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 |
|---|---|---|---|---|
650,866 | <p>So I have the function $$ e^{-2x} $$ and if I derive this I thought that I should get $$ -2xe^{-2x} $$ But the $x$ disappears, why? Is it an inner derivative and because of that, I also have to differentiate the expression $-2x$ when I put it in front of $e$? If that is the case, then $x$ would be 1 and -2 is the only character left.. Am I right?</p>
| Glen O | 67,842 | <p>Go back to the basic definition of derivative. You have this:</p>
<p>\begin{align}
\frac{d}{dx}\left(e^{-2x}\right) &= \lim_{h\to0} \frac{e^{-2(x+h)}-e^{-2x}}{h}\\
&=\lim_{h\to0} \frac{e^{-2x}(e^{-2h}-1)}{h}\\
&=e^{-2x}\lim_{h\to0} \frac{e^{-2h}-1}{h}
\end{align}
As you can see, there is no $x$ outside of the $e^{-2x}$. And you will find that the remaining limit is equal to $-2$.</p>
|
2,419,485 | <blockquote>
<p>In a certain family four girls take turns at washing dishes. Out of a total of four breakages, three were caused by the youngest girl, and she was thereafter called clumsy. Was she justified in attributing the frequency of her breakages to chance?</p>
</blockquote>
<p>I'm not sure how to solve the following question. The answer is <strong>13/64</strong> for a girl if the breakage was random and <strong>13/256</strong> for the youngest girl that broke 3 dishes.</p>
| Ilia Vatahov | 761,820 | <p>There is a mistake in the book and it's not just a typo. It seems that the provided answers are for another task.</p>
<ul>
<li><p>A: three <strong>or</strong> four breakages are caused by one girl.</p>
</li>
<li><p>B: three <strong>or</strong> four breakages are caused by the youngest girl.</p>
</li>
</ul>
<p>If we think about the problem as placing four balls (the broken dishes) into four boxes (the girls) then:</p>
<ul>
<li><span class="math-container">$4^4$</span> - all possible arrangements.</li>
<li><span class="math-container">$\binom{4}{3}\cdot3$</span> - а pre-selected girl breaks three dishes.</li>
<li><span class="math-container">$\binom{4}{4}$</span> - а pre-selected girl breaks four dishes.</li>
</ul>
<p>and:</p>
<ul>
<li><p><span class="math-container">$p(A)=\dfrac{4\cdot\binom{4}{3}\cdot3+4\cdot\binom{4}{4}}{4^4} = \dfrac{13}{64}$</span></p>
</li>
<li><p><span class="math-container">$p(B)=\dfrac{\binom{4}{3}\cdot3+\binom{4}{4}}{4^4} = \dfrac{13}{256}$</span></p>
</li>
</ul>
|
479,594 | <p>I was wandering which is the best way to generate various combinations of $x_i$ such that $$\sum\limits_{i=1}^7 x_i = 1.0$$</p>
<p>where $ x_i \in \{0.0, 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0\}$</p>
<p>I can generate these using brute-force, i.e checking through all $ 11^7$ combinations and only taking those which satisfies our constraint, however I am interested to know if there is another approach for this. Any ideas?</p>
| obataku | 54,050 | <p>By inspection we determine a particular solution to $2a_n-a_{n-1}=1/2$ is given by $a_n=1/2$ trivially -- try an ansatz of the form $a_n=k$ and thus we get $k=2k-k=1/2$.</p>
<p>Considering the homogeneous case, $2a_n-a_{n-1}=0$, let $a_n=\lambda^n$ hence:$$2\lambda^n-\lambda^{n-1}=0\\2\lambda-1=0\\\lambda=\frac12$$... and so it follows that $a_n=(1/2)^n=1/2^n$ is a solution to our general equation and further so is any scalar multiple (since our equation is linear) i.e. $a_n=C/2^n$. Adding our particular equation to the mix we get a solution of the form $a_n=C/2^n+1/2$. Impose your initial conditions to determine $C$.</p>
|
1,259,853 | <p>Why the derivative of $n^{1/n} = \sqrt[n]{n}$ is $n^{1/n} \left( \frac{1}{n^2} - \frac{\log(n)}{n^2}\right)$ (according to Maxima and other tools online)?</p>
<p>I have tried to applied the chain rule, but it comes something completely different:</p>
<p>$$\frac{1}{n} n^{\frac{1}{n} - 1} \cdot 1 = \frac{1}{n} n^\frac{1}{n}n^{-1} = \frac{1}{n^2} n^\frac{1}{n} = \frac{\sqrt[n]{n}}{n^2}$$</p>
<p>Sincerely, I am not seeing where that $\log$ and the rest of the stuff comes from. I have a more difficult problem that is similar and whose solution contains a $\log$ somewhere, but I am not seeing where it comes from.</p>
| robjohn | 13,854 | <p>Write
$$
n^{1/n}=\exp\left(\frac1n\log(n)\right)
$$
Then the chain rule, followed by the product rule, says
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}n}n^{1/n}
&=\exp\left(\frac1n\log(n)\right)\frac{\mathrm{d}}{\mathrm{d}n}\left(\frac1n\log(n)\right)\\
&=n^{1/n}\left(\frac1{n^2}-\frac{\log(n)}{n^2}\right)
\end{align}
$$</p>
|
894,476 | <p>I don't have a strong background in probability/statistics and I'm trying to understand the example at <a href="http://rationalwiki.org/wiki/Extraordinary_claims_require_extraordinary_evidence#Probability_theory" rel="nofollow">http://rationalwiki.org/wiki/Extraordinary_claims_require_extraordinary_evidence#Probability_theory</a></p>
<p>Is that the correct framework to explain the principle? Their $P(A)$ there seems arbitrary, and using other values and the Bayes' formula they employ one gets values for $P(A|B)$ greater than one. </p>
<p>I've been trying to work out a more general example in which the experiment consists of tossing $N$ times and the event $B$ would be guessing right $n$ times, but again it seems to me that there must be something limiting $P(A)$ or else one can get values of $P(A|B)$ higher than one.</p>
<p>But again, I don't really know how to make sense of the example in a completely rigorous way or even if that's the correct approach to illustrate the principle.</p>
<p>Thanks!</p>
| André Nicolas | 6,312 | <p>You can apply it to Simpson's Rule. However, Simpson's Rule is obtained by one extrapolation step from the Trapezoidal Rule, so it makes no difference. </p>
|
201,999 | <p>Prove that the given sequence ${a_n}$ diverges to infinity.</p>
<p>$a_n=\frac{n^3+5}{-n^2+8n}$</p>
<p>I believe that the sequence diverges to -infinity. And I have this for my proof so far:</p>
<p>Let $M>0$ and let $N=$ ?. Then $n>N$ implies... I am confused on how to solve for the N. I believe I have to make the numerator larger and denominator smaller, but I have a hard time visualizing this. </p>
| Community | -1 | <p>Working directly from the definition, you're trying to show that for $n$ sufficiently large, you can make $a_n < -M$ for any $M \in \mathbb{N} .$ It helps to rewrite the general term as $\frac{n + 5/n^2}{-1 + 8/n}.$ The end behavior is suggested by the leading terms in the numerator and the denominator, so there are essentially two items you need to control first - the $\frac{5}{n^2} $ and the $\frac{8}{n}.$ It's easy to see that the first item is going to zero, so we know that at a cetain point ($n \ge 3$), the numerator is bounded by a certain natural number. To control the denominator, we want to make it smaller - just note that $\frac{8}{n}$ is always positive. Below is a concise proof.</p>
<p>Let $M > 1$ so that for $n > N = \max (2, M-1),$ we have $\frac{n^3+5}{-n^2+8n} = \frac{n + 5/n^2}{-1 + 8/n} < - (n+1) < -M .$ </p>
|
38,439 | <p>I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.</p>
<p>This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?</p>
<p>I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).</p>
<p>If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.</p>
| Dmitri Pavlov | 402 | <p>Here is my own favorite construction of the (Lebesgue) integral.</p>
<p>Suppose M is an arbitrary smooth manifold.
Denote by Or(M) the orientation line bundle of M.
This bundle is equipped with a canonical Riemannian metric.
Vectors of length 1 in the fiber of Or(M) over a point p∈M correspond canonically to the two orientations
of the tangent space at the point p.
The manifold M is orientable if and only if the bundle Or(M) is trivializable.
Choosing an orientation of M amounts to choosing an isometric trivialization of Or(M).</p>
<p>The bundle Or(M) together with its natural metric is flat.
Hence we can twist the de Rham complex Ω^0(M)→⋯→Ω^n(M)
by Or(M) and obtain the following <em>twisted de Rham complex</em>: Ω^0(M)⊗Or(M)→⋯→Ω^n(M)⊗Or(M).
(Here by a complex I mean a complex of sheaves.)
The line bundle Ω^n(M)⊗Or(M) is called the <em>bundle of densities</em> and is denoted by Dens(M).
This bundle has a canonical orientation (hence it is trivializable), but does not have a canonical
metric or a canonical trivialization.</p>
<p>The cohomology of the twisted de Rham complex (with compact support) is called the <em>twisted de Rham cohomology</em> (with compact support).
We have a canonical map C^∞_cs(Dens(M))→H^n_cs(M,Or(M)).
Here C^∞_cs is the space of global sections of a vector bundle with compact support
and H^n_cs denotes the nth cohomology with compact support.</p>
<p>The Poincaré duality gives us a canonical isomorphism H^n_cs(M,Or(M))→H_0(M).
Finally, the map from M to the point induces a map in homology H_0(M)→H_0(∙)=R.</p>
<p>The composition of maps C^∞_cs(Dens(M))→H^n_cs(M,Or(M))→H_0(M)→H_0(∙)=R
gives us a map ∫: C^∞_cs(Dens(M))→R, which is the integration map.
Note that the actual integration (over each connected component) happens
in the first map.
The second map is an isomorphism and the third map simply sums integrals over individual connected components.</p>
<p>The map f∈C^∞_cs(Dens(M))→∫|f|∈[0,∞) is a norm on C^∞_cs(Dens(M)).
Completing C^∞_cs(Dens(M)) in this norm yields L_1(M)(=L^1(M)), which
can be identified with the space of finite complex-valued measures on M.</p>
<p>The space of bounded measurable functions on M (=L_0(M)=L^∞(M)) can be constructed by completing C^∞(M) in the σ-weak topology induced by L_1(M).
Other L_p spaces can be constructed in a similar way to L_1(M) by completing
sections of the bundle of p-densities instead of the bundle of 1-densities Dens(M).</p>
<p>The development of the remainder of measure theory in this approach largely parallels
the one explained in <a href="https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820">one of my previous answers</a>.</p>
<p>I want to stress that these constructions do not rely on any existing integration theory.
In fact, they can be used to build integration theory on smooth manifolds from scratch
without ever referring to the usual measure theory with its lengthy and technical proofs.</p>
<p><strong>Added later:</strong> In the middle of the proof, we relied on Poincaré duality, which can be most easily established using sheaf cohomology.
To this end, one must show that the de Rham cohomology is isomorphic to the sheaf cohomology with real coefficients.
This boils down immediately to the Poincaré lemma.</p>
<p>The simplest way to establish the Poincaré lemma is as follows.
The de Rham complex of a finite-dimensional smooth manifold M
is the free C^∞-dg-ring on the C^∞-ring C^∞(M).
If M is the underlying smooth manifold of a finite-dimensional
real vector space V, then C^∞(M) is the free C^∞-ring on
the vector space V* (the real dual of V).
Thus, the de Rham complex of a finite-dimensional real vector space V
is the free C^∞-dg-ring on the vector space V*.
This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain
complex on the vector space V*.
The latter cochain complex is simply V*→V* with the identity differential.
It is cochain homotopy equivalent to the zero cochain complex,
and the free functor from cochain complexes to C^∞-dg-rings
preserves cochain homotopy equivalences.
Thus, the de Rham complex of the smooth manifold V
is cochain homotopy equivalent to the free C^∞-dg-ring on
the zero cochain complex, i.e., <strong>R</strong> in degree 0.</p>
|
3,493,519 | <p>Can I get a verification if this is the right way to approach this problem?</p>
<blockquote>
<p>Give an example of a linear map <span class="math-container">$T$</span> such that <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
</blockquote>
<p>By the fundamental theorem of linear maps,
<span class="math-container">$$\dim V = \dim \operatorname{range}T + \dim\operatorname{null}T,$$</span>
thus <span class="math-container">$\dim V=5$</span>. Let <span class="math-container">$e_1,e_2,e_3,e_4,e_5$</span> be a basis for <span class="math-container">$\mathbb{R}^5$</span>.
Let <span class="math-container">$f_1,f_2$</span> be a basis for <span class="math-container">$\mathbb{R}^2$</span>. Define a linear map <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^5,\mathbb{R}^2)$</span> by <span class="math-container">$$T(a_1e_1+a_2e_2+a_3e_3+a_4e_4+a_5e_5)=a_1f_1+a_2f_2.$$</span></p>
<p>Thus <span class="math-container">$\dim(\operatorname{null}T) = 3$</span> and <span class="math-container">$\dim(\operatorname{range}T) = 2$</span>.</p>
| Arthur | 15,500 | <p>This is completely correct. This will give a linear map with the properties you're asked for.</p>
<p>I think that it is a bit too general to actually be "an example". I think it would be better if you <em>actually pick</em> a concrete basis. But that's a personal aesthetic belief, and one would have to be pretty pedantic about it to say that that makes you wrong.</p>
<p>One objection with a bit more substance is that you haven't actually <em>proven</em> that your claims about the kernel and the image actually holds. You don't need much, but if this were on a test or an assignment and I was correcting it, I would want you to spend a sentence or two on each of them. For instance</p>
<blockquote>
<p><span class="math-container">$\dim \operatorname{im} T=2$</span> because <span class="math-container">$T$</span> is clearly surjective and <span class="math-container">$\dim \Bbb R^2=2$</span>. Then by the rank nullity theorem, we also get <span class="math-container">$\dim\ker T=3$</span>.</p>
</blockquote>
|
1,682,818 | <p>$L(G)$ is an undirected graph without parallel edges and loops such that:</p>
<p><strong>1. every edge in $G$ is an vertex in $L(G)$<br>
2. two vertices in $L(G)$ are connected by edge only if their edges in $G$ have a common vertex.</strong></p>
<p>The mission is to <strong>express the number of edges in $L(G)$ as a function of the vertices' degrees in $G$.</strong></p>
<p>I have the solution, but I don't understand one thing in it.</p>
<p>Let's look at a vertex $v$ in $G$. Every pair of edges connected to $v$ gives an edge to $L(G)$. I don't understand why $\binom{d(v)}2$ is the number of edges which $v$ creates; why it isn't $\frac{d\left(v\right)}{2}$, if every two edges connected to $v$ add an edge?</p>
<p>Can some one explain it to me, maybe by drawing an example, if it needed, because I tried and still don't see it.</p>
| Ove Ahlman | 222,450 | <p>You get $\binom{d(v)}2$ since you need to calculate the number of possible, unordered, pairs when you have $d(v)$ elements. Note especially that, if we call one of the edges in $v$ by the name $a$ then $a$ has $d(v)-1$ different other edges to be paired together with. Thus the answer $d(v)/2$ far from the truth.</p>
|
1,682,818 | <p>$L(G)$ is an undirected graph without parallel edges and loops such that:</p>
<p><strong>1. every edge in $G$ is an vertex in $L(G)$<br>
2. two vertices in $L(G)$ are connected by edge only if their edges in $G$ have a common vertex.</strong></p>
<p>The mission is to <strong>express the number of edges in $L(G)$ as a function of the vertices' degrees in $G$.</strong></p>
<p>I have the solution, but I don't understand one thing in it.</p>
<p>Let's look at a vertex $v$ in $G$. Every pair of edges connected to $v$ gives an edge to $L(G)$. I don't understand why $\binom{d(v)}2$ is the number of edges which $v$ creates; why it isn't $\frac{d\left(v\right)}{2}$, if every two edges connected to $v$ add an edge?</p>
<p>Can some one explain it to me, maybe by drawing an example, if it needed, because I tried and still don't see it.</p>
| Brian M. Scott | 12,042 | <p>Suppose that $G$ is the following graph:</p>
<pre><code> a
|
b
/ \
c d
</code></pre>
<p>It has edges $ab,bc$, and $bd$, so $L(G)$ has three vertices for which I will use the labels $v_{ab},v_{bc}$, and $v_{bd}$.. The edges $ab$ and $bc$ have a vertex in common, so the vertices $v_{ab}$ and $v_{bc}$ are connected by an edge in $L(G)$. In fact, each of the three pairs of edges of $G$ have vertex $b$ in common, so the corresponding vertices in $L(G)$ are connected by edges. Thus, $L(G)$ looks like this:</p>
<pre><code> v_ab
/ \
1/ \2
/ \
v_bc------v_bd
3
</code></pre>
<p>(Here I’ve written <code>v_ab</code> instead of $v_{ab}$, since I can’t write true subscripts in preformatted code.</p>
<p>Now let’s compare this with the argument in question. Specifically, we’ll look at vertex $b$ of $G$. Every <strong>pair</strong> of edges incident at $b$ create an edge in $L(G)$. Those pairs of edges are the pairs $\{ab,bc\}$, $\{ab,bd\}$, and $\{bc,bd\}$. the first of these pairs created the edge labelled $1$ in $L(G)$; the second, the edge labelled $2$; and the third, the edge labelled $3$.</p>
<p>The number of pairs of elements of an $n$-element set is $\binom{n}2=\frac{n(n-1)}2$; in this case $n=d(b)=3$, so we have</p>
<p>$$\binom32=\frac{3\cdot2}2=3$$</p>
<p>pairs of edges incident at vertex $b$. Note that we could not possibly get $\frac{d(b)}2=\frac32$ edges of $L(G)$ from the vertex $b$: that isn’t even an integer.</p>
|
3,575,417 | <p>I have to prove the non-existence of a continuous bijection between <span class="math-container">$[0,1)$</span> and <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>My attempt</strong>:</p>
<p>Since <span class="math-container">$\mathbb{R}$</span> is homeomorphic to <span class="math-container">$(0, 1)$</span>, I tried to prove the non-existence of a continuous bijection <span class="math-container">$g: [0, 1) \rightarrow (0, 1)$</span>. So, assume by contradiction that such a map exists. I wanted to find a contradiction using the Intermediate Value Theorem, but without success.</p>
<p>Any suggestions? Thanks in advance!</p>
| Community | -1 | <p>If a map <span class="math-container">$f:[0,1)\to (0,1)$</span> satifies the intermediate value property, then <span class="math-container">$f\left[(0,1)\right]=I$</span> is a subinterval of <span class="math-container">$(0,1)$</span>. If <span class="math-container">$f$</span> is bijective, then <span class="math-container">$I=(0,1)\setminus\{f(1)\}$</span>. However, <span class="math-container">$(0,1)\setminus \{c\}$</span> is never an interval forany <span class="math-container">$c\in (0,1)$</span>.</p>
|
3,575,417 | <p>I have to prove the non-existence of a continuous bijection between <span class="math-container">$[0,1)$</span> and <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>My attempt</strong>:</p>
<p>Since <span class="math-container">$\mathbb{R}$</span> is homeomorphic to <span class="math-container">$(0, 1)$</span>, I tried to prove the non-existence of a continuous bijection <span class="math-container">$g: [0, 1) \rightarrow (0, 1)$</span>. So, assume by contradiction that such a map exists. I wanted to find a contradiction using the Intermediate Value Theorem, but without success.</p>
<p>Any suggestions? Thanks in advance!</p>
| Henno Brandsma | 4,280 | <p>Suppose <span class="math-container">$f:[0,1) \to \Bbb R$</span> is a continuous bijection.
Then <span class="math-container">$f[(0,1)]=\Bbb R \setminus \{f(0)\}$</span> which is a contradiction, as <span class="math-container">$(0,1)$</span> is connected and so its continuous image is too, while <span class="math-container">$\Bbb R\setminus \{p\}$</span> is disconnected for any <span class="math-container">$p \in \Bbb R$</span>.</p>
|
1,246,356 | <p>Let $A,B \in {M_n}$ . suppose $A$ is normal matrix and has distinct eigenvalue, and $AB=0$. why $B$ is normal matrix?</p>
| Prasun Biswas | 215,900 | <p>$$\sum_{k=5}^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\approx \int\limits_5^{\sqrt n}\frac{\ln(\ln k)}{k\ln k}\,\mathrm dk\stackrel{t=\ln k}=\int\limits_{\ln 5}^{0.5\ln n}\frac{\ln t}{t}\,\mathrm dt\stackrel{u=\ln t}=\int\limits_{\ln(\ln 5)}^{\ln(0.5\ln n)}u\,\mathrm du$$</p>
<p>Can you take it from here?</p>
|
3,454,725 | <p>I am reading real analysis book and encountered this symbol <span class="math-container">$\wedge$</span> and <span class="math-container">$\vee.$</span></p>
<p>The author says following:</p>
<ol>
<li><span class="math-container">$f\wedge g=\frac{1}{2}(f+g-
|f-g|)$</span>,</li>
<li><span class="math-container">$f\vee g=\frac{1}{2}(f+g+|f-g|)$</span>.</li>
</ol>
<p>What are the meanings of <span class="math-container">$\wedge$</span> and <span class="math-container">$\vee$</span>? For example, I want to know what <span class="math-container">$f\wedge g$</span> means like <span class="math-container">$f^{+}$</span> means the positive part of the function.</p>
<p>Edit: he also asserts that <span class="math-container">$\chi_{A\cap B}=\chi_A\wedge \chi_B$</span> and <span class="math-container">$\chi_{A\cup B}=\chi_A\vee\chi_B.$</span></p>
| fleablood | 280,126 | <p>The way I've seen it done is to specify <span class="math-container">$\{i_1, ....., i_k\}\subset \{1,...., |\phi_M|\}$</span> and <span class="math-container">$A = \{x_{i_j}|x_{i_j} \in \phi_M\}$</span>.</p>
<p>In fact just <span class="math-container">$A=\{x_{i_j}\}\subset \phi_M$</span> is usually understood that the <span class="math-container">$j$</span> is is the index for <span class="math-container">$A$</span> and runs <span class="math-container">$1,.... , |A|$</span> and that <span class="math-container">$i_j $</span> is an index element from the indexes of <span class="math-container">$\phi_M$</span>.</p>
<p>Example: <span class="math-container">$\phi_M = \{x_1,....., x_m\}=\{x_i\}$</span> and <span class="math-container">$A = \{x_2,x_3, x_5, x_{13}\} = \{x_{i_1},x_{i_2},x_{i_3},x_{i_4}\}=\{x_{i_j}\}$</span> where <span class="math-container">$i_1 = 2; i_2 =3; i_3 = 5, i_4=13$</span>.</p>
|
601,951 | <p><em><strong>2</strong> + <strong>5</strong> + <strong>8</strong> + . . . + <strong>(6n-1)</strong> = <strong>n(6n+1</strong>)</em></p>
<p>This is what I have so far. </p>
<p>The <strong>sum</strong> of <strong>(3j-1)</strong> from <strong>j=1</strong> to <em>something I`m not sure of</em>.</p>
| ccorn | 75,794 | <p>If you are tired of trying to find an ingenious proof,
here is a computer-aided procedure for proving the identity.</p>
<p>The nomenclature follows that in <a href="http://www.math.upenn.edu/%7Ewilf/AeqB.pdf" rel="nofollow noreferrer">Petkovšek, Wilf, Zeilberger (1997): <span class="math-container">$A=B$</span></a>.
If you are impatient, just read the introduction to chapter 6.
Set
<span class="math-container">$$\begin{align}
F(n,k) &= \binom{2n+1}{2k+1}\binom{m+k}{2n}
\\ f(n) &= \sum_{k\in\mathbb{Z}} F(n,k)
\end{align}$$</span>
Here we use <span class="math-container">$\binom{n}{k}=0$</span> for <span class="math-container">$k<0$</span> as well as for <span class="math-container">$k>n$</span>.
