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 |
|---|---|---|---|---|
139,487 | <p>I have a not-so-complicated piecewise cubic function, shown below as the yellow curve on the right. It's derivative is on the left; the blue lines are references. Please see the code below where I call it <strong>myF</strong> (and its derivative myf).</p>
<p>To my surprise, the integration with a parameter $u$ and sine in the argument gives an ugly result involving the complex $i$.</p>
<blockquote>
<p>Is the result of $\int$myF$(u\sin x)$d$x$ (as shown in the screenshot below) correct? Or is it some kind of failure due to my erroneous way of using <code>Integrate</code> coupled with<code>Piecewise</code>? </p>
</blockquote>
<p>As just the area under the curve modulated by sine, I for now don't see how this can be mathematically true to have involved $\sqrt{-1}$.</p>
<p><a href="https://i.stack.imgur.com/qphqj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qphqj.png" alt="enter image description here"></a></p>
<p>Screen shot below:</p>
<p><a href="https://i.stack.imgur.com/P17ew.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P17ew.png" alt="enter image description here"></a></p>
<p>The code to define the curves, generate the plots, and do the integration is below. When the parameter $u < \frac12$ of course there's no problem since the integration doesn't go over the `splitting point' (where is differentiable).</p>
<p>I tried various ways to code the integration with assumptions, but the outcomes are basically the same. Some hint or confirmation would be greatly appreciated. Thank you.</p>
<pre><code>ClearAll[myF, myf, u, v, ImgSz];
myF[v_] := Piecewise[{
{0, v < 0},
{4/3 v^3, 0 <= v < 1/2},
{1/3 (1 - 6 v + 12 v^2 - 4 v^3), 1/2 <= v < 1},
{1, v >= 1}
}];
myf = D[myF[u], u] /. {u -> v};
ImgSz = 250;
Row[{ Plot[{ 1, (* symmetry reference *)
myf
}, {v, -.3, 1.3}, PlotRange -> {-0.1, 2}, ImageSize -> ImgSz],
Plot[{ v, (* linear reference *)
myF[v]
}, {v, -.3, 1.3}, PlotRange -> {-0.1, 1}, ImageSize -> ImgSz]
}]
Assuming[ 1/2 < u < 1 ,
Integrate[ myF[ u Sin[x]], {x, 0, Pi}] ]
</code></pre>
<hr>
<p><strong>Summarizing Edit</strong></p>
<p>Okay, so I couldn't recognize the expression as real-disguised-in-complex-conjugate, neither did I know reliable ways to test that. I'm glad to have learned from both answer posts.</p>
<p>In my case, defining the function with <code>UnitStep</code> doesn't allow Mathematica to evaluate differently like in another post (which I cannot find right now). Nor does <code>PrincipalValue -> True</code> or change of variables apply here as <a href="https://mathematica.stackexchange.com/questions/127301">it does sometimes</a>. I have accepted that in my case I need to post-process the expression the way I see fit. Nonetheless, for the record, there are some known bugs, old and <a href="https://mathematica.stackexchange.com/questions/99395">new</a>, <a href="https://mathematica.stackexchange.com/questions/48437">solved</a> and <a href="https://mathematica.stackexchange.com/questions/127448">unsolved</a>. </p>
| Himalaya Senapati | 57,938 | <p>Inside If, separate statements with a semicolon, not &&, the code does something. But, it is wrong, and I don't yet know why it is giving wrong output. </p>
<pre><code> IC = {x[0] == 1, x'[0] == 0, a[0] == 10, b[0] == 1/10};
sol = NDSolve[{x''[t] + b[t] x'[t] + a[t] x[t] == 0,
WhenEvent[x[t] == 0,
If[x'[t] < 0, a[t] -> 3 - x[t]; b[t] -> 0.1 - x'[t]]], IC}, {x,
a, b}, {t, 0, 5}, DiscreteVariables -> {a, b}];
Plot[{x[t], a[t]} /. sol, {t, 0, 5}]
Plot[{x'[t], b[t]} /. sol, {t, 0, 5}]
</code></pre>
<p>It looks like a sequence of actions written within If command doesn't work well, thus using two WhenEvent makes it work.</p>
<pre><code>IC = {x[0] == 1, x'[0] == 0, a[0] == 10, b[0] == 1/10};
sol = NDSolve[{x''[t] + b[t] x'[t] + a[t] x[t] == 0,
WhenEvent[x[t] == 0, If[x'[t] < 0, b[t] -> 0.1 - x'[t]]],
WhenEvent[x[t] == 0, If[x'[t] < 0, a[t] -> 3 - x[t]]], IC}, {x, a,
b}, {t, 0, 5}, DiscreteVariables -> {a, b}];
Plot[{x[t], a[t]} /. sol, {t, 0, 5}]
Plot[{x'[t], b[t]} /. sol, {t, 0, 5}]
</code></pre>
|
238,627 | <p>Kindly consider it soft question as I am a software engineer.and I know only software but I have a doubt in my mind that there would be something like null in mathmatics as well. </p>
<p>If Mathematics <code>NULL</code> IS Equivalent to <code>ZERO</code>?</p>
| Dan Brumleve | 1,284 | <p>The distinction between the empty set $\emptyset$ and the number $0$ is similar to that between <code>NULL</code> and <code>ZERO</code>. For example, the set of real solutions (or informally "the solution") to $x^2=-1$ is $\emptyset$, but the solution to $x^2=0$ is $0$.</p>
|
4,617,031 | <p>How would I order <span class="math-container">$x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $</span> without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.</p>
<p>What I tried to so far, because they are all positive numbers, is to square <span class="math-container">$x,y,z$</span> but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that <span class="math-container">$x<z$</span> is easy and so is <span class="math-container">$y<z$</span>, but I'm stuck at <span class="math-container">$x < y \text{ or } x>y$</span>.</p>
| boojum | 882,145 | <p>A visual demonstration of sorts:</p>
<p>Starting from <span class="math-container">$ \ x \ = \ \sqrt3 - 1 \ \ , \ $</span> we have<br />
<span class="math-container">$$ x \ + \ 1 \ \ = \ \ \sqrt3 \ \ \Rightarrow \ \ x^2 \ + \ 2x \ + \ 1 \ \ = \ \ 3 \ \ . \ $$</span> So <span class="math-container">$ \ \sqrt3 - 1 \ $</span> is the larger of the two zeroes of <span class="math-container">$ \ x^2 + 2x - 2 \ \ . \ $</span> As for <span class="math-container">$ \ x \ = \ \sqrt5 \ - \ \sqrt2 \ \ , \ $</span> we may write
<span class="math-container">$$ x \ + \ \sqrt2 \ \ = \ \ \sqrt5 \ \ \Rightarrow \ \ x^2 \ + \ 2 \sqrt2 · x \ + \ 2 \ \ = \ \ 5 \ \ , $$</span>
making <span class="math-container">$ \ \sqrt5 \ - \ \sqrt2 \ $</span> the larger of the two zeroes of the polynomial <span class="math-container">$ \ x^2 + 2 \sqrt2 · x - 3 \ \ . \ $</span> A graph of the two polynomials (the first in red, the second in blue) indicates the relative positions of these two zeroes.</p>
<p>If one mistrusts a graph (a not unreasonable attitude), we can show that the curve for <span class="math-container">$ \ x^2 + 2x - 2 \ $</span> lies "above" that of <span class="math-container">$ \ x^2 + 2 \sqrt2 · x - 3 \ \ $</span> [vertex at <span class="math-container">$ \ (-1 \ , \ -3) \ $</span> <em>versus</em> <span class="math-container">$ \ (-\sqrt2 \ , \ -5) \ \ , \ $</span> "higher" <span class="math-container">$ \ y-$</span> intercept; also, <span class="math-container">$$ x^2 + 2x - 2 \ > \ x^2 + 2 \sqrt2 x - 3 \ \ \Rightarrow \ \ 1 \ > \ 2·(\sqrt2 - 1)·x $$</span> <span class="math-container">$$ \Rightarrow \ \ \frac{1}{2·(\sqrt2 - 1)} \ = \ \frac12·(1 + \sqrt2) \ > \ x \ \ . \ ] $$</span> until the curves intersect at <span class="math-container">$ \ x \ = \ \frac{1}{2 \sqrt2 \ - \ 1} \ = \ \frac12 + \frac{\sqrt2}{2} \ \ , $</span> <span class="math-container">$ y \ = \ \frac32 \sqrt2 - \frac14 \ > \ 0 \ \ . \ $</span> Since this intersection is "above" the <span class="math-container">$ \ x-$</span>axis, the positive <span class="math-container">$ \ x-$</span>intercept of <span class="math-container">$ \ x^2 + 2x - 2 \ $</span> is smaller than that of <span class="math-container">$ \ x^2 + 2 \sqrt2 · x - 3 \ \ . $</span></p>
<p><a href="https://i.stack.imgur.com/ygbuq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ygbuq.png" alt="enter image description here" /></a></p>
|
113,843 | <p>Here comes some sample data</p>
<pre><code>data = {{0, 0.7, 0.4}, {1, 0.831177, 0.51854}, {2, 1.11106, 0.463533},
{3, 1.84226, -0.642571}, {4, 0.677049, -0.327877},
{5, 0.77886, -0.451322}, {6, 0.965874, -0.508772},
{7, 1.34397, -0.202473}, {8, 1.01761, -0.717013},
{9, -0.0507992, -1.3864}, {10, -0.102145, -0.957957},
{11, -0.00228078, -0.861489}}
</code></pre>
<p>The first integer is a counter, while the other two reals are the $(x,y)$ coordinates. </p>
<p>Let's plot these points</p>
<pre><code>d0 = data[[All, {2, 3}]];
L0 = ListPlot[d0, Joined -> True, Mesh -> All,
PlotStyle -> {Black, Thickness[0.002], PointSize[0.012]},
Frame -> True, Axes -> False, PlotRange -> All, ImageSize -> 500]
</code></pre>
<p>Now I want the customize the plot as follows:</p>
<p>(a). Add arrows showing the evolution of the points, like</p>
<p><a href="https://i.stack.imgur.com/RX1Kp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RX1Kp.jpg" alt="enter image description here"></a> </p>
<p>(b). The first point with counter 0 should be plotted in red, while the last point should be plotted in blue.</p>
<p>(c). Add labels near to the points indicating the corresponding counter (e.g, 2, 3, etc). The label of the first point should be $P_0$, instead of 0, while for the last point the label should be $P_f$.</p>
<p>Any suggestions? </p>
| Jason B. | 9,490 | <p>Use a <code>Graph</code> with directed edges</p>
<pre><code>labels = Thread[
Range[12] -> (Placed[#, Above] & /@
Join[{Subscript[x, 0]}, Range[10], {Subscript[x, f]}])];
Graph[# \[DirectedEdge] # + 1 & /@ Range[11], VertexCoordinates -> d0,
VertexLabels -> labels, VertexStyle -> {1 -> Red, 12 -> Blue}]
</code></pre>
<p><img src="https://i.stack.imgur.com/lo1yv.png" alt="Mathematica graphics"></p>
<p>Or, if you need to have it look like the plot above,</p>
<pre><code>labels = Thread[
Range[12] -> (Placed[#, Above] & /@
Join[{Subscript["x", 0]}, Range[10], {Subscript["x", "f"]}])];
labels = MapAt[Style[#, 15, FontFamily -> "Times New Roman"] &,
labels, {All, 2, 1}];
Show[
Graph[# \[DirectedEdge] # + 1 & /@ Range[11],
VertexCoordinates -> d0, VertexLabels -> labels,
VertexStyle -> {1 -> Red, 12 -> Blue}],
Frame -> True, FrameTicks -> Automatic, ImageSize -> 500,
PlotRangePadding -> .1]
</code></pre>
<p><img src="https://i.stack.imgur.com/J17yr.png" alt="Mathematica graphics"></p>
|
3,130,195 | <p>I'm interested of finding a closed formula for the fundamental matrix to the system
<span class="math-container">$$\eqalign{
& y'(t) = a(t)z(t) \cr
& z'(t) = \delta a(t)y(t) \cr} $$</span>
<span class="math-container">$$(y(0),z(0)) = ({y_0},{z_0})$$</span>
where <span class="math-container">$\delta$</span> is some constant and <span class="math-container">$a$</span> is a regular function. Thank you. </p>
| SchrodingersCat | 278,967 | <p>The given equation <span class="math-container">$\displaystyle \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$</span>
can be simplified as
<span class="math-container">$$\displaystyle \sqrt{\frac{x}{a}} = 1 - \sqrt{\frac{y}{b}}$$</span>
Squarring both sides,
<span class="math-container">$$\displaystyle \frac{x}{a}= 1 - 2\sqrt{\frac{y}{b}} + \frac{y}{b}$$</span>
<span class="math-container">$$\displaystyle \frac{x}{a}-1+ \frac{y}{b}=- 2\sqrt{\frac{y}{b}}$$</span>
Squarring again,
<span class="math-container">$$\displaystyle \left(\frac{x}{a}-1+ \frac{y}{b}\right)^2=4\frac{y}{b}$$</span>
<span class="math-container">$$\displaystyle \left(X\right)^2=4Y$$</span>
Where
<span class="math-container">$$X=\frac{x}{a}-1+ \frac{y}{b}$$</span>
and <span class="math-container">$$Y=\frac{y}{b}$$</span>
Now compare with the standard form, apply the formulae and find the required properties.</p>
<p>Hope this helps you.</p>
|
4,073,821 | <p>If <span class="math-container">$\sum a_n$</span> converges, then does <span class="math-container">$\sum |a_n|$</span> converge as well? This is the same as "absolute convergence", where if <span class="math-container">$\sum a_n$</span> converges, then <span class="math-container">$\sum|a_n|$</span> might actually diverge, for example, <span class="math-container">$\sum\frac{(-1)^n}{n}$</span>. So this would serve as a counterexample to the question, and prove it to be wrong, correct?</p>
| S.Farr | 683,130 | <p>Yes, you are correct. If <span class="math-container">$\sum a_n$</span> converges, <span class="math-container">$\sum |a_n|$</span> does not have to converge as well.</p>
|
19,373 | <p>I posted this question earlier today on the Mathematics site (<a href="https://math.stackexchange.com/q/3988907/96384">https://math.stackexchange.com/q/3988907/96384</a>), but was advised it would be better here.</p>
<p>I had a heated argument with someone online who claimed to be a school mathematics teacher of many years standing. The question which spurred this discussion was something along the lines of:</p>
<p>"A horseman was travelling from (location A) along a path through a forest to (location B) during the American War of Independence. The journey was of 22 miles. How far was it in kilometres?"</p>
<p>To my mind, the answer is trivially obtained by multiplying 22 by 1.6 to get 35.2 km, which can be rounded appropriately to 35 km.</p>
<p>I was roundly scolded by this ancient mathematics teacher for a) not using the official conversion factor of 1.60934 km per mile and b) not reporting the correct value as 35.405598 km.</p>
<p>Now I have serious difficulties with this analysis. My argument is: this is a man riding on horseback through a forest in a pre-industrial age. It would be impractical and impossible to measure such a distance to any greater precision than (at best) to the nearest 20 metres or so, even in this day and age. Yet the answer demanded was accurate to the nearest millimetre.</p>
<p>But when I argued this, I was told that it was not my business to round the numbers. I was to perform the conversion task given the numbers I was quoted, and report the result for the person asking the question to decide how accurately the numbers are to be interpreted.</p>
<p>Is that the way of things in school? As a trained engineer, my attitude is that it is part of the purview of anybody studying mathematics to be able to estimate and report appropriate limits of accuracy, otherwise you get laughably ridiculous results like this one.</p>
<p>I confess I have never had a good relationship with teachers, apart from my A-level physics teacher whom I adored, so I expect I will be given a hard time over my inability to understand the basics of what I have failed to learn during the course of the above.</p>
| leftaroundabout | 6,902 | <p>Just to play the <del>devil's</del> teacher's advocate here: one <em>can</em> make a point that rounding should be generally avoided but measurement uncertainty instead be expressed explicitly. Specifically, rounding errors should always be <em>much smaller</em> than measurement errors. Now, if you have a figure of 22 miles, I'd interpret this as <span class="math-container">$(22\pm0.5)\mathrm{mi} = (35.4\pm0.8)\mathrm{km}$</span>. I specified one more digit, but not only did I represent the center value better (which in your rounding adds a whopping 50% error), I also captured that the inaccuracy of that result is even <em>bigger</em> than simply <span class="math-container">$35\:\mathrm{km}$</span> would suggest. In particular, <span class="math-container">$36\:\mathrm{km}$</span> is also within the range!</p>
<p>How many digits to write out is then uncritical; in physics convention is to write two non-significant digits in both the value and uncertainty figure. One is usually enough, but when <em>completely</em> omitting non-significant digits you do introduce excessive extra error. If the numbers are just stored in a computer, you should typically keep <em>all</em> the digits of the binary number representation – with double precision that means you keep a rather absurd 16 decimals! It doesn't really increase the precision, but it also doesn't really cost anything or suggest too high precision (because uncertainty is stored separately), and it makes sure that rounding really will have no contribution to the error of the final result.</p>
|
1,561,716 | <p>I have spent the last couple weeks in my Fourier Analysis course to solve PDEs with the method of separations of variables. However, I have come up with something that annoys me and I can't really explain it. Let me show an example.</p>
<p>I have this problem here</p>
<p><span class="math-container">$$u_{xx}+u_{yy}=0 \\
u(0,y) = u(1,y) = 0 \\
u(x,0) = u(x,1) = \frac{x^3-x}{6}$$</span></p>
<p>So I separate the variables and get these ODEs</p>
<p><span class="math-container">$$
X''(x)+ \lambda^2 X(x) = 0 \\
Y''(y)- \lambda^2 Y(y) = 0
$$</span></p>
<p>and these are simple to solve. The first one is just <span class="math-container">$X(x)= A\cos(\lambda x)+B \sin(\lambda x)$</span> and now is the confusing part. For me the solution to the second equation have always been <span class="math-container">$Y(y)=Ce^{- \lambda y}+ De^{ \lambda y}$</span> but the book have suddenly started to use the hyperbolic equations <span class="math-container">$\cosh$</span> and <span class="math-container">$\sinh$</span>, why is that? Is it something I am missing?</p>
| Xoque55 | 88,545 | <p>The awesome thing about hyberbolic (trig) functions is how they can be represented as sums of exponentials (and vice versa). Recall that <span class="math-container">$\cosh(u) = \dfrac{e^u + e^{-u}}{2}$</span> and <span class="math-container">$\sinh(u) = \dfrac{e^u - e^{-u}}{2}$</span>. It's just a matter of how you want to indicate the constants that come about from your initial conditions.</p>
<p>For example, since you're used to the solution <span class="math-container">$Y(y) = Ce^{\lambda y} + De^{-\lambda y}$</span>. Here's how you could express the same solution with hyperbolic trig functions (namely <span class="math-container">$\cosh$</span> and <span class="math-container">$\sinh$</span>). Let <span class="math-container">$C = \frac{A+B}{2}$</span> and <span class="math-container">$D=\frac{A-B}{2}$</span>. Then</p>
<p><span class="math-container">$$
Ce^{\lambda y} + De^{-\lambda y} \\
= \frac{A+B}{2}e^{\lambda y} + \frac{A-B}{2}e^{-\lambda y} \\
= \frac{A}{2}e^{\lambda y} + \frac{B}{2}e^{\lambda y} + \frac{A}{2}e^{-\lambda y} - \frac{B}{2}e^{-\lambda y} \\
= \frac{A}{2}(e^{\lambda y} + e^{-\lambda y}) + \frac{B}{2}(e^{\lambda y} - e^{-\lambda y}) \\
= A \frac{e^{\lambda y} + e^{-\lambda y}}{2} + B \frac{e^{\lambda y} - e^{-\lambda y}}{2} \\
= A\cosh(\lambda y) + B \sinh(\lambda y).
