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 |
|---|---|---|---|---|
1,071,866 | <p>I am confused by the statement of Sylow's Fourth Theorem:<br>
Let $G$ be a finite group, $p$ a prime. The Sylow $p$-subgroups of $G$ form a single conjugacy class of subgroups.<br>
In particular, I do not understand what it means for the subgroups to form a single conjugacy class?
Thanks!</p>
| Kez | 201,782 | <p>Given a group $G$, you can define an equivalence relation $\sim$ on the set of all subgroups of $G$, where if $H$ and $K$ are subgroups of $G$,
$$H\sim K\iff \text{there exists some }g\in G\text{ such that }K=gHg^{-1}$$
In other words, $H\sim K$ if and only if $K$ is a conjugate of $H$.</p>
<p>Given a subgroup $H$ of $G$, the conjugacy class of $H$ is just the equivalence class of $H$, i.e.,
$$\{\text{subgroups }K\text{ of }G\mid H\sim K\}$$
So to say that the Sylow $p$-subgroups of $G$ form a single conjugacy class of subgroups of $G$ just means that there is some $H$ such that
$$\{\text{Sylow $p$-subgroups of $G$}\}=\{\text{subgroups }K\text{ of }G\mid H\sim K\}$$
It is a general fact/definition that</p>
<ul>
<li>everything equivalent to a given element of an equivalence class is again an element of the equivalence class</li>
<li>every element of an equivalence class is equivalent to every other element.</li>
</ul>
<p>Thus, every Sylow $p$-subgroup of $G$ is conjugate to every other Sylow $p$-subgroup of $G$, and moreover every conjugate of a given Sylow $p$-subgroup is another Sylow $p$-subgroup.</p>
|
2,273,506 | <p>I was able to simplify a boolean expression $$(\neg a*\neg b*c)+(a*\neg b*\neg c)+(a*b*\neg c)+(a*b*c)$$into the form $$\neg b*(a\oplus c)+a*b$$ where $*$ is the logical and, $+$ is the logical or, and $\oplus$ is the logical XOR.</p>
<p>Apparently, from Wolfram Alpha, this expression can be simplified to $$\left(a\oplus c\right)\oplus(b*c)$$</p>
<p>I would like to know how Wolfram Alpha was able to simplify this.</p>
| Fabio Somenzi | 123,852 | <p>I'm not privy to the internals of Wolfram Alpha, but here's a way to derive the expression it produces.</p>
<p>First of all, it's not unreasonable to assume that if the input contains an exclusive OR, WA may decide that the user wants to see exclusive ORs in the output, or is at least OK with them.</p>
<p>Let $f = \neg b * (a \oplus c) + (a * b)$. Applying <a href="https://en.wikipedia.org/wiki/Reed%E2%80%93Muller_expansion" rel="nofollow noreferrer">Reed-Muller expansion</a> with respect to $b$, we get cofactors $f_b = a$, $f_{\neg b} = a \oplus c$, Boolean difference $\frac{\partial f}{\partial b} = c$, and hence</p>
<p>$$ f = f_{\neg b} \oplus b * \frac{\partial f}{\partial b} = (a \oplus c) \oplus (b * c) \enspace. $$</p>
<p>The choice of $b$ for the expansion is enticing when computing by hand because the cofactors and the Boolean difference are easy to find. Is this what WA does? Expansion w.r.t. $a$ produces</p>
<p>$$a \oplus (\neg b * c) \enspace. $$</p>
<p>Expansion w.r.t. $c$ also produces this result. This expression is more concise than the previous one, but not in Reed-Muller canonical form, which may be what WA is shooting for. Conversion produces</p>
<p>$$ a \oplus ((1 \oplus b) * c) = a \oplus c \oplus (b*c) \enspace. $$</p>
<p>So, no matter how we do the expansion, we end up with the expected result if we shoot for Reed-Muller canonical form. The parentheses in the expression produced by WA suggest the expansion is actually carried out w.r.t. $b$. </p>
|
2,984,918 | <p>How can I prove this? </p>
<blockquote>
<p>Prove that for any two positive integers <span class="math-container">$a,b$</span> there are two positive integers <span class="math-container">$x,y$</span> satisfying the following equation:
<span class="math-container">$$\binom{x+y}{2}=ax+by$$</span></p>
</blockquote>
<p>My idea was that <span class="math-container">$\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$</span> and choose <span class="math-container">$x,y$</span>, such that <span class="math-container">$2a=x+2y-1, 2b=y(y-1)$</span>, but using this idea, <span class="math-container">$x,y$</span> won’t be always positive. </p>
| G.Kós | 141,614 | <p>Consider the equation <span class="math-container">$f(x,y)=x^2+2xy+y^2-(2a+1)x-(2b+1)y=0$</span>.
If <span class="math-container">$a=b$</span> then this is satisfied along the parallel lines <span class="math-container">$x+y=0$</span> and <span class="math-container">$x=y+2a+1$</span>, so we can choose say <span class="math-container">$x=1$</span>, <span class="math-container">$y=2a$</span>.</p>
<p>From now on we assume <span class="math-container">$a>b$</span>. Then the equation is satisfied along a parabola, passing through points <span class="math-container">$O=(0,0)$</span>, <span class="math-container">$A=(2a+1,0)$</span> and <span class="math-container">$B=(0,2b+1)$</span> and having its vertex in the second quarter plane.</p>
<p>Starting from the origin, we can perform a sequence of Viéte jumps:
<span class="math-container">$$
(x,y)
\mapsto \big(x,2b+1-2x-y\big)
\mapsto \big(2a+1-2(2b+1-2x-y)-x,2b+1-2x-y\big)
\mapsto \ldots
$$</span>
All these lattice points lie on the parabola. We continue this sequence until we get onto the arc <span class="math-container">$AB$</span>.</p>
<p><a href="https://i.stack.imgur.com/SNQH8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SNQH8.png" alt="enter image description here"></a></p>
<p>More formally, let <span class="math-container">$P_n=(x_n,y_n)$</span> with
<span class="math-container">$$
x_0=y_0=0, \quad
x_{n+1}=2a-4b-1+3x_n+2y_n,
\quad y_{n+1}=2b+1-2x_n-y_n.
$$</span>
Notice that <span class="math-container">$x_{n+1}+y_{n+1}=x_n+y_n+2(a-b)$</span>, so <span class="math-container">$x_n+y_n=2(a-b)n$</span>. Hence, we must take the unique index <span class="math-container">$n$</span> with <span class="math-container">$2b+1<2(a-b)n<2a+1$</span> in order to find <span class="math-container">$P_n$</span> inside arc <span class="math-container">$AB$</span>. Solving the equations <span class="math-container">$f(x,y)=0$</span>, <span class="math-container">$x+y=2(a-b)n$</span>, the coordinates are
<span class="math-container">$$
x_n = \big(2(a-b)n-(2b+1)\big)n, \quad
y_n = \big((2a+1)-2(a-b)n\big)n.
$$</span></p>
<p>(The picture shows the case <span class="math-container">$a=8$</span>, <span class="math-container">$b=6$</span>.)</p>
|
4,498,296 | <p>Is there any subtle way to compute the following integral?</p>
<p><span class="math-container">$$\int \frac{\sqrt{u^2+1}}{u^2-1}~ \mathrm{d}u$$</span></p>
<p>The solution i had in mind was substituting <span class="math-container">$u=\tan (\theta)$</span>,then after a few calculations the integral became <span class="math-container">$$\int \sec (\theta)
~ \mathrm{d}\theta+2\int \frac{\sec (\theta)}{\sec^2 (\theta) -2} ~\mathrm{d}\theta$$</span> I think we can formulate the last integral as <span class="math-container">$$\frac{1}{2}\int \left(\frac{1}{\sec (\theta) -\sqrt{2}}+\frac{1}{\sec (\theta)+\sqrt{2}} \right)\mathrm{d}\theta$$</span> But it still seems to be a daunting task and i think it will require further substitutions.</p>
<p>So could anyone please provide a <em>out of the blue</em> kind of solution or a clever approach to this?Or is it possible to go along my approach shortening the calculations?</p>
| Lai | 732,917 | <p>Letting <span class="math-container">$u=\tan \theta$</span> transforms the integral into
<span class="math-container">$$
\begin{aligned}
\int \frac{\sqrt{u^{2}+1}}{u^{2}-1} d u&=\int \frac{\sec \theta \sec ^{2} \theta d \theta}{\tan ^{2} \theta-1} \\
&=\int \frac{d\theta}{\cos \theta-2 \cos ^{3} \theta} \\
&=\int \left(\frac{1}{\cos \theta}+\frac{2 \cos {\theta}}{1-2 \cos ^{2} \theta}\right) d \theta\\&=\ln |\sec \theta+\tan \theta|+2 \int \frac{d(\sin \theta)}{2 \sin ^{2} \theta-1} \\&=\ln |\sec \theta+\tan \theta|+\sqrt{2} \ln \left|\frac{\sqrt{2} \sin \theta-1}{\sqrt{2} \sin \theta+1}\right|+C\\&=\ln \left|u+\sqrt{1+u^{2}}\right|+\sqrt{2}\ln \left| \frac{\sqrt{2} u-\sqrt{1+u^{2}}}{\sqrt{2} u+\sqrt{1+u^{2}} }\right|+C
\end{aligned}
$$</span></p>
|
1,566,215 | <p>Can someone explain to me the difference between joint probability distribution and conditional probability distribution?</p>
| Jay Kim | 564,425 | <p>How can you apply the first equation in the above, i.e. $P(F \, \text{and} \, L)=P(F)*P(L)$? $P(F)=12/25$, $P(L)=7/20$, then the product is $28/125$ which is different from $23/100$ ? What mistake did I make here?</p>
|
1,548,159 | <p>This is a question asked in India's CAT exam: <a href="http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html" rel="nofollow noreferrer">http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html</a> </p>
<blockquote>
<p>How many numbers with distinct digits are possible product of whose digits is 28?</p>
<p>A. 6</p>
<p>B. 4</p>
<p>C. 8</p>
<p>D. 12 </p>
</blockquote>
<p>Firstly, I couldn't even understand the question because the English seems grammatically incorrect. </p>
<p>Secondly, I couldn't understand how the answer was arrived at either.</p>
<pre><code>Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
</code></pre>
<p>If you can't have three digits, then how can four digits even be considered? And how on earth did they eventually reach 8 numbers? What does this even mean? </p>
<p>ps: I considered asking on <a href="https://puzzling.stackexchange.com/">puzzling.stackexchange</a>, but felt that it'd be more appropriate in a math forum.</p>
| kccu | 255,727 | <p>They arrived at $8$ by counting $2$ possible two-digit numbers (47 and 74) plus $3!=6$ possible three-digit numbers (147, 174, 417, 471, 714, 741). $3!$ counts the number of ways to arrange the three digits $1, 4, 7$ that multiply to $28$.</p>
<p>And you <em>can</em> have three digits, as was demonstrated with 147, etc. That is why they considered four-digit numbers and noted that from that point onward, digits would have to be repeated.</p>
|
1,503,958 | <p>In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?</p>
<p>I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!</p>
| Jack D'Aurizio | 44,121 | <p>Set $x=z^2+1$. Then:
$$ \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+3-8\sqrt{x-1}} = \sqrt{(z-2)^2}+\sqrt{(z-3)^2} = |z-2|+|z-3| $$
equals one for every $z\in[2,3]$, hence for every $x\in[5,10]$.</p>
|
211,903 | <p>Supposing I have a $0/1$ or $\pm1$ matrix $A$ of size $m\times n$, is there a minimum $d$ (that works for every $m\times n$ $A$) such that there exists $m$ lines $r_1,\dots,r_m$, $n$ lines $s_1,\dots,s_n$ in $\Bbb R^d$ such that $r_i$ intersects $s_j$ iff there are some permutations $\Pi_1,\Pi_2$ such that $\widetilde{A}[i,j]=1$ where $\widetilde{A}=\Pi_1A\Pi_2$?</p>
<p>If so what is $d$ as function of $m,n,r$?</p>
<p>It is clear this is possible for special $A$ with $d=2$. Is it possible for every $A$ at some $d$?</p>
<p>Related: <a href="https://mathoverflow.net/questions/132516/when-is-a-0-1-matrix-a-one-intersection-incidence-matrix?rq=1">When is a 0-1 matrix a one-intersection incidence matrix?</a></p>
| Joseph O'Rourke | 6,094 | <p>As Will Sawin points out, this represents only a partial answer.</p>
<p>Theorem 5 of this paper</p>
<blockquote>
<p>Laison, Joshua D., and Yulan Qing. "Subspace intersection graphs." <em>Discrete Mathematics</em> 310, no. 23 (2010): 3413-3416.
<a href="http://www.sciencedirect.com/science/article/pii/S0012365X10002566" rel="nofollow noreferrer">Journal link</a></p>
</blockquote>
<p>proves that this graph
<hr />
<a href="https://i.stack.imgur.com/TbK4m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TbK4m.png" alt="Fig.1"></a></p>
<hr />
<p>cannot be realized as the intersection graph of lines in $\mathbb{R}^d$
for any $d \ge 2$.
The corresponding incidence matrix is shown below,
using the ordering $(a,b,c,i,j,k)$.
$$
\left(
\begin{array}{cccccc}
0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 \\
\end{array}
\right)
$$
The logic of their proof is as follows (paraphrased from the paper).
Because $i$ and $j$ intersect (say at point $p$), they determine a plane $\pi$.
Since $a,b,c$ are mutually disjoint,
at most one could contain $p$.
Hence two of them are also contained in $\pi$,
as they intersect both $i$ and $j$; let these two be $a,b$.
Because $a,b$ are disjoint and lie in $\pi$, they must be parallel.
Now $k$ intersects both $a$ and $b$, so $k$ also lies in $\pi$.
But $k$ is disjoint from $i$ and $j$ (which cross in $\pi$),
which is impossible.</p>
|
211,903 | <p>Supposing I have a $0/1$ or $\pm1$ matrix $A$ of size $m\times n$, is there a minimum $d$ (that works for every $m\times n$ $A$) such that there exists $m$ lines $r_1,\dots,r_m$, $n$ lines $s_1,\dots,s_n$ in $\Bbb R^d$ such that $r_i$ intersects $s_j$ iff there are some permutations $\Pi_1,\Pi_2$ such that $\widetilde{A}[i,j]=1$ where $\widetilde{A}=\Pi_1A\Pi_2$?</p>
<p>If so what is $d$ as function of $m,n,r$?</p>
<p>It is clear this is possible for special $A$ with $d=2$. Is it possible for every $A$ at some $d$?</p>
<p>Related: <a href="https://mathoverflow.net/questions/132516/when-is-a-0-1-matrix-a-one-intersection-incidence-matrix?rq=1">When is a 0-1 matrix a one-intersection incidence matrix?</a></p>
| Noam D. Elkies | 14,830 | <p>It seems that the matrix
$$
A = \left(\begin{array}{cccccc}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 0 \\
\end{array}\right)
$$
is not the incidence matrix for any distinct lines $r_1,\ldots,r_6$ and
$s_1,\ldots,s_6$ for any dimension $d$, over $\bf R$ or any other field,
and thus that there is no such $d(m,n)$ once $m,n \geq 6$.</p>
<p>The $5\times 5$ matrix $A_0$ formed by the first five rows and columns of $A$
is not enough: let $R,S$ be points on a plane $\Pi$;
let $r_1,r_2,r_3 \subset \Pi$ be lines containing $R$ but not $S$,
and $s_1,s_2,s_3 \subset \Pi$ lines containing $S$ but not $R$;
and then choose points $P,Q$ off $\Pi$ and collinear with neither $R$ not $S$,
and set
$$
r_4 = \overline{PS}, \quad
r_5 = \overline{QS}, \quad
s_4 = \overline{QR}, \quad
s_5 = \overline{PR}.
$$
But this seems to be the only way to realize $A_0$, and it does not extend to
$6+6$ lines with the intersections demanded by $A$ itself.</p>
<p>The key point is suggested by <strong>Boris Bukh</strong>'s comment:
of any three $r_i$, two must intersect. Indeed if three pairwise skew lines
$l,l',l''$ have more than one transversal then they determine a quadric surface
$Q$ with two rulings, one of which includes $l,l',l''$ and the other consists
of all the transversals. Then $s_1,s_2,s_3$ are in the other ruling,
and are thus pairwise skew. Then $r_1,r_2,r_3$ are in the first ruling,
and it follows that all five $r_i$ are in one ruling and all five $s_j$
in the other. Hence $r_i$ meets $s_j$ for every $i,j$ $-$ contradiction.</p>
<p>Likewise, of any three $s_j$, two must intersect.</p>
<p>On the other hand, if any three of the $r_i$ meet pairwise,
but not at the same point, then they span a plane $\Pi$ and $s_1,s_2,s_3$
must be on $\Pi$; and then they must also coincide, else every $r_i \subset \Pi$
and we soon reach a contradiction ($s_4$ meets $r_1,r_2,r_3$ but not $r_4$,
so $r_1,r_2,r_3$ meet at some point $R$ and $s_4$ contains $R$ but is not
in $\Pi$; likewise the same is true of $s_5$; but then $r_4,r_5$ also
go through $R$, which contradicts both the assumption at the start of
this paragraph and the $A_0$ condition). Likewise if any three $r_i$
meet at a point $R$ but are not coplanar then $s_1,s_2,s_3$ go through $R$
and <em>are</em> coplanar, else every $r_i$ contains $R$ and again a contradiction
soon ensues.</p>
<p>From here it should be just a matter of hunting down which configurations
of skew/coincident/coplanar lines are possible among the $r_i$ and $s_j$
until only the one configuration described earlier survives…</p>
<p>Another candidate counterexample, with $m=n=7$, is a "double seven"
with $r_i \cap s_j = \emptyset$ <em>iff</em> $i=j$. Even $(m,n)=(6,7)$
would suffice if the only way to get $(m,n)=(6,6)$ is the famous
"double six" configuration on a smooth cubic surface, on which
no two $r_i$ meet but the $r_i$ are not all on the same quadric,
and likewise for the $s_j$.</p>
|
3,847,209 | <p>I'm currently trying to solve the question below.</p>
<h2>Abed is sitting in front of a large screen tv. He thinks he gets the best view when the screen takes up the maximum angle in his field of view.
What is the optimal point where he sits(to get the largest angle in his view)?</h2>
<p>I think it has something to do with a geometric inequality, but I'm stuck at the first step, I can't find any way to progress. I've drawn a few constructions, but I still can't find the optimal point.</p>
<p>EDIT:</p>
<p>I forgot to clarify. We treat the TV as a line, and Abed is looking at it. He cannot leave the 'floor' so he can only move from left to right, not up and down.</p>
| Narasimham | 95,860 | <p>Assuming the TV surface to be a flat rectangle, the normal to the screen at the center point of the rectangle maximizes planar and solid angles subtended by the viewing eye.</p>
<p>This is a consequence of symmetry of the field being viewed.</p>
<p>Answer after your EDIT:</p>
<p>Assuming the TV to be the red horizontal line the normal to the line at the center point of the line maximizes planar angle subtended by the viewing eye moving parallel to it on the ground as shown.</p>
<p>Let length of line TV be <span class="math-container">$2L$</span>, point on considered on it from center at distance on ground <span class="math-container">$x$</span>,central slant distance <span class="math-container">$a$</span> then total subtended angle</p>
<p><span class="math-container">$$ \theta= \tan^{-1}\dfrac{L-x}{a}+\tan^{-1}\dfrac{L+x}{a} $$</span></p>
<p>Differentiate w.r.t. <span class="math-container">$x$</span> and simplify, resulting in the center point</p>
<p><span class="math-container">$$ x=0 $$</span></p>
<p><a href="https://i.stack.imgur.com/1GZHM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1GZHM.png" alt="enter image description here" /></a></p>
|
1,179,195 | <p>Good day everyone. </p>
<p>I need to know automata theory. Can you advice me the best way to study math?
What themes will I need to know to understand automata theory. What a sequence of study? What level will I need to study intermediate themes? Maybe can you say something yet, what can help me quickly learn automata theory?</p>
| Alan | 175,602 | <p>Yes, a contraction has a unique fixed point. Proof: Let $d(\cdot,\cdot)$ be our distance function (which is absolute value on the real numbers, but works in more general settings).</p>
<p>Assume $\phi$ is a contraction and $a,b$ are both fixed points. Let $0<k<1$ be our contraction constant. Then $d(g(a),g(b))<k\cdot d(a,b)$, since it's a contraction, but $g(a)=a,g(b)=b$, so $d(g(a),g(b))=d(a,b)$, hence $d(a,b)<k\cdot d(a,b)$, which is only possible if $d(a,b)=0$, hence $a=b$.</p>
|
1,349,654 | <p>Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$</p>
<p>Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .</p>
<p>Thank you for any kind of help </p>
| Marco Cantarini | 171,547 | <p>We have $$\int_{\pi/4}^{\pi/3}\frac{\sqrt{\tan\left(x\right)}}{\sin\left(x\right)}dx=\int_{\pi/4}^{\pi/3}\frac{1}{\sqrt{\cos\left(x\right)\sin\left(x\right)}}dx=\sqrt{2}\int_{\pi/4}^{\pi/3}\frac{1}{\sqrt{\sin\left(2x\right)}}dx$$ $$=\frac{1}{\sqrt{2}}\int_{\pi/2}^{2\pi/3}\frac{1}{\sqrt{\sin\left(t\right)}}dt
$$ and if we put $$\sin\left(t\right)=\cos\left(2u\right)
$$ we get $$=\sqrt{2}\int_{0}^{\pi/12}\frac{1}{\sqrt{\cos\left(2u\right)}}dt=\sqrt{2}\int_{0}^{\left(\sqrt{3}-1\right)/2\sqrt{2}}\frac{1}{\sqrt{1-2v^{2}}\sqrt{1-v^{2}}}dt=\sqrt{2}F\left(\frac{\pi}{12}\mid2\right)
$$ where $F\left(x\mid k^{2}\right)
$ is the <a href="http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html" rel="nofollow">elliptic integral of the first kind</a>.</p>
|
2,065,639 | <p>$\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$</p>
<p>$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ </p>
<p>Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?</p>
| mickep | 97,236 | <p>One way is to use <a href="https://en.wikipedia.org/wiki/Simpson's_rule" rel="nofollow noreferrer">Simpson's rule</a>.</p>
<p>Without it, one could argue what is faster, but: </p>
<p>If $A=(a+b)/2$ and $D=(b-a)/2$, then your first integrand is
$$
(x-A+D)(x-A-D)=(x-A)^2-D^2.
$$
Thus
$$
\begin{aligned}
\int_a^b (x-a)(x-b)\,dx &=\frac{1}{3}\bigl((b-A)^3-(a-A)^3\bigr)-2D^3\\
&=\frac{1}{3}\bigl(D^3+D^3)-2D^3=-\frac{1}{6}(b-a)^3.
