qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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1,278,719 | <p>This is a problem from Artin's book "Algebra". In the fifth miscellaneous problem of the chapter "Vector spaces", he has asked to prove that:</p>
<p>If $\alpha$ is a cube root of $2$, then the real numbers $a+b\alpha +c\alpha ^2$ with $a,b,c \in \mathbb{Q}$ form a field.</p>
<p>I am stuck at proving this. For example, what would be the inverse of $1+\alpha +\alpha^2$? In the previous subpart, we were asked to prove that $1,\alpha, \alpha^2$ are linearly independent over $\mathbb{Q}$, and that went well. Using this result seems to give me there is no inverse of $1+\alpha + \alpha^2$ in the above set, which can't happen in a field.</p>
<p>Am I missing something? Thanks.</p>
| zhw. | 228,045 | <p>The series converges uniformly on $[0,\infty).$ Proof: Let $S_n(x) = \sum_{k=1}^{n}\frac{(-1)^k}{k+x}.$ Then </p>
<p>$$S_{2n}(x) = \sum_{k=1}^{n} \left ( \frac{1}{2k + x}-\frac{1}{2k-1 + x}\right ) = \sum_{k=1}^{n} \frac{-1}{(2k + x)(2k-1+x)}.$$</p>
<p>In absolute value, the terms in the last sum are $\le \frac{1}{2k(2k-1)}.$ (We used the fact that $x\ge 0$ here.) By the Weierstrass M test, we see $S_{2n}$ converges to some $f$ uniformly on $[0,\infty).$ Since </p>
<p>$$ |S_{2n+1}(x) - f(x)| \le |S_{2n}(x) - f(x)| +\left |\frac{1}{x+2n+1}\right | \le |S_{2n}(x) - f(x)| + \frac{1}{2n+1},$$</p>
<p>we see $S_{2n+1} \to f$ uniformly on $[0,\infty)$ as well. Therefore the full series converges uniformly to $f$ on $[0,\infty).$</p>
<p>What about the rest of $\mathbb {R}$? There is something to worry about with negative $x$ of course: $\frac{1}{x+n}$ blows up at $-n.$ Still I think the following is true: For any $a<0,$ the series converges uniformly on $[a,0]\setminus \{-1,-2,\cdots \}.$ Try to prove this!</p>
|
1,969,996 | <p>I'm new to probability and I'm struggling to figure how to approach this question.</p>
<p>Airport A handles 40% of all airline traffic, and airports B and C handle 40% and 20% respectively. The detection rate for weapons at the three airports are 0.9, 0.5 and 0.4 respectively. If a passenger is found to be carrying a weapon what is the probability that he is using airport A?</p>
<p>Sorry if it's very simple.</p>
| Hagen von Eitzen | 39,174 | <p>No.</p>
<p>The presumably shortest proof for the case you consider is this</p>
<p>Assume $$n^3+(n+1)^3=(n+2)^3. $$
Then
$$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$
We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$</p>
|
1,969,996 | <p>I'm new to probability and I'm struggling to figure how to approach this question.</p>
<p>Airport A handles 40% of all airline traffic, and airports B and C handle 40% and 20% respectively. The detection rate for weapons at the three airports are 0.9, 0.5 and 0.4 respectively. If a passenger is found to be carrying a weapon what is the probability that he is using airport A?</p>
<p>Sorry if it's very simple.</p>
| fleablood | 280,126 | <p>$n^3 + (n+1)^2 = (n+2)^3 \implies $</p>
<p>$(m-1)^3 +m^3=(m+1)^3; m=n+1$</p>
<p>$2m^3-3m^2+3m-1=m^3+3m^2+3m+1$</p>
<p>$m^3=6m^2+2$</p>
<p>$m=6 + 2/m^2$ so $m^2|2$ and $m > 6$. So $m=1>6$. Clearly impossible.</p>
<p>Of course, the shortest proof would be "this is a single cubic equation with one variable and can easily be confirmed to have no integer solution".</p>
<p>This is simply far too specific a case to be of any significance.</p>
|
633,985 | <p>How can I find the values of $a$ for which the following function $f:\{0,1,\dots,m-1\} \rightarrow \mathbb{Z}_m$ is bijective for a fixed $m$?
$$f(n) = \sum_{k=0}^n a^k$$</p>
| Michael Hardy | 11,667 | <p>Maybe the quickest way to do this is to take the number of triangles including degenerate ones, and subtract the number of degenerate triangles.</p>
<p>The number of ways to choose three points out of 25 is
<span class="math-container">$$
\binom{25}{3} = 2300.
$$</span></p>
<p>Now consider how to count the number of degenerate triangles, i.e. those in which the three vertices are collinear. The possible slopes of the lines are <span class="math-container">$0,\infty,\pm1,\pm2,\pm\frac12$</span>. The quickest way to see that is just staring at the grid. For example, with slope <span class="math-container">$1/3$</span>, only two points on the grid can be found on the line.</p>
<p>With slope <span class="math-container">$0$</span>, there are five possible lines and from one of those we must choose three points, so we get
<span class="math-container">$$
5\cdot\binom 5 3 = 50
$$</span>
degenerate triangles. And the same with slope <span class="math-container">$\infty$</span>.</p>
<p>With slope <span class="math-container">$1$</span>, we have two lines of length <span class="math-container">$1$</span>, two of length <span class="math-container">$2$</span>, two of length <span class="math-container">$3$</span>, two of length <span class="math-container">$4$</span>, and just one of length <span class="math-container">$5$</span>. Hence, the number of degenerate triangles is
<span class="math-container">$$
2\binom 1 3 + 2\binom 2 3 + 2\binom 3 3 + 2\binom 4 3 + 2\binom 5 3 = 30.
$$</span>
And the same with slope <span class="math-container">$-1$</span>.</p>
<p>Now on to slope <span class="math-container">$1/2$</span>. There are only three degenerate triangles with slope <span class="math-container">$1/2$</span>. Call them
<span class="math-container">$$
\{(1,1),(3,2),(5,3)\} + (1,k) \text{ for }k=0,1,2.
$$</span>
Look at the grid and you'll see this. Similarly, there are three with slope <span class="math-container">$-1/2$</span> and three with each of the slopes <span class="math-container">$\pm2$</span>.</p>
<p>Now add them up:
<span class="math-container">$$
50+50+30+30+3+3+3+3 = 172.
$$</span></p>
<p>Hence,
<span class="math-container">$$
2300 - 172 = 2128.
$$</span>
non-degenerate triangles.</p>
<p>But you should closely check the details above since I haven't.</p>
|
633,985 | <p>How can I find the values of $a$ for which the following function $f:\{0,1,\dots,m-1\} \rightarrow \mathbb{Z}_m$ is bijective for a fixed $m$?
$$f(n) = \sum_{k=0}^n a^k$$</p>
| Hunter | 120,472 | <p>It is as easy as it seems to be. This problem can be solved by combination and permutation techniques. What you need is that create as many as triangles as can be without creating a line (three collinear points will make a line out of a triangle). A triangle is formed with three noncollinear points. But there can be two collinear (here by collinear I mean being in the same row, column or diagonal) points. Just as combination you have 25 points (5x5 grid gives 25 slots) so 1st point could be from any of these i.e. 1st point can be placed in any of these 25 points. For the 2nd point there is no condition of any point being collinear or noncollinear (but let this pt be collinear with the 1st one although two pts are always collinear) so the second point has 24 points excluding the 1st one. So 2nd point can take any of these 24 points. For the third one, this point cannot be collinear to 1st and 2nd (if 1st and 2nd are collinear and if they are not then this can be collinear to either 1st or second and then consider this point to be the second one and the second point to be third point so that the three pts be non collinear) points both. So this must deduce some points from the remaining 23 slots. This deduce 3 points from the grid so the remaining slots are 20. This formula can be devised as for n*n grid Total no. of triangles are (n*n)*((n*n)-1)*(((n*n)-2)-(n-2)) = (n*n)*((n*n)-1)*((n*n)-n) . This formula can be used to calculate no. of triangles for n*n no. of grid.</p>
<p>So with the above formula and as said above the no. of triangles in a 5*5 grid becomes
(5*5)*((5*5)-1)*((5*5)-5) = 25*24*20 just as the no. of slots I deduced above which comes out to be 12000 no. of triangles.</p>
<p>Also the number of triangles that comes out with this formula consists of too many repeated (by repeated I mean the triangles formed by the same points so the congruent triangles are excluded in this type) triangles. So how to remove these triangles which have repeated and also how many times have they be repeated. This is a simple trick which can be solved by factorialisation. We know that how many points a triangles consists of, obviously only 3 points (vertices). On a grid these three points will repeat themselves 3! (3 factorial) no. of times to form one triangle (6 triangles in all but all are same). So with this we can deduce that in every 6 triangles there are 5 fake ones and 1 real one. So the real : fake ratio becomes 1:5 . So the total no. of triangles omitting the fake ones becomes 1/6th the no. of triangles we got by above formula. So for a 5*5 grid total no. of triangles becomes 12000/6 = 2000. Yes 2000 different triangles but there can be and there are many congruent triangles.</p>
<p>This formula can be applied to other polygons to (with some modifications) and can be used.
Also if someone fails to understand this answer I can post images here (although I don't know how as there is no image posting symbol like in SU).</p>
|
4,572,505 | <p>There are many approximations of <span class="math-container">$\pi$</span> using trigonometric and rational numbers. But I created this one: <span class="math-container">$$\pi \approx \sqrt[11]{294204}$$</span> Which is correct to almost <span class="math-container">$8$</span> decimal places. Are there any other approximations of <span class="math-container">$\pi$</span> using radicals? I know of <span class="math-container">$\sqrt{10}$</span>, <span class="math-container">$\sqrt[3]{31}$</span>, <span class="math-container">$\sqrt[4]{97}$</span>, and so on. But are there more beautiful ways (like using <span class="math-container">$\varphi$</span> since it is composed of radicals)?</p>
| Oscar Lanzi | 248,217 | <p>There are lots of radical approximations to <span class="math-container">$\pi$</span>. A classical example is given by <a href="https://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formula" rel="nofollow noreferrer">Viete's formula</a>, in which a series of increasingly complicated nested radicals may be used to render <span class="math-container">$2/\pi$</span>:</p>
<p><span class="math-container">$2/\pi=\Pi_{n=1}^\infty(a_n/2)$</span></p>
<p><span class="math-container">$a_1=\sqrt2, a_n=\sqrt{2+a_{n-1}}.$</span></p>
<p>Many readers are aware that this series is derived by considering the perimeters of inscribed regular polygons having <span class="math-container">$2^k$</span> sides, these approaching the circumference of a circle as <span class="math-container">$k\to\infty$</span>.</p>
<p>Just for fun, let's explore a method based on the Taylor series for the cosine function. Pick a value of <span class="math-container">$n$</span> for which <span class="math-container">$\cos(\pi/n)$</span> has a convenient radical expression and plug in the Taylor series approximation</p>
<p><span class="math-container">$\cos(\pi/n)\approx 1-[(\pi/n)^2/2]+[(\pi/n)^4/24].$</span></p>
<p>Then substitute the radical expression for <span class="math-container">$\cos(\pi/n)$</span> on the left and solve the biquadratic polynomial by standard methods for "<span class="math-container">$\pi/n$</span>".</p>
<p>With <span class="math-container">$n=5,\cos(\pi/5)=(\sqrt5+1)/4$</span> this gives</p>
<p><span class="math-container">$\pi\approx5\sqrt{6-\sqrt3×(\sqrt5+1)}=\color{blue}{3.142}3...$</span></p>
<p>(correct to the significant digits indicated in blue.)</p>
|
3,653,148 | <p>Let <span class="math-container">$w$</span> be a primitive 5th root of unity. Then the difference equation <span class="math-container">$$x_nx_{n+2}=x_n-(w^2+w^3)x_{n+1}+x_{n+2}$$</span> generates a cycle of period 5 for general initial values:
<span class="math-container">$$u,v,\frac{u-(w^2+w^3)v}{u-1},\frac{uv-(w^2+w^3)(u+v)}{(u-1)(v-1)},\frac{v-(w^2+w^3)u}{v-1},u,v, ...$$</span> </p>
<p>For equations of the form <span class="math-container">$$x_nx_{n+2}=w^{a+b}x_n-(w^a+w^b)x_{n+1}+x_{n+2},\text{ for }w^a+w^b\ne 0$$</span> with the same globally periodic property, I can show that the only possible periods are 5,8,12,18 and 30.</p>
<p>Curiously, the 'same' equation works for all these periods:
<span class="math-container">$$x_nx_{n+2}=w^5x_n-(w^2+w^3)x_{n+1}+x_{n+2},$$</span>
where <span class="math-container">$w$</span> is a primitive <span class="math-container">$p$</span>th root of unity for <span class="math-container">$p=5,8,12,18,30$</span>. Is this a fluke or is there a way of seeing why this 'family' of equations always generate cycles? </p>
| Pavel Kozlov | 143,912 | <p>The substitution <span class="math-container">$z_n=x_n−1$</span> really helps us, it leads to the equation
<span class="math-container">$$z_nz_{n+2}=az_{n+1}+b,$$</span>
where <span class="math-container">$a=-\omega^2-\omega^3, b=1-\omega^2-\omega^3.$</span>
We can easily see that if <span class="math-container">$z_n=u, z_{n+1}=v$</span> then
<span class="math-container">$$z_{n-1}=\frac{au+b}{v}, z_{n+2}=\frac{av+b}{u},$$</span>
and
<span class="math-container">$$z_{n-2}=\frac{a^2u+bv+ab}{uv}, z_{n+3}=\frac{a^2v+bu+ab}{uv},$$</span>
so if <span class="math-container">$a^2=b$</span> then <span class="math-container">$z_n$</span> is <span class="math-container">$5$</span>-periodic, so and <span class="math-container">$x_n$</span>.
It is true because
<span class="math-container">$$a^2=(\omega^2+\omega^3)^2=2+\omega+\omega^4=1+(1+\omega+\omega^4)=1-\omega^2-\omega^3=b.$$</span></p>
|
1,949,966 | <h2>Q 1a</h2>
<p>Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?</p>
<p>I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.</p>
<p>To be more explicit, I want as many properties to hold as possible, e.g. $|a|\times|b|=|a\times b|$ and $|a|=|-a|$, as some properties that seem to hold for all different types of numbers (or in some analogous way).</p>
<hr>
<h2>Q 1b</h2>
<p>If we let the solution to $|x|=-1$ be $x=z_1$, and we allow the multiplicativeness property,</p>
<p>$$|(z_1)^2|=1$$</p>
<p>Or, further,</p>
<p>$$|(z_1)^{2n}|=1\tag{$n\in\mathbb N$}$$</p>
<p>Note that this does not mean $z_1$ is any such real, complex, or any other type of number. We used to think $|x|=1$ had two solutions, $x=1,-1$, but now we can give it the solution $x=e^{i\theta}$ for $\theta\in[0,2\pi)$. Adding in the solution $(z_1)^{2n}$ is no problem as far as I can see.</p>
<p>However, there result in some problems I simply cannot quite see so clearly, for example,</p>
<p>$$|z_1+3|=?$$</p>
<p>There exists no such way to define such values at the moment.</p>
<p>Similarly, let $z_2$ be the number that satisfies the following:</p>
<p>$$|z_2|=z_1$$</p>
<p>As far as I see it, it is not possible to create $z_2$, given $z_1$ and $z_0\in\mathbb C$.</p>
<p>The following has a solution, in case you were wondering.</p>
<p>$$|\sqrt{z_1}|=i$$</p>
<p>so no, I did not forget to consider such cases.</p>
<p>But, more generally, I wish to define the following numbers in a recursive sort of way.</p>
<p>$$|z_{n+1}|=z_n$$</p>
<p>since, as far as I can tell, $z_{n+1}$ is not representable using $z_k$ for $k\le n$. In this way, the nature of $z_n$ goes on forever, unlike $i$, which has the solution $\sqrt i=\frac1{\sqrt2}(1+i)$.</p>
<p>So, my second question is to ask if anyone can discern some properties about $z_n$, defining them as we did above? And what is $|z_1+3|=?$</p>
<hr>
<h2>Q 2a</h2>
<p>This part is important, so I truly want you guys (and girls) to consider this:</p>
<blockquote>
<p>Can you construct a problem such that $|x|=-1$ will be required in a step as you solve the problem, but such that the final solution is a real/complex/anything already well known. This is similar to <em><a href="https://en.wikipedia.org/wiki/Casus_irreducibilis" rel="nofollow noreferrer">Casus irreducibilis</a></em>, which basically forced $i$ to exist by establishing its need to exist.</p>
</blockquote>
<p>I am willing to give a large rep bounty for anyone able to create such a scenario/problem. </p>
<hr>
<h2>Q 2b</h2>
<p>And if it is truly impossible, why? Why is it not possible to define some 'thing' the solution to the problem, keep a basic set of properties of the absolute value, and carry on? What's so different between $|x|=-1$ and $x^2=-1$, for example?</p>
<hr>
<h2>Thoughts to consider:</h2>
<p>Now, <a href="https://math.stackexchange.com/a/1345391/272831">Lucian</a> has pointed out that there are plenty of things we do not yet understand, like $z_i\in\mathbb R^a$ for $a\in\mathbb Q_+^\star\setminus\mathbb N$. There may very well exist such a number, but in a field we fail to understand so far.</p>
<p>Similarly, the triangle inequality clearly cannot coexist with such numbers as it is. For the triangle inequality to exist, someone has to figure out how to make triangles with non-positive/real lengths.</p>
<p>As for the <a href="https://en.wikipedia.org/wiki/Norm_(mathematics)#Definition" rel="nofollow noreferrer">properties/axioms of the norm</a> I want:</p>
<p>$$p(v)=0\implies v=0$$</p>
<p>$$p(av)=|a|p(v)$$</p>
| Anixx | 2,513 | <p>When doing operations on complex or split-complex numbers, one can encounter a result which has negative modulus. In that case we consider it equal to a number with positive modulus but argument shifted by <span class="math-container">$2\pi$</span>.
It is also of note that in split-complex numbers (tessarines) the modulus can be imaginary: <span class="math-container">$|a+bj|=\sqrt{a^2-b^2}$</span>.
Thus, <span class="math-container">$|j|=i$</span>.</p>
|
288,340 | <p>I am having difficulty understanding the recursive definition of a language. The problem asked how to write this non recursively. But I want to understand just how a recursive definition of a language works.</p>
<p>Recursive definition of a subset of L of $\{a,b\}^*$.</p>
<p>Basis : $a\in L$ </p>
<p>Recursive Definition : for any $x\in L$, $ax$ and $xb$ are in $L$.</p>
<p>Below is my attempt at explaining the recursive definition.</p>
<p>Starting with the basis $a$ saying that $ax$ is in it means that all strings formed such as $\{a,aa,aaa,...\}$ are present.</p>
<p>Defining all $xb$ represents $\{ab,abb,abbbb,...\}$</p>
<p>The answer I have so far is $\{a\}^*\{b\}^*$ but again it is understanding it that I am after.</p>
| Clive Newstead | 19,542 | <p>Pretend you're a computer.</p>
<p><strong>Step 0.</strong> You start with $\{ a \}$.</p>
<p>Apply the recursive definitions to each word you have so far. This gives you $aa$ and $ab$, so add them to your list.</p>
<p><strong>Step 1.</strong> You now have $\{ a, aa, ab \}$.</p>
<p>Applying the recursion again to each word gives you $aa,ab,aaa,aab,aab,abb$. There's some duplication going on here, but that doesn't matter.</p>
<p><strong>Step 2.</strong> You now have $\{ a, aa, ab, aaa, aab, abb \}$.</p>
<p>Do it again. This gives you
$$aa,ab,aaa,aab,aab,abb,aaaa,aaab,aaab,aabb,aabb,abbb$$
so append these to your list, ignoring duplicates as before.</p>
<p><strong>Step 3.</strong> You now have $\{ a,aa,ab,aaa,aab,abb,aaaa,aaab,aabb,abbb \}$.</p>
<p>Do you see a pattern emerging? Try and guess the general form of a word in this language and then prove by induction on the 'step' above (a.k.a. structural induction) that your guess is correct.</p>
|
2,602,799 | <p>This is exactly what is written in Walter Rudin chapter 2, Theorem 2.41:</p>
<p>If $E$ is not closed, then there is a point $\mathbf{x}_o \in \mathbb{R}^k$ which is a limit point of $E$ but not a point of $E$. For $n = 1,2,3, \dots $ there are points $\mathbf{x}_n \in E$ such that $|\mathbf{x}_n-\mathbf{x}_o| < \frac{1}{n}$. Let $S$ be the set of these points $\mathbf{x}_n$. Then $S$ is infinite (otherwise $|\mathbf{x}_n-\mathbf{x}_o|$ would have a constant positive value, for infinitely many $n$), $S$ has $\mathbf{x}_o$ as a limit point, and $S$ has no other limit point in $\mathbb{R}^k$. For if $\mathbf{y} \in \mathbb{R}^k, \mathbf{y} \neq \mathbf{x}_o$, then
\begin{align}
|\mathbf{x}_n-\mathbf{y}| \geq{} &|\mathbf{x}_o-\mathbf{y}| - |\mathbf{x}_n-\mathbf{x}_o|\\
\geq {} & |\mathbf{x}_o-\mathbf{y}| - \dfrac{1}{n} \geq \dfrac{1}{2} |\mathbf{x}_o-\mathbf{y}|
\end{align}
for all but finitely many $n$. This shows that $\mathbf{y}$ is not a limit point of $S$.</p>
<p>The question:</p>
<p>I'm stuck in understanding the reason behind why $S$ is infinite.
