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 |
|---|---|---|---|---|
66,187 | <p>I would like to remove facets from an <a href="http://mathworld.wolfram.com/Octahedron3-Compound.html" rel="nofollow noreferrer">octahedron 3-compound</a> - like in the picture below.</p>
<p><img src="https://i.stack.imgur.com/xvvj4.png" alt="octahedron 3-compound with holes"></p>
<p>I tried to combine these two graphics: </p>
<pre><code>PolyhedronData["OctahedronThreeCompound"]
</code></pre>
<p>and</p>
<pre><code>Graphics3D[{Opacity[0], EdgeForm[Thickness[.03]],
{PolyhedronData["OctahedronThreeCompound", "Faces"]}}]
</code></pre>
<p>but I failed - any tips on how to get the desired result?</p>
| Silvia | 17 | <p>You can't get what you want with <code>EdgeForm[Thickness[.03]]</code>, that's not what <a href="http://reference.wolfram.com/language/ref/EdgeForm.html" rel="nofollow noreferrer"><code>EdgeForm</code></a> is meant for (that is, styles for the <em>1-dimensional</em> edges).</p>
<p>Here is a quick solution based on post-processing:</p>
<pre><code>Manipulate[
Normal[PolyhedronData["OctahedronThreeCompound", "Faces"]] /.
p : Polygon[__] :> (
p /. pts : {{_?NumericQ, _, _}, __} :>
Module[{cent, innervtx},
cent = Mean[pts];
innervtx = {1 - δ, δ}.{#, cent} & /@ pts;
Join[pts, pts[[{1}]], innervtx, innervtx[[{1}]]]
]
) // Graphics3D[{FaceForm[White], EdgeForm[Blue], #}] &,
{{δ, .1}, .01, .99}]
</code></pre>
<p><img src="https://i.stack.imgur.com/1eLJu.png" alt="hollowed OctahedronThreeCompound"></p>
|
2,753,812 | <p>So far I have been looking for a possible divisor such that the sum of $7$ seventh powers will never leave a certain remainder but there seems not to be such a divisor below $400$ and the search is too computationally heavy to continue.</p>
| Will Jagy | 10,400 | <p>If you mean non-negative seventh powers, the answer is yes, infinitely many misses. Indeed, the volume of the region
$x_i \geq 0$ and $\sum_{i=1}^7 x_i^7 \leq 1$ is a constant $C$ smaller than $1.$ the volume of $$\sum_{i=1}^7 x_i^7 \leq N$$
is roughly $CN.$ This volume is a good approximation of the (maximum) count of numbers represented as the sum of seven non-negative seventh powers. </p>
<p>Here $$ C = \Gamma \left( \frac{8}{7} \right)^7 \approx 0.6267634 $$</p>
<p>which means that, in the long run, at least 37% of numbers fail to have a representation. </p>
<p>This is usually written $G(7) \geq 8,$ see <a href="https://en.wikipedia.org/wiki/Waring%27s_problem#Lower_bounds_for_G(k)" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Waring%27s_problem#Lower_bounds_for_G(k)</a></p>
<p>Note $\Gamma(1+z) = z \, \Gamma(z)$</p>
<p><a href="https://i.stack.imgur.com/oT3YP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oT3YP.png" alt="enter image description here"></a></p>
|
2,254,672 | <p>I'm trying to solve a differential equation for the modeling of an ultrasonic horn in wich his form is a catenary, and his differential equation for a wave involves a hyperbolic tangent. I have the solution for the differential equation, but I need to do the "steps by steps". I had transformed the trigonometric coefficient into a polynomial but I keep having problems with the series in the development of the solution (I'm using series for solving this) so if someone can give me a hand, because I don't know if I'm doing a good approach to the equation, I would be very grateful. Here is the equation:
$$\ v''+\frac{2k}{r}tanh(\frac{kx}{r})v'+\frac{w^2}{c^2}v=0$$</p>
<p>and here is transformed into a polynomial</p>
<p>$$ \frac{k^2}{r^2}(1+u)^2 v''+\frac{w^2}{c^2}v=0 $$</p>
<p>thanks in advance and I wait for your answers if someone knows. ($k, r, w$ and c are constants)</p>
| Michael Seifert | 248,639 | <p>Define $y(x)$ such that $v(x) = y(x)/\cosh(kx/r)$. (<strong>EDIT:</strong> see below for why you might want to do this.) Then we have
$$
v'(x) = \frac{1}{\cosh (kx/r)} y'(x) - \frac{k}{r} \frac{\sinh (kx/r)}{\cosh^2 (kx/r)} y(x) = \frac{1}{\cosh (kx/r)} \left[ y'(x) - \frac{k}{r} \tanh \left( \frac{kx}{r} \right) y(x)\right],
$$
and (after a fair amount of algebra),
$$
v''(x) = \frac{1}{\cosh (kx/r)} \left[ y''(x) - \frac{2k}{r} \tanh \left( \frac{kx}{r} \right) y'(x) + \frac{k^2}{r^2} \left( 1 - 2 \,\mathrm{sech}^2 (kx/r) \right) y(x)\right].
$$
Thus,
$$
v''(x) + \frac{2k}{r} \tanh \left(\frac{2k}{r} \right) v'(x) = \frac{1}{\cosh(kx/r)} \left[ y''(x) + \frac{k^2}{r^2} \left( 1 - 2 \,\mathrm{sech}^2 (kx/r) - 2 \tanh^2 (kx/r) \right) y(x)\right] \\
= \frac{1}{\cosh(kx/r)} \left[ y''(x) - \frac{k^2}{r^2} y(x)\right],
$$
since $\tanh^2(x) + \mathrm{sech}^2(x) = 1$. Plugging this into the original ODE we have
$$
v''+\frac{2k}{r}\tanh\left(\frac{kx}{r}\right)v'+\frac{w^2}{c^2}v = \frac{1}{\cosh(kx/r)} \left[ y''(x) + \left( \frac{w^2}{c^2} - \frac{k^2}{r^2} \right) y(x) \right] = 0.
$$
Since $\cosh(kx/r) \neq 0$, we can conclude that
$$
y''(x) + \left( \frac{w^2}{c^2} - \frac{k^2}{r^2} \right) y(x) = 0,
$$
which is an ODE that you can probably solve in your sleep. </p>
<hr>
<p>In general, if you have an ODE of the form
$$
y'' + p(x) y' + q(x) y = 0,
$$
you can always make a substitution of the form $y(x) = f(x) z(x)$ to rewrite it in terms of another function $z$. After some algebra you get
$$
z'' + \left( p' + 2\frac{f'}{f} \right) z' + \left( \frac{f''}{f} + p \frac{f'}{f} + q \right) z = 0.
$$
A reasonable guess for $f(x)$ would be that function that causes the coefficient of $z'$ to vanish, i.e.,
$$
\frac{2f'}{f} = - p(x) \quad \rightarrow \quad f = \exp \left[ - \frac{1}{2} \int p(x) dx \right].
$$
Note that if we plug in $p(x) = \frac{2k}{r} \tanh (2kx/r)$, we get $f(x) = 1/\cosh(kx/r)$. So that's nice.</p>
<p>In general, the remaining term proportional to $z$ will still be present in the transformed ODE, and will probably be some really ugly function of $x$. We lucked out in this particular equation in that the quantity
$$
\frac{f''}{f} + \frac{f'}{f} p + q
$$
reduced down to a constant. So this trick won't always work; still, it's probably worth a shot if you have no idea how to solve a second-order ODE.</p>
<p>Note that this trick is similar to the use of <a href="http://mathworld.wolfram.com/IntegratingFactor.html" rel="nofollow noreferrer">integrating factors</a> for first-order ODEs; in that case, it's guaranteed to work because you don't have another term in the ODE to worry about.</p>
|
43,650 | <p>Consider the following problem:</p>
<blockquote>
<p>Let $f:{\mathbb R}^3 \to{\mathbb R}$ be $$f(x,y,z)=x+4z$$ where $$x^2+y^2+z^2\leq 2.$$ Find the minimum of $f$. </p>
</blockquote>
<p>This is similar to the question <a href="https://math.stackexchange.com/q/41385/9464">here</a>. However, since this is not an analytic function with complex variable, one may not be able to use the "Maximum modulus principle". </p>
<p>What I think is that one may rewrite the inequality constraint $x^2+y^2+z^2\leq 2$ as
$$x^2+y^2+z^2=2-\delta\qquad \delta\in[0,2]$$
then one can use the "Lagrange Multiplier Method" with the parameter $\delta$. Or one can do it on the xOz plane with the geometric meaning of $C$ in $C=x+4z$.</p>
<p>Is there any other way better than the idea above to solve this problem?</p>
| Luboš Motl | 10,599 | <p>The function $x+4z$ increases most quickly (measured by the increase per unit length) in the direction of the vector $(1,0,4)$ - the coefficients of the monomials. The inequality $x^2+y^2+z^2 \leq 2$ is a solid ball with radius $\sqrt{2}$. So the minimum value of the linear function is on the boundary of the ball in the direction $(1,0,4)$ from the center which has to be a multiple that belongs to the surface, namely
$$-\sqrt{\frac{2}{17}} (1,0,4)$$
The minus sign was added to make it a minimum. The opposite point with the plus sign is a maximum. The minimum itself is $-\sqrt{34}$.</p>
|
43,650 | <p>Consider the following problem:</p>
<blockquote>
<p>Let $f:{\mathbb R}^3 \to{\mathbb R}$ be $$f(x,y,z)=x+4z$$ where $$x^2+y^2+z^2\leq 2.$$ Find the minimum of $f$. </p>
</blockquote>
<p>This is similar to the question <a href="https://math.stackexchange.com/q/41385/9464">here</a>. However, since this is not an analytic function with complex variable, one may not be able to use the "Maximum modulus principle". </p>
<p>What I think is that one may rewrite the inequality constraint $x^2+y^2+z^2\leq 2$ as
$$x^2+y^2+z^2=2-\delta\qquad \delta\in[0,2]$$
then one can use the "Lagrange Multiplier Method" with the parameter $\delta$. Or one can do it on the xOz plane with the geometric meaning of $C$ in $C=x+4z$.</p>
<p>Is there any other way better than the idea above to solve this problem?</p>
| Per Alexandersson | 934 | <p>Here is an idea:
Consider the plane $x+4z=c$, for different values of c.
Changing c will shift the plane, and clearly, $c$ is the variable we want to minimize. Thus, we seek the minimal $c$ such that the plane intersects the sphere with radius 2. Therefore, the minimizing plane must be a tangent plane to the sphere, otherwise, we decrease c a little and still intersect the sphere.</p>
<p>The normal of the plane is the vector $(1,0,4),$ and thus, the ray $t(1,0,4)$ must hit the point of tangency. Therefore, the minimal/maximal value
is given when $x=t,z=4t$ and $t^2 + (4t)^2 = 2$. The second equation gives $t = \pm \sqrt{2/17}.$ Hence $c = -\sqrt{2/17}-4(4\sqrt{2/17}) = -17\sqrt{2/17} = -\sqrt{34},$ which is the minimum.</p>
|
89,176 | <p>The Hahn–Banach theorem states that: Given a sublinear functional $S: V \rightarrow \mathbb R$, if $T: U \rightarrow \mathbb R$ is a linear functional on a linear subspace $U \subseteq V$ that is dominated by $S$ on $U$, then there exists a linear extension of $T$ to $V$ that is dominated by $S$ on $V$.</p>
<p>Now, let us consider a symmetric multisublinear (positively homogeneous and subadditive in every component) continous functional $S: V\times\cdots\times V \rightarrow\mathbb R$ satisfying some good additionnal assumptions and a symmetric multilinear continous functional $T: U\times\cdots\times U \rightarrow \mathbb R$ that is dominated by $S$ on $U\times\cdots\times U$. Does there exist an extension of $T$ to $V\times\cdots\times V$ that is dominated by $S$ on $V\times\cdots\times V$?</p>
| Matthias Ludewig | 16,702 | <p>In this answer I made more or less the same mistake over and over again. It turns out that sublinear functions are not as nice as I believed. I thought it would be best to delete it.</p>
|
3,567,228 | <p>I am really struggling and have spent hours proving this theorem so I would greatly appreciate some help. I think I have nearly proved the theorem except for showing that <span class="math-container">$f^{-1}(-\infty,0]$</span> is a regular sublevel set of <span class="math-container">$f$</span>. </p>
<p>My attempt so far is this : Let <span class="math-container">$\mathscr{U}= \{(U_p,\phi_p)\}_{p \in D}$</span> be a collection of interior or boundary slice charts for <span class="math-container">$D$</span> in <span class="math-container">$M$</span>. Then <span class="math-container">$\mathscr{U} \cap \{M \backslash D\}$</span> covers <span class="math-container">$M$</span>. Now take the partition of unity <span class="math-container">$\{\psi_\alpha\}$</span> subordinate to this cover and we can construct functions <span class="math-container">$\{f_p \}_{p \in D} \cup \{f_0\}$</span> as follows : Given <span class="math-container">$p \in Int D$</span>, take an interior slice chart <span class="math-container">$U_p$</span> and <span class="math-container">$f_p \equiv -1$</span>, and extend <span class="math-container">$\psi_p f_p$</span> to <span class="math-container">$0$</span> on <span class="math-container">$M \backslash supp \psi_p$</span>. Given <span class="math-container">$p \in \partial D$</span>, take a boundary slice chart <span class="math-container">$U_p$</span> and <span class="math-container">$f_p(x^1,\dots ,x^n ) =-x^n$</span> on <span class="math-container">$U_p$</span>. And extend <span class="math-container">$\psi_p f_p $</span> to <span class="math-container">$0$</span> outside of the support of <span class="math-container">$\psi_p$</span>. </p>
<p>Define <span class="math-container">$f := \sum_\alpha \psi_\alpha f_\alpha$</span>. Then <span class="math-container">$f$</span> is smooth and since if <span class="math-container">$p \in D$</span>, we have <span class="math-container">$f_\alpha (p) = -1$</span> or <span class="math-container">$f_\alpha(p) \le 0$</span>, and <span class="math-container">$\psi_0(p)=0$</span>, we get <span class="math-container">$f(p) \le 0$</span>. </p>
<p>We show that <span class="math-container">$D = f^{-1}((-\infty, 0])$</span>. </p>
<p>Finally if <span class="math-container">$p \notin D$</span>, then <span class="math-container">$f_0\equiv1$</span>. Note that we can restrict the interior domains <span class="math-container">$U_p$</span> to <span class="math-container">$U_p \cap Int D$</span>. Here, from Proposition 5.46 in the text, the topological interior and manifold interior of a regular domain <span class="math-container">$D$</span> are the same so <span class="math-container">$U_p$</span> remains open in <span class="math-container">$M$</span> and we get <span class="math-container">$U_p \cap (M \backslash D) = \emptyset$</span>. Hence for all <span class="math-container">$p \in M \backslash D$</span>, <span class="math-container">$\psi_\alpha(p)=0$</span> if <span class="math-container">$\alpha$</span> belongs to an interior domain. </p>
<p>Hence for <span class="math-container">$p \notin D$</span>, <span class="math-container">$\psi_\alpha f_\alpha (p)=0$</span> for <span class="math-container">$\alpha$</span> belonging to the interior domain and <span class="math-container">$f_\alpha (p)>0$</span> for <span class="math-container">$\alpha$</span> belonging to boundary domain since <span class="math-container">$p \in U_\alpha \cap D^c = \{(x^1, \dots, x^n ) \in U: x^n < 0\}$</span> by definition of slice boundary chart. </p>
<p>Therefore, we have <span class="math-container">$f(p)=\sum_{\alpha \in \text{Boundary} \cup \{0\}} \psi_\alpha f_\alpha(p) \ge 0$</span>, and indeed it is positive since if <span class="math-container">$f(p)=0$</span> then we would have <span class="math-container">$\psi_0(p)=0$</span> and <span class="math-container">$\psi_\alpha(p)=0$</span> for all <span class="math-container">$\alpha \in \text{Boundary}$</span>, so <span class="math-container">$\sum_\beta \psi_\beta (p)=1$</span>, for <span class="math-container">$\beta \in \text{Interior}$</span>. But interior charts were chosen so that <span class="math-container">$\psi_\beta(p)=0$</span> if <span class="math-container">$p \notin D$</span>. Hence we have <span class="math-container">$D= f^{-1}(-\infty,0]$</span>. </p>
<p>Finally, we need to conclude the proof by showing that <span class="math-container">$0$</span> is a regular value of <span class="math-container">$f$</span>. This follows from Proposition 5.47 which states that for each regular value <span class="math-container">$b$</span> of <span class="math-container">$f$</span>, <span class="math-container">$f^{-1}(-\infty , b]$</span> is a regular domain. </p>
<p>However, this I cannot show. And how may we extend this to an exhaustion function if <span class="math-container">$D$</span> is compact?</p>
<p><a href="https://i.stack.imgur.com/exA4j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/exA4j.png" alt="enter image description here"></a></p>
| Li-Chung Wang | 743,599 | <p>The fact that <span class="math-container">$0$</span> is a regular value of <span class="math-container">$f$</span> follows from the formula given in Lee's Smooth Manifolds, p.119, l.5.
The case when <span class="math-container">$D$</span> is compact:
By Lee's Smooth Manifolds, p.46, Proposition 2.28, we may obtain an exhausting function <span class="math-container">$f^0\ge 1$</span> for the open submanifold <span class="math-container">$M\setminus D$</span>. Then replace <span class="math-container">$f_0$</span> with <span class="math-container">$f^0$</span>. If one checks the proofs of the results for the old <span class="math-container">$f$</span> when <span class="math-container">$D$</span> is not restricted to the compact case, one will find these proofs remain valid for the new <span class="math-container">$f$</span> when <span class="math-container">$D$</span> is compact except that trivial changes are required.</p>
|
1,775,166 | <p>If $$T:V\rightarrow V$$ is a bijective linear transformation then can $T^2$ also be a bijective linear transformation? Can $T^n$ also be bijective linear transformation?</p>
| Patrick Abraham | 337,503 | <p>To start an easy example:
$$I_n = I_n^2 = I_n^k$$</p>
<p>But the most easy way to think about this question is, if you look at the determinant and the determinant of matrix products.</p>
<p>$$det(A*B)=det(A)*det(B)$$</p>
<p>So if you have a bijective (non-singular,$\ det(T)\neq 0$) Transformation T follows:</p>
<p>$$det(T^n)=det(T)*det(T^{n-1})=det(T)*...*det(T)\neq 0$$</p>
<p>Thus $T^n$ is non-singular.</p>
<p>Else you would get
$$det(T^n)=det(T)*...*det(T)=0\quad => \quad det(T)=0$$
Thus T wouldn't be invertible (non-singular, bijective)</p>
<hr>
<p>If $T$ is non-singular $T^n$ is non-singular.</p>
|
1,775,166 | <p>If $$T:V\rightarrow V$$ is a bijective linear transformation then can $T^2$ also be a bijective linear transformation? Can $T^n$ also be bijective linear transformation?</p>
| Henry Swanson | 55,540 | <p>In fact, $T^n$ <em>must</em> be a bijective linear transformation.</p>
<p>A function has an inverse iff it's bijective. Since $T$ is bijective, it has an inverse, $T^{-1}$. So consider $(T^{-1})^n$ as a candidate for the inverse of $T^n$.
