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 |
|---|---|---|---|---|
2,328,677 | <p>I wish to find the area between the curves:</p>
<p>$y=\sqrt{x}$</p>
<p>and </p>
<p>$y=x^{2}-3x-2$</p>
<p>and between $x=1, x=9$</p>
<p>Now, I think that the first thing to do is to find the intersection between the two curves, and then to find which curve is the upper one before and after the intersection point. At the end, an integral will follow.</p>
<p>The problem is, how do you find the intersection point? If you compare the curves, you get an equation which is find to solve. Is there a trick I am missing? How do you solve it when you have a square root and a power of 2?</p>
<p>Thank you !</p>
| Michael Rozenberg | 190,319 | <p>We need to solve
$$\sqrt{x}=x^2-3x-2.$$
We have $x\geq0$ and $x^2-3x-2\geq0$, which gives $x\geq\frac{3+\sqrt{17}}{2}$.</p>
<p>Thus, squaring gives
$$(x-4)(x^3-2x^2-3x-1)=0$$
and $x=4$ because for $x\geq\frac{3+\sqrt{17}}{2}$ we see that $$x^3-2x^2-3x-1=x(x^2-3x-2)+x^2-x-1=$$
$$=x(x^2-3x-2)+x^2-3x-2+2x+1>0.$$</p>
<p>Now, we need to calculate two integrals:
$$\int_{1}^4(\sqrt{x}-x^2+3x+2)dx+\int_{4}^9(x^2-3x-2-\sqrt{x})dx.$$</p>
|
3,427,640 | <p>This is a problem that came up as I was learning Hermitian/skew-Hermitian transformations:</p>
<hr>
<p>Let <span class="math-container">$T: V\rightarrow E$</span> be a linear transformation where <span class="math-container">$V$</span> is a subspace of a complex Euclidean space <span class="math-container">$E$</span> and define a scalar-valued function <span class="math-container">$Q$</span> on <span class="math-container">$V$</span> such that <span class="math-container">$\forall x \in V$</span>:</p>
<p><span class="math-container">$$Q(x) = (T(x),x)$$</span></p>
<p>where <span class="math-container">$()$</span> denotes inner product. </p>
<p>Show that if <span class="math-container">$Q(x)$</span> is real for all <span class="math-container">$x$</span>, then <span class="math-container">$T$</span> is Hermitian i.e. <span class="math-container">$(T(x),y) = (x,T(y))$</span></p>
<hr>
<p><strong>My work:</strong></p>
<p>Since <span class="math-container">$Q(x)$</span> is real, we know that <span class="math-container">$\forall x \in V$</span>, <span class="math-container">$Q(x) = \overline{Q(x)}$</span></p>
<p><span class="math-container">$Q(x+ty) = (T(x)+tT(y), x+ty) = Q(x) + t(T(y),x) + \bar{t}(T(x),y) + t\bar{t}Q(y)$</span></p>
<p><span class="math-container">$\overline{Q(x+ty)} = (x+ty, T(x)+tT(y)) = (x,T(x)) + t(y,T(x)) + \bar{t}t\bar{t}(x,T(y))+(y,T(y)) = Q(x) + t(y,T(x)) + \bar{t}(x,T(y)) + t\bar{t}Q(y)$</span></p>
<p>Putting the two equations together, I get: <span class="math-container">$$t(T(y),x) + \bar{t}(T(x),y) = t(y,T(x)) + \bar{t}(x,T(y))$$</span></p>
<p>I feel like this is close, but I'm not sure how to continue. Does anyone have any ideas?</p>
| jazhang | 701,790 | <p>I wasn't fully convinced by comparing coefficients. Here's another way:</p>
<p>Substituting <span class="math-container">$i$</span> and <span class="math-container">$1$</span> for <span class="math-container">$t$</span> in the equation gets:</p>
<p><span class="math-container">$(T(y),x)+(T(x),y) = (y,T(x))+(x,T(y))$</span> and</p>
<p><span class="math-container">$i(T(y),x)+(-i)(T(x),y) = i(y,T(x))+(-i)(x,T(y))$</span> or </p>
<p><span class="math-container">$(T(y),x)+(-1)(T(x),y) = (y,T(x))+(-1)(x,T(y))$</span></p>
<p>Adding, you get:</p>
<p><span class="math-container">$(T(y),x) = (x,T(y))$</span></p>
|
220,170 | <p>My question is rather philosophical : without using advanced tools as Perlman-Thurston's geometrisation, how can we get convinced that the class of closed oriented $3$-manifolds is large and that simple invariants as Betti number are not even close to classify ?</p>
<p>For example i would start with :</p>
<ol>
<li><p>If $S_g$ is the closed oriented surface of genus $g$, the family $S_g \times S^1$ gives an infinite number of non pairwise homeomorphic $3$-manifolds.</p></li>
<li><p>Mapping tori of fiber $S_g$ gives as much as non-diffeomorphic $3$-manifolds as conjugacy classes in the mapping class group of $S_g$ which can be shown to be large using the symplectic representation for instance.</p></li>
</ol>
<p>I think that I would like also say that Heegaard splittings give rise to a lot of different $3$-manifolds which are essentially different, but I don't know any way to do this. </p>
<p>So if you know a nice construction which would help understanding the combinatorial complexity of three manifolds, please share it :) </p>
| Louis Deaett | 2,502 | <p>I'm not sure if this is what you have in mind by "without using advanced tools" or "a nice construction which would help understanding the combinatorial complexity of three manifolds," but how about identifying pairs of faces of different polyhedra? I guess this is what convinced Poincaré...</p>
|
2,015,717 | <p>I know that the O notation tells me to find $n_0$ natural and $c>0$ real, so the
$$2^{(n^2)} \le c*2^{2n} $$
Only step that I can think about is this:
$$2^{(n^2)} \le c*2^{n+n} $$
$$2^{(n^2)} \le c*2^n*2^n $$
But I have no clue what to do next. (Some estimate?)</p>
<p>Do you have some hints please?</p>
<p>EDIT: sorry it should be $2^{(n^2)}$</p>
| mfl | 148,513 | <p>Your claim is false because $$\dfrac{2^{n^2}}{2^{2n}}=2^{n^2-2n}\to \infty$$ as $n\to\infty.$</p>
|
674,620 | <p>What is the slope of the tangent line to the function $$g(x)=x^2 \frac{\cos x}{1+x^3}$$ when $x=\pi/2$?
When $x=a$?</p>
| Heisenberg | 127,460 | <p>The derivative is calculated using the quotient rule,</p>
<p>$$y=\frac{u}{v} \\ y'=\frac{u'v - v'u}{v^2}$$
$$g'(x)=\frac{(1+x^3)(2x\cos x-x^2\sin x)-(x^2\cos x)(3x^2)}{(1+x^3)^2}$$
The slope of the tangent line in calculus is the limit of the rise over run of the tangent at the tangent point - i.e. $dy/dx$ which is the derivative of the function. </p>
<p>The slopes at $x=\frac{\pi}{2}$ would be,</p>
<p>$$g'(\frac{\pi}{2})=\frac{-(\frac{\pi}{2})^2}{1+(\frac{\pi}{2})^3}$$</p>
|
2,559,623 | <p>The question is as follows:<p>
Calculate $\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx$.<p></p>
<p>$\textbf{Some ideas:}$<p>
We can use the fact that $\sin(\frac{x}{n}) \simeq \frac{x}{n} $.But then we find that <p>
$\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx \simeq \lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \times \frac{x}{n}}{1 + x^2} = \lim_{n \to +\infty} \int_{0}^{n} \frac{ x }{1 + x^2} dx $<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int_{0}^{n} \frac{ 2x }{1 + x^2} dx $<p>
$ \hspace{9.1cm} \text{take } x^2=y$<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int \frac{ dy }{1 + y} $<p>
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(y)}{2} $<p>
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(x^2)}{2} \mid_{0}^{n}$<p>
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(n^2)}{2} = +\infty$<p>
But someone said me that the final result should be $\frac{\pi}{2}$?<p>
Can you please let me know where is my mistake?<p>
Thanks!</p>
| Parcly Taxel | 357,390 | <p>There is a neat geometric approach that treats the portion of $C_1$ with $0<|z|\le2\sqrt3$ as one side of a regular hexagon whose centre is $2\sqrt3i$, also the centre of the circle $C_2$.
<img src="https://i.stack.imgur.com/TCBKE.jpg" alt="">
For the two curves to intersect in precisely two points, the exclusive upper bound for $r$ is attained when $C_2$ circumscribes the hexagon, i.e. has a radius of $2\sqrt3$. The exclusive lower bound is attained when $C_2$ is inscribed in the hexagon, i.e. has a radius of 3. Thus the admissible range of $r$ is $3<r<2\sqrt3$.</p>
|
2,743,827 | <p>This came up on page 40 of Hubbard's <em>Teichmuller Theory</em> (Vol 1), in the context of parametrizing geodesics in the upper half.</p>
<p>Where it is essentially implied that
$$\frac{\sinh(t)+i}{\cosh(t)}$$
is an arc-length parametrization of the unit circle.</p>
<p><strong>This is my first time seeing such a parametrization and I was wondering there was a (preferably visual or geometric) way to see this.</strong></p>
| user | 505,767 | <p>Note that</p>
<p>$$\frac{\sinh t+i}{\cosh t}=\frac{\sinh t}{\cosh t}+i\frac{1}{\cosh t}$$</p>
<p>and</p>
<p>$$\left(\frac{\sinh t}{\cosh t}\right)^2+\left(\frac{1}{\cosh t}\right)^2=\frac{\sinh^2t+1}{\cosh^2 t}=1$$</p>
|
2,743,827 | <p>This came up on page 40 of Hubbard's <em>Teichmuller Theory</em> (Vol 1), in the context of parametrizing geodesics in the upper half.</p>
<p>Where it is essentially implied that
$$\frac{\sinh(t)+i}{\cosh(t)}$$
is an arc-length parametrization of the unit circle.</p>
<p><strong>This is my first time seeing such a parametrization and I was wondering there was a (preferably visual or geometric) way to see this.</strong></p>
| Rob Arthan | 23,171 | <p>At least the first part of the following argument is geometric.</p>
<p>Here are two facts about the <a href="https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model" rel="nofollow noreferrer">Poincaré half-plane model</a> with $d$ denoting the metric:</p>
<ol>
<li>if $x$ and $y$ are positive reals, $d(ix, iy) = |\ln(x) - \ln(y)|$</li>
<li>The metric is invariant under linear fractional transformations with real coefficients.</li>
</ol>
<p>If you don't know these facts, see the Wikipedia link above for 1 (which is easy to prove once you know the integral you have to calculate) and see the Wikipedia page on the <a href="https://en.wikipedia.org/wiki/Poincar%C3%A9_metric#Metric_and_volume_element_on_the_Poincar%C3%A9_plane" rel="nofollow noreferrer">Poincaré metric tensor</a> for a proof of 2.</p>
<p>It follows from fact 1 that we can define an arc-length parametrization of the upper half of the imaginary axis by:
$$
u(t) = ie^t
$$
The linear fractional transformation:
$$
f(z) = \frac{z-1}{z+1}
$$
maps $0, i, \infty$ to $-1, i, 1$ in that order. Hence (using fact 2), $f$ is an isometry from the upper half of the imaginary axis to the upper half of the unit circle. It follows that:
$$
v(t) = f(u(t))
$$
gives an arc-length parametrization of the upper half of the unit circle. </p>
<p>The rest of the proof is algebra. For real $t$, we have:
$$
\begin{align}
v(t) &= \frac{ie^t - 1}{ie^t + 1}\\
&= \frac{(ie^t - 1)(-ie^t + 1)}{e^{2t}+1} \tag{A}\\
&= \frac{(ie^t - 1)(-i + e^{-t})}{e^t + e^{-t}} \tag{B} \\
&= \frac{e^t -e^{-t} +2i}{e^t + e^{-t}} \tag{C}\\
&= \frac{\sinh(t) + i}{\cosh(t)} \tag{D}
\end{align}
$$
where (A) follows by calculating the reciprocal of the denominator, (B) follows by multiplying top and bottom by $e^{-t}$, (C) follows by multiplying out and (D) follows from the definitions of $\sinh$ and $\cosh$.</p>
|
175,664 | <p>How can I have a region plot with the logarithmic axis? In the following, I have brought the original case which I want to be logarithmic. In this case, I have a density plot and a region plot that shows the realm of validity of the theory. I want to show this figure in logarithmic scales in order to be more clear.</p>
<pre><code>η4=0.123663
con14=-0.1 + 8 E^8 (-((5 μ)/(3 E^(40/3))) + (
8 π η (BesselI[0, 4] BesselK[0, 4] -
BesselI[1, 4] BesselK[1, 4]))/E^8)
con24=-12.4276 + μ
con34=-((5 μ)/(3 E^(40/3))) + (
8 π η (BesselI[0, 4] BesselK[0, 4] -
BesselI[1, 4] BesselK[1, 4]))/E^8
con44=(35 μ)/(36 E^(16/3)) +
4 π η ((2 BesselI[0, 4] + 4 BesselI[1, 4]) BesselK[0,
4] - (4 BesselI[0, 4] + BesselI[1, 4]) BesselK[1, 4])
pnrhs4=(1/(η Sqrt[
161.08 η +
35. μ]))(-22.1002 η^(3/2) Sqrt[67.0519 η - 5. μ]
Sqrt[μ] +
2.78971 μ^(
5/2) + η^2 (-255.832 Sqrt[μ] -
0.764016 Sqrt[
161.08 η + 35. μ]) + η (-240.094 μ^(3/2) +
1. Sqrt[161.08 η + 35. μ] +
0.00911946 μ Sqrt[161.08 η + 35. μ]))
CPN4=(1/(η Sqrt[
161.08 η +
35. μ]))(-22.1002 η^(3/2) Sqrt[67.0519 η - 5. μ]
Sqrt[μ] +
2.78971 μ^(
5/2) + η^2 (-255.832 Sqrt[μ] -
0.764016 Sqrt[
161.08 η + 35. μ]) + η (-240.094 μ^(3/2) +
0.00911946 μ Sqrt[161.08 η + 35. μ]))
QNN4=(60.1862 Sqrt[0.0216024 η + 0.00469384 μ] Sqrt[μ])/η
Area4 = RegionPlot[{con14 < 0 && con24 < 0 && con34 > 0 && con44 > 0 &&
Abs[CPN4] < 1 && QNN4 > 0}, {μ, 0, 0.028}, {η,
0, η4}, BoundaryStyle -> {Green}, PlotStyle -> {None},
FrameLabel -> Automatic]
P4 = Show[
DensityPlot[(QNN4/pnrhs4), {μ, 0, 0.028}, {η, 0, η4},
PlotRange -> {0, 75}, PlotPoints -> 300,
PlotRangePadding -> {Automatic, 0.00015},
FrameLabel -> {Style[μ, FontSize -> 14, Blue],
Style[η, FontSize -> 14, Blue]},
BaseStyle -> {FontWeight -> Bold, FontSize -> 17},
ColorFunction -> "SunsetColors",
PlotLegends ->
BarLegend[Automatic, LegendMarkerSize -> 230, LegendMargins -> 5,
LegendLabel -> Style["Q/(1+C)", FontSize -> 16],
LabelStyle -> {Bold, FontSize -> 14}],
FrameTicks -> {{{0, 0.04, 0.08, 0.12},
None}, {{0, 0.006, 0.012, 0.018, 0.024}, None}}], Area4]
</code></pre>
<p>Thank you for your help.</p>
| kglr | 125 | <p>You can post-process <code>RegionPlot</code> output to rescale the desired coordinate of graphics primitives:</p>
<pre><code>rp = RegionPlot[Sin[x] Sin[y] > 1/10, {x, Pi, 4 Pi}, {y, Pi, 4 Pi},
ColorFunction -> "DarkRainbow", ImageSize -> 300];
Row[{rp,
Show[rp /. GraphicsComplex[c_, prims___] :> GraphicsComplex[{#, Log @ #2}& @@@ c, prims],
FrameTicks -> {{Charting`ScaledTicks[{Log, Exp}],
Charting`ScaledFrameTicks[{Log, Exp}]}, {Automatic, Automatic}} ,
PlotRange -> All]}, Spacer[10]]
</code></pre>
<p><a href="https://i.stack.imgur.com/OVFFQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OVFFQ.png" alt="enter image description here"></a></p>
<p>Replace <code>{#, Log @ #2}& @@@ c</code> with <code>{Log @ #, #2} & @@@ c</code> to use <code>Log</code> scale on the horizontal axis, and with <code>{Log @ #, Log @ #2} & @@@ c</code> to have both axes in <code>Log</code> scale. With appropriate modifications of the <code>FrameTicks</code> settings we get</p>
<p><a href="https://i.stack.imgur.com/ttN03.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ttN03.png" alt="enter image description here"></a></p>
<p>Another example:</p>
<pre><code>rp = RegionPlot[2 < Abs[ ((x -3 + I (y/10-5))/10 - 2)/( 2(x-3 + I (y/10-5)) /10- 1)] <5,
{x, 3, 20}, {y, 1, 200},
ColorFunction -> "Rainbow", ImageSize -> 300, PlotRange -> All, PlotPoints -> 50]
</code></pre>
<p>we get</p>
<p><a href="https://i.stack.imgur.com/NTNd6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NTNd6.png" alt="enter image description here"></a></p>
<p><strong>Update:</strong> Adding the option <code>ScalingFunctions -> {"Log", "Log"}</code> to OP's <code>DensityPlot</code> and using post-processing idea on OP's <code>Area4</code>:</p>
<pre><code>dp2 = DensityPlot[(QNN4/pnrhs4), {μ, 0, 0.028}, {η, 0, η4},
PlotRange -> {0, 75}, PlotPoints -> 100,
PlotRangePadding -> {Automatic, 0.00015},
BaseStyle -> {FontWeight -> Bold, FontSize -> 17},
ColorFunction -> "SunsetColors",
FrameTicks -> {{{0, 0.04, 0.08, 0.12}, None}, {{0, 0.006, 0.012, 0.018, 0.024}, None}},
ScalingFunctions -> {"Log", "Log"}];
Show[dp2, Area4 /. GraphicsComplex[c_, prims___] :>
GraphicsComplex[{Log@#, Log@#2} & @@@ c, {Thick, prims}], ImageSize -> 600]
</code></pre>
<p><a href="https://i.stack.imgur.com/9Viyx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Viyx.png" alt="enter image description here"></a></p>
<p>To confirm that post-processing gives the correct picture, compare the above result to the one obtained using <code>ContourPlot</code> with <code>ScalingFunctions-> {"Log", "Log"}</code> with <code>Boole[con14 <= 0 && con24 <= 0 && con34 >= 0 && ...]</code> as the first argument:</p>
<pre><code>Area4b = ContourPlot[Boole[con14 <= 0 && con24 <= 0 && con34 >= 0 && con44 >= 0 &&
Abs[CPN4] <= 1 && QNN4 >= 0],
{μ, 0.00001, 0.028}, {η, 0.00001, η4}, ContourStyle -> Green,
Contours -> {1}, PlotPoints -> 50, ScalingFunctions -> {"Log", "Log"},
ContourShading -> {None, Opacity[.5, Yellow]}];
bdg = BoundaryDiscretizeGraphics[Area4b,
MeshCellStyle -> {2 -> None, 1 -> Directive[Thick, Green]}];
Show[dp2, bdg, ImageSize -> 400]
</code></pre>
<p><a href="https://i.stack.imgur.com/aYcN4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aYcN4.png" alt="enter image description here"></a></p>
<p> </p>
|
1,014,614 | <p>That Theorem 3.2 says: Every finite orthogonal set of nonzero vectors is linearly independent.</p>
<p>The proof is simple, but it seems to me that the finiteness is redundant, for the argument in the proof applies to an infinite set. Am I right?</p>
<p>The proof runs as follows: If $k > 0$ is an integer, if $a_{1}, \dots, a_{k}$ are reals, if $v_{1}, \dots, v_{k}$ are vectors,
and if
$$\sum_{1}^{k}a_{j}v_{j} = 0,$$
then, by taking inner product with some $v_{i}$
we have
$$a_{i}(v_{i} \cdot v_{i}) = 0,$$
so that
$$a_{i} = 0.$$
Since this argument holds for any $1 \leq i \leq k,$
qed.</p>
<p>To me, the proof above holds for any $k$.</p>
| Hagen von Eitzen | 39,174 | <p><em>Hint:</em> Pick any $\epsilon>0$. Then pick $\delta>0$ according to the definition of unifom continuity. Then what can you say by induction about $f(\pm k\frac\delta2)-f(0)$ with $k\in\mathbb N$? What about the values inbetween?</p>
|
1,576,561 | <p>I'd like to show that $$\sum\limits_{n = 1}^\infty {{{{x^{n + 1}}} \over {n(n + 1)}}} $$ absolutely converges for $|x| < 1$</p>
| SchrodingersCat | 278,967 | <p>By <a href="https://en.wikipedia.org/wiki/Ratio_test" rel="nofollow noreferrer">Ratio test</a>, we can write for the given series,
<span class="math-container">$$\lim_{n\to\infty} \frac{u_{n+1}}{u_n}=\lim_{n\to\infty} \frac{\frac{x^{n + 2}} {(n+1)(n+2)}}{\frac{x^{n + 1}}{n(n + 1)}}$$</span>