Also I need clarification why the last inequality holds. May someone help, please?</p>
| Michael Hardy | 11,667 | <p>Suppose $S$ is finite. Then the set $\{\, |\mathbf x - \mathbf x_o| : \mathbf x \in S \,\}$ is a finite set of strictly positive numbers. Thus it has a smallest member, which is a positive number. Choose $m$ so large that $1/m$ is less than that smallest number. Then no points of $E$ are within a distance $1/m$ of $\mathbf x_o,$ so $\mathbf x_o$ would not be a limit point.</p>
<p>Let's parse the part about having "constant positive value for infinitely many $n.$"</p>
<p>If a set is finite, then a sequence of members of that set has "constant positive value for infinitely many" terms. For example, consider the digits $0,1,2,3,4,5,6,7,8,9.$ The decimal expansion of an irrational number cannot contain only finitely many of each of those.</p>
<p>If a particular positive number occurs infinitely many times in a sequence, then the sequence cannot approach $0.$ But the sequence of distances $( |\mathbf x_n - \mathbf x_o| )_{n=1}^\infty$ approaches $0.$</p>
<p>That last inequality holds if $n$ is large enough, since in that case $1/n$ is small.</p>
|
1,940,448 | <p>I am stuck on this question. Could someone help me?</p>
<p>$$ \text{Find value of } S = \displaystyle\sum_{n=0}^{\infty} \cfrac{1}{n!(n+2)} $$</p>
<p>I am supposed to show that $ S = 1 $ in two ways: <br /><br />
1) Integrate the taylor series of $ xe^x $ <br />
2) Differentiate the taylor series of $ \frac{e^{x-1}}{x} $</p>
<p>For (1), I tried to using the fact that the taylor series of
$ e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} $</p>
<p>Now, multiplying $ x $ into the taylor series gives:
$$ xe^x = \displaystyle\sum_{n=0}^{\infty} \cfrac{x^{n+1}}{n!} $$</p>
<p>Integrating this yields the following:</p>
<p>$$ \begin{align} \int_0^x xe^x &= \displaystyle\int_0^x\sum_{n=0}^{\infty} \cfrac{x^{n+1}}{n!} \\ &= \displaystyle\sum_{n=0}^{\infty} \cfrac{x^{n+2}}{n!(n+2)} \end{align} $$</p>
<p>I am not sure how to cary on from here.</p>
<p>For (2), I am not sure how to find the taylor series for $ \frac{(e^x - 1)}{x} $</p>
<p>Is anyone able to assist me?</p>
| Jacky Chong | 369,395 | <p>Observe
\begin{align}
\int^1_0 xe^x\ dx = \sum^\infty_{n=0} \int^1_0\frac{x^{n+1}}{n!}\ dx = \sum^\infty_{n=0} \frac{1}{n!(n+2)}
\end{align}</p>
|
4,263,629 | <blockquote>
<p>Let <span class="math-container">$A=\{(x,y) \in \Bbb R^2 \mid x \ge 1, 0<y<\frac{1}{x^2}\}$</span>. Show that <span class="math-container">$m_2(A) < \infty$</span> where <span class="math-container">$m$</span> is the Lebesgue measure.</p>
</blockquote>
<p>I now that the integral <span class="math-container">$\int_{1}^\infty \frac{1}{x^2} dx = 1$</span> is it true that <span class="math-container">$m_2(A)$</span> is less than or equal to the Riemann integral?</p>
<p>Another approach would be to cover the set <span class="math-container">$A$</span>. For this I considered <span class="math-container">$I_k=(k,k+\varepsilon) \times (0, \frac{1}{\varepsilon^2})$</span>. Now <span class="math-container">$\ell(I_k) =\frac{k}{\varepsilon^2} + \frac{1}{\varepsilon}$</span> but the sum of these isn't very nice so I guess I need to find another cover?</p>
| principal-ideal-domain | 131,887 | <p>If you want to cover it simply use rectangles of width <span class="math-container">$1$</span> and height of the maximum of the function for that inverval of length <span class="math-container">$1$</span>. So you have
<span class="math-container">$$m_2(A) \le \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$</span></p>
|
2,961,796 | <p>I was trying to solve the inequality <span class="math-container">$$a-\sqrt[3]{a^3-c\cdot a^2}<b-\sqrt[3]{b^3-c\cdot b^2}$$</span> where <span class="math-container">$a>b>0$</span> and <span class="math-container">$c>0$</span>. I managed to pack the part inside the cube root: <span class="math-container">$$a-\sqrt[3]{a^2(a-c)}<b-\sqrt[3]{b^2(b-c)}$$</span> but I'm stuck after that. Can anybody help prove/disprove the inequality?</p>
| Martin R | 42,969 | <p>For fixed <span class="math-container">$c > 0$</span>, the function <span class="math-container">$f(x) = x-\sqrt[3]{x^3-c x^2}$</span> satisfies</p>
<ul>
<li><span class="math-container">$f(0) = 0$</span>,</li>
<li><span class="math-container">$f(c) = c$</span>,</li>
<li><span class="math-container">$\lim_{x \to \infty} f(x) = \frac c3$</span>,</li>
</ul>
<p>therefore <span class="math-container">$f$</span> is not monotonic on <span class="math-container">$(0, \infty)$</span>.</p>
|
156,179 | <p>Let $A$ be a closed subset of $\mathbb{R}^{n}$. Can the quotient space $\mathbb{R}^{n}/A$ be embedded in some Euclidean space $\mathbb R^{m}$? In particular, assume that $A$ is an algebraic variety of degree $k$, can we control $m$ in term of $n$ and $k$? </p>
| Joseph Van Name | 22,277 | <p>I shall prove the case for $S^{n}$ rather than $\mathbb{R}^{n}$ since $S^{n}$ works better being compact (for closed non-compact sets of $\mathbb{R}^{n}$ it does not work; see Daniele Zuddas's answer). Suppose that $A$ is a closed subset of $S^{n}\subseteq\mathbb{R}^{n+1}$. Then define a mapping $f:S^{n}\rightarrow\mathbb{R}^{n+1}$ by letting $f(x)=d(x,A)\cdot x$ where $d(x,A)$ denotes the distance from $x$ to $A$. Then clearly $f$ is a continuous map. Since $S^{n}$ is compact, the mapping $f$ is a quotient map and $f(x)=f(y)$ if and only if $x=y$ or $x,y\in A$. Therefore $S^{n}/A\simeq f[S^{n}]$.</p>
|
19,598 | <p>I have two independent ODE systems. </p>
<pre><code>A = NDsolve[..., {x, y}, {t, 0, 10}];
B = NDsolve[..., {a, b}, {t, 0, 10}];
</code></pre>
<p>I can draw a <code>ParametricPlot</code> from one ODE. That is, </p>
<pre><code>ParametricPlot[Evaluate[{x[t], y[t]} /. A], {t, 0, 10}]
</code></pre>
<p>I wonder if I can draw a <code>ParametricPlot</code> from the two independent ODE systems. That is a <code>ParametricPlot</code> of <code>x[t]</code> taken from A and <code>a[t]</code> taken from B.</p>
| ssch | 1,517 | <p>Sure you can:</p>
<pre><code>sol1 = NDSolve[{x'[t] == Sin[t], x[0] == 1}, x, {t, 0, 10}];
sol2 = NDSolve[{a'[t] == Cos[t], a[0] == 1}, a, {t, 0, 10}];
ParametricPlot[Evaluate[{x[t], a[t]} /. Flatten@{sol1, sol2}], {t, 0, 10}]
</code></pre>
<p>The <code>Flatten</code> is there for the following reason:</p>
<pre><code>{x[t], a[t]} /. {sol1, sol2}
(* {{{InterpolatingFunction[{{0.,10.}},<>][t],a[t]}},
{{x[t],InterpolatingFunction[{{0.,10.}},<>][t]}}} *)
{x[t], a[t]} /. Flatten@{sol1,sol2}
(* {InterpolatingFunction[{{0.,10.}},<>][t],
InterpolatingFunction[{{0.,10.}},<>][t]} *)
</code></pre>
<p>You can also do:</p>
<pre><code>ParametricPlot[Evaluate[{x[t] /. First@sol1, a[t] /. First@sol2}], {t, 0, 10}]
</code></pre>
|
185,061 | <p>Let $p$ be a prime number. For which finite $p$-groups $H$ is there a finite $p$-group $G$ such that $[G,G] \cong H$?</p>
| Yakov | 83,645 | <p>(Essentially, Burnside) If $H$ is a $p$-group containing a nonabelian characteristic subgroup with cyclic center, then there is no $p$-group $G$ such that $H$ is a $G$-invariant subgroup of $\Phi(G)$. In particular, $H\ne G'$, $H\ne \Phi(G)$. Next, if a two-generator $H$ is $G$-invariant subgroup of $\Phi(G)$, then $H$ is metacyclic.</p>
|
2,728,248 | <blockquote>
<p>Let $K=\mathbb{Q}(\sqrt{-2})$. Show that $\mathcal{O}(K)$ is a principal ideal domain. Deduce that every prime $p\equiv 1, 3$ (mod 8) can be written as $p = x^2 + 2y^2$ with $x, y \in \mathbb{Z}$.</p>
</blockquote>
<p>As $−2$ is squarefree $6\equiv 1$ (mod 4) we have $\mathcal{O}(K) = \mathbb{Z}[
\sqrt{2}$]. The
discriminant is $\Delta$ = −8. The degree $n = 2$. The signature is $(0, 2)$. Thus the Minkowski bound is</p>
<p>$$ B_k = \frac{2!}{2^2}=\frac{4}{\pi}\times \sqrt{8} = \frac{4\sqrt{2}}{\pi}<2$$</p>
<p>Hence $Cl(K)$ is generated by the empty set of ideal classes and so $Cl(K) = \{1\}$. So this means $\mathcal{O}(K)$ is a principal ideal domain I believe...</p>
<p>Ok, now if we let $p \equiv 1$ or $3$ (mod 8). By quadratic reciprocity, $−2 ≡ \alpha^2$
(mod p) for
some integer $\alpha$. Thus</p>
<p>$$X^2 + 2 ≡ (X + \alpha)(X − \alpha) \quad\text{(mod}~~ p).$$</p>
<p>Ok, now I am slightly stuck, can we apply some theorem here? Not sure if what above is correct to get to the desired result</p>
| Rene Schipperus | 149,912 | <p>OK, to finish what you have started, which I believe is the question, you know that $\mathbb{Z}[\sqrt{-2}]$ is a principal ideal domain and thus a unique factorization domain. So given $p$, is it a prime in $\mathbb{Z}[\sqrt{-2}]$ ?
You have also that for $p\equiv 1,3(\mod 8)$ that
$$a^2\equiv -2(\mod p)$$ meaning that
$$p|a^2+2=(a+\sqrt{-2})(a-\sqrt{-2})$$ however $p\not |(a+\sqrt{-2})$ and
$p\not |(a-\sqrt{-2})$, this contradicts the definition of prime so $p$ is not a prime in $\mathbb{Z}[\sqrt{-2}]$. Thus it has a factorization
$$p=(x+y\sqrt{-2})(x-y\sqrt{-2}).$$</p>
|
2,065,254 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is twice differentiable.</p>
<p>We know that:
$$\lim_{x\to-\infty}\ f(x) = 1$$</p>
<p>$$\lim_{x\to\infty}\ f(x) = 0$$</p>
<p>$$f(0) = \pi$$</p>
<p>We have to prove that there exist at least two points of the function in which $f''(x) = 0$.</p>
<p>How could we do it in a rigorous way? It is pretty intuitive, but in a rigorous way it isn't that simple for me...</p>
| Robert Z | 299,698 | <p>Let $x_0$ be a point where $f$ attains its maximum value (which is $\geq \pi$). At $x_0$ we have that $f''(x_0)\leq 0$. </p>
<p>We claim that there is a point $x_1>x_0$ such that $f''(x_1)=0$. </p>
<p>Assume that $x_1$ does not exist then $f''(x)\not=0$ for $x>x_0$. By <a href="https://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)" rel="nofollow noreferrer">Darboux theorem</a> the sign of $f''(x)$ in $(x_0,+\infty)$ should be always the same. We have two cases:</p>
<p>i) $f''(x)>0$ for all $x>x_0$ that is $f(x)$ is strictly convex in $(x_0,+\infty)$. Hence $f'$ is strictly increasing and $f'(x)>0$ for $x>x_0$ which implies that $f(x)>f(x_0)$ against the fact that $x_0$ is a maximum point.</p>
<p>ii) $f''(x)<0$ for all $x>x_0$ that is $f(x)$ is strictly concave in $(x_0,+\infty)$. Hence $f'$ is strictly decreasing and $f'(x)<0$ for $x>x_0$. Moreover the graph of $f$ remains under the tangent line at some $t>x_0$: for all $x\in (x_0,+\infty)$
$$f(x)\leq f'(t)(x-t)+f(t).$$
Now by taking the limit for $x\to +\infty$ to both sides, we get
$$0\leftarrow f(x)\leq f'(t)(x-t)+f(t)\to -\infty.$$
Contradiction!</p>
<p>By the same argument we get another zero of $f''(x)$ for $x<x_0$.</p>
|
2,645,611 | <blockquote>
<p>Prove that:
<span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span></p>
</blockquote>
<h3>My work so far:</h3>
<p><span class="math-container">$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$</span>
<span class="math-container">$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$</span>
<span class="math-container">$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$</span></p>
<p><em>How to prove the last equality?</em></p>
| farruhota | 425,072 | <p>You can calculate directly:
$$\begin{align} & \frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}=\frac{n!(1+n+1)}{1!}; \\
& \frac{n!(n+2)}{1!}+\frac{(n+2)!}{2!}=\frac{n!(n+2)(2+n+1)}{2!}; \\
& \cdots \\
& \frac{n!(n+2)(n+3)\cdots ((n-1)+n+1)}{(n-1)!}+\frac{(n+n)!}{n!}= \\
& \frac{n!(n+2)(n+3)\cdots (n+n+1)}{n!}=\frac{(2n+1)!}{(n+1)!}.
\end{align}$$</p>
|
2,134,653 | <blockquote>
<p>There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?</p>
</blockquote>
<p>I have tried:</p>
<p>Initally Vessel containing 40 litres of Milk :</p>
<p>4 litres out means -> 36 litres </p>
<p>4 litres of water is poured in - > 4 litres</p>
<p>so Now total quantity is 40 litres </p>
<p>Mixture containing water and Milk in the ratio: 36:4 i.e 9:1</p>
<p>Again 5 litres of Mixture is replaced with the six litres of water</p>
<p>for that:</p>
<p>9x - 9/10*5 : x -1/10*5</p>
<p>Now the Ratio becomes:</p>
<p>90x - 45 : 10x -5 i.e 9x -9:2x -1</p>
<p>six litres of water is added</p>
<p>9x -9 :2x -5</p>
<p>again six litres of Mixture is replaced then</p>
<p>9x -9 - 9/10*6 : 2x -5 -9/10*6</p>
<p>that is </p>
<p>90x -144 :10x -84</p>
<p>after adding six litres of water again we got </p>
<p>90x -144 :10x -78</p>
<p>so Milk containing is </p>
<p>90x-144+10x -78 =41</p>
<p>100x -222 =41</p>
<p>100x = 263</p>
<p>x= 2.63</p>
<pre><code>again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk
total itself 41 litres of Milk Please anyone guide me answer
</code></pre>
<p><strong>what i am doing mistake please anyone guide me for the answer</strong></p>
| Sherlock Watson | 104,084 | <p>Think in the terms of %. In the first iteration, <span class="math-container">$10$</span>% is removed, in the second iteration, <span class="math-container">$12.5$</span>% is removed and in the third iteration, <span class="math-container">$15$</span>% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back. </p>
<p><span class="math-container">$40-10$</span>%<span class="math-container">$=$$36$</span></p>
<p><span class="math-container">$36-12.5$</span>%<span class="math-container">$=$$31.5$</span></p>
<p><span class="math-container">$31.5-15$</span>%<span class="math-container">$=26.775$</span>. </p>
<p>Therefore, after the third iteration, milk will be 26.775 liters. </p>
|
3,898,027 | <p>How do I find the posterior predictive distribution in Bayesian Analysis? I'm not looking for a specific case, I would like the general solution.</p>
| Michael Hardy | 11,667 | <p>Say <span class="math-container">$\Theta$</span> is distributed as <span class="math-container">$p(\theta)\, d\theta,$</span> i.e.
<span class="math-container">$$
\Pr(\Theta\in S) = \int\limits_S p(\theta)\,d\theta
$$</span>
and
<span class="math-container">$$
X_1,\ldots,X_n \mid \Theta=\theta\sim\text{i.i.d. } f_\theta(x)\,dx
$$</span>
i.e.
<span class="math-container">$$
\Pr(X_i\in A\mid \Theta=\theta) = \int\limits_A f_\theta(x)\,dx
$$</span>
and <span class="math-container">$X_1,\ldots,X_n$</span> are conditionally independent given <span class="math-container">$\Theta=\theta.$</span></p>
<p>Then the posterior distribution, i.e. the conditional distribution of <span class="math-container">$\Theta$</span> given <span class="math-container">$X_i=x_i$</span> for <span class="math-container">$i=1,\ldots,n,$</span> is
<span class="math-container">$$
q(\theta)\, d\theta = \text{constant} \times p(\theta) f_\theta(x_1)\cdots f_\theta(x_n)\, d\theta
$$</span>
where the constant is so chosen that <span class="math-container">$\displaystyle \int q(\theta)\,d\theta$</span> (the integral being taken over the whole parameter space) is <span class="math-container">$1.$</span></p>
<p>Then the predictive distribution is given by
<span class="math-container">$$
\Pr(X_{n+1}\in A \mid X_1=x_1,\ldots,X_n=x_n) = \int \left( \, \int\limits_A f_\theta(x)\,dx\right) q(\theta) \, d\theta,
$$</span>
the outer integral again being taken over the whole parameter space.</p>
|
3,898,027 | <p>How do I find the posterior predictive distribution in Bayesian Analysis? I'm not looking for a specific case, I would like the general solution.</p>
| tommik | 791,458 | <p>This answer has been moved from <a href="https://math.stackexchange.com/questions/4148311/help-me-with-the-integral-of-the-posterior-predictive-distribution">here</a></p>
<p>Consider firs that (very easy to prove)</p>
<p><span class="math-container">$$\mathbb{P}[A,C|B]=\mathbb{P}[A|B,C]\cdot\mathbb{P}[C|B]$$</span></p>
<p>Now set</p>
<p><span class="math-container">$A=\{y_{n+1}=\tilde{y}\}$</span></p>
<p><span class="math-container">$B=\{Y_1=y_1,\dots,Y_n=y_n=\mathbf{y}\}$</span></p>
<p><span class="math-container">$C=\theta$</span></p>
<p>and get</p>
<p><span class="math-container">$$\mathbb{P}\{\tilde{y},\theta|\mathbf{y}\}=\mathbb{P}\{\tilde{y}|\theta,\mathbf{y}\}\cdot \mathbb{P}\{\theta|\mathbf{y}\}$$</span></p>
<p>Now to eliminate <span class="math-container">$\theta$</span> in LHS you have to integrate both sides w.r.t. <span class="math-container">$\theta$</span> and thus, considering the conditional independence (subordinated to <span class="math-container">$\theta$</span>) of the observations, you get:</p>
<p><span class="math-container">$$\mathbb{P}\{\tilde{y}|\mathbf{y}\}=\int_{\Theta}\mathbb{P}\{\tilde{y}|\theta\}\cdot \mathbb{P}\{\theta|\mathbf{y}\}d\theta$$</span></p>
<p>Where</p>
<ul>
<li><p><span class="math-container">$\mathbb{P}\{\tilde{y}|\theta\}$</span> is the statistical model</p>
</li>
<li><p><span class="math-container">$\mathbb{P}\{\theta|\mathbf{y}\}$</span> is the posterior</p>
</li>
</ul>
<hr />
<p><a href="https://en.wikipedia.org/wiki/Conditional_independence" rel="nofollow noreferrer">Conditional independence</a> of the observations given <span class="math-container">$\theta$</span></p>
<p><a href="https://i.stack.imgur.com/EcSGP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EcSGP.jpg" alt="enter image description here" /></a></p>
<p>... in your case this means that</p>
<p><span class="math-container">$$\mathbb{P}[\tilde{y}|\mathbf{y},\theta]=\mathbb{P}[\tilde{y}|\theta]$$</span></p>
<hr />
<p><strong>A simple example:</strong></p>
<p>Let's suppose to have the following model (bernulli)</p>
<p><span class="math-container">$$\theta^x(1-\theta)^{1-x}$$</span></p>
<p>with <span class="math-container">$x=0,1$</span> and <span class="math-container">$\theta \in (0;1)$</span></p>
<p>As per we said before, the probability of a "future success" given a certain number of previous results is</p>
<p><span class="math-container">$$\mathbb{P}[\tilde{y}=1|\mathbf{y}]=\int_0^1 \theta p(\theta|\mathbf{y}) d\theta= \mathbb{E}[\theta|\mathbf{y}]$$</span></p>
<p>and similarly</p>
<p><span class="math-container">$$\mathbb{P}[\tilde{y}=0|\mathbf{y}]=\int_0^1 (1-\theta) p(\theta|\mathbf{y}) d\theta= 1-\mathbb{E}[\theta|\mathbf{y}]$$</span></p>
|
1,838,002 | <p>There is famous <a href="https://en.wikipedia.org/wiki/Quillen-Suslin_theorem" rel="nofollow">Quillen-Suslin theorem</a> which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free.</p>
<p>I have never carefully read a proof of this theorem, which is for example in the <strong>Lang's Algebra</strong>. <code>Probably it is based on Quillen's original ideas.</code>- this is not true as it was pointed out in the answer below.</p>
<p><strong>Questions:</strong> Is every finitely generated projective modules over $\mathbb{Z}[x_1,...,x_n]$ free? </p>
<p>If yes, then is the proof modification of the one given in <strong>Lang's Algebra</strong>?</p>
<p>And if yes, then how about polynomial rings over other Dedekind domains or number rings?</p>
| syzygy | 349,357 | <p>Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$.</p>
<p><em>Example.</em> (i) If $A$ is local Noetherian then $(S)$ holds.
(ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $\mathrm{Pic}(A)=0$, which is the case iff $A$ is a pid.</p>
<p><strong>Theorem.</strong>(Grothendieck) If $A$ is a regular ring then we have a canonical isomorphism $K_0(A)\rightarrow K_0(A[(T_i)_{i\in I}])$.</p>
<p>Thus a necessary condition for $(S)$ to hold for $A[T_1,\ldots,T_n]$ is $K_0(A)=\mathbf{Z}$.</p>
<p><em>Corollary.</em> Let $A$ be a Dedekind domain. Then $(S)$ holds for $A[T_1,\ldots,T_n]$ if and only if $\mathrm{Pic}(A)=0$, i.e. iff $A$ is a pid.</p>
<p>(This follows from what we said previously and Quillen's theorem stating that if $A$ is a pid, then $A[T_1,\ldots,T_n]$ satisfies $(S)$.)</p>
<p>(<em>Remark.</em> Quillen's original proof is different from the one given in Lang!)</p>
|
64,406 | <p>It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$. </p>
<p>I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these? </p>
| KCd | 619 | <p>See www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf, which discusses groups of order p^3 for any prime p and treats the case p = 2 first.</p>
|
226,323 | <p>Let $X$ and $Y$ be complex Banach spaces and $B(X,Y)$ be the Banach space
of all bounded operators. An operator $T\in B(X,Y)$ is weakly compact if
$T(\{ x\in X;\; \| x\| \leq 1\})$ is relatively compact in the weak topology
of $Y$. If $X$ or $Y$ is reflexive, then every operator in $B(X,Y)$ is weakly
compact. I guess that the converse holds as well: if every operator in $B(X,Y)$
is weakly compact then either $X$ or $Y$ has to be reflexive. But I cannot
find a satisfactory argument for this. I know about some characterizations of weakly
compact operators: factorization through a reflexive Banach space or continuity with
respect to the right topology in $X$. However it seems to me that there must be a
simple argument which forces that either $X$ or $Y$ is reflexive if every operator
in $B(X,Y)$ is weakly compact. I am asking if someone knows this simple argument
or if there is a paper with an answer to my question.</p>
| Friedrich Philipp | 84,182 | <p>I don't know if the following helps:</p>
<p>Let $B(X,Y) = W(X,Y)$. Then the following hold.</p>
<p>(i) If there exists a surjection $T\in B(X,Y)$ then $Y$ is reflexive.</p>
<p>(ii) If there exists an injection $T\in B(X,Y)$ with closed range then $X$ is reflexive.</p>
<p>Indeed, denote by $B_r$ and $K_r$ the open and closed ball around $0$, respectively. By the open mapping theorem, $TB_1$ is open. Thus, there exists $r > 0$ such that $K_r\subset TB_1\subset TK_1$, i.e., $K_r$ is weakly compact in $Y$, meaning that $Y$ is reflexive. If $T$ is an injection with closed range then a similar argument as above shows that $\operatorname{ran} T$ is reflexive. But as $T : X\to\operatorname{ran} T$ admits a bounded inverse, $X$ is reflexive.</p>
|
641,744 | <p>I am obliged to find rank of matrix $A$ depending on $ p $.</p>
<p>I know how to do this using Gauss elimination method but I would like to try solve this using minors. I know that the rank of matrix is equal to degree of the biggest non-zero minor.</p>
<p>$A=\left(
\begin{array}{ccc}
p-1&p-1&1&1\\
1&p^{2}-1&1&p-1\\
1&p-1&p-1& 1
\end{array}
\right)$</p>
<p>So I am taking this sub-matrix:
$A_{1}=\left(
\begin{array}{ccc}
p-1&1&1\\
1&1&p-1\\
1&p-1& 1
\end{array}
\right)$</p>
<p>I am counting $det(A_{1})=-p^{2}+5p-6$</p>
<p>$-p^{2}+5p-6=0 \iff p=2 \lor p = 3 $</p>
<p>So now I got two values when the rank of matrix is not equal to 3. No I will test rank for $p=2$ and $p=3$ and so on. Do I have to check only one minor? Am I doing this right?</p>
| Robert Israel | 8,508 | <p>Yes, you are right so far. Since the matrix has $3$ rows, its rank is at most $3$,
and you know that for $ p \notin \{2,3\}$ the minor you considered is nonzero, making the rank $3$. Then you look at the cases $p=2$ and $p=3$ (but there you will have to look at other minors).</p>
|
4,347,174 | <p>In the book Superlinear Parabolic Problems Blow-up, Global Existence and Steady States, page 493 this equation appears in which the book says it uses Young's inequality</p>
<p><span class="math-container">$$ |\Omega| u^p \leq ku^q +\epsilon(k)u $$</span></p>
<p>where <span class="math-container">$\epsilon(k) = C(|\Omega|,p,q) k^{{-(p-1)}/{(q-p)}}$</span>.
The only things that are given are that <span class="math-container">$p<q$</span> (<span class="math-container">$p$</span> and <span class="math-container">$q$</span> are not conjugate exponents) and <span class="math-container">$k>0$</span> (<span class="math-container">$k$</span> can be taken large enough). I am failing to see how this is an application of classical <a href="https://en.wikipedia.org/wiki/Young%27s_inequality_for_products" rel="nofollow noreferrer">Young inequality</a>.
Is this really an application of the classic young inequality or is it an application of some modification of it?</p>
<p><strong>attempt:</strong> The best I've achieved so far using Young's inequality was</p>
<p><span class="math-container">$$ u^p = [\left( k \frac{(p-1)}{q} \right)^{-(p-1)/q}u][(\left( k \frac{(p-1)}{q} \right)^{(p-1)/q}u^{p-1}] \leq \dfrac{q}{(p-1)}[\left( k \frac{(p-1)}{q} \right)^{(p-1)/q}u^{p-1}]^{q/(p-1)} + \dfrac{q}{(q-p+1)}[\left( k \frac{(p-1)}{q} \right)^{-(p-1)/q}u]^{q/(q-p+1)} \leq ku^q+Ck^{-(p-1)/(q-p+1)}u^{q/(q-p+1)}.$$</span></p>
| timur | 2,473 | <p>The trick is to use the splitting
<span class="math-container">$$
u^p = u^ru^{p-r}
$$</span>
with
<span class="math-container">$$
r = \frac{p-1}{q-1}q,
$$</span>
and apply the Young inequality with the exponents
<span class="math-container">$$
\frac{q-1}{p-1} \qquad\textrm{and}\qquad \frac{q-1}{q-p}.
$$</span></p>
|
49,339 | <p>Are there any known non-slow methods for solving diophantine systems?</p>
<p>I can't find books of mathematics that appear methods explaining how to solve diophantine systems in a manner "not slow", e.g. not force brute enumeration.</p>
| Felipe Voloch | 2,290 | <p>You must not have looked very hard. This will get you started:</p>
<p>Nigel P. Smart. The Algorithmic Resolution of Diophantine Equations. London Mathematical Society Student Texts 41. Cambridge University Press, 1998.</p>
<p>However, as Igor and zroslav have mentioned, some problems are unsolvable, others believed to be very hard, so don't be surprised that there are no "easy" methods.</p>
|
1,137,336 | <p>This is from Discrete Mathematics and its applications
<img src="https://i.stack.imgur.com/jjJiF.png" alt="enter image description here"></p>
<p>I was able to get sum pretty easy. </p>
<p>I am trying to follow this example in the book to get the product of the two binary numbers
<img src="https://i.stack.imgur.com/uce6O.png" alt="enter image description here"></p>
<p>Here's my work so far <img src="https://i.stack.imgur.com/vpA9Z.jpg" alt="enter image description here"></p>
<p>I got the expected output from my calculator windows application. Does anyone see what the issue is? The problem starts when I don't get 7 zeros separating the first two ones. What would you do with the end result, 8? I am still not sure about that. </p>
| Laurent Hayez | 157,160 | <p>Okay let's take a simple example: suppose you want to add $(111)_2 + (111)_2 + (111)_2$. First of all, all the carries will be written in binary, and not in decimal. Here is how you should do that:</p>
<pre><code> 1
Carry: 1101
111
+111
+111
----
10101
</code></pre>
<p><strong>Explanation:</strong> On the right most column, you first add $1+1+1=11_2$. So you note 1 in the result line and note $1$ as you first carry.<br>
Then you add $1+1+1+1=100_2$, therefore you write one $0$ in the result line, and you have a carry of $10$. To handle that carry, you can simply write $10$ in the carry over two columns (that is the 0 will be above one column, and the one above the column on the left).<br>
The next addition is $0+1+1+1=11_2$ so once again you write $1$ in the result line and carry $1$ (which I wrote above the $1$ from the preceding $10$) and the rest is easy.</p>
|
2,922,881 | <p>I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$</p>
<p>The point $z_0=0$ seems to be a singularity. I'm not sure but I think it's also a non-removable one. I tried using the Taylor expansion of $e^x$ and $\cos{x}$ as that usually helps. </p>
<p>$$e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}z+\frac{1}{2!z^2}+\frac{1}{3!z^3}+...)(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-...)\\=>e^{\frac{1}{z}}\cos{\frac{1}{z}}=(1+\frac{1}{z}+...)$$</p>
<p>It seems like the negative power terms are infinite showing that $z_0=0$ is no pole. If I'm correct, the coefficient of $1/z$, which is $1$, is the residue of the singularity and this leads to the result:$$ \int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz=2\pi i$$ </p>
<p>I don't think I've evaluated other integrals with non-removable singularities and I'm not sure about the whole process..</p>
| Alan Muniz | 289,217 | <p>You are right. It is an essential singularity with residue $1$. You can find a primitive (defined away from zero) for $$ e^{\frac{1}{z}}\cos{\frac{1}{z}} - \frac{1}{z} $$ just using that $\displaystyle\frac{z^{1-n}}{1-n}$ is a primitive for $z^{-n}$, with $n>1$. Hence the
$$
\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z} - \frac{1}{z}}dz= 0
$$
And the only term of $e^{\frac{1}{z}}\cos{\frac{1}{z}}$ that contributes to the integral is $\frac{1}{z}$. This type of argument appears in the proof of the Residue Theorem.</p>
|
3,460,843 | <p>I understand that the way to calculate the cube root of <span class="math-container">$i$</span> is to use Euler's formula and divide <span class="math-container">$\frac{\pi}{2}$</span> by <span class="math-container">$3$</span> and find <span class="math-container">$\frac{\pi}{6}$</span> on the complex plane; however, my question is why the following solution doesn't work. </p>
<p>So <span class="math-container">$(-i)^3 = i$</span>, but why can I not cube root both sides and get <span class="math-container">$-i=(i)^{\frac{1}{3}}$</span>. Is there a rule where equality is not maintained when you root complex numbers or is there something else I am violating and not realizing?</p>
| Saketh Malyala | 250,220 | <p>Yup, <span class="math-container">$\tan^{-1}$</span> is a great start.</p>
<p>So once you have <span class="math-container">$\displaystyle -3\int \frac{1}{x^2+4}\,dx$</span>, you can turn this into <span class="math-container">$\displaystyle -\frac{3}{4}\int\frac{1}{(\frac{x}{2})^2+1}\,dx=-\frac{3}{2}\int\frac{\frac{1}{2}}{(\frac{x}{2})^2+1}\,dx$</span>.</p>
<p>You see the perfect u-substitution yet?</p>
|
1,705,159 | <blockquote>
<p>Find necessary and sufficient conditions for a Mobius transformation <span class="math-container">$T(z)=\frac{az+b}{cz+d}$</span> to map the unit circle to itself. So if <span class="math-container">$\gamma$</span> is a circle, <span class="math-container">$T(\gamma)=\gamma$</span>.</p>
<p>I've worked out the necessary conditions. Namely, if <span class="math-container">$T(\gamma)=\gamma$</span>, then</p>
<ol>
<li><p><span class="math-container">$|a|^2+|b|^2=|c|^2+|d|^2$</span></p>
</li>
<li><p><span class="math-container">$a\bar{b}=\bar{d}c$</span></p>
</li>
<li><p><span class="math-container">$\bar{a}b=d\bar{c}$</span></p>
</li>
</ol>
<p>Source: Conway's Complex Functions of One Variable</p>
</blockquote>
<p>How does one go about showing sufficiency? Should I simply assume conditions 1),2) and 3) and try to prove that <span class="math-container">$T(\gamma)=\gamma$</span>? If so I can simply claim that all the implications I used to get these conditions also work backwards. Or just show that <span class="math-container">$|\frac{az+b}{cz+d}|=1$</span> by these conditions, which is rather simple. Is that all there is to this? I just wish the whole "necessary and sufficient" language was scrapped for some direct notation.</p>
<p>As an aside, I'm wondering if I'm using the words "necessary" and "sufficient" correctly in this context. Is what I showed in the first part the necessary conditions (that's what makes sense to me semantically, because they are necessary once I've assumed the map), or are they the sufficient conditions?</p>
| chin | 901,265 | <p>Consider the mapping of the complex plane G(w) =2w^2 on itself. What happens to the unit circle (consisting of complex numbers such that |z|=1)? How many preimages does one point have with this mapping?</p>
|
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).</p>
<p>Thanks for any help.</p>
| Florian | 1,609 | <p>Although in most cases the words function and mapping can be used interchangeably, in several parts of mathematics differences in emphasis, especially in analysis and differential geometry. I can think of two.</p>
<p>First, especially in differential geometry, "mapping" is the universal word, and the word "function" is used for mappings that map to $\mathbb{R}$ or $\mathbb{C}$. Thus a mapping which maps to $\mathbb{R}^n$ for instance would not be called a function. This convention is not always adhered to (you might occasionally read about "vector-valued functions"), but this is the usual interpretation.</p>
<p>Second, especially in analysis, it is not uncommon to call members of $L^p$ "functions", even though they are actually equivalence classes of mappings. Again the idea is that functions should assign numbers to some objects (e.g. points in some space) in a suitable sense. Thus functions are thought of being objects studied in analysis, whereas "mapping" is thought of being a term from set theory.</p>
|
95,741 | <p>I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).</p>
<p>Thanks for any help.</p>
| Community | -1 | <p>From P216 of Mathematical Proofs by Gary Chartrand:</p>
<blockquote>
<p>By a function f from A to B, written f : A → B, we
mean a relation from A to B with the property that every element a in A is the first
coordinate of exactly one ordered pair in f.
...</p>
<p>If (a, b) ∈ f , then we write b = f (a) and refer to b as the image of a. Sometimes f is
said to map a into b. Indeed, f itself is sometimes called a <strong>mapping</strong>.</p>
</blockquote>
|
270,641 | <p>I want to find the inverse triple Laplace transform of <span class="math-container">$L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right]$</span>. I did
<span class="math-container">\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{1}{a} L^{-1}_{x_{1}} \left[ \frac{a}{s^2_{1} + a^2} \right] \right] \right], \ \ a^2 = s^2_{2} + s^2_{3}
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[\frac{ \sin \left( x_{1} \sqrt{ \left( s^2_{2} + s^2_{3}\right)} \right) }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&= (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ \displaystyle\sum_{k = 0}^{\infty} \frac{(-1)^k \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^{2k+1}}{(2k+1)!} }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \frac{ x_{1} \sqrt{s^2_{2} + s^2_{3}} - \frac{1}{6} \left(x_{1} \sqrt{s^2_{2} + s^2_{3}} \right)^3 }{\sqrt{ \left( s^2_{2} + s^2_{3}\right)}} \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ x_{1} - \frac{1}{6} x_{1}^3 \left( s^2_{2} + s^2_{3} \right) \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) - \frac{1}{6} x_{1}^3 s^2_{3} \right] \right]
\\
&\approx (-1) L^{-1}_{x_{3}} \left[ \left( x_{1} - \frac{1}{6} x_{1}^3 s^2_{3} \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \right]
\\
&\approx (-1) \left( \left( x_{1} \delta(x_{3}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{3}) \right) \delta(x_{2}) - \frac{1}{6} x_{1}^3 \delta^{"}(x_{2}) \delta(x_{3}) \right)
\end{align*}</span>
I am wondering if this solution is correct or not? and if it is incorrect, what should I do to get the correct solution? I would appreciate your help.</p>
| user64494 | 7,152 | <p>Your result seems incorrect in view of</p>
<pre><code>LaplaceTransform[-(x1*DiracDelta[x3]*DiracDelta[x2] -
1/6*x1^3 DiracDelta''[x3]*DiracDelta[x2] -
1/6 x1^3*DiracDelta''[x2]*DiracDelta[x3]), {x1, x2, x3}, {s1, s2, s3}]
</code></pre>
<blockquote>
<p><code>-(1/s1^2) + s2^2/s1^4 + s3^2/s1^4</code></p>
</blockquote>
|
3,318,130 | <p>Let <span class="math-container">$(a_j)_{j \in \mathbb{N}}$</span> and <span class="math-container">$(b_j)_{j \in \mathbb{N}}$</span> two real valued sequences such that <span class="math-container">$a_j \nearrow +\infty$</span> and <span class="math-container">$b_j \nearrow +\infty$</span>.
