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 |
|---|---|---|---|---|
3,639,192 | <p>In their article on the <a href="https://en.wikipedia.org/wiki/Brauer_group#Galois_cohomology" rel="nofollow noreferrer">Brauer group</a> Wikipedia writes:</p>
<blockquote>
<p>Since all central simple algebras over a field <span class="math-container">$K$</span> become isomorphic to the matrix algebra over a separable closure of <span class="math-container">$K$</span>, the set of isomorphism classes of central simple algebras of degree <span class="math-container">$n$</span> over <span class="math-container">$K$</span> can be identified with the Galois cohomology set <span class="math-container">$H^1(K, \mathrm{PGL}(n))$</span>.</p>
</blockquote>
<p>I understand all the words here, but the reasoning goes too quick for me to follow it. Could someone explain why this is true?</p>
<p>I believe <span class="math-container">$H^1(K, \mathrm{PGL}(n))$</span> is really the group cohomology <span class="math-container">$H^1(G, \mathrm{PGL}(n,K))$</span> where <span class="math-container">$G$</span> is the Galois group of the separable closure of the field <span class="math-container">$K$</span>. The group cohomology <span class="math-container">$H^1(G,M)$</span> is explained <a href="https://en.wikipedia.org/wiki/Group_cohomology#H_1" rel="nofollow noreferrer">here</a>. Perhaps it's not explained in sufficient generality, since they seem to define <span class="math-container">$H^1(G,M)$</span> only where <span class="math-container">$M$</span> is an abelian group acted on by a group <span class="math-container">$G$</span>. But I think the same thing should work for any set acted on by <span class="math-container">$G$</span>, and I know how <span class="math-container">$PGL(n,K)$</span> is acted on by the Galois group <span class="math-container">$G$</span>.</p>
<p>I guess I need to see how a</p>
<ul>
<li>central simple algebra <span class="math-container">$A$</span> over a field <span class="math-container">$K$</span> that becomes isomorphic to an <span class="math-container">$n \times n$</span> matrix algebra when tensored with the separable closure of <span class="math-container">$A$</span> </li>
</ul>
<p>gives rise to a</p>
<ul>
<li>1-cocycle <span class="math-container">$c_A \colon G \to M$</span></li>
</ul>
<p>and why isomorphic algebras of this sort give cocycles that differ by a coboundary. (Also how to go back.)</p>
| Awnon Bhowmik | 515,900 | <p><span class="math-container">$$\begin{align}9^x+15^x&=25^x\\\left(\dfrac35\right)^{2x}+\left(\dfrac35\right)^x&=1\\y^2+y&=1\qquad\boxed{\text{Let }y=\left(\dfrac35\right)^x}\\y^2+y-1&=0\\y&=\dfrac{-1\pm\sqrt{5}}2\end{align}$$</span></p>
<p><span class="math-container">$$\begin{equation}\begin{split}\left(\dfrac35\right)^x&=\dfrac{-1+\sqrt5}2\\x\ln\left(\dfrac35\right)&=\ln\left(\dfrac{-1+\sqrt5}2\right)\\x&=\dfrac{\ln(-1+\sqrt5)-\ln2}{\ln3-\ln5}\approx 0.942028\end{split}\qquad\begin{split}\left(\dfrac35\right)^x&=\dfrac{-1-\sqrt5}2\\x\ln\left(\dfrac35\right)&=\ln\left(\dfrac{-1-\sqrt5}2\right)\\x&=\dfrac{\ln(1+\sqrt5)-\ln2+i\pi}{\ln 3-\ln5}\in\mathbb{C}\end{split}\end{equation}$$</span></p>
|
47,246 | <p>I got some text scraps with this <em>structure</em></p>
<pre><code>"Focal Plane: 198' Active Aid to Navigation: Yes *Latitude: 35.250 N *Longitude: -75.529 W"
</code></pre>
<p>But some of them lack of parts like this</p>
<pre><code>"Focal Plane: 198' Active Aid to Navigation: Yes *Longitude: -75.529 W"
</code></pre>
<p>Using regular expression matching, I want to extract all the info available, included getting a tag "NotAvailable" for missing data., </p>
<p>I wrote this code, but it can't tell when a pattern does not match -- should the pattern in the middle not match, my code just skips it and goes on to the next pattern:</p>
<pre><code>ptrFocal = "(?<=\\Focal Plane: )(.*?)(?=\\')";
ptrLat = "(?<=\\*Latitude: )(.*?)(?=[a-zA-Z])";
ptrLon = "(?<=\\*Longitude: )(.*?)(?=[a-zA-Z])";
r =
StringTrim /@ StringCases[text,
RegularExpression[ptrFocal <> "|" <> ptrLat <> "|" <> ptrLon]
];
</code></pre>
<p>I would want some pointers on how to do this.</p>
<p><strong>EDIT:</strong></p>
<p>The expected transformation of the example data I gave above is</p>
<blockquote>
<pre><code>{198, 32.250, -75.529}
{198, "NotAvailable", -75.529}
</code></pre>
</blockquote>
| Murta | 2,266 | <p>As requested, here is one approach using pure <code>RegularExpression</code>.</p>
<p>Another difference is the use of <code>Internal`StringToDouble</code> that is faster then ToExpression (velocity test <a href="https://mathematica.stackexchange.com/a/19631/2266">here</a>).</p>
<pre><code>getFields[str_String]:=Module[{fields,r},
fields = {"Focal Plane: ", "Latitude: ", "Longitude: "};
r=Map[StringCases[str,RegularExpression[#<>"((\\+|-)?\\d+(\\.\\d+)?)"]:>Internal`StringToDouble@"$1"]&,fields];
r=Flatten[r/.{}-> Missing]
]
str1 = "Focal Plane: 198.23' Active Aid to Navigation: Yes *Longitude: -75.529 W";
str2 = "Focal Plane: 198.23' Active Aid to Navigation: Yes *Latitude: 35.250 N *Longitude: -75.529 W";
getFields@str1
getFields@str2
</code></pre>
<blockquote>
<p>{198.23, Missing, -75.529}</p>
<p>{198.23, 35.25`, -75.529}</p>
</blockquote>
|
3,065,572 | <p>Function <span class="math-container">$f: (0, \infty) \to \mathbb{R}$</span> is continuous. For every positive <span class="math-container">$x$</span> we have <span class="math-container">$\lim\limits_{n\to\infty}f\left(\frac{x}{n}\right)=0$</span>. Prove that <span class="math-container">$\lim\limits_{x \to 0}f(x)=0$</span>. I have tried to deduce something from definition of continuity, but with no effect.</p>
| Community | -1 | <p>What you're trying to prove is the transitive relation of subsets. The statement is logically equivalent to if A<span class="math-container">$\nsubseteq$</span>C then A<span class="math-container">$\subseteq$</span>B or B<span class="math-container">$\nsubseteq$</span>C. So assume A<span class="math-container">$\nsubseteq$</span>C and B<span class="math-container">$\subseteq$</span>C, now I must show that A<span class="math-container">$\nsubseteq$</span>B. So Since A<span class="math-container">$\nsubseteq$</span>C, <span class="math-container">$\exists$</span> x<span class="math-container">$\in$</span>A such that x<span class="math-container">$\notin$</span>C, and since B<span class="math-container">$\subseteq$</span>C and x<span class="math-container">$\notin$</span>C, it follows that A<span class="math-container">$\nsubseteq$</span>B.</p>
|
113,349 | <p>I have a lot of old <em>Mathematica</em> version 2.2 <code>.ma</code> files that simply crash the current (10.4.1) version of <em>Mathematica</em> if I try to open them.</p>
<p>Is there some way to convert them to <code>.nb</code> notebook files?</p>
<p>Is 5.2 the latest version that will do the conversion? And if so, will that older version run under: Windows 10? Windows 7? OS X 10.11?</p>
| larry | 38,059 | <p>How to save Older Mathematica Files With The .ma Extension.</p>
<p>In order to use them I needed to convert some older (1995) mathematica files I found in a CD in a used book so that they would run on 11.1. They had extension ".ma". </p>
<p>To do this I did the following.
1) Download the Mathematica CDF program. To find this look under resources in the Mathematica start window.
2) Place the files to be modified in a there own directory.
3) Change windows folder options to show all extensions.
4) Change the ma extension to nb. With my ".ma" files if I did not do this CDF would not read them.
5) Read the files into the CDF program.
6) In the edit drop down choose Select All then Copy.
7) Open a new notebook in Mathematica and paste the text into.
8) Save the file. I added the word reworked at the end of the filename but before the extension. Example Chpt1.ma-->Chpt1.nb->>into CDF->>copied text to new file in Mathematica --> saved as Chpt1reworked.nb.</p>
<p>There maybe problems as some instructions have changed, but this will get you started.</p>
<p>Good Luck</p>
|
361,171 | <p>If I am about to fabricate a bracelet and I can select $24$ pearls out of a total of $500$ pearls ($300$ white, $150$ red and $50$ green) how many possible bracelets can I create?</p>
<p>The (official) solution to this question is
$$ \frac{500!}{476!} = 3.4\cdot10^{64} $$
But how can it be that the colors of the pearls were totally neglected? I mean if there would just be red pearls then this solution would be clear to me. But since there are different colors I am confused...</p>
| timidpueo | 54,230 | <p>Yeah it seems like the official answer is assuming each pearl is distinct, in which case the fact that it provided colors is meaningless. On top of that even if you assume each pearl is distinct, the official answer is wrong. It would be right if the question was asking about putting pearls in a line instead of a bracelet. If the question were asking for (say) putting 24 pearls on a merry-go-round, then you need to divide the official answer by 24, because rotation doesn't matter anymore. Now if it were a bracelet, take the merry-go-round answer and divide that by 2 (you can put on a bracelet 'forwards' or 'backwards').</p>
<p>So to sum up: Assuming all pearls are distinct, picking 24 where order matters is:</p>
<p>$$ {500 \choose 24} = \frac{500!}{24! \cdot 476!}.$$</p>
<p>Where order matters, ie putting the pearls in a line (official answer):</p>
<p>$$ \frac{500!}{24! \cdot 476!} \cdot 24! = \frac{500!}{476!}.$$</p>
<p>Now if you were putting the pearls on a bracelet:</p>
<p>$$ \frac{500!}{476!} \frac{1}{24 \cdot 2} .$$</p>
<p>Seems it's just a very poorly written question....</p>
|
361,171 | <p>If I am about to fabricate a bracelet and I can select $24$ pearls out of a total of $500$ pearls ($300$ white, $150$ red and $50$ green) how many possible bracelets can I create?</p>
<p>The (official) solution to this question is
$$ \frac{500!}{476!} = 3.4\cdot10^{64} $$
But how can it be that the colors of the pearls were totally neglected? I mean if there would just be red pearls then this solution would be clear to me. But since there are different colors I am confused...</p>
| Marko Riedel | 44,883 | <p>The first answer is excellent. I would just like to remark that if we are calculating the cycle index of $D_{24}$ anyway, then why not use the Polya enumeration theorem (PET)?</p>
<p>We have by table lookup and/or basic reasoning that for the dihedral group the cycle index is given by
$$ Z(D_n) = \frac{1}{2} Z(C_n) +
\frac{1}{4} \left(a_1^2 a_2^{(n-2)/2} + a_2^{n/2}\right)$$
and
$$ Z(C_n) = \frac{1}{n} \sum_{d|n} \phi(d) a_d^{n/d}.$$</p>
<p>This implies that
$$ Z(D_{24}) =
1/48\,{a_{{1}}}^{24}+{\frac {13}{48}}\,{a_{{2}}}^{12}+1/24\,{a_{{3}}}^
{8}+1/24\,{a_{{4}}}^{6}+1/24\,{a_{{6}}}^{4}+1/12\,{a_{{8}}}^{3}+1/12\,
{a_{{12}}}^{2}\\+1/6\,a_{{24}}+1/4\,{a_{{1}}}^{2}{a_{{2}}}^{11}.$$</p>
<p>The answer to the problem is then given by $Z(D_n)(x+y+z)_{x=1, y=1, z=1},$ but we have
$$ Z(D_n)(x+y+z) =
1/48\, \left( x+y+z \right) ^{24}+{\frac {13}{48}}\, \left( {x}^{2}+{y
}^{2}+{z}^{2} \right) ^{12}\\+1/24\, \left( {x}^{3}+{y}^{3}+{z}^{3}
\right) ^{8}+1/24\, \left( {x}^{4}+{y}^{4}+{z}^{4} \right) ^{6}+1/24
\, \left( {x}^{6}+{y}^{6}+{z}^{6} \right) ^{4}\\+1/12\, \left( {x}^{8}+{
y}^{8}+{z}^{8} \right) ^{3}+1/12\, \left( {x}^{12}+{y}^{12}+{z}^{12}
\right) ^{2}+1/6\,{x}^{24}+1/6\,{y}^{24}+1/6\,{z}^{24}\\+1/4\, \left( x
+y+z \right) ^{2} \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{11}$$
so that $$Z(D_n)(x+y+z)_{x=1, y=1, z=1} = 5884491500$$
confirming the result from the first answer.</p>
<p>Since this problem gets asked quite frequently I am also posting the code for the solution using the PET in Maple for a necklace with any number of beads. Values on the order of $24$ with a lot of divisors usually require some assistance by a CAS. The routine <b>pet_cycleind</b> produces the cycle index and <b>pet_varinto_cind</b> does the substitution of the repertoire generating function.</p>
<pre>
with(numtheory);
with(group):
with(combinat):
pet_cycleind :=
proc(n)
local d, s, t;
s := 0;
for d in divisors(n) do
s := s + phi(d)*a[d]^(n/d);
od;
if type(n, odd) then
t := n*a[1]*a[2]^((n-1)/2);
else
t := n/2*(a[1]^2*a[2]^((n-2)/2)+a[2]^(n/2));
fi;
(s+t)/2/n;
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
</pre>
|
3,697,062 | <p>How many solutions are there for this equation <span class="math-container">$$x_1+x_2+x_3+x_4 = 49\,,$$</span> where <span class="math-container">$x_i,\;i= 1,2,3,4$</span> is a non negative integer such that:<span class="math-container">$$ 1\le x_1\le 8,\;3\le x_2\le 9,\;10\le x_3\le 20,\ 0\le x_4\,?$$</span></p>
| G Cab | 317,234 | <p>We have that
<span class="math-container">$$
\eqalign{
& \left\{ \matrix{
1 \le x_{\,1} \le 8 \hfill \cr
3 \le x_{\,2} \le 9 \hfill \cr
10 \le x_{\,3} \le 20 \hfill \cr
0 \le x_{\,4} \left( { \le 49} \right) \hfill \cr
x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 49 \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
x_{\,1} = y_{\,1} + 1\; \to \;0 \le y_{\,1} \le 7 \hfill \cr
x_{\,2} = y_{\,2} + 3\; \to 0 \le y_{\,2} \le 6 \hfill \cr
x_{\,3} = y_{\,3} + 10\; \to 0 \le y_{\,3} \le 10 \hfill \cr
x_{\,4} = y_{\,4} \; \to 0 \le y_{\,4} \left( { \le 49} \right) \hfill \cr
y_{\,1} + y_{\,2} + y_{\,3} + y_{\,4} = 49 - 14 = 35 \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
0 \le y_{\,1} \le 7 \hfill \cr
0 \le y_{\,2} \le 6 \hfill \cr
0 \le y_{\,3} \le 10 \hfill \cr
y_{\,1} + y_{\,2} + y_{\,3} \le 35 \hfill \cr} \right. \cr}
$$</span></p>
<p>Now, since the upper limits sum to <span class="math-container">$23 < 35$</span>, then all the points inside
the parallepiped <span class="math-container">$[0,7] \times [0,6] \times [0,10]$</span> respect the last
inequality, thus
<span class="math-container">$$
N = 8 \cdot 7 \cdot 11 = 616
$$</span></p>
|
1,560,539 | <p>I'm trying to show that the partial sums of $\log(j) = n\log(n) - n + \text{O}(\log(n))$</p>
<p>I know that $$\int_1^n\log(x)dx = n\log(n) - n + 1$$</p>
<p>so that this number is pretty close to what I want.</p>
<p>Now I look at the difference between sum and integral of log:</p>
<p>$$\sum_{j=1}^n \log(j) - \int_1^n \log(x)dx$$</p>
<p>My work:</p>
<p>Working out the arithmetic and simplifying as much as possible, I am now at
$$\sum_{j=1}^n \log(j) - \int_1^n \log(x)dx = \log(n) + \sum_{j=1}^{n-1} \big[\log(\large \frac{1}{1+\frac{1}{j}}) - \log(1 + \frac{1}{j})(j) + 1]$$ </p>
<p>Where can I go from here? I think I would want to show that this difference, i.e. the R.H.S., is actually $O(\log(n))$ ...</p>
<p>Thanks,</p>
| Marco Cantarini | 171,547 | <p>It is a direct application of <a href="https://en.wikipedia.org/wiki/Abel%27s_summation_formula" rel="nofollow">Abel's summation</a>. We have $$\sum_{k\leq n}\log\left(k\right)=\sum_{k\leq n}1\cdot\log\left(k\right)=n\log\left(n\right)-\int_{1}^{n}\frac{\left\lfloor t\right\rfloor }{t}dt
$$ where $\left\lfloor t\right\rfloor
$ is the floor function and using $\left\lfloor t\right\rfloor =t+O\left(1\right)
$ we have $$\sum_{k\leq n}\log\left(k\right)=n\log\left(n\right)-n+1+O\left(\int_{1}^{n}\frac{1}{t}dt\right)=n\log\left(n\right)-n+O\left(\log\left(n\right)\right).$$</p>
|
234,477 | <p>In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension?</p>
<p>Is it algebraic? Probably not, right? Transcendental? Normal? Finitely generated? Separable?</p>
<p>For instance, I was thinking: Would it be enough to adjoin an infinitesimal to $\mathbb{R}$ and get $\mathbb{R}^*$? The axioms of hyperreals in Keisler's Foundations of Infinitesimal Calculus seem to suggest so, but I'm not sure. </p>
<p>I don't know how to approach this question, since infinitesimals and such things don't "result" from polynomials (like, say, complex numbers do). </p>
<p>Is $\mathbb{R}^*$ maybe isomorphic to the field of fractions of some polynomial ring? (This occured to me after thinking about $\mathbb{R}(\epsilon)$ and such things, where $\epsilon$ is an infinitesimal).</p>
| Mark S. | 26,369 | <p>I don't think it would be enough to adjoin one infinitestimal. To construct the hyperreals via an ultrapower requires the Axiom of Choice, but you could certainly adjoin one infinitesimal without Choice.</p>
|
234,477 | <p>In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension?</p>
<p>Is it algebraic? Probably not, right? Transcendental? Normal? Finitely generated? Separable?</p>
<p>For instance, I was thinking: Would it be enough to adjoin an infinitesimal to $\mathbb{R}$ and get $\mathbb{R}^*$? The axioms of hyperreals in Keisler's Foundations of Infinitesimal Calculus seem to suggest so, but I'm not sure. </p>
<p>I don't know how to approach this question, since infinitesimals and such things don't "result" from polynomials (like, say, complex numbers do). </p>
<p>Is $\mathbb{R}^*$ maybe isomorphic to the field of fractions of some polynomial ring? (This occured to me after thinking about $\mathbb{R}(\epsilon)$ and such things, where $\epsilon$ is an infinitesimal).</p>
| Mikhail Katz | 72,694 | <p>A small correction to the formulation of the original question: the continuum hypothesis is not needed to construct a hyperreal field extension of the reals. The alternative simpler field extension you are proposing is related to the Levi-Civita field, which was of intense interest to Abraham Robinson. However, it does not have the crucial properties that make the hyperreals useful in analysis.</p>
|
2,861,293 | <p>I found this statement with the proof:</p>
<p><a href="https://i.stack.imgur.com/bGRiZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bGRiZ.jpg" alt="enter image description here" /></a></p>
<p>But I don't understand the proof. Where is the contradiction? We have a nonempty interval <span class="math-container">$J$</span> contained in the nonempty interval <span class="math-container">$(f(a),f(b))$</span>. Where is the problem?</p>
| uniquesolution | 265,735 | <p>This is essentially Paul Frost's answer, in a slighly less verbose form.</p>
<p>The range $f(I)$ contains the interval $(f(a),f(b))$, so every point in the interval $(f(a),f(b))$ is of the form $f(x)$ for some $x\in I$. On the other hand, only one endpoint of the non-empty interval $J$ is a value of $f$ - because $f$ jumps from $y_0^-$ to $y_0^+$. That is the contradiction. </p>
|
97,318 | <p>I use this code but it doesn't work.</p>
<pre><code>ZZ = {
{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}
} ;
ZZ // MatrixForm
K = ConstantArray[0, {4, 4, 4}];
K // MatrixForm
For[ i = 1, i = 4, i++,
For [j = 1, j = 4, j++,
K[[i, j, 1]] = ZZ[[i, 1]] ;
K[[i, j, 2]] = ZZ[[i, 2]] ;
K[[i, j, 3]] = ZZ[[i, 3]] ;
K[[i, j, 4]] = ZZ[[i, 4]] ;
]
];
K[[3, 2, 1]]
</code></pre>
<p>What am I doing wrong?</p>
| Chris Degnen | 363 | <p>Needs <code><=</code> to satisfy the condition.</p>
<pre><code>For[i = 1, i <= 4, i++,
For[j = 1, j <= 4, j++,
K[[i, j, 1]] = ZZ[[i, 1]];
K[[i, j, 2]] = ZZ[[i, 2]];
K[[i, j, 3]] = ZZ[[i, 3]];
K[[i, j, 4]] = ZZ[[i, 4]];]]
</code></pre>
<p>Also <code>K2 = Array[ZZ[[{#, #, #, #}]] &, 4]</code></p>
<pre><code>K == K2
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
3,213,854 | <p>From the definition of a inverse standpoint (<span class="math-container">$f^{-1}(f(x))=f(f^{-1}(x))=x$</span>), why does interchanging variables (<span class="math-container">$x$</span> and <span class="math-container">$y$</span>) work to find the inverse? It seems logical to me but I cannot come up with a coherent argument as to why.</p>
| StAKmod | 640,345 | <p>From your definition, suppose that <span class="math-container">$(x,y)$</span> is on that function, we have</p>
<p><span class="math-container">$x=f^{-1}(f(x))=f^{-1}(y)$</span>.</p>
<p>Now like you said, if we change <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, we obtain <span class="math-container">$y=f^{-1}(x)$</span>, which is the inverse.</p>
|
3,829,377 | <p>We know trace of a matrix is the sum of the eigenvalues of the given matrix. Suppose we know the characteristics polynomial of the matrix, is there any result which gives us the sum of the positive eigenvalues of the matrix?</p>
<p>Note that I need the sum of only the positive eigenvalues...not all eigenvalues.</p>
| Disintegrating By Parts | 112,478 | <p>Let <span class="math-container">$C$</span> be a large positively-oriented square contour with the left side of the contour along the imaginary axis. Then
<span class="math-container">$$
p(\lambda)=(\lambda-\lambda_1)^{r_1}(\lambda-\lambda_2)^{r_2}\cdots(\lambda-\lambda_n)^{r_n} \\
\frac{p'(\lambda)}{p(\lambda)}=\frac{r_1}{\lambda-\lambda_1}+\frac{r_2}{\lambda-\lambda_2}+\cdots+\frac{r_n}{\lambda-\lambda_n} \\
\frac{1}{2\pi i}\oint_{C}\lambda\frac{p'(\lambda)}{p(\lambda)}d\lambda = \sum_{\lambda_j > 0}r_j\lambda_j = \mbox{sum of roots in the right half-plane}
$$</span>
If you know all the roots are real, then you have what you want. If not, then you can integrate close enough to the real axis, and you'll be summing only the roots on the real axis. They will be summed according to multiplicity.</p>
|
1,575,107 | <p>$$L^{-1}\frac{4s}{(s-6)^{3}}$$
$$4L^{-1}\frac{s}{s^{3}}|s=s-6$$
$$4L^{-1}\frac{1}{s^{2}}|s=s-6$$
$$4L^{-1}\frac{1!}{s^{1+1}}|s=s-6$$
$$4te^{6t}$$</p>
<p>Is this correct? symbolab and Wolfram are giving me different answers...</p>
| mrf | 19,440 | <p>If you don't know residues, you can rewrite your function as
$$
\frac{4s}{(s-6)^3} = \frac{4(s-6)}{(s-6)^3} + \frac{24}{(s-6)^3} =
\frac{4}{(s-6)^2} + \frac{24}{(s-6)^3}
$$
and use the "translation rule" to invert the two terms separately.</p>
|
2,482,868 | <p>I am trying to find</p>
<p>$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$</p>
<p>where $p>0$. I have tried to factor out as</p>
<p>$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.</p>
| Guy Fsone | 385,707 | <p><strong>Thanks this answer's here</strong> : <a href="https://math.stackexchange.com/questions/2482681/find-lim-n-to-infty-sqrt3n31-sqrtn21#comment5129295_2482681">Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$</a></p>
<p>Consider
$$
f(x)=[( 1+x^{p+1})^{\frac{1}{p+1}}-(1+x^p)^{\frac1p}]
$$
Then
$$
f'(x)=x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]
$$</p>
<blockquote>
<p>If $p>1$ therefore $f'(0)=0$. Therefore
$$
0=f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.</p>
<p>If $0<p<1$ therefore $f'(0)=-\infty$. Therefore
$$
-\infty =f'(0^+)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.</p>
<p>If $p=1$ therefore $f'(0)=-1 $. Therefore
$$
-1 =f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.</p>
</blockquote>
|
685,642 | <p>I am trying to find a basis for the set of all $n \times n$ matrices with trace $0$. I know that part of that basis will be matrices with $1$ in only one entry and $0$ for all others for entries outside the diagonal, as they are not relevant.</p>
<p>I don't understand though how to generalize for the entries on the diagonal. Maybe just one matrix with $1$ in the $(1, 1)$ position and a $-1$ in all other $n - 1$ positions?</p>
| Sammy Black | 6,509 | <p>The <em>matrix unit</em> <span class="math-container">$E_{ij}$</span> is the matrix with <span class="math-container">$1$</span> in the <span class="math-container">$(i, j)$</span>-entry and <span class="math-container">$0$</span> everywhere else. A basis for your space consists is
<span class="math-container">$$
\{ E_{ij} \; \mid \; i \ne j \} \cup \{ E_{ii} - E_{i+1, i+1} \; \mid \; 1 \le i < n \}.
