a bad style.</p>
<p>There are trickier examples like <code>If[a<5,...]</code>. This looks okay but knowing that <code>a<5</code> stays unevaluated if the comparison cannot be done, it is a source of error because you end up with the unevaluated <code>If</code> expression in your wrong result and debugging might be complicated.</p>
<p>In both examples, wrapping <code>TrueQ</code> around the condition resolves the issue and although there might still be a bug, at least you can be sure your <code>If</code> expression is evaluated to some branch.
Other common sources of error are, e.g. <code>x_?testFunc[#]&</code> or implicit multiplication through linebreaks.</p>
<p><strong>Question:</strong> What are common bugs in your code and could they have been pointed out by a linter? If you like to share your thoughts, please provide one issue per answer, so that others can vote. I'm looking forward to your suggestions and see if I can implement some of them in IntelliJ.</p>
<hr>
<p>Example issue: With the <a href="https://mathematica.stackexchange.com/a/176489/187">alternative layout for packages</a> which was pointed out by Leonid, we can use <em>directives</em> for a static code analyzer to easily export symbols or declare them as package symbols. As Leonid pointed out, the directives need to be on their own source-line with nothing else on it. So for the directives</p>
<pre><code>PackageScope["myFunc"]
PackageExport["MyExportedFunc"]
</code></pre>
<p>I implemented the following rules</p>
<ol>
<li>They need to be on their own source line with nothing else on it</li>
<li>Their string argument must be a valid identifier</li>
</ol>
<p><a href="https://i.stack.imgur.com/3bO61.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3bO61.gif" alt="enter image description here"></a></p>
| Szabolcs | 12 | <blockquote>
<h1>Status Partially Completed</h1>
</blockquote>
<h2>Unused variables</h2>
<pre><code>fun[x_, y_] := x^2
^ y is unused
</code></pre>
<pre><code>Module[{x, y}, <-- y is unused
x+1
]
</code></pre>
<pre><code>y=1; <-- this instance of y is effectively unused
y=2;
</code></pre>
<p>The last one is quite involved and may be difficult to implement, therefore I won't give more examples of this type unless you ask.</p>
<p>Generally, the idea is that if a localized variable (<code>Module</code>, <code>Block</code>, <code>With</code> or function arg) is unreferenced within the local scope, there should be a warning.</p>
<p>Cases to exclude: </p>
<ul>
<li><code>Rule</code>, as it is typically used with explicit pattern matching, where naming subpatterns often aids readability even if their value isn't used.</li>
<li>Patterns like <code>fun[list : {x_, x_}] := ...</code> where a subpattern needed to be named to indicate repetition.</li>
</ul>
<p><strong>UPDATE:</strong></p>
<p>It may be best to add this for <code>Module</code> and <code>With</code> only for the time being. <code>Block</code> has too many reasonable uses where the localized variables do not need to appear explicitly, e.g. <code>Block[{$RecursionLimit = ...}, ...]</code> or setting other flags which are user-defined, but affect the Block body in a similar way.</p>
<hr>
<p><strong>UPDATE 2:</strong></p>
<p>The Workbench has this feature:</p>
<p><a href="https://i.stack.imgur.com/WqgJL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WqgJL.png" alt="enter image description here"></a></p>
<p><strong>Comment halirutan:</strong> This inspection has been implemented and is available since WL Plugin version 2019.1</p>
|
2,435 | <p>I'm not sure we already have something similar, but I'm working on more code inspections for the IntelliJ plugin and it's always a good idea to ask the community. Since it doesn't really fit on main, I'm posting it here on Meta.</p>
<p>Linting is an excellent way to point the developer to probable errors that he might have overlooked. With a dynamic language like the one of Mathematica, we are a bit restricted with what we can do, since we cannot evaluate code and since most things require evaluation to be sure if they are a bug or not. Nevertheless, there are checks we can do. For instance <code>If[a=b, ..]</code> is most likely a bug and even if the developer knew what he did, it is a bad style.</p>
<p>There are trickier examples like <code>If[a<5,...]</code>. This looks okay but knowing that <code>a<5</code> stays unevaluated if the comparison cannot be done, it is a source of error because you end up with the unevaluated <code>If</code> expression in your wrong result and debugging might be complicated.</p>
<p>In both examples, wrapping <code>TrueQ</code> around the condition resolves the issue and although there might still be a bug, at least you can be sure your <code>If</code> expression is evaluated to some branch.
Other common sources of error are, e.g. <code>x_?testFunc[#]&</code> or implicit multiplication through linebreaks.</p>
<p><strong>Question:</strong> What are common bugs in your code and could they have been pointed out by a linter? If you like to share your thoughts, please provide one issue per answer, so that others can vote. I'm looking forward to your suggestions and see if I can implement some of them in IntelliJ.</p>
<hr>
<p>Example issue: With the <a href="https://mathematica.stackexchange.com/a/176489/187">alternative layout for packages</a> which was pointed out by Leonid, we can use <em>directives</em> for a static code analyzer to easily export symbols or declare them as package symbols. As Leonid pointed out, the directives need to be on their own source-line with nothing else on it. So for the directives</p>
<pre><code>PackageScope["myFunc"]
PackageExport["MyExportedFunc"]
</code></pre>
<p>I implemented the following rules</p>
<ol>
<li>They need to be on their own source line with nothing else on it</li>
<li>Their string argument must be a valid identifier</li>
</ol>
<p><a href="https://i.stack.imgur.com/3bO61.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3bO61.gif" alt="enter image description here"></a></p>
| b3m2a1 | 38,205 | <p>Things like </p>
<pre><code>f[a_Association]:=...
</code></pre>
<p>Should ask if you want to convert to </p>
<pre><code>f[a_Association?AssociationQ]:=...
</code></pre>
<p>And similarly</p>
<pre><code>f[a_?AssociationQ]:=...
</code></pre>
<p>Should ask if you want to do the same</p>
<p>The longer pattern is the fastest, most robust one, for most pattern types:</p>
<pre><code>f[a_Association?AssociationQ] := True;
g[a_?AssociationQ] := True;
h[a_Association] := True;
a = Association[1];
b = <||>;
c = Dogs;
RepeatedTiming[f[a];, 1] // First
RepeatedTiming[f[b];, 1] // First
RepeatedTiming[f[c];, 1] // First
1.2*10^-6
1.2*10^-6
4.7*10^-7
RepeatedTiming[g[a];, 1] // First
RepeatedTiming[g[b];, 1] // First
RepeatedTiming[g[c];, 1] // First
1.2*10^-6
1.2*10^-6
9.3*10^-7
RepeatedTiming[h[a];, 1] // First
RepeatedTiming[h[b];, 1] // First
RepeatedTiming[h[c];, 1] // First
7.2*10^-7
6.9*10^-7
4.3*10^-7
</code></pre>
|
3,977,687 | <p>A coin of radius 1 cm is tossed onto a plane surface that has been tessellated by right triangles whose sides are 8 cm, 15 cm, and 17 cm long. Find the probability that the coin lands within a triangle.</p>
<p>I know that this has to do with similarity because the inner triangle that is formed by the area where the coin can land is similar to the outer triangle. Therefore, I know the angles of this triangle, but I am not sure how to find one side of this triangle.</p>
| achille hui | 59,379 | <p>You don't need to compute the sides of the new triangle, you only need to know the change in the inradius.</p>
<p>Given any triangle with area <span class="math-container">$\Delta$</span> and perimeter <span class="math-container">$\ell$</span>, its inradius <span class="math-container">$r$</span> equals to <span class="math-container">$\frac{2\Delta}{\ell}$</span>.</p>
<p>If one parallel shift all three edges inwards for a small distance <span class="math-container">$\rho$</span>, the shrinked triangle will be similar to original triangle, sharing same incenter and has inradius <span class="math-container">$r - \rho$</span>. The area of the shrinked triangle is given by:</p>
<p><span class="math-container">$$\Delta_{\rm shrink} = \Delta \left( 1 - \frac{\rho}{r}\right)^2 = \Delta\left(1 - \frac{\rho\ell}{2\Delta}\right)^2$$</span></p>
<p>For the problem at hand, the area and perimeter of a triangle tile are:
<span class="math-container">$$\Delta = \frac{8\cdot 15}{2} = 60\quad\text{ and }\quad \ell = 8+15+17 = 40$$</span></p>
<p>In order for the coin to falls completely inside the interior of a triangle tile, its center needs to lie inside a shrinked triangle of the tile obtained by a shift with <span class="math-container">$\rho = 1$</span>. This means the probability for the coin to fall complete inside a tile is</p>
<p><span class="math-container">$$\frac{\Delta_{\rm shrink}}{\Delta} = \left(1 - \frac{\rho\ell}{2\Delta}\right)^2 = \left(1 - \frac{1\cdot 40}{2\cdot 60}\right)^2 = \left(1 - \frac13\right)^2 = \frac{4}{9}$$</span></p>
|
86,755 | <p>I need to solve the following integral equations for a problem I'm working on - </p>
<p>$\displaystyle \frac{-i}{2 \pi}$ $\int_{-a}^{a} \mathrm{dt}\,\, \frac{e^{i k t}}{t + i \tau}$ and $\displaystyle \frac{-i}{2 \pi}$ $\int_{a}^{\infty} \mathrm{dt}\,\, \frac{e^{i k t}}{t + i \tau}$</p>
<p>where $\tau, k \in \mathbb{R}$.</p>
<p><em>Mathematica</em> can do these numerically and I get sensible plots as a function of $\tau$ for given $a$ and $k$, where by sensible I mean that the integrals go to 0 as $\tau \to \pm \infty$. </p>
<p>However, I'd like to find an analytic expression for these, even if it is in terms of exponential integrals, etc. When I try to do this, <em>Mathematica</em> spits out an answer that agrees with the numerical integration at small $\tau$, but at large $\tau$ the analytic expressions break down. </p>
<p>For concreteness, suppose I take $\frac{-i}{2 \pi}$ $\int_{a}^{\infty} \mathrm{dt}\,\, \frac{e^{i k t}}{t + i \tau}$ with k = 1 and a = 1. </p>
<p>Doing this integral numerically and plotting the real part vs $\tau$,</p>
<pre><code>Plot[Re[-I/(2 π)
NIntegrate[E^(I t)/(
t + I τ), {t, 1, ∞}]], {τ, -200, 200},
PlotRange -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/rMeKZ.jpg" alt="Numerically_Integrated_Plot"></p>
<p>But now, suppose I do this integral analytically.</p>
<pre><code>Integrate[-I/(2 π) E^(I t)/(t + I τ), {t, 1, ∞}]
</code></pre>
<p>This gives an output</p>
<pre><code> (E^τ (π + 2 I CosIntegral[1 + I τ] + 2 I SinhIntegral[I - τ]))/(4 π)
</code></pre>
<p>and when I plot the Real part of this, the result looks like - </p>
<p><img src="https://i.stack.imgur.com/jDOyd.jpg" alt="Analytically_Generated_Plot"></p>
<p>I can't really understand why this breaks down and would like to have analytic expressions for both those integrals for an values of $\tau, k$ and $a$.</p>
| Aegon | 30,377 | <p>So upon thinking about this more and playing with different assumptions when trying to integrate the above function(s), what I found was the following - </p>
<p>$\int_{a}^{\infty} dt \frac{e^{i k t}}{t + i \tau} = e^{k\tau}\,\Gamma\left(k(-ia + \tau) \right)$, where $\Gamma$ is the incomplete Gamma function. </p>
<p>Now, suppose I try doing this integral with Mathematica, with the assumption that $a>0$. I get</p>
<pre><code>E^(k \[Tau]) (Gamma[0, k (-I a + \[Tau])] - Log[-I k] - Log[a + I \[Tau]] + Log[k (-I a + \[Tau])])
</code></pre>
<p>Now for large values of $\tau$, if you want to plot this function, you have to follow @glance's answer given above. On examining the answer, however, you see that the Log part of this answer is 0, leaving one with just the $\Gamma$-function which Mathematica has no problems evaluating for any $\tau$ or $k$! </p>
<p>The integral $\int_{-a}^a dt \frac{e^{ikt}}{t + i \tau}$ is a little more complicated, but in Mathematica, it evaluates to (with assumptions $\tau \in \text{Reals}, a >0$)</p>
<pre><code>E^(k \[Tau]) (-Gamma[0, k (-I a + \[Tau])] + Gamma[0, k (I a + \[Tau])] - Log[a - I \[Tau]] + Log[a + I \[Tau]] + Log[-I \[Tau]] - Log[I \[Tau]] - Log[k (-I a + \[Tau])] + Log[k (I a + \[Tau])])
</code></pre>
<p>Again, looking at the Log part of this answer, one finds that it is </p>
<p>0 if $k \tau > 0$ ; $2 \pi i$ if $k>0$ and $\tau < 0$ ; $-2 \pi i$ if $k < 0$ and $\tau > 0$. </p>
<p>Then, this function can be built with the $\Gamma$-functions and a piece-wise part, which matches the numerically integrated function and is evaluated by Mathematica for any $\tau$ and $k$ without any issues whatsoever. </p>
|
1,336,506 | <p>We know that the usual $\leq$ is a partial order relation on the group of integers $\mathbb Z$ and $\mathbb Z$ is a totally ordered with this partial order relation. Is there any other partially order relation exist in $\mathbb Z$ which makes $\mathbb Z$ a partially ordered group (or totally ordered group)? </p>
| lhf | 589 | <p>If $0$ and $1$ are related, then the only possible partial orders that make $\mathbb Z$ into a partially ordered group are the usual order $\le$ and its reverse order $\ge$:</p>
<p>If $0 \prec 1$, then $1 = 0 + 1 \prec 1 + 1 = 2$, and so $n \prec n+1$ for all $n$ by induction.</p>
<p>If $0 \succ 1$, then $n \succ n+1$ for all $n$.</p>
<p>If $0$ and $1$ are unrelated, I don't know right now... I think you need to consider all integers that are related to $0$. Then there'll be several linear orders "starting" at $0$ in both directions. For instance, if $0 \prec n$, then there will be arithmetic progressions of step $n$ starting at $k=0, \dots, n-1$. </p>
|
2,300,613 | <p>I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way? </p>
<p>$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$</p>
| Jack D'Aurizio | 44,121 | <p>Since, formally,
$$ \frac{1}{1-z}=1+z+z^2+z^3+\ldots $$
the multiplication by $\frac{1}{1-z}$ brings the power series $a_0 + a_1 z+ a_2 z^2 +\ldots $ into the power series $a_0+(a_0+a_1)z+(a_0+a_1+a_2) z^2+\ldots$. It follows that:</p>
<p>$$ \frac{e^{z}}{1-z}=\sum_{n\geq 0}z^n \left(\sum_{j=0}^{n}\frac{1}{j!}\right).$$</p>
|
3,913,244 | <p>My attempt :</p>
<p><span class="math-container">$A=2^3×5^2×7^3$</span></p>
<p>Let's determine number of numbers primes with A, and
Smaller than A</p>
<p><span class="math-container">$\rho (A) = 2^2 ×4×5×6×7^2 =23520$</span></p>
<p>23520 is a number of numbers primes with A and smaller than A</p>
<p><span class="math-container">$\forall k$</span> <span class="math-container">$ \in {23520} $</span> <span class="math-container">$gcd(A, k)=1$</span></p>
<p>So : number of numbers doesn't prime white A is <span class="math-container">$A-23520=45080$</span></p>
<p><span class="math-container">$\forall k $</span> <span class="math-container">$\in {45080} $</span> <span class="math-container">$gcd(A, k) ≠1$</span></p>
<p>let's search about The number of numbers that divide A</p>
<p><span class="math-container">$A=0[k]$</span>
<span class="math-container">$\Leftrightarrow$</span> <span class="math-container">$68600=0[k]$</span></p>
<p>In the first we can see K can take the following values <span class="math-container">${1,2,4, 5,7,8,10,20,40,50,70,100}$</span></p>
<p>After using this theory <span class="math-container">$(a=b[n] $</span>and <span class="math-container">$a=b[m] $</span> and <span class="math-container">$gcd(m, n) =1$$\Rightarrow$</span> <span class="math-container">$a=b[MN] $</span>)</p>
<p>We can see that</p>
<p><span class="math-container">$K=(1, 2,4,5,7,8,10,20,40,50,70,100,35,56,28,14,140,280,350,700)$</span></p>
<p>So igot 20 value of k</p>
<p>But I know, there are 48 values of k</p>
<p>I want to know how can I continue for find 48 value of k</p>
<p>Note : I don't want any
other solution, I want to develop of my attempt</p>
| Calvin Lin | 54,563 | <p><strong>Claim 1:</strong> For a fixed <span class="math-container">$ 1 \leq k \leq p-2$</span>, <span class="math-container">$p \mid \sum_{i=1}^{p-1} i^k$</span>.<br />
This almost corresponds to <span class="math-container">$ F(x) = x^k$</span> in the question.</p>
<p>Proof: Work mod <span class="math-container">$p$</span>. Let <span class="math-container">$ g$</span> be a primitive root modulo <span class="math-container">$p$</span>.<br />
Then, <span class="math-container">$ \{ 1, 2, \ldots, p-1 \} = \{ g^0, g^1, \ldots g^{p-1}\}$</span>.<br />
So <span class="math-container">$ \sum_{i=1}^{p-1} i^k = \sum_{j=1}^{p-1} g^{jk} = \frac{ g^{(p-1)k } - 1 } { g^ k - 1 }$</span> .<br />
Note that <span class="math-container">$ p \not \mid g^k - 1$</span> and <span class="math-container">$ p \mid g^{(p-1)k } - 1 $</span>, hence <span class="math-container">$p$</span> divides the fraction.</p>
<p><strong>Claim 2:</strong> For a fixed <span class="math-container">$ 0 \leq k \leq p-2$</span>, <span class="math-container">$p \mid \sum_{i=0}^{p-1} i^k$</span>.<br />
This corresponds to <span class="math-container">$ F(x) = x^k$</span> in the question.</p>
<p>Proof: For <span class="math-container">$ k = 0$</span>, the sum is <span class="math-container">$p$</span>, so the claim is true.<br />
For <span class="math-container">$k>0$</span>, this is (almost) the previous claim.</p>
<p>Note: For <span class="math-container">$ k = p - 1$</span>, the statement isn't true since the sum is <span class="math-container">$ \equiv p-1 \pmod{p}$</span>. Which of the above steps fail?</p>
<p><strong>Corollary:</strong> The question is proved.<br />
<span class="math-container">$F(x) = \sum_{k=0}^{p-2} a_k x^k,$</span> and the question is true for each <span class="math-container">$a_k x^k$</span>, so it's true for the sum.</p>
<hr />
<p>Note: We can prove Claim 1 without the machinery of primitive roots.<br />
EG Using Method of Differences to show that for prime <span class="math-container">$n$</span>, <span class="math-container">$ n \mid \sum_{i=1}^{n} i^k$</span> because the denominators of the coefficients <span class="math-container">$\mid (p-1)!$</span> and the numerator is a multiple of <span class="math-container">$p$</span>.<br />
Or appeal to <a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula" rel="nofollow noreferrer">Faulhaber's formula</a>.</p>
|
598,962 | <p>I have to determine the following:</p>
<p>$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)$$</p>
<p>I've got so far:</p>
<p>$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)= \lim_{x \rightarrow 0}\left(\frac{27}{x(x+3)^3}-\frac{1}{x}\right)=\lim_{x \rightarrow 0} \left(\frac{27-(x+3)^3}{x(x+3)^3}\right)=\cdots$$</p>
<p>How to go on? I've got $\frac{\infty}{0}...$</p>
| Tī-Kong n̂g | 114,467 | <p>Just develop the nuumerator and you'll get
$$
\lim_{x \rightarrow 0} \frac{-x^3 - 9x^2 - 27x}{x(x+3)^3}
=\lim_{x \rightarrow 0} \frac{-x^2 - 9x - 27}{(x +3)^3} = -1
$$</p>
|
1,512,549 | <p>As the title says everything.</p>
<p>Thanks,</p>
| Community | -1 | <p>Let $s=\dfrac{S_n}{a_1}$ for conciseness.</p>
<p>You need to solve the polynomial equation</p>
<p>$$s(1-q)=1-q^n,$$
or
$$P_n(q)=q^n-sq+s-1=0.$$</p>
<p>There is no analytical solution for general $n$ and you need to resort to numerical methods. <a href="https://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow">Newton</a> works well for this case. Iterate with</p>
<p>$$q\leftarrow q-\frac{P_n(q)}{P'_n(q)}=\frac{nq^n-s+1}{nq^{n-1}-s}.$$</p>
<p>An intial bracketing can be found from</p>
<p>$$s_1=1+q<s<\frac1{1-q}=s_\infty,$$ so that </p>
<p>$$1-\frac1s<q<s-1.$$</p>
<p>And a better approximation can be obtained as the first step of the secant method, with</p>
<p>$$q\leftarrow \frac{P_n(q_1)q_0-P_n(q_0)q_1}{P_n(q_1)-P_n(q_0)},$$where $q_1=1-\dfrac1s$ and $q_0=s-1$.</p>
|
1,512,549 | <p>As the title says everything.</p>
<p>Thanks,</p>
| Lutz Lehmann | 115,115 | <p>Since $q=1+r$ with some small value $r$ for the interest rate (assuming no usury) one gets $q-1=r$ and $q^n-1=nr+\frac{n(n-1)}2r^2+\frac{n(n-1)(n-2)}{6}r^3+…$ so that approximately
$$
r_0=\frac{S}{na}-1\approx \frac{n-1}2r·\left(1+\frac{n-2}3r\right)
$$
Multiplying with $\left(1-\frac{n-2}3r\right)$ and ignoring higher-order terms results in
$$
r_0=\left(\frac{n-1}2+\frac{n-2}3r_0\right)r
\\ \iff \\
r = \frac{6r_0}{3(n-1)+2(n-2)r_0}
$$
as a good initial first approximation for starting Newtons method or the secant method or ...</p>
|
572,541 | <blockquote>
<p>Let $L$ be the set of all lines in the plane. Prove that $L$ is uncountable, but only countably many of the lines in $L$ contain more than one rational point.</p>
</blockquote>
<p><strong>Attempt</strong>: Well, I was trying to define $L$ using linear combinations of points since a line is a linear combo of two points. So, I wanted to define $L=\{ax+yb:x+y+z=k,k∈Z_+\}$. But, this does not seem right. Anyway, once I define $L$ I would try to find a function from $L$ to some <strong>uncountable</strong> set D that is onto or a function from D to L that is 1-1. Help appreciated thank you.</p>
| user99680 | 99,680 | <p>You can List all lines in the plane this way:</p>
<p>Use the fact that a line is described uniquely once you know its slope, and one of its
intercepts with the axes. The slope is indexed by the Reals, and so is the intercept, say the x-intercept. So we count all the possible pairs ( slope, x-intercept), and show
it is equal to $|\mathbb R|$</p>
<p>1) Consider all lines thru the origin $(0,0)$. These are described uniquely by their
slope, and there are $|\mathbb R|$ of them, since the slope is parametrized by the Reals.</p>
<p>2) From 1) , we can cover all other cases of lines not going thru $(0,0)$ , by considering all possible ( say x-) intercepts of a line thru any point, with fixed slope$m$. For every line
in 1), there are $\mathbb R$ lines not going thru the origin, but with the same slope.
