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 |
|---|---|---|---|---|
395,685 | <p>I recall seeing a quote by William Thurston where he stated that the Geometrization conjecture was almost certain to be true and predicted that it would be proven by curvature flow methods. I don't remember the exact date, but it was from after Hamilton introduced the Ricci flow but well before Perelman's work. Unfortunately, most of the results for Geometrization and Ricci flow are from 2003 or after. Does anyone know if the quote I'm referring to actually exists, and if so, where to find it?</p>
<p>There is a quote from Thurston lauding Perelman's work, which suggests that he thought the Ricci flow was a promising approach, but I thought there was one from before as well.</p>
<blockquote>
<p>That the geometrization conjecture is true is not a surprise. That a
proof like Perelman's could be valid is not a surprise: it has a
certain rightness and inevitability, long dreamed of by many people
(including me). What is surprising, wonderful and amazing is that someone – Perelman – succeeded in rigorously analyzing and controlling this process, despite the many hurdles, challenges and potential pitfalls.</p>
</blockquote>
<p>Thanks in advance.</p>
| Misha | 21,684 | <p>Most likely, this is just misremembering (or misattribution).</p>
<ol>
<li><p>In his math writing Thurston did not make any predictions regarding what approach to the <em>Geometrization Conjecture</em> (GC) will be successful. I did not hear him making such predictions in his math lectures (but, of course, I heard only few), and I would be surprised if he did. What Thurston might have said to somebody privately, I do not know.</p>
</li>
<li><p>However, Thurston was well aware of the power of the Ricci Flow (RF) and in the early 1980s Thurston did write few pages with a sketch of the proof of the <em>Orbifold Theorem</em> which is the geometrization theorem for orientable irreducible orbifolds with nonempty orbifold locus:</p>
</li>
</ol>
<ul>
<li><p>W. P. Thurston, <em>Three-manifolds with symmetry</em>, preprint, Princeton University, 1982, 5 pages.</p>
<p>One of the steps of the proof was an application of Hamilton's 1982 result on manifolds of positive curvature (proven via RF, of course). A proof of the Orbifold Theorem was given later on in a work by Boileau, Leeb and Porti (broadly speaking, their proof follows Thurston's outline), shortly before appearance of Perelman's preprints. A bit different proof was later given in a work of Cooper, Hodgson and Kerckhoff. Both proofs use Hamilton's theorem at some point, as Thurston suggested.</p>
</li>
</ul>
<ol start="3">
<li>In the mid 1990s Thurston was developing his own approach to the part of the GC known as the <em>Hyperbolization Conjecture</em>, based on analyzing laminations/foliations in 3-manifolds:</li>
</ol>
<ul>
<li><em>Three-manifolds, Foliations and Circles, I.</em> arXiv:<a href="https://arxiv.org/abs/math/9712268" rel="nofollow noreferrer">math/9712268</a>.</li>
</ul>
<ol start="4">
<li>In the last chapter of my 2001 book</li>
</ol>
<ul>
<li><p><em>Kapovich, Michael</em>, Hyperbolic manifolds and discrete groups, Progress in Mathematics (Boston, Mass.). 183. Boston, MA: Birkhäuser. xxv, 467 p. (2001). <a href="https://zbmath.org/?q=an:0958.57001" rel="nofollow noreferrer">ZBL0958.57001</a>.</p>
<p>I summarized several known approaches to the GC. Two approaches were differential-geometric (RF and the Mike Anderson's approach), one was via min-volume (a combination of differential and Alexandrov geometry) and one was coarse geometric (to the "hyperbolic part" of the conjecture). At the time, it was quite possible that the GC will be proven in a piece-meal fashion with one proof of the hyperbolization conjecture, another for the Poincare Conjecture and yet another for Spherical Space Forms Conjecture.</p>
</li>
</ul>
<ol start="5">
<li>Regarding Ryan's comment: There was nothing even close to a consensus on the most promising approach to the GC. Hamilton and Yau believed it to be the RF, Mike Anderson had his own approach, Besson, Courtois and Gallot were making progress via the min-volume approach, some people in (or close to) Geometric Group Theory (like myself) believed it was via Cannon's Conjecture (still open) and Coarse Hyperbolization, etc. And there was no such thing as "Thurston's program" for proving the GC. The closest thing to it was the plan for proving the Orbifold Theorem, but by the 1990s it was clear that this approach will not prove the full GC. In this list of successes one should also add the work of Gabai who in 1990s was proving directly some consequences of the Hyperbolization Conjecture, as well as the work of Tukia, Mess, Gabai, Casson and Jungreis, proving Seifert Conjecture.</li>
</ol>
|
3,980,441 | <p>I need to prove this. I need your help to verify that my proof is correct (or not) please.</p>
<blockquote>
<p>Prove that this integral exists: <span class="math-container">\begin{align}
\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}} \end{align}</span></p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>Fist we need to observe that <span class="math-container">$\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{\sqrt{x^{3}}}\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}<\int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$</span></p>
<p>Next, we note that <span class="math-container">$\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1+x^{3}}}}{\frac{1}{\sqrt{x^{3}}}}=1 \Longrightarrow$</span> for <span class="math-container">$f(x)=\frac{1}{\sqrt{1+x^{3}}}$</span>, <span class="math-container">$g(x)=\frac{1}{\sqrt{x^{3}}}$</span> both integrals converges or both diverges.</p>
<p>By last, integrals of the form <span class="math-container">$\int_{1}^{\infty}\frac{dx}{x^{p}}$</span> converges if <span class="math-container">$p>1$</span>, <span class="math-container">$\Longrightarrow \int_{1}^{\infty}\frac{dx}{\sqrt{x^{3}}}$</span> converges <span class="math-container">$\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$</span> converges</p>
<p>That implies that, <span class="math-container">$\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$</span> converges, therefore it exists.</p>
<p>Is it correct? Is there another way to prove it? Thank you very much</p>
| Kosh | 270,689 | <p>Yes, you compute a limit that is not necessary to compute. You just need to bound your integrand by something that is integrable (like you did):</p>
<p><span class="math-container">$$
0\leq \int_2^{+\infty} \frac{1}{\sqrt{1+x^3}} dx \leq \int_2^{+\infty} \frac{1}{x^{3/2}} dx
$$</span>
and the last integral is finite. Therefore also the one in the middle of the inequalities is finite.</p>
|
2,129,086 | <p>I know that the total number of choosing without constraint is </p>
<p>$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13·12}{2} =78$</p>
<p>Then with x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3. </p>
<p>the textbook has the following solution </p>
<p>$\binom{3+5−1}{5}=\binom{7}{5}=21$ I can't figure out where is the 5 coming from?</p>
<p>The reason to choose 5 is because the constraint adds up to 6? so 11 -6 =5?</p>
| Jasser | 170,011 | <p>$$(1,0,10),(2,3,6),(1,2,8),(1,3,7)$$ which satisfies above equation, if rearranged gives 24 solutions and you have even more solutions. So your text book answer is wrong and I dont see any problems in ur method.</p>
|
897,756 | <p>How can I solve the following trigonometric inequation?</p>
<p>$$\sin\left(x\right)\ne \sin\left(y\right)\>,\>x,y\in \mathbb{R}$$</p>
<p>Why I'm asking this question... I was doing my calculus homework, trying to plot the domain of the function $f\left(x,y\right)=\frac{x-y}{sin\left(x\right)-sin\left(y\right)}$ and figured out I'd have to solve the inequation $\sin\left(x\right)\ne\sin\left(y\right)$... I was able to come to the answer $y\ne x +2\cdot k\cdot \pi \>,\>k \in \mathbb{N}$. However, the answer on the textbook also includes $y\ne -x +2\cdot k\cdot \pi + \pi \>,\>k \in \mathbb{N}$, so I thought that I was probably doing something wrong while solving that inequation.</p>
| Guy | 127,574 | <p>Use $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$</p>
|
897,756 | <p>How can I solve the following trigonometric inequation?</p>
<p>$$\sin\left(x\right)\ne \sin\left(y\right)\>,\>x,y\in \mathbb{R}$$</p>
<p>Why I'm asking this question... I was doing my calculus homework, trying to plot the domain of the function $f\left(x,y\right)=\frac{x-y}{sin\left(x\right)-sin\left(y\right)}$ and figured out I'd have to solve the inequation $\sin\left(x\right)\ne\sin\left(y\right)$... I was able to come to the answer $y\ne x +2\cdot k\cdot \pi \>,\>k \in \mathbb{N}$. However, the answer on the textbook also includes $y\ne -x +2\cdot k\cdot \pi + \pi \>,\>k \in \mathbb{N}$, so I thought that I was probably doing something wrong while solving that inequation.</p>
| JimmyK4542 | 155,509 | <p>Using the difference to product identity $\sin A - \sin B = 2\sin \dfrac{A-B}{2}\cos\dfrac{A+B}{2}$, we get: </p>
<p>$\displaystyle\lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h} = \lim_{h \to 0}\dfrac{2\sin\frac{h}{2}\cos(x+\frac{h}{2})}{h} = \lim_{h \to 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \lim_{h \to 0}\cos(x+\tfrac{h}{2}) = $...</p>
|
200,777 | <p>I have a question regarding sums in arrays.</p>
<p>So I have the following array:</p>
<pre><code>list=RandomReal[{0,1},{5,2}]
(*{{0.693551,0.447185},{0.274842,0.637526},{0.745271,0.0288363},{0.894933,0.937219},{0.605447,0.0337067}}*)
</code></pre>
<p>And from that I want to have the splitting for each pair like that</p>
<pre><code>Subsets[Range@Length@list, {2}]
{{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}}
</code></pre>
<p>Lets say every array in the list is a pair of x and y coordinates.
Now I compute the distance between the different points using:</p>
<pre><code>dist = EuclideanDistance @@@ Subsets[list, {2}] .
</code></pre>
<p>But now I want to have the sum of all the distances where particle 1 is occuring, so these are the first 5 entries in the array, for particle 2 its the first and the 5, 6 and 7th etc. </p>
<p>In the end I want to have a list containing 5 arrays with the sum for each particle.</p>
<p>So can anyone help me please?</p>
| kglr | 125 | <p>You can use <a href="https://reference.wolfram.com/language/ref/DistanceMatrix.html" rel="nofollow noreferrer"><code>DistanceMatrix</code></a> and use <code>Total</code> with the upper part of the matrix:</p>
<pre><code>rowtotaldistances = Total[UpperTriangularize[DistanceMatrix[#],1],{2}]&
m = Partition[Range[0,10],2];
TeXForm @ MatrixForm @ m
</code></pre>
<blockquote>
<p><span class="math-container">$\left(
\begin{array}{cc}
0 & 1 \\
2 & 3 \\
4 & 5 \\
6 & 7 \\
8 & 9 \\
\end{array}
\right)$</span></p>
</blockquote>
<pre><code> TeXForm @ MatrixForm[UpperTriangularize[DistanceMatrix[m],1]]
</code></pre>
<blockquote>
<p><span class="math-container">$\left(
\begin{array}{ccccc}
0 & 2 \sqrt{2} & 4 \sqrt{2} & 6 \sqrt{2} & 8 \sqrt{2} \\
0 & 0 & 2 \sqrt{2} & 4 \sqrt{2} & 6 \sqrt{2} \\
0 & 0 & 0 & 2 \sqrt{2} & 4 \sqrt{2} \\
0 & 0 & 0 & 0 & 2 \sqrt{2} \\
0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)$</span></p>
</blockquote>
<pre><code>rowtotaldistances @ m // TeXForm
</code></pre>
<blockquote>
<p><span class="math-container">$\left\{20 \sqrt{2},12 \sqrt{2},6 \sqrt{2},2 \sqrt{2},0\right\}$</span></p>
</blockquote>
<p><strong>Note:</strong> Using <code>{2}</code> as the second argument of <code>Total</code> gives the row totals:</p>
<pre><code>mat = Array[Subscript[a, ##] &, {5, 5}];
TeXForm@MatrixForm[mat]
</code></pre>
<blockquote>
<p><span class="math-container">$\left(
\begin{array}{ccccc}
a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} \\
a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} \\
a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & a_{3,5} \\
a_{4,1} & a_{4,2} & a_{4,3} & a_{4,4} & a_{4,5} \\
a_{5,1} & a_{5,2} & a_{5,3} & a_{5,4} & a_{5,5} \\
\end{array}
\right)$</span></p>
</blockquote>
<pre><code>TeXForm @ MatrixForm[Total[m, {2}]]
</code></pre>
<blockquote>
<p><span class="math-container">$\left(
\begin{array}{c}
a_{1,1}+a_{1,2}+a_{1,3}+a_{1,4}+a_{1,5} \\
a_{2,1}+a_{2,2}+a_{2,3}+a_{2,4}+a_{2,5} \\
a_{3,1}+a_{3,2}+a_{3,3}+a_{3,4}+a_{3,5} \\
a_{4,1}+a_{4,2}+a_{4,3}+a_{4,4}+a_{4,5} \\
a_{5,1}+a_{5,2}+a_{5,3}+a_{5,4}+a_{5,5} \\
\end{array}
\right)$</span></p>
</blockquote>
<p>For column totals leave the second argument out (or use <code>1</code> as the second argument):</p>
<pre><code>TeXForm @ MatrixForm[Total[mat]]
</code></pre>
<blockquote>
<p><span class="math-container">$\left(
\begin{array}{c}
a_{1,1}+a_{2,1}+a_{3,1}+a_{4,1}+a_{5,1} \\
a_{1,2}+a_{2,2}+a_{3,2}+a_{4,2}+a_{5,2} \\
a_{1,3}+a_{2,3}+a_{3,3}+a_{4,3}+a_{5,3} \\
a_{1,4}+a_{2,4}+a_{3,4}+a_{4,4}+a_{5,4} \\
a_{1,5}+a_{2,5}+a_{3,5}+a_{4,5}+a_{5,5} \\
\end{array}
\right)$</span></p>
</blockquote>
|
1,251,914 | <p>I do not understand how to set up the following problem:</p>
<p>"Forces of 20 lb and 32 lb make an angle of 52 degrees with each other. find the magnitude of the resultant force."</p>
<p>An actually picture would really help.</p>
| Timothy | 137,739 | <p>When somebody says a statement about the fourth dimension that you think is clearly false such as the fact that a closed string has only one possible knot, the unknot, the usual way to interpret what they're saying is a statement that is false. You're right that it's technically false to say that a string cannot be knotted. What they're really doing is thinking of a totally different statement and using that English sentence to mean something the sentence doesn't actually mean.</p>
<p>Take <span class="math-container">$\mathbb{R}^4$</span>, the set of all ordered quadruplets of real numbers. Now, we can invent a definition of the distance from any member of <span class="math-container">$\mathbb{R}^4$</span> to any other member of <span class="math-container">$\mathbb{R}^4$</span> as follows. For any points in <span class="math-container">$\mathbb{R}^4$</span>, <span class="math-container">$(w_1, x_1, y_1, z_1)$</span> and <span class="math-container">$(w_2, x_2, y_2, z_2)$</span>, the distance from <span class="math-container">$(w_1, x_1, y_1, z_1)$</span> to <span class="math-container">$(w_2, x_2, y_2, z_2)$</span> is <span class="math-container">$\sqrt{(w_2 - w_1)^2 + (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$</span>. We can invent a binary operation * on that space that functions essentially the same as quaternion multiplication. Let's define a function <span class="math-container">$f$</span> from <span class="math-container">$\mathbb{R}^4$</span> to itself to be an origin rotation if and only if <span class="math-container">$\exists y \in \mathbb{R}^4\exists z \in \mathbb{R}^4$</span> such that distance from the origin to <span class="math-container">$y$</span> and the distance from the origin to <span class="math-container">$z$</span> are 1 and <span class="math-container">$\forall x \in \mathbb{R}^4, f(x) = y * x * z$</span>. Let's define any rotation in <span class="math-container">$\mathbb{R}$</span> to be an operation such that there exists a translation <span class="math-container">$T$</span> and an origin rotation <span class="math-container">$f$</span> such that it can be expressed as <span class="math-container">$T \circ f \circ T^{-1}$</span>. Although this definition of a rotation works only for <span class="math-container">$\mathbb{R}^4$</span>, we can show that it satisfies all the basic intuitive properties that the set of all rotations in the third dimension satisfies.</p>
|
21,141 | <p>Is there a way to extract the arguments of a function? Consider the following example:</p>
<p>I have this sum</p>
<pre><code>g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3]
</code></pre>
<p>Now, what I want to do is exctract the function arguments and apply them to another function <code>func</code> which takes the arguments as a list.</p>
<pre><code>func[{1}] + func[{2}] + func[{3}] + func[{1, 3}] + 3*func[{1, 2}] + 6*func[{1, 2, 3}]
</code></pre>
<p>I know there is <code>Extract[g[1]*g[3], Position[g[1]*g[3], _Integer]]</code> but that does not work if I have a multiplicative constant.</p>
<p>Is there a way to do this?</p>
| Mr.Wizard | 121 | <p>The pattern for a direct replacement proves to be a bit tricky:</p>
<pre><code>start = g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3];
start /. y__g z_g | y__g x_. :> x func[Join @@ List @@@ {y, z}]
</code></pre>
<blockquote>
<pre><code>func[{1}] + func[{2}] + func[{3}] + 3 func[{1, 2}] + func[{1, 3}] + 6 func[{1, 2, 3}]
</code></pre>
</blockquote>
<p>If order doesn't matter, and arguably it shouldn't with <code>Times</code> you could use:</p>
<pre><code>start //. g[x__] g[y__] :> g[x, y] /. g[x__] :> func[{x}]
</code></pre>
<blockquote>
<pre><code>func[{1}] + func[{2}] + func[{3}] + 3 func[{1, 2}] + func[{1, 3}] + 6 func[{3, 1, 2}]
</code></pre>
</blockquote>
<hr>
<p>The OP asked how to extract just the arguments of <code>g</code> and do so preserving the order of the input. This requires wrapping the expression in <code>Hold</code> (or similar) before it is evaluated, to prevent the automatic sorting of <code>Plus</code>. I shall use a variation of kguler's cleaner patten, and <code>Cases</code> since we want only parts of the expression and not the entire thing transformed. I shall scan at <em>level 2</em> (rather than the default <em>level 1</em>) to bypass the additional Head.</p>
<pre><code>start2 = Hold[g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3]]
Cases[start2, (x___ : 1) p__g :> First /@ {p}, {2}]
</code></pre>
<blockquote>
<pre><code>{{1}, {2}, {3}, {1, 3}, {1, 2}, {1, 2, 3}}
</code></pre>
</blockquote>
<p>Let me take this opportunity to show an unusual but potentially useful method I also used for <a href="https://mathematica.stackexchange.com/a/7842/121">Convert head Times to List</a>. You can hold your expression unevaluated by using <code>SetDelayed</code>, but normally it is fully evaluated when it is called. (This has the advantage of letting you use the expression elsewhere without additional effort such as <code>ReleaseHold</code>.) To get around that when doing the extraction you can <code>Block</code> the functions that you do not want to evaluate during the call. Example:</p>
<pre><code>start3 := g[1] + g[2] + g[3] + g[1]*g[3] + 3*g[1]*g[2] + 6*g[1]*g[2]*g[3]
Block[{Plus},
Cases[start3, (x___ : 1) p__g :> First /@ {p}]
]
</code></pre>
<blockquote>
<pre><code>{{1}, {2}, {3}, {1, 3}, {1, 2}, {1, 2, 3}}
</code></pre>
</blockquote>
<p>Alternatively, you could use <a href="https://mathematica.stackexchange.com/a/1447/121">my <code>step</code> evaluator function</a> to get the unevaluated expression wrapped in <code>HoldForm</code>, then use <em>level 2</em> again:</p>
<pre><code>Cases[step @ start3, (x___ : 1) p__g :> First /@ {p}, {2}]
</code></pre>
<blockquote>
<pre><code>{{1}, {2}, {3}, {1, 3}, {1, 2}, {1, 2, 3}}
</code></pre>
</blockquote>
|
3,589,685 | <p>Can you give an example of an isomorphism mapping from <span class="math-container">$\mathbb R^3 \to \mathbb P_2(\mathbb R)$</span>(degree-2 polynomials)?</p>
<p>I understand that to show isomorphism you can show both injectivity and surjectivity, or you could also just show that an inverse matrix exists.</p>
<p>My issue is that I don't think you can represent the transformation with a matrix because of the polynomial space. </p>
<p>How would you come to proving isomorphism without the use of matrix representations?</p>
| José Carlos Santos | 446,262 | <p>What about<span class="math-container">$$\begin{array}{rccc}\psi\colon&\mathbb R^3&\longrightarrow&P_2(\mathbb R)\\&(a,b,c)&\mapsto&a+bx+cx^2?\end{array}$$</span>It is linear, injective and surjective.</p>
|
149,769 | <p>Using the <a href="http://szhorvat.net/pelican/latex-typesetting-in-mathematica.html" rel="nofollow noreferrer">MaTeX</a> package from our colleague <a href="https://mathematica.stackexchange.com/users/12/szabolcs">Szabolcs</a> I had a certain problem.
I would like to highlight the result for bold, but I did not get a result that I liked.
I tried several options that I found: <code>\boldmath</code>, <code>\bm</code>, <code>\mathbf</code> and <code>\textbf</code>.
