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 |
|---|---|---|---|---|
885,450 | <blockquote>
<p>After covering a distance of 30Km with a uniform speed, there got some
defect in train engine and therefore its speed is reduced to 4/5 of
its original speed. Consequently, the train reaches its destination 45
minutes late. If it had happened after covering 18Km of distance, the
train would have reached 9 minutes earlier>Find the speed of the train
and the distance of the journey.</p>
</blockquote>
<p>This is my question. .</p>
<p>now after letting the original speed of train be $x Km/Hr$ and the time taken be y Hr, threfore
distance = $xy$</p>
<p>CASE I:</p>
<p>speed = $x-4x/5$ Km/hr</p>
<p>time = $y + 45/60$</p>
<p>$xy=60xy/300 +45x/300 => +300xy-60xy = 45x => 240xy = 45x$ .............[i]</p>
<p>similarly in CASE II we get the equation:</p>
<p>$240xy = -9x$.........[ii]</p>
<p>but after solving these two equations the answer is coming to 0 which is wrong please tell me the correct solution.</p>
<p>thanks</p>
<p>(fast please)</p>
| MCT | 92,774 | <p>Wolfram can actually find it: <a href="http://www.wolframalpha.com/input/?i=lim%28n+to+infinity%29+prod%28k+%3D+0+to+%282%5E%282%5En+-+1%29+%2B+1%29%29+%281+%2B+1%2F%282%5E2%5En+%2B+2k%29%29" rel="nofollow">http://www.wolframalpha.com/input/?i=lim%28n+to+infinity%29+prod%28k+%3D+0+to+%282%5E%282%5En+-+1%29+%2B+1%29%29+%281+%2B+1%2F%282%5E2%5En+%2B+2k%29%29</a> in all it's glory.</p>
<p>I find that condensing the search formula for Wolfram and not relying on its pattern recognition tends to get you better results. In your search query, it probably thought that the top and bottom was identical when finding a pattern.</p>
|
105,040 | <p>This question in stackExchange remained unanswered. </p>
<p>Let $\mathbb F$ be a finite field. Denote by $M_n(\mathbb F)$ the set of matrices of order $n$ over $\mathbb F$ . For a matrix $A∈M_n(\mathbb F)$ what is the cardinality of $C_{M_n(\mathbb F)} (A)$ , the centralizer of $A$ in $M_n(\mathbb F)$? There are papers about it? </p>
| Alexander Chervov | 10,446 | <p>This is just elementary comment, but may be too long for comment, probably you know it, but just for completeness.</p>
<p>I happened to ask almost (but not exactly) the same question some days ago: </p>
<p><a href="https://mathoverflow.net/questions/104439/conjugcy-classes-in-glf-2-glf-q">Conjugcy classes in GL(F_2) ? GL(F_q)</a></p>
<p>It sounds a little different - size of conjugacy classes in GL(F_q).
But conjugacy class of element "C" is size of GL(F_q)/ size of centralizer of element "C".
Because size of any orbit is |G|/|Stabilizer| and here we have action by conjugation and stabilizer
is centralizer.
So if you know the size of centralizer - you know size of conjugacy class.</p>
<p>The difference is, of course, that you asked about Mat(F_q) while setup above is about non-degenerate matrices GL(F_q).</p>
<hr>
<p>Let me also write down some elementary facts for completeness.</p>
<p>If you consider element $C$ such that its characteristic polynomial is irreducible, (then it automatically minimal), then size of centralizer in Mat_n(F_q) is q^n (if I understand correctly)
and q^n-1 in GL_n(F_q).</p>
<p>One has good way to think about this centralizer: let consiser p(x) - char.pol. of "C".
F_q[x]/p(x) is a field F_q^n.
To any element of the field "a" in F_q^n one can correspond a matrix M(a) - the matrix of multiplication by "a":F_q^n -> F_q^n.</p>
<p>Map "a->M(a)" is clearly homomorphism of algebras so in particular all matrices M(a) commute among themselves. They provide a centralizer of a matrix M(x). </p>
|
3,846,717 | <p>Denote <span class="math-container">$\mathbb{F}=\mathbb{C}$</span> or <span class="math-container">$\mathbb{R}$</span>.</p>
<p><strong>Theorem (Cauchy - Schwarz Inequality).</strong> <em>If <span class="math-container">$\langle\cdot,\cdot\rangle$</span> is a semi-inner product on a vector space <span class="math-container">$H$</span>, then</em> <span class="math-container">$$\lvert\langle x,y\rangle\rvert\le\lVert x\rVert\lVert y\rVert,\quad\textit{for all}\;x,y,\in H.$$</span></p>
<p><em>Proof.</em> If <span class="math-container">$x=0$</span> or <span class="math-container">$y=0$</span>, then there is nothing to prove, so suppose that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are both nonzero.</p>
<p>Given any scalar <span class="math-container">$z\in\mathbb{F}$</span>, there is a scalar <span class="math-container">$\alpha$</span> with modulus <span class="math-container">$\lvert\alpha\rvert=1$</span> such that <span class="math-container">$\alpha z=\lvert z \rvert$</span>. In particular, if we set <span class="math-container">$z=\langle x, y\rangle$</span>, then there is a scalar with <span class="math-container">$\lvert \alpha \rvert=1$</span> such that <span class="math-container">$$\langle x,y\rangle=\alpha\lvert\langle x,y\rangle\rvert.$$</span> Multiplying both sides by <span class="math-container">$\overline{\alpha}$</span>, we see that we also have <span class="math-container">$\overline{\alpha}\langle x, y\rangle=\lvert\langle x, y \rangle \rvert$</span>.</p>
<p>For each <span class="math-container">$t\in\mathbb{R}$</span>, using the Polar Identity and antilinearity in the second variable, we compute that
<span class="math-container">\begin{equation}
\begin{split}
0\le\lVert x-\alpha ty\rVert= &\rVert x\lVert^2-2\Re\big(\langle x,\alpha ty\rangle\big)+t^2\rVert y \lVert^2 \\
= &\rVert x\lVert^2-2t\Re\big(\overline{\alpha}\langle x, y\rangle\big)+t^2\lVert y \lVert^2\\
= & \lVert x \rVert ^2-2t\lvert\langle x, y\rangle\rvert+t^2\lVert y \rVert^2\\
= & at^2+bt+c,
\end{split}
\end{equation}</span>
where <span class="math-container">$a=\lVert y \rVert ^2$</span>, <span class="math-container">$b=-2\lvert \langle x, y\rangle \rvert$</span>, and <span class="math-container">$c=\lVert x \rVert ^2$</span>. This is a rel-valued quadratic polynomial in the variable <span class="math-container">$t$</span>. Since this polynomial is nonnegative, <span class="math-container">$\color{red}{it\;can\;have\;at\;most\;\;one\;real\;root}.$</span></p>
<p><span class="math-container">$\color{blue}{This\;implies\;that\;the\;discriminant\; cannot\;be\;strictly\;positive}.$</span></p>
<blockquote>
<p><strong>Question.</strong> What are the reasons why the red and blue assertions hold? I need the precise details.</p>
</blockquote>
<p>Thanks!</p>
| user | 505,767 | <p>The statements are related to the quadratic equation</p>
<p><span class="math-container">$$ax^2+bx+c=0 \implies x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$</span></p>
<p>since when <span class="math-container">$b^2-4ac>0$</span> then <span class="math-container">$\sqrt{b^2-4ac}$</span> exists and we always have two distinct real solutions.</p>
<p>Moreover for the quadratic function</p>
<p><span class="math-container">$$f(x)=ax^2+bx+c$$</span></p>
<p>which represents a parabola, when <span class="math-container">$f(x)=0$</span> has two solutions then <span class="math-container">$f(x)$</span> takes positive and negative values.</p>
|
300,745 | <p>If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?</p>
<p>($a$ and $b$ being finite numbers).</p>
<p>I tried proving and disproving it. Couldn't find an example for a non-bounded image. </p>
<p>Is there any basic proof or counter example for any of the cases?</p>
<p>Thanks a million!</p>
| Ittay Weiss | 30,953 | <p>Hint: Prove first that a uniformly continuous function on an open interval can be extended to a continuous function on the closure of the interval. </p>
|
300,745 | <p>If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?</p>
<p>($a$ and $b$ being finite numbers).</p>
<p>I tried proving and disproving it. Couldn't find an example for a non-bounded image. </p>
<p>Is there any basic proof or counter example for any of the cases?</p>
<p>Thanks a million!</p>
| David Mitra | 18,986 | <p>If, for instance, $\limsup\limits_{x\rightarrow a^+} f(x)=\infty$, then given any $\delta>0$, one could find $x$ and $y$ with $a<x<\delta$ and $a<y<\delta$ such that $|f(x)-f(y)|>1$. Could $f$ then be uniformly continuous on $(a,b)$?</p>
<p>Note if $f$ is continuous on $(a,b)$ and unbounded, then either $f$ is "unbounded at $a$" or "unbounded at $b$" (aside from the above, there are three other cases to consider). </p>
|
741,436 | <p>I get stuck at the following question:</p>
<p>Consider the matrix<br>
$$A=\begin{bmatrix}
0 & 2 & 0 \\
1 & 1 & -1 \\
-1 & 1 & 1\\
\end{bmatrix}$$</p>
<p>Find $A^{1000}$ by using the Cayley-Hamilton theorem.</p>
<p>I find the characteristic polynomial by $P(A) = -A^{3} + 2A^2 = 0$ (by Cayley-Hamilton) but I don't see how to find $A^{1000}$ by this characteristic polynomial.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$$
\expo{At}=\alpha\pars{t} + \beta\pars{t}A + \gamma\pars{t}A^{2}
$$</p>
<blockquote>
<p>$$\dot{\alpha}\pars{t} + \dot{\beta}\pars{t}A + \dot{\gamma}\pars{t}A^{2}
=A\expo{At}
=\alpha\pars{t}A + \beta\pars{t}A^{2} + \gamma\pars{t}\ \overbrace{A^{3}}^{2A^{2}}\,,
\quad
\left\lbrace%
\begin{array}{l}
\alpha\pars{0} = 1
\\[1mm]
\beta\pars{0} = \gamma\pars{0} = 0
\end{array}\right.
$$</p>
</blockquote>
<p>$$
\dot{\alpha}\pars{t} = 0\,,\quad
\dot{\beta}\pars{t} = \alpha\pars{t}\,,\quad
\dot{\gamma}\pars{t} = \beta\pars{t} + 2\gamma\pars{t}
\quad\imp\quad
\left\lbrace%
\begin{array}{rcl}
\alpha\pars{t} & = & 1
\\[1mm]
\beta\pars{t} & = & t
\\[1mm]
\gamma\pars{t} & = & {\expo{2t} - 2t - 1 \over 4}
\end{array}\right.
$$</p>
<blockquote>
<p>$$
\expo{At} = 1 + tA + {\expo{2t} - 2t - 1 \over 4}\,A^{2}
$$</p>
</blockquote>
<p>\begin{align}
A^{1000} &= \left.\totald[1000]{\pars{\expo{At}}}{t}\right\vert_{t = 0}
=\left. {A^{2} \over 4}\,
\totald[1000]{\bracks{\expo{2t} - 2t - 1}}{t}\right\vert_{t = 0}
={A^{2} \over 4}\,2^{1000}
\end{align}
$$
\boxed{\vphantom{\Huge {A \over B}}\quad\color{#00f}{\large A^{1000} = 2^{998}\ A^{2}}\quad}
$$</p>
|
566,993 | <p>Suppose $f(z)=1/(1+z^2)$ and we want to find the power series in $a=1$. I think we have to write $1/(1+z^2)=1/(1+(z-1)+1)^2=1/(1+(1+(z-1)^2+2(z-1)))$, but I'm stuck here.</p>
| Giorgio Mossa | 11,888 | <p><strong>Edit:</strong> Andreas Caranti have pointed out, this proof work iff the all the elements of the group have finite order. So it doesn't answer completely but
is more a partial result.</p>
<p>Let $x,a \in G$ then by hypothesis we have that there's $b \in G$ such that $ab=x$ then
$$a^{-1}x^2a=a^{-1}(ab)^2a=a^{-1}ababa=baba=(ba)^2=(ab)^2=x^2$$
So $x^2 \in Z(G)$ for all $x \in G$.</p>
<p>So we get the quotient $G/Z(G)$ in which all the elements have order two.</p>
<p>If there's an $x \in G \setminus Z(G)$ then $xZ(G) \ne Z(G)$ should have order $2$, and by properties of homomorphisms of groups the order of $x$ should be divided by $2$. That's absurd since by hypothesis $G$ can't have elements of order $2$ hence it can't have either elements which have order a multiple of $2$.</p>
|
566,993 | <p>Suppose $f(z)=1/(1+z^2)$ and we want to find the power series in $a=1$. I think we have to write $1/(1+z^2)=1/(1+(z-1)+1)^2=1/(1+(1+(z-1)^2+2(z-1)))$, but I'm stuck here.</p>
| dan_fulea | 550,003 | <p>The following solution is the same as in the <a href="https://math.stackexchange.com/questions/4270671/prove-that-a-certain-group-is-abelian/4272460#4272460">duplicate</a>, just made shorter. As in the answer of <a href="https://math.stackexchange.com/users/62565/boris-novikov">Boris Novikov</a>, for every <span class="math-container">$x\in G$</span> the element <span class="math-container">$x^2$</span> is in the centralizer of <span class="math-container">$Z(G)$</span> of <span class="math-container">$G$</span>, by using
<span class="math-container">$(ab)^2=(ba)^2$</span> for <span class="math-container">$b=a^{-1}x$</span>:
<span class="math-container">$$
x^2 =(ab)^2=(ba)^2=(a^{-1}xa)^2=a^{-1}x^2a\ .
$$</span>
Let now <span class="math-container">$s,t\in G$</span> be two elements. We show <span class="math-container">$st=ts$</span>. Consider
<span class="math-container">$$a = s^{-2}t^{-2}\; stst\ .
$$</span>
One easily shows <span class="math-container">$a^2=1$</span> by using <span class="math-container">$s^{\pm 2},t^{\pm 2}\in Z(G)$</span>:
<span class="math-container">$$
\begin{aligned}
a^2
&= s^{-4}t^{-4}\cdot stst\cdot stst\\
&= s^{-4}t^{-4}\cdot stst\cdot tsts\\
&= s^{-4}t^{-4}\cdot sts\cdot t^2\cdot sts\\
&= s^{-4}t^{-2}\cdot st\cdot s^2\cdot ts\\
&= s^{-2}t^{-2}\cdot s\cdot t^2\cdot s\\
&= s^{-2}\cdot s^2\\
&=1\ .
\end{aligned}
$$</span>
From the assumption, <span class="math-container">$a=1$</span>, i.e.
<span class="math-container">$$
1=s^{-2}\; stst\; t^{-2}=s^{-1}\; ts\; t^{-1}\ ,
$$</span>
so after multiplying from left with <span class="math-container">$s$</span>, and from right with <span class="math-container">$t$</span> we get
<span class="math-container">$st=ts$</span>.</p>
<p><span class="math-container">$\square$</span></p>
|
131,179 | <p>Consider the assumptions</p>
<pre><code>$Assumptions = {Element[a,Reals], Element[z,Complexes]}
</code></pre>
<p>I'm looking for a test, to be applied on <code>a</code> and <code>z</code>, that gives <code>True</code> if the argument is a complex number such as <code>a</code> and <code>False</code> if it's real such as <code>z</code>.</p>
<p>The aim is to use this test in a replacement in this way</p>
<pre><code>set = {a, z, x};
set /. (x_ :> img /; test[x])
(* {a, img, x} *)
</code></pre>
<p>An example is</p>
<pre><code>set /. (x_ :> img /; Simplify[NotElement[x, Reals] && Element[x,Complexes]] === (NotElement[x, Reals]))
</code></pre>
<p>It is based on the fact that </p>
<pre><code>Simplify[ Element[z,Reals] ]
</code></pre>
<p>remain unevaluated.</p>
<p>Is there another possible test that doesn't rest on this (and, possibly, simpler and without <code>If</code> and similar)?</p>
| ubpdqn | 1,997 | <p>There are a number of ways, e.g.:</p>
<pre><code>f[x_] := With[{ri = ReIm /@ N[x]},
ri /. {{_, 0} :> False, {_, _} :> True}]
</code></pre>
<p>Testing:</p>
<pre><code>r = RandomReal[1, 10];
c = Complex @@@ RandomReal[1, {10, 2}];
test = Join[r, c][[RandomSample[Range@20]]];
Thread[{test, f[test]}] // Grid
</code></pre>
<p><a href="https://i.stack.imgur.com/cqfCJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cqfCJ.png" alt="enter image description here"></a></p>
|
3,006,022 | <p>I have a radioactive decay system to solve for <span class="math-container">$x(t)$</span> and <span class="math-container">$y(t)$</span> (no need for <span class="math-container">$z(t)$</span>):
<span class="math-container">$$\begin{cases}x'=-\lambda x\\
y'=\lambda x-\mu y\\
z'=\mu y\end{cases}$$</span>
with the initial conditions <span class="math-container">$x(0)=1,y(0)=0,z(0)=0$</span>.</p>
<p>I found <span class="math-container">$x(t) = e^{-\lambda t}$</span>, but <span class="math-container">$y(t)$</span> is proving to be difficult. I have tried subbing in <span class="math-container">$x(t) = e^{-\lambda t}$</span> so that we have a linear DE in <span class="math-container">$y$</span>:
<span class="math-container">$$y'+\mu y=\lambda e^{-\lambda t}$$</span>
but ended up getting a giant solution which seems very wrong. </p>
<p>Is this the right approach or no?</p>
| Robert Z | 299,698 | <p>The differential equation
<span class="math-container">$$y'+\mu y=\lambda e^{-\lambda t}$$</span>
is <a href="https://en.wikipedia.org/wiki/Linear_differential_equation#Non_homogeneous_equation_with_constant_coefficients" rel="nofollow noreferrer">linear non-homogeneous with constant coefficients</a>. The general solution is
<span class="math-container">$$y'(t)=Ce^{-\mu t}+y_p(t)$$</span>
where <span class="math-container">$C$</span> is an arbitrary constant and <span class="math-container">$y_p$</span> is a particular solution. As regards <span class="math-container">$y_p$</span> we are supposed to distinguish two cases 1) <span class="math-container">$\lambda\not=\mu$</span> and 2) <span class="math-container">$\lambda=\mu$</span>.
As a reference see the <a href="https://en.wikipedia.org/wiki/Method_of_undetermined_coefficients#Typical_forms_of_the_particular_integral" rel="nofollow noreferrer">method of undetermined coefficients</a>.</p>
<p>Once we have <span class="math-container">$y$</span>, we can solve <span class="math-container">$z'=\mu y$</span> by a straightforward integration.</p>
<p>Can you take it from here?</p>
|
3,012,416 | <p>I know the answer is obvious: In <span class="math-container">$\mathbb{Z}$</span> the only solutions of <span class="math-container">$xy=-1$</span> are <span class="math-container">$x=-y=1$</span> and <span class="math-container">$x=-y=-1$</span>.