$$ (T^{-1})^n \circ T^n = (T^{-1})^{n-1} \circ T^{-1} \circ T \circ T^{n-1} = (T^{-1})^{n-1} \circ T^{n-1} = \cdots = T^{-1} \circ T = id$$</p>
<p>Similarly, $T^n \circ (T^{-1})^n$ is also the identity map. Since $T^n$ has an inverse, it's bijective.</p>
<p>Also, as a composition of linear functions, it's linear, but that probably was the easier thing to show.</p>
|
314,281 | <p>Let <span class="math-container">$B=-B$</span> be a nowhere dense bounded closed convex set in the Hilbert space <span class="math-container">$\ell_2$</span> such that the linear hull of <span class="math-container">$B$</span> is dense in <span class="math-container">$\ell_2$</span>.</p>
<blockquote>
<p><strong>Question.</strong> Is there a non-compact linear bounded operator <span class="math-container">$T:\ell_2\to \ell_2$</span> such that <span class="math-container">$T(B)$</span> is compact?</p>
</blockquote>
| Nik Weaver | 23,141 | <p>No, take <span class="math-container">$B$</span> to be the set of vectors in the unit ball of <span class="math-container">$l^2$</span> whose first coordinate is zero. If <span class="math-container">$T: l^2 \to l^2$</span> is a bounded operator and <span class="math-container">$T(B)$</span> is compact, then letting <span class="math-container">$C$</span> be the set of vectors in the unit ball which are nonzero only in the first coordinate (plus the zero vector), we have that <span class="math-container">$C$</span> is compact and therefore <span class="math-container">$T(C)$</span> is also compact. Thus <span class="math-container">$T(B+C) = T(B) + T(C)$</span> is compact, and <span class="math-container">$B + C$</span> contains the unit ball, so <span class="math-container">$T$</span> is a compact operator.</p>
|
3,114,873 | <p>Does the function <span class="math-container">$\frac{1-2xy}{x^2 +y^2}$</span> have a max or min value for <span class="math-container">$(x,y)=/=0$</span>?</p>
<p>What I've tried so far is to take the the partial derivatives:</p>
<p><span class="math-container">$$\frac{\partial f}{\partial x} = \frac{2(-x+x^2*y - y^3)}{(x^2 + y^2)^2}$$</span>
<span class="math-container">$$\frac{\partial f}{\partial y} = \frac{2(x^3 -x*y^2 +y)}{(x^2 + y^2)^2}$$</span></p>
<p>However I can't see what satisfy will <span class="math-container">$\nabla f(a,b)=0$</span></p>
<p>It looks like the function has a singular point in <span class="math-container">$(0,0)$</span> since it doesn't exist there, but I am told to ignore that point.</p>
<p>And seeing there is no boundary to f, the max/min can't be there either.</p>
<p>So, how can I then show that this function has a max/min other than in <span class="math-container">$(0,0)$</span>? </p>
| Daniel Schepler | 337,888 | <p><strong>Hint:</strong> Observe that for any <span class="math-container">$\sigma \in S_{n-1}$</span>, we have <span class="math-container">$\sigma(n) = n$</span>. Now, if <span class="math-container">$\sigma \in (1~n) \cdot S_{n-1}$</span>, then in that case what can we say about <span class="math-container">$\sigma(n)$</span>?</p>
|
1,040,846 | <p>I want to prove, that $a_n$ is a null sequence if $$\lim_{n \to \infty}|\frac{a_{n+1}}{a_n}|= c < 1$$</p>
<p>That means that $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |\frac{a_{n+1}}{a_n} - c| < \epsilon$</p>
<p>How can I get rid of the $a_{n+1}$ and the $c$, to show $\forall \epsilon > 0: \exists N \in \mathbb{N}: n \ge N: |a_n| < \epsilon$</p>
<p>?</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Alternative solution: for $n$ large enough
$$0\le|a_{n+1}|\le|a_n|,$$
i.e., $(|a_n|)_{n\ge n_0}$ is bounded and monotonic. Let be $l=\lim_{n\to\infty}|a_n|\ge 0$. If $l\ne 0$, then
$$\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \frac{l}{l} = 1\ne c.$$
Finally,
$$
0 = \lim_{n\to\infty}|a_n|=|\lim_{n\to\infty}a_n|\implies\lim_{n\to\infty}a_n = 0
$$</p>
|
1,665,931 | <p>Let $a,c \in \mathbb{R}$ and $b \in \mathbb{C}$. We consider the equation $$a \bar{z}z+ b\bar{z} + \bar{b}z+c=0.$$ What curve it represents in the complex plane?</p>
<p>I think it a circle, but I am not able to conclude. Any help?</p>
| egreg | 62,967 | <p>If $a\ne0$, write $B=-b/a$ and $C=c/a$, so the equation is
$$
z\bar{z}-B\bar{z}-\bar{B}z+B\bar{B}=B\bar{B}-C
$$
that can be written
$$
|z-B|^2=B\bar{B}-C
$$
which is a circle, with center $B$ and radius $B\bar{B}-C$, if $B\bar{B}-C>0$, the point $B$ if $B\bar{B}-C=0$, the empty set if $B\bar{B}-C<0$.</p>
<p>If $a=0$ and $b=0$, the equation is either the plane (if $c=0$) or the empty set (if $c\ne0$), so we can assume $a=0$ and $b\ne0$. Write $z=x+iy$, so we get
$$
b(x-iy)+\bar{b}(x+iy)+c=0
$$
that becomes
$$
(b+\bar{b})x-i(b-\bar{b})y+c=0
$$
Since $b+\bar{b}=A$ is real and $b-\bar{b}=iB$ is purely imaginary, we get
$$
Ax+By+c=0
$$
and at least one among $A$ and $B$ is nonzero. This is the equation of a straight line.</p>
|
348,276 | <p>I am not an expert, thus I apologize if my question is very naive. Let <span class="math-container">$\mathsf{M}$</span> be a model category (I do not assume any functoriality on the factorization),</p>
<blockquote>
<p><strong>Q1.</strong> Is there a reference where it is proven that (all) homotopy colimits exist?</p>
<p><strong>Q2.</strong> Are homotopy colimits characterized by any universal property as in the non-homotopical case (possibly involving the Homotopy category)?</p>
</blockquote>
<p>Let me invite the reader to not underestimate completely this question, under various assumptions on the model category this theorem is proved in many sources, yet I did not manage to find the most general version (if any exists).</p>
<p>I invite anyone to contribute with an answer about the state of art about this question. There is an incredible amount of sources, coming from different decades, thus it is not completely trivial to gather a global and coherent picture.</p>
| Andrea Gagna | 42,137 | <p>This is mostly a comment to Simone's thorough answer. It addresses a different, arguably more natural, way to deal with categories of functor <span class="math-container">$\mathbf{M}^I$</span>, with <span class="math-container">$\mathbf{M}$</span> a model category; but I have nothing to add with respect to Ivan's questions.</p>
<p>In <a href="http://www.numdam.org/item/BSMF_2010__138_3_317_0/" rel="nofollow noreferrer">Catégories dérivables</a>, Cisinski shows in Proposition 6.21 that the prederivator associated to a left derivable category (1.1) which is homotopy complete (6.1) is in fact a left derivator with domain the 2-category of small categories. The same fact is generalised to <span class="math-container">$\infty$</span>-categories in §7.7 of Cisinski's book <a href="http://www.mathematik.uni-regensburg.de/cisinski/CatLR.pdf" rel="nofollow noreferrer">Higher Categories and Homotopical Algebra</a>, although with a slightly different emphasis and terminology, where left derivable <span class="math-container">$\infty$</span>-categories are called <span class="math-container">$\infty$</span>-categories with weak equivalences and fibrations.</p>
<p>This implies in particular that the homotopy limit functor <span class="math-container">$\text{holim} \colon ho(\mathbf{M}^I) \to ho(\mathbf{M})$</span> exists for any small category <span class="math-container">$I$</span>. Notice that any Quillen model category is homotopy complete as a category with weak equivalences and fibrations, by standard closure arguments. The nice technical side of this theory amounts to the fact that the category <span class="math-container">$\mathbf{M}^I$</span> can be endowed with a canonical structure of category with weak equivalences and fibrations, which are defined component-wise (this is Theorem 6.11 of <a href="http://www.numdam.org/item/BSMF_2010__138_3_317_0/" rel="nofollow noreferrer">Catégories dérivables</a> and was proved by Rădulescu-Banu in his thesis).</p>
|
348,276 | <p>I am not an expert, thus I apologize if my question is very naive. Let <span class="math-container">$\mathsf{M}$</span> be a model category (I do not assume any functoriality on the factorization),</p>
<blockquote>
<p><strong>Q1.</strong> Is there a reference where it is proven that (all) homotopy colimits exist?</p>
<p><strong>Q2.</strong> Are homotopy colimits characterized by any universal property as in the non-homotopical case (possibly involving the Homotopy category)?</p>
</blockquote>
<p>Let me invite the reader to not underestimate completely this question, under various assumptions on the model category this theorem is proved in many sources, yet I did not manage to find the most general version (if any exists).</p>
<p>I invite anyone to contribute with an answer about the state of art about this question. There is an incredible amount of sources, coming from different decades, thus it is not completely trivial to gather a global and coherent picture.</p>
| Ivan Di Liberti | 104,432 | <p>I recently found a very convincing and elegant proof of <strong>Q1</strong> in <em>Balzin</em>'s <strong><a href="https://arxiv.org/abs/1803.00681" rel="nofollow noreferrer">Reedy model structures in families</a></strong>. In the paper, it appears as <strong>Cor. 3.41.</strong></p>
<p>The <em>idea of the proof</em> is that it is enough to show that the infinity category associated to a model category <span class="math-container">$\mathcal M$</span> has <span class="math-container">$\infty$</span>-colimits over direct Reedy diagrams, because they imply the existence of pushouts and coproducts. (Dually for limits). In order to do so he uses that when <span class="math-container">$R$</span> is direct Reedy, <span class="math-container">$\mathcal M^R$</span> has always a model structure (without any additional assumption on <span class="math-container">$\mathcal M$</span>) in order to build the Quillen adjunction, <span class="math-container">$$ \text{colim}: \mathcal M^R \leftrightarrows \mathcal M : c_R.$$</span> To finish he applies one of the main theorems of his machinery to derive the behaviour of the derived functors, <strong>Thm. 3.37</strong> in the paper. </p>
|
2,762,951 | <p>Let $u$ be a $W^{1,p}$-Sobolev function defined on the half disk $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$.</p>
<p>If I define a function $\tilde u$ by
$$ \tilde u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}, $$
is this $\tilde u$ a Sobolev $W^{1,p}$-function on the whole unit disk?</p>
<p>It seems obvious to me that if $\tilde u$ does have generalized derivatives, these have to be simply $\partial_x \tilde u (x,y) = \partial_x u (x,|y|)$ and $\partial_y \tilde u(x,y) = \partial_y u(x,y)$ if $y >0$ and $\partial_y \tilde u(x,y) = - \partial_y u(x,-y)$ if $y <0$. These functions are all $L^p$, so the only question that is left is whether these really are the generalized derivatives of $\tilde u$ and how to check this.</p>
<p>Thank you for any help.</p>
| Klaus Niederkrüger | 333,322 | <p>Thank you Revzora for your answer. I'll just add in a few more details to your solution.</p>
<p>Indeed by a <a href="https://en.wikipedia.org/wiki/Sobolev_space#Absolutely_continuous_on_lines_(ACL)_characterization_of_Sobolev_functions" rel="nofollow noreferrer">theorem due to Nykodim</a> a function $f$ is Sobolev $W^{1,p}(\Omega)$ if and only if its restriction to almost every line parallel to the coordinate directions is absolutely continuous, and if $f$ and $\nabla f$ are both in $L^p(\Omega)$.</p>
<p>If $u\in W^{1,p}(D_+^2)$ with $D^2_+ := \{(x,y)\in \mathbb{R}^2| x^2+y^2 < 1, y > 0\}$ the half-disk, we define $\bar u$ by
$$ \bar u\colon (x,y) \mapsto \begin{cases} u(x,y) & y > 0 \\ u(x,-y) & y < 0\end{cases}. $$</p>
<p>Since $u$ is Sobolev, clearly the restriction of $u$ to almost all lines parallel to the coordinate axis will be absolutely continuous. From this we directly see that the restriction of $\bar u$ is absolutely continuous along almost all lines that are parallel to the $x$-direction, and it only remains to verify the lines in $y$-direction.</p>
<p>The defintion of absolute continuity means for a function $f\colon [a,b] \to \mathbb{R}$ that for every $\epsilon > 0$ there exists a $\delta >0$ with the following property: choose any finite sequence $a \le a_1 < b_1 \le a_2 < b_2 \le a_3 < \dotsb \le a_N < b_N \le b$ with $\sum_{j=1}^N (b_j-a_j) < \delta$, then
$$
\sum_{j=1}^N |f(b_j) - f(a_j)| < \epsilon.
$$</p>
<p>If $x_0\in (-1,1)$ such that $y\mapsto u(x_0,y)$ is absolutely continuous then $y\mapsto \bar u(x_0,y)$ will inherit this property. In fact, it trivially holds if we only use intervals that do not include the point $0$ at which we have mirrored the half-line, and if one of the intervals $(a_j,b_j)$ does contain that point, there is no danger either, because we can split $(a_j, b_j)$ into $(a_j,0)$ and $(0,b_j)$, and for such intervals the statement holds.</p>
<p>It still remains to see that the restriction to the lines is $L^p$, but this is obvious. To see that the derivatives are $L^p$, we can use that the generalized partial derivatives agree almost everywhere with the classical partial derivatives.</p>
|
2,906,282 | <p>I want calculate the reflection from a straight <strong>R</strong> on the surface from the straight <strong>BC</strong>: </p>
<p><a href="https://i.stack.imgur.com/bOINy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bOINy.png" alt="enter image description here"></a></p>
<p>The triangle with the position vectors <strong>A,B,C</strong> are given. As well the position vector <strong>P</strong> and the direction vector <strong>R</strong>.</p>
<p><em>what I can calculate:</em></p>
<p>Now I can calculate the intersection point <strong>D(x,y)</strong> and the reflection vector (as a direction vector <strong>R'</strong>) and I know the <strong>x</strong> coordinate from the point <strong>P'(x,?)</strong> because the x value from the point P and P' are equals. </p>
<p><em>what I cannot calculate:</em></p>
<p>I need from the point <strong>P'</strong> the <strong>y</strong> coordinate. Because I need position vectors for drawing the reflected line with python matplotlib.</p>
<p>How can I get the y coordinate from the point <strong>P'</strong> with the given parameters?</p>
<p>The reflected vector will go from <strong>D(x,y)</strong> to <strong>P'(x,y)</strong></p>
<p><a href="https://i.stack.imgur.com/3Bt1B.png" rel="nofollow noreferrer">matplotlib reflection</a></p>
<p>Thanks you!</p>
| PM. | 416,252 | <p>Each elementary matrix $E_i$ is invertible so $\det(E_i)\ne0$.</p>
<p>As $$\det(E_iE_j)=\det(E_i)\det(E_j)\ne0$$ then $E_iE_j$ is invertible.</p>
|
3,224,765 | <p>The following question was asked on a high school test, where the students were given a few minutes per question, at most:</p>
<blockquote>
<p>Given that,
<span class="math-container">$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$</span>
and,
<span class="math-container">$$Q(x)=x^4+x^3+x^2+x+1$$</span>
what is the remainder of <span class="math-container">$P(x)$</span> divided by <span class="math-container">$Q(x)$</span>?</p>
</blockquote>
<hr>
<p>The given answer was:</p>
<blockquote>
<p>Let <span class="math-container">$Q(x)=0$</span>. Multiplying both sides by <span class="math-container">$x-1$</span>:
<span class="math-container">$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$</span>
Substituting <span class="math-container">$x^5=1$</span> in <span class="math-container">$P(x)$</span> gives <span class="math-container">$x^4+x^3+x^2+x+1$</span>. Thus,
<span class="math-container">$$P(x)\equiv\mathbf0\pmod{Q(x)}$$</span></p>
</blockquote>
<hr>
<p>Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?</p>
| Bill Dubuque | 242 | <p>The key idea employed here is the <strong>method of simpler multiples</strong> - a very widely used technique. Note that <span class="math-container">$\,Q\,$</span> has a "simpler" multiple <span class="math-container">$\,QR = x^5\!-\!1,\,$</span> so we can first reduce <span class="math-container">$P$</span> modulo <span class="math-container">$\,x^{\large 5}\! -\! 1\,$</span> via <span class="math-container">$\!\bmod x^{\large 5}-1\!:\,\ \color{#c00}{x^{\large 5}\equiv 1}\Rightarrow\, \color{#0a0}{x^n} =x^{\large r+5q^{\phantom{|}}}\!\!\equiv x^{\large r}(\color{#c00}{x^{\large 5}})^{\large q}\equiv x^{\large r}\equiv \color{#0a0}{x^{n\bmod 5}},\,$</span> then reduce <span class="math-container">$\!\bmod Q,\,$</span> i.e.</p>
<p><span class="math-container">$$P\bmod Q\, =\, (P\bmod QR)\bmod Q\qquad$$</span></p>
<p>Proof: <span class="math-container">$\: \color{darkorange}{P'}:= P\bmod QR\, =\, P-QRS\,\color{darkorange}{\equiv P}\,\pmod{\!Q},\,$</span> <a href="https://math.stackexchange.com/a/614944/242">thus</a> <span class="math-container">$\:\color{darkorange}{P'}\bmod Q = \color{darkorange}{P}\bmod Q$</span></p>
<p>Therefore, if <span class="math-container">$\,P = x^{\color{#0a0}A}+x^{\color{#90f}B} +x^C + x^D + x^E\,$</span> & <span class="math-container">$\bmod 5\!:\ \color{#0a0}A,\color{#90f}B,C,D,E\equiv \color{#0a0}4,\color{#90f}3,2,1,0\,$</span> then <span class="math-container">$\,P\bmod x^5-1 = x^{\large \color{#0a0}4}+x^{\large \color{#90f}3}+x^{\large 2}\,+x^{\large 1}\,+x^{\large 0} = Q,\,$</span> by <span class="math-container">$\,\color{#0a0}{x^n\equiv x^{n\bmod 5}}$</span> as above, hence <span class="math-container">$P\bmod Q \,=\, (P\bmod X^5\!-\!1)\bmod Q \,=\, Q\bmod Q = 0,\,$</span> generalizing the OP.</p>
<p>This idea is ubiquitous, e.g. we already use it implicitly in grade school in radix <span class="math-container">$10$</span> to determine integer parity: first reduce mod <span class="math-container">$10$</span> to get the units digit, then reduce the units digits mod <span class="math-container">$2,\,$</span> i.e.</p>
<p><span class="math-container">$$N \bmod 2\, = (N\bmod 2\cdot 5)\bmod 2\qquad\ $$</span></p>
<p>i.e. an integer has the same parity (even / oddness) as that of its units digit. Similarly since <span class="math-container">$7\cdot 11\cdot 13 = 10^{\large 3}\!+1$</span> we can compute remainders mod <span class="math-container">$7,11,13$</span> <a href="https://math.stackexchange.com/a/3425355/242">by using</a> <span class="math-container">$\,\color{#c00}{10^{\large 3}\equiv -1},\,$</span> e.g. <span class="math-container">$\bmod 13\!:\,\ d_0+ d_1 \color{#c00}{10^{\large 3}} + d_2 (\color{#c00}{10^{\large 3}})^{\large 2}\!+\cdots\,$</span> <span class="math-container">$ \equiv d_0 \color{#c00}{\bf -} d_1 + d_2+\cdots,\,$</span> and, similar to the OP, by <span class="math-container">$\,9\cdot 41\cdot 271 = 10^{\large 5}\!-1\,$</span> we can compute remainders mod <span class="math-container">$41$</span> and <span class="math-container">$271$</span> by using <span class="math-container">$\,\color{#c00}{10^5\!\equiv 1}$</span></p>
<p><span class="math-container">$$N \bmod 41\, = (N\bmod 10^{\large 5}\!-1)\bmod 41\quad $$</span></p>
<p>for example <span class="math-container">$\bmod 41\!:\ 10000\color{#0a0}200038$</span> <span class="math-container">$ \equiv (\color{#c00}{10^{\large 5}})^{\large 2}\! + \color{#0a0}2\cdot \color{#c00}{10^{\large 5}} + 38\equiv \color{#c00}1+\color{#0a0}2+38\equiv 41\equiv 0$</span></p>
<p>Such "divisibility tests" are frequently encountered in elementary and high-school and provide excellent motivation for this method of "divide first by a simpler multiple of the divisor" or, more simply, "mod first by a simpler multiple of the modulus".</p>
<p>This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously when <a href="https://math.stackexchange.com/a/100201/242">rationalizing denominators</a> and in <a href="https://math.stackexchange.com/a/174687/242">Gauss's algorithm</a> for computing modular inverses.</p>
<p>For example, to divide by an algebraic number we can use as a <em>simpler</em> multiple its <em>rational</em> norm = product of conjugates. Let's examine this for a quadratic algebraic number <span class="math-container">$\,w = a+b\sqrt{n},\,$</span> with norm
<span class="math-container">$\,w\bar w = (a+b\sqrt n)(a-b\sqrt n) = \color{#0a0}{a^2-nb^2 = c}\in\Bbb Q\ (\neq 0\,$</span> by <span class="math-container">$\,\sqrt{n}\not\in\Bbb Q),\,$</span> which reduces division by an <em>algebraic</em> to <em>simpler</em> division by a <em>rational</em>, i.e. <span class="math-container">$\, v/w = v\bar w/(w\bar w),$</span> i.e.</p>
<p><span class="math-container">$$\dfrac{1}{a+b\sqrt n}\, =\, \dfrac{1}{a+b\sqrt n}\, \dfrac{a-b\sqrt n}{a-b\sqrt n}\, =\, \dfrac{a-b\sqrt n}{\color{#0a0}{a^2-nb^2}}\,=\, {\frac{\small 1}{\small \color{#0a0}c}}(a-b\sqrt n),\,\ \color{#0a0}{c}\in\color{#90f}{\Bbb Q}\qquad $$</span></p>
<p>so-called <span class="math-container">$\rm\color{#90f}{rationalizing}\ the\
\color{#0a0}{denominator}$</span>. The same idea <a href="https://math.stackexchange.com/a/3225783/242">works even</a> with <span class="math-container">$\,{\rm\color{#c00}{nilpotents}}$</span></p>
<p><span class="math-container">$$\color{#c00}{t^n = 0}\ \Rightarrow\ \dfrac{1}{a-{ t}}\, =\, \dfrac{a^{n-1}+\cdots + t^{n-1}}{a^n-\color{#c00}{t^n}}\, =\, a^{-n}(a^{n-1}+\cdots + t^{n-1})\qquad$$</span></p>
<p>which simplifies by eliminating <span class="math-container">$\,t\,$</span> from the denominator, i.e. <span class="math-container">$\,a-t\to a^n,\,$</span> reducing the division to division by a simpler <em>constant</em> <span class="math-container">$\,a^n\,$</span> (vs. a simpler <em>rational</em> when <em>rationalizing</em> the denominator).</p>
<p><a href="https://math.stackexchange.com/a/95015/242">More generally,</a> often we can use norms to reduce divisibility and factorization problems on <em>algebraic</em> integers to "simpler" problems involving their norm multiples in <span class="math-container">$\Bbb Z$</span> - analogous to above, where we reduced division by the algebraic integer <span class="math-container">$\,w = a + b\sqrt{n}\,$</span> to division by its norm. The same can be done using norms in any (integral) ring extension (i.e. the base ring need not be <span class="math-container">$\Bbb Z$</span>).</p>
<p>Another example is <a href="https://math.stackexchange.com/questions/2991/not-understanding-simple-modulus-congruency/3230#3230">Gauss' algorithm,</a> where we compute fractions <span class="math-container">$\!\bmod m\,$</span> by iteratively applying this idea of simplifying the denominator by scaling it to a smaller multiple. Here we scale <span class="math-container">$\rm\color{#C00}{\frac{A}B} \to \frac{AN}{BN}\: $</span> by the least <span class="math-container">$\rm\,N\,$</span> so that <span class="math-container">$\rm\, BN \ge m,\, $</span> reduce mod <span class="math-container">$m,\,$</span> then iterate this reduction, e.g.</p>
<p><span class="math-container">$$\rm\\ mod\ 13\!:\,\ \color{#C00}{\frac{7}9} \,\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}\qquad\qquad$$</span></p>
<p>Denominators of the <span class="math-container">$\color{#c00}{\rm reduced}$</span> fractions decrease <span class="math-container">$\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$</span> so reach <span class="math-container">$\color{#C00}{1}\,$</span> (not <span class="math-container">$\,0\,$</span> else the denominator would be a <em>proper</em> factor of the <em>prime</em> modulus; it may fail for <em>composite</em> modulus)</p>
<p>See <a href="https://math.stackexchange.com/a/2012356/242">here</a> and its <span class="math-container">$25$</span> linked questions for more examples similar to the OP (some far less trivial).</p>
<p>Worth mention: there are <a href="https://math.stackexchange.com/a/42869/242">simple
algorithms</a> for
recognizing cyclotomics (and products of such), e.g. it's shown
there that <span class="math-container">$\, x^{16}+x^{14}-x^{10}-x^8-x^6+x^2+1$</span> is cyclotomic (a factor of <span class="math-container">$x^{60}-1),\,$</span> so we can detect when the above methods apply for arbitrarily large degree divisors.</p>
|
3,224,765 | <p>The following question was asked on a high school test, where the students were given a few minutes per question, at most:</p>
<blockquote>
<p>Given that,
<span class="math-container">$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$</span>
and,
<span class="math-container">$$Q(x)=x^4+x^3+x^2+x+1$$</span>
what is the remainder of <span class="math-container">$P(x)$</span> divided by <span class="math-container">$Q(x)$</span>?</p>
</blockquote>
<hr>
<p>The given answer was:</p>
<blockquote>
<p>Let <span class="math-container">$Q(x)=0$</span>. Multiplying both sides by <span class="math-container">$x-1$</span>:
<span class="math-container">$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$</span>
Substituting <span class="math-container">$x^5=1$</span> in <span class="math-container">$P(x)$</span> gives <span class="math-container">$x^4+x^3+x^2+x+1$</span>. Thus,
<span class="math-container">$$P(x)\equiv\mathbf0\pmod{Q(x)}$$</span></p>
</blockquote>
<hr>
<p>Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?</p>
| nonuser | 463,553 | <p>Let <span class="math-container">$a$</span> be zero of <span class="math-container">$x^4+x^3+x^2+x+1=0$</span>. Obviously <span class="math-container">$a\ne 1$</span>. Then <span class="math-container">$$a^4+a^3+a^2+a+1=0$$</span>
so multiply this with <span class="math-container">$a-1$</span> we get <span class="math-container">$$a^5=1$$</span> (You can get this also from geometric series <span class="math-container">$$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\over a-1}$$</span> by putting <span class="math-container">$n=4$</span>).</p>
<p>But then <span class="math-container">\begin{eqnarray} Q(a) &=& a^{100}\cdot a^4+a^{90}\cdot a^3+a^{80}\cdot a^2+a^{70}\cdot a+1\\ &=& a^4+a^3+a^2+a+1\\&=&0\end{eqnarray}</span></p>
<p>So each zero of <span class="math-container">$Q(x)$</span> is also a zero of <span class="math-container">$P(x)$</span> and since all 4 zeroes of <span class="math-container">$Q(x)$</span> are different we have <span class="math-container">$Q(x)\mid P(x)$</span>. </p>
|
101,566 | <p>How do I generalize the equation to be able to plug in any result for $\phi(n)=12$ and find any possible integer that works?</p>
| Pete L. Clark | 299 | <p>Note that $\lim_{n \rightarrow \infty} \varphi(n) = \infty$, so for any $k \in \mathbb{Z}^+$, the equation $\varphi(n) = k$ has at most finitely many solutions. How to find all of them?</p>
<p>Solutions to $\varphi(n) = k$ for very small values of $k$ can easily be worked out by hand and probably should be as a good exercise. For instance if $\varphi(k) \leq 4$ then $n$ can only be divisible by primes $2$, $3$ and $5$, i.e., $n = 2^a 3^b 5^c$ for natural numbers $a,b,c$. Since $\varphi(5^2) = 5(5-1) = 20$, we must have $c \in \{0,1\}$. Since $\varphi(3^2) = 3(3-1) = 6$, we must have $b \in \{0,1\}$. Since $\varphi(2^4) = 2^3(2-1) = 8$, we must have $a \in \{0,1,2,3\}$, which leaves us with $2 \cdot 2 \cdot 4 = 16$ values of $n$ to check. </p>
<p>The above method will work for any value of $k$, but it gets very slow as $k$ grows. You could still do it for $k = 12$ -- the value you asked about -- but I wouldn't want to do it by hand even for $k = 50$, say. </p>
<p>In fact for even moderately large $k$ this a problem to get your computer to solve. However, you should and can do something faster than the method sketched out above. Better is to use one of the known <strong>explicit lower bounds</strong> for $\varphi$. In fact this problem came up in a research project for graduate students that I led at UGA a few years back, and the following simple lower bound was perfectly adequate for our purposes:</p>
<p>(1) $\varphi(n) \geq \sqrt{n}$, for $n > 6$ </p>
<p>(I learned this originally from wikipedia, although its article on the totient function seems no longer to include this result.)</p>
<p>Thus, assuming $k > 4$ (we did the other case above!) so $n > 6$, using this inequality to find all values of $\varphi(n) = k$ you just need to evaluate the $\varphi$ function at all positive integers in the range $[1,k^2]$: no problem for a computer. </p>
<p>What if you are working with really large $k$? Well, there is in fact a much better inequality</p>
<p>(2) $\varphi(n) > \frac {n} {e^\gamma\; \log \log n + \frac {3} {\log \log n}}$, for $n > 2$. </p>
<p>This is extremely close to being sharp: if we omitted the $\frac{3}{\log \log n}$ term in the denominator, the right hand side would get ever so slightly larger, but this is large enough for the inequality to be false for infinitely many $n$. I have not given the matter any deep thought, but if anyone knows a faster way to enumerate the solution set to $\varphi(n) = k$ for large $k$ than by a brute-force search using (2) above, I would be interested to know. </p>
|
10,674 | <p>Let $p$ be a rational prime and $K$ a number field.