What came closest was <code>\mathbf</code>, but Pi did not get bold. Is that so?</p>
<pre><code>Needs["MaTeX`"]
MaTeX["\\int_0^r 2\\pi r\\,dr = 2\\pi \\int_0^r r\\,dr = \
2\\pi\\bigg\vert_0^r \\frac{r^2}{2} = \\bigg(2\\pi \
\\frac{r^2}{2}\\bigg)-\\bigg(2\\pi \\frac{0^2}{2}\\bigg) = \
\\frac{2\\pi r^2}{2} = \\mathbf{\\pi r^2}"]
</code></pre>
<p><a href="https://i.stack.imgur.com/gMSIU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gMSIU.png" alt="enter image description here"></a></p>
| Szabolcs | 12 | <p>@JasonB already posted a good answer. I just wanted to point out two more things:</p>
<ul>
<li><p>This is a LaTeX question really, and not specific to MaTeX/Mathematica. I suggest <a href="http://tex.stackexchange.com/">http://tex.stackexchange.com/</a> for similar future questions. The main thing to know about MaTeX is that it interprets all TeX commands in inline math mode. I.e. when people on TeX.SE ask you to provide a complete minimal example, you simply need to put whatever you passed to MaTeX between a pair of <code>$...$</code> and put that into a minimal LaTeX document. If you want to see the exact LaTeX document that MaTeX creates behind the scenes, then first run <code>ClearMaTeXCache[]</code>, then pass the option <code>"TeXFileFunction" -> Print</code> to the <code>MaTeX</code> function.</p></li>
<li><p><code>\bm</code> does work, but you must load the <code>bm</code> package:</p>
<pre><code>MaTeX["\\bm{\\pi}\\pi", "Preamble" -> {"\\usepackage{bm}"}, Magnification -> 3]
</code></pre>
<p><img src="https://i.stack.imgur.com/k76tI.png" alt="Mathematica graphics"></p></li>
</ul>
|
201,122 | <p>A little bit of <em>motivation</em> (the question starts below the line): I am studying a proper, generically finite map of varieties $X \to Y$, with $X$ and $Y$ smooth. Since the map is proper, we can use the Stein factorization $X \to \hat{X} \to Y$. Since the composition is generically finite, $X \to \hat{X}$ is birational, and therefore a sequence of blowups. I am currently interested in the other map: $\hat{X} \to Y$. I would like to apply Casnati–Ekedahl's techniques from “Covers of algebraic varieties I” (Journal of alg. geom., 1996). For this, I need $\hat{X} \to Y$ to be Gorenstein. (Since $Y$ is Gorenstein (since it is smooth), this is equivalent with $\hat{X}$ being Gorenstein.) When is this true?</p>
<p>Specifically, in my case $X \to Y$ is the albanese morphism of a smooth projective surface: so $Y$ is an abelian surface, and I am in the situation that the albanese morphism is surjective.</p>
<hr>
<p>Let $f \colon X \to Y$ be a proper map between two varieties $X$ and $Y$ over a field $k$. Assume $X$ and $Y$ are smooth (and proper, if you want).</p>
<p>Let $\pi \colon X \to \hat{X}$ and $\hat{f} \colon \hat{X} \to Y$ be the Stein factorization ($f = \hat{f} \circ \pi$). Of course, in general $\hat{X}$ is not smooth. However:</p>
<blockquote>
<p><strong>Q1:</strong> Does $\hat{X}$ have some other nice properties?</p>
</blockquote>
<p>I am thinking in the direction of, e.g., Gorenstein or Cohen–Macaulay. If not, does it help if we assume a bit more on $f$? Or, alternatively:</p>
<blockquote>
<p><strong>Q2:</strong> Under what conditions is $\hat{X}$ Gorenstein?</p>
</blockquote>
| Laurent Moret-Bailly | 7,666 | <p>The only restriction I see is that $\hat{X}$ must be normal (because $X$ is): if $\phi$ is a rational function on (some affine open subscheme of) $\hat{X}$ which is integral over $\mathscr{O}_\hat{X}$, then $\phi\circ\pi$ is integral over $\mathscr{O}_{X}$, hence in $\mathscr{O}_{X}$. In other words, $\phi$ lies in $\pi_*\mathscr{O}_{X}=\mathscr{O}_\hat{X}$.</p>
|
2,424,508 | <p>One textbook exercise asks to prove $$|a|+|b|+|c|-|a+b|-|a+c|-|b+c|+|a+b+c| \geq 0.$$</p>
<p>The textbook's solution is:</p>
<blockquote>
<p>If $a$, $b$ or $c$ is zero, the equality follows. Then, we can assume
$|a| \geq |b| \geq |c| > 0$. </p>
<p>Dividing by $|a|$, the inequality is equivalent
to</p>
<p>$$ 1 + |\frac{b}{a}| + |\frac{c}{a}| - |1+\frac{b}{a}| - |\frac{b}{a}+\frac{c}{a}| - |1+\frac{c}{a}| + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 $$</p>
<p>Since $
\frac{b}{a} \leq 1$ and $\frac{c}{a} \leq 1$, we can
deduce that $|1+\frac{b}{a}| = 1+\frac{b}{a}$ and $|1+\frac{c}{a}| =
> 1+\frac{c}{a}$. </p>
<p>Thus, it is sufficient to prove that</p>
<p>$$ |\frac{b}{a}| + |\frac{c}{a}| - |\frac{b}{a}+\frac{c}{a}| - (1+\frac{b}{a}+\frac{c}{a}) + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 .$$</p>
<p>Now, use the triangle inequality shows that the sum of the first three
terms are positive, and absolute value shows that the sum of last two
terms is also positive.</p>
</blockquote>
<p>There may be more intuitive proofs to this, but how can one in 'some semi-logical way' arrive at this exact one? </p>
| Rob Arthan | 23,171 | <p>The product topology on $P = \{0, 2\}^{\Bbb{N}}$ induced from the discrete topology on $\{0, 2\}$ is <strong>not</strong> the discrete topology. The points in $P$ are not open. The product topology on $P$ is the smallest topology that makes each projection $\pi_i : P \to \{0, 2\}$ continuous. Each open set $X$ in the product topology has $\pi_i[X] = \{0, 2\}$ for all but finitely many $i$ (so $X$ is defined by a finite set of conditions on the values of the $\pi_i$). So the answer to your first question is no and you need to rethink your second and third questions.</p>
|
2,424,508 | <p>One textbook exercise asks to prove $$|a|+|b|+|c|-|a+b|-|a+c|-|b+c|+|a+b+c| \geq 0.$$</p>
<p>The textbook's solution is:</p>
<blockquote>
<p>If $a$, $b$ or $c$ is zero, the equality follows. Then, we can assume
$|a| \geq |b| \geq |c| > 0$. </p>
<p>Dividing by $|a|$, the inequality is equivalent
to</p>
<p>$$ 1 + |\frac{b}{a}| + |\frac{c}{a}| - |1+\frac{b}{a}| - |\frac{b}{a}+\frac{c}{a}| - |1+\frac{c}{a}| + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 $$</p>
<p>Since $
\frac{b}{a} \leq 1$ and $\frac{c}{a} \leq 1$, we can
deduce that $|1+\frac{b}{a}| = 1+\frac{b}{a}$ and $|1+\frac{c}{a}| =
> 1+\frac{c}{a}$. </p>
<p>Thus, it is sufficient to prove that</p>
<p>$$ |\frac{b}{a}| + |\frac{c}{a}| - |\frac{b}{a}+\frac{c}{a}| - (1+\frac{b}{a}+\frac{c}{a}) + |1+\frac{b}{a}+\frac{c}{a}| \geq 0 .$$</p>
<p>Now, use the triangle inequality shows that the sum of the first three
terms are positive, and absolute value shows that the sum of last two
terms is also positive.</p>
</blockquote>
<p>There may be more intuitive proofs to this, but how can one in 'some semi-logical way' arrive at this exact one? </p>
| Xander Henderson | 468,350 | <p>The product topology will not be the discrete topology. The product topology consists of sets of the form
$$ \prod_{j=1}^{\infty} U_j $$
where each set $U_j \in \{ \{0\}, \{2\}, \{0,2\}\}$, <strong>and</strong> only finitely many of the $U_j$ are singletons. This second condition gives us something different from the discrete topology. Note that the <em>box topology</em> on $\{0,2\}^{\mathbb{N}}$ will give the discrete topology, but the box topology is not the product topology.</p>
|
68,563 | <p>I was wondering if there's a formula for the cardinality of the set $A_k=\{(i_1,i_2,\ldots,i_k):1\leq i_1<i_2<\cdots<i_k\leq n\}$ for some $n\in\mathbb{N}$. I calculated that $A_2$ has $n(n-1)/2$ elements, and $A_3=\sum_{j=2}^{n-2}\frac{(n-j)(n-j+1)}{2}$. As you can see, the cardinality of $A_3$ is already represented by a not so nice formula. </p>
<p>Is there a general formula?</p>
| Brian M. Scott | 12,042 | <p>You can also get it by induction using the fairly obvious recurrence $$A_{k+1}(n) = \sum_{i=k}^{n-1}A_k(i):$$ if $A_k(i) = \dbinom{i}{k}$, then $$A_{k+1}(n) = \sum_{i=k}^{n-1}\binom{i}{k} = \binom{n}{k+1}$$ by one of the ‘hockey stick’ identities.</p>
|
23,314 | <p>I have been waiting for <em>Mathematica</em> to give me something for the following integral, errors welcome, but it has been "running" for almost 30 minutes now. </p>
<pre><code> h[s_] := If[1 < s, 1, 0]
Integrate[
Abs[1/(b1^2 + b2^2)
2 E^(-b1 s) ((a1 b1 + a2 b2) E^(
b1 s) - (a1 b1 + a2 b2) Cos[b2 s] + (-a2 b1 + a1 b2) Sin[
b2 s]) - h[s]], {s, 0, Infinity}, Assumptions -> b1 > 0]
</code></pre>
<p>Is there a way to simplify this so that <em>Mathematica</em> would not take this long, or is this an entirely lost case?
Thank you in advance.</p>
| whuber | 91 | <p>Provided $2(a_1 b_1 + a_2 b_2) = b_1^2 + b_2^2$, there exists a closed-form solution that <em>Mathematica</em> can obtain quickly. But you have to do a little analysis along the way :-).</p>
<p><strong>I won't carry out the full solution, but will show how to do the difficult part.</strong> The motivation for the first step is recognizing that expressions like $b_1^2+b_2^2$, $a_1 b_1 + a_2 b_2$, etc., have geometric meanings that can be expressed in terms of the lengths of the vectors $(a_1, a_2)$ and $(b_1, b_2)$ and the angles between them. Accordingly, let's introduce polar coordinates for both:</p>
<pre><code>{a1, a2} = a {Cos[\[Alpha]], Sin[\[Alpha]]};
{b1, b2} = b {Cos[\[Beta]], Sin[\[Beta]]};
</code></pre>
<p>Remember, for future use, that $a\ge 0$ and $b \ge 0$.</p>
<p>Applying <code>FullSimplify</code> to the integrand yields an expression similar to this one, in which I have manually distributed the exponentials into the numerator:</p>
<p>$$\frac{ \left|2 a e^{-\Re(b s \cos (\beta ))}\cos (\alpha -\beta -b s \sin (\beta ))+ (b h(s)-2 a \cos (\alpha -\beta ))\right|}{|b|}.$$</p>
<p><strong>For the integral to converge, the first term must decay</strong>--implying $b\cos(\beta)\gt 0$--and the second term simply cannot be present for arbitrarily large $s$. For such values, $h(s)=1$. We must therefore enforce</p>
<p>$$b - 2 a \cos(\alpha-\beta) = 0.$$</p>
<p>This is equivalent to the condition stated at the beginning of this answer:</p>
<pre><code>(-2 (a1 b1 + a2 b2) + ( b1^2 + b2^2))/b // FullSimplify
</code></pre>
<blockquote>
<p>$$b-2 a \text{Cos}[\alpha -\beta]$$</p>
</blockquote>
<p>When this condition holds, and remembering $a$ and $b$ are nonnegative, the integrand is</p>
<p>$$\frac{2 a e^{-b s \cos (\beta )} |\cos (\alpha -\beta -b s \sin (\beta ))|}{b}$$</p>
<p><strong>To simplify this further, let's re-parameterize it.</strong> It looks like it's in the form of a constant times and exponential times a cosine. Here are the new parameters that will make it so:</p>
<pre><code>rules = {c -> 2 a / b, k -> Cot[\[Beta]], u -> b Sin[\[Beta]], \[Gamma] -> (\[Alpha] - \[Beta])};
</code></pre>
<p>As a check, the calculation</p>
<pre><code>c Exp[-k u s] Abs[Cos[\[Gamma] - u s]] /. rules // TraditionalForm
</code></pre>
<p>reproduces the previous formula.</p>
<p>Now for the analysis. First, it is clear that a <strong>change of variable</strong> from <code>s</code> to <code>u s</code> will help: it eliminates one free parameter <code>u</code> (at the cost merely of dividing the answer by $u$ afterwards). Some care will be needed to distinguish the three cases $u \lt 0$, $u=0$, and $u \gt 0$. I will illustrate the case $u \gt 0$. (The case $u \lt 0$ is almost the same and $u = 0$ is easy because the cosine disappears.) Let's illustrate:</p>
<pre><code>With[{k = 1/8, g = 2},
base = Plot[Exp[-k s] Cos[g + s], {s, 0, 20}, Filling -> Axis,
FillingStyle -> {RGBColor[.9, .6, .6], RGBColor[.6, .6, .9]}, PlotRange -> Full]];
first = Plot[Exp[-k s] Cos[g + s], {s, 0, 2 \[Pi] - \[Pi]/2 - u},
Filling -> Axis, FillingStyle -> GrayLevel[0.9]]];
Show[base, first]]
</code></pre>
<p><img src="https://i.stack.imgur.com/CcgBF.png" alt="Plot"></p>
<p>Here I have plotted the argument <em>without the absolute value.</em> With the absolute value, we are requesting the total (unsigned) area of the gray, red, and blue parts. <strong>The strategy is to compute this area in three pieces according to the colors.</strong></p>
<p>We therefore break the argument of the absolute value into its positive and negative parts. To find out where it is positive--if it's not plainly obvious already--apply <code>Reduce</code>:</p>
<pre><code>Reduce[Cos[\[Gamma] - s] >= 0 , s, Reals]
</code></pre>
<blockquote>
<p>$C[1]\in \text{Integers}\&\&\frac{1}{2} (-\pi +2 \gamma +4 \pi C[1])\leq s\leq \frac{1}{2} (\pi +2 \gamma +4 \pi C[1])$</p>
</blockquote>
<p>This is a countable set of regions indexed by the integer <code>C[1]</code>. Let's perform the integration over a single such region and, afterwards, replace the cosines with $1$ and the sines with $0$:</p>
<pre><code>Integrate[
Exp[-k s] Cos[\[Gamma] - s],
{s, 1/2 (-\[Pi] + 2 \[Gamma] + 4 \[Pi] C[1]), 1/2 (\[Pi] + 2 \[Gamma] + 4 \[Pi] C[1])},
Assumptions -> C[1] \[Element] Integers && k > 0 && \[Gamma] \[Element] Reals]
/. {Cos[2 \[Pi] C[1]] -> 1, Sin[2 \[Pi] C[1]] -> 0, C[1] -> n}
</code></pre>
<blockquote>
<p>$$\frac{e^{-\frac{1}{2} k (\pi +4 n \pi +2 \gamma )} \left(1+e^{k \pi }\right)}{1+k^2}$$</p>
</blockquote>
<p>The same treatment of the regions where the integrand is the absolute value of a negative number yields</p>
<blockquote>
<p>$$\frac{e^{-k \left(\left(\frac{3}{2}+2 n\right) \pi +\gamma \right)} \left(1+e^{k \pi }\right)}{1+k^2}$$</p>
</blockquote>
<p>It remains now only to</p>
<ol>
<li><p>Integrate over the first partial period of $\cos(\gamma - s)$ (the gray lobe in the plot) which has a closed form value that <code>Integrate</code> will handily produce,</p></li>
<li><p>Add to that the sum of the preceding values starting at an appropriate initial value of $n$ (indexing precisely the red and blue lobes in the plot).</p></li>
</ol>
<p>At this juncture I will only demonstrate that <strong>the sum has a closed form</strong> (it is, in fact, merely a geometric series) and leave it to interested (or more dedicated) readers to determine where to start the summation.</p>
<pre><code>Sum[(E^(-(1/2) k (\[Pi] + 4 n \[Pi] + 2 \[Gamma])) (1 + E^(k \[Pi])))/(1 + k^2)
+ (E^(-k ((3/2 + 2 n) \[Pi] + \[Gamma])) (1 + E^(k \[Pi])))/(1 + k^2),
{n, 1, Infinity}] // FullSimplify
</code></pre>
<blockquote>
<p>$$\frac{e^{-2 k (\pi +\gamma )} \left(e^{k \left(\frac{3 \pi }{2}+\gamma \right)}+e^{\frac{1}{2} k (\pi +2 \gamma )}\right)}{\left(-1+e^{k \pi }\right) \left(1+k^2\right)}$$</p>
</blockquote>
<p>To account for the change of variable we must divide this by $u$ and then back-substitute:</p>
<pre><code>%/u /. rules
</code></pre>
<blockquote>
<p>$$\frac{e^{-2 (\pi +\alpha -\beta ) \text{Cot}[\beta ]} \left(e^{\frac{1}{2} (\pi +2 (\alpha -\beta )) \text{Cot}[\beta ]}+e^{\left(\frac{3 \pi }{2}+\alpha -\beta \right) \text{Cot}[\beta ]}\right) \text{Csc}[\beta ]}{b \left(-1+e^{\pi \text{Cot}[\beta ]}\right) \left(1+\text{Cot}[\beta ]^2\right)}$$</p>
</blockquote>
<p>One may go further and re-express this in terms of the $a_i$ and $b_i$, but this is far enough.</p>
|
4,475,082 | <p>Problem:</p>
<ul>
<li>Three-of-a-kind poker hand: Three cards have one rank and the remaining two cards have
two other ranks. e.g. {2♥, 2♠, 2♣, 5♣, K♦}</li>
</ul>
<p>Calculate the probability of drawing this kind of poker hand.</p>
<p>My confusion: When choosing the three ranks, the explanation used <span class="math-container">$13 \choose 1$</span> and <span class="math-container">$12 \choose 2$</span>. I used <span class="math-container">$13 \choose 3$</span> instead which ends up being wrong. I do not know why.</p>
| true blue anil | 22,388 | <p>This mistake is made so often by beginners, that I advise that you consider it as reserving bags for chosen ranks from <span class="math-container">$13$</span> bags.</p>
<p>So one bag needs to be reserved for the triple, and two for the two singles, hence <span class="math-container">$\binom{13}1\binom{12}2$</span>.</p>
<p>The order in which we reserve does not matter, so we could as well write <span class="math-container">$\binom{13}2\binom{11}1$</span>, although the first, of course, will seem the more natural.</p>
|
3,969,943 | <p>It's been a few years since doing any type of trigonometry questions and I've seemed to forgotten everything about it. Below is a question with the solution. You're not supposed to use a calculator.</p>
<p><span class="math-container">$$\begin{align}
&\cos\frac{2\pi}{3}+\tan\frac{7\pi}{4}-\sin\frac{7\pi}{6} \\[4pt]
&=-\cos\frac{\pi}{3}-\tan\frac{\pi}{4}-\left(-\sin\frac{\pi}{6}\right) \\[4pt]
&=-\frac12-1+\frac12 \\[4pt]
&=-1
\end{align}$$</span></p>
<p>Can somebody explain the following to me?</p>
<ul>
<li>How <span class="math-container">$\cos(2\pi/3)$</span> becomes <span class="math-container">$-\cos(\pi/3)$</span></li>
<li>How <span class="math-container">$\tan(7\pi/4)$</span> becomes <span class="math-container">$-\tan(\pi/4)$</span></li>
<li>How <span class="math-container">$-\sin(7\pi/6)$</span> becomes <span class="math-container">$-(-\sin(\pi/6))$</span></li>
</ul>
<p>Thanks</p>
| GEdgar | 442 | <p>For (a) it says "explicit formula". Yours is not. Here it is:
<span class="math-container">$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)
\cdots\left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)
\\
=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{n}{n-1}\cdot\frac{n+1}{n}
= \frac{n+1}{1} = n+1
$$</span>
Each numerator cancels the following denominator.</p>
|
2,981,444 | <p><span class="math-container">$AB$</span> is a chord of a circle and the tangents at <span class="math-container">$A$</span>, <span class="math-container">$B$</span> meet at <span class="math-container">$C$</span>. If <span class="math-container">$P$</span> is any point on the circle and <span class="math-container">$PL$</span>, <span class="math-container">$PM$</span>, <span class="math-container">$PN$</span> are the perpendiculars from <span class="math-container">$P$</span> to <span class="math-container">$AB$</span>, <span class="math-container">$BC$</span>, <span class="math-container">$CA$</span>. Prove that <span class="math-container">$PL^2= PM\cdot PN$</span>.</p>
<p>I tried solving for angles and proving <span class="math-container">$\triangle PLM$</span> and <span class="math-container">$\triangle PNL$</span> similar but was unable to do so.</p>
| Batominovski | 72,152 | <p>Let <span class="math-container">$\Gamma$</span> denote the circle. The line <span class="math-container">$MB$</span> is tangent to <span class="math-container">$\Gamma$</span> at <span class="math-container">$B$</span>, so <span class="math-container">$\angle PAB=\angle PBM$</span>. Since <span class="math-container">$PLBM$</span> is a cyclic quadrilateral, <span class="math-container">$\angle PBM=\angle PLM$</span>, whence
<span class="math-container">$$\angle PAB=\angle PBM=\angle PLM\,.$$</span>
Likewise, <span class="math-container">$PLAN$</span> is a cyclic quadrilateral, so
<span class="math-container">$$\angle PNL=\angle PAL=\angle PAB\,.$$</span>
Consequently,
<span class="math-container">$$\angle PNL=\angle PLM\,.$$</span></p>
<p>Similarly, <span class="math-container">$NA$</span> is tangent to <span class="math-container">$\Gamma$</span> at <span class="math-container">$C$</span> so <span class="math-container">$\angle PBA=\angle PAN$</span>. Since <span class="math-container">$PLBM$</span> is a cyclic quadrilateral, <span class="math-container">$\angle PML=\angle PBL=\angle PBA$</span>, whence
<span class="math-container">$$\angle PML=\angle PBA=\angle PAN\,.$$</span>
Likewise, <span class="math-container">$PLAN$</span> is a cyclic quadrilateral, so
<span class="math-container">$$\angle PLN=\angle PAN\,.$$</span>
Consequently,
<span class="math-container">$$\angle PLN=\angle PML\,.$$</span></p>
<p>Therefore, in the triangles <span class="math-container">$MPL$</span> and <span class="math-container">$LPN$</span>, we have <span class="math-container">$\angle PLM=\angle PNL$</span> and <span class="math-container">$\angle PML=\angle PLN$</span>. Thus, <span class="math-container">$MPL$</span> and <span class="math-container">$LPN$</span> are similar triangles, whence
<span class="math-container">$$\frac{PM}{PL}=\frac{PL}{PN}\,,$$</span>
as desired.</p>
|
2,943,329 | <p>There seem to be six essentially different types of cubic polynomials with real coefficients, giving rise to 1, 2 or 3 real roots in different ways. </p>
<p>Consider <span class="math-container">$f(z) = z^3 + a_2z^2 + a_1z + a_0$</span> and let <span class="math-container">$(a_2,a_1,a_0)$</span> be <span class="math-container">$(0,0,0)$</span>, <span class="math-container">$(0,0,1)$</span>, <span class="math-container">$(0,1,0)$</span>, <span class="math-container">$(2,0,0)$</span>, <span class="math-container">$(2,0,1)$</span>, <span class="math-container">$(2,0,-1)$</span> which yield the following graphs:</p>
<p><a href="https://i.stack.imgur.com/NzmLK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NzmLK.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/ELjV0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ELjV0.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/w88tq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w88tq.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/4pYuD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4pYuD.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/Cqk0f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cqk0f.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/doVBN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/doVBN.png" alt="enter image description here"></a></p>
<p>Is this classification sound? How can it be made precise?</p>
<p>To each of these types there correspond types of curves in the complex plane whose intersections give the complex roots (red for <span class="math-container">$\operatorname{Re}f(z) = 0$</span>, blue for <span class="math-container">$\operatorname{Im}f(z) = 0$</span>, for more details see <a href="https://math.stackexchange.com/questions/2943126/symmetries-in-the-equations-and-graphs-of-complex-valued-polynomials">here</a>):</p>
<p><a href="https://i.stack.imgur.com/m9po5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m9po5.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/KoL1x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KoL1x.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/LD5No.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LD5No.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/apHAx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/apHAx.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/5ZCht.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5ZCht.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/fFqHH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fFqHH.png" alt="enter image description here"></a></p>
<p>But interestingly there are cases where some cubic polynomials look essentially the same as real-valued functions (and are very close in "coefficient space"), while their complex counter parts look quite different as curves. For example for <span class="math-container">$(1,3\frac{1}{9},1), (1,3\frac{2}{9},1) (1,3\frac{3}{9},1)$</span>:</p>
<p><a href="https://i.stack.imgur.com/1fYxs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1fYxs.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/DGfkC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DGfkC.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/zOzK0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zOzK0.png" alt="enter image description here"></a></p>
<p>What does this mean? "There's more structure/complexity in the complex numbers than is visible in the real numbers"?</p>
<p>Another question: Which formula yields the critical value <span class="math-container">$a_1 = 3\frac{2}{9}$</span> as a function of the other two <span class="math-container">$(a_2 = a_0 = 1)$</span>? What's the significance of this number (which happens to be rational)? For arbitrary rational <span class="math-container">$a_2, a_0$</span>, or only when <span class="math-container">$a_2 = a_0$</span>? (It turns out that for <span class="math-container">$a_2 = a_0 = 2$</span> the critical value is <span class="math-container">$a_1 = 3\frac{8}{9}$</span> and for <span class="math-container">$a_2 = a_0 = 0$</span> it's <span class="math-container">$a_1 = 1$</span> - see above.)</p>
<p>If you want to play around with the coefficients, you can do it <a href="https://www.desmos.com/calculator/flznlmfodl" rel="nofollow noreferrer">here</a>.</p>
| LucaMac | 586,942 | <p><span class="math-container">$(1+x+x^2+x^3)^2-x^3 = (1+x+x^2)^2 + 2x^3(1+x+x^2) + x^3(x-1)(1+x+x^2) = (1+x+x^2)(1+x+x^2+x^3+x^4)$</span></p>
|
2,366,610 | <p>Let $U$ be an $n \times n$ unitary matrix and $X$ an $n \times n$ real symmetric matrix. Suppose that $$U^\dagger X U = X$$ for all real symmetric $X$, then what are the allowed unitaries $U$? It seems that the only possible $U$ is some phase multiple of the identity $U=aI$ where $|a|=1$ but I'm not able to show that this is the only allowed unitary.</p>
| robjohn | 13,854 | <p>Using the Mean Value Theorem, and the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\sinh^{-1}(x)=\frac1{\sqrt{1+x^2}}$, we get that
$$
\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}=\frac1{\sqrt{1+\xi^2}}\tag{1}
$$
for some $\xi$ between $\sin(x)$ and $\sinh(x)$.</p>
<p>Therefore, since both $\sinh(x)$ and $\sin(x)$ tend to $0$, the $\xi$ in $(1)$ tends to $0$; that is,
$$
\begin{align}
\lim_{x\to0}\frac{\sinh^{-1}(\sinh(x))-\sinh^{-1}(\sin(x))}{\sinh(x)-\sin(x)}
&=\lim_{\xi\to0}\frac1{\sqrt{1+\xi^2}}\\[3pt]
&=1\tag{2}
\end{align}
$$</p>
|
4,444,669 | <p>I'm unsure about the problem below</p>
<hr>
Under which conditions is the following linear equation system solvable ?
