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 |
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3,608,114 | <p>Consider the example where I have a matrix <span class="math-container">$\mathbf{D}$</span> in <span class="math-container">$-1/1$</span> coding with <span class="math-container">$5$</span> columns,</p>
<p><span class="math-container">$$D = \begin{bmatrix}-1&-1&-1&1&1\\1&-1&-1&-1&1\\-1&1&-1&-1&-1\\1&1&-1&1&-1\\
-1&-1&1&1&-1\\1&-1&1&-1&-1\\-1&1&1&-1&1\\1&1&1&1&1\end{bmatrix}$$</span></p>
<p>We see that the fourth and fifth columns are combinations of the first three columns so that if we label the columns <span class="math-container">$a,b,c,d,e$</span> we can say that <span class="math-container">$d=a*b$</span> and <span class="math-container">$e=b*c$</span>. We can separate the columns in two groups: <span class="math-container">$a,b,c$</span> are <strong>basic columns</strong> and <span class="math-container">$e,d$</span> are <strong>added columns</strong>. Furthermore we can call the relations that define <span class="math-container">$e$</span> and <span class="math-container">$d$</span> as a combination of <span class="math-container">$a,b,c$</span> <strong>defining contrasts</strong>.</p>
<p>My question is this:
Is there a way to determine more generally (for a larger matrix or a matrix with different defining contrasts) which columns are combinations of the others and</p>
| P. Lawrence | 545,558 | <p>For an <span class="math-container">$m \times n$</span> matrix <span class="math-container">$A$</span> let <span class="math-container">$A'$</span> be the unique row-reduced echelon matrixx that is row-equivalent to <span class="math-container">$A.$</span> The relations of linear dependence among the columns of <span class="math-container">$A$</span> are the same as those among the columns of <span class="math-container">$A'.$</span> The columns of <span class="math-container">$A'$</span> in which a leading 1 appears are a basis for the column-space of <span class="math-container">$A'$</span> and the columns of <span class="math-container">$A$</span> in the same positions are a basis for the column-space of <span class="math-container">$A.$</span>Let the <span class="math-container">$t-$</span>th column of <span class="math-container">$A$</span> be <span class="math-container">$\zeta_t$</span> and let the <span class="math-container">$t-$</span>th column of <span class="math-container">$A'$</span> be <span class="math-container">$\xi_t$</span> for <span class="math-container">$1 \le t \le n.$</span>For a column <span class="math-container">$\zeta $</span> of <span class="math-container">$A$</span> let <span class="math-container">$\xi$</span> be the column of of <span class="math-container">$A'$</span> in the same position, let the element in row <span class="math-container">$i$</span> of <span class="math-container">$\xi$</span> be <span class="math-container">$c_i$</span> and, if <span class="math-container">$c_i \ne 0$</span> let the leading 1 in row <span class="math-container">$i$</span> of <span class="math-container">$A'$</span> occur in column <span class="math-container">$\nu_i$</span> of <span class="math-container">$A'$</span> for <span class="math-container">$1 \le i \le m.$</span> If <span class="math-container">$c_i=0,$</span> let
<span class="math-container">$\zeta_{\nu_i}=\zeta_{\nu_i}=\mathbf 0.$</span> Then <span class="math-container">$$\xi=\sum_{i=1}^mc_i\xi_{\nu_i}$$</span> and <span class="math-container">$$\zeta=\sum_{i=1}^mc_i\zeta_{\nu_i}$$</span>.</p>
|
575,513 | <p>Can someone help me find the density function $f_X$ for $X$ and hence find $E(X)$ and $Var(X)$ of the following distribution function $F_X$ given by:</p>
<p>$F_X(x)=\begin{cases}
1-(1+x)e^{-x} & x>0 \\
0 & otherwise.
\end{cases}$</p>
<p>$X$ is a continuous random variable.</p>
<p>From memory, do I have to integrate $1-(1+x)e^{-x}$ or something similar? I can't recall on what to do, I get mixed up with the range in which I must integrate these sort of things (I am not even sure if I must integrate it but I know that when going from the probability density function to the distribution function, I must integrate it).</p>
| tomasz | 30,222 | <p>An example for a commutative ring which is not a domain: $R=\{0,a,1-a,1\}$ with $a^2=a,a+a=1+1=0$ and $A=R$. The elements $a,1-a$ are torsion, but $a+(1-a)=1$ isn't.</p>
|
575,513 | <p>Can someone help me find the density function $f_X$ for $X$ and hence find $E(X)$ and $Var(X)$ of the following distribution function $F_X$ given by:</p>
<p>$F_X(x)=\begin{cases}
1-(1+x)e^{-x} & x>0 \\
0 & otherwise.
\end{cases}$</p>
<p>$X$ is a continuous random variable.</p>
<p>From memory, do I have to integrate $1-(1+x)e^{-x}$ or something similar? I can't recall on what to do, I get mixed up with the range in which I must integrate these sort of things (I am not even sure if I must integrate it but I know that when going from the probability density function to the distribution function, I must integrate it).</p>
| Aufenthaltsraum | 238,486 | <p>actually, the Torsion subset is a submodule for all $R$-modules $M$.
Recall that $m\in M$ is called torsion, if there is $r\in R$ which is regular (i.e. not a zero divisor) such that $r.m=0$.</p>
<p>Assume $m$ and $m'$ are torsion with corresponding regular elements $r$ and $r'$. Then $rr'$ is non-zero and still not a zero divisor and we have $rr'.(m+m')=0$. Hence $m+m'$ is torsion.
It is trivial to see that if $m$ is torsion then the same is true for $s.m$ for all $s\in R$.</p>
<p>Note also that over a finite ring every module is torsion.
To see this, notice that an element of a finite ring is either a unit or a zero-divisor. Hence all regular elements act invertibly on modules and hence cannot kill any non-zero elements.</p>
|
239,863 | <p>I've to study this series:</p>
<p>$$\sum_{n=1}^\infty e^{\sqrt n\,x}$$ </p>
<p>My teacher wrote that with the asymptotic comparison with this series:</p>
<p>$$\sum_{n=1}^\infty\frac{1}{n^2}$$<br>
My series converges for every </p>
<p>$$x<0$$</p>
<p>I don't understand the motivation, hoping for someone to enlighten me!</p>
<p>=) Thanks. Leonardo.</p>
| Norbert | 19,538 | <p>There is a more general result.</p>
<p><strong>Theorem.</strong> Let $E$ be a normed space. Let $\{x_n:n\in\mathbb{N}\}\subset E$ and $x\in E$, then the following conditions are equivalent:</p>
<ul>
<li>$\{x_n:n\in\mathbb{N}\}$ weakly converges to $x\in E$</li>
<li>$\{x_n:n\in\mathbb{N}\}$ is bounded and for all $S\subset E^*$ such that $\mathrm{cl}(\mathrm{span} S)=E^*$ holds $\lim\limits_{n\to\infty} f(x_n)=f(x)$ for all $f\in S$.</li>
</ul>
<p><strong>Proof.</strong> Assume $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$. Then it is well knonw that $\{x_n:\in\mathbb{N}\}$ is bounded. Moreover for all $f\in X^*$ we know that $\lim\limits_{n\to\infty}f(x_n)=f(x)$. Hence for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ we have $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$.</p>
<p>Assume that $\{x_n:n\in\mathbb{N}\}$ is bounded by some conatant $M>0$ and for all $S\subset X^*$ such that $\mathrm{cl}(\mathrm{span}S)=X^*$ holds $\lim\limits_{n\to\infty}f(x_n)=f(x)$ for all $f\in S$. Take arbitrary $g\in\mathrm{span}S$, then $g=\sum\limits_{k=1}^m \alpha_k f_k$. Then it is easy to check that $\lim\limits_{n\to\infty}g(x_n)=g(x)$. So for all $g\in\mathrm{span}S$ we have $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Now take arbitrary $g\in X^*=\mathrm{cl}(\mathrm{span}S)$, then $g=\lim\limits_{k\to\infty} g_k$ for some $\{g_k:k\in\mathbb{N}\}\subset\mathrm{span}S$. Fix arbitrary $\varepsilon>0$. Since $g=\lim\limits_{k\to\infty} g_k$, then there exist $K\in\mathbb{N}$ such that $\Vert g-g_k\Vert\leq\varepsilon$ for all $k>K$. Then
$$
\begin{align}
|g(x_n)-g(x)|
&\leq|g(x_n)-g_k(x_n)|+|g_k(x_n)-g_k(x)|+|g_k(x)-g(x)|\\
&\leq\Vert g-g_k\Vert\Vert x_n\Vert+|g_k(x_n)-g_k(x)|+\Vert g_k-g\Vert\Vert x\Vert\\
&\leq\varepsilon M+|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert
\end{align}
$$
Let's take a limit $n\to\infty$ in this inequality, then
$$
\limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq
\varepsilon M+\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|+\varepsilon\Vert x\Vert
$$
Since $g_k\in\mathrm{span}S$, then $\lim\limits_{n\to\infty}|g_k(x_n)-g_k(x)|=0$ and we get
$$
\limsup\limits_{n\to\infty}|g(x_n)-g(x)|\leq
\varepsilon M+\varepsilon\Vert x\Vert
$$
Since $\varepsilon>0$ is arbitrary we conclude $\limsup\limits_{n\to\infty}|g(x_n)-g(x)|=0$. This is equivalent to $\lim\limits_{n\to\infty}|g(x_n)-g(x)|=0$ i.e. $\lim\limits_{n\to\infty}g(x_n)=g(x)$. Since $g\in X^*$ is arbitrary, then $\{x_n:n\in\mathbb{N}\}$ weakly converges to $x$.</p>
|
64,881 | <p>I am having trouble with this problem from my latest homework.</p>
<p>Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.</p>
<p>Any and all help would be appreciated.</p>
| André Nicolas | 6,312 | <p>Note that
$$\frac{x+y}{2}-\sqrt{xy}=\frac{(\sqrt{x}-\sqrt{y})^2}{2}.$$</p>
|
64,881 | <p>I am having trouble with this problem from my latest homework.</p>
<p>Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.</p>
<p>Any and all help would be appreciated.</p>
| Mongol-genius | 111,192 | <p>$$0\le ({\sqrt x}-{\sqrt y})^{2}$$
$$0\le x-2{\sqrt {xy}}+y$$
$$2{\sqrt {xy}}\le x+y$$
$${\sqrt {xy}}\le {x+y\over2}$$</p>
|
64,881 | <p>I am having trouble with this problem from my latest homework.</p>
<p>Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.</p>
<p>Any and all help would be appreciated.</p>
| Daniel W. Farlow | 191,378 | <p>I am surprised no one has given the following very straightforward algebraic argument:
\begin{align}
0\leq(x-y)^2&\Longleftrightarrow 0\leq x^2-2xy+y^2\tag{expand}\\[0.5em]
&\Longleftrightarrow 4xy\leq x^2+2xy+y^2\tag{add $4xy$ to both sides}\\[0.5em]
&\Longleftrightarrow xy\leq\left(\frac{x+y}{2}\right)^2\tag{div. sides by 4 & factor}\\[0.5em]
&\Longleftrightarrow \sqrt{xy}\leq\frac{x+y}{2}.\tag{since $x,y\in\mathbb{R}^+$}
\end{align}
In regards to equality, notice that $\sqrt{xy}\leq\frac{x+y}{2}\leftrightarrow 2\sqrt{xy}\leq x+y$, and it becomes clear that equality holds if and only if $x=y$. $\blacksquare$</p>
|
69,961 | <p>I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s.</p>
<p>$$S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}$$</p>
<p>I want to check whether that would be:
0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on.</p>
<p>Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set?</p>
| robjohn | 13,854 | <p>This doesn't tell you exactly which numbers can be written as $3k+5j$ with $j,k\ge0$, but it might be the best that can be said in general. These are two Theorems that usually accompany <a href="http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" rel="nofollow">Bezout's Identity</a>.
<hr/>
<strong>Theorem $\boldsymbol{1}$:</strong> Suppose $\operatorname{gcd}(a,b)=1$ and $c \ge (a-1)(b-1)$. Then $ax+by=c$ has a non-negative solution, that is, one in which both $x$ and $y$ are
non-negative integers.</p>
<p><strong>Proof of Theorem $\boldsymbol{1}$:</strong><br>
Since $gcd(a,b)=1$, we have some $(u,v)$ so that $au+bv=1$. The set of all
solutions to $ax+by=c$ is $\{ (cu+bk,cv-ak) : k \in Z \}$. Thus, we have a
non-negative solution $(x,y)$, i.e. one in which both $x$ and $y$ are non-
negative, precisely when there is an integer $k$ so that $k \ge -cu/b$ (so
that $cu+bk \ge 0$) and $k \le cv/a$ (so that $cv-ak \ge 0$). Thus, $ax+by=c$
has a non-negative iff there is an integer in $[-cu/b,cv/a]$.</p>
<p>Suppose there is no integer in this interval. This means that it must
be contained in some interval $(j,j+1)$. Since $cu$ and $cv$ are integers,
we must have
$$
-cu/b-j \ge 1/b\tag{1a}
$$
and
$$
j+1-cv/a \ge 1/a\tag{1b}
$$
Adding $(1a)$ and $(1b)$ and multiplying by $ab$ gives
$$
ab-cau-cbv \ge a+b\tag{1c}
$$
Since $au+bv=1$, $(1c)$ becomes
$$
c \le ab-a-b\tag{1d}
$$
Therefore, if $c \ge ab-a-b+1 = (a-1)(b-1)$, then there is a non-negative
solution $(x,y)$ to $ax+by=c$.
$$\square$$</p>
<hr>
<p><strong>Theorem $\boldsymbol{2}$:</strong> Suppose $\operatorname{gcd}(a,b)=1$, $0 \lt c \lt ab$, and neither $a\mid c$ nor $b\mid c$. Then
one and only one of
$$
ax+by=c\tag{2a}
$$
and
$$
ax+by=ab-c\tag{2b}
$$
has a non-negative solution.</p>
<p><strong>Proof of Theorem $\boldsymbol{2}$:</strong><br>
Note that since neither $a\mid c$ nor $b\mid c$, neither $x$ nor $y$ can be $0$ in any solution. Therefore, any non-negative solution must be a positive
solution, that is, one in which both $x$ and $y$ are positive integers.</p>
<p>Suppose both $as+bt=c$ and $au+bv=ab-c$ are positive solutions. Add them
together to get
$$
a(s+u)+b(t+v) = ab\tag{2c}
$$
Since $gcd(a,b)=1$, $(2c)$ says that $b|s+u$ and $a|t+v$. Since $s$, $t$, $u$, and $v$ are positive integers, we must have that $s+u\ge b$ and $t+v\ge a$. However, then $a(s+u)+b(t+v) \ge 2ab$, which contradicts $(2c)$.</p>
<p>Therefore, we have shown that at most one of $(2a)$ and $(2b)$ can have a
non-negative solution.</p>
<p>Suppose $(2a)$ does not have a non-negative solution. Since $gcd(a,b)=1$,
we have some $(u,v)$ so that $au+bv=1$. The set of all solutions to $ax+by=c$
is then $\{ (cu+bk,cv-ak) : k \in Z \}$. Therefore, we can find an $(s,t)$ so
that $as+bt=c$ and $0 \le s \lt b$.</p>
<p>Since $bt = c-as$, we have that $-ab \lt bt \lt ab$. Since $(2a)$ does not have
a non-negative solution, we must have $-ab \lt bt \lt 0$. Thus, we have the
non-negative solution $a(b-s)+b(-t) = ab-c$.</p>
<p>Therefore, we have shown that at least one of $(2a)$ and $(2b)$ must have a
non-negative solution.
$$\square$$</p>
|
252,767 | <p>I'm looking for a tangible example of a free abelian group whose quotient with a subgroup is not free abelian. There's a theorem that says that every abelian group is a quotient of some free group, but I'm looking for a more exact example.</p>
| Bombyx mori | 32,240 | <p>The confusion maybe because every subgroup of a free abelian group is free abelian, while for the quotient group this is not necessarily true. The canoical example maybe $\mathbb{Z}/p\mathbb{Z}$, where $p$ is prime. Here $\mathbb{Z},p\mathbb{Z}$ are both free but the above group is not free. The wiki article probably can provide more information on this. </p>
|
252,767 | <p>I'm looking for a tangible example of a free abelian group whose quotient with a subgroup is not free abelian. There's a theorem that says that every abelian group is a quotient of some free group, but I'm looking for a more exact example.</p>
| pepa.dvorak | 85,466 | <p>The fact that you mention is a more general fact, i.e. every module is factor of a free module - you can imagine the construction in the following way:</p>
<p>take "enough" generators and create a free module over them, then, since different modules differ in "which elements are the same", i.e. in relations between the elements (therefore between the generators), create a submodule of relations $R$; factoring $R$ away is just like saying "the relations in $R$ are not important, zero" and in this way you can naturally get any module.</p>
<p>The example of $\mathbb{Z}_p$'s is:</p>
<p>take one element $x$ and create free group, so you get $F = \mathbb{Z}x$ (which is isomorphic to $\mathbb{Z}$); now you want to say that if two elements differ in a multiple of $p$, they're the same, so the relation submodule $R$ is $p\mathbb{Z}x$ (it contains $...,-1p, 0x, px, 2px..$) and factoring it out you say that all these elements are equal to each other (so to $0$ as well). If now you have for example $(2p + r)x$ and $(3p + r)x$ two elements of $F$, their difference is $p$, which lies in $R$, i.e. it "is not important", these two are the same in factor and you got the structure of $\mathbb{Z}_p$. </p>
|
90,940 | <p>It seems known that the category of hypergraphs is a topos.
I am looking for any reference here, or just a statement of this in the literature, but can't find anything. There is one paper </p>
<blockquote>
<p>A category-theoretical approach to hypergraphs,
W. Dörfler and D. A. Waller, ARCHIV DER MATHEMATIK, Volume 34, Number 1, 185-192, <a href="https://doi.org/10.1007/BF01224952" rel="nofollow noreferrer">DOI:10.1007/BF01224952</a>, 1980</p>
</blockquote>
<p>which might contain information about that, but I don't have access to this paper (and it might take some time to get a copy, likely a paper-copy).</p>
<p>By a hypergraph I mean here a triple $(V,E,h)$, where $V$, $E$ are arbitrary sets, while $h$ is a map from $E$ to the set of finite subsets of $V$ (so $V$ is the set of vertices, $E$ the set of hyperedge-labels, and $h$ yields the hyperedge of a hyperedge-label). Morphisms are pairs $a: V \rightarrow V'$, $b: E \rightarrow E'$, which fulfill the usual commutativity condition.</p>
| James Cranch | 14,901 | <p>One can reinterpret a hypergraph as a span-shaped diagram of sets where the left leg of the span is a finite map (meaning, all preimages are finite). Indeed, given a hypergraph, consider the span
$$V\leftarrow\lbrace(v,e)\in V\times E\mid v\in h(e)\rbrace\rightarrow E;$$
it is clear that this gives a correspondence.</p>
<p>This seems more natural to work with.</p>
<p>The category of sets is a topos. The category of diagrams of some given shape in a topos is itself a topos, so the category of span-shaped diagrams of this sort is again a topos. Imposing finiteness conditions tends not to destroy the property of being a topos, and one can rapidly check that philosophy in this case.</p>
|
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which can help me. </p>
<p>Thanks :) </p>
| Carl Wienecke | 38,256 | <p>$Topology$ by James Munkres is an excellent book for that sort of thing. </p>
|
1,278,860 | <p>Use the process of implicit differentiation to find $dy/dx$ given that:</p>
<p>$$x^2e^y − y^2e^x=0 $$</p>
<p>I am trying first to find $y$, </p>
<p>$$y^2e^x = x^2e^y$$</p>
<p>$$y^2 = (x^2e^y)/e^x$$</p>
<p>$$y = \sqrt{(x^2e^y)/e^x}$$</p>
<p>Is this correct? I have the feeling it is not.</p>
| architectpianist | 141,199 | <p>The math is right, but if you are using implicit differentiation the point is <em>not</em> to solve for $y$. Instead you would differentiate each term with respect to $x$, assuming that $y$ is some function of $x$ whose derivative is $dy/dx$. For instance, the term $y^2e^x$ yields</p>
<p>$$2y\frac{dy}{dx}e^x+y^2e^x$$</p>
<p>Hopefully this helps as a starting point. (You will need to differentiate the equation term by term as shown, then solve for $dy/dx$.)</p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| N. S. | 9,176 | <p><strong>Hint</strong></p>
<p>$$ (n+1)(n+2)...(n+d-1)(n+d)= \left[(n+1)(n+2)(n+3)...(n+d-1)\right]n + \left[(n+1)(n+2)(n+3)...(n+d-1) \right]d$$</p>
<p>$P(n)$ tells you that the first term on RHS is divisible by $d$, while the second one is clearly divisible by $d$...</p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| Brian M. Scott | 12,042 | <p>Note that $n(n+1)\dots(n+d-1)$ is the product of $d$ consecutive integers. Thus, it suffices to prove that if $n,n+1,\dots,n+d-1$ are any $d$ consecutive integers, then $d$ divides one of these integers. I would prove this by induction on $n$, simultaneously for all $d$.</p>
<p>First, it’s clearly true for $n=1$, since in that case the $d$ integers are $1,2,\dots,d$. Suppose now that it’s true for some $n$, and consider the $d$ consecutive integers $n+1,n+2,\dots,n+d$. By the induction hypothesis one of the integers $n,n+1,\dots,n+d-1$ is a multiple of $d$. If it’s one of the integers $n+1,n+2\dots,n+d-1$, you’re done (why?), and if $n$ is a multiple of $d$, then so is ... ?</p>
|
306,011 | <p>Does anyone have a proof for $$\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.$$
I tried to get it from contour integrating $$\frac{e^{iz^2}-1}{z^2},$$ but failed.