This means there are $|\mathbb R|\times |\mathbb R|=|\mathbb R|$ total lines in the plane.</p>
|
457,977 | <p>I am trying to use residues to compute $$\int_0^\infty\frac{\log x}{(1+x)^3}\,\operatorname d\!x.$$My first attempt involved trying to take a circular contour with the branch cut being the positive real axis, but this ended up cancelling off the term I wanted. I wasn't sure if there was another contour I should use. I also had someone suggest using the substitution $x=e^z$, so the integral becomes $$\int_{-\infty}^\infty\frac{ze^z}{(1+e^z)^3}\,\operatorname d\!z$$so that the poles are the at the odd multiples of $i\pi$. I haven't actually worked this out, but it does not seem like the solution the author was looking for (this question comes from an old preliminary exam).</p>
<p>Any suggestions on how to integrate?</p>
| Ron Gordon | 53,268 | <p>Consider the integral</p>
<p>$$\oint_C dz \frac{\log^2{z}}{(1+z)^3}$$</p>
<p>where $C$ is a keyhole contour in the complex plane, about the positive real axis. This contour integral may be seen to vanish along the outer and inner circular contours about the origin, so the contour integral is simply equal to</p>
<p>$$\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{(1+x)^3} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{(1+x)^3}+4 \pi^2 \int_0^{\infty} dx \frac{1}{(1+x)^3}$$</p>
<p>By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=-1=e^{i \pi}$. In this case, with the triple pole, we have the residue being equal to</p>
<p>$$\frac12 \left [ \frac{d^2}{dz^2} \log^2{z}\right]_{z=e^{i \pi}} = 1-i \pi$$</p>
<p>Thus we have that</p>
<p>$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{(1+x)^3}+4 \pi^2 \frac12 = i 2 \pi + 2 \pi^2$$</p>
<p>which implies that</p>
<p>$$\int_0^{\infty} dx \frac{\log{x}}{(1+x)^3} = -\frac12$$</p>
|
996,052 | <p>A disk of radius <span class="math-container">$5$</span> cm has density <span class="math-container">$10$</span> g/cm<span class="math-container">$^2$</span> at its center, density <span class="math-container">$0$</span> at its edge, and its density is a linear function of the distance from the center. Find the mass of the disk.</p>
<p>my answer: <span class="math-container">$157.08$</span>g</p>
<p><span class="math-container">$D=10-2x$</span></p>
<p>Double integration of <span class="math-container">$D$</span> from <span class="math-container">$y=0$</span> to <span class="math-container">$y=2\pi$</span> and <span class="math-container">$x=0$</span> to <span class="math-container">$x = 5$</span></p>
| gniourf_gniourf | 51,488 | <p>You computed the integral:
$$\iint_{x\in[0,5],\,y\in[0,2\pi]}(10-2x)\,\mathrm{d}x\,\mathrm{d}y=50\pi\approx157.08.$$
Though your computation is correct, the result is not the total mass of the disc.</p>
<p>For a 2-dimensional object $\Sigma$ of density of mass $D$, the total mass $m$ of $\Sigma$ is
$$m=\iint_\Sigma D\,\mathrm{d}A$$
where $\mathrm{d}A$ is the surface element. Since your object is a disc, and the way the density of mass is given, it makes sense to use polar coordinates. That's what you did. But in the process, you used the following <em>wrong fact</em>:
$$\mathrm{d}A=\mathrm{d}r\,\mathrm{d}\theta\qquad\textbf{This is wrong!}$$
where $(r,\theta)$ are the polar coordinates, instead of:
$$\mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta\qquad\text{This is correct}.$$
So the total mass should instead be:
$$m=\iint_{(r,\theta)\in[0,5]\times[0,2\pi]}(10-2r)\,r\,\mathrm{d}r\,\mathrm{d}\theta.$$
I'm sure you can take it from here. The answer is $m=\dfrac{250\pi}3\,\mathrm{g}\approx261.80\,\mathrm{g}$.</p>
|
1,081,021 | <p>In what follows I'm only considering positive real valued functions.</p>
<p>Everywhere I look about the definition of the Lebesgue integral it is required to consider a measurable function. Why do we not define the integral for non-measurable functions? From what I see we require measurablility of the simple functions that approximate f, not f itself. The definition I'm considering is given a measure space $X$ with measure $\mu$ and a measurable function $f$ we define</p>
<p>$$
\int_E f \, \mathrm{d}\mu = \sup_{s \in S} \int_X s \,\mathrm{d}\mu
$$
where $S = \{ s : X \to [0, \infty) \mid 0 \le s \le f, s \text{ is simple, measurable} \}$.</p>
<p>For example consider $\mathbb{R}$ with the sigma algebra $\varnothing, \mathbb{R}$ with measure $\mu$ given by $\mu(\varnothing) = 0, \mu(\mathbb{R}) = 1$ and consider $f = \chi_{[0,1]}$ then why can't we say that
$$
\int_{\mathbb{R}} f \,\mathrm{d} \mu = 0
$$
(since the only measurable simple function such that $0\le s \le f$ is $s = 0$) which would follow the definition above? Is this not well defined? In general I'm struggling to see why measurable functions (other than measurable simple functions) are used.</p>
| GEdgar | 442 | <p>Also, maybe it is useful to show the bad properties of this lower integral when applied to non-measurable funtions. Take your $\{\varnothing, \mathbb R\}$ example. Let $f = \chi_{[0,1]}$ and $g = 1-f$. Then $\int f = \int g = 0$ but $\int(f+g) = 1$. So even simple linearity fails.</p>
|
434,290 | <p>According to the <a href="http://arxiv.org/abs/0910.5922" rel="nofollow">equation 4</a>,
$$\phi(0,t)= \frac{A_0}{(1+\frac{2t^2}{R^4})^{3/4}}\cos \left(\sqrt2 t+ \frac{3}{2}\tan^{-1}\left[\frac{\sqrt2 t}{R^2}\right]\right)\tag{1}$$
what conditions makes, $$\cos \left(\sqrt2 t+ \frac{3}{2}\tan^{-1}\left[\frac{\sqrt2 t}{R^2}\right]\right)=1$$
so the equation (1) will be </p>
<p>$$\phi(0,t)= \frac{A_0}{(1+\frac{2t^2}{R^4})^{3/4}}$$
The author used the <a href="http://arxiv.org/abs/hep-ph/9503217" rel="nofollow">article reference</a> to establish the equation
$$\frac{1}{2} \Gamma_{lin}= \frac{1}{\tau_{linear}} \approx \frac{1.196}{\omega_{mass}} \approx \frac{.846}{R^2}$$
but I didn't get any argument there, can you explain this a bit please.</p>
| Quanto | 686,284 | <p><span class="math-container">\begin{align}
\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx
=&\int_0^1 \frac {(1-x^2)\ln x}{x^4+1}\ dx\\
\overset{ibp}=&-\frac1{2\sqrt2}\int_0^1\frac1x
\ln\frac{x^2+\sqrt2 x+1}{x^2-\sqrt2 x+1}dx\\
=&-\frac1{2\sqrt2}\int_0^1\int_{-\pi/4}^{\pi/4}
\frac{2\cos y}{x^2+2x\sin y+1}dy \ dx\\
=&-\frac1{\sqrt2}\int_{-\pi/4}^{\pi/4} \left(\frac\pi4-\frac y2\right) dy=-\frac{\pi^2 }{8\sqrt2}
\end{align}</span></p>
|
4,609,833 | <p>As far as I can tell using Mathematica, the following identity seems to hold:
<span class="math-container">$$(n+2)^n=(n+1)\sum_{k=0}^n\binom{n}{k}\frac{(k+1)^{k-1}(n-k)^{n-k}}{n+1-k},$$</span>
where we define <span class="math-container">$0^0=1$</span>. However, I am having trouble proving it. I thought that this looked familiar and found the formulas (5.64) and (5.65) in Graham, Knuth, and Patashnik's "Concrete Mathematics", which are
<span class="math-container">$$\sum_{k=0}^n\binom{n}{k}(tk+r)^k(tn-tk+s)^{n-k}\frac{r}{tk+r}=(tn+r+s)^n$$</span>
and
<span class="math-container">$$\sum_{k=0}^n\binom{n}{k}(tk+r)^k(tn-tk+s)^{n-k}\frac{r}{tk+r}\frac{s}{tn-tk+s}=(tn+r+s)^n\frac{r+s}{tn+r+s},$$</span>
but these seem to just barely not work for my purpose. Does anyone know a proof of this result (or perhaps a counterexample)? I suspect I will have to do some playing with generating functions.</p>
| Marko Riedel | 44,883 | <p>We seek to prove that</p>
<p><span class="math-container">$$(n+2)^n = (n+1) \sum_{k=0}^n {n\choose k}
\frac{(k+1)^{k-1} (n-k)^{n-k}}{n+1-k}.$$</span></p>
<p>Observe that the RHS is</p>
<p><span class="math-container">$$\sum_{k=0}^n {n+1\choose k}
(k+1)^{k-1} (n-k)^{n-k}
\\ = (n+2)^n +
\sum_{k=0}^{n+1} {n+1\choose k}
(k+1)^{k-1} (n-k)^{n-k}
\\ = (n+2)^n +
\sum_{k=0}^{n+1} {n+1\choose k}
(k+1)^{k-1} (n+1-k-1)^{n+1-k-1}.$$</span></p>
<p>We are therefore tasked with showing that the sum is zero.
Introduce the tree function <span class="math-container">$T(z)$</span> from combinatorics where <span class="math-container">$T(z) = z
\exp T(z)$</span> and <span class="math-container">$T(z) = - W_0(-z).$</span> Note that we have by Cayley's
theorem that <span class="math-container">$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}.$</span> Note also
that the functional equation is derived from the combinatorial class
equation</p>
<p><span class="math-container">$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T}).$$</span></p>
<p>which might be useful in a purely combinatorial proof.</p>
<p>Observe that</p>
<p><span class="math-container">$$P(z) = \frac{T(z)}{z} = \sum_{n\ge 1} n^{n-1} \frac{z^{n-1}}{n!}
= \sum_{n\ge 1} n^{n-2} \frac{z^{n-1}}{(n-1)!}
\\ = \sum_{n\ge 0} (n+1)^{n-1} \frac{z^n}{n!}$$</span></p>
<p>Next introduce</p>
<p><span class="math-container">$$Q(z) = \frac{z}{T(z)}.$$</span></p>
<p>We have for the coefficients that</p>
<p><span class="math-container">$$n! [z^n] Q(z) =
n! \;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+1}} \frac{z}{T(z)}
= n! \;\underset{z}{\mathrm{res}}\;
\frac{1}{z^n} \frac{1}{T(z)}.$$</span></p>
<p>Now put <span class="math-container">$T(z) = w$</span> so that <span class="math-container">$w = z \exp(w)$</span> or <span class="math-container">$z = w \exp(-w)$</span>
and <span class="math-container">$dz = \exp(-w) (1-w) \; dw$</span> to obtain</p>
<p><span class="math-container">$$n! \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^n} \exp(nw) \frac{1}{w} \exp(-w) (1-w)
\\ = n! \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{n+1}} \exp((n-1)w) (1-w)
= n! \frac{(n-1)^n}{n!}
- n! \frac{(n-1)^{n-1}}{(n-1)!}
\\ = (n-1)^n - n (n-1)^{n-1} = - (n-1)^{n-1}$$</span></p>
<p>so that in fact</p>
<p><span class="math-container">$$Q(z) = - \sum_{n\ge 0} (n-1)^{n-1} \frac{z^n}{n!}.$$</span></p>
<p>We thus require by convolution of EGFs with <span class="math-container">$n\ge 0$</span>
the value of</p>
<p><span class="math-container">$$- (n+1)! [z^{n+1}] P(z) Q(z)$$</span></p>
<p>which is</p>
<p><span class="math-container">$$ - (n+1)! [z^{n+1}] 1 = 0$$</span></p>
<p>and we have the claim.</p>
<p><strong>Remark.</strong> Here we have used the fact that when we
multiply two exponential generating functions of the sequences
<span class="math-container">$\{p_n\}$</span> and <span class="math-container">$\{q_n\}$</span> we get that</p>
<p><span class="math-container">$$ P(z) Q(z) = \sum_{n\ge 0} p_n \frac{z^n}{n!}
\sum_{n\ge 0} q_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} p_k q_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} p_k q_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} p_k q_{n-k}\right)\frac{z^n}{n!}.$$</span></p>
|
4,350,695 | <p>My book was introducing the concept of integrals and wrote this:</p>
<p><span class="math-container">$$\text{Area under the curve of $f(x)$}=\lim_{\Delta x\to0}\sum_{n=1}^{N}f(x)\Delta x\tag{1}$$</span></p>
<p>My problem with <span class="math-container">$(1)$</span> is that there is no <span class="math-container">$n$</span> in the expression <span class="math-container">$f(x)\Delta x$</span>. So, what does <span class="math-container">$n=1$</span> below the summation notation even mean then? To elucidate my point:</p>
<p>Let,</p>
<p><span class="math-container">$y=f(x)$</span></p>
<p>Now,</p>
<p><span class="math-container">$$\text{Area under the curve of $f(x)$ from $y_1$ to $y_N$}=\lim_{\Delta x\to0}\sum_{n=1}^{N}y_n\Delta x\tag{2}$$</span></p>
<p>Now, the usage of <span class="math-container">$n$</span> makes sense because there is an actual <span class="math-container">$n$</span> present in the expression <span class="math-container">$y_n\Delta x$</span>.</p>
<p>So, is my book wrong?</p>
| Saksham Paliwal | 1,012,372 | <p>Yeah actually your book wants to write <span class="math-container">$$\sum_{n=1}^{N}f(x_n)\Delta x_n$$</span></p>
<p><span class="math-container">$\Delta x_n$</span> is the width of the elemental rectangle and <span class="math-container">$f(x_)$</span> is the value of the function at the near neighbor of <span class="math-container">$x_$</span>.</p>
|
1,417,286 | <p>So I'm trying to learn about RSA and have come across various subtopics, including the discrete logarithm problem. </p>
<p>This mentions primitive roots, which I do not understand.</p>
<p>Essentially all I want is an answer in simple terms of what a primitive root actually is.</p>
<p>Thanks</p>
| Peter | 82,961 | <p>Let $n>1$ be a natural number.</p>
<p>Let $S$ be the set of numbers coprime to $n$. A number $m$ is called a primitive
root in $\mathbb Z_n$, if the Set $\{m,m^2,m^3,...,m^{\phi(n)}\}$ modulo $n$ contains
every element of $S$.</p>
<p>$\phi(n)$ is the Euler-Phi-Function : The number of $m's$ with $gcd(m,n)=1$</p>
<p>Example : $n=10$ </p>
<p>Numbers coprime to $10$ : $\{1,3,7,9\}$</p>
<p>The elements $3,3^2,3^3,3^4$ are congruent $3,9,7,1$ modulo $10$, so all the
numbers occur. Hence, $3$ is a primitve root in $\mathbb Z_{10}$</p>
|
1,989,253 | <p>I am trying to evaluate: </p>
<p>$$\lim_{x \to 4}\frac{\sqrt{5-x} - 1}{2-\sqrt{x}}.$$</p>
<p>Even though I tried rationalizing both denominator and numerator, I still end up with the functioning being undefined.</p>
<p>How can I solve this without rationalizing?</p>
| Nicolas FRANCOIS | 288,125 | <p>Rationalizing gives :
$$\frac{\sqrt{5-x}-1}{2-\sqrt x} = \frac{(4-x)(2+\sqrt x)}{(4-x)(\sqrt{5-x}+1)}$$
I think you can find the limit from here :-)</p>
|
1,358,002 | <p>My son did something quite impressive the other day. It was shear luck but I don't think I'll ever see it duplicated again in my lifetime. </p>
<p>I brought my kids to the boardwalk and my son wanted to play an amusement game. It was the arrow spin wheel game. It had 90 different names or possibilities to win. You could either pick 1 name for \$1 or 3 names for \$2. </p>
<p>He chose 3 names for \$2 and won on his very first try. He picked his prize and we walked over to a completely different spinning wheel game with the same amount of names and cost. On his 1st spin, he won again and again he chose his prize and we left. Yet again we went to a 3rd similar wheel with the same amount of names and chances and he wins again on the 1st try. </p>
<p>What are the odds that someone could win on their 1st 3 tries (or 3 in a row) like my son did?