Is it possible to extract subsequences <span class="math-container">$(a_{j_i})_i$</span> and <span class="math-container">$(b_{j_i})_i$</span> such that <span class="math-container">$(\frac{a_{j_i}}{b_{j_i}})_i$</span> converges to a positive finite number? </p>
<p>Any help will be very apreciated! Thanks!!</p>
| Kavi Rama Murthy | 142,385 | <p>No. If <span class="math-container">$a_j >jb_j$</span> then <span class="math-container">$\frac {a_{j(i)}} {b_{j(i)}} >j(i)$</span> and <span class="math-container">$j(i)\to \infty$</span> as <span class="math-container">$ i \to \infty$</span> whatever be the subsequence <span class="math-container">$(j(i))$</span>. </p>
|
3,625,069 | <p>How to solve <span class="math-container">$10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$</span>?</p>
<p>I tried to solve this problem by letting <span class="math-container">$x=10\sqrt{10\sqrt[3]{10\sqrt[4]{10...}}}$</span> to observe the pattern.</p>
<p>Based on the pattern, the result is</p>
<p><span class="math-container">$\dfrac{x^{n!}}{10^{((((1)(2)+1)4+1)...n+1)}}$</span> where <span class="math-container">$n$</span> is a positive integer approaching infinity. </p>
<p>This is where I got stuck.</p>
| Matteo | 686,644 | <p>We can rewrite your expression as:
<span class="math-container">$$S=10\cdot(10\sqrt{...})^{\frac{1}{2}}=10\cdot10^{\frac{1}{2}}\cdot(10\sqrt[3]{...})^{{\frac{1}{2}}\cdot{\frac{1}{3}}}=10^{\frac{1}{2}}\cdot10^{{\frac{1}{2}}\cdot{\frac{1}{3}}}\cdots10^{\frac{1}{n!}}$$</span>
Now, using the rule of exponents, we have:
<span class="math-container">$$S=10^{\sum{i=1}^{\infty}\frac{1}{i!}}=10^{e-1}$$</span>
because <span class="math-container">$$\lim_{\xi\to\infty}\sum_{i=1}^\xi\dfrac1{i!}=e-1$$</span></p>
|
365,808 | <p>Sorry if something like this has already been asked, I searched but I couldn't find anything similar to my question.</p>
<p>I'm a senior undergraduate and currently doing my senior thesis. My senior thesis is not original work, however it's quite demanding and I'm learning a lot of high level topics. I have been lurking around arxiv and started reading "Solved and unsolved problems in Number theory" by Daniel Shanks. My plan is to work on some open problems and play around with them so that I can try to get a publication before I graduate. My main reason for trying to get a publication is to increase my chances to get into a good graduate program (my GPA is not that great and I don't have the money to apply to many programs, so unless I publish something I'll probably only apply to safety schools).</p>
<p>With that being said if I were to do original work, how would I go about publishing? I might end up modifying a problem too much and proving something that might not be interesting, so I feel it'll get rejected from a journal for not being profound. I will also attack problems with all I know, so I might also end up using some heavy tools that aren't part of an undergraduate curriculum so I don't if i would send them to an undergraduate research journal. Maybe I could just upload on dropbox or arxiv, but then it's not a publication.</p>
<p>I have thought about asking my advisors about this, but I'll rather not since I'm aware I'm probably being overly ambitious and should probably focus on my thesis instead. Which I can agree with, hence I'll probably play around with problems on the weekends only or once a week. I'm also aware I might end up not publishing anything all, however in my mind unless I give it a shot I won't know. Either way I'll have fun and end up learning a lot about research so I don't see a downside.</p>
<p>(In case my background is relevant, my senior thesis is about perfectoid spaces. I've already taken a graduate course on commutative algebra, have taken a basic course on p-adic analysis,started learning about point free topology, already know the basics of category theory, still learning more about algebraic geometry, will learn about adic spaces soon/already know a bit about krull valuations, learning about homological algebra through weibel's book, started reading szamuely's galois theory book, will have to learn about etale cohomology soon, will also learn some things from almost mathematics, etc.)</p>
| David White | 11,540 | <p>There are many undergraduate journals that would not be likely to reject your work as "not profound enough." Basically, if it's written while the author was an undergraduate, and contains anything novel at all (at the level one would expect of an undergraduate), then a journal can be found for it. This includes well-written expository accounts of existing texts, especially if they work out some more examples. The most famous undergrad math journal is Involve, which has higher standards than the others (more like what a grad student or professor might do). But there are plenty of others that accept papers that might not make it to Involve:</p>
<ul>
<li><a href="https://unl.libguides.com/c.php?g=51642&p=333916" rel="noreferrer">List of undergrad math journals</a></li>
<li><a href="https://www.carthage.edu/undergraduate-research/journals/" rel="noreferrer">List of general undergrad journals</a></li>
<li><a href="https://www.cur.org/resources/students/journals/" rel="noreferrer">Tons of resources for undergrad research</a></li>
</ul>
<p>Like the rest of us, you should do the research first, and think later about where it can be published. I agree with Sam that this should be guided by your advisor. Don't be afraid to have a conversation with your advisor stating that you are hoping for a journal paper in one of these journals. That can guide how focused the research experience should be, and can guide how you write the thesis. It's not overly ambitious at all.</p>
<p>One last point: I don't think it makes sense to tether your perceptions of admission to grad school to whether or not you have a journal publication. For one thing, even after you finish the research, it'll take you weeks or months to write the paper. Then, there will be 6-12 months while the paper is being refereed. Then a back and forth with the referees. The point is: grad programs in math would be crazy to expect undergrads to already have publications before applying. However, if you have a good draft, you can share that when you apply. I think it would matter less than your recommendation letters (another reason to talk to your advisor often), GPA, and test scores. Having fun with it, as you say, is the best idea. Whether or not you prove new results, your advisor can relay your passion and depth of understanding to graduate programs, and that will matter much more than your publication record at this stage of your career. Good luck!</p>
|
1,534,694 | <p>I tried to solve for the following limit: </p>
<p>$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form! </p>
| Brian M. Scott | 12,042 | <p>I have no idea how you got $\lim\limits_{x\to\infty}\frac{4e^{2x}}{4e^{2x}}$, but as has been pointed out, that limit is easily evaluated: the fraction is identically $1$, so the limit is also $1$.</p>
<p>Let $L=\lim\limits_{x\to\infty}\left(e^{2x}+x\right)^{1/x}$; then </p>
<p>$$\ln L=\ln\lim_{x\to\infty}\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\ln\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\frac{\ln\left(e^{2x}+x\right)}x\;.$$</p>
<p>Now apply l’Hospital’s rule.</p>
|
1,534,694 | <p>I tried to solve for the following limit: </p>
<p>$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form! </p>
| DanielWainfleet | 254,665 | <p>$x>0\implies e^x>x \implies e^2=(e^{2 x})^{1/x}<(e^{2x}+x)^{1/x}< (2 e^{2x})^{1/x} =(e^2) (2^{1/x})$..... And $\lim_{x\to \infty}2^{1/x}=1.$</p>
|
2,878,777 | <p>Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,\cdot)$ is isomorphic to $(B,\oplus,\odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:</p>
<p>Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $\varphi:A\to B$ between them, because, just as the apparently distinct sets $\{1,\, 2,\, 3,\, \cdots\}$ and $\{I,\, II,\, III,\, \cdots\}$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.</p>
<p>One can argue that this interpretation is wrong since there is a bijection between $\mathbb{N}$ and $\mathbb{Q}$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $\mathbb{N}$ and $\mathbb{Q}$ are not simply sets, but an algebraic structures. For example, $\mathbb{N}$ is actually a triple $(\mathbb{N},+,\cdot)$ in which $\mathbb{N}=\{1,\, 2,\,3,\, \cdots\}$ and $+$ and $\cdot$ are binary operations defined in $\mathbb{N}$ with which we can build a well-ordering $\leq $ order in $\mathbb{N}$. So, as algebraic structures, $\mathbb{N}$ and $\mathbb{Q}$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:\mathbb{N}\to\mathbb{Q}$ I can represent all elements of $\mathbb{Q}$ with the symbols $\{1,\, 2,\, 3,\, \cdots\}$. To do this, I can simply represent an element $q\in\mathbb{Q}$ by an $n\in \{1,\, 2,\, 3,\, \cdots\}$ that satisfies $f(n)=q$. </p>
<p>If we think well it is not possible to infer that $\mathbb{N}=\{1,\, 2,\,3,\, \cdots\}$ from the axioms of Peano. What happens is that we use the symbols $\{1,\, 2,\, 3,\, \cdots\}$ to represent the natural numbers which is the same attitude as representing $q\in\mathbb{Q}$ by an $n\in \{1,\, 2,\, 3,\, \cdots\}$ satisfying $f(n)=q$. Therefore, if we consider $\mathbb{N}$ and $\mathbb{Q}$ only as sets (and not as algebraic structures), we can say that $\mathbb{N}=\mathbb{Q}$ since it is possible to represent $\mathbb{Q}$ with the same symbols used to represent the elements of $\mathbb{N}$.</p>
<p><strong>My question is:</strong> can we interpret sets of the same cardinality as <strong>distinct representations of the same set</strong>? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,\cdot)$ and $(L,\odot)$ are isomorphic, then $(G,\cdot)$ and $(L,\odot)$ are distinct representations of the same group.</p>
<p><strong>EDIT:</strong> I think I've found a way to express myself better. I can say that $\mathbb{Q}=\{1,2,3,\cdots\} $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:\mathbb{N}\to\mathbb{Q}$ is a bijection). Therefore, treating $\mathbb{Q}$ and $\mathbb{N}$ only as sets (and not as algebraic structures) I believe it makes sense to say $\mathbb{Q}=\mathbb{N}=\{1,2,3,\cdots\}$.</p>
| Community | -1 | <p>The thing to recognize here is that the mathematical notion of equality has bifuricated. </p>
<p>In addition to the traditional notion of <em>equality</em>, we now recognize the value in considering the distinct notion of having an <em>equivalence</em> between two things.</p>
<p>We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification. </p>
<p>For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.</p>
<p>If you only ever ask questions about equivalence, never about equality, then you will indeed see $\mathbb{N}$ and $\mathbb{Q}$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what <em>you</em> mean by "$\mathbb{N}$ is the same as $\mathbb{Q}$" is a bijection $\mathbb{N} \to \mathbb{Q}$.</p>
<p>Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.</p>
<p>However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.</p>
|
2,818,427 | <p>Let $f \in \mathrm{End} (\mathbb{C^2})$ be defined by its image on the standard basis $(e_1,e_2)$: </p>
<p>$f(e_1)=e_1+e_2$</p>
<p>$f(e_2)=e_2-e_1$</p>
<p>I want to determine all eigenvalues of f and the bases of the associated eigenspaces.</p>
<p>First of all how does the transformation matrix of $f$ look like?
Is it </p>
<p>$\begin{pmatrix}1 &-1 \\1 &1 \end{pmatrix}$?</p>
| Jean-Claude Arbaut | 43,608 | <p>The differential equation must be</p>
<p>$$uy''+vy'+wy=0$$</p>
<p>where $u,v,w$ are functions of the variable $x$.</p>
<p>The solution $y=e^{-x}$ leads to $u-v+w=0$, or $v=u+w$.</p>
<p>The solution $y=x^2$ leads to $2u+2xv+x^2w=0$, which must be true for all $x$.</p>
<p>Replace $v=u+w$, then</p>
<p>$$2u+2x(u+w)+x^2w=0$$</p>
<p>$$2(x+1)u+x(x+2)w=0$$</p>
<p>Now choose functions $u$ and $w$ so that this equality holds. An obvious choice would be</p>
<p>$$u(x)=\frac{1}{x+1}$$
$$w(x)=\frac{-2}{x(x+2)}$$</p>
<p>And then</p>
<p>$$v(x)=\frac{1}{x+1}-\frac{2}{x(x+2)}$$</p>
|
3,367,588 | <p>I've been studying Numerical Linear Algebra, Lloyd, 1997. I've came across the below incomprehensible paragraph.</p>
<blockquote>
<p>"Methods like Householder reflections and Gaussian elimination would
solve linear systems of equations exactly in a finite number of steps
if they could be implemented in exact arithmetic. By contrast, any
eigenvalue solver must be iterative. The goal of an eigenvalue solver
is to produce sequences of numbers that converge rapidly toward
eigenvalues."</p>
</blockquote>
<p>In the above, the author differentiates between "a method with a finite number of steps" and "an iterative method",but it seems to me that these words have similar meaning. Could anyone elaborate difference?</p>
| Sudix | 470,072 | <p>A method/an algorithm with finite number of steps is one, that if executed on a machine which can work with real numbers (and do addition and multiplication of real numbers), at some point stops, and the returns the exact result.</p>
<p>For example determining the determinant of a matrix using LR-decomposition will stop at some point, because there's nothing to do anymore.Now, if you wouldn't have made rounding errors, your result would be exact.</p>
<p>On the other hand, an iterating algorithm is one, which even if the machine would work on the real numbers, doesn't have the guarantee that it will ever find the exact result. <br>
Instead, the algorithm guarantees, that the iterations (from some point on) will get closer and closer to the exact result.</p>
|
124,280 | <p>Show that the sequence ($x_n$) defined by $$x_1=1\quad \text{and}\quad x_{n+1}=\frac{1}{x_n+3} \quad (n=1,2,\ldots)$$ converges and determine its limit ? </p>
<p>I try to show ($x_n$) is a Cauchy sequence or ($x_n$) is decreasing (or increasing) and bounded sequence but I fail every step of all.</p>
| Alexander Thumm | 8,087 | <p>Hint: For $x,y \geq 0$ we have $\left\vert\frac{1}{x+3} - \frac{1}{y+3}\right\vert = \left\vert\frac{y-x}{(x+3)(y+3)}\right\vert \leq \frac{1}{9}\vert x-y\vert$.</p>
|
1,707,853 | <p>To be more precise than the title, the function is actually piecewise</p>
<p>$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$</p>
<p>I checked that the function is continuous at $(0,0)$, so I then calculated the partial derivative with respect to $x$ as</p>
<p>$$
f_1(x,y) = \frac{x^4-2xy^3+3x^2y^2}{(x^2+y^2)^2} \tag{1}
$$</p>
<p>This is undefined at $(0,0)$, so I then tried to find the limit around accumulation points. Let $S_1$ be the points on the $x$ axis</p>
<p>$$
\lim_{x \to 0} f_1(x,0) = \frac{x^4}{(x^2)^2} = 1 \tag{2}
$$</p>
<p>Let $S_2$ be the points on the line $y = x$</p>
<p>$$
\lim_{x \to 0} f_1(x,x) = \frac{x^4-2x^4+3x^4}{(x^2+x^2)^2} = \frac{2x^4}{(2x^2)^2} = \frac{1}{2} \tag{3}
$$</p>
<p>So, the limits are different around different accumulation points. That's where I'm confused because the answer should be $1$.</p>
| Ned | 67,710 | <p>Hint: Compute the limit (as $h$ goes to $0$) of $(f(0+h,0)-f(0,0))/h$ to get the value of the partial derivative at $(0,0)$. </p>
<p>What you are looking at with those path limits is the question of continuity of this partial derivative function at $(0,0)$. </p>
|
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https://en.wikipedia.org/w/index.php?title=Artin%27s_conjecture_on_primitive_roots&oldid=374854868" rel="nofollow noreferrer">Wikipedia</a>.
I need to know if it is up-to-date and if one can say something about the case <span class="math-container">$n=2$</span>.</p>
| Community | -1 | <p>@Igor: The membership in 2-generated subgroups of $F\times F'$ (where $F,F'$ are free) is decidable. Take any 2-generated subgroup $H=\langle (a,b), (c,d)\rangle$ of $F\times F'$. First we may suppose that $H$ is a subdirect product, that is $F$ is generated by $a,c$, $F'$ is generated by $b,d$. By Baumslag-Roseblate, $H$ is a pullback of two homomorphisms $f,g: F\to F/N$ where $N$ is the intersection of $H$ with the $F\times \{1\}$, and the $F$ is identified with $F\times \{1\}$. The membership problem is equivalent to the word problem in $F/N$. The subgroup $N$ is then obtained as follows. Let $L$ be the subgroup of $F'$ consisting of all words $w(x,y)$ such that $w(b,d)=1$. Then $N$ is the image of $L$ under the endomorphism $x\to a, y\to b$. The group $L$ is non-trivial if and only if $b,d$ commute. Hence $b^m=d^n$ for some $m,n$. Thus $L$ is generated as a normal subgroup by two words $[x,y], x^my^{-n}$. Hence $F'/L$ (which is isomorphic to $F/N$) is Abelian (in fact cyclic) and the word problem in $F'/L$ is decidable. Hence the membership problem for $H$ is decidable (in linear time because the distortion of $H$ is equivalent to the Dehn function of $F/N$ by our theorem with Olshanskii). </p>
|
47,974 | <p>I am interested in the following question:</p>
<p>Is it known that <span class="math-container">$2$</span> is a primitive root modulo <span class="math-container">$p$</span> for infinitely many primes <span class="math-container">$p$</span>?</p>
<p>There is some information about Artin's conjecture in <a href="https://en.wikipedia.org/w/index.php?title=Artin%27s_conjecture_on_primitive_roots&oldid=374854868" rel="nofollow noreferrer">Wikipedia</a>.
I need to know if it is up-to-date and if one can say something about the case <span class="math-container">$n=2$</span>.</p>
| Alla Detinko | 60,476 | <p>Just some minor comments on the problem of testing whether a
finitely generated subgroup H of SL(n, Z) equals SL(n, Z), n > 2.
This is a partial case of the arithmeticity testing (AT) problem,
i.e., testing whether H has finite index in SL(n, Z). To our
knowledge it is not known whether AT problem is decidable; but
most likely, in general, the answer is negative (some arguments
for this have been provided by C. F. Miller III). If so, the AT
problem is semidecidable: if H does happen to be arithmetic then
it is possible to verify that, by, e.g., the Todd-Coxeter
procedure. So, if H is known to be arithmetic, one can test
whether H = SL(n, Z). However, the obstacle here is that
Todd-Coxeter is impractical (the index of H in SL(n, Z) could be
arbitrarily large). An alternative approach is based on the
congruence subgroup property, i.e., construction of a principal
congruence subgroup in H (for which Todd-Coxeter is not required).
Note that in the class of solvable algebraic groups, the AT
problem is decidable and a practical algorithm is available.</p>
<p>A very brief (although now somewhat outdated) survey on decidable
problems in the class of finitely generated linear groups is
available in Section 3 of London Math. Soc. Lecture Note Ser. 387
(2011) 256–270. Also, a series of papers by F. Grunewald and D.
Segal on decidable problems for explicitly given arithmetic groups
may be of interest with regard to your question (e.g., see `Some
general algorithms. I. Arithmetic groups', Ann. of Math. (2)
112(3) (1980) 531–583.)</p>
|
912,426 | <p>A bag contains six chips, numbered 1 through 6. If two chips are chosen at random without replacement and the values on those two chips are multiplied, what is the probability that this product will be greater than 20?</p>
<p>I tried to solve by counting the total possibilities (36) and solving for 6 choices that worked, e.g. $4x6, 5x5, 5x6, 6x4,6x5,6x6$... so I thought the probability would be $1/6$. </p>
<p>How is my method incorrect?</p>
| Kelenner | 159,886 | <p>We verify easily that for $k\geq 1$, we have $\displaystyle \frac{2k-1}{2k}\leq \sqrt{\frac{k}{k+1}}$. Hence
$$\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \prod_{k=1}^n \sqrt{\frac{k}{k+1}}= \frac{1}{\sqrt{n+1}}$$
and we are done.</p>
|
478,566 | <p>I'm reading a book about combinatorics. Even though the book is about combinatorics there is a problem in the book that I can think of no solutions to it except by using number theory.</p>
<p>Problem: Is it possible to put $+$ or $-$ signs in such a way that $\pm 1 \pm 2 \pm \cdots \pm 100 = 101$?</p>
<p>My proof is kinda simple. Let's work in mod $2$. We'll have:</p>
<p>$\pm 1 \pm 2 \pm \cdots \pm 100 \equiv 101 \mod 2$
but since $+1 \equiv -1 \mod 2$ and there are exactly $50$ odd numbers and $50$ even numbers from $1$ to $100$ we can write:</p>
<p>$(1 + 0 + \cdots + 1 + 0 \equiv 50\times 1 \equiv 0) \not\equiv (101\equiv 1) \mod 2$ which is contradictory. </p>
<p>Therefore, it's not possible to choose $+$ or $-$ signs in any way to make them equal.</p>
<p>Now is there a combinatorial proof of that fact except what I have in mind?</p>
| marty cohen | 13,079 | <p>Replacing 100 with $n$
and using Brian M. Scott's solution,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.</p>
<p>The sum is
$\frac{(n+1)(n+2)}{2}$,
and if $n=4k$,
this is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.</p>
<p>If $n = 4k+1$,
this is
$(2k+1)(4k+3)$
which is also odd,
and therefore impossible.</p>
<p>If $n = 4k+2$,
this is
$(4k+3)(2k+2)$,
so it is not ruled out,
and each sum must be
$(4k+3)(k+1)$.</p>
<p>if $n = 4k+3$,
this is
$(2k+2)(4k+5)$
which is also not ruled out,
and each sum must be
$(k+1)(4k+5)$.</p>
<p>Now I'll try to find a solution
for the not impossible cases.
(I am working these out as I enter them.)</p>
<p>For the $n=4k+2$ case,
the sum must be
$(4k+3)(k+1)
=(4k+4-1)(k+1)
=4(k+1)^2-(k+1)
=(2k+2)^2-(k+1)
$.
The square there suggests,
to me,
the formula for
the sum of consecutive odd numbers
$1+3+...+(2m-1)=m^2$,
so $1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is odd,
remove it from the sum
so it is
$(2k+2)^2-(k+1)$.
If $k+1$ is even,
both $1$ and $k$ are odd,
so remove them from the sum.
In either case, we have the desired partition.</p>
<p>For the $n=4k+3$ case,
the sum must be
$(4k+5)(k+1)
=(4k+4+1)(k+1)
=4(k+1)^2+(k+1)
=(2k+2)^2+(k+1)
$.
Again,
$1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is even,
add it to the sum
so it is
$(2k+2)^2+(k+1)$.
If $k+1$ is odd,
$k+2$ is even,
so remove $1$
and add $k+2$ to the sum.
