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 |
|---|---|---|---|---|
222,105 | <p>I want to use geometric shapes in Mathematica to build complex shapes and use my raytracing algorithm on it. I have a working example where we can get the intersections from a combination of a <code>Cone[]</code> and <code>Cuboid[]</code>, e.g </p>
<pre><code>shape1 = Cone[];
shape2 = Cuboid[];
(* add shapes in this list to make a more complicated shape *)
shapes = {shape1, shape2};
(* this constains the shapes so the shape is considered as a whole *)
constraints[shapes__] :=
And[## & @@ (Not /@
Through[(RegionMember[RegionIntersection@##] & @@@
Subsets[{shapes}, {2}])@#]),
RegionMember[RegionUnion @@ (RegionBoundary /@ {shapes})]@#] &
direction = {-0.2, -0.2, -1};
point = {0.5, 0.5, 1.5};
line = HalfLine[{point, point + direction}];
intersections[l_, s__] :=
NSolve[# ∈ l && constraints[s][#], #] &@({x, y, z}[[;; RegionEmbeddingDimension[l]]])
(* find intersection *)
intersection = intersections[line, ##] & @@ shapes;
points = Point[{x, y, z}] /. intersection;
Graphics3D[{{Opacity[0.2], shapes}, line, {Red, points}},
PlotRange -> {{-1, 1}, {-1, 1}, {-2, 2}}, Axes -> True]
</code></pre>
<p>This works well, and we get the external intersections as we expect. </p>
<p><a href="https://i.stack.imgur.com/3SCo1m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3SCo1m.png" alt="enter image description here"></a></p>
<p>Now, let us try to take the difference between two shapes, modelling something like </p>
<pre><code>square = Cuboid[];
ball = Ball[{0, 0, 1}, 1];
Region[RegionDifference[square, ball]]
</code></pre>
<p><a href="https://i.stack.imgur.com/tz64fm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tz64fm.png" alt="enter image description here"></a></p>
<pre><code>shapes = {RegionDifference[square, ball]};
direction = {0, 0, -1};
point = {0.5, 0.5, 5};
line = HalfLine[{point, point + direction}];
intersection = intersections[line, ##] & @@ shapes
</code></pre>
<p>Does not work, with an error that the constraints are "<em>not a quantified system of equations and inequalities</em>"...even though the constraints look fine </p>
<pre><code>constraints[shapes]
(* (##1 &) @@
Not /@ Through[
Apply[RegionMember[RegionIntersection[##1]] &,
Subsets[{{BooleanRegion[#1 && ! #2 &, {Cuboid[{0, 0, 0}],
Ball[{0, 0, 1}, 1]}]}}, {2}], {1}][#1]] &&
RegionMember[
RegionUnion @@
RegionBoundary /@ {{BooleanRegion[#1 && ! #2 &, {Cuboid[{0, 0,
0}], Ball[{0, 0, 1}, 1]}]}}][#1] & *)
</code></pre>
| Frei Zhang | 79,635 | <p>This issue could be solved by "sphere tracing", it's similar to Tim's particle advancer" and reasonable fast which will converge in a very small amount of steps.</p>
<pre><code>RayIntersect[ray_, region_, maxIteration_, maxRadius_,
radiusThreshold_] := Module[{ rnf, RegionMarcher, result},
rnf = RegionNearest[region];
RegionMarcher[orig_, radius_, step_] :=
Block[{next, closestPoint, advOp, radiusCurrent},
closestPoint = rnf[orig];
advOp =
If[Dot[closestPoint - orig, dir /. ray] > 0,
Subtract, {a, b} |-> b - a];
radiusCurrent = Max[Abs[advOp @@ {closestPoint , orig}]] // N;
next = orig + radiusCurrent * (dir /. ray);
Sow[{orig, next, Abs[radiusCurrent], advOp, step + 1}];
{next, Abs[radiusCurrent], step + 1}
] ;
result =
NestWhile[RegionMarcher @@ # &, {origin /. ray, 1, 0},
e |-> (e[[2]] < maxRadius && e[[2]] > radiusThreshold &&
e[[3]] < maxIteration)];
If[result[[2]] >= maxRadius || result[[3]] >= maxIteration, None,
result[[1]]]
]
RayIntersect[ray_, region_] :=
RayIntersect[ray, region, 10, 256, 0.01]
</code></pre>
<p>And the result:</p>
<pre><code>Graphics3D[{Opacity[.5], Blue,
DiscretizeRegion[RegionDifference[square, ball]],
Arrow[{point, point + direction}], Red,
Map[(e |-> { Sphere[e[[1]], e[[3]]]}),
Reap[RayIntersect[{origin -> point, dir -> direction},
RegionDifference[square, ball]]] // Rest // First // First]},
Axes -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/WxsS4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WxsS4.png" alt="sphere tarcing" /></a></p>
|
2,237,963 | <p>One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.</p>
<p>Let $X$ be a topological space. Then the One-point compactification of $X$ is a certain compact space $X^*$ together with an open embedding $c : X \to X^*$ such that the complement of $X$ in $X^*$ consists of a single point, typically denoted $\infty$. </p>
<p>Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $S_{\alpha} = \{x \mid x \in X \text{ and }x < \alpha \}$.
It is called the section of $X$ by $\alpha$. </p>
<p>I am finding difficulty to do the problem!</p>
| Michael Rozenberg | 190,319 | <p>Since $\sin$ is a concave function on $[0,\pi]$ and sum of concave functions is a concave function, </p>
<p>we have
$$\min_{[0,\pi]}f=\min\{f(0),f(\pi-1),f(\pi-2),f(\pi)\}=f(\pi-1)=2\sin1>\frac{8}{5}$$</p>
|
183,881 | <p>I want to replace an element of a matrix by a function of its current value. The code I'm using is as follows</p>
<pre><code>T = {{2, 3}, {3, 2}, {1, 4}}
KNeeds = Transpose[T][[2]]
pos = Flatten[{Position[KNeeds, Min[KNeeds]], 2}]
ReplacePart[T, pos -> T[[pos[[1]]]] - Epsilon]]
</code></pre>
<p>I was hoping to obtain <code>{{2, 3}, {3, 2-Epsilon}, {1, 4}},</code> but I get <code>{{2, 3}, {3, {3 - Epsilon, 2 - Epsilon}}, {1, 4}}.</code> </p>
<p>I've also tried <code>ReplacePart[T, pos -> T[[pos]] - Epsilon]],</code> but then I get <code>{{2, 3}, {3, {{3 - Epsilon, 2 - Epsilon}, {3 - Epsilon,
2 - Epsilon}}}, {1, 4}}.</code></p>
| kirma | 3,056 | <p>You can also acquire a symbolic solution to this question with <code>Total</code> of <code>Boole</code> (truth value to 0/1) values of membership on parametric regions:</p>
<pre><code>FullSimplify@
Total[Function[{u, v, w},
Evaluate@
Boole@RegionMember[
RegionIntersection[
Cylinder[{{3, 0, 0}, {3, 0, 2}}, x],
Cylinder[{{7, 0, 0}, {7, 0, 2}}, x]], {u, v, w}]] @@@
Table[{i, 0, 1}, {i, 0, 10}]]
</code></pre>
<blockquote>
<p>Boole[x >= 2] +
2 (Boole[x >= 3] + Boole[x >= 4] + Boole[x >= 5] + Boole[x >= 6] + Boole[x >= 7])</p>
</blockquote>
<p>And plot it:</p>
<pre><code>Plot[FullSimplify@
Total[Function[{u, v, w},
Evaluate@
Boole@RegionMember[
RegionIntersection[
Cylinder[{{3, 0, 0}, {3, 0, 2}}, x],
Cylinder[{{7, 0, 0}, {7, 0, 2}}, x]], {u, v, w}]] @@@
Table[{i, 0, 1}, {i, 0, 10}]], {x, 1, 10}, Evaluated -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/aOY7E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aOY7E.png" alt="enter image description here"></a></p>
|
1,706,310 | <p>Suppose X and Y are compact and they are Hausdorff topological spaces. Let $f : X \rightarrow Y$ be a continuous surjective function. Prove that any $U \subset Y$ is open if and only if $f^{−1}(U)$ is open.</p>
<p>I got stuck here because I don't know how to use compactness. Can someone help please?
Thanks</p>
| carmichael561 | 314,708 | <p>If $U$ is open then $f^{-1}(U)$ is open by the definition of continuity.</p>
<p>On the other hand, if $f^{-1}(U)$ is open, then its complement $X\setminus f^{-1}(U)$ is closed, hence compact because $X$ is compact.</p>
<p>Therefore $f(X\setminus f^{-1}(U))$ is compact because $f$ is continuous, and since $Y$ is Hausdorff this implies that $f(X\setminus f^{-1}(U))$ is closed. Finally $f(X\setminus f^{-1}(U))=Y\setminus U$ since $f$ is surjective, hence $U$ is open.</p>
<p>The reason compactness is important here is that the image of a closed set under a continuous function need not be closed.</p>
|
1,785,044 | <p>Looking at the picture below, it's easy to see why the perimeter of a polygon inscribed in a circle is an underestimation of the circle's perimeter. This follows from the triangle inequality: Any side (say $AB$) of the polygon is shorter than the circular arc with the same endpoints ($\stackrel{\frown}{AB}$). Summing all these inequalities shows the perimeter of the inscribed polygon is indeed smaller than that of the circle.</p>
<p>I'm wondering if there is proof that the perimeter of a circumscribed polygon always overestimates the perimeter of the circle, which is as simple as that of the inscribed polygon case. Thanks!</p>
<p><a href="https://i.stack.imgur.com/iYJuW.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYJuW.gif" alt="enter image description here"></a></p>
| marty cohen | 13,079 | <p>This expands
Yves Daoust's comment.</p>
<p>Call the point that
the tangent from $D$
touches the circle
$P$,
and the point where
$DE$ intersects the circle
$Q$.</p>
<p>Then
$DEQ$ is a right triangle.</p>
<p>Let
$t = \angle DEP$.</p>
<p>Then
$\tan(t)
=\dfrac{DP}{PE}
$
so
$DP
=PE \tan(t)
$.</p>
<p>We also have
the length of
arc $QP= t\,PE$.</p>
<p>Therefore
$\dfrac{DP}{arc\ QP}
=\dfrac{PE \tan(t)}{t\,PE}
=\dfrac{\tan(t)}{t}
\gt 1
$
for
$t > 0$.</p>
<p>Since the same holds on both sides,
the sum of the lengths
of the two tangents
is greater than
the length of arc
$DE$
by a factor
$\dfrac{\tan(t)}{t}
$.</p>
|
4,320,849 | <p>I had this problem in an exam I recently appeared for:</p>
<blockquote>
<p>Find the range of
<span class="math-container">$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$</span></p>
</blockquote>
<p>By randomly assuming the value of <span class="math-container">$x$</span>, I got the lower range of this expression as <span class="math-container">$3/7$</span>. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.</p>
<p>Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.</p>
<p>Another guy I met outside the exam center, told me he used an approach of <span class="math-container">$x$</span> tending to infinity in both cases and got the maximum value of this expression as <span class="math-container">$1/2$</span>. But before I could ask him to explain more on this method, he had to leave for his work.</p>
<p>So, will someone please throw some light on this method of <span class="math-container">$x$</span> tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).</p>
| Spectre | 799,646 | <p>As a follow-up to @NikolaAlfredi's answer:</p>
<p><span class="math-container">$ y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \frac{1}{2}(1-\frac{1}{2x^2+4x+9}) \implies 2y = 1 - \frac{1}{2x^2+4x+9}$</span>. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by <span class="math-container">$2$</span> to get the required extremes(taking half since we get values for <span class="math-container">$2y$</span> and not <span class="math-container">$y$</span>).</p>
|
794,301 | <p>I am trying to find out the sum (I just derived these from 2 + 0.5 + 0.125 + 0.03125 + ...):</p>
<p>$$\sum_{n=0}^{\infty} \frac{5^{2n-1}}{10^{2n -1}}$$</p>
<p>It's confusing me because it doesn't match $${ar}^{n-1}$$ the power by which $r$ is raised. </p>
| nadia-liza | 113,971 | <p>$$\sum_{n=0}^{\infty} \frac{5^{2n-1}}{10^{2n -1}}=\sum_{n=0}^{\infty} (\frac{5}{10})^{2n-1}=2\sum_{n=0}^{\infty} (\frac{1}{2})^{2n}= 2 \sum_{n=0}^{\infty} (\frac14)^n=2 \frac{1}{1-\frac14}$$</p>
|
1,513,078 | <blockquote>
<p>Suppose $b\in\mathbb Z$. Find all possible remainders of $b^3$ divided by $7$. </p>
</blockquote>
<p>I know $b^6=1\bmod7$ which means $(b^2)^3=1\bmod7$ so all square numbers^3 leave $1$ as a remainder, but how to continue?</p>
| Patrick Stevens | 259,262 | <p>Notice that this depends only on the value of $b$ mod $7$, since $(b+7k)^3 = b^3 + (\text{multiple of 7})$. So we can work wlog with $b = 0, 1, \dots, 6$.</p>
<p>Your method is also faulty: just because $b^6 \equiv 1 \pmod{7}$, that doesn't mean $b^2 \equiv 1 \pmod{7}$. For example, $3^2 \equiv 2 \pmod{7}$.</p>
<hr>
<p>There's a really brute-force way to do it, modulo 7. I'll use $=$ instead of $\equiv$ to save a bit of typing:</p>
<p>$0^3 = 0, 1^3 = 1, 2^3 = 8 = 1, 3^3 = 27 = -1, 4^3 = 64 = 1, 5^3 = (-2)^3 = -1, 6^3 = (-1)^3 = -1$</p>
<p>yielding the answers $0, 1, -1$.</p>
<p>Really, you only need to check $0, 1, 2, 3$ (and then negate those answers), because $(7-3)^3 \equiv (-3)^3 = - (3^3)$.</p>
|
265,377 | <p>Let $k$ be a finite field, and let $G$ be the absolute Galois group of $k$, which is isomorphic to $\widehat{\mathbb{Z}}$. Let $\mathcal{C}$ be the category of $G$-modules. Then, we have the following:</p>
<p>For a finite $G$-module $N$, we have
$$
Ext^r_{\mathcal{C}}(N, \mathbb{Z}) \simeq H^{r-1}(G,~N^D),
$$
where $N^D:=\operatorname{Hom}_\mathbb{Z}(N, \mathbb{Q}/\mathbb{Z})$ is the Pontrjagin dual of $N$, with the natural $G$-action by $(g\phi)(x):=g(\phi(g^{-1}(x)))=\phi(g^{-1}(x))$ for $g\in G$, $x \in N$ and $\phi \in N^D$.</p>
<p>Why is this true?</p>
| R. van Dobben de Bruyn | 82,179 | <p>Consider the short exact sequence
$$0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0.$$
Note that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q) = 0$ for all $i$: it is torsion since $N$ is torsion, but multiplication by $n \in \mathbb Z_{>0}$ is an isomorphism since it is so on $\mathbb Q$. Thus, the sequence above induces isomorphisms
$$\operatorname{Ext}^{i-1}_\mathcal C(N,\mathbb Q/\mathbb Z) \stackrel\sim\to \operatorname{Ext}^i_\mathcal C(N,\mathbb Z).$$
Thus, we only have to show that $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z) = H^i(G, N^D)$. I will treat the finite group case, and trust that the OP can carry out the limit argument to treat the profinite group $\hat {\mathbb Z}$ (assuming that you are working with discrete [topological] $G$-modules).</p>
<p>Note that if $P$ is a projective $\mathbb Z[G]$-module, then $\operatorname{Hom}_\mathbb Z(P,\mathbb Q/\mathbb Z)$ is an acyclic $\mathbb Z[G]$-module. Indeed, we prove this in three steps:</p>
<ol>
<li><p>Let $P = \mathbb Z[G]$. Then we get the co-induced module $M^G(\mathbb Q/\mathbb Z)$, which is acyclic by Shapiro's lemma.</p></li>
<li><p>The case of free modules follows by taking sums: $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ turns sums into products, and a product of acyclic modules is acyclic. Indeed (by abstract nonsense): $(-)^G$ has a left adjoint $- \otimes_\mathbb Z \mathbb Z[G]$, hence preserves products (you can also easily prove this by hand). Since products of injectives are injective and products are exact, we can take them out of $H^i(G,-)$ as well.</p></li>
<li><p>The general case follows since any projective module is the summand of a free module.</p></li>
</ol>
<p>Now consider the composition of the functors $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z) \colon \mathcal C^{\operatorname{op}} \to \mathcal C$ and $(-)^G \colon \mathcal C \to \operatorname{\underline{Ab}}$. The above argument shows that $\operatorname{Hom}_\mathbb Z(-,\mathbb Q/\mathbb Z)$ takes projectives to acyclics, hence there is a Grothendieck spectral sequence
$$E_2^{pq} = H^p(G,\operatorname{Ext}^q_\mathbb Z(N,\mathbb Q/\mathbb Z)) \Rightarrow \operatorname{Ext}^{p+q}_{\mathbb Z[G]}(N,\mathbb Q/\mathbb Z).$$
But $\mathbb Q/\mathbb Z$ is injective as $\mathbb Z$-module, hence $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z) = 0$ for $i > 0$. Thus, the spectral sequence collapses on the $E_2$ page, and we conclude that
$$H^i(G,\operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)) = \operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z).$$
$\square$</p>
<p><strong>Remark.</strong> If you don't like spectral sequences, you can also give a more concrete proof of the last part. Indeed, consider a free resolution as $\mathbb Z[G]$-modules
$$\ldots \to P_1 \to P_0 \to N.$$
Since $\mathbb Z[G]$ is a free $\mathbb Z$-module, this resolution is also a free resolution as $\mathbb Z$-modules. Thus, it computes both $\operatorname{Ext}^i_\mathbb Z(N,\mathbb Q/\mathbb Z)$ and $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$. More specifically, the former can be computed as the cohomology of the complex
$$0 \to \operatorname{Hom}_\mathbb Z(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_\mathbb Z(P_1,\mathbb Q/\mathbb Z) \to \ldots,\tag{1}\label{Seq 1}$$
and then the latter can be computed as the cohomology of
$$0 \to \operatorname{Hom}_{\mathbb Z[G]}(P_0,\mathbb Q/\mathbb Z) \to \operatorname{Hom}_{\mathbb Z[G]}(P_1,\mathbb Q/\mathbb Z) \to \ldots.\tag{2}\label{Seq 2}$$
Now (\ref{Seq 1}) is exact since $\mathbb Q/\mathbb Z$ is an injective $\mathbb Z$-module. By what we argued above, it is actually an <em>acyclic</em> resolution $Q^\bullet$ of $N^D = \operatorname{Hom}_\mathbb Z(N,\mathbb Q/\mathbb Z)$, so we can use it to compute group cohomology. Thus, $H^i(G,N^D)$ is computed as the cohomology of the complex
$$0 \to (Q^0)^G \to (Q^1)^G \to (Q^2)^G \to \ldots,$$
which on the other hand computes $\operatorname{Ext}^i_\mathcal C(N,\mathbb Q/\mathbb Z)$ by (\ref{Seq 2}). $\square$</p>
|
1,728,910 | <p>Whenever I get this question, I have a hard time with it. </p>
<p>An example of a problem:</p>
<p>In the fall, the weather in the evening is <em>dry</em> on 40% of the days, <em>rainy</em>
on 58% of days and <em>snowy</em> 2% of the days. </p>
<p>At noon you notice clouds in the sky. </p>
<p>Clouds appear
at noon on 10% of the days that have <strong>dry evenings</strong>, </p>
<p>25% of the days that have <strong>rainy evenings</strong>, </p>
<p>and
35% of the days that have <strong>snowy evenings</strong>.</p>
<p>From that I get:</p>
<p>$P(S)$ = "prob. snows" = $.02$</p>
<p>$P(R)$ = "prob. rains" = $.58$</p>
<p>$P(D)$ = "prob. is dry" = $.40$</p>
<p>$P(C|S)$ = "prob. that it is cloudy given that it snows" = $.35$</p>
<p>$P(C|R)$ = "prob. that it is cloudy given that it rains" = $.25$</p>
<p>$P(C|D)$ = "prob. that it is cloudy given that it is dry" = $.10$</p>
<p>However, the question given is:</p>
<p>Find the probability of snow in the evening given that you saw clouds at noon.</p>
<p>I interpret this as:</p>
<p>$$P(S|C)$$</p>
<p>I can't get it right. Is there a formula to it without having P(C)? </p>
| Τίμων | 319,592 | <p>Maybe you are looking for <a href="https://en.wikipedia.org/wiki/Bayes%27_theorem" rel="nofollow">Bayes' theorem</a>, which states</p>
<p>$$P(S\mid C) = \frac{P(C \mid S) \, P(S)}{P(C)}$$</p>
<p>But you need $P(C)$ too.</p>
|
226,534 | <p>I have the following function for which I want to know the range</p>
<pre><code>FunctionRange[{1/
32 (8 - Sqrt[(-8 - 36 Abs[y] (1 + 2 Sqrt[Abs[z]]) -
27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2)^2 -
64 (1 + Abs[y] + 2 Abs[y] Sqrt[Abs[z]])] +
36 Abs[y] (1 + 2 Sqrt[Abs[z]]) +
27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2),
0 < Abs[y] < 1 && 0 < Abs[z] < 1}, {Abs[y], Abs[z]}, t]
</code></pre>
<p>But once the code starts it just keeps on running with no output. I tried to use MaxValue and MinValue too so as to get the Range but that too doesn't seem to work. Is there any other way to find the range when the above options don't work?</p>
| user64494 | 7,152 | <p>This can be done by the change <code>Abs[y] -> a^2, Abs[z] -> b^2</code> in order to obtain a polynomial in <code>a</code> and <code>b</code> as follows.</p>
<pre><code>Maximize[{1/
32 (8 - Sqrt[(-8 - 36 Abs[y] (1 + 2 Sqrt[Abs[z]]) -
27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2)^2 -
64 (1 + Abs[y] + 2 Abs[y] Sqrt[Abs[z]])] +
36 Abs[y] (1 + 2 Sqrt[Abs[z]]) + 27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2),
0 < Abs[y] < 1 && 0 < Abs[z] < 1 && Sqrt[Abs[y]] > 0 &&
Sqrt[Abs[z]] > 0} /.{Sqrt[Abs[y]]-> a,Sqrt[Abs[z]] -> b,Abs[y]->a^2,Abs[z] -> b^2}, {a, b}]
(*{1/4, {a -> 0, b -> Indeterminate}}*)
</code></pre>
<p>and the warning (not an error)</p>
<blockquote>
<p>Maximize::natt: The maximum is not attained at any point satisfying the given constraints.</p>
</blockquote>
<p>The result means that <code>b</code> may be an arbitrary value (of course, <code>0<b&&b<1</code>.
Making use of <code>Minimize</code> instead of <code>Maximize</code>, one obtains</p>
<pre><code>(*{1/32 (359 - 35 Sqrt[105]), {a -> 1, b -> 1}}*)
</code></pre>
<p>and a similar warning.
It remains to notice that the polynomial, being a continuous function, takes each value between its minimum value and maximum value on a (compact) set <code>a>=0&&a<=1&&b>=0&&b<=1</code>. Numeric calculations with <code>NMaximize</code> and <code>NMinimize</code> with <code>Method->"DifferentialEvolution</code>confirm those results.
Summarizing the above, one may conclude that the range under consideration is <span class="math-container">$(\frac{1}{32} \left(359-35 \sqrt{105}\right),\frac 1 4)$</span>.</p>
|
4,169,445 | <p>I'm getting stuck on perhaps a simple step in the Hille-Yosida theorem from 13.37 in Rudin's functional analysis.
I wonder if someone has had this same difficulty before or knows how to get around it -</p>
<p><strong>Setup:</strong> <span class="math-container">$A$</span> is a densely defined operator with domain <span class="math-container">$\mathcal{D}(A)$</span> in a Banach space <span class="math-container">$X$</span> and there are constants <span class="math-container">$C, \gamma >0$</span> such that for every <span class="math-container">$\lambda > \gamma$</span> and <span class="math-container">$m\in \mathbb{N}$</span>,
<span class="math-container">$$
\| (\lambda I - A)^{-m}\| \leq C(\lambda - \gamma)^{-m}.
$$</span>
The claim is then that <span class="math-container">$A$</span> is the <em>infinitesimal generator</em> of a semi-group of operators.
For small <span class="math-container">$\varepsilon$</span>, the bounded operator <span class="math-container">$S(\varepsilon)$</span> is defined to be <span class="math-container">$(I-\varepsilon A)^{-1}:X \to \mathcal{D}(A)$</span>.
It follows from the definitions that <span class="math-container">$AS(\varepsilon) = \varepsilon^{-1}(S(\varepsilon)-I)$</span> from which one can show that <span class="math-container">$e^{tAS(\varepsilon)}$</span> converges weakly to a bounded <span class="math-container">$Q(t)$</span>. Moreover, <span class="math-container">$\{Q(t) \}$</span> gives a semi-group.
Thus it has an infinitesimal generator <span class="math-container">$\tilde{A}$</span>.
