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 |
|---|---|---|---|---|
839,043 | <p>I have been studying power functions, and started to think about imaginary powers. Take the function $x^i$. Because I don't know how to multiply a number $i$ times, I tried to simplify the equation</p>
<p>$x^i = x^{\sqrt{-1}} = x^{(-1)^{1/2}}$</p>
<p>Then, using the property of exponents that states an exponent to an exponent is the two multiplied, I get</p>
<p>$x^{(-1)^{1/2}} = x^{-1/2} = (\sqrt{x})^{-1} = \frac{1}{\sqrt{x}}$</p>
<p>However, plugging in <a href="http://www.wolframalpha.com/input/?i=%7Bx%5Ei%2C+1%2F%28sqrt%28x%29%29%7D" rel="nofollow">any number</a> shows that</p>
<p>$x^i \neq \frac{1}{\sqrt{x}}$</p>
<p>Where did I go wrong?</p>
| Graham Kemp | 135,106 | <blockquote>
<p>Then, using the property of exponents that states an exponent to an exponent is the two multiplied, </p>
</blockquote>
<p>This is where you go wrong. You are thinking of $\left(a^b\right)^c=a^{(bc)}$, but you have $a^{(b)^c}$. That's not the same thing at all.</p>
<p>Thus $x^{(-1)^{1/2}}\neq x^{-1/2}$ because: $(-1)^{1/2}\neq -1/2$</p>
|
4,501,286 | <blockquote>
<p>If a, b, c are positive real numbers such that <span class="math-container">$a^2+ b^2+ c^2 = 1$</span>
<br>Show that:
<span class="math-container">$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$</span></p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>First , I used Holder's : <span class="math-container">$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{27}{3(a+b+c)}$$</span></p>
<p>similarly:</p>
<p><span class="math-container">$$ a+b+c \geq \frac{(a+b+c)^3}{3(a^2+b^2+c^2)} = \frac{(a+b+c)^3}{3}$$</span></p>
<p>Which gives us:</p>
<p><span class="math-container">$$LHS \geq \frac{27}{3(a+b+c)}+ \frac{(a+b+c)^3}{3}$$</span></p>
<p>By AM-GM:</p>
<p><span class="math-container">$$LHS \geq 2\sqrt{3} \times (a+b+c)$$</span></p>
<p>So all we need to do is prove that <span class="math-container">$$2\sqrt{3} \times (a+b+c)\geq 4\sqrt{3}$$</span></p>
<p>And this means we need to prove <span class="math-container">$$a+b+c \geq 2 $$</span></p>
<p>Which I can't.</p>
<p>I am not looking for a solution , only hints that would guide me through solving it .</p>
<p>Thanks in advance for any communicated help !</p>
| Mengchun Zhang | 420,459 | <p>By your argument, we already have
<span class="math-container">$$\frac1a+\frac1b+\frac1c+a+b+c\geq\frac9{a+b+c}+\frac{(a+b+c)^3}3=\frac9u+\frac{u^3}3\equiv f(u)$$</span>
where <span class="math-container">$u=a+b+c>0$</span>. Try to show that <span class="math-container">$f'(x)=0$</span> only has one root <span class="math-container">$x^*=\sqrt3$</span> when <span class="math-container">$x>0$</span> and <span class="math-container">$f''(x^*)>0$</span>, therefore <span class="math-container">$f(u)\geq\min_{x>0}f(x)=f(x^*)=4\sqrt3$</span>.</p>
|
1,013,346 | <p>Given a box which contains $3$ red balls and $7$ blue balls. A ball is drawn from the box and a ball of the other color is then put into the box. A second ball is drawn from the box, What is the probability that the second ball is blue? </p>
<p>could anyone provide me any hint? </p>
<p>Please, don't offer a complete sketch of the solution, a hint is enough for me as this is a homework problem. </p>
| Marco Cantarini | 171,547 | <p>You can also use the well known summation formula $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{residue of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{ poles}\right\}
$$ which is a consequence of the residue theorem. So it is sufficient to note that $$ \sum_{n\geq1}\frac{1}{\left(2n-1\right)^{4}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{\left(2n-1\right)^{4}}=-\frac{1}{2}\left\{ \textrm{residue of }\frac{\pi\cot\left(\pi z\right)}{\left(2z-1\right)^{4}}\textrm{ at }\frac{1}{2}\right\} =\color{red}{\frac{\pi^{4}}{96}}.$$</p>
|
81,267 | <p>I have the following problem: I have a (a lot)*3 table, meaning that I have 3 columns, say X, Y and Z, with real values. In this table some of the rows have the same (X,Y) values, but with different value of Z. For instance</p>
<pre><code>{{12.123, 4.123, 513.423}, {12.123, 4.123, 33.43}}
</code></pre>
<p>have the same (X,Y) but different Z. This is a case of multiplicity=2, but in principle it could be higher. What I want to do is to take all the unique rows, AND in case they have multiplicity >1 (i.e. repeated (X,Y)), pick the one with minimum Z value. In the previous example it would be the second one.</p>
<p>I hope I have been clear! Thank you very much indeed!</p>
| kglr | 125 | <pre><code>f1 = Composition[DeleteDuplicatesBy[Most], SortBy[Last]]
SeedRandom[1]
data = Append @@@ Transpose[{RandomInteger[2,{10,2}], RandomReal[1,{10}]}];
data2 = f1 @ data;
Row[MatrixForm /@ {data, data2}]
</code></pre>
<p><img src="https://i.stack.imgur.com/TVshL.png" alt="Mathematica graphics"></p>
<p>Alternatively,</p>
<pre><code>f2 = DeleteDuplicatesBy[Most] @ SortBy[Last] @ #&;
f1 @ data == f2 @ data
(* True *)
</code></pre>
|
3,609,906 | <p>I need to compute <span class="math-container">$$\lim_{n \to \infty}\sqrt{n}\int_{0}^{1}(1-x^2)^n dx.$$</span>
I proved that
for <span class="math-container">$n\ge1$</span>,
<span class="math-container">$$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!},$$</span>
but I don't know how to continue from here.</p>
<p>I also need to calculate <span class="math-container">$\int_{0}^{1}(1-x^2)^ndx$</span> for <span class="math-container">$n=50$</span> with a <span class="math-container">$1$</span>% accuracy. I thought about using Taylor series but also failed.</p>
| Ninad Munshi | 698,724 | <p>Let <span class="math-container">$y = \sqrt{n}x$</span>. Then we have that</p>
<p><span class="math-container">$$\lim_{n\to\infty} \int_0^{\sqrt{n}} \left(1-\frac{y^2}{n}\right)^n\:dy \longrightarrow \int_0^\infty e^{-y^2}\:dy = \frac{\sqrt{\pi}}{2}$$</span></p>
<p>by dominated convergence.</p>
<hr>
<p><span class="math-container">$\textbf{EDIT}$</span>: To get the numerical accuracy you desire, we can make use of the following result describing the rate of convergence of the limit:</p>
<p><span class="math-container">$$\left(1+\frac{x}{n}\right)^n = e^x - \frac{x^2e^x}{2n} + O\left(\frac{1}{n^2}\right)$$</span></p>
<p>To only achieve <span class="math-container">$1\%$</span> accuracy we can take a few liberties with the calculation, for example, after using the substitution above:</p>
<p><span class="math-container">$$\frac{1}{\sqrt{n}}\int_0^{\sqrt{n}} \left(1-\frac{y^2}{n}\right)^n\:dy = \frac{1}{\sqrt{n}}\int_0^{\sqrt{n}} e^{-y^2}-\frac{1}{2n}y^4e^{-y^2}\:dy + O\left(n^{-\frac{5}{2}}\right)$$</span></p>
<p><span class="math-container">$$ \approx \frac{1}{\sqrt{n}}\int_0^{\infty} e^{-y^2}\:dy - \frac{1}{2\sqrt{n^3}}\int_0^{\infty}y^4e^{-y^2}\:dy$$</span></p>
<p>where we can throw out the <span class="math-container">$O\left(n^{-\frac{5}{2}}\right)$</span> terms and approximate the integrals to <span class="math-container">$\infty$</span> since for <span class="math-container">$\sqrt{n} = 5\sqrt{2}$</span> the integrals have accumulated most of their area up to several decimal places.</p>
<p>The value of the second integral is given by Feynman's trick</p>
<p><span class="math-container">$$\int_0^\infty y^4 e^{-y^2}\:dy = \Biggr[\frac{d^2}{da^2}\int_0^\infty e^{-ay^2}\:dy \Biggr]_{a=1} = \frac{1}{2}\Biggr[\frac{d^2}{da^2} \sqrt{\frac{\pi}{a}} \Biggr]_{a=1} = \frac{3\sqrt{\pi}}{8}$$</span></p>
<p>This gives us a nice, tidy approximation of </p>
<p><span class="math-container">$$\int_0^1 (1-x^2)^{50}\:dx \approx \frac{397}{4000}\sqrt{\frac{\pi}{2}} \sim 0.12439$$</span></p>
<p>Compare this to the actual value</p>
<p><span class="math-container">$$\int_0^1 (1-x^2)dx \sim 0.12440$$</span></p>
<p>and the zeroth order approximation</p>
<p><span class="math-container">$$\frac{1}{10}\sqrt{\frac{\pi}{2}} \sim 0.12533$$</span></p>
|
255,217 | <p>Let a polynomial with integer coefficients be nice if</p>
<ol>
<li>this polynomial has distinct integer roots;</li>
<li>its derivative has also integer roots.</li>
</ol>
<p>For instance
<span class="math-container">$$p(x)=x(x-9)(x-24),\\
p'(x)=3(x-4)(x-18)$$</span>
is the smallest known nice cubic polynomial. Smallest here means a polynomial with the smallest diameter. The diameter of the polynomial <span class="math-container">$p$</span> is the diameter of the set of the roots of <span class="math-container">$p$</span> in the complex plane. But how to verify with the help of MA that there are no smaller ones?</p>
| Cesareo | 62,129 | <p>We can have an insight about the solutions behavior for each <code>m</code> by doing</p>
<pre><code>tmax = 1000;
solution = ParametricNDSolve[{-((m (1 + m) + 4/(9 (-2/3 + t) t)) y[t]) + 2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[t] == (-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c}, y, {t, 1, tmax}, {m, c}]
m = 1;
Plot3D[Evaluate[y[m, c][t] /. solution], {t, 1, tmax}, {c, -6, 0}]
</code></pre>
|
255,217 | <p>Let a polynomial with integer coefficients be nice if</p>
<ol>
<li>this polynomial has distinct integer roots;</li>
<li>its derivative has also integer roots.</li>
</ol>
<p>For instance
<span class="math-container">$$p(x)=x(x-9)(x-24),\\
p'(x)=3(x-4)(x-18)$$</span>
is the smallest known nice cubic polynomial. Smallest here means a polynomial with the smallest diameter. The diameter of the polynomial <span class="math-container">$p$</span> is the diameter of the set of the roots of <span class="math-container">$p$</span> in the complex plane. But how to verify with the help of MA that there are no smaller ones?</p>
| user64494 | 7,152 | <p>Here is an answer for <code>m==2</code> in version 13.1 on Windows 10 (<code>C</code> is reserved in WL.)</p>
<pre><code>ClearAll[m, c]; $Assumptions = {m \[Element] Integers, c \[Element] Reals}; m = 2;
solution = AsymptoticDSolveValue[{-((m (m + 1) + 4/(9 (-2/3 + t) t)) y[t]) +
2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[t] ==
(-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c}, y[t], {t, Infinity, 1}];
Limit[solution, t -> Infinity]
</code></pre>
<blockquote>
<p><code>\[Infinity] (48-36 Log[3]+c (-68+63 Log[3]))</code></p>
</blockquote>
<p>Now</p>
<pre><code>Solve[(48 - 36 Log[3] + c (-68 + 63 Log[3])) == 0, c]
</code></pre>
<blockquote>
<p><code>{{c -> (-48 + 36 Log[3])/(-68 + 63 Log[3])}}</code></p>
</blockquote>
<p>and then</p>
<pre><code>ClearAll[m, c]; m = 2; c = (-48 + 36 Log[3])/(-68 + 63 Log[3]);
solution = AsymptoticDSolveValue[{-((m (m + 1) + 4/(9 (-2/3 + t) t)) y[t]) +
2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[
t] == (-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c},
y[t], {t, Infinity, 1}];
Limit[solution, t -> Infinity]
</code></pre>
<blockquote>
<p><code>0</code></p>
</blockquote>
<p>A similar behavior for other natural <code>m</code> I tried. In the general case <code>AsymptoticDSolveValue</code> fails.</p>
|
353 | <p>One of the challenges of undergraduate teaching is logical implication. The case by case definition, in particular, is quite disturbing for most students, that have trouble accepting "false implies false" and "false implies true" as true sentences. </p>
<blockquote>
<p><strong>Question:</strong> What are good point of view, methods and tips to help students grasp the concept of logical implication?</p>
</blockquote>
<p>To focus the question, I would like to restrict to math majors, although the question is probably equally interesting for other kind of students.</p>
| Community | -1 | <p>Historically, quantificational logic came before propositional logic.</p>
<p>So maybe start with <strong>quantified</strong> sentences instead. It is reasonable to say both <em>"Every ten-foot-tall person plays basketball"</em> and <em>"No ten-foot-tall person plays basketball"</em>.</p>
<p>This leads to accepting both <em>Ten-Foot-Tall $\rightarrow$ Plays-Basketball</em> and <em>Ten-Foot-Tall $\rightarrow \neg$ Plays-Basketball</em>. </p>
|
353 | <p>One of the challenges of undergraduate teaching is logical implication. The case by case definition, in particular, is quite disturbing for most students, that have trouble accepting "false implies false" and "false implies true" as true sentences. </p>
<blockquote>
<p><strong>Question:</strong> What are good point of view, methods and tips to help students grasp the concept of logical implication?</p>
</blockquote>
<p>To focus the question, I would like to restrict to math majors, although the question is probably equally interesting for other kind of students.</p>
| Benoît Kloeckner | 187 | <p>Let me add the tricks I used already to the interesting ones proposed in the other answers. They are partly, but not completely redundant.</p>
<p>First, implication is often best understood in quantified propositions; <a href="https://matheducators.stackexchange.com/a/368">this has already been pointed out by Matt F.</a>, but I am not too found of allegedly real-life assertions, which usually sound rather artificial. I rather use statements like: "every integer $n$ that is divisible by 6 is also divisible by 3" or "for any integer $n$, if $n^2$ is even then so is $n$". One of the best of this kind is "for any integer $n$, if $n$ is a multiple of 4 then $n^2$ also is a multiple of 4": you have all three cases where implication is true by taking $n=2,3,4$.</p>
<p>The second angle is to give an example where one wants to state and use an implication without knowing the truth value of its terms. Here ironically, an allegedly real-life assertion does the job quite beautifully: assume you are investigating a murder that took place in London Saturday and that Colonel Mustard is a suspect. You can say confidently that "if Colonel Mustard is guilty, then she was in London Saturday" even if you do not know whether she is guilty nor whether she was in London Saturday. What is especially good in this example is that it is a good way to introduce contrapositives: the statement "if Colonel Mustard was not in London Saturday then she is not guilty" shows convincingly that the contrapositive has the same truth value than the original implication. I learned this example from Viviane Durand-Guerrier, a researcher in educational studies in mathematics in Montpellier. </p>
|
353 | <p>One of the challenges of undergraduate teaching is logical implication. The case by case definition, in particular, is quite disturbing for most students, that have trouble accepting "false implies false" and "false implies true" as true sentences. </p>
<blockquote>
<p><strong>Question:</strong> What are good point of view, methods and tips to help students grasp the concept of logical implication?</p>
</blockquote>
<p>To focus the question, I would like to restrict to math majors, although the question is probably equally interesting for other kind of students.</p>
| G. Allen | 4,845 | <p>I use a specific example: If I'm in the rain then I'm wet. I also emphasize that we're talking about whether it's possible for the statement to be true. </p>
<p>Now think about the four cases:</p>
<p><strong>true/true:</strong> If you are in the rain then it's possible for you to be wet so this is a true result.</p>
<p><strong>true/false:</strong> If you are in the rain then it <strong>isn't</strong> possible for you not to be wet so this is a false result.</p>
<p><strong>false/true (which I think is the most confusing):</strong> If you aren't in the rain, is it possible that you're wet? The answer to that is yes because you may have gotten wet by other means. That makes this a true result.</p>
<p><strong>false/false:</strong> If you aren't in the rain it's possible that you aren't wet so this is also a true result.</p>
|
353 | <p>One of the challenges of undergraduate teaching is logical implication. The case by case definition, in particular, is quite disturbing for most students, that have trouble accepting "false implies false" and "false implies true" as true sentences. </p>
<blockquote>
<p><strong>Question:</strong> What are good point of view, methods and tips to help students grasp the concept of logical implication?</p>
</blockquote>
<p>To focus the question, I would like to restrict to math majors, although the question is probably equally interesting for other kind of students.</p>
| riceissa | 9,917 | <p>In my opinion, the truth-table definition of material implication is disturbing because "if ..., then ..." is used in mathematics in two distinct ways (and no similar distinction exists for the other connectives like NOT, AND, and OR):</p>
<ol>
<li>"If <span class="math-container">$p$</span>, then <span class="math-container">$q$</span>" means that you can start out with <span class="math-container">$p$</span>, make some deductions, and end up with <span class="math-container">$q$</span>.</li>
<li>As the truth table definition.</li>
</ol>
<p>These end up being the same in classical logic, of course, but they are <em>conceptually distinct</em>. I think what is necessary is to connect these two concepts. In other words, for the truth table definition to make sense, and for a logical connective to be called an "implication", it has to be shown equivalent to the deductive sense. In mathematical logic we have the distinction between <span class="math-container">$\{p\} \vdash q$</span> and <span class="math-container">$\vdash p\to q$</span>, which are shown to be equivalent by the <a href="https://en.wikipedia.org/wiki/Deduction_theorem" rel="nofollow noreferrer">deduction theorem</a> (and its converse), but I believe it is possible to show this same kind of thing without bringing in all of the scary formal notation from mathematical logic.</p>
<p>I have a <a href="https://riceissa.github.io/material-conditional/" rel="nofollow noreferrer">longer essay</a> that implements the above strategy for explaining the material implication.</p>
|
1,897,538 | <p>Next week I will start teaching Calculus for the first time. I am preparing my notes, and, as pure mathematician, I cannot come up with a good real world example of the following.</p>
<p>Are there good examples of
\begin{equation}
\lim_{x \to c} f(x) \neq f(c),
\end{equation}
or of cases when $c$ is not in the domain of $f(x)$?</p>
<p>The only thing that came to my mind is the study of physics phenomena at temperature $T=0 \,\mathrm{K}$, but I am not very satisfied with it.</p>
<p>Any ideas are more than welcome!</p>
<p><strong>Warning</strong></p>
<p>The more the examples are approachable (e.g. a freshman college student), the more I will be grateful to you! In particular, I would like the examples to come from natural or social sciences. Indeed, in a first class in Calculus it is not clear the importance of indicator functions, etc..</p>
<p><strong>Edit</strong></p>
<p>As B. Goddard pointed out, a very subtle point in calculus is the one of removable singularities. If possible, I would love to have some example of this phenomenon. Indeed, most of the examples from physics are of functions with poles or indeterminacy in the domain.</p>
| Alex Ortiz | 305,215 | <p>One discontinuous phenomenon in physics is the electric field above and below a surface with uniform positive charge density $\sigma$. The field above the sheet is pointing directly upwards with $E_\text{top} = +{\sigma \over 2\varepsilon_0} \hat k$ and that below being $E_\text{below} = -{\sigma \over 2\varepsilon_0} \hat k$. You could forgo the vector notation and simply use $+$ and $-$ to distinguish the two.</p>
<p>Reference: any college-level physics textbook, say Griffiths.</p>
|
1,897,538 | <p>Next week I will start teaching Calculus for the first time. I am preparing my notes, and, as pure mathematician, I cannot come up with a good real world example of the following.</p>
<p>Are there good examples of
\begin{equation}
\lim_{x \to c} f(x) \neq f(c),
\end{equation}
or of cases when $c$ is not in the domain of $f(x)$?</p>
<p>The only thing that came to my mind is the study of physics phenomena at temperature $T=0 \,\mathrm{K}$, but I am not very satisfied with it.</p>
<p>Any ideas are more than welcome!</p>
<p><strong>Warning</strong></p>
<p>The more the examples are approachable (e.g. a freshman college student), the more I will be grateful to you! In particular, I would like the examples to come from natural or social sciences. Indeed, in a first class in Calculus it is not clear the importance of indicator functions, etc..</p>