There will be no need to make the dependency on <span class="math-container">$m$</span> explicit.</p>
<p>The claim is <span class="math-container">$f(n)=\binom{2m}{2n}$</span>
which is equivalent to
<span class="math-container">$$\begin{align}
f(n) &= 0 &&\text{for $n < 0$} \tag{$-$}
\\ f(n) &= 1 &&\text{for $n = 0$} \tag{0}
\\ f(n+1) &= \frac{(m-n)(2m-2n-1)}{(n+1)(2n+1)} f(n)
&&\text{for $n\geq 0$} \tag{1}
\end{align}$$</span>
<span class="math-container">$(-)$</span> and <span class="math-container">$(0)$</span> can be verified immediately.
For <span class="math-container">$(1)$</span> we will use Zeilberger's method.
Note that <span class="math-container">$\frac{F(n+1,k)}{F(n,k)}$</span> and <span class="math-container">$\frac{F(n,k+1)}{F(n,k)}$</span>
are rational functions of <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, therefore we call <span class="math-container">$F(n,k)$</span>
a <em>hypergeometric term</em>.
Zeilberger's method finds another hypergeometric term <span class="math-container">$G(n,k)$</span> and
<span class="math-container">$k$</span>-free polynomials <span class="math-container">$a_0(n),\ldots,a_J(n)$</span> (also depending on <span class="math-container">$m$</span>) such that
<span class="math-container">$$\sum_{j=0}^J a_j(n)\,F(n+j,k) = G(n,k+1) - G(n,k) \tag{2}$$</span>
In fact, we will get
<span class="math-container">$$G(n,k) = R(n,k)\,F(n,k)$$</span>
where <span class="math-container">$R(n,k)$</span> is a rational function in <span class="math-container">$n$</span> and <span class="math-container">$k$</span>.
Consequently. for any given <span class="math-container">$n$</span> and <span class="math-container">$m$</span>, there are finite lower and upper bounds
for those <span class="math-container">$k$</span> for which <span class="math-container">$G(n,k)$</span> can be nonzero.
Therefore, summing <span class="math-container">$(2)$</span> over <span class="math-container">$k\in\mathbb{Z}$</span> allows telescoping to
<span class="math-container">$$\sum_{j=0}^J a_j(n)\,f(n+j) = 0 \tag{3}$$</span>
which is a recurrence relation for <span class="math-container">$f$</span>.
The claim is that this recurrence relation
yields the same sequence <span class="math-container">$f(n)$</span> as <span class="math-container">$(1)$</span>.</p>
<p>Note that verification of the proof essentially amounts to verification of
<span class="math-container">$(2)$</span>, which requires no ingenuity because <span class="math-container">$(2)$</span> is equivalent to
<span class="math-container">$$\sum_{j=0}^J a_j(n)\,\frac{F(n+j,k)}{F(n,k)} =
R(n,k+1) \frac{F(n,k+1)}{F(n,k)} - R(n,k) \tag{4}$$</span>
which consists of rational functions only.</p>
<p>It remains to find <span class="math-container">$R(n,k)$</span> and <span class="math-container">$a_0(n),\ldots,a_J(n)$</span>.
This is best done with a suitable computer algebra system.
For example, in <a href="http://maxima.sourceforge.net/" rel="nofollow noreferrer">Maxima</a>,
or in <a href="http://www.sagemath.org/" rel="nofollow noreferrer">SAGE</a> on <code>maxima.console()</code>,
the lines</p>
<pre><code>load(zeilberger);
Zeilberger(binomial(2*n+1,2*k+1)*binomial(m+k,2*n),k,n);
</code></pre>
<p>would suffice. But let us be a bit more verbose and also verify the result:</p>
<pre><code>load(zeilberger);
F(n,k) := binomial(2*n+1,2*k+1)*binomial(m+k,2*n);
define (Fn(n,k), factcomb(makefact(F(n+1,k)/F(n,k)))), sumsplitfact:false;
define (Fk(n,k), factcomb(makefact(F(n,k+1)/F(n,k)))), sumsplitfact:false;
sols: Zeilberger(F(n,k),k,n);
/* Pick the first (and only) solution */
sol: sols[1];
/* sol has the form [R(n,k), [a_0, ..., a_J]] */
define (R(n,k), sol[1]);
/* Horner for lhs: sum(a_i*F(n+i,k)/F(n,k),i,0,length(a)-1); */
a: sol[2];
lhs: block([s], s: 0, for i: length(a) step -1 thru 1 do
s: s*Fn(n+(i-1),k)+a[i], s);
/* Here length(a)=2, so we have lhs: a[1]+a[2]*Fn(n,k); */
rhs: R(n,k+1)*Fk(n,k)-R(n,k);
ratsimp(lhs-rhs);
</code></pre>
<p>These commands should produce output with last line <code>0</code>.
The <code>Zeilberger</code> results are: <span class="math-container">$J=1$</span> and
<span class="math-container">$$\begin{align}
a_0(n) &= (m-n)(2m-2n-1)
\\ a_1(n) &= -(n+1)(2n+1)
\\ R(n,k) &= \frac{k(2k+1)(2n-m-k)(8n^2-6mn-6kn+10n+4km-5m-3k+3)}
{2(n-k+1)(2n+1)(2n-2k+1)}
\\\therefore\quad
G(n,k) &= -\frac{1}{2}(8n^2-6mn-6kn+10n+4km-5m-3k+3)
\binom{2n+1}{2k-1}\binom{m+k}{2n+1}
\end{align}$$</span>
Note that <span class="math-container">$G(n,k)$</span> has the singularities of <span class="math-container">$R(n,k)$</span> removed, as it should be,
and that <span class="math-container">$(3)$</span> is equivalent to
<span class="math-container">$$f(n+1) = -\frac{a_0(n)}{a_1(n)} f(n) =
\frac{(m-n)(2m-2n-1)}{(n+1)(2n+1)} f(n)$$</span>
which indeed matches <span class="math-container">$(1)$</span>.</p>
<p>We should be done now, but you know,
the first way found is usually not the best one.
You will have noticed that <span class="math-container">$k$</span> is the summation variable which we want to
telescope, but there is no particular reason for switching to <span class="math-container">$n$</span>
instead of <span class="math-container">$m$</span> for the recurrence.
Let us try switching the recurrence to <span class="math-container">$m$</span> instead:</p>
<pre><code>Zeilberger(binomial(2*n+1,2*k+1)*binomial(m+k,2*n),k,m);
</code></pre>
<p>This outputs
<span class="math-container">$$\begin{align}
a_0(m) &= -(m+1)(2m+1)
\\ a_1(m) &= (m-n+1)(2m-2n+1)
\\ R(m,k) &= k(2k+1)
\end{align}$$</span>
which simplifies the proof drastically. And I should have foreseen that.
Well, in the outset I supposed tiredness. Now that is proven too.</p>
|
4,405,145 | <p>Given <span class="math-container">$x_1, x_2, x_3, x_4, x_5$</span> be independent standard normal random variable and <span class="math-container">$\bar x$</span> the sample mean <span class="math-container">$\bar x= (x_1 + x_2 + x_3 + x_4 + x_5)/5$</span>. Then <span class="math-container">$\Pr(\bar x\leqslant c)$</span></p>
<p>What is c?</p>
<p>I tried to rewrite the given information mean which is <span class="math-container">$\bar x=∑_{x\in\{1,2,3,4,5\}}x_i/5 $</span> and <span class="math-container">$x_i\sim\mathcal N(0,1)$</span> then got stuck. What to do next?</p>
| user97357329 | 630,243 | <p><strong>A first solution by Cornel Ioan Valean</strong></p>
<p>One of the possible options here is to start by splitting the main integral and write
<span class="math-container">$$\int_0^1 \frac{\arctan^2(x)}{x}\log\left(\frac{x}{(1-x)^2}\right)\textrm{d}x$$</span>
<span class="math-container">$$=\underbrace{\int_0^1 \frac{\arctan^2(x)\log(x)}{x}\textrm{d}x}_{\displaystyle X}-2\underbrace{\int_0^1 \frac{\arctan^2(x)\log(1-x)}{x}\textrm{d}x}_{\displaystyle Y}.\tag1$$</span></p>
<p>A natural way to go for the integral <span class="math-container">$X$</span> is to consider <strong>Cornel</strong>'s solution to the challenging alternating harmonic series <span class="math-container">$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}$</span> you may find <a href="https://math.stackexchange.com/q/3803762">in this answer</a>, and thus we get
<span class="math-container">$$X=\int_0^1 \frac{\arctan^2(x)\log(x)}{x}\textrm{d}x$$</span>
<span class="math-container">$$=\frac{1}{24}\log^4(2)-\frac{1}{4}\log^2(2)\zeta(2)+\frac{7}{8}\log(2)\zeta(3)-\frac{151}{128}\zeta(4)+\operatorname{Li}_4\left(\frac{1}{2}\right).\tag2$$</span></p>
<p>During the calculation process of the result in <span class="math-container">$(2)$</span>, we circumvent the work with the Tetralogarithm involving a complex argument.</p>
<p>For the integral <span class="math-container">$Y$</span>, we combine two well-known results, that is the Cauchy product <span class="math-container">$\displaystyle \arctan^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}x^{2n} \frac{2H_{2n}-H_n}{n}, \ |x|\le1$</span>, and the logarithmic integral <span class="math-container">$\displaystyle \int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_n}{n}$</span> that gives
<span class="math-container">$$Y=\int_0^1 \frac{\arctan^2(x)\log(1-x)}{x}\textrm{d}x=\frac{1}{2}\int_0^1\log(1-x)\sum_{n=1}^{\infty}(-1)^{n-1}x^{2n-1} \frac{2H_{2n}-H_n}{n}\textrm{d}x$$</span>
<span class="math-container">$$=\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{2H_{2n}-H_n}{n}\int_0^1 x^{2n-1}\log(1-x)\textrm{d}x$$</span>
<span class="math-container">$$=\frac{1}{4}\underbrace{\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n H_{2n}}{n^2}}_{\displaystyle A}-\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}^2}{n^2}}_{\displaystyle B}$$</span>
<span class="math-container">$$=\frac{1}{48}\log^4(2)-\frac{1}{8}\log^2(2)\zeta(2)+\frac{7}{16}\log(2)\zeta(3)-\frac{151}{256}\zeta(4)-\frac{1}{2}G^2+\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right),\tag3$$</span>
where the first resulting series <span class="math-container">$A$</span> is <strong>magically</strong> calculated <a href="https://math.stackexchange.com/q/4407406">here</a> and <a href="https://math.stackexchange.com/q/4407788">here</a>, and for the second series <span class="math-container">$B$</span> we have
<span class="math-container">$$B=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}^2}{n^2}$$</span>
<span class="math-container">$$=2 G^2-\log(2)\pi G+\frac{231}{32}\zeta(4)-\frac{35}{16}\log(2)\zeta(3)+\log^2(2)\zeta(2)-\frac{5}{48}\log^4(2)$$</span>
<span class="math-container">$$-2 \pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-\frac{5}{2}\operatorname{Li}_4\left(\frac{1}{2}\right),$$</span>
which is immediately given by exploiting the generating function
<span class="math-container">$$\sum_{n=1}^{\infty} x^n \frac{H_n^2}{n^2}$$</span>
<span class="math-container">$$=2\zeta(4)-\frac{1}{3}\log(x)\log^3(1-x)-\log^2(1-x)\operatorname{Li}_2(1-x)+\frac{1}{2}(\operatorname{Li}_2(x))^2$$</span>
<span class="math-container">$$+2\log(1-x)\operatorname{Li}_3(1-x)+\operatorname{Li}_4(x)-2\operatorname{Li}_4(1-x), \ |x|\le1 \land \ x\neq1,$$</span>
then the well-known polylogarithmic relations <a href="https://math.stackexchange.com/q/920575">on this page</a>, and the special tetralogarithmic values,
<span class="math-container">$$\Re\biggr \{\operatorname{Li}_4\left(\frac{1\pm i}{2}\right)\biggr\}=\frac{343}{1024}\zeta(4)-\frac{5}{128}\log^2(2)\zeta(2)+\frac{1}{96}\log^4(2)+\frac{5}{16}\operatorname{Li}_4\left(\frac{1}{2}\right);$$</span>
<span class="math-container">$$\Re\{\operatorname{Li}_4(1\pm i)\}=\frac{485}{512}\zeta(4)+\frac{1}{8}\log^2(2)\zeta(2)-\frac{5}{384}\log^4(2)-\frac{5}{16}\operatorname{Li}_4\left(\frac{1}{2}\right).$$</span></p>
<p><strong>An extremely important note:</strong> The main core of extracting the tetralogarithmic values above relies on the calculation of the alternating harmonic series <span class="math-container">$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}$</span> you may find <a href="https://math.stackexchange.com/q/3803762">in this answer</a>. The other steps are trivial as explained <a href="https://math.stackexchange.com/questions/984026/inverse-trigonometric-integrals/984371#984371">in this 2014 answer</a>, that is using the classical generating function <span class="math-container">$\displaystyle \sum_{n=1}^{\infty}x^n \frac{H_n}{n^3}$</span> and the tetralogarithmic inverse relation in order to make the connections between that series and the special tetralogarithmic values. Full details about the extraction process of these awesome results (and other related ones) will be found in the sequel of <strong>(Almost) Impossible Integrals, Sums, and Series</strong>.</p>
<p>Finally, combining <span class="math-container">$(1)$</span>, <span class="math-container">$(2)$</span>, and <span class="math-container">$(3)$</span>, we arrive at the desired value</p>
<blockquote>
<p><span class="math-container">$$\int_0^1 \frac{\arctan^2(x)}{x}\log\left(\frac{x}{(1-x)^2}\right)=G^2.$$</span></p>
</blockquote>
<p><strong>End of story</strong></p>
<p><strong>Notes</strong>:</p>
<p><span class="math-container">$1).$</span> The solution is obviously based on deriving separate challenging results.</p>
<p><span class="math-container">$2).$</span> The series <span class="math-container">$\displaystyle A=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n H_{2n}}{n^2}$</span> is an <strong>extremely difficult</strong> (a beast in every way) to derive harmonic series, and <strong>Cornel</strong> already prepared a new full solution to it that I'll post very soon in its dedicated section.
<strong>UPDATE:</strong> Check the solution <a href="https://math.stackexchange.com/q/4407406">here</a>.</p>
<p><span class="math-container">$3).$</span> Is there hope to find a solution that circumvents such separate challenging results? I strongly believe it is possible. Maybe by exploiting the ideas from <a href="https://math.stackexchange.com/q/4407406">this solution</a>. Moreover, the sum of series <span class="math-container">$\displaystyle \sum _{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}}{n^3}+\sum _{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}^{(2)}}{n^2}=2G^2+\frac{37}{64}\zeta(4),$</span> given in <strong>(Almost) Impossible Integrals, Sums, and Series</strong>, page <span class="math-container">$313$</span>, might play a crucial part - they are derived without calculating each series separately.</p>
<p><span class="math-container">$4).$</span> More details about the results above, but also other results alike will be found in the sequel of <strong>(Almost) Impossible Integrals, Sums, and Series</strong> that will appear soon.</p>
<p><strong>ADDITIONAL NOTES:</strong></p>
<p>a). Another result we might consider in the process of trying to find a simple derivation of the main result, without having to calculate advanced results, is
<span class="math-container">$$\zeta(4)$$</span>
<span class="math-container">$$=\frac{32}{45} \pi \int_0^1 \frac{\arctan(t)\operatorname{arctanh(t)}}{t}\textrm{d}t-\frac{32}{45} \int_0^1 \frac{\arctan^2(t)\operatorname{arctanh}(t)}{t} \textrm{d}t,$$</span>
which appareas as <strong>a key identity</strong> in <a href="https://math.stackexchange.com/q/4407406">this post</a>.</p>
|
3,583,879 | <blockquote>
<p>a) $P_5=11$$</p>
<p>b) <span class="math-container">$P_1+P_2+P_3+P_4+P_5 =26$</span></p>
</blockquote>
<p>For the first part
<span class="math-container">$$\alpha^5+\beta ^5$$</span>
<span class="math-container">$$=(\alpha^3+\beta ^3)^2-2(\alpha \beta )^3$$</span></p>
<p>I found the value of <span class="math-container">$\alpha^3+\beta^3=4$</span></p>
<p>So <span class="math-container">$$16-2(-1)=18$$</span> which doesn’t match.</p>
<p>In the second part depends on the value obtained from part 1, so I need to get that cleared up. </p>
<p>I checked the computation many times, but it might end up being just that. Also, is there a more efficient way to do this?</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>Use <span class="math-container">$$(a^3+b^3)(a^2+b^2)=a^5+b^5+a^2b^2(a+b)$$</span> as we know <span class="math-container">$a+b, ab$</span></p>
|
917,272 | <p>The following is an old exam problem (Calc III). It looks simple and technical, but I end up with a difficult integral and I guess I have a mistake somewhere.</p>
<p>We are given the vector field $F(x,y,z)=(4z+2xy,x^2+z^2,2yz+x)$. We are asked to calculate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$, where $C$ is the intersection of the conic $z=\sqrt{x^2+y^2}$ and the cylinder $x^2+(y-1)^2=1$.</p>
<p>Stokes' Theorem allows us to replace the required integral with $\int_{S} \text{Curl}\vec{F} \cdot \hat{n} dS$, where $S$ is a surface bounded by $C$, and $\hat{n}$ is a normal to that surface. </p>
<p>The curl is $\text{Curl}\vec{F}=(0,3,0)$, so the integral simplifies to $3 \int_{S} (0,1,0) \cdot \hat{n} dS$.</p>
<p>I choose the surface to be $(x,y,\sqrt{x^2+y^2})$ with $x^2+(y-1)^2 \le 1$. I choose the parametrization $x=r\cos\theta, y=1+r\sin\theta$, and ended up with the integral $\int_{0}^{1} \int_{0}^{2\pi} \frac{r^2 \sin\theta +r}{\sqrt{r^2+1+2r\sin\theta}} dr d\theta$. I know how to solve similar integrals but this specific one seems non-elementary.</p>
<p>What am I doing wrong?</p>
| Andrés E. Caicedo | 462 | <p>Let $C_0=X$ and $C_{n+1}=f(C_n)$ for all $n$. By induction, note that $C_n\supseteq C_{n+1}$ for all $n$. Let $C=\bigcap_n C_n$, and note that $C$ is nonempty, being the intersection of a decreasing sequence of nonempty compact sets. </p>
<p>To check that $f(C)=C$, consider a point $x\in C$. Since $x\in C_{n+1}$ for all $n$, then there are points $x_n\in C_n$ with $x=f(x_n)$ for all $n$. By compactness of $X$, there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ that converges to a point in $X$, call it $y$. For each $m$, the tail subsequence $(x_{n_k})_{k\ge m}$ is contained $C_m$ (since the $C_n$ are decreasing), and therefore so is its limit $y$. This means that $y\in \bigcap_n C_n=C$. Finally, since $f$ is continuous and $f(x_{n_k})=x$ for all $k$, then also $f(y)=x$. This proves that $x\in f(C)$, and therefore that $C\subseteq f(C)$. </p>
<p>On the other hand, if $t\in f(C)$, then there is a point $z\in C$ with $t=f(z)$. Since $z\in C_n$ for each $n$, then $t\in C_{n+1}$ for all $n$. Clearly, $t\in C_0=X$ as well, so $t\in C$. This proves that $f(C)\subseteq C$, and therefore that indeed $C=f(C)$, as desired.</p>
<p>Let me conclude with a general observation. A <em>complete lattice</em> is a partially ordered set such that each subset has a supremum and an infimum. If $(P,\le)$ is a partial order, a function $g:P\to P$ is <em>order preserving</em> iff whenever $x\le y$ in $P$, then also $g(x)\le g(y)$. The <a href="http://en.wikipedia.org/wiki/Knaster%E2%80%93Tarski_theorem" rel="nofollow">Knaster-Tarski theorem</a> asserts that if $(L,\le)$ is a complete lattice, and $\pi:L\to L$ is order preserving, then $\pi$ admits a fixed point and, in fact, the set of fixed points of $\pi$ also constitutes a complete lattice. </p>
<p>It is easy to verify that if $X$ is a compact metric space, then the collection $K(X)$ of closed subsets of $X$, partially ordered by inclusion, forms a complete lattice: The infimum of a collection of closed sets is their intersection, and their supremum is the closure of their union. if $f:X\to X$ is continuous, then we can see $f$ as a function mapping closed subsets $D$ of $X$ to their pointwise image $f(D)$, and in this way we can consider $f$ as a map from $K(X)$ to itself, that it is obviously order-preserving. It follows that the collection of closed sets $A$ such that $f(A)=A$ forms a complete lattice (this particular case, of course, can be verified directly without having to go through the proof of the general theorem). Its minimum is the empty set, and its maximum is the set $C$ we built in the first paragraph. </p>
|
917,272 | <p>The following is an old exam problem (Calc III). It looks simple and technical, but I end up with a difficult integral and I guess I have a mistake somewhere.</p>
<p>We are given the vector field $F(x,y,z)=(4z+2xy,x^2+z^2,2yz+x)$. We are asked to calculate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$, where $C$ is the intersection of the conic $z=\sqrt{x^2+y^2}$ and the cylinder $x^2+(y-1)^2=1$.</p>
<p>Stokes' Theorem allows us to replace the required integral with $\int_{S} \text{Curl}\vec{F} \cdot \hat{n} dS$, where $S$ is a surface bounded by $C$, and $\hat{n}$ is a normal to that surface. </p>
<p>The curl is $\text{Curl}\vec{F}=(0,3,0)$, so the integral simplifies to $3 \int_{S} (0,1,0) \cdot \hat{n} dS$.</p>
<p>I choose the surface to be $(x,y,\sqrt{x^2+y^2})$ with $x^2+(y-1)^2 \le 1$. I choose the parametrization $x=r\cos\theta, y=1+r\sin\theta$, and ended up with the integral $\int_{0}^{1} \int_{0}^{2\pi} \frac{r^2 \sin\theta +r}{\sqrt{r^2+1+2r\sin\theta}} dr d\theta$. I know how to solve similar integrals but this specific one seems non-elementary.</p>
<p>What am I doing wrong?</p>
| Marc Bogaerts | 118,955 | <p>This sketch of a proof is based on the "special" definition of compactedness on metric spaces : <a href="http://en.wikipedia.org/wiki/Metric_space#Compact_spaces" rel="nofollow">compact metric spaces</a>. Take any point in $x\in X$ and consider the sequence $f(x), f(f(x)), \ldots$ then there is a subsequence that converges to a point $x_0$ then $A= \{x_0\}$.</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| Mauro ALLEGRANZA | 108,274 | <p>In a proof with multiple <em>assumptions</em> you have to choose one of them to be "blamed" for the contradiction.</p>
<p>Think to your example in terms of <em>assumptions</em>; you start with a couple of them (they can be two <em>Lemmas</em> already proved, or two hypotheses) :</p>
<blockquote>
<p>$P→Q$ and $R→¬Q$.</p>
</blockquote>
<p>Then we proceed "formally" as follows (I'll use the <a href="http://en.wikipedia.org/wiki/Natural_deduction">Natural Deduction</a> proof system; for a good explanation of the rules to be used, see : Ian Chiswell & Wilfrid Hodges, <a href="http://rads.stackoverflow.com/amzn/click/0199215626">Mathematical Logic</a> (2007), <strong>Ch.2 : Informal natural deduction</strong>, page 5-on) :</p>
<p><em>1)</em> $P$ --- assumed</p>
<p><em>2)</em> $Q$ --- from 1) and $P→Q$ by $\rightarrow$-elim (<em>modus ponens</em>)</p>
<p><em>3)</em> $R$ --- assumed</p>
<p><em>4)</em> $\lnot Q$ --- from 3) and $R→¬Q$ by $\rightarrow$-elim (<em>modus ponens</em>)</p>
<p><em>5)</em> $\bot$ --- from 2) and 4) by $\lnot$-elim [i.e. using the rule : "from $\varphi$ and $\lnot \varphi$, infer $\bot$]</p>
<p><em>6)</em> $\lnot R$ --- from 3) and 5) by $\lnot$-intro [i.e. using the rule : "if from $\varphi$ we have derived $\bot$, then infer $\lnot \varphi$], "discharging" temporary assumption 3)</p>
<p><em>7)</em> $P \rightarrow \lnot R$ --- from 1) and 6) by $\rightarrow$-intro, "discharging" temporary assumption 1).</p>
<p>Thus we have proved :</p>
<blockquote>
<p>$P→Q, R→¬Q \vdash P \rightarrow \lnot R$.</p>
</blockquote>
<hr>
<p>As per the above answer, we can apply <em>contraposition</em> : $\varphi \rightarrow \lnot \psi \vdash \psi \rightarrow \lnot \varphi$ to conclude also :</p>
<blockquote>
<p>$P→Q, R→¬Q \vdash R \rightarrow \lnot P$.</p>
</blockquote>
<p>In the previous proof, we have <strong>chosen</strong> the assumptiom $R$ to be "blamed" for the contradiction. We can as well choose $P$.</p>
<p>If you rewrite it introducing $\lnot P$ in step 6) above, you will end exactly with : $R \rightarrow \lnot P$.</p>
<hr>
<p><em>Comment</em></p>
<p>In order to "have a feeling" with the above application of logical rules, modify the above proof using a single <em>assumption</em> $P \land R$.</p>
<p>Due to the fact that :</p>
<p>$P \land R \vdash P$ and $P \land R \vdash R$ [by : $\land$-elim]</p>
<p>we can repeat the same steps until 5) : $\bot$.</p>
<p>In this case, we have only one <em>assumption</em> to be "blamed" : $P \land R$ and we conclude with :</p>
<blockquote>
<p>$\lnot (P \land R)$.</p>
</blockquote>
<p>This means that, in the presence of the two <em>Lemmas</em> or hypotheses : $P \rightarrow Q$ and $R \rightarrow \lnot Q$, we <strong>cannot</strong> "jointly assert" $P$ and $R$.</p>
<p>Thus, one of them must be "removed". Which one ? it's up to us ...</p>
|
916,963 | <p>I am reading an introductory book on mathematical proofs and I don't seem to understand the mechanics of proof by contradiction. Consider the following example. </p>
<p>$\textbf{Theorem:}$ If $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$.</p>
<p>$\textbf{Proof:}$ (by contradiction)
Assume $P$, then it follows that $Q$.