Using the resolvent formulas for <span class="math-container">$AS(\varepsilon)$</span> and <span class="math-container">$\tilde{A}$</span>, we have for all <span class="math-container">$x$</span> and <span class="math-container">$\lambda$</span> sufficiently large,
<span class="math-container">$$
(\lambda I - \tilde{A})^{-1}x = \int_0^\infty e^{-\lambda t} Q(t)x dt
$$</span>
and
<span class="math-container">$$
(\lambda I - AS(\varepsilon))^{-1}x = \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt.
$$</span>
One can easily justify the limit
<span class="math-container">$$
\lim_{\varepsilon \to 0} \int_0^\infty e^{-\lambda t} e^{tAS(\varepsilon)}x dt = \int_0^\infty e^{-\lambda t} Q(t)x dt.
$$</span></p>
<p><strong>Question:</strong> In order to compare <span class="math-container">$\tilde{A}$</span> and <span class="math-container">$A$</span>, how does one see that
<span class="math-container">$$
\lim_{\varepsilon \to 0} (\lambda I - AS(\varepsilon))^{-1}x = (\lambda I - A)^{-1}x?
$$</span></p>
<p>THANK YOU!</p>
| Kman3 | 641,945 | <p>Firstly, you have the wrong formula. The area of the semicircle in Cartesian coordinates would be</p>
<p><span class="math-container">$$\int_{-r}^r \sqrt{r^2-x^2} \ dx$$</span></p>
<p>or, as a double integral,</p>
<p><span class="math-container">$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}} \,dy \ dx$$</span></p>
<p>Secondly, you're not following the process for <a href="https://www.math24.net/change-variables-double-integrals" rel="nofollow noreferrer">changing variables in double integrals</a>, given on the linked source as
<span class="math-container">$$\iint\limits_R f(x,y) \, dx \, dy = \iint\limits_S f[x(u,v),y(u,v)] \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, du\,dv$$</span></p>
<p>which ultimately leads to your confusion.</p>
<p>When you change variables in double integrals, you need to compute the <strong>Jacobian</strong> of your transformation. That means that when you substitute <span class="math-container">$x=r\cos \theta$</span> and <span class="math-container">$y=r\sin \theta$</span>, you need to compute</p>
<p><span class="math-container">$$\left| \frac{
\partial(x,y)}{
\partial(r, \theta)}
\right| =\left| \begin{matrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{matrix} \right|$$</span></p>
<p>In this case, that gives you</p>
<p><span class="math-container">$$\left| \begin{matrix}
\cos \theta & -r\sin\theta \\
\sin \theta & r\cos\theta
\end{matrix} \right| =r(\cos^2\theta+\sin^2\theta)=r$$</span></p>
<p>Once you have this, you can safely convert the integral into polar coordinates to get</p>
<p><span class="math-container">$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}}dy \ dx=\int_0^\pi \int_{0}^r r \ dr \ d\theta=\frac{\pi r^2}{2}$$</span></p>
|
1,348,046 | <p>I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?</p>
| Zain Patel | 161,779 | <p><strong>Hint:</strong></p>
<p>$$\frac{3n-4}{n(n-1)(n-2)} \equiv \frac{1}{n-1} + \frac{1}{n-2} - \frac{2}{n}$$ </p>
|
1,348,046 | <p>I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?</p>
| robjohn | 13,854 | <p>Using the <a href="https://en.wikipedia.org/wiki/Heaviside_cover-up_method" rel="nofollow">Heaviside Method of Partial Fractions</a> to solve
$$
\frac{3n-4}{n(n-1)(n-2)}=\frac{A}{n}+\frac{B}{n-1}+\frac{C}{n-2}
$$
we get</p>
<p>multiply by $n$ and set $n=0\implies\frac{3\cdot\color{#C00000}{0}-4}{(\color{#C00000}{0}-1)(\color{#C00000}{0}-2)}=A\implies A=-2$<br>
multiply by $n-1$ and set $n=1\implies\frac{3\cdot\color{#C00000}{1}-4}{\color{#C00000}{1}(\color{#C00000}{1}-2)}=B\implies B=1$<br>
multiply by $n-2$ and set $n=2\implies\frac{3\cdot\color{#C00000}{2}-4}{\color{#C00000}{2}(\color{#C00000}{2}-1)}=C\implies C=1$</p>
<p>Thus,
$$
\frac{3n-4}{n(n-1)(n-2)}=-\frac2n+\frac1{n-1}+\frac1{n-2}
$$
Then applying <a href="https://en.wikipedia.org/wiki/Telescoping_series" rel="nofollow">Telescoping Series</a> to get
$$
\begin{align}
\sum_{n=3}^N\frac{3n-4}{n(n-1)(n-2)}
&=-2\sum_{n=3}^N\frac1n+\sum_{n=3}^N\frac1{n-1}+\sum_{n=3}^N\frac1{n-2}\\
&=\color{#C00000}{-2\sum_{n=3}^N\frac1n}\color{#00A000}{+\sum_{n=2}^{N-1}\frac1n}\color{#0000F0}{+\sum_{n=1}^{N-2}\frac1n}\\
&=\color{#0000F0}{1+\frac12}\color{#00A000}{+\frac12+\frac1{N-1}}\color{#C00000}{-\frac2{N-1}-\frac2N}
\end{align}
$$
Now take the proper limit.</p>
<p>Note that we can't sum from $n=0$ since the denominators will be $0$, so I assumed we were to start at $n=3$.</p>
|
259,751 | <p>I'm playing this video game where people can get kills, deaths, and assists , and all this is recorded on a stats website. The stats website gives you a rating by directly manipulating these numbers.</p>
<p>In the first entry, I have 26 kills, 5 deaths, and 19 assists. The KDA ratio the website gave me was 29.8. At first thought, I guessed that the formula was ratio = (kills*5 + assists)/deaths.</p>
<p>But then the second entry threw me off. 21.33 Kills, 1.33 deaths, and 4.33 assists. And the ratio outputted here is 24.58.</p>
<p>Is there a good approach to figuring out the formula or function when the inputs and outputs are given?</p>
| Mario Carneiro | 50,776 | <p>I thought I'd give an overview of the prime-generating functions in <a href="http://www.sonoma.edu/math/colloq/primes_sonoma_state_9_24_08.pdf" rel="nofollow">Is there a formula that Generates Prime Numbers?</a> and <a href="http://mathworld.wolfram.com/PrimeFormulas.html" rel="nofollow">MathWorld: Prime Formulas</a>, with attention to the constraints set above.</p>
<p>Wilson's Theorem (1770):</p>
<p>$$n\in\mathbb P\iff\frac{(n-1)!-1}n\in\mathbb Z$$</p>
<p>This one is possibly useful, but is only a prime indicator, not a formula.</p>
<p>Willans (1964):</p>
<p>$$f(n)=p_n=1+\sum_{k=1}^{2^n}\left\lfloor\sqrt[n]\frac n{1+\pi(k)}\right\rfloor$$</p>
<p>This one is plainly useless due to the presence of the <a href="http://en.wikipedia.org/wiki/Prime-counting_function" rel="nofollow">prime-counting function</a> $\pi(k)$. Since $f(n)$ depends on $\pi(k)$ up to $k=2^n$, this cannot even be defined recursively. However, Willans gives a non-recursive definition for $\pi(k)$, using Wilson's theorem:</p>
<p>$$\pi(k)=-1+\sum_{j=1}^k\left\lfloor\cos^2\bigg[\pi\frac{(j-1)!-1}j\bigg]\right\rfloor$$</p>
<p>In asymptotic-analysis terms, if we assume the $\cos^2x$ application is $O(1)$ (which is reasonable, because it is being abused here to act as an <a href="http://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow">Iverson bracket</a>), then the full formula for $f(n)$ is $O(\sum_{k=1}^{2^n}\sum_{j=1}^kj)=O(8^n)$. A better formula to get $f(n)$ from $\pi(k)$ (Martin-Ruiz & Sondow 2002) is</p>
<p>$$f(n)=p_n=1+\sum_{k=1}^{2(\lfloor n\log n\rfloor+1)}\left(1-\left\lfloor\frac{\pi(k)}n\right\rfloor\right)$$</p>
<p>which is $O((n\log n)^3)$, assuming $\pi(k)=O(k^2)$. A better formula for $\pi(k)$ also due to (Martin-Ruiz & Sondow 2002) is</p>
<p>$$\pi(k)=k-1+\sum_{j=1}^k\left\lfloor\frac2j\left(1+\sum_{s=1}^{\lfloor\sqrt j\rfloor} \left(\left\lfloor\frac{j-1}s\right\rfloor-\left\lfloor\frac js\right\rfloor\right) \right)\right\rfloor$$</p>
<p>which is only $O(k^{3/2})$, so that combining these gives $f(n)=O((n\log n)^{5/2})$.</p>
<p>By using a "real" primality test, $f^+(n)=\min(\mathbb P\setminus\{1,\dots,n\})$ can be calculated in $O(n\log^{6+\varepsilon}n)$ (using <a href="http://en.wikipedia.org/wiki/AKS_primality_test" rel="nofollow">AKS</a>), so $f(n)=(f^+)^{(n-1)}(2)$ can be calculated, but the asymptotics of this depend on the distribution of primes. The best known (provable) upper bound is $p_n\leq2^n$, so $f(n)=O(2^n\log^{6+\varepsilon}n)$ this way.</p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| Joel | 85,072 | <p>You can also compute it with the following method. First note that $50 = 32 + 16 + 2$. Let $$A=\left(\begin{array}{cc}2&0\\2&1\end{array}\right).$$</p>
<p>Then $$A^{50} = A^{32} A^{16} A^2$$.</p>
<p>We can compute $A^{2^n}$ easier than $A^{50}$,</p>
<p>$$A^2 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)$$</p>
<p>$$A^4 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)\left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)= \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right)$$</p>
<p>$$A^8 = \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)$$</p>
<p>$$A^{16} = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)\left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)=\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)$$</p>
<p>and</p>
<p>$$A^{32} = \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)=\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right).$$</p>
<p>Finally $$A^{50} =\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right) \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right) \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right).$$</p>
<p>This isn't the best way to go about it, but it is a lot easier than multiplying $A$ to itself 50 times. You should probably persue @science's method to be honest.</p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| vudu vucu | 215,476 | <p>Hint: Find a matrix $P$ such that $P^{-1}AP=D$ where $D=diag(2,1)$. Then $P^{-1}A^{50}P=D^{50}=diag(2^{50},1).$</p>
|
167,891 | <p>I'm working through the James Stewart Calculus text to prep for school. I'm stuck at this particular point. </p>
<p>How would you sketch the graph for the parametric equations:
$x = \cos t$, $y = \sin t$, and $z = \sin 5t$? I understand that if it were the case that $z=t$, I'd merely get a helix around the $z$-axis, as $x$ and $y$ form an ellipse. However, I cannot make the leap to solve more exotic problems such as the problem posed or even the case when $z = \ln(t)$.</p>
<p>Some help and a push in the right direction would be appreciated.</p>
| robjohn | 13,854 | <p><strong>Hint:</strong> Note that the $x$ and $y$ coordinates trace out a circle. As they do, the $z$ coordinate goes through $5$ sinusoidal cycles.</p>
|
2,815,900 | <p>Consider the proposition $∀x (P (x) → Q (x))$, where $P (x)$ and $Q (x)$ are predicates on a domain $U$. Prove or disprove the following statement, justifying the answer.</p>
<p>If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$</p>
<p>How can I solve this exercise?</p>
| Bram28 | 256,001 | <p>It's typically a good strategy to try to come up with a counterexample to the given statement: if you can find a counterexample, then the statement is false, and if you can't, then you probably get some idea why you can;t, and that might translate into a proof why the statement is in fact true.</p>
<p>Now, to find a counterexample to your given statement:</p>
<blockquote>
<p>If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$</p>
</blockquote>
<p>we should try to set $∀x (P (x) → Q (x))$ to FALSE, but $∀x P (x)$ *not to FALSE, i.e. to TRUE.</p>
<p>The latter means that everything is a $P$, and the former means that not every $P$ is a $Q$.</p>
<p>So, can you think of a domain of objects where every object in the domain has some property (which will be $P$), but not everything has some othr property (which will be $Q$)?</p>
|
1,721,055 | <p>This afternoon I've been studying the pythagorean identities & compound angles. I've got a problem with a question working with 2 sets of compound angles:</p>
<blockquote>
<p>Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$ </p>
</blockquote>
<p>I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding? </p>
<p>$$\cos\theta\cos 25^\circ-\sin\theta \sin 25^\circ+\sin\theta\cos 65^\circ+\sin 65^\circ\cos\theta=1$$</p>
<p>$$\cos \theta\;\left(\cos 25^\circ+\sin 65^\circ\right) +\sin\theta\;\left(\cos 65^\circ-\sin 25^\circ\right)=1$$</p>
<p>Could you tell me if I've made a mistake or how I could continue;
thanks</p>
<p><em>coffee is wearing out</em></p>
| Jalejo | 97,487 | <p>"Solve, in the interval 0≤θ≤360, cos(θ+25)+sin(θ+65)=1"</p>
<p>$\cos(\theta+25)+\cos(25-\theta)=1$</p>
<p>∵ $\sin(\theta+65)=\cos(90-(\theta+65))=\cos(25-\theta)$</p>
<p>$\cos(\theta)\cos(25)- \sin(\theta)\sin(25)+\cos(25)\cos(\theta)+\sin(\theta)\sin(25)=1$</p>
<p>$2\cos(\theta)\cos(25)=1$</p>
<p>$\cos(\theta)=\frac{1}{2\cos(25)}$</p>
<p>∴$\theta≈55.6°, 326.5°$</p>
|
89,638 | <p>I'm trying to solve an exercise as it follows:</p>
<p>$\alpha, \beta \in \mathbb{C}$ such that $a^{37}=2=\beta^{17}$. Note that both are prime.</p>
<p>a) Find the minimum polynomials of $\alpha$ and $\beta$ over $\mathbb{Q}$.</p>
<p>Well, I guess that $(x^{37}-2)$ and $(x^{17}-2)$ are irreducible over $\mathbb{Q}$, hence minimum polynomials. But does that follow from 37,17 being prime, or just from the fact that the polynomial has no zero in $\mathbb{Q}$? How do I write this down rigorously?</p>
<p>b) Compute $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}]$, and compute $[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$!</p>
<p>My guess here on the first one is that this is equal to $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\beta)}]*[\mathbb{Q}( \beta):\mathbb{Q}]$ and $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\alpha)}]*[\mathbb{Q}( \alpha):\mathbb{Q}]$, so the first product is $17x$, the second is $37y$, so the dimension of field extension is at least $17\times 37$, but this is also the maximum number (right?), so $17\times 37$ is the answer? and this would be the same answer as for the last part, $\mathbb{Q}(\alpha \beta)$? How do I write this down nicely?</p>
| Dilip Sarwate | 15,941 | <p>Suppose that $\psi(t)$ is the characteristic function corresponding to $f(x)$. Using the properties of Fourier transforms, we have that </p>
<ul>
<li><p>since $f(x)$ is a real-valued function, $\psi(t)$ has <em>conjugate symmetry:</em>
$\psi(-t) = [\psi(t)]^*.$</p></li>
<li><p>the characteristic function corresponding to $f(-x)$, the density of $-X$, is $\psi(-t)$.</p></li>
</ul>
<p>Now, $\dfrac{1}{2}(f(x)+f(-x))$ is a density and the corresponding characteristic function is
$$\frac{1}{2}(\psi(t) + \psi(-t)) = \frac{1}{2}(\psi(t) + [\psi(t)]^*)
= \text{Re}(\psi(t)).$$</p>
|
1,571,083 | <p>Given the 2 terms
$$ \frac{k + a}{k + b}$$
and $$\frac{a}{b}$$
with $a, b, k \in \mathbb{R^+}$ and $a > b$</p>
<p>I want to show, that the first term is always bigger than the second one.</p>
<p><strong>My try</strong>
$$
\frac{k + a}{k + b} > \frac{a}{b} \\
\frac{k + b + (a - b)}{k + b} > \frac{b + (a - b)}{b} \\
1 + \frac{a - b}{k + b}> 1 + \frac{a - b}{b} \\
\frac{1}{k + b} > \frac{1}{b} \\
k + b > b
$$
the problem is, that I think the inequality sign has to change to $<$ in the last step because of calculating the inverse on both sides. That would mean
$$ k + b < b$$
which is nonsense.</p>
| Colm Bhandal | 252,983 | <p>Assuming the events are independent and everyone has the same chance of winning, then the odds of the same person winning twice is:</p>
<p>$$\frac{2}{31} \times \frac{1}{25} = \frac{2}{775}$$</p>
<p>In other words, they have a $2$ in $775$ or roughly $0.26\%$ chance of winning both. Unlikely, but not impossible.</p>
<hr>
<p><strong>Update:</strong> In light of Paw's answer above, I must make it clear here that this is the odds of a <em>specific</em> person winning twice in a row. The odds that <em>someone</em> wins twice in a row, can be calculated in two ways. Paw gives one way above, assuming all $25$ people in raffle two are in raffle one. However, there is another way to calculate this probability, that is arguably more general. First, we note that no two people can win twice in a row- i.e. the events are mutually exclusive. A nice property of mutually exclusive events $A$ and $B$ is that $P(A \vee B) = P(A) + P(B)$ i.e. the probability of one or the other is just their sum. From this, we get that the probability of <em>someone</em> winning twice in a row is:</p>
<p>$$\dfrac{2x}{775}$$</p>
<p>where $x$ is the number of people in <em>both</em> raffle number one <em>and</em> raffle number two. Note the maximum $x=25$, and for this we get:</p>
<p>$$\dfrac{2 \times 25}{775} = \dfrac{50}{775} \approx 0.0645$$</p>
<p>Which seems to agree reasonably well with Paw's answer (perhaps the large exponential caused some rounding errors).</p>
<hr>
<p><strong>Update 2:</strong> As A.S. points out in the comments above, this answer fails to incorporate any model of "rigging" or "close personal relationship". In this update, I will try to address these issues. However, as this is mathematics, I must still make some assumptions, and this answer is by no means perfect.</p>
<p>We start by noting that what we are after is the probability that the raffles were rigged, given the outcome of the close friend winning twice in a row. To begin, we introduce a <em>baseline</em> probability of the raffle being rigged, call it $p$. This baseline rig probability is how likely we think it is that the raffle will be rigged before we make any observations of the outcome. Of course, I must warn the reader that from a psychological point of view, it will be very hard to calculate this <em>after</em> said outcome was observed. This is because humans are notoriously bad for exaggerating patters. However, this is a different subject, so let's just assume as mathematicians that we have some baseline estimate for raffle-rigging $p$.</p>
<p>Now, let's assume that if the raffles <em>weren't</em> rigged, then everybody has an even chance of winning. Furthermore, let's assume that if the raffles <em>were</em> rigged, then the "close friend" of the organiser will definitely win them <em>both</em>. These assumptions are quite callous. E.g. they ignore the likely possibility that the organiser has <em>many</em> friendships, all of varying closeness. Worse still, the organiser mightn't even want to rig the draw for a friend: what if she was rigging it for some other reason? Or what if one raffle was rigged, the other not? The possibilities are endless, so we must draw the line somewhere and make assumptions. Hence the above.</p>
<p>OK, let $q$ be the probability that the organiser's friend won both raffles. Then we have, by our assumptions stated above:</p>
<p>$$q = p + (1 - p)\frac{2}{775}$$</p>
<p>Now let $r$ be the probability of the raffle being rigged <em>given</em> that the friend won both raffles. Now, we note that by our assumptions, the probability of the friend winning both raffles given that the raffles were rigged is $1$ i.e. absolute certainty. Then, by <a href="https://en.wikipedia.org/wiki/Bayesian_inference" rel="nofollow">Bayesian inference</a>, we have that:</p>
<p>$$r = \dfrac{p}{p + (1 - p)\frac{2}{775}} = \dfrac{775p}{773p + 2}$$</p>
<p>Using the formula, we can calculate for example, the conditional probability of a rigged raffle, given that the baseline probability was $0.01$ (or $1\%$):</p>
<p>$$\dfrac{7750.01}{7730.01 + 2} = 0.796$$</p>
<p>This is a substantial increase over the baseline of $0.01$. If the baseline was $0.001$, we still get a substantial conditional rig probability of around $0.28$- an even bigger increase. The percentage gain from initial rig probability to conditional rig probability increases monotonically as $p \rightarrow 0$. The increase is given by:</p>
<p>$$\dfrac{775}{773p + 2}$$</p>
<p>Which is just the above formula for $r$ divided by $p$. Clearly this is a maximum at $p=0$ and the increase is $\frac{775}{2}$- but of course with $p=0$, then the conditional probability is also just $0$, so we never experience this increase. So <em>if</em> the above assumptions approximately hold, which is a big "if", then we should expect a significant increase from the baseline probability of a rig.</p>
|
1,571,083 | <p>Given the 2 terms
$$ \frac{k + a}{k + b}$$
and $$\frac{a}{b}$$
with $a, b, k \in \mathbb{R^+}$ and $a > b$</p>
<p>I want to show, that the first term is always bigger than the second one.</p>
<p><strong>My try</strong>
$$
\frac{k + a}{k + b} > \frac{a}{b} \\
\frac{k + b + (a - b)}{k + b} > \frac{b + (a - b)}{b} \\
1 + \frac{a - b}{k + b}> 1 + \frac{a - b}{b} \\
\frac{1}{k + b} > \frac{1}{b} \\
k + b > b
$$
the problem is, that I think the inequality sign has to change to $<$ in the last step because of calculating the inverse on both sides. That would mean
$$ k + b < b$$
which is nonsense.</p>
| fleablood | 280,126 | <p>The probability of winning the first raffle is: $1/31 + 30/31*1/30 = 2/31$.</p>
<p>The probability of winning the second raffle is: $1/25$.</p>
<p>The probability of her winning both: $2/31 * 1/25 = 2/775$ but that's the same probability of <em>any</em> outcome.</p>
<p>The probability of one of the two winners of the first raffle winning the second raffle is 2/25. Which one shouldn't bet on but is hardly suspicious.</p>
<p>Okay, that the person is a close friend of the boss... Well, as 1/13 probability is hardly shocking, this is well within the realm of possibility.</p>
|
2,389,581 | <p>A town has 2017 houses. Of these 2017 houses, 1820 have a dog, 1651 have a cat, and 1182 have a turtle. If x is the largest possible number of houses that have a dog, a cat, and a turtle, and y is the smallest possible number of houses that have a dog, a cat, and a turtle, then what is the value of x−y.</p>
| fleablood | 280,126 | <p>The trick is to think if the set of houses with all kinds of pets isn't maximal or minimal what we can do to maximize or minimize it and what result that will be.</p>
<p>If there are any houses with only a turtle in it, or only a turtle and one other pet it int we can add the other pets from houses with pets but not turtle. (As there are more dogs and cats than turtles there will always be houses with dogs and houses with cats that don't have turtles). So in order to be maximal all the turtles must be in houses that have all pets. And the maximal number of houses with all pets must be the same as the houses with turtle: $118$2.</p>
<p>If there are any empty houses or houses with only one type of pet, and if there are any houses with all types of pets, we can add a pet from a house with all pets. That way we reduce the number of houses with all types of pets by one. So in order to be minimal either there must be no houses with fewer than one pet , or there are no houses with all types of pets.</p>
<p>If there are no houses with fewer than one pet, let $dc, dt,ct all$ be the number of hoses with dogs and cats, dogs and turtles, cats and turtles, and all three:</p>
<p>Then $dc+ dt+ct+all = 2017; dc+dt +all = 1820; dc + ct + all = 1651; dt + ct + all = 1182$. Solving those we get $ct = 197; DC = 366; dt= 835; all = 619$.</p>
<p>If there are no houses with all types of pets, on the other hand, then $dc + dt \ge 1820; dc + ct \ge 1651; dt+ct \ge 1182$ so $dc + dt + ct \ge \frac {1820+1651+1182}2 > 2017 \ge dc + dt + ct$ so this is clearly not possible.</p>
<p>So the number of houses with all three types of pets is between $619$ and $1182$.</p>
|
186,395 | <p>i want write a module to find the integer combination for a multi variable fomula. For example</p>
<p>$8x + 9y \le 124$</p>
<p>The module will return all possible positive integer for $x$ and $y$.Eg. $x=2$, $y=12$.