\end{aligned}
$$
For the second one you write the integrand as
$$
\bigl((x-A)^2-D^2\bigr)(x-A)=(x-A)^3-D^2(x-A)
$$
and do a similar calculation. I leave that to you.</p>
|
185,867 | <p>I hear that the axiom of choice (AC) derives from
The generalized continuum hypothesis(GCH).
And also hear that both AC and GCH are independent of
Zermelo–Fraenkel set theory(ZF).</p>
<p>So, I'm just curious why don't expert mathematicians use ZF+GCH
instead of ZF+AC(ZFC).</p>
| Asaf Karagila | 7,206 | <p>In some sense $\sf GCH$ is a limiting axiom. While it solves a lot of things, it also means that certain things we are interested in become false or trivialized. And that's no fun.</p>
<p>For example, forcing axioms like $\sf MA$ become trivial assuming even just $\sf CH$, and stronger forcing axioms like $\sf PFA,MM$ and others become false (since they imply $2^{\aleph_0}=\aleph_2$).</p>
<p>So while $\sf GCH$ gives us more power in terms of implications, it also gives us more constraints. On the other hand the axiom of choice gives us a lot of power, since it helps tame infinite objects, but it leaves enough space to other axioms to develop.</p>
<p>And since you're asking, we can take this question one step further. $\sf GCH$ can be derived from the axiom $V=L$ (Godel's axiom of constructibility), so why aren't we just assuming it and that's that? Well, essentially the same answer. While it has a lot of merits, it also imposes a lot of constraints, more than we might want to assume.</p>
<hr>
<p>Here's another (perhaps a better) argument in favor of $\sf AC$ and against $V=L$ and $\sf GCH$.</p>
<p>When working over $\sf ZF$ and you use forcing, then once you force $\sf AC$ to be true, it will remain true in every other generic extension, whereas $V=L$ is immediately false once you add any new sets to the model, and will never again be true if you only allow yourself generic extensions; and $\sf GCH$ is like a switch that we can turn on and off using [possibly class-] generic extensions.</p>
<p>So this is a very good argument in favor of the axiom of choice. It is stable under forcing, which is a lovely technique. On the other hand, limiting yourself to $\sf GCH$ makes forcing more difficult (because you will always have to ensure that the forcing does not violate that), and insisting on $V=L$ will make forcing outright impossible.</p>
|
3,410,850 | <p>The limit law <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)$</span> is true given that <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> both exist. I also know that <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span> can exist even if both <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> don't (i.e. if <span class="math-container">$g=-f$</span>). I was wondering whether <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span> can exist if the limit of one function exists while the limit of the other doesn't, and if not, how I would go about to prove this.</p>
<p>This is for a proof in which I want to make the statement "since I know that <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span>, either <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> both exist or both don't exist. Since we know <span class="math-container">$\lim_{x\to c}f(x)$</span> exists, we therefore know that <span class="math-container">$\lim_{x\to c}g(x)$</span> exists."</p>
| gandalf61 | 424,513 | <p>The “area below the curve” interpretation of an integral only makes intuitive sense for functions of one or two variables.</p>
<p>If you have a function of a single variable <span class="math-container">$f(x)$</span> then you can plot the curve <span class="math-container">$y=f(x)$</span> and think of the integral as the area between the curve and the <span class="math-container">$x$</span> axis. If you have a function of two variables <span class="math-container">$f(x,y)$</span> then you can plot the curve <span class="math-container">$z=f(x,y)$</span> and think of the integral as the area between the curve and the <span class="math-container">$x,y$</span> plane.</p>
<p>But if you have a function of three variables <span class="math-container">$f(x,y,z)$</span> then to plot a curve you would have to introduce a fourth variable <span class="math-container">$w=f(x,y,z)$</span> and plot the curve in four dimensional space. And then the interpretation of the integral as the area “below” the curve becomes much less intuitive and much harder (or impossible) to visualise.</p>
|
3,410,850 | <p>The limit law <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)$</span> is true given that <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> both exist. I also know that <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span> can exist even if both <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> don't (i.e. if <span class="math-container">$g=-f$</span>). I was wondering whether <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span> can exist if the limit of one function exists while the limit of the other doesn't, and if not, how I would go about to prove this.</p>
<p>This is for a proof in which I want to make the statement "since I know that <span class="math-container">$\lim_{x\to c}[f(x)+g(x)]$</span>, either <span class="math-container">$\lim_{x\to c}f(x)$</span> and <span class="math-container">$\lim_{x\to c}g(x)$</span> both exist or both don't exist. Since we know <span class="math-container">$\lim_{x\to c}f(x)$</span> exists, we therefore know that <span class="math-container">$\lim_{x\to c}g(x)$</span> exists."</p>
| Peter Melech | 264,821 | <p>As gandalf61 correctly says You must think of a fourth dimension and the integral could be interpreted as the (signed) area of the "curtain" ( as Khan says) along the curve <span class="math-container">$C$</span> with the height given by values of <span class="math-container">$w=f(x,y,z)$</span> "above" this curve.
In Your case this should give
<span class="math-container">$$\int_Cf(x,y,z)dS=\int_0^1f(0,0,t)\sqrt{0^2 +0^2+1^2}dt=\int_0^1 -t^2dt=-\frac{1}{3}.$$</span></p>
|
2,246,137 | <p>Let
$$\parallel\overrightarrow{a}\parallel =6\text{ and}\parallel\overrightarrow{b}\parallel =3$$</p>
<p>$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\text{ and } k\overrightarrow{a}-\overrightarrow{b}\text{ are parallel}$$</p>
<p>Find all the value(s) of k.</p>
<p>How to get the value(s) of k?</p>
<p>I tried the following case,
$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\cdot k\overrightarrow{a}-\overrightarrow{b}=\parallel2\overrightarrow{a}+(k-3)\overrightarrow{b}\parallel\parallel k\overrightarrow{a}-\overrightarrow{b}\parallel\cos0$$</p>
<p>$$2\overrightarrow{a}+(k-3)\overrightarrow{b}\cdot k\overrightarrow{a}-\overrightarrow{b}=\parallel2\overrightarrow{a}+(k-3)\overrightarrow{b}\parallel\parallel k\overrightarrow{a}-\overrightarrow{b}\parallel\cos\pi$$</p>
<p>but when I reached this point I stopped as I don't know how to continue. Can any one help me?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>since the vectors $$\vec{2a}+(k-3)\vec{b}$$ and $$k\vec{a}-\vec{b}$$ are parallel
then exists a real number $t$ with
$$2\vec{a}+(k-3)\vec{b}=t(k\vec{a}-\vec{b})$$ this gives
$$\vec {a}(2-kt)+\vec{b}(k-3+t)=\vec{0}$$
from here we get the System
$$2-kt=0$$
$$k-3+t=0$$
with $$t=\frac{2}{k}$$
plugging this in the second equation we get
$$k-3+\frac{2}{k}=0$$
can you finish?</p>
|
2,262,661 | <p>The question is in how many ways can we select 20 different items from the empty set?</p>
<h3>ans:</h3>
<p>Obviously in 0 ways since the empty set has no items. I mean, this seem obvious, but maybe there is a trick to this question.</p>
| Jair Taylor | 28,545 | <p>The answer is pretty clear if you formulate this question slightly more precisely:</p>
<p>How many sets are there of size $20$ are there that are subsets of the empty set?</p>
<p>We have $\{A| A \subseteq \emptyset \text{ and } |A| = 20\} = \emptyset$ since $A \subseteq \emptyset \implies A = \emptyset \implies |A| = 0$. So the answer is $0$.</p>
|
478,713 | <p>I have this logic statement:</p>
<pre><code> (A and x) or (B and y) or (not (A and B) and z)
</code></pre>
<p>The problem is that accessing A and B are rather expensive. Therefore I'd like to access them only once each. I can do this with an if-then-else construct:</p>
<pre><code>if A then
if x then
true
endif
else
if B then
if y then
true
endif
else
if z then
true
endif
endif
endif
</code></pre>
<p>Is there a way to express this as a boolean expression? I have "and", "or" (both short-circuit) but no "xor".</p>
<p>I thought this would work:</p>
<pre><code>X and (A or (Y and (B xor Z)))
</code></pre>
<p>But my test program (<a href="http://pastebin.com/EjURvpM4" rel="nofollow">http://pastebin.com/EjURvpM4</a>) shows it doesn't.</p>
| Amzoti | 38,839 | <p>Lets graph it and see what happens.</p>
<p><img src="https://i.stack.imgur.com/ZU7lq.png" alt="enter image description here"></p>
<p>Do you see why $f(x) = 0$ at $x=3$ now?</p>
|
478,713 | <p>I have this logic statement:</p>
<pre><code> (A and x) or (B and y) or (not (A and B) and z)
</code></pre>
<p>The problem is that accessing A and B are rather expensive. Therefore I'd like to access them only once each. I can do this with an if-then-else construct:</p>
<pre><code>if A then
if x then
true
endif
else
if B then
if y then
true
endif
else
if z then
true
endif
endif
endif
</code></pre>
<p>Is there a way to express this as a boolean expression? I have "and", "or" (both short-circuit) but no "xor".</p>
<p>I thought this would work:</p>
<pre><code>X and (A or (Y and (B xor Z)))
</code></pre>
<p>But my test program (<a href="http://pastebin.com/EjURvpM4" rel="nofollow">http://pastebin.com/EjURvpM4</a>) shows it doesn't.</p>
| kiss my armpit | 26,975 | <p>The $x$ intercepts are the abscissa of intersection points of the curve with the $x$ axis. When the curve intersects the $x$ axis, the ordinate is always equal to 0. </p>
<p><img src="https://i.stack.imgur.com/N9Iyb.png" alt="enter image description here"> </p>
<p>The code to generate the graph can be found <a href="http://pastebin.com/5zuET7GB" rel="nofollow noreferrer">here</a>.</p>
<p>Therefore, we have the following equation:</p>
<p>$$
(x-3)^2=0
$$</p>
<p>We have a theorem as follows.</p>
<blockquote>
<p>The solution of $$ A \times B =0 $$</p>
<p>is $A=0$ with any $B$ <em>OR</em> any $A$ with $B=0$.</p>
</blockquote>
<p>Thus, the solution of $(x-3)^2=(x-3)(x-3)=$ is $x-3=0$ or $x=3$.</p>
|
1,953,251 | <blockquote>
<p>For what x does the exponential series $P_c(x) = \sum^\infty_{n=0} (-1)^{n+1}\cdot n\cdot x^n$ converge?</p>
</blockquote>
<p><strong>What I got so far:</strong></p>
<p>$\sum^\infty_{n=0} (-1)^{n+1}\cdot n\cdot x^n = (-1)\sum^\infty_{n=0} (-1)^{n}\cdot (\sqrt[n]n)^n\cdot x^n = (-1)\sum^\infty_{n=0} ((-1)\cdot (\sqrt[n]n)\cdot x)^n$</p>
<p>We know that $|((-1)\cdot (\sqrt[n]n)\cdot x)|$ must be < 1 in order to converge <em>(geometric sum)</em>:</p>
<p>$ |((-1)\cdot (\sqrt[n]n)\cdot x)| < 1$ </p>
<p>Case 1:
$ ((-1)\cdot (\sqrt[n]n)\cdot x) < 1 \Leftrightarrow (-1)\cdot (\sqrt[n]n)\cdot x < 1 \Leftrightarrow (-1)\cdot x < \frac{1}{(\sqrt[n]n)} \Leftrightarrow x > \frac{-1}{(\sqrt[n]n)}$</p>
<p>Case 2:
$ ((\sqrt[n]n)\cdot x) < 1 \Leftrightarrow (\sqrt[n]n)\cdot x < 1 \Leftrightarrow x < \frac{1}{(\sqrt[n]n)}$</p>
<p>We know that $x < |\frac{-1}{(\sqrt[n]n)}| = |\frac{1}{(\sqrt[n]n)}| = \frac{1}{(\sqrt[n]n)} \le \pm 1 $</p>
<p>So $x < (-1) < 1$. <strong>Is this way right</strong>, must $x$ be $< -1$ in order for $P_c(x) $ to converge? Is there a simpler way to get to a solution? Thanks in advance!</p>
| Claude Leibovici | 82,404 | <p>May be, you could consider that $$\sum^\infty_{n=0} (-1)^{n+1} n x^n=x\sum^\infty_{n=0} (-1)^{n+1} n x^{n-1}=x \frac{d}{dx} \left(\sum^\infty_{n=0} (-1)^{n+1} x^n \right)=-x\frac{d}{dx} \left(\sum^\infty_{n=0} (-x)^{n} \right)$$ Now, in brackets, you have a simple geometric sum easy to compute (and analyze, for sure).</p>
|
1,528,501 | <p>Let $P(n)$ be a property for all $n \geq 1$. For the phrase "there is some $N \geq 1$ such that $P(n)$ holds for all $n \geq N$" there are some suggestive, convenient abbreviations such as "$P(n)$ holds for large $n$" or "$P(n)$ holds eventually" and so on.</p>
<p>I wonder if there is in literature a like abbreviation for "$P(n)$ holds for infinitely many $n$"? I am aware that in probability theory some authors would write "$P(n)$ holds infinitely often"; but, in my humble opinion, this abbreviation would be not that useful in contexts other than probability theory.</p>
| Henno Brandsma | 4,280 | <p>In descriptive set theory and logic one sometimes uses $\exists^\ast_n P(n)$ for "there are infinitely many $n$ such that $P(n)$ holds", and $\forall^\ast_n P(n)$ for "all but finitely many $n$ satisfy $P(n)$".</p>
<p>Then at least we have $\lnot \forall^\ast_n P(n) \leftrightarrow \exists^\ast_n \lnot P(n)$, like for normal quantifiers, e.g.</p>
|
1,528,501 | <p>Let $P(n)$ be a property for all $n \geq 1$. For the phrase "there is some $N \geq 1$ such that $P(n)$ holds for all $n \geq N$" there are some suggestive, convenient abbreviations such as "$P(n)$ holds for large $n$" or "$P(n)$ holds eventually" and so on.</p>
<p>I wonder if there is in literature a like abbreviation for "$P(n)$ holds for infinitely many $n$"? I am aware that in probability theory some authors would write "$P(n)$ holds infinitely often"; but, in my humble opinion, this abbreviation would be not that useful in contexts other than probability theory.</p>
| yo' | 43,247 | <p>I'm aware of only one way how to write this clearly -- in symbols:</p>
<p>$$ \exists_\infty n\in\mathbb{N} : P(n). $$</p>
<p>However, unless you need this a lot you shouldn't use this notation. If you need it a lot, you can introduce it:</p>
<blockquote>
<p>We suppose that $P(n)$ holds for infinitely many $n$, i.e., that $\exists_\infty n\in\mathbb{N}:P(n)$.</p>
</blockquote>
<p>Some people use: "$P(n)$ holds infinitely many times," but I don't find this one quite nice; also, it's not even shorter. In contexts when no confusion can happen, one can use: "$P(n)$ holds infinitely often." However, be aware that to some, this means that $P(n)$ holds with a positive density.</p>
|
2,109,197 | <p><strong>Update:</strong><br>
(because of the length of the question, I put an update at the top)<br>
I appreciate recommendations regarding the alternative proofs. However, the main emphasis of my question is about the correctness of the reasoning in the 8th case of the provided proof (with a diagram).</p>
<p><strong>Original question:</strong> </p>
<p>I would like to know, whether the following proof, is a valid way to prove that $a^2 + b^2 \neq 3c^2$ for all $a, b, c \in Z$ (except the trivial case, when $a=b=c=0$). More formally, we have to prove the correctness of the following statement:</p>
<p>$$P: (\forall a,b,c \in Z, a^2 + b^2 \neq 3c^2 \lor (a=b=c=0))$$</p>
<p><strong>Proof.</strong> <em>(by contradiction)</em><br>
For the sake of contradiction let's assume, that there exist such $a, b, c \in Z$, that $a^2 + b^2 = 3c^2$ (and the combination of $a,b,c$ is not a trivial case). More, formally, let's assume that $\neg P$ is <em>true</em>:</p>
<p>$$\neg P: (\exists a,b,c \in Z, a^2 + b^2 = 3c^2 \land \neg (a=b=c=0))$$</p>
<p>There are $2^3$ possible combinations of different parities of $a,b,c$ (8 disjoint cases, which cover entire $Z^3$). So, in order to prove the original statement, we have to consider each case, and show that the <em>true</em>-ness of the $\neg P$ always leads to some sort of contradiction.</p>
<p>Let's consider 8 possible cases (7 of which are simple, <strong>whereas the 8th case looks a bit intricate, and I am not sure regarding its correctness</strong>):</p>
<p><strong>Case 1)</strong> $a$ is odd, $b$ is odd, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z + 1)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 2 \cdot (6z^2 + 6z + 1) + 1 \\
\implies even\ number = odd\ number \\
$$</p>
<p>However, the derived result contradicts to the fact that <em>odd numbers</em> and <em>even numbers</em> can't be equal.
Hence: $(even\ number = odd\ number) \land (even\ number \neq odd\ number)$, or equivalently: $(even\ number = odd\ number) \land \neg (even\ number = odd\ number)$. Contradiction.</p>
<p><strong>Case 2)</strong> $a$ is odd, $b$ is odd, $c$ is even<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = 2z$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 12z^2 \\
\implies 2 \cdot (x^2 + x + y^2 + y) + 1 = 6z^2 \\
\implies odd\ number = even\ number
$$
Contradiction.</p>
<p><strong>Case 3)</strong> $a$ is odd, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y)^2 = 3 \cdot (2z + 1)^2 \\
\implies 4x^2 + 4x + 1 + 4y^2 = 12z^2 + 12z + 3 \\
\implies 4\cdot(x^2 + x + y^2) = 2 \cdot (6z^2 + 6z + 1) \\
\implies 2\cdot(x^2 + x + y^2) = 6z^2 + 6z + 1 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 4)</strong> $a$ is odd, $b$ is even, $c$ is even<br>
The square of an odd number is odd (so, $a^2$ is odd).<br>
The square of an even number is even (so, $b^2$ and $3c^2$ are even).<br>
Fact: <em>the sum of an even number and an odd number is odd</em>.<br>
However, equality: $a^2 + b^2 = 3c^2$ leads to the conclusion, that: $odd\ number + even\ number = even\ number$<br>
Contradiction.</p>
<p><strong>Case 5)</strong> $a$ is even, $b$ is odd, $c$ is odd<br>
Symmetric to the Case 3 (because $a$ and $b$ are mutually exchangeable), which shows the contradiction.</p>
<p><strong>Case 6)</strong> $a$ is even, $b$ is odd, $c$ is even<br>
Symmetric to the Case 4, which shows the contradiction.</p>
<p><strong>Case 7)</strong> $a$ is even, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 12z^2 + 12z + 3 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 8)</strong> $a$ is even, $b$ is even, $c$ is even<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = 2z$ for some $x, y, z \in Z$<br>
So:
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 3 \cdot 4z^2 \\
\implies x^2 + y^2 = 3z^2
$$</p>
<p>Now, we are faced with the similar instance of the problem, <strong>however, the size of the problem is strictly smaller</strong> ($x = {a \over 2}$, $y = {b \over 2}$, $z = {c \over 2}$).<br>
At first glance, it seems that we have to consider again the eight possible parities of $x, y, z$. However, if we analyze all dependencies between the cases of the problem, we will notice that the only possible outcomes are either <em>contradiction</em> or <em>the trivial case</em>:</p>
<p><a href="https://i.stack.imgur.com/KVWUL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVWUL.png" alt="enter image description here"></a></p>
<p>We have shown the contradiction in all cases, hence we have subsequently proved the original statement.