$$</span></p>
<p>Notice that there are <span class="math-container">$n(n-1)$</span> of the off diagonal matrices and <span class="math-container">$n-1$</span> of the diagonal ones, for a total of <span class="math-container">$n^2 - 1$</span> matrices. This is the right size set since your space is the null space of the onto map
<span class="math-container">$$
\operatorname{tr}: \Bbb{R}^{n^2} \to \Bbb{R}.
$$</span></p>
|
815,739 | <p>Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?</p>
| Hanul Jeon | 53,976 | <p>In my personal opinion, if $\Bbb{R}^{1/2}$ and $\Bbb{R}^{-2}$ is exists meaningfully, then we can expect that these two objects satisfies following properties:</p>
<ul>
<li>$\Bbb{R}^2\times \Bbb{R}^{-2} \cong \{0\}$</li>
<li>$(\Bbb{R}^{1/3})^3 \cong \Bbb{R}$</li>
</ul>
<p>However, the cardinality of $\Bbb{R}^2\times X$ is greater or equal than the cardinality of continuum unless $X$ is empty. So I think that $\Bbb{R}^{-2}$ does not exist.</p>
<p>The case of $\Bbb{R}^{1/3}$ is more complicated. However, as I know, there is no topological space $X$ satisfy that $X\times X\cong \Bbb{R}$. I think we can prove the nonexistence of (topological!) cube roots of $\Bbb{R}$ (that is, a space $X$ satisfy $X^3\cong \Bbb{R}$) does not exist.</p>
<p>If you drop the suggested property of $\Bbb{R}^{-2}$ of $\Bbb{R}^{1/3}$, then they may exist. But I don't know they can have the meaning.</p>
<hr>
<p>I will give the outline of the proof of nonexistence of "topological square root" of the real line (that is, there is no $X$ such that $X^2\cong \Bbb{R}$ holds.) </p>
<p>At first, we will prove following lemma:</p>
<blockquote>
<p><em>Lemma.</em> Let $X$ and $Y$ be a path-connected space with $|X|>1$. If $(x,y)\in X\times Y$ then $X\times Y-\{(x,y)\}$ is path-connected.</p>
</blockquote>
<p>The idea of the proof of this lemma is "make a detour". Let $(a,b)$ and $(c,d)$ be distinct points of $X\times Y -\{x,y\}$. Take $e\in X-\{x\}$ and find the path $\pi_1$ between $(a,b)$ and $(e,b)$, and find the path $\pi_2$ between $(e,b)$ to $(e,d)$, and find the path $\pi_3$ between $(e,d)$ and $(c,d)$. Adjoin $\pi_1$, $\pi_2$ and $\pi_3$, then we get a path between $(a,b)$ and $(c,d)$.</p>
<p>It is easy to check that if $X$ is path-connected and $f:X\to Y$ is continuous then $f(X)$ is also path-connected. Let assume that $X\times X\cong \Bbb{R}$. Since $X$ is a projection of $\Bbb{R}$, $X$ is also path-connected. By Lemma 1, $\Bbb{R}-\{x\}$ is path-connected but we know that $\Bbb{R}-\{x\}$ is disconnected, a contradiction.</p>
|
815,739 | <p>Let $f :\mathbb R\to \mathbb R$ be a continuous function such that $f(x+1)=f(x) , \forall x\in \mathbb R$ i.e. $f$ is of period $1$ , then how to prove that $f$ is uniformly continuous on $\mathbb R$ ?</p>
| coolpapa | 151,988 | <p>None that I know of. </p>
<p>You could <strike>stretch things a bit and</strike> say that $\mathbb{R}^{0}$ is the $0$-dimensional vector space over $\mathbb{R}$. However, raising something to the one-third power should mean a cube root in SOME sense - to make sense with the definition of $\mathbb{R}^{3}$, we would need $\mathbb{R}^{1/3} \oplus \mathbb{R}^{1/3} \oplus \mathbb{R}^{1/3} = \mathbb{R}$ in some way or another. But I don't know any categories in which such a statement is true - not in groups, rings, vectors spaces.... Similarly, $\mathbb{R}^{-2}$ should satisfy $\mathbb{R}^{2} \oplus \mathbb{R}^{-2} = \mathbb{R}^{0} = \{0\}$. Again, in the context of vector spaces or groups or rings, that just makes no sense.</p>
<p>I mean, if you really want to push it, you could maybe say that $\mathbb{C}^{1/2} = \mathbb{R}$, because as real vector spaces, $\mathbb{C} \cong \mathbb{R} \oplus \mathbb{R}$. But I wouldn't advise it. I've never seen that notation used, and it doesn't really help, so....</p>
|
1,116,435 | <p>How do I get the value of </p>
<p>$$\lim_{x \rightarrow \frac{1}{4} \pi } \frac{\tan x-\cot x}{x-\frac{1}{4} \pi }?$$ </p>
<p>I need the steps without using L'hospital.</p>
| idm | 167,226 | <p>$$...=\underbrace{\frac{1}{\tan x}}_{\to 1}\underbrace{\frac{\tan^2x-1}{(x-\frac{\pi}{4})}}_{ \to \frac{d}{dx}\tan^2(x)\big|_{x=\pi/4}}$$</p>
|
1,437,073 | <p>Ok guys, I have to solve this ODE</p>
<p>$$
\frac{d^2y}{dx^2}=f(x), \quad
x>0,\quad y\left(0\right) = 0, \quad
\left.\frac{dy}{dx}\right\lvert_{x=0}=0
$$
The solution I should get is in the form of
$$y\left(x\right)=\int_0^x k\left(t\right)\, dt $$
Moreover, I should tell what the function $\,k\left(t\right)\,$ is in a simple form. I have tried by substitution, with $\,u=y',\,$ but I have to integrate two times $\,f\left(x\right)\,$, and this seems not at all like a simple form... The function $\,k\left(t\right)\,$ is undoubtedly an exponential, but in which form? Hope that somebody can help!</p>
| Vlad | 229,317 | <p><strong>HINT</strong>: Use the (Leibniz) <a href="https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign" rel="nofollow">formula for differentiation under the integral with variable limits</a></p>
<p>$$
\frac{d}{dx}\left(
\int_{a\left(x\right)}^{b\left(x\right)} f\left(x,t\right)\, dt \right)
= f\big(x,b\left(x\right)\big) \cdot b'\left(x\right)
- f\big(x,b\left(x\right)\big) \cdot a'\left(x\right)
+ \int_{a\left(x\right)}^{b\left(x\right)}
\frac{\partial}{\partial x}\,f\left(x,t\right)\, dt
$$</p>
<hr>
<p>In your particular case we have
$$
y\left(x\right) = \int_0^x k\left(t\right)\, dt
\implies
\frac{dy}{dx} = k\left( x\right) \cdot 1 - 0\cdot 0 + \int_0^x 0\, dt = k\left( x\right)
\implies
\frac{d^2y}{dx^2} = k'\left( x\right)
$$
Substituting this back into original equation we get
$$
k'\left(x\right)=f\left(x\right), \qquad k\left(0\right)=0
$$
After expressing $\,k\left( x\right)\,$ via $\,f\left( x\right)\,$ you can find solution for $\,y\,$ in terms of $\,f.\,$
Hope you can proceed from here.</p>
|
4,341,297 | <p>Evaluate the given expression <span class="math-container">$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$</span> The given answer is <span class="math-container">$\dfrac{1}{4}$</span>. My attempt:
<span class="math-container">$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$</span> This is as far I as I am able to reach. Thank you!</p>
<p>PS I don't see how one can get <span class="math-container">$\dfrac14$</span>. For that we have to get something like <span class="math-container">$\sqrt[n]{A^n}$</span>.</p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$2^{2n+4}+2^{2n+2}=2^{2n}(2^4+2^2)=20\times4^n$</span>,<span class="math-container">$$\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\frac{20}{20\times4^n}}=\frac14.$$</span></p>
|
700,673 | <p>I saw this notation many times, but I don't understand why the $y$ variable is missing in the first term of the first equation below.</p>
<p>$$
\frac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)
$$</p>
<p>It just mean:</p>
<p>$$
\frac{\mathrm{d}y(x,y)}{\mathrm{d}x} = f(x,y)
$$</p>
<p>?</p>
<hr>
<h2>Tell me if I'm wrong</h2>
<p>In my understanding this is a linear ODE of first order in which $y$ is the unknown function and both of them (the derivate and the unknown one) have just a single variable.</p>
<p>$$
y'(x) = f(x)
$$</p>
<p><em>A simple example</em></p>
<p>$$
\begin{align}
y'(x) &= 2x\\
\int y'(x)\!\mathop{}\mathrm{d}x &= \int 2x\!\mathop{}\mathrm{d}x\\
y(x)&= x^2+\mathrm{C}\\
\end{align}
$$</p>
<p>Is this case:</p>
<p>$$
\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))
$$</p>
<p>would it be something like this?</p>
<p>$$
\begin{align}
y'(x,y) &= 2x + 2y\\
\iint y'(x,y)\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y &= \iint 2x + 2y\!\mathop{}\mathrm{d}x\!\mathop{}\mathrm{d}y\\
y(x,y) &= x^2 + y^2+\mathrm{C}\\
\end{align}
$$</p>
<hr>
<p><em>For Christian Blatter</em></p>
<blockquote>
<p>This is what I would expected to be your exaplanation. Can you tell me
why my version is wrong?</p>
</blockquote>
<p>When dealing with ODEs for the first time we are given a function $f:\ (x,y)\mapsto f(x,y)$ defined in some region $\Omega$ of the $(x,y)$-plane. For each $(x,y)\in\Omega$ the value $f(x,y)$ is to be interpreted as a slope assigned to the point $(x,y)$. Therefore we are given a field of "line elements" of various slopes covering $\Omega$.</p>
<p>Given this "slope field" we are interested in curves $$\gamma:\ x\mapsto \gamma(x)=\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)$$ lying in $\Omega$ that have for each of their points $\bigl(x,\gamma(x)\bigr)$ the slope $f\bigl(x,\gamma(x)\bigr)$ prescribed there. This means that we should have
$$\gamma'(x)\equiv f\bigl(x,\gamma(x)\bigr)\qquad(a<x<b)\ .\tag{1}$$</p>
| Git Gud | 55,235 | <p>The notation $\dfrac{\mathrm dy(x)}{\mathrm dx}$ is short for $\dfrac{\mathrm dy}{\mathrm dx}(x)$ or $y'(x)$, if you prefer.</p>
<p>In this context, the equality $\dfrac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y)$ should be read as $\dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x))$.</p>
<p>As for your last example, you got the wrong idea, $y$ is a function whose domain is a subset of $\mathbb R$, so $y(x,y)$ doesn't make sense.</p>
<p>Correct would be to consider the function $f\colon \mathbb R^2\to \mathbb R, (x,y)\mapsto 2x+2y$ and to consider the ODE $y'(x)=f(x,y(x))$. Note that in $f\colon \mathbb R^2\to \mathbb R, (x,\color{red} y)\mapsto 2x+2\color{red} y$ the variable $\color{red} y$ is bounded, it's not a function, it's a variable. If you wish you can rewrite the function as $f\colon \mathbb R^2\to \mathbb R, (x,u)\mapsto 2x+2u$ and then you can consider the ODE $y'(x)=f(x,y(x))$ without any ambiguity.</p>
|
3,218,662 | <p>Let <span class="math-container">$T: X \to Y$</span> be a linear operator between normed Banach spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>.
The definition of the operator norm
<span class="math-container">$$
\| T \|
:= \sup_{x \neq 0} \frac{\| Tx \|}{\| x \|}
$$</span>
is well known.</p>
<p>Now, let <span class="math-container">$T$</span> be bijective.