<span class="math-container">$$=\lim_{n\to\infty} \frac{x}{1+\frac{2}{n}} = x$$</span></p>
<p>So this is a geometric series with common ratio <span class="math-container">$x$</span> which converges iff <span class="math-container">$|x|<1$</span></p>
|
927,815 | <p>Find an equation of the plane that passes through the points $A(0, 1, 0)$, $B(1, 0, 0)$ and $C(0, 0, 1)$.</p>
| georg | 144,937 | <p>I would say</p>
<p>Sectional plane equation: $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$</p>
<p>In this example p = q = r = 1 --> equation x + y + z =1</p>
|
3,566,083 | <p>I have a game that assigns the probability of finding items in treasure chests by making several independent checks until it fills a quota, with duplicates not allowed. I am trying to figure out how to calculate the resulting probabilities of finding each item - without breaking the bank in terms of calculation brute force.</p>
<p>For an example: The % chance of getting a small chest is 30. For a medium chest it's 15, and for a big chest it's 5. (There is no requirement that these add to 100.)</p>
<p>The algorithm:</p>
<ul>
<li>Roll random against the large's 5%.</li>
<li>If successful, it's a large. If not, roll random against the medium's 15%.</li>
<li>If successful, it's a medium. If not, roll random against the small's 30%.</li>
<li>If successful, it's a small. If not, return to rolling for the large, and repeat the process indefinitely until <em>something</em> is successful.</li>
</ul>
<p>This is then repeated over three layers. The layers are:</p>
<ul>
<li>size of treasure chest</li>
<li>item category (e.g. weapon or armour)</li>
<li>specific item (e.g. for weapons: sword or pike; for armour: helmet, gloves, or boots)</li>
</ul>
<p>So first the game makes this "roll until success" check to decide what chest to get. The chest selected determines which item pool is drawn from and how many items are needed. For each item needed, the game does a "roll until success" check for category, and then another for item. It repeats these two checks until it has the requisite number of items. Duplicates are not allowed; if an item would be a duplicate, the process is restarted. (Which I think is identical to changing the % chance of potential duplicates to 0.)</p>
<p>I am trying to, given the % chances for everything in all 3 layers, calculate the final probability that you will get a chest that has a specific item in it. I'm not looking for a one-time result of any particular example data: I'd like help in figuring out how to formulate the algorithm so I can apply it to any data.</p>
<p>This is giving me a headache for two reasons:</p>
<ul>
<li>It's really close to being a geometric distribution, but because the "success rate" at each step is not identical, it isn't. Also you have to fudge the meaning of "success" because what item you get matters.</li>
<li>Ignoring duplicates is a pain. If item 2 is in a different category than item 1, there's no effect. But if item 2 is in the same category - or if one category has been completely exhausted - the rest of the things in that level all have different rates for item 2.</li>
</ul>
<p>The brute force way of formulating this doesn't seem too hard (e.g. doing the <span class="math-container">$(1-p)^k(p)$</span> thing for each p in order a bunch of times). But I don't want to use brute force, because I have to present the data <a href="https://meta.wikimedia.org/wiki/Help:Calculation" rel="nofollow noreferrer">using MediaWiki</a>, which <em>can</em> do this given the variable and loop extensions but I imagine doing this a ton will not be ideal - taking 2 items from a chest that has 4 categories and 3 things in each category looks like I need 21 iterations of <span class="math-container">$(1-p)^k(p)$</span> (1 for "chance of picking this", 1 for each other option to get "chance of picking this given I previously picked that"). If possible, I'm looking for something more pragmatic.</p>
<hr>
<p>Other notes:</p>
<ul>
<li>Some items can appear in more than one size of chest, with different rates. It should be easy enough to calculate them separately and proportionally add them together.</li>
</ul>
<p>Related-looking questions:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/2805529/algorithm-game-item-drop-probability">Exhibit A</a> - Very similar, but no answer.</li>
<li><a href="https://math.stackexchange.com/questions/1341703/what-is-the-probability-of-an-item-in-a-list-being-chosen-if-both-the-list-and-i">Exhibit B</a> - The same problem in essence (including the part where there's multiple levels of selection), but without the all-important "repeat until something succeeds" condition.</li>
</ul>
<hr>
<p>Examples of possible data:</p>
<pre><code>Set 1
1% large chest (pick 2)
50% weapon
30% big sword
10% claymore
20% armour
50% steel helmet
50% steel gauntlets
50% steel boots
10% jewel
20% ruby
30% sapphire
40% emerald
20% potion
10% red potion
20% blue potion
30% green potion
40% yellow potion
10% medium chest (pick 2)
50% potion
40% water
20% red potion
10% weapon
40% medium sword
20% armour
40% iron helmet
40% iron boots
100% small chest (10% chance of pick 2, 90% chance of pick 1)
100% creature bits
15% tail
70% hair
Set 2
5% large chest (pick 2)
50% weapon
30% claymore
30% big sword
20% giant club
30% armour
100% magic bracelet
10% jewel
50% emerald
100% diamond
30% potion
10% red potion
20% blue potion
30% green potion
40% yellow potion
15% medium chest (pick 2)
40% potion
80% water
10% red potion
5% weapon
40% small sword
20% medium sword
25% armour
40% iron helmet
40% iron boots
100% small chest (10% chance of pick 2, 90% chance of pick 1)
100% creature bits
40% tail
70% hair
</code></pre>
| Robert Israel | 8,508 | <p>The equation
<span class="math-container">$\ln(b^y) = y \ln(b)$</span> is not true for negative <span class="math-container">$b$</span>.</p>
|
214,705 | <p>If $a(x) + b(x) = x^6-1$ and $\gcd(a(x),b(x))=x+1$ then find a pair of polynomials of $a(x)$,$b(x)$.</p>
<p>Prove or disprove, if there exists more than 1 more distinct values of the polynomials.</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ \ $ If $\rm\:a\!+\!b = c\:(x\!+\!1)\:$ then </p>
<p>$$\rm\:x\!+\!1 = (a,b) = (a,c\,(x\!+\!1)\!-\!a) = (a,c\,(x\!+\!1))\iff a = d(x\!+\!1),\,\ (d,c) = 1$$</p>
<p>Thus for $\rm\:c(x\!+\!1) = x^6\!-\!1\:$ it's true $\rm\iff a = d(x\!+\!1),\:$ for $\rm\,d\,$ coprime to $\rm\:c = (x^6\!-\!1)/(x+1),\:$ or, equivalently, $\rm\,d\,$ coprime to $\rm\:x\!-\!1,\ x^2\pm x + 1$</p>
|
975,207 | <p><strong>Question:</strong></p>
<blockquote>
<p>Solve the following system for $a,b,c\in \mathbb{R}$:
$$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$</p>
</blockquote>
<p>I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6)^2=4(2c+6)$$
Then maybe $a=b=c$ is one case.</p>
<p>Thank you.</p>
| Tito Piezas III | 4,781 | <p>Surprisingly, this system <strong><em>does</em></strong> have solutions in $\mathbb{R}$ where $a\neq b\neq c$. Unfortunately, it involves a deg-$54$ equation. Given,</p>
<p>$$ (b^2-6)^2=4(2a+6)\tag1$$
$$(c^2-6)^2=4(2b+6)\tag2$$
$$(a^2-6)^2=4(2c+6)\tag3$$</p>
<p>Do the substitutions:</p>
<p>$$a =\tfrac{1}{2}(-6+x^2)\tag4$$</p>
<p>$$b =\tfrac{1}{8}(12-12c^2+c^4)\tag5$$</p>
<p>$$c =\tfrac{1}{128}(-240 - 288x^2 + 168 x^4 - 24 x^6 + x^8)\tag6$$</p>
<p>This satisfies $(2),(3)$. However, $(1)$ is satisfied if a factorable $2+2+6+54=64$-deg eqn in $x$ is solved. The quadratic roots are $-1\pm\sqrt{3},\,1\pm\sqrt{7}$ which yield $a=b=c$, the sextic has all complex roots, but the $54$-deg has <strong><em>12 real roots</em></strong>. (You can see that monster equation in all its glory here: <a href="http://www.wolframalpha.com/input/?i=Factor%5BResultant%5B-240-288c%5E2%2B168c%5E4-24c%5E6%2Bc%5E8-128x%2C%20-240-128c-288x%5E2%2B168x%5E4-24x%5E6%2Bx%5E8%2Cc%5D%5D" rel="nofollow"><em>WolframAlpha resultant</em></a>.)</p>
<p>From the definitions of $(4),(5),(6)$, if $x$ is real, then $a,b,c$ are also real and $a\neq b\neq c$. We give one root to 50 decimal places,</p>
<p>$$x =-2.9849962763167580809615620439502727527653816890554$$</p>
<p>so,</p>
<p>$$\begin{align}
a &= 1.4551013848124557801133370412617382434937978383740\dots\\
b &= -0.1732265781180354496841227929738397618222596966\dots\\
c &= -1.1155995411074882152529545257291565862139304073\dots\\
\end{align}$$</p>
|
2,842,481 | <p>If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?</p>
<p>I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!</p>
| Maverick | 171,392 | <p>Solving <span class="math-container">$$y=\frac{x^2-3x+c}{x^2+3x+c}$$</span> and <span class="math-container">$$y=7$$</span></p>
<p>we put <span class="math-container">$$\frac{x^2-3x+c}{x^2+3x+c}=7$$</span></p>
<p>to get <span class="math-container">$$x^2-4x+c=0$$</span> which should have exactly one root.</p>
<p>So putting <span class="math-container">$Discriminant=0$</span></p>
<p>we get <span class="math-container">$c=4$</span></p>
|
40,618 | <p>Let $f_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows:
(1) if $x = y$ then $f_{=}(x,y) = 1$;
(2) $f_{x,y} = 0$ otherwise.</p>
<p>I would like to have a proof for / a reference to a textbook proof of the following theorem (if it indeed is a theorem):</p>
<p>$f_{=}$ is uncomputable even if one restricts the domain of $f_{=}$ to a proper subset of $\mathbb{R}^{2}$, viz. the set of the computable real numbers</p>
<p>Thanks!</p>
| Gerald Edgar | 454 | <p>Aren't computable functions of reals automatically continuous? And isn't your function discontinuous? Of course you need a <em>definition</em> of computable in this setting to make sense of this...</p>
|
40,618 | <p>Let $f_{=}$ be a function from $\mathbb{R}^{2}$ be defined as follows:
(1) if $x = y$ then $f_{=}(x,y) = 1$;
(2) $f_{x,y} = 0$ otherwise.</p>
<p>I would like to have a proof for / a reference to a textbook proof of the following theorem (if it indeed is a theorem):</p>
<p>$f_{=}$ is uncomputable even if one restricts the domain of $f_{=}$ to a proper subset of $\mathbb{R}^{2}$, viz. the set of the computable real numbers</p>
<p>Thanks!</p>
| Paul Taylor | 2,733 | <p>It is entirely reasonable to ask for a continuous and computable equality-test function on~$\mathbb R$. The unreasonable thing is to expect it to take values in the <em>discrete</em> two-element space $\mathbf 2$.</p>
<p>The truth-value space that we need to use is the <strong>Sierpinski space</strong>, for which I write $\Sigma$. It has (in the classical interpretation) two elements,</p>
<ul>
<li>$\top$, for which $\lbrace\top\rbrace\subset\Sigma$ is an open subspace, and</li>
<li>$\bot$, for which $\lbrace\bot\rbrace\subset\Sigma$ is a closed subspace.</li>
</ul>
<p>This space has the property that there is a three-way bijection amongst</p>
<ul>
<li>continuous functions $f:X\to\Sigma$;</li>
<li>open subspaces $U=f^{-1}(\top)\subset X$ and</li>
<li>closed subspaces $C=f^{-1}(\bot)\subset X$
for any space $X$.</li>
</ul>
<p>This is like the subobject classifier $\Omega$ in a topos, where "open subspace" becomes "subobject". Indeed, both in set theory and topology the classification properties are rather trivial in the classical case, but become powerful definitions when read constructively.</p>
<p>In particular, a space $X$ is <strong>Hausdorff</strong> iff the diagonal $X\subset X\times X$
is closed. This happens iff there is a continuous function
$$ (\neq):X\times X\to\Sigma $$
for which
$$ (\neq)(x,y) = \bot \iff x=y. $$</p>
<p>In the discrete space $\mathbf 2$ both points are both open and closed and there is a <strong>negation</strong> function $\lnot:\mathbf 2\to\mathbf 2$ that swaps them.</p>
<p>All of this is meaningful in both general topology and recursion theory, replacing
- <em>continuous function</em> by <em>computable function</em> and
- <em>open subspace</em> by <em>recursively enumerable</em> subspace.</p>
<p>In particular, a computable function $f:X\to\Sigma$ is one that may or may not terminate but otherwise has no return value, so it is like $\mathtt{void}$ in $C$ or $\mathtt{unit}$ in ML. Then $\top$ denotes termination and $\bot$ divergence.</p>
<p>There is no negation function on this $\Sigma$ because if there were it would be exactly a solution of the Halting Problem.</p>
<p>In particular, slimton is right in saying that a positive solution to the question as originally posed would be equivalent to the
Halting Problem.</p>
<p>The interpretation of computation in general topology was pioneered by Dana Scott in the 1970s. Conversely, my research programme <a href="http://www.paultaylor.eu/ASD" rel="nofollow"><strong>Abstract Stone Duality</strong></a> is about reformulating topology to make it equivalent to computation.</p>
<p>The best introduction to this programme for the general mathematician is my paper <a href="http://www.paultaylor.eu/ASD/lamcra" rel="nofollow"><em>A Lambda Calculus for Real Analysis</em></a>.</p>
|
265,549 | <p>$$ \left\{ x \in\mathbb{R}\; \middle\vert\; \tfrac{x}{|x| + 1} < \tfrac{1}{3} \right\}$$</p>
<p>What is the supremum and infimum of this set? I thought the supremum is $\frac{1}{3}$. But can we say that for any set $ x < n$ that $n$ is the supremum of the set? And for the infimum I have no idea at all. Also, let us consider this example:</p>
<p>$$ \left\{\tfrac{-1}{n} \;\middle\vert\; n \in \mathbb{N}_0\right\}$$</p>
<p>How can I find the infimum and supremum of this set? It confuses me a lot. I know that as $n$ gets bigger $\frac{-1}{n}$ asymptotically approaches $0$ and if $n$ gets smaller $\frac{-1}{n}$ approaches infinity, but that's about it. </p>
| André Nicolas | 6,312 | <p>Let $x$ be negative. then $\dfrac{x}{|x|+1}$ is negative, and in particular $\lt \dfrac{1}{3}$.</p>
<p>So there is no infimum. (But some people allow the symbol $-\infty$ as an infimum.)</p>
<p>For the supremum, note that there are positive $x$ such that $\dfrac{x}{x+1}\lt \dfrac{1}{3}$.</p>
<p>Since $\dfrac{x}{x+1}=1-\dfrac{1}{x+1}$, our function is increasing for positive $x$. Solve $1-\dfrac{1}{x+1}=\dfrac{1}{3}$. We get $x=\dfrac{1}{2}$. </p>
<p>Thus $\dfrac{1}{2}$ is an upper bound for our set.</p>
<p>However close we are to $\dfrac{1}{2}$, but below $\dfrac{1}{2}$, we will have $\dfrac{x}{x+1}\lt\dfrac{1}{3}$. So there is no cheaper upper bound than $\dfrac{1}{2}$. </p>
<p><strong>Remark:</strong> For your other question, I think you intend to look at $\{-\frac{1}{n}\}$, where $n$ ranges over the positive integers. The smallest element of this set is $-1$. It is therefore the infimum. For the supremum, note that our numbers are all $\lt 0$, but can be made arbitrarily close to $0$ by choosing $n$ large enough. So the supremum is $0$.</p>
|
944,992 | <p>I'm trying to understand each part of this completed proof that my professor did, here is my interpretation in parentheses, please advise as necessary.</p>
<p>Proof: Assume that $2^{1/2}$ is rational, then $2^{1/2}$ = $p/q$ for some integer $p$ and $q$ with $q \neq 0$ and $\gcd(p,q) = 1$</p>
<p>$$(p/q)^2 = 2 = p^2 / q^2 $$ $$\text{ (What was the purpose of squaring √2 and p/q?)}$$</p>
<p>$$p^2 = 2q^2$$ $$ \text{
(Is this basically just canceling out the q^2 on left side, and multiplying the right side?)}$$</p>
<p>2 | p^2 => (2 divides p^2 since it is a factor)</p>
<p>2 | p => (thus 2 divides p)</p>
<p>p = 2k for some integer k => (definition of even)</p>
<p>(2k)^2 = 4k^2 = 2q^2 => (Why are we squaring 2k, and how does 4k^2 = 2q^2?) </p>
<p>q^2 = 2k^2 => (Canceled out 2 on one side, and dividing on other)</p>
<p>2 | q^2 => (2 divides q^2)</p>
<p>2 | q => (thus 2 divides q)</p>
<p>2 | gcd(p,q) => (2 divides gcd of p and q)</p>
<p>2 | 1 Which is a Contradiction (no clue how I got here)</p>
| NicNic8 | 24,205 | <p>Unfortunately not. Physical explanations like that are a great way to develop intuition that might lead to a proof, but it's not a mathematical proof. A mathematical proof is a set of logical statements that lead from axioms or theorems to the desired conclusion.</p>
<p>In fact, physical intuition can lead to some major errors in math. A great example is the axiom of choice. The axiom of choice says something like: if there's a set of stuff than one can choose one element of the stuff. This seems reasonable. But if one accepts this axiom, then one can split a ball into two balls of exactly the same volume. And this, of course, is impossible physically.</p>
|
1,918,885 | <p>I would like to show that $\mathbb{Q}$ is the smallest field between $\mathbb{R}$ and $\mathbb{Z}$. In other words, if there is a field $\mathbb{F}$ where $\mathbb{Z} \subset \mathbb{F} \subset \mathbb{R}$, then it must also be that $\mathbb{Q} \subset \mathbb{F}$.</p>
<p>My proof is by contradiction, assume that it is not the case $\mathbb{Q} \subset \mathbb{F}$. Then, it is possible to find a $y \in \mathbb{Q}$ such that $y \notin \mathbb{F}$. Suppose that $y \in \mathbb{Q}$, then, by definition of rational numbers, we may find a $c,d \in \mathbb{Z}$ where $d \neq 0$ such that $y=\frac{c}{d}$.</p>
<p>BUT, we know from the assumption above that $\mathbb{Z} \subset \mathbb{F}$. This means that there exists $a,b \in \mathbb{Z}$ where $a=c$ and $b=d$. Furthermore, because $\mathbb{Z} \subset \mathbb{F}$, $a,b \in \mathbb{F}$. Now, due to the multiplicative inverse properties of $\mathbb{F}$, we know that $\frac{a}{b}$ exists and that $\frac{a}{b} \in \mathbb{F}$. </p>
<p>Putting it all together, we have that:</p>
<p>$$
y= \frac{c}{d} = \frac{a}{b} \in \mathbb{F}.