Thanks.</p>
| hunminpark | 54,833 | <p><strong>Another solution.</strong> <em>(which does not use complex analysis)</em><br>
Substitute $u=x^2$, then the integral becomes
$$A:=\int_{0}^{\infty}\frac{\sin (x^2)}{x^2}dx=\frac{1}{2}\int_{0}^{\infty}u^{-3/2}\sin u du$$
Now we'll consider more general one;
$$f(p):=\int_{0}^{\infty}\frac{\sin u}{u^p}du\phantom{a} (p\in (1, 2))$$
From the definition of the gamma function, we can easily deduce that
$$\frac{\Gamma(p)}{u^p}=\int_{0}^{\infty}v^{p-1}e^{-uv}dv$$
This formula gives
$$f(p)\Gamma(p)=\int_{0}^{\infty}\sin u\left(\int_{0}^{\infty}v^{p-1}e^{-uv} dv\right)du$$
By Fubini's theorem,
$$f(p)\Gamma(p)=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u dvdu=\int_{0}^{\infty}\int_{0}^{\infty}v^{p-1}e^{-uv}\sin u du dv$$
So
$$f(p)\Gamma(p)=\int_{0}^{\infty}v^{p-1}\left(\int_{0}^{\infty}e^{-uv}\sin udu\right)dv=\int_{0}^{\infty}\frac{v^{p-1}}{v^2+1}dv$$
Substitute $v=\tan \theta$, then
$$\begin{aligned}f(p)\Gamma(p)&=\int_{0}^{\pi /2}\tan^{p-1}\theta d\theta=\int_{0}^{\pi /2}\cos^{2\cdot(1-p/2)-1}\theta\sin^{2\cdot p/2-1}\theta d\theta\\&=\frac{1}{2}\operatorname{B}\left(1-\frac{p}{2},\frac{p}{2}\right)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)\end{aligned}$$
($\operatorname{B}(\phantom{x},\phantom{y})$ denotes Beta function.)<br>
By Euler's reflection formula,
$$f(p)\Gamma(p)=\frac{1}{2}\Gamma\left(1-\frac{p}{2}\right)\Gamma\left(\frac{p}{2}\right)=\frac{1}{2}\frac{\pi}{\sin(\pi p/2)}$$
Therefore
$$f(p)=\int_{0}^{\infty}\frac{\sin u}{u^p}du=\frac{1}{2}\frac{\pi}{\Gamma(p) \sin(\pi p/2)}$$
Substituting $\displaystyle p=\frac{3}{2}$, we obtain
$$A=\frac{1}{2}f\left(\frac{3}{2}\right)=\frac{1}{2}\cdot\frac{1}{2}\frac{\pi}{\sqrt{\pi}/2 \cdot \sqrt{2}/2}=\frac{\sqrt{\pi}}{\sqrt{2}}$$</p>
|
1,458,975 | <p>I'm having a issue with solving this problem. I know that the answer is $ a=3, b=1 $. But i'm not sure how to get to that conclusion.</p>
<p>Given that $(a+i)(2-bi)=7-i$, find the value of $a$ and of $b$, where $a,b \in \mathbb{Z}$.</p>
| Asinomás | 33,907 | <p>If that were possible then we would have $(a+b\sqrt{2})^2)=a^2+2b^2+2ab\sqrt{2}=3$, this would imply $2ab\sqrt{2}\in\mathbb Q$. So we would have $ab=0$.</p>
<p>So you are left with two cases:</p>
<p>$a=0,b=0$.</p>
<p>The first case gives us $a^2=3$ which is clearly imposible in the rationals. The second one gives us $2b^2=3$, this is impossible. Let $b=\frac{p}{q}$ a reduced fraction. Then we have $2p^2=3q^2$ and so $3$ divides $p$. But then $p=3r$ and we have $2(3r)^2=3q^2\implies 2r3=q^2$ and so $q$ is also a multiple of $3$, a contradiction.</p>
|
2,161,294 | <p>I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:</p>
<p>Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infinite? Is it a dense set? Is in countable? Are their members irrational numbers??</p>
<p><strong>Many thanks in advance!!</strong></p>
| Amin235 | 324,087 | <p>If you want to find these numbers, you should search in the limit values of sequence numbers like Fibonacci numbers. For example, Consider the following sequence</p>
<p>\begin{equation}
a_n=
\left\{
\begin{array}{cc}
a_{n-3} & n=1(mod \hspace{1mm}2)~,\\ \\
a_{n-3}+ a_{n-2} & n=0(mod \hspace{1mm}2)~.
\end{array}
\right.
\end{equation}
with the following initial values </p>
<p>$$a_0=0~, \hspace{5mm} a_1=1~, \hspace{5mm} a_2=0~.$$</p>
<p>Now, consider the limit values of the $a_n$ sequence are defined as follows</p>
<p>$$
\lim_{n\rightarrow\infty}\frac{a_{2n}}{a_{2n+1}}=\alpha_1~,\quad
\lim_{n\rightarrow\infty}\frac{a_{2n+1}}{a_{2n+2}}=\alpha_2~.
$$ </p>
<p>with calculation, you can see that</p>
<p>\begin{equation}
\left\{
\begin{array}{ccc}
\alpha_1 &=& 1.4655712318767680267\, , \\
&&\\
\alpha_2 &=& 0.4655712318767680267\, .
\end{array}
\right.
\end{equation}</p>
|
2,406,043 | <p>Let the triangle $\triangle ABC$ have sides $a,b,c$ and be inscribed in a circle with radius $R$. If $p=\frac{a+b+c}{2}$ The radius of the circle can be expressed as</p>
<p>a) $$R=\frac{\sqrt{p(p-a)(p-b)(p-c)}}{4abc}$$</p>
<p>b) $$R=\frac{4\sqrt{p(p-a)(p-b)(p-c)}}{abc}$$</p>
<p>c) $$R=\frac{abc}{4\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<p>d) $$R=\frac{4abc}{\sqrt{p(p-a)(p-b)(p-c)}}$$</p>
<hr>
<p>So clearly Heron's formula is being used here. I know that the radical is an expression of the area of the triangle. Let's denote $A_T=\sqrt{p(p-a)(p-b)(p-c)}$ and solve for it in all four cases:</p>
<p>a) $A_T=4Rabc$</p>
<p>b) $A_T=\frac{Rabc}{4}$</p>
<p>c) $A_T=\frac{abc}{4R}$</p>
<p>d) $A_T=\frac{4abc}{R}$</p>
<p>None of the RHS's even closely resembles to the area of the triangle of the form $A=\frac{base \cdot height}{2}.$ How should I do this? Keep in mind that this is from a test where one is to have about 3 minutes per question, so no complicated solution should be needed.</p>
| Michael Rozenberg | 190,319 | <p>Another way.</p>
<p>We need to prove that
$$a^3+b^3+c^3-3abc\geq0.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.</p>
<p>Hence, our inequality is a linear inequality of $v^2$ because it's third degree.</p>
<p>Thus, it remains to prove our inequality for an extremal value of $v^2$,</p>
<p>which happens for equality case of two variables.</p>
<p>Since the last inequality is homogeneous and symmetric, we can assume $v=c=1$,</p>
<p>which gives $$a^3+2-3a\geq0$$ or
$$(a+2)(a-1)^2\geq0.$$
Done!</p>
|
2,489,498 | <p>A={a,b,c,d}</p>
<p>R={(a,b),(a,c),(c,b)}</p>
<p>According to the definition for transitive relation, if there is (a,b) and (b,c) there should be (a,c)</p>
<p>In the above relation there is (a,c),(c,b) as well as (a,b). Shouldn't it be transitive?</p>
| Jaideep Khare | 421,580 | <p>For quickly finding the limit; apply L'Hospital's rule to $$\lim_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}}$$</p>
<p>To get $$\lim_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}} =\lim_{x \to 0} \frac{\frac 1x}{\frac{-1}{2 x \sqrt x}}=\lim_{x \to 0} \left(-2\sqrt x \right)=0$$</p>
|
2,505,863 | <p>I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2).
I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that.
So I take the formula: x' = a + Ax and started to fill in what we know about. We know that a = (2,2,2) to be able to map Q and we are looking for a matrix that can also transform P to P'. So I filled in P as x, and P' as x'.</p>
<p>$\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix} + \begin{pmatrix}a11+a12+a13\\a21+a22+a23\\a32+a32+a33\end{pmatrix}*\begin{pmatrix}1\\1\\1\end{pmatrix}$</p>
<p>Then I get 3 equations with 9 variables so I'm not sure how to solve it:</p>
<p>I: a11 + a12 + a13 = -3</p>
<p>II: a21 + a22 + a23 = -3</p>
<p>III: a31 + a32 + a33 = -3</p>
<p>Am I on the right way? How can I solve this equations?</p>
| Guy Fsone | 385,707 | <p>Hint use the definition of derivative at x=4 as follows
$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} =
\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{x-4}
\cdot\lim_{x\to 4} \frac{x-4}{\sqrt{x-2}-\sqrt{2}}=(\sqrt{2x+1})'
\frac{1}{(\sqrt{x-2})'}\Big|_{x=4}
= \frac{1}{3}\frac{2\sqrt2}{1}$$</p>
|
3,778,024 | <p>Let <span class="math-container">$(\Omega, \mathcal{F}, P)$</span> be a probability space, <span class="math-container">$X$</span> a random variable and <span class="math-container">$F(x) = P(X^{-1}(]-\infty, x])$</span>. The statement I am trying to prove is</p>
<blockquote>
<p>The distribution function <span class="math-container">$F$</span> of a random variable <span class="math-container">$X$</span> is right continuous, non-decreasing and satisfies <span class="math-container">$\lim_{x \to \infty}F(x) = 1$</span>, <span class="math-container">$\lim_{x \to -\infty} F(x) = 0$</span>.</p>
</blockquote>
<p>As <span class="math-container">$F(x + \delta) = F(x) + P(]x, x + \delta])$</span>, we have that <span class="math-container">$F$</span> is non-decreasing, but is the measure of an interval bounded by its length? In that case we would have right continuity as well.</p>
<p>For the limits, we have <span class="math-container">$F(x) + P(X^{-1}(]x, \infty]) = P(\Omega) = 1$</span>, so <span class="math-container">$F(x) = 1 - P(X^{-1}(]x, \infty])$</span>, so it suffices for <span class="math-container">$P(X^{-1}(]x, \infty])$</span> to get small as <span class="math-container">$x$</span> gets large and to get large as <span class="math-container">$x$</span> gets small. This is not true for general measures, take the Lebesgue measure for example, but maybe because we need <span class="math-container">$P(X^{-1}(\mathbb{R}))$</span> to be <span class="math-container">$1$</span>?</p>
| Michael Hardy | 11,667 | <p>The probability assigned to an interval is certainly not bounded by its length. For example, discrete distributions assign positive probability to intervals of length <span class="math-container">$0.$</span></p>
<p>To prove right-continuity you need countable additivity.</p>
<p><span class="math-container">\begin{align}
F(x) & = \Pr(X\le x) = 1 - \Pr(X>x) \\[8pt]
& = 1 - \Pr(x+1 < X \text{ or } x+\tfrac 1 2 < X\le x+1 \text{ or } x+\tfrac 1 3 < X\le x + \tfrac 1 2 \text{ or } \cdots) \\[8pt]
& = 1 - \big( \Pr(x+1< X) +\Pr(x+\tfrac 1 2 < X\le x+1) + \Pr(x+\tfrac 1 3< X\le x + \tfrac 1 2) + \cdots \\[8pt]
& = 1 - \lim_{N\,\to\,\infty} \sum_{n\,=\,0}^N \Pr( x + \tfrac 1 {n+1} < X \le x + \tfrac 1 n) \\[8pt]
& = \lim_{N\,\to\,\infty} \Pr(X\le x + \tfrac 1 {N+1}) = \lim_{N\,\to\,\infty} F(x + \tfrac 1{N+1}).
\end{align}</span></p>
<p>Given <span class="math-container">$\varepsilon>0,$</span> find <span class="math-container">$N$</span> large enough so that <span class="math-container">$F(x+\tfrac 1{N+1}) < F(x)+\varepsilon, $</span> and then choose <span class="math-container">$\delta= 1/N.$</span> Then for <span class="math-container">$x < w < x+\delta,$</span> you have <span class="math-container">$F(x)\le F(w)< F(x)+\varepsilon.$</span> The point of this paragraph is that it's not just <span class="math-container">$\lim_{N\to\infty} F(x+\tfrac 1 {N+1}) = F(x),$</span> but <span class="math-container">$\lim_{w\,\downarrow\,x} F(w) = F(x).$</span></p>
|
2,243,083 | <p>I'm writing an advanced interface, but I don't yet have a concept of derivatives or integrals, and I don't have an easy way to construct infinite many functions (which could effectively delay or tween their frame's contributing distance [difference between B and A] over the next few frames).</p>
<p>I can store values for a frame, and I can also consume them or previous values and map into A.</p>
<p>For example, each frame could calculate the distance between B and A, then add that distance to A, but they would be perfectly in sync.</p>
<p>I can keep track of the last N frames' distances and constantly shift old distances off, but this would create a delay, not an elastic effect. Somehow, the function that's popping off past-frames' distances needs to adjust for how long it's been for those 10 frames.</p>
<hr>
<p><strong>Is there any function I can rewrite each frame, which picks up the progress from it's predecessor, and contributes it's correct delta amount, adjusting for the new total distance between B and A?</strong></p>
<p>Does this question make sense? How can I achieve behavior where A is constantly catching up to B in a non-linear, exponential way?</p>
<hr>
| Narasimham | 95,860 | <p>Elastic motion obeys a time differential equation representing a dynamic system of order two or higher in which elasticity constants like $m,k$ are fixed. The simplest harmonic motion $ m \ddot x + k x=0 $ enforces distances and you have no further control except on the imposed boundary conditions.</p>
|
3,636,667 | <blockquote>
<p>Evaluate
<span class="math-container">$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$</span> </p>
</blockquote>
<p>Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good ideas yet, so I will appreciate any hints !</p>
| Gary | 83,800 | <p>A lower bound is given by
<span class="math-container">$$
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}} \ge \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {i^p } \\ = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {\frac{i}{p}} \right)^p } = \int_0^1 {x^p dx} = \frac{1}{{p + 1}} .
$$</span>
An upper bound:
<span class="math-container">\begin{align*}
& \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {(i + 1)(i + 2) \cdots (i + p)}
\\ &
\le \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\left( {i + \frac{{p + 1}}{2}} \right)^p } \le \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\int_i^{i + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} }
\\ & =
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\int_1^{n + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} \\ & = \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1} - \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1}
\\ &
= \frac{1}{{p + 1}}.
