qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,864,604 | <p>What's the difference between $f(x)=f(a-x)$ and $f(x)=f(x-a)$ ?</p>
<p>It's a pretty simple question maybe, but I'm unable to understand this one. </p>
| JonesY | 265,330 | <p>The sign "$-$" here refers to a binary operation, lets call it $O(x,y)=x-y$. Writing it like this shows the difference between $O(x,y)$ and $O(y,x)$</p>
<p>So the difference is $O(x,y)\neq O(y,x)$ since a binary operation does not have to be commutative, the result could be different. </p>
<p>There are a lot of examples of non commutative operation:</p>
<ol>
<li>Composition of functions $f\circ g\neq g\circ f$.</li>
</ol>
<p>2.Division $4/2\neq 2/4$.</p>
<ol start="3">
<li>Multiplication of matrices (with at least 2 rows).</li>
</ol>
<p>To read more about non commutative binary operations I would suggest you read about basic Relations in Set Theory and about basic Group Theory.</p>
|
117,500 | <p>How would you go about finding the conjugacy classes of the nonabelian group of order 21, $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$?</p>
| Mariano Suárez-Álvarez | 274 | <p>The group has a normal Sylow $7$-subgroup, generated by $x$, and it is clear from the way $y$ acts on $x$ that the conjugacy relation is generated by $x^i\sim x^{2i}$: this gives two conjugacy classes of elements of order $7$.</p>
<p>Using the Sylow theorems for $p=3$ and the fact that a Sylow $3$-subgroup cannot be normal (for otherwise the group would be abelian) you see that there is $1$ conjugacy class of $7$ cyclic subgroups of order $3$. They must be simply transitively permuted by the Sylow $7$-subgroup, so they give us 2 more conjugacy classes of elements of order three. </p>
<p>Finally, there's the class of $1$. In all, there are five classes then, which we can describe as follows: the classes of $1$, $y$, $y^2$, $x$, $x^3$.</p>
<p>We can check with GAP:</p>
<pre><code>GAP4, Version: 4.4.10 of 02-Oct-2007, i686-pc-linux-gnu-gcc
Components: small 2.1, small2 2.0, small3 2.0, small4 1.0, small5 1.0, small6 1.0, small7 1.0, small8 1.0,
small9 1.0, small10 0.2, id2 3.0, id3 2.1, id4 1.0, id5 1.0, id6 1.0, id9 1.0, id10 0.1, trans 1.0,
prim 2.1 loaded.
Packages: AClib 1.1, Polycyclic 2.2, Alnuth 2.2.5, CrystCat 1.1.2, Cryst 4.1.5, Carat 2.0.2, AutPGrp 1.2,
CRISP 1.3.2, CTblLib 1.1.3, TomLib 1.1.2, FactInt 1.5.2, GAPDoc 1.2, IO 2.3, FGA 1.1.0.1,
IRREDSOL 1.1.2, LAGUNA 3.4, Sophus 1.23, Polenta 1.2.7, ResClasses 2.5.3, EDIM 1.2.3 loaded.
gap> List(AllSmallGroups(21, IsAbelian, false), g -> Length(ConjugacyClasses(g)));
[ 5 ]
gap>
</code></pre>
|
1,041,731 | <p>I want to prove that if $A$ in an infinite set, then the cartesian product of $A$ with 2 (the set whose only elements are 0 and 1) is equipotent to $A$.</p>
<p>I'm allowed to use Zorn's Lemma, but I can't use anything about cardinal numbers or cardinal arithmetic (since we haven't sotten to that topic in the course).</p>
<p>I read a proof of the fact that if $a$ is an infinite cardinal number, then $a+a=a$, which is something similar to what I want to prove. </p>
<p>Any suggestions will be appreciated :)</p>
| Git Gud | 55,235 | <p>The sum of the entries on each row is always $1$, so $\left(1,\begin{bmatrix} 1\\1\\1\end{bmatrix}\right)$ is an eigenpair. </p>
<p>Since $\det(A)=2$ (easily seen by Laplace expansion on the first row) and $\text{tr}(A)=4$, the other 'two' eigenvalues follow easily.</p>
<hr>
<p>Actually answering your question, since $$-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0\iff \lambda^3-4\lambda ^2+5\lambda -2=0$$</p>
<p>(I'd rather work with monic polynomials, this equivalence is unnecessary), the <a href="http://en.wikipedia.org/wiki/Rational_root_theorem" rel="nofollow">rational root theorem</a> says the rational roots, if any, are in $\left\{-2-1,1,2\right\}$, so you can try these and hope for the best.</p>
|
2,798,206 | <p>How do you prove this using the epsilon-delta definition? I'm unsure of using the min = { } function.</p>
<p>$\lim \limits_{x \to \infty}\frac{2x+1}{1-x}$</p>
<p>These are my steps: </p>
<p>$ |f(x) - L| < \epsilon => |\frac{2x+1}{1-x} +2|< \epsilon $</p>
<p>$ \qquad \qquad \; \; \; \; \; =>|\frac{3}{1-x} | < \epsilon $</p>
<p>$ \qquad \qquad \; \; \; \; \; =>|\frac{-3}{x-1} | < \epsilon$</p>
<p>$ \qquad \qquad \; \; \; \; \; =>|-3||\frac{1}{x-1} | < \epsilon$</p>
<p>$ \qquad \qquad \; \; \; \; \; =>|\frac{1}{x-1} | < \frac{\epsilon}{3}$</p>
<p>$ \qquad \qquad \; \; \; \; \; =>|{x-1} | < \frac{\epsilon |x-1|}{3}$</p>
<p>How do I continue from here?</p>
| Theo Bendit | 248,286 | <p>A few pointers:</p>
<ul>
<li>The $\varepsilon$-$\delta$ definition of a limit pertains to when you have a limit as $x \to a$, where $a$ is a finite quantity. For a limit as $x \to \infty$, you need an $\varepsilon$-$M$ definition, along the lines of, for all $\varepsilon > 0$, there exists some $M$ such that
$$x > M \implies |f(x) - L| < \varepsilon.$$</li>
<li>Your last line has an error. You'll want to be reciprocating both sides, to get $|x - 1| > \frac{3}{\varepsilon}$. Note the change in sign, as the function $\frac{1}{x}$ is decreasing for positive $x$.</li>
<li>This means, you are looking for values of $x$ that are a greater distance than $\frac{3}{\varepsilon}$ from $1$. This suggests a that $M = 1 + \frac{3}{\varepsilon}$ is a good choice of $M$. Note that no maximums or minimums are required!</li>
<li>Beware the flow of logic here! You've written in implications $\implies$ between each step. While this is true, what you really need for the definition is backwards implications: $\impliedby$. This is because you're starting with $|f(x) - L| < \varepsilon$, the thing you need to conclude! This is ok, provided each step you perform <strong>implies the previous step</strong>, not the next step as we often tend to do. The other way is to simply rewrite the entire proof backwards, starting with assuming $x > 1 + \frac{3}{\varepsilon}$, and working logically to the desired conclusion.</li>
</ul>
<p>EDIT: Regarding the solution you posted as an answer, it's definitely looking better, though I have a couple more pointers:</p>
<ul>
<li>You only really need the last block of implications. Your entire solution could be presented by simply starting with the line that begins
$$|x - 1| > \frac{3}{\varepsilon} \iff \ldots$$
and going from there. Sure, it doesn't explain where you got $\frac{3}{\varepsilon}$ from, but then again, it doesn't have to! The logic shows that such a choice was a good choice, no matter how you came by it.</li>
<li>You still haven't come from the starting point of $x > M$, which you should, as this is part of the definition of a limit as $x \to \infty$. So, I would just add to your solution,
\begin{align*}
x > 1 + \frac{3}{\varepsilon} &\implies x - 1 > \frac{3}{\varepsilon} > 0 \\
&\implies |x - 1| > \frac{3}{\varepsilon} \\
&\implies \ldots
\end{align*}
Note that the above implication works because $x - 1 \ge 0$, and so $|x - 1| = x - 1$.</li>
</ul>
|
1,083,277 | <p>$a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$</p>
<p>I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next</p>
| Redundant Aunt | 109,899 | <p>As it was said, you can take $x=2^a$, $y=2^b$ and $z=2^c$ to obtain:
$$
x^3+y^3+z^3\ge x+y+z
$$
With $x,y,z>0$ and $xyz=1$. It is equivalent to:
$$
x^2(x-1)+y^2(y-1)+z^2(z-1)\ge 0 \iff \frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge 0
$$
Since $x^2$ and $x-1$ ar equally ordered, we might apply Chebychevs inequality to obtain:
$$
\frac{x^2(x-1)+y^2(y-1)+z^2(z-1)}{3}\ge \frac{x^2+y^2+z^2}{3}\frac{(x-1)+(y-1)+(z-1)}{3}
$$
And to show that the two factors are greater than zero isn't difficult. (AM-GM)</p>
|
134,205 | <blockquote>
<p>Find the expectation of a Geometric distribution using $\mathbb{E}(X)= \sum_{k=1}^\infty P(X \ge k)$. </p>
</blockquote>
<p>Okay I know how to find the expectation using the definition of the geometric distribution $$P(X=k)= p \cdot(1-p)^{k-1}$$ and I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it. </p>
<p>I know the expectation is $\frac{1}{p}$ but I just get $\mathbb E(X)= \frac{1}{p^2}$ using the method specified in the question.</p>
| thepiercingarrow | 274,737 | <p>A simpler way would be to plug in $q=1-p$ and solve it that way using formula for geometric sequences:</p>
<p>\begin{align*}
E(X) &= \sum\limits_{k=1}^\infty kpq^{k-1}\\
&= \frac{p}{q} \sum\limits_{k=1}^\infty kq^{k}\\
&= \frac{p}{q} \frac{q}{(1-q)^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{1}{p}.
\end{align*}</p>
|
650,395 | <p>I am given generating functions $f(x)= \frac{x}{1-x}$ or $f(x)=\frac{1}{1+x^{2}}$ or $f(x)=\frac{1}{x^2-5x+6}$ and I am obliged to write sequence which are generated by this functions. What is the fastest algorithm to solve these problems? I have problem with even starting. I will be glad if anyone would be so nice to explain me algorithm to solve this kind of exercises or post any reference that is related with my problem.</p>
| Alex | 38,873 | <p>Hint: $\frac{1}{1-x}=(1-x)^{-1}=\sum_{k=0}^{\infty} 1 \cdot x^k$, so the sequence this GF generates is $<1,1, \ldots>$</p>
<p>EDIT: 'in general' you need to represent the GF in the form $\sum_{k=0}^{\infty} a_k x^k$, and $\{ a_k \}$ is the sequence that this GF generates. </p>
|
481,952 | <p>Why is a union of infinitely many bounded sets not necessarily bounded, please? In addition, what condition can we add to make this union bounded, please?</p>
| Amitesh Datta | 10,467 | <p>$\mathbb{R}^n=\bigcup_{n=1}^{\infty} \{x\in\mathbb{R}^n:\left\|x\right|\leq n\}$</p>
<p>A finite union of bounded sets is bounded. </p>
<p>I hope this helps!</p>
|
3,203,607 | <p>"Each cell of a 100 × 100 table is painted either black or white and all
the cells adjacent to the border of the table are black. It is known that in every
2 × 2 square there are cells of both colours. Prove that in the table there is 2 × 2
square that is coloured in the chessboard manner."</p>
<p><a href="https://cms.math.ca/crux/v44/n9/OCP_44_9.pdf" rel="nofollow noreferrer">Source of problem</a></p>
<p>How to solve this problem?</p>
| Chathura Gunasekera | 567,103 | <p>Step 1:First reduce the problem to a simpler,but proportional version of the problem
Let's consider a 10 x 10 board where all the border lining boxes are black.
now imagine painting inward as shown in the following figure.
<a href="https://i.stack.imgur.com/QVCRf.jpg" rel="nofollow noreferrer">Click to see image</a> </p>
<p>Then there would be way to colour the square in the middle without leaving one 2x2 square in one colour.Therefore one square(not the corners) should be changed to the opposite colour(i.e black) then this generates a 2x2 square coloured like the chess board when the entire board is coloured <a href="https://i.stack.imgur.com/aUqb9.jpg" rel="nofollow noreferrer">Click here to see picture</a></p>
<p>now returning to the original problem we must compute how many squares are left when we finish painting like bands.(step 1).
As the board is 100 x 100, there are also 100 diagonal squares(just as there are 100 diagonal squares in the 10 x 10 grid) then at the dead-center of the square there will not be a square but set of 4 squares (see above image and compare). [this is because there are even no. of squares.]
Then the same argument for the 10 x 10 square works. </p>
<p>Therefore there should be one 2 x 2 square painted like in chess board.</p>
|
1,079,493 | <blockquote>
<p>Prove that <span class="math-container">$f(x) = x^3 + 3x - 1$</span> is irreducible in <span class="math-container">$\mathbb Q[X]$</span>.<br />
Let <span class="math-container">$\theta$</span> be a root of <span class="math-container">$f(x)$</span>. Compute <span class="math-container">$\frac{1}{\theta}$</span> and <span class="math-container">$(2 + \theta^2)^{-1} $</span> in <span class="math-container">$\mathbb Q[\theta ]$</span>.</p>
</blockquote>
<p><span class="math-container">\begin{array}{l}
f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\
\Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\
\Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\
\end{array}</span></p>
<p><span class="math-container">\begin{array}{l}
\frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\
\Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\
\Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\
\Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\
\end{array}</span></p>
<p>But I can't show that <span class="math-container">$f$</span> is irreducible.</p>
| Peter Woolfitt | 145,826 | <p>Let $$g(x)=f(x+1)=x^3+3x^2+3x+1+3x+3-1=x^3+3x^2+6x+3$$</p>
<p>Now we can apply <a href="http://en.wikipedia.org/wiki/Eisenstein%27s_criterion" rel="noreferrer">Eisenstein's Criterion</a> (with $p=3$) to find that $g(x)$ is irreducible in $\mathbb{Q}[x]$, so $f(x)$ is also irreducible in $\mathbb{Q}[x]$.</p>
|
1,079,493 | <blockquote>
<p>Prove that <span class="math-container">$f(x) = x^3 + 3x - 1$</span> is irreducible in <span class="math-container">$\mathbb Q[X]$</span>.<br />
Let <span class="math-container">$\theta$</span> be a root of <span class="math-container">$f(x)$</span>. Compute <span class="math-container">$\frac{1}{\theta}$</span> and <span class="math-container">$(2 + \theta^2)^{-1} $</span> in <span class="math-container">$\mathbb Q[\theta ]$</span>.</p>
</blockquote>
<p><span class="math-container">\begin{array}{l}
f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\
\Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\
\Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\
\end{array}</span></p>
<p><span class="math-container">\begin{array}{l}
\frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\
\Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\
\Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\
\Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\
\end{array}</span></p>
<p>But I can't show that <span class="math-container">$f$</span> is irreducible.</p>
| Mr.Fry | 68,477 | <p>For polynomials of deg $\leq 3$ it is enough to check that there are no rational roots (i.e a linear factor). Applying the Rational Roots Theorem the set of possible roots are $\pm 1$, which clearly $g(\pm 1) \not = 0$.</p>
|
3,403,255 | <p>I am trying to follow wikipedia's page about matrix rotation and having a hard time understanding where the formula comes from.</p>
<p><a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Rotation_matrix</a> Wiki page about it.</p>
<p>what i have so far:</p>
<p>y<sub>2</sub>=sin(<em>a<sub>1</sub>+a<sub>2</sub></em>)R -> where R is hypotenuse, a1 is current angle and a2 is the angle by which something must rotate.</p>
<p>this how i used to calculate my rotation, but it takes long time to compute and uses up a lot of cpu time for square roots and other heavy stuff due tot he need of finding the initial angle.</p>
<p>So i decided to reduce computation time and found that sin(a1+a2) could be writen as <code>sin(a1)cos(a2)+cos(a1)sin(a2)</code> and from there i got to the point where it is:</p>
<p>y<sub>2</sub>=y<sub>1</sub>cos(<em>a<sub>2</sub></em>)+xsin(<em>a<sub>2</sub></em>)sin(<em>a<sub>1</sub></em>)</p>
<p>But wiki page says that it must b:</p>
<p>y<sub>2</sub>=y<sub>1</sub>cos(<em>a<sub>2</sub></em>)+xsin(<em>a<sub>2</sub></em>)</p>
<p><a href="https://i.stack.imgur.com/vRzo7.png" rel="nofollow noreferrer">My work book</a></p>
| DanLewis3264 | 480,329 | <p>Think: where are the unit vectors <span class="math-container">$(1,0)$</span> and <span class="math-container">$(1,1)$</span> sent to under a rotation by angle <span class="math-container">$\theta$</span>? It should not take too long to convince yourself that the answers are <span class="math-container">$(\cos \theta,\sin \theta)$</span> and <span class="math-container">$(-\sin \theta, \cos\theta)$</span> respectively.</p>
|
2,360,268 | <p>Draw a triangle given $A-B=90$(degree) and length of $AC,BC$.</p>
<p><strong>My attempt</strong>:I thought It would be a good idea to draw a right angle so I made the picture below:</p>
<p><a href="https://i.stack.imgur.com/yikB7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yikB7.png" alt="enter image description here"></a></p>
<p>But I don't know how to find the appropriate point on the circle?</p>
| g.kov | 122,782 | <p>From
\begin{align}
\frac{b}{\sin\beta}
&=
\frac{a}{\sin(90^\circ+\beta)}
=
\frac{a}{\cos\beta}
\end{align}</p>
<p>we have </p>
<p>\begin{align}
\tan\beta&=\frac{b}{a}
,\quad
\sin\beta=\frac{b}{\sqrt{a^2+b^2}}
,\\
2\,R&=\frac{b}{\sin\beta}=\sqrt{a^2+b^2}
.