<p><strong>Edit</strong></p>
<p>As B. Goddard pointed out, a very subtle point in calculus is the one of removable singularities. If possible, I would love to have some example of this phenomenon. Indeed, most of the examples from physics are of functions with poles or indeterminacy in the domain.</p>
| Rob Arthan | 23,171 | <p>Navigation may provide interesting examples: you can fly west forever (if you don't start at one of the poles) but you can't fly north forever (from anywhere on the earth).</p>
|
1,897,538 | <p>Next week I will start teaching Calculus for the first time. I am preparing my notes, and, as pure mathematician, I cannot come up with a good real world example of the following.</p>
<p>Are there good examples of
\begin{equation}
\lim_{x \to c} f(x) \neq f(c),
\end{equation}
or of cases when $c$ is not in the domain of $f(x)$?</p>
<p>The only thing that came to my mind is the study of physics phenomena at temperature $T=0 \,\mathrm{K}$, but I am not very satisfied with it.</p>
<p>Any ideas are more than welcome!</p>
<p><strong>Warning</strong></p>
<p>The more the examples are approachable (e.g. a freshman college student), the more I will be grateful to you! In particular, I would like the examples to come from natural or social sciences. Indeed, in a first class in Calculus it is not clear the importance of indicator functions, etc..</p>
<p><strong>Edit</strong></p>
<p>As B. Goddard pointed out, a very subtle point in calculus is the one of removable singularities. If possible, I would love to have some example of this phenomenon. Indeed, most of the examples from physics are of functions with poles or indeterminacy in the domain.</p>
| Community | -1 | <p>One form of the Kronecker delta function (aka discrete delta function) is the one satisfying</p>
<p>$$ \delta(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} $$</p>
<p>This usually crops up in discrete contexts, but it makes sense in this context too.</p>
<p>Another appearance of the same function is as the indicator function of the set $\{ 0 \}$. Recall that indicator functions are defined by</p>
<p>$$ \chi_S(x) = \begin{cases} 1 & x \in S \\ 0 & x \notin S \end{cases} $$</p>
|
3,283,606 | <p>Good Evening,</p>
<p>I know this is a basic question, but I haven't been able to find a clear explanation for how to simplify the follow equation:
<span class="math-container">$$n\log_2n=10^6$$</span>
Solving this equation is part of the solution for Problem 1-1 from the Intro. to Algorithms book by CLRS:
<a href="http://atekihcan.github.io/CLRS/P01-01/" rel="nofollow noreferrer">http://atekihcan.github.io/CLRS/P01-01/</a></p>
<p>The author there simplifies the above to:
<span class="math-container">$$n=62746$$</span>
But I can't see how to do this. Thank you.</p>
| Claude Leibovici | 82,404 | <p>Starting from Parcly Taxel's answer, the problem is to compute <span class="math-container">$W(x)$</span> for a very large argument. If you do not have access to Lambert function, let me suggest the very fast algorithm presented in <a href="https://arxiv.org/pdf/1504.01964.pdf" rel="nofollow noreferrer">this paper</a>.</p>
<p>For the computation of <span class="math-container">$$y=\log\big(W(e^x)\big)\implies W(e^x)=e^y$$</span> the author proposes the very simple iterative scheme
<span class="math-container">$$y_{n+1}=y_n-\frac{2 \left(e^{y_n}+1\right) \left(e^{y_n}+y_n-x\right)}{2
\left(e^{y_n}+1\right)^2-e^{y_n} \left(e^{y_n}+y_n-x\right)}$$</span> If you need to compute it many time, you could save a lot defining obvious intermediate terms.</p>
<p>The author proposes as starting value <span class="math-container">$y_0=\log(x)$</span>; in fact, it should be better in my humble opinion, to use
<span class="math-container">$$y_0=\log\big(x-\log(x)\big)$$</span></p>
<p>Applied to your case, we should get the following iterates
<span class="math-container">$$\left(
\begin{array}{cc}
n & W\big(10^6\log(2)\big) \\
0 & \color{red} {1}0.85009305921355973225341 \\
1 & \color{red} {11.0468}4846509467698238553 \\
2 & \color{red} {11.046852125521960930}48720 \\
3 & \color{red} {11.04685212552196093051026}
\end{array}
\right)$$</span></p>
|
3,283,606 | <p>Good Evening,</p>
<p>I know this is a basic question, but I haven't been able to find a clear explanation for how to simplify the follow equation:
<span class="math-container">$$n\log_2n=10^6$$</span>
Solving this equation is part of the solution for Problem 1-1 from the Intro. to Algorithms book by CLRS:
<a href="http://atekihcan.github.io/CLRS/P01-01/" rel="nofollow noreferrer">http://atekihcan.github.io/CLRS/P01-01/</a></p>
<p>The author there simplifies the above to:
<span class="math-container">$$n=62746$$</span>
But I can't see how to do this. Thank you.</p>
| Arthur | 15,500 | <p>Note that <span class="math-container">$2^{20}=16\cdot 2^{16}=1\,048\,576$</span> is pretty darn close to one million, so <span class="math-container">$n=2^{16}$</span> is pretty close to the answer. Also note that decreasing <span class="math-container">$n$</span> by a relatively small amount won't make a big change to <span class="math-container">$\log_2(n)$</span>. We want a result that's <span class="math-container">$48\,576$</span> smaller, we can as a first attempt just reduce our <span class="math-container">$n$</span> by <span class="math-container">$48\,576/16=3036$</span> and see what we get:
<span class="math-container">$$
\log_2(2^{16}-3036)\cdot(2^{16}-3036)\approx 995\,723
$$</span>
which is only <span class="math-container">$4277$</span> away. We missed because of course <span class="math-container">$\log_2(2^{16}-3036)\approx 15.93$</span> is a little smaller than <span class="math-container">$16$</span>.</p>
<p>However, we can now repeat this process, noting that increasing <span class="math-container">$n$</span> by <span class="math-container">$4277/15.93\approx 268$</span> ought to give us an even better result, this time missing only overshooting by 380.</p>
<p>There is a better method, though, which in a much better way encompasses how also the logarithm changes: use the derivative. Differentiating <span class="math-container">$f(n)=n\log_2(n)$</span>, we get
<span class="math-container">$$
f'(n)=\log_2(n)+\frac1{\ln(2)}
$$</span>
We only used the first term when deciding to decrease <span class="math-container">$n$</span> by <span class="math-container">$3036$</span> the first time. However, using the full derivative, we're told to decrease <span class="math-container">$n$</span> by <span class="math-container">$48\,576/(16+1/\ln2)\approx 2785$</span>. Let's see where this takes us:
<span class="math-container">$$
\log_2(2^{16}-2785)(2^{16}-2785)\approx 1\,000\,085
$$</span>
and look how close we got immediately, compared to the previous attempt, just by adding <span class="math-container">$\frac1{\ln2}$</span>. One more iteration ought to do it. Next we will decrease <span class="math-container">$n$</span> by <span class="math-container">$85/f'(2^{16}-2785)\approx4.9$</span>, and this gives us
<span class="math-container">$$
\log_2(2^{16}-2789.9)(2^{16}-2789.9)\approx999\,999.5
$$</span>
and I daresay we clearly can't find a better integer approximation to the solution than <span class="math-container">$2^{16}-2790=62\,746$</span>.</p>
<p>This is called Newton's method, and it works remarkably well when you have a decent guess to start with. It is a widely used algorithm for numerical solution of equations because of its simplicity and efficiency.</p>
|
937,912 | <p>I'm looking for a closed form of this integral.</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$</p>
<p>where $\operatorname{Li}_2$ is the <a href="http://mathworld.wolfram.com/Dilogarithm.html" rel="noreferrer">dilogarithm function</a>.</p>
<p>A numerical approximation of it is</p>
<p>$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$</p>
<p>As Lucian said $I$ has the following equivalent forms:</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$</p>
<p>According to <em>Mathematica</em> it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us <a href="https://math.stackexchange.com/a/937926/153012">this form</a>.</p>
<p>With <em>Maple</em> using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's <a href="https://math.stackexchange.com/a/938024/153012">answer</a>, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. </p>
<p>I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.</p>
<p>I hope it exists. Similar integrals are the following.</p>
<p>$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$</p>
<p>Related techniques are in <a href="http://carma.newcastle.edu.au/jon/Preprints/Papers/ArXiv/Zeta21/polylog.pdf" rel="noreferrer">this</a> or in <a href="http://arxiv.org/pdf/1010.6229.pdf" rel="noreferrer">this</a> paper. <a href="http://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/Livres/Connon%20some%20series%20and%20integrals/Vol-3.pdf" rel="noreferrer">This</a> one also could be useful.</p>
| Claude Leibovici | 82,404 | <p>According to a CAS, $$I = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{t} \right)}{2 \, \sqrt{t} \, \sqrt{1-t}} \,dt =\,
_4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right
)+\frac{\pi ^3}{48}-\frac{1}{4} \pi \log ^2(2)$$</p>
<p>Enjoy !</p>
|
937,912 | <p>I'm looking for a closed form of this integral.</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$</p>
<p>where $\operatorname{Li}_2$ is the <a href="http://mathworld.wolfram.com/Dilogarithm.html" rel="noreferrer">dilogarithm function</a>.</p>
<p>A numerical approximation of it is</p>
<p>$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$</p>
<p>As Lucian said $I$ has the following equivalent forms:</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$</p>
<p>According to <em>Mathematica</em> it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us <a href="https://math.stackexchange.com/a/937926/153012">this form</a>.</p>
<p>With <em>Maple</em> using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's <a href="https://math.stackexchange.com/a/938024/153012">answer</a>, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. </p>
<p>I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.</p>
<p>I hope it exists. Similar integrals are the following.</p>
<p>$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$</p>
<p>Related techniques are in <a href="http://carma.newcastle.edu.au/jon/Preprints/Papers/ArXiv/Zeta21/polylog.pdf" rel="noreferrer">this</a> or in <a href="http://arxiv.org/pdf/1010.6229.pdf" rel="noreferrer">this</a> paper. <a href="http://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/Livres/Connon%20some%20series%20and%20integrals/Vol-3.pdf" rel="noreferrer">This</a> one also could be useful.</p>
| Jack D'Aurizio | 44,121 | <p>Following Anastasiya-Romanova's approach, we have:</p>
<p>$$ I = \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}\int_{0}^{\pi/2}\sin^n x\,dx =\frac{\pi}{16}\sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n}+\frac{1}{4}\sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2}\tag{1}$$
where:
$$ S_1 = \sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n} = \zeta(2)-2\log^2 2 \tag{2}$$
and the second sum is the problematic one, leading to a value for a hypergeometric function $\phantom{}_4 F_3$:</p>
<p>$$\begin{eqnarray*}S_2 = \sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2} &=& -\int_{0}^{1}\frac{2\arcsin x}{x\sqrt{1-x^2}}\log x\,dx\\&=&-2\int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\log\sin\theta\,d\theta. \tag{3}\end{eqnarray*}$$</p>
<p>However, since the Fourier cosine series of $\log\sin\theta$ is well-known:
$$\log\sin\theta = -\log 2-\sum_{n\geq 1}\frac{\cos(2n\theta)}{n}\tag{4} $$
we just need to compute the Fourier cosine series of $\frac{\theta}{\sin\theta}$. Since the Fourier sine series of the triangle wave is given by:
$$ \theta = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sin(2n\theta) \tag{5}$$
by exploiting $\frac{\sin(2n\theta)}{\sin\theta}=2\left(\cos\theta+\cos(3\theta)+\ldots+\cos((2n-1)\theta)\right)$ we have:</p>
<p>$$\frac{\theta}{\sin\theta}=2\sum_{k=1}^{+\infty}\left(\sum_{n\geq k}\frac{(-1)^{n+1}}{n}\right)\cos((2k-1)\theta)\tag{6}$$
so, at least in principle, $S_2$ is computable through a Fourier-analytic approach, by exploiting:
$$ \int_{0}^{\pi/2}\cos((2n-1)\theta)\cos(2m\theta)\,d\theta = \frac{(2n-1)(-1)^m}{4m^2-(2n-1)^2}.\tag{7}$$</p>
<p>It is also interesting to notice that the last integral appearing in $(3)$ is very similar to the one appearing in this <a href="https://math.stackexchange.com/questions/1168209/closed-form-of-int-0-pi-2x-cot-leftx-right-cos-leftx-right-log-left/1168221#1168221">related question</a>, but the latter is way easier to compute since in the Fourier cosine series of $\frac{\theta}{\sin\theta}\,\cos^2\theta$ there are only "even cosines".</p>
|
937,912 | <p>I'm looking for a closed form of this integral.</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$</p>
<p>where $\operatorname{Li}_2$ is the <a href="http://mathworld.wolfram.com/Dilogarithm.html" rel="noreferrer">dilogarithm function</a>.</p>
<p>A numerical approximation of it is</p>
<p>$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$</p>
<p>As Lucian said $I$ has the following equivalent forms:</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$</p>
<p>According to <em>Mathematica</em> it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us <a href="https://math.stackexchange.com/a/937926/153012">this form</a>.</p>
<p>With <em>Maple</em> using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's <a href="https://math.stackexchange.com/a/938024/153012">answer</a>, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. </p>
<p>I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.</p>
<p>I hope it exists. Similar integrals are the following.</p>
<p>$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$</p>
<p>Related techniques are in <a href="http://carma.newcastle.edu.au/jon/Preprints/Papers/ArXiv/Zeta21/polylog.pdf" rel="noreferrer">this</a> or in <a href="http://arxiv.org/pdf/1010.6229.pdf" rel="noreferrer">this</a> paper. <a href="http://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/Livres/Connon%20some%20series%20and%20integrals/Vol-3.pdf" rel="noreferrer">This</a> one also could be useful.</p>
| Sangchul Lee | 9,340 | <p><strong>My attempt.</strong> This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.</p>
<p>First, from <a href="http://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm" rel="noreferrer">Landen's identity</a> of the following form</p>
<p>$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$</p>
<p>we observe that</p>
<p>\begin{align*}
I
&= -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \left\{ \mathrm{Li}_2\left(-\frac{x}{1-x}\right) - \frac{1}{2}\log^{2}(1-x) \right\} \, dx \\
&= -\int_{-\infty}^{0} \frac{2\mathrm{Li}_2 (t) + \log^{2}(1-t)}{2\sqrt{1-2t}(1-t)} \, dt
\tag{1}
\end{align*}</p>
<p>By noting that</p>
<p>$$ \frac{d}{dt} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) = \frac{1}{2\sqrt{1-2t}(1-t)}, $$</p>
<p>integration by parts and the substitution $x = (1-2t)^{-1/2}$ shows that (1) is equal to</p>
<p>\begin{align*}
I
&= \int_{-\infty}^{0} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) \frac{2\log(1-t)}{t(t-1)} \, dt \\
&= \int_{0}^{1} \frac{8x \arctan x}{1 - x^4} \log \left( \frac{1+x^2}{2x^2} \right) \, dx \tag{2}
\end{align*}</p>
<p>The following observation</p>
<p>$$ \Re \log \left(\frac{1+ix}{\sqrt{2}} \right) = \frac{1}{2}\log \left( \frac{1+x^2}{2} \right)
\quad \text{and} \quad
\Im \log \left(\frac{1+ix}{\sqrt{2}} \right) = \arctan x $$</p>
<p>somehow seems to suggest complex-analytic approach, but I have not been successful with such approaches so far. Next, from the following simple formula</p>
<p>$$ \log \left( \frac{1+x^2}{2x^2} \right) \, dx = \int_{0}^{1} \frac{d}{dy} \log \left( \frac{y^2+x^2}{y^2 + 1} \right) \, dy $$</p>
<p>the integral (2) can be further decomposed into the following form</p>
<p>$$ I = \int_{0}^{1}\int_{0}^{1} \frac{16xy \arctan x}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy. \tag{3} $$</p>
<p>Simple calculation shows that</p>
<p>$$ \int_{0}^{1}\int_{0}^{1} \frac{16xy}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy = 2\zeta(2), $$</p>
<p>so I suspect that the situation in (3) is not that bad.</p>
|
937,912 | <p>I'm looking for a closed form of this integral.</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$</p>
<p>where $\operatorname{Li}_2$ is the <a href="http://mathworld.wolfram.com/Dilogarithm.html" rel="noreferrer">dilogarithm function</a>.</p>
<p>A numerical approximation of it is</p>
<p>$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$</p>
<p>As Lucian said $I$ has the following equivalent forms:</p>
<p>$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$</p>
<p>According to <em>Mathematica</em> it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us <a href="https://math.stackexchange.com/a/937926/153012">this form</a>.</p>
<p>With <em>Maple</em> using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's <a href="https://math.stackexchange.com/a/938024/153012">answer</a>, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. </p>
<p>I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.</p>
<p>I hope it exists. Similar integrals are the following.</p>
<p>$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$</p>
<p>Related techniques are in <a href="http://carma.newcastle.edu.au/jon/Preprints/Papers/ArXiv/Zeta21/polylog.pdf" rel="noreferrer">this</a> or in <a href="http://arxiv.org/pdf/1010.6229.pdf" rel="noreferrer">this</a> paper. <a href="http://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/Livres/Connon%20some%20series%20and%20integrals/Vol-3.pdf" rel="noreferrer">This</a> one also could be useful.</p>
| Jack D'Aurizio | 44,121 | <p>An alternative approach. As shown by <em>nospoon</em> <a href="https://math.stackexchange.com/questions/1438381">here</a>,
\begin{equation}\label{shalev} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(1- x\sin^2\theta )}}\,dx = \frac{8}{\sin\theta}\left[\frac{\theta^3}{3}-\text{Im}\,\text{Li}_3\left(1-e^{2i\theta}\right)\right]\tag{1}\end{equation}
holds for any $\theta\in\left(0,\frac{\pi}{2}\right)$. This is an istance of a very nice principle, according to which every hypergeometric function of the $\phantom{}_{p+1}F_{p}\left(\frac{1}{2},\frac{1}{2},\ldots;\frac{3}{2},\frac{3}{2},\ldots;z\right)$ kind has a closed form in terms of polylogarithms. We need to evaluate $\phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$, hence our arrival point is already pretty close to the statement of $(1)$.</p>
<p>The evaluation of $(1)$ at $\theta=\frac{\pi}{4}$ leads to
\begin{equation}\label{shalev2} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(2- x )}}\,dx = 8\left[\frac{\pi^3}{192}-\text{Im}\,\text{Li}_3\left(1-i\right)\right]\tag{2}\end{equation}
and the <a href="http://mathworld.wolfram.com/Dilogarithm.html" rel="noreferrer">functional relations for $\text{Li}_2$</a> reduce the original problem to $(2)$. In particular
$$\begin{eqnarray*} \int_{0}^{\pi/2}\text{Li}_2(\sin\theta)\,d\theta &=& \int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{1-t^2}{1+t^2}\right)\,dt\\
\int_{0}^{\pi/2}\text{Li}_2(\cos\theta)\,d\theta&=&\int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{2t}{1+t^2}\right)\,dt\\
&=&\frac{5\pi^3}{24}+4\,\text{Im}\,\text{Li}_3(1-i)\tag{3}\end{eqnarray*}$$
as already shown by Reshetnikov.</p>
|
939,212 | <p>Apologies if this isn't at the level of questions expected here!</p>
<p>I've got two simultaneous equations to solve.</p>
<p>(Equation 1): $ x y = 4 $</p>
<p>(Equation 2): $ x + y = 2 $</p>
<p>They produce the following curves:</p>
<p><img src="https://i.stack.imgur.com/dMrNi.png" alt="enter image description here"></p>
<p><strong>Question:</strong> Whilst it's graphically obvious that they do not make contact, what is the <em>algebraic</em> indicator that these two lines do not meet? How do you prove that?</p>
| Kim Jong Un | 136,641 | <p>Suppose that the curves meet: there are $x,y$ such that $xy=4$ and $x+y=2$. Then, $x$ and $y$ are real roots to the polynomial
$$
P(z)=z^2-2z+4.