My problem is that I want to formally prove it and I don't know how to write it. Where do you even begin for such a trivial statement?</p>
<p><strong>Edit: I would like to prove it viewing <span class="math-container">$\mathbb{Z}$</span> as a ring. This is, just using the sum and product of integers. No order, no absolute value, etc...</strong></p>
| Michael Burr | 86,421 | <p>Suppose that <span class="math-container">$xy=-1$</span>, then <span class="math-container">$-xy=1$</span>, so <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are units (by definition). Since the only units in <span class="math-container">$\mathbb{Z}$</span> are <span class="math-container">$\pm 1$</span>, we can check which combinations result in a product of <span class="math-container">$-1$</span>.</p>
|
419,176 | <p>Let $f:\left\{x\in\mathbb{R}^n\vert\parallel x\parallel<1\right\}\rightarrow\mathbb{R}$ be an one-to-one bounded continuous function.<br>
I want to construct such $f$ which is not uniformly continuous.<br><br>
In this case, I thought I can construct $f$ with a restriction $n=2$.<br>
But I'm confused because $f$ is bounded so I can't use functions like $\frac{1}{x}$ on $(0,1)$.<br>
To top that off, $f$ is even one-to-one so I gave up and now I'm writing this to get some help from you who is smarter than me.<br><br>
Please give me some help to solve this problem.<br>
Thanks.</p>
| Vishal Gupta | 60,810 | <p>At least for $n =1$, any such function will have a limit on the end points $-1,1$ in this case and hence will have an extension on the interval $[-1,1]$ and hence will be uniformly continuous. So, you cannot construct a function you want in the case $n=1$.</p>
<p>For higher dimensions, the main thing is to examine whether the limit exists on the boundary of the closed ball (in the case $n=1$ one-one and continuity implies monotonicity, which gives the limit). If there is a limit, then the function will again be uniformly continuous.</p>
|
2,871,892 | <p><a href="https://i.stack.imgur.com/XLen7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XLen7.png" alt="Q1"></a></p>
<p>Solution is 4. </p>
<p>Original matrix is simply [v1;v2;v3;v4]. It forms an identity matrix. Hence the only alteration of the determinant comes from row 1 operation where v1 is multiplied by 2. Then the determinant will also be multiplied by two so 2*2 =4. </p>
<p>We ignore the rest of row operations since they have no effect on the determinant.</p>
<p>Is this line of reasoning correct?</p>
| hmakholm left over Monica | 14,366 | <p>If you take the equation
$$ \tag{*} f(x) = x^n $$
and "set $x=0$", what you get is <em>not</em> $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).</p>
<p>As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $\text{(*)}$ is <strong>simply not true</strong>. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).</p>
|
204,106 | <p>I would like to know what the definition of a short proof is.</p>
<p>In Lance Fortnow’s article “<a href="http://cacm.acm.org/magazines/2009/9/38904-the-status-of-the-p-versus-np-problem/fulltext" rel="nofollow">The Status of the P Versus NP Problem</a>”, Communications of the ACM, Vol. 52 No. 9, he says,</p>
<blockquote>
<p>If a formula θ is not a tautology, we can give an easy proof of that fact by exhibiting an assignment of the variables that makes θ false. But if θ were indeed a tautology, we don’t expect short proofs. If one could prove there are no short proofs of tautology that would imply P ≠ NP.</p>
</blockquote>
<p>I have tried to find a definition of a “short proof”, but have not been able to.</p>
| zeb | 2,363 | <p>Allow me to muddy the waters a little bit. Before we start making statements about the lengths of proofs, we should first formally define what a proof <em>is</em>. For that, we want the concept of a <em>proof system</em>.</p>
<h2>What is a proof system?</h2>
<p>A proof system is a Turing machine which runs in polynomial time and defines a function $f$ taking ordered pairs $(\varphi,p)$, where $\varphi$ is a formula and $p$ is some binary string purporting to be a proof of $\varphi$, to the set $\{V,I\}$ ($V$ for Valid, $I$ for Invalid). It is <em>sound</em> if it never returns $V$ unless $\varphi$ is actually a tautology, and it is <em>complete</em> if for every tautology $\varphi$ there is a proof $p$ with $f(\varphi,p) = V$.</p>
<h2>How do I check whether a proof system is sound?</h2>
<p>In general you can't, by an easy reduction to the Halting Problem. (Start with a sound proof system, and modify it by first simulating some other program $X$ for $n$ steps, where $n$ is the length of the binary representation of $\varphi$. Output $V$ if $X$ halts within $n$ steps, otherwise run the original sound proof system.)</p>
<h2>Is there a "best" sound proof system?</h2>
<p>Nobody knows? If I am interpreting <a href="http://people.cs.uchicago.edu/~razborov/files/php_survey.pdf" rel="nofollow">this paper</a> correctly, the naive proof system - ordinary proofs in propositional logic - has trouble proving special cases of the Pigeonhole Principle in less than exponential time. (<strong>Edit:</strong> I think I was misinterpreting - the Pigeonhole Principle is hard to prove when you restrict yourself to "bounded-depth Frege proofs".)</p>
<p>A better proof system might be the following: a valid proof $p$ of the statement $\varphi$ consists of an ordered triple $(M,q,r)$, where $M$ is a proof system, $q$ is a binary string encoding an ordinary proof in first order logic that the axioms of ZFC imply $M$ is sound, and $r$ is a proof of $\varphi$ in the proof system defined by $M$. Unfortunately, we can't prove that this proof system is sound (without finding an inconsistency in ZFC): if we could, then we could convert that proof into a proof that ZFC is consistent, violating Gödel's second incompleteness theorem.</p>
<p>Of course, if P = NP, then there is a sound and complete proof system which ignores $p$ entirely...</p>
<h3>This is stupid</h3>
<p>If we now make the statement you quoted precise, it becomes: "If for every sound proof system there is a sequence $(\varphi_i)_{i\in\mathbb{N}}$ of tautologies, such that $\varphi_n$ has length $n$ for every $n$ and such that there is no polynomial $P$ such that for every $n$ $\varphi_n$ has a valid proof of length at most $P(n)$, then P $\ne$ NP."</p>
<p>Since we don't have a practical method for checking whether a proof system is sound, I can't help but think that this statement is completely pointless (like many tautologies).</p>
|
4,463,105 | <p>Write down the radius of convergence of the power series, obtained by the Taylor expansion of the analytic functions about the stated point, in</p>
<p><span class="math-container">$f(z) = \frac{(z+20)(z+21)}{(z-20i)^{21} (z^2 +z+1)}$</span> about <span class="math-container">$z = 0$</span>.</p>
<p>My attempt: Since power series are continuous on the disk of convergence, the radius of convergence is the distance to the nearest point of discontinuity. <span class="math-container">$f(z)$</span> is not analytic at <span class="math-container">$z=20i$</span>, hence the radius of convergence would be <span class="math-container">$|0-20i|= \sqrt{20}$</span>.</p>
<p>Am I correct?</p>
| José Carlos Santos | 446,262 | <p>The radius of convergence is the distance from <span class="math-container">$0$</span> to the nearest singularity. And the nearest singularities are <span class="math-container">$-\frac12\pm\frac{\sqrt3}2i$</span> (the roots of <span class="math-container">$z^2+z+1$</span>), whose distance to <span class="math-container">$0$</span> is <span class="math-container">$1$</span>.</p>
|
667,371 | <p>I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$</p>
<p>So what i did was:</p>
<p>$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$</p>
<p>$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$</p>
<p>$$4*{(x+2)}*(x-3)=x^2+25+10x$$</p>
<p>$$4x^2-4x-24=x^2+25+10x$$</p>
<p>$$3x^2-14x-49$$</p>
<p>But this seems to be wrong! What did i wrong?</p>
| gt6989b | 16,192 | <p>first step on the right hand side should be $3x+5$ not $3x+4$</p>
<p>Then proceeding the same way:
$$
\begin{split}
4(x^2-x-6) &= x^2 + 12x + 36 \\
3x^2 - 16x - 60 &= 0
\end{split}
$$
Since the discriminant of that equation is $16^2 - 4 \cdot 3 \cdot 60 < 0$ there are no solutions.</p>
<p><strong>UPDATE</strong> you changed the problem now, so you get
$4(x^2-x-6) = x^2 + 10x + 25$ which results in a quadratic $3x^2 - 14x - 49 = 0$ so $(x-7)(3x+7)=0$ and $x=7$ or $x = -7/3$. The second solution is impossible (drives roots to be complex) so $x=7$ is the only one.</p>
|
2,076 | <p>I'm attempting for the first time to create a map within <em>Mathematica</em>. In particular, I would like to take an output of points and plot them according to their lat/long values over a geographic map. I have a series of latitude/longitude values like so:</p>
<pre><code> {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046},
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024},
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966},
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47},
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254},
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65},
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092},
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659},
{42.3398, -70.9843}}
</code></pre>
<p>I've tried finding documentation on how I would proceed but I cannot find anything that doesn't assume a certain level of introduction to geospatial data. Does anyone know of a good resource online or is there a simple explanation one can supply here? </p>
| kglr | 125 | <p>data:</p>
<pre><code>latlong = {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046},
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024},
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966},
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47},
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254},
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65},
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092},
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659}, {42.3398, -70.9843}}
</code></pre>
<p>To put the data on latitude-longitude pairs on a map, yo will need to transform your data based on the projection method used by the map.</p>
<p>For example, </p>
<pre><code> coords = CountryData["UnitedStates", "Coordinates"];
</code></pre>
<p>gives the latitude-longitude data for US boundaries.</p>
<p>To use this data to put together a map with a specific projection method (say <code>Mercator</code>),
you need to transform your data </p>
<pre><code> Map[ GeoGridPosition[ GeoPosition[#], "Mercator"][[1]] & , {latlong}, {2}]
</code></pre>
<p>which gives</p>
<blockquote>
<pre><code> {{{1.09884, 0.602677}, {1.18857, 0.778879}, {1.0813,
0.632239}, {1.18707, 0.781777}, {1.08315, 0.632597}, {1.08169,
0.62617}, {1.08057, 0.634228}, {1.00789, 0.704491}, {0.995431,
0.879708}, {1.09687, 0.602482}, {0.993756, 0.874393}, {1.00409,
0.714614}, {1.09785, 0.604149}, {0.998381, 0.868794}, {1.00786,
0.708463}, {0.998538, 0.88544}, {1.00021, 0.947273}, {1.00095,
0.870866}, {1.694, 0.816595}, {1.85624, 0.824365}, {1.01069,
0.958578}, {1.01072, 0.97155}, {1.0015, 0.892771}, {1.00079,
0.970088}, {1.90268, 0.817169}}}
</code></pre>
</blockquote>
<p>Doing this transformation for both your data and the latitude-longitude data for world countries inside <code>Graphics</code>:</p>
<pre><code> Graphics[{Red, Point /@ Map[
GeoGridPosition[ GeoPosition[#],
"Mercator"][[1]] & , {latlong}, {2}], Gray,
Polygon[Map[ GeoGridPosition[ GeoPosition[#], "Mercator"][[1]] & ,
CountryData[#, "Coordinates"], {2}]] & /@
CountryData["Countries"]}]
</code></pre>
<p>you get:</p>
<p><img src="https://i.stack.imgur.com/jDt96.png" alt="world map and lat-long points"></p>
<p>Now I know I can focus on US:</p>
<pre><code> Graphics[{ Gray,
Polygon[Map[ GeoGridPosition[ GeoPosition[#], "Mercator"][[1]] & ,
CountryData["UnitedStates", "Coordinates"], {2}]], Red,
PointSize[.02], Point /@ Map[
GeoGridPosition[ GeoPosition[#],
"Mercator"][[1]] & , {latlong}, {2}]}]
</code></pre>
<p>to get</p>
<p><img src="https://i.stack.imgur.com/AMqsZ.png" alt="us and points"></p>
<p><strong>A simpler method</strong> avoiding <code>GeoPosition</code>, <code>GeoGridPosition</code> ... etc</p>
<p>Get the coordinates of US:</p>
<pre><code> coords = CountryData["UnitedStates", "Coordinates"];
</code></pre>
<p>and use </p>
<pre><code> Graphics[{EdgeForm[Black], Polygon[Reverse /@ First[coords]], Red,
Point /@ Reverse /@ latlong}]
</code></pre>
<p>to get</p>
<p><img src="https://i.stack.imgur.com/QBKi6.png" alt="simpler approach result"></p>
|
2,076 | <p>I'm attempting for the first time to create a map within <em>Mathematica</em>. In particular, I would like to take an output of points and plot them according to their lat/long values over a geographic map. I have a series of latitude/longitude values like so:</p>
<pre><code> {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046},
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024},
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966},
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47},
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254},
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65},
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092},
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659},
{42.3398, -70.9843}}
</code></pre>
<p>I've tried finding documentation on how I would proceed but I cannot find anything that doesn't assume a certain level of introduction to geospatial data. Does anyone know of a good resource online or is there a simple explanation one can supply here? </p>
| Artes | 184 | <p>Given latitude/longitude values: </p>
<pre><code>list = {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046}, \
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024}, \
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966}, \
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47}, \
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254}, \
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65}, \
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092}, \
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659}, \
{42.3398, -70.9843}};
</code></pre>
<p>I make a graphics with <code>Tooltip</code> to show coordinates of positions in the list in <code>DMSString</code> {degree, minute, second} format. </p>
<pre><code>Graphics[{Darker[Green], CountryData["UnitedStates", "Polygon"],
PointSize[Large], Blue, Tooltip[{PointSize[0.005], Point[Reverse@#]},
DMSString[#]] & /@ list}]
</code></pre>
<p><strong>Edit</strong></p>
<p>It would be more useful if we could find two nearest big cities to every specified position in the list. We can fulfill such an expectation with a handy function from Mathematica documentation, like this :</p>
<pre><code>nearLC = Nearest[ CityData[ #, "Coordinates"]
-> # & /@ CityData[{Large, "UnitedStates"}]];
</code></pre>
<p>Now we can adapt this function to the data we are deal with in order to show with <code>Tooltip</code> two nearest big cities (of population over 100000) for every point : </p>
<pre><code>Graphics[{ Lighter[Gray], CountryData["UnitedStates", "Polygon"],
Blue, Tooltip[{PointSize[0.007], Point[Reverse@#]},
Flatten[ nearLC[#, 2] /. {a_, b_, c_} -> {a, b}]] & /@ list}]
</code></pre>
<p><img src="https://i.stack.imgur.com/uoDJ7.gif" alt="enter image description here"></p>
|
2,076 | <p>I'm attempting for the first time to create a map within <em>Mathematica</em>. In particular, I would like to take an output of points and plot them according to their lat/long values over a geographic map. I have a series of latitude/longitude values like so:</p>
<pre><code> {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046},
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024},
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966},
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47},
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254},
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65},
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092},
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659},
{42.3398, -70.9843}}
</code></pre>
<p>I've tried finding documentation on how I would proceed but I cannot find anything that doesn't assume a certain level of introduction to geospatial data. Does anyone know of a good resource online or is there a simple explanation one can supply here? </p>
| Jason B. | 9,490 | <p>Just in case anyone comes across this post in a search, here is a one-liner here for version 10 and after. If <code>latlong</code> is the list from the OP, then you can do this</p>
<pre><code>GeoGraphics[{Red, PointSize[.02], Point@GeoPosition@latlong}]
</code></pre>
<p><img src="https://i.stack.imgur.com/tgU6N.png" alt="Mathematica graphics"></p>
<p>The input for <code>GeoGraphics</code> is structured like that for <code>Graphics</code>, and you can use a whole host of options</p>
<pre><code>GeoGraphics[{Red, PointSize[.01], Point@GeoPosition@latlong},
GeoRange -> "World", GeoProjection -> "Mollweide",
GeoGridLines -> Quantity[5, "AngularDegrees"],
GeoBackground -> "ContourMap"]
</code></pre>
<p><img src="https://i.stack.imgur.com/1HzxF.png" alt="Mathematica graphics"></p>
|
2,355,579 | <blockquote>
<p><strong>Problem:</strong> James has a pile of n stones for some positive integer n ≥ 2. At each step, he
chooses one pile of stones and splits it into two smaller piles and writes the product
of the new pile sizes on the board. He repeats this process until every pile is exactly
one stone.</p>
<p>For example, if James has n = 12 stones, he could split that pile into a pile of size
4 and another pile of size 8. James would then write the number 4 · 8 = 32 on the
board. He then decides to split the pile of 4 stones into a pile with 1 stone and a
pile with 3 stones and writes 1 · 3 = 3 on the board. He continues this way until he
has 12 piles with one stone each.</p>
<p>Prove that no matter how James splits the piles (starting with a single pile of n
stones), the sum of the numbers on the blackboard at the end of the procedure is
always the same.</p>
</blockquote>
<p><em>Hint: First figure out what the formula for the final sum will be. Then prove it
using strong induction.</em></p>
<p>The formula I came up with is $n(n-1) / $2. It could be totally wrong...</p>
<p><strong>My (Partial) Solution:</strong></p>
<p>1) <strong>Base Case:</strong> (n=2) James has one pile of 2 stones. Suppose he takes the top-most stone and puts in one pile A, which now has size 1. He takes the remaining stone and places it into another pile B, which now has size 1. Then the product of the sizes is 1. The sum of all the products on his board is 1. Now, if he were to start again, and take the bottom stone first and put it in pile A, he has a pile of size 1, and then takes the top stone and puts it in pile B, he has a pile of size 1, and the product is 1, with sum of all products on the board is 1. </p>
<p>2) <strong>Inductive Hypothesis (Strong Induction):</strong> Suppose for some $k \geq 2$, $n$ stones can be split in $ 2 \leq n \leq k $ stones and $k-n$ stones.</p>
<p>3) <strong>Inductive Step</strong>: Consider $n=k+1$. What do I do now?</p>
| emma | 406,460 | <p>I think the first step would be to try to figure out what the formula for the sum would be. If James has $n$ stones, one way he could work through the process is to remove one stone from the pile each time (putting it into its own 1-stone pile). So at first he would have two piles, one with $n-1$ stones and the other with 1 stone - so a product of $n-1$. Then in the next step he would have three piles, with $n-2$, 1, and 1 stones - a product of $n-2$. And so on, until the sum of his products is </p>
<p>$$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$$</p>
<p>Then, for your induction step, you would assume that for any $2 \leq k \leq n$, the sum of the products at the end of all the pile dividing is $\frac{(k-1)k}{2}$. You would use that to show that if you have $n+1$ stones, the sum of the products would be $\frac{n(n+1)}{2}$. So, you'd have to think about what your piles would look like after the first division, and what conclusions you could draw from that - can you take it from there?</p>
|
3,182,587 | <p>I think it is simpler if I just focus on the derivation of <span class="math-container">$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$</span>.</p>
<p>I start from the formulae for the complex exponential, assuming <span class="math-container">$z=x+iy$</span>: <span class="math-container">$e^{iz}=e^{-y}*(\cos(x)+i\sin(x))$</span> and <span class="math-container">$e^{-iz}=e^{y}*(\cos(x)-i\sin(x))$</span>.</p>
<p>And here I start running into problems, I can simply not reproduce the given formula! The furthest I obtain is <span class="math-container">$$e^{iz}+e^{iz}=(e^{-y}+e^{y})*\cos(x)+i(e^{y}-e^{-y})*\sin(x)$$</span> and an equivalent formula for the difference. </p>
<p>Please help.</p>
| Kandinskij | 657,309 | <p>Consider <span class="math-container">$e^{ix}$</span> and <span class="math-container">$e^{-ix}$</span> remember that sine is an odd function and cosine is an even function:</p>
<p><span class="math-container">$$e^{ix}=\cos(x)+i\sin(x)$$</span>
<span class="math-container">$$e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)$$</span></p>
<p>Sum both sides:</p>
<p><span class="math-container">$$e^{ix}+e^{-ix}=2\cos(x)$$</span></p>
<p>:)</p>
|
93,499 | <p>Let $P$ be a normal sylow $p$-subgroup of a finite group $G$.</p>
<p>Since $P$ is normal it is the unique sylow $p$-subgroup.</p>
<p>I would like to say if $\phi$ is an automorphism then $\phi(P)$ is also a sylow $p$-subgroup. Then uniqueness would finish the proof. But is that true?</p>
<p>Does an automorphism of a group always send subgroups to subgroups of the same order?</p>
| Community | -1 | <p>Let $\phi:G \to G$ be the automorphism. We'll prove $\phi(P)$ is a $p$-Sylow Subgroup of $G$.</p>
<p><em>$\phi$ takes identity to itself</em>:</p>
<p>For any $e_G \in G$, the identity in G, $\phi(e_G.e_G)=\phi(e_G)\phi(e_G)$
which proves the result.(from cancellation law in a group)</p>
<p><em>$\phi$ takes inverses to inverses</em>:</p>
<p>Use the fact that $gg^{-1}=e_G$ for any $g \in G$. Do a computation similar to the one above.</p>
<p><em>$\phi$ takes subgroups to subgroups</em>:</p>
<p>Let $H \leq G$. We intend to prove $\phi(H)$ is a subgroup. Use the 'lemma' we have proved before and verify the subgroup criterion (that $\phi(H)$ is closed under multiplication and inverses. )</p>
<p>Now, by one of my comments above, (in fact by just using the bijectivity of the map $\phi$, and by looking at its restriction to $H$), we'll prove that $|\phi(H)|=|H|$. </p>
<p>Note that the definition for two sets to be of same cardinality is that there exists a bijection between them. </p>
<p>So, You are through.</p>
|
1,346,039 | <p>Let <span class="math-container">$F$</span> be a field of characteristic prime to <span class="math-container">$n$</span>, and let <span class="math-container">$F^a$</span> be an algebraic closure of <span class="math-container">$F$</span>. Let <span class="math-container">$\zeta$</span> be a primitive <span class="math-container">$n$</span>th root of unity in <span class="math-container">$F^a$</span>. I know that the monic polynomial <span class="math-container">$\Phi_n(X)$</span> factorizes as <span class="math-container">$$\Phi_n(X) = \prod_{m \in (\mathbb{Z}/n)^\times} (X - \zeta^m).$$</span>I also know that there is a canonical injective group homomorphism <span class="math-container">$\chi: \text{Gal}(F(\zeta)/F) \to (\mathbb{Z}/n)^\times$</span> that is characterized by the property that <span class="math-container">$g(\alpha) = \alpha^{\chi(g)}$</span> for any <span class="math-container">$g \in \text{Gal}(F(\zeta)/F)$</span> and any <span class="math-container">$\alpha \in \mu_n$</span>.</p>
<p>What is the easiest and quickest way see that the following properties are true/where can I find a reference to their proofs?</p>
<blockquote>
<p>Let <span class="math-container">$S \subset (\mathbb{Z}/n)^\times$</span> be the image of the above homomorphism <span class="math-container">$\chi$</span>.</p>
<ol>
<li><p>If <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> are two elements of <span class="math-container">$(\mathbb{Z}/n)^\times$</span> then <span class="math-container">$\zeta^{m_1}$</span> and <span class="math-container">$\zeta^{m_2}$</span> are conjugate to another if and only if <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> lie in the same coset of <span class="math-container">$S$</span> in <span class="math-container">$(\mathbb{Z}/n)^\times$</span>.</p>
</li>
<li><p>Let <span class="math-container">$S = S_1, \dots, S_s$</span> denote the cosets of <span class="math-container">$S$</span> in <span class="math-container">$(\mathbb{Z}/n)^\times$</span>. Then for each <span class="math-container">$i = 1, \dots, s$</span>, the polynomial <span class="math-container">$f_i = \prod_{m \in S_i} (X - \zeta^m)$</span> is an irreducible element of <span class="math-container">$F[X]$</span>.</p>
</li>
<li><p><span class="math-container">$\Phi_n(X) = f_1(X) \dots f_s(X)$</span>.</p>
</li>
</ol>
</blockquote>
<p>Much thanks in advance.</p>
| mich95 | 229,072 | <p>I am really not sure of this, but here's what I could do!