$$</p>
<p>But, this contradicts the assumed that $y \notin \mathbb{F}$. </p>
<p>I am not sure if the above proof works, because in a proof by contradiction, I normally contradict something in the assumption area. However, it seems here that I am contradicting the negated result. </p>
<p>In other words, if I let $P$ be the statement: <strong>"if there is a field $\mathbb{F}$ where $\mathbb{Z} \subset \mathbb{F} \subset \mathbb{R}$"</strong></p>
<p>and $Q$ be the statement <strong>"then it must also be that $\mathbb{Q} \subset \mathbb{F}$"</strong>, </p>
<p>then in a direct proof I normally have $P \implies Q$. In a contradiction proof, what I understand is that we assume NOT $Q$, then try to contradict something in $P$. But here, it seems I am not really contradicting anything in $P$, but rather something in $Q$, and so seems almost circular. </p>
<p>Could anyone help me see what is missing? Thanks!</p>
| fleablood | 280,126 | <p>You can be pretty straightforward.</p>
<p>Let $\mathbb Z \subset \mathbb F$ be a field. Then $\{a*(1/b)|a\in \mathbb Z; b \in \mathbb Z; b \ne 0\}\subset \mathbb F$. But $\{a*(1/b)|a\in \mathbb Z; b \in \mathbb Z; b \ne 0\}= \mathbb Q$. So $\mathbb Q \subset \mathbb F$.</p>
<p>That's it. Nothing more needs to be done. </p>
<p>Your proof by contradiction is fine. But you prove the contradiction by proving the statement <em>directly</em> to get the contradiction. That is not <em>circular</em>; it is redundant.</p>
<p>Try to follow this reasoning: We are asked to prove $4n$ is even. We chose to do a contradiction. So we assume $4n$ is odd. So $4n = 2*(2n) = 2k$ for $k = 2n$. So $4n$ is even. This contradicts that $4n$ is odd. So $4n$ is not odd. Therefore it is even. </p>
<p>Note: that is not circular. But it is redundant.</p>
|
657,801 | <p>So this is a simple problem but I'm just getting stumped. The question is:<br><br>
A particle not connected to a spring, moving in a straight line, is subject to a retardation force of magnitude $\beta(\frac{dx}{dt})^n$, with $\beta > 0$.<br>
a) Show that if 0 < n < 1, the particle will come to rest in a finite time. How far will the particle travel, and when will it stop?</p>
<p>So I think this would be the starting equation: $m\frac{d^2x}{dt^2}+\beta (\frac{dx}{dt})^n=0$ and the particle will stop when $\frac{dx}{dt}=0$ but that's all I got..I don't really know what to do from here. The question asks for n in a range so that's kind of throwing me off. Any ideas? Thanks</p>
| Robert Lewis | 67,071 | <p>I too think</p>
<p>$m\dfrac{d^2x}{dt^2} + \beta(\dfrac{dx}{dt})^n = 0, \tag{1}$</p>
<p>is the right equation with which to start. Then observe that (1) implies</p>
<p>$m(\dfrac{dx}{dt})^{-n}\dfrac{d^2x}{dt^2} + \beta = 0, \tag{2}$</p>
<p>and that</p>
<p>$\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} = (1 - n)(\dfrac{dx}{dt})^{- n}\dfrac{d^2x}{dt^2}, \tag{3}$</p>
<p>whence (2) may be written</p>
<p>$\dfrac{m}{1 - n}\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} + \beta = 0 \tag{4}$</p>
<p>or</p>
<p>$\dfrac{m}{1 - n}\dfrac{d}{dt}(\dfrac{dx}{dt})^{1 - n} = -\beta. \tag{5}$</p>
<p>Note that all these calculations are valid for $n$ in the range $0 < n < 1$. It should also be observed that the assumption $dx / dt
\ge 0$ is implicit in (1) and what follows, since $(dx/dt)^n$ only makes sense for the given range of $n$ if this is the case. These things being said, if we integrate (5) 'twixt $t_0$ and $t$, and divide through by $m / (1 - n)$ we find that</p>
<p>$(\dfrac{dx}{dt})^{1 - n}(t) - (\dfrac{dx}{dt})^{1 - n}(t_0) = -\dfrac{1 - n}{m}\beta(t - t_0), \tag{6}$</p>
<p>or</p>
<p>$(\dfrac{dx}{dt})^{1 - n}(t) = (\dfrac{dx}{dt})^{1 - n}(t_0) -\dfrac{1 - n}{m}\beta(t - t_0). \tag{7}$</p>
<p>Suppose the particle has velocity $v_0 = (dx/dt)(t_0)$ at $t = t_0$. Then (7) becomes</p>
<p>$(\dfrac{dx}{dt})^{1 - n}(t) = v_0^{1 - n} -\dfrac{1 - n}{m}\beta(t - t_0), \tag{8}$</p>
<p>and if we set the left-hand side equal to $0$, then we can solve for the value $t_r$ of $t$ at which the particle has zero velocity, i.e, when it "comes to rest":</p>
<p>$v_0^{1 - n} -\dfrac{1 - n}{m}\beta(t_r - t_0) = 0 \Rightarrow t_r = t_0 + \dfrac{ m}{\beta(1 - n)}v_0^{1 - n}. \tag{9}$</p>
<p>The distance $d$ the particle will travel is given by integrating $dx / dt$ between $t_0$ and $t_r$; indeed, from equation (8) we have</p>
<p>$\dfrac{dx}{dt} = (v_0^{1 - n} - \dfrac{1 - n}{m}\beta (t -t_0))^{1 / (1 - n)}, \tag{10}$</p>
<p>and thus</p>
<p>$x_r - x_0 = \int_{t_0}^{t_r} (\dfrac{dx}{dt})dt = \int _{t_0}^{t_r} (v_0^{1- n} - \dfrac{1-n}{m}\beta(t - t_0))^{1 / (1 - n)}dt; \tag{11}$</p>
<p>the right-hand integral occurring in (11) looks at first glance formidable, but becomes quite tractable under the substitution</p>
<p>$u = v_0^{1 - n} - \dfrac{1 -n}{m}\beta(t - t_0), \tag{12}$</p>
<p>for then</p>
<p>$du = -\dfrac{1-n}{m}\beta dt, \tag{13}$</p>
<p>so that</p>
<p>$dt = -\dfrac{m}{\beta(1 - n)}du, \tag{14}$</p>
<p>and the integral on the right-hand side of (11) becomes</p>
<p>$-\dfrac{m}{\beta(1 - n)}\int _{v_0^{1 - n}}^0 u^{1 / (1 - n)}du = -\dfrac{m}{\beta(1 - n)}(\dfrac{1 - n}{2 - n}u^{\frac{2 - n}{1 - n}}\mid_{v_0^{1 - n}}^0 = \dfrac{m}{\beta(2 - n)}v_0^{2 - n}, \tag{15}$</p>
<p>since $u(t_0) = v_0^{1 - n}$ and $u(t_r) = 0$. The distance the particle travels is thus </p>
<p>$d = \dfrac{m}{\beta(2 - n)}v_0^{2 - n}. \tag{16}$</p>
<p>Well, I <em>think</em> I got all these calculations right. Gotta run to work right now, will check it all again later!</p>
<p>Hope this helps! Cheerio,</p>
<p>and as always,</p>
<p><em><strong>Fiat Lux!!!</em></strong></p>
|
657,801 | <p>So this is a simple problem but I'm just getting stumped. The question is:<br><br>
A particle not connected to a spring, moving in a straight line, is subject to a retardation force of magnitude $\beta(\frac{dx}{dt})^n$, with $\beta > 0$.<br>
a) Show that if 0 < n < 1, the particle will come to rest in a finite time. How far will the particle travel, and when will it stop?</p>
<p>So I think this would be the starting equation: $m\frac{d^2x}{dt^2}+\beta (\frac{dx}{dt})^n=0$ and the particle will stop when $\frac{dx}{dt}=0$ but that's all I got..I don't really know what to do from here. The question asks for n in a range so that's kind of throwing me off. Any ideas? Thanks</p>
| Qmechanic | 11,127 | <p>Hint: Split the problem in two:</p>
<ol>
<li><p>First solve the 1st order ODE $m\frac{dv}{dt}=-\beta v^n$ for the velocity $v(t)$ as a function of time $t$. (The exercise formulation doesn't say so, but there are physical reasons to believe that the velocity $v\geq 0$ should be assumed to be non-negative).</p></li>
<li><p>Next integrate the velocity $v(t)$ wrt. $t$ to find the position $x(t)$ as a function of time $t$.</p></li>
</ol>
|
64,653 | <p>Consider the two (inequivalent) $\mathbb{Z}$-representations $\phi,\psi$ of the symmetric group $S=S_3$ given by</p>
<p>$(1,2)^\phi=\left(\begin{array}{rr}0 &-1\\\ -1 & 0\end{array}\right), \qquad
(1,2,3)^\phi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right);$</p>
<p>$(1,2)^\psi=\left(\begin{array}{rr}0 &1\\\ 1 & 0\end{array}\right), \qquad
(1,2,3)^\psi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right).$</p>
<p>Now, let $F=\langle x,y\rangle$ be a free 2-generated group. The representation $\phi$ can be "lifted" to an embedding $\tau:S\to\rm{Aut}(F)$ as follows:</p>
<p>$(1,2)^\tau=[x\mapsto y^{-1};\quad y\mapsto x^{-1}], \qquad
(1,2,3)^\tau=[x\mapsto y;\quad y\mapsto x^{-1}y^{-1}].$</p>
<p><strong>Question.</strong> Can one similarly lift $\psi$?</p>
<p><strong>Remark 1.</strong> By "lifting" a representation $\phi:S\to\rm{GL}_2(\mathbb{Z})$ I mean finding an embedding $\tau:S\to\rm{Aut}(F)$ such that $\phi=\tau\alpha$, where $\alpha:\rm{Aut}(F)\to\rm{GL}_2(\mathbb{Z})$ is the natural epimorphism.</p>
<p><strong>Remark 2.</strong> A naïve attempt to send
$(1,2)\ \mapsto\ [x\mapsto y;\quad y\mapsto x], \qquad
(1,2,3)\ \mapsto\ [x\mapsto y;\quad y\mapsto x^{-1}y^{-1}]$</p>
<p>does <em>not</em> give a lifting of $\psi$.</p>
| DavidLHarden | 12,610 | <p>I have wondered about such lifts myself, and I want to give what I hope is a tantalizing hint of what such lifts may be able to tell us:</p>
<p>The lift you give of $S_{3}$ from $GL_{2}( \mathbb{Z} )$ to $Aut(F_{2})$ also gives an embedding of $C_{3} = A_{3}$ into $Aut(F_{2})$. Why is this useful? It gives a character-free proof of a congruence about conjugacy classes of a finite group (here $c(G)$ denotes the number of conjugacy classes of the group $G$):</p>
<p>Theorem. If $G$ is a finite group with $|G|$ not divisible by $3$, then $|G| \equiv c(G) \mod{3}$. </p>
<p>Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is not a multiple of $3$, these dimensions aren't multiples of $3$. Now reduce modulo $3$ and obtain the congruence. </p>
<p>Proof, without character theory: $|G|(|G| - c(G))$ is the number of non-commuting ordered pairs of elements of $G$. Since $|G|$ is a nonmultiple of $3$, the congruence may be proved by showing that this set has a number of elements which is a multiple of $3$. Now just let the lift of $\tau$ act on it: If $(x,y)$ is a fixed point, then reading the first coordinate gives $x=y$, which trivially implies $xy=yx$. So the action is fixed-point-free, and we are done.</p>
<p>Theorem. If $G$ is a finite group of odd order, then $|G| \equiv c(G) \mod{8}$. </p>
<p>Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is odd, these dimensions are odd. Now reduce modulo $8$ and obtain the congruence. </p>
<p>Proof, without character theory: Instead of lifting $S_{3}$ to $Aut(F_{2})$, lift the dihedral group of order $8$, which is generated by the involutions $(x,y) \to (x^{-1},y)$ and $(x,y) \to (y,x)$. It suffices to check that none of the $5$ involutions of this group has a fixed point in the action on the non-commuting pairs of elements of $G$. This is easy to do, and it involves recalling that, since $|G|$ is odd, $t^{2}=1$ implies $t=1$ for $t \in G$.</p>
<p>This is all well and good, but, it is not the end of the story:</p>
<p>Theorem. The number of rows in the character table of $G$ which are entirely real-valued is the number of conjugacy classes $C$ of $G$ such that $x \in C$ iff $x^{-1} \in C$. </p>
<p>Corollary. If $|G|$ is odd, then the only entirely real-valued character of $G$ is the trivial character. </p>
<p>This leads to a strengthening of the character theory argument used above, so that it now proves that $|G| \equiv c(G) \mod{16}$.<br>
In fact, $16$ is the highest power of $2$ that works here, since for any prime $p \equiv 3 \mod{8}$, we can let $G$ be a nonabelian group of order $p^{3}$. This gives $|G| - c(G) = p^{3} - (p^{2}+p-1) = (p^{2}-1)(p-1) \equiv 16 \mod{32}$. </p>
<p>The really tantalizing thing here is that $(p^{2}-1)(p-1)$ is the $p$-free part of the order of the general linear group $GL_{2}(\mathbb{Z} / (p))$.<br>
I do not know whether $|G|-c(G)$ is always a multiple of $(p^{2}-1)(p-1)$ when $G$ is a $p$-group (though the theorems whose proofs I outlined above establish it when $p = 2$ or $p = 3$), and I do not know if some analogue of the lifting argument works, possibly with $Aut(F_{2})$ replaced by $Aut(B(2,p))$ or $Aut(B(2,p^{k}))$, where $B(r,n)$ denotes the rank $r$, exponent $n$ Burnside group. </p>
<p>Also see Bjorn Poonen's paper:<br>
Congruences relating the order of the group to the number of conjugacy classes, American Mathematical Monthly, 105(1995), 440-442.</p>
|
463,539 | <p>I've transformed the number 11 to:</p>
<p>$11^e = 677.32$</p>
<p>Given the exponent and the transformed value, how can I solve for the original number?</p>
<p>I know that $x = y^z$ and that $z = \log_y(x)$, but I don't know how to solve for $x$? Can anyone explain how I can use the exponent $e$ and $677.32$ to find the $x$ value of 11?</p>
| cuneiformium | 500,021 | <p>This is equation with constant e (2.71) as exponent and no obvious unknown quantities. We can set equation with one unknown from Rasmus's answer (the general solution). However, the terms can be manipulated with the rules of logs and exponents, assuming work uses natural logarithms for base e.</p>
<pre><code> 1. a**b=c definition of exponent
2. x=c**(1/b) Rasmus's answer (from general solution)
3. loading terms, x=677.32**(1/2.71) or 11.0
4. check answer. 11**2.71 approximates 664
</code></pre>
<p>Refs</p>
<pre><code> 1. Exponentiation with arbitrary bases, exponents, Math 152, section
2. 55, Vipul Naik , pdf on internet
3. Wikipedia, exponentiation
</code></pre>
|
1,628,386 | <p>I had following limit of two variables as a problem on my calculus test. How does one show whether the limit below exists or does not exist? I think it does not exist but I was not able to show that rigorously. There was a hint reminding that $\lim_{t\to 0}\sin t / t=1$.</p>
<p>$$\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$$</p>
| Brad A.M. | 296,774 | <p>Consider various paths that approach $(0,0)$. For instance,</p>
<p>$$\lim_{(x,y)\to (0,0)}\frac{3x^2\sin^2y}{2x^4+2\sin(y^4)}$$ along the line $y=0$ becomes</p>
<p>$$\lim_{x\to 0}\frac{3x^2\cdot 0}{2x^4}=0$$</p>
<p>However, along the line $y=x$ the limit is </p>
<p>$$\lim_{x\to 0}\frac{3x^2\sin^2x}{2x^4+2\sin(x^4)}=1$$ I evaluated this limit numerically. This shows that the limit does not exist. You can evaluate it by l'Hopital, careful algebra and the hint.</p>
|
1,407,944 | <p>Can this be proved that $\log(n)$ is irrational for every $n=1,2,3,\dots$ ?</p>
<p>I find that question in my mind in searching for if $\log(x)$ is irrational for every rational number $x\gt0$. </p>
| Ben Sheller | 250,221 | <p>Let $n\in\mathbb{N}$, $n\neq 1$. Suppose that there exist $a,b\in\mathbb{Z}$ such that $\log(n)=\frac{a}{b}$. Then we have that $n=e^{a/b}$ is a trancendental number since $e$ is a trancendental number and $\frac{a}{b}$ is rational. This is a contradiction. Therefore, $\log(n)$ is irrational for every $n=2,3,4,....$</p>
|
2,266,342 | <p>So on my review for my final exam there is this question: </p>
<p>Is there a linear transformation from $P_2$ to $P_2$ with the following properties? In each case, either give an example of such a transformation or prove that no such transformation exist.</p>
<blockquote>
<p>$T(t^2+t+1) = t^2+t+1, T(t^2+2t+3) = 3t^2+2t+1, T(t^2+2t+2) = 2t^2+2t+1$</p>
</blockquote>
<p>So we know that the standard basis for $P_2$ is </p>
<p>{$1,t,t^2$}</p>
<p>My question is how do you even start? Is there a trick? we have never done this kind of question in our class before.</p>
| freshman | 217,694 | <p>Hint: $\{t^2+t+1, t^2+2t+3,t^2+2t+2\}$ is independent in $P_2$. Thus there exists such liner map $T$ and is unique. Find $T(t^2), T(t)$ and $T(1)$ solving the system
$$T(t^2)+T(t)+T(1)=t^2+t+1\\
T(t^2)+2T(t)+3T(1)=3t^2+2t+1\\
T(t^2)+2T(t)+2T(1)=2t^2+2t+1$$</p>
|
925,941 | <p>Solve:</p>
<p>$xy=-30$<br>
$x+y=13$</p>
<p>{15, -2} is a particular solution, but, how would I know if is the only solution, or what would be the way to solve this without "guessing" ?</p>
| Jose Arnaldo Bebita Dris | 28,816 | <p>The sum</p>
<p>$$x + y = 13 = -a$$</p>
<p>and the product</p>
<p>$$xy = -30 = b$$</p>
<p>gives rise to the quadratic equation</p>
<p>$$z^2 + az + b = 0.$$</p>
<p>Consequently, we have to solve the quadratic equation</p>
<p>$$z^2 - 13z - 30 = 0$$</p>
<p>for $z$, in order to get $x$ and $y$.</p>
<p>Since the discriminant $a^2 - 4b = {17}^2$, $z^2 - 13z - 30$ is factorable as:</p>
<p>$$z^2 - 13z - 30 = (z - 15)(z + 2).$$</p>
<p>Setting it equal to zero, we get the roots: $z = 15$ or $z = -2$.</p>
<p>Since the equations</p>
<p>$$x + y = 13$$</p>
<p>and</p>
<p>$$xy = -30$$</p>
<p>are symmetric in $x$ and $y$, we get the solutions:</p>
<p>$$x = 15, y = -2$$</p>
<p>or</p>
<p>$$x = -2, y = 15.$$</p>
|
6,990 | <p>The Fourier transform of periodic function $f$ yields a $l^2$-series of the functions coefficients when represented as countable linear combination of $\sin$ and $\cos$ functions.</p>
<ul>
<li><p>In how far can this be generalized to other countable sets of functions? For example, if we keep our inner product, can we obtain another Schauder basis by an appropiate transform? What can we say about the bases in general?</p></li>
<li><p>Does this generalize to other function spaces, say, periodic functions with one singularity?</p></li>
<li><p>What do these thoughts lead to when considering the continouos FT?</p></li>
</ul>
| David Hansen | 1,464 | <p>If you are interested in more general Fourier transforms, then the two things which spring immediately to my mind are:</p>
<ol>
<li><p>Titchmarsh's book <em>Fourier Integrals</em> contains a detailed treatment of what he calls "generalized kernels", which vaguely are pairs of functions $h(x),k(x)\in L^2(\mathbb{R})$ such that</p>
<p>$\int_{0}^{\infty}k(xy)\int_{0}^{\infty}h(yw)f(w)dwdy=f(x)$.</p></li>
<li><p>There is a lovely theory of "wavelets" due to Daubechies et al, which are described in many places.</p></li>
</ol>
|
2,083,475 | <p>This is the conic $$x^2+6xy+y^2+2x+y+\frac{1}{2}=0$$
the matrices associated with the conic are:
$$
A'=\left(\begin{array}{cccc}
\frac{1}{2} & 1 & \frac{1}{2} \\
1 & 1 & 3 \\
\frac{1}{2} & 3 & 1
\end{array}\right),
$$</p>
<p>$$
A=\left(\begin{array}{cccc}
1 & 3 \\
3 & 1
\end{array}\right),
$$</p>
<p>His characteristic polynomial is: $p_A(\lambda) = \lambda^2-2\lambda-8$<br>
A has eigenvalues discordant $(\lambda = 4, \lambda = -2)$, so it's an Hyperbole.
