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 |
|---|---|---|---|---|
1,373,103 | <p>I was wondering if $|f(x)g(x)| = |f(x)| |(g(x)|$ is true all the time as in the case of real numbers.</p>
<p>I was not convinced enough that that was true.</p>
<p>But I can't think of any counterexample.</p>
<p>Thank you.</p>
| Tucker | 256,305 | <p>For $|x|\leq1$, the Maclaurin series (of cos(x)) is an decreasing alternating series, so the error in your approximation will be no worse then the next non-zero term in the series, namely $|-\frac{x^{6}}{6!}|$. However you will want to review estimates for the remainder in a Taylor series. You will not always have this trick at your disposal. </p>
|
2,441,894 | <p>The matrix $$\pmatrix{100\sqrt{2}&x&0\\-x&0&-x\\0&x&100\sqrt{2}},\quad x>0$$ have two equal eigenvalues. How can I find $x$?
What I tried is this. If $\lambda_1$ is doubly degenerate and $\lambda_2$ the third eigenvalue, then the characteristic equation is $(\lambda-\lambda_1)^2(\lambda-\lambda_2)=0$. Also $2\lambda_1+\lambda_2=200\sqrt{2},\quad \lambda_1^2\lambda_2=200\sqrt{2}x^2$. I do not know how to proceed from here. </p>
| euraad | 443,999 | <p>Here is the answer.</p>
<p>$$A = [A;A](:)$$</p>
<p>Easy!</p>
|
3,118,298 | <p>So I have a formula for arc length <span class="math-container">$$s(t)=\int_{t_0}^t \vert\vert \dot\gamma(u)\vert\vert du$$</span>
I computed that <span class="math-container">$$\vert\vert \dot\gamma(t)\vert\vert=\sqrt{2(1-\cos t)}$$</span></p>
<p>Substituting this into the integral <span class="math-container">$$\int_0^{2\pi} \sqrt{2(1-\cos t)} dt$$</span>
<span class="math-container">$$=2\int_0^{2\pi} \sqrt{1-\cos t} dt$$</span></p>
<p><span class="math-container">$$=\sqrt{2}\int_0^{2\pi} \sqrt{1-\cos t} \frac{\sqrt{1+\cos t}}{\sqrt{1+\cos t}} dt$$</span></p>
<p><span class="math-container">$$=\sqrt{2}\int_0^{2\pi} \frac{\sin t}{\sqrt{1+\cos t}}dt$$</span>
Using subsitution <span class="math-container">$u=1+\cos t$</span>, <span class="math-container">$du=-\sin t dt$</span></p>
<p><span class="math-container">$$=\sqrt{2}\int_0^{2\pi} \frac{-1}{\sqrt{u}} du$$</span>
<span class="math-container">$$=-\sqrt{2} (2\sqrt{1+\cos t}\big\vert_0^{2\pi})=0$$</span></p>
<p>Obviously this has to be wrong, but I'm not sure what I'm doing wrong.</p>
| MAHI | 821,955 | <p><em>My Approach-</em></p>
<p>Let M be the LUB of {<span class="math-container">$x_n$</span>}.
Then <span class="math-container">$x_n \in N_ε(M)$</span> for some <span class="math-container">$n\in S\forall ε>0$</span> ,(i. e.<span class="math-container">$x_n \in N_ε(M)$</span> when <span class="math-container">$K\leq n\leq s^ * $</span> , where <span class="math-container">$K:=inf S , s^ *:=sup S$</span>).Here S is an interval in <span class="math-container">$\Bbb R$</span> ,such that <span class="math-container">$infS,supS\in\Bbb N$</span>.</p>
<p>But let <span class="math-container">$\exists k>s^*=supS$</span> such that <span class="math-container">$x_k \notin N_ε(M).(\therefore k>n)$</span></p>
<p><span class="math-container">$\implies x_n>x_k$</span>,viz. a contradiction. <span class="math-container">$^. .^. (x_n)$</span> is monotonically increasing.</p>
<p><span class="math-container">$\therefore \nexists k\in\Bbb N :k>s^*$</span> for which <span class="math-container">$x_k \notin N_ε(M) \forall ε>0$</span>.</p>
<p><span class="math-container">$\implies \forall ε>0 ,\exists K $</span> such that <span class="math-container">$x_n\in N_ε(M) \forall n\geq K.$</span></p>
<p>Hence {<span class="math-container">$x_n$</span>}<span class="math-container">$\to M$</span>. ■</p>
|
11,457 | <p>In their paper <em><a href="http://arxiv.org/abs/0904.3908">Computing Systems of Hecke Eigenvalues Associated to Hilbert Modular Forms</a></em>, Greenberg and Voight remark that</p>
<p>...it is a folklore conjecture that if one orders totally real fields by their discriminant, then a (substantial) positive proportion of fields will have strict class number 1.</p>
<p>I've tried searching for more details about this, but haven't found anything. </p>
<p>Is this conjecture based solely on calculations, or are there heuristics which explain why this should be true? </p>
| JSE | 431 | <p>Maybe it's worth a word about <em>why</em> Cohen-Lenstra predicts this behavior. Suppose K is a field with r archimedean places. Then Spec O_K can be thought of as analogous to a curve over a finite field k with r punctures, which is an affine scheme Spec R. Write C for the (unpunctured) curve. Then the class group of R is the quotient of Pic(C)(k) by the subgroup generated by the classes of the punctures -- or, what is the same, the quotient of Jac(C)(k) by the subgroup generated by degree-0 divisors supported on the punctures. (This last subgroup is just the image of a natural homomorphim from Z^{r-1} to Jac(C)(k).)</p>
<p>The Cohen-Lenstra philosophy is that these groups and the puncture data are "random" -- that is, you should expect that the p-part of the class group of R looks just like what you would get if you chose a random finite abelian p-group (where a group A is weighted by 1/|Aut(A)|) and mod out by the image of a random homomorphism from Z^{r-1}. (There are various ways in which this description is slightly off the mark but this gives the general point.)</p>
<p>It turns out that when r > 1 the chance is quite good that a random homomorphism from Z^{r-1} to A is surjective. In fact, the probability is close enough to 1 that when you take a product over all p you still get a positive number. In other words, when r > 1 Cohen-Lenstra predicts a positive probability that the class group will have trivial p-part for all p; in other words, it is trivial. (In fact, it predicts a precise probability, which fits experimental data quite well.)</p>
<p>When r = 1, on the other hand, the class group is just A itself, and the probability its p-part is trivial is on order 1-1/p. Now the product over all p is 0, so one does NOT expect to see a positive proportion of trivial class groups. And in fact, when there is just one archimedean place -- i.e. when K is imaginary quadratic -- this is just what happens! </p>
|
107,171 | <p>I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get </p>
<p>$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$</p>
<p>What do I do from here? </p>
<p>Thanks a lot!</p>
| Peđa | 15,660 | <p>According to Maple solution is given by :</p>
<p>$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{(3n+1)^2} = \frac{1}{9} \Psi\left(1,\frac{1}{3}\right)$$</p>
<p>where $\Psi\left(1,\frac{1}{3}\right)$ is <a href="http://mathworld.wolfram.com/PolygammaFunction.html" rel="nofollow">polygamma function</a> .</p>
|
107,171 | <p>I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get </p>
<p>$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$</p>
<p>What do I do from here? </p>
<p>Thanks a lot!</p>
| Riccardo.Alestra | 24,089 | <p>The sum is $$\frac19{\Psi(1,\frac13)}$$</p>
<p>where $\Psi$ is the Poligamma function</p>
|
107,171 | <p>I'm trying to find $$\lim\limits_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4} .$$
After I tried couple of algebraic manipulation, I decided to use the polaric method.
I choose $x=r\cos \theta $ , $y=r\sin \theta$, and $r= \sqrt{x^2+y^2}$, so I get </p>
<p>$$\lim\limits_{r \to 0} \frac{e^{-\frac{1}{r^2}}}{r^4\cos^4 \theta+r^4 \sin^4 \theta } $$</p>
<p>What do I do from here? </p>
<p>Thanks a lot!</p>
| Daoyi Peng | 29,776 | <p>The sum can express
$$\begin{align}
& \sum\limits_{n=0}^{\infty }{\frac{1}{{{(3n+1)}^{2}}}} \\
& =\frac{1}{9}\left[ \frac{{{\Gamma }''}(1/3)}{\Gamma (1/3)}-{{\left( \frac{{\Gamma }'(1/3)}{\Gamma (1/3)} \right)}^{2}} \right] \\
& =1+\frac{1}{9}\int_{0}^{\infty }{\frac{t{{\mathrm{e}}^{-t/3}}}{{{\mathrm{e}}^{t}}-1}\mathrm{d}t} \\
\end{align}$$
But it can't find the specific value.</p>
|
3,005,842 | <blockquote>
<p>Let <span class="math-container">$(X,Y)$</span> be the coordinates of a point uniformly chosen from a
quadrilateral with vertices <span class="math-container">$(0,0)$</span>, <span class="math-container">$(1,0)$</span>, <span class="math-container">$(1,1)$</span>, <span class="math-container">$(0,2)$</span>. Find
the marginal probability density functions of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>.</p>
</blockquote>
<h2>Try</h2>
<p>Well, these points are vertices of a quadriteral whose area is <span class="math-container">$\frac{2}{3}$</span>. Since it is uniformly chosen the points, then</p>
<p><span class="math-container">$$ f_{X,Y}(x,y) = \frac{1}{\text{Area(S)}} = \frac{1}{2/3} = \frac{3}{2} $$</span></p>
<p>Where <span class="math-container">$(x,y) \in S $</span>, the quadrilateral. Hence,</p>
<p><span class="math-container">$$ f_X(x) = \int\limits_0^2 \frac{2}{3} dy = \frac{4}{3} $$</span></p>
<p>and</p>
<p><span class="math-container">$$ f_Y(y) = \int\limits_0^1 \frac{2}{3} = \frac{2}{3} $$</span></p>
<p>Is this correct?</p>
| Graham Kemp | 135,106 | <p>To be correct you need to include the support. This can be expressed in two equivalent ways: <span class="math-container">$$f_{X,Y}(x,y)~{=\tfrac 32\mathbf 1_{0\le x\leq 1, 0\leq y\leq2-x}\\=\tfrac 32\mathbf 1_{0\le y\leq 2, 0\leq x\leq \min(1,2-y)}}$$</span></p>
<p>Then integrate over this support with respect to the appropriate variables.</p>
|
1,576,713 | <p>$X$ and $Y$ are two sets and $f:X\to Y$. If $f(C)=\{f(x):x\in C\}$ for $C\subseteq X$ and $f^{-1}(D)=\{x:f(x)\in D\}$ for $D\subseteq Y$, then the true statement is </p>
<p>(A) $f(f^{-1}(B))=B$</p>
<p>(B) $f^{-1}(f(A))=A$</p>
<p>(C) $f(f^{-1}(B))=B$ only if $B\subseteq f(X)$</p>
<p>(D) $f^{-1}(f(A))=A$ only if $f(X)=Y$</p>
| user 1 | 133,030 | <p><code>(A) is not true</code>.<br>
Let $X=A=\{a\}$, and $Y=B=\{a,b\}$. Define $f:X\to Y; \quad f(a)=a.$ Then $f(f^{-1}(B))=\{a\}\neq B.$ </p>
<hr>
<p><code>(B) is not true</code>.<br>
Let $X=Y=\{a,b\}$, and $A=B=\{a\}$ . Define $f:X\to Y; \quad f(x)=a, \forall x\in X.$ Then $f^{-1}(f(A))=X\neq A.$ </p>
<hr>
<p><code>(D) is not true</code>.<br>
Let $X=\{a,b\}$, $Y=\{a,b,c\}$, and $A=B=\{a\}$. Define $$f:X\to Y; \quad f(x)=x, \forall x\in X.$$
We have $f^{-1}(f(A))=A,$ but $f(X)\neq Y.$</p>
<hr>
<p><code>(C) is true</code>:<br>
Let $f(f^{-1}(B))=B.$ We should prove $B\subseteq f(X).$ If not, we have $b\in B\setminus f(X).$ Hence $b\in B\setminus f(f^{-1}(B)),$ Since $f^{-1}(B)\subseteq X.$ (contradiction!)</p>
|
7,761 | <p>Our undergraduate university department is looking to spruce up our rooms and hallways a bit and has been thinking about finding mathematical posters to put in various spots; hoping possibly to entice students to take more math classes. We've had decent success in finding "How is Math Used in the Real World"-type posters (mostly through AMS), but we've been unable to find what I would call interesting/informative math posters. </p>
<p>For example, I remember seeing a poster once (put out by Mathematica) that basically laid out how to solve general quadratics, cubics, and quartics. Then it had a good overview of proving that no formula existed for quintics. So not only was it pretty to look at, but if you stopped to read it you actually learned something.</p>
<p>Does anyone know of a company or distributor that carries a variety of posters like this? </p>
<p>I've tried searching online but all that comes up is a plethora of posters of math jokes. And even though the application/career-based posters are nice and serve a purpose, I don't feel like you actually gain mathematical knowledge by reading them.</p>
| celeriko | 3,237 | <p>There is a series of posters made by Key Curriculum Press that each detail a different culture and their contribution to mathematics. They are colorful, informative, and have, in my experience, been very engaging for students to look at and read.</p>
<p>Unfortunately, with the McGraw-Hill purchase of Key Curriculum a few years ago, they have been pigeonholed into only working on the GSP program, so i can't find these poster sets on their website. However, the following two links provide contact information that may or may not still be valid...</p>
<p><a href="http://www.maa.org/publications/periodicals/convergence/multicultural-classroom-posters-sets-1-2" rel="noreferrer">http://www.maa.org/publications/periodicals/convergence/multicultural-classroom-posters-sets-1-2</a></p>
<p><a href="http://www.maa.org/publications/periodicals/convergence/multicultural-classroom-posters-sets-3-4" rel="noreferrer">http://www.maa.org/publications/periodicals/convergence/multicultural-classroom-posters-sets-3-4</a></p>
<p>I do not have the time to try and track these down but if you are successful in finding a place to purchase these posters, please let me know in the comments so I can update this answer!</p>
<p>Update: While this is not the specific poster set that I was thinking of, these look nice and are in a similar vein to the Key Curriculum posters described above</p>
<p><a href="http://www.enasco.com/product/TB18084T" rel="noreferrer">http://www.enasco.com/product/TB18084T</a></p>
|
18,459 | <p>Does anyone know of any studies or have personal experience dealing with difficulties (if any) faced by students studying mathematics if they come from countries which use languages written from right-to-left or top-down? </p>
<p>I have been wondering about this recently because I have been working on supplemental study guides. It occured to me that mathematics covering everything from arithmetic to calculus is heavily viewed in terms of a left-to-right perspective (e.g. digits with the highest place value are the leftmost ones in a numeral and are thus read first; the positive x-axis extends to the right; a line with positive slope is seen as one which goes up when viewed from left-to-right)</p>
<p>Then again there are many accomplished mathematicians who come from countries which speak and write these languages so maybe it's not that difficult of a conceptual jump after all. I think I for one would have difficulty though...</p>
| Amy B | 5,321 | <p>I have some personal experience.</p>
<p>I taught in a school that has Israeli students whose families had moved to the US. These children had Hebrew as their native language. Hebrew is written right to left. I never saw these students struggle with math because they learned to write from right to left.</p>
<p>Furthermore all students in my school studied both Hebrew and English, learning to write both right to left and left to right. No one seemed to struggle with direction in math class which was taught in English.</p>
|
3,346,775 | <p>Do there exist non-zero expectation, dependent, uncorrelated random variables <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>? The examples that I have found have at least one of the variables have zero expectation.</p>
| Robert Israel | 8,508 | <p>Take any example and add nonzero constants to <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>. This changes the expectations but does not affect dependency or correlation.</p>
|
2,185,585 | <p>Triangular numbers (See <a href="https://en.wikipedia.org/wiki/Triangular_number" rel="noreferrer">https://en.wikipedia.org/wiki/Triangular_number</a> )</p>
<p>are numbers of the form $$\frac{n(n+1)}{2}$$</p>
<p>In ProofWiki I found three claims about triangular numbers. The three claims are that a triangular number cannot be a cube, not a fourth power and not a fifth power. Unfortunately, neither was a proof given nor did I manage to do it myself. Therefore my qeustions :</p>
<blockquote>
<p>Does someone know a proof that a triangular number cannot be a cube, a fourth power or a fifth power ?</p>
</blockquote>
| Qiaochu Yuan | 232 | <p>Yes, the real $1$-dimensional projective line $\mathbb{RP}^1$ is homeomorphic (in fact diffeomorphic) to the circle $S^1$. It can be thought of as the circle $S^1$ with antipodal points identified, which reflects the fact that the circle is its own double cover: the double cover map is given explicitly by</p>
<p>$$S^1 \ni z \mapsto z^2 \in S^1$$</p>
<p>thinking of $S^1$ as the unit complex numbers. In other words, quotienting by antipodes is the same as quotienting by $180^{\circ}$ rotation; note that this is very much false for the 2-sphere. </p>
|
2,022,700 | <blockquote>
<p>In how many ways can the letters in WONDERING be arranged with exactly
two consecutive vowels</p>
</blockquote>
<p>I solved and got answer as $90720$. But other sites are giving different answers. Please help to understand which is the right answer and why I am going wrong.</p>
<p><strong>My Solution</strong></p>
<p>Arrange 6 consonants $\dfrac{6!}{2!}$<br>
Chose 2 slots from 7 positions $\dbinom{7}{2}$<br>
Chose 1 slot for placing the 2 vowel group $\dbinom{2}{1}$<br>
Arrange the vowels $3!$</p>
<p>Required number of ways:<br>
$\dfrac{6!}{2!}\times \dbinom{7}{2}\times \dbinom{2}{1}\times 3!=90720$</p>
<p><strong>Solution taken from <a href="http://www.sosmath.com/CBB/viewtopic.php?t=6126" rel="nofollow noreferrer">http://www.sosmath.com/CBB/viewtopic.php?t=6126</a>)</strong></p>
<p><a href="https://i.stack.imgur.com/7JNnA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7JNnA.jpg" alt="enter image description here"></a></p>
<p><strong>Solution taken from <a href="http://myassignmentpartners.com/2015/06/20/supplementary-3/" rel="nofollow noreferrer">http://myassignmentpartners.com/2015/06/20/supplementary-3/</a>
<a href="https://i.stack.imgur.com/O1uLK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O1uLK.jpg" alt=""></a></strong></p>
| Community | -1 | <p>The total number of ways of arranging the letters is $\frac{9!}{2!} = 181440$. Of these, let us count the cases where no two vowels are together. This is $$\frac{6!}{2!} \times \binom{7}{3}\times 3! = 75600$$
Again, the number of ways in which all vowels are together is 15120. Thus the number of ways in which exactly two vowels are together is
$$181440 - 75600 - 15120 = 90720$$</p>
|
287,947 | <p>For example, $\sqrt 2 = 2 \cos (\pi/4)$, $\sqrt 3 = 2 \cos(\pi/6)$, and $\sqrt 5 = 4 \cos(\pi/5) + 1$. Is it true that any integer's square root can be expressed as a (rational) linear combinations of the cosines of rational multiples of $\pi$?</p>
<p>Products of linear combinations of cosines of rational multiples of $\pi$ are themselves such linear combinations, so it only needs to be true of primes. But I do not know, for example, a representation of $\sqrt 7$ in this form.</p>
| Johannes Hahn | 3,041 | <p>Yes, that is true. The general case is the Kronecker-Weber theorem as Lucia mentioned in the comments. For square roots one can be more explicit and prove $\mathbb{Q}(\zeta_p) \cap \mathbb{R} = \mathbb{Q}[\sqrt{ (-1)^{(p-1)/2} p }]$ for odd prime numbers $p$ using properties of the Legendre symbol. Therefore you can write $\sqrt{p}$ either as $\mathfrak{Re}(\sqrt{p})$ or $\mathfrak{Im}(\sqrt{-p})$ depending on $p\mod 4$ so that you can write it as rational (in fact: integral) linear combination of real- or imaginary parts of powers of $\zeta_p$, i.e. cosine or sine values.</p>
|
521,589 | <p>In a rectangle $ABCD$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $ABCD$ has equation $2x-y+4=0$. Then the area of the rectangle is:</p>
<p>My work: I found the equations of $AD$ and $BC$ of the rectangle. Taking the points $C$ and $D$ as $(x_1,y_1)$ and $(x_2,y_2)$ I wrote the equation for $AD=BC$. I think that the equation given for the diameter of the circle should pass through the point of intersection of the diagonals of the rectangle and wrote equation for the point of intersection. This gives me two equations and four unknowns. I know there is some problem with my method making the answer to this problem hard. So, please help me do the problem by the right method. </p>
| copper.hat | 27,978 | <p>Here is one ugly way.</p>
<p>Since the diameter lies on the given line, we know the centre of the circle $P=(x,y)$ lies on this line. The distance from $P$ to $A=(1,2)$ and $B=(3,6)$ must be the same so we have the equation:
$(x-1)^2+(y-2)^2 = (x-3)^2+(y-6)^2$. Simplifying gives the equation $2y+x = 10$.