\end{align*}</span>
I first used the inequality between the geometric and the arithmetic mean, then estimated each term by an integral taking into account the monotonicity of the power function. Thus, the limit in question is <span class="math-container">$\frac{1}{p+1}$</span>.</p>
|
1,309,670 | <p>Suppose $D \subset \mathbb{R}$ is open, $f : D \to \mathbb{R}$ is a smooth (not necessarily real analytic) function, $x_0 \in D$, and $T_n$ is the degree $n$ Taylor polynomial of $f$ centered at $x_0$. Let $S=\{ x \in D : f(x)=T_n(x) \}$. It is not hard to see that $S$ is closed and contains $x_0$. What else can be said about it (possibly with additional hypotheses)? I suppose the most general possible question is: can $S$ be an arbitrary closed subset of $D$?</p>
<p>I know that $S$ need not be finite; for instance, the constant approximation of $\sin$ at $\pi/2$ has $S=\pi/2 + 2\pi \mathbb{Z}$. I also know that $S$ need not be discrete, at least if the domain is not compact; for instance, if the domain is $(0,1]$ and $f(x)=\sin(1/x)$ and we take the constant approximation at any point, then $S$ has a limit point at $0$.</p>
<p>Major edit: evidently $D \setminus S$ can at least be any open interval, because $f$ could be a bump function supported interval $I$ and take a point of expansion outside $I$, in which case $T_n$ will be zero and $T_n - f$ will be zero exactly outside $I$. Can we use this to prove that $S$ can be any closed set containing $x_0$?</p>
| zhw. | 228,045 | <p>I add this because I'm not sure if Robert Israel is giving the same answer. Let $E \subset \mathbb {R}$ be closed, with $0\in E.$ Then there exists $g\in C^\infty(\mathbb {R})$ such that $g=0$ on $E$ and $g>0$ on $\mathbb {R}\setminus E.$ Let $P$ be a polynomial of degree $n.$ Define</p>
<p>$$f(x) = e^{-1/x^2}g(x) + P(x).$$</p>
<p>Then $f\in C^\infty(\mathbb {R}),$ the $n$th order Taylor polynomial of $f$ at $0$ is $P,$ and the set where $f=P$ is precisely the set $E.$</p>
|
1,419,897 | <blockquote>
<p><strong>Theorem:</strong> Let $A$ be a bounded infinite subset of $\mathbb{R}^l$. Then
it has a limit point.</p>
</blockquote>
<p>So this is the Euclidean version of the Bolzano-Weierstrass theorem, the thing is that I was trying to prove it by induction, but it doesn't help because in the case $l=1$ we constructed a sequence of intervals, such that each iteration has infinite number of points of the set, but what is the anlogy in $\mathbb{R}^l$?, Can someone help me to prove this please?</p>
<p>Thanks a lot in advance :)</p>
| principal-ideal-domain | 131,887 | <p>The closure of $A$ is compact. In compact metric spaces each sequence has a convergent subsequence. If you may use that result you are done.</p>
<p><strong>Elaboration:</strong> Since $A$ is infinite there is a injective sequence $(a_n)_{n\in\mathbb N}\subseteq A$. Since $\overline{A}$ is compact $(a_n)_{n\in\mathbb N}$ has a convergent subsequence with limit in $\overline{A}$. Since the limit is the limit of an injective sequence of elements of $A$ it is a limit point of $A$.</p>
|
909,228 | <p>I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.</p>
<p>Could you help me with it?</p>
| Cleo | 97,378 | <p>$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}=\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right).$$</p>
|
997,602 | <blockquote>
<p>Prove that the function <span class="math-container">$x \mapsto \dfrac 1{1+ x^2}$</span> is uniformly continuous on <span class="math-container">$\mathbb{R}$</span>.</p>
</blockquote>
<p>Attempt: By definition a function <span class="math-container">$f: E →\Bbb R$</span> is uniformly continuous iff for every <span class="math-container">$ε > 0$</span>, there is a <span class="math-container">$δ > 0$</span> such that <span class="math-container">$|x-a| < δ$</span> and <span class="math-container">$x,a$</span> are elements of <span class="math-container">$E$</span> implies <span class="math-container">$|f(x) - f(a)| < ε.$</span></p>
<p>Then suppose <span class="math-container">$x, a$</span> are elements of <span class="math-container">$\Bbb R. $</span>
Now
<span class="math-container">\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}</span></p>
<p>I don't know how to simplify more. Can someone please help me finish? Thank very much.</p>
| Tom | 103,715 | <p>Hint: $$\frac{|x|}{(1+x^2)(1+a^2)} \leq \frac{|x|}{1+x^2} < 1$$
and
$$\frac{|a|}{(1+x^2)(1+a^2)} \leq \frac{|a|}{1+a^2} < 1$$</p>
|
86,800 | <p>I am curious about how the Heegaard genus changes after a finite covering. </p>
<p>Is there anyone constructing an closed hyperbolic 3-manifold $N$ such that </p>
<p>the Heegaard genus of a finite covering of $N$ is less than the Heegaard genus of $N$? </p>
<p>Thank you!</p>
<p>Note: Heegaard genus of a 3-manifold means the minimal genus of all Heegaard splittings.</p>
| Yo'av Rieck | 22,631 | <p>Hyam Rubinstein and me have results about the behavior of the Heegaard genus under double covers for non-Haken manifolds, see <a href="http://arxiv.org/abs/math/0607145">http://arxiv.org/abs/math/0607145</a>. Essentially, we show that the Heegaard genera of the two manifolds bound each other linearly. (The statement is a little more complicated for branched covers.)</p>
|
2,745,570 | <p>Use the mathematical Induction show that $H_{2^n}\le n+1$</p>
<p>here $H$ is harmonic numbers ie. $H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}$</p>
<p><strong>my idea</strong></p>
<p>so for $n=0$ L.H.S=R.H.S</p>
<p>Suppose this is true for $n$</p>
<p>we prove for $n+1$</p>
<p>So $H_{2^{n+1}}=1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{2^n}+\frac{1}{2^n+1}+....\frac{1}{2^{n+1}}\\
=H_{2^n}+\frac{1}{2^n+1}+.......\frac{1}{2^{n+1}}$</p>
<p>How to continue from here</p>
| user061703 | 515,578 | <p>We have already assumed that $$H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}\le n+1$$</p>
<p>We need to prove that:</p>
<p>$$H_{n+1}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n}+\frac{1}{2^n+1}+...+\frac{1}{2^n+2^n}\le n+2$$</p>
<p>This is true because $$\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n}<\frac{1}{2^n+1}+\frac{1}{2^n+1}+...+\frac{1}{2^n+1}\text{(there are $2^n$ numbers)}$$</p>
<p>$$\Rightarrow \frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n}<\frac{2^n}{2^n+1}<1$$</p>
<p>Combined with: $$H_n=1+\frac{1}{2}+\frac{1}{3}+.....\frac{1}{2^n}\le n+1$$ to finish the proof.</p>
|
801,081 | <p>I was doing some school work and got bored so I started messing with k-gonal numbers. I started with the triangular numbers, square numbers and looked for patterns. I noticed something.</p>
<p>Let $n^{(k)}$ denote the $n$-th $k$-gonal number. For example, $3^{(3)}$ is the third triangular number, 6.</p>
<p>I found that there was an easy way to compute the formula for each $k$-gonal and noticed that the
$$n^{(k)}=n^{(k-1)}+(n-1)^{(3)}$$</p>
<p>So to find the formula for the $n$-th pentagonal number,
$$n^{(5)}=n^{(4)}+(n-1)^{(3)}$$
$$n^{(5)}=n^2+\frac{n(n-1)}{2}$$
$$n^{(5)}=\frac{n(3n-1)}{2}$$</p>
<p>So after doing this a bunch of times, I think I found the pattern...</p>
<p>Let $f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ such that
$$f(n,k)=\frac{n[(k-2)n-(k-4)]}{2}$$
Is this the formula for the $n$-th $k$-gonal number? Are there any other intersting formulas that come out of the 2 x 2 array of these numbers like the one I derived above?</p>
| Janaka Rodrigo | 1,043,137 | <p>Let <span class="math-container">$n$</span> th <span class="math-container">$r$</span>-gonal number be <span class="math-container">$u(r,n)$</span></p>
<p>By the patterns of terms up
to heptagonal numbers it can be observed that,
<span class="math-container">$$u(r,n) = u(r-1,n) + u(3,n-1)$$</span></p>
<p>That is, <span class="math-container">$$u(r,n) - u(r-1,n) = u(3,n-1)$$</span></p>
<p><span class="math-container">$r=4$</span> gives <span class="math-container">$u(4,n) - u(3,n) = u(3,n-1)$</span></p>
<p><span class="math-container">$r=5$</span> gives <span class="math-container">$u(5,n) - u(4,n) = u(3,n-1)$</span></p>
<p>and</p>
<p><span class="math-container">$r=k$</span> gives <span class="math-container">$u(k,n) - u(k-1,n) = u(3,n-1)$</span>.</p>
<p>By adding these <span class="math-container">$k-3$</span> equations,</p>
<p><span class="math-container">\begin{align}
u(k,n)- u(3,n) &= (k-3)u(3,n-1)\\
u(k,n) &= u(3,n) + (k-3)u(3,n-1)\\
u(k,n) &= \frac{n(n+1)}2 + (k-3)\frac{(n-1)n}2\\
u(k,n) &= \frac{n}2\cdot ( k(n-1) -2n +4 )
\end{align}</span></p>
|
644,163 | <p>The question asks:
Find the line through $(3,1,-2)$ that intersects and is perpendicular to</p>
<p>$$x = -1 + t, y = -2 + t, z = -1 + t.$$</p>
<p>My thoughts:
Say the point of intersection is $(x_0,y_0,z_0)$, then my line can be of the form</p>
<p>$$L(s) = (3,1,-2) + (x_0- 3,y_0- 1,z_0+ 2)s$$</p>
<p>Then I tried setting up a system of equations at the intersection like this:</p>
<p>$$-1 + t = 3 + (x_0- 3)s$$</p>
<p>$$-2 + t = 1 + (y_0- 1)s$$</p>
<p>$$-1 + t = -2 + (z_0+ 2)s$$</p>
<p>And tried finding the point $(x_0,y_0,z_0)$, but I feel that I'm not on the right track. Could someone explain to me how to appropriately tackle this problem?</p>
| David Park | 99,469 | <p>These kind of problems are especially appropriate to Grassmann algebra so, if I may, I would like to show this approach using the Mathematica code of John Browne. First, some nomenclature:</p>
<p><img src="https://i.stack.imgur.com/TqphR.png" alt="enter image description here"></p>
<p>Load the package, define a 3-dimensional space with an origin, define t as a scalar.</p>
<pre><code><< GrassmannCalculus`
SetPreferences["Grassmann3Space", "Vector"]
DeclareExtraScalarSymbols[t]
</code></pre>
<p>We will use the idea that when the vector from the fixed point to the line is a minimum length, the vector must be perpendicular to the line.</p>
<p>Define the fixed point, a parameterized line, and a parameterized vector from the fixed point to a point on the line. </p>
<p><img src="https://i.stack.imgur.com/dj0uc.png" alt="enter image description here"></p>
<p>Giving for the point on the line and the vector from the fixed point to the line point:</p>
<p><img src="https://i.stack.imgur.com/85IIl.png" alt="enter image description here"> </p>
<p>We calculate the length squared of the vector from the point to the line as a function of t and minimize by solving for when the derivative is zero. </p>
<p><img src="https://i.stack.imgur.com/TdAbP.png" alt="enter image description here"></p>
<p>Giving:</p>
<p><img src="https://i.stack.imgur.com/wGF3n.png" alt="enter image description here"> </p>
<p>The resulting vector from the fixed point to the line is:</p>
<p><img src="https://i.stack.imgur.com/ehrjS.png" alt="enter image description here"> </p>
<p>If the answer is correct, the scalar product of the vector12 and the direction of the line should be zero. </p>
<p><img src="https://i.stack.imgur.com/JimGm.png" alt="enter image description here"></p>
|
4,442,223 | <p>How does one show this?
<span class="math-container">$$
\exp(-x) \sum_{k=0}^\infty x^k \frac{(k+m)!}{(k!)^2} = L_m(-x) m!,
$$</span> where <span class="math-container">$m$</span> is a positive integer, and <span class="math-container">$L_{m}(x)$</span> is the <span class="math-container">$m$</span>th order Laguerre polynomial.</p>
| Lee Mosher | 26,501 | <p>What you heard is correct: for any simplicial complex <span class="math-container">$X$</span> endowed with the simplicial topology, if we let <span class="math-container">$X^{(0)}$</span> denote its set of vertices endowed with the subspace topology, then <span class="math-container">$X^{(0)}$</span> is indeed a discrete topological space. This is an almost immediate consequence of the definition of the simplicial topology.</p>
<p>Popular accounts of simplicial complexes, such as the one in that wolfram link you gave in a comment, might not contain any formal definition of the simplicial topology. You can find a description of the simplicial topology at this <a href="https://en.wikipedia.org/wiki/Abstract_simplicial_complex#Geometric_realization" rel="nofollow noreferrer">wikipedia link</a>, although perhaps a good textbook description would be better.</p>
<p>Proving that <span class="math-container">$X^{(0)}$</span> is discrete should be a straightforward exercise once one understand that definition. Here's a brief outline. In the simplicial topology on <span class="math-container">$X$</span>, a subset of <span class="math-container">$X$</span> is open if and only it its intersection with each individual simplex is an open subset of that simplex. Using this, it's not hard to directly construct, for each <span class="math-container">$x \in X^{(0)}$</span>, an open subset <span class="math-container">$U_x \subset X$</span> such that <span class="math-container">$X^{(0)} \cap U_x = \{x\}$</span>. It follows that <span class="math-container">$\{x\}$</span> is an open subset of the subspace topology on <span class="math-container">$X^{(0)}$</span>.</p>
|
1,690,715 | <p>I have this space $E=\mathcal{C}([0,1],\mathbb{R})$ and the inner product $d(f,g)=\int_0^1 |f(x)-g(x)|\,{\rm d}x$.</p>
<p>Who have an idea about a simple sequence $\{f_n\}_{n=1}^\infty$ which is Cauchy but not convergent in $(E,d)$?</p>
| nullUser | 17,459 | <p>Take $f= 1_{[1/2,1]}$. Approximate it by $f_n$ which agrees with $f$ outside of $[1/2-1/n,1/2+1/n]$ and interpolates linearly inbetween. By dominated convergence $f_n \to f \in L^1([0,1])$ and hence $f_n$ is a Cauchy sequence in the $d$-metric. Now let $g \in \mathcal{C}([0,1],\mathbb{R})$ and assume for contradiction that $d(f_n,g) \to 0$. Since $g \in L^1$ and limits in $L^1$ are unique, we find $g=f$ almost everywhere. Since $f,g$ are both right continuous we find $g=f$, and thus $g \notin \mathcal{C}([0,1],\mathbb{R})$, a contradiction.</p>
|
3,631,648 | <p>Suppose <span class="math-container">$X_1, ..., X_n \stackrel{iid}{\sim}$</span> Exponential(rate = <span class="math-container">$\lambda$</span>) independent of <span class="math-container">$Y_1, ..., Y_n \stackrel{iid}{\sim}$</span> Exponential<span class="math-container">$(1)$</span>. </p>
<p>Define <span class="math-container">$Z_i \equiv \min\{X_i, Y_i\}$</span></p>
<p>I want to find the maximum likelihood estimator for <span class="math-container">$\lambda$</span> in the following scenario: I observe <span class="math-container">$Z_1, ..., Z_n$</span> and <span class="math-container">$Y_1, ..., Y_n$</span> but NOT any of the <span class="math-container">$X_i$</span>.</p>
<p>First I need to determine the likelihood and then maximize it over <span class="math-container">$\theta > 0$</span>, but I'm not really sure of the right approach. I calculate the joint cdf as follows:</p>
<p><span class="math-container">$$P(Z_i \leq z, Y_i \leq y) = \begin{cases} P(Y_i \leq y), & y \leq z \\ P(Y_i \leq z, Y_i \leq X_i) + P(Y_i \leq y, X_i \leq z, X_i < Y_i), & y > z\end{cases} \\
= \begin{cases} 1- e^{-y}, & y \leq z \\
1-e^{-z} + (e^{-z}-e^{-y})(1-e^{-\lambda z}), & y > z \end{cases}$$</span></p>
<p>This is because <span class="math-container">$Z_i \leq Y_i$</span> always. Would the likelihood function therefore be:</p>
<p><span class="math-container">$$L(\lambda |Y_i, Z_i, i \in \{1,...n\}) = \prod_{\{i : Y_i = Z_i\}} (1-e^{-Y_i}) \prod_{\{i:Y_i > Z_i\}} \lambda e^{-Y_i}e^{-\lambda Z_i}$$</span></p>
<p>splitting into the "discrete" and "continuous" parts? Or am I getting this wrong? Or should I be doing something like <a href="https://math.stackexchange.com/questions/3512290/exam-prep-maximum-likelihood-estimator">here</a> or <a href="https://math.stackexchange.com/questions/3156467/how-to-find-the-mle-of-these-parameters-given-distribution?noredirect=1&lq=1">here</a>? I should note my scenario is different than theirs, as intuitively at least, observing the magnitude of the difference between the minimum and the maximum (in the cases where <span class="math-container">$Z_i$</span> and <span class="math-container">$Y_i$</span> differ) should give us more information about <span class="math-container">$\lambda$</span>, right?</p>
| heropup | 118,193 | <p>If you observe both <span class="math-container">$Z_i$</span> and <span class="math-container">$Y_i$</span>, then when they are equal, you know <span class="math-container">$X_i > Y_i$</span>. When they are not, you know <span class="math-container">$X_i = Z_i$</span>. Therefore, your likelihood function is <span class="math-container">$$\begin{align*}\mathcal L(\lambda \mid \boldsymbol z, \boldsymbol y) &= \prod_{i=1}^n \left(f_X(z_i) \mathbb 1 (z_i \ne y_i) + (1 - F_X(y_i)) \mathbb 1 (z_i = y_i) \right) \\
&= \prod_{i=1}^n \left(\lambda e^{-\lambda z_i} \mathbb 1 (z_i \ne y_i) + e^{-\lambda y_i} \mathbb 1 (z_i = y_i) \right) \\
&= \lambda^{\sum_{i=1}^n \mathbb 1(z_i \ne y_i)} \prod_{i=1}^n e^{-\lambda z_i} \\
&= \lambda^{\sum_{i=1}^n \mathbb 1(z_i \ne y_1)} e^{-\lambda n \bar z}. \end{align*}$$</span> Notice here that the density and survival functions we choose are for <span class="math-container">$X$</span>, not on <span class="math-container">$Y$</span> or <span class="math-container">$Z$</span>! Then the log-likelihood is <span class="math-container">$$\ell (\lambda \mid \boldsymbol z, \boldsymbol y) = ( \log \lambda ) \sum_{i=1}^n \mathbb 1 (z_i \ne y_i) - \lambda n \bar z,$$</span> and we solve for the extremum as usual, giving <span class="math-container">$$\hat \lambda = \frac{\sum_{i=1}^n \mathbb 1(z_i \ne y_i)}{n \bar z},$$</span> where the numerator counts the number of paired observations that are not equal, and the denominator is the sample total of <span class="math-container">$z$</span>.</p>
<p>Simulation of this is straightforward and I invite you to try it out to confirm the estimator works. Here is code in <em>Mathematica</em> to perform the estimation based on a sample of size <span class="math-container">$n$</span> and any <span class="math-container">$\lambda = t$</span>:</p>
<pre><code>F[n_, t_] := RandomVariate[TransformedDistribution[{Min[x, y], y},
{Distributed[x, ExponentialDistribution[t]],
Distributed[y, ExponentialDistribution[1]]}], n]
T[d_] := Length[Select[d, #[[1]] != #[[2]] &]]/Total[First /@ d]
T[F[10^6, Pi]]
</code></pre>
<p>The last expression evaluates <span class="math-container">$\hat \lambda$</span> for <span class="math-container">$n = 10^6$</span> and <span class="math-container">$\lambda = \pi$</span>. I got <span class="math-container">$3.14452$</span> when I ran it.</p>
|
172,292 | <p>I am trying to find the residue of the function $$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.
I tried the following in Mathematica</p>
<pre><code>Residue[(z+1)^2*Exp[3/z^2],{z,0}]
</code></pre>
<p>which remains unevaluated. Computing this by hand gives the value of $6$. What is going on?</p>
<p>I’ve noticed that Mathematica has a problem with the Laurent series of $e^{3/z^2}$ at $z=0$.</p>
| Carl Woll | 45,431 | <p>You could use <a href="http://reference.wolfram.com/language/ref/SeriesCoefficient" rel="nofollow noreferrer"><code>SeriesCoefficient</code></a> instead:</p>
<pre><code>SeriesCoefficient[(z+1)^2 Exp[3/z^2], {z, 0, -1}]
</code></pre>
<blockquote>
<p>6</p>
</blockquote>
<p><strong>Addendum</strong></p>
<p>Another possibility is to note that the residue at 0 and the residue at infinity must sum to zero, since they are the only singularities of the function. Hence we can do:</p>
<pre><code>- Residue[(z + 1)^2 Exp[3/z^2], {z, Infinity}]
</code></pre>
<blockquote>
<p>6</p>
</blockquote>
<p>which is the same answer as before.</p>
|
208,802 | <p>Is there a continuous increasing function $ f : [0, \pi] \to [0, e] $ such that $ f(0) = 0, f(\pi) = e $ and $ f (q ) \in \mathbb{Q} $ for $ q \in \mathbb{Q} $ and $ f (q ) \in \mathbb{Q}^c $ for $ q \in \mathbb{Q}^c $? I think there should be, but I am unable to construct one. </p>
| Hagen von Eitzen | 39,174 | <p>Find a suitable strictly ascending sequence $(c_n)_n$ of rationals and define $f(c_1)=0$ and recursively for $x\in[c_{n-1},c_n]$ by $f(c_n)=f(c_{n-1})+\frac12 (x-c_{n-1})$ if $n$ is even and $f(c_n)=f(c_{n-1})+c_n-c_{n-1}$ if $n$ is odd.
Let $d_n=c_{n+1}-c_n$.
In order to make this an example you are looking for, we have to make sure</p>
<ul>
<li>that $c_1=0$ and $c_n\to \pi$ (so that $f$ is defined on all of $[0,\pi]$)</li>
<li>that $\sum_n d_n = \pi$ (which implies $c_n\to c_0+\pi)$</li>
<li>that $\sum_{n\text{ even}}d_n+\frac12\sum_{n\text{ odd}}d_n$ = e</li>
</ul>
<p>This can be accomplished by selecting positive rational numbers $d_{2n-1}$ such that their sum converges to $2(\pi-e)$ and positive rational numbers $d_{2n}$ such that theri sum converges to $2e-\pi$.</p>
<p>That $f$ maps rational to rational and irrational to irrational follows from the fact that each segemnt is of the form $f(x)=ax+b$ with positive rational $a$ and rational $b$.</p>
|
1,144,695 | <p>I'm currently trying to solve this problem. </p>
<blockquote>
<p>Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.</p>
</blockquote>
<p>I solved the ideal part. </p>
<p>Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$. </p>
<p>Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$. </p>
<p>Therefore $f(P)$ is an ideal. </p>
<p>The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime. </p>
<p>I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.</p>
<p>Any help would be appreciated.</p>
<p>Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.</p>
| MooS | 211,913 | <p>You should show $R/P \cong S/f(P)$. To that extend, consider the composition $$R \to S \to S/f(P).$$</p>
|
1,144,695 | <p>I'm currently trying to solve this problem. </p>
<blockquote>
<p>Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.</p>
</blockquote>
<p>I solved the ideal part. </p>
<p>Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$. </p>
<p>Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$. </p>
<p>Therefore $f(P)$ is an ideal. </p>
<p>The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime. </p>
<p>I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.</p>
<p>Any help would be appreciated.</p>
<p>Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.</p>
| Slade | 33,433 | <p>$f:R\to S$ induces an isomorphism $\overline{f}:R/K \to S$. So it is enough to show that $P/K$ is a (completely) prime ideal of $R/K$.</p>
<p>Every ideal of $R/K$ can be written uniquely in the form $I/K$ for some ideal $I\supset K$ of $R$ (take the preimage under the projection $R\to R/K$). But if $A/K\cdot B/K \subset P/K$, then $AB \subset P+K = P$, and the conclusion follows shortly thereafter.</p>
|
1,722,964 | <p>Expression :$$(p\rightarrow q)\leftrightarrow(\neg q\rightarrow \neg p)$$
What does the symbol $\leftrightarrow$ mean ? Please explain by drawing the truth table for this expression and also with other examples if possible. <strong>I'm in a desperate situation so I'd really appreciate a quick response !</strong></p>
| Matthias | 164,923 | <p>$$A\leftrightarrow B$$ is the same as </p>
<p>$$(A\rightarrow B) \land (B\rightarrow A)$$</p>
|
2,516,123 | <p>Problem 11985, by Donald Knuth, <em>American Mathematical Monthly</em>, June-July, 2017:</p>
<blockquote>
<p>For fixed $s,t \in \mathbb{N}$. with $s\leq t$. let $a_{n}=\sum\limits_{k=s}^{t}$ $ {n}\choose{k}$. Prove that this sequence is log-concave, namely that $a_{n}^{2}\geq a_{n-1}a_{n+1} \ \forall n\geq 1$. </p>
</blockquote>
<p>The submission deadline for this problem was over on 31st October.