\end{align}</p>
<p>$\triangle ABC$ and $\triangle A_1BC$
are inscribed in the same circle with the radius $R$:</p>
<p><a href="https://i.stack.imgur.com/J0vLd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J0vLd.png" alt="enter image description here"></a></p>
|
3,397,548 | <p>For a sequence <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span>, define <span class="math-container">$$\Delta x_n:=x_{n+1}-x_n,~\Delta^2 x_n:=\Delta x_{n+1}-\Delta x_n,~(n=1,2,\ldots)$$</span> which are named <strong>1-order</strong> and <strong>2-order difference</strong>, respectively. </p>
<p>The problem is stated as follows:</p>
<blockquote>
<p>Let <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span> be <strong>bounded</strong> , and satisfy
<span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0$</span>. Prove or disprove
<span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n=0.$</span></p>
</blockquote>
<p>By intuiton, the conclusion is likely to be true. According to <span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0,$</span> we can estimate <span class="math-container">$\Delta x_n$</span> almost equal with an increasing <span class="math-container">$n$</span>. Thus, <span class="math-container">$\{x_n\}$</span> looks like an <strong>arithmetic sequence</strong>. If <span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n \neq 0$</span>, then <span class="math-container">$\{x_n\}$</span> can not be bounded.</p>
<p>But how to prove it rigidly?</p>
| Daniel Fischer | 83,702 | <p>Suppose that <span class="math-container">$\Delta x_n \not\to 0$</span>. Then there is - without loss of generality, replace <span class="math-container">$x_n$</span> with <span class="math-container">$-x_n$</span> if necessary - a <span class="math-container">$c > 0$</span> such that <span class="math-container">$\Delta x_n > 2c$</span> for infinitely many <span class="math-container">$n$</span>. Now for every <span class="math-container">$\varepsilon > 0$</span> there exists an <span class="math-container">$N_{\varepsilon}$</span> such that <span class="math-container">$\lvert \Delta^2 x_n\rvert < \varepsilon$</span> for all <span class="math-container">$n \geqslant N_{\varepsilon}$</span>. Pick an <span class="math-container">$n_1 \geqslant N_{\varepsilon}$</span> with <span class="math-container">$\Delta x_{n_1} > 2c$</span>. Then
<span class="math-container">$$\Delta x_{n_1 + k} \geqslant \Delta x_{n_1} - k\varepsilon > c$$</span>
for <span class="math-container">$0 \leqslant k \leqslant c/\varepsilon$</span>. It follows that
<span class="math-container">$$x_{n_1+k+1} - x_{n_1} = \sum_{\kappa = 0}^k \Delta x_{n_1 + \kappa} > (k+1)\cdot c$$</span>
for <span class="math-container">$0 \leqslant k \leqslant c/\varepsilon$</span>, and thus</p>
<p><span class="math-container">$$\sup_n x_n - \inf_n x_n \geqslant \frac{c^2}{\varepsilon}$$</span>
for every <span class="math-container">$\varepsilon > 0$</span>, which says that <span class="math-container">$x_n$</span> is unbounded.</p>
|
3,811,753 | <p>Show that the equation:</p>
<p><span class="math-container">$$
y’ = \frac{2-xy^3}{3x^2y^2}
$$</span></p>
<p>Has an integration factor that depends on <span class="math-container">$x$</span> And solve it that way.</p>
<hr />
<p>Already we got to:</p>
<p><span class="math-container">$$
y’ + \frac{xy^3}{3x^2y^2} = \frac{2}{3x^2y^2}
$$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$$
y’ + \frac{1}{3x}y = \frac{2}{3x^2}y^{-2}
$$</span></p>
<p>But, in order to get an integration factor, shouldn’t we have a linear equation? Of the form:</p>
<p><span class="math-container">$$
y‘ + p(x)y = g(x)
$$</span></p>
<p>That way getting the integration factor:</p>
<p><span class="math-container">$$
\mu = ke^{\int p(x)}, k \in R
$$</span></p>
<p>But what we have is a non-linear equation, so how could an integration factor exists?</p>
<p>Thanks.</p>
| Peanut | 144,455 | <p>You can manipulate the expression and obtain <span class="math-container">$(xy^3)' = 2/x$</span></p>
|
230,504 | <p>Again, this question is related (**) to a <a href="https://mathoverflow.net/questions/101700/large-cardinals-without-the-ambient-set-theory?rq=1">previous one</a>:</p>
<p>in standard books on basic set theory, after stating the axioms of ZFC, ordinal numbers are introduced early on. Afterwards cardinals appear: they are special ordinals which are minimal with respect to equinumerosity. </p>
<p>Ordinals and cardinals live happily within standard set theory.
Nothing wrong with that, but <strong>what about axiomatizing them <em>directly</em>, ie without the underlying set theory</strong>? </p>
<p>What I mean is: develop a first-order theory of some number system (the intended ordinals), such that they are totally ordered, there is an initial limit ordinal, and satisfy induction with respect to formulas in the language of ordinal arithmetic (here, of course, one will have to adjust the induction schema to accommodate the limit case).</p>
<p>The cardinals could be introduced by adding an order-preserving operator $K$ on the ordinal numbers mimicking their definition in ZFC: cardinals would then be the fixed points of $K$.</p>
<ol>
<li>has some direct axiomatization along these lines been fully developed? I would suspect that the answer is in the affirmative, but I have no refs. </li>
<li>the induction schema would be limited to first order formulae, so, assuming that the answer to 1 is yes, is there a theory of non-standard models of Ordinal Arithmetic? </li>
<li>Assuming 1 AND 2, what about weaker induction schemas for ordinals (*)? </li>
</ol>
<p>(*) I am thinking again of formal arithmetics and the various sub-systems of Peano</p>
<p>(**) it is not the same, though: here I am asking for a direct axiomatization of ordinals, and indirectly of cardinals, via the operator $k$</p>
| Thomas Benjamin | 20,597 | <p>You might consider taking a look at a paper by Athanassios Tzouvaras titled "Cardinality without enumeration" (look under title on the Web), especially at Definition 3.1. Since it is short, I will quote it verbatim:</p>
<p>"Definition 3.1 Let $M$ be a model of $ZF$. A <em>notion of cardinality</em> for $M$ is a mapping $C$$\subset$$M$ such that:</p>
<p>(1) <em>dom</em>($C$)=$M$ and <em>r ng</em>($C$)=<em>Card</em></p>
<p>(2) $C$($\kappa$)=$\kappa$ for every $\kappa$$\in$_Card_</p>
<p>(3) For any disjoint sets $x$, $y$ $C$($x$$\cup$$y$)=$C$($x$)$+$$C$($y$)</p>
<p>(4) For any $x$, $y$ $C$($x$$\times$$y$)=$C$($x$) $\cdot$ $C$($y$)</p>
<p>(5) If $f$ : $x$$\rightarrow$$y$ is an injective mapping, then $C$($x$)$\le$$C$($y$)</p>
<p>A cardinality notion $C$ is said to be <em>standard</em> if in addition the converse of (5) holds, i.e. if $C$($x$)$\le$$C$($y$) implies that there is an injective $f$ from $x$ to $y$."</p>
<p>An especially interesting consequence of the cardinal notion
$C$ being standard is Remark 3.2(ii):</p>
<p>"If $C$ is standard, then $C$($x$)=$C$($y$) implies that there is an injective $f$ from $x$ onto $y$. So the existence of a standard notion of cardinality implies $AC$ [and the Kunen inconsistency as well--my comment]. Moreover in the presence of $AC$ there is a unique notion of cardinality, the standard one. Thus in order to depart from the standard notion of cardinality, we must drop $AC$." </p>
<p>Is Definition 3.1 more along the lines of what you are looking for, at least regarding cardinality? </p>
|
231,187 | <p>I'm wondering if there's an efficient way of checking to see if two context free grammars are equivalent, besides working out "test cases" by hand (ie, just trying to see if both grammars can generate the same things, and only the same things, by trial and error).</p>
<p>Thanks!</p>
| Micah | 30,836 | <p>There is not. In fact, there isn't even an inefficient way!</p>
<p>That is, the problem of determining whether two given CFGs represent the same language is undecidable. In fact, an even stronger statement is true: the problem of determining whether a given CFG accepts all strings on its alphabet is undecidable.</p>
<p>The proof of this can be found in chapter 5 of Sipser's <em><a href="http://rads.stackoverflow.com/amzn/click/113318779X">Introduction to the Theory of Computation</a></em>. The basic idea is that, for any Turing machine $M$, we can obtain a context-free grammar which accepts all strings that do not encode a proof that $M$ halts (under some specific encoding that's more complicated than I really want to get into here). So determining whether this grammar accepts all strings is equivalent to solving the halting problem for $M$.</p>
|
487,123 | <p>How to evaluate the following limit?
$$\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}$$</p>
<p>For this problem I have two methods. But I'd like to know if there are better methods.</p>
<p><strong>My solution 1:</strong></p>
<p>Using Stolz-Cesaro Theorem, we have
$$\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}=\lim_{n\to\infty}\dfrac{n!}{n!-(n-1)!}=\lim_{n\to\infty}\dfrac{n}{n-1}=1$$</p>
<p><strong>My solution 2:</strong></p>
<p>$$1=\dfrac{n!}{n!}<\dfrac{1!+2!+\cdots+n!}{n!}<\dfrac{(n-2)(n-2)!+(n-1)!+n!}{n!}=\dfrac{n-2}{n(n-1)}+\dfrac{1}{n}+1$$</p>
| Chris | 164,598 | <p>Perhaps you might like the following argument:</p>
<p>Notice that <span class="math-container">$$\frac{\sum_{k=1}^{n} i!}{n!} = 1 + \frac{1}{n}\frac{\sum_{k=1}^{n-1} i!}{n-1!};$$</span>
from this, we get the recurrence</p>
<p><span class="math-container">$$\frac{\sum_{k=1}^{2} i!}{2!} = 1 + \frac{1}{2},$$</span>
<span class="math-container">$$\frac{\sum_{k=1}^{3} i!}{3!} = 1 + \frac{1}{3}(1 + \frac{1}{2}),$$</span></p>
<p>and in general</p>
<p><span class="math-container">$$\frac{\sum_{k=1}^{n} i!}{n!} = 1 + \frac{1}{n}(1 + \frac{1}{n-1}(...(1 + \frac{1}{3}(1+ \frac{1}{2}))...)) \\<
1 + \frac{1}{n}(1 + \frac{1}{2}(...(1 + \frac{1}{2}(1+\frac{1}{2}))...)) \\ < 1 + \frac{1}{n}(2)$$</span></p>
<p>and so your sequence is bounded above by one converging to <span class="math-container">$1$</span>. Since it is also trivially bounded below by the constant sequence of <span class="math-container">$1$</span>, your sequence thus converges to <span class="math-container">$1$</span>.</p>
|
1,302,932 | <p>Given $$A=\begin{pmatrix}
2 & 0 & 0\\
a & 2& 0\\
a+3 & a &-1
\end{pmatrix}$$<br>
For which values of $a$ can $A$ be diagonal?<br>
I found that $p_A(x)=(x-2)^2(x+1)$ and tried to find the eigen subspace of 2, to see if the geomtric multiplicity of the eigenvalue $2$ is $2$.<br>
I got a set of equations:$$2x=2x ; ax+2y=2y ; (a+3)x+ay-z=2z$$<br>
But I could not understand how to extract the relevant information from it.</p>
| Anurag A | 68,092 | <p>Assuming you got the correct set of equations, from the first equation we get $x$ can be anything, the second equation gives, $y$ can be anything but $ax=0$. From here either we have $a=0$ or $x=0$. </p>
<p>If $a=0$, then the third equation gives $z=0$. But since $x$ and $y$ have no restrictions, therefore you can generate two independent eigen vectors by choosing $x=1, y=0$ and $x=0,y=1$. Hence diagonalizable.</p>
<p>If $a \neq 0$, then $x$ has to be $0$, $y$ can still be anything and from the third equation we get $ay+3z=0$. Now ask can you generate two independent vectors in this case?</p>
|
1,302,932 | <p>Given $$A=\begin{pmatrix}
2 & 0 & 0\\
a & 2& 0\\
a+3 & a &-1
\end{pmatrix}$$<br>
For which values of $a$ can $A$ be diagonal?<br>
I found that $p_A(x)=(x-2)^2(x+1)$ and tried to find the eigen subspace of 2, to see if the geomtric multiplicity of the eigenvalue $2$ is $2$.<br>
I got a set of equations:$$2x=2x ; ax+2y=2y ; (a+3)x+ay-z=2z$$<br>
But I could not understand how to extract the relevant information from it.</p>
| MathNewbie | 24,672 | <p>You'll get for $a \neq 0$, that the nullspace of $A-2I$ is less than $2$. To see why, you have that </p>
<p>\begin{align*} A - 2I = \left(\begin{matrix} 0 & 0 & 0 \\ a & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*}<br>
is row equivalent to </p>
<p>\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ a+3 & a & - 3 \end{matrix}\right) \end{align*}</p>
<p>since $a \neq 0$. This is row equivalent to </p>
<p>\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & a & -3 \end{matrix}\right)\end{align*}</p>
<p>regardless if $a = -3$ or not. So that this is row equivalent to</p>
<p>\begin{align*}\left(\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -\frac{3}{a} \end{matrix}\right)\end{align*}</p>
<p>since $a \neq 0$. I think you can extract the relevent information from there.</p>
|
3,350,021 | <blockquote>
<p>We have the following quadratic equation:</p>
<p><span class="math-container">$2x^2-\sqrt{3}x-1=0$</span> with roots <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>.</p>
<p>I have to find <span class="math-container">$x_1^2+x_2^2$</span> and <span class="math-container">$|x_1-x_2|$</span>.</p>
</blockquote>
<p>First we have: <span class="math-container">$x_1+x_2=\dfrac{\sqrt{3}}{2}$</span> and <span class="math-container">$x_1x_2=-\dfrac{1}{2}$</span></p>
<p>So <span class="math-container">$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$</span></p>
<p>Can someone help me with the second one?</p>
<p>I forgot to tell that solving the equation is not an option in my case.</p>
| Saketh Malyala | 250,220 | <p>Well, you know that, by the Quadratic Formula, <span class="math-container">$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$</span>, so the difference between the two roots is <span class="math-container">$\frac{1}{a}\sqrt{b^2-4ac}=\frac{1}{a}\sqrt{3+4(2)(1)}=\frac{1}{2}\sqrt{11}$</span></p>
|
3,350,021 | <blockquote>
<p>We have the following quadratic equation:</p>
<p><span class="math-container">$2x^2-\sqrt{3}x-1=0$</span> with roots <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>.</p>
<p>I have to find <span class="math-container">$x_1^2+x_2^2$</span> and <span class="math-container">$|x_1-x_2|$</span>.</p>
</blockquote>
<p>First we have: <span class="math-container">$x_1+x_2=\dfrac{\sqrt{3}}{2}$</span> and <span class="math-container">$x_1x_2=-\dfrac{1}{2}$</span></p>
<p>So <span class="math-container">$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$</span></p>
<p>Can someone help me with the second one?</p>
<p>I forgot to tell that solving the equation is not an option in my case.</p>
| MPW | 113,214 | <p><strong>Hint:</strong> What is <span class="math-container">$\tfrac{-b+\sqrt{b^2-4ac}}{2a} - \frac{-b-\sqrt{b^2-4ac}}{2a}$</span> ?</p>
|
1,560,411 | <p>If $B_1$ and $B_2$ are the bases of two integer lattices $L_1$ and $L_2$, i.e.</p>
<p>$L_1=\{B_1n:n\in\mathbb Z^d\}$ and $L_2=\{B_2n:n\in\mathbb Z^d\}$,</p>
<p>is there an easy way to determine a basis for $L_1\cap L_2$? Answers of the form "Plug the matrices into a computer and ask for Hermite Normal Form, etc" are perfectly acceptable as this is a practical problem and the matrices of integers $B_1$ and $B_2$ are known, but I need some algorithmic way because the procedure will be repeated many times.</p>
| Steven Stadnicki | 785 | <p>A set of lecture notes up at <a href="http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf" rel="nofollow">http://cseweb.ucsd.edu/classes/wi10/cse206a/lec2.pdf</a> suggests computing the dual bases $D_1$ and $D_2$ of your lattices, then getting the HNF/orthogonalization of the concatenated matrix $[D_1\mid D_2]$ and then computing the dual of that. You'll want to be slightly careful with your algorithm for computing normal form to keep the intermediate coefficients from blowing up, but that shouldn't be too hard.</p>
|
227,109 | <p>I keep mixing them up, because they are very similar.</p>
<p>Some contrapositives resemble some contradictions.</p>
| Robert Mastragostino | 28,869 | <p>A contrapositive has truth value equivalent to the original statement:</p>
<p>$$\text{It is raining}\implies\text{I have an umbrella}$$
has a contrapositive (and is equivalent to)
$$\text{I do not have an umbrella}\implies\text{it is not raining}$$</p>
<p>Proving the contrapositive is equivalent to proving the original statement, and can sometimes be cleaner. Note that we're dealing with (hopefully) true statements this way. We're trying to prove <em>truth</em>, not falsity.</p>
<p>A proof by contradiction proves a statement true that can be proven false (typically is already known to be false) by other means, meaning that the logic being used is inconsistent. Rather than working with a statement directly, it assumes its <em>negation</em> and derives an absurdity. The goal here is to reach a falsehood, not truth. While this works, and can sometimes be nice, working in an inconsistent theory means that intermediate results proven or needed along the way are useless. If a proof by contradiction suggests (or can be easily turned into) a more direct proof of similar difficulty, the direct proof is typically preferred.</p>
|
892,114 | <p>i have three number
1 2 3 which will always be in this order {123}, i want to find out number of cases can be made,
like {1},{2},{23},{13},{12},{123}{3},{}. but each number has two states like "a" "b", i.e, each one will become different entity,like 2a,2b,3a,3b,1a,
with only exception i.e. 1 will have only one state 1a.</p>
<p>please tel me step wise using formulas, so that i can understand, also, any link will be helpfull.
yours sincerly</p>
| Maman | 167,819 | <p>$x^{2}-4x-8$=$(x-2)^{2}-4-8$=$(x-2)^{2}-12$</p>
|
1,474,867 | <p>I was trying to prove </p>
<p>$$\left|\int_{0}^{a}{\frac{1-\cos{x}}{x^2}}dx-\frac{\pi}{2}\right|\leq \frac{3}{a}$$ or $\leq \frac{2}{a}$. My work: I would like to use Fubini's theorem to prove it. </p>
<p>I notice that $\frac{1}{x^2}=\int^{\infty}_{0}{ue^{-xu}}du$. </p>
<p>Then, I got $\int_{0}^{a}{\frac{1-\cos{x}}{x^2}}dx=\int_{0}^{\infty}u\int_{0}^{a}{(1-\cos{x})e^{-xu}}dxdu$. </p>
<p>Then, I got $\int_{0}^{a}{(1-\cos{x})e^{-xu}}dx=-e^{-au}u+\frac{1}{u+u^3}+e^{-au}\frac{u^2\cos{a}-u\sin{a}}{u+u^3}$.</p>
<p>Then, $\int_{0}^{a}{\frac{1-\cos{x}}{x^2}}dx=\int_0^{\infty}u(\frac{e^{au}-1}{u}+\frac{u-e^{au}(u\cos{a}+\sin{a})}{1+u^2})du\\=\int_0^{\infty}({e^{au}+\frac{-ue^{au}(u\cos{a}+\sin{a}-2)}{1+u^2}})du+\frac{\pi}{2}.$ </p>
<p>I was trying to show $|\int_0^{\infty}({e^{au}+\frac{-ue^{au}(u\cos{a}+\sin{a}-2)}{1+u^2}})du|\leq\frac{3}{a}$ or $\frac{2}{a}$. </p>
<p>But I do not have a clue. Can some give me hints?</p>
| Jack D'Aurizio | 44,121 | <p>Since we have:
$$ \frac{\pi}{2}-\int_{0}^{a}\frac{1-\cos x}{x^2}\,dx = \int_{a}^{+\infty}\frac{1-\cos x}{x^2}\,dx \tag{1}$$
it trivially follows that:
$$\left|\frac{\pi}{2}-\int_{0}^{a}\frac{1-\cos x}{x^2}\,dx\right|\leq \int_{a}^{+\infty}\frac{2}{x^2}\,dx = \frac{2}{a}.\tag{2}$$
If we use integration by parts, from:
$$\int_{a}^{+\infty}\frac{1-\cos x}{x^2}\,dx = \frac{\sin(a)+a}{a^2}-2\int_{a}^{+\infty}\frac{\sin x}{x^3}\,dx \tag{3}$$
we also have an improved upper bound:</p>
<blockquote>
<p>$$\left|\frac{\pi}{2}-\int_{0}^{a}\frac{1-\cos x}{x^2}\,dx\right|\leq \frac{a+2}{a^2}.\tag{4}$$</p>
</blockquote>
|
1,108,832 | <p>Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?</p>
| DeepSea | 101,504 | <p><strong>Hint:</strong> Bowlers: $5$</p>
<p>Keepers: $2$</p>
<p>Others: $8$</p>
<p>$A = A_{41} + A_{51} + A_{42} + A_{52}$</p>
<p>$A_{41} = \binom{5}{4}\binom{2}{1}\binom{8}{6}$</p>
<p>$A_{51} = \binom{5}{5}\binom{2}{1}\binom{8}{5}$, etc..</p>
|
2,510,322 | <p>$\left( f(x) \right ) =\min_{t<x}\left(t^2\right)$</p>
<p>How do I sketch this function for all real x? I don't get what minimum means in this context how do I sketch such a function when t is in the function but x isn't the square term? </p>
| Siong Thye Goh | 306,553 | <p>Guide:</p>
<p>Suppose $f(x) = \min_{t\le x} t^2$.</p>
<p>Let's illustrate how to evaluate this function value at $-1$ and $1$.</p>
<p>$f(-1) =\min_{t \leq {-1}}t^2$, since $t^2$ is a decreasing function for non-positive value, the minimimum value occur at $-1$, hence $$f(1) =\min_{t \leq {-1}}t^2=(-1)^2=1$$</p>
<p>$f(1) = \min_{t \leq 1} t^2$, we can differentaite the function $t^2$, set the derivative to be equal to $0$ and verify that the minimum value is $0$. Hence $f(1)=0$.</p>
<p>Another hint: The graph that you plot can be described in the following form</p>
<p>$$f(x)=\begin{cases} g(x) & ,x \leq a \\ h(x) &, x>a\end{cases}$$</p>
<p>where I am sure you have seen $g(x)$ and $h(x)$ before.</p>
|
439,302 | <p>@HansEngler Left the following response to <a href="https://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it">this question</a> regarding "bad math" that works,</p>
<blockquote>
<p>Here's another classical freshman calculus example: </p>
<p><strong>Find $\frac{d}{dx}x^x$.</strong> </p>
<p>Alice says "this is like $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$."