Dedekind's discriminant theorem tells us that
$p$ ramifies in $K$ $\iff$ $p$ divides the discriminant of $K$.
Hence if $p$ does not divide discriminant of $K$,
$(p)$ either splits, i.e., </p>
<p>(i) $(p)=P_1 \cdots P_g$ for $P_i \neq P_j$ and $g \geq 2$
or </p>
<p>(ii) $(p)$ remains prime. </p>
<p>Now, my question is: what are some criteria which can tell if $p$ will split or remain prime?</p>
| tkr | 389 | <p>Adding a few more words to Pete's answer...In the Galois case, at least, the inertia degrees $f$ and ramifications indices $e$ of each prime $P_i$ lying over $p$ are the same. If $p$ is unramified then $e = 1$ and so the only question is what is $f$ and what is $g$? The common degree $f$ is the order of any (and each) of the Frobenius elements $(P_i | K/\mathbb Q)$ which map to the generator of the local Galois group $G(k(P_i)/\mathbb F_p)$. I always recommend Milne's online notes <a href="http://www.jmilne.org/math/CourseNotes/" rel="nofollow"><a href="http://www.jmilne.org/math/CourseNotes" rel="nofollow">http://www.jmilne.org/math/CourseNotes</a></a>, both the algebraic number theory and class field theory, for this kind of question. I also remember well reading this material in the Janusz book "Algebraic Number Fields". In the latter, I believe Chapter 1 and Chapter 3 are most relevant, but I do not have a copy with me.</p>
|
2,692,957 | <p>I am interested in the isometries of the hyperbolic plane in the Beltrami–Klein disk model. (<a href="https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model</a>) The Wikipedia article does not say anything about the structure of the isometries in this model.</p>
<p>Since the isometries in the upper half-plane model are well-known, I did a change of variables to go from upper half-plane to the Beltrami–Klein disk. After some calculations, I determined that every isometry in the Beltrami-Klein disk model must be a projective transformation $\mathbb R \mathbb P^2 \to \mathbb R \mathbb P^2$ which maps the "unit disk" $\{ [x:y:1] \in \mathbb R \mathbb P^2 ~|~ x^2 + y^2 < 1 \}$ to itself. (The unit disk is, after all, the disk in the Beltrami-Klein disk model of hyperbolic space.)</p>
<p>Some questions I have:</p>
<ol>
<li>Is every projective transformation of this form an isometry of the hyperbolic plane? (Is there an easy way to see this?)</li>
<li>The set of such projective transformations is a subgroup of $SL(3,\mathbb R)$. Is there a nice characterization of this group?</li>
</ol>
<p>Update: I just realized that every projective transformation which maps the unit disk to itself must be an element of $SO(2,1)$. Here is the reason: such a projective transformation must map the unit circle to itself, so the corresponding linear transformation $\mathbb R^3 \to \mathbb R^3$ must preserve the quadratic form $x^2+y^2-z^2$. Therefore, if the answer to Question 1 is "yes," then the answer to Question 2 should be $SO(2,1)$.</p>
| Alan C | 5,894 | <p>In addition to Lee Mosher's very nice answer, here are some texts that some of my friends have recommended:</p>
<ul>
<li>John Ratcliffe: <em>Foundations of Hyperbolic Manifolds</em> </li>
<li>Cannon, Floyd,
Kenyon, Parry: <em>Hyperbolic Geometry</em></li>
</ul>
|
29,530 | <p>On the questions tab can we have a filter feature (I mean like frequented votes etc) of watched tags. I am aware that the tags that I have watched are demarked in a different colour. But I would like to discuss the pros and cons of getting a dedicated watched tags filter because I feel most people are well versed in certain specific areas of mathematics and would like to answer questions on that domain. This may not be so useful for asking questions as exploring newer areas of mathematics is generally encouraged upon. On the other hand its good for everybody that they answer questions on domains that they have command over. Any suggestions in this regard are welcome. </p>
| Andrew T. | 174,736 | <p><a href="https://meta.stackexchange.com/questions/330326/custom-filters-release-announcement">Custom Question Lists</a> is now available on all SE sites.</p>
<p>One of the features is a filter to list only watched tags:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/rKGxt.png" alt="Custom Question Lists' filter"></p>
</blockquote>
|
960,026 | <p>I'm having some trouble finding out how to calculate the probability of the following problem:</p>
<p><img src="https://i.stack.imgur.com/1VdCX.png" alt="enter image description here"></p>
<p>I'm confused as to how I should be calculating the probability of B. I'm not sure how to factor in the possibility of A winning in one state but B winning in another. </p>
| Graham Kemp | 135,106 | <p>Let $K$ be the event that party A wins in Kentucky. Let $T$ be the event that party A wins in Tennessee. The events of B winning in each state is the complementary events $K^c, T^c$</p>
<p>We are given: $\mathsf P(K)=0.50,\, \mathsf P(T)=0.50,\, \mathsf P(T\mid K)=0.72$</p>
<p>We wish to find the probability that party B wins in at least one of the states. This event is the complement of party A winning in both states. We can then apply the product rule for the conjunction of conditionally dependent events.</p>
<p>$$\begin{align}\mathsf P(K^c\cup T^c) & = 1-\mathsf P(K\cap T) \\ & = 1 - \mathsf P(K)\cdot\mathsf P(T\mid K)\\ & = 1-0.5\cdot 0.72 \\ & = 0.64\end{align}$$</p>
|
718,590 | <p>Is it true that in general complete metric space $(M,d)$, a closed ball of radius $r$ centered at $p\in M$ is always compact? That is, the ball is the set of all points $\left\{x:d(x,p)\leq r\right\}$.</p>
| Vincent Boelens | 94,696 | <p>For a Banach space $X$ the closed unit ball is compact iff dim $X<\infty$.</p>
|
2,940,072 | <p><span class="math-container">$$\left(\frac{f}{g}\right)'(x_{0})=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$$</span></p>
<p>So, <span class="math-container">$\frac{1}{g}.f=\frac{f}{g}$</span>, then <span class="math-container">$$\frac{f}{g}'(x_0)=\frac{f(x)\frac{1}{g(x)}-f(x_0)\frac{1}{g(x_0)}}{x-x_0}=f(x)\frac{\frac{1}{g(x)}-\frac{1}{g(x_0)}}{x-x_0}-\frac{1}{g(x_0)}\frac{f(x)-f(x_0)}{x-x_0}$$</span> I took the limit of everything as <span class="math-container">$x \to x_0$</span> <span class="math-container">$$=f(x_0)\frac{1}{g'(x_0)}-\frac{1}{g(x_0)}f'(x_0)=\frac{f(x_0)}{g'(x_0)}-\frac{f'(x_0)}{g(x_0)}=\frac{g(x_0)f(x_0)-f'(x_0)g'(x_0)}{g'(x_0)g(x_0)}$$</span> which clearly it's fake.</p>
<p>Where did I go wrong?
Thank you!</p>
| Neeyanth Kopparapu | 591,204 | <p>You are close. Note that instead of cancelling the term <span class="math-container">$\frac{\frac{f(x)}{g(x_0)}}{x-x_0}$</span> you actually subtract it twice. Second, note that the <span class="math-container">$$\lim_{x \rightarrow x_0} \frac{\frac{1}{g(x)} - \frac{1}{g(x_0)}}{x-x_0} = \left(\frac{1}{g(x_0)}\right)'\neq \frac{1}{g'(x_0)}.$$</span></p>
<p>Instead of doing it this way, I like to prove the quotient rule using the product rule: <span class="math-container">$\frac{f}{g} = f \cdot g^{-1}$</span>. <span class="math-container">$\frac{d}{dx} (fg^{-1}) = f'g^{-1} - fg'g^{-2} = \frac{f'g - fg'}{g^2}.$</span></p>
|
2,298,359 | <p>Which of the following has an element that is less than any another element in that set?</p>
<p>I. The set of positive rational numbers</p>
<p>II. The set of positive rational numbers $r$ such that $r^2$ is greater than or equal to $2$</p>
<p>III. The set of positive rational numbers such that $r^2 > 4$. </p>
<p>The answer is None. </p>
<p>I am not very clear why. Can anyone please explain intuitively? I know that there is no such thing as the smallest positive rational number. </p>
| Toby Mak | 285,313 | <p>HINT:</p>
<p>An important point for these types of questions are understanding the properties of rational numbers. For example, by contradiction, question I is not true. Let us suppose there is a smallest positive rational number $\frac{p}{q}$. We can multiply by $\frac{1}{2}$ to get $\frac{p}{2q}$. However, this is smaller than $\frac{p}{q}$, so there is no such thing as
"the smallest positive rational number" (that is larger or smaller than a given value).</p>
<p>Can you use a similar type of argument for questions II and III with the appropriate bounds, which are $r>\sqrt2$ for question II and $r>2$ for question III?</p>
<p>Note: The negative root is not required because the question specifies the numbers have to be positive.</p>
|
1,174,433 | <p>I'm trying to prove $\mathbb Z_p^*$ is a group if and only if $p$ is prime. I know that if $p$ is prime $\mathbb Z_p^*$ is a group, but how can I do the converse? In another words, if the equation $ax\equiv 1 (\mod m)$ has solution for every integer $a$ which is not divisible by $m$, then $m$ is prime.</p>
<p>I'm trying to solve this exercise:</p>
<p><img src="https://i.stack.imgur.com/u2Hfc.png" alt=""></p>
<p>Thanks </p>
| Asinomás | 33,907 | <p>The equation in $x$ : $ax\equiv 1 \bmod n$ has a solution if and only if $(a,n)=1$. If $n$ is not prime take $a$ a proper divisor of $n$. Then $a$ is not divisible by $n$ , while $ax\equiv 1$ has no solution.</p>
<p>Although note that $\mathbb Z_p^*$ is the group in which the elements are the congruence classes relatively prime to $p$. Not those that are not multiples of $p$.</p>
|
1,984,190 | <p>$X \in \mathbb{R}$ is the number of coin tosses it takes to get the first head.</p>
<p><strong>Question</strong>: find the probability of getting the first head at an even number of tosses.</p>
<p><strong>Official solution</strong>:
$P(X_{even}) = \frac{1}{2^{2}} + \frac{1}{2^{4}} + ... = \frac{1}{3}$ (sum of a geometric progression).</p>
<p><strong>Why is this not correct solution</strong>:
$P(X)=1$; $X_{odd} = X_{even}$; $P(X_{even})=\frac{n(X_{even})}{n(X)}=\frac{1}{2}$</p>
| martini | 15,379 | <p>Because $X_{\rm odd} \ne X_{\rm even}$. The most probable result is to get a head with the first toss (we have $P(X = 1) = \frac 12$), and $1$ is odd.</p>
|
1,984,190 | <p>$X \in \mathbb{R}$ is the number of coin tosses it takes to get the first head.</p>
<p><strong>Question</strong>: find the probability of getting the first head at an even number of tosses.</p>
<p><strong>Official solution</strong>:
$P(X_{even}) = \frac{1}{2^{2}} + \frac{1}{2^{4}} + ... = \frac{1}{3}$ (sum of a geometric progression).</p>
<p><strong>Why is this not correct solution</strong>:
$P(X)=1$; $X_{odd} = X_{even}$; $P(X_{even})=\frac{n(X_{even})}{n(X)}=\frac{1}{2}$</p>
| Mees de Vries | 75,429 | <p>Another way of looking it is as follows: suppose that you throw pairs of coins, one for odd, one for even, the odd one first, until at least one of them comes up heads. Then, when your experiment has ended, the outcome of the last two coins is either HH, HT, or TH. In two out of those three, the odd one gets the first heads (HH, HT); and in one of three, the even coin does (TH).</p>
|
2,379,881 | <p>I want to know if this integral converges or not : $\int _{1}^{\infty }\!{\frac {\sin \left( \cos \left( x \right) +\sin\left( x\sqrt {3} \right) \right) }{x}}{dx}.$ I tried to integrate by parts or to use dirichlet's test, but it seems impossible to prove that $\int _{1}^{x}\!\sin \left( \cos \left( t \right) +\sin \left( t\sqrt {3} \right) \right) {dt}$ is bounded. Do you have any idea how to solve this problem?</p>
| cdamle | 352,854 | <p>Here is my idea to solve this problem: </p>
<p>Note that $\cos x$ has positive values on $I_n = (\frac{2n-1}{2} \pi, \frac{2n+1}{2} \pi)$ for every $n \in \mathbb{Z}$ and that $\sin (\sqrt{3} x)$ has positive values on $J_k = (\frac{2k}{\sqrt{3}} \pi, \frac{2k+1}{\sqrt{3}} \pi)$ for every $k \in \mathbb{Z}$. </p>
<p>For each fixed $n \in \mathbb{Z}$, we can find $k \in \mathbb{Z}$ such that
$$\left(\frac{2n-1}{2}\right) \pi < \frac{2k}{\sqrt{3}} \pi < \left(\frac{2n+1}{2} \right) \pi,$$
which implies that for each $I_n$, there exists $J_k$ such that $I_n \cap J_k \neq \emptyset$. </p>
<p>Here, we can choose a small subinterval $L \subseteq I_n \cap J_k$ that makes $\cos x + \sin (\sqrt{3} x) > \alpha $ for some $0< \alpha < 1$. </p>
<p>Thus $$\frac{\sin(\cos x + \sin(\sqrt{3} x) )}{x} > \frac{\beta}{x}$$</p>
<p>on some sub-subinterval $\tilde{L} \subseteq L$ for some $0<\beta<\alpha$ ($\sin x$ is continuous). </p>
<p>Since there are infinitely many such disjoint subintervals $L \subseteq (1,\infty)$, <em>so the integral diverges if the lengths of $L$ and $\tilde{L}$ and the number $\beta >0$ are chosen appropriately.</em></p>
|
353,630 | <p>Consider a uniform random tournament with <span class="math-container">$n$</span> vertices. (Between any two vertices <span class="math-container">$x,y$</span>, with probability <span class="math-container">$0.5$</span> draw an edge from <span class="math-container">$x$</span> to <span class="math-container">$y$</span>; otherwise draw an edge from <span class="math-container">$y$</span> to <span class="math-container">$x$</span>.) Let <span class="math-container">$S$</span> be the set of all out-degrees. Let <span class="math-container">$s_1$</span> be the largest element of <span class="math-container">$S$</span>, and <span class="math-container">$s_2$</span> the next largest. (If <span class="math-container">$S$</span> is a singleton, let <span class="math-container">$s_2=s_1$</span>.) </p>
<p>Let <span class="math-container">$c\in (0,1)$</span> be a constant. What is <span class="math-container">$\lim_{n\rightarrow\infty}\text{Pr}[s_1-s_2<cn]$</span>?</p>
<p>My guess is that the limit should go to <span class="math-container">$1$</span>, that is, the two largest out-degrees are close to each other compared to the size of the tournament.</p>
| Seva | 9,924 | <p>The outdegree of every fixed vertex is distributed binomialy with the mean <span class="math-container">$n/2$</span> and the variance <span class="math-container">$n/4$</span>. Hence, the probability that the outdegree deviates from <span class="math-container">$n/2$</span> by <span class="math-container">$cn/3$</span> at least is extremely small, and by the union bound so is the probability that there is at least one vertex with the outdegree deviating from <span class="math-container">$n/2$</span> by <span class="math-container">$cn/3$</span>. This means that with probability <span class="math-container">$1-o(1)$</span>, the outdegrees of all vertices are in the range <span class="math-container">$(n/2-cn/3,n/2+cn/3)$</span>, confirming your conjecture.</p>
|
1,499,822 | <p>So we were shown this problem from our first functional analysis lecture and I was wondering if someone could help or give a hint:</p>
<blockquote>
<p>Show that the unit sphere $\mathbb{S}^{n-1} := \{ x \in \mathbb{R}^n: \| x \| = 1 \}$ is a complete metric space equipped with $d(x,y):= \arccos \langle x,y \rangle_{\mathbb{R}^n}$ where $\langle x,y \rangle_{\mathbb{R}^n}$ denotes the standard dot product. </p>
</blockquote>
<p>We're having no trouble showing positivity and symmetry but could use help showing completeness and the triangle inequality.</p>
<p>Kind regards</p>
<p>Edit:</p>
<p>So with MatiasHeikkilä's help, what's left to show is that $\theta_{x,y} \leq \theta_{x,z} + \theta_{y,z}$ if I'm not mistaken. Do I need to make a case-by-case proof to show this inequality or is there an easier way? I would argue something along the lines of: "$z$ lies between $x$ and $y$ implies the equality", "$z$ does not lie on the (smallest) path between $x$ and $y$ but $z$ lies on the half-sphere with $x$ and $y$ on it implies the inequality (since $\theta_{x,y} < \theta_{x,z}$) and lastly $z$ does neither lie on the (smallest) path between $x$ and $y$ nor does $z$ lie on the half-sphere implies the inequality (since $\theta_{x,z} + \theta_{y,z} = 2 \pi - \theta_{x,y} \geq \pi $)</p>
| Christian Blatter | 1,303 | <p>For $x$, $y\in S^{n-1}$ write $\arccos\langle x,y\rangle=:\theta_{xy}$. Then $$d(x,y):=\theta_{xy}\in[0,\pi]$$ is nothing else but the usual spherical distance on $S^{n-1}$. Using elementary geometry one immediately verifies that
$$|x-y|=2\sin{\theta_{xy}\over 2}\qquad(x, \>y\in S^{n-1})\ .\tag{1}$$
In particular, $\theta_{xy}$ is a strictly increasing function of the eucliden distance $|x-y|$.</p>
<p>In order to prove the triangle inequality it is sufficient to consider three points $x$, $y$, $z\in S^2$. If $\theta_{xy}+\theta_{yz}\geq\pi$ we immediately obtain $$d(x,z)\leq\pi\leq d(x,y)+d(y,z)\ .$$ Therefore assume $\theta_{xy}+\theta_{yz}<\pi$. Consider $y$ as north pole, and rotate the point $z$ around the axis $Oy$ to a position $z'$ such that $z'$ and $x$ are on opposite meridians. We then have
$$|x-z|\leq |x-z'|$$
and therefore
$$d(x,z)\leq d(x,z')=d(x,y)+d(y,z')=d(x,y)+d(y,z)\ .$$</p>
<p>The completeness of $(S^{n-1},d)$ is seen as follows: From $${2\over \pi}\>t\leq \sin t\leq t\qquad\left(0\leq t\leq{\pi\over2}\right)$$
and $(1)$ one concludes that
$$|x-y|\leq d(x,y)\leq{\pi\over2}\>|x-y|\qquad(x, \>y\in S^{n-1})\ .$$
This immediately implies that the euclidean metric $|\cdot|$ and $d$ are equivalent: A sequence on $S^{n-1}$ that converges with respect to $|\cdot|$ converges with respect to $d$ as well, and vice versa.
But more than that: A sequence that is Cauchy with respect to $|\cdot|$ is Cauchy as well with respect to $d$, and vice versa. This allows to conclude that the metric space $(S^{n-1},d)$ is indeed complete.</p>
|
89,987 | <p>In the sequence: </p>
<p>$$\lim _{n\rightarrow \infty }{\frac {n+1}{2\,n+3}}\neq 3$$</p>
<p>I know how to prove that the limit is actually $1/2$, but is there another way to prove that 1 is NOT the limit? </p>
<p>I tried to prove by negation showing that if 1 is the limit I can't find an $N$ that for every epsilon etc etc but got a bit lost. I know this is trivial but would appreciate the help.