In either case, we have the desired partition.</p>
<p>I do not know if these partitions
are unique.</p>
|
1,479,095 | <blockquote>
<p>For $f:[0,1]\to \mathbb{R}$ let $E\subset\left\{x \mid f'(x) \text{exists}\right\}$. Prove that if $|E|=0$, then $|f(E)|=0$.</p>
</blockquote>
<p>My attempt:</p>
<p>Let $E_{nk}=\left\{x\in [0,1]|\frac{|f(x+h)-f(x)|}{h}\leq n, |h|< \frac{1}{k} \right\}$. </p>
<p>I am not sure where to go from here, but I think that $E\subset \bigcup E_{nk}$, but I am not sure.</p>
<p>Any hints? Thank you.</p>
| Community | -1 | <p>Hint: Note that if $f$ is differentiable at $x$, then the limit </p>
<p>$$ \frac{f(x+ h) - f(x)}{h}$$</p>
<p>exists as $h \to 0$. So whenever $|h|$ is small, the above expression is bounded by a constant. </p>
|
4,253,640 | <p>For example:
suppose we need to find <strong>x</strong> given that <strong>x mod 7 = 5</strong> and <strong>x mod 13 = 8</strong>.</p>
<p><strong>x = 47</strong> is a solution but needs hit and trial.</p>
<p>Is there any shortcut to calculate such number?</p>
| kbx12 | 436,239 | <p><span class="math-container">$x=\frac{5*13+8*7}{7+13}=121/20=121*41=47\mod{(7*13)}$</span></p>
<p><span class="math-container">$x=m_1\mod{n_1}$</span><br>
<span class="math-container">$x=m_2\mod{n_2}$</span><br>
<span class="math-container">$x=m_3\mod{n_3}$</span><br>
...<br>
<span class="math-container">$x=m_n\mod{n_n}$</span><br></p>
<p><span class="math-container">$$x=\frac{m_1/n_1+m_2/n_2+m_3/n_3+...m_n/n_n}{1/n_1+1/n_2+1/n_3+...1/n_n}\mod{(n_1*n_2*n_3...n_n)}$$</span></p>
|
1,216,983 | <p>Let $f(x)$ be a polynomial with complex coefficients such that $\exists n_0 \in \mathbb Z^+$ such that $f(n) \in \mathbb Z , \forall n \ge n_0$, then is it true that $f(n) \in \mathbb Z , \forall n \in \mathbb Z$ ?</p>
| vonbrand | 43,946 | <p>Erdös et al, "The Asymptotic Behavior of a Family of Sequences" (Pacific Journal of Mathematics 126:2, pp 227-241, dec 1987, <a href="http://www.dtc.umn.edu/~odlyzko/doc/arch/sequence.family.pdf" rel="nofollow noreferrer">here</a> a more readable version than the one in the journal) discuss exactly this type of sequences.</p>
|
3,768,333 | <blockquote>
<p>Let <span class="math-container">$f: [a,b] \to R$</span> be a differentiable function of one variable such that <span class="math-container">$|f'(x)| \le 1$</span> for all <span class="math-container">$x\in [a,b]$</span>. Prove that <span class="math-container">$f$</span> is a contraction. (Hint: use MVT.) If in addition <span class="math-container">$|f'(x)| < 1$</span> for all <span class="math-container">$x \in [a,b]$</span> and <span class="math-container">$f'$</span> is continuous, show that <span class="math-container">$f$</span> is a strict contraction.</p>
</blockquote>
<p>Using MVT, <span class="math-container">$|f(x) - f(y)| = |f'(c)(x-y)| \le |x-y|$</span> for <span class="math-container">$c$</span> between <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>I don't know the proof for a strict contraction. I guess that I need to use the continuity of <span class="math-container">$f'$</span>, but I am not sure how to use it. Any help would be appreciated.</p>
| Michael Hardy | 11,667 | <p>Besides the mean-value theorem, another "MVT" is the "maximum-value theorem": A continuous function on a closed bounded interval has an absolute maximum. If <span class="math-container">$|f'|$</span> is continuous, then there is some point <span class="math-container">$c\in[a,b]$</span> for which <span class="math-container">$|f'|$</span> is at least as big as it is at any point in that interval. And that value must be less than <span class="math-container">$1,$</span> by hypothesis. Then apply the mean value theorem again.</p>
<p>(You probably won't see that other theoem called the "MVT", for "maximum-value theorem", since it is instead called the extreme-value theorem, applying to both maxima and minima.)</p>
|
1,640,678 | <p>How do you find the maximum of a quadratic function? Specifically,
$R(x) = -4x^2 + 4000x$</p>
| KonKan | 195,021 | <p>Find the roots of $$-4x^2 + 4000x=0$$ These are $x=0$ and $x=1000$. The maximum of the quadratic function appears at the midpoint between the roots i.e. at $x=500$ and its value is $$R(500)=-4*500^2+4000*500=1000000$$</p>
|
3,325,658 | <blockquote>
<p>Count the number of 5 cards such that there's exactly 2 suits</p>
</blockquote>
<p>Suppose we draw five cards from a standard deck of 52 cards. I want to count the number of ways I can draw five cards such that the hand contains exactly 2 suits.</p>
<p>Here's my intuition:<br/>
There are two cases, one case involves a single card with a different suit from the other 4, and the other case involves two cards with the same suit, and the other three have a different suit.</p>
<p>Case 1: <span class="math-container">${4 \choose 1} {13 \choose 1} \cdot {3 \choose 1} {13 \choose 4}$</span></p>
<p>Case 2: <span class="math-container">${4 \choose 1} {13 \choose 2} \cdot {3 \choose 1} {13 \choose 3}$</span></p>
<p>Am I correct with the cases? I'm confused on the coefficients for choosing the suits for the first group of cards since it will limit the number of suits to choose for the next group of cards. Should I multiply each case by <span class="math-container">$2$</span> ?</p>
| David Popović | 549,692 | <p>Your cases 1 and 2 are correct, so the total number of hands satisfying the property in question is <span class="math-container">$\binom{4}{1} \binom{13}{1} \binom{3}{1} \binom{13}{4} + \binom{4}{1} \binom{13}{2} \binom{3}{1} \binom{13}{3}$</span>.</p>
<p>There is no need to multiply anything by 2. It is true that the selections of the suits of the large and small groups are not independent, but whatever you choose for the first one (in 4 possible ways), there will be 3 choices for the second one.</p>
|
199,549 | <p>By using the Löwenheim–Skolem theorem & Mostowski collapse, in every model $V$ of $ZF+Con(ZF)$ there is a countable transitive set $M$ such that $(M,\in_M) \models ZF$. Is the following "converse" true?</p>
<blockquote>
<p>In every model $V$ of $ZF$ and every transitive set $M \in V$ such that $(M,\in_M) \models ZF$, there exists a transitive set $N \in V$ such that:</p>
<ol>
<li><p>$M \in N$</p></li>
<li><p>$(N,\in_N) \models ZF$</p></li>
<li><p>$M$ is countable inside $N$</p></li>
</ol>
</blockquote>
| Asaf Karagila | 7,206 | <p>While the original question has been answered, remember that Skolem's paradox says that if $M$ is a countable model of set theory, then it knows only about countably many real numbers.</p>
<p>The reverse Skolem paradox, if so, is the following situation that can occur:</p>
<blockquote>
<p>There $(M,E)\models\sf ZF$ such that $\{x \mid M\models x\text{ is an integer}\}$ is uncountable. (Assuming that there is a model of $\sf ZF$ to begin with, but let's dispense of that issue for now.)</p>
</blockquote>
<p>The point is that $M$ thinks that an uncountable set is in fact countable. This $M$ cannot be transitive (or well-founded), and obtaining it is a simple exercise in compactness or ultrapowers. This shows that not only that "uncountable" is relative and internal to the model, but also "countable", and in fact in this model $M$, even "finite" is relative. </p>
<p>Since $M$ will think about certain infinite set that they are finite (e.g. various bounded sets of "natural numbers" in $M$ will contain infinitely many objects, but $M$ will still think they are finite because $M$ has a twisted view as for what is finite and countable).</p>
|
4,196,583 | <p>More precisely:</p>
<blockquote>
<p><strong>Definition.</strong><br />
A subset <span class="math-container">$S \subset \Bbb R$</span> is called <em>good</em> if the following hold:</p>
<ol>
<li>if <span class="math-container">$x, y \in S$</span>, then <span class="math-container">$x + y \in S,$</span> and</li>
<li>if <span class="math-container">$(x_n)_{n = 1}^\infty \subset S$</span> is a sequence in <span class="math-container">$S$</span> and <span class="math-container">$\sum_{n = 1}^\infty x_n$</span> converges, then <span class="math-container">$\sum_{n = 1}^\infty x_n \in S$</span>.</li>
</ol>
</blockquote>
<p>In other words, a good subset is closed under finite sums and countable sums whenever the sum does exist.</p>
<blockquote>
<p><strong>Question:</strong> What are all the good subsets of <span class="math-container">$\Bbb R$</span>?</p>
</blockquote>
<hr />
<p><strong>Origin</strong></p>
<p><a href="https://math.stackexchange.com/questions/4194005/">This question</a> was asked recently and <a href="https://math.stackexchange.com/users/152568/conifold">Conifold</a> had <a href="https://math.stackexchange.com/questions/4194005/does-1-frac14-frac19-frac116-cdots-frac-pi26-imply-that-the-rational#comment8699081_4194005">commented</a> how the only subsets of <span class="math-container">$\Bbb R$</span> closed under countable summations are <span class="math-container">$\varnothing$</span> and <span class="math-container">$\{0\}$</span>. It was then natural to ask "closed under countable summation, assuming it exists".</p>
<hr />
<p><strong>My thoughts</strong></p>
<p>Here are some examples of familiar sets which are good: <span class="math-container">$\varnothing$</span>, <span class="math-container">$\{0\}$</span>, <span class="math-container">$\Bbb Z_{\geq 0}$</span>, <span class="math-container">$\Bbb Z_{> 0}$</span>, <span class="math-container">$\Bbb Z$</span>, <span class="math-container">$n\Bbb Z$</span>, <span class="math-container">$\Bbb R$</span>.<br />
We even have the following:<br />
<span class="math-container">$$r \Bbb Z := \{rn : n \in \Bbb Z\},$$</span>
where <span class="math-container">$r$</span> is any real number.</p>
<p>But the examples apart from <span class="math-container">$\Bbb R$</span> are good for a trivial reason: Those are sets that are closed under finite summation and have the property that they are discrete enough so that the only convergent sums are those where the terms are eventually <span class="math-container">$0$</span>. (In the case of <span class="math-container">$\Bbb Z_{> 0}$</span>, there is no such sum.)</p>
<p>On the same note, intervals of the form <span class="math-container">$[a, \infty)$</span> and <span class="math-container">$(a, \infty)$</span> are good for <span class="math-container">$a > 0$</span>.</p>
<p>In general, suppose that <span class="math-container">$S$</span> satisfies the following: <span class="math-container">$S$</span> is closed under finite sums and there exists <span class="math-container">$\epsilon > 0$</span> such that <span class="math-container">$|s| > \epsilon$</span> for all <span class="math-container">$s \in S$</span>. Then, <span class="math-container">$S$</span> and <span class="math-container">$S \cup \{0\}$</span> are good.</p>
<p>Another example: <span class="math-container">$[0, \infty)$</span> and <span class="math-container">$(0, \infty)$</span> are good and do not follow the criteria above.</p>
<hr />
<p>The following are some examples of <em>not</em> good sets: <span class="math-container">$\Bbb Q$</span>, <span class="math-container">$\Bbb R \setminus \Bbb Q$</span>, <span class="math-container">$\Bbb R \setminus \Bbb Z$</span>, a proper cofinite subset of <span class="math-container">$\Bbb R$</span>, any bounded set apart from <span class="math-container">$\{0\}$</span> and <span class="math-container">$\varnothing$</span>. In fact, excluding <span class="math-container">$\Bbb Q$</span>, the other ones are not even closed under finite sums.<br />
In the same vein as <span class="math-container">$\Bbb Q$</span>, we also have the set of real algebraic numbers which is not good (but is indeed closed under finite sums).</p>
<p>Here's a nontrivial one: Consider the set <span class="math-container">$$B = \left\{\frac{1}{2^k} : k \in \Bbb Z_{> 0}\right\}.$$</span> Then, <em>any</em> countable subset of <span class="math-container">$\Bbb R$</span> that contains <span class="math-container">$B$</span> is not good.<br />
<em>Proof.</em> <span class="math-container">$(0, 1]$</span> is uncountable and every element in it can be written as a sum of elements of <span class="math-container">$B$</span>. (Binary expansions.) <span class="math-container">$\Box$</span></p>
<p>A bit more thought actually shows that more is true: Since <span class="math-container">$(0, 1]$</span> is contained in the set of all possible sums, any good set containing <span class="math-container">$B$</span> must contain all of <span class="math-container">$(0, \infty)$</span>.</p>
<p>Another one: let <span class="math-container">$(a_n)_{n \ge 1}$</span> be any real sequence such that <span class="math-container">$\sum a_n$</span> converges conditionally. Then, the only good subset containing <span class="math-container">$\{a_n\}_{n \ge 1}$</span> is <span class="math-container">$\Bbb R$</span>, by the Riemann rearrangement theorem.</p>
<hr />
<p><strong>Additional comments</strong></p>
<p>There are some variants that come to mind. Not sure if any of them are any more interesting. But I'd be happy with an answer that answers only one of the following variants as well.</p>
<ol>
<li>What if I exclude point 1. from my definition? Let's call such a set nice.<br />
In that case, the set <span class="math-container">$\{1\}$</span> is nice but not good. (Of course, if <span class="math-container">$0 \in S$</span>, then nice is equivalent to good.) What are the nice subsets of <span class="math-container">$\Bbb R$</span>? What are the nice subsets which are not good?</li>
<li>What if I consider only those sums which have converge absolutely?</li>
</ol>
<hr />
<p><strong>More observations</strong><br />
(These are edits, which I'm adding later)</p>
<p>Here are some additional observations:</p>
<ol>
<li>Arbitrary intersection of good sets is good and <span class="math-container">$\Bbb R$</span> is good. Thus, it makes sense to talk about the smallest good set containing a given subset of <span class="math-container">$\Bbb R$</span>.<br />
So, given a subset <span class="math-container">$A \subset \Bbb R$</span>, let us call this smallest good set to be the <em>good set generated by <span class="math-container">$A$</span></em> and notationally denote it as <span class="math-container">$\langle A \rangle$</span>.<br />
(In particular, <span class="math-container">$A$</span> is good iff <span class="math-container">$A = \langle A \rangle$</span>.)</li>
<li><span class="math-container">$A \subset B \implies \langle A \rangle \subset \langle B \rangle$</span>.</li>
<li><span class="math-container">$\langle (0, \epsilon) \rangle = (0, \infty)$</span> and similar symmetric results.</li>
<li>Suppose <span class="math-container">$S \subset (0, \infty)$</span> is dense in <span class="math-container">$(0, \infty)$</span>, then <span class="math-container">$\langle S \rangle = (0, \infty)$</span>.<br />
Indeed, pick <span class="math-container">$a_0 \in (0, \infty)$</span>.<br />
Then, there exists <span class="math-container">$s_1 \in (a_0/2, a_0)$</span>.<br />
Put <span class="math-container">$a_1 := a - s_1$</span>. Then, <span class="math-container">$a_1 \in (0, a_0/2)$</span>.<br />
Now pick <span class="math-container">$s_2 \in (a_1/2, a_1)$</span> and so on. Then, <span class="math-container">$\sum s_n = a_0$</span>.<br />
Symmetric results apply. In particular, the only dense good subset of <span class="math-container">$\Bbb R$</span> (or <span class="math-container">$\Bbb R^+$</span> or <span class="math-container">$\Bbb R^-$</span>) is the whole set.</li>
<li>Suppose <span class="math-container">$S \subset (0, \infty)$</span> contains arbitrarily small elements (i.e., <span class="math-container">$S \cap (0, \epsilon) \neq \varnothing$</span> for all <span class="math-container">$\epsilon > 0)$</span>), then <span class="math-container">$\langle S \rangle = (0, \infty)$</span>.<br />
To see this, let <span class="math-container">$a > 0$</span> be arbitrary. Pick <span class="math-container">$s_1 \in (0, a)$</span>.<br />
Let <span class="math-container">$n_1$</span> be the largest positive integer such that <span class="math-container">$n_1 s_1 < a$</span>.<br />
Then pick <span class="math-container">$s_2 \in (0, a - n_1s_1)$</span>. Let <span class="math-container">$n_2$</span> be the largest such <span class="math-container">$n_2s_2 < a - n_1s_1$</span> and so on.<br />
Then, <span class="math-container">$$\underbrace{s_1 + \cdots + s_1}_{n_1} + \underbrace{s_2 + \cdots + s_2}_{n_2} + \cdots = a.$$</span></li>
</ol>
| Kr Dpk | 499,096 | <p>Maybe this'll add to the classification.</p>
<p>Let <span class="math-container">$S$</span> be a good set such that there exist atleast one series <span class="math-container">$\sum\limits_{n=1}^{\infty} x_n$</span> which converges.</p>
<p>As mentioned in the above answers, if <span class="math-container">$\sum x_n$</span> is conditionally convergent, then by Riemann's rearranging theorem , <span class="math-container">$S = \mathbb{R}$</span>.<br />
So, lets consider positive series only (equivalently negative series may be considered).<br />
We prove that under the given properties of <span class="math-container">$S$</span>, we can make sure that every positive real belongs to <span class="math-container">$S$</span>. WLOG, lets prove that <span class="math-container">$1 \in S$</span>, and same procedure can be used for any positive real.<br />
Let <span class="math-container">$\sum\limits_{n=1}^{\infty} x_n$</span> be convergent to some positive value. This implies that we can chose the tail of the sequence, <span class="math-container">$\sum\limits_{n=k}^{\infty} x_n $</span>(let be <span class="math-container">$\epsilon$</span>), whose value is less than <span class="math-container">$\frac{1}{2^a}$</span>, where <span class="math-container">$a$</span> is a positive integer. Since <span class="math-container">$\epsilon$</span> is less then 1, we can add <span class="math-container">$\epsilon$</span> some finite '<span class="math-container">$l$</span>' times, so that <span class="math-container">$0 \leq 1- l*\epsilon \leq \frac{1}{2^a}$</span>. Now, from our original series, we chose a tail whose value is less than <span class="math-container">$ \frac{1}{2^b}$</span> ( <span class="math-container">$< 1 - l*\epsilon$</span>) for some positive integer <span class="math-container">$'b'$</span>. Let the value of the tail be <span class="math-container">$\epsilon$</span>'. Adding <span class="math-container">$\epsilon$</span>' some finite <span class="math-container">$l'$</span> times, we can get <span class="math-container">$0 \leq 1 - l*\epsilon - l' *\epsilon ' \leq \frac{1}{2^b}$</span>. In this way, we keep adding terms to get as close to <span class="math-container">$1$</span> as we want. The series that we produce in this way, sums to <span class="math-container">$1$</span>.<br />
Thus, <span class="math-container">$1 \in S$</span>. But we could have done this procedure for any positive real. Hence, if any positive series converges in <span class="math-container">$S$</span>, then <span class="math-container">$(0,\infty) \subseteq S$</span>.</p>
<p>Now, suppose that we could find one such convergent series of positive reals. If <span class="math-container">$S$</span> contains atleast one negative real, then adding the multiples of that one negative value with <span class="math-container">$(0,\infty)$</span>, we can generate the whole <span class="math-container">$\mathbb{R}$</span>.</p>
<p>Hence, if the good set <span class="math-container">$S$</span>, do indeed follow the second property (and exhibits at least one such series in itself), then <span class="math-container">$S$</span> is : <span class="math-container">$\mathbb{R}$</span>, <span class="math-container">$(0,\infty), (-\infty,0), [0,\infty), (-\infty,0]$</span>.</p>
<p>Otherwise, <span class="math-container">$S$</span> can be considered closed under addition only. And then it is a semi group with addition.</p>
|
4,117,409 | <blockquote>
<p>Prove or disprove: if for every <span class="math-container">$n\in\Bbb{N}, |a_{n+1}-a_n|<\frac{1}{n^2}$</span> then <span class="math-container">$a_n$</span> converges.</p>
</blockquote>
<p>I think this is true, and tried using Cauchy's theorem - I take some <span class="math-container">$\varepsilon > 0$</span>. There exists such <span class="math-container">$N$</span> s.t <span class="math-container">$\varepsilon > \frac{1}{N^2}$</span>. So, for every <span class="math-container">$m,n>N$</span>, we get that:
<span class="math-container">$|a_m-a_n|\leq|a_m-a_{m-1}|+...+|a_{n+1}-a_n|<\frac{1}{m^2}+...+\frac{1}{n^2}<\frac{m-n}{N^2}$</span>. Now I think the righthand side tends to <span class="math-container">$0$</span>, but this feels like cheating. Am I correct or is there something I'm missing?</p>
| RRL | 148,510 | <p>Note that for all sufficiently large <span class="math-container">$n$</span> and all <span class="math-container">$m > n$</span>, we have</p>
<p><span class="math-container">$$|a_m-a_n| \leqslant \sum_{k=n}^m \frac{1}{k^2} \leqslant \int_{n-1}^m \frac{dx}{x^2} = \frac{1}{n-1} - \frac{1}{m} < \frac{1}{n-1} < \epsilon$$</span></p>
|
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| Asaf Karagila | 622 | <p>You cannot do that because a countable set can have an uncountably many limit points. The points in the Cantor set are limit points of these endpoints.</p>
<p>For example, the real numbers are all limit points of the rational numbers. If between every two real numbers there is a rational number, but we still can't establish that the real numbers are countable.</p>
|
210,735 | <p>The Cantor set is closed, so its complement is open. So the complement can be written as a countable union of disjoint open intervals. Why can we not just enumerate all endpoints of the countably many intervals, and conclude the Cantor set is countable?</p>
| AnotherPerson | 185,237 | <p>There are a couple other ways to conclude that the Cantor set is uncountable, which you can read more about in Charles Pugh's "Real Mathematical Analysis" 2nd edition. Since the arguments clarified a lot for me I figured it would be good to put them in here for others to see since they have not been mentioned already.</p>
<p>While this theorem does not give much intuition about the points included in the Cantor set which make it uncountable, we can use the fact that every complete, perfect, and nonempty metric space is uncountable, and show that the Cantor set meets these criteria to show that it is uncountable. </p>
<p>Now for a more illuminating explanation that relates to the expression in base $3$ argument, we create an address for each point in the Cantor set which consists of an infinite string of $0$'s and $2$'s which is determined as follows.</p>
<p>We know that $C=\bigcap_{n=1}^{\infty}C^n$ where each $C^n$ consists of the $2^n$ subdivisions of $[0,1]$ each of length $\frac{1}{3^n}$. So we create an address system which indicates to which of the $2^n$ subintervals a given point $p$ belongs, where the string up to the $n$-th digit will tell us the specific subinterval in $C^n$. The general idea is that a $0$ indicates belonging to the left subinterval after the removal of the middle third , while a $2$ indicates belonging to the right subinterval after the removal of the middle third of a given interval of $C^n$. </p>
<p>Let's consider the example with $p=\frac{1}{4}$. We know that the first splitting of the interval $[0,1]$ in $C^1$ consists of the two subintervals $C_0= [0, \frac{1}{3}]$ and $C_2=[\frac{2}{3}, 1]$. and that $\frac{1}{4} \in [0, \frac{1}{3}]$. Therefore the first number of the address string for $\frac{1}{4}$ is a $0$. </p>
<p>Now for $C^2$, we know this consists of $4$ subintervals of $[0,1]$, namely $C_{00} =[0, \frac{1}{9}]$, $C_{02}=[\frac{2}{9}, \frac{1}{3}]$, $C_{20}=[\frac{2}{3}, \frac{7}{9}]$, and $C_{22}=[\frac{8}{9},1]$. We know that $\frac{2}{9}<\frac{1}{4}< \frac{1}{3}$, we have that $\frac{1}{4}\in C_{02}=[\frac{2}{9}, \frac{1}{3}]$, making our address for $\frac{1}{4}$ is $02....$. Continuing on to $C^3$, $C^4$,... we get an infinite address string for $\frac{1}{4}$.</p>
<p>The set of all of these infinite address strings can be shown to be uncountable, and it is in bijection with the Cantor set. By the continuum hypothesis, any uncountable subset of $\mathbb{R}$ has a cardinality equal to $\mathbb{R}$. Therefore the Cantor set is uncountable.</p>
|
715,706 | <p>I try to find a partial fraction expansion of $\dfrac{1}{\prod_{k=0}^n (x+k)}$ (to calculate its integral).