Now, assume $R$, then it follows that $\neg Q$. Contradiction, we have $Q$ and $\neg Q$ at the same time. Hence, $\neg R$.
Therefore, if $P \rightarrow Q$ and $R \rightarrow \neg Q$, then $P \rightarrow \neg R$, as desired.</p>
<p>What I don't understand in this proof, is that why having arrived at contradiction, we decide that our assumption that $R$ is necessarily false? It also could have been that our first assumption, namely, $P$, was false. Or both of them could be false.</p>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
| David | 119,775 | <blockquote>
<p>So my question is: in general, when proving by contradiction, how do we know which assumption exactly is false? And how do we know that exactly one assumption must be wrong in order to proceed with the proof?</p>
</blockquote>
<p>If you make more than one assumption and get a contradiction, you cannot without further argument know which assumption(s) is(are) false.</p>
<p>However IMHO the point (which I cannot see clearly stated in any previous answers) is that if you set up the problem carefully you need only make one assumption: namely, assume that the result <strong>as a whole</strong> is false. In this case, assume that</p>
<blockquote>
<p>If $P\to Q$ and $R\to\neg Q$, then $P\to\neg R$</p>
</blockquote>
<p>is false. By standard propositional logic equivalences, this means that</p>
<blockquote>
<p>$P\to Q$ and $R\to\neg Q$ are both true, and $P\to\neg R$ is false,</p>
</blockquote>
<p>that is,</p>
<blockquote>
<p>$P\to Q$ and $R\to\neg Q$ are both true, and $P$ is true and $R$ is true.</p>
</blockquote>
<p>From this you easily deduce that $Q$ is true and therefore $R$ is false; but $R$ is also true, which is a contradiction.</p>
<p>Therefore, <strong>the theorem as a whole</strong> cannot be false and must be true.</p>
|
3,478,098 | <p><a href="https://i.stack.imgur.com/dcxhi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dcxhi.jpg" alt="enter image description here" /></a></p>
<p>Guys, Why is this weird statement true? It seems counterintuitive to me I cannot understand or lack creativity understanding it can you help me explain it? Guys please if possible make it visual.</p>
| David K | 139,123 | <p>It all comes from the three lines before the part you highlighted.</p>
<p>You might be thinking that no matter how small we make <span class="math-container">$\delta,$</span> there is always a rational number in the interval <span class="math-container">$(a - \delta, a + \delta).$</span>
If so, you would be right.
In fact there are infinitely many rational numbers there.
But they might all have very large denominators.</p>
<p>Consider the closest integer to <span class="math-container">$a.$</span> There is only one.
(Really only one; since <span class="math-container">$a$</span> is irrational, it can't be exactly halfway between two integers.)
Now consider the closest rational number with denominator <span class="math-container">$2.$</span>
Next, the the closest rational number with denominator <span class="math-container">$3.$</span>
Then <span class="math-container">$4.$</span> Then <span class="math-container">$5.$</span></p>
<p>Keep on going like that until you have the closest rational number with denominator <span class="math-container">$N.$</span>
Now you have <span class="math-container">$N$</span> rational numbers and <span class="math-container">$N$</span> distances from each of those numbers to <span class="math-container">$a.$</span>
Pick the number that is closest to <span class="math-container">$a$</span>. Make <span class="math-container">$\delta$</span> less than the distance between that number and <span class="math-container">$a.$</span></p>
<p>Now which of those <span class="math-container">$N$</span> rational numbers can be in <span class="math-container">$(a - \delta, a + \delta)$</span>?
None of them: we chose an interval that all of them are outside of.</p>
<p>What about all the other rational numbers with denominators <span class="math-container">$1,$</span> <span class="math-container">$2,$</span> <span class="math-container">$3,\ldots, N$</span>?
Also outside <span class="math-container">$(a - \delta, a + \delta)$</span>, because we made sure we had already looked at the closest rational number with each denominator. All the others are further.</p>
<p>This is what the passage is saying, although I gave a lot of unnecessary detail.
We know that only finitely many rationals with denominators <span class="math-container">$1,$</span> <span class="math-container">$2,$</span> <span class="math-container">$3,\ldots, N$</span>
can exist in the interval <span class="math-container">$[0,1]$</span> where we found <span class="math-container">$a.$</span>
One of those numbers has to be the closest one to <span class="math-container">$a.$</span> We just need to make
<span class="math-container">$a - \delta$</span> and <span class="math-container">$a + \delta$</span> closer.</p>
|
462,397 | <p>So, I read the John Baez essay "Lectures on n-categories and cohomology" and I understand the notion of a (-1)-category" and a (-2)-category" and how to derive them. However, I'm not totally clear on what a (-1)-morphism is.</p>
<p>At nLab at <a href="http://ncatlab.org/nlab/show/k-morphism" rel="nofollow noreferrer">http://ncatlab.org/nlab/show/k-morphism</a>, I found:</p>
<blockquote>For the purposes of negative thinking, it may be useful to recognise that every ∞-category has a (−1)-morphism, which is the source and target of every object.</blockquote>
<p>I have an idea of how this works, but I'm not sure if it's correct. Basically, my thinking is that if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are objects in a 1-category <span class="math-container">$C$</span>, and <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are both parallel morphisms that map <span class="math-container">$A\to B$</span>, and we take <span class="math-container">$S = hom_C(A,B)$</span> then we get a 0-category (a set) that contains <span class="math-container">$f$</span> and <span class="math-container">$g$</span> as elements (and their identities as morphisms). Both <span class="math-container">$hom_S(f,g)$</span> and <span class="math-container">$hom_S(f,g)$</span> each give us an empty (-1)-category (obviously isomorphic). Then <span class="math-container">$hom_S(f,f)$</span> and <span class="math-container">$hom_S(g,g)$</span> each give us a (-1)-category with a single object (also obviously isomorphic). Now, we treat the isomorphic (-1)-categories as equivalent, so there are only the two: the empty one named <span class="math-container">$False$</span>; and, the non-empty one named <span class="math-container">$True$</span>. So, since all non-empty (-1)-categories are equivalent, we decide that their morphisms are also equivalent. </p>
<p>So, here's my question that I'm not sure about: </p>
<ul>
<li><b>Question 1: </b> Am I correct in my understanding that the identity morphism of an object in a (-1)-category is a (-1)-morphism? It's a 0-morphism if the (-1)-category were treated as a 1-category, but if we take into account that the (-1)-category is derived from some higher category, then does that effectively make the single 1-morphism in the (-1)-category a (-1) morphism in the higher-category? </li>
</ul>
<p>This seems like it would make sense of the assertion that the single (-1)-morphism is the source and target of every 0-morphism/object.</p>
<p>I also came across another passage meant to explain the (-1)-morphism that threw me off a little bit:</p>
<blockquote>
Also note that every k-morphism has k+1 identity (k+1)-morphisms, which just happen to all be the same (which can be made a result of the Eckmann–Hilton argument). Thus, the (−1)-morphism has 0 identity 0-morphisms, so we don't need any object. (This confused me once.)
</blockquote>
<ul>
<li><b>Question 2:</b> Does this mean that an empty category still has a (-1)-morphism? If so, where does it come from?</li>
<li><b>Question 3:</b> Does this mean that a 1-morphism has 2 identity 2-morphisms? My understanding of an "identity" necessitates uniqueness. Having 2 identities for the same thing seems to cause a contradiction. How can I reconcile this?</li>
</ul>
| user062295 | 305,314 | <p>For an example, take the $\infty$ category $Kom$, i.e. the nerve of the category of chain complexes over a field. A $-1$ morphism is a map from one chain complex to another chain complex of degree $-1$.</p>
|
1,016,682 | <p>is my proof correct?</p>
<p>Definition:</p>
<p>Let $X\subset\mathbb R$ and let $x'\in\mathbb R$, we say that $x'$ is an adherent point of $X$ iff $\forall\epsilon>0\exists x\in X \text{ s.t. }d(x′,x)≤ε$. the closure of X is denoted as $\overline X$ and is defined to be the set of all the adherent points of $X$.</p>
<p>show: $\overline{\overline X} = \overline X$</p>
<p>suppose $\exists z \in \overline{\overline X}~~and~~z \notin \overline X$</p>
<p>then, $\exists x' \in \overline X ~~s.t.~~ |z-x'|\leq \epsilon$</p>
<p>$|z-x'|\leq \frac{\epsilon}{2}$</p>
<p>but we also know that $\exists x \in X s.t. |x'-x|\leq \frac{\epsilon}{2}$</p>
<p>hence, $z-x+x-x' \leq \epsilon/2$</p>
<p>$z-x\leq0$</p>
<p>hence z is an adherent point of X ($z\in \overline X$). but this is a contradiction with the above condition on $z$
hence, $\overline{\overline X} = \overline X$</p>
| idm | 167,226 | <p>An other way to prove:
$\overline{\bar X}$ is the smallest close set that contain $\bar X$. But $\bar X$ is close then the smallest close set that contain $\bar X$ is $\bar X$, therefore $$\overline{\bar X}=\bar X$$</p>
|
86,536 | <p>Considering we have a an association:</p>
<pre><code>assc = <|"A" -> <|"a" -> 1, "aa" -> 2|>, "B" -> 0, "C" -> 5,"D" -> <|"d" -> 2, "dd" -> 12|>|>
</code></pre>
<p>Let's also consider we have 2 known lists and one list for nested keys
(**
- <em>how to create this list?</em>
**):</p>
<pre><code>bigKeys = {"A", "D"}
lst1 = {"a", "dd", "B"}
lst2 = {"aa", "d", "C"}
</code></pre>
<p>I would like to loop over <code>assc</code> and assign the values for the keys that are from <code>lst1</code> and are <strong>not</strong> <code>bigKeys</code> - <strong>0</strong>, and to the values for the keys that are from <code>lst2</code> and again are <strong>not</strong> <code>bigKeys</code> - <strong>1</strong>. </p>
<p>As a result to have: </p>
<pre><code>asscNew = <|"A" -> <|"a" -> 0, "aa" -> 1|>, "B" -> 0, "C" -> 1, "D" -> <|"d" -> 1, "dd" -> 0|>|>
</code></pre>
<p>Thanks!</p>
| C. E. | 731 | <p>This solution will traverse all nested associations regardless of depth and replace each value with the value that corresponds to its key in <code>replacements</code>.</p>
<pre><code>f[Rule[key_, assoc_Association]] := Rule[key, AssociationMap[f, assoc]]
f[Rule[key_, val_]] := Rule[key, key /. replacements]
replacements = {
"a" -> 1, "dd" -> 1, "B" -> 1,
"aa" -> 0, "d" -> 0, "C" -> 0
};
assoc = <|"A" -> <|"a" -> 1, "aa" -> 2|>, "B" -> 0, "C" -> 5,
"D" -> <|"d" -> 2, "dd" -> 12|>|>;
AssociationMap[f, assoc]
(* Out: <|"A" -> <|"a" -> 1, "aa" -> 0|>, "B" -> 1, "C" -> 0,
"D" -> <|"d" -> 0, "dd" -> 1|>|> *)
</code></pre>
<p><code>key /. replacements</code> can be replaced by a more general transformation <code>func[key, val]</code>.</p>
|
3,300,469 | <p>I have a problem counting all the possible ways of "pairing" two datasets of size n and m, including partial pairing. </p>
<p>Example:
Assume we have two sets <span class="math-container">$\{A,B\}$</span> and <span class="math-container">$\{1,2,3\}$</span>. My aim is to find all ways of pairing letters with numbers, including the consideration of situations, where only a fraction of possible pairs is actually present, including a case of "no pairing at all".</p>
<p>In this case I would get the following ways of pairing:<br>
<span class="math-container">$\{A,1\},\{B,2\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,1\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B,3\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,3\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,2\},\{B\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,3\},\{B\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,1\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,2\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,3\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B\},\{1\},\{2\},\{3\}$</span> (zero pairs out of two coexisting pairs possible) </p>
<p>How can I generalize it to bigger sets of size n and m?</p>
| SlipEternal | 156,808 | <p>Assume your sets are <span class="math-container">$A,B$</span> with <span class="math-container">$|A|\le |B|$</span> and <span class="math-container">$A\cap B = \emptyset$</span>. Then you are looking for the sum of the ways for each <span class="math-container">$X\subset A$</span>, the number of injections <span class="math-container">$f:X\hookrightarrow B$</span>. This is counted by <span class="math-container">$X$</span>-permutations of <span class="math-container">$B$</span>.</p>
<p>So, the number of ways is given by:</p>
<p><span class="math-container">$$\sum_{X \in P(A)} (|B|)_{|X|}$$</span></p>
<p>where <span class="math-container">$(|B|)_{|X|}$</span> is the <a href="https://en.wikipedia.org/wiki/Falling_and_rising_factorials" rel="nofollow noreferrer">Descending factorial</a>.</p>
|
1,515,823 | <p>I am doing my research in Functional Analysis, especially in "Generalized inverse of Linear Maps".</p>
<p>I have come across Probability by studying only the methods or Distributions(like Binomial, poisson, normal,etc)</p>
<p>Now I wish to study the mathematical background and intuitive way of looking on it.</p>
<p>Can you suggest some probability text book which explains,</p>
<blockquote>
<p><span class="math-container">$\bullet$</span> Geometrical ideas of the Probability Distributions. May be using diagrams, graphs, etc.</p>
<p><span class="math-container">$\bullet$</span> Proofs for the distribution functions.</p>
<p><span class="math-container">$\bullet$</span> Problems with natural solutions and then generalizations
Like, for a binomial distribution(how they are giving the probability mass function).</p>
</blockquote>
| Trajan | 119,537 | <p>I think your best bet would be <strong>"Probability And Random Processes" by Grimmett and Stirzacker</strong>, see <a href="http://www.amazon.co.uk/Probability-Random-Processes-Geoffrey-Grimmett/dp/0198572220" rel="nofollow">http://www.amazon.co.uk/Probability-Random-Processes-Geoffrey-Grimmett/dp/0198572220</a>. It provides a concise yet introductory treatment of probability theory, free of measure theory. The latter chapters may not be of use to you but the earlier chapters should be. This would be the quickest way to get the basics. </p>
<p>I would also recommend, <strong>"A Course in Probability Theory" by Chung</strong>, <a href="http://www.amazon.co.uk/Course-Probability-Theory-Third-Edition/dp/0121741516" rel="nofollow">http://www.amazon.co.uk/Course-Probability-Theory-Third-Edition/dp/0121741516</a>, its a more rigorous treatment than others mentioned so far. If you are researching functional analysis, this should be quite accessible to you.</p>
<p>I would not recommend Ross' book at is it too elementary, too slow and not suitable for a postgraduate. It is aimed at first year mathematics students, many of whom are making the transition from high school. The only reason it is recommended is because when people who did one course in probability as a undergrad, this is the text that would have been recommended.</p>
|
45,771 | <p>Hi, it seems like a big field and I'm having trouble getting some solid/classic references to get me started.</p>
<p>If $U \subset \mathbb{R}^d$ is a bounded domain with, say, $C^2$-boundary $\partial U$ and $(S(t),t \ge 0)$ is the Dirichlet heat semigroup on $L^p(U)$ then $(S(t) f)(x) = \int_U G_U(t,x,y) f(y)\,dy$ for $f \in L^p(U)$ where $G_U$ is the Dirichlet heat kernel.</p>
<p>I would like to find a good reference for bounds of the type:
$$
\left|\frac{\partial G_U}{\partial \nu_y} (t,x,y)\right| \le C_1 t^{-(d + k)/2} \exp\left(- \frac{|x-y|^2}{C_2 t}\right)
$$
It seems like it should be classic result? I found Aronson's 1968 paper but it only contains estimates for the kernel and not the 'derivatives'. I can find lots of recent papers on manifolds and such but I am just looking for a solid and accessible reference for my simple case.</p>
<p>Further, if one has a semigroup $(T(t), t \ge 0)$ with a kernel $k(t,x,y)$ that has a pointwise Gaussian estimate $|k(t,x,y)| \le c_1 e^{\omega t} t^{-d/2} e^{-|x-y|^2/(c_2 t)}$, do the estimates for the 'derivatives' as above follow readily? or are more assumptions needed? Again, a reference to point me in the right direction would be greatly appreciated.</p>
<p>Thanks.</p>
| Piero D'Ancona | 7,294 | <p>I would like to know a complete answer to this question myself, since this is useful in a number of situations. From what I know, contrary to estimates for the kernel which are a quite general phenomenon (see the classical book by Davies for your case), estimates for the derivatives are much more subtle and are connected with several properties of the generator such as <span class="math-container">$L^p$</span> boundedness for the associated Riesz operator when <span class="math-container">$p>2$</span> (while when <span class="math-container">$p<2$</span> the estimates for the derivatives are not needed).</p>
<p>The paper "On second-order periodic elliptic operators in divergence form" by A.F.M. ter Elst, Derek W. Robinson and Adam Sikora gives some info in the case of elliptic operators, and on manifold this is studied in a 2004 paper by Auscher, Coulhon, Duong and Hofmann (Riesz transform on manifolds and heat kernel regularity) which should be on arXiv.</p>
|
1,398,956 | <p>I saw from literature that the expected value of a random variable $f(X)$ is either $E f(X)$, $E(f(X))$ or $E[f(X)]$. Is there a standard which one notation should one use? Is the expected value a function $f(X)\to\mathbb R$?</p>
| BruceET | 221,800 | <p>It is totally up to the author. Some authors use a different
typeface for the $E$ and they tend to be the ones who avoid
parentheses or brackets: $\mathbf{E}X,$ $\mathbb{E}X,$ $\mathsf{E}X$, and so on (including a script E, which I've forgotten how to make).</p>
<p>It is also common to use $\mu$ when there is only one random
variable under discussion or $\mu_X$ and $\mu_Y$ when there are
several. This make is convenient to write expressions such as
$E(X - \mu_X)^2$ without accumulating too many brackets or parentheses.
as in $E\{[X - E(X)]^2\}.$ </p>
<p>As for your second question: A random variable is a function
from the sample space to the real numbers, sometimes written as
$\Omega \stackrel{X}{\rightarrow} \mathbb{R}.$ Moreover, if $f$ is a function
from the real numbers to the real numbers then $f(X)$ is another
random variable. Then we might write $\Omega \stackrel{X}{\rightarrow} \mathbb{R} \stackrel{f}{\rightarrow}\mathbb{R}$ or
$\Omega \stackrel{f(X)}{\rightarrow}\mathbb{R}.$ Expectation itself is not a function (because it yields a constant);
the word 'operator' is often used for the process of producing
that constant.</p>
<p>I see that another Answer by @Karl has appeared with some entirely
different notations for expected values. Outside the mainstream
of mathematical statistics, there are many more such notations.</p>
<p>One potentially confusing notation for math stat people is common in queueing
theory where $L$ can stand for the AVERAGE number of people
in a queueing system, $L_Q$ for the average number of
people waiting to be served, $W_Q$ for the average time
waiting to be served, and so on. The random variables themselves
are seldom mentioned and have a variety of notations other than
capital letters when they are mentioned. </p>
<p>Adding to the diversity is the routine use of small Greek letters (such
as $\xi$ and $\eta$) or small Roman letters (such as $\text{a}$ amd $\text{b}$ for random variables, common in some European and Asian
countries. </p>
|
2,637,337 | <p>$ABC$ is a triangle and $A_1, B_1, C_1$ are points on $BC, CA, AB$ such that $$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\lambda$$</p>
<p>If $A_2, B_2, C_2$ are points on $B_1C_1, C_1A_1$, and $A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1}=\frac{C_1B_2}{B_2A_1}=\frac{A_1C_2}{C_2B_1}=\frac{1}{\lambda}$$</p>
<p>prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of simlitude.</p>
<p>I'm missing something obvious, but I don't know what.</p>
| dezdichado | 152,744 | <p>A more thorough hint: </p>
<p>Use the theorem of sines on triangles $BB_2C_1$ and $BB_2A_1$, to conclude that:
$$\dfrac{\sin\angle B_2BA}{\sin\angle B_2BC} = \dfrac{A_1C}{BC_1}.$$</p>
<p>Then the sine version of Ceva's theorem will let you conclude that $AA_2, BB_2, CC_2$ are concurrent. This will greatly help you prove the $A_2B_2//AB$, for example.</p>
<p>Let me know if you still have trouble after this. </p>
|
2,637,337 | <p>$ABC$ is a triangle and $A_1, B_1, C_1$ are points on $BC, CA, AB$ such that $$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\lambda$$</p>
<p>If $A_2, B_2, C_2$ are points on $B_1C_1, C_1A_1$, and $A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1}=\frac{C_1B_2}{B_2A_1}=\frac{A_1C_2}{C_2B_1}=\frac{1}{\lambda}$$</p>
<p>prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of simlitude.</p>
<p>I'm missing something obvious, but I don't know what.</p>
| Michael Rozenberg | 190,319 | <p>$$\vec{A_2B_2}=\vec{A_2C_1}+\vec{C_1B_2}=\frac{1}{1+\frac{1}{\lambda}}\vec{B_1C_1}+\frac{\frac{1}{\lambda}}{1+\frac{1}{\lambda}}\vec{C_1A_2}=$$
$$=\frac{\lambda}{1+\lambda}\vec{B_1C_1}+\frac{1}{1+\lambda}\vec{C_1A_2}=\frac{\lambda}{1+\lambda}\left(\vec{B_1A}+\vec{AC_1}\right)+\frac{1}{1+\lambda}\left(\vec{C_1B}+\vec{BA_1}\right)=$$
$$=\frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{CA}+\frac{\lambda}{1+\lambda}\vec{AB}\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{AB}+\frac{\lambda}{1+\lambda}\vec{BC}\right)=$$
$$=\frac{\lambda}{(1+\lambda)^2}\left(\vec{BC}+\vec{CA}\right)+\frac{1+\lambda^2}{(1+\lambda)^2}\vec{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}\vec{AB},$$
which says $A_2B_2||AB.$</p>
<p>By the same way we can prove that $A_2C_2||AC$ and $B_2C_2||BC,$</p>
<p>which says that $$\Delta ABC\sim\Delta A_2B_2C_2$$ and
$$\frac{A_2B_2}{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}.$$</p>
|
325,588 | <p>Is there an analytic function $f$ in $\mathbb{C}\backslash \{0\}$ s.t. for every $z\ne0$: $$|f(z)|\ge\frac{1}{\sqrt{|z|}}\, ?$$</p>
| Hagen von Eitzen | 39,174 | <p>If $0$ is an essential singuarity of $f$, then by the <a href="http://en.wikipedia.org/wiki/Picard_theorem" rel="nofollow">Big Picard theorem</a>, $f(z)$ leaves out at most one value in every punctured neighbourhood of $0$.