Does this statement follow from some well known results?</p>
| Dap | 467,147 | <p>This follows from the log-concavity of binomial coefficients. Using the identity $\binom nk=\binom{n-1}{k-1}+\binom{n-1}{k}$ we can express the desired inequality $a_n^2\geq a_{n-1}a_{n+1}$ in terms of binomial coefficients of $n-1:$ we need to show</p>
<p>$$\sum_{i=s}^t\sum_{j=s-2}^{t-2}\binom{n-1}{i}\binom{n-1}{j}\leq \sum_{i=s-1}^{t-1}\sum_{j=s-1}^{t-1}\binom{n-1}{i}\binom{n-1}{j}.$$</p>
<p>The only terms that don't cancel here are those with $i=t$ or $j=s-2$ on the left-hand-side, and the terms with $i=s-1$ or $j=t-1$ on the right-hand-side. For these we can use
$$\binom{n-1}{t}\binom{n-1}{j}\leq \binom{n-1}{j+1}\binom{n-1}{t-1}\qquad(j\leq t-2)$$
$$\binom{n-1}{i}\binom{n-1}{s-2}\leq \binom{n-1}{s-1}\binom{n-1}{i-1}\qquad(i\geq s)$$
which are essentially the log-concavity of $\binom{n-1}{k}$ as $k$ varies i.e. $\binom{n-1}{k}/\binom{n-1}{k-1}=\frac{n-k}k$ is non-increasing in $k.$</p>
|
2,516,123 | <p>Problem 11985, by Donald Knuth, <em>American Mathematical Monthly</em>, June-July, 2017:</p>
<blockquote>
<p>For fixed $s,t \in \mathbb{N}$. with $s\leq t$. let $a_{n}=\sum\limits_{k=s}^{t}$ $ {n}\choose{k}$. Prove that this sequence is log-concave, namely that $a_{n}^{2}\geq a_{n-1}a_{n+1} \ \forall n\geq 1$. </p>
</blockquote>
<p>The submission deadline for this problem was over on 31st October.
Does this statement follow from some well known results?</p>
| Sil | 290,240 | <p>Solution by Roberto Tauraso <a href="http://www.mat.uniroma2.it/~tauraso/AMM/AMM11985.pdf" rel="nofollow noreferrer">http://www.mat.uniroma2.it/~tauraso/AMM/AMM11985.pdf</a> (who by the way has solutions to many of AMM's problems):</p>
<blockquote>
<p>Let $$F_n(x):=\sum_{k=s}^{t}\binom{n}{k}x^k.$$</p>
<p>Then $$F_n(x)=\sum_{k=s}^{t}\left(\binom{n-1}{k-1}+\binom{n-1}{k}\right)x^k=\binom{n-1}{s-1}x^s-\binom{n-1}{t}x^{t+1}+(x+1)F_{n-1}(x).$$</p>
<p>Let $P_n(x):=F^2_n(x)-F_{n-1}(x)F_{n+1}(x).$ Then</p>
<p>\begin{align}
P_n(x)&=F_n(x)\left(\binom{n-1}{s-1}x^s-\binom{n-1}{t}x^{t+1}+(x+1)F_{n-1}
(x)\right)\\
&\ -F_{n-1}(x)\left(\binom{n}{s-1}x^s-\binom{n}{t}x^{t+1}+(x+1)F_{n}
(x)\right)\\
&=\left(\binom{n-1}{s-1}F_n(x)-\binom{n}{s-1}F_{n-1}(x)\right)x^s+\left(\binom{n}{t}F_{n-1}(x)-\binom{n-1}{t}F_{n}(x)\right)x^{t+1}\\
&= \sum_{k=s}^{t}\left(\binom{n-1}{s-1}\binom{n}{k}-\binom{n}{s-1}\binom{n-1}{k}\right)x^{k+s}+\sum_{k=s}^{t}\left(\binom{n}{t}\binom{n-1}{k}-\binom{n-1}{t}\binom{n}{k}\right)x^{k+t+1}.
\end{align}</p>
<p>Since $s \leq k \leq t$, it is easy to verify that</p>
<p>$$
\binom{n-1}{s-1}\binom{n}{k} \geq \binom{n}{s-1}\binom{n-1}{k}\ \ \mbox{ and }\ \ \ \binom{n}{t}\binom{n-1}{k} \geq \binom{n-1}{t}\binom{n}{k}.
$$</p>
<p>Hence the polynomial $P_n$ has non-negative coefficients which implies that</p>
<p>$$
P_n(1)=F^2_n(1)-F_{n-1}(1)F_{n+1}(1)=a^2_n-a_{n-1}a_{n+1} \geq 0
$$</p>
<p>and the sequence $(a_n)_n$ is log-concave.</p>
</blockquote>
|
3,611,072 | <p>Show that if a prime <span class="math-container">$p ≠ 3$</span> is such that <span class="math-container">$p≡1$</span> (mod 3) then p can be written as <span class="math-container">$a^2-ab+b^2$</span> where a and b are integers. </p>
<p>I have no idea how to approach this question, so any help much appreciated! This is in the context of algebraic number theory, so I'm not sure if it's helpful to consider rings of integers or anything like that. </p>
| Piquito | 219,998 | <p>COMMENT.- I fear it is a problem not too elementary. It can be solved using the theory of representation of integers by quadratic forms. In short, consider the discriminant of the form <span class="math-container">$x^2-xy+y^2$</span> which is equal to <span class="math-container">$\Delta=-3$</span>. </p>
<p>One can use the following result: The integer <span class="math-container">$n$</span> is represented by a quadratic form of discriminant <span class="math-container">$\Delta$</span> if and only if <span class="math-container">$\Delta$</span> is a square modulo <span class="math-container">$4n$</span>.</p>
<p>Here four examples.</p>
<p>►For <span class="math-container">$p=7$</span> one has <span class="math-container">$x^2=-3\pmod{28}$</span> since <span class="math-container">$-3=25$</span>, obvio.</p>
<p>►For <span class="math-container">$p=13$</span> one has <span class="math-container">$-3$</span> is a square modulo <span class="math-container">$52$</span> because <span class="math-container">$-3=49$</span> obvio.</p>
<p>►For <span class="math-container">$p=19$</span> one has <span class="math-container">$-3=x^2\pmod{76}$</span> has, for example, the solution <span class="math-container">$x=15$</span>(there are other solutions).</p>
<p>►For <span class="math-container">$p=31$</span> the equation <span class="math-container">$-3=x^2\pmod{124}$</span> has, for example,the solution <span class="math-container">$x=11$</span>. </p>
|
2,710,681 | <p>If I have a function of three variables and I want to create a new function in which it equals the other function squared, could I literally just square the other function or does this violate any rules? Would this also mean its gradient vector is just squared at a certain point?</p>
| gt6989b | 16,192 | <p><strong>HINT</strong></p>
<p>Say you have $f(x,y,z)$ and you would like to define $$g(x,y,z) = f(x,y,z)^2.$$ Then,
$$
\vec{\nabla} g =
\begin{pmatrix}
\partial g/\partial x \\
\partial g/\partial y \\
\partial g/\partial z
\end{pmatrix}
$$
For example, chain rule implies
$$
\frac{\partial g(x,y,z)}{\partial x}
= \frac{\partial f(x,y,z)^2}{\partial x}
= 2 f(x,y,z) \frac{\partial f(x,y,z)}{\partial x}.
$$</p>
|
76,683 | <p>How do I force mathematica to display the below expression as a sum <code>a+b</code> with a scaling factor of <code>1/r</code>.</p>
<p>(a+b)/r</p>
<p>I would like Mathematica to display (1/r) (a+b), ie. I want it to show 1/r as a scaling factor. </p>
<p>currently, it shows (a+b)/r , with r as a common denominator. </p>
| Nasser | 70 | <pre><code>expr = (a + b)/r
1/Denominator[expr]
</code></pre>
<p><img src="https://i.stack.imgur.com/LNEqO.png" alt="Mathematica graphics"></p>
|
1,621,269 | <p>I have tried everything in my knowledge and no, I cannot state it.
I have tried a factorizor online which tells me that it is not factorizable hence irreducible. But I cannot reason why.</p>
<p>I looked at Eisenstein's criteria but obviously, there is no prime $q$ that fits the criteria so this is useless.</p>
<p>I then tried reducibility via modulo reduction, and this should give me the options to test irreducibility up to mod $8$ since that is the largest coefficient in the polynomial...yes? But every mod arrives at the polynomial being reducible...so it basically fails to tell me that it is irreducible.
For mod 2, I get 0 as a solution to the reduced polynomial so that means I can factor it out with $x$.
Similarly, mod 3 says 1 is a solution so $(x-1)$ should be a solution.
In a similar fashion, I get mod 4,5,6,7 to have solutions 0,3,4,6.</p>
<p>Am I missing anything? This is the best I can do, nothing more assures me irreducibility at all. Ideas please...?</p>
| lulu | 252,071 | <p>Let $\phi(x)$ denote your polynomial. Then we note that $$\phi(x+1)=x^5+3x^2+9x+3$$ and we can invoke Eisenstein's criterion.</p>
|
1,621,269 | <p>I have tried everything in my knowledge and no, I cannot state it.
I have tried a factorizor online which tells me that it is not factorizable hence irreducible. But I cannot reason why.</p>
<p>I looked at Eisenstein's criteria but obviously, there is no prime $q$ that fits the criteria so this is useless.</p>
<p>I then tried reducibility via modulo reduction, and this should give me the options to test irreducibility up to mod $8$ since that is the largest coefficient in the polynomial...yes? But every mod arrives at the polynomial being reducible...so it basically fails to tell me that it is irreducible.
For mod 2, I get 0 as a solution to the reduced polynomial so that means I can factor it out with $x$.
Similarly, mod 3 says 1 is a solution so $(x-1)$ should be a solution.
In a similar fashion, I get mod 4,5,6,7 to have solutions 0,3,4,6.</p>
<p>Am I missing anything? This is the best I can do, nothing more assures me irreducibility at all. Ideas please...?</p>
| Travis Willse | 155,629 | <p>One option is to reduce the given polynomial modulo $11$, in which case it factors (over $\Bbb F_{11}$) as
$$(x - 5)(x^4 - x^2 - x - 3).$$
So, if the polynomial is reducible over $\Bbb Q$, it has one linear factor and one irreducible quartic factor there.</p>
<p>On the other hand, checking the short list, $\pm 1, \pm 2, \pm 4$, of candidates given by the Rational Root Test shows that the polynomial has no rational roots and hence no linear factors. (In fact, since the signs of the polynomial are alternating, all of its real roots are positive, so we need only check $+1, +2, +4$.) Thus, the polynomial is irreducible.</p>
<p>(Alternatively, the given polynomial is irreducible altogether modulo $17$, but this is surely even less pleasant to check by hand than the irreducibility modulo $11$ of the above quartic.)</p>
|
476,899 | <p>Does someone know a proof that $\{1,e,e^2,e^3\}$ is linearly independent over $\mathbb{Q}$?</p>
<p>The proof should not use that $e$ is transcendental.</p>
<p>$e:$ Euler's number.</p>
<p><a href="http://paramanands.blogspot.com/2013/03/proof-that-e-is-not-a-quadratic-irrationality.html#.Uhv87tJFUnl">$\{1,e,e^2\}$ is linearly independent over $\mathbb{Q}$</a></p>
<p>Any hints would be appreciated.</p>
| Paramanand Singh | 72,031 | <p>I thought to add an answer instead of giving long comments.</p>
<p>From <a href="http://en.wikipedia.org/wiki/Proof_that_e_is_irrational" rel="nofollow noreferrer">Wikipedia</a> we have the following quote</p>
<p>"In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that $e$ is not a root of a third degree polynomial with rational coefficients. In particular, $e^{3}$ is irrational."</p>
<p>The reference quoted is Hurwitz, Adolf (1933) [1891]. "Über die Kettenbruchentwicklung der Zahl $e$".</p>
<p>Luckily after much searching I was able to find this reference in <a href="https://archive.org/download/schriftenderphys3132phys/schriftenderphys3132phys.djvu" rel="nofollow noreferrer">an old journal available on internet archive</a>. Here Hurwitz analyzes the simple continued fractions of numbers related with $e$ and notices that most of them have terms in an arithmetic progression (after a certain point).</p>
<p>He then proves the following theorem:</p>
<p><em>If simple continued fractions of two positive numbers $x, y$ have terms which are in arithmetic progression (after a certain point) then we can't have a non-trivial bi-linear relation of the form $$y = \frac{Ax + B}{Cx + D}$$ with integer coefficients $A, B, C, D$ unless the terms in their continued fraction belong to the same arithmetic progression.</em></p>
<p>Then Hurwitz notes that $$x = \frac{e - 1}{2} = [0, 1, 6, 10, 14, 18,\ldots]$$ and $$y = \frac{e^{2} - 1}{2} = [3, 5, 7, 9, \ldots]$$ where notation $$[a_{0}, a_{1}, a_{2}, \ldots]$$ represents the continued fraction $$a_{0} + \dfrac{1}{a_{1} + \dfrac{1}{a_{2} + \dfrac{1}{a_{3} + \cdots}}}$$ And clearly both of them have terms belonging to arithmetic progressions ($6, 10, 14, \ldots$ and $3, 5, 7, 9, \ldots$ respectively) but these are not the terms belonging to same AP and hence there is no non-trivial bi-linear relation of type $$y = \frac{Ax + B}{Cx + D}$$ with integer coefficients $A, B, C, D$.</p>
<p>Now it follows easily that $1, e, e^{2}, e^{3}$ are linearly independent over $\mathbb{Q}$. If it was not the case then we have integers $a, b, c, d$ not all $0$ such that $$ae^{3} + be^{2} + ce + d = 0$$ Using $e^{2} = 2y + 1$ and $e = 2x + 1$ we get $$a(2x + 1)(2y + 1) + b(2y + 1) + c(2x + 1) + d = 0$$ which leads to $$Axy + Bx + Cy + D = 0$$ with $A, B, C, D$ as integers or $$y = -\frac{Bx + D}{Ax + C}$$ and this is not allowed by the theorem of Hurwitz mentioned above.</p>
<p>Unfortunately I could not understand the proof of his theorem on continued fractions (because the whole paper/journal is in German). With reasonable effort and Google Translate I was able to understand the gist of the paper and I have presented the same in this answer. I have <a href="https://math.stackexchange.com/questions/1391869/bi-linear-relation-between-two-continued-fractions">asked for the proof of Huzwitz theorem on MSE</a>.</p>
|
4,064,760 | <p><a href="https://i.stack.imgur.com/tDP8G.png" rel="nofollow noreferrer">image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x</a></p>
<p>I solved this <span class="math-container">$y'' - 4y' + 5y = e^{-x}$</span> equation with the guess of:</p>
<p><span class="math-container">$Ae^{-x}$</span>, such that A ended up being <span class="math-container">$A = 1/2$</span>;</p>
<p>The solution given by math calculator websites was the one of the image —</p>
<p><span class="math-container">$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$</span></p>
<p>but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?</p>
| 19aksh | 668,124 | <p>The auxiliary equation of the given ODE is,</p>
<p><span class="math-container">$m^2-4m+5 = 0 \Rightarrow (m-2)^2 +1 = 0 \Rightarrow \boxed{m = 2 \pm i}$</span></p>
<p>So the solution (complementary function) will be,</p>
<p><span class="math-container">$$y_{CF}(x) = k_1 e^{(2 + i)x} + k_2 e^{(2 - i)x} =e^{2x}(k_1 e^{ix} + k_2 e^{-ix}) \equiv \boxed{e^{2x}(C_1 \sin x + C_2\cos x)}$$</span></p>
<p>Now you have to add this with the particular solution to get the complete solution.</p>
|
4,064,760 | <p><a href="https://i.stack.imgur.com/tDP8G.png" rel="nofollow noreferrer">image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x</a></p>
<p>I solved this <span class="math-container">$y'' - 4y' + 5y = e^{-x}$</span> equation with the guess of:</p>
<p><span class="math-container">$Ae^{-x}$</span>, such that A ended up being <span class="math-container">$A = 1/2$</span>;</p>
<p>The solution given by math calculator websites was the one of the image —</p>
<p><span class="math-container">$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$</span></p>
<p>but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?</p>
| Henry Lee | 541,220 | <p>okay lets solve it, first:
<span class="math-container">$$y''-4y'+5y=0$$</span>
lets make an educated guess that the solutions will be of the form:
<span class="math-container">$$y=Ae^{\lambda x}$$</span>
now sub in:
<span class="math-container">$$Ae^{\lambda x}(\lambda^2-4\lambda+5)=0$$</span>
solving for <span class="math-container">$\lambda$</span> the non-trivial solutions are going to be in the quadratic:
<span class="math-container">$$\lambda^2-4\lambda+5=0$$</span>
<span class="math-container">$$\lambda=2\pm i$$</span>
now we add these two solutions together:
<span class="math-container">$$y=Ae^{(2+i)x}+Be^{(2-i)x}=e^{2x}(Ae^{ix}+Be^{-ix})$$</span>
now notice that this will just form sines and cosines so lets redefine our constants and we arrive at:
<span class="math-container">$$y=e^{2x}\left(C_1\sin x+C_2 \cos x\right)$$</span>
This part forms the "complementary function" so now you just need to find the particular solution. Hope this helps</p>
|
4,064,760 | <p><a href="https://i.stack.imgur.com/tDP8G.png" rel="nofollow noreferrer">image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x</a></p>
<p>I solved this <span class="math-container">$y'' - 4y' + 5y = e^{-x}$</span> equation with the guess of:</p>
<p><span class="math-container">$Ae^{-x}$</span>, such that A ended up being <span class="math-container">$A = 1/2$</span>;</p>
<p>The solution given by math calculator websites was the one of the image —</p>
<p><span class="math-container">$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$</span></p>
<p>but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?</p>
| Community | -1 | <p>Let us pretend that we know nothing about the linear ODE with constant coefficients nor complex exponentials.</p>
<p>We will try by factoring <span class="math-container">$y$</span> and get some simplification.</p>
<p><span class="math-container">$$y=zh,\\y'=z'h+zh',\\y''=z''h+2z'h'+zh''.$$</span></p>
<p>We plug this in the ODE:</p>
<p><span class="math-container">$$z''h+2z'(h'-2h)+z(h''-4h'+5h)=e^{-x}$$</span> and let the coefficient of <span class="math-container">$z'$</span> vanish.</p>
<p>From <span class="math-container">$h'-2h=0$</span>, which is separable,</p>
<p><span class="math-container">$$\frac{h'}h=(\log(h))'=2$$</span> and <span class="math-container">$h=e^{2x}$</span> is a solution.</p>
<p>Plugged in the ODE gain,</p>
<p><span class="math-container">$$z''e^{2x}+ze^{2x}=e^{-x}$$</span> or</p>
<p><span class="math-container">$$z''+z=e^{-3x}.$$</span></p>
<p>Now if we try a solution of the form <span class="math-container">$e^{-3x}$</span> by educated guess, we get that</p>
<p><span class="math-container">$$z''+z=10\,e^{-3x}$$</span> and it suffices to correct with a suitable factor.</p>
<p>So, using linearity of the derivative, we can move the RHS as</p>
<p><span class="math-container">$$z''-\frac9{10}e^{-3x}+z-\frac1{10}e^{-3x}=0$$</span> or</p>
<p><span class="math-container">$$\left(z-\frac1{10}e^{-3x}\right)''+\left(z-\frac1{10}e^{-3x}\right)=0$$</span></p>
<p>which is of the form <span class="math-container">$$w''+w=0.$$</span> We make it integrable by multiplying by <span class="math-container">$2w'$</span>:</p>
<p><span class="math-container">$$2w'w''+2ww'=(w'^2+w^2)'=0$$</span> and for some (positive) constant <span class="math-container">$a^2$</span>,</p>
<p><span class="math-container">$$w'^2=a^2-w^2$$</span> then</p>
<p><span class="math-container">$$\frac{w'}{\sqrt{a^2-w^2}}=\pm1.$$</span></p>
<p>After integration,</p>
<p><span class="math-container">$$\arccos\frac wa=\pm t+c$$</span></p>
<p>or</p>
<p><span class="math-container">$$w=a\cos(\pm t+c)=c_c\cos t+c_s\sin t.$$</span></p>
<p>Finally,</p>
<p><span class="math-container">$$y=ze^{2x}=\left(w+\frac{e^{-3x}}{10}\right)e^{2x}=\left(c_c\cos t+c_s\sin t+\frac{e^{-3x}}{10}\right)e^{2x}
\\=\left(c_c\cos t+c_s\sin t\right)e^{2x}+\frac{e^{-x}}{10}$$</span> as claimed.</p>
|
2,960,501 | <p><span class="math-container">$(0^n 1)^* \ \ , n\geq 0 $</span></p>
<p>According to wiki</p>
<blockquote>
<p>If V is a set of strings, then V* is defined as the smallest superset of V that contains the empty string ε and is closed under the string concatenation operation</p>
<p>If V is a set of symbols or characters, then V* is the set of all strings over symbols in V, including the empty string ε.</p>
</blockquote>
<p>So this language accepts all strings over <span class="math-container">$\Sigma^*$</span> which must be regular.