Then i found that the center of the conic is: $(-\frac{1}{16}, -\frac{5}{16})$<br>
Then with the eigenvalues i found the two lines passing through the center:
$$4x-4y-1=0$$
$$8x+8y+3=0$$</p>
<p><a href="https://i.stack.imgur.com/OlKbQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OlKbQ.png" alt="enter image description here"></a></p>
<p>Now i want to find the focus and the asymptotes but i have no idea how to do it.There is a way to find These two things through the data I have now? or do i need the canonical form of the conical? Thanks</p>
| MvG | 35,416 | <p>It seems you're homogenizing by introducing a first coordinate equal to one, instead of a last coordinate <a href="http://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections" rel="nofollow noreferrer">conventionally used</a>. So your conic is essentially described as</p>
<p>$$
(1,x,y)\cdot\begin{pmatrix}
\tfrac12 & 1 & \tfrac12 \\
1 & 1 & 3 \\
\tfrac12 & 3 & 1
\end{pmatrix}\cdot\begin{pmatrix}1\\x\\y\end{pmatrix} = 0
$$</p>
<p>I'll stick to your notation for this post, but readers should be careful if translating between this and a different convention.</p>
<h2>Tangents</h2>
<p>The asymptotes are the tangents at the points at infinity. So first you intersect the conic with the line at infinity, e.g. by plugging a generic point at infinity into the equation:</p>
<p>$$
0 = (0,x,y)\cdot\begin{pmatrix}
\tfrac12 & 1 & \tfrac12 \\
1 & 1 & 3 \\
\tfrac12 & 3 & 1
\end{pmatrix}\cdot\begin{pmatrix}0\\x\\y\end{pmatrix} = x^2+6xy+y^2
$$</p>
<p>You can check that neither $x=0$ nor $y=0$ is a solution. As the result is only defined up to a scalar factor, you may assume $y=1$ and solve for $x$:</p>
<p>$$x_{1,2} = \pm\sqrt8-3$$</p>
<p>Now you compute the tangent, which is simply the matrix times the point at infinity.</p>
<p>$$g_{1,2} =
\begin{pmatrix}
\tfrac12 & 1 & \tfrac12 \\
1 & 1 & 3 \\
\tfrac12 & 3 & 1
\end{pmatrix}\cdot\begin{pmatrix}0\\\pm\sqrt8-3\\1\end{pmatrix}
=\begin{pmatrix}\pm\sqrt8-\tfrac52\\\pm\sqrt8\\\pm\sqrt{72}-8\end{pmatrix}
$$</p>
<p>Now you can read this as the equations of a pair of lines:</p>
<p>$$(\pm\sqrt8-\tfrac52) + (\pm\sqrt8)x + (\pm\sqrt{72}-8)y = 0$$</p>
<p>Note that the principal axes (the ones you computed using eigenvectors) are the perpendicular bisectors of these asymptotes. Computing the axes from the asymptotes is possible, but the converse is impossible as the axes alone don't contain sufficient information.</p>
<h2>Foci</h2>
<p>For foci you can do the following:</p>
<ol>
<li>Compute the <a href="https://en.wikipedia.org/wiki/Pole_and_polar" rel="nofollow noreferrer">polar lines</a> of the ideal circle points which are $(0,1,\pm i)$ in your coordinates. That's the imaginary unit $i$ there, so you need to perform the computation using complex numbers. The polar line is matrix times the coordinates of a point, so in the special case of a point that lies on the conic the polar will be the tangent.</li>
<li>Intersect these polar lines with the conic. That requires a conic-line intersection similar to the one above, except that you can't simply assume that one of the coordinates is zero. But it is still just a quadratic equation in a single parameter. Or you can follow <a href="https://math.stackexchange.com/a/867428/35416">this approach</a> instead.</li>
<li>Connect the points of intersection with the circle point from which you started, to get the tangents from this point. So here you have two ideal circle points, and from each of them you have two tangents to the conic.</li>
<li>Intersect these lines, taking one line for each of the ideal circle points. There are four ways to intersect them, resulting in four foci. Two of them will be real, two complex. Finding the pairings is simple: if you have one real focus, then switching to the other tangent for each of the circle points will result in the other real focus.</li>
</ol>
<p>This approach follows J. Richter-Gebert's <a href="http://rads.stackoverflow.com/amzn/click/3642172857" rel="nofollow noreferrer"><em>Perspectives on Projective Geometry</em></a> Section 19.4.</p>
<p>For your input you get</p>
<ol>
<li>For $I_1$ = $(0,1,i)$ you get $h_1:(1+\tfrac12i)+(1+3i)x+(3+1i)y=0$<br>
and for $I_2$ = $(0,1,-i)$ you get $h_2:(1-\tfrac12i)+(1-3i)x+(3-1i)y=0$.</li>
<li>For $h_1$ you get $P_{11}=(32,-2-(1-3i)\sqrt{6i},-10+(3-i)\sqrt{6i})$<br>
and $P_{12}=(32,-2+(1-3i)\sqrt{6i},-10-(3-i)\sqrt{6i})$.<br>
For $h_2$ you get $P_{21}=\overline{P_{11}}$ and $P_{22}=\overline{P_{12}}$ since $h_2=\overline{h_1}$.</li>
<li>Connecting $I_a$ to $P_{ab}$ gives $t_{ab}$ like this:<br>
$\begin{array}{rr}
t_{11}:& (10-2i-6\sqrt{6i}) - 32i\,x + 32\,y = 0 \\
t_{12}:& (10-2i+6\sqrt{6i}) - 32i\,x + 32\,y = 0 \\
t_{21}:& (10+2i-6\sqrt{-6i}) + 32i\,x + 32\,y = 0 \\
t_{22}:& (10+2i+6\sqrt{-6i}) + 32i\,x + 32\,y = 0
\end{array}$</li>
<li>Intersecting $t_{11}$ with $t_{21}$ yields one real focus at
$F_1=(16,-1-3\sqrt3,-5+3\sqrt3)$,<br>
so $t_{12}$ with $t_{22}$ yield the other at $F_2=(16,-1+3\sqrt3,-5-3\sqrt3)$.<br>
The other foci form a pair of complex conjugate points at $F_{3,4}=(16,-1\pm3i\sqrt3,-5\pm3i\sqrt3)$.</li>
</ol>
<p>Note that I just wrote $16$ in the first coordinate to avoid fractions. If you don't like working with homogeneous coordinates, simply divide by this number and you get plain coordinates $x$ and $y$ in the second and third coordinate. The results are $F_1\approx(-0.387,0.012)$ and $F_2\approx(0.262,-0.637)$.</p>
<p><img src="https://i.stack.imgur.com/r11Yq.png" width="500" alt="Figure"></p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| J. M. ain't a mathematician | 498 | <p>Two related integrals:</p>
<p>$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$</p>
<p>$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Douglas S. Stones | 139 | <p>\[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| J126 | 2,838 | <p>$$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| 2'5 9'2 | 11,123 | <p>\begin{align}
\frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\text{Power Rule?}&\ \text{False}\\
\frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\ \text{False}\\
\frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\
\end{align}</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Noam D. Elkies | 93,983 | <p>$$ 71 = \sqrt{7! + 1}. $$</p>
<p>Besides the amusement of reusing the decimal digits $7$ and $1$,
this is conjectured to be the last solution of $n!+1 = x^2$ in integers.
($n=4$ and $n=5$ also work.) Even finiteness of the set of solutions
is not known except using the ABC conjecture.</p>
|
8,814 | <p>Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?</p>
| Feynman | 196,748 | <p>I have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.</p>
<p>Let us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\phi$, the Euler-Mascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that</p>
<p>$$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$$</p>
|
2,505,971 | <blockquote>
<p>Let $\lim_{x\to \infty }x^n f(x)=0$ for any $n \in \mathbb{N}$. Then find the symptote
$$g(x)=\dfrac{x^2+f(x)}{x+1+2f(x)}.$$</p>
</blockquote>
<hr>
<p>My try :
$$\lim_{x \to \infty }\dfrac{x^2+f(x)}{x+1+2f(x)}=\lim_{x \to \infty }\dfrac{x^{n+2}+x^nf(x)}{x^{n+1}+2x^nf(x)}=\infty?$$
Please help me.</p>
| Marcus Andrews | 97,648 | <p>As a general rule, when you have a recurrence $S(n)$ that is a summation of itself, it usually helps to try $S(n)-S(n-1)$ so you can cancel out most of the summation.</p>
<p>$S(n)=\sum\limits_{i=1}^{n-1}iS(i)$</p>
<p>$S(n)-S(n-1)=\sum\limits_{i=1}^{n-1}iS(i) - \sum\limits_{i=1}^{n-2}iS(i) = (n-1)S(n-1)$</p>
<p>Now we are dealing with the recurrence $S(n) - S(n-1) = (n-1)S(n-1)$ which is a little easier to work with. Add $S(n-1)$ to both sides:</p>
<p>$S(n) = nS(n-1)$</p>
<p>And since $S(1) = 1$, we can see that $S(n) = n \cdot (n-1) \cdot (n-2) \cdot \cdots\cdot 2 \cdot 1 = n!$</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| Charles Matthews | 6,153 | <p>Not sure I agree with the whole post in detail. Distinguish "pure algebra" from "applied algebra"; and within "pure algebra" distinguish "structural" issues from "combinatorial" ones such as the Burnside problem. Remembering that "abstract algebra" is the modern term for what used to be called "modern algebra", we should probably drop the "abstract" to get a more reasonable view (the scope of "old" or 19th century algebra being that of Chrystal's <em>Algebra</em> say, some would now count as other branches of mathematics, such as numerical methods).</p>
<p>So which questions are worth studying? Not just one kind, surely. Algebraic geometry, algebraic topology, algebraic number theory all do ask serious and interesting algebraic questions. See for example the <a href="https://en.wikipedia.org/wiki/Golod%E2%80%93Shafarevich_theorem" rel="nofollow noreferrer">Golod–Shafarevich theorem</a> which is pure algebra to start with. Parts of algebra come across as "general" compared to mathematics as a whole, but this is somewhat subjective criterion these days. There are both general-structural and general-combinatorial parts of algebra. There do need to be some criteria operating in, say, infinite group theory and infinite-dimensional Lie algebra theory. Generality in the sense of category theory is rather 1960s in feel; derived categories are "abstract" but I wonder who these days would argue that they are too "general"? I suppose the general module over the general ring still looks troublesome as a setting for research.</p>
<p>Well, I think "follow the masters" is probably the best advice,</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| David Corfield | 447 | <p>There are surely no hard and fast rules as to assessing the importance of a generalization of a concept. I once took a <a href="http://books.google.co.uk/books?id=s37rhfWs73QC">look</a> (chap. 9) at debates surrounding the move from groups to groupoids. One important step up for a concept is being deemed essential rather than merely useful. To achieve this it must find its place in an array of good storylines.</p>
|
39,828 | <p>Dear MO-community, I am not sure how mature my view on this is and I might say some things that are controversial. I welcome contradicting views. In any case, I find it important to clarify this in my head and hope that this community can help me doing that.</p>
<p>So after this longish introduction, here goes: Many of us routinely use algebraic techniques in our research. Some of us study questions in abstract algebra for their own sake. However, historically, most algebraic concepts were introduced with a specific goal, which more often than not lies outside abstract algebra. Here are a few examples:</p>
<ul>
<li>Galois developed some basic notions in group theory in order to study polynomial equations. Ultimately, the concept of a normal subgroup and, by extension, the concept of a simple group was kicked off by Galois. It would never have occurred to anyone to define the notion of a simple group and to start classifying those beasts, had it not been for their use in solving polynomial equations.</li>
<li>The theory of ideals, UFDs and PIDs was developed by Kummer and Dedekind to solve Diophantine equations. Now, people study all these concepts for their own sake.</li>
<li>Cohomology was first introduced by topologists to assign discrete invariants to topological spaces. Later, geometers and number theorists started using the concept with great effect. Now, cohomology is part of what people call "commutative algebra" and it has a life of its own.</li>
</ul>
<p>The list goes on and on. The axiom underlying my question is that you don't just invent an algebraic structure and study it for its own sake, if it hasn't appeared in front of you in some "real life situation" (whatever this means). Please feel free to dispute the axiom itself.</p>
<p>Now, the actual question. Suppose that you have some algebraic concept which has proved useful somewhere. You can think of a natural generalisation, which you personally consider interesting.</p>
<blockquote>
<p>How do you decide whether a generalisation (that you find natural) of an established algebraic concept is worth studying? How often does it happen (e.g., how often has it happened to you or to your colleagues or to people you have heard of) that you undertake a study of an algebraic concept and when you try to publish your results, people wonder "so what on earth is this for?" and don't find your results interesting? How convincing does the heuristic "well, X naturally generalises Y and we all know how useful Y is" sound to you?</p>
</blockquote>
<p>Arguably, the most important motivation for studying a question in pure mathematics is curiosity. Now, you don't have to explain to your colleagues why you want to classify knots or to solve a Diophantine equation. But might you have to explain to someone, why you would want to study ideals if he doesn't know any of their applications (and if you are not interested in the applications yourself)? How do you motivate that you want to study some strange condition on some obscure groups?</p>
<p>Just to clarify this, I have absolutely no difficulties motivating myself and I know what curiosity means subjectively. But I would like to understand, how a consensus on such things is established in the mathematical community, since our understanding of this consensus ultimately reflects our choice of problems to study.</p>
<p>I could formulate this question much more widely about motivation in pure mathematics, but I would rather keep it focused on a particular area. But one broad question behind my specific one is</p>
<blockquote>
<p>How much would you subscribe to the statement that
EDIT: "studying questions for the only reason that one finds them interesting is something established mathematicians do, while younger ones are better off studying questions that they know for sure the rest of the community also finds interesting"?</p>
</blockquote>
<p>Sorry about this long post! I hope I have been able to more or less express myself. I am sure that this question is of relevance to lots of people here and I hope that it is phrased appropriately for MO.</p>
<hr>
<p>Edit: just to clarify, this question addresses the status quo and the prevalent consensus of the mathematical community on the issues concerned (if such a thing exists), rather than what you would like to be true.</p>
<hr>
<p>Edit 2: I received some excellent answers that helped me clarify the situation, for which I am very grateful! I have chosen to accept Minhyong's answer, as that's the one that comes closest to giving examples of the sort I had in mind and also convincingly addresses the more general question at the end. But I am still very grateful to everyone who took the time to think about the question and I realise that for other people who find the question relevant, another answer might be "the correct one".</p>
| Gerhard Paseman | 3,402 | <p>I think something is worth studying if it helps one of:</p>
<ul>
<li><p>solving a problem I know about,</p></li>
<li><p>giving a new perspective on something I know, or</p></li>
<li><p>raising interesting questions, some of which are easy to solve and
some of which aren't.</p></li>
</ul>
<p>Especially, I study it if it gives me some degree of gratification.
Here are a couple of examples of things that I hope to pursue after
my current interests wane:</p>
<p>Recursive clone theory: A class of functions on a set which is closed
under composition and having projections is called a clone; the notion is a part of
basic general algebra. Something that should be mentioned in basic
recursion theory classes but is not is that various definitions are
specializations of clones: primitive recursive functions, partial
recursive functions, total recursive functions. I think it would be useful to
blend the ongoing research in clone theory with a computational component
that can answer how complex a class can be.</p>
<p>Transforming Shelah's classification theory: In determining how many
inequivalent models of cardinality kappa exist for a first order theory,
Saharon Shelah came up with conditions on the theory which (loosely and
inaccurately speaking) sometimes dealt with whether a theory could
encode a particular order or a certain simpler theory. I think the
ideas can be moved into the domain of computation over finite structures.
In particular, languages that are members of some complexity class (oh, say, NP)
could be shown to satisify properties analogous to what Shelah developed
for first order theories. I think that this would be a promising route
to find a language in NP - P .</p>
<p>Granted, these are not generalizations so much as taking tools, trying
them on a new kind of widget, and then retooling the tool to work on the
widget. The justifications for working on them should be the same and
(I think) apply to your questions.</p>
<p>Gerhard "Ask Me About System Design" Paseman, 2010.09.24</p>
|
3,843,559 | <p>Can someone please give me a hint why for metric spaces we have</p>
<p><span class="math-container">$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\subset \{x|d_1(x,y)<\varepsilon\}$</span></p>
<p>I have expected the opposite:</p>
<p><span class="math-container">$d_1(x,y)<d(x,y)\Rightarrow \{x|d(x,y)<\varepsilon\}\supset \{x|d_1(x,y)<\varepsilon\}$</span></p>
| José Carlos Santos | 446,262 | <p>Because if <span class="math-container">$d(x,y)<\varepsilon$</span>, then <span class="math-container">$d_1(x,y)<d(x,y)<\varepsilon$</span>.</p>
|
2,454,455 | <p>I know this is a soft and opinion based question and I risk that this question get's closed/downvoted but I still wanted to know what other persons, who are interested in mathematics, think about my question.</p>
<p>Whenever people are talking about the most beautiful equation/identity Euler's identity is cited in this fashion:</p>
<p>$$e^{i\pi}+1=0.$$</p>
<p>While I would agree that this is a beautiful identity (see my avatar) I personally always wondered why not </p>
<p>$$e^{2i\pi}-1 = 0$$</p>
<p>is the most beautiful identity. It has $e$, $i$, $\pi$, $0$ and the number $2$ in it. I prefer it because the number $2$ is the first and at the same time the only even odd prime number. Having the prime numbers, which are in some way the atoms of mathematics, included makes this formula even more pleasant for me. The minus sign seems a little bit "negative" but the good part is that it is displaying the principle of inversion.</p>
<blockquote>
<p>So my question is, why is this not the form in which it is most often
presented?</p>
</blockquote>
| CopyPasteIt | 432,081 | <p>Looking at the answers and comments I feel compelled to offer this as a compromise,</p>
<h1><span class="math-container">$\frac {e^{i \pi} + e^{-i \pi}}{2} = -1$</span></h1>
|
1,729,893 | <p>Let $\nu_1, \nu_2, \nu_3 \in \mathbb{R}$ not all be zero.</p>
<p>I wish to show
$$\nu_1^2 + \nu_1\nu_2 + \nu_2^2+\nu_2\nu_3 + \nu_3^2 > 0\text{.}$$</p>
<p><a href="http://www.wolframalpha.com/input/?i=x%5E2%2Bx*y%2By%5E2%2By*z%2Bz%5E2+%3E+0" rel="nofollow">Wolfram</a> seems to suggest splitting this into cases, but I'm wondering if there's a shorter way to approach this. This expression seems very similar to the binomial expansion, minus the fact that we have three terms that are squared and two cross-terms occurring (rather than one cross-term multiplied by $2$).</p>
<p>As for my work, if any one of these are $0$, (I think) this is a trivial exercise. If any two of these are $0$, this is a trivial exercise (you're left with a squared non-zero term). But if all three are non-zero? Then I'm at a loss on how to pursue this, because there isn't a clean way to deal with three variables (or is there?). I've tried seeing if Wolfram could perhaps factor the above. It can't, but maybe, I thought, we could try working with
$$(\nu_1 + \nu_2 + \nu_3)^2 = (\nu_1^2 + \nu_1\nu_2 + \nu_2^2+\nu_2\nu_3 + \nu_3^2) +2\nu_1\nu_3+\nu_1\nu_2+\nu_2\nu_3$$
but there is no guarantee that this is $> 0$ either (take $\nu_3 = -(\nu_1 + \nu_2)$, for example).</p>
| Will Jagy | 10,400 | <p>$$
\frac{1}{2}
\left(
\begin{array}{ccc}
2 & 1 & 0 \\
1 & 2 & 1 \\
0 & 1 & 2
\end{array}
\right)
$$
is positive definite. If you leave off the $1/2$ factor, you have the Hessian matrix of second partial derivatives of the form. </p>
<p>Meanwhile, if we allow $x,y,z$ to take only integer values, $$ x^2 + y^2 + z^2 + yz + xy $$ takes all integer values other than $4^k (16n + 14).$ Note that we need to allow $x,y,z$ to be positive, negative, or zero to guarantee getting all the promised (positive) integers. </p>
|
3,151,143 | <p>I want to find a symmetric matrix <span class="math-container">$A$</span>, whose eigenvalues are <span class="math-container">$4$</span> and <span class="math-container">$-1$</span>. One of the eigenvectors corresponding to the eigenvalue <span class="math-container">$4$</span> is <span class="math-container">$(2,3)$</span>. I want to find an eigenvector corresponding to the eigenvalue <span class="math-container">$-1$</span> and then find the matrix <span class="math-container">$A$</span>.</p>
| gt6989b | 16,192 | <p><strong>HINT</strong></p>
<p>Pick your favorite eigenvector , say <span class="math-container">$(1, 0)^T$</span> and remember that the diagonalized form of <span class="math-container">$A$</span> looks like <span class="math-container">$A= V^{-1}DV$</span> for special <span class="math-container">$D$</span> and <span class="math-container">$V$</span>. Can you reconstruct <span class="math-container">$A$</span>?</p>
|
356,940 | <p>According to Wolfie:</p>
<p>$2^{-1} \bmod 5 = 3$</p>
<p><a href="http://www.wolframalpha.com/input/?i=2%5E-1+mod+5" rel="nofollow">http://www.wolframalpha.com/input/?i=2%5E-1+mod+5</a></p>
<p>Why is that?</p>
| Alfonso Fernandez | 54,227 | <p>Because $2\cdot3=1\pmod5$ - in this sense, $3$ is the multiplicative inverse of $2$ modulo $5$, and this is formally expressed as $2^{-1}=3\pmod5$</p>
|
147,338 | <p>I'm helping a student through a course in mathematics. In the course text, we came across the following problem concerning the Lagrange multiplier technique.