Thanks!</p>
| Tryss | 216,059 | <p>With 3 names, the probability to win a game on the first try is $\frac{1}{30}$</p>
<p>So the probability to win 3 games in a row is<br>
$$\frac{1}{30} \times \frac{1}{30} \times \frac{1}{30} = \frac{1}{27000} \simeq 0.003\%$$</p>
|
646,109 | <p>For function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies $f\left(x+y\right)=f\left(x\right)f\left(y\right)$
and is not the zero-function I can prove that $f\left(1\right)>0$
and $f\left(x\right)=f\left(1\right)^{x}$ for each $x\in\mathbb{Q}$.
Is there a way to prove that for $x\in\mathbb{R}$?</p>
<p>This question has been marked to be a duplicate of the question whether $f(xy)=f(x)f(y)$ leads to $f(x)=x^p$ for some $p$. I disagree on that. Both questions are answered by means of construction of a function $g$ that suffices $g(x+y)=g(x)+g(y)$. In this question: $g(x)=\log f(x)$ and in the other $g(x)=\log f(e^x)$. So the answers are alike, but both questions definitely have another startpoint.</p>
| Andreas Caranti | 58,401 | <p>No, because if $f$ is any of your functions, you may take any additive function $g : \mathbb{R} \to \mathbb{R}$ (that is, a function such that $g(x+y) = g(x) + g(y)$), and $f \circ g$ will still satisfy your assumption, as $f \circ g (x + y) = f(g(x+y)) = f(g(x) + g(y)) = f(g(x)) f(g(y)) = f \circ g(x) f \circ g(y)$.</p>
<p>And there are plenty such $g$, see under <a href="http://mathworld.wolfram.com/HamelBasis.html" rel="nofollow">Hamel basis</a>.</p>
|
1,178,080 | <p>How to calculate the number of solutions of the equation $x_1 + x_2 + x_3 = 9$ when $x_1$, $x_2$ and $x_3$ are integers which can only range from <code>1</code> to <code>6</code>.</p>
| Mayank Jha | 220,999 | <p>A very simple way first count no.of
Positive solutions by $(n-1)C(r-1)$.
I hope you understand the meaning of $C$.Then if $x_1$ is $7$ then no. of solutions is $1$ similarly for $x_2$ & $x_3$
is also $1$, so number of solution is $28-3=25$.</p>
|
48,077 | <p>First, I'm a beginner.</p>
<p>I can compute the sum of roots with the follwing:</p>
<pre><code>Roots[x^7 + 5 x^6 + x^5 + x + 1 == 0, x]
Plus @@ (x /. {ToRules[%]}) // Simplify
</code></pre>
<p>Of course I get, except the sign, the coefficient of x^6.</p>
<p>Now, is there a way to compute more elaborate symmetric functions, for example the sum of xi/xj for all i,j ?</p>
| Wouter | 7,680 | <p>It is not clear if this is what you were asking, but it might help:
from the link to <a href="https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial" rel="nofollow">https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial</a> we extract </p>
<pre><code>Product[\[Lambda]+Subscript[x, k],{k,7}]==Sum[\[Lambda]^k ee[7-k,7],{k,0,7}]
</code></pre>
<p>with </p>
<pre><code> ee[k_,n_]:=SymmetricPolynomial[k, Thread[Subscript[x, Range[n]]]]
</code></pre>
<p>Now, with your definitions:</p>
<pre><code>(x/.{ToRules[Roots[x^7+5 x^6+x^5+x+1==0,x]]})//Simplify ;
ru = Thread[Thread[Subscript[x, Range[7]]] -> %];
Sum[\[Lambda]^k ee[7 - k, 7], {k, 0, 7}]/. ru // FullSimplify
</code></pre>
<p>returns (* -1+[Lambda]+[Lambda]^5-5 [Lambda]^6+[Lambda]^7 *)</p>
<p>The mathematics of this are obviously clear to you, maybe you had not yet located the function 'SymmetricPolynomial'?</p>
|
944,948 | <p>$\textbf{QUESTION-}$
Let $P$ be a p-group with $|P:Z(P)|\leq p^n$. Show that $|P'| \leq p^{n(n-1)/2}$.</p>
<p>If $P=Z(P)$ it is true. Now let $n > 1$, then</p>
<p>If I see $P$ as a nilpotent group and construct its upper central series, it will end , so let it be,</p>
<p>$e=Z_0<Z_1<Z_2<......<Z_r=P$ </p>
<p>Now as $Z_{i+1}/Z_i=Z(P/Z_i)$, so if if I take some $x\in Z_2$\ $Z_1$ then $N$={$[x,y]|y\in P$} $\leq Z_1(P)$ and $N \triangleleft P $, so $P/N$ is a group with order $\leq p^{n-1}$.</p>
<p>Now if I let $H=P/N$ then obviously |$H/Z(H)$|$\leq p^{n-1}$.</p>
<p>Now $H'\cong P'N/N \cong P'/(P' \cap N)$ so from here I could finally bring $P'$ atleast into the picture, now |$P'$|=$|H'|.|P'\cap N|$ so $|P'|\leq |H'||N|$. </p>
<p>This is where I am $\textbf{STUCK}$</p>
<p>Now , from here how can I calculate or find some power $p$ bounds on $|H'|$ and $|N|$ so i could get my result.</p>
| egreg | 62,967 | <p>You can write your equation as
$$
f(x+y)=f(x)+x^2y+xy^2+f(y)
$$
so
$$
f'(x)=\lim_{y\to0}\frac{f(x+y)-f(x)}{y}=
\lim_{y\to0}\frac{f(x)+x^2y+xy^2+f(y)-f(x)}{y}=
\lim_{y\to0}\frac{x^2y+xy^2+f(y)}{y}
$$
I bet you can take on from here.</p>
|
590,817 | <p>(I'm a software developer so excuse me)</p>
<p>I'm building an application for a client and one of the formulas that has been provided in the spec is <code>value1 = value2 * (1 + 5%)</code>. When I asked about it I was told that it's some kind of notation for <code>value1 = value2 * 0.15</code>.</p>
<p>Also, they said that <code>value1 = value2 * (1 + -5%)</code> is the same as <code>value1 = value2 * 0.95</code>.</p>
<p>Could someone tell me what this means? Where can I read about what is actually going on with these calculations?</p>
<p>Another question: How can a percentage be negative?</p>
<p><strong>edit:</strong> I could be wrong about the first example. Client might have said 1.05 and not 0.15.</p>
| alexjo | 103,399 | <p>Note that $ a\%$ means $a/100$, so $5\%=5/100=0.05$ and $1+5\%=1+0.05=1.05$ and $1-5\%=1-0.05=0.95$. Thus value1=value2*(1+5%) is the same as value1=value2*1.05 and value1=value2*(1+ -5%) is the same as value1=value2*0.95.</p>
<p>For more explanations you can read <a href="http://en.wikipedia.org/wiki/Percentage" rel="nofollow">http://en.wikipedia.org/wiki/Percentage</a></p>
|
1,211,287 | <p>Given that the angles between the consecutive lateral edges AB, AC & AD meeting at the vertex A of a tetrahedron ABCD are $ α, β, γ$ (as shown in the diagram below). Is there any set-formula to find out the solid angle subtended by the tetrahedron at the same vertex? </p>
<p>Note: A tetrahedron is a solid having 4 triangular faces, 6 edges & 4 vertices. Three triangular faces meet together at each of four vertices & each of six edges is shared (common) by two adjacent triangular faces. </p>
<p><img src="https://i.stack.imgur.com/CdCvt.jpg" alt="Tetrahedron"></p>
| AB Balbuena | 147,730 | <p>Denote the solid angle by $\omega$ and let $v_1,v_2,v_3$ be the vectors from vertex $A$ along the edges $AB, AD, AC$. Then we have (using the usual cross product, dot product, and Euclidean norm):</p>
<p>$$(4 \pi)\omega + \pi = \cos ^{-1} \left( \frac{ (v_1 \times v_2) \cdot (v_1 \times v_3)}{||v_1 \times v_2|| ||v_1 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_2 \times v_1) \cdot (v_2 \times v_3)}{||v_2 \times v_1|| ||v_2 \times v_3||}\right) + \cos ^{-1} \left( \frac{ (v_3 \times v_1) \cdot (v_3 \times v_2)}{||v_3 \times v_1|| ||v_3 \times v_2||}\right)$$</p>
|
1,878,573 | <p><a href="https://i.stack.imgur.com/3iZQ8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3iZQ8.png" alt="enter image description here"></a></p>
<p>I cannot get the $f'(0)$ by using L'Hôpital's rule, because it appears recurrence item. Can you help me?</p>
| DanielWainfleet | 254,665 | <p>For $x\ne 0$ we have $|\sin \frac {1}{x^3}|\leq 1.$ We have $e^x=1+x +O(x^2)$ as $x\to 0$. So $e^{x^2}-e^{-x^2}=2x^2 +O(x^4)$ as $x\to 0.$ So for $x\ne 0$ we have $$|\frac {f(x)-f(0)}{x-0}|=|\frac {f(x)}{x}|\leq |\frac {e^{x^2}-e^{-x^2}}{x}|=|\frac {2x^2+O(x^4)}{x}|=$$ $$=|2x+O(x^3)|.$$ So $f'(0)=0 .$</p>
|
3,227,215 | <p><a href="https://i.stack.imgur.com/7pJ4t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7pJ4t.png" alt="enter image description here" /></a></p>
<blockquote>
<p><span class="math-container">$(O, R)$</span> is the circumscribed circle of <span class="math-container">$\triangle ABC$</span>. <span class="math-container">$I \in \triangle ABC$</span>. <span class="math-container">$AI$</span>, <span class="math-container">$BI$</span> and <span class="math-container">$CI$</span> intersects <span class="math-container">$AB$</span>, <span class="math-container">$BC$</span> and <span class="math-container">$CA$</span> respectively at <span class="math-container">$M$</span>, <span class="math-container">$N$</span> and <span class="math-container">$P$</span>. Prove that <span class="math-container">$$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$$</span></p>
</blockquote>
<p>I have provided my own solution and I would be greatly appreciated if there are any other solutions, perhaps one involving trigonometry. I deeply apologise for the misunderstanding.</p>
| Lê Thành Đạt | 654,749 | <p>Firstly,</p>
<p><span class="math-container">$$\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP} = \frac{A_{CIB}}{A_{CAB}} + \frac{A_{AIC}}{A_{ABC}} + \frac{A_{BIA}}{A_{BCA}} = 1$$</span></p>
<p>where <span class="math-container">$A_m$</span> denotes the area of shape <span class="math-container">$m$</span>.</p>
<p><span class="math-container">$$\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP} = 3 - \left(\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP}\right) = 2$$</span></p>
<p>We have that <span class="math-container">$$\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{1}{3}\left(\frac{1}{AM} + \frac{1}{BN} + \frac{1}{CP}\right)^2$$</span></p>
<p><span class="math-container">$$\le \frac{1}{3}\left[\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2\right]\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right)$$</span></p>
<p>(according to the Cauchy-Schwarz inequality)</p>
<p>However </p>
<p><span class="math-container">$$\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2 \ge \dfrac{1}{3}\left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)^2 = \frac{4}{3}$$</span></p>
<p>Now I just need to prove that</p>
<p><span class="math-container">$$\frac{1}{3}\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right) \le \frac{1}{(R - OI)^2}$$</span></p>
<p>which can be easily proven because</p>
<p><span class="math-container">$$\left\{ \begin{align} AI \ge |AO - OI|\\ BI \ge |BO - OI|\\ CI \ge |CO - OI|\\ \end{align} \right. = |R - OI|$$</span></p>
|
821,875 | <p>A school director must randomly select 6 teachers to participate in a training session. There are 30 teachers at the school. In how many different ways can these teachers be selected, if the order of selection does not matter?</p>
| Asinomás | 33,907 | <p>If the order mattered it would be $30\cdot29\cdot28\cdot27\cdot26\cdot25$ since there would be $30$ option for the first to be picked, $29$ for the second to be piecked and so on...</p>
<p>However if order does not matter each unordered selection appears $6\cdot5\cdot4\cdot3\cdot2$ in the ordered selections.</p>
<p>So the answer you are looking for is $\frac{30\cdot29\cdot28\cdot27\cdot26\cdot25}{6\cdot5\cdot4\cdot3\cdot2}$</p>
<p>Using factorial notation this is $\frac{30!}{6!(30-6)!}$</p>
<p>using binomial coefficient notation this is $\binom{30}{6}$ or $30\text C6$</p>
|
120,067 | <p>The <em>theta function</em> is the analytic function $\theta:U\to\mathbb{C}$ defined on the (open) right half-plane $U\subset\mathbb{C}$ by $\theta(\tau)=\sum_{n\in\mathbb{Z}}e^{-\pi n^2 \tau}$. It has the following important transformation property.</p>
<blockquote>
<p><strong>Theta reciprocity</strong>: $\theta(\tau)=\frac{1}{\sqrt{\tau}}\theta\left(\frac{1}{\tau}\right)$.</p>
</blockquote>
<p>This theorem, while fundamentally analytic—the proof is just Poisson summation coupled with the fact that a Gaussian is its own Fourier transform—has serious arithmetic significance.</p>
<ul>
<li><p>It is the key ingredient in the proof of the functional equation of the Riemann zeta function.</p></li>
<li><p>It expresses the <em>automorphy</em> of the theta function.</p></li>
</ul>
<p>Theta reciprocity also provides an analytic proof (actually, the <em>only</em> proof, as far as I know) of the Landsberg-Schaar relation</p>
<p>$$\frac{1}{\sqrt{p}}\sum_{n=0}^{p-1}\exp\left(\frac{2\pi i n^2 q}{p}\right)=\frac{e^{\pi i/4}}{\sqrt{2q}}\sum_{n=0}^{2q-1}\exp\left(-\frac{\pi i n^2 p}{2q}\right)$$</p>
<p>where $p$ and $q$ are arbitrary positive integers. To prove it, apply theta reciprocity to $\tau=2iq/p+\epsilon$, $\epsilon>0$, and then let $\epsilon\to 0$.</p>
<p>This reduces to the formula for the quadratic Gauss sum when $q=1$:</p>
<p>$$\sum_{n=0}^{p-1} e^{2 \pi i n^2 / p} =
\begin{cases}
\sqrt{p} & \textrm{if } \; p\equiv 1\mod 4 \\\
i\sqrt{p} & \textrm{if } \; p\equiv 3\mod 4
\end{cases}$$</p>
<p>(where $p$ is an odd prime). From this, it's not hard to deduce Gauss's "golden theorem".</p>
<blockquote>
<p><strong>Quadratic reciprocity</strong>: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)(q-1)/4}$ for odd primes $p$ and $q$.</p>
</blockquote>
<p>For reference, this is worked out in detail in the paper "<a href="http://www.math.kth.se/~akarl/langmemorial.pdf">Applications of heat kernels on abelian groups: $\zeta(2n)$, quadratic reciprocity, Bessel integrals</a>" by Anders Karlsson.</p>
<hr>
<p>I feel like there is some deep mathematics going on behind the scenes here, but I don't know what.</p>
<blockquote>
<p>Why should we expect theta reciprocity to be related to quadratic reciprocity? Is there a high-concept explanation of this phenomenon? If there is, can it be generalized to other reciprocity laws (like Artin reciprocity)?</p>
</blockquote>
<p>Hopefully some wise number theorist can shed some light on this!</p>
| Jonah Sinick | 683 | <p>Hecke generalized the argument that you mention to prove quadratic reciprocity relative to any given number field $K$ (see, e.g. his <a href="http://books.google.com/books/about/Lectures_on_the_Theory_of_Algebraic_Numb.html?id=QEocn1kbfxAC" rel="nofollow noreferrer">Lectures on the Theory of Algebraic Numbers</a>).</p>
<p>In <a href="https://rads.stackoverflow.com/amzn/click/0471358304" rel="nofollow noreferrer">The Fourier-Analytic Proof of Quadratic Reciprocity</a> Michael C. Berg describes the subsequent development of this line of research. Quoting from the book's summary:</p>
<blockquote>
<p>The relative quadratic case was first
settled by Hecke in 1923, then recast
by Weil in 1964 into the language of
unitary group representations. The
analytic proof of the general n-th
order case is still an open problem
today, going back to the end of
Hecke's famous treatise of 1923.</p>
</blockquote>
|
3,950,463 | <blockquote>
<p>What is <span class="math-container">$100$</span>th derivative of <span class="math-container">$y=\ln(2x-x^2)$</span> at <span class="math-container">$x=1$</span>?</p>
<p><span class="math-container">$a)2\times99!$</span></p>
<p><span class="math-container">$b)-2\times99!$</span></p>
<p><span class="math-container">$c)2\times101!$</span></p>
<p><span class="math-container">$d)-2\times101!$</span></p>
</blockquote>
<p>I tried to find a pattern by calculating first derivatives of <span class="math-container">$y$</span>:</p>
<p><span class="math-container">$$y'=\cfrac{2-2x}{2x-x^2} ,\quad y'(1)=0$$</span></p>
<p><span class="math-container">$$y''=\cfrac{-2(2x-x^2)-(2-2x)^2}{(2x-x^2)^2},\quad y''(1)=-2$$</span></p>
<p>And <span class="math-container">$y'''$</span>is going to get really ugly. so I couldn't find any pattern.</p>
| dezdichado | 152,744 | <p>You split them first:</p>
<p><span class="math-container">$$y = \ln(2-x) + \ln(x)\implies y' = \frac{1}{x-2}+\frac 1x\dots{}{} $$</span></p>
|
10,505 | <p>I've only done a few questions here but already it's grinding on me. Why can't we have the writing-answer panel and the preview panel side by side, rather than below, this means for big answers I can't make use of the preview! It'd be great if side by side, two scroll-bars, or even a pop out (I have a window manager that gives me an always-on-top button :)) but this may not help those 'less fortunate' when it comes to OS.</p>
<p>It'd also be great if the preview wasn't trying to update EVERY keystroke, I'd love to have (the option, as I can see others disagreeing) an update button, or only when I pressed enter. Maybe only try and update everything after the last parsed expression (rather than the entire messages) </p>
<p>It can get QUITE laggy. So please please do this, or have like a profile option to do it. </p>
<p>Thanks</p>
| Antonio Vargas | 5,531 | <p>I've been contributing to a userscript which adds side-by-side editing functionality to Stack Exchange. It can be toggled on and off by clicking a new button in the editor and works when writing new as well as editing existing questions and answers.</p>
<h2>Screenshot</h2>
<p><img src="https://i.stack.imgur.com/RSP8a.png" alt="editor on the left, preview on the right" /></p>
<p><strong><a href="https://stackapps.com/q/6461/36263">More info can be found over at StackApps SE.</a></strong></p>
|
35,463 | <p>I was asked the following vector calculus problem:</p>
<blockquote>
<p>Let <span class="math-container">$D$</span> be the unit ball and let <span class="math-container">$S$</span> be the unit sphere in <span class="math-container">$\mathbb{R}^3$</span>. Suppose that <span class="math-container">$F:\mathbb{R}^3\rightarrow \mathbb{R}^3$</span> is a <span class="math-container">$C^1$</span> vector field on some open neighborhood of <span class="math-container">$D$</span> which satisfies:</p>
<p><span class="math-container">$(i) \nabla\times F=0$</span></p>
<p><span class="math-container">$(ii) \nabla\cdot F=0$</span></p>
<p><span class="math-container">$(iii)$</span> On <span class="math-container">$S$</span>, <span class="math-container">$F$</span> is orthogonal to the radial vector.</p>
<p>Prove that <span class="math-container">$F=0$</span> on all of <span class="math-container">$D$</span>.</p>
</blockquote>
<p>Conditions <span class="math-container">$(i)$</span> and <span class="math-container">$(ii)$</span> imply that <span class="math-container">$F=\nabla g$</span> for some <span class="math-container">$g:\mathbb{R}^3\rightarrow \mathbb{R}$</span> where <span class="math-container">$g$</span> must be harmonic as well.</p>
<p>I know one solution (see end), however my initial instinct was to try to use the max/min property of harmonic functions, and I couldn't get it to work. Since the gradient is always orthogonal to the sphere, there must be a point on the sphere where it is <span class="math-container">$0$</span>. (Hairy ball) If that was a local max or min in <span class="math-container">$\mathbb{R}^3$</span> we would be done, by taking a small neighborhood around it. If it is a saddle point this doesn't work. (We know that it must be a local max/min on <span class="math-container">$S$</span> since it is harmonic)</p>
<p><strong>My question is:</strong> Is there any way to modify this approach, and solve the problem?</p>
<p>Thanks!</p>
<p><strong>Other Solution:</strong> Here is one solution that first uses the fact that the radial vector is orthogonal, and then applies Gauss's Divergence theorem to the function <span class="math-container">$gF$</span>. (<span class="math-container">$\nabla g=F$</span>) That is <span class="math-container">$$0=\iint_S (gF\cdot n)dS=\iiint_D \nabla\cdot (gF)dV=\iiint_D \|F\|^2dV,$$</span> and since the integrand on the right hand side is non-negative, continuous and integrates to give zero, it must be zero.</p>
| Hans Engler | 9,787 | <p>Your second proof is a good one and entirely appropriate within vector calculus. The first attempt has a gap (I think). </p>
<p>A third proof relies on Hopf's Lemma (commonly taught in graduate level classes on partial differential equations) which implies here that if a function $u$ satisfying $\Delta u \le 0$ in $D$ attains a strict minimum at $z \in \partial D = S$, then the outer normal derivative at that point satisfies $\nu \cdot \nabla u(z) < 0$. Applying this with $g = u$, it follows that $g$ must attain its minimum in the interior of $D$ and hence must be constant. </p>
|
2,317,867 | <p>It is said that asymmetry implies antisymmetry, but how come so?</p>
<p>If $aRb \Rightarrow \neg(bRa)$, isn't this also the case for $a = b$, which indicates $aRa \Rightarrow \neg(aRa)$, which is a contradiction.</p>
| user454135 | 454,135 | <p>Asymmetry says that given $aRb$, we can't have $bRa$.
