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 |
|---|---|---|---|---|
160,801 | <p>Here is a vector </p>
<p>$$\begin{pmatrix}i\\7i\\-2\end{pmatrix}$$</p>
<p>Here is a matrix</p>
<p>$$\begin{pmatrix}2& i&0\\-i&1&1\\0 &1&0\end{pmatrix}$$</p>
<p>Is there a simple way to determine whether the vector is an eigenvector of this matrix?</p>
<p>Here is some code for your convenience.</p>
<pre><code>h = {{2, I, 0 },
{-I, 1, 1},
{0, 1, 0}};
y = {I, 7 I, -2};
</code></pre>
| David G. Stork | 9,735 | <pre><code>MemberQ[myeigens = Normalize/@Eigenvectors[h], Normalize[y]]|| MemberQ[myeigens, -Normalize[y]]
</code></pre>
<p>(* False *)</p>
|
160,801 | <p>Here is a vector </p>
<p>$$\begin{pmatrix}i\\7i\\-2\end{pmatrix}$$</p>
<p>Here is a matrix</p>
<p>$$\begin{pmatrix}2& i&0\\-i&1&1\\0 &1&0\end{pmatrix}$$</p>
<p>Is there a simple way to determine whether the vector is an eigenvector of this matrix?</p>
<p>Here is some code for your convenience.</p>
<pre><code>h = {{2, I, 0 },
{-I, 1, 1},
{0, 1, 0}};
y = {I, 7 I, -2};
</code></pre>
| AccidentalFourierTransform | 34,893 | <p>You already have several good answers. An alternative is to use a <a href="https://en.wikipedia.org/wiki/Rayleigh_quotient" rel="noreferrer">Rayleigh quotient</a>,</p>
<pre><code>r = First[y.h.ConjugateTranspose[{y}]/Norm[y]];
</code></pre>
<p>The vector <code>y</code> is an eigenvector of <code>h</code> if and only if the matrix
$$
h-r1_{3\times3}
$$
is singular:</p>
<pre><code>MatrixRank[h - IdentityMatrix[Length[y]] R]<Length[y]
(*False*)
</code></pre>
<p>or</p>
<pre><code>Det[h - IdentityMatrix[Length[y]] R] == 0
(*False*)
</code></pre>
<p>If you are using floating point numbers, you should change this condition into</p>
<pre><code>MatrixRank[h - IdentityMatrix[Length[y]] R,Tolerance->epsilon]<Length[y]
</code></pre>
<p>or</p>
<pre><code>Abs[Det[h - IdentityMatrix[Length[y]] R]] < epsilon
</code></pre>
<p>where <code>epsilon</code> is some small number.</p>
<p>As a matter of fact, the <code>MatrixRank</code> method is slightly faster than the <code>Det</code> one. It seems to me that it is also faster than the methods suggested by other users, but confirming this would require a more thorough analysis.</p>
|
2,080,042 | <blockquote>
<p>I am interested of finding examples of non-zero homomorphisms $f:R\to S$ of rings with unity such that $f(1_R)\neq 1_S$.</p>
</blockquote>
<p>I will provide one example and I will be glad if others can also give examples. </p>
| Juniven Acapulco | 44,376 | <p>Let $R=\mathbb{Z}$ and $S=\mathbb{Z}\times \mathbb{Z}$. Define $f:R\to S$ by $f(n)=(n,0)$ for all $n\in\mathbb{Z}$. Then $f$ is a nonzero homomorphism but
$$f(1_R)=f(1)=(1,0)\neq (1,1)=1_S.$$</p>
|
2,354,036 | <p>$\require{AMScd}\def\colim{\text{colim}}$I need this result in less generality, but I'd be happy to know this stronger version holds.</p>
<p>Let $\{\alpha_c : Fc \to Gc\}$ be arrows in a category $D$, indexed by the objects of a category $C$, for two functors $F,G: C\to D$. </p>
<p>Let $A\subseteq C$ be a dense subcategory (meaning that $i : A \hookrightarrow C$ is a dense functor). Then $\alpha$ is a natural transformation <em>if and only if it is natural on the components $c\in A$ (and with respect to morphisms of $A$)</em>.</p>
<p><strong>Edit</strong>: This is not what I wanted to know (and false, see below): instead, I'm trying to prove that there is a unique extension of a natural transformation $Fi\to Gi$ to a natural transformation $F\to G$.</p>
<p>Example: for a functor $K: [I°,Set] \to [I°,Set]$, a natural transformation $\alpha : 1_{[I°,Set]}\to K$ is such if and only if it is natural when restricted to representables.</p>
<p>It is easy to see that the right universal property gives that if $P = \colim y(X_i)$ is a colimit of representables, then there is
$$
P \cong \colim\; y(X_i) \xrightarrow{\colim \alpha_{X_i}} \colim\; Ky(X_i) \to K\big(\colim\; y(X_i)\big) \cong KP
$$
It seems then that all boils down to the fact that the morphism $\colim\; KX\to K(\colim\; X)$ is "natural", i.e.
$$
\begin{CD}
KX @>>> K(\colim_I X_i)\\
@VVV @VVV\\
KY @>>> K(\colim_J Y_j)
\end{CD}
$$
commutes (but how do you induce the right-vertical arrow?).</p>
| Younesse Kaddar | 581,427 | <p>$$
\newcommand{\colim}{\mathop{\rm colim}\nolimits}
\newcommand{\Nat}{\mathop{\rm Nat}\nolimits}
\newcommand{\cancom}[1]{\big((i/#1)\stackrel{\mathrm{pr}_}{→} \stackrel{i}{→} ℂ\big)}
$$</p>
<p>First, I thank <a href="http://www.cs.ox.ac.uk/people/ohad.kammar/main.html" rel="nofollow noreferrer">Ohad Kammar</a>, <a href="https://math.stackexchange.com/users/525482/mathieu-huot">Mathieu Huot</a> and <a href="https://www.univ.ox.ac.uk/academics/sean-moss/" rel="nofollow noreferrer">Sean Moss</a> who helped me to come up with this answer.</p>
<p>To answer your <strong>EDIT</strong> in a nutshell: the result <em>doesn't stand</em> in general when $F$ is not cocontinuous.</p>
<p>Here are two counter-examples to the existence and unicity of your natural transformation extension:</p>
<h2>Counter-example to existence</h2>
<p>Let</p>
<ul>
<li>$ℂ = \, ≝ \, X ⟶ Y$ (category with two objects and one non-trivial morphism between them)</li>
<li>$ \, ≝ \, \mathbb{1}$ (terminal category)</li>
<li>$i: \begin{cases}
↪︎ ℂ \\
\ast ⟼ Y
\end{cases}$</li>
<li>$F \, ≝ \, \mathrm{const}_Y ∈ [ℂ, ]$ (constant functor sending all the $ℂ$-objects to $Y$)</li>
<li>$G \, ≝ \, \mathrm{Id}_{ℂ} ∈ [ℂ, ]$</li>
</ul>
<p>then there is a natural transformation between $Fi: \begin{cases}
\mathbb{1} ⟶ \\
\ast ⟼ Y
\end{cases}$ and $Gi: \begin{cases}
\mathbb{1} ⟶ \\
\ast ⟼ Y
\end{cases}$ (the identity), but no natural transformation from $F$ to $G$, as there is no morphism from $F(X) = Y$ and $G(X) = X$.</p>
<h2>Counter-example to uniqueness</h2>
<ul>
<li>As before, let's set $ℂ ≝ \, X ⟶ Y, \; \, ≝ \, \mathbb{1}, \; i: \begin{cases}
↪︎ ℂ \\
\ast ⟼ Y
\end{cases}$</li>
<li>Let $$ be the category given by
<ul>
<li>two object $X'$ and $Y'$</li>
<li>two non-trivial morphisms: $g: X' ⟶ Y'$ and $f: X' ⟶ X'$ such that $$gf = g \qquad \text{ and} \qquad f^2 = \mathrm{id}_{X'}$$</li>
</ul></li>
<li>$F = G: \begin{cases}
ℂ ⟶ \\
X ⟼ X' \\
Y ⟼ Y'
\end{cases}$</li>
</ul>
<p>then there is a natural transformation between $Fi: \begin{cases}
\mathbb{1} ⟶ \\
\ast ⟼ Y'
\end{cases}$ and $Gi: \begin{cases}
\mathbb{1} ⟶ \\
\ast ⟼ Y'
\end{cases}$ (the identity), but two distinct natural transformations from $F$ to $G$ (the identity and the one whose component at $X$ is $f$).</p>
<h2>Fixed statement</h2>
<p>Your assertion is true if $F$ is assumed to be <strong>cocontinuous</strong> though:</p>
<blockquote>
<p>Let $F: ℂ ⟶ $ be a <strong>cocontinuous</strong> functor, $G: ℂ ⟶ $ a functor and $ \overset{i}{\hookrightarrow} ℂ$ a dense subcategory. There is a unique extension of every natural transformation $Fi ⇒ Gi$ to a natural transformation $F ⇒ G$.</p>
</blockquote>
<p>The proof is as shown in the diagram below: for every $c \cong \colim\cancom{c}, \, c' \cong \colim\cancom{c} ∈ ℂ, \, f: c ⟶ c'$:</p>
<ul>
<li><p>there is a cocone from $\cancom{c}\stackrel{F}{→} $ to $G(c)$ (resp. from $\cancom{c'}\stackrel{F}{→} $ to $G(c')$)</p>
<ul>
<li><em>indeed</em>: the green subdiagram commutes (since $Gϕ_1 = G ϕ_2 \circ Gg$ (cocone property) and $Gg \circ α_{ia_1} = α_{ia_2} \circ Fg$ (naturality))</li>
</ul>
<p>so there exists a unique corresponding cocone morphism $α_c: F(c) ⟶ G(c)$ (resp. $α_c': F(c') ⟶ G(c')$), due to $F$ being concontinuous</p></li>
<li><p>the two cocones $\cancom{c}\stackrel{F}{→} $ to $G(c)$ obtained by post-composing by $α_c \circ Ff$ and $Gf \circ α_{c'}$ are equal</p>
<ul>
<li><em>indeed</em>: the red subdiagram commutes since $Gϕ_1' \circ α_{ia_1'} = α_{c'} \circ F ϕ_1'$ (colimit property of $F(c')$) and $α_c \circ Ff \circ Fϕ_1' = Gf \circ Gϕ_1' \circ α_{ia_1'}$ (colimit property of $F(c)$))</li>
</ul>
<p>so there is a unique cocone morphism from $\cancom{c}\stackrel{F}{→} $ to $G(c)$ for these cocones, and the red square commutes: $Gf \circ α_{c'} = α_c \circ Ff$, which ends the proof.</p></li>
</ul>
<p><img src="https://i.stack.imgur.com/BdMMU.png" width="900"></p>
<p><a href="http://younesse.net/assets/StackExchange/naturality-can-be-detected-on-a-dense-subcategory/svg/extension.svg" rel="nofollow noreferrer">Link to the svg picture</a></p>
<p>By duality, one can also prove that:</p>
<blockquote>
<p>Let $F: ℂ ⟶ $ be a functor, $G: ℂ ⟶ $ a <strong>continuous</strong> functor and $ \overset{i}{\hookrightarrow} ℂ$ a co-dense subcategory. There is a unique extension of every natural transformation $Fi ⇒ Gi$ to a natural transformation $F ⇒ G$.</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/umKsY.png" width="900"></p>
<p><a href="http://younesse.net/assets/StackExchange/naturality-can-be-detected-on-a-dense-subcategory/svg/extension_dual.svg" rel="nofollow noreferrer">Link to the svg picture</a></p>
<h2>Application: Nerve functor fully faithful</h2>
<p>With the dual version of the lemma, we get this well-known result as a corollary:</p>
<blockquote>
<p><strong>Corollary</strong>: If $ℂ \stackrel{i}{\hookrightarrow} $ is dense, the nerve functor
$$N_i: \begin{cases}
&⟶ \widehat ℂ \\
d &⟼ (i(-), d)
\end{cases}$$
is fully faithful.</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/9OIAT.png"></p>
<p><a href="http://younesse.net/assets/StackExchange/naturality-can-be-detected-on-a-dense-subcategory/svg/nerve.svg" rel="nofollow noreferrer">Link to the svg picture</a></p>
<p><em>Proof</em>: We want to prove that for all $d, d' ∈ $:
$$\Nat\Big(\underbrace{(i(-), d)}_{\mathbb{y}_(d)i}, \, \underbrace{(i(-), d')}_{\mathbb{y}_(d')i}\Big) \cong \Nat(d, \, d') \overset{\scriptsize\text{Yoneda lemma}}{\cong} \Nat\Big(\underbrace{(-, d)}_{\mathbb{y}_(d)}, \, \underbrace{(-, d')}_{\mathbb{y}_(d')}\Big)$$</p>
<p>The result follows by applying the dual version of the lemma with $i^{\rm op}$ ($i$ post-composed with the $\rm op$ functor) and $F ≝ \mathbb{y}_(d), \; G ≝ \mathbb{y}_(d')$:</p>
<p><img src="https://i.stack.imgur.com/MEorj.png"></p>
<p><a href="http://younesse.net/assets/StackExchange/naturality-can-be-detected-on-a-dense-subcategory/svg/two_arrows.svg" rel="nofollow noreferrer">Link to the svg picture</a></p>
|
438,166 | <p>Let $X$ be a bounded connected open subset of the $n$-dimensional real Euclidean space. Consider the Laplace operator defined on the space of infinitely differentiable functions with compact support in $X$. </p>
<p>Does the closure of this operator generate a strongly continuous semigroup on $C_0(X)$ endowed with the supremum norm?</p>
<p>I think it is equivalent to the following question: </p>
<blockquote>
<p>Is the space of infinitely differentiable functions with compact support in $X$ a core for the Dirichlet Laplacian on $X$?</p>
</blockquote>
| Community | -1 | <p>The closure of the Laplacian on $C^\infty_0(X)$
cannot generate a strongly continuous semigroup on $C(X)$,
since there are two distinct (closed!) generators that extend $(\Delta, C^\infty_0(X))$; namely
the Dirichlet Laplacian and the Neumann Laplacian. Probabilistically,
these correspond to absorbing and reflecting Brownian motion, respectively. </p>
<p>This follows from the basic fact that generators of semigroups cannot strictly extend one another. </p>
<p><strong>Lemma:</strong>
Suppose that $(T_t)_{t\geq 0}$ is a strongly continuous semigroup with generator $(L,D(L))$, and that
$(S_t)_{t\geq 0}$ is a strongly continuous semigroup with generator $(K,D(K))$.
Suppose also that $L$ extends $K$, that is, $D(K)\subseteq D(L)$ and $Ku=Lu$ for all
$u\in D(K)$. Then $T_t=S_t$ for all $t\geq 0$. </p>
<p><strong>Proof.</strong> Fix $u\in D(K)$ and $t>0$, and consider the continuous map from $[0,t]$ to $C(X)$ given by $s\to T_{t-s} S_su$. This map is differentiable on $(0,t)$ with
$$\biggl({d\over{ds}}\biggr) T_{t-s} S_su= -L T_{t-s} S_su+ T_{t-s} K S_su=
T_{t-s} (K-L)(S_su)=0.$$
It follows that the value of this map at $s=0$ and at $s=t$ must be equal,
that is $T_tu=S_tu$.
Since this equation is true on the dense set $D(K)$ and since
$T_t$ and $S_t$ are bounded operators, it follows that $T_t=S_t$.</p>
<p>Therefore, the extension
$D(K)\subseteq D(L)$ is not genuine; that is, $D(L)=D(K)$.
So if any two generators extend one another, they in fact must coincide. </p>
|
148,185 | <p>Let $ X = \mathbb R^3 \setminus A$, where $A$ is a circle. I'd like to calculate $\pi_1(X)$, using van Kampen. I don't know how to approach this at all - I can't see an open/NDR pair $C,D$ such that $X = C \cup D$ and $C \cap D$ is path connected on which to use van Kampen. </p>
<p>Any help would be appreciated. Thanks</p>
| Simon Markett | 30,357 | <p>I am not sure whether there is a nicer choice but this is how I think about it. Intuitively the fundamental group should be $\mathbb Z$ - a path may jump through the hoop a couple of times or not. I choose the open sets to model this somewhat. One open set is the interiour of a filled torus with the circle lying on the surface. The other set is the whole of $\mathbb R^3$ with the closed disk (bounded by the circle) removed. Then the first set contracts to a circle, the second set contracts to a sphere and the intersection is contractible. </p>
<p>Edit: To make the sets more precise:
$$U=\mathbb R^3-D^2\simeq S^2$$
such that $$A=\partial D^2\subseteq D^2$$
and
$$V=int(S^1\times D^2)\simeq S^1$$
such that $$A=\ast\times \partial D^2\subseteq S^1\times D^2.$$
Then $$U\cap V=int(S^1\times D^2-\ast\times D^2)\cong int(I\times D^2)\simeq\ast$$</p>
|
23,471 | <p>I'm trying to find an explanation for the different sizes I'm seeing for fonts added to graphics in different ways, and haven't yet located an easy to understand explanation. Here's a minimal example:</p>
<pre><code>Graphics[
{LightGray,
Rectangle[{0, 0}, {72, 72}],
Red,
Style[
Text["Hig", {0, 0}, {-1, -1}], 72,
FontFamily -> "Times New Roman"],
Black,
First[
First[
ImportString[
ExportString[
Style["Hig",
FontFamily -> "Times New Roman",
FontSize -> 72], "PDF"],If[$VersionNumber>=13,{"PDF","PageGraphics"},"PDF"],
"TextMode" -> "Outlines"]]]
},
PlotRange -> {{0, 100}, {0, 100}},
Axes -> True,
Ticks -> {Table[x, {x, 0, 100, 10}], Table[x, {x, 0, 100, 10}]},
Epilog -> {Text["72", {200, 75}], Line[{{0, 72}, {200, 72}}]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/l2wT7.png" alt="fonts" /></p>
<p>The red text doesn't change size when you resize the graphic, although everything else changes. The black text resizes along with everything else. But neither text seems to be the result of specifying 72.</p>
<h2>Update</h2>
<p>After changing the screen resolution to Automatic following @Sjoerd's suggestion, I can see how the red text is basically displaying at a fixed size that's independent of <em>Mathematica</em>. In this picture, the left image shows 72 dpi, the red font is using 70ish pixels of vertical space. The right image is at 'Automatic' (so presumably 133.51 for my machine, according to <a href="http://members.ping.de/%7Esven/dpi.html" rel="nofollow noreferrer">this site</a>, and similarly uses up 70ish pixels.</p>
<p><img src="https://i.stack.imgur.com/Qjbuv.png" alt="resolutions" /></p>
<p>I'm still puzzled by the size of the black font, which doesn't seem to be related to the specified font size or to the screen resolution. Perhaps the PDF translation introduces another scaling factor.</p>
| Alexey Popkov | 280 | <p>As it is pointed out in my old answer <a href="https://stackoverflow.com/a/6124065">here</a>, graphics and text are displayed inside of the FrontEnd in the style environment defined by the <code>ScreenStyleEvironment</code> option while are <code>Export</code>ed into <code>"PDF"</code> in the style environment defined by the <code>PrintingStyleEnvironment</code> option:</p>
<pre><code>Options[$FrontEnd, {ScreenStyleEnvironment,
PrintingStyleEnvironment}]
</code></pre>
<blockquote>
<pre><code>{PrintingStyleEnvironment -> "Printout",
ScreenStyleEnvironment -> "Working"}
</code></pre>
</blockquote>
<p>The <code>"Working"</code> and <code>"Printout"</code> style environments are defined in the default Core.nb stylesheet and differ in the value of <code>Magnification</code> option which is set to <code>1</code> in the <code>"Working"</code> and to <code>0.8</code> in the <code>"Printout"</code> style environment correspondingly:</p>
<p><img src="https://i.stack.imgur.com/06ahx.png" alt="screenshot of the Core.nb stylesheet"></p>
<p>It is possible to override the value of this option by setting it explicitly via <code>Style</code>:</p>
<pre><code>text = Style["Hig", 72, FontFamily -> "Times New Roman",
Magnification -> 1]
textOutlines =
First[ImportString[ExportString[text, "PDF"],
"TextMode" -> "Outlines"]];
Style[textOutlines, Magnification -> 1]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/CU5bu.png" alt="screenshot"></p>
</blockquote>
<p>We can check that the outlines and the text have identical sizes:</p>
<pre><code>Style[Show[textOutlines,
Prolog -> {Red, Text[text, {0, 0}, {-1, -1}]}], Magnification -> 1]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/o3FxA.png" alt="screenshot2"></p>
</blockquote>
<p>A step-by-step explanation HOW it works:</p>
<ol>
<li>As Sjoerd correctly points out in his <a href="https://mathematica.stackexchange.com/a/23481/280">answer</a>, the glyphs are displayed on the screen with <code>FontProperties -> "ScreenResolution" -> 72</code> by default. In practice it means that each displayed pixel on the screen corresponds to 1/72 of inch (when <code>Magnification</code> is set to <code>1</code>) and so a glyph of height 72 printer's points will have height 72 pixels on the screen (even if on the level of operating system is set other resolution for rendering of fonts).</li>
<li>The internal coordinate system of PDF file is defined in printer's points (one printer's point = 1/72 of inch). It means that if we <code>Export</code> a glyph of height 72 printer's points to PDF <strong>with <code>Magnification -> 1</code></strong> then the glyph will have height 72 units in the internal coordinate system of the PDF file. So when we convert the glyphs to outlines and <code>Import</code> the outlines we get a plot which internal coordinate system is defined in printer's points.</li>
<li><p>The <code>Graphics</code> object created from the <code>Import</code>ed PDF figure has <code>AspectRatio -> Automatic</code> and <code>ImageSize</code> set to the size of the imported figure in printer's points without any extra paddings:</p>
<pre><code>textOutlines // Options
</code></pre>
<blockquote>
<pre><code>{ImageSize -> {108., 81.},
PlotRange -> {{0., 108.}, {0., 81.}}, AspectRatio -> Automatic}
</code></pre>
</blockquote></li>
</ol>
<p>The combination of these factors gives us the situation when each unit on the plot corresponds to one pixel on the screen and is equal to one printer's point of displayed glyph (<strong>when <code>Magnification</code> is set to <code>1</code></strong>). So we get identical sizes of the glyphs and outlines. But when we turn on the <code>Frame</code> we add extra paddings and the coordinates no longer correspond to printer's points:</p>
<pre><code>Style[Show[textOutlines,
Prolog -> {Red, Text[text, {0, 0}, {-1, -1}]}, Frame -> True],
Magnification -> 1]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/dgDvS.png" alt="screenshot3"></p>
</blockquote>
<p>For obtaining correct coordinate scalings we need to carefully define <code>PlotRange</code>, <code>ImagePadding</code> and <code>ImageSize</code>:</p>
<pre><code>plotSize = {120, 100};(* in arbitrary units *)
imagePadding = 20;(* in printer's points *)
imageSize = plotSize + 2*imagePadding;(* in printer's points; now the internal
coordinate system of the plot will be in printer's points *)
plot = Style[
Show[textOutlines, Frame -> True,
FrameTicks -> {{Automatic, {72}}, {Automatic, None}},
GridLines -> {None, {72}}, ImagePadding -> imagePadding,
PlotRange -> {{0, plotSize[[1]]}, {0, plotSize[[2]]}},
ImageSize -> imageSize,
Prolog -> {Red, Text[text, {0, 0}, {-1, -1}]}], Magnification -> 1]
</code></pre>
<blockquote>
<p><img src="https://i.stack.imgur.com/bCDUy.png" alt="screenshot4"></p>
</blockquote>
|
3,760,253 | <blockquote>
<p>The diagram shows the line <span class="math-container">$y=\frac{3x}{5\pi}$</span> and the curve <span class="math-container">$y=\sin$</span>
<span class="math-container">$x$</span> for <span class="math-container">$0\le x\le \pi$</span>.</p>
<p>Find (as an exact value) the enclosed area shown shaded in the diagram.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/fTGUr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fTGUr.png" alt="enter image description here" /></a></p>
<p>I'm not sure where I've made an error, but this is my working out so far:</p>
<p>Area under <span class="math-container">$y=\frac{3x}{5\pi}$</span> from <span class="math-container">$0$</span> to <span class="math-container">$\frac{5\pi}{6}$</span> is:</p>
<p><span class="math-container">$$\int_{0}^{5\pi/6} \frac{3x}{5\pi}dx=\frac{5\pi}{24}$$</span></p>
<p>The remaining <strong>unshaded</strong> "triangle" area from <span class="math-container">$\frac{5\pi}{6}$</span> to <span class="math-container">$\pi$</span>:</p>
<p><span class="math-container">$$\frac{1}{2} \cdot \frac{\pi}{6} \cdot \frac{1}{2} = \frac{\pi}{24}$$</span></p>
<p>Area under <span class="math-container">$y=\sin x$</span> for <span class="math-container">$0\le x\le\pi$</span>:</p>
<p><span class="math-container">$$\int_{0}^{\pi} \sin{x}dx=2$$</span></p>
<p>Hence, shaded area = <span class="math-container">$$2-\frac{5\pi}{24} - \frac{\pi}{24} = \frac{-\pi}{4} + 2$$</span></p>
<p>The <strong>correct</strong> answer is:</p>
<p><span class="math-container">$$1+\frac{\sqrt3}{2} - \frac{5\pi}{24} units^2$$</span></p>
<p>If someone could explain how to correctly solve this, it would be greatly appreciated!</p>
| Henry | 6,460 | <p>If <span class="math-container">$L=1-x+{x}^2-\cdots$</span> converges for <span class="math-container">$|x|<1$</span></p>
<p>then <span class="math-container">$xL= x-x^2+{x}^3-\cdots$</span> also converges for <span class="math-container">$|x|<1$</span></p>
<p>and <span class="math-container">$L+xL = 1$</span></p>
<p>so <span class="math-container">$L=\frac{1}{1+x}$</span> for <span class="math-container">$|x|<1$</span></p>
|
2,604,093 | <p>I would like to study the convergence of the series:</p>
<p>$$\sum_{n=1}^\infty \frac{\log n}{n^2}$$</p>
<p>I could compare the generic element $\frac{\log n}{n^2}$ with $\frac{1}{n^2}$ and say that
$$\frac{1}{n^2}<\frac{\log n}{n^2}$$ and $\frac{1}{n^2}$ converges but nothing more about.</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Let use <a href="https://en.wikipedia.org/wiki/Convergence_tests#Limit_comparison_test" rel="nofollow noreferrer">Limit comparison test</a> test with $$\frac{1}{n^{\frac32}}$$</p>
<p>Related OP with a more general discussion <a href="https://math.stackexchange.com/questions/2586107/determine-if-sum-k-1-infty-frac-lnkk-sqrtk-converges-diverges/2586110#2586110">Determine if $\sum_{k=1}^{\infty}\frac{\ln{k}}{k\sqrt{k}}$ converges/diverges.</a></p>
|
3,968,905 | <p>I am trying to prove this:</p>
<p><span class="math-container">$\bullet$</span> Prove that <span class="math-container">$\Delta(\varrho_\epsilon \star u) = \varrho_\epsilon \star f $</span> in the sense of distributions, if <span class="math-container">$\Delta u = f$</span> in the sense of distributions, <span class="math-container">$ u \in L^1_{loc}(\mathbb{R}^n)$</span>.</p>
<p>Can anybody help me?