After checking some values of $n$, I noticed that it seems to be true that $\dfrac{n!}{\prod_{k=0}^n (x+k)}=\sum_{k=0}^n\dfrac{(-1)^k{n \choose k}}{x+k}$. However, I can't think of a way to prove it. Can somebody please help me?</p>
| Community | -1 | <p>Since every $k,\; k=-n,\ldots, 0$ is a simple pole of the given fraction then its decomposition take the form</p>
<p>$$\frac{1}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{a_k}{x+k}$$
and we have
$$a_k=\lim_{x \to -k}\sum_{i=0}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=0}^n\frac{a_i}{x+i}$$
$$= \lim_{x \to -k} \frac{x+k}{x(x+1)(x+2)...(x+n)}=\frac{1}{-k(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^k}{k!(n-k)!}$$
so yes it's true that
$$\frac{n!}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{(-1)^k{n\choose k}}{x+k}$$</p>
|
1,600,428 | <p>Find the equation of a line that passes through the origin, with positive slope, and its tangent to the parabola given by :$ y = x^2 - 2x + 2$</p>
<p>My approach to this problem was to differentiate the equation of the parabola, so I can et an expression, that determines the tangent line anywhere on the parabola. I then got $2x - 2$, and I replaced x with a point (p), which is the second point of the line, where the first point is the origin $(0,0)$, and the second point being (p) which is tangent to the parabola, I then used $2p - 2$ as the slope of , my line, and the y intercept of the line is zero, because the line goes through the origin. The equation of the then becomes : $y = x (2p-2)$, and now I set the line and the parabola equal to each other, and at that point is where I get stuck, because I cant solve for $(p)$. I am pretty sure that my approach is wrong, and this is why I needed help solving this problem. </p>
<p>Could somebody, who solves this problem, provide a full solution below? Thank You !</p>
| Ahmed S. Attaalla | 229,023 | <p>At the point $x=x_0$ the slope of the line,as you said, is :</p>
<p>$$2x_0-2$$</p>
<p>And so the line is of the form:</p>
<p>$$y=(2x_0-2)x+b$$</p>
<p>What is $b$? </p>
<p>You did say it passes through the origin where $x=0$ and $y=0$. So with some basic algebra we can come up with:</p>
<p>$$b=0$$</p>
<p>$$y=(2x_0-2)x$$</p>
<p>Again, this you correctly executed.Now all that is left is to conclude what $x_0$ and the corresponding slope(s) may be. How do we do this?</p>
<p>We incorporate our final bit out information. That our line must pass through the point:</p>
<p>$$(x_0,x_0^2-2x_0+2)$$</p>
<p>Because it is tangent to our function at $x=x_0$.</p>
<p>And we get,</p>
<p>$$x_0^2-2x_0+2=(2x_0-2)x_0$$</p>
<p>$$x_0=\sqrt{2},x_0=-\sqrt{2}$$</p>
<p>$$y=(2\sqrt{2}-2)x,y=(-2\sqrt{2}-2)x$$</p>
<p>But you only want the one with a positive slope which is:</p>
<p>$$y=(2\sqrt{2}-2)x$$</p>
<p>So really the trick is to plug in $x=p$ and $y=p^2-2p+2$ into your linear equation. </p>
|
3,005,965 | <p>I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.</p>
<p>I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.</p>
<p>So here is the link to the my earlier post containing the question and my concerns</p>
<p><a href="https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal">https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal</a></p>
<p>if you don't want to look at the link then I can put the same photo here and show the direct question also</p>
<p><a href="https://i.stack.imgur.com/K5VYs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K5VYs.jpg" alt="enter image description here"></a></p>
<p>I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?</p>
| Théophile | 26,091 | <p>You've gone too far. After the first step, when you have <span class="math-container">$5z-4$</span>, you should stop because the exponent in <span class="math-container">$z^2$</span> is too high for it to go into <span class="math-container">$5z$</span>. You simply have a remainder of <span class="math-container">$5z-4$</span>, thus:</p>
<p><span class="math-container">$$\frac{4z^2-5z}{2z^2-5z+2} = 2 + \frac{5z-4}{2z^2-5z+2}$$</span></p>
<p>Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.</p>
|
30,220 | <p>Jeremy Avigad and Erich Reck claim that one factor leading to abstract mathematics in the late 19th century (as opposed to concrete mathematics or hard analysis) was <em>the use of more abstract notions to obtain the same results with fewer calculations.</em></p>
<p>Let me quote them from their remarkable historical <a href="https://www.andrew.cmu.edu/user/avigad/Papers/infinite.pdf" rel="nofollow noreferrer">paper</a> "Clarifying the nature of the infinite: the development of metamathematics and proof theory".</p>
<blockquote>
<p>The gradual rise of the opposing viewpoint, with its emphasis on conceptual
reasoning and abstract characterization, is elegantly chronicled by <a href="http://mcps.umn.edu/philosophy/11_10Stein.pdf" rel="nofollow noreferrer">Stein</a> (<a href="https://web.archive.org/web/20140415224643/http://mcps.umn.edu/philosophy/11_10Stein.pdf" rel="nofollow noreferrer">Wayback Machine</a>),
as part and parcel of what he refers to as the “second birth” of mathematics.
The following quote, from Dedekind, makes the difference of opinion very clear:</p>
</blockquote>
<blockquote>
<blockquote>
<p>A theory based upon calculation would, as it seems to me, not offer
the highest degree of perfection; it is preferable, as in the modern
theory of functions, to seek to draw the demonstrations no longer
from calculations, but directly from the characteristic fundamental
concepts, and to construct the theory in such a way that it will, on
the contrary, be in a position to predict the results of the calculation
(for example, the decomposable forms of a degree).</p>
</blockquote>
</blockquote>
<blockquote>
<p>In other words, from the Cantor-Dedekind point of view, abstract conceptual
investigation is to be preferred over calculation.</p>
</blockquote>
<p><strong>What are concrete examples from concrete fields avoiding calculations by the use of abstract notions?</strong> (Here "calculation" means any type of routine technicality.) Category theory and topoi may provide some examples.</p>
<p>Thanks in advance.</p>
| Carl Mummert | 5,442 | <p>In computability theory, it is often necessary to prove some particular function is a "computable function". Until the 1960s, this was most commonly done by actually demonstrating a formal algorithm for the function in a kind of pseudocode, or giving a set of recursion equations. Needless to say this style of presentation was heavily symbolic and conveyed little intuition about why the function was defined the way it was. </p>
<p>The more modern style of presentation relies instead on having a good sense of the closure properties of computable functions, and identifying a large class of basic computable functions (the "primitive recursive functions"). So one can simply explain how to obtain the function at hand from primitive recursive functions using operations that preserve computability. This style of proof allows for much more detailed exposition of the intuition behind the definition of a computable function. Everyone in the field understands how, in principle, to take this kind of proof and obtain a formal algorithm, if it is ever necessary. </p>
|
1,414,316 | <p>I am trying to optimize distance from point to plane using Lagrange multiplier.</p>
<p>Usually for such problems you are given specific point like (1,2,3) in 3D, and then an exact plane which is just the subject of Lagrange. But what I have here doesn't specify values for point and plane.</p>
<p>It says problem happens in a D-dimensional space. It denotes the point as X, the plane as (wT)x+b=0, where wT is the transpose of w (which is just a n-by-1 matrix I suppose), and requires me to use Lagrange to optimize the distance. The final result should be expressed with w, b and X. Without specific values I am really lost in how to approach such kind of problem. Any suggestions?</p>
| Asaf Karagila | 622 | <p>No, not necessarily.</p>
<p>Suppose $\kappa$ is measurable in $W$, let $M$ be $W[G]$ the model obtained after forcing with $\operatorname{Col}(\omega,<\kappa)$. If $j\colon W\to W'$ is an ultrapower embedding with critical point $\kappa$, then we can force over $M$ to obtain a model in which $j$ can be extended from $W$ to $M$.</p>
<p>Another important example is the stationary tower forcing, which can produce from a Woodin cardinal in $V$, an embedding [definable in $V[G]$] from $V\to M$, with critical point $\omega_1$.</p>
<p>Somewhere to look for this could be Matt Foreman's chapter in the Handbook "Ideals and Generic Elementary Embeddings". Although admittedly I just learned these facts from my advisor (I don't have notes to distribute, though, sorry).</p>
|
344,166 | <p>I was for some time curious about William Feller's probability tract (first volume); luckily, I could lay my hands on it recently and I find it of super qualities. It provides a complete exposition of elementary(no measures) probability. The book is rigorous "hard" math but doesn't escape from giving a solid intuitive feeling. The author discusses a topic, mentions an example, proposes different scenarios that gives back more math. His first chapter on "nature of probability" is essential. It gives a good feeling for what statistical probability means, and why/how it was defined as it is. </p>
<p>Question: I'm looking for other math books on fundamental mathematics(algebra, real analysis, etc...)- essential mathematics that is not very advanced(algebraic geometry for example) - of high qualities like Feller's probability text. Feller might not be used anymore, but its full of exercises that would make it a working textbook written by a master.</p>
<p>To be specific and not too general. I'm looking exactly for inspiring Feller style books in real analysis and abstract algebra. Rudin is good, but its not a master book. I don't know much about abstract algebra available textbooks/master expositions. </p>
| Asinomás | 33,907 | <p>I don't know if this is exactly what you mean but the book <a href="http://rads.stackoverflow.com/amzn/click/088385757X">visual group theory</a> is a great way to develop intuition in abstract algebra. </p>
<p>Another great book is <a href="http://www.amazon.com/s/ref=nb_sb_ss_i_0_17?url=search-alias%3Dstripbooks&field-keywords=adventures+in+group+theory&sprefix=adventures+in+gro%2Cstripbooks%2C226">adventures in group theory</a> where they use mathematical toys to give an insight of group theory </p>
<p>Finally, for a serious text I would recommend <a href="http://rads.stackoverflow.com/amzn/click/0821847813">Paolo Aluffi</a></p>
|
189,744 | <p>The following procedure can be easily aborted using Evaluation > Abort Evaluation menu item:</p>
<pre><code>Do[j + k, {j, 1, 10000}, {k, 1, 10000}]
</code></pre>
<p>But it is not possible if <code>NotebookEvaluate</code> was used:</p>
<pre><code>nb = CreateDocument[ ExpressionCell[Defer[Plot[Sin[x], {x, 0, 2 Pi}]], "Input"]]
NotebookEvaluate[nb, InsertResults -> True];
</code></pre>
<p>Now run this again:</p>
<pre><code>Do[j+k,{j,1,10000},{k,1,10000}]
</code></pre>
<p>and try to abort. It does not work and the loop is on.</p>
<p>Anyone knows why? Is there any alternative to <code>NotebookEvaluate</code>?</p>
| Jerry | 47,524 | <pre><code>Short/@Partition[Range[10^6], 1000]
Shallow/@Partition[Range[10^6], 1000]
</code></pre>
|
2,807,356 | <blockquote>
<p><strong>If $z_1,z_2$ are two complex numbers such that $\vert
z_1+z_2\vert=\vert z_1\vert+\vert z_2\vert$,then it is necessary that</strong> </p>
<p>$1)$$z_1=z_2$</p>
<p>$2)$$z_2=0$</p>
<p>$3)$$z_1=\lambda z_2$for some real number $\lambda.$</p>
<p>$4)$$z_1z_2=0$ or $z_1=\lambda z_2$ for some real number $\lambda.$</p>
</blockquote>
<p>From Booloean logic we know that if $p\implies q$ then $q$ is necessary for $p$.</p>
<p>For $1)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $1\neq 2$.So,$(1)$ is false.</p>
<p>For $2)$taking $z_1=1$ and $z_2=2$ then $\vert 1+2 \vert=\vert 1\vert+\vert 2\vert$ but $2\neq 0$.So,$(2)$ is false.</p>
<p>I'm not getting how to prove or disprove options $(3)$ and $(4)?$</p>
<p>Need help</p>
| Mundron Schmidt | 448,151 | <p><strong>Hint:</strong></p>
<p>3) Consider $\lambda=-1$ for some $z_1\neq 0$.</p>
<p>4) Complex numbers forms a field. So $z_1z_2=0$ implies $z_1=\ldots$ or $z_2=\ldots$. Hence, 4) is the same as $\ldots$ or 3)</p>
<p>Nevertheless, 3) would be true, if $\lambda$ were a real and positive number.</p>
|
1,434,630 | <p>Does $O(n\cdot n!) = O(n!)$?</p>
<p>I know that $n*n! < (n+1)!$, and in Big Oh we usually throw out constants, so it seems like we could make this conclusion. However I am not sure how to show this mathematically.</p>
| Brian M. Scott | 12,042 | <p>Look at the ratio: </p>
<p>$$\lim_{n\to\infty}\frac{n\cdot n!}{n!}=\lim_{n\to\infty}n=\infty\;,$$</p>
<p>so $n\cdot n!$ cannot be $O(n!)$.</p>
|
1,434,630 | <p>Does $O(n\cdot n!) = O(n!)$?</p>
<p>I know that $n*n! < (n+1)!$, and in Big Oh we usually throw out constants, so it seems like we could make this conclusion. However I am not sure how to show this mathematically.</p>
| Ivan Neretin | 269,518 | <p>We don't throw out constants. See, suppose we have $O(n^2)$; now what is that little figure <strong>2</strong>? A variable? Definitely not. A constant? Surely! Can't we just throw it away?</p>
<p>We may throw out additive constants, but that's another story.</p>
|
191,984 | <p>In this context composition series means the same thing as defined <a href="http://en.wikipedia.org/wiki/Composition_series#For_groups" rel="noreferrer">here.</a></p>
<p>As the title says given a finite group <span class="math-container">$G$</span> and <span class="math-container">$H \unlhd G$</span> I would like to show there is a composition series containing <span class="math-container">$H.$</span></p>
<p>Following is my attempt at it.</p>
<p>The main argument of the claim is showing the following.</p>
<blockquote>
<p><strong>Lemma.</strong> If <span class="math-container">$H \unlhd G$</span> and <span class="math-container">$G/H$</span> is not simple then there exist a subgroup <span class="math-container">$I$</span> such that <span class="math-container">$H \unlhd I \unlhd G.$</span></p>
</blockquote>
<p>The proof follows from the 4th isomorphism theorem since if <span class="math-container">$G/H$</span> is not simple then there is a normal subgroup <span class="math-container">$\overline{I} \unlhd G/H$</span> of the form <span class="math-container">$\overline{I} = H/I.$</span></p>
<p>Suppose now that <span class="math-container">$G/H$</span> is not simple. Using the above lemma we deduce that there exist a finite chain of groups (since <span class="math-container">$G$</span> is finite) such that <span class="math-container">$$H \unlhd I_1 \unlhd \cdots \unlhd I_k \unlhd G$$</span> and <span class="math-container">$G/I_k$</span> is simple. Now one has to repeat this process for all other pairs <span class="math-container">$I_{i+1}/I_{i}$</span> and for <span class="math-container">$I_1/H$</span> until the quotients are simple groups. This is all fine since all the subgroups are finite as well.</p>
<p>Now if <span class="math-container">$H$</span> is simple we are done otherwise there is a group <span class="math-container">$J \unlhd H$</span> and we inductively construct the composition for <span class="math-container">$H.$</span></p>
<p>Is the above "proof" correct? If so, is there a way to make it less messy?</p>
| rschwieb | 29,335 | <p>Your approach has all the elements of a proof, but I wanted to offer suggestions for a "less messy" version.</p>
<p>First of all, as long as you are using that "any finite group has a composition series", you shouldn't need your lemma.</p>
<p>You could go about it this way. Let <span class="math-container">$H$</span> be a normal subgroup of <span class="math-container">$G$</span>. Then <span class="math-container">$H$</span> has a composition series <span class="math-container">$1=H_0<\dots<H_k=H$</span>. The group <span class="math-container">$G/H$</span> also has a composition series, which we will enumerate this way:</p>
<p><span class="math-container">$$
H/H=H_{k}/H<\dots <H_n/H=G/H
$$</span></p>
<p>By an isomorphism theorem, the fact that <span class="math-container">$(H_{j}/H)/(H_{j-1}/H)$</span> is simple is equivalent to <span class="math-container">$H_j/H_{j-1}$</span> being simple.</p>
<p>Putting these two chains together, you have that <span class="math-container">$H_0<\dots <H_n$</span> is a composition series for <span class="math-container">$G$</span> through <span class="math-container">$H$</span>.</p>
|
3,458,406 | <p>Define <span class="math-container">$f(n)=(2n)^2 + 1,n \in \mathbb{N}$</span></p>
<p>From <span class="math-container">$1$</span> to <span class="math-container">$10^7$</span> there's <span class="math-container">$15$</span> numbers that <span class="math-container">$f(n)$</span> is prime, <span class="math-container">$f(f(n)), f(f(f(n)))$</span> and <span class="math-container">$f(f(f(f(n))))$</span> are also primes.</p>
<p>The <span class="math-container">$15$</span> numbers are:</p>
<pre><code>625678, 704613, 717718, 1182168, 3147353,
4869813, 5339178, 5363578, 5411562, 846777,
7848283, 7970403, 8152962, 9220303, 9727978
</code></pre>
<p>Is that normal <span class="math-container">$4$</span> numbers in this <span class="math-container">$15$</span> numbers end in <span class="math-container">$78$</span>?</p>
| Community | -1 | <p>To expand on my comment, here are the first 15 values of f(n):</p>
<p><span class="math-container">$$5,17,37,65,101,145,197,257,325,401,485,577,677,785,901$$</span> Because of symmetry for certain primes (those congruent to <span class="math-container">$1 \bmod 4$</span>), this can test those up as high as 29 (possibly higher) if you do mod these by those primes (<span class="math-container">$5,13,17,29$</span>) you get the following sequences (respectively):<span class="math-container">$$0,2,2,0,1,0,2,2,0,1,0,2,2,0,1$$</span> <span class="math-container">$$5,4,11,0,10,2,2,10,0,11,4,5,1,5,4$$</span> <span class="math-container">$$5,0,3,14,16,9,10,2,2,10,9,16,14,3,0$$</span> <span class="math-container">$$5,17,8,7,14,0,23,25,6, 24, 21, 26, 10, 2, 2$$</span> disallowing: <span class="math-container">$$1,4\bmod 5$$</span><span class="math-container">$$4,9\bmod 13$$</span><span class="math-container">$$2,15\bmod 17$$</span><span class="math-container">$$6,23\bmod 29$$</span> as start values( in fact, at best end iterate values if 1 more than a quadratic residue), continuing we get: <span class="math-container">$$0\bmod 5$$</span><span class="math-container">$$2,11\bmod 13$$</span><span class="math-container">$$8,9\bmod 17$$</span><span class="math-container">$$ 8,10,19,21\bmod 29$$</span> as illegal prior to the second last iterate : <span class="math-container">$$1,4\bmod 5$$</span><span class="math-container">$$3,6,7,10\bmod 13$$</span><span class="math-container">$$6,11\bmod 17$$</span><span class="math-container">$$3,11,13,16,18,26\bmod 29$$</span> for third last iterate. Taking the complement, of the union,for each prime; we get:<span class="math-container">$$2,3\bmod 5$$</span><span class="math-container">$$0,1,5,8,12\bmod 13$$</span><span class="math-container">$$0,1,3,4,5,7,10,12,13,14,16\bmod 17$$</span><span class="math-container">$$0,1,2,4,5,7,9,12,14,15,17,20,22,24,25,27,28\bmod 29$$</span> Long story short is just 2 residues mod 5, 5 residues mod 13, 11 residues mod 17, and 17 residues mod 29 survive as possible start residues, for 1870 residues ( fixed calculation error in the edit, aka under 5.9% of residue classes ) mod their product (32045) and 20 times their product is a multiple of 100 so you can go from there.</p>
<p><strong>EDIT</strong></p>
<p>Turns out it's just as likely as any other ending via this, you just seem to be getting roughly 1 every multiple of 640900. might have to do with placement in multiples of 640900 if any of these numbers where divisible by a prime that is of form <span class="math-container">$4x+3$</span>, we get it divisible by an even number of them. Any ending that is 0 mod 3, shows up only if 640900 isn't multiplied by a multiple of 3, and non-zero residue two digit endings don't happen when 640900 multiplied by their additive inverse mod 3. Might explain some of why they are spread out. maybe some multipliers are just denser in working values, when you consider other factors.</p>
|
2,476,717 | <p>Let $f: \mathbb R^n \rightarrow \mathbb R^n$ with arbitrary norm $\|\cdot\|$. It exists a $x_0 \in \mathbb R^n$ and a number $r \gt 0 $ with </p>
<p>$(1)$ on $B_r(x_0)=$ {$x\in \mathbb R^n: \|x-x_0\| \leq r$} $f$ is a contraction with Lipschitz constant L</p>
<p>$(2)$ it applies $\|f(x_0)-x_0\| \le (1-L)r$</p>
<p>The sequence $(x_k)_{k\in\mathbb N}$ is defined by $x_{k+1}=f(x_k).$</p>
<p>How do I show that $x_k \in B_r(x_0) \forall k \in \mathbb N$?</p>
<p>How do I show that $f$ has a unique Fixed point $x_f$ with $x_f = \lim_{k \rightarrow\infty} x_k$?</p>
<p>I know this has something to do with Banach but I am totally clueless on how to prove this. Any help is welcome. Thanks. </p>
| Lutz Lehmann | 115,115 | <p>As for the unique fixed point, one shows that the iteration sequence is related to a geometric sequence with factor $L$ and uses that to show that it is a Cauchy sequence. Completeness of the space gives existence, contractivity the uniqueness of the fixed point.</p>
|
2,662,554 | <p>I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even. </p>
<p>Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$. </p>
<p>What I have done so far:</p>
<p>\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}</p>
<p>Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake? </p>
<p>Thank you.</p>
| Mr Pie | 477,343 | <p>If $n + 1$ is even, then $n$ must be odd. Let $n = 2k + 1$ since it is in the form of an odd number. Now substitute into the quadratic expression: $$\begin{align} n^2 - 2n + 7 &= (2k+1)^2 - 2(2k+1) + 7 \\ &= (2k)^2 + 1 + 4k - 4k - 2 + 7 \\ &= 4k^2 + 6 \\ &= 2(2k^2 + 3).\end{align}$$ Since the quadratic is an even number, this also completes the proof. For more sensibility, since $n + 1$ is even, we can let $n + 1 = 2r$ and then let $n = 2r - 1$, but the same is implied anyway.</p>
|
2,396,561 | <blockquote>
<p>Solve the differential equation
$$x^2y''+xy'-y=\frac{x^2}{2+x}$$</p>
</blockquote>
<p>My attempt: put $x=e^z$ then $z=\log x$</p>
<p>and equation reduced to $\theta^2-1=\frac{e^{2z}}{2+e^z}$</p>
<p>so $y_c=c_1x+c_2\frac{1}{x}$ </p>
<p>how to find particular integral </p>
| Guy Fsone | 385,707 | <p>Using successively the change of variables $x = \sqrt{a}t$, $u = t^2$ and $ s =\frac{1}{1+u}$ we obtain</p>
<p>$$ I_{n,a} = 2\sqrt{a}\int_{0}^{\infty} \frac{dt}{(1+t^2)^n} = \sqrt{a}\int_{0}^{\infty} \frac{ u^{-1/2}du}{(1+u )^n} = \sqrt{a}\int_{0}^{1}s^{n-1/2-1} (1-s)^{-1/2} ds $$ </p>
<p>Then we obviously see that $0\le I_{n,a}\le I_{n-1,a}$ i.e a nondecreasing and bounded sequence this give the convergence of $(I_{n,a})_n$.