Let $\tau$ be in the Galois group. $\tau$ is uniquely determined by what it does to $\zeta$. And $\tau_{k}(\zeta)=\zeta^{k}$. Now if $m_{1}$ and $m_{2}$ lie in the same coset of $S$, then $m_{1}m_{2}^{-1} \in S$, So $m_{1}m_{2}^{-1}=t$ for some $t$ in $S$. But we know that we can find an elememt $\tau_{t}$ in the Galois group such that $\tau_{t}(\zeta)=\zeta^{t}$, so $\tau_{t}(\zeta^{m_{2}})=\zeta^{m_{2}t}=\zeta^{m_{1}}$.
Assume they're conjugate, then we can find $\sigma$ in the Galois group such that $\sigma(\zeta^{m_{1}})=\zeta^{m_{2}}$, and hence we can find $t \in S$ such that $\sigma(\zeta)=\zeta^{k}$ so $\zeta^{km_{1}}=\zeta^{m_{2}}$ and since $\zeta$ is primitive, $m_{1}k=m_{2}$ so $m_{2}m_{1}^{-1}=k \in S$.
For the second, your polynomial indeed belongs to $F[X]$ as it's invariant under the action of the Galois group, by using what we proved in part 1. Now, if we have a proper irreducible factor $g_{i}(x)$ of $f_{i}(x)$, then $g_{i}(x)=\prod\limits_{k \in T_{i}}(x-\zeta^{k})$, where $T_{i}$ is a proper subset of $S_{i}$. But, also using the first part, you can prove that this polynomial does not belong to $F[X]$ as you can produce an automorphism that does not fix it!.
Now for third part, I think you can use the fact that the cosets of $S$ partition $(\mathbb{Z}/n\mathbb{Z})^{\star}$</p>
|
2,806,432 | <p>Let $(\mathbb{R}^N,\tau)$ a topological space, where $\tau$ is the usual topology.
Let $A\subset\mathbb{R}^N$ a compact. If $(A_n)_n$ is a family of open such that
\begin{equation}
\bigcup_nA_n\supset A,
\end{equation}
then, from compact definition
\begin{equation}
\bigcup_{i=1}^{k}A_i\supset A
\end{equation}
Now, if I find a family of open $(B_n)_n$ such that
\begin{equation}
cl\bigg(\bigcup_n B_n\bigg)\supset A,
\end{equation}
can I say that
\begin{equation}
cl\bigg(\bigcup_{i=1}^kB_i\bigg)\supset A\quad?
\end{equation}
Thanks for the attention!</p>
| Riccardo Sven Risuleo | 419,249 | <p>The first thing to notice is that $B$ throws 50 coins and makes 225 points.
If $n_H$ and $n_T$ are the numbers of heads and tails, 50 throws make it so that
$$ n_H + n_T = 50.$$
Given that a heads gives 1 point and a tails 8 points, we can set up the equation
$$ 1 \cdot n_H + 8 \cdot n_T = 225.$$</p>
<p>Solving the two equations together gives $n_t = 25$; so we are after the probability tha 25 tails pop up in 50 throws. You can either do it with combinations and permutations (as your teacher did), or plainly use the pmf for the <a href="https://en.wikipedia.org/wiki/Binomial_distribution" rel="nofollow noreferrer">binomial distribution</a> that represents your situation solving for 25 successes in 50 trials.</p>
<p>If you want the graphics, below is the plot of the distribution $B(50,0.5)$
<a href="https://i.stack.imgur.com/LlT6r.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LlT6r.png" alt="enter image description here"></a>
with a line through $0.11227...$ which is the required probability.</p>
<p>While it is true that the binomial distribution tends to a normal distribution, you cannot interpret the values of a probability <em>density</em> function as the probability of an event.</p>
|
1,369,409 | <p>I have a bit of an advanced combination problem that has left me stumped for a few days. Essentially my question is if you have n sets of items, and you can select a different number of items from each set, how do you compute the combinations without first creating new sets.</p>
<p>An example in pictures:
I have three sets:
<br>Set A has elements: <br>
Dog<br>
Cat<br>
Rhino<br><br>
Set B has elements:
<br>Pig
<br>Horse
<br>Cow
<br><br>Set C has elements:
<br>Lizard
<br>Snake
<br>Crocodile
<br>Alligator
<br><br>
And now I would like to compute all of the combinations with the criteria that 2 elements be selected from set a, 1 element is selected from set
B, and 2 elements are selected from set C.
<br><br>
The end result would contain all the unique combinations with those specifications.<br>
A current way I am using is to take Set A and turn Set A into all of the combinations of Set A that has 2 elements and storing it in a different set, Set D, then doing the same for Set C, storing in Set E, and then selecting one item from each Set B, Set D, and Set E to get all the unique combinations but I was wondering if there was a better solution.</p>
<p><br>EDIT: By compute I am referring to generating a list of all the possible sets, NOT figure out the count or number of items. That being said, this implies that each set must be unique (Dog, Cat, Pig, Lizard, Snake) is the same as (Cat, Dog, Pig, Snake, Lizard). </p>
| Graham Kemp | 135,106 | <p>Assuming "compute" means "count".</p>
<blockquote>
<p>And now I would like to compute all of the combinations with the criteria that 2 elements be selected from set A (3 elements), 1 element is selected from set B (3 elements), and 2 elements are selected from set C (4 elements). </p>
</blockquote>
<p>$${^{3}{\rm C}_{2}}{^{3}{\rm C}_{1}}{^{4}{\rm C}_{2}} = \frac{3!}{2!1!}\cdot\frac{3!}{1!2!}\cdot\frac{4!}{2!^2} = 3\cdot 3\cdot 6 = 54$$</p>
|
1,100,812 | <p>Here is the statement:
($\bf{Tonelli}$) If $f\in L^+(X,Y)$, then $\displaystyle g:x\mapsto\int_Yf_xd\nu$ is $\mathcal{M}$-measurable,\
$\displaystyle h:y\mapsto \int_Xf^yd\mu$ is $\mathcal{N}$-measurable (so $g\in L^+(X)$ and $h\in L^+(Y)$). And $$\displaystyle \int_{X\times Y}fd\mu \times\nu=\int_Xgd\mu=\int_Yhd\nu.$$
That is, $$\displaystyle \int_{X\times Y}fd\mu \times\nu=\int_X\left(\int_Yf_xd\nu\right)d\mu(x)=\int_Y\left(\int_Xf^yd\mu\right)d\nu(y)$$</p>
<p>($\bf{Fubini}$) If $f\in L^1(X\times Y)$, then $f_x\in L^1(Y,\nu)$ for a.e. $x\in X$ and $f^y\in L^1(X,\mu)$ for a.e. $y\in Y$. The a.e. defined functions $g$ and $h$ above are $\mathcal{M}$-measurable and $\mathcal{N}$-measurable respectively and the conclusion from above holds. </p>
<p>($\bf{Fubini}$) Since $f\in L^1(X\times Y)$, then $|f|\in L^+$. That is $(\int_{X\times Y})|f|<\infty$. Then
$\Rightarrow$ $\displaystyle\int_{X\times Y}|f|d\mu \times\nu=\int_Y\left(\int_X|f^y|d\mu\right)d\nu=\int_X\left(\int_Y|f_x|d\nu\right)d\mu$
$\Rightarrow \displaystyle \int_X|f^y|d\mu<\infty$ a.e. $y$ and $\displaystyle \int_Y|f_x|d\nu<\infty$ a.e. $x$.
$\Rightarrow f^y\in L^1(\mu)$ a.e. $y$ and $f_x\in L^1(\nu)$ a.e. $x$.
Let $f=f^+-f^-$, then $f_x=(f_x)^+-(f_x)^-=(f^+)_x-(f^-)_x$. So
\begin{eqnarray*}
\int_{X\times Y}f&=&\int_{X\times Y}f^+-\int_{X\times Y}f^-\\
&=&\int_X\left[\int_Y(f^+)_x\right]-\int_X\left[\int_Y(f^-)_x\right]\\
&=&\int_X\left[\int_Y(f^+)_x-(f_x)^-\right]\\
&=&\int_X\int_Yf_x.
\end{eqnarray*}
Similar for $f^y$.</p>
<p>My question is about the a.e. part of this proof. I'm not sure exactly why line 2 implies the a.e. conclusion. Should we not show this rigorously or is there a way of looking at these types of a.e. statements and seeing it automatically? Thank you! </p>
| pre-kidney | 34,662 | <p>Here is where the a.e. is coming from. Suppose we have a measurable function $g$, and we know that
$$
\int |g|\ d\mu<\infty.
$$
Then it is intuitively clear that $|g|<\infty$, because integrating $\infty$s would give you $\infty$. However, we have to hedge our statement because integration w.r.t. a measure "forgets" about null sets. Therefore all we can say with certainty is that $|g|<\infty$ $\mu$-almost everywhere.</p>
<p>It's trivial to come up with counterexamples, for example:
$$
g(x)=\begin{cases}
\infty & x=0\\
0 & x\not=0
\end{cases}
$$
Then $\int_{\mathbb{R}} g(x)\ dx=0$ (where $dx$ denotes Lebesgue measure), however $g$ takes on the value $\infty$.</p>
<p><strong>Edit:</strong></p>
<p>In line 2 of your proof, the idea I've described happens twice, with different $g$ functions. First it happens with $g(y)=\int_X |f^y|\ d\mu$, then immediately afterwards we fix $x$ and apply the argument to $g(y)=f_x(y)$.</p>
|
2,589,019 | <p>I am aware that there is a result saying that if $L_1, L_2$ are two algebraically closed fields of the same characteristic, then either $L_1$ is isomorphic to the subfield of $L_2$, or $L_2$ is isomorphic to a subfield of $L_1$.</p>
<p>I am unsure however of how to prove this result. I am quite surprised that this is true, because I feel like these algebraic subfields can be so large that there's no reason why their structures have to have similarities like this. Could anyone offer some insight please? </p>
| Eric Wofsey | 86,856 | <p>Let $k$ be the prime field of the common characteristic of $L_1$ and $L_2$, let $B_1$ be a transcendence basis for $L_1$ over $k$, and let $B_2$ be a transcendence basis for $L_2$ over $k$. Then either $|B_1|\leq |B_2|$ or $|B_2|\leq |B_1|$; suppose WLOG that $|B_1|\leq |B_2|$. Choose an injection $i:B_1\to B_2$; this then gives a homomorphism $j:k(B_1)\to k(B_2)\subset L_2$. Since $L_1$ is algebraic over $k(B_1)$ and $L_2$ is algebraically closed, we can extend $j$ to a homomorphism $L_1\to L_2$. Thus $L_1$ embeds in $L_2$.</p>
|
381,566 | <p>I know practically nothing about fractional calculus so I apologize in advance if the following is a silly question. I already tried on math.stackexchange.</p>
<p>I just wanted to ask if there is a notion of fractional derivative that is linear and satisfy the following property <span class="math-container">$D^u((f)^n) = \alpha D^u(f)f^{(n-1)}$</span> where <span class="math-container">$\alpha$</span> is a scalar. In the case of standard derivatives we would have <span class="math-container">$\alpha = n$</span>.</p>
<p>Thank you very much.</p>
| Terry Tao | 766 | <p>There are basically no interesting solutions to this equation beyond first and zeroth order operators, even if one only imposes the stated constraint for <span class="math-container">$n=2$</span>.</p>
<p>First, we can <a href="https://en.wikipedia.org/wiki/Polarization_identity" rel="nofollow noreferrer">depolarise</a> the hypothesis
<span class="math-container">$$ D^u(f^2) = \alpha_2 D^u(f) f \quad (1)$$</span>
by replacing <span class="math-container">$f$</span> with <span class="math-container">$f+g, f-g$</span> for arbitrary functions <span class="math-container">$f,g$</span> and subtracting (and then dividing by <span class="math-container">$4$</span>) to obtain the more flexible Leibniz type identity
<span class="math-container">$$ D^u(fg) = \frac{\alpha_2}{2}( D^u(f) g + f D^u(g) ). \quad (2)$$</span></p>
<p>There are now three cases, depending on the value of <span class="math-container">$\alpha_2$</span>:</p>
<ol>
<li><span class="math-container">$\alpha_2 \neq 1,2$</span>. Applying (2) with <span class="math-container">$f=g=1$</span> we then conclude that <span class="math-container">$D^u(1)=0$</span>, and then applying (2) again with just <span class="math-container">$g=1$</span> we get <span class="math-container">$D^u(f)=0$</span>. So we have the trivial solution <span class="math-container">$D^u=0$</span> in this case.</li>
<li><span class="math-container">$\alpha_2=2$</span>. Then <span class="math-container">$D^u$</span> is a <a href="https://en.wikipedia.org/wiki/Derivation_(differential_algebra)" rel="nofollow noreferrer">derivation</a> and by induction we have <span class="math-container">$D^u(f^n) = n D^u(f) f^{n-1}$</span>, just as with the ordinary derivative, so we just have <span class="math-container">$\alpha_n=n$</span> for all <span class="math-container">$n$</span> with no fractional behavior.</li>
<li><span class="math-container">$\alpha_2=1$</span>. Applying (2) with <span class="math-container">$g=1$</span> we obtain (after a little bit of algebra) <span class="math-container">$D^u(f) = mf$</span> where <span class="math-container">$m := D^u(1)$</span>. Thus <span class="math-container">$D^u$</span> is just a multiplier operator, which obeys <span class="math-container">$D^u(f^n) = D^u(f) f^{n-1}$</span>, thus <span class="math-container">$\alpha_n=1$</span> for all <span class="math-container">$n$</span>.</li>
</ol>
<p>Thus there are no linear solutions to your equation other than the usual derivations (e.g., <span class="math-container">$D^u(f) = a(x) \frac{d}{dx} f$</span> for any smooth symbol <span class="math-container">$a$</span>) and multiplier operators <span class="math-container">$D^u(f) = mf$</span>, i.e., first order and zeroth order operators.</p>
<p>On the other hand, fractional derivatives <span class="math-container">$D^u$</span> tend to obey a "fractional chain rule"
<span class="math-container">$$ D^u( F(f) ) = D^u(f) F'(f) + E$$</span>
for various smooth functions <span class="math-container">$F,f$</span>, where the error <span class="math-container">$E$</span> obeys better estimates in various Sobolev spaces than the other two terms in this equation. In particular, for <span class="math-container">$F(t) = t^n$</span>, we would have
<span class="math-container">$$ D^u(f^n) = n D^u(f) f^{n-1} + E$$</span>
for a "good" error term <span class="math-container">$E$</span>. For instance, taking <span class="math-container">$u=n=2$</span> with <span class="math-container">$D$</span> the usual derivative, we have
<span class="math-container">$$ D^2(f^2) = 2 D^2(f) f + E \quad (3)$$</span>
with <span class="math-container">$E$</span> the "<a href="https://mathscinet.ams.org/mathscinet-getitem?mr=521767" rel="nofollow noreferrer">carré du champ</a>" operator
<span class="math-container">$$ E := 2 (Df)^2.$$</span>
Note that the error <span class="math-container">$E$</span> is controlled uniformly by the <span class="math-container">$C^1$</span> norm of <span class="math-container">$f$</span> but the other two terms in (3) are not. See my previous MathOverflow answer at <a href="https://mathoverflow.net/a/94039/766">https://mathoverflow.net/a/94039/766</a> for some references and further discussion.</p>
|
1,700,590 | <p>Let $A=\{a_1\ldots,a_m \}$ be a set of linearly independent vectors. Suppose that each $a_j$ $(j=1,\ldots,m)$ can be written as a linear combination of vectors in the set $B=\{b_1,\ldots,b_n\}$.</p>
<p>Then how to show that $m \le n$?</p>
<p>I have tried as follows:</p>
<p>Since $a_j \in A$, we have $a_j \in \operatorname{span} (B)$.</p>
<p>Then we have
$$ A \subseteq \operatorname{span} (B).$$</p>
<p>Therefore, can we say that $m \le n$?</p>
<p>Please give me the correct answer for this!!</p>
| mordecai iwazuki | 167,818 | <p>You are almost there. Consider for the sake of contradiction that $m>n$ but you still have $A\subseteq$span$(B)$. Then, you have a set of $m>n$ vectors in a space spanned by $n$ vectors so the set $\{a_j\}$ must be linearly dependent, a contradiction to our assumptions so it must be that $m\leq n$.</p>
|
200,278 | <p>Say I have two TimeSeries:</p>
<pre><code>x = TimeSeries[{2, 4, 1, 10}, {{1, 2, 4, 5}}]
y = TimeSeries[{6, 2, 6, 3, 9}, {{1, 2, 3, 4, 5}}]
</code></pre>
<p>x has a value at times: 1,2,4,5</p>
<p>y has a value at times: 1,2,3,4,5</p>
<p>I would like to build a list of pairs {<span class="math-container">$x_i$</span>, <span class="math-container">$y_i$</span>} which would not include missing elements (in this case element x element at time 3 is missing)</p>
<p>The desired result would be: </p>
<pre><code>{{2,6}, {4,2}, {1,3}, {10,9}}
</code></pre>
<p>I have a feeling that this should be simple and perhaps I'm not using right tools.</p>
| Sjoerd Smit | 43,522 | <p><code>TimeSeriesThread</code> is probably the tool for the job when you specify the right options:</p>
<pre><code>x = TimeSeries[{2, 4, 1, 10}, {{1, 2, 4, 5}}];
y = TimeSeries[{6, 2, 6, 3, 9}, {{1, 2, 3, 4, 5}}];
DeleteMissing[
TimeSeriesThread[Identity, {x, y}, ResamplingMethod -> None]["Values"],
1, 1
]
</code></pre>
<blockquote>
<p>{{2, 6}, {4, 2}, {1, 3}, {10, 9}}</p>
</blockquote>
|
1,942,641 | <p>Recall that a function $f: A \rightarrow B$ is continuous means that for any open set $\beta \in B$, $f^{-1}(\beta)$ is open, where $f^{-1}$ is the preimage set.</p>
<p>Say that $f$ is not continuous. Then, that means that there exists some open set $\beta$ in $B$ such that $f^{-1}(\beta)$ is...not open?</p>
<p>What does it mean to be not open? Closed? But open and closed are not necessarily mutually exclusive...</p>
<p>So, what does it mean for a function to not be continuous, properly speaking?</p>
| Thomas Andrews | 7,933 | <p>It's really as simple as noting that $a+b\geq 2\sqrt{ab}$, and similarly for $a+c$ and $b+c$, then multiplying it out.</p>
|
1,942,641 | <p>Recall that a function $f: A \rightarrow B$ is continuous means that for any open set $\beta \in B$, $f^{-1}(\beta)$ is open, where $f^{-1}$ is the preimage set.</p>
<p>Say that $f$ is not continuous. Then, that means that there exists some open set $\beta$ in $B$ such that $f^{-1}(\beta)$ is...not open?</p>
<p>What does it mean to be not open? Closed? But open and closed are not necessarily mutually exclusive...</p>
<p>So, what does it mean for a function to not be continuous, properly speaking?</p>
| ADAM | 312,147 | <p>Using AM-GM, we get that
$$\frac{a+b}{2} \geq \sqrt{ab} $$
$$\Rightarrow a+b \geq 2 \sqrt{ab}$$</p>
<p>Similar manipulations show that<br>
$$ b+c \geq 2 \sqrt{bc}$$
$$ a+c \geq 2 \sqrt{ac}$$</p>
<p>Multiplying all these inequalities together ( this is allowed here, since all numbers involved are positive), we get
$$ \begin{align}
(a+b)(b+c)(a+c) &\geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac}) \\
&= 8 \sqrt{ab.bc.ac} \\
&= 8 \sqrt{a^{2} b^{2} c^{2}} \\
&= 8abc
\end{align}
$$
as required. </p>
|
9,484 | <p>Let <span class="math-container">$F(k,n)$</span> be the number of permutations of an n-element set that fix exactly <span class="math-container">$k$</span> elements.</p>
<p>We know:</p>
<ol>
<li><p><span class="math-container">$F(n,n) = 1$</span></p>
</li>
<li><p><span class="math-container">$F(n-1,n) = 0$</span></p>
</li>
<li><p><span class="math-container">$F(n-2,n) = \binom {n} {2}$</span></p>
<p>...</p>
</li>
<li><p><span class="math-container">$F(0,n) = n! \cdot \sum_{k=0}^n \frac {(-1)^k}{k!}$</span> (the subfactorial)</p>
</li>
</ol>
<p>The summation formula is obviously</p>
<p><span class="math-container">$\displaystyle\sum_{k=0}^n F(k,n) = n!$</span></p>
<p>A recursive definition of <span class="math-container">$F(k,n)$</span> is (my claim):</p>
<p><span class="math-container">$$F(k,n) = \binom {n} {k} \cdot \Big( k! - \displaystyle\sum_{i=0}^{k-1} F(i,k) \Big)$$</span></p>
<p>Question 1: Is there a common name for the "generalized factorial" <span class="math-container">$F(k,n)$</span>?</p>
<p>Question 2: Does anyone know a closed form for <span class="math-container">$F(k,n)$</span> or have an idea how to get it from the recursive definition? (generating function?)</p>
| Hans-Peter Stricker | 2,672 | <p>see <a href="https://oeis.org/A000166" rel="nofollow noreferrer">Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points</a></p>
|
9,484 | <p>Let <span class="math-container">$F(k,n)$</span> be the number of permutations of an n-element set that fix exactly <span class="math-container">$k$</span> elements.</p>
<p>We know:</p>
<ol>
<li><p><span class="math-container">$F(n,n) = 1$</span></p>
</li>
<li><p><span class="math-container">$F(n-1,n) = 0$</span></p>
</li>
<li><p><span class="math-container">$F(n-2,n) = \binom {n} {2}$</span></p>
<p>...</p>
</li>
<li><p><span class="math-container">$F(0,n) = n! \cdot \sum_{k=0}^n \frac {(-1)^k}{k!}$</span> (the subfactorial)</p>
</li>
</ol>
<p>The summation formula is obviously</p>
<p><span class="math-container">$\displaystyle\sum_{k=0}^n F(k,n) = n!$</span></p>
<p>A recursive definition of <span class="math-container">$F(k,n)$</span> is (my claim):</p>
<p><span class="math-container">$$F(k,n) = \binom {n} {k} \cdot \Big( k! - \displaystyle\sum_{i=0}^{k-1} F(i,k) \Big)$$</span></p>
<p>Question 1: Is there a common name for the "generalized factorial" <span class="math-container">$F(k,n)$</span>?</p>
<p>Question 2: Does anyone know a closed form for <span class="math-container">$F(k,n)$</span> or have an idea how to get it from the recursive definition? (generating function?)</p>
| Sam OT | 59,264 | <p>Various links on this answer have expired, so I thought I would add an answer.</p>
<p>One can use inclusion--exclusion. First, note (as in @ReidBarton's answer) that
<span class="math-container">$$ F(k,n) = \binom kn F(0,n-k). $$</span>
So it is sufficient to only study permutations with no fixed points.