Thank you in advance and I take advantage of the situation to wish you a happy new year :D</p>
| dan_fulea | 550,003 | <p>We write the given equation equivalently:
<span class="math-container">$$
\begin{aligned}
0 &= x^2y^2 - 2x^2y + 2xy^2 + x^2 - 4xy + y^2 + 2x - 2y - 3\ ,\\
0 &= x^2(y-1)^2 + 2x(y-1)^2 + (y-1)^2-4\ ,\\
4 &= (x+1)^2(y-1)^2\ .
\end{aligned}
$$</span>
Now consider all possible ways to write <span class="math-container">$4$</span> as a product of two perfect squares.</p>
|
76,163 | <p>Represent the position of a unit-length, oriented segment $s$ in the plane
by the location $a$ of its <em>basepoint</em> and
an orientation $\theta$: $s = (a,\theta)$. So $s$ can be
viewed as a point in $\mathbb{R^2} \times \mathbb{S^1}$.
Now I'll define a metric on this space.
Define the distance $d(s_1,s_2)$ between two positions of
unit-length segments as the average distance between their corresponding
points:
<br />
<img src="https://i.stack.imgur.com/4G9o9.jpg" alt="Segments" />
<br />
Above the distances are about 0.31, 0.61, and 0.53, left-to-right.</p>
<p>So if the endpoints of $s_i$ are $a_i$ and $b_i$,
then $d(s_1,s_2)$ is the average of the Euclidean distances
between $(1-t) a_1 + t b_1$ and $(1-t) a_2 + t b_2$ as $t$ varies in $[0,1]$.
This is indeed a metric, I believe, because the triangle inequality
holds between corresponding points in three positions of the segment.
This metric is intended to capture the intuitive notion of how much
work is required to move $s_1$ to $s_2$.</p>
<p>My question is: What are the geodesics in this space under this metric?
Certainly a pure translation of $s$ is a geodesic.
It seems that a pure rotation by at most $\pi$ of $s$ about a
point $p \in s$ should also be a geodesic, but even this is
not so clear to me. Certainly a rotation about a point not on $s$
is (generally) not a geodesic. Of course the main interest would be
in geodesics that mix translation and rotation, showing (locally) optimal
repositioning paths.</p>
<p>I investigated this long ago when working on motion-planning algorithms
("moving a ladder"), but got quite blocked on this natural question.
This superficially seems related to the
<a href="http://en.wikipedia.org/wiki/Kakeya_set#Kakeya_needle_problem" rel="nofollow noreferrer">Kakeya needle problem</a>,
but the metric I propose does not measure swept area.
Perhaps it has been studied in some guise previously.
If so, a pointer would be appreciated. Thanks!</p>
<p><b>Addenda</b>. (<em>26Sep11</em>.) I just ran across this book, by V. A. Dubovit͡s︡kiĭ,
which seems relevant:
<em>The Ulam problem of optimal motion of line segments</em>,
Translation Series in Mathematics and Engineering, Optimization Software, 1985.
It may take some time for me to locate a copy...</p>
<p>(<em>11Nov11</em>). I finally have this book in my hands.
The Preface by Hestenes says,</p>
<blockquote>
<p>Dubovitskij has succeeded in solving in closed form a generalization of a problem
of S[.] Ulam..:
Among all continuous motions of an oriented line segment $S$ in $\mathbb{E}^n$ from one
position to another, which preserves its length [...], find one for which the sum
of the lengths of the paths swept by its endpoints is minimal.</p>
</blockquote>
<p>The concentration here on the motion of the endpoints—in contrast to
the average distance metric I proposed—seems to render these results as not directly relevant, although nevertheless <em>quite</em> interesting.</p>
| Anton Petrunin | 1,441 | <p>Let us start with the metric on $\mathbb R^4=\mathbb R^2\times \mathbb R^2$
defined by the norm $\|{*}\|$ defined by
$$\|(x,y)\|=\int_0^1|t\cdot x+(1-t)\cdot y|\,dt,$$
where $|{ * }|$ denotes the Euclidean norm on $\mathbb R^2$.
This norm is not strongly convex, so you should expect many geodesics between close points.</p>
<p>Now, your metric is the intrinsic metric induced on the hypersurface $\Sigma$ described by $|x-y|=1$.
So, you have a Finsler metric on $\Sigma=\mathbb S^1\times \mathbb R^2$.
The unit ball in the tangent plane is isometric the intersection of the ball in the above norm with the 3-dimensional subspace in one special direction.
It seems that this intersection is strongly convex.
The metric is smooth (since there is a transitive isometric group action on $\mathbb S^1\times \mathbb R^2$).
It remains to write a differential equation for the geodesics;
this should be in any book on Finsler geometry.
(It should be a pain, but it might help.)</p>
|
76,163 | <p>Represent the position of a unit-length, oriented segment $s$ in the plane
by the location $a$ of its <em>basepoint</em> and
an orientation $\theta$: $s = (a,\theta)$. So $s$ can be
viewed as a point in $\mathbb{R^2} \times \mathbb{S^1}$.
Now I'll define a metric on this space.
Define the distance $d(s_1,s_2)$ between two positions of
unit-length segments as the average distance between their corresponding
points:
<br />
<img src="https://i.stack.imgur.com/4G9o9.jpg" alt="Segments" />
<br />
Above the distances are about 0.31, 0.61, and 0.53, left-to-right.</p>
<p>So if the endpoints of $s_i$ are $a_i$ and $b_i$,
then $d(s_1,s_2)$ is the average of the Euclidean distances
between $(1-t) a_1 + t b_1$ and $(1-t) a_2 + t b_2$ as $t$ varies in $[0,1]$.
This is indeed a metric, I believe, because the triangle inequality
holds between corresponding points in three positions of the segment.
This metric is intended to capture the intuitive notion of how much
work is required to move $s_1$ to $s_2$.</p>
<p>My question is: What are the geodesics in this space under this metric?
Certainly a pure translation of $s$ is a geodesic.
It seems that a pure rotation by at most $\pi$ of $s$ about a
point $p \in s$ should also be a geodesic, but even this is
not so clear to me. Certainly a rotation about a point not on $s$
is (generally) not a geodesic. Of course the main interest would be
in geodesics that mix translation and rotation, showing (locally) optimal
repositioning paths.</p>
<p>I investigated this long ago when working on motion-planning algorithms
("moving a ladder"), but got quite blocked on this natural question.
This superficially seems related to the
<a href="http://en.wikipedia.org/wiki/Kakeya_set#Kakeya_needle_problem" rel="nofollow noreferrer">Kakeya needle problem</a>,
but the metric I propose does not measure swept area.
Perhaps it has been studied in some guise previously.
If so, a pointer would be appreciated. Thanks!</p>
<p><b>Addenda</b>. (<em>26Sep11</em>.) I just ran across this book, by V. A. Dubovit͡s︡kiĭ,
which seems relevant:
<em>The Ulam problem of optimal motion of line segments</em>,
Translation Series in Mathematics and Engineering, Optimization Software, 1985.
It may take some time for me to locate a copy...</p>
<p>(<em>11Nov11</em>). I finally have this book in my hands.
The Preface by Hestenes says,</p>
<blockquote>
<p>Dubovitskij has succeeded in solving in closed form a generalization of a problem
of S[.] Ulam..:
Among all continuous motions of an oriented line segment $S$ in $\mathbb{E}^n$ from one
position to another, which preserves its length [...], find one for which the sum
of the lengths of the paths swept by its endpoints is minimal.</p>
</blockquote>
<p>The concentration here on the motion of the endpoints—in contrast to
the average distance metric I proposed—seems to render these results as not directly relevant, although nevertheless <em>quite</em> interesting.</p>
| Sergei Ivanov | 4,354 | <p>In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation.</p>
<p>Since the metric has three-dimensional group of isometries (generated by rigid motions of the plane), the corresponding <a href="http://en.wikipedia.org/wiki/Noether%2527s_theorem" rel="noreferrer">Noether's integrals</a> make the geodesic flow <a href="http://en.wikipedia.org/wiki/Integrable_system" rel="noreferrer">completely integrable</a>. More precisely, the first derivative is uniquely determined by the position and the values of the integrals. Actually a two-dimensional group is sufficient because for the third integral one can always take the speed (= the norm of the velocity vector) of a geodesic.</p>
<p>So let's use invariance under parallel translations only. We have a manifold of all unit segments in the plane, and a tangent vector to that manifold can be thought of as the vector field along a segment (representing velocities of all its points). This vector field has a form
$$
v(t) = tx + (1-t)y, \qquad t\in[0,1]
$$
where $x,y\in\mathbb R^2$ are such that the vector $x-y$ is orthogonal to the segment. The norm of this tangent vector is $\int |v(t)| dt$ (as in Anton's answer), and the Noether integral corresponding to translations (if my quick computation is correct) boils down to the following: if $v$ is a velocity vector of a geodesic represented in the above form, then
$$
\int_0^1 \frac{v(t)}{|v(t)|} \ dt = V_0
$$
where $V_0$ is a constant (for each geodesic) vector in the plane. This vector $V_0$ (and, say, the assumption that the geodesic is unit-speed) determines the parameters $x$ and $y$ in the formula for $v(t)$ uniquely (as a function of the segment's current direction), so the geodesic equation is reduced to a 1st order ODE. Furthermore, the above integral can be found as an explicit (but weird) function of $x$ and $y$, so the equation becomes easy to solve, at least numerically.</p>
<p>By the way, I second Jean-Marc Schlenker's proposal to consider quadratic mean of the distances (i.e. $L^2$ norm rather than $L^1$), especially if you have any physics-related application in mind. In this case the metric is Riemannian, the energy is just the standard kinetic energy of the moving segment, so the Noether intergrals are just the standard conservation laws: the linear momentum and the angular momentum. And the geodesics are very simple: the segment rotates at a fixed angular speed while its midpoint (the barycenter) moves along a straight line with a constant speed.</p>
|
813,825 | <p>In strong Induction for the induction hypothesis you assume for all K, p(k) for k
<p>If for example I am working with trees and not natural numbers can I still use this style of proof?</p>
<p>For example if I want my induction hypothesis to be that p(k) for k < n where n is a node in the tree and everything smaller than/bellow it (The nodes children,k) is assumed to be true.</p>
<p>In the proof would I have to define what the < operator does for two nodes?</p>
| vadim123 | 73,324 | <p>Induction, and strong induction, are used to prove statements that are indexed by the natural numbers $\mathbb{N}=\{1,2,3,\ldots\}$. $k$ and $n$ need to be natural numbers, with a minimum value ($0$ or $1$ are the usual choices). The <em>statements</em> need not be natural numbers, for example, the first statement $p(1)$ could be that "node number 1 satisfies property x" and the second statement $p(2)$ could be that "nodes number 1 and 2 satisfy property x" etc.</p>
|
1,177,988 | <p>Comparing the equation
$$x^4+3x+20=0$$<br>
With the equation
$$(x^2+\lambda)^2-(mx+n)^2=0$$
we get </p>
<p>$m^2=2\lambda,$</p>
<p>$-2mn=3,$<br>
$n^2=\lambda^2-20$ </p>
<p>Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$. </p>
<p>How can I find easily the values of $\lambda$ from the above equation.</p>
<p>Please suggest me.</p>
| Luigi D. | 164,401 | <p>The cubic equation is in depressed form (i.e., its quadratic coefficient is 0). To find the first root, you can use Cardano's formula $$\lambda_1=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$$ where $p = -20$ and $q = -\frac{9}{8}$.</p>
|
1,177,988 | <p>Comparing the equation
$$x^4+3x+20=0$$<br>
With the equation
$$(x^2+\lambda)^2-(mx+n)^2=0$$
we get </p>
<p>$m^2=2\lambda,$</p>
<p>$-2mn=3,$<br>
$n^2=\lambda^2-20$ </p>
<p>Now, $4m^2n^2=9\Rightarrow 4(2\lambda)(\lambda^2-20)=9\Rightarrow 8\lambda^3-160\lambda-9=0$. </p>
<p>How can I find easily the values of $\lambda$ from the above equation.</p>
<p>Please suggest me.</p>
| Bernard | 202,857 | <p>You can't find them easily, as the discriminant of the equation is standard form:
$$x^3-20x+\frac 98=0$$
is $ 4\cdot (-20)^3+ 27\cdot\Bigl(\dfrac 98\Bigr)<0$.</p>
<p>In such a case, we know the equation has $3$ real toots, but Cardano's formula requires using complex numbers since the square root is that of a negative number, then computing the cubic roots of a complex number.</p>
<p>A trigonometric method is more direct: set $x=A\cos t,\enspace A>0$. The equation becomes
$$A^3\cos^3 t -20A\cos t+\frac 98=0.$$
We the choose $A>0$ so that $\dfrac{A^3}4=\dfrac{-20A}3$. Let's call $k$ this common ratio. We get:
$$k(4\cos^3t-3\cos t)=k\cos 3t=-\frac 98$$
whence
$$3t\equiv \pm\arccos\Bigl(-\frac 9{8k}\Bigr)\mod 2\pi\iff t\equiv\pm\frac 13\arccos\Bigl(-\frac 9{8k}\Bigr)\mod\frac{2\pi}3. $$
These $6$ solutions are equal in pairs, and there are $3$ roots in all, of course.</p>
|
183,243 | <p>Gödel's incompleteness theorem states that: "<em>if a system is consistent, it is not complete.</em>" And it's well known that there are unprovable statements in ZF, e.g. GCH, AC, etc.</p>
<p>However, why does this mean that ZF is consistent? What does "<em>relatively consistent</em>" actually mean?</p>
| user642796 | 8,348 | <p>We have to hedge quite a bit in what you said. First, there are many complete theories: the theory of all groups of order 7 is complete, as an example. However, Gödel showed that any "nice enough" (read: "recursively enumerable") consistent first-order theory capable of encoding basic arithmetic is incomplete. (There is a complete theory of number theory, usually denoted $\mathrm{Th}(\mathbb{N})$, which consists of all sentences that are true in the standard model of number theory; unfortunately, this is not a "nice enough" theory — it's impossible to program a computer to print out all and only the axioms of this theory!)</p>
<p>Also, these statements that you gave, $\mathsf{(G)CH}$, $\mathsf{AC}$, are not provably unprovable in $\mathsf{ZF}$. However, it is provable that they are unprovable in $\mathsf{ZF}$ only if $\mathsf{ZF}$ is itself inconsistent.</p>
<p>This is what a relative consistency result means: it is a result that says the consistency of one theory implies the consistency of the other. Kurt Gödel proved that if $\mathsf{ZF}$ is consistent, then so if $\mathsf{ZFC}+\mathbf{V}=\mathbf{L}$ (and also that $\mathsf{ZFC}+\mathbf{V}=\mathbf{L}$ implies $\mathsf{GCH}$). Paul Cohen proved that if $\mathsf{ZF}$ is consistent, then both $\mathsf{ZF}+\neg\mathsf{AC}$ and $\mathsf{ZFC}+\neg\mathsf{CH}$ are consistent. More elementarily, with $\mathsf{ZF}$ one can construct the natural numbers, and so the consistency of $\mathsf{ZF}$ implies the consistency of $\mathsf{PA}$. These are <em>fundamentally</em> relative consistency results, and cannot be improved to straight consistency results. This stems from <a href="https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem" rel="nofollow noreferrer">Gödel's Second Incompleteness Theorem</a>. Because of this the acceptance of the consistency of $\mathsf{ZF(C)}$ is an article of faith (though one I'm happy to believe in).</p>
<p>I recommend that you look at the first page of George Boolos's <a href="http://www.jstor.org/stable/2253954" rel="nofollow noreferrer"><em>Gödel's second incompleteness theorem explained in words of one syllable</em></a> (Mind, vol.103, pp.1-3). It is a quite entertaining look at the meaning of Gödel's Second Incompleteness Theorem.</p>
|
481,173 | <p>The most common way to find inverse matrix is $M^{-1}=\frac1{\det(M)}\mathrm{adj}(M)$. However it is very trouble to find when the matrix is large.</p>
<p>I found a very interesting way to get inverse matrix and I want to know why it can be done like this. For example if you want to find the inverse of $$M=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$$</p>
<p>First, write an identity matrix on the right hand side and carry out some steps:</p>
<p>$$\begin{bmatrix}1 & 2 &1 &0 \\ 3 & 4&0&1\end{bmatrix}\to\begin{bmatrix}1 & 2 &1 &0 \\ 3/2 & 2&0&1/2\end{bmatrix}\to\begin{bmatrix}1/2 & 0 &-1 &1/2 \\ 3/2 & 2&0&1/2\end{bmatrix}\to\begin{bmatrix}3/2 & 0 &-3 &3/2 \\ 3/2 & 2&0&1/2\end{bmatrix}$$
$$\to\begin{bmatrix}3/2 & 0 &-3 &3/2 \\ 0 & 2&3&-1\end{bmatrix}\to\begin{bmatrix}1 & 0 &-2 &1 \\ 0 & 2&3&-1\end{bmatrix}\to\begin{bmatrix}1 & 0 &-2 &1 \\ 0 & 1&3/2&-1/2\end{bmatrix}$$</p>
<p>You can 1. swap any two row of the matrix 2. multiply a constant in any row 3. add one row to the other row. Just like you are doing Gaussian elimination. when the identical matrix shift to the left, the right hand side become</p>
<p>$$M^{-1}=\begin{bmatrix}-2 &1 \\3/2&-1/2\end{bmatrix}$$</p>
<p>How to prove this method work?</p>
| Agustí Roig | 664 | <p>Because when you are performing Gauss reduction and elimination to</p>
<p>$$
(A \ \vert \ I) \ ,
$$</p>
<p>what your are doing in fact is finding the solutions of all these linear systems:</p>
<p>$$
AX_1 = e_1 , \dots , AX_n = e_n \ .
$$</p>
<p>Aren't you?</p>
<p>So, the matrix formed by the columns $X_1, \dots , X_n$ is the inverse of $A$:</p>
<p>$$
A^{-1} = (X_1 \dots X_n) \ .
$$</p>
<p>If you did it one by one, the solution of every system would appear on the right at the end of the process:</p>
<p>$$
(A \ \vert \ e_i) \longrightarrow \dots \longrightarrow (I \ \vert \ X_i) \ .
$$</p>
<p>Hence, if you do it in one go,</p>
<p>$$
(A \ \vert \ I) = (A \ \vert \ e_1 \dots e_n) \longrightarrow \dots \longrightarrow (I \ \vert\ X_1\dots X_n) = (I \ \vert \ A^{-1}) \ ,
$$</p>
<p>the inverse of $A$ appears where it should.</p>
|
488,258 | <p>What are the last two digits of $11^{25}$ to be solved by binomial theorem like $(1+10)^{25}$?
If there is any other way to solve this it would help if that is shown too.</p>
| Mher | 80,548 | <p>$$11^{25} = (1+10)^{25} = \dbinom{25}{0} + 10\cdot\dbinom{25}{1} + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$
$$ 1 + 5\cdot10 + 2\cdot 10^2 + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$
$$\underline{1} + \underline{5}\cdot10 + 10^2\cdot \left(2+\sum_{k=2}^{25}\dbinom{25}{k}\cdot10^{k-2}\right), $$
hence the last digits are 5 and 1:</p>
|
640,554 | <p>For the system
$$
\left\{
\begin{array}{rcrcrcr}
x &+ &3y &- &z &= &-4 \\
4x &- &y &+ &2z &= &3 \\
2x &- &y &- &3z &= &1
\end{array}
\right.
$$
what is the condition to determine if there is no solution or unique solution or infinite solution? </p>
<p>Thank you!</p>
| shariva | 431,031 | <p>There is an easier way to determine whether a system of equations has unique, infinite or no solution. It is as follows: calculate determinant <span class="math-container">$D$</span> of the coefficients of the three variables in three equations, then calculate <span class="math-container">$Dx$</span>, where the x coefficients with the constant terms in the determinant <span class="math-container">$D$</span>. Similarly calculate <span class="math-container">$Dy$</span> and <span class="math-container">$Dz$</span>. Then, if <span class="math-container">$D$</span> is not equal to zero, the system has a unique solution. This is called consistent. If <span class="math-container">$D$</span> not equal to <span class="math-container">$0$</span>, at least one of <span class="math-container">$Dx$</span>, <span class="math-container">$Dy$</span>, <span class="math-container">$Dz$</span> are not equal to zero, the system has no solution and this is called inconsistent. If <span class="math-container">$D=Dx=Dy=Dz=0$</span>, the system has infinite solutions.</p>
|
244,492 | <p>Find $m \in \mathbb R$ for which the equation $|x-1|+|x+1|=mx+1$ has only one unique solution. When does a absolute value equation have only 1 solution?</p>
<p>I solved for $x$ in all 4 cases and got $x=\frac{1}{-m-2},x=\frac{1}{2-m},x=\frac{1}{m},x=-\frac{3}{m}$</p>
| Martin Argerami | 22,857 | <p>You have
<span class="math-container">$$
|x-1|+|x+1|=\begin{cases} 2x,&\text{ if }x\geq1 \\2,&\text{ if }x\in(-1,1)\\ -2x,&\text{ if }x<-1\end{cases}
$$</span>
so
<span class="math-container">$$
g(x)=|x-1|+|x+1|-mx-1=\begin{cases} 2x-mx-1,&\text{ if }x\geq1 \\2-mx-1,&\text{ if }x\in(-1,1)\\ -2x-mx-1,&\text{ if }x<-1\end{cases}
$$</span>
For <span class="math-container">$g(x)=0$</span> we need:</p>
<ol>
<li>if <span class="math-container">$x\geq1$</span>, <span class="math-container">$(2-m)x-1=0$</span> requires <span class="math-container">$1/(2-m)\geq1$</span> to have a zero, i.e. <span class="math-container">$1\leq m<2$</span>.</li>
<li>if <span class="math-container">$x\in(-1,1)$</span>, <span class="math-container">$2-mx-1=0$</span>, i.e. <span class="math-container">$mx=1$</span>. This is <span class="math-container">$x=1/m\in(-1,1)$</span>, so <span class="math-container">$|m|>1$</span>.</li>
<li>if <span class="math-container">$x\leq-1$</span>, <span class="math-container">$(-2-m)x-1=0$</span> is <span class="math-container">$x=1/(-2-m)$</span>; as <span class="math-container">$x\leq-1$</span>, this is <span class="math-container">$1/(-2-m)\geq1$</span>, i.e. <span class="math-container">$-2<m\leq-1$</span>.</li>
</ol>
<p>In conclusion: if <span class="math-container">$|m|<1$</span>, there is no possible solution. If <span class="math-container">$m=1$</span>, there is a single solution (from case 1); if <span class="math-container">$m=-1$</span>, there is a single solution (from case 3); if <span class="math-container">$|m|\in(1,2)$</span>, there is a solution from case 2 and another one from either cases 1 and 3.</p>
<p>In short, single solution if and only if <span class="math-container">$|m|=1$</span>.</p>
<p><em>(it is a lot easier to understand all this if you draw a picture of <span class="math-container">$|x-1|+|x+1|$</span> and of <span class="math-container">$mx+1$</span>)</em></p>
|
90,263 | <p>Let $\mathcal{E} = \lbrace v^1 ,v^2, \dotsm, v^m \rbrace$ be the set of right
eigenvectors of $P$ and let $\mathcal{E^*} = \lbrace \omega^1 ,\omega^2,
\dotsm, \omega^m \rbrace$ be the set of left eigenvectors of $P.$ Given any two
vectors $v \in \mathcal{E}$ and $ \omega \in \mathcal{E^*}$ which correspond to
the eigenvalues $\lambda_1$ and $\lambda_2$ respectively. If $\lambda_1 \neq
\lambda_2$ then $\langle v, \omega^\tau\rangle = 0.$ </p>
<p>Proof. For any eigenvector $v\in \mathcal{E}$ and $ \omega \in
\mathcal{E^*}$ which correspond to the eigenvalues $\lambda_1$ and $\lambda_2$
where $\lambda_1 \neq \lambda_2$ we have, \begin{equation*} \begin{split}\langle\omega,v\rangle = \frac{1}{\lambda_2}\langle \lambda_2 \omega, v\rangle = \frac{1}{\lambda_2} \langle P^ \tau \omega ,v\rangle = \frac{1}{\lambda_2}\langle\omega,P v\rangle = \frac{1}{\lambda_2} \langle \omega,\lambda_1v\rangle = \frac{\lambda_1}{\lambda_2}\langle \omega,v\rangle .\end{split}
\end{equation*} This implies $(\frac{\lambda_1}{\lambda_2} - 1)\langle\omega,v\rangle =
0.$ If $\lambda_1 \neq \lambda_2$ then $\langle \omega,v\rangle = 0.$</p>
<p>My question: what if $ \lambda_2 = 0 \neq \lambda_1,$ how can I include this case in my proof.</p>
| deinst | 943 | <p>Mathematics may be universal, but learning styles are not. The relationship between teacher and student vary vastly between (and even within) cultures. Your job is to get knowledge from your brain to theirs. If this were a one size fits all problem, you would be replaced by a You-tube video.</p>
|
129,295 | <p>$$\int{\sqrt{x^2 - 2x}}$$</p>
<p>I think I should be doing trig substitution, but which? I completed the square giving </p>
<p>$$\int{\sqrt{(x-1)^2 -1}}$$</p>
<p>But the closest I found is for</p>
<p>$$\frac{1}{\sqrt{a^2 - (x+b)^2}}$$ </p>
<p>So I must add a $-$, but how? </p>
| Mike | 17,976 | <p>The main obstacle here is the square root. It is likely that eliminating it will allow us to proceed. So we want something squared minus 1 is the square of something. That leaves secant (or cosecant) as the best option.</p>
<p>As N3buchadnezzar states, it is not the only option. Consider the formulas</p>
<p>$(m+n)^2=m^2+2mn+n^2,(m-n)^2=m^2-2mn+n^2$</p>
<p>It is possible to do a substitution such that </p>
<p>$(m+n)^2-1=(m-n)^2$</p>
<p>$(m+n)^2-(m-n)^2=1$</p>
<p>$4mn=1$</p>
<p>So any substitution $x-1=\frac{f(t)}2+\frac1{2f(t)}$ will work, though back substitution may be messy and will likely require the same properties to undo. For example, let's do</p>
<p>$x-1=\frac12(e^t+e^{-t}),dx=\frac12(e^t-e^{-t})dt$</p>
<p>$\int\sqrt{(x-1)^2-1}dx=\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}+2)-1}dt$=</p>
<p>$\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}+2-4)}dt=$</p>
<p>$\int\frac12(e^t-e^{-t})\sqrt{\frac14(e^{2t}+e^{-2t}-2)}dt=\int(\sqrt{\frac14(e^{2t}+e^{-2t}-2)})^2dt=$</p>
<p>$\frac14\int e^{2t}+e^{-2t}-2dt=\frac18e^{2t}-\frac18e^{-2t}-\frac t2+C$</p>
<p>Now the problem of back-substitution. How do we get back a function in terms of x? Well, we have</p>
<p>$x-1=\frac12(e^t+e^{-t}),\sqrt{(x-1)^2-1}=\frac12(e^t-e^{-t})$</p>
<p>$x-1+\sqrt{x^2-2x}=e^t,t=\ln(x-1+\sqrt{x^2-2x})$</p>
<p>That you'll need for the $\frac t2$ term. For the rest, you'd want to use</p>
<p>$\frac18e^{2t}-\frac18e^{-2t}=\frac12(\frac{e^t+e^{-t}}2)(\frac{e^t-e^{-t}}2)=\frac12(x-1)\sqrt{x^2-2x}$</p>
<p>So, like I said, secant is the best option. :)</p>
|
84,711 | <p>This is a homework question I was asked to do</p>
<p>Of a twice differentiable function $ f : \mathbb{R} \to \mathbb{R} $ it is given that $f(2) = 3, f'(2) = 1$ and $f''(x) = \frac{e^{-x}}{x^2+1}$ . Now I have to prove that $$ \frac{7}{2} \leq f\left(\frac{5}{2}\right) \leq \frac{7}{2} + \frac{e^{-2}}{40} . $$ I tried this by computing the third Taylor polynomial of $f$ near $a=2$, setting $x = \frac{5}{2}$, which gave me $$f(5/2) \approx 7/2 + \frac{e^{-2}}{40} - \frac{ - e^{-5/2}}{48} $$, but now I don't know what to do next. I guess one has to do something with finding the error of the first and second order Taylor polynomials, but I'm not sure how to do so.