$$
But $P(z)=(z-1)^2+3$ clearly has no real roots, so contradiction. (Ref: <a href="http://en.wikipedia.org/wiki/Vieta%27s_formulas" rel="nofollow">Vieta's formulas</a>.)</p>
|
1,979,876 | <p>Currently I started studying about ray-casting when I came across this following problem based on ray-triangle intersection. The problem was:</p>
<p>You are provided with a triangle with vertices ,<strong>(x1,y1,z1)</strong>, <strong>(x2,y2,z2)</strong> and <strong>(x3,y3,z3)</strong>. A ray with origin <strong>(a1,b1,c1)</strong> and direction <strong>(a2,b2,c2)</strong> is also given. Your task is to find: </p>
<p>1)Whether or not the ray intersects the triangle</p>
<p>2) If the ray intersects the triangle, what's the point of intersection and also find the distance of theat point from the origin of the ray</p>
<p>Examples:</p>
<p>1) For <strong>(x1,y1,z1)</strong>=<strong>(-2,2,6)</strong>, <strong>(x2,y2,z2)</strong>=<strong>(2,2,6)</strong>, <strong>(x3,y3,z3)</strong>=<strong>(0,-4,6)</strong> , <strong>(a1,b1,c1)</strong>=<strong>(1,0,0)</strong>,<strong>(a2,b2,c2)</strong>=<strong>(-0.2,0,1)</strong> the answer is: coordinates of intersection:<strong>(-0.2,0,6)</strong> and the distance of the origin of the ray from the point of intersection is <strong>6.12</strong></p>
<p>2) For <strong>(x1,y1,z1)</strong>=<strong>(-2,2,1)</strong>, <strong>(x2,y2,z2)</strong>=<strong>(2,2,1)</strong>, <strong>(x3,y3,z3)</strong>=<strong>(0,-4,1)</strong> , <strong>(a1,b1,c1)</strong>=<strong>(0,0,0)</strong>,<strong>(a2,b2,c2)</strong>=<strong>(0,0,1)</strong> the answer is: coordinates of intersection:<strong>(0,0,1)</strong> and the distance of the origin of the ray from the point of intersection is <strong>1</strong> </p>
<p>3) For <strong>(x1,y1,z1)</strong>=<strong>(-10,-2.3,0)</strong>, <strong>(x2,y2,z2)</strong>=<strong>(4.4,20.3,9.5)</strong>, <strong>(x3,y3,z3)</strong>=<strong>(9.8,-10,0)</strong> , <strong>(a1,b1,c1)</strong>=<strong>(0,0,0)</strong>,<strong>(a2,b2,c2)</strong>=<strong>(0.68,-1.14,1.82)</strong> the answer is: coordinates of intersection:<strong>(0.67, -1.12, 1.79)</strong> and the distace of the origin of the ray from the point of intersection is <strong>2.22</strong> </p>
<p>All I need is to understand the general equations to solve this question. I need to know how the equations are formed and how the problem is solved. This seems a pretty hard question for me.</p>
| K. Miller | 264,375 | <p>The ray can be parameterized by the equation $\mathbf{y}(t) = (a_1,b_1,c_1) + t(a_2,b_2,c_2)$ for all $t\geq 0$. The boundaries of the triangle may be parameterized by</p>
<p>\begin{align}
\mathbf{s}_1(\tau) &= (x_1,y_1,z_1) + \tau(x_2-x_1,y_2-y_1,z_2-z_1)\\
\mathbf{s}_2(\tau) &= (x_2,y_2,z_2) + \tau(x_3 - x_2,y_3-y_2,z_3-z_2)\\
\mathbf{s}_3(\tau) &= (x_3,y_3,z_3) + \tau(x_1 - x_3,y_1 - y_3,z_1-z_3)
\end{align}
for all $\tau\in[0,1]$. The triangle lies in a plane with normal vector </p>
<p>$$
\mathbf{n} = (x_2-x_1,y_2-y_1,z_2-z_1)\times(x_2 - x_3,y_2 - y_3,z_2-z_3)
$$</p>
<p>If $(a_2,b_2,c_2) \cdot \mathbf{n} \neq 0$, then the line intersects the plane for some $t\in\mathbb{R}$. The intersection point with the ray, if it exists, is given by solving $\mathbf{n}\cdot(\mathbf{y}(t) - (x_1,y_1,z_1)) = 0$ for $t\geq 0$. Lastly, you have to check if the intersection point lies on the triangle. To do this let $\mathbf{y}$ be the intersection point and suppose it does not lie on a boundary of the triangle (if it does then you are done). Define the line $\mathbf{x}(t) = \mathbf{y} + t(\mathbf{s}_1(0.5) - \mathbf{y})$ for all $t\geq 0$. If $\mathbf{x}(t) = \mathbf{s}_2(\tau)$ or $\mathbf{x}(t) = \mathbf{s}_3(\tau)$ have a solution for $t \geq 0$ and $\tau\in[0,1]$, then the point is exterior to the triangle, otherwise it is interior. The distance from the intersection point to the origin of the ray is simply $\|\mathbf{y} - (a_1,b_1,c_1)\|$.</p>
|
3,404,673 | <p>Let <span class="math-container">$G$</span> be a unipotent connected linear algebraic group over a field <span class="math-container">$F$</span>. Then <span class="math-container">$G$</span> is called <em>split</em> if there is a series of closed subgroup schemes <span class="math-container">$1 = G_0 \subset G_1 \subset \cdots \subset G_t =G$</span> with each <span class="math-container">$G_i$</span> normal in <span class="math-container">$G_{i+1}$</span> and <span class="math-container">$G_i/G_{i+1} \cong_F \mathbb G_a$</span>.</p>
<p>I have read that for <span class="math-container">$F$</span> perfect, every such <span class="math-container">$G$</span> is split, provided it is connected. And <span class="math-container">$F$</span> of characteristic zero, every such <span class="math-container">$G$</span> is moreover automatically connected.</p>
<p>I was trying to verify this in the case of <span class="math-container">$G = \operatorname{Res}_{E/F} \mathbb G_a$</span>, where <span class="math-container">$E/F$</span> is a quadratic extension. Does <span class="math-container">$G$</span> really have such a composition series?</p>
<p>I tried considering the diagonal embedding <span class="math-container">$\Delta$</span> of <span class="math-container">$\mathbb G_a$</span> into <span class="math-container">$G$</span> and looking at the quotient <span class="math-container">$G/\Delta$</span>. It should be the case that <span class="math-container">$G/\Delta \cong \mathbb G_a$</span>, but I'm not able to verify this. The natural map <span class="math-container">$G/\Delta \rightarrow \mathbb G_a$</span> given by <span class="math-container">$(x,y)\Delta \mapsto x- y $</span> is not defined over <span class="math-container">$F$</span>. </p>
| WhatsUp | 256,378 | <p>It's not that complicated. I believe you have all the needed knowledge, but just have to unpack the definitions.</p>
<p>By fixing a basis of <span class="math-container">$E/F$</span>, we know that <span class="math-container">$E$</span> is isomorphic to <span class="math-container">$F^2$</span> as <span class="math-container">$F$</span>-vector spaces, hence for any <span class="math-container">$F$</span>-algebra <span class="math-container">$R$</span>, one has <span class="math-container">$\mathbb{G}_a(E\otimes_F R) = E\otimes_F R \simeq R^2 = \mathbb{G}_a^2(F)$</span>, where the middle isomorphism is functorial in <span class="math-container">$R$</span>.</p>
<p>Hence by the definition of restriction of scalars, we see that <span class="math-container">$\operatorname{Res}_{E/F}\mathbb{G}_a \simeq \mathbb{G}_a^2$</span>.</p>
|
623,703 | <blockquote>
<p>Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator. </p>
</blockquote>
<p>I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$</p>
<p>So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$. </p>
<p>The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you. </p>
| Zhoe | 99,231 | <p>$\hskip2in$ <img src="https://i.stack.imgur.com/fYbLE.gif" alt="enter image description here"></p>
<p>Using the triangle above...& the fact that $$\tan x = \frac{\text{opp}}{\text{adj}}, \space \tan \left(\frac{\pi}{4}\right)=...$$</p>
|
623,703 | <blockquote>
<p>Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator. </p>
</blockquote>
<p>I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$</p>
<p>So, $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )=\tan\left( \dfrac{\pi}{4}\right)$. </p>
<p>The answer is $1$. Can you show how to solve $\tan\left( \dfrac{\pi}{4}\right)$ to get $1$? Thank you. </p>
| N74 | 288,459 | <p>You have</p>
<p><span class="math-container">$
\sin x=\dfrac{\sqrt{2}}{2}
$</span></p>
<p>so</p>
<p><span class="math-container">$
\cos x=\sqrt{1-(\sin x)^{2}}=\dfrac{\sqrt{2}}{2}
$</span></p>
<p>and</p>
<p><span class="math-container">$
\tan x =\dfrac{\sin x}{\cos x}\ = 1
$</span></p>
|
309,395 | <p>I am currently learning about <em>Direct Proofs</em>. I am struggling trying to find a starting point to prove the Statement: <em>For all real numbers $c$, if $c$ is a root of a polynomial with rational coefficients, then c is a root of a polynomial with integer coefficients</em>. Based on a definition given by the book and the professor: A number is $c$ is called a root of a polynomial $p(x)$ <em>if, and only if,</em> $p(c) = 0$. But how can I prove the veracity of this statement using the given definition?</p>
| half-integer fan | 54,125 | <p>Any polynomial with rational coefficients can be made into a polynomial with integer coefficients by multiplying through by the product of the denominators of the coefficients (or the LCM of those denominators). All that remains, then, is for you to show that if a number $c$ is the root of a polynomial, then it is also the root of any polynomial formed by multiplying that polynomial by a constant.</p>
|
1,816,807 | <blockquote>
<p>If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$</p>
</blockquote>
<p>$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$</p>
<p>So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$</p>
<p>Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then</p>
<p>$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$</p>
<p>So we get $\displaystyle t^3-4t^2-5t-c=0$</p>
<p>Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$</p>
<p>and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$</p>
<p>So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$</p>
<p>Now How can I solve it after that, Help required, Thanks</p>
| Ángel Mario Gallegos | 67,622 | <p>By CS inequality we get
$$|x^3+y^3+z^3|\le\sqrt{x^2+y^2+z^2}\sqrt{x^4+y^4+z^4}\tag{1}$$
Where
\begin{align*}
x^4+y^4+z^4&=(x^2+y^2+z^2)^2-2[(xy+yz+xz)^2-2xy^2z-2xyz^2-2x^2yz]\\
&=36-2[5^2-2xyz(4)]\\
x^4+y^4+z^4&=16xyz-14\tag{2}
\end{align*}
Now, from the identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$
follows
\begin{align*}
x^3+y^3+z^3-3xyz&=(4)(6-5)\\
x^3+y^3+z^3&=3xyz+4\tag{3}
\end{align*}
$(1)$, $(2)$ and $(3)$ imply
\begin{align*}
|3xyz+4|\le\sqrt{6}\sqrt{16xyz-14}
\end{align*}
By setting $t=xyz$ and squaring the last inequality we get
$$9t^2+24t+16\le 96t-84\quad\iff\quad 9t^2-72t+100\le0\quad\iff\quad(3t-12)^2-44\le0$$
So
$$\boxed{\color{blue}{4-\frac23\sqrt{11}\le xyz\le4+\frac23\sqrt{11}}}$$</p>
|
1,816,807 | <blockquote>
<p>If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$</p>
</blockquote>
<p>$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$</p>
<p>So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$</p>
<p>Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then</p>
<p>$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$</p>
<p>So we get $\displaystyle t^3-4t^2-5t-c=0$</p>
<p>Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$</p>
<p>and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$</p>
<p>So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$</p>
<p>Now How can I solve it after that, Help required, Thanks</p>
| Samrat Mukhopadhyay | 83,973 | <p>Note that $$c=t((t-2)^2+1),\ t\in \mathbb{R}$$, so according to the calculations, $c\in [a,b]$ where $$a=t((t-2)^2+1)|_{t=5/3}=\frac{50}{27},\ b=t((t-2)^2+1)|_{t=1}=2$$</p>
|
3,874,020 | <p>As it says in the title, any help would be greatly appreciated. I've tried some things but just can't get it to click.</p>
| Tryst with Freedom | 688,539 | <p><span class="math-container">$$ \vec{r(t) } = \langle1 , 0 ,-1\rangle +\langle 3 \cos t , 4 \cos t , 5 \sin t\rangle$$</span></p>
<p>And,</p>
<p><span class="math-container">$$ |\langle 3 \cos t , 4 \cos t + 5 \sin t\rangle | = \sqrt{ ( 3 \cos t)^2 + (4 \cos t)^2 + (5 \sin t)^2} = 5$$</span></p>
<p>So we have that no matter what the value of <span class="math-container">$t$</span> is the second term is at a distance of <span class="math-container">$5$</span> from the point <span class="math-container">$\langle1,0,-1\rangle$</span>.</p>
<p>Can you take it from here?</p>
|
3,874,020 | <p>As it says in the title, any help would be greatly appreciated. I've tried some things but just can't get it to click.</p>
| Rubén Colomina Citoler | 37,795 | <p><span class="math-container">$r(t)$</span> defines a curve in <span class="math-container">$3$</span>-dimensions as a functions of every real <span class="math-container">$t$</span> from <span class="math-container">$\mathbb R$</span> onto <span class="math-container">$\mathbb R^3$</span>. We will have to get rid of its parameter <span class="math-container">$t$</span> solving the following system of equations:</p>
<p><span class="math-container">$ x = 1+3\cos(t) $</span></p>
<p><span class="math-container">$ y = 4\cos(t)$</span></p>
<p><span class="math-container">$ z = -1 + 5\sin(t)$</span></p>
<p>Do squares over all the equations:</p>
<p><span class="math-container">$ (x -1)^2 = 9 \cos^2(t) $</span></p>
<p><span class="math-container">$ y^2 = 16 \cos^2(t)$</span></p>
<p><span class="math-container">$ (z+1)^2 = 25 \sin^2(t)$</span></p>
<p>Now add the three of them together:</p>
<p><span class="math-container">$ (x -1)^2 + y^2 + (z+1)^2 = 9 \cos^2(t) + 16 \cos^2(t) + 25 \sin^2(t)$</span></p>
<p>Applying <span class="math-container">$\cos^2(u) + \sin^2(u) =1 $</span> for every real number.</p>
<p><span class="math-container">$(x-1)^2 + y^2 + (z+1)^2 = 25$</span></p>
<p>The last equation defines a sphere of radius <span class="math-container">$5$</span> and center <span class="math-container">$(1,0,-1)$</span> in <span class="math-container">$3$</span>D space.</p>
|
712,697 | <p>What is the radius of convergence?</p>
<p>$$\sum_{n=0}^{\infty} n^3 (5x+10)^n$$</p>
| Makoto Kato | 28,422 | <p>Since $X$ is quasi-compact and locally noetherian, there exists a finite open cover $V_1,\ldots,V_n$ each of which is noetherian.
Let $\mathcal V$ be any non-empy family of open sets.
Write $\mathcal V_i=\{U\cap V_i\mid U\in\mathcal V\}$.</p>
<p>We define inductively an element $U_i \in \mathcal V_i$ and a non-empty subset $\mathcal U_i$ of $\mathcal V$ for each $i$.
Since $V_1$ is noetherian, there exists a maximal element $U_1$ of $\mathcal V_1$.
Let $\mathcal U_1 = \{U \in \mathcal V\mid U\cap V_1 = U_1\}$.
Let $i$ be an integer such that $1 \lt i \le n$.
Suppose $U_j$ and $\mathcal U_j$ are defined for all $j$ such that $1 \le j \lt i$
Let $U_i$ be a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$.
Let $\mathcal U_i = \{U \in \mathcal U_{i-1}\mid U\cap V_i = U_i\}$.
Notice that $\mathcal U_i$ is not empty since $V_i$ is noetherian.</p>
<p>Now that $U_i$ and $\mathcal U_i$ are defined for all $i(1 \le i \le n)$, it suffices to prove that any element $W$ of $\mathcal U_n$ is a maximal element of $\mathcal V$.
Let $V$ be an element of $\mathcal V$ such that $W \subset V$.
It suffices to prove that $W = V$.
We first prove by induction that $V \in \mathcal U_i$ for all $i = 1, \cdots n$.
Since $\mathcal U_n \subset \mathcal U_{n-1} \subset \cdots \subset \mathcal U_1 \subset \mathcal V$, $W \in \mathcal U_i$ for all $i = 1, \cdots, n$.
Since $W\in \mathcal U_1$, $W\cap V_1 = U_1$.
Since $U_1$ is a maximal element of $\mathcal V_1$ and $V\in \mathcal V$,
$W\cap V_1 = V\cap V_1 = U_1$.
Hence $V\in \mathcal U_1$.
Let $i$ be an integer such that $1 \lt i \le n$.
Suppose $V \in \mathcal U_j$ for all $j$ such that $1 \le j \lt i$.
Since $W \in \mathcal U_i, W\cap V_i = U_i$.
Since $U_i$ is a maximal element of $\{U\cap V_i\mid U\in\mathcal U_{i-1}\}$ and $V \in \mathcal U_{i-1}$, $V\cap V_i = U_i$.
Hence $V \in \mathcal U_i$</p>
<p>Therefore $W\cap V_i = V \cap V_i = U_i$ for each $i$.
Hence $W = V$ as desired.
<strong>QED</strong></p>
|
3,337,382 | <p>I want to find how many roots of the equation <span class="math-container">$z^4+z^3+1=0$</span> lies in the first quadrant.</p>
<p>Using Rouche's Theorem how to find ?</p>
| Jack D'Aurizio | 44,121 | <p>All the roots of the given polynomial lie inside <span class="math-container">$|z|\leq 2$</span>: if we assume <span class="math-container">$|z|>2$</span> and <span class="math-container">$z^4=-(z^3+1)$</span>, we get <span class="math-container">$2<|z|=|z^{-3}+1|<\frac{9}{8}$</span> which is a contradiction. Additionally it is simple to check that the roots are simple and there are no purely real or purely imaginary roots, so the number of roots in the first quadrant is given by
<span class="math-container">$$ \frac{1}{2\pi i}\oint_{\partial Q}\frac{4z^3+3z^2}{z^4+z^3+1}\,dz $$</span>
where <span class="math-container">$Q$</span> is the square with vertices in <span class="math-container">$(0,0),(2,0),(2,2),(0,2)$</span>. It is enough to approximate the (imaginary parts of the) four integrals associated to the sides of <span class="math-container">$Q$</span> with decent accuracy (for instance, through <a href="https://en.wikipedia.org/wiki/Newton-Cotes_formulas" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Newton-Cotes_formulas</a>) to discover that there is exactly one root per quadrant.</p>
|
3,337,382 | <p>I want to find how many roots of the equation <span class="math-container">$z^4+z^3+1=0$</span> lies in the first quadrant.</p>
<p>Using Rouche's Theorem how to find ?</p>
| Andrés Martínez | 1,103,687 | <p>For completeness, here I present a more standard solution to this problem:</p>
<p>Let <span class="math-container">$p(z):=z^4+z^3+1$</span>. Consider the following path</p>
<p><a href="https://i.stack.imgur.com/VDgt4m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VDgt4m.png" alt="enter image description here" /></a></p>
<p>Since <span class="math-container">$p(x)>0$</span> and <span class="math-container">$p(ix)=x^4+1-ix^3\neq 0$</span> for <span class="math-container">$x\geq 0$</span>, we deduce that <span class="math-container">$p$</span> has no roots on <span class="math-container">$\gamma_R$</span>. Therefore, we can apply the Argument Principle to obtain that the number of roots of <span class="math-container">$p$</span> in the first cuadrant is the winding number of <span class="math-container">$0$</span> around <span class="math-container">$p\circ \gamma_R$</span>, for sufficiently big <span class="math-container">$R$</span>.</p>
<p>Besides, for big <span class="math-container">$R$</span> the argument variation of <span class="math-container">$p\circ \text{quarter circle}$</span> is approximately <span class="math-container">$\Delta \arg z^4=4\frac{\pi}{2}=2\pi$</span>. Therefore, this fact combined with <span class="math-container">$p(x)>0$</span>, <span class="math-container">$p(ix)\in 4\text{th quadrant}$</span> tells us that <strong><span class="math-container">$p$</span> has exactly one root in the first quadrant</strong>.</p>
|
635,301 | <p>I need some help with the following problem: </p>
<blockquote>
<p>Let $f:\Bbb C \to \Bbb C$ be continuous satisfying that $f(\Bbb C)$ is an open set and that $|f(z)| \to \infty$ as $z\to \infty$. Prove that $f(\Bbb C)=\Bbb C$. </p>
</blockquote>
<p>My idea on this one is to prove by contradiction and assume that $S=f(\Bbb C)\ne\Bbb C$ to get some contradiction with the given two properties of the function. But I have no idea on how to proceed next. </p>
<p>Thanks in advance. </p>
| Igor Rivin | 109,865 | <p>By your condition, the image of $f$ is closed (why? take a convergent sequence of points in the image, and their preimages. The preimages do not diverge to infinity, by assumption, so they have a limit point, which, by continuity, goes to the limit you want). So, $f(\mathbb{C})$ is open (by assumption) and closed. So, since $\mathbb{C}$ is connected, the image is all of $\mathbb{C}.$</p>
|
465,001 | <p>$\mathbf{h}_i\in\mathbb{C}^{M}$ are column vectors $\forall i=\{1, 2, \cdots, K\}$.</p>
<p>$q_i\in\mathbb{R}_+$ are scalars $\forall i=\{1, 2, \cdots, K\}$</p>
<p>$\lvert\bullet\rvert$ denotes determinant of a square matrix or Euclidean norm of a vector according to the context. </p>
<p>From Sylvester's theorem, it's trivial to show that $\lvert\mathbf{I}_M+\mathbf{h}_1q_1\mathbf{h}_1^{\text{H}}\rvert=1+q_i\left|\mathbf{h}_1\right|^2$. Is it possible to extend the theorem to simplify $\lvert\mathbf{I}_M+\sum_{i=1}^K\mathbf{h}_iq_i\mathbf{h}_i^{\text{H}}\rvert$?</p>
<p>P. S. It's part of a problem involving multiple access channels and I am not sure if a definite solution exists. </p>
| Ben Grossmann | 81,360 | <p>You are correct in saying that since $\cos\left(\frac1x\right)$ is an even function: if the limit as $x\to0$ from one direction exists, it must be equal to the limit from the other direction. However, neither $\lim_{x\to0^+}\cos\left(\frac1x\right)$ nor $\lim_{x\to0^-}\cos\left(\frac1x\right)$ exist. In fact, these limits don't exist for the same reason that $\lim_{x\to0^+}\sin\left(\frac1x\right)$ doesn't exist, and comparing the right-side and left-side limits doesn't give you the whole picture in any of these cases.</p>
<hr>
<p>Here's another way of thinking about this question to hopefully put you on the right track: what is the limit $\lim_{x\to\infty}\sin(x)$? Does it exist? If so, what is it? If not, then why not? Why is this limit the same as your limit?</p>
|
8,013 | <p>Given a set $S$, one can easily find a set with greater cardinality -- just take the power set of $S$. In this way, one can construct a sequence of sets, each with greater cardinality than the last. Hence there are at least countably infinite many orders of infinity.</p>
<p><strong>But do there exist uncountably infinite orders of infinity?</strong></p>
<p>To be precise, does there exist an uncountable set of sets whose elements all have distinct cardinalities?</p>
<p>The first answer to <a href="https://math.stackexchange.com/questions/5378/types-of-infinity">Types of infinity</a> suggests the answer is "yes", but only establishes a countable number of cardinalities (which, to be fair, was what the question was asking about).</p>
<p>I've been exposed to enough mathematical logic to realize that I'm walking in a minefield; let me know if I've already mis-stepped.</p>
| morgan | 317 | <p>I can't believe this question is not closed.</p>
<p>As near as I can tell, Chloe solved the problem which has stymied Dr. Rush by successfully integrating a constant. Apparently Rush's expertise in ancient technology does not cover first year calculus. </p>
<p><img src="https://i.stack.imgur.com/kv6EZ.jpg" alt="Before"></p>
<p><img src="https://i.stack.imgur.com/Z4lZ7.jpg" alt="After"></p>
<p><strong>Update</strong>: Actually, looking over it, I might be mistaken. I interpreted the question as $-\int_0^t g \sin \gamma dt$ and Chloe's answer as $-\ddot{r} t \sin \gamma$, which is integrating a constant and making a gratuitous substitution of $\ddot{r}$ for $g$. I can't make any sense at all of the middle line in Chloe's answer.</p>
|
8,013 | <p>Given a set $S$, one can easily find a set with greater cardinality -- just take the power set of $S$. In this way, one can construct a sequence of sets, each with greater cardinality than the last. Hence there are at least countably infinite many orders of infinity.</p>
<p><strong>But do there exist uncountably infinite orders of infinity?</strong></p>
<p>To be precise, does there exist an uncountable set of sets whose elements all have distinct cardinalities?</p>
<p>The first answer to <a href="https://math.stackexchange.com/questions/5378/types-of-infinity">Types of infinity</a> suggests the answer is "yes", but only establishes a countable number of cardinalities (which, to be fair, was what the question was asking about).</p>