When can we say that
<span class="math-container">$$
\| T^{-1} \|
= \inf_{x \neq 0} \frac{\| x \|}{\| T x \|}
$$</span>
holds?</p>
| operatorerror | 210,391 | <p>This fails in finite dimensions. </p>
<p>Take
<span class="math-container">$$
T=\begin{pmatrix} 2&0\\0&3 \end{pmatrix}
$$</span>
So <span class="math-container">$||T||=3$</span> and
<span class="math-container">$$
T^{-1}=\begin{pmatrix} 1/2&0\\0&1/3 \end{pmatrix}
$$</span>
and <span class="math-container">$||T^{-1}||=1/2$</span>. </p>
<p>But,
<span class="math-container">$$
\inf_{x\ne0}\frac{||x||}{||Tx||}\geq\frac{1}{||T||}=\frac13
$$</span>
while <span class="math-container">$x=(0,1)$</span> achieves this lower bound, showing that
<span class="math-container">$$
\inf_{x\ne0}\frac{||x||}{||Tx||}=\min_{x\ne 0}\frac{||x||}{||Tx||}=\frac13\ne \frac12
$$</span> </p>
|
1,590,817 | <p>I was hoping someone could explain to me how to prove a sequence is Cauchy. I've been given two definitions of a Cauchy sequence:</p>
<p>$\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n,m> N$ $\Rightarrow |a_n - a_m| ≤ \epsilon$</p>
<p>and equivalently $\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n> N$ $\Rightarrow |a_{n+p} - a_n| ≤ \epsilon$, $\forall p \in \mathbb{N}$</p>
<p>I understand that proving a sequence is Cauchy also proves it is convergent and the usefulness of this property, however, it was never explicitly explained how to prove a sequence is Cauchy using either of these two definitions. I'd appreciate it if someone could explain me how to prove a sequence is Cauchy perhaps $a_n = \sqrt{n+1} - \sqrt{n}$ ? or another example just for me to grasp the concept.</p>
| Pedro M. | 21,628 | <p>For the particular example you chose, it is very easy to show directly that it converges to zero, because
$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.$$
Nevertheless, this same identity allows you to show that it is Cauchy, since
$$\left|\frac{1}{\sqrt{m+1} + \sqrt{m}} - \frac{1}{\sqrt{n+1} + \sqrt{n}}\right| \leq \left| \frac{1}{2\sqrt{m}} - \frac{1}{2\sqrt{n}} \right| \leq \frac{1}{2\sqrt{\min(m,n)}}< \epsilon$$
whenever $\min(m,n) \geq N > \frac{1}{4\epsilon^2}$.</p>
|
677,859 | <p>$f(x)= f(x+1)+3$ and $f(2)= 5$, determine the value of $f(8)$.</p>
<p>I don't understand how $f(x)$ can equal $f(x+1)+3$</p>
| Ashot | 51,491 | <p>It means that $f(x+1) = f(x) - 3$. Now you can find values of $f(x)$ consequently.</p>
|
203,827 | <p>Suppose I have the following lists: </p>
<pre><code>prod = {{"x1", {"a", "b", "c", "d"}}, {"x2", {"e", "f",
"g"}}, {"x3", {"h", "i", "j", "k", "l"}}, {"x4", {"m",
"n"}}, {"x5", {"o", "p", "q", "r"}}}
</code></pre>
<p>and </p>
<pre><code>sub = {{"m", "n"}, {"o", "p", "r", "q"}, {"g", "f", "e"}};
</code></pre>
<p>for each element in <code>sub</code> I want to go through <code>prod</code> and select if the element exist such that I get the following output, </p>
<pre><code> {{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q","r"}}}
</code></pre>
<p>I tried doing: </p>
<pre><code>Table[Select[
prod[[All, 2]][[i]], # == ContainsAny[Map[Sort, sub]][[i]] &], {i,
Length[sub]}]
</code></pre>
<p>yet it doesn't work, am I missing something? </p>
| C. E. | 731 | <p>A rule-based approach:</p>
<pre><code>patt = List /@ OrderlessPatternSequence @@@ Alternatives @@ sub;
Cases[prod, {x_, y : patt} :> {x, y}]
</code></pre>
<blockquote>
<p><code>{{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q", "r"}}}</code></p>
</blockquote>
<p>And a functional approach:</p>
<pre><code>sel = Outer[ContainsAll, Last /@ prod, sub, 1];
Pick[prod, Or @@@ sel]
</code></pre>
<blockquote>
<p><code>{{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q", "r"}}}</code></p>
</blockquote>
<p>A simpler one:</p>
<pre><code>select[{label_, values_}] := MemberQ[sub, {OrderlessPatternSequence @@ values}]
Pick[prod, select /@ prod]
</code></pre>
<blockquote>
<p><code>{{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q", "r"}}}</code></p>
</blockquote>
<p>You can use the same approach with <code>Select</code>:</p>
<pre><code>Select[prod, select]
</code></pre>
<blockquote>
<p><code>{{"x2", {"e", "f", "g"}}, {"x4", {"m", "n"}}, {"x5", {"o", "p", "q", "r"}}}</code></p>
</blockquote>
|
1,997,513 | <p>I spend so much time for proving this triangle and i still don't know. </p>
<p>Question :</p>
<p>Given Triangle ABC, AD and BE are altitudes of the triangle. Prove that Triangle DEC similarity with triangle ABC</p>
| arberavdullahu | 377,862 | <p>Since $AD=AC\cdot cos C$ and $BE=BC\cdot cos C$ we have that
$$\frac{AD}{BE}=\frac{AC}{BE}$$
And since the angle that form those two pairs are equal ($\angle C$) the result follows</p>
|
115,385 | <p>I heard that computation results can be very sensitive to choice of random number generator. </p>
<ol>
<li><p>I wonder whether it is relevant to program own Mersenne-Twister or other pseudo-random routines to get a good number generator. Also, I don't see why I should not trust native or library generators as random.uniform() in numpy, rand() in C++. I understand that I can build generators on my own for distributions other than uniform (inverse repartition function methor, polar method). But is it evil to use one built-in generator for uniform sampling? </p></li>
<li><p>What is wrong with the default 'time' seed? Should one re-seed and how frequently in a code sample (and why)?</p></li>
<li><p>Maybe you have some good links on these topics!</p></li>
</ol>
<p>--edit
More precisely, I need random numbers for multistart optimization routines, and for uniform space sample to initialize some other optimization routine parameters. I also need random numbers for Monte Carlo methods (sensibility analysis). I hope the precisions help figure out the scope of question.</p>
<p>Kind regards</p>
| Yuval Filmus | 1,277 | <p>Let's take your points one by one.</p>
<ol>
<li><p>Is it relevant to program your own PRNG? Not really. Someone already did it for you. There is no more reason to program a PRNG than it is to write your own hash function; that is, only if you're not satisfies with the performance of the library function. In the case of the PRNG, there can be two reasons. The first one is that the PRNG isn't random enough. This is the case for some really old standard C functions, whose pseudorandomness might actually be noticeable by your Monte Carlo simulation. But if you use anything reasonably new, you're fine in this regard. The second one is that the PRNG is too slow. That's not usually a problem since modern PRNG are pretty fast, and the processing that you do far outweighs the time required to generate the numbers. But if you do ever find yourself in this situation, ask here, and either someone will give you a faster PRNG, or explain why you don't really need it.</p></li>
<li><p>Should you use time to generate the seed? When doing Monte Carlo simulations, you want your experiments to be reproducible. Also, you sometimes want to test several variants of your code on the same random data. But it's not a sin to generate the random seeds (once and for all) using whatever method. In Unix you have a /dev/random you can use.</p></li>
<li><p>Any links? Lot of links, but I don't think they're relevant. It's better for you to concentrate on the actual thing that you're doing, rather on this aspect, which perhaps interests you, but is marginal (or should be) from your point of view. However, it's always good to be inquisitive, so if you feel like, explore. Just don't think that it has any bearing on your actual simulations.</p></li>
</ol>
|
90,712 | <p>How many <em>unique</em> pairs of integers between $1$ and $100$ (inclusive) have a sum that is even? The solution I got was</p>
<p>$${100 \choose 1}{99 \choose 49}$$</p>
<p>I don't have a way to verify it, but I figured you pick one card from the 100, then you can pick 49 of the other cards (if the first card is even the other has to be even and if the first card is odd the other has to be odd as well).</p>
| Arturo Magidin | 742 | <p>This doesn't make much sense: $\binom{100}{1}\binom{99}{49}$ counts the number of ways of picking one out of 100 possibilities <em>and</em> 49 out of 99 possibilities. That is not what you want: you don't want to pick 49 <em>other</em> cards, you just want to pick one out of the 49 that are still "in play".</p>
<p>That is, I <em>think</em> what you meant was:</p>
<blockquote>
<p>I can pick the first number arbitrarily, and there are $\binom{100}{1}$ ways of doing that; once I pick that, there are only $49$ other numbers that I can pick (those with the same parity as the one I picked), and that means $\binom{49}{1}$ of them. Multiplying we have $\binom{100}{1}\binom{49}{1}$.</p>
</blockquote>
<p>(Note the $\binom{49}{1}$ rather than $\binom{99}{49}$). </p>
<p>That's <em>almost</em> right, but you are overcounting: you count the pair $(1,99)$ once when you first pick $1$ and <em>then</em> pick $99$; and you count it again when you first pick $99$ and <em>then</em> pick $1$. So, what's the easy way of taking care of that?</p>
<p>(To see that your answer cannot possibly be correct, note that there are only
$$\binom{100}{2} = \frac{100(99)}{2} = 4950$$
pairs of numbers between $1$ and $100$; but $\binom{100}{1}\binom{99}{49} \approx 5\times 10^{30}$, which is <em>way</em> too many.)</p>
|
28,456 | <p>I built a <code>Graph</code> based on the permutations of city's connections from :</p>
<pre><code>largUSCities =
Select[CityData[{All, "USA"}], CityData[#, "Population"] > 600000 &];
uScityCoords = CityData[#, "Coordinates"] & /@ largUSCities;
Graph[#[[1]] -> #[[2]] & /@ Permutations[largUSCities, {2}] ,
VertexCoordinates -> Reverse[uScityCoords, 2], VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},ImageSize -> 650]
</code></pre>
<p>It looks like this:
<img src="https://i.stack.imgur.com/1ea6c.png" alt="graph"></p>
<p>My question, is there any way to have the Graph like this?
<img src="https://i.stack.imgur.com/ajwOT.png" alt="enter image description here"></p>
| Simon Woods | 862 | <p>I think what you are after is a Delaunay triangulation of the city coordinates. For example:</p>
<pre><code>Graphics`Mesh`MeshInit[];
Graph[
Range[Length[uScityCoords]],
UndirectedEdge @@@ Delaunay[Reverse[uScityCoords, 2]]["Edges"],
VertexCoordinates -> Reverse[uScityCoords, 2],
VertexStyle -> Red,
Prolog -> {LightBrown, CountryData["USA", "FullPolygon"]},
ImageSize -> 650
]
</code></pre>
<p><img src="https://i.stack.imgur.com/dFMqE.png" alt="enter image description here"></p>
|
1,723,718 | <p>Knowing $f(x,y) = 2x^2 +3y^2 -7x +15y$, one simply proves $$|f(x,y)|\leq 5(x^2+y^2)+22 \sqrt{x^2 + y^2}$$ How can I use this info to compute
$$ \lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}}\;\;\; ?$$</p>
<p>Thanks!</p>
| Anonymous | 327,815 | <p>Approach limit with $y = mx$
then
$$
\lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}} =
$$
$$
\lim_{x\to0} \frac{x(2x +3m^2x -7 +15m)}{x^{1/2}(1+m^2)^{1/4}} - 2 =
$$
$$
\lim_{x\to0} \frac{x^{1/2}( -7 +15m)}{(1+m^2)^{1/4}} - 2 = -2
$$</p>
|
984,852 | <p>Let $p$ be a univariate polynomial over a field $F$, and let $K$ be an extension of $F$. </p>
<p>If $p(x) = 0$ for all $x \in F$, does this imply that $p(x) = 0$ for all $x \in K$? How about if $p$ is multivariate?</p>
<p>For context, I'm trying to understand if doing Schwartz-Zippel-style arithmetic circuit identity-testing over a large enough extension field gives the right answer when the degree of the expression may be high.</p>
| ajotatxe | 132,456 | <p>No. Take $p(X)=X(X+1)$, $F=\Bbb F_2$, $K=\Bbb F_4$.</p>
<p>Indeed, let $\beta\in\Bbb F_4-\Bbb F_2$. We know that $\beta^2+\beta=1$, that is, $p(\beta)=1$. And $p(0)=p(1)=0$.</p>
|
3,540,045 | <p>The definite integral, <span class="math-container">$$\int_0^{\pi}3\sin^2t\cos^4t\:dt$$</span></p>
<p><strong>My question</strong>: for the trigonometric integral above the answer is <span class="math-container">$\frac{3\pi}{16}$</span>. What I want to know is how can I compute these integrals easily. Is there more than one way to solve it? If so, is the key to solving these integrals, just recognizing some trig identities and using u-sub until it looks like a simpler integral?</p>
<p><strong>Here's what I tried (Why doesn't it work!):</strong></p>
<p>I rewrote the integrand as: <span class="math-container">$3(1-\cos^2t)\cos^4t\:dt$</span> then foiled it in,</p>
<p><span class="math-container">$3\cos^4t-3\cos^6t dt$</span>
, then I used the power rule and multipied through by chain rule and then did </p>
<p><span class="math-container">$F(\pi)-F(0)$</span> and got the answer: <span class="math-container">$\frac{6}{35}$</span></p>
<p>Why does this not work?!?</p>
| Vedant Chourey | 638,765 | <p><strong>Note</strong> :-The solution is specifically for the problem in question</p>
<p>The given function is <strong>even</strong></p>
<p>So let <span class="math-container">$$I = 3\int_0^{\pi} \sin^{2}t \cos^{4}t dt$$</span>
And it is equal to
<span class="math-container">$$I = 6\int_0^{\pi/2} \sin^{2}t \cos^{4}t dt\tag{1}$$</span></p>
<p>Now, we know that
<span class="math-container">$$\int_a^b f(x) dx = \int_a^b f(b+a-x) dx$$</span></p>
<p>Therefore
We write the integral as
<span class="math-container">$$I = 6\int_0^{\pi/2} \sin^{2}(\frac{\pi}{2}-t) \cos^{4}(\frac{\pi}{2}-t) $$</span></p>
<p>That is equal to
<span class="math-container">$$I = 6\int_0^{\pi/2} \cos^{2}t \sin^{4}t dt\tag{2}$$</span></p>
<p>adding equation (1) and (2)
we get
<span class="math-container">$$2I =6 \int_0^{\pi/2} \sin^{2}t \cos^{4}t + \cos^{2}t \sin^{4}t dt$$</span>
Further
<span class="math-container">$$2I = 6\int_0^{\pi/2} \cos^{2}t \sin^{2}t[\sin^{2}t+\cos^{2}t] $$</span>
<span class="math-container">$$I = 3\int_0^{\pi/2} \sin^{2}t \cos^{2}t dt$$</span>
<span class="math-container">$$I = \frac{3}{4} \int_0^{\pi/2} (2\cos t \sin t)^2 dt $$</span></p>
<p>Integrating,
<span class="math-container">$$I = \frac{3}{4} [( \frac{\pi}{4})-(\frac{1}{2} -\frac{1}{2})]$$</span></p>
<p>So,</p>
<p><span class="math-container">$$I= \frac {3\pi}{16}$$</span></p>
|
3,130,059 | <p>I have faced this differential problem: <span class="math-container">$(y'(x))^3 = 1/x^4$</span>. </p>
<p>From the fundamental theorem of algebra i know there exist 3 solutions <span class="math-container">$y_1$</span>, <span class="math-container">$y_2$</span>, <span class="math-container">$y_3$</span>, but formally how can I procede to deduce that? </p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Write <span class="math-container">$$\frac{dy}{dx}=x^{-4/3}$$</span>
Use the factorization of <span class="math-container">$$a^3-b^3=(a-b)(a^2+ab+b^2)$$</span>
This is
<span class="math-container">$$(y'(x)-x^{-4/3})(y'(x)^2+y'(x)x^{-4/3}+x^{-8/3})$$</span></p>
|
3,130,059 | <p>I have faced this differential problem: <span class="math-container">$(y'(x))^3 = 1/x^4$</span>. </p>
<p>From the fundamental theorem of algebra i know there exist 3 solutions <span class="math-container">$y_1$</span>, <span class="math-container">$y_2$</span>, <span class="math-container">$y_3$</span>, but formally how can I procede to deduce that? </p>
| user247327 | 247,327 | <p>Every number has three cube roots. You only used one of them.</p>
<p>In particular, if a is real number then the three cube roots are <span class="math-container">$a^{1/3}$</span>, a real number, <span class="math-container">$a^{1/3}(cos(2\pi/3)+ isin(2\pi/3))$</span>, and <span class="math-container">$a^{1/3}(cos(4\pi/3)+ isin(4\pi/3))$</span>.</p>
<p>Given <span class="math-container">$(y'(x))^3= \frac{1}{x^4}= x^{-4}$</span>, we have
<span class="math-container">$y'(x)= x^{-4/3}$</span>, <span class="math-container">$y'(x)= x^{-4/3}(cos(2\pi/3)+ isin(2\pi/3))$</span>, and <span class="math-container">$y'(x)= x^{-4/3}(cos(4\pi/3)+ isin(4\pi/3))$</span>.</p>
|
1,950,077 | <blockquote>
<p>A standard deck of cards consists of 26 red cards (hearts and diamonds) and 26 black cards (clubs and spades). Suppose you shuffle such a deck and draw three cards at random without replacement. Let <span class="math-container">$A_i =$</span> the event that the <span class="math-container">$i$</span>th card is a red card, for <span class="math-container">$i = 1, 2, 3$</span>. Mark each of the following statements as TRUE or FALSE.</p>
<p>(a) <span class="math-container">$P(A_2)> P(A_1)$</span></p>
<p>(b) <span class="math-container">$A_1$</span> and <span class="math-container">$A_3$</span> are independent.</p>
<p>(c) <span class="math-container">$P(A_1|A_3)< P(A_1)$</span></p>
</blockquote>
<p>I'm having a lot of trouble with part a. Obviously <span class="math-container">$P(A_1)$</span> is <span class="math-container">$26/52$</span>, but <span class="math-container">$P(A_2)$</span> is either (<span class="math-container">$26/51$</span>) or (<span class="math-container">$25/51$</span>) depending on what happened in <span class="math-container">$A_1$</span>.</p>
| copper.hat | 27,978 | <p>Each permutation of the deck is equally likely. When you select the three cards, each one is equally likely to be red or black from which the probablity
follows (${1 \over 2}$).</p>
<p>Hence $pA_1 p A_3 = {1 \over 4}$, however $p (A_1 \text{ and } A_3) = { \binom{26}{2}\over \binom{52}{2}} = {25 \cdot 26 \over 51 \cdot 52 } = {25 \over 102} < {1 \over 4}$.</p>
<p>Intuitively, if you know that $A_3$ occurred, there are fewer red cards than black cards remaining, hence the chance of $A_1$ being red is lessened slightly. It is easy to compute $p [A_1 | A_3] = { p (A_1 \text{ and } A_3) \over p A_3} = {50 \over 102} < {1 \over 2}$</p>
|
2,275,679 | <p>Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \ \ \text{for} \ \ x, a,b,c > 0 \ .
$$
To do so, I figured it is sufficient to show that
$$
f(x) = \sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)
$$</p>
<p>is monotonically increasing for $x > 0$.
Of course, I took the derivative $f'(x)$ and proceeded with the demand
$$
\frac{\frac{b}{a}x}{\sqrt{a \cdot b + \frac{b}{a}x^2}} \arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) - \frac{c \frac{b}{a} x}{a \cdot b + \frac{b}{a} x^2 + c^2} > 0 \ .
$$
In the end, I got stuck with
$$
\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) > \frac{c \sqrt{a \cdot b + \frac{b}{a}x^2}}{a \cdot b + \frac{b}{a} x^2 + c^2} \ .
$$</p>
<p>Inserting values for a,b, and c seems to work perfectly, but I can't manage to analytically solve the inequation.
Does anyone have an idea of how to approach this problem? </p>
<p>Edit:
I came across the Shafer-Fink inequality stating
$$
\frac{3y}{1+2\sqrt{1 + y^2}} < \arctan y < \frac{\pi y}{1 + 2\sqrt{1 + y^2}} \ .
$$
Can I substitute
$$
y = \frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}
$$
and therefore receive
$$
\arctan (y) > \frac{3y}{1+2\sqrt{1 + y^2}} > \frac{y}{1 + y^2} ?
$$
Is that a proper way?</p>
| Cye Waldman | 424,641 | <p>You are quite correct in your supposition that there should be some symmetry in the surface area and volume of a body of revolution. Your only error is that in the surface are we do not want the area under $ydx$, but rather along $yds$.</p>
<p>These ideas are expressed quite clearly in Pappus's Centroid Theorems:</p>
<p>Pappus's $(1^{st})$ Centroid Theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid (<a href="https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem" rel="nofollow noreferrer">Pappus's centroid theorem</a>). Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by</p>
<p>$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$</p>
<p>Pappus's $(2^{nd})$ Centroid Theorem says the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$. The centroid of a volume is given by</p>
<p>$$\mathbf{R}=\frac{\int_A \mathbf{r}dA}{\int_A dA}=\frac{1}{A} \int_A \mathbf{r}dA$$</p>
<p>Thus we can say for your cases that</p>
<p>$$
S=2\pi\int y\ ds \\
V=\pi\int y^2\ dx
$$</p>
<p>Notice that here, the length, $L$ and the volume, $V$ have cancelled out, thus disguising the true nature of the equations.</p>
|
175,971 | <p>Let's $F$ be a field. What is $\operatorname{Spec}(F)$? I know that $\operatorname{Spec}(R)$ for ring $R$ is the set of prime ideals of $R$. But field doesn't have any non-trivial ideals.</p>
<p>Thanks a lot!</p>
| LinAlgMan | 49,785 | <p>I know that the spectrum of a field in the Zariski/commutative-algebra/algebraic-geometry meaning is simply $\{ (0) \}$ but it is too trivial, so I <strong>suspect</strong> there is another meaning for this term applied to fields. In a paper by <em>Yuval Flicker et al.</em> there is a related term: "real spectrum of a field" $\Omega_k$ which is the topological space of all orderings $\xi$ of $k$ ($k$ is a field). Its topology is generated by the subsets
$$ \{ \xi \in \Omega_k \mid a > 0 \mbox{ at } \xi \}, \quad a \in k \ .$$
So I ask: is there a generalization of this which is called spectrum of a field?</p>
<p>See: Yuval Z. Flicker, Claus Scheiderer, R. Sujatha, <em>Grothendieck's Theorem on Non-Abelian $H^2$ and Local-Global Principles</em>, Journal of the American Mathematical Society, 1998*.</p>
|
3,669,700 | <blockquote>
<p>Let <span class="math-container">$f(x)$</span> be a monic, cubic polynomial with <span class="math-container">$f(0)=-2$</span> and <span class="math-container">$f(1)=−5$</span>. If the sum of all solutions to <span class="math-container">$f(x+1)=0$</span> and to <span class="math-container">$f\big(\frac1x\big)=0$</span> are the same, what is <span class="math-container">$f(2)$</span>?</p>
</blockquote>
<p>From <span class="math-container">$f(0)$</span> I got that <span class="math-container">$f(x)=x^3+ax^2+bx-2$</span> and from <span class="math-container">$f(1)=-5$</span> that <span class="math-container">$a+b = -4$</span> however I'm not sure how to use the info about the transformations to find <span class="math-container">$f(2).$</span> It seems that <span class="math-container">$(x+1)$</span> is a root for <span class="math-container">$f(x+1)$</span> and the same logic applies for <span class="math-container">$f\big(\frac1x\big)$</span>?</p>
<p>Should I use Vieta's here or what's the appropriate way to go?</p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>If <span class="math-container">$\alpha$</span> is a root of <span class="math-container">$f(x)$</span> then <span class="math-container">$\alpha - 1$</span> is a root of <span class="math-container">$f(x+1)$</span> (and vice-versa). So the sum of roots of <span class="math-container">$f(x+1)$</span> is the sum of roots of <span class="math-container">$f(x)$</span>, minus <span class="math-container">$3$</span> which by Vieta' formula is <span class="math-container">$-a$</span> minus <span class="math-container">$3 =-a-3$</span>.</p>
<p>Now, the roots of <span class="math-container">$f(\frac 1x) = \frac{-2x^3+bx^2+ax+1}{x^3}$</span> are also the roots of <span class="math-container">$-2x^3+bx^2+ax+1$</span>, the <em>reciprocal polynomial</em> of <span class="math-container">$f$</span>. The sum of roots of this polynomial is <span class="math-container">$\frac b2$</span> by Vieta's formula.</p>
<p>Finally, <span class="math-container">$-a-3= \frac b2$</span>. Conclude.</p>
|
9,010 | <p>I hear people use these words relatively interchangeably. I'd believe that any differentiable manifold can also be made into a variety (which data, if I understand correctly, implicitly includes an ambient space?), but it's unclear to me whether the only non-varietable manifolds should be those that don't admit smooth structures. I'd hope there's more to it than that.</p>
<p>I've heard too that affine schemes are to schemes local coordinates are to manifolds, so maybe my question should be about schemes instead -- I don't even know enough to know...</p>
| Matt | 2,000 | <p>A variety is usually defined to be the zero set of some polynomials. I don't know much about this, but I think people usually call manifolds that can be realized as varieties "algebraic manifolds". Take for example the polynomial $x^2+y^2-1$, then the zero set of this is $x^2+y^2=1$, or the circle $S^1$. So that is an algebraic manifold.</p>
<p>There can be varieties that are not manifolds, for instance, $y^2-x^2(x+1)=0$ is a "nodal cubic" and so it has a singularity at $(0,0)$. It can't be a manifold because it looks like "X" a cross at the origin so is not homeomorphic locally to $\mathbb{R}$. </p>
<p>There is a very useful theorem, sometimes called the Regular Level Set Theorem, which says that if you have a smooth map $M\to N$, then the level set of a regular value is a smooth manifold. So in this particular case, our polynomial $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth map, and we care about whether or not zero is a regular value. </p>
<p>This amounts to checking whether or not the matrix $(\frac{\partial f}{\partial x_1}(c), \ldots , \frac{\partial f}{\partial x_n}(c))$ has rank 1 at all $c$ such that $f(c)=0$. i.e. if at least one of the partial derivatives is non-zero. This can easily be extended to work if your variety is defined by a bunch of polynomials $\mathbb{R}^n\to\mathbb{R}^m$.</p>
<p>This has a name, called the Jacobian criterion, and for varieties it says that if the Jacobian has full rank, then the variety is non-singular. But these two things are equivalent (Jacobian criterion for non-singular variety and the level set theorem to be a smooth manifold). Thus every non-singular variety is a smooth manifold (Just to be clear, this was considering "variety" to mean a zero set of a collection of polynomials over $\mathbb{R}$)</p>
<p>So to reiterate, some varieties are manifolds (if the defining polynomials satisfy a certain condition on partial derivatives) and some are not. Is every smooth manifold a variety? I think not, but it seems harder. A quick Google search turns up a result that seems to say every smooth compact manifold is algebraic.</p>
|
2,557,520 | <p>PS: Before posting it, one tried to grasp <a href="https://math.stackexchange.com/questions/1996141/if-ffx-x2-x1-what-is-f0">this</a> (although didn't understand mfl's answer completely either).</p>
<p>I was shown the way to solve it: if I set $$\frac{3x-2}{2}=0 \Rightarrow x=\frac{2}{3} \Rightarrow x^2-x-1=-\frac{11}{9}$$ that answer will also be the one of $f(0)$ (i.e. $f(0)=-11/9$), which is true, but I don't quite get it: how the answer of $f(\frac{3x-2}{2})$ (i.e. $-11/9$) is also the answer of $f(0)$? why the answer of the quad equ. is also the answer of $f(0)$ (that it is $f(0)=-11/9$)? I asked the person who solved it for me if there was a proof for this, but she couldn't bring it. So, I would be very happy and obliged if someone here would show the proof that it is $f(0)=-11/9$.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>substituting $$t=\frac{3x-2}{3}$$ then we get
$$x=\frac{2t+2}{3}$$ then we get
$$f(t)=\left(\frac{2t+2}{3}\right)^2-\frac{2+2t}{3}-1$$ then we get ...?</p>
|
50,547 | <p>let $X$ be a $n$-manifold. let $A=\{(x,y,z) \, |\,x=y\}$. I want to see if $A$ is a submanifold of $X^3$.</p>
<p>Consider the map $\Delta\times 1:X\times X\rightarrow X \times X\times X;\, (x,y)\mapsto (x,x,y)$. </p>
<p>If $U_x$ and $U_y$ are neighborhoods of $x$ and $y$ in $X$ then $\Delta\times 1 (U_x\times U_y)=\Delta(U_x)\times U_y$ is a neighborhood of
$\Delta(x,y)=(x,x,y)$ in $X^3$, and we can see that $\Delta(U_x)\times U_y$ is in $A$. So this is a neighborhood of $(x,x,y)$ in $A$. </p>
<p>Now $U_x\cong \mathbb R^n$ and $\Delta(U_x)\cong U_x\cong \mathbb R^n$ and so $A$ is an $2n$-manifold. </p>
<p>Is my argument correct?and does it matter that the diagonal $\Delta(U_x)$ is a closed set, I mean don't we always want a neighborhood to be open? </p>
| Marko Riedel | 44,883 | <p>Suppose we are interested in
$$I = \int_0^\infty \frac{x}{1+x^4} dx$$</p>
<p>and evaluate it by integrating
$$f(z) = \frac{z}{1+z^4}$$</p>
<p>around a pizze slice contour with the horizontal side $\Gamma_1$ of
the slice on the positive real axis and the slanted side $\Gamma_3$
parameterized by $z= \exp(2\pi i/4) t = \exp(\pi i/2) t = it$ with the
two connected by a circular arc $\Gamma_2$ parameterized by $z= R
\exp(it)$ with $0\le t\le \pi i/2.$ We let $R$ go to infinity.</p>
<p><P> The integral along $\Gamma_1$ is $I$ in the limit. Furthermore
we have in the limit
$$\int_{\Gamma_3} f(z) dz
= - \int_0^\infty
\frac{\exp(\pi i/2) t}{1+t^4 \exp(2\pi i)} \exp(\pi i/2) dt
\\ = - \exp(\pi i) \int_0^\infty
\frac{t}{1+t^4} dt = I.$$</p>
<p>Furthermore
$$\int_{\Gamma_2} f(z) dz \rightarrow 0$$
by the ML bound which yields
$$\lim_{R\rightarrow \infty} \pi i/2 R \frac{R}{R^4-1} = 0.$$</p>
<p>The four poles are at
$$\rho_k = \exp(\pi i/4 + 2\pi i k/4).$$</p>
<p>Considering the one pole $\rho_0$ inside the slice ($\rho_1 = \exp(\pi
i/4 + \pi i/2) = \exp(3\pi i/4)$ so it is not inside the contour) we
thus obtain</p>
<p>$$2 I = 2\pi i \mathrm{Res}_{z=\rho_0} f(z)$$</p>
<p>or</p>
<p>$$I = \pi i \frac{\rho_0}{4\rho_0^3}
= \pi i \frac{\rho_0^2}{4\rho_0^4}
= -\frac{1}{4} \pi i \exp(\pi i/2)
= \frac{\pi}{4}.$$</p>
<p><strong>Remark.</strong> We see that the pizza slice is in fact a quarter slice and
the parameterization may use the rotation $it$ by $\pi/2$ throughout.