By assumption, $f(0)\ne 0$ for all $z$, hence we find $z$ with $0<|z|<1$ and $f(z)=1$, so for this $z$ we have $|f(z)|<\frac1{\sqrt{|z|}}$.</p>
<p>Therefore $f$ cannot have an essential singularity at $0$, and (as clearly $f\ne0$) it can be written as $f(z)=z^kg(z)$ where $k\in\mathbb Z$ and $g$ is an entire function with $g(0)\ne0$.</p>
<p>If $k\ge0$, then $f$ is bounded in a neighbourhood of $0$, say $|f(z)|<M$ for $|z|<\epsilon$. If additionally $|z|<\frac1{M^2}$, we obtain $|f(z)|<\frac1{\sqrt{|z|}}$.</p>
<p>If $k<0$, then $g(0)\ne 0$ implies $|g(z)|>a$ for $|z|<\epsilon$ for some $\epsilon>0$, $a>0$.
By assumption, $|g(z)|\ge|z|^{-k-1/2}$ for $z\ne0$, hence $|g(z)|\ge \epsilon^{-k-1/2}$ for $|z|\ge \epsilon$.
This shows that $z\mapsto \frac1{g(z)}$ is an entire function that is bounded by $\max\{a^{-1},\epsilon^{k+1/2}\} $, hence constant.
Then $f(z)=cz^{k}$ for some $c\ne 0$.
As soon as $|z|>|c|^{-k-1/2}$, we get $|f(z)|<\frac1{\sqrt{|z|}}$.</p>
<p>In all cases, we exhibited some $z$ with $|f(z)|<\frac1{\sqrt{|z|}}$, hence no analytic function $\mathbb C^\times\to\mathbb C$ with $|f(z)|\ge\frac1{\sqrt{|z|}}$ for all $z\ne 0$ can exist.</p>
|
1,992,256 | <p>I have to prove</p>
<p>$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le \frac{2}{3}*(n+1)\sqrt{n+1}$</p>
<p>by using math induction. </p>
<p>First step is to prove that it works for n = 1 , which is true.
Next step is to prove it for n + 1. We can rewrite the formula using</p>
<p>$\sum_{i=1}^{n+1} \sqrt{i}= \sum_{i=1}^{n}\sqrt{i} + \sqrt{i+1}$</p>
<p>and we can substitute sum </p>
<p>$\frac{2}{3}(n+1)\sqrt{n+1} +\sqrt{n+1} \le \frac{2}{3}(n+2)\sqrt{n+2}$</p>
<p>we can transform the left side into </p>
<p>$\sqrt{n+1}(\frac{2}{3}(n+1)+1)$</p>
<p>but how to I further transform the formula in order to find if the sentence is true?</p>
<p>Thanks for all help!</p>
| Martin Sleziak | 8,297 | <p>To finish your proof by induction we could use the fact the following inequalities are equivalent to each other
\begin{align*}
\frac23(n+1)^{3/2}+\sqrt{n+1} &\le \frac23(n+2)^{3/2}\\
\frac32\sqrt{n+1} &\le (n+2)^{3/2} - (n+1)^{3/2}\\
\frac32\sqrt{n+1} &\le (\sqrt{n+2} - \sqrt{n+1}) (n+2 + \sqrt{(n+2)(n+1)} + n+1)\\
\frac32\sqrt{n+1}(\sqrt{n+2} + \sqrt{n+1}) &\le n+2 + \sqrt{(n+2)(n+1)} + n+1\\
\frac32(n+1 + \sqrt{(n+2)(n+1)}) &\le 2n+3 + \sqrt{(n+2)(n+1)}\\
\frac12\sqrt{(n+2)(n+1)} &\le \frac{n+3}2
\end{align*}
I have:<br>
Used $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=\sqrt{n+2}$ $b=\sqrt{n+1}$<br>
Multiplied both sides by $\sqrt{n+2}+\sqrt{n+1}$ to use that $(\sqrt{n+2} - \sqrt{n+1})(\sqrt{n+2} + \sqrt{n+1}) = (n+2)-(n+1)=1$.</p>
<hr>
<p>If you already know how to integrate, you can simply use that
$$\newcommand{\dx}{\; \mathrm{d}x}\sqrt k \le \int_{k}^{k+1} \sqrt x \dx.$$
Simply by adding these inequalities together for $k=0$ to $n$ you get
$$ \sum_{k=0}^n \sqrt n \le \int_0^{n+1} \sqrt x \dx = \frac{(n+1)^{3/2}}{3/2} = \frac23 (n+1)\sqrt{n+1}.$$</p>
<p>This can be also visualized by noticing that one side is are of rectangle under the curve $y=\sqrt x$ and the other is the whole area under the curve.</p>
<p>I will add the following - somewhat similar - picture which might help (it was taken from <a href="https://math.stackexchange.com/questions/1496171/how-do-i-find-the-upper-and-lower-sum-of-the-area-of-the-graph-by-summation">this post</a>):</p>
<p><a href="https://i.stack.imgur.com/CnHQF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CnHQF.png" alt="lower and upper integral sum for sqrt(x)"></a></p>
|
252,820 | <p><code>Sound[]</code> generates a visual representation of notes. I would like to extract that image, only notes, without controls and borders. How can I do it?</p>
<p>Take this example</p>
<pre><code>Sound[SoundNote @@@ Transpose @ {
{"E5", "D5", "F#4", "G#4", "C#5", "B4", "D4", "E4", "B4", "A4", "C#4", "E4", "A4"},
{1/8, 1/8, 1/4, 1/4, 1/8, 1/8, 1/4, 1/4, 1/8, 1/8, 1/4, 1/4, 1/2 }
}]
</code></pre>
<p><a href="https://i.stack.imgur.com/v3qNP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/v3qNP.png" alt="enter image description here" /></a></p>
<p>So far, I found that I can <code>Rasterize</code> the whole output.
Also, if I press <code>Ctrl Shift E</code> on the output it seems like there is something, but the code is not simple and very low level. I'm not sure how to translate that into graphics primitives. This method requires manual intervention anyway, and I would like to get the image automatically.</p>
| LouisB | 22,158 | <p>This example returns a <code>SparseArray</code>. Use <code>Normal</code> to make it a nested list, if you prefer.</p>
<pre><code>mat = Block[{b, n = 6},
b = Riffle[ConstantArray[1, n/2], 0];
SparseArray[{Band[{1, 2}] -> b, Band[{2, 1}] -> -b}, {n, n}]
];
MatrixForm @ mat
</code></pre>
<p><span class="math-container">$$\left(
\begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & -1 & 0 \\
\end{array}
\right)$$</span></p>
<p>Alternatively,</p>
<pre><code>mat = With[{n = 6},
KroneckerProduct[IdentityMatrix[n/2], {{0, 1}, {-1, 0}}]]
</code></pre>
|
2,804,495 | <p>I was asked to solve this double integral:
Compute the area between $y=2x^2$ and $y=x^2$ and the hyperbolae $xy=1$ and $xy=2$ in </p>
<p>$$ \iint dx \,dy$$</p>
<p>I tried to solve it starting with considering that </p>
<p>$$x^2 \leq y \leq 2x^2 $$</p>
<p>suitabile for integration interval in $y$, obtaining the incomplete form</p>
<p>$$ \int^{x^2}_{2x^2} \int_\ldots^\ldots dx \,dy$$</p>
<p>but I also have
$$1 \leq xy \leq 2$$ and I would obtain a result in which I still have one independent variabile. </p>
<p>Please, can anyone help me? Thanks in advance.</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Let consider the change of variables</p>
<ul>
<li>$u=x^2\implies 1\le u\le 2$</li>
<li>$v=xy\implies 1\le v\le 2$</li>
</ul>
<p>and</p>
<p>$$dudv=|J|dxdy=\begin{vmatrix}2x&0\\y&x\end{vmatrix}dxdy=2x^2dxdy\implies dxdy=\frac1{2u}dudv$$</p>
|
2,995,327 | <p>Suppose a,b ∈ Z. If 4 | <span class="math-container">$(a^2 + b^2)$</span> then a and b are not both odd.</p>
<p>So, assuming that 4 | <span class="math-container">$(a^2 + b^2)$</span> and <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are odd</p>
<p>this gives <span class="math-container">$4k=(2l+1)^2+(2u+1)^2$</span> for some <span class="math-container">$k,l,u\in z$</span></p>
<p>eventually leading to <span class="math-container">$4k=4(l^2+l+u)+2(u^2+1)$</span></p>
<p>The RHS is not a multiple of 4 when <span class="math-container">$u=2$</span> contradiction.</p>
<p>Is this valid, thanks.</p>
| Nodt Greenish | 614,664 | <p>If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.</p>
|
1,199,304 | <p>Let $M\neq \{0\}$ be a semi-simple left $R$ module .Prove that it contains a simple sub-module.</p>
<p>An $R-$ module $M$ is said to be semi-simple if every submodule of $M$ is a direct summand of M
<strong>My solution</strong></p>
<p>Since $M\neq \{0\}$; $\exists m\in M$ such that $m\neq 0$.Then I can consider the left $R-$ module $Rm$ ;By hypothesis it is a direct summand of $M$.Thus $M=N+Rm$ where $N$ is a sub module of $M$ </p>
<p>How to proceed next?</p>
| egreg | 62,967 | <p>First show that every submodule of a semisimple module is also semisimple.</p>
<p>Let $M$ be semisimple, $N$ a submodule of $M$ and $L$ a submodule of $N$. Since $M$ is semisimple, we have $N\oplus N'=M$ and also $L\oplus L'=M$. Consider then $L''=L'\cap N$. Then $L\cap N=L\cap L'\cap N=\{0\}$. If $x\in N$, we have $x=y+z$, with $y\in L$ and $z\in L'$. But $z=x-y$, so $z\in L'\cap N=L''$. Therefore $L\oplus L''=N$.</p>
<p>So, let $M\ne0$ be semisimple and take $x\ne0$, $x\in M$. Then $Rx$ is semisimple and it's sufficient to show that $Rx$ has a simple submodule.</p>
<p>Consider $I=\operatorname{Ann}_R(x)=\{r\in R:rx=0\}$. Then $Rx\cong R/I$ and $I\ne R$. Let $M$ be a maximal left ideal containing $I$; then $L=Mx$ is a direct summand of $Rx$ and $L/Mx\cong (R/I)/(M/I)\cong R/M$ is simple, so a complement of $L$ is a simple submodule of $Rx$.</p>
|
2,545,516 | <p>So I have to assess the convergence of $$\displaystyle\sum_{n=1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right).$$</p>
<p>I'm told that it diverges, but can't really see why.</p>
<p>The divergence test doesn't really help, because
$\lim\limits_{x\to\infty}\displaystyle\frac{1}{\sqrt{n}}=0$, so</p>
<p>$\lim\limits_{x\to\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right)=0$, which doesn't conclude its divergence.</p>
<p>I doubt the ratio test would be much of use in this situation.</p>
<p>I can't imagine using the integral comparison test, as I wouldn't know where to start with $\displaystyle\int_{1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right) \mathrm dx$.</p>
| qualcuno | 362,866 | <p>You can do a limit comparison with <span class="math-container">$\displaystyle\sum\limits_{n\geq1}\dfrac{1}{\sqrt{n}}$</span>, since</p>
<p><span class="math-container">$$
\lim_{n\to\infty}\sqrt{n}\sin\left(\frac{1}{\sqrt{n}}\right) = 1
$$</span></p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Mozibur Ullah | 35,706 | <p>Personally, I think the best way to illustrate curvature is to start from the simplest case. This is how Hilbert illustrated it in his book, <em>Geometry and the Imagination</em>.</p>
<blockquote>
<ol>
<li>The simplest curve is the the straight line and its curvature obviously should be zero.</li>
</ol>
</blockquote>
<blockquote>
<ol start="2">
<li>The next simplest curve is the circle. Since the only invariant of a circle is the radius, the curvature must involve this. The larger the circle, the less curved it is. The simplest way to describe this is to say that curvature is the reciprocal of the radius.</li>
</ol>
</blockquote>
<blockquote>
<ol start="3">
<li>For any other curve we measure its curvature by fitting a circle to it and calling its radius, the radius of curvature.</li>
</ol>
</blockquote>
<p>Of course this leaves unexplained how to actually find this fitting circle. Both Newton and Liebniz provided different but equivalent method: Basically, take two points on either side of the chosen point and draw the tangent there. Then draw the normals. Where they intersect is the approximate centre of this circle. Then by taking the limits of both points approaching the chosen point we find the true centre and hence radius of curvature.</p>
<p>I like this because it directly connects our intuitive notion of curvature to its formalisation. Hilbert doesn't describe how to generalise this to higher dimensions, but presumably it's a question of fitting ellipsoids...</p>
<p>For example, on a surface, we would get a fitting ellipse whose major and semi-major axes will be the principal axes of curvature. Gauss then showed that the product of these the two radii, the Gaussian curvature, was independent of the embedding of the surface in Euclidean space - his <em>theorem egregium</em>, his 'remarkable theorem'. This inaugurated the era of <em>intrinsic</em> geometry as opposed to <em>extrinsic</em> geometry where we consider a geometrical object embedded in another.</p>
<p>It's probably worthwhile to add that this product is - upto a constant of proportionality - the area of the bounding rectangle. And this hints at the usefulness of tensors in dealing with curvature as tensors geometrically represent volumes (that is, upto rotations and shears).</p>
<p>Of course it is a mistake to think intrinsic geometry is the final word on geometry. After all, the theory of bundles (and of cobundles), as used in gauge theory and homology/cohomology shows that it is equally important to keep the extrinsic view in mind.</p>
<p>More practically, we can see this with the Möbius strip. Intrinsically, it can only have two twists; the trivial one and the half-twist; extrinsically, there is a natural numbers worth of full twists and of half twists. This is important in the theory of spin in physics.</p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Nitin Nitsure | 148,928 | <p>Curvature can be very easily pictured using `geodesic quadrilateral gaps',
which can be more generally used to recover the torsion tensor, and if the torsion is
identically zero, then the curvature tensor, for a manifold equipped with an affine connection.</p>
<p>In the special case of an oriented Riemannian surface <span class="math-container">$(M,g)$</span> with its Riemannian connection <span class="math-container">$\nabla$</span>,
this works as follows to pictorially give us
the Gaussian curvature <span class="math-container">$\kappa(P_0)$</span> at any point <span class="math-container">$P_0 \in M$</span>.
Travel along a geodesic from <span class="math-container">$P_0$</span> in the starting direction given by a unit vector <span class="math-container">$u\in T_{P_0}M$</span>,
and take the point <span class="math-container">$P_1$</span> on it at a small distance
<span class="math-container">$s$</span> from <span class="math-container">$P_0$</span>. Turn left in <span class="math-container">$90$</span> degrees, and follow the geodesic in that direction for
the same distance <span class="math-container">$s$</span> to
arrive at a point <span class="math-container">$P_2$</span>. Iterate the left turn and the travel along the geodesic for
distance <span class="math-container">$s$</span> twice more, to successively arrive at points <span class="math-container">$P_3$</span> and <span class="math-container">$P_4$</span>.
If the surface was flat, and <span class="math-container">$s$</span> small enough, then we would have
traveled along a closed geodesic quadrilateral and arrived back at the starting point, that is,
<span class="math-container">$P_4 = P_0$</span>. But if the curvature is non zero, then the vector <span class="math-container">$P_4 - P_0$</span> (which you can
define in terms of a local smooth embedding of <span class="math-container">$M$</span> in a higher dimensional vector space)
is non-zero, and satisfies the following formula. Let <span class="math-container">$v\in T_{P_0}M$</span> be the
vector such that <span class="math-container">$(u,v)$</span> is a right-handed
orthonormal basis for <span class="math-container">$T_{P_0}M$</span>. Then
<span class="math-container">$$\lim_{s\to 0}\, {P_4 - P_0 \over s^3} = {\kappa(P_0) \over 2}(u - v)$$</span></p>
<p>More generally, let there be give a pair <span class="math-container">$(M,\nabla)$</span>
where <span class="math-container">$M$</span> is a smooth manifold and <span class="math-container">$\nabla$</span> is a connection on <span class="math-container">$TM$</span>.
Consider any <span class="math-container">$P\in M$</span> and a pair of vectors <span class="math-container">$u,v \in T_PM$</span>. From the triple
<span class="math-container">$(P,u,v)$</span> and a small real number <span class="math-container">$s$</span>, we can make a new triple <span class="math-container">$(P',u',v')$</span>
as follows. Take the geodesic from <span class="math-container">$P$</span> with starting tangent vector <span class="math-container">$u$</span>,
and let <span class="math-container">$P'$</span> be the point on it where the affine parameter takes the value <span class="math-container">$s$</span>
(where the parameter has value <span class="math-container">$0$</span> at <span class="math-container">$P$</span>).
Let <span class="math-container">$u',v' \in T_{P'}M$</span> where <span class="math-container">$u'$</span> is parallel transport of <span class="math-container">$v$</span> and <span class="math-container">$v'$</span> is <span class="math-container">$(-1)$</span>-times
the parallel transport of <span class="math-container">$u$</span> along this geodesic.
Starting with a triple <span class="math-container">$(P,u,v)$</span> for which <span class="math-container">$P = P_0$</span>, and iterating the
above, we get an open geodesic quadrilateral with vertices <span class="math-container">$P_0$</span>, <span class="math-container">$P_1 = (P_0)'$</span>,
<span class="math-container">$P_2 = (P_1)'$</span>, <span class="math-container">$P_3 = (P_2)'$</span> and <span class="math-container">$P_4= (P_3)'$</span>.
The quadrilateral is closed if <span class="math-container">$P_4 = P_0$</span>. But in general, we have the formula</p>
<p><span class="math-container">$$\lim_{s\to 0}\,{P_4 - P_0\over s^2} = - T(u,v)$$</span></p>
<p>where <span class="math-container">$T(u,v) = \nabla_uv - \nabla_vu - [u,v]$</span> is the torsion tensor.
If the torsion tensor <span class="math-container">$T$</span> is identically zero on <span class="math-container">$M$</span>, then the gap <span class="math-container">$P_4 - P_0$</span> is given
in terms of the Riemann curvature tensor by the formula</p>
<p><span class="math-container">$$\lim_{s\to 0}\,{P_4 - P_0\over s^3} = {1\over 2}R(u,v)(u+v)$$</span></p>
<p>where by definition
<span class="math-container">$R(u,v)(w) = \nabla_u\nabla_vw - \nabla_v\nabla_uw - \nabla_{[u,v]}w$</span>.
The above formula can be `inverted' to recover the curvature tensor
when the torsion is identically zero, as the tensor <span class="math-container">$R(u,v)(w)$</span> can be recovered uniquely from
the tensor <span class="math-container">$R(u,v)(u+v)$</span> using the symmetries of <span class="math-container">$R(u,v)(w)$</span>.</p>
<p>The above results are proved in arXiv:1910.06615, which is written in an expository style.</p>
|
275,539 | <p>Kind of leading on from my other question, how would I solve for $i$? Or how would I check that it is possible to have such an $i$?</p>
<p>First I had to check for all $2^i$ and clearly this doesn't happen as all $2^i$ are even and so I will just get even $x's$ such that $2^i \equiv x \mod 28$. So the next one I go onto is $3$.</p>
<p>Now how do I go about doing this? </p>
| lab bhattacharjee | 33,337 | <p><a href="http://mathworld.wolfram.com/CarmichaelFunction.html" rel="nofollow">Carmichael Function</a> is more useful than <a href="http://mathworld.wolfram.com/TotientFunction.html" rel="nofollow">Totient Function</a>, while dealing with composite numbers like $28$.</p>
<p>For
$\phi(28)=\phi(7)\cdot\phi(4)=6\cdot2=12\implies 3^{12}\equiv1\pmod{28}$</p>
<p>$\lambda(28)=lcm(\lambda(7),\lambda(4))=lcm(6,2)=6\implies 3^6\equiv1\pmod{28}$</p>
<p>So, if $ord_{28}3=d,d\mid 6\implies d $ can be $1,2,3$ or $6$</p>
<p>$3^1=3\not\equiv 1\pmod{28},3^2=9\not\equiv 1\pmod{28},3^3=27\not\equiv 1\pmod{28}\implies ord_{28}3=6$</p>
|
1,368,988 | <p>I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.</p>
<p><img src="https://i.stack.imgur.com/vueMQ.png" alt="An example for n=8."></p>
<p>Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$</p>
<p>We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$</p>
<p>Now for my question: <strong>How would you solve the opposite problem?</strong> To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?</p>
| Bhaskara-III | 246,676 | <blockquote>
<p>Just recall double angle trig. identity: $\cos 2x=1-2\sin^2x$,</p>
</blockquote>
<p>Starting from,
$$\begin{align}
\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(2\pi/n\right)}\\=\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2+4\sin^2\left(\pi/n\right)}\\
=\lim_{n\to \infty}\frac{n}{2}\cdot 2\sin\left(\pi/n\right)\\
=\pi\cdot \lim_{n\to \infty}\frac{\sin(\pi/n)}{\pi/n}\\
=\pi\cdot \lim_{t\to 0}\frac{\sin(t)}{t}\\
=\pi\cdot 1\\=\pi
\end{align}$$</p>
|
103,675 | <p>I have defined a recursive sequence</p>
<pre><code>a[0] := 1
a[n_] := Sqrt[3] + 1/2 a[n - 1]
</code></pre>
<p>because I want to calculate the <code>Limit</code> for this sequence when n tends towards infinity.</p>
<p>Unfortunately I get a <code>recursion exceeded</code> error when doing:</p>
<pre><code>Limit[a[n], n -> Infinity]
</code></pre>
<p>How can I calculate the <code>Limit</code> for this sequence using Mathematica?</p>
| bbgodfrey | 1,063 | <p>Alternatively, and perhaps more directly, use</p>
<pre><code>RSolve[{a[n] == Sqrt[3] + 1/2 a[n - 1], a[0] == 1}, a[n], n]
(* {{a[n] -> 2^-n (1 - 2 Sqrt[3] + 2^(1 + n) Sqrt[3])}} *)
Limit[a[n] /. %[[1]], n -> Infinity]
(* 2 Sqrt[3] *)
</code></pre>
|
1,903,473 | <p>Given three permutations $p_1,p_2,p_3$ of $\{1,2,\ldots,n^3+1\}$, prove that two of them have a common subsequence of length $n+1$.</p>
<p>I have tried to solve this using the pigenhole principle but I didnt progress too much, any help would be appreciated</p>
<p>edit: when I say subsequence I mean that there are $1<=r<q<=3$
and $1<=i(1)<i(2)<...<i(n+1)<=n^3+1$ and $1<=j(1)<j(2)<...<j(n+1)<=n^3+1$ so that $p_r(i1)=p_q(j1)$, $p_r(i2)=p_q(j2)$ and so on, not necessarily consecutive</p>
| openspace | 243,510 | <p>That's not true.