Bob says "no, this is like $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$."
Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".</p>
<p>The answer $\frac{d}{dx}x^x = (1 + \log x)x^x $ turns out to be correct.</p>
</blockquote>
<p>In <a href="https://math.stackexchange.com/questions/260656/cant-argue-with-success-looking-for-bad-math-that-gets-away-with-it#comment571410_261057">this comment</a>, @joriki asserts that this is not "bad math" but rather a legitimate technique,</p>
<blockquote>
<p>You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding other instances constant.</p>
</blockquote>
<p>I had never previously seen such a technique so naturally I tested it on a few examples, including $\frac{d}{dx} \left( x^{ \sin x}\right)$, etc. and it provided the correct result. The following three questions arose,</p>
<p>$ \ \ $ <strong>1. What is the proof of its is validity?</strong> <br>
$ \ \ $ <strong>2. Are there any examples where this technique outshines standard methods?</strong></p>
| nbubis | 28,743 | <p>Look at $y = f(u(x),v(x))$:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}$$
Now, note that $x^{\sin x}$ can be written as:
$$y=x^{\sin x} = f(x,\sin x), \ f(u,v) = u^v$$
So that:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}=
vu^{v-1} \cdot\frac{du}{dx} + \log u \cdot u^v \cdot\frac{dv}{dx}$$
In fact, this holds for any number of terms (since so does the chain rule), where each time we only look at the derivative by $x$ of only one of the terms, multiply those by the "internal" derivative, and sum them all up.</p>
|
596,374 | <p>I solved this , but I am not sure if I did in the right way.</p>
<p>$$2^{2x + 1} - 2^{x + 2} + 8 = 0$$</p>
<p>$$2^{x + 2} - 2^{2x + 2} = 8$$</p>
<p>$$\log_22^{x + 2} - \log_22^{2x + 2} = \log_28$$</p>
<p>$$x + 2- 2x - 2 = 3$$</p>
<p>solving for $x$:</p>
<p>$$x = -2$$</p>
<p>any feedback would be appreciated.</p>
| abkds | 112,225 | <p>You can do it without including logarithms . Just take $t = 2^x$. It will become a quadratic in $t$ solve for $t$ . If $t$ has any negative value , neglect it beacuse $t> 0$ . </p>
<p>$t^2-2t+4 = 0 $
which has complex roots and hence there is no solution for $x$ in the real domain .</p>
|
3,850,422 | <p>For a few days now I've been trying to find a closed form expression for the determinant of the following <span class="math-container">$n\times n$</span> tridiagonal matrix</p>
<p><span class="math-container">$$\begin{pmatrix}c_1+b_1+a_1 & b_1 & 0 & \ddots & 0 \\ c_2 & c_2+b_2+a_2 & b_2 & \ddots & 0 \\ 0 & c_3 & c_3+b_3+a_3 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & b_{n-1}\\ 0 & ... & ... & c_{n} & c_{n}+b_n +a_n\end{pmatrix}$$</span></p>
<p>For the sequences <span class="math-container">$c_n$</span>, <span class="math-container">$b_n$</span>, and <span class="math-container">$a_n$</span>. I've figured out closed form expression for special cases. Namely, when <span class="math-container">$a_n=0$</span>, the determinant is
<span class="math-container">$$\Big(\prod_{i=1}^nb_i\Big)\sum_{l=0}^n\prod_{k=1}^l\frac{c_{k}}{b_k}$$</span>
When <span class="math-container">$l=0$</span> in the product series, that returns a <span class="math-container">$1$</span>. Additionally, if <span class="math-container">$c_1=0$</span>, then the determinant is simply <span class="math-container">$$\prod_{i=1}^nb_i.$$</span></p>
<p>I would really like to find an analogous formula in the case where <span class="math-container">$a_n \neq 0$</span>. For your benefit I will list the first few determinants for small <span class="math-container">$n$</span>
<span class="math-container">$$n=1:\quad\quad c_1+b_1+a_1$$</span>
<span class="math-container">$$n=2:\quad\quad a_1a_2+b_1a_2+a_1b_2+b_1b_2+c_1a_2+c_1b_2+a_1c_2+c_1c_2$$</span>
<span class="math-container">$$n=3:\quad\quad a_1a_2a_3+b_1a_2a_3+a_1b_2a_3+b_1b_2a_3+a_1a_2b_3+b_1a_2b_3+a_1b_2b_3+b_1b_2b_3+c_1a_2a_3+c_1b_2a_3+c_1a_2b_3+c_1b_2b_3+a_1c_2a_3+a_1c_2b_3+c_1c_2a_3+c_1c_2b_3+a_1a_2c_3+b_1a_2c_3+c_1a_2c_3+a_1c_2c_3+c_1c_2c_3$$</span></p>
<p>When you look at this, you may suspect that it is just the sum of every <span class="math-container">$n$</span>th order product of <span class="math-container">$a$</span>'s <span class="math-container">$b$</span>'s and <span class="math-container">$c$</span>'s with no subscript repeated, however this is not the case. For instance, <span class="math-container">$b_1c_2$</span> does not appear in the <span class="math-container">$n=2$</span> formula. Similarly there are <span class="math-container">$6$</span> terms which do not appear in the <span class="math-container">$n=3$</span> formula.</p>
<p>I would really appreciate anyones input on this!</p>
| Servaes | 30,382 | <p>Your matrix is a general tridiagonal matrix, with <span class="math-container">$d_i:=a_i+b_i+c_i$</span> along the diagonal. If we denote the determinant of the <span class="math-container">$n\times n$</span>-matrix by <span class="math-container">$f_n$</span>, then we have the recurrence relation
<span class="math-container">$$f_n=d_nf_{n-1}-b_{n-1}c_{n-1}f_{n-2}.$$</span>
Not much more can be said for general sequences <span class="math-container">$b_n$</span>, <span class="math-container">$c_n$</span> and <span class="math-container">$d_n$</span>. For more information see <a href="https://en.wikipedia.org/wiki/Tridiagonal_matrix#Determinant" rel="nofollow noreferrer">Wikipedia</a>.</p>
|
2,637,983 | <p>I was working on a program to carry out some computations, and ran into an issue of needing to compare some algebraic numbers, but not having enough precision to do it without exact arithmetic, and not knowing how to do it with exact arithmetic.</p>
<p>A little algebra shows that the statement
$$a+b\sqrt{n}>0$$
is equivalent to asking that either $a^2>nb^2$ and $a>0$ or $nb^2>a^2$ and $b>0$. In particular, this means that we can easily compute the order on $\mathbb Q(\sqrt{5})$ using only rational arithmetic on the coefficients of polynomials in $\sqrt{5}$.</p>
<p>However, it seems not so clear how to generalize this reasoning even to an example like deciding whether $a+b\sqrt[3]{n}+c\sqrt[3]{n}^2$ is positive.</p>
<p>In general, suppose that $f$ is an irreducible polynomial in $\mathbb Q[x]$ and has some real root $\alpha$. Let $F=\mathbb Q[x]/(f)\cong \mathbb Q(\alpha)$ be the corresponding field extension. This field clearly can be ordered, as it is identified with a subfield of $\mathbb R$.</p>
<p>Is it possible to compute an explicit order* on $F$ using only rational arithmetic? I feel that this must be possible, but can't figure out how.</p>
<p>I'm most interested in whether, for each fixed field extension $F$, there exists an algorithm taking as input a polynomial in $\alpha$ of degree less than $\deg f$ and deciding whether it is positive or not, using a bounded number of operations. I want this primarily for field extensions of low degree, so I'm less interested in how the complexity grows as $F$ becomes more complex than in how algorithms tailored to a single $F$ fare.</p>
<p>(*Obviously, I'm most interested in being able to compute the order on $\mathbb Q(\alpha)$ inherited from $\mathbb R$, but given that this field is isomorphic to $\mathbb Q(\alpha')$ for any other root of $f$, there are probably multiple orders - any of which would be interesting to compute)</p>
| Dap | 467,147 | <p>Yes.</p>
<p>For the sledgehammer approach, note that the set of $(c_0,\dots,c_{\deg(\alpha)-1})$ such that $\sum_{n=0}^{\deg(\alpha)-1} c_n\alpha^n>0$ can be defined in the language of real-closed fields, so by the Tarski-Seidenberg theorem can be expressed by polynomial inequalities.</p>
<p>This is sort of circular (or at least overkill), since the proof of that theorem relies on Sturm theory which already gives such an algorithm more directly using polynomial remainder sequences; more efficient algorithms use subresultants/Sturm-Habicht sequences. Yap's "Fundamental Problems in Algorithmic Algebra" puts it this way:</p>
<blockquote>
<p>(Generalized Sturm) Let $A$ dominate $B$ and let $α < β$ so that $A(α)A(β) \neq 0.$ Then
$$\mathrm{Var}_{A,B}[α, β] = \sum_{γ,r,s}\mathrm{sign}(A^{(r)}(γ)B^{(s)}(γ))$$
where $γ$ ranges over all roots of $A$ in $[α, β]$ of multiplicity $r ≥ 1$ and $B$ has multiplicity $s$ at $γ,$ and $r + s$ is odd.</p>
</blockquote>
<p>Here $\mathrm{Var}$ is the difference in the number of sign variations in a generalized Sturm sequence defined by $A$ and $B$ when evaluated at $\alpha$ and $\beta,$ and "dominate" is means that if $\gamma$ is an $r$-fold root of $A,$ it is at most an $r$-fold root of $B.$</p>
<p>My $\gamma$ will be your $\alpha,$ sorry. Taking $A\in\mathbb Q[X]$ to the minimal polynomial of $\gamma,$ and taking $[\alpha,\beta]$ to be a rational interval isolating $\gamma$ as a root of $A,$ we can determine the sign of a rational combination $B(\gamma)=\sum_{n=0}^{\deg(A)-1} c_n\gamma^n$ by the generalized Sturm theorem with $r=1$ and $s=0.$ The value of $\mathrm{sign}(A^{(1)}(\gamma))$ can be baked into the algorithm.</p>
|
1,138,789 | <p><a href="http://en.wikipedia.org/wiki/Free_object" rel="nofollow noreferrer">Wikipedia</a> defines free objects as follows:</p>
<blockquote>
<p>Let <span class="math-container">$(\mathcal{C},F)$</span> be a concrete category (i.e. <span class="math-container">$F : \mathcal{C} \to {\rm \bf{Set}}$</span> is a faithful functor), let <span class="math-container">$X$</span> be a set (called <em>basis</em>), <span class="math-container">$A \in \mathcal{C}$</span> an object, and <span class="math-container">$i: X \to F(A)$</span> a map between sets (called <em>canonical injection</em>). We say that <span class="math-container">$A$</span> is the <em>free object</em> on <span class="math-container">$X$</span> (with respect to <span class="math-container">$i$</span>) if and only if they satisfy this universal property:</p>
<p>for any object <span class="math-container">$B$</span> and any map between sets <span class="math-container">$f : X \to F(B)$</span>, there exists a unique morphism <span class="math-container">$\tilde{f} : A \to B$</span> such that <span class="math-container">$f = F(\tilde{f})\circ i$</span>. That is, the following diagram commutes:</p>
<p><img src="https://i.stack.imgur.com/F53eE.png" alt="I took this image from Wikipedia." /></p>
</blockquote>
<p>I just can't seem to understand the categorical definition of free objects in terms of commutative diagrams, so I would appreciate any help in grasping the intuition behind the way free objects are defined in categorical setting.</p>
| individ | 128,505 | <p>For the equation:</p>
<p>$$3x^2+xy+5y^2=z^3$$</p>
<p>write the formula so that it was easier to go through. To facilitate calculations will make the replacement.</p>
<p>$$p=15k^2+3q^2-21n^2-11qk$$</p>
<p>$$s=25n^2+15k^2+3q^2+10qn-11qk-38kn$$</p>
<p>$$a=42n^2-30k^2-46qn+60qk-16q^2$$</p>
<p>$$t=42n^2-6q^2+8k^2-46nk+12qk$$</p>
<p>$$b=3p^2+6ps-17s^2$$</p>
<p>$$j=p^2+6ps-3s^2$$</p>
<p>Then decisions can be recorded and they are.</p>
<p>$$x=ab-5tj$$</p>
<p>$$y=(a-3t)b+3(2t-a)j$$</p>
<p>$$z=3p^2+4ps+21s^2$$</p>
<p>$q,k,n$ - integers, which we ask.</p>
|
18 | <p>Some teachers make memorizing formulas, definitions and others things obligatory, and forbid "aids" in any form during tests and exams. Other allow for writing down more complicated expressions, sometimes anything on paper (books, tables, solutions to previously solved problems) and in yet another setting students are expected to take questions home, study the problems in any way they want and then submit solutions a few days later.</p>
<p>Naturally, the memory-oriented problem sets are relatively easier (modulo time limit), encourage less understanding and more proficiency (in the sense that the student has to be efficient in his approach). As the mathematics is in big part thinking, I think that it is beneficial to students to let them focus on problem solving rather than recalling and calculating (i.e. designing a solution rather than modifying a known one). There is a huge difference between work in time-constrained environment (e.g. medical teams, lawyers during trials, etc.) where the cost of "external knowledge" is much higher and good memory is essential.
However, math is, in general (things like high-frequency trading are only a small part math-related professions), slow.</p>
<p>On the other hand, memory-oriented teaching is far from being a relic of the past. Why is this so? As this is a broad topic, I will make it more specific:</p>
<p><strong>What are the advantages of memory-oriented teaching?</strong></p>
<p><strong>What are the disadvantages of allowing aids during tests/exams?</strong></p>
| Sue VanHattum | 60 | <p>I allow notes on tests, because math is less about memory than about understanding, and I don't want students to focus on the memory part. I don't allow notes on quizzes, because they are on just one problem type, and I want students to be ready to think it through. </p>
<p>You may find this blog post helpful: <a href="http://exzuberant.blogspot.com/2012/07/monkey-and-mathematician-learn-calculus.html">http://exzuberant.blogspot.com/2012/07/monkey-and-mathematician-learn-calculus.html</a></p>
<p>Getting more into particulars:
One disadvantage of allowing notes for tests is that students will spend too much time hunting for the answer in their notes. To offset that, I allow only a 3x5 card.</p>
<p>One advantage of memory-oriented teaching is that it can help students retain. But if that retention is bought at the expense of deeper understanding, you've made a serious mistake. In pre-calc, I ask student to show the end behavior of a function with their arms. This is a memory technique. My hope is that it complements the understanding of <em>why</em> odd-degree functions have "one arm up, one arm down". </p>
|
18 | <p>Some teachers make memorizing formulas, definitions and others things obligatory, and forbid "aids" in any form during tests and exams. Other allow for writing down more complicated expressions, sometimes anything on paper (books, tables, solutions to previously solved problems) and in yet another setting students are expected to take questions home, study the problems in any way they want and then submit solutions a few days later.</p>
<p>Naturally, the memory-oriented problem sets are relatively easier (modulo time limit), encourage less understanding and more proficiency (in the sense that the student has to be efficient in his approach). As the mathematics is in big part thinking, I think that it is beneficial to students to let them focus on problem solving rather than recalling and calculating (i.e. designing a solution rather than modifying a known one). There is a huge difference between work in time-constrained environment (e.g. medical teams, lawyers during trials, etc.) where the cost of "external knowledge" is much higher and good memory is essential.
However, math is, in general (things like high-frequency trading are only a small part math-related professions), slow.</p>
<p>On the other hand, memory-oriented teaching is far from being a relic of the past. Why is this so? As this is a broad topic, I will make it more specific:</p>
<p><strong>What are the advantages of memory-oriented teaching?</strong></p>
<p><strong>What are the disadvantages of allowing aids during tests/exams?</strong></p>
| Geoff | 685 | <p>I’m a high school maths teacher and I use a way to test where crib sheets are allowed; yet pupils are rewarded for not using it. </p>
<p>At the start of the test, the pupils use black ink and are not allowed to use the test aids (such as crib sheet and calculator). When they want to, they indicate to me that they want to use their aids, at which point they must now continue their exam in blue ink. They can only do this swap once, the black pen is removed from their desk. Solutions written in black carry more weight than solutions written just in blue (typically indicated on the exam sheet).</p>
<p>There are many extra advantages. The pupils like to know that at some point they can check their work. They can even correct their work if they find they have made a mistake (which gains credit). Because of this, students learn to present their work better, since they themselves have to come back to their work to fill in the gaps at a later time in the exam. Students who are algebraically strong get their due credit. Weaker students can let the calculator do the algebra, so they can still solve the questions but knowingly obtain fewer points. If a student knows a formula, they can get credit for it. Finally, it is easier for me to write the tests as I don’t need to write two parts: one without aids and one with.</p>
<p>I admit, it takes one test to get the students to appreciate how the system works and how they can benefit from it. The feedback I have had is mostly positive.</p>
|
18 | <p>Some teachers make memorizing formulas, definitions and others things obligatory, and forbid "aids" in any form during tests and exams. Other allow for writing down more complicated expressions, sometimes anything on paper (books, tables, solutions to previously solved problems) and in yet another setting students are expected to take questions home, study the problems in any way they want and then submit solutions a few days later.</p>
<p>Naturally, the memory-oriented problem sets are relatively easier (modulo time limit), encourage less understanding and more proficiency (in the sense that the student has to be efficient in his approach). As the mathematics is in big part thinking, I think that it is beneficial to students to let them focus on problem solving rather than recalling and calculating (i.e. designing a solution rather than modifying a known one). There is a huge difference between work in time-constrained environment (e.g. medical teams, lawyers during trials, etc.) where the cost of "external knowledge" is much higher and good memory is essential.