$$</span></p>
<p>Again, notice that the only thing that changed was really how you defined the constants with a relation from <span class="math-container">$C,D$</span> to <span class="math-container">$A,B$</span>. I stress that these are direct results from your initial (or given) conditions.</p>
<hr />
<p>As far as your question from the comments, "Why write it in that way instead of just the exponentials?", it is really a matter of conveniently denoting the properties of a solution. In an analogous fashion, even your familiar sines and cosines are merely just one way of expressing solutions to almost identical differential equations.</p>
<p>Since <span class="math-container">$\cos(u) = \dfrac{e^{iu}+e^{-iu}}{2}$</span> and <span class="math-container">$\sin(u) = \dfrac{e^{iu}+-e^{-iu}}{2i}$</span> (where <span class="math-container">$i^2 = -1$</span>), it depends on the context of your problem or possibly just your preference of where to write a solution as <span class="math-container">$P\cos(x)+Q\sin(x)$</span> or in the form <span class="math-container">$Ue^{ix}+Ve^{-ix}$</span>. It is straightforward to relate the constants <span class="math-container">$P,Q$</span> to <span class="math-container">$U,V$</span> in the same way that <span class="math-container">$A,B$</span> and <span class="math-container">$C,D$</span> were related for the hyperbolic trig case.</p>
<hr />
<p>There are several examples that I'd love to present to show why expressing your solutions as a sum of hyperbolic trig functions. There are properties of hyperbolic trig functions that are so closely related to the familiar circular trig functions that manipulating them can be very natural (as opposed to working with a sum of 2 exponentials).</p>
<p>For example, let's say your solution was expressed as <span class="math-container">$y(x) = C\cosh(mx)+D\sinh(mx)$</span>. How would you find the <span class="math-container">$n$</span>-th derivative of <span class="math-container">$y(x)$</span>? Since <span class="math-container">$\frac{d}{dx}[\cosh{mx}] = m\sinh(mx)$</span> and <span class="math-container">$\frac{d}{dx}[\sinh{mx}] = m\cosh(mx)$</span>, it's very easy to express the derivatives of <span class="math-container">$y(x)$</span>:</p>
<p><span class="math-container">$$
y(x) = y^{(0)}(x) = C\cosh(mx)+D\sinh(mx) \\
y'(x) = y^{(1)}(x) = Cm\sinh(mx) + Dm\cosh(mx) \\
y''(x) = y^{(2)}(x) = Cm^2\cosh(mx)+Dm^2\sinh(mx) \\
y'''(x) = y^{(3)}(x) = Cm^3\sinh(mx) + Dm^3\cosh(mx) \\
\vdots
$$</span></p>
|
1,561,716 | <p>I have spent the last couple weeks in my Fourier Analysis course to solve PDEs with the method of separations of variables. However, I have come up with something that annoys me and I can't really explain it. Let me show an example.</p>
<p>I have this problem here</p>
<p><span class="math-container">$$u_{xx}+u_{yy}=0 \\
u(0,y) = u(1,y) = 0 \\
u(x,0) = u(x,1) = \frac{x^3-x}{6}$$</span></p>
<p>So I separate the variables and get these ODEs</p>
<p><span class="math-container">$$
X''(x)+ \lambda^2 X(x) = 0 \\
Y''(y)- \lambda^2 Y(y) = 0
$$</span></p>
<p>and these are simple to solve. The first one is just <span class="math-container">$X(x)= A\cos(\lambda x)+B \sin(\lambda x)$</span> and now is the confusing part. For me the solution to the second equation have always been <span class="math-container">$Y(y)=Ce^{- \lambda y}+ De^{ \lambda y}$</span> but the book have suddenly started to use the hyperbolic equations <span class="math-container">$\cosh$</span> and <span class="math-container">$\sinh$</span>, why is that? Is it something I am missing?</p>
| marty cohen | 13,079 | <p>I think that
the reason that
hyperbolic functions are used
is that they satisfy
many identities similar to the ones
satisfied by the trig functions.</p>
<p>An example is
$\cosh^2(x)-\sinh^2(x)
= 1
$.</p>
|
102,721 | <p>This is probably a very simple question, but I couldn't find a duplicate.</p>
<p>As everybody knows, <code>{x, y} + v</code> gives <code>{x + v, y + v}</code>. But if I intend <code>v</code> to represent a vector, for example if I am going to substitute <code>v -> {vx, vy}</code> in the future, then the result <code>{x + v, y + v}</code> is meaningless.</p>
<p>How I can indicate to Mathematica that <code>v</code> is not a scalar and functions like <code>Plus</code> should not treat it as one? I tried setting <code>$Assumptions = v ∈ Vectors[2]</code> but that didn't help.</p>
| eldo | 14,254 | <p>You can use <code>Inactivate</code></p>
<pre><code>pl = Inactivate[{x, y} + v, Plus]
</code></pre>
<p><a href="https://i.stack.imgur.com/b7qQz.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/b7qQz.jpg" alt="enter image description here"></a></p>
<p>As an indication to not being "active" the plus sign is dimmed</p>
<p>You can still use a (partially) inactivated expression</p>
<pre><code>ti = pl * 8
</code></pre>
<p><a href="https://i.stack.imgur.com/lhuCK.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/lhuCK.jpg" alt="enter image description here"></a></p>
<p>and <code>Activate</code> it whenever you want</p>
<pre><code>Activate[ti]
</code></pre>
<p><a href="https://i.stack.imgur.com/hBrZV.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/hBrZV.jpg" alt="enter image description here"></a></p>
|
3,703,495 | <p><a href="https://en.m.wikipedia.org/wiki/Doomsday_argument" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Doomsday_argument</a></p>
<p>Suppose each new human born has the knowledge of the total number of humans born so far. So in their life, each human multiplies that number by 20 to get the upper bound for the total number of humans.</p>
<p>Assuming humans will eventually go extinct, 95% of the humans will calculate the upper bound correctly using the 'multiply by 20' rule.</p>
<p>But still, does that really imply that the number that I calculate right now would be correct with 95% accuracy?</p>
<p>What if we repeat this experiment to check if my calculation really works 95% of the time?</p>
<p>For each experiment, God randomly chooses a natural number N (as the total number of humans to be born). Assuming there is only one point in time when I can be born (which seems like a valid assumption), in each experiment, I'm always born as the <span class="math-container">$n$</span>th human. So the upper bound I calculate is always <span class="math-container">$20n$</span>.</p>
<p>Now it's easy to see that my calculation will be correct in 0% of these experiments, instead of 95.</p>
<p>So what is the correct probability? 0% or 95%?</p>
| Ryder Rude | 445,404 | <p>I think the problem is that 60B (current population) is not a randomly selected number out of the set {1,2,3,4.....N}, where N is the total number of humans ever going to exist.</p>
<p>Assuming we know nothing about <span class="math-container">$N$</span>, and each natural number value of <span class="math-container">$N$</span> is equally likely, the law of total probability says that there's a 0% chance that 60B is at least 5% of <span class="math-container">$N$</span></p>
<p><span class="math-container">$$P(60B\geq 0.05N)=\lim_{N\rightarrow \infty} \sum_{n=60B}^{1200B} \frac{1}{N} P(60B\geq 0.05N | N=n) $$</span></p>
<p><span class="math-container">$$=\lim_{N\rightarrow \infty} \sum_{n=60B}^{1200B} \frac{1}{N}=0$$</span> </p>
<p>If <span class="math-container">$60B$</span> had been randomly selected out of the set {1,2,3....N}, then we'd have gained some information about <span class="math-container">$N$</span> (larger values of <span class="math-container">$N$</span> would've been less likely), and hence the above calculation wouldn't apply in that case.</p>
|
2,072,059 | <blockquote>
<p>Let $x$ be an element of the group $G$. Prove that $x^2 = 1$ <em>iff</em> $|x|$ is either $1$ or $2$</p>
</blockquote>
<p>Now I know that $G$ is implicitly a group under multiplication, therefore $G = (A, \ \cdot \ )$, for some underlying set $A$.</p>
<p>The problem I'm facing is that $A$ could be any set. So now if we take $A = \mathbb{R}$, then $x^2 = 1 \implies |x| = 1$, but it certainly does not imply that $|x| = 2$. So how is this problem even well defined?</p>
<p>This problem is taken from <em>Chapter 1: Group Theory - Exercise 16 in 'Abstarct Algebra' by Dummit and Foote</em></p>
<p>Is there a mistake in this problem? If not how would one proceed to prove that $x^2 = 1$ <em>iff</em> $|x|$ is either $1$ or $2$, despite the counterexample I gave above?</p>
| q.Then | 222,237 | <p>You need not attempt to deduce properties of this specific group or set. This is true because of the group properties in general, not of $G = (A, \cdot )$. You may also want to read your textbook further, by conventional context in group theory, $|a|$ is by definition the order, which is the lowest positive integer such that $a^n = 1 \implies |a| = n$. If $|a| = 1$, then $a^1, a^2, a^3, ....$ is always going to be $1$ since any positive integer divides $1$. If $|a| = 2$, you immediately have $a^2 = 1$</p>
|
951 | <p>I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?</p>
| Michael Lugo | 173 | <p>One way that functions from C to C can be represented is to show the image of a grid. That is, plot the images of the lines x = constant and y = constant under your function, where z = x + yi. </p>
<p>Another is what Wikipedia calls <a href="http://en.wikipedia.org/wiki/Domain_coloring">domain coloring</a> (see <a href="http://www.mai.liu.se/~halun/complex/domain_coloring-unicode.html">this article by Hans Lundmark</a> for a more detailed exposition. The idea here is to color the range of the complex function -- since color space is three-dimensional there are multiple ways to do this -- and then color each point in the domain by the color of the corresponding point in the range.</p>
|
951 | <p>I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?</p>
| Steve Bosman | 167 | <p>Open two browser windows side-by-side and use <a href="http://www.wolframalpha.com/" rel="nofollow">wolfram alpha</a>. Recast your function f(z) as f(x+iy) and plot the <a href="http://bit.ly/bEZgsA" rel="nofollow">real part</a> in one window and the <a href="http://bit.ly/bfeLyS" rel="nofollow">imaginary part</a> in another. My example links provide plots for z<sup>3</sup>.</p>
|
65,059 | <p>I have two points ($P_1$ & $P_2$) with their coordinates given in two different frames of reference ($A$ & $B$). Given these, what I'd like to do is derive the transformation to be able to transform any point $P$ ssfrom one to the other.</p>
<p>There is no third point, but there <em>is</em> an extra constraint, which is that the y axis of Frame $B$ is parallel to the $X$-$Y$ plane of Frame $A$ (see sketch below). I <em>believe</em> that is enough information to be able to do the transformation.</p>
<p><img src="https://i.stack.imgur.com/2d6QH.png" alt="Two frames"></p>
<p>Also:</p>
<ul>
<li>The points are the same distance apart in both frames (no scaling).</li>
<li>The points don't coincide.</li>
<li>The origins don't necessarily coincide.</li>
</ul>
<p>As you may have gathered, I'm <em>not</em> a mathematician (ultimately this will end up as code), so please be gentle...</p>
<p><sub>I've seen this question (<a href="https://math.stackexchange.com/questions/23197/finding-a-rotation-transformation-from-two-coordinate-frames-in-3-space">Finding a Rotation Transformation from two Coordinate Frames in 3-Space</a>), but it's not <em>quite</em> the same as my problem, and unfortunately I'm not good enough at math to extrapolate from that case to mine.</sub></p>
<p><strong>EDIT</strong> I've updated the diagram, which makes it a bit cluttered, but (I hope) shows all the 'bits': $P3_B$ is what I'm trying to calculate...</p>
| joriki | 6,622 | <p>The problem is to find a rotation matrix $R$ and a translation vector $\vec t$ such that</p>
<p>$$R\vec p_{1B}+\vec t=\vec p_{1A}\;,\tag1$$
$$R\vec p_{2B}+\vec t=\vec p_{2A}\;.\tag2$$</p>
<p>Subtracting these yields</p>
<p>$$R\left(\vec p_{1B}-\vec p_{2B}\right)=\vec p_{1A}-\vec p_{2A}\;.\tag3$$</p>
<p>Since the $y$ axis of system $B$ is parallel to the $x$-$y$ plane of system $A$, we can obtain the rotation by first rotating around the $y$ axis and then rotating around the $z$ axis, so $R$ must have the form</p>
<p>$$
\begin{pmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\\&&1\end{pmatrix}
\begin{pmatrix}\cos\beta&&\sin\beta\\&1&\\-\sin\beta&&\cos\beta\end{pmatrix}
=
\begin{pmatrix}
\cos\alpha\cos\beta&\sin\alpha&\cos\alpha\sin\beta\\
-\sin\alpha\cos\beta&\cos\alpha&-\sin\alpha\sin\beta\\
-\sin\beta&&\cos\beta
\end{pmatrix}
\;.
$$</p>
<p>So the third row depends only on $\beta$, and writing out the corresponding row of $(3)$ yields a trigonometric equation for $\beta$:</p>
<p>$$-\sin\beta(p_{1Bx}-p_{2Bx})+\cos\beta(p_{1Bz}-p_{2Bz})=p_{1Az}-p_{2Az}\;.$$</p>
<p>Since we'll get another equation of the form</p>
<p>$$a\sin\phi+b\cos\phi=c\tag4$$</p>
<p>shortly, I'll solve it in that general form, and you can substitute</p>
<p>$$
\begin{eqnarray}
a&=&p_{2Bx}-p_{1Bx}\;,\\
b&=&p_{1Bz}-p_{2Bz}\;,\\
c&=&p_{1Az}-p_{2Az}
\end{eqnarray}
$$</p>
<p>to solve this one. Writing $a$ and $b$ in polar form, $a=r\cos\xi$, $b=r\sin\xi$, leads to</p>
<p>$$r\cos\xi\sin\phi+r\sin\xi\cos\phi=c\;,$$</p>
<p>$$\sin(\xi+\phi)=\frac cr\;.$$</p>
<p>You can get one solution from</p>
<p>$$\phi_1=\arcsin\frac cr-\xi\;,$$</p>
<p>but note that in general there's a second one, $\phi_2=\pi-\phi_1$, which makes sense, since there are in general two different angles through which you can turn a vector around the $y$ axis to give it a certain $z$ component.</p>
<p>You can determine $r$ and $\xi$ using $r=\sqrt{a^2+b^2}$ and $\xi=\text{atan}(b,a)$, where $\text{atan}$ is the two-argument arctangent function found in many programming environments, which takes into account the signs of both arguments to disambiguate the arctangent on the full range of angles.</p>
<p>Now you have two values of $\beta$, and you can substitute them into the rotation matrix. For instance, substituting into the first row of $(3)$ yields</p>
<p>$$
\cos\alpha\cos\beta(p_{1Bx}-p_{2Bx})+\sin\alpha(p_{1By}-p_{2By})+\cos\alpha\sin\beta(p_{1Bz}-p_{2Bz})=p_{1Ax}-p_{2Ax}\;,
$$</p>
<p>which is again a trigonometric equation for $\alpha$ of the form $(4)$, with</p>
<p>$$
\begin{eqnarray}
a
&=&
p_{1By}-p_{2By}\;,
\\
b
&=&
\cos\beta(p_{1Bx}-p_{2Bx})+\sin\beta(p_{1Bz}-p_{2Bz})
\;,
\\
c
&=&
p_{1Ax}-p_{2Ax}
\;.
\end{eqnarray}
$$</p>
<p>You can solve it just like the other one, to obtain two values of $\alpha$ for each of the two values of $\beta$. Again, this makes sense, since we've only used the first row of $(3)$ so far and there are two different angles through which you can turn a vector around the $z$ axis to give it a certain $x$ component. However, only one of each of these two pairs of values of $\alpha$ will also solve the second row of $(3)$ (that is, the other one would produce a wrong $y$ component), so in the end you obtain two sets of solutions for $\alpha$ and $\beta$. You can substitute each of these into either $(1)$ or $(2)$ to get the corresponding value of $\vec t$.</p>
<p>So in general you get two different affine transformations for valid point coordinates, though sometimes, e. g. if $a=b=c=0$ in either trigonometric equation, the solution will be a lot more underdetermined than that, and if you substitute invalid point coordinates (e. g. for points at different distances from each other, or such that $|c|>r$ in either trigonometric equation), there will be no solutions at all.</p>
<p>I wrote the affine transform for transforming from $B$ to $A$ to simplify the calculations, but you can easily obtain the transform in the other direction as</p>
<p>$$\vec p_B=R^T(\vec p_A-\vec t)\;.$$</p>
|
3,340,093 | <p>Is the following statement true?</p>
<blockquote>
<p>Two real numbers a and b are equal iff for every ε > 0, |a − b| < ε.</p>
</blockquote>
<p>I got that if a and b are equal then |a-b|=0 which is less than ε.
But I'm not sure if the converse also holds.</p>
| drhab | 75,923 | <p>The statement is correct. Here is a proof of the converse.</p>
<p>Suppose that <span class="math-container">$a\neq b$</span>. </p>
<p>Then <span class="math-container">$\epsilon:=|a-b|>0$</span> but we do not have <span class="math-container">$|a-b|<\epsilon$</span>.</p>
|
1,757,556 | <p>Solve the recursion $p_n = p \cdot(1 - p_{n-1}) + (1-p)p_{n-1}$</p>
<p>$p_n = p \cdot(1 - p_{n-1}) + (1-p)p_{n-1}$</p>
<p>$= p + (1-2p)p_{n-1}$ I can see that this step simply rearranges the expression, but what's the point of it? What are we trying to accomplish here? Is it to combine the $p_{n-1}$ terms?</p>
<p>$p_n - \frac{1}{2} = (1-2p)(p_{n-1} - \frac{1}{2})$ Again, I'm not quite sure why the solutions rearranged the expression this way</p>
<p>$p_n = \frac{1}{2} + C(1-2p)^n$ Here I assume $C = (p_{n-1} - \frac{1}{2})$?</p>
<p>And as $p_0 = 1$, $C = \frac{1}{2}$, and </p>
<p>$p_n = \frac{1}{2} + \frac{1}{2}(1- 2p)^n$</p>
| Christopher Carl Heckman | 261,187 | <p>$$p_n = p \cdot(1 - p_{n-1}) + (1-p)p_{n-1}= p + (1-2p)p_{n-1}$$
<em>I can see that this step simply rearranges the expression, but what's the point of it? What are we trying to accomplish here? Is it to combine the $p_{n-1}$ terms?</em></p>
<p>Yes.</p>
<p>$$p_n - \frac{1}{2} = (1-2p)\left(p_{n-1} - \frac{1}{2}\right)$$
<em>Again, I'm not quite sure why the solutions rearranged the expression this way.</em></p>
<p>This looks like a cooked-up example. It appears that, instead of solving for $p_n$, the person who wrote the solution is looking for a a relation that $\displaystyle q_n=p_n-{1\over2}$ satisfies. That would turn the equation into
$$q_n = (1-2p) q_{n-1}.$$
The solution to this will be simply $q_n = (1-2p)^n \cdot C$, for some constant $C$ (which also equals $q_0$):
$$\eqalign{
q_1 &= (1-2p) q_0\cr
q_2 &= (1-2p) q_1 = (1-2p)^2 q_0\cr
q_3 &= (1-2p) q_2 = (1-2p)^3 q_0\cr
}$$</p>
<p>And as $p_0 = 1$, $\displaystyle q_0=1-{1\over2}$, $\displaystyle C = q_0=\frac{1}{2}$,
$$q_n = {1\over2}(1-2p)^n$$
so $\displaystyle p_n = {1\over2} + q_n = {1\over2}+{1\over2}(1-2p)^n$.</p>
|
1,199,746 | <p>Let $f : (−\infty,0) → \mathbb{R}$ be the function given by $f(x) = \frac{x}{|x|}$. Use the $\epsilon -\delta$ definition of a $\lim\limits_{x \to 0^-} f(x) = -1.$</p>
<p>Workings:</p>
<p>Informal Thinking:
We want $|f(x) - L| < \epsilon$</p>
<p>$\left|\frac{x}{|x|} - -1\right| < \epsilon$</p>
<p>$\left|\frac{x}{|x|} + 1 \right| < \epsilon$</p>
<p>$\left|\frac{x + |x|}{|x|} \right| < \epsilon$</p>
<p>$x + |x| < |x|\epsilon$</p>
<p>$|x| < |x| \epsilon - x$</p>
<p>Take $\delta = |x|\epsilon - x$</p>
<p>Proof:</p>
<p>Suppose $\epsilon > 0$ and let $\delta = |x|\epsilon - x$</p>
<p>So $0 < |x| < \delta = |x| \epsilon - x$</p>
<p>I'm wondering if what I did so far is correct and what I should do next. Any help will be appreciated.</p>
| abcdef | 223,597 | <p>Take some $\epsilon >0$. We need to schow there exists a $\delta >0$ such that $\forall x: 0>x>-\delta \Rightarrow |f(x)-(-1)| <\epsilon$. <br/>
Hint: If $x<0$ what does it say about $f(x)$ what is $|f(x)+1|$ in this case?</p>
|
3,700,938 | <p>I know that the equations are equivalent by doing the math with the same value for x, but I don't understand the rules for changing orders or operations.<br>
When it is not the first addition or subtraction happening in the equation, parentheses make the addition subtraction and vice versa? Are there any other rules?</p>
<p><span class="math-container">$x^2+x-x-1 = (x^2+x)-(x+1)$</span>? </p>
<p>What if you put the parentheses around the two <span class="math-container">$x$</span>s in the middle?<br>
<span class="math-container">$x^2+(x-x)+1$</span>? Should that be (x+x) in the middle?</p>
| Community | -1 | <p>The sign before the parenthesis distributes to the content:</p>
<p><span class="math-container">$$\begin{align}+(a+b)&=a+b\\-(a+b)&=-a-b\\+(a-b)&=a-b\\-(a-b)&=-a+b.\end{align}$$</span></p>
|
2,561,968 | <p>Is there a sequence of integers such that for ∀ k it contains an arithmetic subsequence of length k but it does have no infinitely long arithmetic subsequence?</p>
| Misha Lavrov | 383,078 | <p>There's two approaches to constructing a sequence like this.</p>
<p>First, we might try to space out blocks of the sequence sufficiently far. For example,
$$
1, \qquad 3,4, \qquad 9,10,11, \qquad 23,24,25,26, \qquad 53,54,55,56,57, \dots
$$
where, whenever a block ends at $x$, the next block begins at $2x+1$.</p>
<p>In particular, the gap between the $k^{\text{th}}$ block and the $(k+1)^{\text{th}}$ is wider than the range spanned by the first $k$ blocks. If an arithmetic progression includes at least two terms from the first $k$ blocks, its difference is less than that range, so it can never jump the gap. So there is no infinite arithmetic progression, but there are obviously finite ones of arbitrary length (each block by itself).</p>
<p>Second, we can apply a <a href="https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/" rel="nofollow noreferrer">"just-do-it" construction</a>. For simplicity, we'll consider only increasing sequences of integers.</p>
<p>Enumerate <em>all</em> the infinite arithmetic progressions: this is possible, because there are only countably many pairs $(a,d)$ where $a$ is the first term and $d$ is the difference. Now, to generate our sequence, we alternate the following steps:</p>
<ol>
<li>Find a term, $x$, of the $k^{\text{th}}$ infinite arithmetic progression which is larger than any term of our current sequence, and write down $x+1$ (ensuring that we will never write down $x$).</li>
<li>Write down $x+2, x+3, \dots, x+k$, ensuring that we have a finite arithmetic progression of length $k$.</li>
</ol>
<p>For each infinite arithmetic progression, eventually we will skip one of its terms, and therefore no infinite progression is a subsequence.</p>
|
337,655 | <p>I am seeking for some intuition why norm (for any reasonable norm on functions) of a function is smaller if the function is smoother.</p>
| 40 votes | 85,506 | <p>spitespike gave a good explanation already, but here is a another point of view.
Suppose $\|\cdot\|$ is a translation-invariant norm on some space of functions on $\mathbb R$. Translation-invariant means $\|f\|=\|f(\cdot-h)\|$ for all $h\in\mathbb R$. Consider the convolution $f*\varphi$ with some <a href="http://en.wikipedia.org/wiki/Mollifier" rel="nofollow">mollifier</a> $\varphi$. In functional terms, this is a weighted average of translates of $f$. By assumption, all of these translates belong to the closed ball of radius $\|f\|$. Therefore, the average is in this ball as well:
$$\|f*\varphi\| \le \|f\| \tag1$$
We see that making $f$ smoother does not increase its norm. If the norm is strictly convex, then strict inequality holds in (1).</p>
|
243,336 | <p>This question was asked to me in an interview, I still cannot think of its solution. Can anyone help? Following is the question:</p>
<blockquote>
<p>Given an infinite number of ropes of length $R$, you have to calculate the
minimum number of cuts required to make $N$ ropes of length $C$. You can
append ropes one after the other to make ropes of length $2R, 3R, 4R, \ldots$</p>
</blockquote>
| P.. | 39,722 | <ul>
<li><p>Suppose there are $k,m\in \mathbb{N}$ such that $k\cdot R=m \cdot C$
and the minimum $m$ is $m_0$.</p>
<ol>
<li>For $N=r\cdot m_0, \ \ r \in
\mathbb{N}$ the minimum number of cuts is $r\cdot (m_0-1)$ (Make
$r$ ropes of length $m_0\cdot C$ and with $m_0-1$ cuts in each rope
you can produce $m_0$ ropes of length $C$).</li>
<li>If $N\neq r\cdot m_0, \ \forall r \in\mathbb{N}$ then you can write $N=r\cdot m_0 +N_0, \text{ for some } \ r,N_0 \in\mathbb{N} \cup \{0\}$ with $0<N_0<m_0$. Use 1. to make $r\cdot m_0$ ropes of length $C$ and 3. to make $N_0$ ropes of length $C$.</li>
<li>If $N<m_0$ make a rope of length $>N\cdot C$ and use $N$ cuts to make $N$ ropes of length $C$.</li>
</ol></li>
<li><p>If $k\cdot R\neq m \cdot C$ for any $k,m\in \mathbb{N}$ then make a rope of length $>N\cdot C$ and use $N$ cuts to make $N$ ropes of length $C$.</p></li>
</ul>
|
1,802,020 | <p>Let $S$ be a set such that if $A,B\in S$ then $A\cap B,A\triangle B\in S,$ where $\triangle$ denotes the symmetric difference operator. I would like to show that if $S$ contains $A$ and $B$, then it also contains $A\cup B, A\setminus B$.</p>
<p>The difference was easy to find, but I am not succeeding with the union. I was able to show that each of the sets $$\emptyset,A\setminus B, B\setminus A,(A\cap B)\cup(A\setminus B),(A\cap B)\cup(B\setminus A),A\cup(A\triangle B),B\cup(A\triangle B),\\(A\cap B)\cup(B\setminus A)\cup(A\triangle B)$$ is an element of $S$. Combining these I could go on producing other elements, and probably I would eventually find $A\cup B$. But I have already spent too long on the problem, there must be a cleverer approach, right?</p>
| joriki | 6,622 | <p>$$
\left(A\triangle B\right)\triangle\left(A\cap B\right)=A\cup B\;.
$$</p>
|
1,750,592 | <p>Can anyone help with this? I got a wrong answer. </p>
<p>Problem: Joe's French poodle, FooFoo, is tied to the corner of the barn which measures 40 x 30. FooFoo's rope is 50 long. In terms of π, over how many square feet can FooFoo wander?</p>
<p><a href="https://i.stack.imgur.com/Nmrc4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nmrc4.jpg" alt="enter image description here"></a></p>
| user19405892 | 282,223 | <p><a href="https://i.stack.imgur.com/dc2ps.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dc2ps.png" alt="enter image description here"></a></p>
<p>The picture above shows the limits that FooFoo can turn. It is then easy to compute our answer </p>
<p>$$\dfrac{3}{4} 50^2 \pi+\dfrac{1}{4}10^2 \pi + \dfrac{1}{4}20^2 \pi = 2000 \pi.$$</p>
|
300,900 | <p>I am learning random matrix theory. Unfortunately I do not like combinatorics, and have never really been good at it. But I found that random matrix theory has heavily relied on combinatorics, particularly in finding the limiting spectral distribution, at least for symmetric matrices, where the moment method is the most popular one. It may have an elementary appeal but I do not really find any beauty in that and I am also aware of its limitations.</p>
<p>I am a probabilist. Unfortunately in all the problems in random matrix theory I have seen, there is almost zero probability and a lot of counting involving graphs and trees. It is not doing any justice to my knowledge of probability.</p>
<p>I am studying from Bai and Silverstein's book "Spectral Analysis of Large Dimensional Random Matrices". It is a good book in terms of the results it has collected and some interesting ways in which it has proved them. But then I have been terribly disappointed in the combinatorial way it has approached the subject.</p>
<p>I am looking for questions in random matrix theory that are not combinatorial in nature. I like analysis a lot more, so if you can suggest that I look into an area that has both random matrix theory and analysis, I would be thrilled. I feel that I am not getting any insight while applying these counting principles, in that I am just solving a problem but not really understanding the structure of a random matrix. I would like to understand a random matrix. Please help me find an answer.</p>
<p>I searched online but was surprised there is no result on non combinatorial random matrix theory. It is strange that everybody studying random matrix theory is very happy with the combinatorial jargon thrown at them!</p>
<p>NOTE: I am aware of an invariance principle developed by Chatterjee, although I have not studied that in depth. In case you are aware of that, does it provide a different perspective into the universality of random matrices?</p>
| Bazin | 21,907 | <p>Using Fourier transformation, your question is about the existence of a subspace $X$ of $L^2(\mathbb R^d)$ such that
$$
\lim_{t\rightarrow 0_+}\left\{\sup_{v\in X, \Vert v\Vert=1}\int(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2 d\xi\right\}=0.
$$
Of course, as you noted, taking $X=L^2(\mathbb R^d)$ does not work since
$
\Vert 1-e^{-t\vert \xi\vert^2}\Vert_{L^\infty(\mathbb R^d)}=1.