Can you help me?</p>
<p>Thanks in advance,</p>
| Arturo Magidin | 742 | <p>Using a local linear approximation (that is, a degree 1 Taylor polynomial approximation), we have that
$$f(x) \approx f(2) + f'(2)(x-2) = x+1.$$</p>
<p>Using the Lagrange Error Bound (with $n=1$) we have that
$$\left| f(x) - (x+1)\right| \leq \frac{M}{2!}|x-2|^2$$
where $\max|f''(x)|\leq M$ on the interval between $2$ and $x$.</p>
<p>On the interval $[2,2.5]$, the function $f''(x) = \frac{e^{-x}}{x^2+1}$ is decreasing (and positive), since the derivative is
$$-\frac{(x^2+1)e^{-x} + 2xe^{-x}}{(x^2+1)^2},$$
so we can take $M=f''(2) = \frac{e^{-2}}{5}$. Thus, the bound at $x=\frac{5}{2}$ is
$$\frac{M}{2}\left(\frac{1}{2}\right)^2 = \frac{e^{-2}}{10}\left(\frac{1}{2}\right)^2.$$</p>
<p>Plugging into the Lagrange Error Bound and resolving the absolute value gives:
$$-\left(\frac{e^{-2}}{10}\right)\left(\frac{1}{2}\right)^2 \leq f\left(\frac{5}{2}\right) - \frac{7}{2}\leq \frac{e^{-2}}{10}\left(\frac{1}{2}\right)^2$$
from which you should be able to deduce what you want.</p>
|
1,422,990 | <p>How to show that $(2^n-1)^{1/n}$ is irrational for all integer $n\ge 2$?</p>
<p>If $(2^n-1)^{1/n}=q\in\Bbb Q$ then $q^n=2^n-1$ which doesn't seem right, but I don't get how to prove it.</p>
| Spenser | 39,285 | <p>If $(2^n-1)^{1/n}=a/b$ for $a,b\in\Bbb Z$ and $n\ge 3$, then
$$a^n+b^n=(2b)^n,$$
contradicting <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="nofollow">Fermat's Last Theorem</a>.</p>
|
1,892 | <p>Although whether $$ P = NP $$ is important from theoretical computer science point of view, but I fail to see any practical implication of it.</p>
<p>Suppose that we can prove all questions that can be verified in polynomial time have polynomial time solutions, it won't help us in finding the actual solutions. Conversely, if we can prove that $$ P \ne NP,$$ then this doesn't mean that our current NP-hard problems have no polynomial time solutions. </p>
<p>From practical point of view ( practical as in the sense that we can immediately use the solution to real world scenario), it shouldn't bother me whether P vs NP is proved or disproved any more than whether my <strong><em>current problem</em></strong> has a polynomial time solution. </p>
<p>Am I right?</p>
| T.. | 467 | <p>If $P = NP$, computational revolution (once a specific algorithm is identified for an NP-hard problem, with explicit asymptotic runtime bounds).</p>
<p>If $P < NP$ <em>and one can prove it</em>, secure (classical) cryptography provably exists, and a huge missing piece in our understanding of computation is filled in. The first already has significant implications for daily life, and developing the second would have much larger implications.</p>
<p>You should also understand that after 40 years of research, today P=NP carries a host of related ideas like: easy-to-hard phase transition in combinatorial problems; quantifiable boundaries between easy and hard approximate versions of specific NP-complete problems (so getting within 7/8 of the optimal solution is easy but anything closer is NP-complete); counting and randomly sampling combinatorial objects are the same problem; zero-knowledge proofs "that reveal nothing but their own validity" (unforgeable ID cards). It's a very rich universe of ideas and it doesn't run out of questions once you know the answer to P=NP.</p>
|
4,277,616 | <blockquote>
<p>In how many different ways can we arrange <span class="math-container">$120$</span> students into <span class="math-container">$6$</span> groups for <span class="math-container">$6$</span> different classes so that the largest group has at most <span class="math-container">$2$</span> members more than the smallest group?</p>
</blockquote>
<p>My initial plan was to use a generating function, but I stumbled across a problem. Let's mark the groups with numbers <span class="math-container">$1$</span> to <span class="math-container">$6$</span> and let <span class="math-container">$n_i,i\in\{1,\ldots,6\}$</span> denote the number of members of the <span class="math-container">$i-$</span> th group in some arrangment. To see where this would lead me, for a moment, I assumed <span class="math-container">$n_1\le n_2\le\cdots\le n_6\le n_1+2$</span> in hope to find some range <span class="math-container">$\{m,\ldots M\}$</span> for <span class="math-container">$n_i$</span>'s and use a generating function <span class="math-container">$f(x)=(x^m+\cdots+x^M)^6$</span> and find <span class="math-container">$\langle x^{120}\rangle-$</span> the coefficient in front of <span class="math-container">$x^{120}$</span>, however students are distinct entities and <span class="math-container">$m$</span> and <span class="math-container">$M$</span> still remained misterious. I then tried figuring out if I was on the somewhat right track by, again taking <span class="math-container">$m=\min\{n_1,\ldots, n_6\}$</span> and write <span class="math-container">$n_i=m+j_i, j_i\in\{0,1,2\}.$</span> I believe, an arrangement with <span class="math-container">$2$</span> groups of <span class="math-container">$19, 2$</span> groups of <span class="math-container">$20$</span> and <span class="math-container">$2$</span> groups of <span class="math-container">$21$</span> people suggests there should be at least <span class="math-container">$19$</span> people in each group.</p>
<p>I also had a look at this problem:</p>
<p><a href="https://math.stackexchange.com/q/2386814/942722">The number of the partition of the set $A$ into $k$ bounded blocks.</a></p>
<p>but, again, I don't have any bounds on the blocks.</p>
<p>How should I proceed?</p>
| Asinomás | 33,907 | <p>Let <span class="math-container">$l$</span> be the size of the smallest group, we get <span class="math-container">$6l \leq 120 \leq 6l+5\times 2$</span>. Hence we must have <span class="math-container">$6l = 120$</span> or <span class="math-container">$6l= 114$</span>.</p>
<p>If <span class="math-container">$6l = 120$</span> then there is <span class="math-container">$\binom{120}{20,20,20,20,20,20}/6!$</span> solutions as all groups have size <span class="math-container">$20$</span>. See <a href="https://math.stackexchange.com/questions/3578443/the-number-of-partitions-of-n-to-k-equal-sized-sets">this related question</a>.</p>
<p>So lets move to the other case. We must see the possibilities for the group sizes, each group has size at least <span class="math-container">$19$</span> but there is <span class="math-container">$6$</span> extra people if they all have size <span class="math-container">$19$</span>, so we must find all partitions of <span class="math-container">$6$</span> into at most <span class="math-container">$5$</span> summands, such that each is at most <span class="math-container">$2$</span>.</p>
<p>There are only <span class="math-container">$3$</span> ways:</p>
<p><span class="math-container">$2,2,2$</span>. This corresponds to groups of sizes <span class="math-container">$19,19,19,21,21,21$</span> for which there are <span class="math-container">$\binom{120}{19,19,19,21,21,21}/(3!3!)$</span> partitions.</p>
<p><span class="math-container">$1,1,2,2$</span>. This corresponds to groups of sizes <span class="math-container">$19,19,20,20,21,21$</span> for which there are <span class="math-container">$\binom{120}{19,19,20,20,21,21}/(2!^3)$</span> partitions.</p>
<p><span class="math-container">$1,1,1,1,2$</span>. This corresponds to groups of sizes <span class="math-container">$19,20,20,20,20,21$</span> for which there are <span class="math-container">$\binom{120}{19,20,20,20,20,21}/{4!}$</span> partitions.</p>
|
4,277,616 | <blockquote>
<p>In how many different ways can we arrange <span class="math-container">$120$</span> students into <span class="math-container">$6$</span> groups for <span class="math-container">$6$</span> different classes so that the largest group has at most <span class="math-container">$2$</span> members more than the smallest group?</p>
</blockquote>
<p>My initial plan was to use a generating function, but I stumbled across a problem. Let's mark the groups with numbers <span class="math-container">$1$</span> to <span class="math-container">$6$</span> and let <span class="math-container">$n_i,i\in\{1,\ldots,6\}$</span> denote the number of members of the <span class="math-container">$i-$</span> th group in some arrangment. To see where this would lead me, for a moment, I assumed <span class="math-container">$n_1\le n_2\le\cdots\le n_6\le n_1+2$</span> in hope to find some range <span class="math-container">$\{m,\ldots M\}$</span> for <span class="math-container">$n_i$</span>'s and use a generating function <span class="math-container">$f(x)=(x^m+\cdots+x^M)^6$</span> and find <span class="math-container">$\langle x^{120}\rangle-$</span> the coefficient in front of <span class="math-container">$x^{120}$</span>, however students are distinct entities and <span class="math-container">$m$</span> and <span class="math-container">$M$</span> still remained misterious. I then tried figuring out if I was on the somewhat right track by, again taking <span class="math-container">$m=\min\{n_1,\ldots, n_6\}$</span> and write <span class="math-container">$n_i=m+j_i, j_i\in\{0,1,2\}.$</span> I believe, an arrangement with <span class="math-container">$2$</span> groups of <span class="math-container">$19, 2$</span> groups of <span class="math-container">$20$</span> and <span class="math-container">$2$</span> groups of <span class="math-container">$21$</span> people suggests there should be at least <span class="math-container">$19$</span> people in each group.</p>
<p>I also had a look at this problem:</p>
<p><a href="https://math.stackexchange.com/q/2386814/942722">The number of the partition of the set $A$ into $k$ bounded blocks.</a></p>
<p>but, again, I don't have any bounds on the blocks.</p>
<p>How should I proceed?</p>
| Robert Shore | 640,080 | <p>You can do this a smidgen more simply (at least, it's simpler to me).</p>
<p>First, the average size of a group is <span class="math-container">$20$</span>. If any group is smaller than average, then some other group must be larger than average, which means that there must be a group of size at least <span class="math-container">$21$</span>. Thus, the smallest group must have size at least <span class="math-container">$19$</span>.</p>
<p>Distribute <span class="math-container">$19$</span> people into each group. This can be done in <span class="math-container">$x=\binom{114}{19, 19, 19, 19, 19, 19}$</span> ways.</p>
<p>Now count the ways to distribute the remaining <span class="math-container">$6$</span> people across the groups with the constraint that no more than <span class="math-container">$2$</span> of these people can go into any group and multiply the result by <span class="math-container">$x$</span>. The unconstrained number of solutions of <span class="math-container">$x_1 + \cdots + x_6=6$</span> is <span class="math-container">$\binom{11}{6}$</span>. Using the principal of inclusion and exclusion, you must subtract the number of solutions where <span class="math-container">$x_i \geq 3$</span>. For any <span class="math-container">$i$</span>, there are <span class="math-container">$\binom 83$</span> such solutions, so you must subtract <span class="math-container">$6 \binom 83$</span>. But you also must add back the double-counted solutions where <span class="math-container">$x_i, x_j \geq 3$</span>, of which there are <span class="math-container">$\binom 62$</span>.</p>
<p>So the answer is:</p>
<p><span class="math-container">$$\left ( \binom{11}{6}-6 \binom 83 + \binom 62 \right) x.$$</span></p>
|
2,995,408 | <blockquote>
<p><span class="math-container">$$
\lim_{x\to 2^-}\frac{x(x-2)}{|(x+1)(x-2)|}=
\lim_{x\to 2^-}\left(\frac{x}{|x+1|}\cdot \frac{x-2}{|x-2|}\right)
$$</span></p>
</blockquote>
<p>So as the title says, is it okay to separate function under absolute value like this (i.e In form of Products) as shown in the denominator?</p>
| Ross Millikan | 1,827 | <p>Yes, in general <span class="math-container">$|ab|=|a||b|$</span>. You can check all the sign combinations to justify it.</p>
|
1,281,627 | <p>Today I completed the chapter of '**Limits **' in my school, and I found this chapter very fascinating. But the only problem I have with limits and Derivatives is that I don't know How can I use it in my daily life. (Any Book Recommendation?)</p>
| drawnonward | 51,530 | <p>A great way to see applications of the derivative is to consider real life functions, and look at the units you get when you apply Newton's Quotient.</p>
<p>As an example, lets say you have a velocity function based on time, then if you apply Newton's Quotient, you will see you are left with $m/s^2$(acceleration), while a position function will yield a velocity in $m/s$</p>
<p>As for books, I suppose it depends on where your interests lie.</p>
|
1,281,627 | <p>Today I completed the chapter of '**Limits **' in my school, and I found this chapter very fascinating. But the only problem I have with limits and Derivatives is that I don't know How can I use it in my daily life. (Any Book Recommendation?)</p>
| Jesse P Francis | 45,937 | <p>You might have calculated $\lim\limits_{n\to\infty}\frac{1}{n}$ during your course work ($=0$)- what does it mean? Does it mean $\frac{1}{n}=0$ for some n?</p>
<p>In other words, $\lim\limits_{n\to\infty}\frac{1}{n}=0$ is a neat and accurate way of saying, as the value of n gets bigger, $\frac{1}{n}$ is almost near 0 (and never equals 0).</p>
<hr>
<p>Have you heard of <a href="http://en.wikipedia.org/wiki/Zeno%27s_paradoxes#Dichotomy_paradox" rel="nofollow">Zeno's Paradox</a>? Mathematically, we will say, he will never reach the target (which is the real solution). But in reality, doesn't it sound a bit absurd? He reaches almost there, but not exactly there - do you think it has something to do with limits as I explained above?</p>
<hr>
<p>Also, it may be hard to find a direct application of a mathematical concept. But it may be a tool used in one of the concepts which has direct application in real life. Limit for example, is one such. Derivatives are widely used. You might have already studied that <a href="http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfDerivative.aspx" rel="nofollow">derivatives are defined using limits</a> in analysis. You might have learned integration, but have you ever wondered <a href="http://tutorial.math.lamar.edu/Classes/CalcI/DefnofDefiniteIntegral.aspx" rel="nofollow">how integration is defined</a>?</p>
<p>Once you get into higher classes, for nearly every other thing, limits are going to haunt you!</p>
|
110,722 | <p>1) Many Mathematics departments ask to send a "list of publications" while applying for research postdoctoral jobs. My question is: how important is it to post my papers in arXiv. I know, posting on arXiv is always good, because people might search for the arXiv -ed papers, but how much difference is publication on arXiv going to make ? What if I prepare a publication list (in .pdf) of accepted paper(s), and submitted paper(s) and paper(s) in preparation and send it to the employers ? Will that be sufficient or as good as putting them on arXiv ? (I will also send them a link to my site containing the publication list, including the downloadable links).</p>
<p>The situation is: I can put one of my papers in arXiv, but the others are either in collboration or deal with problems that stem from a question raised by collaborator(s) and the collaborator(s) did't agree to put on arXiv right away. Hence I am asking.</p>
<p>2) Also what exactly should I write in the "list of publications", the name of the paper, author(s) and its status(accepted/submitted/in preparation), that's all or or should I briefly also describe its content/abstract (to make it more informational) ?</p>
<p>Thank you.</p>
| Per Alexandersson | 1,056 | <p>You should put the preprints on arxiv. This allows for other researchers to find your work, and to cite it.
In extreme cases, it can take <strong>2-3 years to get a a referee report</strong> after submission, and then perhaps another year before finally getting published.
I am not kidding - I am speaking of personal experience.
In this time-frame, you might write several follow-up papers based on this initial paper. In order to get a follow-up paper refereed, you more or less must have all works it relies on available.</p>
|
2,178,714 | <p>Let $ f: (-1,1) \rightarrow \mathbb{R}$ be a bounded and continuous function . Prove that the function $ g(x)=(x^{2}-1)f(x) $ is uniformly continuous on $ (-1,1)$ . $$ $$ My little approach is, Since $f$ is bounded on $(-1,1)$ , there is positive $M \in \mathbb{R}$ such that </p>
<p>$$\forall x \in (-1,1)\,|f(x)| \leq M$$</p>
<p>Now, $ |g(x)-g(y)|=|(x^{2}-1)f(x)-(y^{2}-1)f(y)|=|x^{2}f(x)-y^{2}f(y)-(f(x)-f(y))|$</p>
| Adren | 405,819 | <p>Given $\epsilon>0$, there exists $a\in(-1,0)$ and $b\in(0,1)$ such that :</p>
<p>$$\forall x\in(-1,a],\,|g(x)|\le\epsilon$$</p>
<p>and similarly :</p>
<p>$$\forall x\in[b,1),\,|g(x)|\le\epsilon$$</p>
<p>(this is because $\lim_{x\to\pm 1}g(x)=0$)</p>
<p>By Heine theorem, $g$ is uniformly continuous on $[a,b]$ : there exists $\delta>0$ such that for every $(x,y)\in[a,b]$, if $x-y<\delta$ then $|g(x)-g(y)|\le\epsilon$.</p>
<p>It is now easy to show, with the triangle inequality, that $|g(x)-g(y)|\le2\epsilon$, for any $(x,y)\in(-1,1)^2$,</p>
|
173,131 | <p>Let's suppose that for the following expression:</p>
<p>$\qquad \alpha\,\beta +\alpha+\beta$</p>
<p>I know that $\alpha$ and $\beta$ are of small magnitude (e.g., 0 < $\alpha$ < 0.02 and 0 < $\beta$ < 0.02). Therefore, the magnitude of $\alpha\,\beta$ is negligible, i.e., the original expression can be approximated by</p>
<p>$\qquad \alpha+\beta$</p>
<p>Is there any command in Mathematica to do such an operation?</p>
<hr>
<p>Reminder from original question: if and $\alpha$ and $\beta$ are of small magnitude, we may approximate the original equation by disregarding nonlinear terms, such as $\alpha^n$, $\beta^n$,n=2,3,..., and $\alpha\beta$</p>
<p>My observation is that by applying the good code suggestion of @Henrik Schumacher in a fraction it seems to not generate a proper result to the whole fraction. For instance,if </p>
<p>$\text{numerator}=\alpha \beta +\alpha +\beta ^2+\beta -\lambda \epsilon q(t)+\text{$\beta $q}(t)$ </p>
<pre><code>numerator = α*β + β^2 + α + β + β q[t] - ϵ*λ*q[t]
f = numerator;
(f /. {α -> 0, β -> 0}) + (D[f, {{α, β}, 1}] /. {α -> 0, β -> 0}) . {α, β}
</code></pre>
<p>The code generates the correct elimination on numerator :$\alpha +\beta -\lambda \epsilon q(t)+\text{$\beta $q}(t)$</p>
<p>and</p>
<p>$\text{denominator}=\alpha +\alpha (-\beta ) \text{LD}(t)+\alpha \beta q(t)+\beta q(t)+\epsilon +1$:</p>
<pre><code>denominator= 1 +α +ϵ -α*β*LD[t] +β*q[t] +α*β*q[t]
f = denominator;
(f /. {α -> 0, β -> 0}) + (D[f, {{α, β}, 1}] /. {α -> 0, β -> 0}) . {α, β}
</code></pre>
<p>The code generates the correct elimination on denominator: $\alpha +\beta q(t)+\epsilon +1$ </p>
<p>Nevertheless, when applying the suggested first order Taylor Series expansion in the fraction as a whole: </p>
<pre><code>f = (numerator /denominator);
(f /. {α -> 0, β -> 0}) + (D[
f, {{α, β}, 1}] /. {α -> 0, β ->
0}).{α, β} // Simplify
</code></pre>
<p>The result generated is incorrect.
$\frac{(\epsilon +1) (\alpha +\beta )-q(t) (\lambda \epsilon (-\alpha +\epsilon +1)+\beta \text{$\beta $q}(t))+\beta \lambda \epsilon q(t)^2+(-\alpha +\epsilon +1) \text{$\beta $q}(t)}{(\epsilon +1)^2}$</p>
<p>That can be observed by, for instant, noticing that in the output above (i) the numerator has $\beta^2$, (ii) the code generated $q[t]^2$ that did not exist in the original equation.</p>
<p>I hope that this time I could express my concern in a proper format.