Since $P$ lies on the given line, we must also have $2x-y+4 = 0$ as well, and solving gives the rather ugly $P=\frac{1}{5}(2, 24)$.</p>
<p>Now we can compute $C$ by reflecting through $P$, to get $C = P+(P-A) = \frac{1}{5}(-1, 38)$, then the area is given by $|A-B||B-C| = \sqrt{(2^2+4^2)\frac{1}{25}(16^2+8^2)} = 16$.</p>
|
50,521 | <p>I would like to know if there are some open mathematical problems in General Relativity, that are important from the point of view of Physics. </p>
<p>Is there something that still needs to be justified mathematically in order to have solid foundations? </p>
| Willie Wong | 1,543 | <p>Let me sneak in an answer before this gets closed, because the answer to this question is likely to be very very different depending if you ask a mathematician or a physicist. <code>:)</code> So here is a long list of questions that a <em>mathematician</em> considers to be important open problems in general relativity, and why they are physically relevant. </p>
<p><strong>Cosmic censorship</strong></p>
<p>Let me first get the biggie out of the way. There are actually <em>two</em> different cosmic censorship hypotheses, which are minimally related to each other. The Weak Cosmic Censorship (WCC) hypothesis states that (a fairly modern formulation)</p>
<blockquote>
<p>For generic initial data to the evolution problem in general relativity, there cannot be naked singularities. </p>
</blockquote>
<p>Unfortunately, if you ask me to define the terms in that hypothesis, I can just shrug and tell you, "no can do". This question is so open as an open problem that we don't even know the correct formulation of the statement! If you are interested, you may want to consult Christodoulou, D. "On the global initial value problem and the issue of singularities", <em>Classical Quant. Grav.</em>, 1999, 16, A23. Let me give just a short digression on what the difficulty of formulating the problem is. </p>
<p>The first is the problem of <em>what do you mean by a naked singularity</em>? In certain situations, the definition of a naked singularity is very clear cut (for example, if we assume mathematically that the space-time is spherically symmetric). But in general, the problem of what a naked singularity is is very difficult to define accurately in mathematics. The main problem being this: the equations of motion given by the Einstein's equations are hyperbolic. As such, we cannot necessarily meaningfully extend the notion of a solution to points which can "see" the singularity. (Sort of a catch 22: morally speaking, if you can see the singularity, you are at a (possibly different) singular event, so it doesn't make much sense to use the physics definition of a naked singularity being one that can be observed by a far away observer.) (This is not to say this is completely impossible: there <em>are</em> singular solutions which the solution can extend continuously to the singular boundary, and so technically you can exist all the way up to the moment you see the singularity, and then poof, you are gone.) So we have to probe at the notion of a naked singularity using alternative definitions that are more stable and can be mathematically formulated. And that is hard. (Once physical intuition fails, things get hard.)</p>
<p>The second problem is generic. It is already known that there are examples of solutions in general relativity which contains a naked singularity for all reasonable definitions of "naked singularity". So it is in fact impossible to prove a statement that says "naked singularity cannot exists. Period.". Any statement you prove must be a statement that is true away from some exceptional set. The hope is that the bad set is positive codimension in the space of all initial data. But we don't know. </p>
<p>The other cosmic censorship hypothesis is the Strong Cosmic Censorship (SCC). Despite the name it is independent of WCC. The SCC states that</p>
<blockquote>
<p>A generic solution to the Einstein's equation cannot be continued beyond the Cauchy horizon.</p>
</blockquote>
<p>Again, this is from the view of an evolution equation. A funny thing about general relativity as an evolution equation is that you build your manifold <em>and</em> your solution on the manifold at the same time, which causes an interesting problem. Roughly speaking, usually when you solve an nonlinear wave equation, and you extended your solution as much as you can, then you either exhausted the entire space-time, or you ran into a singularity. But in general relativity, "running out of space-time" is not a hard obstruction to extending the solution further. (This goes back to the hyperbolicity of the equation.) Basically, you can "run out of space-time that can be completely predicted by your initial data", while still have a possible extension of your solution "to infinity and beyond". The classic examples of this are the Reissner-Nordstrom and the Kerr-Newman black hole solutions. If you solve it as an initial value problem in partial differential equations, you get a black hole solution. But if you start from this black hole solution, you can actually analytically continue the solution (similar to how analytic continuation is done in complex analysis) to beyond what can be defineable from the inital data. The continuation is in some sense "outside" our universe, so the existence of this type of solutions causes some, at the very least, epistemological difficulties. The SCC posits that these type of solutions are rare. Just like the case of the WCC, the correct mathematical statement for this problem is also not known, and therefore it is hard to say what techniques of mathematics will be required to resolve it.</p>
<p><strong>Final state conjecture</strong></p>
<p>The final state conjecture in classical relativity states that (assuming the universe is not expanding) if you let the universe run long enough, because gravity is an attractive force, and is the strongest force over cosmological distances, we expect the entire universe to either collapse into a black hole, or become "mostly" empty (things drifting away from each other). Or slightly more jargony: the end state of the universe is either a single black hole or vacuum. </p>
<p>Mathematically this requires deriving the asymptotic behaviour of solutions to the Einstein's equations in general relativity. The Einstein equations are a highly nonlinear system of partial differential equations, and it is very difficult to study these things asymptoticly. There are however, some special cases that have either been worked out or are receiving a lot of specialist attention. </p>
<p><em>Stability of Minkowski space</em></p>
<p>One of the first things to consider, in terms of the final state conjecture, is that whether, if we start with a universe that is already very sparse, we can guarantee that this will end up asymptotic to Minkowski space. The answer to this, as it turns out, is "yes". The first work was due to D. Christodoulou and S. Klainerman, with extensions of the result and improvements by H. Lindblad and I. Rodnianski, L. Bieri, N. Zipser, J. Speck, etc. </p>
<p><em>Uniqueness of black holes</em></p>
<p>Now, if we posit that the final state of the universe is either Minkowski space or a black hole, we better show that a black hole is the only possible stationary solution! (Any other stationary solution would be another steady state!) The problem of black hole uniqueness is almost, but not completely, resolved. </p>
<p>In the case where we assume some additional symmetry (that the universe is either axially symmetric or rotationally symmetric), the uniqueness of black holes is known. In the case where we assume that all the objects we are dealing with are <em>real analytic</em>, we also know that black holes are unique. </p>
<p>The remaining case, which is expected to be true, is when we relax the regularity assumption to just infinitely differentiable, but not real analytic. In this case we only have some partial results, recently due to A. Ionescu, S. Alexakis, S. Klainerman, plus some additional contributions by P. Yu and yours truly. So far we have two major class of results: one is that if we allow only small perturbations of a stationary black hole, then there are no other stationary solutions that are approximately a known black hole solution without being one. The other is that if you assume certain special structures on the event horizon (roughly speaking, assume that the event horizon looks like the one for a known black hole solution), you cannot have other stationary exteriors. </p>
<p><em>Stability of Kerr-Newman black holes</em></p>
<p>Now, assuming we accept that the known Kerr-Newman family of black holes form the unique stationary state of general relativity, the next problem is to prove that they are actually stable under perturbations. That is, if we start out with initial data very similar to a Kerr-Newman black hole, is the evolution going to "track" a Kerr-Newman black hole? So far, there are quite some progress made in the linearised problem, due to contributions from (in no particular order) I. Rodnianski, M. Dafermos, P. Blue, A. Soffer, W. Schlag, R. Donniger, D. Tataru, J. Luk, J. Sterbenz, G. Holzegel, and many others. (It is quite an active field at the moment.) But the way forward for the full nonlinear problem is still somewhat elusive. </p>
<p><strong>Penrose inequality and friends</strong></p>
<p>Recently between H. Bray and the team of Huisken and Illmanen, what is known as the Riemannian Penrose inequality was proved. The Riemannian Penrose inequality sits in a large class of <em>mass inequalities</em> for space-time manifolds. These types of inequalities all take the form </p>
<blockquote>
<p>Mass of the space-time is bounded below by Blah.</p>
</blockquote>
<p>The first, and most famous, of these type of results is the positive mass theorem of Schoen and Yau, which states that the mass is bounded below by 0. The Riemannian Penrose Inequality states that the mass is bounded below by number related to the size of the black hole (when there is one). In other words, a black hole must contribute a certain amount of mass to the universe, the amount depending on its surface area. </p>
<p>There are further refinements of the known statements conjectured. One is the version called the Gibbons-Penrose inequality, which gives some improved lower bounds when there are multiple black holes (the attracting force between the multiple black holes will creating some contribution to the mass at infinity). And there are also other versions which also factors in electromagnetic forces and so forth. Most of them are open questions. </p>
<p>A somewhat related problem is the <em>Hoop Conjecture</em> of Kip Thorne. The conjecture roughly states that </p>
<blockquote>
<p>If you try to squeeze lots of mass into a small space-time volume, what determines whether the mass will collapse and form a black hole is the "circumference" of the space-time region. </p>
</blockquote>
<p>This problem is being actively pursued by M. Khuri and others, and is still open to my knowledge. </p>
<hr>
<p>Some other open problems:</p>
<p><strong>Low regularity evolution</strong> Can we solve the equations of motions assuming very rough (in the sense of having very few derivatives) initial data? These types of question are important because (a) when comparing to physical measurements, the rougher the data the theory can accomodate, the less sensitive the solution is to small perturbations, and so the better we can figure out the "error bar" due to theory (b) the problem of shockwaves. Physical systems can exhibit shock waves, and it is important to have a theory that can deal with that. </p>
<p><strong>Higher dimensional cases</strong> In some cosmological/string theories, one is lead to consider space-time manifolds of higher dimensions. On the one hand, radiative decay of solutions is stronger on higher dimensions, and it would enhance stability. On the other, there are more degrees of freedom, which will increase possibilities of instability. There are numerical evidence to suggest that certain types of higher dimensional black holes are in fact unstable, and could be poked and perturbed into other types of black holes. </p>
<p><strong>Evolution of apparent horizon</strong> Under the banner of dynamical horizon is a collection of topics related to how the apparent horizon (the "boundary of a black hole") will evolve in general relativity. It is important to note that a lot of the physics literature on the evolution of apparent horizon (such as black hole evaporation) is based on a linear analysis, which to first order assumes that the horizons do not move. (Yes, by assuming that the horizons do not move (much), they show that quantum effects will make the horizons shrink.) The true nonlinear versions of the evolution is not yet well understood.</p>
<p><strong>Geodesic hypothesis</strong> One of the postulates in studying general relativity is that <em>point particles</em> with negligible mass will travel along geodesics of the space-time. This question has a surprisingly long history (it was first considered by Einstein in the 1920s), and is still not completely resolved. The main problem is how to make the process of "taking the negligible mass limit" rigorous. For a physical object in general relativity, when it moves, its motion will cause "ripples" in the space-time caused by gravitational backreaction of its own presence. While the three body problem is difficult in classical mechanics, even the two body problem in full generality is still unresolved in general relativity. </p>
<hr>
<p>I hope the above provides some keywords or phrases for you to look up. </p>
<hr>
<p><em>[Addendum: 18.09.2013]</em></p>
<p>Here are some more open problems in mathematical relativity that concern <em>Cosmology</em>. To start with, let me first quickly say what I mean by cosmology here. To me, cosmology is the study of the <em>large scale</em> behaviour of the universe. In other words, it is the "blurry" picture of what the universe would look like if we ignore the small-scale structures such as stars and galaxies. As such, in the study of cosmology we often assume the <em>Cosmological Principle</em> holds. The cosmological principle is a strengthening of the <a href="http://en.wikipedia.org/wiki/Copernican_principle">Copernican principle</a>, and says that </p>
<blockquote>
<p>On the large scale, the universe can be modeled by a solution to Einstein's equations which admits a foliation by Cauchy hypersurfaces each of which is <em>homogeneous</em> and <em>isotropic</em>. </p>
</blockquote>
<p>In layman's words, it says that we can pick a (possibly preferred) notion of cosmological time, such that at every instant of cosmological time, space looks identical in all directions, and you cannot distinguish between any two points. (If we factor in the small-scale structure, this is quite obviously false: we live on a planet we call Earth orbiting a star we call Sol, that quite clearly distinguish us from many other points in the universe. But remember that the cosmological principle is only supposed to apply "in the large".) One can get into an argument about whether this principle holds in practice (cf. the cosmic microwave background anisotropy), but it is under this principle most of the computations in cosmology are made. </p>
<p>This leads to some immediate open problems:</p>
<p><strong>Stability of Cosmological Solutions</strong></p>
<p>Part of the cosmological principle assumes that the universe is well-modeled by the spatially homogeneous and isotropic cosmological solutions. In particular, this means that we are assuming the cosmological solutions are <em>stable</em> (that if we perturb the cosmological solutions a tiny bit, say, by now factoring in the small-scale structure, the evolution as governed by Einstein's equations will be, in the large, not very different from the evolution of the cosmological solutions). </p>
<p>Since in general this requires studying a large and complicated set of PDEs around a special solution, this is generally tackled on a case-by-case basis and there are still many interesting cosmological models to be studied. </p>
<p><strong>Cosmological principle as a derived consequence</strong></p>
<p>It would be much more satisfying, however, if the cosmological principle can be derived as a consequence of general relativity (under suitable assumptions on the cosmological constant and what not), rather than something assumed <em>a priori</em>. Namely, we ask the question:</p>
<blockquote>
<p>Can we find some conditions such that the evolution of arbitrary data in the cosmological setting will eventually converge to one for which the cosmological principle holds? </p>
</blockquote>
<p>Mathematically, we are asking whether <em>homogenisation</em> and <em>isotropisation</em> can be derived as a consequence of Einstein's equations. </p>
<p>We remark here that it is generally accepted that the sufficient conditions will involve the space-time being <em>expanding</em>. Heuristically speaking if the space-time is expanding then the relative strength of the small-scale structure should get weaker as time goes along (since they get relatively smaller and smaller) and so one should converge toward a cosmological space-time. (On the other hand, homogenisation and isotropisation are not expected to hold in contracting space-times [either toward the big-bang singularity in the past or possibly toward a big-crunch singularity in the future], because if space-time contracts, the relative size of the small-scale features get bigger...) Luckily for us, the recent astrophysical observations seem to suggest that our universe is in a regime of accelerated expansion, so it is okay to not consider the other alternatives. <code>:-)</code></p>
<p><strong>Geometrisation</strong></p>
<p>One can conjecture that something even stronger than homogenisation and isotropisation should hold. This conjecture is backed up by all the currently known results in this direction. </p>
<p>From a <a href="http://en.wikipedia.org/wiki/Geometrization_conjecture">powerful result in three dimensional topology</a>, we expect that spatial slices of the universe (assuming that the universe as a compact topology) can be written as the sum of some <a href="http://en.wikipedia.org/wiki/Seifert_fiber_space">graph manifolds</a> with some hyperbolic three-manifolds. </p>
<p>Then the conjecture is the following:</p>
<blockquote>
<p>In the expanding direction, as time goes to infinity, the limits of the spatial slices is such that we can rescale the metric in such a way that (a) the rescaled metric on the hyperbolic pieces converge to the standard complete hyperbolic metric and (b) the rescaled metric on the graph manifold parts collapse.</p>
</blockquote>
<p>In special situations or under special assumptions, this conjecture has been confirmed by</p>
<ul>
<li>Lars Andersson and Vincent Moncrief by showing the stability of the <a href="http://en.wikipedia.org/wiki/Milne_model">Milne model</a></li>
<li>Vincent Moncrief and Yvonne Choquet-Bruhat in their work studying the $U(1)$ symmetric space-times (under their assumptions the spatial slices must be Seifert fibered spaces, and they do in fact show the desired [rescaled] collapse). </li>
<li>The work of Michael Anderson, Martin Reiris, and possibly others, which show the desired conclusion if one were to assume some strong a priori bounds on the geometry. </li>
</ul>
<hr>
<p><em>[Added 23.01.2014]</em></p>
<p>There are, of course, many, many interesting problems concerning the <em>initial data</em> for Einstein's equations. Just a quick review: the strong ties to geometry of Einstein's equation in general relativity means that the system of equations is somehow simultaneously over- and underdetermined. The latter problem can be cured by "gauge fixing"; but the former problem manifests at the level of initial data. </p>
<p>One is used to be able to freely specify "initial conditions" when attacking an evolution equation. In the case of Einstein's equations (and many other geometric partial differential equations), the initial data must satisfy certain initial constraints to be compatible with the global geometry: here we take the naive view that the initial data should contain the metric tensor on the initial slice and also its first time derivative. (<a href="http://arxiv.org/abs/gr-qc/0405092">This review by Bartnik and Isenberg</a> gives a pretty high-powered review if you are already familiar with Lorentzian geometry; if you are not then I suggest looking at standard textbooks such as Wald's <em>General Relativity</em>.) </p>
<p>Several natural questions arise concerning these <strong>constraint equations</strong>.</p>
<p><strong>How rigid are the constraints</strong>? </p>
<p>By writing down the constraint equations (which can be expressed as 4 partial differential equations), we immediately see that relative to the number of degrees of freedom we have, the constraint equations are under-determined. Therefore we expect "lots" of solutions. But natural questions like "how many solutions are there given a set of prescribed asymptotics" still need to be answered. This is especially so since as part of the proof of the <a href="http://en.wikipedia.org/wiki/Positive_energy_theorem">Positive Mass Theorem</a> is included the following rigidity statement:</p>
<blockquote>
<p>... the ADM mass of a complete asymptotically flat initial data set under assumption of dominant energy vanishes if and only if the initial data is a space-like initial data slice of Minkowski space.</p>
</blockquote>
<p>Therefore we know that certain asymptotics for the initial data are <em>very</em> rigid. </p>
<p>An interesting development is the gluing technique due to Justin Corvino and Rick Schoen from a few years back. A corollary to the gluing technique is that once the ADM mass is allowed to be positive, the asymptotics become a lot less rigid for restricting the initial data. In particular, they were able to construct many initial data sets with asymptotics virtually identical to that of the Schwarzschild and Kerr black holes, but with different topology. The full force of this gluing technique and its implication is still under investigation today. </p>
<p>On the other hand, the success of the positive mass theorem (which is essentially a result on scalar curvature comparison) has led to many current research topics concerning various aspects of the rigidity of asymptotically flat manifolds with non-negative scalar curvature. These questions are mostly Riemannian geometric in nature and practice, and in addition to the Riemannian version of the Penrose inequality mentioned above, just to give a taste the kind of results obtained and questions asked (since I am not really too much of an expert in this direction), one can check out <a href="http://arxiv.org/abs/1303.3545">this recent result of Michael Eichmair and Simon Brendle</a> which does not admit an easy physical interpretation. </p>
<p><strong>How to parametrise the initial data?</strong></p>
<p>Giving that the initial data sets need to satisfy the constraint equations, one is led to ask whether there exists an alternative way of parametrising the initial data (as opposed to the naive initial metric + first time derivative splitting) such that</p>
<ol>
<li>The data can be freely prescribed. For this we need a method to start with some tensors and a method of transformations which guarantees us to recover geometric data that satisfy the constraint equations. </li>
<li>Is exhaustive; that is, every solution of the constraint equations can be generated from free data using this method. </li>
</ol>
<p>This dream dates back to at least Andre Lichnerowicz and is still not entirely settled. A part of the story is the immense success of his <em>conformal method</em> in constructing and parametrising initial data sets of general relativity; it turns out that however, this conformal method is very well suited for a certain situation (data that has constant mean curvature or almost constant mean curvature), but may (the understanding on this issue is a bit muddy at the moment; but this is a post on open questions after all) be very much less well suited for other situations (data that has strongly varying mean curvature). If you are interested, I would highly recommend the <a href="http://www.msri.org/workshops/691">two videos of David Maxwell's lectures</a> from September of last year. </p>
|
394,085 | <p>How is it possible to establish proof for the following statement?</p>
<p>$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$</p>
<p>Where $n,x,k$ are $\text{integers}$.</p>
<hr>
<p>To be more verbose:</p>
<p>I conjecture that;</p>
<p>If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$</p>
<p>How could one prove this?</p>
| Warren Moore | 63,412 | <p>What is $x$? I assume that it is some natural number: then the conjecture is false. If you let $x=0$, then $n=2$, but $n\equiv 2\text{ (mod }10)\not\equiv 7\text{ (mod 10})$.
<hr />
For $n$ to be an integer, we must have $x$ even, and since $n\ne 2$, we must have $n$ odd. So in particular, we cannot have $5x+4$ even. This forces $x=4k+2$, hence
$$
\frac{1}{2}(5x+4)=\frac{1}{2}(5(4k+2)+4)=\frac{1}{2}(20k+14)=10k+7
$$</p>
|
3,256,646 | <p>I find it really hard to find the range. I usually substitute the x's with y and then solve for y, but it does not always work for me. Do you have any advice?</p>
<p>Function in question: </p>
<p><span class="math-container">$$f(x) = \frac{e^{-2x}}{x}$$</span></p>
| cmk | 671,645 | <p>First, <span class="math-container">$A$</span> and <span class="math-container">$B$</span> should be open.</p>
<p>Suppose that we can take <span class="math-container">$x\in A,\ y\in B.$</span> Then, by assumption, there exists a connected subset <span class="math-container">$C'$</span> of <span class="math-container">$C$</span> so that <span class="math-container">$x,y\in C'.$</span> Note that <span class="math-container">$C'\subset A\cup B.$</span> Can you finish from here? </p>
<p>EDIT: Details for remainder of the proof are in comments below. I'll add them here, as a spoiler:</p>
<blockquote class="spoiler">
<p> Consider <span class="math-container">$A'=A\cap C'$</span> and <span class="math-container">$B'=B\cap C'$</span>. These are disjoint, relatively open, non-empty (<span class="math-container">$x\in A', y\in B'$</span>), and have union <span class="math-container">$C'$</span>. This forms a disconnection of the connected set <span class="math-container">$C',$</span> a contradiction. So, we cannot have arbitrary elements <span class="math-container">$x,y$</span> with <span class="math-container">$x\in A$</span> and <span class="math-container">$y\in B.$</span> Hence, every element of <span class="math-container">$C$</span> is contained in either <span class="math-container">$A$</span> or <span class="math-container">$B$</span>, contradicting that both are non-empty.</p>
</blockquote>
|
227,562 | <p>Let $K\subset \mathbb{R}^n$ be a compact convex set of full dimension. Assume that $0\in \partial K$. </p>
<p><strong>Question 1.</strong> Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K\cap \varepsilon S^{n-1}$ is contractible? Here $\varepsilon S^{n-1}$ is the unit sphere centered at 0 of radius $\varepsilon$.</p>
<p>If Question 1 has a positive answer I would like to generalize it a little bit. Under the above assumptions, assume in addition that a sequence $\{K_i\}$ of compact convex sets converges in the Hausdorff metric to $K$.</p>
<p><strong>Question 2.</strong> Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K_i\cap \varepsilon S^{n-1}$ is contractible for $i>i(\varepsilon)$?</p>
<p>A reference would be helpful.</p>
| Mohammad Ghomi | 68,969 | <p>The answer to Question 1 is yes, which is precisely Lemma 3.6 in the paper:</p>
<p><a href="http://people.math.gatech.edu/~ghomi/Papers/torsionrq.pdf" rel="nofollow noreferrer">Boundary torsion and convex caps of locally convex surfaces,
J. Differential Geom., 105 (2017), 427-486.</a></p>
<p>Although the lemma is stated and proved for $R^3$, the same proof works in $R^n$. The proof also indicates how to give a positive answer to Question 2.</p>
<p>The proof is elementary and proceeds as follows: a convex surface can be represented locally as the graph of a convex function over a convex domain in a support plane, which we may identify with $R^{n-1}$. Then the upper half of the sphere corresponds to the graph of a concave function over the same convex domain, assuming that the radius is sufficiently small and after we readjust the domain. Now the portion of the surface cut off by the sphere is the set of points where the convex graph lies below the concave graph. It is easy to check, using the standard inequalities for the convex and concave functions, that this portion projects onto a convex domain. Hence it is a disk.</p>
<p>In the setting of Question 2, we have a sequence of convex functions converging to the convex function mentioned in the previous paragraph. So eventually they will be defined over the same convex domain and lie below the concave function corresponding to the hemisphere, whence again we may conclude that the upper hemisphere cuts out a disk from their graphs. So the answer to Question 2 is yes as well.</p>
|
1,084,041 | <p>Introduction: I've been studying integrals of the form $$\int_0^\infty \frac{x^a}{(e^x-1)^b}dx$$ where a and b are real parameters. I've been able to find closed forms for the integral in terms of the Riemann Zeta function, the Gamma function and the Polygamma functions provided the integral converges when at least one of the parameters is a positive integer.</p>
<p>I then thought of generalizing this to the case where both a and b are non-integers. As a start I considered the integral $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx$$ Using substitution methods and integration by parts I deduced that $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx=\int_1^\infty \frac{\sqrt{\ln x}}{x\sqrt{x-1}}dx=\int_0^1 \sqrt\frac{-\ln x}{x(1-x)} dx=2\sqrt2\int_0^1 \sqrt\frac{-\ln x}{1-x^2}dx$$ $$=\sqrt2\int_0^1 \frac{\arcsin x}{x\sqrt{-\ln x}}dx=\sqrt2\int_0^\infty \frac{\arcsin(e^{-x})}{\sqrt x} dx=2\sqrt2\int_0^\infty \arcsin(e^{-x^2})dx$$ $$=2\sqrt2\int_0^\frac{\pi}{2} \sqrt{-\ln(\sin x)} dx$$</p>
<p>I tried different methods on each integral but pretty much nothing worked. With each method not shown here the integral got much more complicated. The last integral looks similar to the Clausen's integral but I wasn't able to establish a relation between them. It is of course possible to represent the value of the integral as an infinite series but my question is that does anybody know a method to evaluate this integral in terms of well-known special functions or mathematical constants? Thanks in advance.</p>
| Olivier Oloa | 118,798 | <p>A partial answer.</p>
<p>I'm sure you already know the Lerch transcendent, a <strong>special function</strong> which may initially be defined as
$$\Phi(z,s,a):=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}, \quad a>0,\Re s>1,|z|<1.$$
It admits the following integral representation, which you obtain by expanding the integrand as a powers series of $z$:
$$
\Phi(z,s,a)=\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-ze^{-x}}{\rm d}x.
$$ By differentiation with respect to $z$, you get
$$
\partial_z^r\Phi(z,s,a)=(-1)^r\int_0^{\infty}\frac{x^{s-1}e^{-(a+r)x}}{(1-ze^{-x})^{r+1}}{\rm d}x.\tag1
$$
This shows the <strong>level of complexity</strong> of your family of integrals: <a href="http://en.wikipedia.org/wiki/Fractional_calculus" rel="noreferrer">fractional calculus</a>.</p>
<p>For example, from $(1)$, you have</p>
<blockquote>
<p>$$
\int_0^\infty \sqrt\frac{x}{e^x-1}dx= \color{blue}{\left(-\partial_z\right)^{\large \! -1/2}\Phi\left(1,3/2,1\right). }
$$ </p>
</blockquote>
<p>I would be very surprised to learn that we have more than the above expression.</p>
|
1,084,041 | <p>Introduction: I've been studying integrals of the form $$\int_0^\infty \frac{x^a}{(e^x-1)^b}dx$$ where a and b are real parameters. I've been able to find closed forms for the integral in terms of the Riemann Zeta function, the Gamma function and the Polygamma functions provided the integral converges when at least one of the parameters is a positive integer.</p>
<p>I then thought of generalizing this to the case where both a and b are non-integers. As a start I considered the integral $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx$$ Using substitution methods and integration by parts I deduced that $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx=\int_1^\infty \frac{\sqrt{\ln x}}{x\sqrt{x-1}}dx=\int_0^1 \sqrt\frac{-\ln x}{x(1-x)} dx=2\sqrt2\int_0^1 \sqrt\frac{-\ln x}{1-x^2}dx$$ $$=\sqrt2\int_0^1 \frac{\arcsin x}{x\sqrt{-\ln x}}dx=\sqrt2\int_0^\infty \frac{\arcsin(e^{-x})}{\sqrt x} dx=2\sqrt2\int_0^\infty \arcsin(e^{-x^2})dx$$ $$=2\sqrt2\int_0^\frac{\pi}{2} \sqrt{-\ln(\sin x)} dx$$</p>
<p>I tried different methods on each integral but pretty much nothing worked. With each method not shown here the integral got much more complicated. The last integral looks similar to the Clausen's integral but I wasn't able to establish a relation between them. It is of course possible to represent the value of the integral as an infinite series but my question is that does anybody know a method to evaluate this integral in terms of well-known special functions or mathematical constants? Thanks in advance.</p>
| Lucian | 93,448 | <p>Starting from the general <span class="math-container">$~\displaystyle\int_0^\infty\frac{x^a}{e^x-u}dx~=~\frac{\Gamma(a+1)\cdot\text{Li}_{a+1}(u)}u,~$</span> we can then </p>
<p>deduce, by way of repeated differentiation under the integral sign with regard to <em>u</em>, </p>
<p>that <span class="math-container">$~\displaystyle\int_0^\infty\frac{x^a}{(e^x-u)^b}dx~=~(-1)^b\cdot\Gamma(a+1)\cdot{\large\partial}^b_u~\bigg[\dfrac{\text{Li}_{a+1}(u)}u\bigg],~$</span> for all <strong>real</strong> <span class="math-container">$a>-1$</span> </p>
<p>and <span class="math-container">$u\in\mathbb C$</span>, with <span class="math-container">$b\in\mathbb N.~$</span> However, in this case, <span class="math-container">$~b=\dfrac12~$</span> is fractional, so, if you are </p>
<p>looking for a closed form, you will have to appeal to the concept of <a href="http://en.wikipedia.org/wiki/Fractional_calculus" rel="nofollow noreferrer">fractional calculus</a>, </p>
<p>which has already been mentioned above, though in relation to a lesser-known special </p>
<p>function.</p>
|
1,181,631 | <p>Let $f : \mathbb R \to \mathbb R$ continuous. Prove that graph $G = \{(x, f(x)) \mid x \in \mathbb R\}$ is closed.</p>
<p>I'm a little confused on how to prove $G$ is closed. I get the general strategy is to show that every arbitrary convergent sequence in $G$ converges to a point in $G$.</p>
<p>Here is what I tried so far:</p>
<ol>
<li>Let $x_k$ be a sequence which converges to $x$.</li>
<li>Since $f$ is continuous, this implies that $f(x_k)$ converges to $f(x)$.</li>
<li>At this point, can you say every $(x_k, f(x_k))$ converges to a $(x, f(x))$, so $G$ is closed?</li>
</ol>
| Marm | 159,661 | <p>Define at first $F:\mathbb R^2 \rightarrow \mathbb R$,$F(x,y)=f(x)-y$.</p>
<p>Next note that $F$ is continuous (because $f$ and "+" are continuous)</p>
<p>Then the graph is exactly the inverse image of $\{0\}\in \mathbb R$, hence</p>
<p>the graph is closed as an inverse image of a closed set under a continuous function.</p>
|
1,887,536 | <p>Howdy just a simple question,</p>
<p>I know when A is diagonalizable, the eigenvalues of F(A) are just simply $F(\lambda_i)$ where $\lambda_i \exists \sigma (A)$</p>
<p>I'm interested in the case of when A is not diagonalizable. I look at A as a Jordan form, but I cannot seem to show that when $A$ is not diagonalizable, that the eigenvalues of F(A) are $F(\lambda_i$). I'm okay with no proof, just want to know if the eigenvalues of F(A) are $F(\lambda_i$) when A is not diagonalizable. </p>
| imranfat | 64,546 | <p>As hinted in a comment, we assume here that the coefficients $a,b,c,d$ are real.