The four poles are centered on the diagonals of the four quadrants.</p>
|
2,610,501 | <p>A five digit number has to be formed by using the digits $1,2,3,4$ and $5$ without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.</p>
<p>This question came in my test where you literally get $2$ minutes to solve one problem. I want to how to solve this problem more "mathematicaly" instead of listing all the $36$ cases. </p>
| Travis Willse | 155,629 | <p><strong>Hint</strong> If you subtract a number satisfying the criterion from $66666$, you get a different number satisfying the criterion.</p>
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2,991,719 | <p>How to prove/disprove</p>
<blockquote>
<p>If <span class="math-container">$f : (0, \infty) \to \mathbb{R}^n$</span> is continuous on <span class="math-container">$[a, \infty) , \forall a>0$</span> then <span class="math-container">$f$</span> is continuous on <span class="math-container">$(0, \infty)$</span></p>
</blockquote>
<hr>
<p><strong>My try</strong></p>
<p>Assume it is not continuous on <span class="math-container">$(0, \infty)$</span>.</p>
<p>Let's denote a discontinuity point <span class="math-container">$x_0 \in (0,\infty)$</span>. <span class="math-container">$\exists \epsilon >0$</span> s.t. <span class="math-container">$\forall \delta >0$</span>, <span class="math-container">$|x - x_0| < \delta$</span> and <span class="math-container">$|f(x) - f(x_0)| \ge \epsilon $</span>. </p>
<p>But since <span class="math-container">$f$</span> is continuous on <span class="math-container">$[x_0/2, \infty)$</span>, it is contradiction.</p>
<p>This proof seems doubtful for me, but I cannot specify the doubts.</p>
<p>Is there anyone to clarify this?</p>
| user | 505,767 | <p>Yes that's correct, indeed by the <a href="https://en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant" rel="nofollow noreferrer"><strong>properties of the determinant</strong></a> we have that</p>
<p><span class="math-container">$$\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=\det\begin{bmatrix}a&b&c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=\\=\frac13 \det\begin{bmatrix}3a&3b&3c\\d&e&f\\g-a&h-b&i-c\end{bmatrix}=-\frac13 \det\begin{bmatrix}3a&3b&3c\\-d&-e&-f\\g-a&h-b&i-c\end{bmatrix}$$</span></p>
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1,440,106 | <p>I am currently studying how to prove the Fibonacci Identity by Simple Induction, shown <a href="http://mathforum.org/library/drmath/view/52718.html">here</a>, however I do not understand how $-(-1)^n$ becomes $(-1)^{n+1}$. Can anybody explain to me the logic behind this?</p>
| MrYouMath | 262,304 | <p>$-(-1)^n=(-1)^1(-1)^n=(-1)^{n+1}$</p>
|
1,440,106 | <p>I am currently studying how to prove the Fibonacci Identity by Simple Induction, shown <a href="http://mathforum.org/library/drmath/view/52718.html">here</a>, however I do not understand how $-(-1)^n$ becomes $(-1)^{n+1}$. Can anybody explain to me the logic behind this?</p>
| Uri Goren | 203,575 | <p>$$-x=(-1)x$$
Now substitute $x=(-1)^n$
$$-(-1)^n=(-1)(-1)^n=(-1)^{n+1}$$</p>
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3,066,913 | <p>Given a linear operator <span class="math-container">$T$</span> on a finite-dimensional vector space <span class="math-container">$V$</span>, satisfying <span class="math-container">$T^2 = T$</span> answer the following:</p>
<p>(a) Using the dimension theorem, show that <span class="math-container">$N(T) \bigoplus R(T) = V$</span>;</p>
<p>(b) Identify all eigenvalues of <span class="math-container">$T$</span> and the corresponding eigenspaces and show that <span class="math-container">$T$</span> is diagonalizable.</p>
<p>I have already proven part (a). Basically define <span class="math-container">$Tv = w$</span> and the rest follows.</p>
<p>Part (b) however I am honestly confused on where to start. </p>
| user3482749 | 226,174 | <p>You've rather overcomplicated the issue: you know that <span class="math-container">$T^{-1}S^{-1}$</span> is a linear map that exists (since <span class="math-container">$S$</span> and <span class="math-container">$T$</span> have inverses), so all you need to is check that it's an inverse for ST. As a hint: this result is true in all kinds of more general spaces (all the way up to an arbitrary associative algebra), so there's no need to be messing around with evaluating your functions. </p>
<p>Also, <span class="math-container">$S(X)T(Y)$</span> is <em>not</em> equal to <span class="math-container">$ST(XY)$</span>, and <span class="math-container">$T(X) = 0$</span> implying <span class="math-container">$X = 0$</span> is <em>not</em> sufficient for <span class="math-container">$T$</span> to be invertible (it also needs to be surjective). </p>
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804,532 | <p>I a working my way through some old exam papers but have come up with a problem. One question on sequences and induction goes:</p>
<p>A sequence of integers $x_1, x_2,\cdots, x_k,\cdots$ is defind recursively as follows: $x_1 = 2$ and $x_{k+1} = 5x_k,$ for $k \geq 1.$</p>
<p>i) calculate $x_2, x_3, x_4$</p>
<p>ii) deduce a formula for the $n$th term i.e. $x_n$ in terms of $n$ and then prove its validity, using the principles of mathematical induction.</p>
<p>It is the last part that is giving me trouble. I think $x_2, x_3$ and $x_4$ are $10, 50$ and $250$ respectively. I also think I managed to work out the formula, it is $f(n) = 2 \cdot 5^{n-1}.$</p>
<p>However I'm not sure how I'm supposed to prove this using induction... induction is only used when you're adding the numbers in a sequence I thought? I've looked everywhere and can't find any answer specific enough to this question to help. Any help appreciated. Thanks.</p>
| vonbrand | 43,946 | <p>Induction is aplicable in many cases, not just sums. In this case you have a expression for $x_n$, prove (base) that it is valid for $n = 1$ and (induction) if it is valid for $n$ it is valid for $n + 1$.</p>
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1,593,282 | <p>Say we have the function $f:A \rightarrow B$ which is pictured below.<a href="https://i.stack.imgur.com/WfCxs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WfCxs.jpg" alt="enter image description here"></a></p>
<p>This function is not bijective, so the inverse function $f^{-1}: B \rightarrow A$ does not exist. However, we can see that elements $d$ and $e$ in set $B$ are each mapped to by a single element in $A$, so they are kind of "nice" in this sense, almost like a bijective function.</p>
<p>Would it be acceptable to use the notation $f^{-1}(d)=6$ and $f^{-1}(e)=7$ here, even though not all the elements of $B$ have a single inverse? Or does the use of $f^{-1}$ always imply that the entire function $f$ has an inverse?</p>
| Alex M. | 164,025 | <p>No, but it does make sense to say that $f^{-1} (\{d\}) = \{6\}$ (which reads as: the preimage of the set $\{d\}$ is $\{6\}$). This is standard mathematical notation. This also allows you to say that $f^{-1} (\{c\}) = \emptyset$, so it's a very convenient notation.</p>
<p>Note that with this notation, you must now view the formula $f^{-1} (B)$ as an indecomposable formula, i.e. you are <em>not</em> talking about the function $f^{-1}$ applied to some subset $B$, because $f^{-1}$ does not exist.</p>
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3,336,135 | <p>I was looking at sequences <span class="math-container">$x_n=\{n\alpha\}$</span> with <span class="math-container">$n=1,2,\cdots$</span> and <span class="math-container">$\alpha\in [0, 1]$</span> an irrational number. Such sequences are known to be equidistributed, so they get arbitrarily close to any number <span class="math-container">$x\in[0, 1]$</span>, time and over. The brackets represent the fractional part function.</p>
<p>There is a subsequence <span class="math-container">$t_n$</span> of indices such that <span class="math-container">$x_{t_n}$</span> is the <span class="math-container">$n$</span>-th record: it corresponds to the largest value observed sequentially, that is, <span class="math-container">$t_n$</span> is the smallest index such that <span class="math-container">$x_{t_n} > x_{t_{n-1}}$</span>. Here <span class="math-container">$x_{t_{n-1}}$</span> is the previous record. In short, <span class="math-container">$t_n$</span> is the arrival time (index) of the <span class="math-container">$n$</span>-th record. The sequence <span class="math-container">$t_n$</span> is strictly increasing and never ends. For <span class="math-container">$\alpha = \sqrt{2}/2$</span> (our focus in this question) we have:</p>
<p><a href="https://i.stack.imgur.com/X5fAn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/X5fAn.png" alt="enter image description here"></a></p>
<p>I did not expect to see such a pattern. What's more,</p>
<p><span class="math-container">$$t_{2n+1} = \Big\lfloor \frac{\sqrt{3+2\sqrt{2}}}{2}\Big(3+2\sqrt{2}\Big)^{n}\Big\rfloor,$$</span></p>
<p><span class="math-container">$$t_{2n} = \Big\lfloor \frac{\sqrt{2}}{2}\Big(3+2\sqrt{2}\Big)^{n}\Big\rfloor.$$</span></p>
<p>Let us denote as <span class="math-container">$u_{2n+1}$</span> and <span class="math-container">$u_{2n}$</span> the expression inside the brackets <span class="math-container">$\lfloor \rfloor$</span> in the above twin formulas for <span class="math-container">$t_{2n+1}$</span> and <span class="math-container">$t_{2n}$</span>. We have: <span class="math-container">$u_n - t_n \rightarrow 0$</span> (whether <span class="math-container">$n$</span> is even or odd) and the convergence is very fast.</p>
<p>Anyone interested in proving this result? It is based solely on empirical evidence. A similar result seems achievable for <span class="math-container">$\sqrt{3}/2$</span>. Now for non-algebraic number, the kind of periodicity observed here is lost, but we get very intriguing results, with more mysterious patterns. Below is the data for <span class="math-container">$\alpha = \log_2 3$</span>:</p>
<p><a href="https://i.stack.imgur.com/3S47m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3S47m.png" alt="enter image description here"></a></p>
<p>The nice pattern seems to end abruptly beyond the data being displayed here, and indeed the next maximum take places at index 190,537 and 10,781,274. I could not find the next one, maybe due to machine precision or due to a massive gap. </p>
<p>I am very interested in some asymptotic value for <span class="math-container">$t_n$</span> especially when <span class="math-container">$\alpha$</span> is not an algebraic number. I am surprised that the arrival times <span class="math-container">$t_n$</span> of the successive records behave so much differently than if the underlying sequence <span class="math-container">$\{n \alpha\}$</span> was truly random (I studied the random case <a href="https://www.datasciencecentral.com/profiles/blogs/distribution-of-arrival-times-of-extreme-events" rel="noreferrer">here</a>.) </p>
<p><strong>Update</strong></p>
<p>If <span class="math-container">$\alpha = \pi$</span>, we get the following: <span class="math-container">$(t_5,t_6,t_7,t_8,t_9)=$</span> <span class="math-container">$(7,113,$</span> <span class="math-container">$33215,99532,364913)$</span>. Define <span class="math-container">$\Delta t_n = t_n-t_{n-1}$</span>. Then
<span class="math-container">$(\Delta t_6, \Delta t_7,\Delta t_8,\Delta t_9)=(106, 33102, 66317,265381)$</span>. </p>
<p>Now look at <span class="math-container">$\pi$</span> convergents (skipping the first ones): <span class="math-container">$22/7$</span>, <span class="math-container">$333/106$</span>, <span class="math-container">$355/113$</span>, <span class="math-container">$103993/33102$</span>, <span class="math-container">$104348/33215$</span>, <span class="math-container">$208341/66317$</span>, <span class="math-container">$312689/99532$</span>, <span class="math-container">$833719/265381$</span>, <span class="math-container">$11464108/364913$</span>. The denominators in odd positions corresponds to the successive <span class="math-container">$\Delta t_n$</span>. The denominators in even positions corresponds to the successive <span class="math-container">$t_n$</span>. </p>
<p>This works for all irrationals (need to check) and it gives an alternate way to compute the convergents. For instance, take <span class="math-container">$t_8=99532$</span>. Then the fraction <span class="math-container">$m/99532$</span> closest to <span class="math-container">$\pi$</span>, with <span class="math-container">$m$</span> an integer, is a convergent. Using this fact, you can easily obtain <span class="math-container">$m=312689$</span>. </p>
<p><strong>Background</strong></p>
<p>My interest in this problem stems from the following. The sequence <span class="math-container">$z_n =2^{\{-n \log_2 3\}}$</span> consists of rational numbers, all having a period (in their binary expansion) equal to <span class="math-container">$2\cdot 3^{n-1}$</span>. All have exactly 50% zeros, 50% ones in their binary expansion. The median, which is one of them if <span class="math-container">$n$</span> is odd, tends to <span class="math-container">$\sqrt{2}/2$</span>. Does it mean that <span class="math-container">$\sqrt{2}/2$</span> has 50% zeros too? No, you need to remove "extreme" numbers from this sequence. For instance all numbers <span class="math-container">$z_{t_n}$</span> with <span class="math-container">$t_n$</span> listed in the above table for <span class="math-container">$\alpha = \log_2 3$</span>. What other numbers should I remove? I don't know yet, but this is a starting point. It depends on how fast/slowly the "extreme" numbers are building up as <span class="math-container">$n$</span> increases. See discussion on this topic <a href="https://math.stackexchange.com/questions/3326227/limit-associated-with-a-recursion">here</a>, look at the section entitled <em>A different approach</em>. </p>
<p><strong>Potential machine learning applications of continued fractions</strong> </p>
<p>We found a connection between records (maxima) of some sequences, and continued fractions. Doing the same thing with minima will complete this analysis. Extreme events in business or economics settings are typically modeled using statistical distributions (Gumbel and so on), which did not live to their expectations in many instances. </p>
<p>The theory outlined here provides an alternative, with each number <span class="math-container">$\alpha$</span> serving as a particular model, with its own features distinct from all other numbers. In particular, the numerators and denominators of convergents of <span class="math-container">$\alpha$</span> can be used to model the arrival times of extreme events. While the records here are getting closer either to zero or to one, using a logarithm transformation allows you to model phenomena that are not bounded. </p>
<p>Note that the sequence <span class="math-container">$\{n\alpha\}$</span> has strong, long-range auto-correlations that are well studied (see appendix B in <a href="https://www.datasciencecentral.com/profiles/blogs/free-book-statistics-new-foundations-toolbox-and-machine-learning" rel="noreferrer">this book</a>). One could try with the sequence <span class="math-container">$\{b^n \alpha\}$</span> instead (<span class="math-container">$b > 1$</span>) as it exhibits exponentially decaying auto-correlations, also studied in the same book. Even cross-correlations between two sequences have been studied (say <span class="math-container">$\{\alpha n\}$</span> and <span class="math-container">$\{\beta n\}$</span>), allowing you to model multivariate time series and their cross-dependencies. Other sequences are also being investigated. </p>
| Daniel Castle | 398,608 | <p>For the case of <span class="math-container">$\frac{\sqrt{2}}{2}$</span>, we look to find <span class="math-container">$k\in\mathbb{N}$</span> such that <span class="math-container">$\frac{\sqrt{2}}{2}k\pmod 1$</span> is as large as possible.</p>
<p>First, consider the sequence of "best possible" rational over-estimates of <span class="math-container">$\frac{\sqrt{2}}{2}$</span>. We can do this using finite iterations of the continued fraction <span class="math-container">$[0;1,\bar 2]$</span> evaluated with the last term as <span class="math-container">$0$</span>, giving the sequence:
<span class="math-container">$$S_n=\frac{1}{1},\frac{3}{4},\frac{5}{7},\frac{17}{24},\frac{29}{41}...\space\space\space\space$$</span></p>
<p>The (strictly increasing) sequence of denominators is then:</p>
<p><span class="math-container">$$D_n=1,4,7,24,41...$$</span></p>
<p>Since we know <span class="math-container">$\frac{\sqrt{2}}{2}<S_n$</span> for all <span class="math-container">$n$</span>, we have that:
<span class="math-container">$$\frac{\sqrt{2}}{2}D_n<S_nD_n=k,\space k\in\mathbb{N}$$</span></p>
<p>This implies:</p>
<p><span class="math-container">$$\left(\frac{\sqrt{2}}{2}-S_n\right)D_n\equiv\frac{\sqrt{2}}{2}D_n\pmod 1\tag{1}$$</span></p>
<p>Since <span class="math-container">$S_n$</span> is a "best possible" approximation for <span class="math-container">$\frac{\sqrt{2}}{2}$</span>, we have that:</p>
<p><span class="math-container">$$\left|\frac{\sqrt{2}}{2}-\frac{a}{b}\right|<\left|\frac{\sqrt{2}}{2}-S_n\right|\implies b>D_n$$</span></p>
<p>Hence:</p>
<p><span class="math-container">$$\left|\frac{\sqrt{2}}{2}-S_n\right|D_n=\left(S_n-\frac{\sqrt{2}}{2}\right)D_n$$</span></p>
<p>Is minimal. Multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$-1$</span>, we find that:</p>
<p><span class="math-container">$$\left(S_n-\frac{\sqrt{2}}{2}\right)D_n\equiv -\frac{\sqrt{2}}{2}D_n\equiv 1-\frac{\sqrt{2}}{2}D_n \pmod 1$$</span></p>
<p>Must also be minimal. Therefore:</p>
<p><span class="math-container">$$\frac{\sqrt{2}}{2}k\pmod 1$$</span></p>
<p>Is maximal for some <span class="math-container">$k\in\mathbb{N}\iff k=D_n$</span></p>
<h2>Note:</h2>
<p>On further inspection, I realised that (provided my proof is correct), substituting any irrational number <span class="math-container">$\alpha$</span> (perhaps rational too?) for <span class="math-container">$\frac{\sqrt{2}}{2}$</span> would lead to the same result:</p>
<p><span class="math-container">$\alpha D_n\pmod 1>\alpha k\pmod 1$</span> for all <span class="math-container">$k\in\mathbb{N}, k<D_n \iff D_n$</span> represents the sequence of over-estimate convergent denominators from the continued fraction of <span class="math-container">$\alpha$</span>.</p>
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3,098,245 | <p><span class="math-container">$$ \frac{d}{dx}e^x =\frac{d}{dx} \sum_{n=0}^{ \infty} \frac{x^n}{n!}$$</span>
<span class="math-container">$$ \sum_{n=0}^{ \infty} \frac{nx^{n-1}}{n!}$$</span>
<span class="math-container">$$ \sum_{n=0}^{ \infty} \frac{x^{n-1}}{(n-1)!}$$</span>
This isn't as straightforward as I thought it would be. I'm assuming there is some way to shift indices to complete the proof, but I'm not sure exactly how. </p>
| Theo Bendit | 248,286 | <p>Well, first of all, careful with your first term, since you are dividing by <span class="math-container">$(-1)!$</span>. The first term is constant and differentiates to <span class="math-container">$0$</span>, so your sum should start from <span class="math-container">$n = 1$</span>.</p>