Given a differentiable function with continuous partial derivatives $f:\mathbb{R}\to\mathbb{R}:(x,y)\mapsto f(x,y)$ that has to be extremized and a constraint given by an implicit relation $g(x,y)=0$ with $g$ likewise having continuous partial derivatives. The Lagrange multiplier technique looks for points $(x^*,y^*,\lambda^*)$ such that</p>
<p>$$\nabla f(x^*,y^*)=\lambda^* \nabla g(x^*,y^*)$$</p>
<p>$\lambda$ is the so-called Lagrange multiplier.</p>
<p>The technique rests upon the fact that the gradient of the constraint is different from the zero vector: $\nabla g(x^*,y^*) \neq 0$. Then, <strong>the text says that if that condition is not fulfilled, it is possible to not have solutions of the system of equations resulting from the technique while there can actually be a constrained extremum.</strong> </p>
<p>I have tried to construct such an example, but until now unsuccessfully as every attempt I make with a $\nabla g(x^*,y^*) = 0$ turns out to have a solution. Does anyone know of a nice counterexample? </p>
<p>And does one know of a counterexample when there are $n$ variables and $m$ constraints with $n>m>1$ ? For the latter, the condition becomes that the gradients of the constraints have to be linearly independent. So the question is: can we find an example where there is no solution for the Lagrange multiplier method, but there is an extremum and the gradients of the constraints are linearly dependent? </p>
| Stefan Smith | 55,689 | <p>In the Lagrange multiplier technique, what you really need is a point $(x^*,y^*)$ where $\nabla f(x^*,y^*)$ and $\nabla g(x^*,y^*)$ are linearly dependent. If the text says otherwise, they are wrong. This is slightly different from what you wrote down. Try $g(x,y)=x^2+y^2$ and $f(x,y) = (x-1)^2=0$, which has a constrained minimum at $(1,0)$. This doesn't satisfy $n > m >1$, but it shouldn't be to hard to modify it so it does.</p>
|
1,243,750 | <p>$A \in \mathbb{R}^{n\times n}$, with $A^2 = 1$ and $A\ne\pm1$</p>
<p>Show that the only eigenvalues of $A$ are $1$ and $-1$.</p>
| Nescrio | 230,074 | <p>$$
Av = \lambda v \Rightarrow A^2v = \lambda Av = v
$$</p>
<p>So
$$
\lambda Av = v \Rightarrow \lambda \lambda v = \lambda ^2v = v
$$</p>
<p>This shows $\lambda = 1$ or $\lambda = -1$.</p>
|
587,484 | <p>Show that $\left(\displaystyle\frac{\partial (u,v)}{\partial (x,y)}\right)\left(\displaystyle\frac{\partial (x,y)}{\partial (r,s)}\right)=\displaystyle\frac{\partial (u,v)}{\partial (r,s)}$. Thus, prove that $\displaystyle\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{\displaystyle\frac{\partial (x,y)}{\partial (u,v)}}$ </p>
<hr>
<p>I have no idea on how to get started with this problem. I tried attacking it down with Jacobians, but to no avail. I'm not even sure how did they define the systems in this case.</p>
| aaaaa | 112,182 | <p>Let $$u_1=u, u_2=v, x_1=x, x_2=y, r_1=r, r_2=s$$ Here I'll use the symbol you use for the determinant for the matrix itself, that is:
$$\left(\frac{\partial (a_1, a_2)}{\partial (b_1,b_2)}\right)_{ij}:=\frac{\partial a_j}{\partial b_i}$$
Then
$$\left(\left(\frac{\partial (x,y)}{\partial (r,s)}\right)\left(\frac{\partial (u,v)}{\partial (x,y)}\right)\right)_{ij} = \sum_{k=1}^2\left(\frac{\partial x_k}{\partial r_i}\right)\left(\frac{\partial u_j}{\partial x_k}\right) = \left(\frac{\partial u_j}{\partial r_i}\right)=\left(\frac{\partial (u,v)}{\partial (r,s)}\right)_{ij}$$
so that there is an equality of the Jacobian matrices:
$$\left(\frac{\partial (x,y)}{\partial (r,s)}\right)\left(\frac{\partial (u,v)}{\partial (x,y)}\right)=\left(\frac{\partial (u,v)}{\partial (r,s)}\right)$$
Notice the order of multiplication. Our matrix could be the transposed of the one I defined, which would change the order. Nevertheless, take the deteterminant from both sides and you get the first result. Take $$r=x, s=y$$ and you get the second</p>
|
1,888,732 | <p>When I had the calculus class about the limit, one of my classmate felt confused about this limit:</p>
<blockquote>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$</span></p>
</blockquote>
<p>What he thought that since <span class="math-container">$x^2 > x$</span> and <span class="math-container">$x^2 > 3x$</span> when <span class="math-container">$x \to \infty$</span> so the first square root must be <span class="math-container">$x$</span> and same for the second. Hence, the limit must be <span class="math-container">$0$</span>.</p>
<p>It is obviously problematic.</p>
<p>And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.</p>
<p>After perfect-squaring, <span class="math-container">$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$</span> I assert that since there is a perfect square and a square root. As <span class="math-container">$x \to \infty$</span>, the constant does not matter. So</p>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$</span></p>
<p>But when I conquer this limit: <a href="https://math.stackexchange.com/questions/1888153/better-way-to-find-lim-x-to-infty-frac-sqrtx2-2x3-sqrt4x25x-6x">this post</a>.</p>
<p>Here is my argument:</p>
<blockquote>
<p><span class="math-container">$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$</span> when <span class="math-container">$x \to \infty$</span></p>
<p><span class="math-container">$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$</span></p>
</blockquote>
<p>Very concise get this answer but it gets downvoted.</p>
<p>I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:</p>
<blockquote>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$</span></p>
</blockquote>
<p>My solution is:</p>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$</span></p>
<p>So the limit becomes: <span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$</span></p>
<p><a href="http://www.wolframalpha.com/input/?i=lim%20%5B(x%5E3%2B6x%5E2%2B9x%2B1)%5E(1%2F3)-(x%5E3%2B5x%5E2%2Bx%2B1)%5E(1%2F3),%20x-%3Einfinity%5D" rel="nofollow noreferrer">This result gets verified by wolframalpha</a>.</p>
<p>To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.</p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>Let our problem be of the form $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)}$, where $p$ and $q$ are $n$th degree monic(coefficient of $x^n$ is $1$) polynomials. First of all, without loss of generality, we will rewrite $p$ and $q$ in the forms:
$$
p = (x + a)^n + \sum_{i=0}^{n-1} a_ix^i, q = (x + b)^n + \sum_{i=0}^{n-1} b_ix^i
$$ </p>
<p>where $a_i$ and $b_i$ are real constants which may be zero.(This we did in the cubic and quadratic case).</p>
<p>Now, note that $x+a \leq \sqrt[n]{(x + a)^n + \sum_{i=0}^{n-1} a_ix^i} \leq \sqrt[n]{(x + a)^n} + \sqrt[n]{\sum_{i=0}^{n-1} a_ix^i}$.Note that $$
\sqrt[n]{\sum_{i=0}^{n-1} a_ix^i} \leq \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}
$$
However, the term on the left goes to $0$ as $x \to \infty$, because $i < n, $ so $x^i \to 0$. A similar logic follows for $q$. To complete the argument,</p>
<p>However, this is better expressed by: $x+a - ((x+b) + \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) \leq \sqrt[n]{p(x)} - \sqrt[n]{q(x)} \leq (x+a +\sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) - (x+b))$.</p>
<p>On taking limits on both sides, we get $a-b$ on both sides, and using squeeze theorem, we get the result: $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)} = a-b$.</p>
|
1,888,732 | <p>When I had the calculus class about the limit, one of my classmate felt confused about this limit:</p>
<blockquote>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$</span></p>
</blockquote>
<p>What he thought that since <span class="math-container">$x^2 > x$</span> and <span class="math-container">$x^2 > 3x$</span> when <span class="math-container">$x \to \infty$</span> so the first square root must be <span class="math-container">$x$</span> and same for the second. Hence, the limit must be <span class="math-container">$0$</span>.</p>
<p>It is obviously problematic.</p>
<p>And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.</p>
<p>After perfect-squaring, <span class="math-container">$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$</span> I assert that since there is a perfect square and a square root. As <span class="math-container">$x \to \infty$</span>, the constant does not matter. So</p>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$</span></p>
<p>But when I conquer this limit: <a href="https://math.stackexchange.com/questions/1888153/better-way-to-find-lim-x-to-infty-frac-sqrtx2-2x3-sqrt4x25x-6x">this post</a>.</p>
<p>Here is my argument:</p>
<blockquote>
<p><span class="math-container">$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$</span> when <span class="math-container">$x \to \infty$</span></p>
<p><span class="math-container">$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$</span></p>
</blockquote>
<p>Very concise get this answer but it gets downvoted.</p>
<p>I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:</p>
<blockquote>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$</span></p>
</blockquote>
<p>My solution is:</p>
<p><span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$</span></p>
<p>So the limit becomes: <span class="math-container">$$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$</span></p>
<p><a href="http://www.wolframalpha.com/input/?i=lim%20%5B(x%5E3%2B6x%5E2%2B9x%2B1)%5E(1%2F3)-(x%5E3%2B5x%5E2%2Bx%2B1)%5E(1%2F3),%20x-%3Einfinity%5D" rel="nofollow noreferrer">This result gets verified by wolframalpha</a>.</p>
<p>To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.</p>
| Claude Leibovici | 82,404 | <p>Consider the general case $$A=\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}=x\left(\sqrt{1+\frac a x+\frac b {x^2}}-\sqrt{1+\frac c x+\frac d {x^2}}\right)$$ and use the fact that, for small $y$, using the generalized binomial theorem or Taylor series, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ For the first radical, replace $y$ by $\frac a x+\frac b {x^2}$ and by $\frac c x+\frac d {x^2}$ for the second radical. You then obtain $$A=x\left(1+\frac{a}{2
x}+\frac{\frac{b}{2}-\frac{a^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)-\left(1+\frac{c}{2
x}+\frac{\frac{d}{2}-\frac{c^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)\right)\right)$$ $$A=\frac{a-c}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.</p>
<p>Doing the same with $$B=\sqrt[3]{x^3+ax^2+bx+c}-\sqrt[3]{x^3+dx^2+ex+f}$$ and using $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ you should arrive to $$B=\frac{a-d}{3}+\frac{-a^2+3 b+d^2-3 e}{9 x}+O\left(\frac{1}{x^2}\right)$$
Doing the same with $$C=\sqrt[p]{x^p+a_1x^{p-1}+a_2x^{p-2}+\cdots}-\sqrt[p]{x^p+b_1x^{p-1}+b_2x^{p-2}+\cdots}$$ the limit will just be $$\frac{a_1-b_1} p$$</p>
<p>Your strategy works well for the limit because you just ignore the terms of degrees lower than $p-1$.</p>
|
2,665,539 | <p>How to solve the following system of equations</p>
<blockquote>
<p>$$
\begin{cases}
a+c=12\\
b+ac+d=86\\
bc+ad=300\\
bd=625\\
\end{cases}
$$</p>
</blockquote>
| Community | -1 | <p>Well, $b=\cfrac {625}{d}$ from your last equation, and $a=-c+12$ from your first equation.</p>
<p>Substitute $b$ and $a$ into your second equation to get $\cfrac{625}{d}+(-c+12)(c)+d=86$</p>
<p>Do the same for your third equation.</p>
|
2,348,290 | <p>The question and its answer is given in the following pictures:<a href="https://i.stack.imgur.com/xYhqD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xYhqD.png" alt="enter image description here"></a></p>
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<p>The first line in the solution is not clear for me, my questions are:</p>
<p>1- where is the interval from 0 to 1 ?</p>
<p>2- why we changed the x inside the floor function to n and did not change the x in the power of e ?</p>
<p>Could anyone illustrate this for me please? </p>
| J.G. | 56,861 | <p>In answer to your first question, on that interval the integrand vanishes, so there is no contribution to the integral. In answer to your second question, on each of the length-$1$ intervals considered the floor factor is constant but $e^{-x}$ is not.</p>
|
2,348,290 | <p>The question and its answer is given in the following pictures:<a href="https://i.stack.imgur.com/xYhqD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xYhqD.png" alt="enter image description here"></a></p>
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<p>The first line in the solution is not clear for me, my questions are:</p>
<p>1- where is the interval from 0 to 1 ?</p>
<p>2- why we changed the x inside the floor function to n and did not change the x in the power of e ?</p>
<p>Could anyone illustrate this for me please? </p>
| stity | 285,341 | <p>$$ \forall x \in [n, n+1[, \lfloor x \rfloor = n$$</p>
<p>and $$\int_0^1{\lfloor x \rfloor e^{-x}dx} = \int_0^1{0 e^{-x}dx} =0$$</p>
|
1,241,639 | <p>How do I prove that $\lim_{(x,y)\to (0,0)} \frac{x^2 y }{x^2 + y^2} = 0$?</p>
<p>I can prove this by notifying $x=rcos\theta$ and $y=rsin\theta$, but I remember that it could also be proven by squeeze theorem.</p>
<p>How do I prove this using squeeze theorem?</p>
| agha | 118,032 | <p>By AM-GM inequality we have:</p>
<p>$$2|xy|\leq x^2+y^2$$</p>
<p>So:</p>
<p>$$0<\frac{|xy|}{x^2+y^2}\leq\frac{1}{2}$$</p>
<p>Multiply both sides by $|x|$:</p>
<p>$$0\leq \frac{|x^2y|}{x^2+y^2} \leq \frac{|x|}{2}$$</p>
<p>Now $\frac{|x|}{2}$ goes to zero when $(x,y) \to 0$, so $\frac{|x^2y|}{x^2+y^2}$ also goes to zero.</p>
|
1,090,974 | <p>We have two functions of time $f(t)$ and $g(t)$, for which convolution and correlation are defined as following:</p>
<p>Convolution: $(f(t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f(t)g(\tau-t)dt}$</p>
<p>Correlation: $(f(t)\star g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(t)g(\tau+t)dt}$</p>
<p>In the english wikipedia and in other sources I found that the following relationship should hold:</p>
<p>$(f(t)\star g(t))(\tau) = (f^\ast(-t)\ast g(t))(\tau)$</p>
<p>Is this correct? If so, how can i prove this? Usually, i would try substitution, but how to change the $g(\tau+t)$ to $g(\tau-t)$?</p>
| Timitry | 157,781 | <p>I figured out the answer while writing down the question. Here it is:</p>
<p>$(f^\ast(-t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(-t)g(\tau-t)dt} = -\int_{\infty}^{-\infty}{f^\ast(t)g(\tau+t)dt} = \int_{-\infty}^{\infty}{f^\ast(t)g(\tau+t)dt} = (f(t)\star g(t))(\tau)$</p>
<p>In the second step, the substitution $t\to -t$ took place.</p>
|
341,621 | <p>Is there an inequality such as
$$(a+b)^2 \leq 2(a^2 + b^2)$$
for higher powers of $k$
$$(a+b)^k \leq C(a^k + b^k)?$$</p>
| Samuel | 405 | <p>The <a href="http://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality">generalized mean inequality</a> states that
$$\dfrac{a+b}2\leq \left(\dfrac{a^k+b^k}2\right)^{1/k},$$
with equality if and only if $a=b$, from which it follows that
$$(a+b)^k\leq 2^{k-1}(a^k+b^k),$$
with equality if and only if $a=b$. Thus $C=2^{k-1}$ works and no smaller $C$ works.</p>
|
2,941,579 | <p>In Taylor's series, to determine the number of terms needed to obtain the desired accuracy, sometimes one needs to solve inequalities of the form <span class="math-container">$$\frac{a^n}{n!}<b,$$</span>
where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are fixed positive numbers. In most textbooks in calculus, the only introduced method to solve <span class="math-container">$\frac{a^n}{n!}<b$</span> for <span class="math-container">$n$</span> is trial and error. While this method works well in many cases, I feel that it is inefficient when <span class="math-container">$a$</span> is large and <span class="math-container">$b$</span> is small. (For example, how about solving <span class="math-container">$\frac{1000^n}{n!}<0.01$</span>?) </p>
<p>My Question: Apart from using brutal force, is there another method to solve the inequality <span class="math-container">$\frac{a^n}{n!}<b$</span> for <span class="math-container">$n$</span>?</p>
| Andrei | 331,661 | <p>Because you started with <span class="math-container">$$0\lt|x-3|\le\delta$$</span>
Multiply <strong>every</strong> term in this inequality by <span class="math-container">$4$</span> you get <span class="math-container">$$4\cdot 0\lt4\cdot |x-3|\le4\cdot\delta$$</span>
or, just using the last two terms <span class="math-container">$$|4x-12|\lt 4\delta$$</span></p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| Community | -1 | <p>The answer is $42$.</p>
<p>$69$ is also the answer.</p>
<p>"Purple feelings" is also an answer.</p>
<p>The truth of each of these is, of course, vacuous. :)</p>
<hr>
<p>If the question is posed as something other than multiplication, then it is the fault of the questioner for miscommunicating.</p>
<p>Although, one could arguably blame the person trying to solve this problem for not doing enough to extract enough requirements from the 'customer' to be able to provide a solution. In some settings, this is an <em>extremely</em> important skill.</p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| hrkrshnn | 63,095 | <p>$$p(x)=$$</p>
<p>$$-\frac{1486263915627335609976345925580307452480}{198824918770116952269605821139049374259}-\frac{23535858736574459335924875719051524464677 x}{1789424268931052570426452390251444368331}+\frac{1532186339457747628597246965489647712097745599 x^2}{742539494635629574624160683858739355082631760}-\frac{5300973178829466500668773673899060773511329723 x^3}{62373317549392884268429497444134105826941067840}+\frac{425139989729581169917246837619141657974952401 x^4}{374239905296357305610576984664804634961646407040}-\frac{15160892592292573821061148160317799661783 x^5}{7128379148502043916391942565043897808793264896}+\frac{2379833487879115598578638026951579913181 x^6}{1496959621185429222442307938659218539846585628160}-\frac{133849478325585275186149006837381343 x^7}{249493270197571537073717989776536423307764271360}+\frac{9291465647310545015926219743101 x^8}{136087238289584474767482539878110776349689602560}$$</p>
<p>Then<br>
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
x & 12\color{grey}{(3\times 4)} & 20\color{grey}{(4\times 5)} & 30\color{grey}{(5\times 6)} &42\color{grey}{(6\times 7)}&56\color{grey}{(7\times 8)}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016}\\ \hline
p(x)& 8 & 50 & 30 &49&\color{red}{224}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016} \\
\hline
\end{array}$$</p>
<p>To learn to play this 'game', <a href="https://math.stackexchange.com/questions/819101/how-to-solve-this-sequence-165-195-255-285-345-x/819382#819382">read me.</a></p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| gnasher729 | 137,175 | <p>The answer stares you right in the face.</p>
<p>7 x 8 is a question mark.</p>
<p>Now I should add that one moderator apparently believed this was a joke answer. It should be obvious that it isn't. If I wanted to make a joke, I would have added a comment. Mathematics is about the manipulation of symbols, and this is an example of symbol manipulation creating a riddle with the answer hidden in plain sight. </p>
<p>The riddle equates various symbols resembling products with other symbols in a rather pointless way. The question of the riddle is what the last symbol "7 x 8" equates. It obviously is meant to equate whatever symbol is to the right of the "=" sign. </p>
<p>There is one answer here by maddog2k that I would consider better (that 7 x 8 = 56, since we shouldn't care about all the wrong answers given in the riddle but just give the correct answer). </p>
|
866,921 | <p>On the <a href="http://chat.stackexchange.com/rooms/36/mathematics">Mathematics chat</a> we were recently talking about the following problem <a href="https://math.stackexchange.com/users/32016/chriss-sis">@Chris'ssis</a> had to solve during an interview :</p>
<p>$$3\times 4=8$$
$$4\times 5=50$$
$$5\times 6=30$$
$$6\times 7=49$$
$$7\times 8=?$$</p>
<p>We have not managed to solve it so far, all we know is the solution (which was given <strong>after</strong> we had given up) :</p>
<blockquote class="spoiler">
<p> $224$</p>
</blockquote>
<p>How do we find this solution ?</p>
| Kal S. | 117,757 | <p>We rewrite the riddle using prime decomposition.</p>
<p><strong>Notation:</strong> <span class="math-container">$2^{x_1}3^{x_2}5^{x_3}7^{x_4}\leftrightarrow (x_1,x_2,x_3,x_4)$</span>
We have:
<span class="math-container">$$
(a_1,a_2,a_3,a_4) \odot (b_1,b_2,b_3,b_4) = (c_1,c_2,c_3,c_4)\\
(0,1,0,0) \odot (2,0,0,0) = (3,0,0,0)\\
(2,0,0,0) \odot (0,0,1,0) = (1,0,2,0)\\
(0,0,1,0) \odot (1,1,0,0) = (1,1,1,0)\\
(1,1,0,0) \odot (0,0,0,1) = (0,0,0,2)\\
(0,0,0,1) \odot (3,0,0,0) = ?