$\blacksquare$</p>
<p>So, I would like to know, if there is any problem with reasoning in the 8th case?</p>
| pjs36 | 120,540 | <p>Here is another approach that relies on the idea that its easy to detect if three complex numbers lie on a line through the origin, and we attempt to reduce the original problem to this simpler one. </p>
<p>To this end, we might hope that translating our complex numbers (in this case, subtracting $z_1$ from each) doesn't affect whether the function $f(z_1, z_2, z_3) = \overline{z_1}z_2 + \overline{z_2}{z_3} + \overline{z_3}{z_1}$ takes a real value or not. </p>
<p>And indeed, we'll see that</p>
<p>$$\operatorname{Im}\Big(f(z_1, z_2, z_3)\Big) = \operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big).$$</p>
<p>Observe that</p>
<p>\begin{align*}
\operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big)
&= \operatorname{Im}\Big(\overline{z_2 - z_1}(z_3 - z_1)\Big) \\[7pt]
&= \operatorname{Im}\Big(\overline{z_2}z_3 - \overline{z_1}z_3 - \overline{z_2}z_1 + \overline{z_1}{z_1}\Big) \\[7pt]
&= \operatorname{Im}\big(\overline{z_2}z_3) - \operatorname{Im}\big(\overline{z_1}z_3\big) - \operatorname{Im}\big(\overline{z_2}z_1\big) + \underbrace{\operatorname{Im}\big(|z_1|^2\big)}_{=0}
\end{align*}</p>
<p>where $\overline{z_1}{z_3} = \overline{z_1\overline{z_3}}$ so the two have opposite imaginary parts; that is, $\operatorname{Im}\big(z_1\overline{z_3}) = - \operatorname{Im}(\overline{z_1}z_3)$. Now picking up where we left off,</p>
<p>\begin{align*}
\operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big)
&= \operatorname{Im}\big(\overline{z_2}z_3) - \operatorname{Im}\big(\overline{z_1}z_3\big) - \operatorname{Im}\big(\overline{z_2}z_3\big) \\[7pt]
&= \operatorname{Im}\big(\overline{z_2}z_3) + \operatorname{Im}\big(\overline{z_3}z_1\big) + \operatorname{Im}\big(\overline{z_1}z_2\big)\\
&= \operatorname{Im}\Big(f(z_1, z_2, z_3)\Big)
\end{align*}</p>
<p>Now we simply need to show that $z_1, z_2, z_3$ are collinear if and only if $f(0, z_2 - z_1, z_3 - z_1) \in \Bbb R$.</p>
<p>Letting $w_2 = z_2 - z_1$ and $w_3 = z_3 - z_1$, this is equivalent to showing that $0, w_1, w_2$ are collinear if and only if $f(0, w_2, w_3) \in \Bbb R$. </p>
<p>But since $f(0, w_2, w_3) = \overline{w_2}w_3$, this is just the well-known fact that $\overline{w_2}w_3 \in \Bbb R$ if and only if $w_3 = c w_2$ for some real scalar $c$.</p>
|
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>
| fleablood | 280,126 | <p>If <span class="math-container">$a^2 + ab = b^2$</span> then <span class="math-container">$a^2 + ab -b^2 =0$</span> and you can solve for <span class="math-container">$a$</span> in terms of <span class="math-container">$b$</span> or <span class="math-container">$b$</span> in terms of <span class="math-container">$a$</span>.</p>
<p><span class="math-container">$a = \frac {-b \pm\sqrt{b^2 +4b^2}}2=\frac {(-1\pm \sqrt 5)b}2$</span></p>
<p>so </p>
<p><span class="math-container">$3^x = 5^x\frac {(-1\pm \sqrt 5)}2$</span></p>
<p><span class="math-container">$(\frac 35)^x = \frac {(-1\pm \sqrt 5)}2$</span>. As <span class="math-container">$(\frac 35)^x > 0$</span> we have</p>
<p><span class="math-container">$x = \log_{\frac 35} \frac {(-1+ \sqrt 5)}2$</span></p>
|
1,247,185 | <p>I already know, and so ask NOT about, the proof of: <a href="https://math.stackexchange.com/a/463407/53259">$A$ only if $B$ = $A \Longrightarrow B$</a>.<br>
Because I ask only for intuition, please do NOT prove this or use truth tables. </p>
<p><strong>My problem:</strong> I try to avoid memorisation. So whenever I see this statement, I always need to pause for 5 minutes to remember my linked post above, in order to determine the meaning.<br>
This pause reveals deficiency in my knowledge that stifles me. So please help me dig deeper. How can this statement be naturalised? </p>
| hmakholm left over Monica | 14,366 | <p>I understand "A only if B" as "A can be true <em>only</em> in those possible worlds where B is true also".</p>
|
4,116,134 | <p>Find angle between <span class="math-container">$y=\sin x$</span> and <span class="math-container">$y=\cos x$</span> at their intersection point.</p>
<p>Intersection points are <span class="math-container">$\frac{\pi}{4}+\pi k$</span> and to find angle between them we need to compute derivatives at intersection points but then I can't combine them to get an answer which is <span class="math-container">$\arctan2\sqrt2$</span>. Will be thankful for your help.</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>If <span class="math-container">$u$</span> is the angle <span class="math-container">$$\tan u=\left|\dfrac{\cos x-(-\sin x)}{1+\cos x(-\sin x)}\right|$$</span></p>
<p>At the point of intersection,</p>
<p><span class="math-container">$$\cos x+\sin x=\sqrt2\sin(x+\pi/4)=\cdots=\sqrt2(-1)^n$$</span></p>
<p><span class="math-container">$$\sin x\cos x=\dfrac{\sin2(n\pi+\pi/4)}2=\dfrac12$$</span></p>
|
866,808 | <p>In my lecture notes:</p>
<p>Let $m,n\in \mathbb{N}$ be relatively prime. The fundamental theorem of arithmetic implies that each divisor of $mn$ is the product of two unique positive relatively prime integers $d_1|m$ and $d_2|n$.</p>
<p>Please could someone help me understand how this is implied? I have no idea</p>
<p>thanks!</p>
| TonyK | 1,508 | <p>$d_1$ and $d_2$ are not, in general, prime numbers. They are <em>relatively</em> prime (to each other); that means that the highest common factor of $d_1$ and $d_2$ is $1$.</p>
|
4,151,381 | <p>I am a nube just getting into mathematics and set theory.</p>
<p>I am learning about how we can produce the list of ordinal numbers by purely using the null set, with 0 standing for Ø, 1 standing for {Ø}, 2 standing for {Ø, {Ø}} and so forth. What I am confused about is the operation at play here to produce the larger sets with more elements. It seems to me to be the power set axiom being applied to create a new larger set. But elsewhere I have seen this called the axiom of subsets. Is this the same thing? Or am I confused?</p>
<p>Thanks so much :)
A</p>
| HFKy | 534,596 | <p>By contradiction, suppose <span class="math-container">$h(\Omega(f))\subsetneq \Omega(g)$</span>. We can assume <span class="math-container">$Y = \Omega(g)$</span> since <span class="math-container">$K = h^{-1}(\Omega(g))$</span> is compact and <span class="math-container">$f-$</span>invariant, so we have <span class="math-container">$h:K\to \Omega(g)$</span> surjective continuous function which satisfies <span class="math-container">$hf = gh$</span>.</p>
<p>Now since <span class="math-container">$h(\Omega(f))$</span> is closed in <span class="math-container">$X$</span> we can find an element <span class="math-container">$w$</span> and an open set <span class="math-container">$U$</span> containing <span class="math-container">$w$</span> which does not intersect <span class="math-container">$h(\Omega(f))$</span>. Since there is a dense subset of recurrent points (because <span class="math-container">$Y = \Omega(g)$</span>), this shows we can find an element <span class="math-container">$y \in U$</span> such that <span class="math-container">$y$</span> is recurrent, i.e, there exists <span class="math-container">$n_k \to +\infty$</span> such that <span class="math-container">$g^{n_k}(y) \to y$</span>. Take <span class="math-container">$x$</span> any preimage of <span class="math-container">$y$</span>. Taking subsequences we can assume that <span class="math-container">$f^{n_k}(x)$</span> is convergent to some element <span class="math-container">$z$</span>. Since <span class="math-container">$h$</span> is a semi-conjugacy then <span class="math-container">$h(z)=y$</span>. But <span class="math-container">$z \in \omega(x)$</span>, then <span class="math-container">$z \in \Omega(f)$</span> which is absurd because <span class="math-container">$y \notin h(\Omega(f))$</span>.</p>
|
4,277,924 | <p><strong>1. p ∧ ¬q = T</strong>
<br><br>
<strong>2. (q ∧ p) → r = T</strong>
<br><br>
<strong>3.¬p → ¬r = T</strong>
<br><br>
<strong>4.(¬q ∧ p) → r = T</strong>
<br><br>
From <strong>Eq 1</strong>, we got <strong>p = T</strong> and <strong>q = F</strong>
<br>
Now Apply value of <strong>P</strong> in <strong>Eq 3</strong>, we get:
<br>
<span class="math-container">$$\begin{array}{cc}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$</span></p>
<p><strong>Now there are two possibilities when ¬p→r is T, and ¬p is F but the r has two separate values.</strong></p>
<p><strong>Is this System consistent or inconsistent?</strong></p>
| mohottnad | 955,538 | <p>It's perfectly ok that your set of sentences is consistent if you have 2 different models satisfying your set of sentences (a theory) since consistency has nothing to do with uniqueness of model as referenced <a href="https://en.wikipedia.org/wiki/Consistency" rel="nofollow noreferrer">here</a></p>
<blockquote>
<p>a consistent theory is one that does not lead to a logical contradiction. The lack of contradiction can be defined in either semantic or syntactic terms. The semantic definition states that a theory is consistent if it has a model, i.e., there exists an interpretation under which all formulas in the theory are true.</p>
</blockquote>
<p>In fact any set of tautological sentences such as {<span class="math-container">$(p=p), (q=q), (r=r)$</span>} can always have different truth values for any propositional sentence <span class="math-container">$p,q,r$</span> to stay to be consistent.</p>
<p>But look further about your set of particular sentences, <span class="math-container">$r$</span> can only be true from the constraint of your last sentence 4 since the antecedent of your material conditional is true then <span class="math-container">$r$</span> has to be true...</p>
|
4,277,924 | <p><strong>1. p ∧ ¬q = T</strong>
<br><br>
<strong>2. (q ∧ p) → r = T</strong>
<br><br>
<strong>3.¬p → ¬r = T</strong>
<br><br>
<strong>4.(¬q ∧ p) → r = T</strong>
<br><br>
From <strong>Eq 1</strong>, we got <strong>p = T</strong> and <strong>q = F</strong>
<br>
Now Apply value of <strong>P</strong> in <strong>Eq 3</strong>, we get:
<br>
<span class="math-container">$$\begin{array}{cc}
p&¬p&r&¬r&¬p\to r\\ \hline
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{T}}&\color{red}{\text{F}}&\color{red}{\text{T}}\\
\color{red}{\text{T}}&\color{red}{\text{F}}&\color{blue}{\text{F}}&\color{red}{\text{T}}&\color{red}{\text{T}}\\
\text{F}&\text{T}&\text{T}&\text{F}&\text{F}\\
\text{F}&\text{T}&\text{F}&\text{T}&\text{T}
\end{array}$$</span></p>
<p><strong>Now there are two possibilities when ¬p→r is T, and ¬p is F but the r has two separate values.</strong></p>
<p><strong>Is this System consistent or inconsistent?</strong></p>
| ryang | 21,813 | <p>A set of sentences as consistent iff their conjunction is satisfiable.</p>
<p>(Informally: a consistent system is one whose premises/axioms are coherent in <em>some</em> universe.)</p>
<p>So, in propositional logic, an inconsistent system is one whose conjunction is a contradiction, i.e., whose conjunction is false regardless of the combination of truth values of its atomic propositions.</p>
<p>So, in your exercise, the system is inconsistent iff <span class="math-container">$$(1 ∧ 2 ∧ 3 ∧ 4) \equiv\bot,$$</span> i.e., regardless of <span class="math-container">$(p,q,r)$</span>'s value, <span class="math-container">$(1 ∧ 2 ∧ 3 ∧ 4)=$</span> F,</p>
<p>i.e., every row of <span class="math-container">$(1 ∧ 2 ∧ 3 ∧ 4)$</span>'s truth table has a False <em>main connective</em>.</p>
<p>Because the main connective <span class="math-container">$\to$</span> in your simplified truth table of <span class="math-container">$(1 ∧ 2 ∧ 3 ∧ 4)$</span> is True thrice, your system is consistent.</p>
|
4,238,241 | <blockquote>
<p>Let tangents <span class="math-container">$PA$</span> and <span class="math-container">$PB$</span> on hyperbola from any point <span class="math-container">$P$</span> on the Director Circle of hyperbola such that <span class="math-container">$d(P,AB).d(C,AB)=4d(S_1,PA).d(S_2,PA)$</span> and <span class="math-container">$|AS_1-AS_2|=4$</span> where <span class="math-container">$d(P,AB)$</span> denotes distance of point <span class="math-container">$P$</span> to the line <span class="math-container">$AB$</span> and <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> are the foci and <span class="math-container">$C$</span> is centre of the hyperbola. If asymptotes of hyperbola pass through the point <span class="math-container">$(1,2)$</span> and a line <span class="math-container">$y=4$</span> intersects the branch of hyperbola which lies entirely on the first quadrant only at a point then find the length of traverse axis of the hyperbola and its equation.</p>
</blockquote>
<p>This is originally a question i had encountered in a test I had given, and i do know however that the director circle is the circle from which a pair of perpendicular tangents can be drawn and from the question it becomes clear that <span class="math-container">$2a=4$</span> but I am unable to make any use of the equation <span class="math-container">$d(P,AB).d(C,AB)=4d(S_1,PA).d(S_2,PA)$</span> maybe it is some property of hyperbola I am unaware of, in the given 'solution' they had directly stated the value of <span class="math-container">$b$</span> (of hyperbola) but I have no idea how they got that and I would highly appreciate any help with this question as it had been bugging me for quite a while now, thanks!</p>
<p><a href="https://i.stack.imgur.com/1twDX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1twDX.jpg" alt="enter image description here" /></a></p>
| Math Lover | 801,574 | <p>It is not a horizontal or vertical hyperbola - it has one branch which is entirely in the first quadrant and with other information provided, this fact will become clear as we go through the solution steps.</p>
<p>There are properties of hyperbola that would hold regardless of whether you rotated or shifted the coordinate axes. So let's work through them, first considering a hyperbola in a coordinate system such that,</p>
<p><span class="math-container">$\displaystyle \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$</span></p>
<p>We are given <span class="math-container">$|AS_1 - AS_2| = 4 \implies a = 2$</span> (by a property of hyperbola that you seem to be aware of)</p>
<p>Now I will use two other properties of hyperbola</p>
<p>First: <em>The product of perpendicular distances from foci to any tangent is <span class="math-container">$b^2$</span></em>.</p>
<p>So, <span class="math-container">$d(S_1,PA) \cdot d(S_2,PA) = b^2 \tag1$</span></p>
<p>Second: <em>The product of perpendicular distances from a point <span class="math-container">$P$</span> on the director circle and the center of the hyperbola / director circle to the chord of contact formed by tangents from point <span class="math-container">$P$</span> is given by <span class="math-container">$\frac{a^2b^2}{a^2-b^2}$</span></em>.</p>
<p>So,
<span class="math-container">$d(P,AB) \cdot d(C,AB) = \frac{a^2b^2}{a^2-b^2} \tag2$</span></p>
<p>I came to this on my own while working on this problem. I am sure there is literature available showing it but I have not been able to search and find this on the internet yet. See at the end of the answer for further details and how to go about a proof.</p>
<p>It is given that <span class="math-container">$d(P,AB).d(C,AB)=4d(S_1,PA).d(S_2,PA)$</span></p>
<p><span class="math-container">$\implies b = \sqrt3$</span></p>
<p>So we have the equation of hyperbola as,</p>
<p><span class="math-container">$\displaystyle \frac{X^2}{4}-\frac{Y^2}{3} = 1$</span></p>
<p>Now to find the equation of hyperbola in the original coordinate system, we look at a few other information provided in the question.</p>
<p>One of the branches of hyperbola is entirely in first quadrant and <span class="math-container">$y = 4$</span> intersects it only once. Then one of the asymptotes has to be parallel to x-axis given that the center of the hyperbola is <span class="math-container">$(1, 2)$</span>. As asymptote passes through the center, we conclude that <span class="math-container">$y = 2$</span> is one of the asymptotes. This asymptote got transformed to <span class="math-container">$Y = - \frac{\sqrt3}{2}X$</span> in the new coordinates system <span class="math-container">$XY$</span>.</p>
<p>From here on, it is just series of boring steps using rotation and translation back to original coordinate system to finally obtain the equation of hyperbola as,</p>
<p><span class="math-container">$ \large (y-2) (4 \sqrt3 x - y - 4 \sqrt3 + 2) = 12$</span></p>
<p>See below for additional info.</p>
<hr />
<p>Here is a plot and certain info using desmos and WolframAlpha.</p>
<p><a href="https://i.stack.imgur.com/90Od7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/90Od7.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/carNm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/carNm.png" alt="enter image description here" /></a></p>
<p>Now what if you are not convinced on <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> and want to prove it. By the way, there is enough literature easily available that shows <span class="math-container">$(1)$</span>.</p>
<p>For <span class="math-container">$(1)$</span>, take equation of a tangent to hyperbola at point <span class="math-container">$(X_0, Y_0)$</span> which is given by <span class="math-container">$\displaystyle \frac{X X_0}{a^2} - \frac{YY_0}{b^2} = 1$</span> and find perpendicular distance to this line from points <span class="math-container">$(\pm ae, 0)$</span> and multiply it. Using the fact that <span class="math-container">$\displaystyle \frac{X_0^2}{a^2} - \frac{Y_0^2}{b^2} = 1$</span>, you should be able to show it.</p>
<p>For <span class="math-container">$(2)$</span>, you know that director of circle is given by <span class="math-container">$X^2 + Y^2 = a^2 - b^2$</span>. Take a point <span class="math-container">$P (X_1, Y_1)$</span> satisfying this equation and draw pair of tangents and then chord of contacts <span class="math-container">$AB$</span> to hyperbola from external point <span class="math-container">$P$</span> is given by,</p>
<p><span class="math-container">$\displaystyle \frac{X X_1}{a^2} - \frac{YY_1}{b^2} = 1$</span></p>
<p>Find distance to <span class="math-container">$AB$</span> from <span class="math-container">$(X_1, Y_1)$</span> and <span class="math-container">$(0, 0)$</span> and take their product. Using the fact that <span class="math-container">$(X_1, Y_1)$</span> satisfies <span class="math-container">$X_1^2 + Y_1^2 = a^2 - b^2$</span>, you should be able to prove it.</p>
|
124,060 | <p>I have a small dark object in the upper left corner of an <code>image</code>.</p>
<p>How can I separate it from the noisy rest and determine its <code>IntensityCentroid</code>?</p>
<p><a href="https://i.stack.imgur.com/D7pHv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D7pHv.png" alt="enter image description here" /></a></p>
<p>The problem is that the objects intensity is not significantly lower than the intensity in the lower part of the image:</p>
<pre><code>ListPlot3D[ImageData[image] // N]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/GoLaS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GoLaS.png" alt="enter image description here" /></a></p>
</blockquote>
<p><strong>UPDATE</strong>:</p>
<p>Following the question <a href="https://mathematica.stackexchange.com/questions/109658/finding-objects-in-images-with-inhomogeneous-background">Finding objects in images with inhomogeneous background</a> I tried:</p>
<pre><code>imageN = ColorNegate[image];
background = ImageConvolve[imageN, GaussianMatrix[7]];
subImage = ImageSubtract[imageN, background];
t = FindThreshold[subImage, Method -> "Entropy"];
binImg = DeleteSmallComponents[Binarize[subImage, t], 5];
pts = ComponentMeasurements[ImageMultiply[image, binImg],
"IntensityCentroid"]
Show[image, Graphics[{Red, Point[pts[[All, 2]]]}]]
</code></pre>
<p>which gives:</p>
<blockquote>
<pre><code>{1 -> {77.2279, 397.997}}
</code></pre>
<p><a href="https://i.stack.imgur.com/8RCob.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8RCob.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Unfortunately the other solutions for the mentioned question cannot detect the object here.</p>
<p>As Rahul mentioned also <code>background = MedianFilter[imagen, 10]</code> can be used to detect the object. But this is much slower:</p>
<pre><code>background = ImageConvolve[imageN, GaussianMatrix[7]]; // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{0.005, Null}
</code></pre>
</blockquote>
<pre><code>background = MedianFilter[imageN, 20]; // RepeatedTiming
</code></pre>
<blockquote>
<pre><code>{22.8, Null}
</code></pre>
</blockquote>
| Svyatoslav Korneev | 42,293 | <p>Let me give you my general solution for the scientific image processing. The most important is to try to use less subjective parameters, like a threshold value. First, let's assume that the image is taken from charge-coupled device (CCD) or CMOS matrix. These detectors are widely used in most of the modern digital electronics: photo camera, scanner etc.</p>
<p>Second, plot the image histogram.</p>
<pre><code>im = Import["image.png"]
imdata = ImageData[im];
data = Flatten[ImageData[im]];
SmoothHistogram[data, PlotRange -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/nwwcu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nwwcu.png" alt="enter image description here"></a></p>
<p>You can see that the histogram looks like combination of two Gaussian PDF. It has Gaussian shape due to the photon-shot noise. Because the number of photon counts is finite. In general, the noise has Poisson distribution, but for the large number of photon counts Poisson PDF can be approached by Gaussian. Also from the image histogram, you can see that Gaussian curves are very close to each others, and the image has low contrast. To increase the contrast, you have to separate distributions.