Thank you all for your support!</p>
<hr>
<p>Questions related to @Akku14's code suggestion</p>
<p><strong>Question 1</strong>:
@Akku14, I was trying to use your code suggestion with the slight modification in the original purpose(instead of eliminating α and β, now eliminating α, β and LD[t]), but I had no success. I think the reason is because I could not find a way of writing LD[t] as a parameter of function g: </p>
<p>for the following equation: </p>
<pre><code>\[CapitalDelta]p[t] = (α*β + β^2 + α + β + β*q[t] - ϵ*λ*q[t])/(1 + α + ϵ -α*LD[t] + β*q[t] + α*β*q[t])
</code></pre>
<p>$\text{$\Delta $p}(t)=\frac{\alpha \beta +\alpha +\beta ^2+\beta +\beta q(t)-\lambda \epsilon q(t)}{\alpha +\alpha (-\text{LD}(t))+\alpha \beta q(t)+\beta q(t)+\epsilon +1}$</p>
<pre><code>g[α_, β_, LD[t] _] = \[CapitalDelta]p[t]
</code></pre>
<p>By using @Akku14's suggestion:</p>
<pre><code>ser = (Series[g[α eps, β eps, LD[t] eps], {eps, 0, 1}] //
Normal) /. eps -> 1 // Simplify
</code></pre>
<p>I get the following output:</p>
<p>$\frac{(\epsilon +1) (\alpha +\beta )+q(t) (\lambda \epsilon (\alpha -\epsilon -1)+\beta (\epsilon +1)-\alpha \lambda \epsilon LD[t]+\beta \lambda \epsilon q(t)^2}{(\epsilon +1)^2}$</p>
<p>which is incorrect since $\alpha \lambda \epsilon LD[t]$ is present in the numerator of <code>ser</code>.</p>
<p>Again, I think that the problem is that my <code>g</code> is not recognizing LD[t] as a parameter; would any of you know how to approach this issue?</p>
<p><strong>Question 2.1</strong>: in case I wanted second order Taylor Series for <code>g[α, β, LD[t]]</code>, changing <code>{eps, 0, 1}</code> to <code>{eps, 0, 2}</code> at <code>ser</code> would be enough to get the correct result? Like:</p>
<pre><code>ser = (Series[g[α eps, β eps, LD[t] eps], {eps, 0, 2}] //
Normal) /. eps -> 1 // Simplify
</code></pre>
<p><strong>Question 2.2</strong>: in case I wanted n order Taylor Series for <code>g[α, β, LD[t]]</code>, changing <code>{eps, 0, 1}</code> to <code>{eps, 0, n}</code> at <code>ser</code> would be enough to get the correct result?</p>
| Henrik Schumacher | 38,178 | <p>You could use first order Taylor expansion, e.g. with</p>
<pre><code>f = α β + α + β;
(f /. {α -> 0, β -> 0}) + (D[f, {{α, β}, 1}] /. {α -> 0, β ->0}).{α, β}
</code></pre>
<blockquote>
<p>$\alpha +\beta$</p>
</blockquote>
<p>For your second example (with typos fixed), I obtain</p>
<pre><code>numerator = α β + β^2 + α + β + β q[t] - ϵ λ q[t];
denominator = 1 + α + ϵ - α β LD[t] + β q[t] + α β q[t];
f = numerator/denominator;
(f /. {α -> 0, β -> 0}) + (D[ f, {{α, β}, 1}] /. {α -> 0, β -> 0}).{α, β}
</code></pre>
<blockquote>
<p>$$\alpha \left(\frac{\lambda \epsilon q(t)}{(\epsilon
+1)^2}+\frac{1}{\epsilon +1}\right)+\beta \left(\frac{\lambda
\epsilon q(t)^2}{(\epsilon +1)^2}+\frac{q(t)+1}{\epsilon
+1}\right)-\frac{\lambda \epsilon q(t)}{\epsilon +1}$$</p>
</blockquote>
|
3,765,398 | <p>The range of <span class="math-container">$\alpha$</span> for which all the points of local extrema of the function <span class="math-container">$f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$</span> lie in the interval (–2, 4), is</p>
<p>(A) (–1, 3)</p>
<p>(B) (3, 4)</p>
<p>(C) (–4, –2)</p>
<p>(D) (–2, –1)</p>
<p>My approach is as follow</p>
<p><span class="math-container">$f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$</span></p>
<p><span class="math-container">$f'\left( x \right) = 3{x^2} - 6\alpha x + 3\left( {{\alpha ^2} - 1} \right) = 0$</span></p>
<p><span class="math-container">$D = 36{\alpha ^2} - 36\left( {{\alpha ^2} - 1} \right) \geqslant 0$</span></p>
<p>I can easily solve this problem if I can get the roots of x and I can get the value at which x is at the extremum position, but as I am not getting the value I am struck in this problem.</p>
| Orenio | 783,418 | <p>Solve the quadratic equation with the parameter <span class="math-container">$\alpha$</span>:</p>
<p><span class="math-container">$x_{1,2}=\frac{6\alpha\pm\sqrt{({36\alpha^2-36\alpha^2+36})}}{6}$</span></p>
<p>Meaning <span class="math-container">$x_1=\alpha+1, x_2=\alpha-1$</span></p>
<p>Can you finish solving from here?</p>
|
3,765,398 | <p>The range of <span class="math-container">$\alpha$</span> for which all the points of local extrema of the function <span class="math-container">$f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$</span> lie in the interval (–2, 4), is</p>
<p>(A) (–1, 3)</p>
<p>(B) (3, 4)</p>
<p>(C) (–4, –2)</p>
<p>(D) (–2, –1)</p>
<p>My approach is as follow</p>
<p><span class="math-container">$f\left( x \right) = {x^3} - 3\alpha {x^2} + 3\left( {{\alpha ^2} - 1} \right)x + 1$</span></p>
<p><span class="math-container">$f'\left( x \right) = 3{x^2} - 6\alpha x + 3\left( {{\alpha ^2} - 1} \right) = 0$</span></p>
<p><span class="math-container">$D = 36{\alpha ^2} - 36\left( {{\alpha ^2} - 1} \right) \geqslant 0$</span></p>
<p>I can easily solve this problem if I can get the roots of x and I can get the value at which x is at the extremum position, but as I am not getting the value I am struck in this problem.</p>
| Quanto | 686,284 | <p>Note that</p>
<p><span class="math-container">\begin{align}
f'( x ) & = 3{x^2} - 6\alpha x + 3\left( {{\alpha ^2} - 1} \right)\\
&= 3[x^2 -2ax + (a+1)(a-1)] \\
& = 3[x-(a+1)][x-(a-1)]=0
\end{align}</span></p>
<p>which leads to the roots <span class="math-container">$x=a\pm 1$</span>. Then, solve <span class="math-container">$-2<a\pm 1< 4$</span> to obtain <span class="math-container">$a\in (-1,3)$</span>.</p>
|
2,214,236 | <p>The question:</p>
<blockquote>
<p>An object is dropped from a cliff. How far does the object fall in the 3rd second?"</p>
</blockquote>
<p>I calculated that a ball dropped from rest from a cliff will fall $45\text{ m}$ in $3 \text{ s}$, assuming $g$ is $10\text{ m/s}^2$.</p>
<p>$$s = (0 \times 3) + \frac{1}{2}\cdot 10\cdot (3\times 3) = 45\text{ m}$$</p>
<p>But my teacher is telling me $25\text{ m}$! </p>
<p>EDITS: His reasoning was that from $t=0$ to $t=1$, $s=10\text{ m}$, and from $t=1$ to $t =2$, $s=20$...</p>
<p>The mark scheme also says $25\text{ m}$</p>
| Narasimham | 95,860 | <p>Your teacher is correct. Question asks how much distance is covered between $t= 2$ and $ t= 3.$ Time lapse is 1 second, that is, in the <em>third second of duration</em>. In meters, distance travelled =</p>
<p>$$ s = \frac12 \cdot 10\cdot (3^2-2^2) = 25, $$ </p>
<p>and, if you draw the parabola graph, $s_2-s_1 = a (t_2^2-t_1^2)/2. $</p>
|
236,933 | <p>When I use ListPlot, I want to show the labels, such as</p>
<pre><code>ListPlot[Callout[#, #, Above] & /@ Range[10], Joined -> True, Mesh -> All]
</code></pre>
<p>Now all positions are <strong>Above</strong>, but sometimes the labels will over other text, so I want to set some of them <strong>Below</strong>, for example, the second and third point labels are <strong>Below</strong>, as the image shows.</p>
<p><a href="https://i.stack.imgur.com/xZ06X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xZ06X.png" alt="enter image description here" /></a></p>
| cvgmt | 72,111 | <p>One way is use <code>MapAt</code></p>
<pre><code>data = Range[10];
mapBelow[n_] := MapAt[Callout[#, #, Below] &, n];
mapAbove[n_] := MapAt[Callout[#, #, Above] &, n];
below = Table[mapBelow[n], {n, {2, 3, 8}}];
above = Table[mapAbove[n], {n, Complement[Range[10], {2, 3, 8}]}];
ListPlot[Composition[Sequence @@ Flatten[{above, below}]][data],
Joined -> True, Mesh -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/eWhRb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eWhRb.png" alt="enter image description here" /></a></p>
|
30,918 | <p>Judging by some of the posts on meta<sup>1</sup> and comments posted there it seems that there are users who try to improve the posts by correcting spelling mistakes. Of course, there are other ways to improve the posts via editing, some of them probably more important than grammar and spelling.<sup>2</sup> Still this is certainly an improvement, a post with correct spelling is easier to read and using English correctly also increases the professional appearance of this site. Moreover, it helps with searching. (Posts which contain only the misspelled form of the words are less likely to be found when searching.)</p>
<p>Maybe it could be useful to <em>collect some commonly misspelled words</em>. This could make it easier for users, who have some spare time and want to do a few such edits, to find posts which have this problem.</p>
<p>For this reason it might be useful also include link to a search (or a <a href="https://math.meta.stackexchange.com/tags/data-explorer/info">SEDE query</a>) which returns those posts. Some minor comments on searching:</p>
<ul>
<li>For some words, the search also returns posts which contain minor variants of the words. Since we are searching for misspellings, in such cases adding quotation marks might help to find the misspelled ones. (For example, searching for <a href="https://math.stackexchange.com/search?q=occurence">occurence</a> returns many results, but we are more interested in the ones which are found when we search for <a href="https://math.stackexchange.com/search?q="occurence"">"occurence"</a> and <a href="https://math.stackexchange.com/search?q="occurences"">"occurences"</a>.)</li>
<li>A simple SEDE query searching for some string can return words which contain this string as a some part. So sometimes we might want to look only at results containing exactly the given word. (Compare the results <a href="https://data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=banch?num=5000&text=teh" rel="nofollow noreferrer">searching for teh</a> with the results where <a href="https://data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=teh" rel="nofollow noreferrer">teh is preceded and followed by a non-letter</a>).</li>
<li>It is better to use case insensitive version of SEDE queries. (So that we catch also misspelled word at the beginning of a sentence. This is especially important for names.)</li>
<li>With SEDE you can find several words which contain the same substring. For example, words containing "anali" might often be typos (such as analisis, analise, analitic, etc.} and you can search for <a href="https://data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=anali" rel="nofollow noreferrer">anali</a>.</li>
</ul>
<p>A few comments about editing the misspellings. (Some of them apply to editing in general.)</p>
<ul>
<li>It is better to avoid <a href="https://math.meta.stackexchange.com/questions/5068/how-much-bumping-is-too-much">bumping too many old posts at the same time</a>. (Frontpage is considered a precious commodity.)</li>
<li>For this reason it is better to concentrate on questions which are new or have been recently bumped for some other reason.</li>
<li>If you edit some post, try to check whether there are also other improvements to do. (Since the post was bumped by correcting the misspelled word, it is better to improve also other things which are worth editing than to bump the same post by additional edits later.)</li>
<li>Difference between British/American spelling should not be considered a misspelling, as discussed before: <a href="https://math.meta.stackexchange.com/q/30335">Is English (US) vs English (UK) grounds for an edit?</a></li>
</ul>
<p>Some names of mathematicians which are commonly misspelled can also be found in this answer: <a href="https://math.meta.stackexchange.com/q/8698#8709">Searching for accented characters is too strict</a>. (Although that answer is primarily about alternative spellings.) This might also help when creating list of common misspellings.</p>
<p><sup>1</sup>Posts such as: <a href="https://math.meta.stackexchange.com/q/13730">Edit session for wrong spelling of mathematicians and mathematical concepts</a> or
<a href="https://math.meta.stackexchange.com/q/30912">Wrong spelling of “occurrence”</a>. Some users have mentioned in chat that they do this kind of edits, for example, <a href="https://chat.stackexchange.com/transcript/message/2443651#2443651">Srivatsan</a>.</p>
<p><sup>2</sup>Just to name a few common problems with posts on this site which could be helped by editing: <a href="https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question#10144">non-descriptive titles</a>, incorrectly tagged question, various problems with MathJax in the post or <a href="https://math.meta.stackexchange.com/questions/9687/guidelines-for-good-use-of-rm-latex-in-question-titles">in the title</a>, etc. (Still, I'd guess that posts with misspellings are slightly more likely to have also some other problems - so perhaps checking them from time to time might be useful also for other purposes.)</p>
<p><sup>3</sup>These SEDE queries are more for people who are curious (since they do not actually help with editing), but they still might be interesting anyway. You can find <a href="https://data.stackexchange.com/math/query/1157683/posts-which-contained-the-text-in-some-revision-and-it-was-later-removed-case-in?word=haussdorff" rel="nofollow noreferrer">posts where a specific word has already been corrected</a>. You can also search in comments <a href="https://data.stackexchange.com/math/query/556789/comments-containing-given-keyword-with-text-and-author?Word=occurence" rel="nofollow noreferrer">for some string</a> or for <a href="https://data.stackexchange.com/math/query/1157639/comments-containing-exact-string-with-text-and-author?text=teh" rel="nofollow noreferrer">an exact word</a>.</p>
| Martin Sleziak | 8,297 | <p>Here are some words which have been misspelled in some posts on the site, also some SEDE queries are included.<sup>1</sup></p>
<ul>
<li>alegbra; <a href="//math.stackexchange.com/search?tab=active&q=alegbra">search</a></li>
<li>analisis; <a href="//math.stackexchange.com/search?tab=active&q=analisis">search</a></li>
<li>analyis; <a href="//math.stackexchange.com/search?tab=active&q=analyis">search</a></li>
<li>aplication; <a href="//math.stackexchange.com/search?q=aplication">search</a></li>
<li>asyptotic; <a href="//math.stackexchange.com/search?tab=active&q=asyptotic">search</a></li>
<li>Banch; <a href="//math.stackexchange.com/search?tab=active&q=banch">search</a></li>
<li>biyective; <a href="//math.stackexchange.com/search?tab=active&q=biyective">search</a></li>
<li>boudary; <a href="//math.stackexchange.com/search?tab=active&q=boudary">search</a></li>
<li>Caley; <a href="//math.stackexchange.com/search?tab=active&q=caley">search</a></li>
<li>caracteristic; <a href="//math.stackexchange.com/search?tab=active&q=caracteristic">search</a></li>
<li>Cauchy+Swartz; <a href="//math.stackexchange.com/search?tab=active&q=cauchy+swartz">search</a>, <a href="//data.stackexchange.com/math/query/1157669/posts-containing-two-given-words-case-insensitive?text1=cauchy&text2=swartz" rel="nofollow noreferrer">SEDE</a></li>
<li>Cauchy+Shwartz; <a href="//math.stackexchange.com/search?tab=active&q=cauchy+shwartz">search</a>, <a href="//data.stackexchange.com/math/query/1157669/posts-containing-two-given-words-case-insensitive?text1=cauchy&text2=shwartz" rel="nofollow noreferrer">SEDE</a></li>
<li>Cauchy+Schwartz; <a href="//math.stackexchange.com/search?tab=active&q=cauchy+schwartz">search</a>, <a href="//data.stackexchange.com/math/query/1157669/posts-containing-two-given-words-case-insensitive?text1=cauchy&text2=schwartz" rel="nofollow noreferrer">SEDE</a></li>
<li>choosen; <a href="//math.stackexchange.com/search?tab=active&q=choosen">search</a></li>
<li>compacness; <a href="//math.stackexchange.com/search?q=compacness">search</a></li>
<li>conjecutre; <a href="//math.stackexchange.com/search?tab=active&q=conjecutre">search</a></li>
<li>continity; <a href="//math.stackexchange.com/search?tab=active&q=continity">search</a></li>
<li>continuty; <a href="//math.stackexchange.com/search?tab=active&q=continuty">search</a></li>
<li>continous; <a href="//math.stackexchange.com/search?tab=active&q=continous">search</a></li>
<li>continus; <a href="//math.stackexchange.com/search?tab=active&q=continus">search</a></li>
<li>continuos; <a href="//math.stackexchange.com/search?tab=active&q=continuos">search</a></li>
<li>Cuachy; <a href="//math.stackexchange.com/search?tab=active&q=cuachy">search</a></li>
<li>definiton; <a href="//math.stackexchange.com/search?tab=active&q=definiton">search</a></li>
<li>defintion; <a href="//math.stackexchange.com/search?tab=active&q=defintion">search</a></li>
<li>defition; <a href="//math.stackexchange.com/search?tab=active&q=defition">search</a></li>
<li>defnition; <a href="//math.stackexchange.com/search?tab=active&q=defnition">search</a></li>
<li>deriative; <a href="//math.stackexchange.com/search?tab=active&q=deriative">search</a></li>
<li>deriavtive; <a href="//math.stackexchange.com/search?tab=active&q=deriavtive">search</a></li>
<li>deriviative; <a href="//math.stackexchange.com/search?tab=active&q=deriviative">search</a></li>
<li>derteminant; <a href="//math.stackexchange.com/search?tab=active&q=derteminant">search</a></li>
<li>derterminant; <a href="//math.stackexchange.com/search?tab=active&q=derterminant">search</a></li>
<li>dervative; <a href="//math.stackexchange.com/search?q=dervative">search</a></li>
<li>diferent; <a href="//math.stackexchange.com/search?tab=active&q=diferent">search</a></li>
<li>diffrent; <a href="//math.stackexchange.com/search?tab=active&q=diffrent">search</a></li>
<li>Eluer; <a href="//math.stackexchange.com/search?tab=active&q=eluer">search</a></li>
<li>exercice; <a href="//math.stackexchange.com/search?tab=active&q=exercice">search</a></li>
<li>finte; <a href="//math.stackexchange.com/search?tab=active&q=finte">search</a></li>
<li>fintely; <a href="//math.stackexchange.com/search?tab=active&q=fintely">search</a></li>
<li>fucntion; <a href="//math.stackexchange.com/search?tab=active&q=fucntion">search</a></li>
<li>funcion; <a href="//math.stackexchange.com/search?tab=active&q=funcion">search</a></li>
<li>fundemental; <a href="//math.stackexchange.com/search?tab=active&q=fundemental">search</a></li>
<li>funtion; <a href="//math.stackexchange.com/search?tab=active&q=funtion">search</a></li>
<li>Haussdorff; <a href="//math.stackexchange.com/search?tab=active&q=haussdorff">search</a></li>
<li>Hausdorf; <a href="//math.stackexchange.com/search?tab=active&q=hausdorf">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=hausdorf" rel="nofollow noreferrer">SEDE</a></li>
<li>inegral; <a href="//math.stackexchange.com/search?tab=active&q=inegral">search</a></li>
<li>inegration; <a href="//math.stackexchange.com/search?tab=active&q=inegration">search</a></li>
<li>ineqality; <a href="//math.stackexchange.com/search?tab=active&q=ineqality">search</a></li>
<li>ineqaulity; <a href="//math.stackexchange.com/search?tab=active&q=ineqaulity">search</a></li>
<li>infnity; <a href="//math.stackexchange.com/search?tab=active&q=infnity">search</a></li>
<li>integal; <a href="//math.stackexchange.com/search?tab=active&q=integal">search</a></li>
<li>integation; <a href="//math.stackexchange.com/search?tab=active&q=integation">search</a></li>
<li>integtal; <a href="//math.stackexchange.com/search?tab=active&q=integtal">search</a></li>
<li>intergal; <a href="//math.stackexchange.com/search?tab=active&q=intergal">search</a></li>
<li>intergral; <a href="//math.stackexchange.com/search?tab=active&q=intergral">search</a></li>
<li>intergate; <a href="//math.stackexchange.com/search?tab=active&q=intergate">search</a></li>
<li>intergrate; <a href="//math.stackexchange.com/search?tab=active&q=intergrate">search</a></li>
<li>intervall; <a href="//math.stackexchange.com/search?tab=active&q=intervall">search</a></li>
<li>invaraint; <a href="//math.stackexchange.com/search?tab=active&q=invaraint">search</a></li>
<li>inyective; <a href="//math.stackexchange.com/search?tab=active&q=inyective">search</a></li>
<li>Kirchoff; <a href="//math.stackexchange.com/search?tab=active&q=kirchoff">search</a></li>
<li>Lapace; <a href="//math.stackexchange.com/search?tab=active&q=lapace">search</a></li>
<li>Laplce; <a href="//math.stackexchange.com/search?tab=active&q=laplce">search</a></li>
<li>Lebegue; <a href="//math.stackexchange.com/search?tab=active&q=lebegue">search</a></li>
<li>Lebesque; <a href="//math.stackexchange.com/search?tab=active&q=lebesque">search</a></li>
<li>Lebesue; <a href="//math.stackexchange.com/search?tab=active&q=lebesue">search</a></li>
<li>mathemaic; <a href="//math.stackexchange.com/search?tab=active&q=mathemaic">search</a></li>
<li>mathemaical; <a href="//math.stackexchange.com/search?tab=active&q=mathemaical">search</a></li>
<li>matrice; <a href="//math.stackexchange.com/search?tab=active&q=%22matrice%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=matrice" rel="nofollow noreferrer">SEDE</a></li>
<li>Mclaurin; <a href="//math.stackexchange.com/search?tab=active&q=mclaurin">search</a> </li>
<li>Minkowsky; <a href="//math.stackexchange.com/search?tab=active&q=minkowsky">search</a></li>
<li>miminum; <a href="//math.stackexchange.com/search?tab=active&q=miminum">search</a></li>
<li>minimun; <a href="//math.stackexchange.com/search?tab=active&q=minimun">search</a></li>
<li>mininum; <a href="//math.stackexchange.com/search?tab=active&q=mininum">search</a></li>
<li>miniscule; <a href="//math.stackexchange.com/search?q=miniscule">search</a></li>
<li>minumum; <a href="//math.stackexchange.com/search?tab=active&q=minumum">search</a></li>
<li>Neuman; <a href="//math.stackexchange.com/search?tab=active&q=neuman">search</a></li>
<li>notrivial; <a href="//math.stackexchange.com/search?tab=active&q=notrivial">search</a></li>
<li>occurence; <a href="//math.stackexchange.com/search?tab=active&q=%22occurence%22">search</a></li>
<li>ocurrence; <a href="//math.stackexchange.com/search?tab=active&q=%22ocurrence%22">search</a></li>
<li>paralel; <a href="//math.stackexchange.com/search?tab=active&q=%22paralel%22">search</a></li>
<li>patrial; <a href="//math.stackexchange.com/search?tab=active&q=patrial">search</a></li>
<li>polinomial; <a href="//math.stackexchange.com/search?tab=active&q=polinomial">search</a></li>
<li>polinomyal; <a href="//math.stackexchange.com/search?tab=active&q=polinomyal">search</a></li>
<li>poynomial; <a href="//math.stackexchange.com/search?tab=active&q=poynomial">search</a></li>
<li>proble; <a href="//math.stackexchange.com/search?tab=active&q=proble">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=proble" rel="nofollow noreferrer">SEDE</a></li>
<li>proofed; <a href="//math.stackexchange.com/search?tab=active&q=%22proofed%22">search</a></li>
<li>quare; <a href="//math.stackexchange.com/search?tab=active&q=quare">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=quare" rel="nofollow noreferrer">SEDE</a></li>
<li>represetnation; <a href="//math.stackexchange.com/search?tab=active&q=represetnation">search</a></li>
<li>Reiman; <a href="//math.stackexchange.com/search?tab=active&q=reiman">search</a></li>
<li>Reimann; <a href="//math.stackexchange.com/search?tab=active&q=reimann">search</a></li>
<li>Rieman; <a href="//math.stackexchange.com/search?tab=active&q=rieman">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=rieman" rel="nofollow noreferrer">SEDE</a></li>
<li>Shwartz; <a href="//math.stackexchange.com/search?tab=active&q=shwartz">search</a></li>
<li>subsitution; <a href="//math.stackexchange.com/search?tab=active&q=subsitution">search</a></li>
<li>substract; <a href="//math.stackexchange.com/search?tab=active&q=substract">search</a></li>
<li>suryective; <a href="//math.stackexchange.com/search?tab=active&q=suryective">search</a></li>
<li>teh; <a href="//math.stackexchange.com/search?tab=active&q=teh">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=teh" rel="nofollow noreferrer">SEDE</a></li>
<li>theorm; <a href="//math.stackexchange.com/search?tab=active&q=theorm">search</a></li>
<li>thinnk; <a href="//math.stackexchange.com/search?tab=active&q=thinnk">search</a></li>
<li>toplogy; <a href="//math.stackexchange.com/search?tab=active&q=toplogy">search</a></li>
<li>ultrafiler; <a href="//math.stackexchange.com/search?tab=active&q=ultrafiler">search</a></li>
<li>ultrafiter; <a href="//math.stackexchange.com/search?tab=active&q=ultrafiter">search</a></li>
<li>utrafilter; <a href="//math.stackexchange.com/search?tab=active&q=utrafilter">search</a></li>
<li>varition; <a href="//math.stackexchange.com/search?tab=active&q=varition">search</a></li>
<li>vertice; <a href="//math.stackexchange.com/search?tab=active&q=%22vertice%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=vertice" rel="nofollow noreferrer">SEDE</a></li>
<li>Weirstrass; <a href="//math.stackexchange.com/search?tab=active&q=%22Weirstrass%22">search</a></li>
<li>Wierstrass; <a href="//math.stackexchange.com/search?tab=active&q=%22Wierstrass%22">search</a></li>