<p>Then, just make a substitution <span class="math-container">$m = n - 1$</span>. When <span class="math-container">$n = 1$</span>, we get <span class="math-container">$m = 0$</span>, and so</p>
<p><span class="math-container">$$\frac{\mathrm{d}}{\mathrm{d}x} e^x = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^m}{m!}.$$</span></p>
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3,596,514 | <p>Let <span class="math-container">$0\le x \le 1$</span>
find the maximum value of <span class="math-container">$x(9\sqrt{1+x^2}+13\sqrt{1-x^2})$</span></p>
<p>I try to use am-gm inequality to solve this because it's similar to CMIMC2020 team prob.12 but i don't know how to do next. </p>
| Matteo | 686,644 | <p>You can compute the first derivate of <span class="math-container">$f(x)$</span> which is:
<span class="math-container">$$f'(x)=x\left (\frac{9x}{\sqrt{x^2+1}}-\frac{13x}{\sqrt{1-x^2}}\right )+9\sqrt{x^2+1}+13\sqrt{1-x^2}$$</span>
Now you have to study <span class="math-container">$f'(x)=0$</span> and so:
<span class="math-container">$$x=\frac{2}{\sqrt{5}}$$</span>
There is another solutions, bit with this we find the local maximum. Now to find the maximim, we compute:
<span class="math-container">$$f\left (\frac{2}{\sqrt{5}}\right)=16$$</span></p>
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738,122 | <p>I have been looking at Kolmogorov's book "Introductory Real Analysis" and have become stuck at the problem 4a) on page 301.</p>
<p>In this problem we are given $f$ a nonnegative and integrable function on $A$, a set of finite measure. We are asked to show that</p>
<p>$\int_A f(x) d \mu \ge 0$</p>
<p>However, how would we prove this? In this text the Lebesgue integral of a function $f$ is defined to be (when the limit exists);</p>
<p>$\int_A f(x) d\mu = \lim\limits_{n \rightarrow \infty} \int_{A} f_n(x) d\mu$</p>
<p>where $f_n$ is a sequence of integrable simple functions that converges uniformly to $f$.</p>
<p>I have attempted to show this by showing that the sequence of simple functions converging to a nonnegative function, is a sequence of nonnegative functions. However this isn't true, take $f_n = -1/n$ and $f = 0$. How else could I go about this? </p>
<p>Thanks.</p>
| AlexR | 86,940 | <p>Hint: First show the result for non-negative simple functions. Then use that you can find $f_n \ge 0$ simple with $f_n \nearrow f$ uniformly to see that
$$\int_A f d\mu = \lim_{n\to\infty} \underbrace{\int_A f_n d\mu}_{\ge 0} \ge 0$$
For simple functions, the result is fairly easy to prove but depends slightly on the concrete definition of a "simple function" in your book.</p>
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352,305 | <p>For <span class="math-container">$\mathcal{S}$</span> the <span class="math-container">$(\infty,1)$</span>-category of spaces <a href="https://mathoverflow.net/questions/239383/the-homotopy-category-is-not-complete-nor-cocomplete">its homotopy category <span class="math-container">$h\mathcal{S}$</span> does not have pushouts or pullbacks</a>. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in <span class="math-container">$\mathcal{S}$</span>.</p>
<p>Generally, filtered colimits are much better behaved than general ones. For example <a href="https://mathoverflow.net/questions/235526/when-do-colimits-agree-with-homotopy-colimits">filtered colimits in the (Quillen) model category <span class="math-container">$\mathit{sSet}$</span> compute homotopy colimits in <span class="math-container">$\mathcal{S}$</span>.</a>
In light of this I was wondering the following:</p>
<blockquote>
<p>Does <span class="math-container">$h\mathcal{S}$</span> have filtered colimits and, if so, does the functor <span class="math-container">$\mathcal{S} \to h\mathcal{S}$</span> preserve them?</p>
</blockquote>
<p>More generally one could ask the same for any presentable <span class="math-container">$(\infty,1)$</span>-category <span class="math-container">$\mathcal{C}$</span>; I particularly care about the case of the derived category <span class="math-container">$\mathcal{D}(R)$</span> of a ring <span class="math-container">$R$</span>.</p>
| user152131 | 152,131 | <p>No, for a diagram <span class="math-container">$X: I \to \mathcal{S} \to h\mathcal{S}$</span> the colimit in <span class="math-container">$h\mathcal{S}$</span> would satisfy <span class="math-container">$[\mathrm{colim} X(i),Y] \cong \lim [X(i),Y]$</span> where brackets denote morphisms in <span class="math-container">$h\mathcal{S}$</span>. Informally, the set of maps in <span class="math-container">$h\mathcal{S}$</span> from the (homotopy) colimit in <span class="math-container">$\mathcal{S}$</span> should involve also higher derived limits.</p>
<p>For a concrete example, you could take <span class="math-container">$I = (\mathbb{N},<)$</span>, <span class="math-container">$X(i) \simeq S^1$</span> for all <span class="math-container">$i$</span>, <span class="math-container">$X(i) \to X(i+1)$</span> a map of degree 2. Then <span class="math-container">$\mathrm{colim}(X(i)) \simeq K(\mathbb{Z}[1/2],1)$</span> is not the colimit in <span class="math-container">$h\mathcal{S}$</span> since it has distinct morphisms to <span class="math-container">$Y = K(\mathbb{Z},2)$</span> which become equal upon restricting to each <span class="math-container">$X(i)$</span>.</p>
<p>Similar examples exist in <span class="math-container">$D(\mathbb{Z})$</span>.</p>
|
352,305 | <p>For <span class="math-container">$\mathcal{S}$</span> the <span class="math-container">$(\infty,1)$</span>-category of spaces <a href="https://mathoverflow.net/questions/239383/the-homotopy-category-is-not-complete-nor-cocomplete">its homotopy category <span class="math-container">$h\mathcal{S}$</span> does not have pushouts or pullbacks</a>. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in <span class="math-container">$\mathcal{S}$</span>.</p>
<p>Generally, filtered colimits are much better behaved than general ones. For example <a href="https://mathoverflow.net/questions/235526/when-do-colimits-agree-with-homotopy-colimits">filtered colimits in the (Quillen) model category <span class="math-container">$\mathit{sSet}$</span> compute homotopy colimits in <span class="math-container">$\mathcal{S}$</span>.</a>
In light of this I was wondering the following:</p>
<blockquote>
<p>Does <span class="math-container">$h\mathcal{S}$</span> have filtered colimits and, if so, does the functor <span class="math-container">$\mathcal{S} \to h\mathcal{S}$</span> preserve them?</p>
</blockquote>
<p>More generally one could ask the same for any presentable <span class="math-container">$(\infty,1)$</span>-category <span class="math-container">$\mathcal{C}$</span>; I particularly care about the case of the derived category <span class="math-container">$\mathcal{D}(R)$</span> of a ring <span class="math-container">$R$</span>.</p>
| Tim Campion | 2,362 | <p>Ironically, I was wondering something similar earlier this week (the irony is that I was sitting next to the OP while doing my wondering). Here's another reason why this can't be the case. In general, if <span class="math-container">$C\subseteq D$</span> is a full subcategory and every object in <span class="math-container">$D$</span> is a colimit of a diagram in <span class="math-container">$C$</span>, then it follows that <span class="math-container">$C$</span> is a strong generator for <span class="math-container">$D$</span> -- i.e. the hom functors <span class="math-container">$\{Hom(c,-): D \to Set\}_{c \in C}$</span> are jointly faithful and conservative. If <span class="math-container">$C$</span> is essentially small, then one can take a product (or coproduct) of these functors to obtain a faithful, conservative functor <span class="math-container">$D \to Set$</span>.</p>
<p>At the <span class="math-container">$\infty$</span>-categorical level, every space <span class="math-container">$X$</span> is the filtered colimit <span class="math-container">$X = \varinjlim_i X_i$</span> of finite complexes <span class="math-container">$X_i$</span>. If this colimit were also a colimit in the homotopy category, then <span class="math-container">$X$</span> would be a colimit of finite complexes in the homotopy category. Since the homotopy category of finite complexes is essentially small, we would get a faithful, conservative functor <span class="math-container">$hS \to Set$</span> by the above.</p>
<p>But as <a href="http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html" rel="nofollow noreferrer">Freyd famously showed</a>, there does not exist a faithful functor <span class="math-container">$hS \to Set$</span>! <a href="https://mathoverflow.net/questions/351653/does-the-homotopy-category-of-spaces-admit-a-weak-generating-set">I learned recently</a> that <a href="https://arxiv.org/abs/1910.04141" rel="nofollow noreferrer">Carlson and Christensen</a> have shown there also does not exist a set of spaces such that the corresponding functor <span class="math-container">$hS \to Set$</span> is conservative!</p>
<hr>
<p>We can conclude something stronger from this: there does not exist a regular cardinal <span class="math-container">$\lambda$</span> such that <span class="math-container">$\lambda$</span>-filtered colimits in <span class="math-container">$S$</span> can be computed in the homotopy category. More generally, let <span class="math-container">$A$</span> be a <span class="math-container">$\lambda$</span>-accessible <span class="math-container">$\infty$</span>-category, and suppose that <span class="math-container">$A \to hA$</span> preserves <span class="math-container">$\lambda$</span>-filtered colimits. Then, since <span class="math-container">$A \to hA$</span> is obtained by applying the filtered-colimit-preserving functor <span class="math-container">$\pi_0: Spaces \to Sets$</span> to the homsets, one sees that any <span class="math-container">$\lambda$</span>-presentable object in <span class="math-container">$A$</span> is also <span class="math-container">$\lambda$</span>-presentable in <span class="math-container">$hA$</span>. Moreover, every object is a <span class="math-container">$\lambda$</span>-filtered colimit of such objects. It follows that the functor <span class="math-container">$hA \to Ind_\lambda(hA_\lambda)$</span> is fully faithful (here <span class="math-container">$A_\lambda \subseteq A$</span> is the full subcategory of <span class="math-container">$\lambda$</span>-presentable objects); by Rosicky's theorem (see below), this functor is also essentially surjective. So <span class="math-container">$hA = Ind_\lambda(hA_\lambda)$</span> is also <span class="math-container">$\lambda$</span>-accessible! (As an alternative to Rosicky's theorem, if we assume that <span class="math-container">$hA$</span> has <span class="math-container">$\lambda$</span>-filtered colimits, then it since every object is a <span class="math-container">$\lambda$</span>-filtered colimit of <span class="math-container">$\lambda$</span>-presentable objects, it follows directly that <span class="math-container">$hA$</span> is accessible) That is, if you have an accessible <span class="math-container">$\infty$</span>-category and you can compute sufficiently-filtered colimits in the homotopy category, then the homotopy category is also accessible. Beyond the existence of classic counterexamples like Spaces, my sense is that this situation is comparatively rare in practice (except when <span class="math-container">$A$</span> was a 1-category to start with). In particular, I think it's rare for the homotopy derived category of a ring to be accessible. I don't know about the <em>existence</em> of filtered colimits in these homotopy categories though, but I suspect they typically don't exist.</p>
<p>A notable exception would be the derived category of field <span class="math-container">$D(k)$</span>. In this case, the homotopy category is the category of graded <span class="math-container">$k$</span>-vector spaces, which <em>is</em> accessible (indeed, it's locally presentable -- though not all colimits are preserved by the forgetful functor) and the forgetful functor can be identified with homology, which <em>does</em> preserve filtered colimits.</p>
<p>I haven't thought about this much -- it's possible that there are a lot of rings <span class="math-container">$R$</span> for which the homotopy category of <span class="math-container">$D(R)$</span> is accessible...</p>
<hr>
<p>I think there's a lot more to be said about understanding (highly) filtered colimits by looking at the homotopy category.</p>
<ul>
<li><p>One commonly-used trick is that (co)products can be computed in the homotopy category, so if you have a triangulated category with countable coproducts, then you can compute a sequential colimit <span class="math-container">$\varinjlim_n X_n$</span> as the cofiber (using the triangulated structure) of the map <span class="math-container">$\oplus_n X_n \to \oplus_n X_n$</span> given by the difference of the identity and the coproduct of the maps <span class="math-container">$X_n \to X_{n+1}$</span>. This allows you to compute countable filtered colimits using only the homotopy category + triangulated structure, by passing to a cofinal chain. But I don't think this generalizes to larger filtered colimits.</p></li>
<li><p>There are Brown representability-type results which look at the functor <span class="math-container">$ho(Ind_\lambda(C)) \to Ind_\lambda(ho(C))$</span>. Rosicky <a href="http://www.kurims.kyoto-u.ac.jp/EMIS/journals/TAC/volumes/14/19/14-19.pdf" rel="nofollow noreferrer">showed</a> that this functor is always essentially surjective (perhaps after passing to a higher <span class="math-container">$\lambda$</span>?). In some cases, Adams apparently showed that this functor is also full, and I think there's more work on this in the triangulated categories literature. But maybe this is not so close to the original question.</p></li>
<li><p>Another tangentially related point -- not every idempotent in the homotopy category splits! (This is also a theorem of Freyd). The splitting of an idempotent is an example of a colimit of a diagram which is <span class="math-container">$\lambda$</span>-filtered for every <span class="math-container">$\lambda$</span>. This is only tangentially related, because if you know that your idempotent arises from a genuine homotopy-coherent idempotent (i.e. an idempotent in the <span class="math-container">$\infty$</span>-category of spaces) then it <em>does</em> split; in particular the splitting in the homotopy category exists and agrees with the <span class="math-container">$\infty$</span>-categorical splitting.</p></li>
<li><p>I think in the triangulated categories literature, people deal with this issue by various "weak colimit" and "minimal weak colimit" constructions, but I don't know exactly how these work. Perhaps an expert can comment.</p></li>
<li><p>Maybe it's implicit in things I've said above, but with these sorts of questions there can be a big difference between <span class="math-container">$\infty$</span>-categories like <span class="math-container">$Spaces$</span> and <span class="math-container">$\infty$</span>-categories like <span class="math-container">$D(R)$</span>. For example, there <em>does</em> exist a conservative functor <span class="math-container">$D(R) \to Set$</span> (even already there is a conservative functor <span class="math-container">$h(pointed spaces) \to Set$</span> by Whitehead's theorem) -- although I think still not a faithful functor.</p></li>
</ul>
|
744,787 | <p>I need the equation of the line passing through a given point $A(2,3,1)$ perpendicular to the given line
$$
\frac{x+1}{2}= \frac{y}{-1} = \frac{z-2}{3}.
$$</p>
<p>I think there must bee some kind of rule to do this, but I can't find it anyway.</p>
| JJacquelin | 108,514 | <p>Your wording (x+1)/2= y/-1 = (z-2)/3 looks strange. Why not (x+1)/2= -y = (z-2)/3 ? Is there not a typing mistake ? If it's OK, an analylical method is shown below :</p>
<p><img src="https://i.stack.imgur.com/0dOWt.jpg" alt="enter image description here"></p>
|
285,719 | <p>Why is $(-3)^4 =81$ and $-3^4 =-81 $?This might be the most stupidest question that you might have encountered,but unfortunately i'am unable to understand this.</p>
| Artem | 28,379 | <p>By definition</p>
<p>$(-3)^4 = (-3)\cdot(-3)\cdot(-3)\cdot(-3) = 9\cdot 9 = 81$</p>
<p>$-3^4 = -(3^4) = -(3\cdot 3\cdot 3\cdot 3) = - 81$</p>
|
645,579 | <p>Let $n$ be a positive integer. Find a general expression for $$\int x^n\cos(x)~dx$$ None of the standard integration techniques or the standard tricks I've seen for difficult integrals seem to apply to this one. I guess it is some type of reduction, but how to get a closed form?</p>
| Peter | 82,961 | <p>Use repeated integration by parts. Take the antiderivate of the cos(x) or sin(x)-function and differentiate the polynomial term. The degree decreases in each step.</p>
|
645,579 | <p>Let $n$ be a positive integer. Find a general expression for $$\int x^n\cos(x)~dx$$ None of the standard integration techniques or the standard tricks I've seen for difficult integrals seem to apply to this one. I guess it is some type of reduction, but how to get a closed form?</p>
| doraemonpaul | 30,938 | <p><a href="http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_cosine" rel="nofollow">http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions#Integrands_involving_only_cosine</a> can find the reduction formula and also the final result.</p>
|
182,101 | <p>With respect to assignments/definitions, when is it appropriate to use $\equiv$ as in </p>
<blockquote>
<p>$$M \equiv \max\{b_1, b_2, \dots, b_n\}$$</p>
</blockquote>
<p>which I encountered in my analysis textbook as opposed to the "colon equals" sign, where this example is taken from Terence Tao's <a href="http://terrytao.wordpress.com/">blog</a> :</p>
<blockquote>
<p>$$S(x, \alpha):= \sum_{p\le x} e(\alpha p) $$</p>
</blockquote>
<p>Is it user-background dependent, or are there certain circumstances in which one is more appropriate than the other?</p>
| Carl Mummert | 630 | <p>It's entirely up to the whim of the author. Other symbols that can mean the same thing are $\triangleq$ and $=_{def}$. I think that only a minority of authors use any special notation, however; the majority just use a regular equals sign. </p>
|
182,101 | <p>With respect to assignments/definitions, when is it appropriate to use $\equiv$ as in </p>
<blockquote>
<p>$$M \equiv \max\{b_1, b_2, \dots, b_n\}$$</p>
</blockquote>
<p>which I encountered in my analysis textbook as opposed to the "colon equals" sign, where this example is taken from Terence Tao's <a href="http://terrytao.wordpress.com/">blog</a> :</p>
<blockquote>
<p>$$S(x, \alpha):= \sum_{p\le x} e(\alpha p) $$</p>
</blockquote>
<p>Is it user-background dependent, or are there certain circumstances in which one is more appropriate than the other?</p>
| nullUser | 17,459 | <p>The notation $x:= y$ is preferred as $\equiv$ has another meaning in modular arithmetic (though it is almost always clear from context as to which is meant). However, there is one big advantage to using the $:=$. That is, it is not graphically symmetric and hence allows for strings such as
$$
y:= f(x) \leq g(x) =: L
$$
where here we are defining both $y$ and $L$. This statement would be much more cumbersome using $\equiv$, and it would not make sense if one simply wrote
$$
y \equiv f(x) \leq g(x) \equiv L.