$
On the other hand, one crude way of doing this would be assume that the regularity of $u$ is slightly better, e.g. $u\in H^s$, $s \in(0,1]$. We get then for $\lambda>0, v=\hat u$,
\begin{multline*}
\int(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2 d\xi
\\\le
\int_{\vert \xi\vert\le \lambda}(1-e^{-t\vert \xi\vert^2})\vert v(\xi)\vert^2
d\xi
+t^s\int_{\vert \xi\vert> \lambda}
\underbrace{(t\vert \xi\vert^2)^{-s}(1-e^{-t\vert \xi\vert^2})}_{\le C_s}\vert v(\xi)\vert^2 \vert\xi\vert^{2s}
d\xi
\\
\le(1-e^{-t\lambda^2})\Vert u\Vert^2_{L^2}+C_st^s\Vert u\Vert^2_{H^s},
\end{multline*}
providing for $s\in (0,1]$,
$
\lim_{t\rightarrow 0_+}\Bigl(\sup_{\Vert u\Vert_{H^s}=1}
\Vert u-e^{t\Delta}u\Vert_{L^2}\Bigr)=0.
$
Some variations could be made by choosing $\lambda$ dependent of $t$ and $\Vert u\Vert_{H^s}.$</p>
|
4,623,590 | <blockquote>
<p>Find all positive integers <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> satisfying <span class="math-container">$(x+y+z)(xy+yz+zx)=12xyz$</span>.</p>
</blockquote>
<p>Firstly, I think of <span class="math-container">$\textsf{Vieta Jumping}$</span>, but the quadratic coefficient is not a number, so I abandon this thought.</p>
<p>Then I wanted to do something with this homogeneous condition, so I divide the equation by <span class="math-container">$x$</span>, then we have to find <span class="math-container">$b\overset{\rm WLOG}\ge c\in\Bbb Q_+$</span>, satisfying
<span class="math-container">$$(1+b+c)(b+c+bc)=12bc.$$</span>
It seems tempting to do find some algebraic properties. By Cauchy inequality,
<span class="math-container">$$12bc\ge\left(\sqrt{bc}+\sqrt b+\sqrt c\right)^2.$$</span>
So <span class="math-container">$12\ge\left(1+\sqrt{\dfrac bc}+\sqrt{\dfrac cb}\right)^2$</span>, then we can solve out the range of <span class="math-container">$\dfrac bc$</span>.</p>
<p>But none of these is making any fundamental progress. How to solve this?</p>
| achille hui | 59,379 | <p>There are no positive integer solutions for the equation</p>
<p><span class="math-container">$$(x+y+z)(xy + yz + zx) = 12xyz$$</span></p>
<p>The way I do it might be an overkill. I transform the problem to one finding rational points on an elliptic curve. It turns out the elliptic curve has rank zero, so it has finitely many rational points. One can verify none of them give us a positive integer solution.</p>
<hr />
<p>Since <span class="math-container">$$(x+y+z)(xy+yz+xz) = (x+y)(y+z)(z+x) + xyz$$</span>
the question at hand is equivalent to finding solutions of
<span class="math-container">$$(x+y)(y+z)(z+x) = 11xyz$$</span>
for <span class="math-container">$x,y,z \in \mathbb{Z}_{+}$</span>. Since this equation is homogeneous, it is equivalent to finding solutions for <span class="math-container">$x,y,z \in \mathbb{Q}_{+}$</span>. This is because any rational solution can be converted to an integer one by clearing the denominators.</p>
<p>We will "normalized" the equation by assuming <span class="math-container">$x + y + z = 1$</span>. Under this assumption, the equation becomes
<span class="math-container">$$\left(\frac1x - 1\right)\left(\frac1y - 1\right)\left(\frac1z - 1\right) = 11$$</span>
Call the <span class="math-container">$3$</span> factors on LHS as <span class="math-container">$u,v,w$</span>, we have <span class="math-container">$(x,y,z) = \left(\frac{1}{1+u},\frac{1}{1+v},\frac{1}{1+w}\right)$</span> and the problem is equivalent to finding solutions for
<span class="math-container">$$uvw = 11\quad\text{ and }\quad \frac1{1+u} + \frac{1}{1+v} + \frac{1}{1+w} = 1$$</span>
for <span class="math-container">$u,v,w\in \mathbb{Q}_{+}$</span>.
Set <span class="math-container">$u = \frac{11}{vw}$</span>, substitute that in <span class="math-container">$2^{nd}$</span> equation and rearrange, we obtain:
<span class="math-container">$$vw(9-(v+w)) = 11$$</span>
Let <span class="math-container">$p = v+w$</span> and <span class="math-container">$q = v-w$</span>, this becomes
<span class="math-container">$$(p^2 - q^2)(9 - p) = 44
\iff q^2(9-p) = p^2(9-p) - 44$$</span>
Multiply both sides by <span class="math-container">$\frac{(-44)^2}{(9-p)^3}$</span>, this leads
to
<span class="math-container">$$\left(-\frac{44q}{9-p}\right)^2 =
\left(-\frac{44}{9-p}\right)^3 + \left(\frac{44p}{9-p}\right)^2$$</span></p>
<p>Change variable to <span class="math-container">$(X,Y) = \left( -\frac{44}{9-p}, \frac{-44q}{9-p} \right) \iff (p,q) = \left( 9 + \frac{44}{X}, \frac{Y}{X} \right)$</span>, the problem transforms to finding rational solution on elliptic curve
<span class="math-container">$$Y^2 = X^3 + (9X+44)^2$$</span>
Throwing following commands to the <a href="http://magma.maths.usyd.edu.au/calc/" rel="noreferrer">online magma calculator</a>,</p>
<pre><code>Q<x> := PolynomialRing(Rationals());
E00:=EllipticCurve(x^3+(9*x+44)^2);
MordellWeilShaInformation(E00);
P := Generators(E00)[1];
for i := 1 to 5 do
print i, i*P;
end for;
</code></pre>
<p>one find this elliptic curve has rank <span class="math-container">$0$</span> and Torsion subgroup <span class="math-container">$\mathbb{Z}/6\mathbb{Z}$</span> consists of 6 elements:</p>
<p><span class="math-container">$$(X,Y) = (44,-528),(0,-44),(-4,0),(0,44),(44,528),(\infty,\infty)$$</span></p>
<p>None of these give us a solution with all <span class="math-container">$u,v,w > 0$</span>, so the original equation doesn't have any positive rational and hence positive integer solutions.</p>
|
4,623,590 | <blockquote>
<p>Find all positive integers <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> satisfying <span class="math-container">$(x+y+z)(xy+yz+zx)=12xyz$</span>.</p>
</blockquote>
<p>Firstly, I think of <span class="math-container">$\textsf{Vieta Jumping}$</span>, but the quadratic coefficient is not a number, so I abandon this thought.</p>
<p>Then I wanted to do something with this homogeneous condition, so I divide the equation by <span class="math-container">$x$</span>, then we have to find <span class="math-container">$b\overset{\rm WLOG}\ge c\in\Bbb Q_+$</span>, satisfying
<span class="math-container">$$(1+b+c)(b+c+bc)=12bc.$$</span>
It seems tempting to do find some algebraic properties. By Cauchy inequality,
<span class="math-container">$$12bc\ge\left(\sqrt{bc}+\sqrt b+\sqrt c\right)^2.$$</span>
So <span class="math-container">$12\ge\left(1+\sqrt{\dfrac bc}+\sqrt{\dfrac cb}\right)^2$</span>, then we can solve out the range of <span class="math-container">$\dfrac bc$</span>.</p>
<p>But none of these is making any fundamental progress. How to solve this?</p>
| Steffen Jaeschke | 629,541 | <p>The given equation can be rewritten as:</p>
<p><span class="math-container">$x^2 y+x y^2+x^2 z-9 x y z+y^2 z+x z^2+y z^2 = 0$</span></p>
<p>The biggest problem of the task is that this is a volume and volumes of that kind are not visibly nice to humans. That has impact an the solutions methodologies. Most of them attempt to give <span class="math-container">$z(x,y)$</span> or even make the solution have only two free parameters.</p>
<p>I interpret the task that like the most trivial solution <span class="math-container">$(0,0,0)$</span> seeks for triples only. Lacking a constant makes this type of equation not homogenous.</p>
<p><a href="https://i.stack.imgur.com/kN3qK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kN3qK.png" alt="picture one" /></a></p>
<p><a href="https://i.stack.imgur.com/RwRRd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RwRRd.png" alt="picture two" /></a></p>
<p><a href="https://i.stack.imgur.com/W8GME.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W8GME.png" alt="picture three" /></a></p>
<p>My pictures show a present day solution from a bolide symbolic mathematics solver of the task. This is more that 2 and quarter of pages long and just the solution not the solutions paths.</p>
<p>As could have been expected this heavily shows the <span class="math-container">$9 x y z$</span> term. The root symbol is simply square rooting, the Root function determines sets of roots. There is a case for <span class="math-container">$x$</span> being less or equal <span class="math-container">$-1$</span>,<span class="math-container">$0$</span> or bigger equal <span class="math-container">$1$</span> by the solutions method start with <span class="math-container">$x$</span>.</p>
<p>Sorry for such small picture size. The layout enforcements of math.stackexchange.com make my pictures smaller than they really are. Click on the picture or download them to properly ready the details. Even better get yourself access to wolframcloud.com and produce the solution yourself for verification purposes. This is thereby the 13.2 version of Mathematica calculating the result.</p>
<p>Setting <span class="math-container">$x=$</span> gives pretty easy looking solutions worth for a closer look with simpler methods.</p>
<p><a href="https://www.wolframalpha.com/input?i=Reduce%5B%28x%2By%2B%20z%29%28x%20y%2By%20z%2B%20z%20x%29%3D%3D12x%20y%20z%2C%7Bx%2Cy%2Cz%7D%2CIntegers%5D&key=tave0" rel="nofollow noreferrer">Reduce[(x+y+ z)(x y+y z+ z x)==12x y z,{x,y,z},Integers]</a></p>
<p>on wolframalpha.com gives some more visual result.</p>
<p><a href="https://i.stack.imgur.com/isEd4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/isEd4.png" alt="enter image description here" /></a></p>
|
3,105,546 | <p>Suppose we have two functions <span class="math-container">$g:\mathbb{R}\rightarrow [0,1]$</span> and <span class="math-container">$f:\mathbb{R}\rightarrow [0,1]$</span> such that
<span class="math-container">$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}g(y)=\lim\limits_{y\rightarrow -\infty}f(y)=0 \tag{1}$$</span>
Moreover, we know that
<span class="math-container">$$\hspace{1cm}\lim\limits_{y\rightarrow -\infty}\frac{g(y)}{f(y)}=0 \tag{2}$$</span>
which I interpret as saying "g goes to zero faster than <span class="math-container">$f$</span>".</p>
<p>Do (1), (2) imply that
<span class="math-container">$$
\lim\limits_{y\rightarrow -\infty}\frac{f(y)}{g(y)}
$$</span>
cannot be a finite number? If this is not true, can you make a counterexample? </p>
| feynhat | 359,886 | <p>Your assertion is true.</p>
<p><span class="math-container">$\lim_{y\to-\infty} \frac{f(y)}{g(y)} = 0$</span> means that for any <span class="math-container">$\epsilon > 0$</span>, <span class="math-container">$\exists M > 0$</span> such that <span class="math-container">$\left|\frac{f(y)}{g(y)}\right| < \epsilon$</span> whenever <span class="math-container">$y < -M$</span>.</p>
<p>So, it follows that <span class="math-container">$\left|\frac{g(y)}{f(y)}\right| > \frac{1}{\epsilon}$</span>, whenever <span class="math-container">$y < -M$</span>. Thus, <span class="math-container">$\lim_{y\to-\infty}\left|\frac{g(y)}{f(y)}\right| = \infty$</span>.</p>
|
1,255,376 | <blockquote>
<p>For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ so that $q + n = 271$.</p>
</blockquote>
<p>This is true? Because both $q$ and $n$ are rational numbers and $271$ is an integer thus it's a rational number? </p>
<p>Also,</p>
<blockquote>
<p>For all $n \in \Bbb Z$ there exists $q \in \Bbb Q$ so that $q + n = 271$. </p>
</blockquote>
<p>Isn't it the same thing? One should be true and one should be false. Any help? </p>
| Jonathan Hebert | 120,932 | <p>No, it's not the same thing.</p>
<p>Note that $\frac{1}{2}$, a rational number, cannot be added to any integer to get 271. The first one is false.</p>
<p>The second is true. For any integer, there is some integer I can add to it to yield 271, and all integers are also rational numbers.</p>
|
1,507,076 | <p>I'm supposed to give a constructive proof of the following claim, but to get some hints on what to do I tried to prove the inverse and see where the proof breaks down. I have the following.</p>
<p>Claim: Define $X * Y = \{ xy \mid x \in X, y \in Y\}$. There is a ring $R$ with ideal $U$ such that $R / U$ is not closed under *.</p>
<p>What I've done:</p>
<p>Observe $S := (a + U) * (b + U) = \{ xy \mid x \in a + U, y \in b + U\}
= \{(a + u_1)(b + u_2) \mid u_1, u_2 \in U\}
= \{ ab + au_2 + u_1b + u_1u_2 \mid u_1, u_2 \in U\}$
Since $U$ is an ideal, $au_2 + u_1b + u_1u_2 \in U$. Therefore, $(a + U)*(b + U)$ could only possibly equal $ab + U$. Thus if I can find a $u \in U$ not expressible as $au_2 + u_1b + u_1u_2$, then that completes the claim, as it would make $(a + U)*(b + U)$ a strict subset of $ab + U$. But at this point I'm not sure how to proceed.</p>
| TJ Combs | 285,776 | <p>Let $R=M_2(\mathbb{F}_2)$, the ring of $2 \times 2$ matrices with entries in $\mathbb{F}_2$.</p>
<p>Let $ U = \{ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \}$, the set of upper triangular matrices of $R$.</p>
<p>I claim that $R / U$ is not closed under $*$.</p>
<p>You can show this by showing $U * \left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} +U \right)\notin R/U$.</p>
<p>$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} +U = \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \}$</p>
<p>$U * \left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} +U \right) = \{ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\} \notin U$</p>
|
1,842,365 | <p>I recently got acquainted with a theorem:</p>
<p>If $f(x)$ is a periodic function with period $P$, then $f(ax+b)$ is periodic with period $\dfrac{P}{a}$ , $a>0$.</p>
<p>I am having a difficulty in understanding this theorem. Does this theorem mean that $f(ax+b)=f(ax+b+ \dfrac{P}{a})$? </p>
<p>If the above meaning is true, then how can a function, initially having a single period, acquire another period by just changing the arguments?</p>
| Jochen | 38,982 | <p>In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$. </p>
<p>Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of
$\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \mathbb P^{Z_2}$ (here independence is used) under the transformation $(u,v)\mapsto (u+v,v)$. </p>
|
24,550 | <p>Let <span class="math-container">$H$</span> and <span class="math-container">$K$</span> be affine group schemes over a field <span class="math-container">$k$</span> of characteristic zero. Let <span class="math-container">$\varphi:H\to Aut(K)$</span> be a group action. Then we can form the semi-direct product <span class="math-container">$G = K\ltimes H$</span>.</p>
<p><strong>Problem</strong>: Describe the tannakian category <span class="math-container">$Rep_k(G)$</span> in terms of <span class="math-container">$Rep_k(K)$</span>, <span class="math-container">$Rep_k(H)$</span> and <span class="math-container">$\varphi$</span>.</p>
<p>If <span class="math-container">$\varphi$</span> is the trivial action, <span class="math-container">$G$</span> is the direct product <span class="math-container">$K\times H$</span>. In this case, <span class="math-container">$Rep_k(G) = Rep_k(K) \boxtimes Rep_k(G)$</span> is just the Deligne tensor product of the tannakian categories.</p>
<p><strong>Question</strong>: What happens in the opposite scenario where the action <span class="math-container">$\varphi$</span> is faithful? Can one characterize this situation in terms of Ext groups of simple objects in <span class="math-container">$Rep_k(K)$</span> and <span class="math-container">$Rep_k(H)$</span>?</p>
<p>I'm mostly insterested in the case where <span class="math-container">$K$</span> and <span class="math-container">$H$</span> are extensions of <span class="math-container">$\mathbb{G}_m$</span> by a pro-unipotent group but I have no idea where to start.</p>
| Victor Ostrik | 4,158 | <p>This is just a reiteration of David's answer: in this situation the group $H$ acts on the category
$Rep(K)$ (an element $h\in H$ sends a representation of $K$ into its twist by $\phi(h)$; I should add that the notion of an action of affine group on abelian category is a bit subtle..). In this language, the category $Rep(G)$ is just an "equivariantization" of the category $Rep(K)$ (in other words the objects of $Rep(G)$ are just $H-$equivariant objects of $Rep(K)$).</p>
|
2,178,428 | <p><strong>Goal</strong>: I need help to intuitively understand division by a fraction. </p>
<p><strong>Background</strong>: I've read <a href="https://math.stackexchange.com/questions/248385/what-is-the-physical-explanation-of-a-division-by-a-fraction">this</a> and <a href="http://www.mathsisfun.com/division.html" rel="nofollow noreferrer">this</a>. I understand that division is "even distribution" so, with integers, $\frac 6 2 = 3$ could mean that 6 friends get 2 chocolates each.</p>
<p><strong>Observation</strong>: When the denominator is >0, the solution is less than the numerator (e.g., $\frac 3 4$ and $\frac 4 3$), but when 0 < denominator < 1, the solution is larger than the numerator (e.g., $\frac {1} {\frac 1 2}$ = 2). <a href="https://www.google.com/search?hl=en&q=y%3D1%2Fx" rel="nofollow noreferrer">Graph of y=$\frac 1 x$</a>.</p>
<p><strong>Problem</strong>: "Dividing" (as understood by "into smaller parts") by a fraction doesn't seem possible: you end up multiplying, never dividing--and yes, I understand that division by a fraction is multiplication by that fraction's multiplicative inverse <a href="https://www.youtube.com/watch?v=f3ySpxX9oeM" rel="nofollow noreferrer">(Khan Academy)</a>.</p>
<p><strong>Question</strong>: Since 3--2 (three minus negative two) = 3+(-(-2))--i.e., addition not subtraction (<a href="https://math.stackexchange.com/questions/851987/subtraction-of-a-negative-number">source</a>)--is division by a fraction similar? In other words, is division by a fraction <strong>not division</strong> at all, but multiplication?</p>
| mlchristians | 681,917 | <p>First of all, a <em>fraction</em> is any real number of the form <span class="math-container">$\frac{a}{b}$</span>, where <span class="math-container">$a, b$</span> are integers and <span class="math-container">$b \neq 0$</span>. So to answer your second question in the title, <em>yes</em>---you can only divide by nonzero integers; for otherwise, to say that this is possible:</p>
<p><span class="math-container">$$\frac{a}{0}$$</span> </p>
<p>is to say that <span class="math-container">$0$</span> divides into <span class="math-container">$a$</span> a certain number of times, say <span class="math-container">$x$</span>; that is, <span class="math-container">$x$</span> is that number by which</p>
<p><span class="math-container">$$ 0 x = a;$$</span></p>
<p>But as we can see, no such number <span class="math-container">$x$</span>; and so, division by <span class="math-container">$0$</span> is not permitted.</p>
<p>Now, getting to the first part of what you ask in the title---</p>
<p>Any rational number is expressible as a fraction; e.g., <span class="math-container">$7$</span> is a rational number, and so, can be written exactly for example, as </p>
<p><span class="math-container">$$\frac{7}{1};$$</span></p>
<p>We can also express <span class="math-container">$7$</span> as </p>
<p><span class="math-container">$$\frac{1}{\frac{1}{7}}$$</span> </p>
<p>although in this form it is not a fraction, but a rational number equivalent to <span class="math-container">$7$</span>. </p>
<p>To see this, I now want to multiple the last expression by a number equivalent to <span class="math-container">$1$</span> but expresses such that when I multiply <span class="math-container">$\frac{1}{1/7}$</span> by it, the denominator becomes <span class="math-container">$1$</span>.</p>
<p>The number I look for is <span class="math-container">$\frac{7}{7}$</span>; and so,</p>
<p><span class="math-container">$$ \frac{1}{\frac{1}{7}} \cdot \frac{7}{7} = \frac{(1)(7)}{\left(\frac{1}{7}\right)(7)} = \frac{7}{1} = 7.$$</span></p>
<p>We usually think of division as being an arithmetic operation separate from multiplication; but this is not true. Just as subtraction is a form of addition, so is division a form of multiplication.</p>
<p>Again, take the fraction <span class="math-container">$1/7$</span>. We think we are dividing <span class="math-container">$7$</span> into <span class="math-container">$1$</span> (which we are), but equivalently, we are multiplying <span class="math-container">$1$</span> by the reciprocal of the denominator; that is,</p>
<p><span class="math-container">$$ \frac{1}{7} = 1 \left(\frac{1}{7}\right).$$</span></p>
<p>I hope this helps a little. </p>
|
549,159 | <p>How to simplify this:</p>
<p>$$(5-\sqrt{3}) \sqrt{7+\frac{5\sqrt{3}}{2}}$$</p>
<p>Dont know how to minimize to 11.</p>
<p>Thanks in advance!</p>
| njguliyev | 90,209 | <p><em>Hint:</em> $7+\dfrac{5\sqrt{3}}{2} = \dfrac{1}{2^2}\left(5^2+\left(\sqrt{3}\right)^2+2\cdot5\cdot\sqrt{3}\right)$.</p>
|
549,159 | <p>How to simplify this:</p>
<p>$$(5-\sqrt{3}) \sqrt{7+\frac{5\sqrt{3}}{2}}$$</p>
<p>Dont know how to minimize to 11.</p>
<p>Thanks in advance!</p>
| André Nicolas | 6,312 | <p>It helps to know the answer! (One might suspect it, for example, from a numerical calculation. )</p>
<p>Equivalently, we need to ask whether
$$\frac{11}{5-\sqrt{3}}\overset{??}{=}\sqrt{7+\frac{5\sqrt{3}}{2}}.\tag{1}$$</p>
<p>To simplify the left side of (1), rationalize the denominator by multiplying top and bottom by $5+\sqrt{3}$. We get
$$\frac{5-\sqrt{3}}{2}.\tag{2}$$
The square of this is equal to
$$\frac{14-\sqrt{3}}{2}.$$
It is immediate that the square of the right-hand side of (1) is also equal to $\frac{14-\sqrt{3}}{2}$, so we are finished. </p>
|
416,253 | <p>I am playing around a bit with <span class="math-container">$W^*$</span>-algebras, and I'm trying to come up with a definition for the <span class="math-container">$W^*$</span>-algebraic tensor product. Here is my first attempt:</p>
<p><a href="https://i.stack.imgur.com/WE797.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WE797.png" alt="enter image description here" /></a></p>
<p>It is easy to show that such an object exists. Simply represent the <span class="math-container">$W^*$</span>-algebras as concrete von Neumann algebras on Hilbert spaces and consider the usual von Neumann algebra tensor product (i.e. the weak closure of the algebraic tensor product <span class="math-container">$M \odot N$</span>).</p>
<p>However, I fail to show that the above definition determines the <span class="math-container">$W^*$</span>-algebra <span class="math-container">$M \overline{\otimes} N$</span> uniquely (up to (normal) <span class="math-container">$*$</span>-isomorphism). Is this true? If not, can we still modify the above universal property to get a well-working definition?</p>
<p>Basically, I want my universal property to convey the intuition that the <span class="math-container">$W^*$</span>-tensor product is simply the usual von Neumann tensor product when we appropriately represent the spaces involved.</p>
<p>Thanks in advance for any help!</p>
| Nik Weaver | 23,141 | <p>You can see this is false by taking <span class="math-container">$N = \mathbb{C}$</span>. Then, given a von Neumann algebra <span class="math-container">$M$</span>, you are asking for a von Neumann algebra <span class="math-container">$\widetilde{M}$</span> and a weak* dense embedding <span class="math-container">$\iota: M \to \widetilde{M}$</span>, such that there is a weak* continuous homomorphism <span class="math-container">$\pi: \widetilde{M} \to M$</span> with <span class="math-container">$\pi\circ\iota = {\rm id}_M$</span>. Your proposed solution is <span class="math-container">$\widetilde{M} = M$</span>, but you could also take <span class="math-container">$\widetilde{M} = M^{**}$</span>, for instance. Here <span class="math-container">$\iota: M \to M^{**}$</span> would be the natural embedding and <span class="math-container">$\pi: M^{**} \to M$</span> would be the adjoint of the natural embedding from <span class="math-container">$M_*$</span> into <span class="math-container">$M^*$</span>.</p>
<p>To bring this a little more down to earth, consider <span class="math-container">$M = l^\infty \cong C(\beta \mathbb{N})$</span> and take <span class="math-container">$\widetilde{M} = l^\infty(\beta\mathbb{N})$</span> (which is not all of <span class="math-container">$M^{**}$</span>, only part of it). Here <span class="math-container">$\iota$</span> is the identity map from <span class="math-container">$C(\beta\mathbb{N})$</span> into <span class="math-container">$l^\infty(\beta\mathbb{N})$</span> and <span class="math-container">$\pi$</span> is the restriction map from <span class="math-container">$l^\infty(\beta\mathbb{N})$</span> to <span class="math-container">$l^\infty(\mathbb{N})$</span>.</p>
|
416,253 | <p>I am playing around a bit with <span class="math-container">$W^*$</span>-algebras, and I'm trying to come up with a definition for the <span class="math-container">$W^*$</span>-algebraic tensor product. Here is my first attempt:</p>
<p><a href="https://i.stack.imgur.com/WE797.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WE797.png" alt="enter image description here" /></a></p>
<p>It is easy to show that such an object exists. Simply represent the <span class="math-container">$W^*$</span>-algebras as concrete von Neumann algebras on Hilbert spaces and consider the usual von Neumann algebra tensor product (i.e. the weak closure of the algebraic tensor product <span class="math-container">$M \odot N$</span>).</p>
<p>However, I fail to show that the above definition determines the <span class="math-container">$W^*$</span>-algebra <span class="math-container">$M \overline{\otimes} N$</span> uniquely (up to (normal) <span class="math-container">$*$</span>-isomorphism). Is this true? If not, can we still modify the above universal property to get a well-working definition?</p>
<p>Basically, I want my universal property to convey the intuition that the <span class="math-container">$W^*$</span>-tensor product is simply the usual von Neumann tensor product when we appropriately represent the spaces involved.</p>
<p>Thanks in advance for any help!</p>
| bathalf15320 | 159,073 | <p>Let me add that the basic flaw in your attempt is that you are working with a l.c. topology which is not complete. This is a problem common to the array of standard topologies which are used in von Neumann algebras--weak, strong, strong <span class="math-container">$\ast$</span>, ultraweak, ultrastrong (not, of course for the norm but it is useless here). However, this is not a "remedy is none" situation. There is a standard ploy (the mixed or strict topology) which saves the day in each case. One replaces each of the first three by the so-called <span class="math-container">$\beta$</span> version--the finest l.c. topology which agrees with it on the unit ball. These are complete. They are described in the chapter on von Neumann algebras of the monograph "Saks Spaces and Applications to Functional Analysis". Carrying out an analogous construction with one of these topologies validates your argument.</p>
|
4,380,748 | <p>I suppose by contradiction that <span class="math-container">$x+a$</span> is a factor of <span class="math-container">$x^n-a^n$</span> for all odd <span class="math-container">$n$</span>. In particular for <span class="math-container">$n=1$</span>, we have that <span class="math-container">$x+a$</span> is a factor of <span class="math-container">$x-a$</span>, but that is not possible. So that would be a contradiction. If my proof is correct?</p>
| Misha Lavrov | 383,078 | <p>Your analysis is good. Here is another argument.</p>
<p>The three statements contradict each other, so at least two of them are false, and there are at least two werewolves.</p>
<p>If there are three werewolves, all three statements must be false - but one of the statements <em>says</em> that there are three werewolves, which would be true. So this case is impossible.</p>
<p>Therefore there are exactly two werewolves, and the speaker that says so is the only human.</p>
|
2,859,463 | <blockquote>
<p>Prove or disprove. All four vertices of every regular tetrahedron in $ \mathbb{R}^3$ have at least two irrational coordinates.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/hYrWv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYrWv.png" alt="enter image description here"></a></p>
<p>This question arose from my inability to construct a tetrahedron in $\mathbb{R}^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y \;\mathrm
{and}\; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers. </p>
| S. Piérard | 317,824 | <p>False.</p>
<p>Proof.