If $x=1+4i $ is a root, then also its conjugate $x=1-4i$. Now here is where I differ.
It is better to write $x-1=4i$, then square which gives $x^2-2x+1=-16. $ Do you recognize that this move captures the conjugate root as well? So we have $x^2-2x+17$ as one part of the polynomial. The other root $x=2$ comes from the factor $x-2$ and so we arrive at $f(x)=a(x-2)(x^2-2x+17). $ Now substituting $(1,-32)$ for $x$ and $y$ gives you access to find $a$. At this point you can work out the brackets if needed, but that's basic algebra. </p>
|
365,631 | <p>Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.</p>
<blockquote>
<p><strong>What is your favourite example of a proof of this type?</strong></p>
</blockquote>
<p>Here are some examples:</p>
<ul>
<li><p><strong>The probabilistic method in combinatorics</strong> As I understand it, a
typical pattern of argument is as follows. We have a set <span class="math-container">$X$</span> and want to
show that at least one element of <span class="math-container">$X$</span> has property <span class="math-container">$P$</span>. We choose some
function <span class="math-container">$f: X \to \{0, 1, \ldots\}$</span> such that <span class="math-container">$f(x) = 0$</span> iff <span class="math-container">$x$</span> satisfies
<span class="math-container">$P$</span>, and we choose a probability measure on <span class="math-container">$X$</span>. Then we show that
with respect to that measure, <span class="math-container">$\mathbb{E}(f) < 1$</span>. It follows that
<span class="math-container">$f^{-1}\{0\}$</span> has positive measure, and is therefore nonempty.</p>
</li>
<li><p><strong>Real analysis</strong> One example is <a href="http://www.artsci.kyushu-u.ac.jp/%7Essaito/eng/maths/Cauchy.pdf" rel="noreferrer">Banach's
proof</a>
that any measurable function <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> satisfying
Cauchy's functional equation <span class="math-container">$f(x + y) = f(x) + f(y)$</span> is linear.
Sketch: it's enough to show that <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span>, since then
it follows from additivity that <span class="math-container">$f$</span> is continuous everywhere, which makes
it easy. To show continuity at <span class="math-container">$0$</span>, let <span class="math-container">$\varepsilon > 0$</span>. An
argument using Lusin's theorem shows that for all sufficiently small
<span class="math-container">$x$</span>, the set <span class="math-container">$\{y: |f(x + y) - f(y)| < \varepsilon\}$</span> has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives <span class="math-container">$|f(x)| < \varepsilon$</span>.</p>
<p>Another example is the existence of real numbers that are
<a href="https://en.wikipedia.org/wiki/Normal_number" rel="noreferrer">normal</a> (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.</p>
</li>
<li><p><strong>Set theory</strong> Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set <span class="math-container">$X$</span> being a finitely additive <span class="math-container">$\{0, 1\}$</span>-valued
probability measure defined on the full <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$P(X)$</span>. Some
existence proofs work by proving that the subset of elements with the
desired property has measure <span class="math-container">$1$</span> in the ultrafilter, and is therefore nonempty.</p>
<p>One example is a proof that for every measurable cardinal
<span class="math-container">$\kappa$</span>, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a <span class="math-container">$\kappa$</span>-complete ultrafilter on <span class="math-container">$\kappa$</span>. Make an inspired choice of function <span class="math-container">$\kappa \to \{\text{cardinals } <
\kappa \}$</span>. Push the ultrafilter forwards along this function to give
an ultrafilter on <span class="math-container">$\{\text{cardinals } < \kappa\}$</span>. Then prove that the set
of inaccessible cardinals <span class="math-container">$< \kappa$</span> belongs to that ultrafilter ("has
measure <span class="math-container">$1$</span>") and conclude that, in particular, it's nonempty.</p>
<p>(Although it has a similar flavour, I would <em>not</em> include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an <em>explicit construction</em>
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)</p>
</li>
</ul>
<p>(Mathematicians being mathematicians, someone will probably observe that
<em>any</em> existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)</p>
<p><strong>PS</strong> I'm aware of the earlier question <a href="https://mathoverflow.net/questions/34390">On proving that a certain set is
not empty by proving that it is actually
large</a>. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on <em>positive measure</em>, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.</p>
| Terry Tao | 766 | <p><a href="https://en.wikipedia.org/wiki/Szemer%C3%A9di%27s_theorem" rel="noreferrer">Szemerédi's theorem</a> asserts that every set <span class="math-container">$A$</span> of integers of positive upper density (thus <span class="math-container">$\limsup_{N \to \infty} \frac{|A \cap [-N,N]|}{|[-N,N]|} > 0$</span>) contains arbitrarily long arithmetic progressions. One of the shortest (but not the most elementary) proofs of this remarkably deep theorem deduces it from a result in ergodic theory:</p>
<blockquote>
<p><strong>Furstenberg recurrence theorem</strong>: Let <span class="math-container">$E$</span> be a subset of a probability space <span class="math-container">$(X,\mu)$</span> of positive measure, and let <span class="math-container">$T: X \to X$</span> be an invertible measure-preserving shift. Then for any <span class="math-container">$k \geq 1$</span> there exists a positive integer <span class="math-container">$n$</span> such that <span class="math-container">$E \cap T^n E \cap T^{2n} E \cap \dots \cap T^{(k-1) n} E$</span> has positive measure.</p>
</blockquote>
<p>The case <span class="math-container">$k=1$</span> is trivial, and the case <span class="math-container">$k=2$</span> is the classical Poincare recurrence theorem. The general case was established in</p>
<p><em>Furstenberg, Harry</em>, <a href="http://dx.doi.org/10.1007/BF02813304" rel="noreferrer"><strong>Ergodic behavior of diagonal measures and a theorem of Szemeredi on arithmetic progressions</strong></a>, J. Anal. Math. 31, 204-256 (1977). <a href="https://zbmath.org/?q=an:0347.28016" rel="noreferrer">ZBL0347.28016</a>.</p>
<p>Roughly speaking the deduction of Szemerédi's theorem from Furstenberg's theorem is as follows. By hypothesis, there is a sequence <span class="math-container">$N_j \to \infty$</span> such that <span class="math-container">$\frac{|A \cap [-N_j,N_j]|}{|[-N_j,N_j]|}$</span> converges to a positive limit. One can define a generalised density of subsets <span class="math-container">$B \subset {\bf Z}$</span> by the formula <span class="math-container">$\mu(B) := \tilde \lim_{j \to \infty} \frac{|B \cap [-N_j,N_j]|}{|[-N_j,N_j]|}$</span> where <span class="math-container">$\tilde \lim$</span> is an extension of the limit functional <span class="math-container">$\lim$</span> to bounded sequences (this can be constructed using the Hahn-Banach theorem or using an ultrafilter). Morally speaking, this turns the integers <span class="math-container">${\bf Z}$</span> into a probability space <span class="math-container">$({\bf Z},\mu)$</span> in which <span class="math-container">$A$</span> has positive measure and the shift <span class="math-container">$T: n \mapsto n-1$</span> is measure-preserving. Then by the Furstenberg recurrence theorem, for every <span class="math-container">$k$</span>, there is a positive integer <span class="math-container">$n$</span> such that <span class="math-container">$A \cap T^n A \cap \dots \cap T^{(k-1) n} A$</span> has positive measure, hence non-empty, hence <span class="math-container">$A$</span> contains arbitrarily long arithmetic progressions.</p>
<p>(I cheated a little because <span class="math-container">$\mu$</span> is only a finitely additive measure rather than countably additive, but one can massage the finitely additive probability space <span class="math-container">$({\bf Z},\mu)$</span> constructed here into a countably additive model <span class="math-container">$(X, \tilde \mu)$</span> by a little bit of measure-theoretic trickery which I will not detail here.)</p>
|
365,631 | <p>Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.</p>
<blockquote>
<p><strong>What is your favourite example of a proof of this type?</strong></p>
</blockquote>
<p>Here are some examples:</p>
<ul>
<li><p><strong>The probabilistic method in combinatorics</strong> As I understand it, a
typical pattern of argument is as follows. We have a set <span class="math-container">$X$</span> and want to
show that at least one element of <span class="math-container">$X$</span> has property <span class="math-container">$P$</span>. We choose some
function <span class="math-container">$f: X \to \{0, 1, \ldots\}$</span> such that <span class="math-container">$f(x) = 0$</span> iff <span class="math-container">$x$</span> satisfies
<span class="math-container">$P$</span>, and we choose a probability measure on <span class="math-container">$X$</span>. Then we show that
with respect to that measure, <span class="math-container">$\mathbb{E}(f) < 1$</span>. It follows that
<span class="math-container">$f^{-1}\{0\}$</span> has positive measure, and is therefore nonempty.</p>
</li>
<li><p><strong>Real analysis</strong> One example is <a href="http://www.artsci.kyushu-u.ac.jp/%7Essaito/eng/maths/Cauchy.pdf" rel="noreferrer">Banach's
proof</a>
that any measurable function <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> satisfying
Cauchy's functional equation <span class="math-container">$f(x + y) = f(x) + f(y)$</span> is linear.
Sketch: it's enough to show that <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span>, since then
it follows from additivity that <span class="math-container">$f$</span> is continuous everywhere, which makes
it easy. To show continuity at <span class="math-container">$0$</span>, let <span class="math-container">$\varepsilon > 0$</span>. An
argument using Lusin's theorem shows that for all sufficiently small
<span class="math-container">$x$</span>, the set <span class="math-container">$\{y: |f(x + y) - f(y)| < \varepsilon\}$</span> has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives <span class="math-container">$|f(x)| < \varepsilon$</span>.</p>
<p>Another example is the existence of real numbers that are
<a href="https://en.wikipedia.org/wiki/Normal_number" rel="noreferrer">normal</a> (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.</p>
</li>
<li><p><strong>Set theory</strong> Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set <span class="math-container">$X$</span> being a finitely additive <span class="math-container">$\{0, 1\}$</span>-valued
probability measure defined on the full <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$P(X)$</span>. Some
existence proofs work by proving that the subset of elements with the
desired property has measure <span class="math-container">$1$</span> in the ultrafilter, and is therefore nonempty.</p>
<p>One example is a proof that for every measurable cardinal
<span class="math-container">$\kappa$</span>, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a <span class="math-container">$\kappa$</span>-complete ultrafilter on <span class="math-container">$\kappa$</span>. Make an inspired choice of function <span class="math-container">$\kappa \to \{\text{cardinals } <
\kappa \}$</span>. Push the ultrafilter forwards along this function to give
an ultrafilter on <span class="math-container">$\{\text{cardinals } < \kappa\}$</span>. Then prove that the set
of inaccessible cardinals <span class="math-container">$< \kappa$</span> belongs to that ultrafilter ("has
measure <span class="math-container">$1$</span>") and conclude that, in particular, it's nonempty.</p>
<p>(Although it has a similar flavour, I would <em>not</em> include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an <em>explicit construction</em>
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)</p>
</li>
</ul>
<p>(Mathematicians being mathematicians, someone will probably observe that
<em>any</em> existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)</p>
<p><strong>PS</strong> I'm aware of the earlier question <a href="https://mathoverflow.net/questions/34390">On proving that a certain set is
not empty by proving that it is actually
large</a>. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on <em>positive measure</em>, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.</p>
| Stanley Yao Xiao | 10,898 | <p>In general, the probabilistic method of Erdos follows exactly this philosophy: prove that an object with a certain property of number theoretic interest exists by showing that the probability a random set satisfies the desired property with positive probability (usually the probability is one!)</p>
<p>Example: a subset <span class="math-container">$S \subset \mathbb{N}$</span> is an <em>asymptotic additive basis of order <span class="math-container">$k$</span></em> if there exists <span class="math-container">$N_0 > 0$</span> such that for all <span class="math-container">$N > N_0$</span>, there exists <span class="math-container">$x_1, \cdots, x_k \in S$</span> (not necessarily distinct) such that <span class="math-container">$N = x_1 + \cdots + x_k$</span>. In other words, every sufficiently large positive integer is the sum of <span class="math-container">$k$</span> elements of <span class="math-container">$S$</span> (with possible repetition).</p>
<p>If we define <span class="math-container">$r_S^k(n) = \# \{(x_1, \cdots, x_k) \in S^k : n = x_1 + \cdots + x_k\}$</span> to be the <em>representation function of order <span class="math-container">$k$</span></em> with respect to <span class="math-container">$S$</span>, then the average size of <span class="math-container">$r_S^k(n)$</span> is a measure of how "optimal" the set <span class="math-container">$S$</span> as an additive basis. For instance it is known that the set <span class="math-container">$\mathcal{S}$</span> of square integer is an additive basis of order <span class="math-container">$4$</span> (Lagrange's theorem), but it is hardly optimal since <span class="math-container">$r_\mathcal{S}^4(n) \gg n$</span> for all <span class="math-container">$n$</span>. How small can <span class="math-container">$r_S^k(n)$</span> be on average provided that it is positive for all sufficiently large <span class="math-container">$n$</span>?</p>
<p>Erdos and Fuchs gave a "lower bound" for this average: <span class="math-container">$r_S^k(n)$</span> cannot be constant on average. Further, Erdos and Turan made the following conjecture: if <span class="math-container">$S$</span> is an asymptotic additive basis of order <span class="math-container">$k$</span>, then <span class="math-container">$\liminf_{n \rightarrow \infty} r_S^k(n) = \infty$</span>.</p>
<p>Erdos further refined this conjecture to assert that the lower bound ought to be of order <span class="math-container">$\log n$</span>. To show that such optimal additive bases exist he used the probabilistic method. The case <span class="math-container">$k = 2$</span> is due to Erdos and the general case due to Erdos and Tetali.</p>
|
438,336 | <p>This a two part question:</p>
<p>$1$: If three cards are selected at random without replacement. What is the probability that all three are Kings? In a deck of $52$ cards.</p>
<p>$2$: Can you please explain to me in lay man terms what is the difference between with and without replacement.</p>
<p>Thanks guys!</p>
| W_D | 85,348 | <p>No, no, dear MethodManX, while computing probabilities, addition refers to "or", multiplication - to "and". Here you have "and": the first card is a King AND the second is a King AND the fird is a King, so it's rather $\frac{4}{52}\cdot\frac{3}{52}\cdot\frac{2}{52}$.</p>
|
918,788 | <p>How to do this integral</p>
<p>$$\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x$$</p>
<p>for any $k > 0$ ?.</p>
<p>I tried to use gamma function, but sometimes the series doesn't converge.</p>
| Jack D'Aurizio | 44,121 | <p>The integral equals $\sqrt{\pi}e^{-k^2/4}$. To show this, just consider that:
$$ I =\Re\int_{-\infty}^{+\infty}e^{ikx-x^2}\,dx = e^{-k^2/4}\cdot \Re\int_{-\infty}^{+\infty}e^{-(x-ik/2)^2}\,dx $$
and prove that the complex shift does not affect the value of the integral:
$$\int_{-\infty}^{+\infty}e^{-(x-ik/2)^2}\,dx = \int_{-\infty-ik/2}^{+\infty-ik/2}e^{-x^2}\,dx = \int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}.$$
This happens because $e^{-z^2}$ is an entire function whose absolute value when $|\Re(z)|\to +\infty$ and $\Im(z)$ stays bounded goes to zero really fast.</p>
|
240,700 | <p>How can I prove that every maximal ideal of $B= \mathbb{Z} [(1+\sqrt{5})/2] $ is a principal?</p>
<p>I know if I show that B has division with remainder, that means it is a Euclidean domain. It follows that B is PID, and then every maximal ideal is principal ideal in PID. </p>
<p>However, I haven't been able to show that $B$ has division with remainder.</p>
| Davide Giraudo | 9,849 | <p>Consider $e^n$ the sequence whose unique non-zero entry is the $n$-th (which is $1$), and the sequence $x_n=\sum_{j=1}^n2^{-j}e^j\in\Bbb R^{\infty}$. This sequence is Cauchy for the $\ell^p$ norm for all $p$, but doesn't converge to an element of $\Bbb R^{\infty}$ for the $\ell^p$ norm. Indeed, this converge implies componentwise convergence, so the potential limit is $\sum_{j=1}^{+\infty}2^{-j}e^j$, which is not finitely supported.</p>
<p>Actually, no norm can make $\Bbb R^\infty$ complete by Baire category theorem (but we don't need this argument if we work with a specified well-behaved norm).</p>
|
2,882,985 | <p>Let $f_n(x)=\frac{1}{n}\boldsymbol 1_{[0,n]}(x)$. This sequence is bounded in $L^1(\mathbb R)$ since $\|f_n\|_{L^1}=1$. But why is there no subsequence that convergent weakly ? I know that if such subsequence exist (still denote $f_n$), then $\|f_n\|_{L^1}=1$ Let denote $f$ it's limit. Then, since $f_n\to 0$ pointwise then $f=0$ because $$\lim_{n\to \infty }\int_{\mathbb R}f_n\varphi=\int_{\mathbb R}\lim_{n\to \infty }f_n\varphi=0.$$</p>
<p>Therefore $\|f\|=0$. Now, for me there is no reason to have $$\lim_{n\to \infty }\|f_n\|=0.$$ Indeed, using semi lower continuity of the norm, we have that $$1=\liminf_{n\to \infty }\|f_n\|\geq \|f\|=0,$$
and thus, there is no contradiction. But I know that I should have a contradiction, but where ?</p>
| Dzoooks | 403,583 | <p>Let $\{n_k\}_{k \geq 1}$ be any subsequence. Then we readily compute $$\lim_{k \to \infty} \int_{\mathbb{R}} 1 \cdot \left(f_{n_k}(x) - 0\right)dx = \lim_{k \to \infty} 1 = 1 \neq 0.$$</p>
|
1,196,424 | <p>So I'm reviewing my notes and I just realized that I can't think of how to show that a particular integer mod group is abelian. I know how to do it with symmetric but not with integers themselves.</p>
<p>For example, lets say I was asked to show $\mathbb{Z_5}$ is abelian.</p>
<p>I know for symmetric groups, lets say $S_5$ i can pick two elements in $S_5$, for example (123),(23) and if (123)(23)=(23)(123) then I know it is abelian but how would I go about the integers?</p>
| David Wheeler | 23,285 | <p>What you really want to show, for any cyclic group $G = \langle a\rangle$, is that:</p>
<p>$a^ka^m = a^ma^k$ for any integers $k,m$.</p>
<p>This takes care of $\Bbb Z$ <em>and</em> $\Bbb Z_n$ (for any $n$) "all at once".</p>
<p>Can you show how to leverage the "rules of exponents" and commutativity of addition in the integers, here?</p>
|
3,207,767 | <p>What is the general solution of differential equation <span class="math-container">$y\frac{d^{2}y}{dx^2} - (\frac{dy}{dx})^2 = y^2 log(y)$</span>.</p>
<p>The answer to this DE is <span class="math-container">$log(y) = c_1 e^x + c_2 e^{-x}$</span></p>
<p>I don't know the method to solve differential equation with degree more than 1. Please tell me how to solve these types of equations.</p>
| Michael Rozenberg | 190,319 | <p>Another way for the proof of the inequality <span class="math-container">$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$</span></p>
<p>We obtain: <span class="math-container">$$(x^2+3)(y^2+3)=x^2y^2+3(x^2+y^2)+9=$$</span>
<span class="math-container">$$=(xy-1)^2+(x-y)^2+2(x+y)^2+8\geq2((x+y)^2+4).$$</span>
By the same way <span class="math-container">$$(z^2+3)(t^2+3)\geq2((z+t)^2+4).$$</span>
Id est, by C-S <span class="math-container">$$\prod_{cyc}(x^2+3)\geq4((x+y)^2+4)(4+(y+z)^2)\geq$$</span>
<span class="math-container">$$\geq4(2x+2y+2z+2t)^2=16(x+y+z+t)^2.$$</span></p>
|
1,134,145 | <p>A set S is bounded if every point in S lies inside some circle |z| = R other it is unbound. Without appealing to any limit laws, theorems, or tools from calculus, prove or disprove that the set {$\frac{z}{z^2 + 1}$; z in R} is bounded.</p>
<p>I imagine that it's simple, but I have no clue where to start due to the restrictions. Thanks</p>
| user2566092 | 87,313 | <p>Hint: First assume $|z| < 1$ and prove that the set is contained in some bounded interval, say $[-1,1]$. Then assume $|z| \geq 1$ and prove the set is contained in the interval $[-1,1]$.</p>
|
1,600,307 | <p>Let $n$ be an integer greater than 1, $\alpha$ be a real number, and consider the quadratic form $Q_{\alpha}$ given by: </p>
<p>for every $(x_1, ... , x_n) \in R^n$, </p>
<p>$$Q_{\alpha}(x_1,...,x_n)= \sum_{i=1}^n x_i^2 - \alpha(\sum_{i=1}^n x_i)^2$$</p>
<p>Find all the eigenvalues of $Q_{\alpha}$ in terms of $\alpha$ and $n$. What is the condition on $\alpha$ for $Q_{\alpha}$ to be positive-definite?</p>
<p>EDIT: I am currently expanding out every term, and then regrouping the common $x_ix_j$ terms to see what I can come up with and to see whether I can spot a matrix that I can put together...</p>
<p>Any ideas are welcome. </p>
<p>Thanks,</p>
| K. Miller | 264,375 | <p>Here is a hint. Let $x = (x_1,\ldots,x_n)^T$ and let $e$ denote the $n$-vector of all ones. Then</p>
<p>$$
Q_\alpha(x) = x^Tx - \alpha (x^Te)(e^Tx) = x^T(I - \alpha ee^T)x
$$</p>
<p>So you need to find the eigenvalues of the matrix $A_\alpha = I - \alpha ee^T$.</p>
|
1,220,923 | <p>Find the value of the integral
$$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx.$$
I tried the substitution $x=t^5$ to obtain
$$\int_0^\infty \frac{5t^6}{1+t^{10}}dt.$$
Now we can factor the denominator to polynomials of degree two (because we can easily find all roots of polynomial occured in the denominator of the former integral by using complex numbers) and then by using partial fraction decomposition method find the integral!</p>
<p>Is there any simple method to find the integral value??!!</p>
| Chappers | 221,811 | <p>Your integral is
$$ \int_0^{\infty} \frac{x^{7/5}}{1+x^2} \frac{dx}{x}. $$
Substituting $u=x^2$, $du/u=2dx/x$, the integral becomes
$$ \frac{1}{2}\int_0^{\infty} \frac{u^{7/10}}{1+u} \, du $$
Now,
$$ \frac{1}{1+u}= \int_0^{\infty} e^{-(1+u)\alpha} \, d\alpha, $$
and interchanging the order of integration gives
$$ \frac{1}{2}\int_0^{\infty} e^{-\alpha} \left( \int_0^{\infty} u^{7/10-1} e^{-\alpha u} \, du \right) \, d\alpha $$
Changing variables in the inner integral shows it has value $\alpha^{-7/10}\Gamma(7/10)$. We then do
$$ \frac{\Gamma(7/10)}{2} \int_0^{\infty} \alpha^{3/10-1} e^{-\alpha} \, d\alpha = \frac{\Gamma(7/10)}{2}\Gamma(3/10) = \frac{\pi}{2\sin{(3\pi/10)}}, $$
which you do by whatever arcane trigonometry you have to hand.</p>
<hr>
<p>On the other hand, there is a more direct way of getting the relation
$$ \int_0^{\infty} \frac{x^{s-1}}{1+x} \, dx = \frac{\pi}{\sin{\pi s}}, \quad 0<\Re(s)<1. $$
Split the integral in two at $x=1$:
$$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_1^{\infty} \frac{x^{s-1}}{1+x} \, dx. $$
Now set $u=1/x$ in the second integral, and we find
$$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_0^{1} \frac{u^{1-s-1}}{1+u} \, du = \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx. $$
And now we use a trick I saw in some of G.H. Hardy's work: expand the denominator in a power series and change the order of integration (we can check this is legal easily enough):
$$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} (x^{n+s-1}+x^{n-s}) \, dx $$
Doing the integrals gives
$$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{n+s} + \frac{1}{n-s+1} \right) $$
Shifting the second terms by one (which you can check is okay by trucating the sums, we end up with
$$ \int_0^{\infty} \frac{x^s}{1+x} \, dx = \frac{1}{s} + \sum_{n=1}^{\infty} (-1)^n \left(\frac{1}{s+n}+\frac{1}{s-n} \right), $$
which is a well-known expression for $\pi\csc{\pi s}$. (Which is actually what Hardy says. You can get it using Fourier series, and so avoid both complex analysis and the Gamma function entirely.)</p>
|
1,611,078 | <p>If we have the function $f : \mathbb{R}\rightarrow \mathbb{R} : x \mapsto x^2 + \frac{x}{3}$ and the sequence $(a_n)_{n \in \mathbb{N}}$ which is recursively specified for $n \in \mathbb{N_+}$:</p>
<p>$a_n =_{def} f(a_{n-1})$</p>
<p>(So the sequence is fixed by $a_0$) </p>
<p>How to determine all real numbers $x \in \mathbb{R}$ for which the sequence $(a_n)_{n \in \mathbb{N}}$ with $a_0 = x $ converges and the associated limits? </p>
| Hagen von Eitzen | 39,174 | <p>Finding the limits is easy: If $a$ is the limit of such a sequence, then certainly $f(a)=a$. This leads to $a=a^2+\frac13a$, i.e., $a=0$ or $a=\frac23$.</p>
<p>If $x>\frac23$, then $f(x)=x^2+\frac13x>\frac23x+\frac13x=x$, i.e., $a_0>\frac23$ produces a strictly increasing sequence. This could only converge to a fixpoint of $f$ that is $>\frac23$. As there is no such fixpoint, sequences starting with $a_0>\frac23$ do not converge.</p>
<p>If $a_0=\frac23$, the sequence is constant (and obviously the limit is $\frac23$).</p>
<p>If $0<x<\frac23$ then $f(x)=x^2+\frac13x>0$ and $f(x)=x^2+\frac13x<\frac23x+\frac13x=x$, so $0<a_0<\frac23$ leads to a strictly decreasing sequence. As it is bounded from below, it must converge to a fixpoint of $f$ that is $<\frac23$, i.e., the sequence converges to $0$.</p>
<p>If $a_0=0$, the sequence is constant and converges to $0$, obviously.</p>
<p>If $-\frac16\le x<0$ then $f(x)=x^2+\frac13x>\frac13 x>x$ and $f(x)=x^2+\frac13x\ļe -\frac16x+\frac13x=\frac16x<0$, so a sequence starng with $-\frac16\le a_0<0$ is strictly increasingm bounded from above by $0$, hence converges to a limit $\le 0$ and in fact converges to $0$ as that is the only fixpoint $\le 0$.</p>
<p>If $a_0< \frac16$ then the sequence is the same (except for tha initial term) as that starting with $-\frac13-a_0$. Using this symmetry we conclude</p>
<blockquote>
<p>We have $\lim_{n\to\infty} a_n=\frac23$ if $a_0\in\{-1,\frac23\}$. We have $\lim_{n\to\infty}a_n=0$ if $-1<a_0<\frac23$. In all other cases, $a_n$ diverges.</p>
</blockquote>
|
42,787 | <p>I am using <code>ListPlot</code> to display from 5 to 12 lines of busy data. The individual time series in my data are not easy to distinguish visually, as may be evident below, because the colors are not sufficiently different.</p>
<p><img src="https://i.stack.imgur.com/PiMMh.png" alt="enter image description here"></p>
<p>I have been trying to use <code>PlotStyle</code>, <code>ColorData</code>and related functions to get better colors. I would rather not have to specify a specific list of colors because the number of plot items varies from test to test. I created a toy plot to experiment with - the problem is illustrated by lines "F" and "G", which seem to be almost the same color. <code>PlotStyle</code> -> <code>ColorData</code> doesn't seem to work. Is there a simple way to do this?</p>
<pre><code>ListPlot[Table[i*Range[0, 10], {i, 1, 5, 0.5}]
, Frame -> True, Joined -> True, PlotRange -> All
, PlotLegends -> SwatchLegend[Characters["ABCDEFGHIGKLMNOP"]]
, PlotStyle -> ColorData["TemperatureMap"] ]
</code></pre>
<p><img src="https://i.stack.imgur.com/Wdi0O.png" alt="enter image description here"></p>
<p>It looks like </p>
<pre><code>ListLinePlot[Table[data2*i, {i, k}], PlotStyle -> Thick,
ColorFunction -> Function[{x1, x2}, ColorData[c1][x2]]]
</code></pre>
<p>from another <a href="https://mathematica.stackexchange.com/questions/27131/plotstyle-in-listplot-change-color-scheme-manually-choose-color-of-first-plot?rq=1">question</a> may be the answer. I didn't see that before. I'll try it out. I don't think I really understand <code>ColorData</code>. Meanwhile, if anyone has generally enlightening comments, I would appreciate them.</p>
| Mike Honeychurch | 77 | <p>I just use <code>"ColorList"</code></p>
<pre><code>ListPlot[Table[i*Range[0, 10], {i, 1, 5, 0.5}], Frame -> True,
Joined -> True, PlotRange -> All,
PlotStyle -> ColorData[3, "ColorList"]]
</code></pre>
<p><img src="https://i.stack.imgur.com/NuzKE.png" alt="enter image description here"></p>
<p>You can add some interactivity and choose between the 60+ colour data numbers until you find your preferred set.</p>
<p>From the docs:</p>
<p><img src="https://i.stack.imgur.com/h1UHM.png" alt="enter image description here"></p>
|
3,213,464 | <p>Does 22.449 approximate to 22 or 23?