<p>I've been exposed to enough mathematical logic to realize that I'm walking in a minefield; let me know if I've already mis-stepped.</p>
| SpaceMika | 55,118 | <p>Awww, if you want screenshots of my Crazy Hall, all you have to do is ask! Yes, it's all legit math, and all internally consistant throughout the seasons. The core science is usually from peer-reviewed publications, just blended together in ways that don't typically make sense in the Real World (when would you ever feed the energy of a solar flare into a black hole, except with Stargate?)</p>
<p>I frequently use the established alien alphabets for variables, making it a bit more squiggle-loving, and anything that's going to be completed-by-an-actor is always simplified down to "write the last 1-3 characters," with the assumption that we're catching them at the end of a big scribble-fest. Thus, it wasn't really "Chloe knows how to integrate & Rush doesn't," as it was, "Chloe has thought of a new approach to this problem that Rush didn't." Each character has different and distinct handwriting associated with them (usually based off writing samples from the actor), but admittedly that'd almost impossible to spot in the fuzzy darkness that is Destiny's hallways.</p>
<p>For SGU specifically, the Crazy Hall evolved over time based on what the crew had encountered. For example, when going through the "How do we collect Lt. Scott from that shuttlecraft before we jump?!", an entire wall on orbital dynamics, with particular focus on energy needs for intersecting orbits, got added in.</p>
|
2,793,077 | <blockquote>
<p>Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.</p>
</blockquote>
<p>From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.</p>
<p>Under addition a ring is a commutative group. </p>
<p>$a + b = b + a$ (commutative)</p>
<p>$(a + b) + c = a + (b + c)$ (associative)</p>
<p>$a + (-a) = 0$ (inverse exists for every element)</p>
<p>$a + 0 = a$ (identity exists)</p>
<p>Where $a,b,c \in R$</p>
<p>But I'm not really sure how to proceed with the proof from here. Any idea?</p>
| hmakholm left over Monica | 14,366 | <p><strong>Hint.</strong> Compute $(a+1)^2$ in two ways -- once by the Boolean ring property, another time by using the distributive law that works in all rings to multiply out $(a+1)(a+1)$. Then apply $a^2=a$ once more and cancel terms that appear in both results.</p>
|
34,204 | <p>I have several contour lines and one point. How can I find a point in one of those contour lines which is nearest to the given point?</p>
<pre><code>(*Create the implicit curves*)
Data={{10,20,1},{10,40,2},{10,60,3},{10,80,4},{20,25,2},{20,45,3},{20,65,4},{30,30,3},{30,50,4},{40,35,4},{40,55,5},{50,20,4},{50,40,5},{60,25,5}};
U=NonlinearModelFit[Data,a x^b (y^(1-b))+c,{a,b,c},{x,y}];
L={ContourPlot[U[x,y]=={1},{x,0,100},{y,0,100},ContourStyle->Red],ContourPlot[U[x,y]=={2},{x,0,100},{y,0,100},ContourStyle->Magenta],ContourPlot[U[x,y]=={3},{x,0,100},{y,0,100},ContourStyle->Brown],ContourPlot[U[x,y]=={4},{x,0,100},{y,0,100},ContourStyle->Blue],ContourPlot[U[x,y]=={5},{x,0,100},{y,0,100},ContourStyle->Green]};
(*Point nearest to which we need to find the points on the curves*)
pt={30,50};
(*Graphic*)
Show[L,Graphics[{PointSize[Large],Blue,Point[pt]}],FrameLabel->{"X","Y"}]
</code></pre>
<p><img src="https://i.stack.imgur.com/kElkt.jpg" alt="enter image description here"></p>
| cormullion | 61 | <p>One way you could try is to examine the points of the plot. If <code>cp</code> is the plot:</p>
<pre><code>points = Cases[cp, {x_?NumericQ, y_?NumericQ} :> {x, y}, Infinity];
nearest = Last@Nearest[points, First[pt], 2]
</code></pre>
<blockquote>
<pre><code>{36.7666, 51.0523}
</code></pre>
</blockquote>
<p>This is not the nearest point to <code>pt</code> - that's <code>pt</code> itself - but the next nearest: </p>
<pre><code>Show[
cp,
Graphics[{Red, Disk[nearest]}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/kcN29.png" alt="plot"></p>
<p>As yves points out in the comments, this only works as well as the original plot allows:</p>
<p><img src="https://i.stack.imgur.com/QrK4X.png" alt="blow up"></p>
<p>so technically I haven't found the nearest possible point, just the nearest sampled point.</p>
|
1,878,806 | <p>I am a graduate school freshman.</p>
<p>I did not take a probability lecture.</p>
<p>So I don't have anything about Probability.</p>
<p>Could you suggest Probability book No matter What book level?</p>
| EHH | 133,303 | <p>This is a standard and good introduction to probability:
<a href="https://www.amazon.co.uk/Probability-Introduction-Oxford-Science-Publications/dp/0198532644" rel="nofollow">https://www.amazon.co.uk/Probability-Introduction-Oxford-Science-Publications/dp/0198532644</a></p>
<p>I would also recommend avoiding the approach recommended by Justin in the comments. He is right that measure theory is important in advanced probability theory, however, going straight to this without having first studied introductory level probability I believe would be confusing, unhelpful and just not necessary. Furthermore measure theory itself is a topic which first requires a good foundation in dealing with abstract maths and so not a good place to start.</p>
|
848,415 | <p>If the limit of one sequence $\{a_n\}$ is zero and the limit of another sequence $\{b_n\}$ is also zero does that mean that $\displaystyle\lim_{n\to\infty}(a_n/b_n) = 1$?</p>
| Jika | 143,855 | <p>Of course not.</p>
<ul>
<li>$a_n=\dfrac{(-1)^n}{n}$,</li>
<li>$b_n=\dfrac{1}{n}$, </li>
<li>Hence, $r_n=\dfrac{a_n}{b_n}=(-1)^n$ and the limit of $r_n$ does not even exist.</li>
</ul>
|
848,415 | <p>If the limit of one sequence $\{a_n\}$ is zero and the limit of another sequence $\{b_n\}$ is also zero does that mean that $\displaystyle\lim_{n\to\infty}(a_n/b_n) = 1$?</p>
| Tom | 103,715 | <p>If we want an example such that the limit exists: Let $r \in \Bbb{R}$ be your favorite number, suppose that $b_n \to 0$, and let $a_n = r b_n$. Then $a_n \to 0$ and $a_n/b_n \to r$. </p>
<p>If we want an example such that the limit does not exist: Let $b_n \to 0$ and $\{c_n\}$ be your favorite <em>bounded</em> sequence such that $\lim c_n$ does not exist. Let $a_n = c_n b_n$, then $a_n \to 0$ and $a_n/b_n = c_n$, and the limit does not exist.</p>
<p>But, don't consider this a disappointing result! We would be in a much different world if this was not the case; for example, imagine how boring L'Hopital's rule would be.</p>
|
136,996 | <p>What is known about $f(k)=\sum_{n=0}^{k-1} \frac{k^n}{n!}$ for large $k$?</p>
<p>Obviously it is is a partial sum of the series for $e^k$ -- but this partial sum doesn't reach close to $e^k$ itself because we're cutting off the series right at the largest terms. In the full series, the $(k+i-1)$th term is always at least as large as the $(k-i)$th term for $1\le i\le k$, so $f(k)< e^k/2$. Can we estimate more precisely how much smaller than $e^k$ the function is?</p>
<p>It would look very nice and pleasing if, say, $f(k)\sim e^{k-1}$ for large $k$, but I have no real evidence for that hypothesis.</p>
<p>(Inspired by <a href="https://math.stackexchange.com/questions/136932/">this question</a> and my answer thereto).</p>
| Robert Israel | 8,508 | <p>$e^{-k} f(k) = P(X_k < k)$ where $X_k$ is a Poisson random variable with parameter $k$. Since the sum of $k$ independent Poisson random variables with parameter $1$ is Poisson with parameter $k$, the Central Limit Theorem says the probability distribution of $Z_k = (X_k - k)/\sqrt{k}$ approaches a standard normal distribution. Thus $P(X_k < k) = P(Z_k < 0) \to 1/2$ as $k \to \infty$. </p>
|
2,536,185 | <p><strong>Lemma 3.21</strong>. </p>
<p>Let $S$ be a subset of $R$. Then $\bar{S}$ is a closed set. $\bar{S}$ denotes the closure of $S$.</p>
<p>The following is a proof of Lemma 3.21:</p>
<p><em>Proof</em>. </p>
<blockquote>
<p>By Corollary 3.16 it is enough to show that $\bar{S^c}$
is open.</p>
</blockquote>
<p>I understand that if we show that $\bar{S^c}$ is open, then $\bar{S}$ is closed, since Corollary 3.16 states that "a nonempty set $E$ is closed iff $E^c$
is open."</p>
<blockquote>
<p>We will show that
$\bar{S^c}$
is open using Lemma 3.15 (c) with $S$ replaced by $\bar{S^c}$
. So let $x \in \bar{S^c}$, if we can
show that $x$ has a neighborhood $U$ which is contained in $\bar{S^c}$
then we will be done (by that Lemma).</p>
</blockquote>
<p>Lemma 3.15 (c) states that "for every point $x \in S$, there exists $\epsilon > 0$ s.t. $N(x,\epsilon) ⊂ S$." So, I see why we replace $S$ with $\bar{S^c}$. Lemma 3.15 (a) states that "$S$ is open" and Lemma 3.15 (a) and (c) are equivalent. So, essentially, we would show that $\bar{S^c}$ is open.</p>
<blockquote>
<p>By the contrapositive of Lemma 3.20, there is a neighborhood $N$ of $x$ which is contained in $\bar{S^c}$.</p>
</blockquote>
<p>I understand that this is the contrapositive of Lemma 3.20. Lemma 3.20 states that "if $S$ is a nonempty set of real numbers then $x \in \bar{S}$ iff every neighborhood of $x$ contains at least one point $\in S$."</p>
<blockquote>
<p>If $y \in U$, then since $N$ is open, by the contrapositive of the last assertion in Lemma 3.20 with $x$ replaced by $y$, we see that $y \notin \bar{S}$. </p>
</blockquote>
<p>I understand how we arrive at the contrapositive of the statement. But, why do we replace $x$ with $y$? And is $N$ referring to an arbitrary neighborhood from Lemma 3.20?</p>
<blockquote>
<p>So $N ⊂ \bar{S^c}$
. Hence by Lemma 3.15 (c), $U = \bar{S^c}$
is open.</p>
</blockquote>
<p>Since the contrapositive states that $y \notin \bar{S}$, I get why $y \in \bar{S^c}$. But, I don't follow why $N ⊂ \bar{S^c}$ and why $U = \bar{S^c}$.</p>
| Mastrem | 253,433 | <p>Let $m,n\ge 3$ be given and suppose that $m\neq n$. Without loss of generality, we can say that $n>m$.</p>
<p>Assume there are primes between $m$ and $n$. Let $p$ be the largest prime number with $n>p\ge m$. It is clear that $p\not\in P_m$. If $p\not\in P_n$, we must have $p\mid n$, so $n\ge 2p$. However, Betrand's postulate tells us there is at least one prime inbetween $p$ and $2p$, contradicting the definition of $p$. We conclude that $p\in P_n$ and $P_n\neq P_m$.</p>
<p>Now, say there are no primes between $n$ and $m$. Now, we have $P_n=P_m$ if and only if $rad(n)=rad(m)$. Say $rad(n)=r$. So now we have $k,l\in\mathbb{N}$ with $n=kr$ and $m=lr$. This is the hard part. (In fact it's very hard, as shown in an answer to the repost on Mathoverflow)</p>
|
4,085,635 | <p>What is the volume of an n-dimension cube? Consider the length of each side to be <span class="math-container">$a$</span>. How to solve this problem?</p>
| Jaap Scherphuis | 362,967 | <p>If there are 4 or more sides, you can always reduce the number of sides by joining together two adjacent ones into one long edge (essentially setting the vertex angle to 180). If that would result in a side longer than all the other sides put together, simply change your mind and choose an adjacent pair of those other sides instead (which when joined together clearly cannot form a side that is longer than the sum of the others). Repeat till you have only 3 sides left.</p>
<p>So if you can see why it is sufficient for a triangle, it is clear that it works for the original sequence of side lengths, as they form a triangle with some extra vertices on the sides of the triangle.</p>
|
2,681,451 | <p>For the following question I'm getting stuck on a proof. Below I've just written out all the things/steps I've tried (even if they might be wrong). Could someone steer me in the right direction?</p>
<blockquote>
<p>Suppose $f$ is a real-valued function $f:\mathbb{R}\to\mathbb{R}$,
which is continuous at $0$, with the property
$\forall_{x,y\in\mathbb{R}}:f(x+y)=f(x)+f(y)$. Show that
$\exists_{\lambda\in\mathbb{R}}\forall_{x\in\mathbb{R}}:f(x)=\lambda
x$.</p>
</blockquote>
<p>So far I've found that $f(0)=0$, as $f(x+0)=f(x)+f(0)=f(x)$.
As $f$ is continuous at $0$:
$$\forall_{\epsilon >0}\exists_{\delta >0}:|x-0|<\delta\implies|f(x)-f(0)|<\epsilon \ \ \iff \ \ \ |x|<\delta\implies|f(x)|<\epsilon$$</p>
<p>I suspect that $f|x|=|f(x)|$, which I have yet to prove or find. I'm also afraid the next step I tried is illegal: $|x|<\delta\implies f|x|<f|\delta|$, cause this way, you can compare it to $\epsilon$ by $|f(x)|=f|x|$. This didn't really take me anywhere but I thought I'd just write out everything I tried so far.</p>
<p>I also though that $f$ is continuous everywhere: $f(x-y)=f(x)-f(y)$ gives:
$$|x|<\delta\implies|f(x)|<\epsilon \ \ \iff \ \ \ |a-b|<\delta\implies |f(a-b)|=|f(a)-f(b)|<\epsilon$$
as we substitute $x=a-b$ for arbitrary $a,b\in\mathbb{R}$.</p>
| Arian | 172,588 | <p>It is clear that any function of the form $f(x)=\lambda x$ satisfies $f(x+y)=f(x)+f(y)$ and $\lim_{x\to 0}f(x)=f(0)$. Now let $f:\mathbb{R}\to\mathbb{R}$ be such that these two conditions hold. It is not difficult to see that: $f(0+0)=f(0)+f(0)\Rightarrow f(0)=0$ and that for $x+y=0$ we have $f(x+y)=f(0)=f(x)+f(-x)\Rightarrow f(-x)=-f(x)$ hence $f(x)$ is an odd function. Moreover $$f(2x)=f(x+x)=f(x)+f(x)=2f(x)$$
and in general one can show by induction that $$f(nx)=f(x)+f((n-1)x)=f(x)+(n-1)f(x)=nf(x)$$
whenever $n\in\mathbb{N}$. Analogue for negative integers using that $f(x)$ is odd function by induction then
$$f(-nx)=f(-x)+f(-(n-1)x)=-f(x)-(n-1)f(x)=-nf(x)$$
We can extend this identity $f(nx)=nf(x)$ for $n\in\mathbb{Z}$ to rationals as well. Just let $x=y/n$ to obtain
$$f(ny/n)=nf(y/n)\Rightarrow f(y)=nf(y/n)\Rightarrow \frac{m}{n}f(y)=mf(\frac{y}{n})=f(\frac{m}{n}y)$$
where $m\in\mathbb{Z}$ and the last equality follows from the main identity $f(nx)=nx$ when $n\in\mathbb{Z}$. But $m/n\in\mathbb{Q}$ was arbitrary so indeed we have $f(qx)=qx$ for all $q\in\mathbb{Q}$. But $\mathbb{Q}$ is dense in $\mathbb{R}$. Let $r\in\mathbb{R}$ then there exists a sequence $(q_k)\subset\mathbb{Q}$ such that $\lim_k q_k=r$. But $f(q_kx)=q_kx$ for all $k$ so $\lim_kf(q_k x)=\lim_k(q_kx)=x\lim_kq_k=rx$. Continuity of $f$ implies $\lim_kf(q_kx)=f(\lim_kq_kx)=f(rx)$ hence $f(rx)=rf(x)$ for all $r\in\mathbb{R}$. Together with the condition $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$ implies that indeed $f$ is a linear function (often these two conditions are used as a definition for linear functions, operators etc.). So there exists some $\lambda\in\mathbb{R}$ such that $f(x)=\lambda x$ and $\lambda=f(1)$.</p>
|
4,063,337 | <p>In an exercise I'm asked the following:</p>
<blockquote>
<p>a) Find a formula for <span class="math-container">$\int (1-x^2)^n dx$</span>, for any <span class="math-container">$n \in \mathbb N$</span>.</p>
<p>b) Prove that, for all <span class="math-container">$n \in \mathbb N$</span>: <span class="math-container">$$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$</span></p>
</blockquote>
<p>I used the binomial theorem in <span class="math-container">$a$</span> and got:</p>
<p><span class="math-container">$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$</span></p>
<p>and so in part (b) i got:</p>
<p><span class="math-container">$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$</span></p>
<p>I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?</p>
| PierreCarre | 639,238 | <p>Part b) can be obtained just by induction. The result is immediate for <span class="math-container">$n=1$</span> and
<span class="math-container">\begin{align*}
\int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!}-\frac{1}{2n+2}\int_0^1(1-x^2)^{n+1} dx
\end{align*}</span></p>
<p>This implies that
<span class="math-container">$$
\int_0^1 (1-x^2)^{n+1}dx = \frac{2n+2}{2n+3} \frac{2^{2n}(n!)^2}{(2n+1)!}=\frac{(2n+2)^2 2^{2n} (n!)^2}{(2n+1)!(2n+2)(2n+3)}=\dfrac{2^{2n+2}((n+1)!)^2}{(2n+3)!}.
$$</span></p>
<hr />
<p><span class="math-container">\begin{align*}
\int_0^1 x (-2x) (1-x^2)^n dx = &\left[x \dfrac{(1-x^2)^{n+1}}{n+1} \right]_0^1 - \frac{1}{n+1}\int_0^1 (1-x^2)^{n+1}
\end{align*}</span></p>
|
4,063,337 | <p>In an exercise I'm asked the following:</p>
<blockquote>
<p>a) Find a formula for <span class="math-container">$\int (1-x^2)^n dx$</span>, for any <span class="math-container">$n \in \mathbb N$</span>.</p>
<p>b) Prove that, for all <span class="math-container">$n \in \mathbb N$</span>: <span class="math-container">$$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$</span></p>
</blockquote>
<p>I used the binomial theorem in <span class="math-container">$a$</span> and got:</p>
<p><span class="math-container">$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$</span></p>
<p>and so in part (b) i got:</p>
<p><span class="math-container">$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$</span></p>
<p>I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?</p>
| mathstackuser12 | 361,383 | <p>Here is an approach based on your attempt using a series</p>
<p><span class="math-container">$$F\left( n \right)=\int\limits_{0}^{1}{{{\left( 1-{{x}^{2}} \right)}^{n}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
n \\
k \\
\end{matrix} \right){{\left( -1 \right)}^{k}}}\int\limits_{0}^{1}{{{x}^{2k}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
n \\
k \\
\end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}}{2k+1}}$$</span>
Now note
<span class="math-container">$$\left( \begin{matrix}
n \\
k \\
\end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$</span>
And so
<span class="math-container">$$F\left( n \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{2k+1}\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}}=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\frac{{{\left( 1+z \right)}^{n}}\arctan \left( {{z}^{-1/2}} \right)}{\sqrt{z}}dz}$$</span></p>
<p>However
<span class="math-container">$$\arctan \left( z \right)=\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{{{z}^{2k+1}}}{{{\left( 1+{{z}^{2}} \right)}^{k+1}}}}$$</span>
which yields
<span class="math-container">$$F\left( n \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{1}{{{\left( z+1 \right)}^{k-n+1}}}}dz}$$</span>
and the residue therefore provides the result
<span class="math-container">$$F\left( n \right)=\frac{{{2}^{2n}}{{\left( n! \right)}^{2}}}{\left( 2n+1 \right)!}$$</span></p>
|
4,063,337 | <p>In an exercise I'm asked the following:</p>
<blockquote>
<p>a) Find a formula for <span class="math-container">$\int (1-x^2)^n dx$</span>, for any <span class="math-container">$n \in \mathbb N$</span>.</p>
<p>b) Prove that, for all <span class="math-container">$n \in \mathbb N$</span>: <span class="math-container">$$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$</span></p>
</blockquote>
<p>I used the binomial theorem in <span class="math-container">$a$</span> and got:</p>
<p><span class="math-container">$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$</span></p>
<p>and so in part (b) i got:</p>
<p><span class="math-container">$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$</span></p>
<p>I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?</p>
| Quanto | 686,284 | <p>Integrate by parts</p>
<p><span class="math-container">$$I_n= \int_0^1 (1-x^2)^n dx=\int_0^1 \frac{(1-x^2)^n}{x^{2n}}d(\frac{x^{2n+1}}{2n+1})=\frac{2n}{2n+1}I_{n-1},\>\>\>I_0=1
$$</span>
which leads to <span class="math-container">$I_n= \frac{2^{2n}(n!)^2}{(2n + 1)!}$</span>.</p>
|
4,063,337 | <p>In an exercise I'm asked the following:</p>
<blockquote>
<p>a) Find a formula for <span class="math-container">$\int (1-x^2)^n dx$</span>, for any <span class="math-container">$n \in \mathbb N$</span>.</p>
<p>b) Prove that, for all <span class="math-container">$n \in \mathbb N$</span>: <span class="math-container">$$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$</span></p>
</blockquote>
<p>I used the binomial theorem in <span class="math-container">$a$</span> and got:</p>
<p><span class="math-container">$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$</span></p>
<p>and so in part (b) i got:</p>
<p><span class="math-container">$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$</span></p>
<p>I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?</p>
| robjohn | 13,854 | <p><strong>Beta Integral</strong>
<span class="math-container">$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\frac12\int_0^1\left(1-t\right)^nt^{-1/2}\mathrm{d}t\tag{1a}\\
&=\frac{\Gamma(n+1)\Gamma(1/2)}{2\Gamma(n+3/2)}\tag{1b}\\
&=2^{2n}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\tag{1c}\\
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{1d}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(1a)}$</span>: substitute <span class="math-container">$t\mapsto t^{1/2}$</span><br />
<span class="math-container">$\text{(1b)}$</span>: <a href="https://en.wikipedia.org/wiki/Beta_function" rel="nofollow noreferrer">Beta Integral</a><br />
<span class="math-container">$\text{(1c)}$</span>: <a href="https://en.wikipedia.org/wiki/Multiplication_theorem" rel="nofollow noreferrer">Gauss' Multiplication Formula</a><br />
<span class="math-container">$\text{(1d)}$</span>: convert to <a href="https://en.wikipedia.org/wiki/Binomial_coefficient" rel="nofollow noreferrer">Binomial Coefficients</a> via <a href="https://en.wikipedia.org/wiki/Factorial#The_gamma_and_pi_functions" rel="nofollow noreferrer">Factorials</a></p>
<hr />
<p><strong>Integration by Parts</strong>
<span class="math-container">$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\int_0^12nt^2\left(1-t^2\right)^{n-1}\,\mathrm{d}t\tag{2a}\\
&=\frac{2n}{2n+1}\int_0^1\left(1-t^2\right)^{n-1}\mathrm{d}t\tag{2b}\\
&=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{2c}\\
&=\color{#090}{\frac1{2n+1}\frac{2^nn!}{(2n)!}}\color{#C00}{2^nn!}\tag{2d}\\[3pt]
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{2e}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(2a)}$</span>: integrate by parts<br />
<span class="math-container">$\text{(2b)}$</span>: add <span class="math-container">$\frac{2n}{2n+1}$</span> of the left side of <span class="math-container">$\text{(2a)}$</span> to <span class="math-container">$\frac1{2n+1}$</span> of the right side<br />
<span class="math-container">$\text{(2c)}$</span>: induction<br />
<span class="math-container">$\text{(2d)}$</span>: evaluate the products<br />
<span class="math-container">$\text{(2e)}$</span>: simplify</p>
|
16,290 | <p>Hi I am new here and have a calculus question that came up at work.</p>
<p>Suppose you have a $4' \times 8'$ piece of plywood. You need 3 circular pieces all equal diameter. What is the maximum size of circles you can cut from this piece of material?
I would have expected I could write a function for the area of the 3 circles in terms of $x$ and $y$, then differentiate it, find a point of maxima/minima and go from there.</p>
<p>My coworker did cut three $33''$ circles and that solved the real-world problem. But my passion would be to find the mathematical answer to this. I hope that my new stackexchange.com friends have the same passion, and can help me find the answer to this in general terms. </p>
<p>What I mean by that is someone says I have a piece of material Q units by 2Q units, what are the three circles of maximum size?... I hope you understand what I am asking.