<span class="math-container">$$x_1 + 2x_2 - 3x_3 = a$$</span>
<span class="math-container">$$3x_1 - x_2 + 2x_3 = b$$</span>
<span class="math-container">$$x_1 - 5x_2 + 8x_3 = c$$</span>
<hr>
<p>We set up our matrix</p>
<p><span class="math-container">$$\begin{bmatrix}
1 & 2 & -3 & | a \\
3 & -1 & 2 & | b \\
1 & -5 & 8 & | c \\
\end{bmatrix}$$</span></p>
<p>We apply -3 first row to second row and -1 first row to third row. Then we add -1 second row to third row. We get</p>
<p><span class="math-container">$$\begin{bmatrix}
1 & 2 & -3 & |a\\
0 & -7 & 11 & |b - 3a\\
0 & 0 & 0 & |2a - b + c\\
\end{bmatrix}$$</span></p>
<p>So <span class="math-container">$2a - b + c = 0$</span> for the system to be solvable. Is this correct ? I fear that there are other conditions that I forgot ?</p>
| Hagen von Eitzen | 39,174 | <p>I suppose we are allowed to use the following:</p>
<blockquote>
<p>If <span class="math-container">$S$</span> is a non-empty subset of <span class="math-container">$\Bbb R$</span> and is bounded from below, then there exists a real number denoted as <span class="math-container">$\inf S$</span> such that</p>
<ul>
<li><span class="math-container">$s\ge \inf S$</span> for all <span class="math-container">$s\in S$</span></li>
<li>If <span class="math-container">$a>\inf S$</span>, then there exists <span class="math-container">$s\in S$</span> with <span class="math-container">$s<a$</span>.</li>
</ul>
</blockquote>
<p>The set <span class="math-container">$S:=\{\,\frac1n\mid n\in\Bbb N\,\}$</span> is non-empty and bounded from below by <span class="math-container">$0$</span>, hence <span class="math-container">$\inf S$</span> exists.
Clearly, <span class="math-container">$\inf S\ge 0$</span> (as otherwise the second bullet point fails for <span class="math-container">$a=0$</span>).
Suppose <span class="math-container">$\inf S>0$</span>. Then with <span class="math-container">$a:=2\inf S$</span>, the second bullet point tells us that there exists <span class="math-container">$n\in\Bbb N$</span> with <span class="math-container">$\frac1n<2\inf S$</span>. But then <span class="math-container">$\frac1{2n}<\inf S$</span>, contradicting the first bullet point.</p>
|
43,095 | <p>The continuous max flow problem is posed as follows : </p>
<p>sup $\int_\Omega p_s(x)dx$</p>
<p>subject to : </p>
<p>$|p(x)| \le C(x); \forall x \in \Omega $</p>
<p>$p_s(x) \le C_s(x); \forall x \in \Omega $</p>
<p>$p_t(x) \le C_t(x); \forall x \in \Omega $</p>
<p>$\nabla \cdot p(x) - p_s(x) + p_t(x) = 0; \forall x \in \Omega $</p>
<p>Here $p(x)$ is a field vector and is analogous to the flow in the discrete domain. $\nabla \cdot p$ is the divergence of the field p.</p>
<p>How do i find out the dual of this maximization problem using the lagrangian dual technique, i.e. the equivalent min cut formulation of the problem in the continuous domain.</p>
| Jim Belk | 1,726 | <p>Given arbitrary continuous functions $\lambda,\mu,\nu\colon\Omega\to [0,\infty)$ and $\varphi\colon\Omega\to\mathbb{R}$, define
$$
G(\lambda,\mu,\nu,\varphi) \;=\; \sup_{(p_s,p_t,p)}\int_\Omega \Bigl(p_s + (C-|p|)\lambda+(C_s - p_s)\mu + (C_t - p_t)\nu + (\nabla\cdot p - p_s + p_t)\varphi \Bigr)
$$</p>
<p>Then clearly $\sup_{(p_s,p_t,p)} \int_\Omega p_s \leq G(\lambda,\mu,\nu,\varphi)$ for any $\lambda,\mu,\nu,\varphi$. I would conjecture that
$$
\inf_{(\lambda,\mu,\nu,\varphi)} G(\lambda,\mu,\nu,\varphi) \;=\; \sup_{(p_s,p_t,p)} \int_\Omega p_s\text{,}
$$
but I do not know enough about dual problems to determine whether this is necessarily the case.</p>
|
2,761,509 | <p>I hope it's not a duplicate but I've been searching about this problem for some time on this site and I couldn't find anything. My problem is why a number $\in(-1,0)$ raised to $\infty$ is $0$. For example let's take
$$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n$$
Which is equivalent to
$$\left(\frac{-1}{2}\right)^\infty=(-1)^\infty\left(\frac{1}{2}\right)^\infty=0(-1)^\infty$$
But if a sequence converges all its subsequences converge to the same limit.
And $(-1)^{2n}$ is a subsequence of $(-1)^n$ that converges to $1$ when $(-1)^{2n + 1}$ is a subsequence of $(-1)^n$ that converges to $-1$. So $(-1)^\infty$ does not exist. It remains that$$\lim_{n\to \infty} \left(\frac{-1}{2}\right)^n=0(DNE)$$ </p>
| Delta-u | 550,182 | <p><strong>Hint</strong>:</p>
<p>Indeed $(-1)^\infty$ does not exists. To show that $(-1/2)^n$ goes to $0$, an idea is to show that:
$$ \lim_{n \to \infty}\left|\left(\frac{-1}{2} \right)^n-0 \right|=0$$ </p>
|
3,766,585 | <p>Let <span class="math-container">$X_1,X_2,...,X_n$</span> be random sample from a DF <span class="math-container">$F$</span>, and let <span class="math-container">$F_n^* (x)$</span> be the sample distribution function.</p>
<p>We have to find <span class="math-container">$\operatorname{Cov}(F_n^* (x), F_n^* (y))$</span> for fixed real numbers <span class="math-container">$x, y$</span> where <span class="math-container">$F_n^* (x)$</span> is a sample distribution.</p>
<p>My approach:</p>
<p><span class="math-container">$$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}[(F_n^* (x) - \mathbb{E}[F_n^* (y)])(F_n^* (y) - \mathbb{E}[F_n^* (y)])]$$</span></p>
<p><span class="math-container">$$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}[F_n^* (x) .F_n^* (y)] - \mathbb{E}[F_n^* (x)]\mathbb{E}[F_n^* (y)]$$</span></p>
<p><span class="math-container">$$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}\bigg[\bigg(F_n^* (\min(x, y))\bigg) \bigg(F_n^* (\min(x, y)) + \int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] - \mathbb{E}[F_n^* (x)]\mathbb{E}[F_n^* (y)]$$</span></p>
<p>where <span class="math-container">$f_n^*(x)$</span> is a probability density function of the sample.</p>
<p><span class="math-container">$$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}\bigg[\bigg(F_n^* (\min(x, y))\bigg)^2\bigg]+\mathbb{E}\bigg[ \bigg(F_n^* (\min(x, y)).\int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] - \frac{F_n (x)F_n (y)}{n^2}$$</span></p>
<p><span class="math-container">$$\text{Cov}(F_n^* (x), F_n^* (x)) = \frac{(F_n (\min(x, y))^2}{n} - \frac{F_n (x)F_n (y)}{n^2} +\mathbb{E}\bigg[ \bigg(F_n^* (\min(x, y)).\int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] $$</span></p>
<p>How can I proceed from here?</p>
| Bernard | 202,857 | <p>First observe the absolute value isn't necessary, as <span class="math-container">$1+x^3>0$</span> if <span class="math-container">$-1<x<1$</span>. You can find the limit very with <em>equivalents</em> near <span class="math-container">$0$</span>: we know that <span class="math-container">$\sin\sim_0 x$</span> and <span class="math-container">$\ln(1+u)\sim_0 u $</span>, whence
<span class="math-container">$$\frac{\ln(1+x^3)}{\sin^3x}\sim_0 \frac{x^3}{x^3}=1.$$</span></p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| Gerhard Paseman | 3,468 | <p>I don't know that there is such a line to walk. Justification usually requires some specific context, and an answer that is out of context assumed by the asker usually does not convince the asker.</p>
<p>I would not dare to convince someone that study in mathematics should have higher priority than study in physics, or biochemistry, or psychology, or architecture, or art. Nor would I ask a funding organization that mathematics education should deserve a greater share of funds than educational efforts in other disciplines.</p>
<p>I would note that if they attempted a variety of studies without mathematics or a mathematical viewpoint, they would be missing out on an opportunity to help better understand much of their world. I would point out that trying to understand music without ratios, to understand realism in painting without some knowledge of geometry, to understand social and economic forces without some notion of function and rate of change, to arrange and prepare parties and supplies without basic notions of recipe/algorithm and algebra, much of it would have to be invented, arranged, and taught as some class called "Ratiogeomfunctionrecipalgebrology". Fortunately, older and wiser heads already chose to group such things under "Mathematics".</p>
<p>Gerhard "Head Getting Older At Least" Paseman, 2015.05.11</p>
|
8,052 | <p>I wonder how you teachers walk the line between justifying mathematics because of
its many—and sometimes surprising—applications, and justifying it as the study
of one of the great intellectual and creative achievements of humankind?</p>
<p>I have quoted to my students G.H. Hardy's famous line,</p>
<blockquote>
<p>The Theory of Numbers has always been regarded as one of the most obviously useless branches of Pure Mathematics.</p>
</blockquote>
<p>and then contrasted this with the role number theory plays in contemporary
cryptography.
But I feel slightly guilty in doing so, because I believe that even without
the applications to cryptography that Hardy could not foresee—if in fact
number theory were completely "useless"—it
would nevertheless be well-worth studying for anyone.</p>
<p>One provocation is
Andrew Hacker's influential article in
the <em>NYTimes</em>,
<a href="http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html?_r=0" rel="noreferrer">Is Algebra Necessary?</a>
I believe your cultural education is not complete unless you understand
something of the achievements of mathematics, even if pure and useless.
But this is a difficult argument to make when you are teaching students how
to factor quadratic polynomials, e.g.,
<a href="https://matheducators.stackexchange.com/q/8020/511">The sum - product problem</a>.</p>
<p>So, to repeat, how do you walk this line?</p>
| Adam Rubinson | 12,691 | <p>Mathematics is often thought of as a bunch of clever people doing weird things with symbols for reasons that are difficult to relate to reality. Whilst this is not entirely false, it is for the most part (almost humorously so in my opinion), the antithesis of the actual role of mathematics in today's world.</p>
<p>I challenge anyone to seriously tackle the world's major problems without first being an expert in mathematics.</p>
<p>Our planet is on the verge of human extinction due to climate change. There will be wars over water. There is currently a global pandemic. There are billions of people living in poverty. There is famine, disease, etc...</p>
<p>Has any of the progress made (towards solving these problems) in the past been done without rigorous problem-solving? Without collecting and analysing data (statistics)?</p>
<p>Ah, <em>rigorous problem solving</em>. This is, in my opinion, the most accurate definition of mathematics.</p>
<p>Does anyone seriously think we can tackle any of these problems without greatly advancing technology for example machine learning or <a href="https://en.wikipedia.org/wiki/Computer_simulation" rel="nofollow noreferrer">computer simulation</a>, in which being good at mathematics is a prerequisite to even understanding to what is going on, let alone being able to work towards solutions?</p>
<p><em>Why would / how can you not learn mathematics?</em></p>
<p>Even if you don't care about the world's serious problems, but just want to further your own career, you may want to advertise your product or service to make it arrive at the top of a google search. In which case, it may help to understand how <a href="https://en.wikipedia.org/wiki/PageRank" rel="nofollow noreferrer">google's pagerank algorithm</a> works and how you can exploit it (although there are already people who are specialist in this type of work- but clearly <em>they</em> need to be good at maths.)</p>
<p>Then there are Scientists who must use maths to both gather and analyse data and also to help solve problems e.g. <a href="https://en.wikipedia.org/wiki/Mathematical_model" rel="nofollow noreferrer">Mathematical modelling</a>.</p>
<p>What about trying to come up to solutions to economic problems. If you take a look at <a href="https://economics.stackexchange.com/">economics stackexchange</a>. Take a look at some of the questions on there and notice the high proportion of questions with some reasonably heavy mathematical machinery in them.</p>
<p>I could easily give another thousand examples, but you can find lists of the applications of mathematics if you do a quick google search.</p>
<p>Again, in response to the original question of, "Why do I need to learn mathematics?". I don't see how you can do without mathematics.</p>
<p>Not to the extent of air and water, I'm not delusional. For example, you <em>can avoid</em> mathematics and pursue a career of art or music or entertainment or "social media influencer" and whilst I don't think these people are entirely useless to society, the fact remains that they are not the problem-solvers.</p>
|
2,798,847 | <p>I would like to prove that $\|e^A-e^B\| \leq \|A-B\|e^{max\{\|A\|,\|B\|\}}$, where $A,B \in \mathbb{R}^{n \times n}$.</p>
<p>So far I was able to create the first difference term, but I have no idea how to incorporate the max norm.
I've read <a href="https://math.stackexchange.com/questions/2262000/inequality-norm-of-difference-in-exponential-of-matrices?newreg=15d63d96ac0f4ad08a10eb1a97a1a7d4">this</a> post, where the Fréchet calculus was mentioned, but I'm still stuck.</p>
<p>Any help would be appreciated.