Consider three permutations :
$$p_1 = (1, \dots, n^3+ 1)$$
$$p_2 = (n^3+1, \dots, 1)$$
$$p_3 = (1, n^3, 3, n^3-2, \dots)$$
where $p_3[i] = p_{i \bmod 2}[i]$</p>
|
1,903,473 | <p>Given three permutations $p_1,p_2,p_3$ of $\{1,2,\ldots,n^3+1\}$, prove that two of them have a common subsequence of length $n+1$.</p>
<p>I have tried to solve this using the pigenhole principle but I didnt progress too much, any help would be appreciated</p>
<p>edit: when I say subsequence I mean that there are $1<=r<q<=3$
and $1<=i(1)<i(2)<...<i(n+1)<=n^3+1$ and $1<=j(1)<j(2)<...<j(n+1)<=n^3+1$ so that $p_r(i1)=p_q(j1)$, $p_r(i2)=p_q(j2)$ and so on, not necessarily consecutive</p>
| arghbleargh | 362,140 | <p>You can prove this with ideas that are used in proving the Erdos-Szekeres theorem. Without loss of generality, we can assume $p_1$ is the identity permutation (otherwise just relabel numbers). Then, if either of $p_2$ or $p_3$ has an increasing subsequence of length $n + 1$, we are done. So assume their longest increasing subsequences are both at most $n$.</p>
<p>For each $k \in \{1, 2, \ldots , n^3 + 1\}$, define $f_i(k)$ to be the length of the longest increasing subsequence of $p_i$ whose last term is $k$. Our assumption says that $1 \le f_i(k) \le n$.</p>
<p>Note that there are only $n^2$ possible values for the pair $(f_2(k), f_3(k))$. Thus, by the pigeonhole principle, there must exist $n + 1$ integers</p>
<p>$$k_1 > k_2 > \cdots > k_{n + 1}$$</p>
<p>such that $(f_2(k_i), f_3(k_i)) = (f_2(k_j), f_3(k_j))$ for all $i$ and $j$. </p>
<p>We claim now that $(k_1, k_2, \ldots , k_{n + 1})$ is a subsequence of both $p_2$ and $p_3$. Indeed, suppose instead that for some $i$, $k_i$ appears after $k_{i + 1}$ in permutation $p_2$. Then, any increasing subsequence ending at $k_{i + 1}$ can be extended one longer by adding $k_i$, which contradicts the equality $f_2(k_i) = f_2(k_{i + 1})$. The same argument applies for $p_3$, so we have obtained our common subsequence of length $n + 1$.</p>
|
3,497,420 | <p>Consider the function <span class="math-container">$$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$</span> The point <span class="math-container">$(0,0)$</span> is a critical point. Observe,
<span class="math-container">\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}</span></p>
<p>So, in order to determine the nature of the above critical point, we need to check the Hessian at <span class="math-container">$(0,0)$</span> which is <span class="math-container">$0$</span> and hence the test is inconclusive. <span class="math-container">$$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$</span>So, I tried to see the function on slices like <span class="math-container">$y=0$</span> and <span class="math-container">$y=x$</span> but nothing worked. So please suggest me how do I find the nature of the critical point in this case?</p>
| bjorn93 | 570,684 | <p>You have <span class="math-container">$f(0,0)=0$</span>. You can find both positive and negative values of <span class="math-container">$f(x,y)$</span> in any region around <span class="math-container">$(0,0)$</span>, which means that you don't have a local extremum at this point:
<span class="math-container">$$f(x,x^3)= -x^5(1-x)(2-x)<0\quad\text{for}\,\, |x|<1, x\ne 0$$</span>
and
<span class="math-container">$$f(x,0)=x^6>0\quad\text{for}\,\, x\ne 0 $$</span></p>
|
2,235,427 | <p>I am trying to characterize the continuum $\mathcal{C}$ using only the notion of midpoint, i.e. the operation $\mu : \mathcal{C}\times\mathcal{C} \to \mathcal{C}$ assigning to each pair of points the midpoint between them. I thought of defining “unbounded interval” as a set $I \subseteq \mathcal{C}$ such that $I$ as well as $\mathcal{C}\backslash I$ are closed under $\mu$. But is that even true? </p>
<p>My thoughts so far: If $I$ has the above property but is not an interval, then both $I$ and $\mathcal{C}\backslash I$ are dense in $\mathcal{C}$ and noncountable (it is easy to find an injection from one into the other)...</p>
<p>Note: The more general question whether it is possible to define "interval" or order in the above setting has now been answered <a href="https://math.stackexchange.com/questions/2235490/characterizing-the-continuum-using-only-the-notion-of-midpoint">here (in the comment by Matt F.)</a> and the answer is no.</p>
| Misha Lavrov | 383,078 | <p>Working with the usual definition of $\mathbb R$ (and assuming choice), we can define a set $I$ that's not an unbounded interval, but for which both $I$ and $\mathbb R \setminus I$ are closed under taking midpoints. (I use $\mathbb R$ to refer to the usual notion of realnumberness to distinguish from $\mathcal C$ which you want to build from the ground up.)</p>
<p>Choose a Hamel basis for $\mathbb R$ as a vector space over $\mathbb Q$, and let $b$ be an element of that basis. Then define $I$ to be the set of all elements of $\mathbb R$ so that, when represented in this basis, the coefficient of $b$ is nonnegative.</p>
<p>If we do this, then both $I$ and $\mathbb R\setminus I$ are closed under midpoint. If $x, y \in \mathbb R$ and have $b$-coefficients $x_b, y_b$, then $\frac{x+y}{2}$ has $b$-coefficient $\frac{x_b+y_b}{2}$. So if $x_b, y_b \ge 0$, then $\frac{x_b+y_b}{2} \ge 0$, and if $x_b, y_b < 0$, then $\frac{x_b + y_b}{2} <0$.</p>
<p>It's also worth noting that if $I$ and $\mathbb R\setminus I$ are closed under midpoints, and $I$ is bounded from above, <em>then</em> $I$ is an unbounded interval. Let $x = \sup I$. For any $\epsilon>0$, there is some $y \in I$ with $y > x-\epsilon$. If we now take $z < x-2\epsilon$, then $2y-z > x$, so $2y-z \notin I$. So we must have $z \in I$, or else $\frac{z + (2y-z)}{2} = y \notin I$ would be a contradiction. Therefore $(-\infty, x-2\epsilon) \subseteq I$ for all $\epsilon$, and therefore their union, $(-\infty, x)$, is contained in $I$. So $I$ is either $(-\infty,x]$ or $(-\infty,x)$ depending on whether $x \in I$ or not.</p>
|
2,235,427 | <p>I am trying to characterize the continuum $\mathcal{C}$ using only the notion of midpoint, i.e. the operation $\mu : \mathcal{C}\times\mathcal{C} \to \mathcal{C}$ assigning to each pair of points the midpoint between them. I thought of defining “unbounded interval” as a set $I \subseteq \mathcal{C}$ such that $I$ as well as $\mathcal{C}\backslash I$ are closed under $\mu$. But is that even true? </p>
<p>My thoughts so far: If $I$ has the above property but is not an interval, then both $I$ and $\mathcal{C}\backslash I$ are dense in $\mathcal{C}$ and noncountable (it is easy to find an injection from one into the other)...</p>
<p>Note: The more general question whether it is possible to define "interval" or order in the above setting has now been answered <a href="https://math.stackexchange.com/questions/2235490/characterizing-the-continuum-using-only-the-notion-of-midpoint">here (in the comment by Matt F.)</a> and the answer is no.</p>
| William Balthes | 231,063 | <p>Are you putting metric structure on these midpoints; such as as a weakened version of midpoint convexity.</p>
<p>Or just that the midpoint functions values exist in that appropriates, F(c/2+c1/2) exists in the structure somewhere between the c1 and c2 c1
|
1,572,045 | <p>This is maybe a stupid question, but I want to find the roots of:</p>
<blockquote>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$</p>
</blockquote>
<p>What that I did:</p>
<p>$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$</p>
<p>So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ </p>
<p>My questions:</p>
<p>$1)$ Is there an easy way to see that $x=-8$ is a root too?</p>
<p>$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$</p>
| user236182 | 236,182 | <p>Use the Distributive Property.</p>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2$$</p>
<p>$$=\underbrace{(x+2)(x-1)^2}_{\text{common factor}}\left(2(x-1)\right)-\underbrace{(x+2)(x-1)^2}_{\text{common factor}}(3(x+2))$$</p>
<p>$$=(x+2)(x-1)^2\left(2(x-1)-3(x+2)\right)$$
$$=(x+2)(x-1)^2(-(x+8))$$</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| mounir ben salem | 244,326 | <p>you can use: $$ \cos(x)= \sqrt{1-\sin(x)^{2}}$$ $$ \text{then let}\qquad \sin(x)=t\\
\sqrt{1-t^{2}}+t-1=0$$ now you can continue</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| Bernard | 202,857 | <p>$\cos x+\sin x=\sqrt2\cos\Bigl(x-\dfrac\pi4\Bigr)$, hence the equation is equivalent to:
$$\cos\Bigl(x-\frac\pi4\Bigr)=\frac1{\sqrt2}\iff x-\frac\pi4\equiv\pm\frac\pi4\mod 2\pi\iff x\equiv 0,\,\frac\pi2\mod2\pi.$$</p>
|
713,521 | <p>There are so many notations for differentiation. Some of them are:
$$
f^\prime(x) \qquad \frac{d}{dx}(f(x))\qquad \frac{dy}{dx}\qquad \frac{df}{dx}\qquad
D f(x)\qquad y^\prime\qquad D_x f(x)
$$
Why are there so many ways to say "the derivative of $f(x)$"? Is there a specific use for each notation? What is the difference between $\dfrac{d}{dx}$ and $\dfrac{dy}{dx}$? I am only asking this because I am worried that I might use the wrong notation sometimes. For example, I don't know when I should use $\dfrac{dy}{dx}$ instead of $D_xf(x)$, or vice versa. I thank you in advance for your answers.</p>
| Slade | 33,433 | <p>For the most part, the things you've written are equivalent, and the reason there are so many is partly historical, partly practical (e.g. $D_x$ is better notation when one is using the language of operators or partial derivatives, $y'$ saves space when it's unambiguous, etc.).</p>
<p>But there are two really huge points here. First of all, the difference between $\frac{df}{dx}$ and $\frac{dy}{dx}$. It's very important not to confuse these—put simply: $f$ and $y$ are different letters! They might mean the same thing in certain problems, but they might not—e.g. $y=f(x)$ is a curve in $\mathbb{R}^2$, where $f$ and $y$ mean mostly the same thing, $z=f(x,y)$ is a surface in $\mathbb{R}^3$, where $f$ and $y$ mean totally different things. Usually, $f$ denotes a function $f(x)$, and $y$ denotes a coordinate, but you should always look at the notation of the specific problem before you make assumptions—you should be able to handle a question about $(q,w)$-plane instead of the $(x,y)$ plane, without getting confused.</p>
<p>The other point is the difference between $\frac{d}{dx}$ and $\frac{df}{dx}$. It's also very important not to confuse these! $\frac{df}{dx}$ is the derivative of $f$ with respect to $x$, and it's a function of $x$. $\frac{d}{dx}$ is just the derivative with respect to $x$, and it's not a function at all—it eats functions and spits out their derivatives: $\frac{d}{dx} (x^3+3x) = 3x^2+3$, $\frac{d}{dx} (e^y+f(x)) = e^y \frac{dy}{dx} + f'(x)$, and so on. When we write $\frac{dy}{dx}$, we just mean $\frac{d}{dx} (y)$. Keep these concepts separate—this is the same as the comparison $+1$ versus $y+1$, or $\sqrt{\phantom{1}}$ versus $\sqrt{y}$.</p>
|
846,108 | <p>How do you solve this equation: $2x+8=6x-12$ by using the guess and check method?</p>
<p>I divide $2x+8$ and I get $4$ then I divide $6x-12$ and I get $-2$ but I don't know what to do next or is it wrong?</p>
| Joel | 85,072 | <p>Your instructor wants you to try plugging in a bunch of numbers to guess the correct value. Just by eyeballing this, we might try 4 or 5. If you plug in $5$ on the left you get $$2(5)+8=18$$ and on the right you get $$6(5)-12 = 30 - 12 = 18.$$ This tells us that $x=5$ is a solution to this equation.</p>
<p>Remember that the old methods still work here. So if you add 12 to both sides and subtract $2x$ from both sides you get $$20=4x$$ and then you can solve for $x$ by dividing both sides by 4.</p>
|
2,353,142 | <p>Solve:
$$(\cot^{-1} (x))^2 - 4\cot^{-1} (x) + 3 \geq 0$$.</p>
<p>My Attempt:
$$(\cot^{-1} (x))^2 - 4\cot^{-1} (x) + 3 \geq 0$$.
Let $\cot^{-1} (x)=t$. then</p>
<p>$$t^2-4t+3\geq 0$$
$$(t-3)(t-1)\geq 0$$
Either, $\cot^{-1} (x) \leq 1$</p>
<p>Or, $\cot^{-1} (x) \geq 3$</p>
<p>I solved till here, but couldn't get the answer given in book. The answer in book is $x \in (-\infty, \cot (3)] \cup [\cot (3), \infty )$.</p>
| Michael Rozenberg | 190,319 | <p>It's true for $p=2$ only because for real $p$ we need $x>0$ by definition of $x^p$ and our inequality can not hold for all reals $x$ and $y$.</p>
<p>If $p>2$ and odd then $x=y=-1$ gets a contradiction.</p>
<p>If $p>2$ and even then $x=y\rightarrow0^+$ gets a contradiction again. </p>
|
691,494 | <p>Suppose I am given a state space $S=\{0,1,2,3\}$ with transition probability matrix </p>
<p>$\mathbf{P}= \begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3} & 0\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & 0 & 1
\end{bmatrix}$</p>
<p>and I want the expected number of steps from states $0 \rightarrow 3$ which I will denote $E_0(N(3))$.</p>
<p>Attempt at solving: First I write the transient states $\{0,1,2\}$ and recurrent state $\{3\}$ which I got from drawing the chain. I now want to write $\mathbf{P}$ in canonical form, i.e. with state space $S=\{3,0,1,2\}$ as so:</p>
<p>$\mathbf{P}=\begin{bmatrix}
1 & 0 & 0 & 0\\[0.3em]
0 & \frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
0 & \frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
\frac{2}{3} & 0 & 0 & \frac{1}{3}
\end{bmatrix}$</p>
<p>It's clear that the transient matrix is</p>
<p>$\mathbf{Q}=
\begin{bmatrix}
\frac{2}{3} & \frac{1}{3} & 0 \\[0.3em]
\frac{2}{3} & 0 & \frac{1}{3}\\[0.3em]
0 & 0 & \frac{1}{3}
\end{bmatrix}$</p>
<p>Now I can get the matrix I want for computing expected steps (calculated with Mathematica):</p>
<p>$\mathbf{M}=(\mathbf{I}-\mathbf{Q})^{-1}=\begin{bmatrix}
9 & 3 & 3\\[0.3em]
6 & 3 & 3\\[0.3em]
0 & 0 & \frac{3}{2}
\end{bmatrix}$</p>
<p>From this, we get $E_0(N(3))=9+3+3=15$. Is this correct? I am sort of weak in finding the "canonical form" of a matrix.
Note: although this looks like a homework question, it's simply a preparation problem for an upcoming exam, so a complete solution/correction of my work is appreciated.</p>
| Rodrigo Ribeiro | 44,681 | <p>Usually, my approach to this kind of question is to solve a very simple recurrence.</p>
<p>Just look to $ E_{0} N(3)$ but conditioned in each of the two possible first steps to get (I will write $N$ instead of $N(3)$</p>
<p>$$E_0 N = \frac{2}{3}E_0N + \frac{1}{3}E_1N +1$$</p>
<p>The term +1 shows up because once you had walked one step you have to add this step to the account.
(if you want a rigorous proof of this equation, use the simple property of Markov)</p>
<p>If you repeat this process to $E_1N$ and $E_2N$ and replace what you find in the first equation you will find that $E_0N = 39$ :(</p>
<p>Now I'm curious about the results... I made the calculations twice, but it is possible that I forgot something. </p>
|
1,066,484 | <p>Given a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $k\ge 2$. Prove that there is no cut vertex exist in $G$. </p>
<p>Cut vertex $v$ here is a vertex which make the graph induced have number of connected component $>1$ when $v$ is removed.</p>
<p>I have tried to prove by contradiction but i have no clue about what contradiction can be obtained. I am quite curious about what the point the graph has to be bipartite is here. I have not come across any place to adopt the property of a bipartite graph so far. Any hints on tackling this problem would be appreciated. Last but not least, thanks for reading.</p>
<p>Edited: i have found something </p>
| Leen Droogendijk | 95,972 | <p>Hint:</p>
<p>Show that the number of edges in the original graph is a multiple of $k$.</p>
<p>Show that the removal of one vertex removes $k$ edges.</p>
<p>Show that the number of edges in each component (after the removal) still is a multiple of $k$.</p>
<p>Conclude that the original graph must already have been disconnected.</p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| Clement C. | 75,808 | <p>Taylor, first order, would do. $\arctan u = u+o(u)$, $\ln(1+u) = u+o(u)$, so than $\ln(1+\arctan u) = u+o(u)$ when $u\to 0$. So your limit is $e^{\frac{1}{u}\ln(1+\arctan u)} = e^{1+o(1)}$, with $u=\frac{x}{2}$; $$e^{\frac{1}{u}\ln(1+\arctan u)} \xrightarrow[u\to0]{} e^1 = e.$$</p>
<hr>
<p><em>Remark: there may be simpler. But this works, and takes roughly one minute if you know the expansions at $0$ (to very low order, here first) of some standard functions.</em></p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| Paramanand Singh | 72,031 | <p>First and foremost you can get rid of the $x/2$ (it is a pain to type fractions) by putting $t = x/2$ and using the fact that as $x \to 0$ the variable $t \to 0$. We thus need to evaluate $$L = \lim_{t \to 0}(1 + \arctan t)^{1/t}\tag{1}$$ We can now proceed in your manner
\begin{align}
\log L &= \log\left\{\lim_{t \to 0}(1 + \arctan t)^{1/t}\right\}\notag\\
&= \lim_{t \to 0}\log(1 + \arctan t)^{1/t}\text{ (via continuity of log)}\notag\\
&= \lim_{t \to 0}\frac{\log(1 + \arctan t)}{t}\notag\\
&= \lim_{t \to 0}\frac{\log(1 + \arctan t)}{\arctan t}\cdot\frac{\arctan t}{t}\notag\\
&= 1\cdot 1\notag\\
&= 1 \notag
\end{align}
Hence $L = e$.</p>
|
872,889 | <p>What determinant is zero? What equation does this give for the plane?</p>
<p>I need some help here, am pretty stuck</p>
| Steven Stadnicki | 785 | <p>Expanding my comments into an answer: by distributing the divisions by $a_0$, $a_1$, $\ldots$ successively you can rewrite such an upward continued fraction in the equivalent form $\frac1{a_0}+\frac1{a_0a_1}+\frac1{a_0a_1a_2}+\cdots$. This is known as the <em><a href="http://en.wikipedia.org/wiki/Engel_expansion" rel="nofollow">Engel expansion</a></em> of the number, and their coefficients have some interesting limiting properties (in particular, for almost all real numbers the coefficients grow exponentially); the Wikipedia article on them should offer up several good pointers for more information.</p>
|
2,024,997 | <blockquote>
<p>$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$</p>
</blockquote>
<p>I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?</p>
| Caleb Stanford | 68,107 | <p>Since the limit of both the top and bottom is $\infty$ alone, l'hopital's rule gives us
\begin{align*}
\lim_{x \to \infty}\frac{\log_{1.1} x}{x}
&= \lim_{x \to \infty}\frac{\ln x}{(\ln 1.1) x} \\
&= \lim_{x \to \infty}\frac{(1/x)}{(\ln 1.1)} \\
&= 0.
\end{align*}</p>
|
2,024,997 | <blockquote>
<p>$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$</p>
</blockquote>
<p>I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?</p>
| Bernard | 202,857 | <p>Use the definition: $\;\log_{1.1}x=\dfrac{\ln x}{\ln 1.1}$, so
$$\frac{\log_{1.1}x}{x}=\frac 1{\ln 1.1}\dfrac{\ln x}{x}\xrightarrow[x\to+\infty]{}\frac 1{\ln 1.1}\cdot 0=0.$$</p>
|
1,919,912 | <blockquote>
<p>Let $D$ be the Integral Domain with characteristic $m>0$. Prove that $m$ is prime.</p>
</blockquote>
<p>My Proof: </p>
<p>Since the characteristic of $D$ is $m$, $m\cdot b=0$ for all $b\in D$ and if $n\cdot b=0$ for all $b\in D$, then $m\leq n$. </p>
<p>Assume that $m$ is composite number. Then $m=n_1n_2$ where $n_1,n_2>1$.</p>
<p>Let $a\in D$. Then $m\cdot a=0$ i.e., $n_1(n_2\cdot a)=0$. Take $c=n_2\cdot a$, then $n_1\cdot c=0$. Since $a$ is an arbitrary element of $D$, $c$ is also an arbitrary element of $D$. So, we have $n_1\cdot c=0$ for all $c\in D$ and $m> n_1$. This is a contradiction. So, our assumption that $m$ is composite number, is false. Hence $m$ is a prime number.</p>
| Exit path | 161,569 | <p>To simplify things, let $m$ be the characteristic of $D$. Let $m=nk$ be a factorization of $m$. Then $m\cdot 1=(nk)\cdot 1=(n \cdot 1)(k\cdot 1)=0$. Since $D$ is an integral domain, either $n \cdot 1=0$ or $k \cdot 1=0$. But $m$ is the least integer for which $m \cdot 1=0$, implying $n=1$ or $k=1$, so $m$ is prime. </p>
|
2,068,951 | <p>I'm interested in proving the following claim:</p>
<p>There exists a sequence of natural numbers $\left(a_{n}\right)_{n=1}^{\infty}$ such that
$$\lim_{n\to\infty}\left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\frac{1}{2}
$$</p>
<p>I've studied a fair amount of calculus and algebra, yet I've never encountered such a problem before.