It does not necessary be exactly $124$, could be any number less or equal to $124$. Must be as close as possible to $124$ if no exact solution could be found.</p>
<p>I do not want to solve with brute force as the number of variable could be any...$(5,10,100,...n)$</p>
<p>Any algorithm could solve this?</p>
| David Nehme | 16,013 | <p>You are trying to solve</p>
<p>$${\rm maximize} 8x + 9y$$
subject to
$$8x + 9y \le 124$$
$$ x, y \in \mathbb{Z}^+$$</p>
<p>Which is essentially a special case of the <a href="http://en.wikipedia.org/wiki/Knapsack_problem#Multiple_Knapsack_Problem" rel="nofollow">knapsack problem</a>.
$${\rm maximize} \sum_i a_i x_i \le b$$
subject to
$$\sum_i a_i x_i \le b$$
$$ x_i \in \mathbb{Z}^+$$
This can be solved with a MIP <a href="http://www.or-exchange.com/questions/26/open-sourcefree-mip-solver" rel="nofollow">solver</a>.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| angryavian | 43,949 | <p>Hint: if $n$ is odd, then is $n^2$ odd or even? what about $7n$? Ask the same questions if $n$ is even.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| Cameron Buie | 28,900 | <p><strong>Hint</strong>: You should be able to fill in the the following blanks and justify each claim made. We start by noting that $$n^2+7n=n(n+7).$$ If $n$ is an integer, then either $n$ or $n+7$ must be _____ and the other must be _____. Hence, $n(n+7)$ must be _____, and so....</p>
|
850,162 | <p>How can i prove that there exist some real $a >0$ such that $\tan{a} = a$ ? </p>
<p>I tried compute $$\lim_{x\to\frac{\pi}{2}^{+}}\tan x=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\cos x}$$ </p>
<p>We have the situation " $\frac{1}{0}$ " which leads us " $\infty$ " </p>
<p>$$\lim_{x\to\frac{\pi}{2}^{-}}\tan x=\lim_{x\to\frac{\pi}{2}^{-}}\frac{\sin x}{\cos x}$$ </p>
<p>We have the situation " - $\frac{1}{0}$" which tells us " $- \infty$" </p>
<p>This means for any real number $y$ there exists $x_1$ and $x_2$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that </p>
<p>$f(x_1)<y<f(x_2)$. </p>
<p>Remember the Intermediate Value Theorem. If f is a continuous function and $f(a)<0<f(b)$ then there exist $x \in (a,b)$ such that $f(x) = 0$ </p>
<p>So $f(x_1)<y<f(x_2)$ is equivalent to </p>
<p>f(x1) - y < 0 < f(x2) - y (I just subtracted y from each part) </p>
<p>Now we can use the Intermediate Value Theorem (applied to $(f - y)$ to say there exists an $x \in (x_1 , x_2)$ such that $f(x) - y = 0 $</p>
<p>or $f(x) = y$ or $\tan x = y$ </p>
<p>We know $x_1 < x < x_2$ and $-\frac{\pi}{2}<x_1<x_2<\frac{\pi}{2}$ </p>
<p>So we know that $-\frac{\pi}{2}<x<\frac{\pi}{2}$, then $f(x) - y = 0$.</p>
<p>But how can i prove it for $a >0$ such that $\tan{a} = a$ ?</p>
| Adam Hughes | 58,831 | <p>Use the intermediate value theorem on $\tan x -x$ after plugging in $x={3\pi\over 4}$ and $x={35\pi\over 24}$.</p>
<p>I chose those specific values because (in theory) you can compute them by hand using half-angle formulae and it's more constructive, but one can use the definition of unboundedness near $3\pi/2$ and the boundedness of the identity function on any compact set to make things slicker.</p>
|
301,318 | <p>As the title stated , what is the meaning of infinitely many ? When we say a set contains infinitely many elements, does this mean we cannot finish counting all the elements in the set ? Does infinitely many same as $\forall$ ? </p>
| Sigur | 31,682 | <p><strong>Infinitely many</strong> means that you can find a subset and a bijection with $\mathbb{N}$, that is, you have an infinite elements.</p>
|
3,767,421 | <p>This is a rather soft question.</p>
<p><strong>My understanding:</strong></p>
<p>Suppose we have <span class="math-container">$x \in ℝ$</span> and <span class="math-container">$x^2 = -1$</span> [in the normal interpretation].</p>
<p>Then the statement "<em>there exists <span class="math-container">$r \in ℝ$</span> such that <span class="math-container">$r^2 = -1$</span></em>" is true.</p>
<p>This is because <span class="math-container">$x \in ℝ$</span> and <span class="math-container">$x^2 = -1$</span> form a contradiction, and under contradictory settings any statement follows. That is, in an inconsistent system, any statement is true. [UPDATE: this should be "any statement can be proved" as pointed out in the following answers/comments]</p>
<p><strong>My question:</strong></p>
<p>So, is that still valid to say that the negation of a true statement in an inconsistent system is false? If yes, then we would have any statement in an inconsistent system is simultaneously true and false. [UPDATE: this implication is actually wrong and has been corrected in the following answers/comments]</p>
<p>Or do we rather leave <em>false</em> to be <em>undefined</em> in inconsistent systems? (since I think the definition of <em>false</em> is to some extent redundant in such systems)</p>
<p><strong>Motivation:</strong></p>
<p>I am thinking about what does it actually mean when we say some statement is <em>true</em>.</p>
<p>In a vacuous implication, we say that the premise is <em>false</em>. However, for example, when we are using proof by contradiction to <em>test</em> if a statement is false, we actually treat the statement as if it is a true statement <em>until</em> we hit a contradiction, and then conclude that the statement is false, under the given settings. In other words, a statement is not necessary to be false if we don’t expect consistent system in the first place.</p>
| lemontree | 344,246 | <p>First off, "There exist <span class="math-container">$x \in \mathbb{R}$</span> such that <span class="math-container">$x^2 = -1$</span>" is by itself not at all contradictory. It is just not true in the actual world by our usual understanding of the symbols <span class="math-container">$\mathbb{R}, -x$</span> etc, in which there just so happen to be no negative squares. <strong>A contradiction only arises if the theory additionally <em>proves</em></strong> that there are no negative squares, in which case the theory proves both the statement and its negation. This is what I will assume in the following.</p>
<p>Remember that statements aren't just true or false by themselves: <strong>Truth is defined relative to interpretations.</strong> So what exactly is it that you're asking? In which structures would you like the statements to be false?</p>
<blockquote>
<p>Are there any theorems that are false in all models of the inconsistent theory?</p>
</blockquote>
<p>In a consistent theory, the answer would be "no, trivially", because the models of a theory are <em>defined</em> as those structures in which all theorems hold, i.e. in which no statement of the theory is false.<br />
But an inconsistent theory has no model: There is no structure in which a contradiction is true. So the answer to this question is: <strong>Yes, vacuously</strong>, because there are no models to begin with, so in particular there are none in which there are <em>not</em> any statements of the theory that are false in it.</p>
<p>Instead, we may ask:</p>
<blockquote>
<p>Are there any theorems that are false in any conceivable structure whatsoever?</p>
</blockquote>
<p>In classical logic, with the principle of explosion, an inconsistent theory proves everything. This means in particular that it proves <span class="math-container">$\phi$</span> and <span class="math-container">$\neg \phi$</span> for any statement <span class="math-container">$\phi$</span>. But although both may be provable, <span class="math-container">$\phi$</span> and <span class="math-container">$\neg \phi$</span> can never be simultaneously <em>true</em> under a given interpretation. So in any conceivable structure, for all the infinitely many sentences <span class="math-container">$\phi$</span>, either <span class="math-container">$\phi$</span> is true but <span class="math-container">$\neg \phi$</span> false in that structure or vice versa, whereas both of them are theorems. So here the answer is: <strong>Yes</strong>, there are infinitely many such structures in which infinitely many statements of the theory are false.</p>
<p>In the context of theories, truth is often understood as <strong>truth in the standard model with the "intended interpretation" for the non-logical symbols</strong>: By saying "<span class="math-container">$s(0) + s(0) = s(s(0))$</span> is true" we mean that it is true in the structure of the natural numbers with the successor function and addition defined as usual.<br />
But again: Since an inconsistent theory doesn't have any models, it doesn't have a standard model either. So the question</p>
<blockquote>
<p>Are there any theorems that are false in the standard model of the inconsistent theory?</p>
</blockquote>
<p>can not be answered.</p>
<p>But the idea of a standard model is that it is a formalization of the real world. So we may ask:</p>
<blockquote>
<p>Are there any theorems that are false in the real world?</p>
</blockquote>
<p>Again, for every of the infinitely many provable pairs of statements <span class="math-container">$\phi, \neg \phi$</span>, one of them must be false under each interpretation, such as the real world. So the answer here is again <strong>yes</strong>: An inconsistent theory proves statements that are false in the real world, namely those whose negation is true in the real world.</p>
<p>This is a crucial point to understand in symbolic logic: Truth exists only relative to interpretations, and the real world/standard model with the intended meaning of the symbols is just one of them. We very well can also have non-standard interpretations in which we, say, take the symbol "<span class="math-container">$\_^2$</span>" to mean "square root", that yield different truth values for the same sentences. When asking about truth, you have to specify which interpretation you are talking about.</p>
<p>In any given interpretation, any given statement takes exactly one of the truth values "true" or "false". An inconsistent theory is inconsistent precisely because it has no models, i.e. no structure that makes all statements of the theory come true: <strong>There can be no possible interpretation</strong> in which a statement is both true and false.</p>
|
1,283,541 | <p>Find the least value of $f(x)=3^{-x+1} + e^{-x-1}$.</p>
<p>I tried to use the maxima/minima concept but it was of no use. Please help.</p>
| Barry Cipra | 86,747 | <p>Trigonometric identities and other such properties of trig functions are independent of whether angles are measured in radians, degrees, or what have you. But the squeeze theorem for $\sin x\over x$ and other such relations <em>do</em> depend on $x$ being measured in radians, so the underlying geometry is inescapable.</p>
|
89,622 | <p>I'm having a hard time trying to understand a proof of the Principle of Local Reflexivity. I'm following the proofs from </p>
<p>1) Topics in Banach Space Theory (by Fernando Albiac, Nigel J. Kalton)
2) <a href="http://www.math.tamu.edu/~schlump/sofar.pdf" rel="nofollow">http://www.math.tamu.edu/~schlump/sofar.pdf</a> (some notes from Professor Schlump)</p>
<p>The proof in 2) is basically the same as in 1) , but with further details. I don't quite get why ker $S$ is contained in Ker $S_1$ . And I don't get why that implies that there is a $T$ such that $S_1 = TS$. </p>
<p>Can someone enlight me ? I would really appreciate it!</p>
| Keaton Hamm | 22,500 | <p>I cannot comment above, so I have to write another answer. To clarify for you Rafael why such an $i$ exists in Bill's comment. Take the same setup, $\xi\in\mathbb{K}^N$ such that
\begin{equation}\underset{j=1}{\overset{N}{\sum}}\xi_jx_j^{\*\*}=0 ;\quad\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\neq0 \end{equation}<br>
(i.e. $\xi\in\ker(S)\setminus\ker(S_1)$). But recall that the $x_i$'s were chosen in $S_{X^*}$ so that $$\langle x_j^{**},x_j^*\rangle=\langle x_j,x_j^*\rangle=\langle x_j^*,x_j\rangle$$</p>
<p>Now then by assumption, we have that for all $i$,
$$0=\langle\underset{j=1}{\overset{N}{\sum}}\xi_jx_j^{\*\*},x_i^*\rangle=\langle\underset{j=1}{\overset{N}{\sum}}\xi_jx_j,x_i^*\rangle=\langle x_i^*,\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\rangle$$
So if the right hand side is equal to 0, but $\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\neq0$ as we assumed, then we must have that for some $i\in\{1,2,\dots,N\}$,
$$\langle x_i^*,\underset{j=1}{\overset{N}{\sum}}\xi_jx_j\rangle\neq0$$
which gives a contradiction.</p>
|
214,520 | <p>I'm trying to make tables like in the image to calculate max and min of the expressions.
Whenever I change the max and min of <code>a</code> and <code>b</code> in the upper table then the <code>MixEx</code>and <code>MaxEx</code> are updated automatically.</p>
<p>How can I do that in Mathematica?
<a href="https://i.stack.imgur.com/pwvP4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pwvP4.jpg" alt="enter image description here"></a></p>
| kglr | 125 | <p>A few customizations for the <code>IntervalSlider</code> and <code>InputField</code> controls:</p>
<pre><code>ClearAll[thumb, intSlider, inpField]
thumb = Graphics[{#, Text[Style["▲", #, 16], Offset[{0, -20}, {0, 0}]],
Text[Style[#2, 12], Offset[{0, -35}, {0, 0}]]}, ImageSize -> 20] &;
intSlider[Dynamic[{x_, y_}], range_, opts___ : OptionsPattern[]] :=
IntervalSlider[Dynamic[{x, y}], range, Method -> "Stop",
Appearance -> {"ThumbAppearance" -> {thumb[Red, Dynamic[x]], None,
thumb[Blue, Dynamic[y]]}}, ImageSize -> {400, 50}, opts]
inpField = InputField[#, Appearance -> "Frameless", FieldSize -> 5,
Alignment -> Center] &;
</code></pre>
<p>Given a list of input functions, we only need the functions <code>Interval</code> and <code>MinMax</code> to find the extrema of the input expressions: </p>
<pre><code>ClearAll[functions, minMax]
functions = {# + #2 &, -# &, # + 2 #2 &, -#2 &, #- #2&, #2 - #&, 1/(2 # - #2) &};
minMax[f_][x_, y_] := MinMax@f[Interval@x, Interval@y]
</code></pre>
<p>The <code>IntervalSlider</code>s and <code>InputField</code>s are used to specify the input. </p>
<pre><code>DynamicModule[{a = {20, 50}, b = {35, 75}}, Dynamic @
Grid[{{Grid[{{Labeled[intSlider[Dynamic@{a[[1]], a[[2]]}, {0, 100, 1}],
Style["a", 16], Left], SpanFromLeft, SpanFromLeft, SpanFromLeft},
{Labeled[intSlider[Dynamic@{b[[1]], b[[2]]}, {0, 100, 1}],
Style["b", 16], Left], SpanFromLeft, SpanFromLeft, SpanFromLeft}},
Dividers -> All, ItemSize -> 10]},
{Grid[{{"a", SpanFromLeft, "b", SpanFromLeft} /. s_String :>
Item[s, Background -> LightGray],
Item[#, Background -> LightBlue] & /@ {"min", "max", "min", "max"},
Item[#, Background -> LightYellow] & /@
{inpField[Dynamic[a[[1]]]], inpField[Dynamic[a[[2]]]],
inpField[Dynamic[b[[1]]]], inpField[Dynamic[b[[2]]]]}},
Frame -> {All, All}, ColumnsEqual -> True, ItemSize -> 10, Alignment -> Center]},
{Dynamic@ Grid[{{"expr", SpanFromLeft, "min (expr)", "max (expr)"} /.
s_String :> Item[s, Background -> LightMagenta],
## & @@ Table[{foo["a", "b"], SpanFromLeft, ##& @@ minMax[foo][a, b]},
{foo, functions}]},
Dividers -> All, ColumnsEqual -> True, ItemSize -> 10, Alignment -> Center]}}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/yoN4S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yoN4S.png" alt="![enter image description here"></a></p>
<p><strong>Note:</strong> You can also define <code>functions</code> as</p>
<pre><code>functions = Function[{x, y}, #] & /@
{x + y, -x, x + 2 y, -y, x - y, y - x, 1/(2 x - y)}
</code></pre>
<p><strong>Update:</strong> Dealing with cases where some input expression are lists, as in,for example,</p>
<pre><code>expList = {x + y, 2 x, {x + y, 2 x}, -x, x + 2 y, -y, x - y, y - x, 1/(2 x - y)}
</code></pre>
<p>We need to modify <code>minMax</code> to handle lists:</p>
<pre><code>ClearAll[minMax, functions]
minMax[f_][x_, y_] := MinMax@f[Interval@x, Interval@y]
minMax[f_List][x_, y_] := MinMax@Transpose[minMax[#][x, y] & /@ f]
</code></pre>
<p>Second, we need to transform the input list to a list of functions:</p>
<pre><code>functions = Block[{bar},
Map[bar, expList, 1] /. bar[l_List] :> bar /@ l /. bar -> (Function[{x, y}, #] &)]
</code></pre>
<p>Finally, we need to change the first argument of <code>Table</code> in the last grid to handle lists in the <code>functions</code> list properly.</p>
<p>With these changes (removing the second grid with input fields):</p>
<pre><code>DynamicModule[{a = {20, 50}, b = {35, 75}},
Dynamic@Grid[{{Grid[{{Labeled[intSlider[Dynamic@{a[[1]], a[[2]]}, {0, 100, 1}],
Style["a", 16], Left], SpanFromLeft, SpanFromLeft, SpanFromLeft},
{Labeled[intSlider[Dynamic@{b[[1]], b[[2]]}, {0, 100, 1}],
Style["b", 16], Left], SpanFromLeft, SpanFromLeft, SpanFromLeft}},
Dividers -> All, ItemSize -> 10]},
{Dynamic@ Grid[{{"expr", SpanFromLeft, "min (expr)", "max (expr)"} /.
s_String :> Item[s, Background -> LightMagenta],
## & @@ Table[{If[Head[foo] === List, Through@foo["a", "b"], foo["a", "b"]],
SpanFromLeft, ## & @@ minMax[foo][a, b]},
{foo, functions}]},
Dividers -> All, ColumnsEqual -> True, ItemSize -> 10, Alignment -> Center]}}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/LZMAz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LZMAz.png" alt="enter image description here"></a></p>
|
3,004,767 | <p>If <span class="math-container">$\displaystyle\lambda_n = \int_{0}^{1} \frac{dt}{(1+t)^n}$</span> for <span class="math-container">$n \in \mathbb{N}$</span>. Then prove that <span class="math-container">$\lim_{n \to \infty} (\lambda_{n})^{1/n}=1.$</span></p>
<p><span class="math-container">$$\lambda_n=\int_{0}^{1} \frac{dt}{(1+t)^n}= \frac{2^{1-n}}{1-n}-\frac{1}{1-n}$$</span></p>
<p>Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.</p>
| Jack D'Aurizio | 44,121 | <p>In general, if <span class="math-container">$f(x)$</span> is a continuous and non-negative function on <span class="math-container">$[0,1]$</span>, </p>
<p><span class="math-container">$$ \lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}f(x)^n\,dx} = \max_{x\in[0,1]}f(x) $$</span>
by the inequality between means.</p>
|
2,118,266 | <p>In Order Theory, what is the exact definition of a dual statement? And what is the duality principle for posets/lattices? I haven't been able to find an exact statement or definition in this regard.</p>
| Hagen von Eitzen | 39,174 | <p>If $(X,\le)$ is a poset, then so is $(X,\preceq)$ with $a\preceq b\iff b\le a$. Any statement about the former translates into a statement of the latter</p>
|
2,345,142 | <p>Can you use Skolemization to reduce a formula to the variables you want it to be about? I was trying to think of a nice algorithmic way to do it but only ended up having problems. </p>
<p>Say you have a formula $G$ in prenex normal form, i.e. something looking like this:</p>
<p>$$G \equiv Q_1x_1...Q_nx_n\;F$$</p>
<p>where $Q_i$ are quantifiers binding variables $x_i$ and occuring exclusively on the left of formula $F$ (which, consequently does not contain quantifiers itself). It is known that $G$ can be transformed into Skolem normal form, i. e. all existential quantifiers can be eliminated by substituting the variables they are binding by functions. But can <i>any</i> variable be substituted so that in the end the formula expresses something about the variables you wish to keep?</p>
<p>The possibly cheapest way to do it involves bracketing $F$ with its nearest quantifier. </p>
<p><em>[EDIT: as you can read in the comments, this manoeuvre involves a major flaw, which shall stay undeleted for the record.]</em></p>
<p>If the quantifier nearest to $F$ is $\exists$, it would go like this:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;\exists x_n\;F$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\exists x_n\;F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(F[x_n/c_1])$$</p>
<p>...where $F[x_n/c_1]$ expresses that $x_n$ has been substituted by constant $c_1$ in $F$.</p>
<p>If the quantifier nearest to $F$ is $\forall$, the procedure takes some extra legwork and two negations:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall x_n F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\neg \exists x_n\;\neg F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;\neg (\exists x_n\;\neg F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\neg F[x_n/c_1])$$</p>
<p>Iterating the process for all bound variables would leave you with a formula with no quantifiers but many constants you cannot assume much about.</p>
<p>To sustain "expessive power" regarding the variables designated to stay, there ought to be a way to substitute the rest of the variables by functions with arities higher than 0. I was thinking of (if required, repeatedly) putting the formula in the following form:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall y_{k} ... \forall y_n\;\exists x_n F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall y_{k} ... \forall y_n\;F[x_n/f(y_{k},..., y_n)])$$</p>
<p>...where $y_i$ shall be the variables preserved in the final form and $x_i$ the variables to be eliminated.</p>
<p>Two problems arise:</p>
<p>1) As any quantifiers can bind $y_i$ in the original prenex normal form, the existential quantifiers binding $y_i$ must first be rephrased into universal ones for the Skolemization to yield functions of $y_i$. This does not always lead to a chain of universal quantifiers with no negations in between. It seems like it's not always possible to get rid of the negation in $\forall y_i \; \neg \forall y_{i-1}$.</p>
<p>2) Consequently, opportunities to rephrase $Q_ix_i\; Q_{i+1}x_{i+1}$ as $Q_{i+1}x_{i+1}\;Q_ix_i$ are limited.</p>
<p>So far, I managed to fabricate formulas containing only the variables I picked as survivors in the initial stage. Anyway, I could not avoid constants in every case. Can you think of a procedure with rules that would prevent constants appearing in any resulting formula?</p>
| Noah Schweber | 28,111 | <p>Your version of Skolemization is incorrect, and we can see this at the very first step. "$\forall xL(x)$" is <strong>not</strong> equivalent to "$\neg L(x/c)$:" while the former implies the latter, the latter doesn't imply the former (how do we know $c$ wasn't "chosen badly"?) Similarly, "$\exists x L(x)$" is not equivalent to "$L(x/c)$:" while the latter implies the former,the former doesn't imply the latter (for the same reason as above. So the way you've introduced constants is broken.</p>
<p>Skolemization doesn't introduce new functions - rather, it introduces new <em>existential quantifiers over functions</em>. E.g. the Skolemized version of "$\forall x\exists yL(x, y)$" is "$\exists F^1\forall x(L(x, F(x)))$." (The "$^1$" indicates that $F$ is a <em>unary</em> function symbol.) </p>
<p>Things get clearer once we have more than one existential quantifier. The Skolemized version of "$\exists x\forall y\exists z L(x, y, z)$" is "$\exists F^0\exists G^1\forall y(L(F^0, y, G(y)))$." Note two things here:</p>
<ul>
<li><p>All the existential function quantifiers are together, at the front. </p></li>
<li><p>We've proceeded <em>from the outside in</em>: the $F^0$ corresponds to $x$, and the $G^1$ corresponds to $y$.</p></li>
</ul>
<p>Basically, the Skolemized version of a formula "$\overline{Q}\overline{x}L(\overline{x})$" is "There are functions picking witnesses for the existential quantifiers in $\overline{Q}$ appropriately."</p>
<hr>
<p>By the way, we can also perform "partial" Skolemization: we can Skolemize away any subset of the first-order existential quantifiers while leaving the rest. For instance, $$\forall x\exists y\forall z\exists w\forall a\exists b(L(x, y, z,w, a, b))$$ is equivalent to $$\exists F^1\exists G^2\exists H^3\forall x \forall z\forall a(L(x, F(x), z, G(x, z), a, H(x, z, a)))$$ (this is the fully Skolemized version), but it's also equivalent to the <em>partially</em> Skolemized sentence $$\exists F^1\exists H^3\forall x\forall z\exists w\forall a(L(x, F(x), z, w, a, H(x, z, a))).$$ <em>Note that when we leave an existential quantifier alone it stays where it was - it doesn't move to the front.</em></p>
|
804,963 | <p>Let $f,g:\mathbb{R}\longrightarrow \mathbb{R}$ be Lebesgue measurable. If $f$ is Borel measurable, then $f\circ g$ is Lebesgue mesuarable. In general, $f\circ g$ is not necessarily Lebesgue measurable. Is there any counterexample?</p>
| Etienne | 80,469 | <p>Let $g:\mathbb R\to \mathbb R$ be any Borel function with the following property: there exists a Lebesgue-measurable set $B$ such that $g^{-1}(B)$ is not Lebesgue-measurable. Such functions do exist (see below).</p>
<p>Now take $f:=\mathbf 1_{B}$. Then $f$ is Lebesgue-measurable because $B$ is, but $f\circ g=\mathbf 1_{g^{-1}(B)}$ is not Lebesgue-measurable.</p>
<p><em>Existence of the function $g$</em>. Take any one-to-one Borel function $g:\mathbb R\to\mathbb R$such that $g^{-1}(C)$ does not have measure $0$, where $C$ is the usual Cantor ternary set. There exist even continuous functions with this property (see e.g. the link given by FisiaiLusia). If you don't require continuity, just take a one-to-one Borel function $g_0$ such that $g_0(\mathbb R) =C$. Let us show that $g$ has the required property.</p>
<p>Since $g^{-1}(C)$ does not have measure $0$, it contains a set $A$ which is not Lebesgue-measurable. (If you take $g=g_0$, you just need to know that there exists at least one non-measurable set in $\mathbb R$). Set $B:=g(A)$. Then $B$ is Lebesgue-measurable because it is contained in the measure $0$ set $C$; and $A=g^{-1}(B)$ because $g$ is one-to-one. So $g^{-1}(B)$ is indeed non-measurable.</p>
|
1,317,974 | <p>Consider the affine toric variety $V \subset k^{5}$ parametrized by $$\Phi(s,t,u) = (s^{4},t^{4},u^{4},s^{8}u,t^{12}u^{3}) \in k^{5}$$ where k is an algebraically closed field of characteristic 2. This is problem 1.1.8 from Cox, Little, and Schenck, but my question regards a more general notion: How exactly does one determine the generators of the toric ideals. Just by looking at the parametrization, I can deduce the generators $x_{4}x_{5} - x_{1}^{2}x_{2}^{3}x_{3}, x_{1}^{8}x_{3} - x_{4}^{4}$, and $x_{5}^{4} - x_{2}^{12}x_{3}^{3}$ </p>
<p>Would these then generate the toric ideal? In general, is there a way to determine whether the toric ideal is the correct one other than just looking at the relations?</p>
<p>Thanks!</p>
| Community | -1 | <p>The following items illustrate an algorithm to compute saturation of ideals. Let $I\subset S := k[x_1,\dots,x_n]$ be an ideal and fix $f\in S$. Then the <strong>saturation</strong> of $I$ w.r.t. $f$ is the set
$$I:f^\infty=\{g\in S|f^m g\in I \text{ for some } m>0\}.$$</p>
<ul>
<li>$I:f^\infty$ is an ideal.</li>
<li>We have the ascending chain of ideals
$I:f \subset I:f^2\subset I:f^3 \subset \dots.$</li>
<li>There is an integer $N$ for which $I:f^N=I:f^{N+1}=I:f^\infty$.</li>
<li>$I:f^\infty=I:f^m$ iff $I:f^m=I:f^{m+1}$.</li>
<li>$I:f=\frac{1}{f}(I\cap \langle f \rangle )$.</li>
</ul>
<p>Let $L\subset \mathbb{Z}^m$ be a sublattice and $\ell^1, \dots, \ell^r$ be a basis for $L$. Then the lattice ideal $I_L$ can be computed as the saturation
$$I_L = \langle x^{\ell^i_{+}}-x^{\ell^i_{-}}|i=1,\dots,r\rangle:\langle x_1\cdots x_m\rangle^\infty,$$
where $\ell_{+}=\sum_{\ell_i>0}\ell_{i}e_i$ and $\ell_{-}=-\sum_{\ell_i<0}\ell_{i} e_i$,
both in $\mathbb{N}^m$.</p>
<p>In the given exercise we define the <strong>matrix of parametrization</strong> by </p>
<p>$$A:=\begin{pmatrix}
4 & 0 &0 \\
0& 4& 0 \\
0& 0& 4\\
8& 0& 1\\
0& 12& 3\\
\end{pmatrix}.$$</p>
<p>If $V$ is a toric affine variety given by a parametrization matrix $A\in \mathbb{Z}^{s\times r}$, then the toric ideal of $V$ is the lattice ideal $I_L$, where $L$ is the kernel of the map $\mathbb{Z}^s \rightarrow \mathbb{Z}^r; e_i \mapsto i\mbox{'th row of } A$, i.e., the basis of $L$ is the set of rows of the matrix of row-syzygies of $A$.</p>
<pre><code>gap> LoadPackage( "RingsForHomalg" );
true
gap> ZZ := HomalgRingOfIntegers();
Z
gap> A := HomalgMatrix( [[4,0,0],[0,4,0],[0,0,4],[8,0,1],[0,12,3]],5,3,ZZ );
<A 5 x 3 matrix over an internal ring>
gap> L := SyzygiesOfRows(A);
<A non-zero 2 x 5 matrix over an internal ring>
gap> Display( L );
[ [ -8, 0, -1, 4, 0 ],
[ 6, -3, 0, -3, 1 ] ]
</code></pre>
<p>Hence $I_L=\langle x_1^8 x_3 - x_4^4,
x_1^6 x_5 - x_2^3 x_4^3 \rangle:(x_1 x_2 x_3 x_4 x_5)^\infty$
which can easily be computed as described above.</p>
<p>In Gap the toric ideal can also be computed using 4ti2</p>
<pre><code>gap> LoadPackage( "4ti2Interface" );
true
gap> A := [ [ 4, 0, 0 ], [ 0, 4, 0 ], [ 0, 0, 4 ], [ 8, 0, 1 ], [ 0, 12, 3 ] ];;
gap> 4ti2Interface_groebner_matrix(A);
[ [ -8, 0, -1, 4, 0 ], [ -6, 3, 0, 3, -1 ], [ -4, 6, 1, 2, -2 ], [ -2, -3, -1, 1, 1 ], [ -2, 9, 2, 1, -3 ], [ 0, -12, -3, 0, 4 ] ]
</code></pre>
|
43,640 | <p>Consider the following problem:</p>
<p>Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?</p>
<p>A. If $A$ is open, then $A={\mathbb R}$</p>
<p>B. If $A$ is closed, then $A={\mathbb R}$</p>
<p>Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer. </p>
<p>Here are my questions:</p>
<blockquote>
<p>Why A is not necessarily true? What can be a counterexample?</p>
</blockquote>
| Jonas Meyer | 1,424 | <p>The question boils down to whether there are nonempty subsets of $\mathbb{R}\setminus \mathbb{Q}$ that are closed in $\mathbb{R}$. The easiest examples are finite sets, as <a href="https://math.stackexchange.com/questions/43640/open-set-in-bf-r/43642#43642">Luboš Motl noted</a>. An easy infinite example is $\sqrt{2}+\mathbb{Z}$. <a href="https://math.stackexchange.com/questions/43640/open-set-in-bf-r/43643#43643">Theo Buehler showed</a> that there are positive measure examples, which is much stronger and closely related to the question at <a href="https://math.stackexchange.com/questions/21094/is-there-a-compact-subset-of-the-irrationals-with-positive-lebesgue-measure/">this link</a>.</p>
<p>Another direction to strengthen the result is to show that there are perfect examples, which is the subject of the question at <a href="https://math.stackexchange.com/questions/1064/perfect-set-without-rationals">this link</a>.</p>
|
43,640 | <p>Consider the following problem:</p>
<p>Let ${\mathbb Q} \subset A\subset {\mathbb R}$, which of the following must be true?</p>
<p>A. If $A$ is open, then $A={\mathbb R}$</p>
<p>B. If $A$ is closed, then $A={\mathbb R}$</p>
<p>Since $\overline{\mathbb Q}={\mathbb R}$, one can immediately get that B is the answer. </p>
<p>Here are my questions:</p>
<blockquote>
<p>Why A is not necessarily true? What can be a counterexample?</p>
</blockquote>
| Michael Hardy | 11,667 | <p>"Why A is not necessarily true? What can be a counterexample?"</p>
<p>I'm surprised at the complexity of some answers given to this. Here's a counterexample:
$$
(-\infty,\pi)\cup (\pi,\infty).