$$</span>
We note that the following definition of <span class="math-container">$(c_1,c_2,c_3,c_4)$</span> satisfy the given relations:
<span class="math-container">$$
c_1=a_2+a_3+b_1+b_3-b_2-b_4+2a_4\\
c_2=a_3\\
c_3=a_1+b_2-b_4\\
c_4=a_4+2b_4\\
$$</span></p>
<p>We then calculate:
<span class="math-container">$$
(0,0,0,1) \odot (3,0,0,0)=(5,0,0,1) \leftrightarrow 2^57=224
$$</span></p>
<p>We note though that <span class="math-container">$a_4$</span> can appear anywhere and satisfy the given relations namely the riddle can have multiple answers. Knowing the answer is <span class="math-container">$224$</span> we can include <span class="math-container">$a_4$</span> in the defining relation of <span class="math-container">$c_1,c_4$</span>.</p>
|
2,125,136 | <p>I understand the usual motivation behind the truth table for the logical connective <span class="math-container">$\to$</span>.</p>
<p>However, I would like to know if there is a more fundamental reason for that truth table. Something that would have to do with arguments and validity.</p>
<p>A.G.Hamilton writes in <em>Logic for Mathematicians</em> that "the significance of the conditional statement <span class="math-container">$A\to B$</span> is that its truth enables the truth of <span class="math-container">$B$</span> to be inferred from the truth of <span class="math-container">$A$</span>, and nothing in particular to be inferred from the falsity of <span class="math-container">$A$</span>".</p>
<blockquote>
<p><strong>Question:</strong> What does Hamilton mean precisely? Something like <span class="math-container">$\to$</span> is the only binary truth function such that</p>
<ul>
<li><p>The argument form <span class="math-container">$(p\to q),p;\therefore q$</span> is valid</p>
</li>
<li><p>The argument form <span class="math-container">$(p\to q),\sim p;\mathcal{A}$</span> is invalid unless <span class="math-container">$\mathcal{A}$</span> is a statement form logically equivalent to <span class="math-container">$\sim p$</span> ?</p>
</li>
</ul>
</blockquote>
| Asinomás | 33,907 | <p>assuming you counted correctly:</p>
<p>The probability that he misses the first attempt is $\dfrac{4319}{4320}$, the probability he misses the second is $ \dfrac{4318}{4319}$ and so on.</p>
<p>The probability he doesn't guess is the product of all these fractions, almost everything cancels and you get $\dfrac{4310}{4320}$. Therefore the probability he guesses is $\dfrac{10}{4320}=\dfrac{1}{432}$</p>
|
163,043 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/160847/polynomials-irreducible-over-mathbbq-but-reducible-over-mathbbf-p-for">Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$</a> </p>
</blockquote>
<p>Anyone know an $f\in \mathbb{Z}[x]$ that is irreducible over $\mathbb{Q}$ but whose reduction mod $p$ is reducible over the first three positive primes?</p>
| KReiser | 21,412 | <p>Yes. The polynomial $x^4+1$ does the trick. </p>
|
163,043 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/160847/polynomials-irreducible-over-mathbbq-but-reducible-over-mathbbf-p-for">Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$</a> </p>
</blockquote>
<p>Anyone know an $f\in \mathbb{Z}[x]$ that is irreducible over $\mathbb{Q}$ but whose reduction mod $p$ is reducible over the first three positive primes?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\rm\ \ x^2 +\, 2\cdot3\cdot 5 $</p>
|
63,641 | <p>Here is the problem: Two mathematicians meet at a bar. They like each other and tend to collaborate. But it is not so clear what problems could be of common interest to both of them. Of course, the traditional way is they keep describing they work or their field in general so that hopefully they catch something at the end. But is there any reference, graph, table or whatever that they can use to help them? This, of course, makes sense only when such a reference keeps updated based on the continuous production in mathematics.</p>
| Tim Dokchitser | 3,132 | <p>No, such an isomorphism does not always exist, and the smallest counterexample is $G=C_5\rtimes C_4$ with $C_4$ acting faithfully. It is not hard to see that the only automorphisms of $G$ are inner, and that they cannot map an element of order 4 to its inverse.</p>
|
859,493 | <p>Given that any fixed integer $n>0$, let $S=\{1,2,3,4,...,n\}$. Now a Red-Blue subset of $S$ is called $T$. Every element of $T$ is given a colour (either red or blue). For instance $\{17 (\text{red})\}, \{1 (\text{red}), 5 (\text{red})\}$ and $\{1(\text{red}),5(\text{blue})\}$ are $3$ different subsets of $S$.</p>
<p>Determine the number of different RB-coloured subsets of $S$.</p>
<p>First thing I did was look at the problem without the colour restrictions to find out how many different subsets I can have. The number of different I calculated was $2^n - 1$. The $-1$ is from subtracting the null set. </p>
<p>I'm not sure how to number of possibilities when the colour restrictions are introduced.</p>
<p>Also how would I go about determining the number of different subsets of size $i$, where $0<i<n$?</p>
| kingW3 | 130,953 | <p>By the binomial theorem $$(2x^2+(x-3))^8=\sum_{k=0}^8{8\choose k}2^{8-k}x^{16-2k}(x-3)^{k}$$
for $k<8$ term power of $x$ is bigger than $1$,so $k=8$ than it simplifies to find coefficient of $x$ in $(x-3)^8$ which is $8\cdot(-3)^7=-17496$</p>
|
1,920,781 | <p>Let's say I have a basic equation (here a linear combination):
$$ax_1 + bx_2=y$$</p>
<p>Is there any way to get the value of the sum $x_1+x_2$ if $a$, $b$ and $y$ is known?</p>
| Bobson Dugnutt | 259,085 | <p>If there are no further restrictions on $x_1$ and $x_2$, then we cannot <em>in general</em> (see <em>Lacklub</em>s comment below) say anything about $x_1+x_2$, as the system is <a href="https://en.wikipedia.org/wiki/Underdetermined_system" rel="nofollow">underdetermined</a>, i.e., there are more variables than equations. </p>
|
3,609,791 | <p><span class="math-container">$f(x) = \frac{x}{(1+x^2)}$</span></p>
<p>I need to find the range of function.</p>
<p>My method-</p>
<p>Let <span class="math-container">$f(x)=y$</span> </p>
<p>Then <span class="math-container">$x²y-x+y=0$</span></p>
<p><span class="math-container">$$x=1\pm\frac{\sqrt{(1-4y²)}}{2y}$$</span></p>
<p>For <span class="math-container">$x$</span> to be real , </p>
<p><span class="math-container">$1-4y^2\ge0$</span> <strong>and</strong> <span class="math-container">$y\neq 0$</span></p>
<p><span class="math-container">$(2y+1)(2y-1)\ge0$</span> <strong>and</strong> <span class="math-container">$y\neq0$</span></p>
<p>Hence <span class="math-container">$y \in [-1/2,1/2] -\{0\}$</span></p>
<p>So far so good.</p>
<p>But if I put <span class="math-container">$0$</span> in the function,</p>
<p>Then <span class="math-container">$f(0)= 0/1+0 =0$</span></p>
<p>While my solution says that <span class="math-container">$y$</span> cannot be zero. </p>
<p>Where am I going wrong?</p>
| joeb | 362,915 | <p>Short answer ...</p>
<p>In your efforts to describe the solution set of the statement <span class="math-container">$$y = x/(1+x^2),$$</span> you manipulated algebraically, and arrived at the statement <span class="math-container">$$x = (1 + \sqrt{1-4y^2})/2y \qquad \text{or} \qquad x = (1 + \sqrt{1-4y^2})/2y,$$</span> and focused your attention on describing its solution set, which appeared to be the same as that of the first statement. It is <em>almost</em> the same - when you manipulated algebraically you altered the original solution set by dropping the point <span class="math-container">$(0,0)$</span>.</p>
|
2,407,700 | <p>If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.</p>
<p>This equation becomes $x^4+3x^3-26x^2+3x+1=0$ which has four solutions. One of the solutions is $2+\sqrt3$ which has the form $a+\sqrt b$. So, $a+b=2+3=5$. How does this look?</p>
| GAVD | 255,061 | <p>Let $t = x + \frac{1}{x}$, then the equation is $$t^2 - 2 + 3t = 26,$$ or $$t^2 + 3t -28 = 0.$$</p>
<p>Then, we have $t = 4$ or $t = -7$. </p>
<p>If $t=4$, then $x + \frac{1}{x} = 4$, then $x = 2 \pm \sqrt{3}$.</p>
<p>If $t=-7$, then $x + \frac{1}{x} = -7$, then $x = \frac{1}{2}(-7\pm \sqrt{45})$.</p>
<p>Because $a$ and $b$ are positive integers, then $x = 2 + \sqrt{3}$. And yes, $a+b=5$.</p>
|
2,407,700 | <p>If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.</p>
<p>This equation becomes $x^4+3x^3-26x^2+3x+1=0$ which has four solutions. One of the solutions is $2+\sqrt3$ which has the form $a+\sqrt b$. So, $a+b=2+3=5$. How does this look?</p>
| Davood | 477,916 | <p>We can factor this equation as follows:
$$x^4+3x^3-26x^2+3x+1 = (x^2-4x+1) (x^2+7x+1).$$</p>
<hr>
<p>One can check that $2\pm \sqrt3$ are solutions to the $(x^2-4x+1)$,<br>
alse we can check that $\dfrac{-7\pm 3\sqrt5}{2}$ are solutions to the $(x^2+7x+1)$.</p>
|
65,892 | <p>Hello!</p>
<p>Let $M$ be an almost complex manifold. Let $TM$ denote its tangent bundle. Then we have the decomposition $TM\otimes\mathbb{C}=T^{1,0}M\oplus T^{0,1}M$ corresponding to the eigenvalues of the almost complex structure. This decomposition yields the decomposition:
$$
\Lambda^r(T^\star M\otimes\mathbb{C})=\Lambda^r(T^{1,0}M^\star\oplus T^{0,1}M^\star)=\bigoplus_{p+q=r}\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star)
$$
Now take a section $\omega$ of the complex vector bundle
$$
\Lambda^{p,q}:=\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star)
$$
$\omega$ is called a complex differential form of type $(p,q)$. Consider a complex $(p,q)$-form $\omega$ and take its differential. Its differential $\mathrm{d}\omega$ is a section of:
$$
\Lambda^{p+q+1}(T^\star M\otimes\mathbb{C})=\bigoplus_{m+n=p+q+1}\Lambda^{m,n}
$$
Therefore $\mathrm{d}\omega$ can be decomposed in a sum of complex differential forms of type $(m,n)$ with $m+n=p+q+1$. However I have read that there are only four terms. My second question is:</p>
<p><em>How do we prove that in fact $\mathrm{d}\omega$ is a section of: $$\Lambda^{p+2,q-1}\oplus\Lambda^{p+1,q}\oplus\Lambda^{p,q+1}\oplus\Lambda^{p-1,q+2}$$ only?</em></p>
<p>I am aware that in the case where the almost complex structure is integrable we get only two terms such that finally we have $\mathrm{d}=\partial+\bar{\partial}$. But in fact it seems that in the almost complex case already we do not have so many terms (namely we have only 4 as above). I think this has something to do with the graduation of the algebra of differential forms and the nilpotence of the differential itself but I am not able to prove it.</p>
<p>At last, since I am interesting in the same kind of question concerning Lie and Courant algebroids, I was wondering if this fact could be recast in the language of homotopical algebras (by which I vaguely mean that usual identities on brackets hold up to something else)? This is because the algebra of differential forms is a supercommutative algebra and that we can reformulate $\mathrm{d}^2=0$ by $[\mathrm{d},\mathrm{d}]=0$. Could somebody point me toward an article?</p>
<p>Thank you very much!</p>
| diverietti | 9,871 | <p>Call $C^{\infty}_{p,q}(M)$ the space of smooth complex sections of the bundle $\Lambda^{p,q}T^*_M$ and let $2n$ be the real dimension of $M$. </p>
<p>The fact that
$$
dC^{\infty}_{p,q}(M)\subset C^{\infty}_{p+2,q-1}(M)+C^{\infty}_{p+1,q}(M)+C^{\infty}_{p,q+1}(M)+C^{\infty}_{p-1,q+2}(M)
$$
follows immediately from the two following facts:</p>
<ol>
<li>The (bigraded) algebra $C^{\infty}_{\bullet,\bullet}(M)=\bigoplus_{p,q=0}^n C^{\infty}_{p,q}(M)$ is locally generated by $C^{\infty}_{0,0}(M)$, $C^{\infty}_{1,0}(M)$ and $C^{\infty}_{0,1}(M)$;</li>
<li>There are (obvious) inclusions
$$
dC^{\infty}_{0,0}(M)\subset C^{\infty}_{1,0}(M)+C^{\infty}_{0,1}(M),
$$
$$
dC^{\infty}_{1,0}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M),
$$
$$
dC^{\infty}_{0,1}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M).
$$</li>
</ol>
<p>Moreover, for an almost complex manifold $M$ with complex structure $J$, the following facts are equivalent:</p>
<ul>
<li>$J$ has no torsion (and thus, by Newlander-Nirenberg theorem $J$ is a true complex structure and $M$ a complex analytic manifold);</li>
<li>$dC^{\infty}_{1,0}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)$
and $dC^{\infty}_{0,1}(M)\subset C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M)$;</li>
<li>$dC^{\infty}_{p,q}(M)\subset C^{\infty}_{p+1,q}(M)+C^{\infty}_{p,q+1}(M)$ for all $p,q=0,1,\dots,n$.</li>
</ul>
|
3,636,040 | <p>I am reading a paper titled "Cross-Calibration for Data Fusion of EO-1/Hyperion and Terra/ASTER" which mentions a particular minimization problem
<img src="https://i.stack.imgur.com/UgZrB.png" alt="Frobenius norm of y(I)-r(I)x"></p>
<p>I am not able to comprehend how this equation is converted (during implementation, in the code) to a quadratic optimization problem specified in the form of <img src="https://i.stack.imgur.com/x0eWT.png" alt="generic form of quadratic optimization equation"> and how would the matrices <span class="math-container">$H$</span> and <span class="math-container">$f$</span> be computed. </p>
<p>I know the question is a bit ambiguous and might require more information. But I'm just looking for some guidance as to how to begin with the conversion between the two forms (Frobenius norm and Quadratic optimization)</p>
| dohmatob | 168,758 | <p><strong>Note.</strong> For ease of writing, everything is transposed in my answer, but should change anything in the implementation...</p>
<hr>
<p>So you have <span class="math-container">$L_m$</span> constrained problems of minimization of Frobenius norm. The objective function of the <span class="math-container">$i$</span>th problem can be written as <span class="math-container">$
\|y_i - Xr_i\|^2 = \|y_i\|^2 - 2y_i^TXr_i + \|Xr_i\|^2$</span>. Discarding the constant term <span class="math-container">$\|y_i\|^2$</span> (it doesn't depend on the optimization variable <span class="math-container">$r_i$</span>), we can further rewrite the objective as <span class="math-container">$\frac{1}{2}r_i^T H r_i + f^Tr_i$</span>, where <span class="math-container">$H := 2X^TX$</span> and <span class="math-container">$f_i := -2y_i^TX$</span>.</p>
<p>As for the constraint <span class="math-container">$\|r_i - r_i^*\|_\infty \le \epsilon r_i^*$</span>, well, as you've been told in (15), they can be rewritten as <span class="math-container">$Ar_i \le b$</span>, where <span class="math-container">$A :=\begin{bmatrix}-I\\I\end{bmatrix}$</span> and <span class="math-container">$b_i=\begin{bmatrix}-(1-\epsilon)r_i^*\\(1+\epsilon)r^*_i\end{bmatrix}$</span>.</p>
<p>Thus, all in all, your <span class="math-container">$L_m$</span> problems are QP problems of the form</p>
<p><span class="math-container">$$
\text{minimize}\; \frac{1}{2}r_i^T H r_i + f_i^Tr_i\text{ subject to }Ar_i \le b_i,\; r_i \ge 0, \tag{$P_i$}
$$</span></p>
<p>with the <span class="math-container">$H$</span>, <span class="math-container">$f_i$</span>, <span class="math-container">$A$</span>, and <span class="math-container">$b_i$</span> as given above.</p>
|
2,706,061 | <p>This can be rewritten as $A\vec{x}=0$, where $A=(1, -2, 3), \vec{x}=(x, y, z)^T$.