Antisymmetry has both $aRb$ and $bRa$ as hypothesis, so in this case it is true by vacuity, since both hypothesis can never be satisfied simultaneously.</p>
|
2,317,867 | <p>It is said that asymmetry implies antisymmetry, but how come so?</p>
<p>If $aRb \Rightarrow \neg(bRa)$, isn't this also the case for $a = b$, which indicates $aRa \Rightarrow \neg(aRa)$, which is a contradiction.</p>
| luifrancgom | 768,068 | <p>First some definitions:</p>
<ul>
<li><p><span class="math-container">$R \text{ is antisymmetric in } A \iff (\forall x)(\forall y)((x \in A \land y \in A \land xRy \land yRx) \implies x = y)$</span></p></li>
<li><p><span class="math-container">$R \text{ is asymmetric in } A \iff (\forall x)(\forall y)((x \in A \land y \in A \land xRy) \implies \neg(yRx))$</span></p></li>
</ul>
<p>Second, as you mention if <span class="math-container">$R$</span> is asymmetric then <span class="math-container">$(x \in A \land y \in A \land xRx) \implies \neg(xRx)$</span> is a contradiction. Therefore if <span class="math-container">$R$</span> is asymmetric in <span class="math-container">$A$</span>, <span class="math-container">$x \in A$</span>, <span class="math-container">$y \in A$</span> and <span class="math-container">$x = y$</span> then by <em>reductio ad absurdum</em><span class="math-container">$^1$</span> <span class="math-container">$\neg(xRy)$</span></p>
<ul>
<li>In particular if <span class="math-container">$R$</span> is asymmetric and <span class="math-container">$x \in A$</span> then <span class="math-container">$\neg(xRx)$</span></li>
</ul>
<p>In that sense, if <span class="math-container">$R$</span> is asymmetric in <span class="math-container">$A$</span>, <span class="math-container">$x \in A$</span> and <span class="math-container">$y \in A$</span> then <span class="math-container">$x \neq y \iff xRy \lor yRx$</span> <span class="math-container">$^2$</span> </p>
<p>Third, we can rewrite the definition of antisymmetric as:</p>
<ul>
<li><span class="math-container">$R \text{ is antisymmetric in } A \iff (\forall x)(\forall y)(x \neq y \implies (x \notin A \lor y \notin A \lor \neg(xRy) \lor \neg(yRx)))$</span></li>
</ul>
<p>Fourth, we have that if <span class="math-container">$R$</span> is asymmetric in <span class="math-container">$A$</span>, <span class="math-container">$x \in A$</span>, <span class="math-container">$y \in A$</span> then <span class="math-container">$x \neq y \iff xRy \lor xRy$</span>. Let us assume that <span class="math-container">$xRy$</span> without loss of generality. Since <span class="math-container">$R$</span> is asymmetric <span class="math-container">$xRy \implies \neg(yRx)$</span>. Therefore we have that</p>
<ul>
<li><span class="math-container">$R$</span> is asymmetric in <span class="math-container">$A$</span>, <span class="math-container">$x \in A$</span>, <span class="math-container">$y \in A$</span> and <span class="math-container">$xRy$</span> then <span class="math-container">$x \neq y \implies \neg(yRx)$</span></li>
</ul>
<p>So <span class="math-container">$R$</span> is also antisymmetric <span class="math-container">$\square$</span> </p>
<p>If you want to learn more about relations check out in your local library the book:</p>
<ul>
<li>Mendelson, Elliott. 2008. Number Systems and the Foundations of Analysis. Mineola, N.Y: Dover Publications. ISBN-13: 978-0486457925</li>
</ul>
<p>In particular check out: <strong>Chapter 1 Basic Facts and notions of logic and set theory > 1.16 Reflexivity, symmetry and transitivity > Exercises > 2 > b</strong></p>
<p><span class="math-container">$^1$</span> This means that if <span class="math-container">$\neg\alpha \implies (\beta \land \neg\beta)$</span> is true then <span class="math-container">$\alpha$</span> is true. In our case assuming <span class="math-container">$R$</span> is asymmetric and <span class="math-container">$xRx$</span> implies <span class="math-container">$\neg(xRx)$</span> meaning that <span class="math-container">$xRx \land \neg(xRx)$</span> so <span class="math-container">$xRx$</span> can't be true.</p>
<p><span class="math-container">$^2$</span> To be more precise <span class="math-container">$x \neq y \iff xRy \oplus yRx$</span> where <span class="math-container">$\oplus$</span> refers to the exclusive or. </p>
|
3,985,177 | <p>When it comes to proving that two sets are equal, say <span class="math-container">$A = B$</span>, we're usually told that we have to prove that <span class="math-container">$A \subset B$</span> and <span class="math-container">$B \subset A$</span>. However, I'm under the impression that this strategy isn't unique. Two sets are equal if they have the same elements, and this commonly used strategy is just one way to prove that.</p>
<p>A better way to prove that two sets are equal, at least in my opinion, is to use set notation. However, because I never see this being used, I'm unsure if this method is correct. Example:</p>
<p><em>Prove that <span class="math-container">$ (A \cup B) \times C = (A \times C) \cup (B \times C) $</span></em>.</p>
<p><em>Proof</em>. We have,
<span class="math-container">\begin{align}
(A \cup B) \times C &= \{x; x \in A \operatorname{or} B \} \times \{y; y \in C \} \\
&= \{(x, y); x \in A \operatorname{or} B, y \in C \} \\
&= \{(x, y); x \in A, y \in C \} \cup \{(x, y); x \in B, y \in C \} \\
&= (A \times C) \cup (B \times C).
\end{align}</span></p>
<p>So the question is, is this method correct? Thanks in advance.</p>
| Strichcoder | 655,660 | <p>Yes, Mobius transformations send generalized circles to generalized circles. A generalized circle is a cirlce or a line. If there was a Mobius transformation sending <span class="math-container">$\mathbb{C}$</span> to the disk, then its inverse would also be a Mobius transformation and would send the unit circle to some other generalized circle, exactly as you say. Since Mobius transformations are continuous the unit circle will be mapped on one of the sides of the generalized circle.</p>
|
3,064,501 | <p>So I was trying to find the <strong>time complexity</strong> of an algorithm to find the <span class="math-container">$N$</span>th prime number (where <span class="math-container">$N$</span> could be any positive integer).</p>
<p>So is there any way to exactly determine how far <span class="math-container">$(N+1)$</span>th prime number will be if <span class="math-container">$N$</span>th prime number is already known ?</p>
| S Spring | 622,021 | <p>A good method for the operation of a prime number program is to output only the primes between an upper-bound and the upper-bound - k . Then the program just appears to be jumping from one section of a list to the other on each computation. Of course required consecutive primes are obviously found within the range of the output.</p>
<p>Here is my previous post:</p>
<p>Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.</p>
<p>Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.</p>
<p>Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuing and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.</p>
<p>A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - k. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.</p>
|
2,298,665 | <p>*prior to the body, note that title might be insufficient or inappropriate. Please edit it if it's needed. </p>
<p>I am proving the claim below: </p>
<p>Let $f: [a,b] \to \Bbb R$ be of bounded variation. </p>
<p>$f(x) \ge c \gt 0$ for all $x \in [a, b]$ where $c$ is a constant</p>
<p>$\Rightarrow$ $h(x)$ = $1 \over f(x)$ is of bounded variation on $[a, b]$.</p>
<p>To prove it, I had made up inequality such as - $\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le \mid {1 \over h(x_i)}\mid + \mid {1 \over h(x_{i-1)}}\mid \le {2 \over c}$</p>
<p>Then I want to derive from above, the fact that $\sum_{i=1}^{n}\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le {2 \over c}n$(*)</p>
<p>but from (*) to derive the fact that $\sum_{i=1}^{n}\mid {1 \over h(x_i)} - {1 \over h(x_{i-1})} \mid \le {2 \over c}n \lt\infty$, I need a guarantee that partition of definition of Bounded of Variation is $\lt \infty$. </p>
<p>so Is it true for the defnition of Bounded Variation, it only requires the finite partition?</p>
| Arthur | 15,500 | <p>For instance, for any $p, q\in \Bbb Q$ with $p\neq 0$, $\pi\mapsto p\pi+q$ is an automorphism.</p>
|
268,676 | <p>It is not hard to check that the three roots of $x^3-2=0$ is $\sqrt[3]{2}, \sqrt[3]{2}\zeta_3, \sqrt[3]{2}\zeta_3^{2}$, hence the splitting field for $x^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}[\sqrt[3]{2}, \sqrt[3]{2}\zeta_3, \sqrt[3]{2}\zeta_3^{2}]$. However, since $\sqrt[3]{2}\zeta_3^{2}$ can be compute through $\sqrt[3]{2}, \sqrt[3]{2}\zeta_3$ then the splitting field is $\mathbb{Q}[\sqrt[3]{2}, \sqrt[3]{2}\zeta_3]$.</p>
<p>In the case $x^5-2=0$, in the book Galois theory by J.S.Milne, the author said that the splitting field is $\mathbb{Q}[\sqrt[5]{2}, \zeta_5]$. </p>
<p>My question is : </p>
<ol>
<li>How can the other roots of $x^5-2$ be represented in term of $\sqrt[5]{2}, \zeta_5$, so that he can write the splitting field is$\mathbb{Q}[\sqrt[5]{2}, \zeta_5] $ ?</li>
<li>Is the splitting field for $x^n -a$ over $\mathbb{Q}$ is $\mathbb{Q}[\alpha,\zeta_n]$, where $\alpha$ is the real $n$-th root of $a$ ?</li>
</ol>
| pritam | 33,736 | <p>Note that the roots of $x^n-a$ are $\alpha ,\alpha\zeta_n,\alpha\zeta_n^2,\ldots,\alpha\zeta_n^{n-1}$ where $\alpha$ is as you said. Then the splitting field of $x^n-a$ is $F:=\mathbb{Q}(\alpha ,\alpha\zeta_n,\ldots,\alpha\zeta_n^{n-1})$. Now $\alpha\in F$ and $\zeta_n=\alpha^{-1}(\alpha\zeta_n)\in F$, hence $\mathbb{Q(\alpha ,\zeta_n})\subseteq F$. Now $\alpha\zeta_n^i\in \mathbb{Q}(\alpha ,\alpha\zeta_n)$ for each $i$, hence $F\subseteq \mathbb{Q}(\alpha ,\alpha\zeta_n)$. Therefore $F=\mathbb{Q}(\alpha ,\alpha\zeta_n).$</p>
|
1,930,901 | <p>I need to prove that:</p>
<p>$$f(z) = \frac{Re(z)}{|z|}$$</p>
<p>and </p>
<p>$$g(z) = \frac{Im(z^2)}{|z^2|}$$</p>
<p>both have limit at $z=0$</p>
<p>If I see $z$ as $z = x+iy$ I have:</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{x}{\sqrt{x^2+y^2}}$$</p>
<p>but if I take this limit at $y = x$ we have:</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{x}{\sqrt{2}\sqrt{x^2}}$$</p>
<p>won't that depend on the signal of $x$? So wouldn't this limit be inexistent?</p>
<p>For $g$ we should have:</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{2xy}{\sqrt{(x^2-y^2)^2+(2xy)^2}} = \lim_{(x,y)\to (0,0)}\frac{2xy}{\sqrt{x^4-2x^2y^2+y^4+4x^2y^2}} = $$</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{2xy}{x^2+y^2}$$</p>
<p>doesn't this go to infinity as well?</p>
| Olivier Moschetta | 369,174 | <p>Let $\varepsilon>0$. Write first:
$$\left|\dfrac{1}{j}\sum_{k=1}^j a_k-l\right|\leq\dfrac{1}{j}\sum_{k=1}^j |a_k-l|=\dfrac{1}{j}\sum_{k=1}^{N}|a_k-l|+\dfrac{1}{j}\sum_{k=N+1}^{j}|a_k-l|\,(\ast)$$
where $1\leq N\leq j$. Since $a_j\rightarrow l$ we can choose $N$ such that
$$|a_k-l|<\dfrac{\varepsilon}{2}\text{ if }k>N$$
Next choose $j$ large enough so that
$$\dfrac{1}{j}\sum_{k=1}^{N}|a_k-l|<\dfrac{\varepsilon}{2}$$
This is possible since $N$ has been fixed, so $\sum_{k=1}^{N}|a_k-l|$ is a fixed constant. For this choice of $N$ and if $j$ is large enough we obtain from $(\ast)$:
$$\left|\dfrac{1}{j}\sum_{k=1}^j a_k-l\right|<\dfrac{\varepsilon}{2}+\dfrac{1}{j}\sum_{k=N+1}^j\dfrac{\varepsilon}{2}<\varepsilon$$
since there are fewer than $j$ terms in that latter sum. This shows that $m_j\rightarrow l$.</p>
|
706,546 | <p>If we have a non-zero real $n$ by $m$ matrix $M$, then there may exist a non-zero unit vector $v$ of $m$ elements so that $Mv = 0$. I understand we can't call this an eigenvector with eigenvalue $0$. </p>
<blockquote>
<p>Why is this not a sensible definition of an eigenvector of a rectangular matrix?</p>
</blockquote>
<p>If $m > n$ then $M$ must have rank less than $m$. </p>
<blockquote>
<p>Can such a matrix still have no non-zero unit vectors $v$ of $m$
elements so that $Mv = 0$?</p>
</blockquote>
<p>Note that the $0$ term on the right hand side is $n$-dimensional and $v$ is $m$-dimensional.</p>
| Gabriel Romon | 66,096 | <p>Let <span class="math-container">$F=\mathrm{span}({I_n,A,A^2,...})$</span>.