Let's find distributions property. It can be done by various ways, but I use Expectation-Maximization algorithm for Gaussian Mixture Models. I gave the example of the algorithm for other <a href="https://mathematica.stackexchange.com/questions/28429/how-to-make-a-multi-peak-histogram-fitting/123814#123814">question</a>. The results of the algorithm you can find at the image below. Please note, it is important to choose the the starting point as better as you can, otherwise it may converge to single Gaussian only.</p>
<p>The result of the algorithm, where the red curve shows the recovered distribution, where two Gaussian functions where used.</p>
<p><a href="https://i.stack.imgur.com/IaYqs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IaYqs.png" alt="enter image description here"></a></p>
<p>And the left Gaussian has mean value </p>
<pre><code>μ1 = 0.588906
</code></pre>
<p>and the right</p>
<pre><code>μ2 = 0.623881
</code></pre>
<p>Let's define linear mapping to increase the contrast.</p>
<pre><code>df[gl_] := If[gl < μ2, (gl - μ2)/(μ1 - μ2), 0];
</code></pre>
<p>For the left maximum it gives 1 and for the right 0. Now let's map the initial image to the high-contrast image.</p>
<pre><code>{nRow, nCol} = Dimensions[imdata]
imdf = Table[df[imdata[[i]][[j]]], {i, 1, nRow}, {j, 1, nCol}];
</code></pre>
<p>And plot.</p>
<pre><code>Image[imdf]
</code></pre>
<p>As you can see, the spot and other features have higher contrast now.</p>
<p><a href="https://i.stack.imgur.com/kNu2a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNu2a.png" alt="enter image description here"></a></p>
<p>Now, you perform the simple threshold-based segmentation, where the threshold value is between pikes 0.5.</p>
<pre><code>segment =
Table[If[imdf[[i]][[j]] > 0.5, 1, 0], {i, 1, nRow}, {j, 1, nCol}];
binary = Image[segment]
</code></pre>
<p><a href="https://i.stack.imgur.com/r59vx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r59vx.png" alt="enter image description here"></a></p>
<p>Delete small and border components</p>
<pre><code>binnew = DeleteSmallComponents[DeleteBorderComponents[binary], 100]
</code></pre>
<p><a href="https://i.stack.imgur.com/D18yj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D18yj.png" alt="enter image description here"></a></p>
<p>And find your centroid</p>
<pre><code>ComponentMeasurements[binnew, "IntensityCentroid"]
{1 -> {77.2571, 397.8}}
</code></pre>
<p>UPDATE:</p>
<p>Actually, Otsu's algorithm gives similar result directly for the low-contrast image. As far as I see, the idea behind the Otsu's algorithm is similar to what I write here. But now it is clear how to extend the segmentation to arbitrary number of Gaussian.</p>
<p>Segmentation with Otsu's algorithm of the low-contrast image.</p>
<pre><code>ColorNegate[Binarize[im]]
</code></pre>
<p><a href="https://i.stack.imgur.com/NNoJ7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NNoJ7.png" alt="enter image description here"></a></p>
|
1,509,007 | <blockquote>
<p>Consider an unweighted and undirected graph $G=(V,E)$, where the vertices $V$ of $G$ lie on the unit n-sphere. If we choose a normal vector uniformly at random on this $n$-sphere, then the corresponding hyperplane (shifted to go through the origin) splits the vertices into two disjoint sets $A$ and $B$, and the expected number of edges cut by this hyperplane is
$$
\mathbb{E}[Cut(A,B)] =
\sum_{(i,j)\in E}\frac{\arccos(v_i\cdot v_j)}{\pi}\; .
$$</p>
</blockquote>
<p>So I understand that $\frac{1}{\pi}\arccos(v_i\cdot v_j)$ is basically the probability that a randomly chosen hyperplane which goes through the origin will separate $v_i$ and $v_j$. But I can't grasp how this sum is equal to the expectation described above.</p>
<p>Also, can this sum be extended to work for weighted graphs?</p>
| Tenno | 286,106 | <p>It isn't clear to me exactly what you are asking the community to provide, but you can compare this to previously established bounds to get a sense for how well you are doing. In particular, a mathematician named Robin showed that
$$
\sigma(n) < e^{\gamma}n\log(\log(n)) + \frac{.65n}{\log(\log(n))}
$$
holds for all $n\geq 3$. The constant $.65$ isn't quite right, but I don't have my references handy (maybe someone can comment/edit one in), but already this beats your bound for large $n$. So you do not have a "sharp" inequality, but it may have merits in that the proof might be easier than Robin's result. You will probably have to do the computational aspects of your problem yourself. I recommend using Python or Sage. </p>
|
1,030,274 | <p>$A,B,X,Y$ are four invertible matrices. If $AYB=XY$, can I express matrix $X$ in the terms of $A$ and $B$?</p>
| Andrea Mori | 688 | <p>Consider the ideal $I=(2,X)$ in $\Bbb Z[X]$.</p>
<p>Concretely, it consists of polynomials with integer coefficients of the form $a_0+a_1X+\cdots$ with $a_0$ even.</p>
<p>It cannot be generated by a single element $P(x)$ because you can never get $2$ and $X$ both multiples of a single polynomial.</p>
<p>But it is also false that $I=(2)\oplus(X)$ because $(2)\cap(X)$ contains non-zero elements (i.e. $2X$). A similar argument works if you attempt in any other way to generate $I$ with more than one generator.</p>
<p>The situation can be replicated almost verbatim for any ring which admits non principal ideals.</p>
|
267,051 | <p>Games appear in pure mathematics, for example, <a href="https://en.wikipedia.org/wiki/Ehrenfeucht%E2%80%93Fra%C3%AFss%C3%A9_game" rel="noreferrer">Ehrenfeucht–Fraïssé game</a> (in mathematical logic) and <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_game" rel="noreferrer">Banach–Mazur game</a> (in topology).</p>
<p>But the Game Theory behind those applications is not so deep, and we don't need to know some fundamental theorems for them. Maybe except Zarmelo Theorem.</p>
<p>Are there applications of <strong>advanced</strong> (anything behind the basic definitions) game theory ideas in pure mathematics?</p>
<p>Thanks!</p>
| Eilon | 64,609 | <p>You can read about the application of Blackwell's approachability theory, a nontrivial area in game theory, to normal numbers in <a href="https://www.jstor.org/stable/30035680?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">https://www.jstor.org/stable/30035680?seq=1#page_scan_tab_contents</a></p>
|
1,833,854 | <p>We have $M$ Binomial random variables, where $X_0 \sim $ Bin$(n,p)$ and $X_i \sim $ Bin$(n,1/2)$. </p>
<p>Suppose $p > 1/2$. I'm interested in the probability that $\mathbb{P}(\max \{X_1,\dots,X_M\} \geq X_0)$. Is this tractable? </p>
<p>If not, is it tightly boundable/approximable? If this is a very difficult question, I'd accept a reference providing some insight on the problem as well.</p>
| Jason | 164,082 | <p>$P( \max \{X_1,...,X_M \} \geq X_0)=P(X_i \geq X_0$ for some i) = $1-P( \text{each }X_i < X_0)$.
<br>
Now $P( \text{each }X_i < X_0)=\sum_{k=0}^n P(\text{each }X_i<X_0 \vert X_0=k) \cdot P(X_0 =k)$ =$\sum_{k=0}^n P( \text{each } X_i<k \text{ and } X_0=k)$ = $\sum_{k=1}^n \big[ P(X_1<k) \big]^M \cdot P(X_0=k)$. Note the k=0 term is 0.</p>
|
2,748,495 | <p>I have a numeric table for artillery operations (Royal Italian Army, year 1940), in the instructions it refers to a measure of a planar angle as $32.00^{\circ\circ}$ and it seems to me that this angle is equivalent to $\pi$ radian. I had a look at <a href="https://en.wikipedia.org/wiki/Gradian" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gradian</a> but it states that $\pi$ radian is equal to 200 gradian.</p>
<p>Referring to the following figure, given the angles $\alpha$, $\beta$ and the distance $b$, the numeric table will give the distance $X$.</p>
<p><a href="https://i.stack.imgur.com/mkwD7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mkwD7.jpg" alt="enter image description here"></a></p>
<p>My rough translation of the interesting part of the instruction is this:</p>
<blockquote>
<p>The angles $\alpha$ and $\beta$ will be measured by means of
goniometers. So in each end-point of the baseline you will set up a
goniometer, the direction of the goniometer is the direction of the
baseline $SD$ and the orientation of the goniometer will be from $D$
to $S$. So the angle of the line-of-sight from $D$ to $S$ will be set
up, in the goniometer in $D$, equal to $0.00^{\circ\circ}$ and the
angle of the line-of-sight from $S$ to $D$ will be set up, in the
goniometer in $S$, equal to $32.00^{\circ\circ}$.</p>
</blockquote>
<p><strong>Edit</strong></p>
<p>After the <a href="https://math.stackexchange.com/a/2748607/10799">answer</a> by <a href="https://math.stackexchange.com/users/1303/christian-blatter">Christian Blatter</a> I have found some nice pictures of the Swiss Army Compass with the "Art. ‰" scale:</p>
<p><a href="https://image.jimcdn.com/app/cms/image/transf/none/path/s904213d06cc35930/image/ifba6fe2fc60a4024/version/1424088065/image.jpg" rel="nofollow noreferrer"><img src="https://image.jimcdn.com/app/cms/image/transf/none/path/s904213d06cc35930/image/ifba6fe2fc60a4024/version/1424088065/image.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/QulF9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QulF9.jpg" alt="enter image description here"></a></p>
| Christian Blatter | 1,303 | <p>The full circle is subdivided into $6400$ <em>artillery promilles</em>, whereby "pro mille" is latin for ${1\over1000}$. The idea behind this angle measure is that a circle of radius $1000$ m has a circumference of $$2\pi\cdot 1000\ {\rm m}=6283.2\ {\rm m}\ \approx 6400\ {\rm m}\ ,$$ so that an angle of $1$ artillery promille corresponds to $1$ m circumference at a distance of $1000$ m. One then can say that also the tangent of this angle is about ${1\over1000}$, or $1$ per thousand. This means that you can easily convert promilles deviation measured with your binoculars into distances orthogonal to the line of sight, if you know the distance to the target, or conversely, can convert the (small) angle under which you see, e.g., a telephone pole into the distance to this pole.</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>
| Alexey Popkov | 280 | <p><a href="http://forums.wolfram.com/mathgroup/archive/2012/May/msg00371.html" rel="nofollow">This</a> <a href="https://groups.google.com/forum/#!forum/comp.soft-sys.math.mathematica" rel="nofollow">MathGroup</a> discussion should answer your question, so I'll cite it here:</p>
<blockquote>
<p>On Tue, 29 May 2012 05:47:52 -0400 (EDT), JCW wrote:</p>
<blockquote>
<p>Please forgive my dragging up ancient history: I have been using
Mathematica
from version 2.2 through 7.0. I remember at least one (maybe two?) format
conversions that were necessary to update old notebooks to the newer
formats. I also remember at least one batch-notebook-conversion command.
Unfortunately I cannot remember the details:</p>
<p>1) Between which versions did these format changes occur? (I'm pretty
sure there was one from the vs. 2.2 x.ma and x.mb format to the newer
x.nb format. Was there another format change within x.nb versions?)</p>
<p>2) What was the name of the batch converter (or converters) mentioned
above? (I know I have it in my archives, but it's difficult to find
without a name!)</p>
<p>Thanks in advance for any help!</p>
</blockquote>
<p>I'm always one for ancient history.</p>
<p>There was only the one switch from 2.2 .ma/.mb files to 3.0 .nb files.
Possibly, you might be remembering v6 when pre-v6 notebooks were
opened with an compatibility tool added to the top, but the tool had
nothing to do with the file format...it was merely an aid to updating
some of the code inside the notebook.</p>
<p>Also, .nb files are not wholly backward compatible. They'll always
open in older versions, but as new features get added to Mathematica,
inputs/outputs depending upon those features may not render or
function correctly in older versions. The most radical such change
would be in v6, when we switched from using PostScript for graphics
to more native constructs in the notebooks. Which means that pretty
much any v6 or later notebook is going to produce graphics which are
wholly unusable by v5 or earlier (but the notebooks <em>do</em> open, and
textual content comes through just fine).</p>
<p>The converter for .ma->.nb is <code>Developer`NotebookConvert</code>. It's still
documented in the latest shipping Mathematica.</p>
<p>Sincerely,</p>
<p>John Fultz</p>
<p>jfultz at wolfram.com</p>
<p>User Interface Group</p>
<p>Wolfram Research, Inc.</p>
</blockquote>
|
52,802 | <p>This is partly a programming and partly a combinatorics question.</p>
<p>I'm working in a language that unfortunately doesn't support array structures. I've run into a problem where I need to sort my variables in increasing order.</p>
<p>Since the language has functions for the minimum and maximum of two inputs (but the language does not allow me to nest them, e.g. <code>min(a, min(b, c))</code> is disallowed), I thought this might be one way towards my problem.</p>
<p>If, for instance, I have two variables $a$ and $b$, I only need one temporary variable so that $a$ ends up being less than or equal to $b$:</p>
<pre><code>t = min(a, b);
b = max(a, b);
a = t;
</code></pre>
<p>for three variables $a,b,c$, the situation is a little more complicated, but only one temporary variable still suffices so that $a \leq b \leq c$:</p>
<pre><code>a = min(a, b);
t = max(a, b);
c = max(t, c);
t = min(t, c);
b = max(a, t);
a = min(a, t);
</code></pre>
<p>Not having a strong combinatorics background, however, I don't know how to generalize the above constructions if I have $n$ variables in general. In particular, is there a way to figure out how many temporary variables I would need to sort out $n$ variables, and to figure out what is the minimum number of assignment statements needed for sorting?</p>
<p>Thanks in advance!</p>
| James | 68,019 | <p>Following Sleziak's suggestion, it might be useful to point out that under the bijection $A\mapsto\chi_{A}$, the symmetric difference is realized as </p>
<p>\[ A\triangle B\sim\left|\chi_{A}-\chi_{B}\right|^{2}.\]</p>
<p>It is then straightforward to check that multiplication is distributive: </p>
<p>\[ C\cap\left(A\triangle B\right)
\sim
\chi_{C}\left|\chi_{A}-\chi_{B}\right|^{2}=\left|\chi_{C\cap A}-\chi_{C\cap B}\right|^{2}
\sim
\left(C\cap A\right)\triangle\left(C\cap B\right).\]</p>
|
2,439,744 | <p>How to prove that $\lim_{x\to 0,y\to 0}{\frac {\sqrt {a+x^2y^2} -1} {x^2+y^2}} (a>0)$ doesn't exist while a $\ne 1$?</p>
<p>I already calculated that when a = 1 by multiplying $\sqrt {a+x^2y^2} + 1$ on both denominator and numerator and using the fact $x^2y^2<(x^2+y^2)^2/4$.</p>
<p>Any help will be appreciated.</p>
| Parcly Taxel | 357,390 | <p>Evaluating on the path $x=y$:
$$\lim_{x,y\to0}{\frac{\sqrt{a+x^2y^2}-1}{x^2+y^2}}=\lim_{x\to0}{\frac{\sqrt{a+x^4}-1}{2x^2}}=\lim_{x\to0}{\frac{\sqrt a-1}{2x^2}}$$
If $a\ne1$ then $\sqrt a-1\ne0$ and a singularity exists at $x=y=0$, so the limit does not exist.</p>
|
2,662,605 | <p>Problem: If ${F_n}$ is a sequence of bounded functions from a set $D \subset \mathbb R^p$ into $ \mathbb R^q$ and if ${F_n}$ converges uniformly to $F$ on $D$, then $F$ is also bounded. </p>
<p>Proof(Attempt): Let $\epsilon >0$. Since ${F_n}$ converges uniformly to $F$ on $D$, then there is an $N \in\mathbb R$ such that $||F(x)-F_n(x)||< \epsilon $ whenever $x \in D$ and $n\ge N$. </p>
<p>I'm using Joseph Taylor's Foundation of Analysis textbook.
Since ${F_n}$ is bounded we know that $||F_n|| \le M $ for every $x\in D$.<br>
At this point, I'm thinking that I need to use a trick that gets $||F|| \le M$ from the inequality $||F(x)-F_n(x)||< \epsilon $. </p>
<p>I would appreciate advise and hints that will help guide me. </p>
| spaceisdarkgreen | 397,125 | <p><strong>HINT</strong></p>
<p>We have $$ \| F(x) \| = \|F(x)-F_n(x) + F_n(x)\| \le \|F(x)-F_n(x)\| + \|F_n(x)\|.$$</p>
|
251,559 | <p>First of all, this is a very silly question, but I finished high school a long time ago and I really don't remember much about some basic stuff.</p>
<p>I have:</p>
<p>$$T(x) = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>$T$ = temperature</p>
<p>$x$ = hour</p>
<p>And they ask me to find the hour in which the temperature reached zero. So I know I must replace:</p>
<p>$$0 = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>It's very easy, I know, but I really don't remember how to clear/free the $x$ in this case.
Could you guys help me?</p>
| Jebruho | 40,030 | <p>Use the quadratic formula so that the zeroes of the function occur at $\frac{-b \pm \sqrt {b^{2}-4ac}}{2}$, where $a=\frac{-1}{10},b=\frac{24}{10}x,c=\frac{-44}{10}$. In general, polynomials of up to degree 4 have solutions in roots, but solutions of degree 5 or higher do not.</p>
|
251,559 | <p>First of all, this is a very silly question, but I finished high school a long time ago and I really don't remember much about some basic stuff.</p>
<p>I have:</p>
<p>$$T(x) = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>$T$ = temperature</p>
<p>$x$ = hour</p>
<p>And they ask me to find the hour in which the temperature reached zero. So I know I must replace:</p>
<p>$$0 = -\frac1{10}x^2 + \frac{24}{10}x - \frac{44}{10}$$</p>
<p>It's very easy, I know, but I really don't remember how to clear/free the $x$ in this case.
Could you guys help me?</p>
| Thomas | 26,188 | <p>You want to solve the quadratic equation:</p>
<p>$$
0 = -\frac{1}{10}x^2 + \frac{24}{10}x - \frac{44}{10}
$$
You can indeed just apply the quadratic formula, but first you can make things a tiny bit easier by multiplying by -10 on both sides to get
$$
0 = x^2 - 24x + 44.
$$
Now if $a$ and $b$ are the solutions you can also factor the polynomial as
$$
(x-a)(x-b) = x^2 - (a+b) + ab.
$$
So if you wanted to try and guess the solutions, you are trying to find two numbers $a$ and $b$ such that $-(a + b) = -24$ and $ab = 44$. Maybe it is not too hard too see that $a = 22$ and $b = 2$. So in all you get the two solutions to the original equation:
$$x = 22 \quad \text{and} \quad x = 2.
$$</p>
|
4,198,496 | <p>If the value of <span class="math-container">$\int_{1}^2{e^{x^{2}}}dx$</span> is <span class="math-container">$\alpha$</span>, then what is <span class="math-container">$$\int_{e}^{e^{4}}{\sqrt{\log x}}dx$$</span></p>
<p>It just seems that substitution of any sort does not help. Is there some other way in which these two integrals are related?</p>
<p>The answer is <span class="math-container">$2e^4 - e - \alpha$</span></p>
| Tito Eliatron | 84,972 | <p>Take <span class="math-container">$t=\sqrt{\log x}$</span> then <span class="math-container">$x=e^{t^2}$</span> and <span class="math-container">$dx=2te^{t^2}$</span>. So
<span class="math-container">$$I=\int_e^{e^4}\sqrt{\log x}dx=\int_1^22t^2e^{t^2}dt.
$$</span>
Using integration by parts: <span class="math-container">$u=t$</span> <span class="math-container">$dv=2te^{t^2}dt$</span>, you'll get that <span class="math-container">$v=e^{t^2}$</span> and
<span class="math-container">$$I=te^{t^2}\bigg|_1^2-\int_1^2e^{t^2}dt=2e^4-e-\alpha
$$</span></p>
|
1,225,359 | <p>Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:</p>
<p>What I tried to do was make pairs of two terms and the expand it as a binomial expression and then again expand the binomial in the resulting series which gave me an expression with lot of unknowns and I got stuck. </p>
<p>Any help would be appreciated.