I understand that one can find vectors perpendicular to this place by finding the basis for the null space, as the null space is perpendicular to the row space.</p>
<p>This basis is $(2, 1, 0), (-3, 0, 1)$. I believe that the space describing all perpendicular vectors can be written as $a(2, 1, 0)^T+b (-3, 0, 1)^T$? In order to find the basis for this space, we solve for the null space of
$$
B=
\left[ {\begin{array}{cc}
2, 1, 0 \\
3, 0, 1 \\
\end{array} } \right]
$$</p>
<p>This gives you $(1, -2, 3)^T$ which makes sense as this described the original plane. But I don't understand why we combine the two solutions for the null space and transpose them to find the basis for all vectors $a(2, 1, 0)^T+b (-3, 0, 1)^T$. Could somebody explain this to me? Thank you.</p>
<p>Note: <a href="https://math.stackexchange.com/questions/2681592/find-a-basis-for-all-vectors-perpendicular-to-x-2y3z-0">This question</a> is about the same problem but has a different misunderstanding.</p>
| Tal-Botvinnik | 331,471 | <p>This is because of linearity of the dot product. First, write your $A$ matrix as a vector $\vec y$, it will be simpler in a second, and note that what you mean by $A\vec x$ is nothing more than $\vec y\cdot \vec x$. I.e. your plane consists in all vectors perpendicular to the vector $y$. I guess you know this but I'm just settling notation.</p>
<p>Now, you have found two vectors $\vec u,\vec v$ such that $\vec y\cdot\vec u=0$ and $\vec y\cdot \vec v=0$. This means you can write a whole family of other vectors verifying the same. For instance, $2\vec u$ is also perpendicular to $y$. Now, go further and take any two real or complex values $a,b$. Then we have
$$\vec y \cdot (a\vec u+b\vec v)= a\vec y\cdot\vec u+b\vec y\cdot\vec v= 0.$$</p>
<p>Actually this family of vectors (one for each pair $(a,b)$) contains all vectors in the plane. To see why this happens, see @José Carlos Santos answer.</p>
|
1,621,989 | <p>If we have any two co-prime positive integers <em>x</em> and <em>y</em>, does there always exist a positive integer <em>C</em> such that all integers greater than <em>C</em> can be expressed as <em>Ax</em>+<em>By</em> where <em>A</em> and <em>B</em> are also non-negative integers?</p>
<p>Do we have a formula to calculate the largest non-expressable integer (i.e. <em>C</em>-<em>1</em>) in such a case?</p>
<p>EDIT: <em>A</em> and <em>B</em> are non-negative, not necessarily positive. Either one of them can be 0.</p>
| David | 119,775 | <p>I'm going to use more conventional notation otherwise my head will explode ;-)</p>
<p><strong>Theorem</strong>. Let $a,b$ be coprime positive integers, let $c$ be an integer, and consider the equation
$$ax+by=c\ .$$</p>
<ul>
<li>If $c=ab$ then the equation has no solutions in positive integers $x,y$.</li>
<li>If $c>ab$ then the equation has solutions in positive integers $x,y$.</li>
</ul>
<p><strong>Proof</strong>. For the first result we consider
$$ax+by=ab\ ,$$
and suppose that $x,y$ are positive integers. From this equation we have
$$b\mid ax\quad\hbox{and}\quad a\mid by\ ,$$
and since $a,b$ are coprime this gives
$$b\mid x\quad\hbox{and}\quad a\mid y\ .$$
Since $x,y$ are positive we have $x\ge b$, $y\ge a$ and so $ax+by\ge2ab$, which contradicts the given equation.</p>
<p>For the second, consider
$$ax+by=c\ ,$$
where $c>ab$. The general integer solution of this is
$$x=x_0+bt\ ,\quad y=y_0-at\ ,\quad t\in{\Bbb Z}\ ,$$
where $(x_0,y_0)$ is a specific solution, that is, $ax_0+by_0=c$. There will be a positive solution if we can find an integer $t$ such that
$$-\frac{x_0}b<t<\frac{y_0}a\ .$$
But the interval from $-x_0/b$ to $y_0/a$ has length
$$\frac{y_0}a-\Bigl(-\frac{x_0}b\Bigr)=\frac{ax_0+by_0}{ab}=\frac{c}{ab}>1\ ,$$
so it must contain an integer. This completes the proof.</p>
|
470,081 | <p>There is a question from an old topology prelim that is somewhat giving me a hard time. Consider the cylinder $X= S^1 \times [-1,1]$. Now we define an equivalence relation $\sim$ as follows: For points $v,v' \in S^1$, we have $(v,-1) \sim (v',-1)$ and $(v,1) \sim (v',1)$. I am asked to show that the quotient space $X^{*}= S^1 \times [-1,1]/\sim$ is homeomorphic to the unit sphere $S^2$. The problem is I can't off the top of my head come up with a decent continuous bijection from the quotient space onto $S^2$. What might work here? </p>
<p>Suppose I had some sort of continuous bijection $h: X^{*} \rightarrow S^2$. Now the quotient map $p: X \rightarrow X^{*}$ is continuous and surjective, and since $X$ is compact, so is $X^{*}$. We also know that $S^2$ being a topological manifold is Hausdorff. Recall that if there is a continuous bijection between the compact space $X^{*}$ (any compact space for that matter) and the Hausdorff space $S^2$ (or any Hausdorff space), then that continuous bijection is a homeomorphism. This is what I intended to do, but I still can't come up with such a continuous bijection. Also, perhaps I am a bit confused in trying to visualize the quotient space. I would really appreciate some input on this, and any ideas that may prove useful.</p>
| Stefan Hamcke | 41,672 | <p>Try to find for a point $((a,b),z)\in S^1\times [-1,1]$ a point $(x,y,z)$ in $S^2$ with the same $z$-coordinate and $x=\lambda a,\ y=\lambda b$ for a $\lambda$ which is a function in $z$.</p>
<p>Then show that the map $f:((a,b),z)\mapsto (x,y,z)$ is surjective and continuous, and that it induces a map $\tilde f:X^*\to S^2$. Is $\tilde f$ injective?</p>
|
4,453,356 | <p>I had been having trouble understanding a proof of the irrational nature of √2.</p>
<p>I found this proof in the first page of the foreward to <a href="https://www.cs.ru.nl/%7Efreek/comparison/comparison.pdf" rel="nofollow noreferrer"><em>17 theorem provers of the world</em></a> where a 'geometrical proof' (is that a right term?) of the irrationality of √2 is mentioned (page 2 of the pdf).</p>
<p>It goes like this:</p>
<blockquote>
<p>Call the original triangle ABC, with the right angle at C. Let the hypothenuse AB = p, and let the legs AC = BC = q. As remarked, p² = 2q².</p>
<p>Reflect ABC around AC obtaining the congruent copy ADC. On AB position E so that BE = q. Thus AE = p − q. On CD position F so that BF = p. Thus DF = 2q − p. The triangle BF E is congruent to the original triangle ABC. EF is perpendicular to AB, the lines EF and AD are parallel.</p>
<p>Now, position G on AD so that AG = EF = q. Since AEFG is a rectangle, we find AG = q. Thus, DG = F G = AE = p − q. So, the triangle DFG is an isosceles right triangle with a leg = p − q and hypothenuse = 2q − p.</p>
<p>If there were commensurability of p and q, we could find an example with integer lengths of sides and with the perimeter p + 2q a minimum. But we just constructed another example with a smaller perimeter p, where the sides are also obviously integers. Thus, assuming commensurability leads to a contradiction.</p>
<p><a href="https://i.stack.imgur.com/nEkgA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nEkgA.png" alt="enter image description here" /></a></p>
</blockquote>
<p>What I understood was, we assume that we have got the smallest possible isosceles right triangle: ABC.</p>
<p>It has hypotenuse p and leg q, where p,q ∈ ℤ.</p>
<p>ie, p and q are smallest pair integers for which ABC would be a right isosceles triangle.</p>
<p>That means p² = 2q².</p>
<p>Then we derive a smaller isosceles right triangle (DFG) with hypotenuse 2q-p and leg p-q.</p>
<p>Thus, the assumption that ABC was the smallest right isosceles triangle was wrong.</p>
<p>But how can that be used to say that p and q are not commensurable?</p>
<p>(p and q being commensurable means p/q is rational, right?)</p>
<p>My level of math knowledge is very basic.</p>
<p>Could you help me understand this?</p>
| TonyK | 1,508 | <p>If <span class="math-container">$\sqrt 2$</span> is rational, it can be expressed as <span class="math-container">$p/q$</span> where <span class="math-container">$p,q$</span> are <em>integers</em>. (Is this what you were missing?) The proof then goes on to construct a rational representation of <span class="math-container">$\sqrt 2$</span> with smaller integers in the numerator and denominator. But if this were possible, you could repeat this process indefinitely, which would lead to an infinite sequence of decreasing positive integers -- a contradiction.</p>
|
688,577 | <p>A beautiful polyhedron with 20 hexagons and 60 pentagons can be seen here: <a href="http://robertlovespi.wordpress.com/2013/11/03/a-polyhedron-with-80-faces/" rel="nofollow">http://robertlovespi.wordpress.com/2013/11/03/a-polyhedron-with-80-faces/</a> . Euler formula and the corresponding Diophantine equation give a smaller possible combination: 7 hexagons and 20 pentagons adjacent by two to hexagons' vertices and by five in the pentagonal vertices. Does such a polyhedron really exist? I doubt but my only argument is "I was not able to compose it". At the same time I do know that the non-existence of polyhedra permitted by Euler equation is not elementary (cf. the not-existing polyhedron with 12 pentagons and 1 hexagon only). </p>
| vonbrand | 43,946 | <p>There are 13 <a href="http://en.wikipedia.org/wiki/Archimedean_solid" rel="nofollow">Archimedean solids</a>, none of them with your recipe.</p>
|
688,577 | <p>A beautiful polyhedron with 20 hexagons and 60 pentagons can be seen here: <a href="http://robertlovespi.wordpress.com/2013/11/03/a-polyhedron-with-80-faces/" rel="nofollow">http://robertlovespi.wordpress.com/2013/11/03/a-polyhedron-with-80-faces/</a> . Euler formula and the corresponding Diophantine equation give a smaller possible combination: 7 hexagons and 20 pentagons adjacent by two to hexagons' vertices and by five in the pentagonal vertices. Does such a polyhedron really exist? I doubt but my only argument is "I was not able to compose it". At the same time I do know that the non-existence of polyhedra permitted by Euler equation is not elementary (cf. the not-existing polyhedron with 12 pentagons and 1 hexagon only). </p>
| John Joy | 140,156 | <p>I'm not sure if this leads anywhere, but I would ignore the hexagons and just deal with the region bounded by the planes that the pentagons lie on. This should result in a polyhedron with 60 triangles. Try to show that if these 60 triangles were edge-connected, that they would all lie on the same plane (and thus they do not form a polyhedron at all).</p>
|
2,003,660 | <p>I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:</p>
<p>Suppose x is a real and $0 < x < 4$, it follows that,</p>
<p>\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow 4 < (x - 2)^2 < 4\\
&\Rightarrow 0 \leq (x - 2)^2 < 4
\end{align*}</p>
<p>Since, $x(4 - x) = 4x - x^2 = 4 - (x - 2)^2$, then</p>
<p>$$\dfrac{4}{x(4 - x)} = \dfrac{4}{4 - (x - 2)^2}.$$</p>
<p>This expression is greater or equal to $1$ for
$0 \leq (x - 2)^2 < 4$. Thus,</p>
<p>$$\dfrac{4}{x(4 - x)} \geq 1.$$</p>
<p>I'm quite new to proof technique and I'm using this book to self-learn logic and proofing writing. My question is: is the solution stated above logically sound? Would my arguments be considered sufficient to prove that $P \Rightarrow Q$?</p>
| Darío G | 27,454 | <p>We can do this proof by contradiction: We want to prove $P\rightarrow Q$ and we prove it by assuming $P$ and $\neg Q$, and getting a contradiction.</p>
<p>Suppose $0<x<4$ with $x\in\mathbb{R}$, and the inequality $\dfrac{4}{x(x-4)}\geq 1$ does not hold. </p>
<p>Then we have $0<x,0<4-x$ (both follow from $0<x<4$) and $\dfrac{4}{x(x-4)}<1$, and we have:</p>
<p>\begin{align*}
\dfrac{4}{x(4-x)}&<1 & &\\
4&<x(4-x) & &\text{[multiplying both sides by $x(x-4)>0$]}\\
0&<x(4-x)-4\\
0&<-x^2+4x-4\\
x^2-4x+4&<0\\
(x-2)^2&<0
\end{align*}</p>
<p>which is absurd because every squared real number is positive. </p>
<p>This contradiction allows us to conclude that $\dfrac{4}{(x-4)}\geq 1$.</p>
|
1,493,783 | <p>We have $4$ pockets each containing $4$ placement positions, we want to obtain an estimate on the number of possible arrangements, when the exact set of allowed elements is given:
$$ls=[a,a,a,a,b,b,c,c,d,d,x,y] \tag{1}$$
and numbers [1:4] are used for empty slots. For example if one pocket is of the form -,a,-,- then it corresponds to $1$ empty slot then letter a then $2$ empty slots.</p>
<p>Only real restriction is that each pocket can only contain 4 elements at most, and the size is always 4. Two different examples to clarify the size discussion: a,b,c,d has $4$ letters, thus size $4$, another would be 1,a,2 which has $1$ letter thus 1+1+2 = 4. Another clarification, for example a,b,c,d is distinct from b,a,c,d, else we would be counting combinations.</p>
<p>I'm really interested to learn how one should go about doing the count of arrangements of all four pockets, when there are restrictions such as above on pockets. For simplicity, let us assume for the moment that all items in $(1)$ without exception should be inserted into pockets. One example of an arrangement would be: $$a,b,c,d - c,b,a,a - d,a,x,y - 4$$</p>
<ul>
<li>Essentially the question is: how many possible arrangements are there like the above?</li>
<li>The main difficulty to me, lies on the one hand in the fact that we have restrictions, and that once one pocket is filled and we move to the next, the set of elements for the remaining pockets has changed. If we had only one pocket then the problem would translate to: Permutation of $n$ objects taken $4$ at a time.</li>
</ul>
<hr>
<p>As pointed out by Ian Miller, the problem can effectively be formulated in terms of number of permutations of n elements with repetition. So as an additional sub-question: how to include restrictions such as the a's only allowed to be in certain number of positions (e.g. from 1 to 8 i.e. first 2 pockets, or 8 to 12 etc).</p>
| Ian Miller | 278,461 | <p>16! ways to place all the items (include 4 null objects).
The four $a$'s are identical so 4! ways to interchange those.
The four $-$'s are identical so 4! ways to interchange those.
The two $b$'s are identical so 2! ways to interchange those.
The two $c$'s are identical so 2! ways to interchange those.</p>
<p>Answer: $$\frac{16!}{4!4!2!2!}=9081072000$$</p>
<p>(If you only want combination of pockets then its just another divide by $4!$ arrangements of pockets.)</p>
|
483,533 | <p>How to prove that the following linear map :</p>
<p>$T : \Bbb R^2 \to \Bbb R^2$ defined by $$T(e_1)=e_2$$
$$T(e_2)=0$$ </p>
<p>can't be diagonalized with respect to any basis. Here $e_i$s are standard basis.</p>
| Vishal Gupta | 60,810 | <p>Any map is diagonalizable if and only if it has the same number of linearly independent eigenvectors as the dimension of the space.</p>
<p>In your case, there is only one eigenvalue i.e. $0$ and the corresponding eigenspace is one dimensional. Hence, it does not have "enough" number of linearly independent eigenvectors and therefore cannot be diagonalized.</p>
<p>To see why there is no other eigenvalue, let</p>
<p>$$ T(ae_{1} + be_{2}) = \lambda(ae_{1} + be_{2})$$</p>
<p>which implies</p>
<p>$$ ae_{2} = \lambda a e_{1} + \lambda b e_{2}$$</p>
<p>which immediately gives $\lambda = 0$.</p>
|
483,533 | <p>How to prove that the following linear map :</p>
<p>$T : \Bbb R^2 \to \Bbb R^2$ defined by $$T(e_1)=e_2$$
$$T(e_2)=0$$ </p>
<p>can't be diagonalized with respect to any basis. Here $e_i$s are standard basis.</p>
| Robert Lewis | 67,071 | <p>One explanation by Yours Truly can be found in my answer to this question:</p>
<p><a href="https://math.stackexchange.com/questions/482688/proof-of-a-matrix-is-positive-semidefinite-iff-it-can-be-written-in-the-form-x">Proof of a matrix is positive semidefinite iff it can be written in the form $X'X$</a></p>
<p>Look between equations (10) and (12)!</p>
|
160,765 | <p>I have a functor $F:C \to D$ between poset-enriched categories, and I'd like to show that the induced map on classifying spaces is a homotopy-equivalence. To this end, I am trying to establish the presence of initial objects in all the fibers $d\setminus F$ and use the 2-categorical version of Quillen's Theorem A due to Bullejos and Cegarra (see their pdf <a href="http://hera.ugr.es/doi/14976262.pdf">here</a>). It'd be nice if these fibers had initial elements, so</p>
<blockquote>
<p>What is an initial object $i$ in a 2-category $C$, and where can I find a reference in the literature?</p>
</blockquote>
<p>I imagine that instead of having a unique morphism $1 \to c$ for every object $c$ of $C$ like one does for a 1-categorical initial object, we'd now want the category of all morphisms from $1$ to $c$ in $C$ to be contractible. So if $C$ is poset-enriched it would suffice for the poset $C(i,c)$ to have a minimal element for all objects $c$. Is this accurate, and if so, what can I cite as a reference? </p>
| Jonathan Chiche | 5,587 | <p>Dimitri Ara has just brought this question to my attention. Perhaps you will find the following useful. </p>
<p>Let us say that an object $z$ of a $2$-category $\mathcal{A}$ has a terminal object if, for every object $x$ of $\mathcal{A}$, the category $Hom_{\mathcal{A}}(x,z)$ has a terminal object. (In particular, this category is non-empty.) </p>
<p>The terminology comes from the fact that a category has a terminal object in this sense, viewed as an object of the $2$-category of categories, if and only if it has a terminal object in the usual sense. (This terminology was suggested to me by Bénabou because of that. See also Michael Barr's relevant answer to my question <a href="https://mathoverflow.net/questions/108397/is-there-a-standard-name-for-a-2-category-which-has-an-object-z-such-that-for-e">Is there a standard name for a 2-category which has an object z such that, for every object x, the category Hom(x,z) has a terminal object?</a>) </p>
<p>There are three dual definitions obtained by replacing $Hom_{\mathcal{A}}(x,z)$ by $Hom_{\mathcal{A}}(z,x)$ or "terminal" by "initial". The result below could of course be dualized accordingly. </p>
<p>Denote by $e$ the trivial $2$-category. If a small $2$-category $\mathcal{A}$ has an object which has a terminal object, then the geometric realization of $\mathcal{A} \to e$ is a homotopy equivalence. This is Lemme 2.27 of <a href="http://www.math.jussieu.fr/~chiche/Maths/TheoremA.pdf" rel="nofollow noreferrer">http://www.math.jussieu.fr/~chiche/Maths/TheoremA.pdf</a>. I still have to upload the final version on Arxiv. This is to be published in TAC.</p>
<p>Not sure whether it answers the question you were asking but hope this helps nevertheless.</p>
|
3,563,419 | <p>What I did was writing <span class="math-container">$(1\, 2)(3\, 4)(1\, 3)(2\, 4)$</span> as <span class="math-container">$\bigl(\begin{smallmatrix}
1& 2 &3 & 4\\
2& 1 &4 & 3
\end{smallmatrix}\bigr)\bigl(\begin{smallmatrix}
1& 2 &3 &4 \\
3& 4& 1 & 2
\end{smallmatrix}\bigr)$</span> which is equal to <span class="math-container">$(1 4)(2 3)$</span></p>
<p>Is what I'm doing right ? And isn't there any other method for it ?</p>
| Luca Goldoni Ph.D. | 264,269 | <p>A rigorous proof could be the following:</p>
<ol>
<li><span class="math-container">$$
\begin{gathered}
\cos \left( {\frac{{\pi n^2 }}
{{n + 1}}} \right) = \cos \left[ {\pi \left( {n - 1} \right) + \frac{\pi }
{{n + 1}}} \right] = \hfill \\
\hfill \\
= \cos \left[ {\pi \left( {n - 1} \right)} \right]\cos \left( {\frac{\pi }
{{n + 1}}} \right) - \sin \left[ {\pi \left( {n - 1} \right)} \right]\sin \left( {\frac{\pi }
{{n + 1}}} \right) = \hfill \\
\hfill \\
= \left( { - 1} \right)^{n - 1} \cos \left( {\frac{\pi }
{{n + 1}}} \right) \hfill \\
\end{gathered}
$$</span></li>
<li><span class="math-container">$$
\sum\limits_{n = 2}^{ + \infty } {\frac{{\cos \left( {\frac{{\pi n^2 }}
{{n + 1}}} \right)}}
{{\log ^2 n}}} = \sum\limits_{n = 2}^{ + \infty } {\left( { - 1} \right)^{n - 1} \frac{{\cos \left( {\frac{\pi }
{{n + 1}}} \right)}}
{{\log ^2 n}}}
$$</span></li>
<li>We prove now that
<span class="math-container">$$
a_n = \frac{{\cos \left( {\frac{\pi }
{{n + 1}}} \right)}}
{{\log ^2 n}} \downarrow 0\,\,\,\,\,\,n \to + \infty
$$</span></li>
<li>Let be
<span class="math-container">$$
f(x) = \frac{{\cos \left( {\frac{\pi }
{{x + 1}}} \right)}}
{{\log ^2 x}}
$$</span>
with <span class="math-container">$x \geq 2$</span>. Then
<span class="math-container">$$
f'(x) = - \frac{{2\cos \left( {\frac{\pi }
{{x + 1}}} \right)}}
{{x\log ^3 (x)}} + \frac{{\pi \sin \left( {\frac{\pi }
{{x + 1}}} \right)}}
{{(x + 1)^2 \log ^2 (x)}}
$$</span>
thus
<span class="math-container">$$
f'(x) = - \frac{{2\cos \left( {\frac{\pi }
{{x + 1}}} \right)}}
{{x\log ^3 (x)}} + o\left( {\frac{1}
{{x\log ^3 (x)}}} \right)
$$</span>
as <span class="math-container">$x \to +\infty$</span>. This means that eventually <span class="math-container">$f'(x)<0$</span> and therefore <span class="math-container">$f(x)$</span> is monotonically decreasing as well as <span class="math-container">$a_n$</span>. Now the series is convergent by Leibniz Test.</li>
</ol>
|
3,432,861 | <p><span class="math-container">$y'$</span> and <span class="math-container">$y$</span> occur linearly in the ODE
<span class="math-container">$$y'=\frac{2x+y}{y-x}.$$</span>
yet it is a first order non-linear ODE. I can find a family solutions of this homogeneous ODE by using <span class="math-container">$y=vx \implies \frac{dy}{dx}= v+x \frac{dv}{dx}$</span>. I want to know if there is(are) singular solution(s) of this equation.</p>
| vic165 | 723,908 | <p><span class="math-container">$$ y'=\frac{x(2+\frac{y}{x})}{x(\frac{y}{x}-1)} $$</span>
then you can make a replacement <span class="math-container">$$ t=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{dt}{dx}x+t \Rightarrow \frac{dt}{dx}x+t=\frac{2+t}{t-1} $$</span>
It remains only to divide the variables and take the integral :)</p>
|
3,432,861 | <p><span class="math-container">$y'$</span> and <span class="math-container">$y$</span> occur linearly in the ODE
<span class="math-container">$$y'=\frac{2x+y}{y-x}.$$</span>
yet it is a first order non-linear ODE. I can find a family solutions of this homogeneous ODE by using <span class="math-container">$y=vx \implies \frac{dy}{dx}= v+x \frac{dv}{dx}$</span>. I want to know if there is(are) singular solution(s) of this equation.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Write <span class="math-container">$$y'=\frac{2+\frac{y}{x}}{-1+\frac{y}{x}}$$</span> and substitute <span class="math-container">$$u=\frac{y}{x}$$</span></p>
|
3,432,861 | <p><span class="math-container">$y'$</span> and <span class="math-container">$y$</span> occur linearly in the ODE
<span class="math-container">$$y'=\frac{2x+y}{y-x}.$$</span>
yet it is a first order non-linear ODE. I can find a family solutions of this homogeneous ODE by using <span class="math-container">$y=vx \implies \frac{dy}{dx}= v+x \frac{dv}{dx}$</span>. I want to know if there is(are) singular solution(s) of this equation.</p>
| Z Ahmed | 671,540 | <p>You can get the singular solution of this equation by puttin <span class="math-container">$y=vx$</span>, but treat <span class="math-container">$v$</span> as constant for a change. then
<span class="math-container">$$v=\frac{2+v}{v-1} \implies v=1\pm \sqrt{3} ~~real~~.$$</span> so the singular solutions are
<span class="math-container">$y=(1+\sqrt{3})x, ~ y=(1-\sqrt{3}) x.$</span> These are essential solutions free from any constant. These solutions occur if one gets real value(s) of <span class="math-container">$v$</span> as in this examples. If so, the origin is either a node (stable /unstable) or a saddle point. These singular(essential) solutions are apart from the usual solution which will depend on a given initial condition.</p>
|
307,822 | <p>Excuse my incorrect use of terminology, I hope my question is clear:</p>
<p>I am coding a Python module which tests whether a given number is a member of the Fibonacci series. No problem with that. Additionally, should a number not be a member of the series, I would like to test whether it is significantly close to its nearest Fibonacci neighbor. Here an increasing deviation margin of what is considered "close" is needed, along the lines of:</p>
<pre><code>deviation margin (x) increases as given number (n) increases
</code></pre>
<p>For my purposes 9 is significantly close to Fibonacci number 8 but 10 is not. 1600 is significantly close to Fibonacci number 1597 but 1610 is not, etc. So, the test for "significant closeness" is applying a deviation margin of 1 at lower numbers and an increasing deviation margin as the series increments up to infinity.</p>
<p>I figured a logical candidate for inclusion in the test would be Standard Deviation. So I have calculated a coarse margin as follows:</p>
<pre><code>margin = (StdDev / ( n + closestfibneighbor)) * StdDev
</code></pre>
<p>This does not give me good control over the margin rate of increase and I am sure there is a more appropriate function to express the growth in margin caused by increments in x. Please feel free to elaborate on the mathematics of this - I am seeking a general solution and not a Python-specific function.</p>
| venzen | 62,930 | <p>A suitable solution (suggested by <a href="https://math.stackexchange.com/users/856/rahul-narain">Rahul Narain</a>) is to simply calculate the margin as a fixed percentage of each successive Fibonacci number in the series:</p>
<pre><code>margin = 1 + 0.01n
</code></pre>
<p>This gives the desired outcome of allowing a margin of ~1 at the lowest Fib numbers and an increasing margin as the series increments. In the spirit of the Fib series I calculate the margin as 13% of n instead of 10%.</p>
<p>Using standard deviation as a variable in this specific context is not elegant since a percentage of Std Dev will have to be applied in a working solution anyway.</p>
|
2,060,633 | <p>Let $R=(\Bbb Z/5\Bbb Z$)[X]/ $\langle X^n \rangle $ with $n\in\Bbb N_+$. Show that $f\in R$ is a zero divisor if and only if $f_0=0$. And how many zero dividers exist?</p>
<p>I know that zero divispr means there exists non-zero $m$ such that $fm=0$ in $R$ and that $\Bbb Z/5\Bbb Z$ has no zero divisors, meaning $f_0$ or $m_0$ are equal to $0$ if $fm=0$.</p>
<p>How can i show that if $m_0\neq0$ then $f_0=0$?</p>
| sami ullah | 976,897 | <p>Yeah, this question gave me a headache for a week but then it was easy once I realized how to look at the question. The first catch is that the probability is being calculated for each group. Do not confuse with a sequence of events.