$$
You can construct lots of others similar to that but more complicated if need be.</p>
|
3,346,543 | <p>I am aware this is a pretty big topic, but the attempts at layman's explanations I have seen either barely provide commentary on the formal proofs, or fail to provide an explanation (e.g "it gets too complex" does not really say anything)</p>
<p>Is there a good intuitive explanation as to why we fail to obtain a general solution for a 5th+ degree polynomial, and why this happens at the 5th degree and not below or above?</p>
| NBeneparte | 709,516 | <p>Responding to Dmitry Ezhov's comment above: the "finite combination" is necessary, since with an infinite number of operations one <em>can</em> solve an arbitrary quintic. For example, consider <span class="math-container">$x^5 - x - 1$</span>. Galois theory tells us the roots of this cannot be expressed in terms of a finite combination of radicals and field operations, but using infinitely many it's not too bad: if <span class="math-container">$x^5 - x - 1 = 0$</span>, then <span class="math-container">$x^5 = x+1$</span>, so <span class="math-container">$x = \sqrt[5]{1+x}$</span>. Plugging this back into itself and iterating yields a solution <span class="math-container">$x = \sqrt[5]{1+\sqrt[5]{1+\cdots}}$</span> to the original quintic.</p>
|
497,687 | <p>How would I integrate the following.</p>
<p>$$
\int_{\ln2}^{\ln3}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}\,dx
$$ </p>
<p>I think I have to use the $\arcsin(x)$ formula.</p>
<p>Which means would I use $y=u^2$ with $u=e^{-x}$ but that does not seem to work.</p>
| Mercy King | 23,304 | <p>Setting
$$
t=e^{-x},\ t=\sin\theta
$$
we have
\begin{eqnarray}
\int_{\ln2}^{\ln3}\frac{e^{-x}}{\sqrt{1-e^{-2x}}}\,dx
&=&\int_{\frac13}^{\frac12}\frac{1}{\sqrt{1-t^2}}\,dt=\int_{\arcsin\frac13}^{\arcsin\frac12}\frac{\cos\theta}{\sqrt{1-\sin^2\theta}}\,d\theta\\
&=&\int_{\arcsin\frac13}^{\frac\pi6}\,d\theta=\frac\pi6-\arcsin\frac13.
\end{eqnarray}</p>
|
1,630,831 | <p>I just woke up this and I was extremely curious if there were any subspaces of a vector space such that $X\cap (Y + Z) \neq (X \cap Y) + (X \cap Z)$?</p>
<hr>
<p>Just a curious person. Thanks in advances. </p>
| DylanSp | 308,461 | <p>$5 \frac{(x +5)(6x - 1)}{(7 - x)(8x + 2)} = \frac{5[(x +5)(6x - 1)]}{(7 - x)(8x + 2)}$. Just use FOIL to multiply the binomials in the numerator, then use the distributive law to multiply the resulting trinomial by 5.</p>
|
1,630,831 | <p>I just woke up this and I was extremely curious if there were any subspaces of a vector space such that $X\cap (Y + Z) \neq (X \cap Y) + (X \cap Z)$?</p>
<hr>
<p>Just a curious person. Thanks in advances. </p>
| Narasimham | 95,860 | <p>No need to FOIL. Divide numerator and denominator with $x^2,$ and let x go to infinity.</p>
<p>$$ 5 \dfrac{(x +5)(6x - 1)}{(7 - x)(8x + 2)} = \dfrac{5\,(1 +5/x)(6 - 1/x)}{(-1 +7/x)(8 + 2/x)} = \frac{30}{-8} = - 3.75 $$</p>
|
1,587,007 | <p>I have the following matrix $\mathbf{U}$ which is in echelon form. The strange to me is that I havent met a matrix with first column zero. </p>
<p>$$\mathbf{U} = \begin{bmatrix}
0 & 5 & 4 & 3 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$</p>
<p>According to theory columns with pivots define a basis. Also according to theory, a matrix can have more that one basis. So, according to these facts, we may say that the basis are:</p>
<p>$$B_1= \begin{Bmatrix}\begin{bmatrix} 5 \\ 0 \\ 0 \\0 \\ \end{bmatrix},
\begin{bmatrix} 4 \\ 2 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix} ,~ B_2= \begin{Bmatrix}\begin{bmatrix} 5 \\ 0 \\ 0 \\0 \\ \end{bmatrix},
\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix} \text{and}~B_3= \begin{Bmatrix}\begin{bmatrix} 4 \\ 2 \\ 0 \\0 \\ \end{bmatrix},
\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}\end{Bmatrix}$$</p>
<p>Taking one of these bases ($B_1$ or $B_2$ or $B_3$) we may represent the non basis column as a combination of the basis columns. I think this is by definition of the column space basis. However, the last hold only when taking $B_3$ as column space basis.</p>
<p><strong>Can you please correct possible flaws on my logic?</strong> </p>
<p>Thanks!!</p>
| janmarqz | 74,166 | <p>The three sets $B_1,B_2,B_3$ serve as a basis for a unique two dimensional subspace of $\Bbb R^4$ because each set has two linearly independent vectors.</p>
|
382,935 | <p>I reading my textbook and I don't understand the concept of distributions or number of solutions to an equation. It's explained that this problem is 1/4 types of sampling/distributions problems. An example is provided to illustrate:</p>
<blockquote>
<p>In how many ways can 4 identical jobs (indistinguishable balls) be distributed among 26 members (urns) without exclusion (since one member can do multiple jobs)?</p>
<p>A sample outcome might be:<br>
$\text{_____________}$<br>
| A | B | C |...| Z | <br>
$\text{--------------------}$<br></p>
<p>$\text{_____________}$<br>
| o | oo| | | ||| o | <br>
$\text{_____________}$<br>
| A | B | C |... | Z | <br>
$\text{--------------------}$<br></p>
<p>Thus, the question is reduced to, "How many $(26-1+4)$ letter words are there consisting of four circles and $(26-1)$ vertical lines?" Therefore, the solution is: $\binom{26-1+4}{4}$</p>
</blockquote>
<p>I really don't understand why its $(26-1+4)$. There's only 26 different spots or "urns" to place the 4 jobs. Can someone please explain? </p>
<p>I looked through another text to try and understand and I found it explained as such: </p>
<blockquote>
<p>There are $\binom{n+r-1}{r-1}$ distinct nonnegative integer valued vectors $(x_1,...,x_r$ satisfying the equation $(x_1 + ... + x_r = x_n$ for $x\ge0$. $\spadesuit$</p>
</blockquote>
<p>How in the world are they deriving this? For <em>distinct positive integers</em> I understand:</p>
<p>Assume I have 8 balls (n=8) and I have 3 urns (r=3):
o^o^o^o^o^o^o^o, where o represents a ball and ^ represents a place holder where an urn could be placed. For this scenario:</p>
<blockquote>
<p>There are $\binom{n-1}{r-1}$ distinct positive integer valued vectors $(x_1,...,x_n)$ satisfying the equation: $x_1 + ... + x_n = n, x_i>0, i=1,..,r$ $\clubsuit$</p>
</blockquote>
<p>It's clear that I could have this specific case ooo|ooo|oo. Here the bar represents a divide for the urn and you see I have 3 sections. So that case is clear. Can anyone please explain this problem to me? I don't understand the nonnegative integer case. </p>
<p>Also, people who post tend to be crazy smart and explain things very in a complicated manner. I'd appreciate it if it could be explained in layman's terms as much as possible. </p>
<p>Thank you!!!</p>
| Alex Wertheim | 73,817 | <p>Like so many problems in combinatorics, the key to this problem is understanding precisely what kind of things you are counting. You confusion seems to be arising, at least in part, from the fact that you are trying to count distributions of jobs to urns, and your authors are counting something seemingly different and abstruse to you. I will try to explain why they are in fact the same thing.</p>
<p>Consider the task of distributing 4 jobs among 26 urns as posed in your problem. If you were to naively count distributions of jobs to urns, you might start building distributions as follows in terms of sequences, where the value at the $i$th index in this list represents the number of jobs distributed to the $i$th urn: </p>
<p>$(1, 1, 1, 1, 0, 0, \ldots, 0)$
$(1, 1, 1, 0, 1, 0, \ldots, 0)$
$(1, 1, 1, 0, 0, 1, \ldots, 0)$</p>
<p>...</p>
<p>For a problem this large, you would get frustrated pretty quickly. While it's a nice idea, this type of solution doesn't seem to get us anywhere. So we have to conceptualize the problem in a different way, which leads me to your authors suggestion. Instead of building sequences, a common technique in combinatorics, why not instead pictorally represent the problem with o's and |'s, where o's represent jobs, and |'s (or dividers) represent divisions into urns. That is to say, the first sequence posed would look something like:</p>
<p>o|o|o|o|||||||$\ldots$|</p>
<p>where the two parallel dashes with no 'o's in between would represent an urn with no jobs in it. </p>
<p>The reason this technique is powerful is because it allows us to now count the different kinds of symbolic sequences we can build in a way that represents exactly what we're looking for: distributions of jobs to urns. But how do we count these kinds of distributions?</p>
<p>Well, we can see that as there are 26 urns, and 4 jobs to distribute, there are going to be a total of 29 symbols in our sequence (not 30, because we only need 25 (i.e., 26- 1) dividers to represent division into 26 total urns). Since we require that four of these symbols are 'o's, i.e., jobs, then we see that all possible sequences are formed by "choosing" four of the symbols in the sequence to be 'o's, and allocating the rest to be dividers. These build all the possible distributions we are looking for, and for this particular problem, we see that the number of possible distributions is thus ${26 - 1 + 4 \choose 4}$.</p>
<p>You might rightfully protest, "Why are we choosing 'o's? Shouldn't this method work if we instead chose 25 of the symbols to be dividers?" Well, that's a perfectly reasonable objection; but by the symmetry of $n C r$, we in fact have that ${29 \choose 4} = {29 \choose 25}$. You can see that this type of thinking quickly generalizes to distributions of $n$ indistinguishable things to $k$ distinguishable people; there will be $n$ objects, $k-1$ dividers, and we choose $n$ of the symbols to be objects, so we get ${n + k - 1 \choose n}$ possible distributions.</p>
<p>This explanation might seem a bit overkill (it probably is!), but this type of thinking can help you solve a lot of distribution problems of this kind. The analogue in the case of non-negative vector solutions to $x_{1} + \cdots + x_{k} = n$ is in realizing that this is the same as distributing $n$ indistinguishable things among $k$ distinguishable people, where $x_{i}$ simply represents the number of "things" distributed to the $i$th person. </p>
<p>I hope this helps!</p>
<p>Edit: in response to your next question, I'm not sure why your authors would present it this way (typically, the flow of logic is reversed!) but I can certainly explain.</p>
<p>The key is in noting that strictly positive solutions require $y_{i} > 0$ for all $i$, i.e., $y_{i} \geq 1$. One way to ensure this is to simply add one to each element of the non-negative solutions; that is, set each $y_{i}$ equal to $x_{i} + 1$. Since each $x_{i} \geq 0$, we then have that each $y_{i} \geq 1$ as desired. However, if we wish to leave the number of distributions unchanged, we cannot change what we are doing combinatorially; hence, for each $y_{i}$, if we set $y_{i} = x_{i} + 1$, then we must add 1 to the right hand side of the expression $x_{1} + \cdots + x_{r} = n$ for each $i \in \{1, 2, \cdots, r\}$. Then we have $y_{1} + \cdots + y_{r} = n + 1 + \cdots + 1 = n + 1*r = n + r$. </p>
<p>As a sort of combinatorial exercise, one way you can convince yourself that the number of distributions is unchanged is to think of it as follows. Imagine instead of distributing $n$ objects to $r$ people, accepting distributions where some people get zero objects, we instead consider distributing one object to each of the $r$ people a priori in order to ensure that each person gets at least one object. But to leave the distribution unchanged, that would mean we would distribute $n+r$ objects, allocating one object to each of the $r$ people from the start - hence why in "equation form" we would set $y_{i} = x_{i} + 1$.</p>
<p>Hope this helps! Feel free to post if you need further clarification.</p>
|
1,627,381 | <p>There is a subset of sigma field $G_2$, say $G_1 \subset G_2$. $G_1$ is proven to be a sigma field.
Does this necessarily imply that $G_1 = G_2$?</p>
| BrianO | 277,043 | <p>Not at all: the Borel sets of $\Bbb R$ form a $\sigma$-field $\mathcal{B}$, and the power set $\mathcal{P}(\Bbb R)$ of $\Bbb R$ is a $\sigma$-field, but $\mathcal{B} \subsetneqq \mathcal{P}(\Bbb R)$, as not every set of reals is Borel.</p>
|
30,789 | <p>I'm trying to identify the frequencies in my time history samples, and I can see a frequency in the time history, but can't see it in its Fourier transform.
Here it is :</p>
<p>the sample data:</p>
<pre><code>dt = 0.01;(*0.01 second per sample*)
ls={7.18384,9.08503,7.13301,9.03243,7.23692,8.82911,7.48153,8.50053,7.8291,8.09453,8.22514,7.67123,8.60473,7.29656,8.90489,7.02926,9.07228,6.91356,9.07469,6.96968,8.90404,7.19156,8.58122,7.54573,8.15062,7.97689,7.67665,8.41549,7.23096,8.78898,6.88426,9.03305,6.69172,9.10231,6.68698,8.97821,6.87272,8.6724,7.22283,8.22643,7.68225,7.70488,8.17857,7.18711,8.62957,6.75303,8.9597,6.47266,9.10889,6.39174,9.04594,6.52704,8.77139,6.85901,8.32088,7.33824,7.7584,7.88776,7.16829,8.41967,6.64141,8.84366,6.2609,9.0868,6.08919,9.10045,6.15615,8.87392,6.45525,8.43236,6.94108,7.83828,7.53853,7.17843,8.15053,6.55427,8.67686,6.06288,9.02603,5.78403,9.13458,5.76455,8.97352,6.01177,8.55858,6.48966,7.94428,7.12458,7.22101,7.8152,6.49678,8.4485,5.88528,8.91739,5.48272,9.13786,5.35709,9.06375,5.53107,8.69409,5.98259,8.07629,6.64157,7.29757,7.40537,6.47512,8.14882,5.73434,8.74894,5.19349,9.09985,4.93916,9.13459,5.01757,8.83338,5.41929,8.23096,6.08644,7.41062,6.91285,6.49317,7.76769,5.61863,8.50777,4.92325,9.00806,4.51966,9.1747,4.47503,8.96814,4.80297,8.40434,5.455,7.55984,6.33198,6.55581,7.29271,5.54458,8.18165,4.6821,8.847,4.10633,9.17184,3.91187,9.08704,4.13559,8.59066,4.74753,7.74309,5.65489,6.66645,6.71462,5.51928,7.75486,4.47976,8.6024,3.70977,9.10911,3.33653,9.17844,3.42289,8.77952,3.96388,7.95795,4.87716,6.82561,6.02223,5.55073,7.21349,4.32513,8.25592,3.34263,8.96977,2.75835,9.22521,2.67336,8.96055,3.10556,8.19668,3.99626,7.03496,5.20525,5.6427,6.54315,4.23034,7.78851,3.01522,8.73391,2.19205,9.20894,1.89461,9.11841,2.17992,8.45091,3.00859,7.29079,4.25733,5.80068,5.727,4.20335,7.18224,2.74288,8.37806,1.64996,9.10926,1.10137,9.23383,1.19232,8.70843,1.91736,7.5871,3.1692,6.02681,4.75307,4.25335,6.41448,2.53923,7.8804,1.14864,8.89922,0.307694,9.28655,0.154088,8.95041,0.725381,7.91675,1.93765,6.31868,3.60713,4.38985,5.46625,2.41628,7.21338,0.707136,8.55287,-0.470752,9.2482,-0.919297,9.15831,-0.561298,8.26386,0.562301,6.67532,2.27715,4.61492,4.31759,2.39053,6.34975,0.340327,8.03907,-1.21045,9.08947,-2.01384,9.30485,-1.9283,8.61208,-0.958302,7.08603,0.756819,4.93354,2.94639,2.47094,5.26353,0.0706556,7.32281,-1.89254,8.77714,-3.10504,9.358,-3.36416,8.93794,-2.61487,7.53742,-0.963082,5.34406,1.33854,2.66827,3.9233,-0.0836731,6.37173,-2.49189,8.2693,-4.17005,9.28401,-4.84777,9.20861,-4.3986,8.01348,-2.88195,5.83756,-0.523299,2.99222,2.304,-0.10749,5.14496,-2.98081,7.52648,-5.18357,9.04878,-6.32959,9.4304,-6.18952,8.56459,-4.73628,6.56344,-2.13332,3.70414,1.31944,0.401559,5.19872,-2.85418,9.0291,-5.5357,12.3241,-7.17454,14.6869,-7.4465,15.8834,-6.29832,15.9227,-3.97558,15.044,-0.986409,13.6463,2.07958,12.1451,4.72194,10.8576,6.6742,9.93404,7.90659,9.37837,8.56294,9.10124,8.84598,8.99267,8.93815,8.96116,8.95636,8.95544,8.95677,8.95391,8.95455,8.95365,8.95341,8.95371,8.95173,8.95356,8.95088,8.95381,8.9499,8.95321,8.94933,8.95282,8.94887,8.9513,8.94838,8.95018,8.94857,8.94853};
</code></pre>
<p>Fourier transform function:</p>
<pre><code>DFT[A_, ht_] := RotateRight[ht/Sqrt[2 \[Pi]]*Fourier[RotateLeft[A, Length[A]/2 - 1],
FourierParameters -> {1, 1}], Length[A]/2 - 1];
(*shift the zero frequency to the center*)
</code></pre>
<p>plot the time history:</p>
<pre><code>ListPlot[ls, PlotRange -> All, Joined -> True, DataRange -> {0, dt*Length[ls]}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/4hnHe.png" alt="enter image description here"></p>
<p>We can see there are a fast frequency with period about 0.02s and a slow frequency with period about 1.5s, but in the Fourier transform we only see the fast frequency(except the zero frequency)</p>
<pre><code>ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {- 1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/3qthH.png" alt="enter image description here"></p>
<p>So where is the low frequency?</p>
<hr>
<p><strong>Update</strong></p>
<p>As Simon and bill suggest, the slow oscillation is the beating of two close high frequencies.
Since the Fourier transform resolution is 2Pi/(N*dt), where N is the number of sample points, so if I increase the resolution by increasing the number of sample points I should see two separated peaks.
So I tried to increase the number of sample points, but I can only see one peak all the time.
Here is how I did it:</p>
<pre><code>w = 5.0; dt = 0.66125;
f[x_] := Sin[w x]
ls = Table[f[x], {x, dt, 200 dt, dt}];
</code></pre>
<p>we see beating in the plot</p>
<pre><code>ListPlot[ls, PlotRange -> All, Joined -> True, DataRange -> {dt, 200 dt}]
</code></pre>
<p><img src="https://i.stack.imgur.com/h30HL.png" alt="enter image description here"></p>
<p>but one on peak in the Fourier transform:</p>
<pre><code>ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/U8r7E.png" alt="enter image description here"></p>
<p>If we increase the number of sample points, we still only see one peak:</p>
<pre><code>ls2 = Table[f[x], {x, dt, 800 dt, dt}];
ListPlot[Abs[DFT[ls2, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/ynyfB.png" alt="enter image description here"></p>
<p>So where is the problem?</p>
| bill s | 1,783 | <p>What is happening in your second example (with the single sine wave giving the "beating") is that you have exceeded the <a href="http://en.wikipedia.org/wiki/Nyquist_frequency" rel="nofollow noreferrer">Nyquist frequency</a>: what you are seeing is called <a href="http://en.wikipedia.org/wiki/Aliasing" rel="nofollow noreferrer">aliasing</a>. Here's a simple way to explore this (using your DFT function):</p>
<pre><code>dt = 0.66125;
f[w_, x_] := Sin[w x];
Manipulate[ls = Table[f[w, x], {x, dt, 200 dt, dt}];
Column[{ListPlot[ls, PlotRange -> All, Joined -> True, DataRange -> {dt, 200 dt}],
ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True,
DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]}], {w, 0, 10}]
</code></pre>
<p><img src="https://i.stack.imgur.com/XElWn.png" alt="enter image description here"></p>
<p>If you play with the slider, you'll see a single sine wave up to the Nyquist frequency (in both the time and frequency plots). When you get higher than the Nyquist frequency, the signal can do many things, among them show the beating that you noticed. Observe that the frequency line begins to descend as the frequency increases (just after you pass Nyquist). When this has certain relationships to the frequency of the sine wave, you get the beating effect.</p>
<p>So -- in your first question, where we could not see the source of the data (you just gave us a list of values) we had assumed you sampled correctly (i.e., below Nyquist). Simon's guess about beating was a good one. In the second question, you have created a sine wave and are sampling it improperly... there are many things that can go wrong once you do this. </p>
|
1,232,439 | <p><strong>EDIT:</strong> This is for a production scheduling problem with quadratic production and linear inventory costs. </p>
<p>The goal is to \begin{equation*}
\max_{u} \int_{0}^{T} -(c_{1}u^{2} + c_{2}y) dt
\end{equation*}
subject to
\begin{align*}
y' &= u,\\
y(0) &= 0,\\
y(T) &=B.\\
\end{align*}</p>
<p>So the Bellman equation is: \begin{equation*}
\max_{u} {-c_{1}u^{2} + c_{2}y + \frac{\partial V}{\partial y}u + \frac{\partial V}{\partial y}} =0.
\end{equation*}</p>
<p>By first-order condition, we obtain $u^{*} = \frac{1}{2c_{1}}\left ( \frac{\partial V}{\partial y} \right )$.</p>
<p>We insert this $u^{*}$ into the Bellman equation to obtain:</p>
<p>$$0 = -c_{1}\left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial y} \left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right ) + \frac{\partial V}{\partial t}$$ </p>
<p>And then I obtain the following PDE: $$0 = \frac{1}{4c_{1}}\left (\frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial t}$$</p>
<p>I have to solve for $V(y,t)$.</p>
<p>My work: $V = e^{ay + bt}$. </p>
<p>Then I find $V_{y} = ae^{ay+bt}$, and $V_{t} = be^{ay+bt}$.</p>
<p>Then I put it values of $V_{y}$ and $V_{t}$ in the PDE to solve for $b$: </p>
<p>$$\frac{1}{4c_{1}}(ae^{ay+bt})^{2} + be^{ay+bt} = 0$$ </p>
<p>$$\Rightarrow a^{2}e^{2(ay+bt)} + be^{ay+bt} = 0$$</p>
<p>$$\Rightarrow a^{2}e^{2} e^{(ay+bt)} = -be^{ay+bt}$$</p>
<p>Assuming that $e^{ay+bt} \neq 0$, we obtain $b = -a^{2}e^{2}$.</p>
<p>Now, substitute the value of $b$ back in the equation for $V$, to obtain $$V(y,t) = e^{ay - a^{2} e^{2}t} = e^{a(y-aAt)},$$ where $A = e^{2}$.</p>
<p>Does this look correct?</p>
<p>Also, the boundary conditions are given with $y(0)=0$, and $y(T)=B$. </p>
| Anurag A | 68,092 | <p>Let $n$ be the total number of people at the meeting before Mr. L arrived. Then the number of handshakes will be $\dfrac{n(n-1)}{2}$. Suppose $m$ people left before Mr. L arrived .So he shakes $n-m$ hands. Then
\begin{align*}
\frac{n(n-1)}{2}+n-m & =100\\
n^2+n-2m & = 200\\
m & = \frac{n(n+1)}{2}-100.
\end{align*}<br>
But $0 < m < n$ (assuming that at least one person left). So we want
\begin{align*}
0 & < \frac{n(n+1)}{2}-100 < n\\
200 & < n(n+1) < 2n+\color{red}{200}.