If we see it one way
<span class="math-container">$22.449≈22$</span>
But on the other hand
<span class="math-container">$22.449≈22.45≈22.5≈23$</span>
Which one is correct?</p>
| Wrzlprmft | 65,502 | <p>Your problem seems to be that you implicitly expected that rounding behaves like equality, “less than”, and similar.
Speaking in notation, you seem to assume that the <span class="math-container">$≈$</span> relation (meaning something like “the left-hand side rounded yields the right-hand side”) behaves like the <span class="math-container">$=$</span> relation (equality). I assume that from this you concluded:</p>
<p><span class="math-container">$$22.449≈22.45≈22.5≈23 \qquad ⇒ \qquad 22.449 ≈ 23.$$</span></p>
<p>While you can make such a conclusion for other relation operators such as <span class="math-container">$=$</span>, <span class="math-container">$≥$</span>, <span class="math-container">$⊂$</span>, this is not valid for <span class="math-container">$≈$</span> (in your sense).
Since the symbol <span class="math-container">$≈$</span> suggests this, I do not consider it a good choice of notation here.
Another argument against using <span class="math-container">$≈$</span> here is that it has a direction, even though its symbol suggests that it doesn’t.
Finally, you use the same symbol for rounding to different digits.</p>
<p>Consider the alternative notation:</p>
<p><span class="math-container">$$22.449
~\stackrel{0.01}{⇝}~ 22.45
~\stackrel{0.1 }{⇝}~ 22.5
~\stackrel{1 }{⇝}~ 23;
\qquad 22.449
~\stackrel{1 }{⇝}~ 23.$$</span></p>
<p>Would you still assume that the left statement implies the right statement? And most importantly, would you still consider the right statement correct?</p>
<p>More abstractly speaking, the symbol <span class="math-container">$≈$</span> suggests an <a href="https://en.wikipedia.org/wiki/Equivalence_relation" rel="nofollow noreferrer">equivalence relation</a>, though you used it for something that doesn’t fulfil the requirements of such a relation, in particular <a href="https://en.wikipedia.org/wiki/Transitive_relation" rel="nofollow noreferrer">transitivity</a>. (The others are not fulfilled either, but that’s not the cause of your problem.)</p>
<p>As much as we like all sorts of relations to have the nice properties of equivalence relations, rounding simply doesn’t have them.
This sometimes leads to seemingly paradoxical behaviour – usually if you wrap everything up in a manner that suggests that rounding behaves like equality.</p>
|
1,716,656 | <p>I am having trouble solving this problem</p>
<blockquote>
<p>Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?</p>
</blockquote>
<p>My attempt:</p>
<p>I first want to find the deposit per month.</p>
<p>I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,</p>
<p>$D*100(Ia_{30|0.08}) = 100,000$</p>
<p>However, the $D$ I got was 8.12, which is clearly not right.</p>
<p>Can someone help?</p>
| Martín Vacas Vignolo | 297,060 | <p>Total possible results: $6\times6=36$</p>
<p>Favorable results: $1-3,2-4,3-5,4-6$ and opposites, $8$.</p>
<p>Then the probability is $8/36=2/9$.</p>
|
2,426,263 | <p>Nowadays, the most widely-taught model of computation (at least in the English-speaking world) is that of Turing Machines, however, it wasn't the first Turing-Complete model out there: μ-recursive functions came a few years earlier, and λ-calculus came a year earlier. Why is it that Turing machines are so popular today. Intuitive appeal? The depth or notoriety of Turing's work? Or is the model perceived to be "natural" in some sense, and if so, why?</p>
| Bram28 | 256,001 | <p>If you read Turing's original 1936 paper, you will see how he tried to get at the fundamental ingredients of a human following some systematic process of symbol manipulation. The word 'effective' is sometimes used in this context: a computation or algorithm is 'effective' when a human can perform it. E.g Long division is an effective algorithm because we can do it. In fact, when we are performing long division, we are the 'computer'. Indeed, in Turing's time a 'computer' was often still understood as a human performing some calculation. The Manhattan project, for example, used many human computers, and the recent movie Hidden Figures showed the human computers employed at NASA at the time of the Apollo Project.</p>
<p>Turing's conceptual path that led him to his 'Turing Machines' is actually fairly straightforward. Any computer, Turing said, manipulates a bunch of symbols, which we can think of written down on a sheet of paper. Now, while computing, the computer takes discrete steps, where each step taken is dependent on what <em>stage</em> or <em>state</em> the process is at, and what symbols the computer sees in front of it. </p>
<p>Going back to the addition of numbers, for example, we may at some point find ourselves in the 'state' of having to add two specific digits, followed by the 'state' of having to write down the result, etc. Now, Turing said, while our natural language description of these states may use words like 'adding two digits' or 'writing down the result', all that is really important is that the computer is able to <em>distinguish</em> between the different states. That is, for all that the computer cares, the different states could simply be numbered $1$, $2$, $3$, etc. and all that the computer needs to know in order to make its next move, is to know whether we're in state $1$, $2$, or whichever other state there may be there. </p>
<p>A similar story goes for the symbols, Turing argued. That is, while we are used to the decimal representation of numbers, we could of course use any other kind of representational scheme. Again, Turing argued, all that matters is that we are able to <em>distinguish</em> one symbol from a different symbol. Or, a little more technical, if we use a set of symbols $s_1$, $s_2$, $s_3$, etc. then all that should matter to the computer is that the computer be able to recognize that some written symbol is, say, symbol $s_2$, rather than any of the other symbols. In short, the computer simply needs to have an alphabet of symbols that the computer is able to read and write.</p>
<p>How are the symbols organized? Turing writes:</p>
<blockquote>
<p>Computing is normally done by writing certain symbols on paper. We may suppose this paper to be divided into squares like a child’s arithmetic book. In elementary arithmetic the 2-dimensional character of the paper is sometimes used. But such use is always avoidable, and I think it will be agreed that the two-dimensional character of paper is no essential of computation. I assume then that the computation is carried out on one-dimensional paper, on a tape divided into squares. </p>
</blockquote>
<p>Turing here makes the point that whatever we can compute using a two-dimensional (or even higher-dimensional, as when we use multiple sheets of paper) layout for the symbols, we can always simulate using a one-dimensional layout, since we can always 'string' together all the lines and make it into one long string of symbols, possibly using special symbols to demarcate what would normally be line, page, or other breaks. </p>
<p>Also, since it is not predictable how much 'scratch' work one has to do while performing the computation, Turing decided that we can think of having a tape with an infinite number of squares on it, and that symbols can be placed in each of those squares. Or, maybe more realistically, that one always has the ability extend the tape by adding squares on either direction, so that one is guaranteed of any amount of 'working space' as needed.</p>
<p>How many states and symbols can there be? Turing said that there could be any number, as long as it is finite. Turing argued for both to be finite saying that the human mind and human memory was limited. Also, assuming a finite size of the squares on the tape in which to place our symbols , there can only be finitely many different symbols, or else the differences between the symbols would be come infinitely small, and we would not be able to recognize and discriminate the one symbol from the next:</p>
<blockquote>
<p>I shall also suppose that the number of symbols which may be printed is finite. If we were to allow an infinity of symbols, then there would be symbols differing to an arbitrarily small extent.</p>
</blockquote>
<p>Finally, as a computer we must be able to go to any arbitrary place in this one long string of symbols. But for that, Turing argued, all that is required is an ability to go either left or right one square relative from the square one is currently looking at: wherever one wants to go, eventually one should be able to get there. Thus, Turing imagined that one would have a 'read/write' head that at any particular point in time would be 'at' some particular square on the tape, and that this 'head' can move left or right, one square at a time.</p>
<p>And there you have it: all the components that we now call a Turing machine!</p>
<p>Moreover, this very argument makes the Turing-Church Thesis plausible that any computation can be performed using Turing-computation.... which is why it is important that instructors covering Turing-machines should relate Turing's analysis to the students, rather than immediately starting with a definition of Turing-machines as unfortunately so many instructors do.</p>
|
449,631 | <p>Again a root problem..
$\sqrt{2x+5}+\sqrt{5x+6}=\sqrt{12x+25}$</p>
<p>Isn't there any standardized way to solve root problems..Can u plz help by giving some tips and stategies for root problems??</p>
| Pedro | 23,350 | <p>I think some things can be written in a clearer manner. First, I would change </p>
<blockquote>
<p>This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$.</p>
</blockquote>
<p>For</p>
<blockquote>
<p>This implies we can write $A=\{x_n:n\geqslant 1\}$ </p>
</blockquote>
<p>or </p>
<blockquote>
<p>Let $\{x_n:n\geqslant 1\}$ be an enumeration of $A$.</p>
</blockquote>
<p>Then, I would say</p>
<blockquote>
<p>If $S$ is a subset of the natural numbers, let $\min S$ denote least element of $S$. Define $S_1=\{k:x_k\in E\}$. $S_2=S_1\setminus \{\min S_1\}$ and in general $$S_{n+1}=S_{n}\setminus\{\min S_1,\ldots,\min S_{n}\}$$ Then define $n_k=\mathscr \min S_k$</p>
</blockquote>
<p>I guess the idea is clear: consider the set of subscripts such that $x_k\in E$. By the well ordering of the natural numbers, we can extract a sequence $n_k$ such that $$n_1<n_2<\cdots\\E=\{x_{n_k}:k\geqslant 1\}$$</p>
<p>by considering the first subscript with $x_k\in E$ removing this one from the list and looking at the new first subscript (our second in the list) and so on. Some details should be addressed</p>
<p>$(1)$ The set $S_{n}$ is never empty. <em>Reason:</em> Since $E\subseteq A$; $S_1$ is not empty. Moreover, $E$ is by assumption infinite, thus removing one element every time cannot exhaust it.</p>
<p>$(2)$ The construction exhausts the elements of $E$ -- that is, it is a surjection. <em>Reason:</em> Pick $m$ such that $x_m\in E$. We need to find $k$ such that $n_k=m$. Consider the finite set $\{x_1,\dots,x_m\}$. Keeping the order, remove all elements such that $x_i\notin E$. We're left with a finite set, and it must be the case $\{x_{n_1},\dots,x_{n_k}\}$ for some $k$, and $n_k=m$, by definition of the $n_k$. </p>
<p>$(3)$ The construction is an injection. <em>Reason:</em> By construction, $n_k\neq n_j$ if $j\neq k$ for if $j>k$ then $n_k\notin S_{j}$.</p>
<p><em>Conclusion</em> We obtain an bijection of $E$ with an infinite subset $F$ of $\Bbb N$. Thus $E\simeq F\simeq \Bbb N$, that is $E\simeq \Bbb N$.</p>
|
449,631 | <p>Again a root problem..
$\sqrt{2x+5}+\sqrt{5x+6}=\sqrt{12x+25}$</p>
<p>Isn't there any standardized way to solve root problems..Can u plz help by giving some tips and stategies for root problems??</p>
| CopyPasteIt | 432,081 | <blockquote>
<p>Let <span class="math-container">$A$</span> be a countably infinite set and <span class="math-container">$E\subset A$</span> be an infinite subset. Then <span class="math-container">$E$</span> is countably infinite.</p>
</blockquote>
<p>The OP is using the 'can be bijectively enumerated' definition for a countably infinite set. Their proof concludes by writing <span class="math-container">$E$</span> as a union of a family of singletons indexed by <span class="math-container">$\Bbb N$</span>.</p>
<p>Writing out a precise/formal proof that builds the OP's subsequence can be broken into parts, each of which is easily digestible.</p>
<p>Let <span class="math-container">$(x_{n})_{\, n\in\mathbb{N}}$</span> be a bijective enumeration of <span class="math-container">$A$</span>.</p>
<p>Lemma 1: We can associate to any finite subset <span class="math-container">$F$</span> of <span class="math-container">$E$</span> with <span class="math-container">$k$</span> elements another subset <span class="math-container">$\Gamma(F)$</span> of <span class="math-container">$E$</span> satisfying</p>
<p><span class="math-container">$\tag 1 F \subset \Gamma(F)$</span>
<span class="math-container">$\tag 2 |\Gamma(F)| = k + 1$</span></p>
<p>Proof<br>
Since <span class="math-container">$F$</span> is finite and <span class="math-container">$E$</span> is infinite, the set <span class="math-container">$S = \{m \in \mathbb N \, | \, x_m \notin F\}$</span> is nonempty. Let <span class="math-container">$\gamma(F)$</span> be the least integer in <span class="math-container">$S$</span> and define</p>
<p><span class="math-container">$$ \Gamma(F) = F \cup \{ x_{ \gamma(F) } \}$$</span>
<span class="math-container">$\blacksquare$</span></p>
<p>Using recursion and lemma 1 we can prove the following:</p>
<p>Proposition 2: There exists a family of sets <span class="math-container">$(F_n)_{\, n \ge 1}$</span> satisfying for every integer <span class="math-container">$n \ge 1$</span>,</p>
<p><span class="math-container">$\tag 3 F_n \subset E$</span>
<span class="math-container">$\tag 4 F_n \subset F_{n+1}$</span>
<span class="math-container">$\tag 5 |F_n| = n$</span>
<span class="math-container">$\tag 6 E = \bigcup_{n \ge 1} F_n$</span></p>
<p>The OP can now 'take those singleton accretions' to define the bijective enumeration <span class="math-container">$(y_{n})_{\, n \ge 1}$</span> of <span class="math-container">$E$</span>.</p>
|
2,720,694 | <p>I am facing difficulty to calculate the second variation to the following functional.</p>
<p>Define $J: W_{0}^{1,p}(\Omega)\to\mathbb{R}$ by
$J(u)=\frac{1}{p}\int_{\Omega}|\nabla u|^p\,dx$ where $p>1$.</p>
<p>I am able to calculate the first variation as follows:
$J'(u)\phi=\int_{\Omega}\,|\nabla u|^{p-2}\nabla u\cdot\nabla\phi\,dx$
which I have got by using the functional $E:\mathbb{R}\to\mathbb{R}$ defined by $E(t)=J(u+t\phi)$.</p>
<p>But I am unable to calculate the second variation. </p>
<p>Any type of help is very much appreciated.</p>
<p>Thanks.</p>
| user284331 | 284,331 | <p>\begin{align*}
\int_{0}^{2\pi}\sin(nx)\sin(mx)dx&=-\dfrac{1}{4}\int_{0}^{2\pi}(e^{inx}-e^{-inx})(e^{imx}-e^{-imx})dx\\
&=-\dfrac{1}{4}\int_{0}^{2\pi}(e^{i(m+n)x}-e^{-i(m+n)x}-e^{-i(n-m)x}-e^{-i(m-n)x})dx\\
&=-\dfrac{1}{4}\int_{0}^{2\pi}(-e^{-i(n-m)x}-e^{-i(m-n)x})dx\\
&=-\dfrac{1}{4}\delta_{m,n}(-2\pi-2\pi)\\
&=\delta_{m,n}\pi.
\end{align*}</p>
|
2,720,694 | <p>I am facing difficulty to calculate the second variation to the following functional.</p>
<p>Define $J: W_{0}^{1,p}(\Omega)\to\mathbb{R}$ by
$J(u)=\frac{1}{p}\int_{\Omega}|\nabla u|^p\,dx$ where $p>1$.</p>
<p>I am able to calculate the first variation as follows:
$J'(u)\phi=\int_{\Omega}\,|\nabla u|^{p-2}\nabla u\cdot\nabla\phi\,dx$
which I have got by using the functional $E:\mathbb{R}\to\mathbb{R}$ defined by $E(t)=J(u+t\phi)$.</p>
<p>But I am unable to calculate the second variation. </p>
<p>Any type of help is very much appreciated.</p>
<p>Thanks.</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>The trick for trigonometric integrals:
$$
z = e^{ix}\implies\sin(nx) =
\frac12(e^{inx} − e^{−inx}) = \frac12(z^n - z^{-n})
\qquad dz = iz\,dx
$$
$$
\int_{0}^{2\pi}\sin(nx)\sin(mx)\,dx =
\frac14\int_{|z|=1}(z^n - z^{-n})(z^m - z^{-m})\frac1{iz}\,dz =
$$
$$
\frac1{4i}\int_{|z|=1}(z^{m-n-1} + z^{n-m-1} - z^{m+n-1} - z^{-m-n-1})\,dz = \cdots
$$
In your problematic case:
$$
m = n\ne 0\implies
z^{m-n-1} = z^{n-m-1} = z^{-1}, z^{m+n-1}\ne z^{-1},
z^{-m-n-1}\ne z^{-1},
$$
and the integral is $\frac{2\cdot2\pi i}{4i} = \pi$.</p>
|
122,471 | <p>Can anyone explain how I can prove that either $\phi(t) = \left|\cos (t)\right|$ is characteristic function or not? And which random variable has this characteristic function? Thanks in advance.</p>
| Sasha | 11,069 | <p>Since $\phi(t) = | \cos(t) |$ is periodic with period $\pi$ and even and if it is valid, it should correspond to a symmetric discrete random variable. </p>
<p>It is not hard to establish that:
$$
| \cos(t) | = \frac{2}{\pi} + \frac{4}{\pi} \sum_{m=1}^\infty \frac{(-1)^{m-1}}{4 m^2-1} \cos(2 m t)
$$
<img src="https://i.stack.imgur.com/JrSnk.png" alt="enter image description here"></p>
<p>Comparing this to $\phi_X(t) = \sum_{m=-\infty}^\infty c_m \mathrm{e}^{i t m}$ we see that $c_4 = - \frac{2}{\pi} \cdot \frac{1}{15}$ is negative, thus can not be a probability of any random variable. </p>
|
3,430,812 | <p>Consider the set of integers, <span class="math-container">$\Bbb{Z}$</span>. Now consider the sequence of sets which we get as we divide each of the integers by <span class="math-container">$2, 3, 4, \ldots$</span>.</p>
<p>Obviously, as we increase the divisor, the elements of the resulting sets will get closer and closer.</p>
<p><strong>Question:</strong> In the limit as <span class="math-container">$\text{divisor}\to\infty$</span>, what will the "limiting" set be?
(I don't think it could be <span class="math-container">$\Bbb{R}$</span>.)</p>
| Robert Z | 299,698 | <p>Let <span class="math-container">$A_n=\{x/n:x\in\mathbb{Z}\}$</span> with <span class="math-container">$n$</span> any integer greater than <span class="math-container">$1$</span> then it is easy to see that <span class="math-container">$\mathbb{Z}\subset A_n\subset \mathbb{Q}$</span>. We claim that
<span class="math-container">$$\limsup_n A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)=\mathbb{Q}$$</span> and <span class="math-container">$$\liminf_n A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)=\mathbb{Z}.$$</span>
See <a href="https://en.wikipedia.org/wiki/Set-theoretic_limit" rel="nofollow noreferrer">set-theoretic limit</a> for more details about the limit of a sequence of sets.</p>
|
1,395,619 | <p>One of my friend asked this doubt.Even in lower class we use both as synonyms,he says that these two concepts have difference.Empty set $\{ \}$ is a set which does not contain any elements,while null set ,$\emptyset$ says about a set which does not contain any elements.</p>
<p>I could not make out that...is his argument correct ? if so how ?</p>
| Christoph | 86,801 | <p>In measure theory, a <a href="https://en.wikipedia.org/wiki/Null_set" rel="nofollow noreferrer">null set</a> refers to a set of measure zero. For example, in the reals, <span class="math-container">$\mathbb R$</span> with its standard measure (Lebesgue measure), the set of rationals <span class="math-container">$\mathbb Q$</span> has measure <span class="math-container">$0$</span>, so <span class="math-container">$\mathbb Q$</span> is a null set in <span class="math-container">$\mathbb R$</span>. Actually, all finite and countably infinite subsets of <span class="math-container">$\mathbb R$</span> have measure <span class="math-container">$0$</span>. In contrast, the empty set <em>always</em> refers to the unique set with no elements, denoted <span class="math-container">$\left\{ \right\}$</span>, <span class="math-container">$\varnothing$</span> or <span class="math-container">$\emptyset$</span>.</p>
|
320,452 | <p>For any positive integer <span class="math-container">$n\in\mathbb{N}$</span> let <span class="math-container">$S_n$</span> denote the set of all bijective maps <span class="math-container">$\pi:\{1,\ldots,n\}\to\{1,\ldots,n\}$</span>. For <span class="math-container">$n>1$</span> and <span class="math-container">$\pi\in S_n$</span> define the <em>neighboring number</em> <span class="math-container">$N_n(\pi)$</span> as the minimum distance of <span class="math-container">$\pi$</span>-neighbors, or more formally: <span class="math-container">$$N_n(\pi) = \min \big(\big\{|\pi(k)-\pi(k+1)|:k\in\{1,\ldots,n-1\}\big\}\cup \big\{|\pi(1) - \pi(n)|\big\}\big).$$</span></p>
<p>For <span class="math-container">$n>1$</span> let <span class="math-container">$E_n$</span> be the expected value of the neighboring number of a member of <span class="math-container">$S_n$</span>.</p>
<p><strong>Question.</strong> Do we have <span class="math-container">$\lim\sup_{n\to\infty}\frac{E_n}{n} > 0$</span>?</p>
| Wojowu | 30,186 | <p><span class="math-container">$\newcommand{\e}{\varepsilon}$</span>
For any fixed <span class="math-container">$\e>0$</span>, permutations with <span class="math-container">$N_n(\pi)>\e n$</span> asymptotically have density zero. Indeed, consider values <span class="math-container">$\pi(1),\dots,\pi(k)$</span>. There are <span class="math-container">$n$</span> choices for <span class="math-container">$\pi(1)$</span>. For <span class="math-container">$\pi(2)$</span>, we have at most <span class="math-container">$(1-\e)n$</span> choices, since it has to be at distance at least <span class="math-container">$\e n$</span> from <span class="math-container">$\pi(1)$</span>. Similarly, for <span class="math-container">$\pi(3),\dots,\pi(k)$</span> we have at most <span class="math-container">$(1-\e)n$</span> choices. For remaining <span class="math-container">$\pi(j),j>k$</span> we just take the estimate <span class="math-container">$n-j+1$</span>. It follows the ratio of permutations satisfying <span class="math-container">$N_n(\pi)>\e n$</span> is at most
<span class="math-container">$$\frac{(1-\e)^{k-1}n^k}{n(n-1)\dots(n-k+1)}=\frac{(1-\e)^{k-1}}{(1-1/n)(1-2/n)\dots(1-k/n)}<\frac{(1-\e)^{k-1}}{(1-k/n)^k}.$$</span>
Taking <span class="math-container">$k=\e n/2$</span> this bound goes to zero exponentially.</p>
<p>Since for large enough <span class="math-container">$n$</span> all but <span class="math-container">$\e n!$</span> permutations have <span class="math-container">$N_n(\pi)<\e n$</span> and remaining ones have <span class="math-container">$N_n(\pi)<n$</span>, it follows <span class="math-container">$E_n\leq 2\e n$</span>, so <span class="math-container">$E_n/n\to 0$</span>. </p>
|
909,741 | <blockquote>
<p><strong>ALREADY ANSWERED</strong></p>
</blockquote>
<p>I was trying to prove the result that the OP of <a href="https://math.stackexchange.com/questions/909712/evaluate-int-0-frac-pi2-ln1-cos-x-dx"><strong><em>this</em></strong></a> question is given as a hint.</p>
<p>That is to say: <em>imagine that you are not given the hint and you need to evaluate</em>:</p>
<blockquote>
<p>$$I = \int^{\pi/2}_0 \log{\cos{x}} \, \mathrm{d}x \color{red}{\overset{?}{=} }\frac{\pi}{2} \log{\frac{1}{2}} \tag{1}$$ </p>
</blockquote>
<p><em>How would you proceed?</em></p>
<hr>
<p>Well, I tried the following steps and, despite it seems that I am almost there, I have found some troubles:</p>
<ul>
<li>Taking advantage of the fact: $$\cos{x} = \frac{e^{ix}+e^{-ix}}{2}, \quad \forall x \in \mathbb{R}$$</li>
<li>Plugging this into the integral and performing the change of variable $z = e^{ix}$, so the line integral becomes a contour integral over <em>a quarter of circumference of unity radius centered at $z=0$</em>, i.e.:
$$ I = \frac{1}{4i} \oint_{|z|=1}\left[ \log{ \left(z+\frac{1}{z}\right)} - \log{2} \right] \, \frac{\mathrm{d}z }{z}$$</li>
</ul>
<blockquote>
<p>$\color{red}{\text{We cannot do this because the integrand is not holomorphic on } |z| = 1 }$</p>
</blockquote>
<ul>
<li>Note that the integrand has only one pole lying in the region enclosed by the curve $\gamma : |z|=1$ and it is holomorphic (is it?) almost everywhere (except in $z =0$), so the residue theorem tells us that:</li>
</ul>
<p>$$I = \frac{1}{4i} \times 2\pi i \times \lim_{z\to0} \color{red}{z} \frac{1}{\color{red}{z}} \left[ \underbrace{ \log{ \left(z+\frac{1}{z}\right)} }_{L} - \log{2} \right] $$</p>
<ul>
<li>As I said before, it seems that I am almost there, since the result given by eq. (1) follows iff $L = 0$, which is not true (I have tried L'Hôpital and some algebraic manipulations).</li>
</ul>
<p>Where did my reasoning fail? Any helping hand?</p>
<p>Thank you in advance, cheers!</p>
<hr>
<p>Please note that I'm not much of an expert in either complex analysis or complex integration so please forgive me if this is trivial.</p>
<hr>
<p>Notation: $\log{x}$ means $\ln{x}$.</p>
<hr>
<p>A graph of the function $f(z) = \log{(z+1/z)}$ helps to understand the difficulties:</p>
<p><img src="https://i.stack.imgur.com/jAjTP.png" alt="enter image description here"></p>
<p>where $|f(z)|$, $z = x+i y$ is plotted and the white path shows where $f$ is not holomorphic.</p>
| dustin | 78,317 | <p>Since this post was tagged with complex analysis, I can provide a contour integration solution as well. Again, I will exploit the identity given to you by @idm.