<li>Weierstrauss; <a href="//math.stackexchange.com/search?tab=active&q=%22Weierstrauss%22">search</a></li>
<li>whit; <a href="//math.stackexchange.com/search?tab=active&q=whit">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=whit" rel="nofollow noreferrer">SEDE</a></li>
</ul>
<p><sup>1</sup>There is character limit on the length of the post, to save space the SEDE queries are left only in the cases where <em>exact</em> search is needed and in cases where you search for <em>multiple</em> strings. You can easily modify SEDE query by changing parameter in the link, here is query for <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=alegbra" rel="nofollow noreferrer">algebra</a>. Notice that this query simply searches whether the post contains the given string - in cases when the word can be a substring of a different word, you should use <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=rieman" rel="nofollow noreferrer">a different query</a>. (You want to find posts containing the word Rieman, but not the word Riemann - which would be among the search results if you used the first query.)</p>
|
3,960,282 | <p>I am trying to find <span class="math-container">$z$</span> such that
<span class="math-container">$$\dot{z} = -1 + e^{-iz^*},$$</span>
where <span class="math-container">$*$</span> denotes complex conjugate and the dots represent derivatives with respect to time. The time dependence of the dependent variables is suppressed for clarity of presentation. Letting <span class="math-container">$z=x+iy$</span>,
<span class="math-container">$$\dot{x} = -1 - e^{-y}\cos x, \quad \dot{y}= e^{-y}\sin x,$$</span>
together with the constraint that
<span class="math-container">$$A = -y+ e^{-y}\cos x$$</span>
is a constant, i.e. <span class="math-container">$\dot{A}=0$</span>.</p>
<p>The normal way I would proceed would be to eliminate <span class="math-container">$y$</span> in favor of <span class="math-container">$x$</span>, which yields product log (ie Lambert W) functions, but the resulting ODE is complicated, and I cannot solve it in closed form. Note, if we take <span class="math-container">$z = i \log \zeta$</span>, we can equate this to a problem involving a point vortex in a uniform stream, but I'm not sure this identification helps.</p>
| JJacquelin | 108,514 | <p><span class="math-container">$$\begin{cases}
\frac{dx}{dt}=-1-e^{-y}\cos(x) \\
\frac{dy}{dt}=e^{-y}\sin(x)
\end{cases}\quad\implies\quad
\frac{dy}{dx}=\frac{e^{-y}\sin(x)}{-1-e^{-y}\cos(x)}$$</span>
<span class="math-container">$$\left(1+e^{-y}\cos(x) \right)dy+e^{-y}\sin(x)dx=0$$</span>
<span class="math-container">$$dy+d\left(-e^{-y}\cos(x) \right)=0$$</span>
<span class="math-container">$$y-e^{-y}\cos(x)=\text{constant}$$</span>
This is equivalent to the so called "constrain" :
<span class="math-container">$$-y+e^{-y}\cos(x)=A$$</span>
Thus this is not realy a constrain. This is the equation of trajectory corresponding to the solution of the original system of ODEs.</p>
<p>The above implicit equation can be explicitly written on various forms :
<span class="math-container">$$x(y)=\cos^{-1}\left( (y+A)e^y\right)$$</span>
<span class="math-container">$$y(x)=-A+\text{W}\left(e^A\cos(x) \right)$$</span>
W denotes the Lambert W function.</p>
<p><span class="math-container">$$\frac{dy}{dt}=e^{-y}\sin(x)=\pm e^{-y}\sqrt{1-\cos^2(x)}=\pm e^{-y}\sqrt{1-(y+A)^2e^{2y}}$$</span>
<span class="math-container">$$\frac{dy}{dt}=\pm\sqrt{e^{-2y}-(y+A)^2}$$</span>
<span class="math-container">$$t=\pm\int \frac{dy}{\sqrt{e^{-2y}-(y+A)^2}}$$</span>
This gives the analytic solution <span class="math-container">$t(y)$</span>.</p>
<p>As far as I know, this integral cannot be expressed with a finite number of standard functions. A-fortiori don't expect the inverse function <span class="math-container">$y(t)$</span> on explicit form. Better use numerical calculus.</p>
|
1,940,446 | <p>I'm working through Bona's "A Walk Through Combinatorics" and I came across this problem:</p>
<blockquote>
<p>A company has $20$ employees, $12$ male and $8$ female. How many ways are
there to form a $5$ person committee that contains at least one male and
at least one female?</p>
</blockquote>
<p>I realise that this question has been asked here before multiple times and that the standard way to solve such problems is to enumerate all the possible committees and then subtract the ones that do not satisfy the constraint, i.e., consist entirely of the same sex; thus giving the answer: $20C5-12C5-8C5 = 14656$</p>
<p>However, I first tried to solve it like this:
A pair of a man and a woman can be first chosen in $12*8$ ways, thus leaving $3$ open slots on the committee which can be filled in $18C3$ ways; thus giving an answer $12*8*18C3 = 78336$</p>
<p>What is the error in reasoning here?</p>
| barak manos | 131,263 | <p>The error in the other method is double-counting.</p>
<p>For example, the following combinations are essentially identical:</p>
<ul>
<li>$M_1+W_1,M_2+M_3+M_4$</li>
<li>$M_2+W_1,M_1+M_3+M_4$</li>
<li>$M_3+W_1,M_1+M_2+M_4$</li>
<li>$M_4+W_1,M_1+M_2+M_3$</li>
</ul>
|
2,530,820 | <p>Let $1\leq p<\infty$ and $q$ be the conjugate exponent of $p$. Suppose that $\lbrace a^n \rbrace_{n=1}^{\infty} \subset \ell^q$ in a sequence in $\ell^q$ such that $f_{a^n}(x) \mapsto 0~( n \mapsto \infty)$ for all $x \in \ell^p$ where $f_{a^n}(x)=\sum_{i=1}^{\infty} a_i^{(n)}x_i$. Show that the sequence $\lbrace a^n \rbrace_{n=1}^{\infty}$ is bounded, i.e., there exists $M\geq0$ such that $\lVert a^n\rVert_q\leq M$ for all $n$. </p>
<p>I get across with a theorem in Bryan P. Rynne Book which shows that $f_{a^n}$ defines a linear functional $f_{a^n} \in (\ell^q)'$ with $\lVert f_{a^n}\rVert=\lVert a^n\rVert_q$. I think it will help but I am still confused how to prove the statement above</p>
| Martin Argerami | 22,857 | <p>You have a sequence that converges weakly to zero, and you want to conclude that it is bounded. What you need is to use the <a href="https://en.wikipedia.org/wiki/Uniform_boundedness_principle" rel="nofollow noreferrer">Uniform Boundedness Principle</a>. </p>
|
1,232,420 | <p>Consider the ideal $I = (13x+16y, 11x+13y)$ in the ring R = $\mathbb{Z}[x,y].$</p>
<p>Prove that $I=(x-2y, 3x+y)$ by using mutual inclusion.</p>
<p>I'm confused on how to start...do I begin by multiplying the elements in the ideal by a general element of R?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ Use basis transformations to triangularize the basis, i.e. to the form $\,(x+cy, dy)$</p>
<p>$(a,b) = (x\!-\!2y,3x\!+\!y) = (x\!-\!2y,\overbrace{7y}^{b-3a})$</p>
<p>$(13x\!+\!16y,11x\!+\!13y) = (\underbrace{2x\!+\!3y}_{a-b},11x\!+\!13y) = (2x\!+\!3y,\underbrace{x\!-\!2y}_{b-5a}) = (\underbrace{7y}_{a-2b},x\!-\!2y)$</p>
<p><strong>Remark</strong> $\ $ As in the extended Euclidean alorithm you can augment the reductions to keep track of the generators in terms of the original basis if so desired.</p>
|
15,480 | <p>Say I have two lists,</p>
<pre><code>list1 = {a, b, c}
list2 = {x, y, z}
</code></pre>
<p>and I want to map a function f over them to produce</p>
<pre><code>{f[a,x], f[a,y], f[a,z], f[b,x], f[b,y], f[b,z], f[c,x], f[c,y], f[c,d]}
</code></pre>
<p>I would assume I map the function over the first list to produce a "list of functions", which then run over the 2nd list, something like:</p>
<pre><code>Map[Map[f[#1, #2]&,list1]&, list2]
</code></pre>
<p>but I can't figure out how to leave #2 "empty" until the 2nd map kicks in. How can I separate them to generate all combinations of arguments?</p>
| Sascha | 4,597 | <p>Or you could use <code>Tuples</code>, which appears a bit more natural to me.</p>
<pre><code>Tuples[{{a, b, c}, {x, y, z}}]
</code></pre>
<p>creates</p>
<pre><code>{{a, x}, {a, y}, {a, z}, {b, x}, {b, y}, {b, z}, {c, x}, {c, y}, {c, z}}
</code></pre>
<p>Afterwards <code>Apply</code> can be used to apply your function to the sublist</p>
<pre><code>Apply[f , Tuples[{{a, b, c}, {x, y, z}}], {1}]
</code></pre>
<p>creates:</p>
<pre><code>{f[a, x], f[a, y], f[a, z], f[b, x], f[b, y], f[b, z], f[c, x], f[c, y], f[c, z]}
</code></pre>
<h2>Addition: comparison of speed:</h2>
<p>create some random data:</p>
<pre><code>list1 = RandomReal[1, 10^3];
list2 = RandomReal[1, 10^3];
</code></pre>
<p>Usage of a pure function to summarize the arguments <code>(#1 + #2) &</code></p>
<pre><code> Apply[(#1 + #2) &, Tuples[{list1, list2}], {1}]; // AbsoluteTiming
</code></pre>
<p>yields <code>{0.944316, Null}</code></p>
<pre><code>Outer[(#1 + #2) &, list1, list2]; // AbsoluteTiming
</code></pre>
<p>yields <code>{0.506706, Null}</code></p>
<hr>
|
2,223,039 | <p>Consider an urn containing 90 balls numbered from 1 to 90, plus 3 balls attached with 3 distinct (known) numbers still from 1 to 90, say 1,2,3. I'm trying to find the probability that at least two equal numbers are extracted after 5 extractions without replacement.
I've tried two approaches which give different answers, and I don't really know which is correct (mistakes aside).</p>
<p>Firstly, with $\Omega_0=\{1,2,\cdots,90,1a,2a,3a \}$ I would define as a sample space $\Omega=\{\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5)\in\Omega_0^5\},$ as a $\sigma$-algebra $F=P(\Omega)$ and as a probability measure $\mathbb{P}$ such that $\mathbb{P}(\{\omega\})= 1/ \binom {93}{5} .$</p>
<p>Let $A $ be the event of two equal numbers being extracted. My first approach is to consider $A^c=\{\omega \in \Omega :\omega_i \ne\omega_j \text{ for } i \ne j\};$ I think $\lvert A^c\rvert =\binom{90}{5}$, which gives $\mathbb{P}(A)=1-\frac{88\cdot87\cdot86}{93\cdot92\cdot91}.$</p>
<p>Second approach: $\mathbb{P}(A)=\mathbb{P}(``\text{two 1's" or }`\text{two 2's" or }``\text{two 3's}")=3\mathbb{P}(``\text{two 1's}") -3\mathbb{P}(``\text{two 1's or two 2's}")$. Then from $$\mathbb{P}(``\text{two 1's}")=\frac{\binom{2}{2}\binom{91}{3}}{{93}\choose{5}}$$ and $$\mathbb{P}(``\text{two 1's and two 2's}")=\frac{\binom{2}{2} {2\choose2} {{89}\choose1}}{{93}\choose{5}}$$ I proceed to the result but it doesn't equal what I got earlier. Where do I go wrong?</p>
| Stella Biderman | 123,230 | <p>You should approach this using the following formula, which is valid for all complex numbers $z$</p>
<p>$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$</p>
<p>We also have the cooresponding formula for sine</p>
<p>$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$</p>
<p>These can both be proven by using $e^{iz}=\cos(z)+i\sin(x)$ and solving for the respective quantities.</p>
<p>As pointed out by Rob Israel, these two formulas are sometimes used to <em>define</em> the sine and claims functions on the complex plane. The other way that they are sometimes defined is via their Taylor polynomials.</p>
|
2,223,039 | <p>Consider an urn containing 90 balls numbered from 1 to 90, plus 3 balls attached with 3 distinct (known) numbers still from 1 to 90, say 1,2,3. I'm trying to find the probability that at least two equal numbers are extracted after 5 extractions without replacement.
I've tried two approaches which give different answers, and I don't really know which is correct (mistakes aside).</p>
<p>Firstly, with $\Omega_0=\{1,2,\cdots,90,1a,2a,3a \}$ I would define as a sample space $\Omega=\{\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5)\in\Omega_0^5\},$ as a $\sigma$-algebra $F=P(\Omega)$ and as a probability measure $\mathbb{P}$ such that $\mathbb{P}(\{\omega\})= 1/ \binom {93}{5} .$</p>
<p>Let $A $ be the event of two equal numbers being extracted. My first approach is to consider $A^c=\{\omega \in \Omega :\omega_i \ne\omega_j \text{ for } i \ne j\};$ I think $\lvert A^c\rvert =\binom{90}{5}$, which gives $\mathbb{P}(A)=1-\frac{88\cdot87\cdot86}{93\cdot92\cdot91}.$</p>
<p>Second approach: $\mathbb{P}(A)=\mathbb{P}(``\text{two 1's" or }`\text{two 2's" or }``\text{two 3's}")=3\mathbb{P}(``\text{two 1's}") -3\mathbb{P}(``\text{two 1's or two 2's}")$. Then from $$\mathbb{P}(``\text{two 1's}")=\frac{\binom{2}{2}\binom{91}{3}}{{93}\choose{5}}$$ and $$\mathbb{P}(``\text{two 1's and two 2's}")=\frac{\binom{2}{2} {2\choose2} {{89}\choose1}}{{93}\choose{5}}$$ I proceed to the result but it doesn't equal what I got earlier. Where do I go wrong?</p>
| The_Sympathizer | 11,172 | <p>To explain the deeper answer to the "why" of why it will happen is first off to point out that one <em>cannot</em> and <em>should not</em> expect that the behavior of at least analytic complex functions should mirror that on the real number line.</p>
<p>In particular, one very important theorem of complex analysis is that <em>any non-constant entire function is unbounded</em> (more often-stated converse: a bounded entire function is constant), where "entire" means it is complex differentiable everywhere. The same is <em>not</em> true for real-valued functions, as $\cos$ and $\sin$ illustrate.</p>
<p>(There is another interesting theorem in the same vein which says that an <em>injective entire function is linear</em>, which means that almost all the "nicest" complex functions are <em>not invertible</em>!)</p>
<p>Thus if we are to extend $\cos$ and $\sin$ in <em>any</em> way to the complex plane that preserves their differentiability they <em>must be unbounded</em> and thus there <em>must</em> be some way to approach $\infty$ under them following some suitable path in the plane (which can be discretized to a sequence). It does not matter how we do it -- as long as we want to be able to take a derivative everywhere this <em>must</em> hold. It's even better than that -- if we want to take $\cos$ and $\sin$ as agreeing with their real number equivalents but are willing to admit only an <em>isolated</em> set of nondifferentiability points (i.e. just a few scattered points), we will actually get them to be differentiable everywhere and so they will still be unbounded.</p>
|
2,462,297 | <p>Let $(x_1,...,x_n)$ be real numbers and M be an $n \times n$ matrix whose its column is given by the entries $x_i,x_i^2, x_i^3,...x_i^n$. Compute the determinant of M.</p>
<p>I computed the formula for the determinant of M in terms of $x_1...x_n$ but I wonder if I can find a specific value.</p>
| Tsemo Aristide | 280,301 | <p>Hint: use this determinant formula</p>
<p><a href="https://en.wikipedia.org/wiki/Vandermonde_matrix" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Vandermonde_matrix</a></p>
<p>take $x_{n+1}=0$ and develop the Vandermonde $(n+1)\times (n+1)$-determinant relatively to the last line.</p>
|
899,249 | <p>I got this problem from my friend. I have been doing it for hours.</p>
<p>$a_1 = 2$</p>
<p>$a_{n+1} = 2a^2_n+1$</p>
<p>$a_n = ?$</p>
<p>Could you please tell me how to solve this? Thanks!</p>
<p>BTW: I failed to solve it by using Mathematica <code>RSolve[{a[1] == 2, a[n + 1] == 2 a[n]^2 + 1}, a[n], n]</code></p>
| Claude Leibovici | 82,404 | <p>Assuming, as Will Jagy suspected, that the problem is $${ a_1 = 2, \; \; a_{n+1} = 2 a_n^2 - 1}$$ using a CAS I obtained, after some manipulations and simplifications, the surprizing form $$a_n=\cosh\Big(2^{n-1}\log(2+\sqrt 3)\Big)$$ or $$a_n=\frac{1}{2} \left(2+\sqrt{3}\right)^{-2^{n-1}}
\left(1+\left(2+\sqrt{3}\right)^{2^n}\right)$$</p>
|
529,861 | <p>If $m,n$ are coprime positive integers and $m-n$ is odd, then $(m-n),(m+n),m,n,$ are coprime each other?</p>
<p>How do I prove it?</p>
<p>Especially how do I prove $(m-n), (m+n)$ are coprime?</p>
| Prahlad Vaidyanathan | 89,789 | <p>Hints :</p>
<ol>
<li><p>Show that it is a decreasing sequence bounded below.</p></li>
<li><p>Notice that $a_0 = x$, and $a_k = f(a_{k-1})$ where $f(y) = xy$. By (1), we know that $L = \lim a_k$ exists. Now
$$
a_k \to L \Rightarrow f(a_k) \to f(L)
$$
since $f$ is continuous. However, $f(a_k) = a_{k+1}$, and so the the sequences $\{f(a_k)\}$ and $\{a_k\}$ are actually <em>the same sequence.</em> Since a sequence cannot converge to two different points,
$$
L = f(L) = xL
$$</p></li>
</ol>
|
1,606,978 | <p>What is the value for $\lim \limits _{x\to\infty} \frac{\sin x} x$? </p>
<p>I solved it by expanding $\sin x$ as</p>
<p>$$\sin x = x - \frac {x^3} {3!} \dotsc$$</p>
<p>So $\lim \limits _{x\to\infty} \frac {\sin x} x = 1 -\infty = - \infty$,</p>
<p>but the answer is $0$. Why? What I am doing wrong?</p>
| Community | -1 | <p>Yes , the answer is $0$ . </p>
<p>One way to see this is by using the inequality :</p>
<p>$$\left |\frac{\sin x}{x}\right | \leq \frac{1}{x}$$ when $x>0$ (this happens because $|\sin x\ | \leq 1$ )</p>
<p>When $x \to \infty $ we have $\frac{1}{x} \to 0$ so the limit must be $0$ .</p>
|
1,606,978 | <p>What is the value for $\lim \limits _{x\to\infty} \frac{\sin x} x$? </p>
<p>I solved it by expanding $\sin x$ as</p>
<p>$$\sin x = x - \frac {x^3} {3!} \dotsc$$</p>
<p>So $\lim \limits _{x\to\infty} \frac {\sin x} x = 1 -\infty = - \infty$,</p>
<p>but the answer is $0$. Why? What I am doing wrong?</p>
| johnnyb | 298,360 | <p>The range of $\sin(x)$ will always be a value between -1 and 1, no matter what the input. However, there is no such restriction on the denominator. Therefore, if your numerator is restricted to a finite value, and your denominator is not, as the denominator goes to infinity the value of the whole expression will go to zero.</p>
|
2,986,515 | <p>Can anyone help me with this problem? </p>
<p>Prove that for any real number <span class="math-container">$x > 0$</span> and for any <span class="math-container">$M > 0$</span> there is <span class="math-container">$N ∈ \mathbb N$</span> so that if <span class="math-container">$n > N$</span> then <span class="math-container">$(1 + x)^n > M.$</span></p>
<p>A sequence <span class="math-container">${a_n}$</span> diverges to <span class="math-container">$+\infty$</span> if for any <span class="math-container">$M > 0$</span> there is <span class="math-container">$N ∈\mathbb N$</span> so that if <span class="math-container">$n > N$</span> then <span class="math-container">$a_n > M$</span>. Suppose <span class="math-container">${a_n}$</span> diverges to <span class="math-container">$+\infty$</span> and <span class="math-container">$a_n \ne 0$</span> for all <span class="math-container">$n$</span>. Prove that in this case <span class="math-container">$\frac{1}{a_n}$</span> converges to <span class="math-container">$0$</span>.</p>
<p>How do I approach these proofs?</p>
| Seth | 610,132 | <p>If <span class="math-container">$\cfrac{3n^2-9n+6}{n^3+5n^2+8n+4}=0$</span>, then <span class="math-container">$3n^2-9n+6=0$</span>, or <span class="math-container">$3(n-2)(n-1)=0$</span>, so since the denominator is always positive and the poynomial is positive for <span class="math-container">$x \notin (1,2)$</span>, it is greater than zero for all integers.</p>
<p>If <span class="math-container">$\cfrac{3n^2-9n+6}{n^3+5n^2+8n+4}=1$</span>, then <span class="math-container">$n^3+5n^2+8n+4=3n^2-9n+6$</span>, or <span class="math-container">$n^3+2n^2+17n-2=0$</span>, but you can see from any graphics calculator that that cubic is positive for all <span class="math-container">$n \geq 1$</span>, and the denominator is more than <span class="math-container">$1$</span> for <span class="math-container">$n$</span> positive, so the function is less than one for all natural numbers</p>
|
1,087,015 | <p>I'm looking for a polynomial $P(x)$ with the following properties:</p>
<ol>
<li>$P(0) = 0$.</li>
<li>$P\left(\frac13\right) = 1$</li>
<li>$P\left(\frac23\right) = 0$</li>
<li>$P'\left(\frac13\right) = 0$</li>
<li>$P'\left(\frac23\right) = 0$</li>
</ol>
<p>From 1 and 3 we know that $P(x) = x\left(x - \frac23\right)Q(x)$. From 4 and 5 we know that $P'(x) = \alpha\left(x - \frac13\right)\left(x - \frac23\right)$.
$$
\begin{align}
P(x) & = \alpha\int\left(x - \frac13\right)\left(x - \frac23\right)\text{d}x \\
& = \alpha\left(\frac13x^3 - \frac12x^2 + \frac29x + \text{C}\right)
\end{align}
$$
Now 1 implies that the constant is $\text{C} = 0$, but 3 implies that the constant is $\text{C} = -\frac{2}{81}$. Am I doing something wrong or does this polynomial not exist? </p>
<p>If it doesn't exist, how close could I get to making a polynomial that satisfies these 5 conditions?</p>
| Lucian | 93,448 | <p><strong>Hint:</strong> $P'(a)=0$ implies $\Big(a,P(a)\Big)$ is either among the extrema, or an inflection point. My advice for you would be to draw a simple graphic. It is clear that we area dealing with something that is at least a cubic. Also, the addition of any other root or maxima or inflection points does not contradict the existence of the ones we already know about, so it is clear that an infinity of such polynomials exist for any degree greater than three. But the cubic is uniquely determined. Why ? Because $\dfrac23$ is a double root. So $P(x)=ax\bigg(x-\dfrac23\bigg)^2$. Now, what is the value of <em>a</em> ?</p>
|
1,087,015 | <p>I'm looking for a polynomial $P(x)$ with the following properties:</p>
<ol>
<li>$P(0) = 0$.</li>
<li>$P\left(\frac13\right) = 1$</li>
<li>$P\left(\frac23\right) = 0$</li>
<li>$P'\left(\frac13\right) = 0$</li>
<li>$P'\left(\frac23\right) = 0$</li>
</ol>
<p>From 1 and 3 we know that $P(x) = x\left(x - \frac23\right)Q(x)$. From 4 and 5 we know that $P'(x) = \alpha\left(x - \frac13\right)\left(x - \frac23\right)$.
$$
\begin{align}
P(x) & = \alpha\int\left(x - \frac13\right)\left(x - \frac23\right)\text{d}x \\
& = \alpha\left(\frac13x^3 - \frac12x^2 + \frac29x + \text{C}\right)
\end{align}
$$
Now 1 implies that the constant is $\text{C} = 0$, but 3 implies that the constant is $\text{C} = -\frac{2}{81}$. Am I doing something wrong or does this polynomial not exist? </p>
<p>If it doesn't exist, how close could I get to making a polynomial that satisfies these 5 conditions?</p>
| Brady Gilg | 188,927 | <p>This is the same solution as Ross's, but from a more linear algebra focused perspective.</p>
<p>You have five linearly independent conditions, so a polynomial with five parameters should work.</p>
<p>$P(x) = ax^4 + bx^3 + cx^2 + dx + e$</p>
<p>Now your five conditions can be written as such:</p>
<ol>
<li>$e = 0$</li>
<li>$a(\frac{1}{3})^4 + b(\frac{1}{3})^3 + c(\frac{1}{3})^2 + d(\frac{1}{3}) + e = 1$</li>
<li>$a(\frac{2}{3})^4 + b(\frac{2}{3})^3 + c(\frac{2}{3})^2 + d(\frac{2}{3}) + e = 0$</li>
<li>$4a(\frac{1}{3})^3 + 3b(\frac{1}{3})^2 + 2c(\frac{1}{3}) + d = 0$</li>
<li>$4a(\frac{2}{3})^3 + 3b(\frac{2}{3})^2 + 2c(\frac{2}{3}) + d = 0$</li>
</ol>
<p>We can group these five equations together in matrix-vector form as </p>
<p>$\left( \begin{array}{ccccc}
0 & 0 & 0 & 0 & 1 \\
\frac{1}{81} & \frac{1}{27} & \frac{1}{9} & \frac{1}{3} & 1 \\
\frac{16}{81} & \frac{8}{27} & \frac{4}{9} & \frac{2}{3} & 1 \\
\frac{4}{27} & \frac{3}{9} & \frac{2}{3} & 1 & 0 \\
\frac{32}{27} & \frac{12}{9} & \frac{4}{3} & 1 & 0 \end{array} \right) \cdot \left( \begin{array}{c} a \\ b \\ c \\ d \\ e \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array} \right) $</p>
<p>Use a computerized linear algebra solver to find</p>
<p>$\left( \begin{array}{c} a \\ b \\ c \\ d \\ e \end{array} \right) = \left( \begin{array}{c} 81 \\ -108 \\ 36 \\ 0 \\ 0 \end{array} \right)$</p>
<p>So your polynomial is $P(x) = 81x^4 - 108x^3 + 36x^2$. This is the same one Ross Millikan found.</p>
<p>The advantage of this method is that it is more flexible. You could start with $P(x) = ax^{50} + bx^{42} + cx^9 + dx^2 + e$ if you wanted a find a degree 50 polynomial that would also work. You could even start with a function that wasn't polynomial at all. The disadvantage is that if you have more than 4 parameters or so then solving the system will be too cumbersome to do by hand, and you'll have to rely on a computer.</p>
|
2,900,454 | <p>There are so many different methods I've found on SE and through Matlab, and they're all giving me different results.</p>
<p>Specifically, I have {v1} = (1,2,1) and {v2} = (2,1,0) in set S. What is the method to find {v3} vectors that are orthogonal to both v1 and v2?</p>
<p>I'm preparing for a final and I'm trying to find a flexible method for many cases. The answer I got for above was v3 = {1,-2,3} but different methods are returning different results.</p>
| Siong Thye Goh | 306,553 | <p>Guide:</p>
<p>Your answer should be a non-zero scalar multiple of the $v_3$ that you provided since $v_1$ and $v_2$ are not parallel to each other.</p>
<p>Method $1$:</p>
<ul>
<li>Compute the cross product of $v_1$ and $v_2$, that will give you a valid solution.</li>
</ul>
<p>Method $2$:</p>
<ul>
<li>Solve the linear system $v_1^Tx=0$ and $v_2^Tx=0$ by reducing the system say to REF. You will obtain multiple solution, of which all of them are scalar multiple of $v_3$.</li>
</ul>
<p>Either method should be fine.</p>
|
744,034 | <p>How do I show that for all integers $n$, $n^3+(n+1)^3+(n+2)^3$ is a multiple of $9$?