\end{align*}</p>
<p><strong>edit:</strong>
I had made a typo because of which my initial conclusion was incorrect:</p>
<p>The first inequality suggests that $n \geq 14$ but the second inequality suggests $n \leq 14$. So $n=14$ is the answer. This yields $m=5$.</p>
|
1,160,454 | <p>I missed two classes in calculus and we're on a subject that I do not understand at all. If someone could just walk me through this problem I could probably begin to comprehend the rest.</p>
<p>The base of a solid elliptical cylinder is given by $ (x/5)^2 + (y/3)^2 = 1.$ A solid is formed by cutting off or removing some material such that the cross-sections perpendicular to the x-axis are all squares. Find the volume of such a solid. </p>
| MonkeysUncle | 217,283 | <p>First imagine how you'd do this problem in 2 dimensions. You need to set up a double integral over some x interval and then some y interval. The equation you're working with is in the form of an ellipse centered at the origin. Sections perpendicular to the x axis have constant x value, so you want x to be your last integral. So slice up your ellipse from left to right in segments of dx thickness from -5 to 5 and length $l$.</p>
<p>As you move from left to right, the length $l$ of the infinitesimal segments you are summing up changes. You can find this change just by looking at your equation. Solve for y and you get $y= \pm3\sqrt{1-\frac{x^2}{25}}$. So for some specific x value, these are the ranges y goes from. In two dimensions you would be able to set up your integral at this point.</p>
<p>In this case the integral is over 3 dimensions. but we're told each cross-section perpendicular to the x axis is a square. Therefore, if y goes from some -a to a, then z goes from the same exact -a to a. In other words, the bounds on z will be exactly the same as the bounds on y. This allows us to set up the integral.</p>
<p>$$ Volume=\int_{-5}^5\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}dzdydx $$</p>
<p>First step is to integrate over z. $\int dz = z$, so the integral becomes</p>
<p>$$ Volume = \int_{-5}^5\int_{-3\sqrt{1-\frac{x^2}{25}}}^{3\sqrt{1-\frac{x^2}{25}}}dydx. $$</p>
<p>Now integrate over y. No y variables have been introduced into the equation yet, so this is also just a simple $\int dy=y$, and we have
$$ Volume = \int^5_{-5} 4 * 9 (1-\frac{x^2}{25})dx .$$</p>
<p>The last integral is a piece of cake.</p>
|
527,799 | <p>In Hatcher's Algebraic Topology section 1.3, Cayley complexes are explained. The book states that we get a Cayley complex out of a Cayley graph by attaching a 2-cell to each loop. There is an example showing the Cayley complex for $\mathbb{Z}\times\mathbb{Z}$ (the fundamental group of the torus). We attach one 2-cell to each loop and we get $\mathbb{R}^{2}$ with vertical and horizontal tiling. I understand this.</p>
<p>The book then says (example 1.47) that the Cayley complex of a cyclic group of order $ n $ is $n$ disks with boundaries identified. I can't for the life of me figure out where the $n$ disks come from. In the Cayley graph, we have one loop $e \to x \to x^2 \to \cdots \to x^n = e$. I guess the relation $x^n = e$ somehow generates $n$ loops, but I don't understand why.</p>
<p>The next example is for $\mathbb{Z}_2*\mathbb{Z}_2$ in which two 2-cells are attached to each loop. I also don't understand why two.</p>
<p>I'm looking for a canonical description of the algorithm to build Cayley complexes, and the application of the algorithm to build Cayley complexes for finite cyclic groups and $\mathbb{ℤ}_2*\mathbb{ℤ}_2$.</p>
<p>Thank you.</p>
| Benjamin Steinberg | 85,907 | <p>This is essentially the same answer as user32240 but I will try to explain it differently. Hatcher's description is a bit sloppy. The correct thing to say is that if $R$ is the set of defining relators for $G$, then each element $r\in R$ labels a loop based at every vertex of the Cayley graph. To each of these based loops, you add a 2-cell. </p>
<p>The reason for this is you want the group $G$ to act freely on the Cayley complex. Now if you have a relator of the form $r=s^n$ where $s$ is not a proper power, then each loop labeled by $s^n$ in the Cayley graph can be read from $n$ different starting vertices and so you need a 2-cell for each one.</p>
<p>So, for example, if $G=\mathbb Z_2$ with presentation $\langle a\mid a^2=1\rangle$, then you want to have 2 2-cells and have $\mathbb Z_2$ permute them so that you have a free action. The two 2-cells come from the loop labelled $a^2$ at 1 and the looped labeled by $a^2$ at $a$. You can think of the 2 2-cells as the northern and southern hemisphere of a sphere.</p>
<p>If you attached only one $2$-cell, you would get a disk. The group $G$ would fix the center of this disk and so the action is not free. Although the projective plane is a disk with antipodal points identified, the quotient map is not a covering map. By attaching 2 disks you get a covering map.</p>
<p>Incidentally, this issue is not handled properly in the book of Lyndon and Schupp if memory recalls. Cohen makes a big point of this in his book and on the necessity of using $n$ disks for relators $r^n$.</p>
|
2,745,436 | <p>For some reason I have been struggling with this problem for the past couple hours.</p>
<p><a href="https://i.stack.imgur.com/diFLe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/diFLe.png" alt="enter image description here"></a></p>
<p>I believe I have solved part a.</p>
<p>Since there are 6 states (assuming a standard die and the die is fair), then there is an equal chance of landing in any of the 6 states for every dice throw. Therefore I concluded that the transition matrix looks as such:</p>
<p><code>[1 0 0 0 0 0]
[0 1 0 0 0 0]
[0 0 1 0 0 0]
[0 0 0 1 0 0]
[0 0 0 0 1 0]
[0 0 0 0 0 1]</code></p>
<p>For part b I setup my starting matrix as:</p>
<p>s1 = </p>
<p><code>[1]
[0]
[0]
[0]
[0]
[0]</code></p>
<p>I multiplied my transition matrix by <code>s1</code> 5 times and arrived at the same vector <code>s1</code>. Arithmetically, this makes sense. However, logically it does not.</p>
<p>Can somebody help explain where I went wrong?</p>
<p><strong>Edit:</strong> Based on the answers I believe the transition matrix would be:</p>
<p><code>[1/6 0 0 0 0 0]
[1/6 2/6 0 0 0 0]
[1/6 1/6 3/6 0 0 0]
[1/6 1/6 1/6 4/6 0 0]
[1/6 1/6 1/6 1/6 5/6 0]
[1/6 1/6 1/6 1/6 1/6 1]</code></p>
| Michael Hardy | 11,667 | <p>From state $s_1$ you have $1/6$ chance of remaining in that state and a $1/6$ chance of transitioning to any of the other states.</p>
<p>From state $s_2$ you have a $2/6$ chance of remaining there (by getting either a $1$ or a $2$) anad a $1/6$ chance of transitioning to each of $3,4,5,6.$</p>
<p>From state $s_3$ you have a $3/6$ chance of remaining there (by getting $1,$ $2,$ or $3$) and a $1/6$ chance of transitioning to each of $4,5,6.$</p>
<p>From state $s_4$ you have a $4/6$ chance of remaining there (by getting $1,$ $2,$ $3,$ or $4$) and a $1/6$ chance of transitioning to each of $5,6.$</p>
<p>From state $s_5$ you have a $5/6$ chance of remaining there (by getting $1,$ $2,$ $3,$ $4,$ or $5$) and a $1/6$ chance of transitioning to $6.$</p>
<p>From state $s_6$ you can only remain there.</p>
|
2,343,216 | <p>Let $\operatorname{gd}(x)$ the Gudermannian function, defined as in this <a href="http://mathworld.wolfram.com/Gudermannian.html" rel="nofollow noreferrer">MathWorld's article</a>, and implemented in Wolfram Language as <em>Gudermannian[x]</em>. </p>
<p>This afternoon I've spent an hour playing with <a href="https://www.wolframalpha.com" rel="nofollow noreferrer">Wolfram Alpha online calculator</a> about integrals, for instance, like this $$I=\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx.$$
Notice that the plot in previous article of MathWorld tell us that for $0<x<1$ one has $0<\operatorname{gd}(x)<1$. </p>
<p>My calculations to get an approximation were $$I\approx\int_0^1 \operatorname{gd}(x)\,dx=-\frac{\pi}{2}+i\left(\operatorname{Li_2}(-ie)-\operatorname{Li_2}(ie)\right)-2C,$$ where I've combined with the indefinite integral of the Gudermannian functions that tell us MathWorld. Here $C$ is Catalan's constant. But this quantity is about $\approx 0.464065$.</p>
<p>But WA calculated $I$ as $0.365619$, play this code:
<code>integrate log(1+(gd(x)))dx, from x=0 to x=1</code></p>
<blockquote>
<p><strong>Question.</strong> Can you provide us a better approximation, than mine, for our integral $$\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx?$$
Please provide also the more important justifications in your calculations. Also if your approach uses the numerical analysis and/or calculations with your computer provide us some detail of your calculation/method. <strong>Thanks in advance.</strong></p>
</blockquote>
| Raffaele | 83,382 | <p>You considered $\log(1+x)\approx x$ forgetting that gd(x) goes from 0 to 0.6 circa on the given interval, that is too much and gives a huge error</p>
<p>Compare $\log(1+0.4)\approx 0.33;\;\log(1+0.5)\approx 0.4$ and $\log(1+0.6)\approx 0.47$ </p>
<p>This causes your approximation being too high. Taking integration limit lower than 1, like 0.2 or 0.3 gives good approximation.</p>
|
4,373,262 | <p>Let G be a graph with order 9 so that the degree of each vertex is either 5 or 6. Prove
that there are either at least 5 vertices of degree 6 or at least 6 vertices of degree 5</p>
| Community | -1 | <p>Let the graph have <span class="math-container">$m$</span> vertices of degree <span class="math-container">$6$</span> and <span class="math-container">$n$</span> vertices of degree <span class="math-container">$5$</span>.</p>
<p>Then <span class="math-container">$6m+5n$</span> is twice the number of edges and so <span class="math-container">$n$</span> is even. If <span class="math-container">$n<6$</span> then we know <span class="math-container">$n\le 4$</span>.</p>
<p>Since <span class="math-container">$m+n=9$</span>, we therefore have <span class="math-container">$m\ge 5$</span>.</p>
|
1,665,107 | <p>I must use a generating function to solve this question: </p>
<p>In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the other two people each give $0$, $1$, or $5$ dollars?</p>
<p>Here is what I have so far.</p>
<p>The generating function is $(1+x)^6(1+x+x^5)^2$. Then let $f(x)= (1+x)^6$, and $g(x) = (1+x+x^5)^2$. We want the coefficient of $x^6$. Then, we note f(x) has the expansion:</p>
<p>$(1+x)^6$ = $1$ + $6 \choose 1$$x$ + $6 \choose 2$$x^2$ + ... + $6 \choose 6$$x^6$</p>
<p>But here is where I get stuck. We can let $h(x) = f(x)g(x)$ where $a_0b_0 + (a_1b_0+a_0b_1)x + (a_2b_0+ a_1b_1 + a_0b_2)x^2 + ... + (a_rb_0+ a_{r-1}b_1 + a_{r-2}b_2 + ... + a_0b_r)x^r$,</p>
<p>But I can't figure out the expansion of $g(x) = $$(1+x+x^5)^2$.</p>
<p>Can someone help me find the expansion of $g(x)$, and help me solve the problem?</p>
| Archis Welankar | 275,884 | <p>Hint its multinomial theorem where $n=x_1+x_2+...x_n$ and coefficient is given by ${n\choose x_1,x_2,x_3..x_n}$</p>
|
1,665,107 | <p>I must use a generating function to solve this question: </p>
<p>In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the other two people each give $0$, $1$, or $5$ dollars?</p>
<p>Here is what I have so far.</p>
<p>The generating function is $(1+x)^6(1+x+x^5)^2$. Then let $f(x)= (1+x)^6$, and $g(x) = (1+x+x^5)^2$. We want the coefficient of $x^6$. Then, we note f(x) has the expansion:</p>
<p>$(1+x)^6$ = $1$ + $6 \choose 1$$x$ + $6 \choose 2$$x^2$ + ... + $6 \choose 6$$x^6$</p>
<p>But here is where I get stuck. We can let $h(x) = f(x)g(x)$ where $a_0b_0 + (a_1b_0+a_0b_1)x + (a_2b_0+ a_1b_1 + a_0b_2)x^2 + ... + (a_rb_0+ a_{r-1}b_1 + a_{r-2}b_2 + ... + a_0b_r)x^r$,</p>
<p>But I can't figure out the expansion of $g(x) = $$(1+x+x^5)^2$.</p>
<p>Can someone help me find the expansion of $g(x)$, and help me solve the problem?</p>
| choco_addicted | 310,026 | <p>In general,
$$
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
$$
Thus
\begin{align}
(1+x+x^5)^2&=1+x^2+x^{10}+2(x+x^6+x^5)\\
&=1+2x+x^2+2x^5+2x^6+2x^{10}
\end{align}
and so the coefficient of $x^6$ is
\begin{align}
\binom{6}{6}+2\binom{6}{5}+\binom{6}{4}+2\binom{6}{1}+\binom{6}{0}&=1+2\cdot 6+15+ 2\cdot 6+1\\
&=41.
\end{align}</p>
|
3,517,722 | <p>Let <span class="math-container">$\alpha:= \sqrt[7]{2},\omega:= e^{\frac{2\pi i }{7}}\in \Bbb C$</span>. We set <span class="math-container">$E:=\Bbb Q(\alpha,\omega)$</span> and <span class="math-container">$B:=\Bbb Q(\omega+\omega^2+\omega^4)\leq E$</span>. </p>
<p>Thus, we have the Tower of Fields
<span class="math-container">$$\Bbb Q \leq B \leq E.$$</span>
We can prove that <span class="math-container">$|\mathrm{Aut}(E/\Bbb Q)|=42$</span> and that <span class="math-container">$\theta,\sigma\in \mathrm{Aut}(E/\Bbb Q)$</span>, given by
<span class="math-container">$$\theta:\alpha \longmapsto \alpha \omega,\ \omega \longmapsto \omega$$</span>
<span class="math-container">$$\sigma: \alpha \longmapsto \alpha,\ \omega \longmapsto \omega^3.$$</span></p>
<p><strong><em>Question:</em></strong> How can we prove that <span class="math-container">$\mathrm{Aut}(E/B)=\langle \theta,\sigma^2\rangle$</span>?</p>
<p>A first thought is to compute the Galois group <span class="math-container">$\mathrm{Aut}(E/B)$</span>, but this seems to be extremely hard, because of <span class="math-container">$B$</span>. Another might be to use somehow the Fundamental Theorem of Galois Theory, so to exploit the relations <span class="math-container">$[E:B]=|\mathrm{Aut}(E/B)|$</span> and then find the order of the subgroup <span class="math-container">$\langle \theta,\sigma^2 \rangle$</span>.</p>
<p>Note that <span class="math-container">$\theta^i(\alpha)=\alpha\omega^i$</span> and <span class="math-container">$\sigma^j(\omega)=\omega^{3^j}$</span>.</p>
| reuns | 276,986 | <p>You meant <span class="math-container">$\alpha=\sqrt[7]{2}$</span>. Let <span class="math-container">$K=\Bbb{Q}(\omega)$</span>.</p>
<p>We get <span class="math-container">$E/K/B/\Bbb{Q}$</span> where <span class="math-container">$E/\Bbb{Q}$</span> and <span class="math-container">$K/B$</span> are Galois.</p>
<p>We know that <span class="math-container">$Aut(K/\Bbb{Q})=\Bbb{Z/7Z}^\times=\langle \sigma\rangle,Aut(E/K)=\Bbb{Z/7Z}$</span>. It is quite immediate that both mix together to give <span class="math-container">$Gal(E/\Bbb{Q})=Aff(\Bbb{Z/7Z})$</span> (the group of affine transformations <span class="math-container">$x\to ax+b$</span> which corresponds to <span class="math-container">$\omega^x\alpha\to \omega^{ax+b}\alpha$</span>).</p>
<p><span class="math-container">$B$</span> is a subfield of <span class="math-container">$K$</span> fixed by <span class="math-container">$\sigma^2$</span>, and it is not <span class="math-container">$\Bbb{Q}$</span>, so it is the subfield fixed by <span class="math-container">$\sigma^2$</span> and hence <span class="math-container">$ Aut(K/B)=\langle \sigma^2\rangle$</span>. </p>
<p><span class="math-container">$K$</span> is the subfield of <span class="math-container">$E$</span> fixed by <span class="math-container">$\theta$</span>.</p>
<p>Extend <span class="math-container">$\sigma\in Aut(K/\Bbb{Q})$</span> to an element <span class="math-container">$\sigma'\in Aut(E/\Bbb{Q})$</span>. </p>
<blockquote>
<p>Then it is immediate that <span class="math-container">$B$</span> is the subfield of <span class="math-container">$E$</span> fixed by <span class="math-container">$(\sigma')^2$</span> and <span class="math-container">$\theta$</span> which means that <span class="math-container">$E/B$</span> is Galois with Galois group <span class="math-container">$Aut(E/B)=\langle(\sigma')^2,\theta\rangle$</span>.</p>
</blockquote>
<p><span class="math-container">$[E:B]=[E:K][K:B]=21$</span> and <span class="math-container">$Aut(E/B)= \{x\to a^2 x+b\}$</span>.</p>
|
2,851,609 | <p>I need to find the solution to the inequality $(x - y)(x + y -1) > z$, where $x,y,z \geq 0$ and $x,y,z \leq 1$. As $z$ is positive, then the inequality holds whenever (i) $x - y > 0$ and $x + y - 1 > 0$ OR (ii) $x - y < 0$ and $x + y - 1 < 0$. I can solve the cases (i) and (ii) on their own, but I don't know how to relate them to the value of $z$. Probably the solution is simple, but I am not an expert in math so any help would be appreciated. </p>
| Luca Bressan | 89,879 | <p>If we multiply and rearrange the terms in the inequality, we get:
<span class="math-container">$$z < x^2 - y^2 - x + y$$</span>
which is equivalent to:
<span class="math-container">$$z < \left (x - \frac 1 2 \right)^2 - \left (y - \frac 1 2 \right)^2.$$</span>
The associated equation represents a hyperbolic paraboloid having its saddle point at <span class="math-container">$\left ( \frac 1 2 , \frac 1 2, 0 \right )$</span>.</p>
<p>If we rewrite the inequality as the system:
<span class="math-container">$$\begin{cases} \left (x - \frac 1 2 \right)^2 - \left (y - \frac 1 2 \right)^2 > k \\ z = k \end{cases} \qquad (k \in \mathbb R)$$</span>
we see that the inequality represents the set <span class="math-container">$S$</span> of all points that, at a given altitude <span class="math-container">$k$</span>, lie outside the corresponding hyperbola.</p>
<p>Now, if we also require that <span class="math-container">$0 \le x, y, z \le 1$</span>, we are just intersecting <span class="math-container">$S$</span> with the cube centered at <span class="math-container">$\left ( \frac 1 2, \frac 1 2, \frac 1 2 \right)$</span> having side of length <span class="math-container">$1$</span>. This means that we consider only the points having altitude <span class="math-container">$k$</span> with <span class="math-container">$0 \le k \le 1$</span> that lie on the inside of the square having vertices <span class="math-container">$(0, 0, k)$</span>, <span class="math-container">$(1, 0, k)$</span>, <span class="math-container">$(1, 1, k)$</span>, <span class="math-container">$(0, 1, k)$</span> and on the outside of the corresponding hyperbola.</p>
<p>We can look at some of those points for given altitudes.</p>
<p>For <span class="math-container">$k = 0$</span> we get:</p>
<p><a href="https://i.stack.imgur.com/SqZAj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SqZAj.png" alt="k=0" /></a></p>
<p>For <span class="math-container">$k = \frac 1 {20}$</span>, for instance, we get:</p>
<p><a href="https://i.stack.imgur.com/6Qml2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Qml2.png" alt="k=0.05" /></a></p>
<p>and so on, until for <span class="math-container">$\frac 1 4 \le k \le 1$</span> we get:</p>
<p><a href="https://i.stack.imgur.com/vePg4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vePg4.png" alt="k=0.25" /></a></p>
<p>because for <span class="math-container">$k = \frac 1 4$</span> the hyperbola has its vertices at <span class="math-container">$(0, \frac 1 2, k)$</span> and <span class="math-container">$(1, \frac 1 2, k)$</span>.</p>
|
223,509 | <p>Let $\ulcorner \cdot \urcorner$ be a fixed encoding of formulas by numbers, e.g.
let $\ulcorner \varphi \urcorner$ denote the <a href="https://en.wikipedia.org/wiki/G%C3%B6del_numbering" rel="nofollow">Godel number</a> of $\varphi$.
Let $T$ be a first-order arithmetic theory, e.g. PA.
Let $\Phi$ be a class of closed first-order formulas e.g. $\Sigma^0_1$.
Let $Tr$ be a first-order formula defining
the truth predicate for the formulas in $\Phi$ in the standard model:</p>
<blockquote>
<p>for all $\varphi \in \Phi$, $\mathbb{N} \vDash \varphi$ iff
$\mathbb{N} \vDash Tr(\ulcorner \varphi \urcorner)$</p>
</blockquote>
<p>E.g.
we have:
$\mathbb{N} \vDash Tr(\ulcorner \top \urcorner)$,
$\mathbb{N} \nvDash Tr(\ulcorner \bot \urcorner)$,
$\mathbb{N} \nvDash Tr(\ulcorner \exists x \ (x + 1 = 0)\urcorner)$,
etc.</p>
<p>Consider the following property: </p>
<blockquote>
<p>for all $\varphi \in \Phi$,
$T \vdash Tr(\ulcorner \varphi \urcorner) \to \varphi$.</p>
</blockquote>
<p><strong>Does this property have a standard name?</strong></p>
<hr>
<p>The reflection is similar to the property above
with <em>truth</em> replaced with <em>provability</em>:
let $\square$ denote the provability in theory $T'$.
Then reflection property is:</p>
<blockquote>
<p>for all $\varphi \in \Phi$, $T \vdash \square(\ulcorner \varphi \urcorner) \to \varphi$.</p>
</blockquote>
| Bjørn Kjos-Hanssen | 4,600 | <p>At the risk of stating the obvious, in modal logic the axiom schema
$$\Box\varphi\rightarrow\varphi$$
is called the schema T.</p>
|
8,756 | <p>I'd like to suggest that after a Question receives an Accepted Answer, some consideration be given to revising the title (if appropriate) to reflect what the real issue turned out to be.</p>
<p>It seems to me users often pick titles when first posting a question that are uninformative and are worth revisiting once the actually point is identified. Improved subject lines will help in searches, and by extension help with identifying duplicates.</p>
<p>High rep users, it seems to me, are not excessively shy about making changes to tags, which helps with later searches somewhat. But I think a good title trumps all that, and the SE search engine can use all the help we can give it.</p>
<p>Discussion?</p>
| Brett Frankel | 22,405 | <p>I think we should be more conservative about changing titles than retagging.</p>
<p>It's reasonable to suppose that someone searching for an answer to a similar question will be thinking along the lines of the OP's original wording, and not necessarily in the language of the answer. I can't see why anybody would object to revising a title after some discussion has clarified the language of the original question; retitling questions to reflect answers, however, may make it harder for users to search for answers to questions they are about to ask.</p>
<p>We already have too many users who post questions without first searching to see if their questions have already been answered. To keep the search process as easy and effective as possible, I think we should avoid revising titles to the point that they contain substantial language outside of the original question.</p>
|
8,756 | <p>I'd like to suggest that after a Question receives an Accepted Answer, some consideration be given to revising the title (if appropriate) to reflect what the real issue turned out to be.</p>
<p>It seems to me users often pick titles when first posting a question that are uninformative and are worth revisiting once the actually point is identified. Improved subject lines will help in searches, and by extension help with identifying duplicates.</p>
<p>High rep users, it seems to me, are not excessively shy about making changes to tags, which helps with later searches somewhat. But I think a good title trumps all that, and the SE search engine can use all the help we can give it.</p>
<p>Discussion?</p>
| Alexander Gruber | 12,952 | <p>I'm for it. I have been making an effort lately to do this myself. I am sick of seeing titles like "<a href="https://math.stackexchange.com/q/328718/12952">Proof about a semigroup</a>" (an example taken from the front page right at this moment). This provides absolutely no information, and there's no reason why the title can't be "In a semigroup, $ab=ba\Rightarrow (ab)^k=a^kb^k$." In fact, I don't see any reason to wait for an answer to be accepted to change it.</p>
<p>I agree with Brett's question about revising original wording, though. I think we should not change the title if it includes mistakes or bad wording <em>which are descriptive of the question's content</em>. As he points out, the whole point is to make the search better.</p>
|
2,037,428 | <p>Let $D$ be a compact, connected $Jordan$ domain in $R^n$ with positive volume, and suppose that the fuction $f:D →R$ is continuous. Show that there is a point $x$ in $D$ in which
$f(x)=(1/volD)$$\int$$f$ over $D$.
(Mean Value Property for integral)</p>
<p>I can prove it when $D$ is pathwise connected</p>
<p>But i have no idea when $D$ is just connected</p>
<p>Is it true compact, connected subset of $R^n$ is pathwise connected?</p>
<p>Or is there any other ways to solve it?</p>
| Julián Aguirre | 4,791 | <p>No. The standard example is the <em>Topologist sine curve</em>:
$$
\Bigl\{\Bigl(x,\sin\frac{1}{x}\Bigr):0<x\le1\Bigr\}\bigcup\{0\}\times[-1,1].
$$
To prove the MVT all you need is that the continuous image of a connected set is connected.</p>
|
991,377 | <p>I have to evalute this integral.