I am looking to be a friend and contributor
BD</p>
| Isaac | 72 | <p>I believe the optimal configuration is as shown below, with the circles tangent to the sides of the rectangle and to one another.</p>
<p><img src="https://i.stack.imgur.com/nT3kG.png" alt="diagram"></p>
<p>Let the rectangle have sides of length $Q$ and $2Q$, as you specified, and let the radius of all three circles be $r$. Consider the right triangle with the segment joining the centers of the left and middle circles as its hypotenuse and with legs parallel to the sides of the rectangle.</p>
<p><img src="https://i.stack.imgur.com/4lORT.png" alt="diagram with triangle"></p>
<p>The hypotenuse has length $2r$, and the legs have lengths $Q-r$ and $Q-2r$, so by the Pythagorean theorem, $(2r)^2=(Q-r)^2+(Q-2r)^2$. Solving for $r$ in terms of $Q$ gives $r=(3\pm\sqrt{7})Q$, but since $r<Q$, $r=(3-\sqrt{7})Q\approx 0.354249Q$. For your 4-by-8 board, that gives an optimal radius of approximately $0.354249\cdot 48\approx 17.0039$ inches, so a diameter of just over 34 inches.</p>
|
1,435,603 | <blockquote>
<p>Let $S$ be a subset of $\mathbb{R}$ with strictly positive Lebesgue measure. Prove that almost every (with respect to Lebesgue measure) real number can be written as the sum of an element of $S$ and an element of $\mathbb{Q}$. </p>
</blockquote>
<p>So, I remember proving a long time ago that $S-S$ contains a neighborhood of 0, and I thought that it would be possible to use this to prove the problem, but I was unable to do so.<br>
I know that $S + \mathbb{Q}$ is dense in the reals and has infinite measure, but I can't figure out why it must be $\mathbb{R}$ minus a measure zero set.<br>
Thanks in advance :).</p>
| maibaita | 159,452 | <p>Let $R=\mathbb{R}-(S+\mathbb{Q})$. Assume its measure is strictly positive. Then there are $a,b\in\mathbb{R}$ with $a<b$ such that $(a,b)\subset{}R$. Since the measure of $S$ is strictly positive as well, the same is true for $S$, so $(c,d)\subset{}S$. That clearly contradicts $R=\mathbb{R}-(S+\mathbb{Q})$, since it's easy to find a rational $q$ for which $\gamma+q=\alpha$ with $\alpha\in(a,b)$ and $\gamma\in(c,d)$.</p>
|
724,045 | <p>I believe I'm not correctly understanding the concept of unique factorization and irreducibles.</p>
<p>Consider $R = \mathbb{F}_7$ and $h \in R[x]$ where $h = x^4 + 4x^3 + 3x^2 + 5x + 6$. Now $h$ has the following factorizations:</p>
<p>$h_1 = (3x^2 + 3x + 4)(5x^2 + x + 5)$</p>
<p>and</p>
<p>$h_2 = (x^2 + x + 6)(x^2 + 3x + 1)$</p>
<p>All four of those quadratics are irreducible in $R[x]$.</p>
<p>So factorization does not appear to be unique in $R[x]$. But I read on wikipedia (which may be wrong), that if $R$ is a UFD, then $R[x]$ is a UFD.</p>
<p>I must have a misunderstanding because $R = \mathbb{F}_7$ is a UFD, but we have shown two distinct factorizations of a polynomial in $R[x]$. Where am I going wrong?</p>
| Martin Argerami | 22,857 | <p>Consider the following sequence of elements in the space $V$ of finite sequences:
$$
u_1=(1,0,0,\ldots),\ \ u_2=(0,\frac12,0,\ldots),\ \ u_3=(0,0,\frac13,0,\ldots)
$$
Then
$$
\sum_{k=1}^nu_k=(1,\frac12,\frac13,\ldots,\frac1n,0,\ldots)
$$
Now consider these two norms on $u=(a_1,a_2,\ldots)$:
$$
\|u\|_1=\sum_{k=1}^\infty|a_k|,\ \ \ \|u\|_2=\left(\sum_{k=1}^\infty|a_k|^2\right)^{1/2}
$$
Then, for the $u_n$ defined above,
$$
\left\|\sum_{k=M}^nu_k\right\|_1=\sum_{k=M}^N\frac1k,\ \ \ \left\|\sum_{k=M}^Nu_k\right\|_2=\sum_{k=M}^N\frac1{k^2}
$$
So, in $\|\cdot\|_1$, the tails of the series $\sum_{k=1}^\infty u_k$ are unbounded, which means that the series diverges; while in $\|\cdot\|_2$, the tails go to zero, so the series converges. </p>
|
3,163,317 | <p>I am given the question "Find the equation of the line in standard form with slope <span class="math-container">$m = -3$</span> and passing through the point <span class="math-container">$(1, \frac{1}{3})$</span>"</p>
<p>The solution is provided as <span class="math-container">$x + 3y = 2$</span></p>
<p>I arrived at <span class="math-container">$3y + 3x =4$</span></p>
<p>Here is my working:</p>
<p>point slope formula:
<span class="math-container">$y - y1 = m(x - x1)$</span></p>
<p>My given slope m is <span class="math-container">$-\frac{1}{3}$</span> and point is <span class="math-container">$(1, \frac{1}{3})$</span></p>
<p>So:</p>
<p><span class="math-container">$y - \frac{1}{3} = -\frac{1}{3}(x - 1)$</span></p>
<p>I then multiplied both sides by 3 to get rid of the fractions:</p>
<p><span class="math-container">$3(y - \frac{1}{3}) = 3(-\frac{1}{3}(x - 1))$</span></p>
<p><span class="math-container">$3y - 1 = -3(x - 1)$</span></p>
<p><span class="math-container">$3y - 1 = -3x + 3$</span></p>
<p><span class="math-container">$3y = -3x + 4$</span></p>
<p><span class="math-container">$3y + 3x = 4$</span></p>
<p>Where did I go wrong and how can I arrive at <span class="math-container">$x + 3y = 2$</span>?</p>
| J. W. Tanner | 615,567 | <p>You made a mistake: <span class="math-container">$ 3(-\frac{1}{3}(x - 1))= -(x-1),$</span>
not
<span class="math-container">$ -3(x - 1).$</span></p>
|
575,232 | <p>Consider all strings whose letters belong to the set:</p>
<p>$A = \{ a, b, c, d, e\}$</p>
<p>How many strings of length $6$ are there that contain exactly one $a$?</p>
<p>Attempt:</p>
<p>Since we are only using $\frac{4}{5}$ letters for the rest of the string, </p>
<p>There are $1* 4^5$ strings that contain exactly one $a$.</p>
<p>Book answer: $6 * 4^5$</p>
<p>What am I doing wrong? Where does the six come from?</p>
| Daniel Robert-Nicoud | 60,713 | <p>Ok, a couple of months have passed since I posted this question, and I have begun this project on quantum groups. I am now in the condition of answering a couple of the questions I posed.</p>
<ol>
<li>There are many different definitions of quantum groups, all of which are related somehow. A possible definition is: a (Drinfeld-Jimbo) quantum group is a $1$-parameter deformation of the universal enveloping algebra $U\mathfrak{g}$ of a Lie algebra $\mathfrak{g}$ which is always a Hopf algebra.</li>
<li>The subject is interesting because it relates to the subject of <a href="http://en.wikipedia.org/wiki/Integrable_system#Quantum_integrable_systems" rel="noreferrer">quantum integrable systems</a> in quantum mechanics. There is an ongoing effort trying to geometrize the notion of quantum groups (as stated at page $28$ of the paper of Maulik and Okounkov).</li>
<li>See the wonderful and exhaustive <a href="https://math.stackexchange.com/a/589904/60713">answer</a> by Will Orrick.</li>
<li>To learn quantum groups you will need some knowledge of physics (to have a motivation for the subject and maybe to get some intuition), Lie groups and Lie algebras, and representation theory.</li>
</ol>
<p>For the questions added in the edits:</p>
<ol>
<li>I have just begun to read the paper, maybe in a couple of months, when I'll have a clearer idea of what is happening, I will update this answer to give an overview. From what I have been told by my professor, it should be about finding a geometrical approach to quantum groups.</li>
<li>To understand the paper easily you will need a lot of algebraic geometry (you should at least know what a Nakajima variety is, and know stuff about Gromov-Witten invariants) and at least some K-theory. I have to admit I am struggling, but I am also learning a lot.</li>
</ol>
<hr>
<p><strong>Edit:</strong> There are a few more subjects you will need to know to approach the paper and hope to undertand the majority of it. I would advise to take a look at:</p>
<ul>
<li>Symplectic geometry (for example <a href="http://www.math.ist.utl.pt/~acannas/Books/lsg.pdf" rel="noreferrer">here</a>, or any of the other references you can easily find online), in particular you should know at least what a moment map is.</li>
<li>Geometric Invariant Theory (GIT), which you can look up <a href="http://www-fourier.ujf-grenoble.fr/sites/ifmaquette.ujf-grenoble.fr/files/ginzburg_rev.pdf" rel="noreferrer">here</a>.</li>
<li>Equivariant cohomology, if you have access to Springer through an university, <a href="http://link.springer.com/chapter/10.1007/978-94-011-4542-8_3#page-1" rel="noreferrer">Bott's introduction</a> is really clear, else you can probably find something online.</li>
</ul>
<p>I will update if I find other useful stuff.</p>
|
3,336,742 | <p>Can we evaluate <span class="math-container">$\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$</span> ?</p>
<p>where <span class="math-container">$H_n=\sum_{k=1}^n\frac1n$</span> is the harmonic number.</p>
<p>A related integral is <span class="math-container">$\displaystyle\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2\left(\frac x2\right)}{x}dx$</span>.</p>
<p>where <span class="math-container">$\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$</span> is the dilogarithmic function.</p>
<hr>
<p>Here is how the integral and the sum are related:</p>
<p>From <a href="https://math.stackexchange.com/questions/3263927/prove-frac-partial-partial-m-textbn-m-textbn-m-sum-k-0n-1-f/3271177#3271177">here</a> we have </p>
<p><span class="math-container">$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$</span></p>
<p>Divide both sides by <span class="math-container">$n^22^n$</span> then sum up we get</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{H_n^2+H_n^{(2)}}{n^32^n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{x^n}{n^22^n}dx=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$</span></p>
| Ali Shadhar | 432,085 | <p><strong>Different Solution with a big bonus:</strong></p>
<p>First lets define the following:</p>
<p><span class="math-container">$$M=\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$$</span></p>
<p><span class="math-container">$$N=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}$$</span></p>
<p>I am going to establish two relations of <span class="math-container">$M$</span> and <span class="math-container">$N$</span> and solve them as a system of equations.</p>
<hr>
<p><strong>The first relation:</strong></p>
<p>In the question body, we reached</p>
<p><span class="math-container">$$M+N=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$</span></p>
<p>and by using <span class="math-container">$\quad\displaystyle\frac{\operatorname{Li}_2(x)}{x}=-\int_0^1\frac{\ln y}{1-xy}dy\quad$</span> we can write</p>
<p><span class="math-container">$$M+N=-\frac12\int_0^1\ln y\left[\int_0^1\frac{\ln^2(1-x)}{1-(y/2)x}dx\right]dy$$</span></p>
<p><span class="math-container">$$=2\int_0^1\frac{\ln y}{y}\operatorname{Li}_3\left(\frac{y}{y-2}\right)dy\overset{IBP}{=}-\int_0^1\frac{\operatorname{Li}_2(-y)}{y}\ln^2\left(\frac{2y}{1+y}\right)dy$$</span></p>
<p><span class="math-container">$$\small{=-\int_0^1\frac{\operatorname{Li}_2(-y)}{y}\left[\ln^22+2\ln2\ln y+\ln^2y-2\ln2\ln(1+y)-2\ln y\ln(1+y)+\ln^2(1+y)\right]dy}$$</span></p>
<p><span class="math-container">$$=-\ln^22\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)}{y}dy}_{I_1}-2\ln2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln y}{y}dy}_{I_2}-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln^2y}{y}dy}_{I_3}\\+2\ln2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln(1+y)}{y}dy}_{I_4}+2\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln y\ln(1+y)}{y}dy}_{I_5}-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-y)\ln^2(1+y)}{y}dy}_{I_6}$$</span></p>
<p><span class="math-container">$$I_1=\operatorname{Li}_3(-1)=-\frac34\zeta(3)$$</span></p>
<p><span class="math-container">$$I_2\overset{IBP}{=}-\int_0^1\frac{\operatorname{Li}_3(-y)}{y}dy=-\operatorname{Li}_4(-1)=\frac78\zeta(4)$$</span></p>
<p><span class="math-container">$$I_3\overset{IBP}{=}-2\int_0^1\frac{\operatorname{Li}_3(-y)\ln y}{y}dy\overset{IBP}{=}2\int_0^1\frac{\operatorname{Li}_4(-y)}{y}dy=2\operatorname{Li}_5(-1)=-\frac{15}8\zeta(5)$$</span></p>
<p><span class="math-container">$$I_4=-\frac12\operatorname{Li}_2^2(-1)=-\frac5{16}\zeta(4)$$</span></p>
<p><span class="math-container">$$I_5\overset{IBP}{=}\frac12\int_0^1\frac{\operatorname{Li}_2^2(-y)}{y}dy=\frac38\zeta(2)\zeta(3)-\frac{17}{32}\zeta(5)$$</span></p>
<p>where the last result follows from <a href="https://math.stackexchange.com/q/3236084">this solution</a>, check Eq<span class="math-container">$(3)$</span></p>
<p><span class="math-container">$I_6$</span> is elegantly evaluated by Cornel <a href="https://www.researchgate.net/publication/336378340_The_calculation_of_a_harmonic_series_with_a_weight_5_structure_involving_the_product_of_harmonic_numbers_H_n_H_2_2n" rel="nofollow noreferrer">here</a> in page <span class="math-container">$5$</span> and by me <a href="https://math.stackexchange.com/q/3462815">here</a>,</p>
<p><span class="math-container">$$\small{I_6=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)\zeta(3)+\frac{7}{4}\ln^22\zeta(3)-\frac2{3}\ln^32\zeta(2)+\frac{2}{15}\ln^52}$$</span></p>
<p>Combine these results to obtain </p>
<blockquote>
<p><span class="math-container">$$M+N=-4\operatorname{Li}_5\left(\frac12\right)-4\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{131}{32}\zeta(5)$$</span>
<span class="math-container">$$-\frac{19}{8}\ln2\zeta(4)+\frac{7}{8}\zeta(2)\zeta(3)-\ln^22\zeta(3)+\frac2{3}\ln^32\zeta(2)-\frac{2}{15}\ln^52\tag1$$</span></p>
</blockquote>
<hr>
<p><strong>The second relation:</strong></p>
<p>From <a href="https://math.stackexchange.com/questions/3366039/a-group-of-important-generating-functions-involving-harmonic-number">here</a></p>
<p><span class="math-container">$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)\tag2$$</span></p>
<p>multiply both sides by <span class="math-container">$\frac{\ln^2x}{x}$</span> then integrate from <span class="math-container">$x=0$</span> to <span class="math-container">$1/2$</span>
we have</p>
<p><span class="math-container">\begin{align}
I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^{1/2}x^{n-1}\ln^2x\ dx\\
&=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\left(\frac{\ln^22}{n2^n}+\frac{2\ln2}{n^22^n}+\frac{2}{n^32^n}\right)\\
&=\ln^22\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n2^n}+2\ln2\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}+2\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}\\
&=\ln^22S_1+2\ln2S_2+2M-2N
\end{align}</span></p>
<p>Or </p>
<p><span class="math-container">$$M-N=\frac12I-\frac12\ln^22S_1-\ln2S_2$$</span></p>
<p><strong>Evaluation of <span class="math-container">$I:$</span></strong></p>
<p><span class="math-container">\begin{align}
I&=\int_0^{1/2}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\overset{1-x\mapsto x}{=}\int_{1/2}^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\
2I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\underbrace{\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{1-x\mapsto x}\\
I&=\int_0^{1}\frac{\ln^2(1-x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1x^n\ln^2x\ dx\\
&=4\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=4\sum_{n=1}^\infty\frac{H_n}{n^4}-4\zeta(5)=8\zeta(5)-4\zeta(2)\zeta(3)
\end{align}</span></p>
<p><strong>Evaluation of <span class="math-container">$S_1$</span>:</strong></p>
<p>Divide both sides of (2) by <span class="math-container">$x$</span> then integrate from <span class="math-container">$x=0$</span> to <span class="math-container">$1/2$</span> and use the fact that <span class="math-container">$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$</span></p>
<p><span class="math-container">\begin{align}
S_1&=\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n2^n}=\int_0^{1/2}\frac{\ln^2(1-x)}{x(1-x)}\ dx\\
&=\int_{1/2}^{1}\frac{\ln^2x}{x(1-x)}\ dx=\sum_{n=0}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\
&=\frac13\ln^32+\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^2x\ dx\\
&=\frac13\ln^32+\sum_{n=1}^\infty\left(\frac2{n^3}-\frac{\ln^22}{n2^n}-\frac{2\ln2}{n^22^n}-\frac{2}{n^32^n}\right)\\
&=\frac13\ln^32+2\zeta(3)-\ln^32-2\ln2\operatorname{Li}_2\left(\frac12\right)-2\operatorname{Li}_3\left(\frac12\right)=\frac14\zeta(3)
\end{align}</span></p>
<p>where we used <span class="math-container">$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$</span> and <span class="math-container">$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$</span></p>
<p><strong>Evaluation of <span class="math-container">$S_2$</span>:</strong></p>
<p>integrate both sides of (2) from <span class="math-container">$x=0$</span> to <span class="math-container">$x$</span> to have</p>
<p><span class="math-container">$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)\tag{3}$$</span></p>
<p>Now divide both sides of (3) by <span class="math-container">$x$</span> then integrate from <span class="math-container">$x=0$</span> to <span class="math-container">$1/2$</span> and use the fact that <span class="math-container">$\int_0^{1/2}x^{n-1}=\frac1{n2^n}$</span></p>
<p><span class="math-container">$$-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx=\sum_{n=1}^\infty\frac{1}{n^22^n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$</span></p>
<p>Rearranging the terms</p>
<p><span class="math-container">$$S_2=\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^22^n}=2\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\frac13\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx-2\operatorname{Li}_4\left(\frac12\right)$$</span></p>
<p><a href="https://math.stackexchange.com/q/3195207">Substitute</a> </p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42$$</span></p>
<p>and </p>
<p><span class="math-container">\begin{align}
\int_0^{1/2}\frac{\ln^3(1-x)}{x}\ dx&=\int_{1/2}^{1}\frac{\ln^3x}{1-x}\ dx\\
&=\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\
&=\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\
&=\ln^42+3\ln^32\operatorname{Li}_2\left(\frac12\right)+6\ln2\operatorname{Li}_3\left(\frac12\right)+6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)\\
&=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}4\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac12\ln^42
\end{align}</span></p>
<p>we get </p>
<p><span class="math-container">$$S_2=-2\operatorname{Li}_4\left(\frac12\right)+\frac94\zeta(4)-2\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42$$</span></p>
<p>collect the results of <span class="math-container">$I$</span>, <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> we get</p>
<blockquote>
<p><span class="math-container">$$M-N=2\ln2\operatorname{Li}_4\left(\frac12\right)+4\zeta(5)$$</span>
<span class="math-container">$$-\frac{9}{4}\ln2\zeta(4)-2\zeta(2)\zeta(3)+\frac{15}8\ln^22\zeta(3)-\frac1{2}\ln^32\zeta(2)+\frac{1}{12}\ln^52\tag4$$</span></p>
</blockquote>
<hr>
<p>Now we are ready to calculate the two sums:</p>
<p><span class="math-container">$$M=\frac{(1)+(4)}{2}=-2\operatorname{Li}_5\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{64}\zeta(5)-\frac{37}{16}\ln2\zeta(4)-\frac{9}{16}\zeta(2)\zeta(3)\\+\frac{7}{16}\ln^22\zeta(3)+\frac1{12}\ln^32\zeta(2)-\frac{1}{40}\ln^52$$</span></p>
<p><span class="math-container">$$N=\frac{(1)-(4)}{2}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$</span></p>
|
983,200 | <p>Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:</p>
<p>$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$</p>
<p>would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$?</p>
| heropup | 118,193 | <p>Something to notice is that $$x^4 - 6x^2 + 9 = (x^2 - 3)^2$$ and consequently, $$f(x) = 2 (x^3 - 4x + 6) \log (x^2 - 3).$$ This makes differentiation slightly easier: $$f'(x) = 2\left( (3x^2 - 4)\log(x^2 - 3) + \frac{x^3-4x+6}{x^2-3} \cdot 2x \right).$$</p>
|
1,523,427 | <p>Is it possible to cover all of $\mathbb{R}^2$ using balls $\{ B(x_n,n^{-1/2})\}_{n=1}^\infty$ of decreasing radius $n^{-1/2}$? I know that if we chose e.g. radius $n^{-1}$ it could never work because $\sum \pi (n^{-1})^2 < \infty$. But in this case the balls cover an infinite amount of area, so it seems that it may be possible to construct.</p>
<p>Edit: It can definitely be done with balls of radius $n^{-1/2+\epsilon}$ for any $\epsilon > 0$.</p>
| David C. Ullrich | 248,223 | <p>It's true for any $r_n$ with $\sum r_n^2=\infty$.</p>
<p>Say $S(x,r)$ is the square with center $x$ and side length $r$.