How should I approach this claim, or rather what tools should I read about?</p>
<p>Thanks!</p>
| Barry Cipra | 86,747 | <p>Try </p>
<p>$$a_n=\lfloor2^n\ln2\rfloor$$</p>
<p>and use the inequalities $2^n\ln2-1\lt a_n\le2^n\ln2$ in the Squeeze Theorem: Since $1-{1\over2^n}\lt1$, we have</p>
<p>$$\left(1-{1\over2^n}\right)^{2^n\ln2}\le\left(1-{1\over2^n}\right)^{a_n}\lt\left(1-{1\over2^n}\right)^{2^n\ln2-1}$$</p>
<p>The left- and right-hand expressions are easily seen to tend to $(e^{-1})^{\ln2}={1\over e^{\ln2}}={1\over2}$ as $n\to\infty$.</p>
|
671,407 | <p>I have problem with equation: $4^x-3^x=1$. </p>
<p>So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions? </p>
| Blaubaer | 126,764 | <p>Show that $4^x$ grows much faster than $3^x$.</p>
|
3,057,198 | <p><span class="math-container">$$\iint_{G}\!x^2\,\mathrm{d}x\mathrm{d}y$$</span></p>
<p>where <span class="math-container">$G := \left\{(x,y)\in\mathbb{R}^{2}\,;\,|x|+|y| \le 1\right\}$</span></p>
<p>How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).</p>
<p>Oh and also Happy New Year in advance!</p>
| Rafa Budría | 362,604 | <p>You have to solve the inequalities for the four cases:</p>
<p><span class="math-container">$x\geq0\,\land\,y\geq0\,\land x+y\leq1$</span>, the triangle with vertices at <span class="math-container">$(0,0),(0,1),(1,0)$</span>, limited by the function <span class="math-container">$y=1-x$</span></p>
<p><span class="math-container">$x\geq0\,\land\,y<0\,\land x-y\leq1$</span>, the triangle with vertices at <span class="math-container">$(0,0),(0,-1),(1,0)$</span> with <span class="math-container">$y=x-1$</span></p>
<p><span class="math-container">$x<0\,\land\,y\geq0\,\land -x+y\leq1$</span>, the triangle with vertices at <span class="math-container">$(0,0),(0,1),(-1,0)$</span>, with <span class="math-container">$y=1+x$</span></p>
<p><span class="math-container">$x<0\,\land\,y<0\,\land -x-y\leq1$</span>, the triangle with vertices at <span class="math-container">$(0,0),(0,-1),(-1,0)$</span>, with <span class="math-container">$y=-x-1$</span></p>
<p>So, <span class="math-container">$x$</span> runs between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> or <span class="math-container">$-1$</span>. The integral is better solved splitting it in two or four for a clearer calculation.</p>
|
3,057,198 | <p><span class="math-container">$$\iint_{G}\!x^2\,\mathrm{d}x\mathrm{d}y$$</span></p>
<p>where <span class="math-container">$G := \left\{(x,y)\in\mathbb{R}^{2}\,;\,|x|+|y| \le 1\right\}$</span></p>
<p>How does one go about finding the boundaries of these types of integrals? I did fail at searching for examples like this as I don't even know their name(if they do have a specific one).</p>
<p>Oh and also Happy New Year in advance!</p>
| StubbornAtom | 321,264 | <p>Sketch the region <span class="math-container">$G$</span>. </p>
<p><img src="https://i.stack.imgur.com/Z8wxO.jpg" alt="enter image description here"></p>
<p>It is now clear that <span class="math-container">$$G=\left\{(x,y)\in\mathbb R^2: |y|\le 1-|x|\,,\,|x|\le 1\right\}$$</span></p>
<p>So, </p>
<p><span class="math-container">\begin{align}
\iint_G x^2 \,dx\,dy&=\int_{-1}^{1}x^2\int_{-1+|x|}^{1-|x|}\,dy \,dx
\\&=\int_{-1}^1 2x^2 (1-|x|)\,dx
\\&=2\int_0^1 2x^2(1-x)\,dx
\end{align}</span></p>
|
3,031,290 | <p>Can you choose <span class="math-container">$11$</span> different numbers among them so that the numbers <span class="math-container">$|a_1-a_2|, |a_2-a_3|,\ldots,|a_{10}-a_{11}|,|a_{11}-a_{1}|$</span> are all different. The smartest thing that my dumbest mind could accomplish is that all those differences are <span class="math-container">$1,2,3,...,11$</span>. From this, there are two ways - construct an example which is quite painful - but I did it for <span class="math-container">$n=4,5$</span> and tried for <span class="math-container">$6$</span> so I couldn't find example for <span class="math-container">$3$</span>, so may be there something about <span class="math-container">$6$</span> and its multiplicators? another way - prove by something, may be algebra or properties of numbers of <span class="math-container">$1,2,3,\ldots,11$</span> (sum of squares so we could get rid of modulus), try to prove that there always will be at least two equal differences and so on. Can you give me some hint? </p>
| Jam | 161,490 | <p><strong>@Saulspatz</strong> has shown an example but I've found a methodical approach to finding one that could be done by hand in a few minutes to give you an answer in a competition. It could also be easily computerised.</p>
<p>Start by laying out a grid with columns, <span class="math-container">$D=1,2,\ldots11$</span>, representing the particular differences you can have between successive elements in a circular permutation. Then lay out the rows, <span class="math-container">$B=12,11,\ldots1$</span> and entries <span class="math-container">$A$</span>, representing successive pairs of elements which would give you the particular <span class="math-container">$B-A=D$</span>. Then do this again for another <span class="math-container">$B=12,11,\ldots1$</span> but representing the successive pairs of elements that give you <span class="math-container">$B-A=-D$</span>. It should look like the following figure.</p>
<p><a href="https://i.stack.imgur.com/vNT0w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vNT0w.jpg" alt="enter image description here"></a></p>
<p>I'll be highlighting chosen cells in yellow, and highlighting impossible cells in red. Now we can introduce the rules. We want one of each <span class="math-container">$D$</span>, so each column needs exactly <span class="math-container">$1$</span> yellow. No <span class="math-container">$B$</span> can be followed by more than one <span class="math-container">$A$</span> (but must be followed by something) so exactly one yellow per row. No <span class="math-container">$A$</span> can follow more than one <span class="math-container">$B$</span> so once an <span class="math-container">$A$</span> is highlighted, all other equal <span class="math-container">$A$</span>'s are removed from the grid. We are also only picking <span class="math-container">$11$</span> elements so we have to remove all <span class="math-container">$B$</span>'s and <span class="math-container">$A$</span>'s for a particular number (I've chosen <span class="math-container">$6$</span>). Let's begin by removing all <span class="math-container">$B=6$</span> rows and <span class="math-container">$A=6$</span> cells.</p>
<p><a href="https://i.stack.imgur.com/LL8yT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LL8yT.jpg" alt="enter image description here"></a></p>
<p>Since there are multiple choices that thin out at the right, I'll be going for the topmost possible cell while moving left to right. We can then pick the cell <span class="math-container">$B=12,A=1$</span> such that <span class="math-container">$D=11$</span>. This means we have to remove all other <span class="math-container">$B=12$</span>'s, all <span class="math-container">$A=1$</span>'s and all <span class="math-container">$D=11$</span>'s. Hence:</p>
<p><a href="https://i.stack.imgur.com/e51Y8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e51Y8.jpg" alt="enter image description here"></a></p>
<p>We can continue with this approach and generate the sequences:</p>
<p><span class="math-container">$$\begin{aligned}(&12,1)\\(2,&12,1)\\(11,2,&12,1)\\(3,11,2,&12,1)\\(10,3,11,2,&12,1)\\(4,10,3,11,2,&12,1)\\(9,4,10,3,11,2,&12,1)\end{aligned}$$</span></p>
<p>But now we run into a slight problem shown in the next figure. After we've chosen that <span class="math-container">$4$</span> follows <span class="math-container">$9$</span>, we're left with the following chart. For <span class="math-container">$D=4$</span>, if we pick <span class="math-container">$B=5$</span>, <span class="math-container">$A=9$</span>, we will have to remove <span class="math-container">$B=1$</span>, <span class="math-container">$A=5$</span> since it is also <span class="math-container">$D=4$</span>. In which case we will fail, since we will have no cell for <span class="math-container">$B=1$</span>. Hence, we must pick <span class="math-container">$B=1$</span>, <span class="math-container">$A=5$</span>.</p>
<p><a href="https://i.stack.imgur.com/UxpdE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UxpdE.jpg" alt="enter image description here"></a></p>
<p>We can then proceed as normal and get the following chart. This gives us the sequence <span class="math-container">$(9,4,10,3,11,2,12,1,5,8,7)$</span>, which proves the result. </p>
<p><a href="https://i.stack.imgur.com/Wojb5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wojb5.jpg" alt="enter image description here"></a></p>
|
4,651,596 | <p>I know the proof of the "<a href="https://en.wikipedia.org/wiki/Doubling_the_cube" rel="nofollow noreferrer">Doubling the cube problem</a>". What is used there is the fact that if a number <span class="math-container">$a$</span> is constructible then <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span>.</p>
<p>I found in a German textbook the remark that the inversion is not correct:
If <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span>, then <span class="math-container">$a$</span> is not necessarily constructible.</p>
<p>Do you know an example for an <span class="math-container">$a$</span> where <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]$</span> is a power of <span class="math-container">$2$</span> and which is not constructible – or a textbook with an example?</p>
| Michael Weiss | 79,741 | <p>Cox's <em>Galois Theory</em>, Example 10.1.13 (p.263) gives the example <span class="math-container">$f(x)=x^4-4x^2+x+1$</span>. It is irreducible, so if <span class="math-container">$f(a)=0$</span> then <span class="math-container">$[\mathbb{Q}(a):\mathbb{Q}]=4$</span>. However, the splitting field of <span class="math-container">$f$</span> over <span class="math-container">$\mathbb{Q}$</span> has degree 24.</p>
<p>Cox also shows that <span class="math-container">$a$</span> is constructible if and only if the splitting field of <span class="math-container">$f$</span> over <span class="math-container">$\mathbb{Q}$</span> has degree a power of 2, where <span class="math-container">$f$</span> is the minimal polynomial of <span class="math-container">$a$</span>. (Theorem 10.1.12, p.262). Therefore the above <span class="math-container">$a$</span> is not constructible.</p>
|
4,177,639 | <p>I have an object with known coordinates in in 3D but on the ground (<code>z=0</code>). The object has a direction vector. My goal is to move this object on the ground (so <code>z</code> stays <code>0</code>) using its direction vector and via randomly-generated velocity vectors with one condition: I want to ensure that the generate velocity vector can only move the object within a "valid arc", defined in degrees with respect to the direction vector of the object. More specifically, the way I determine valid range is by ensuring that the new position is within 45 degrees of the old object's direction vector (-45 degrees to the left and +45 degrees to the right). Can someone write a pseudocode on how I can achieve this?</p>
<p>Here's my attempt to do this but this doesn't seem to be the correct way to help me achieve what I want:</p>
<pre><code>object_dir = object_position # the direction could be the same as the object's coortinates
while True:
vel_vec=[uniform(-max_vel, max_vel), uniform(-max_vel, max_vel)] # generate a random velocity vector
new_pos = object_dir + vel_vec # compute a new position (and/or object direction vector) for the object
if (compute_angle(new_pos, object_dir) < 45 or compute_angle(new_pos, object_dir) > 315):
break
</code></pre>
| IV_ | 292,527 | <p><span class="math-container">$$a^x+b^x=1$$</span></p>
<p>Substitute <span class="math-container">$x=\frac{\ln(t)}{\ln(a)}$</span>:</p>
<p><span class="math-container">$$a^{\frac{\ln(t)}{\ln(a)}}+b^{\frac{\ln(t)}{\ln(a)}}=1$$</span></p>
<p><span class="math-container">$$t+t^{\frac{\ln(b)}{\ln(a)}}=1$$</span></p>
<p>Substitute <span class="math-container">$\frac{\ln(b)}{\ln(a)}=\alpha$</span>:</p>
<p><span class="math-container">$$t+t^\alpha=1$$</span></p>
<p>For rational <span class="math-container">$\alpha$</span>, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations. For <a href="https://www.wolframalpha.com/input?i=solve+ln%28b%29%2Fln%28a%29%3Dbeta%2Cb" rel="nofollow noreferrer">certain <span class="math-container">$a,\beta$</span> and <span class="math-container">$b=a^\beta$</span> (WolframAlpha)</a>, <span class="math-container">$\alpha$</span> is rational.<br />
For rational <span class="math-container">$\alpha\neq 0,1$</span>, the equation is related to a <a href="https://en.wikipedia.org/wiki/Trinomial" rel="nofollow noreferrer">trinomial equation</a>.<br />
For real or complex <span class="math-container">$\alpha\neq 0,1$</span>, the equation is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent <a href="https://en.wikipedia.org/wiki/Fox%E2%80%93Wright_function" rel="nofollow noreferrer">Fox-Wright Function</a> <span class="math-container">$\ _1\Psi_1$</span> therefore.</p>
<p>see also: <a href="https://math.stackexchange.com/questions/4109555/how-to-isolate-x-in-ax-bx-c-for-use-in-medical-statistics/4496115#4496115">How to isolate $x$ in $a^x + b^x = c$? (For use in medical statistics)</a></p>
<p><a href="https://www.researchgate.net/publication/225959964_On_the_roots_of_the_trinomial_equation" rel="nofollow noreferrer">Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104</a></p>
<p><a href="https://link.springer.com/article/10.1007/s10910-018-0985-3" rel="nofollow noreferrer">Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106</a><br />
<span class="math-container">$\ $</span></p>
<p>If <span class="math-container">$a,b,x\in\mathbb{N}$</span>, your equation is a <a href="https://en.wikipedia.org/wiki/Diophantine_equation" rel="nofollow noreferrer">Diophantine equation</a>.</p>
|
87,437 | <p>Let <span class="math-container">$R$</span> be a rectangular region of the integer lattice <span class="math-container">$\mathbb{Z}^2$</span>,
each of whose unit squares is labeled with a number
in <span class="math-container">$\lbrace 1, 2, 3, 4, 5, 6 \rbrace$</span>.
Say that such a labeled <span class="math-container">$R$</span> is <em>die-rolling Hamiltonian</em>,
or simply <em>rollable</em>,
if there is a Hamiltonian cycle obtained by rolling a unit die
cube over its edges so that, for each square <span class="math-container">$s \in R$</span>,
the cube lands on <span class="math-container">$s$</span> precisely once, and when it does so,
the top face of the cube matches the number in <span class="math-container">$s$</span>.
For example, the <span class="math-container">$4 \times 4$</span> "board" shown below
is rollable.
<br />
<img src="https://i.stack.imgur.com/6aR0a.jpg" alt="Dice Rolling" />
<br /></p>
<blockquote>
<p><b>Q</b>. Is it true that, if <span class="math-container">$R$</span> is die-rolling Hamiltonian, then the
Hamiltonian cycle is unique, i.e., there are never two distinct
die-rolling Hamiltonian cycles on <span class="math-container">$R$</span>?</p>
</blockquote>
<p>This "unique-rollability"
question arose out of a problem I posed in 2005, and was largely
solved two years later, in a paper entitled,
"On rolling cube puzzles" (complete citation below;
the <span class="math-container">$4 \times 4$</span> example above is from Fig. 17 of that paper).
Although the original question involved computational complexity,
the possible uniqueness of Hamiltonian cycles is independent
of those computational issues, so I thought it might be useful
to expose it to a different community, who might bring
different tools to bear.
It is known to hold for <span class="math-container">$R$</span> with side lengths at most 8.
If not every cell of <span class="math-container">$R$</span> is labeled, and unlabeled cells are forbidden
to the die, then there are examples with more than one Hamiltonian cycle.</p>
<p><b>Edit1</b>. Rolling a regular tetrahedron on the equilateral triangular (hexagonal) lattice
is not as interesting. See the Trigg article cited below.</p>
<p><b>Edit2</b>.
Serendipitously, <em>gordon-royle</em> posted a perhaps(?) relevantly related
question:
"<a href="https://mathoverflow.net/questions/87496/">Uniquely Hamiltonian graphs with minimum degree 4</a>."</p>
<hr />
<ul>
<li> The computational version is
<a href="http://cs.smith.edu/~jorourke/TOPP/P68.html#Problem.68" rel="nofollow noreferrer">
Open Problem 68</a> at
<a href="http://cs.smith.edu/~jorourke/TOPP/" rel="nofollow noreferrer">The Open Problems Project</a>.
</li>
<li>
"On rolling cube puzzles."
Buchin, Buchin, Demaine, Demaine, El-Khechen, Fekete, Knauer, Schulz, Taslakian.
<em>Proceedings of the 19th Canadian Conference on Computational Geometry</em>, Pages 141–144, 2007.
<a href="http://people.csail.mit.edu/schulz/papers/RollingFull.pdf" rel="nofollow noreferrer">PDF download.</a>
</li>
<li>
Charles W. Trigg. "Tetrahedron rolled onto a plane." <em>J. Recreational Mathematics</em>, 3(2):82–87, 1970.
</li>
</ul>
| Marzio De Biasi | 35,419 | <p>While browsing the recreational-mathematics tag I noticed this old question again.</p>
<p>In 2012 I published on my blog a draft paper in which I give another example of a (bigger) board with two distinct Hamiltonian cycles (but domotorp was quicker to answer).</p>
<p>In the paper I also prove that the <em>Rolling Cube Puzzle in labeled boards without free cells and with blocked cells is NP-complete</em> (an open problem).</p>
<p><a href="http://www.nearly42.org/cstheory/rolling-a-cube-can-be-tricky/" rel="nofollow noreferrer">http://www.nearly42.org/cstheory/rolling-a-cube-can-be-tricky/</a></p>
<p>... I never had the time to clean it up and submit it to a journal for a serious review (though I tested the gadgets with Mathematica).</p>
<p>If someone wants to give it a look ...</p>
|
3,691,255 | <p>Pierre runs a game at a fair, where each player is guaranteed to win $10. </p>
<p>Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs. </p>
<p>When a ‘6’ occurs, Pierre gives the player $10 and the game concludes. </p>
<p>On average, Pierre wishes to make a profit of $2 per game. How much does he need to charge for each roll of the die?</p>
| gnasher729 | 137,175 | <p>In the long term, one out of six throws gives a six, costing Pierre 10 dollars and ending a game. Pierre wants to make 2 dollars profit per game, so he must ask for 12 dollars for each six thrown, so he charges 2 dollars per throw. No sums needed. </p>
|
1,908,923 | <p>Let $X$ be a Riemannian manifold*, and $S$ a compact submanifold of $X$. </p>
<p>Assume there exists an <strong>open, dense</strong> subset $Y$ of $\,X$, such that for any element $y \in Y$, there exists a unique element in $S$ closest to $y$; i.e there is a function $\tilde s:Y \to S$ such that $$ d(y,\tilde s(y))=d_S(y)=\inf\{d(s,y)|\, \,s \in S\},$$</p>
<p>and $\tilde s(y)$ is the only element in $S$ satisfying the above equality.</p>
<p><strong>Question:</strong></p>
<p>Let $x \in X$, and let $\tilde s \in S$ be a closest element to $x$ in $S$. (such an element exists by compactness of $S$). </p>
<p>Is it true that there exists a sequence $y_n \in Y$, $y_n \to x$, such that $\tilde s(y_n) \to \tilde s$?</p>
<p>Note that I do <strong>not</strong> assume $x$ has a unique closest point on $S$. (If this were true the question would become trivial; On any subset consisting of elements which have unique closest points, the "closest point mapping" $\tilde s$ is continuous).</p>
<hr>
<p>*(Actually, one could ask the question in a more general setting when $X$
is a metric space, and $S$ is a subset of it, but I am not sure which "strange pathologies" can arise then)</p>
| Asaf Shachar | 104,576 | <p>I am adding some details to the answer of HK Lee:</p>
<p>We assume $X$ is geodesically convex, i.e there is a minimizing geodesic between every two points. (In particular, our argument holds for any complete manifold).</p>
<p>Let $x \in X$, and suppose $s_0$ is a closest point to $x$ in $S$. Let $c:[0,1] \to X$ be a minimizing geodesic from $x$ to $S$, i.e $c(0)=x,c(1)=s_0$.</p>
<p>We begin with the following observation:</p>
<p><strong>Lemma (1):</strong> For every $0<t_0<1$, $s_0$ is the a <strong>unique</strong> closest point to $c(t_0)$ on $S$.</p>
<p><strong>Proof Lemma (1):</strong></p>
<p>The fact $s_0$ is a closest point to $c(t_0)$ is trivial.
Assume $s_1 \neq s_0$ satisfies $$d(c(t_0),s_1) = d(c(t_0),S) = d(c(t_0),s_0),\ s_1\in S$$</p>
<p>Let $\gamma$ be a minimizing geodesic from $c(t_0)$ to $s_1$.
Look at the concatentation $\gamma * c|_{[0,t_0]}$; $$L(\gamma)=d(c(t_0),s_1)= d(c(t_0),s_0)=L(c|_{[t_0,1]}),$$ so</p>
<p>$$L(\gamma * c|_{[0,t_0]})=L(c|_{[0,t_0]}) + L(\gamma)=L(c)=d(x,s_0)=d(x,S)$$</p>
<p>$\gamma * c|_{[0,t_0]}$ is therefore a <strong>broken</strong> path (not smooth at $t_0$) connecting $x$ to $s_1$. Since a minimizing path between every two points must be smooth, it follows that $ \gamma * c|_{[0,t_0]}$ is not a minimizer, i.e $d(x,s_1) < L(\gamma * c|_{[0,t_0]})=d(x,S)$ which is a contradiction.</p>
<hr>
<p><strong>Lemma (2):</strong> For every $0<t<1$, there exists a sequence $y_n \in Y$ such that $y_n \to c(t), \tilde s(y_n) \to s_0$.</p>
<p><strong>Proof Lemma (2):</strong>
Let $0<t<1$, and let $y_n \in Y, y_n \to c(t)$. $$(1) \lim_{n \to \infty} d(y_n,\tilde s(y_n))=\lim_{n \to \infty} d(y_n,S)=d(c(t),S)$$</p>
<p>Since $S$ is compact, we can assume W.L.O.G that $\tilde s(y_n)$ is converging to some $\tilde s \in S$. Thus,</p>
<p>$$ (2) \lim_{n \to \infty} d(y_n,\tilde s(y_n))=d(c(t),\tilde s) $$</p>
<p>$(1),(2)$ imlpies $d(c(t),S)=d(c(t),\tilde s)$</p>
<p>Lemma (1) implies $\tilde s=s_0$.</p>
<hr>
<p><strong>Proof of the main proposition:</strong></p>
<p>Define $t_n=\frac{1}{n}$. By Lemma (2) (applied for each $t_n$), there exists $y_n \in Y$ such that $d(y_n,c(t_n))<\frac{1}{n}$, and $d(\tilde s(y_n),s_0) < \frac{1}{n}$. Thus, $y_n \to x, \tilde s(y_n) \to s_0$, as required.</p>
|
1,257,598 | <p>Suppose A is a family of subsets of R with the property that the intersection of any two sets in A is finite. Show that $|A|\leq 2^{\aleph_0}$.</p>
<p>I was told that choosing a countable $D \subset B$ for all $B \in A$ would be helpful. I'm just really not sure where to go with this. Any hints would be appreciated! </p>
| Camilo Arosemena-Serrato | 33,495 | <p><strong>Hint:</strong> Prove that the set $[\Bbb R]^{<\omega}$, the set of all finite subsets of $\Bbb R$, has size $2^{\aleph_0}$.
Fix $C_0\in A$, then prove that for each $C\in [\Bbb R]^{<\omega}$, the set $A_C:=\{B\in A:B\cap C_0=C \}$ has size $\leq2^{\aleph_0}$. Notice that $\{A_C:C\in [\Bbb R]^{<\omega}\}$ is a partition of $A-\{C_0\}$, and thus $|A|\leq 2^{\aleph_0}$.</p>
|
1,297,690 | <p>If I have a program that creates, let's say, one billion integers, with each having a pure $50 - 50$ chance to be one or zero,</p>
<p>what is the chance of finding $x$ zeros in a row?</p>
<p>for brownie points, instead of the program creating a set billion numbers, what would the equation be with $z$ numbers?</p>
| alkabary | 96,332 | <p>consider with just 10 integers $$\large{x_1x_2x_3x_4x_5x_6x_7x_8x_9x_{10}}$$ where all $\large{x_i}$ is either $0$ or $1$.</p>
<p>What is the chance of finding $\large{x_1x_2x_3x_4}$ all zeros ?</p>
<p>It is just $\large{(\frac{1}{2})^4 = \frac{1}{16}}$. Now you just apply this to your question</p>
|
222,596 | <p>I would like to find a temperature by knowing the enthalpy, is this possible?