However, math is, in general (things like high-frequency trading are only a small part math-related professions), slow.</p>
<p>On the other hand, memory-oriented teaching is far from being a relic of the past. Why is this so? As this is a broad topic, I will make it more specific:</p>
<p><strong>What are the advantages of memory-oriented teaching?</strong></p>
<p><strong>What are the disadvantages of allowing aids during tests/exams?</strong></p>
| ncr | 1,537 | <p>In many of the classes I've taught, I've allowed students to have a sheet of notes (e.g., a course on differential equations where some of the recipes they're asked to apply can be easily mixed up) but with the following two features:</p>
<ol>
<li>I talk a lot about how, like others mention above, if they're going to rely on what they've written down during the exam, they probably won't have time to finish the exam. I sell it more as a way for them to organize their own understanding of the material.</li>
<li>I ask them to turn it in to me a few days before the exam and I return it to them in class, on the day of the exam. This helps them not only to start the studying process earlier, but also helps them to not rely on it in the last few days before the exam.</li>
</ol>
|
2,257,365 | <p>In a semicircle of diameter $CD$ there's a chord $AB$ of length 7, and it's parallel to the diameter. There's also a small semicircle that is tangent to $AB$ and its diameter is a segment in $CD$ . Find the area of the semicircle without the small semicircle.</p>
<p>I'm pretty curious about this problem, i've tried many things, like drawing triangles that are similar and also rectangles, so i tries with pithagorean theorem but i got nothing. I'd really like to know how to solve it, thanks.</p>
| egreg | 62,967 | <p>You have
$$
\tan\frac{\alpha}{2}=\frac{1/2}{1}=\frac{1}{2}
$$
Then use
$$
\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}
$$
With $\beta=\alpha/2$, we get
$$
\tan\alpha=\frac{1}{1-1/4}=\frac{4}{3}
$$</p>
<hr>
<p>Using your tools it can be done as well; set $r=AM$, for simplicity. Then
$$
\frac{1}{2}=\frac{1}{2}r^2\sin\alpha
$$
from the area and
$$
1=r^2+r^2-2r^2\cos\alpha
$$
from the cosine law.</p>
<p>Thus
$$
\sin\alpha=\frac{1}{r^2}\qquad \cos\alpha=\frac{2r^2-1}{2r^2}
$$
Finally
$$
\tan\alpha=\frac{1}{r^2}\frac{2r^2}{2r^2-1}=\frac{2}{2r^2-1}
$$</p>
<p>Since $r=\sqrt{1+\frac{1}{4}}=\sqrt{5}/2$, we have
$$
\tan\alpha=\frac{2}{5/2-1}=\frac{4}{3}
$$</p>
|
2,257,365 | <p>In a semicircle of diameter $CD$ there's a chord $AB$ of length 7, and it's parallel to the diameter. There's also a small semicircle that is tangent to $AB$ and its diameter is a segment in $CD$ . Find the area of the semicircle without the small semicircle.</p>
<p>I'm pretty curious about this problem, i've tried many things, like drawing triangles that are similar and also rectangles, so i tries with pithagorean theorem but i got nothing. I'd really like to know how to solve it, thanks.</p>
| Steven Alexis Gregory | 75,410 | <p>You've already found the lengths shown below.</p>
<p><a href="https://i.stack.imgur.com/pZykI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pZykI.jpg" alt="enter image description here" /></a></p>
<p>Hence
<span class="math-container">\begin{align}
u^2 &= \left(\frac{\sqrt 5}{2}\right)^2 -\left(\frac{2}{\sqrt 5}\right)^2 \\
&= \frac 54 - \frac 45 \\
&= \frac{9}{20} \\
u &= \frac{3}{2\sqrt 5}
\end{align}</span></p>
<p>It follows that
<span class="math-container">$ \tan \alpha = \dfrac{\frac{2}{\sqrt 5}}{\frac{3}{2\sqrt 5}} = \frac 43$</span></p>
|
743,227 | <p>I've this question:</p>
<blockquote>
<p>Find the area of the intersection between the sphere $x^2 + y^2 + z^2 = 1$ and the cylinder $x^2 + y^2 - y = 0$.</p>
</blockquote>
<p>Is this second equation even a closed shape? If one were to plot points satisfying that equation, one gets things like $(2, \sqrt{-2})$, $(3, \sqrt{-6})$ and all that.</p>
<p><strong>Edit:</strong> I understand the equation for a circle and such, and have (with the help of everyone who answered) found my issue.</p>
<p>I was plugging in (whole) numbers that weren't in the codomain of the cylinder, similar to having, say, the equation of a circle $x^2 + y^2 = 16$ and plugging in $25$ for $x$—you will get a complex number for $y$. If one plugs in only numbers not in the domain/codomain, then the equation will not <em>seem</em> like the shape it should be. </p>
<p>Sorry for my shortsightedness, and thanks everyone for replying so promptly. :)</p>
| Guy | 127,574 | <p>Yes it is. </p>
<p>Consider this equation only in the $xy$ ,i.e, $(z=0)$ plane.</p>
<p>Clearly it is a circle(why? Prove)</p>
<p>Now since it is independent of $z$, this equation will form a circle for any plane $z\in \Bbb R$</p>
<p>Do you see why that is a cylinder?</p>
|
2,926,270 | <p>The base step is pretty obvious: <span class="math-container">$1 \geq \frac{2}{3}$</span>.</p>
<p>Then we assume that <span class="math-container">$P(k)$</span> is true for some <span class="math-container">$k \in \mathbb{Z}^{+}$</span> and try to prove <span class="math-container">$P(k+1)$</span>. So I have</p>
<p><span class="math-container">$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$</span> </p>
<p>by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than <span class="math-container">$\frac{2}{3}(k+1)\sqrt{k+1}$</span>.</p>
<p>Would appreciate any help!</p>
| Hagen von Eitzen | 39,174 | <p>You hope to have
<span class="math-container">$$\frac23k\sqrt k+\sqrt{k+1}\stackrel?\ge \frac23(k+1)\sqrt{k+1} $$</span>
or equivalently after simple transformations,
<span class="math-container">$$\frac23k\sqrt k+\sqrt{k+1}\stackrel?\ge \frac23k\sqrt{k+1} +\frac23\sqrt{k+1},$$</span>
<span class="math-container">$$\frac13\sqrt{k+1}\stackrel?\ge \frac23k\sqrt{k+1}-\frac23k\sqrt k,$$</span>
<span class="math-container">$$\sqrt{k+1}\stackrel?\ge 2k(\sqrt{k+1}-\sqrt k).$$</span>
A good trick when seeing differences of square roots is often to multiply with their sum, so here
<span class="math-container">$$\sqrt{k+1}(\sqrt{k+1}+\sqrt k)\stackrel?\ge 2k(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)=2k((k+1)-k)=2k.$$</span>
And now the claim is clear as we indeed have
<span class="math-container">$$2k=\sqrt k(\sqrt k+\sqrt k)<\sqrt{k+1}(\sqrt{k+1}+\sqrt k). $$</span></p>
|
1,433,456 | <p>Given matrics $A = \begin{bmatrix}1 && 5 && 2 \\ -1 && 0 && 1 \\ 3 && 2 && 4\end{bmatrix}$ and $B = \begin{bmatrix}6 && 1 && 3 \\ -1 && 1 && 2 \\ 4 && 1 && 3\end{bmatrix}$, find $-3\mathop{Tr}(A-3B)$.</p>
<p>I am not exactly sure what is $-3\mathop{Tr}(\cdot)$, but I did till $(A-3B)$ part.</p>
<p>Where $3B = \begin{bmatrix}18 && 3 && 9 \\ -3 && 3 && 6 \\ 12 && 3 && 9\end{bmatrix}$, and $A-3B$ is $\begin{bmatrix}-17 && 2 && -7 \\ 2 && -3 && -5 \\ -9 && -1 && -5\end{bmatrix}$</p>
<p>How do I apply the $-3\mathop{Tr}(\cdot)$ to $(A-3B)$?</p>
<p>Thanks for the help!</p>
| Hirshy | 247,843 | <p>Hint: for a matrix $A\in \mathbb R^{n\times n}$ the trace of $A$ is a real number. Thus $3\operatorname{trace}(A-3B)$ is just...can you take it from here?</p>
|
1,433,456 | <p>Given matrics $A = \begin{bmatrix}1 && 5 && 2 \\ -1 && 0 && 1 \\ 3 && 2 && 4\end{bmatrix}$ and $B = \begin{bmatrix}6 && 1 && 3 \\ -1 && 1 && 2 \\ 4 && 1 && 3\end{bmatrix}$, find $-3\mathop{Tr}(A-3B)$.</p>
<p>I am not exactly sure what is $-3\mathop{Tr}(\cdot)$, but I did till $(A-3B)$ part.</p>
<p>Where $3B = \begin{bmatrix}18 && 3 && 9 \\ -3 && 3 && 6 \\ 12 && 3 && 9\end{bmatrix}$, and $A-3B$ is $\begin{bmatrix}-17 && 2 && -7 \\ 2 && -3 && -5 \\ -9 && -1 && -5\end{bmatrix}$</p>
<p>How do I apply the $-3\mathop{Tr}(\cdot)$ to $(A-3B)$?</p>
<p>Thanks for the help!</p>
| Emilio Novati | 187,568 | <p>I suppose that you know that the <a href="https://en.wikipedia.org/wiki/Trace_(linear_algebra)" rel="nofollow">trace</a> of a matrix is the sum of the elements on the principal diagonal. You have correctly found $A-3B$, so: trace$(A-3B)=-25$ and $-3 \mbox{trace} (A-3B)= -3\times (-25)$ .</p>
|
663,736 | <p>For a very large number n, how many divisibility tests are required to establish if its prime?</p>
<p>I know this has something to do with the Golden Number, but I can't figure out what. I did try searching for an answer but not much luck.</p>
<hr>
<p>!!EDIT!!
(It wont let me answer my own question for upto 8hours)</p>
<p>I found something posted by someone else on the primality test by golden ratio, although just like Fermat's probability test, it also fails at times. </p>
<blockquote>
<p>There is a primality test by Golden ratio that is used in conjunction with the Lucas N+1 primality test. It is based on the relation between Lucas numbers and Fibonacci numbers. Primality test by Golden ratio states that if </p>
<p>$g^p+(1-g)^p \equiv 1\mod p$ , where g is golden ration, is true then p is prime. In other words, if</p>
<p>$\frac{g^p+(1-g)^p-1}{p} $ divides wholly then p is prime. The expression </p>
<p>$g^p+(1-g)^p$ is a formula for the p-th Lucas number, i.e. </p>
<p>$g^p+(1-g)^p = L_p$. As a result, we can say that if p-th Lucas number minus 1 divides by p wholly then p is prime, i.e.
$ \forall p \in \mathbb{N}, \frac{L_p-1}{p}=a$ where a $\in \mathbb{N} \Rightarrow $ p is prime.</p>
<p>Aaaand it is not true. If you check a composite number 705 which is equal to 3*5*47:</p>
<p>$ \frac{L_{705}-1}{705} = \frac{g^{705} +(1-g)^{705}}{705} = 3.031556 * 10^{144}$</p>
<p>$3.031556 *10^{144}$ is a whole number and the test fails. Fermat's primality test suffers from a similar problem. </p>
</blockquote>
| Newb | 98,587 | <p>To test if some $x$ is prime, we generally have to do divisibility tests only up to and including $\sqrt{x}$. </p>
<p>That's because if some $y > \sqrt{x}$ were a factor of $x$, then there would have to be some $z$ such that $zy = x$. And $z < \sqrt{x}$ because if $z > \sqrt{x}$, then clearly $zy > x$ (as both $z$ and $y$ would be greater than $\sqrt{x}$). But if $z < \sqrt{x}$, then we've already tested $z$ in going up to $\sqrt{x}$!</p>
<p>And we don't have to do trial division for <strong>every</strong> integer up to $\sqrt{x}$. Using the <a href="http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes" rel="nofollow noreferrer">Sieve of Eratosthenes</a>, for example, after testing some $n$, we 'cross out' the integers that are multiples of $n$, because if some multiple of $n$ divides $x$, then $n$ has to divide $x$. So we only need to check the case for $n$. </p>
<p>In that spirit, there are lots of heuristics we can use to make prime-testing more efficient. However, finding large primes remains computationally very difficult.</p>
<p>Finally, I cannot immediately find an application of the golden ratio to this. You may be misremembering. I could dig up <a href="https://i.imgur.com/stLnVYk.jpg" rel="nofollow noreferrer">Prime Number Spirals</a>, but I don't see how they'd be related. </p>
|
2,654,538 | <p>If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?</p>
<p>My attempt:</p>
<p>Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$</p>
<p>The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$</p>
<p>For having $7$ distinct solutions, $n$ can have value = 0,1,2,3</p>
<p>So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>simplifying the given equation we get
$$-5\cos(x)^2-5\cos(x)+4=0$$
with $$\cos(x)=t$$ we get the quadratic equation
$$-5t^2-5t+4=0$$
solving this we get
$$t_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{105}}{10}$$
can you finish?</p>
|
2,205,950 | <p>If $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies $f'(a) \neq 0$ for all $a \in \mathbb{R}$, show that $f$ is one-to-one for all $a\in \mathbb{R}$.</p>
<h2>My attempt</h2>
<p>We know that $f(a)$ is not a constant because $f'(a)\neq 0$.Define $f$ by $f(a)=bx$. $f'(a)=x\neq 0$</p>
<p>If $f(x)=f(v)$ then $$bx=bv$$</p>
<p>$$x=v$$</p>
<p>Thus, $f$ is one-to-one.</p>
| szw1710 | 130,298 | <p>The intermediate value property of a derivative is not needed to get our conclusion. Simply apply the Mean Value Theorem: if $x\ne y$ then there exists an intermmediate point $a$ between $x$ and $y$ s.t. $f(y)-f(x)=f'(a)(y-x)\ne 0$ which means that $f(x)\ne f(y)$.</p>
|
334,701 | <p>The question is really simple, its just terminology.</p>
<p>For simplicity we work on smooth algebraic surfaces and we consider the intersection form on curves on the surface.</p>
<p>So let $S$ be a surface and $D \in \operatorname{Pic}(S)$ a divisor. Then $D$ is said to be nef if
$$
D.C \geq 0
$$
for all curves $C$ on $S$ (wasn't it Miles Reid who introduced this term?).</p>
<p>See <a href="http://en.wikipedia.org/wiki/Nef_line_bundle">http://en.wikipedia.org/wiki/Nef_line_bundle</a>.</p>
<p>My question is now that with this definition, an effective curve need not be nef! This happens exactly when it has negative self-intersection, as is very well possible.