It suffices to prove that <span class="math-container">$(I_n, A,..., A^{n-1})$</span> is a family with cardinality <span class="math-container">$n$</span> that spans <span class="math-container">$F$</span>.</p>
<p>By Cayley-Hamilton, one can write <span class="math-container">$A^n=a_{n-1}A^{n-1}+\ldots+a_0I_n$</span>, hence <span class="math-container">$$A^n\in \text{span}(I_n, A,..., A^{n-1}).$$</span></p>
<p>More generally, it is proved by induction on <span class="math-container">$m$</span> that <span class="math-container">$\forall m \geq n, A^m \in \operatorname{span}(I_n, A,..., A^{n-1}) $</span>.</p>
<p>Hence <span class="math-container">$(I_n, A,..., A^{n-1})$</span> spans <span class="math-container">$F$</span>, thus <span class="math-container">$\dim F\leq n$</span>.</p>
|
5,711 | <p>I am teaching abroad to non-native English speakers with a large variance of language skills. </p>
<p>I teach both pre-calculus and AP calculus (AB & BC). For both of those classes I define the new terms. I use word problems, have them read from the textbook, take notes in English during class, have the students speak English in class, and teach the class in English. </p>
<p>I am asking this question because my CP told me I need to embed more English into my lesson, so the students learn English. </p>
<p>So what are some techniques to embed English into a mathematics lesson, and what skills and lessons have you learned from teaching mathematics in an ESL environment?</p>
| Pablo B. | 3,260 | <p>Well, apart from what you are already doing, there are not too many things you can do. You teach mathematics in English during Math class. And your students learn the grammar, general vocabulary, etc. during English class. And this is something your principal should understand. Depending on the level of your students, it will be also allowed to use their native language to explain some misunderstandings.</p>
<p>However, you could satisfy your principal if using, for example, short video fragments taken from films, TV series, Youtube, etc. Take a look to these sites, they are just a sample of what you can find:</p>
<p><a href="http://www.qedcat.com/moviemath/index.html" rel="nofollow">Mathematical Movie Database</a> </p>
<p><a href="http://numb3rs.wolfram.com" rel="nofollow">The Math Behind Numb3rs</a></p>
<p><a href="http://mathsci2.appstate.edu/~sjg/simpsonsmath/" rel="nofollow">Simpsons Math</a></p>
<p>The main idea is to create a problem or situation with a real-life context, or easily imaginated by the students. That is where the fragments help. They provide a visual anchor where develop further mathematics. Sometimes an explicit problem will also be set from an scene (it's a classic the movie <em>Die Hard: With a Vengeance</em>, where some problems are proposed to Bruce Willis). </p>
<p>By this way you will not only satisfy your principal, but also improving your teaching practice. </p>
|
207,515 | <p>Suppose I have the following list, </p>
<pre><code>l = {{"b", "c", "d"}, {"e", "b"}, {"a", "b", "d", "e"}}
</code></pre>
<p>and further suppose I have the following association, </p>
<pre><code>l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
</code></pre>
<p>I wonder how can I replace the keys into my list such that I get, </p>
<pre><code>{{2, 3, 4}, {5, 2}, {1, 2, 4, 5}}
</code></pre>
| kglr | 125 | <p>For the specific numbering in OP, you can also use <a href="https://reference.wolfram.com/language/ref/LetterNumber.html" rel="noreferrer"><code>LetterNumber</code></a>:</p>
<pre><code>LetterNumber[l]
</code></pre>
<blockquote>
<p>{{2, 3, 4}, {5, 2}, {1, 2, 4, 5}} </p>
</blockquote>
|
2,354,467 | <p>I am trying to evaluate the following
\begin{equation}
I(a,b) = \int_{a}^{\frac{a+b}{2}} (x-a)^{\alpha-1} \, x^n \, dx + \int_{\frac{a+b}{2}}^{b} (b-x)^{\alpha-1} \, x^n \, dx,
\end{equation}
where $0<\alpha<1$. Wolfram alpha gives no solution. I tried integration by parts without success. My problem is that I don't understand well the evaluation of the limit of the upper limit and this integrand.</p>
| JJacquelin | 108,514 | <p>\begin{equation}
I(a,b) = \int_{a}^{\frac{a+b}{2}} (x-a)^{p-1} \, x^n \, dx + \int_{\frac{a+b}{2}}^{b} (x-a)^{p-1} \, x^n \, dx \qquad 0<p<1.
\end{equation}</p>
<p>I wonder if where is a typo in the equation : Why you don't simply wrote
\begin{equation}
I(a,b) = \int_{a}^{b} (x-a)^{p-1} \, x^n \, dx \qquad 0<p<1.
\end{equation}</p>
<p>Supposing that $\quad b>a\quad$ so that $\quad (x-a)^{p-1}\quad$ be real on the range $\quad a<x<b$</p>
<p>Since $\quad p-1>-1\quad$ the integral is convergent at the lower bound :
$$I(a,b\to a)\sim a^n\frac{(b-a)^p}{p}$$
This is easy to prove with change of variable $\quad x=a+\epsilon$</p>
<p>Moreover, I cannot see any problem of convergence at the upper bound insofar $b>a$.</p>
<p>For further calculus, HINT : The integral involves the Incomplete Beta function.</p>
<p>ATTENTION :</p>
<p>As suspected, there was a mistake in the initial wording of the question.</p>
<p>The typo was corrected by Math. So, My answer to the initial question is no longer consistent with the new wording.</p>
<p>Anyways, the HINT remains valid : The integrals involve the Incomplete Beta function. </p>
|
62,539 | <p>I am using two books for my calculus refresher.</p>
<ol>
<li>Thomas' Calculus </li>
<li>Higher Math for Beginners by Ya. B. Zeldovich</li>
</ol>
<p><strong>My question is :</strong> When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.</p>
<p>Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??</p>
<p>Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.</p>
<p>In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.</p>
| joriki | 6,622 | <p>I like this question; what you're trying to understand is important to understand.</p>
<p>In this answer I'll be talking loosely about infinitesimal quantities linear or quadratic in $\mathrm dx$; I think this is the best way to get a feel for this sort of thing, but similar arguments could also be presented more rigorously.</p>
<p>Basically, the reason is that in the case of the surface area, the effect from the slope is linear in $\mathrm dx$, whereas in the case of the volume, it's quadratic in $\mathrm dx$. Thus we can neglect it in the limit $\mathrm dx\to0$ in the latter case but not in the former.</p>
<p>Let's see what happens if we take the slope into account in adding up the volume of slices of the solid of revolution generated by a function $f(x)$ rotated around the $x$ axis. As you say, after the cylindrical volume $\pi f^2\mathrm dx$ the next order of approximation would be a cone, or more precisely a conical frustum, corresponding to a linear approximation to the function. The volume of such a frustum between $x$ and $x+\mathrm dx$ would be</p>
<p>$$\begin{eqnarray}
\frac13\pi\mathrm dx\left(f(x)^2+f(x)f(x+\mathrm dx)+f(x+\mathrm dx)^2\right)
&\approx&
\frac13\pi\mathrm dx\left(3f(x)^2+3f(x)\mathrm dx\right)
\\
&=&
\pi\mathrm dx\left(f(x)^2+f(x)\mathrm dx\right)\;,
\end{eqnarray}
$$</p>
<p>which differs from the cylindrical volume by the second term, which contains one more factor of $\mathrm dx$ than the first one and therefore vanishes in the limit.</p>
<p>By contrast, for the surface area, taking into account the slope leads to a surface element $2\pi f(x)\sqrt{1+f'(x)^2}$, whereas not taking it into account would lead to just $2\pi f(x)$, the surface area of a cylindrical slice. Here we don't have two terms with one negligible and dominated by the other, but an additional factor that survives the limit.</p>
<p>You can also try to picture this geometrically. Think of a conical slice and the corresponding cylindrical slice, and imagine shrinking their width. As you shrink, the portion of volume in that little extra bit on the boundary becomes negligible compared to the bulk of the slice -- whereas the bulk only shrinks with the width, the extra bit shrinks both with the width and with the vertical deviation, which is the slope times the width, so it shrinks quadratically while the bulk shrinks linearly. For the surface, there's no such effect, since there's no "bulk" of the surface; all of the surface is at the boundary, and tilting it by the slope makes all of it larger, not just a small portion that becomes negligible in the limit.</p>
|
1,300,853 | <p>Could somebody tell me the exact value of this series?
$$
\sum_{k=1}^{\infty} (-1)^k\frac{H_k^{(5)}}{k}
$$
where
$$
H_k^{(n)}=\sum_{i=1}^{k}\frac{1}{i^n}
$$</p>
<p>Thanks!</p>
| Przemo | 99,778 | <p>Denote ${\bf H}^{(q)}_n(t) := \sum\limits_{m=1}^\infty H_m^{(q)}/m^n \cdot t^m$. By using the formula on the bottom of the answer to <a href="https://math.stackexchange.com/questions/2169507/calculating-alternating-euler-sums-of-odd-powers/2251637#2251637">Calculating alternating Euler sums of odd powers</a> we have:
\begin{equation}
{\bf H}^{(5)}_1(t) = Li_6(t) + Li_1(t) Li_5(t) - Li_2(t) Li_4(t) + \frac{1}{2} [Li_3(t)]^2
\end{equation}
Since $Li_n(-1) = (1-2^{n-1})/2^{n-1} \zeta(n)$ the result is expressed through zeta functions only.</p>
|
24,873 | <p>It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.</p>
<p>Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.</p>
<p>However, the general result that $\mathbb{R^n}$ isn't homeomorphic to $\mathbb{R^m}$ for $n\neq m$, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.</p>
<p>Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?</p>
| M. Mueger | 75,007 | <p>I second Brandsma's answer. Still such proofs tend to be 10 pages long if everything is spelled out. (Engelking spends 7 on proving Brouwer's fixed point theorem using Sperner's lemma, then several pages on dimension theory.)</p>
<p>I wrote down a fairly simple self-contained proof of invariance of dimension (for cubes, not ${\mathbb R^n}$) in four pages. See <a href="http://arxiv.org/abs/1310.8090" rel="nofollow">http://arxiv.org/abs/1310.8090</a></p>
<p>The proof consists of (i) proving a cubical version of Sperner's lemma, (ii) using that to prove that the $n$-cube $I^n$ is $n$-connected in the following sense:</p>
<p>Let $I_i^-=\{ x\in I^n\ | \ x_i=0\},\ I_i^+=\{ x\in I^n\ | \ x_i=1\}$ be the faces of $I^n$.
For $i=1,\ldots,n$, let $H_i^+, H_i^-\subset I^n$ be closed sets such that for
all $i$ one has $I_i^\pm\subset H_i^\pm$ and $H_i^-\cup H_i^+=I^n$. Then $\bigcap_i(H_i^-\cap H_i^+)\ne\emptyset$. </p>
<p>In step (iii) one shows that this implies dim$(I^n)\ge n$ for a somewhat ad-hoc, but convenient, notion of dimension (defined in the same spirit as the above definition of $n$-connectedness), and (iv) one invokes a simple but clever proposition of van Mill showing dim$(I^n)\le n$. Now the claim follows from the (obvious) fact that dim is homeomorphism invariant. </p>
<p>Note: the proofs of (i) and (ii) are due to Kulpa and replace barycentric subdivision by a simple compactness argument.</p>
|
1,858,297 | <p>Suppose the diameter of a nonempty set $A$ is defined as </p>
<p>$$\sigma(A) := \sup_{x,y \in A} d(x,y)$$</p>
<p>where $d(x,y)$ is a metric.</p>
<p>Is $\sigma(.)$ a 'measurement'? I.e., how do I prove the countable additivity for this particular case?</p>
| chi | 207,328 | <p>Let $A$ be a circle of diameter 1, and let $B$ a circle of diameter 2, having the same center of $A$. Note that $A \subseteq B$.</p>
<p>Now, $\sigma(B) = 2$, yet $\sigma(B \setminus A) + \sigma(A) = 2 + 1 = 3$.</p>
|
3,717,932 | <p>How can this identity convolution be shown?</p>
<p><span class="math-container">$$\int^\infty_{-\infty} f(\tau)\delta(t-\tau)d\tau=f(t)$$</span></p>
<p>I keep getting stuck in traps when trying to show this and need a bit of assistance</p>
| Daniel Robert-Nicoud | 60,713 | <p>First of all, one needs to understand that the Dirac delta is <em>not</em> a function, but a distribution, ie an element of the dual space of some space of test functions. Thus, one <em>defines</em> the Dirac delta by its action on smooth, compactly supported function
<span class="math-container">$$\delta(f):=f(0)\ ,$$</span>
and one writes (this is just notation!)
<span class="math-container">$$\delta(f) = \int_{\mathbb{R}}\delta(t)f(t)dt\ .$$</span>
One then <em>defines</em> <span class="math-container">$\delta(t-\tau)$</span> by a formal change of variable <span class="math-container">$u = t-\tau$</span>:
<span class="math-container">$$\int_{\mathbb{R}}\delta(t-\tau)f(t)dt = \int_{\mathbb{R}}\delta(u)f(u+\tau)dt = f(\tau)\ .$$</span></p>
<p><em>Note:</em> Sorry, I switched <span class="math-container">$t$</span> and <span class="math-container">$\tau$</span> with respect to the OP.</p>
|
280,393 | <p>I want to show that $(3, \sqrt 15)$ is not a principal ideal in the ring $ R = \mathbb{Z}[\sqrt{15}]$ with norm $N(a + b \sqrt 15) = a^2 - 15b^2$.</p>
<p>My attempt:</p>
<p>Suppose $(3, \sqrt 15) = (x) $</p>
<p>Then $3 = x * r1$ and $\sqrt 15 = x * r2$ , $r1,r2 \in R$.</p>
<p>$N(3) = 9 = N(x) N(r1)$ and $N(\sqrt 15) = -15 = N(x)N(r2)$ so $N(x) = 3$ or -3 or 1 or -1.</p>
<p>Any ideas on how to continue? Thanks.</p>
| Community | -1 | <p>Here is a proof using some results from algebraic number theory. Suppose your ideal $I = (3,\sqrt{15})$ is a principal generated by some $(\alpha)$ with $\alpha = a + b\sqrt{15}$. Then firstly your ideal $I$ is prime since
$$\begin{eqnarray*} R/I &\cong& \Bbb{Z}[x]/(x^2 - 15)/(3,x)/(x^2 - 15)\\
&\cong& \Bbb{Z}[x]/(3,x) \\
&\cong&\Bbb{Z}/3\Bbb{Z}.\end{eqnarray*}$$</p>
<p>Consequently we also have the <a href="http://en.wikipedia.org/wiki/Ideal_norm" rel="nofollow">absolute norm</a> of $I$ denoted $||I||$ being equal to $3$. On the other hand it is well known that $||(\alpha)|| = |N_{\Bbb{Q}(\alpha)/\Bbb{Q}}(\alpha)| = |a^2 -15b^2|.$</p>
<p>Thus it will now suffice to understand why $a^2 - 15b^2 \neq \pm 3$. Now the square of every integer mod $5$ is $0,1$ or $4$. What happens when you reduce mod $5$?</p>
|
765,404 | <p>Can anyone explain the partial derivative below:</p>
<p>$\frac{\partial a^tX^{-1}b}{\partial X} = -X^{-t}ab^tX^{-t}$</p>
<p>I was trying to derive this equation using the below formula, but failed.</p>
<p><img src="https://i.stack.imgur.com/apR2q.png" alt="enter image description here"></p>
| Hyperplane | 99,220 | <p>A totally mechanical approach. By the chain rule:</p>
<p><span class="math-container">$$\frac{∂a^⊤ X^{-1} b}{∂ X} = \frac{∂a^⊤ X^{-1} b}{∂ X^{-1}}∘\frac{∂X^{-1} }{∂ X}$$</span></p>
<p>Consider the first term <span class="math-container">$\frac{∂a^⊤ X^{-1} b}{∂ X^{-1}}$</span>. Note that the nominator is linear in <span class="math-container">$X^{-1}$</span>, therefore its derivative is found directly by bringing it to the standard form of a linear function "<span class="math-container">$x↦A⋅x$</span>":</p>
<p><span class="math-container">$$a^⊤ X^{-1} b = ⟨ab^⊤∣X^{-1}⟩ ⟹ \frac{∂a^⊤ X^{-1} b}{∂ X^{-1}} = ab^⊤$$</span></p>
<p>Secondly, let's figure out <span class="math-container">$\frac{∂X^{-1}}{∂ X}$</span> first. Note that</p>
<p><span class="math-container">$$ X⋅X^{-1} = ⟹ \frac{d}{dX}(X⋅X^{-1}) =0$$</span></p>
<p>Apply product rule:</p>
<p><span class="math-container">$$\begin{aligned}
0 = \frac{d}{dX}(X⋅X^{-1})
&= \frac{∂\, Y⋅Z}{∂(Y, Z)}\Bigg|_{\begin{aligned}Y&=X\\ Z&=X^{-1}\end{aligned}} \cdot
\frac{∂(X, X^{-1})}{∂X}
\\&= \begin{bmatrix}⊗X^{-⊤},\, X⊗\end{bmatrix}⋅\begin{bmatrix}⊗\\ \frac{∂X^{-1}}{∂ X}\end{bmatrix}
\\&= (⊗X^{-⊤}) + (X⊗)\frac{∂X^{-1}}{∂ X}
\\⟹ \frac{∂X^{-1}}{∂ X} &= -(X⊗)^{-1}(⊗X^{-⊤})
\\&= -(X^{-1}⊗)(⊗X^{-⊤}) = -X^{-1}⊗X^{-⊤}
\end{aligned}$$</span></p>
<p>That is, <span class="math-container">$\frac{∂X^{-1}}{∂ X}$</span> is the linear map <span class="math-container">$V↦ (-X^{-1}⊗X^{-⊤})⋅V = -X^{-1}VX^{-1}$</span></p>
<hr />
<p>Putting both together we have:</p>
<p><span class="math-container">$$\begin{aligned}
\frac{∂a^⊤ X^{-1} b}{∂ X^{-1}}∘\frac{∂X^{-1} }{∂ X}
&= (V↦ ⟨ab^⊤∣V⟩) ∘ (V↦ -X^{-1}VX^{-1})
\\ &= (V↦ ⟨ab^⊤∣-X^{-1}VX^{-1}⟩)
\\ &= (V↦ ⟨-X^{-⊤}ab^⊤X^{-⊤}∣V⟩)
\end{aligned}$$</span></p>
|
650,710 | <p>How would I go about simplifying $4(a-2(b-c)-(a-(b-2)))$. Show working out and steps please.</p>
<p>I'd show my working out but I'm not really sure where to start. Firstly, I would want to get rid of the 4 so I'd times everything else by 4 right? No idea. </p>
| Karolis Juodelė | 30,701 | <p>Basic algebra is all about term rewriting. You need to identify complete parts of the expression, extract them, solve them and place the results back where they were taken from.</p>
<p>You can start by separating $4(a-2(b-c)-(a-(b-2)))$ into $4(a+?-(a-(b-2)))$ and $? = -2(b-c)$. Then solve $? = -2(a-c) = -2a+2c$ and put it back as $4(a-2a+2c-(a-(b-2)))$.</p>
<p>Also, from the other side, you could start by separating $4(a-2(b-c)-(a-(b-2)))$ into $? = 2(b-c)$, $! = a-(b-2)$ and $4(a-?-!)$. Then solve $4(a-?-!) = 4a - 4\cdot ? - 4\cdot!$ and put the extracted bits back as $4a - 4\cdot 2(b-c) - 4\cdot(a-(b-2))$.</p>
|
1,913,873 | <p>If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$</p>
<p>I don't know how to begin. What should be the approach?</p>
| Joey Zou | 260,918 | <p>Note that $x^2-4(x-1) = (x-2)^2\ge 0$ for all $x\in\mathbb{R}$, and hence $x^2\ge 4(x-1)$ for all $x\in\mathbb{R}$. As such, we have