Here is one possible solution:</p>
<ul>
<li>vertex 1: (1 0 0)</li>
<li>vertex 2: (0 0 1)</li>
<li>vertex 3: (0 1 0)</li>
<li>vertex 4: (1 1 1)</li>
</ul>
<p>You can check that:</p>
<ul>
<li>all coordinates of all vertices are rational;</li>
<li>and that all distances from one vertex to another are equal (thus this is a regular tetrahedron).</li>
</ul>
|
1,184,327 | <p>Find $\lim_{x\rightarrow 0}\exp(-1/(x^2+y^2))$.</p>
<p>I've tried multiplying the power out by the conjugate, which gives me a power of $y^2-x^2/x^4-y^4$, but I still can't figure out how to whittle this down. The most confusing thing to me is that the answer is supposedly $0$. How can $e^\text{anything}$ limit to zero?</p>
<p>EDIT: I should add: I understand the general concept of what the graph looks like, and as the exponent shrinks, the number stretches towards infinity in the x direction. The issue is that I don't know how to actually formally solve this question other than a verbal explanation.</p>
| Julián Aguirre | 4,791 | <p>The equation is in separated variables:
$$
\frac{dP}{k\,P \left (1- \dfrac P {P_\infty} \right)-H}=dt.
$$</p>
|
1,184,327 | <p>Find $\lim_{x\rightarrow 0}\exp(-1/(x^2+y^2))$.</p>
<p>I've tried multiplying the power out by the conjugate, which gives me a power of $y^2-x^2/x^4-y^4$, but I still can't figure out how to whittle this down. The most confusing thing to me is that the answer is supposedly $0$. How can $e^\text{anything}$ limit to zero?</p>
<p>EDIT: I should add: I understand the general concept of what the graph looks like, and as the exponent shrinks, the number stretches towards infinity in the x direction. The issue is that I don't know how to actually formally solve this question other than a verbal explanation.</p>
| boojum | 882,145 | <p>There are in fact three cases to consider for the logistic model with harvesting. Multiplying out the expression on the right side of the differential equation produces <span class="math-container">$$ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} \ P^2 \ + \ k P \ - \ H \ \ , $$</span></p>
<p>where I am replacing <span class="math-container">$ \ P_{\infty} \ $</span> with <span class="math-container">$ \ M \ \ , $</span> since the former label can be misleading in the harvesting model, as we shall see. We may solve the quadratic polynomial for its zeroes, which will give us the equilibrium solutions of the differential equation (if they exist). For <span class="math-container">$ \ P^2 \ - \ MP \ + \ \frac{HM}{k} \ = \ 0 \ \ , $</span> we obtain
<span class="math-container">$$ P \ \ = \ \ \frac{M \ \pm \ \sqrt{M^2 \ - \ \frac{4HM}{k}}}{2} \ \ . $$</span></p>
<p>How we factor the right side of the differential equation then depends upon the "size" of the harvesting term.</p>
<p><span class="math-container">$ \mathbf{{M^2 - \frac{4HM}{k}} \ > \ 0 \ \Rightarrow \ H < \frac{kM}{4} \ \ - \ } $</span> ("subcritical harvesting"): we have two equilibrium solutions smaller than <span class="math-container">$ \ M \ $</span> , the "carrying capacity" in the standard logistic model, that we could call <span class="math-container">$ \ M_{+} \ $</span> and <span class="math-container">$ \ M_{-} \ $</span> ; the differential equation becomes <span class="math-container">$ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · (P - M_{+}) · (P - M_{-}) \ \ , $</span> which is separable and can be solved by using partial-fraction decomposition; we in turn have three cases for the initial condition, <span class="math-container">$ \ P_0 < M_{-} \ \ , \ \ M_{-} < P_0 < M_{+} \ \ , \ $</span> and <span class="math-container">$ \ P_0 > M_{+} \ \ ; $</span> we find that <span class="math-container">$ \ M_{+} \ $</span> is the "attractor" (stable) equilibrium solution and <span class="math-container">$ \ M_{-} \ $</span> is the "repeller" (unstable) equilibrium</p>
<p><span class="math-container">$ \mathbf{ H > \frac{kM}{4} \ \ - \ } $</span> ("over-harvesting"): the polynomial <span class="math-container">$ \ P^2 \ - \ MP \ + \ \frac{HM}{k} \ $</span> cannot be factored in real numbers, so there is <strong>no</strong> equilibrium solution; the equation is separable, but we must use "completion-of-squares" to write <span class="math-container">$ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · \left(P - \frac{M}{2} \right)^2 \ - \ \left(\frac{4H \ - \ kM}{4} \right) \ \ ; $</span> we will have an "arctanh-type" solution, with the population falling to zero ("and beyond") regardless of the initial condition</p>
<p><span class="math-container">$ \mathbf{ H = \frac{kM}{4} \ \ - \ } $</span> ("critical harvesting"): the differential equation is again separable, and can be written as <span class="math-container">$ \frac{dP}{dt} \ \ = \ \ -\frac{k}{M} · \left(P - \frac{M}{2} \right)^2 \ \ ; $</span> there is a single critical solution which is a peculiar equilibrium solution sometimes called "semi-stable": <span class="math-container">$ \frac{M}{2} \ $</span> is an attractor for <span class="math-container">$ \ P_0 > \frac{M}{2} \ $</span> , but is a repeller for <span class="math-container">$ \ P_0 < \frac{M}{2} \ \ ; $</span> this can be thought of as the "merger" of the two equilibria in the subcritical case</p>
<p>Note that in all three of these models, a population at a sufficiently low level (or any over-harvested population) will "crash" to zero, since <span class="math-container">$ \ \frac{dP}{dt} < 0 \ $</span> at <span class="math-container">$ \ P = 0 \ $</span> ; the population will become <em>abruptly</em> extinct, rather than disappearing gradually.</p>
|
265,635 | <p>I'm trying to create a user-defined function that computes the equivalent resistance of <span class="math-container">$n$</span> resistors in parallel.</p>
<p>As we know, such formula is:</p>
<p><span class="math-container">$R_\text{eq.p} = \dfrac{1}{\displaystyle\sum_{k=1}^{n} \dfrac{1}{R_k}} = \left( \displaystyle\sum_{k=1}^{n} R_k^{-1} \right)^{-1} \tag*{}$</span></p>
<p>The code would seem straight forward. I tried:</p>
<pre><code>Rp[list_] := 1/Sum[1/list[[k]], {k, Length[list]}];
Attributes[Rp] = {Listable};
</code></pre>
<p>where I'm using <a href="https://reference.wolfram.com/language/ref/Listable.html" rel="noreferrer"><code>Listable</code></a> because the input of the function is a list/vector. To test it, I created the list <code>test = {1, 2, 3}</code>, yet when I enter <code>Rp[test]</code> I get the error <em>Power: Infinite expression 1/0 encountered</em>. Why isn't this working?</p>
| bmf | 85,558 | <pre><code>Quit[]
Rp[list_] := 1/Sum[1/list[[k]], {k, Length[list]}];
test = {1, 2, 3}
</code></pre>
<p>and then either</p>
<pre><code>Rp[test]
</code></pre>
<p>or my preference</p>
<pre><code>Rp@test
</code></pre>
<p>to get</p>
<blockquote>
<p><code>6/11</code></p>
</blockquote>
<p>which is the right result; see</p>
<pre><code>(1/1 + 1/2 + 1/3)^-1
</code></pre>
<p>by a direct application of the formula.</p>
|
265,635 | <p>I'm trying to create a user-defined function that computes the equivalent resistance of <span class="math-container">$n$</span> resistors in parallel.</p>
<p>As we know, such formula is:</p>
<p><span class="math-container">$R_\text{eq.p} = \dfrac{1}{\displaystyle\sum_{k=1}^{n} \dfrac{1}{R_k}} = \left( \displaystyle\sum_{k=1}^{n} R_k^{-1} \right)^{-1} \tag*{}$</span></p>
<p>The code would seem straight forward. I tried:</p>
<pre><code>Rp[list_] := 1/Sum[1/list[[k]], {k, Length[list]}];
Attributes[Rp] = {Listable};
</code></pre>
<p>where I'm using <a href="https://reference.wolfram.com/language/ref/Listable.html" rel="noreferrer"><code>Listable</code></a> because the input of the function is a list/vector. To test it, I created the list <code>test = {1, 2, 3}</code>, yet when I enter <code>Rp[test]</code> I get the error <em>Power: Infinite expression 1/0 encountered</em>. Why isn't this working?</p>
| John Doty | 27,989 | <p><code>Sum</code> is really for symbolic sums. It's clumsy here. I suggest:</p>
<pre><code>Rp[r_List] := 1/Total[1/r]
</code></pre>
|
1,691,605 | <p>I've been given the following definition:</p>
<blockquote>
<p>For a THMC with one step transition matrix $\mathbf{P}$, the row vector $\mathbf{\pi}$ with elements $(\pi_{i})_{i \in S}$ (where $S$ is the state space) is a stationary distribution iff $\mathbf{\pi \; P} = \mathbf{\pi}$</p>
</blockquote>
<p>However, I also know that many THMCs will have multiple stationary distributions. </p>
<p>This leaves me with the following questions:</p>
<ol>
<li>How can you tell how many stationary distributions a THMC has?</li>
<li>How can I show that a THMC has only one stationary distribution?</li>
<li>The equation $\mathbf{\pi \; P} = \mathbf{\pi}$ looks like it should only have one solution, so how is it possible to have multiple stationary distributions?</li>
</ol>
| Paramanand Singh | 72,031 | <p>The expression $$\frac{a^{a^x} - a^{x^a}} {a^x-x^a} $$ can be written as $$\frac{\exp(a^x\log a) - \exp(x^{a} \log a) }{a^x-x^a}=\exp(x^a\log a) \cdot\frac{\exp((a^x-x^a) \log a) - 1} {(a^x-x^a)\log a} \cdot\log a$$ and the above clearly tends to $\exp(a^a\log a) \log a=a^{a^a} \log a$. </p>
|
3,567,563 | <p>We have the polynomial <span class="math-container">$ f= X^4+X^3+X^2+X+2$</span> with <span class="math-container">$f\in \Bbb C[X] $</span>, it asks to determine the quotient of the division of the polynomial <span class="math-container">$f$</span> by the polynomial <span class="math-container">$g$</span>, <span class="math-container">$g(X)=X-\cos(\alpha)+i \sin(\alpha) \in \Bbb C[x] $</span>, <span class="math-container">$α \in(0,π/2)$</span>, knowing that <span class="math-container">$r$</span>(remainder)<span class="math-container">${}=1+i(1+\sqrt{2})$</span>. Now, what I've tried is doing long division but it seems like that might not be the first step to start with. So I was looking for a solution.</p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>As a complement, you can use <em>synthetic division</em> by <span class="math-container">$X-\cos\alpha+i\sin\alpha=X-\mathrm e^{-i\alpha}$</span>. </p>
<p>You obtain for the quotient:
<span class="math-container">$$q(X)=X^3+(1+\mathrm e^{-i\alpha})X^2+(1+\mathrm e^{-i\alpha}+\mathrm e^{-2i\alpha})X+(1+\mathrm e^{-i\alpha}+\mathrm e^{-2i\alpha}+\mathrm e^{-3i\alpha}),$$</span>
whic after some computations with the geometric series can be written as
<span class="math-container">$$\frac1{\sin\tfrac\alpha 2\,\mathrm e^{\tfrac{3i\alpha}2}}\biggl(\sin\tfrac\alpha 2\Bigl(\mathrm e^{\tfrac{i\alpha}2}X\Bigr)^3+\sin\alpha\Bigl(\mathrm e^{\tfrac{i\alpha}2}X\Bigr)^2+\sin\tfrac{3\alpha} 2\,\mathrm e^{\tfrac{i\alpha}2}X+\sin2\alpha\biggr)$$</span></p>
|
2,417,542 | <p>$$\sum_{i,j}{n\brack i+j}\binom{i+j}i$$
Does this have a combinatorial interpretation? I don't see how to use Stirling numbers of the first kind in interpretations. I know that the answer is $(n+1)!$ , but the original question didn't provide it.</p>
| Dan Robertson | 243,692 | <p>The question is asking for the angle through which a vector (perpendicular to the axis of rotation) is rotated by the rotation matrix. Consider the effect of the matrix on a vector and work out how much it is rotated by. You should find the answer to be $\frac{3\pi}2$ anti-clockwise.</p>
<p>For more general matrices I find the best way is to look at the eigenvalues of the matrix. First consider the eigenvalues of your matrix, which one can determine by solving
$$\lambda^2+1=0.$$
This gives $\lambda=\pm\mathrm i = \mathrm e^{\pm \frac{\mathrm i \pi}2}$, corresponding to a rotation of $\pm\frac\pi2,$ with the sign depending on the direction of rotation.</p>
<p>For a 3-dimensional rotation matrix you will find that one eigenvalue is 1. The corresponding eigenvector is the axis of the rotation.</p>
|
1,218,354 | <p>I've read in some textbooks that $\vdash$ and $\vDash$ are symbols present only in metalanguage. From this, I infer that their use in object language is unacceptable.</p>
<p>I would like to know why. Can't we define them as relation symbols in a structure? Or introduce them in statements for the sake of formal proofs?</p>
| tomasz | 30,222 | <p>You can use whatever symbols you want to denote whatever you want. For example, you can use ${\mathfrak w}_\$^7$ as a binary function symbol for addition of natural numbers. The utility of that is questionable, though.</p>
<p>More seriously, you certainly can introduce them if the structure in question is rich enough to intepret those in a meaningful way: for example, we can certainly interpret $\vdash$ and $\vDash$ in models of set theory.</p>
<p>This is actually how Gödel's incompleteness theorems are proved: by interpreting $\vdash$ in a theory of arithmetic. The problem with that is this is merely an interpretation, and depending on the model in which it is considered (which is why the second incompleteness theorem is possible).</p>
<p>Those interpretations are distinct from the "true" $\vdash$ and $\vDash$ (though they will, of course, coincide in case of "true" natural numbers and "true" set-theoretic universe).</p>
<p>For an illustration of how such an interpretation can stray from being "true", consider the natural numbers $({\bf N},<)$ with standard ordering. This is a nice structure. It is isomorphic to $({\bf N},<_1)$ where $n<_1m$ iff $m=0$ and $n=1$ or $n<m$. So now you have an interpretation of $<$ which is quite different from the "true" $<$. Worse yet, there are (countable!) structures elementarily equivalent to $({\bf N},<)$ with infinite descending chains. In those structures, the interpretation of $<$ is (in some ways) <em>very</em> different from the standard interpretation.</p>
<p>The same can happen whenever you try to express $\vdash,\vDash$ in a formal language.</p>
|
4,035,300 | <p>If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!</p>
<p>'''</p>
<p>When a company bids for contracts it estimates the probability of winning each contract is <span class="math-container">$0.18$</span>, independent on whether other contracts have been won or lost.</p>
<p>(a) If the company bids for 5 contracts, what is the probability it wins:</p>
<p>i) at least one contract</p>
<p>ii) at least two contracts</p>
<p>'''</p>
<p>i)I used the OR rule for this question so if there's a <span class="math-container">$0.18$</span> chance then surely it's <span class="math-container">$0.18$</span> OR <span class="math-container">$0.18$</span>, etc which is just <span class="math-container">$0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$</span></p>
<p>ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did <span class="math-container">$0.18 * 0.18 = 0.0324$</span></p>
<p>If someone could just correct me if I took my stupid pills this morning I would be very grateful!</p>
| Luke Collins | 301,095 | <p>Let <span class="math-container">$\alpha$</span> denote the real root of <span class="math-container">$X^3+X^2-1$</span>. Then <span class="math-container">$\alpha^3 = 1-\alpha^2$</span>. If we plug in <span class="math-container">$\alpha^2$</span> into <span class="math-container">$X^3-X^2+2X-1$</span>, we obtain
<span class="math-container">\begin{align*}
(\alpha^2)^3-(\alpha^2)^2+2(\alpha^2)-1 &= (1-\alpha^2)^2-\alpha^4+2\alpha^2-1\\
&=(1-\alpha^2)^2-(1-\alpha^2)^2=0,
\end{align*}</span>
so <span class="math-container">$\alpha^2$</span> is a root of <span class="math-container">$X^3-X^2+2X-1$</span>.</p>
|
2,373,175 | <p>It is probably an easy solution to this problem but I am either too overwhelmed already or not smart enough. Please help me out with this problem from the last year test</p>
<p><a href="https://i.stack.imgur.com/rzQWv.jpg" rel="nofollow noreferrer">Here is the problem</a></p>
| Adarsh Sharma | 471,162 | <p>First, let's have a look at how we can find a pattern to get the number of perfect squares in the range $R = 1, 2, 3 ... n$:</p>
<p>Let $k$ be the number of perfect squares in $R$. We need to find $k$.</p>
<p>Difference between consecutive perfect squares is $1 + 2i$, where $i = 1, 2, 3 ... k-1$. Upon subtle observation, you can discover that any perfect square can be written by the following formula:</p>
<p>Perfect Square (PS) = $1+i+2*(1+2+3+ ... i)$, where $i = 1,2,3 ... k-1$</p>
<p>Last perfect square has to be less than or equal to $n$. Therefore, putting $k-1$ in 'PS,' and equating it with $n$, we get</p>
<p>$1+k-1+2*(1+2+3+ ... k-1) <= n$</p>
<p>$\Rightarrow 1+k-1+2*\frac{(k-1)*k}{2} <= n$</p>
<p>$\Rightarrow k+ k^{2} - k <= n$</p>
<p>$\Rightarrow k <= \sqrt{n}$</p>
<p>Since, $k$ has to be positive, I neglected the negative part.</p>
<p>Now, the probability of selecting a perfect square from $R$ = $\frac{\sqrt{n}}{n}$</p>
<p>Note: If you find anything ambiguous, then please let me know in the comments below. </p>
|
3,229,789 | <p>For proving <span class="math-container">$$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$</span></p>
<p>I tried to use that:</p>
<p><span class="math-container">\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}</span></p>
<p>but i can't try more</p>
| lab bhattacharjee | 33,337 | <p><span class="math-container">$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$</span></p>
<p><span class="math-container">$\iff c^4+s^2=c^2+s^4=P$</span>(say) where <span class="math-container">$c=\cos x,s=\sin x$</span></p>
<p><span class="math-container">$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-\dfrac{\sin^22x}2=2-\dfrac{1-\cos4x}4=?$</span></p>
<p>using <span class="math-container">$\sin2x=2\sin x\cos x,\cos2y=2-2\sin^2y$</span></p>
<p>We need <span class="math-container">$$\dfrac1P+\dfrac2P=?$$</span></p>
|
27 | <p>Burt Totaro has a result that for a certain class of algebraic stacks, having affine diagonal is equivalent to the stabilizers at closed points begin affine. Is there an example of this equivalence failing in general?</p>
| David Rydh | 40 | <p>The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.</p>
<p>There are (non-separated) <em>schemes</em> with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.</p>
<p>Also see the related <a href="http://www.mathoverflow.net/questions/76/" rel="nofollow">question</a>.</p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| fedja | 1,131 | <p>Here is another model from Gil Kalai last idea:</p>
<p>We have the moving compressible fluid in which the particles can teleportate, diffuse, and organize the motion, while contamination can only diffuse. Each particle can be contaminated or not. You see contamination levels but not individual contamination. How to move? </p>
<p>The answer is that unorganized teleportation should be excluded (uncontrolled private cars), diffusion should be reduced (Israeli 100 meters from home rule for walking out), the movement of all particles should be <em>towards</em> the higher contamination density until the maximum allowed density of the fluid is reached there (meaning sending goods only from warehouses in less contaminated areas to the shops in more contaminated areas, but <em>never</em> the other way around, sending the hospital cases to the hospitals in the highest contamination areas until full capacity is reached there, the doctors working in the <em>nearest to home</em> hospitals whenever the switch is possible, etc, etc. Has this been implemented <em>anywhere</em>?</p>
<p>Now the next question: how to let China return to the normal mode and provide goods for the rest of the world? That would require more than hemotaxis in the model: we need a flow. It cannot be maintained forever, of course, but it is possible for quite a long time. What would that mean in practical terms? First, no one can enter China (they believe their cases now are just "imported", but why do we need any import? BTW, no question is rhetorical now). No shipments of goods should be sent to China; all hospital cases should be moved to the closest hospital to where they get most cases at the moment (by military helicopters or planes if it is not 911 and they can afford now to test every symptomatic case and move it), and the areas that should be opened to the relaxed life should be furthest from the areas with cases. In other words, if the cases are imported, they should stay at the border, the doctors treating them should stay at the borders, and somewhere along the borderline a deserted area should be created, so China will split itself into productive inland and defensive outskirts. In the fluid terminology, in addition to hemotaxis on the patch boundary towards the contamination, there should be hemotaxis away from the boundary inside until the patch separates into the outer layer and the inner bulk with minimal density in between (transit transport only, no real inhabitance).