On other hand integration by path give </p>
<p>$$ I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$.</p>
<p>Also we can from the given formula above that </p>
<p>$$I_{n,a}=\sqrt{a} B\left(1/2,n-1/2\right)=\sqrt{a}\frac{\Gamma\left(1/2\right)\Gamma\left(n-1/2\right)}{\Gamma\left(n\right)}$$</p>
<p>Using the following famous Stirling formula: Given $x>0$
$$ \lim_{x\to +\infty} \frac{\Gamma(x+1)}{\left(\frac{x}{e}\right)^x \sqrt{2\pi x} }=1. $$
Where $\Gamma$ is the Gamma function of Euler and $n! =\Gamma(n+1)$ for $n\in \mathbb{N}$ .</p>
<p>we find out that,
\begin{split}
\lim_{n\to \infty} I_{n+1,n+1} &=&\Gamma\left(1/2\right)\lim_{n\to \infty} \sqrt{n+1}\frac{\Gamma\left(n+1-1/2\right)}{\Gamma\left(n+1\right)}\\
&= & \Gamma\left(1/2\right)\lim_{n\to \infty} \sqrt{n+1}\left(\frac{e}{n}\right)^n\left(\frac{n-1/2}{e}\right)^{n-1/2} \\
&=& \Gamma\left(1/2\right)\lim_{n\to \infty} \sqrt{\frac{n+1}{n}}\left(\frac{n-1/2}{n}\right)^{n-1/2}e^{1/2} \\
&=& e^{1/2} \Gamma\left(1/2\right)\lim_{n\to \infty} \left(\frac{n-1/2}{n}\right)^{n-1/2}
\end{split}</p>
<p>But (easy tocheck)</p>
<p>$$\lim_{n\to \infty} \left(\frac{n-1/2}{n}\right)^{n-1/2} =e^{-1/2} $$</p>
<p>Finally we have $$ \lim_{n\to \infty} I_{n,n} = \Gamma(1/2) := \int_0^{\infty} e^{-t} t^{1/2-1} dt = \int_{-\infty}^{\infty}e^{-x^2}dx$$</p>
|
4,605,888 | <p>Let's say I have a uniformly distributed random number sequence whose values are in the range [1, <strong>m</strong>]. Each value has a chance of <strong>p</strong> = 1/<strong>m</strong> appearing. Take a sample of size <strong>s</strong> from that sequence. For a given value in the sample, let <strong>n</strong> be the number of duplicates in the rest of the sample. Let <strong>c</strong> be the count of numbers in the sample that have the same <strong>n</strong>.</p>
<p>For example:</p>
<ul>
<li>if <strong>n</strong> = 0, then <strong>c</strong> is the number of unique values in the sample.</li>
<li>if <strong>n</strong> = 1, then <strong>c</strong> is the number of values that have one same value in the rest of the sample.</li>
</ul>
<p>So for a given <strong>n</strong>, how to calculate the expected <strong>c</strong>?</p>
<p>Below are some simulated results.</p>
<p>for <strong>m</strong> = 65536, <strong>s</strong> = 1000</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>n</th>
<th>c</th>
</tr>
</thead>
<tbody>
<tr>
<td>0</td>
<td>978</td>
</tr>
<tr>
<td>1</td>
<td>22</td>
</tr>
<tr>
<td>2</td>
<td>0</td>
</tr>
</tbody>
</table>
</div>
<p>for <strong>m</strong> = 65536, <strong>s</strong> = 10,000</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>n</th>
<th>c</th>
</tr>
</thead>
<tbody>
<tr>
<td>0</td>
<td>8598</td>
</tr>
<tr>
<td>1</td>
<td>1284</td>
</tr>
<tr>
<td>2</td>
<td>114</td>
</tr>
<tr>
<td>3</td>
<td>4</td>
</tr>
<tr>
<td>4</td>
<td>0</td>
</tr>
</tbody>
</table>
</div>
<p>for <strong>m</strong> = 65536, <strong>s</strong> = 100,000</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>n</th>
<th>c</th>
</tr>
</thead>
<tbody>
<tr>
<td>0</td>
<td>21806</td>
</tr>
<tr>
<td>1</td>
<td>33326</td>
</tr>
<tr>
<td>2</td>
<td>24915</td>
</tr>
<tr>
<td>3</td>
<td>13076</td>
</tr>
<tr>
<td>4</td>
<td>4865</td>
</tr>
<tr>
<td>5</td>
<td>1566</td>
</tr>
<tr>
<td>6</td>
<td>350</td>
</tr>
<tr>
<td>7</td>
<td>96</td>
</tr>
<tr>
<td>8</td>
<td>0</td>
</tr>
</tbody>
</table>
</div> | pre-kidney | 34,662 | <p>The mathematical content behind this question has already been answered here:
<a href="https://math.stackexchange.com/questions/1056540">Sampling with replacement - Expected number of duplicates, triplicates, ..., n-tuples</a></p>
<p>However, the phrasing is slightly different, so let me help translate. In this problem, we have a sequence of random variables <span class="math-container">$X_1,\ldots,X_s$</span>, where each <span class="math-container">$X_i$</span> is uniformly distributed in <span class="math-container">$\{1,\ldots,m\}$</span>, and we wish to compute the expected number of <span class="math-container">$X_i$</span>'s which are unique (<span class="math-container">$n=0$</span>), repeat exactly twice (<span class="math-container">$n=1$</span>), and so on, for all <span class="math-container">$n$</span>.</p>
<p>In the question I linked to above, the names of the variables are different but it is the same mathematical setup. The <span class="math-container">$m$</span> in our problem here, is the genome size in the other question. The <span class="math-container">$s$</span> in our problem here, is the sample size of the number of genes in the other question. Rephrasing the answer of <a href="https://math.stackexchange.com/users/6460/henry">https://math.stackexchange.com/users/6460/henry</a> in this notation, the probability of seeing a particular value of <span class="math-container">$X_i$</span> exactly <span class="math-container">$k$</span> times is
<span class="math-container">$$
\binom{s}{k}\frac{(m-1)^{s-k}}{m^s}
$$</span>
and therefore the expected number of indices <span class="math-container">$i$</span> such that the value <span class="math-container">$X_i$</span> appears exactly <span class="math-container">$k$</span> times is
<span class="math-container">$$
k\binom{s}{k}\frac{(m-1)^{s-k}}{m^{s-1}}.
$$</span>
A value that appears exactly <span class="math-container">$k$</span> times corresponds to an <span class="math-container">$n$</span> value of <span class="math-container">$k-1$</span> in the notation of this question, so we obtain the formula:
<span class="math-container">$$
\mathbb E[c_n]=(n+1)\binom{s}{n+1}\frac{(m-1)^{s-n-1}}{m^{s-1}}
$$</span></p>
|
3,806,122 | <p>I tried using Chinese remainder theorem but I kept getting 19 instead of 9.</p>
<p>Here are my steps</p>
<p><span class="math-container">$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$</span></p>
| Jonathan Gai | 814,615 | <p>You used the inverse of <span class="math-container">$x_i$</span> instead of the inverse of <span class="math-container">$\frac{M}{m_i}$</span>. So for example,
<span class="math-container">$$
9 \cdot \frac{88}{11} \cdot 5 + 1 \cdot \frac{88}{8} \cdot 1 \equiv 0 + 1 \cdot 11 \cdot 1 \equiv 3 \not \equiv 1\pmod{8}.
$$</span>
If you use the inverse of <span class="math-container">$\frac{M}{m_i}$</span> instead, you would have
<span class="math-container">$$
x_1 \cdot \frac{88}{11} \cdot 8^{-1} + x_2 \cdot \frac{88}{8} \cdot 11^{-1} \equiv 0 + 1 \cdot 11 \cdot 11^{-1} \equiv 1\pmod{8}.
$$</span></p>
|
1,494,022 | <p>Find a closed expression in terms of $n$.
$$\sum_{k=1}^n(k!)(k^2+k+1); n=1,2,3...$$<br>
Any idea about how to do this.. I'm a new to this so a little explanation would be helpful. Thanks in advance!</p>
| Ron Gordon | 53,268 | <p>Write $k^2+k+1 = (k+1)^2-k$ so that</p>
<p>$$(k!)(k^2+k+1) = (k+1)(k+1)! - k \, k!$$</p>
<p>Now you have a telescoping sum.</p>
|
713,104 | <p>Are there any combinatorial games whose order (in the usual addition of combinatorial games) is finite but neither $1$ nor $2$?</p>
<p>Finding examples of games of order $2$ is easy (for example any impartial game), but I have not been able to think up an example with finite order where the order did not come from some sort of symmetry (for example even though Domineering is not impartial, it is easy to see that any square board will give a game of order $1$ or $2$), and such a symmetry only gives $1$ or $2$ as the possible orders.</p>
| Gottfried Helms | 1,714 | <p>I think the initial steps towards a formula look a bit simpler than seen so far here.</p>
<p>Beginning with
<span class="math-container">$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$</span>
we count the number of occurences of the <span class="math-container">$1$</span>, of the <span class="math-container">$2$</span>
<span class="math-container">$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$</span>
and rewrite a bit
<span class="math-container">$$ \begin{align} H^{(2)}(n) &= 1\cdot ((n+1)-1) + 2 \cdot ((n+1)-2) + 3\cdot ((n+1)-3) + ... + n \cdot ((n+1)-n) \\ \qquad \\
&= (1+2+3+...+n)(n+1) - (1^2+2^2+3^2+...+n^2) \\ \qquad \\
&= {n(n+1) \over 2}(n+1) - {2n^3 + 3n^2 + 1n\over 6} \end{align}$$</span>
The last steps stems from the faulhaber formulae for sums-of-like-powers.</p>
<hr />
<p>Final step, expanding & collecting this gives the same result as derived by other answerers:
<span class="math-container">$$ H^{(2)}(n) = {3n(n+1)(n+1)-2n^3 - 3n^2 - 1n \over 6} \\
= {n^3+3n^2+2n \over 6} \\
= {n(n+1)(n+2) \over 3!} $$</span>
For <span class="math-container">$n=5$</span> we get the sum <span class="math-container">$H^{(2)}(5)=35$</span></p>
|
135,252 | <p>Evaluate $\displaystyle \lim_{n \to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$. What are the ways of counting such things? My last topic in school was Riemann integral, can I use it here?</p>
| Ragib Zaman | 14,657 | <p>You sure can!</p>
<p>$$ \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{(k/n)} .$$</p>
<p>The second version of writing the sum makes it clearer that it is the Riemann sum of $f(x) = 1/x $ obtained by dividing $[1,2]$ into $n$ pieces and setting up rectangles over those intervals. As such, your limit is $ \int^2_1 \frac{1}{x} dx = \log 2.$</p>
|
4,234,095 | <p>I need to show that <span class="math-container">$[\mathbb{Q}(2^{1/4},2^{1/6}):\mathbb{Q}]$</span> is a field extension of degree <span class="math-container">$12$</span>. It is possible to show that the degree is at least <span class="math-container">$12$</span> because it is divisible by <span class="math-container">$6$</span> and <span class="math-container">$4$</span> by finding the minimal polynomial of the simple field extensions of <span class="math-container">$2^{1/4}$</span> and <span class="math-container">$2^{1/6}$</span>, but I am not sure how to bound the inequality in the other direction.</p>
<p>Another way to approach this problem might just be to explicitly find the basis, but I think there should be a way to find a bound on the inequality.</p>
| ndhanson3 | 808,617 | <p>Degrees of field extensions are multiplicative. Can you find <span class="math-container">$[\mathbb{Q}(2^{1/4}):\mathbb{Q}]$</span> by explicitly producing the minimal polynomial of <span class="math-container">$2^{1/4}$</span> over <span class="math-container">$\mathbb{Q}$</span>? Not too hard. You'll find the degree of this extension is <span class="math-container">$4$</span>.</p>
<p>Then can you find <span class="math-container">$[\mathbb{Q}(2^{1/4},2^{1/6}):\mathbb{Q}(2^{1/4})]$</span> by finding the minimal polynomial of <span class="math-container">$2^{1/6}$</span> over <span class="math-container">$\mathbb{Q}(2^{1/4})$</span>? Since you know the degree of the full extension should be <span class="math-container">$12$</span>, the degree of this extension should be <span class="math-container">$3$</span>. So perhaps a polynomial of degree <span class="math-container">$3$</span>. To show that the polynomial you get is irreducible over <span class="math-container">$\mathbb{Q}(2^{1/4})$</span>, simply find its roots in <span class="math-container">$\mathbb{C}$</span> and note that they do not lie in <span class="math-container">$\mathbb{Q}(2^{1/4})$</span>.</p>
|
3,562,294 | <p>I got <span class="math-container">$x^2+y^2$</span> could factorized by <span class="math-container">$(x+yi)(x-yi)$</span></p>
<p>But Could we get factorization of <span class="math-container">$x^2+y^2+1$</span></p>
<p>I tried <span class="math-container">$(x+yi+i)(x-yi-i)$</span> but i couldn't guess it.</p>
<p>By FTA, It is possible but I couldn't gess it....</p>
| Robert Israel | 8,508 | <p>It is not possible to factor <span class="math-container">$x^2 + y^2 + 1$</span> into polynomials in <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, i.e. it is irreducible over <span class="math-container">$\mathbb C[x,y]$</span>. Of course you could factor into polynomials in <span class="math-container">$x$</span> as <span class="math-container">$(x + \sqrt{-1-y^2})(x - \sqrt{-1-y^2})$</span>, but these are not polynomials in <span class="math-container">$y$</span>.</p>
|
4,235,480 | <p><span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~~ \leftarrow~~ \text{The set of integers each of which is greater or equal than zero} ~$</span></p>
<p>I want to know or create the alternative(s) of set of <span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~$</span></p>
<p>As I write <span class="math-container">$~ \mathbb{N} \cup \left\{ 0 \right\} ~$</span> using a pen in a paper, then a symbol of <span class="math-container">$~ \cup ~$</span> may sometimes be seen as letter U and the ambiguity seems happens.</p>
<p>Does anyone know it?</p>
<p><strong>Add</strong></p>
<p>Thought that <span class="math-container">$~ \mathbb Z_{\geq 0} ~$</span> may be the one of the answers .</p>
| ancient mathematician | 414,424 | <p>(a)
At the outset we have (if we add the trivial character)
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\\
\\
\\
\end{matrix}.
$$</span></p>
<p>(b) Recall that the complex conjugate of an irreducible character is also an irreducible character and get
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\\
1 & &\zeta^2\\
\\
\end{matrix}.
$$</span></p>
<p>(c) Use the fact that <span class="math-container">$|G|=1+3+4+4$</span> to get
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\\
1 & &\zeta^2\\
3& \\
\end{matrix}.
$$</span></p>
<p>(d) Use the fact that <span class="math-container">$|C_3|=4$</span> to see that the length of the 3rd column is <span class="math-container">$12/4=3$</span> and so get
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\\
1 & &\zeta^2\\
3& &0 &\\
\end{matrix}.
$$</span></p>
<p>(e) Note that as the second character is not real on <span class="math-container">$C_3$</span> the elements of <span class="math-container">$C_3$</span> are not conjugate to their inverses, so their inverses form a distinct class, say <span class="math-container">$C_4$</span>, and each character on <span class="math-container">$C_4$</span> is the complex conjugate of the values on <span class="math-container">$C_3$</span>, since the eigenvalues of the representation on <span class="math-container">$C_4$</span> are the inverses of the eigenvalues of the representation on <span class="math-container">$C_3$</span>. So we have
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\zeta^2\\
1 & &\zeta^2&\zeta\\
3& &0 &0\\
\end{matrix}.
$$</span></p>
<p>(f) Use the fact that the second column is orthoogonal to each of the other columns, to see first that it is orthogonal to the sum of the other three columns, and so get
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& & \zeta&\zeta^2\\
1 & &\zeta^2&\zeta\\
3&-1 &0 &0\\
\end{matrix}.
$$</span></p>
<p>(g) Use whatever orthogonality relations you like to complete the table and get
<span class="math-container">$$
\begin{matrix}
1 & 1 &1 &1\\
1& 1& \zeta&\zeta^2\\
1 &1 &\zeta^2&\zeta\\
3&-1 &0 &0\\
\end{matrix}.
$$</span></p>
<p><em>Note. I think it is possible to do this without the hypotheses that the four classes have orders <span class="math-container">$1,3,4,4$</span> but it's a bit harder.</em></p>
|
2,209,034 | <blockquote>
<p>The question asks to compute:
<span class="math-container">$$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$</span>
where <span class="math-container">$\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$</span> are the <span class="math-container">$n$</span>-th roots of unity.</p>
</blockquote>
<p>I started off by simplifiyng and got it as:</p>
<p><span class="math-container">$$=-n+2\left(\sum_{k=0}^{n-1} \dfrac{1}{2-\alpha_k}\right)$$</span></p>
<p>Now I was stuck. I can rationalise the denominator, but we know <span class="math-container">$\alpha_k$</span> has both real and complex components, so it can't be simplified by rationalising. What else can be done?</p>
| jonsno | 310,635 | <p>I think the following answers the question using the method that <strong>Jyrki</strong> posted <a href="https://math.stackexchange.com/a/1909368/310635">here</a>.</p>
<p>Since <span class="math-container">$\alpha_k$</span> are nth roots, so they satisfy <span class="math-container">$$f(x)=x^n-1=\prod_{k=0}^{n-1}(x-\alpha_k)$$</span></p>
<p>Putting in logarithm and derivating,</p>
<p><span class="math-container">$$f'(x)/f(x)=\dfrac{nx^{n-1}}{x^n-1}=\sum_{k=0}^{n-1}\dfrac{1}{x-\alpha_k}$$</span></p>
<p>Thus <span class="math-container">$$f'(2)/f(2) = \sum_{k=0}^{n-1}\dfrac{1}{2-\alpha_k} = \dfrac{n\cdot 2^{n-1}}{2^n-1}$$</span></p>
<p>Thus the required answer is given as:</p>
<p><span class="math-container">$$-n+ 2\left(\dfrac{n\cdot 2^{n-1}}{2^n-1}\right)$$</span>
<span class="math-container">$$=\dfrac{n}{2^n-1}$$</span></p>
|
2,209,034 | <blockquote>
<p>The question asks to compute:
<span class="math-container">$$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$</span>
where <span class="math-container">$\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$</span> are the <span class="math-container">$n$</span>-th roots of unity.</p>
</blockquote>
<p>I started off by simplifiyng and got it as:</p>
<p><span class="math-container">$$=-n+2\left(\sum_{k=0}^{n-1} \dfrac{1}{2-\alpha_k}\right)$$</span></p>
<p>Now I was stuck. I can rationalise the denominator, but we know <span class="math-container">$\alpha_k$</span> has both real and complex components, so it can't be simplified by rationalising. What else can be done?</p>
| Marko Riedel | 44,883 | <p>For future reference here is a solution using residues. We have that
with <span class="math-container">$\zeta_k = \exp(2\pi i k/n)$</span> so that <span class="math-container">$\zeta_k^n = 1$</span></p>
<p><span class="math-container">$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k}
\frac{1}{2-z} \frac{n}{z^n-1}
= \sum_{k=0}^{n-1} \left.
\frac{1}{2-z} \frac{n}{nz^{n-1}} \right|_{z=\zeta_k}
\\ = \sum_{k=0}^{n-1} \left.