Thank you in advance!</p>
| Keerthana Gurushankar | 647,133 | <p>You can first get <span class="math-container">$\|B^k-A^k\|\leq k\|B-A\|(\max(\|A\|,\|B\|))^{k-1}$</span> as follows</p>
<p><span class="math-container">\begin{align*}
\|B^k-A^k\| &= \left\lVert\sum_{l=0}^{k-1}B^{l}(B-A)A^{k-1-l}\right\rVert
\leq \sum_{l=0}^{k-1}\left\lVert B^{l}(B-A)A^{k-1-l}\right\rVert\\
&\leq \sum_{l=0}^{k-1}\|B\|^l\|B-A\|\|A\|^{k-1-l}
\leq k\|B-A\|\left(\max(\|A\|,\|B\|)\right)^{k-1}
\end{align*}</span></p>
<p>Then<span class="math-container">\begin{align*}\|e^B-e^A\|=\left\lVert \sum_{k=0}^\infty \frac{B^k-A^k}{k!}\right\rVert\leq \sum_{k=0}^\infty \frac{k\left(\max(\|A\|,\|B\|)\right)^{k-1}\|B-A\|}{k!}=\left|B-A\right|e^{\max(\|A\|,\|B\|)}\end{align*}</span></p>
|
1,687,714 | <p><a href="https://i.stack.imgur.com/nZEAy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nZEAy.jpg" alt=""></a></p>
<p>I am given a problem in my textbook and I am left to determine the Laplace transform of a function given its graph (see the attached photo) - a square wave - using the theorem that $$F(s) = \frac{1}{1-e^{-ps}} \int_0^p e^{-st}f(t)dt$$ where $f(t)$ is a periodic function with period $p$. From the graph and the information in the theorem, I deduce that the Laplace transform of the function can be calculated as follows: $$F(s) = \frac{1}{1-e^{-2as}} \int_0^{2a} e^{-st}f(t)dt = \frac{1}{1-e^{-2as}} \int_0^a e^{-st}dt\qquad (1)$$ because $f(t) = 1$ for $0 \le t \le a$, and $f(t) = 0$ for $a \le t \le 2a$. However, the book gets $$F(s) = \frac{1}{s(1+e^{-as})}\qquad (2)$$ </p>
<p>Could someone lend me a hand with this problem?</p>
<p>Numbers added for convenience, thank you in advance mates. Fair winds.</p>
| John B | 301,742 | <p>$$
\frac{1}{1-e^{-2as}} \int_0^a e^{-st}dt=\frac{1-e^{-as}}{s(1-e^{-2as})}=\frac{1}{s(1+e^{-as})}
$$</p>
|
412,482 | <p>I have a few (semi-)related questions regarding certain Hilbert space representations of locally compact groups that come up in the theory of automorphic forms. </p>
<p>Let $G$ be a unimodular locally compact Hausdorff group, $\Gamma$ a discrete (hence closed) subgroup of $G$, and $Z$ a closed subgroup of the center of $G$. The example of interest to me is:</p>
<p>(*) $G$ is the group of adelic points of a connected reductive $\mathbf{Q}$-group $\mathscr{G}$, $\Gamma=\mathscr{G}(\mathbf{Q})$, and $Z$ is the group of adelic points of the maximal split torus in the center of $\mathscr{G}$. </p>
<p>I would be happy to know the answer to the following question just in case (*).</p>
<p><strong>Is the group $Z\Gamma$ closed in $G$, and if so, is it unimodular?</strong></p>
<p>The answer to the first part of the question would be yes if one of $Z$, $\Gamma$ were compact, but this isn't so in many cases of interest to me (e.g. $G=\mathrm{GL}_n$). I suspect the answer to both parts of the question are "yes" because of the following definition which appears in the literature (I will assume $Z\Gamma$ is closed and unimodular to make the definition, although I'm hoping, at least in the case of interest mentioned above, that this is unnecessary):</p>
<p>For a unitary character $\omega$ of $Z$, let $L^2(Z\Gamma\setminus G,\omega)$ be the space of equivalence classes (the equivalence being almost every equality relative to the unique-up-to-scaling $G$-invariant measure on $\Gamma\setminus G$) of Borel measurable $f:\Gamma\setminus G\rightarrow\mathbf{C}$ with the following properties: for each $z\in Z$, $f(\Gamma zg)=\omega(z)f(\Gamma g)$ for almost every $\Gamma g\in\Gamma\setminus G$, $\vert f\vert$ is Borel measurable on $Z\Gamma\setminus G$ (this makes sense because of the unitary of $\omega$), and $\int_{Z\Gamma\setminus G}\vert f\vert^2<\infty$ (the integral being taken with respect to the unique-up-to-scaling $G$-invariant measure on $Z\Gamma\setminus G$). Assuming I've got this definition right, it will be a Hilbert space with the natural inner product.</p>
<p>My second question concerns the definition of the so-called cuspidal subspace $L_0^2(Z\Gamma/G,\omega)$, and I'm only interested in the case of (*) with $\mathscr{G}=\mathrm{GL}_n$, so $G=\mathrm{GL}_n(\mathbf{A}_\mathbf{Q})$, $\Gamma=\mathrm{GL}_n(\mathbf{Q})$, and $Z=\mathbf{A}_\mathbf{Q}^\times$ is the center of $G$. The (apparently) standard definition of $L_0^2(Z\Gamma\setminus G,\omega)$ is: the subspace of $L^2(Z\Gamma\setminus G,\omega)$ consisting of equivalence classes containing a function $f$ such that $\int_{\mathbf{A}_\mathbf{Q}/\mathbf{Q}}f\big(\big(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\big)g\big)dx=0$ for almost every $g\in G$ ($dx$ means Haar measure on $\mathbf{A}_\mathbf{Q}/\mathbf{Q}$). There are some things to check to make precise sense of this definition, but the important thing is that it should be a closed subspace of the full $L^2$-space. This is taken for granted in, e.g., Bump's book on automorphic representations and Godement's notes on Jacquet-Langlands. The closedness doesn't seem completely obvious to me, and I worry about it because some other authors, notably Lang in his book on $\mathrm{SL}_2$, define the cuspidal subspace as the Hilbert space completion of the space of bounded continuous functions on $\Gamma\setminus G$ with the appropriate properties. I would hope these definitions yield the same space but it isn't clear to me.</p>
<p><strong>Assuming the definition I've given of $L_0^2(Z\Gamma\setminus G,\omega)$ is technically correct (i.e. I haven't left out any conditions on the functions under consideration), is it closed in $L^2(Z\Gamma\setminus G,\omega)$?</strong></p>
| TheBridge | 4,437 | <p>You need to apply this <a href="http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions" rel="noreferrer">theorem</a> about how on conditioning a Gaussian vectors.</p>
<p>You are in the situation where you have this 3-d Gaussian vector $V=(X_s,X_u,X_t)$.</p>
<p>You will have to compute the mean vector and covariance matrix of $V$, and then applying the theorem to your situation. </p>
<p>Regards</p>
|
1,572,954 | <p>What is the only ordered pair of numbers $(x,y)$ which, for all $a$ and $b$, satisfies </p>
<p>$$x^a y^b=\left(\frac34\right)^{a-b} \text{and } x^b y^a=\left(\frac34\right)^{b-a}$$</p>
<p>I started off with the trivial cases, $a=0$ and $b=0$ and you get $1=1$ on both sides, so that works.</p>
<p>I can't seem to find anymore cases. Any ideas?</p>
| Kay K. | 292,333 | <p>$$\left(x^ay^b\right)\times\left(x^by^a\right)=\left(\frac34\right)^{a-b}\times\left(\frac34\right)^{b-a}$$
$$(xy)^{a+b}=1$$
$$xy=1$$
$$x^ax^{-b}=\left(\frac34\right)^{a-b}$$
$$\therefore x=\frac34,\space y=\frac43$$</p>
|
418,026 | <p>I am asked to find the power series of the function $f(x)=\arctan(\frac{x}{\sqrt{2}})$. I first found the derivative of this function which is: $f'(x)=\frac{\sqrt{2}}{2+x^{2}}$. Then I found the power series of $f'(x)$ which is: $\sum_{n=0}^{\infty }\frac{1}{\sqrt{2}}(-1)^{n}\left ( \frac{1}{2} \right )^{n}x^{2n}$. Then to get the power series of $f(x)$, I just integrated the previous power series term by term to get: $\sum_{n=0}^{\infty }\frac{1}{\sqrt{2}}(-1)^{n}\left ( \frac{1}{2} \right )^{n}\frac{x^{2n+1}}{2n+1}$. Now to find the radius of convergence, I found $-\sqrt{2}\leqslant x\leqslant \sqrt{2}$. However, when I entered this answer in the software, it doesn't accept this answer as a valid answer. Can anybody tell me where the problem is? Because I checked and I found that the series is convergent at the end points of my interval of convergence. Thanks!</p>
| Branimir Ćaćić | 49,610 | <p>First, since $G$ is symmetric positive definite, you know in particular that it is indeed invertible.</p>
<p>Now, let $T := \sum_{j=1}^n u_j u_j^T$. The question is, what is $TGe_k$ for each $k$?</p>
|
1,201,942 | <p>I apologize if this question gets down-voted ahead of time.</p>
<p>I've been working on the Collatz Conjecture all day with Python, because that is the language I'm most familiar with (I'm not a CS student, just majoring in math). Below is the function I'm using for your reference during my comments:</p>
<p>\begin{align}
T\left(n\right)&=\begin{cases}1 & \text{if}\;n=1\\
T\left(\frac{n}{2}\right) & \text{if}\;n\;\text{is even}\\
T\left(3n+1\right) & \text{if}\;n\;\text{is odd}\end{cases}\tag{1}
\end{align}</p>
<p>I've created graphs of the number of iterations vs. the integers up to $10^7$, one of which I've shown below for integers up to $2\cdot 10^6$:</p>
<p><img src="https://i.stack.imgur.com/8eZBA.png" alt="Collatz Conjecture"></p>
<p>I've read on <a href="http://mathworld.wolfram.com/CollatzProblem.html" rel="noreferrer">Wolfram's Mathworld</a> that the conjecture has been tested via computers for numbers up to $\approx 5.48\cdot 10^{18}$, which is quite impressive, although I was thinking ahead of time it would have been tested by this time for incredibly large integers given we've had computers for >50 years now.</p>
<p>Now to my simple question: would it be true that, given such an integer actually does exist that does not satisfy the conjecture, the smallest such number must be odd because if it is even there must exist a smaller such integer that does not satisfy the conjecture? </p>
<p>Further, suppose we did find such an integer. In my writings I've noticed that all integers within the sequences generated by this function eventually break down to a repeat of an earlier sequence. For instance, the following two sequences:</p>
<blockquote>
<p>$\color{red}{7}$-22-11-34-17-52-26-13-40-20-10-5-16-8-4-2-1</p>
<p>14-$\color{red}{7}$</p>
</blockquote>
<p>These are simple, but even more lengthy sequences of numbers I've found break down to prior sequences that have already been computed, and must if we are to end up at $1$ unless it is defining a new sequence to be called by a later one (I wonder what the frequency of this is). But if we find such a number and it is odd, then all other numbers contained within this sequence must also not satisfy the conjecture. But is this true? Or am I thinking of "not satisfying" the conjecture in the wrong way?</p>
<p>Thank you for your time,</p>
| fulges | 53,662 | <p>You are correct, in both your points. Suppose that the conjecture is false.</p>
<p>If $N$ is a counterexample and $M$ is an element of the sequence generated by $N$, then $M$ is a counterexample as well, since if the sequence generated by $M$ ended to $1$, then also the sequence generated by $N$ ends at $1$.</p>
<p>For the same reason, the minimal counterexample has to be odd, because if $N$ is an even counterexample, then $N/2$ is a counterexample as well, as it is the first term generated by the sequence starting with $N$.</p>
|
3,168,662 | <p>How do you evaluate <span class="math-container">$\int_{|z|=1} \frac{\sin(z)}{z^2+(3-i)z-3i}dz$</span> ? </p>
<p>Here is my thought process: </p>
<p>I want to use <a href="http://mathworld.wolfram.com/CauchyIntegralFormula.html" rel="nofollow noreferrer">Cauchy's Integral Formula</a>, but in order to use it I need to find the poles and make sure one of them is in the interior of the unit circle (the contour we are supposed to integrate over).</p>
<p>First, I need to set the denominator in the integral to <span class="math-container">$0$</span> and solve the <a href="http://mathworld.wolfram.com/QuadraticEquation.html" rel="nofollow noreferrer">quadratic equation</a>: <span class="math-container">$z^2+(3-i)z-3i=0$</span></p>
<p>Using Wolfram, here: <a href="https://www.wolframalpha.com/input/?i=solve+z%5E2+%2B+(3-i)z+-3i+%3D+0" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=solve+z%5E2+%2B+(3-i)z+-3i+%3D+0</a></p>
<p>I get <span class="math-container">$z=-3$</span> and <span class="math-container">$z=i$</span></p>
<p>A requirement for Cauchy's Integral Formula is one of these poles must be in the intereior of the curve we are integrating over (in this case the unit circle). However, <span class="math-container">$z=-3$</span> is outside the unit circle and <span class="math-container">$z=i$</span> is on the unit circle itself, not in its interior. So I cannot use Cauchy's Integral Formula. </p>
<p>What other options do I have?</p>
| Mark Viola | 218,419 | <p>First, we can write the integrand as </p>
<p><span class="math-container">$$\frac{\sin(z)}{(z+3)(z-i)}=\underbrace{\frac1{3+i}\left(\frac{\sin(z)-\sin(i)}{z-i}-\frac{\sin(z)}{z+3}\right)}_{\text{Analytic for }|z|\le1}+\underbrace{\frac1{3+i}\left(\frac{\sin(i)}{z-i}\right)}_{\text{Has simple pole at }z=i}$$</span></p>
<p>The key observation is that the integral <span class="math-container">$\oint_{|z|=1}\frac1{z-i}\,dz$</span> is <strong><em>not</em></strong> defined since the path of integration actually contains the simple pole singularity point <span class="math-container">$z=i$</span>. </p>
<p>However, if we exclude a "small" neighborhood of <span class="math-container">$i$</span> from the unit circle and integrate over the path parametrically described by <span class="math-container">$z=e^{i\phi}$</span>, <span class="math-container">$\phi\in [\pi/2+\Delta \phi,\pi/2+2\pi-\Delta \phi]$</span>, for <span class="math-container">$\Delta \phi>0$</span>, then we find </p>
<p><span class="math-container">$$\begin{align}
\int_{\pi/2+\Delta \phi}^{\pi/2+2\pi-\Delta \phi}\frac{ie^{i\phi}}{e^{i\phi}-i}\,d\phi&=i\int_{\Delta \phi}^{2\pi-\Delta \phi}\frac{1-\cos(\phi)-i\sin(\phi)}{2(1-\cos(\phi))}\,d\phi\\\\
&=i(\pi+\Delta \phi)
\end{align}$$</span></p>
<p>Letting <span class="math-container">$\Delta \phi\to0$</span>, we have </p>
<p><span class="math-container">$$\text{PV}\left(\oint_{|z|=1}\frac{\sin(z)}{z^2+(3-i)z}\right)=\frac{i\pi \sin(i)}{3+i}=-\frac1{10}(3-i)\pi\sinh(1)$$</span></p>
<p>where <span class="math-container">$\text{PV}$</span> denotes the Cauchy principal value of the contour integral. </p>
<p>Interestingly, this result is exactly the arithmetic average of the integral over the circular contours <span class="math-container">$|z|=1+\epsilon$</span> and <span class="math-container">$|z|=1-\epsilon$</span>, <span class="math-container">$0<\epsilon<2$</span>. The former would yield <span class="math-container">$2\pi i \text{Res}\left(\frac{\sin(z)}{(z+3)(z-i)}, z=i\right)=-\frac2{10}(3-i)\pi \sinh(1) $</span>, while the latter would yield <span class="math-container">$0$</span>.</p>
|
3,168,662 | <p>How do you evaluate <span class="math-container">$\int_{|z|=1} \frac{\sin(z)}{z^2+(3-i)z-3i}dz$</span> ? </p>
<p>Here is my thought process: </p>
<p>I want to use <a href="http://mathworld.wolfram.com/CauchyIntegralFormula.html" rel="nofollow noreferrer">Cauchy's Integral Formula</a>, but in order to use it I need to find the poles and make sure one of them is in the interior of the unit circle (the contour we are supposed to integrate over).</p>
<p>First, I need to set the denominator in the integral to <span class="math-container">$0$</span> and solve the <a href="http://mathworld.wolfram.com/QuadraticEquation.html" rel="nofollow noreferrer">quadratic equation</a>: <span class="math-container">$z^2+(3-i)z-3i=0$</span></p>
<p>Using Wolfram, here: <a href="https://www.wolframalpha.com/input/?i=solve+z%5E2+%2B+(3-i)z+-3i+%3D+0" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=solve+z%5E2+%2B+(3-i)z+-3i+%3D+0</a></p>
<p>I get <span class="math-container">$z=-3$</span> and <span class="math-container">$z=i$</span></p>
<p>A requirement for Cauchy's Integral Formula is one of these poles must be in the intereior of the curve we are integrating over (in this case the unit circle). However, <span class="math-container">$z=-3$</span> is outside the unit circle and <span class="math-container">$z=i$</span> is on the unit circle itself, not in its interior. So I cannot use Cauchy's Integral Formula. </p>
<p>What other options do I have?</p>
| Community | -1 | <p>You must be in Dr. Gowda's Complex Analysis class as well. See the new email from Dr. Gowda that says to integrate over <span class="math-container">$\left\lvert z\right\rvert=2$</span> instead. We cannot use Cauchy's integral formula here because <span class="math-container">$z=i$</span> lies on the boundary of <span class="math-container">$\left\lvert z\right\rvert=1$</span>, so he updated it to the contour <span class="math-container">$\left\lvert z\right\rvert=2$</span>. With the updated version, consider <span class="math-container">$f\left(z\right)=\frac{\sin z}{z+3}$</span>, then using Cauchy's integral formula, we have
<span class="math-container">$$\int_{\left\lvert z\right\rvert=2}\frac{\sin z}{\left(z-i\right)\left(z+3\right)}\text{d}z=2\pi i\left(\frac{1}{2\pi i}\int_{\left\lvert z\right\rvert=2}\frac{\sin z}{\left(z-i\right)\left(z+3\right)}\text{d}z\right)=2\pi i\frac{\sin\left(i\right)}{i+3},$$</span>
which can be changed into
<span class="math-container">$$\frac{\pi\left(1-e^{2}\right)}{10e}\left(3-i\right)$$</span>
using the fact that <span class="math-container">$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$</span>.</p>
|
1,725,337 | <p>How does the following definition of Taylor polynomials:</p>
<p>$f(x_0 + h)= f(x_0) + f'(x_0)\cdot h + \frac{f''(x)}{2!}h^2+ ... +\frac{f^(k)(x_0)}{k!}\cdot h^k+R_k(x_0,h),$ </p>
<p>where $R_k(x_0,h)=\int^{x_0+h}_{x_0} \frac{(x_0+h-\tau)^k}{k!}f^{k+1}(\tau) d\tau$</p>
<p>where I guess $\lim_{h\to 0} \frac{R_k(x_0,h}{h^k}=0$</p>
<p>differ from </p>
<p>$f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots +\frac {f^k(a)}{k!} (x-a)^k + R(x) $</p>
<p>where $R(x)$ is the corresponding error function.</p>
<p>I understand the intuition of the second definition and how it is derived but how does the first definition approximate the function $f$? <em>Can you please show how to derive the definition or give an intuitive explanation</em> in the way Tom Apostol does for the first definition: </p>
<p><a href="https://i.stack.imgur.com/aL0Rz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aL0Rz.png" alt="enter image description here"></a></p>
<p>I know a similar question is asked at <a href="https://math.stackexchange.com/questions/648540/two-definitions-of-taylor-polynomials">Two definitions of Taylor polynomials</a> but it isn't quite the same. </p>
| Anonymous | 327,815 | <p>$$
g(x) = \frac{1}{(8+x)} = \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{8^{n+1}}
$$
Taking the derivative in both sides: (n=0 is constant)
$$
g'(x) = \frac{-1}{(8+x)^2} = \sum_{n=1}^{\infty} \frac{n(-1)^nx^{n-1}}{8^{n+1}}
$$
$$
-f(x) = \frac{-1}{(8+x)^2} = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n+1}x^{n}}{8^{n+2}}
$$
$$
f(x) = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n}x^{n}}{8^{n+2}}
$$</p>
|
39,551 | <p>How can I use <em>Mathematica</em> to equate coefficients in a non-power-series equation?</p>
<p>For example, I would like to take an equation like the following:
$$af_x+\frac{b}{2}f_xf_y+chf_x=f_x+e^af_x+3f_xf_y+2bhf_x$$
and produce the following system:
$$a=1+e^a$$
$$\frac{b}{2}=3$$
$$c=2b$$
<strong>EDIT:</strong> This is a rather small example. If possible, I would prefer a solution that requires minimal human inspection of the original equation. The equations I will be working with will have many, perhaps hundreds of partial derivative terms, and it would be unfeasible to do things like individually pick them out. Ideally, I would like to specify only the unknowns I am interested in (in this case, {a, b, c}) and let <em>Mathematica</em> take it from there.</p>
| Nasser | 70 | <p>Probably (ok, most likely) not the most elegant way. But here we go: May be you can let $x=f_x$ and $y=f_y$ and use <code>CoefficientList</code> to help.</p>
<pre><code>ClearAll[x, y, a, b, c, h, r, lhs, rhs]
fx = D[f[x, y], x];
fy = D[f[x, y], y];
expr = a fx + b/2 fx fy + c h fx == fx + Exp[a] fx + 3 fx fy + 2 b h fx
</code></pre>
<p><img src="https://i.stack.imgur.com/uusbR.png" alt="Mathematica graphics"></p>
<pre><code>expr = expr /. {fx -> x, fy -> y}
</code></pre>
<p><img src="https://i.stack.imgur.com/PIzY7.png" alt="Mathematica graphics"></p>
<pre><code>lhs = Last@CoefficientList[expr /. (lhs_ == rhs_) :> lhs, {x, y}];
rhs = Last@CoefficientList[expr /. (lhs_ == rhs_) :> rhs, {x, y}];
(r = MapThread[Equal, {lhs, rhs}]) // TableForm
</code></pre>
<p><img src="https://i.stack.imgur.com/xYXGS.png" alt="Mathematica graphics"></p>
<p>One small part left</p>
<pre><code>r2 = r[[1]] /. (h any_ + __) :> h any;
r2 /. (any1_ h == any2_ h) :> any1 == any2
Inner[Subtract, r[[1]], r2, Equal]
</code></pre>
<p><img src="https://i.stack.imgur.com/gBc1l.png" alt="Mathematica graphics"></p>
|
1,578,783 | <p>A friend of mine found a book in which the author said that the dual space of $L^\infty$ is $L^1$, of course not with the norm topology but with the weak-* topology. Does anyone know where I can find this result? Thanks.</p>
| Tomasz Kania | 17,929 | <p>For any $C(K)$-space we have $C(K)^*\cong L_1(\mu)$ for some usually humongous measure $\mu$. See the proof of Proposition 4.3.8(iii) in </p>
<blockquote>
<p>F. Albiac, N.J. Kalton, <em>Topics in Banach Space Theory</em>, Grad. Texts in Math. 233, Springer, 2006.</p>
</blockquote>
<p>Of course, $L_\infty(\nu)\cong C(K)$ for some compact, Hausdorff space $K$. However, there is no clear relation between the measures $\mu$ and $\nu$. In fact, if $L_\infty(\nu)$ is infinite-dimensional, then $\mu$ is not even $\sigma$-finite.</p>
|
127,108 | <p>If you take a subtraction-free rational identity like $(xxx+yyy)/(x+y)+xy=xx+yy$ and replace $\times$,$/$,$+$,$1$ by $+$,$-$,min,$0$, do you always get a valid min,plus,minus identity like min(min($x+x+x,y+y+y$)$-$min($x,y$),$\:x+y$)$\ =\ $min($x+x,y+y$)?</p>
| Colin McQuillan | 408 | <p>It suffices to show that whenever $F$ is a function $\mathbb R_{\geq 0}^k\to\mathbb R_{\geq 0}$ defined using $\times,/,+,1$, and $f,g$ is the corresponding tropicalization $\mathbb R^k\to\mathbb R$, for all real $x_1,\dots,x_k$ we have $$F(\exp(-\beta x_1),\dots,\exp(-\beta x_k))^{1/\beta}\to \exp(-f(x_1,\dots,x_k)).$$
as $\beta\to+\infty$.</p>
<p>This follows by structural induction on the formula defining $f$, using the following lemma.</p>
<p><strong>Lemma:</strong> Let $F$ be one of the operations $\times,/,+$, let $f$ be the corresponding tropical operation, and let $G_1$ and $G_2$ be functions $\mathbb R_{> 0}\to\mathbb R_{> 0}$ such that
$G_1(\beta)^{1/\beta}$ tends to some limit $e^{-x_1}$ as $\beta\to\infty$, and likewise $G_2(\beta)^{1/\beta}\to e^{-x_2}$. Then
$$F(G_1(\beta),G_2(\beta))^{1/\beta}\to \exp(- f(x_1,x_2)).$$
as $\beta\to+\infty$.</p>
<p><strong>Proof:</strong> The operations $\times$ and $/$ are easy - the only non-trivial step is
$$(G_1(\beta)+G_2(\beta))^{1/\beta}\to e^{-\beta \min(x_1,x_2)}.$$</p>
|
1,907,743 | <p>I'm having trouble with a step in a paper which I believe boils down to the following inequality:
$$
\left\| \sum_{k\in\mathbb{Z}} f(\cdot+k) \right\|_{L^2(0,1)}
\leq c \|f\|_{L^2(\mathbb{R})}.
$$
I haven't come up with many ideas. Hitting the left-hand side with Minkowski, for example, produces something which can exceed $\|f\|_{L^2(\mathbb{R})}$.</p>
<p>I also put a bit of effort this afternoon into falsifying the above inequality (it may be that I'm misunderstanding the omitted steps in the original paper). Begin with a power series $g(x)=\sum a_kx^k$ which has a local $L^2$ singularity. It's then not too hard to use this representation to construct $f$ satisfying
$$
\sum_{k\in\mathbb{Z}} f(x+k) = g(x).
$$
However, the few times I attempted this did not result in an $L^2(\mathbb{R})$ function.</p>
<p>Any help one way or the other is appreciated.</p>
| Community | -1 | <p>The inequality is not true: counterexample: </p>
<p>$$f(x) = \sum_{n=1}^\infty \frac 1n \chi_{[n, n+1]}.$$</p>
<p>Then $f\in L^2(\mathbb R)$, but </p>
<p>$$g(x):= \sum_{k\in \mathbb Z} f(\cdot + k) = \sum_{n=1}^\infty \frac 1n = \infty$$</p>
<p>is not in $L^2(0,1)$. </p>
|
2,603,799 | <p>Good morning, i need help with this exercise.</p>
<blockquote>
<p>Prove all tangent plane to the cone $x^2+y^2=z^2$ goes through the origin</p>
</blockquote>
<p><strong>My work:</strong></p>
<p>Let $f:\mathbb{R}^3\rightarrow\mathbb{R}$ defined by $f(x,y,z)=x^2+y^2-z^2$</p>
<p>Then,</p>
<p>$\nabla f(x,y,z)=(2x,2y,-2z)$</p>
<p>Let $(a,b,c)\in\mathbb{R}^3$ then
$\nabla f(a,b,c)=(2a,2b,-2c)$</p>
<p>By definition, the equation of the tangent plane is</p>
<p>\begin{eqnarray}
\langle(2a,2b,-2c),(x-a,y-b,z-c)\rangle &=& 2a(x-a)+2b(y-b)+2c(z-c)\\
&=&2ax-2a^2+2by-2b^2+2cz-2c^2 \\
&=&0
\end{eqnarray}</p>
<p>In this step i'm stuck, can someone help me?</p>
| Community | -1 | <p>Actually you want $(a, b, c)$ on the cone... so $a^2+b^2-c^2=0$...</p>
<p>Now the equation of the tangent plane is always satisfied by the origin $(0, 0, 0) $ </p>
<p>Note: you lost a minus sign in your equation (the $z $ -term). .. </p>
<p>Substitute $(x, y, z)=(0, 0, 0) $ to find that the origin satisfies the equation of the plane...</p>
|
2,590,068 | <p>$$\epsilon^\epsilon=?$$
Where $\epsilon^2=0$, $\epsilon\notin\mathbb R$.
There is a formula for exponentiation of dual numbers, namely:
$$(a+b\epsilon)^{c+d\epsilon}=a^c+\epsilon(bca^{c-1}+da^c\ln a)$$
However, this formula breaks down in multiple places for $\epsilon^\epsilon$, yielding many undefined expressions like $0^0$ and $\ln 0$. So, here's my question: what is $\epsilon^\epsilon$ equal to for <a href="https://en.wikipedia.org/wiki/Dual_number" rel="noreferrer">dual numbers</a>?</p>
| Anixx | 2,513 | <p>Since <span class="math-container">$f(\varepsilon)=f(0)+f'(0)\varepsilon$</span>,</p>
<p>formally we have</p>
<p><span class="math-container">$\varepsilon^\varepsilon=1+\varepsilon+\varepsilon\ln 0$</span></p>
<p>The quantity <span class="math-container">$\ln 0$</span> is infinite with negative sign. We could stop here, but if you want you can write it as a divergent integral:</p>
<p><span class="math-container">$\varepsilon^\varepsilon=1+\varepsilon-\varepsilon\int_0^1\frac{dx}x=1+\varepsilon-\varepsilon\gamma -\varepsilon\int_1^\infty\frac{dx}x$</span></p>
<p>I have no ideas where we can advance further.</p>
|
1,701,935 | <p>I've been experimenting with recursive sequences lately and I've come up with this problem:</p>
<blockquote>
<p>Let $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$
<br><br>Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$ </p>
</blockquote>
<p><br><br> I know of $3$ ways of examining whether a set is dense:
<br><br>$i)$The definition, that is, whether its closure is the set on which it is dense, in our case this means if: $\bar L=[-1,1]$.