How much of that is feasible is another question. But does it make sense?</p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| Martin Rubey | 3,032 | <p>I have one rather practical answer, which does not answer the question in the body, but possibly in the title.</p>
<p>I was very surprised to learn that a (small) hospital in my home town was doing their scheduling ("rostering" might be the correct term, but I don't know) by hand.</p>
<p>They were clever enough to ask for help just in time. Again surprisingly: informally via a mathematician they knew privately. Thus, they now have a professional university team developing a plan that fits their new needs - minimal service, minimizing the probability that a large portion of their staff simultaneously drops out.</p>
<p>I guess that such a strategy might also make sense for other community services, who might not even be aware of this fact.</p>
<p>(Disclaimer: I am not helping with the implementation, but I was lucky to know someone who knew someone who was a professional in this field, who luckily agreed to do it.)</p>
|
2,467,228 | <p>I attempted to start with the $L_p$ norm and raise it to the power of $p$ but got stuck because I realized that I have no idea how to eliminate the integrand.</p>
<hr>
<p><strong>$L_p$ norm:</strong>
$||f||_p = ||f||_{L_p[a,b]} = (\int_{a}^{b}~|f(x)|^p~~dx)^{\frac{1}{p}}$</p>
<p>$\\$</p>
<hr>
<p><strong>C-norm:</strong>
<a href="https://i.stack.imgur.com/OiHHj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OiHHj.png" alt="enter image description here"></a></p>
| Learnmore | 294,365 | <p>Note that $(\int _a^b|f_n-f|^p)\le \max_{x\in [a,b]} |f_n(x)-f(x)|^p\times (b-a)^p$</p>
<p>Choose $\epsilon<1$. </p>
<p>Now since $f_n\to f$ in $C-$ Norm </p>
<p>$\implies ||f_n-f||=\max_{x\in [a,b]} |f_n(x)-f(x)|<\epsilon(<1)\implies |f_n(x)-f(x)|<\epsilon$</p>
<p>$\implies |f_n(x)-f(x)|^p<\epsilon$</p>
<p>$\implies \max_{x\in [a,b]} |f_n(x)-f(x)|^p< \epsilon$ for arbitrary $\epsilon>0$.</p>
<p>Hence $||f_n-f||_p=(\int _a^b|f_n-f|^p)^{\frac{1}{p}}< \epsilon (b-a)$</p>
|
2,477,107 | <p>Okay so I have to prove this. I can write that if 2 divides n and 7 divides n, then there must be integers k and m such that
$2*k=n$
and
$7*m=n$</p>
<p>So $14*k*m=n^2$</p>
<p>But what to do after that?</p>
<p>If I say that then 14 divides $n^2$, I get bit of a circular argument, but if I write that n divides $14*k*m$, then I don't know what to do next.</p>
<p>Any help/suggestions?</p>
| nonuser | 463,553 | <p>You can say much more: <strong>Say $a$ and $b$ are relatively prime. If $a|n$ and $b|n$ then $ab|n$.</strong> </p>
<p>Proof: Since $a|n$ we can write $n=ak$. Now since $b|ak$ we have, by Euclid lemma $b|k$, so $k=bl$. Thus $n=abl$ and so $ab|n$. </p>
|
2,477,107 | <p>Okay so I have to prove this. I can write that if 2 divides n and 7 divides n, then there must be integers k and m such that
$2*k=n$
and
$7*m=n$</p>
<p>So $14*k*m=n^2$</p>
<p>But what to do after that?</p>
<p>If I say that then 14 divides $n^2$, I get bit of a circular argument, but if I write that n divides $14*k*m$, then I don't know what to do next.</p>
<p>Any help/suggestions?</p>
| orangeskid | 168,051 | <p>Write $n = 7 m$. If $m$ were odd, $7m$ would also be odd, contradiction. So $m$ is even, $m=2k$, and $n = 14k$.</p>
|
71,822 | <p><em>I have moved this question here from MSE, because I did not receive any answers as of yet over there.</em></p>
<p>I know that there are statements that are neither provable nor disprovable within some set of axioms, and I also know that such statements are called undecidable. Please allow me to call these statements to be undecidable to the first order. These statements belong to $U_1$.</p>
<p>I was wondering if there is some kind of generalization of this concept. </p>
<blockquote>
<p>Question 1. Are there any conjectures/statements of which we can prove that we cannot prove whether it is decidable or not? Related: how would such a proof look like?</p>
</blockquote>
<p>Such a statement/conjecture would be undecidable to the second order, or belong to $U_2$.
Generalizing even further
(Question 2):</p>
<blockquote>
<p>What about statements that are in $U_\infty$ ?</p>
</blockquote>
| Ron Maimon | 14,689 | <p>In questions like these, I think it is important to say right off the bat: Theorems are only undecidable in a given axiom system. For all we know, all meaningful mathematical questions, including all questions about the halting/nonhalting of any computer program, are settled from a strong enough axiom of higher infinity. It's just that you need stronger axioms for stronger theorems, and you can't use a fixed computer program to list all the the axioms.</p>
<p>Anyway, given some consistent axiomatic system, like ZF, consider the Godel statement, which is constructed as follows:</p>
<p>Write computer program GODEL to do the following:</p>
<ol>
<li><p>print its own code into a variable R</p></li>
<li><p>deduce all consequences of ZF, looking for the theorem "R does not halt"</p></li>
<li><p>if it finds this theorem, halt.</p></li>
</ol>
<p>A little reflection shows that GODEL halts if and only if ZF proves "GODEL does not halt". That is, if you call the sentence "GODEL does not halt" by the name G, then G is true if only if ZF does not prove G.</p>
<p>This sentence is unprovable in ZF, so it is order 0.</p>
<p>But it is also order 1: ZF can't prove that it is unprovable, because of the fact that it is a pi-1-0 statement, it is a statement about the non-halting of a computer programs. If program P halts, then ZF proves that this program halts (and ZF can also prove this fact about itself).</p>
<p>A little reflection shows that G is also of order 2, and order 3, of any order really, because the whole point of the Godel sentence is that ZF can't prove it, can't prove it can't prove it, can't prove that it can't prove that it can't prove it, etc. So the Godel sentence answers your question in a technical sense, but probably not the way you intended.</p>
<p>But colloquially, one says that "G is provably unproveable" when you show that G is equivalent to "consistency of ZF", so if you prove " (ZF is consistent) iff (G)", you have proved G is unprovable in common parlance, because the common intuition is that the consistency of an axiom system is to be taken for granted. But G is provable in a stronger system where the consistency of ZF is explicitly stated as a new axiom.</p>
<p>So in order to get a non-vacuous hierarchy, one could interpret your levels as follows:</p>
<p>level 0: "ZF proves that con(ZF) is equivalent to G." (So G is provably unprovable)
level 1: "ZF proves that con(ZF+con(ZF)) is equivalent to G'. (G is unprovable, and it is provably not equivalent to con(ZF), so it is provably unprovably unprovable)
level 2: "ZF proves that con(ZF+con(ZF+con(ZF))" (this is harder to state in English, but it's level 2)</p>
<p>etc, etc.</p>
<p>In this case, you can define the levels up to n, and higher levels by transfinite induction. This is studied in ordinal proof theory.</p>
|
347,214 | <p>We have to find a continuous model for a curved path which you then solve. A woman is running in the positive y-direction starting at x=50 (50,0) which is orthogonal to the x axis. At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman. We need to use the dogs position the woman's position and the gradient of the dogs location to find the model. How would I go about doing this? </p>
| johannesvalks | 155,865 | <p>You ask:</p>
<p><em>At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman.</em></p>
<hr>
<p>Path length is given by</p>
<p>$$
s = \int_0^T ds = \int_0^T \frac{ds}{dt} dt = \int_0^T v(t) dt = \Big[ v(t) t \Big]_0^T - \int_0^T v'(t) t dt = v(T)T - \int_0^T v'(t)tdt,
$$</p>
<p>and if the speed is constant we obtain</p>
<p>$$
s = v T.
$$</p>
<p>As the speed is constant for both the dog and the woman, we obtain</p>
<p>$$
s_w = v_w T
$$</p>
<p>and</p>
<p>$$
s_d = v_d T,
$$</p>
<p>therefore</p>
<p>$$
s_d = \frac{v_d}{v_w} s_w,
$$</p>
<p>i.e. the ratios of the path length is the ratios of the speeds, assuming the $v_d > v_w$.</p>
<p>If you want to know the path of the dog - the solution is more complex.</p>
<hr>
<h1>The path</h1>
<p>The situation is desribed as</p>
<p>$$
\dot{x} = \alpha (50 - x)\\
\dot{y} = \alpha (v_wt - y)
$$</p>
<p>And constant speed implies that</p>
<p>$$
\dot{x}^2 + \dot{y}^2 = v_d^2,
$$</p>
<p>so</p>
<p>$$
\alpha = \frac{v_d}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}.
$$</p>
<p>Whence</p>
<p>$$
\dot{x} = v_d \frac{50 - x}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}\\
\dot{y} = v_d \frac{v_wt - y}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}
$$</p>
<p>and I wonder if we can solve this...</p>
|
2,875 | <p>I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group <code>G</code> being <code>SL(2, R)</code>, can be completely described and that there is a discrete and continuous part of the spectrum of <code>L^2(G)</code>.</p>
<ol>
<li>How are those representations described?</li>
<li>Do all unitary representations come from <code>L^2(G)</code>?</li>
<li>How are those related to representation of compact <code>SO(3, R)</code>? </li>
<li>What happens in the flat limit between <code>SL(2, R)</code> and <code>SO(3, R)</code>?</li>
</ol>
<p>Also, is it possible to answer the questions above simultaneously for all Lie groups, not just <code>SL(2, R)</code>?</p>
| Rob Harron | 1,021 | <p>I strongly recommend you read the article "Representations of semisimple Lie groups" by Knapp and Trapa in the park city/ias proceedings "Representation theory of Lie groups". It's a very nice introduction to the problem of describing the "unitary dual" (which is what you are asking about) which focusses on SL(2,R). For example, page 9 says "the irreducible unitary representations that appear in L^2(G) do not nearly exhaust the unitary dual" for general semisimple Lie groups (thus answering you question 2). For more info, you can check out knapp's book "Representation theory of semisimple groups: an overview based on examples". For example, sections II.4 and II.5 describe the unitary duals of SL(2,C) and SL(2,R) respectively. The unitary duals of GL(n,C) and GL(n,R) were described by Vogan. Some other unitary duals are known, but in general, I don't think anything else is known. One approach is via Langlands' parametrization of irreducible admissible representations of reductive groups. This result is known for all groups and unitary representations are admissible, so the problem would be to identify which admissible representations are unitary (the knapp-trapa article talks about this). As for 3), every irreducible unitary representation of a compact group is finite-dimensional, so you don't get any of the infinite dimensional representations you get for SL(2,R). I don't know what you mean by 4).</p>
<p>For a full answer to 1) you can check out <a href="http://en.wikipedia.org/wiki/Representation_theory_of_SL2%28R%29#Unitary_representations">link text</a>.</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Simply Beautiful Art | 272,831 | <p>I too tried the same thing:</p>
<p>$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{x^2}x\stackrel{L'H}=2\lim_{x\to\infty}x$$</p>
<p>Thus,</p>
<p>$$\lim_{x\to\infty}x=2\lim_{x\to\infty}x$$</p>
<p>And as you have said,</p>
<p>$$\lim_{x\to\infty}x=0$$</p>
<p>QED (?)</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Barry Cipra | 86,747 | <p>What you have (very cleverly!) shown is that <em>if the limit $\lim_{x\to\infty}{1\over x}$ exists</em>, then, by L'Hopital, it can only equal $0$. Simply Beautiful Art's answer establishes the same result for $\lim_{x\to\infty}x$. The difference is, in your case the limit actually does exist, while in SBA's case it doesn't. That was SBA's implicit message: You haven't proven the limit is $0$, you've only proven a conditional statement; it remains to show that the limit exists.</p>
<p>One possible way to show that the limit exists without explicitly computing it would be to invoke (or prove) a theorem saying that a monotonically decreasing function that's bounded below necessarily has a limit as $x$ tends to infinity.</p>
<p>In essence, you've done the second step of a two-step process. There are other MSE questions where assuming the limit exists allows you to compute it; when I have more time I'll try to provide some links. This is the first time I can think of, though, where I've seen L'Hopital's rule used as part of the derivation.</p>
|
228,224 | <p>I am faced with the following expression</p>
<p><span class="math-container">$$
-\frac{(1 - a x^{2})^{b/2}}{b} {{}_2F_1} (1, \frac{b}{2}; \frac{c}{2}; 1 - a x^{2}) = - p t
$$</span></p>
<p>where <span class="math-container">$ a, b, c, p $</span> are constant values. Also, <span class="math-container">$ {{}_2F_1} $</span> is a hypergeometric function. Now, I am trying to plot <span class="math-container">$ x $</span> versus <span class="math-container">$ t $</span>. I am thankful for any suggestions.</p>
<hr />
<p>I don't know why the code doesn't work in my machine. I am trying a different way by this</p>
<p><code>f[a_, b_, c_, p_] := -((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2] == -p t; Plot[Evaluate@Table[f[1, 2, 3, p], {p, 1, 2, 3}], {t, -1, 2}, {x, -2, 2}]</code></p>
<p>is that correct?</p>
| Αλέξανδρος Ζεγγ | 12,924 | <p>I think <a href="https://reference.wolfram.com/language/ref/ContourPlot.html" rel="nofollow noreferrer"><code>ContourPlot</code></a> works for your case because it can find the curves represented by equations. I do not know what the values of your those constants so I make wild choices in the code below:</p>
<pre><code>With[{a = 1, b = 2, c = 3, p = 4},
ContourPlot[
-((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2] == -p t,
{t, -1, 2}, {x, -2, 2}, AspectRatio -> Automatic, FrameLabel -> {t, x}
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/6AIQe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6AIQe.png" alt="enter image description here" /></a></p>
<hr />
<p><strong>Update</strong></p>
<p>Suppose that <code>p</code> has several values,</p>
<pre><code>Block[{a = 1, b = 2, c = 3, ps = {1, 2, 3}, eqns},
eqns[p_] := -((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2] == -p t;
ContourPlot[Evaluate[eqns /@ ps], {t, -1, 2}, {x, -2, 2},
AspectRatio -> Automatic, FrameLabel -> {t, x},
PlotLegends -> StringTemplate["p = ``"] /@ ps
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/LnZGR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LnZGR.png" alt="enter image description here" /></a></p>
<hr />
<p>How about this unpacked version:</p>
<pre><code>Clear[a, b, c, ps, eqns]
a = 1; b = 2; c = 3; ps = {1, 2, 3};
eqns[p_] := -((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2] == -p t;
ContourPlot[Evaluate[eqns /@ ps], {t, -1, 2}, {x, -2, 2},
AspectRatio -> Automatic, FrameLabel -> {t, x},
PlotLegends -> (StringTemplate["p = ``"] /@ ps)
]
</code></pre>
<p><a href="https://i.stack.imgur.com/rX3ik.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rX3ik.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/1qAQX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1qAQX.png" alt="enter image description here" /></a></p>
|
3,890,064 | <p>When Max is 8 m from a lamp post which is 6 m high his shadow is 2 m long. When Max is 3 m from the lamp post, what is the length of his shadow?</p>
| poetasis | 546,655 | <p>When he is <span class="math-container">$8m$</span> from the lamp post, the end of the shadow is <span class="math-container">$10m$</span> from the lamp post meaning we have two legs of a triangle with <span class="math-container">$6$</span> and <span class="math-container">$10$</span> respectively.</p>
<p>By proportion, then, Max is X-high where <span class="math-container">$$\frac{X}{2}=\frac{6}{10}\implies X=\frac{12}{10}=1.2\space m=3.927'\quad \text{not very tall}$$</span></p>
<p>If we now move him closer (<span class="math-container">$3\space m$</span>) from the post, we can work out the proportions the other way where here, the shadow will be <span class="math-container">$Y$</span>.
<span class="math-container">$$\frac{X}{Y}=\frac{6}{3+Y}\implies 3X+XY=6Y\implies Y = -\frac{3 X}{X - 6}\land X\ne6$$</span></p>
<p><span class="math-container">$$Y = -\frac{3(1.2)}{(1.2 - 6)}=-\frac{3.6}{-4.8}=0.75\space m$$</span></p>
|
3,890,064 | <p>When Max is 8 m from a lamp post which is 6 m high his shadow is 2 m long. When Max is 3 m from the lamp post, what is the length of his shadow?</p>
| Toby Mak | 285,313 | <p>Alternately, from the gradient of <span class="math-container">$AH'$</span> being <span class="math-container">$\frac{h-6}{3}$</span>, we can work out <span class="math-container">$h$</span> using similarity: <span class="math-container">$\text{small $\Delta$ ratio} = \text{big $\Delta$ ratio} \Rightarrow \frac{h}{2} = \frac{6}{10} \Rightarrow h = 1.2$</span>, so the gradient of <span class="math-container">$AH'$</span> is <span class="math-container">$\frac{1.2-6}{3} = -1.6$</span>.</p>
<p>Then the light ray cast when Max is <span class="math-container">$3$</span> m from the lamppost has the equation <span class="math-container">$y = -1.6x + 6$</span> (<span class="math-container">$y$</span>-intercept must be <span class="math-container">$6$</span>). When the light hits the ground at <span class="math-container">$y = 0, -6 = -1.6x \Rightarrow x = \frac{15}{4}$</span>, so his shadow is <span class="math-container">$\frac{15}{4} - 3 = \frac{3}{4}$</span> m long.</p>
|
1,465,167 | <p>Given 2 points $p_1=(x_1^1, x_2^1, ..., x_n^1)$ and $p_2=(x_1^2, x_2^2, ..., x_n^2)$ in $n$-dimensional Euclidean space, how would you define the straight-line from $p_1$ to $p_2$ with these 2 points being the endpoints of the line.</p>
<p>There are a few of things I've been completely unable to figure out. One is the slope of the line, and the other is how to define the line such that it has endpoints.</p>
<p>For now, I've only attempted to do this as a Cartesian equation with a number of inequality conditions which I feel could be more complex than necessary. If there are alternative methods of defining a line I will be happy to see that.</p>
| Arpit Kansal | 175,006 | <p>Note that for $a\gt 0$,
$$a^b = e^{b\ln a}.$$
So
$$\lim_{n\to\infty}x^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln x}.$$
Since the exponential is continuous, we have
$$\lim_{n\to\infty}e^{\frac{1}{n}\ln x} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln x}.$$</p>
<p>Can you conclude now?</p>
<p><strong>OR</strong> Use a sledgehammer,</p>
<p>We use the following theorem:</p>
<p>If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$</p>
|
1,465,167 | <p>Given 2 points $p_1=(x_1^1, x_2^1, ..., x_n^1)$ and $p_2=(x_1^2, x_2^2, ..., x_n^2)$ in $n$-dimensional Euclidean space, how would you define the straight-line from $p_1$ to $p_2$ with these 2 points being the endpoints of the line.</p>
<p>There are a few of things I've been completely unable to figure out. One is the slope of the line, and the other is how to define the line such that it has endpoints.</p>
<p>For now, I've only attempted to do this as a Cartesian equation with a number of inequality conditions which I feel could be more complex than necessary. If there are alternative methods of defining a line I will be happy to see that.</p>
| Claudius | 218,931 | <p>First assume $x\ge 1$. Write $x_n = \sqrt[n]{x} - 1$. Notice that $x_n\ge0$, because $\sqrt[n]x\ge1$ if $x\ge1$. Using Bernoulli inequality we get
$$
x = (1+x_n)^n \ge 1 + n\cdot x_n
$$
for all $n$. This yields $0\le x_n\le \frac{x-1}n$ for all $n$. This shows $\lim_{n\to\infty} x_n = 0$ or $\lim_{n\to\infty} \sqrt[n]x = 1$.</p>
<p>If $0<x<1$, then we have $\lim_{n\to\infty}\sqrt[n]{1/x} = 1$ by the above and hence
$$\lim_{n\to\infty}\sqrt[n]{x} = \frac1{\lim_{n\to\infty}\sqrt[n]{1/x}} = \frac11 = 1.