Thanks.</p>
| lab bhattacharjee | 33,337 | <p>$$(x+y+z+w)^6=\{y+(x+w+z)\}^6=\cdots+(x+w+z)^6$$</p>
<p>$$(x+w+z)^6=\sum_{r=0}^6\binom6rx^{6-r}(w+z)^r$$</p>
<p>We need $r\ne6$</p>
<p>The sum of the reuqired coefficients should be $\sum_{r=0}^5\binom6r(1+1)^r$ (setting $x=w=z=1$)</p>
<p>$$=\sum_{r=0}^6\binom6r2^r-\binom662^6=(1+2)^6-2^6$$</p>
|
3,146,161 | <blockquote>
<p>Find particular solution of <span class="math-container">$\dfrac{dx}{dy} +x\cot y =y\cot y$</span> given <span class="math-container">$x=0$</span> when <span class="math-container">$y= π/2$</span>.</p>
</blockquote>
<p>Please help me figure this out
I can't separate them out in order to integrate</p>
| Henry Lee | 541,220 | <p>For an integral to be convergent:
<span class="math-container">$$\int_0^\infty f(x)dx=I$$</span>
where <span class="math-container">$I$</span> is a real number. We need the function to decrease such that:
<span class="math-container">$$\lim_{x\to\infty}f(x)=0$$</span>
however for this function,
<span class="math-container">$$\lim_{x\to\infty}e^{x^2}\to\infty$$</span>
so the integral cannot converge</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>
| user236182 | 236,182 | <p>The question says $p>0$. Let $x=\frac{1}{t}$. $$\lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}]=$$</p>
<p>$$=\lim_{t\to 0^+}\frac{(t^{p+1}+1)^{\frac{1}{p+1}}-1}{t}-\lim_{t\to 0^+}\frac{(t^p+1)^{\frac{1}{p}}-1}{t}=$$</p>
<p>$$=((t^{p+1}+1)^{\frac{1}{p+1}})'|_{t=0}-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=$$</p>
<p>$$=\left(\frac{1}{p+1}(t^{p+1}+1)^{-\frac{p}{p+1}}(p+1)t^p\right)|_{t=0}-$$</p>
<p>$$-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=0-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}=$$</p>
<p>$$=-((t^p+1)^{\frac{1}{p}})'|_{t=0^+}$$</p>
<p>If $p>1$, then</p>
<p>$$=-\left(\frac{1}{p}(t^p+1)^{\frac{1-p}{p}}pt^{p-1}\right)|_{t=0^+}=-0=0$$</p>
<p>If $0<p<1$, then</p>
<p>$$=-\left(\frac{1}{p}(t^p+1)^{\frac{1-p}{p}}pt^{p-1}\right)|_{t=0^+}=-\infty$$</p>
<p>If $p=1$, then</p>
<p>$$=-(t+1)'|_{t=0^+}=-1$$</p>
|
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>
| user44197 | 117,158 | <p>Note that trace equals zero says that the $n,n$ term is given by the remaining. So start with the usual basis with
$$
A_{k,m}=1, ~~~ (k,m) \ne (n,n),\\ A_{i, j} = 0 , ~~~~(i,j) \ne (k, m), \\
A_{n,n} = -\sum_{i=1}^{n-1} A_{i,i}$$</p>
<p>Thus the basis has $n^2-1$ elements.</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>
| obataku | 54,050 | <p>The Green's function for $\dfrac{d^2}{dx^2}$ is the ramp function $x\theta(x)$, where $\theta(x)$ is the <a href="http://en.wikipedia.org/wiki/Heaviside_step_function" rel="nofollow">Heaviside step function</a>. We can write the solution for the general inhomogeneous problem as follows. $$\begin{align*}y(x)&=\int_{-\infty}^\infty (x-\chi)\theta(x-\chi)f(\chi)d\chi\\&=\int_{-\infty}^x (x-\chi) f(\chi)\, d\chi\end{align*}$$for suitable conditions on $f$.</p>
|
509,928 | <p>There is a square cake. It contains N toppings - N disjoint axis-aligned rectangles. The toppings may have different widths and heights, and they do not necessarily cover the entire cake.</p>
<p>I want to divide the cake into 2 non-empty rectangular pieces, by either a horizontal or a vertical cut, such that the number of toppings I destroy (i.e. cross in the interior) is minimized.</p>
<p>What is the number of toppings I will have to destroy, in the worst case, as a function of N?</p>
<p>CURRENT BOUNDS:</p>
<p><strong>Upper bound $N/2$</strong>:</p>
<p>Take any horizontal cut. If it crosses no more than N/2 toppings, then we are done. Otherwise, make a vertical cut between two of the crossed rectangles. This vertical cut does not cross any rectangle crossed by the horizontal cut, therefore it crosses at most N/2 toppings.</p>
<p><strong>Lower bound $N/4$</strong>:</p>
<p>In the following cake, with 4 toppings, every cut must cross at least 1 topping:</p>
<pre><code>aaaaaaaa bb
aaaaaaaa bb
cc ..... bb
cc ..... bb
cc ..... bb
cc dddddddd
cc dddddddd
</code></pre>
<p>As MvG suggested, it is possible to cut each rectangle into $N/4$ parallel strips, forcing a cut to destroy at least $⌊N/4⌋$ toppings.</p>
<p>NOTE: I just found out that this problem is related to the topic of <a href="https://en.wikipedia.org/wiki/Geometric_separator" rel="nofollow">Geometric separators</a>. The lower bound example and the upper bound proofs are given in Section 4 of <a href="http://users.monash.edu.au/~nwormald/papers/geomsep.ps.gz" rel="nofollow">Smith and Wormald (1998)</a>. There is still a gap between the lower and upper bound.</p>
| Jeff Snider | 119,951 | <p>Supposing the cake is open, and the toppings are open, we can position them so that a cut between edge and topping is impossible, and cuts between adjacent toppings are possible.</p>
<p>It appears that $\lfloor \frac{N}{4} \rfloor$ is the maximal case for a given $N>1$ on your two-dimensional cake, as demonstrated by your $N=4$ layout. For $N=8$ put two parallel toppings where each one of yours is. For $N=4k$ put $k$ parallel toppings.</p>
<p>For $N=1$ we just use a single topping covering the whole surface.</p>
<p>Proof that for $N=2$ or $N=3$ there is no layout requiring a cut topping is a first step to proving $\lfloor \frac{N}{4} \rfloor$ is the correct formula.</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>
| Mastrem | 253,433 | <p>Using the identity <span class="math-container">$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$</span>, we find that
<span class="math-container">$$
\begin{align*}
\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}} &= \sqrt[n]{\frac1{2^n}}\cdot\sqrt[n]{\frac{20}{2^4+2^2}}\\
&= \sqrt[n]{\frac1{4^n}}\cdot\sqrt[n]{\frac{20}{16+4}}\\
&=\sqrt[n]{\left(\frac14\right)^n}\cdot\sqrt[n]{\frac{20}{20}}\\
&=\frac14.
\end{align*}
$$</span></p>
|
2,685,424 | <p>What I don't understand is that why can't we find the general solution of non homogeneous differential equation from the non homogeneous one itself. Currently we use the homogeneous equation also. </p>
<p>Why isn't it that general solution is not available from the non-homogeneous equation itself?</p>
| symplectomorphic | 23,611 | <p>The whole point is that if $x_1$ and $x_2$ solve $L(x)=y$, where $L$ is a linear differential operator, then $$0=y-y=L(x_1)-L(x_2)=L(x_1-x_2)$$ so $x_1$ and $x_2$ differ by a homogeneous solution. Note that linearity is crucial here.</p>
<p>(You seem to think the inhomogeneous equation has a <em>unique</em> particular solution. But this is not true. When you solve an inhomogeneous equation, you find <em>a</em> particular solution of the infinitely many that are available. The other solutions to the equation differ from the one you find by a homogeneous solution, that is, by an element of the kernel of the linear operator.)</p>
|
501,250 | <p>I want to say that $|\textbf{x}-\textbf y|<\delta$ implies $|x_1- y_1|<\delta$ and $|x_2- y_2|<\delta$ for a proof I am working on. This is assuming that $\textbf{x}=(x_1,x_2) \in \text R^2$ and $\textbf{y}=(y_1,y_2) \in \text R^2$. If true, I'd also like to extend this to $\textbf{x} \in \text R^{n_1+n_2}$ and $\textbf{x} \in \text R^{n_1+n_2}$ where $n_1 , n_2 \in \text N$.</p>
<p>I am wondering if this is obvious enough to state or would it be better to prove it? If I should prove it then what would be the best way? My guess would be to use the distance formula. </p>
| Prahlad Vaidyanathan | 89,789 | <p>$C$ is not closed because $0$ is a limit point of $C$, but $O \notin C$.</p>
<p>$C$ is not open because, for any $x = 1/n \in C$, and any $\delta > 0$, there is a point $z\in \mathbb{R}\setminus C$ such that $x-\delta < z < x+\delta$</p>
|
63,348 | <p>This question arises from a discussion with my friends on a commonly encountered IQ test questions: "What's the next number in this series 2,6,12,20,...". Here a "number" usually means an integer. I was wondering whether there is a systematical way to solve such problems.Let us call a point on a plane integer point if all the components of it are integers. I want to know the following:</p>
<p>Give a finite set of integer points, Can we always find a corresponding polynomial that passes all these points and maps integers to integers? </p>
| Robert Israel | 13,650 | <p>Let the points be $(x_j, y_j), j=1\ldots n$. If the $x_j$ are consecutive, the Lagrange interpolating polynomial will take integers to integers: easy proof by induction, using the difference operator $\Delta(p)(x) = p(x+1) - p(x)$. If not, choose arbitrary integers for the $y$ values to fill in the gaps. </p>
|
1,710,517 | <blockquote>
<p>Show that 13 is the largest prime which divides two consecutive terms of $n^2 + 3$.</p>
</blockquote>
<p>The integers are $39$ and $52$. First of all, I set the variable for the number as $k$. So, $k|n^2 +3$ and $k|n^2 + 2n+ 4$ which imply that $k|2n+1$. $n=6$ over here. And the fact that 13 is the largest 'prime' makes me feel it is hard to prove. That's all I have managed to get. I need a few hints to set me in the right direction. Thanks.</p>
| paw88789 | 147,810 | <p>If $k\mid n^2+3$ and $k\mid n^2+2n+4$, then as you noted, $k\mid 2n+1$.</p>
<p>But then also from $k\mid n^2+3$ we have $k\mid 2n^2+6$, and from $k\mid 2n+1$ we have $k\mid 2n^2+n$.</p>
<p>Hence $k\mid (2n^2+n)-(2n^2+6))=n-6$. And so $k\mid 2n-12$.</p>
<p>From $k\mid 2n-12$ and $k\mid 2n+1$, we obtain $k\mid 13$.</p>
|
4,292,815 | <p>Compute line integral <span class="math-container">$\int_a^b (y^2z^3dx + 2xyz^3dy + 3xy^2z^2dz)$</span> where <span class="math-container">$a = (1,1,1)$</span> and <span class="math-container">$b = (2,2,2)$</span></p>
<p>What I have done:</p>
<p>To find <span class="math-container">$t$</span> I used the calculation for slope:</p>
<p><span class="math-container">$\frac{x-1}{2-1}=t, \frac{y-1}{2-1}=t, \frac{z-1}{2-1}=t$</span> and then re-arranged for x,y,z to calculate the derivate to find that <span class="math-container">$dx =1, dy=1, dz=1$</span> and <span class="math-container">$t+1 = (x, y, z)$</span></p>
<p>Plugging this back into the integral:
<span class="math-container">$$6\int_a^b (t+1)^5 dt$$</span></p>
<p>However, what do I do with the integral bounds? Do I set them from <span class="math-container">$t \in [1,2]$</span>?</p>
| Community | -1 | <p>The <a href="https://math.stackexchange.com/questions/4292815/compute-line-integral-int-ab-y2z3dx-2xyz3dy-3xy2z2dz/4292824#comment8938860_4292815">hint</a> you mentioned implies that you can use the <a href="https://en.wikipedia.org/wiki/Gradient_theorem" rel="nofollow noreferrer">fundamental theorem of calculus</a>. Note that
<span class="math-container">$$
(y^2z^3,2xyz^3,3xy^2z^2)=\nabla f(x,y,z)
$$</span>
where <span class="math-container">$f(x,y,z)=xy^2z^3$</span>. So the integral is path independent, and you don't really need to parametrize your path:
<span class="math-container">$$
\int_A^B \nabla f(x,y,z)\cdot dr=f(B)-f(A)
$$</span>
where <span class="math-container">$A=(1,1,1)$</span>, <span class="math-container">$B=(2,2,2)$</span>.</p>
<p>Usually, the notation <span class="math-container">$\int_A^B$</span> is only used for path integrals that are path independent. Otherwise, one usually uses <span class="math-container">$\int_\gamma$</span> instead and describes <span class="math-container">$\gamma$</span> in words.</p>
|
858,353 | <p>$$x^2(y')^2+3xyy'+2y^2=0$$
I have no idea how to start, I probably need to do some tricky substitution but as of now I cant see any options.</p>
| Community | -1 | <p>It factors as $$(xy'+2y)(xy'+y)=0.$$ Now solve each equation separately.</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>
| kglr | 125 | <p>You can also turn <code>prod</code> into an <code>Association</code> using sorted second elements as keys and then use <a href="https://reference.wolfram.com/language/ref/Lookup.html" rel="nofollow noreferrer"><code>Lookup</code></a>:</p>
<pre><code>Lookup[Sort/@sub] @ GroupBy[ Sort@*Last] @ prod
</code></pre>
<blockquote>
<p>{{{"x4", {"m", "n"}}}, {{"x5", {"o", "p", "q", "r"}}}, {{"x2", {"e", "f", "g"}}}} </p>
</blockquote>
<p>Make it a function:</p>
<pre><code>ClearAll[lookUp]
lookUp[keys_] := Lookup[Sort /@ keys] @* GroupBy[Sort@*Last]
lookUp[sub] @ prod
</code></pre>
<blockquote>
<p>{{{"x4", {"m", "n"}}}, {{"x5", {"o", "p", "q", "r"}}}, {{"x2", {"e",
"f", "g"}}}}</p>
</blockquote>
|
162,611 | <p>I am working with the square-roots of square symmetric matrices. The answers are to be binary symmetric matrices.</p>
<p>If we take the matrix $$M = \begin{pmatrix}1&1&1&0&0&0&0\\1&0&0&1&1&0&0\\1&0&0&0&0&1&1\\0&1&0&0&1&0&1\\0&1&0&1&0&1&0\\0&0&1&0&1&1&0\\0&0&1&1&0&0&1 \end{pmatrix},$$ then $M^2$ will yield a $7 \times 7$ matrix with $3$ down the main diagonal and $1$ elsewhere.Once we have the matrix $M^2$ then we want to take the square root of it in order to get M.Now this is the simplest of all examples that I can give. M is the incidence matrix of a symmetric balanced incomplete block design. This design has 7 varieties, 7 blocks,each variety occurs in 3 blocks,each block contains 3 varieties, each pair of varieties occurs in exactly one block. If you are familiar with block designs then the parameters are (v,b,r,k,l)=(7,7,3,3,1) known as the fano plane.</p>
| Will Jagy | 10,400 | <p>Gerry Myerson gave a recipe for finding a symmetric solution, also providing a solution in a comment. I wanted to implement that while preserving the structure of the Don Giles matrix for order 2 (Fano plane). That is, I wanted the first four rows and first four columns to be the analogous pattern to the 7 by 7. Gerry's method amounts to this: take all 13 lines in $\mathbb F_3^3,$ where $\mathbb F_3$ is the field with three elements. The 13 lines can be parametrized by triples of $1,0,-1,$ where the first nonzero entry is $+1.$ There is a $1$ in the matrix for a pair of vectors out of the 13 if their dot product is $0 \pmod 3.$ I decided to label the 13 points using capital letters. Notice that $A \cdot A \equiv G \cdot G \equiv J \cdot J \equiv M \cdot M \equiv 0 \pmod 3.$</p>
<p>LABELS:
$$
\begin{array}{rrrr}
0 & 0 & 1 & K \\
0 & 1 & -1 & B \\
0 & 1 & 0 & H \\
0 & 1 & 1 & F \\
1 & -1 & -1 & G \\
1 & -1 & 0 & D \\
1 & -1 & 1 & J \\
1 & 0 & -1 & C \\
1 & 0 & 0 & E \\
1 & 0 & 1 & I \\
1 & 1 & -1 & M \\
1 & 1 & 0 & L \\
1 & 1 & 1 & A
\end{array}
$$</p>
<p>The order of both the rows and the columns of the matrix is alphabetical, $ABCDEFGHIJKLM.$ As I said, there is a $1$ in row $X,$ column $Y$ (and the reverse) when $X \cdot Y \equiv 0 \pmod 3.$ </p>
<p>MATRIX:
$$ P \; = \;
\left( \begin{array}{cccc|ccc|ccc|ccc}
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ \hline
0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\ \hline
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ \hline
0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1
\end{array}
\right).
$$</p>
<p>The result is that $P$ is symmetric and $P^2$ is the matrix with all 4's on the main diagonal and 1's everywhere else.</p>
|
839,431 | <p>Can someone be kind enough to show me the steps to integrate this, I'm sure it's by parts but how do I split up the sin part?
$$x\sin(1+2x)$$</p>
| user1729 | 10,513 | <p>I would start with an integration by substitution first. So, let $z=1+2x$ and note that this implies $x=\frac{z-1}{2}$, so we obtain the following. $$\frac14\int(z-1)\sin z\operatorname{d}z=\frac14\int z\sin\operatorname{d}z-\frac14\int \sin z\operatorname{d}z$$
Then, integrating $\sin z$ is standard, while $z\sin z$ is a standard easy example of integration by parts (and is probably in your notes!).</p>
|
253,746 | <p>I am currently aware of the following two versions of the global Cauchy Theorem. Which one is stronger?</p>
<p>1.)If the region $U$ is simply connected, then for every closed curve contained therein, the integral of the holomorphic function $f$ defined on $U$ over the curve is zero.</p>
<p>2.) If the domain $D$ is an arbitrary open set, then for every closed curve contained therein with the property that the index is zero for points outside $D$, the integral of the holomorphic function $f$ defined on $D$ is zero.
Thanks in advance.</p>
| Julián Aguirre | 4,791 | <p>The second one is stronger, since any closed curve in a simply connected domain satisfies the condition imposed on the second version of Cauchy's theorem.</p>
<p>Somewhat more general versions of the theorem are called the homotopical version (corresponding to your first theorem) and the homological version (corresponding to your second one.) The homological version is stronger, since any null-homotopic curve is also null-homlogous.</p>
|
13,989 | <p>Suppose $E_1$ and $E_2$ are elliptic curves defined over $\mathbb{Q}$.
Now we know that both curves are isomorphic over $\mathbb{C}$ iff
they have the same $j$-invariant.</p>
<p>But $E_1$ and $E_2$ could also be isomorphic over a subfield of $\mathbb{C}$.
As is the case for $E$ and its quadratic twist $E_d$. Now the question general is.</p>
<blockquote>
<p>$E_1$ and $E_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$
the smallest subfield of $\mathbb{C}$ such that $E_1$ and $E_2$ become isomorphic over $K$.
What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be
said about the extension $K|\mathbb{Q}$.</p>
</blockquote>
<p>My second question is something goes something like in the opposite direction. I start again with
quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension
$\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$)
which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E_d$.
Again I would like to know what happens if we take a larger extension.</p>
<blockquote>
<p>Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension.
Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic
to $E$ over K.</p>
</blockquote>
<p>I have no idea what is the right context to answer such questions.</p>
| H. Hasson | 2,665 | <p>Hopefully I'll have some time later to ellaborate, but for now - here is a great reference I wish somebody had shown me when I started out, for how to attack questions of this type:
<a href="http://books.google.com/books?id=l0DgAIx_djoC&printsec=frontcover&dq=waterhouse+affine&source=bl&ots=nup7qU4Aln&sig=VXgsPcQNgAkCbNsH8KG-VAakjio&hl=en" rel="nofollow">http://books.google.com/books?id=l0DgAIx_djoC&printsec=frontcover&dq=waterhouse+affine&source=bl&ots=nup7qU4Aln&sig=VXgsPcQNgAkCbNsH8KG-VAakjio&hl=en</a></p>
<p>look at chapter 17 (and 18).</p>
|
497,546 | <p>Let $A$ be an infinite set.</p>
<p>Then, we can construct an injective function $f:\omega \rightarrow A$. </p>
<p>But how do i construct this via orginal statement of $AC_\omega$? (i.e. $\forall countable X, [\emptyset \notin X \Rightarrow \exists f:X\rightarrow \bigcup X \forall A\in X, f(A)\in A$)</p>
<p>So my question is:</p>
<p>(1) how do i prove that every infinite set has a countable subset?</p>
<p>(2) how do i write down a situation such as: "after choosing $x_1,...,x_k$, choosing $x_k$ satisfying a given condition. Continue this and form a sequence"</p>
| You_Don't_Know_Who | 484,838 | <p>Hint: Actually, one can simply use binomial theorem to get the expansion of <span class="math-container">$(1-q)^{1/2}$</span>. Then, substitute <span class="math-container">$q=2xt-t^2$</span>. Again, use binomial expansion on the powers of <span class="math-container">$q$</span>. Combining the two expansions, you can show that the power <span class="math-container">$t^n$</span> has the coefficient <span class="math-container">$P_n(x)$</span>. Then the proof completes.</p>
|
2,759,407 | <p>Suppose there's a exam with 5 questions.