Thanks</p>
| GEdgar | 442 | <p>For example, show the limit is $\le 1/2$ like this:</p>
<p>$$
\frac{n+1}{2n+3} < \frac{n+1}{2n+2} = \frac{1}{2}
$$</p>
|
4,374,739 | <p>Spivak's chapter 6, question 1 asks: for which of the following functions <span class="math-container">$f$</span> is there a continuous function <span class="math-container">$F$</span> with domain <span class="math-container">$\mathbb{R}$</span> such that <span class="math-container">$F(x) = f(x)$</span> for all <span class="math-container">$x$</span> in the domain of <span class="math-container">$f$</span>. Part (iv) asks us to consider the function <span class="math-container">$f(x) = 1/q, x = p/q$</span> rational in lowest terms.</p>
<p>The answer book says: No, because it would have to be that <span class="math-container">$F(x) = 0$</span> for irrational <span class="math-container">$x$</span>, but then <span class="math-container">$F$</span> would not be continuous at rational numbers. I understand the second part (that <span class="math-container">$F$</span> would not be continuous at rational <span class="math-container">$x$</span> if <span class="math-container">$F(x) = 0$</span> for irrational <span class="math-container">$x$</span>). But I don't know how to prove the first part. If there were an <span class="math-container">$F$</span> such that <span class="math-container">$F(x) = f(x)$</span> for all <span class="math-container">$x$</span> in the domain of <span class="math-container">$f$</span>, why does it follow that <span class="math-container">$F(x) = 0$</span> for irrational <span class="math-container">$x$</span>?</p>
| Cameron Buie | 28,900 | <p>The general idea is to pick some positive <span class="math-container">$\varepsilon$</span> and some irrational <span class="math-container">$z,$</span> then show that there is an open interval containing <span class="math-container">$z$</span> such that for every rational <span class="math-container">$x$</span> (in lowest terms) in that interval, the denominator is large enough that <span class="math-container">$f(x)<\varepsilon.$</span> Then you have to show that if <span class="math-container">$y<0,$</span> then there is some positive <span class="math-container">$\varepsilon$</span> such that no matter what <span class="math-container">$x\in\Bbb Q$</span> you choose, <span class="math-container">$|f(x)-y|>\varepsilon.$</span></p>
|
1,398,464 | <p>Folks - Please help. What's the gradient for the cost function below?</p>
<p>$ D(Y||AX)=\frac{1}{2} ||Y-AX||^2_F $</p>
<p>Additional info - </p>
<p>-need to get the derivative of that with respect to A.
-Multiplicative NMF Algorithms Based on the Squared Euclidean Distance</p>
| john316 | 262,158 | <p>Let $M=(AX-Y)$, then the function and its differential can be expressed in terms of the Frobenius (:) product as
$$\eqalign{
f &= \frac{1}{2}\,M:M \cr\cr
df &= M:dM \cr
&= (AX-Y):dA\,X \cr
&= (AX-Y)X^T:dA \cr
}$$
Since $df=(\frac{\partial f}{\partial A}):dA\,\,$ the gradient must be
$$\eqalign{
\frac{\partial f}{\partial A} &= (AX-Y)X^T \cr
}$$
If you dislike Frobenius products, you can replace them with traces, e.g. $\,{\rm tr}(A^TB)=A:B$</p>
|
4,610,313 | <p>I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:</p>
<blockquote>
<p>Given that <span class="math-container">$\log_{4n} 40\sqrt{3} = \log_{3n} 45$</span>, find <span class="math-container">$n^3$</span> (MA<span class="math-container">$\Theta$</span> 1991).</p>
</blockquote>
<p>My approach is to isolate <span class="math-container">$n$</span> and then cube it. Observe:
<span class="math-container">\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}</span></p>
<p>Dividing through and putting the coefficients as powers, we have:
<span class="math-container">\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}</span></p>
<p>which shows that</p>
<p><span class="math-container">\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}</span></p>
<p>Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?</p>
| lone student | 460,967 | <p>In this answer, we provide <em>another</em> method that uses the <em>definition</em> and <em>basic properties</em> of logarithm.</p>
<p>The original equation states :</p>
<p><span class="math-container">$$
\begin{align}&\log_{4n}{40}\sqrt{3}=\log_{3n}{45}\\
\implies &{40}\sqrt 3=\left(4n\right)^{\log_{3n}45}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=\left(3n\cdot \frac 43\right)^{\log_{3n}{45}}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=45\cdot \left(\frac 43\right)^{\log_{3n}{45}}\end{align}
$$</span></p>
<p>This leads to the following :</p>
<p><span class="math-container">$$
\begin{align}\left(\frac {2}{\sqrt 3}\right)^{2\log_{3n}{45}}&=\frac {40\sqrt 3}{45}=\left(\frac {2}{\sqrt 3}\right)^3\end{align}
$$</span></p>
<p>Finally, again using the definition of logarithm, we get the desired result:</p>
<p><span class="math-container">$$
\begin{align}&\log_{3n}{45^2}=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace .\end{align}
$$</span></p>
|
2,618,813 | <p>What is the image of the set {$ { {z ∈ C : z = x + iy, x ≥ 0, y ≥ 0} } $} under
the mapping $ z \to z^2$</p>
<p>my answer : $f(z) = z^2 =(x+iy)^2 = x^2-y^2 +2ixy$,</p>
<p>here I get $u=x^2-y^2$ and $v=2xy$. After that I am confused that how can I find the image of the set.</p>
<p>Please help me, as any Hints can be appreciated or if u have time u can tell me the solution. I would be more thankful.</p>
| Sahiba Arora | 266,110 | <p><strong>Hint:</strong> The map $z \mapsto z^2$ is just the map $$re^{ i\theta}\mapsto r^2e^{ i2\theta}.$$</p>
|
1,481,627 | <p>I have been given homework in which the $2\times 2$ matrices of determinant $1$ are equipped with the subspace topology of $\mathbb{R}^4$. However, $\mathbb{R}^4$ is a space of 4-tuples, while 2x2 matrices are not n-tuples. How do I get the open sets of this topology?</p>
<p>How is this set of matrices a subset of $\mathbb{R}^4$?</p>
| Community | -1 | <p>Any matrix
$$\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\in\mbox{Mat}^{2\times 2}(\mathbb{Z})$$
can be interpreted as the touple $(a,b,c,d)\in\mathbb{R}^4$, so $\mbox{Mat}^{2\times 2}(\mathbb{Z})$ can be thought of a subspace of $\mathbb{R}^4$ and be given the subspace topology.</p>
|
2,798,338 | <p>Let $G$ be a group of order 20, Prove that $G$ has a normal subgroup of order 5.</p>
<p>Obviously by Sylow theorem there is a subgroup of order 5, and since all Sylow p-subgroups are conjugate the only problem is to show that there is only one sungroup of order 5.</p>
<p>any help would be appreciated</p>
| alper akyuz | 483,006 | <p>Or we can say just use the fact that there is only 1 group of order 5 which is cyclic group of order 5 and it is normal, we also know that all p subgroups are conjugate, but cyclic group of order 5 is normal so it only conjugates with itself.</p>
|
3,271,507 | <p>Given a ring <span class="math-container">$R$</span> and a multiplicatively closed system <span class="math-container">$S\subset R$</span>, we define the ring <span class="math-container">$S^{-1}R=\{\frac rs | r\in R, s\in S\}$</span>. However if I have <span class="math-container">$R=\Bbb Z$</span> and <span class="math-container">$S=\Bbb Z \backslash 3\Bbb Z=\{0,1,2\}$</span>, do the denominators in <span class="math-container">$S^{-1}R$</span> contain only <span class="math-container">$1$</span> and <span class="math-container">$2$</span> or do they contain all the non multiples of <span class="math-container">$3$</span> in <span class="math-container">$\Bbb Z $</span>? </p>
| Alekos Robotis | 252,284 | <p>Be careful: <span class="math-container">$3\mathbb{Z}=\{m\in \mathbb{Z}: m=3k\:\text{for}\:k\in \mathbb{Z}\}.$</span> <span class="math-container">$S=\mathbb{Z}\setminus 3\mathbb{Z}$</span> is indeed a multiplicative system, but actually consists of all integers not divisible by <span class="math-container">$3$</span>. Then <span class="math-container">$S^{-1}R$</span> in your notation is the set of fractions <span class="math-container">$\frac{r}{s}$</span> so that <span class="math-container">$r\in \mathbb{Z}$</span> and <span class="math-container">$3\nmid s$</span>. For instance <span class="math-container">$\frac{3}{2}\in S^{-1}R$</span> as is <span class="math-container">$\frac{79}{14}$</span>.</p>
|
1,915,273 | <p>$$\int_0^r\sqrt{x-x^2}.dx$$</p>
<p>I only have basic calculus and would like to know how would one go about integrating an expression of this form.
I have tried substituting say $u=x-x^2$ but I'm still left with an $x$.</p>
<p>The method is not in my book and I can't find a similar example anywhere.</p>
<p>A hint in the right direction is sufficient.</p>
| egreg | 62,967 | <p>First of all, the function is defined for $0\le x\le 1$, so we have to assume the same for $r$.</p>
<p>Consider the curve $y=\sqrt{x-x^2}$ that we can square leading to $x^2+y^2-x=0$, which is the equation of a circle with center at $(1/2,0)$ and radius $1/2$. Thus a substitution that can work is
$$
x-\frac{1}{2}=\frac{1}{2}\cos t
$$
with $t\in[0,\pi]$ so $\sin t\ge0$.</p>
<p>With this substitution we have $\sqrt{x-x^2}=\frac{1}{2}\sin t$ “by design”, or, if you prefer to work it out,
$$
x-x^2=\frac{1}{2}+\frac{1}{2}\cos t-\frac{1}{4}+\frac{1}{2}\cos t+\frac{1}{4}\cos^2t=\frac{1}{4}\sin^2t
$$
Now the integral becomes, after noting that $dx=-\frac{1}{2}\sin t\,dt$,
$$
-\frac{1}{4}\int_{\pi}^{\arccos(2r-1)}\sin^2t\,dt=
\frac{1}{4}\int_{\arccos(2r-1)}^{\pi}\sin^2t\,dt
$$
An antiderivative of $\sin^2t$ is well known to be
$$
\frac{1}{2}(t-\sin t\cos t)
$$
so the integral is
$$
\frac{1}{8}\Bigl[t-\sin t\cos t\Bigr]_{\arccos(2r-1)}^{\pi}=
\frac{1}{8}\Bigl(\pi-\arccos(2r-1)+(2r-1)\sqrt{1-(2r-1)^2}\,\Bigr)
$$</p>
|
3,636,563 | <p>How to find the values of <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span> if
<span class="math-container">$3x²-3(1126)x=96y²+24(124)y=8z²-4(734)z $</span>?</p>
<p>I dont have any idea!! I think we can have many values of <span class="math-container">$x,y$</span> and <span class="math-container">$z$</span> at a time or it is a no solution??</p>
| manifold | 363,892 | <p>let partial sum be
<span class="math-container">$$S_n = \sum_{k=2}^n \int_{\frac{1}{k}}^{\frac{1}{k-1}}f$$</span> then we have to show that <span class="math-container">$\lim_{n \rightarrow \infty} S_n = \int_0^1 f$</span> and key point we will use is that <span class="math-container">$ \exists M>0 \,\,\, |f| < M$</span> since <span class="math-container">$f$</span> is Riemann Integrable (i.e it is bounded)</p>
<p><span class="math-container">$$ \Big|\sum_{k=2}^n \int_{\frac{1}{k}}^{\frac{1}{k-1}} f - \int_{0}^1 f \Big| = |\int_{1/n}^1 f - \int_{0}^1f | = \Big|\int_0^{\frac{1}{n}} f \Big| \leq \int_0^{\frac{1}{n}} |f| \leq \frac{M}{n} $$</span> </p>
<p>which implies that <span class="math-container">$n>\frac{M}{\epsilon}$</span> suffices and then we have limit</p>
<p>Edit:@Aditya proof is also good but existence of antiderivative is not necessary and it is doesn't imply from Integrability look at examples here : <a href="https://math.stackexchange.com/questions/239324/is-integrability-equivalent-to-having-antiderivative">Is "integrability" equivalent to "having antiderivative"?</a></p>
|
406,966 | <p>Define a graph with vertex set <span class="math-container">$\mathbb{R}^2$</span> and connect two vertices if they are unit distance apart. The famous <a href="http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem" rel="nofollow noreferrer">Hadwiger-Nelson problem</a> is to determine the <a href="http://en.wikipedia.org/wiki/Graph_coloring#Vertex_coloring" rel="nofollow noreferrer">chromatic number</a> <span class="math-container">$\chi$</span> of this graph. For the problem as stated (there are many variations), the best known bounds are <span class="math-container">$4 \leq \chi \leq 7$</span>. The lower bound comes from the existence of a clever subgraph on just seven vertices that is readily seen to require four colors. The upper bound comes from a straightforward tiling of the plane by monochromatic hexagons that admits a proper <span class="math-container">$7$</span>-coloring.</p>
<p>I like to show this problem to young students because they are always fascinated by the fact that we can cover "everything" that is known about it in just a few minutes. What are some other problems of this type?</p>
<blockquote>
<p>What are some famous problems for which the best known results are fairly obvious or elementary?</p>
</blockquote>
<p>Update: Aubrey de Grey has recently improved the <a href="https://arxiv.org/abs/1804.02385" rel="nofollow noreferrer">lower bound</a> to 5 using an elaborate subgraph, so we now know slightly more than the elementary.</p>
| MJD | 25,554 | <p>Let $S$ be a set of integers, each greater than 1, and $m(S)$ be the largest positive integer that is <em>not</em> expressible as a sum of elements of $S$, if it exists. (Elements of $S$ may be used more than once, of course, so that $7 = 2+2+3$ is considered to be a sum of 2's and 3's.) <a href="https://en.wikipedia.org/wiki/McNugget_number#McNugget_numbers">A well-known example</a> is $m(\{6, 9 ,20\}) = 43$.</p>
<p>J.J. Sylvester showed (1884) that $m(S)$ exists if and only if $\gcd(S) = 1$, and also that $m(\{s_1, s_2\})$, when it exists, is equal to $$s_1s_2 - s_1 - s_2.$$ Both of these could certainly be conjectured, and probably proved, in an hour by a bright high-school student.</p>
<p>But for $|S| > 2$ the problem is wide open.</p>
|
118,185 | <blockquote>
<p>Given a polynomial f with only real coefficients, the zeros of f are conjugates of zeros of f.</p>
</blockquote>
<p>(Sorry my English left me, when trying to formulate this theorem.) </p>
<p>So I would like to know how to refer to this theorem in a scientific text.</p>
| Stefan Hansen | 25,632 | <p>There is an equivalent definition of uniform integrability which goes: Let $(\Omega,\mathcal{F},P)$ be a probability space. A set $\mathcal{H}$ of random variables is uniformly integrable if and only if
$$
\sup_{X\in \mathcal{H}}\int_\Omega |X| dP < \infty
$$
and
$$
\forall \epsilon >0\, \exists \delta >0\;\forall A\in \mathcal{F}: P(A)\leq \delta \Rightarrow \sup_{X\in\mathcal{H}} \int_A |X| dP\leq \epsilon.
$$
Maybe this it what you are looking for.</p>
|
118,185 | <blockquote>
<p>Given a polynomial f with only real coefficients, the zeros of f are conjugates of zeros of f.</p>
</blockquote>
<p>(Sorry my English left me, when trying to formulate this theorem.) </p>
<p>So I would like to know how to refer to this theorem in a scientific text.</p>
| Did | 6,179 | <p>Three conditions are involved:</p>
<blockquote>
<p>(C) For every <span class="math-container">$\varepsilon\gt0$</span>, there exists a finite <span class="math-container">$c$</span> such that, for every <span class="math-container">$X$</span> in <span class="math-container">$\mathcal H$</span>, <span class="math-container">$\mathrm E(|X|:|X|\geqslant c)\leqslant\varepsilon$</span>.</p>
<p>(C1) There exists a finite <span class="math-container">$C$</span> such that, for every <span class="math-container">$X$</span> in <span class="math-container">$\mathcal H$</span>, <span class="math-container">$\mathrm E(|X|)\leqslant C$</span>.</p>
<p>(C2) For every <span class="math-container">$\varepsilon\gt0$</span> there exists <span class="math-container">$\delta\gt0$</span> such that, for every measurable <span class="math-container">$A$</span> such that <span class="math-container">$\mathrm P(A)\leqslant\delta$</span> and every <span class="math-container">$X$</span> in <span class="math-container">$\mathcal H$</span>, <span class="math-container">$\mathrm E(|X|:A)\leqslant\varepsilon$</span>.</p>
</blockquote>
<p>Then (C) is equivalent to (C1) and (C2). To wit:</p>
<ul>
<li><p>(C) implies (C1) since <span class="math-container">$\mathrm E(|X|)\leqslant c+\mathrm E(|X|:|X|\geqslant c)$</span>.</p>
</li>
<li><p>(C) implies (C2) since <span class="math-container">$\mathrm E(|X|:A)\leqslant c\mathrm P(A)+\mathrm E(|X|:|X|\geqslant c)$</span>.</p>
</li>
<li><p>(C1) and (C2) imply (C) since <span class="math-container">$\mathrm P(|X|\geqslant c)\leqslant\mathrm E(|X|)/c$</span>.</p>
</li>
</ul>
<p>The intuition might lie in the decomposition <span class="math-container">$A=(A\cap[|X|\lt c])\cup(A\cap[|X|\geqslant c])$</span> used to show that (C) implies (C2): for any given <span class="math-container">$c$</span>, the expectation of <span class="math-container">$|X|$</span> on the first part is controlled uniformly over <span class="math-container">$X$</span> by something of the order of <span class="math-container">$\mathrm P(A)$</span> and the expectation of <span class="math-container">$|X|$</span> on the second part is controlled uniformly over <span class="math-container">$A$</span> thanks to condition (C).</p>
|
1,861,292 | <p>I'm reading Osbourne's Basic Homological Algebra, and on page 18 he has this situation where we've got a ring $R$ and a right-ideal $I$, and some left $R$-module $B$. He says $I\otimes B$ is not a subgroup of $R\otimes B$, but he doesn't give an example, just states "the free abelian groups and equivalence relations defining them are different". Well, sure, but the free abelian group on $I\times B$ is certainly a subgroup of the one on $R\times B$, and I cannot see how restricting the equivalence relations on $R\times B$ to just elements in $I\times B$ is going to make some new group that doesn't exist inside $R\otimes B$. </p>
<p>I've tried to make examples, but failed to find one which demonstrated this. </p>
<p>I've seen examples of tensor products which do this with more relaxed conditions, such as $\mathbb{Z}\otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z})$ is not a subgroup of $\mathbb{Q}\otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z})$ since the latter is 0 while the former is not, although obviously $\mathbb{Z}$ is a subgroup of $\mathbb{Q}$ - but it is not an ideal. </p>
<p>I would appreciate some examples, and some idea of what causes it to happen. </p>
| Rob Arthan | 23,171 | <p>One example is the case $R = \Bbb{Z}_4$, $B = I = \{0, 2\}$ that Osbourne is talking about at that point in the book. In this example, you find that $I \otimes B \simeq R \otimes B \simeq \Bbb{Z}_2$, but the natural mapping from $I\otimes B$ to $R \times B$ is not an injection. In fact, $2 \otimes 2$ generates $I \otimes B$ but maps to zero in $R \otimes B$, because in $R \otimes B$, we have $2 \otimes 2 = 4(1 \otimes 1) = 0$, but $1 \otimes 1$ isn't in the free abelian group on $I \times B$, and in $I \otimes B$ we don't have this equation. ($R \otimes B$ is generated by $1 \otimes 2$.)</p>
|
325,890 | <p>To evaluate this integral I have got to use Parseval's Theorem and the fourier transform of</p>
<p>$$s(x)=\begin{cases}
0 & x\leq -a \\
1 & -a<x<a \\
0 & x \geq a.\end{cases}$$</p>
<p>This works out to be</p>
<p>$$\tilde{s}(k)=\frac{e^{ika}}{ik}-\frac{e^{-ika}}{ik}$$</p>
<p>I have then changed this into sin and cos and got to the stage of</p>
<p>$$|\tilde{s}(k)|^{2}=\left|\frac{-4\sin^{2}(ka)}{-k^{2}}\right|$$ </p>
<p>but this is as far as I have got and I can't figure out what to do from here.</p>
| TCL | 3,249 | <p>Take $a=1$ in your function $s(x)$. The Fourier transform of $s$ is $$\frac{1}{\sqrt{2\pi}}\int_{-1}^1 e^{-iwx}\,dx=\sqrt{\frac{2}{\pi}}\frac{\sin w}{w} $$ Then Parseval (also called Plancherel) relation says $$\int_{-\infty}^{\infty}\frac{2}{\pi}\frac{\sin^2w}{w^2}\,dw = \int_{-1}^1 1^2\,dx=2.$$</p>
|
177,643 | <p>Let $I=\{0, 1, \ldots \}$ be the multiplicative semigroup of non-negative integers. It is possible to find a ring $R$ such that the multiplicative semigroup of $R$ is isomorphic (as a semigroup) to $I$?</p>
| robjohn | 13,854 | <p>With all these simple solutions available, how about a complex one?</p>
<ol>
<li>The argument of the product of two complex numbers is the sum of their arguments.</li>
<li>The imaginary part of a complex number divided by its real part is the tangent of its argument.</li>
</ol>
<p>Consider the product
$$
\begin{align}
&(1+i\tan(A))(1+i\tan(B))(1+i\tan(Y))\\
&=1-\tan(A)\tan(B)-\tan(B)\tan(Y)-\tan(Y)\tan(A)\\
&+i\,(\tan(A)+\tan(B)+\tan(Y)-\tan(A)\tan(B)\tan(Y))\tag{1}
\end{align}
$$
Using 1. and 2., $(1)$ says that
$$
\tan(A+B+Y)=\frac{\tan(A)+\tan(B)+\tan(Y)-\tan(A)\tan(B)\tan(Y)}{1-\tan(A)\tan(B)-\tan(B)\tan(Y)-\tan(Y)\tan(A)}\tag{2}
$$</p>
|
3,603,066 | <p>My question: How does one distinguish between two embeddings of the same graph on the plane? </p>
<p>For instance, are two such embeddings considered the same if the degree sequence of their faces are the same? Or does even the slightest change, such as drawing an edge more crooked in one embedding than the other, constitute a different embedding? (Or perhaps none of these?)</p>
| User49582934 | 106,207 | <p>This can be solved by taking log on both sides.</p>
<p>Growth formula is given by
<span class="math-container">$$ P = P_0 e^{rt} $$</span></p>
<p>Here, <span class="math-container">$P_0$</span> and <span class="math-container">$P$</span> are initial and final length of the tree, respectively, <span class="math-container">$r$</span> is continuously compounded rate of growth, <span class="math-container">$t$</span> is time in years.</p>
<p><span class="math-container">$$15 = 3 e^{r \times 12}$$</span>
<span class="math-container">$$\log_e(15 / 3) = r \times 12 $$</span></p>
<p><span class="math-container">$$r \approx 0.1341$$</span></p>
<p>Hence the growth rate is 13.41% per year.</p>
|
2,586,466 | <p>Write the value of $n$ if the sum of n terms of the series $1+3+5+7...n =n^2$.</p>
<p>I'm not getting the right value if I proceed with the general formula for finding sum of n terms of a arithmetic series. The general summation formula for arithmetic series is $\frac{n(2a+(n-1)d)}{2}$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.</p>
| Community | -1 | <p>Note that the sum of an arithmetic progression with $n$ terms having a starting term $a$ and common difference $d$ is: $$S_n = \frac{n}2 \left[2a + (n-1)d\right]$$ Now, if $a=1, d=2$, what do you get?</p>
<p>$$S_n = \frac{n}{2}\left[2+(n-1)2\right] = \frac{n}{2}\left[2n \right] = n^2$$</p>
|
2,759,112 | <p>Are there cases where two events have high probability (i.e. each of them has high probability) but, at the same time, the probability that they both obtain is low? If yes, could you please provide a simple, daily-life example?</p>
| trancelocation | 467,003 | <p>There is always a restriction to the probability of $A \cap B$:</p>
<p>$$1 \geq P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow$$
$$P(A \cap B) \geq P(A) + P(B) -1$$
For example:</p>
<ul>
<li>$P(A) = 0.9$ and $P(B) = 0.9 \Rightarrow P(A \cap B) \geq 0.9+0.9-1 = 0.8$</li>
</ul>
<p>So, the higher the probability of $A$ and $B$, the higher the lower bound for the probability of $A \cap B$.</p>
|
2,738,679 | <p>I am having trouble to prove this exercise for a Real Analysis course. I tried two approaches. However, I am insecure about both of them.</p>
<p>This is the question:</p>
<p>If $X$ is a partition of ℕ* such that:</p>
<p>i) $1 ∈ X$;</p>
<p>ii) For all $n ∈ ℕ$*, if $n ∈ X$, then $2n ∈ X$;</p>
<p>iii) For all $n ∈ ℕ$*, if $n+1 ∈ X$, then $n ∈ X$;</p>
<p>Show that $ X = ℕ$*</p>
<hr>
<p>First attempt:</p>
<p>I tried using the definition of set partition and a family of subsets. This is my try: </p>
<p>Let X be a part of $ℕ$* </p>
<p>Hence, we have a family of subsets with:</p>
<p>${ X_i : i ∈ I}$</p>
<p>$ X_1 = {1} $</p>
<p>$ X_2 = {2n} $ For all $n ∈ ℕ$* </p>
<p>$ X_3 = {n+1} $ For all $n ∈ ℕ$* and $n ∉ X_2 $ </p>
<p>All properties of a partition hold:</p>
<p>$ X_1, X_2 ,X_3 ≠ ø $ </p>
<p>$ X_1 ∪ X_2 ∪ X_3 = X$ </p>
<p>$ X_1 ∩ X_2 ∩ X_3 = ø $</p>
<p>$ X_1 ∩ X_2 = ø $</p>
<p>$ X_1 ∩ X_3 = ø $</p>
<p>$X_2 ∩ X_3 = ø $</p>
<p>Finally, $ ⋃ X_i$ = X = $ℕ$*</p>
<hr>
<p>The second approach is more intuitive and I don't know how to express it mathematically!</p>
<p>The second rule on the question states that:</p>
<p>ii) For all $n ∈ ℕ$*, if $n ∈ X$, then $2n ∈ X$;</p>
<p>So, every number expressed as $2^k$ belongs to X.</p>
<p>And, by the third rule: </p>
<p>iii) For all $n ∈ ℕ$*, if $n+1 ∈ X$, then $n ∈ X$;</p>
<p>I can get from $2^k$ to any number between $2^k$ and 1, by decreasing $2^k$</p>
<hr>
<p>Thanks!</p>
| Rai | 548,890 | <p>Your second approach is correct.<br>
You want to prove that $X = \mathbb{N}$<br>
For being $X$ a subset of $\mathbb{N}$ , then $X \subset\mathbb{N}$<br>
Now we have to proof that $\mathbb{N}\subset X$<br>
So let be $n \in\mathbb{N}$, then it exist some $k \in\mathbb{N}$ with $n=2^{k+1}-p$, where p$\leq k$<br>
For i): $1 \in X \Rightarrow 2 \in X$$\;$ for ii)<br>