<br><br>ii)$(\forall x \in [-1,1])(\forall \epsilon>0)(\exists b \in L):|x-b|<\epsilon$
<br><br>$iii)$ $(\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow
x$ </p>
<p>So far I haven't been able to use these to answer the question. I tried plugging in different values of $a_0$ and see where that leads but I have not found any corresponding promising "pattern" for $a_n$. Any ideas on how to approach this?</p>
| rtybase | 22,583 | <p>Ok, if you study recursive sequences, then you probably heard about <a href="https://books.google.co.uk/books?id=Bud_HAemxR8C&pg=PA318&lpg=PA318&dq=Lamere%20ladder&source=bl&ots=86YJxwbZFt&sig=HLADUXex5F9Y-ZQ_-XUet3KJ_-M&hl=en&sa=X&ei=KZVEVe_KHY7KaL6CgIgN&ved=0CCMQ6AEwAA#v=onepage&q=Lamere%20ladder&f=false" rel="nofollow">"Lamere Ladder"</a>.</p>
<p>According to Lamere Ladder for $\cos(x)$ (which is also a <a href="https://en.wikipedia.org/wiki/Banach_fixed-point_theorem" rel="nofollow">contraction</a> because of <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow">MVT</a>), this function has a <a href="https://www.wolframalpha.com/input/?i=cos(x)%3Dx" rel="nofollow">fixed</a> (stationary) point. So, regardless of $a_0$, $a_1$ will end in between $[-1, 1]$ and from there on, the sequence $\{a_n\}$ will tend to the fixed point of $\cos(x)$. Which makes $L$ a converging sequence, so $L$ can't be dense, because the only point satisfying ii) (in your question) is its limit.</p>
<p>On another note, <a href="http://mathworld.wolfram.com/KroneckersApproximationTheorem.html" rel="nofollow">Kronecker's approximation theorem</a> is quite an useful tool too. For example $\{n+ m \cdot 2 \cdot \pi \space | \space m,n \in \mathbb{Z} \}$ is dense on $\mathbb{R}$ and $cos(x)$ is a continuous function, making $\{cos(n)\}_{n \in \mathbb{Z}}$ dense on $[-1,1]$. </p>
|
3,319,629 | <p>The question is from a practice exam I am currently trying to do:
<a href="https://i.stack.imgur.com/VWQhs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VWQhs.png" alt="enter image description here"></a></p>
<p>I am really not sure how to go about this one. In essence, I'd imagine that the idea is to find the area of the greater curve, and then from that subtract the calculated area of the smaller one. How do I go about doing this? I figure you can do this with one iterated integral?</p>
<p>Further, if possible, as an aside to this question, I am also wondering how to find the area of just <strong>one</strong> of these curves with iterated integrals?</p>
<p>Thank you very much!</p>
| David G. Stork | 210,401 | <p><a href="https://i.stack.imgur.com/f9HL8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f9HL8.png" alt="enter image description here"></a></p>
<p><span class="math-container">$$\int\limits_{\theta = 0}^{2 \pi} \int\limits_{r = 2 + \sin (3 \theta)}^{4 - \cos (3 \theta)} r\ dr\ d\theta = 12 \pi$$</span></p>
|
2,033,485 | <p>I have an equilateral triangle with each point being a known distance of N units from the center of the triangle.</p>
<p>What formula would I need to use to determine the length of any side of the triangle?</p>
| Kunal Chawla | 309,858 | <p>First draw the lines from all three points to the center point of this triangle, then notice that two if these lines make an isosceles triangle with angles $2\pi/3$, $\pi/6$, and $\pi/6$. Since we're given the length of two of the sides of this triangle, and the angles are known, we can calculate the length of the third side </p>
<p>Let $a$ denote the length of the inradius, and $b$ the sidelength if the triangle. Then, by the cosine law, we have</p>
<p>$b^2 = a^2 + a^2 - 2aa\cos(2\pi/3)$</p>
<p>$= 2a^2(1 - \cos(2\pi/3)$</p>
<p>$= 2a^2(3/2) $</p>
<p>So we have $b = \sqrt(3a^2)$ or $\sqrt3a$</p>
|
3,853,351 | <p>Given an n-dimensional ellipsoid in <span class="math-container">$\mathbb{R}^n$</span>, is any orthogonal projection of it to a subspace also an ellipsoid? Here, an ellipsoid is defined as</p>
<p><span class="math-container">$$\Delta_{A, c}=\{x\in \Bbb R^n\,:\, x^TAx\le c\}$$</span></p>
<p>where <span class="math-container">$A$</span> is a symmetric positive definite n by n matrix, and <span class="math-container">$c > 0$</span>.</p>
<p>I'm just thinking about this because it gives a nice visual way to think about least-norm regression.</p>
<p>I note that SVD proves immediately that any linear image (not just an orthogonal projection) of an ellipsoid is also an ellipsoid, however there might be a more geometrically clever proof when the linear map is an orthogonal projection.</p>
| alphacapture | 334,625 | <p>Yes. An ellipsoid is a linear transformation of a spherical ball, and orthogonal projection is also a linear transformation, so it suffices to show that any linear transformation whose image is a subspace sends a spherical ball to an ellipsoid in that space.</p>
<p>A linear transformation can be decomposed into orthogonal projection by its kernel followed by some invertible linear transformation. Orthogonal projection sends a spherical ball to a spherical ball in the subspace, so we are done.</p>
|
2,674,853 | <blockquote>
<p>Suppose $E \subset \mathbb R^d$ has measure $0$ and $f: \mathbb R^d \longrightarrow \mathbb R$ is measurable. Does $f (E)$ necessarily have measure $0$?</p>
</blockquote>
<p>I tried to find a counter-example though I failed.It is clear that countable subset will not work for otherwise the image of it is again countable in $\mathbb R$ and hence has measure $0$. So we need to find an uncountable set of measure $0$ which maps to some set in $\mathbb R$ of positive measure under some measurable function in order to find a counter-example.How to find it?</p>
<p>Please help me in finding this example.Then it will really help me.</p>
<p>Thank you in advance.</p>
| Calvin Khor | 80,734 | <p>Suppose $d>1$. $f(x_1,\dots,x_d) := x_1$ is a measurable map $\mathbb R^d\to\mathbb R$. Furthermore, the set $$E := \{x\in \mathbb R^d \mid x_2 = x_3 = \dots =x_d = 0 \}$$ is null, but $f(E) = \mathbb R$.</p>
|
318,983 | <p>$$\int_{-\infty}^{\infty} \frac{x^2}{x^6+9}dx$$ I'm a bit puzzled as how to go about solving this integral. I can see that it isn't undefined on -infinity to infinity. But I just need maybe a hint on how to go about solving the problem.</p>
| André Nicolas | 6,312 | <p>It is natural to let $u=x^3$. But note that the substitution $x^3=3u$ is more efficient. </p>
|
318,983 | <p>$$\int_{-\infty}^{\infty} \frac{x^2}{x^6+9}dx$$ I'm a bit puzzled as how to go about solving this integral. I can see that it isn't undefined on -infinity to infinity. But I just need maybe a hint on how to go about solving the problem.</p>
| DonAntonio | 31,254 | <p>With complex analysis: define the complex function</p>
<p>$$f(z)=\frac{z^2}{z^6+9}\;\;,\;\;\text{with simple poles at}\,\,\,z_k:=\sqrt[6] 9\,\,e^{\frac{\pi i}{6}(1+2k)}\;\;,\;k=0,1,2,3,4,5$$</p>
<p>Note that the only poles on the upper half plane are the first three ones $\,z_0,\,z_1,\,z_2\,$ , with residues</p>
<p>$$Res_{z=z_0}(f)=Res_{z=z_2}(f)=-\frac{i}{18}\;\;,\;\;\;Res_{z=z_1}(f)=\frac{i}{18}$$</p>
<p>Choosing the contour</p>
<p>$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;|z|=R\,,\,\Im (z)\ge 0\}$$</p>
<p>we get, by Cauchy's Integral Theorem, that</p>
<p>$$(**)\;\;\;\;\;\;\;2\pi i\left(\sum_{k=0}^2 Res_{z=z_k}(f)\right)=\frac{\pi}{9}=\oint_{C_R}f(z)\,dz=\int\limits_{-R}^Rf(x)\,dx+\int\limits_{\gamma_R} f(z)\,dz$$</p>
<p>But</p>
<p>$$\left|\int\limits_{\gamma_R} f(z)\,dz\right|\le\max_{z\in \gamma_R}\frac{|z|^2}{|z^6+9|}\pi R\le\frac{\pi R^3}{R^6-9}\xrightarrow[R\to\infty]{}0$$</p>
<p>Thus passing to the limit in (**) above we get</p>
<p>$$\int\limits_{-\infty}^\infty\frac{x^2}{x^6+9}dx=\frac{\pi}{9}$$</p>
|
213,639 | <blockquote>
<p>Where is axiom of regularity actually used? Why is it important? Are there some proofs, which are substantially simpler thanks to this axiom?</p>
</blockquote>
<p>This question was to some extent provoked by <a href="https://math.stackexchange.com/users/3515/dan-christensen">Dan Christensen</a>'s <a href="https://math.stackexchange.com/questions/204865/axiom-of-regularity#comment467198_204866">comment</a>: <em>Would regularity ever be used in a formal development of, say, number theory or real analysis? I can't imagine it.</em></p>
<p>I have to admit that I do not know other use of this axiom than the proof that every set has <a href="http://en.wikipedia.org/wiki/Von_Neumann_universe" rel="noreferrer">rank</a> in cumulative hierarchy, and a few easy consequences of this axiom, which are mentioned in <a href="http://en.wikipedia.org/wiki/Axiom_of_regularity#Elementary_implications_of_regularity" rel="noreferrer">Wikipedia article</a>.</p>
<p>I remember seeing an introductory book in axiomatic set theory, which did not even mention this axiom. (And that book went through plenty of stuff, such as introducing ordinals, transfinite induction, construction of natural numbers.)</p>
<p>Wikipedia article
on <a href="http://en.wikipedia.org/wiki/Non-well-founded_set_theory" rel="noreferrer">Non-well-founded set theory</a> links to Metamath page for <a href="http://us.metamath.org/mpegif/axreg.html" rel="noreferrer">Axiom of regularity</a> and says: <em>Scroll to the bottom to see how few Metamath theorems invoke this axiom.</em></p>
<p>Based on the above, it seems that quite a lot of stuff can be done without this axiom.</p>
<p>Of course, it's quite possible that this axiom becomes important in some areas of set theory which are not familiar to me, such as forcing or working without Axiom of Choice. (It might be difficult to define cardinality without AC and regularity, as mentioned <a href="https://math.stackexchange.com/a/53771/">here</a>.) But even if the this axiom is important only for some advanced stuff - which I will probably never get to - I'd be glad to know that.</p>
| Asaf Karagila | 622 | <p><em>Basic</em> mathematics was done long before set theory. Its users couldn't care less whether or not ZF is the underlying theory of the universe or some other theory. As long as it works fine.</p>
<p>There are a few things to point out:</p>
<ol>
<li><p>The natural numbers (standard model of PA), on which classical number theory is developed, are well-ordered regardless to their construction. This is one of their defining properties, after all. Therefore all the inductions we use on them hold regardless to whether or not regularity holds in general.</p></li>
<li><p>Classical analysis can be developed, again, without regularity to hold. Most basic theorems (i.e. freshman calculus) are theorems about continuous functions or sequences of real numbers.</p></li>
<li><p>Even if the axiom of regularity fails, it still holds in an inner model (granted the other axioms of ZF hold). Therefore even without regularity we can still insist on working with sets for which regularity holds.</p></li>
</ol>
<p>The main idea behind having set theory as a foundational theory is that we can carry out the "non set-theoretic" constructions as sets. There is, at least in the basic level of mathematics, little to none interaction between the mathematics and the foundational theories. This is the same reason we don't see proofs assuming CH during our first two-three years of mathematics, these assumptions while interesting are irrelevant for basic mathematics.</p>
<p>Within set theory, the rank function and the Mostowski collapse are already enough reason for assuming it, and these are used a lot.</p>
|
3,536,061 | <p>Find the number of ways you can invite <span class="math-container">$3$</span> of your friends on <span class="math-container">$5$</span> consecutive days, exactly one friend a day, such that no friend is invited on more than two days. </p>
<p>My approach: Let <span class="math-container">$d_A,d_B$</span> and <span class="math-container">$d_C$</span> denote the total number of days <span class="math-container">$A, B$</span> and <span class="math-container">$C$</span> were invited respectively. According to the question we must have <span class="math-container">$0\le d_A,d_B,d_C\le 2.$</span> Also, we must have <span class="math-container">$$d_A+d_B+d_C=5.$$</span> </p>
<p>Now let <span class="math-container">$d_A+c_A=2, d_B+c_B=2, d_C+c_C=2,$</span> for some <span class="math-container">$c_A, c_B, c_C\ge 0$</span>. </p>
<p>This implies that <span class="math-container">$c_A+c_B+c_C=1$</span>. </p>
<p>Therefore the problem translates to finding the number of non-negative integer solutions to the equation <span class="math-container">$$c_A+c_B+c_C=1.$$</span> </p>
<p>By the stars and bars method the total number of required solutions is equal to <span class="math-container">$$\dbinom{1+3-1}{3-1}=3.$$</span></p>
<p>But the number of ways to invite the friends will be higher than this, since the friends are distinguishable and we have assumed them to be indistinguishable while applying the stars and bars method. </p>
<p>How to proceed after this?</p>
| WaveX | 323,744 | <p>The fact that the only way to achieve this is by inviting one friend over once and the other two twice will make this problem simpler.</p>
<p>How many ways are there to pick the one friend (from <span class="math-container">$3$</span>) that will be only visiting one day instead of two?</p>
<p>Now if we assume the days are Monday through Friday, how many ways are there to pick the day that the one-day friend will visit?</p>
<p>Finally, for the remaining four days, how many ways are there to pick two of them? This will be the number of ways to arrange the remaining two friends.</p>
<p>Now multiply the three of these numbers together for the final answer </p>
<hr>
<p>A side proof as to why we must have it with one friend one day and the other two with <span class="math-container">$2$</span>: We obviously can it have any friend arrive three days out of the five, as per given in the question. If all three come twice, this would require <span class="math-container">$6$</span> days so this isn't possible either. If two friends only come one day each, then we aren't using up all <span class="math-container">$5$</span> days.</p>
|
251,182 | <p>Is 13 a quadratic residue of 257? Note that 257 is prime.</p>
<p>I have tried doing it. My study guide says it is true. But I keep getting false. </p>
| Oscar Lanzi | 248,217 | <p>Quadratic reciprocity (QR) is the way to go, but a little ingenuity simplifies the solution.</p>
<p>Start with $(13|257)=(257|13)=(10|13)$. We can't directly apply QR again because $10$ is even. But, since $13$ is a prime one greater than a multiple of $4$, $(10|13)=(-10|13)=(3|13)$. Now we apply QR again: $(3|13)=(13|3)=(1|3)$.</p>
<p>So $(13|257)=(10|13)=(1|3)$. Through a great deal of arduous effort we find that $(1|3)=1$ so $(13|257)=1$ rendering $13$ a quadratic residue $\bmod 257$.</p>
|
1,006,562 | <p>So I am trying to figure out the limit</p>
<p>$$\lim_{x\to 0} \tan x \csc (2x)$$</p>
<p>I am not sure what action needs to be done to solve this and would appreciate any help to solving this. </p>
| Crostul | 160,300 | <p>Note that $\csc (2x) = \frac{1}{\sin(2x)} = \frac{1}{2 \sin x \cos x}$, and $\tan x = \frac{\sin x}{\cos x}$.</p>
<p>So $$\lim_{x \to 0} \tan x \csc (2x) = \lim_{x \to 0} \frac{1}{2 \sin x \cos x} \frac{\sin x}{\cos x} = \lim_{x \to 0} \frac{1}{2 \cos^2 x} = \frac{1}{2}$$</p>
|
2,428,243 | <p>How can I evalute this product??</p>
<p>$$\prod_{i=1}^{\infty} {(n^{-i})}^{n^{-i}}$$</p>
<p>Unfortunately, I have no idea.</p>
| neonpokharkar | 477,567 | <p>Taking log
$$\ln\Big( \prod_{i=1}^{\infty} (n^{-i})^{n^{-i}}\Big)$$$$$$
$$\sum_{i=1}^{\infty} \frac{i}{n^i} \ln n$$$$$$
$$\ln n \sum_{i=1}^{\infty} \frac{i}{n^i} $$$$$$</p>
<blockquote>
<p>Let$$$$$$S= \sum_{i=1}^{k} \frac{i}{n^i}$$ $$$$
$$S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\cdots+\frac{k}{n^k}$$$$$$
Multiplying by $\frac{1}{n}$$$$$
$$\frac{S}{n}=\frac{1}{n^2}+\frac{2}{n^3}+\frac{3}{n^4}+\cdots+\frac{k}{n^{k+1}}$$$$$$
Now $S-\frac{S}{n}$$$$$
$$\frac{n-1}{n} S =\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\cdots+\frac{1}{n^k}-\frac{k}{n^{k+1}}$$$$$$
Using GP formula,
$$\frac{n-1}{n} S =\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}$$$$$$
$$S=\frac{n}{n-1} \Biggr(\frac{1}{n}\frac{\Big(\frac{1}{n}\Big)^k-1}{\frac{1}{n}-1} -\frac{k}{n^{k+1}}\Biggr)$$$$$$
For $k=\infty$ if $n\gt 1$$$$$
$$S=\frac{n}{n-1}\frac{1}{n} \Big(\frac{n}{n-1}\Big)$$$$$$
$$S=n\Big(\frac{1}{n-1}\Big)^2$$$$$$</p>
</blockquote>
<p>Then substitute it back</p>
<blockquote class="spoiler">
<p>$$n \ln n \Big(\frac{1}{n-1}\Big)^2$$Taking antilog,$$n^{n\Big(\frac{1}{n-1}\Big)^2}$$</p>
</blockquote>
<p>This is only for $n\gt 1$</p>
|
3,265,403 | <p>While trying to compute the line integral along a path K on a function, I need to parametrize my path K in terms of a single variable, let's say this single variable will be <span class="math-container">$t$</span>.
My path is defined by the following ensemble: <span class="math-container">$$K=\{(x,y)\in(0,\infty)\times[-42,42]|x^2-y^2=1600\}$$</span> I know how to calculate the line integral, that is not my issue. My problem is to parametrize <span class="math-container">$x^2-y^2=1600$</span>. I tried using the identities: <span class="math-container">$$\sin^2(t)+\cos^2(t)=1$$</span> <span class="math-container">$$\sec^2(t)-\tan^2(t)=1$$</span>
But I did not get anywhere with my parametrization (see below for my poor try into parametrizing). I would welcome any help/hints and if you happen to know some good reading to learn more about parametrization, I am also interested. <span class="math-container">$$r(t)=1600\sec^2(t)-1600\tan^2(t)=1600$$</span>
for <span class="math-container">$$x=40\sec(t) \land y=40\tan(t)$$</span></p>
| TurlocTheRed | 397,318 | <p><span class="math-container">$$x^2-y^2=1600$$</span>
<span class="math-container">$$(\frac{x}{40})^2 - (\frac{y}{40})^2=1$$</span></p>
<p>Let <span class="math-container">$x=40\cdot(e^t+e^-t)/2=40\cdot\cosh{t}$</span>.</p>
<p>Let <span class="math-container">$y=40\cdot (e^t-e^{-t})/2=40 \cdot\sinh{t}$</span></p>
<p>Use positive values of t if you need positive values of <span class="math-container">$x$</span>. </p>
<p><span class="math-container">$d/dt(\cosh{t})=\sinh{t}$</span></p>
<p><span class="math-container">$d/dt(\sinh{t})=\cosh{t}$</span></p>
|
3,245,270 | <p>From Statistical Inference by Casella and Berger:</p>
<blockquote>
<p>Let <span class="math-container">$X_1, \dots X_n$</span> be a random sample from a discrete distribution
with <span class="math-container">$f_X(x_i) = p_i$</span>, where <span class="math-container">$x_1 \lt x_2 \lt \dots$</span> are the possible
values of <span class="math-container">$X$</span> in ascending order. Let <span class="math-container">$X_{(1)}, \dots, X_{(n)}$</span>
denote the order statistics from the sample. Define <span class="math-container">$Y_i$</span> as the number of <span class="math-container">$X_j$</span> that are less than or equal to
<span class="math-container">$x_i$</span>. Let <span class="math-container">$P_0 = 0, P_1 = p_1, \dots, P_i = p_1 + p_2 + \dots + p_i$</span>.</p>
</blockquote>
<p>If <span class="math-container">$\{X_j \le x_i\}$</span> is a "success" and <span class="math-container">$\{X_j \gt x_i\}$</span> is a "failure", then <span class="math-container">$Y_i$</span> is binomial with parameters <span class="math-container">$(n, P_i)$</span>.</p>
<p><strong>Then the event <span class="math-container">$\{X_{(j)} \le x_i\}$</span> is equivalent to the event <span class="math-container">$\{Y_i \ge j\}$</span></strong></p>
<p>Can someone explain why these two are equivalent?</p>
<p><span class="math-container">$\{X_{(j)} \le x_i\} = \{s \in \text{dom}(X_{(j)}) : X_{(j)}(s) \le x_i\}$</span></p>
<p><span class="math-container">$\{Y_i \ge j\} = \{s' \in \text{dom}(Y_i) : Y_i(s') \ge j\}$</span></p>
<p>I'm having trouble understanding how these random variable functions show this equivalence.</p>
| Henno Brandsma | 4,280 | <p>A pseudometric space is symmetric (also called <a href="https://en.wikipedia.org/wiki/T1_space" rel="nofollow noreferrer"><span class="math-container">$R_0$</span></a>): if <span class="math-container">$x \in \overline{\{y\}}$</span> then <span class="math-container">$y \in \overline{\{x\}}$</span> (basically because <span class="math-container">$d(x,y)=0$</span> implies <span class="math-container">$d(y,x)=0$</span> too, also in pseudometric spaces). </p>
<p>Sierpinski space (<span class="math-container">$X=\{0,1\}$</span> with topology <span class="math-container">$\{\emptyset,\{0\},X\}$</span>) is not symmetric so not pseudometrisable. (Because <span class="math-container">$1 \in \overline{\{0\}}$</span> but not the other way around). This is in a way the simplest example, certainly the smallest one.</p>
<p>If <span class="math-container">$X$</span> is <span class="math-container">$T_1$</span> then <span class="math-container">$X$</span> is metrisable iff <span class="math-container">$X$</span> is pseudometrisable. (the <span class="math-container">$T_1$</span> ensures that <span class="math-container">$X$</span> is also <span class="math-container">$R_0$</span> and so the non-existence of any points <span class="math-container">$x,y$</span> with <span class="math-container">$x \neq y$</span> but <span class="math-container">$d(x,y)=0$</span>. So the pseudometric for <span class="math-container">$X$</span> on the right is then a metric.)</p>
<p>So spaces like the cofinite topology on <span class="math-container">$\mathbb{N}$</span> is not pseudometrisable, as it's not metrisable (not Hausdorff to start with...)