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 4\left(\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\right). $$
So it suffices to show
$$\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-1}\ge 5.$$
Can you see how to finish?</p>
|
1,913,873 | <p>If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$</p>
<p>I don't know how to begin. What should be the approach?</p>
| zyx | 14,120 | <p>The functions $x^2$ and $(x-1)$ are both increasing in $x$. Therefore, by the rearrangement inequality, the sum is $ \geq \sum \frac{a^2}{a-1}$ which is easier to control, in fact each term of this sum is $\geq 4$.</p>
|
1,913,873 | <p>If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$</p>
<p>I don't know how to begin. What should be the approach?</p>
| Landon Carter | 136,523 | <p>By Cauchy Schwarz, $($given expression$)(a+b+c+d+e-5)\geq(a+b+c+d+e)^2$</p>
<p>Now this shows $($given expression$)\geq \dfrac{(a+b+c+d+e)^2}{a+b+c+d+e-5}$</p>
<p>Now $(a+b+c+d+e)^2=(a+b+c+d+e-5)^2+2\times(a+b+c+d+e-5)\times 5+25$ hence on division by $a+b+c+d+e-5$ we get $a+b+c+d+e-5+10+\dfrac{25}{a+b+c+d+e-5}=(a+b+c+d+e-5)+\dfrac{25}{a+b+c+d+e-5}+10$</p>
<p>Letting $x=a+b+c+d+e-5$ we note $x\geq0$ and the above is $x+\dfrac{25}{x}+10\geq 10+10=20$ by AM-GM.</p>
|
3,027,528 | <p>I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:</p>
<p>Suppose <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, and <span class="math-container">$P$</span> are <span class="math-container">$R$</span>-modules, and <span class="math-container">$f:A \rightarrow B$</span> and <span class="math-container">$g:B\rightarrow C$</span> are both <span class="math-container">$R$</span>-module morphisms.</p>
<p>1) <span class="math-container">$\forall \phi : C \rightarrow P$</span> morphism, if <span class="math-container">$\phi \circ g = 0 \Rightarrow \phi = 0$</span>, for a morphism <span class="math-container">$\phi : C \rightarrow P$</span>, does this imply that <span class="math-container">$g$</span> is surjective? Why?</p>
<p>2) If <span class="math-container">$\phi \circ g \circ f = 0$</span> <span class="math-container">$ \forall \phi : C \rightarrow P$</span> morphism does this mean that <span class="math-container">$g \circ f = 0$</span>? Why?</p>
| Servaes | 30,382 | <p>For the first point consider for <span class="math-container">$\phi$</span> the quotient morphism <span class="math-container">$\pi:\ C\ \longrightarrow\ \operatorname{coker}g$</span>.</p>
<p>For the second point consider for <span class="math-container">$\phi$</span> the identity morphism <span class="math-container">$\operatorname{id}:\ C\ \longrightarrow\ C$</span>.</p>
|
1,230,112 | <p>Given a vector $\mathbf{x} \in \mathbb{R}^n$, a scalar $r\gt 0$ and an invertible matrix $\mathbf{A} \in \mathbb{R}^{n\times n}$, I'd like to maximize one of the components $x_\alpha$ constrained by $\mathbf{x}^T\mathbf{A}^T\mathbf{A}\mathbf{x}=r^2$.</p>
<p>I tried to do this with tensor algebra but I'm pretty new to that and while I did get a result, I'm almost certain that it is wrong. I used some steps which I think are dodgy at best, so I have a rough idea where the problem(s) might lie, however I do not know how to fix them. Here is what I have thus far:</p>
<ol>
<li>rewriting $\mathbf{x}^T\mathbf{A}^T\mathbf{A}\mathbf{x}=r^2$ in Tensor notation:</li>
</ol>
<p>$$x_i a_{ji} a_{jk} x_k = r^2$$</p>
<ol start="2">
<li>Building a Lagrangian:</li>
</ol>
<p>$$L\left(\mathbf{x},\lambda\right)=x_\alpha + \lambda \left(x_i a_{ji} a_{jk} x_k-r^2\right)$$</p>
<p>The $x_\alpha$ here is the component I want to eventually maximize. This is where the dodginess begins: The indices aren't balanced. I add a component of a vector to a scalar. However, since I actually want a component of a vector (effectively a scalar) in the end, and not a full vector, it seemed ok to me in this case.</p>
<ol start="3">
<li>taking the gradient of $L$:</li>
</ol>
<p>$$ \frac{\partial L}{\partial x_l} = \delta_{l\alpha}+\lambda a_{ji} a_{jk} \left( \delta_{l k} x_i + \delta_{l i} x_k \right) \\
\frac{\partial L}{\partial x_l} = \delta_{l\alpha} +\lambda a_{j l} \left( a_{j i} x_i + a_{j k} x_k \right) = 0$$</p>
<ol start="4">
<li>Solving for $x_m$ (while ramping up the dodginess):</li>
</ol>
<p>$$ \delta_{l\alpha} +\lambda a_{j l} \left( a_{j i} \delta_{i m} x_m + a_{j k} \delta_{k m} x_m \right) = 0\\
\delta_{l\alpha} +\lambda a_{j l} \left( a_{j m} + a_{j m} \right) x_m = 0 \\
x_m = -\frac{\delta_{l\alpha}}{2\lambda a_{jl} a_{jm}}\\
x_m = -\frac{1}{2\lambda a_{j\alpha} a_{j m}}$$</p>
<p>Am I actually allowed to divide by those matrix components like that? What about zero-components? And then, am I allowed to simplify $\frac{\delta_{i j}}{a_{j k}}=\frac{1}{a_{i k}}$?</p>
<ol start="5">
<li>Plugging that into my constraint and solving for $\lambda$:</li>
</ol>
<p>$$ x_i a_{j i} a_{j k} x_k = r^2 \\
\left(-\frac{1}{2\lambda a_{l\alpha}a_{l i}}\right)\left(-\frac{1}{2\lambda a_{m \alpha} a_{m k}}\right)a_{ji}a_{jk}=r^2\\
\frac{a_{j i}a_{j k}}{4 \lambda^2 a_{l\alpha} a_{m\alpha} a_{l i} a_{m k}}=r^2\\
\lambda=\pm\frac{1}{2r}\sqrt{\frac{a_{ji}a_{jk}}{a_{l\alpha}a_{m\alpha}a_{li}a_{mk}}}$$</p>
<ol start="6">
<li>Plugging that back into $x_\beta$</li>
</ol>
<p>$$x_\beta = -\frac{1}{\pm\frac{1}{r}\sqrt{\frac{a_{ji}a_{jk}}{a_{l\alpha}a_{m\alpha}a_{li}a_{mk}}} a_{n\alpha} a_{n\beta}}\\
x_\beta=\mp\frac{r}{a_{n\alpha} a_{n\beta}}\sqrt{\frac{a_{l\alpha}a_{m\alpha}a_{li}a_{mk}}{a_{ji}a_{jk}}}$$</p>
<ol start="7">
<li>Sanity check, using $\alpha = \beta$ and $\mathbf{A}= \delta_{ij}$:</li>
</ol>
<p>$$x_\alpha=\mp\frac{r}{\delta_{n\alpha}^2}\sqrt{\frac{\delta_{l\alpha}\delta_{m\alpha}\delta_{li}\delta_{mk}}{\delta_{ji}\delta_{jk}}}\\
x_\alpha=\mp\frac{r}{\delta_{nn}}\sqrt{\frac{\delta_{lm}\delta_{li}\delta_{mk}}{\delta_{ik}}}\\
x_\alpha=\mp\frac{r}{\delta_{nn}}\sqrt{\frac{\delta_{ik}}{\delta_{ik}}}\\
x_\alpha=\mp\frac{r}{\delta_{nn}}$$</p>
<p>This result is obviously incorrect: if $\mathbf{A}$ is the identity matrix, the result should simply be $x_\alpha=\pm r$ (and the other coordinates should all be $0$). Though it's so close (just incorrect by a factor dependent on the dimension of the involved vectorspace) that I must assume what I did isn't complete nonsense. So, where did I go off-rails and what would be the correct way to do this?</p>
<p>Also, with all those indices, I have a hard time turning that result back into usual matrix notation. If that's possible, how would that look like then?</p>
| mrp | 134,447 | <p>As 5xum mentioned in the comments, union starts with a u and has a symbol $\cup$ that looks very much like a $u$. Then intersection is simply the same symbol flipped $\cap$.</p>
<p>As for what they mean, you can think of union $A \cup B$ as a cup (indeed, the LaTeX command for it is <code>\cup</code>) in which you pour <em>all</em> of the elements of both $A$ and $B$, whereas $A \cap B$ has two legs, one in $A$ and one in $B$, so it contains only those elements that are in both $A$ and $B$.</p>
|
19,865 | <p>What does this set look like? $A = \{ (x, \frac{1}{x}) |  x \in \mathbb{R} \backslash 0 \}$, where $A$ is a subset of $\mathbb{R}^2$ with the Euclidean topology.</p>
<p>I thought $A = (-\infty, 0] \cup [0, \infty)$ but I know that its projection $(x,y) \rightarrow x$ is not closed, so this must be wrong.</p>
<p>Thanks for your help.</p>
| Willie Wong | 1,543 | <p>Interpreting your notation to mean that $A$ is a certain subset of $\mathbb{R}^2$...</p>
<p><strong>Hint</strong> $A$ can be treated as the graph of some function.</p>
|
19,865 | <p>What does this set look like? $A = \{ (x, \frac{1}{x}) |  x \in \mathbb{R} \backslash 0 \}$, where $A$ is a subset of $\mathbb{R}^2$ with the Euclidean topology.</p>
<p>I thought $A = (-\infty, 0] \cup [0, \infty)$ but I know that its projection $(x,y) \rightarrow x$ is not closed, so this must be wrong.</p>
<p>Thanks for your help.</p>
| user6495 | 6,495 | <p>@Matt: The set $A$ is closed so it is equal to its own closure. To show it is closed you can use the characterization of a closed set (in metric spaces) in terms of sequences.</p>
|
2,439,340 | <p>How would one proceed to prove this statement?</p>
<blockquote>
<p>The set of the strictly increasing sequences of natural numbers is not enumerable.</p>
</blockquote>
<p>I've been trying to solve this for quite a while, however I don't even know where to start.</p>
| paw88789 | 147,810 | <p>Hint: Could you uniquely associate to a sequence of $1$s and $2$s a strictly increasing sequence of natural numbers?</p>
|
2,439,340 | <p>How would one proceed to prove this statement?</p>
<blockquote>
<p>The set of the strictly increasing sequences of natural numbers is not enumerable.</p>
</blockquote>
<p>I've been trying to solve this for quite a while, however I don't even know where to start.</p>
| Siong Thye Goh | 306,553 | <p>Given a sequence $(a_0, a_1, \ldots, )$, map it to the sequence $(a_1-a_0, a_2-a_1, \ldots)$.</p>
<p>The image of such map would be the sequence of all natural numbers . Indeed the map is surjective since given a sequence of natural number $(b_0, b_1, \ldots)$, we can find a preimage $(a_0, a_1, \ldots)$ which satisfy $a_0 =1$, $a_{i+1} = b_i + a_i , \forall i \geq 0$. That is the cardinality of the set of increasing sequence of natural number is at least as big as the set of sequence of natural number.</p>
<p>By Cantor diagonalization, we know the set of all sequence of natural number is not countable.</p>
|
2,439,340 | <p>How would one proceed to prove this statement?</p>
<blockquote>
<p>The set of the strictly increasing sequences of natural numbers is not enumerable.</p>
</blockquote>
<p>I've been trying to solve this for quite a while, however I don't even know where to start.</p>
| fleablood | 280,126 | <p>Let $\{{s_i}_j\}$ be a countable list of strictly increasing sequences; define $\{c_i\}$ via $c_i = \max (c_{i-1}, {s_i}_i)+1$ and .... presto, Cantor!</p>
<p>But it could be simpler (depending on one's idea of simple) to reduce to things we already know are uncountable.</p>
<p>For simplicity, we can consider the sequence of differences between terms and not have to worry about the terms being increasing. i.e $a_{i+1} > a_i$ so $b_{i+1} = a_{i+1} - a_i > 0$ so $b_{i+1} \in \mathbb N$ and if $b_0 = a_0$. We have a one-to-one correspondence between $\{$ all increasing sequence of natural numbers $\}$ and $\{$ all sequences of natural numbers $\}$.</p>
<p>$\{$ all sequences of natural numbers $\} \supset \{$ all sequences of 1.... 10$\} \cong \{$ all sequences of 0....9 $\} \cong \{$all real numbers between 0 and 1$\}$ which is uncountable.</p>
|
3,425,373 | <p>Consider:</p>
<p><span class="math-container">$$ 1+1/2^2+2/3^2+1/4^2+2/5^2+1/6^2+...$$</span></p>
<p>Does this sum have a closed form?</p>
<p>If all the numerators are <span class="math-container">$1$</span> then it does have a closed form. </p>
| Calvin Lin | 54,563 | <p>Yes it does.</p>
<p><strong>Hint:</strong> <span class="math-container">$\sum \frac{1}{n^2} = \frac{\pi^2}{6}$</span></p>
<p><strong>Hint:</strong> <span class="math-container">$\sum \frac{1}{(2n)^2} = \frac{\pi^2}{24}$</span></p>
|
872,493 | <p>$$\lim_{x\to 0^+} \sin(x)^\frac{1}{\ln(x)} = ... =
\exp \left(\lim_{x\to 0^+} \frac{\ln(\frac{\sin x}{x}) + \ln(x)}{\ln(x)}\right)$$</p>
<p>Now, from continuity we can evaluate each term separately.</p>
<p>$\lim_{x\to 0^+} \ln(\frac{\sin x}{x}) = 0$</p>
<p>Therefore, we have:<br>
$$\exp \left(\lim_{x\to 0^+} \frac{\ln(x)}{\ln(x)}\right)$$</p>
<p>Technically, we have in the exponent $\frac{t}{t} = 1$, where $t=\ln(x)$.<br>
$e$ is indeed the answer, but are we allowed to do this direct arithmetic operation?<br>
I mean, shouldn't it be solved using L'Hôpital's rule if we want to be rigorous? </p>
<p>I'm asking that because my instructor marked $\checkmark$ on this and didn't mention something is wrong about this.</p>
<p><strong>So, I'm actually asking:</strong> When are we allowed to do this kind of operation? </p>
<p><strong>Update:</strong><br>
Isn't it like claiming that $$\frac{-\infty}{-\infty} = 1$$</p>
<p>Thanks. </p>
| Mark | 24,958 | <p>The L'Hôpital's rule isn't more rigorous than others, nay, mathematicians generally prefer not to use this rule because it is seen as a mechanical way to solve a limit. Furthermore the rule, if it is used mechanically, not always allows to solve a limit. For example, with the limit:
$$\lim_{x\rightarrow\infty}\frac{e^x}{\sqrt{x^2+1}}$$
this rule fails.</p>
|
4,164,650 | <p>The outline of the exercise is that: Fix <span class="math-container">$b > 1, y > 0$</span> and show that there is a unique real <span class="math-container">$x$</span> such that <span class="math-container">$b^x = y$</span>. (For further specification see e.g. <a href="https://math.stackexchange.com/questions/598890/baby-rudin-excercie-1-7-existence-of-logarithms">this post</a>) Part f.) requires to show that <span class="math-container">$b^x = y$</span> where <span class="math-container">$x = \sup A, A = \{w \in \mathbb{R}\mid b^w < y\}$</span>. It is very well possible that I am too pedantic, but I am struggling to prove the part since I can't even show that <span class="math-container">$A$</span> is necessarily 1.) non-empty 2.) bounded above.</p>
<p>The three cases to consider are that i.) <span class="math-container">$b < y$</span>, ii.) <span class="math-container">$b = y$</span>, iii.) <span class="math-container">$b > y$</span>. In the second case the claim is straightforward to prove. But how can you show in the i.) and iii.) that <span class="math-container">$A$</span> is bounded above and non-empty, respectively? Specifically I am stuck at the thought that since we are proving the existence of the logarithm, and at this point in the book the author hasn't covered sequences, series and limits, you can't really make an argument that you can multiply/divide <span class="math-container">$b$</span> enough times by itself to get a value greater/less than <span class="math-container">$y$</span>.</p>
| Jan Pulmann | 760,426 | <p>Let's set <span class="math-container">$\theta_0=0$</span>, since rotations preserve the quadratic form for conic sections. Then, you can see this quickly by rewriting the equation to
<span class="math-container">$$ r = \frac{1}{1+ex/r}$$</span>
which gives
<span class="math-container">$$ r+ex=1.$$</span>
Separating <span class="math-container">$r$</span> and squaring gives
<span class="math-container">$$ x^2 +y^2 = 1 - 2ex + e^2 x^2$$</span>
i.e.
<span class="math-container">$$ (1-e^2) x^2 + y^2+2ex+1=0.$$</span>
The discriminant is <span class="math-container">$4(e^2-1)$</span>, which recovers the classification of conic sections.</p>
<p>Note that the squaring also allows for the other branch of the hyperbola.</p>
<p><strong>Edit:</strong> of course, this does not explain the fact that the focus is in the origin.</p>
|
2,995,643 | <p>Here is a thought experiment I have. </p>
<p>Say we flip a unique coin where we have a 99.99999999999% chance of it landing on heads, and a .000000000001% chance of it landing on tails (the two possibilities equal to 100%).</p>
<p>And say we have an <em>infinite</em> number of coins flipped all at once (and only one time).</p>
<p>Is it possible that none of the trials will experience the coin land on tails?</p>
| Bram28 | 256,001 | <p>Sure! In fact, this would be possible with a fair coin already. In fact, it would be possible with a coin that is largely biased to come up with tails: one that has a 99.99999999999% chance of it landing on tails, and a .000000000001% chance of it landing on heads: it is <em>always</em> possible that you get no tails at all.</p>
<p>But, the probability of that would be infinitely small. In fact, mathematically, it would just be <span class="math-container">$0$</span>.</p>
|
366,687 | <p>I am interested in the status of the conjecture about the minimum number of edge crossings <span class="math-container">$cr(K_{m,n})$</span> in a drawing of the complete bipartite graph <span class="math-container">$K_{m,n}$</span>.</p>
<p>The Wikipedia article <a href="https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_brick_factory_problem" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_brick_factory_problem</a> led me to study the original papers of Zarankiewicz (<em>On a problem of P. Turan concerning graphs</em>) from 1954 and of Urbanik (<em>Solution du problème posé par P. Turán</em>) from 1955.</p>
<p>I wondered whether someone could tell me whether an asymptotic approach has been successfully attempted (letting <span class="math-container">$n\to\infty$</span>). If so, I would be very interested in any references for that.</p>
| ZsuzsannaMath | 161,836 | <p>It is a fascinating conjecture. The following might be a good reference for you: In 1997, Richter & Thomassen showed that
<span class="math-container">$$\lim_{n\to\infty}cr(K_{n,n})\left(\begin{array}{c} n \\ 2 \end{array}\right)^{-2}$$</span>
exists and is at most <span class="math-container">$1/4$</span>. If the conjecture is true, the value of this limit is exactly <span class="math-container">$1/4$</span>.