If the probability that Student $1$ correctly answers question $i$ is $P_{1.i}$, then</p>
<p>$P_{1.1} = 0.3$ , $P_{1.2} = 0.4$ , $P_{1.3} = 0.9$, $P_{1.4} = 0.7$ , $P_{1.5} = 0.1$</p>
<p>For Student $2$, </p>
<p>$P_{2.1} = 0.4$ , $P_{2.2} = 0.5$ , $P_{2.3} = 0.2$, $P_{2.4} = 0.8$ , $P_{2.5} = 0.1$</p>
<p>What is the probability that Student $1$ performs better than Student $2$ ?</p>
<p>How to solve something like that? I want an expression to do this.</p>
| Henno Brandsma | 4,280 | <p>Compute all probabilities $P(C_i = j)$ where $C_i$ is the number of correct questions scored by student $i$. These are just sums of products of the $P_{i,j}$ really. Then sum all $P(C_1 = j)P(C_2 = k)$ over all pairs $j > k$ that are allowed. This is allowed by independence of the scores of both students.</p>
<p>E.g. to compute the probability that student 1 has 2 correct answers, consider all combinations of two answers out of the five. Say one of these is $\{i,j\}$, which means that student $1$ answers questions $i$ and $j$ correctly and the others incorrectly. The chance of exactly that happening is $P_{1,i} \times P_{1,j}$ times all factors $(1-P_{1,n})$ where $n \neq i,j$. And this we have to do for all $\binom{5}{2}$ sets of possible $\{i,j\}$ to compute $P(C_1 = 2)$.</p>
<p>This is quite tedious to do by hand, so I wrote a <a href="https://gist.github.com/HendrikSikke/f81cf3c60de6f138f797241fb0189700" rel="nofollow noreferrer">program</a>:</p>
<p>The results were:</p>
<p><code>P student 1 has 0 correct = 0.011340
P student 2 has 0 correct = 0.043200
P student 1 has 1 correct = 0.142200
P student 2 has 1 correct = 0.260400
P student 1 has 2 correct = 0.397800
P student 2 has 2 correct = 0.406400
P student 1 has 3 correct = 0.340000
P student 2 has 3 correct = 0.236400
P student 1 has 4 correct = 0.101100
P student 2 has 4 correct = 0.050400
P student 1 has 5 correct = 0.007560
P student 2 has 5 correct = 0.003200
1 does better with probability 0.471532
2 does better with probability 0.243788
equal score with probability 0.284680</code></p>
|
2,292,096 | <p>I'm learning inequalities for the first time, and except a paragraph by Paul Zeitz in his book Art and Craft of problem solving, none actually give much motivation of why should I care about inequalities.</p>
<p>The example given by Paul Zeitz was that to prove $b^2-b+1$ is never a perfect square for integer $b$. Well- that kinda motivates a tiny bit, but the inequality used is triviality obvious; I want much deeper.</p>
<blockquote>
<p>What are some easy-to-state, moderately hard to solve number theoretic or combinatorics problem which requires applying a nontrivial inequality ?</p>
</blockquote>
| DanielWainfleet | 254,665 | <p>(I). Many Q's are readily answered by the following: Let $J$ be a (bounded or unbounded) interval of $\mathbb R.$ Let $f:J\to \mathbb R$ with $f''(x)<0$ for all $x\in J.$ Then when $w_1,...,w_n$ are non-negative with $\sum_{i=1}^nw_i=1,$ and $x_1,...,x_n \in J$ then $$(*)\quad f\left(\sum_{i=1}^nw_ix_i\right)\geq \sum_{i=1}^nw_if(x_i)$$ with equality iff $x_i=x_j$ whenever $w_i\ne 0\ne w_j.$ </p>
<p>Example: $J=(0,\infty)$ and $f(x)=\ln x.$ Then $\log (\sum_{i=1}^nw_ix_i)\geq \sum_{i=1}^nw_i\log x_i,\;$ which is equivalent to $\sum_{i=1}^n w_ix_i\geq \prod_{i=1}^nx_i^{w_i}.$ </p>
<p>In particular when $w_i=1/n$ for each $n$ we have $(\sum_{i=1}^nx_i)/n\geq (\prod_{i=1}^nx_i)^n,$ which is the AGM inequality.</p>
<p>If, instead, we have $f''(x)>0$ for all $x\in J$ then the inequality in (*) is reversed. An example Q that was on this site this month (May 2017) was to minimize $\sum_{i=1}^nix_i^2\;$ given that $\sum_{i=1}^nix_i=1,$ for $x_1,...x_i\in \mathbb R.$ With $f(x)=x^2$ and $J=\mathbb R,$ let $w_i=i/((n^2+n)/2)$ for each $i$, and the answer is obtained immediately. </p>
<p>(II). On a different subject we have : Let $x\in \mathbb R.$ If for every $r>0$ there exist $a,b\in \mathbb Z$ such that $0<|x-\frac {a}{b}|<\frac {r}{|b|}$ then $x\not \in \mathbb Q.$</p>
<p>PROOF: If $x=c/d$ with $c,d\in \mathbb Z$ then for $a,b \in \mathbb Z$ we have $$0<|x-\frac {a}{b}|\iff 0<|cb-ad| \iff 1\leq |cb-ad| \iff$$ $$\iff \frac {1/|d|}{|b|}\leq |\frac {c}{d}-\frac {a}{b}|=|x-\frac {a}{b}|$$ so we cannot have $0<|x-\frac {a}{b}|<\frac {r}{|b|}$ unless $r\geq \frac {1}{|d|}.$</p>
<p>Note how the inequality $0<|cb-ad|$ strengthens to $1\leq |cb-ad|$ because $cb-ad\in \mathbb Z.$</p>
|
4,258,226 | <p>Suppose a group <span class="math-container">$G$</span> splits as a semidirect product <span class="math-container">$N\rtimes\mathbb{Z}_2$</span>, and let <span class="math-container">$\phi:G\to\mathbb{Z}_2$</span> the the associated quotient map. If I have a subset of elements <span class="math-container">$\{g_1,\dots,g_n,h\}$</span> of <span class="math-container">$G$</span> such that <span class="math-container">$g_i\in N$</span> for <span class="math-container">$1\le i\le n$</span>, and <span class="math-container">$\phi(h)=[1]_2$</span> which I know <em>a priori</em> generates <span class="math-container">$G$</span>, <strong>can I conclude that <span class="math-container">$\{g_1,\dots,g_n\}$</span> generates <span class="math-container">$N$</span> as a subgroup of <span class="math-container">$G$</span>?</strong></p>
<p>I am particularly interested in the case the <span class="math-container">$G$</span> is a Coxeter group, so <span class="math-container">$N$</span> can be thought of (morally at any rate) as the subgroup of orientation preserving transformations. I can't see how I might prove this, but I also haven't been able to think of a counter example.</p>
| Derek Holt | 2,820 | <p>A counterexample is the Coxeter Group <span class="math-container">$$G =W(A_3) = \langle x,y,z \mid x^2,y^2,z^2,(xy)^3,(yz)^3,(xz)^2 \rangle.$$</span></p>
<p>Then <span class="math-container">$G$</span> is isomorphic to <span class="math-container">$S_4$</span> with <span class="math-container">$x,y,z \mapsto (1,2),(2,3),(3,4)$</span>, and the orientation preserving subgroup <span class="math-container">$N$</span> corresponds to <span class="math-container">$A_4$</span>.</p>
<p>Now <span class="math-container">$G = \langle x,yz \rangle$</span> (because <span class="math-container">$S_4$</span> is generated by <span class="math-container">$(1,2)$</span> and <span class="math-container">$(2,3,4)$</span>), but <span class="math-container">$N$</span> is not generated by <span class="math-container">$yz$</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>
| Ross Millikan | 1,827 | <p>The formal answer is that the first card is red or black with probability $\frac 12$. The probability the second card is red is then $\frac{25}{51} \cdot \frac 12$ (first card is red) + $\frac{26}{51} \cdot \frac 12$ (first card is black)$=\frac 12$, so it is false. The intuitive answer is to draw the two cards. Clearly $P(A_1)=\frac 12$. Before you look at them, swap them. Now $P(A_2)=\frac 12$. Done.</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>
| Community | -1 | <p>Here's an explanation that mixes varying amounts of intuition and slight rigor.</p>
<hr>
<p><strong>Volume of a solid of revolution:</strong></p>
<p>When using the disk method the idea is that we're adding up the volumes of a massive amount of extremely thin disks between $x=a$ and $x=b$ in order to get the volume of the solid. The disks each have radius given by $y(x)$ and thickness given by $\Delta x$. So the volume of each disk is $\pi [y(x)]^2 \Delta x$. Then we push this through the limit process so that we're using integration to add up the volumes of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$ and $\Delta x$ becomes $dx$ in the integral, and we have $\displaystyle V = \pi \int_a^b [y(x)]^2 \, dx$.</p>
<p>The "slant" of the function is completely irrelevant here.</p>
<hr>
<p><strong>Surface area of a solid of revolution:</strong></p>
<p>To find the surface area, you want to add up the surface areas of the boundaries of a massive amount of extremely tiny approximate disks. (My use of the word "approximate" will be explained shortly, and until then I'll just keep saying disk and I'll also stop specifying that we only want the surface areas of the boundaries.) Each disk has radius given by $y(x)$. But this time, we can't take $\Delta x$ to be the thickness. Why not? Because $\Delta x$ is a good approximation for the thickness of the <em>interior</em> of the disk, but not for the <em>boundary</em> of the disk. And it's the boundary we care about for surface area, because surface area depends on circumference, and the circumference depends on what's happening on the boundary.</p>
<p>How can we get the thickness (or length) of the boundary? We need to approximate the boundary with tiny line segments. Each line segment can be viewed as the hypotenuse of a tiny right triangle whose legs are parallel to the $x$- and $y$-axes. The leg parallel to the $x$-axis has length $\Delta x$ and the leg parallel to the $y$-axis has length $\Delta y$. If we let $\Delta s$ denote the length of our tiny line segment then we get $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}.$</p>
<p>Here's a picture illustrating this. Imagine this is some function that we very closely zoomed in on:</p>
<p><a href="https://i.stack.imgur.com/og5hs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/og5hs.png" alt="enter image description here"></a></p>
<p>So, each of our little tiny disks has surface area approximately given by $2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2}$. Note that our disks actually have slanted boundaries. (The line segments whose lengths are $\Delta s$ are the boundaries of our disks.) This is why I said "approximate disks" earlier: Our tiny disks for surface area are not perfect disks (like we had for volume) because they're actually disks with slanted boundaries. And this whole slanted boundary thing is necessary here because we must accommodate for the slant of the function, because that slant affects the boundaries of our disks (but not their interiors, which is why we didn't care about slant for the volume).</p>
<p>Anyway, we can rewrite the approximate surface area of each slanted-boundary disk as:
$$ 2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2} = 2\pi y(x) \sqrt{(\Delta x)^2 \left[\frac{(\Delta x)^2}{(\Delta x)^2} + \frac{(\Delta y)^2}{(\Delta x)^2}\right]} = 2\pi y(x) \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \, \Delta x$$</p>
<p>Then we push this through the limit process so that we're using integration to add up the surface areas of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$, and $\Delta y$ becomes $dy$ and $\Delta x$ becomes $dx$. So we get $\displaystyle S = 2\pi \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$.</p>
<hr>
<p><strong>Summary:</strong></p>
<p>For volume, we start by adding up the volumes of a massive amount of extremely thin disks. Since we're adding volumes, it's the <em>interiors</em> of the disks that we care about. And $\Delta x$ is a good enough approximation for the thickness of each disk if we're looking only at the interiors of the disks.</p>
<p>For surface area, we start by adding up the surface areas of a massive amount of extremely thin disks. Since we're adding surface areas, it's the <em>boundaries</em> of the disks that we care about. But $\Delta x$ is not a good enough approximation for the thickness of each disk, because it doesn't account for the slant of the function. This is why $\Delta s$ is necessary. And the slant of the function is important for this because the function <strong>is</strong> the boundary of our solid of revolution, and we need to use the thickness of that boundary to get the circumferences of our "disks" so that we can get a working approximation of the surface areas.</p>
|
2,904,276 | <p>I have this, what seems basic triangle problem, and not certain if the given information is sufficient to solve the problem of finding an angle.
We have one main isoceles and another one inside of it.
I have attached here a diagram, and we wish to find the angle in red:</p>
<p><a href="https://i.stack.imgur.com/moRH9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/moRH9.png" alt="enter image description here"></a></p>
<p>The labels blue means : lengths PQ = QR
The label green means : lengths QS = QT
The given angle PQS = 24 (deg)</p>
<p>We wish to find angle in red , angle RST.</p>
<p>I am trying to figure this out only using geometric principle without trying to use a system of algebraic equations. I tried to create parallel lines to help, but it did not, I tried to make use of principle of exterior angle theorem, but still could not get it.
Hope someone here can help me on this.</p>
| Donald Splutterwit | 404,247 | <p>Let $\widehat{QRP}= \widehat{RPQ}=x$, so $\widehat{RQP}=180-2x$, so $\widehat{RQS}=156-2x$, so the sum of the angles $\widehat{QST}$ and $ \widehat{QTS}$ is $24+2x$, so they are both $x+12$, so $\widehat{RTS}=168-x$, so $\widehat{RST}=12$.</p>
|
2,904,276 | <p>I have this, what seems basic triangle problem, and not certain if the given information is sufficient to solve the problem of finding an angle.
We have one main isoceles and another one inside of it.
I have attached here a diagram, and we wish to find the angle in red:</p>
<p><a href="https://i.stack.imgur.com/moRH9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/moRH9.png" alt="enter image description here"></a></p>
<p>The labels blue means : lengths PQ = QR
The label green means : lengths QS = QT
The given angle PQS = 24 (deg)</p>
<p>We wish to find angle in red , angle RST.</p>
<p>I am trying to figure this out only using geometric principle without trying to use a system of algebraic equations. I tried to create parallel lines to help, but it did not, I tried to make use of principle of exterior angle theorem, but still could not get it.
Hope someone here can help me on this.</p>
| Edward Porcella | 403,946 | <p>Although the solution of @Donald Splutterwit might seem like algebra and equations, it is really straight-forward geometry, using the principle that the three angles of a triangle sum to $180^o$.</p>
<p>Perhaps the following may be another helpful way to exhibit the geometry of the situation.
<a href="https://i.stack.imgur.com/owilG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/owilG.png" alt="isosceles within isosceles"></a>
Since $PRQ$ and $STQ$ are isosceles and share vertex $Q$, let the circle with center $Q$ and radius $QT$ intersect $PQ$ at $U$ and $PR$ at $V$, and join $TU$. Thus, since$$\frac{QU}{QP}=\frac{QT}{QR}$$then$$TU\parallel RP$$and arcs$$\ VT= SU$$so that$$\angle VST=\angle STU$$But$$\angle SQU=2\angle STU$$Therefore$$\angle SQU=2\angle VST$$that is$$\angle SQP=2\angle RST$$Thus under the given conditions, the angle we are looking for is always half the given angle. </p>
|
2,041,484 | <p>Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$</p>
<p>I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.<br>
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5}}\le y \le {\frac 45}$}</p>
<p>I don't know what to do next.</p>
| ILoveMath | 42,344 | <p>Hint: Notice $x,y \neq 0 $</p>
<p>$$ \frac{12}{5x} + \frac{4}{5y} = 2 $$</p>
|
4,019,748 | <p>Let <span class="math-container">$H_1=(H_1, (\cdot, \cdot )_1)$</span> and <span class="math-container">$H_2=(H_2, (\cdot, \cdot )_2)$</span> be Hilbert spaces. Suppose that <span class="math-container">$H_1$</span> is continuously and densely embedded in <span class="math-container">$H_2$</span>. Simbolically, <span class="math-container">$H_1 \stackrel d{\hookrightarrow} H_2$</span>. Let <span class="math-container">$X \subset H_1$</span> and <span class="math-container">$Y \subset H_2$</span> be closed subspaces such that <span class="math-container">$X \subset Y$</span>, with <span class="math-container">$X,Y \neq \emptyset$</span>. I know that
<span class="math-container">$$
X=(X, (\cdot, \cdot )_1) \quad \text{and} \quad Y=(Y, (\cdot, \cdot )_2) \tag{1}
$$</span>
are Hilbert spaces.</p>
<p><strong>Question.</strong> Is <span class="math-container">$X \stackrel d{\hookrightarrow} Y$</span>?</p>
<p>Its clear for me that, since <span class="math-container">$X \subset Y$</span> and due to <span class="math-container">$(1)$</span> that <span class="math-container">$X$</span> is continuously embedded in <span class="math-container">$Y$</span>. But I don't know (and I couldn't prove) that <span class="math-container">$X$</span> is dense in <span class="math-container">$Y$</span>. I tried using the Hahn-Banach theorem, but without success.</p>
| JLMF | 551,373 | <p>This doesn't seem true to me without more restrictions.
For example, take <span class="math-container">$X$</span> the zero vector space in <span class="math-container">$H_1$</span>, then <span class="math-container">$X$</span> is not dense in <span class="math-container">$Y$</span>.</p>
<p>Edit: What if we consider a separable Hilbert space <span class="math-container">$H_2$</span> with basis <span class="math-container">$\beta$</span>, then let the space generated by <span class="math-container">$\beta$</span> be <span class="math-container">$H_1$</span>.</p>
<p>Taking a non-trivial subspace <span class="math-container">$Y \subset H_2$</span> and take a sub-basis <span class="math-container">$\beta_Y$</span> for <span class="math-container">$Y$</span>, then remove one or more vectors from the said basis <span class="math-container">$\beta_Y$</span> (by construction or by a unilateral shift). Take the space generated by this shrunken basis to be <span class="math-container">$X$</span>, it is indeed a subspace of <span class="math-container">$H_1$</span>, and is a subspace of <span class="math-container">$Y$</span>, but it is not dense.</p>
|
204,043 | <p>I am looking at the following optimization problem</p>
<p><span class="math-container">$$
\begin{align*}
\max\ & 1000 r_1 + \frac{1}{2}r_2 + \frac{1}{3}r_3\\
\text{s.t. }& 1000^2 r_1 + \frac{1}{4}r_2 + \frac{1}{9}r_3 = \frac{1}{9},\\
& 1000^2 p_1 + \frac{1}{4}p_2 + \frac{1}{9}p_3 = \frac{1}{9},\\
& r_1 + r_2 + r_3 + r_4 = 2\\
& p_1 + p_2 + p_3 + p_4 = 2\\
& 0\leq p_i\leq 1,\quad i = 1,\dots,4\\
& r_i^2\leq p_i,\quad i = 1,\dots,4
\end{align*}
$$</span>
with the following Mathematica code (it is clear that <span class="math-container">$r_1=p_1=0$</span>, <span class="math-container">$r_2=p_2=0$</span>, <span class="math-container">$r_3=p_3=1$</span>, <span class="math-container">$r_4=p_4=1$</span> is a feasible solution so I picked it to be the initial solution)</p>
<pre><code>FindMaximum[{1000 r1 + 1/2 r2 + 1/3 r3,
{1000^2 r1 + 1/4 r2 + 1/9 r3 == 1/9,
r1 + r2 + r3 + r4 == 2,
r1^2 <= p1, r2^2 <= p2, r3^2 <= p3, r4^2 <= p4,
1000^2 p1 + 1/4 p2 + 1/9 p3 == 1/9,
p1 + p2 + p3 + p4 == 2,
0 <= p1 <= 1, 0 <= p2 <= 1, 0 <= p3 <= 1, 0 <= p4 <= 1}
},
{{r1, 0}, {r2, 0}, {r3, 1}, {r4, 1}, {p1, 0}, {p2, 0}, {p3, 1}, {p4, 1}}, AccuracyGoal -> 10]
</code></pre>
<p>Mathematica returns the following answer</p>
<pre><code>{0.45521, {r1 -> -6.30073*10^-8, r2 -> 0.268328, r3 -> 0.963328,
r4 -> 0.768345, p1 -> 0., p2 -> 0.0719999, p3 -> 0.928, p4 -> 1.}}
</code></pre>
<p>If you look at the constraint that <span class="math-container">$1000^2 p_1 + \frac{1}{4}p_2 + \frac{1}{9}p_3 = \frac{1}{9}$</span>, you can see that in the solution Mathematica returns,
the left-hand side evaluates to <span class="math-container">$0.12111\cdots$</span>, so the constraint is not satisfied.</p>
<p>Is this a bug or just a numerical inaccuracy? Just the value seems too large compared with <span class="math-container">$1/9$</span> -- the additive error is not at the scale of <span class="math-container">$10^{-4}$</span> or <span class="math-container">$10^{-5}$</span>.</p>
| A.G. | 7,060 | <p>It looks like a bug. If you define the <code>error</code> variable like this</p>
<pre><code>sol = FindMaximum[{
1000 r1 + 1/2 r2 + 1/3 r3,
{
1000^2 r1 + 1/4 r2 + 1/9 r3 == 1/9,
r1 + r2 + r3 + r4 == 2,
r1^2 <= p1,
r2^2 <= p2,
r3^2 <= p3,
r4^2 <= p4,
1000^2 p1 + 1/4 p2 + 1/9 p3 == 1/9,
error == 1000^2 p1 + 1/4 p2 + 1/9 p3 - 1/9, (* just change this *)
p1 + p2 + p3 + p4 == 2,
0 <= p1 <= 1, 0 <= p2 <= 1, 0 <= p3 <= 1, 0 <= p4 <= 1}
}, {{r1, 0}, {r2, 0}, {r3, 1}, {r4, 1}, {p1, 0}, {p2, 0}, {p3,
1}, {p4, 1}, {error, 0}}, AccuracyGoal -> 10]
</code></pre>
<p>you get the same solution:</p>
<pre><code>{0.45521, {r1 -> -6.30073*10^-8, r2 -> 0.268328, r3 -> 0.963328,
r4 -> 0.768345, p1 -> 0., p2 -> 0.0719999, p3 -> 0.928, p4 -> 1.,
error -> 5.75352*10^-18}}
</code></pre>
<p>however</p>
<pre><code>1000^2 p1 + 1/4 p2 + 1/9 p3 - 1/9 /. Last[sol]
error /. Last[sol]
</code></pre>
<p>yields</p>
<pre><code>0.00999999
5.75352*10^-18
</code></pre>
<p>although clearly the two figures should be identical. I suggest you contact Wolfram.</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| zyx | 14,120 | <p>$x \to x^p$ is an automorphism sending $r$ to $r-a$ for any root $r$ of the polynomial. This operation is cyclic of order $p$, so that one can get from any root to any other by applying the automorphism several times. The Galois group thus acts transitively on the roots, which is equivalent to irreducibility.</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| Jyrki Lahtonen | 11,619 | <p>Greg Martin and zyx have given you IMHO very good answers, but they rely on a few basic facts from Galois theory and/or group actions. Here is a more elementary but also a longer approach. </p>
<p>Because we are in a field with $p$ elements, we know that $p$ is the characteristic of our field. Hence, the polynomial $g(x)=x^p-x$ has the property $$g(x_1+x_2)=g(x_1)+g(x_2)$$ whenever $x_1$ and $x_2$ are two elements of an extension field of $\mathbb{F}_p$. By little Fermat we know that $g(k)=k^p-k=0$ for all $k\in \Bbb{F}_p$. Therefore, if $r$ is one of the roots of $f(x)=x^p-x+a$, then $$f(r+k)=g(r+k)+a=g(r)+g(k)+a=f(r)+g(k)=0,$$ so all the elements $r+k$ with $k \in \Bbb{F}_p$ are roots of $f(x)$, and as there are $p$ of them, they must be all the roots. It sounds like you have already shown that $r$ cannot be an element of $\Bbb{F}_p$.</p>
<p>Now assume that $f(x)=f_1(x)f_2(x)$, where both factors $f_1(x),f_2(x)\in \Bbb{F}_p[x]$. From the above consideration we can deduce that
$$
f_1(x)=\prod_{k\in S}(x-(r+k)),
$$
where $S$ is some subset of the field $\Bbb{F}_p$. Write $\ell=|S|=\deg f_1(x)$. Expanding the product we see that
$$
f_1(x)=x^\ell-x^{\ell-1}\sum_{k\in S}(r+k)+\text{lower degree terms}.