Also, the Sorgenfrey line, the Michael line, double arrow space etc etc.</p>
|
3,743,673 | <p>Using calculus to find the minima:</p>
<p><span class="math-container">$$y(x) = x^x$$</span></p>
<p><span class="math-container">$$ln(y) = x*ln(x)$$</span></p>
<p><span class="math-container">$$(1/y)*\frac{dy}{dx} = ln(x) + x*\left(\frac{1}{x}\right) = ln(x) + 1$$</span></p>
<p><span class="math-container">$$\frac{dy}{dx} = y*(ln(x) + 1)$$</span></p>
<p><span class="math-container">$$\frac{dy}{dx} = (x^x)*(ln(x) + 1)$$</span></p>
<p>Though arriving at this next step, one can assume from looking at it graphically, that <span class="math-container">$x^x$</span> will never be <span class="math-container">$0$</span>, thus <span class="math-container">$(ln(x) + 1) = 0$</span>, however how can it be <strong>shown</strong> that <span class="math-container">$(x^x)$</span> is never <span class="math-container">$0$</span>, instead of making a bold assumption?</p>
<p><span class="math-container">$$0 = (x^x)*(ln(x) + 1)$$</span></p>
<p><span class="math-container">$$ln(x) = -1$$</span></p>
<p><span class="math-container">$$x = exp(-1) = \frac{1}{e}$$</span></p>
<p><span class="math-container">$$y = \left(\frac{1}{e}\right)^{\left(\frac{1}{e}\right)} ~= 0.6922$$</span></p>
| Arthur | 15,500 | <p>I don't think that's a bold assumption. For instance, <span class="math-container">$x^x=e^{x\ln x}$</span> is never zero, as <span class="math-container">$e$</span> raised to the power of any real number is strictly positive.</p>
<p>Alternatively, there are no strictly positive <span class="math-container">$a,b$</span> that makes <span class="math-container">$a^b$</span> zero, and restricting our attention to the special case of <span class="math-container">$a=b$</span> doesn't change that.</p>
<p>As for exactly what happens at <span class="math-container">$x=0$</span>, that's a matter of definitions, not a matter of calculation. I think <span class="math-container">$0^0=1$</span> is most sensible, but others may disagree.</p>
<p>The rest of your proof seems good.</p>
|
3,743,673 | <p>Using calculus to find the minima:</p>
<p><span class="math-container">$$y(x) = x^x$$</span></p>
<p><span class="math-container">$$ln(y) = x*ln(x)$$</span></p>
<p><span class="math-container">$$(1/y)*\frac{dy}{dx} = ln(x) + x*\left(\frac{1}{x}\right) = ln(x) + 1$$</span></p>
<p><span class="math-container">$$\frac{dy}{dx} = y*(ln(x) + 1)$$</span></p>
<p><span class="math-container">$$\frac{dy}{dx} = (x^x)*(ln(x) + 1)$$</span></p>
<p>Though arriving at this next step, one can assume from looking at it graphically, that <span class="math-container">$x^x$</span> will never be <span class="math-container">$0$</span>, thus <span class="math-container">$(ln(x) + 1) = 0$</span>, however how can it be <strong>shown</strong> that <span class="math-container">$(x^x)$</span> is never <span class="math-container">$0$</span>, instead of making a bold assumption?</p>
<p><span class="math-container">$$0 = (x^x)*(ln(x) + 1)$$</span></p>
<p><span class="math-container">$$ln(x) = -1$$</span></p>
<p><span class="math-container">$$x = exp(-1) = \frac{1}{e}$$</span></p>
<p><span class="math-container">$$y = \left(\frac{1}{e}\right)^{\left(\frac{1}{e}\right)} ~= 0.6922$$</span></p>
| Sameer Baheti | 567,070 | <blockquote>
<p>Show that <span class="math-container">$x^x = a$</span> has no real solutions when <span class="math-container">$a < \left(\frac1e\right)^{\frac1e}$</span>.</p>
</blockquote>
<p>Enough to visualize that</p>
<ul>
<li><span class="math-container">$\displaystyle\lim_{x\to 0}x^x=\displaystyle\lim_{x\to 0}e^{x\ln x}=e^{\displaystyle\lim_{x\to 0}x\ln x}=e^{\displaystyle\lim_{x\to 0}\frac{\ln x}{\frac1x}}=e^{\displaystyle\lim_{x\to 0}\frac{\frac1x}{\frac{-1}{x^2}}}=e^{-\displaystyle\lim_{x\to 0}x}=1$</span></li>
<li><span class="math-container">$\displaystyle\lim_{x\to +\infty}x^x\to+\infty$</span></li>
<li><span class="math-container">$f(\frac1e)=\left(\frac1e\right)^{\frac1e}$</span></li>
<li><span class="math-container">$$f^\prime(x)=x^x(1+\ln x)=
\begin{cases}<0&x\in(0,\frac1e)\\0&x=\frac1e\\>0&x\in(\frac1e,+\infty)
\end{cases}$$</span>
So, <span class="math-container">$f(x)=x^x$</span> is tangential to the line <span class="math-container">$y=\left(\frac1e\right)^{\frac1e}$</span>.</li>
</ul>
<p><a href="https://i.stack.imgur.com/UJTGf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UJTGf.png" alt="enter image description here" /></a></p>
|
1,085,511 | <p>What would be the irrational number $\dfrac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers given by this expression:
$$
\left(
\begin{array}{@{}c@{}}2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\dotsb}}}\end{array}
\right)^{1/8}
$$</p>
| Alex Silva | 172,564 | <p><strong>Hint:</strong></p>
<p>$$x = 2207 - \frac{1}{x},$$ where</p>
<p>$$x = 2207-\dfrac{1}{2207-\dfrac{1}{2207-\dfrac{1}{2207...}}}.$$</p>
|
555,446 | <p>Given this shape: <img src="https://i.stack.imgur.com/1rRsC.png" alt="diagram showing a 4000 unit wide cyan square with a 400 unit wide red square in the middle"></p>
<h1>Is it possible to divide the cyan area into 5 equal area shapes</h1>
<p>such that:</p>
<ol>
<li>Each shape is the same</li>
<li>Each shape has an edge touching the red square</li>
<li>Each shape has an edge touching the outside.</li>
<li>No diagonal lines.</li>
</ol>
<p>Its reasonably easy to perform #2 and #3 as long as you violate #1</p>
<p>And I don't believe its possible to actually satisfy #1, and the very fact its contained within a square suggests #1 is not satisfiable given the presence of #2.</p>
<p>Though its just a bit of fun really =).</p>
<p>Context: Was simply devising a town plan for a guild oriented town for a minecraft world, and it became possible that we might want 5 guilds, and the fun of giving each guild a fair equal area region , in conjunction with all 5 guilds having a shared space in the middle.</p>
<p>Diagonals are unwanted as it makes dividing the land fairly and applying region controls onerous, as most things are rectangular, regions included.</p>
<p>So while you can roughly approximate a diagonal with sufficiently many rectangles, the less steps, the better.</p>
<h1>If it is not possible</h1>
<p>Please provide reasoning as to why not.</p>
<h1>Additionally</h1>
<p>It would be interesting to see what sort of alternatives people can come up with, perhaps there is an optimal shape that results in all the shapes being highly similar geometrically, despite not being identical.</p>
| RicardoCruz | 36,340 | <p>What about the shape below made in geogebra?
<img src="https://i.stack.imgur.com/qoYrh.jpg" alt="enter image description here"></p>
|
2,713,873 | <p>We know that if a real valued function $f$ is continuous over an interval $[a,b]$ then the following integral $$\int_a^bf(x)dx$$ represents the area between horizontally the line $y=0$ and the curve of $f$, vertically between the lines $x=a$ and $x=b$. So what represent the following $$\int_{[a,b]\times [c,d]}g(x,y)dxdy$$ and $$\int_{[a,b]\times [c,d]\times [e,f]}h(x,y,z)dxdydz$$ where $g$ and $h$ are two continuous real valued 2d and 3d functions.
Thanks</p>
| HBR | 396,575 | <p>The double integral, represents the sum of the infinite areas under the curves $g(x,y)$ with $x=constant$ defined as:
$$A(y)=\int{g(x,y)\,dx}$$
each one between $y$ and $y+dy$, which gives you a volume. (see <a href="https://en.m.wikipedia.org/wiki/Fubini%27s_theorem" rel="nofollow noreferrer">Fubini</a> 's explanation)</p>
<p>The last integral gives you the infinite sum of volumes $V(z)$ defined as
$$V(z)=\int\int{h(x,y,z)\,dxdy}$$
which gives you a hypervolume.</p>
<p>Imagine that if the function $h$ is a sphere the number $\int V(z) \,dz$ will be the sum of all the volumes of the infinite spheres of radius $z$ between the integration interval $[z_1,z_2]$</p>
|
4,310,003 | <p>Suppose you have a non empty set <span class="math-container">$X$</span>, and suppose that for every function <span class="math-container">$f : X \rightarrow X$</span>, if <span class="math-container">$f$</span> is surjective, then it is also injective. Does it necessarily follow that <span class="math-container">$X$</span> is finite ?</p>
<p>Every example I've been able to think of leads me to believe this is true. Is it ? Or could anyone provide a counterexample?</p>
| 2 is even prime | 994,639 | <p><span class="math-container">$$\int_{-\infty}^\infty \frac{e^x}{1 + e^{4x}}dx$$</span>
Split the integral at <span class="math-container">$0$</span> and bring them between same bounds i.e <span class="math-container">$[0,\infty]$</span>.
<span class="math-container">$$\int_{0}^{\infty}\frac{e^{3x}+e^{x}}{e^{4x}+1} dx$$</span>
Substitute <span class="math-container">$u=e^{x}\implies dx=\frac{du}{u}$</span>.
<span class="math-container">$$\int_{1}^{\infty}\frac{u^2+1}{u^{4}+1} du$$</span>
Substitute <span class="math-container">$k=u^2\implies du=\frac{dk}{2\sqrt{k}}$</span>.
<span class="math-container">$$\frac{1}{2}\int_{1}^{\infty}\frac{\sqrt{k}+\frac{1}{\sqrt{k}}}{k^2+1} dk$$</span>
Put, <span class="math-container">$ k=\frac{1}{t} \implies dk=\frac{-dt}{t^2} $</span>
<span class="math-container">$$\int_{0}^{1}\frac{\sqrt{t}+\frac{1}{\sqrt{t}}}{t^{2}+1} dt$$</span>
This step allows us to use formula for infinite geometric sums but the problem is at the boundary condition. To solve that take a limit to <span class="math-container">$1$</span> from negative as follows,
<span class="math-container">$$\lim_{a\to 1^{-}} \int_{0}^{a}\frac{\sqrt{t}+\frac{1}{\sqrt{t}}}{t^{2}+1} dt$$</span>
Write, <span class="math-container">$\frac{1}{t^2+1}=\sum_{n\ge 0}t^{2n}$</span>.
<span class="math-container">$$\lim_{a\to 1^{-}} \sum_{n\ge 0}\int_{0}^{a} t^{2n-\frac{1}{2}}+t^{2n+\frac{1}{2}} dt$$</span>
I think you can proceed from here.</p>
|
151,430 | <p>Let $Y\subset X$ be a codimension $k$ proper inclusion of submanifolds. If we choose a coorientation of $Y$ inside of $X$ (that is, an orientation of the normal bundle), then we get a class $[Y]\in H^k(X)$. If $X$ and $Y$ are oriented, then $[Y]$ may be defined as the fundamental class of $Y$ in the Borel-Moore homology of $X$, which is isomorphic to the cohomology of $X$. What is the simplest definition of $[Y]$ in the general case (where $X$ and $Y$ are not necessarily oriented)?</p>
<p>Note that a simple generalization of this question would be to ask how to define the pushforward in cohomology along a proper oriented map. (Then $[Y]$ would simply be the pushforward of $1\in H^0(Y)$.) I would be happy to know the answer to this more general question, but I asked the simpler version to be as concrete as possible.</p>
| Dan Petersen | 1,310 | <p>The easiest definition is via the Pontrjagin--Thom construction (I think). Let $N$ be a tubular neighbourhood of $Y$, isomorphic to the normal bundle. Let $X'$ be the space obtained from $X$ by collapsing the complement of $N$ to a point. Then $X'$ is isomorphic to the Thom space of the normal bundle, and if the normal bundle is oriented then there is the Thom isomorphism:
$$ H^\bullet(Y,\mathbf Z) \cong H^{\bullet+\dim X - \dim Y}(X',\mathbf Z).$$
Composing with the natural $H^\bullet(X',\mathbf Z) \to H^\bullet(X, \mathbf Z)$ gives the map you want.</p>
|
151,430 | <p>Let $Y\subset X$ be a codimension $k$ proper inclusion of submanifolds. If we choose a coorientation of $Y$ inside of $X$ (that is, an orientation of the normal bundle), then we get a class $[Y]\in H^k(X)$. If $X$ and $Y$ are oriented, then $[Y]$ may be defined as the fundamental class of $Y$ in the Borel-Moore homology of $X$, which is isomorphic to the cohomology of $X$. What is the simplest definition of $[Y]$ in the general case (where $X$ and $Y$ are not necessarily oriented)?</p>
<p>Note that a simple generalization of this question would be to ask how to define the pushforward in cohomology along a proper oriented map. (Then $[Y]$ would simply be the pushforward of $1\in H^0(Y)$.) I would be happy to know the answer to this more general question, but I asked the simpler version to be as concrete as possible.</p>
| John Pardon | 35,353 | <p>Your construction for the oriented case extends to the unoriented case just by using twisted coefficients.</p>
<p>Every manifold $Y$ has a fundamental class $[Y]\in H_\ast^{\mathrm{lf}}(Y;\mathfrak o_Y)$ (where $\mathfrak o_Y$ is the orientation sheaf of $Y$).
The coorientation of $Y\subseteq X$ gives an identification of orientation sheaves $\mathfrak o_Y\xrightarrow\sim\mathfrak o_X$ over $Y$, and thus there is a pushforward map:
$$H_\ast^{\mathrm{lf}}(Y;\mathfrak o_Y)\to H_\ast^{\mathrm{lf}}(Y;\mathfrak o_X)\to H_\ast^{\mathrm{lf}}(X;\mathfrak o_X)$$
Now use the Poincare duality isomorphism $H^\ast(X;\mathbb Z)\xrightarrow{\cap[X]}H_\ast^{\mathrm{lf}}(X;\mathfrak o_X)$.</p>
|
1,787,806 | <p>I've recently had this problem in an exam and couldn't solve it.</p>
<p>Find the remainder of the following sum when dividing by 7 and determine if the quotient is even or odd:</p>
<p>$$\sum_{i=0}^{99} 2^{i^2}$$</p>
<p>I know the basic modular arithmetic properties but this escapes my capabilities. In our algebra course we've seen congruence, divisibility, division algorithm... how could I approach it?</p>
| user90369 | 332,823 | <p>E.g. use </p>
<p>$\sum\limits_{i=0}^{99}2^{i^2}=\sum\limits_{i=0}^{32}(2^{(3i)^2}+2^{(3i+1)^2}+2^{(3i+2)^2})+2^{3(3(33^2))}$ </p>
<p>and $2^{3n}=(7+1)^n$ divided by $7$ has the remainder $1$ for any integer $n$. </p>
<p>The rest should be clear.</p>
|
4,064,684 | <p>I'm working through some notes for my signal processing class and there's something elementary that baffles me. We spent lots of hours and dozens of pages setting up the entire theory of Hilbert spaces in order to define the Fourier series of a square integrable periodic function in terms of the orthogonal basis of exponentials <span class="math-container">$ e_n(t):[0,2\pi] \to \mathbb{C}: t \to e^{int} $</span>.</p>
<p>Then all of the sudden, out of the blue, the notes assault me with a seemingly unrelated theorem about pointwise convergence of the series, and I find out that pointwise convergence isn't guaranteed by <span class="math-container">$L^2$</span>-convergence. So my (probably naive) question is: what was all that work good for? What use is <span class="math-container">$L^2$</span>-convergence if it doesn't guarantee pointwise convergence?</p>
| paul garrett | 12,291 | <p>A good question! First, even though <span class="math-container">$L^2$</span> convergence does not imply pointwise convergence, Parseval's theorem says that the Fourier series converges in that sense, whether or not pointwise. The standard elementary curriculum does seem to urge us to think that the only reasonable notion of convergence of a sequence or infinite sum of functions is pointwise, but that is toooo naive/optimistic.</p>
<p>So, one point is that various manipulations of Fourier series do indeed give correct results with or without pointwise convergence.</p>
<p>"Not all is lost", also, because "Sobolev space theory" is an <span class="math-container">$L^2$</span>-based extension of the basic <span class="math-container">$L^2$</span> theory which <em>does</em> capture pointwise convergence, if one wants.</p>
<p>But/and a very useful technical virtue of <span class="math-container">$L^2$</span> is that it is a Hilbert space, in which a (true) "Dirichlet/minimum" principle holds (as opposed to Banach spaces, such as <span class="math-container">$C^o[0,1]$</span>, which might seem more natural, but in which the Dirichlet/minimum principle fails violently...) Thus, the <span class="math-container">$L^2$</span>-Sobolev theory can recover pointwise features (continuity, for example), within a Hilbert space context. A good thing.</p>
<p>But/and, again, yes, it was a shock to me years ago when it finally got through to me that <span class="math-container">$L^2$</span> convergence and pointwise convergence (e.g., of Fourier series) were not literally/immediately/provably the same thing. Crazy! Who knew? :)</p>
<p>EDIT: while we're here, as @leslietownes rightly comments, there <em>are</em> results about getting pointwise convergence from <span class="math-container">$L^2$</span>, namely, Lennart Carleson's result that the Fourier series of an <span class="math-container">$L^2$</span> function does converge to it almost everywhere. This is a very difficult theorem.</p>
<p>And, while we're here, Kolmogorov constructed a rather tricky <span class="math-container">$L^1$</span> function whose Fourier series diverges everywhere. His example is so crazy that it is not possible to succinctly describe it, and I can't even remember the key mechanism. Something number theoretic, in the vein of Diophantine approximation, if I remember at all correctly... Fun stuff, but not easy.</p>
|
4,064,684 | <p>I'm working through some notes for my signal processing class and there's something elementary that baffles me. We spent lots of hours and dozens of pages setting up the entire theory of Hilbert spaces in order to define the Fourier series of a square integrable periodic function in terms of the orthogonal basis of exponentials <span class="math-container">$ e_n(t):[0,2\pi] \to \mathbb{C}: t \to e^{int} $</span>.</p>
<p>Then all of the sudden, out of the blue, the notes assault me with a seemingly unrelated theorem about pointwise convergence of the series, and I find out that pointwise convergence isn't guaranteed by <span class="math-container">$L^2$</span>-convergence. So my (probably naive) question is: what was all that work good for? What use is <span class="math-container">$L^2$</span>-convergence if it doesn't guarantee pointwise convergence?</p>
| Alex R. | 22,064 | <p>In addition to paul's answer, one thing to keep in mind is that Hilbert spaces are typically meaningless when it comes to function values at individual points. An often glanced over fact is that if you take a function <span class="math-container">$f\in L^2$</span> and modify it at a single point <span class="math-container">$x_0$</span> calling the result <span class="math-container">$g$</span>, then <span class="math-container">$\|f-g\|_2=0$</span>. More generally, the elements of <span class="math-container">$L^2$</span> are <em>representatives</em> of the equivalence relation <span class="math-container">$f\equiv g$</span> defined by <span class="math-container">$\|f-g\|_2=0$</span>. This can also be expressed in terms of measure theory, in that <span class="math-container">$\|f-g\|_2=0$</span> iff <span class="math-container">$f$</span> and <span class="math-container">$g$</span> differ on a set of measure 0. So, when speaking about convergence in these spaces, typically the best you can do is almost-everywhere convergence, and not point-wise.</p>
|
2,637,914 | <p>I would like to teach students about the pertinence of the Axiom of Infinity. Are there any high school-level theorems of arithmetic, algebra, or calculus, whose proof depends on the Axiom of Infinity? If there are no such examples, what would be the simplest theorem which demands the Axiom of Infinity?</p>
<p>It seems we can still generate endless numbers without the Axiom of Infinity, but this axiom lets us treat infinite sets as a whole -- is this true?</p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>The base case $n=1$ is clear.</p>
<p>For the induction case, start with what you know is true, whch is the statement for $n$ : $\sum_{i=1}^n \frac{i}{i+1} \leq \frac{n^2}{n+1}$. If we show that each time the left hand side's increase is slower than the right hand side then we are done.</p>
<p>When we put $n+1$ in place of $n$, the left hand side increases by $\frac{n+1}{n+2} = 1 - \frac{1}{n+2}$. The right hand side, on the other hand, increases by $\frac{(n+1)^2}{n+2} - \frac {n^2}{n+1} = 1-\frac{1}{(n+1)(n+2)}$.</p>
<p>Now, clearly, $\frac{1}{n+2} > \frac{1}{(n+1)(n+2)}$. Therefore, the right hand side has increased more than the left hand side. Adding this to the inequality given by the case for $n$
gives you the proof.</p>
|
1,896,024 | <p><span class="math-container">$f(n) = 2n^2 + n$</span></p>
<p><span class="math-container">$g(n) = O(n^2)$</span></p>
<p>The question is to find the mistake in the following process:</p>
<blockquote>
<p><span class="math-container">$f(n) = O(n^2) + O(n)$</span></p>
<p><span class="math-container">$f(n) - g(n) = O(n^2) + O(n) - O(n^2)$</span></p>
<p><span class="math-container">$f(n)-g(n) = O(n)$</span></p>
</blockquote>
<p>From how I understand it, Big-Oh represents the upper bound on the number of operations (when <span class="math-container">$n$</span> tends to very large value). So, the difference between an order of <span class="math-container">$n^2$</span> minus an order of <span class="math-container">$n^2$</span> should be negligible if <span class="math-container">$n$</span> is very large.</p>
<p>But the individual steps seem correct. It seems to me that the mistake is that when doing the minus with large values, the <span class="math-container">$O(n)$</span> will also get consumed.</p>
<p>I need clarification on whether I'm correct. If I'm not, then where is the mistake?</p>
| Paolo Leonetti | 45,736 | <p>When you write $O(h(x))$ (as $x \to \infty$, for some function $h$) you have to imagine that it is written "something (in modulo) which is at most $h(x)$ (up to a constant)", giving you an upper bound for the growth of your function.</p>
<p>This is why, if you write $O(h(x))-O(h(x))$, then it does <em>not</em> simplify to $0$, but it is still $O(h(x))$. You may obtain something "smaller" than $O(h(x))$ only if you have some additional information on what is your first $O(h(x))$ and your second $O(h(x))$.</p>
|