This is known as the number of <em>derangements</em>.
Various proofs using inclusion--exclusion can be found on Wikipedia:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Derangement#Counting_derangements" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Derangement#Counting_derangements</a></li>
<li><a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#Counting_derangements_2" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#Counting_derangements_2</a></li>
</ul>
|
156,479 | <p>Let $S$ be a compact oriented surface of genus at least $2$ (possibly with boundary). Let $X$ be a connected component of the space of embeddings of $S^1$ into $S$.</p>
<p>Question : what is the fundamental group of $X$? My guess is that the answer is $\mathbb{Z}$ with generator the loop of embeddings obtained by precomposing the base embedding with a sequence of rotations of $S^1$.</p>
<p>I'm also interested in the higher homotopy groups of $X$, which I would guess are trivial.</p>
<hr>
<p>Edit: In response to Sam Nead's question, I'm most interested in the smooth category, but am also interested in the topological category. There are technical issues in giving an appropriate topology to mapping spaces in the PL category, so the question doesn't really make sense there.</p>
| Sam Nead | 1,650 | <p>Edit - </p>
<p>As pointed out by Igor below, my "proof" of the first bullet point is incomplete. I'll leave the rest of the post here: perhaps some kind soul will fix the gap. </p>
<p>Original - </p>
<p>Let's restrict attention to the case where all embeddings are smooth and where $S$ has negative Euler characteristic, or is an annulus or Möbius band. </p>
<ul>
<li>If $X$ contains a trivial curve, then $X$ is homotopy equivalent to the unit tangent bundle to the surface $S$. </li>
<li>If $X$ contains an essential curve, then $X$ is homotopy equivalent to a circle.</li>
</ul>
<p>Both of these results follow from Matt Grayson's curve-shortening flow (Annals, 1989). If $S$ is the torus or Klein bottle, then first bullet still holds, but the second does not. Instead $X$ is homotopy equivalent to (a cover of) $S$. Finally, if $S$ is the sphere or projective plane, then I believe that $X$ has nontrivial higher homotopy groups. </p>
<p>If the embeddings are only continuous, well, that seems tricky. </p>
|
216,532 | <p>How do I find the limit of something like</p>
<p>$$ \lim_{x\to \infty} \frac{2\cdot3^{5x}+5}{3^{5x}+2^{5x}} $$</p>
<p>?</p>
| Hagen von Eitzen | 39,174 | <p><strong>Hint:</strong> $2\cdot 3^{5x}+5=3^{5x}\cdot (2+\frac5{3^{5x}})$ and $3^{5x}+ 2^{5x}=3^{5x}\cdot(1+(\frac23)^{5x})$.</p>
|
216,532 | <p>How do I find the limit of something like</p>
<p>$$ \lim_{x\to \infty} \frac{2\cdot3^{5x}+5}{3^{5x}+2^{5x}} $$</p>
<p>?</p>
| Cameron Buie | 28,900 | <p>More generally, if you're trying to determine limiting behavior of a function of form <span class="math-container">$$\frac{f(x)}{g(x)}$$</span> as <span class="math-container">$x\to\infty$</span>, and the limit is of form "<span class="math-container">$\pm\frac\infty\infty$</span>", then you can look for a dominating term--that is, a term that (eventually) grows more rapidly in size than any other as <span class="math-container">$x$</span> gets sufficiently large--and then divide top and bottom by that dominating term (as filmor and Hagen demonstrated in their answers), so that all but a few terms vanish in the limit.</p>
<p>Some guidelines:</p>
<blockquote>
<p>If <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are linear combinations of nonnegative powers of <span class="math-container">$x$</span>, then the dominating term is the highest power of <span class="math-container">$x$</span> that appears in the expression.</p>
<p>If <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are linear combinations of exponential expressions, then the dominating term will be the one with the largest base (and in case of a tie, the one with the fastest growing power).</p>
<p>Exponential terms (with bases greater than <span class="math-container">$1$</span>) dominate powers of <span class="math-container">$x$</span>, and both of these dominate logarithmic terms.</p>
</blockquote>
|
2,782,658 | <p>Here's what I have to show:</p>
<blockquote>
<p>If $x\le y+z$ for every $z>0$ then $x\le y$.</p>
</blockquote>
<p>I tried proof by contradiction but it does not seem to work. Suppose $y<x$. Then $x-y>0$. Let $z_{0}=x-y$. Then $x\le y+z_{0} \Rightarrow x\le x$ which is true. Had $x$ been strictly less than $y+z$, it would have worked.</p>
<p>Can I get hints?</p>
| fleablood | 280,126 | <p>"Had x been strictly less than y+z, it would have worked."</p>
<p>So instead of making $z = x-y$ make $0 < z < x-y$.</p>
<p>Then $x \le y + z < y + x-y = x$.</p>
|
2,782,658 | <p>Here's what I have to show:</p>
<blockquote>
<p>If $x\le y+z$ for every $z>0$ then $x\le y$.</p>
</blockquote>
<p>I tried proof by contradiction but it does not seem to work. Suppose $y<x$. Then $x-y>0$. Let $z_{0}=x-y$. Then $x\le y+z_{0} \Rightarrow x\le x$ which is true. Had $x$ been strictly less than $y+z$, it would have worked.</p>
<p>Can I get hints?</p>
| robjohn | 13,854 | <p>You are almost there. Assume that $x\gt y$, then set $z=\frac{x-y}2\gt0$. That would mean that $x=y+2z\gt y+z$ and $z\gt0$, which is a contradiction.</p>
|
3,405,622 | <p>Let <span class="math-container">$\int_{0}^{1}fg \text{ }d\mathbb{P}=0$</span>, for all <span class="math-container">$f$</span> <span class="math-container">$\in$</span> <span class="math-container">$L^{\infty}([0,1],\mathbb{P})$</span> and <span class="math-container">$g$</span> be a fixed function in <span class="math-container">$L^{1}([0,1],\mathbb{P})$</span> and where <span class="math-container">$\mathbb{P}$</span> is a probability measure in <span class="math-container">$[0,1]$</span>. Is it true <span class="math-container">$g=0$</span> almost everywhere?
Attempt: I was trying to use dominated convergence theorem to show <span class="math-container">$\int|g|^{2}\text{ }d\mathbb{P}=0$</span>, just want to find <span class="math-container">$f_{n}\rightarrow \overline{g}$</span> bounded sequence, that I can do using density of <span class="math-container">$L^{\infty}$</span> in <span class="math-container">$L^{1}$</span>, but the problem is <span class="math-container">$f_{n}$</span> need not dominated by <span class="math-container">$L^{1}$</span> function for applying DCT.</p>
| Marios Gretsas | 359,315 | <p>Yes since the dual of <span class="math-container">$L^1$</span> is <span class="math-container">$L^{\infty}$</span> </p>
<p><span class="math-container">$T_g(f):=\int_0^1fg$</span> is a bounded linear functional <span class="math-container">$T:L^{\infty} \to \Bbb{C}$</span></p>
<p>and <span class="math-container">$T_g(f)=0,\forall f \Longrightarrow g=0$</span></p>
<p><span class="math-container">$|T_g(f)| \leq ||f||_{\infty}||g||_1$</span></p>
|
2,857,769 | <blockquote>
<p>Find $t$ such that $$\lim_{n\to\infty} \frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)}=\frac 45.$$</p>
</blockquote>
<p>At first sight this question scared the hell out of me. I tried using the general known formulas like $$\sum_{r=1}^n r^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{6}$$ and $$\sum_{r=1}^n r^5=\frac {n^2(n+1)^2(2n^2+2n-1)}{12}.$$</p>
<p>But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it. </p>
<p>Any help would be greatly appreciated. Thanks. </p>
| Robert Z | 299,698 | <p>Hint. Note that for $a>0$
$$F(a):=\lim_{n\to\infty} \frac{1}{n^{a+1}}\left(\sum_{r=1}^n r^a\right)=
\lim_{n\to\infty} \frac{1}{n}\left(\sum_{r=1}^n \left(\frac{r}{n}\right)^a\right)\to \int_0^1x^a dx=\frac{1}{a+1}.$$
Then, as $n\to\infty$,
$$\frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)}
=\frac{\frac{1}{n^5}\left(\sum_{r=1}^n r^4\right)\cdot\frac{1}{n^6}\left(\sum_{r=1}^n r^5\right)}{\frac{1}{n^{t+1}}\left(\sum_{r=1}^n r^t\right)\cdot\frac{1}{n^{10-t}}\left(\sum_{r=1}^n r^{9-t}\right)}
\to\frac{F(4)\cdot F(5)}{F(t)\cdot F(9-t)}.$$</p>
|
2,857,769 | <blockquote>
<p>Find $t$ such that $$\lim_{n\to\infty} \frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)}=\frac 45.$$</p>
</blockquote>
<p>At first sight this question scared the hell out of me. I tried using the general known formulas like $$\sum_{r=1}^n r^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{6}$$ and $$\sum_{r=1}^n r^5=\frac {n^2(n+1)^2(2n^2+2n-1)}{12}.$$</p>
<p>But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it. </p>
<p>Any help would be greatly appreciated. Thanks. </p>
| Paramanand Singh | 72,031 | <p>The key here is Cesaro-Stolz. Using Cesaro-Stolz we have $$f(t) =\lim_{n\to\infty} \frac{1}{n^{t+1}}\sum_{r=1}^{n}r^t=\lim_{n\to \infty} \frac{n^t} {n^{t+1}-(n-1)^{t+1}}=\frac{1}{t+1}$$ Dividing the numerator and denominator of the given expression by $n^{11}$ we can see that the desired limit is $$\frac{f(4)f(5)}{f(t)f(9-t)}$$ and now you can equate this to $4/5$ and get $t$. </p>
|
1,613,171 | <p>On page $61$ of the book <a href="http://solmu.math.helsinki.fi/2010/algebra.pdf" rel="nofollow">Algebra</a> by Tauno Metsänkylä, Marjatta Näätänen, it states</p>
<blockquote>
<p>$\langle \emptyset \rangle =\{1\},\langle 1 \rangle =\{1\}. H\leq G \implies \langle H \rangle =H$</p>
</blockquote>
<p>where $H \leq G$ means that H is the subgroup of G.</p>
<p>Now assumme $H=\emptyset$ so $\langle \emptyset \rangle = \emptyset \not = \{1\}$, contradiction. Please explain p.61 of the book that is the line in orange above.</p>
| 5xum | 112,884 | <p>$\emptyset$ is not a subgroup of $G$, because it is not a group. Because $\langle X\rangle$ is <strong>by definition</strong></p>
<blockquote>
<p>The smallest subgroup of $H$ that includes $X$</p>
</blockquote>
<p>then if $X=\emptyset$, it is equal to $\{1\}$</p>
|
4,086,995 | <p>Let <span class="math-container">$ABCD$</span> be a rectangle.</p>
<p>Given:</p>
<p><span class="math-container">$A(2;1)$</span></p>
<p><span class="math-container">$C(5;7)$</span></p>
<p><span class="math-container">$\overline{BC}=2\overline{AB}$</span>.</p>
<p>I tried to solve it, but after using the Pythagoras theorem I got that <span class="math-container">$\overline{AB}=\overline{DC}=3$</span> and <span class="math-container">$\overline{AD}=\overline{BC}=6$</span> but I don't know what I do from here.</p>
<p>How can I get points <span class="math-container">$B$</span> and <span class="math-container">$D$</span>? (There are two answers)</p>
| Arturo Magidin | 742 | <p>Jose Carlos Santos has explained why the automorphism group of <span class="math-container">$\mathbb{Z}$</span> is very constrained, while that of <span class="math-container">$\mathbb{Z}_n$</span> is not as restricted, leading to more possibilities.</p>
<p>Paraphrasing US Supreme Court justices, I write separately to emphasize that I don’t think it is conceptually a good idea to think of <span class="math-container">$\mathbb{Z}$</span> as being a sort of “limit” of <span class="math-container">$\mathbb{Z}_n$</span> as <span class="math-container">$n$</span> gets larger. It just doesn’t.</p>
<p>First, if you think of <span class="math-container">$\mathbb{Z}_n$</span> as <span class="math-container">$\mathbb{Z}/n\mathbb{Z}$</span>, then <span class="math-container">$\mathbb{Z}\cong\mathbb{Z}/0\mathbb{Z}$</span>, so in fact <span class="math-container">$\mathbb{Z}$</span> isn’t what happens when “<span class="math-container">$n$</span> gets larger”, it’s what happened when <span class="math-container">$n$</span> was a small as possible.</p>
<p>Second, you don’t have (nontrivial) maps from <span class="math-container">$\mathbb{Z}_n$</span> into <span class="math-container">$\mathbb{Z}$</span>: you only have maps going the other way. If you think of group morphisms as going “left to right”, like most of our function arrows, the <span class="math-container">$\mathbb{Z}$</span> is the <em>leftmost</em> group among the cyclic groups, not the rightmost. So there is not good way to think of <span class="math-container">$\mathbb{Z}$</span> as the “limiting group” of the <span class="math-container">$\mathbb{Z}_n$</span>.</p>
<p>Thirdly, you want to be careful with how you think of the <span class="math-container">$\mathbb{Z}_n$</span> “fitting inside each other”. Note that <span class="math-container">$\mathbb{Z}_m$</span> has a (unique) subgroup isomorphic to <span class="math-container">$\mathbb{Z}_n$</span> if and only if <span class="math-container">$n|m$</span>. So it’s not a nice straight progression, but rather a somewhat complicated arrangement of these groups as they fit inside each other; what you get is what is called a “directed partially ordered set”.</p>
<p>Now, there is a way to try to figure out what the “limiting group” of the <span class="math-container">$\mathbb{Z}_n$</span> is; it’s called a <a href="https://en.wikipedia.org/wiki/Direct_limit" rel="noreferrer">direct limit</a>. If you do that to the finite cyclic groups, you don’t get <span class="math-container">$\mathbb{Z}$</span>; instead, <a href="https://math.stackexchange.com/questions/1016124/union-of-all-finite-cyclic-groups">you get <span class="math-container">$\mathbb{Q}/\mathbb{Z}$</span></a>, a very different group. And this group <em>does</em> have <em>lots</em> of automorphisms! You can decompose it into a direct product of Prüfer <span class="math-container">$p$</span>-groups, and each of those has at least one automorphism for each unit of the <span class="math-container">$p$</span>-adic integers (see <a href="https://math.stackexchange.com/questions/1048116/automorphisms-of-the-pr%c3%bcfer-group">here</a>).</p>
|
4,086,995 | <p>Let <span class="math-container">$ABCD$</span> be a rectangle.</p>
<p>Given:</p>
<p><span class="math-container">$A(2;1)$</span></p>
<p><span class="math-container">$C(5;7)$</span></p>
<p><span class="math-container">$\overline{BC}=2\overline{AB}$</span>.</p>
<p>I tried to solve it, but after using the Pythagoras theorem I got that <span class="math-container">$\overline{AB}=\overline{DC}=3$</span> and <span class="math-container">$\overline{AD}=\overline{BC}=6$</span> but I don't know what I do from here.</p>
<p>How can I get points <span class="math-container">$B$</span> and <span class="math-container">$D$</span>? (There are two answers)</p>
| user909807 | 909,807 | <p>All these groups are 1-generated. Thus, the image of a generator must be a generator. (Isomorphisms preserve such things.) But
<span class="math-container">$\Bbb Z$</span> has only two generators, whereas <span class="math-container">$\Bbb Z_n$</span> can have many. In fact, the number of automorphisms of <span class="math-container">$\Bbb Z_n$</span> equals the number of its generators.</p>
|
1,844,374 | <p>Why does the "$\times$" used in arithmetic change to a "$\cdot$" as we progress through education? The symbol seems to only be ambiguous because of the variable $x$; however, we wouldn't have chosen the variable $x$ unless we were already removing $\times$ as the symbol for multiplication. So why do we? I am very curious. It seems like $\times$ is already quite sufficient as a descriptive symbol.</p>
| Allie | 315,111 | <p>This is primarily done to emphasize different multiplication operations in terms of vector and multidimensional calculus. In particular, this is to emphasize that the dot product $\cdot$ is mechanically different from the cross product $\times$, although in operations on objects of one dimension, they are virtually the same.</p>
|
1,844,374 | <p>Why does the "$\times$" used in arithmetic change to a "$\cdot$" as we progress through education? The symbol seems to only be ambiguous because of the variable $x$; however, we wouldn't have chosen the variable $x$ unless we were already removing $\times$ as the symbol for multiplication. So why do we? I am very curious. It seems like $\times$ is already quite sufficient as a descriptive symbol.</p>
| DpS | 350,927 | <p>It's because . stands for any binary operation which might look like 'multiplication' in some particular setting, or might be a substitute for multiplication. Hence, it is more general in nature.</p>
|
1,844,374 | <p>Why does the "$\times$" used in arithmetic change to a "$\cdot$" as we progress through education? The symbol seems to only be ambiguous because of the variable $x$; however, we wouldn't have chosen the variable $x$ unless we were already removing $\times$ as the symbol for multiplication. So why do we? I am very curious. It seems like $\times$ is already quite sufficient as a descriptive symbol.</p>
| Robert Soupe | 149,436 | <p>The lowercase letter $x$ and the multiplication cross $\times$ (<code>\times</code> in TeX, <code>&times;</code> in HTML) are very different symbols. One can be used to represent a variable, as you have already halfway surmised, but it shouldn't be used to denote any kind of multiplication, whereas the other can be used to denote multiplication but should not be used to represent a variable. In some contexts you will see $\otimes$ used for extra clarity.</p>
<p>The big problem here is that these two symbols that have different etymologies and different uses look so much alike, and nowhere was this problem felt more acutely than during the early history of computer programming languages in the time of ASCII.</p>
<p>$x \times y$ is clear enough, at least for us with good enough eyes, but the multiplication cross is not in the ASCII character set, and so <code>x x y</code> would be hellishly ambiguous. And so the asterisk was co-opted for the multiplication operator, thus $x \times y$ becomes <code>x * y</code>, something you still see in Javascript and Mathematica. In C++ there are some operators that are actually words, but in general, in computer programming, the arithmetic operators are not alphanumeric symbols.</p>
<p><strong>EDIT:</strong> Doing some research after posting this answer, I came across <a href="http://mathserver.neu.edu/~bridger/U201/History_of_Math_Mathematical_Symbols.htm" rel="nofollow">a page from Northeastern University on math symbols</a>. William Oughtred, a 16th century mathematician, came up with a cross with vertical serifs as a multiplication symbol. Oughtred was rebuked by his now more famous contemporary Leibniz, who wrote:</p>