Also regular languages are closed under kleene star.</p>
<p>But again on wiki</p>
<blockquote>
<p><span class="math-container">$V^* = \bigcup\limits_{i\geq 0}^{} V_i = {\epsilon} \ \cup \ V_1 \ \cup V_2 \ \cup \ V_3 \ \cup .....$</span></p>
</blockquote>
<p>Now according to this definition strings such as <span class="math-container">$01001$</span> cannot be a part of given language so <span class="math-container">$0$</span>'s prior of every 1 are compared within a string, so this can't be regular.</p>
<p>But according to the former definition <span class="math-container">$01001$</span> is a part of language because it can be formed with symbols <span class="math-container">$01$</span> and <span class="math-container">$001$</span> both are part of <span class="math-container">$0^n 1$</span>.</p>
<p>Can someone help me in identifying the class of these types of languages</p>
| J.-E. Pin | 89,374 | <p>If I understand correctly (and no, your definition is neither clear nor correct since
<span class="math-container">$\{(0^n1)^* \mid n \geqslant 0\}$</span> does not make any sense), your language is
<span class="math-container">$\{0^n1 \mid n \geqslant 0\}^*$</span>, which can be rewritten as <span class="math-container">$(0^*1)^*$</span>, which is indeed a regular language.</p>
|
2,960,501 | <p><span class="math-container">$(0^n 1)^* \ \ , n\geq 0 $</span></p>
<p>According to wiki</p>
<blockquote>
<p>If V is a set of strings, then V* is defined as the smallest superset of V that contains the empty string ε and is closed under the string concatenation operation</p>
<p>If V is a set of symbols or characters, then V* is the set of all strings over symbols in V, including the empty string ε.</p>
</blockquote>
<p>So this language accepts all strings over <span class="math-container">$\Sigma^*$</span> which must be regular.
Also regular languages are closed under kleene star.</p>
<p>But again on wiki</p>
<blockquote>
<p><span class="math-container">$V^* = \bigcup\limits_{i\geq 0}^{} V_i = {\epsilon} \ \cup \ V_1 \ \cup V_2 \ \cup \ V_3 \ \cup .....$</span></p>
</blockquote>
<p>Now according to this definition strings such as <span class="math-container">$01001$</span> cannot be a part of given language so <span class="math-container">$0$</span>'s prior of every 1 are compared within a string, so this can't be regular.</p>
<p>But according to the former definition <span class="math-container">$01001$</span> is a part of language because it can be formed with symbols <span class="math-container">$01$</span> and <span class="math-container">$001$</span> both are part of <span class="math-container">$0^n 1$</span>.</p>
<p>Can someone help me in identifying the class of these types of languages</p>
| rici | 59,314 | <p>My reading of this question (which I think is the natural reading, notwithstanding other possibilities) is that the language being defined is:</p>
<p><span class="math-container">$$L = \bigcup\limits_{n\geq 0}^{} (0^n1)^*$$</span></p>
<p>which is, roughly speaking, the language of all strings in <span class="math-container">$\{0,1\}^*$</span> ending in <span class="math-container">$1$</span> in which the <span class="math-container">$1$</span>s are evenly-spaced. (In other words, there is an implicit "union over all <span class="math-container">$n$</span>".) That language is not regular, which is easy to prove with the pumping lemma. (Take the string <span class="math-container">$(0^{p+1}1)^5$</span>.)</p>
<p>I don't see any natural interpretation of <span class="math-container">$(0^n1)^*$</span> in which the <span class="math-container">$n$</span> is not fixed. It seems unlikely that the intent was <span class="math-container">$\left(\bigcup\limits_{n\geq 0}0^n1\right)^*$</span>, since that would naturally be written <span class="math-container">$(0^*1)^*$</span>, not <span class="math-container">$(0^n1)^*$</span>. That language is regular, as you know, but I don't think it is relevant to this question. </p>
|
3,527,919 | <p>I've tried to prove this property of Bessel function but I don't seem to be going anywhere</p>
<p><span class="math-container">$$\sqrt{\frac 12 \pi x} J_\frac 32 (x) = \cfrac{\sin x}{x} - \cos x$$</span></p>
<p>I have tried substituting <span class="math-container">$\frac 32$</span> for <span class="math-container">$J_n (x)$</span> and then manipulating with the product but it doesn't seem to give me something similar with the series on my LHS. I don't know if there is another different approach which I must follow. </p>
| Gary | 83,800 | <p>Using the series expansion of <span class="math-container">$J_{3/2}(x)$</span> and the Legendre duplication formula for the gamma function, we find
<span class="math-container">$$
\sqrt {\frac{{\pi x}}{2}} J_{3/2} (x) = \sqrt {\frac{{\pi x}}{2}} \left( {\tfrac{1}{2}x} \right)^{3/2} \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\left( {\frac{1}{4}x^2 } \right)^n }}{{n!\Gamma \left( {n + \frac{5}{2}} \right)}}}
\\
= \sqrt \pi \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{x^{2n + 2} }}{{2^{2n + 2} n!\Gamma \left( {n + \frac{5}{2}} \right)}}} = \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{(2n + 2)}}{{(2n + 3)!}}x^{2n + 2} }
\\
= \sum\limits_{n = 1}^\infty {( - 1)^{n + 1} \frac{{2n}}{{(2n + 1)!}}x^{2n} } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} } - \sum\limits_{n = 1}^\infty {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}}
\\
= \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(2n + 1)!}}x^{2n} } - \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{x^{2n} }}{{(2n)!}}} = \frac{{\sin x}}{x} - \cos x.
$$</span></p>
|
1,023,575 | <p>How would one factor a number, say $9+4\sqrt{2}$ in $\mathbb{Z}[\sqrt{2}]$?</p>
<p>This is what I've attemped to do:
$$(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2}) $$
$$a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2$$
Thus,
\begin{eqnarray}
a_1a_2+2b_1b_2&=&9 \\
a_1b_2+a_2b_1 &=& 4.
\end{eqnarray}</p>
<p>But this results in 4 variables and only 2 equations.</p>
| Kevin Arlin | 31,228 | <p>The point is, of course, that you want to factor into <em>primes</em>. The norm in $\mathbb{Z}[\sqrt{2}]$ is $N(a+b\sqrt 2)=a^2-2b^2$, so $N(9+4\sqrt 2)=49$ and we only have to worry about primes of norm $\pm 7$. So, when does $a^2-2b^2=\pm 7$ with $a,b$ integers? Well, $(3,1)$ looks tempting, but doesn't work. So we simply take the conjugate of $3+\sqrt 2$, which must be in the other prime dividing $7$. And indeed, $(3-\sqrt 2)(5+3\sqrt 2)$ is the desired factorization. Note $3-\sqrt 2$ must be an associate of $1+2\sqrt 2$, which figured in the other suggested factorization.</p>
|
1,931,754 | <p>I am trying to show that the interval $[0,1)$ is a closed subset of $(-1,1)$ by using the definition that a closed subset contains all of its limit points.
So for a convergent sequence $\{x_n\}$ in $[0,1)$ we have that $0 \leq x_{n} < 1$ for all $n \in \mathbf{N}$. How can I show that $\lim_{n \rightarrow \infty}x_{n} = x$ implies that $x \in [0,1)$? I understand that 1 cannot be a limit point of $[0,1)$ since $1 \not\in (-1,1)$ but I'm having a hard time saying that in a formal way?</p>
| Mark Viola | 218,419 | <p>Herein, we present a way forward that does not rely on differential calculus, but rather uses an elementary pair of inequalities and the squeeze theorem. To that end we proceed.</p>
<blockquote>
<p><strong>PRIMER:</strong></p>
<p>In <a href="https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263">THIS ANSWER</a>, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities</p>
<p><span class="math-container">$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag1$$</span></p>
<p>for <span class="math-container">$x>0$</span>.</p>
</blockquote>
<p>Using <span class="math-container">$(1)$</span> with <span class="math-container">$x=\frac{n}{n-1}$</span>, we find that</p>
<p><span class="math-container">$$\frac{n+1}{n}\le 1+\log\left(\frac{n}{n-1}\right)\le \frac{n}{n-1} \tag 2$$</span></p>
<p>Finally, from <span class="math-container">$(2)$</span> we find that</p>
<p><span class="math-container">$$\left(1+\frac1n\right)^n\le \left(1+\log\left(\frac{n}{n-1}\right)\right)^n\le \frac{1}{\left(1-\frac1n\right)^n} \tag 3$$</span></p>
<p>whereupon applying the squeeze theorem to <span class="math-container">$(3)$</span> yields the coveted limit</p>
<p><span class="math-container">$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty} \left(1+\log\left(\frac{n}{n-1}\right)\right)^n=e}$$</span></p>
|
1,745,136 | <p>Show that among every consecutive 5 integers one is coprime to the others<br>
I considered these 5 numbers as: $5k,5k+1,5k+2,5k+3,5k+4$<br>
It's seen that for example $5k+1$ is coprime to $5k$ and $5k+2$,now it remains to show $5k+1$ is coprime to $5k+3,5k+4$<br>
Let $\gcd(5k+1,5k+3)=d\Rightarrow\ d|2\Rightarrow\ d=1\ or\ 2$<br>
If $d=2$ then $5k+1$ must be even , so does $5k+3$ , so $k$ must be odd.<br>
I stopped here!!</p>
| user133281 | 133,281 | <p>Among any $6$ consecutive integers, there are two that are coprime to $6$. So among any $5$ consecutive integers, there is at least one that is coprime to $6$. This number if also coprime to the others, because the only possible common prime divisors are $2$ and $3$.</p>
|
1,745,136 | <p>Show that among every consecutive 5 integers one is coprime to the others<br>
I considered these 5 numbers as: $5k,5k+1,5k+2,5k+3,5k+4$<br>
It's seen that for example $5k+1$ is coprime to $5k$ and $5k+2$,now it remains to show $5k+1$ is coprime to $5k+3,5k+4$<br>
Let $\gcd(5k+1,5k+3)=d\Rightarrow\ d|2\Rightarrow\ d=1\ or\ 2$<br>
If $d=2$ then $5k+1$ must be even , so does $5k+3$ , so $k$ must be odd.<br>
I stopped here!!</p>
| Patrick Da Silva | 10,704 | <p>Suppose the prime $p$ divides two of the integers $5k,\cdots,5k+4$. Then there are two integers $i,j \in \{0,\cdots,4\}$ such that $i \equiv j \pmod p$, so that $p = 2$ or $p=3$. The integer $6$ can obviously never divide two integers whose distance is less than $5$, so the gcd of two such integers is either $1,2$, $3$ or $4$. </p>
<p>Now it is not hard to check that at most $3$ integers out of those $5$ are divisible by $2$ and at most $2$ integers out of those $5$ are divisible by $3$. But it can't be that all five are divisible by either $2$ or $3$ since when three are divisible by $2$ and two are divisible by $3$, one of the five is divisible by $6$ (check this!). </p>
<p>Another way of saying this is that the units of $\mathbb Z/6\mathbb Z$ are $\overline 1$ and $\overline 5$. Any set of five consecutive integers, when reduced modulo $6$, contains at least one of these two. </p>
<p>Hope that helps,</p>
|
1,745,136 | <p>Show that among every consecutive 5 integers one is coprime to the others<br>
I considered these 5 numbers as: $5k,5k+1,5k+2,5k+3,5k+4$<br>
It's seen that for example $5k+1$ is coprime to $5k$ and $5k+2$,now it remains to show $5k+1$ is coprime to $5k+3,5k+4$<br>
Let $\gcd(5k+1,5k+3)=d\Rightarrow\ d|2\Rightarrow\ d=1\ or\ 2$<br>
If $d=2$ then $5k+1$ must be even , so does $5k+3$ , so $k$ must be odd.<br>
I stopped here!!</p>
| MathWiz | 323,681 | <p>Let $n$ be a natural number. Consider five consecutive numbers $(n-2),(n-1),(n),(n+1),(n+2)$.</p>
<ul>
<li><p>If $n$ is even then $n-1$ and $n+1$ only can be coprime to all others.
since these two are consecutive odd numbers, thus they are coprime .
now the largest odd number less than $ 5$ is $3$.
if $n-1$ is a multiple of $3$ then so will $n+2$ be , in which case $n+1$ definately
would be coprime to all others.</p></li>
<li><p>If $n$ is odd then
($n-2$, $n$),($n$, $n+2$) being pairs of consecutive odds will be coprime.
if $n-2$ is a multiple of $3$ then so will $n+1$ be. but $n$ and $n+2$ wont be multiples
of $3$, in which case both $n$ and $n+2$ will be coprime to all others.
note: if the first odd number in each case (i.e. n being even or odd) is not a
multiple of $3$ then the reasoning is simpler.</p></li>
</ul>
|
1,355,133 | <p>A while ago I asked a question about probability here <a href="https://math.stackexchange.com/questions/1353044/why-is-binomial-probability-used-here/">Why is binomial probability used here?</a></p>
<p>I get that you can find how many ways of choosing the $6$ correct out of $10$ questions.</p>
<p>But why do we <strong>multiply</strong> by $\binom{n}{k}$? </p>
<p>I thought it is simple casework that:</p>
<p>$$P(\text{total probability}) = P(\text{Q 1->6 right and 7-10 wrong} + P(\text{Q 1->5 right, 6 wrong, 7 right, 8-10 wrong}) + ...$$</p>
<p>What is the idea? </p>
| Conrado Costa | 226,425 | <p>How many ways can you get 6 questions right?</p>
<p>1->6 right and 7->10 wrong is an event. But you need to count the others. For instance 1->3 wrong and 4->10 right. </p>
<p>How many ways can you get 6 questions out of 10 right? Choose $6$ out of $10$ to get right: ${10\choose 6}$. The rest will follow if you understand this.</p>
|
1,951 | <p>In <a href="https://matheducators.stackexchange.com/a/1949/704">this answer</a>, user <a href="https://matheducators.stackexchange.com/users/942/robert-talbert">Robert Talbert</a> stated that</p>
<blockquote>
<p>There are some amazing things you can do pedagogically with clickers.</p>
</blockquote>
<p>I'd like to see some examples. (Not that I'm eager to try that myself, but I'm just curious.) As I stated in one of the comments to the linked question,</p>
<blockquote>
<p>I can't see much point in using such devices, at least in the context of teaching.</p>
</blockquote>
<p>– but I'm probably mistaken. Please enlighten me, and at the same time give MESE users some neat ideas! :)</p>
| Adrienne | 1,207 | <p>Warning. Biologist is answering.</p>
<p>Our instructors are often in very large, very sleepy lecture halls. Clickers provide a stimulus for student discussion and trigger learning through testing effects. </p>
<p>Common uses:</p>
<ol>
<li><p>The instructor is about to begin a new subject. She opens a clicker
question that contains a common misconception as one of the answer
options. Students answer incorrectly. The instructor can now address
the misconception and correct it. Students are given additional, new
problems to answer, either individually or in groups, and then click
in.</p></li>
<li><p>The instructor asks students to rank a list of procedures, processes, steps in order. Can be done individually, and then in groups. </p></li>
<li><p>If the instructor has assigned pre-class reading, students can take a 5-minute clicker quiz when they walk in to reinforce the assignment and remind them that the instructor will not be covering the basics.</p></li>
</ol>
<p>Newer, web-based audience response systems can require students to draw a curve on a graph or type in a short answer. Some (Learning Catalytics) can determine which students answered correctly or incorrectly, and tell Marcos to "turn to Jane on your left and discuss your answer," maximizing student teaching.</p>
<p>Attached is an example of the research on the subject in biology:</p>
<p>Smith, M. K., Wood, W. B., Krauter, K., & Knight, J. K. (2011). <a href="http://www.lifescied.org/content/10/1/55.full">Combining peer discussion with instructor explanation increases student learning from in-class concept questions.</a> CBE-Life Sciences Education, 10(1), 55-63.</p>
|
2,440,802 | <p>The number of positive integers that $n$ can take in between the range $100$ to $200$.</p>
<p>I tried a lot using the prime factorization method but no use. </p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>$$n^2-n-2=(n-2)(n+1)$$</p>
<p>$$n^2+2n-3=(n+3)(n-1)$$</p>
<p>As $n+1-(n-2)=3,n-2,n+1$ are of opposite parity, exactly one of them must be divisible by $8$</p>
<p>As $n+3-(n-1)=4,$ exactly one of them must be divisible by $27$</p>
<p>Now use <a href="http://mathworld.wolfram.com/ChineseRemainderTheorem.html" rel="noreferrer">CRT</a> for all $2\cdot2$ possible cases.</p>
|
1,860,459 | <blockquote>
<p>Prove that $4k < 2^k$ by induction.</p>
</blockquote>
<p>It holds for $k = 5$. Assume $ k = n + 1 $. Then</p>
<p>$4(n+1) < 2^{(n+1)}$</p>
<p>$4n + 4 < 2^n * 2$</p>
<p>$2n + 2 \leq 2^n$</p>
<p>Now I just need to show that</p>
<p>$2n + 2 \leq 4n$</p>
<p>$n + 1 \leq 2n$</p>
<p>$1 \leq n$</p>
<p>And because I chose $n = 5$ which is greater than $1$, this should prove that the formula holds for $n \geq 5$.</p>
<p>Is this correct?</p>
| Daniel W. Farlow | 191,378 | <p>Consider the following (see if you can determine how one step relates to another):
\begin{align}
4(k+1)&=4k+4\\[1em]
&< 2^k+4\tag{why?}\\[1em]
&< 2^k+2^k\tag{why?}\\[1em]
&= 2\cdot2^k\\[1em]
&=2^{k+1}.
\end{align}</p>
|
2,801,936 | <p>To me, it seems obvious that the binary quadratic form $x^2+8y^2$ does not properly represent 3. However, I have managed to prove that it does so I think I must be doing something stupid.
I have used the following:</p>
<p><strong>Let f be a a binary quadratic form and n an integer. We say that f <em>properly represents</em> n if there exists [x,y]∈$\mathbb Z^2$ such that (x,y)=1 and f(x,y)=n. (x,y) is defined as the greatest common divisor of x and y.</strong></p>
<p><strong>Lemma 5.3 (iii) Some form of discriminant d properly represents n if and only if $u^2\equiv d\pmod {4n}$</strong></p>
<p>Then here is my 'proof':</p>
<p>Let $f(x,y)=x^2+8y^2$. Then, by Lemma 5.3(iii), $f(x,y)$ properly represents n=3 if and only if there is a solution to $u^2\equiv d\pmod{4\cdot3}$</p>
<p>$d=b^2-4ac=0^2-4\cdot1\cdot8=-32$ so</p>
<p>$u^2\equiv -32\pmod{12}$ which gives us</p>
<p>$u^2\equiv 4\pmod{12}$</p>
<p>which clearly has the solution $u=2$ so $f(x,y)$ properly represents n=3.</p>
<p>I know that this is wrong and it's very likely wrong for a stupid reason but I can't figure out what that is so any help would be appreciated.</p>
| user328442 | 328,442 | <p>A little more general:</p>
<p>Theorem: Suppose $f$ is a multiplicative function. Then $$\sum_{d|n} \mu(d) f(d) = \prod_{p|n} (1-f(p)).$$</p>
<p>Proof: Let $$g(n) = \sum_{d|n} \mu(d) f(d).$$ Then $g$ is multiplicative (the product $\mu f$ is obviously multiplicative so the Direchlet convolution $\mu f * u$ where $u(n) = 1$ for every $n$ is multiplicative). This implies that $g$ is completely determined by its behavior on powers of primes. </p>
<p>That is, $$g(p^k) = \sum_{d|p^k} \mu(d) f(d) = \mu(1) f(1) + \mu(p) f(p) = 1-f(p).$$ Therefore, $$g(n) = \prod_{p|n}g(p^k) = \prod(1-f(p)).$$</p>
<p>Note: Try to deal with the prime powers individually rather than dealing with all of them at once. That is, try handling the case where $n = p^k$ first and then apply the multiplicative property to attain the general case. </p>
|
1,221,158 | <p>I'm interested in knowing whether $a^0 = 1$ ('$a$' not zero) is a definition.