But doesn't nef stand for numerically effective? That suggests that it is weaker then effective..</p>
<p>Please enlighten me, i guess a single line answer will do. Thanks!</p>
| Community | -1 | <p>Yes, this is confusing. Let me try to clarify.</p>
<p>The term "nef" was indeed coined by Reid. But it was <strong>not</strong> supposed to be an abbreviation of "numerically effective"! Rather, it was meant as an acronym standing for "numerically eventually free". (This is why sometimes, in older references, you might see it written as "NEF".)</p>
<p>The idea is that if $D$ is an effective Cartier divisor with the property that, for a sufficiently large natural number $m$, the linear system $|mD|$ has no basepoints, then it must be the case that $D \cdot C \geq 0$ for all curves $C$. So the nef condition is supposed to encapsulate the numerical behaviour of "eventually free" divisors.</p>
<p>Unfortunately, this doesn't really make things any less confusing, because a nef divisor certainly doesn't have to be eventually free (or numerically equivalent to an eventually free divisor.)</p>
<p>There is a quote due to Kollár somewhere, saying that nef means nef and nothing else --- in other words, one should forget where the word came from, and just learn the definition. This seems like a sensible idea to me! </p>
|
4,590,677 | <p>This is perhaps a silly question related to calculating with surds. I was working out the area of a regular pentagon ABCDE of side length 1 today and I ended up with the following expression :</p>
<p><span class="math-container">$$\frac{\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}}}{4}$$</span></p>
<p>obtained by summing the areas of the triangles ABC, ACD and ADE.</p>
<p>I checked my solution with Wolfram Alpha which gave me the following equivalent expression :</p>
<p><span class="math-container">$$\frac{\sqrt{25+10\sqrt{5}}}{4}$$</span></p>
<p>I was able to show that these two expressions are equivalent by squaring the numerator in my expression, which gave me</p>
<p><span class="math-container">$$15+4\sqrt5+2\sqrt{70+30\sqrt5},$$</span></p>
<p>and then "noticing" that</p>
<p><span class="math-container">$$\sqrt{70+30\sqrt5}=\sqrt{25+30\sqrt5+45}=5+3\sqrt5.$$</span>
My question is the following : how could I have known beforehand that my sum of surds could be expressed as a single surd, and is there a way to systematize this type of calculation ? I would have liked to find the final, simplest expression on my own without the help of a computer.</p>
<p>Thanks in advance !</p>
| Hirofumi Ryo | 300,531 | <p>Yes. There is a way to formalize this particular type of sum of square roots, similar to the way a determinant is developed for quadratic equations.</p>
<p>We can write the generic form of the expression in the first place as follows.</p>
<p><span class="math-container">$\sqrt{a+b\sqrt{s}}+\sqrt{c+d\sqrt{s}}=\sqrt{x+y\sqrt{s}}$</span></p>
<p>Note that</p>
<ol>
<li><span class="math-container">$a, b, c, d, s$</span> are given and rational. We aim to express <span class="math-container">$x,y$</span> in terms of them.</li>
<li>The value in the double surds are identical, which are all <span class="math-container">$s$</span>.</li>
<li>Only square roots are considered.</li>
</ol>
<p>Then,</p>
<p><span class="math-container">$(\sqrt{a+b\sqrt{s}}+\sqrt{c+d\sqrt{s}})^2=(\sqrt{x+y\sqrt{s}})^2$</span></p>
<p><span class="math-container">$(a+c)+(b+d)\sqrt{s}+2\sqrt{(a+b\sqrt{s})(c+d\sqrt{s})}=x+y\sqrt{s}$</span></p>
<p><span class="math-container">$2\sqrt{(a+c+bds)+(ad+bc)\sqrt{s}}=(x-a-c)+(y-b-d)\sqrt{s}$</span></p>
<p><span class="math-container">$4(a+c+bds)+4(ad+bc)\sqrt{s}=(x-a-c)^2+(y-b-d)^2s+2(x-a-c)(y-b-d)\sqrt{s}$</span></p>
<p>Let <span class="math-container">$p=x-a-c,\ q=y-b-d$</span>.</p>
<p>By comparing the coefficients in the rational and irrational terms,</p>
<p><span class="math-container">$\begin{cases}
p^2+q^2s=4(a+c+bds)\\
pq=2(ad+bc)
\end{cases}$</span></p>
<p>By substituting <span class="math-container">$q=\frac{2(ad+bc)}{p}$</span> and <span class="math-container">$p=\frac{2(ad+bc)}{q}$</span> from the second equation to the first equation, we can get a formula for each of <span class="math-container">$p$</span> and <span class="math-container">$q$</span>.</p>
<p><span class="math-container">$\begin{cases}
p^4-4(ac+bds)p^2+4(ad+bc)^2s=0\\
sq^4-4(ac+bds)q^2+4(ad+bc)^2=0
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
p^2=2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]\\
q^2=\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
p=\pm\sqrt{2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}\\
q=\pm\sqrt{\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
x=\pm\sqrt{2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}+(a+c)\\
y=\pm\sqrt{\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}+(b+d)
\end{cases}$</span></p>
<p>As a result, there are three determinants to look at:</p>
<ol>
<li><span class="math-container">$D_1=(ac+bds)^2-(ad+bc)^2s$</span> has a rational square root.</li>
<li><span class="math-container">$D_2=2(ac+bds\pm\sqrt{D_1})$</span> has a rational square root.</li>
<li><span class="math-container">$D_3=\dfrac{D_2}{s}$</span> has a rational square root.</li>
</ol>
<hr />
<p>We can use the two surds that you dealt with as an example.</p>
<p><span class="math-container">$a=5, b=2, c=10, d=2, s=5$</span></p>
<p><span class="math-container">$D_1=(ac+bds)^2-(ad+bc)^2s=400$</span>, which is a perfect square.</p>
<p><span class="math-container">$D_2=2(ac+bds\pm\sqrt{D_1})=100\text{ or }180$</span>. 100 is a perfect square.</p>
<p><span class="math-container">$D_3=\dfrac{D_2}{s}=36\text{ or }20$</span>. 36 is a perfect square.</p>
<p>Therefore, the two surds can be summed to a single surd.</p>
<p>As seen above, the determinants are much more complex than the quadratic formula. I wonder if anyone would memorize them like <span class="math-container">$(b^2-4ac)$</span>.</p>
|
2,006,118 | <p>When defining the notion of a measurable cardinal using the ultrafilter definition, why do we require the ultrafilter to be $\kappa$-complete? I know this makes our definition of a measurable cardinal more restrictive (by requiring more sets to be in the ultrafilter) but intuitively speaking, what effect does this condition have on the "size" of kappa? What would happen to our definition if we left it out?</p>
| Stefan Mesken | 217,623 | <p>Your question leaves room for interpretations and may therefore not have a definitive answer. Thus, instead of trying to come up with one, let me highlight two factors such an answer would have to take into account.</p>
<p>First, there is a historical reason. The definition of a measurable cardinal is ultimately due to Stanislaw Ulam, who at the age of 21 proved in <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm16/fm16114.pdf" rel="nofollow noreferrer">Zur Masstheorie der allgemeinen Mengenlehre</a> (in German) the following fact.</p>
<p><strong>Theorem</strong> [Ulam]. If there is a $\sigma$-additive nontrivial measure on $X$ then there is a a two-valued measure on $X$ and $\operatorname{card}(X)$ is greater than or equal to the least inaccessible or there is an atomless measure on $2^{\aleph_0}$ and $2^{\aleph_0}$ is greater than or equal to the least weakly inaccessible.</p>
<p>Combine this with the the following two facts:</p>
<p><strong>Lemma</strong>. If $\kappa$ admits a two valued measure $\mu$, then there is a $\sigma$-complete ultrafilter $U_\mu$ given by
$$
X \in U :\iff \mu(X) = 1
$$
and</p>
<p><strong>Lemma</strong>. Let $\kappa$ be the least cardinal for which there is a $\sigma$-complete ultrafilter $U$ on $\kappa$. Then there is a $\kappa$-complete ultrafilter on $\kappa$.</p>
<p>and you see that the least candidate for a measurable cardinal will in fact admit a $\kappa$-closed ultrafilter.</p>
<p>Second, let $\lambda$ be an inaccessible cardinal and let $U$ be an ultrafilter on $\lambda$. Suppose that $U$ is at least $\sigma$-closed and let $\kappa$ be least such that $U$ is not $\kappa^+$-closed, i.e. there is a sequence $(X_\alpha \mid \alpha < \kappa) \in ^\kappa U$ such that $\bigcap_{\alpha < \kappa} X_\alpha \not \in U$. (Note that $\kappa \le \lambda$.) We may form an elementary embedding
$$
j_U \colon V \to \operatorname{Ult}(V;U)
$$
such that $\operatorname{Ult}(V;U)$ is transitive and one can show $\kappa$ is the least ordinal $\alpha$ with $j_U(\alpha) \neq \alpha$. We say that $\kappa$ is the critical point of $j_U$. This allows us to define a $\kappa$-closed ultrafilter $W$ on $\kappa$ by
$$
X \in W \iff \kappa \in j_U(X).
$$
In particular, we may now form
$$
j_W \colon V \to \operatorname{Ult}(V; W).
$$
But from $j_W$ we can recover a $\kappa$-closed ultrafilter $\bar{U}$ on $\lambda$ by letting
$$
X \in \bar{U} \iff \kappa \in j_W(\bar{U}).
$$</p>
<p>Thus, if $\kappa < \lambda$, then the fact that $\lambda$ admits a $\kappa$-closed ultrafilter really doesn't have anything to do with $\lambda$. Instead it is an consequence of the fact that $\kappa$ admits a $\kappa$-closed ultrafilter. A slight variations of this also shows that any $\mu > \kappa$ admits a $\kappa$-closed ultrafilter and therefore the reason that all of these $\mu > \kappa$ have a $\kappa$-closed ultrafilter on them is really due to the <em>measurability</em> of $\kappa$.</p>
<p>You can find a proof of all of the above facts and much more about measurable cardinals in Jech's wonderful textbook 'Set Theory (The 3rd millenium edition)'.</p>
|
2,006,118 | <p>When defining the notion of a measurable cardinal using the ultrafilter definition, why do we require the ultrafilter to be $\kappa$-complete? I know this makes our definition of a measurable cardinal more restrictive (by requiring more sets to be in the ultrafilter) but intuitively speaking, what effect does this condition have on the "size" of kappa? What would happen to our definition if we left it out?</p>
| Mitchell Spector | 350,214 | <blockquote>
<p>"When defining the notion of a measurable cardinal using the ultrafilter definition, why do we require the ultrafilter to be $\kappa$-complete? I know this makes our definition of a measurable cardinal more restrictive (by requiring more sets to be in the ultrafilter)..."</p>
</blockquote>
<p>It's not correct to say that $\kappa$-completeness requires more sets to be the in the ultrafilter. It's true that $\kappa$-completeness means that the intersection of fewer than $\kappa$ many sets in the ultrafilter is also in the ultrafilter — but you could just as easily say that $\kappa$-completeness requires more sets <em>not</em> to be in the ultrafilter because the union of fewer than $\kappa$ many sets not in the ultrafilter is also not in the ultrafilter.</p>
<p>It's best to think of $\kappa$-completeness as simply a structural requirement on the ultrafilter as a whole; it doesn't mean that more sets or fewer sets are in the ultrafilter than in a non-$\kappa$-complete ultrafilter, just <em>different</em> sets.</p>
<blockquote>
<p>"... but intuitively speaking, what effect does this condition have on the "size" of kappa? What would happen to our definition if we left it out?"</p>
</blockquote>
<p>If the alternative is just to require $\kappa$ to be an uncountable cardinal that carries a non-principal ultrafilter (with no completeness requirement at all), then <em>every</em> uncountable cardinal satisfies the property. So this isn't a large cardinal property at all (or even a newly interesting property).</p>
<p>If you replace $\kappa$-completeness by countable completeness, then the cardinals satisfying this property are precisely those cardinals greater than or equal to the first measurable cardinal. So, for $\kappa$ greater than the first measurable cardinal, this doesn't imply that $\kappa$ is dramatically larger than the cardinals less than $\kappa.$</p>
<p>With the standard definition of measurability, every measurable cardinal is <em>much</em> larger than the cardinals less than it. That's why it has been adopted as the definition of measurability.</p>
|
681,543 | <p>So I have a function </p>
<p>$$r= ( x^2 + y^2)^{1/2}$$</p>
<p>and I want to show that </p>
<p>$$\operatorname{grad} f(r) = f'(r)(\operatorname{grad} r).$$</p>
<p>I don't really know where to begin do you say that $f(r) = (f \circ r)(x,y)$ and then use the definition of gradient to work it out. Please give a relatively basic answer as I'm new to multi-variable calculus, thanks. </p>
| Nitish | 61,574 | <p>Let's break it down:</p>
<p><span class="math-container">\begin{align}
f &: \Re^+\to \Re\\f(z) &=\sqrt z\\f'(z)&=\frac{1}{2\sqrt{z}}\\
\\g&:\Re^2\to\Re\\ g(x,y)&=x^2+y^2\\\nabla g(x,y)&=\binom{2x}{2y}
\\\mbox{So,}
\\f'(g(x,y))&=f'(g(x,y))\cdot\nabla g(x,y)\\
&=\frac{1}{2\sqrt{x^2+y^2}}\cdot\binom{2x}{2y}
\end{align}</span></p>
|
3,043,846 | <p>I want to rewrite a question not so well written on this site and clarified by Mr. Lahtonen (thank you again).</p>
<p>So here the question:</p>
<blockquote>
<p>Let the extention <span class="math-container">$GF(p^m) \supset GF(p)$</span> that contains roots of
<span class="math-container">$p(x)=x^{p^{m}}-1$</span>. Show that those roots are distinct and that forms
a field</p>
</blockquote>
<p>I know that the roots of <span class="math-container">$p(x)=x^{p^{}}-1$</span> are contained in <span class="math-container">$p(x)=x^{p^{m}}-1$</span>, but then?</p>
<p>edit: probably the correct exercise was <span class="math-container">$p(x)=x^{p^{m}-1}$</span></p>
| user1729 | 10,513 | <p>As another example, both the trivial group <span class="math-container">$Id$</span> and the cyclic group of order two <span class="math-container">$C_2$</span> have trivial automorphism group:
<span class="math-container">$$\operatorname{Aut}(Id)\cong Id\cong\operatorname{Aut(C_2)}$$</span></p>
<p>This is the smallest possible example...</p>
<p>(These are the only two groups <span class="math-container">$G$</span> with <span class="math-container">$\operatorname{Aut}(G)\cong Id$</span>. See <a href="https://groupprops.subwiki.org/wiki/Trivial_automorphism_group_implies_trivial_or_order_two" rel="nofollow noreferrer">here</a> for a proof.)</p>
|
895,759 | <p>Is there some sorts of Krull's theorem (that every ring has maximal ideal) for rings that do not have multiplicative identity (unit)? So I know that non-unital rings do not satisfy Krull's theorem, but for some types of non-unital rings, theorem does get satisfied. So what is it?</p>
<p>Edit: Wikipedia seems to mention the case with regular ideal, but does not explain it.</p>
| Xam | 133,781 | <p>I found a theorem given in David Burton's book: "A first course in rings and ideals" where he proves a "non-unitary" version of Krull theorem. It says that every nonzero finitely generated ring $R$ has a maximal ideal. The proof is essentially the same than for the usual version of Krull theorem.
So this result works for non-unitary rings, but actually this is just a particular case of Krull theorem for modules: <a href="https://math.stackexchange.com/questions/110845/maximal-submodule-in-a-finitely-generated-module-over-a-ring">Maximal submodule in a finitely generated module over a ring</a></p>
|
825,318 | <p>Can someone please help me with these True and False questions? I've tried them myself, but I'm not very good at discrete math... Thank you in advance!</p>
<ol>
<li><p>Any set $A$ and $B$ with $B\subseteq A$ and $f: B \to A$ be $1$-$1$ and onto, then $B = A$</p>
<p>False?</p></li>
<li><p>Let $A$ and $B$ be nonempty sets and $f:A \to B$ be a $1$-$1$ function. Then $f(X\cap Y) = f(X)\cap f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$</p>
<p>True?</p></li>
<li><p>Let $A$ and $B$ be nonempty sets and $f:A \to B$ be a function. Then if $f(X\cap Y) = f(X)\cap f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$, then $f$ must be $1$-$1$.</p>
<p>False?</p></li>
<li><p>There is no one-to-one correspondence between the set of all positive integers and the set of all odd positive integers because the second set is a proper subset of the first.</p>
<p>False?</p></li>
<li><p>if $(A \cup B\subset A \cup C)$ then $B\subset C$</p>
<p>False?</p></li>
<li><p>If $A$, $B$, and $C$ are three sets, then the only way that $A \cup C$ can equal $B \cup C$
is $A = B$.</p>
<p>False?</p></li>
<li><p>If the product $A \times B$ of two sets $A$ and $B$ is the empty set , then both $A$ and
$B$ have to be empty set.</p>
<p>False?</p></li>
</ol>
| André Nicolas | 6,312 | <p>It is an unfortunate fact that when one sees, for example, $N(20.6,0.8)$, one does not immediately know whether $0.8$ is the variance or the standard deviation. Each convention is used. I suspect that in "applied" settings, the second parameter is more frequently the standard deviation. </p>
<p>There is a similar issue with the
exponential distribution: the parameter used might be the mean, or the reciprocal of the mean. Things get worse with the gamma distribution. </p>
<p>And there are issues with the geometric distribution (and the negative binomial)). In the geometric, the random variable might be the number of trials until the first success, or the number of failures until the first success. </p>
|
50,227 | <p>The problem I'm having is mapping a 3D triangle into 2 dimensions. I have three points in $(x,y,z)$ form, and want to map them onto the plane described by the normal of the triangle, such that I end up with three points in $(x,y)$ form.</p>
<p>My guess would be it'd assign an arbitrary up vector and then doing something? Finding the distance traveled along the plane from one vertex to another? What do I do, and how do I do it?</p>