For ii) again: $2 \in X \Rightarrow 2^{k+1} \in X \Rightarrow 2^{k+1}-p \in X$ $\;$(because of iii)<br>
So $\mathbb{N}\subset X$<br>
And then $X = \mathbb{N}$</p>
|
6,155 | <p>I have a category $C$, which is equipped with a symmetric monoidal structure (tensor product $\otimes$, unit object $1$). My category also has finite coproducts (I'll write them using $\oplus$, and write $0$ for the initial object), and $\otimes$ distributes over $\oplus$.</p>
<p>By an <em>exponential monad</em>, I mean a monad $(T,\eta,\mu)$ on $C$, where the functor $T:C\to C$ is equipped with some structure maps of the form
$$\nu \colon 1 \to T(0)$$
and
$$\alpha\colon T(X)\otimes T(Y) \to T(X\oplus Y).$$
The structure maps are isomorphisms, and are suitably "coherent" with respect to the two monoidal structures $\otimes$ and $\oplus$.</p>
<p>The simplest example is: $C$ is the category of $k$-vector spaces, and $T=\mathrm{Sym}$ is the commutative $k$-algebra monad (i.e., $\mathrm{Sym}(X)$ is the symmetric algebra $\bigoplus \mathrm{Sym}^q(X)$).</p>
<p>Now, I'm sure I can work out all the formalism that I need for this, if I have to. My question is: is there a convenient place in the literature I can refer to for this? Alternately, is there suitable categorical language which makes this concept easy to talk about?</p>
<p>I'd also like to have a good formalism for talking about a "grading" on $T$. This means a decomposition of the functor $T=\bigoplus T^q$, where $T^q\colon C\to C$ are functors, which have "nice" properties (for instance, $T^m(X\oplus Y)$ is a sum of $T^p(X)\otimes T^{m-p}(Y)$). The motivating example again comes from the symmetric algebra: $\mathrm{Sym}=\bigoplus \mathrm{Sym}^q$.</p>
| Peter Teichner | 4,625 | <p>There is a beautiful explanation of the Chern character that my student Fei Han proved in his thesis: The Chern character is given by the map that "crosses with the circle". The hard part is to explain why domain and range of this map are K-theory respectively de Rham cohomology. This uses isomorphisms</p>
<p>$K^0(X) \cong 1|1-EFT[X]$ and $H^{ev}_{dR}(X) \cong 0|1-EFT[X]$</p>
<p>where $d|1-EFT[X]$ are concordance classes of $d|1$-dimensional Euclidean field theories over the manifold $X$. Since the circle of length one is a Euclidean 1-manifold, it is not hard to believe, modulo the precise definitions, that crossing with it gives a map as required. </p>
<p>In fact, his result works even before taking concordance classes, where the left hand side is replace by vector bundles with connection and the right hand side becomes (even closed) differential forms.</p>
|
23,639 | <p>Prove that for every positive integer $x$ of exactly four digits, if the sum of digits is divisible by $3$, then $x$ itself is divisible by 3 (i.e., consider $x = 6132$, the sum of digits of $x$ is $6+1+3+2 = 12$, which is divisible by 3, so $x$ is also divisible by $3$.)</p>
<p>How could I approach this proof? I'm not sure where I would even begin. </p>
| Qiang Li | 6,016 | <p>Actually, this is true for an integral number with any digits. The proof is quite easy. Let's denote the integral number by $\overline{a_n a_{n-1} \ldots a_1}$. If the sum of its digits $\sum_{i=1}^n{a_i}$ is divisible by 3, then $\sum_{i=1}^n{(1+\overline{9...9}_{i-1})*a_i}$ is too. Here $\overline{9...9}_{i-1}$ denotes the integer with $i-1$ 9's. But this second sum is just the original number $\overline{a_n a_{n-1} \ldots a_1}$ expanded.</p>
|
179,322 | <p>I'm interested in classes C of $R^1$-valued random variables which possess the following properties:</p>
<p>1) the sum of two independent random variables from class C belongs to class C;</p>
<p>2) for any $\lambda \in R^1$, $\xi \in C$ we have $\lambda\xi \in C$; </p>
<p>3) any random variable from class $C$ has tails which are heavier than the Gaussian tails;</p>
<p>4) for any $\xi \in C$ we have $\mathrm{E}\xi=0, \mathrm{E}\xi^2<\infty$. </p>
<p>If we omit the restriction $\mathrm{E}\xi=0$ and restriction No. 2 then there is an obvious example:
a class of gamma-distributed random variables with one parameter of the distribution fixed. Can anyone suggest other examples? </p>
<p>Thanks in advance. </p>
<p>Best wishes, Ievgen. </p>
| Vivek Shende | 4,707 | <p>I think at some level this is a convention having to do with the fact that our exponential is $x \mapsto e^x$ rather than $x \mapsto e^{2\pi i x}$. But having made this choice, here's one way that factor appears:</p>
<p>Over $\mathbb{C}$, you have the exponential sequence </p>
<p>$$0 \to 2\pi i \mathbb{Z} \to \mathbb{G}_a \to \mathbb{G}_m \to 1$$</p>
<p>Line bundles are classified by $H^1(\,\cdot\,, \mathbb{G}_m)$, and the first Chern class is the natural map </p>
<p>$$c_1: H^1(\, \cdot \, , \mathbb{G}_m) \to H^2(\,\cdot \,, 2\pi i\mathbb{Z})$$</p>
<p>There's also the $\ell$-th power map. If $\mu_\ell$ is the group of $\ell$-th roots of unity, then you get a sequence:</p>
<p>$$1 \to \mu_\ell \to \mathbb{G}_m \to \mathbb{G}_m \to 1$$ </p>
<p>Taking etale cohomology gives a notion of Chern class sensible over any field
(... of characteristic prime to $\ell$):
$$c_1: H^1_{et}(\, \cdot \, ,\mathbb{G}_m)
\to H^2_{et}(\,\cdot \,, \mu_\ell)$$ </p>
<p>The first sequence maps to the second by sending the first two terms via $\mathrm{exp}(\,\cdot\, /\ell)$, and the third by the identity. The second sequence for $\mu_{\ell^2}$ maps to that for $\mu_\ell$ by the $\ell$-th power map on the first two terms and the identity on the third. By definition we have $\lim \mu_{\ell^n} = \mathbb{Z}_\ell(1)$, so taking an inverse limit gives</p>
<p>$$c_1: H^1_{et}(\, \cdot \, ,\mathbb{G}_m)
\to H^2_{et}(\,\cdot \,, \mathbb{Z}_\ell(1))$$ </p>
<p>So if you want your Chern classes to line up, it's nice to identify the Tate twist $(1)$ in etale cohomology with $\cdot 2\pi i$ in the singular theory. </p>
|
50,923 | <p>I have two questions:</p>
<ol>
<li><p>How can I calculate compound interest when I want to reinvest a percentage of the sum of my principal and interest periodically?</p>
<p>To make it clearer,</p>
<p>Usually I use this formula for calculating compound interest: <span class="math-container">$M = P( 1 + i )^n$</span>.</p>
<p>This time I want to compound only a percentage of the sum of the principal and interest</p>
<p><strong>EDIT</strong></p>
<p>This is what I am looking at:</p>
<blockquote>
<p>I invest <span class="math-container">$1000$</span> for a month, at the end of the month I get <span class="math-container">$1100$</span> (principal plus interest), and I want to re-invest <span class="math-container">$60%$</span> of the total amount for <span class="math-container">$n$</span> periods.</p>
</blockquote>
</li>
<li><p>I want to compound the interest on the principal daily till the investment matures, i.e. if am investing for one year, I want the daily interest to be compounded to the principal.</p>
</li>
</ol>
| Ross Millikan | 1,827 | <p>For 2, you just regard each day as a separate term. So the interest rate is the rate per day and you have 365 (or 366 or 360!?) terms. For 365, if the interest rate were 10%/year, it would be 10/365$\approx $0.0274%/day, and the final value would be $1.000274^{365}$ times the original value, or about 1.105 times the original value.</p>
<p>For 1, what do you want to calculate? The final balance after N terms when at the end of each term the principal is reduced? The amount of interest received under those conditions? Probably it is easiest to make a spreadsheet for what you want.</p>
|
50,923 | <p>I have two questions:</p>
<ol>
<li><p>How can I calculate compound interest when I want to reinvest a percentage of the sum of my principal and interest periodically?</p>
<p>To make it clearer,</p>
<p>Usually I use this formula for calculating compound interest: <span class="math-container">$M = P( 1 + i )^n$</span>.</p>
<p>This time I want to compound only a percentage of the sum of the principal and interest</p>
<p><strong>EDIT</strong></p>
<p>This is what I am looking at:</p>
<blockquote>
<p>I invest <span class="math-container">$1000$</span> for a month, at the end of the month I get <span class="math-container">$1100$</span> (principal plus interest), and I want to re-invest <span class="math-container">$60%$</span> of the total amount for <span class="math-container">$n$</span> periods.</p>
</blockquote>
</li>
<li><p>I want to compound the interest on the principal daily till the investment matures, i.e. if am investing for one year, I want the daily interest to be compounded to the principal.</p>
</li>
</ol>
| AndyS | 13,294 | <p><strong>1.</strong> For question 1 you just have to redefine the amount you want to invest after the end of every period. If $x$ is the daily interest rate (if, for example, the daily interest rate is 5% then $x=0.05$), then after 30 days (1 month) your initial capital, $c$, has become
$$
c_1=(1+x)^{30} c.
$$</p>
<p>At that point you only want to invest a percentage of $c_1$, given, say, by $p$ (in your example $p=0.6$). All you do is multiply with $p$:
$$
c_{1,\text{inv}}=(1+x)^{30} pc.
$$</p>
<p>At the end of the second month you have
$$
c_2=(1+x)^{30}c_{1,\text{inv}}=(1+x)^{60}pc.
$$
After $n$ months this becomes
$$
c_n=(1+x)^{30n}p^{n-1}c.
$$</p>
<p><strong>2.</strong> For question 2 just take the yearly interest, $X$, divide it by the number of days in the year, and apply it every day. With initial capital $C$ you get, at the end of the year,
$$
C_1=\left(1+\frac{X}{365}\right)^{365}C
$$</p>
|
187,486 | <p>Let $SL(2,{\mathbb R})$ act on ${\mathbb R}^2$ by matrix multiplication.</p>
<p>What is known about group cohomology $H^*(SL(2,{\mathbb R}),{\mathbb R}^2)$?</p>
<p>And about $$H^*(\Gamma,{\mathbb R}^2)$$ for a cocompact lattice $$\Gamma\subset SL(2,{\mathbb R})?$$</p>
| Oscar Randal-Williams | 318 | <p>It is zero. This is an application of the "centre kills" trick, which I will state in homology.</p>
<p><strong>Trick.</strong> Let $M$ be a $G$-module for which there is an element $z$ in the centre of $G$ which acts as $-1$ on $M$. Then $2H_*(G;M)=0$.</p>
<p>In your situation the homology is a real vector space, so if multiplication by 2 kills it then it is already dead. You can then use Universal Coefficients to get the result in cohomology, or just work out the details of the trick in cohomology.</p>
<p>The proof is as follows. For any element $z \in G$, the map
$$m \vert g_1 \vert g_2 \vert \cdots \vert g_n \longmapsto m \cdot z \vert g_1^z \vert g_2^z \vert \cdots \vert g_n^z$$
on the bar complex is i) a chain map, and ii) chain homotopic to the identity, so induces the identity map on homology. But if $z$ has the properties described then this map is equal to $-1$.</p>
|
200,298 | <p>How to find the sum of the absolute values for the roots of this equation:</p>
<p>$$x^4-4x^3-4x^2+16x-8=0$$</p>
| Clive Newstead | 19,542 | <p>Notice that
$$\begin{align}x^4-4x^3-4x^2+16x-8 &= (x-1)^4 - 10(x-1)^2 + 1 \\
&= ((x-1)^2-5)^2-24 \end{align}$$
so you can actually calculate the roots explicitly and sum their absolute values.</p>
|
1,687,206 | <p>Do I multiply $4$ to the power of $2$ or do I multiply $8$ to the power of $2$? What is the answer? I'm thinking it's $76$.</p>
| Soham | 242,402 | <p>$$V=3b+2b^2$$</p>
<p>So, when $b=4$, we get $$V=3\times4+2\times4^2=12+(2\times16)=12+32=44$$</p>
|
1,702,885 | <blockquote>
<blockquote>
<p>Question: Prove $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$</p>
</blockquote>
</blockquote>
<p>RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$</p>
<p>$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)}{\sin(x)}-2\cos^2(x)}{1-\frac{\cos^2(x)}{\sin^2(x)}} $$</p>
<p>$$ ⇔ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)\sin(x)}{\sin^2(x)}-\frac{2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$</p>
<p>$$ ⇔ \frac{\frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$</p>
<p>$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)} \times {\frac{\sin^2(x)}{\sin^2(x)-\cos^2(x)}} $$</p>
<p>$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)-\cos^2(x)} $$</p>
<p>$$ ⇔ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$</p>
<p>Now I am stuck</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Writing $\cos x\sin x=u$</p>
<p>$$1-u-2u^2=(1-2u)(1+u)$$</p>
<p>$1-2u=(\sin x-\cos x)^2$</p>
|
925,005 | <p>Show that $$\|a+b+c\|^2\leq 3\|a\|^2+3\|b\|^2+3\|c\|^2$$
where $a,b,c$ are in some Hilbert space $(H,\langle\cdot,\cdot \rangle)$?</p>
<p>I see that we have $$\|a+b\|^2\leq2\|a\|^2 +2 \|b\|^2$$ due to the parallelogram law. What about the first inequality?</p>
| Kim Jong Un | 136,641 | <p>You can expand the LHS and use $\left<x,y\right>+\left<y,x\right>\leq ||x||^2+||y||^2$ 3 times. This latter inequality follows by expanding the non-negative entity $\left<x-y,x-y\right>$.</p>
|
3,490,198 | <p>For all <span class="math-container">$a,b \in \mathbb{Z}$</span> and for all <span class="math-container">$m,n \in \mathbb{N}\setminus \left\lbrace0\right\rbrace$</span>,</p>
<p>is <span class="math-container">$a^{48m+1}+b^{48n+1} \equiv 0 \pmod{39} \iff a+b \equiv 0 \pmod{39}$</span>?</p>
<p>I think the answer is yes, but I can't prove it. Is there somebody who can help me? Thank you.</p>
| nonuser | 463,553 | <p>Hint: Use Euler theorem <span class="math-container">$$a^{\varphi(n)}\equiv 1\pmod n$$</span>
if <span class="math-container">$\gcd(a,n )= 1$</span>. Notice that <span class="math-container">$$\varphi (39) = 2\cdot 12 =24$$</span></p>
<hr>
<p><strong>Edit: 12. 29. 2019</strong></p>
<ul>
<li>a) If <span class="math-container">$(a,39) =1$</span> then clearly also <span class="math-container">$(b,39)=1$</span> so we have <span class="math-container">$a^{24} \equiv 1$</span> and <span class="math-container">$b^{24} \equiv 1$</span> so <span class="math-container">$a^{48m}a \equiv a$</span> and <span class="math-container">$b^{48n}b \equiv b$</span> and thus: <span class="math-container">$$a^{48m+1}+b^{48n+1} \equiv a+b$$</span> and the equivalence follows. </li>
<li><p>b) If <span class="math-container">$(a,39) \ne 1$</span> then <span class="math-container">$3\mid a$</span> or <span class="math-container">$13 \mid a$</span> or <span class="math-container">$39\mid a$</span>. </p>
<ul>
<li><p>If <span class="math-container">$39\mid a$</span> then also <span class="math-container">$39\mid b$</span> and claim follows</p></li>
<li><p>If <span class="math-container">$3\mid a$</span> and <span class="math-container">$13\nmid a$</span> then if <span class="math-container">$3\mid a^{48m+1}+b^{48n+1} $</span> we have also <span class="math-container">$3\mid b$</span> so <span class="math-container">$3\mid a+b$</span> and vice versa. And for <span class="math-container">$13$</span> you can repeat a process in case a). </p></li>
<li><p>If <span class="math-container">$13\mid a$</span> and <span class="math-container">$3\nmid a$</span> then repeat a process from previous case</p></li>
</ul></li>
</ul>
|
1,281,142 | <p>I am trying to find all the closed points of $\mathbb{A}_{\mathbb{R}}^2$.</p>
<p>After a quick google research, I found that $\mathbb{A}_{\mathbb{R}}^2 = \operatorname{Spec}(\mathbb{R}[x,y])$ and then all we need to find is the maximal ideals of $\mathbb{R}[x,y]$.</p>
<p>However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.</p>
<p>And more than that, how do I find the maximal ideals of $\mathbb{R}[x,y]$? Seems hard.</p>
<p>Any help would be appreciated.</p>
<p>Thanks!</p>
| Daniel McLaury | 3,296 | <p>Re: </p>
<blockquote>
<p>However, set-theoretically this equalty does not make sense, as we are talking about points and prime ideals.</p>
</blockquote>
<p>The points of a scheme are*, by definition, prime ideals of a ring. If this is new to you, you're probably not ready to be looking at this question until you go back and learn what a scheme is.</p>
<p>Once you understand the question, Slade's answer shows you a good way of working the problem.</p>
<p>*Or, if you want to be incredibly pedantic, <em>correspond to</em> the ideals of a ring, since a scheme is a locally ringed space that <em>happens to be</em> locally isomorphic to the spectrum of a ring.</p>
|
243,849 | <p>I was working on an examples from my textbook concerning transforming formulae into disjunctive-normal form (DNF) until I found an expression that I cannot solve. I hope somebody can help me transform the following statement into DNF:</p>
<p>$$ (\lnot q \lor r) \land ( q \lor \lnot r)$$</p>
| Dominik | 50,527 | <p>By the distibution law one obtains $(\lnot q \lor r) \land ( q \lor \lnot r) \leftrightarrow ((\lnot q \lor r) \land q) \lor ((\lnot q \lor r) \land \lnot r) \leftrightarrow (\lnot q \land q) \lor (r \land q)\lor (\lnot q \land \lnot r) \lor (r \land \lnot r)\leftrightarrow (r \land q)\lor (\lnot q \land \lnot r)$</p>
|
1,368,180 | <p>Here is a tricky compass and straightedge construction problem. </p>
<blockquote>
<p>Given triangle $\triangle ABC$ and point $D$ on segment $\overline{AB}$, construct point $P$ on line $\overleftrightarrow{CD}$ such that $\angle APB = \angle BAC$. </p>
</blockquote>
<p>This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.</p>
| Tae Hyung Kim | 94,401 | <p>I misread the question the first time I answered this, so let me try again. </p>
<p>The condition that you gave $\angle APB = \angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $\triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$\angle APB = 180^\circ - \angle AID = \angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$\triangle ABC \sim \triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $\angle B$ onto side $AC$), and everything can be reversed to find $P$. </p>
|
1,368,180 | <p>Here is a tricky compass and straightedge construction problem. </p>
<blockquote>
<p>Given triangle $\triangle ABC$ and point $D$ on segment $\overline{AB}$, construct point $P$ on line $\overleftrightarrow{CD}$ such that $\angle APB = \angle BAC$. </p>
</blockquote>
<p>This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.</p>
| PRITHU PURKAIT | 635,014 | <p>Construct line <span class="math-container">$AK$</span> with <span class="math-container">$K$</span> on <span class="math-container">$BC$</span> (produced, if necessary) and <span class="math-container">$\angle BAK = \angle ACB$</span>. The circumcircle of <span class="math-container">$\triangle ABK$</span> intersects <span class="math-container">$CD$</span> (again, produced if needed) at two points <span class="math-container">$P$</span> and <span class="math-container">$P1$</span>. Without loss of generality, let <span class="math-container">$P$</span> be on the same side of <span class="math-container">$AB$</span> as <span class="math-container">$K$</span>. Then, <span class="math-container">$\angle APB = \angle AKB$</span> and by construction <span class="math-container">$\angle AKB = \angle BAC$</span>. (since<span class="math-container">$\angle ABC + \angle BCA + \angle BAC = \angle ABK + \angle BAK + \angle AKB) $</span>.<br>
Thus we are done</p>
|
698,460 | <p>I've literally tried everything including proofs by cases. If I were to try a an odd integer, then I would get $8k^3 + 4k^2 + 8k^2 + 4k + 1$, which obviously couldn't convert to a $\mod 7$, and I'm honestly not sure what I can do! Does anyone here have any tips?</p>
<p>Note, this is a soft question.</p>
| Sangchul Lee | 9,340 | <p>Euler's identity will help you expand $\cos^{2N}\theta$ in terms of trigonometric polynomial. Then use the identity back to simplify them. For example, if $N = 1$, then</p>
<p>$$ \cos^{2}\theta = \left( \frac{e^{i\theta} + e^{-i\theta}}{2} \right)^{2} = \frac{e^{2i\theta} + 2 + e^{-2i\theta}}{4}= \frac{1}{2} + \frac{\cos2\theta}{2}, $$</p>
<p>which is the Fourier expansion of $\cos^{2}\theta$.</p>
|
432,262 | <p>Let <span class="math-container">$x_0=1$</span> and
<span class="math-container">$$x_{k+1} = (1-a_k)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x_k}\right)$$</span>
where <span class="math-container">$a_n$</span> is a known sequence satisfying that <span class="math-container">$a_k\in(0,1)$</span> for all <span class="math-container">$k$</span> and <span class="math-container">$a_k\to 0$</span> as <span class="math-container">$k\to\infty$</span>. How to prove that <span class="math-container">$x_k\to 1$</span> as <span class="math-container">$k\to\infty$</span>?</p>
<hr />
<p>The difficulty here is that</p>
<ol>
<li>It is not known how fast <span class="math-container">$a_k$</span> converges to zero, and I don't know how it affect the convergence of <span class="math-container">$x_k$</span>;</li>
<li><span class="math-container">$x_k$</span> may change sign and is not monotone, so I don't know how to prove <span class="math-container">$x_k$</span> even converges;</li>
<li>Furthermore, if we assume <span class="math-container">$x_k$</span> do converge to some limit <span class="math-container">$x^*$</span>, then by taking the limit,
<span class="math-container">$$x^*=(1-0)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x^*}\right)$$</span>
I find there are two possible solution <span class="math-container">$x^*=1/2$</span> or <span class="math-container">$x^*=1$</span>. How to exclude the case that <span class="math-container">$x^*=1/2$</span>?</li>
</ol>
| Alapan Das | 156,029 | <p>Taking <span class="math-container">$b_k=(1-a_k), y_k=\sqrt{2}x_k, c=\frac{3}{\sqrt{2}}$</span> we have <span class="math-container">$y_{k+1}=b_k(c-\frac{1}{y_k})$</span>. Now as <span class="math-container">$b_k \rightarrow 1$</span> assuming it's variation to be sufficiently small for large <span class="math-container">$k$</span>, we take <span class="math-container">$b_k=1 \forall k\geq k_0$</span>.</p>
<p>The recurrence relation becomes <span class="math-container">$y_{k+1}=c-\frac{1}{y_k},k≥k_0$</span>. The constancy points of this recurrence is at <span class="math-container">$\frac{1}{\sqrt{2}}$</span> and <span class="math-container">$\sqrt{2}$</span>. So, we have <span class="math-container">$\Delta y=-\frac{(y-1/\sqrt{2})(y-\sqrt{2})}{y}$</span>,which is positive when <span class="math-container">$y \in [1/\sqrt{2},\sqrt{2}]$</span>, with the maximum being at <span class="math-container">$1$</span>. Now, for some <span class="math-container">$y_k$</span> in this interval (<span class="math-container">$I$</span>); <span class="math-container">$(\sqrt{2}-y)-\Delta y_k=\frac{1-y/\sqrt{2}}{y} \geq 0$</span> implying that <span class="math-container">$y_{k+1} \in I; y_{k+1}>y_k$</span>.</p>
<p>So, if <span class="math-container">$y_{k_0} \in I$</span>, the sequence converges to <span class="math-container">$\sqrt{2}$</span> from left. While if
<span class="math-container">$y_{k}>\sqrt{2}$</span>, <span class="math-container">$y_{k+1}=\frac{3}{\sqrt{2}}-\frac{1}{y}\in (\sqrt{2}, \frac{3}{\sqrt{2}})$</span>, hence <span class="math-container">$y_k$</span> converges to <span class="math-container">$\sqrt{2}$</span> from right.</p>
<p>Similarly for <span class="math-container">$y_{k}<0$</span>, <span class="math-container">$y_{k+1}=\frac{3}{\sqrt{2}}+\frac{1}{|y_k|} >\sqrt{2}$</span>. The secuence converges.</p>
<p>Lastly, when <span class="math-container">$0<y<\frac{1}{\sqrt{2}}, \Delta y_k<0$</span> and decreasing very fast towards zero. So, <span class="math-container">$y_{k+r}<0$</span> for some <span class="math-container">$r>0$</span> which then converges to <span class="math-container">$\sqrt{2}$</span> as previously mentioned. So, the sequence <span class="math-container">$x_k$</span> converges to <span class="math-container">$1$</span>.</p>
|
718,664 | <p>I'm trying to find a small primitive root modulo $p^k$, where $p$ is prime. My strategy is to test small numbers $g=2,3,\ldots$ until I find a primitive root modulo $p$. That is, until $ord_pg=\phi(p)=p-1$. There are results that suggest such a search won't take long.</p>
<p>To go from modulo $p$ to modulo $p^k$, there is the following <a href="http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots" rel="noreferrer">well-known</a> theorem:</p>
<blockquote>
<p>Thm: If $g$ is a primitive root modulo $p$, then $$\begin{cases} g+p & g^{p-1}\equiv 1\pmod{p^2}\\g & g^{p-1}\not\equiv 1\pmod{p^2}\end{cases}$$
is a primitive root modulo $p^k$, for all $k\in\mathbb{N}.$</p>
</blockquote>
<p>However, I was unable to find an example that falls into the first case (except $p=2$, trivially). Hence, I have two related questions:</p>
<ol>
<li>Are there any odd primes $p$, and primitive roots $g$, such that $g^{p-1}\equiv 1\pmod{p^2}$?</li>
<li>Are there any odd primes $p$, and minimal primitive roots $g$, such that $g^{p-1}\equiv 1\pmod{p^2}$?</li>
</ol>
| mercio | 17,445 | <p>With computer search I found that $g=5$ for $p = 40487$ is an example of (2) .