If not, can anyone please help me with proving this?</p>
| Daniel | 150,142 | <p>It's a definition. A convenient definition. </p>
<p>We know from our early encounter with mathematics that $a^m\times a^n=a^{m+n}$ if $m,n\in \mathbb{N}$ (not including zero) because it's very natural: "Multiplying $m$ times and multiplying $n$ times, and then multiplying those values should be the same as multiplying $m+n$ times". </p>
<p>What if $m$ or $n$ is $0$? We'd like that formula to still be true, i.e. we'd like </p>
<p>$$a^0\times a^n = a^{0+n}=a^n$$</p>
<p>However, the only number that satisfies this is $1$ so it's necessary to define $a^0:=1$ in order to keep this property.</p>
<p>This is exactly the same reason why we define</p>
<p>$$a^n:=\frac{1}{a^{-n}}$$</p>
<p>if $n$ is a negative integer, because we want laws as </p>
<p>$$\frac{a^m}{a^n}=a^{m-n}$$</p>
<p>to hold even if $m$ is not greater than $n$.</p>
|
1,221,158 | <p>I'm interested in knowing whether $a^0 = 1$ ('$a$' not zero) is a definition.
If not, can anyone please help me with proving this?</p>
| Chenkodan | 146,844 | <p>$${a^b\over a^c} = a^{b-c} $$and vice versa. {Index rule}</p>
<p>Therefore, $$a^0 = a^{x-x}$$ for any x</p>
<p>$$= {a^x\over a^x} ,$$ using the aforementioned index rule.</p>
<p>$$= 1 $${since any thing divided by itself is 1 except 0)</p>
|
551,662 | <p>I am reading "What Is Mathematics? An Elementary Approach to Ideas and Methods"
And I am stuck here, I don't get it. I have posted a screen shot underlining what my doubt is..</p>
<p>I dont get it when the author says while the pythagoras theorem is : $a^2 + b^2 = c^2$
and then he says $x=a/c$ and $y=b/c$
and then the equation should be according to me , $ax+by=c$.. right??
but the author writes y^2 = (1-x)(1+x)</p>
<p>Which I think may have came from something like
$x^2 + y^2 = 1$
$y^2= 1^2 + x^2$
$y^2= (1+x)(1-x)$ (since $a^2 + b^2 = (a+b)(a-b)$ )</p>
<p>but I dont get it where did that x^2 + y^2 = 1 came from ??
Is it that author assumed that x=a/c and y=b/c and then stoped talking about pythagoras theorem and started talking on x^2 + y^2 = 1???</p>
<p>and then further he introduces a number t.. i don't get it how it got converted into y=t(1+x) and (1-x)=ty ???</p>
<p>Can please someone help?</p>
<p>also , while i was writing the question it clicked me that if:
x=a/c (i.e. opposite upon hypotenuse means x is <strong><em>sin</em></strong> )
y=b/c (i.e. adjacent upon hypotenuse means y is <strong><em>cos</em></strong>)
therefore x^2 + y^2 = 1 (sin^2 + cos^2 = 1)</p>
<p>but then if i assume i am right about the sin cos thing then how come the last step is derived ?? (the one in blue color underline and box)
also that as per me x is sin , but the formula in that book is of cos2x? when t=tanx??</p>
<p>so m lil confused .. if you want any more clean way of me asking my doubt then do tell me i will rephrase the entire question .. :)</p>
<p><img src="https://i.stack.imgur.com/QQIVM.png" alt="Image of my question asked above,, screenshot of the book"></p>
| Olivier | 45,622 | <p>For the first part of your question: the author divides both sides of the Pythagorean equation by $c^2$. This yields: $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, he defines $x = a/c$ and $y = b/c$. The equation can than be rewritten to: $x^2 + y^2 = 1$. Substracting $x^2$ from both sides yields $y^2 = 1 - x^2$, or equivalent: $y^2 = (1-x)(1+x)$. </p>
<p>He then devides both sides by $y$ and $(1+x)$. This gives $y/(1+x) = (1-x)/y$. His next step is to say that the expression is actually $t = t$, where $t$ is a rational. So, $ y/(1+x) = t \rightarrow y =t(1+x)$ and, in the same way, $(1-x)/y = t \rightarrow (1-x) = ty$. </p>
<p>Working this out gives the two equations $tx - y = -t$ and $x + ty = 1$. Now you have two unknowns in two equations. </p>
<p>This can be easily solved:</p>
<p>$tx - y = -t \rightarrow x = \frac{y-t}{t}$. Substitute this into the second equation. That gives us$\frac{y-t}{t} + ty = 1 \rightarrow (y-t) + t^2y = t$. Now, solve for y:
$y + t^2y = 2t \rightarrow y(1+t^2) = 2t \rightarrow y = \frac{2t}{1 + t^2}$. </p>
<p>Substitute this value for y back into the first to get the expression for $x$.</p>
|
391,333 | <p>It is well known that $\sum_{k = 1}^{n}k^3 =\Big [\sum_{k=1}^{n}k^1\Big]^2$. My question is very simple.</p>
<blockquote>
<p>There are $3$-tuples $(p, q, \alpha) \in
\mathbb{N}\times\mathbb{N}\times\mathbb{N}$, in addition to $(3,1,2)$,
such that $\alpha\geq 2$ and $$\sum_{k = 1}^{n}k^{\,p} =\Big [\sum_{k=1}^{n}k^{\,q}\Big]^\alpha, \quad \forall n \in \mathbb{N}{\;\Large ?}$$ </p>
</blockquote>
| Ivan Loh | 61,044 | <p>Let $n=2$, so $(1+2^p)=(1+2^q)^{\alpha}$, so $(1+2^q)^{\alpha}-2^p=1$. If $p=1$, then $q=\alpha=1$, contradicting $\alpha \geq 2$. Otherwise $1+2^q, \alpha, 2, p>1$, so by <a href="http://en.wikipedia.org/wiki/Catalan%27s_conjecture" rel="nofollow">Mihailescu's theorem</a> $1+2^q=3, \alpha=2, p=3$. This gives $(3, 1, 2)$, which is indeed a solution. </p>
|
847 | <p>Apologies in advance if this is obvious.</p>
| Ben Webster | 66 | <p>By the way, <a href="http://www.ams.org/mathscinet-getitem?mr=1155753" rel="nofollow">this paper</a> may be of interest. It shows that for solvable groups, one doesn't have to do the Hilbert class extension moonface suggests, but for some non-solvable ones you do. Also <a href="http://www.ams.org/mathscinet-getitem?mr=2381795" rel="nofollow">this one</a> has more examples.</p>
|
982,780 | <p>I have the following system of <span class="math-container">$M$</span> linear equations in <span class="math-container">$N$</span> unknowns.</p>
<p><span class="math-container">$$
\begin{bmatrix}
3 & 0 & 1 & 0 & -1 & -3 & 2\\
1 & 2 & 0 & 4 & 0 & 0 & -1\\
1 & 1 & 0 & 0 & -1 & -1 & -2\\
0 & 0 & 1 & 0 & -3 & -1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4} \\
x_{5} \\
x_{6} \\
x_{7} \\
\end{bmatrix} =
\begin{bmatrix}
1\\
0\\
0\\
-1\\
\end{bmatrix}$$</span></p>
<p>Is there any algorithm for finding answers of this equations that <span class="math-container">${x_{i} \ge 0}$</span>?</p>
<p><strong>Comment</strong>: I just want that <span class="math-container">$x_i \ge 0$</span>.</p>
<p>It can change to</p>
<p><span class="math-container">$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 2/3 & -2/3 & 1/3 & 2/3\\
0 & 1 & 0 & 0 & -5/3 & -1/3 & -7/3 & -2/3 \\
0 & 0 & 1 & 0 & -3 & -1 & 1 & -1 \\
0 & 0 & 0 & 1 & 2/3 & 1/3 & 5/6 & 1/6 \\
\end{bmatrix}
$$</span></p>
| Henno Brandsma | 4,280 | <p>Prove by induction on $\beta$ that $L(\beta) = \{x: x \le \beta \}$ is compact, for all ordinals $\beta$.</p>
<p>This is clear for $\beta = 0$, where $L(0) = \{0\}$ and if $\beta+1$ is a successor, then $L(\beta+1) = L(\beta) \cup \{\beta+1\}$, so if $L(\beta)$ is compact, so is $L(\beta+1)$.</p>
<p>So assume $L(\alpha)$ is compact for all $\alpha < \beta$ and $\beta$ is a limit ordinal.
Let $(U_i), i \in I$ be an open cover of $L(\beta)$. Some $U_j$ covers the point $\beta \in L(\beta)$, and so for some $\alpha < \beta$ we have that $(\alpha, \beta] \subset U_j$. Then as $L(\alpha)$ is compact, it is covered by finitely many $U_i$, and adding $U_j$ gives us the required finite subcover. So $L(\beta)$ is compact.</p>
|
1,282,419 | <p>Let $\Delta\subset\Bbb C$ be the open unitary disk. Let $\varphi:\Delta\to\Bbb R$ defined as follows: $\varphi(z)=1$ if $\Re z\ge0$, $\varphi(z)=0$ otherwise. So $\varphi$ is upper semicontinous.</p>
<p>In order to prove $\varphi$ is NOT subharmonic, I've to find a compact subset $K\Subset\Delta$ and a real valued function $h\in\mathcal{C}^0(K)\cap\mathcal H(\stackrel{\circ}{K})$ such that</p>
<ul>
<li>$\varphi\leq h$ on $\partial K$</li>
<li>$\varphi\nleq h$ on $\stackrel{\circ}{K}$</li>
</ul>
<p>Here $\mathcal H(\stackrel{\circ}{K})$ is the set of harmonic functions on the interior of $K$.</p>
<p>I tried for hours, but I did't find an answer.</p>
<p>The only thing which gave me a chance for one second was $h(z):=\log\left(\frac{|z|}{a}\right)+1$ on $K:=\bar\Delta_{0,a}$ which is the closure of the disk centered in $0$ of radius $a\in]0,1[$. This function would be right, the problem is it's not defined in $0$. So I tried cutting out a small circle around zero, but we get other obvious problems in satisfying the given conditions.</p>
| Daniel Fischer | 83,702 | <p>Take a disk $\lvert z\rvert \leqslant r$ for some $0 < r < 1$, and define the boundary values by</p>
<p>$$h_0(re^{i\vartheta}) = l(\vartheta),$$</p>
<p>where</p>
<p>$$l(t) = \begin{cases} \frac{2}{\pi}(t+\pi) &, -\pi \leqslant t \leqslant -\frac{\pi}{2}\\ \quad 1 &, -\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\\ \frac{2}{\pi}(\pi-t) &, \frac{\pi}{2} \leqslant t \leqslant \pi.\end{cases}$$</p>
<p>Then let $h$ be the solution to the Dirichlet problem with boundary values $h_0$. Since $h_0$ is not constant, $h$ is not constant, and by the maximum principle, $h(z) < 1$ for $\lvert z\rvert < r$. But then $\varphi(r/2) = 1 > h(r/2)$.</p>
|
3,995,492 | <p>I have no clue how to do this, I manage to get I get that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$</span> but I can't get anywhere from there.</p>
| Olivier Roche | 649,615 | <p>All you need to know for this is that, since <span class="math-container">$13$</span> is a prime number, <span class="math-container">$\mathbb{Z} / 13 \mathbb{Z}$</span> is a <strong>field</strong>.</p>
<p>In particular, every non zero element has a unique inverse for multiplication. Constating that
<span class="math-container">$$11 \times 6 = 66 \equiv 1 \hspace{0.1cm} \text{mod} (13)$$</span>
it turns out that the multiplicative inverse of <span class="math-container">$11$</span> is <span class="math-container">$6$</span> : <span class="math-container">$\boxed{11^{-1} \equiv 6}$</span>.</p>
<p>Now, since you know that <span class="math-container">$11^{36} \equiv 1 \hspace{0.1cm} \text{mod} (13)$</span>, it turns out that :</p>
<p><span class="math-container">$$11^{35} \equiv 11^{36} \times 11^{-1} \equiv 1 \times 6 \equiv 6 \hspace{0.1cm} \text{mod} (13)$$</span></p>
|
2,476,973 | <p>A fair six-sided die carries $1$ on one face, $2$ on two of its faces, and<br>
$3$ on the remaining three faces. </p>
<p>Suppose the die is rolled twice, and let $X$ be the random variable ’total score'. Find the probability distribution of $X$.</p>
| A. M. | 123,356 | <p>For example, $P(T=2)$ is the probability that we get $1$ at both tries. Since the trials are independent, this means:</p>
<p>$P(T=2) = P(1~on~first~try)\times P(1~on~second~try)=1/6\times 1/6=1/36$.
You can work through all other situations as it is suggested.</p>
|
3,695,127 | <p>Before the moderators close my question, I cant think of any starting approach to the question. </p>
<p>Another question of the similar type I am having trouble with is: 12 balls are distributed at random among 3 boxes. What is the probability that the first box will contain 3 balls?
For the second question I can figure out the exhaustive number of outcomes will be 3 raised to the power 12 since each ball has 12 options.</p>
| Kevin.S | 724,407 | <p>Geometrically, <span class="math-container">$S^1\times\{1\}\cup\{1\}\times S^1\cong S^1\vee S^1$</span>, and <span class="math-container">$S^1\times S^1\setminus\{(-1,-1)\}$</span> is the punctured torus (<span class="math-container">$=T^2\setminus \{p\}\cong([-1,1]^2/\sim)\setminus\{p\}$</span>).</p>
<p>I think this is what you want:</p>
<p>Let <span class="math-container">$p=(-1,-1)$</span>, <span class="math-container">$I=[-1,1]$</span> and <span class="math-container">$(s_1,s_2)\in (I^2/\sim)\setminus\{p\}$</span> then we construct the deformation retraction <span class="math-container">$H:(I^2/\sim)\setminus\{p\}\times[0,1]\to (s_1\times\{1\}\cup \{1\}\times s_2)/\sim$</span> by:
<span class="math-container">$$
H(s_1,s_2,t)=
\begin{cases}
(t(\frac{2(s_1+1)}{|s_2+1|}-1-s_1)+s_1, (1-s_2)t+s_2)& \text{if }s_2\ge s_1\\
((1-s_1)t+s_1, t(\frac{2(s_2+1)}{|s_1+1|}-1-s_2)+s_2)& \text{if }s_2\le s_1
\end{cases}
$$</span></p>
<p>The map is well-defined and continuous at both branches.</p>
<p>A way to visualize this map is to think about the square under the quotient <span class="math-container">$\sim$</span>, because in this case, <span class="math-container">$\frac{s_1\times\{1\}\cup \{1\}\times s_2}{\sim}\cong S^1\times\{1\}\cup\{1\}\times S^1$</span>. <span class="math-container">$H$</span> expands the hole created by <span class="math-container">$\{p\}$</span> (in the left corner) and maps every point in <span class="math-container">$T^2\setminus\{p\}$</span> to the two edges that have one point in common by sliding each point linearly through the line between <span class="math-container">$(-1,-1)$</span> and the two edges which is obviously <strong>equal</strong> to the space indicated in the first sentence. ("the two edges" means <span class="math-container">$\frac{s_1\times\{1\}\cup \{1\}\times s_2}{\sim}$</span>) </p>
<p>The space <span class="math-container">$\frac{s_1\times\{1\}\cup \{1\}\times s_2}{\sim}$</span> is the same as <span class="math-container">$S^1\times\{1\}\cup\{1\}\times S^1$</span> because <span class="math-container">$\sim$</span> generates the equivalence relation which states that <span class="math-container">$(s_1,1)\sim(s_1,-1),(1,s_2)\sim(-1,s_2)\implies (-1,1)\sim (1,1)\sim (1,-1)$</span> which implies that the two edges are actually two circles. After identification, we see that only <span class="math-container">$(1,1)$</span> belongs to both circles simultaneously so this is also equivalent to the wedge sum of two circles, namely <span class="math-container">$S^1\vee S^1$</span>. (hope this makes it clearer...)</p>
<hr>
<p>If there is anything ambiguous, please tell me and I'll try to help you. :)</p>
|
134,987 | <blockquote>
<p>$$3x^2 + 2y^4 = z^4$$</p>
</blockquote>
<p><em>How do I solve this??</em> I would like to use so-called "elementary number theory", not abstract algebra (e.g. $\mathbb{Z} ( \sqrt d)$) or elliptic curves.</p>
<p>Note: I'm not asking <em>what</em> the solutions are, but rather <em>how</em> to find them.</p>
<p>My instincts are: </p>
<ul>
<li>search the internet (I compared this equation with the ~280 here on MSE, and tried a variety of similar searches on uniquation.com ...)</li>
<li>search the 3 number theory books that I have</li>
<li>try to find solutions "by inspection" (possibly after reducing the order of the variables)</li>
<li>do some magic with modular arithmetic </li>
<li>use <a href="http://www.alpertron.com.ar/QUAD.HTM" rel="nofollow">Alpern's solver</a> - which seemed to indicate that there are no solutions (though I might have made an illegal substitution, so to speak) </li>
</ul>
<p>I was able to identify $A = 6, B = 3, C = 6$ as solutions of $ \ 3A + 2B \ ^2 = C \ ^2$, but those aren't squares!</p>
<blockquote>
<p>What is the number-theoretic approach to such problems? Is there a general method?</p>
</blockquote>
| Will Jagy | 10,400 | <p>Maybe what you need is Legendre's Theorem. Certainly it covers this situation. It tells you exactly what needs to be checked. It is presented in Ireland and Rosen, A Classical Introduction to Modern Number Theory, chapter 17, section 3. A very similar treatment is in <a href="http://alpha.math.uga.edu/%7Epete/4400rationalqf.pdf" rel="nofollow noreferrer">PETE</a>, pages 5-8.</p>
<p>Anyway, if <span class="math-container">$a,b,c$</span> are integers, not all positive and not all negative, but all nonzero, they are all squarefree, <span class="math-container">$\gcd(b,c) = \gcd(c,a) = \gcd(a,b) = 1,$</span>, then
<span class="math-container">$$ a x^2 + b y^2 + c z^2 = 0 $$</span> has a nontrivial solution in integers if and only if all three of these are true:</p>
<p>(i) <span class="math-container">$-bc$</span> is a square <span class="math-container">$\pmod a,$</span></p>
<p>(ii) <span class="math-container">$-ca$</span> is a square <span class="math-container">$\pmod b,$</span></p>
<p>(iii) <span class="math-container">$-ab$</span> is a square <span class="math-container">$\pmod c.$</span></p>
<p>In these we include <span class="math-container">$0$</span> as a square.</p>
<p>This result is also done in any book on quadratic forms. I like <a href="http://store.doverpublications.com/0486466701.html" rel="nofollow noreferrer">Rational Quadratic Forms</a> by J. W. S. Cassels.</p>
<p>Anyhoo, you have <span class="math-container">$$ 3 U^2 + 2 V^2 - W^2 = 0,$$</span> so <span class="math-container">$a=3, \; b=2, \; c = -1.$</span> Condition (iii) asks if <span class="math-container">$-6$</span> is a square <span class="math-container">$\pmod {-1},$</span> the answer is yes as it is a multiple. Condition (ii), is <span class="math-container">$3$</span> a square <span class="math-container">$\pmod 2,$</span> again the answer is yes. However, condition (i), <span class="math-container">$2$</span> is not a square <span class="math-container">$\pmod 3.$</span> So there you are.</p>
<p>Leonard Eugene Dickson devoted a whole chapter to this in <em>Introduction to the Theory of Numbers</em> (1929). He still gave a section on it in <em>Modern Elementary Theory of Numbers</em> (1939).</p>
<p>NOTE: so this problem can be done without the fourth powers. The traditional one where the fourth powers really matter is <span class="math-container">$X^4 + Y^4 = Z^2,$</span> which is part of the proof of Fermat's Last Theorem, and is not just an application of Legendre's Theorem (although not difficult).</p>
|
2,332,750 | <p>At the end of chapter 5 of stein's book <a href="http://wstein.org/books/ant/ant.pdf" rel="nofollow noreferrer">A Computational Introduction to Algebraic Number Theory</a> he proves proposition 5.2.4 which states that:</p>
<p>Given a prime ideal $\mathfrak{p}$ in a Dedekind domain $R$ we have the isomorphism
$$
\frac{\mathfrak{p}^n}{\mathfrak{p}^{n+1}} \cong \frac{R}{\mathfrak{p}}
$$
of $R$-modules for any $n \geq 0$.</p>
<p>What is the point of including this proposition, other than the fact it can be proved using the Chinese Remainder Theorem? I don't see why this is important.</p>
| nguyen quang do | 300,700 | <p>I don't know much about effective calculation in ANT, but it seems to me that your question is rather a general one about the "purpose" of the natural isomorphisms ¤ $\mathfrak p^n /\mathfrak p^{n+1} \cong R/\mathfrak p := k_\mathfrak p$ (the residue field at $\mathfrak p$, viewed as an additive group). Moreover it appears to be of a "local" nature. More precisely, let $R$ be the ring of integers of a number field $K$, $K_\mathfrak p$ = the completion of $K$ w.r.t. the $\mathfrak p$-adic valuation, with uniformizer $\pi$, and $R_\mathfrak p$ the ring of integers of $K_\mathfrak p$. Introduce the group of $\mathfrak p$-adic units $U_0$ and the decreasing sequence of subgroups of principal units $U_n=1 + (\pi^n)$. In addition to the isomorphisms ¤ recalled above, one has isomorphisms $U_0/U_1 \cong k_\mathfrak p ^{*}$ (obvious) and $U_n/U_{n+1} \cong \mathfrak p^n /\mathfrak p^{n+1}\cong k_\mathfrak p$ (coming obviously from ¤, or the other way around). </p>
<p>The principal direct (this means : without CFT) applications of these isomorphisms, I think, concern ramification , see e.g. Serre's "Local Fields", chapters IV and V. From now on, for convenience, we drop the subscript $\mathfrak p$ in the notations for $\mathfrak p$-adic local fields. For a finite Galois extension $L/K$ of such fields, with Galois group $G$, the decreasing sequence of ramification subgroups $G_i$ is defined as follows : for any integer $i\ge 0$, $G_0$ is the inertia subgroup, and $s\in G_0$ belongs to $G_i$ iff $s(\pi_L)/\pi_L\in U_i (L)$. The goal is to "unscrew" the Galois group $G$ by means of the successive quotients $G_i/G_{i+1}$. Using the aforementioned isomorphisms, one can show that the map $s\in G_i \to s(\pi_L)/\pi_L$ induces an isomorphism of $G_i/G_{i+1}$ into $U_i (L)/U_{i+1} (L)$, whose image, for certain adequate values of $i$, can be determined as the kernel of a map induced by the norm of $L/K_{nr}$, where $K_{nr}$ denotes the maximal unramified subextension of $L/K$. For the numerous information thus obtained on the filtration of the $G_i$'s, see <em>op. cit.</em></p>
|
2,612,134 | <p>One of the exercise in Artin's algebra gives an eigenvector of an element of $SO(3)$, in one possible case. Namely, it is asked to show that </p>
<blockquote>
<p>If $A=[a_{ij}]$ is a rotation in $SO(3)$, then the vector
$$v=\begin{bmatrix} (a_{23}+a_{32})^{-1}\\ (a_{13}+a_{31})^{-1} \\ (a_{12}+a_{21})^{-1}\end{bmatrix}$$
is an eigenvector of $A$, whenever these entries of the vector are well-defined in $\mathbb{R}$. </p>
</blockquote>
<p>I couldn't get any way to proceed to prove this. What I tried was, consider $A$ as sum of symmetric and skew-symmetric- $\frac{1}{2}(A+A^t)+\frac{1}{2}(A-A^t)$ and show that $v$ is an eigenvector of both with eigenvalues $1$ and $0$ respectively; but, this was not working beyond long algebraic/symbolic computations. Any hint for this?</p>
| Widawensen | 334,463 | <p>Let me use letter $R$ instead of $A$ for the considered matrix. </p>
<p>Generally for a rotation matrix $R(v,\theta)$ you can calculate the axis unit vector from the formula:
$v= {\dfrac {1}{2sin(\theta)}}\begin{bmatrix}
r_{32}-r_{23} \\
r_{13}-r_{31} \\
r_{21} -r_{12}
\end{bmatrix}$ where $r_{ij}$ are appropriate entries of $R$ matrix.</p>
<p>Now the problem is equivalent to the checking whether cross product<br>
$c=\begin{bmatrix}
r_{32}-r_{23} \\
r_{13}-r_{31} \\
r_{21} -r_{12}
\end{bmatrix} \times \begin{bmatrix} (r_{32}+r_{23})^{-1} \\
(r_{13}+r_{31})^{-1} \\
(r_{21} +r_{12})^{-1})
\end{bmatrix}$ is really a zero vector what is equivalent to the condition for the vectors to be parallel.</p>
<p>Calculations can be checked with WolphramAlpha.<br>
Use the command </p>
<blockquote>