$\displaystyle\iint\limits_{D}(2+x^2y^3 - y^2\sin x)\,dA$
$$D=\left \{ (x, y):\left | x \right |+\left | y \right | \leq 1\right \}$$</p>
<p>At first, I evaluated simply by putting $-1\leq x\leq 1, -1\leq y\leq 1$, thus making
$$
\int_{-1}^{1}\int_{-1}^{1}(2+x^2y^3 - y^2\sin x)\,dx\,dy,
$$
but the answer I got was 8, not 4, which my answersheed requires me. </p>
| MvG | 35,416 | <p>This is a coordinate-based approach, making heavy use of tools from projective geometry.</p>
<p>Without loss of generality, you can choose the coordinate system in such a way that the inscribed circle is the unit circle. On that you can use a rational parametrization, i.e. choose $a,b,c\in\mathbb R$ such that $A'=(1-a^2,2a)/(1+a^2)$ and likewise for $B'$ and $C'$. Or, even better, use homogeneous coordinates for these:</p>
<p>$$
A'=\begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix}\qquad
B'=\begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix}\qquad
C'=\begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix}
$$</p>
<p>From that, everything else follows. The tangent at a point on the circle is simply its polar line, which you get by multiplication with the matrix of the unit circle, namely the matrix</p>
<p>$$U=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$</p>
<p>You can join points using lines, and intersect lines to obtain points, simply by computing the cross product.</p>
<p>$$A=(U\cdot B')\times(U\cdot C')$$</p>
<p>and likewise for $B$ and $C$, the other two corners of the triangle. Then you get</p>
<p>$$G=(A\times A')\times(B\times B')$$</p>
<p>A circle through three points can be constructed as a conic through these points and the special points $I=(1,i,0)$ and $J=(1,-i,0)$ which have complex coordinates and lie on the line at infinity. Since they also lie on every circle, they are often called the ideal circle points. So we construct the matrix of the circle $\bigcirc GA'B'$ in several steps:</p>
<p>\begin{align*}
M_1 &= (A'\times I)\cdot(B'\times J)^T &
M_2 &= (A'\times J)\cdot(B'\times I)^T \\
M_3 &= M_1 + M_1^T & M_4 &= M_2 + M_2^T \\
M_5 &= (G\cdot M_3\cdot G)M_4 - (G\cdot M_4\cdot G)M_3 &
M_C &= iM_5
\end{align*}</p>
<p>$M_1$ and $M_2$ describe degenerate conics through $A',B',I,J$. $M_3$ and $M_4$ are the same except using symmetric matrices. $M_5$ is a linear combination which also passes through $G$, so that's the circle. $M_C$ is a real matrix describing the same circle, avoiding all the purely imaginary entries of $M_5$.</p>
<p>To intersect that circle with $AC$ (which is the same line as $B'C$) you compute</p>
<p>$$C_A=(C^T\cdot M_C\cdot C)B'-2(B'^T\cdot M_C\cdot C)C$$</p>
<p>This is still a homogeneous coordinate vector of a point, e.g. some $(x,y,z)$. Dehomogenize that to $(x/z, y/z)$ then take the norm of that:</p>
<p>\begin{align*}
\lVert C_A\rVert =& \frac{\sqrt s}{t} \\
s =& \phantom+ (a^4b^4 + a^4c^4 + b^4c^4) \\&
- 2\,abc\,(a^3b^2 + a^2b^3 + a^3c^2 + b^3c^2 + a^2c^3 + b^2c^3) \\&
+ 3\,a^2b^2c^2\,(a^2 + b^2 + c^2) \\&
+ 11\,(a^4b^2 + a^2b^4 + a^4c^2 + b^4c^2 + a^2c^4 + b^2c^4) \\&
+ 16\,abc\,(a^2b + ab^2 + a^2c + b^2c + ac^2 + bc^2) \\&
- 20\,(a^3b^3 + a^3c^3 + b^3c^3) \\&
- 20\,abc\,(a^3 + b^3 + c^3) \\&
- 42\,a^2b^2c^2 \\&
- 2\,(a^3b + ab^3 + a^3c + b^3c + ac^3 + bc^3) \\&
+ 3\,(a^2b^2 + a^2c^2 + b^2c^2) \\&
+ (a^4 + b^4 + c^4) \\
t =&\phantom+ (a^2 + b^2 + c^2) \\&
- (ab + ac + bc) \\&
+ (a^2b^2 + a^2c^2 + b^2c^2) \\&
- abc\,(a + b + c)
\end{align*}</p>
<p>This is the radius for one of the six points of your claimed circle. The other five can be obtained using the same computation, starting from a permutation of the three initial points. So the result will be the same except for a permutation of the parameters $a,b,c$. But the formula stated above is invariant under such a permutation, therefore all six points lie on a circle as claimed. Its center is the center of the coordinate system, i.e. the incenter of the triangle. Its radius will be the fraction described by the lengthy expressions stated above.</p>
|
2,282,359 | <p>I am trying to calculate this limit:
$$\lim_{x \to \infty} x^2(\ln x-\ln (x-1))-x$$
The answer is $1/2$ but I am trying to verify this through proper means. I have tried L'Hospital's Rule by factoring out an $x$ and putting that as $\frac{1}{x}$ in the denominator (indeterminate form) but it becomes hopeless afterwards. Also I am a little hesitant about series involving the natural log because of restricted interval of convergence as $x$ is going to infinity. Is there a different approach how to do evaluate this limit? Thanks. </p>
| CY Aries | 268,334 | <p>\begin{align*}
\lim_{x \to \infty} [ x^2(\ln x-\ln (x-1))-x]&=\lim_{x \to \infty} \frac{\displaystyle\ln x-\ln (x-1)-\frac{1}{x}}{\displaystyle\frac{1}{x^2}}\\
&=\lim_{x \to \infty} \frac{\displaystyle\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x^2}}{\displaystyle\frac{-2}{x^3}}\quad (\text{ L'Hopital Rule})\\
&=\lim_{x \to \infty} \frac{\displaystyle\frac{-1}{x^2(x-1)}}{\displaystyle\frac{-2}{x^3}}\\
&=\frac{1}{2}
\end{align*}</p>
|
390,718 | <p>I am trying to prove something that seemed simple to me at first sight but apparently it is giving me a hard time. Here is <a href="https://math.stackexchange.com/questions/4108555/bounded-sequence-in-spatial-tensor-product-and-boundedness-of-simple-tensor-summ">the same question on MSE</a>.</p>
<p>Let <span class="math-container">$E\subset A$</span> be a finite dimensional operator system with Hamel basis <span class="math-container">$\{x_1,\dots,x_n\}$</span> and let <span class="math-container">$B$</span> be an arbitrary <span class="math-container">$C^*$</span>-algebra. I am trying to show that, if <span class="math-container">$\|\cdot\|$</span> is the minimal norm on <span class="math-container">$E\odot B$</span> (i.e. the spatial norm) and <span class="math-container">$E\otimes B:=\overline{E\odot B}^{\|\cdot\|}$</span>, then every element of <span class="math-container">$E\otimes B$</span> is written uniquely as <span class="math-container">$\sum_{i=1}^nx_i\otimes b_i$</span>. This is of course equivalent to showing that <span class="math-container">$E\odot B$</span> is complete with the minimal norm.</p>
<p>It is trivial to verify the claim for the elements of the algebraic tensor product <span class="math-container">$E\odot B$</span>, by the following lemma:</p>
<blockquote>
<p>If <span class="math-container">$X,Y$</span> are vector spaces and <span class="math-container">$\{a_1,\dots,a_n\}\subset X$</span> are linearly independent, then
<span class="math-container">$$\sum_{i=1}^na_i\otimes b_i=0\implies b_1=\dots=b_n=0.$$</span></p>
</blockquote>
<p>This lemma also takes care of uniqueness for us. But what about elements of <span class="math-container">$E\otimes B$</span> in general? Of course, the ideal thing would be to establish an inequality of the form <span class="math-container">$\|x_j\otimes b_j\|\leq C\|\sum_{i=1}^nx_i\otimes b_i\|$</span>, but how?</p>
<p>I tried to consider <span class="math-container">$B\oplus\dots\oplus B\to (E\odot B,\|\cdot\|)$</span>, <span class="math-container">$(b_1,\dots,b_n)\mapsto\sum_{i=1}^nx_i\otimes b_i$</span> and observed that this is bounded and bijective. The annoying thing: if I knew that <span class="math-container">$E\odot B$</span> is complete (which is what I want to show), then I could apply the open mapping theorem to deduce that the inverse is bounded, hence obtain the desired estimate.</p>
<p>I also tried an induction argument on <span class="math-container">$\dim(E)$</span> but it didn't work out.</p>
| Mateusz Wasilewski | 24,953 | <p>One way to see it is to use the dual basis of <span class="math-container">$\{x_1,\dots,x_n\}$</span>. Let <span class="math-container">$\{\varphi_1,\dots,\varphi_n\} \subset E^{\ast}$</span> be functionals such that <span class="math-container">$\varphi_i(x_j)=\delta_{ij}$</span>. Because functionals are automatically completely bounded, we get bounded maps <span class="math-container">$\Phi_i:= \varphi_i \otimes Id : E\otimes B \to B$</span>. On the subspace <span class="math-container">$E\odot B$</span> we have <span class="math-container">$\Phi_i(\sum_{j=1}^{n} x_{j} \otimes b_{j}) = b_{i}$</span>, so <span class="math-container">$\|b_{i}\| \leqslant \|\Phi_i\|\cdot \|\sum_{j=1}^{n} x_{j} \otimes b_{j}\|$</span>. As a consequence, if a sequence <span class="math-container">$y(m):= \sum_{i=1}^{n} x_{i} \otimes b_i(m) \in E\odot B$</span> converges in <span class="math-container">$E \otimes B$</span> then for each <span class="math-container">$i\in \{1,\dots, n\}$</span> the sequence <span class="math-container">$b_i(m)$</span> converges in <span class="math-container">$B$</span>. Let <span class="math-container">$B_i = \lim_{m\to\infty} b_i(m)$</span>. Then <span class="math-container">$\lim_{m\to \infty} \sum_{i=1}^{n} x_i \otimes b_i(m) = \sum_{i=1}^{n} x_i \otimes B_i \in E \odot B$</span>, hence <span class="math-container">$E\odot B$</span> is equal to <span class="math-container">$E\otimes B$</span>.</p>
|
390,718 | <p>I am trying to prove something that seemed simple to me at first sight but apparently it is giving me a hard time. Here is <a href="https://math.stackexchange.com/questions/4108555/bounded-sequence-in-spatial-tensor-product-and-boundedness-of-simple-tensor-summ">the same question on MSE</a>.</p>
<p>Let <span class="math-container">$E\subset A$</span> be a finite dimensional operator system with Hamel basis <span class="math-container">$\{x_1,\dots,x_n\}$</span> and let <span class="math-container">$B$</span> be an arbitrary <span class="math-container">$C^*$</span>-algebra. I am trying to show that, if <span class="math-container">$\|\cdot\|$</span> is the minimal norm on <span class="math-container">$E\odot B$</span> (i.e. the spatial norm) and <span class="math-container">$E\otimes B:=\overline{E\odot B}^{\|\cdot\|}$</span>, then every element of <span class="math-container">$E\otimes B$</span> is written uniquely as <span class="math-container">$\sum_{i=1}^nx_i\otimes b_i$</span>. This is of course equivalent to showing that <span class="math-container">$E\odot B$</span> is complete with the minimal norm.</p>
<p>It is trivial to verify the claim for the elements of the algebraic tensor product <span class="math-container">$E\odot B$</span>, by the following lemma:</p>
<blockquote>
<p>If <span class="math-container">$X,Y$</span> are vector spaces and <span class="math-container">$\{a_1,\dots,a_n\}\subset X$</span> are linearly independent, then
<span class="math-container">$$\sum_{i=1}^na_i\otimes b_i=0\implies b_1=\dots=b_n=0.$$</span></p>
</blockquote>
<p>This lemma also takes care of uniqueness for us. But what about elements of <span class="math-container">$E\otimes B$</span> in general? Of course, the ideal thing would be to establish an inequality of the form <span class="math-container">$\|x_j\otimes b_j\|\leq C\|\sum_{i=1}^nx_i\otimes b_i\|$</span>, but how?</p>
<p>I tried to consider <span class="math-container">$B\oplus\dots\oplus B\to (E\odot B,\|\cdot\|)$</span>, <span class="math-container">$(b_1,\dots,b_n)\mapsto\sum_{i=1}^nx_i\otimes b_i$</span> and observed that this is bounded and bijective. The annoying thing: if I knew that <span class="math-container">$E\odot B$</span> is complete (which is what I want to show), then I could apply the open mapping theorem to deduce that the inverse is bounded, hence obtain the desired estimate.</p>
<p>I also tried an induction argument on <span class="math-container">$\dim(E)$</span> but it didn't work out.</p>
| Matthew Daws | 406 | <p>Here's an approach motivated by considerations of Banach space tensor products; it owes a debt to the approach taken by Takesaki in his book, Volume 1, Chapter IV, Sections 2 and 4. Another good book is <a href="https://link.springer.com/book/10.1007/978-1-4471-3903-4" rel="nofollow noreferrer">Ryan's book</a>, "Introduction to Tensor Products of Banach Spaces"</p>
<p>For Banach spaces <span class="math-container">$E,F$</span> a norm on <span class="math-container">$E\odot F$</span> is a <em>cross-norm</em> if <span class="math-container">$\|x\otimes y\| = \|x\| \|y\|$</span>. There are two natural cross-norms: the <em>injective tensor norm</em> defined by the map <span class="math-container">$E\odot F\rightarrow B(E^*,F)$</span>, so
<span class="math-container">$$ \lambda\Big(\sum_{i=1}^n x_i\otimes y_i\Big) = \sup\Big\{ \Big\|\sum_{i=1}^n f(x_i) y_i \Big\|: f\in E^*, \|f\|\leq 1 \Big\}; $$</span>
and the <em>projective tensor norm</em>,
<span class="math-container">$$ \pi(u) = \inf\Big\{ \sum_{i=1}^n \|x_i\| \|y_i\| : u = \sum_{i=1}^n x_i\otimes y_i \Big\}. $$</span>
These are cross-norms. Given a cross-norm <span class="math-container">$\beta$</span>, there is a norm <span class="math-container">$\beta^*$</span> on <span class="math-container">$E^*\odot F^*$</span> given by the natural dual-pairing between <span class="math-container">$E\odot F$</span> and <span class="math-container">$E^*\odot F^*$</span>. Then <span class="math-container">$\beta^*$</span> is a cross-norm if and only if <span class="math-container">$\lambda \leq \beta \leq \pi$</span>. (So such <span class="math-container">$\beta$</span> are in some sense well-behaved).</p>
<hr/>
<p>If <span class="math-container">$E$</span> is finite-dimensional, with basis <span class="math-container">$(x_i)_{i=1}^n$</span> then (much as Mateusz argues) we can use the dual basis <span class="math-container">$(x_i^*)$</span> to see that
<span class="math-container">$$ \lambda\Big(\sum_{i=1}^n x_i\otimes y_i\Big) \geq K^{-1} \|y_j\| $$</span>
for any <span class="math-container">$j$</span>, where <span class="math-container">$K=\max_i \|x_i^*\|$</span>. Clearly also
<span class="math-container">$$ \pi(\sum_{i=1}^n x_i\otimes y_i) \leq \sum_i \|x_i\| \|y_i\|
\leq L \max_i \|y_i\| $$</span>
where <span class="math-container">$L=\sum_i \|x_i\|$</span>. Thus both <span class="math-container">$\lambda$</span> and <span class="math-container">$\pi$</span> are equivalent to the max-norm <span class="math-container">$\max_i \|y_i\|$</span>, and so any "nice" cross-norm on <span class="math-container">$E\odot F$</span> gives a norm equivalent to the direct sum of <span class="math-container">$n$</span> copies of <span class="math-container">$F$</span>.</p>
<hr/>
<p>Now, it turns out that any <span class="math-container">$C^*$</span>-tensor norm on <span class="math-container">$A\odot B$</span> is a "nice" cross-norm: this is shown by Takesaki on pages 207--208. As the theory of <span class="math-container">$C^*$</span>-tensor norms is quite intricate, it's hard to say exactly <em>why</em> this is so. However, we then see that the dual norm on <span class="math-container">$A^*\odot B^*$</span> is a cross-norm (in fact, and harder to show, the dual norm on <span class="math-container">$A^*\odot B^*$</span> is always the same, independent of the <span class="math-container">$C^*$</span>-tensor norm on <span class="math-container">$A\odot B$</span>.)</p>
<p>The result follows.</p>
<p>In fact, if you are happy to use that <span class="math-container">$\|x\otimes b\| = \|x\| \|b\|$</span> on <span class="math-container">$E\odot B$</span>, then already <span class="math-container">$\|\cdot\|\leq\pi$</span>, and so it remains to show that <span class="math-container">$\lambda\leq\|\cdot\|$</span>. We need only show that if <span class="math-container">$f\in E^*, g\in B^*$</span> then <span class="math-container">$f\otimes g$</span> induces a functional on <span class="math-container">$E\odot B$</span> of norm at most <span class="math-container">$\|f\| \|g\|$</span>. Hahn-Banach <span class="math-container">$f$</span> to a member of <span class="math-container">$A^*$</span> and hence work on <span class="math-container">$A\odot B$</span>. The result follows from a polar-decomposition argument and then a GNS argument, using the definition of the spatial <span class="math-container">$C^*$</span>-tensor norm: this is exactly how Takesaki's book proceeds.</p>
|
11,315 | <p>I just answered this question:</p>
<p><a href="https://math.stackexchange.com/questions/528880/boolean-formula">Boolean formula over 64 Boolean variables X</a></p>
<p>By the time I had posted my answer, another user had edited the question so as to remove all the mathematical interest.</p>
<p>What is going on here? I feel very demotivated after spending time crafting an answer to a question and then seeing a sensible question trashed.</p>
<p>Regards,</p>
<p>Rob.</p>
| Rob Arthan | 23,171 | <p>Thanks to everyone for their comments and to Andres Caicedo for undoing the irresponsible edit. I conclude that I was bamboozled into solving a homework question. When I gave an answer the student or students tried to cover up what had happened by conspiring to delete the mathematical content of the question.</p>
<p>Or maybe I should apply <a href="http://en.wikipedia.org/wiki/Hanlon%27s_razor" rel="nofollow">Hanlon's Razor</a> and conclude it was all down to incompetence rather than malice.</p>
|
48,237 | <p>I have found by a numerical experiment that first such primes are:
$2,5,13,17,29,37,41$. But I cannot work out the general formula for it.<br>
Please share any your ideas on the subject.</p>
| GEdgar | 442 | <p>In a number theory course, you would probably come to the <em>Legendre symbol</em> after a while... <a href="http://en.wikipedia.org/wiki/Legendre_symbol" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Legendre_symbol</a> </p>
|
4,415,037 | <p>I would like to prove upper and lower bounds on <span class="math-container">$|\cos(x) - \cos(y)|$</span> in terms of <span class="math-container">$|x-y|$</span>. I was able to show that <span class="math-container">$|\cos(x) - \cos(y)| \leq |x - y|$</span>. I'm stuck on the lower bound. Does anyone know how to approach this?</p>
<p>Update: Over the interval <span class="math-container">$[0,\pi/2]$</span>, I was able to show that <span class="math-container">$|\cos(x) - \cos(y)| \geq \frac{2 \min(x,y)}{\pi}|x-y|$</span>. But I would like a lower bound that holds for any interval.</p>
| jjagmath | 571,433 | <p>Assume there exists a function <span class="math-container">$f:\Bbb R^+ \to \Bbb R$</span> such that <span class="math-container">$\left|\cos x - \cos y\right| \ge f(\left|x-y\right|)$</span> for all <span class="math-container">$x, y$</span>.</p>
<p>Choose any <span class="math-container">$a \ge 0$</span>.</p>
<p>Set <span class="math-container">$x = \frac{a}{2}$</span> and <span class="math-container">$y=-\frac{a}{2}$</span>. By the assumption <span class="math-container">$$0 =\left|\cos (\tfrac{a}{2}) - \cos(-\tfrac{a}{2})\right| \ge f(\left|\tfrac{a}{2}-(-\tfrac{a}{2})\right|) = f(a)$$</span></p>
<p>So <span class="math-container">$f(a)$</span> is non-positive for all the values of <span class="math-container">$a$</span>. That proves that the best lower bound for <span class="math-container">$\left|\cos x - \cos y\right|$</span> depending only on <span class="math-container">$|x-y|$</span> is <span class="math-container">$0$</span>.</p>
|
873,183 | <p>Are variables logical or non-logical symbols in a logic system?
I understand constants are 0-ary logical operation symbols. I think variables are non-logical symbols.</p>
<p>But here are two contrary examples:</p>
<p>It seems that variables are logical symbols in a propositional logic system, according to <a href="http://en.wikipedia.org/wiki/First-order_logic#Logical_symbols" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/First-order_logic#Logical_symbols</a></p>
<blockquote>
<p><strong>Logical symbols</strong></p>
<p>An infinite set of <strong>variables</strong>, often denoted by lowercase letters at
the end of the alphabet x, y, z, … . Subscripts are often used to
distinguish variables: x0, x1, x2, … .</p>
</blockquote>
<p>It seems that variables are non-logical symbols in a propositional logic system, according to <a href="http://en.wikipedia.org/wiki/Propositional_logic#Generic_description_of_a_propositional_calculus" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Propositional_logic#Generic_description_of_a_propositional_calculus</a></p>
<blockquote>
<ul>
<li><p>The alpha set is a finite set of elements called <strong>proposition symbols or propositional variables</strong>. Syntactically
speaking, these are the most basic elements of the formal language
<span class="math-container">$\mathcal{L}$</span>, otherwise referred to as atomic formulæ or terminal
elements. In the examples to follow, the elements of are
typically the letters <span class="math-container">$p, q, r$</span>, and so on.</p>
</li>
<li><p>The omega set <span class="math-container">$\Omega$</span> is a finite set of elements called <strong>operator symbols or logical connectives</strong>.</p>
</li>
</ul>
</blockquote>
<p>Thanks.</p>
| Mauro ALLEGRANZA | 108,274 | <p>We can agree that there is some "variability" in the practice, regarding the definition (if any) of <em>logical symbols</em> in <em>first-order</em> logic.</p>
<p>According to the definition in Herbert Enderton, <a href="https://books.google.it/books?id=dVncCl_EtUkC&pg=PA69" rel="nofollow">A Mathematical Introduction to Logic</a> (2nd - 2001), of <strong>First-Order Languages</strong> [page 69], we have :</p>
<blockquote>
<p><strong>A. Logical symbols</strong></p>
<p>$0$. Parentheses: $(,)$. </p>
<p>$1$. Sentential connective symbols: $\rightarrow, \lnot$. </p>
<p>$2$. Variables (one for each positive integer $n$): </p>
<blockquote>
<p>$v_\, v_2, \ldots$.</p>
</blockquote>
<p>$3$. Equality symbol (optional): $=$. </p>
<p><strong>B. Parameters</strong> </p>
<p>$0$. Quantifier symbol: $\forall$.</p>
<p>$1$. Predicate symbols: For each positive integer $n$, some set (possibly empty) of symbols, called $n$-place predicate symbols. </p>
<p>[...]</p>
</blockquote>
<p>Compare with Dirk van Dalen, <a href="https://books.google.it/books?id=u0wlXPHATDcC&pg=PA56" rel="nofollow">Logic and Structure</a> (5th ed - 2013), page 56 :</p>
<blockquote>
<p>The alphabet consists of the following symbols:</p>
<ol>
<li><p>Predicate symbols: $P_1, \ldots, P_n, =$</p></li>
<li><p>Function symbols: $f_1, \ldots, f_m$</p></li>
<li><p>Constant symbols: $c_i$ for $i \in I$</p></li>
<li><p>Variables: $x_0, x_1, x_2, \ldots$ (countably many)</p></li>
<li><p>Connectives: $∨,∧,→,¬,↔,⊥,∀,∃$</p></li>
<li><p>Auxiliary symbols: $(, )$.</p></li>
</ol>
</blockquote>
<p>Note that the expression "logical symbols" has been avoided.</p>
<p>The issue is related to that of <a href="http://plato.stanford.edu/entries/logical-form/" rel="nofollow">Logical Form</a> and <a href="http://plato.stanford.edu/entries/logical-constants/" rel="nofollow">Logical Constants</a>. Traditionally :</p>
<blockquote>
<p>The most venerable approach to demarcating the logical constants identifies them with the language's <em>syncategorematic</em> signs: signs that signify nothing by themselves, but serve to indicate how independently meaningful terms are combined. </p>
</blockquote>
<p>Roughly speaking, <a href="http://en.wikipedia.org/wiki/Syncategorematic_term" rel="nofollow">syncategorematic signs</a> are the <em>logical constants</em> or <em>logical symbols</em>, i.e. those symbols (like <em>conncetives</em>) which are <strong>not</strong> interpreted or, according to modern semantics of first-order logic, do not "change meaning" when we vary the interpretation of the language.</p>
<p>According to this criteria, <em>variables</em> are ambiguos, because they have no "fixed meaning"; but also, their meaning is not fixed by the interpretation.</p>
<p>They are <em>placeholders</em>; in principle (see Frege) we can dispense with them. We can index argument places writing, instead of $Q(x, y)$ :</p>
<blockquote>
<p>$Q[( \quad )_i , ( \quad )_j]$.</p>
</blockquote>
<p>Thus, we can say that variables are another category of auxiliary symbols.</p>
|
2,707,744 | <h2>The Problem</h2>
<p>I am really struggling to wrap my head around the definition of the integral in measure theory. According to <a href="http://measure.axler.net/" rel="nofollow noreferrer">Axler</a>, the integral of a nonnegative function is defined as follows:</p>
<blockquote>
<p>Suppose $(X, \mathcal S,\mu)$ is a measure space and $f : X \to [0,\infty]$ a measurable function.
The integral of $f$ with respect to $\mu$, denoted $\int f\, \mathrm d \mu$ is defined by
\begin{multline}
\int f\,\mathrm d\mu = \sup \bigg\{\sum_{k=1}^{m} a_k \mu(A_k) : \sum_{k=1}^{m} a_k\chi_{A_k} \leq f,\quad m\in\mathbb Z^+,\quad a_1,\ldots,a_m > 0\quad \text{and} \quad A_1,\ldots, A_m\in\mathcal S\bigg\}.
\end{multline}</p>
</blockquote>
<p>I've been asked to prove Theorems 3.10 and 3.12 in the linked book,
that the integral is order-preserving, as in if $f \leq g$, then
\begin{equation}
\int f\,\mathrm d \mu \leq \int g \,\mathrm d \mu
\end{equation}
and that the integral is positively homogeneous,
as in if $c\in\mathbb R : c \geq 0$, then
\begin{equation}
\int cf\,\mathrm d\mu = c\int f\,\mathrm d\mu\,.
\end{equation}
The book gives hints in both cases,
namely that</p>
<blockquote>
<h2>3.10</h2>
<p>The supremum defining $\int f\,\mathrm d\mu$ is taken over a subset of the corresponding set for the supremum defining $\int g\,\mathrm d\mu$.</p>
</blockquote>
<p>and in the case of positive homogeneity:</p>
<blockquote>
<h2>3.12</h2>
<p>The supremum defining $\int cf\,\mathrm d\mu$ is taken over a set consisting of $c$ times the set, whose supremum defines $\int f\,\mathrm d\mu$.</p>
</blockquote>
<p>You would think that this was enough of a hint,
but I am having trouble justifying the steps in the "calculations".</p>
<h2>3.10</h2>
<p>Let $f,g : X\to[0,\infty]$ be measurable,
meaning that
\begin{equation}
F = \bigcup_{k=1}^{m} F_k = f^{-1}([0,\infty]) \in\mathcal S
\quad \text{and}\quad
G = \bigcup_{k=1}^{m} G_k = g^{-1}([0,\infty]) \in\mathcal S\,.
\end{equation}
If we now take the hint,
choose the sets $F\subseteq G \subseteq X$ so that each of the sums in the definition of the integral reach their supremums.