<a href="http://www.wolframalpha.com/input/?i=5%5E40486+mod+1639197169" rel="nofollow">http://www.wolframalpha.com/input/?i=5%5E40486+mod+1639197169</a></p>
<p>I would guess that this should happen with probability $1/p$.
$\sum_{2 < p \le 40487} \frac 1p = 2.1235$ only. In retrospect it makes sense that the examples are rare.</p>
|
2,760,605 | <p>$||x-1|-2|=|x-3|$.
Find the value of $x$.<br>
In my attempt I got the critical values of the expression as $1$ and $3$.
But I’m not sure is we can just not consider the $2$ in the LHS.</p>
<p>My steps are Case 1: when $x>3$, $x-1-2=x-3$, $0=0$</p>
<p>Case 2: $1<x<3$, $X-1-2=-x+3 \implies 2x=6 \implies x= 3$</p>
<p>Case 3: $x<1$, $-x+1-2=-x+3$. Not possible.</p>
| Cesareo | 397,348 | <p>Hint.</p>
<p>Use the Stirling asymptotic approximation formula</p>
<p>$$
n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+O(1/n))
$$</p>
<p>because</p>
<p>$$
\frac{n}{(n!)^{1/n}}= \frac{n}{(2\pi n)^{1/(2n)}(n/e)}(1+O(1/n)) = \frac{e}{(2\pi n)^{1/(2n)}}(1+O(1/n))
$$</p>
<p>hence </p>
<p>$$
\lim_{n\rightarrow \infty}\frac{n}{(n!)^{1/n}} = e
$$</p>
|
85,984 | <p>I was doing good at school in plane geometry and trigonometry - especially in geometric proofs like proving the equality of two line segments or two angles - more than I was doing in analytic geometry.</p>
<p>I am considering doing research in mathematics to be my career (and my life) someday. and I am wondering about the most interesting research area for me.</p>
<p>What I am asking about is :</p>
<blockquote>
<p>Is there still open research fields in plane geometry or Euclidean geometry? Or this area is considered to be fundamentals that are already investigated enough?</p>
</blockquote>
<p>Thank you!</p>
| Paul | 16,158 | <p>I am not sure this is something you are looking for. But I find it interesting: it is easy to understand and is kind of related to plane geometry: </p>
<p>A graph is planar if it can be drawn on the plane without any crossing. It is well known that every planar graph can be drawn on the plane such that every edge is drawn as a straight line segment. Kemnitz and Harborth made the following stronger conjecture:</p>
<p><strong>Conjecture.</strong> Every planar graph can be drawn on the plane such that every edge is a straight line segment with integer edge lengths.</p>
<p>For the case when the graph is cubic (that is, every vertex has exactly three edges connecting to it), it was known to be true. See <a href="http://onlinelibrary.wiley.com/doi/10.1002/jgt.20304/abstract" rel="noreferrer">journal link</a> or <a href="http://www.math.uwaterloo.ca/~jfgeelen/publications/planar.pdf" rel="noreferrer">paper</a>. But the general case is still open. You can understand the problem more by reading the paper. </p>
|
31,544 | <p>For example, how to make Mathematica display only the lines beginning with <code>In[some number]</code>?</p>
<pre><code>In[8]:= m1 = {{1, 1, 1}, {8, 4, 2}, {64, 16, 4}}
</code></pre>
<blockquote>
<pre><code>Out[8]= {{1, 1, 1}, {8, 4, 2}, {64, 16, 4}}
</code></pre>
</blockquote>
<pre><code>In[9]:= m1.{b1, b2, b3} == {1, 1, 1}
</code></pre>
<blockquote>
<pre><code>Out[9]= {b1 + b2 + b3, 8 b1 + 4 b2 + 2 b3, 64 b1 + 16 b2 + 4 b3} == {1, 1, 1}
</code></pre>
</blockquote>
| Szabolcs | 12 | <p>I see three ways:</p>
<ol>
<li><p>Copy and paste everything into a new notebook (to ensure we don't destroy the output for good). Then select Cell -> Delete All Output from the menu.</p>
</li>
<li><p><kbd>alt</kbd>-click (<kbd>⌥</kbd>-click on a Mac) on an input cell's bracket. All input cells will get selected. Now copy them and paste them into a new notebook.</p>
</li>
<li><p><kbd>alt</kbd>-click (<kbd>⌥</kbd>-click on a Mac) on an input cell's bracket. All input cells will get selected. Now press <kbd>ctrl</kbd>-<kbd>'</kbd> (<kbd>⌘</kbd>-<kbd>'</kbd> on a Mac). All outputs will be hidden, but they won't be deleted.</p>
<p>(If you need to hide the input and keep the output, you can do the same but start with alt-clicking an output cell.)</p>
</li>
</ol>
|
1,350,030 | <blockquote>
<p>Let's consider a field $F$ and $\alpha,\beta\in\overline{F}-F$ (where $\overline{F}$ is an algebraic closure). If $F(\alpha)=F(\beta)$, is it true that $\alpha$, $\beta$ have the same minimal polynomial over $F$? </p>
</blockquote>
<p>For reference, I explain my way. (From here forward, all field isomorphisms are the identity on $F$.)</p>
<p>There exist polynomials $f,g\in F[x]$ such that $f(\beta)=\alpha$, $g(\alpha)=\beta$. Let $p$ and $q$ be the minimal polynomials of $\alpha$ and $\beta$ over $F$. There are field isomorphisms
$$\sigma:F[x]/(p)\to F(\alpha),\qquad \tau:F[x]/(q)\to F(\beta)$$
(these two isomorphisms are constructed in the proof of the fundamental theorem of field theory), and the trivial isomorphism $i:F(\alpha)\to F(\beta)$ (<em>I hope that there is isomorphism $j$ with $j(\alpha)=\beta$, but I can't prove it. How do you think about this?</em>) Then
$$\rho:=(\tau^{-1}\circ i\circ\sigma)$$
is also an isomorphism with
$$\rho(x+(p))=f(x)+(q),\qquad \rho^{-1}(x+(q))=g(x)+(p).$$
From this, I got $q\mid f\circ p$ but not $q\mid p$. Is this way right? If it is, please proceed this way.</p>
| Zev Chonoles | 264 | <p>Let $a$ and $b$ be any two distinct elements of $F$. Then we have $F(a)=F(b)=F$ but the minimal polynomial of $a$ over $F$ is $x-a$, whereas the minimal polynomial for $b$ over $F$ is $x-b$, which are different.</p>
<p>This is just a trivial <em>example</em> of the general behavior. Let $F$ be <strong>any</strong> infinite field, and let $a$ be <strong>anything</strong> that's algebraic over $F$ (regardless of whether it's an element of $F$ or not). Then there are infinitely many $b$'s that are algebraic over $F$ such that $F(a)=F(b)$ — consider $b=a+c$ as $c$ ranges over all elements of $F$ — but only finitely many roots of the minimal polynomial of $a$ (since a polynomial of degree $n$ over a field has at most $n$ distinct roots). Thus, all but finitely many of the $b$'s such that $F(a)=F(b)$ <strong>do not</strong> share a minimal polynomial with $a$.</p>
<p>If you really want a concrete example, observe that $\mathbb{Q}(\sqrt{2})=\mathbb{Q}(1+\sqrt{2})$ but
$$\text{min poly of }\sqrt{2} = x^2-2,\qquad \text{min poly of }1+\sqrt{2}=x^2-2x-1$$</p>
|
80,242 | <p>Let $K_0$ be a bounded convex set in $\mathbb{R}^n$ within which lie two sets $K_1$ and $K_2$. $K_0,K_1,K_2$ have nonempty interior. Assume that,</p>
<ol>
<li>$K_1\cup K_2=K_0$ and $K_1\cap K_2=\emptyset$.</li>
<li>The boundary between $K_1$ and $K_2$ is unknown. (To avoid the trivial case, I assume that the boundary is not a hyperplane.)</li>
<li>Either $K_1$ or $K_2$ is a convex set, but we don't know which one is.</li>
<li>We have two initial points $\mathbf{x},\mathbf{y}$ on hand, where $\mathbf{x}\in K_1$ and $\mathbf{y}\in K_2$.</li>
</ol>
<p>Essentially, $K_0$ can be viewed as a black box. Further assume that one can query any point in $K_0$ with a <em>membership oracle</em>, namely a procedure that given a point $\mathbf{p}\in K_0$, reports the set contains $\mathbf{p}$. </p>
<p><strong>The goal is to determine which set is convex using membership queries.</strong></p>
<p>My questions:</p>
<ol>
<li>Can this be done with <em>finite</em> number of queries?</li>
<li>What is the complexity class of this problem?</li>
</ol>
<p>Thanks.</p>
| Joel David Hamkins | 1,946 | <p>You haven't said precisely what you mean by an algorithm here, and this is actually a nontrivial issue since your question is outside the usual finitary context of computability theory, so it isn't clear what you might mean, and there are choices to be made about it. </p>
<p>One way to make the concept precise would be to work with $\mathbb{Q}^n$ rather than $\mathbb{R}^n$ (or some other countable dense set), that is, to finitize the problem by regarding all the relevant reals as rational. In this case, we can use a standard Turing machine concept of computability to make the question precise. For this manner of formalization, we regard the "input" as fixing the oracles for the $K_i$, and the question is whether there is a Turing machine program that can detect the answer correctly regardless of the oracle (subject to them satisfying your hypotheses). Let me assume first that the hypotheses are that exactly one of the $K_i$ are non-convex. </p>
<p><b>Theorem.</b> With the $\mathbb{Q}^n$ formalization just described, the answer is yes, there is an algorithm.</p>
<p>Proof. Consider the algorithm that systematically produces all triples $(p,q,r)$ of rational points in $\mathbb{Q}^n$, such that $q$ is on the line from $p$ to $r$, and checks membership in each $K_i$. Note that any instance of non-convexity will therefore be revealed. Thus, we will at some finite time in the algorithm, discover which of the $K_i$ is not convex, and thus produce the correct answer in finitely many steps. QED</p>
<p>The previous algorithm relied on the fact that instances of non-convexity can be discovered in finitely many steps. </p>
<p><b>Theorem.</b> In the $\mathbb{Q}^n$ formalization, in the context where the input sets $K_1$ and $K_2$ might <em>both</em> be convex (and in such a case only, we allow the boundary to be a hyperplane), then there is no computable procedure that will always work.</p>
<p>Proof. Suppose that there were such a procedure in the case $n=1$. Let $K_0=(0,2)$, $K_1=(0,1)$ and $K_2=[1,2)$. The algorithm will halt in finitely many steps, having made queries about reals $q_0$, $q_1$, ... $q_n$, and stating that one of the sets, lets suppose $K_1$, is convex, which is true. Consider now the modification of the sets by letting $r$ be the largest $q_i$ below $2$ (or $\frac32$, whichever is larger), and using $K_1'=(0,1)\cup (r,2)$ and $K_2'=[1,r]$. Since the answers to the queries about the $q_i$ are exactly the same for this configuration as in the original problem, the algorithm will again say that $K_1'$ is convex, but in this case, it would be incorrect. QED</p>
<p>Since the $K_i$ input is infinitary in nature, it isn't immediately clear what you might mean by "complexity class". However, the problem does have an inherent semi-decidable nature, since instances of non-convexity are revealed by a single instance. In the case where one side or the other must be non-convex, this semi-decidable nature becomes decidable, since we can know which is convex by discovering the other set to be non-convex. </p>
<p>Let us now try to consider the general case of $\mathbb{R}^n$. We don't have a standard formal notion of computability here (there are several distinct notions of computability on the reals, such as BSS machines, computable analysis, descriptive set-theoretic notions of "computability", infinite time computability). Note that the input to the algorithm is the <em>sets</em> $K_i$, rather than a real number. </p>
<p>But let's try to be flexible and allow a more generous concept of algorithm, subject to the following properties: an algorithm is a deterministic procedure that (somehow) produces points $p$ in $\mathbb{R}^n$, makes inquiries about whether they are in $K_0$, $K_1$ and $K_2$, and then uses the resulting yes/no answers to those queries to produce additional real numbers about which queries may be made. Eventually, based on the result of the queries, the algorithm is supposed to give an output by specifying whether $K_1$ or $K_2$ is convex. In particular, in this set-up, the same algorithm can be used with many different $K_0$, $K_1$ and $K_2$, and the reals produced depend only on the answers to the queries, rather than on the oracle sets themselves. We give the "input" of $K_i$ to the algorithm by attaching the black boxes, without providing any additional information about the $K_i$ sets, except that they satisfy the assumptions you identified. In this case, it doesn't matter whether one insist that exactly one or at most one $K_i$ is non-convex. </p>
<p><b>Theorem.</b> With the $\mathbb{R}^n$ manner of formalization just described, there can be no algorithm correctly identifying a convex $K_i$. </p>
<p>Proof. Suppose that there were such an algorithm in the case $n=1$. Call the algorithm $e$. Consider the set $Q$ of all real numbers that will ever be produced by the algorithm $e$ for a membership query for any combination of sets $K_0$, $K_1$ and $K_2$. Our assumptions on the nature of the algorithm ensure that $Q$ is a countable set, since any real produced during the course of any computation is produced as the result of a finite sequence of yes/no answers to the previous queries, and there are only countably many such possible patterns of membership. Let $r\lt s\lt t$ be real numbers not in $Q$, and consider the two possible inquiries:</p>
<ul>
<li>$K_0=[r,t]$, $K_1=\{r\}\cup[s,t]$, $K_2=(r,s)$. </li>
<li>$K_0=[r,t]$, $K_1=[s,t)$, $K_2=[r,s)\cup\{t\}$. </li>
</ul>
<p>Because these two input configurations agree on $Q$, the algorithm must give the same output for both of them. But in the first case, it $K_2$ that is convex, while in the second, it is $K_1$ that is convex. So there can be no such algorithm. QED</p>
|
855,686 | <p>In a triangle, the sum of two sides is $x$ and the product of the same two sides is $y$.
If $x^2 - c^2=y$ where c is the third side, then what is the ratio of the inradius to the circumradius of the triangle?</p>
<p>I guess I have found half of it: if the two sides of the triangle are $a$ and $b$, then $x=a+b$ and $y=ab$. Therefore
$x^2 - c^2 = y \Rightarrow (a+b)^2 - c^2 = ab \Rightarrow
a^2 + b^2 +ab = c^2.$
But
$a^2 + b^2 - 2ab\cos\theta=c^2$ (as c is the third side),
therefore $\theta= 120^{\circ} = \frac{2\pi}{3}$.</p>
<p>OK, so what about the inradius and circumradius ?</p>
| Peter Woolfitt | 145,826 | <p>We have equations for the inradius $r$ and the circumradius $R$, namely
$r=\dfrac{\triangle}{s}$, and $R=\dfrac{abc}{4rs}$ where $\triangle$ denotes the area of the triangle and $s$ denotes the semiperimeter. Hence we have
$$\frac{r}{R}=\frac{4r^2s}{abc}=\frac{4\triangle^2}{sabc}$$
We have $c=\sqrt{x^2-y}$, $s=\dfrac{1}{2}(a+b+c)=\dfrac{1}{2}(x+\sqrt{x^2-y})$, and $\triangle=\dfrac{1}{2}ab\sin\theta=\dfrac{\sqrt{3}}{4}y$. Plugging everything in, we get</p>
<p>$$\frac{r}{R}=\frac{3y^2}{4(\frac{1}{2}(x+\sqrt{x^2-y})y(\sqrt{x^2-y}))}
=\frac{3y}{2(x\sqrt{x^2-y}+x^2-y)}$$</p>
|
1,920,624 | <p>Prove that if 1 is a linear combination of two integers $a, b$, then $a$ and $b$
are relatively prime. Use only basic divisibility properties. (Do not use GCDLC).</p>
<p>I can only figure out how to prove this using the GCDLC.</p>
| Fly by Night | 38,495 | <p>Assume that $ma+nb=1$, and that $a$ and $b$ are not coprime, i.e. there is an integer $d>1$ which divides both $a$ and $b$.</p>
<p>If $d$ divides $a$ and $b$ then there exist integers $s$ and $t$ for which $a=ds$ and $b=dt$. In that case</p>
<p>$$ma+nb=1 \iff m(ds)+n(dt)=1 \iff d(ms+nt)=1$$</p>
<p>It follows that $d$ divides $1$. The only positive number that divides $1$ is $1$ itself, meaning that $d=1$. </p>
<p>Hence, the only positive number dividing both $a$ and $b$ is $1$, i.e. $a$ and $b$ are coprime.</p>
|
1,920,624 | <p>Prove that if 1 is a linear combination of two integers $a, b$, then $a$ and $b$
are relatively prime. Use only basic divisibility properties. (Do not use GCDLC).</p>
<p>I can only figure out how to prove this using the GCDLC.</p>
| Vidyanshu Mishra | 363,566 | <p>the combination possible is a-b(for any two consecutive integers with a>b).
a-b=1 so a=b+1, which means that a and b are consecutive integers.then one must be even and other must be odd, since odd and even numbers are coprime thus so are a and b.</p>
|
2,785,464 | <p>It's clear that every orthonormal basis is a frame, but is <strong>every</strong> basis a frame? In particular, is a non-orthogonal basis always a frame? If yes, why?</p>
<p>I'm asking about general Hilbert spaces, in particular infinite dimensional spaces.</p>
| pcp | 430,515 | <p>Let $(e_n)_{n\in \mathbb N}$ be an ONB for a Hilbert space. $(\frac{1}{n}e_n)_{n\in \mathbb N}$ is spanning and representation is unique, i.e., it is a basis. But it is not a frame, since
$$
\sum_{n\in \mathbb N} \left|\langle e_k, \frac{1}{n}e_n \rangle\right|^2 \leq \frac{1}{k^2}.
$$</p>
|
356,594 | <p>The roots of the quadratic equation $ax^ 2-bx+c=0,$ $a>0$, both lie within the interval $[2,\frac{12}{5}]$. Prove that:
(a) $a \leq b \leq c <a+b$.