<blockquote>
<p>"I do not like (the cross) as a symbol for multiplication, as it is easily confounded with x; .... often I simply relate two quantities by an interposed dot and indicate multiplication by ZC.LM."</p>
</blockquote>
<p>This reinforces my point about how $x$ and $\times$ have different origins but are confusingly similar in appearance.</p>
<hr>
<p>And by the way, don't ever use $\Sigma$ (uppercase Sigma) as a cheap way to write E when you want to give something a Greek flavor. That Greek letter was chosen as the summation operator because it is an S sound.</p>
|
1,844,374 | <p>Why does the "$\times$" used in arithmetic change to a "$\cdot$" as we progress through education? The symbol seems to only be ambiguous because of the variable $x$; however, we wouldn't have chosen the variable $x$ unless we were already removing $\times$ as the symbol for multiplication. So why do we? I am very curious. It seems like $\times$ is already quite sufficient as a descriptive symbol.</p>
| Community | -1 | <p>$\times$ is clearly visible.</p>
<p>$\cdot$ is more discrete and is often left implicit.</p>
<p>With maturity, we become able to supply the operator where required.</p>
|
250,687 | <p>I'm doing a sanity check of the following equation:
<span class="math-container">$$\sum_{j=2}^\infty \frac{(-x)^j}{j!}\zeta(j) \approx x(\log x + 2 \gamma -1)$$</span></p>
<p>Naive comparison of the two shows a bad match but I suspect one of the graphs is incorrect.</p>
<ol>
<li>Why isn't there a warning?</li>
<li>How do I compute this sum correctly?</li>
</ol>
<pre><code>katsurda[x_] := NSum[(-x)^j/j! Zeta[j], {j, 2, Infinity}];
katsurdaApprox[x_] := x (Log[x] + 2 EulerGamma - 1) - Zeta[0];
plot1 = DiscretePlot[katsurda[x], {x, 0, 40, 2}];
plot2 = Plot[katsurdaApprox[x], {x, 0, 40}];
Show[plot1, plot2]
</code></pre>
<p><a href="https://i.stack.imgur.com/pBmVX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pBmVX.png" alt="enter image description here" /></a></p>
<ol start="3">
<li><strong>meta</strong> How do I avoid being mislead by incorrect numeric results? Would using <code>NIntegrate</code> instead of <code>NSum</code> give better guarantees? My usual approach of a avoiding machine precision, checking <code>Precision</code> of the answer and minding warnings fails in the example below</li>
</ol>
<pre><code>katsurda[x_] :=
NSum[(-x)^j/j! Zeta[j], {j, 2, Infinity}, WorkingPrecision -> 32,
NSumTerms -> 2.5 x];
katsurdaApprox[x_] := x (Log[x] + 2 EulerGamma - 1) - Zeta[0];
Print["Precision: ", Precision@katsurda[100]] (* 13.9729 *)
Print["Discrepancy: ", katsurda[100] - katsurdaApprox[100]] (* 94.65088290385, but should be <1 *)
</code></pre>
<p><strong>Background:</strong> the expression comes from "Power series with the Riemann zeta-function in the coefficients" by Katsurada M (<a href="https://projecteuclid.org/journals/proceedings-of-the-japan-academy-series-a-mathematical-sciences/volume-72/issue-3/Power-series-with-the-Riemann-zeta-function-in-the-coefficients/10.3792/pjaa.72.61.pdf" rel="nofollow noreferrer">paper</a>)</p>
| Mariusz Iwaniuk | 26,828 | <p>Another way using:</p>
<p><span class="math-container">$$\sum _{n=1}^{\infty } \left(\exp \left(-\frac{x}{n}\right)-1+\frac{x}{n}\right)=\int_0^{\infty } \left(\frac{x}{-1+e^z}-\frac{\sqrt{x}
J_1\left(2 \sqrt{x} \sqrt{z}\right)}{\left(-1+e^z\right) \sqrt{z}}\right) \, dz$$</span></p>
<pre><code> f[x_] := NIntegrate[InverseLaplaceTransform[Exp[-x/n] - 1 + x/n, n, z]/(Exp[z] - 1), {z, 0, Infinity}, Method -> "LocalAdaptive"];
fApprox[x_] := x (Log[x] + 2 EulerGamma - 1) + 1/2;
plot1 = DiscretePlot[f[x], {x, Round[(E - 1)^Range[10, 17]]}];
plot2 = Plot[fApprox[x], {x, 50, 10000}];
Show[plot1, plot2, PlotRange -> All]
DiscretePlot[f[x] - fApprox[x], {x, Round[(E - 1)^Range[10, 17]]}]
</code></pre>
|
3,124,412 | <p>I'm having a hard time coming up with a formal proof by cases method for this set of premises and conclusion. Note that <span class="math-container">$\neg$</span> refers to negation and <span class="math-container">$\wedge$</span> denotes AND. | Denotes subproof</p>
<ol>
<li><p><span class="math-container">$\neg(A\wedge B)$</span></p></li>
<li><p><span class="math-container">$\neg(A\wedge\neg B)$</span></p></li>
</ol>
<p>Thus,</p>
<ol start="2">
<li><span class="math-container">$\neg A$</span></li>
</ol>
<p>Use # for contradiction; justify subproof assumptions with Assume; always drop outer parentheses; no spaces in PROP.</p>
<ol>
<li>~(A&B) Premise</li>
<li>~(A&~B) Premise</li>
<li>| B Assume</li>
<li>|| ~B Assume</li>
<li>|| B Reit;3</li>
<li>|| # #Intro;4,5</li>
<li>| A #Elim;6</li>
<li>| A&~B &Intro;4,7</li>
<li>| # #Intro;2,8</li>
<li>~A #Elim;9</li>
</ol>
<p>This is what I have currently, but am not sure if the steps I took are okay.</p>
| Lutz Lehmann | 115,115 | <p>Insert the Taylor expansion
<span class="math-container">$$
f(x_k)=f(ξ+e_k)=0+f'(ξ)e_k+\frac12f''(ξ)e_k^2+...
$$</span>
However, apart from not knowing what <span class="math-container">$p_2$</span> is relative to <span class="math-container">$p_1$</span>, it also does not make systematic use of <span class="math-container">$p_2(x_{n+1})=0$</span> as suggested.</p>
|
3,124,412 | <p>I'm having a hard time coming up with a formal proof by cases method for this set of premises and conclusion. Note that <span class="math-container">$\neg$</span> refers to negation and <span class="math-container">$\wedge$</span> denotes AND. | Denotes subproof</p>
<ol>
<li><p><span class="math-container">$\neg(A\wedge B)$</span></p></li>
<li><p><span class="math-container">$\neg(A\wedge\neg B)$</span></p></li>
</ol>
<p>Thus,</p>
<ol start="2">
<li><span class="math-container">$\neg A$</span></li>
</ol>
<p>Use # for contradiction; justify subproof assumptions with Assume; always drop outer parentheses; no spaces in PROP.</p>
<ol>
<li>~(A&B) Premise</li>
<li>~(A&~B) Premise</li>
<li>| B Assume</li>
<li>|| ~B Assume</li>
<li>|| B Reit;3</li>
<li>|| # #Intro;4,5</li>
<li>| A #Elim;6</li>
<li>| A&~B &Intro;4,7</li>
<li>| # #Intro;2,8</li>
<li>~A #Elim;9</li>
</ol>
<p>This is what I have currently, but am not sure if the steps I took are okay.</p>
| Carl Christian | 307,944 | <p>Let <span class="math-container">$\xi$</span> denote the root and let <span class="math-container">$x_0 \not = x_1$</span> denote the initial approximation. We assume that <span class="math-container">$f(x_1) \not = f(x_2)$</span> such that the secant step
<span class="math-container">$$x_2 = x_1 - \frac{x_1 -x_0}{f(x_1)-f(x_0)} f(x_1) = x_1 - \frac{f(x_1)}{f[x_0,x_1]}$$</span> is well-defined. Let <span class="math-container">$e_i = \xi - x_i$</span> denote the error at the <span class="math-container">$i$</span>th iteration. Our goal is develop an error formula which connects <span class="math-container">$e_2$</span> to <span class="math-container">$e_1$</span> and <span class="math-container">$e_0$</span>. We have
<span class="math-container">\begin{align}
e_2 &= \xi - x_2 = \xi - x_1 + \frac{f(x_1)}{f[x_0,x_1]} = e_1 - \frac{f(\xi)-f(x_1)}{f[x_0,x_1]} = e_1 - \frac{f[x_1,\xi]e_1}{f[x_0,x_1]} \\ &= e_1 \left(1 - \frac{f[x_1,\xi]}{f[x_0,x_1]}\right) = - e_1 \left( \frac{f[x_1,\xi]-f[x_0,x_1]}{f[x_0,x_1]} \right) = -e_1 e_0 \frac{f[x_0,x_1,\xi]}{f[x_0,x_1]}
\end{align}</span>
<hr>A sanity check is in order before we proceed. In the limit where <span class="math-container">$x_0 \rightarrow x_1$</span>, the secant step becomes a Newton step and <span class="math-container">$e_1 \rightarrow e_0$</span>. Our formula reduces to
<span class="math-container">$$ e_2 = - e_0^2 \frac{f[x_0,x_0,\xi]}{f'(x_0)} = -\frac{1}{2} e_0^2 \frac{f''(\theta)}{f'(x_0)},$$</span>
for at least one value of <span class="math-container">$\theta$</span>. In other words, we recover the error formula for Newton's method.
<hr>
In general, we have
<span class="math-container">$$e_{n+1} = -e_n e_{n-1} \frac{f[x_{n-1},x_n,\xi]}{f[x_{n-1},x_n]}.$$</span>
In the event of convergence, we have
<span class="math-container">$$ f[x_{n-1},x_n] \rightarrow f'(\xi), \quad n\rightarrow \infty, \quad n \in \mathbb{N} $$</span> and <span class="math-container">$$ f[x_{n-1},x_n,\xi] \rightarrow \frac{1}{2} f''(\xi), \quad n\rightarrow \infty. \quad n \in \mathbb{N}$$</span>
We make the natural assumptions that <span class="math-container">$f'(\xi) \not = 0$</span> and <span class="math-container">$f''(\xi) \not = 0$</span> and define <span class="math-container">$$A = - \frac{1}{2}\frac{f''(\xi)}{f'(\xi)}$$</span> This definition allows us to write
<span class="math-container">$$ e_{n+1} \approx A e_{n} e_{n-1}$$</span>
but we <em>cannot</em> expect equality as this is only an approximation. Many texts will ignore this distinction and simply proceed from <span class="math-container">$$e_{n+1} = A e_{n} e_{n-1}.$$</span> We shall not make this mistake. Instead, we exploit the fact that
<span class="math-container">$$ \left| \frac{f[x_{n-1},x_n,\xi]}{f[x_{n-1},x_n]} \right| \rightarrow |A|, \quad n\rightarrow \infty, \quad n \in \mathbb{N}.$$</span>
Given <span class="math-container">$\epsilon > 0$</span>, there exists <span class="math-container">$N = N(\epsilon)$</span> such for all <span class="math-container">$n \ge N$</span> we have
<span class="math-container">$$ \left| \frac{f[x_{n-1},x_n,\xi]}{f[x_{n-1},x_n]} \right| \leq |A| + \epsilon =: C(\epsilon).$$</span>
In reality, there is no harm in assuming that <span class="math-container">$N=1$</span>. If necessary, we simply discard the first finitely many approximations and renumber the rest. From our error formula we deduce
<span class="math-container">$$ |e_{n+1}| \leq |e_n||e_{n-1}| C(\epsilon).$$</span>
It follows that
<span class="math-container">$$ C(\epsilon) |e_{n+1}| \leq (C(\epsilon) |e_n|) (C(\epsilon) |e_{n-1}|).$$</span>
At this point we pause to recognize that we must have
<span class="math-container">$$ C(\epsilon) |e_n| \leq g_n$$</span>
where
<span class="math-container">$$ g_0 = C(\epsilon) |e_0|, \quad g_1 = C(\epsilon) |e_1|, \quad g_{n+1} = g_{n} g_{n-1}.$$</span>
We now impose the vital conditions that <span class="math-container">$$g_0 < 1, \quad g_1 < 1$$</span> to ensure <span class="math-container">$$g_n \rightarrow 0_+, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.$$</span>
This is what we will mean when we say that <span class="math-container">$x_0$</span> and <span class="math-container">$x_1$</span> are <em>sufficiently</em> close to <span class="math-container">$\xi$</span>.</p>
<p>By applying the logarithm we find the Fibonacci sequence
<span class="math-container">$$ \log g_{n+1} = \log g_{n} + \log g_{n-1},$$</span>
We have
<span class="math-container">$$ \log g_n = a_0 \lambda^n + a_1 \mu^n $$</span>
where
<span class="math-container">$$ \lambda = \frac{1+\sqrt{5}}{2}, \quad \mu= \frac{1-\sqrt{5}}{2} $$</span>
and
<span class="math-container">$$ \begin{bmatrix} 1 & 1 \\ \lambda & \mu \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \end{bmatrix} = \begin{bmatrix} \log g_0 \\ \log g_1 \end{bmatrix}. $$</span>
We have
<span class="math-container">$$
\begin{bmatrix} a_0 \\ a_1 \end{bmatrix} = \frac{1}{\mu - \lambda} \begin{bmatrix} \mu & -1 \\ -\lambda & 1 \end{bmatrix} \begin{bmatrix} \log g_0 \\ \log g_1 \end{bmatrix}.
$$</span>
so <span class="math-container">$a_0 < 0$</span>. We stress that <span class="math-container">$a_0 < 0$</span> also follows from <span class="math-container">$g_n \rightarrow 0_+$</span>, because <span class="math-container">$\lambda > 1$</span> and <span class="math-container">$|\mu| < 1$</span>. We can now conclude
<span class="math-container">$$ |e_n| \leq \frac{1}{C(\epsilon)} \exp( a_0 \lambda^n + a_1 \mu^n) $$</span>
At this point, we are very nearly done. We consider the right hand side expression and define
<span class="math-container">$$ y_n = \frac{1}{C(\epsilon)} \exp( a_0 \lambda^n + a_1 \mu^n).$$</span>
Then <span class="math-container">$$\frac{y_{n+1}}{y_n^\lambda} = C(\epsilon)^{\lambda-1} \frac{\exp( a_0 \lambda^{n+1} + a_1 \mu^{n+1})}{\exp( a_0 \lambda^{n+1} + a_1 \lambda \mu^n).} \rightarrow C(\epsilon)^{\lambda-1} \not = 0, \quad n \rightarrow \infty$$</span>
This shows that the auxiliary sequence <span class="math-container">$y_n$</span> converges to zero and the order is <span class="math-container">$\lambda$</span>. By definition, it follows that <span class="math-container">$|e_n|$</span> converges to zero with order at least <span class="math-container">$\lambda$</span>.
<hr> I have a few comments regarding the original exercise which prompted the question above. I have no idea what <span class="math-container">$p_2$</span> is. The text contains a few oddities and it is plausible that <span class="math-container">$p_2$</span> is simply a typographical error. Subquestion (c) asks for a conclusion which is not entirely correct. I have supplied the correct answer. Subquestion (d) asks for a conclusion which is not entirely correct. I have pushed the analysis as far as I can. The underlying issue is fundamental. The mathematical definition of order <span class="math-container">$p$</span> convergence is beautiful, but essentially useless for the analysis of some algorithms such as the secant method and even bisection. Newton's method is the only exception which comes to mind. Why? Normally, we work with upper bounds for the error and we can show that the upper bounds decay to zero with order <span class="math-container">$p$</span>. In such cases, we say that the original errors tend to zero with an order that is at least <span class="math-container">$p$</span>.</p>
|
70,976 | <blockquote>
<p>I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.</p>
</blockquote>
<p>I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.</p>
| Chris Eagle | 5,203 | <p>If $n$ is even, then $2$ divides $\sqrt{-n}^2=-n$ but does not divide $\sqrt{-n}$, so $2$ is a nonprime irreducible. In a UFD, all irreducibles are prime, so this shows $\mathbb{Z}[\sqrt{-n}]$ is not a UFD.</p>
<p>Similarly, if $n$ is odd, then $2$ divides $(1+\sqrt{-n})(1-\sqrt{-n})=1+n$ without dividing either of the factors, so again $2$ is a nonprime irreducible.</p>
<p>This argument works equally well for $n=3$, but fails for $n=1,2$, and in fact $\mathbb{Z}[\sqrt{-1}]$ and $\mathbb{Z}[\sqrt{-2}]$ are UFDs.</p>
|
3,580,068 | <p>Let <span class="math-container">$f:[0,1] \rightarrow \mathbb{R}$</span> continuous such that <span class="math-container">$\int_0^1 xf(x)\,dx=\frac{\pi}{4}$</span>. Prove that there is <span class="math-container">$c\in (0,1)$</span> such that <span class="math-container">$c^3f(c)+cf(c)-1=0$</span>.</p>
<p>Here is what I think, MVT for integrals gives that there is some <span class="math-container">$c\in(0,1)$</span> such that <span class="math-container">$cf(c)=\frac{\pi}{4}$</span>, but I doubt this is relatable with the condition to prove. Maybe someone can give me a hint?</p>
| LHF | 744,207 | <p>Since <span class="math-container">$$\displaystyle\frac{\pi}{4}=\int_0^1\frac{1}{1+x^2}\, dx$$</span> </p>
<p>we can write the condition as:</p>
<p><span class="math-container">$$\int_0^1\left(xf(x)-\frac{1}{1+x^2}\right)\,dx=0$$</span></p>
<p>and from the mean value theorem, there exists some <span class="math-container">$c\in (0,1)$</span> such that:</p>
<p><span class="math-container">$$cf(c)-\frac{1}{1+c^2}=0$$</span></p>
<p>This is equivalent with the equality to prove.</p>
|
3,580,068 | <p>Let <span class="math-container">$f:[0,1] \rightarrow \mathbb{R}$</span> continuous such that <span class="math-container">$\int_0^1 xf(x)\,dx=\frac{\pi}{4}$</span>. Prove that there is <span class="math-container">$c\in (0,1)$</span> such that <span class="math-container">$c^3f(c)+cf(c)-1=0$</span>.</p>
<p>Here is what I think, MVT for integrals gives that there is some <span class="math-container">$c\in(0,1)$</span> such that <span class="math-container">$cf(c)=\frac{\pi}{4}$</span>, but I doubt this is relatable with the condition to prove. Maybe someone can give me a hint?</p>
| Milo Brandt | 174,927 | <p>You can get this somewhat easily by examining cases: since <span class="math-container">$f$</span> is continuous, so is the function taking <span class="math-container">$c$</span> to <span class="math-container">$c^3f(c)+cf(c)-1$</span>, meaning that if this quantity is never zero, it must either always be negative or positive. So there are two cases: If this quantity is always negative we get
<span class="math-container">$$c^3f(c)+cf(c)-1<0$$</span>
which, for <span class="math-container">$c\in (0,1]$</span> rearranges to
<span class="math-container">$$cf(c)<\frac{1}{c^2+1}.$$</span>
The other case would rearrange to
<span class="math-container">$$cf(c)>\frac{1}{c^2+1}.$$</span>
However, if you integrate both sides of either inequality over <span class="math-container">$[0,1]$</span>, you get that the integral on the right is <span class="math-container">$\int_{0}^1\frac{1}{c^2+1}\,dc=\frac{\pi}4$</span> and that the integral on the left is either greater than or less than this - but cannot be equal because the inequality of functions is strict everywhere. This is basically an application of the intermediate value theorem.</p>
|
260,516 | <p>I was inspired by <a href="https://math.stackexchange.com/questions/2062960/there-exist-infinite-many-n-in-mathbbn-such-that-s-n-s-n-frac1n2?noredirect=1#comment4336226_2062960">this</a> topic on Math.SE.<br>
Suppose that $H_n = \sum\limits_{k=1}^n \frac{1}{k}$ - $n$th harmonic number. Then</p>
<h2>Conjecture</h2>
<blockquote>
<p>Let $M$ be a set of all $n$ such that
$$H_n - \lfloor{H_n\rfloor} < \frac{1}{n^{1+\epsilon}}.$$
Then
$$\forall\epsilon>0 : |M| = \bar\eta(\epsilon) < \infty.$$</p>
</blockquote>
<p>Picture below illustrates my conjecture, where I have checked this conjecture for $n < 10^6$ for each $\epsilon\in(0,1.1)$ with step 0.01
<a href="https://i.stack.imgur.com/ArpSz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ArpSz.png" alt="enter image description here"></a> </p>
<p>The following picture based on data provided by @GottfriedHelms for $n \approx 10^{100}$ (see answer below).