Then
\begin{align}
\int_F f\,\mathrm d\mu
&= \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{F_k} \leq f \bigg\}\\
&\leq \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(G_k) : \sum_{k=1}^{m} a_k \chi_{G_k} \leq g \bigg\}\\
&= \int_G g\,\mathrm d\mu,
\end{align}
where the inequality follows from the fact that measure preserves order, as in if $F \subseteq G$, then $\mu(F) \leq \mu(G)$.
This seems to be enough of a proof in my head, but I am not totally convinced.
As for the next theorem...</p>
<h2>3.12</h2>
<p>Let $f$ be a measurable function and $c \geq 0$,
and once again choose the pre-image of $f$, $F$, so that the sum in the definition of the integral reaches its supremum and
also define the set
\begin{equation}
cF = \{ca : a\in F\} = c\bigcup_{k=1}^{m} F_k = \bigcup_{k=1}^{m} cF_k\,.
\end{equation}
Then
\begin{align}
\int_{cF} cf\,\mathrm d\mu
&= \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(cF_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\
&= \sup\bigg\{ \sum_{k=1}^{m} a_k c\mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\
&= \sup\bigg\{c \sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\
&= c\sup\bigg\{\sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\
&= c\int_F f\,\mathrm d\mu\,.
\end{align}
This is where I'm not getting it.
The proof works if
\begin{equation}
\mu(cF_k) = c\mu(F_k),
\end{equation}
but I don't remember such a result from the book or our exercise sessions.
What I'm especially interested in is, is my understanding of the idea behind the proof correct, or have I totally just missed the target?</p>
| nonuser | 463,553 | <p>You can not do that since $L$ is not linear transformation. </p>
<p>If $L$ is linear, then $L(\vec{0}) = \vec{0}$ but in your case $L(\vec{0}) = a\ne \vec{0}$</p>
|
3,968,429 | <p>I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e - problem 29 in section 1.7).</p>
<p>I am confused by particularly by hint #3 which says:</p>
<p>Why is <span class="math-container">$\lvert x^2 -4x + 4 \rvert = \lvert x + 2 \rvert \lvert x - 2 \rvert$</span></p>
<p>I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see <a href="https://www.stewartcalculus.com/media/10_inside_homework_hints.php?show_cat=2&book=7e&hint_id=h00570" rel="nofollow noreferrer">here</a> (I would have attached the image but I cannot since I have not posted enough).</p>
<p>Left to my own device I would have figured that <span class="math-container">$\lvert x^2 -4x + 4\rvert \lt \epsilon $</span> where the left hand sides could be reworked as <span class="math-container">$ \lvert (x -2)^2 \rvert \lt \epsilon $</span> and furthermore that <span class="math-container">$ x - 2 = \sqrt{\epsilon}$</span></p>
<p>Not sure where I am going wrong here, can somebody please help?</p>
| user0102 | 322,814 | <p><strong>Solution</strong></p>
<p>Suppose that <span class="math-container">$|x - 2| \leq \delta$</span>. Then one has that
<span class="math-container">\begin{align*}
|f(x) - f(2)| & = |x^{2} - 4x + 5 - 1|\\\\
& = |x^{2} - 4x + 4|\\\\
& = |x-2|^{2} \leq \delta^{2} = \varepsilon
\end{align*}</span></p>
<p>Thus we can conclude that for every <span class="math-container">$\varepsilon > 0$</span>, there corresponds <span class="math-container">$\delta = \sqrt{\varepsilon}$</span> such that
<span class="math-container">\begin{align*}
|x - 2| \leq \delta \Rightarrow |f(x) - f(2)| \leq \varepsilon
\end{align*}</span></p>
<p>and we are done.</p>
<p>If you still have any doubts, please let me know.</p>
<p><strong>Comments</strong></p>
<p>The proposed identity is not correct. Indeed, <span class="math-container">$|x^{2} - 4x + 4| \neq |x-2||x+2|$</span>.</p>
|
3,968,429 | <p>I am working through the Stewart Calculus text independently and am stuck on one of the practice problems (edition 7e - problem 29 in section 1.7).</p>
<p>I am confused by particularly by hint #3 which says:</p>
<p>Why is <span class="math-container">$\lvert x^2 -4x + 4 \rvert = \lvert x + 2 \rvert \lvert x - 2 \rvert$</span></p>
<p>I am quite confused by this and I cannot figure out how these two sides of the equation are equal. For full context of the homework hints please see <a href="https://www.stewartcalculus.com/media/10_inside_homework_hints.php?show_cat=2&book=7e&hint_id=h00570" rel="nofollow noreferrer">here</a> (I would have attached the image but I cannot since I have not posted enough).</p>
<p>Left to my own device I would have figured that <span class="math-container">$\lvert x^2 -4x + 4\rvert \lt \epsilon $</span> where the left hand sides could be reworked as <span class="math-container">$ \lvert (x -2)^2 \rvert \lt \epsilon $</span> and furthermore that <span class="math-container">$ x - 2 = \sqrt{\epsilon}$</span></p>
<p>Not sure where I am going wrong here, can somebody please help?</p>
| VIVID | 752,069 | <p>Regarding your question, I think there was a typo in that hint #3; that should be <span class="math-container">$$|x^2-4x+4| = |x\color{red}-2||x-2|$$</span> instead.</p>
|
3,104,210 | <p>Is there a commonly accepted notation for a vector whose first <span class="math-container">$k$</span> entries are <span class="math-container">$1$</span>'s, with <span class="math-container">$0$</span>'s afterwards? I have seen <span class="math-container">$\mathbf{e}_i$</span> for a vector with a <span class="math-container">$1$</span> in the <span class="math-container">$i$</span>th entry, and <span class="math-container">$\mathbf{e}$</span> for a vector of all ones, but not this particular case.</p>
| J.G. | 56,861 | <p>Given any unary predicate <span class="math-container">$\phi(u)$</span>, comprehension implies <span class="math-container">$\exists p\forall u (u\in p\iff u\in z\land\phi(u))$</span>, with <span class="math-container">$z$</span> defined as per your usage of pair. For this inference to coincide with pair#, it suffices by <em>modus ponens</em> to take <span class="math-container">$\phi(u)$</span> to be <span class="math-container">$u=x\lor u=y$</span>. Of course, any <span class="math-container">$\phi$</span> that suffices could be replaced with <span class="math-container">$u\in z\land\phi(u)$</span>, but this is an unnecessary complication. This is what I imagine you intended when you wrote, "Let <span class="math-container">$\phi$</span> be the set property that <span class="math-container">$\{u\in z:u=x\lor u=y\}$</span>" (i.e. I think you were defining <span class="math-container">$\phi(u)$</span> as <span class="math-container">$u\in z\land (u=x\lor u=y)$</span>).</p>
|
1,833,406 | <p>Now I have a topological space $X$ that is $C_2$ and $T_4$, and $U$ is an open set in it, I want to show that $U$ can be expressed as $\cup_{i\in\Bbb Z_+} F_i$ where $F_i$ are closed sets, <strong>without the aid of any metrisation theorem</strong> (but Uryshon's theorem about normal spaces and the equivalent Tietze extension theorem are usable, if needed). </p>
<p>How can I proceed then? I totally have no clue. The only thing I know is that according to $C_2$, $U$ is a countable union of base sets $B_n$. I think the next thing to do is to find a closed subset for each $B_n$ that is "saturated" enough. By $T_4$ we can find an ascending chain of open subsets $G_1\subset G_2\subset \cdots G_k\subset \cdots\subset B_n$ such that $\bar G_k\subset G_{k+1}$, but even so, there seems to be no way to guarantee that $G_k$ can literally "approach" $B_n$. </p>
<p>Of course, an appeal to Uryshon metrisation theorem makes this problem somewhat trivial, but is apparently an overkill here. </p>
| Empy2 | 81,790 | <p>Simply multiply each term in $\arctan(x)$ by $x^2+1$, then collect like terms.</p>
|
1,833,406 | <p>Now I have a topological space $X$ that is $C_2$ and $T_4$, and $U$ is an open set in it, I want to show that $U$ can be expressed as $\cup_{i\in\Bbb Z_+} F_i$ where $F_i$ are closed sets, <strong>without the aid of any metrisation theorem</strong> (but Uryshon's theorem about normal spaces and the equivalent Tietze extension theorem are usable, if needed). </p>
<p>How can I proceed then? I totally have no clue. The only thing I know is that according to $C_2$, $U$ is a countable union of base sets $B_n$. I think the next thing to do is to find a closed subset for each $B_n$ that is "saturated" enough. By $T_4$ we can find an ascending chain of open subsets $G_1\subset G_2\subset \cdots G_k\subset \cdots\subset B_n$ such that $\bar G_k\subset G_{k+1}$, but even so, there seems to be no way to guarantee that $G_k$ can literally "approach" $B_n$. </p>
<p>Of course, an appeal to Uryshon metrisation theorem makes this problem somewhat trivial, but is apparently an overkill here. </p>
| C. Dubussy | 310,801 | <p>The key here is that you want the same power of $x$ for all the terms of the sum.</p>
<p>\begin{align*}
(x^2+1)\arctan(x) & = (x^2+1) \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=1}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}+x\\
& =\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}-\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+3}+x\\
& =x+\sum_{k=0}^{+\infty}(-1)^k \left(\frac{1}{2k+1}-\frac{1}{2k+3}\right)x^{2k+3}
\end{align*}</p>
|
1,955,509 | <p>There's this exercise in Hubbard's book:</p>
<blockquote>
<p>Let $ h:\Bbb R \to \Bbb R $ be a $C^1$ function, periodic of period $2\pi$, and define the function $ f:\Bbb R^2 \to \Bbb R $ by
$$f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=rh(\theta)$$</p>
<p>a. Show that $f$ is a continuous real-valued function on $\Bbb R^2$.</p>
<p>b. Show that $f$ is differentiable on $\Bbb R^2 - \{\mathbf 0\}$.</p>
<p>c. Show that all directional derivatives of $f$ exist at $\mathbf 0$ if and only if</p>
<p>$$ h(\theta) = -h(\theta + \pi) \ \text{ for all } \theta $$</p>
<p>d. Show that $f$ is differentiable at $ \mathbf 0 $ if an only if $h(\theta)=a \cos \theta + b \sin \theta$ for some number $a$ and $b$. </p>
</blockquote>
<p>I can't find how to prove $ f $ is continuous, I tried to prove
$$ \lim_{\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix} \to \begin{pmatrix}s\cos\phi\\s\sin\phi \end{pmatrix}} f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=s\ h(\phi) $$ for all $s$ and $\phi$.
But I can't do much else.</p>
| Olivier Oloa | 118,798 | <p>One may start with the standard <strong>finite</strong> evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by differentiating once more one gets
$$
2\times 1+3\times 2 x^2+...+n\times (n-1)x^{n-2}=\frac{2-x^{n-1}\left(n+n^2 (1-x)^2+2x-nx^2\right)}{(1-x)^3},\tag3
$$ then <strong>by making</strong> $n \to +\infty$ in $(3)$, <strong>using $|x|<1$</strong>, one obtains </p>
<blockquote>
<p>$$
\sum_{n=0}^\infty(n+1)\cdot n \cdot x^n=\frac{2 x}{(1-x)^3},\quad |x|<1, \tag4
$$ </p>
</blockquote>
<p>from which one gets the desired series by putting $x:=\dfrac14.$</p>
|
1,955,509 | <p>There's this exercise in Hubbard's book:</p>
<blockquote>
<p>Let $ h:\Bbb R \to \Bbb R $ be a $C^1$ function, periodic of period $2\pi$, and define the function $ f:\Bbb R^2 \to \Bbb R $ by
$$f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=rh(\theta)$$</p>
<p>a. Show that $f$ is a continuous real-valued function on $\Bbb R^2$.</p>
<p>b. Show that $f$ is differentiable on $\Bbb R^2 - \{\mathbf 0\}$.</p>
<p>c. Show that all directional derivatives of $f$ exist at $\mathbf 0$ if and only if</p>
<p>$$ h(\theta) = -h(\theta + \pi) \ \text{ for all } \theta $$</p>
<p>d. Show that $f$ is differentiable at $ \mathbf 0 $ if an only if $h(\theta)=a \cos \theta + b \sin \theta$ for some number $a$ and $b$. </p>
</blockquote>
<p>I can't find how to prove $ f $ is continuous, I tried to prove
$$ \lim_{\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix} \to \begin{pmatrix}s\cos\phi\\s\sin\phi \end{pmatrix}} f\begin{pmatrix}r\cos\theta\\r\sin\theta \end{pmatrix}=s\ h(\phi) $$ for all $s$ and $\phi$.
But I can't do much else.</p>
| robjohn | 13,854 | <p>Using the formula for the sum of a geometric series:
$$
\sum_{k=0}^\infty x^k=\frac1{1-x}\tag{1}
$$
Taking the derivative of $(1)$ and tossing the terms which are $0$:
$$
\sum_{k=1}^\infty kx^{k-1}=\frac1{(1-x)^2}\tag{2}
$$
Taking the derivative of $(2)$ and tossing the terms which are $0$:
$$
\sum_{k=2}^\infty k(k-1)x^{k-2}=\frac2{(1-x)^3}\tag{3}
$$
Reindexing the sum in $(3)$:
$$
\sum_{k=0}^\infty(k+2)(k+1)x^k=\frac2{(1-x)^3}\tag{4}
$$
Plug in $x=\frac14$:
$$
\begin{align}
\sum_{k=0}^\infty\frac{(k+2)(k+1)}{4^k}
&=\frac2{\left(\frac34\right)^3}\\[6pt]
&=\frac{128}{27}\tag{5}
\end{align}
$$</p>
|
2,253,768 | <p>I am currently working on a small optimization problem in which I need to find an optimal number of servers for an objective function that incorporates the Erlang Loss formula. To this end, I have been searching for an expression for the first order difference of the Erlang Loss formula with respect to the number of servers, i.e. $B(E,m+1)-B(E,m)$, where <em>m</em> is the number of servers and $B(E,m)$ is given by:</p>
<p>$B(E,m)={\dfrac {{\dfrac {E^{m}}{m!}}}{\sum _{{i=0}}^{m}{\dfrac {E^{i}}{i!}}}}$</p>
<p>Unfortunately, until now I can't derive or find such an expression (if it exists) and was wondering whether one on this forum could help me out? </p>
<p>Many thanks in advance! </p>
| egreg | 62,967 | <p>Consider the homomorphism
$$
R[x]\to R/M,\qquad f(x)\mapsto f(0)+M
$$
and compute its kernel and image.</p>
|
923,871 | <p>Let $Y=\{(a,b)∈ \Bbb R\times \Bbb R∣ a≠0\}$. Given $(a,b),(c,d)\in Y$, define $(a,b)*(c,d)=(ac,ad+b)$. Prove that $Y$ is a group with the operation $*$.</p>
<p>I already do the proof of ∗ is an operation on Y. And proof is associative like this:
$$\{(A,B)*(C,D)\}*(E,F)=(A,B)*\{(C,D)\}*(E,F)\}$$</p>
<p>$$(AC,AD+B)(E,F)=(A,B)(CE,CF+D)$$</p>
<p>$$(ACE,ACF+AD+B)=(ACE,ACF+AD+B)$$</p>
<p>but I get stock trying the identity and the inverse proof. dont know how to start</p>
| Community | -1 | <p><strong>Hint</strong></p>
<p>Prove the associativity of $*$, find the neutral element and prove that each element has an inverse.</p>
|
3,373,576 | <p>What are the non-trivial solutions of
<span class="math-container">$$\tan x = \arctan x$$</span></p>
<p>Can these solutions be expressed e.g. in terms of <span class="math-container">$\pi$</span> or in radicals?
I mean are they some "nice" numbers?</p>
<p>E.g. do we know if these solutions are irrational, or rational, some rational multiple of <span class="math-container">$\pi$</span>, or say something like e.g. <span class="math-container">$\frac{\sqrt{2}}{7}$</span>?</p>
<p>What about <span class="math-container">$\cot x = \text{arccot} (x)$</span>? I became curious about these problems after graphing the four functions.</p>
| hamam_Abdallah | 369,188 | <p>Put
<span class="math-container">$$a=x^2-1=(x-1)(x+1)$$</span>
and
<span class="math-container">$$b=x^2-4x+3=(x-1)(x-3).$$</span></p>
<p>observe that when <span class="math-container">$ x \to 1 $</span>, <span class="math-container">$ a $</span> and <span class="math-container">$b $</span> go to zero. so</p>
<p><span class="math-container">$$\lim_{x\to 1}\frac{\tan(b)}{b}\frac{a}{\sin(a)}\frac{b}{a}$$</span></p>
<p><span class="math-container">$$=\lim_{x\to 1}\frac ba=\lim_{x\to 1}\frac{x-3}{x+1}=-1$$</span></p>
|
1,361,948 | <p>$\frac{df}{dx} = 2xe^{y^2-x^2}(1-x^2-y^2) = 0.$</p>
<p>$\frac{df}{dy} = 2ye^{y^2-x^2}(1+x^2+y^2) = 0.$</p>
<p>So, $2xe^{y^2-x^2}(1-x^2-y^2) = 2ye^{y^2-x^2}(1+x^2+y^2)$.</p>
<p>$x(1-x^2-y^2) = y(1+x^2+y^2)$</p>
<p>$x-x^3-xy^2 = y + x^2y + y^3$</p>
<p>Is the guessing the values of the variables the only way of solving this? With $y = 0$ and I can already figure out that $x = 0$, $1$ or $-1$ but It's bothering me how I had to randomly guess that. </p>
| Community | -1 | <p><strong>Without computing</strong>:</p>
<p>The denominator of $z$ has non-trivial roots (when the argument of the exponential is $i2k\pi$); not the denominator of $\Re(z)$.</p>
|
2,319,119 | <p>Been given a question and find it to be too vague to know what's going on. </p>
<p>The question is: </p>
<p>$f(x) = 2x + 2$. Define $f(x)$ recursively.</p>
<p>I'm just quite puzzled as there is no $f(0)$, $f(1)$ or $f(x-1)$ function to go by other than the original function.</p>
<p>Supposed to be in the form of $f(x-1)$.</p>
<p>Any help appreciated thanks.</p>
| DanielWainfleet | 254,665 | <p>(1). If $X$ is a normed linear space and $S\subset X$ then $Y$ is dense in $X$ iff $v+S=\{v+y:y\in S\}$ is dense in $X$ for every $v\in X.$</p>
<p>(2). Let $v_1(x)=-5$ for all $x\in [0,1].$ So $S$ is dense in $C[0,1]$ iff $v_1+S=\{f\in C[0,1]:\int_0^1 f(x)dx=0\}$ is dense.</p>
<p>(3). The function $\psi(f)=\int_0^1f(x)dx$ is Lipschitz-continous on $C[0,1].$ because $|\psi(f-g)|\leq \int_0^1|f(x)-g(x)|dx\leq \int_0^1\|f-g\|dx=\|f-g\|.$</p>
<p>(4). Referring to (2), the set $v_1+S=\psi^{-1}\{0\}$ is closed in $C[0,1]$ because $\psi:C[0,1]\to \mathbb R $ is continuous. If $v_1+S$ were closed and dense it would be equal to $C[0,1]$. But obviously $\psi^{-1}\{0\} \ne C[0,1]$.</p>
|
4,095,715 | <p>I know how to do these in a very tedious way using a binomial distribution, but is there a clever way to solve this without doing 31 binomial coefficients (with some equivalents)?</p>
| Ankit Saha | 876,128 | <p>You can try using the multinomial theorem
<span class="math-container">$$(1+x^2+x^4+x^6)^n = \sum_{0 \le a,b,c,d \le n} \dfrac{n!}{a!~b!~c!~d!}~1^a~x^{2b}~x^{4c}~x^{6d} $$</span>
<span class="math-container">$$ a+b+c+d=n ~~~~;~~~ a,b,c,d \in \mathbb{W}$$</span>
For the coefficient of <span class="math-container">$x^{12}$</span>,
<span class="math-container">$$ 2b+4c+6d = 12$$</span>
<span class="math-container">$$ b +2c+3d=6$$</span>
Also, <span class="math-container">$$0 \le a,b,c,d \le n $$</span>
<span class="math-container">$$0 \le b \le6 $$</span>
<span class="math-container">$$0 \le c \le3 $$</span>
<span class="math-container">$$0 \le d \le2 $$</span>
Now you just have to find all solutions that satisfy the above criteria.</p>
|
313,298 | <p>For example, let's say that I have some sequence $$\left\{c_n\right\} = \left\{\frac{n^2 + 10}{2n^2}\right\}$$ How can I prove that $\{c_n\}$ approaches $\frac{1}{2}$ as $n\rightarrow\infty$?</p>
<p>I'm using the Buchanan textbook, but I'm not understanding their proofs at all.</p>
| Tom Oldfield | 45,760 | <p>In general, we say that a sequence $x_n \rightarrow x$ if $|x_n - x| \rightarrow 0$. This means that for all $\epsilon > 0$ there exists some $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_n - x| < \epsilon$.</p>
<p>We can also use results about sequences that will be proved in any good introductory texts, such as if $x_n \rightarrow x$ and $y_n \rightarrow y$, $\lambda \in \mathbb{R}$: </p>
<p>$i) x_n + y_n \rightarrow x+y$</p>
<p>$ii) x_ny_n \rightarrow xy$</p>
<p>$iii)\lambda x_n \rightarrow \lambda x$</p>
<p>$iv)$ If $x \not= 0$ then $\frac{y_n}{x_n} \rightarrow \frac{y}{x}$</p>
<p>Using these results we can see that $\displaystyle c_n = \frac{n^2 + 10}{2n^2} = \frac{1+\frac{10}{n^2}}{2}$. So if $a_n = \frac{10}{n^2} \rightarrow a$ then $c_n \rightarrow \frac{1+a}{2}$. Now we can prove that $a_n \rightarrow 0$ using epsilon delta techniques, so it follows that $c_n \rightarrow \frac12$.</p>
|
53,989 | <p>The broad, generic and badly posed question may be formulated in this way:</p>
<blockquote>
<p>Let <span class="math-container">$X$</span> be a compact Kälher manifold (or even a projective one). If one considers the Hodge decomposition <span class="math-container">$H^k(X, \mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X)$</span> and interprets the left hand side as Betti cohomology <span class="math-container">$\text{Hom}(\pi_1(X), \mathbb{C})$</span>, which homomorphisms fit into which direct summand?</p>
</blockquote>
<p>I'm sure this is well studied, but couldn't find any reference. Without a precise question there is probably not much to say, so let us look at a few motivation and give more precise statements below.</p>
<h3>Motivations</h3>
<p>I was trying to study the easy cases of Simpson's non-abelian cohomology (cfr. for example his paper "Moduli of representations of the fundamental group of a smooth projective variety"), and the first possible idea is to reduce oneself to the abelian case. So some questions which are of interest in the non-abelian case should become easy in the abelian context, and the above provides the basics for one of them: Simpson defines Betti and Dolbeaut non-abelian cohomology, then proves that the constructed spaces are in fact naturally homeomorphic, and studies interactions between objects defined in these "different worlds". In the abelian context this boils down to interpreting the cohomology as singular cohomology on one side and as the direct sum of the Dolbeaut cohomology on the other.</p>
<h3>More specific questions</h3>
<p>One can in particular consider any given subgroup <span class="math-container">$G \subset \mathbb{C}$</span>, and seek only those homomorphisms which take values in this subgroup (if <span class="math-container">$G$</span> is a subfield this means considering <span class="math-container">$H^k(X, G)$</span>, otherwise there may be some torsion in this cohomology group which is ignored if we just take the subgroup of <span class="math-container">$H^k(X, \mathbb{C})$</span> as above). Then, can one say whether this subgroup, call it <span class="math-container">$H_G \subset H^k(X, \mathbb{C})$</span>, is contained in some pure part of the Hodge decomposition <span class="math-container">$H^{p,q}(X)$</span>? I would have thought that this depended entirely on the nature of <span class="math-container">$X$</span>, but on second thought it seems that something can always be said: for example, if <span class="math-container">$k = 1$</span>, then <span class="math-container">$H_{\mathbb{R}}$</span> is never contained in any pure part of some weight (since they are interchanged by conjugation), while some work I did would lead one to think that if <span class="math-container">$G = \lbrace k \pi i \rbrace_{k \in \mathbb{Z}}$</span>, then <span class="math-container">$H_G \subset H^{0,1}(X)$</span>, at least for <span class="math-container">$X$</span> projective such that <span class="math-container">$H_1(X, \mathbb{Z})$</span> has no torsion (I'm absolutely not sure of the rightness of this proof, and if this fact was true there should really be a more straight idea then the one I thought of). Furthermore, for <span class="math-container">$k = 2$</span>, then always <span class="math-container">$H_{\mathbb{R}} \cap H^{1,1}(X) \neq \emptyset $</span> since there is the Kähler form, while the question if <span class="math-container">$H_{\mathbb{Z}}$</span> is a subset of <span class="math-container">$H^{1,1}(X)$</span> is by Kodaira embedding theorem equivalent to <span class="math-container">$X$</span> being projective.</p>
<p>Hence a slightly more focused question could be:</p>
<blockquote>
<p>For which classes of compact Kähler manifolds and which (in case cyclic) subgroups <span class="math-container">$G \subset \mathbb{C}$</span> one knows the Dolbeaut type of the elements of <span class="math-container">$ H_G $</span> inside some <span class="math-container">$H^k(X, \mathbb{C})$</span>?</p>
</blockquote>
<p>The question may become more interesting (hence, probably won't have any sensitive answer) if one allows (in case small) variations of the Hodge structure, by considering a family <span class="math-container">$X \to S$</span>. I do not know much about variations of Hodge structures, so this could be well known. In this case, what is surely known is that the question for <span class="math-container">$k = 2$</span> and <span class="math-container">$G = \mathbb{Z}$</span> is false (small deformations of projective varieties need not be projective) and for <span class="math-container">$G = \mathbb{R}$</span> is true (small deformation of Kähler manifolds are Kähler). In this case one may want to drop the smoothness hypothesis, or, on the other hand, require the morphism <span class="math-container">$X \to S$</span> to be both projective and smooth.</p>
| Donu Arapura | 4,144 | <p>Marco,</p>
<p>Since I don't fully understand what you are asking, let me instead discuss
what you say is motivating you. If it isn't relevant, you can say so.
When $X$ is a compact Kaehler manifold
then what Simpson calls the Betti moduli space $M_B(X)$
is the set of semisimple representations in $Hom(\pi_1(X),GL_n(\mathbb{C}))/conjugacy$.