This is what i've tried so far:</p>
<pre><code>V1 = 150;
V2 = 4;
T1 = 15 + 273;
Enthalpy
h[T_] := QuantityMagnitude[
ThermodynamicData["Air",
"Enthalpy", {"Temperature" -> Quantity[T, "Kelvins"]}]]
h1 = h[T1]
sol = Solve[h1 + V1^2/2 == h2 + V2^2/2]
{{h2 -> 425466.}}
FindRoot[h[T2] == h2 /. sol, {T2, 300}]
During evaluation of ThermodynamicData::quant: T2 is not a real number.
During evaluation of FindRoot::jsing: Encountered a singular Jacobian at the point {T2} = {300.}. Try perturbing the initial point(s).
{T2 -> 300.}
</code></pre>
| flinty | 72,682 | <p><em>Mathematica</em> is missing enthalpy data below 60K. Also evaluating the <code>ThermodynamicData</code> inside the <code>Solve</code> is slow. Try this:</p>
<pre><code>enthalpy[t_?NumericQ] :=
QuantityMagnitude[
QuantityMagnitude[
ThermodynamicData["Air",
"Enthalpy", {"Temperature" -> Quantity[t, "Kelvins"]}]]]
h2 = With[{V1 = 150, V2 = 4, T1 = 15 + 273},
x /. Last@
NMinimize[{((enthalpy[T1] + V1^2/2) - (x + V2^2/2))^2, x > 60}, x,
MaxIterations -> 5]]
NMinimize[{(enthalpy[t] - h2)^2 , t > 60}, t, MaxIterations -> 5]
</code></pre>
<p>It's still a bit slow, and will be even slower with more iterations, but I get 299.173K.</p>
|
4,177,829 | <p>Given angles <span class="math-container">$0<\theta_{ij}<\pi$</span> for <span class="math-container">$1\leq i<j\leq k$</span>, what conditions are there on the angles to ensure that there exists <span class="math-container">$k$</span> unit vector <span class="math-container">$v_i\in \mathbb R^k$</span> so that the angle between <span class="math-container">$v_i$</span> and <span class="math-container">$v_j$</span> is <span class="math-container">$\theta_{ij}$</span>?</p>
<p>There are clearly problems when <span class="math-container">$\theta_{12},\theta_{13},\theta_{23}$</span> are all close to <span class="math-container">$\pi$</span>. What if I can ensure, for a given <span class="math-container">$\epsilon>0$</span> that <span class="math-container">$|\theta_{ij}-\frac{\pi}2|<\epsilon?$</span></p>
<p>Intuitively, this is true for <span class="math-container">$k=3$</span> and the angles close to <span class="math-container">$\frac{\pi}2$</span>. We can easily pick <span class="math-container">$v_1,v_2$</span> and the locus of points for <span class="math-container">$v_3$</span> meeting the angle requirement with <span class="math-container">$v_1$</span> is a near-great circle in the unit sphere. With <span class="math-container">$v_2$</span>, the same. And these two near-great circles are near-perpendicular. We need them to intersect for <span class="math-container">$v_3$</span> to be found.</p>
<p>But I can’t prove it, and my intuition for <span class="math-container">$k>3$</span> spheres is negligible.</p>
<p>This is a possible solution for <a href="https://math.stackexchange.com/q/4177715">this question</a>. (In fact, I only need <span class="math-container">$k>3$</span> to complete that question - I’ve got an entirely different solution for <span class="math-container">$k=3$</span> in that question.)</p>
<hr />
<p>I think a minimum necessary condition is, for all distinct <span class="math-container">$i,j,k$</span>:
<span class="math-container">$$\theta_{ij}+\theta_{jk}\geq \theta_{ik}.\tag 1$$</span> (Here we need <span class="math-container">$\theta_{ij}=\theta_{ji}$</span> for <span class="math-container">$i>j$</span> to include all the correct cases in (1).)</p>
<p>Also, probably:
<span class="math-container">$$\theta_{ij}+\theta_{jk}+\theta_{ik}\leq 2\pi\tag 2$$</span></p>
<p>It seems like, when <span class="math-container">$k=3$</span>, (1) and (2) should be enough.</p>
| NN2 | 195,378 | <p>Let us represent the vectors in question as directed line segments <span class="math-container">$\{\overrightarrow{OA}_i\}_{1 \le i \le k} $</span> from the origin <span class="math-container">$O$</span> to the points <span class="math-container">$A_i$</span> on the unit sphere <span class="math-container">$\mathcal{B}(O,1)$</span>.</p>
<p><strong>Remark</strong>: perhaps it may be necessary to create apoint <span class="math-container">$A_0$</span> to be able to apply the Cayley-Menger determinant as we need <span class="math-container">$k+1$</span> points in the space <span class="math-container">$\Bbb R^k$</span>. If yes, the point <span class="math-container">$A_0$</span> can be created as the symmetric point of <span class="math-container">$A_1$</span>. The angle between <span class="math-container">$\overrightarrow{OA}_0$</span> and <span class="math-container">$\overrightarrow{OA}_i$</span> (for <span class="math-container">$1 \le i \le k$</span>) is <span class="math-container">$\theta_{0i} = \pi-\theta_{0i}$</span> for <span class="math-container">$i=1,...,k$</span>.</p>
<p>The distance between two points <span class="math-container">$A_i$</span> and <span class="math-container">$A_j$</span> is then determined by the angle <span class="math-container">$\theta_{ij}$</span> via the law of cosines. I call it <span class="math-container">$a_{ij}$</span>:
<span class="math-container">$$a_{ij} =|A_iA_j| = \sqrt{2-2\cos(\theta_{ij})} = 2\sin\left(\frac{\theta_{ij}}{2} \right) \qquad \text{for } 0 \le i< j \le k\tag{1}$$</span></p>
<p>Of course, the distance from <span class="math-container">$O$</span> to each <span class="math-container">$A_i$</span> is <em>also</em> fixed at <span class="math-container">$1$</span>. Your question now stands thus: can we construct points in <span class="math-container">$\mathbb{R}^k$</span> with specified distances (to each other, and to <span class="math-container">$O$</span>)?</p>
<p>The answer is entirely determined by the <a href="https://en.wikipedia.org/wiki/Cayley%E2%80%93Menger_determinant" rel="nofollow noreferrer">Cayley-Menger determinant</a> <span class="math-container">$\Gamma$</span>, which calculates the area of the simplex marked out by those points. <a href="http://www.maia.ub.edu/%7Esombra/publications/cayley/cayley.pdf" rel="nofollow noreferrer">This nice exposition</a> mentions that it can be used on page <span class="math-container">$2$</span>, citing Theorem 9.7.3.4 of <a href="http://library.lol/main/5CA1D714B834C99E7F92B4A0D1A50503" rel="nofollow noreferrer">Berger's <em>Geometry I</em></a> :</p>
<p><a href="https://i.stack.imgur.com/rfSBb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rfSBb.png" alt="Screenshot of theorem 9.7.3.4 from Berger's Geometry I" /></a></p>
<p>To summarize the argument:</p>
<ul>
<li>Work by induction on <span class="math-container">$k$</span> with two base cases.</li>
<li>Construct <span class="math-container">$k-2$</span> points in a <span class="math-container">$(k-2)$</span>-dimensional subspace by hypothesis. Henceforth, I'm going to pretend the <span class="math-container">$k-2$</span> points are just the single point <span class="math-container">$O$</span>, and quotient out by their <span class="math-container">$(k-2)$</span>-dimensional subspace. I'm also going to choose an arbitrary splitting of the complementary subspace <span class="math-container">$\mathbb{R}^2=Y\oplus W$</span>.</li>
<li>Adding in one more point is easy: the inductive hypothesis tells us we can do it in any <span class="math-container">$(k-1)$</span>-dimensional subspace. So, put <span class="math-container">$A_{k-1}\in Y$</span>.</li>
<li>Now, for any codimension-<span class="math-container">$1$</span> subspace ("line") <span class="math-container">$L$</span>, there are two possible places for <span class="math-container">$A_k$</span> (ignoring our requirement on <span class="math-container">$|A_{k-1}A_k|$</span>). In fact, as we rotate <span class="math-container">$L$</span>, we get a circle.</li>
<li>The distance <span class="math-container">$x=|A_{k-1}A_k|$</span> still isn't right, but now we can fix it. The two examples when <span class="math-container">$L=Y$</span> are extremal: any distance between these two values of <span class="math-container">$x$</span> is feasible by finding the correct point on the circle. Those examples are also zeros of <span class="math-container">$\Gamma$</span>.</li>
<li><span class="math-container">$\Gamma$</span> is a quadratic in <span class="math-container">$x$</span>. So <span class="math-container">$\Gamma$</span> has a specific sign at values of <span class="math-container">$x$</span> between the extreme examples. The sign condition given in the theorem statement ensures we are in that region.</li>
</ul>
<p>Q.E.D.</p>
<p>From this theorem, you have necessary and sufficient conditions for the existence of the <span class="math-container">$k$</span> vectors in terms of distances, and from (1) you have necessary and sufficient conditions in terms of the angles.</p>
|
95,964 | <p>On the page 43 of <em>Real Analysis</em> by H.L. Royden (1st Edition) we read: "(Ideally) we should like $m$ (the measure function) to have the following properties:</p>
<ol>
<li>$m(E)$ is defined for each subset $E$ of real numbers.</li>
<li>For an interval $I$, $m(I) = l(I)$ (the length of $I$).</li>
<li>If $\{E_n\}$ is a sequence of disjoint sets (for which $m$ is defined),
$m(\bigcup E_n)= \sum m (E_n)$."</li>
</ol>
<p>Then at the end of page 44 we read : "If we assume the Continuum Hypothesis (that every non countable set of real numbers can be put in one to one correspondence with the set of all real numbers) then such a measure is impossible," and no more explanation was given.</p>
<p>Now assuming the Continuum Hypothesis I am not able to see why such a measure is not possible. Would you be kind enough to help me?</p>
| Jonas Meyer | 1,424 | <p>I refer you to the MathOverflow question "<a href="https://mathoverflow.net/questions/45784/does-pointwise-convergence-imply-uniform-convergence-on-a-large-subset">Does pointwise convergence imply uniform convergence on a large subset?</a>" for proofs and references for the following:</p>
<blockquote>
<p>Assuming the Continuum Hypothesis, there exists a sequence of real-valued functions on $\mathbb R$ that converges pointwise to a function on $\mathbb R$ but does not converge uniformly on any uncountable set.</p>
</blockquote>
<p>Suppose there were a function $m:\mathcal{P}(\mathbb R)\to[0,\infty]$ satisfying the three named conditions. Then $m$ is a measure on $\mathbb R$ for which every subset of $\mathbb R$ is measurable. Let $(f_n)$ be a sequence of real-valued functions on $[0,1]$ that converges pointwise to some function. By Egorov's theorem, there is a set $E\subseteq [0,1]$ with $m(E)>0$ such that $(f_n)$ converges uniformly on $E$. The fact that $m(E)>0$ implies that $E$ is uncountable. Since $(f_n)$ was arbitrary, this implies by the above cited result that the Continuum Hypothesis does not hold.</p>
<hr>
<p>In light of Michael's answer, I want to mention that the hypothesis 2, that $m(I)=l(I)$ for each interval $I$, could be replaced by the following two properties:</p>
<p>(a) For all $x\in\mathbb R$, $m(\{x\})=0$.</p>
<p>(b) There exists $X\subseteq\mathbb R$ such that $0<m(X)<\infty$.</p>
<p>Then the same argument as above would apply with such an $X$ in place of $[0,1]$.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Blue | 409 | <p>A couple of my previous answers to similar questions:</p>
<p>Find a rubber chicken(!) and introduce the kids to $\pi$:
<a href="https://math.stackexchange.com/a/395933/409">https://math.stackexchange.com/a/395933/409</a> (It won't take a half-hour, but it's a good way to warm-up the crowd. :)</p>
<p>This exercise is described with a focus on conceptualizing logarithms, but the exponential aspect alone could be engaging: <a href="https://math.stackexchange.com/a/129246/409">https://math.stackexchange.com/a/129246/409</a></p>
<hr>
<p>Also, you could take a sheet of paper and <a href="http://www.kidzone.ws/magic/walkthrough.htm" rel="nofollow noreferrer">cut a hole big enough to walk through</a>. While this is usually described as a simple trick (or bar bet), I like to look at it as a demonstration that (<strong>very</strong> loosely speaking) even a finite area "contains" infinite length ... a concept that becomes clearer as one investigates <a href="http://en.wikipedia.org/wiki/Space-filling_curve" rel="nofollow noreferrer">space-filling curves</a>.</p>
<p>You might also confront the kids with <a href="http://en.wikipedia.org/wiki/Zeno%27s_paradoxes" rel="nofollow noreferrer">Zeno's tortoise paradox</a>. Kids love animal stories! :) Instead of just telling the story, though, you could have a couple of kids re-enact the parts. Relatedly, you could have kids do the "walk half-way to the door, then half-of-halfway, then half-of-half-of-halfway" thing as a first brush with limits. (You'll never get <em>beyond</em> the door that way, but at some step you'll pass any particular point <em>before</em> it; therefore, the journey must "converge" at the door.) </p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| KCd | 619 | <ol>
<li><p>Discuss the birthday paradox if there are at least 23 people in the room. In fact, ask the teacher in advance if he/she knows from student records if two students share the same birthday (an illustration with the students in the room won't go over well if you try it and nobody shares a birthday).</p></li>
<li><p>Look at the sum of odd numbers: 1, 1+3, 1+3+5, 1+3+5+7,... until someone notices a pattern. This is basically your idea of the sum of an arithmetic progression, but made concrete in an easily grasped way.</p></li>
<li><p>The 3x+1 problem. This will show them an example of an elementary unsolved problem that they can experiment with.</p></li>
<li><p>Examples of patterns that break down: if a sequence starts off as 1, 2, 4, 8, 16, what is the next term? I would hope some of the students will recognize these as powers of 2. The point is to show them several examples of counting problems that all start off this way but the next term is not 32. See examples 1, 5, and 6 in my answer at <a href="https://mathoverflow.net/questions/52101/longest-coinciding-pair-of-integer-sequences-known">https://mathoverflow.net/questions/52101/longest-coinciding-pair-of-integer-sequences-known</a>.</p></li>
</ol>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Loki Clock | 58,287 | <p>Describe multiplication on a 12-hour clock, pointing out that 3*4=12 makes it unlike regular multiplication. Ask them which clocks don't have two numbers that can be multiplied to get the number of hours on the clock.</p>
<p>Taking (topological) dual polyhedra by letting new points be face centers and connecting centers of faces that share an edge.</p>
<p>What happens when you place a black square on the top-left corner of a grid, then make any square on the next line black (repeating this line by line) when there's a black square either above or above and to the left, but not both.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Eric Jablow | 70,913 | <ol>
<li>Make a bagatelle and drop balls down it. Use this to demonstrate the central limit theorem, suitably simplified to "Large amounts of random anything look like a bell curve." Then, measure your students' heights and compare.</li>
<li>Try logic puzzles and games.
<ol>
<li>Nim</li>
<li>The Monty Hall Problem</li>
<li>Dice probabilities. Don't be afraid of teachers who don't want gambling; the students play <em>Monopoly</em> and <em>Settlers</em>, don't they?</li>
</ol></li>
<li>Show R(3, 3) = 6. They go to parties too.</li>
</ol>
|
366,311 | <blockquote>
<p>Show that the sequence $\displaystyle (x_n)=\left( \sum_{i=1}^n\frac 1 i\right)$ diverge by epsilon delta definition.</p>
</blockquote>
<p>I'm not familiar with proving divergent sequence. Do anyone have any des? Thank you.</p>
| Community | -1 | <p>For $\epsilon=\frac{1}{2}$ and $\forall n , \displaystyle|x_{2n}-x_n|=\sum_{k=n+1}^{2n}\frac{1}{k}\geq n\times\frac{1}{2n}=\epsilon$ hence the sequence $(x_n)$ is divergent since it's not a Cauchy sequence.</p>
|
1,005,193 | <p>My problem is as follows:</p>
<p>I have a point $A$ and a circle with center $B$ and radius $R$. Points $A$ and $B$ are fixed, also $A$ is outside of the circle. A random point $C$ is picked with uniform distribution in the area of disk $B$. My question is how to calculate the expected value of $AC^{-4}$. I am working with the path loss in Wireless Communication so $AC^{-4}$ measures how much energy is dissipated along the distance $AC$</p>
<p>My approach is to first denote $\theta$ as the angle between AB and BC then $\theta$ is uniformly distributed between $[0,2\pi]$. Denote $r$ as the distance of BC then distribution of $r$ in $[0,R]$ is $\frac{2r}{R^2}$. Using the formula $AC^2 = AB^2 + BC^2 - 2AB\times BC \times \cos\theta$ , I have</p>
<p>\begin{align}
E[AC^{-4}] & = \int_0^{2\pi}\int_0^R (AB^2 + BC^2 - 2AB\times BC \times
\cos\theta)^{-2} f_\theta f_r \, dr \, d\theta \\
& = \int_0^{2\pi}\int_0^R (AB^2 + r^2 - 2AB\times r \times \cos\theta)^{-2} \frac{1}{2\pi} \frac{2r}{R^2} \, dr \, d\theta
\end{align}</p>
<p>However, I am unable to solve this integration. I want to ask if anyone know any method that can give me the closed-form of the above expected value. If not, then maybe an approximation method that can give a closed-form is also good. Thanks in advance.</p>
| Did | 6,179 | <blockquote>
<p>I don't know what limits to use.</p>
</blockquote>
<p>Note that $x=w/u$, $y=u$, $z=\sqrt{v}$ with $0\leqslant x,y,z\leqslant1$ hence the domain of integration is $$0\leqslant w/u,u,\sqrt{v}\leqslant1,$$ or, equivalently, $$0\leqslant w\leqslant u\leqslant1,\qquad0\leqslant v\leqslant1.$$</p>
<blockquote>
<p>Find the joint pdf of $W:=XY$ and $V:=Z^2$.</p>
</blockquote>
<p>This can be simplified by noting that $W$ and $V$ are independent hence their marginal densities suffice to solve the question.</p>
|
34,215 | <p>How do professional mathematicians learn new things? How do they expand their comfort zone? By talking to colleagues? </p>
| Pietro Majer | 6,101 | <p>Well, understanding mathematics has different levels: Understanding is, mainly, pretending to understand. If you are able to cheat your professors, then your students, and your colleagues, it's OK. If you succeed to cheat yourself, then it means you went a bit too far. </p>
|
188,087 | <p>Is there a function that can extract a list of variables in an expression?