$$
\int_0^{\pi/2}\ln(\sin(\theta))d\theta = \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta
$$
Consider $1 - e^{2iz} = -2ie^{iz}\sin(z)$. We can write $1 - e^{2iz}$ as
$$
1 - e^{-2y}(\cos(2x) + i\sin(2x)) < 0\text{ when } x=\pi n, \ y\leq 0
$$
Now let's consider the contour from $0$ to $\pi$ to $\pi + iA$ to $iA$ where we take a quarter of a circle around $0$ and $\pi$ with radius $\epsilon$. From the periodicity of the function, the vertical line segments cancel each other since they have opposite signs. Additionally, as $A\to\infty$, the top integral of the top line goes to zero and as $\epsilon\to 0$, the integral around $0$ and $\pi$ go to zero.
\begin{align}
\ln(-2ie^{ix}\sin(z)) &= \ln(-2i) + \ln(e^{ix}) + \ln(\sin(\theta))\\
&= \ln|-2i| + i\arg(-2i) + ix + \ln(\sin(\theta))\\
&= \ln(2) - i\frac{\pi}{2} + \ln(\sin(\theta)) + i\frac{\pi}{2}
\end{align}
where $\ln(2i) = \ln(2) + i\arg(-2i)$ and we take the principle argument to be $-\frac{\pi}{2}$ and the imaginary part of $ix$ is between $0$ and $\pi$. Since there are no poles in them contour, by the Cauchy integral formula, the integral is equal to zero.
\begin{alignat}{2}
\int_0^{\pi/2}\ln(\sin(\theta))d\theta &=\frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta\\
&= \frac{\ln(2)}{2}\int_0^{\pi}d\theta - \frac{i\pi}{4}\int_0^{\pi}d\theta + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi}{4}\int_0^{\pi}d\theta &&{}= 0\\
&= \frac{\pi\ln(2)}{2} - \frac{i\pi^2}{4} + \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta + \frac{i\pi^2}{4} &&{}=0\\
\frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2}\\
\int_0^{\pi/2}\ln(\sin(\theta))d\theta &= -\frac{\pi\ln(2)}{2}
\end{alignat}</p>
|
3,834,894 | <p>I understand a continuous function may not be differentiable. But does every continuous function have directional derivative at every point? Thanks!</p>
| Marius S.L. | 760,240 | <p>There is no need to distinguish between one and more directions. If a continuous function like <span class="math-container">$x\longmapsto |x|$</span> isn't differentiable at <span class="math-container">$x=0,$</span> then there is no directional derivative in <span class="math-container">$x$</span>-direction. <br />
The contrary is the case: every differentiation is always a directional differentiation. We have that <span class="math-container">$f$</span> is differentiable at <span class="math-container">$x_0$</span> if there is a linear function <span class="math-container">$\mathbf{J}$</span> and a remainder function <span class="math-container">$\mathbf{r}$</span> such that
<span class="math-container">\begin{equation}\mathbf{f(x_{0}+v)=f(x_{0})+J(v)+r(v)} \end{equation}</span>
where <span class="math-container">$\mathbf{v}$</span> is the direction where we consider a change of slope. Partial derivatives are along the coordinate axis, the total derivative is a linear combination of partial derivatives, but all are in some direction <span class="math-container">$\mathbf{v}$</span>.</p>
|
2,153,743 | <p>I have the relation $R = \{(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)\}$ on the set $\{1,2,3,4\}$. I have to find [1] and [4] but I don't really understand what that means. </p>
<p>I get that $[a]$ is the set of all elements of $A$ related (by $R$) to $a$ so $[a]=\{x\in A : x$ $R$ $a\}$ right? But I don't get the significance of the number 1 or 4 inside of the brackets ([1] and [4]).</p>
| ConMan | 82,793 | <p>$[1]$ is the set of all things that are equivalent to $1$, given the relation. So, since $1R1$ and $1R2$ are the only relations given that involve $1$, $[1] = \{1, 2\}$ (and in particular note that $[1] = [2]$). You can do the same thing for $[4]$.</p>
|
706,980 | <p>If I know that $f(z)$ is differentiable at $z_0$, $z_0 = x_0 + iy_0$.
How do I prove that $g(z) = \overline{f(\overline{z})}$ is differentiable at $\overline z_0$?</p>
| Christian Blatter | 1,303 | <p>The function $f$ is defined in a neighborhood of $z_0$. Therefore the function $g$ is defined in a neighborhood of $\bar z_0$.</p>
<p>By assumption there is a number $a\in{\mathbb C}$ such that
$$f(z_0+\zeta)=f(z_0)+ a\zeta +o(\zeta)\qquad(\zeta\to0)\ .$$</p>
<p>Using the definition of $g$ we therefore have
$$g(\bar z_0+\zeta)=\overline{f(z_0+\bar\zeta)}= \overline{f(z_0)}+\overline{a\bar\zeta}+\overline{o(\bar\zeta)}=\overline{f(z_0)}+\bar a\>\zeta+o(\zeta)\qquad(\zeta\to0)\ .$$
This shows that $g'(\bar z_0)=\bar a$.</p>
|
3,288,815 | <p>I'm reading theorem 3.11 in Rudin's RCA what says <span class="math-container">$L^p$</span> space is a complete metric space.
At the end of the proof, Rudin says that "Then <span class="math-container">$\mu(E)=0$</span>, and on the complement of <span class="math-container">$E$</span> the sequence <span class="math-container">${f_n}$</span> converges uniformly to a bounded function <span class="math-container">$f$</span>".
Why <span class="math-container">${f_n}$</span> converges uniformly?
I see <span class="math-container">$$E^c=\{f_n : |f_n (x)|\leq \|f_n\|_\infty \text{ and } |f_n-f_m|\leq \|f_n-f_m\|_\infty \}$$</span></p>
<p><img src="https://i.stack.imgur.com/EqFah.jpg" alt=""></p>
| mechanodroid | 144,766 | <p>By definition of <span class="math-container">$\|\cdot\|_\infty$</span>, the sets
<span class="math-container">$$A_k=\{x \in X : |f_k(x)| > \|f_k\|_\infty\},\quad B_{m,n}=\{x \in X : |f_n(x)-f_m(x)| > \|f_n-f_m\|_\infty\}$$</span>
are <span class="math-container">$\mu$</span>-null sets so <span class="math-container">$E$</span> is a <span class="math-container">$\mu$</span>-null sets as a countable union of such sets.</p>
<p>For every <span class="math-container">$x \in E^c$</span> we have
<span class="math-container">$$|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty \xrightarrow{m,n\to\infty} 0 \quad \text{ uniformly in } x \in E^c$$</span>
so <span class="math-container">$(f_n|_{E^c})_n$</span> is uniformly Cauchy. Also since <span class="math-container">$|f_n(x)| \le \|f_n\|_{\infty}, \forall x \in E^c$</span> we have that <span class="math-container">$f_n|_{E^c}$</span> are bounded functions.</p>
<p>Now <a href="https://math.stackexchange.com/q/3239854/144766">recall</a> that the space <span class="math-container">$B(E^c, \mathbb{R})$</span> of bounded functions <span class="math-container">$E^c \to \mathbb{R}$</span> equipped with the metric <span class="math-container">$d(f,g) = \sup_{x \in E^c}|f(x)-g(x)|$</span> is a complete metric space. Since <span class="math-container">$(f_n|_{E^c})_n$</span> is a Cauchy sequence in <span class="math-container">$B(E^c, \mathbb{R})$</span>, there exists <span class="math-container">$f \in B(E^c, \mathbb{R})$</span> such that <span class="math-container">$f_n|_{E^c} \to f$</span> uniformly on <span class="math-container">$E^c$</span>.</p>
<p>In particular we have <span class="math-container">$f_n|_{E^c} \to f$</span> pointwise so <span class="math-container">$f$</span> is a measurable function as a limit of measurable functions.</p>
|
490,802 | <p>Is $(x,3,5)$ a plane, for $x\in\mathbb{R}$?</p>
<p>I know that if two of the coordinates are "arbitrary", like $(x,y,4)$or $(3,y,z)$, then it creates a plane (for $x,y,z\in \mathbb{R}).$</p>
<p>Is there a way to tell if it would create a plane in $\mathbb{R}^3?$</p>
| Mark Bennet | 2,906 | <p>The equation $y=3$ defines a line in $\mathbb R^2$, but a plane in $\mathbb R^3$. Likewise $z=5$ defines a plane in $\mathbb R^3$. Each equation can be seen as constraining a point in one dimension, leaving it free to be located in a two-dimensional space (in this case a plane).</p>
<p>The two equations $y=3, z=5$ taken together define the intersection between the two planes, which is a line (this always happens with two planes in $\mathbb R^3$, except in the case where the planes are parallel). A third equation would identify a particular point on the line (except in special cases).</p>
<p>This is very much akin to the way in which two straight lines in the plane intersect in a single point unless they are parallel.</p>
<p>Note that each equation reduces the dimension by $1$ - we start with $\mathbb R^3$ - with one equation we get to a plane, two equations gives us a line and three give a point. This counting-down principle also applies in greater numbers of dimensions (noting always that we have to exclude special cases).</p>
|
1,182,953 | <p>Does anyone know the provenance of or the answer to
the following integral</p>
<p>$$\int_0^\infty\ \frac{\ln|\cos(x)|}{x^2} dx $$</p>
<p>Thanks.</p>
| Lucian | 93,448 | <p><strong>Hint:</strong> Let $I(n)=\displaystyle\int_0^\infty\frac{1-\cos^{2n}x}{x^2}~dx.~$ Prove first that $I(n)=n\pi~\dfrac{\displaystyle{2n\choose n}}{4^n}~,~$ then evaluate $I'(0)$.</p>
|
4,344,571 | <p>In a previous exam assignment, there is a problem that asks for a proof that <span class="math-container">$\mathbb{Z}_{24}$</span> and <span class="math-container">$\mathbb{Z}_{4}\times\mathbb{Z}_6$</span> are <strong>not</strong> isomorphic.</p>
<p>We have <span class="math-container">$\mathbb{Z}_{24}$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span> if there exists a bijective function <span class="math-container">$f∶ \mathbb{Z}_{24}\rightarrow\mathbb{Z}_{4}\times\mathbb{Z}_6$</span> such that <span class="math-container">$f(a+b)=f(a)+f(b)$</span> and <span class="math-container">$f(ab)=f(a)f(b) \forall a,b\in R$</span>. Since there are exactly <span class="math-container">$24$</span> unique elements in both <span class="math-container">$\mathbb{Z}_{24}$</span> and <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span>, we can construct a bijective function <span class="math-container">$f∶ \mathbb{Z}_{24}\rightarrow\mathbb{Z}_{4}\times\mathbb{Z}_6$</span>. Consider then
<span class="math-container">$$\begin{aligned}
&f\left([a]_{24}+[b]_{24}\right) \\
&=f\left([a+b]_{24}\right) \\
&=\left([a+b]_{4},[a+b]_{6}\right) \\
&=\left([a]_{4}+[b]_{4},[a]_{6}+[b]_{6}\right) \\
&=\left([a]_{4},[a]_{6}\right)+\left([b]_{4},[b]_{6}\right) \\
&=f\left([a]_{24}\right)+f\left([b]_{24}\right)
\end{aligned}$$</span>
and
<span class="math-container">$$\begin{aligned}
&f\left([a]_{24}[b]_{24}\right) \\
&=f\left([a b]_{24}\right) \\
&=\left([a b]_{4},[a b]_{6}\right) \\
&=\left([a]_{4}[b]_{4},[a]_{6}[b]_{6}\right) \\
&=\left([a]_{4},[a]_{6}\right)\left([b]_{4},[b]_{6}\right) \\
&=f\left([a]_{24}\right) f\left([b]_{24}\right).
\end{aligned}$$</span>
It therefore seems to me that this function shows that <span class="math-container">$\mathbb{Z}_{24}$</span> <em>is</em> isomorphic to <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span>.</p>
<p>Can someone tell me where I go wrong with this "proof", and tell me how I can show that the rings are <em>not</em> isomorphic?</p>
| Mark | 470,733 | <p>I guess you defined <span class="math-container">$f([a]_{24})=([a]_4, [a]_6)$</span>? This is clearly not a bijection. For example, <span class="math-container">$[0]_{24}$</span> and <span class="math-container">$[12]_{24}$</span> are mapped to the same element.</p>
<p>The rings are not isomorphic because the additive group of <span class="math-container">$\mathbb{Z_{24}}$</span> is cyclic, while the additive group of the other ring isn't. So there can't even be a bijection which only satisfies <span class="math-container">$f(a+b)=f(a)+f(b)$</span>.</p>
|
311,677 | <p>The problem from the book. </p>
<blockquote>
<p>$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 -y$ </p>
</blockquote>
<p>I understand the solution till this part. </p>
<p>$\ln \vert 6 - y \vert = x + C$ </p>
<p>The solution in the book is $6 - Ce^{-x}$ </p>
<p>My issue this that this solution, from the book, doesn't seem to resolve the issue of the abs value of $\vert 6 - y\vert$ </p>
| David Mitra | 18,986 | <p>You should have, as your general solution,
$$
-\ln|6-y|=x+C\ \quad\iff\quad |6-y|=e^C e^{-x} .
$$</p>
<p>If $y-6>0$, you have the solution
$$y-6= e^Ce^{-x}\ \quad\iff\quad y=6+ e^Ce^{-x} .
$$</p>
<p>If $y-6<0$, you have the solution
$$6-y= e^Ce^{-x}\ \quad\iff\quad y=6- e^Ce^{-x} .
$$</p>
<p>In either case, the solution can be written as $y=6- Ce^{-x} $, for some constant $C$ (different from the $C$ above).</p>
|
904,041 | <p>$$tx'(x'+2)=x$$
First I multiplied it:
$$t(x')^2+2tx'=x$$
Then differentiated both sides:
$$(x')^2+2tx'x''+2tx''+x'=0$$
substituted $p=x'$ and rewrote it as a multiplication
$$(2p't+p)(p+1)=0$$
So either $(2p't+p)=0$ or $p+1=0$</p>
<p>The first one gives $p=\frac{C}{\sqrt{T}}$
The second one gives $p=-1$. My question is how do I take the antidervative of this in order to get the answer for the actual equation?</p>
| Mary Star | 80,708 | <p><strong>EDIT:</strong></p>
<p>$x=t(x')^2+2tx'$</p>
<p>$p=x'$</p>
<p>$x=tp^2+2tp$</p>
<p>We differentiate in respect to $t$:</p>
<p>$p=p^2+t2pp'+2p+2tp' \Rightarrow p'(2tp+2t)=(p-p^2-2p) \Rightarrow p'(p+1)2t=-(p^2+p) \Rightarrow p'(p+1)2t=-p(p+1) \Rightarrow p'(p+1)2t+p(p+1)=0 \Rightarrow (p+1)(2tp'+p)=0 \\ \Rightarrow p+1=0 \text{ or } 2tp'+p=0 \\ \Rightarrow p=-1 \text{ or } p'=-\frac{1}{2t}p \\ \Rightarrow p=-1 \text{ or } \frac{p'}{p}=-\frac{1}{2t} \\ \Rightarrow p=-1 \text{ or } \ln{p}=-\frac{1}{2}\ln{t} +c\\ \Rightarrow p=-1 \text{ or } \ln{p}=\ln{t^{-\frac{1}{2}}} +c\\ \Rightarrow p=-1 \text{ or } p= \pm e^c \frac{1}{\sqrt{t}} \\ \Rightarrow x'=-1 \text{ or } x'= \pm e^c\frac{1}{\sqrt{t}} \\ \Rightarrow x(t)=-t+c_1 \text{ or } x(t)=2 C\sqrt{t}+c_2, \text{ where } C= \pm e^c$.</p>
<p>$$$$</p>
<ul>
<li>$x(t)=-t+c_1 \Rightarrow x'=-1$</li>
</ul>
<p>Replacing this at the initial equation we get:</p>
<p>$x=t(x')^2+2tx' \Rightarrow -t+c_1=t-2t \Rightarrow -t+c_1=-t \Rightarrow c_1=0$</p>
<p>Therefore, $x(t)=-t$.</p>
<p>$$$$</p>
<ul>
<li>$x(t)=2 C\sqrt{t}+c_2 \Rightarrow x'(t)=\frac{C}{\sqrt{t}} $</li>
</ul>
<p>Replacing this at the initial equation we get:</p>
<p>$x=t(x')^2+2tx' \Rightarrow 2 C\sqrt{t}+c_2=t\frac{C^2}{t}+2t\frac{C}{\sqrt{t}} \Rightarrow 2C \sqrt{t}+c_2=C^2+\frac{2Ct}{\sqrt{t}} \\ \Rightarrow 2Ct+c_2\sqrt{t}=C^2\sqrt{t}+2Ct \Rightarrow c_2\sqrt{t}=C^2\sqrt{t} \Rightarrow c_2=C^2$</p>
<p>Therefore, $x(t)=2 C\sqrt{t}+C^2$.</p>
|
3,819,202 | <p>Can anyone explain to solve the identity posted by my friend <span class="math-container">$$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$</span> which is an infinite nested square roots of 2. <strong>(Pattern <span class="math-container">$++--$</span> repeating infinitely)</strong></p>
<p>Converging to finite nested radical of <span class="math-container">$2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$</span></p>
<p>The finite nested radical, I was able to derive <span class="math-container">$\cos12° = \cos(30-18)°$</span> as follows</p>
<p><span class="math-container">$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$</span>
<span class="math-container">$$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$</span>
Where <span class="math-container">$\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$</span> (by Half angle cosine formula) and
<span class="math-container">$\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$</span> (solving again by half angle cosine formula)
<span class="math-container">$2\cos36° =\frac{ \sqrt5 +1}{2}$</span> which is golden ratio</p>
<p><span class="math-container">$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$</span></p>
<p>Further steps finally lead to the finite nested radical</p>
<p>Method actually I tried to solve infinite nested square roots of 2 is as follows.</p>
<p><span class="math-container">$2\cos\theta = \sqrt{2+2\cos2\theta}$</span> and
<span class="math-container">$2\sin\theta = \sqrt{2-2\cos2\theta}$</span></p>
<p>Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical
<span class="math-container">$$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$</span></p>
<p>Simplifying step by step as follows</p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$</span> then</p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$</span> (by <span class="math-container">$\sin\theta = \cos(90-\theta)$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$</span></p>
<p><span class="math-container">$2\cos12° = \sqrt{2+2\cos24°}$</span></p>
<p><span class="math-container">$2\cos12° = 2\cos12°$</span></p>
<p>We are back to <span class="math-container">$\sqrt1$</span></p>
<p>Actually this is how I got stuck!</p>
<p>But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.</p>
<p>Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots</p>
<p>Thanks in advance.</p>
| saulspatz | 235,128 | <p>If the value of the radical is <span class="math-container">$x$</span>, then we have <span class="math-container">$$x=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}}\tag1$$</span> Repeated squaring gives
<span class="math-container">$$\left(\left(\left(x^2-2\right)^2-2\right)^2-2\right)^2=2-x\tag2$$</span></p>
<p>Now, <span class="math-container">$(2)$</span> has <span class="math-container">$8$</span> solutions, and notice for all choices of the first three signs in <span class="math-container">$(1)$</span>, repeated squaring gives <span class="math-container">$(2)$</span>. Thus, the solutions of <span class="math-container">$(2)$</span> are the eight solutions to <span class="math-container">$$x=\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\sqrt{2-x}}}}$$</span></p>
<p>So, we must first show that <span class="math-container">$2\cos12^\circ$</span> satisfies <span class="math-container">$(2)$</span>, and then to show that it is the root given by the choice of signs in the question.</p>
<p>To verify that <span class="math-container">$2\cos12^\circ$</span>, we use the formula <span class="math-container">$$(2\cos\theta)^2-2 = 2(2\cos^2\theta-1)=2\cos2\theta\tag3$$</span>
Then setting <span class="math-container">$x=2\cos12^\circ$</span>, <span class="math-container">$(3)$</span> gives
<span class="math-container">$$\begin{align}
x^2-2&=2\cos24^\circ\\
(x^2-2)^2-2&=2\cos48^\circ\\
((x^2-2)^2-2)^2-2&=2\cos96^\circ\\
(((x^2-2)^2-2)^2-2)-2&=2\cos192^\circ=-2\cos12^\circ=-x\\
\end{align}$$</span>
as required.</p>
<p><strong>ADDENDUM</strong></p>
<p>Since <span class="math-container">$0\leq x\leq 2$</span>, there is a value <span class="math-container">$0\leq\theta\leq\frac\pi2$</span> such that <span class="math-container">$x=2\cos\theta$</span>. The argument above gives <span class="math-container">$2\cos16\theta=-2\cos\theta$</span> so either <span class="math-container">$$16\theta=(2n+1)\pi+\theta$$</span>or<span class="math-container">$$16\theta=(2n+1)\pi-\theta$$</span> The condition <span class="math-container">$0\leq\theta\leq\frac\pi2$</span> gives <span class="math-container">$8$</span> possibilities for <span class="math-container">$\theta$</span>: either<span class="math-container">$$\theta=\frac{(2n+1)\pi}{15},\ n=0,1,2,3$$</span> or
<span class="math-container">$$\theta=\frac{(2n+1)\pi}{17},\ n=0,1,2,3$$</span>
so if you can sort the values of the nested radicals for the <span class="math-container">$8$</span> choices of sign in decreasing order, you not only evaluate the given infinite nested radical, but <span class="math-container">$7$</span> more. For example, it seem clear that choosing the choosing the <span class="math-container">$3$</span> plus signs would give the larges value, which would mean that the value of that radical would be <span class="math-container">$2\cos\frac\pi{17}$</span>. If the second largest value comes from choosing the first <span class="math-container">$2$</span> signs as <span class="math-container">$+$</span> and the third as <span class="math-container">$-$</span>, that would complete the proof for the original question.</p>
<p>Actually, there's still something missing. For this argument to work, we also have to show that the infinite nested radical converges for all <span class="math-container">$8$</span> sign choices. I'm having trouble coming up with an economical way to do that.</p>
<p>I carried out the numerical work, and found that
<span class="math-container">$$\begin{align}
2\cos\frac{7\pi}{15}&=
\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{7\pi}{17}&=
\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{5\pi}{15}&=
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{5\pi}{17}&=
\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{3\pi}{15}&=
\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\
2\cos\frac{3\pi}{17}&=
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{\pi}{15}&=
\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\
2\cos\frac{\pi}{17}&=
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\
\end{align}$$</span></p>
<p>Still haven't found a nice argument for convergence, though it's clear numerically that all sequences converge rapidly.</p>
|
1,436,867 | <p>I don´t know an example wich $ \rho (Ax,Ay)< \rho (x,y) $ $ \forall x\neq y $ is not sufficient for the existence of a fixed point .