This can be made into a variety (actually a scheme). When $n=1$, this is simply
$$Hom(\pi_1(X),\mathbb{C}^*)=H^1(X,\mathbb{C}^*)$$
On the other side, the "Dolbeault moduli space" $M_{Dol}(X)$ is the moduli space of Higgs bundles (satisfying appropriate
conditions). Again, in the rank one case, this simply the cotangent bundle
of the Picard torus $T^*Pic^0(X)= Pic^0(X)\times H^0(X,\Omega_X^1)$. The correspondence
in this case can be deduced from standard Hodge theory; it
is sketched on page 21 of Simpson's <em>Higgs bundles and local systems</em>, and it
is worthwhile to fill in the details yourself.</p>
<p>One thing you will notice is that this story doesn't really have
much to do with cohomology beyond $H^1$. Actually, this isn't quite true.
Even though the FAQ discourages discussion of open problems, let me mention one
that may have some relevance to your question:</p>
<blockquote>
<p>Is the image of the natural map
$H^*(\pi_1(X),\mathbb{C})\to H^*(X,\mathbb{C})$ a sub Hodge structure?</p>
</blockquote>
|
53,989 | <p>The broad, generic and badly posed question may be formulated in this way:</p>
<blockquote>
<p>Let <span class="math-container">$X$</span> be a compact Kälher manifold (or even a projective one). If one considers the Hodge decomposition <span class="math-container">$H^k(X, \mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X)$</span> and interprets the left hand side as Betti cohomology <span class="math-container">$\text{Hom}(\pi_1(X), \mathbb{C})$</span>, which homomorphisms fit into which direct summand?</p>
</blockquote>
<p>I'm sure this is well studied, but couldn't find any reference. Without a precise question there is probably not much to say, so let us look at a few motivation and give more precise statements below.</p>
<h3>Motivations</h3>
<p>I was trying to study the easy cases of Simpson's non-abelian cohomology (cfr. for example his paper "Moduli of representations of the fundamental group of a smooth projective variety"), and the first possible idea is to reduce oneself to the abelian case. So some questions which are of interest in the non-abelian case should become easy in the abelian context, and the above provides the basics for one of them: Simpson defines Betti and Dolbeaut non-abelian cohomology, then proves that the constructed spaces are in fact naturally homeomorphic, and studies interactions between objects defined in these "different worlds". In the abelian context this boils down to interpreting the cohomology as singular cohomology on one side and as the direct sum of the Dolbeaut cohomology on the other.</p>
<h3>More specific questions</h3>
<p>One can in particular consider any given subgroup <span class="math-container">$G \subset \mathbb{C}$</span>, and seek only those homomorphisms which take values in this subgroup (if <span class="math-container">$G$</span> is a subfield this means considering <span class="math-container">$H^k(X, G)$</span>, otherwise there may be some torsion in this cohomology group which is ignored if we just take the subgroup of <span class="math-container">$H^k(X, \mathbb{C})$</span> as above). Then, can one say whether this subgroup, call it <span class="math-container">$H_G \subset H^k(X, \mathbb{C})$</span>, is contained in some pure part of the Hodge decomposition <span class="math-container">$H^{p,q}(X)$</span>? I would have thought that this depended entirely on the nature of <span class="math-container">$X$</span>, but on second thought it seems that something can always be said: for example, if <span class="math-container">$k = 1$</span>, then <span class="math-container">$H_{\mathbb{R}}$</span> is never contained in any pure part of some weight (since they are interchanged by conjugation), while some work I did would lead one to think that if <span class="math-container">$G = \lbrace k \pi i \rbrace_{k \in \mathbb{Z}}$</span>, then <span class="math-container">$H_G \subset H^{0,1}(X)$</span>, at least for <span class="math-container">$X$</span> projective such that <span class="math-container">$H_1(X, \mathbb{Z})$</span> has no torsion (I'm absolutely not sure of the rightness of this proof, and if this fact was true there should really be a more straight idea then the one I thought of). Furthermore, for <span class="math-container">$k = 2$</span>, then always <span class="math-container">$H_{\mathbb{R}} \cap H^{1,1}(X) \neq \emptyset $</span> since there is the Kähler form, while the question if <span class="math-container">$H_{\mathbb{Z}}$</span> is a subset of <span class="math-container">$H^{1,1}(X)$</span> is by Kodaira embedding theorem equivalent to <span class="math-container">$X$</span> being projective.</p>
<p>Hence a slightly more focused question could be:</p>
<blockquote>
<p>For which classes of compact Kähler manifolds and which (in case cyclic) subgroups <span class="math-container">$G \subset \mathbb{C}$</span> one knows the Dolbeaut type of the elements of <span class="math-container">$ H_G $</span> inside some <span class="math-container">$H^k(X, \mathbb{C})$</span>?</p>
</blockquote>
<p>The question may become more interesting (hence, probably won't have any sensitive answer) if one allows (in case small) variations of the Hodge structure, by considering a family <span class="math-container">$X \to S$</span>. I do not know much about variations of Hodge structures, so this could be well known. In this case, what is surely known is that the question for <span class="math-container">$k = 2$</span> and <span class="math-container">$G = \mathbb{Z}$</span> is false (small deformations of projective varieties need not be projective) and for <span class="math-container">$G = \mathbb{R}$</span> is true (small deformation of Kähler manifolds are Kähler). In this case one may want to drop the smoothness hypothesis, or, on the other hand, require the morphism <span class="math-container">$X \to S$</span> to be both projective and smooth.</p>
| David E Speyer | 297 | <p>This is really a comment on Donu Arapura's answer, but it seemed large enough to deserve it's own post. Working again in the case of $GL_1$, Simpson considers three spaces:</p>
<p>$M_{betti}$: The space of $\mathbb{C}^*$-local systems on $X$. If you like, you can think of this as smooth complex line bundles with an integrable connection.</p>
<p>$M_{DR}$: The space of holomorphic line bundles $L$ on $X$, equipped with an integrable holomorphic connection. (Being compatible with a holomorphic connection forces $c_1(L)$ to be $0$ in $H^2(X)$.)</p>
<p>$M_{Dol}$: The space of holomorphic line bundles $L$ on $X$, with $c_1(L)= 0$ in $H^2(X)$, and equipped with an $\mathrm{End}(L)$-valued $1$-form. $\mathrm{End}(L)$ is naturally isomorphic to $\mathcal{O}$, so this is just a $1$-form, but it is the $\mathrm{End}(L)$ version which generalizes to higher $GL_n$.</p>
<p>The first space is $\mathrm{Hom}(\pi_1(X), \mathbb{C}^*) = H^1(X, \mathbb{Z}) \otimes \mathbb{C}^*$. In this latter form, it has a natural algebraic structure, as a multiplicative algebraic group.</p>
<p>The second space is an affine bundle over $\mathrm{Pic}^0(X)$. Each fiber is a torsor for $H^0(X, \Omega^1)$, so we can describe this space by giving a class in $H^1(\mathrm{Pic}^0(X), \mathcal{O}) \otimes H^0(X, \Omega^1)$. By GAGA, this cohomology group on $\mathrm{Pic}$ is the same algebraically or analytically; viewing it algebraically, we get an algebraic structure on $M_{DR}$.</p>
<p>The third space is simply $\mathrm{Pic}^0(X) \times H^0(X, \Omega^1)$ (for larger $n$, this vector bundle can be nontrivial). For obvious reasons, this has an algebraic structure.</p>
<p>The relations between these spaces are the following: All three are diffeomorphic. $M_{betti}$ and $M_{DR}$ are isomorphic as complex analytic varieties, but have different algebraic structure. $M_{Dol}$ and $M_{DR}$ are not isomorphic as complex analytic varieties, rather, $M_{Dol}$ is the vector bundle for which the affine bundle $M_{DR}$ is a torsor.</p>
<p>You might enjoy writing this all down in coordinates for $X$ an elliptic curve. As smooth manifolds, all three spaces should be $(\mathbb{C}^*)^2$.</p>
|
523,376 | <p>Suppose $a_i$ is a sequence of positive integers. Define $a_1 = 1$, $a_2 = 2$ and $a_{n+1} = 2a_n + a_{n-1}$. Does it follow that </p>
<p>$$ \gcd(a_{2n+1} , 4 ) = 1 $$ ???</p>
<p>Im trying to see this by induction assuming above holds, we need to see that $\gcd(a_{2n+3} , 4 ) = 1$.</p>
<p>But, $\gcd(a_{2n+3} , 4 ) = n_0(2a_{2n+1} + a_{2n-1}) + 4n_1$ for integers $n_0, n_1$. But this quantity does not seem to give me $1$. Can someone help me with this problem? thanks</p>
| Michael Hoppe | 93,935 | <p>The normal vector field has to be differentiable, in this case it's not even continuous.</p>
|
106,708 | <p><strong>My fragile attempt:</strong> Note that if $1987^k-1$ ends with 1987 zeros, that means $1987^k$ has last digit 1 (and 1986 "next" ones are zeros). For this to be satisfied, $k$ has to be in form $k=4n$, where $n\in N$. This means out number can be written in form </p>
<p>$$ [(1987^n)^2+1][1987^n+1][1987^n-1]. $$</p>
<p>This number has to be dividable by $10^{1987}$ if there is such a number that is asked for in question.</p>
<p>Now, I believe that the fact 1987 is a prime is very important here. There are probably some theorems from number theory about primes and their powers. For example, if $p$ is a prime (distinct from 2 if needed), are there any important things about number such as $p^2-1$? </p>
<p>If I'm going at the right direction with this, I'd appreciate a hint. Please don't use too advanced techniques if possible. Thanks.</p>
| Michael Sink | 26,102 | <p>Hint: You've already done the hard part, you have identified how the formula can be factored. Don't look for powers of 10, as all 3 terms are divisible by 2. Focus on powers of 5. </p>
|
738,435 | <p>I am having a problem understanding how to determine if a function is one to one.</p>
<p>The problem is: Show that the function f(x) = 3x+4 is one-to-one.</p>
<p>Also, I'm being thrown off by the notation x[subset 1] = x[subset 2], what does that mean, loosely speaking? "In my eyes" that would mean two different values are being represented. </p>
| Brian J. Fink | 125,131 | <p>Actually, the question of whether a function is one-to-one is answered by whether there is one and only one value of every $f(x)$ for every $x$. Since it can be demonstrated that $f(x)=3x+4$ is linear and continuous (there are no gaps), there is both one and only one value of $3x+4$ for every $x \in \mathbb{R}$. Therefore, it is one-to-one. Another way is to rearrange the function so that the $x$ is on the left side, and solve for x in terms of $f(x)$. The new equation will be the inverse function. If there is one and only one value of $x$ for every $f(x)$, then the function is indeed one-to-one.</p>
|
2,107,990 | <p>I'm looking for suggestions for textbooks covering multivariable calculus. I am looking for two textbooks, one which covers the theory and the other which covers the computational aspects. I have already taken a (not so taught well) first course in multivariable calculus, but I'd ideally like to to keep a computational/intuitive text with me.</p>
<p>I have covered a first course analysis with a focus on point set topology. After revising a bit of the theory of integration and covering Lebesgue integrals which I haven't done thus far, I'd like to move on to cover the aforementioned texts. </p>
<p>My aim is the following: over the semester, I'll be enrolled in a course in advanced calculus. From what I gather, the course will cover manifolds etc. but it'll cover them from a very not so rigorous/differential geometric point of view. With that course and a course in GR, I'd like to get introduced to the rudiments of manifolds etc. With my out-of-class work on multivariable calculus, I hope to build up on the material of the courses in a few month's time -- before the term ends -- and use the material covered on both these fronts to start off with basic differential geometry. By then, I hope to have covered some algebra, topology etc. on my own as well.</p>
<p>With my motivation in mind, it'd be great if you could recommend some texts.</p>
| Batominovski | 72,152 | <p>If $a$, $b$, and $c$ are required to only be positive integers and some of them is $1$, then we have a unique solution $(a,b,c)=(1,1,1)$. For solutions with $a,b,c>1$, note that
$$(a-1)(b-1)(c-1)=abc-bc-ca-ab+a+b+c-1=a+b+c-3\,.$$
Set $x:=a-1$, $y:=b-1$, and $z:=c-1$. Therefore,
$$xyz=x+y+z\,.$$
Without loss of generality, suppose that $x\leq y\leq z$ (noting that they are positive integers). Now, we have
$$y+z=x(yz-1)\geq yz-1\,,\text{or }(y-1)(z-1)\leq 2\,.$$
If $(y-1)(z-1)=0$, then $y=1$, making $x=1$ as well, so $$z=xyz=x+y+z=2+z\,$$ which is absurd. If $(y-1)(z-1)=1$, then $y=2$ and $z=2$, so $$4x=xyz=x+y+z=x+4\,,$$ leading to $x\notin\mathbb{Z}$, which is a contradiction. Thus, $(y-1)(z-1)=2$, and so $y=2$ and $z=3$. Ergo, $$6x=xyz=x+y+z=x+5\,,\text{ or }x=1\,.$$
Consequently, $(a,b,c)=(2,3,4)$ is the only solution, up to permutation.</p>
|
1,758,194 | <p>Consider the function f: {-1, +1} -> R defined by</p>
<p>$f(x)= \arcsin (\frac{1+x^2}{2x})$.</p>
<p>Due to the following two inequalities :</p>
<p>(i) $1+x^2 \geq 2x$</p>
<p>(ii)$1+x^2 \geq -2x$ , </p>
<p>the function can only be defined at $x=1$ and $x=-1$. I have learnt that the epsilon delta definition only includes those values of $x$ which are in the domain of $f(x)$. But in this case, the function isn't defined on <em>either</em> side of x=1.</p>
<p>So this is my question: Is it correct to say that the limit as $x$ approaches $1$ of $f(x$) is $\frac{\pi}{2}$ ?</p>
<p>Can the above question be given a <strong>definitive</strong> "yes" or "no" answer, or must it(unfortunately) vary from person to person? </p>
<p>If the latter, is the "precise" definition of a limit not precise enough?</p>
<p>How can the answer be proved or disproved using the epsilon delta definition?</p>
<p>I have also read that functions are by default continuous at isolated points. Can I conclude from the definition of continuity (the limit equals the value of the function evaluated at the point) that the limit must exist?</p>
<p><strong>Note :</strong> Forgive my ignorance but I do not know a thing about topology. I'm looking for a detailed answer but in simple terms, preferably written in the language of calculus.</p>
<p>Thanks for the help. </p>
| Christian Blatter | 1,303 | <p>It seems that $x$ is tacitly assumed to be real. In this case the given expression is defined only on the two-element set $S=\{-1,1\}\subset{\mathbb R}$. The set $S$ inherits from ${\mathbb R}$ the metric $d(x,y):=|x-y|$, and as $|1-(-1)|=2$ it follows that $S$ is a discrete metric (resp., topological) space. It follows that any function on $S$, in particular the function considered in the question, is continuous.</p>
<p>As for the limit $\lim_{x\to1} f(x)$ I think that one should not even consider such a limit. Formally, for each $\lambda\in{\mathbb R}$, and any given $\epsilon>0$, one (trivially) has $\bigl|f(x)-\lambda\bigr|<\epsilon$ as soon as $0<|x-1|<1$.</p>
|
1,668,653 | <blockquote>
<p>If $\sum a_n$ is convergent and $a_n>0$, then $\sum (-1)^n a_n$ is convergent.</p>
</blockquote>
<p>So far I've tried convergence/divergence tests and also I tried to prove this using partial sums. But tests do not work because the latter series <em>includes negative terms</em>. Maybe the Alternating Series Test could have worked but "$a_n$ is decreasing" is not an initial condition. I know the proof will be very easy but I am stuck here. Could you provide me with a little <strong>hint</strong>?</p>
| frosh | 211,697 | <blockquote>
<p><strong>Theorem:</strong> If $\sum |a_n|$ is convergent, then $\sum a_n$ is convergent.</p>
</blockquote>
<p><strong>Proof:</strong> Let $b_n=|a_n|+a_n \Rightarrow b_n=0 \: \: \mbox{or} \: \: b_n=2|a_n| \Rightarrow 0\le b_n \le 2|a_n| \stackrel{\mbox{DCT}}{\Rightarrow} \sum b_n$ is convergent. Therefore, $\sum b_n - |a_n|=\sum a_n$ is convergent.</p>
<p>This answers the question I asked.</p>
<p><strong>Remark:</strong> Actually, the result is known as Absolute Convergence Test. I am learning the subject this semester. My apologies for asking the question early.</p>
|
4,079,842 | <p>Consider the below function.
<span class="math-container">$$f(x,y) =
\begin{cases}
1 & xy \neq 0 \\
0 & xy = 0
\end{cases}$$</span>
Suppose i want to calculate the below repeated limit.
<span class="math-container">$$\lim_{x\to0}\lim_{y\to0}f(x,y)$$</span>.
In general textbooks in India for BS courses,the above repeated limit computation is given to be <span class="math-container">$1$</span> which is fine if we know that the point <span class="math-container">$(x,y)\neq 0$</span>. However , if we be really precise, then ,this limit computation is<span class="math-container">$$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$$</span>.
The inner limit is unknown to us as we donot know about the nature of <span class="math-container">$x$</span>. So my question is , in repeated limit calculation , do we assume that there is some distance left that we still have to move some in <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to approach a given point?</p>
| José Carlos Santos | 446,262 | <p>The limit <span class="math-container">$\lim_{y\to0}f(x,y)$</span> is <span class="math-container">$0$</span> when <span class="math-container">$x=0$</span> and <span class="math-container">$1$</span> otherwise. But when you compute<span class="math-container">$$\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right),\tag1$$</span>what happens when <span class="math-container">$x=0$</span> does not matter. So, <span class="math-container">$(1)=\lim_{x\to0}1=1$</span>.</p>
|
492,125 | <p>How many different equivalence relations can be defined on a set of five elements?</p>
| azarel | 20,998 | <p>Hint: In how many ways can you partition a five element set?</p>
|
384,090 | <p>Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$</p>
<p>I have tried to fiddle with it as follows: </p>
<p>$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$</p>
<p>Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.</p>
| Community | -1 | <p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$6(8^x+27^x)=7(12^x+18^x)$</span></p>
<p>Divide by<span class="math-container">$12^x$</span></p>
<p><span class="math-container">$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$</span></p>
<p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$</span></p>
<p>We put :<span class="math-container">$t=(\frac{3}{2}) ^x$</span></p>
<p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$</span></p>
<p>Multiply by <span class="math-container">$t$</span></p>
<p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$</span></p>
<p>We can see <span class="math-container">$-1$</span> is a solution of the equation</p>
<p>So:after division by <span class="math-container">$t+1$</span> we see</p>
<p><span class="math-container">$ t^2 - \frac{13}{6}x+1=0$</span></p>
<p><span class="math-container">$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $</span></p>
<p>So :</p>
<p><span class="math-container">$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $</span></p>
<p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$x_1=1$</span></p>
<p>And</p>
<p><span class="math-container">$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $</span></p>
<p><span class="math-container">$\Rightarrow $</span> <span class="math-container">$x_2=-1$</span></p>
<p>Finally</p>
<p><span class="math-container">$x=-1$</span> or <span class="math-container">$ x= 1$</span></p>
|
4,381,737 | <p><span class="math-container">$\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\}$</span> are the Gaussian integers. I want to show that there is no ideal <span class="math-container">$I \subseteq \mathbb{Z}[i]$</span> such that the quotient <span class="math-container">$\mathbb{Z}[i]/I$</span> is a field of size <span class="math-container">$3$</span> (i.e, is <span class="math-container">$\mathbb{F}_3$</span>).</p>
<p>Here's my approach:</p>
<p>Since <span class="math-container">$\mathbb{Z}[i]/I$</span> is a field, this would make <span class="math-container">$I$</span> a prime/maximal ideal in <span class="math-container">$\mathbb{Z}[i]$</span>. Moreover, in the reduction <span class="math-container">$\mathbb{Z}[i] \rightarrow \mathbb{Z}[i]/I = F_3$</span>, since <span class="math-container">$1 \mapsto 1$</span>, we would use homomorphism properties to see that <span class="math-container">$3 \mapsto 0$</span> which is just a complicated way of saying that <span class="math-container">$3 \in I$</span>.</p>
<p>Now <span class="math-container">$(3) \subseteq I$</span>. But we know that through Fermat's sum of squares theorem, since <span class="math-container">$3 \equiv 3$</span> mod <span class="math-container">$4$</span>, we know that <span class="math-container">$3$</span> is a prime in <span class="math-container">$\mathbb{Z}[i]$</span> and thus maximal as well. This would mean that <span class="math-container">$I = (3)$</span>.</p>
<p>Now, we obtain a contradiction in that <span class="math-container">$\mathbb{Z}[i]/(3)$</span> has size <span class="math-container">$9$</span>, not <span class="math-container">$3$</span> (i.e, all elements of the form <span class="math-container">$a + bi$</span> were <span class="math-container">$a, b \leq 3$</span>).</p>
<p>This concludes the proof.</p>
<p>Does this look correct? Is there another way to do it without wielding heavy results like Fermat's sum of squares?</p>
| Claudius | 218,931 | <p>Since you asked whether there are other solutions:</p>
<p>If there existed a ring homomorphism <span class="math-container">$\mathbb{Z}[i] \to \mathbb{F}_3$</span>, then the image of <span class="math-container">$i$</span> in <span class="math-container">$\mathbb{F}_3$</span> would be a square root of <span class="math-container">$-1$</span> in <span class="math-container">$\mathbb{F}_3$</span>, which is clearly not possible.</p>
<p>In particular, there is no ideal <span class="math-container">$I\subseteq \mathbb{Z}[i]$</span> such that <span class="math-container">$\mathbb{Z}[i]/I \cong \mathbb{F}_3$</span>.</p>
|
3,397,834 | <blockquote>
<h2><span class="math-container">$$49y″−98y′+48y= 0 \quad\quad\quad \,\, y(2)=3,y′(2)=9.$$</span></h2>
</blockquote>
<p>When I solved, I got that my <span class="math-container">$r_1= \frac67$</span> and <span class="math-container">$r_2= \frac87.$</span> Then I got that <span class="math-container">$y= C_1e^{(\frac67)}t +C_2e^{\frac87}t.$</span> I’m not sure how to proceed from here in regards to substituting <span class="math-container">$y(2)= 3$</span> and <span class="math-container">$y’(3)=9$</span> to get <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span> values. </p>
| Robert Lewis | 67,071 | <p>Starting with</p>
<p><span class="math-container">$y^5 + x^5 = 8, \tag 1$</span></p>
<p>we differentiate once to obtain</p>
<p><span class="math-container">$5y^4 y' + 5x^4 = 0, \tag 2$</span></p>
<p>from which we may isolate <span class="math-container">$y'$</span>:</p>
<p><span class="math-container">$y' = -\dfrac{x^4}{y^4}; \tag 3$</span></p>
<p>differentiating this expression using the quotient rule yields</p>
<p><span class="math-container">$y'' = -\dfrac{4x^3y^4 - 4x^4y^3y'}{y^8} = -\dfrac{4x^3y - 4x^4y'}{y^5}; \tag 4$</span></p>
<p>substitute (3) into (4):</p>
<p><span class="math-container">$y'' = -\dfrac{4x^3y - 4x^4(-x^4/y^4)}{y^5} = -\dfrac{4x^3y^5 + 4x^8}{y^9}; \tag 5$</span></p>
<p>we can now in fact eliminate <span class="math-container">$y$</span> from the numerator using (1) in the form</p>
<p><span class="math-container">$y^5 = 8 - x^5, \tag 6$</span></p>
<p>that is</p>
<p><span class="math-container">$y'' = -\dfrac{4x^3y^5 + 4x^8}{y^9}$</span>
<span class="math-container">$= -\dfrac{4x^3(8 - x^5) + 4x^8}{y^9} = -\dfrac{32x^3 - 4x^8 + 4x^8}{y^9} = -\dfrac{32x^3}{y^9}; \tag 7$</span></p>
<p>it is <em>possible</em> to use (6) once again to reduce the power of <span class="math-container">$y$</span> in the denominator, <em>viz</em>.</p>
<p><span class="math-container">$y'' = -\dfrac{32x^3}{y^9} = -\dfrac{32x^3}{(8 - x^5)y^4}; \tag 8$</span></p>
<p>an expression which may have some utility, though I am somewhat skeptical as to its genuine advantage over, say, (7).</p>
|
3,791,350 | <p>An example question is:</p>
<p>In radian measure, what is <span class="math-container">$\arcsin \left(\frac{1}{2}\right)$</span>?</p>
<p>Select one:</p>
<p>a. <span class="math-container">$0$</span></p>
<p>b. <span class="math-container">$\frac{\pi}{6}$</span></p>
<p>c. <span class="math-container">$\frac{\pi}{4}$</span></p>
<p>d. <span class="math-container">$\frac{\pi}{3}$</span></p>
<p>e. <span class="math-container">$\frac{\pi}{2}$</span></p>
<hr />
<p>So, in the exam, I will be given only four function calculator. And is it possible to calculate this kind of trigo function? Or, do I have to memorise common values of trigo functions? Is there any tricks and tips for this problem?</p>
| Air Mike | 802,358 | <p>The function <span class="math-container">$\arcsin$</span> is the inverse of <span class="math-container">$\sin$</span>.</p>
<p>So to compute <span class="math-container">$\arcsin(\frac{1}{2})$</span> we have to see “where” does <span class="math-container">$\sin$</span> of some angle equals <span class="math-container">$\frac{1}{2}$</span>.</p>
<p>And that would be <span class="math-container">$\frac{\pi}{6}$</span>. So the correct answer is <strong>option b</strong>.</p>
<hr />
<p>It will help you all the time to know the values of trigonometry functions at some angles (for instance, at <span class="math-container">$0$</span>, <span class="math-container">$\frac{\pi}{3}$</span>, <span class="math-container">$\frac{\pi}{4}$</span>, <span class="math-container">$\frac{\pi}{6}$</span>...)</p>
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507,494 | <p>$\begin{align*}\frac{n!}{e}-!n&=n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}\\&=\frac{(-1)^{n+1}}{n+1}\left(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\cdots\right)\end{align*}$</p>
<p>and hence</p>
<p>$$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$</p>
<p>I can understand how$$ \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$</p>
<p>is but can't understand how the inequality $$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| $$ is true.</p>
<p>Please HELP.</p>
<p>Thank you.</p>
| Silent | 94,817 | <p>I think $$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right|$$ is because $$\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}=\frac{1}{n+2}$$ and there are many many terms to add in $\frac{1}{n+2}$ so it is less than $\left|\frac{n!}{e}-!n\right|$.</p>
<p>Is my answer correct?</p>
<p>Please verify!</p>
<p>Thank you.</p>
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