$$ </p>
|
2,909,626 | <p>I'm having trouble remembering how to solve when you have an equation such as </p>
<p>$$0.3=(1-0.63x)^{5.26}$$</p>
<p>any help would be appreciated.</p>
| David C. Ullrich | 248,223 | <p>The sentence you say you're concerned about has more or less nothing to do with "<em>extension</em> by continuity". A careful justification (assuming my guesses about the meaning of the notation are correct):</p>
<p>Say $S\subset L^p$ is the set of simple functions. Define $\Lambda:L^p\to\Bbb R$ by $$\Lambda g=Fg-\int fg.$$Since $\Lambda$ is continuous, $\Lambda g=0$ for all $g\in S$ and $S$ is dense in $L^p$ it follows that $\Lambda g=0$ for all $g\in L^p$.</p>
<p>This uses the BFADS:</p>
<blockquote>
<blockquote>
<p><strong>Basic Fact About Dense Subsets</strong> Suppose $X$ is a metric space, $D\subset X$ is dense, $f:X\to\Bbb R$ is continuous, and $f(x)=0$ for all $x\in D$. Then $f(x)=0$ for all $x\in X$.</p>
</blockquote>
</blockquote>
<p>Proof: Suppose $x\in X$. Let $\epsilon>0$. Choose $\delta>0$ so that $d(x,y)<\delta$ implies $|f(y)-f(x)|<\epsilon$. Since $D$ is dense, there exists $y\in D$ with $d(y,x)<\delta$. Hence $|f(x)|=|f(x)-f(y)|<\epsilon$.</p>
<p>And so $|f(x)|<\epsilon$ for every $\epsilon>0$, hence $f(x)=0$.</p>
|
1,597,247 | <p>Give the continued fraction expansion of two real numbers $a,b \in \mathbb R$, is there an "easy" way to get the continued fraction expansion of $a+b$ or $a\cdot b$?</p>
<p>If $a,b$ are rational it is easy as you can easily conver the back to the 'rational' form, add or multiply and then conver them back to continued fraction form. But is there a way that requres <em>no conversation</em>?</p>
<p>Other than that I found no clues whether there is an "easy" way to do it for irrational numbers.</p>
| Gerry Myerson | 8,269 | <p>Gosper found efficient ways to do arithmetic with continued fractions (without converting them to ordinary fractions or decimals). <a href="http://perl.plover.com/yak/cftalk/" rel="noreferrer">Here</a> is a page with links to Gosper's work, but also with an exposition of Gosper's methods. </p>
<p>See also this older m.se question, <a href="https://math.stackexchange.com/questions/232422/faster-arithmetic-with-finite-continued-fractions">Faster arithmetic with finite continued fractions</a></p>
|
3,535,316 | <p><span class="math-container">$$\int_{0}^{\pi}e^{x}\cos^{3}(x)dx$$</span></p>
<p>I tried to solve it by parts.I took <span class="math-container">$f(x)=\cos^{3}(x)$</span> so <span class="math-container">$f'(x)=-3\cos^{2}x\sin x$</span> and <span class="math-container">$g'(x)=e^{x}$</span> and I got"</p>
<p><span class="math-container">$$\int_{0}^{\pi}e^{x}\cos^{3}(x)dx=e^{x}\cos(x)+3\int_{0}^{\pi}e^{x}\cos^{2}(x)\sin(x)dx.$$</span></p>
<p>How to approach the second integral?</p>
| Community | -1 | <p>Your sum is:</p>
<p><span class="math-container">$$\sum_{n\ge1}^{}\frac{n}{\left(n-1\right)!}$$</span>
Setting <span class="math-container">$n-1 \mapsto n$</span> gives:</p>
<p><span class="math-container">$$=\sum_{n\ge0}^{}\frac{n+1}{n!}=1+\sum_{n\ge1}^{}\frac{n+1}{n!}=1+\sum_{n\ge1}^{}\frac{1}{n!}+\sum_{n\ge1}^{}\frac{1}{\left(n-1\right)!}=\sum_{n\ge0}^{}\frac{1}{n!}+\sum_{n\ge0}^{}\frac{1}{n!}=\color{red}{2e}$$</span></p>
|
3,166,419 | <blockquote>
<p>A fair coin is tossed three times in succession. If at least one of
the tosses has resulted in Heads, what is the probability that at
least one of the tosses resulted in Tails?</p>
</blockquote>
<p>My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is <span class="math-container">$ \frac 12 + \frac 12 - \frac 14 = \frac 34 $</span>. </p>
<p>The given answer: <span class="math-container">$ \frac 67 $</span>. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is <span class="math-container">$ \frac 67 $</span>.</p>
<p>Why is my answer wrong? What am I missing?</p>
| LuuBluum | 321,835 | <p>Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or <span class="math-container">$P(T>0|X>0)$</span> Using the definition of conditional probability, we can say that <span class="math-container">$P(T>0|X>0)=P(T>0,X>0)/P(X>0)$</span>. Then, using some identities, we have that <span class="math-container">$P(T>0,X>0)=1-P(T=0)-P(X=0)$</span> and <span class="math-container">$P(X>0)=1-P(X=0)$</span>.</p>
<p>To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.</p>
<p>Then, doing the remaining math:</p>
<p><span class="math-container">$$\frac{1-P(T=0)-P(X=0)}{1-P(X=0)}=\frac{1-\frac{1}{8}-\frac{1}{8}}{1-\frac{1}{8}}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$$</span></p>
<p>Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.</p>
|
3,090,879 | <p>Is there prime number of the form <span class="math-container">$1+11+111+1111+11111+...$</span>. I've checked it up to first 2000 repunits, but i found none. If <span class="math-container">$R_1=1$</span>, <span class="math-container">$R_2=1+11$</span>, <span class="math-container">$R_3=1+11+111$</span>, <span class="math-container">$R_n=1+11+111+...+$</span>nth repunit. What is the smallest n such that <span class="math-container">$R_n$</span> is prime ?</p>
| Roman | 632,893 | <p>The first one is at <span class="math-container">$n=2497$</span>, you just missed it!</p>
<pre><code>Do[If[PrimeQ[(10^(n+1)-9n-10)/81], Print[n]], {n, 10^4}]
(* 2497 *)
(* 3301 *)
...
</code></pre>
<p>The direct formula for the <span class="math-container">$n^{\text{th}}$</span> term is from</p>
<pre><code>Sum[10^j, {i, 0, n - 1}, {j, 0, i}]
(* 1/81 (-10 + 10^(1 + n) - 9 n) *)
</code></pre>
|
3,090,879 | <p>Is there prime number of the form <span class="math-container">$1+11+111+1111+11111+...$</span>. I've checked it up to first 2000 repunits, but i found none. If <span class="math-container">$R_1=1$</span>, <span class="math-container">$R_2=1+11$</span>, <span class="math-container">$R_3=1+11+111$</span>, <span class="math-container">$R_n=1+11+111+...+$</span>nth repunit. What is the smallest n such that <span class="math-container">$R_n$</span> is prime ?</p>
| DanaJ | 117,584 | <p>To give some more examples of solutions, using Roman's direct formula for conciseness (alternately you can do something like <code>r = 10*r+1; N += r</code> in the loop).</p>
<p>Pari/GP (test is AES BPSW):</p>
<pre><code>? for(n=1,6000,ispseudoprime((10^(n+1)-9*n-10)/81) && print(n))
2497
3301
time = 2min, 5,916 ms.
</code></pre>
<p>Perl/ntheory (v0.74) (test is ES BPSW):</p>
<pre><code><span class="math-container">$ time perl -Mntheory=:all -E 'is_prob_prime(divint(powint(10,$</span>_+1)-9*$_-10,81)) and say for 1..6000;'
2497
3301
real 0m44.473s
</code></pre>
<p>PFGW (test is base 3 Fermat):</p>
<pre><code><span class="math-container">$ echo -e "ABC2 (10^(\$a+1)-9*\$a-10)/81\na: from 1 to 6000" > repsum.txt
$</span> time ./pfgw64 -k -u0 -f50 repsum.txt
(10^(2497+1)-9*2497-10)/81 is 3-PRP! (0.0793s+0.0362s)
(10^(3301+1)-9*3301-10)/81 is 3-PRP! (0.1333s+0.0615s)
real 0m42.963s
</code></pre>
<p>As the size gets larger, PFGW will get much faster relative to the others. E.g. times on Macbook Pro (2015) for checking n from 6001 to 7000 the times are 88.9s for Pari/GP, 26.0s for Perl/ntheory, and 13.4s for PFGW. Albeit for any discovered values we'd want to run another more strict test such as BPSW.</p>
|
3,148,094 | <p>In class, my professor computed the density of a gamma random variable by taking the derivative of its CDF, but he skipped many steps. I am trying to go through the derivation carefully but cannot reproduce his final result.</p>
<p>Let <span class="math-container">$k$</span> be the shape and <span class="math-container">$\mu$</span> be the scale. Then the CDF for a gamma random variable <span class="math-container">$T$</span> is</p>
<p><span class="math-container">$$
F(t) = 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!}
$$</span></p>
<p>Using the product rule, I get</p>
<p><span class="math-container">$$
\begin{align}
f(t)
&= \frac{\partial}{\partial t} F(t)
\\
&= \frac{\partial}{\partial t} \Big( 1 - \sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{\partial}{\partial t} \Big( \frac{e^{-\mu t} (\mu t)^{i}}{i!} \Big)
\\
&= - \sum_{i=0}^{k-1} \frac{1}{i!} \frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big)
\end{align}
$$</span></p>
<p>where</p>
<p><span class="math-container">$$
\frac{\partial}{\partial t} \Big( e^{-\mu t} (\mu t)^{i} \Big) = e^{-\mu t} (-\mu)(\mu t)^i + e^{-\mu t} i (\mu t)^{i-1}
$$</span></p>
<p>Putting everything together, we get</p>
<p><span class="math-container">$$
f(t)
=
\sum_{i=0}^{k-1} \frac{\mu e^{-\mu t} (\mu t)^i}{i!}
-
\sum_{i=0}^{k-1} \frac{e^{-\mu t} (\mu t)^{i-1}}{(i-1)!}
$$</span></p>
<p>But here I am stuck. I know that</p>
<p><span class="math-container">$$
f(t) = \frac{\mu e^{-\mu t} (\mu t)^{k-1}}{(k-1)!}
$$</span></p>
<p>but am not sure how to get there.</p>
| Peter Szilas | 408,605 | <p>Set <span class="math-container">$A=$</span>{<span class="math-container">$x \in \mathbb{R} | x^2 \le 9$</span> and <span class="math-container">$x^2>4$</span>}.</p>
<p>1) <span class="math-container">$x^2 \le 9$</span> , i.e. <span class="math-container">$-3 \le x \le 3$</span>.</p>
<p>2) <span class="math-container">$x^2 >4$</span>, i.e. <span class="math-container">$x <-2$</span> or <span class="math-container">$x >2$</span>.</p>
<p>1) and 2) : <span class="math-container">$x \in [-3,-2)$</span> or <span class="math-container">$x \in (2,3]$</span>.</p>
<p>Rewriting:</p>
<p><span class="math-container">$A=$</span>{<span class="math-container">$x| x \in [-3,-2)$</span> or <span class="math-container">$x \in (2,3]$</span>}$.</p>
<p><span class="math-container">$B=(2,3]$</span>, definition.</p>
<p>Hence <span class="math-container">$B \subset A$</span> (Why proper?) </p>
|
241,210 | <p>I am confused with the concept of topology base. Which are the properties a base has to have?</p>
<p>Having the next two examples for $X=\{a,b,c\}$:</p>
<p>1) $(X,\mathcal{T})$ is a topological space where $\mathcal{T}=\{\emptyset,X,\{a\},\{b\},\{a,b\}\}$. Which is the general procedure to follow in order to get a base for $(X,\mathcal{T})$? Can $\mathcal{B}=\{\{a\},\{b\}\}$ be a base?</p>
<p>2) Having just $X$ and no topology $\mathcal{T}$ defined for $X$, is $\mathcal{A}=\{X,\{a\},\{c\}\}$ a base? What is the topology that it generates? </p>
<p>Thank you very much.</p>
| Rudy the Reindeer | 5,798 | <p>A base is a collection $B$ of sets such that every set in the topology can be written as a union of sets in $B$.</p>
<p>1) Can you write every set in $T$ as a union of $\{a\}$ and/or $\{b\}$?</p>
<p>2) Produce all the unions then you'll see what the topology is.</p>
|
417,356 | <p>I'd like to show that the function,</p>
<p>$$f(x) = \begin{cases} 0 & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$$</p>
<p>is discontinuous via using the sequence-limit-criterion:</p>
<p>And I'd like to show:</p>
<p>"There is a sequence $(a_n)$ that contains only rational numbers and converges to $x_0$. Now $f(a_n) = 1$ but $f(a_n)−>f(x_0)=0$ is not possible$</p>
<p>I wonder which $" (a_n) "$ this could be. Which sequence can be chosen?</p>
| Tim | 74,128 | <p>Hint:</p>
<p>Let $k_n$ be the largest integer such that $k_n^2<n^2$. Then $\dfrac {k_n}n\to\sqrt 2$ as $n\to\infty$.</p>
|
417,356 | <p>I'd like to show that the function,</p>
<p>$$f(x) = \begin{cases} 0 & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$$</p>
<p>is discontinuous via using the sequence-limit-criterion:</p>
<p>And I'd like to show:</p>
<p>"There is a sequence $(a_n)$ that contains only rational numbers and converges to $x_0$. Now $f(a_n) = 1$ but $f(a_n)−>f(x_0)=0$ is not possible$</p>
<p>I wonder which $" (a_n) "$ this could be. Which sequence can be chosen?</p>
| DonAntonio | 31,254 | <p>Choose for example</p>
<p>$$x_0:=3\;,\;x_1:=3.1\;,\;....,x_n=3.141592...a_n\;,\;a_n:=\text{the n-th decimal digit of }\;\;\pi$$</p>
<p>Now, the above isn't very nice since one usually does <em>not</em> know what that $\,a_n\,$ is, in particular for big values of $\,n\,$, yet it is possible to calculate as many digits of $\,\pi\,$ by means fo power series, say.</p>
<p><strong>Added:</strong> An easier and nicer example can be given the other way around: choose the irrational sequence</p>
<p>$$\left\{\;a_n:=\frac{\pi n+1}{\pi n}\;\right\}\xrightarrow[n\to\infty]{}1$$</p>
|
417,356 | <p>I'd like to show that the function,</p>
<p>$$f(x) = \begin{cases} 0 & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$$</p>
<p>is discontinuous via using the sequence-limit-criterion:</p>
<p>And I'd like to show:</p>
<p>"There is a sequence $(a_n)$ that contains only rational numbers and converges to $x_0$. Now $f(a_n) = 1$ but $f(a_n)−>f(x_0)=0$ is not possible$</p>
<p>I wonder which $" (a_n) "$ this could be. Which sequence can be chosen?</p>
| Hagen von Eitzen | 39,174 | <p>Let $a_n=\frac 1n\lfloor nx_0\rfloor$ (where $\lfloor x\rfloor$ denotes "greatest integer $\le x$"). Then $\mathbb Q\ni a_n\to x_0$. If $x_0$ is irrational, this shows that $f$ is not continuous at $x_0$.
If $x_0$ is rational, consider $a_n=\frac 1n\lfloor n(x_0-\sqrt 2)\rfloor+\sqrt 2$.</p>
|
1,557,039 | <h2>Background</h2>
<p>I am a software engineer and I have been picking up combinatorics as I go along. I am going through a combinatorics book for self study and this chapter is absolutely destroying me. Sadly, I confess it makes little sense to me. I don't care if I look stupid, I want to understand how to solve these problems. </p>
<p>I am studying counting with repetition. That is, generalized binomial coefficients and generating functions. </p>
<h2>Problem</h2>
<p>Suppose that an unlimited amount of jelly beans is available in each of the five different colors: red, green, yellow, white, and black. </p>
<ul>
<li>How many ways are there to select from twenty jelly beans? </li>
<li>How many ways are there to select twenty jelly beans if we must select at least two jelly beans of each color? </li>
</ul>
<h2>Attempted Solution</h2>
<ul>
<li><p>How many ways are there to select from twenty jelly beans?</p>
<p>$5^{20}$</p></li>
<li><p>How many ways are there to select twenty jelly beans if we must select at least two jelly beans of each color? </p></li>
</ul>
<p>I keep thinking of applying the hypergeometric distribution here, but I think this is dead wrong. The entire chapter is on series, so I am confused as to what these series are and why they are being applied to solve these problems? The above solutions (some I am too embarrassed to share) didn't pass the smell test at all :( </p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: we have $3^{2(n+1)}=3^{2n}\cdot 3^2$</p>
|
1,583,747 | <p>I just started a course on queue theory, yet equations are given for granted without any demonstrations, which is very frustrating... Thus</p>
<ol>
<li>Why is the mean number of people in a queue system following an $M/M/1$ system</li>
</ol>
<p>$$E(L)=\frac{\rho}{1-\rho}$$</p>
<p>with $\rho=\frac{\lambda}{\mu}$ with $\lambda$ the clients arrival rate and $\mu$ the service rate.</p>
<ol start="2">
<li>And the mean number of people waiting in the queue:</li>
</ol>
<p>$$E(L^q)=\frac{\rho^2}{1-\rho}$$</p>
| Jan van der Vegt | 298,403 | <p>The $M/M/1$ queue can be modelled as a so called birth/death process. Since the arrivals are a Poisson process (the interarrival times are exponential) and the departures (service times) are exponentially distributed this will lead to some nice properties.</p>
<p><strong>The birth/death process</strong></p>
<p>If we call $X$ the state of the system, with the corresponding value being the number of customers in the system, we can start working towards long-term probabiltiies of the system being in a specific state. Once we have this we can use this to determine all kinds of metrics like $\mathbb{E}(L)$ and $\mathbb{E}(L^q)$.</p>
<p>The idea is that customers arrive and leave one at a time, this means the system can only change by one. Once this is established we can use the fact that between every two neighbour states the rate in has to be the same as the rate out. If we name the steady-state probabilities $\pi_i$ where $i$ is the number of customers, $\pi_i$ is the expected fraction of time the system is in state $i$. Using the rate in/rate out theorem we can get to a system of equations:</p>
<p>$\lambda\pi_0 = \mu\pi_1$</p>
<p>$\lambda\pi_1 = \mu\pi_2$</p>
<p>...</p>
<p>Rewriting this gives us:</p>
<p>$\pi_1 = \frac{\lambda}{\mu}\pi_0 = \rho\pi_0$</p>
<p>$\pi_2 = \frac{\lambda}{\mu}\pi_1 = \frac{\lambda^2}{\mu^2}\pi_0 = \rho^2\pi_0$</p>
<p>...</p>
<p>Now we have all the steady-state probabilities in terms of $\pi_0$, we also know that these have to add up to 1 since they are probabilities. This means:</p>
<p>$\pi_0 = \{\sum_{i=0}^\infty\rho^i\}^{-1}$</p>
<p>You can find a picture of this kind of system at <a href="https://en.wikipedia.org/wiki/Birth%E2%80%93death_process" rel="nofollow">Wikipedia</a> (I don't know if I can post that here)</p>
<p>With these probabilities you can calculate all kind of expected values.</p>
|
1,278,442 | <p>I was wondering if the singular homology functor preserve injectivity and surjectivity? I've been trying to figure out a proof or counterexample for ages now and I just can't.</p>
<p>This came up when I was looking at the reduced homology $H_p(S^{n},S^{n-1})$. To calculate it, I have looked at the canonical injection $$\iota: S^{n-1} \longrightarrow S^{n}.$$ I'm viewing $S^{n-1}$ as the equator of $S^n$, in this case does the functor preserve injectivity? That is, I want to say that $\iota_*$ is injective. Is this true? Thanksarinos</p>
| Alex Fok | 223,498 | <p>Your example already answers your question. $\iota_*: H_*(S^{n-1})\to H_*(S^n)$ is not injective. Also, $\pi: S^2\to\mathbb{RP}^2$ is surjective but the induced map on homology is not.</p>
|
3,424,720 | <p>I want to calculate the above limit. Using sage math, I already know that the solution is going to be <span class="math-container">$-\sin(\alpha)$</span>, however, I fail to see how to get to this conclusion.</p>
<h2>My ideas</h2>
<p>I've tried transforming the term in such a way that the limit is easier to find:
<span class="math-container">\begin{align}
\frac{\cos(\alpha + x) - \cos(\alpha)}{x}
&= \frac{\cos(x)\cos(\alpha)-\sin(x)\sin(\alpha)-\cos(\alpha)}{x} & (1) \\
&= \frac{\cos(x)\cos(\alpha)-\cos(x)}{x} - \frac{\sin(x)\sin(\alpha)}{x} & (2) \\
&= \frac{\cos(\alpha)(\cos(x)-1)}{x} - \sin(\alpha) & (3) \\
\end{align}</span></p>
<p>However, I'm not sure whether term <span class="math-container">$(3)$</span> is enough to solve the problem. Surely, for <span class="math-container">$x \to 0$</span>, this evaluates to <span class="math-container">$\frac{0}{0} - \sin(\alpha)$</span>, which is not enough to determine the limit.</p>
<p>Another try was to transform the term in such a way that I can use <span class="math-container">$\lim_{x \to 0} \frac{\sin x}{x} = 1$</span>. For instance, I found that
<span class="math-container">$$
\frac{\cos(\alpha + x) - \cos(\alpha)}{x} = \frac{\sin(\alpha - 90^° + x) - \sin(\alpha-90^°)}{x} \qquad (4)
$$</span></p>
<p>However, this only seems to lead to more complicated term manipulations that I did not manage to bring to a useful point.</p>
<p>What can I do to solve this problem?</p>
| Vasili | 469,083 | <p>Use that <span class="math-container">$\cos(\alpha + x) - \cos(\alpha)=-2\sin(\alpha+\frac{x}{2})\sin(\frac{x}{2})$</span> to get <span class="math-container">$$\lim_{x \rightarrow 0} -\sin(\alpha+\frac{x}{2})\frac{\sin\frac{x}{2}}{\frac{x}{2}}$$</span></p>
|
2,113,387 | <p><strong>An Ordered Primitive Pythagorean Triple</strong> $(a,b,c)$ is one in which $a \le b \le c$ are coprime and $a^2+b^2 = c^2$.</p>
<p>$f(n) = |\{(a,b,c)~|~ a^2+b^2=c^2,a\le b\le c,~c \le n\}|$.</p>
<p>Function $f(n)$ defines the number of all distinct Ordered Primitive Pythagorean Triple $(a,b,c)$ with $c \le n$. For example, </p>
<p>$f(4) = 0$ </p>
<p>$f(5) = 1$ with triple $(3,4,5)$</p>
<p><strong>Conjecture</strong>: for any $\epsilon > 0$, there exists $N_0$ such that
$\forall n \ge N_0, \frac{n}{f(n)} \in (2 \pi - \epsilon, 2 \pi + \epsilon)$.</p>
<p><strong>Question</strong>:I'd like to ask is there anyone has proposed such conjecture or is there any proof (true or false) for this conjecture?</p>
<p>Here are the first several triples</p>
<p>n f(n) n/f(n)</p>
<p>5 1 5</p>
<p>13 2 6.5</p>
<p>17 3 5.66666666666667</p>
<p>25 4 6.25</p>
<p>[3, 4, 5]</p>
<p>[5, 12, 13]</p>
<p>[8, 15, 17]</p>
<p>[7, 24, 25]</p>
| Michael Lugo | 173 | <p>Wolfram MathWorld, in its <a href="http://mathworld.wolfram.com/PythagoreanTriple.html" rel="nofollow noreferrer">article on Pythagorean triples,</a> says,</p>
<blockquote>
<p>Lehmer (1900) proved that the number of primitive solutions with
hypotenuse less than N satisfies</p>
<p>$$\lim_{n \to \infty} {\Delta_p(N) \over N} = {1 \over 2\pi} = 0.1591549\cdots$$</p>
</blockquote>
<p>and taking the reciprocal gives your result. This paper can be found on <a href="https://books.google.com/books?id=GSBLAAAAYAAJ&pg=PA303&lpg=PA303&dq=Asymptotic%20Evaluation%20of%20Certain%20Totient%20Sums&source=bl&ots=V7nm9Na7YK&sig=20vftnJRVnjIl2IqJonXQK_OqO4&hl=en&sa=X&ved=0ahUKEwjVr6uxx93RAhUNxmMKHYb9DZwQ6AEISDAJ#v=onepage&q=Asymptotic%20Evaluation%20of%20Certain%20Totient%20Sums&f=false" rel="nofollow noreferrer">Google Books</a>, and your conjecture is given in the discussion on pp. 327-328. The proof seems to rest on a lot of theoretical apparatus, though; I don't know of a "simple" proof.</p>
|
2,113,387 | <p><strong>An Ordered Primitive Pythagorean Triple</strong> $(a,b,c)$ is one in which $a \le b \le c$ are coprime and $a^2+b^2 = c^2$.</p>
<p>$f(n) = |\{(a,b,c)~|~ a^2+b^2=c^2,a\le b\le c,~c \le n\}|$.</p>
<p>Function $f(n)$ defines the number of all distinct Ordered Primitive Pythagorean Triple $(a,b,c)$ with $c \le n$. For example, </p>
<p>$f(4) = 0$ </p>
<p>$f(5) = 1$ with triple $(3,4,5)$</p>
<p><strong>Conjecture</strong>: for any $\epsilon > 0$, there exists $N_0$ such that
$\forall n \ge N_0, \frac{n}{f(n)} \in (2 \pi - \epsilon, 2 \pi + \epsilon)$.</p>
<p><strong>Question</strong>:I'd like to ask is there anyone has proposed such conjecture or is there any proof (true or false) for this conjecture?</p>
<p>Here are the first several triples</p>
<p>n f(n) n/f(n)</p>
<p>5 1 5</p>
<p>13 2 6.5</p>
<p>17 3 5.66666666666667</p>
<p>25 4 6.25</p>
<p>[3, 4, 5]</p>
<p>[5, 12, 13]</p>
<p>[8, 15, 17]</p>
<p>[7, 24, 25]</p>
| Thomas Andrews | 7,933 | <p>A loose heuristic argument uses the sort-of-result:</p>
<blockquote>
<p>Picking two random integer, the probability that they are relatively prime is $\frac{6}{\pi^2}$</p>
</blockquote>
<p>This is a "sort-of" result because we are actually talking about density, not probability.</p>
<p>Every primitive triple can be written $u^2-v^2,2uv,u^2+v^2$ with $\gcd(u,v)=1$, $u-v$ is odd.</p>
<p>So $f(n)$ can be seen as the number of pairs $v<u$ with $u^2+v^2\leq n$ and $\gcd(u,v)$ and $u-v$ is odd.</p>
<p>But then $g(n)$, the number of pairs $u^2+v^2\leq n$ is roughly $\frac{\pi n}{4}$, since it is the number of lattice points in a quadrant of a circle of radius $\sqrt{n}$. Of those, approximately $\frac{6}{\pi^2} g(n)$ have $\gcd(u,v)=1$, and of those, about $\frac{2}{3}$ have $u-v$ odd (because we've already excluded the cases where $u,v$ are both even.) And half of those have $v<u$.</p>
<p>So, altogether, you get: $$f(n)\approx \frac{1}{2}\frac{2}{3}\frac{6}{\pi^2}\frac{\pi n}{4}=\frac{1}{2\pi} n$$</p>
<p>Making this all rigorous would require an effort to get bounds in the various estimates, but this is the fundamentals of why this $\frac{n}{f(n)}$ converges to $2\pi$ - these estimates turn out to be "good enough."</p>
|
3,237,242 | <p>I have the following problem:</p>
<p>I need to prove that given the following integral</p>
<p><span class="math-container">$\int_{0}^{1}{c(k,l)x^k(1-x)^l}dx = 1$</span>,</p>
<p>we the constant <span class="math-container">$c(k,l) = (k+l+1) {{k+l}\choose{k}} = \frac{(k+l+1)!}{k!l!}$</span>,</p>
<p>with the use of two dimensional mathematical induction on <span class="math-container">$min(k,l)$</span>.