| Ross Millikan | 1,827 | <p>You have not specified the problem well enough. Do you have three points $(x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3)$ to map to two dimensional points? The simplest is to ignore the third coordinate. This is not as stupid as it sounds-you are projecting the triangle on the $xy$ plane. If you want to project onto another plane, how is it defined?</p>
|
50,227 | <p>The problem I'm having is mapping a 3D triangle into 2 dimensions. I have three points in $(x,y,z)$ form, and want to map them onto the plane described by the normal of the triangle, such that I end up with three points in $(x,y)$ form.</p>
<p>My guess would be it'd assign an arbitrary up vector and then doing something? Finding the distance traveled along the plane from one vertex to another? What do I do, and how do I do it?</p>
| Glorious Nathalie | 948,761 | <p>Suppose you have three points A(x_1, y_1, z_1) , B(x_2, y_2, z_2), C(x_3, y_3, z_3). You can have the axes of the coordinate system attached to the plane containing <span class="math-container">$A,B,C$</span> anywhere on this plane, and the orientation of the its <span class="math-container">$x', y'$</span> axes can be oriented freely as well. So to specify the problem further, let's assume that you want to place point <span class="math-container">$A$</span> at the origin of the <span class="math-container">$O'x'y'$</span> plane (which is coincident with the plane of the triangle). Further, assume that the <span class="math-container">$x'$</span> axis is chosen to be along the vector <span class="math-container">$\vec{AB}$</span>. These two assumptions make the coordinate system <span class="math-container">$O'x'y'$</span> unique. It is specified in steps as follows:</p>
<p>Step 1: Find the normal vector to the plane of the triangle, this is given by</p>
<p><span class="math-container">$ N = ( B - A) \times (C - A ) $</span></p>
<p>Step 2: Normalize vector <span class="math-container">$N$</span> by defining the unit vector along its direction by</p>
<p><span class="math-container">$\hat{n} = \dfrac{N}{\| N \| } $</span></p>
<p>Step 3: Choose the <span class="math-container">$x'$</span> axis to point in the direction of <span class="math-container">$\vec{AB} = B - A $</span>, therefore, the unit vector along the <span class="math-container">$x'$</span> axis will be</p>
<p><span class="math-container">$ \hat{x'}= \dfrac{ B - A}{\| B - A \|} $</span></p>
<p>Step 4: Now for the right handed system, the <span class="math-container">$y'$</span> axis unit vector is given by</p>
<p><span class="math-container">$ \hat{y'} = \hat{n} \times \hat{x'} $</span></p>
<p>Step 5: Place the three vectors <span class="math-container">$\hat{x'}, \hat{y'}, \hat{n} $</span> are the columns of a <span class="math-container">$3 \times 3 $</span> matrix R $ in this order.</p>
<p>Step 6: The coordinate system <span class="math-container">$O' x' y' n$</span> and the world coordinate system are related by</p>
<p><span class="math-container">$ \large {P = A + R Q} $</span></p>
<p>where <span class="math-container">$P = [x, y, z]^T $</span> is the coordinate vector of a point expressed in the world coordinate frame, and <span class="math-container">$Q$</span> is the coordinate vector of the same point expressed in the frame what we just constructed.</p>
<p>From Step 6. we have</p>
<p><span class="math-container">$Q = R^T (P - A) $</span></p>
<p>So for our triangle, we have</p>
<p><span class="math-container">$A' = R^T (A - A) = 0 $</span></p>
<p><span class="math-container">$B' = R^T (B - A) $</span></p>
<p><span class="math-container">$C' = R^T (C - A) $</span></p>
<p>Points <span class="math-container">$A', B', C'$</span> are the coordinate vectors in the coordinate system <span class="math-container">$Ox'y'z'$</span> (the <span class="math-container">$z'$</span> axis is the axis along <span class="math-container">$\hat{n}$</span> ). Note that the <span class="math-container">$z'$</span> coordinate of <span class="math-container">$A', B', C'$</span> is zero, therefore, we effectively transformed the <span class="math-container">$3D$</span> world coordinates of <span class="math-container">$A,B,C$</span> to <span class="math-container">$2D$</span> coordinates in the reference frame <span class="math-container">$O'x'y'z'$</span> that we constructed in steps 1. through 6.</p>
|
2,157,914 | <p>I am struggling with the next exercise of my HW:</p>
<p>How many conjugacy classes are in $GL_3(\mathbb{F}_p)$? And how many in $SL_2(\mathbb{F}_p)$?</p>
<p>It's on the topic of Frobenius normal form of finitely generated modules over $\mathbb{F}_p$.</p>
<p>I'd appreciate any idea.</p>
| N. S. | 9,176 | <p><strong>Hint</strong> $$k-\frac{1}{2} \leq \sqrt{n} < k+\frac{1}{2} \Leftrightarrow k^2-k+\frac{1}{4} \leq n <k^2+k+\frac{1}{4}$$</p>
<p>There are $2k$ integers for which $f(n)=\frac{1}{k}$.</p>
|
1,814,823 | <p>In this question , multiple concepts of graphical transformations are involved. I am facing problems in applying all of them in a single question.</p>
| Spenser | 39,285 | <p>The <a href="https://en.wikipedia.org/wiki/Graph_of_a_function" rel="nofollow">graph</a> of your function is by definition the set
$$\left\{\left(x,\frac{x}{1+|x|}\right):x\in\Bbb R\right\}\subseteq\Bbb R^2.$$</p>
|
353,947 | <p>Problem. Let $f:[a,b]\to\mathbb{R}$ be a function such that $ f\in C^3([a,b])$ and $f(a)=f(b)$. Prove that $$ \left|\int\limits_{a}^{\frac{a+b}{2}}f(x)dx-\int\limits_{\frac{a+b}{2}}^{b}f(x)dx\right|\leq\frac{(b-a)^4}{192}\max_{x\in [a,b]}|f'''(x)|.$$
Any idea are welcome.</p>
| r9m | 129,017 | <p>Idea : Making use of successive Integration by parts ,</p>
<p>$\int P(x)f^{(3)}(x)\,dx = P(x)f^{(2)}(x)-P^{(1)}(x)f^{(1)}(x)+P^{(2)}(x)f(x)-\int P^{(3)}(x)f(x)\,dx$</p>
<p>Consider the two third degree monic polynomials, $P_1(x)$ and $P_2(x)$.</p>
<p>Now, we compute the difference of the definite integrals: </p>
<p>$\int\limits_{\frac{a+b}{2}}^b P_2(x)f^{(3)}(x)\,dx-\int\limits_a^{\frac{a+b}{2}} P_1(x)f^{(3)}(x)\,dx$ </p>
<p>$=[P_2(b)f^{(2)}(b)+P_1(a)f^{(2)}(a)-P_2(\frac{a+b}{2})f^{(2)}(\frac{a+b}{2})-P_1(\frac{a+b}{2})f^{(2)}(\frac{a+b}{2})] - [P_2'(b)f^{(1)}(b)+P_1'(a)f^{(1)}(a)-P_2'(\frac{a+b}{2})f^{(1)}(\frac{a+b}{2})-P_1'(\frac{a+b}{2})f^{(1)}(\frac{a+b}{2})] + [P_2''(b)f(b)+P_1''(a)f(a)-P_2''(\frac{a+b}{2})f(\frac{a+b}{2})-P_1''(\frac{a+b}{2})f(\frac{a+b}{2})] - [\int\limits_{\frac{a+b}{2}}^b P_2'''(x)f(x)\,dx - \int\limits_a^{\frac{a+b}{2}} P_1'''(x)f(x)\,dx]$</p>
<p>We want our polynomials to satisfy the relations:</p>
<p>$P_2(b)=0,P_1(a)=0,P_2(\frac{a+b}{2})+P_1(\frac{a+b}{2})=0$ ... (<strong>i</strong>)</p>
<p>$P_2'(b)=0,P_1'(a)=0,P_2'(\frac{a+b}{2})+P_1'(\frac{a+b}{2})=0$ ... (<strong>ii</strong>)</p>
<p>$P_2''(b)+P_1''(a)=0,P_2''(\frac{a+b}{2})+P_1''(\frac{a+b}{2})=0$ ... (<strong>iii</strong>)</p>
<p>(In (<strong>iii</strong>) we made use of $f(a)=f(b)$)</p>
<p>From, (<strong>ii</strong>) and (<strong>iii</strong>), it follows that $P_1(x)=(x-a)^2(x-c_1)$ and $P_2(x)=(x-b)^2(x-c_2)$, plugging in the other conditions we get $c_1=\frac{a+3b}{4}$ and $c_2=\frac{b+3a}{4}$</p>
<p>Therefore,</p>
<p>$\int\limits_a^{\frac{a+b}{2}} P_1(x)f^{(3)}(x)\,dx-\int\limits_{\frac{a+b}{2}}^b P_2(x)f^{(3)}(x)\,dx$</p>
<p>$=6\int\limits_{\frac{a+b}{2}}^b f(x)\,dx-6\int\limits_a^{\frac{a+b}{2}}f(x)\,dx$</p>
<p>Making use of the fact $P_1(x)\le0$ in the interval $[a,\frac{a+b}{2}]$ (Since, $\frac{a+3b}{4} \ge \frac{a+b}{2}$) </p>
<p>and $P_2(x)\ge 0$ in the interval $[\frac{a+b}{2},b]$ (Since, $\frac{b+3a}{4} \le \frac{a+b}{2}$)</p>
<p>and, $\int\limits_a^{\frac{a+b}{2}}(x-a)^2(x-\frac{a+3b}{4})=-\frac{1}{64}(b-a)^4$</p>
<p>$\int\limits_{\frac{a+b}{2}}^b (x-b)^2(x-\frac{b+3a}{4})=\frac{1}{64}(b-a)^4$</p>
<p>Therefore, $\left|\int\limits_{\frac{a+b}{2}}^b f(x)\,dx-\int\limits_a^{\frac{a+b}{2}}f(x)\,dx\right| \le \frac{2}{64\times 6}(b-a)^4\max_{x\in [a,b]}|f'''(x)|=\frac{1}{192}(b-a)^4\max_{x\in [a,b]}|f'''(x)|$</p>
|
4,158,431 | <p>Say we have the set of real numbers. Can we construct a different ring than the usual ring of real numbers?</p>
<p>I am trying to wrap my head around the idea of rings and I couldn't find two other operations that will hold distributive property other than usual addition and multiplication.</p>
<p>If we can find another ring with the set of real numbers, is there a set with only one ring structure? If we can't find another ring with the set of real numbers, is there a set with different ring structures?</p>
| Kavi Rama Murthy | 142,385 | <p>(31) shows that <span class="math-container">$\phi_n(t)\to \phi (t)$</span> for each <span class="math-container">$t\neq x$</span>. Now let <span class="math-container">$m \to \infty$</span> in the inequality <span class="math-container">$$|\phi_n(t)-\phi_m (t)| \leq \frac {\epsilon} {b-a}$$</span> to see that <span class="math-container">$$|\phi_n(t)-\phi (t)| \leq \frac {\epsilon} {b-a}$$</span> for <span class="math-container">$n \geq N$</span> for <span class="math-container">$t \neq x$</span>.</p>
|
4,158,431 | <p>Say we have the set of real numbers. Can we construct a different ring than the usual ring of real numbers?</p>
<p>I am trying to wrap my head around the idea of rings and I couldn't find two other operations that will hold distributive property other than usual addition and multiplication.</p>
<p>If we can find another ring with the set of real numbers, is there a set with only one ring structure? If we can't find another ring with the set of real numbers, is there a set with different ring structures?</p>
| user912011 | 912,011 | <p>Since <span class="math-container">$f_n(t) \rightarrow f(t)$</span> for any <span class="math-container">$a \leq t \leq b$</span>, by theorem 3.3,
<span class="math-container">$$ f_n(t)-f_n(x) \rightarrow f(t)-f(x) $$</span>
and hence
<span class="math-container">$$ \frac{f_n(x)-f_n(t)}{t-x} \rightarrow \frac{f(t)-f(x)}{t-x} $$</span>
pointwise.</p>
|
2,261,927 | <p>How to get alternative form from equation 1)</p>
<p>$$ 1) -a^2 + a + b^2 -b $$</p>
<p>to equation 2)</p>
<p>$$ 2) (a-b)(a+b-1)$$</p>
| StackTD | 159,845 | <p>Group terms and factor, e.g. by first using $\color{blue}{a^2-b^2=(a-b)(a+b)}$ and then putting the common term $\color{red}{a-b}$ up front:
$$\begin{align}-a^2 + a + b^2 -b
& = -\left(\color{blue}{a^2-b^2}\right)+a-b \\[4 pt]
& = -\color{blue}{\left(a-b\right)\left(a+b\right)}+a-b \\[4 pt]
& = -\color{red}{\left( a-b \right)} \left( \left(a+b\right)+1\right) \\[4 pt]
& = -\left( a-b \right) \left( a+b+1\right)
\end{align}$$
Note that there's an extra minus (missing in your answer).</p>
|
1,407,714 | <p>Why is:
$$\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}\,=\emptyset\quad?$$
Indeed,
$$\forall n\in\mathbb N, 0\in\mathopen{]}0,e^{-n}\mathclose{[},$$
and thus $0\in\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}$. So, what's wrong here ? </p>
| Augustin | 241,520 | <blockquote>
<p>$$\forall n\in\mathbb N, 0\in ]0,e^{-n}[$$</p>
</blockquote>
<p>This is wrong. $0\notin ]0,e^{-n}[$.</p>
|
1,407,714 | <p>Why is:
$$\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}\,=\emptyset\quad?$$
Indeed,
$$\forall n\in\mathbb N, 0\in\mathopen{]}0,e^{-n}\mathclose{[},$$
and thus $0\in\bigcap_{n=0}^\infty\,\mathopen{]}0,e^{-n}\mathclose{[}$. So, what's wrong here ? </p>
| Scientifica | 164,983 | <p><strong>Hint</strong>: Try to show that if $\bigcap_{n=0}^\infty\neq \emptyset$ which means $\exists x\in\mathbb{R},\,x\in\bigcap_{n=0}^\infty]0,\,e^{-n}[$ then $\exists m\in\mathbb{N},\,e^{-m}<x$ and so $x\notin]0,\,e^{-m}[$ which leads to a contradiction. Note that since $x\in\bigcap_{n=0}^\infty]0,\,e^{-n}[$ then in particular $x\in ]0,1[$ and so $x>0$. Use the fact that $\lim\limits_{n\to +\infty}e^{-n}=0$.</p>
|
1,581,545 | <p>I am looking for a proof of Euclid's Lemma, i.e if a prime number divides a product of two numbers then it must at least divide one of them.</p>
<p>I am coding this proof in Coq, and i'm doing it over <em>natural numbers</em>. I aim to prove the uniqueness of prime factorization (So I cannot use this lemma!). However, I can use the existence of a prime factorization, which I already proved.</p>
<p>I do not want to use the gcd algorithm as that would involve coding it in Coq and proving it is correct which may be difficult. The idea is to use this proof in a computer science course, so I do not want to overcomplicate things.</p>
<p>Is there any proof of this lemma that does not use gcd, or Bezout's lemma, or the uniqueness of prime factorization? Maybe something using induction?</p>
<p>Thank you in advance.</p>
<p>EDIT: The Proof should be on NATURAL NUMBERS. No answer did the proof in N.</p>
| CopyPasteIt | 432,081 | <p>The OP's stated goal is to prove the uniqueness of prime factorization using elementary/foundational techniques, perhaps by proving Euclid's lemma in a different way.</p>
<p>Since the end goal is to prove uniqueness in the FTA, of interest here are two similar elementary proofs.</p>
<p>(1) The wikipedia article: <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Elementary_proof_of_uniqueness" rel="nofollow noreferrer">Elementary proof of uniqueness</a></p>
<p>which uses the distributivity law a couple of times.</p>
<p>(2) Bill Dubuque's: <a href="https://math.stackexchange.com/a/117416/432081">Elementary proof of the FTA</a></p>
<p>The 'hook' used there: every <span class="math-container">$n \gt 1$</span> has a least (unique) factor <span class="math-container">$p \ne 1$</span>. The argument works around that to get the FTA. </p>
<p>I think that the techniques found in the above arguments are beautiful, although those who have not invested the time most likely would not agree.</p>
<p>In conclusion, a budding <span class="math-container">$\text{AI MATH BOT}$</span> might enjoy chewing on this elementary theory!</p>
|
987,895 | <blockquote>
<p>Let $(G,\cdot)$ be a group, $g \in G$.</p>
<p>For $a,b \in G$ define $a * b := a \cdot g^{-1} b$. Show that $(G,*)$ is a group with the neutral element $g$ and $f : (G,*) \rightarrow (G,\cdot), a \mapsto a \cdot g^{-1}$ is a group isomorphism.</p>
</blockquote>
<p>In order to show that $(G,*)$ is a group, I need to show that $*$ is closed and associative, g is the neutral element and an inverse element exists for each $x \in G$.</p>
<p>(* is closed, associativity)</p>
<p>$*$ is closed because $\cdot$ is, $\forall x,y \in G: x * y = x \cdot g^{-1} \cdot y$, which is closed because $(G,\cdot)$ is a group. The same applies for associativity: $(x * y) * z = (x \cdot g^{-1} \cdot y) \cdot g^{-1} \cdot z = x \cdot (g^{-1} \cdot y) \cdot g^{-1} \cdot z) = x * (y * z)$. </p>
<p>(neutral element)</p>
<p>$x * g = x \cdot g^{-1} g = x \cdot e = x, g * x = g \cdot g^{-1} x = e \cdot x = x$</p>
<p>Is this right?</p>
<p>How do I show that there is an inverse element, so that $\forall x \in G \; \exists x': x * x' = g$?</p>
<p>In order to show that f is a group isomorphism, I need to show that it is a group homomorphism, and f is a bijection, right?</p>
<p>I already figured out how to show that f is surjective, but how do I show that f is injective? </p>
| Hagen von Eitzen | 39,174 | <p>It is best to start from the end, i.e. first to show that $f$ has the isomorphism property (even if we don't know yet that $(G,*)$ is a group), that is
$$\tag1 f(a*b)=f(a)\cdot f(b)\qquad\text{for all $a,b\in G$}$$
and
$$\tag2 f\text{ is bijective}.$$
Once you have shown $(1)$ and $(2)$, the group properties are straightforward:
$$ f((a*b)*c)=f(a*b) f(c)=(f(a) f(b)) f(c)=f(a)(f(b) f(c))=f(a) f(b*c)=f(a*(b*c))$$
and hence by injectivity of $f$, $(a*b)*c=a*(b*c)$.</p>
<p>Or $a*f^{-1}(e)=f^{-1}(e)*a=a$ for all $a\in G$ because after application of $f$, we have $f(a)\cdot e=e\cdot f(a)=f(a)$. (And you can readily compute $f^{-1}(e)$ explicitly).</p>
<p>Moreover, the inverse is what it should be accoring to $f$, i.e. $a*a'=a'*a=f^{-1}(e)$ is equivalent to $f(a)f(a')=f(a*)f(a)=e$, i.e. we must have $a'=f^{-1}(f(a)^{-1})$.</p>
|
215,835 | <p>According to Willard,</p>
<p>If $(X,\tau)$ is a topological space, a base for $\tau$ is a collection $\mathscr{B} \subset \tau$ such that $\tau=\{ \bigcup_{B \in \mathscr C} : \mathscr C \subset \mathscr B\}$. Evidently, $\mathscr B$ is a base for $X$ iff whenever $G$ is an open set in $X$ and $p \in G$ there is some $B \in \mathscr B$ such that $p \in B \subset G$.</p>
<p>Question 1: Is it safe to assume that in the sentence beginning with "Evidently" it is assumed that $\mathscr B \subset \tau$, for otherwise the iff statement is not true.</p>
<p>Question 2: I've been told that not all basic sets are open, but it seems by the above definition that they are defined to be open.</p>
<p>Comment on Question 2: There is also a definition of being a base for "a" topology. Is this what was meant by not all basic sets are open? Is this just a semantic issue, i.e. the basic sets are open in the topology that the base is a base for but not open in general? Or can there be a base for a topology where the basic sets are not open in that same topology?</p>
| Rudy the Reindeer | 5,798 | <p>Question 1: Yes. It says so in the first sentence. Unless I am misunderstanding your question.</p>
<p>Question 2: Every base is a subset of the topology it generates. By definition, all sets in a topology are open. Hence all sets in a base also have to be open. Hence "no" to your last question in the comment to question 2.</p>
|
1,859,719 | <blockquote>
<p>Let be $U (x,y) = x^\alpha y^\beta$. Find the maximum of the function $U(x,y)$ subject to the equality constraint $I = px + qy$.</p>
</blockquote>
<p>I have tried to use the Lagrangian function to find the solution for the problem, with the equation</p>
<p>$$\nabla\mathscr{L}=\vec{0}$$</p>
<p>where $\mathscr{L}$ is the Lagrangian function and $\vec{0}=\pmatrix{0,0}$.