Do I use induction for showing this? If not what do I use and how? And is this question asking me to prove it or show it? How do I show it? </p>
| DeepSea | 101,504 | <p>Use the identity: $a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc$,</p>
<p>and substituting $a = n$, $b = n + 1$, and $c = n + 2$ into the equation to get:</p>
<p>$n^3 + (n+1)^3 + (n+2)^3 = (3n + 3)(n^2 + (n+1)^2 + (n+2)^2 - n(n+1) - (n+1)(n+2) - n(n+2)) + 3n(n+1)(n+2) = 3(n+1)(n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 - n^2 - n - n^2 - 3n - 2 - n^2 - 2n) + 3n(n+1)(n+2) = 3(n+1)(3n+3) + 3n(n+1)(n+2) = 9(n+1)^2 + 3n(n+1)(n+2)$. Observe that $n(n+1)(n+2$) is divisible by $3! = 6$. So $3n(n+1)(n+2)$ is divisible by $18$ and inturn divisible by $9$. So we're done.</p>
|
3,623,368 | <p>for example this equation: 505x-673y=1 . x=4 and y = 3. but how can I find them with mathematics. What would be the approach here?</p>
| Jean Marie | 305,862 | <p>This was a comment, but I think deserves to be an answer.</p>
<p>As your question is asking for a method, here it is.</p>
<ul>
<li><p>1) First, divide by the common factor <span class="math-container">$d=GCD(a,b,c)$</span>.</p></li>
<li><p>2) Then apply "extended Euclidean algorithm" which will give you coefficients <span class="math-container">$x',y'$</span> called "Bezout coefficients". </p></li>
<li><p>3) The final coefficients are : <span class="math-container">$x'=dx, y'=dy$</span>.</p></li>
</ul>
<p>See the details of point 2) here.</p>
|
2,901,783 | <p>I am having trouble solving a multi part question.</p>
<p>Express $ \frac x{x^2-3x + 2} $ in the partial fraction form.</p>
<p>The answer I got was $\frac2{x-2}-\frac1{x-1}$ .</p>
<p>The problem comes when they asked:</p>
<p>Show that, if $x$ is so small that $x^3$ and higher powers of $x$ can be neglected, then:</p>
<p>$$\frac x{x^2-3x +2}\approx\frac12x+\frac34x^2$$</p>
<p>I understand that I had to expand the partial fraction using Generalized Binomial Theorem such that I needed to expand: $-\left[1+\left(-\frac12x\right)\right]^{-1}+\left[1+\left(-x\right)\right]^{-1}$ since I needed to manipulate the equation into $\left(1+a\right)^n$ format.</p>
<p>I got $$\frac12x+\frac34x^2+\frac78x^3+\frac{15}{16}x^4+\cdots$$</p>
<p>but I am not sure how to continue to <strong>show that , if $x$ is so small that $x^3$ and higher powers of $x$ can be neglected</strong>. Do i just sub in numbers ranging from -1 to 1 and that's it or is there a structured way to show?</p>
| R zu | 587,462 | <p>Just cross out terms with $x^3$, $x^4$, ... in the approximation.</p>
<p>The question assumes these terms are too small, and we don't care.</p>
<p>For example, if I know x is between 0 and 0.1. Then $x^3$ is at most 0.001. $x^4$ is even smaller. </p>
<p>I think the sum of the terms $x^3$, $x^4$, ... are at least smaller than $x^2$, which is at most 0.01. </p>
<p>I forget the exact formula for the errors. I just use sum of a geometric series and approximate the coefficient of all terms as 1.</p>
<p>Let's say an error of 0.02 is acceptable in the approximation.</p>
<p>I would use just the $x$ and $x^2$ terms in that case.</p>
<p>Less terms means a computer program can approximate the thing faster.</p>
|
574,041 | <p>Consider a set of linear equations described by $A\vec{X}=\vec{B}$ is given, where $A$ is an $n\times n$ matrix and $\vec{X}$ and $\vec{B}$ are n-row vectors. Also suppose that this system of equations have a unique solution and this solution is given.</p>
<p>Imagine a new set of linear equations $A'\vec{X}=\vec{B}$, where all elements of $A'$ is equal to those of $A$ but one element $A_{ij}$ which is increased by $k$. I am interested to know if I could somehow relate the solution of the first problem to the second problem by knowing the value of $k$ and $A$. In other words, I would like to derive the new solution without resolving the set of linear equations.</p>
| Community | -1 | <p>Changing a single element corresponds to a special rank one update of the form
$$(a'(i,j)-a(i,j))e_i e_j^T$$where $e_k$ is a vector with zero except at location $k$, which is one. You can use <a href="http://en.wikipedia.org/wiki/Woodbury_matrix_identity" rel="nofollow">Sherman-Morrison-Woodbury</a> to construct the solution for low-rank updates.</p>
|
3,878,174 | <p>a) Prove that = {<span class="math-container">$_{,}$</span> | , ∈ ℝ, ≠ 0} is a group, where the operation is composition.<br />
Let 1 = {<span class="math-container">$\big(\begin{smallmatrix}
a & 0\\
b & 1
\end{smallmatrix}\big)$</span> ∶ , ∈ ℝ, ≠ 0}, where the operation is matrix multiplication, and let
be the group in part a).\</p>
<p>b.)Prove that the map : → 1 defined as (,) = <span class="math-container">$\big(\begin{smallmatrix}
a & 0\\
b & 1
\end{smallmatrix}\big)$</span> for all <span class="math-container">$_{,}$</span> ∈ , is an isomorphism.</p>
<p>Proof:<br />
a) the associative property is always true for composition of functions.
I am not sure how to do the other two properties for group to show that there is an inverse and identity exist. This is preventing me to do part b if someone can help guide me through this proof. I need help formulating the rest of the proof.</p>
| User203940 | 333,294 | <p>We need to show that <span class="math-container">$G$</span> is closed under composition. Let <span class="math-container">$f_{a,b}, f_{c,d} \in G$</span>. Show that <span class="math-container">$f_{a,b} \circ f_{c,d} \in G$</span>.</p>
<p><em>Hint:</em> We see <span class="math-container">$f_{a,b} \circ f_{c,d}(x) = f_{a,b}(cx + d) = a(cx+d)+b = \cdots.$</span></p>
<p>We need to show that there is an identity. The identity for composition of functions is a function such that <span class="math-container">$f_{a,b}(x) = x$</span> for all <span class="math-container">$x$</span> (why?). Determined <span class="math-container">$a$</span> and <span class="math-container">$b$</span> which will give us this.</p>
<p><em>Hint:</em> We want <span class="math-container">$f_{a,b}(x) = ax+b = x,$</span> so<span class="math-container">$\ldots$</span></p>
<p><em>Hint:</em> To see this is an identity, notice <span class="math-container">$f_{a,b} \circ f_{c,d}(x) = \cdots$</span> for the above <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.</p>
<p>We need to show existence of inverse. Fix <span class="math-container">$f_{a,b} \in G$</span>. In the first part, we calculated <span class="math-container">$f_{a,b} \circ f_{c,d}$</span>. What conditions do we need on <span class="math-container">$c$</span> and <span class="math-container">$d$</span> so that <span class="math-container">$f_{a,b} \circ f_{c,d}(x) = x$</span> for all <span class="math-container">$x$</span>? For the same <span class="math-container">$c$</span> and <span class="math-container">$d$</span>, do we have <span class="math-container">$f_{c,d} \circ f_{a,b}(x) = x$</span>?</p>
<p>This shows that <span class="math-container">$G$</span> is a group. (Assuming you know associativity)</p>
|
3,878,174 | <p>a) Prove that = {<span class="math-container">$_{,}$</span> | , ∈ ℝ, ≠ 0} is a group, where the operation is composition.<br />
Let 1 = {<span class="math-container">$\big(\begin{smallmatrix}
a & 0\\
b & 1
\end{smallmatrix}\big)$</span> ∶ , ∈ ℝ, ≠ 0}, where the operation is matrix multiplication, and let
be the group in part a).\</p>
<p>b.)Prove that the map : → 1 defined as (,) = <span class="math-container">$\big(\begin{smallmatrix}
a & 0\\
b & 1
\end{smallmatrix}\big)$</span> for all <span class="math-container">$_{,}$</span> ∈ , is an isomorphism.</p>
<p>Proof:<br />
a) the associative property is always true for composition of functions.
I am not sure how to do the other two properties for group to show that there is an inverse and identity exist. This is preventing me to do part b if someone can help guide me through this proof. I need help formulating the rest of the proof.</p>
| Mateus Figueiredo | 674,120 | <p>Let <span class="math-container">$a,b,c,d\in\mathbb{R}$</span> be arbitrary elements, <span class="math-container">$a,d\neq 0$</span>. Observe that for all <span class="math-container">$x\in \mathbb{R}$</span>, <span class="math-container">$$f_{a,b}\circ f_{c,d}(x)=f_{a,b}(f_{c,d}(x))=f_{a,b}(cx+d)=a(cx+d)+b=(ac)x+ad+b=f_{ac,~ad+b}(x),$$</span> i.e., <span class="math-container">$f_{a,b}\circ f_{c,d}=f_{ac,~ad+b}$</span>, this means that the composition is closed on <span class="math-container">$G$</span>. Now, <span class="math-container">$f_{1,0}$</span> is the identity of <span class="math-container">$\circ$</span>. In fact, <span class="math-container">$$f_{a,b}\circ f_{1,0}(x)=f_{a,b}(f_{1,0}(x))=f_{a,b}(x)=ax+b=f_{1,0}(ax+b)=f_{1,0}(f_{a,b}(x))=f_{1,0}\circ f_{a,b}(x),$$</span> for all <span class="math-container">$x\in \mathbb{R}$</span>. Now we will calculate <span class="math-container">$f_{a,b}^{-1}$</span>. We want to found a map <span class="math-container">$f_{r,s}\in G$</span> such that <span class="math-container">$$f_{a,b}\circ f_{r,s}=f_{r,s}\circ f_{a,b}=f_{1,0}.$$</span> For all <span class="math-container">$x\in \mathbb{R}$</span>, <span class="math-container">$f_{a,b}\circ f_{r,s}(x)=f_{a,b}(rx+s)=a(rx+s)+b=arx+s+b=f_{1,0}(x)=x$</span> if and only if <span class="math-container">$ar=1$</span> and <span class="math-container">$s+b=0$</span>, that is, <span class="math-container">$r=1/a$</span> and <span class="math-container">$s=-b$</span>. It means that <span class="math-container">$f_{1/a,~-b}$</span> is a right inverse to <span class="math-container">$f_{a,b}$</span> on <span class="math-container">$G$</span>. You can check that <span class="math-container">$f_{1/a,~-b}$</span> is a left inverse to <span class="math-container">$f_{a,b}$</span> on <span class="math-container">$G$</span> and so it is an inverse to <span class="math-container">$f_{a,b}$</span>. In particular, every element of <span class="math-container">$G$</span> has an inverse because <span class="math-container">$a,b\in \mathbb{R}$</span>, <span class="math-container">$a\neq 0$</span>, are arbitrary. Finally, as you said, the composition is always associative. Therefore, <span class="math-container">$G$</span> is a group.</p>
<p>Your map <span class="math-container">$\phi$</span> cannot by an isomorphism because it is not even a homomorphism: it is easy to check that <span class="math-container">$\phi(f_{1,2}\circ f_{2,1})\neq \phi(f_{1,2})\phi(f_{2,1})$</span>. Now, if you define <span class="math-container">$$G_{1}=\{A=[a_{ij}]\in M_{2\times 2}(\mathbb{R});~a_{11}\neq 0,~a_{21}=0~\text{and}~a_{22}=1\}$$</span> we will show that <span class="math-container">$G$</span> and <span class="math-container">$G_{1}$</span> are isomorphics. Consider <span class="math-container">$\phi:G\longrightarrow G_1$</span> defined by <span class="math-container">$\phi(f_{a,b})=\left[ {\begin{array}{cc}
a & b \\
0 & 1 \\
\end{array} } \right].$</span>
Observe that <span class="math-container">$$\phi(f_{a,b}\circ f_{c,d})=\phi(f_{ac,ad+b})=\left[ {\begin{array}{cc}
ac & ad+b \\
0 & 1 \\
\end{array} } \right]=\left[ {\begin{array}{cc}
a & b \\
0 & 1 \\
\end{array} } \right]\left[ {\begin{array}{cc}
c & d \\
0 & 1 \\
\end{array} } \right]=\phi(f_{a,b})\phi(f_{c,d}),$$</span> and so <span class="math-container">$\phi$</span> is a homomorphism. Clearly <span class="math-container">$\phi$</span> is a surjective map and we just have to show <span class="math-container">$\phi$</span> is injective. But, if <span class="math-container">$f_{a,b}=1_{G_{1}}$</span>, then <span class="math-container">$a=1$</span> and <span class="math-container">$b=0$</span>, i.e., <span class="math-container">$f_{a,b}=f_{1,0}$</span>. It means that <span class="math-container">$\phi$</span> is injective and so it is a isomorphism between <span class="math-container">$G$</span> and <span class="math-container">$G_1$</span>.</p>
|
2,945,367 | <p><a href="https://i.stack.imgur.com/MGzHc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MGzHc.png" alt="enter image description here"></a></p>
<p>We were given a couple formulas, but the one that immediately stood out to me was the Vfinal = Vinitial + at</p>
<p>so we know the patrol will constantly accelerate "until he pulls next to the speeding car" so Vfinal = 30m/s and Vinitial = 0(cop at rest) and the acceleration is a constant 3m/s^2</p>
<p>so 30 = 3t, t = 10. However when I continue through this lecture, it turns out t = 20. he uses the formula x = xo + v0t + 1/2at^2 and somehow gets a different time than me. I don't understand how it's possible that we are both solving for time but getting different results. </p>
<p>Is it safe to assume that when xo = 0 and v0 = 0 that</p>
<p>sqrt(2x/a) = Vfinal / a </p>
<p>since we cancel out terms and they both equal t?</p>
| Paramanand Singh | 72,031 | <p>You just need to consider the Riemann sum of <span class="math-container">$f$</span> over partition <span class="math-container">$P''=\{x_i, y, x_{i+1}\} $</span> of <span class="math-container">$[x_i, x_{i+1}]$</span>. By <span class="math-container">$(a) $</span> we have <span class="math-container">$$U(f, P'')\leq M_{i+1}(x_{i+1}-x_i)=f(\xi_{i+1})(x_{i+1}-x_i) $$</span> and adding terms corresponding to intervals <span class="math-container">$$[x_0,x_1],\dots,[x_{i-1},x_i],[x_{i+1},x_{i+2}],\dots, [x_{n-1},x_n]$$</span> to the above inequality we get <span class="math-container">$U(f, P') \leq U(f, P) $</span>.</p>
<p>I have deliberately stayed away from the notation used in your image as it is not standard. </p>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| Hong Liu | 13,033 | <p>I think fibre bundles should be introduced to give a modern viewpoint of tensor analysis.</p>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| Deane Yang | 613 | <p>I'm not sure if the original question is about a one semester or year course.</p>
<p>If this is the first course the students have ever had in differential geometry, then I still agree with Anton that at least the first semester should be about only 2-dimensional manifolds embedded in $R^3$ and Gauss-Bonnet. The point here is that everything can be understood visually, but you learn how to deploy linear algebra and calculus to prove what seems obvious visually. The full power of differential geometry is displayed very nicely. Guillemin and Pollack provides a nice textbook to base the course on. I also like O'Neill's elementary differential geometry textbook.</p>
<p>I would not introduce the more abstract machinery until the second semester, and even then try to be selective about what is discussed because there is just too much. It seems best to focus on basic Riemannian geometry and what, say, sectional curvature means (this builds nicely on what was done in the first semester). It is of course important to introduce many different examples. Although the basic abstract definitions and properties of Lie groups and algebras could be introduced, I believe the focus should be on how to build interesting geometric spaces from standard matrix groups ($GL(n)$, $SL(n)$, $SO(n)$, $SU(n)$).</p>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| Spencer | 4,281 | <p>This is in agreement with Igor's comment on Anton's answer, but became too long.</p>
<p>I'd say whatever approach you ultimately take, for a first-year grad course it surely has to be done 'properly', i.e. starting from intrinsic definition of a smooth manifold and using the 'modern' language and general definitions of tensor bundles, connections etc. </p>
<p>Absolutely crucially (and here's what inspired this comment), the course simply <em>has</em> to teach people that there is more to manifolds than 2D surfaces because that's why the theory is quite so useful and so prominent in modern mathematics. The whole point is surely the sheer diversity of objects amenable to geometric thought (whatever that means). The job of the teacher would then be to maintain the intuition of "surfaces in R^3" while using general definitions. I believe this can be done. If it cannot, then what on Earth are we all doing?</p>
<p>By the look of the books mentioned in the question, it certainly looks like a course on what I would call "Differential Topology". Sure, there is nothing wrong with a good course on Differential Topology! However, it doesn't seem to me to be synonymous with "A First Course on Smooth Manifolds". My go to book for the latter is John Lee's Introduction to Smooth Manifolds. </p>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| John Klein | 8,032 | <p>I do have one addition to make to the above. At our university we usually use a combination of Guillemin and Pollack and Milnor. There is another approach at a first course which some have found useful: Bott and Tu's book, </p>
<pre><code> Differential forms in algebraic topology
</code></pre>
<p>This text covers an alternative set of topics that overlap both manifold theory and algebraic topology.</p>
<p>Disclaimers: (1) I have never used the text myself, but several colleagues have said in the past that it is a good book to use---and I am personally a big fan of Bott's approach to mathematical writing.</p>
<p>(2) If one uses Bott and Tu, then one has to sacrifice</p>
<pre><code> transversality.
</code></pre>
<p>Andrew Ranicki once told me that transversality counts as one of the most important gems
of 20th century mathematics.</p>
|
195,832 | <p>I want to download the content of the website(contains text) and only a few lines from the content (from a specific number of the line up to the last line minus specific offset). Unfortunately, I do not know how to get the number of line in the content. For example, I want to replace 59 with the length of the content minus specific offset.</p>
<pre><code>data1 = Import["https://en.wikipedia.org/wiki/Segment_tree"];
Snippet[data1, 51 ;; 59]
</code></pre>
| amator2357 | 61,985 | <pre><code>length = ToExpression[ToString[StringCount[#,"\n"] & /@ FullForm[data1]]]+1
Snippet[data1,138;;length]
</code></pre>
<blockquote>
<p>This page was last edited on 27 January 2019, at 05:26 (UTC) . Text \
is available under the Creative Commons Attribution-ShareAlike \
License ; additional terms may apply. By using this site, you agree \
to the Terms of Use and Privacy Policy . Wikipedia® is a registered \
trademark of the Wikimedia Foundation, Inc. , a non-profit \
organization.<br>
Privacy policy About Wikipedia Disclaimers Contact Wikipedia \
Developers Cookie statement Mobile view</p>
</blockquote>
|
294,519 | <p>The problem I am working on is:</p>
<p>Translate these statements into English, where C(x) is “x is a comedian” and F(x) is “x is funny” and the domain consists of all people.</p>
<p>a) $∀x(C(x)→F(x))$ </p>
<p>b)$∀x(C(x)∧F(x))$</p>
<p>c) $∃x(C(x)→F(x))$ </p>
<p>d)$∃x(C(x)∧F(x))$</p>
<h2>-----------------------------------------------------------------------------------------</h2>
<p>Here are my answers:</p>
<p>For a): For every person, if they are a comedian, then they are funny.</p>
<p>For b): For every person, they are both a comedian and funny.</p>
<p>For c): There exists a person who, if he is funny, is a comedian</p>
<p>For d): There exists a person who is funny and is a comedian.</p>
<p>Here are the books answers:</p>
<p>a)Every comedian is funny.
b)Every person is a funny comedian.
c)There exists a person such that if she or he is a comedian, then she or he is funny.