<p>[r_32-r_23,r_13-r_31,r_21-r_12] x [1/(r_32+r_23),1/(r_31+r_13),1/(r_21+r_12)]</p>
</blockquote>
<p>$c=[\dfrac{r_{12}^2}{(r_{12} + r_{21}) (r_{13} + r_{31})} + \dfrac {r_{13}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31})} - \dfrac {r_{21}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31})} - \dfrac {r_{31}^2}{ (r_{12} + r_{21}) (r_{13} + r_{31}) } \ , - \dfrac {r_{12}^2}{ (r_{12} + r_{21}) (r_{23} + r_{32}) } + \dfrac{ r_{21}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32}) } + \dfrac {r_{23}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32} ) } - \dfrac{ r_{32}^2 }{ (r_{12} + r_{21}) (r_{23} + r_{32}) } \ ,- \dfrac{r_{13}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } - \dfrac{ r_{23}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } + \dfrac {r_{31}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32}) } + \dfrac {r_{32}^2 }{ (r_{13} + r_{31}) (r_{23} + r_{32})) }]^T$</p>
<p>what really will give the searched <strong>zero vector</strong> if we additionally take into consideration that norms of column and row vectors for rotation matrix are equal to $1$. </p>
<p>(for example:<br>
$ \ r_{11}^2+r_{12}^2+r_{13}^2=1$, $ \ r_{11}^2+r_{21}^2+r_{31}^2=1$ from this $ \ r_{12}^2+r_{13}^2-(r_{21}^2+r_{31}^2)=0$, etc..)</p>
|
2,612,134 | <p>One of the exercise in Artin's algebra gives an eigenvector of an element of $SO(3)$, in one possible case. Namely, it is asked to show that </p>
<blockquote>
<p>If $A=[a_{ij}]$ is a rotation in $SO(3)$, then the vector
$$v=\begin{bmatrix} (a_{23}+a_{32})^{-1}\\ (a_{13}+a_{31})^{-1} \\ (a_{12}+a_{21})^{-1}\end{bmatrix}$$
is an eigenvector of $A$, whenever these entries of the vector are well-defined in $\mathbb{R}$. </p>
</blockquote>
<p>I couldn't get any way to proceed to prove this. What I tried was, consider $A$ as sum of symmetric and skew-symmetric- $\frac{1}{2}(A+A^t)+\frac{1}{2}(A-A^t)$ and show that $v$ is an eigenvector of both with eigenvalues $1$ and $0$ respectively; but, this was not working beyond long algebraic/symbolic computations. Any hint for this?</p>
| Widawensen | 334,463 | <p>I would like to propose another approach to the problem. As it is substantially other that given above it will be presented as a separate answer.</p>
<p>Let $R(u,\theta)$ be rotation matrix with the unit axis vector $u=[x,y,z]^T$ and rotation angle $\theta$.<br>
In this situation we have <a href="https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula#Matrix_notation" rel="nofollow noreferrer">Rodrigues formula</a> for generating the exact form of this matrix.</p>
<p>$R(u,\theta)=I+\sin(\theta)K+(1-\cos(\theta))K^2$</p>
<p>where skew-symetric matrix $ K=
\begin{bmatrix} 0 & -z & y \\
z & 0 &-x \\ -y & x & 0 \end{bmatrix} $</p>
<p>It can be checked additionally that $K^2$ is symmetric matrix and </p>
<p>$K^2=uu^T-I = \begin{bmatrix} x^2 -1 & xy & xz \\ xy & y^2-1 & yz \\ xz & yz & z^2-1 \end{bmatrix} $ </p>
<p>Now we can easily calculate the given vector $v$</p>
<p>$v=\begin{bmatrix} { (r_{32}+r_{23})^{-1} \\(r_{13}+r_{31})^{-1} \\ (r_{21}+r_{12})^{-1} }\end{bmatrix}= (1-\cos(\theta))^{-1}\begin{bmatrix} { (2yz)^{-1} \\(2xz)^{-1} \\ (2xy)^{-1} }\end{bmatrix} \ \ \ (*)$.</p>
<p>It is visible that only symmetric part of rotation matrix influences on the form of vector $v$.</p>
<p>The generated vector $v$ is parallel to $u$ because it's sufficient to multiply $v$ by a scalar $c=2(1-\cos(\theta))xyz \ \ $ to obtain $ \ \ u=cv \ \ $ so really the vector $v$ is the eigenvector of $R$ like vector $u$.</p>
<p>The approach is valid for all $\theta$ except of course $\theta=0$ when corresponding matrix becomes identity matrix. Additionally $x,y,z$ must be different from $0$ what is the condition for real components of $v$.</p>
<p>As procedure above seems to be the easiest way of proving that $v$ is an eigenvector of $R$, the Rodrigues form can be also used for proving that $0$ is eigenvalue for skew-symmetric part of $R$ ( i.e. $\sin(\theta)K$) and $1$ is eigenvalue for symmetric part (i.e. $I+ (1-\cos(\theta))K^2)$ with the given eigenvector $v$. It's sufficient to use form (*) for $v$ in appropriate calculations.</p>
|
114,122 | <p>I am trying to figure out the maximum possible combinations of a (HEX) string, with the following rules:</p>
<ul>
<li>All characters in uppercase hex (ABCDEF0123456789)</li>
<li>The output string must be exactly 10 characters long</li>
<li>The string must contain at least 1 letter</li>
<li>The string must contain at least 1 number</li>
<li>A number or letter can not be represented more than 2 times</li>
</ul>
<p>I am thinking the easy way to go here (I am most likely wrong, so feel free to correct me):</p>
<ol>
<li>Total possible combinations: $16^{10} = 1,099,511,627,776$</li>
<li>Minus all combinations with just numbers: $10^{10} = 10,000,000,000$</li>
<li>Minus all combinations with just letters: $6^{10} = 60,466,176$</li>
<li>etc...</li>
</ol>
<p>Can someone could tell me if this is the right way to go and if so, how to get the total amount of possible combinations where a letter or a number occur more than twice.</p>
<p>Any input or help would be highly appreciated!</p>
<p>Muchos thanks!</p>
<p>PS.</p>
<p>I don't know if I tagged this question right, sorry :(</p>
<p>DS.</p>
| J. Kyle | 221,466 | <p>I had a similar question. Basically I think I figured out how to do this.</p>
<p>First of all, if you are dealing with only numbers and you have a base 10 number system, it's pretty easy to figure out the number of combinations. If you have a 3 digit code and only use numbers, you have 999 possible combinations, right? Well, 10^3 gets you there. And I'm using 10, because there are 10 possible combinations of each digit (i.e. 0,1,2,3,4,5,6,7,8,9).
Now if you add the hex letters a,b,c,d,e,f then now there are 16 possible combinations for each digit (the 10 numbers + 6 letters). So I think it stands to reason that you take the number of possible combinations for a digit, 16, and raise it by the number of digits in your code. So 16^8 for an 8 digit code, 16^10 for a 10 digit code, 16^16 for a 16 digit code, etc.</p>
<p>I think my thought process is sound, but I am not a mathematician. Maybe someone can verify that this is correct. Of course this assumes perfectly random codes, not within your restrictions of must contain 1 letter, must contain 1 number, can't use any letter or number or letter more than twice. That should significantly reduce the number of possible combinations.<br>
I think though to eliminate number-only codes (the ones that don't have letters) you would just subtract the total number of possibilities, obviously 9,999,999,999. Then subtract the number of letter only codes, which I believe is 6^10. For no letter or number used more than twice, you would have to find how many possible codes fit that condition, and subtract those as well.</p>
|
784,753 | <p>In spherical coordinates, we have</p>
<p>$ x = r \sin \theta \cos \phi $;</p>
<p>$ y = r \sin \theta \sin \phi $; and </p>
<p>$z = r \cos \theta $; so that</p>
<p>$dx = \sin \theta \cos \phi\, dr + r \cos \phi \cos \theta \,d\theta – r \sin \theta \sin \phi \,d\phi$;</p>
<p>$dy = \sin \theta \sin \phi \,dr + r \sin \phi \cos \theta \,d\theta + r \sin \theta \cos \phi \,d\phi$; and</p>
<p>$dz = \cos \theta\, dr – r \sin \theta\, d\theta$</p>
<p>The above is obtained by applying the chain rule of partial differentiation.</p>
<p>But in a physics book I’m reading, the authors define a volume element $dv = dx\, dy\, dz$, which when converted to spherical coordinates, equals $r \,dr\, d\theta r \sin\theta \,d\phi$. How do the authors obtain this form?</p>
| Omish | 486,107 | <p>After 3.5 year there needs to be an answer to this for searchers :D
First of all there's no need for complicated calculations. You can obtain that expressions just by looking at the picture of a spherical coordinate system.
The only thing you have to notice is that there are two definitions for unit vectors of spherical coordinate system. The only difference between these two definitions is that theta and phi angles are replaced by eachother. This common form of element volume you mentioned is based on the uncommon form of coordinate set in which theta is the angle between z axis and the point.like this :</p>
<p><a href="https://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/illustr/spherical_coordinates.gif" rel="nofollow noreferrer">https://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/illustr/spherical_coordinates.gif</a></p>
<p>While in common mathematics we say theta is the angle in xy plane. The rest should be easy, if there are any more problems please let me know I'd be happy to help with the detail if any need. :)</p>
|
2,206,247 | <p><strong>Question:</strong> Consider the following non linear recurrence relation defined for $n \in \mathbb{N}$:</p>
<p>$$a_1=1, \ \ \ a_{n}=na_0+(n-1)a_1+(n-2)a_2+\cdots+2a_{n-2}+a_{n-1}$$</p>
<p>a) Calculate $a_1,a_2,a_3,a_4.$</p>
<p>b) Use induction to prove for all positive integers that:</p>
<p>$$a_n=\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{3+\sqrt{5}}{2}\right)^n-\left(\dfrac{3-\sqrt{5}}{2}\right)^n\right]$$
Hi all! I'm having trouble solving this problem. I have no problem with part (a), but I'm having lots of troubles with part (b). I proved the base case (which is quite trivial), but I'm having trouble for the inductive step (proving k->k+1).</p>
<p><a href="https://i.stack.imgur.com/JgGfv.png" rel="nofollow noreferrer">Attempt</a></p>
<p>I don't know what to do from this point. Thank you! </p>
| marty cohen | 13,079 | <p>Just square both sides
and see what matches on each side.</p>
|
1,238,210 | <p>How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?</p>
<p>P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D</p>
| egreg | 62,967 | <p>Since
$$
e^{x^2}=1+x^2+\frac{x^4}{2!}+\dotsb
$$
we have that, for $x\ge0$, $e^{x^2}\ge1+x^2$. So
$$
\int_{0}^{x}e^{t^2}\,dt\ge\int_{0}^x(1+t^2)\,dt=x+\frac{x^3}{3}
$$
Can you finish?</p>
|
1,238,210 | <p>How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?</p>
<p>P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D</p>
| zhw. | 228,045 | <p>Or just use $e^{x^2} \ge 1$ on $[0,\infty)$ to see $\int_0^x e^{t^2}dt \ge x \to \infty.$</p>
|
64,643 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number">$a^{1/2}$ is either an integer or an irrational number</a> </p>
</blockquote>
<p>I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?</p>
| Ragib Zaman | 14,657 | <p>Suppose $\sqrt{3} = a/b$ where $a$ and $b$ have no common factor (and note $b\neq 1$). Then $ 3 = a^2/b^2$, but $a^2$ and $b^2$ no common factors to cancel to produce an integer, so we have a contradiction.</p>
|
64,643 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/4467/a1-2-is-either-an-integer-or-an-irrational-number">$a^{1/2}$ is either an integer or an irrational number</a> </p>
</blockquote>
<p>I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?</p>
| marty cohen | 13,079 | <p>Another variation on a theme:</p>
<p>If $\sqrt 3 = m/n$, where $n$ is as small as possible, then
$$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1}
= \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$
and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).</p>
<p>This can be used to show (IIRC) that $\sqrt k$ is irrational for any non-square k by multiplying $\sqrt k$ by $\frac{\sqrt k - j}{\sqrt k - j}$ where $j = \lfloor \sqrt k \rfloor$.</p>
|
12,057 | <p>Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?</p>
| Alex B. | 3,212 | <p>The answer is clearly no: take $p=\alpha$, $q=\beta$. Then $p\beta+q\alpha=2pq$ and $q\beta=pq$.</p>
<p>If however $\beta$ is co-prime to $q$, then the answer is equally clearly yes, since no divisor of $q$ divides $p\beta$ and no divisor of $\beta$ divides $q\alpha$.</p>
|
12,057 | <p>Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?</p>
| TCL | 3,249 | <p>No, e.g. $p=3,q=5,\alpha=3,\beta=5$ is a counterexample.</p>
|
4,519,106 | <p>After I learned about the existence of such a concept as a contrapositive, I always try to translate any statements into a contrapositive. And every time I fail. I haven't found a general technique for this yet. I think that if I know the statement and its contrapositive form, it will give me a better understanding.</p>
<p>The statement :<span class="math-container">$$A\subset B \implies \min A \geq \min B.$$</span></p>
<p>Its contrapositive form must look like <span class="math-container">$$\neg (\min A\geq \min B)\implies \neg (A\subset B).$$</span></p>
<p>The first thing which comes to mind is <span class="math-container">$$\min A<\min B \implies B\subseteq A.$$</span></p>
<p>On the other hand I think I must write down what I mean by <span class="math-container">$A\subset B$</span> and I get</p>
<p><span class="math-container">$$\neg (\min A\geq \min B)\implies \neg (x\in A \implies x\in B ).$$</span></p>
<p>And then I get <span class="math-container">$$\min A<\min B \implies x\in A, x\notin B$$</span> which looks very strange.</p>
<p><strong>Question</strong>: What is the correct way to do this?</p>
| ryang | 21,813 | <p>Elaborating on Nitin’s answer: <span class="math-container">$A$</span> isn’t a proper subset of <span class="math-container">$B$</span> means precisely that<br><em>EITHER</em> some element of <span class="math-container">$A$</span> isn’t in <span class="math-container">$B\;$</span> <em>OR</em> <span class="math-container">$\;A=B,$</span> that is, <span class="math-container">$$⊄\;\iff \\\big(A=B\quad\lor\quad \exists x{\in}A\; x\not\in B \big). $$</span></p>
|
308,329 | <p>I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.</p>
<p>My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?</p>
<p>I know that the answer should be:</p>
<p>$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$</p>
<p>Please explain how to do this.</p>
<p>Thank you!</p>
| muzzlator | 60,855 | <p>$$\begin{align}\sin^4 \theta &= (\sin^2\theta)^2\\ &= \left(\frac12-\frac12\cos(2\theta)\right)^2\\ &= \frac14 \left(1 - \cos(2\theta)\right)^2\\ &= \frac14\left(1 - 2 \cos(2\theta) + \cos^2(2 \theta)\right)\\ &= \frac14\left(1 - 2 \cos(2 \theta) + \frac12(\cos (4\theta) + 1)\right)\\ &= \frac14\left(\frac32 - 2\cos(2\theta) + \frac12\cos(4 \theta)\right)\\ &= \frac38 - \frac12\cos(2\theta) + \frac18\cos(4\theta).\end{align}$$</p>
|
308,329 | <p>I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.</p>
<p>My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?</p>
<p>I know that the answer should be:</p>
<p>$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$</p>
<p>Please explain how to do this.</p>
<p>Thank you!</p>
| Barbara Osofsky | 59,437 | <p>Hint: Start by noting $\sin^4 (\theta)=\left(\sin^2(\theta)\right)\cdot\left(\sin^2(\theta)\right)$. Then use the double angle formula derived by $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)=1-2\sin^2 (\theta)$ so $$\sin^2(\theta)={1\over 2}\cdot\left(1-\cos(2\theta)\right)$$ and you now plug into the factors of $\sin^4 (\theta) $, and use the double angle formula for $\cos^2(2\theta) $ to get the required answer.</p>
|
1,807,479 | <blockquote>
<p>I recently took a test and was confused about a question. I feel that
the answer is B. Could anyone please elucidate it. Thanks!</p>
</blockquote>
<p>The point $(−4, 3)$ is on the terminal side of angle $\theta$ as sketched below. Find $\cos\theta$.</p>
<p><a href="https://i.stack.imgur.com/BiOiI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BiOiI.png" alt="terminal_side_of_angle"></a></p>
<p>A. $-(4/5)$</p>
<p>B. $-(3/4)$</p>
<p>C. $-(\sqrt{2}/2)$</p>
<p>D. $(\sqrt{3}/2)$</p>
<p>E. $(4/5)$</p>
| N. F. Taussig | 173,070 | <p>The number you computed is $\tan\theta = -\frac{3}{4}$. If an angle is in standard position (vertex at the origin, initial side on the positive $x$-axis) and $(x, y)$ is the point where the terminal side of the angle intersects the circle with radius $r$ with center at the origin, then
\begin{align*}
\sin\theta & = \frac{y}{r} & \csc\theta & = \frac{r}{y}\\[2mm]
\cos\theta & = \frac{x}{r} & \sec\theta & = \frac{r}{x}\\[2mm]
\tan\theta & = \frac{y}{x} & \cot\theta & = \frac{x}{y}
\end{align*}</p>
<p>With that in mind, consider the figure below.</p>
<p><a href="https://i.stack.imgur.com/t6A6n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t6A6n.jpg" alt="terminal_side_of_second-quadrant_angle_intersects_circle_with_radius_5"></a></p>
<p>By the Pythagorean Theorem, the point $(-4, 3)$ lies on a circle with radius $5$ and center at the origin. Hence, we have $x = -4$, $y = 3$, and $r = 5$. Therefore,
$$\cos\theta = \frac{x}{r} = -\frac{4}{5}$$</p>
|
28,877 | <p>Since I self-study mathematical analysis without <em>formal</em> teacher, I can only appeal to help from out site most of the time. It's obvious that to grasp the underlying concepts in mathematics, we must roll the sleeves and solve problems.</p>
<p>It's clear that there are actually mistakes and misunderstanding that are too subtle for me to recognize, so it's very natural for me to ask for proof verifications.</p>
<p>Even though I tried to write my proofs as detailed and clear as possible, they seem to attract little attention from other users. It seems to me that proof checking is a boring and tedious job, but it is essential for me (and possibly for all of us) to know whether and where I get wrong.</p>
<p>How can I make my post for proof verification more attractive and consequently attract more attention?</p>
<p>Below are questions that i have not received any answer. Most of them are related to Cantor-Bernstein-Schröder theorem. It would be great if someone can help me improve them so that they get an answer. Thank you so much!</p>
<p><a href="https://math.stackexchange.com/questions/2813526/is-this-a-mistake-in-the-proof-of-halls-marriage-theorem-from-https-proofwiki">Is this a mistake in the proof of Hall's Marriage Theorem from https://proofwiki.org?</a></p>
<p><a href="https://math.stackexchange.com/questions/2748266/top-down-and-bottom-up-proofs-of-a-lemma-used-to-prove-cantor-bernstein-schr%C3%B6der">Top-down and Bottom-up proofs of a lemma used to prove Cantor-Bernstein-Schröder theorem and their connection</a> (this is the question that i would like to receive answer most)</p>
<p><a href="https://math.stackexchange.com/questions/2759971/is-my-proof-of-cantor-bernstein-schr%C3%B6der-theorem-correct">Is my proof of Cantor-Bernstein-Schröder theorem correct?</a></p>
<p><a href="https://math.stackexchange.com/questions/2751389/bottom-up-proof-of-a-lemma-used-to-prove-bernstein-schr%C3%B6der-theorem">Bottom-up proof of a lemma used to prove Bernstein-Schröder theorem</a></p>
<p><a href="https://math.stackexchange.com/questions/2749527/julius-k%C3%B6nigs-proof-of-schr%C3%B6der-bernstein-theorem">Julius König's proof of Schröder–Bernstein theorem</a></p>
| Arnaud Mortier | 480,423 | <p>Sometimes how attractive a question is is also a matter of luck, of who is connected at the time you ask. But in general here are a few tips:</p>
<ul>
<li>Avoid asking ten questions in one (I've seen that)</li>
<li>Try to <em>emphasize the point</em> - even if the context requires some terminology and perhaps an unavoidable long text, you can always <em>give a clear and short introduction</em> where people will see immediately if they can answer or not.</li>
<li><p>If it <em>has</em> to be long, try to be as structured as possible, using <code>- bullet point lists</code>, <code>**bold**</code> and <code>*emphasized text*</code>,
<code>## Titles ##</code>, <code>> quotes</code>
etc.</p></li>
<li><p>Offer a bounty (if the question really needs efforts from the answerer).</p></li>
<li><p>Of course, explain why/where what you have tried has failed.</p></li>
<li><p>Explain briefly the points that you have understood (I've seen someone say that they had understood why the probability of something was 13, which was quite helpful in answering)</p></li>
</ul>
|
2,464,890 | <p>Here is link to some limit questions:</p>
<p><a href="https://i.stack.imgur.com/2rM9f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2rM9f.png" alt="Example" /></a>
Can anyone explain how has answers were derived? In (a), how can we cancel out <span class="math-container">$(x-2)$</span>? And how can answer be 0? When <span class="math-container">$x\to 2$</span>, <span class="math-container">$x-2\to 0$</span> and the answer should be infinity. Similarly in (b) the answer should be infinity. Can anyone explain?</p>
| Tanmay | 687,415 | <p>If there be a given limit like the one given underneath:</p>
<p><span class="math-container">$$\lim_{x \to 2}~ \frac{x^2-4}{x-2}$$</span></p>
<p>As it can be seen that as the denominator tends to <span class="math-container">$~0~$</span> the limit approaches to infinity,and therefore we cancel out both <span class="math-container">$~(x-2)~$</span> from numerator as well as from denominator.</p>
<p>The final result comes like this:</p>
<p><span class="math-container">$$\lim_{x \to 2} (x+2)$$</span></p>
<p>Which can be easily simplified by putting the limit value,i.e.,<span class="math-container">$~2~$</span></p>
<p>But my question arises out from the point that "is it correct to cancel out <span class="math-container">$~(x-2)~$</span> from both numerator and denominator ?", as
<span class="math-container">$$x-2\implies 2-2\implies 0$$</span>
and we can't cancel out <span class="math-container">$~\left(\frac{0}{0}\right)~$</span> as it is an intermediate form. And if this be the case then we can also prove that <span class="math-container">$~\left(\frac{0}{0}\right)~$</span> is <span class="math-container">$~\left(\frac{2}{3}\right)~$</span>,<span class="math-container">$~2~$</span>,<span class="math-container">$~\left(\frac{1}{3}\right)~$</span> etc.