$$</p>
|
2,155,740 | <p>This is a homework problem for which I think I've missed the point or have incorrectly done the proof (or both). There are two parts to the problem: Let $(a_n)$ be a sequence with $a_n \ge 0,$ for all $n$.</p>
<p>Part 1:<br>
Suppose that $a_n \rightarrow 0$. Show that $\sqrt{a_n}\rightarrow 0$. I recognized that these are actually two different sequences. Here is my proof.<br>
<strong>Proof</strong><br>
Let $(b_n)=\sqrt{a_n}$ and $\forall a\in (a_n), a\ge 0$.<br>
By assumption, $(a_n) \rightarrow 0$.<br>
By definition of <em>Square Root</em>, $\forall b_n, b_n < a_n$.<br>
So, since $b_n < a_n$ and $(a_n) \rightarrow 0, (b_n) \rightarrow 0 \square$.</p>
<p>Part 2:<br>
Suppose that $a_n \rightarrow L$. Show that $\sqrt{a_n}\rightarrow \sqrt{L}$.</p>
<p>Assuming I did things correctly in part 1, shouldn't this be a virtual <em>ditto</em>? My teacher gave this <em>hint</em> which makes me think I'm way off base:<br>
<em>you can assume that $L\ne 0$. Use that $\sqrt{x} - \sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}$</em>.</p>
<p>Since my proof in part 1 didn't use anything like this, I'm assuming I'm off in the weeds. Since this homework hasn't yet been graded, I'll need hints rather than solutions. Thanks.</p>
| Squirtle | 29,507 | <p>For the first one it is NOT true that $\sqrt{x_n} < x_n$ but the limit is still zero all the same. Are you allowed to use the fact that you can "pass a limit through a continuous function"? If so... then
$$\lim_{n\to \infty} g(x_n) = g(\lim_{n\to \infty} x_n) = g(0)$$
where $g(x)=\sqrt{x}$ and we know that $g(0)=0$. Please let me know if you are not aware of this result or cannot use it.</p>
|
2,570,008 | <p>For what value of a will this equation have only one real root: </p>
<p>$$(2a−5)x^2−2(a−1)x+3=0$$</p>
<p>Note: $x$ is a variable</p>
<p>If found that $a=4$ works, but there seems to be another solution. Any help?</p>
| Tanuj | 438,329 | <p>A quadratic equation has equal roots or repeated roots , when it fulfills the condition that </p>
<p>$D=0$ </p>
<p>Here $D$ is the discriminant of the quadratic given by $D=b^2-4a\cdot c$</p>
<p>Here , $a$ is the coefficient of $x^2$ , $b$ is the coefficient of $x$ and $c$ is the constant term .</p>
<p>Can you take it on from here ?</p>
|
4,157,472 | <p>Let <span class="math-container">$V$</span> be a vector space, and <span class="math-container">$T:V→V$</span> a linear transformation such that:</p>
<p><span class="math-container">$T(2v_1 + 3v_2) = -5v_1 - 4v_2$</span> and <span class="math-container">$T(3v_1 + 5v_2) = 3v_1 -2v_2$</span></p>
<p>Then:</p>
<p>T(v<sub>1</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>T(v<sub>2</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>T(4v<sub>1</sub>+2v<sub>2</sub>)= ? v<sub>1</sub>+ ? v<sub>2</sub></p>
<p>I cannot solve this problem and have been at it for hours. I found a similar question here: <a href="https://math.stackexchange.com/questions/1223337/finding-the-basis-of-a-vector-space">Finding the basis of a vector space</a>. I tried applying the same operations, but do not understand how they got to the final solution.</p>
| herb steinberg | 501,262 | <p>Since <span class="math-container">$||v||=\alpha$</span>, there is an <span class="math-container">$n$</span> such that <span class="math-container">$|v_n|=\alpha$</span> Let <span class="math-container">$u_n=-\alpha$</span> and <span class="math-container">$u_k=0$</span> for <span class="math-container">$k\ne n$</span>. Then <span class="math-container">$(u,v)=-\alpha^2=-\alpha||u||_1$</span>.</p>
|
279,064 | <p>I find this question interesting, but need to get it out of my system: is the space of connections (modulo gauge) on a compact four-manifold paracompact, in the Sobolev topology?</p>
<p>If so, I believe it would admit partitions of unity, which would surely make life easier in gauge theory. But I haven't seen the experts make use of such a fact. I have also heard that spaces of curves (as used in symplectic geometry) do not always admit partitions of unity.</p>
<p>The question occurred to me while reading about the first of the "five gaps" described by the late Abbas Bahri:
<a href="http://sites.math.rutgers.edu/~abahri/papers/five%20gaps.pdf" rel="noreferrer">http://sites.math.rutgers.edu/~abahri/papers/five%20gaps.pdf</a></p>
<p>As Bahri himself points out, this so-called "gap" has been filled in several different ways (via the Freed-Uhlenbeck theorem or via holonomy perturbations; that is, if the original approach is indeed flawed). Still, I think it would be useful for the younger generation to understand the core of the difficulty. I had wondered if it can be summed up in a negative answer to this question.</p>
| Tobias Diez | 17,047 | <p>Yes, the space of gauge orbits of connections is paracompact (even when you the use Fréchet topology).</p>
<p>First, the space of all connections is paracompact since it is an affine space modelled on a nuclear Fréchet space (and/or it is metrisable). Narasimhan & Ramadas (Geometry of SU(2) Gauge Fields) showed that the action of the group of gauge transformation is proper (they actually only consider SU(2) gauge theories over $S^3$ but their argument generalizes to arbitrary structure groups and arbitrary compact base manifolds). For proper group actions, the orbit space is Hausdorff and the canonical projection map is closed. Since the <a href="https://ncatlab.org/nlab/show/Michael%27s+theorem" rel="noreferrer">image of a paracompact Hausdorff space under a closed continuous map is also paracompact</a>, we conclude that the orbit space is paracompact.</p>
<p>Moreover, the gauge orbit space is stratified by smooth (Fréchet) manifolds (it's the usual orbit type stratification, the top stratum being the space of irreducible connections, see for example <a href="http://dx.doi.org/10.1016/0393-0440(89)90025-9" rel="noreferrer">The orbit space of the action of gauge transformation group on connections</a> or <a href="http://iopscience.iop.org/article/10.1088/0305-4470/35/28/201/meta" rel="noreferrer">On the gauge orbit space stratification: a review</a>). Since every stratum is modelled on a nulear Fréchet space, they are even <em>smoothly</em> paracompact (see Convenient Setting of Global Analysis by Kriegl & Michor for further details, especially chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality)).</p>
|
2,889,929 | <p>I have following function:</p>
<p>$$f(x)=x^2\cdot\left({\sin{\frac 1 x}}\right)^2$$</p>
<p>I want to find the limit of the function for $x\rightarrow0^\pm$. First I analyze $\frac 1 x$:</p>
<ul>
<li>$\frac {1}{x}\rightarrow +\infty$ for $x\rightarrow0^+$</li>
</ul>
<p>but the $\sin$ of infinity does not exist. Then I use the comparison theorem (I don't know how it's called in English) and conclude that, because</p>
<p>$$\left|{x^2\left({\sin{\frac 1 x}}\right)}^2 \right| \le \frac{1}{x^2}\rightarrow0^+$$</p>
<p>therefore the initial function tends to $0$. Is this reasoning correct? Are there better ways? </p>
| prog_SAHIL | 307,383 | <p><strong>HINT:</strong></p>
<p>$\text{P(Required)= P(A and B hits, C misses)+P(A and C hits, B misses) +}$</p>
<p>$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{P(C and B hits, A misses)+P(A,B,C all hit)}$</p>
<p>$\text{P(A)=A hits
, P(A')=A misses= 1- P(A)}$</p>
<p>$\text{P(B)=B hits
, P(B')=B misses= 1- P(B)}$</p>
<p>$\text{P(C)=C hits
, P(C')=C misses= 1- P(C)}$</p>
<p>So you can write,</p>
<p>$\text{P(Required)= P(A)P(B)P(C')+P(A)P(B')P(C) +P(A')P(B)P(C)+P(A)P(B)P(C)}$</p>
<p>$$=\frac{1}{2}\cdot\frac{1}{3}\cdot\bigg(1-\frac{1}{4}\bigg)+\frac{1}{2}\cdot\frac{1}{4}\cdot\bigg(1-\frac{1}{3}\bigg)+\frac{1}{4}\cdot\frac{1}{3}\cdot\bigg(1-\frac{1}{2}\bigg)+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}$$</p>
<p>$$=\frac{3}{24}+\frac{2}{24}+\frac{1}{24}+\frac{1}{24}$$</p>
<p>$$=\frac{7}{24}$$</p>
<p>Comment if you have any more doubts.</p>
|
1,925,182 | <p>I am trying to understand the representation theory of $GL_2(\mathbb{R})$, and I have a few questions. I know that it is generally inadvisable to put a few questions in the same topic, but since they are quite inter-related I will give it a shot like this.</p>
<p>I can understand the idea of classifying irreducible unitary $GL_2(\mathbb{R})$-representations by first classifying $(gl_2(\mathbb{R}), O(2))$-modules, then picking out the unitary ones from there. However,</p>
<ol>
<li><p>To classify $(gl_2(\mathbb{R}), O(2))$-modules, one is led to study the parabolic induction. But why induce from parabolics? Is it because flag manifolds are well-studied and we can compute a lot of things about the cohomology of its vector bundles?</p></li>
<li><p>A more confusing point, is the study of irreducibility of parabolic induction. How does one study its (ir)reducibility? My main concern is that Schur's lemma fails in this case. Several places I looked at are very unsatisfying,</p>
<ul>
<li>Bump first calculated the necessary conditions on irreducible admissible $(gl_2(\mathbb{R}, O(2))$-modules, then show that they all arise from parabolic induction, by actually calculating the $K$-types and the infinitesimal character. Such approach does not look very generalizable to arbitrary reductive groups, mainly because the necessary conditions on irreducible $({\frak{g}},K)$-modules in general cannot be written as easily I imagine.</li>
<li>Jacquet-Langlands directly quoted Harish-Chandra's paper, oh well..</li>
<li>Moeglin's exposition "Representations of $GL(n,\mathbb{R})$" (on Speh-Vogan's method) seems to rely on Langlands classification, and I really want to know if there are direct ways to study the reducibility.</li>
<li>Some other places focus on the unitary representations and thus use Schur's lemma. I can accept this, but what if I just want to look at $({\frak{g}}, K)$-modules in general?</li>
</ul></li>
</ol>
<p>I know that there are a lot of questions here, and I would be immensely grateful if someone can at least tell me where to look. Thanks!</p>
| Kimball | 11,323 | <p>Let me just give a very basic answer to 1 (which may be too simple for you, I don't know). In general, the basic way to study representations of a group $G$ is to try to reduce the problem to the case of simpler subgroups $H$. For $G=GL(n)$, one of the simplest kinds of subgroups you can consider are the (proper) Levi subgroups $M$, which are direct products of $GL(m)$'s for smaller $m$'s. So you can try to construct representations of $G$ by inducing from $M$, but if you directly induce your representation will be too big. So you first extend representations of $M$ to an appropriate parabolic $P \supset M$, and induce from $P$, which gives you parabolic induction.</p>
<p>For $GL(2,\mathbb R)$, there is only one interesting Levi (up to isomorphism), $M = \mathbb R^\times \times \mathbb R^\times$. Thus parabolic induction lets you study representations of the very non-abelian group $GL(2,\mathbb R)$ by reducing to the abelian case of pairs of characters of $\mathbb R^\times$. Here we luck out because you can describe all irreducible unitary representations of $GL(2, \mathbb R)$ from parabolic induction. In the case of $GL(2, \mathbb Q_p)$ parabolic induction does not suffice, in some sense because there are a bunch of other interesting subgroups, and one also need to consider inducing from non-split tori or open compact subgroups like $GL(2, \mathbb Z_p)$ (which gives the so-called supercuspidal representations).</p>
|
1,925,182 | <p>I am trying to understand the representation theory of $GL_2(\mathbb{R})$, and I have a few questions. I know that it is generally inadvisable to put a few questions in the same topic, but since they are quite inter-related I will give it a shot like this.</p>
<p>I can understand the idea of classifying irreducible unitary $GL_2(\mathbb{R})$-representations by first classifying $(gl_2(\mathbb{R}), O(2))$-modules, then picking out the unitary ones from there. However,</p>
<ol>
<li><p>To classify $(gl_2(\mathbb{R}), O(2))$-modules, one is led to study the parabolic induction. But why induce from parabolics? Is it because flag manifolds are well-studied and we can compute a lot of things about the cohomology of its vector bundles?</p></li>
<li><p>A more confusing point, is the study of irreducibility of parabolic induction. How does one study its (ir)reducibility? My main concern is that Schur's lemma fails in this case. Several places I looked at are very unsatisfying,</p>
<ul>
<li>Bump first calculated the necessary conditions on irreducible admissible $(gl_2(\mathbb{R}, O(2))$-modules, then show that they all arise from parabolic induction, by actually calculating the $K$-types and the infinitesimal character. Such approach does not look very generalizable to arbitrary reductive groups, mainly because the necessary conditions on irreducible $({\frak{g}},K)$-modules in general cannot be written as easily I imagine.</li>
<li>Jacquet-Langlands directly quoted Harish-Chandra's paper, oh well..</li>
<li>Moeglin's exposition "Representations of $GL(n,\mathbb{R})$" (on Speh-Vogan's method) seems to rely on Langlands classification, and I really want to know if there are direct ways to study the reducibility.</li>
<li>Some other places focus on the unitary representations and thus use Schur's lemma. I can accept this, but what if I just want to look at $({\frak{g}}, K)$-modules in general?</li>
</ul></li>
</ol>
<p>I know that there are a lot of questions here, and I would be immensely grateful if someone can at least tell me where to look. Thanks!</p>
| tracing | 200,415 | <p>Regarding your question 1.,
there is an important underlying fact at work here, which you may not know:</p>
<p>It is a theorem of Casselman (the <em>subrepresentation theorem</em>) that for a real reductive Lie group $G$, any irreducible admissible $(\mathfrak g, K)$-module can be embedded as a subrepresentation of a parabolically induced representation. (This generalizes an earlier result of Harish Chandra, which was the slightly weaker statement in which subrepresentation is replaced by <em>subquotient</em>. For more discussion, you could read <a href="http://www.mathunion.org/ICM/ICM1978.2/Main/icm1978.2.0557.0564.ocr.pdf" rel="nofollow noreferrer">Casselman's 1978 ICM talk</a>.)</p>
<p>Given this general fact, it certainly makes sense to study parabolic inductions and their reducibility in order to classify irreducible admissible $(\mathfrak gl_2, O_2)$-modules (and hence also unitary representations of $\mathrm{GL}_2(\mathbb R)$).</p>
<p>As to how Harish Chandra and Casselman discovered their theorems, or (more-or-less equivalently) how they realized that it was useful/important to focus on parabolic inductions, I don't know if there is simple answer. Ceratinly it is related to the geometry of flag manifolds, as well as analogies with finite dimensional representations and Borel--Weil--Bott theory. </p>
<p>From a structural/algebraic point of view, the Iwasawa decomposition
$G = KAN$ leads to a tensor decomposition of the enveloping algebra
$U\mathfrak g = U\mathfrak k \otimes U\mathfrak a \otimes U\mathfrak n.$ Now on an admissible representation $U\mathfrak k$ acts through finite dimensional quotients, while $U\mathfrak a$ is closely related to the centre of $U\mathfrak g$ via the Harish Chandra isomorphism, and so also acts through a finite quotient on an irreducible representation. The upshot is that an irreducible (or more generally a f.g.) $(\mathfrak g, K)$-module will actually be
f.g. over $U\mathfrak n$ (this is the so-called <em>Lemma
of Osborne</em>). This suggests a possible relationship to parabolically induced representations.</p>
<p>From a geometric perspective, one knows that f.g. $(\mathfrak g, K)$-modules
can be described as $K$-equivariant coherent $\mathcal D$-modules on the flag variety (this is Beilinson--Bernstein theory), while $\mathfrak g$-modules
in so-called category $\mathcal O$ (so Verma-type modules, which are closely related to parabolically induced representations) can be described
as $N$-equivariant $\mathcal D$-modules on the flag variety. Geometrically, one can relate the two geometric situations by using a certain flow on the flag variety to move the $K$-orbits to the $N$-orbits, and so give a geometric proof of Casselman's theorem. This is done in a paper of Emerton--Nadler--Vilonen (see e.g. <a href="https://arxiv.org/abs/math/0311339" rel="nofollow noreferrer">here</a>).</p>
<hr>
<p>Regarding 2, for general groups, the reducibility of principal series can be subtle, and even for tempered reps. is the subject of a general theory,
due to Knapp--Zuckerman (see e.g. <a href="https://publications.ias.edu/node/17" rel="nofollow noreferrer">Langlands notes</a>). </p>
<p>But in the case of $\mathfrak gl_2$ --- or, more-or-less equivalently --- of $\mathfrak sl_2$, the analysis is pretty easy, and very similar to the finite-dimensional situation. The induction of a character from the Borel is a direct sum of weight spaces for $SO(2)$, each of dimension one, and each composite $X^+ X^-$ and $X^- X^+$ (lowering and then raising, or raising and then lowering) acts on a given weight space by a scalar depending just on the weight and on the Casimir eigenvalue. With this information computed, it is then easy to determine any reducibility's, just by looking for weights where $X^+ X^- =0$ or $X^- X^+ =0$).</p>
|
1,925,182 | <p>I am trying to understand the representation theory of $GL_2(\mathbb{R})$, and I have a few questions. I know that it is generally inadvisable to put a few questions in the same topic, but since they are quite inter-related I will give it a shot like this.</p>
<p>I can understand the idea of classifying irreducible unitary $GL_2(\mathbb{R})$-representations by first classifying $(gl_2(\mathbb{R}), O(2))$-modules, then picking out the unitary ones from there. However,</p>
<ol>
<li><p>To classify $(gl_2(\mathbb{R}), O(2))$-modules, one is led to study the parabolic induction. But why induce from parabolics? Is it because flag manifolds are well-studied and we can compute a lot of things about the cohomology of its vector bundles?</p></li>
<li><p>A more confusing point, is the study of irreducibility of parabolic induction. How does one study its (ir)reducibility? My main concern is that Schur's lemma fails in this case. Several places I looked at are very unsatisfying,</p>
<ul>
<li>Bump first calculated the necessary conditions on irreducible admissible $(gl_2(\mathbb{R}, O(2))$-modules, then show that they all arise from parabolic induction, by actually calculating the $K$-types and the infinitesimal character. Such approach does not look very generalizable to arbitrary reductive groups, mainly because the necessary conditions on irreducible $({\frak{g}},K)$-modules in general cannot be written as easily I imagine.</li>
<li>Jacquet-Langlands directly quoted Harish-Chandra's paper, oh well..</li>
<li>Moeglin's exposition "Representations of $GL(n,\mathbb{R})$" (on Speh-Vogan's method) seems to rely on Langlands classification, and I really want to know if there are direct ways to study the reducibility.</li>
<li>Some other places focus on the unitary representations and thus use Schur's lemma. I can accept this, but what if I just want to look at $({\frak{g}}, K)$-modules in general?</li>
</ul></li>
</ol>
<p>I know that there are a lot of questions here, and I would be immensely grateful if someone can at least tell me where to look. Thanks!</p>
| Devadatta Hegde | 628,020 | <p>For almost all that I am going to say below the reference is V.S. Varadarajan, An introduction to Harmonic analysis on semisimple Lie groups. </p>
<p>The concept of parabolic induction is due to Gelfand and Naimark from their monograph "Unitary representations of classical groups" (~1950). They used the principal series representations (induced from minimal parabolic) to obtain a Plancherel formula for <span class="math-container">$SL(n,C)$</span>. They were inspired to look at <em>smooth</em> sections of <em>real analytic</em> line bundles instead of <em>holomorphic</em> sections of <em>complex analytic</em> ones. </p>
<p>Irreducibility is, as far as I know, highly non-trivial. Let us assume we are only interested in inductions which give unitary representations. The following statements should illustrate that there is <em>some</em> subtlety. </p>
<ol>
<li><p>Induction from Borel for <em>complex Lie groups</em> is always irreducible. (This is much easier than others below )</p></li>
<li><p>For real Lie groups, induction from parabolics with non-discrete series data on the Levi-components is not necessarily irreducible. </p></li>
<li><p>It was Harish-Chandra who showed that with discrete series data on the Levi the induced representations are <em>generically</em> irreducible. ( Generically here means ,for example, if the inductions are parametrized by the real line then except for finitely many as in the case of <span class="math-container">$SL(2,R)$</span>)</p></li>
<li><p>If induced from a parabolic subgroup that is associated to a Cartan subgroup which is <em>fundamental</em> (A techinical condition that has relevance for Plancherel theroem) with discrete series data on the Levi then the induced representation is <em>always</em> irreducible. </p></li>
</ol>
<p>A place which I think, although I have not read, has a proof of these assertions is "On the transverse symbol of vectorial distributions and some applications to harmonic analysis" by Varadarajan and Kolk. You can find it by googling. </p>
|
2,936,608 | <blockquote>
<p>Let <span class="math-container">$a_n$</span> where <span class="math-container">$n \in \mathbb {N}$</span> be a sequence of rational numbers converging to <span class="math-container">$a$</span>. Suppose <span class="math-container">$a \neq 0$</span>, for <span class="math-container">$k = 1, 2, ...$</span> let
<span class="math-container">$$b_k=\begin{cases} 0 & \text{if}\;a_k=0\\\\\frac{1}{a_k} &\text{if}\;a_k \neq 0\end{cases}$$</span>
Prove that <span class="math-container">$b_n$</span> converges to <span class="math-container">$\frac{1}{a}$</span>.</p>
</blockquote>
<p>I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?</p>
<p>Thanks</p>
<p>edit: is it possible to solve this in terms of Cauchy Sequence?</p>
| 1123581321 | 482,390 | <p><strong>Hint</strong></p>
<p>Because <span class="math-container">$a\not=0$</span> (let's say <span class="math-container">$a>0$</span>), show that <span class="math-container">$\exists n_0\in\mathbb{N}: \forall n\geq n_0\quad a_n>0$</span> </p>
|
3,855,736 | <p>Mathematics is not my primary discipline, but I know enough about both it and academics in general to know that many to most mathematical researchers do what they do because they enjoy doing it. This would seem to make "recreational mathematics" a rhetorical tautology, yet the term is used as if it were a subdiscipline in its own right. Universities offer courses on recreational mathematics. There are academic journals on recreational mathematics. There's a tag right here on SE Mathematics that reads <code>recreational-mathematics</code>. It's defined as "Mathematics done just for fun, often disjoint from typical school mathematics curriculum." Yet it does seem to be applied to a certain specific, albeit rather eclectic category of mathematical problems.</p>
<p>A lot of math that is called recreational falls quite neatly under some other category of math, often logic or combinatorics. In some cases, a kind of recreational math seems to be characterized as such only because some other formulation of the same ideas "got there first": it's relaxing to draw shapes with a compass and straightedge in much the same way that folding paper is relaxing, and the mathematics behind origami apprehend most if not all of the same mathematical truths that Euclidean geometry does, but mathematical origami is considered recreational while Euclidean geometry generally isn't.</p>
<p>With that in mind, what <em>is</em> recreational math? "Math that's done for fun" doesn't seem to make sense, because, again, that applies in some capacity to all math. Is it math that's done casually, with less concern for rigorous proof? Is it math that has not (yet) found an application in engineering or the empirical sciences? Is it the mathematical counterpart to popular science? What <em>is</em> it?</p>
<p>It's good math, but I don't understand this label attached to it. What makes a particular bit of math recreational?</p>
| Robert Israel | 8,508 | <p>Recreational math tends to comprise topics that are fun, and easily accessible to people without a lot of specialized training, but, for one reason or another, are not usually topics of serious mathematical research. Factors in that lack of serious mathematical research may include lack of applications of the topic, lack of mathematical tools
to apply to the topic, a feeling that the topic itself is "silly" or arbitrary, or simply outside the mathematical fashion of the day.</p>
<p>You might look at <a href="https://www.jstor.org/stable/2689642?seq=1" rel="noreferrer">"What is recreational mathematics?"</a> by Charles Trigg, Mathematics Magazine 51(1), 18-21. doi:10.2307/2689642</p>
|
4,380,274 | <p>Indefinite integral is pretty easy to solve, I did it by substitution and I'm pretty sure it can be done relatively easy via integration by parts.
The problem are boundaries.</p>
<p>After substitution <span class="math-container">$arcsin x=t$</span> we get</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{(\sin t)^3}dt$$</span>
so we have</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{\cos t}{(\sin t)^2} dt+ \int_0^\frac{\pi}{2} \frac{t \cos t}{(\sin t)^3}dt$$</span></p>
<p>Now for example first integral is easy to compute and I get <span class="math-container">$\frac{-1}{\sin t}$</span> from 0 to <span class="math-container">$\frac{\pi}{2}$</span>.But in 0 the value is <span class="math-container">$\infty$</span>.
The second integral can also be solved using partial integration with <span class="math-container">$t$</span> and <span class="math-container">$\frac{cost}{(sint)^3}$</span>.</p>
<p>The only method to bypass this that I know of is linear substitution (I'm not sure how you call it) but that doesn't give anything useful.</p>
<p>Any hints?</p>
| Svyatoslav | 869,237 | <p>We can also try to evaluate the integral simply integrating by part.