It's enough to show that if $\sum r_n^2>4$ then there exist $x_n$ such
that $S(x_n,r_n)$ covers $[0,1]^2$.</p>
<p>For every $n$ there exists $j$ with $2^{-j}\le r_n<2^{-j+1}$. Replace $r_n$ with $2^{-j}$:</p>
<p>It's enough to show that if each $r_n$ is of the form $2^{-j}$ and $\sum r_n>1$ then there exist $x_n$ such that $S(x_n,r_n)$ cover the unit square.</p>
<p>A <em>dyadic interval</em> is an interval $[j2^{-n},(j+1)2^{-n}]$. A <em>dyadic square</em> is the product of two dyadic intervals with the same side length.</p>
<p>Start: If there exists $n$ with $r_n\ge1$ we're done. Suppose not. Divide $[0,1]^2$ into four dyadic squares of side length $1/2$. Call them $Q$'s.</p>
<p>If there are four $n$ such that $r_n=1/2$ we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ with $r_n=1/2$.</p>
<p>Discard the $r_n$'s we just used. Note we used <em>all</em> the $r_n$ that equal $1/2$. Discard the $Q$'s we just covered. Divide each remaining $Q$ into four subsquares of side length $1/4$.</p>
<p>If there are enough $n$ with $r_n=1/4$ to cover all the reminaing $Q$'s we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ where $r_n=1/4$. Note we've used <em>all</em> the $r_n$ that equal $1/4$. Discard those $r_n$, discard the $Q$'s we just covered. Divide each remaining $Q$ into four suubsquares of side length $1/8$.</p>
<p>Etc.</p>
<p>If at some point we're done then we're done. Suppose not. Then we've used up all the $r_n$ covering essentially disjoint subsquares of $[0,1]^2$; hence $\sum r_n^2\le 1$.</p>
|
26,259 | <p>I've been reading <em>generationgfunctionology</em> by Herbert S. Wilf (you can find a copy of the second edition on the author's page <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">here</a>).</p>
<p>On page 33 he does something I find weird. He wants to shuffle the index forward and does so like this:
\begin{align*}
(f_{n+1})_{n\in N_0} &= \frac{(f(X)-f(0))}{X}\\
(f_{n})_{n\in N_0} &= \sum_{n\in N_0} f_nx^n
\end{align*}</p>
<p>Why is this allowed? Namely $X$ has a noninvertible (0) constant term, so how is this division (multiplication with the reciprocal) defined within the ring of formal power series?</p>
| ypercubeᵀᴹ | 7,367 | <p>Are you allowed to write the following?</p>
<p>\begin{align*}
X = \frac{X^2}{X}\
\end{align*}</p>
|
26,259 | <p>I've been reading <em>generationgfunctionology</em> by Herbert S. Wilf (you can find a copy of the second edition on the author's page <a href="http://www.math.upenn.edu/~wilf/DownldGF.html" rel="nofollow">here</a>).</p>
<p>On page 33 he does something I find weird. He wants to shuffle the index forward and does so like this:
\begin{align*}
(f_{n+1})_{n\in N_0} &= \frac{(f(X)-f(0))}{X}\\
(f_{n})_{n\in N_0} &= \sum_{n\in N_0} f_nx^n
\end{align*}</p>
<p>Why is this allowed? Namely $X$ has a noninvertible (0) constant term, so how is this division (multiplication with the reciprocal) defined within the ring of formal power series?</p>
| Bill Dubuque | 242 | <p>The ring of formal power series over a domain is itself a domain, hence nonzero elements are cancellable, quotients are unique, etc. But you don't need these results for the whole ring - only for the element $\rm\:X\:,\: $ and this case can be proved quite simply. It is obvious that $\rm\:X\:$ is not a zero-divisor, i.e. $\rm\ X\ f\:(X) = 0\ \Rightarrow\ f\:(X)=0\:,\: $ from which closely related results follow immediately, viz.</p>
<p><strong>THEOREM</strong> $\ \ $ TFAE for $\rm\ a \in R\ $ any ring</p>
<p>$(1)\quad\ \rm a\: b\ =\ a\: c\ \ \Rightarrow\ \ b\ =\ c\quad\quad\quad\quad\quad\quad\ $ ($\rm\:a\:$ is cancellable)</p>
<p>$(2)\quad\rm\ a\:b\ =\ \ 0\ \ \ \Rightarrow\ \ b\ =\ 0\quad\quad\quad\quad\quad\quad\ $ ($\rm\:a\:$ is not a zero-divisor)</p>
<p>$(3)\quad $ roots of $\rm\ a\ x\ =\ b\ $ are unique $\quad\ \ $ (uniqueness of division by $\rm\:a\:$)</p>
<p><strong>Proof</strong> $\rm\ \ \ (1\Rightarrow 2)\ \ $ Specialize $\rm\: c = 0$</p>
<p>$\rm (2\Rightarrow 3)\ \ \ a\ x = b = a\ x'\ \Rightarrow\ \ a\ (x-x') = 0\ \ \Rightarrow\ \ x = x'$</p>
<p>$\rm (3 \Rightarrow 1)\ \ \ a\ x = a\: c\ $ has roots $\rm\ b,c\ $ so $\rm\ b = c$ </p>
|
1,511,733 | <p>B = matrix given below. I is identity matrix.</p>
<pre><code> [1 2 3 4]
[3 2 4 3]
[1 3 2 4]
[5 4 3 7]
</code></pre>
<p>So What will be the relation between the matrices A and C if AB = I and BC = I?</p>
<p>I think that A = C because both AB and BC have B in common and both of their product is an identity matrix but I'm not sure. </p>
| Peter | 82,961 | <p>Hint : Multiply $AB=I$ from right with $B^{-1}$ and $BC=I$ from left with $B^{-1}$. What do you get ?</p>
|
4,274,085 | <p>A linear equation in one variable <span class="math-container">$x$</span>, <span class="math-container">$$ax+b=k$$</span> has only one non-negative integer solution. For example, <span class="math-container">$2x+3=85$</span> has a solution 41.</p>
<p>How to find the number of non-negative integer solutions of a linear equation in two variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, for example, <span class="math-container">$5x+3y=100$</span> and in general, <span class="math-container">$$ax+by=k$$</span></p>
<p>By brute-force search, I found 7 non-negative integer solutions of <span class="math-container">$5x+3y=100$</span>. They are <span class="math-container">$(2, 30), (5, 25), (8, 20), (11, 15), (14, 10), (17, 5), \text{ and }(20, 0)$</span>.</p>
<p><em>P.S.</em> I came to know that a linear equation in two variables is also know as linear Diophantine equation. I also came across <a href="https://en.wikipedia.org/wiki/Ku%E1%B9%AD%E1%B9%ADaka" rel="nofollow noreferrer">Kuṭṭaka</a>, Aryabhata's algorithm for solving linear Diophantine equations in two variables.</p>
| Robert Shore | 640,080 | <p>Assume <span class="math-container">$\gcd(a, b)=1$</span>. If there are any such solutions at all in non-negative integers, there will be a unique solution where <span class="math-container">$0 \leq x \lt b$</span>. Start there. Increment <span class="math-container">$x$</span> by <span class="math-container">$b$</span> while simultaneously decrementing <span class="math-container">$y$</span> by <span class="math-container">$a$</span>. When this process results in <span class="math-container">$y \lt 0$</span>, you're done. In your example, you start with <span class="math-container">$(2, 30)$</span> (note that <span class="math-container">$0 \leq 2 \lt 3$</span>) and then increase the first coordinate by <span class="math-container">$3$</span> while simultaneously decreasing the second coordinate by <span class="math-container">$5$</span>.</p>
<p>If <span class="math-container">$\gcd(a, b) \neq 1$</span>, then <span class="math-container">$\gcd(a, b) \mid k$</span> (or there would be no solutions in integers) so divide <span class="math-container">$a, b, k$</span> by <span class="math-container">$\gcd(a, b)$</span> and apply the first paragraph.</p>
|
4,274,085 | <p>A linear equation in one variable <span class="math-container">$x$</span>, <span class="math-container">$$ax+b=k$$</span> has only one non-negative integer solution. For example, <span class="math-container">$2x+3=85$</span> has a solution 41.</p>
<p>How to find the number of non-negative integer solutions of a linear equation in two variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, for example, <span class="math-container">$5x+3y=100$</span> and in general, <span class="math-container">$$ax+by=k$$</span></p>
<p>By brute-force search, I found 7 non-negative integer solutions of <span class="math-container">$5x+3y=100$</span>. They are <span class="math-container">$(2, 30), (5, 25), (8, 20), (11, 15), (14, 10), (17, 5), \text{ and }(20, 0)$</span>.</p>
<p><em>P.S.</em> I came to know that a linear equation in two variables is also know as linear Diophantine equation. I also came across <a href="https://en.wikipedia.org/wiki/Ku%E1%B9%AD%E1%B9%ADaka" rel="nofollow noreferrer">Kuṭṭaka</a>, Aryabhata's algorithm for solving linear Diophantine equations in two variables.</p>
| Thomas Andrews | 7,933 | <p>One way to solve this, when <span class="math-container">$(a,b)=1,$</span> is to use partial fractions:</p>
<p><span class="math-container">$$\frac1{(1-x^a)(1-x^b)}=\frac{\alpha}{(1-x)^2}+\frac\beta{1-x}+\frac{p(x)}{1-x^a}+\frac{q(x)}{1-x^b}$$</span> where <span class="math-container">$\deg p<a, \deg q<b,$</span> and <span class="math-container">$p(1)=q(1)=0.$</span></p>
<p>If <span class="math-container">$p(x)=\sum_i p_i x^i, q(x)=\sum_{i} q_i x^i,$</span> then the coefficient of <span class="math-container">$x^k$</span> is <span class="math-container">$$\alpha(k+1)+\beta +p_{k\bmod a}+q_{k\bmod b}$$</span></p>
<p>It’s easy to calculate the dominant term, <span class="math-container">$\alpha=\frac1{ab}.$</span></p>
<p>So if <span class="math-container">$C_k$</span> is the count of solutions, <span class="math-container">$C_k\sim \frac{k}{ab}.$</span> We also get <span class="math-container">$C_{k+ab}=C_k+1.$</span></p>
<p>For <span class="math-container">$k<ab,$</span> <span class="math-container">$c_{k}=0,1,$</span> so this means the <span class="math-container">$\beta+p_{k\bmod a}+q_{k\bmod b}$</span> is always of absolute value <span class="math-container">$<1.$</span></p>
<p>For <span class="math-container">$a=3,p=5,$</span> Wolfram Alpha gives:</p>
<p><span class="math-container">$\beta=\frac15,$</span> <span class="math-container">$p_*=\left(\frac13,-\frac13,0\right),$</span> <span class="math-container">$q_*=\left(\frac25,0,-\frac25,\frac15,-\frac15\right).$</span></p>
<hr />
<p>The general question, given relatively prime <span class="math-container">$a_1,\cdots,a_n$</span> of <span class="math-container">$$a_1x_1+\cdots+a_nx_n=k,$$</span> the partial fraction formula is similar if the <span class="math-container">$a_i$</span> are pairwise relatively prime. It gets more complicated with common factors.</p>
<p>In either event, the dominant term is still simple: <span class="math-container">$$\frac{\binom{k+n-1}{n-1}}{a_1a_2\cdots a_n}$$</span></p>
|
380,530 | <p>It's easy to show that there's a function such that $\int_1^\infty f $ diverges, but $\int_1^\infty |f|$ converges, such as $f = 1/(-1+x)$. </p>
<p>But is there a function such that $\int_1^\infty f $ converges, but $\int_1^\infty |f| $ diverges?</p>
| Community | -1 | <p>First I recall that
$$\text {if}\quad \int_1^\infty |f|\quad\text{ converges then so does }\quad \int_1^\infty f$$</p>
<p>The integral
$$\int_1^\infty\frac{ \sin x}{x}dx$$
is convergent but
$$\int_1^\infty\frac{ |\sin x|}{x}dx$$
is divergent.</p>
|
4,016,921 | <p>Let <span class="math-container">$X, Y, Z$</span> be three random variables. Is the equality
<span class="math-container">$$
P(X=x, Y=y |Z=z) = P(X=x|Y=y,Z=z) P(Y=y|Z=z)
$$</span>
true?
Note that <span class="math-container">$P(X,Y)$</span> denotes the joint probability of random variables <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>.</p>
| user | 293,846 | <p>First of all we can cancel one parameter:
<span class="math-container">$$\mbox{$_2$}F_1(a,b;a; x)=\mbox{$_1$}F_0(b;;x)$$</span></p>
<p>Further:
<span class="math-container">$$
\mbox{$_1$}F_0\left(-\frac12;;x\right)=\sum_{n\ge0}\frac{\left(-\frac12\right)^\overline{n}}{n!}x^n=\sum_{n\ge0}\binom{\frac12}n(-x)^n=\sqrt{1-x}.
$$</span></p>
|
32,997 | <p>I have made a rather obvious yet peculiar observation while calculating with quadratic inequalities. Take a simple quadratic inequality like the one below</p>
<p>$\frac{x^2+1}{x}>1$</p>
<p>by multiplying both sides by $x$, then subtracting $x$ from both sides we get</p>
<p>$x^2-x+1>0$</p>
<p>Hence, both the above inequalities are one and the same in theory. However, the solution to the first inequality is $x>0$, while the second inequality is satisfied for all $x\in\mathbb R$.</p>
<p>It is not difficult to explain this simple case, seeing that the LHS of the first inequality will turn negative for negative $x$, but my question is more of a general nature, if you like. I wonder, why do two identical (but paraphrased/rewritten) inequalities have different solutions?</p>
| Rasmus | 367 | <p>If you multiply an inequality with a negative number, you have to exchange the inequality signs $<$ and $>$.</p>
<p>For instance, multiplying $1<2$ with $(-1)$ you get
$$-1>-2.$$</p>
|
32,997 | <p>I have made a rather obvious yet peculiar observation while calculating with quadratic inequalities. Take a simple quadratic inequality like the one below</p>
<p>$\frac{x^2+1}{x}>1$</p>
<p>by multiplying both sides by $x$, then subtracting $x$ from both sides we get</p>
<p>$x^2-x+1>0$</p>
<p>Hence, both the above inequalities are one and the same in theory. However, the solution to the first inequality is $x>0$, while the second inequality is satisfied for all $x\in\mathbb R$.</p>
<p>It is not difficult to explain this simple case, seeing that the LHS of the first inequality will turn negative for negative $x$, but my question is more of a general nature, if you like. I wonder, why do two identical (but paraphrased/rewritten) inequalities have different solutions?</p>
| Joseph Malkevitch | 1,369 | <p>Another way of thinking about of some of the previous answers and comments is to use a geometric point of view. Draw a graph of the function y = f(x) =((x^2+ 1)/x) - 1. Often this can be done without specialized algebraic knowledge. When the graph is above the line y = 0 you can read off the answer from what you have graphed.</p>
|
2,784,697 | <p>Find the solutions to:$\displaystyle\frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2$. </p>
<p>I got the following solutions:-</p>
<p>$\left(\frac{dy}{dx}\right)=0\Rightarrow y=c_1$ is a solution</p>
<p>$\left(\frac{dy}{dx}\right)=1\Rightarrow y=x+c_2$ is another solution </p>
<p>Are there any other solutions?</p>
<p>I dont have any idea about how to solve a $2^{nd}$ order non linear DE. As far as i Know , a $2^{nd}$ order linear DE could be solved with the help of auxillary equations , Is there any such similar methods applicable to this problem </p>
| Bill O'Haran | 551,590 | <p><strong>Hint:</strong>
Let $p(x) = \frac{dy}{dx}$. Then:
$$
p'(x) = p(x)^2
$$
is a first order DE.</p>
|
1,610,616 | <blockquote>
<p>$6$ letters are to be posted in three letter boxes.The number of ways
of posting the letters when no letter box remains empty is?</p>
</blockquote>
<p>I solved the sum like dividing into possibilities $(4,1,1),(3,2,1)$ and $(2,2,2)$ and calculated the three cases separately getting $90,360$ and $90$ respectively.</p>
<p>What I wanted to know is there any shortcut method to solve this problem faster?Can stars and bars be used?How?</p>
<p>Or can the method of coefficients be used like say finding coefficient of $t^6$ in $(t+t^2+t^3+t^4)^3$.Will that method work here?</p>
| Empy2 | 81,790 | <p>Inclusion-Exclusion. </p>
<p>$3^6$ ways total.<br>
$2^6$ ways if you only mail to $1\&2,2\&3$ or $1\&3$, subtract them off (3 alike cases).<br>
$1^6$ way if you only mail to $1$, or to $2$, or to $3$, add them back in because they were subtracted twice (agai, 3 cases all alike).</p>
<p>$3^6 - 3 \cdot 2^6 + 3 \cdot 1^6$</p>
|
245,921 | <p>I'm new to Mathematica (trying it out for an electromagnetism course) and struggling with plotting magnetic field lines using a <strong>line</strong> of current. Not a torturous, not a cylinder, not a circle. A line. Almost every example I see is <em>far</em> more complex and it's beyond what I'm trying to do... I know I have to use VectorPlot, but I'm feeling kind of lost. Any help on this would be welcome!</p>
| Daniel Huber | 46,318 | <p>in your course you will have learned that the magnetic field at r of a a current infinitesimal dI at position r0 is proportional to:</p>
<pre><code>Cross[dI,(r-r0)]/Norm[r-r0]^3
</code></pre>
<p>Now if we consider a current of 1 along the x ax at {x0,0,0}: dI= {1,0,0} dx0 and the field at {x,y,z};</p>
<pre><code>dField= Cross[{1,0,0},({x,y,z}-{x0,0,0})]/ Norm[{x,y,z}-{x0,0,0}] dx0
</code></pre>
<p>If we integrate this from -Infinity to Infinity we get a field that is proportional to H or B. As we integrate from -Infinity to Infinity the integral can not depend on x, we may therefore calculate the integral for x==0:</p>
<pre><code>field[y_,z_]= -Integrate[Cross[{1,0,0},{x0,y,z}]/ Norm[{x0,y,z}]^3,{x0,-Infinity,Infinity},Assumptions -> {y > 0, z > 0}]
</code></pre>
<p>We can plot this field using "VectorPlot3D", but note: by default all arrows have the same length, field strength is indicated by color. To get different length arrows, we use the options <code>, VectorScaling -> Automatic, VectorSizes -> {0, 1}]</code>:</p>
<pre><code> d = 1;
VectorPlot3D[field[y, z], {x, -d, d}, {y, -d, d}, {z, -d, d},
VectorScaling -> Automatic, VectorSizes -> {0, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/aQ8no.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aQ8no.png" alt="enter image description here" /></a></p>
|
245,921 | <p>I'm new to Mathematica (trying it out for an electromagnetism course) and struggling with plotting magnetic field lines using a <strong>line</strong> of current. Not a torturous, not a cylinder, not a circle. A line. Almost every example I see is <em>far</em> more complex and it's beyond what I'm trying to do... I know I have to use VectorPlot, but I'm feeling kind of lost. Any help on this would be welcome!</p>
| Alex Trounev | 58,388 | <p>We can use 3D graphics as follows</p>
<pre><code>Graphics3D[{{Red, Arrowheads[0.025],
Arrow[Line[{{0, 0, -4}, {0, 0, 4}}]]},
Table[{Blue, Arrowheads[0.025],
Arrow[BSplineCurve@
Table[{r Cos[t], r Sin[t], z0}, {t, 0, 2 Pi, Pi/50}]]}, {r, {1,
2, 3}}, {z0, {-3, 0, 3}}]}, Boxed -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/y3DZu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y3DZu.png" alt="Figure 1" /></a></p>
|
388,561 | <p>I am trying to do this question in Bredon's <em>Topology and geometry</em> about using the transversality theorem to show that the intersection of two manifolds is a manifold.</p>
<p>Now it goes as follows:</p>
<p>Let $f(x,y,z)=(2-(x^2+y^2)^{1/2})+z^2$ on $\mathbb{R}^3 - (0,0,z)$. Then one can show that $M=f^{-1}(1)$ is a manifold.</p>
<p>Now let $N = \{ (x,y,z) \in \mathbb{R}^3 | x^2+y^2=4 \}$. I need to show that $M \cap N$ is a manifolds by using the transversality theorem, but I'm not quite sure how to do this.</p>
<p>I thought about finding the tanget planes to each to these manifolds and then showing that there a points in one of them that aren't in the other, but I dont know if this will work. So I was wondering if I could get some hints on this.</p>
<p>Thank you</p>
| Ted Shifrin | 71,348 | <p>I couldn't fit my response in the character limit of the comment box.</p>
<p>No, your first function needs to be the function $f$ you started with :) You're correct that I was guiding you just to use the regular value set-up. Of course, this is a special case of the transversality theorem. If you try to study the function $h(x,y,z)=x^2+y^2$ on $M$, you end up with a messy local-coordinates computation, so I always discourage my students from trying to show "directly" that the function $h\colon M\to\mathbb R$ has $4$ as a regular value. You could try to see that the inclusion map $\iota\colon M\to\mathbb R^3$ is transverse to the cylinder $N$: As you surmised, you just need to see that at any point $P\in M\cap N$, $T_PM+T_PN = \mathbb R^3$. The way I was leading you is doing this geometrically, by showing that $\nabla f(P)$ (which is the normal vector to $M$) and $\nabla h(P)$ (which is the normal vector to $N$) are linearly independent at every $P\in M\cap N$.</p>
|
1,170,602 | <p>How to evaluate the integral </p>
<p>$$\int \sqrt{\sec x} \, dx$$</p>
<p>I read that its not defined.<br>
But why is it so ? Does it contradict some basic rules ?