(R.B. Richter, C. Thomassen, "Relations between crossing numbers of complete and complete bipartite graphs" Amer. Math. Monthly , 104 (1997) pp. 131–137)</p>
<p>As of 2017, the exact value of this limit is still not known, see for example Gethner, Hogben, Lidick, Pfender, Ruiz, Young: „Crossing numbers of complete tripartite and balanced complete multipartite graphs“" Journal of Graph Theory 84, no. 4 (2017): 552-565. DOI: 10.1002/jgt.22041</p>
|
1,502,676 | <p>I'm aware of this similiar question:</p>
<p><a href="https://math.stackexchange.com/questions/1249308/what-is-the-difference-between-an-elliptical-and-circular-paraboloid-3d">what is the difference between an elliptical and circular paraboloid? (3D)</a></p>
<p>But I need help in a different way. In my calculus exam, I was asked to name the following surface:</p>
<p>$$(x-2)^2+(y-4)^2+1=z$$</p>
<p>I named it as an <strong>elliptic paraboloid</strong>, and according to my teacher, it is wrong because the correct answer is <strong>circular paraboloid</strong>.</p>
<p>What I was thinking at that time was that it can't be wrong to call elliptic paraboloid since horizontal planes would give a circumference which is a specific case of a ellipse.</p>
<p>Later I questioned why my question was wrong and explained the above. She gave me an explanation that I really couldn't understand at the time, but it was short and not convincing.</p>
<p>So I'm searching for definitions of a elliptic paraboloid and the "best" thing I found was this from a dictionary (not good I know, that's why I need your help, please):</p>
<p><a href="http://www.thefreedictionary.com/Elliptic+paraboloid" rel="nofollow noreferrer">http://www.thefreedictionary.com/Elliptic+paraboloid</a></p>
<p>Is it going to be convincing? How can I argue in my case? </p>
<p>Calculus is really important to my course and I want to do good. Thanks in advance, sorry my english.</p>
| Community | -1 | <p>A circle is a special case of an ellipse; a circular paraboloid is a special case of an elliptic paraboloid. This is how I use these terms, and so do many others. </p>
<p>But it's possible that your teacher used them differently, when introducing these terms <em>in the context of the class</em>. That's her decision, and it's natural that she expects answers to be consistent with the terminology adopted for the class.</p>
<p>Also, even when circular is considered a special case of elliptical paraboloid, there may still be an expectation that a student will pick the answer that is <em>the best</em> description of the object. The more specific description "circular" conveys more information and is to be preferred.</p>
|
1,850,258 | <p>From where can I learn mathematics from the basic blocks up? I feel like I have a lot of holes in the mathematics that I know and I would like to see where all those concepts come from. I would like to see what are the ideas that are took from granted, as foundation, and which ideas are made from this foundation.</p>
| Peter | 82,961 | <p>$1$ It can be viewed as a lattice in the $xy$-plane.</p>
<p>$2$ Addition and multiplication is like in $\mathbb C$</p>
<p>$3$ Yes, $1+i$ , for example. The definition is analogue to the usual
primes. A Gaussian number is prime if it cannot be written as
a product of two Gaussian numbers, both of which are neither $\pm1$ nor $\pm i$.</p>
<p>$4$ Yes, prime factorization is unique.</p>
<p>Addition : $$(a+bi) + (c+di) = (a+c) + (b+d)i$$</p>
<p>Multiplication : $$(a+bi)\cdot (c+di)=(ac-bd) + (ad+bc)i$$</p>
<p>Units : $1,i,-1,-i$</p>
<p>If $a^2+b^2$ is prime (in $\mathbb N$) , then $a+bi$ is prime as well. The converse is not necessarily true.</p>
|
3,057,278 | <blockquote>
<p><strong>Question:</strong> Can we show that <span class="math-container">$$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}=\frac{\pi}{8} $$</span> ?</p>
</blockquote>
<hr>
<p>According to <a href="https://www.wolframalpha.com/input/?i=sum+(-1)%5E(n%2B1)+%5B(2n-3)!!%2F(2n%2B3)!!%5D+from+n%3D+0+to+infinity" rel="noreferrer">wolfram alpha this result is true</a>.</p>
<p>Just amateur curiosity, not sure of a starting point to show if this true.</p>
| Claude Leibovici | 82,404 | <p><em>Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.</em></p>
<p>Let
<span class="math-container">$$S_p(x)=\sum_{n=0}^p(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}x^p$$</span> It write
<span class="math-container">$$S_p(x)=\frac{\left(8 p^3+36 p^2+46 p+15\right) \,
_2F_1\left(-\frac{1}{2},1;\frac{5}{2};-x\right)+3 (-x)^{p+1} \,
_2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-x\right)}{3 \left(8 p^3+36 p^2+46
p+15\right)}$$</span> Using
<span class="math-container">$$\,
_2F_1\left(-\frac{1}{2},1;\frac{5}{2};-1\right)=\frac{3 \pi }{8}$$</span> then
<span class="math-container">$$S_p(1)=\frac{\pi \left(8 p^3+36 p^2+46 p+15\right)-8 (-1)^p \,
_2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)}{8 \left(8 p^3+36 p^2+46
p+15\right)}$$</span> that it to say
<span class="math-container">$$S_p(1)=\frac \pi 8-\frac{(-1)^p }{8 p^3+36
p^2+46 p+15}\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$$</span> and <span class="math-container">$\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$</span> looks like an hyperbolic function going asymptotically to <span class="math-container">$\frac 12$</span>.</p>
<p>Then <span class="math-container">$S_\infty(1)=\frac{ \pi }{8}$</span>.</p>
|
1,700,246 | <p>Let $F=\mathbb{F}_{q}$, where $q$ is an odd prime power. Let $e,f,d$ be a standard basis for the $3$-dimensional orthogonal space $V$, i.e. $(e,e)=(f,f)=(e,d)=(f,d)$ and $(e,f)=(d,d)=1$. I have an element $g\in SO_{3}(q)$ defined by: $g: e\mapsto -e$, $f\mapsto \frac{1}{2}e -f +d$, $d\mapsto e+d$. I would like to determine the spinor norm of this element using Proposition 1.6.11 in the book 'The Maximal Subgroups of the Low-Dimensional Finite Classical Groups' by Bray, Holt and Roney-Dougal. </p>
<p>The proposition is quite long to state so it would be handy if someone who can help already has a copy of the book to refer to. If not, then please let me know and I can post what the proposition says. </p>
<p>I have followed the proposition and have that the matrices
$$A:=I_{3}-g=\left( \begin{array}{ccc}
2 & -\frac{1}{2} & -1 \\
0 & 2 & 0 \\
0 & -1 & 0 \end{array} \right)$$
$$F= \textrm{matrix of invariant symmetric bilinear form =}\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \end{array} \right).$$</p>
<p>If $k$ is the rank of $A$, the proposition says to let $B$ be a $k\times 3$ matrix whose rows form a basis of a complement of the nullspace of $A$. I have that $ker\, A=<(1,0,2)^{T}>$. Now by the way the proposition is stated it seems as if it does not make a difference as to what complement is taken. However, I have tried 3 different complements and I get contradictory answers each time.</p>
<p>Try 1) Orthogonal complement of $Ker\, A$, where $B=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 1 \end{array} \right)$. This gives me $det(BAFB^{T})=-25$.</p>
<p>Try 2) $B=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=-4$.</p>
<p>Try 3) $B=\left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$. This gives me $det(BAFB^{T})=0$.</p>
<p>The problem is that whatever complement I take the determinants should all be non-zero squares at the same time, but this is not the case. </p>
<p>I am not sure if I have misunderstood how to use the proposition. Any help will be appreciated. Thanks.</p>
| Andres Mejia | 297,998 | <p>This in general is not true. This is because $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^2$ is not a surjective function. More specifically, for all $r<0$, this statement won't be true for some neighborhood around $r$.</p>
<p>For a more concrete example, pick any $x<0$ and try to find a squared rational between $x$ and $0$.</p>
|
1,182,684 | <p>Let $\mathbf{F}(x,y,z) = y \hat{i} + x \hat{j} + z^2 \hat{k}$ be a vector field. Determine if its conservative, and find a potential if it is.</p>
<p><strong>Attempt at solution:</strong></p>
<p>We have that $\frac{\partial F_1}{\partial y} = 1 = \frac{\partial F_2}{\partial x} $, $\frac{\partial F_1}{\partial z} = 0 = \frac{\partial F_3}{\partial x}$, $\frac{\partial F_2}{\partial z} = 0 = \frac{\partial F_3}{\partial y}$, so the potential might exist.</p>
<p>Now we need to find a function $f$ such that $\nabla f = \mathbf{F}$.</p>
<p>For the first component, this means that $\frac{\partial f(x,y,z)}{\partial x} = y $, or after integrating, $f(x,y,z) = yx + C(y,z)$. Now I don't know how to determine the constant of integration $C(y,z)$, and I don't understand if I should add another constant when I integrate the second component.</p>
<p>For the second component, we have that $f(x,y,z) = xy + D(x,z)$, and for the third $f(x,y,z) = \frac{z^3}{3} + E(x,y)$. What now?</p>
<p>Any help please? In my textbook this is explained really in a terrible way.</p>
| kobe | 190,421 | <p>The function $\phi(x,y,z) = xy + \frac{z^3}{3}$ is a potential for $\mathbf{F}$ since $$\operatorname{grad} \phi = \phi_x \mathbf{i} + \phi_y \mathbf{j} + \phi_z \mathbf{k} = y\mathbf{i} + x \mathbf{j} + z^2\mathbf{k} = \mathbf{F}.$$</p>
<p>To actually derive $\phi$, we solve $\phi_x = F_1, \phi_y = F_2, \phi_z = F_3$. Since $\phi_x = F_1 = y$, by integration $\phi(x,y,z) = xy + u(y,z)$. Now $\phi_y = x + u_y$, so from $\phi_y = F_2 = x$, we have $x + u_y = x$, or $u_y = 0$. By integration, $u(y,z) = v(z)$. Thus $\phi(x,y,z) = xy + v(z)$. Then $\phi_z = v'(z)$, so from $\phi_z = F_3 = \frac{z^3}{3}$, we get $v'(z) = \frac{z^3}{3}$, which by integration yields $v(z) = z^2 + C$, where $C$ is a constant independent of $x,y,z$. This gives the general solution $\phi(x,y,z) = xy + \frac{z^3}{3} + C$. For convenience we set $C = 0$, giving the particular solution $\phi(x,y,z) = xy + \frac{z^3}{3}$.</p>
|
621,461 | <p>I'm having trouble understanding division when the divisor is greater than the dividend, for ex 1/4.</p>
<p>I think of division as "how many times can the divisor fit into the dividend evenly". </p>
<p>Intuitively, when I see 1/4 in the context of slices of pizza, I think of it as 1 "out of" 4, but I can't seem to grasp it in terms of "how many times does 4 fit into 1" if that makes any sense.</p>
<p>In other words my question could be why do we use division to represent "one out of 4"?.</p>
<p>If it helps you guys understand, this came about as I was trying to find the percentage representation of two populations, as in "there are 1253 A's and 747 B's, what is the proportion of each in % ?".</p>
<p>Conceptually I understand that I need to add up those two populations and then find the proportion they represent of that total. However, when I got to that second part, I couldn't reason through whether I needed to divide the total by a population, or a population by the total in order to find the desired result. </p>
<p>Obviously I eventually found the right way to do it, but it still doesn't make sense to me.</p>
<p>Sorry if it's very vague, this is really bothering me; I can't seem to reconcile those two ways of thinking about division.</p>
| user110219 | 110,219 | <p>Suppose you have 1 bread and you want to distribute it among 4 people equally. The only solution is to cut the single bread into 4 equal pieces.</p>
<p>For the population suppose that you have collected 8 from A and B such that their donations are 5 and 3 respectively. So if someone asks you that what is the contribution of A or B in the donation in terms of the percentage. Then what should you do? </p>
|
621,461 | <p>I'm having trouble understanding division when the divisor is greater than the dividend, for ex 1/4.</p>
<p>I think of division as "how many times can the divisor fit into the dividend evenly". </p>
<p>Intuitively, when I see 1/4 in the context of slices of pizza, I think of it as 1 "out of" 4, but I can't seem to grasp it in terms of "how many times does 4 fit into 1" if that makes any sense.</p>
<p>In other words my question could be why do we use division to represent "one out of 4"?.</p>
<p>If it helps you guys understand, this came about as I was trying to find the percentage representation of two populations, as in "there are 1253 A's and 747 B's, what is the proportion of each in % ?".</p>
<p>Conceptually I understand that I need to add up those two populations and then find the proportion they represent of that total. However, when I got to that second part, I couldn't reason through whether I needed to divide the total by a population, or a population by the total in order to find the desired result. </p>
<p>Obviously I eventually found the right way to do it, but it still doesn't make sense to me.</p>
<p>Sorry if it's very vague, this is really bothering me; I can't seem to reconcile those two ways of thinking about division.</p>
| Zhoe | 99,231 | <p>Division is often represented algebraically as the <strong>fraction</strong> $\frac{a}{b}$ where "$a$ is divided by $b$". $a$ is the numerator (dividend) and $b$ the denominator (divisor). The numerator represents the number of equal parts and the denominator indicates how many of those parts make up the whole. Take for example $\frac{3}{4}$; the numerator indicates that the fraction represents $3$ equal parts while the denominator indicates that $4$ parts make up the whole.</p>
<p>You can still think of division as "how many times can the divisor fit into the dividend evenly" in instances where the divisor is larger than the dividend; but you should realize that $\frac{1}{4}$ is just <strong>one way</strong> of expressing division. What might help you see this is considering writing the fraction as its decimal equivalent:
$$\frac{1}{4} = 4 \div 1 = 0.25$$</p>
<p>And so, $1$ is divided by $4$ evenly $0.25$ times.</p>
<hr>
<p>Following the definition of a fraction: <em>a representation of a part of a whole</em>, it should now be easier for you to see which (either the total of $A$/$B$'s population <strong>or</strong> the total population) goes on top and which goes on bottom. Using above, the numerator represents the number of equal parts, in this case $A=1253$ and $B=747$ and the denominator would be how many parts make up the whole, i.e. $A+B=1253+747=2000$. </p>
<p>Hence the numerator would be either $1253$ (for $A$'s percentage) or $747$ (for $B$'s percentage) and the denominator for both would the be total, $2000$.</p>
|
1,762,001 | <p>I recently watched a <a href="https://www.youtube.com/watch?v=SrU9YDoXE88" rel="noreferrer">video about different infinities</a>. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \ldots, \omega_\omega$, etc..</p>
<p>I can't find myself in all of this. Why there are so many infinities, and why even bother to classify infinity, when infinity is just... infinity? <strong>Why do we use all of these symbols? What does even each type of infinity mean?</strong></p>
| Ben Millwood | 29,966 | <p>One natural use for smallish infinite ordinals, which makes clear that different infinite ordinals have distinct properties, is the length of games.</p>
<p>I'm playing a chess-like (turn-taking, deterministic, complete information) board game of some kind against an opponent. My teammate is also playing a separate game against my opponent's teammate. Moves are made in both games at the same time.</p>
<p>Sadly, I'm losing my game, and can no longer do anything to avoid an eventual loss. Luckily, my teammate is winning, and the rules are such that as long as I can delay my loss for long enough for my teammate to force their win, we're OK.</p>
<p>Borrowing chess terminology, my opponent's position can be described as <em>win in $n$</em> inductively, by saying that they win in $0$ if they've already won, and they win in $k+1$ if no matter my next move, they will be able to guarantee ending up in a win-in-$k$ position.</p>
<p>I'm losing, so my opponent has win-in-$p$ for some $p$. My teammate has win-in-$q$ for some $q$. I'm OK exactly if $p > q$ (what happens if $p = q$ isn't important for what follows).</p>
<p>So far, it seems like we can measure the length of winning sequences using natural numbers. But then I spot something interesting: I actually have infinitely many moves available to me, and I can see a move that leaves me $1$ step from defeat, one which loses in $2$, one which loses in $3$, and so on. Every move seems to lose ultimately, each in a fixed finite number of steps, but I can at least choose how long it takes for me to lose. So if my teammate is in a win-in-$k$ position, I just need to pick the move that loses in $k+1$ moves and I'm all right!</p>
<p>In some sense, then, the winning sequence (losing sequence) that I'm facing is "longer" than any finite sequence my teammate might need, despite the fact that in all circumstances I lose in a finite number of moves. For this particular situation, we might say I will lose in $\omega$ moves, where $\omega$ is some kind of special number (spoiler: ordinal) that is bigger than any natural number.</p>
<p>However, I notice my teammate is actually in the same exact situation: they will win, but their opponent will be able to choose, on their next move, how long it takes. Once they've made that choice, their fate is sealed, but if they make their choice at the same time as I make mine, I can't ensure that they won't pick a longer finishing sequence than me. So I need to find a move such that I can put off my decision until my next turn, see what my teammate's opponent picks, and just pick a longer losing sequence. If you remember our inductive definition of win-in-$k$ from before, you might say that I'm looking for a lose-in-$(\omega + 1)$ situation, i.e. for a way to ensure that I'm in a lose-in-$\omega$ situation next turn.</p>
<p>Once you can imagine that, you can perhaps imagine losing in $\omega + 7$ moves, or even $\omega + \omega$, $\omega + \omega + \omega$, even $\omega^2$ and things more exotic than that. You can come up with a natural comparison ordering of these game lengths that precisely corresponds to whether or not I'll be able to guarantee that my teammate wins before I lose. If you do that, you have a game length for every natural number, then you have a game length which is larger than any natural number, then you have many more, still larger, game lengths. We can consider each of these long game lengths as a variety of infinite numbers that are each distinct and comparable with each other in meaningful ways, and as you might've guessed there's a meaningful addition (or at least concatenation) concept on these lengths which satisfies some (but not all) of the properties of addition on natural numbers. These are (some of) the ordinal numbers.</p>
<p>All this to say, there isn't simply one infinity that sits at the top of all numbers. There are many objects that don't seem to have this property or characteristic of "finiteness", indeed as rich and diverse a variety of non-finite objects as there are finite ones.</p>