The book has skipped a step as well. let's assume the initial state. Four groups and each group have 4 location that is valid for a grad student to occupy. The probability that a group will have a grad student is <span class="math-container">$\frac{4}{16}$</span> and overall probability of grad student in any of the group can be calculated as <span class="math-container">$\frac{4}{16} + \frac{4}{16} +\frac{4}{16} +\frac{4}{16} =\frac{16}{16} = 1 $</span>. So the probability of having a grad in any valid position is one. Now we have occupied the first group with a valid position whats the probability of the second grad to have a valid position</p>
<p>Group 1: X _ _ _
Group 2: _ _ _ _
Group 3: _ _ _ _
Group 4: _ _ _ _</p>
<p>we have in total of 15 vacant positions out of which 3 are in group 1 that we can not be used and 12 are valid positions for second grad and the probability is <span class="math-container">$\frac{12}{15}$</span>. Now we assume that second grad occupied valid positions we have the following
Group 1: X _ _ _
Group 2: X _ _ _
Group 3: _ _ _ _
Group 4: _ _ _ _</p>
<p>now for 3rd grad student to be in a valid position we have 8 locations and the total slots are 14 so the probability is <span class="math-container">$\frac{8}{14}$</span>. After the 3rd valid allocation, the final situation is as below
Group 1: X _ _ _
Group 2: X _ _ _
Group 3: X _ _ _
Group 4: _ _ _ _</p>
<p>Four valid poistion in group 4 out of 13 so the probability of that group is <span class="math-container">$\frac{4}{13}$</span>. Now multiply all. <span class="math-container">$1.\frac{12}{15}.\frac{8}{14}.\frac{4}{13}$</span></p>
|
1,574,290 | <p>How do I prove this? </p>
<p>For the Fibonacci numbers defined by $f_1=1$, $f_2=1$, and $f_n = f_{n-1} + f_{n-2}$ for $n ≥ 3$, prove that $f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n$ for all $n≥ 1$.</p>
| David Holden | 79,543 | <p>the inductive step:
$$
f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n \Leftrightarrow \\
f_{n+1}f_{n-1} - f_n^2 =(-1)^n \Leftrightarrow \\
(f_{n+2}-f_n)(f_{n+1}-f_n) -f_n^2 = (-1)^n \Leftrightarrow \\
f_{n+2}f_{n+1}-f_nf_{n+3} = (-1)^n \Leftrightarrow \\
f_{n+1}^2+f_nf_{n+1}-f_nf_{n+1}-f_nf_{n+2}=(-1)^n \Leftrightarrow \\
f_{n+2}f_n-f_{n+1}^2 = (-1)^{n+1} \Leftrightarrow \\
f_{n+2}^2-f_{n+2}f_{n+1}-f_{n+1}^2 =(-1)^{n+1}
$$</p>
|
2,423,086 | <p>Show, by listing the elements or how we can list the elements, that </p>
<p>$\mathbb{N}^3$ = $\mathbb{N×N×N}$ is a countable set.</p>
<p><strong>Attempt at a solution:</strong></p>
<p>I was thinking about making use of Cantor's diagonal argument. If this were $\mathbb{N×N}$ I could just do:</p>
<p>(1,1), (1,2), (1,3),...</p>
<p>(2,1), (2,2), (2,3)...</p>
<p>(3,1), (3,2), (3, 3)...</p>
<p>and snake my way through it.</p>
<p>However, I am having trouble finding a way to list $\mathbb{N×N×N}$ such that I can implement the diagonal argument.</p>
| Hagen von Eitzen | 39,174 | <p>For an explicit listing of the elements of $\Bbb N^3$, you might start with all $(x,y,z)$ with $x+y+z=3$, then those with $x+y+z=4$, etc.
For each sum $s$, there are only finitely many choices of $x,y,z$ (namely $s-1\choose 2$), and these can be ordered lexically, i.e., sort by $x$, then $y$ (then $z$, but that's irrelevant). With patience, you could develop explicit formulae from this procedure to compute the $n$ triple.</p>
|
2,642,547 | <p>If a subset A of $\mathbb{R}^n$ has no interior, must it be closed?</p>
<p>Can I prove this using the example of a subset A that consists of a single point, so A has no interior yet it is closed?</p>
| user87690 | 87,690 | <p>In general, every countable subset has empty interior, but not every countable subset is closed.</p>
|
2,642,547 | <p>If a subset A of $\mathbb{R}^n$ has no interior, must it be closed?</p>
<p>Can I prove this using the example of a subset A that consists of a single point, so A has no interior yet it is closed?</p>
| Asaf Karagila | 622 | <p>You have received numerous examples of countable sets which are with no interior, but not closed either. Let me note that the irrational numbers are not closed and have an empty interior. So it is not the countability which hinders your statement.</p>
<p>Let me address your secondary question. No, you cannot prove something of the form "Every such and such is such and such" by providing an example. Example are used to <em>disprove</em> that (or prove the opposite "There exists such and such with such and such properties).</p>
<p>When in doubt, always try to recast your question as a preposterous hyperbole. For example, "Can I prove that all numbers are $0$ using an example that $0$ is $0$?" or "Can I prove that all sets are empty using $\varnothing$ as an example of an empty set?", the answer to both is silly, since $1\neq 0$, and $\{1\}$ is not empty.</p>
|
11,652 | <p>Last August, so-called "review audits" <a href="https://math.meta.stackexchange.com/q/10740/43351">were introduced</a> on MSE.</p>
<p>These are fake review samples added to the review queues, designed to catch those users who (in a dash to get a badge) practice mindless button mashing instead of careful consideration while reviewing.</p>
<p>Prior to the introduction, people were enthusiastic -- me included.</p>
<hr>
<p>However, in three months, we have generated:</p>
<ul>
<li><a href="https://math.meta.stackexchange.com/q/10836/43351">STOP! Look and Listen.</a> (+58)</li>
<li><a href="https://math.meta.stackexchange.com/q/10924/43351">Examples of poor review audits</a> (+26)</li>
<li><a href="https://math.meta.stackexchange.com/q/11641/43351">Audits: this is getting ridiculous.</a> (+27)</li>
<li><a href="https://math.meta.stackexchange.com/q/11307/43351">Try to comment: Fail review audit</a> (+13)</li>
</ul>
<p>in which threads, more and more people start stating their resentment of the whole auditing thing, with wordings like <a href="https://math.meta.stackexchange.com/a/11322/43351">Vedran Šego's</a>:</p>
<blockquote>
<p>The first time I got an audit, I thought it was a good idea. Well, not anymore.</p>
</blockquote>
<hr>
<p>It is time to evaluate. Some valid concerns have been raised, but not addressed by the SE developers. A tour around Meta.StackOverflow reveals a similar picture. The developers are unwilling to invest the time to do something about at least the most-heard complaints about the audit system.</p>
<p>However, the current implementation has a <strong>far too high</strong> "false negative" statistic: Too many conscientious reviewers are being shouted at by the audit system for trying to be helpful to their best judgement.</p>
<blockquote>
<p>Therefore, pending improvements to the audit system to reduce the "false negatives", I want the audits to be disabled on Math.StackExchange. Do you agree?</p>
</blockquote>
<p>Please find two polling answers below -- you can use an upvote to indicate your preference.</p>
<hr>
<p>Seeing as (to the best of my knowledge) the audit system is standardised SE-network-wide, the relevant feature requests for improving the audit system should be filed at <a href="http://meta.stackoverflow.com">Meta.StackOverflow</a>.</p>
<hr>
<p>The vote tallies have more or less crystallised. Disregarding downvotes, and taking into account my own, the final result is:</p>
<blockquote>
<ul>
<li>Disable audits: 24 votes</li>
<li>Keep audits: 15 votes</li>
</ul>
</blockquote>
<p>This falls short (and has consistently fallen short throughout the poll) of the "qualified majority" (two thirds) I had in mind for requesting the termination of the audits. That means the audits remain, if I am to decide.</p>
| Lord_Farin | 43,351 | <p>No.</p>
<p>Please keep the review audits.</p>
|
1,809,871 | <p>I am trying to solve Poisson equation using FFT. The issue appears at wavenumber $k = 0$ when I want to get inverse Laplacian which means division by zero.</p>
<p>We have </p>
<p>${\nabla ^2}\phi = f$ </p>
<p>Taking FFT from both side we get:</p>
<p>$-k^2\hat\phi = \hat f $</p>
<p>or </p>
<p>$\hat\phi = \frac{\hat f}{-k^2} $</p>
<p>Assuming that we want to solve this equation in periodic domain and using DFT using FFTW package, at $k=0$ we have a division by zero. Anybody knows how to deal with this singularity?</p>
| Vitality | 55,395 | <p><strong>THE PROBLEM</strong></p>
<p>Let me first restate the problem in 2 dimensions:</p>
<p>Solve the Poisson equation</p>
<p><span class="math-container">$\nabla^2\phi(x,y)=f(x,y)\;\;\;(x,y)\in[0, L_x]\times[0, L_y]$</span></p>
<p>subject to periodic boundary conditions, namely</p>
<p><span class="math-container">$\phi(x,0)=\phi(x,L_y) \;\;\; x\in[0, L_x]$</span></p>
<p><span class="math-container">$\phi(0,y)=\phi(L_x,y) \;\;\; y\in[0, L_y]$</span></p>
<p><span class="math-container">$\frac{\partial \phi}{\partial x}(0, y) = \frac{\partial \phi}{\partial x}(L_x, y)\;\;\; y\in[0, L_y]$</span></p>
<p><span class="math-container">$\frac{\partial \phi}{\partial y}(x, 0) = \frac{\partial \phi}{\partial y}(x, L_y)\;\;\; x\in[0, L_x]$</span></p>
<p><strong>THE FFT-BASED SOLUTION APPROACH</strong></p>
<p>Let us expand <span class="math-container">$\phi(x,y)$</span> in the Fourier series</p>
<p><span class="math-container">$$\phi(x,y)=\sum_{p,q=-\infty}^{\infty}\hat{\phi}_{pq}e^{-j2\pi p\frac{x}{L_x}}e^{-j2\pi q\frac{y}{L_y}}$$</span></p>
<p>where the <span class="math-container">$\hat{\phi}_{pq}$</span>'s are the expansion coefficients, and let us do the same for <span class="math-container">$f(x,y)$</span> with expansion coefficients <span class="math-container">$\hat{f}_{pq}$</span>. On substituting the Fourier series expansion into the Poisson equation we have</p>
<p><span class="math-container">$$\left[\left(-j2\pi \frac{p}{L_x}\right)^2+\left(-j2\pi \frac{q}{L_y}\right)^2\right]\hat{\phi}_{pq}=\hat{f}_{pq}$$</span></p>
<p>Solving the Poisson equation amounts at solving such an equation for the <span class="math-container">$\hat{\phi}_{pq}$</span>'s.</p>
<p>On specializing the Fourier series of <span class="math-container">$\phi(x,y)$</span> at the points <span class="math-container">$(x_m, y_n) = (m \Delta x, n \Delta y)$</span>, with <span class="math-container">$m = 0,1,\ldots,M$</span> and <span class="math-container">$n=0,1,\ldots,N$</span>, we have</p>
<p><span class="math-container">$$\phi_{mn}=\phi(m\Delta x,n\Delta y)=\sum_{p,q=-\infty}^{\infty}\hat{\phi}_{pq}e^{-j2\pi p\frac{m\Delta x}{L_x}}e^{-j2\pi q\frac{n\Delta y}{L_y}}$$</span></p>
<p>Finally, if we choose</p>
<p><span class="math-container">$\Delta x = \frac{L_x}{M}$</span> and <span class="math-container">$\Delta y = \frac{L_y}{N}$</span> and truncate the Fourier series to <span class="math-container">$(M+1)\times(N+1)$</span> terms, then</p>
<p><span class="math-container">$$\phi_{mn}=\sum_{p=0}^{M}\sum_{q=0}^{N}\hat{\phi}_{pq}e^{-j2\pi p\frac{m\Delta x}{L_x}}e^{-j2\pi q\frac{n\Delta y}{L_y}}$$</span></p>
<p>which is the expression of a Discrete Fourier Transform (DFT). The same holds true for <span class="math-container">$f(x,y)$</span>, i.e.</p>
<p><span class="math-container">$$f_{mn}=\sum_{p=0}^{M}\sum_{q=0}^{N}\hat{f}_{pq}e^{-j2\pi p\frac{m\Delta x}{L_x}}e^{-j2\pi q\frac{n\Delta y}{L_y}}$$</span></p>
<p>where <span class="math-container">$f_{mn}=f(m \Delta x, n\Delta y)$</span>. The approach is then the following:</p>
<ol>
<li>Sample <span class="math-container">$f(x,y)$</span> at the points <span class="math-container">$(m \Delta x, n\Delta y)$</span>;</li>
<li>Perform the FFT of <span class="math-container">$f_{mn}$</span>;</li>
<li>Divide <span class="math-container">$\hat{f}_{pq}$</span> by <span class="math-container">$\left[\left(-j2\pi \frac{p}{L_x}\right)^2+\left(-j2\pi \frac{q}{L_y}\right)^2\right]$</span> to compute <span class="math-container">$\hat{\phi}_{pq}$</span>;</li>
<li>Perform the IFFT of <span class="math-container">$\hat{\phi}_{pq}$</span>.</li>
</ol>
<p><strong>THE ISSUE</strong></p>
<p>The above equation can be solved for any <span class="math-container">$(p,q)$</span>, except for <span class="math-container">$p=q=0$</span>, giving an ambiguity for the determination of the zero frequency Fourier coefficient <span class="math-container">$\hat{\phi}_{00}$</span>.</p>
<p><strong>THE SOLUTION</strong></p>
<p>Seeing the issue from a different perspective, the boundary conditions define the solution of the above Poisson equation only
up to an additive constant: if <span class="math-container">$\phi(x,y)$</span> solves this equation and obeys the above periodic boundary conditions, then so does <span class="math-container">$\phi(x,y)+c$</span> for any real constant <span class="math-container">$c$</span>. Accordingly, we need another condition to pin the constant <span class="math-container">$c$</span> down. The standard stipulation is that</p>
<p><span class="math-container">$$\int_{0}^{L_x}\int_{0}^{L_y}\phi(x,y)dxdy = 0$$</span></p>
<p>Note that, on using the periodic boundary conditions and on integrating the Poisson equation, it can be easily shown that the forcing term <span class="math-container">$f(x,y)$</span> must also obey the condition</p>
<p><span class="math-container">$$\int_{0}^{L_x}\int_{0}^{L_y}f(x,y)dxdy = 0$$</span></p>
<p><strong>REFERENCE</strong></p>
<p>Arieh Iserles, <em>A first course in the numerical analysis of differential equations</em>, Section 10.4, Cambridge Texts in Applied Mathematics.</p>
|
2,276,429 | <p>I've tried multiple variations of polygons but can't find any that work. Do they exist?</p>
<p>Is it possible to draw a polygon on a grid paper and divide it into two equal parts by a cut of the shape shown on the Figure (a)? </p>
<p>Solve the same problem for a cut shown on Figure (b).</p>
<p>Solve the same problem for a cut shown on Figure (c).</p>
<p><a href="https://i.stack.imgur.com/iYSlV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYSlV.jpg" alt="enter image description here"></a></p>
<p>(In every problem a cut is inside the polygon, with the ends lying on the boundary. The sides of the polygons and the cuts must lie on the grid lines. The small links of the cuts are twice as short as the large ones)</p>
| tomi | 215,986 | <p><a href="https://i.stack.imgur.com/plHyU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/plHyU.jpg" alt="enter image description here"></a></p>
<p>So your polygon can be a rectangle?</p>
<p>You need $a^2=(a+b)(a+c)-a^2$</p>
<p>$a^2=a^2+ac+ab+bc-a^2$</p>
<p>$a^2=ac+ab+bc$</p>
|
2,276,429 | <p>I've tried multiple variations of polygons but can't find any that work. Do they exist?</p>
<p>Is it possible to draw a polygon on a grid paper and divide it into two equal parts by a cut of the shape shown on the Figure (a)? </p>
<p>Solve the same problem for a cut shown on Figure (b).</p>
<p>Solve the same problem for a cut shown on Figure (c).</p>
<p><a href="https://i.stack.imgur.com/iYSlV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iYSlV.jpg" alt="enter image description here"></a></p>
<p>(In every problem a cut is inside the polygon, with the ends lying on the boundary. The sides of the polygons and the cuts must lie on the grid lines. The small links of the cuts are twice as short as the large ones)</p>
| Andrei | 331,661 | <p>Here is something for (a) and (b)</p>
<p><a href="https://i.stack.imgur.com/722dH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/722dH.png" alt="enter image description here"></a></p>
|
92,613 | <p>This is a question in Pinter's <em>A Book of Abstract Algebra</em>.</p>
<blockquote>
<p>Let $S=\{g\in G\mid \operatorname{ord}(g)=p\}$. Prove the order of $S$ is a multiple of $p-1$.</p>
</blockquote>
<p>In his solution Pinter says $a \in S$ implies that $a$ generates a subgroup with $p-1$ elements. Shouldn't there be $p$ elements $\{1,a^1,\dots,a^{p-1}\}$? Or is it typical to only count the non-trivial elements in a subgroup?</p>
| Justin Campbell | 4,842 | <p>You are correct, an element of order $n$ generates a subgroup with $n$ elements. Perhaps Pinter means that the subgroup generated by an element of order $p$ contains $p-1$ elements of order $p$ (namely, the nontrivial ones).</p>
|
1,405,809 | <p>So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.<br>
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.</p>
| Patrick Stevens | 259,262 | <p>It is precisely $f'(0)$, where $f(r) = a^r$.</p>
|
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