Please clarify it .</p>
| kryomaxim | 212,743 | <p>$$\int \sqrt{\sec x} \,dx = \int \frac{1}{\sqrt{\cos x}}\, dx$$; Now substituting $u = \cos x$ then it follows $$du = - \sin x \, dx = - \sqrt{1-u^2} dx.$$ Hence:</p>
<p>$$\int \sqrt{\sec x} \, dx = - \int \frac{1}{\sqrt{u(1-u^2)}}du$$ This integral is an <a href="http://mathworld.wolfram.com/EllipticIntegral.html" rel="nofollow">elliptic integral</a>.</p>
|
646,010 | <p>So i kinda think i have figured this out, i'm not very good at math, and need a formula to figure out some stats for a game i'm playing.</p>
<p>I have a Weapon with a reload speed of X sec.. however, i also have a modifier attached, that will make the weapon reload faster by +Y%</p>
<p>i made this formula, mostly by guessing, as i got no clue what i am doing.</p>
<pre><code>100/(100+Y)*X
</code></pre>
<p>the results i am getting looks right to me, but is the formula ok?</p>
| T.J. Gaffney | 34,380 | <p>$\sin(x^2)>.5$ on $(\sqrt{\pi/6+2k\pi},\sqrt{5\pi/6+2k\pi}),k\in\mathbb{N}$. When $x>e$, $\log^2(x)>1$, so we can see that the positive part of the integral is greater than $.5\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)$ for any $K\in\mathbb{N}\setminus\{1\}$. Now see that </p>
<p>$$
\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)=\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}+\sqrt{\pi/6+2k\pi}}\right)\geq\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}}\right)
$$</p>
<p>which we see is unbounded. We conclude that the positive part of the integral is unbounded. Similarly, we can conclude that the negative part of the integral is unbounded. Therefore it is not integrable.</p>
|
2,869,031 | <p>Let $A$ a domain, i.e. $ab\in A\implies a=0$ or $b=0$. It's written that all domains are commutative. Is it by definition, or can we prove that domains are commutative? I mean, do we only consider domaind for commutative rings, or is a ring that is a domain then commutative?</p>
| José Carlos Santos | 446,262 | <p>That's a mistake: <a href="https://en.wikipedia.org/wiki/Domain_(ring_theory)" rel="noreferrer">a domain doesn't have to be commutative</a>. When it is, we call it an <em>integral domain</em>.</p>
|
2,869,031 | <p>Let $A$ a domain, i.e. $ab\in A\implies a=0$ or $b=0$. It's written that all domains are commutative. Is it by definition, or can we prove that domains are commutative? I mean, do we only consider domaind for commutative rings, or is a ring that is a domain then commutative?</p>
| rschwieb | 29,335 | <p>Some authors do allow "domain" to refer to a noncommutative ring. But nearly universally "integral domain" will imply commutativity.</p>
<p>Here are <a href="http://ringtheory.herokuapp.com/search/results/?H=8&L=1" rel="noreferrer">nine examples of noncommutative rings with no nonzero zero divisors</a>.</p>
<p>I'll call out at least two of them, the <a href="https://en.wikipedia.org/wiki/Hurwitz_quaternion" rel="noreferrer">Hurtwitz and Lipschitz quaternions</a>.</p>
|
3,050,235 | <p>A polynomial is exactly divided by <span class="math-container">$x+1$</span>, and when it is divided by <span class="math-container">$3x-1$</span>, the remainder is <span class="math-container">$4$</span>. Given that the polynomial gives a remainder <span class="math-container">$hx+k$</span> when divided by <span class="math-container">$3x^2+2x-1$</span>, find <span class="math-container">$h$</span> and <span class="math-container">$k$</span></p>
<p>I've been having a little bit of trouble with this question because I'm not entirely sure what to do when the polynomial isn't given.</p>
| Deepak | 151,732 | <p>Givens:</p>
<p><span class="math-container">$P(x) = Q_1(x)\cdot (x+1)$</span> [note here that <span class="math-container">$P(-1) = 0$</span>]</p>
<p>and</p>
<p><span class="math-container">$P(x) = Q_2(x)\cdot (3x-1) + 4$</span> [note here that <span class="math-container">$P(\frac 13) = 4$</span>]</p>
<p>Need to determine <span class="math-container">$h,k$</span> where:</p>
<p><span class="math-container">$P(x) = Q_3(x)\cdot (x+1)(3x-1) + hx + k$</span></p>
<p><span class="math-container">$P(-1) = 0 \implies -h + k = 0 \implies h = k$</span>.</p>
<p><span class="math-container">$P(\frac 13)= 4 \implies \frac h3 + k = 4 \implies \frac {4h}3 = 4 \implies h = k = 3$</span>.</p>
|
3,050,235 | <p>A polynomial is exactly divided by <span class="math-container">$x+1$</span>, and when it is divided by <span class="math-container">$3x-1$</span>, the remainder is <span class="math-container">$4$</span>. Given that the polynomial gives a remainder <span class="math-container">$hx+k$</span> when divided by <span class="math-container">$3x^2+2x-1$</span>, find <span class="math-container">$h$</span> and <span class="math-container">$k$</span></p>
<p>I've been having a little bit of trouble with this question because I'm not entirely sure what to do when the polynomial isn't given.</p>
| Bill Dubuque | 242 | <p><span class="math-container">$\ \overbrace{f\!-\!4\bmod (3x\!-\!1)(x\!+\!1)}^{\Large {\rm factor\ out}\,\ 3x\,-1\ \mid\ f-4\ \ \ \rightarrow\!\!\!\!\!\!\!} =\, (3x\!-\!1)\!\!\underbrace{\overbrace{\left[\dfrac{\color{#c00}f\!-\!4}{\color{#0a0}{3x}\!-\!1}\bmod x\!+\!1\right]}^{\Large\ \color{#c00}f\bmod x+1\ =\ \color{#c00}0\ \ \,\rightarrow\!\!\!\!\!\!\!\!\!\!\!}}_{\Large \color{#0a0}g\bmod x+1\, =\, \color{#0a0}{g(-1)}\ \ \rightarrow\!\!\!\!\!\!}\!\! =\, (3x\!-\!1)\left[\dfrac{\ \ \color{#c00}0\!-\!4}{\color{#0a0}{-3}\!-\!1}\right] =\, 3x\!-\!1$</span></p>
|
307,529 | <p>I am trying to prove that if $L/K$ is an algebraic extension and if $\alpha \in L$, then </p>
<ul>
<li><p>$\alpha$ is separable over $K$ if $\mathrm{char}(K)=0$. This is clear because $K$ is perfect which in turn implies that $L/K$ is seperable . </p></li>
<li><p>Now if $\mathrm{char}(K)=p$ is prime, then the statement is: $\alpha$ is separable if and only if $K(\alpha) =K(\alpha^p)$.
This problem is from a past examination and looks very interesting, but somehow I am not able to connect what's happening with the extension of $K$ with $\alpha$ and $\alpha^p$ .</p></li>
</ul>
<p>I need some help here. </p>
<p>This somehow makes me think that the minimal polynomial of $\alpha$ and $\alpha^p$ have different roots and $\alpha$ is given by some root of minimal polynomial of $\alpha^p$ . </p>
<p>Note that I am using the definition of separability as follows : </p>
<blockquote>
<p>An extension $L/K$ is called separable if for every $\alpha \in L$ the minimal polynomial of $\alpha$ has distinct roots in $L$ . </p>
</blockquote>
<p>Thanks for helping. </p>
| Barbara Osofsky | 59,437 | <p>In characteristic $p$, the Frobenius map $\Phi:K\rightarrow K$ with $\Phi(\kappa)=\kappa^p$ is additive (that is, $\Phi(\kappa +\lambda)=\Phi(\kappa)+\Phi(\lambda)$ because all the binomial coefficients that arise in $(\kappa+\lambda)^p$ are divisible by $p$. Taking $p$ powers is always multiplicative so $\Phi$ must be a field endomorphism of the base field $K$. Field homomorphisms are always monomorphisms since the only ideals of a field $K$ are $0$ and $K$. </p>
<p>Now let $Q(X)=X^m+\sum_{i=0}^{m-1} c_iX^i$ be the minimal polynomial for $\alpha$ over $K$ . Now use calculus to show that $Q$ has a multiple root in $K[\alpha]$ if and only if $Q$ is a polynomial in $X^p$. Also, because $Q$ is irreducible so not a $p^{th}$ power, some coefficient of $Q(X^p)$ must not be in $\Phi(K)$ to avoid $Q(X)=\left(\Phi^{-1}(Q)\right)^p$ over $K$. Now note that from $Q(\alpha)=0$ we get $\Phi(Q(X))$ has root $\alpha^p$ so eventually some power $\alpha^{p^k}$ will be a root of a separable factor of $\Phi^{k}(Q[X])$, and use this to show $K[\alpha]=K[\alpha^p]=\cdots =K[\alpha^{p^k}]$ if and only if $Q$ is separable. </p>
|
3,278,797 | <p>I tried to solve it and I got answer '3'. But that is just my intuition.I don't have concrete method to prove my answer .I did like this, in order to maximize the fraction, we need to minimize the denominator .So if plug in '1' in expression, denominator becomes '1'.Now denominator is minimalized,the result of expression is '3'. Thats how I got to conclusion.But I can't prove that " 3 is the solution" mathematically.
Can anyone show me how to prove it properly</p>
| nonuser | 463,553 | <p>Hint: <span class="math-container">$$y= \frac{x^2+x+1}{x^2-x+1}=\frac{x+1+{1\over x}}{x-1+{1\over x}} = {t+2\over t} = 1+{2\over t}$$</span></p>
<p>where <span class="math-container">$t=x-1+{1\over x}$</span>. Since for positive <span class="math-container">$x$</span> have <span class="math-container">$$x+{1\over x}\geq 2 \implies t\geq 1 \implies y\leq 3$$</span> and for negative <span class="math-container">$x$</span> we have <span class="math-container">$$x+{1\over x}\leq -2 \implies t\leq -3 \implies y\geq -1$$</span>. </p>
|
3,278,797 | <p>I tried to solve it and I got answer '3'. But that is just my intuition.I don't have concrete method to prove my answer .I did like this, in order to maximize the fraction, we need to minimize the denominator .So if plug in '1' in expression, denominator becomes '1'.Now denominator is minimalized,the result of expression is '3'. Thats how I got to conclusion.But I can't prove that " 3 is the solution" mathematically.
Can anyone show me how to prove it properly</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: It is <span class="math-container">$$\frac{x^2+x+1}{x^2-x+1}\le 3$$</span> if <span class="math-container">$$x^2+x+1\le 3x^2-3x+3$$</span> and this is true since <span class="math-container">$$0\le x^2-2x+1$$</span> . This is <span class="math-container">$$(x-1)^2\geq 0$$</span> and the equal sign holds if <span class="math-container">$$x=1$$</span></p>
|
130,806 | <p><strong>Qusestion:</strong> Let f be a continuous and differentiable function on $[0, \infty[$, with $f(0) = 0$ and such that $f'$ is an increasing function on $[0, \infty[$. Show that the function g, defined on $[0, \infty[$ by $$g(x) = \begin{cases} \frac{f(x)}{x}, x\gt0& \text{is an increasing function.}\\ f'(0), x=0 \end{cases}$$</p>
<p>I have tried to solve this problem but I don't know whether I have done it right or not. </p>
<p><strong>Solution:</strong> I have applied mean value theorem on the interval $[0, x]$. Then, $$\frac{f(x)}{x} =f'(c) , 0\lt c \lt x$$ </p>
<p>It is given that $f'$ is an increasing function. So I deduce that $\frac{f(x)}{x}$ is also increasing.</p>
<p>Further, $$g(x) = f'(c) \text {such that } 0<c<x$$ Therefore, $$g(0) =f'(c) \text{such that} 0<c<0$$
So, $c=0$</p>
<p>Thus $g(x) = f'(0)$ at $x=0$</p>
| noname1014 | 27,743 | <p>Base on @chemeng $g(x)$is increasing in$(0,\infty)$.when $x=0$,
$$\lim_{x \to 0^+} g(x)=\lim\frac{f(x)-f(0)}{x-0}=f'(0)=g(0)
$$,so $g(x)$ is contentious at $0$,last g(x) is an increasing function in
$[0,\infty)$</p>
|
3,336,506 | <p>Let <span class="math-container">$V=\mathbb R^3$</span> be an inner product space with the standard inner product (that means <span class="math-container">$\langle(x_1,y_1,z_1),(x_2,y_2,z_2)\rangle=x_1y_1+x_2y_2+x_3y_3$</span> ).<br>
<span class="math-container">$U=span\{(1,2,3),(1,2,1)\}\subseteq V$</span> </p>
<p>a) Find an orthonormal basis for <span class="math-container">$U$</span><br>
b) Find an explicit formula for the orthogonal projection <span class="math-container">$P_U: V\rightarrow U$</span></p>
<p>So I found an orthonormal basis for <span class="math-container">$V$</span>, denote that basis <span class="math-container">$B=(b_1,b_2)=\bigg(\frac{1}{\sqrt6}(1,2,1),\frac{1}{\sqrt{30}}(-1,-2,5)\bigg)$</span>, and I'm totally sure it's an orthonormal basis, so no problem with that. On the other hand, what I'm having trouble with is finding the explicit formula. What I did was to use the following formula for the orthogonal projection:</p>
<p>Let <span class="math-container">$W\subseteq V$</span> be a vector subspace, where <span class="math-container">$V$</span> is an inner product vector space of finite dimension, and let <span class="math-container">$S=\{s_1,\dots,s_k\}$</span> be an orthonormal basis for <span class="math-container">$W$</span>, then the orthogonal projection upon <span class="math-container">$W$</span> is given by: <span class="math-container">$P_W(u)=\sum_{i=1}^k\langle u,s_i\rangle\cdot s_i$</span></p>
<p>This is the way I used it:<br>
<span class="math-container">$P_U(x,y,z)=\langle(x,y,z),b_1\rangle\cdot b_1+\langle(x,y,z),b_2\rangle\cdot b_2=\dots=\frac{1}{5}(x+2y,2x+4y,5z)$</span></p>
<p>The problem is that I got a projection upon <span class="math-container">$\mathbb R^3$</span>, instead of only upon <span class="math-container">$U$</span> (I took vectors<span class="math-container">$\in\mathbb R^3\backslash U$</span> to find that out), and I dont get what is wrong with what I did.</p>
<p>Any help would be appreciated.</p>
<p>Edit: I found this similar post: <a href="https://math.stackexchange.com/questions/1809883/orthogonal-projection-formula-question?rq=1">orthogonal projection formula question</a> , but since we didn't learned cross products I can't use the solution given there.</p>
| Ninad Munshi | 698,724 | <p>As you noted, unless <span class="math-container">$\phi(x) = 0$</span> this distribution isn't really a distribution since distributions need to be continuous linear functionals. But we can modify it by taking the action <span class="math-container">$$T_x(\phi) = \text{p.v.}\int_{\mathbb{R}^n} \frac{\phi(y)-\phi(x)}{|x-y|^n}dy$$</span> which we can consider as a family of distributions that are shifted by x from <span class="math-container">$$T_0(\phi) = \text{p.v.}\int_{\mathbb{R}^n} \frac{\phi(y)-\phi(0)}{|y|^n}dy$$</span> <span class="math-container">$T_0$</span> has some special properties, namely it is a homogeneous radial distribution of degree <span class="math-container">$-n$</span>. In <span class="math-container">$\mathbb{R}^n$</span> another radial distribution with that homogeneity is the Dirac delta <span class="math-container">$\delta_0$</span> in <span class="math-container">$n$</span> dimensions. In some sense, <span class="math-container">$T_0$</span> can tell you what is happening with the test function everywhere except <span class="math-container">$0$</span>, but <span class="math-container">$\delta_0$</span> can only tell you what is happening at <span class="math-container">$0$</span>. In fact, any radial distribution with homogeneity <span class="math-container">$-n$</span> has to be some linear combination of <span class="math-container">$T_0$</span> and <span class="math-container">$\delta_0$</span></p>
|
2,097,557 | <blockquote>
<p>If $0< \alpha, \beta< \pi$ and $\cos\alpha + \cos\beta-\cos (\alpha + \beta) =3/2$ then prove $\alpha = \beta= \pi/3$</p>
</blockquote>
<p>How do I solve for $\alpha$ and $\beta$ when only one equation is given? By simplification I came up with something like
$$
\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \cos \frac {\alpha +\beta}{2}=\frac{1}{8}.
$$
I don't know if this helps. How to do this?</p>
| egreg | 62,967 | <p>Set $p=\cos\beta$ and $q=\sin\beta$; then your equality can be written
$$
\cos\alpha+p-p\cos\alpha+q\sin\alpha=\frac{3}{2}
$$
or
$$
(1-p)\cos\alpha+q\sin\alpha=\frac{3}{2}-p
$$
Set $X=\cos\alpha$ and $Y=\sin\alpha$; then the equation becomes
$$
\begin{cases}
(1-p)X+qY=\frac{3}{2}-p\\[6px]
X^2+Y^2=1
\end{cases}
$$
and the distance of the line from the origin should be $\le1$, or the line and the circle wouldn't meet:
$$
\frac{(\frac{3}{2}-p)^2}{(1-p)^2+q^2}\le1
$$
that becomes
$$
\frac{9}{4}-3p+p^2\le 1-2p+p^2+q^2
$$
that is, recalling that $p^2+q^2=1$,
$$
p^2-p+\frac{1}{4}\le0
$$
Can you finish up?</p>
|
2,485,276 | <p>Is the statement true? if it is, how to prove it?</p>
<p>If $\binom{p}{k} \mod p=0 $ for $k=1,2,..,p−1$ then $p$ is prime.</p>
| user19405892 | 282,223 | <p>Suppose that $N = ab$ is composite with positive integers $a,b > 1$ and $a$ is the smallest prime factor of $N$. Then $$\binom{N}{a} = \binom{ab}{a} = \dfrac{ab(ab-1) \cdots (ab-(a-1))}{a!} = \dfrac{b(ab-1) \cdots (ab-(a-1))}{(a-1)!}.$$ Then $(ab-1)(ab-2) \cdots (ab-(a-1)) \equiv \pm (a-1)!\pmod{ab}$. Thus, $\binom{ab}{a} \equiv b \pmod{ab}$.</p>
|
1,102,668 | <p><a href="https://math.stackexchange.com/q/67994/198434">This question</a> shows how dividing both sides of an equation by some $f(x)$ may eliminate some solutions, namely $f(x)=0$. Naturally, all examples admit $f(x)=0$ as a solution to prove the point.</p>
<p>I tried to find a simple example of an equation that could be solved by dividing both sides by some $f(x)$ but where $f(x)=0$ was not a solution, and failed miserably. Sure, I can divide both sides of, let's say, $x^2-1=0$ by $x$ ($x=0$ is not a solution), but that doesn't help me solve the equation.</p>
<p>I started wondering if actually the equations that can be solved (or at least simplified) by dividing both sides by some $f(x)$ were precisely those where $f(x)=0$ is a solution: by removing a solution, the division reduces the equation to a simpler form. This is particularly obvious with this example given in that question's <a href="https://math.stackexchange.com/a/68001/198434">accepted answer</a>:</p>
<p>$$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=0.$$</p>
<p>By successively dividing by $x-1$, $x-2$ and so on, the equation becomes simpler as the solutions are removed, until there's no solution left ($1=0$).</p>
<p>However, both the accepted answer and the quote in the question itself say that $f(x)=0$ <em>may</em> be a solution, which I also understand as it <em>may not</em> be one.</p>
<p>So, are there equations where dividing by some $f(x)$ significantly improves the equation resolution, without $f(x)=0$ being a solution?</p>
| Peter Woolfitt | 145,826 | <p>Well, assuming that we are looking for solutions over the reals, we can solve say $$(x^2+1)(x+1)=0$$ by dividing both sides by $x^2+1$ to find that the only solution is $x=-1$. However, we still have to check that $f(x)=x^2+1=0$ has no solutions in the reals (easy to see because $f(x)\ge1$ for every $x$). Is this the sort of thing you are looking for?</p>
<p>For a slightly more complicated example, we can take the equation</p>
<p>$$\sin(2x)=0$$
to be solved on the interval $(0,\pi)$. Since $\sin(2x)=2\sin(x)\cos(x)$, we can divide both sides by $2\sin(x)$ (since $\sin(x)$ is positive on the interval in question) to find the solutions are exactly where $\cos(x)=0$. This would give us exactly one solution at $x=\frac{\pi}{2}$.</p>
|
2,210,531 | <blockquote>
<p>Prove that the limit is zero:
$$ F(x, y)= \frac{x^2y^3}{3x^2+ 2y^3}$$</p>
</blockquote>
<p>Definition1. Let $ U ⊂ R^n$ be an open set and letf : $U→R^m$ beafunction with domain U. Let x0 be a vector in U or on the boundary of U. Let b∈Rm. We say that the limit of f as x approaches x0 is b, written
lim f(x) = b, x→x0
provided that for every ε > 0 there is a δ > 0 so that ||f(x)−b|| < ε whenever 0 < ||x − x0|| < δ and x ∈ U.</p>
<p>First:</p>
<p>$ 0\le\sqrt {x^2+y^2} <\delta $ and $ |F(x,y) -0| < \epsilon$</p>
<p>$|\frac{x^2y^3}{3x^2+ 2y^3}| = \frac{x^2|y^3|}{|3x^2+ 2y^3|} = ...$ how can i solve this? Please, help me.</p>
| Barry Cipra | 86,747 | <p>The limit actually does not exist. Here's one way to see why:</p>
<p>Let $(x,y)\to(0,0)$ along the curve defined by $2y^3=kx^4-3x^2$. Then</p>
<p>$${x^2y^3\over3x^2+2y^3}={kx^6-3x^4\over2kx^4}={1\over2}x^2-{3\over2k}\to-{3\over2k}$$</p>
<p>Since we can use any value we like for $k$ (except $0$, of course), we can get any value we like in the limit.</p>
<p>The key here is that you can take cube roots of negative numbers, so that $y=\sqrt[3]{(kx^4-3x^2)/2}$ is defined for all $x$.</p>
|
280,156 | <p>I have a code as below:</p>
<pre><code>countpar = 10;
randomA = RandomReal[{1, 10}, {countpar, countpar}];
randomconst = RandomInteger[{0, 1}, {countpar, 1}];
For[i = 1, i < countpar + 1, i++,
If[randomconst[[i, 1]] != 0,
randomA[[All, i]] = 0.; randomA[[i, All]] = 0.;
randomA[[i, i]] = 1;
];
];
</code></pre>
<p>The problem is when I have changed countpar, lets say, i.e, countpar=1000, then the computational time of the for loop increases dramatically. Is there a way to decrease this time from an expert eye ?</p>
<p>Best Regards,</p>
<p>Ahmet</p>
| bill s | 1,783 | <p>The formula is a recursion, and so can be written:</p>
<pre><code>Clear[y]; x = 2; n = 2; y[1] = 1;
y[k_] := y[k] = (y[k - 1]^(x - 1) + n)/(y[k - 1]^(x - 1) + y[k - 1]^(x - 2))
</code></pre>
<p>Then you can get individual values as</p>
<pre><code>{y[1], y[2], y[3]}
</code></pre>
<p>or get 10 at once:</p>
<pre><code>y /@ Range[10]
</code></pre>
|
178,342 | <p>This is an exercise from Kunen's book.</p>
<p>Write a formula expressing $z=\langle \langle x,y\rangle, \langle u,v\rangle \rangle$ using just $\in$ and $=$.</p>
<p>What I've tried: because $\langle x,y\rangle= \{\{x\},\{x,y\}\}$ and $\langle u,v\rangle= \{\{u\},\{u,v\}\}$, and hence $z=\{\{\langle x,y\rangle\},\{\langle x,y\rangle,\langle u,v\rangle\}\}= \{\{\{\{x\},\{x,y\}\}\} ,\{\{\{x\},\{x,y\}\},\{\{u\},\{u,v\}\}\}\}$. </p>
<blockquote>
<p>So the formula expressing is this: $z_1\in z$, then $z_1= \{\{x\},\{x,y\}\}$ or $z_1= \{\{u\},\{u,v\}\}$</p>
</blockquote>
<p>Am I right? Thanks ahead:)</p>
| Thanaton | 22,787 | <p>Although I am not an expert in this field (merely a beginning student), maybe I can help you out.</p>
<p>We will have to find a formula only using <span class="math-container">$\in$</span> and <span class="math-container">$=$</span> to describe <span class="math-container">$z = \langle \langle x, y \rangle, \langle u, v \rangle \rangle$</span>. This essentially means we want to say that: '<span class="math-container">$z$</span> is an ordered pair of two ordered pairs <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. The ordered pair <span class="math-container">$a$</span> consists of the elements <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, the ordered pair <span class="math-container">$b$</span> consists of the elements <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.'</p>
<p>--</p>
<p>To describe that <span class="math-container">$z$</span> is an ordered pair, you can use the formula:</p>
<p><span class="math-container">$$(\exists a \in z)(\exists b \in z)(\exists d \in a)(\exists f \in b) \left[ (\forall c \in z)(c = a \vee c = b) \wedge (\forall c \in a)(c = d) \wedge (d \in b) \wedge (\forall c \in b)(c = d \vee c = f) \right]$$</span></p>
<p>Lets call this formula <span class="math-container">$o(z)$</span>.</p>
<p>--</p>
<p>Given that <span class="math-container">$z$</span> is an ordered pair, we can describe that <span class="math-container">$g$</span> is the first coördinate of <span class="math-container">$z$</span> with:</p>
<p><span class="math-container">$$(\forall a \in z)(g \in a)$$</span></p>
<p>Lets call this formula <span class="math-container">$p(z, g)$</span>.</p>
<p>--</p>
<p>Given that <span class="math-container">$z$</span> is an ordered pair, we can describe that <span class="math-container">$i$</span> is the second coördinate of <span class="math-container">$z$</span> with:</p>
<p><span class="math-container">$$(\exists a \in z)(i \in a) \wedge (\forall b \in z)(\forall c \in z)(b \neq c \Rightarrow (i \notin b \vee i \notin c))$$</span></p>
<p>Call this formula <span class="math-container">$q(z, i)$</span>.</p>
<p>--</p>
<p>By combining these formulas, we can describe <span class="math-container">$z = \langle \langle x,y \rangle, \langle u,v \rangle \rangle$</span> as follows:</p>
<p><span class="math-container">$$o(z) \wedge (\exists a)(\exists b)(o(a) \wedge o(b) \wedge p(z, a) \wedge q(z, b) \wedge p(a, x) \wedge q(a, y) \wedge p(b, u) \wedge p(b, v))$$</span></p>
<p>--</p>
<p>EDIT (NOT APPLICABLE ANYMORE)</p>
<blockquote>
<p>As I do not have enough reputation to post a comment on other answers yet (:-s), I will comment on Asaf Karagila's answer in this way. The formula <span class="math-container">$\phi(u,v,z)$</span> you use is not correct, because <span class="math-container">$\phi(u,v,z)$</span> does not hold if and only if <span class="math-container">$z = \langle u, v \rangle$</span>. It is for example possible that <span class="math-container">$z = \{ \{ u \} , \{ v \} \}$</span>, or <span class="math-container">$z = \{ \{ u \} \}$</span> with <span class="math-container">$u \neq v$</span>.</p>
<p>You really need to use the definition of ordered pairs with first and second coördinate I use in my answer, elsewise you won't have an 'if and only if':</p>
<p><span class="math-container">$$z = \langle u, v \rangle \mbox{ if and only if } o(z) \wedge p(z, u) \wedge q(z, v)$$</span></p>
</blockquote>
|
1,279,328 | <p>Can someone please explain to me if a 2/3D Poisson's equation is separable or non separable?