<a href="https://i.stack.imgur.com/ySVSZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ySVSZ.png" alt="enter image description here"></a></p>
| Gerhard Paseman | 3,402 | <p>I am letting $A = H_n - \lfloor H_n \rfloor$ and computing it in a dumb way, and then computing $1/nA$ when $A \lt 1/n $. For $n=83$ I get $1/nA \approx 5.825$, which is the largest such value I find so far, and which means 83 is one of three values of $n$ which are allowed when $\epsilon$ is 0.3. A smarter way to compute the values of $n$ for which $1/nA$ is worth computing is to use $\log(n+1/2) + \gamma$ for just those $n$ for which the expression is very near an integer, for $H_n$ is within $1/24n^2$ of this value. $\log(1/nA )/\log(n)$ then serves as an upper bound on $\epsilon$.</p>
<p>My calculations show $1 \lt 1/nA \lt 2$ much of the time, which means $\epsilon \lt 1/\log n$ for admissible $n$. I expect the conjecture to hold and when proved to reveal some of the nature of Euler's $\gamma$.</p>
<p>Gerhard "Still Computing Oh So Slowly" Paseman, 2017.01.25.</p>
|
3,308,291 | <p>I have an array of numbers (a column in excel). I calculated the half of the set's total and now I need the minimum number of set's values that the sum of them would be greater or equal to the half of the total. </p>
<p>Example:</p>
<pre><code>The set: 5, 5, 3, 3, 2, 1, 1, 1, 1
Half of the total is: 11
The least amount of set values that need to be added to get 11 is 3
</code></pre>
<p>What is the formula to get '3'?</p>
<p>It's probably something basic but I have not used calculus in a bit hence may have forgotten just forgotten it.</p>
<p>Normally I would use a simple while loop with a sort but I am in excel so I was wondering is there a more elegant solution. </p>
<p>P.S. I have the values sorted in descending order to make things easier.</p>
<p>EDIT: Example</p>
| Sayan Goswami | 595,808 | <p>1+1=2 and 4+4=3 is not in your set. So it is not a semigroup.
Ok, now you can restrict your operation as, you will not alow a*a, i.e, if you say that the operation between same element is not allowed, i.e., 1+1 and 4+4 is not defined then {0,1,4} is closed. But it is not so much interesting.</p>
|
2,582,046 | <p>My professor showed the following false proof, which showed that complex numbers do not exist. We were told to find the point where an incorrect step was taken, but I could not find it. Here is the proof: (Complex numbers are of the form <span class="math-container">$\rho e^{i\theta}$</span>, so the proof begins there) <span class="math-container">$$\large\rho e^{i\theta} = \rho e^{\frac{2 \pi i \theta}{2\pi}} = \rho (e^{2\pi i})^{\frac{\theta}{2\pi}} = \rho (1)^{\frac{\theta}{2\pi}} = \rho$$</span>
<span class="math-container">$$Note: e^{i\pi} = -1, e^{2\pi i} = (-1)^2 = 1$$</span>
Since we started with the general form of a complex number and simplified it to a real number (namely, <span class="math-container">$\rho$</span>), the proof can claim that only real numbers exist and complex numbers do not. My suspicion is that the error occurs in step <span class="math-container">$4$</span> to <span class="math-container">$5$</span> , but I am not sure if that really is the case.</p>
| ArsenBerk | 505,611 | <p>Because for complex numbers, $e^{zc}\ne(e^z)^c$ for $z,c \in \mathbb{C}$.</p>
|
1,641,255 | <p>I am having difficulty find the centroid of the region that is bound by the surfaces $x^2+y^2+z^2-2az=0$ and $3x^2+3y^2-z^2=0$ (lying above $xy$ plane, consider the inner region). I know the first surface is a sphere, while the second is an infinite cone.</p>
<p>I just dont know how to even approach it as I dont know how to even visualize these surfaces.</p>
<p>Can anyone help?</p>
<p>My only thoughts would be to probably use spherical coordinates.</p>
<p>My possible thoughts( I have no clue if this will work, just some thoughts)</p>
<p>if we had</p>
<p>$$\rho^{2}=x^2+y^2+z^2$$</p>
<p>$$x=\rho\sin(\psi)\cos(\theta)$$</p>
<p>$$y=\rho\sin(\psi)\sin(\theta)$$</p>
<p>$$z=\rho\sin(\psi)$$</p>
<p>We could do some rearranging </p>
<p>$\rho=2acos(\psi)$</p>
<p>Since we are only considering above xy plane, maybe then $\theta$ would only run though fourth quadrant?</p>
<p>In regard to the centroid aspect, I know my answer will need to be of the form $C=(X_{c},Y_{c},Z_{c})$ where each component is obtained by evaluating the respective integral and dividing by the total mass.</p>
<p>But I am not sure, so looking for help.</p>
<p>Update: A suggestion was made to use cylindrical coordinates instead.</p>
<p>If I were to try this, I would have</p>
<p>$$x=r\cos\theta$$</p>
<p>$$y=r\sin\theta$$</p>
<p>$$x^2+y^2=r^2$$</p>
<p>$$z=z$$</p>
<p>And from rearranging I would have</p>
<p>$r^2+(z-a)^2=a^2$
and
$3r^2=z^2$</p>
<p>$$r=\pm \sqrt{\frac{z^2}{3}}$$</p>
<p>But I am still left confused on how to tie it all together. I am still wondering this even after all this time. I really would like some advice and help. Anyone? I am tried all I know and tried the current answer, but I am making no progress.</p>
| David G. Stork | 210,401 | <p>For the first equation, complete the square in $z$... that is, convert the equation to the form $x^2 + y^2 + (z - q)^2 = r^2$. That will give you a sphere centered at $\{0,0,q\}$ of radius $r$. You must express $q$ and $r$ in terms of your $a$.</p>
<p>For the second equation, note that the region is rotationally symmetric about the $z$ axis because the $x$-$y$ dependence is of the form $\alpha (x^2 + y^2)$ or $\alpha r^2$. That radius $r$ depends upon $z$. Hence this region is cone shaped.</p>
<p>To visualize, why not just use some simple computer plotting routine?</p>
<p>Note that both regions are symmetric around the $z$ axis, and thus the centroid must have coordinates $\{ 0, 0, z_0 \}$ for some $z_0$. Just perform the integral to find the center of mass in the $z$ direction.</p>
<p>Exploit the <em>cylindrical</em> symmetry of the problem, not the <em>spherical</em> (non) symmetry: Express everything in terms of $r$ (distance from vertical axis) and $z$. Only the $z$ integral is non-trivial.</p>
|
1,778,440 | <p>I'm new to combinatorics, Although I understood most of the concepts this one baffles me.</p>
<blockquote>
<p>How many words exist that have exactly $5$ distinct consonants and $2$ identical vowels?</p>
</blockquote>
<p>The Answer is $$\binom{21}{5} \binom{5}{1} \frac{7!}{2!}$$</p>
<p>My doubt is:
Why do we write $\binom{5}{1}$ when we have to select 2 vowels?(Answers with examples appreciated).</p>
| lomot | 338,275 | <p>Since we have 2 <strong>identical</strong> vowels, we should to choose only one "kind" of vowels to put it in two places.</p>
|
1,778,440 | <p>I'm new to combinatorics, Although I understood most of the concepts this one baffles me.</p>
<blockquote>
<p>How many words exist that have exactly $5$ distinct consonants and $2$ identical vowels?</p>
</blockquote>
<p>The Answer is $$\binom{21}{5} \binom{5}{1} \frac{7!}{2!}$$</p>
<p>My doubt is:
Why do we write $\binom{5}{1}$ when we have to select 2 vowels?(Answers with examples appreciated).</p>
| gt6989b | 16,192 | <p>You are selecting 1 vowel (since your problem requires 2 <em>identical</em> ones) and then choosing where to place consonants and vowels in $7!/2!$ ways</p>
|
3,309,511 | <p>Prove that there exists infinitely many pairs of positive real numbers <span class="math-container">$x$</span> and <span class="math-container">$y$</span> such that <span class="math-container">$x\neq y$</span> but <span class="math-container">$ x^x=y^y$</span>.</p>
<p>For example <span class="math-container">$\tfrac{1}{4} \neq \tfrac{1}{2}$</span> but </p>
<p><span class="math-container">$$\left( \frac{1}{4} \right)^{1/4} = \left( \frac{1}{2} \right)^{1/2}$$</span></p>
<p>I am confused how to approach the problem. I think we have to find all the sloutions in a certain interval, probably <span class="math-container">$(0,1]$</span>.</p>
| Especially Lime | 341,019 | <p>In fact is is true for any <span class="math-container">$a<b$</span> (in this case, <span class="math-container">$a=1/4, b=1/2$</span>) and any function <span class="math-container">$f$</span> satisfying</p>
<ul>
<li><span class="math-container">$f(a)=f(b)$</span> and</li>
<li><span class="math-container">$f$</span> is continuous on <span class="math-container">$[a,b]$</span></li>
</ul>
<p>that there are infinitely many pairs <span class="math-container">$x<y$</span> satisfying <span class="math-container">$f(x)=f(y)$</span>. </p>
<p>If <span class="math-container">$f$</span> is constant on <span class="math-container">$[a,b]$</span> this is trivially true; if not there is some <span class="math-container">$c\in [a,b]$</span> with <span class="math-container">$f(c)\neq f(a)$</span>, and then the result follows by applying the intermediate value theorem to <span class="math-container">$[a,c]$</span> and <span class="math-container">$[c,b]$</span>.</p>
|
94,525 | <p>I am trying to solve the equation
$$z^n = 1.$$</p>
<p>Taking $\log$ on both sides I get $n\log(z) = \log(1) = 0$.</p>
<p>$\implies$ $n = 0$ or $\log(z) = 0$</p>
<p>$\implies$ $n = 0$ or $z = 1$.</p>
<p>But I clearly missed out $(-1)^{\text{even numbers}}$ which is equal to $1$.</p>
<p>How do I solve this equation algebraically?</p>
| David Mitra | 18,986 | <p>Working in the reals, take the $n$th root of both sides. </p>
<p>If $n$ is even then you'd write $(z^n)^{1/n}=|z|$.</p>
<p>For example $z^2=1 \iff (z^2)^{1/2}=1^{1/2} \iff |z|=1$.</p>
<p>For odd powers, you could say, for example:
$z^3=1 \iff (z^3)^{1/3}=1^{1/3} \iff z =1$.</p>
|
94,525 | <p>I am trying to solve the equation
$$z^n = 1.$$</p>
<p>Taking $\log$ on both sides I get $n\log(z) = \log(1) = 0$.</p>
<p>$\implies$ $n = 0$ or $\log(z) = 0$</p>
<p>$\implies$ $n = 0$ or $z = 1$.</p>
<p>But I clearly missed out $(-1)^{\text{even numbers}}$ which is equal to $1$.</p>
<p>How do I solve this equation algebraically?</p>
| Robert Israel | 8,508 | <p>If you want to do this properly you need complex numbers. The logarithm is a multivalued function; $z^n = 1$ is equivalent not to $n \log z = \log 1$ but to
$n \log z = 2 \pi i m$ where $m$ is an arbitrary integer. If $n \ne 0$ this says
$\log z = (2 \pi i m)/n$ and $z = e^{2 \pi i m/n}$. In particular with $n = 2 m$, $z = e^{\pi i} = -1$.</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Charles Siegel | 622 | <p>As above, a groupoid is an object where every morphism is an isomorphism, and generalizes groups. As for why to get excited about them, they're useful in classification of things. Like, say you want to understand vector bundles on a space. One method of doing so is constructing the "stack" of vector bundles on that space, which, to each open set (actually, a bit more generally, but thinking concretely here rather than going to grothendieck topologies) associates the groupoid consisting of vector bundles on that open set along with isomorphisms, so that the set of vector bundles is the set of isomorphism classes in this groupoid. The stack made up this way has the property that any family of bundles (or whatever, groupoids work for many things) over some space T is equivalent to giving a morphism from T to the stack.</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| David Zureick-Brown | 2 | <p>A set is an example of a groupoid, and I care about groupoids as a generalization of <strong>sets</strong> as opposed to groups. My most fundamental tool is Yoneda's lemma, which says that one can think of a category C as being embedded in the category C-hat of presheaves (C-hat := Hom(C,Sets)); this is a really useful way to think for instance about the category of Schemes (which is special because instead of presheaves you actually get sheaves). Similarly, if you want to think about things like algebraic groups, it is extremely useful to consider Hom(Schemes,Grps) instead. </p>
<p>A stack is a generalization of the notion of a scheme, which one would like to think of as a functor from schemes to groupoids; this doesn't quite work (one only gets a `pseudofunctor') and the notion of a stack is a gadget that makes this work. Just as with algebraic groups, sometime you want the target of the functor that your geometric object represents to have the type of extra structure that your geometric object has; groupoids come up for me in moduli theory, where they keep track of extra automorphisms, or when trying to construct quotients by group actions, where you can keep track of stabilizers.</p>
<p>So not a very precise answer, but also not a technical one, so maybe it will be useful.</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Qiaochu Yuan | 290 | <p>Personally, the reason I'm interested in groupoids is something called <a href="http://math.ucr.edu/home/baez/counting/" rel="noreferrer">groupoid cardinality</a> and some other related ideas (the link contains a lot of other links). A motivating idea here is that certain sets X of algebraic objects have the property that ($\sum_{x \in X} 1$) is ugly but ($\sum_{x \in X} \frac 1{Aut(x)}$) is much nicer, and so we should think of this is the "true" cardinality of the set (which is actually a groupoid). </p>
<p>Interesting combinatorial stuff happens when you take this philosophy seriously: for example, the cardinality of the groupoid of finite sets and bijections between them is e. Why is this interesting? It suggests that one reason exponential generating functions are important is that the denominator of n! is an indication that what you're really working with is some kind of structure defined over the groupoid of finite sets and bijections. And indeed, there's an approach to combinatorics called species theory that defines a combinatorial species, such as "binary trees," as a functor from this groupoid to itself. From this information one can extract a generating function, but the really important point here is that constructions such as the sum of generating functions are seen to be "decategorifications" of more fundamental combinatorial constructions, so one can avoid the machinery of working with generating functions by working directly with species instead. A good reference here is <a href="http://books.google.com/books?hl=en&lr=&id=83odtWY4eogC&oi=fnd&pg=PR5&dq=trees+and+species+flajolet&ots=m_b5VIIqYZ&sig=pd6dsgaTPANLVBgm0Uvr1OrOLuo#v=onepage&q=trees%20and%20species%20flajolet&f=false" rel="noreferrer">Bergeron, Labelle, and Laroux</a>.</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Kevin H. Lin | 83 | <p>Let me expand a bit on what Dave said.</p>
<p>The Yoneda lemma tells us that given an object $X$ of a category $\mathcal C$, the (covariant, contravariant, whatever) functor $h_X : \mathcal C \to \mathsf{Set}$, which sends an object $Y$ to the set $\mathsf{Hom}(Y,X)$, can be thought of as the "same" as the object $X$. There are many situations in which we are interested in a functor $F : \mathcal C \to \mathsf{Set}$, and we might like to know whether $F$ is isomorphic to $h_M$ for some object $M$, because that reduces the study of $F$ to the study of a single object $M$. In such a case we say that $F$ is represented by $M$. The letter $M$ here, suggestively, stands for "moduli".</p>
<p>Example: Given a group $G$, the functor $BG' : \mathsf{Top} \to \mathsf{Set}$ is the functor which sends a topological space $\mathcal X$ to the set of isomorphism classes of principal $G$-bundles over $\mathcal X$. (You can also do the analogous thing for schemes.)</p>
<p>Example: The functor $M_g' : \mathsf{Sch} \to \mathsf{Set}$ is the functor which sends a scheme $X$ to the set of isomorphism classes of flat families of genus $g$ curves over $X$.</p>
<p>In both of the above examples, there is no object $M$ for which $h_M$ is isomorphic to the functor. So this is perhaps not so nice. But, without getting into too many details, there is a natural "fix", namely we can instead consider the functor $BG : \mathsf{Top} \to \mathsf{Groupoid}$ (resp. $M_g : \mathsf{Sch} \to \mathsf{Groupoid}$) which sends a topological space (resp. a scheme) to the <em>groupoid</em> of $G$-bundles (resp. flat families of genus $g$ curves). This groupoid has objects $G$-bundles and morphisms isomorphisms of $G$-bundles (resp. the obvious analogous thing). The original set-valued functor is just the composition of this functor with the functor $\mathsf{Groupoid}$ to $\mathsf{Set}$ which takes a groupoid and returns the set of isomorphism classes of objects in the groupoid.</p>
<p>Anyway, despite the fact that the set-valued functors are not so "geometric", since they are not represented by a "geometric" object (topological space and scheme, respectively), the groupoid-valued functors are more "geometric". In the case of $M_g$, the "geometric" structure we get is that of a "Deligne-Mumford stack", which essentially means that we can for practical purposes pretend that it is represented by a scheme with only some slightly "weird" properties. In the case of $BG$ (the topological one) you can take a "geometric realization" and recover the classifying space $BG$ that we know and love.</p>
<p>Another very important reason for studying groupoids and another very important class of groupoids comes from, as others have already mentioned, group actions. When a group acts on a manifold or a variety, the naive quotient may be badly behaved, for example it may no longer be a manifold (e.g. it might not be smooth, or it might not be Hausdorff) or respectively a variety (or it may not even be clear how to take the quotient at all!), which makes it harder to study geometric properties of the alleged "quotient". However, the groupoid viewpoint allows us to get a better handle on the quotient and its geometry. More precisely, if $G$ is a group acting on a space (manifold, scheme, variety, whatever) $X$, then the "correct" quotient is actually the functor $X/G : \mathcal C \to \mathsf{Groupoid}$ (where $\mathcal C$ is the category of manifolds, schemes, whatever) which sends an object $Y$ to the groupoid of pairs ($G$-bundles $E$ over $Y$, $G$-equivariant morphism from the total space of $E$ to $X$). The functor $BG$ is a special case of this; it's $\mathrm{pt}/G$.</p>
<p>There's some further discussion on this sort of stuff at the nLab: </p>
<p><a href="http://ncatlab.org/nlab/show/moduli+space" rel="nofollow noreferrer">http://ncatlab.org/nlab/show/moduli+space</a></p>
<p><a href="http://ncatlab.org/nlab/show/classifying+space" rel="nofollow noreferrer">http://ncatlab.org/nlab/show/classifying+space</a></p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Ari | 673 | <p>While the categorical definition of groupoid is the most concise, you can also think of a groupoid as being like a group, except where multiplication is only partially defined, rather than being defined for any pair of elements. Here are a few of my favorite examples:</p>
<ul>
<li><p>Given a vector bundle E, the general linear groupoid GL(E) is the groupoid of linear isomorphisms between fibers. Given a map from Ex -> Ey, and another from Ey' -> Ez, we can only compose them if y=y'. When E is just a vector space over a single point, then this is the usual general linear group. In differential geometry, this gives a very natural way to think about frames and G-structures on a differentiable manifold M: just look at the general linear groupoid GL(TM). G-structures can be understood as subgroupoids: for example, a Riemannian structure corresponds to the orthogonal subgroupoid O(TM), consisting of elements of GL(TM) which are also isometries.</p></li>
<li><p>More generally, given a principal G-bundle, the gauge groupoid consists of G-equivariant maps between fibers. This is useful for talking about connections, holonomy, etc., without having to fix a particular gauge.</p></li>
<li><p>Given a directed graph, one can construct the free groupoid generated by the edges. As a special case, the free group on n elements is generated by the graph with one vertex and n self-loops. (There is also a forgetful functor from groupoids to graphs, which is adjoint to the free functor.)</p></li>
</ul>
<p>The first two examples happen to be <em>Lie groupoids</em>, and they have corresponding <em>Lie algebroids</em>, which generalizes the relationship between Lie groups and Lie algebras. Whereas a Lie algebra is a vector space with a bracket between elements, a Lie algebroid is a vector bundle with a bracket between sections (as well as an additional structure called the anchor map). For example, if Q is a principal G-bundle, then the gauge groupoid is (Q x Q)/G, while the corresponding gauge algebroid is TQ/G. This comes in handy in geometric mechanics, particularly in reduction theory. If we have a Lagrangian L: TQ -> R, which is invariant with respect to the action of a Lie group G, then is useful to look at the reduced Lagrangian \ell: TQ/G -> R. There are some subtleties arising from the fact that TQ/G is not a tangent bundle, but it <em>is</em> still a Lie algebroid, so this has motivated the study of mechanics on Lie algebroids.</p>
|
744,952 | <p>Is it true that a map between ${\bf T1}$ topological spaces $f:X \to Y$ is surjective iff the induced geometric morphism $f:Sh(Y) \to Sh(X)$ is a surjection (i.e. its inverse image part $f^*$ is faithful)?</p>
<p>In "Sheaves in Geometry and Logic" a proof is given, but the the "if" part leaves me a bit unsatisfied (and, btw, the "only if" part holds for any topological space).