For example, assume we have an expression</p>
<pre><code>x^2+y^3+z
</code></pre>
<p>This expression has variables x, y and z. The result should be</p>
<pre><code>{x, y, z}
</code></pre>
<p>. Is there a way to get this?</p>
| Xminer | 61,541 | <p>I like the following approach x):</p>
<pre><code>expr = x^2 + y^3 + z;
Select[DeleteDuplicates@Level[expr, Depth@expr], Head[#] == Symbol &]
</code></pre>
<p>the result is:</p>
<pre><code>{x, y, z}
</code></pre>
|
1,319,476 | <p>This is a question related to another posted question:</p>
<p>The answer to the following question "Find all solutions to: $e^{ix}=i$" is as follows: </p>
<p>"Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$,</p>
<p>so: $ \cos x+i\sin x=0+1⋅i$</p>
<p>compare real and imaginary parts
$\sin(x)=1$
and
$\cos(x)=0$</p>
<p>$x=\frac{(4n+1)π}2$, $n∈$
(W stands for set of whole number W={0,1,2,3,.......,n})."</p>
<p>My question: Where does $x=\frac{(4n+1)π}2$, $n∈$ come from? </p>
<p>My steps: </p>
<ol>
<li><p>$\cos(x) + i\sin(x) = 0 + i(1)$</p></li>
<li><p>$\cos(x) = i(1 - \sin(x))$</p></li>
<li><p>... </p></li>
<li><p>how does $x=\frac{(4n+1)π}2$ follow? </p></li>
</ol>
| Deepak | 151,732 | <p>I always find it easier to use a fixed method, and I thought you might find this explanation easier, so I'm posting it. </p>
<p>Start by putting everything into exponential form. Now $i = e^{\frac{i\pi}{2}}$. You can derive this from $e^{i\pi} = -1$ and taking square roots on both sides.</p>
<p>Now note that for <em>any</em> $\theta$, $e^{i\theta} = e^{i(\theta + 2k\pi)}, k \in \mathbb{Z}$, and this is because $e^{2k\pi i} = 1$. Essentially, this can be viewed as the periodicity of the exponential form. To compute general solutions or roots, you would be well-advised to include this term so that you don't miss any solutions. </p>
<p>Hence you can now write $i = e^{i(\frac{\pi}{2} + 2k\pi)} = e^{i\pi\frac{4k+1}{2}}$</p>
<p>Note that the final step is just an algebraic rearrangement of the exponent.</p>
<p>You can now immediately solve the equation by taking logs of both sides, i.e.</p>
<p>$e^{ix} = e^{i\pi\frac{4k+1}{2}}\\ \implies x = \pi\frac{4k+1}{2}$ which is essentially the required form.</p>
|
1,913,320 | <blockquote>
<p>Let <span class="math-container">$A=(a_{ij})_{n\times n}$</span> and <span class="math-container">$A=(a_{ij})_{n\times n}$</span> be two upper triangular matrices, i.e. <span class="math-container">$a_{ij}=b_{ij}=0$</span> whenever <span class="math-container">$i>j$</span>.</p>
<p><span class="math-container">$(a)$</span> Show that the <span class="math-container">$(i,j)$</span>-entry of <span class="math-container">$AB$</span> is <span class="math-container">$0$</span> if <span class="math-container">$i>j$</span>, i.e <span class="math-container">$AB$</span> is an upper triangular matrix.</p>
<p><span class="math-container">$(b)$</span> Find the <span class="math-container">$(i,i)$</span>-entry of <span class="math-container">$AB$</span>.</p>
</blockquote>
<p>I have already proven part (a). How do I go about finding part (b)? Any help would be greatly appreciated! Thank you so much!</p>
| DonAntonio | 31,254 | <p>$$2\cdot3\cdot5\cdot7\cdot11\cdot13+1=59\cdot509$$</p>
|
2,745,918 | <p>The numbers $1,2, \ldots, n$ are written in a board, with $n \in \mathbb{N}$. In every move, we can choose two numbers of the board, find their $\rm lcm$, and replace the two numbers with it. After $k$ moves, we find the sum of the numbers in the board, and we name it $S$. Find the minimum and the maximum value of $n$, such that there is $k$ such that $S=2017$.</p>
<p>This is an exercise I show at a Mathematical Forum, posted on 2017, and I have been trying this for a long time actually, and the things I have proven aren't something important, so I would prefer full answers.</p>
<p>As far as I now, it isn't from an exam or a Math Olympiad. </p>
| Joffan | 206,402 | <p>Some hints (all numbers are $\in \Bbb N$).</p>
<ul>
<li>for any $k$, ${\rm lcm}(k,1)=k$.</li>
<li>for any $j,k,$ ${\rm lcm}(j,jk)=jk$.</li>
<li>the $\rm lcm$ of a prime number $p$ and any other distinct number $k$ is never less than the prime number. If $k>1,$ ${\rm lcm}(p,k)>p$.</li>
<li>if we have a set of powers of distinct prime numbers, their least common multiple $({\rm lcm})$ is the product of those numbers.</li>
<li>the product of a set of prime powers is greater than their sum.</li>
</ul>
<p>Implications for the puzzle, to achieve $S=2017$: </p>
<ul>
<li>the sum of the maximum powers of prime numbers less than $n$ must be less than $2017$</li>
<li>numbers that have multiples in the set can be removed (or retained) (including $1$).</li>
</ul>
<p>A couple of example solutions, just for checking:</p>
<ul>
<li><p>For $n=63, $ the initial $S_0= 2016$. We can take $2,3$ as our first $\rm lcm$ pair, which are replaced with $6$ to give $S_1=2017$</p></li>
<li><p>For $n=64,$ initial $S_0 = 2080$ and for a target of $2017$ we need to lose $63$. We can feed the powers of $2$ into the $\rm lcm$ calculation to replace all of $\{1,2,4,8,16,32,64\}$ with $64,$ achieving the target sum at $k=6$. Or we could use $\rm lcm$ pairs $(32,64)$ and $(31,62)$ to achieve it at $k=2$.</p></li>
</ul>
<hr>
<p>The highest value of $n$ is likely determined by the case where none of the remaining values is a multiple of any other; that is, the last set where $\lfloor n/2 \rfloor {+}1$ to $n$ sums to less that $2017$. This happens for $\fbox{$n=72$}$, when taking $(k,2k)\to 2k$ for every $k$ up to $36$ except for $k=19$ will give the required sum.</p>
<p>The lowest value of $n$ is at least $9$, since below that ${\rm lcm}(1..n)$ is too far below $2017$ - in particular ${\rm lcm}(1..8) = 840$. It's relatively easy to solve this, for example, for $n=24$ by assembling ${\rm lcm}$s from the original numbers in sequential pairs $\{(1,2),(3,4),(5,6),\ldots\}$ for a total of $2444$ and then disassembling some pairs and creating others appropriately to get the target value: $\{1, (2,5),$ $(3,4), 6,$ $(7,8),$ $9, 10,$ $(11,12),$ $(13,14),$ $(15,16),$ $(17,18),$ $19, 20,$ $(21,22),$ $(23,24)\}$. There will be solutions for smaller values of $n$ which use more than two numbers in ${\rm lcm}$ calculations - my smallest is $\fbox{$n=12$}$ which is achieved with $(9,10,11,12)\to 1980,$ $(1,4,6)\to 12,$ and $\{2,3,5,7,8\}$ adding in unchanged.</p>
|
1,005,154 | <p>I don't know how to advance in the following <em><strong>problem</strong></em>:</p>
<p>Let $X$, $Y$ and $Z$ independent random variables equally distributed with uniform distribution over $[0,1]$.</p>
<ul>
<li>Find the joint pdf of $W:=XY$ and $V:=Z^2$.</li>
</ul>
<hr>
<p><em><strong>I tried to</strong></em> answer this problem by declaring a new random variable $U:= Y$ (to my opinion necessary to get the transformation).</p>
<p>Then:<br>
$w=xy,$<br>
$v=z^2,$<br>
$u=y.$</p>
<p>We can see that dividing the first equation by the second one:<br>
$x=\dfrac{w}{u},$<br>
$y=u,$<br>
$z=\sqrt{v}.$</p>
<p>Consider the transformation $h(x(w,v,u),y(w,v,u),z(w,v,u))=\left(\dfrac{w}{u},u,\sqrt{v}\right)$ gives us</p>
<p>$$f_{WVU}(w,v,u)=|\boldsymbol{J(h)}|f_{XYZ}(h(x,t,z))=\frac{1}{2u\sqrt{v}}.$$</p>
<p>To find the pdf of $W,V$:
$$f_{WV}(w,v)=\int_u\frac{1}{2u\sqrt{v}}du.$$
However I don't know what limits to use. Any help is appreciated.</p>
| Vladimir Vargas | 187,578 | <p>Notice that:</p>
<p>$$f_{WVU}(w,v,u)=|\boldsymbol{J(h)}|f_{XYZ}(h(x,t,z))=|\boldsymbol{J(h)}|f_X\left(\dfrac{w}{u}\right)\chi_{[0,1]}(w)f_Y(u)\chi_{[w,1]}(u)f_Z(\sqrt{v})\chi_{[0,1]}(v).$$</p>
|
3,145,896 | <h1>Solve for <span class="math-container">$x$</span></h1>
<p>I have an equation that I have been working on solving; I know the solution, but I cannot get to it myself. Almost every simplification I do reverts back to a previous step. Can anyone show me how to solve for <span class="math-container">$x$</span> in this equation?</p>
<h3>Equation:</h3>
<p><span class="math-container">$$\log_6(2x-3)+\log_6(x+5)=\log_3x$$</span></p>
<h3>Solution:</h3>
<p><span class="math-container">$$x ≅ \frac{3347}{2000} ≅ 1.6735$$</span>
<br>
<strong>Note:</strong> upon further analysis of the answer, while close, it does not seem to be the <em>exact</em> solution.</p>
<hr>
<h3>What I Have Tried So Far</h3>
<p><span class="math-container">$$\log_6(2x-3) + \log_6(x + 5) = \log_3x$$</span>
<span class="math-container">$$\frac{\log(2x-3)}{\log6} + \frac{\log(x + 5)}{\log6} = \frac{\log x}{\log3}$$</span>
<span class="math-container">$$\log3 \cdot \log(2x-3) + \log3 \cdot \log(x + 5) = \log6 \cdot \log x$$</span>
<span class="math-container">$$\log3 \cdot \log \left[(2x - 3)(x + 5)\right] = \log6 \cdot \log x$$</span>
<span class="math-container">$$\frac{\log \left[(2x - 3)(x + 5)\right]}{\log_3 10} = \frac{\log6}{\log_x10}$$</span>
<span class="math-container">$$\log_x10 \cdot \log \left[(2x - 3)(x + 5)\right] = \log_3 10 \cdot \log6$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right] = \log_3 6$$</span>
<span class="math-container">$$\log_x3 \cdot \log_x \left[(2x - 3)(x + 5)\right] = \frac{\log_3 6}{\log_3 x}$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right]^{\ \log_x3} = \log_x 6$$</span>
<span class="math-container">$$\left[(2x - 3)(x + 5)\right]^{\ \log_x3} = 6$$</span>
<span class="math-container">$$(2x - 3)(x + 5) = x^{\log_3 6}$$</span></p>
<p><br>
I know these steps aren't really working towards the solution at points; I was sort of just playing around with the equation. Regardless, I really don't know how to go about moving forward from here.</p>
<hr>
| Claude Leibovici | 82,404 | <p><em>Just for the fun of it !</em></p>
<p>Since heropup already gave the answer, let us do the same using <strong>one single</strong> iteration using <a href="http://numbers.computation.free.fr/Constants/Algorithms/newton.html" rel="nofollow noreferrer">high order methods</a> with <span class="math-container">$x_0 = \frac{3347}{2000}$</span></p>
<p><span class="math-container">$$\left(
\begin{array}{ccc}
n & x_1 & \text{Method} \\
1 & \color{blue}{1.67351617}525914512770936502001715480227532437 & \text{Newton} \\
2 & \color{blue}{1.6735161761242}2229623976832154151388262101881 & \text{Halley}\\
3 & \color{blue}{1.67351617612426023}725117282565297065627253993 & \text{Householder}\\
4 & \color{blue}{1.6735161761242602388483}2362357774862967541409 & \text{no name}\\
5 & \color{blue}{1.6735161761242602388483916}1928419708807692011 & \text{no name}\\
6 & \color{blue}{1.673516176124260238848391622220}46176652833341 & \text{no name}\\
7 & \color{blue}{1.67351617612426023884839162222058963}357352924 & \text{no name}\\
8 & \color{blue}{1.673516176124260238848391622220589639170}05729 & \text{no name}
\end{array}
\right)$$</span></p>
|
2,194,190 | <p>I need to check the irreducibility of $p(x) \in F[x]$, where $F$ is a finite field.
I have read and checked on several exercises on the internet. Their solutions are as follows:</p>
<p>For instance, let $p(x)$ an arbitrary polynomial in $\mathbb{Z}_5[x]$. </p>
<p>If $p(x)$ has no zeros in $\mathbb{Z}_5$, then they say that $p(x)$ is an irreducible polynomial in $\mathbb{Z}_5[x]$.</p>
<p><em>I am confused at this point:</em> The polynomial $p(x)=(x^2+2)(x^2+3)$ has no zeros in $\mathbb{Z}_5[x]$, but it is reducible? Where is my mistake?</p>
| Akash Patalwanshi | 168,676 | <p>Let $D$ be an integral domain with unity, a polynomial $f(x) ∈ D[x]$ such that $deg(f(x)) ≥ 1$ is irreducible polynomial in $D[x]$, if whenever $f(x) = g(x) • h(x)$ then either $deg((g(x)) = 0$ or $deg((h(x)) = 0$. </p>
<p>There is one more thing which is irreducible element in Integral domains. Let $R$ be a CRU (commutative ring with unity) then an element $ a∈ R $ is called irreducible element, if </p>
<p>(i) $a ≠ 0$ and a is non-unit</p>
<p>(ii) whenever $ a= bc$ for some $b, c ∈ R$ then either $b$ is unit or $c$ is unit in $R$</p>
<p>For example:
Consider,
$f(x) = 2x^2 + 4∈Z[x] $
then $f(x) = 2(x^2 + 2) = g(x) • h(x)$ (say)
Then we saw that, $deg(g(x)) = 0$ so that $f(x)$ is irreducible polynomial in $Z[x]$ but it not an irreducible element in $Z[x]$, since neither $2$ nor $ (x^2 + 2)$ are units in $Z[x]$. </p>
<p>So "irreducible polynomial need not be an irreducible element"
But when $F$ is field then "every irreducible polynomial in $F[x]$ is an irreducible element of $F[x]$ and conversely"! </p>
<p>In fact you can check if you take
$f(x) = 2x+ 2 ∈ Z_3 [x]$ then $f(x)$ has root in $Z_3$ but $f(x)$ is irreducible polynomial over $Z_3$</p>
|
493,104 | <p>I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating </p>
<p>$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$</p>
<p>So we should be able to find the area of the ellipse by </p>
<p>$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$</p>
<p>$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$</p>
<p>$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$</p>
<p>$$=\pi\frac{a^2+b^2}{2}.$$</p>
<p>First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as <a href="http://www.wolframalpha.com/input/?i=1%2F2+*+integral+from+0+to+2*pi+of+%28cos%28x%29%29%5E2+%2B+2*+%28sin%28x%29%29%5E2" rel="nofollow">Wolfram Alpha tells me</a>.</p>
<p>What am I doing wrong?</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Putting $x=r\cos\theta,y=r\sin\theta$</p>
<p>$$\frac {x^2}{a^2}+\frac{y^2}{b^2}=1,$$</p>
<p>$$r^2=\frac{a^2b^2}{b^2\cos^2\theta+a^2\sin^2\theta}=b^2\frac{\sec^2\theta}{\frac{b^2}{a^2}+\tan^2\theta}$$</p>
|
164,152 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/33215/what-is-48293">What is 48÷2(9+3)?</a> </p>
</blockquote>
<p>Please, look at the picture?</p>
<p><img src="https://i.stack.imgur.com/3Tauh.jpg" alt="http://s16.radikal.ru/i190/1206/3b/7411739c6d9a.jpg"></p>
<p>Which calculator shows true result and how to prove achieved result?</p>
| Jorge Leitao | 25,780 | <p>The reason why calculators are giving different results is because one is calculating,</p>
<p>$$\frac{6}{2\cdot(2+1)} = 1$$</p>
<p>while the other is calculating,</p>
<p>$$\frac{6}{2}\cdot(2+1) = 9$$</p>
<p>Depending on what you want to calculate, you could obtain different answers.</p>
|
1,957,453 | <p>Can you please help me with this question?</p>
<p>Question: Find the radius of curvature, and the equation of the osculating circle, for the following curve for <span class="math-container">$t\geq0$</span>.</p>
<p><span class="math-container">$r(t) = \sin(\sqrt{e^t+1}) \hat{i} - \cos(\sqrt{e^t+1}) \hat{j} + 0 \hat{k}$</span></p>
<p>Attempt:
v = (e^t cos(sqrt(e^t+1))/(2sqrt(e^t+1) i + (e^t sin(sqrt(e^t+1))/(2sqrt(e^t+1)j
|v| = 0.5e^t sqrt(1/e^t+1)</p>
<p>I am not able to calculate the curvature and the equation of the osculating circle</p>
| Michael Hoppe | 93,935 | <p>It is much easier to tackle the problem in common. The curve $r$ in question is a curve in the Euclidian plane. Its curvature $\kappa_r$ is the normal component of the acceleration divided by the square of its velocity. Define $J(a,b):=(-b,a)$ to be the rotation by $\pi/2$ as usual. So
$$\kappa_r=\frac{\langle \ddot r,\frac{J\dot r}{\|\dot r\|}\rangle}{\|\dot r\|^2}=\frac{\langle\ddot r,J\dot r\rangle}{\|\dot r\|^3}.$$
That's for the facts.</p>
<p>Let $r\circ\varphi$ be a $C^2$ reparametrization of $r$ with $\varphi'>0$.</p>
<p>In our case $r(t)=\bigl(\sin(t),-\cos(t)\bigr)$ and $\varphi=\sqrt{e^t+1}$; surely $\varphi'>0$. Let's compute $\kappa_{r\circ\varphi}$. The first derivative of $r\circ\varphi$ is $\dot r(\varphi)\cdot\varphi'$ and the second one is $\ddot r(\varphi)\cdot \varphi'^2+\dot r(\varphi)\cdot\varphi''$. Hence
$$\kappa_{r\circ\varphi}=\frac{\langle\ddot r(\varphi)\cdot \varphi'^2+\dot r(\varphi)\cdot\varphi'',J(\dot r(\varphi)\cdot\varphi' \rangle}{\|\dot r(\varphi)\cdot\varphi'\||^3}.$$
Since $v\perp Jv$ and $J$ is linear we have
$$\kappa_{r\circ\varphi}=\frac{\langle\ddot r(\varphi)\cdot \varphi'^2,J(\dot r(\varphi)\cdot\varphi' \rangle}{\|\dot r(\varphi)\cdot\varphi'\||^3}$$
and as $\varphi'>0$ we arrive in </p>
<p>$$\kappa_{r\circ\varphi}=\frac{\varphi'^3\cdot\langle\ddot r(\varphi),J(\dot r(\varphi)\rangle}{\varphi'^3\cdot\|\dot r(\varphi)\|^3}=\frac{\langle\ddot r(\varphi),J(\dot r(\varphi)\rangle}{\|\dot r(\varphi)\|^3}=\kappa_r\circ\varphi,$$
i.e., <em>reparametrization which preserves orientation doesn't change curvature.</em> (If $\varphi'<0$ we had arrive in $\kappa_{r\circ\varphi}=-\kappa_r\circ\varphi$.)</p>
<p>From here we're done as $r$ in our case represents the unit circle $S^1$, so $\kappa_r=1$.</p>
|
3,440,093 | <p>The problem is minimize over all <span class="math-container">$\theta \in \mathbb{R}^n$</span></p>
<p><span class="math-container">$$\frac{1}{2} ||Y - \theta||^2$$</span> subject to <span class="math-container">$A \theta = 0$</span> where <span class="math-container">$A$</span> is <span class="math-container">$m \times n$</span>. </p>
<p>Let <span class="math-container">$\theta^*$</span> be the solution. My notes seem to suggest <span class="math-container">$$\theta^* = (I - A'(AA')^{-1} A)Y = (I - P)Y$$</span></p>
<p>where <span class="math-container">$P = A'(AA')^{-1} A $</span> is being referred to as the projection matrix. </p>
<p>If <span class="math-container">$P$</span> is projection into, maybe <span class="math-container">$\ker$</span> <span class="math-container">$A$</span>, then <span class="math-container">$\theta^*$</span> should be <span class="math-container">$PY$</span>. But since <span class="math-container">$\theta^* = Y - YP$</span>, this suggests <span class="math-container">$P$</span> is projection into <span class="math-container">$(\ker A)^\perp$</span>. </p>
<p>Can someone prove this? Or in general prove why <span class="math-container">$\theta^* = (I - P)Y$</span>?</p>
| pre-kidney | 34,662 | <p>Another way to phrase the question is to find which vector in the subspace <span class="math-container">$V=\ker A$</span> is closest to the vector <span class="math-container">$Y\in \mathbb R^n$</span>.</p>
<p>Suppose we can find <span class="math-container">$v\in V$</span> such that <span class="math-container">$Y-v$</span> is orthogonal to the subspace <span class="math-container">$V$</span>. Then by the pythagorean theorem, for any vector <span class="math-container">$w\in V$</span> we would have
<span class="math-container">$$
\|w-Y\|^2=\|(w-v)+(v-Y)\|^2=\|w-v\|^2+\|v-Y\|^2\geq \|v-Y\|^2,
$$</span>
and since equality is attained when <span class="math-container">$w=v$</span> it follows that <span class="math-container">$v$</span> would be the closest vector to <span class="math-container">$Y$</span> that lies in <span class="math-container">$V$</span>.</p>
<p>The mapping sending <span class="math-container">$Y$</span> to <span class="math-container">$v$</span> is exactly what is meant by "orthogonal projection": it is a linear operation, and thus can be represented by a matrix, with the explicit formulas you have written down. If you are looking for a proof of such formulas, it can be achieved using a judicious chose of basis.</p>
|
2,210,893 | <p>A lot of times when proving for example inequalities like $$x \leq y$$
for real numbers $x,y$ the argument looks like
$$x \leq y + \varepsilon$$
for all $\varepsilon > 0$, hence $x \leq y$. </p>
<p>Now this is obviously very intuitive, but is there a "proof" that this conclusion is correct? And is it always sufficient in order to proof $x \leq y$ to show $x \leq y + \epsilon$ for all $\varepsilon > 0$? </p>
<p>I'd appreciate any explanations!</p>
<p><strong>NOTE:</strong> I know that these kinds of arguments are correct when dealing with sequences. But here we have no sequences so I wanted to understand this too. </p>
| fleablood | 280,126 | <p>A silly direct proof (for those who don't) like proofs by contradiction).</p>
<p>$x\le y +\epsilon$ for all $\epsilon > 0$.</p>
<p>Let $\tau =y-x $.</p>
<p>$y= x +\tau $ so $\tau \le 0$. </p>
<p>So $y \le x $</p>
<p>$-\tau \ge 0$. So as $x\ge y $ we have $x+(-\tau) \ge y+(-\tau) =y-y+x = x $. So $-\tau \le 0$.</p>
<p>Well, I guess technically that is a proof by contradict. Still... I thought it was cute.</p>
<p>So $\tau = y-x=0$. And $x=y $.</p>
|
2,359,700 | <p>Given the vector space, $ C(-\infty,\infty)$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{a^x\mid a \ge 1 \}$, a subspace of the given vector space?</p>
<p>As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it). </p>
<p>My argument so far is that the set $U$ is a subset of the set of all differentiable functions, which itself is a subset of $C(-\infty,\infty)$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).</p>
| StackTD | 159,845 | <p>If you're going over these kind of limits, you have probably heard of the <em>path test</em>.</p>
<p>If you can find two different paths leading to $(0,0)$ but where the limits along those paths are different, then the initial, two-variable limit does not exist.</p>
<p><strong>Hint</strong>: looking at the powers of $x$ and $y$, try $y=x$ and $y=x^2$. What do you get?</p>
|
3,099,815 | <p>I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone has an approach example to post I would be very grateful:</p>
<p>"Consider a relation <span class="math-container">$R$</span> defined on the set of integers. Determine for the following if the relation is reflexive, symmetric, and transitive: <span class="math-container">$R = \{(x, y)|x = 2y \}.$</span>"</p>
| MJD | 25,554 | <p>Let's ask if <span class="math-container">$R$</span> is symmetric. For <span class="math-container">$R$</span> to be symmetric, we would have to have <span class="math-container">$$y\ R\ x\qquad\text{whenever}\qquad x\ R\ y.$$</span> </p>
<p>According to the definition of <span class="math-container">$R$</span> that you gave, <span class="math-container">$x\ R \ y$</span> just means <span class="math-container">$x=2y$</span>. So what we are really asking about is whether it's true that <span class="math-container">$$y=2x\qquad\text{whenever}\qquad x=2y.$$</span></p>
<p>But simple algebra tells us this only happens for <span class="math-container">$x=0, y=0$</span>. In particular, when <span class="math-container">$x=2, y=1$</span> we have <span class="math-container">$x\ R\ y$</span> but not <span class="math-container">$y\ R\ x$</span>. So <span class="math-container">$R$</span> is <em>not</em> symmetric.</p>
<p>Now you try the other two.</p>
|
4,380,475 | <p>I'm trying to differentiate <span class="math-container">$x\sqrt{4-x^2}$</span> using the definition of derivative.</p>
<p>So it would be something like</p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$</span></p>
<p>I was trying to solve and I just can end up with something like</p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$</span></p>
<p><span class="math-container">$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$</span></p>
<p>Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like <span class="math-container">$h(2x^2+4)$</span> that would cancel up.</p>
<p>I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".</p>
<p>The final result shall be
<span class="math-container">$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$</span></p>
| Andrew D. Hwang | 86,418 | <p>As Spivak points out in his <em>Calculus</em>, the proofs of limit theorems are strategies for implementing the definition of a limit.</p>
<p>Here, guided by the trick
<span class="math-container">\begin{align*}
\frac{f(x + h)g(x + h) - f(x)g(x)}{h}
&= \frac{f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)}{h} \\
&= f(x + h)\frac{g(x + h) - g(x)}{h} + \frac{f(x + h) - f(x)}{h}g(x)
\end{align*}</span>
in the proof of the power rule, we can add and subtract from the numerator either of
<span class="math-container">$$
f(x + h)g(x) = (x + h)\sqrt{4 - x^{2}},\qquad
f(x)g(x + h) = x\sqrt{4 - (x^{2} + 2xh + h^{2})}.
$$</span>
(Alex's (+1) answer is another approach, but the one here applies more generally.)</p>
|
1,988,419 | <p>Any hint for proving this? If $Y$ is a subspace of $X$, what I am able to find is a closed subset $V$ in $Y$, hence $\mbox{cl}_Y(V)$ is compact, whose closure is contained in a neighborhood of a point $x$, by regularity of $Y$. Restricted to $X$, this $V_x=V \cap X$ is closed. I dont see any way to prove that $V_x$ is compact in $X$ but this is the obvious path. Thanks a lot for any suggestion.</p>
| DanielWainfleet | 254,665 | <p>Let $X=[0,1]$ and $Y=X\cap \mathbb Q.$ Suppose $p\in U\subset Y$ where $U$ is a nbhd ,in $Y,$ of $p.$ Let $U\supset U'\cap Y$ where $U'$ is open in $X$ and $p\in U'.$ There exists $[a,b]\subset X$ with $a<b$ and $p\in [a,b]\cap Y$ and $[a,b]\subset U'.$ Now $$Cl_Y(U)=Y\cap Cl_X(U)\supset Y\cap Cl_X(U'\cap Y)\supset Y\cap Cl_X([a,b]\cap Y)=Y\cap [a,b].$$ Let $(x_n)_{n\in \mathbb N}$ be a strictly increasing sequence of irrationals and let $(y_n)_{n \in \mathbb N}$ be a strictly decreasing sequence of irrationals, with $a<x_1<y_1<b,$ and with $\lim_{n\to \infty}x_n=\lim_{n\to \infty}y_n\not \in \mathbb Q.$ Then $$\{[0,x_1)\cap Y, (y_1,1]\cap Y \}\;\cup \{(x_n,x_{n+1})\cap Y: n\in \mathbb N\}\;\cup \{(y_{n+1},y_n)\cap Y\}$$ is an open cover, in $Y,$ of $Cl_Y(U) ,$ without a finite sub-cover.</p>
<p>So no nbhd in $Y$ of any $p\in Y$ has compact closure in $Y.$</p>
|
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