Thank you</p>
| Snehasish | 241,303 | <p>May be what you are looking for is, solving poission's equation using separation of variable method. Here is a link,<a href="http://farside.ph.utexas.edu/teaching/em/lectures/node66.html" rel="nofollow">http://farside.ph.utexas.edu/teaching/em/lectures/node66.html</a>, to explore more about it.</p>
|
3,916,130 | <p>Assume the weight of a person follows a normal distribution N(71,7). What is the probability of 4 people weighing more than 300kg?</p>
<p>I tried solving this by multiplying the values by 4, so it'd be N(284,28). I converted that into <span class="math-container">$x=284+28z$</span> which lead to <span class="math-container">$z=\frac{x-284}{24}$</span>.
To solve for <span class="math-container">$P(x>300)$</span>, I converted it around to <span class="math-container">$ 1-P(x<300)$</span> <span class="math-container">$$ 1-P\left(\frac{300-284}{24}\right) $$</span>
<span class="math-container">$$ 1-P\left(z<\frac{2}{3}\right) $$</span>
Then i used the normal distribution table and got the following result: <span class="math-container">$1-0.2546 = 0.745$</span>, however I have a suspicion I started from the wrong track and this result isn't correct.</p>
<p>This was an exam question at a college statistics class</p>
| BruceET | 221,800 | <p>I assume heights are independently normally distributed with mean <span class="math-container">$\mu = 71$</span> and <em>standard deviation</em> <span class="math-container">$\sigma = 7.$</span> [Sometimes the second parameter is the variance and sometime the variance, you should always say which convention you are using.]</p>
<p>Then <span class="math-container">$S = X_1+X_2+X_3+X_4$</span> has <span class="math-container">$E(S) = E(X_1)+\cdots+E(X_4) = 4(71) = 284.$</span>
and (because of independence) <span class="math-container">$V(S) = V(X_1)+\cdots+V(X_4) = 4(7^2) = 196.$</span></p>
<p>So <span class="math-container">$P(S > 300) = 1 - P(S_4 \le 300 = 0.1265.$</span> Computation in R (where the second parameter is the SD). You can get approximately the same answer by standardizing and using printed tables of the standard normal CDF (which
involves some rounding.).</p>
<pre><code>1 - pnorm(300, 284, sqrt(196))
[1] 0.126549
</code></pre>
<p>So an elevator with safe weight load 300, serving a population with
weights distributed <span class="math-container">$\mathsf{Norm}(\mu = 71, \sigma=7)$</span> should have
a limit of three (not four) passengers.</p>
<p>Simulation: With a million sums of four we get <span class="math-container">$0.1266\pm 0.0007.$</span></p>
<pre><code>set.seed(1120)
x1 = rnorm(10^6, 71, 7)
x2 = rnorm(10^6, 71, 7)
x3 = rnorm(10^6, 71, 7)
x4 = rnorm(10^6, 71, 7)
s = x1 + x2 + x3 + x4
summary(s); sd(s)
Min. 1st Qu. Median Mean 3rd Qu. Max.
217.9 274.6 284.0 284.0 293.4 351.7
[1] 13.99282
mean(s > 300)
[1] 0.126589 # aprx P(S > 300) = 0.126549
2*sd(s > 300)/1000
[1] 0.0006650243 # aprs 95% margin of simulation err
</code></pre>
<p><a href="https://i.stack.imgur.com/eUqD4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eUqD4.png" alt="enter image description here" /></a></p>
<pre><code>hist(s, prob=T, br=30, col="skyblue2", main="Sim Dist'n Sum of Four")
curve(dnorm(x, 284, 14), add=T, col="red", lwd=2)
abline(v=300, lty="dotted", lwd=2)
</code></pre>
<p>You can get a good approximation by standardizing and using printed tables of
the standard normal CDF (a proxess which requires some rounding).</p>
|
167,848 | <p>I am trying to solve numerically an equation and generate some results. I use the following code </p>
<pre><code>u[c_] := (c^(1 - σ) - 1)/(1 - σ)
f[s_] := g s (1 - s/sbar1)
h[s_] := (2 hbar)/(1 + Exp[η (s/sbar - 1)])
co[a_] := ϕ (a^2)/2
ψ[k_] := wbar (ω + (1 - ω) Exp[-γ k])
</code></pre>
<p>The equation I try to solve is the following</p>
<pre><code>adap[k_, s_] := (ρ + δ) u'[f[s] - priceadap δ k] co'[δ k] + ψ'[k] h[s]
</code></pre>
<p>I have the following constant parameter set </p>
<pre><code>paramFinal2 = {σ -> 1.7, ρ -> 0.025, g -> 0.05, sbar -> 10, η -> 11, hbar -> 0.5, priceadap -> 0.0006, γ -> 0.6, χ -> 1000, ϕ -> 0.05, ω -> 0.35, β -> 0.8, δ -> 0.065, sbar1 -> 10, wbar -> 1000};
</code></pre>
<p>So, for different values of $s$, I try to generate the corresponding values of $k$.</p>
<p>For this, I use the following code </p>
<pre><code>tmax1 = 10;
solK[i_] := Solve[adap[k, i] == 0 /. paramFinal2, k];
Table[solK[i], {i, 1, tmax1}];
</code></pre>
<p>Unfortunately, this does not give any result. Mathematica is always on mode "Running...".</p>
<p>P.S I am using Mathematica 9.0</p>
| Hugh | 12,558 | <p>I always find it a good idea to plot the functions you are investigating. Thus</p>
<pre><code>tmax1 = 10;
Column@Table[
Plot[Evaluate[adap[k, i] /. paramFinal2], {k, 0, 20}], {i, 1 tmax1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/Thz8L.png" alt="Mathematica graphics"></p>
<p>They all look fine except for the last one. The zero is at about k = 10 for most of the functions.
The equation is not a polynomial and thus may not be solvable using Solve. I don't know what range of equations Solve can deal with but some functions are just too complicated. The solution for this is to use FindRoot. </p>
<pre><code>rts = Table[
FindRoot[adap[k, i] /. paramFinal2, {k, 10}], {i, 1, tmax1}]
</code></pre>
<p>which gives</p>
<pre><code>{{k -> 10.7363}, {k -> 12.1674}, {k -> 12.8497}, {k -> 13.185}, {k ->
13.2841}, {k -> 13.169}, {k -> 12.797}, {k -> 12.0139}, {k ->
10.3238}, {k -> 3.43075*10^39 - 3.28305*10^36 I}}
</code></pre>
<p>The last root is clearly wrong and is complex which gives us a clue. If we do </p>
<pre><code>adap[k, 10] /. paramFinal2 /. k -> 1
</code></pre>
<p>We get</p>
<pre><code>5269.92 + 7400.72 I
</code></pre>
<p>So this looks like a complex function. Thus </p>
<pre><code>Plot[Evaluate[ReIm[adap[k, 10] /. paramFinal2]], {k, 0, 20}]
</code></pre>
<p><img src="https://i.stack.imgur.com/Fn2jo.png" alt="Mathematica graphics"></p>
<p>This suggests that you might not have a real root for this case. </p>
<p>Hope that helps. </p>
|
3,399,586 | <p>Suppose that <span class="math-container">$f$$:$$\mathbb{R}\to\mathbb{R}$</span> is differentiable at every point and that </p>
<p><span class="math-container">$$f’(x) = x^2$$</span></p>
<p>for all <span class="math-container">$x$</span>. Prove that </p>
<p><span class="math-container">$$f(x) = \frac{x^3}{3} + C$$</span></p>
<p>where <span class="math-container">$C$</span> is a constant. </p>
<p>This has to be done without integrating, I have only been taught differential calculus and this question only assumes knowledge of that.</p>
<p>I tried applying mean value theorem and taylor’s approximation but could not come up with the proof. Can someone please provide the solution?</p>
| wonderman | 715,632 | <p>Let <span class="math-container">$g(x) = f(x) - \frac{x^3}{3}$</span>. What can you say about the derivative of <span class="math-container">$g(x)$</span>?</p>
|
165,560 | <p>To find the volume of the following region:</p>
<pre><code>fn[x_, y_, z_]:= Abs[0.7*x*Exp[I*y] + 0.3*Sqrt[x^2 + 8*10^-5]
+ Sqrt[x^2 + 3*10^-3]*0.02*Exp[I*z]]
R = ImplicitRegion[fn[x, y, z]<=3*10^-3, {{x, 0, 0.015}, {y, 0, 2*Pi}, {z, 0, 2*Pi}}]
RegionPlot3D[
fn[x, y, z] <= 3*10^-3, {x, 0, 0.015}, {y, 0, 2*Pi}, {z, 0, 2*Pi},
PlotPoints -> 50, AxesLabel -> {x, y, z},
PlotStyle -> Directive[Yellow, Opacity[0.5]], Mesh -> None]
</code></pre>
<p><a href="https://i.stack.imgur.com/DXSbo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DXSbo.jpg" alt="Plot of region of interest"></a></p>
<p>Using <code>Volume</code>:</p>
<pre><code>Volume[R]
</code></pre>
<blockquote>
<p>Volume::nmet: Unable to compute the volume of region</p>
</blockquote>
<p>Using <code>NIntegrate</code>:</p>
<pre><code>NIntegrate[
Boole[fn[x, y, z] <= 3*10^-3], {x, 0, 0.015}, {y, 0, 2*Pi}, {z, 0,
2*Pi}, Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> 50000},
PrecisionGoal -> 3] // Timing
</code></pre>
<blockquote>
<p>NIntegrate::slwcon:
Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.</p>
<p>NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 50000 times. The global error is expected to decrease monotonically after a number of integrand evaluations... NIntegrate obtained 0.12219642052793653 and 0.007515502934285486 for the integral and error estimates.</p>
<p>{19.25, 0.122196}</p>
</blockquote>
<p>The result does not converge.</p>
<p>Using <code>MonteCarlo</code>:</p>
<pre><code>NIntegrate[
Boole[fn[x, y, z] <= 3*10^-3], {x, 0, 0.015}, {y, 0, 2*Pi}, {z, 0,
2*Pi}, Method -> {"MonteCarlo", "MaxPoints" -> 10^12,
"RandomSeed" -> 9}, PrecisionGoal -> 3] // Timing
</code></pre>
<blockquote>
<p>{8.39063, 0.121479}</p>
</blockquote>
<p>This is slow, and if <code>PrecisionGoal -> 5</code> is set,</p>
<pre><code>NIntegrate[
Boole[fn[x, y, z] <= 3*10^-3], {x, 0, 0.015}, {y, 0, 2*Pi}, {z, 0,
2*Pi}, Method -> {"MonteCarlo", "MaxPoints" -> 10^12,
"RandomSeed" -> 9}, PrecisionGoal -> 5]
</code></pre>
<p>It does not give a result even after running for more than 2 minutes.</p>
<p><strong>I hope to find a better way to calculate the volume of this region with:</strong></p>
<ol>
<li>High accuracy: at least <code>PrecisionGoal -> 5</code></li>
<li>Good error estimate</li>
<li>Fast</li>
</ol>
| José Antonio Díaz Navas | 1,309 | <p>I have tried this with MMA 11.2 (macOs 10.13.3) in my old Intel I3. First, I have simplified a bit your <code>fn</code> as it involves <code>Abs</code>, and by setting the definition of <code>ImplicitFucntion</code> in a standard form:</p>
<pre><code>fn[x_, y_, z_] := 8.4*10^-6 + 0.5804 x^2 + 0.00375659 x Sqrt[1 + 12500 x^2] Cos[y] +
0.000885438 Sqrt[3 + 1000 x^2] (1. x Cos[y - z] + 0.00383326 Sqrt[1 + 12500 x^2] Cos[z])
R = ImplicitRegion[fn[x, y, z] <= 9/1000000 &&
0 <= x <= 0.015 && 0 <= y <= 2 Pi && 0 <= z <= 2 Pi, {x, y, z}];
NIntegrate[1, {x, y, z} \[Element] R,Method -> {"MonteCarlo", "MaxPoints" -> 10^6},
PrecisionGoal -> 5]
</code></pre>
<p>The high number of points used is to provide the required <code>PrecisionGoal</code>, and after 11.5 sec:</p>
<pre><code>NIntegrate::maxp: The integral failed to converge after 1000100 integrand evaluations.
NIntegrate obtained 0.12154680553578552` and 0.00023916058934647544` for the integral
and error estimates.
(* 0.121547 *)
</code></pre>
<p>which is slow, and not enough. </p>
<p>However, by using <code>BoundaryDiscretizeRegion</code>, we can use <code>Volume</code>. This is quite slow, as we need to represent the region with enough accuracy (see also <a href="https://mathematica.stackexchange.com/questions/158971/volume-of-solids-of-revolution-through-two-equations/158988#158988">this</a>), but provides a reference value for the volume calculated by using <code>NIntegrate</code>: </p>
<pre><code>R2 = BoundaryDiscretizeRegion[R, BoxRatios -> {1, 1, 1}, MaxCellMeasure -> 0.000075];
Volume[R2]
(* 0.121502 *)
</code></pre>
<p><a href="https://i.stack.imgur.com/t6iZl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t6iZl.png" alt="enter image description here"></a></p>
<p>For <code>MaxCellMeasure->0.00006</code> the volume is 0.121509.</p>
<p>By means of:</p>
<pre><code>NIntegrate[Boole[fn[x, y, z] <= 9/1000000], {x, 0, 0.005, 0.01, 0.015},
{y, 0, 2/3 Pi, 4/3 Pi, 2*Pi}, {z, 0, 2/3 Pi, 4/3 Pi, 2*Pi},
Method -> {"MonteCarlo", "MaxPoints" -> 10^7, "RandomSeed" -> 9},
PrecisionGoal -> 5] // AbsoluteTiming
</code></pre>
<p>I get (again, very slow):</p>
<pre><code>NIntegrate::maxp: The integral failed to converge after 10000100 integrand evaluations.
NIntegrate obtained 0.12149430710916392` and 0.000041400109821011765` for the integral
and error estimates.
(* {20.1802, 0.121494} *)
</code></pre>
<p>Well, It seems that your problem is not so easy to make it more efficient. I always used in this case the <code>MonteCarlo</code> method as the faster one with <code>NIntegrate</code>. However, an expert on numerical integration can explore the multiple options to obtain a proper combination of them to speed up the calculation.</p>
<p>The use of <code>Volume</code> seems to be the faster as long as the process in obtaining the <code>BoundaryDiscretizeRegion</code> will be also fast. It was so to achieve the required <code>PrecisionGoal</code> in my computer, compared to <code>NIntegrate</code> .</p>
<p>Hope this helps.</p>
|
3,346,676 | <blockquote>
<p><strong>Question.</strong> Find a divergent sequence <span class="math-container">$\{X_n\}$</span> in <span class="math-container">$\mathbb{R}$</span> such that for any <span class="math-container">$m\in\mathbb{N}$</span>,
<span class="math-container">$$\lim_{n\to\infty}|X_{n+m}-X_n|=0$$</span></p>
</blockquote>
<p>I don't really know, if someone could walk me through this it'd be really appreciated.
Edit: I'm dumb af ignore what I said before I deleted it. lmao</p>
| copper.hat | 27,978 | <p>Let <span class="math-container">$x_n= 1+ {1 \over 2} + \cdots + {1 \over n}$</span>, then <span class="math-container">$x_n \to \infty$</span>.</p>
<p>For any <span class="math-container">$m$</span>, we see that
<span class="math-container">$x_{n+m} - x_n = {1 \over n+1} + \cdots + {1 \over n+m} \le {m \over n}$</span>, hence
<span class="math-container">$\lim_{n \to \infty} x_{n+m} - x_n = 0$</span>.</p>
|
767,304 | <p>Prove that there are no real numbers $x$ such that</p>
<p>$$\sum_{n\,=\,0}^\infty \frac {(-1)^{n + 1}} {n^x} = 0$$</p>
<p>Can I have a hint please?</p>
| Bruno Joyal | 12,507 | <p>This is an observation which is too long for a comment, but which someone will probably appreciate. Remark that</p>
<p>$$\sum_{n=1}^\infty \frac{(-1)^n}{n^s} = \zeta(s) - 2\times 2^{-s}\zeta(s) = (1-2^{1-s})\zeta(s).$$</p>
<p>The Riemann zeta function is equal to the Euler product $\prod_p (1-p^{-s})^{-1}$ in the right half-plane $\Re s>1$. The Euler product converges in the sense of <a href="http://en.wikipedia.org/wiki/Infinite_product" rel="nofollow">convergence for infinite products</a>. A convergent infinite product is never zero, so $\zeta(s)\neq 0$ for $\Re s > 1$. Moreover, it's easy to see that $1-2^{1-s} \neq 0$ for $\Re s> 1$. So we get your result for $\Re s >1$. </p>
<p>For $0<\Re s \leq 1$, it's a little tricky because the Euler product no longer converges. However, your sum is still equal to $(1-2^{1-s})\zeta(s)$ when $0<\Re s \leq 1$, essentially by analytic continuation. At $s=1$, $(1-2^{1-s})\zeta(s) \neq 0$ because the zero of $1-2^{1-s}$ is cancelled by the simple pole of $\zeta(s)$.</p>
<p>Thus, the proofs that Felix and Daniel have given show that the Riemann zeta function doesn't vanish for real $s$ between $0$ and $1$! </p>
|
2,978,605 | <p>Is there a sophisticate way to proof that:</p>
<p><span class="math-container">$$\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{1}{{N\choose n}}$$</span></p>
<p>where <span class="math-container">${N\choose n}$</span> denotes combinations.</p>
<p>When replacing <span class="math-container">$N$</span> and <span class="math-container">$n$</span> with values, both produce the same result.</p>
| KM101 | 596,598 | <p>Simplify the numerator and denominator.</p>
<p><span class="math-container">$$\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{n!}{\frac{N!}{(N-n)!}}$$</span></p>
<p>Simplify by <span class="math-container">$n!$</span>.</p>
<p><span class="math-container">$$\frac{1}{\frac{N!}{n!(N-n)!}}$$</span></p>
<p>The denominator is in the form <span class="math-container">$n$</span> choose <span class="math-container">$r$</span>.</p>
<p><span class="math-container">$$\frac{1}{N \choose n}$$</span></p>
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2,558,988 | <blockquote>
<p>Let us consider a function $f(x,y)=4x^2-xy+4y^2+x^3y+xy^3-4$. Then find the maximum and minimum value of $f$.</p>
</blockquote>
<p>My attempt. $f_x=0$ implies $8x-y+3x^2y+y^3=0$ and $f_y=0$ implies $-x+8y+x^3+3xy²=0$ and $f_{xy}=3x^2+3y^2-1$. Now $f_x+f_y=0$ implies $(x+y)((x+y)^2+7)=0$ implies $x=-y$ as $x$ and $y$ are reals. Now putting it in $f_x=0$ get the three values of $x$ as $0$, $(-3+3\sqrt{5})/2$ and $(-3-3\sqrt{5})/2$. And then $f_{xy}(0,0)<0$ implies $f$ has maximum at $(0,0)$. But at other two points $f_{xy}$ gives the positive value. So how can I proceed to solve the problem.
Please help me to solve it.</p>
| Bart Michels | 43,288 | <p>Yes. It is the image of the connected space $X\times \mathbb R_{>0}$ under the continuous map $$(x, t)\mapsto (x, f(x)-t)$$</p>
|
843,634 | <p>I am wondering whether for any two lines $\mathfrak{L}, \mathfrak{L'}$ and any point $\mathfrak{P}$ in $\mathbf{P}^3$ there is a line having nonempty intersection with all of $\mathfrak{L}, \mathfrak{L'}$, $\mathfrak{P}$. I don't really know how to approach this, because I was never taught thinking about such a problem, not even related ones. I think the answer should be no, but have no means to justify it. Perhaps use the Klein representation of lines in 3 space? Could someone also recommend a book where similar problems are solved or at least posed as exercises? Also feel free to give an algebraic geometry perspective, but I don't really know how to approach this with algebraic geometry.</p>
| lab bhattacharjee | 33,337 | <p>Do you know the difference between Congruence Equation and Algebraic equation which obeys <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra" rel="nofollow">Fundamental theorem of algebra</a> unlike the previous one?</p>
<p>For example $$x^2\equiv2\pmod3$$ is not solvable unlike $$x^2=2$$ </p>
<p>Again, $$a^p\equiv a\pmod p$$ is basically Identity Congruence Equation i.e., any integer satisfies this equation just like $$\prod_{r=0}^{p-1}(x-r)\equiv0\pmod p$$</p>
|
1,584,594 | <p>Find all $n$ for which $n^8 + n + 1$ is prime. I can do this by writing it as a linear product, but it took me a lot of time. Is there any other way to solve this? The answer is $n = 1$.</p>
| Paolo Leonetti | 45,736 | <p>Since $n^2+n+1$ divides $n^8+n+1$ and $1<n^2+n+1<n^8+n+1$ for $n>1$, then $n=1$ is the unique solution (which indeed gives a prime).</p>
|
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