can anybody help me? please</p>
| Community | -1 | <p>I was about to answer this question <a href="https://math.stackexchange.com/questions/1444036/dtx-ty-dx-y-does-not-guarantee-existence-of-fixed-point">here</a>, but it got marked as a duplicate of this one. Thus, I will answer it on this thread.</p>
<blockquote>
<p>Let $T$ be a mapping of a complete metric space $X$ into itself. How do I show that the condition $d(Tx, Ty) < d(x, y)$, where $x \neq y$, does not suffice for the existence of a fixed point for $T$?</p>
</blockquote>
<p>Let $$F(y) = \int_{-\infty}^y e^{-\pi x^2} dx.$$The relevant properties of $F$ are that $F$ is strictly positive, strictly increasing in $y$, and$$0 < F(y_1) - f(y_2) \le y_1 - y_2$$whenever $y_1 > y_2$. (Convergence of this integral is evident by comparing to $e^{-|x|}$ on $|x| > 1$, and monotonicity because the integrand is positive. The last part follows from the Mean Value Theorem because $F'(y) = e^{-\pi y^2}$ and $0 < e^{-\pi y^2} \le 1$.) We then claim that $T(y) = y- F(y)$ is strictly decreasing: if $y_1 > y_2$, then the inequalities$$0 < F(y_1) - F(y_2) \le y_1 - y_2$$are equivalent to $$y_1 - y_2 > y_1 - y_2 - (F(y_1) - F(y_2)) \ge 0.$$The middle quantity above is $T(y_1) - T(y_2)$ so $T$ decreases distances on $\mathbb{R}$, but clearly $T$ has no fixed points because $T(y) < y$ for all $y$.</p>
|
2,038,189 | <p>(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)</p>
<p>I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.</p>
<p>However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?</p>
<p>I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".</p>
<p>I have a theory:</p>
<ul>
<li>Invert both sides and then multiply both sides by 2;</li>
</ul>
<p>Is this correct?</p>
| qwr | 122,489 | <p>Multiply by $x$: $$5x=2$$</p>
<p>Divide by $5$: $$\frac{2}{5} = x$$</p>
<p>More generally, $$\frac{x}{b} = a \iff x=ab \iff \frac{x}{a} = b$$</p>
|
2,038,189 | <p>(Note: I didn't learn how to solve equations the conventional way; instead I was just taught to "move numbers from side to side", inverting the sign or the operation accordingly. I am learning the conventional way though because I think it makes the process of solving equations clearer. That being said, I apologize if this question is too "basic".)</p>
<p>I know that when I have an equality such as $5 = \frac{x}{2}$ I have to multiply both sides by 2 to get the answer.</p>
<p>However, what is the process behind $5 = \frac{2}{x} \Leftrightarrow \frac{2}{5} = x$ ?</p>
<p>I know that when I have an equation which the variable is in the denominator I have to move the numerator to the other side and make it the numerator and the number that's already there the denominator, but I don't really know why that is or how that's done "mathematically".</p>
<p>I have a theory:</p>
<ul>
<li>Invert both sides and then multiply both sides by 2;</li>
</ul>
<p>Is this correct?</p>
| Frank | 332,250 | <p>From $5=\frac 2x$ to get $x=\frac 25$, you're multiplying both sides by $x$ to get $$5x=2$$
And dividing by $5$, we get $x=\frac 25$.</p>
|
2,334,215 | <p><a href="https://math.stackexchange.com/questions/20223/getting-an-x-for-chinese-remainder-theorem-crt">Getting an X for Chinese Remainder Theorem (CRT)</a></p>
<p>In the "Easy CRT" part of the answer to this problem, the author demonstrates that (-3/77) mod 65 is equal to 16. I don't understand - how is this accurate? I sort of understand the steps, but wouldn't the answer just be 62/77?</p>
<p>Thanks, and I apologize if I've missed something obvious!</p>
| Bernard | 202,857 | <p>In the ring of integers mod. $n$, there is no ‘real’ fraction. Some elements have reciprocals, others don't, depending whether they're coprime with the modulus or not.</p>
<p>For instance, modulo $14$, $5$ is a unit, since $3\cdot 5\equiv 1\pmod{14}$, and $5^{-1}=3$, so one can be tempted to write, say, $\dfrac 45$ instead of $4\cdot 5^{-1}=4\cdot 3=12\pmod{14}$. This way of writing is very confusional, from my point of view.</p>
<p>In the present case, we have $77\equiv 12\pmod{65}$ and one computes with the <em>extended Euclidean algorithm</em> that $12^{-1}\equiv -27$, so that
$$-\frac 3{77}=(-3)(-27)=81\equiv 16\pmod{65}.$$</p>
|
2,334,215 | <p><a href="https://math.stackexchange.com/questions/20223/getting-an-x-for-chinese-remainder-theorem-crt">Getting an X for Chinese Remainder Theorem (CRT)</a></p>
<p>In the "Easy CRT" part of the answer to this problem, the author demonstrates that (-3/77) mod 65 is equal to 16. I don't understand - how is this accurate? I sort of understand the steps, but wouldn't the answer just be 62/77?</p>
<p>Thanks, and I apologize if I've missed something obvious!</p>
| Trevor Gunn | 437,127 | <p>The definition of $\frac{1}{x}$ is that $\frac{1}{x}$ is the quantity such that $x \cdot \frac1x = 1$ (which may or may not exist). Therefore
$$ \frac{-3}{77} \equiv 16 \pmod {65} \text{ if and only if } -3 \equiv 77 \cdot 16 \pmod {65} $$
This happens if and only if
$$77 \cdot 16 + 3\equiv 0 \pmod {65}$$
which by definition of congruence mod $65$ says that
$$65 \mid (77 \cdot 16 + 3)$$
which is true because $$ 77 \cdot 16 + 3 = 1235 = 65 \cdot 19. $$</p>
|
3,031,460 | <blockquote>
<p>Give an example of an assertion which is not true for any positive
integer, yet for which the induction step holds.</p>
</blockquote>
<p>First of all, definition.</p>
<blockquote>
<p>In <strong>inductive step</strong>, we suppose that <span class="math-container">$P(k)$</span> is true for some positive
integer <span class="math-container">$k$</span> and then we prove that <span class="math-container">$P(k + 1)$</span> is true.</p>
</blockquote>
<p><strong>My thoughts</strong> : Since the assertion has to satisfy the induction step, it must be false for <span class="math-container">$n=1$</span>. </p>
<p>But I cannot think of any such assertion.
Does anyone have any hints? Thank you.</p>
<p><strong>Edit</strong>: Just realised that my thoughts don't make much sense, as the question demands the assertion to be false for every positive integer. </p>
| Shubham Johri | 551,962 | <p><span class="math-container">$P(k):k$</span> is irrational</p>
<p><span class="math-container">$P(k): \{k\}>0$</span>, where <span class="math-container">$\{k\}$</span> denotes the fractional part of <span class="math-container">$k$</span></p>
|
3,031,460 | <blockquote>
<p>Give an example of an assertion which is not true for any positive
integer, yet for which the induction step holds.</p>
</blockquote>
<p>First of all, definition.</p>
<blockquote>
<p>In <strong>inductive step</strong>, we suppose that <span class="math-container">$P(k)$</span> is true for some positive
integer <span class="math-container">$k$</span> and then we prove that <span class="math-container">$P(k + 1)$</span> is true.</p>
</blockquote>
<p><strong>My thoughts</strong> : Since the assertion has to satisfy the induction step, it must be false for <span class="math-container">$n=1$</span>. </p>
<p>But I cannot think of any such assertion.
Does anyone have any hints? Thank you.</p>
<p><strong>Edit</strong>: Just realised that my thoughts don't make much sense, as the question demands the assertion to be false for every positive integer. </p>
| fleablood | 280,126 | <p>It's easier than you think:</p>
<p><span class="math-container">$n \le n-2$</span></p>
<p>Then <span class="math-container">$n+1 \le (n-2) + 1 = n-1 = (n+1) -2$</span>.</p>
<p><span class="math-container">$n$</span> is not an integer.</p>
<p><span class="math-container">$n + 1$</span> is not an integer.</p>
<p><span class="math-container">$n$</span> is infinite.</p>
<p>If <span class="math-container">$n+1$</span> were finite then <span class="math-container">$n = (n+1) - 1=n$</span> would be finite which is a contradiction.</p>
<p><span class="math-container">$n = n-1= 6$</span></p>
<p><span class="math-container">$n+1 = (n-1)+1 = n = n-1 = 6$</span> </p>
<p>and so on....</p>
|
3,630,421 | <p>If <span class="math-container">$x+y = 5$</span>, <span class="math-container">$xy = 1$</span> and <span class="math-container">$x > y$</span>, then <span class="math-container">$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= ?$</span> The answer key gives for the asnwer <span class="math-container">$\frac{\sqrt{21}}{3}$</span>, <span class="math-container">$\frac{7}{\sqrt{21}}$</span>, <span class="math-container">$\frac{\sqrt{7}}{\sqrt{3}}$</span>, <span class="math-container">$\frac{7}{3}$</span>.
<span class="math-container">$$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} * \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}= \frac{x+2\sqrt{x}\sqrt{y}+y}{x-y}= \frac{5+2\sqrt{1}}{x-y}= \frac{7}{x-y}$$</span>
Once <span class="math-container">$$x=5-y $$</span>
<span class="math-container">$$(5-y)y=1 $$</span>
<span class="math-container">$$y^2-5y+1 $$</span>
<span class="math-container">$$y=\frac{5\pm \:\sqrt{21}}{2} $$</span>
Considering <span class="math-container">$x=5-y$</span>, in order to <span class="math-container">$x>y$</span>, the value of y should be: <span class="math-container">$y=\frac{5-\sqrt{21}}{2} $</span>. Therefore, x will be <span class="math-container">$y=\frac{5+\sqrt{21}}{2}.$</span> Now solving <span class="math-container">$x-y:$</span></p>
<p><span class="math-container">$$x-y\Rightarrow \frac{5+\sqrt{21}}{2}- \frac{5-\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y\Rightarrow \frac{5-5+\sqrt{21}+\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y\Rightarrow \frac{2\sqrt{21}}{2}$$</span>
<span class="math-container">$$x-y= \sqrt{21}$$</span>
Back to <span class="math-container">$\frac{7}{x-y}$</span>
<span class="math-container">$$\frac{7}{x-y}= \frac{7}{\sqrt{21}} $$</span>
<span class="math-container">$$∴\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}= \frac{7}{\sqrt{21}} $$</span></p>
<p>I did solve it but it wasn't a nice solution. Can anyone give a cleaver and more interesting solution? I am studying for a test and wish there were a quick way to solve this kind of question, I would appreciate some interesting ideas.</p>
| Soyeb Jim | 774,258 | <p>If you want to find the volume using integration you can integrate in <span class="math-container">$z$</span> axis by summing the trapezoidal cross sections. [Assuming the base of the pyramid is a square]</p>
<p>Suppose highest point of the pyramid is <span class="math-container">$(0,0,0)$</span>. Assume the pyramid is upside down. Now we can find a function which gives us the length of the base of trapezoidal cross section. As at <span class="math-container">$y=0, x=0$</span> and <span class="math-container">$y=h, x=b$</span> the function would be <span class="math-container">$y=\frac{b}{h}x$</span></p>
<p>Now the volume of the cross section of <span class="math-container">$dx$</span> thickness would be
<span class="math-container">$$ \left( \frac{b}{h}x \right)^2 dx $$</span></p>
<p>Now the area of the pyramid would be-
<span class="math-container">$$\int_{0}^{h} \left( \frac{b}{h}x \right)^2 dx = \frac{b^2}{h^2} \int_{0}^{h}x^2 dx = \frac{b^2}{h^2} \frac{x^3}{3}=\frac{b^2 \times h^3}{3 h^2} = \frac{b^2 h}{3}$$</span> </p>
<p><a href="https://i.stack.imgur.com/xa4TX.pngD" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xa4TX.pngD" alt="enter image description here"></a></p>
<p>Now why did your method gave a wrong answer? If I am not wrong summing the prisms will make a 3D shape with many prism side by side each having the same <span class="math-container">$dx$</span> base and height would increase due to <span class="math-container">$y=\frac{2h}{b}x$</span>. The face of the 3D same will look kinda like this.
<a href="https://i.stack.imgur.com/n7BN3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n7BN3.png" alt="enter image description here"></a></p>
<p>There for the value you get is half of the volume of a prism with a face area <span class="math-container">$\frac{1}{2} h \times b$</span> and thickness <span class="math-container">$b$</span> </p>
|
22,753 | <p>I've learned the process of orthogonal diagonalisation in an algebra course I'm taking...but I just realised I have no idea what the point of it is.</p>
<p>The definition is basically this: "A matrix <span class="math-container">$A$</span> is orthogonally diagonalisable if there exists a matrix <span class="math-container">$P$</span> which is orthogonal and <span class="math-container">$D = P^tAP$</span> where <span class="math-container">$D$</span> is diagonal". I don't understand the significance of this though...what is special/important about this relationship?</p>
| Yuval Filmus | 1,277 | <p><b>If a matrix $A$ is unitarily diagonalizable, then one can define a "Fourier transform" for which $A$ is a "convolution" matrix.</b></p>
<p>Here is an example. We have a family $F$ of subsets of some finite set $S$, i.e. $F \subset 2^S$, such that any two sets in $F$ agree in some coordinate, i.e. for any two $A,B \in F$, there is some element $x \in S$ that either belongs to both $A,B$ or to neither; we call such a family $F$ <i>agreeing</i>.</p>
<p>How big can an agreeing family $F$ be? Clearly, it can contain at most half the sets, since $A$ and $S \setminus A$ cannot both belong to $F$. On the other hand, for any $x \in S$, the family $$F = \{ A \subset S : x \in A \}$$ is an agreeing family containing exactly half the sets.</p>
<p>Here is a different proof. Let $F$ be an agreeing family, and $f$ its characteristic function. i.e. $f(A) = 1$ iff $A \in F$. Its Fourier transform is defined by $$ \hat{f}(B) = 2^{-|S|} \sum_A f(A) (-1)^{|A \cap B|}. $$ The Fourier transform is really presenting the function $f$ in the orthonormal basis $$\chi_B(A) = (-1)^{|A\cap B|},$$ which is orthonormal with respect to the inner product $$\langle g,h \rangle = 2^{-|S|} \sum_A g(A) h(A). $$ The Fourier transform is defined so that $$f = \sum_B \hat{f}(B) \chi_B, $$ and so the formula for the transform can also be written $$\hat{f}(B) = \langle f,\chi_B \rangle;$$ this works since the basis is orthonormal.</p>
<p>An easy calculation gives that $\hat{f}(\emptyset) = |F|/2^{|S|}$. Moreover, $$\langle f,f \rangle = \sum_{B,C} \langle \hat{f}(B) \chi_B, \hat{f}(C) \chi_C \rangle = \sum_B \hat{f}(B)^2,$$ again from orthonormality. Note that $$\langle f,f\rangle = \langle f^2, 1 \rangle = \langle f,\chi_\emptyset \rangle = \hat{f}(\emptyset),$$ where we used the fact that $f$ is $\{0,1\}$-valued.</p>
<p>Consider now the operator $X$ which corresponds to complementation, i.e. $$Xe_A = e_{S \setminus A},$$ where $e_A$ is the vector which is $1$ only for the set $A$. Another way to look at the operator $X$ is that it is convolution with $S$, where the group operation is symmetric difference (i.e. $A \triangle S = S \setminus A$).</p>
<p>A straightforward computation shows that the eigenvectors of $X$ are exactly the Fourier basis vectors $\chi_B$:
$$ X \chi_B(A) = (-1)^{|(S\setminus A)\cap B|} = (-1)^{|S \cap B|} (-1)^{|A \cap B|} = (-1)^{|B|} \chi_B(A). $$
This is not surprising since $X$ is a convolution operator for the group $\mathbb{Z}_2^{|S|}$, and the Fourier basis is a basis of characters for that abelian group.</p>
<p>Since the family $f$ is agreeing, $f(A)f(S\setminus A) = 0$, and so $$0 = \langle f,Xf \rangle = \sum_B (-1)^{|B|} \hat{f}(B)^2,$$ where we again used orthonormality of the Fourier basis. Denoting $|f| = |F|/2^{|S|}$, we have seen above that $$\sum_B \hat{f}(B)^2 = |f|, \quad \hat{f}(\emptyset)^2 = |f|^2.$$ So the even and odd squared Fourier coefficients balance; their total weight is $|f|$, and the weight of one of them is $|f|^2$. Evidently, $|f|^2$ can be at most half the total weight, and so $|f| \leq 1/2$.</p>
<hr>
<p>This proof might seem silly (since we presented a "one-line" proof preceding it), but the same method can be used to prove much more difficult theorems. For example, we can look at a family of "colored" sets, i.e. generalize the two colors above (corresponding to "not in the set" and "in the set") to an arbitrary number of colors. The largest family is still obtained by fixing one coordinate, but the combinatorial proof is more difficult (takes about a page); the Fourier proof is almost the same.</p>
<p>Here are some more difficult examples:</p>
<ul>
<li>Families of graphs over a fixed vertex set, the intersection of any two of which contains a triangle. The "best" family is obtained by taking all supergraphs of a fixed triangle. The only known proof is using very similar Fourier methods.</li>
<li>Families of permutations, any two of which have a common "input/output" pair. The "best" family is in general obtained by fixing the image of some element (all permutations taking $i$ to $j$). There is a simple combinatorial proof of that along the lines above. But it's much harder to prove that these are the unique optimal families, whereas it follows relatively easily from the "Fourier" proof. Moreover, the latter can be extended to the case where the permutations are required to have $t$ matching input/output pairs. The "best" families take $t$ fixed inputs to $t$ fixed outputs. The only known proof is spectral (it requires the representation theory of $S_n$).</li>
<li>Families of subsets of $S$ of size $k < |S|/2$, any two of which intersect. The maximal families are the same as above (supersets of a fixed elements). This is the celebrated Erdős-Ko-Rado theorem. There is a Fourier proof with some extra benefits over some of the combinatorial proofs, viz. it can describe the structure of almost optimal families.</li>
</ul>
<p>The Fourier proof of the last example actually optimizes some skewed measure of a general family of subsets, in other words, instead of limiting the sets to be of size $k$, we just give more "relevance" to sets of size roughly $k$. The proof goes as follows:</p>
<ol>
<li>Find some inner product so that $\langle f,1 \rangle_p$ is the required skewed measure (which is $\mu_p(A) = p^{|A|} (1-p)^{|S\setminus A|}$ for $p \approx k/n$).</li>
<li>Find some "convolution operator" $X$ such that $\langle f,Xf \rangle_p = 0$ for every intersecting family, and moreover the eigenvectors of $X$ are orthogonal with respect to the inner product.</li>
<li>Follow the same steps as above to conclude that $\mu_p(f) \leq p$.</li>
</ol>
<p>The construction of the convolution operator $X$ crucially uses the fact that symmetric matrices are unitarily diagonalizable (the symmetric matrix in question is obtained from a reversible Markov chain). The operator $X$ is not strictly unitarily diagonalizable, since its eigenvectors are only orthogonal with respect to the skewed inner product (in which the stationary distribution of the Markov chain crops up), but it is obtained from such a matrix through scaling of rows.</p>
<hr>
<p>The material is taken from a series of papers by Ehud Friedgut and friends. I will refer to them by their current numbers on his <a href="http://www.ma.huji.ac.il/~ehudf/PPP" rel="nofollow">list</a>.</p>
<ol>
<li>The general method (including the bound on "agreeing families of colored sets") is #16.</li>
<li>The application to permutations is #2 (there are some follow-up papers by David Ellis).</li>
<li>The application to graphs is #1.</li>
<li>The spectral proof of Erdős-Ko-Rado is #7.</li>
<li>The general construction (using the crucial property that symmetric matrices are unitarily diagonalizable) is #6.</li>
<li>The connection between #7 and #6 is explained in the last section of #5.</li>
</ol>
|
1,452,943 | <p>I'm working on problem where I want to use the continuity of $f'$ to assert that $f'(x)$ cannot be zero ("bounded away from zero"?) near $x = 0$. We know that $(f'(0))^2 >3$.</p>
<p>So, I think that what I really want to ask is this: if $f'$ is cts, must $f'$-squared also be continuous? </p>
<p>Can I use the epsilon-delta definition?:</p>
<p>Since $f'$ is continuous, for every $\epsilon>0$, there exists $\delta>0$ such that:</p>
<p>$$|x-0|<\delta \implies|f'(x)-f'(0)|<\epsilon$$
$$ \implies -\epsilon < f'(x) - f'(0) < \epsilon $$
$$ \implies f'(0)-\epsilon < f'(x) < f'(0)+\epsilon $$</p>
<p>From here I'm not sure how to use the fact that $(f'(0))^2>3$.</p>
<p>Thanks,</p>
| Cameron Buie | 28,900 | <p>Well, $x\mapsto f'(x)$ is continuous, and $y\mapsto y^2$ is also continuous, so what can you say about $x\mapsto\left( f'(x)\right)^2$?</p>
|
1,452,943 | <p>I'm working on problem where I want to use the continuity of $f'$ to assert that $f'(x)$ cannot be zero ("bounded away from zero"?) near $x = 0$. We know that $(f'(0))^2 >3$.</p>
<p>So, I think that what I really want to ask is this: if $f'$ is cts, must $f'$-squared also be continuous? </p>
<p>Can I use the epsilon-delta definition?:</p>
<p>Since $f'$ is continuous, for every $\epsilon>0$, there exists $\delta>0$ such that:</p>
<p>$$|x-0|<\delta \implies|f'(x)-f'(0)|<\epsilon$$
$$ \implies -\epsilon < f'(x) - f'(0) < \epsilon $$
$$ \implies f'(0)-\epsilon < f'(x) < f'(0)+\epsilon $$</p>
<p>From here I'm not sure how to use the fact that $(f'(0))^2>3$.</p>
<p>Thanks,</p>
| Hasan Saad | 62,977 | <p>$f'(0)^2>3\implies |f'(0)|>\sqrt{3}$</p>
<p>However, the function is continuous at $x=0$, so for some $\delta$, we have $|f'(x)-f'(0)|<\sqrt{3}$ whenever $|x|<\delta$.</p>
<p>However, this asserts that, $|f'(0)-f'(x)|<\sqrt{3}\implies |f'(0)|-|f'(x)|<\sqrt{3}$ by using the triangle inequality.</p>
<p>Thus, we have,</p>
<p>$|f'(x)|>|f'(0)|-\sqrt{3}>\sqrt{3}-\sqrt{3}=0$ whenever $|x|<\delta$.</p>
<p>Thus, we reach the conclusion that $f$ is bounded away from $0$ near $x=0$.</p>
|
19,495 | <p>I was told that one of the most efficient tools (e.g. in terms of computations relevant to physics, but also in terms of guessing heuristically mathematical facts) that physicists use is the so called "Feynman path integral", which, as far as I understand, means "integrating" a functional (action) on some infinite-dimentional space of configurations (fields) of a system.</p>
<p>Unfortunately, it seems that, except for some few instances like Gaussian-type integrals, the quotation marks cannot be eliminated in the term "integration", cause a mathematically sound integration theory on infinite-dimensional spaces — I was told — has not been invented yet.</p>
<p>I would like to know the state of the art of the attempts to make this "path integral" into a well-defined mathematical entity.</p>
<p>Difficulties of analytical nature are certainly present, but I read somewhere that perhaps the true nature of path integral would be hidden in some combinatorial or higher-categorical structures which are not yet understood...</p>
<p>Edit: I should be more precise about the kind of answer that I expected to this question. I was not asking about reference for books/articles in which the path integral is treated at length and in detail. I'd have just liked to have some "fresh", (relatively) concise and not too-specialistic account of the situation; something like: "Essentially the problems are due to this and this, and there have been approaches X, Y, Z that focus on A, B, C; some progress have been made in ... but problems remain in ...".</p>
| user1504 | 35,508 | <p>It's not accurate to say that no theory of integration on infinite-dimensional spaces exists. The Euclidean-signature Feynman measure has been constructed -- as a measure on a space of distributions -- in a number of non-trivial cases, mainly by the Constructive QFT school in the 70s. </p>
<p>The mathematical constructions reflect the physical ideas of effective quantum field theory: One obtains the measure on the space of field histories as the limit of a sequence/net of "regularized" integrals, which encode how the effective "long distance" degrees of freedom interact with each other after one averages out the short distance degrees of freedom in various ways. (You can imagine here that long/short distance refers to some wavelet basis, and that we get the sequence of regularized integrals by varying the way we divide the wavelet basis into short distance and long distance components.)</p>
<p>I don't think the main problem in the subject is that we need some new notion of integration. The Feynman measures we mathematicians can construct exhibit all the richness of the "higher categories" axioms, and moreover, the numerical computations in lattice gauge theory and in statistical physics indicates that the existing framework is at the least a very good approximation. </p>
<p>The problem, rather, is that we need a better way of constructing examples. At the moment, you have to <em>guess</em> which family of regularized integrals you ought to study when you try to construct any particular example. (In Glimm & Jaffe's book, for example, they simply replace the interaction Lagrangian with the corresponding "normally ordered" Lagrangian. In lattice gauge theory, they use short-distance continuum perturbation theory to figure out what the lattice action should be.) </p>
<p>Then -- and this is the really hard and physically interesting part -- you have to have enough analytic control on the family to say which observables (functions on the space of distributions) are integrable with respect to the limiting continuum measure. This is where you earn the million dollars, so to speak.</p>
|
115,483 | <p>Edited:</p>
<p>I guess </p>
<p>$$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$</p>
<p>We know that if $\operatorname{Supp} H^i_I(M)\subseteq V(I)\cap \operatorname{Supp}(M)$, then
$$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=\lbrace(x,y) \rbrace\cup \lbrace(x,y,p) \rbrace$$ where $p$ is prime number.</p>
<p>If we could show that for every $P\in V((x,y))$, $$\left(H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)\right)_P=0$$</p>
<p>then it is done.</p>
<p><strong>background:</strong>
$H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$.
$V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $\operatorname{Supp}(M)=\lbrace P\in \operatorname{Spec}(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$. Furthermore $\operatorname{Supp}(R/I)=V(I)$.</p>
| user26857 | 23,950 | <p>$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology
$$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$
But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).</p>
|
1,874,634 | <blockquote>
<p>Corollary (of Schur's Lemma): Every irreducible complex representation of a finite abelian group G is one-dimensional.</p>
</blockquote>
<p>My question is now, why has the group to be abelian? As far as I know, we want the representation $\rho(g)$ to be a $Hom_G(V,V)$, where $V$ is the representation space. Isn't this always the case (i.e. even if the $\rho(g)$ is not abelian) as it is by definition a function $G \rightarrow GL(V)$?</p>
| Maik Pickl | 317,129 | <p>The trick here is, that for a abelian group every element is a intertwining operator. This means let $h \in G$, then $\rho(h)\rho(g)\rho(h^{-1})=\rho(g)$ for all $g$ and therefore by Schur's lemma $\rho(h)=\lambda id$. Since your representation was assumed to be irreducible it follows that it is one dimensional. Note that we used the commutativity of the group here in an essential way. This is no longer true for non abelian groups. </p>
|
1,779,068 | <p>Let $H$ and $K$ be two subgroups of a group $G$ such that $[G : H]=2$ and $K$ is not a subgroup of $H$. Then show that $HK=G$.