|
3,792,683 | <p>Where <span class="math-container">$\alpha$</span> is a real constant, consider the sequence {<span class="math-container">$z_n$</span>} defined by <span class="math-container">$z_n=\frac{1}{n^\alpha}$</span>. For which value of <span class="math-container">$\alpha$</span> is {<span class="math-container">$z_n$</span>} a bounded sequence?</p>
<p>How do I start with this kind of question? I think that <span class="math-container">$\forall\space \alpha\in\Bbb{R}_{\geq0}$</span> the sequence is convergent and therefore bounded, but how do I write it out?</p>
| Bloodpolyhydron | 606,626 | <p><span class="math-container">$$\begin{array}\ \int^{+\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx - \int^{+\infty}_0 \frac{x^2e^{-x^2}}{(x^2+\frac{1}{2})^2} dx\bigg]\\ &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx + \frac{1}{2}\int^{+\infty}_0xe^{-x^2}d(\frac{1}{x^2+\frac{1}{2}})\bigg]\\ &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx - \frac{1}{2}\int^{+\infty}_0\frac{1-2x^2}{x^2+\frac{1}{2}}e^{-x^2}dx\bigg]\\ &= 2\int^{+\infty}_0e^{-x^2}dx\\ &= \sqrt{\pi} \end{array}$$</span></p>
|
91,645 | <p>I asked a similar question previously, though this is more specific and directed. In the writing of mathematics research papers, when is information cited, such as definitions? I have read that if it is fairly recent, then cite it. But what is "fairly recent?" Also, should books from whence a definition came be cited? In other words, what deems something cite-worthy? I have read several articles on the matter and it all was very vague.</p>
| JRN | 18,398 | <p>Definitions that "every mathematician knows" need no citation. For example, if you define $\mathbb{N}$ as the set of positive integers (or, if you like, the set of non-negative integers), you do not need a citation.</p>
<p>"Fairly recent" means that there is currently no consensus, that is, other people have different definitions, terms, or notation for the same concept. I would say "fairly recent" covers the past few decades.</p>
|
91,645 | <p>I asked a similar question previously, though this is more specific and directed. In the writing of mathematics research papers, when is information cited, such as definitions? I have read that if it is fairly recent, then cite it. But what is "fairly recent?" Also, should books from whence a definition came be cited? In other words, what deems something cite-worthy? I have read several articles on the matter and it all was very vague.</p>
| Carl Mummert | 630 | <p>The standard for citation in mathematics papers is very different than in (say) humanities. The <a href="http://www.ams.org/about-us/governance/policy-statements/sec-ethics">AMS Ethical Guidelines</a> say</p>
<blockquote>
<p>The correct attribution of mathematical results is essential, both because it encourages creativity, by benefiting the creator whose career may depend on the recognition of the work and because it informs the community of when, where, and sometimes how original ideas entered into the chain of mathematical thought.</p>
</blockquote>
<p>As the quote implies, we are often interested in attribution to the original discoverer, rather than the most recent person to use the definition. </p>
<p>At the same time, there is a question of who the audience for your paper is. In general, a high-level research paper will be read (or skimmed) by other researchers and read by some of their graduate students. So the author has to ask herself what citations these readers would appreciate seeing. </p>
<p>If a definition is new enough that it does not appear in any graduate textbooks yet, I think it's good practice to include a citation to some paper, because otherwise people may mistakenly think you made up the definition. As a general rule of thumb, if you are going to say "This is a definition of Smith", you might as well include a reference to some paper by Smith, and this will prevent some poor graduate student from an afternoon of trying to locate the definition in the wrong paper. </p>
<p>For definitions that are in the standard textbook(s), it may be better to just cite one of the textbooks, if you think readers may not be familiar with the definition yet, or to just indicate that the definition is standard, if most readers will already know it, or say nothing, if the definition is completely standard. </p>
|
91,645 | <p>I asked a similar question previously, though this is more specific and directed. In the writing of mathematics research papers, when is information cited, such as definitions? I have read that if it is fairly recent, then cite it. But what is "fairly recent?" Also, should books from whence a definition came be cited? In other words, what deems something cite-worthy? I have read several articles on the matter and it all was very vague.</p>
| Stella | 117,795 | <p>I've always been told to use the APA citation style.</p>
<p>APA citation style refers to the rules and conventions established by the American Psychological Association for documenting sources used in a research paper
<a href="http://www.apastyle.org/" rel="nofollow">http://www.apastyle.org/</a></p>
|
3,701,582 | <p>I still struggle mighty with basic conceptions of truth and proof. </p>
<p>For example: The Continuum Hypothesis (CH) is either true or false, i.e. either CH or ~CH holds. Now, Goedel and Cohen proved that CH/~CH are independent from ZFC, so ZFC + CH and ZFC + ~CH are consistent (in case ZFC is consistent but mathematicians assume that anyway). But since we know that CH or ~CH must be false, how can that be? One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false).</p>
<p>Another example is the parallel axiom (P) in euclidean geometry. P is true or false, i.e. P or ~P. That would mean that either the euclidean or non-euclidean geometry system has to be inconsistent (= no model).</p>
<p>Can somebody explain where I make a mistake?</p>
| Noah Schweber | 28,111 | <p>As in <a href="https://math.stackexchange.com/q/3925373/28111">this other question of yours</a>, the issue comes from conflating <strong>truth/falsity</strong> (which are relevant to <strong>structures</strong>) with <strong>provability/disprovability</strong> (which are relevant to <strong>theories</strong>).</p>
<p>More specifically, you're trying to apply <strong>bivalence</strong> - "everything is either correct or incorrect" - in a way it can't be applied. The satisfaction relation for individual structures is indeed bivalent: if <span class="math-container">$\mathcal{A}$</span> is a structure and <span class="math-container">$\varphi$</span> is a sentence (in the same language), then either <span class="math-container">$\mathcal{A}\models\varphi$</span> xor <span class="math-container">$\mathcal{A}\models\neg\varphi$</span>. However, bivalence does <strong>not</strong> hold for provability: e.g. there is no sense in which "<span class="math-container">$\mathsf{CH}$</span> is either true in <span class="math-container">$\mathsf{ZFC}$</span> or false in <span class="math-container">$\mathsf{ZFC}$</span>."</p>
<p>So when you invoke bivalence to claim</p>
<blockquote>
<p>The Continuum Hypothesis (CH) is either true or false, i.e. either CH or ~CH holds,</p>
</blockquote>
<p>that's correct <strong>iff</strong> we're tacitly referring to some specific structure <span class="math-container">$\mathcal{S}$</span>. However, your later claim</p>
<blockquote>
<p>One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false)</p>
</blockquote>
<p>is then mixing up provability and truth. We can see this clearly if we make explicit the tacit assumption above:</p>
<blockquote>
<p>One of those axiom systems must be inconsistent since it has no models (because one of its axioms is false <span class="math-container">$\color{red}{\mbox{in the structure $\mathcal{S}$}}$</span>).</p>
</blockquote>
<p>It should be clear that this is unwarranted: just because a theory is not satisfied in <em>some particular</em> structure doesn't mean it's not satisfied in <em>any</em> structure.</p>
<hr />
<p>To avoid exactly this sort of confusion, we should be careful - at least until the fundamentals are mastered - to only use <strong>true/false</strong>/etc. when talking about <em>structures</em>, and use <strong>provable/disprovable</strong>/etc. when talking about <em>theories</em>. This way we won't be tempted to improperly use our intuitions about truth/falsity when talking about provability/disprovability (or at least, we'll be <em>less</em> tempted).</p>
|
2,544,755 | <p>How would I start to prove this set is bounded above and below? A hint would be appreciated!</p>
<p>$A:=[0,1)\cup\{2-\frac{1}{n^5}:n\in \mathbb{N}\}$</p>
| fleablood | 280,126 | <p>If $x \in [0,1)$ then $x \ge 0$. If $x \in \{2-\frac 1{n^5}\}$ then $x \ge 1 > 0$. (Because $\frac 1{n^5} \le 1$ for all $n$ so if $x= 2 - \frac 1{n^5}$ then $x \ge 2 -1 =1$.). So if $x$ is in $A$ then $0 \le x$.</p>
<p>So $0$ is a lower bound of $A$.</p>
<p>If $x \in [0,1)$ then $x < 1$. If $x \in \{2-\frac 1{n^5}\}$ then $x < 2$ (Because $\frac 1{n^5} > 0$ for all $n$ so if $n=2-\frac 1{n^5}$ then $x < 2-0 =2$.). So if $x \in A$ then either $x < 1 < 2$ or $x < 2$. So either way, $x < 2$.</p>
<p>So $2$ is an upper bound of $A$. </p>
|
238,392 | <p>Do you know interesting examples of purely geometric or topological results which can be proved using group theory? To make precise what I have in mind, let us consider the two following examples:</p>
<blockquote>
<p>There does not exist any Riemannian metric on the torus whose sectional curvature is $<0$.</p>
</blockquote>
<p>This is a consequence of Milnor's article <em>A note on curvature and fundamental group</em>, where he proves that the fundamental group of a negatively-curved Riemannian manifold has exponential growth. On the other hand, the fundamental group of the torus, namely $\mathbb{Z}^2$, has quadratic growth.</p>
<blockquote>
<p>Any compact Riemannian manifold whose sectional curvature is $\equiv 0$ has a torus as a finite cover.</p>
</blockquote>
<p>This is a consequence of Bieberbach theorem. </p>
<p>More recently, showing that quasiconvex subgroups of hyperbolic cubulable groups are separable was the key point in the proof of the virtual Haken's conjecture. However, this is more technical.</p>
| Steve Huntsman | 1,847 | <p>Determining whether or not two simplicial complexes are homeomorphic is undecidable. Markov <em>fils</em> showed this in 1958 by reduction to the word problem for groups.</p>
|
3,961,131 | <p>Suppose we are given 2 predicates
<span class="math-container">$A(x)$</span> and <span class="math-container">$B(x)$</span> with domain <span class="math-container">$M$</span>.</p>
<p>Suppose next we are given the following predicate
<span class="math-container">$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$</span>
which we know is true, so
<span class="math-container">$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x))) = 1$$</span></p>
<p>The question is how does it restrict the truth sets of <span class="math-container">$A(x)$</span> and <span class="math-container">$B(x)?$</span></p>
<p>It is obvious that we have
<span class="math-container">$$\neg (A(x) \land B(x)) = 1 \\
A(x) \land B(x) = 0\\
A(x) = 0 \lor B(x) = 0$$</span>
So from that we get that either truth set for <span class="math-container">$A(x)$</span> is <span class="math-container">$E_A \neq M$</span> or truth set for <span class="math-container">$B(x)$</span> is <span class="math-container">$E_B \neq M$</span>.</p>
<p>But knowing that
<span class="math-container">$$\forall x(A(x) \rightarrow B(x)) = 1\\
\forall x(\neg A(x) \lor B(x)) = 1$$</span>
I have no idea how to link it to useful information on truth sets of <span class="math-container">$A(x)$</span> and <span class="math-container">$B(x)$</span>, any suggestions?</p>
| yugikaiba | 865,997 | <p>1
<span class="math-container">$$\neg (A(x) \land B(x)) \land (\forall x(A(x) \rightarrow B(x)))$$</span></p>
<p>2
<span class="math-container">$$\neg (A(x) \land B(x)) \equiv (\neg A(x)) \lor (\neg B(x))$$</span></p>
<p><span class="math-container">$$(\forall x(A(x) \rightarrow B(x))) \equiv (\neg A(x)) \lor B(x)$$</span></p>
<p>3
<span class="math-container">$$(\neg A(x) \lor \neg B(x)) \land (\neg A(x) \lor B(x)) \equiv (\neg A(x))$$</span></p>
|
401,389 | <p>I worked through some examples of Bayes' Theorem and now was reading the proof.</p>
<p>Bayes' Theorem states the following:</p>
<blockquote>
<p>Suppose that the sample space S is partitioned into disjoint subsets <span class="math-container">$B_1, B_2,...,B_n$</span>. That is, <span class="math-container">$S = B_1 \cup B_2 \cup \cdots \cup B_n$</span>, <span class="math-container">$\Pr(B_i) > 0$</span> <span class="math-container">$\forall i=1,2,...,n$</span> and <span class="math-container">$B_i \cap B_j = \varnothing$</span> <span class="math-container">$\forall i\ne j$</span>. Then for an event A,</p>
<p><span class="math-container">$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$</span></p>
</blockquote>
<p><img src="https://i.stack.imgur.com/nnbZU.png" alt="enter image description here" /></p>
<p>The numerator is just from definition of conditional probability in multiplicative form.</p>
<p>For the denominator, I read the following:</p>
<p><span class="math-container">$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$</span></p>
<p>Now this is what I don't understand:</p>
<blockquote>
<p><strong>The sets <span class="math-container">$A \cup B_i$</span> are disjoint because the sets <span class="math-container">$B_1, B_2, ..., B_n$</span> form a partition.<span class="math-container">$\tag{$\clubsuit$}$</span></strong></p>
</blockquote>
<p>I don't see how that is inferred or why that is the case. What does B forming a partition have anything to do with it being disjoint with A. Can someone please explain this conceptually or via an example?</p>
<p>I worked one example where you had 3 coolers and in each cooler you had either root beer or soda. So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda. But I don't see why these would be disjoint. If anything, I would say they weren't disjoint because each cooler contains both types of drinks.</p>
<p>Thank you in advance! :)</p>
| André Nicolas | 6,312 | <p>It is the sets $A\cap B_i$ that are pairwise disjoint, and that is precisely what you need to calculate the probability in the denominator.</p>
|
1,647,673 | <p>Prove or disprove that $$\left|a_1\right|+\left|a_2\right|+\ldots+\left|a_n\right|\leq n\sqrt{a_1^2+\ldots+a_n^2}$$</p>
<p>Where $a_1,\ldots,a_n\in\mathbb{R}$ and $n\in\mathbb{N}$.</p>
<p>EDIT: I was hoping there is a way without using a known inequality, ie to prove that $RHS-LHS\geq 0$</p>
| vrugtehagel | 304,329 | <p>Notice that $|a_i|^2=a_i^2$. Using the <a href="https://en.wikipedia.org/wiki/Generalized_mean" rel="nofollow">Generalized Mean Inequality</a>, we see $$\frac{|a_1|+\cdots+|a_n|}{n}\leq \sqrt{\frac{|a_1|^2+\cdots+|a_n|^2}{n}}$$
which we can rewrite to
$$|a_1|+\cdots+|a_n|\leq \sqrt{n}\sqrt{a_1^2+\cdots+a_n^2}$$
and since $\sqrt{n}\leq n$ we get $$|a_1|+\cdots+|a_n|\leq n\sqrt{a_1^2+\cdots+a_n^2}$$
<hr>
Alternatively, we can use <a href="https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality" rel="nofollow">Cauchy-Schwarz</a>:
$$(|a_1|\cdot |1|+\cdots |a_n|\cdot |1|)^2\leq (|a_1|^2+\cdots |a_n|^2)(|1|^2+\cdots+|1|^2)$$ which we can also rewrite to $$|a_1|+\cdots+|a_n|\leq \sqrt{n}\sqrt{a_1^2+\cdots+a_n^2}$$ and finishing the proof the same way.</p>
|
869,506 | <p>In a paper I am reading, there is a step that seems to come from the following inequality:
$$(1+x)^\alpha \le 1+2^\alpha x,$$
where $0<x<1$. (Also, $3\le \alpha \le 9/2$ in the context of the paper, but the above probably holds for more general $\alpha$, say, $\alpha\ge 1$.) It is stated with no explanation, and I feel that there is a slick solution, but I am unable to prove it without calculus.</p>
<hr>
<p>I was unsuccessful with the binomial expansion due to the generalized binomial coefficients.</p>
<p>I tried a calculus approach that assumes $0<x<1$ and $\alpha \ge 1$, and I think I was successful.
Consider $f(x):=1+2^\alpha x - (1+x)^\alpha$, and note that $f(0)=0$ and $f(1)=1$. Then, $$f'(x)=2^\alpha-\alpha(1+x)^{\alpha-1}.$$
Since $f'(0)=2^\alpha-\alpha>0$, we know $f$ is increasing at $0$. If we find zero or one critical point in $[0,1]$, we are finished.</p>
<p>Setting $f'(x^*)=0$ gives
\begin{align*}
2^\alpha &= \alpha(1+x^*)^{\alpha-1}\\
x^* &=\left(\frac{2^\alpha}{\alpha}\right)^{1/(\alpha-1)}-1 \ge 0\\
\end{align*}
(because $2^\alpha>\alpha$), so we have one critical point in the positive reals, and we are finished.</p>
<p>I'm a little uncertain about these last few steps involving the critical point; is it correct?</p>
<hr>
<p><strong>Question</strong>: Is there an easier way to prove the inequality, and does it hold for more general $x$ and $\alpha$?</p>
| Semiclassical | 137,524 | <p>Here's a simple convexity argument. Let $g(x)=(1+x)^\alpha$, and note that $\alpha>1$ implies $g''(x)=\alpha(\alpha-1)(1+x)^{\alpha-2}>0$ for $0<x<1$; so $g(x)$ is concave up. In that case $gf(x)$ will be bounded above on the interval $[0,1]$ by $g(0)(1-x)+x g(1)$, i.e. the line segment connecting the endpoints of $g(x)$ on this interval. Since $g(0)=1$ and $g(1)=2^\alpha+1$, one has $g(0<x<1) \leq 1+(2^\alpha-1)x$. Note that this is a stronger bound than the one stated since $2^\alpha-1<2^\alpha$.</p>
|
614,962 | <blockquote>
<p>We have a continuous function <span class="math-container">$f:(a,b)\to \mathbb R$</span></p>
<p>Prove that: <span class="math-container">$\forall n: x_1...x_n\in(a,b):\exists x\in(a,b)$</span> such that:</p>
<p><span class="math-container">$$f(x)=\frac1n ( f(x_1)+...+f(x_n) ) $$</span></p>
</blockquote>
<p>Experience tells me that it may be possible with induction but I have no clue on how to begin, I don't even see how is that possible.</p>
<p>Help please ?</p>
| mathlove | 78,967 | <p>HINT : You can use $\lim_{x\to 0}\frac{\sin(2x)}{2x}=1.$</p>
|
614,962 | <blockquote>
<p>We have a continuous function <span class="math-container">$f:(a,b)\to \mathbb R$</span></p>
<p>Prove that: <span class="math-container">$\forall n: x_1...x_n\in(a,b):\exists x\in(a,b)$</span> such that:</p>
<p><span class="math-container">$$f(x)=\frac1n ( f(x_1)+...+f(x_n) ) $$</span></p>
</blockquote>
<p>Experience tells me that it may be possible with induction but I have no clue on how to begin, I don't even see how is that possible.</p>
<p>Help please ?</p>
| Claude Leibovici | 82,404 | <p>The Taylor expansion of Csc[y] built around y=0 is 1/y + y/6. You may forget the denominator and arrive to x (1/(2x) + (x/3)) = 1/2</p>
|
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