<span class="math-container">$$I=\int_0^1 \frac{x-\arcsin x}{x^3}dx=-\frac{1}{2}\int_0^1 (x-\arcsin x)d\Big(\frac{1}{x^2}\Big)$$</span>
<span class="math-container">$$=-\frac{x-\arcsin x}{2x^2}\Big|_0^1+\frac{1}{2}\int_0^1\frac{dx}{x^2}\Big(1-\frac{1}{\sqrt{1-x^2}}\Big)$$</span>
Making the substitution <span class="math-container">$t=x^2$</span> in the second integral
<span class="math-container">$$I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\int_0^1\Big(t^{-\frac{1}{2}-1}-t^{-\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1}\Big)dt$$</span>
Now we can use the analytical continuation of Beta-function <span class="math-container">$\Bigl (B(\gamma,\alpha)=\int_0^1s^{\gamma-1}(1-s)^{\alpha-1}ds$</span>, if <span class="math-container">$\gamma, \alpha >0\,\Bigr)$</span> for negative <span class="math-container">$\gamma\in(-1,0)$</span>:
<span class="math-container">$$B(\gamma,\alpha)=-\frac{1}{(\exp(2\pi{i}\alpha)-1)(\exp(2\pi{i}\gamma)-1)}\oint_Ps^{\gamma-1}(1-s)^{\alpha-1}ds$$</span></p>
<p>where <span class="math-container">$P$</span> is Pochhammer contour in the complex plane.</p>
<p>It can be shown (<a href="https://math.stackexchange.com/questions/4119506/integrate-int-0-infty-big-y-x-lambda-1-y-lambda-1-bi/4121834#4121834">for example, here</a>) that for <span class="math-container">$\gamma\in(-1;0)$</span> and <span class="math-container">$\alpha>0$</span> <span class="math-container">$B(\gamma,\alpha)=\lim_{r\to0}(\int_r^1s^{\gamma-1}(1-s)^{\alpha-1}ds+\frac{r^\gamma}{\gamma})$</span>,
so <span class="math-container">$$J(\gamma,\alpha,\alpha')=\int_0^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds$$</span><span class="math-container">$$=\lim_{r\to0}\int_r^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds=\lim_{r\to0}\left(B(\gamma,\alpha)-\frac{r^\gamma}{\gamma}-B(\gamma,\alpha')+\frac{r^\gamma}{\gamma}\right)$$</span> <span class="math-container">$$J(\gamma,\alpha,\alpha')=B(\gamma,\alpha)-B(\gamma,\alpha')\,,\,\gamma\in(-1;0)\,,\,\, \alpha,\alpha'>0 $$</span></p>
<p>It can also be proved that analytically continued Beta-function is expressed in the usual way in terms of Gamma-function: <span class="math-container">$B(\gamma,\alpha)=\frac{\Gamma(\gamma)\Gamma(\alpha)}{\Gamma(\gamma+\alpha)}$</span>. This expression is valid for all complex <span class="math-container">$\alpha, \gamma$</span>.</p>
<p>Coming back to the initial integral
<span class="math-container">$$I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\Big(B\big(-\frac{1}{2};1\big)-B\big(-\frac{1}{2};-\frac{1}{2}\big)\Big)$$</span>
<span class="math-container">$$=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\bigg(\frac{\Gamma\big(-\frac{1}{2}\big)\Gamma(1)}{\Gamma\big(\frac{1}{2}\big)}-\frac{\Gamma\big(-\frac{1}{2}\big)\Gamma\big(\frac{1}{2}\big)}{\Gamma (0)}\bigg)$$</span>
Given than <span class="math-container">$\Gamma(0)=\infty$</span> and <span class="math-container">$\Gamma\big(-\frac{1}{2}\big)=-2\Gamma\big(\frac{1}{2}\big)$</span>
<span class="math-container">$$I=\frac{\pi}{4}-\frac{1}{2}-\frac{1}{2}=\frac{\pi}{4}-1$$</span>
WolframAlpha gives the same result.</p>
|
4,146,858 | <blockquote>
<p>Q) For every twice differentiable function <span class="math-container">$f:\mathbb{R}\longrightarrow [-2,2] $</span> with <span class="math-container">$[f(0)]^2+[f'(0)]^2=85$</span> , which of the following statement(s) is(are) TRUE?</p>
</blockquote>
<blockquote>
<p>(A) There exists <span class="math-container">$r,s \in\mathbb{R}$</span> , where <span class="math-container">$r<s$</span> , such that f is one-one on the open interval <span class="math-container">$(r,s)$</span></p>
</blockquote>
<blockquote>
<p>(B) There exists <span class="math-container">$x_0 \in (-4,0)$</span> such that <span class="math-container">$|f'(x_0)|\leq 1$</span></p>
</blockquote>
<blockquote>
<p>(C) <span class="math-container">$lim_{x\to \infty}f(x)=1$</span></p>
</blockquote>
<blockquote>
<p>(D) There exists <span class="math-container">$\alpha \in (-4,4)$</span> such that <span class="math-container">$f(\alpha)+f"(\alpha)=0 and f'(\alpha)=0$</span></p>
</blockquote>
<p>I have problem with option B. Here's the given solution-
<a href="https://i.stack.imgur.com/DV1ND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DV1ND.jpg" alt="enter image description here" /></a></p>
<p>But I think this is wrong because <span class="math-container">$f(0)$</span> cannot be equal to <span class="math-container">$2$</span> as that gives us the value of <span class="math-container">$f'(0)=9$</span> (from the condition given in the question). If the slope is positive at <span class="math-container">$x=0$</span> and the function is achieving its highest value there then for the points in the right neighbourhood of <span class="math-container">$x=0$</span> the slope would have to abruptly change to <span class="math-container">$0$</span> otherwise the function would obtain values greater than <span class="math-container">$2$</span> which are not in its co-domain. But it cannot abruptly change either coz it's given to be a twice differentiable function.</p>
<p>And I took <span class="math-container">$f(x)=9$</span> only but of course similar argument can be made with the negative value and a similar argument can be made if they took <span class="math-container">$f(0)=-2$</span> and <span class="math-container">$f(-4)=2$</span> in the solution.</p>
<p>My question-</p>
<ol>
<li>Is the given solution wrong because of what I just said? Or did I go wrong somewhere? If not then can this be put into some more concrete words or if there's a theorem related to this?</li>
<li>Can an alternate solution be proposed for option B?</li>
</ol>
<p>I'm under confident about this because this is a JEE Advanced 2018 question and no objections were made against this that I'm aware of.</p>
| Shinrin-Yoku | 789,929 | <p>You can derive the quotient rule by considering <span class="math-container">$\dfrac{f(x)}{g(x)}$</span> as <span class="math-container">$f(x)\cdot(g(x))^{-1}$</span> and then using the product and chain rule.<br><br>The quotient rule gives a <em>formula</em> (under the right conditions) for evaluating the derivative of a quotient without using the product and chain rule <em>each time.</em></p>
|
4,146,858 | <blockquote>
<p>Q) For every twice differentiable function <span class="math-container">$f:\mathbb{R}\longrightarrow [-2,2] $</span> with <span class="math-container">$[f(0)]^2+[f'(0)]^2=85$</span> , which of the following statement(s) is(are) TRUE?</p>
</blockquote>
<blockquote>
<p>(A) There exists <span class="math-container">$r,s \in\mathbb{R}$</span> , where <span class="math-container">$r<s$</span> , such that f is one-one on the open interval <span class="math-container">$(r,s)$</span></p>
</blockquote>
<blockquote>
<p>(B) There exists <span class="math-container">$x_0 \in (-4,0)$</span> such that <span class="math-container">$|f'(x_0)|\leq 1$</span></p>
</blockquote>
<blockquote>
<p>(C) <span class="math-container">$lim_{x\to \infty}f(x)=1$</span></p>
</blockquote>
<blockquote>
<p>(D) There exists <span class="math-container">$\alpha \in (-4,4)$</span> such that <span class="math-container">$f(\alpha)+f"(\alpha)=0 and f'(\alpha)=0$</span></p>
</blockquote>
<p>I have problem with option B. Here's the given solution-
<a href="https://i.stack.imgur.com/DV1ND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DV1ND.jpg" alt="enter image description here" /></a></p>
<p>But I think this is wrong because <span class="math-container">$f(0)$</span> cannot be equal to <span class="math-container">$2$</span> as that gives us the value of <span class="math-container">$f'(0)=9$</span> (from the condition given in the question). If the slope is positive at <span class="math-container">$x=0$</span> and the function is achieving its highest value there then for the points in the right neighbourhood of <span class="math-container">$x=0$</span> the slope would have to abruptly change to <span class="math-container">$0$</span> otherwise the function would obtain values greater than <span class="math-container">$2$</span> which are not in its co-domain. But it cannot abruptly change either coz it's given to be a twice differentiable function.</p>
<p>And I took <span class="math-container">$f(x)=9$</span> only but of course similar argument can be made with the negative value and a similar argument can be made if they took <span class="math-container">$f(0)=-2$</span> and <span class="math-container">$f(-4)=2$</span> in the solution.</p>
<p>My question-</p>
<ol>
<li>Is the given solution wrong because of what I just said? Or did I go wrong somewhere? If not then can this be put into some more concrete words or if there's a theorem related to this?</li>
<li>Can an alternate solution be proposed for option B?</li>
</ol>
<p>I'm under confident about this because this is a JEE Advanced 2018 question and no objections were made against this that I'm aware of.</p>
| marty cohen | 13,079 | <p>I like using
<span class="math-container">$(\ln(f))'
=\dfrac{f'}{f}
$</span>
for general
products and quotients.</p>
<p>If
<span class="math-container">$f = \dfrac{\prod u_k}{\prod v_k}
$</span>
then
<span class="math-container">$\ln(f)
= \sum \ln(u_k)-\sum \ln(v_k)
$</span>
so
<span class="math-container">$(\ln(f))'
= \sum \dfrac{u_k'}{u_k}-\sum \dfrac{v_k'}{v_k}
=\dfrac{f'}{f}
$</span>
so
<span class="math-container">$f'
= f\left(\sum \dfrac{u_k'}{u_k}-\sum \dfrac{v_k'}{v_k}\right)
$</span>.</p>
<p>From this
all the product and quotient rules
are special cases.</p>
<p>For example,
if <span class="math-container">$f = \dfrac{u}{v}
$</span>
then</p>
<p><span class="math-container">$\begin{array}\\
f'
&=f(\dfrac{u'}{u}-\dfrac{v'}{v})\\
&=\dfrac{u}{v}(\dfrac{u'v-v'u}{uv})\\
&=\dfrac{u'v-v'u}{v^2}\\
\end{array}
$</span></p>
<p>Since this is formal,
I don't worry about the sign
of the <span class="math-container">$u_k$</span> and <span class="math-container">$v_k$</span>.</p>
|
2,984,918 | <p>How can I prove this? </p>
<blockquote>
<p>Prove that for any two positive integers <span class="math-container">$a,b$</span> there are two positive integers <span class="math-container">$x,y$</span> satisfying the following equation:
<span class="math-container">$$\binom{x+y}{2}=ax+by$$</span></p>
</blockquote>
<p>My idea was that <span class="math-container">$\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$</span> and choose <span class="math-container">$x,y$</span>, such that <span class="math-container">$2a=x+2y-1, 2b=y(y-1)$</span>, but using this idea, <span class="math-container">$x,y$</span> won’t be always positive. </p>
| Wolfram | 397,123 | <p>Case <span class="math-container">$a=b$</span> is trivial after dividing both sides by <span class="math-container">$x+y$</span>, <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are symmetric in the problem, so in the following text WLOG I assume <span class="math-container">$a < b$</span>.</p>
<p>One can visualize the problem as follows. Denote <span class="math-container">$z=x+y$</span>, then the left-hand side is <span class="math-container">$\frac{z(z-1)}{2}$</span>. Let's subdivide the interval <span class="math-container">$(az, bz)$</span> into <span class="math-container">$z$</span> equal parts: then the right-hand side can be any point of subdivision (not including the ends) if you choose appropriate <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. So the problem is equivalent to finding <span class="math-container">$z$</span> such that <span class="math-container">$\frac{z(z-1)}{2}$</span> is a point of the subdivision of interval <span class="math-container">$(az, bz)$</span> into <span class="math-container">$z$</span> equal parts. If we multiply everything (<span class="math-container">$\frac{z(z-1)}{2}$</span>, interval ends and points of the subdivision) in this condition by <span class="math-container">$2/z$</span>, it becomes:</p>
<p><strong>Condition.</strong> <span class="math-container">$(z-1)$</span> is a point of the subdivision of interval <span class="math-container">$(2a, 2b)$</span> into <span class="math-container">$z$</span> equal parts.</p>
<p>Now the simplest try of finding such <span class="math-container">$z$</span> seems to be: make all integer points between <span class="math-container">$2a$</span> and <span class="math-container">$2b$</span> the points of subdivision into <span class="math-container">$z$</span> equal parts, while <span class="math-container">$2a < z-1 < 2b$</span>, making <span class="math-container">$z-1$</span> a point of such subdivision, because it is an integer. All integer points on <span class="math-container">$(2a, 2b)$</span> are in subdivision into <span class="math-container">$z$</span> equal parts iff <span class="math-container">$z$</span> is a multiple of <span class="math-container">$d=2b-2a$</span>. Let's consider as possible values <span class="math-container">$(z-1)=-1,(-1+d),(-1+2d),(-1+3d),\cdots$</span> This is a progression with step <span class="math-container">$d$</span> which will definitely hit the segment <span class="math-container">$[2a, 2b]$</span> of length <span class="math-container">$d$</span>, and because all its members are odd, it will actually hit the interval <span class="math-container">$(2a, 2b)$</span>. The <span class="math-container">$z$</span> for which this happens satisfies both <span class="math-container">$2a < z-1 < 2b$</span> and <span class="math-container">$d \mid z$</span> conditions, as required.</p>
|
90,812 | <p>How do I use the Edmonds-Karp algorithm to calculate the maximum flow? I don't understand this algorithm $100\%$. What I need to know is about flow with minus arrow. Here is my graph: </p>
<p><img src="https://i.stack.imgur.com/nep5F.jpg" alt="the graph">. </p>
<p>Our $1-6-11-12$, the flow is $4$. On the next iteration $1-2-4-11-6-7-9-12$, the flow on $6-11$ decrease on $3$, on other $+3$
on how do next? $1-3-5-11-6-8-10-12$? What will be with $11-6$? We must take $-3$, we will get negative flow on $6-11$ or what? Help me please.</p>
| Gerry Myerson | 8,269 | <p>You are correct that the next step is to use the path $P$ given by 1-3-5-11-6-8-10-12, but since the flow from 6 to 11 is only 1, you can only send 1 unit of flow along $P$ (which is the solution rattle found). The flow in each edge must be non-negative at all times. </p>
|
4,498,296 | <p>Is there any subtle way to compute the following integral?</p>
<p><span class="math-container">$$\int \frac{\sqrt{u^2+1}}{u^2-1}~ \mathrm{d}u$$</span></p>
<p>The solution i had in mind was substituting <span class="math-container">$u=\tan (\theta)$</span>,then after a few calculations the integral became <span class="math-container">$$\int \sec (\theta)
~ \mathrm{d}\theta+2\int \frac{\sec (\theta)}{\sec^2 (\theta) -2} ~\mathrm{d}\theta$$</span> I think we can formulate the last integral as <span class="math-container">$$\frac{1}{2}\int \left(\frac{1}{\sec (\theta) -\sqrt{2}}+\frac{1}{\sec (\theta)+\sqrt{2}} \right)\mathrm{d}\theta$$</span> But it still seems to be a daunting task and i think it will require further substitutions.</p>
<p>So could anyone please provide a <em>out of the blue</em> kind of solution or a clever approach to this?Or is it possible to go along my approach shortening the calculations?</p>
| Accelerator | 1,065,595 | <p>Here are a couple of ideas.</p>
<p>By using <span class="math-container">$\cos^{2}\left(\theta\right)\ =\ 1-\sin^{2}\left(\theta\right)$</span> and <span class="math-container">$\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$</span>, we can rewrite one of your integrals as</p>
<p><span class="math-container">$$\int\frac{\sec\left(\theta\right)}{\sec^{2}\left(\theta\right)-2}d\theta = \int\frac{\cos\left(\theta\right)}{2\sin^{2}\left(\theta\right)-1}d\theta.$$</span></p>
<p>Let <span class="math-container">$v = \sin{(\theta)}$</span> and do some partial fractions. (I think you can take it from there.)</p>
<p>For the original integral, you can let <span class="math-container">$u = \sinh{(a)}$</span>, do some algebra, and then make another substitution like <span class="math-container">$v = \tanh{(a)}$</span>, but you'd still have to do some partial fractions.</p>
<p>I'm not sure if you consider these substitutions out-of-the-blue or subtle like what you're interested in, but these are some ideas I had in mind and wanted to share.</p>
|
3,861,319 | <p>I have 3 features <code>age</code>, <code>income</code> and <code>rating</code>.</p>
<p>In case of age I have 3 buckets.</p>
<p>for income I have 4 buckets.</p>
<p>and for rating I have 2 buckets.</p>
<p>If one could filter data where a person could select 1 or more than 1 bucket from each feature what would be the total number of combinations that can be selected?</p>
<p>E.g</p>
<pre><code>Age = [less than 18, 18 < age <= 30, greater than 30]
Rating = [Low, High]
Combination_1 = [less than 18, Low]
Combination_2 = [18 < age <= 30, Low]
Combination_3 = [greater than 30, Low]
Combination_4 = [less than 18, High]
Combination_5 = [18 < age <= 30, High]
Combination_6 = [greater than 30, High]
Combination_7 = [less than 18, None]
Combination_8 = [18 < age <= 30, None]
Combination_9 = [greater than 30, None]
Combination_10 = [None, Low]
Combination_11 = [None, High]
Combination_12 = [None,None]
</code></pre>
<p>In this example there will be 12 combinations. What would be a generalized formula to achieve this?
Here <code>None</code> represents nothing is filtered from the feature.</p>
<p>Edit:
<code>None</code> here is not a feature but just a representation to tell it can be possible that nothing is selected.</p>
<pre><code>Combination_1 = [less than 18, Low]
Combination_2 = [18 < age <= 30, Low]
Combination_3 = [greater than 30, Low]
Combination_4 = [less than 18, High]
Combination_5 = [18 < age <= 30, High]
Combination_6 = [greater than 30, High]
Combination_7 = [less than 18, None]
Combination_8 = [18 < age <= 30, None]
Combination_9 = [greater than 30, None]
Combination_10 = [None, Low]
Combination_11 = [None, High]
Combination_12 = [None,None]
</code></pre>
<p>As one could select more than 1 bucket from each feature following combinations are also possible.</p>
<pre><code>Combination_13 = [[less than 18,18 < age <= 30], Low]
Combination_14 = [[less than 18,18 < age <= 30], high]
Combination_15 = [[less than 18,18 < age <= 30], None]
Combination_16 = [[less than 18,18 < age <= 30, greater than 30], Low]
Combination_17 = [[less than 18,18 < age <= 30, greater than 30], high]
Combination_18 = [[less than 18,18 < age <= 30, greater than 30], None]
Combination_19 = [less than 18, [High,Low]]
Combination_20 = [18 < age <= 30, [High,Low]]
Combination_21 = [greater than 30, [High,Low]]
Combination_22 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_23 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_24 = [[less than 18,18 < age <= 30], [High,Low]]
Combination_25 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
Combination_26 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
Combination_27 = [[less than 18,18 < age <= 30, greater than 30], [High,Low]]
</code></pre>
| Parcly Taxel | 357,390 | <p>The outermost quantifier is <span class="math-container">$\exists$</span>, so you're supposed to come up with one <span class="math-container">$x$</span> that satisfies <span class="math-container">$\forall y(2\mid5x+4y)$</span>. We can take <span class="math-container">$x=0$</span>. I will let you continue from here.</p>
|
1,566,215 | <p>Can someone explain to me the difference between joint probability distribution and conditional probability distribution?</p>
| BLAZE | 144,533 | <p>I will answer this question with a typical example as it might be easier to understand the difference, but first for some definitions:</p>
<p><code>Definition 1:</code></p>
<p>Given two independent events $A$ and $B$ $$P(A \space\mathrm{and}\space B)=P(A)\times P(B)=\space$$The probability of event $A$ <em>and</em> event $B$ happening simultaneously. This is known as $\color{#180}{\fbox{Joint Probability}}$</p>
<hr>
<p><code>Definition 2:</code></p>
<p>$$P(A\mid B)=\cfrac{P(A \space\mathrm{and}\space B)}{P(B)}=\space$$The probability that event $A$ occurs <strong>given</strong> that event $B$ has occurred. This is $\color{#F80}{\fbox{Conditional Probability}}$.</p>
<blockquote>
<p>The following table gives data on the type of car, grouped by petrol
consumption, owned by $100$ people. </p>
<p>$$\begin{array}{|c|l:l|}\hline & \mathsf{Low} & \,\mathsf{Medium} &
\,\mathsf{High}& \,\mathbf{Total} \\ \hline \mathsf{Male} & 12 & 33 &
7 \\ \hdashline \mathsf{Female} & 23 & 21 & 4 \\ \hdashline & &
& & 100 \\ \hline \end{array}$$</p>
<p>One person is selected at random.</p>
<p>Let $L$ be the event "the person owns a low rated car"</p>
<p>Let $F$ be the event "a female is chosen".</p>
</blockquote>
<p><code>Questions:</code></p>
<p>Find </p>
<p>i). $P(L)=\space$The probability of selecting someone with a low rated car,</p>
<p>ii). $P(F \space\mathrm{and}\space L)=\space$The probability of selecting a female with a low rated car.</p>
<p>iii). $P(F\mid L)=\cfrac{P(F \space\mathrm{and}\space L)}{P(L)}=\space$The probability of selecting a female given the car is low rated. </p>
<p><code>Solutions:</code></p>
<p>i).</p>
<p>$$\begin{array}{|c|l:l|}\hline
& \mathsf{Low} & \,\mathsf{Medium} & \,\mathsf{High}& \,\mathbf{Total} \\
\hline
\mathsf{Male} & \color{blue}{12} & 33 & 7 \\ \hdashline
\mathsf{Female} & \color{blue}{23} & 21 & 4
\\ \hdashline
& \color{red}{35}& & & \color{red}{100} \\ \hline
\end{array}$$
$$P(L)=\frac{12+23}{100}=\frac{35}{100}=\frac{7}{20}$$</p>
<p>ii). $$\begin{array}{|c|l:l|}\hline
& \mathsf{Low} & \,\mathsf{Medium} & \,\mathsf{High}& \,\mathbf{Total} \\
\hline
\mathsf{Male} & 12 & 33 & 7 \\ \hdashline
\mathsf{Female} & \color{red}{23} & 21 & 4
\\ \hdashline
& & & & \color{red}{100} \\ \hline
\end{array}$$
$$P(F \space\mathrm{and}\space L) = \cfrac{23}{100}$$ This is the joint probability and is the probability that two events happen simultaneously, note that this is sometimes written $P(F \cap L)$ which is known as the intersection of events $F$ and $L$.</p>
<p>iii). We need to be careful with the denominators in ii). and iii). Here we are given that the car is low rated; so we want the total of that column: $$\begin{array}{|c|l:l|}\hline
& \mathsf{Low} & \,\mathsf{Medium} & \,\mathsf{High}& \,\mathbf{Total} \\
\hline
\mathsf{Male} & \color{blue}{12} & 33 & 7 \\ \hdashline
\mathsf{Female} & \color{blue}{23} & 21 & 4
\\ \hdashline
& \color{red}{35} & & & 100 \\ \hline
\end{array}$$</p>
<p>$P(F\mid L)=\cfrac{23}{35}$ </p>
<p>It's helpful to notice that from the definition of conditional probability: $P(L)\times P(F\mid L)=\cfrac{7}{20}\times\cfrac{23}{35}=\cfrac{23}{100}=P(F \space\mathrm{and}\space L)$ as calculated in part ii).</p>
<hr>
<p>Just one small note: When I put "events $A$ and $B$ happen simultaneously" the word 'simultaneously' is not essentially describing time; it is just a word we use to state that both events happen at once.</p>
|
1,179,195 | <p>Good day everyone. </p>
<p>I need to know automata theory. Can you advice me the best way to study math?