Here <span class="math-container">$k$</span> and <span class="math-container">$l$</span> are two nonnegative integers.</p>
<p>(THUS: I need to proof that <span class="math-container">$c(k,l)$</span> is equal to <span class="math-container">$(k+l+1) {{k+l}\choose{k}}$</span>)</p>
<p>For the base step I have proved that <span class="math-container">$c(k, 0) = c(0, k) = k + 1$</span> for all <span class="math-container">$k$</span>.</p>
<p>I am given a hint that for the induction step I could try using integration by parts to show <span class="math-container">$c(k,l) = \frac{k+1}{l} c(k+1,l−1)$</span>.</p>
<p>By integrating the following by parts I indeed managed to show the latter:</p>
<p><span class="math-container">$\int_{0}^{1}{c(k+1,l-1)x^{k+1}(1-x)^{l-1}dx}=1$</span>.</p>
<p>However, I don't really see how this helps me to complete my proof, since I don't really get the idea of two dimensional induction.</p>
<p>Can someone maybe clarify this a bit for me, and help me further with my proof?</p>
| Seewoo Lee | 350,772 | <p>In some sense, I think we don't need two-dimensional induction here. Assume that you showed the recurrence relation
<span class="math-container">$$
c(k, l) = \frac{k+1}{l}c(k+1, l-1)
$$</span>
by using the integration by parts. Also, we know base case: <span class="math-container">$c(k, 0) = c(0, k) = k+1$</span>. Then
<span class="math-container">$$
c(k, l) = \frac{k+1}{l} c(k+1, l-1) = \frac{k+1}{l} \frac{k+2}{l-1} c(k+2, l-2) \\= \cdots = \frac{(k+1)(k+2)\cdots(k+l)}{l(l-1)\cdots1}c(k+l, 0)= \frac{(k+1)(k+2)\cdots (k+l+1)}{l(l-1)\cdots 2\cdot 1} = \frac{(k+l+1)!}{k!l!}
$$</span></p>
<hr>
<p>If you want to use induction, you can write in this way: First, we know <span class="math-container">$c(k, 0) = k+1$</span> for all <span class="math-container">$k$</span>. (This is a base step)
Assume that <span class="math-container">$c(k, l) =\frac{(k+l+1)!}{k!l!}$</span> is true for all <span class="math-container">$k$</span> and <span class="math-container">$l \leq L$</span>. For <span class="math-container">$l = L+1$</span>, we have
<span class="math-container">$$
c(k, L+1) = \frac{k+1}{L+1} c(k+1, L) = \frac{k+1}{L+1} \frac{(k+L+2)!}{(k+1)!L!} = \frac{(k+L+2)!}{k!(L+1)!}
$$</span>
so it is also true for <span class="math-container">$l = L+1$</span>. </p>
|
109,961 | <p>Suppose $x^2\equiv x\pmod p$ where $p$ is a prime, then is it generally true that $x^2\equiv x\pmod {p^n}$ for any natural number $n$? And are they the only solutions?</p>
| Zev Chonoles | 264 | <p>No, it is not generally true. For example,
$$6^2=36\equiv 6\bmod 5$$
but
$$6^2=36\not\equiv 6\bmod 25.$$</p>
<hr>
<p>Suppose we are given an integer $x$. Note that, for any prime power $p^n$,
$$x^2\equiv x\pmod {p^n}\iff p^n\mid x(x-1)\iff p^n\mid x\;\; \text{ or }\;\;p^n\mid (x-1)$$
because $x$ and $x-1$ are relatively prime. Therefore,
$$x^2\equiv x\pmod {p^n}$$
is true if and only if $$n\leq\max\{v_p(x),v_p(x-1)\}$$
where $v_p(d)$ denotes the number of times $p$ goes into $d$. </p>
|
621,109 | <p>I need to find all the numbers that are coprime to a given $N$ and less than $N$.
Note that $N$ can be as large as $10^9.$ For example, numbers coprime to $5$ are $1,2,3,4$.</p>
<p>I want an efficient algorithm to do it. Can anyone help? </p>
| user44197 | 117,158 | <p>I assume you know how to implement sets. We will only add numbers $<N$ to the set so any bit representation is okay.</p>
<p>If you know that the primes dividing $N$ then the following works:</p>
<pre><code>S={} // Empty sets
for p in primes dividing $N$
add p,2p,3p... to S
end for
</code></pre>
<p>If you do not know the factors of $N$ then here is one not very efficient (but not terribly inefficient) way to do it.</p>
<pre><code>S={} // Empty set
for k=2 to N
if k is not in the set S then
if gcd(k,N) > 1 then
Add the following to S: k, 2k, 3k, 4k
end if
end if
end for
</code></pre>
<p>Now all the elements in $S$ are <em>not</em> co prime to N, so its complement is the numbers you want</p>
|
4,222,110 | <p>I am following course on topology that is kind of lack luster (not made for mathematicians). The course starts off with predicate logic and axiomatic set theory (ZFC). Now, I reached a point where the author defined the partition of unity and used the set of all continuous functions between 2 sets. But at the starts of the course, we learned about the principle of restricted comperhension, which requires us to state a set, say <span class="math-container">$D$</span>, in order for <span class="math-container">$\{\phi\in D|\phi:M\rightarrow N\,continuous\}$</span> to be a set. So, my question is: in what set are the maps from <span class="math-container">$M$</span> to <span class="math-container">$N$</span> found?</p>
| Spectre | 799,646 | <p>Well, as per your request, I'd add an answer so the comments section is clear (sorry again for the inconvenience).</p>
<p>Notice that you can rearrange the equation as <span class="math-container">$p = (y + 3x)(y-3x)$</span>.
The least bound for <span class="math-container">$p$</span> is <span class="math-container">$10^9$</span> as per your question, so I'd always suggest starting an exhaustive search (due to my being a rookie/ newbie at such big problems) from there so that you can get the least possible <span class="math-container">$p, x$</span> and <span class="math-container">$y$</span> that satisfies your equation. From there, tactically plan how you can manipulate the value of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> and still obtain values for <span class="math-container">$p$</span>. For example, you can add a <span class="math-container">$9a^2 \pm 6ax$</span> or <span class="math-container">$a^2 \pm 2ay$</span> (I'd suggest going for <span class="math-container">$9a^2 + 18ax$</span> or <span class="math-container">$a^2+ 2ay$</span> as long as you don't override a solution during the exhaustive search) to both sides of the equation and see the changes yourself.</p>
<p>If you're looking for multiple solutions for the same <span class="math-container">$p$</span>, you can exchange things around.</p>
<h4>Edit : OmG's answer is worth a special mention, please go check it above.</h4>
<h4>Edit: Another great contribution by Servaes; please go check below</h4>
|
4,222,110 | <p>I am following course on topology that is kind of lack luster (not made for mathematicians). The course starts off with predicate logic and axiomatic set theory (ZFC). Now, I reached a point where the author defined the partition of unity and used the set of all continuous functions between 2 sets. But at the starts of the course, we learned about the principle of restricted comperhension, which requires us to state a set, say <span class="math-container">$D$</span>, in order for <span class="math-container">$\{\phi\in D|\phi:M\rightarrow N\,continuous\}$</span> to be a set. So, my question is: in what set are the maps from <span class="math-container">$M$</span> to <span class="math-container">$N$</span> found?</p>
| Servaes | 30,382 | <p><strong>Solution:</strong> Compute the remainder of <span class="math-container">$p$</span> modulo <span class="math-container">$36$</span>.</p>
<ol>
<li>If <span class="math-container">$p\equiv2,3,5,6,8,10,11,12,14,15,17,18,20,21,22,23,24,26,29,30,32,33,34,35\pmod{36}$</span>, then there are no solutions.</li>
<li>If <span class="math-container">$p\equiv9,27\pmod{36}$</span> then for every factorization <span class="math-container">$p=9uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> take
<span class="math-container">$$x=\frac{u-v}{2}\qquad\text{ and }\qquad y=3\frac{u+v}{2}.$$</span></li>
<li>If <span class="math-container">$p\equiv0\pmod{36}$</span> then for every factorization <span class="math-container">$p=36uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> take
<span class="math-container">$$x=u-v\qquad\text{ and }\qquad y=3(u+v).$$</span></li>
<li>If <span class="math-container">$p\equiv1,7,13,19,25,31\pmod{36}$</span> then for every factorization <span class="math-container">$p=uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> take
<span class="math-container">$$x=\frac{u-v}{6}\qquad\text{ and }\qquad y=\frac{u+v}{2}.$$</span></li>
<li>If <span class="math-container">$p\equiv4,16,28\pmod{36}$</span>, all solutions <span class="math-container">$(x,y)$</span> are of the form <span class="math-container">$(2x',2y')$</span> where <span class="math-container">$(x',y')$</span> is a solution for <span class="math-container">$p'=\tfrac p4$</span>.</li>
</ol>
<p><strong>Explanation:</strong></p>
<p>Let <span class="math-container">$p$</span> be a large integer. We want to find all positive integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span> such that
<span class="math-container">$$9x^2+p=y^2.\tag{1}$$</span>
First note that there are no solutions if <span class="math-container">$p\equiv2\pmod{3}$</span> or <span class="math-container">$p\equiv2\pmod{4}$</span>.</p>
<p>If <span class="math-container">$p$</span> is divisible by <span class="math-container">$3$</span>, which is easy to check given the decimal representation of <span class="math-container">$p$</span>, then also <span class="math-container">$y$</span> is divisible by <span class="math-container">$3$</span>. Then <span class="math-container">$y=3z$</span> for some positive integer <span class="math-container">$z$</span> and so
<span class="math-container">$$p=y^2-9x^2=9z^2-9x^2=9(x^2-z^2),$$</span>
which shows that <span class="math-container">$p$</span> must be divisible by <span class="math-container">$9$</span>. That is to say, if <span class="math-container">$p\equiv3,6\pmod{9}$</span> then there are no solutions. Write <span class="math-container">$p=9q$</span> to see that
<span class="math-container">$$q=z^2-x^2=(z+x)(z-x),\tag{2}$$</span>
and so the integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span> in <span class="math-container">$(1)$</span> yield a factorization of <span class="math-container">$q$</span> into two factors that are congruent mod <span class="math-container">$2$</span>. This is of course impossible if <span class="math-container">$q\equiv2\pmod{4}$</span>, which shows that there are no solutions if <span class="math-container">$p\equiv6\pmod{12}$</span>. Otherwise, if <span class="math-container">$q$</span> is odd then for every factorization <span class="math-container">$q=uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> the pair
<span class="math-container">$$z=\frac{u+v}{2}\qquad\text{ and }\qquad x=\frac{u-v}{2},$$</span>
yields a solution to <span class="math-container">$(2)$</span>. Similarly, if <span class="math-container">$q\equiv0\pmod{4}$</span> then for every factorization <span class="math-container">$q=4uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> the pair
<span class="math-container">$$z=u+v\qquad\text{ and }\qquad x=u-v,$$</span>
yields a solution to <span class="math-container">$(2)$</span>. Either way we see that:</p>
<ol>
<li>If <span class="math-container">$p\equiv2\pmod{3}$</span> or <span class="math-container">$p\equiv3,6\pmod{9}$</span> or <span class="math-container">$p\equiv 6\pmod{12}$</span>, or equivalently <span class="math-container">$$p\equiv2,3,5,6,8,11,12,14,15,17,18,20,21,23,24,26,29,30,32,33,35\pmod{36},$$</span> then there are no solutions.</li>
<li>If <span class="math-container">$p\equiv0,9,27\pmod{36}$</span> then finding all solutions is equivalent to factoring <span class="math-container">$p$</span>.</li>
</ol>
<p>On the other hand, if <span class="math-container">$p\equiv1\pmod{3}$</span> and <span class="math-container">$p\not\equiv2\pmod{4}$</span> then <span class="math-container">$p\equiv1,4,7\pmod{12}$</span>.</p>
<p>If <span class="math-container">$p\equiv4\pmod{12}$</span> then <span class="math-container">$p\equiv4\pmod{6}$</span> and from
<span class="math-container">$$p=y^2-9x^2=(y+3x)(y-3x),$$</span>
we get a factorization of <span class="math-container">$p$</span> into two factors that are congruent modulo <span class="math-container">$6$</span>. It follows that
<span class="math-container">$$y+3x,y-3x\equiv2\pmod{6},$$</span>
and so both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are even, say <span class="math-container">$x=2x'$</span> and <span class="math-container">$y=2y'$</span>. Then <span class="math-container">$p$</span> is divisible by <span class="math-container">$4$</span>, say <span class="math-container">$p=4p'$</span>, and we find that
<span class="math-container">$$9(x')^2+p'=(y')^2,$$</span>
where of course <span class="math-container">$p'\equiv1\pmod3$</span> because <span class="math-container">$p\equiv4\pmod{12}$</span>. This shows that every solution <span class="math-container">$(x,y)$</span> for <span class="math-container">$p$</span> is of the form <span class="math-container">$(2x',2y')$</span> where <span class="math-container">$(x',y')$</span> is a solution for <span class="math-container">$p'=\tfrac p4$</span>.</p>
<p>What remains are the cases <span class="math-container">$p\equiv1,7\pmod{12}$</span>, or equivalently the case <span class="math-container">$p\equiv1\pmod{6}$</span>. If <span class="math-container">$p\equiv1\pmod{6}$</span> then for every factorization <span class="math-container">$p=uv$</span> into positive integers <span class="math-container">$u$</span> and <span class="math-container">$v$</span> with <span class="math-container">$u>v$</span> we have <span class="math-container">$u\equiv v\pmod{6}$</span>. Then setting
<span class="math-container">$$y=\frac{u+v}{2}\qquad\text{ and }\qquad x=\frac{u-v}{6},$$</span>
yields a solution to <span class="math-container">$(1)$</span>. Conversely, every solution <span class="math-container">$(x,y)$</span> to <span class="math-container">$(1)$</span> yields a factorization
<span class="math-container">$$p=(y+3x)(y-3x),$$</span>
into distinct factors that are congruent modulo <span class="math-container">$6$</span>. So in this case the problem is again equivalent to factoring <span class="math-container">$p$</span>.</p>
|
1,042,375 | <p><strong>Question:</strong></p>
<blockquote>
<p>Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Prove that $\{ x \in \mathbb{R} \mid f(x) > 0\}$ is an open subset of $\mathbb{R}$.</p>
</blockquote>
<p>At first I thought this was quite obvious, but then I came up with a counterexample. What if $f(x)=1$ for all $x$? Then the set becomes $\{ x \in \mathbb{R} \mid f(x) = 1\}$, a proper subset of the set in question, which is a closed subset.</p>
| Nishant | 100,692 | <p>You are asked to show that the preimage of $(0, \infty)$ is open. But since $f$ is continuous, the preimage of an open set is open; $(0, \infty)$ is open, therefore its preimage is also open.</p>
|
320,228 | <p>Today, in my lesson, I was introduced to partial derivatives. One of the things that confuses me is the notation. I hope that I am wrong and hope the community can contribute to my learning. In single-variable calculus, we know that, given a function $y =f(x)$, the derivative of $y$ is denoted as $\frac {dy}{dx}$. I understand this as the relative change in $y$, $\delta y$ given a small change in $x$, $\delta x$.</p>
<p>However, in today's lesson on partial derivative, my professor constantly used this notation. </p>
<p>Given a function $z = f(x,y)$, the first derivative with respected to $x$ is written as</p>
<p>$$ \frac{\partial z}{\partial x} $$</p>
<p>So, for example</p>
<p>$$
z = 5x+3y\\
\frac{\partial z}{\partial x} = 5
$$</p>
<p>Why can't I just write it as
$$
z = 5x+3y\\
\frac{d z}{d x} = 5
$$</p>
<p>Is it some convention or am I not understanding something in the notation?</p>
| Community | -1 | <p>A nicer notion is that of the differential:</p>
<p>$$ \text{If} \qquad z = 5x + 3y \qquad \text{then} \qquad dz = 5\, dx + 3\,dy $$</p>
<p>Then if you decide to hold $y$ constant, that makes $dy = 0$, and you have $dz = 5 \, dx$.</p>
<hr>
<p>Another notation that works well with function notation is that if we define</p>
<p>$$ f(x,y) = 5x + 3y$$</p>
<p>then $f_i$ means derivative of $f$ with respect to the $i$-th entry; that is</p>
<p>$$ f_1(x,y) = 5 \qquad \qquad f_2(x,y) = 3 $$</p>
<p>This doesn't work well with a common <em>abuse</em> of notation, though; sometimes people write $f(r,\theta)$ when they really mean "evaluate $f$ at the $(x,y)$ pair whose polar coordinates are $(r, \theta)$" rather than the 'correct' meaning of that expression "evaluate $f$ at $(r, \theta)$". So if you're in the habit of doing that, don't try to indicate derivatives by their position.</p>
<hr>
<p>I confess I really dislike partial derivative notation; when one writes $\partial/\partial x$, one "secretly" means that they intend to hold $y$ constant, then when one passes it through the differential, one gets</p>
<p>$$ \frac{\partial z}{\partial x} = 5 \frac{\partial x}{\partial x} + 3 \frac{\partial y}{\partial x} = 5 \cdot 1 + 3 \cdot 0 = 5$$</p>
<p>However, the suggestive form of Leibniz notation starts becoming very misleading at this point; for example, let's compute other partial derivatives.</p>
<ul>
<li>$\partial z / \partial x = 5$, holding $y$ constant as the notation suggests</li>
<li>$\partial x / \partial y = -3/5$, holding $z$ constant as the notation suggests</li>
<li>$\partial y / \partial z = 1/3$, holding $x$ constant as the notation suggests</li>
</ul>
<p>Then putting it together,</p>
<p>$$ \frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = 5 \cdot \left(-\frac{3}{5}\right) \cdot \frac{1}{3} = -1 $$</p>
<p>This is a big surprise if you expect partial derivatives to behave similarly to fractions as their notation suggests!!!</p>
|
3,927,845 | <p>n is a natural number. Prove <span class="math-container">$6^n \geq n3^n$</span> holds for every natural number.</p>
<p><span class="math-container">$n = 1:$</span></p>
<p><span class="math-container">$$6 \geq 3 $$</span></p>
<p><span class="math-container">$n \rightarrow n + 1:$</span></p>
<p><span class="math-container">$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$</span></p>
<p>Is my proof correct?</p>
<p>EDIT:
<span class="math-container">$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$</span></p>
| Raffaele | 83,382 | <p>Suppose <span class="math-container">$6^n\ge n\cdot 3^n$</span> is true for <span class="math-container">$n$</span> and let's prove it for <span class="math-container">$(n+1)$</span>.</p>
<p><span class="math-container">$6^{n+1}=6\cdot 6^n\ge 6\left(n\cdot 3^n\right)=2\cdot 3\left(n\cdot 3^n\right)=2 n\cdot 3^{n+1}\ge (n+1)3^{n+1}$</span></p>