$$
This polynomial was assumed to have coefficients in the field $\Bbb{F}_p$. From the above expansion we read that the coefficient of degree $\ell-1$ is $|S|\cdot r+\sum_{k\in S}k$. This is an element of $\Bbb{F}_p$, if and only if the term $|S|\cdot r\in\Bbb{F}_p$. Because $r\notin \Bbb{F}_p$, this can only happen if $|S|\cdot1_{\Bbb{F}_p}=0_{\Bbb{F}_p}$. In other words $f_1(x)$ must be either of degree zero or of degree $p$.</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| Utkarsh | 39,872 | <p>$x^p-x+a$ divides $x^{p^p}-x$. If $f$ is an irreducible divisor of $x^p-x+a$ of degree $d$ then $\mathbf{Z}_p[x]/f$ will be a subfield of the field with $p^p$ elements so $p^p = (p^d)^e$ and so $d=1$ or $e=1$. since $x^p-x+a$ has no roots $e=p.$</p>
|
81,588 | <p>A certain function contains points $(-3,5)$ and $(5,2)$. We are asked to find this function,of course this will be simplest if we consider slope form equation </p>
<p>$$y-y_1=m(x-x_1)$$</p>
<p>but could we find for general form of equation? for example quadratic? cubic?</p>
| Community | -1 | <p>The supposition of Greg Martin is truth, if a polinomyal $f$ with $deg(f)=n$ satisfies the property, then $n\ge p$, by contradiction argument, just write the expansion with the Newton's formula and analyse the coeficient of $x^{n-1}$ term, you get $\binom n 1a_{n}+a_{n-1}=a_{n-1}$, if $n\lt p$, this equation is an absurd.</p>
|
467 | <p>Moderators have started incorporating the old faq material in the new <a href="https://mathoverflow.net/help">help system</a>. It wasn't a perfect fit, a lot of stuff is no longer relevant, redundant, missing or broken. You can help by going through the <a href="https://mathoverflow.net/help">help center</a> and post anything that needs fixing here.</p>
<p>In case substantial editing is needed it would help to add a proposal; we may not use it verbatim but it will nevertheless save a lot of effort on our end.</p>
<p>Note that this is not the right place for policy discussion. If you think we should be wearing shirts instead of wearing pants, start a new question with your proposal...</p>
| Community | -1 | <p>Below a suggestion for the 'What about open problems' part of What topics can I ask about here, ie presently the old FAQs. It is based on various contributions and input for details see <a href="https://meta.mathoverflow.net/questions/360/what-should-be-the-policy-on-open-problems-on-mo">What should be the policy on "open problems" on MO?</a> Main goals were to sharpen what is meant with 'open problem' also the descirption of the usage of the open-problem tag was modified with the dual goal of simplfying it and making it closer to the usage that happens in practise. </p>
<hr>
<p><strong>What about open problems?</strong></p>
<p>It <em>depends</em> on what is meant by an open problem: here, we mean a problem which is well-known to be unsolved and very hard. Questions for which you simply do not know the answer are not the issue, as those are the norm on MathOverflow.</p>
<p>On the one hand, a question which simply states a famous open problem will often not be well received. If you mainly want to record some open problem, this site is not the appropriate place: you may do so at <a href="http://www.openproblemgarden.org/" rel="nofollow noreferrer">the Open Problem Garden</a>, for example.</p>
<p>On the other hand, questions that <em>touch upon</em> a well-known open problem are certainly welcome, as long as they follow the general guidelines for asking on MathOverflow [[comment: perhaps place a link here to the guidelines, or a 'see above' or whatever seems practical in the actual final layout]]. Nevertheless, when asking about an open problem, you should mention that it is known to be unsolved. In this case, please use the <a href="https://mathoverflow.net/questions/tagged/open-problem" class="post-tag" title="show questions tagged 'open-problem'" rel="tag">open-problem</a> tag, in addition to other appropriate tags.</p>
<p>Most importantly, please remember that MathOverflow is a question and answer site: the prototypical question is posed under the assumption that some other user will be able to answer it. This principle also applies to questions about open problems. For example, the community will usually welcome inquiries on partial progress, results in special cases, etc.</p>
<p><strong>What if my question turns out to be a well-known open problem?</strong></p>
<p>If you ask a question which is widely known to be unsolved, someone will typically point that out and provide references to the relevant literature. In most such cases, this is the best answer one can expect. At this time, you can add the <a href="https://mathoverflow.net/questions/tagged/open-problem" class="post-tag" title="show questions tagged 'open-problem'" rel="tag">open-problem</a> tag to your question. If you then have more detailed follow-up questions, feel free to ask them in new posts.</p>
|
467 | <p>Moderators have started incorporating the old faq material in the new <a href="https://mathoverflow.net/help">help system</a>. It wasn't a perfect fit, a lot of stuff is no longer relevant, redundant, missing or broken. You can help by going through the <a href="https://mathoverflow.net/help">help center</a> and post anything that needs fixing here.</p>
<p>In case substantial editing is needed it would help to add a proposal; we may not use it verbatim but it will nevertheless save a lot of effort on our end.</p>
<p>Note that this is not the right place for policy discussion. If you think we should be wearing shirts instead of wearing pants, start a new question with your proposal...</p>
| Danny Ruberman | 3,460 | <p>Two small things I noticed on "<a href="https://mathoverflow.net/help/on-topic">on-topic</a>": The link <a href="https://meta.mathoverflow.net/questions/882/how-to-write-a-good-mathoverflow-question">How to ask a good MathOverflow question</a> seems to be broken. Also, in the subtopic "MathOverflow is not a discussion forum" the word discussion is mis-spelled:...Stack Exchange software is bad for <strong>disscusions</strong> ...</p>
|
369,589 | <p>I am not a math student, and only kind of picking up something whenever I need it. After emerged in the field of machine learning, probability, measure theory and functional analysis seem to be quite intriguing. I am considering learning stochastic calculus myself, but do not quite know what kind of prerequisites should I have. Anyone got some suggestions about the things I need to check before I challenge myself in stochastic calculus?</p>
| Alexander Sokol | 28,924 | <p>Stochastic calculus relies heavily on martingales and measure theory, so you should definitely have a basic knowledge of that before learning stochastic calculus. Basic analysis also figures prominently, both in stochastic calculus itself (where limit procedures of various kinds appear, as well as the occasional Hilbert or $L^p$ space argument) and in martingale theory itself.</p>
<p>Summing up, it would be beneficial for you to first familiarize yourself with elementary mathematical tools such as:</p>
<p>-Real analysis (e. g., Carothers "Real analysis" or Rudin's "Real and complex analysis")</p>
<p>-Measure theory (e. g. Dudley's "Real analysis and probability", or Ash and Doleans-Dade's "Probability and measure theroy")</p>
<p>and furthermore learn basic probability theory such as</p>
<p>-Discrete-time martingale theory</p>
<p>-Theories of convergence of stochastic processes</p>
<p>-Theory of continuous-time stochastic processes, Brownian motion in particular</p>
<p>This is all covered in volume one of Rogers and Williams' "Diffusions, Markov processes and martingales", and also sporadically in the two probability-related books above by Dudley and Ash and Doleans-Dade.</p>
<p>With a background like that, you should be well prepared to learn stochastic calculus, which you can do from volume two of Rogers and Williams' "Diffusions, Markov processes and martingales", or Karatzas and Shreve's "Brownian motion and stochastic calculus".</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>
| Arnaud Mortier | 480,423 | <p>Choose a position for the $2$. How many of these numbers have the $2$ in that position? There are two options for the $4$ and $3!$ options for the remaining $3$ numbers, that gives you $12$ numbers. Hence the $2$ brings $12\times 20202$ to the sum.
Similarly the $4$ brings $12\times 40404$.</p>
<p>Now if you fix an odd number at an even place there are $6$ ways to put the even numbers and then $2$ ways to put the other two odd ones. So $1$ brings $12\times 1010$, $3$ brings $12\times 3030$ and $5$ brings $12\times 5050$.</p>
<p>Finally if you fix an odd number at an odd place there are $2$ options for the other odd ones and $2$ options for the even ones. </p>
<p>Altogether $$S=12\times 69696+4\times 90909$$</p>
|
2,485,425 | <p>If $A$ is a non empty subset of the reals and $f$ is a bounded function from $A$ to the reals, how can we show that:</p>
<blockquote>
<p>$\sup|f(x)| - \inf|f(x)| \le \sup(f(x)) - \inf(f(x))$?</p>
</blockquote>
<p>I started by stating that since $f$ is bounded, $\inf(f(x)) \le f(x) \le \sup(f(x))$. And then $|f(x)| \le \max\{|\sup(f(x))|,|\inf(f(x))|\}$.</p>
<p>So $\sup|f(x)| = \max\{|\sup(f(x))|,|\inf(f(x))|\}$ and $\inf|f(x)| = \max\{\min\{|\sup(f(x))|,|\inf(f(x))|\},0\}$.</p>
<p>I feel like my $\inf|f(x)|$ is wrong though, and I don't know where to get it.</p>
<p>Any help would be really appreciated!</p>
| Brethlosze | 386,077 | <p>From <a href="https://en.wikipedia.org/wiki/Line_(geometry)" rel="nofollow noreferrer">wikipedia</a> we can extract:</p>
<blockquote>
<p>Thus in differential geometry a line <em>may</em> be interpreted as a geodesic
(shortest path between points), while in some projective geometries a
line is a 2-dimensional vector space (all linear combinations of two
independent vectors). This flexibility also extends <em>beyond mathematics</em>
and, for example, permits physicists to think of the path of a light
ray as being a line.</p>
</blockquote>
<p>Hence the Equator can be interpreted as a line in the geodesic sense.</p>
|
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>
| Vincent Granville | 574,948 | <p>So here is how it works. No proof here, but I am showing the pattern for all irrational numbers. </p>
<p><a href="https://i.stack.imgur.com/Bdpq5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bdpq5.png" alt="enter image description here"></a></p>
<p>The above table is for the number <span class="math-container">$e$</span> but it works the same way for all irrational numbers. Follow the red path! The convergents of <span class="math-container">$e$</span> are listed below (source: <a href="https://oeis.org/A007676/internal" rel="nofollow noreferrer">here</a>). Look at the denominator of each convergent to identify the pattern: </p>
<p>8/3, </p>
<p>11/4, </p>
<p>19/7, </p>
<p>87/32, </p>
<p>106/39, </p>
<p>193/71, </p>
<p>1264/465, </p>
<p>1457/536, </p>
<p>2721/1001, </p>
<p>23225/8544, </p>
<p>25946/9545, </p>
<p>49171/18089, </p>
<p>517656/190435, </p>
<p>566827/208524, </p>
<p>1084483/398959, </p>
<p>13580623/4996032, </p>
<p>14665106/5394991, </p>
<p>28245729/10391023, </p>
<p>410105312/150869313, </p>
<p>438351041/161260336, </p>
<p>848456353/312129649,... </p>
|
3,323,170 | <p>This question was originally posted <a href="https://crypto.stackexchange.com/q/72456/62225">here</a> on Crypto StackExchange. As suggested by an answer I am posting it here to help get a better perspective on the math side.</p>
<blockquote>
<p>Public-key cryptography was not invented until the 1970's. Apart from the idea not existing earlier (<a href="https://crypto.stackexchange.com/a/59542/62225">as talked about here</a>), is there any reason it could not have been used earlier? For example, are there forms that are easy enough to perform by hand but complicated enough to not be solved (easily) by hand?</p>
</blockquote>
| Wuestenfux | 417,848 | <p>My hypothesis is a bit different. From Euclid's ''Elements'' enough number theory (gcd, lcm) has been available to introduce (with some research) the theory of the RSA algorithm. The necessity to use cryptography can be seen from the Caesar cipher. Some handcrafting could have led to some larger primes and encrypting/decrypting via modular arithmetic would have not been a big deal due to successive multiplication and reduction.</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>
| Spock | 108,632 | <p>This integral is explicitly explained <a href="http://en.wikipedia.org/wiki/Integration_by_reduction_formulae" rel="nofollow">here</a>.</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>
| Claude Leibovici | 82,404 | <p>Probably as you did, integrating twice by parts, you arrived to a simple recurrence relation between I(n) and I(n-2). You can check your results at<br>
<a href="http://en.wikipedia.org/wiki/Integration_by_reduction_formulae" rel="nofollow">http://en.wikipedia.org/wiki/Integration_by_reduction_formulae</a><br>
as mentioned by Spock. </p>
<p>In practice, it is always a good idea for this kind of integrals, to use the form using cos(x) and the form using sin(x). </p>
<p>These recurrence relations are sufficient for almost any purpose I know, but I agree, they do not give you closed forms ... which exist; however, I am not sure you would enjoy them since, for your integral, the result is </p>
<p>-x^(1 + n) (ExpIntegralE[-n, -I x] + ExpIntegralE[-n, I x]) / 2</p>
|
214,475 | <p>Function:
<a href="https://i.stack.imgur.com/sH7mh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sH7mh.png" alt="enter image description here"></a></p>
<p>I am to solve for <span class="math-container">$T_{12}(4.8), T_{24}(1.2)$</span>, using <strong>If</strong> and <strong>Which</strong> functions.</p>
<p>I started with this function and keep getting a recursion limit error:</p>
<pre><code>t[n_] := (7/2 x) t[n - 1] - (7/2) t[n + 1]
</code></pre>
| Cesareo | 62,129 | <p>Try</p>
<pre><code>Clear[T, n, x]
Tn = T[n] /. RSolve[{T[n + 1] == 1/x T[n - 1] - 2/7 T[n], T[0] == 1, T[1] == x}, T, n][[1]]
Tn /. {n -> 12, x -> 4.8}
(* -0.0153785 *)
</code></pre>
|
2,071,332 | <p>I am working on implementation of a machine learning method that in part of the algorithm I need to calculate the value of $\Gamma (\alpha) / \Gamma (\beta) $. $\alpha$ and $\beta$ are quite large numbers (i.e. bigger than 200) and it causes the python $gamma$ function to overflow. However, as the difference of $\alpha$ and $\beta$ is relatively small (e.g. $|\alpha-\beta|<5$), the final result is not such a big number and can be used for later purposes. So, I am trying to calculate (or approximate) the value of $\Gamma (\alpha) / \Gamma (\beta) $ without going through the calculation of $\Gamma (\alpha)$ and $\Gamma (\beta)$ directly. If $\alpha$ and $\beta$ were integers, the result would be simple equal to $\alpha . \alpha+2. \alpha+3... \beta-1$, But I can not imagine how this formula will be changed if we let $\alpha$ and $\beta$ to be real numbers.</p>
| Mark | 310,244 | <p>From <a href="https://math.stackexchange.com/questions/98348/how-do-you-prove-gautschis-inequality-for-the-gamma-function">this answer</a> we have Gautschi's inequality:</p>
<p>$$x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s},\quad x>0,0<s<1$$</p>
<p>We can combine this with the functional equation for the gamma function:
$$\Gamma(z+1) = z\Gamma(z)$$
which holds for all $z$ where $\Gamma$ is defined.</p>
<p>Now, assume that $\alpha\leq\beta$, and that $|\beta-\alpha|\leq 5$. Now, let $k$ be the greatest integer such that $\alpha+k\leq\beta$ (so clearly, $k\in\{0,1,2,3,4,5\}$. It follows that $\alpha+k+s = \beta$ where $0\leq s<1$. </p>
<p>Now, we have that $\Gamma(\beta) = \Gamma(\alpha+k+s)$, and we can apply the functional equation $k+1$ times to get that $\Gamma(\beta) =\prod_{i = 0}^{k+1} (\alpha+k+s)\Gamma(\alpha-1+s)$. It follows that:
$$\frac{\Gamma(\alpha)}{\Gamma(\beta)} = \frac{\Gamma(\alpha)}{\Gamma(\alpha+k+s)} = \frac{\Gamma(\alpha)}{\prod_{i = 0}^k (\alpha+k+s)\Gamma(\alpha-1+s)}$$
Now, if we substitue in $\alpha-1 = x$, we see that:
$$\frac{\Gamma(\alpha)}{\Gamma(\beta)} = \frac{1}{\prod_{i = 0}^{k+1}((\alpha+k+s)}\frac{\Gamma(x+1)}{\Gamma(x+s)}$$
This part in the front is just some constant $c$, so we can rewrite this as:
$$\frac{\Gamma(\alpha)}{\Gamma(\beta)} = c\frac{\Gamma((\alpha-1)+1)}{\Gamma((\alpha-1)+s)}$$
From Gautschi's inequality, we have that:
$$(\alpha-1)^{1-s}<\frac{\Gamma(\alpha-1+1)}{\Gamma(\alpha-1+s)}<\alpha^{1-s}$$
Now, as you said that $\alpha\gg 5$, we have that $c\geq 0$, so we can multiply through by $c$ to get that:
$$c(\alpha-1)^{1-s}<\frac{\Gamma(\alpha)}{\Gamma(\beta)}<c\alpha^{1-s}$$
The above calculation implicitly assume that $s\neq 0,1$, but if this is true then we can get the result just from the functional equation for the gamma. Moreover, $0<1-s<1$, so while $\alpha$ may be large, $\alpha^{1-s}$ is hopefully manageable.</p>
<p>This is all to try to give you another option.</p>
|
4,058,319 | <p>My background in mathematical logic, model theory, etc. is patchy, so I'm looking for a clearer way to think about this. (Edit: I'm not asking what an isomorphism is, I'm asking how to formalize the idea of "preserving all logical properties" in order to state the principle described below in its full power.)</p>
<p>Motivating example: we know that <span class="math-container">$\Bbb N$</span> and <span class="math-container">$\Bbb Z$</span> are not order-isomorphic, because <span class="math-container">$\Bbb N$</span> has a least element whereas <span class="math-container">$\Bbb Z$</span> does not. The principle underlying this argument seems to be that an order isomorphism preserves the truth values of (first-order) sentences: if <span class="math-container">$(A,<_A)\cong(B,<_B)$</span> then <span class="math-container">$\operatorname{Th}(A,<_A)=\operatorname{Th}(B,<_B)$</span>. Since the property "there is a least element" corresponds to the sentence <span class="math-container">$\exists x\forall y. x\le y$</span>, the principle applies. The principle also seems to apply to higher-order properties. For example, we can conclude that <span class="math-container">$(\Bbb R\setminus\{0\},<)\not\cong(\Bbb R,<)$</span> by checking the least-upper-bound property, which is second-order.</p>
<p>More generally (if we think of set theory as modeling higher-order logic in first-order logic), I'm tempted to claim that the whole "set theory of <" (I'm not sure what this is exactly - a two-sorted first-order logic with relation symbols <span class="math-container">$<,\in$</span>, maybe?) is preserved by isomorphisms. I think I can convince myself of this fact by observing that when <span class="math-container">$f:A\to B$</span> is an isomorphism, certain substitution rules involving <span class="math-container">$f$</span> in various syntactic contexts are valid, allowing us to prove equivalences between propositions about <span class="math-container">$<_A$</span> and the corresponding propositions about <span class="math-container">$<_B$</span>.</p>
<p>I'm calling this a "principle" instead of a "theorem" because it seems easier to apply in specific cases (by reasoning directly about the elements and the isomorphism, without formalizing the notion of sentences) than to express in its full generality. My claim that "the set theory of <span class="math-container">$<$</span> is preserved" seems closely related to the fact that set theory is my metatheory (if that's the right word), so some kind of circularity or self-reference is appearing when I attempt to state the principle, and I wonder whether any formal version of the principle can fully eliminate the need to apply the principle directly.</p>
<p>So I guess my questions are:</p>
<ul>
<li>In what sense (if any) is there a most general version of this principle, applying to all properties preserved by order isomorphisms?</li>
<li>Can it be expressed as a theorem in the deductive system where we want to apply it, or is there a theoretical obstacle to this?</li>
<li>Am I getting anything wrong? Or missing a clarifying perspective?</li>
</ul>
<p>Thanks!</p>
| Noah Schweber | 28,111 | <p>Here is an outline of a theorem which, I think, addresses the situation pretty well (incidentally, <a href="https://mathoverflow.net/questions/336191/cauchy-reals-and-dedekind-reals-satisfy-the-same-mathematical-theorems">this MO question</a> is related):</p>
<p>Most logic books will include a proof that first-order logic is isomorphism-invariant. But what about <em>non-first-order</em> properties? For example, the isomorphism-invariance of (say) Archimedeanness in the context of ordered fields is not a mere case of the isomorphism-invariance of first-order logic, even though "obviously" it holds for exactly the same reason.</p>
<p>There is in fact a framework for extending isomorphism-invariance from first-order logic to all "reasonable" logics, in a precise sense. The logics beyond first-order logic that we actually consider are really just "first-order logic in a superstructure." E.g. saying that a relation <span class="math-container">$R$</span> is definable over a structure <span class="math-container">$\mathcal{A}$</span> in monadic second-order logic is the same as saying that it is definable over <span class="math-container">$\mathcal{A}\sqcup\mathcal{P}(\mathcal{A})$</span> (being a bit vague about what that is, in the interest of brevity) in first-order logic.</p>
<p>This motivates the following idea. Working in <span class="math-container">$\mathsf{ZFC}$</span>, let <span class="math-container">$V$</span> as usual denote the whole set-theoretic universe. There is a standard machinery for developing, first-order-definably inside <span class="math-container">$V$</span>, a "version of <span class="math-container">$V$</span> with urelements" where the urelements correspond to elements of a given structure; the details are a bit tedious, but see e.g. <a href="https://projecteuclid.org/ebooks/perspectives-in-logic/admissible-sets-and-structures-an-approach-to-definability-theory/toc/pl/1235418470" rel="nofollow noreferrer">Barwise's book</a>. Call the resulting object "<span class="math-container">$V_\mathcal{A}$</span>." Being a bit sloppy about set/class issues, we can prove in <span class="math-container">$\mathsf{ZFC}$</span> that <span class="math-container">$\mathcal{A}\cong\mathcal{B}$</span> implies <span class="math-container">$V_\mathcal{A}\cong V_\mathcal{B}$</span>. This, together with the fact that isomorphism preserves <em>first-order</em> truth, gives the following result:</p>