(b) $\frac{a}{a+c}+\frac{b}{b+a}>\frac{c}{c+b}$</p>
<p>So we can use the quadractic formula and obtain
$2 \leq \frac{b \pm \sqrt{b^2-4ac}}{2a} \leq \frac{12}{5}$</p>
<p>and as $a>0$, this implies that</p>
<p>$4a \leq b \pm \sqrt{b^2-4ac} \leq \frac{24a}{5}$.</p>
<p>Then we can divide this into two cases (plus or minus) --- But how can we manipulate it to yield the required result?</p>
<p>Another idea is to make use of the fact that $b^2-4ac$ is non-negative</p>
| 23rd | 46,120 | <p>As shown in other answers, let $\alpha,\beta$ be the two real roots of $ax^2-bx+c=0$, i.e. $$ax^2-bx+c=a(x-\alpha)(x-\beta)\iff \alpha+\beta=\frac{b}{a},\ \alpha\beta=\frac{c}{a}.$$</p>
<p>Since $\alpha,\beta\in[2,2.4]$ and $a>0$,
$$\frac{b}{a}=\alpha+\beta\ge 4\Rightarrow a\le \frac{b}{4}<b;$$
$$\frac{b}{c}=\frac{\alpha+\beta}{\alpha\beta}=\frac{1}{\alpha}+\frac{1}{\beta}\le 1\Rightarrow b\le c;$$
and
$$(\alpha-1)(\beta-1)<2\Rightarrow 1+\frac{b}{a}=\alpha+\beta+1>\alpha\beta=\frac{c}{a}\Rightarrow c< a+b.$$
Therefore,
$$\frac{a}{a+c}+\frac{b}{a+b}\ge \frac{a}{a+c}+\frac{b}{a+c}=\frac{a+b}{a+c}>\frac{c}{a+c},$$
where the first inequality is due to $a,b>0$ and $0\le b\le c$ and the second one is due to $a+b>c$ and $a+c>0$.</p>
|
2,916,600 | <p>So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:</p>
<p>Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$. </p>
<p>Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector? </p>
<p>I have included a picture just to make it more clear.</p>
<p>Thank you in advance :)</p>
<p><a href="https://i.stack.imgur.com/LLoj4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LLoj4.png" alt="enter image description here"></a></p>
| user | 505,767 | <p>Here we are using the property of <a href="https://en.wikipedia.org/wiki/Cross_product" rel="nofollow noreferrer"><strong>cross product</strong></a> which is defined only for $v\in \mathbb{R^3}$.</p>
<p><a href="https://i.stack.imgur.com/ACugs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ACugs.png" alt="enter image description here"></a></p>
<p>The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.</p>
|
2,916,600 | <p>So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:</p>
<p>Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$. </p>
<p>Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector? </p>
<p>I have included a picture just to make it more clear.</p>
<p>Thank you in advance :)</p>
<p><a href="https://i.stack.imgur.com/LLoj4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LLoj4.png" alt="enter image description here"></a></p>
| José Carlos Santos | 446,262 | <p>The cross-product $u_1\times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$\lVert u_1\times u_2\rVert=\lVert u_1\rVert.\lVert u_2\rVert.\sin\theta,$$where $\theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1\times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).</p>
|
4,188,147 | <p>I'm trying to show the momentum operator <span class="math-container">$P : \mathcal D(P)\ \dot{=}\ C^\infty_0(\Bbb R)\to L^2(\Bbb R)$</span> with <span class="math-container">$P=-i\hbar\partial_x$</span> is essentially self-adjoint, which is equivalent to saying that it is symmetric (densely defined, which is true because <span class="math-container">$\overline{C^\infty_0}(\Bbb R)=L^2(\Bbb R)$</span>, and Hermitian) and <span class="math-container">$\ker(P^*\pm i 1) = \{0\}$</span>. The momentum is symmetric because
<span class="math-container">$$\require{cancel}(\varphi|P\psi) = \int_\Bbb R \overline{\varphi(x)}(-i\hbar\partial_x\psi)(x)\ dx = \cancel{[-i\hbar \overline \varphi \psi]_{-\infty}^{+\infty}} + \int_\Bbb R \overline{-i\hbar \text{w-}\partial_x\varphi(x)}\psi(x)\ dx, $$</span>
where <span class="math-container">$\text{w-}\partial_x$</span> indicates the weak derivative, and the equation holds for all <span class="math-container">$\psi \in C^\infty_0(\Bbb R)$</span> and <span class="math-container">$\varphi \in \mathcal D(P^*)$</span>, which can (and should) be taken to be the maximal domain of definition, i.e. the Sobolev space
<span class="math-container">$$\mathcal D(P^*) = W^{1,2}(\Bbb R)=\{\varphi \in L^2(\Bbb R)\ |\ \text{w-}\partial_x\varphi \in L^2(\Bbb R)\}. $$</span>
Now, to show the kernels of <span class="math-container">$P^*\pm i 1$</span> are trivial, I can take <span class="math-container">$\psi_\pm\in\ker(P^*\pm i1)$</span>, which means that
<span class="math-container">$$\text{w-}\partial_x \psi_\pm(x) = \pm \frac 1 \hbar \psi_\pm(x). $$</span>
At this point, I could proceed to ignore the <span class="math-container">$\text{w-}$</span> symbol and just treat the derivative as a strong derivative, so I can solve the ODE via <span class="math-container">$\psi_\pm(x) = a_\pm e^{\pm x/\hbar}$</span>, which is <span class="math-container">$\in L^2$</span> iff <span class="math-container">$a_\pm = 0$</span> as needed. But <em>why</em> can I do this? The <a href="https://en.wikipedia.org/wiki/Sobolev_inequality?oldformat=true#k_%3E_n/p" rel="nofollow noreferrer">Sobolev embedding theorem</a> only seems to guarantee that <span class="math-container">$\psi_\pm \in C^0(\Bbb R)$</span>, which wouldn't be enough to argue that <span class="math-container">$\psi_\pm$</span> can be strongly differentiated.</p>
| Michał Miśkiewicz | 350,803 | <p>Jose27's approach is probably the simplest, but I suppose someone may find another approach useful too. The main point I'd like to stress is: there's a number of ways in which one manipulate weak derivatives in the same way as strong derivatives.</p>
<p><span class="math-container">$\newcommand{\R}{\mathbb{R}}$</span>
Let's say <span class="math-container">$\psi \in W^{1,2}(\R)$</span> solves the ODE
<span class="math-container">$$
\psi' = \dfrac{\psi}{h}
$$</span>
in the weak sense (I'm dropping the <span class="math-container">$\text{w-}\partial_x$</span> notation here). What would be your first instinct if this was a classical ODE? Chances are, you would consider the function <span class="math-container">$\gamma(x) := e^{-x/h} \psi(x)$</span> and argue that it is constant. The same argument works here.</p>
<p>The first step is showing that
<span class="math-container">$$
\gamma'(x) = -\frac 1h e^{-x/h} \psi(x) + e^{-x/h} \psi'(x)
= -\frac 1h e^{-x/h} \psi(x) + e^{-x/h} \frac{\psi'(x)}{h} = 0,
$$</span>
again, in the weak sense. This manipulation is quite standard, but this is how you justify it. For any <span class="math-container">$\eta \in C_c^\infty(\R)$</span> we have
<span class="math-container">\begin{align*}
\int \gamma(x) \eta'(x)
& = \int e^{-x/h} \psi(x) \eta'(x) \\
& = \int \psi(x) (e^{-x/h} \eta(x))'
+ \int \psi(x) \cdot \frac 1h e^{-x/h} \eta(x) \\
\text{(by our weak ODE)} \quad
& = - \int \frac{\psi(x)}{h} \cdot e^{-x/h} \eta(x)
+ \int \psi(x) \cdot \frac 1h e^{-x/h} \eta(x) \\
& = 0.
\end{align*}</span></p>
<p>Now that we know <span class="math-container">$\gamma \in W^{1,2}_{loc}(\R)$</span> has <span class="math-container">$\gamma' \equiv 0$</span> (in the weak sense), it has to be constant: <span class="math-container">$\gamma \equiv c$</span>. In consequence, <span class="math-container">$\psi(x) = c e^{-x/h}$</span>, as required. This step is again standard. One of the ways to see it is to consider mollifications <span class="math-container">$\gamma_\varepsilon := \gamma * \eta_\varepsilon$</span> with suitably chosen smooth <span class="math-container">$\eta_\varepsilon$</span>. One checks that <span class="math-container">$\gamma_\varepsilon$</span> is smooth (as a convolution with a smooth function) and that <span class="math-container">$\gamma'_\varepsilon = \gamma' * \eta_\varepsilon = 0$</span> (this follows from the definition of weak derivatives, just as before). Then <span class="math-container">$\gamma_\varepsilon$</span> has to be constant (as a smooth function with zero derivative), but <span class="math-container">$\gamma_\varepsilon \to \gamma$</span> locally in <span class="math-container">$L^1$</span>, hence <span class="math-container">$\gamma$</span> is also constant.</p>
<hr />
<p><em>Side remark.</em> It's possible that for many people the first instinct would be to consider the function <span class="math-container">$\kappa(x) := \log \psi(x)$</span>, as this is what <em>separation of variables</em> yields. If only <span class="math-container">$\psi>0$</span> (so that the logarithm makes sense), then
<span class="math-container">$$
\kappa'(x) = \frac{\psi'(x)}{\psi(x)} = \frac 1h,
$$</span>
and in consequence <span class="math-container">$\kappa(x) = c + \frac{x}{h}$</span> for some <span class="math-container">$c$</span>, and finally <span class="math-container">$\psi(x) = e^c e^{x/h}$</span>. To justify these calculations in the weak setting, one needs to put in some extra work to deal with the logarithm (what if <span class="math-container">$\psi(x)=0$</span>? what if <span class="math-container">$\psi(x)<0$</span>?), but the same can be said about the strong-derivative case, right?</p>
|
4,224,303 | <p><strong>Question</strong></p>
<p>Let <span class="math-container">$X_0, X_1, X_2, \dots$</span> be independent and identically distributed continuous random variables with density <span class="math-container">$f(x)$</span>. Let <span class="math-container">$N$</span> be the first index <span class="math-container">$k$</span> such that <span class="math-container">$X_k > X_0$</span>. For example, <span class="math-container">$N = 1$</span> if <span class="math-container">$X_1 > X_0$</span>, <span class="math-container">$N = 2$</span> if <span class="math-container">$X_2 > X_0$</span> and <span class="math-container">$X_1 \leq X_0$</span> and so on. Determine the probability mass function of <span class="math-container">$N$</span>.</p>
<p><strong>My thoughts</strong></p>
<p>I am thinking that <span class="math-container">$N$</span> should follow a geometric distribution, since we are somewhat modelling the probability of the first success (i.e. the first <span class="math-container">$X_k$</span> exceeding <span class="math-container">$X_0$</span>) but if so, then what should the parameter for my geometric distribution be? Any intuitive explanations will be greatly appreciated :)</p>
<p><strong>Edit</strong></p>
<p>Following some help from an answer below and hints from my professor, I now know that my original thought process was not entire correct!</p>
| tommik | 791,458 | <blockquote>
<p>N should follow a geometric distribution... what should the parameter for my geometric distribution be?</p>
</blockquote>
<p>That's correct.</p>
<p>To get the parameter observe that</p>
<p><span class="math-container">$$P(Y>X)=\int_{-\infty}^{\infty}f(x)\left[ \int_{x}^{\infty}f(y)dy\right]dx=$$</span></p>
<p><span class="math-container">$$=\int_{-\infty}^{\infty}f(x)\left[F_Y(y) \right]_{x}^{\infty}dx=\int_{-\infty}^{\infty}f(x)dx-\int_{-\infty}^{\infty}f(x)F(x)dx=1-I$$</span></p>
<p><span class="math-container">$$I=\int_{-\infty}^{\infty}f\times F dx=\left[F\times F \right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}f\times F dx$$</span></p>
<p>That is</p>
<p><span class="math-container">$$2\int_{-\infty}^{\infty}f\times F dx=1$$</span></p>
<p>and thus</p>
<p><span class="math-container">$$I=\frac{1}{2}$$</span></p>
<p>Conclunding: your geometric parameter is 0.5</p>
<hr />
<p>EDIT: reasoning using the hint of your professor I realized that my answer is not correct. The idea I had now is the following.</p>
<p>The support of N is <span class="math-container">$n=1,2,3,\dots$</span></p>
<p>N=1</p>
<p>...easy: <span class="math-container">$P(N=1)=P(X_1>X_0)=\frac{1}{2}$</span></p>
<p>N=2</p>
<p><span class="math-container">$X_2>X_0>X_1$</span></p>
<p>This probability is <span class="math-container">$\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}$</span></p>
<p>N=3</p>
<p>We must have</p>
<p><span class="math-container">$X_0=\max(X_0,X_1,X_2)\rightarrow p=\frac{1}{3}$</span></p>
<p>AND</p>
<p><span class="math-container">$X_3=\max(X_0,X_1,X_2,X_3)\rightarrow p=\frac{1}{4}$</span></p>
<p>Thus <span class="math-container">$P(N=3)=\frac{1}{3}\cdot\frac{1}{4}$</span></p>
<p>by induction...</p>
<p><span class="math-container">$$P(N=n)=\frac{1}{n(n+1)}$$</span></p>
<p>where <span class="math-container">$n=1,2,3,\dots$</span></p>
|
706,250 | <p>For all $n\ge3\in \mathbb N$, if $n \equiv 3 \pmod{4}$ then ${3^n} \equiv 2 \pmod{5}$.</p>
<p>I tried to set $n = 3+4k$ but it doesn't help.</p>
<p>Any hints first please?</p>
| William Chang | 133,204 | <p>Simply: $3^n=3^{3+4k}$ for some $k \in \mathbb{Z}$, so $3^n=27*3^{4k} \equiv 2*(3^4)^k \equiv 2*1^k \equiv2 \pmod{5}.$</p>
<p>We use Fermat's Little Theorem, $a^{p-1} \equiv 1 \pmod{p}$ for any prime number $p$.</p>
|
262,509 | <p>I have a list of data <code>{{x1,c11}, {x1,c12}, {x2,c21},{x2,c22}, {x3,c31},{x3,c32}, {x4,c4}, {x5,c51},{x5,c52}, ...}</code>, in which <code>x</code> is real value and <code>c</code> is complex value. For some <code>x</code>, there is only a <code>c</code> value, for example, <code>{x4,c4}</code> pair, while for the most <code>x</code> there are a pair of distinct <code>c</code>, e.g. <code>{x2,c21},{x2,c22}</code>.</p>
<p>I want to plot <code>x</code> versus the real and imaginary parts of <code>c</code>, respectively, and render the <code>x</code>-<code>Re[c]</code> and <code>x</code>-<code>Im[c]</code> curves with a same color for the associated <code>Re[c]</code> and <code>Im[c]</code>. In other words, in both <code>x</code>-<code>Re[c]</code> and <code>x</code>-<code>Im[c]</code> plots, there will be two curves, I need to show the 4 curves in two different colors with the same color means the values of <code>Re[c]</code> and <code>Im[c]</code> are from the same <code>c</code>.</p>
<p>For example, <code>c21=Re[c21]+i*Im[c21]</code> and <code>c22=Re[c22]+i*Im[c22]</code>, the points <code>{x2,Re[c21]}</code> and <code>{x2,Im[c21]}</code> should be in a color, while the points <code>{x2,Re[c22]}</code> and <code>{x2,Im[c22]}</code> should use another color.</p>
<p>The key point of the question could be how to separate the interleaved data, in which most values of <code>x</code> have a pair of <code>k</code> but with some exceptions. <strong>I need a general method to handle such data with the above-mentioned features.</strong> Thank you very much!</p>
<p>Here is the <a href="https://www63.zippyshare.com/v/PbdGojjD/file.html" rel="nofollow noreferrer">sample data</a> for testing.</p>
<pre><code>test = ToExpression /@ Import["Documents\\testdata.csv"];
xci = test /. {x_, c_} -> {x, Im[c]};
xcr = test /. {x_, c_} -> {x, Re[c]};
{ListPlot[xci, PlotStyle -> Blue, PlotRange -> All, Frame -> True],
ListPlot[xcr, PlotStyle -> Red, PlotRange -> All, Frame -> True]}
</code></pre>
<p><a href="https://i.stack.imgur.com/MpoQo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MpoQo.png" alt="enter image description here" /></a></p>
<p><strong>Update</strong>: The problem can be converted to plot two curves in 3D with different colors. As can be seen, the two curves are well separated in the <code>{Re[c], Im[c], x}</code> space, thus this way might be easier.</p>
<pre><code>xc3D = test /. {x_, c_} -> {Re[c], Im[c], x};
ListPointPlot3D[xc3D, PlotStyle -> Red, AxesLabel -> {cr, ci, x}]
</code></pre>
<p><a href="https://i.stack.imgur.com/NI6y0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NI6y0.png" alt="enter image description here" /></a></p>
| yarchik | 9,469 | <p>Let <code>d</code> be your data. Then</p>
<pre><code>d1 = Cases[d, {x_, y_} /; Im[y] > -5]
d2 = Cases[d, {x_, y_} /; Im[y] < -5]
ListLinePlot[{Re[d1], Re[d2]}]
</code></pre>
<p>It happens that the <code>Im[y] > -5</code> criterion works just fine for the data in the OP. In more complicated situations, a more complicated criterion will be needed, but the syntax remains.</p>
<p><a href="https://i.stack.imgur.com/kLfKy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kLfKy.png" alt="enter image description here" /></a></p>
<p><strong>P.S.</strong></p>
<p>My answer concerns with the initial version of the question, which however has been updated recently. As the data has been changed quantitatively but not qualitatively, the same approach still applies by changing the threshold value and names of the variables.</p>
|
262,509 | <p>I have a list of data <code>{{x1,c11}, {x1,c12}, {x2,c21},{x2,c22}, {x3,c31},{x3,c32}, {x4,c4}, {x5,c51},{x5,c52}, ...}</code>, in which <code>x</code> is real value and <code>c</code> is complex value. For some <code>x</code>, there is only a <code>c</code> value, for example, <code>{x4,c4}</code> pair, while for the most <code>x</code> there are a pair of distinct <code>c</code>, e.g. <code>{x2,c21},{x2,c22}</code>.</p>
<p>I want to plot <code>x</code> versus the real and imaginary parts of <code>c</code>, respectively, and render the <code>x</code>-<code>Re[c]</code> and <code>x</code>-<code>Im[c]</code> curves with a same color for the associated <code>Re[c]</code> and <code>Im[c]</code>. In other words, in both <code>x</code>-<code>Re[c]</code> and <code>x</code>-<code>Im[c]</code> plots, there will be two curves, I need to show the 4 curves in two different colors with the same color means the values of <code>Re[c]</code> and <code>Im[c]</code> are from the same <code>c</code>.</p>
<p>For example, <code>c21=Re[c21]+i*Im[c21]</code> and <code>c22=Re[c22]+i*Im[c22]</code>, the points <code>{x2,Re[c21]}</code> and <code>{x2,Im[c21]}</code> should be in a color, while the points <code>{x2,Re[c22]}</code> and <code>{x2,Im[c22]}</code> should use another color.</p>
<p>The key point of the question could be how to separate the interleaved data, in which most values of <code>x</code> have a pair of <code>k</code> but with some exceptions. <strong>I need a general method to handle such data with the above-mentioned features.</strong> Thank you very much!</p>
<p>Here is the <a href="https://www63.zippyshare.com/v/PbdGojjD/file.html" rel="nofollow noreferrer">sample data</a> for testing.</p>
<pre><code>test = ToExpression /@ Import["Documents\\testdata.csv"];
xci = test /. {x_, c_} -> {x, Im[c]};
xcr = test /. {x_, c_} -> {x, Re[c]};
{ListPlot[xci, PlotStyle -> Blue, PlotRange -> All, Frame -> True],
ListPlot[xcr, PlotStyle -> Red, PlotRange -> All, Frame -> True]}
</code></pre>
<p><a href="https://i.stack.imgur.com/MpoQo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MpoQo.png" alt="enter image description here" /></a></p>
<p><strong>Update</strong>: The problem can be converted to plot two curves in 3D with different colors. As can be seen, the two curves are well separated in the <code>{Re[c], Im[c], x}</code> space, thus this way might be easier.</p>
<pre><code>xc3D = test /. {x_, c_} -> {Re[c], Im[c], x};
ListPointPlot3D[xc3D, PlotStyle -> Red, AxesLabel -> {cr, ci, x}]
</code></pre>
<p><a href="https://i.stack.imgur.com/NI6y0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NI6y0.png" alt="enter image description here" /></a></p>
| Edmund | 19,542 | <p>You may use <a href="https://reference.wolfram.com/language/ref/FindClusters.html" rel="nofollow noreferrer"><code>FindClusters</code></a> to separate the <code>Complex</code> curves. Then display as needed.</p>
<p>Using file in OP</p>
<pre><code>test = ToExpression /@ Import[FileNameJoin[{$HomeDirectory, "Downloads", "testdata.csv"}]];
</code></pre>
<p><code>FindClusters</code> using the complex number of the pairs.</p>
<pre><code>temp = FindClusters[Last@# -> # & /@ test];
</code></pre>
<p>The curves are separated as shown below.</p>
<pre><code>ComplexListPlot[temp[[All, All, -1]]
, AspectRatio -> 1/GoldenRatio
, PlotRange -> All
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/PiD88.png" alt="Mathematica graphics" /></p>
</blockquote>
<p>Next format the curves' data to plot verses x values.</p>
<pre><code>curves= Apply[Outer[List, {#}, ReIm@#2] &, temp, {2}];
curves= Transpose@Flatten[#, 1] & /@ curves;
</code></pre>
<p>Can <a href="https://reference.wolfram.com/language/ref/ListPlot.html" rel="nofollow noreferrer"><code>ListPlot</code></a> the <code>ReIm</code> verses the x values separately for each curve</p>
<pre><code>Block[{i = 1}
, ListPlot[#
, PlotLegends -> {"Re", "Im"}
, PlotLabel -> StringTemplate["Curve``"][i++]
, PlotRange -> All
] & /@ curves
]
</code></pre>
<p><img src="https://i.stack.imgur.com/rpe3M.png" alt="Mathematica graphics" /></p>
<p>Or each of real and imaginary jointly for all curves.</p>
<pre><code>Block[{i = 1}
, ListPlot[#
, PlotLegends -> StringTemplate["Curve``"] /@ Range@Length@curves
, PlotLabel -> Switch[i++, 1, "Real", 2, "Imaginary"]
, PlotRange -> All
] & /@ Transpose@curves
]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/xomE3.png" alt="Mathematica graphics" /></p>
</blockquote>
<p>Hope this helps.</p>
|
147,917 | <blockquote>
<p>Let $A,B\in M_{n\times n} (\mathbb{C})$. Is it possible that $ABA-BAB=I$?</p>
</blockquote>
<p>I came across this interesting problem as I was studying for an exam. I guess in the case when $A$ and $B$ commute we have $A(A-B)B=I$ and I am not sure if it can happen.</p>
<p>Any ideas?</p>
| Robert Israel | 8,508 | <p>Indeed, you could have $A = I$ and then you need $B - B^2 - I = 0$, which is true e.g. for $B = \pmatrix{0 & 1\cr -1 & 1\cr}$.</p>
<p>EDIT: For a non-commuting example, try $A = \pmatrix{2 & 0\cr 0 & 1\cr}$, $B = \pmatrix{-1/2 & 1\cr -7/2 & 3\cr}$</p>
|
2,529,990 | <p>Find the area of the shape surrounded by $y = \sin(x)$, $y = -\cos(x)$, $x = 0$, $x = π/4$.