Now, since $HK$ is a subset of $G$ we need only to show that $G$ is a subset of $HK$. But how can I show it? Please help me. Thank you in advance.</p>
| Alex Wertheim | 73,817 | <p>Since $[G:H]$, $H$ has two distinct right cosets of $H$ in $G$: one is $H$, and the other can be written as $Hk$ for any $k \notin H, k \in K$. The right cosets of $H$ in $G$ partition $G$, so what can you say about $HK$, which obviously contains $H \cup Hk$? </p>
|
207,778 | <p>I want to save expressions as well as their names in a file.</p>
<pre><code> func[i_] := i;
Do[func[i] >>> out.m,{i,1,3}];
</code></pre>
<p>The output is </p>
<pre><code> cat out.m
1
2
3
</code></pre>
<p>However the desired output is</p>
<pre><code> cat out.m
func[1] = 1;
func[2] = 2;
func[3] = 3;
</code></pre>
<p><code>Save</code> does not save here.</p>
| CA Trevillian | 63,039 | <p>You can use <a href="https://reference.wolfram.com/language/ref/Save.html" rel="noreferrer"><code>Save</code></a> in this way:</p>
<pre><code>func[i_] := func[i] = i;
Do[func[i], {i, 1, 3}];
FullDefinition@func
(*
func[1] = 1
func[2] = 2
func[3] = 3
func[i_] := func[i] = i
*)
Save["out.m",func];
ClearAll[func]; Get["out.m"]; FullDefinition@func
(*
func[1] = 1
func[2] = 2
func[3] = 3
func[i_] := func[i] = i
*)
</code></pre>
<p>So, here, we show that memoization can be used as a method to record the defined expressions when one is going to save them externally.</p>
<p>Your suggested method of using <code>Save</code> was not exactly proper, which is why it would not work for you. Here, this plays out as expected. While the memoized func name cannot be avoided in this procedure, it seems to do as you would like.</p>
<p>Does this make sense?</p>
|
1,660,289 | <p>I want to find the line that passes through $(3,1,-2)$ and intersects at a right angle the line $x=-1+t, y=-2+t, z=-1+t$. </p>
<p>The line that passes through $(3,1,-2)$ is of the form $l(t)=(3,1,-2)+ \lambda u, \lambda \in \mathbb{R}$ where $u$ is a parallel vector to the line. </p>
<p>There will be a $\lambda \in \mathbb{R}$ such that $3+ \lambda u_1=-1+t, 1+ \lambda u_2=-2+t, -2+ \lambda u_3=-1+t$. </p>
<p>Is it right so far? How can we continue?</p>
| Chris Culter | 87,023 | <p>Starting from a prime $p_n$, the first candidate for a twin prime that the pattern generates is $3p_n-4$. If this formula does indeed produce prime numbers, then by iterating it, we have a simple way to generate arbitrarily large primes. That would be a big deal all by itself, as no such generator is known today. So we should be skeptical that the formula works...</p>
<p>Well, let's try it!</p>
<p>$5 \times 3 - 4 = 11$<br>
$11 \times 3 - 4 = 29$<br>
$29 \times 3 - 4 = 83$<br>
$83 \times 3 - 4 = 245$</p>
<p>but $245$ is not prime.</p>
|
2,821,323 | <blockquote>
<p>How to show that a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$? For example, I try to show this polynomial $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c)(*)$$ is irreducible, where $a,b,c\in \mathbb{Q}$.</p>
</blockquote>
<p>The related problem is <a href="https://math.stackexchange.com/questions/2779545/ask-for-the-rational-roots-of-fracabc-fracbac-fraccab-4">Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$</a>. Could I consider the points $(*)$ intersect with $L_{\infty}$ are three? $L_{\infty}$ is the infinity line in a projective space $\mathbb{C}P^2$.</p>
| Will Jagy | 10,400 | <p>To ask about complete (three linear factor) reducibility over the complexes, we take the Hessian matrix of second partials. The entries are linear in the named variables. Next, let $\Delta$ be the determinant of the Hessian. This $\Delta$ is once again a cubic form. The original cubic (homogeneous) ternary form factors completely if and only if $\Delta $ is a constant multiple of it. I have finished this first test, your cubic does not factor completely.</p>
<p>More difficult if the cubic might be a linear times an irreducible quadratic. In that case, there is still a conclusive test:</p>
<p><a href="https://i.stack.imgur.com/JxISF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JxISF.jpg" alt="enter image description here"></a></p>
<p>For your problem, the coefficients in order from $0$ to $9$ are
$$ 1,-3,-3,-3,-5,-3,1,-3,-3,1 $$
where the best looking diagram is a triangle as in bowling. The $-5$ refers to $-5abc \; ,$ the $1$s refer to $a^3,b^3,c^3 \; .$ </p>
<p>Alright, I wote a little program to correctly type the 8 by ten matrix, then put that into gp-Pari. The original polynomial is irreducible over the complex numbers.</p>
<hr>
<pre><code>?
? sch = [ 0, 3, 0, -6, -6, 0, -3, -5, -3, 0; 0, 0, 3, 0, -6, -6, 0, -3, -5, -3; -3, -6, -5, 3, -6, -3, 0, 0, 0, 0; 0, 0, -3, 0, -6, -5, 0, 3, -6, -3; -3, -5, -6, -3, -6, 3, 0, 0, 0, 0; 0, -3, 0, -5, -6, 0, -3, -6, 3, 0; 1, 0, -3, 3, 0, -3, -2, 3, 0, 1; 1, -3, 0, -3, 0, 3, 1, 0, 3, -2]
%25 =
[ 0 3 0 -6 -6 0 -3 -5 -3 0]
[ 0 0 3 0 -6 -6 0 -3 -5 -3]
[-3 -6 -5 3 -6 -3 0 0 0 0]
[ 0 0 -3 0 -6 -5 0 3 -6 -3]
[-3 -5 -6 -3 -6 3 0 0 0 0]
[ 0 -3 0 -5 -6 0 -3 -6 3 0]
[ 1 0 -3 3 0 -3 -2 3 0 1]
[ 1 -3 0 -3 0 3 1 0 3 -2]
? matrank(sch)
%26 = 8
?
?
?
</code></pre>
<hr>
|
1,942,364 | <p>How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.</p>
<p>How can I go about counting the number of squares of each size?</p>
| marty cohen | 13,079 | <p>Another way of counting:</p>
<p>For each
$(i, j, k)$
with
$1 \le i, j \le n$
and
$0 \le k \le \min(n-i, n-j)$
there is a square
with upper left corner
at $(i, j)$
and lower right corner at
$(i+k, j+k)$.</p>
<p>Therefore the total is</p>
<p>$\begin{array}\\
s(n)
&=\sum_{i=1}^n \sum_{j=1}^n (1+\min(n-i, n-j))\\
&=\sum_{i=0}^{n-1} \sum_{j=0}^{n-1} (1+\min(i, j))\\
&=n^2+\sum_{i=0}^{n-1} \sum_{j=0}^{n-1} \min(i, j)\\
&=n^2+\sum_{i=0}^{n-1} (\sum_{j=0}^{i-1} \min(i, j)+\sum_{j=i} \min(i, j)+\sum_{j=i+1}^{n-1} \min(i, j))\\
&=n^2+\sum_{i=0}^{n-1} (\sum_{j=0}^{i-1} j+i+\sum_{j=i+1}^{n-1} i)\\
&=n^2+\sum_{i=0}^{n-1} (\frac12i(i-1)+i+i(n-1-(i+1)+1))\\
&=n^2+\sum_{i=0}^{n-1} (\binom{i}{2}+i+i(n-i-1))\\
&=n^2+\binom{n}{3}+\frac12 n(n-1)+\sum_{i=0}^{n-1} (ni-i(i+1))\\
&=n^2+\binom{n}{3}+\frac12 n(n-1)+\frac12 n^2(n-1)-\sum_{i=1}^{n} i(i-1)\\
&=n^2+\binom{n}{3}+\frac12 n(n-1)+\frac12 n^2(n-1)-2\binom{n+1}{3}\\
&=\frac13(n+1)n(n-1)\\
\end{array}
$</p>
|
2,626,506 | <p><strong>Proof: There is no other prime triple then $3,5,7$</strong></p>
<p>There are already lots of questions about this proof, but I can't find the answer to my question.</p>
<p>The complete the proof, we consider mod $3$ so $p=3k; p=3k+1; p=3k+2$ </p>
<p>But why do we look at divisibility by $3$?</p>
<p>Do we look at mod $4$ for prime quads?</p>
| robjohn | 13,854 | <p>$$
\begin{align}
p(p+2)(p+4)
&=p^3+6p^2+8p\\
&=3\left[5\binom{p}{1}+6\binom{p}{2}+2\binom{p}{3}\right]
\end{align}
$$
is divisible by $3$, so unless one of the factors <em>is</em> $3$, one of the factors is not prime.</p>
|
2,626,506 | <p><strong>Proof: There is no other prime triple then $3,5,7$</strong></p>
<p>There are already lots of questions about this proof, but I can't find the answer to my question.</p>
<p>The complete the proof, we consider mod $3$ so $p=3k; p=3k+1; p=3k+2$ </p>
<p>But why do we look at divisibility by $3$?</p>
<p>Do we look at mod $4$ for prime quads?</p>
| Adam Bailey | 22,062 | <p>Suppose $p,p+2,p+4$ are all prime. </p>
<p>Consider $p+2,p+3,p+4$. Since these are three consecutive integers, one of them must be divisible by $3$ (that's why we consider divisibility by $3$ and not by some other number). It can't be $p+2$ or $p+4$ because they are primes and not equal to $3$ (otherwise $p$ would be $1$ or $-1$ which are not primes. So $3 | p+3$ and therefore $3 | p$. Hence the only such triple is $3,5,7$.</p>
|
210,655 | <p>The lower density of $A\subseteq\mathbb{N}$ is defined to be $\lambda(A)=\lim\text{inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$. We set $${\cal C} = \{A\subseteq \mathbb{N}: \lambda(\mathbb{N}\setminus A) = 1 - \lambda(A)\}.$$</p>
<p>Do both ${\cal C}$ and ${\cal P}(\mathbb{N})\setminus {\cal C}$ have cardinality $2^{\aleph_0}$?</p>
| Salvo Tringali | 16,537 | <p>Let $\alpha \in {]0,1]}$. On the one hand, the set $\{\lfloor \alpha^{-1}n \rfloor: n \in \mathbf N\}$ has (natural) density equal to $\alpha$, so it belongs to $\mathcal C$. On the other hand, the set $\bigcup_{n \ge 1} [\![(2n-1)!\,\alpha + (2n)!\,(1-\alpha) + 1, (2n)!+1]\!]$ has lower (natural) density equal to $0$ and upper (natural) density equal to $\alpha$ (as it follows from <a href="https://mathoverflow.net/questions/207522/reference-for-a-lemma-on-the-asymptotic-upper-density-of-special-sets-with-large">here</a>), so it belongs to $\mathcal P(\mathbf N) \setminus \mathcal C$.</p>
<p><strong>Edit (July 05, 2015).</strong> For those who may care, here is a complement to Stefan Kohl's and my own answer, to highlight some differences between the two approaches (and suggest a couple of questions that someone else will hopefully find interesting). </p>
<p>To start with, let an upper quasi-density (on $\mathbf N$) be a function $\mu^\ast: \mathcal P(\mathbf N) \to \mathbf R$ such that, for all $X,Y \subseteq \mathbf N$ and $h,k \in \mathbf N^+$, the following hold:</p>
<ol>
<li>$\mu^\ast(\mathbf N) = 1$;</li>
<li>$\mu^\ast(X \cup Y) \le \mu^\ast(X) + \mu^\ast(Y)$;</li>
<li>$\mu^\ast(k \cdot X + h) = \frac{1}{k} \mu^\ast(X)$;</li>
<li>$\mu^\ast(X) \le 1$.</li>
</ol>
<p>If, in addition, $\mu^\ast$ is monotonic, viz. $\mu^\ast(X) \le \mu^\ast(Y)$ whenever $X \subseteq Y \subseteq \mathbf N$, then we call $\mu^\ast$ an <em>upper density (on $\mathbf N$)</em>, in which case condition 4 is implied by the others.</p>
<p>Given an upper quasi-density $\mu^\ast$, we define its lower dual $\mu_\ast$ as the function
$$
\mathcal P(\mathbf N) \to \mathbf R: X \mapsto 1 - \mu^\ast(\mathbf N \setminus X),
$$
and accordingly we let the quasi-density induced by $\mu^\ast$ be the <em>partial</em> function
$$
\mu: \mathcal P(\mathbf N) \rightharpoonup \mathbf R: X \mapsto \mu^\ast(X)
$$
whose domain is the set
$$
\mathcal D_\mu := \{X \in \mathcal P(\mathbf N): \mu_\ast(X) = \mu^\ast(X)\}.
$$
Since the upper asymptotic density is an upper density in the sense of the definitions above (the same is also true, among the others, for the upper Banach density, upper logarithmic density, upper analytic density, and Buck's measure density), we see that the OP is asking whether $\mathcal D_\mu$ and $\mathcal P(\mathbf N) \setminus \mathcal D_\mu$ have both the same cardinality of $\mathbf R$.</p>
<p>In fact, it can be proved, see Theorem 2 <a href="http://arxiv.org/abs/1506.04664" rel="nofollow noreferrer">here</a>, that for every $\alpha \in [0,1]$ there exists $X \subseteq \mathbf N$ such that $\mu(X) = \alpha$, which yields that $|\mathcal D_\mu| = |\mathbf R|$: This is based on the explicit construction of such a set $X$, and can't be proved by Stefan Kohl's argument, insofar as we don't know whether $\mu^\ast(X) = \mu^\ast(X \cup Y)$ for every finite set $Y \subseteq \mathbf H$ (this is Question 7 in the linked preprint). </p>
<p>On the other hand, if $\mu^\ast$ is an upper density (i.e., is monotonic), then the "invariance-under-union-with-finite-sets property" is true, and Stefan Kohl's reasoning can be adapted to prove that $|\mathcal P(\mathbf N) \setminus \mathcal D_\mu| = |\mathbf R|$ if $\mathcal D_\mu \subsetneq \mathcal P(\mathbf N)$, namely $\mu_\ast(X) < \mu^\ast(X)$ for some $X \subseteq \mathbf N$ (see the note below), whereas my argument can't, insofar as we don't know whether it is true, again under the (necessary) assumption that $\mathcal D_\mu \subsetneq \mathcal P(\mathbf N)$, if for every $\alpha \in [0,1]$ there exists $X \subseteq \mathbf N$ such that $\mu_\ast(X) = 0$ and $\mu^\ast(X) = \alpha$ (which is a special case of Question 3 in the same linked preprint).</p>
<p><em>Note.</em> In ZFC, say, there exist upper densities on $\mathbf N$ for which the inequality is false, see <a href="https://mathoverflow.net/questions/209706/unicity-of-additive-1-homogeneous-and-shift-invariant-probability-measure">Question 209706: Unicity of additive, $(−1)$-homogeneous, and shift invariant probability measures on $\mathbf N^+$</a>.</p>
|
3,624,524 | <p>I want to figure out the process for showing why the function <span class="math-container">$\cos(1-\frac{1}{z})$</span> has an essential singularity at <span class="math-container">$z=0$</span> without using knowledge of the Laurent expansion. I know the process should be to rule out the possibility of removable singularities or poles, but do not know how to do this for this function.</p>
<p>Attempt
I was thinking I would show since <span class="math-container">$$\lim_{z\to 0} |\cos(1-\frac{1}{z})| \text{ DNE } $$</span> since the function oscillates between <span class="math-container">$1$</span> and <span class="math-container">$-1$</span> for <span class="math-container">$z$</span> near zero for positive values, this rules out the possibility of a pole since the limit is not <span class="math-container">$\infty$</span> and the singularity is not removable since the limit is not finite.Is this the correct approach? What are some other ways, to show that zero is an essential singularity?</p>
| Mnifldz | 210,719 | <p>Yes, it is true for all tangent bundles on all manifolds. This stems from the fact that tangent bundles are specific examples of vector bundles. Vector bundles are defined by having locally trivial neighborhoods.</p>
|
3,055,208 | <p>I am trying to compute the below limit through Taylor series:
<span class="math-container">$\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$</span></p>
<p>What I have already tried is first of all change the variable x to
<span class="math-container">$x=1/t$</span> and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change <span class="math-container">$e$</span> to exponent polynomial up to t=6 + <span class="math-container">$O(X^6)$</span>
however, I don`t know what can I do with square root. </p>
| hamam_Abdallah | 369,188 | <p><strong>hint</strong></p>
<p>The square root becomes</p>
<p><span class="math-container">$$\frac{\sqrt{1+3t^6}}{|t^3|}=$$</span></p>
<p><span class="math-container">$$\frac{1}{|t^3|}\Bigl(1+\frac{3t^6}{2}-\frac 98t^{12}+t^{12}\epsilon(t)\Bigr)$$</span></p>
|
3,055,208 | <p>I am trying to compute the below limit through Taylor series:
<span class="math-container">$\lim \limits_{x\to \infty} ((2x^3-2x^2+x)e^{1/x}-\sqrt{x^6+3})$</span></p>
<p>What I have already tried is first of all change the variable x to
<span class="math-container">$x=1/t$</span> and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change <span class="math-container">$e$</span> to exponent polynomial up to t=6 + <span class="math-container">$O(X^6)$</span>
however, I don`t know what can I do with square root. </p>
| Doug M | 317,162 | <p><span class="math-container">$y = \frac 1x$</span></p>
<p>then we have</p>
<p><span class="math-container">$\lim_\limits{y\to 0^+} \frac {(2 -2y+ y^2)e^y - \sqrt {1+3y^6}}{y^3}$</span></p>
<p>Now if you want to do a Taylor expansion...</p>
<p><span class="math-container">$\lim_\limits{y\to 0^+} \frac {(2 -2y+ y^2)(1+y+\frac 12 y^2+\cdots) - (1+\frac 32 y^6 - \cdots )}{y^3}$</span></p>
<p><span class="math-container">$\lim_\limits{y\to 0^+} \frac {1 + O(y)}{y^3} = \infty$</span></p>
|
3,413,261 | <p>I know this was answered before but I'm having one particular problem on the proof that I'm not getting.</p>
<p>My Understanding of the distribution law on the absorption law is making me nuts, by the answers of the proof it should be like this.</p>
<p>A∨(A∧B)=(A∧T)∨(A∧B)=A∧(T∨B)=A∧T=A</p>
<p>This should prove the Absoption Law but on the Step (A ^ (T v B)), I'm not getting how they get to it.</p>
<p>If (A ^ T) v (A ^ B) will be distributed by my understand of this, the following is the answer
(A v A) ^ (A v B) ^ (T v A) ^ (T v B)
That we can go to
A ^ (A v B) ^ T
I'm getting lost on something here, because it looks to me we will enter a loop on it as:
A ^ T ^ (A v B) will be distributed again and I will go back to A ^ B if I distribute with A but if I distribute with T it will be B ^ T, nothing usefull also is it?</p>
<p>Can anyone help on this? Thanks in advance.</p>
| Mirko | 188,367 | <p><a href="https://en.wikipedia.org/wiki/Lower_limit_topology" rel="noreferrer">The Sorgenfrey line</a>, also called the lower-limit topology on the real line. It has a basis of intervals <span class="math-container">$[a,b)$</span> (or some authors prefer <span class="math-container">$(a,b]$</span>, upper-limit topology). </p>
<p>It is hereditarily Lindelof (and hereditarily separable, i.e. every subspace has a countable dense subset), but it is not second countable. </p>
<p>Real line with discrete topology is not Lindelof. </p>
<p><span class="math-container">$(0,1)$</span> with the usual topology is Lindelof, it is homeomorphic to the real line.
The real line is <span class="math-container">$\sigma$</span>-compact, that is union of countably many compact subspaces. The real line is the union <span class="math-container">$\cup_n[-n,n]$</span>, clearly <span class="math-container">$\sigma$</span>-compact.
The interval <span class="math-container">$(0,1)=\cup_n [\frac1n,1-\frac1n]$</span> is <span class="math-container">$\sigma$</span>-compact. </p>
<p>Every <span class="math-container">$\sigma$</span>-compact space is Lindelof (easy to prove). </p>
<p>In particular, clearly every countable space is Lindelof. But not every countable space is second countable (even with nice separation axioms). For example, the countable sequential fan is not second countable. It is obtained by taking a disjoint family of countably many convergent sequences, and
"gluing" their limit points into one limit point, via a quotient map. The result is not first countable at that point. (I couldn't find a suitable online reference to the countable sequential fan, but it has similar properties to the quotient space <span class="math-container">$\Bbb R/\Bbb N$</span>, which is also not first countable, and likely discussed in most topology books.) </p>
<p>There is an online searchable database (called <span class="math-container">$\pi$</span>-base), you can make a query asking for Lindelof, not second countable spaces. For many more examples see </p>
<p><a href="https://topology.jdabbs.com/spaces?q=lindelof%20%2B%20~Second%20Countable" rel="noreferrer">https://topology.jdabbs.com/spaces?q=lindelof%20%2B%20~Second%20Countable</a></p>
<p>Every second countable space is Lindelof (you may need to assume some separation axioms, often included in the definitions). </p>
<p>One of the examples at <span class="math-container">$\pi$</span>-base is the one-point Lidelofication of uncountable discrete space. Take any uncountable set, and a point <span class="math-container">$p$</span>, and isolate all point but <span class="math-container">$p$</span>. The neighborhoods of <span class="math-container">$p$</span> are co-countable (that is, they have a countable complement). The definition easily implies that the result is a Lindelof space. But it is not first-countable at <span class="math-container">$p$</span>, and hence not second countable. Often (in the case when "uncountable" is taken to be the first uncountable cardinal) this space is described as the set of all countable ordinals <span class="math-container">$\omega_1=\{\alpha:\alpha<\omega_1\}$</span> together with the first uncountable ordinal <span class="math-container">$\omega_1$</span>, so <span class="math-container">$X=[0,\omega_1]=\{\alpha:\alpha\le\omega_1\}$</span>, with all countable ordinals <span class="math-container">$\alpha<\omega_1$</span> isolated, and with basic neighborhoods of <span class="math-container">$\omega_1$</span>
of the form <span class="math-container">$(\alpha,\omega_1]$</span>, with <span class="math-container">$\alpha<\omega_1$</span>. This is, in addition, an example of a Linearly Ordered Topological Space (LOTS), with only one non-isolated point. (It is a LOTS under a somewhat different order, one may insert a decreasing sequence in front of every limit ordinal.) </p>
<p>The last (and some of the previous examples) are not second countable, because they are not even first countable (and, for the one-point Lindelofication the point <span class="math-container">$p$</span> is not even a <span class="math-container">$G_\delta$</span> point, that is, it is not the intersection of any countable family of open sets, and the space is not <span class="math-container">$\sigma$</span>-compact). On the other hand the Sorgenfrey line is perfectly normal: Every closed set (and in particular every point) is the intersection of a countable family of open sets (the proof that the Sorgenfrey line is hereditarily Lindelof uses this). But it is not second countable, since for every basis, and for every <span class="math-container">$x$</span> there must be a basic element <span class="math-container">$B_x$</span> with <span class="math-container">$x\in B_x\subseteq[x,\infty)$</span> and clearly if <span class="math-container">$x\neq y$</span> then <span class="math-container">$B_x\neq B_y$</span>. </p>
<p>An example of a space that is separable but not second countable and not Lindelof is the <a href="https://en.wikipedia.org/wiki/Moore_plane" rel="noreferrer">Moore, or Niemytzki plane</a> (also called tangent-disk space, usually available in topology texts). </p>
<p>There are also compact spaces (which of course is stronger than Lindelof) that are not second countable. One such example closely related to the Sorgenfrey line is the <a href="https://en.wikipedia.org/wiki/Split_interval" rel="noreferrer">Alexandrov double arrow space</a>, also called split interval.