Thanks in advance!</p>
| E.P. | 30,935 | <p>It is obvious that this cannot hold unless $f$ is continuous at $0$, because otherwise the value $f(0)$ becomes decoupled to the continuum of values that determine the integral, and the result is meaningless. Therefore, whatever route you take must be based upon this continuity condition.</p>
<p>So, how do you do this? First of all, you unpack the definition. Let $\newcommand{\eps}{\varepsilon}\eps$ be some positive number which we'll fix later. Then we know that there exists $\delta=\delta_\eps$ such that
$$
|x|<\delta\Longrightarrow |f(x)-f(0)|<\eps.
$$
How does this relate to your integral? Your transformed form,
$$
I=\int_0^\infty f(t/a) \frac{\sin(t)}{t}\mathrm dt,
$$
involves values of $x=t/a$ which are both below and above the $\delta$ threshold. The ones below, i.e. those $t$ such that $|t|<\delta a$, have $f(t/a)$ close to $f(0)$, which means that they will end up contributing the most, whereas the other ones are unconnected to $f(0)$ and must therefore decouple in some way. To make this change in behaviour explicit, you can split up your integral:
$$
I=J+K=
\int_{0}^{\delta a} f(t/a) \frac{\sin(t)}{t}\mathrm dt
+
\int_{\delta a}^\infty
f(t/a) \frac{\sin(t)}{t}\mathrm dt.
$$</p>
<hr>
<p>Now, it is tempting to deal with the two chunks at the same time, but the truth is that they require different handling. Let's deal with the messy, unimportant one first.
$$
K=K(a)=\int_{\delta a}^\infty f(t/a) \frac{\sin(t)}{t}\mathrm dt
$$
goes to zero as $a\to\infty$. This is semi-plausible: you are taking the tails of the integral, which ought to vanish, but your function $f(t/a)$ rescales itself with $a$ to match the outgoing lower limit, so that it always equals $f(\delta)$ there. It is therefore the decay of the $\mathrm{sinc}$ function that drives the limit.</p>
<p>One way to make this precise is to use the Cauchy-Schwarz inequality in what is effectively an $L_2$ dot product:
$$
|K(a)|^2
\leq
\int_{\delta a}^\infty f(t/a)^2\mathrm dt
\int_{\delta a}^\infty \frac{\sin^2(t)}{t^2}\mathrm dt.
$$
The first one is constant, equal to $\int_{\delta}^\infty f(t)^2\mathrm dt$ and bounded by $\int_{-\infty}^\infty f(t)^2\mathrm dt$, while the second one converges to zero. Thus $\lim_{a\to\infty}K(a)=0$.</p>
<hr>
<p>The interesting integral, on the other hand,
$$J=J(a)=\int_{0}^{\delta a} f(t/a) \frac{\sin(t)}{t}\mathrm dt,$$
needs to converge to $\propto f(0)$, so we need to use the closeness of $f(t/a)$ to $f(0)$. You can thus unpack $f(x)$ into $f(0)+(f(x)-f(0))$ to get
$$
J=
f(0)\int_{0}^{\delta a} \frac{\sin(t)}{t}\mathrm dt
+
\int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt.
$$
The second term is essentially bounded by $\eps$. You could try to say that
$$
\left|\int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt\right|
<
\eps\int_{0}^{\delta a} \left|\frac{\sin(t)}{t}\right|\mathrm dt
$$
but the second integral diverges. Instead, one repeats the Cauchy-Schwarz trick:
$$
\begin{align}
\left|\int_{0}^{\delta a} \left[f(t/a)-f(0)\right] \frac{\sin(t)}{t}\mathrm dt\right|^2
&\leq
\int_{0}^{\delta a} \left|f(t/a)-f(0)\right|^2\mathrm dt\int_{0}^{\delta a} \left|\frac{\sin(t)}{t}\right|^2\mathrm dt
\\ & <
\delta a \eps^2\int_{-\infty}^{\infty} \left|\frac{\sin(t)}{t}\right|^2\mathrm dt.
\end{align}
$$
You now have enough ingredients to get the limit, and you can get the nice number by doing the Dirichlet integral
$$
\int_{0}^\infty \frac{\sin(t)}{t}\mathrm dt=\frac\pi2.
$$
The limiting procedure takes a bit of care, because you have a fair few quantities involved. If you package all the results so far, you can say that
$$
|I-\frac\pi2 f(0)|
<
f(0)\left|\int_{\delta a}^\infty \frac{\sin(t)}{t}\mathrm dt\right|
+
\eps\sqrt{\delta a} \|\mathrm{sinc}\|
+
\|f\|\sqrt{\int_{\delta a}^\infty \left|\frac{\sin(t)}{t} \right|^2 \mathrm dt},
$$
and you want to bound this arbitrarily close to zero. This is a bit of a balancing act, because you need $\delta a$ to be large and $\eps\sqrt{\delta a}$ to be small, but it can be done. To do it, take $A$ such that the first and third terms are small when $\delta a=A$; then, choose $\eps$ such that the second term is small enough, for $\delta a=A$ fixed; this determines the necessary $\delta$; and finally take $a=A/\delta$. This is enough to guarantee the limit.
$$\tag*{$\blacksquare$}$$</p>
<hr>
<p>Finally, let me put in a word or two about why I went all this way instead of cracking the problem with some huge theorem. Essentially, trying to apply some form of dominated convergence theorem will never work, because there <em>is</em> no dominating function. In physicist language, the Dirichlet kernel converges to a Dirac delta function:
$$\lim_{a\to\infty}\frac{\sin(ax)}{x}=\frac\pi2 \delta(x).$$
This is loose language but its precise statement is exactly the integral you want to prove.</p>
<p>In other words, this integral breaks exactly the same dominated convergence schemes that you're trying to use, and the limit - in whatever sense it exists (which it does) - is outside of the function space you're looking in. This is why, unless you have fancier tools available, you need to work it from the ground up.</p>
|
2,930,292 | <p>I'm currently learning the unit circle definition of trigonometry. I have seen a graphical representation of all the trig functions at <a href="https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/unit-circle-definition-of-trig-functions/a/trig-unit-circle-review" rel="nofollow noreferrer">khan academy</a>. </p>
<p><img src="https://i.stack.imgur.com/zp5WB.png" alt="enter image description here"></p>
<p>I understand how to calculate all the trig functions and what they represent. Graphically, I only understand why sin and cos is drawn the way it is. I'm having trouble understand why tangent, cotangent, secant, cosecant are drawn the way they are. </p>
<p>Can someone please provide me with some intuitions.</p>
| Kurt Schwanda | 592,591 | <p>At least in the first quadrant of the unit circle, the tangent of the angle is equal to the length of the tangent segment connecting the point on the circle to the x-axis.</p>
<p>So in your image, imagine a radius being drawn from the origin to the green point on the circle's circumference (let's call it <span class="math-container">$P$</span>). Call the angle between this radius and the positive x-axis <span class="math-container">$\theta$</span>.</p>
<p>In terms of the x and y coordinates of <span class="math-container">$P$</span>, <span class="math-container">$\tan \theta = \frac y x$</span>.</p>
<p>If you think more in terms of right triangle trig and "SOH CAH TOA", then <span class="math-container">$\tan \theta = \frac{\text{ opposite}}{\text{ adjacent}} = \frac{\text{ length of tangent segment }}{\text{ length of radius }} = \frac{\text{ length of tangent segment }}{1} = \text{ length of tangent segment }$</span>.</p>
|
2,845,085 | <p>Find $f(5)$, if the graph of the quadratic function $f(x)=ax^2+bx+c$ intersects the ordinate axis at point $(0;3)$ and its vertex is at point $(2;0)$</p>
<p>So I used the vertex form, $y=(x-2)^2+3$, got the quadratic equation and then put $5$ instead of $x$ to get the answer, but it's wrong. I think I shouldn't have added $3$ in the vertex form but I don't know how else I can solve this</p>
| Rory Daulton | 161,807 | <p>You are right that one solution is to use the vertex form</p>
<p>$$y=a(x-h)^2+k$$</p>
<p>but you have $k$ wrong. If the vertex is at $(2,0)$ then $h=2$ (which you have) and $k=0$ (which you got wrong). So the equation is now</p>
<p>$$y=a(x-2)^2+0$$
or
$$y=a(x-2)^2$$</p>
<p>Now use the fact that $(0,3)$ is on the parabola, so $x=0,y=3$ satisfies the equation:</p>
<p>$$3=a(0-2)^2$$</p>
<p>So $a=\frac 34$. Thus the full equation is</p>
<p>$$y=\frac 34(x-2)^2$$</p>
<p>Finally, we want the value of $y$ when $x=5$, which is</p>
<p>$$y=\frac 34(5-2)^2$$
or
$$y=\frac{27}4$$</p>
|
514,922 | <p>I need to prove the following affirmation: If $ \lim x_{2n} = a $ and $ \lim x_{2n-1} = a $, prove that $\lim x_n = a $ (in $ \mathbb{R} $ )</p>
<p>It is a simple proof but I am having problems how to write it. I'm not sure it is the right way to write, for example, that the limit of $(x_{2n})$ converges to a:</p>
<p>$ \forall \epsilon > 0 \; \exists n_1 \in \mathbb{N} $ such that if $n \in \mathbb{N}$ and $n \geq n_1$, then (*) $ | x_{2n} - a| < \epsilon $</p>
<p>About the (*) step, is that correct to write $ x_{2n} $? Or should I use another notation?</p>
<p>Thanks for the help! </p>
| kedrigern | 97,299 | <p>$$\forall\varepsilon>0 \; \exists n_1\in\mathbb{N} \; \forall k>n_1 |a_{2k} - a|<\varepsilon
\\ \forall\varepsilon>0 \; \exists n_2\in\mathbb{N} \; \forall k>n_2 |a_{2k+1} - a|<\varepsilon$$
Therefore if we set $n'=\max\{n_1,n_2\}$ we get
$$\forall\varepsilon>0 \; \exists n'\in\mathbb{N} \; \forall k>n' \;|a_{2k} - a|<\varepsilon
\\ \forall\varepsilon>0 \; \exists n'\in\mathbb{N} \; \forall k>n' \; |a_{2k+1} - a|<\varepsilon$$
As $2k>2n', 2k+1>2n'+1$ we can set $n=2n'$ and get
$$\forall\varepsilon>0 \; \exists n\in\mathbb{N} \; \forall k>n \;|a_k - a|<\varepsilon$$</p>
|
2,918,091 | <p>Suppose I want to find the locus of the point $z$ satisfying $|z+1| = |z-1|$</p>
<p>Let $z = x+iy$</p>
<p>$\Rightarrow \sqrt{(x+1)^2 + y^2} = \sqrt{(x-1)^2 + y^2}$ <br/>
$\Rightarrow (x+1)^2 = (x-1)^2$ <br/>
$\Rightarrow x+1 = x-1$ <br/>
$\Rightarrow 1= -1$ <br/>
$\Rightarrow$ Loucus does not exist</p>
<p>Is my approach incorrect? The answer I was given was that the y-axis describes the locus.</p>
<p>Any help would be appreciated.</p>
| cansomeonehelpmeout | 413,677 | <p>Instead of taking the square-root of both sides, try to expand to get $$(x-1)^2=(x+1)^2\\x^2-2x+1=x^2+2x+1\\-2x=2x\\x=0$$</p>
|
1,928,149 | <p>I have the following general question about geodesics.
I know the following equation for a geodesic $\sigma$ on a manifold $M\subset R^n$ of dimension $m$, written in local coordinates:
$${\sigma^k}^{''} (t) + \Gamma_{i,j}^k {\sigma^{i}}'{\sigma^{j}}'=0,$$</p>
<p>for $i,j,k=1, \dots, m$.</p>
<p>Now, if I have a curve $\gamma(t)=(\gamma_1(t), \dots, \gamma_n(t))$ in $M$, how can I check that such a curve is a geodesic?</p>
<p>More precisely, how can I write my curve in local coordinates, in order to check if it satisfies my equation? I am stuck. Examples are really welcomed too.</p>
<p>Thank you.</p>
| janmarqz | 74,166 | <p>With a parametrization $\Phi$ from an open set $U$ of $\Bbb R^m$ and a curve $\sigma$ in $U$ you will get $\gamma=\Phi\circ\sigma$. For that $\sigma$ you ought to get your $\sigma^k$ in the geodesic equations.</p>
|
1,168,968 | <p>So I'm doing some cryptography assignment and I'm dealing with a modular arithmetic in hexadecimal. Basically I have the values for $n$ and the remainder $x$, but I need to find the original number $m$, e.g.</p>
<p>$$m \mod 0x6e678181e5be3ef34ca7 = 0x3a22341b02ad1d53117b.$$</p>
<p>I just need a formula to calculate $m$.</p>
<p>Edit: ok, let's put it this way, $x = K^e \mod n$, I know the values for $x$, $e$ and $n$. Does that help?</p>
<p>Ok, I realized I was approaching the problem in a wrong way, basically I had the RSA public key and I should have used RSA problem to decrypt the file without having the private key. Sorry for the stupid question.</p>
| user26486 | 107,671 | <p><strong>Hint:</strong></p>
<p>$$\frac{5+\sqrt{x^2+5}}{x-6}=\frac{\frac{5}{x}+\sqrt{\frac{x^2}{x^2}+\frac{5}{x^2}}}{\frac{x}{x}-\frac{6}{x}}=\frac{\frac{5}{x}+\sqrt{1+\frac{5}{x^2}}}{1-\frac{6}{x}}$$</p>
<p>$$\lim_{x\to\infty}\frac{\frac{5}{x}+\sqrt{1+\frac{5}{x^2}}}{1-\frac{6}{x}}=\frac{0+\sqrt{1+0}}{1-0}=1$$</p>
|
2,743,288 | <p>I need help with this exponential equation: $5^{x+2}\ 2^{4-x} = 1000 $</p>
<p>We know that $ 1000 = 10^3$, so:</p>
<p>$$\ln(5^{x+2}\cdot2^{4-x}) = \ln10^3 \implies\ln(5^{x+2}) + \ln(2^{4-x}) = \ln10^3$$</p>
<p>In the next step I use that: $\ln(a^x) = x\ln(a)$</p>
<p>$$(x+2)\ln 5 + (4-x)\ln 2 = 3\ln 10$$</p>
<p>And I'm stuck here. </p>
| Michael Hoppe | 93,935 | <p>$1000=5^{x+2}2^{4-x}=5^32^35^{x-1}2^{1-x}\iff 1=(5/2)^{x-1}\iff x=1$</p>
|
2,743,288 | <p>I need help with this exponential equation: $5^{x+2}\ 2^{4-x} = 1000 $</p>
<p>We know that $ 1000 = 10^3$, so:</p>
<p>$$\ln(5^{x+2}\cdot2^{4-x}) = \ln10^3 \implies\ln(5^{x+2}) + \ln(2^{4-x}) = \ln10^3$$</p>
<p>In the next step I use that: $\ln(a^x) = x\ln(a)$</p>
<p>$$(x+2)\ln 5 + (4-x)\ln 2 = 3\ln 10$$</p>
<p>And I'm stuck here. </p>
| danimal | 202,026 | <p>Remember that $\ln 5$, $\ln2$ etc are just (real) numbers, so they can be used like real numbers, so:
$$(x+2)\ln(5) + (4-x)\ln(2) = 3\ln(10)$$
expanding the brackets and factoring the $x$ s:
$$x(\ln5 -\ln2) = 3\ln10-2\ln5-4\ln2$$
and combining the logs
$$x\ln(5/2) = \ln1000-\ln25-\ln16 = \ln\left({1000\over 16\times25}\right)=\ln(2.5) $$
i.e.
$$x\ln(2.5)=\ln(2.5)$$
and so $$x=1$$</p>
|
1,836,190 | <p>I've been working on a problem and got to a point where I need the closed form of </p>
<blockquote>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$</p>
</blockquote>
<p>I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this: </p>
<p>$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2)n+1)}{(m+2) (m+3)}\binom{m+n+1}{ m+1}.$$</p>
<p>That's quite the nasty formula. Can anyone provide some insight or justification for that answer? </p>
| Leucippus | 148,155 | <p>The series can also be seen as the following.
\begin{align}
\sum_{k=0}^{n} k \, \binom{m+k}{m+1} &= \frac{1}{m+1} \, \sum_{k=0}^{n} k \, \frac{(m+1)_{k}}{k!} \\
&= \frac{1}{m+1} \, \left[ \sum_{k=0}^{n-2} \frac{(m+1)_{k+2}}{k!} + \sum_{k=0}^{n-1} \frac{(m+1)_{k+1}}{k!} \right] \\
&= (m+2) \, \frac{(n-1) \, \Gamma(m+n+2)}{\Gamma(n) \, \Gamma(m+4)} + \frac{ \Gamma(m+n+2)}{\Gamma(n) \, \Gamma(m+3)} \\
&= \frac{(m+n+1)!}{(n-1)! \, (m+3)!} \, (m \, n + 2n + 1) \\
&= \frac{m \, n + 2n + 1}{m+3} \, \binom{m+n+1}{n-1}.
\end{align}
Rearranging terms yields the form presented in the proposed problem. The notation used in Pochhammer's notation, namely,
\begin{align}
(x)_{k} = \frac{\Gamma(x+k)}{\Gamma(x)}.