\frac{1}{2-z} \frac{z}{z^{n}} \right|_{z=\zeta_k}
= \sum_{k=0}^{n-1} \frac{\zeta_k}{2-\zeta_k}$$</span></p>
<p>which is our sum <span class="math-container">$S_n.$</span></p>
<p>Now observe that</p>
<p><span class="math-container">$$\mathrm{Res}_{z=2}
\frac{1}{2-z} \frac{n}{z^n-1}
= -\frac{n}{2^n-1}.$$</span></p>
<p>Furthermore the residue at infinity</p>
<p><span class="math-container">$$\mathrm{Res}_{z=\infty}
\frac{1}{2-z} \frac{n}{z^n-1} = 0$$</span></p>
<p>since we have the bound <span class="math-container">$2\pi n R / R /R^n = 2\pi n / R^n \rightarrow
0$</span> as <span class="math-container">$R\rightarrow\infty.$</span> Residues sum to zero and we get</p>
<p><span class="math-container">$$S_n - \frac{n}{2^n-1} = 0
\quad\text{or}\quad
S_n = \frac{n}{2^n-1}$$</span></p>
<p>as claimed.</p>
|
2,213,626 | <p>How can you prove that if the gcd(a,b) = 1 then gcd(a,bi) = 1 in the Gaussian integers? I know that $i$ is a unit in the ring, but how can you rigorously prove this?</p>
| Jean Marie | 305,862 | <p>Hints (I do not pretend to have a complete solution) </p>
<p>1) You can transform your equation in order to have a single parameter instead of 2:</p>
<p>Look for solutions of the form $y=\sqrt[4]{a}z$.</p>
<p>In this way, the given differential equation becomes:</p>
<p>$$\tag{1} \ z''+\dfrac{1}{z^3}=c \ \ \ \text{where} \ \ \ c:=\tfrac{b}{\sqrt[4]{a}} >0$$</p>
<p>2) A solution to the homogeneous equation associated with (1) (RHS = $0$) is:</p>
<p>$$z=\sqrt{2x}$$</p>
|
2,571,395 | <p>I recently reached got a nice answer from my <a href="https://math.stackexchange.com/questions/2567486/integrating-int-x-1x-2-sqrt12at-ax-1x-22-dt">previous question</a> but I quickly that the problem would be unreasonable unless $x_1$ is not a variable and always holds some value, preferably 0, which simplifies the problem significantly leading to the above $L=\frac{1}{a}\int_0^{ax_2}\sqrt{1+t^2}dt$. I know that you can take the easy first step to get $La=\int_0^{ax_2}\sqrt{1+t^2}dt$ and then take a derivative to get rid of the integral, but I am unsure what variable(s) it would make sense to take the derivative in respect to, would I take the derivative in terms of $ax_2$? And then if that was the case, how would I take the derivative of $La$ in respect to $ax_2$?</p>
| Community | -1 | <p>Note that evaluating $I = \int \sqrt{1+t^2}\,\, dt$ has already been asked before <a href="https://math.stackexchange.com/questions/431143/evaluating-int-sqrt1-t2-dt">on MSE</a>.</p>
<p>Now, calculating the definite integral from $0$ to $ax_2$ gives us the answer as $$\boxed{L = \frac{1}{2a}[\sinh^{-1}(ax_2)+ax_2\sqrt{1+a^2x_2^2}]}$$</p>
|
4,286,296 | <p>I am trying to prove the following claim:</p>
<blockquote>
<p>Let <span class="math-container">$ 0\leq n \in \Bbb Z$</span> and suppose that there exists a <span class="math-container">$k \in \Bbb Z$</span> such that <span class="math-container">$n=4k+3$</span>.
Prove or disprove: <span class="math-container">$\sqrt n \notin \Bbb Q$</span> .</p>
</blockquote>
<p>The problem I am having is that I am trying to assume by contradiction that <span class="math-container">$\sqrt n \in \Bbb Q$</span> and then I say that there are <span class="math-container">$a,b \in \Bbb Z$</span> such that <span class="math-container">$n=\sqrt {4k+3}=\frac ab$</span>. I finally get to a point where <span class="math-container">$k=\frac {a^2-3b^2}{4b^2}$</span>. Yet I can't find any <span class="math-container">$a,b \in \Bbb Z$</span> that will help me show that the claim is false, nor show a contradiction that will cause the claim to be true.
Any help will be welcomed.</p>
| Servaes | 30,382 | <p>It is given that <span class="math-container">$n=4k+3$</span>, so I think you mean to say that there exist <span class="math-container">$a,b\in\Bbb{Z}$</span> such that
<span class="math-container">$$\sqrt{n}=\sqrt{4k+3}=\tfrac ab,$$</span>
in the hopes of reaching a contradiction. Indeed some algebra then leads to
<span class="math-container">$$k=\frac{a^2-3b^2}{4b^2},$$</span>
which means that <span class="math-container">$4b^2$</span> should divide <span class="math-container">$a^2-3b^2$</span> because <span class="math-container">$k$</span> is an integer. In particular <span class="math-container">$4$</span> should divide <span class="math-container">$a^2-3b^2$</span>. This implies that <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are both even [prove this!], say <span class="math-container">$a=2A$</span> and <span class="math-container">$b=2B$</span>. Plugging this in then gives
<span class="math-container">$$k=\frac{(2A)^2-3(2B)^2}{4(2B)^2}=\frac{4A^2-12B^2}{16B^2}=\frac{A^2-3B^2}{4B^2},$$</span>
and so by the same argument <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are again both even. Can you see the contradiction from here?</p>
|
1,643,013 | <p>I have just started learning about differential equations, as a result I started to think about this question but couldn't get anywhere. So I googled and wasn't able to find any particularly helpful results. I am more interested in the reason or method rather than the actual answer. Also I do not know if there even is a solution to this but if there isn't I am just as interested to hear why not.</p>
<p>Is there a solution to the differential equation:</p>
<p>$$f(x)=\sum_{n=1}^\infty f^{(n)}(x)$$</p>
| s.harp | 152,424 | <p>$f(x)=\exp(\frac{1}{2}x)$ is such a function, since $f^{(n)}=2^{-n} f(x)$, you have</p>
<p>$$\sum_{n=1}^\infty f^{(n)}(x)=\sum_{n=1}^\infty 2^{-n}f(x)=(2-1)f(x)=f(x)$$</p>
<p>This is the only function (up to a constant prefactor) for which $\sum_{n}f^{(n)}$ and its derivatives converge uniformly (on compacta), as $$f'=\sum_{n=1}^\infty f^{(n+1)}=f-f'$$ follows from this assumption. But this is the same as $f-2 f'=0$, of which the only (real) solutions are $f(x)= C \exp{\frac{x}{2}}$ for some $C \in \mathbb R$.</p>
|
396,297 | <p>Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?</p>
<p>I've calculated that the recurrence relation for this integral is:</p>
<p>$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$</p>
<p>So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.</p>
<p>Can this be useful here?</p>
<p>I would appreciate all your help.</p>
| Hu Zhengtang | 53,845 | <p>Here is an alternative argument. Note that for every $a\in [0,1)$,
$$\frac{(1-a^{n+1})e^a}{n+1}=e^a\int_a^1x^ndx\le\int_a^1x^ne^xdx\le\int_0^1x^ne^xdx\le e \int_0^1x^ndx=\frac{e}{n+1}.$$
Multiplying the inequality above by $2n+1$ and letting $n\to\infty$, it follows that:
$$2e^a\le\liminf_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le\limsup_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le 2e.$$
Since $a\in [0,1)$ is arbitrary, we can conclude that
$$\lim_{n\to\infty}(2n+1)\int_0^1x^ne^xdx=2e.$$</p>
|
129,261 | <p>I need to prove several inequalities trivially. (i.e. without using AM-GM, re-arrangement etc). I just keep hitting a blank. Could anyone help?</p>
<p>$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$</p>
| Tomarinator | 21,832 | <p>we know that,
$$ \frac{x^{4}+y^{4}+z^{4}}{3}\geq (\frac{x+y+z}{3})^{4} $$</p>
<p>$$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)^{4} $$</p>
<p>$\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$</p>
<p>$(x+y+z)^3 \geq 27xyz$</p>
<p>hence
$$ x^{4}+y^{4}+z^{4}\geq \frac{1}{27}(x+y+z)(27xyz) $$</p>
<p>$$x^{4}+y^{4}+z^{4}\geq x^{2}yz+xy^{2}z+xyz^{2}$$</p>
|
1,978,935 | <p>Let $f,g : [a,b] \rightarrow \mathbb{R}$ be continuous. We know that $f$ and $g$ have maximal values, as they are continuous on a closed interval. Let $M_f$ be the maximal value of $f$, and $M_g$ the maximal value of $g$. Show that if $M_f$ = $M_g$, then there exists $\psi \in [a,b]$ with $f(\psi) = g(\psi)$</p>
<p>Would it suffice to show that $\psi$ = maximal values, and show that this is an example which shows the exist of such a $\psi$?</p>
| Prahlad Vaidyanathan | 89,789 | <p>Suppose that $f(x_1) = M_f$ and $g(x_2) = M_g$ and assume without loss of generality that $x_1 < x_2$. Now consider $h:=f-g$ restricted to the interval $[x_1,x_2]$. Note that
$$
h(x_1) = f(x_1) - g(x_1) = M_g - g(x_1) \geq 0
$$
and
$$
h(x_2) = f(x_2) - M_f \leq 0
$$
So if $h(x_1) = 0$ or $h(x_2) = 0$ we are done. Else, the intermediate value theorem applies.</p>
|
2,872,701 | <blockquote>
<p>Let $K\subset N\subset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.</p>
</blockquote>
<p>It is easy to show that </p>
<blockquote>
<p>If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.</p>
</blockquote>
<p>I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.</p>
<blockquote>
<p>Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/K\oplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/K\oplus N'/K=M/K$.</p>
</blockquote>
<p>I want to say something like $N\oplus N'=M$ which is not true since $N'\cap N\supset K\neq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=N\oplus B$ but $B$ is not a submodule.</p>
| Julian Kuelshammer | 15,416 | <p>The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=\langle e_3\rangle \subset N=\langle e_2, e_3\rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.</p>
|
3,555,084 | <blockquote>
<p>Let
<span class="math-container">$$f(z) = e^z (1+\cos\sqrt{z} ) $$</span>
<span class="math-container">$\Omega=\{z\in\Bbb C: |z|\gt r\}$</span>, <span class="math-container">$r\gt 0$</span>. What is <span class="math-container">$f(\Omega)$</span>?</p>
<p>where <span class="math-container">$\sqrt{z}=\exp{(\text{Log }z/2)}, \text{Log }z=\log|z|+i\arg z,\arg z\in(-\pi,\pi]$</span></p>
</blockquote>
<p><span class="math-container">$\infty$</span> is an essential singularity. Picard's Great theorem , <span class="math-container">$\Bbb C\setminus f(\Omega) $</span> contains at most one point. <span class="math-container">$f(\Omega)$</span> is <span class="math-container">$\Bbb C$</span> ? </p>
<p>When <span class="math-container">$ z\in\Bbb R $</span>, <span class="math-container">$f(z)\geq0$</span>. According to <a href="https://en.wikipedia.org/wiki/Schwarz_reflection_principle" rel="nofollow noreferrer">Schwarz reflection principle</a> and Picard's Great theorem, all <span class="math-container">$ x+iy(y\ne0), \in\Bbb f(\Omega) $</span>. How to show that all negative real numbers belong to <span class="math-container">$f(\Omega)$</span>? thanks a lot!</p>
<p><strong>Is there a simple, more Elementary Proof</strong> than Conard's ? </p>
| IrbidMath | 255,977 | <p><strong>Hint:</strong> Write the terms as a difference <span class="math-container">$a-b$</span>, then multiply with the conjugate <span class="math-container">$a+b$</span> </p>
|
3,555,084 | <blockquote>
<p>Let
<span class="math-container">$$f(z) = e^z (1+\cos\sqrt{z} ) $$</span>
<span class="math-container">$\Omega=\{z\in\Bbb C: |z|\gt r\}$</span>, <span class="math-container">$r\gt 0$</span>. What is <span class="math-container">$f(\Omega)$</span>?</p>
<p>where <span class="math-container">$\sqrt{z}=\exp{(\text{Log }z/2)}, \text{Log }z=\log|z|+i\arg z,\arg z\in(-\pi,\pi]$</span></p>
</blockquote>
<p><span class="math-container">$\infty$</span> is an essential singularity. Picard's Great theorem , <span class="math-container">$\Bbb C\setminus f(\Omega) $</span> contains at most one point. <span class="math-container">$f(\Omega)$</span> is <span class="math-container">$\Bbb C$</span> ? </p>
<p>When <span class="math-container">$ z\in\Bbb R $</span>, <span class="math-container">$f(z)\geq0$</span>. According to <a href="https://en.wikipedia.org/wiki/Schwarz_reflection_principle" rel="nofollow noreferrer">Schwarz reflection principle</a> and Picard's Great theorem, all <span class="math-container">$ x+iy(y\ne0), \in\Bbb f(\Omega) $</span>. How to show that all negative real numbers belong to <span class="math-container">$f(\Omega)$</span>? thanks a lot!</p>
<p><strong>Is there a simple, more Elementary Proof</strong> than Conard's ? </p>
| Peter Szilas | 408,605 | <p>x>0;</p>
<p><span class="math-container">$f(x):=\sqrt{x^2+6}= x\sqrt{1+6/x^2}=$</span></p>
<p><span class="math-container">$x(1+3/x^2+$</span></p>
<p><span class="math-container">$(1/2)(-1/2)(1/2!)(6/x^2)^2+ O(1/x^6))=$</span></p>
<p><span class="math-container">$x(1+3/x^2-(6^2/8)/x^4+O(1/x^6));$</span></p>
<p><span class="math-container">$x^2(x^2-xf(x)+3)=$</span></p>
<p><span class="math-container">$x^2(- 3+3+(9/2)/x^2+O(1/x^4))$</span>;</p>
<p>Take the limit.</p>
|
3,336,870 | <p>Suppose <span class="math-container">$S= \{x_1+x_5\}$</span> is a vector space in <span class="math-container">$R^5$</span>.</p>
<p>Then what is the orthogonal complement for <span class="math-container">$S$</span>?</p>
<p><em>My interpretation:</em></p>
<p>We can represent as <span class="math-container">$[1, 0, 0, 0, 1] [x_1, x_2, x_3, x_4, x_5]^{T} = 0$</span></p>
<p>So, the solutions for plane lies in null space and hence <span class="math-container">$S^{\bot}$</span> is the row space i.e., <span class="math-container">$c[1, 0, 0, 0, 1]$</span> for all real <span class="math-container">$c$</span>.</p>
<p>Am I correct?</p>
| parsiad | 64,601 | <p><strong>Assumption</strong>. Apples are identical but people are not.</p>
<ol>
<li>Due to the assumption that apples are indistinguishable, we can give each person 2 apples so that we have <span class="math-container">$10-3\times2=4$</span> apples left. Therefore, the problem is reduced to counting the number of ways to distribute 4 apples among 3 people.</li>
<li>The number of ways to distribute 4 apples among 3 people is <span class="math-container">$\binom{4+3-1}{3-1}=15.$</span> The argument used to establish this is often referred to as <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Examples" rel="nofollow noreferrer">stars and bars</a>, which says that there are <span class="math-container">$\binom{n+k-1}{k-1}$</span> ways to place <span class="math-container">$n$</span> identical stars into <span class="math-container">$k$</span> non-identical bins. In our case, the stars are the <span class="math-container">$n=4$</span> apples and the bins are the <span class="math-container">$k=3$</span> people.</li>
</ol>
|
164,629 | <p>Probably this is well known to those who know it.</p>
<p>Got an argument and numerical support that over
number fields elliptic curves in minimal models
might have unbounded number of integral points,
the number depending on the degree of the field.</p>
<p>Set $f(x)=x^3+ax+b$ and consider the curve
$E: y^2=f(x)$.</p>
<p>Chose $x_1 \ldots x_n$ such that $f(x_n)$ is
prime and work in $K=\mathbb{Q}[\sqrt{f(x_1)},\ldots\,\sqrt{f(x_n)}]$.</p>
<p>$E$ has the obvious $n$ points $(x_n,\sqrt{f(x_n)})$.</p>
<p>Experimentally for $f(x)=x^3-x+1$ over
$\mathbb{Q}[\sqrt{7},\sqrt{61},\sqrt{211},\sqrt{337},\sqrt{991}]$ the five points are linearly independent according to sage so the
rank is at least $5$.</p>
<p>Computing the absolute field is not efficient for me.</p>
<p>Over the rationals there is a conjecture relating
the number of integral points to the rank,
is there a similar conjecture for number fields?</p>
<p>Is there an example (with few primes) when in
this construction the points are linearly dependent?</p>
<p>The same argument works for higher genus.</p>
| Joe Silverman | 11,926 | <p>joro asks: "Over the rationals there is a conjecture relating the number of integral points to the rank, is there a similar conjecture for number fields?"</p>
<p>Yes, the conjecture is that for a given field $K$, on a quasi-minimal Weierstrass equation for $E/K$, the number of $S$-integral points satisfies
$$
\# E(R_S) \le C(K)^{1+\#S+\text{rank} E(K)}.
$$
This is known to be a consequence of the $abc$-conjecture for $K$; see [1]. It is also known unconditionally in the weaker form
$$
\# E(R_S) \le C(K,\nu(j_E))^{1+\#S+\text{rank} E(K)},
$$
where $\nu(j_E)$ is the number of primes $\mathfrak{p}$ of $R_K$ such that $\text{ord}_{\mathfrak{p}}(j_E)<0$; see [2].</p>
<p>[1] M. Hindry, J.H. Silverman, The canonical height and integral points on
elliptic curves, <em>Invent. Math.</em> <strong>93</strong> (1988), 419-450.</p>
<p>[2] J.H. Silverman, A quantitative version of Siegel's theorem: Integral
points on elliptic curves and Catalan curves, <em>J. Reine
Angew. Math.</em> <strong>378</strong> (1987), 60-100.</p>
|
331,710 | <p>I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.</p>
<p>If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.</p>
<p>It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something. </p>
| amWhy | 9,003 | <p>Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$</p>
<p>The area of a parallelogram is also equal to the magnitutude of the cross product of the component vectors $\vec u, \vec v$ = $$|\vec u\times \vec v| = |\vec u|\,|\vec v|\sin \theta$$</p>
<p>where $\theta$ is the measure of the angle formed by the component vectors $\vec u, \vec v$.</p>
<p>Use your points to determine the component vectors $\vec u, \vec v$. </p>
<p>For parallelogram formed by $p_1, p_2, p_3, p_4$, put $\vec u = p_2 - p_1$, $\vec v = p_3 - p_1$ (where $p_4$ is the point opposite $p_1$, $p_2$ the point opposite $p_3$):
$$\vec u = \langle 4 -(- 1), 1-(-1)\rangle = \langle 5, 2\rangle$$
$$\vec v = \langle 5 - (-1), 3 - (-1)\rangle = \langle 6, 4\rangle$$</p>
<p>So compute $$A = \det \begin{pmatrix} 5 & 6 \\ 2 & 4 \end{pmatrix},\;\;\text{or}\;\; A = \vert \vec u\times \vec v\vert$$</p>
<p><br>See the parallelogram whose area can be determined by component vectors $\vec u, \vec v\;\;$ (left of the image):</p>
<p><img src="https://i.stack.imgur.com/oy5pu.gif" alt="enter image description here"></p>
|
3,413,837 | <p>Jerry the mouse is hungry and according to some confidential information, there is a tempting piece of cheese at the end of one of the three paths after the junction he just found himself!</p>
<p>Fortunately, Tom is standing right there and Jerry hopes he can get some useful information as to which path he must get; most importantly because Spike and Tyke, the dogs, are at the end of the other two paths!</p>
<p>The only problem is that Tom gives true and false replies in alternating order. Furthermore, he has no way of knowing which will be first, the truth or the lie!</p>
<p>He is only allowed to ask Tom 2 questions that can be answered by a “yes” and a “no”.</p>
<p>What must be the two questions he must ask?</p>
<hr>
<p>No matter how hard I tried, I can't figure out anything...
I have seen several variations for the 2 doors problem but this one is different!</p>
| Rushabh Mehta | 537,349 | <p>Think about a question of the following form:</p>
<blockquote>
<p>If I were to ask you whether door X is the right door in the next question, what would you respond?</p>
</blockquote>
<p>Tom will always lie when responding to this. Can you solve from here?</p>
|
2,170,501 | <p>How do you prove the sum of two monotone sequences is also monotone? </p>
<p>Here is my thought process: </p>
<p>Let $a_n$ and $b_n$ be two monotone increasing sequences. Then $\forall n \in N$, $a_n \leq a_{n+1}$ and $b_n \leq b_{n+1}$. Adding both inequalities you get $a_n + b_n \leq a_{n+1} + b_{n+1}$. Therefore in this specific case of both sequences being monotone increasing I have proven their sum is monotone increasing, and so it is also monotone. </p>
<p>But I'm having trouble figuring out how to apply similar logic for an $a_n$ is monotone increasing but $b_n$ is monotone decreasing. I also tried to do proof by contradiction (like supposing the sum is not monotone), but that also lead to nowhere. </p>
<p>How do I prove the other cases?</p>
| Kenny Wong | 301,805 | <p>The result is not true when $a_n$ is monotone increasing and $b_n$ is monotone decreasing.</p>
<p>For example,
$$a_n = 0, +1, +1, +2, +2, +3, +3, +4, +4, +5, +5, \dots$$
$$b_n = 0, \ \ 0, -1, -1, -2, -2, -3, -3, -4, -4, -5, \dots $$
gives
$$ a_n + b_n = 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, \dots$$</p>
|
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