Another is the <a href="https://math.stackexchange.com/q/1331117">Alexandroff double circle</a>. Note that every second countable space in hereditarily Lindelof (since every subspace is second countable, and hence Lindelof). The Alexandroff double circle is compact (and hence Lindelof), but has an uncountable discrete subspace, which of course is not Lindelof. Hence the Alexandroff double circle is not second countable. </p>
|
2,467,327 | <p>How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.<br>
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.</p>
| Michael Rozenberg | 190,319 | <p>If $a^2+b^2$ is divisible by $3$ then $a$ and $b$ are divisible by $3$ because
$$x^2\equiv0,1\pmod3.$$
Let $A=\{0,1\}$.</p>
<p>Thus, $3\not\in A+A$ and $0\in A+A$ for $0=0+0$ only.</p>
<p>Similarly, </p>
<p>if $a^2+b^2$ is divisible by $7$ then $a$ and $b$ are divisible by $7$ because
$$x^2\equiv0,1,2,4\pmod7.$$</p>
|
2,132,936 | <p>How do you simplify this problem?
$$ \frac {\mathrm d}{\mathrm dx}\left[(3x+1)^3\sqrt{x}\right] $$
$$= \frac {(3x+1)^3}{2\sqrt {x}} + 9\sqrt{x} (3x+1)^2 $$
$$\frac{(3x+1)^2(21x+1)}{2\sqrt x} $$</p>
| spaceisdarkgreen | 397,125 | <p>Multiplying top and bottom of the term $9\sqrt x(3x+1)^2$ by $2\sqrt x$ gives $$ \frac{18x(3x+1)^2}{2\sqrt x}.$$</p>
<p>Now can combine with the first term to get $$\frac{(3x+1)^3 + 18x(3x+1)^2}{2\sqrt x}.$$</p>
<p>Then we can factor a $(3x+1)^2$ out of the numerator, giving $$\frac{(3x+1)^2((3x+1)+18x)}{2\sqrt x} = \frac{(3x+1)^2(21x+1)}{2\sqrt x} $$</p>
|
3,960,404 | <p>Let <span class="math-container">$T$</span> be a tree.</p>
<p>Suppose that <span class="math-container">$T$</span> doesn’t have a perfect matching and let <span class="math-container">$M$</span> be a matching of <span class="math-container">$T$</span>, <span class="math-container">$|M| = k$</span>. Prove
that there exists a matching of the same cardinality <span class="math-container">$k$</span> which exposes at least a pendant vertex
of <span class="math-container">$T$</span>.</p>
<p>My idea:</p>
<p>If <span class="math-container">$T$</span> has not a perfect matching it means that <span class="math-container">$\exists x$</span> in <span class="math-container">$V(T)$</span> and <span class="math-container">$x$</span> is exposed. I don't know how to use this in symmetric difference to find another matching <span class="math-container">$M_1$</span>, <span class="math-container">$|M_1|=k$</span> which exposes at least a pendant vertex of <span class="math-container">$T$</span>.</p>
| Misha Lavrov | 383,078 | <p>First, here are some general ideas. In any graph, not just a tree, given a matching <span class="math-container">$M$</span> that is not perfect, we can try to find an augmenting path to improve the matching.</p>
<p>This is a path <span class="math-container">$v_0 v_1 v_2 \dots v_{2k+1}$</span> with the property that <span class="math-container">$v_1v_2, v_3 v_4, \dots, v_{2k-1} v_{2k}$</span> are edges of <span class="math-container">$M$</span>, and <span class="math-container">$v_0$</span> and <span class="math-container">$v_{2k+1}$</span> are exposed. By deleting those <span class="math-container">$k$</span> edges of <span class="math-container">$M$</span>, and adding the <span class="math-container">$k+1$</span> edges <span class="math-container">$v_0v_1, v_2v_3, \dots, v_{2k} v_{2k+1}$</span> instead, we get a bigger matching.</p>
<p>How might we find an augmenting path? Here is a procedure that may or may not work:</p>
<ol>
<li>Start with any exposed vertex <span class="math-container">$v_0$</span>.</li>
<li>Having found a vertex <span class="math-container">$v_{2i}$</span>, if it has any exposed neighbor, let <span class="math-container">$v_{2i+1}$</span> be that neighbor, and stop.</li>
<li>Otherwise, if <span class="math-container">$v_{2i}$</span> has <em>any</em> neighbors whatsoever we haven't seen yet, let <span class="math-container">$v_{2i+1}$</span> be one of them, let <span class="math-container">$v_{2i+2}$</span> be the vertex it is matched to, and continue from <span class="math-container">$v_{2i+2}$</span>.</li>
<li>Otherwise, if all of <span class="math-container">$v_{2i}$</span>'s neighbors are already in the path, we give up. (Fancier algorithms do more here, but we don't need anything fancy for this problem.)</li>
</ol>
<hr />
<p>So let's say that <span class="math-container">$M$</span> is a matching in a <em>tree</em> <span class="math-container">$T$</span>, and it leaves at least one vertex exposed. What can happen if we perform the procedure above on <span class="math-container">$T$</span>?</p>
<ul>
<li>It could succeed in finding a bigger matching <span class="math-container">$M'$</span> with <span class="math-container">$|M'|=|M|+1$</span>. We get the new matching we want by deleting an edge from <span class="math-container">$M'$</span>.</li>
<li>It could fail, stopping at a vertex <span class="math-container">$v_{2i}$</span> whose neighbors are all already on the path.</li>
</ul>
<p>I claim that in the second case, <span class="math-container">$v_{2i}$</span> is a pendant vertex (prove this!) and we can use the path <span class="math-container">$v_0 v_1 v_2 \dots v_{2i}$</span> to make a new matching, of the same size as <span class="math-container">$M$</span>, that leaves <span class="math-container">$v_{2i}$</span> exposed.</p>
|
3,043,780 | <p><a href="https://i.stack.imgur.com/h1M7D.png" rel="nofollow noreferrer">the image shows right-angled triangles in semi-circle</a></p>
<p>In Definite Integration, we know that area can be found by adding up the total area of each small divided parts.</p>
<p>So, base on the Definite Integration, we may say the area of circle is equal to
<span class="math-container">$\sum^{n}_{k=1}(h_k)$</span></p>
<p>And we also know that <span class="math-container">$A_k=2r(h_k)(1/2)=rh_k$</span> while <span class="math-container">$A$</span> is refer to the right-angled triangle's area.</p>
<p>so that, <span class="math-container">$A_k(1/r)=h_k$</span></p>
<p>As a result this equation comes out with fixed position of diameter:</p>
<p><span class="math-container">$\sum^{n}_{k=1}(h_k)=\pi r^2$</span></p>
<p><span class="math-container">$\sum^{n}_{k=1}(A_k)(1/r)=\pi r^2$</span></p>
<p><span class="math-container">$\sum^{n}_{k=1}(A_k)=\pi r^3$</span></p>
<p>I wish to know whether I am correct or not, thanks.</p>
| Oldboy | 401,277 | <p>Your method is totally inefficient and involves <span class="math-container">$\arctan$</span> function which is expensive from the computation point of view. </p>
<p>It's better to check angle orientations:</p>
<p><a href="https://i.stack.imgur.com/XNznF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XNznF.jpg" alt="enter image description here"></a></p>
<p>If the origin is in the triangle, angles <span class="math-container">$\angle AOB$</span>, <span class="math-container">$\angle BOC$</span>, <span class="math-container">$\angle COA$</span> have the same orientation.</p>
<p><a href="https://i.stack.imgur.com/SczWa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SczWa.jpg" alt="enter image description here"></a></p>
<p>If the origin is outside of the triangle, angles <span class="math-container">$\angle AOB$</span>, <span class="math-container">$\angle BOC$</span>, <span class="math-container">$\angle COA$</span> do not have the same orientation.</p>
<p>Angle orientation between vectors is reflected in the cross product. In this particular case:</p>
<p><span class="math-container">$$\vec{OA}\times\vec{OB}=(x_Ay_B-x_By_A)\vec k$$</span>
<span class="math-container">$$\vec{OB}\times\vec{OC}=(x_By_C-x_Cy_B)\vec k$$</span>
<span class="math-container">$$\vec{OC}\times\vec{OA}=(x_Cy_A-x_Ay_C)\vec k$$</span></p>
<p>If the value between brackets is positive, the cross product is pointing upwards and the angle has counterclockwise orientation. It's exactly the opposite if the value in brackets is negative.</p>
<p>The algorithm would look like this:</p>
<p>1) Calclulate the following values:</p>
<p><span class="math-container">$$z_1=x_Ay_B-x_By_A$$</span>
<span class="math-container">$$z_2=x_By_C-x_Cy_B$$</span>
<span class="math-container">$$z_3=x_Cy_A-x_Ay_C$$</span></p>
<p>2) If <span class="math-container">$z_1=0$</span> or <span class="math-container">$z_2=0$</span> or <span class="math-container">$z_3=0$</span>, the origin lies on some triangle edge</p>
<p>3) If <span class="math-container">$z_1,z_2,z_3$</span> are all of the same sign, the origin is inside the triangle, outside of it otherwise.</p>
<p>You can probably fix your method and force it to work, but it's still too complicated, totally inefficient and hard to follow. </p>
<p>Also functions like <span class="math-container">$\arctan$</span> are for mathematicians. If you have to compute the value of it in your computer code, you should use something like atan2(). That function takes two arguments (<span class="math-container">$x$</span> and <span class="math-container">$y$</span> values) and takes care of the quadrant in question.</p>
|
1,115,222 | <blockquote>
<p>Suppose <span class="math-container">$f$</span> is a continuous, strictly increasing function defined on a closed interval <span class="math-container">$[a,b]$</span> such that <span class="math-container">$f^{-1}$</span> is the inverse function of <span class="math-container">$f$</span>. Prove that,
<span class="math-container">$$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$</span></p>
</blockquote>
<p>A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.</p>
<p>However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with <span class="math-container">$f$</span> continuous, I cannot apply change of variables technique.</p>
<p>Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.</p>
<p>The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given <span class="math-container">$f$</span> is differentiable. However, this seems to hold for non-differentiable functions also.</p>
<p>I would really want some help. Pictorial proofs and/or area arguments are invalid.</p>
| kobe | 190,421 | <p>Let $C$ be the graph of $y = f(x)$ over the interval $[a,b]$. Then $\int_a^b f(x)\, dx$ is the line integral $\int_C y\, dx$, and $\int_{f(a)}^{f(b)} f^{-1}(y)\, dy$ is the line integral $\int_C x\, dy$. Thus $$\int_a^b f(x)\, dx + \int_{f(a)}^{f(b)} f^{-1}(y)\, dy = \int_C x\, dy + y\, dx = \int_C d(xy) = bf(b) - af(a).$$</p>
|
223,385 | <p>I would like to recreate the following picture in Mathematica. I know how to draw a tree with GraphLayout. But I don't know how to create the shape of nodes as below. A bit hints about where to start will be appreciated!</p>
<p><a href="https://i.stack.imgur.com/PtK1F.png" rel="noreferrer"><img src="https://i.stack.imgur.com/PtK1F.png" alt="enter image description here"></a></p>
| C. E. | 731 | <pre><code>vertexShape[n_] := Graphics[{
EdgeForm[Black],
FaceForm[Lighter@Gray],
Disk[{0, 0}],
White,
Disk[0.5 #, 0.2] & /@ CirclePoints[n]
}]
shapes = Thread[Range[0, 11] -> Table[
vertexShape@RandomInteger[4],
12]];
SeedRandom[110]
g = TreeGraph[
RandomInteger[#] \[UndirectedEdge] # + 1 & /@ Range[0, 10],
EdgeStyle -> Black,
VertexSize -> 0.5,
VertexShape -> shapes
]
</code></pre>
<p><a href="https://i.stack.imgur.com/d0DVp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d0DVp.png" alt="Example"></a></p>
<p>Here's an example of how you can set internal nodes to blank and set the other ones according to some other rules:</p>
<pre><code>shapes = Cases[
Thread[Range[0, Length@VertexList[g] - 1] -> VertexOutDegree[g]],
(x_ -> 1) :> (x -> vertexShape@RandomInteger[{1, 4}])
];
shapes = Prepend[shapes, _ -> vertexShape[0]];
</code></pre>
|
3,979,674 | <p>Let
<span class="math-container">$$I=\int\frac{dx}{\sqrt{ax^2+bx+c}}$$</span>
I know this can be either
<span class="math-container">$$\displaystyle I=\frac{1}{\sqrt{a}}\ln\left({2\sqrt{a}\sqrt{ax^2+bx+c}+2ax+b}\right)+C$$</span>
<span class="math-container">$$\displaystyle I=-\frac{1}{\sqrt{-a}}\arcsin{\left(\frac{2ax+b}{\sqrt{b^2-4ac}}\right)}+C$$</span>
Or
<span class="math-container">$$I=\frac1{\sqrt{a}}\mathrm{arcsinh}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+C$$</span></p>
<p>Can you tell me the difference between these three solutions?</p>
<p>Thanks a lot</p>
| GEdgar | 442 | <p>Suppose <span class="math-container">$a>0$</span>. Complete the square. You get one of these cases:
<span class="math-container">$$
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2}},\qquad \beta\in \mathbb R,\\
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2+\gamma^2}},\qquad \beta\in \mathbb R, \gamma > 0,\\
\frac{1}{\sqrt{a}}\int\frac{dx}{\sqrt{(x-\beta)^2-\gamma^2}},\qquad \beta\in \mathbb R, \gamma > 0 .
$$</span>
The first one has a "<span class="math-container">$\log$</span>" solution, the second one has an "<span class="math-container">$\arcsin$</span>" solution, the third one has an "<span class="math-container">$\text{asinh}$</span>" solution.</p>
<p>There are three similar cases when <span class="math-container">$a<0$</span>.</p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| KCd | 619 | <p>To rationalize denominators that are not quadratic is a nice application of linear algebra over fields. For example, if asked to rewrite
$$
\frac{1}{1-5\sqrt[3]{2}}
$$
with a rational denominator, the best way to approach the problem is to work in the field ${\mathbf Q}(\sqrt[3]{2})$ with ${\mathbf Q}$-basis $1,\sqrt[3]{2},\sqrt[3]{4}$.</p>
|
217,291 | <p>I am trying to recreate the following image in latex (pgfplots), but in order to do so I need to figure out the mathematical expressions for the functions</p>
<p><img src="https://i.stack.imgur.com/jYGNP.png" alt="wavepacket"></p>
<p>So far I am sure that the gray line is $\sin x$, and that
the redline is some version of $\sin x / x$.
Whereas the green line is some linear combination of sine and cosine functions.</p>
<p>Anyone know a good way to find these functions? </p>
| kjetil b halvorsen | 32,967 | <p>Group theory is useful in statistics, in experimental design: Look at the book by Box, Hunter & Hunter: "Statistics for Experimenters" (maybe the best book you will ever find about statistics, so you should have a look!) The first chapters is about $2^p$ designs and $2^{n-p}$ fractional designs. Everything they show can be seen as examples of group theory (but they do not do it that way). But if you can see it is group theory, maybe you can learn it faster.</p>
|
453,295 | <p>I wanna show that the non-zero elements of $\mathbb Z_p$ ($p$ prime) form a group of order $p-1$ under multiplication, i.e., the elements of this group are $\{\overline1,\ldots,\overline{p-1}\}$. I'm trying to prove that every element is invertible in the following manner:</p>
<blockquote>
<p><strong>Proof (a)</strong></p>
<p>By Bézout's lemma given $\bar a\in\mathbb Z_p$, there are $x,y \in
\mathbb Z$ such that </p>
<p>$ax+py=1\implies\overline {ax+py}=\overline 1\implies \overline a
\overline x+\overline p\overline y=\overline1\implies \overline a
\overline x+\overline 0=\overline1\implies \overline a\overline
x=\overline1$</p>
</blockquote>
<p>There are two problems with this proof, first $\overline 0$ is not defined, because $\overline 0$ is not defined because isn't in $\{\overline 1,\ldots \overline {p-1}\}$ secondly the sum is not defined, because I'm using multiplication in this case.</p>
<blockquote>
<p><strong>Proof (b)</strong></p>
<p>By Fermat's little theorem, for each $\overline a \in \mathbb Z_p$, we
have:</p>
<p>$a^{p-1}\equiv 1$ (mod $p$), then $\overline {a^{p-2}}$ is an inverse
to $\overline a$. </p>
</blockquote>
<p>My problem with this proof is why $\overline {a^{p-2}}\in
\{\overline1,\ldots,\overline{p-1}\}$.</p>
<p>I've already known these proofs, but with a little bit more experience, I found these lacks of rigor.</p>
<p>Thanks in advance.</p>
| davidlowryduda | 9,754 | <p>With respect to (a) first:</p>
<p>Why isn't $\bar0$ defined? It seems you're working in $\mathbb{Z}/p\mathbb{Z}$ and not only the group of units; thus $\bar{0}$ is defined just fine. Secondly, it seems you are deliberately forgetting that $\mathbb{Z}/p\mathbb{Z}$ is a ring and not just a group, i.e. that you have both addition and multiplication. Perhaps you thought in both cases that you were working in the group of units instead of the ring of integers mod $p$?</p>
<p>What you should do is work in the ring, realize $a$ has a (non-zero unital) inverse, and conclude that $a$ is in the group of units. This angle is the one espoused by (a), and is what makes it fine.</p>
<p>With respect to (b):</p>
<p>I like this one a lot less because standard proofs of Fermat's little theorem use (often explicitly) the size of the group of units (and Lagrange's theorem). Or they use the size of the group of units and then multiply them all together, do a little cancellation (maybe with Wilson's theorem), and get the result. But in both cases, you use the size of the group of units, so there is danger of self-referentiality.</p>
<p>But there are ways of proving Fermat's little theorem avoiding knowing the size of the group of units; so let's assume we've done that.</p>
<p>A priori, you only know that $a^{p-2} \in \mathbb{Z}/p\mathbb{Z}$, i.e. that it could be $0$. Well, is it $0$? If it were $0$, this means that $p \mid a^{p-2}$. But $p$ is prime, so if $p \nmid a$ then $p \nmid a^{x}$. So $a^{p-2} \neq 0$, and thus is in $1, \ldots, p-1$. (I stopped using bars, but every number here is in $\mathbb{Z}/p\mathbb{Z}$).</p>
|
1,511,078 | <p><strong>Show that the product of two upper (lower) triangular matrices is again upper (lower) triangular.</strong></p>
<p>I have problems in formulating proofs - although I am not 100% sure if this text requires one, as it uses the verb "show" instead of "prove". However, I have found on the internet the proof below but my problem is not just that I can't do one by myself, but also that it happens that I don't understand a proof which is already written - thing which let me think that I should quit these studies.</p>
<p><strong>My first question is: how to choose the right strategy for proving something, in this case as in others?</strong>. I guess it is also matter of interest...Interest for numbers and their properties. I've never had such interest, frankly - although, perhaps, it is not nice to say this here. The reason for which I started studying math and some physics over an year and a half ago is because I dreamed about and still dreaming to do a kind of research for which I am interested and I need scientific background.</p>
<p>Sorry for the digression but I am so upset and I need some general advice too. Coming back to the topic of the present proof, <strong>my second question is: could you explain me in plain English the idea and logic behind the following proof, please? Is this proof general enough to cover also my case and what changes do I need to add for it?</strong> - I ask this because the text to which I refer speaks about upper and lower triagular matrices, while the text of the proof I report here speaks about upper triangular matrices only. I also notice that the following proof considers only square matrices but also rectangular matrices can be upper/lower triangular.</p>
<p>The only thing I know about all this is what upper/lower triangular matrices are and how to perform multiplication between matrices. This is the proof found on the internet:</p>
<p>"Suppose that U and V are two upper triangular <em>n × n</em> matrices. By the row-column rule for matrix multiplication we know that the <em>(i,j)-th</em> entry of the product UV is $ui1v1j + ui2v2j +···+ uinvnj$. We need to show that if i > j then this expression evaluates to 0. In fact, we will show that every term $uikvkj$ of this expression evaluates to 0. To prove this, we consider two cases: • If i > k then $uik = 0$ since U is upper triangular. Hence $uikvkj = 0$. • If k > j then $vkj = 0$ since V is upper triangular. Hence $uikvkj = 0$. Since i > j, for every k weeither have $i > k$ or $k > j$ (possibly both) so the set wo cases cover all possibilities for k."</p>
| Bernard | 202,857 | <p>I would say this:</p>
<p>In $\displaystyle\sum_{k=1}^nu_{ik}v_{kj}\enspace(i>j)$, then</p>
<ul>
<li>either $i>k$ and $u_{ik}=0$, hence $u_{ik}v_{kj}=0$,</li>
<li>or $k\ge i>j$ and $v_{kj}=0$, hence $u_{ik}v_{kj}=0$.</li>
</ul>
<p>In plain English, it says that in each term of the sum at least one of $u_{ik}$, $v_{kj}$ is $0$, hence the sum is $0$.</p>
|
1,077,504 | <p>Evaluate:</p>
<p>$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$</p>
<p>Without <strong>the use of complex-analysis.</strong></p>
<p>With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?</p>
| lab bhattacharjee | 33,337 | <p>Setting $x=\dfrac1y,$</p>
<p>$$I=\int_0^\infty\frac1{1+x^6}dx=\cdots=\int_0^\infty\frac{x^4}{1+x^6}dx$$</p>
<p>$$2I=\int_0^\infty\frac{1+x^4}{1+x^6}dx=\int_0^\infty\frac{(1+x^2)^2-2x^2}{1+x^6}dx$$</p>
<p>$$=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx-\frac23\int_0^\infty\frac{3x^2}{1+x^6}dx$$</p>
<p>$$I_1=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx=\int_0^\infty\frac{1/x^2+1}{1/x^2-1+x^2}dx$$</p>
<p>$$=\int_0^\infty\frac{1/x^2+1}{\left(x-\dfrac1x\right)^2-3}dx =\int_{-\infty}^\infty\frac{dz}{z^2-3}$$</p>
<p>$$=\frac1{2\sqrt3}\ln\frac{z-\sqrt3}{z+\sqrt3}|_{-\infty}^\infty=\frac1{2\sqrt3}\ln\frac{1-\sqrt3/z}{1+\sqrt3/z}|_{-\infty}^\infty=\ln1-\ln1=0$$</p>
<p>$$I_2=\int_0^\infty\frac{3x^2}{1+x^6}dx=\arctan(1+x^3)|_0^{\infty}=\frac\pi2-\frac\pi4=\frac\pi4$$</p>
|
3,935,811 | <p>While solving a bigger problem, I stumbeled upon a system of parametric equations
<span class="math-container">$$
\left\{
\begin{array}{ll}
\dfrac{x-a}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{x-c}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0\\
\dfrac{y-b}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}} + \dfrac{y-d}{\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}} = 0
\end{array}
\right.
$$</span>
I need to solve it. I don't actually need all of the solutions to this system, one is enough.</p>
<p>Thanks, in advance.</p>
| Quanto | 686,284 | <p>Note that <span class="math-container">$d_1=\sqrt{\left(x-a\right)^2+\left(y-b\right)^2}$</span> and <span class="math-container">$d_2=\sqrt{\left(x-c\right)^2+\left(y-d\right)^2}$</span> are the distances from
the point <span class="math-container">$(x,y)$</span> to the points <span class="math-container">$(a,b)$</span> and <span class="math-container">$(c,d)$</span>, respectively. Rewrite the given system as</p>
<p><span class="math-container">$$\frac{x-a}{c-x}= \frac{y-b}{d-y}= \frac{d_1}{d_2}$$</span></p>
<p>So, geometrically, <span class="math-container">$(x,y)$</span> represent all the points on the line connecting <span class="math-container">$(a,b)$</span> and <span class="math-container">$(c,d)$</span>.</p>
|
762,651 | <p>I have to prove that "any straight line $\alpha$ contained on a surface $S$ is an asymptotic curve and geodesic (modulo parametrization) of that surface $S$". Can I have hints at tackling this problem? It seems so general that I am not sure even how to formulate it well, let alone prove it. Intuitively, I imagine that the normal $n_{\alpha}$ to the line/curve is perpendicular to the normal vector $N_{S}$ to the surface $S$, thus resulting in the asymptoticity; alternatively, a straight line has curvature $k = 0$ everywhere, and so the result follows. Is this reasoning adequate for a proof of the first part? I also realize that both geodesics and straight lines are the paths of shortest distance between two points on given surfaces (here, both $S$), thus the straight line must be a geodesic of any surface which contains it; should I quantify this statement, though?</p>
<p>Let $\mathbb{H}^2 = \{(x, y) \in \mathbb{R}^2: y>0 \}$ be the hyperbolic plane with the Riemannian metric $(\mathbb{d}s)^2 =\frac{(\mathbb{d}x)^2+(\mathbb{d}y)^2}{y^2}$. Consider a "square" $P = \{ (x, y) \in \mathbb{H}^2: 1 \leq x,y \leq 2 \}$. I need to calculate the geodesic curvature of the sides of $P$ and, for Gaussian curvature $K$ of $P$, I have to calculate $\int_{P} (K) \mathbb{d}\sigma$, where $\mathbb{d}\sigma$ is the area measure element of $\mathbb{H}^2$. Just hints as to how to start would be helpful. (I see that I have the first fundamental form, from which I can derive the coefficients $E$, $F$, and $G$ and thereby (hopefully easily) Christoffel symbols and an expression for area, but I do not see how any of this takes the actual square into account. Only the coordinates at which I evaluate these quantities seem to come come from the square! But I would still like detailed examples of even these things, please.)</p>
| Karl | 437,524 | <p>"Likewise when proving: If $0\leq a<b$, and $a^2<b^2$, then $a<b$. Why isn't taking the square root of both sides done?"</p>
<p>There is an intermediate step to this.</p>
<p>$$a^2<b^2$$</p>
<p>$$\sqrt{a^2}<\sqrt{b^2}$$
... notice that the square is inside the square root. Recall that this is equal to the absolute value.
<a href="https://math.stackexchange.com/questions/59630/square-root-of-a-number-squared-is-equal-to-the-absolute-value-of-that-number">Square root of a number squared is equal to the absolute value of that number</a></p>
<p>$$ |a| < |b| $$</p>
<p>... This is where it gets hairy, and depends case-to-case.</p>
<p>a) Here, both values are non-negative,
... the abs of which are themselves: $|a| = a, |b| = b$</p>
<p>$$ a < b $$</p>
<p>b) If both are non-positive, $|a| = -a, |b| = -b$, The abs resolves to:
$$ -a < -b $$</p>
<p>$$ b < a $$ ... I prefer not to "multiply by -1"
, but rather transpose a to the right and b to the left.</p>
<p>$$ a > b $$</p>
<p>c,d) If they have different signs, The abs conditionally resolves to</p>
<p>$$ a < -b \quad if \quad a\geq 0,b\leq 0 $$</p>
<p>$$ -a < b \quad if \quad a\leq 0,b\geq 0 $$</p>
<p>WHICH DO NOT GIVE ANY RELATIONSHIPS BETWEEN a AND b,</p>
<p>...but rather with the inverse of the other.</p>
<p>I always live by my math professor's advice:
If you can't specify what rule you're using on your equations, don't use it.</p>
<p>CONCLUSION:
I hope this answers your question, as to "why you cant",</p>
<p>... and better yet, "why you can in some cases."</p>
|
3,407,489 | <p><span class="math-container">$\neg\left (\neg{\left (A\setminus A \right )}\setminus A \right )$</span></p>
<p><span class="math-container">$A\setminus A $</span> is simply empty set and <span class="math-container">$\neg$</span> of that is again empty set. Empty set <span class="math-container">$\setminus$</span> A is empty set or? But every empty set is included in every set?</p>
<p>I am confused when it comes to this..</p>
| lonza leggiera | 632,373 | <p>Just use the laws of operator precedence to evaluate the result of each binary or unary operation in the proper order:
<span class="math-container">\begin{align}
A\setminus A&=\emptyset\\
\therefore \neg(A\setminus A)&=\Omega\ \ \ \text{(universal set)}\\
\therefore \neg(A\setminus A)\setminus A&=\Omega\setminus A\\
\therefore \neg(\neg(A\setminus A)\setminus A)&=\neg(\Omega\setminus A)\\
&=\neg\Omega\cup A\ \ \ \text{(De Morgan)}\\
&=\emptyset\cup A\\
&=A
\end{align}</span></p>
|
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