47,792 | <p>I think this may be a silly question, but here goes. Let $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\overline{\mathbf{F}_p})$ be a representation; say $\rho$ <em>is of S-type</em> if it is continuous, unramified almost everywhere, and the determinant of complex conjugation is $-1$. Serre's conjecture, now a theorem of Khare-Wintenberger, states that every $\rho$ of S-type arises from some modular form $f=\sum a_n e(nz)$ in the sense that $\mathrm{tr}\rho(\mathrm{Frob}_l)=a_l\;( \mathrm{mod}\;p)$ for (almost all) primes $l$.</p>
<blockquote>
<p>Question: Are S-type representations $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\mathbf{Z}/p^n\mathbf{Z})$ for $n\geq 2$ also expected/known to be modular? </p>
</blockquote>
| Emerton | 2,874 | <p>In what sense? If you mean "come from the reduction of $\rho_f$ for some Hecke eigenform $f$'', no, they are not.</p>
<p>If you mean "come from the reduction of $\rho$ where $\rho:G_{\mathbb Q} \to
GL_2(\mathbb T)$ is the Galois rep'n attached to the Hecke algebra $\mathbb T$ acting on
modular forms of some sufficiently large level, then the answer is known to be yes in most
cases (i.e. with comparitively minor technical restrictions on $\rho$). This is the content
of so-called big $R = $ big $\mathbb T$ theorems, due to Gouvea--Mazur, Boeckle, and others
(combined with Serre's conjecture to know that $\overline{\rho}$ is modular).</p>
|
47,792 | <p>I think this may be a silly question, but here goes. Let $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\overline{\mathbf{F}_p})$ be a representation; say $\rho$ <em>is of S-type</em> if it is continuous, unramified almost everywhere, and the determinant of complex conjugation is $-1$. Serre's conjecture, now a theorem of Khare-Wintenberger, states that every $\rho$ of S-type arises from some modular form $f=\sum a_n e(nz)$ in the sense that $\mathrm{tr}\rho(\mathrm{Frob}_l)=a_l\;( \mathrm{mod}\;p)$ for (almost all) primes $l$.</p>
<blockquote>
<p>Question: Are S-type representations $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\mathbf{Z}/p^n\mathbf{Z})$ for $n\geq 2$ also expected/known to be modular? </p>
</blockquote>
| Community | -1 | <p>To elaborate on Emerton's answer, an arbitrary (finitely ramified and odd) representation <span class="math-container">$\rho_n:G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{Z}/p^n\mathbb{Z})$</span> can't lift to one coming from an eigenform because of some reasons which may be explained locally. Let <span class="math-container">$\bar{\rho}$</span> denote the mod <span class="math-container">$p$</span> representation and <span class="math-container">$Ad^0\bar{\rho}$</span> the <span class="math-container">$\mathbb{F}_p[G_{\mathbb{Q}}]$</span>-module of trace zero <span class="math-container">$2\times 2$</span> matrices over <span class="math-container">$\mathbb{F}_p$</span> equipped with the adjoint action, i.e.<span class="math-container">$g\in G_{\mathbb{Q}}$</span> acts on a matrix <span class="math-container">$A$</span> by conjugation <span class="math-container">$g\cdot A:=\bar{\rho}(g) A\bar{\rho}(g)^{-1}$</span>. Let <span class="math-container">$l$</span> be a prime at which <span class="math-container">$H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})\neq 0$</span> (here <span class="math-container">$G_{\mathbb{Q}_l}$</span> is a decomposition subgroup at <span class="math-container">$l$</span>). The obstruction-class associated with <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span> is a cohomology class <span class="math-container">$O_l({\rho_n}_{\restriction G_{\mathbb{Q}_l}})\in H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})$</span> which is non-zero if there is no lift of the local representation <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span> to <span class="math-container">$GL_2(\mathbb{Z}/p^{n+1}\mathbb{Z})$</span>. This local obstructedness phenomenon is an issue when <span class="math-container">$n\geq 2$</span> though not so for <span class="math-container">$n=1$</span>. This can be further explained as follows, at each prime <span class="math-container">$l$</span> there is a smooth subscheme in the scheme parametrizing local deformations of <span class="math-container">$\bar{\rho}_{\restriction G_{\mathbb{Q}_l}}$</span>, if for <span class="math-container">$n\geq 2$</span> at some prime <span class="math-container">$l$</span> where <span class="math-container">$H^2(G_{\mathbb{Q}_l}, Ad^0\bar{\rho})\neq 0$</span> it is possible that <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span> does not lie on this smooth subscheme in the space of all deformations of <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span>. In this case, <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span> cannot lift one more step (and thus the global representation <span class="math-container">$\rho_n$</span> will not lift one more step either). The construction of these smooth schemes in the spaces of local deformations (dubbed smooth/liftable local deformation conditions) is carried out in the paper of Ramakrishna: "Deforming Galois representations and the Conjectures of Serre and Fontaine-Mazur" and at the prime <span class="math-container">$p$</span> this was previously carried out in his paper "On a Variation of Mazur's Deformation Functor".</p>
<p>There is now another issue, one does require that the lift to characteristic zero must satisfy a <span class="math-container">$p$</span> adic Hodge theoretic condition at <span class="math-container">$p$</span>, this can be done when <span class="math-container">$n=1$</span> but there can be issues when <span class="math-container">$n\geq 2$</span>.</p>
<p>It would be of interest to know if in case the local representations <span class="math-container">${\rho_n}_{\restriction G_{\mathbb{Q}_l}}$</span> are all unobstructed (i.e. their obstruction classes are trivial, i.e. they lift to <span class="math-container">$GL_2(\mathbb{Z}/p^{n+1}\mathbb{Z})$</span>)) at the finitely many primes at which it is unramified, and if <span class="math-container">$\rho_{\restriction G_{\mathbb{Q}_p}}$</span> satisfies a further condition, then if indeed it lifts to the representation associated to a cuspidal eigenform/ big Hecke algebra?</p>
|
1,743,935 | <p>Not sure if I have done this correctly, seems too straight forward, any help is very appreciated. </p>
<blockquote>
<p>QUESTION:<br>
Find the real and imaginary parts of $f(z) = \cos(z)$.</p>
</blockquote>
<p>ATTEMPT:<br>
$\cos(z) = \cos(x+iy) = \cos x\cos(iy) − \sin x\sin(iy) =
\cos x\cosh y − i\sin x\sinh y$</p>
<p>Is that correct? </p>
| Brian | 331,755 | <p>It is simple, but tis the beauty of the trig/exponential functions! You're $correct$!</p>
|
1,061,077 | <p>SO, I drop a piece of bread and jam repeatedly. It lands either jam face-up or jam face-down and I know that jam side down probability is <span class="math-container">$P(Down)=p$</span></p>
<p>I continue to drop the bread until it falls jam side up for the first time. What is the expression for the expected number of drops? By considering the binomial expansion of <span class="math-container">$(1-p)^{-2}$</span>, can you that the expected number is <span class="math-container">$1/(1-p)$</span>?</p>
<p>Also, please include a rough sketch of the probability distribution function for the number of times it falls jam side down for N drops, where N is large</p>
<h3>My Progress</h3>
<p>The probability that jam side is down for n drops: <span class="math-container">$p^n$</span></p>
<p>The probability that jam side is up for the first time at the <span class="math-container">$n^{th}$</span> drop is: <span class="math-container">$p^{n-1}(1-p)$</span></p>
<p>How do I continue from there?</p>
<p>Thank you very much! :)</p>
| Frank Mayer | 197,167 | <p>Average number of throws to get first up results = $\Sigma N Q_{N}$ where $Q_{N}$ is the probability of getting first $N-1$ throws down, then an up. </p>
<p>$Q_{N} = p^{N-1} (1 - p)$ So $\Sigma N Q_{N} = (1 - p) \Sigma Np^{N-1} = \frac{1 - p}{(1 - p)^2}= \frac{1}{(1 - p)}$</p>
<p>The probability distribution function for getting N consecutive ups from the start is $p^N$. For large $N$ this drops swiftly with p.</p>
|
3,189,173 | <p>What will be the remainder when <span class="math-container">$2^{87} -1$</span> is divided by <span class="math-container">$89$</span>?</p>
<p>I tried it solving by Euler's remainder theorem by separating terms:</p>
<p><span class="math-container">$$ \frac {2^{87}}{89} - \frac{1}{89}$$</span></p>
<p><span class="math-container">$\phi (89) =88 $</span></p>
<p>remainder <span class="math-container">$\dfrac{{87}}{88} = 87;$</span></p>
<p>this led me to the point from where I started.</p>
| J. W. Tanner | 615,567 | <p><span class="math-container">$89$</span> is prime, so by Fermat's little theorem <span class="math-container">$2^{88}\equiv 1\pmod {89},$</span> so <span class="math-container">$2^{87}\equiv 2^{-1} \pmod {89}$</span>.</p>
<p>Now <span class="math-container">$45\times 2 = 90 \equiv 1 \pmod {89},$</span> so this means <span class="math-container">$2^{87}\equiv 45 \pmod {89},$</span> so <span class="math-container">$2^{87}-1\equiv44 \pmod {89}$</span>.</p>
|
3,073,832 | <p>I need to understand the meaning of this mathematical concept: "undecided/undecidable". </p>
<p>I know what it means in the English dictionary. But, I don't know what it means mathematically.</p>
<p>If You answer this question with possible mathematical examples, it will be very helpful to understand this issue.</p>
<p>Thank you very much!</p>
| user247327 | 247,327 | <p>Saying that a statement is "undecidable" means that there can be no proof, even theoretically, that the true nor can there be a proof that it is false.</p>
|
2,913,017 | <p>Imagine I have a real random variable $X$ with some distribution (continuous, discrete or continuous with atoms)</p>
<p>Now Imagine I have i.i.d. copies $X_1,...,X_n$, all independently and equally distributed as $X$</p>
<p>My claim is:</p>
<p>$$\mathbb{P}(X_2>X_1)=\mathbb{P}(X_2<X_1)$$
My secondy claim is the following:If I order them by size, so that $X_{(1)}<X_{(2)}<\ldots<X_{(n)}$ and I define the interval $I_n=[X_{(1)},X_{(n)}]$; Then I claim:</p>
<p>$$\mathbb{P}(X_{n+1}<X_{(1)})=\mathbb{P}(X_{n+1}>X_{(n)})$$</p>
<p>So the probability that the $n+1$-th number exceeds the interval on the left equals the probability it exceeds on the right</p>
<p>I guess the first one is true, but the second one not;</p>
<p>E.g. Assume X can take the value 0 and 1; and assume $n=3$; Then</p>
<p>$$\mathbb{P}(X_3>{X_1,X_2})=\mathbb P (X_3=1)\mathbb P (X_2=0)\mathbb P (X_1=0)=\mathbb P (X=1)\mathbb P (X=0)\mathbb P (X=0)$$
but also
$$\mathbb{P}(X_3<{X_1,X_2})=\mathbb P (X_3=0)\mathbb P (X_2=1)\mathbb P (X_1=1)=\mathbb P (X=0)\mathbb P (X=1)\mathbb P (X=1)$$</p>
<p>which is gernerally not the same; But What I am wondering is if there are simply conditions that it would become true</p>
| user97678 | 591,449 | <p>You're asking whether $P(X > \max({X_1, \ldots, X_n})) = P(X < \min({X_1, \ldots, X_n}))$, where $X_1, \ldots, X_n$ and $X$ are all independent and distributed the same.</p>
<p>This is true in some distributions and false in others. For example if the $X_i$s are sampled uniformly at random from $[0,1]$ then the probabilities would be the same (due to symmetry).</p>
<p>In general, for any distribution over $[a,b]$ which is symmetric about $(a+b)/2$, you would expect this to be true.</p>
<p>Also, I believe this is true for any continuous distribution over $(a,b)$ (i.e. such that the probability of any two variables receiving exactly the same value is 0). To see this, note that $P(X > \max({X_1, \ldots, X_n})) = \frac{1}{n+1} = P(X < \min({X_1, \ldots, X_n}))$. This is because we can first generate $X, X_1, \ldots, X_n$, then order $X_1, \ldots, X_n$, without affecting the probabilities, and any of these are equally likely to be the maximum or the minimum.</p>
<p>To see when this could be false, suppose that the $X_i$s are chosen as follows: with probability $1/2$, set $X_i = 10$, otherwise sample $X_i$ uniformly from $[0,1]$. Then $P(X > \max({X_1, \ldots, X_n}))$ will approach $0$ exponentially fast, since it is impossible for $P(X > 10)$ to happen, but $P(X < \min({X_1, \ldots, X_n}))$ is roughly proportional to $1/n$. </p>
|
3,262,714 | <p>So, I need an exponential function on the form <span class="math-container">$e^{-ax}$</span> that is 1 at <span class="math-container">$x=0$</span> and approaches <span class="math-container">$0.3$</span> as <span class="math-container">$x \rightarrow \infty$</span>. I tried doing <span class="math-container">$e^{-ax} + 0.3$</span>, but that only lead to the function starting at <span class="math-container">$1.3$</span> (although it did approach <span class="math-container">$0.3$</span> as <span class="math-container">$x \rightarrow \infty$</span>)</p>
<p>The answer is probably really simple but I can't seem to figure it out.</p>
| Community | -1 | <p>Multiply <span class="math-container">$e^{-ax}$</span> by <span class="math-container">$0.7$</span>. So, consider <span class="math-container">$0.7e^{-ax}+ 0.3$</span>.</p>
|
3,464,615 | <p>A novel process of manufacturing laptop screens is under test. In recent tests, it is found that 75% of the screens are acceptable. What is the most probable number of acceptable screens in the next batch of 10 screens and what is the probability?</p>
<p>Does that mean 7 screens out of 10 will pass with a probability of 0.75?
I was skeptical about using binomial or geometric probability law because there is a huge difference in each calculation.
Any help would be appreciated!</p>
| David G. Stork | 210,401 | <p>Here's the relevant binomial distribution:</p>
<p><a href="https://i.stack.imgur.com/qyWR1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qyWR1.png" alt="enter image description here"></a></p>
<p>If you're forced to choose an integer, then <span class="math-container">$n=8$</span>, and <span class="math-container">$p = 0.2815684$</span>.</p>
|
416,153 | <p>Show that $x^2+y^2=p$ has a solution in $\mathbb{Z}$ if and only if $ p≡1 \mod 4$. Thnx, if someone can help</p>
| Zach L. | 43,128 | <p>My favorite way of doing this:</p>
<p>$\mathbb{Z}[i]$ is a UFD, so this has a solution iff $\mathbb{Z}[i]/(p)$ is not an integral domain. But $\mathbb{Z}[i]$ is constructed via a quotient, so you can show it's isomorphic to $\mathbb{Z}_p[t]/(x^2 + 1)$, where I'm using $\mathbb{Z}_p$ to denote $\mathbb{Z}/p\mathbb{Z}$. Now, if this new ring is not an integral domain, then $x^2 + 1$ must factor into linear factors over $\mathbb{Z}/p\mathbb{Z}$. So, there is $x$ in the finite field such that $x^2 = -1$. So why is this being solvable equivalent to $p \equiv 1 \pmod 4$?</p>
<p>$(\Rightarrow)$ What does $x^4 = 1$ say about the group $(\mathbb{Z}/p\mathbb{Z})^*$?</p>
<p>$(\Leftarrow)$ This direction is harder! But note that since $x^2 - 1 = 0$ can only have two solutions, $-1$ is the only order 2 element of $(\mathbb{Z}/p\mathbb{Z})^*$ (The other is, of course 1. Unless $-1 = 1$, in which case $p =2$ and this can be handled separately.)</p>
|
3,042,802 | <p>For each <span class="math-container">$n ≥ 1$</span>, let <span class="math-container">$T_n = \{x ∈ l_2(N) : ||x||_1 ≤ n \}$</span>.</p>
<p>For <span class="math-container">$n ≥ 1$</span>, is <span class="math-container">$T_n$</span> an absorbing subset of <span class="math-container">$l_2(N) $</span>, but why?
I would like to show that <span class="math-container">$T_n$</span> has empty interior, for all <span class="math-container">$n ≥ 1. $</span> and that <span class="math-container">$T_n$</span> is closed in <span class="math-container">$l_2(N)$</span>, for all <span class="math-container">$n ≥ 1 $</span>. </p>
| David C. Ullrich | 248,223 | <p>Assuming <span class="math-container">$N=\Bbb N$</span>: <strong>Hint:</strong> If <span class="math-container">$x=???$</span> then <span class="math-container">$x\in\ell_2(N)$</span> but <span class="math-container">$||x||_1=\infty$</span>.</p>
|
358,102 | <p>How would I go about doing this?</p>
<p>I assume it is some integral I have to solve, but I have no idea what.</p>
<p>(Note:Not a physicist so please excuse incompetence with regard standard notation.)</p>
<p>Context is I want to estime the energy of N point particles spread over the unit sphere. This is an equation of the for $E(N)=N^2/2 - aN^{3/2}$. I know the $N^2/2$ comes from the uniform charge density on the sphere, and have been told that the $N^{3/2}$ comes from "recovering the energy of distribution of point charges, therefore subtracting the self energies of a et of N uniformly charges disks, which can be shown to be proportional to $N^{3/2}$''.</p>
<p>I get what it is doing, but I don't know how to show it is proportional to $N^{3/2}$. I have already managed to show the $N^2/2$ term though. Also I don't need to find a, I just need to show the power.</p>
| Douglas S. Stones | 139 | <p>We can show that belonging to the same strongly connected component is an <a href="http://en.wikipedia.org/wiki/Equivalence_relation" rel="nofollow">equivalence relation</a>. The strongly connected components thus form the equivalence classes under this relation. We can also show that in any equivalence relation no element belongs to two distinct equivalence classes.</p>
|
187,974 | <p>If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$? </p>
<p>the answer is given as $-1$ in my book, but how do you arrive at this conclusion?</p>
| lab bhattacharjee | 33,337 | <p>Taking $x=\cot a$, $x+\frac{1}{x}=1\implies x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\frac{1}{x}=1-2=-1$</p>
<p>Alternatively, $x+\frac{1}{x}=1\implies x^2-x+1=0$</p>
<p>$x^2-x+1=0\implies x^3+1=0$</p>
<p>So,</p>
<p>$x^{3m}+(\frac{1}{x})^{3m}=(x^3)^m+\frac{1}{(x^3)^m}=2(-1)^m$</p>
<p>$x^{3m+1}+(\frac{1}{x})^{3m+1}=(x^3)^m\cdot x+\frac{1}{(x^3)^m\cdot x}=(-1)^m(x+\frac{1}{x})=(-1)^m$</p>
<p>$x^{3m+2}+(\frac{1}{x})^{3m+2}=(x^3)^m\cdot x^2+\frac{1}{(x^3)^m\cdot x^2}=(-1)^m(x^2+\frac{1}{x^2})=(-1)^m(-\frac{1}{x}-x)$ as $x^3=-1\implies x^2=-\frac{1}{x}$ and $\frac{1}{x^2}=x$</p>
<p>So,$x^{3m+2}+(\frac{1}{x})^{3m+2}=(-1)^{m+1}\cdot (x+\frac{1}{x})=(-1)^{m+1}$</p>
<p>If we put $m=0$, $x^2+\frac{1}{x^2}=(-1)^1=-1$</p>
|
54,506 | <p><a href="http://www.hardocp.com/news/2011/07/29/batman_equation/" rel="noreferrer">HardOCP</a> has an image with an equation which apparently draws the Batman logo. Is this for real?</p>
<p><img src="https://i.stack.imgur.com/VYKfg.jpg" alt="Batman logo"></p>
<p><strong>Batman Equation in text form:</strong>
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}</p>
| stoicfury | 13,615 | <p>Here's the equations typed out if you want save time with writing it yourself.</p>
<pre><code>(x/7)^2*SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\*SQRT(ABS(y+3*SQRT(33)/7)/(y+3*SQRT(33)/7))-1=0
ABS(x/2)-((3*SQRT(33)-7)/112)*x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0
9*SQRT(ABS((ABS(x)-1)*(ABS(x)-0.75))/((1-ABS(x))*(ABS(x)-0.75)))-8*ABS(x)-y=0
3*ABS(x)+0.75*SQRT(ABS((ABS(x)-0.75)*(ABS(x)-0.5))/((0.75-ABS(x))*(ABS(x)-0.5)))-y=0
2.25*SQRT(ABS((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x)))-y=0
(6*SQRT(10))/7+(1.5-0.5*ABS(x))*SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6*SQRT(10))/14)*SQRT(4-(ABS(x)-1)^2)-y=0
</code></pre>
<p>Also: <a href="http://pastebin.com/x9T3DSDp">http://pastebin.com/x9T3DSDp</a></p>
|
54,506 | <p><a href="http://www.hardocp.com/news/2011/07/29/batman_equation/" rel="noreferrer">HardOCP</a> has an image with an equation which apparently draws the Batman logo. Is this for real?</p>
<p><img src="https://i.stack.imgur.com/VYKfg.jpg" alt="Batman logo"></p>
<p><strong>Batman Equation in text form:</strong>
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}</p>
| Community | -1 | <p>You may be able to see more easily the correspondences between the equations and the graph through the following picture <s>which is from the <a href="http://everythingnew.net/wp-content/uploads/2011/07/Batman-Equation-solved.png" rel="noreferrer">link</a> I got after a curious search on Google</s>(link broken now):</p>
<p><img src="https://i.stack.imgur.com/EHNR8.png" alt="enter image description here"> </p>
|
914,936 | <p>Does anyone know where I can find the posthumously published (I think) chapter 8 of Gauss's Disquisitiones Arithmaticae?</p>
| Maths Nerd | 209,035 | <p>I have read some information that it may have been published as 'General Investigations of Curved Surfaces' however I am having some difficulty confirming.</p>
|
1,943,328 | <p>I know about $S_n$, $D_n$ and $A_n$. And from my limited understanding there seem to be many more. I would like to know whether there is some kind of relation that links a small set of non Abelian groups to create the other ones. Something like with the Abelian groups and the Fundamental Theorem of Abelian Groups.</p>
| quid | 85,306 | <p>In general there is a huge number of ways how to combine certain groups.
The relatively most extreme being when the order of the groups are all powers of two.</p>
<p>For example, already for groups with $32$ elements there are $51$ different groups. And it only gets much worse, even for a relatively modest number like $1024$ there are $49487365422$ different groups of that order, and for $2048$ no one knows the exact answer.</p>
<p>Thus while there are unifying construction principles as discussed in other answers what they'll give exactly is hard to predict. </p>
|
1,946,824 | <p>In his book "Analysis 1", Terence Tao writes:</p>
<blockquote>
<p>A logical argument should not contain any ill-formed
statements, thus for instance if an argument uses a statement such
as x/y = z, it needs to first ensure that y is not equal to zero.