\end{align}</p>
|
512,768 | <p>I am trying to intuitively understand why the solution to the following problem is $-2$. $$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$
$$\lim_{x\to\infty}(\sqrt{x^2-4x}-x)\frac{\sqrt{x^2-4x}+x}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{x^2-4x-x^2}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{-4x}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{-4}{\sqrt{1-\frac{4}{x}}+1}$$
$$\frac{-4}{\sqrt{1-0}+1}$$
$$\frac{-4}{2}$$
$$-2$$
I can understand the process that results in the answer being $-2$. However, I expected the result to be $0$. I have learned that when dealing with a limit approaching $\infty$, only the highest degree term matters because the others will not be as significant. For this reason, I thought that the $4x$ would be ignored, resulting in:
$$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$
$$\lim_{x\to\infty}\sqrt{x^2}-x$$
$$\lim_{x\to\infty}x-x$$
$$\lim_{x\to\infty}0$$
$$0$$
Why is the above process incorrect?</p>
| Paramanand Singh | 72,031 | <p>Neglecting lower order terms while dealing with certain limits is a sort of "thumb rule" which applies in certain circumstances and its proper justification in those circumstances is based on the standard rules of limits. You are trying to assume that this thumb rule is more fundamental than the rules of limits and can be used in any circumstances. After gaining reasonable experience with limits an expert is able to figure out where and how to apply this thumb rule correctly. But for a beginner it makes sense to apply the rules of limits rather than rely on these non-fool-proof thumb rules.</p>
<p>If you have a look at the comment from Daniel Fischer you can see how he is able to use the thumb rule effectively without any problem by transforming the expression under square root properly and then ignoring the constant term. The fundamental issue in applying the thumb rule is to figure out which terms to ignore and which ones to keep.</p>
|
57,232 | <p>Given a Heegaard splitting of genus $n$, and two distinct orientation preserving homeomorphisms, elements of the mapping class group of the genus $n$ torus, is there a method which shows whether or not these homeomorphisms, when used to identify the boundaries of the pair of handlebodies, will produce the same $3$-manifold?</p>
| John Sidles | 11,394 | <p>The high level of abstraction that number theorists sustain is a continual source of amazement to me ... doesn't anyone want to see what $\hat\mu(k,n)$ concretely <i>looks</i> like? </p>
<p>So let's do it! Mainly for fun (and as a gesture of respect for Gil), here is a density plot of $n^{1/2} \hat\mu(k,n)$ for all values $(n,k) \in 1,1024$ (note that a normalizing factor $n^{1/2}$ is included; this scaling yields near-uniform luminance in the plot):</p>
<p><a href="http://faculty.washington.edu/sidles/Moebius/Moebius__1024_canonical.png" rel="noreferrer">(canonical Moebius plot) http://faculty.washington.edu/sidles/Moebius/Moebius__1024_canonical.png</a></p>
<p>As usual, the argument (phase) of $\hat\mu(k,n)$ is encoded as hue, and the magnitude as saturation and value. For details, see the <a href="http://faculty.washington.edu/sidles/Moebius/Moebius_Mathematica.m" rel="noreferrer">Mathematica code</a> that produced the above plot (for many people, the most interesting aspect of the code will be the idiom for exporting Mathematica graphics to png files; certainly this is by far the longest part of the code, and the toughest to debug).</p>
<p>Such codes code encourages us to do numerical experiments ...</p>
<p>What happens if we randomize the sign of the Möbius function?</p>
<p><a href="http://faculty.washington.edu/sidles/Moebius/Moebius__1024_randomized.png" rel="noreferrer">(canonical Moebius plot) http://faculty.washington.edu/sidles/Moebius/Moebius__1024_randomized.png</a></p>
<p>Hmmm ... when we randomize the sign of the Möbius function, $\hat\mu(k,n)$ doesn't look much different, does it? </p>
<p>On the other hand, we know (from other numerical experiments) that when we compute a pseudo-Riemann function $\zeta^\prime(s) = (\sum_{r=1}^\infty \mu(r)/r^s)^{-1}$ using a randomized-sign Möbius function, then the resulting $\zeta^\prime(s)$ has a distribution of zeros that is grossly different from the distribution of zeros in $\zeta(s)$ (to see this, just try it).</p>
<p>In summary, the special properties of the distribution of primes (relative to randomized distributions) that are so plainly evident when we "look" at the zeros of $\zeta(s)$, are not immediately evident when we "look" at the magnitude and phase of $\hat\mu(k,n)$.</p>
<p>As for what this observation means (if anything at all) ... well ... that is for the number theorists to comment upon, not me! :)</p>
|
4,411,247 | <blockquote>
<p>If <span class="math-container">$G$</span> is finite group, how to prove that <span class="math-container">$f(g)=ag$</span>, <span class="math-container">$a \in G$</span>, is a bijection for all <span class="math-container">$g \in G$</span>? Here <span class="math-container">$ag$</span> is <span class="math-container">$a \cdot g$</span>, where <span class="math-container">$\cdot$</span> is the operator from the group <span class="math-container">$G$</span>.</p>
</blockquote>
<p>This is what I've tried so far:</p>
<p><span class="math-container">$f(g)=ag
\\
a^{-1}g(g)=g.
$</span></p>
<p>Since the inverse has been dotted so <span class="math-container">$f$</span> is a bijection, I would like to know if there's a misstep on that.</p>
| Berci | 41,488 | <p>The inverse of <span class="math-container">$g\mapsto ag$</span> is simply <span class="math-container">$h\mapsto a^{-1}h$</span>.</p>
|
2,713,311 | <p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $</p>
<p><strong>Answer:</strong></p>
<p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Rightarrow 2x-2ax-a-b=0 \ \ (?) $ </p>
<p>Comparing both sides , we get </p>
<p>$ 2-2a=0 \\ a+b=0 \ $</p>
<p>Solving , we get </p>
<p>$ a=1 , \ b=-1 \ $</p>
<p>But I am not sure about the above line where question mark is there.</p>
<p><strong>Can you help me?</strong></p>
| user | 505,767 | <p>Note that</p>
<p>$$\frac{x^2+1}{x+1}-ax-b=\frac{x^2+1-ax^2-bx-ax-b}{x+1}=\frac{x^2(1-a)-x(a+b)-b+1}{x+1}$$</p>
<p>and in order to have limi zero we need</p>
<ul>
<li>$(1-a)=0 \implies a=1$</li>
<li>$(a+b)=0\implies b=-1$</li>
</ul>
<p>indeed</p>
<p>$$\frac{x^2(1-1)-x(1-1)-(-1)+1}{x+1}=\frac{2}{x+1}\to 0$$</p>
|
2,401,281 | <blockquote>
<p>Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. </p>
</blockquote>
<p>This question is from a math olympiad contest. </p>
<p>I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.</p>
<p><strong>Note</strong>: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. </p>
| Dr. Sonnhard Graubner | 175,066 | <p>plugging $$c=-a-b$$ in the term $$2(a^4+b^4+c^4)$$ we get $$4\, \left( {a}^{2}+ab+{b}^{2} \right) ^{2}$$ and this is a perfect square.</p>
|
2,401,281 | <blockquote>
<p>Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. </p>
</blockquote>
<p>This question is from a math olympiad contest. </p>
<p>I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.</p>
<p><strong>Note</strong>: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. </p>
| lab bhattacharjee | 33,337 | <p>$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=?$$</p>
<p>Now $$(a^2)^2+(b^2)^2+(c^2)^2=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$</p>
<p>$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-abc(a+b+c)=?$$</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| javier | 914 | <p>The "Touching Soap Films" series by Springer. about minimal surfaces. Some excerpts of the video are available here:
<a href="http://page.mi.fu-berlin.de/polthier/video/Touching/Scenes.html" rel="nofollow">http://page.mi.fu-berlin.de/polthier/video/Touching/Scenes.html</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Charles Siegel | 622 | <p>Not lecture videos or anything, but the stuff from <a href="https://www.youtube.com/results?search_query=oliver+labs&search_type=&aq=f" rel="nofollow noreferrer">Oliver Labs</a> is very good for just illustrating geometric stuff, like blowups and dual curves.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| bhwang | 239 | <p>An excellent (and very lively) overview of basic one-variable calculus: Calculus I in 20 minutes: <a href="http://www.youtube.com/watch?v=EX_is9LzFSY" rel="nofollow">Part I</a>, <a href="http://www.youtube.com/watch?v=Q9OkFTDG4fY" rel="nofollow">Part II</a>.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Andy Putman | 317 | <p>Most of the talks at MSRI are videotaped and placed on the web here:</p>
<p><a href="http://www.msri.org/communications/vmath/index_html">http://www.msri.org/communications/vmath/index_html</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Ryan Budney | 1,465 | <p>This isn't purely a math video, it's an interview with Peter Woit and it is something of a summary of the main issues discussed on his blog and in his book. He talks about math vs. physics culture, especially the string theory community.</p>
<p>edit: the link appears to have changed.</p>
<p><a href="https://bigthink.com/u/peterwoit" rel="nofollow noreferrer">https://bigthink.com/u/peterwoit</a></p>
<p>Discussion here:</p>
<p><a href="http://www.math.columbia.edu/%7Ewoit/wordpress/?p=2670" rel="nofollow noreferrer">http://www.math.columbia.edu/~woit/wordpress/?p=2670</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Daniel Pape | 11,176 | <p>Some talks on history by some leading mathematicians (mostly in French):</p>
<p><a href="http://www.archivesaudiovisuelles.fr/FR/_LibraryThemas.asp?thema=541" rel="nofollow">http://www.archivesaudiovisuelles.fr/FR/_LibraryThemas.asp?thema=541</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Jesus Martinez Garcia | 1,887 | <p>David Cox's lectures in toric varieties at <a href="http://www.msri.org/web/msri/scientific/workshops/summer-graduate-workshops/show/-/event/Wm463" rel="nofollow">MSRI</a></p>
<p>Something really good to end the evening with :)</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| jfm314 | 25,910 | <p>I have compiled a list (1500+) of math videos at <a href="http://pinterest.com/mathematicsprof/">http://pinterest.com/mathematicsprof/</a> . If anyone is aware of others, please send them to me. </p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| user48365 | 48,365 | <p>So much maths video in <a href="http://nptel.ac.in/" rel="nofollow">http://nptel.ac.in/</a> National Programme on Technology Enhanced Learning</p>
|
480,828 | <p>We can suppose that we will create a new number system with essentially two imaginaries that do not interact. (Besides this, all quantities are taken to be integers) For example, we have an $i_1$ and an $i_2$. Then we could say</p>
<p>$$(a+b i_1)(c+d i_1) = ac + (ad + bc)i_1-bd$$</p>
<p>and, similarly for $i_2$:</p>
<p>$$(a+b i_2)(c+d i_2) = ac + (ad + bc)i_2-bd$$</p>
<p><strong>However</strong>, for a system with $i_1$ AND $i_2$:</p>
<p>$$(a+b i_1+c i_2)(d + e i_1 + f i_2)=$$<br>
$$(ad + ae i_1 + af i_2) + (bd i_1 - be + 0i_1i_2) + (cd i_2 + 0i_1i_2 -cf)$$</p>
<p>Above, the key thing to note is that $i_1\cdot i_2 = 0$.</p>
<p><strong>QUESTIONS</strong></p>
<p>I'm wondering if there is any idea or statement in math that says that I simply cannot do this. Without having worked in general systems of numbers, I'm wondering what ideas I should know about when I try to create a system like this.</p>
<p>Can I create this system if my main purposes are to carry out addition, subtraction, and multiplication with these numbers? Also, if I pose the additional constraint that all of these calculations are carried out modulo a prime, will this affect the system?</p>
| hmakholm left over Monica | 14,366 | <p><strong>BEWARE. This answer is as wrong as it is possible for it to be without actually containing any technically false statements. See my other answer.</strong> I'm leaving this up because the references to ring theory and quaternions are probably still helpful.</p>
<hr>
<p>If $i_1 i_2=0$, then $i_1$ and $i_2$ are <em>zero divisors</em>, and neither of them can have multiplicative inverses (assuming other usual rules such as associativity and $0x=0$ hold), so what you get isn't a field.</p>
<p>You get even more zero divisors from the fact that $i_1$ and $i_2$ have the same square without being each other's negatives: $(i_1-i_2)(i_1+i_2)=i_1^2-i_2^2=0$.</p>
<p>It is, however, a <em>ring</em>, which allows you to do addition, subtraction and multiplication as you please. Division won't always be possible, like in the integers. There's a quite rich and well developed theory about such structures (look for "ring theory" or "commutative algebra"); what you have is a particular instance of one of the most common way of producing new rings, namely as quotients of polynomial rings. More specifically, in standard notation your ring can be constructed as $\mathbb R[X,Y]/\langle X^2+1, Y^2+1, XY\rangle$.</p>
<p>You can certainly use modular arithmetic as a base instead of $\mathbb R$. In standard notation that will give you $\mathbb Z[X,Y]/\langle p, X^2+1, Y^2+1, XY\rangle$. Note that you may get <em>yet more</em> zero divisors if $-1$ is already a square in $\mathbb Z$ modulo $p$.</p>
<hr>
<p>(Ah, I just noticed that the question already specified that the base was the integers rather than $\mathbb R$. Then of course it is not a particular problem that you cannot always divide, because you can't do that even in $\mathbb Z$ itself. Note, however, that even <em>division with remainder</em> (which works well in $\mathbb Z$) won't have any nice generalization to any setting where there are zero divisors).</p>
<hr>
<p>You might also look into <em>quaternions</em> which manages to be <em>almost</em> a field by having <em>three</em> imaginary units $i$, $j$, $k$, where the product of any two is either the third one or its negative. Then inverses always exist, but on the other hand multiplication is no longer commutative.</p>
|
480,828 | <p>We can suppose that we will create a new number system with essentially two imaginaries that do not interact. (Besides this, all quantities are taken to be integers) For example, we have an $i_1$ and an $i_2$. Then we could say</p>
<p>$$(a+b i_1)(c+d i_1) = ac + (ad + bc)i_1-bd$$</p>
<p>and, similarly for $i_2$:</p>
<p>$$(a+b i_2)(c+d i_2) = ac + (ad + bc)i_2-bd$$</p>
<p><strong>However</strong>, for a system with $i_1$ AND $i_2$:</p>
<p>$$(a+b i_1+c i_2)(d + e i_1 + f i_2)=$$<br>
$$(ad + ae i_1 + af i_2) + (bd i_1 - be + 0i_1i_2) + (cd i_2 + 0i_1i_2 -cf)$$</p>
<p>Above, the key thing to note is that $i_1\cdot i_2 = 0$.</p>
<p><strong>QUESTIONS</strong></p>
<p>I'm wondering if there is any idea or statement in math that says that I simply cannot do this. Without having worked in general systems of numbers, I'm wondering what ideas I should know about when I try to create a system like this.</p>
<p>Can I create this system if my main purposes are to carry out addition, subtraction, and multiplication with these numbers? Also, if I pose the additional constraint that all of these calculations are carried out modulo a prime, will this affect the system?</p>
| hmakholm left over Monica | 14,366 | <p>Um, scratch the conclusions of my earlier answer. You get into more trouble than mere zero divisors:</p>
<p>$$1 = (-1)\cdot (-1) = i_1^2\cdot i_2^2 = (i_1 i_2)^2 = 0^2 = 0 $$</p>
<p>so everything collapses!</p>
<p>My observation that what you get is a quotient ring was technically right, but I failed to notice that the ideal $\langle X^2+1,Y^2+1,XY\rangle$ is just $\langle 1\rangle$ because $$1 = 1\cdot(X^2+1) - X^2\cdot(Y^2+1) + XY\cdot(XY) $$so the quotient ring you get is actually the zero ring.</p>
<hr>
<p>If you drop the rule that $i_1\cdot i_2=0$ and instead just accept that your objects can also contain some multiple of $i_1 i_2$, then you stand a better chance.</p>
<p>In fact, if one decides that multiplication is not commutative and $i_1 i_2 = -i_2i_1$ but ordinary numbers commute with everything, then what one gets is exactly the quaternions!</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| bof | 111,012 | <p>The definition of <em>set</em> depends on what kind of set theory you're using. Here are two examples.</p>
<p>In the kind of set theory described in the appendix of <a href="https://en.wikipedia.org/wiki/John_L._Kelley" rel="nofollow">John L. Kelley</a>'s <em>General Topology</em> (available at the <a href="https://archive.org/details/GeneralTopology" rel="nofollow">Internet Archive</a>), the so-called <a href="https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory" rel="nofollow">Morse-Kelley set theory</a>, a <em>set</em> is is something which is an element of something else. (<em>Element</em> is undefined; you can't define everything.) In symbols:
$$x\text{ is a set }\iff\exists y\ (x\in y)$$</p>
<p>In the very popular <a href="https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory" rel="nofollow">Zermelo–Fraenkel set theory</a>, <em>set</em> is not only not defined, it is not even an undefined term; there is no need to speak of sets because everything is a set. If you insist on defining such a useless term, you could use the same definition as in Morse-Kelley set theory, or more simply:
$$x\text{ is a set }\iff x=x$$</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| Vlad | 229,317 | <p>I disagree with people who say that <em>set</em> is undefined.
I am also not very fond of definitions which describe <em>set</em> as an <em>"object"</em>, a <em>"list"</em>, or a <em>"collection"</em> of something.
Each of above approaches raise somewhat difficult questions, with <em>"collection"</em> being the least problematic of all, in my opinion. </p>
<ul>
<li>If we try to define set as an <em>object</em>, we inevitably run into discussion of what is <em>object</em>.
We end up defining <em>set</em> in terms of something, which is probably even more difficult to define.</li>
<li>We cannot define set as <em>list</em> because we cannot list or even comprehend/recognize/operate every member of many sets arising in mathematics.</li>
<li>Although the least contradictory one from my prospective, an attempt to define <em>set</em> as a <em>collection</em> of something has it disadvantages as well.
Dictionaries define <em>"collection"</em> either as a set, in which case we fall into a vicious circle, or as the act of <em>collecting</em> something, which is not very convenient in context of mathematics.
When talking about set of numbers, we do not mean that they "lie somewhere collected in a pile", or "gathered together" somehow.</li>
</ul>
<p>We are on the right track with the last interpretation though.</p>
<hr>
<p>According to Mathematical Analysis lecturer in my undergrad, </p>
<blockquote>
<p><em><strong>set</strong> is the way our brain groups abstract objects</em>.</p>
</blockquote>
<p>It is not an actual "group" (or "collection") of physical objects or even mental images, but rather a <strong>pattern</strong> in one's brain which connects (or puts together) certain ideas/images (of <em>elements</em> of set).</p>
<p>In a sense, <em>set</em> consisting of some "elements" is nothing more but a <strong>boundary</strong> our consciousness draws between the mental image of these "elements" and the rest of object in the minds.
This boundary can be based on anything our brains can operate – common features, resemblance, random assignment, etc.
Ultimately, my bottom line is:</p>
<blockquote>
<p><strong><em>set</strong> is a <strong>property</strong> of mental image(s) in someone's brain.</em></p>
</blockquote>
<p>I know this definition is not mathy/rigorous at all, and perhaps contradicts some definitions you can find in the literature.
But this is the best answer to the question I kept asking myself for the longest time:
<em>"what <strong>exactly</strong> do they mean when they say 'set'?"</em></p>
|
1,893,280 | <p>How to show $\frac{c}{n} \leq \log(1+\frac{c}{n-c})$ for any positive constant $c$ such that $0 < c < n$?</p>
<p>I'm considering the Taylor expansion, but it does not work...</p>
| kobe | 190,421 | <p>By the mean value theorem, $e^{c/n} = 1 + e^{\alpha}(c/n)$ for some $\alpha\in (0, c/n)$. Since $e^{\alpha} \le e^{c/n}$, then $e^{c/n} \le 1 + e^{c/n}(c/n)$. So $(1 - c/n)e^{c/n} \le 1$, or </p>
<p>$$e^{c/n} \le \frac{1}{1 - c/n} = \frac{n}{n-c} = 1 + \frac{c}{n-c}$$</p>
<p>Now take logarithms.</p>
|
971,160 | <p>So, this is actually 2 questions in 1. I apologize if that is bad practice, but I didn't want to write 2 questions when they're a word different. So, I have</p>
<ol>
<li>Prove or disprove that if $a|(sb+tc), \forall s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.</li>
</ol>
<p>and then,</p>
<ol start="2">
<li>Prove or disprove that if $a|(sb+tc)$ for some $s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.</li>
</ol>
<p>I know how to prove if $a|b$ and $a|c$, then $a|(sb+tc), \forall s,t \in\mathbb{Z}$, but I'm certain I can't just write the proof backwards, and show that it works. So, how would I tackle these two problems, especially the second one?</p>
| N. S. | 9,176 | <ol>
<li><p>What happens when one of $s,t$ is one and the other one is zero?</p></li>
<li><p>$2014| 2014 \cdot 1 + 2014 \cdot 3$ but $2014$ doesn't divide $1$ or $3$...</p></li>
</ol>
|
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