d)Some comedians are funny. </p>
<p>Does the meaning of my answers seem to be in harmony with the meaning of the answers given in the solution manual? The reason I ask is because part a), for instance, is a implication, and "Every comedian is funny," does not appear to be an implication.</p>
| Damien L | 59,825 | <p>This is because the mathematical language is more accurate than the usual language. But your answers are right and they are the same as the book.</p>
|
857,801 | <p>"If in the obvious equalities $(k+1)^3−k^3=3k^2+3k+1$, for the different values $k=1,2,…,n−1$, we add the left and the right sides separately, we obtain the equation $n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$, where $σ_n=1^2+2^2+…+(n−1)^2$."</p>
<p>I'm stuck trying to understand what the author has done in the paragraph below. Is it possible to explain it using high school-level math? For instance, what is meant with "for the different values $k=1,2,…,n−1$, we add the left and the right sides separately"? I don't understand what he does after that either.</p>
| Sebastian Garrido | 62,734 | <p>He performed a summation over the given value of k on each side. You may want to look into series to get a clearer idea.</p>
|
857,801 | <p>"If in the obvious equalities $(k+1)^3−k^3=3k^2+3k+1$, for the different values $k=1,2,…,n−1$, we add the left and the right sides separately, we obtain the equation $n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$, where $σ_n=1^2+2^2+…+(n−1)^2$."</p>
<p>I'm stuck trying to understand what the author has done in the paragraph below. Is it possible to explain it using high school-level math? For instance, what is meant with "for the different values $k=1,2,…,n−1$, we add the left and the right sides separately"? I don't understand what he does after that either.</p>
| Mathmo123 | 154,802 | <p>It's a concise way of explaining the following:</p>
<p>We wish to calculate $\displaystyle \sum_{i=1}^ni^2$ by considering $S_n =\displaystyle \sum_{i=1}^n i^3$, and we know (by evaluating) that $(k+1)^3 - k^3 = 3k^2 + 3k+ 1$</p>
<p>So $$S_n - S_{n-1}= 1+\sum_{i=1}^{n-1} (i+1)^3 - i^3 = \sum_{i=1}^{n-1} 3i^2+ 3i +1 = 3\sigma_n + \frac{3n(n-1)}2 + n$$</p>
<p>But equally, $S_n - S_{n-1} = n^3$ by cancelling terms:$$S_n - S_{n-1} = 1+\sum_{i=1}^{n-1} (i+1)^3 - i^3 = 1+n^3 - \left ( (n-1)^3 - (n-1)^3\right ) + \ldots + \left (2 - 2\right) - 1=n^3$$</p>
|
947,358 | <p>Okay $g(x)= \sqrt{x^2-9}$</p>
<p>thus, $x^2 -9 \ge 0$</p>
<p>equals $x \ge +3$ and $x \ge -3$</p>
<p>thus the domains should be $[3,+\infty) \cup [-3,\infty)$ how come the answer key in my book is stating $(−\infty, −3] \cup[3,\infty)$. </p>
| Timbuc | 118,527 | <p>You can try the following argument:</p>
<p>$$x^2-9=(x-3)(x+3)\ge 0\iff x\le -3\;\;\text{or}\;\;x\ge 3$$</p>
<p>You can see the above easily and geometrically: the function $\;f(x)=x^2-9\;$ is a parabolla opening upwards, and if you draw it it is non-negative exactly when $\;x\le -3\;$ or $\;x\ge 3\;$ .</p>
|
3,366,569 | <p>I am trying to solve the following problem;</p>
<p>Write all elements of the following set: <span class="math-container">$ A=\left \{ x\in\mathbb{R}; \sqrt{8-t+\sqrt{2-t}}\in\mathbb{R}, t\in\mathbb{R} \right \}$</span> .</p>
<p>My assumption is that the solution is <span class="math-container">$\mathbb{R}$</span> and we don't need to solve when are the square roots defined, because of the <span class="math-container">$x$</span>. Am I correct?</p>
<p>Thanks</p>
| Oscar Lanzi | 248,217 | <p>Render <span class="math-container">$y=u/x$</span> where <span class="math-container">$1/x$</span> is the homogeneous solution with the lower power of <span class="math-container">$x$</span> (using the other solution could complicate the integral you eventually get by forcing unneeded fractions in the integrand). Terms containing <span class="math-container">$u$</span> with no derivative cancel out and then</p>
<p><span class="math-container">$x\dfrac{d^2u}{dx^2}-\dfrac{du}{dx}=x^2e^{2x}$</span></p>
<p><span class="math-container">$\color{blue}{\text{When you "vary the coefficient" on a homogeneous solution,}}$</span></p>
<p><span class="math-container">$\color{blue}{\text{adding a constant just adds multiples of the homogeneous solution.}}$</span></p>
<p><span class="math-container">$\color{blue}{\text{So only derivatives of your variable coefficient will have any effect.}}$</span></p>
<p>Introduce an integrating factor <span class="math-container">$1/x^2$</span> which matches the left side with the product rule for derivatives:</p>
<p><span class="math-container">$\dfrac{1}{x}\dfrac{d^2u}{dx^2}-\dfrac{1}{x^2}\dfrac{du}{dx}=e^{2x}$</span></p>
<p><span class="math-container">$\dfrac{d}{dx}(\dfrac{1}{x}\dfrac{du}{dx})=e^{2x}$</span></p>
<p><span class="math-container">$\color{blue}{\text{The ratio of homogeneous solutions is }x^{-1}/x=x^{-2}.}$</span></p>
<p><span class="math-container">$\color{blue}{\text{That will become the coefficient of the first derivative in the exact differential.}}$</span></p>
<p><span class="math-container">$\color{blue}{\text{We choose our integrating factor accordingly.}}$</span></p>
<p>Integrating (and leaving out constants since only a particular solution is needed):</p>
<p><span class="math-container">$\dfrac{du}{dx}=\dfrac{xe^{2x}}{2}$</span></p>
<p><span class="math-container">$u=\dfrac{(x-2)e^{2x}}{4}$</span> with integration by parts</p>
<p>Back substituting:</p>
<p><span class="math-container">$\color{blue}{y=\dfrac{(x-2)e^{2x}}{4x}}$</span></p>
|
1,261,825 | <p>How can I find the inverse function of $f(x) = x^x$? I cannot seem to find the inverse of this function, or any function in which there is both an $x$ in the exponent as well as the base. I have tried using logs, differentiating, etc, etc, but to no avail. </p>
| Renato Faraone | 217,700 | <p>To find the inverse of the function $ y=x^x$ (where it is well defined) you first have to switch $y$ with $x$ and viceversa and now you proceed this way:</p>
<p>$x=y^y$</p>
<p>$\ln x=y\ln y$</p>
<p>$\ln x=e^{\ln y}\ln y$</p>
<p>Now we apply Lambert's W function defined as:</p>
<p>$$W(z)e^{W(z)}=z$$</p>
<p>$\ln y=W(\ln x)$</p>
<p>$$y=e^{W(\ln x)}$$</p>
|
21,201 | <p>Next Monday, I'll have an interview at Siemens for an internship where I have to know about fluid dynamics/computational fluid dynamics. I'm not a physicist, so does somebody have a suggestion for a good book where I can read about some basics? Thank you very much.</p>
| Bugs Bunny | 5,301 | <p>You may be able to salvage something by looking at the Bursnide ring of your group. The isomorphism class of the permutation representation of a set $X$ is governed by products $[X]e$ with some of the idempotents and you need to include missing idempotents (of the Burnside ring) to have a version of Brauer's permutation lemma for any group...</p>
|
1,445,702 | <p>i'm a little confused. </p>
<p>1)Which axis is which in 3 Dimensional system?</p>
<p>2)Does it matter if I switch the x-axis to y-axis?</p>
| uniquesolution | 265,735 | <p>1) Traditionally, the Cartesian coordinate system in $\mathbb{R}^3$ uses the notation $(x,y,z)$ to denote a general point in three-space. The reason is that these letters appear in this order in the alphabet. Now the $x$ axis is obtained by putting zeros for the $z$ and $y$, so that you are left with points of the form $(x,0,0)$. Thus, $(1,0,0), (-2,0,0)$ are points on the $x$-axis. Similarly, the $y$ and the $z$-axis are obtained by putting zeros for the other two, respectively.</p>
<p>2) Once you fix a coordinate system, you have to be careful when you switch axes. This is because representation of points and functions on them usually depends on the choice of a particular coordinate system. What is, after all, the meaning of the point $(1,0,0)$? If you assume the natural coordinate system (the one described in 1)), this is a point lying one unit of distance to the perfect east of the origin $(0,0,0)$. Note that this latter description, (one unit of distance to the east), used the notion of distance and direction (namely, "east"). If you were to switch the axes now, and decide that points to the east (and west) of the origin are the $y$-axis, then the same point would have to be written as $(0,1,0)$.</p>
|
2,585,265 | <p>I understand that $\cos(\theta) = \sin(\pi/2 - \theta)$ holds true. But, </p>
<blockquote>
<p>Does $\cos(\theta) = \sin(\pi/2 +\theta)$ always hold true?</p>
</blockquote>
<p>I am asking this question because I encountered the following question in my workbook.</p>
<p>If $h(x) = \cos x$, $g(x) = \sin x$, and $h(x) = g(f(x))$, which of the following can be $f(x)$?</p>
<p>(a) $-x$</p>
<p>(b) $\pi/2 + x$</p>
<p>(c) $\pi - x$</p>
<p>(d) $3\pi/2 - x$</p>
<p>(e) $3\pi/2 + x$</p>
<p>My book says the correct answer is (b), and I am a bit baffled by this.</p>
<p>I can see that this holds true by plugging in certain values for $x$. But is there a mathematical proof for $\cos(\theta) = \sin(\pi/2 + \theta)$?</p>
| lab bhattacharjee | 33,337 | <p>Let $$\sqrt{\dfrac{x+36}{x-36}}=n\implies x=\dfrac{36(n^2-1+2)}{n^2-1}=36+\dfrac{72}{n^2-1}$$</p>
<p>So $n^2-1(\ge-1)$ must divide $72$ and</p>
<p>$$n^2-1\le72\implies2\le n\le8$$</p>
|
366,401 | <p>Let <span class="math-container">$\nu$</span> be the uniform measure on the unit circle <span class="math-container">$\mathbb{S}^1 \subset \mathbb{R}^2$</span>, normalised so that <span class="math-container">$\nu(\mathbb{S}^1) = 1$</span>. Suppose <span class="math-container">$\mu$</span> is a Borel probability measure on <span class="math-container">$\mathbb{S}^1$</span> which is absolutely continuous w.r.t. <span class="math-container">$\nu$</span>, that is <span class="math-container">$\mu \ll \nu$</span>. Let <span class="math-container">$\{f_n\}_{n\geq 1}$</span> be an orthonormal basis for <span class="math-container">$L^2(\mathbb{S}^1,\mu)$</span>. Is it true that for <span class="math-container">$g \in C^k(\mathbb{S}^1)$</span>
<span class="math-container">$$
\int_{[0,2\pi]} f_n(\theta) g(\theta)d\mu(\theta) = o(1/n^k).
$$</span>
Or is it possible to <em>choose</em> an ONB such that the above holds? My question is motivated by the case when <span class="math-container">$\mu = \nu$</span> and the ONB is <span class="math-container">$\{1,z,\overline{z},z^2,\overline{z^2},\ldots\}$</span>, where it is known to be true (See <a href="https://math.stackexchange.com/questions/52260/asymptotic-estimate-for-riemann-lebesgue-lemma">this</a>.)</p>
<p>Thanks!</p>
| Piotr Hajlasz | 121,665 | <p>In general it is not true. Let <span class="math-container">$\{f_n\}_{n\geq 1}=\{1,z,\overline{z},z^2,\overline{z^2},\ldots\}$</span>, then as the OP pointed out <span class="math-container">$a_n=o(n^{-k})$</span>.
However, with a suitable permutation <span class="math-container">$\sigma$</span> of the basis <span class="math-container">$\{f_n\}_{n\geq 1}$</span>, we will have that coefficients in this new basis satisfy <span class="math-container">$\tilde{a}_n=a_{\sigma(n)}$</span>.
We can choose <span class="math-container">$\sigma$</span> so that for infinitely many <span class="math-container">$n$</span>, <span class="math-container">$\sigma(n)\gg n$</span>.Then it might happen for such <span class="math-container">$n$</span> that
<span class="math-container">$$
\frac{\tilde{a}_n}{n^k}=\frac{a_{\sigma(n)}}{\sigma(n)^k}\left(\frac{\sigma(n)}{n}\right)^k\to \infty.
$$</span>
Indeed, although <span class="math-container">$a_{\sigma(n)}/\sigma(n)^k$</span> is small, <span class="math-container">$(\sigma(n)/n)^k$</span> might be very large.
Providing a more explicit example from this sketch is now a simple exercise.</p>
|
812,778 | <p>Prove that $(4/5)^{\frac{4}{5}}$ is irrational.</p>
<p><strong>My proof so far:</strong></p>
<p>Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.</p>
<p>Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.</p>
<p>Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$</p>
<p>$\therefore$ $4^4q^5=5^4p^5$</p>
<p>I've got to this point and now I don't know where to go from here.</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $ $ By unique factorization, comparing powers of $\,2\,$ as below yields a contradiction $\,5\mid 8$ </p>
<p>$$\begin{align}2^{\large\color{#c00}8}\, Q^{\large\color{#0a0}5}\! &=\, P^{\large\color{#f70}5}\, 5^{\large 4}\\
\Rightarrow\ \color{#c00}8\! +\! \color{#0a0}5q &=\, \color{#f80}5p\\
\Rightarrow\qquad\, 8 &=\, 5(p\!-\!q)\end{align}$$</p>
<p><strong>Remark</strong> $\ $ This is a generalization of the classical proof of the irrationality of $\,\sqrt 2\,$ by comparing the parity of powers of $\,2\,$ in $\,P^2\! = 2Q^2.\,$ It compares powers mod $5$ (vs. $2)$ since it involves $5$'th
vs. $2$nd powers.
The proof does not require the full power of unique prime factorization, only that each natural may be written uniquely in the form $\,2^j n,\,\ n\,$ odd (which has very a simple inductive proof).</p>
|
1,304,529 | <p>I have come across these while studying the limsup & liminf of sequence of subset of a set. In order to understand that, I have to understand what least upper bound & greatest lower bound of a sequence of subset mean. I would be grateful if anyone helps me comprehend this concept intuitively as I am new & novice to this topic.</p>
| paw88789 | 147,810 | <p>$x$ is in the lim sup of a sequence of sets if and only if it is in infinitely many of the sets. </p>
<p>And $x$ is in the lim inf of a sequence of sets if and only if it is in all but finitely many of the sets (or equivalently if it is in all the sets from some point on).</p>
|
88,145 | <p>A couple of recent questions on MO have involved the characters or the orders of specific finite groups of the form $G(\mathbb{Z}/n\mathbb{Z})$ for a familiar algebraic group $G$ defined over $\mathbb{Z}$ (implicitly as a group scheme) especially when $n$ is a prime power: <a href="https://mathoverflow.net/questions/87254/">here</a> and
<a href="https://mathoverflow.net/questions/87904/">here</a>.</p>
<p>It's reasonable to ask in what generality questions of the second kind can be treated uniformly. Here for example the question doesn't seem to require any knowledge of the structure of symplectic matrices, once one knows the orders of the finite symplectic groups over fields. (These have been written down in numerous papers and books.) In particular:</p>
<blockquote>
<p>For group schemes $G$ over $\mathbb{Z}$, in what natural generality is there a uniform procedure for computation of the orders of finite groups $G(\mathbb{Z}/p^k\mathbb{Z})$ as $p$ ranges over the prime numbrs?</p>
</blockquote>
<p>An old computation having a similar flavor occurs for the multiplicative group scheme (which is reductive but not simple) in the determination of the group of units in the ring $\mathbb{Z}_p$ of $p$-adic integers, written as an inverse limit of finite groups with successive quotients of order $p-1$ or $p$. Here the additive group scheme (viewed as the Lie algebra of the multiplicative group scheme) enters the picture, providing an iterated extension of the group of units of the residue field $\mathbb{F}_p$. See for example the book by Serre <em>A Course in Arithmetic</em>, II, section 3.</p>
<p>As George McNinch points out in his comment on Scott Carnahan's direct computation of the order for finite symplectic groups, a uniform approach is suggested by a paper of Serre, "Exemples de plongements des groupes PSL$_2(\mathbb{F}_p)$ dans des groupes de Lie simples", <em>Invent. Math.</em> 124 (1996), 3.1. Here <em>G</em> is a connected simple algebraic group over an algebraically closed field, essentially treated as a group scheme. Serre refers in turn to Demazure-Gabriel, <em>Groupes Algebriques</em> (1970), II, section 4, no. 3, a treatise influenced by the earlier Demazure-Grothendieck seminar SGA3. </p>
<p>In a Bourbaki talk, Chevalley showed how to view the simple adjoint groups, which he had constructed in a uniform way in 1955, as group schemes over $\mathbb{Z}$, a theme refined further by Kostant, and more recently by Lusztig in <em>J. Amer. Math. Soc.</em> 22 (2009). That seems to be a good setting for the question I've raised, though perhaps one can go further in the direction of reductive groups?</p>
<p>ADDED: The answers and comments are very interesting, but while I think further about them and the literature I should clarify that I'm taking for granted the standard (though nontrivial) formulas for the group orders over finite fields. And while it's natural (at least for Chevalley groups) to start over the ring of integers, there is certainly a passage to local rings implied here. The specific prime stays in the background, since unlike many questions in Lie theory this kind of computation doesn't distinguish "good" and "bad" primes: whatever is done should apply uniformly to all primes. Meanwhile a result in Appendix A.5 of the recent book by Conrad-Gabber-Prasad on pseudo-reductive groups has been pointed out to me. This suggests that groups aren't so essential to my question, but only a well-behaved class of schemes (again assuming that one can already count their points over finite fields). </p>
| Marty | 3,545 | <p>The places to look are: </p>
<ol>
<li>Steinberg, "Endomorphisms of linear algebraic groups." Memoir AMS 80, (1968), and </li>
<li>Gross, "The motive of a reductive group" Invent. math. 130, 287 ± 313 (1997).</li>
</ol>
<p>(I learned about the former from the latter).</p>
<p>To any (quasi-split) group $G$ with maximal torus $T$, there is an associated Artin-Tate motive $M$ (i.e., a Galois representation). </p>
<p>The point-count over finite fields is related to the action of Frobenius on the twisted dual motive: equation (3.1) of Gross's article (citing Steinberg) gives us:
$$ \vert G(k) \vert / q^{dim(G)} = \prod_{d \geq 1} det(1 - Fr \cdot q^{-d} : V_d ),$$
where $Fr$ denotes the Frobenius, $q$ the order of the finite field $k$, and $V = \bigoplus V_d$ a graded ${\mathbb Q}$ vector space as follows: $V = R / R^+$ where $R$ is the graded algebra $Sym^\bullet(E)^W$, where $E$ is the character lattice of $T$ tensored with ${\mathbb Q}$ (section 1 of Gross).</p>
|
88,145 | <p>A couple of recent questions on MO have involved the characters or the orders of specific finite groups of the form $G(\mathbb{Z}/n\mathbb{Z})$ for a familiar algebraic group $G$ defined over $\mathbb{Z}$ (implicitly as a group scheme) especially when $n$ is a prime power: <a href="https://mathoverflow.net/questions/87254/">here</a> and
<a href="https://mathoverflow.net/questions/87904/">here</a>.</p>
<p>It's reasonable to ask in what generality questions of the second kind can be treated uniformly. Here for example the question doesn't seem to require any knowledge of the structure of symplectic matrices, once one knows the orders of the finite symplectic groups over fields. (These have been written down in numerous papers and books.) In particular:</p>
<blockquote>
<p>For group schemes $G$ over $\mathbb{Z}$, in what natural generality is there a uniform procedure for computation of the orders of finite groups $G(\mathbb{Z}/p^k\mathbb{Z})$ as $p$ ranges over the prime numbrs?</p>
</blockquote>
<p>An old computation having a similar flavor occurs for the multiplicative group scheme (which is reductive but not simple) in the determination of the group of units in the ring $\mathbb{Z}_p$ of $p$-adic integers, written as an inverse limit of finite groups with successive quotients of order $p-1$ or $p$. Here the additive group scheme (viewed as the Lie algebra of the multiplicative group scheme) enters the picture, providing an iterated extension of the group of units of the residue field $\mathbb{F}_p$. See for example the book by Serre <em>A Course in Arithmetic</em>, II, section 3.</p>
<p>As George McNinch points out in his comment on Scott Carnahan's direct computation of the order for finite symplectic groups, a uniform approach is suggested by a paper of Serre, "Exemples de plongements des groupes PSL$_2(\mathbb{F}_p)$ dans des groupes de Lie simples", <em>Invent. Math.</em> 124 (1996), 3.1. Here <em>G</em> is a connected simple algebraic group over an algebraically closed field, essentially treated as a group scheme. Serre refers in turn to Demazure-Gabriel, <em>Groupes Algebriques</em> (1970), II, section 4, no. 3, a treatise influenced by the earlier Demazure-Grothendieck seminar SGA3. </p>
<p>In a Bourbaki talk, Chevalley showed how to view the simple adjoint groups, which he had constructed in a uniform way in 1955, as group schemes over $\mathbb{Z}$, a theme refined further by Kostant, and more recently by Lusztig in <em>J. Amer. Math. Soc.</em> 22 (2009). That seems to be a good setting for the question I've raised, though perhaps one can go further in the direction of reductive groups?</p>
<p>ADDED: The answers and comments are very interesting, but while I think further about them and the literature I should clarify that I'm taking for granted the standard (though nontrivial) formulas for the group orders over finite fields. And while it's natural (at least for Chevalley groups) to start over the ring of integers, there is certainly a passage to local rings implied here. The specific prime stays in the background, since unlike many questions in Lie theory this kind of computation doesn't distinguish "good" and "bad" primes: whatever is done should apply uniformly to all primes. Meanwhile a result in Appendix A.5 of the recent book by Conrad-Gabber-Prasad on pseudo-reductive groups has been pointed out to me. This suggests that groups aren't so essential to my question, but only a well-behaved class of schemes (again assuming that one can already count their points over finite fields). </p>
| George McNinch | 4,653 | <p>I think the assumptions needed on a group scheme $G$ over $\mathbf{Z}$ are that $G$
should be smooth, affine, and of finite type over $\mathbf{Z}$. Actually,
I'm going to take the point of view that $p$ is fixed and so I'll suppose that $G$ is smooth, affine, and of finite type over the local ring $\mathbf{Z}_{(p)}$.</p>
<p>I'm going to write $\Lambda = \Lambda_r = \mathbf{Z}/p^r\mathbf{Z}$ and $F = \mathbf{F}_p = \Lambda/p\Lambda$.</p>
<p>To give a formula for $|G(\Lambda)|$, you should have a formula for the number of points of the special fiber -- i.e. for $|G(\mathbf{F}_p)|$ (which is OK e.g. if the special fiber is reductive by well-known order formulas as mentioned in Marty's answer).</p>
<p>Let $H$ be an affine group scheme over $\Lambda$ (say, arising by base-change from
the group scheme $G$ over $\mathbf{Z}_{(p)}$).</p>
<p>The result in Demazure-Gabriel to which Serre refers in the article "Exemples de plongements..." (as discussed in the question) implies that the kernel $I$
of the natural mapping $H(\Lambda_k) \to H(\Lambda_{r-1})$
identifies naturally with the $F$-vector space </p>
<p>$$Hom_{\Lambda}(\omega_{H/\Lambda},p^{r-1}\Lambda)$$ </p>
<p>where $\omega_{H/\Lambda}$ is the module of Kahler differentials of $H$ over $\Lambda$.</p>
<p>In particular, $|I| = p^d$ where $d = \dim_F (\omega_{H/\Lambda} \otimes_{\Lambda} F)$.</p>
<p>If $H$ is smooth and of finite type over $\Lambda$ (e.g. if $H = G_{/\Lambda}$ for
$G$ smooth over $\mathbf{Z}_{(p)}$) then the natural mapping
$H(\Lambda_r) \to H(\Lambda_{r-1})$ is surjective (say, by [SGA1, Corr. III.5.3])
and $\omega_{H/\Lambda}$ is a free $\Lambda$-module of rank $\dim H_{/F}$.
And in this case, one gets the formula for $|H(\Lambda)|$ given in Stasinski's answer.</p>
<p>And if $H = G_{/\Lambda}$ then of course $H(\Lambda) = G(\Lambda)$.</p>
<p>Remarks:
(1) It seems to me (?) that the result just quoted from Demazure-Gabriel answers
the conjecture from the end of Greenberg's 1963 paper mentioned in Stasinski's answer.</p>
<p>(2) Let $k$ be (say) an algebraically closed field, and let $W = W_r$ be the
ring of Witt vectors of length $r$ over $k$. Then our smooth group scheme $G$ as before determines a group scheme
$H = G_{/W}$ over $W$. By "Greenberg's functor" we can view the group
of points $G(W) = H(W)$ as a linear algebraic group over $k$. </p>
<p>If say $r=2$, then there is a strictly exact sequence of linear algebraic groups
$$ (*) \quad 0 \to V \to H(W) \to G_{/k} \to 1$$
where $V = $ Lie$(G_{/k})^{[1]}$ is the vector group on which $G_{/k}$ acts via the <em>first Frobenius twist</em> of the adjoint representation Lie$(G_{/k})$.</p>
<p>I mention this because one somehow doesn't "see" the Frobenius twist when the residue
field is $\mathbf{F}_p$, but the Frobenius twist
is important in general.</p>
<p>Indeed, if say $G_{/k}$ is a simple alegbraic group, the exact sequence $(*)$ is not split -- it corresonds
to a non-zero class in $H^2(G_{/k},$Lie$(G_{/k})^{[1]})$.
So the Frobenius
twist is crucial since most of the time $H^2(G_{/k},$Lie$(G_{/k})) = 0$.</p>
|
746 | <p>There have been a number of questions in the Close part of Review lately which were basically asking for help creating an algorithm to do some mundane task (see <a href="https://mathoverflow.net/questions/140585/how-to-perform-divide-step-of-in-place-quicksort#comment362909_140585">here</a>, <a href="https://mathoverflow.net/questions/139681/using-the-affine-maxima-package">here</a>, <a href="https://mathoverflow.net/questions/140705/numerical-solutions-of-ode-by-free-parameters-with-matlab">here</a> for example). I wonder if some of these people could be helped by migrating them over to <a href="https://softwareengineering.stackexchange.com/">https://softwareengineering.stackexchange.com/</a>. Even if you don't think these questions in particular should go there, I'll bet that those who are okay with migration in general would be okay with the option to migrate to the programmers website. Hence my question: are others in favor of adding this feature? If so, can someone make a request to the appropriate powers?</p>
| Scott Morrison | 3 | <p>I don't really like our setup at the moment. Anna said we need to establish precedents of migrations before the open migration paths, but for now only moderators can actually do these migrations. Because we're actually a long established site where moderators area relatively rarely involved in closing questions, we're sort of stuck. </p>
<p>It would be helpful, I guess, if posts that should be migrated get flagged, even if they are already closed. We can reopen then migrate them. </p>
|
1,177,493 | <p>If $p$ is a prime and $p \equiv 1 \bmod 4$, how many ways are there to write $p$ as a sum of two squares? Is there an explicit formulation for this?</p>
<p>There's a theorem that says that $p = 1 \bmod 4$ if and only if $p$ is a sum of two squares so this number must be at least 1. There's also the Sum of Two Squares Theorem for the prime factorization of integers and the Pythagorean Hypotenuse Proposition which says that a number $c$ is a hypotenuse if and only if it's a factor of $1 \bmod 4$ primes. All of these theorems only assert the existence of $1 \bmod 4$ primes as the sum of two squares. How do I (perhaps use these altogether) to find the exact number of different ways to write such prime as a sum of two squares?</p>
| user2566092 | 87,313 | <p>Here is a reference that gives an explicit formula for your problem, even extending to all integers (instead of just primes) <a href="http://mathworld.wolfram.com/SumofSquaresFunction.html" rel="nofollow">http://mathworld.wolfram.com/SumofSquaresFunction.html</a></p>
|
1,738,153 | <p>I know the definition is given as follows:</p>
<p>A map $p: G \rightarrow GL(V)$ such that $p(g_1g_2)=p(g_1)p(g_2)$ but I still do not really understand what this means</p>
<p>Can someone help me gain some intuition for this - perhaps a basic example?</p>
<p>Thanks</p>
| snulty | 128,967 | <p>If you know a little bit of group theory your familiar with the idea of homomorphism. </p>
<p>Now the idea of a linear representation is to take a group and find a homomorphism to a group of linear operators on a vector space. If you've fixed a basis then you're finding a homomorphism to a matrix group. If the vector space is $n$ dimensional, then the matrices are $n\times n$. </p>
<p>To be honest the simplest representation I can think of other than of the trivial group, is of $S_2$, the permutations of two numbers $\{1,2\}$. This is the group $\Bbb Z_2$ but I think permutations are more intuitive.</p>
<p>So if you want to think intuitively (for me at least) think of $S_2$ permuting the fixed basis say $\{e_1,e_2\}$ of a two dimensional space. </p>
<p>Then the $id$ element should leave the basis alone and so is just mapped to the identity matrix under our map $\rho$. Now for a non trivial representation the other permutation should swap the two basis elements and so should map to $$\left( \begin{array}{cc}0 & 1 \\
1 & 0 \end{array} \right).$$</p>
<p>Then in fact there is a natural representation of $S_n$ on an $n$ dim vector space, where fixing a basis $\{e_1,\ldots,e_n\}$, where for $\sigma\in S_n$, let $\rho(\sigma) (e_i)= e_{\sigma(i)}$.</p>
|
1,961,727 | <p>As far as I understood <a href="https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process">Gram–Schmidt orthogonalization</a> starts with a set of linearly independent vectors and produces a set of mutually orthonormal vectors that spans the same space that starting vectors did.</p>
<p>I have no problem understanding the algorithm, but here is a thing I fail to get. Why do I need to do all these calculations? For example, instead of doing the calculations provided in that wiki page in example section, why can't I just grab two basis vectors $w_1 = (1, 0)'$ and $w_2 = (0, 1)'$? They are clearly orthonormal and span the same subspace as the original vectors $v_1 = (3, 1)'$, $v_2 = (2, 2)'$.</p>
<p>It is clear that I'm missing something important, but I can't see what exactly.</p>
| user376902 | 376,902 | <p>Your choice of $w_1 = (1,0)$ and $w_2 = (0,1)$ fails one of the basic purposes of the Gram-Schmidt process: the result of the algorithm would not only have $\mathrm{Span}(v_1,v_2) = \mathrm{Span}(w_1,w_2)$ but also $\mathrm{Span}(v_1) = \mathrm{Span}(w_1).$</p>
<p>There are a few things I should mention:</p>
<p>(1) There are finite-dimensional vector spaces for which it is not easy to guess an orthonormal basis (or indeed any basis). Modular forms with their Petersson scalar product are an intimidating example of this.</p>
<p>(2) The Gram-Schmidt process is smooth in an appropriate sense, which makes it possible to use the Gram-Schmidt process to orthogonalize sections of a Euclidean bundle (a vector bundle with scalar product) and in particular to define things like the orthogonal complement of subbundles. This turns out to be important.</p>
<p>(3) There are also concrete and relatively elementary situations such as QR-factorization for why Gram-Schmidt orthogonalization is useful. Maybe you are already familiar with this.</p>
|
1,961,727 | <p>As far as I understood <a href="https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process">Gram–Schmidt orthogonalization</a> starts with a set of linearly independent vectors and produces a set of mutually orthonormal vectors that spans the same space that starting vectors did.</p>
<p>I have no problem understanding the algorithm, but here is a thing I fail to get. Why do I need to do all these calculations? For example, instead of doing the calculations provided in that wiki page in example section, why can't I just grab two basis vectors $w_1 = (1, 0)'$ and $w_2 = (0, 1)'$? They are clearly orthonormal and span the same subspace as the original vectors $v_1 = (3, 1)'$, $v_2 = (2, 2)'$.</p>
<p>It is clear that I'm missing something important, but I can't see what exactly.</p>
| Community | -1 | <p>If you want to represent a subspace, the standard basis will no do.</p>
<p>For example, assume a plane defined by two arbitrary vectors in 3D: G-S will define an orthonormal base that spans the plane.</p>
|
1,913,689 | <blockquote>
<p>Let $f: X \rightarrow Y$ be a function. $A \subset X$ and $B \subset Y$.