<blockquote>
<p>Isomorphisms preserve truth with respect to every "set-theoretically-definable" logic.</p>
</blockquote>
<p>Here the idea is that each sentence <span class="math-container">$\varphi$</span> in a set-theoretically-definable logic should correspond to a first-order sentence <span class="math-container">$\hat{\varphi}$</span> such that <span class="math-container">$$\mathcal{A}\models\varphi\quad\iff\quad V_\mathcal{A}\models\hat{\varphi}$$</span> (or at worst, a first-order sentence <span class="math-container">$\hat{\varphi}$</span> with a parameter from <span class="math-container">$V$</span> - note that <span class="math-container">$V$</span> sits inside <span class="math-container">$V_\mathcal{A}$</span> in a canonical way, so a single <span class="math-container">$V$</span>-parameter makes sense across all <span class="math-container">$V_\mathcal{A}$</span>s independently of <span class="math-container">$\mathcal{A}$</span>; we need to allow parameters to handle logics with "too many sentences," such as infinitary logic).</p>
<p>So what we have here is a way to "bootstrap" the isomorphism-invariance of first-order truth to all logics <em>which are set-theoretically definable in a precise sense</em>.</p>
<p>I'm skipping a <strong>lot</strong> of detail in the above, since quite frankly it's rather tedious. If you're interested, let me know and I'll expand on this. But I think it is worth mentioning, even if only very briefly, the existence of a precise theorem of the above type.</p>
|
886,935 | <p>Let $A_1,A_2,A_3,\dots$ be a sequence of sequences where each
$$A_i = a_{i,1},a_{i,2},a_{i,3},\dots$$</p>
<p>Each sequence $A_i$ converges and in particular as $t \rightarrow \infty$, $a_{i,t} \rightarrow L_i$ for every $i$. We also have that in the limit as $i \rightarrow \infty$ the limits of these sequences converge to 1. I.e. as $i \rightarrow \infty$, $L_i \rightarrow 1$.</p>
<p>QUESTION: When does the following occur?</p>
<p>In the limit as $t \rightarrow \infty$
$$\prod_{i=1}^{\infty} a_{i,t} \rightarrow \prod_{i=1}^{\infty}L_i $$</p>
<p>I know how to prove that the limit of the product of two (convergent) sequences is the product of the limit of those sequences. I cannot see whether this argument extends to infinite products. Also I am not sure how to use the fact that the $L_i$ converge to 1.</p>
| Ian | 83,396 | <p>Taking the theorem about limits of finite products for granted, you are asking the following: given $a_{i,j} \to L_i$ as $j \to \infty$ and $L_i \to 1$ as $i \to \infty$, when do we have</p>
<p>$$\lim_{j \to \infty} \lim_{k \to \infty} \prod_{i=1}^k a_{i,j} = \lim_{k \to \infty} \lim_{j \to \infty} \prod_{i=1}^k a_{i,j}?$$</p>
<p>This is a question about interchange of limits. The generic answer is that you cannot do it: some additional hypotheses are required. The usual approach for infinite products is to convert to a problem about infinite sums by taking logarithms:</p>
<p>$$\ln \left ( \lim_{j \to \infty} \lim_{k \to \infty} \prod_{i=1}^k a_{i,j} \right ) = \lim_{j \to \infty} \lim_{k \to \infty} \sum_{i=1}^k \ln \left ( a_{i,j} \right )$$</p>
<p>using continuity of $\ln$ and the usual logarithm properties. Now if we can interchange the limits on the right side, then we can "undo" the logarithm procedure (that is, pass the $\ln$ back through the limits and apply $\exp$) to get the result. </p>
<p>The "advanced calculus criterion" for being able to interchange these limits is that the sum converges uniformly in $j$, that is, for any $\varepsilon > 0$ there exists $N > 0$ so that for all $n \geq N$ and all $j$</p>
<p>$$\left | \sum_{i=n}^\infty \ln \left ( a_{i,j} \right ) \right | < \varepsilon$$</p>
<p>One can prove that this condition is necessary. The cleanest proof I can think of rephrases summation in terms of integration with respect to the counting measure and then applies the Vitali convergence theorem. Then uniform integrability is trivial for summation, and tightness is equivalent to uniform convergence of the sums.</p>
<p>The condition that $L_i \to 1$ is required for us to use this procedure, because an infinite product can "diverge to zero" if all but finitely many terms are bounded a fixed distance below $1$. We say that it diverges to zero because in this case the infinite sum of the logarithms diverges to $-\infty$. In this case the sequence of partial products converges but the sequence of partial sums of logarithms does not. By assuming $L_i \to 1$ we remove this issue.</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>
| Noah Caplinger | 581,608 | <p><span class="math-container">$A$</span> is a proper subset of <span class="math-container">$B$</span> if <span class="math-container">$B \subseteq A$</span> and there exists an <span class="math-container">$x \in A$</span> such that <span class="math-container">$x \notin B$</span>.</p>
<p>To show <span class="math-container">$B \subseteq A$</span>: Let <span class="math-container">$x \in B$</span>. Then, <span class="math-container">$x \leq 3 $</span> and <span class="math-container">$ x > 2$</span>. Squaring both sides and moving everything to the left we have <span class="math-container">$x^2 - 9 \leq 0$</span> and <span class="math-container">$x^2 -2>0$</span>, so <span class="math-container">$x \in A$</span>. Thus <span class="math-container">$B \subseteq A$</span>. </p>
<p>Note that <span class="math-container">$(-3)^2 - 9 \leq 0$</span> and <span class="math-container">$(-3)^2 - 4 > 0$</span>, so <span class="math-container">$-3 \in A$</span>, but clearly <span class="math-container">$-3 \notin B$</span>.</p>
<p>Putting these together, we can see that <span class="math-container">$A$</span> is a proper subset of <span class="math-container">$B$</span></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>
| lhf | 589 | <p>Actually, $3^{2n}\equiv 1\bmod 8$.</p>
<p>Indeed, by the binomial theorem, $3^{2n} = 9^n = (8+1)^n = 8a + 1$.</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>
| André Nicolas | 6,312 | <p><strong>Way 1:</strong> Let us compute a bit. Let $p=2$. Note that $x^2\equiv x \pmod 2$ for any $x$. Is it true that always $x^2\equiv x\pmod 4$? No, let $x=2$.</p>
<p><strong>Way 2:</strong> We do more work, but will get a lot more information. Rewrite the congruence $x^2\equiv x \pmod p$ as $x^2-x\equiv 0 \pmod p$, and then as $x(x-1)\equiv 0\pmod p$.</p>
<p>This says that the product of $x$ and $x-1$ is divisible by $p$. A product $ab$ is divisible by the <strong>prime</strong> $p$ if and only if $p$ divides $a$ or $p$ divides $b$, or both. </p>
<p>So $x(x-1)$ is divisible by $p$ if and only if $x$ is divisible by $p$ or $x-1$ is divisible by $p$, that is, if and only if $x\equiv 0 \pmod p$ or $x\equiv 1 \pmod p$.</p>
<p>Now does this force $x(x-1) \equiv 0 \pmod {p^2}$? No, for example we could take $x=p$. Certainly it is not true that $p(p-1)$ is divisible by $p^2$. We could also take $x=2p$, or $x=3p$, and so on up to $x=(p-1)p$. Or we could put the badness in the $x-1$ part, by putting $x=1+p$, or $x=1+2p$, and so on up to $x=1+(p-1)p$. We get a total of $2(p-1)$ numbers $x$ between $0$ and $p^2-1$ such that $x^2\equiv x \pmod p$ but $x^2\not\equiv x \pmod{p^2}$. </p>
<p>We can similarly identify the numbers $x$ between $0$ and $p^n-1$ such that $x(x-1)\equiv 0 \pmod p$ but $x(x-1)\not \equiv 0 \pmod{p^n}$. Again, they are of two types: (i) the numbers $p$, $2p$, and so on up to $p(p^{n-1}-1)$ and (ii) the numbers obtained by adding $1$ to numbers in list (i).</p>
|
102,624 | <p>I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one. </p>
<p>Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder
1) for which lattices this conjecture had been conjectured?
2) what is the motivation for this conjecture?
3) to whom this conjecture is due?
4) is it published somewhere?
5) is it proven in some cases?</p>
| Peter Humphries | 3,803 | <p>At the risk of blowing my own trumpet, I feel like I ought to mention a <a href="http://arxiv.org/abs/1502.06885" rel="nofollow">recent preprint of mine</a> that addresses this question.</p>
<p>Marc Palm answered your question 2, 3, and 4 reasonably well. (For question 4, the reference for where it is published is Cartier's paper.) For question 1, I know that Bolte and Johansson certainly expect the conjecture to be true for $\Gamma_0(q) \backslash \mathbb{H}$ with $q$ squarefree, provided one first removes all the Maaß oldforms (those coming from lower level), as they will always give rise to spectral multiplicity at least $2$. So for $q > 1$, the conjecture should be modified to only be about the eigenvalues of Maaß newforms.</p>
<p>Bolte and Johansson (and later Strömbergsson) describe a spectral correspondence between the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$, $q$ squarefree, and eigenvalues of the automorphic Laplacian for the group of units of norm one in a maximal order in an indefinite quaternion division algebra over $\mathbb{Q}$. Bolte and Johansson conjecture that the spectrum of this automorphic Laplacian is simple (<a href="http://dx.doi.org/10.1007/978-1-4612-1544-8_2" rel="nofollow">see the Hypothesis on p.61</a>), and hence that the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$ are simple when $q$ is squarefree. Here we are looking at the congruence subgroups
\[\Gamma_0(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} * & * \\\ 0 & * \end{pmatrix} \pmod{q}\right\},\]
\[\Gamma_1(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & * \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\},\]
\[\Gamma(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\}.\]</p>
<p>On the other hand, I show in my preprint that if $q$ is odd but not squarefree, then the new part of the spectrum of the Laplacian on $\Gamma_0(q) \backslash \mathbb{H}$ is <em>never</em> simple; there is always a positive proportion of eigenvalues $\lambda \leq T$ for which the corresponding eigenspaces of Maaß newforms are at least two-dimensional. For $\Gamma_1(q) \backslash \mathbb{H}$, the situation is even worse: there is spectral multiplicity even if $q$ is squarefree (provided $q \neq \{1,2,3,6\}$), and the dimension of an eigenspace can be proven to grow with $m$ if $q = p^m$ for an odd prime $p$. The proof doesn't use anything about the eigenvalue $1/4$, as in GH from MO's answer, but rather looks at twists of newforms that have the same level $q$ after twisting; this gives rise to spectral multiplicity.</p>
<p>There is one remaining case where I know that spectral multiplicity must occur; that of $\Gamma(p) \backslash \mathbb{H}$, with $p$ an odd prime; <a href="http://dx.doi.org/10.1090/S0002-9939-1981-0593486-1" rel="nofollow">Randol</a> shows that there must be infinitely many eigenvalues with multiplicity at least $\frac{1}{2} \left(p + (-1)^{(p - 1)/2}\right)$. I don't think anything is known for noncongruence subgroups.</p>
<p>I still think that for any congruence subgroup $\Gamma$, the multiplicity of an eigenvalue of the Laplacian on $\Gamma \backslash \mathbb{H}$ ought to be uniformly bounded as $\lambda \to \infty$, with the bound depending on $\Gamma$. Unfortunately our only tool for attacking this question (finding an upper bound on the multiplicity) seems to be via getting better error terms in the Weyl law, and we are very far off making that approach be useful.</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>
| sirous | 346,566 | <p>An experimental approach:</p>
<p>Here I find one solution for this equation which a family of solutions can be based on:</p>
<p><span class="math-container">$9x^2=y^2-p=(y-\sqrt p)(y+\sqrt p)$</span></p>
<p>we can can construct following system of equations:</p>
<p><span class="math-container">$\begin{cases}y-\sqrt p=9\\y+\sqrt p=x^2\end{cases}$</span></p>
<p>Subtracting first equation from second one we get:</p>
<p><span class="math-container">$2\sqrt p=x^2-9=(x-3)(x+3)$</span></p>
<p>Suppose:</p>
<p><span class="math-container">$x-3=2\Rightarrow x=5$</span></p>
<p><span class="math-container">$x+3=\sqrt p$</span></p>
<p>subtracting first fro, second we get:</p>
<p><span class="math-container">$\sqrt p-2=6\Rightarrow \sqrt p=8\rightarrow p=64$</span></p>
<p>Now we have:</p>
<p><span class="math-container">$9x^2=y^2-64=(y-8)(y+8)$</span></p>
<p>Let <span class="math-container">$y-8=9\rightarrow y=17$</span></p>
<p><span class="math-container">$x^2=y+8=25$</span></p>
<p>hence :</p>
<p><span class="math-container">$x=5$</span>, <span class="math-container">$y=17$</span> and <span class="math-container">$p=64$</span></p>
<p><span class="math-container">$9\times5^2+64=17^2\space\space\space\space\space (1)$</span></p>
<p><span class="math-container">$10^{12}\div 64=15625\times10^6>10^9$</span></p>
<p>So we can construct following relation by multiplying both sides of the relation (1) by <span class="math-container">$15625\times 10^6=(125\times 10^3)^2$</span>:</p>
<p><span class="math-container">$9(5\times 125000)^2+15625\times 10^6=(17\times125000)^2$</span></p>
<p>that is one solution of equation under condition <span class="math-container">$p=15625000000>10^9$</span> is:</p>
<p><span class="math-container">$x=625\times10^3$</span></p>
<p><span class="math-container">$y=2125\times 10^3$</span></p>
<p>This experiment shows that only with particular values of p the equation may have integer solutions.</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>
| poetasis | 546,655 | <p>If we let <span class="math-container">$\quad A^2+B^2=C^2
\qquad A=3x\quad B=\sqrt{p}\quad C=y\qquad$</span>
we can use Euclid's formula for Pythagorean triples to find infinite solutions based on the value of <span class="math-container">$(x)$</span>. We start with the formula</p>
<p><span class="math-container">$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$</span>
and solve the <span class="math-container">$A$</span>-function for <span class="math-container">$(k).\quad$</span>
Then we test a defined range of <span class="math-container">$m$</span>-values to see which yield integers.</p>
<p><span class="math-container">\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}</span>
The lower limit ensures <span class="math-container">$k\in\mathbb{N}$</span> and the upper limit ensures <span class="math-container">$m> k.$</span></p>
<p>There are Pythagorean triples for all <span class="math-container">$A=2n+1$</span> but A=<span class="math-container">$3x$</span>, so we are limited to odd muliples of <span class="math-container">$3:$</span>
<span class="math-container">$$ x\in\big\{1,3,5,7,\cdots\big\}\implies
A\in\big\{3,9,15,21,\cdots\big\}.\quad $$</span></p>
<p>Here, we will use <span class="math-container">$x=5\implies A=15$</span></p>
<p><span class="math-container">$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\\
\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$</span>
<span class="math-container">$$F(4,1)=(15,8,17)\implies (x,p,y)=(5,8^2,17)\\
F(8,7)=(15,112,113)\implies (x,p,y)=(5,112^2,113) $$</span></p>
|
96,657 | <p>I'm trying to understand an alternative proof of the idea that if $E$ is a dense subset of a metric space $X$, and $f\colon E\to\mathbb{R}$ is uniformly continuous, then $f$ has a uniform continuous extension to $X$.</p>
<p>I think I know how to do this using Cauchy sequences, but there is this suggested alternative. For each $p\in X$, let $V_n(p)$ be the set of $q\in E$ such that $d(p,q)<\frac{1}{n}$. Then prove that the intersections of the closures
$$
A=\bigcap_{n=1}^\infty\overline{f(V_n(p))}
$$
consists of a single point, $g(p)$, and so $g$ is the desired continuous extension of $f$. Why is this intersection a single point, and why is $g$ continuous?</p>
<hr>
<p>This is what I did so far. Since $f$ is uniformly continuous, for given $\epsilon>0$, there is $\delta>0$ such that $\text{diam }f(V)<\epsilon$ whenever $\text{diam }V<\delta$. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply
$$
\text{diam }f(V_n(p))=\text{diam }\overline{f(V_n(p))}<\epsilon
$$
So I think $\lim_{n\to\infty}\text{diam }\overline{f(V_n(p))}=0$, which would imply $A$ consists of at most one point. I noticed that the closures form a descending sequence of closed sets, but I couldn't tell if they are bounded since $X$ is an arbitrary metric space, in order to conclude that the intersection is nonempty, and hence a single point.</p>
<p>Lastly, why is $g$ continuous at points $p\in X\setminus E$? I was trying to think of an argument with sequences converging to $p$ since $p$ is a limit point of $E$, but got stumping on how to show $g$ is actually continuous. Thanks.</p>
| Community | -1 | <p>Here is a perhaps an alternative way of looking at what happens before you take the infinite intersection. We know that $\textrm{diam} V_n(p) < \frac{2}{n}$. So by uniform continuity of $f$, for all $\epsilon > 0$ there exists $\delta > 0$ (and hence an $n\in \Bbb{N}$ such that $0 < \frac{1}{2n} < \delta$ by the Archimedean property) such that </p>
<p>$$\textrm{diam} f(V_n(p)) < \epsilon.$$</p>
<p>In other words we have that $\textrm{diam} f(V_n(p)) \rightarrow 0$. Now Theorem 3.10(a) of Rudin tells you that for any subset $E$ of a metric space $X$, $\textrm{diam} E = \textrm{diam} \overline{E}$. Applying it here we have $\textrm{diam} \overline{f(V_n(p))} \rightarrow 0$. Now we have a nested sequence of closed sets</p>
<p>$$\overline{f(V_1(p))} \supset \overline{f(V_2(p))} \supset \ldots $$</p>
<p>Instead of using compactness, you can pick for each $n$ an $x_n \in \overline{f(V_n(p))}$. It is easy to see that $x_n$ is a cauchy sequence so by completeness of $\Bbb{R}$ must convergence to some $x$. This $x$ is easily seen to be in the infinite intersection and the diameter going to zero means that this is the only element left in it.</p>
|
41,302 | <p>This is a two part question, and for that I apologize.. but they're related!</p>
<p>Here's what I'm working with:</p>
<pre><code>d1 = Import["file.CSV", "List"]
size = Length[d1]
dis1 = RandomChoice[{d1}, {100, size}]
</code></pre>
<ul>
<li><p><strong>Q1</strong>: Length views <code>d1</code> as $300,000$ individual elements but <code>RandomChoice</code> views it as a single element. Thus when I execute I get $100$ repetitions of <code>d1</code> in the exact order <code>d1</code> is presented in. </p>
<p>Have I made an annotation error in assigning the <code>RandomChoice</code> number pool as <code>{d1}</code>? If I replace <code>{d1}</code> with a hand written list of values it executes perfectly, so I assume it is a presentation issue.. but I can't tell what I've done wrong from the documentation of the function.</p></li>
<li><p><strong>Q2</strong>: Asking to make $100$ random lists of <code>size</code> ~ $300,000$ elements requires a lot of memory. The thing is I don't need to save each <code>RandomChoice</code> output, I just need the mean/median/SD/SEM for each of the $100$ sets I've tasked <code>RandomChoice</code> with. </p>
<p>Is there a way to tell the program to spit out those end-point values and dump the accumulated list after each one is generated?</p></li>
</ul>
| DumpsterDoofus | 9,697 | <p>For Q1, your syntax mistake is an extra pair of brackets. The following code should work properly:</p>
<pre><code>d1 = Import["file.CSV", "List"]
size = Length[d1]
dis1 = RandomChoice[d1, {100, size}]
</code></pre>
|
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>
| joriki | 6,622 | <p>First, rest assured that you're not the only one who's confused by the standard notation for partial derivatives. See <a href="https://math.stackexchange.com/a/101011">this answer</a> for a collection of answers I've written in response to such confusions.</p>
<p>The problem is that the standard notation doesn't indicate which variables are being held constant. It assumes that you've defined a function of a certain set of variables, and that everyone remembers what these are. That's fine if you only introduce a single function and write its partial derivatives as</p>
<p>$$
\frac{\partial f(x,y,z)}{\partial x}
$$</p>
<p>and the like, since the arguments for the function evaluation make up for what the notation for the partial derivative is missing, but it becomes a problem when you start writing things like</p>
<p>$$
\frac{\partial f}{\partial x}
$$</p>
<p>and especially when you have lots of things like $x,y,z$ floating around that all look like variables and the notation doesn't contain the slightest clue which of these are being treated as functions and which as independent variables being held constant.</p>
<p>In a certain sense, you're right that you could always regard $\dfrac{\partial f}{\partial x}$ as $\dfrac{\mathrm df}{\mathrm dx}$ of a univariate function, namely by regarding all other variables as parameters. That is, given a function $f(x,y)$ of two variables, you can regard $y$ as a fixed parameter and write $g(x)=f(x,y)$, and then $\dfrac{\mathrm dg}{\mathrm dx}=\dfrac{\partial f}{\partial x}$. Then if you feel things aren't quite confusing enough already, you can instead call this new univariate function by the same name as the multivariate function $f$ and write $\vphantom{\dfrac{\partial f}{\partial x}}f(x)=f(x,y)$, and then indeed $\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\partial f}{\partial x}$, but you need to remember what you mean by that: $\dfrac{\mathrm df(x)}{\mathrm dx}=\dfrac{\partial f(x,y)}{\partial x}$, with two different uses of the symbol $f$.</p>
<p>However, this view is rarely very helpful, since the variables of a multivariate function are usually variables on an equal footing for good reason, and one would usually have introduced them as fixed parameters in the first place if that were the natural way to think of them. So usually it's better to view the univariate function that you get by keeping all but one variable fixed as a more temporary construct that's used only for defining and thinking about the partial derivative, but not as something that should appear in the notation as a univariate function in its own right.</p>
|
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