Many purported proofs of “0=1” or other false statements rely on
overlooking this “statements must be well-formed” criterion.</p>
</blockquote>
<p>Can you give an example of such a proof of "0=1"?</p>
| Patrick Stevens | 259,262 | <p>It does make sense.</p>
<p>Define $S_n = \{ F_0, F_1, F_2, \dots, F_n \}$. (I think you'll agree that this set exists for every $n$.)</p>
<p>Then let $$S = \bigcup_{n \in \mathbb{N}} S_n$$</p>
<p>Since each $S_n$ exists, and $\mathbb{N}$ certainly exists, the union does as well (we have an axiom, the Axiom of Unions, to tell us that).</p>
|
257,562 | <p>Given a piecewise function, such as</p>
<p>$$f(t) =
\begin{cases}
2, & \text{if }t \lt a \\
t^2, & \text{if }t \geq a
\end{cases}$$</p>
<p>Or some other piecewise function, how can we write it in the form $u(t-a)f(t-a)$ for $$u(t-a) =
\begin{cases}
0, & \text{if }t \lt a \\
1, & \text{if }t \geq a
\end{cases}$$</p>
<p>A simple explanation that doesn't involve too much math knowledge is preferable, but I would still like to generalize this across other functions that I may come across.</p>
<p>Thank you.</p>
| Artem | 29,547 | <p>\begin{align}
f(t)&=2(u(t)-u(t-a))+t^2u(t-a)\\
&=2u(t)-2u(t-a)+(t-a+a)^2u(t-a)=\ldots
\end{align}</p>
|
257,562 | <p>Given a piecewise function, such as</p>
<p>$$f(t) =
\begin{cases}
2, & \text{if }t \lt a \\
t^2, & \text{if }t \geq a
\end{cases}$$</p>
<p>Or some other piecewise function, how can we write it in the form $u(t-a)f(t-a)$ for $$u(t-a) =
\begin{cases}
0, & \text{if }t \lt a \\
1, & \text{if }t \geq a
\end{cases}$$</p>
<p>A simple explanation that doesn't involve too much math knowledge is preferable, but I would still like to generalize this across other functions that I may come across.</p>
<p>Thank you.</p>
| JohnD | 52,893 | <p>Think of $u(t-a)$ as simply a "switch" that is "off" for $t<a$ and so has a value of $0$, but turns on at $t=a$ with a value of $1$ and remains "on" for $t\ge a$. This is why the function is called the <em>unit step function</em> (activated at $t=a$): <em>unit</em> because it has a value of one and <em>step</em> because it instantaneously steps up from a value of zero to a value of $1$ at $t=a$. Many students find it helpful to think about unit step functions as switches. </p>
<p>Hopefully this helps you see how a piecewise defined function can be written as a linear combination of products involving unit step functions. In your example, the function has the value $2$ for $t<a$ but at $t=a$ a switch is flipped which results in a value of $t^2$ for $t\ge a$. This last part is represented as $u(t-a)(-2+t^2)$ or if you prefer, $u(t-a)(t^2-2)$. Thus, we can write the original function as $$f(t)=2+u(t-a)(t^2-2).$$</p>
<p>(Note how we had to include a $-2$ there in order to eradicate the value of 2 that was "on" by default.) You might find the examples <a href="http://tutorial.math.lamar.edu/Classes/DE/StepFunctions.aspx" rel="nofollow">here</a> helpful.</p>
<p>Techniques like this are useful when you want to rewrite a piecewise defined function as a unit step times another function, e.g., when you want to perform Laplace transforms in a differential equations course.</p>
|
2,220,738 | <p>How do I go about finding the dimension of the subspace:
$$$$S:={p($x$) ∈ $P_4$: p($x$)= 2p($x$) for all $x\in\mathbb{R}$} of $P_4$</p>
<p>My textbook says $dim(P_n)=n+1$, but this does not give me the correct answer. All help is appreciated.</p>
| Dude | 490,091 | <p>I'm extremely confused by the equivalence drawn in linear algebra between vectors in Rn and polynomials. Can someone please point out the flaws in my understanding?</p>
<p>For Rn, a vector (1, 2, 3) has components in three different dimensions (along three different axes), so adding these together is impossible, right?</p>
<p>And when we express a system of linear equations in matrix form, say</p>
<p>2x + 3y + 10z = 45</p>
<p>we treat the variables x, y, z as living in 3 different dimensions that cannot be directly added together, just like the different dimensions ("axes") of the vector (1, 2, 3) in Rn, right?</p>
<p>My understanding (please correct me if I'm wrong) is that x y z somehow correspond to DIFFERENT dimensions, just as 1, 2, 3 lie on three different axes, so you can't just add them together like (1,2,3) = 1+2+3 = 6, nor can you do x+y+z = 3x, for example.</p>
<p>But in linear algebra I see people treat the <strong>same variable x*</strong> raised to different degrees, as different variables, exactly as in the previous case. So for</p>
<p>2x^3 + 3x^2 + 10x = 45</p>
<p>you would make a matrix with three columns, corresponding to the x^3, x^2, and x, exactly as if it had been</p>
<p>2x + 3y + 10z = 45</p>
<p>This feels <strong>totally</strong> wrong to me, because x^3 and x^2 all come from x!</p>
<p>I assume that since you can't just add variables of a different power together in a <strong>linear</strong> way, that's why they can be slotted into the matrix in this way? </p>
<p>i.e.)</p>
<p>2x^2 + 3x^2 can be comined into 5x^2 but 2x^2 + 3x^3 cannot be combined.</p>
<p>But it still feels like a stretch to me and is completely baffling conceptually. How can you say that the <strong>same variable</strong> raised to different degrees lives in different dimensions, in the same way as the components of a vector in Rn??? For one, x^2 can be calculated if you know x, but a value living in dimension 2 of Rn, can <strong>never</strong> be calculated by just knowing a value in dimension 1 - they have totally different meanings.</p>
<p>If you graphed y = x and y = x^2, the two different graphs would look kind of different, but still be in the same dimension on the graph paper, right? One wouldn't be 3D or sticking out of the paper, right? So how do you draw this equivalency, that the 3 dimensional vector (1,2,3) corresponds to (x, x^2, x^3)?</p>
|
182,785 | <p>I haved plot a graph from two functions:</p>
<pre><code>η = 52;
h = 0.5682;
dpdx = -4.092*10^(-2);
Fg = dpdx;
Fl = dpdx/η;
Bl = ((Fg - Fl) h^2 - Fg)/(2 h - 2 η*h + 2 η);
Cg = -Fg/2 - η*Bl;
Bg = η*Bl;
Ut1[y_] := Fg*y^2/2 + Bg*y + Cg;
Ut2[y_] := Fl*y^2/2 + Bl*y;
Plot1 = Plot[Ut1[y]*1000, {y, h, 1}];
Plot2 = Plot[Ut2[y]*1000, {y, 0, h}, PlotStyle -> Orange];
Show[{Plot1, Plot2}, PlotRange -> All, AxesLabel -> {"y", "U"},
AxesStyle -> FontSize -> 14]
</code></pre>
<p>The result:</p>
<p><a href="https://i.stack.imgur.com/9BNfN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9BNfN.png" alt="enter image description here"></a></p>
<p>How to flip and transform the graph to this way?</p>
<p><a href="https://i.stack.imgur.com/0JvMC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0JvMC.jpg" alt="enter image description here"></a></p>
<p>PS: The numbers at the axes has to be legible. </p>
| David G. Stork | 9,735 | <p>Define myFigure as your figure and then:</p>
<pre><code>Rotate[myFigure, -90 Degree]
</code></pre>
<p><a href="https://i.stack.imgur.com/wkR7A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wkR7A.png" alt="enter image description here"></a></p>
|
2,801,433 | <p>I have made the following conjecture, and I do not know if this is true.</p>
<blockquote>
<blockquote>
<p><strong>Conjecture:</strong></p>
</blockquote>
<p><span class="math-container">\begin{equation*}\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}\stackrel{k\to\infty}{\longrightarrow}2\verb| such that we denote by | p_n\verb| the | n^\text{th} \verb| prime.|\end{equation*}</span></p>
</blockquote>
<p>Is my conjecture true? It seems like it, according to a plot made by Wolfram|Alpha, but if it does, then it converges.... <em>very</em>.... <em>very</em>, slowly. In fact, let <span class="math-container">$k=5000$</span>, then the sum is approximately equal to <span class="math-container">$1.97$</span>, which just proves how slow it would be.</p>
<p>Is there a way of showing whether or not this is indeed convergent? For any other higher values of <span class="math-container">$k$</span>, it seems that it is just too much for Wolfram|Alpha to calculate, and it does not give me a result when I let <span class="math-container">$k=\infty$</span>. Also, for users who might not understand the notation, we can similarly write that <span class="math-container">$$\sum_{n=1}^\infty\frac{1}{\pi^{1/n}p_n}=2\qquad\text{ or }\qquad\lim_{k\to\infty}\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}=2.$$</span> Also, without Wolfram|Alpha, I have <em>no idea</em> how to approach this problem in terms of proving it or disproving it. Does the sum even converge <em>at all</em>? If so, to what value? Any help would be much appreciated.</p>
<hr />
<p>Thank you in advance.</p>
<p><strong>Edit:</strong></p>
<p>I looked at <a href="https://math.stackexchange.com/questions/2070991/is-sum-limits-n-1-infty-frac1nk1-frac12-for-k-to-infty?rq=1">this post</a> to see if I could rewrite my conjecture as something else in order to help myself out. Consequently, I wrote that <span class="math-container">$$\sum_{n=1}^k\frac{1}{\pi^{1/n}p_n}\stackrel{k\to\infty}{\longleftrightarrow}4\sum_{n=1}^\infty\frac{1}{n^k+1}\tag{$\text{LHS}=2$}$$</span> since both sums look very similar. Could <em>this</em> be of use?</p>
| Akash Roy | 545,756 | <p>$\sum_{n=1}^{\infty} \frac{1}{s_n}$is a divergent series therefore if you tend the limit to $\infty$ ;
$\lim_{n\to \infty}\pi^{1/n}=1$
You can see that the above term tends to $1$.
The remaining term is similar to above mentioned divergent series. </p>
|
1,355,509 | <p>In my mathematical travels, I've stumbled upon the implicit formula $y^2+x^2+\frac{y}{x}=1$ and found that every graphing program I've plugged it in to seems to believe that there is large set of points which satisfy the equation $(y^2+x^2+\frac{y}{x})^{-1}=1$ which do not satisfy the original equation and this has me quite perplexed. I suspect that this is simply a glitch in the software and this question might therefore be better suited in the CS forum but I figured I would post it here first in the event that someone may have a mathematical explanation for this bizarre behavior. Any and all insights are welcome!</p>
| Tobia | 66,257 | <p>Here is a color plot (aka. scalar field) of $$x^2+y^2+{y\over x}$$
White is around zero, gray is positive and red is negative. The thick line is $0$ and the thin line is $1$, your original curve.</p>
<p><img src="https://i.stack.imgur.com/cOun9.png" alt="color plot"></p>
<p>It's easy to see that by taking the inverse, the thin line will be preserved, but you will get a discontinuity at the thick line, because one side will go to $+\infty$ and the other to $-\infty$. </p>
<p>Now, from a CS point of view, plotting an implicit formula as a <em>line drawing</em> is not a trivial matter. It is usually done by sampling the implicit formula in many places, and then "closing in" around the desired value.</p>
<p>Many pieces of software are not sophisticated enough to deal with discontinuities, especially when plotting implicit formulas. Therefore when they see a negative value on one side and a positive on the other side, they will conclude it must pass from $1$ in between, where in fact it goes to $+\infty$ and comes back from $-\infty$.</p>
<p>Software is stupid like that :-)</p>
<p><img src="https://i.stack.imgur.com/yynmF.png" alt="inverse plot"></p>
<p>Plots done with Grapher, a great plotting app that comes free with every Mac.</p>
|
211,533 | <p>I am trying to automate the process of running a Mathematica notebook that takes as an input a .txt file and exports another .txt file. This notebook was not written by me, and I don't fully understand its inner workings, but I think all I need to know is the following parts of the code:</p>
<p>The input statement is:</p>
<pre><code>my_input_file = Import["my_file_name.txt", "Table"];
</code></pre>
<p>The export statement is:</p>
<pre><code>Export["exported_file_name.txt", my_input_file_transformed, "Table"]
</code></pre>
<p>I would like to run notebook for many (100) different <code>my_file_name.txt</code> files which I could name however I want to facilitate the import process, e.g. <code>my_file_name_1.txt</code>, <code>my_file_name_2.txt</code>, etc., and put them in the same folder as the <code>my_mathematica_notebook.nb</code>.</p>
<p>Similarly, I would like to save the 100 different exported .txt files, in the same form I input them in order to be able to reconcile each imported-exported pair of files, e.g. something like <code>exported_file_name_1.txt</code>, <code>exported_file_name_2.txt</code></p>
<p>I guess my question is how to run the notebook many times for similarly named files and how to export the files as described above. I have read answers on related questions that call a notebook from another notebook, but none that automates the import of .txt files.</p>
<p>My knowledge of Mathematica is very basic, especially that related to file management, so I am unable to translate the answers I found into a solution for my problem. If it helps, I am working on a Linux machine and could use the command line to automate the process if needed.</p>
<p>Any help with this would be very appreciated. Many thanks.</p>
| kickert | 54,320 | <p>Pure functions will be helpful here. Start by using <code>Table</code> to automatically create the 100 files names you plan on entering.</p>
<pre><code>importnames=Table[StringJoin["my_file_name_", ToString[i], ".txt"], {i, 1, 100}]
</code></pre>
<p>Since this will all be text, you can Import them all at once and keep them in a single variable to alter later.</p>
<pre><code>imported=Import[#]&/@importnames
</code></pre>
<p>Now you can manipulate your text. Let's pretend you want it all upper case.</p>
<pre><code>manipulated=ToUpperCase/@imported
</code></pre>
<p>Next create all your export names in the same manner as above.</p>
<pre><code>exportnames=Table[StringJoin["exported_file_name_", ToString[i], ".txt"], {i, 1, 100}]
</code></pre>
<p>An easy way to automate the export process is create a list of lists where each entry contains the export name and the file contents. You can do that with <code>Transpose</code>.</p>
<pre><code>exportdata=Transpose[{exportnames,manipulated}]
</code></pre>
<p>Then run all the exports at the same time.</p>
<pre><code>Export[#[[1]],#[[2]]]&/@exportdata
</code></pre>
<p>If you aren't familiar with slots and mapping, you can read up on <a href="https://reference.wolfram.com/language/howto/WorkWithPureFunctions.html" rel="nofollow noreferrer">Pure Functions</a>.</p>
|
211,533 | <p>I am trying to automate the process of running a Mathematica notebook that takes as an input a .txt file and exports another .txt file. This notebook was not written by me, and I don't fully understand its inner workings, but I think all I need to know is the following parts of the code:</p>
<p>The input statement is:</p>
<pre><code>my_input_file = Import["my_file_name.txt", "Table"];
</code></pre>
<p>The export statement is:</p>
<pre><code>Export["exported_file_name.txt", my_input_file_transformed, "Table"]
</code></pre>
<p>I would like to run notebook for many (100) different <code>my_file_name.txt</code> files which I could name however I want to facilitate the import process, e.g. <code>my_file_name_1.txt</code>, <code>my_file_name_2.txt</code>, etc., and put them in the same folder as the <code>my_mathematica_notebook.nb</code>.</p>
<p>Similarly, I would like to save the 100 different exported .txt files, in the same form I input them in order to be able to reconcile each imported-exported pair of files, e.g. something like <code>exported_file_name_1.txt</code>, <code>exported_file_name_2.txt</code></p>
<p>I guess my question is how to run the notebook many times for similarly named files and how to export the files as described above. I have read answers on related questions that call a notebook from another notebook, but none that automates the import of .txt files.</p>
<p>My knowledge of Mathematica is very basic, especially that related to file management, so I am unable to translate the answers I found into a solution for my problem. If it helps, I am working on a Linux machine and could use the command line to automate the process if needed.</p>
<p>Any help with this would be very appreciated. Many thanks.</p>
| OpticsMan | 58,442 | <p>Not knowing all of the details, I suggest using an additional Notebook to automate the operation of "my_mathematica_notebook.nb"</p>
<p>In the new Notebook</p>
<pre><code> names = FileNames["*.txt", {".", "*"}, 1]
num = Dimensions[names][[1]]
(* finds all txt files *)
process[index_]:=Module[{}
CopyFile[names[[index]],"my_file_name.txt"];
(* copy the file to the placeholder filename *)
NotebookEvaluate["my_mathematica_notebook.nb", InsertResults -> False];
(* Run the Notebook *)
CopyFile["exported_file_name.txt",StringJoin["exported",names[[index]]]];
(* Copy the default export to the desired filename *)
]
(* "loop over" all txt files located *)
Table[process[indx],{indx,1,num,1}]
</code></pre>
|
798,897 | <p>In our lecture we ran out of time, so our prof told us a few properties about measure: He said that a measure is $\sigma$-additive iff it has a right-side continuous function that it creates. And he was not only referring to probability measures.
After going through my lecture notes, I thought that this would imply that there can be no other measures than ones having a right-side continuous function (I think they are called Lebesgue-Stieltjes measures) as $\sigma$-additivity is a prerequisite to be a measure. So somehow, this does not fit together. Does anybody know what he could have meant here? Or was he only referring to probability measures?</p>
<p>Is anything unclear about my question?</p>
| C-star-W-star | 79,762 | <p><strong>The statement is certainly wrong by consistency:</strong></p>
<p>Right-side continuous functions are defined/only make sense for functions whos domain is the real line (actually a right-closed subset of the real line).
Thus it gives rise at most for measures over the reals (or over a right-closed subset of the reals).</p>
<p>However, when restricting to measures over the reals the statement remains still wrong as pointed out above.</p>
<p>Besides, depending on the range this yields a real measure or more generally a vector measure...</p>
|
4,314,162 | <p>Assume k is a finite field with n elements, how many elements are in the projective line <span class="math-container">$\mathbb{P}^{1}(k)$</span> and how do I work this out?</p>
<p>I know that an element of <span class="math-container">$\mathbb{P}^{1}(k)$</span> is represented by <span class="math-container">$[a, b]$</span>, where <span class="math-container">$a, b \in k$</span>, not both of the coordinates are 0, and two elements <span class="math-container">$[a, b]$</span> and <span class="math-container">$[c, d]$</span> are equal if for some <span class="math-container">$\lambda \in k^{*}$</span> we have <span class="math-container">$a=\lambda c, b=\lambda d$</span></p>
<p>However, I’m not sure how I can use this to work out the number of elements?</p>
<p>Likewise how would I advance this to work out the number of elements in <span class="math-container">$\mathbb{P}^{2}(k)$</span> where the elements are the triples [a,b,c] ?</p>
| Wuestenfux | 417,848 | <p>The elements are (in homogeneous coordinates):</p>
<p><span class="math-container">$$(0,1),(1,1),\ldots, (q-1,1), (1,0),$$</span></p>
<p>where <span class="math-container">$k = \{0,1,\ldots,q-1\}$</span> has <span class="math-container">$q$</span> elements. So the number of elements is <span class="math-container">$q+1$</span>.</p>
<p>In the first <span class="math-container">$q$</span> elements, the 2nd coordinate is normalized to <span class="math-container">$1$</span>. In the last element, the 2nd coordinate is <span class="math-container">$0$</span> and the first coordinate is normalized to <span class="math-container">$1$</span>.</p>
<p>Using normalization, the projective plane can be described similarly.</p>
|
2,958,135 | <p>A "standard" example of Bayes Theorem goes something like the following:</p>
<blockquote>
<p>In any given year, 1% of the population will get disease <em>X</em>. A particular test will detect the disease in 90% of individuals who have the disease but has a 5% false positive rate. If you have a family history of <em>X</em>, your chances of getting the disease are 10% higher than they would have been otherwise.</p>
</blockquote>
<p>Virtually all explanations I've seen of Bayes' Theorem will include all of those facts in their formulation of the probability. It makes perfect sense to me to account for patient-specific factors like family history, and it also makes perfect sense to me to include information on the overall reliability of the test. I'm struggling to understand the relevance of the fact that 1% of the population will get disease <em>X</em>, though. In particular, that fact is presumably true for all patients who receive the test; that being the case, wouldn't Bayes' Theorem imply that the <em>actual</em> probability of a false positive is much higher than 5% (and that one of the numbers is therefore wrong)?</p>
<p>Alternatively, why doesn't the 5% figure already account for that fact? Given that the 5% figure was presumably calculated directly from the data, wouldn't Bayes' Theorem effectively be contradicting the data in this case?</p>
| CyclotomicField | 464,974 | <p>I believe it's commonly included because it's counterintuitive. You would expect a test with a high degree of accuracy to be right most of the time but this isn't actually the case and requires more evidence. To address this I think of it as the "error of one sample" fallacy which is to say you can't do an experiment one time and make strong conclusions, even if the experiment is well-designed.</p>
|
333,467 | <p>I was reading in my analysis textbook that the map $ f: {\mathbf{GL}_{n}}(\mathbb{R}) \to {\mathbf{GL}_{n}}(\mathbb{R}) $ defined by $ f(A) := A^{-1} $ is a continuous map. I also saw that $ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $. My question is:</p>
<blockquote>
<p>What is the unique extension of $ f $ to $ {\mathbf{M}_{n}}(\mathbb{R}) $?</p>
</blockquote>
| Elchanan Solomon | 647 | <p>The matrices</p>
<p>$$ A_{n} = \begin{pmatrix} \frac{1}{n} & 0\\ 0 & \frac{1}{n} \end{pmatrix}$$</p>
<p>converge to the zero matrix as $n \to \infty$. Their inverses are</p>
<p>$$ A_{n} = n \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$</p>
<p>And this goes to infinity in matrix norm as $n\to \infty$. Thus there is no way to extend the inverse to even the zero matrix.</p>
<p>This example is analogous to trying to extend $f(x) = x^{-1}$ from $(0,1)$ to $[0,1]$.</p>
|
2,362,354 | <p>I have seen a proof that B is row equivalent to A iff exist invertible matrix C such that :
B = CA , because C is elementary matrix , but i cant find the next step . can B "used" as the elementary matrix that lead from B to BA without changing the row space of A , so A and BA have same rowspace? </p>
| Siong Thye Goh | 306,553 | <p>if $B$ is an invertible matrix, its reduced row echelon form is the identity matrix.</p>
<p>That is I can write I can elementary matrices $E_i$ such that </p>
<p>$$E_m \ldots E_1 B = I$$</p>
<p>$$B=E_1^{-1}\ldots E_m^{-1}$$</p>
<p>Hence I can write $B$ as a product of elemntary matrices.</p>
<p>$$BA=E_1^{-1}\ldots E_m^{-1}A$$</p>
<p>Note that elementary matrices doesn't change the row space.</p>
|
2,362,354 | <p>I have seen a proof that B is row equivalent to A iff exist invertible matrix C such that :
B = CA , because C is elementary matrix , but i cant find the next step . can B "used" as the elementary matrix that lead from B to BA without changing the row space of A , so A and BA have same rowspace? </p>
| Vim | 191,404 | <p>Well, since $B=CA$, you see that rows of $B$ can be written as linear combinations of rows of $A$. (To be more explicit, write
$$B=\begin{bmatrix}\mathbf{b_1\\b_2\\ \vdots \\ b_m}\end{bmatrix},\quad A=\begin{bmatrix}\mathbf{a_1\\a_2\\ \vdots, \\ a_m}\end{bmatrix},\quad C=\begin{bmatrix}c_{11} &\cdots & c_{1m}\\
\vdots & \ddots&\vdots\\ c_{m1} & \cdots & c_{mm}
\end{bmatrix}$$
and perform partitioned matrix multiplication. Here letters in boldface denote row vectors comprising matrices.)</p>
<p>Likewise we note $A=C^{-1}B$.</p>
|
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