What themes will I need to know to understand automata theory. What a sequence of study? What level will I need to study intermediate themes? Maybe can you say something yet, what can help me quickly learn automata theory?</p>
| tomasz | 30,222 | <p>A contraction mapping can never have more than one fixed point: if $a,b$ are both fixed points, then $d(a,b)=d(f(a),f(b))\leq \lambda d(a,b)$. This is only possible if $d(a,b)=0$.</p>
<p>On the other hand, by Banach fixed-point theorem, any contraction mapping of a complete metric space into itself has a fixed point. The proof is fairly straightforward: for any $x$, successive applications of the contraction to $x$ yield a Cauchy sequence, which converges by completeness, and by continuity the limit must be a fixed point.</p>
<p>The hypothesis of completeness is necessary, for example consider the map $x\mapsto x/2$ on $(0,1)$ with standard metric. It is a contraction mapping without any fixed points. Of course, there is a fixed point in the completion, namely $0$.</p>
|
123,587 | <p>I always wonder how many inbuilt functions does Mathematica have (of course you can google for it) and how they are connected with each other! So I tried this (v10.1).</p>
<pre><code>SetDirectory[$InstallationDirectory<>"/Documentation/English/System/ReferencePages/Symbols"]
comms = FileNames[];
ncomms = Length[comms]
</code></pre>
<blockquote>
<p>4613</p>
</blockquote>
<p>I believe this is the total number of documented functions. Now the second part where I am stuck in. <strong>How to find the connections among them?</strong></p>
<p>What I am thinking is to get the list of functions from the <code>See Also</code> section of an example. Here I do one manually. Let's say I start with <code>Plot</code>. <code>See Also</code> in <code>Plot.nb</code> contains ({DiscretePlot, ListLinePlot, ParametricPlot, PolarPlot, Plot3D, ContourPlot, Graphics, Show}). So I store them.</p>
<pre><code>link["Plot"] = {"DiscretePlot", "ListLinePlot", "ParametricPlot" ,
"PolarPlot", "Plot3D", "ContourPlot", "Graphics", "Show"};
</code></pre>
<p>Now I want to scan each element of the list. For example, I take <code>Plot3D</code> and <code>Graphics</code></p>
<pre><code>link["Plot3D"] = {"ListPlot3D", "ContourPlot", "DensityPlot", "ParametricPlot3D",
"Graphics3D", "ListSurfacePlot3D", "Plot", "Show"};
link["Graphics"] = {"Plot", "ListPlot", "ListLinePlot", "ParametricPlot", "DensityPlot",
"ArrayPlot", "RegionPlot", "ContourPlot", "Show", "Graphics3D",
"Image", "Import", "Sound"};
</code></pre>
<p>Now combine and map them</p>
<pre><code>map = Join[DirectedEdge[num["Plot"], num[#]] & /@ link["Plot"],
DirectedEdge[num["Plot3D"], num[#]] & /@ link["Plot3D"],
DirectedEdge[num["Graphics"], num[#]] & /@ link["Graphics"]];
Graph[map, VertexLabels -> Table[n -> sets[[n]], {n, nsets}], ImagePadding -> 10]
</code></pre>
<p><a href="https://i.stack.imgur.com/zgODy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zgODy.png" alt="enter image description here"></a></p>
<p><strong>How can I put this whole thing in a Mathematica code?</strong></p>
| Wjx | 6,084 | <p>There is a way for you to find out the relations without internet, and I think it fit your need more as they are actually extracted directly from those "See Also" links.</p>
<p>I'm currently using v11, 5000+ Functions, it takes about 7 min before it finishes the main evaluation, and another 10 min or so for plotting the community graph out, but it's totally internet free, thus can be a bit more helpful when under bad net conditions.</p>
<pre><code>dir = FileNameJoin[{$InstallationDirectory,
"Documentation\\English\\System\\ReferencePages\\Symbols"}];
files = Import[dir];
rules = Flatten[Function[{name}, Thread[StringSplit[name, "."][[1]] ->
Block[{f = Import[FileNameJoin[{dir, name}]], p},
p = Position[f, "See Also"];
If[p != {}, Cases[#, TextData[s_String]:>s] &@
Extract[f, Drop[p[[-1]], -5]], {}]]]] /@ files];
Graph@rules
CommunityGraphPlot@%
</code></pre>
<p>This method follows how we humans know about the links between functions-----via checking the "See Also" part of the documentation of course. So we can extract those part out one by one, make rules out of them, then use <code>Graph</code> and <code>CommunityGraphPlot</code> to visualize them.</p>
<p>The result is stylish(If we view in the Abstract Painting's style :P) and the structure is clear(Though a bit messy): Mathematica's core functions are closely related!!!</p>
<p><a href="https://i.stack.imgur.com/VxmGL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VxmGL.jpg" alt="graph"></a></p>
<p><a href="https://i.stack.imgur.com/wkw8v.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wkw8v.jpg" alt="community graph"></a></p>
|
1,349,654 | <p>Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$</p>
<p>Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .</p>
<p>Thank you for any kind of help </p>
| robjohn | 13,854 | <p>$$
\begin{align}
\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\int_{\pi/4}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\frac{\mathrm{d}\tan(x)}{\sec^2(x)}\tag{1}\\
&=\int_1^\infty\frac{\sqrt{u}}{\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}\tag{2}\\
&=\frac12\int_0^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\tag{3}\\
&=\frac14\int_0^\infty\frac{t^{-3/4}}{(1+t)^{1/2}}\,\mathrm{d}t\tag{4}\\[6pt]
&=\tfrac14\mathrm{B}\left(\tfrac14,\tfrac14\right)\tag{5}\\[6pt]
&=\frac{\frac14\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\tag{6}\\
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)}{\sqrt\pi}\tag{7}\\
\end{align}
$$
Explanation:<br>
$(1)$: $\mathrm{d}\tan(x)=\sec^2(x)\,\mathrm{d}x$<br>
$(2)$: $u=\tan(x)$<br>
$(3)$: substitution $u\mapsto\frac1u$ leaves the integral alone<br>
$(4)$: $u^2=t$<br>
$(5)$: definition for Beta function<br>
$(6)$: $\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$<br>
$(7)$: $x\Gamma(x)=\Gamma(x+1)$</p>
<p>Note that
$$
\begin{align}
\int_{\pi/3}^{\pi/2}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\int_{\sqrt3}^\infty\frac{\mathrm{d}u}{\sqrt{u(1+u^2)}}\\
&=2\int_0^{3^{-1/4}}\frac{\mathrm{d}v}{\sqrt{1+v^4}}\tag{8}
\end{align}
$$
Using the Binomial Theorem to get a series, integrating, and considering the ratios of the terms, we get
$$
\begin{align}
\int\frac1{\sqrt{1+v^4}}\,\mathrm{d}v
&=\sum_{k=0}^\infty\int\binom{2k}{k}\left(-\frac{v^4}4\right)^k\,\mathrm{d}v\\
&=\sum_{k=0}^\infty(-1)^k\frac{\binom{2k}{k}}{4^k}\frac{v^{4k+1}}{4k+1}\\
&=\vphantom{\mathrm{F}}_2\mathrm{F}_1\left(\frac14,\frac12;\frac54;-v^4\right)v\tag{9}
\end{align}
$$
At $v=3^{-1/4}$, the series above converges more than $0.477$ digits per term.</p>
<p>Thus, we get
$$
\begin{align}
\int_{\pi/4}^{\pi/3}\frac{\sqrt{\tan(x)}}{\sin(x)}\,\mathrm{d}x
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac54\right)}{\sqrt\pi}-\frac2{3^{1/4}}\,\vphantom{\mathrm{F}}_2\mathrm{F}_1\left(\frac14,\frac12;\frac54;-\frac13\right)\\[6pt]
&\doteq0.37913313649500221817\tag{10}
\end{align}
$$
This matches numerically the answer Mathematica gives, although the form is a bit different.</p>
|
2,065,639 | <p>$\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$</p>
<p>$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ </p>
<p>Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. One may use the fact that
$$
4uv=(u+v)^2-(u-v)^2 \tag1
$$ giving
$$
\begin{align}
4\int_a^b (x-a)(x-b)\:dx=4\int_a^b \left(x-\frac{a+b}2\right)^2\:dx-\int_a^b (b-a)^2\:dx \tag2
\end{align}
$$ which is <em>easier</em> to evaluate.</p>
<p>Then, one has $$
\begin{align}
&4\int_a^b (x-a)(x-b)\left(x-\frac{a+b}2\right)\:dx
\\\\&=\int_a^b \left[4\left(x-\frac{a+b}2\right)^2- (b-a)^2\right]\left(x-\frac{a+b}2\right)\:dx
\\\\&=4\int_a^b \left(x-\frac{a+b}2\right)^3\:dx- (b-a)^2\int_a^b\left(x-\frac{a+b}2\right)\:dx\tag3
\end{align}
$$ which is <em>easier</em> to evaluate.</p>
|
2,065,639 | <p>$\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$</p>
<p>$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ </p>
<p>Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?</p>
| Community | -1 | <p>If you're looking for elementary methods there's not going to be a faster one than expanding the integral or using integration by parts. For another solution you could remember the formula, and I think the following <strong>heuristic</strong> is a good mnemonic for that:</p>
<p>We know that $\int_a^b(x-a)(x-b)\,dx$ is a polynomial of degree $3$ in "the variables $a$ and $b$", so if we fix $b$ we might think of $\int_a^b(x-a)(x-b)\,dx$ as a polynomial $p(a)$ of degree $3$. To determine $p(a)$ notice that for $a$ close to $b$ we have
$$
\int_a^b(x-a)(x-b)\,dx\approx (b-a)(b-a)(a-b).
$$
This is a terrible estimate, but it tells us that $p(a)$ must have a zero of order $3$ in $b$, so $p(a)=C(b-a)^3$ for some $C$, which we can determine by calculating $p(a_0)$ for a suitable choice of $a_0$. For example
$$
Cb^3=p(0)=\int_0^bx(x-b)\,dx=\frac{b^3}{3}-\frac{b^3}{2}=-\frac{b^3}{6}.
$$</p>
<p>For the second integral the same procedure works. In that case this other terrible estimate
$$
\int_a^{\frac{a+b}{2}}(x-a)\left(x-\frac{a+b}{2}\right)(x-b)\,dx\approx\left(a-\frac{a+b}{2}\right)(b-a)\left(a-\frac{a+b}{2}\right)(a-b)
$$
tells us that $p(a)$ must have a zero of order $4$ in $b$, and calculating $p(0)$ (for example) again determines $p(a)$.</p>
|
185,867 | <p>I hear that the axiom of choice (AC) derives from
The generalized continuum hypothesis(GCH).
And also hear that both AC and GCH are independent of
Zermelo–Fraenkel set theory(ZF).</p>
<p>So, I'm just curious why don't expert mathematicians use ZF+GCH
instead of ZF+AC(ZFC).</p>
| GH from MO | 11,919 | <p>Warning: naive answer follows. I think AC is more intuitively true than GCH. Also, I believe, AC is more indispensable for mathematics than GCH. For example, there are many equivalent forms of AC within ZF that come up naturally, such as the Zorn lemma or the well-ordering principle.</p>
|
387,202 | <p>What is the smallest 3-regular graph to have a unique perfect matching?</p>
<p>With a large enough number of nodes, it is possible for a 3-regular graph to have no perfect matching (example can be seen in this question <a href="https://mathoverflow.net/questions/98385/cubic-graphs-without-a-perfect-matching-and-a-vertex-incident-to-three-bridges">Cubic graphs without a perfect matching and a vertex incident to three bridges</a> ). So I believe 3-regular graphs with a unique matching likely exists, but I am unsure how to go about constructing and proving what the smallest one is. Likely there is no better answer than to brute force check all the possibilities, so I am hoping someone happens to know what this graph looks like.</p>
<p>Even better: Does anyone know of an online searchable graph database that allows searching for small graphs with certain properties?</p>
| PickupSticks | 176,479 | <p>There is no such graph.</p>
<p>I have some reading to do as my intuition is off, but the details and a related question are available here:
<a href="https://mathoverflow.net/questions/349421/does-there-exist-an-r-regular-graph-r%E2%89%A52-with-a-unique-maximum-matching">Does there exist an r-regular graph (r≥2) with a unique maximum matching?</a></p>
<p>Akbari, Ghodrati, Hosseinzadeh (2017), <em>On the structure of graphs having a unique k-factor</em>, Aust. J. Combin. (<a href="https://ajc.maths.uq.edu.au/pdf/69/ajc_v69_p063.pdf" rel="noreferrer">pdf</a>) show:</p>
<blockquote>
<p>... we prove that there is no r-regular graph (r≥2) with a unique perfect matching.</p>
</blockquote>
|
947,626 | <p>What are the conditions under which the center of a group will have a cyclic subgroup? (with proof, of course)</p>
| Timbuc | 118,527 | <p>Always, certainly: if $\;Z(G)=1\;$ there's nothing to prove, and otherwise take any $\;z\in Z(G)\;$ and then $\;\langle z\rangle\;$ is a cyclic subgroup of the center.</p>
<p>In fact, we don't even need the first case $\;Z(G)=1\;$ .</p>
|
1,476,982 | <p>I'm trying to understand why the volume of a parallelepiped whos sides are $s,u,w$ is $ V = s \cdot(u \times w)$.</p>
<p>Even the units of measurement don't add up. The length of the vectors $s,u,w$ is measured in centimeters, the volume is measured in cubic cm.</p>
<p>$u\times w$ is a vector. It is a vector that is orthogonal to $u$ and $w$, but still a vector, so its length is again measure in cms. So overall $V=s \cdot(u \times w)$ means that $V$ is equal to the product of $2$ vectors, so the unit of measurement for $V$ is squared centimeters, not cubed.</p>
<p>I'm struggling to understand how can $|u\times w|$ be equal to the area of a parallelogram. That is equivalent to saying "The time it takes for me to solve a problem is the distance between New York and London."</p>
| Narasimham | 95,860 | <p>The cross product has dimensions of area $ |A|\cdot|B| \sin \theta $. Just because it is a vector, it need not be a straight line always. We say an <em>area</em> is directed quantity. An area or even a volume can be a vector. </p>
<p>How do we accept angles, angular velocities, angular accelerations as vectors? Forces as vectors? Current and voltage as vectors? Each has a physical magnitude and direction.</p>
<p>The triple product is scalar with no direction.</p>
<p>The dimensions of volume of a parallelopiped is 2+1 = 3.</p>
|
4,173 | <p>I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- </p>
<p>Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?</p>
<p>$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$</p>
<p>Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the <a href="http://en.wikipedia.org/wiki/Householder_transformation" rel="nofollow">Householder matrix</a>, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.</p>
<p>Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n</p>
<p>Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of <a href="http://yaroslavvb.com/upload/bigmatrix.png" rel="nofollow">this</a> form orthonormal basis for d=20</p>
<p>$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$</p>
<p>$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$</p>
<p>$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$</p>
<p>$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$</p>
<p>$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$</p>
<p><b>Update 09/08</b>
I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where</p>
<p>$$M_{ij}=(-1)^{x_i \cdot x_j}$$</p>
| ronaf | 1,547 | <p>a slight retouching of unkz's idea is the following orthogonal basis for $v^\perp$:</p>
<p>$$u_1 = (1,-1,0,\dots,0),\
u_2 = (1,1,-2,0,\dots,0),\
\dots, \
u_{d-1} = (1,\dots,1,-d+1)$$</p>
<p>in this form it is pretty easy to see that the $\{u_i\}$ are in $v^\perp$ and are mutually orthogonal.</p>
<p>i don't quite understand the remark that normalizing the $\{u_i\}$ to get an ONB is the hardest part - since $d$ is unknown. what sort of expression "in terms of [the] unknown $d$" <em>would</em> be suitable?</p>
<p>ps: i would post this as a comment - if this @!*&*%#! site would let me.</p>
|
2,000,013 | <p>On an NFA, how can the empty set ∅ and {ϵ} be considered regular languages? Does it make sense that a machine that accepts no symbols or a machine that takes the empty symbol exist? I could think of a machine (laptop) that is in off mode, where no entries (symbols) are accepted, but, is that of interest? or could it still be considered as a machine? the same model could apply for a rock for example.</p>
<p>thanks!</p>
| Brian M. Scott | 12,042 | <p>Let $M$ be a machine with any alphabet, one state, $q_0$, that is the initial state, and no transitions. If $q_0$ is not an acceptor state, $M$ accepts $\varnothing$, and if $q_0$ is an acceptor state, $M$ accepts $\Sigma^*$, where $\Sigma$ is the alphabet.</p>
<p>Now let $M$ have any alphabet and two states, the initial state $q_0$ and one other state $q_1$; $q_0$ is an acceptor state, and $q_1$ is not. The transitions are simple: in state $q_0$ every input sends $M$ to $q_1$, and in $q_1$ every input also sends $M$ to $q_1$. This $M$ accepts the empty word and nothing else.</p>
|
108,594 | <p>I would like to know if one can weaken conditions of Proposition 2.8 in
<a href="http://www.jmilne.org/math/xnotes/CA.pdf" rel="nofollow">http://www.jmilne.org/math/xnotes/CA.pdf</a></p>
<p>The proposition says that if an ideal $a$ in a ring $A$ is contained in the union of ideals $p_1,...,p_r$ with $p_2,...,p_r$ prime, then $a$ is contained in one of $p_i$. </p>
<p>Why do we need to require that $p_2,...,p_r$ are all prime? What would be the simplest example where some of $p_i$ are non-prime and the proposition does not hold?</p>
<p>It seems to me at least that if $A$ is a polynomial ring $\mathbb C[x_1,...,x_n]$ then one does not need to require that $p_i$ are prime. Am I wrong?</p>
| Mohan | 9,502 | <p>For an infinite base field $k$, you are right. Assume $a\subset\cup_{i=1}^r a_i$ and choose $r$ the least such. Then we can pick $f_i\in a$ not in the union of the the $a_j$'s with $j\neq i$. Consider the $k$-vector space $V$ generated by the $f_i$'s and let $V_i=a_i\cap V$. Then $V\subset \cup V_i$, but $V_i$s are proper subspaces of $V$, a contradiction, since $k$ is infinite.</p>
|
4,380,124 | <p>I understand that the double integral is
<a href="https://i.stack.imgur.com/29e8B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/29e8B.png" alt="enter image description here" /></a>
However what confuses me is when I try to visualize why this formula only accounts for the region inside the bounds and not the whole rectangular region.
<a href="https://i.stack.imgur.com/Zs2t1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zs2t1.jpg" alt="enter image description here" /></a></p>
<p>My guess is it has to do with one of the bounds being (x) but I have a hard time visualizing that. Could someone maybe give me the 2 variable analog to this equation of the single integral, and show how it only looks at points within the specified region
<a href="https://i.stack.imgur.com/KrnII.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KrnII.png" alt="enter image description here" /></a></p>
| CatsOnAir | 974,003 | <p><a href="https://i.stack.imgur.com/PkLBr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkLBr.png" alt="enter image description here" /></a>
The equation works for non rectangular regions because dy or dx can be a function of x and y itself meaning the height and width of your rectangles dydx changes based on were you evaluate your point, thus only accounting for the area inside the the region as n goes to infinty.</p>
|
4,380,124 | <p>I understand that the double integral is
<a href="https://i.stack.imgur.com/29e8B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/29e8B.png" alt="enter image description here" /></a>
However what confuses me is when I try to visualize why this formula only accounts for the region inside the bounds and not the whole rectangular region.
<a href="https://i.stack.imgur.com/Zs2t1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zs2t1.jpg" alt="enter image description here" /></a></p>
<p>My guess is it has to do with one of the bounds being (x) but I have a hard time visualizing that. Could someone maybe give me the 2 variable analog to this equation of the single integral, and show how it only looks at points within the specified region
<a href="https://i.stack.imgur.com/KrnII.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KrnII.png" alt="enter image description here" /></a></p>
| Andrew D. Hwang | 86,418 | <p>To illustrate Hans' (+1) answer:</p>
<p><a href="https://i.stack.imgur.com/E8J4X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E8J4X.png" alt="Riemann sums for a non-rectangular plane region" /></a></p>
<p>Extending the integrand by <span class="math-container">$0$</span> outside the triangle of integration gives us a function whose Riemann sums we can compute. The darker gray squares represent the contributions where the integrand is given by a piecewise formula (i.e., not by <span class="math-container">$f(x, y)$</span> or by <span class="math-container">$0$</span>). The total contribution from these squares is negligible in the limit as the mesh of the partition goes to <span class="math-container">$0$</span>.</p>
<p>As for the variable limits of integration, the terms of the Riemann sum where <span class="math-container">$x < y$</span> (i.e., the lighter-shaded squares) are zero. We can think of including those terms in the Riemann sum (because <span class="math-container">$f(x_{i}, y_{j}) = 0$</span>), or we can omit those terms by summing only over the unshaded and darker-shaded squares. Either way, in the limit as the mesh decreases we obtain the stated iterated integral. For each <span class="math-container">$x$</span>, the integral over <span class="math-container">$0 \leq y \leq x$</span> is approximated by the Riemann sum over the blue slice.</p>
|
1,528,501 | <p>Let $P(n)$ be a property for all $n \geq 1$. For the phrase "there is some $N \geq 1$ such that $P(n)$ holds for all $n \geq N$" there are some suggestive, convenient abbreviations such as "$P(n)$ holds for large $n$" or "$P(n)$ holds eventually" and so on.</p>
<p>I wonder if there is in literature a like abbreviation for "$P(n)$ holds for infinitely many $n$"? I am aware that in probability theory some authors would write "$P(n)$ holds infinitely often"; but, in my humble opinion, this abbreviation would be not that useful in contexts other than probability theory.</p>
| BrianO | 277,043 | <p>In some contexts (set theory, order theory, point set topology, though probably never in probability) you can say <em>cofinally</em>, or <em>cofinally many</em>, <em>cofinally often</em>. Given a preorder $(A,\preceq)$, a subset $X\subseteq A$ is <em>cofinal</em> in $A \Leftrightarrow$ for every $a\in A$ there is $x\in X$ with $a\preceq x$. A predicate $\varphi(x)$ holds <em>cofinally often</em>, and is true for <em>cofinally many</em> $x$, iff $\{x\in A\mid \varphi(x)\}$ is cofinal in $A$. For $\Bbb N$ with the usual order, this is exactly the same notion as "infinitely often".</p>
<p>"Eventually" is often used in such a setting (vis a vis a preorder) to mean: $eventually_x, \varphi(x) \!\stackrel{def}\iff\!$ there is $a\in A$ such that for all $x\succeq a, \varphi(x)$. In that case, "frequently" can be and has been used rather than "cofinally often".</p>
<p>Note that these two notions are quantifiers, duals of each other:
$$
\text{frequently$_x \varphi(x) \iff \neg$ eventually$_x \neg\, \varphi(x)$. }
$$
In his book <a href="https://ia600501.us.archive.org/16/items/GeneralTopology/Kelley-GeneralTopology.pdf" rel="nofollow">General Topology</a> [p.65], Kelley uses the terms "eventually" and "frequently", for special cases of these notions: given a directed preorder $(A,\preceq)$, a function $f\colon A\to B$ is <em>frequently</em> in a subset $Y\subseteq B$ <em>iff</em> $\{x\in A\mid f(x)\in Y\}$ is cofinal in $A$. Similarly for the dual notion "$f$ is <em>eventually</em> in <em>Y</em>". A net $f$ <em>converges</em> to a point $b\in B \!\iff\!$ for every neighborhood $U$ of $b$, $f$ is eventually in $U$.</p>
|
2,679,153 | <p>Let $A\in\mathbb{R}^{n\times n}$ be a generic <em>lower triangular</em> matrix and let $P\in\mathbb{R}^{n\times n}$ be a symmetric <em>positive definite</em> matrix.</p>
<blockquote>
<p><strong>True or false.</strong> Does $AP + PA^\top=0$ imply $AP=0$? </p>
</blockquote>
| Doug M | 317,162 | <p>The ratio of sides is 1:1.6</p>
<p>or lengths $x$ and $1.6x$ </p>
<p>Area $= \frac 12 bh\\ 0.8 x^2 = 51.2\\x^2=64\\x = 8$</p>
<p>Two of the sides then are $8$ and $12.8$</p>
<p>The 3rd side you can find using the Pythagorean theorem.</p>
|
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