Do I subtract $S_2$ from the $S_1$? How do I find the shape's area?
<a href="https://i.stack.imgur.com/67G7B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/67G7B.png" alt="Picture of the graph: "></a></p>
| John Wayland Bales | 246,513 | <p>If you added $1$ to both the upper and the lower function, that would move both graphs one unit higher so that the region between them would lie entirely above the $x$-axis. Yet the area of the region would remain the same. So the position of the $x$-axis makes no difference. You merely integrate the difference of the upper and lower bounds: upper minus lower.</p>
|
4,518,339 | <p>Before I start, please excuse my jargon and any preconceived notions. I am trying to fix how I think about this. It feels as if I am missing something.</p>
<p>So I understand that matrix multiplication linearly maps a vector or matrix to that basis. It describes a linear transformation, as if the coordinate plane scaled and changed direction.</p>
<p>Let's imagine we have basis vectors and some additional vector.</p>
<p><span class="math-container">$$
\hat{i} =
\begin{bmatrix}
2 \\
0 \\
\end{bmatrix},
\hat{j} =
\begin{bmatrix}
1 \\
4 \\
\end{bmatrix},
V =
\begin{bmatrix}
3 \\
2 \\
\end{bmatrix}
$$</span></p>
<p>The following is the corresponding matrix of those basis vectors. This defines our vector space.</p>
<p><span class="math-container">$$
A =
\begin{bmatrix}
2 & 1 \\
0 & 4 \\
\end{bmatrix}
$$</span></p>
<p>If we want to linearly map V to the vector space described by A, we multiply. This convention will scale <span class="math-container">$\hat{i}$</span>'s magnitude by V's x component, and <span class="math-container">$\hat{j}$</span>'s magnitude by V's y component. Then, these products will be added, creating linearly mapped vector V.</p>
<p><span class="math-container">$$
1. \space AV =
\begin{bmatrix}
2 & 1 \\
0 & 4 \\
\end{bmatrix}
\begin{bmatrix}
3 \\
2 \\
\end{bmatrix}
$$</span>
<span class="math-container">$$
2. \space AV =
3
\begin{bmatrix}
2 \\
0 \\
\end{bmatrix}
+2
\begin{bmatrix}
1 \\
4 \\
\end{bmatrix}
$$</span>
<span class="math-container">$$
3. \space AV =
\begin{bmatrix}
6 \\
0 \\
\end{bmatrix}
+
\begin{bmatrix}
2 \\
8 \\
\end{bmatrix}
$$</span>
<span class="math-container">$$
4. \space AV =
\begin{bmatrix}
8 \\
8 \\
\end{bmatrix}
$$</span></p>
<p>The scaling part in step #2 I do not get. Why is the x component of <span class="math-container">$\hat{j}$</span> relevant to the scaling of V's y component? Maybe I'm starting to understand as I'm writing... A linear mapping is almost like a new coordinate plane. Maybe we can consider <span class="math-container">$\hat{j}$</span>, "new y plane". If I tried to plot a point on the <span class="math-container">$\hat{j}$</span> axis, its new point is a linear mapping that would be described in our terms (Euclidean space): (2, 8) as it was found in step three. This seems correct.</p>
<p>So is this simply a convention to transform linear spaces or just vectors, such that, it can be expressed in Euclidean space terms?</p>
<p>But there is also the addition, step #3 that I am confused about. So I mapped V to both Vectors. OH but I guess the vectors are written in Euclidean space, but represent an x axis and y axis inside the linear mapping. So I can just add the x and y components to get the transformed coordinates. This is because we have the points on a seemingly different 'plane'. I've been stuck on this for days I suppose the Feynman technique really does work (assuming I'm right o_o).</p>
<p>Visual Aide</p>
<p><a href="https://i.stack.imgur.com/zSjo7.png" rel="nofollow noreferrer">Graph representing my math above.</a></p>
<p>Please, if you can answer my questions or guide me in a way to break my preconceived notions, that would be much appreciated.</p>
| Aiden Chow | 743,034 | <p>Let's say that you choose the starting number to be some number <span class="math-container">$k$</span>, and the difference to be <span class="math-container">$d$</span>. Then, the sum of the <span class="math-container">$n$</span> numbers will be:
<span class="math-container">$$k+(k+T_1d)+(k+T_2d)+\cdots+(k+T_{n-1}d)=nk+\frac{n(n-1)(n+1)}6d$$</span>
where <span class="math-container">$T_i$</span> is the <span class="math-container">$i$</span>'th triangular number.</p>
<p>Expanded, this would be:
<span class="math-container">$$nk+\frac{n(n-1)(n+1)}6d=nk+\frac{dn^3-dn}6=\frac d6n^3+\left(k-\frac d6\right)n$$</span>
This sum is equal to <span class="math-container">$m$</span>, so we have the equation:
<span class="math-container">$$\frac d6n^3+\left(k-\frac d6\right)n=m$$</span>
There could be multiple choices for <span class="math-container">$k$</span> and <span class="math-container">$d$</span> at this point, but to simplify the calculations, we can choose <span class="math-container">$k=d/6$</span>:
<span class="math-container">$$\frac d6n^3=m\implies d=\frac{6m}{n^3}$$</span>
Which means <span class="math-container">$$k=\frac m{n^3}$$</span>
(Note that <span class="math-container">$k$</span> and <span class="math-container">$d$</span> are not guaranteed to be integers.)</p>
<h3>Example</h3>
<p>Let's take <span class="math-container">$n=5$</span> and <span class="math-container">$m=250$</span>.</p>
<p>We get:
<span class="math-container">$$d=\frac{6m}{n^3}=\frac{1500}{125}=12$$</span>
So:
<span class="math-container">$$k=\frac d6=\frac{12}6=2$$</span>
Which means that the numbers generated are:
<span class="math-container">$$k,k+d,k+3d,k+6d,k+10d\implies2,14,38,74,122$$</span>
To confirm the answer, we can first take the consecutive difference, which gives:
<span class="math-container">$$12,24,36,48$$</span>
We can see that the difference between each consecutive term is increasing by <span class="math-container">$d=12$</span> each time, as expected.</p>
<p>Now let's take the sum of all the terms:
<span class="math-container">$$2+14+38+74+122=250$$</span>
<span class="math-container">$m=250$</span> so this is correct.</p>
<hr />
<p>Here is some example python code which generates the <span class="math-container">$n$</span> numbers based on what we did above (takes in <span class="math-container">$n$</span> and <span class="math-container">$m$</span> as space separated integers): <a href="https://tio.run/##FY3NCsIwEITv@xRzTGqxorRIwSeRHgLxJyTZLjEefPq4YQ4zHzMw8qvvnS9XKa3xiIwbshMTuI6B5VuNPX4kBXVLXstlyJPhQWUpKvtpIUoa7hu5zvTcCwICozh@PQzblZBw0EXcCLEnR3DdPZEUvTLJtjbjPJ/@" rel="nofollow noreferrer">code</a></p>
|
3,291,975 | <p>Transpose this formula to make <span class="math-container">$y$</span> the subject.</p>
<p><span class="math-container">$$x=\sqrt{x^2y^2+1-y}$$</span></p>
<p>My try:</p>
<p><span class="math-container">$$x^2=x^2y^2+1-y$$</span></p>
<p><span class="math-container">$$x^2-x^2y^2=1-y$$</span></p>
<p><span class="math-container">$$x^2(1-y^2)=1-y$$</span></p>
<p>Here I got 2 <span class="math-container">$y$</span> terms, but I am not sure what to do next.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>You can write <span class="math-container">$$0=y^2-\frac{1}{x^2}y+\frac{1-x^2}{x^2}$$</span> using the quadratic formula we obtain
<span class="math-container">$$y_{1,2}=\frac{1}{2x^2}\pm\sqrt{\left(\frac{1}{2x^2}\right)^2-\frac{1-x^2}{x^2}}$$</span></p>
|
3,246,533 | <p>ABC is a triangle such that AB = AC. Let D be the foot of the perpendicular from C to AB and E the foot of the perpendicular from B to AC. Then</p>
<p>(a) <span class="math-container">$BC^3 < BD^3 + BE^3$</span></p>
<p>(b) <span class="math-container">$BC^3 = BD^3 + BE^3$</span></p>
<p>(c) <span class="math-container">$BC^3 > BD^3 + BE^3$</span></p>
<p>(d) None of the foregoing statements need always be true.</p>
<p>I drew the diagram and showed that the triangles BEC and BDC are similar and:
<span class="math-container">$CE/BD=BE/CD=BC/BC=1$</span>, or, <span class="math-container">$CE=BC$</span> and <span class="math-container">$BE=CD$</span></p>
<p>Then, from the same triangles, we know,
<span class="math-container">$BD+CD>BC$</span>, or, <span class="math-container">$BD+BE>BC$</span> (1) since <span class="math-container">$CD=BE$</span></p>
<p>We also know that <span class="math-container">$(BD-BE)^2>0$</span></p>
<p>or, <span class="math-container">$BD^2 -BE.BD+BE^2> BE.BD>0$</span> (2)</p>
<p>therefore from (1) and (2), I concluded that,
<span class="math-container">$BE^3+BD^3>BC$</span></p>
<p>How do I proceed further?</p>
| amitava | 566,278 | <p>Note that, triangles BDC and CEB are congruent as @SS_C4 mentioned. Then <span class="math-container">$CD = BE$</span>, Now</p>
<p><span class="math-container">$BC^2 = BD^2+CD^2$</span>, </p>
<p>So <span class="math-container">$BC^3 = BD^{2}BC+CD^{2}BC = BD^{2}(BD+d_1)+CD^{2}(CD+d_2)\\
= BD^3+CD^3+BD^{2}d_1 + CD^{2}d_2 $</span></p>
<p>so <span class="math-container">$BC^3 > BD^3+CD^3$</span></p>
|
23,306 | <p>I am trying to perform a simple arithmetic in complex algebra.</p>
<pre><code> a := Subscript[a, 1] + I*Subscript[a, 2];
b := Subscript[b, 1] + I*Subscript[b, 2];
r[z_] := a/(b - z) + a\[Conjugate]/(b\[Conjugate] - z);
</code></pre>
<p>when I try to evaluate r[0], mathematica gives me the following message</p>
<pre><code> $RecursionLimit::reclim: Recursion depth of 256 exceeded.
</code></pre>
<p>I was expecting something like the following</p>
<pre><code> 2*Re[a/b] = 2 (a_1 * b_1 + a_2 * b_2)/(b_1 ^2 + b_2 ^2)
</code></pre>
<p>Could someone please point out where I did mess up?</p>
| amr | 950 | <p>I did this and it seems to work:</p>
<pre><code>SetAttributes[Subscript, HoldAll];
a := Subscript[a, 1] + I*Subscript[a, 2];
b := Subscript[b, 1] + I*Subscript[b, 2];
r[z_] := a/(b - z) + a\[Conjugate]/(b\[Conjugate] - z);
r[0] // ComplexExpand // Simplify
</code></pre>
<p>I'm not really sure if this will work in a more general setting though.</p>
|
23,306 | <p>I am trying to perform a simple arithmetic in complex algebra.</p>
<pre><code> a := Subscript[a, 1] + I*Subscript[a, 2];
b := Subscript[b, 1] + I*Subscript[b, 2];
r[z_] := a/(b - z) + a\[Conjugate]/(b\[Conjugate] - z);
</code></pre>
<p>when I try to evaluate r[0], mathematica gives me the following message</p>
<pre><code> $RecursionLimit::reclim: Recursion depth of 256 exceeded.
</code></pre>
<p>I was expecting something like the following</p>
<pre><code> 2*Re[a/b] = 2 (a_1 * b_1 + a_2 * b_2)/(b_1 ^2 + b_2 ^2)
</code></pre>
<p>Could someone please point out where I did mess up?</p>
| m_goldberg | 3,066 | <p>Attributes are powerful, but so is <code>With</code>, which is what I would use here.</p>
<pre><code>With[{a = Subscript[a, 1] + I*Subscript[a, 2],
b = Subscript[b, 1] + I*Subscript[b, 2]},
r[z_] := a/(b - z) + a\[Conjugate]/(b\[Conjugate] - z)]
r[0] // ComplexExpand // Simplify
</code></pre>
<blockquote>
<pre><code>(2*(Subscript[a, 1]*Subscript[b, 1] + Subscript[a, 2]*Subscript[b, 2])) /
(Subscript[b, 1]^2 + Subscript[b, 2]^2)
</code></pre>
</blockquote>
<p>This works because the <code>a</code> on the lhs of assignment is now distinct from the <code>a</code> on the rhs. Same goes for <code>b</code>.</p>
|
3,446,691 | <p>[Please excuse ignorance, I'm not a mathematician, though I do data analysis.]</p>
<p>The question is strictly for <span class="math-container">$2\times2$</span> real non-symmetric matrix <span class="math-container">$\bf A$</span>. Let us assume, having real eigenvalues.</p>
<p>I've learned how to compute the eigenvalues and the eigenvectors for such a matrix, so that <span class="math-container">$\bf V D V^{-1} = A$</span>, where <span class="math-container">$\bf D$</span> is the diagonal matrix of eigenvalues <span class="math-container">$\lambda$</span>. Note that <span class="math-container">$\bf V$</span> are usually non-orthogonal eigenvectors.</p>
<p>But instead of this I want to find <em>orthogonal</em> "eigenvectors" <span class="math-container">$\bf U$</span> and the upper <em>triangular</em> <span class="math-container">$\bf T$</span> (<span class="math-container">$\lambda$</span> being its diagonal) so that <span class="math-container">$\bf U T U' = A$</span>.</p>
<p>I know this is what eigen functions usually do via QR iterations of some sort. (I've read a bit about it and can program such iterations.) It is called Schur form, if I'm correct.</p>
<p>My <strong>question</strong>: Is it possible to obtain <span class="math-container">$\bf T$</span> and <span class="math-container">$\bf U$</span> for a <span class="math-container">$2\times2$</span> case in a closed form - without iterations? Is that easy to do "by hand"? How?</p>
<p>Illustration</p>
<pre><code>A
-5 4
3 6
Eivenvalues are -6 an 7
Eigenvectors:
V
-.9701425001 .3162277660
.2425356250 .9486832981
V'V
1.000000000 -.076696499
-.076696499 1.000000000
(not orthogonal)
Restore A:
V * -6 0 * inv(V)
0 7
yields A
-------------------------------------
But,
U * -6 1 * t(U)
0 7
(T is in the middle) also yields A
where U
.9701425001 .2425356250
-.2425356250 .9701425001
is orthogonal:
U'U
1.000000000 .000000000
.000000000 1.000000000
Need to find T (that is, its element [1,2]) and U.
</code></pre>
| Devansh Bisla | 686,238 | <p>Looks like Schur decomposition to me. We can do it by hand at least for 2x2 matrices here is an example:</p>
<p>For any arbitrary matrix, A, Schur decomposition is defined as <span class="math-container">$A = UTU^{*}$</span> where T is upper triangular, U is a unitary matrix and <span class="math-container">$*$</span> is conjugate transpose.</p>
<p>Let,<br>
<span class="math-container">$A = \begin{bmatrix} 7 & -2\\ 12 & -3\end{bmatrix}$</span></p>
<p>First, we find eigenvectors and eigenvalues of the matrix such that: <span class="math-container">$A = V \Lambda V^{-1}$</span>. For our matrix
<span class="math-container">$\Lambda = \begin{bmatrix} 1 & 0\\ 0 & 3\end{bmatrix}$</span> and <span class="math-container">$V = \begin{bmatrix} 1 & 1\\ 3 & 2\end{bmatrix}$</span>.</p>
<p>Next, we can take any one of the eigenvectors and find its orthogonal vector. Lets take <span class="math-container">$v_{1} = \begin{bmatrix} 1 \\ 3\end{bmatrix}$</span> therefore <span class="math-container">$\{v_{1}\}^{\perp} = \begin{bmatrix} -3 \\ 1\end{bmatrix}$</span>. Normalize them and create a unitary matrix <span class="math-container">$U = \frac{1}{\sqrt{10}}\begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix}$</span></p>
<p>Now, from Schur decomposition we know, <span class="math-container">$A = UTU^{*}$</span>, we have found our unitary matrix <span class="math-container">$U$</span> we can get T as <span class="math-container">$T = U^{*}AU$</span>. Thus,
<span class="math-container">$T = \begin{bmatrix} 1 & -14 \\ 0 & 3 \end{bmatrix}$</span></p>
<p>Further reading:
<a href="http://www.home.uni-osnabrueck.de/mfrankland/Math416/Math416_SchurDecomposition.pdf" rel="nofollow noreferrer">http://www.home.uni-osnabrueck.de/mfrankland/Math416/Math416_SchurDecomposition.pdf</a></p>
|
2,055,546 | <p>So I stumbled across this math problem:</p>
<p>$1015$ USD is to be shared between $3$ people: Person A, Person B and Person C.
Person A gets double of what Person B gets and Person B gets $100$ USD more than Person C</p>
<p>Basically:</p>
<p>$A = 2B$</p>
<p>$B = C + 100$</p>
<p>$A + B + C = 1015$</p>
<p>Can someone please tell me how it's solved, rather than telling me the outcome?</p>
<p>UPDATE: Thank you to people who helped me solve it. What I realized was, that I used the equation C = B - 100 to try to solve the problem. You made me realize that. So thank you.</p>
| Community | -1 | <p>Basically, $A=2B$ and $B=C+100$. We have $$A+B+C =1015$$ $$\Rightarrow 2B+B+(B-100) =1015$$ because $B+C=100$. Hence, $$4B-100 =1015 \Rightarrow B=\frac{1115}{4}$$ Then we can find $A=\frac{1115}{2}$ and $C=\frac{715}{4}$. Hope it helps.</p>
|
647 | <p>For example, I gave an exam earlier today with a problem that ended in the sentence</p>
<blockquote>
<p>Use the chain rule to find $(f\circ g)'(3)$.</p>
</blockquote>
<p>During the exam, one of the students asked me what the circle between the $f$ and $g$ means, and I answered that it represents the composition of functions.</p>
<p>Later on, another student was working on a question that contained the phrase</p>
<blockquote>
<p>. . . where $V$ represents the volume of water in the tank.</p>
</blockquote>
<p>and also the phrase</p>
<blockquote>
<p>Given that water initially drains from the tank at a rate of $2.0\;\mathrm{L}/\mathrm{min}$ . . .</p>
</blockquote>
<p>The student asked whether $2.0\;\mathrm{L}/\mathrm{min}$ is the value of $\dfrac{dV}{dt}$, and I said I couldn't help with that.</p>
<p>Would you have answered these students' questions? How do you decide what to do in these situations? I usually handle these on a case-by-case basis, but I'm curious if anyone has any good general rules for deciding what information to give.</p>
| Dave L Renfro | 745 | <p>For questions like "what does the circle between $f$ and $g$ mean" in a calculus course, I would tell the student and then write on the board (for everyone else) something like:</p>
<blockquote>
<p>In problem 5, $(f \circ g)(x)$ stands for $f(g(x))$ (composition of functions).</p>
</blockquote>
<p>I wouldn't do this in a precalculus course, however, where this notation is one of the things being tested.</p>
<p>As for questions that I don't think are appropriate to answer, I usually make an assessment of how much I think knowing the answer is worth for the problem at hand. For example, for a 10-point applied related rates problem in which your second question came up, I might tell the student that I'll provide the answer for a 3-point penalty, which they can then choose to accept or reject. I explain this sort of bargaining in advance, when reviewing the syllabus the first day of class and/or when reviewing for the test before the test is given. Personally, I'd much rather do this (and mark on their test, in ink and in my handwriting, "$-3$ for hint") than to later deal with the much more time consuming task of deciding how to grade a problem that I think they could have worked if they knew such-and-such, but it's difficult to tell for sure because what they don't know and what they do know is mixed together in their solution.</p>
|
1,063,382 | <p>Is there any proper way to know whether a function has the same domain and range $[a,b]$ where $a,b<\infty$ i.e. $f:[a,b] \to [a,b]$ ?</p>
<p>For example:
$$
f(x) = e^{−x} ,\qquad [\ln(1.1), \ln(3)]
$$</p>
| Ahaan S. Rungta | 85,039 | <p>So, first, you realize that the equation of the circle must be of the form $$ (x-h)^2 + (y-k)^2 = r^2. $$Then, since $(1,1)$ is on the circle, we get $$ (1-h)^2 + (1-k)^2 = r^2. $$Then, we note that $(2,3)$ is the <em>only</em> intersection of $$ (x+2)^2 + (y-3)^2 = 16 $$ and $$ (x-h)^2 + (y-k)^2 = r^2. $$So let us say we were given $h,k,r$. Then, we can solve for the intersection(s) of the two circles. We want $(2,3)$ to be the only one. One way to finish from here is to subtract the two equations, solve for $y$ in terms of $x$, substitute back in, and keep bashing. </p>
<p>However, there's an easier way. We want the two circles to be tangent at $(2,3)$, so that already brings it down to two circles $-$ one that is internally tangent and one that is externally tangent. In both cases, it is clear that we want the points $(-2,3)$, $(2,3)$, and $(h,k)$ collinear. This is just because the centers of the circle and the tangency point must be collinear.</p>
<p>Now, you can finish. </p>
|
82,070 | <p>I reading on Sums and I am reading about the difference between using a generalized Sigma notation and the delimited form. Ok, I understand that the generalized form is more expressive. </p>
<p>But I found the following example confusing to me:<br>
It says: </p>
<blockquote>
<p>In the following we can change the index variable from $k$ to $k+1$ and easily do the substitution:
$$
\sum_{1 \leqslant k \leqslant n} a_k = \sum_{1 \leqslant k+1 \leqslant n} a_{k+1}.
$$</p>
<p>But with the delimited form we have:<br>
$$
\sum_{k=1}^n a_k = \sum_{k=0}^{n-1} a_{k+1};
$$</p>
<p>and is harder to make a mistake. </p>
</blockquote>
<p>My question is the following:<br>
In the generalized form the index is expressed to be $k+1$ in the notation so we have $k+1=1$, $k+1=2$. </p>
<p>But why is the comparison done like this?<br>
I mean shouldn't we have in the right hand of the delimited form $k+1=1$ instead?<br>
I.e. </p>
<blockquote>
<p>$$
\sum_{k=1}^n a_k = \sum_{{\Large \mathbf{k+1=1}}}^n a_{k+1}.
$$</p>
</blockquote>
<p>Or is it not allowed to use a complex index in the delimited form? </p>
| Phira | 9,325 | <p>A complex index in delimited form is never used, the main reason being that this is highly problematic for non-linear variable changes if you think of the delimited form as running through values 1 to n and the non-delimited form as a system of conditions that happen to be inequalities in your case.</p>
<p>If the variable change is linear, your proposed notation will not lead to problems and will certainly be understood, but I would not advice to use it.</p>
|
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