Using this method I have a system of $3$ equations with $3$ variables, but I can't simplify this system:</p>
<p>$$ax^{\alpha-1}y^\beta-p\lambda=0$$
$$\beta y^{\beta-1}x^\alpha-q\lambda=0$$
$$I=px+qx$$</p>
| Jashaszun | 258,167 | <p>You can certainly simplify that system!</p>
<p>$$\lambda = \frac{\alpha x^{\alpha - 1}y^\beta} p$$
$$\lambda = \frac{\beta y^{\beta - 1}x^\alpha} q$$</p>
<p>Thus</p>
<p>$$\frac{\alpha x^{\alpha - 1}y^\beta} p = \frac{\beta y^{\beta - 1}x^\alpha} q$$</p>
<p>and</p>
<p>$$q\left(\alpha x^{\alpha - 1}y^\beta\right) = p\left(\beta y^{\beta - 1}x^\alpha\right)$$</p>
<p>You can reduce powers:
$$q\alpha \cdot y^\beta = p\beta \cdot xy^{\beta-1}$$
$$q\alpha \cdot y = p\beta \cdot x$$</p>
<p>So now you have two equations:</p>
<p>$$qy\alpha = px\beta$$
$$I = px + qy$$</p>
<p>We then have $$qy = px\frac\beta\alpha$$
$$I = px + px\frac\beta\alpha = x\cdot p\left(1+\frac\beta\alpha\right)$$</p>
<p>So $$x = \frac I {p\left(1+\frac\beta\alpha\right)}$$</p>
<p>Do similar computations for $y$, and you get $$(x, y) = \left(\frac I {p\left(1+\frac\beta\alpha\right)}, \frac I {q\left(1+\frac\alpha\beta\right)}\right)$$</p>
<p>You can then continue with Lagrange multipliers.</p>
|
2,781,801 | <p>When asked to evaluate $g$ at the point specified above we would get $\dfrac{1}{e} \cdot \log_e(\frac{1}{\sqrt e})$ and that evaluates to some -0.18393... but the correct answer is -1/2e. How does it get simplified to that?</p>
| Claude Leibovici | 82,404 | <p>$$g(x)=x^2 \log(x) \implies g\left(\frac{1}{\sqrt{e}}\right)=\left(\frac 1 {{\sqrt{e}}}\right)^2 \log\left(\frac{1}{\sqrt{e}}\right)=-\frac 1 e\times \log(\sqrt{e})=-\frac 1 e\times\frac 12 \log(e)$$ and, by definition $\log(e)=1$.</p>
|
108,253 | <p>I would like to assign 'x' individuals to 'y' groups, randomly. For example, I would like to divide 50 individuals into 100 groups randomly. Of course, with more groups than individuals many of the groups will have zero individuals, while some groups will have multiple individuals. That is fine. With random assignment, the distribution of the number of individuals per group should fit a Poisson distribution.</p>
<p>I feel like there should be a simple function in Mathematica for partitioning X things into Y groups randomly. I have searched and haven't found anything to do this. Please help!</p>
| Bob Hanlon | 9,362 | <pre><code>randChoice[
individuals_Integer?NonNegative,
groups_Integer?Positive] :=
RandomChoice[Range[groups], individuals];
choices = randChoice[50, 100]
(* {58, 83, 34, 28, 25, 97, 6, 73, 28, 91, 6, 42, 93, 48, 56, 64, 20, \
88, 73, 11, 79, 65, 34, 3, 16, 18, 4, 18, 53, 30, 20, 97, 79, 30, 91, \
35, 35, 49, 31, 29, 80, 91, 87, 2, 23, 94, 27, 29, 87, 36} *)
Tally[choices] // Sort
(* {{2, 1}, {3, 1}, {4, 1}, {6, 2}, {11, 1}, {16, 1}, {18, 2}, {20,
2}, {23, 1}, {25, 1}, {27, 1}, {28, 2}, {29, 2}, {30, 2}, {31,
1}, {34, 2}, {35, 2}, {36, 1}, {42, 1}, {48, 1}, {49, 1}, {53,
1}, {56, 1}, {58, 1}, {64, 1}, {65, 1}, {73, 2}, {79, 2}, {80,
1}, {83, 1}, {87, 2}, {88, 1}, {91, 3}, {93, 1}, {94, 1}, {97, 2}} *)
</code></pre>
|
3,306,571 | <p>I know that the function <span class="math-container">$f(x) = \frac{x}{x}$</span> is not differentiable at <span class="math-container">$x = 0$</span>, but according to the definition of differentiable functions:</p>
<blockquote>
<p>A differentiable function of one real variable is a function whose derivative exists at each point in its domain</p>
</blockquote>
<p>since <span class="math-container">$x = 0$</span> is not in the domain of <span class="math-container">$f$</span>, it doesn't have to be differentiable at that point for the function to be differentiable. This suggests that <span class="math-container">$f$</span> is differentiable as every other points in the domain has a derivative of <span class="math-container">$0$</span>.</p>
<p>However, some say that a function must be continuous if it's differentiable. This disproves the fact that <span class="math-container">$f$</span> is differentiable since it's not a continuous function.</p>
<p>Then is it really a differentiable function?</p>
| fleablood | 280,126 | <blockquote>
<p>"I know that the function <span class="math-container">$f(x)=\frac xx$</span> is not differentiable at <span class="math-container">$x=0$</span>"</p>
</blockquote>
<p>The function isn't <em>ANYTHING</em> at <span class="math-container">$x=0$</span>. <span class="math-container">$0$</span> is not in the domain.</p>
<blockquote>
<p>since x=0 is not in the domain of f, it doesn't have to be differentiable at that point for the function to be differentiable. This suggests that f is differentiable as every other points in the domain has a derivative of 0.</p>
</blockquote>
<p>Absolutely correct.</p>
<blockquote>
<p>However, some say that a function must be continuous if it's differentiable. This disproves the fact that f is differentiable since it's not a continuous function.</p>
</blockquote>
<p>But <span class="math-container">$f$</span> <em>is</em> a continuous function. It's continuous at every point of its domain.... which doesn't include <span class="math-container">$0$</span>.</p>
|
336,196 | <p>Can anyone help me with the following SDE?</p>
<p>Solve the following stochastic differential equation:
$$dY_t=aY_tdt+(b(t)+cY_t)dB_t$$
with $Y_0=0$.</p>
<p>Hint: Try a solution of the form $Z_tH_t$ where $Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$ and $dH_t=F(t)dt+G(t)dB_t$ for some adapted process F and G which need to be determined.</p>
<p>Thanks gt6989b for your input!</p>
<p>Please correct me if I'm wrong anywhere.</p>
<p>Since, $$Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$$
Hence we get: $dZ_t = Z_t(cdB_t-\frac{1}{2}c^2)dt$</p>
<p>and we also have $dH_t=F(t)dt+G(t)dB_t$</p>
<p>Then we apply Ito's Lemma on $Z_tH_t$, which yields
\begin{eqnarray*}
d(Z_tH_t) &=& Z_tdH_t+H_tdZ_t+dH_tdZ_t \\
&=& Z_t(F(t)dt+G(t)dB_t)+Z_tH_t(cdB_t-\frac{1}{2}c^2dt)+Z_tcG(t)dt \\
&=& Z_t(F(t)-\frac{1}{2}c^2H_t+cG(t))dt+(Z_tG(t)+cZ_tH_t)dB_t
\end{eqnarray*}</p>
<p>By letting $Y_t = Z_tH_t$, we compare between the expressions $dY_t$ and $d(Z_tH_t)$ in the $dt$ and $dB_t$ terms respectively. And we get the following:</p>
<p>\begin{eqnarray}
Y_t &=& Z_tH_t \\
G(t) &=& \frac{Z_t}{b(t)} \\
F(t) &=& (a+\frac{1}{2}c^2)H_t - c\frac{Z_t}{b(t)}
\end{eqnarray}</p>
<p>Note: $F(t) = P$ and $G(t) = Q$ for your P and Q respectively.</p>
<p>But now, I don't really understand how would this result help me in solving the original $dY_t$ SDE. </p>
| gt6989b | 16,192 | <p><strong>Here is an idea to get you started.</strong></p>
<p>Note that
$$
dZ_t = cZ_t dB_t + \frac{c^2}{2}Z_t dt = Z_t \left(cdB_t + \frac{c^2}{2}dt\right)
$$</p>
<p>and let $dH_t = P dt + Q dB_t$.</p>
<p>Consider $A_t = Z_t H_t$. Then, by Ito's Lemma, and expansion for $dZ_t$,</p>
<p>$$\begin{split}
dA_t &= Z_t dH_t + H_t dZ_t + dZ_t dH_t \\
&= Z_t (P dt + Q dB_t) + H_t Z_t \left(cdB_t + \frac{c^2}{2}dt\right) + cQdt \\
&= dt\left[ ??? \right] + dB_t \left[ ??? \right]
\end{split}
$$</p>
<p>Now match this to $dY_t$, what can you say about $P,Q$?</p>
|
2,567,332 | <p>A Greek urn contains a red, blue, yellow, and orange ball. A ball is drawn from the urn at random and then replaced. If one does this $4$ times, what is the probability that all $4$ colors were selected?</p>
<p>I approached this questions by doing $(1/4)^4$ because there's always a $1/4$ chance of selected a specific color ball if it's replaced. I also tried doing if not the correct ball was selected; so I did $(3/4)^4$ but that didn't work either. What am I doing wrong?</p>
| Austin Weaver | 480,825 | <p>The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.</p>
<p>The first draw yields a new ball, guaranteed:
$$P(\text{ball 1 new})=1$$</p>
<p>For the second draw, there are $3$ possible new balls and $4$ total balls, so:
$$P(\text{ball 2 new})=\frac34$$</p>
<p>For the third, there are $2$ possible new balls and $4$ total balls, so:
$$P(\text{ball 3 new})=\frac24=\frac12$$</p>
<p>For the fourth, there is one new ball and there are $4$ total balls, so:
$$P(\text{ball 4 new}) = \frac14$$</p>
<hr>
<p>Thus, the answer is: $$\prod P = 1\cdot\frac34\cdot\frac12\cdot\frac14 = \frac3{32}=0.09375\text{ chance.}$$</p>
|
690,331 | <p>Does it make a different when you parametrize a counterclockwise full circle and a clockwise circle in the complex plane? </p>
<p>For example, I am looking at computing an integral $\int_\gamma {1\over{z+4}}dz$ where $\gamma$ is the circle of radius $1$, centered at $-4$, oriented <strong>counterclockwise.</strong></p>
<p>My parametrization look like this: $\gamma(t)=p+Re^{it}=-4+e^{it}, 0\leq t\leq 2\pi$. Would the parametrization look the same as well if the circle oriented clockwise? </p>
<p>I have the final answer for integral as $2\pi i$, which makes sense, would it be the same regardless? </p>
| Bill Cook | 16,423 | <p>No. This doesn't work. You have to check that $E_i \cap \left(E_1+E_2+\cdots+E_{i-1}+E_{i+1}+\cdots+E_p\right) = \{0\}$ for each $i$. </p>
<p>Just checking that the subspaces don't pairwise intersect is not enough.</p>
<p>Consider the example: $U = \{ (0,y) \;|\; y \in \mathbb{R} \}$, $V =\{ (x,0)\;|\; x \in \mathbb{R} \}$, and $W = \{ (x,x) \;|\; x \in \mathbb{R} \}$. </p>
<p>Each pair trivially intersect: $U \cap V = U \cap W = V \cap W = \{ (0,0) \}$. But this definitely is not a direct sum: $(-1,0)+(0,-1)+(1,1)=(0,0)$.</p>
<p>[Notice that $W \cap (U+V) = W \cap \mathbb{R}^2 = W \not= \{(0,0)\}$.] </p>
|
1,672,131 | <p>A card game is played with a deck whose cards can be one of 6 suits, one of the suits being hearts, and one of 11 ranks. A hand is a subset of 3 cards. What is the probability that a hand has exactly two hearts given that it has the 2 of hearts? Please explain.</p>
| Hagen von Eitzen | 39,174 | <p><strong>Hint:</strong> $10^n$ has $n+1$ digits.</p>
|
2,619,185 | <p>Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural. </p>
<p>I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$. </p>
<p>Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.</p>
| Yash Jain | 522,158 | <p>Let us restate the value of $P$ first.</p>
<p>$P = (X+2)^m+(X+2)^{2m+3}+1^{2m+3} = (X+2)^m+(X+2)^{2m+3}+1 =$</p>
<p>$(X+2)^m+(X+2)^{2m+3}+(X+2)^0 = (X+2)^{3m+3} = (X+2)^{(m+1)^{3}}$</p>
<p>And, Let us restate the value of $Q$.</p>
<p>$Q = X^2+4X+4+X+3 = X^2+4X+4+X+2+1 = (X+2)^2+(X+2)^1+(X+2)^0$</p>
<p>Do you notice the relationship between $P$ and $Q$?</p>
|
4,265,001 | <p>I'm a bit confused about expanding out the notation of product of matrices, in the context of quadratic forms.</p>
<p>If <span class="math-container">$x \in \mathbb{R}^n, \, \, A \in \mathbb{R}^{n \times n}$</span> then</p>
<p><span class="math-container">$$x^TAx = \sum_{i,j=1}^na_{ij}x_ix_j$$</span></p>
<p>But then if I consider a <strong>matrix</strong> <span class="math-container">$X \in \mathbb{R}^{n \times n}$</span> how should I write the expanded form of</p>
<p><span class="math-container">$$X^TAX = \, \,...\, \, ?$$</span></p>
<p>This time the result will be a <strong>matrix</strong>.. will it be something like</p>
<p><span class="math-container">$$X^TAX = \sum_{i,j=1}^n a_{ij}x_ix_j^T$$</span></p>
<p>and if yes, why?</p>
<p>Sorry if this is pretty straightforward but it always happens to get a little bit stuck with matrix notation.</p>
<p>Many thanks,</p>
<p>James</p>
| Mas | 973,301 | <p>The <span class="math-container">$(i,j)$</span>-coefficient of the matrix <span class="math-container">$X^TAX$</span> is given by
<span class="math-container">\begin{equation}
\sum_{k,l=1}^nx_{ki}a_{kl}x_{lj},
\end{equation}</span>
where <span class="math-container">$x_{ij}$</span> (resp. <span class="math-container">$a_{ij}$</span>) is the coefficient of <span class="math-container">$X$</span> (resp. <span class="math-container">$A$</span>).</p>
|
2,060,156 | <p>First thing I want to mention is that this is not a topic about why $1+2+3+... = -1/12$ but rather the connection between this summation and $\zeta$.</p>
<p>I perfectly understand that the definition using the summation $\sum_{k=1}^\infty k^{-s}$ of the zeta function is only valid for $Re(s) > 1$ and that the function is then extrapolated through analytic continuation in the whole complex plan.</p>
<p>However some details bother me : Why can we manipulate the sum and still obtain correct final answer.
$$
S_1 = 1-1+1-1+1-1+... = 1-(1-1+1-1+1-...)= 1-S_1 \implies S_1 = \frac{1}{2} \\
S_2 = 1-2+3-4+5-... \implies S_2 - S_1 = 0-1+2-3+4-5... = -S_2 \implies S_2 = \frac{1}{4} \\
S = 1+2+3+4+5+... \implies S-S_2 = 4(1+2+3+4+...) = 4S \implies S = -\frac{1}{12} \\
S "=" \zeta(-1)
$$
Clearly these manipulations are not legal since we're dealing with infinite non-converging sums. But it works ! Why ?
Is there a real connection between the analytic continuation which yields the "true" value $\zeta(-1) = -1/12$ and these "forbidden manipulations" ? Could we somehow consider these manipulations as "continuation of non-converging sums" ? If so, is there a well-defined framework with defined rules because it is clear that we must be careful when playing with non-converging sums if we don't want to break the mathematics ! (For example <a href="https://en.wikipedia.org/wiki/Riemann_series_theorem" rel="nofollow noreferrer"> Riemann rearrangement theorem</a>)</p>
<p>And since it seems that these illegal operations can be used to compute some value of zeta in the extended domain $Re(s) < 1$, are there other examples of such derivations, for example $0 = \zeta(-2) "=" 1^2 + 2^2 + 3^2 + 4^2 + ...$ ?</p>
<p>Hopefully this is not an umpteenth vague question about zeta and $1+2+3+4...$ I did some research about it but couldn't find any satisfying answer. Thanks !</p>
| snulty | 128,967 | <p>$$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}$$</p>
<p>$$28kl-20l-21k+15=30kl-18l-5k+3$$</p>
<p>$$2kl+2l+16k-12=0$$</p>
<p>$$kl+l+8k-6=0$$</p>
<p>either:</p>
<p>$$l(1+k)=2(3-4k)$$
so
$$l=2\frac{3-4k}{1+k}$$</p>
<p>or:</p>
<p>$$k(l+8)=6-l$$</p>
<p>so</p>
<p>$$k=\frac{6-l}{l+8}$$</p>
<p>Let's go with this second one to complement the other answer.</p>
<p>Then $$k=-\frac{l-6}{l+8}=-\left(\frac{l+8-14}{l+8}\right)=-\left(1-\frac{14}{l+8}\right)$$</p>
<p>We want $l+8=\pm(1,2,7,14)$</p>
<p>So $l=-7,-9,-6,-10,-1,-15,6,-22$ or more nicely ordered $$l=-22,-15,-10,-9,-7,-6,-1,6$$</p>
<p>The pairs are then $(l,k)=$ $(-22,-2)$, $(-15,-3)$, $(-10,-8)$, $(-9,-15)$, $(-7,13)$, $(-6,6)$, $(-1,1)$, $(-22,-2)$, $(6,0)$,
or that the fractions are</p>
<p>$$\frac{19}{13},\frac{13}{9},\frac{61}{43},\frac{55}{39},\frac{43}{31},\frac{37}{27},1,\frac{5}{3}.$$</p>
|
64,544 | <blockquote>
<p>Please let me know what is the standard notation for group action.</p>
</blockquote>
<p>I saw the following three notations for group action.
(All the images obtained as <code>G\acts X</code> for different deinitions of <code>\acts</code>.) </p>
<p>(1) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6Q736oVI/AAAAAAAAACU/7li7VA1-FTc/s144/B.png" alt="alt text"> </p>
<p>I saw this one most, but only in handwriting and I like it. But I did not find a better way to write it in LaTeX.</p>
<pre><code>\usepackage{mathabx,epsfig}
\def\acts{\mathrel{\reflectbox{$\righttoleftarrow$}}}
</code></pre>
<p>(2) <img src="https://lh5.googleusercontent.com/_7jyZyirE1is/TcM6XymimxI/AAAAAAAAACY/byg0FDFv7wM/s144/C.png" alt="alt text"> </p>
<p>It is almost as good as 1, but in handwriting this arrow can be taken as $G$.</p>
<pre><code>\usepackage{mathabx}
\def\acts{\lefttorightarrow}
</code></pre>
<p>(3) <img src="https://lh6.googleusercontent.com/_7jyZyirE1is/TcM6JeflvjI/AAAAAAAAACQ/n_dfYCfeQEc/s800/A.png" alt="alt text"> </p>
<p>I saw this one in print, I guess it is used since there is no better symbol in "amssymb".</p>
<pre><code>\usepackage{amssymb}
\def\acts{\curvearrowright}
</code></pre>
| Stefan Waldmann | 12,482 | <p>IN addition to what has been said, I would like to mention the following aspect: a group action usually does not come alone. When $G$ acts on a set $X$ you will almost for sure be interested also in the induced action of $G$ on certain kind of functions on $X$: think of a smooth manifold $M$ and of smooth functions, or of smooth tensor fields on $M$, or what not. Then the induced group action can be either viewed as a <em>right</em> action when $G$ acts from the left on $X$ or you have to plug in a $^{-1}$ to turn things again into a left action. In various situations (say equivariant maps in the dual of a Lie algebra etc) it might become a notational desaster if you want to specify all kind of actions with a separate symbol. However, it might be important to keep track of whether you have a left or a right action.</p>
<p>So my habit is to denote the action of a group element $g$ on some object $x \in X$ by $g \triangleright x$ if it is a left action and by $g \triangleleft x$ or better $x \triangleleft g$ if it is a right action. Then you don't have to bother whether you have already included the $^{-1}$ in the coadjoint action to make it a left action or just stay with a right action :)</p>
|
3,115,168 | <p>I've converted <span class="math-container">$\cos^3(x)$</span> into <span class="math-container">$\cos^2(x)\cos(x)$</span> but still have not gotten the answer. </p>
<p>The answer is <span class="math-container">$\dfrac{\sin(x)(3\cos^2x + 2\sin^2x)}{3}$</span></p>
<p>My answer was the same except I did not have a <span class="math-container">$3$</span> infront of <span class="math-container">$x$</span> and my <span class="math-container">$2\sin^2x$</span> was not squared.</p>
<p>Help! </p>
| José Carlos Santos | 446,262 | <p>Since <span class="math-container">$\cos^3x=(1-\sin^2 x)\cos x$</span>, you can do <span class="math-container">$\sin x=t$</span> and <span class="math-container">$\cos x\,\mathrm dx=\mathrm dt$</span>, thereby getting<span class="math-container">$$\int1-t^2\,\mathrm dt.$$</span></p>
|
3,115,168 | <p>I've converted <span class="math-container">$\cos^3(x)$</span> into <span class="math-container">$\cos^2(x)\cos(x)$</span> but still have not gotten the answer. </p>
<p>The answer is <span class="math-container">$\dfrac{\sin(x)(3\cos^2x + 2\sin^2x)}{3}$</span></p>
<p>My answer was the same except I did not have a <span class="math-container">$3$</span> infront of <span class="math-container">$x$</span> and my <span class="math-container">$2\sin^2x$</span> was not squared.</p>
<p>Help! </p>
| Sri-Amirthan Theivendran | 302,692 | <p>Note that
<span class="math-container">$$
\int\cos^3x\, dx=\int\cos^2x\, d(\sin x)=\cos^2x\sin x+\int2\sin^2 x\cos x\, dx
$$</span>
To compute the last integral make the substitution <span class="math-container">$u=\sin x$</span>.</p>
|
777,186 | <p>My equation is the following, and I would like to find which $k$ can make it a circle.</p>
<p>$$x^2+y^2+4x-6y+k=0$$</p>
<p>My naive approach is to have $k$ to be $-4x+6y+c$ where c is any number, so that I can have any circle that is in 0. However k is a parameter and I can't really figure that out if I am missing something. Any advice?</p>
| Fly by Night | 38,495 | <p><strong>HINT</strong></p>
<p>Complete the square on $x^2+4x$ and $y^2-6y$ separately.</p>
<p>Take all of the numbers, and the $k$, over to the right hand side.</p>
<p>You will have something like $(x-a)^2 + (y-b)^2 = t$, where $a$ and $b$ are numbers and $t$ is a mixture of numbers and the letter $k$.</p>
<p>You need $t > 0$ for a circle. </p>
<p>What range of values of $k$ ensures that $t>0$?</p>
|
1,594,722 | <p>The ODE is</p>
<p>($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$</p>
<p>I have tried everything like integrating factor,it is not homogenous and not linear differential equation..What should be done now?</p>
| SchrodingersCat | 278,967 | <p>Re-arranging your differential equation, we have
$$\frac{dx}{dy}=xy^3+x^2y^7$$
$$\frac{dx}{dy}-xy^3=x^2y^7$$
$$\frac{1}{x^2}\cdot \frac{dx}{dy}-\frac{1}{x}\cdot y^3=y^7$$
$$-\frac{d}{dy}\left(\frac{1}{x}\right)-y^3\cdot \left(\frac{1}{x}\right) =y^7$$</p>
<p>Put $u=\frac{1}{x}$</p>
<p>You get $$\frac{du}{dy}+uy^3=y^7$$</p>
<p>The integrating factor comes out to be $e^\frac{y^4}{4}$.</p>
<p>Then we have $$\frac{d}{dy}\left(ue^\frac{y^4}{4}\right)=y^7e^\frac{y^4}{4}$$</p>
<p>Now $$\int y^7e^\frac{y^4}{4} dy = \int y^4 e^\frac{y^4}{4} \cdot y^3 dy$$
$$=\int 4ze^z dz$$
where $z=\frac{y^4}{4}$.</p>
<p>Can you complete the integration now?</p>
|
1,969,169 | <p>We have to do the following integral.
$$\int_1^{\frac{1+\sqrt{5}}{2}}\frac{x^2+1}{x^4-x^2+1}\ln\left(1+x-\frac{1}{x}\right)dx$$
I tried it a lot.