Prove $A \subset f^{-1}(f(A))$.</p>
</blockquote>
<p>Here is my approach. </p>
<p>Let $x \in A$. Then there exists some $y \in f(A)$ such that $y = f(x)$. By the definition of inverse function, $f^{-1}(f(x)) = \{ x \in X$ such that $y = f(x) \}$. Thus $x \in f^{-1}(f(A)).$</p>
<p>Does this look OK, and how can I improve it?</p>
| Caleb Stanford | 68,107 | <p>Your proof looks pretty good. The only thing to point out is when you said:</p>
<blockquote>
<p>By the definition of inverse function, $f^{-1}(f(x)) = \{ x \in X$ such that $y = f(x) \}$. Thus $ x \in f^{-1}(f(A)).$</p>
</blockquote>
<p>Two comments on this:</p>
<ol>
<li><p>This isn't usually called the <em>inverse function</em> -- we reserve that for when $f$ is invertible, and has a function $f^{-1}: Y \to X$. Instead, $f^{-1}$ here is called the <em>inverse image</em>, which is not a function $Y \to X$ but instead it takes subsets of $Y$ to subsets of $X$. It's confusing, I know, that they have the same symbol $f^{-1}$ for both.</p></li>
<li><p>Your description of $f^{-1}(f(x))$ is a bit muddled, even though the reasoning is correct. It should be, $\{x' \in X \text{ such that } f(x') = f(x)\}$. That's the definition of $f^{-1}(a)$, where I put in $a = f(x)$. And I used $x'$ instead of $x$ because you don't want to mix up your two different $x$s.</p></li>
</ol>
|
2,007,373 | <p>At some point in your life you were explained how to understand the dimensions of a line, a point, a plane, and a n-dimensional object. </p>
<p>For me the first instance that comes to memory was in 7th grade in a inner city USA school district. </p>
<p>Getting to the point, my geometry teacher taught,</p>
<p>"a point has no length width or depth in any dimensions, if you take a string of points and line them up for "x" distance you have a line, the line has "x" length and zero height, when you stack the lines on top of each other for "y" distance you get a plane"</p>
<p>Meanwhile I'm experiencing cognitive dissonance, how can anything with zero length or width be stacked on top of itself and build itself into something with width of length?</p>
<p>I quit math. </p>
<p>Cut to a few years after high school, I'm deep in the math's. </p>
<p>I rationalized geometry with my own theory which didn't conflict with any of geometry or trigonometry. </p>
<p>I theorized that a point in space was infinitely small space in every dimension such that you can add them together to get a line, or add the lines to get a plane. </p>
<p>Now you can say that the line has infinitely small height approaching zero but not zero.</p>
<p>What really triggered me is a Linear Algebra professor at my school said that lines have zero height and didn't listen to my argument. . . </p>
<p>I don't know if my intuition is any better than hers . . . if I'm wrong, if she's wrong . . . </p>
<p>I would very much appreciate some advice on how to deal with these sorts of things. </p>
| Ben Grossmann | 81,360 | <p>I don't know if your post has much to do with "life advice", but the question of whether there should be an "infinitely small but non-zero width" is something that bears answering.</p>
<p>The way math is done (with the standard set of axioms), it is indeed taken as fact that a point has <strong>exactly zero</strong> length and that a line, which has non-zero length, has <strong>exactly zero</strong> thickness. With an understanding of measure theory, one can see that putting "uncountably many" "measure-zero" things together can produce something with non-zero measure, even if that seems a bit counterintuitive. </p>
<p>However, "non-standard analysis" (which, as its name suggests, is not standard) does allow for a notion of "infinitely small but non-zero" quantities. I don't know enough about this, however, to judge how well the usual formulation of those ideas lines up with your intuition.</p>
<p>As far as life-advice goes, I'd say don't be afraid of not knowing, even if you're the only one in the room who appears to be struggling with the idea. Most of the time, other people in your situation are struggling with the same idea, but these things are hard to express and potentially embarrassing to ask about. </p>
<p>You are certainly not the only one to struggle with the notion of "of course there are infinitely small things!". For example, you'll find that the question of <em>why does $0.999\dots = 1$</em> has been asked an re-asked many times over on this site.</p>
|
2,007,373 | <p>At some point in your life you were explained how to understand the dimensions of a line, a point, a plane, and a n-dimensional object. </p>
<p>For me the first instance that comes to memory was in 7th grade in a inner city USA school district. </p>
<p>Getting to the point, my geometry teacher taught,</p>
<p>"a point has no length width or depth in any dimensions, if you take a string of points and line them up for "x" distance you have a line, the line has "x" length and zero height, when you stack the lines on top of each other for "y" distance you get a plane"</p>
<p>Meanwhile I'm experiencing cognitive dissonance, how can anything with zero length or width be stacked on top of itself and build itself into something with width of length?</p>
<p>I quit math. </p>
<p>Cut to a few years after high school, I'm deep in the math's. </p>
<p>I rationalized geometry with my own theory which didn't conflict with any of geometry or trigonometry. </p>
<p>I theorized that a point in space was infinitely small space in every dimension such that you can add them together to get a line, or add the lines to get a plane. </p>
<p>Now you can say that the line has infinitely small height approaching zero but not zero.</p>
<p>What really triggered me is a Linear Algebra professor at my school said that lines have zero height and didn't listen to my argument. . . </p>
<p>I don't know if my intuition is any better than hers . . . if I'm wrong, if she's wrong . . . </p>
<p>I would very much appreciate some advice on how to deal with these sorts of things. </p>
| Mikhail Katz | 72,694 | <p>After Abraham Robinson died in 1974 there was a bit of a pogrom against both the framework with infinitesimals that he developed, and against his students who had great trouble finding jobs. Most of the critics were less than well-informed, as richly illustrated in the current literature. </p>
<p>Thus, Alain Connes' flawed critique is analyzed in <a href="http://dx.doi.org/10.1007/s10699-012-9316-5" rel="nofollow noreferrer">this article</a>. Errett Bishop's flawed critique is analyzed in <a href="http://dx.doi.org/10.1007/s10699-013-9340-0" rel="nofollow noreferrer">this article</a>.</p>
<p>In recent years Robinson's framework got an eloquent advocate in the person of Fields medalist Terry Tao, who has been publishing books and writing blogs about this. Robinson's framework with infinitesimals remains the best approach to geometry, calculus, and analysis, and in particular answers your question about infinitesimal width.</p>
|
1,017,707 | <p>Are there any proofs of this equality online? I'm just looking for something very simply that I can self-verify. My textbook uses the result without a proof, and I want to see what a proof would look like here.</p>
| Andreas H. | 44,523 | <p>I am not sure which proof you are looking for.</p>
<p>Your polynomial $Q_n$ is called the Chebyshev polynomial of degree $N$. </p>
<p>Probably the wikipedia page (<a href="http://en.wikipedia.org/wiki/Chebyshev_polynomials" rel="nofollow">http://en.wikipedia.org/wiki/Chebyshev_polynomials</a>) will help you out in finding what you are looking for. </p>
|
1,976,382 | <p>Hölder's inequality for finite sums is given by
$$\sum_{k=0}^n|a_kb_k|\leq\left(\sum_{k=0}^n|a_k|^p\right)^{1/p}\left(\sum_{k=0}^n|b_k|^q\right)^{1/q},$$
where $1/p+1/q=1$, $p,q\in(1,\infty)$.</p>
<p>Is there a "similar" inequality which gives a lower bound for the left hand sum? I have searched, but found nothing so far.</p>
| Jacky Chong | 369,395 | <p>Just consider Cauchy-Schwarz. We see that
\begin{align}
|a_1b_1 + a_2b_2| \leq \sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}.
\end{align}
If you are hoping for estimates of the nature
\begin{align}
f\left(\sqrt{a_1^2+a_2^2}, \sqrt{b_1^2+b_2^2}\right) \leq |a_1b_1+a_2b_2|
\end{align}
for some function $f(x, y)$ which is nonnegative on $\{x \geq 0, y\geq 0\}$, then you better hope if $(a_1, a_2)\perp (b_1, b_2)$ you have
\begin{align}
f\left(\sqrt{a_1^2+a_2^2}, \sqrt{b_1^2+b_2^2}\right) \leq 0.
\end{align}
Well. Guess $f$ has to be trivial. </p>
|
3,982,103 | <p>Let <span class="math-container">$(X,\tau)$</span> be a topological space. Prove that <span class="math-container">$\tau$</span> is the finite-closed topology on <span class="math-container">$X$</span> if and only if (i)<span class="math-container">$(X,\tau)$</span> is a <span class="math-container">$T_1$</span>-space, and (ii) every infinite subset of <span class="math-container">$X$</span> is dense in <span class="math-container">$X$</span>.</p>
<p>I already proved the forward direction but I'm stuck on the backward direction. We know that everyone singleton is closed because of i) and from ii) every open set intersects any infinite set non trivially. Now I need to figure out how to show that every open set are just infinite sets with countably finite points removed, thus the topology is finite-closed.</p>
| Kavi Rama Murthy | 142,385 | <p>Let <span class="math-container">$U$</span> be a non-empty open set. If <span class="math-container">$X \setminus U$</span> has an infinite number of points then the infinite set <span class="math-container">$X \setminus U$</span> cannot be dense. This is because it does not intersect <span class="math-container">$U$</span>. This contradiction proves that <span class="math-container">$X \setminus U$</span> is finite whenever <span class="math-container">$U$</span> is a non-empty open set.</p>
|
3,982,103 | <p>Let <span class="math-container">$(X,\tau)$</span> be a topological space. Prove that <span class="math-container">$\tau$</span> is the finite-closed topology on <span class="math-container">$X$</span> if and only if (i)<span class="math-container">$(X,\tau)$</span> is a <span class="math-container">$T_1$</span>-space, and (ii) every infinite subset of <span class="math-container">$X$</span> is dense in <span class="math-container">$X$</span>.</p>
<p>I already proved the forward direction but I'm stuck on the backward direction. We know that everyone singleton is closed because of i) and from ii) every open set intersects any infinite set non trivially. Now I need to figure out how to show that every open set are just infinite sets with countably finite points removed, thus the topology is finite-closed.</p>
| user126154 | 126,154 | <p>You have to prove that, under hypothesis <span class="math-container">$(i),(ii)$</span>, a non-trivial set (i.e. non-empnty and not <span class="math-container">$X$</span>) is closed if and only if it is finite.</p>
<p>Any closed set different from <span class="math-container">$X$</span> is not dense (because it coincides with its closure). Then by <span class="math-container">$(ii)$</span> it is not infinite, hence it is finite. Conversely, any finite set is finite union of points, that are closed because of <span class="math-container">$(i)$</span>, so it is closed.</p>
|
9,934 | <p>I have been given some code with the following line</p>
<pre><code>PeriodicExtension[g_, x_] := If[Abs[x] < Pi, g[x], PeriodicExtension[g, x - 2 Sign[x] Pi]]
</code></pre>
<p>I do not understand the syntax. I would appreciate if someone can explain what this
code does for different values of <code>x</code>.</p>
| stevenvh | 1,450 | <p>I agree with J.M.: don't use recursion if you don't need it. I was just running a benchmark when he posted his answer.</p>
<pre><code>PeriodicExtension[g_, x_] := If[Abs[x] < Pi, g[x], PeriodicExtension[g, x - 2 Sign[x] Pi]];
g[x_] := x^3
Timing[Plot[PeriodicExtension[g, x], {x, 0, 1000 Pi}]]
</code></pre>
<blockquote>
<p>{ 9.922 }</p>
</blockquote>
<pre><code>PeriodicExtension[g_, x_] := g[Mod[x + Pi, 2 Pi] - Pi];
g[x_] := x^3
Timing[Plot[PeriodicExtension[g, x], {x, 0, 1000 Pi}]]
</code></pre>
<blockquote>
<p>{ 0.234 }</p>
</blockquote>
<p>The non-recursive version runs 42 times faster! (For an interval of 10 000 Pi even 100 times.) What's worse is that the recursion fails for large intervals. The recursive version for an interval [0, 100 000] gives this plot: </p>
<p><img src="https://i.stack.imgur.com/9RJg8.png" alt="enter image description here"> </p>
<p>whereas the non-recursive one gives </p>
<p><img src="https://i.stack.imgur.com/Jh5E2.png" alt="enter image description here"></p>
<p>still within 0.4 seconds.</p>
|
3,133,695 | <p>A spotlight on the ground shines on a wal 12 m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s, how fast is the length of his shadow on the building decreasing when he is 4m from the building.</p>
<p>How do you solve this word problem. I have drawn a picture to figure out the solution but I have failed to come up with anything.</p>
<p><a href="https://i.stack.imgur.com/PFCDG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PFCDG.png" alt="enter image description here"></a></p>
<p>Let x be the distance between the man and the spotlight, the distance between him and the building shall be <span class="math-container">$12-x$</span>. <span class="math-container">$12-x=4$</span> means <span class="math-container">$x=8$</span></p>
<p>At this point I don't know how to proceed.</p>
<p>Please help</p>
| Anders Kaseorg | 38,671 | <p>Yes. There are two degrees of freedom in the solution space, so we can add two extra constraints to make the computation easier. Here’s a solution with a 90° angle and a 45° angle:</p>
<p><a href="https://i.stack.imgur.com/uS9nN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uS9nN.png" alt="diagram" /></a></p>
<p>The side lengths, from longest to shortest, are
<span class="math-container">$$\begin{align*}
1&, \\
α &\approx 0.838045, \\
(2 + \sqrt2) + (6 + 4\sqrt2) α + \sqrt2 α^2 - (12 + 8\sqrt2) α^3 &\approx 0.454531, \\
(3 + 3\sqrt2) + (9 + 7\sqrt2) α - (2 + 2\sqrt2) α^2 - (16 + 12\sqrt2) α^3 &\approx 0.284498,
\end{align*}$$</span></p>
<p>where <span class="math-container">$α$</span> is the positive root of</p>
<p><span class="math-container">$$1 - \sqrt2 - 2α - 2α^2 + (4 - \sqrt2)α^3 + 4α^4 = 0.$$</span></p>
|
2,305,656 | <p>I solved this problem on my own, months ago, but the solution seems to me completely forgotten, a little help on it would be appreciated:</p>
<p>Suppose $\alpha= \alpha(t)$ on an interval $I$ is a smooth (of class C$^1$) parametric representation of the curve $C$, and for any $t \in I$ we have $\space\space\frac{d}{dt}\alpha(t).v=0$, where $v$ is a constant vector, furthermore $\alpha(0)$ is perpendicular to $v$.
Then $\forall t\in I: \space \alpha(t).v=0$.</p>
| Chandramauli Chakraborty | 375,555 | <p>We know that $r=4R\sin{(\dfrac{A}{2})} \sin{(\dfrac{B}{2})} \sin{(\dfrac{C}{2})}.$ where A, B and C are the angles of the triangles.
And $\dfrac{a}{c}=\tan{A}$, where A is the equal angles or base angles which can be proofed using sine rule.</p>
<p>Now $\frac{r}{R}=k=8\sin{(\dfrac{A}{2})}\cos{(\dfrac{A}{2})}$,
$\therefore k=4\sin{A}$.
Now you should express $\sin{A}$in terms of $\tan{A}$, and then find the values of $\tan{A}$ in terms of k and it is done.</p>
|
846,797 | <p>I encountered this calculation in a problem $\dfrac{\sin 150^o\times\sin 20^o}{\sin 80^o\times\sin 10^o}$ and calculated that it equals 1.</p>
<p>Is it just a coincidence or is there any identity that says $\sin 150^o\times\sin 20^o=\sin 80^o\times\sin 10^o$?</p>
<p>I am trying to use the addition formulae and<br>
$\sin \phi\sin \theta\equiv\dfrac{\cos (\phi-\theta)-\cos (\phi+\theta)}{2}$,<br>
which reduces to showing $\cos 130^0\cos 170^0\equiv\cos 70^0\cos 90^0$, but still unable to explain why.</p>
<p>Any help is really appreciated. Many thanks!</p>
| puru | 151,916 | <p>$\sin 150^0=\sin 30^0=1/2$ </p>
<p>Hence LHS=$\frac{\sin 20^0}{2}$ </p>
<p>$(\sin 2\theta)/2= \sin \theta \cos \theta\dots(1)$</p>
<p>$\sin 80^0=\cos 10^0 \implies$ RHS= $\sin 10^0 \cos 10^0$ = LHS from $(1)$</p>
<p>Hope this helps! </p>
|
846,797 | <p>I encountered this calculation in a problem $\dfrac{\sin 150^o\times\sin 20^o}{\sin 80^o\times\sin 10^o}$ and calculated that it equals 1.</p>
<p>Is it just a coincidence or is there any identity that says $\sin 150^o\times\sin 20^o=\sin 80^o\times\sin 10^o$?</p>
<p>I am trying to use the addition formulae and<br>
$\sin \phi\sin \theta\equiv\dfrac{\cos (\phi-\theta)-\cos (\phi+\theta)}{2}$,<br>
which reduces to showing $\cos 130^0\cos 170^0\equiv\cos 70^0\cos 90^0$, but still unable to explain why.</p>
<p>Any help is really appreciated. Many thanks!</p>
| Anonymous Computer | 128,641 | <p>You should know that $\sin 150^\circ=\sin 30^\circ=\frac 12$.</p>
<p>Now, $\sin 80^\circ=\cos 10^\circ$, because of the property $\sin x =\cos(90-x)$. </p>
<p>Now we have:
$$\frac 12 \sin 20^\circ=\cos 10^\circ \times \sin 10^\circ$$
Recall the double angle identity $\sin(2x)=2\sin x\cos x$. This means that $\frac{\sin(2x)}{2}=\sin x\cos x$. Let's rewrite our equality:
$$\frac{\sin 2(10)^\circ}{2}=\sin 10^\circ \times \cos 10^\circ$$
We can see that this is a double angle identity, where $x=10^\circ$. The LHS does indeed equal the RHS.</p>
<p>That is why $\sin 150^\circ \times \sin 20^\circ=\sin 80^\circ\times\sin 10^\circ$.</p>
|
4,594 | <p>I would like to open an Excel file and manipulate it as a COM object. While I'm able to open an instance of excel with</p>
<pre><code>Needs["NETLink`"]
InstallNET[]
excel = CreateCOMObject["Excel.Application"]
</code></pre>
<p>This doesn't work for me:</p>
<pre><code> wb = excel@Workbooks@Open["D:\\prices.csv"]
</code></pre>
<p>Producing these errors:</p>
<pre>NET::netexcptn: A .NET exception occurred: System.Runtime.InteropServices.COMException (0x80028018): Old format or invalid type library. (Exception from HRESULT: 0x80028018 (TYPE_E_INVDATAREAD))
at Microsoft.Office.Interop.Excel.Workbooks.Open(String Filename, Object UpdateLinks, Object ReadOnly, Object Format, Object Password, Object WriteResPassword, Object IgnoreReadOnlyRecommended, Object Origin, Object Delimiter, Object Editable, Object Notify, Object Converter, Object AddToMru, Object Local, Object CorruptLoad).</pre>
<p>Is this a known problem? I would very much appreciate any ideas on how to open an excel file with Mathematica as a COM object.</p>
| Chris Degnen | 363 | <p>You don't need the initial <code>InstallNET[]</code>. That should come after <code>Needs["NETLink"]</code>.</p>
<p>I made a post on this topic a while back, here: <a href="http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00386.html" rel="noreferrer">http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00386.html</a></p>
<p>Some code to illustrate the method:</p>
<pre><code>Needs["NETLink`"]
ReadFromExcel[file_String, cell_String, rows_Integer, cols_Integer] :=
Module[{excel, workbook, worksheet, srcRange, data},
NETBlock[InstallNET[];
excel = CreateCOMObject["Excel.Application"];
If[! NETObjectQ[excel], Return[$Failed],
excel[Visible] = True;
workbook = excel@Workbooks@Open[file];
worksheet = workbook@Worksheets@Item[1];
srcRange = worksheet@Range[cell]@Resize[rows, cols];
data = srcRange@Value;
workbook@Close[False];
excel@Quit[];
]];
LoadNETType["System.GC"];
GC`Collect[];
data];
ReadFromExcel["testdata.xlsx", "B1", 2, 3]
</code></pre>
<blockquote>
<p>{{1., 2., 3.}, {4., 5., 6.}}</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/yQ6Qd.png" alt="enter image description here"></p>
|
3,356,544 | <p>A lot of calculators actually agree with me saying that it is defined and the result equals 1, which makes sense to me because:</p>
<p><span class="math-container">$$ (-1)^{2.16} = (-1)^2 \cdot (-1)^{0.16} = (-1)^2\cdot\sqrt[100]{(-1)^{16}}\\
= (-1)^2 \cdot \sqrt[100]{1} = (-1)^2 \cdot 1 = 1$$</span></p>
<p>However, there are certain calculators (WolframAlpha among them) which contest this answer, and instead claim it is equal to:</p>
<p><a href="https://i.stack.imgur.com/XB8nG.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XB8nG.png" alt="enter image description here"></a></p>
<p>Graphing this as an exponential function was not possible.</p>
<p>What's going on?</p>
| Brandon O. Salazar | 480,071 | <p>I don't think you are allowed to split negative numbers up like that. Since <span class="math-container">$-1=e^{i\pi}$</span> then <span class="math-container">$$(-1)^{2.16}=e^{i\pi\cdot 2.16}$$</span> which is not a real number.</p>
|
2,281,932 | <p>If Peano axioms uniquely determine the natural numbers, doesn't this mean that Peano axioms are categorical and hence complete?</p>
<p>If above is true, how is it explained by Goedel's incompleteness theorem?</p>
| Noah Schweber | 28,111 | <p>It depends what you mean by "Peano axioms".</p>
<p>M.Winter's answer assumes you mean the <em>first-order</em> Peano axioms. By the compactness theorem, no infinite structure can <em>ever</em> be captured up to isomorphism by a first-order theory, and no countable rigid structure (like $\mathbb{N}$) can ever be captured up to isomorphism <em>amongst countable structures</em> by a first-order theory. Moreover, by Godel's incompleteness theorem, PA isn't even complete: there are sentences which are true in some models of PA and false in others.</p>
<p>So first-order PA does a very bad job of capturing PA. The second-order axioms, meanwhile, are categorical: $\mathbb{N}$ is their only model (up to isomorphism). However, working in second-order logic has drawbacks: the semantics for second-order logic depend highly on set theory, to the point that there is a second-order sentence which is a validity if and only if the Continuum Hypothesis is true, and second-order logic has no complete and sound proof system. Of course, if we embrace set theory then things get somewhat better: any model $M$ of the first-order(!) theory ZFC contains a unique up-to-internal-isomorphism structure which it thinks is a model of the second-order Peano axioms. However, different models can disagree about what exactly these axioms entail, so this ultimately just pushes back the question to the incompleteness of the theory ZFC (and also is really a thinly-veiled transformation of second-order PA into something embedded in a first-order theory, so dodging the really deep second-order questions). Ultimately there's no escaping the logical commitments needed for second-order logic to work properly in this context.</p>
<p>So the version of the PA axioms which do capture $\mathbb{N}$, do so by invoking a very messy logical framework. </p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.