<p>proved.</p>
|
59,965 | <p>If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?</p>
<p>Thanks.</p>
| anon | 11,763 | <p>If a multivariable function $f(x_1,\dots,x_n)$ is separable then the <a href="http://en.wikipedia.org/wiki/Hessian_matrix" rel="nofollow">Hessian</a> $H\ln f$ is diagonal, which isn't usually the case. Moreover, if $f$ decomposes into $g_1(x_1)\cdots g_n(x_n)$, then</p>
<p>$$\int \frac{\partial\ln f}{\partial x_i} dx_i=g_i(x_i)+C$$</p>
<p>where the integration is done in a single-variable sense.</p>
<hr>
<p>Can someone prove a $C^2$ function $f$ is always separable when $H\ln f$ is diagonal?</p>
|
59,965 | <p>If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?</p>
<p>Thanks.</p>
| Qiaochu Yuan | 232 | <p>It <strong>is</strong> true that linear combinations of such functions are dense. Depending on exactly what space of functions you're working with, this should follow from the locally compact form of the <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Locally_compact_version" rel="nofollow">Stone-Weierstrass theorem</a>. Thus it is not unreasonable to expect that we can find solutions to a linear PDE by taking the closure of the space of linear combinations of separable solutions. </p>
|
2,496,114 | <p>Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $</p>
<p>Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $</p>
<p><strong>Answer:</strong></p>
<p>$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \Rightarrow a^{2^{3-2}}=a^{2^1} \equiv 1 \pmod{2^3} $</p>
<p><strong>Am I right?</strong></p>
| Gabriel Romon | 66,096 | <p>Since $$ \frac{\sin(x)}{x} = \int_{0}^{1} \cos(tx) dt$$</p>
<p>with proper justifications (differentiation under the integral sign) you may derive </p>
<p>$$\left(\frac{\sin(x)}{x}\right)^{(n)} = \int_{0}^{1} t^n \cos(tx+n\pi/2) dt$$ which immediately yields the wanted estimate.</p>
|
1,173,002 | <p>Do I have to use the diagonalization of A?</p>
| Laars Helenius | 112,790 | <p>There are 49 $2\times 2$ boards.</p>
<p>A single domino can cover 6 $2\times 2$ boards.</p>
<p>Thus you need at least 9 dominoes.</p>
<p>On the other hand we can cover your squares claim with 11 dominoes using a greedy type algorithm. I can deploy a maximum of 6 dominoes that cover 6 $2\times 2$ squares per domino. Place dominoes at $(B7,C7)$, $(E7,F7)$, $(B5,C5)$, $(E5,F5)$, $(B3,C3)$, and $(E3,F3)$. The remaining $2\times 2$ all lie along the perimeter of the board and I can deploy 4 dominoes that cover 12 more $2\times 2$ squares. Place dominoes at $(B1,C1)$, $(E1,F1)$, $(H7,H6)$, and $(H4,H3)$. Leaving the final square to be covered with a domino at $(H1,H2)$.</p>
<p>UPDATE: In order to use 9 dominoes, 8 of the dominoes would have to be placed so that they could each cover 6 $2\times 2$ squares, with no overlap. This is impossible. Thus the minimum is at least 10 and I have found a way to do that, but the trick is to allow multiple dominoes cover some squares previously covered by other dominoes.</p>
|
449,296 | <p>I am trying to deduce how mathematicians decide on what axioms to use and why not other axioms, I mean surely there is an infinite amount of available axioms. What I am trying to get at is surely if mathematicians choose the axioms then we are inventing our own maths, surely that is what it is but as it has been practiced and built on for so long then it is too late to change all this so we have had to adapt and create new rules?? I'm not sure how clear what I am asking is or whether it is even understandable but would appreciate any answers or comments, thanks.</p>
| Carl Mummert | 630 | <p>The axioms of mathematics come, in many cases, from the mathematical objects that are being axiomatized.</p>
<p><strong>Example 1: Euclidean Geometry</strong> Many ancient cultures needed to work with geometric ideas in order to build things. Euclid famously wrote a system of axioms for plane geometry based on his existing intuition about the plane. Much later, in the 1800s, it was realized that Euclid's axioms were problematic (e.g. Euclid used <a href="https://en.wikipedia.org/wiki/Pasch%27s_axiom" rel="nofollow">Pasch's axiom</a> implicitly but his axioms cannot prove it). Hilbert developed a different axiomatic system for geometry that did not have the problems that they ten realized Euclid's axioms had.</p>
<p>So how did Hilbert come up with those axioms? He already understood plane geometry! The goal was to isolate some basic properties that would be able to develop the theory he wanted. </p>
<p><strong>Example 2: Topology</strong> According to the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Topology#History" rel="nofollow">topology</a>: </p>
<blockquote>
<p>Unifying the work on function spaces of Georg Cantor, Vito Volterra, Cesare Arzelà, Jacques Hadamard, Giulio Ascoli and others, Maurice Fréchet introduced the metric space in 1906. A metric space is now considered a special case of a general topological space. In 1914, Felix Hausdorff coined the term "topological space" and gave the definition for what is now called a Hausdorff space. Currently, a topological space is a slight generalization of Hausdorff spaces, given in 1922 by Kazimierz Kuratowski.</p>
</blockquote>
<p>Again, these researchers had many examples of mathematical objects at hand, and the goal was to isolate certain properties of those objects and write down axioms describing just those properties. </p>
<p>There are many more examples that can be given, including group theory, Zermelo-Fraenkel set theory, graph theory, matroid theory, computable functions, and more. None of these was invented "out of thin air" as a set of axioms - they each developed over time as more examples were considered. </p>
<p>One way that progress in mathematics happens is that we see a phenomenon that happens in a particular example, and we ask what properties of the example are needed for the phenomenon to happen. If the class of all objects that have those properties turns out to be interesting, then the properties may become axioms for a new class of objects. Studying those axioms may show us new phenomena, leading to new systems of axioms, and so on. </p>
|
238,970 | <p>Recently I've come across 'tetration' in my studies of math, and I've become intrigued how they can be evaluated when the "tetration number" is not whole. For those who do not know, tetrations are the next in sequence of iteration functions. (The first three being addition, multiplication, and exponentiation, while the proceeding iteration function is pentation) </p>
<p>As an example, 2 with a tetration number of 2 is equal to $$2^2$$ 3 with a tetration number of 3 is equal to $$3^{3^3}$$ and so forth.</p>
<p>My question is simply, or maybe not so simply, what is the value of a number "raised" to a fractional tetration number. What would the value of 3 with a tetration number of 4/3 be?</p>
<p>Thanks for anyone's insight</p>
| Simply Beautiful Art | 272,831 | <p>Disclaimer: My proofs of the following facts may be faulty, but numerical computations suggest otherwise.</p>
<p>The factorial can be uniquely extended by enforcing the condition that it grows at a certain asymptotic rate, namely that <span class="math-container">$(n+x)!\sim n^x\cdot n!$</span> as <span class="math-container">$n\to\infty$</span> by inducting on the fact that <span class="math-container">$(n+1)!\sim n\cdot n!$</span>.</p>
<p>Similarly, tetration can be uniquely extended by enforcing a condition on its asymptotic rate of growth when it converges monotonically. With the help of some calculus, one can show that for <span class="math-container">$1<a<e^{1/e}$</span> we have</p>
<p><span class="math-container">$$^{n+1}a-{}^\infty a\sim({}^na-{}^\infty a)\ln({}^\infty a)$$</span></p>
<p>and hence the natural condition is</p>
<p><span class="math-container">$${}^{n+x}a-{}^\infty a\sim({}^na-{}^\infty a)\ln({}^\infty a)^x$$</span></p>
<p>which, when rearranged in terms of <span class="math-container">${}^xa$</span> gives us the definition:</p>
<p><span class="math-container">$$^xa=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a+({}^na-{}^\infty a)\ln({}^\infty a)^x)\tag{$\star$}$$</span></p>
<p>where <span class="math-container">$\log^{\circ n}$</span> is the <span class="math-container">$n$</span>-fold logarithm e.g. <span class="math-container">$\log^{\circ2}(x)=\log(\log(x))$</span>.</p>
<p>A <a href="https://math.stackexchange.com/q/3490579/272831">chain of posts</a> discusses this definition, including proofs that this does in fact give a tetration, as it satisfies the basic properties:</p>
<ul>
<li><p><span class="math-container">$^0a=1$</span></p></li>
<li><p><span class="math-container">$^{x+1}a=a^{({}^xa)}$</span></p></li>
</ul>
<p>as well as having the additional advantages that it is unique, satisfies nice asymptotics, and is analytic. Due to the last property, it is possible to analytically continue this definition outside of the initial interval, though the computation is largely difficult for two main reasons:</p>
<ul>
<li><p>The definition itself requires large amounts of precision, since <span class="math-container">$^na-{}^\infty a$</span> involves large amounts of cancellation and loss of significant figures. This can be somewhat avoided, however, with the use of acceleration techniques such as Aitken's delta squared method.</p></li>
<li><p>Analytic continuation is a fairly difficult task to perform numerically, and I've yet to find an alternative representation of this tetration that converges further out. This can be somewhat circumvented using modern techniques, such as conformal mappings (e.g. expanding <span class="math-container">${}^x(a+t/(1-t))$</span> about <span class="math-container">$t=0$</span>), to allow for convergence even on the entire interval <span class="math-container">$[1,\infty)$</span>.</p></li>
</ul>
|
563,927 | <p>Show that $\mathbb{R}$
is not a simple extension of $\mathbb{Q}$
as follow:</p>
<p>a. $\mathbb{Q}$
is countable.</p>
<p>b. Any simple extension of a countable field is countable.</p>
<p>c. $\mathbb{R}$
is not countable.</p>
<p>I 've done a. and c. Can anyone help me a hint to prove b.?</p>
| DonAntonio | 31,254 | <p>To simplify things let us divide in two cases:</p>
<p><strong>The extension of $\;\Bbb Q\;$ is algebraic:</strong> Then we have $\;\Bbb Q(\alpha)/\Bbb Q\;$ of degree $\;n\;$ , and every element in the extension is of the form $\;a_0+a_1\alpha+\ldots+a_{n-1}\alpha^{n-1}\;,\;\;a_i\in\Bbb Q\;$ ...</p>
<p><strong>The extension of $\;\Bbb Q\;$ is transcendental:</strong> Then $\;\Bbb Q(\omega)/\Bbb Q\;$ isn't finite, but in fact $\;\Bbb Q(\omega)\;$ is the fractions field of $\;\Bbb Q[\omega]\cong\Bbb Q[x]=\;$the polynomial ring in one variable over the rationals, and then we're almost done since</p>
<p>$$\left|\Bbb Q(\omega)\right|:=\left|\left\{\frac ab\;;\;a,b\in\Bbb Q[\omega]\;,\;b\neq 0\right\}\right|\le\left|\Bbb Q[x]\times\Bbb Q[x]\right|=\aleph_0\cdot\aleph_0=\aleph_0$$</p>
|
1,072,639 | <p>A function is <b>bijective</b> if it is both <b>surjective</b> and <strong>injective</strong>. Is there a term for when a function is both <strong>not surjective</strong> and <strong>not injective</strong>?</p>
| Mister Benjamin Dover | 196,215 | <p>Of course not. This would be a useless notion (so there is no need for such terminology). Most functions are not surjective or injective (in the sense that if you take a "random" one, then you cannot expect it to be surjective or injective), mind you. We are mostly interested in the special class of mappings called injections and surjections, and for this reason the terminology "injective" and "surjetive" is useful.</p>
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4,057,255 | <p>Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.</p>
<p>Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)</p>
<p>1-(6<em>P(1 specific empty bin) – 15</em>P(2 specific empty bins) + 20<em>P(3 specific empty bins) – 15</em>P(4 specific empty bins) +6*P(5 specific empty bins)</p>
<p>This is what I have so far but I don't know how to determine the individual probabilities above.</p>
| Math Lover | 801,574 | <p>It can be written as,</p>
<p><span class="math-container">$\displaystyle \small 1 - 6 \cdot \bigg(\frac{5}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{4}{6}\bigg)^{10} - 20 \cdot \bigg(\frac{3}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{2}{6}\bigg)^{10} - 6 \cdot \bigg(\frac{1}{6}\bigg)^{10}$</span></p>
<p><span class="math-container">$\displaystyle \small \frac{5}{6}$</span> represents the probability that an item will land into one of the five specific bins out of six (i.e. a specific bin is empty). Similarly <span class="math-container">$\displaystyle \small \frac{4}{6}$</span> represents the probability that an item will land into one of the four specific bins and so on.</p>
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3,827,449 | <p>For example, <span class="math-container">$(1,1,1)$</span> is such a point. The sphere must contain all points that satisfy the condition.</p>
<p>So, I've been milling over this question on and off for the past few days and just can't seem to figure it out. I think I would have to use the distance formula from some point to (0,0,0) or (3,3,3) but can not wrap my head around how to set it up, or what point to use. Any help would be greatly appreciated.</p>
| JonathanZ supports MonicaC | 275,313 | <p>The standard way to do this is</p>
<ul>
<li><p>let <span class="math-container">$(x, y, z)$</span> be an arbitrary point on your surface,</p>
</li>
<li><p>using your geometry toolkit write the formulas for the various distances you're interested in,</p>
</li>
<li><p>set the appropriate quantities equal to each other,</p>
</li>
<li><p>and finally clean up the algebra by possibly squaring thing and collecting like terms.</p>
</li>
</ul>
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3,359,436 | <p>So I remember as a child when I was taught: <span class="math-container">$ . \bar9 =1 $</span>
The proof was taught as:</p>
<p><span class="math-container">$$x = 0.\bar{9} \\
10x = 9.\bar{9} \\
10x - x = 9.\bar{9} - 0.\bar{9} \\
9x = 9 \\
x = 1 \\
\therefore 0.\bar{9} = 1$$</span></p>
<p>I was found the whole thing quite counter-intuitive and created my own "analog proof" of why it "must" be an absurd statement:</p>
<p>Let us "assume" <span class="math-container">$\bar 9$</span> can exist the same way we assumed <span class="math-container">$. \bar 9$</span> can exist.</p>
<p><span class="math-container">$$x = \bar{9} \\
\frac{x}{10} = \bar9.{9} \\
x - \frac{x}{10} = \bar{9} - \bar{9}.9 \\
.9 x = -.9 \\
x = -1 \\
\therefore \bar{9} = - 1$$</span></p>
<p>Hence, by the same set of logic if <span class="math-container">$. \bar 9 = 1$</span> then <span class="math-container">$\bar 9 = -1$</span>. I remember the maths teacher being really frustrated with me because of these kind of "stunts." I kind of sympathize with him that it would be really difficult to explain to a child without using "radius of convergence", etc. </p>
<h2>Question</h2>
<p>Is it possible to make sense to the "child version" of me without using the words "radius of convergence"?</p>
| 5xum | 112,884 | <p>A sum of logs is easily simplified as the log of the product, i.e.</p>
<p><span class="math-container">$$\sum_{i=1}^n \log a_n = \log\left(\prod_{i=1}^n a_n\right).$$</span></p>
<p>There is no such conversion for the log of a sum.</p>
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3,359,436 | <p>So I remember as a child when I was taught: <span class="math-container">$ . \bar9 =1 $</span>
The proof was taught as:</p>
<p><span class="math-container">$$x = 0.\bar{9} \\
10x = 9.\bar{9} \\
10x - x = 9.\bar{9} - 0.\bar{9} \\
9x = 9 \\
x = 1 \\
\therefore 0.\bar{9} = 1$$</span></p>
<p>I was found the whole thing quite counter-intuitive and created my own "analog proof" of why it "must" be an absurd statement:</p>
<p>Let us "assume" <span class="math-container">$\bar 9$</span> can exist the same way we assumed <span class="math-container">$. \bar 9$</span> can exist.</p>
<p><span class="math-container">$$x = \bar{9} \\
\frac{x}{10} = \bar9.{9} \\
x - \frac{x}{10} = \bar{9} - \bar{9}.9 \\
.9 x = -.9 \\
x = -1 \\
\therefore \bar{9} = - 1$$</span></p>
<p>Hence, by the same set of logic if <span class="math-container">$. \bar 9 = 1$</span> then <span class="math-container">$\bar 9 = -1$</span>. I remember the maths teacher being really frustrated with me because of these kind of "stunts." I kind of sympathize with him that it would be really difficult to explain to a child without using "radius of convergence", etc. </p>
<h2>Question</h2>
<p>Is it possible to make sense to the "child version" of me without using the words "radius of convergence"?</p>
| Henry | 6,460 | <p>Logarithms are strictly increasing functions and assuming that the logarithm is a real number, i.e. the sum is positive (or at least non-negative, as it should be if this is a sum of probabilities or densities or likelihoods or similar), </p>
<p>then the maximum of the logarithm of a sum is equal to the logarithm of the maximum of the sum and occurs for the same value of the variables </p>
<p>So <span class="math-container">$$\max_\theta {\log{\sum_z{p(x_i, z)}}} = \log \left(\max_\theta {{\sum_z{p(x_i, z)}}}\right)$$</span> is as easy to find as finding the maximum of <span class="math-container">$\sum\limits_z{p(x_i, z)}$</span> </p>
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2,995,495 | <p>I'm trying to prove that, for every <span class="math-container">$x \geq 1$</span>:</p>
<p><span class="math-container">$$\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}.$$</span> </p>
<p>I could do it graphically on <span class="math-container">$\Bbb R$</span>, but how to make a formal algebraic proof?</p>
| user | 505,767 | <p>We have that for <span class="math-container">$x\ge 1$</span></p>
<p><span class="math-container">$$\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\le 0 \implies \left|\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\right| =\frac{π}{4}+\frac{(x-1)}{2}-\arctan x$$</span></p>
<p>then consider</p>
<p><span class="math-container">$$f(x)=\frac{(x-1)^2}{2}+\arctan x-\frac{π}{4}-\frac{(x-1)}{2}\implies f'(x)=\frac1{x^2+1}+x-\frac32\ge 0$$</span></p>
<p>and since <span class="math-container">$f(1)=0$</span> we have that <span class="math-container">$f(x)\ge 0$</span> and the inequality is proved.</p>
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