<span class="math-container">$$\frac{10(10-10)}{30(10-10)}\implies \frac{10}{30}\implies \frac{1}{3}$$</span>
Here <span class="math-container">$~(10-10)~$</span> gets cancelled out both from both numerator as well as from the denominator just like the previous problem where <span class="math-container">$~\frac{x-2}{x-2}~$</span> got cancelled out from the numerator as well as from the denominator.</p>
<p>Note:I am just a class <span class="math-container">$11$</span> student,and that also a beginner,so I might be wrong at somewhere,so please help me to figure it out and let me rectify my mistake...</p>
<p>My question </p>
|
3,865,954 | <p>Suppose that I have the following sum:
<span class="math-container">$\sum_{m=0}^{\infty}(e^{it}(1-p))^{m}$</span>, where <span class="math-container">$i^2 = -1$</span>.</p>
<p>This is a geometric series, but involving the complex number <span class="math-container">$i$</span>. Can I just apply the geometric series formula and conclude that the sum if <span class="math-container">$\frac{1}{1-e^{it}(1-p)}$</span>?</p>
<p>Thank you,</p>
| J.G. | 56,861 | <p>To prove that if <span class="math-container">$|z|<1$</span> then <span class="math-container">$\sum_{m\ge0}z^m=\tfrac{1}{1-z}$</span>, note that <span class="math-container">$\tfrac{1}{1-z}-\sum_{m=0}^{n-1}z^m=\frac{z^n}{1-z}$</span> has <span class="math-container">$n\to\infty$</span> limit <span class="math-container">$0$</span>. (By contrast, if <span class="math-container">$|z|\ge1$</span> then this <span class="math-container">$n\to\infty$</span> behaviour of the error term is lost, so the series diverges.) Note nothing about this reasoning cares whether <span class="math-container">$z$</span> is real or complex.</p>
|
1,807,456 | <p>I don't know asymptotic behaviour of the integral $$\int_{0}^{\infty}\frac{du}{\sqrt{4\pi u^{3}}}\left(1-\frac{e^{-\Omega u}}{\sqrt{\frac{1-\exp\left(-2u\right)}{2u}}}\right),$$
when I read a physics paper. It says that the integral have asymptotic behaviour $\log\left(\pi\Omega B\right)/\sqrt{2\pi}$, when $\Omega\to 0$. But most of paper about asymptotic behaviour are talking about the Laplace’s method, which is not match this integral. So I want to ask to get the asymptotic behaviour of the integral.</p>
<p>Thank you for your reading.</p>
| tired | 101,233 | <p>Please view this as a supplement to @Jack's Answer which avoids the use of special functions and adds another constant contribution which might explain the differences between numerics and former asymptotic calculations.</p>
<p>To make the analysis simpler, let us rescale $\Omega u=x$ to obtain (we drop the $4\pi$ troughout and set $\Omega>0$)</p>
<p>$$
I(\Omega)=\sqrt{\Omega}\int_0^{\infty}\frac{1}{x^{3/2}}\left[1-\sqrt{\frac{2 x}{\Omega}}\frac{e^{-x}}{\sqrt{1-e^{-2x/\Omega}}}\right]
$$</p>
<p>the only lengthscale in this problem is determined by $x_1=\Omega/2$ so it is reasonable to splint the range of integration at $x_1$. We call the resulting Integrals $I_<$ and $I_>$. </p>
<p>Starting with the first, we may replace the exponents by their Taylor approximants. We get </p>
<p>$$
I_<(\Omega)\sim \sqrt{2} \int_0^{\Omega/2}\sum_{n\geq 0}c_n\frac{x^{n-1/2}}{\Omega^{n+1/2}}+\mathcal{O}\left(\sqrt{\Omega}\right)
$$</p>
<p>with fastly decaying coefficents $c_n$ ($c_0=1/2,c_1=1/24,c_2=1/48,c_3=1/640$...). Here and throughout we mean by $\sim$ asymptotically equal in the limit $\Omega \rightarrow 0_+$.</p>
<p>Obviously upon integration we get a constant contribution (C is defined through the series above)</p>
<blockquote>
<p>$$
\color{\green}{I_<=C+\mathcal{O}\left(\sqrt{\Omega}\right)}
$$</p>
</blockquote>
<p>Now turning to $I_>$ we may expand ind powers of $e^{-2x/\Omega}$, yielding</p>
<p>$$
I_>(\Omega)\sim \sqrt{\Omega}\int_{\Omega/2}^{\infty}\left[\frac{1}{x^{3/2}}-\sqrt{\frac{2 }{\Omega}}\frac{e^{-x}}{x}\left(1+\mathcal{O}\left(e^{-2 x/\Omega}\right)\right)\right]
$$</p>
<p>the first part of this integral gives a constant contribution namely $2\sqrt{2}$. Furthermore it holds that</p>
<p>$$
-\int_{\Omega/2}^{\infty}\frac{e^{-x}}{x}\sim\log(\Omega/2)+\gamma+\mathcal{O}(\Omega)
$$</p>
<p>which can be shown by integrating the Taylor series of $e^{-x}/x$ and using the representation $\gamma = \int_0^{\infty}dx\left[\frac{1}{e^x x}-\frac{1}{e^x -1}\right]$ of the Euler constant (Note that this series not converge in a classical sense, it is a truely asymptotic power series). Therefore</p>
<blockquote>
<p>$$
\color{\red}{I_>\sim \sqrt{2}\log(\Omega)+\sqrt{2}( \gamma-\log(2)+2)+\mathcal{O}(\Omega)}
$$</p>
</blockquote>
<p>and last but not least </p>
<blockquote>
<p>$$
I(\Omega)=\color{\red}{I_>}+\color{\green}{I_<}\sim \sqrt{2} \log( \Omega)+C'+\mathcal{O}(\sqrt{\Omega})
$$</p>
</blockquote>
<p>with $C'=C+\sqrt{2}( \gamma-\log(2)+2)$. Note that this seems to match the other answer expect that we included another constant contribution from the region of integration around zero</p>
<p>from which we can obtain an value for your constant $B$. Since $C\approx 1/2$, $e^C\approx 1.6$ which seems to close the gap a bit between Jack's answer and numerical predictions</p>
<p>PS: Because <em>Mathematica</em> is not avaiable at the moment, i can't crosscheck my results properly so please forgive me some minor errors!</p>
|
2,612,308 | <p>Obviously we can rearrange for <span class="math-container">$x$</span> in a polynomial of degree 2. </p>
<p>Let <span class="math-container">$y=ax^2+bx+c$</span></p>
<p>then </p>
<p><span class="math-container">$x=\frac{-b\pm\sqrt{b^2-4ac+4ay}}{2a}$</span></p>
<p>Similarly, for <span class="math-container">$y=ax^3+bx^2+cx+d$</span>, although it is very difficult and long, there apparently also exists a way to make <span class="math-container">$x$</span> the subject. </p>
<p>Now I'm wondering whether it is always possible to make <span class="math-container">$x$</span> the subject when <span class="math-container">$y=p_n(x)$</span>, where <span class="math-container">$p_n(x)$</span> is any polynomial of degree <span class="math-container">$n$</span>.</p>
<p>If so, is it always possible to make <span class="math-container">$x$</span> the subject when <span class="math-container">$y=f(x)$</span>, where <span class="math-container">$f(x)$</span> is any function of <span class="math-container">$x$</span>. </p>
<p>And lastly, is there always an exact way to get a desired expression on one side of an equation, obviously still being equivalent to the initial one. If this sounds vague, here is the equation that got me thinking about this:</p>
<p><span class="math-container">$y^3+x^3=3xy$</span></p>
<p>is there a way to make <span class="math-container">$x$</span> the subject?</p>
| Martin Argerami | 22,857 | <p>You cannot expect in general to be able to solve for $x$. For instance, consider
$$
x^5-4x+2=0.
$$
One can easily show, using calculus (or by just plotting) that it has three real roots. One can, however, <strong>prove</strong> using <a href="https://en.wikipedia.org/wiki/Galois_theory" rel="nofollow noreferrer">Galois Theory</a> that no formula exists (that is sums, products, power, roots) that expresses the roots in terms of the coefficients. </p>
<p>And the above is only for polynomial equations. There are many other equations, like <em>trascendental</em> ones, where no closed form solution exists. Example: $x=e^x$. </p>
|
2,647,194 | <p>show that $p(x)=x^3-x^2-4x+5$ is irreducible in $\mathbb{Q}[x]$ </p>
<p>How do we decide if a polynomial $p (x)$ in $\mathbb{Q}[x]$ is irreducible?</p>
| Dietrich Burde | 83,966 | <p>For a non-constant polynomial $f(x)$ of degree $n\le 3$ over a field we know that $f(x)$ is irreducible if and only if it has no root. This can be decided by the <a href="https://en.wikipedia.org/wiki/Rational_root_theorem" rel="nofollow noreferrer">rational root test</a>. The divisors of $5$ are $\pm 1$ and $\pm 5$, and so it is easy to see that $f(x)$ is irreducible.</p>
|
1,234,093 | <p>Given that $ e= \frac{a^2-b^2}{b^2} $ , and $L$ is the length of the perimeter, which equals $4aE(e, \pi/2)$, find the length of the perimeter up to $e^2$ in terms of $a$ and $b$.</p>
<p>How does one begin this?</p>
| Jack D'Aurizio | 44,121 | <p>If our ellipse is given by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, its length is given by:
$$ L(a,b)=4a\int_{0}^{\pi/2}\sqrt{1+e^2\sin^2\theta}\,d\theta. \tag{1}$$
Since:
$$ \int_{0}^{\pi}\sqrt{1+x^2+2x\cos(2\theta)}\,d\theta = \pi\sum_{n\geq 0}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2 x^{2n}\tag{2} $$
if we define the elliptic parameter $\lambda$ as $\frac{a-b}{a+b}$ it follows that:
$$ L(a,b) = \pi(a+b)\left(1+\frac{1}{4}\sum_{n\geq 0}\left(\frac{1}{(n+1)4^n}\binom{2n}{n}\right)^2 \lambda^{2n+2}\right).\tag{3}$$
If we define the mean of order $\alpha>0$ between $a$ and $b$ as:
$$ M_{\alpha}(a,b)=\left(\frac{a^\alpha+b^{\alpha}}{2}\right)^{1/\alpha} $$
two classical inequalities about the ellipse length are given by Muir (1883) and Alzer-Qiu (2004):</p>
<blockquote>
<p>$$ M_{\frac{3}{2}}(a,b)\leq \frac{L(a,b)}{2\pi}\leq M_{\frac{\log 2}{\log(\pi/2)}}(a,b)\tag{4} $$</p>
</blockquote>
<p>and since $\frac{\log 2}{\log(\pi/2)}=1.5349285\ldots$ is pretty close to $\frac{3}{2}$, for low-eccentricity ellipses $2\pi$ times the $\frac{3}{2}$-th mean of $a,b$ is an eccellent approximation for the length.</p>
|
2,138,916 | <p>My question read: </p>
<p>Show that $S_{10}$ contains elements of orders $10,20,30$. Does it contain an element of order $40$? </p>
<p>I am not too sure what the question is asking. Would I have to explicitly write out all the permutations in $S_{10}$ first and then find the orders for all of them? </p>
<p>Update: I understand I need to only show a few examples of disjoint cycles, but I am not sure how to show if order 40 is possible.</p>
| Ron Gordon | 53,268 | <p>Suppose $a \gt b$ for now. Consider the contour integral in the complex plane</p>
<p>$$\oint_C dz \frac{\log{\left ( z^2+a^2 \right )}}{z^2+b^2} $$</p>
<p>where $C$ is a semicircle of radius $R$ in the upper half-plane with a detour down and up the imaginary axis about the branch point $z=i a$. In the limit as $R \to \infty$, the contour integral is equal to</p>
<p>$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} + i \int_{\infty}^a dy \frac{\log{\left ( y^2-a^2 \right )}+i \pi}{b^2-y^2} + i \int_a^{\infty} dy \frac{\log{\left ( y^2-a^2 \right )}-i \pi}{b^2-y^2}$$</p>
<p>Note that the log terms in the latter two integrals vanish. Now, the contour integral is also equal to the residue of the pole of the integrand at $z=i b$. Thus</p>
<p>$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} - 2 \pi \int_a^{\infty} \frac{dy}{y^2-b^2} = i 2 \pi \frac{\log{\left ( a^2-b^2 \right )}}{i 2 b} $$</p>
<p>Accordingly, after doing out that second integral and performing a little algebra, we get...</p>
<blockquote>
<p>$$\frac12 \int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} = \frac{\pi}{b} \log{\left ( a+b \right )} $$</p>
</blockquote>
<p><strong>ADDENDUM</strong></p>
<p>For $a \lt b$, the answer is the same as above but the contour is altered. This time, the contour $C$ must detour about the pole at $z=i b$ along each side of the branch cut with a semicircle of radius $\epsilon$. The contour integral is this equal to</p>
<p>$$\int_{-\infty}^{\infty} dx \frac{\log{\left ( x^2+a^2 \right )}}{x^2+b^2} + i PV \int_{\infty}^a dy \frac{\log{\left ( y^2-a^2 \right )}+i \pi}{b^2-y^2} + i PV \int_a^{\infty} dy \frac{\log{\left ( y^2-a^2 \right )}-i \pi}{b^2-y^2} \\ + i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{\log{\left [- \left (i b + \epsilon e^{i \phi} \right )^2-a^2 \right ]}+i \pi}{\left (i b + \epsilon e^{i \phi} \right )^2+b^2}+ i \epsilon \int_{3 \pi/2}^{\pi/2} d\phi \, e^{i \phi} \frac{\log{\left [ -\left (i b + \epsilon e^{i \phi} \right )^2-a^2 \right ]}-i \pi}{\left (i b + \epsilon e^{i \phi} \right )^2+b^2}$$</p>
<p>Note that the sum of the two final integrals - the pieces that go around the pole - is equal to the residue of the pole at $z=i b$ in the limit as $\epsilon \to 0$. The $\pm i \pi$ pieces cancel. Also note that the principal value integrals are the same as the corresponding integrals above for $a \gt b$. Thus, the result for $a \lt b$ is the same as that for $a \gt b$.</p>
|
1,454,344 | <p>Does span=(2,-1,1,2), (-2,1,-1,-2) represent a line, plane or hyperplane in R4?</p>
<p>We haven't learned matrices yet either </p>
| jimbo | 115,363 | <p>How $(2,-1,1,2)=-(-2,1,-1,-2)$ are dependent, then represent a line in $\mathbb{R}^4$</p>
|
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