I substitute $t=1+x-(1/x)$, $dt=1+(1/x^2)$</p>
<p>But then I stuck at
$$\int\limits_{1}^{2} \frac{\ln(t)}{(t-1)^{2} + 1} \mathrm{d}t$$</p>
<p>But now how to proceed.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mc{J} & = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\dd x
\\[5mm] & =
\int_{1}^{\pars{1 + \root{5}}/2}{1 \over x^{2} - 1 + 1/x^{2}}\,
\ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x
\\[5mm] & =
\int_{1}^{\pars{1 + \root{5}}/2}{1 \over \pars{x - 1/x}^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x
\\[5mm] \stackrel{t\ =\ x - 1/x}{=} &\
\int_{0}^{1}{\ln\pars{1 + t} \over t^{2} + 1}\,\dd t
\,\,\,\stackrel{t\ =\ \tan\pars{\theta}}{=}\,\,\,\
\underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta}_{\ds{=\ \mc{J}}}\ =\
\int_{-\pi/4}^{0}\ln\pars{1 + {\tan\pars{\theta} + 1 \over 1 - \tan\pars{\theta}}}\,\dd\theta
\\[5mm] & =
{1 \over 4}\,\pi\ln\pars{2} -
\int_{-\pi/4}^{0}\ln\pars{1 - \tan\pars{\theta}}\,\dd\theta =
\color{#f00}{{1 \over 4}\,\pi\ln\pars{2}} -
\underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta}
_{\ds{=\ \mc{J}}}
\end{align}
<hr>
$$
\mc{J} = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\dd x =
{\mc{J} + \mc{J} \over 2} = {\pi\ln\pars{2}/4 \over 2} =
\bbx{\ds{{1 \over 8}\,\pi\ln\pars{2}}}
$$</p>
|
3,981,809 | <p>Imagine that we have two pairs of integers <span class="math-container">$(a_1,b_1)$</span> and <span class="math-container">$(a_2, b_2)$</span> where</p>
<p><span class="math-container">$$ a_1b_1\equiv 0,\,\ a_2b_2\equiv 0,\,\ a_1b_2+a_2b_1\equiv 0\pmod n$$</span></p>
<p>Does this imply that
<span class="math-container">$$ a_1 b_2 \equiv 0\pmod n $$</span></p>
<p>I assume that <span class="math-container">$a_1 b_1 \equiv 0\pmod n$</span> is only true if <span class="math-container">$a_1$</span> and <span class="math-container">$b_1$</span> are divisors of <span class="math-container">$n$</span>. Using that, I've checked <span class="math-container">$a_1b_2 = 0$</span> mod <span class="math-container">$n$</span> numerically for a large number of values of <span class="math-container">$n$</span> and it seems to hold, but I am not sure how to prove it.</p>
| Bill Dubuque | 242 | <p>Hypotheses <span class="math-container">$\Rightarrow \dfrac{a_1b_2}n$</span> & <span class="math-container">$\dfrac{a_2b_1}n\,$</span> have sum & product <span class="math-container">$\in\Bbb Z\,$</span> <a href="https://math.stackexchange.com/a/3900209/242">thus</a> both are <span class="math-container">$\in\Bbb Z,\,$</span> <a href="https://math.stackexchange.com/a/3900209/242.">by Rational Root Test</a></p>
|
3,470,208 | <p><span class="math-container">$$f(x)=\begin{cases}
\dfrac{x}{\sin x}, & x>0\\
2-x, & x\le0
\end{cases}$$</span></p>
<p><span class="math-container">$$g(x)=\begin{cases}
x+3, &x<1\\
x^2-2x-2, &1\le x<2\\
x-5, & x\ge2
\end{cases}$$</span></p>
<p>Find left hand limit and right hand limit of <span class="math-container">$g(f(x))$</span> at <span class="math-container">$x = 0$</span> and hence find
<span class="math-container">$\lim_{x\to0}g(f(x))$</span></p>
<p>My attempt is as follows:-</p>
<p><span class="math-container">$$g(f(x))=\begin{cases}
5-x, &x\le0\\
\dfrac{x}{\sin x}+3 &0<x<1\\
\left(\dfrac{x}{\sin x}\right)^2-\dfrac{2x}{\sin x}-2, & 1\le x<2\\
\dfrac{x}{\sin x}-5 &x\ge2
\end{cases}$$</span></p>
<p>Let's find left hand limit</p>
<p><span class="math-container">$$l=\lim_{x\to0^{-}}g(f(x))$$</span>
<span class="math-container">$$l=\lim_{x\to0^{-}}5-x$$</span>
<span class="math-container">$$l=5$$</span></p>
<p>Let's find right hand limit</p>
<p><span class="math-container">$$r=\lim_{x\to0^{+}}g(f(x))$$</span>
<span class="math-container">$$r=\lim_{x\to0^{+}}\dfrac{x}{\sin x}+3$$</span>
<span class="math-container">$$r=4$$</span></p>
<p><span class="math-container">$$l\ne r$$</span>
<span class="math-container">$$\lim_{x\to0}g(f(x)) \text { doesn't exist }$$</span> </p>
<p>But actual answer is following:</p>
<p><span class="math-container">$$l=-3,r=-3,\lim_{x\to0}g(f(x))=-3$$</span></p>
<p>What mistake am I making here? I tried to find it but didn't get any breakthrough.</p>
| user | 505,767 | <p>We have that since <span class="math-container">$x>\sin x$</span> for <span class="math-container">$x\neq 0$</span></p>
<ul>
<li>as <span class="math-container">$x\to0^+, y=f(x)\to1^+ \implies \lim_{x\to0^+}
g(f(x))=\lim_{y\to1^+} g(y)=-3$</span></li>
</ul>
<p>and</p>
<ul>
<li>as <span class="math-container">$x\to0^-, y=f(x)\to2^+\implies \lim_{x\to0^-}
g(f(x))=\lim_{y\to2^+} g(y)=-3$</span></li>
</ul>
<p>therefore the limit exists.</p>
|
149,558 | <p>I always use <code>InputForm</code> to check the result object,such as <code>Dataset</code> or <code>Graphics</code> or other objects.But if you are in the result of <code>InputForm</code>,you cannot use the Front-End function of balance the bracket. Note this gif</p>
<p><a href="https://i.stack.imgur.com/51OYd.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/51OYd.gif" alt="enter image description here" /></a></p>
<p>When I double click the input line, I will select all content just in this bracket.But when I'm in result of <code>InputForm</code>,I will select all line. Of course I can copy the output of <code>InputForm</code> as another new input,but which will make the notebook more mess.</p>
<p>Any method can make the output of <code>InputForm</code> support the function of balance the bracket?</p>
| Carl Woll | 45,431 | <p>Per my comment, I like using <code>expr //InputForm //SequenceForm</code>, but another similar possibility is to use a custom head with a custom <a href="http://reference.wolfram.com/language/ref/MakeBoxes" rel="noreferrer"><code>MakeBoxes</code></a> rule. For instance, let's call the custom form <code>myInputForm</code>. Then, we can define:</p>
<pre><code>myInputForm /: MakeBoxes[myInputForm[expr_], StandardForm] := MakeBoxes[InputForm[expr]]
</code></pre>
<p>We also need to add <code>myInputForm</code> to $OutputForms:</p>
<pre><code>Unprotect[$OutputForms];
AppendTo[$OutputForms, myInputForm];
Protect[$OutputForms];
</code></pre>
<p>Now, <code>myInputForm</code> gets stripped before being stored in <a href="http://reference.wolfram.com/language/ref/Out" rel="noreferrer"><code>Out</code></a>. For instance:</p>
<pre><code>Interpolation[{1,2,3,4}]
% //myInputForm
% //Head
</code></pre>
<blockquote>
<p>InterpolatingFunction[{{1, 4}}, <>]</p>
<p>InterpolatingFunction[{{1, 4}}, {
5, 3, 0, {4}, {4}, 0, 0, 0, 0, Automatic, {}, {}, False}, {{1, 2, 3, 4}}, {{
1}, {2}, {3}, {4}}, {Automatic}]</p>
<p>InterpolatingFunction</p>
</blockquote>
|
2,755,143 | <p>Find Number of integers satisfying $$\left[\frac{x}{100}\left[\frac{x}{100}\right]\right]=5$$ where $[.]$ is Floor function.</p>
<p>I assumed $$x=100q+r$$ where $0 \le r \le 99$</p>
<p>Then we have </p>
<p>$$\left[\left(q+\frac{r}{100}\right)q\right]=5$$ $\implies$</p>
<p>$$q^2+\left[\frac{rq}{100}\right]=5$$</p>
<p>Since $rq$ is an integer we have $$rq=100p+r_1$$ where $0 \le r_1 \le 99$</p>
<p>Then we have</p>
<p>$$q^2+p+\left[\frac{r_1}{100}\right]=5$$ $\implies$</p>
<p>$$q^2+p=5$$ so the possible ordered pairs $(p,q)$ are</p>
<p>$(1,2)$, $(1,-2)$, $(-4, 3)$ i am getting infinite pairs.</p>
<p>How to proceed?</p>
| xpaul | 66,420 | <p>Using
$$ [x]\le x<[x]+1 $$
one has
$$ 5\le\frac{x}{100}\left[\frac{x}{100}\right]<6. \tag{1}$$
Let $x=100q+r$, where $q$ and $r$ are an integer and $r\in[0,99]$. Then (1) becomes
$$ 500\le (100q+r)q<600. \tag{2}$$
Noting that
$$ 100q^2 \le (100q+r)q<600$$
implies $q^2<6$, one has $q=\pm1,\pm2$. If $q=\pm1$, it is easy to see that (2) can't hold.
So $q=\pm2$. If $q=-2$, (2) becomes
$$500\le-2(-200+ r)<600 $$
which gives
$$ 50\le -2r<100$$
which is impossible since $r\ge0$.
If $q=2$, (2) becomes
$$500\le2(200+ r)<600 $$
which gives
$$ 50\le r<100.$$
Thus $r=50,51,\cdots99$ and hence
$$ x=250,251,\cdots,299. $$</p>
|
3,519,515 | <p>Here, I wonder what is a good way to use the epsilon delta definition or converging sequences to show that the set S containing quotients on [0,1] have/does not have volume 0, (i.e. whether there exist a <strong>finite</strong> number of intervals which union contain all of S such that the <strong>sum</strong> of length of all intervals is less than any <span class="math-container">$\epsilon > 0$</span> you fix). I believe it more likely does not have volume 0 from my intuition . I am lost on where to start the proof. </p>
<p>Does the idea of closure of S play a part in this proof? and how?</p>
<p>Also, is it possible to prove this using pigeonhole principle involving infinite rationals in one interval?</p>
| AZM | 440,612 | <p>In order to provide an effective answer, just think of the volume of the complementary set, which is the whole interval but a countable union of zero-measure sets, hence the complementary set has measure one (i.e., the set you are looking for has measure zero).</p>
|
237,031 | <p>The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the strongest large cardinal not known to be inconsistent with choice, as I understand)? This is implicit in the ordering of things on <a href="http://cantorsattic.info/Upper_attic" rel="noreferrer">Cantor's Attic</a>, for example, but I've been unable to find a proof (granted, I don't necessarily have the best nose for where to look!).</p>
<p>One thing that worries me is that when there <em>is</em> a ZFC analog of a ZF statement, many equivalent formulations of the ZFC statement may become inequivalent in ZFC. So we don't have much assurance that the usual definition of a Reinhardt cardinal is "correct" in the absence of choice.</p>
<p>I think it should be clear that Con(ZF + Reinhardt) implies Con(ZF + I0). But again, it's not clear that ZF+I0 is equiconsistent with ZFC+I0.</p>
<p>It's apparently not possible to formulate Reinhardt cardinals in a first-order way, so I should really talk about NBG + Reinhardt, or maybe ZF($j$) + Reinhardt, where ZF($j$) has separation and replacement for formulas involving the function symbol $j$.</p>
<p><strong>EDIT</strong></p>
<p>Since this question has attracted a bounty from Joseph Van Name, maybe it's appropriate to update it a bit. Now, I'm not actually a set theorist, but it's not even clear to me that Con(ZF + Reinhardt) implies Con(ZFC + an inaccessible). So perhaps the question should really be: what large cardinal strength, if any, can we extract from the theory ZF + Reinhardt?</p>
| Mohammad Golshani | 11,115 | <p>The answer to your question is (almost) yes (almost is because of the addition of DC to the statement).</p>
<p>Recently Gabriel Goldberg has proved </p>
<blockquote>
<p>''Con(NBG+DC+Reinhardt)<span class="math-container">$ \implies$</span> Con(ZFC+I0)''. </p>
</blockquote>
<p>See the abstract of the talk by Gabriel Goldberg <a href="http://www.crm.cat/en/Activities/Curs_2016-2017/Documents/abstract-Goldberg.pdf" rel="noreferrer">Choiceless cardinals and I0</a>.</p>
<p>(Thanks to <a href="https://mathoverflow.net/users/38866/rahman-m">Rahman</a> for pointing this to me).</p>
<hr>
<p><strong>Edit.</strong> The result of Goldberg is now available,where indeed something stronger is proved. See <a href="https://arxiv.org/abs/2006.01084" rel="noreferrer">Even ordinals and the Kunen inconsistency</a>. It is shown, assuming DC, the existence of an elementary embedding from <span class="math-container">$V_{λ+3}$</span> to <span class="math-container">$V_{λ+3}$</span> implies the consistency of ZFC + <span class="math-container">$I_0$</span>, while by a recent result of Schlutzenberg, an elementary embedding from <span class="math-container">$V_{λ+2}$</span> to <span class="math-container">$V_{λ+2}$</span> does not suffice.
The paper of Schlutzenberg is
<a href="https://arxiv.org/abs/2006.01077" rel="noreferrer">On the consistency of ZF with an elementary embedding <span class="math-container">$j : V_{λ+2} → V_{λ+2}$</span></a>.</p>
|
533,855 | <p>I need to show that $\{x_{n}\}$ is Cauchy given that there exists $0<C<1$ s.t. $|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}|$. Intuitively, that statement clearly implies $\{x_{n}\}$ is Cauchy, since it implies the sequence terms become arbitrarily close. But how to make it precise? </p>
<p>Couldn't it also be said from the given information that the sequence is either monotone increasing or monotone decreasing and bounded? Then it would converge, which means it is Cauchy. </p>
<p>Thanks for any assistance! </p>
| André Nicolas | 6,312 | <p>We change the question, to what is more likely to appear first, TT or HT. If We get TT at the beginning (probability $\frac{1}{4}$, then TT wins. In all other cases, HT wins.</p>
<p>(For the original question, TT and TH are equally likely to be first.) </p>
|
2,316,159 | <p>I'm interested in the differences in the groups but also in the Lie algebra associated. I know that two groups can have the same lie algebra if they differ from discrete elements, for instance: $SO(n)$ and $O(n)$ should have the same algebra. But then if I have a group $O(2,2)$, what is the associated Lie algebra? Does $O(2)\times O(2)$ have the same associated Lie algebra that $SO(2)\times SO(2)$ does?</p>
| Travis Willse | 155,629 | <p>No, the groups $O(2, 2)$ and $O(2) \times O(2)$ (and the corresponding algebras) are different---in fact, they have different dimensions as Lie groups.</p>
<p>The group $O(2, 2)$ is the group preserving an inner product of signature $(2, 2)$ on a $4$-dimensional real vector space $\Bbb V$. Concretely, if $g$ is the usual (definite-signature) inner product on $\Bbb R^2$, then we may identify $O(2, 2)$ is the group preserving $g \oplus -g$ on $\Bbb R^2 \oplus \Bbb R^2$. (The matrix representation of this bilinear form w.r.t. the standard basis is the matrix given in Tsemo Aristide's excellent answer. Thus, the explicit matrix representations of $O(2, 2)$ and $\mathfrak{so}(2, 2)$ for this choice are the ones given there.) It has dimension $\dim O(2, 2) = \frac{1}{2} (4) (4 - 1) = 6$.</p>
<p>The Lie algebra $\mathfrak{so}(2, 2)$ of $O(2, 2)$ is actually decomposable: $\mathfrak{so}(2, 2) \cong \mathfrak{sl}(2, \Bbb R) \oplus \mathfrak{sl}(2, \Bbb R)$. Better yet, this lifts to an isomorphism $SO_0(2, 2) \cong SL(2, \Bbb R) \times SL(2, \Bbb R)$ (see <a href="https://physics.stackexchange.com/a/143561">this Physics SE answer</a>); here, $SO_0(2, 2)$ just denotes the identity component of $O(2, 2)$.</p>
<p>On the other hand, we may identify $O(2)$ with the group preserving the inner product $g$ itself, and it has dimension $1$, so it has abelian Lie algebra $\mathfrak{so}(2, \Bbb R) \cong \Bbb R$. Thus, $O(2) \times O(2)$ has dimension $2$ and abelian Lie algebra $\mathfrak{so}(2, \Bbb R) \oplus \mathfrak{so}(2, \Bbb R) \cong \Bbb R^2$.</p>
<p>We <em>can</em> view $O(2) \times O(2)$ naturally, however, as a subgroup of $O(2, 2)$ in terms of our above realization: It is exactly the subgroup that preserves the vector space decomposition $\Bbb R^2 \oplus \Bbb R^2$.</p>
<p>Alternatively, the inclusion $O(2) \hookrightarrow SL(2, \Bbb R)$ gives a different way to view $O(2) \times O(2)$ as a subgroup of $O(2, 2)$: $$O(2) \times O(2) \hookrightarrow SL(2, \Bbb R) \times SL(2, \Bbb R) \cong SO_0(2, 2) \leq O(2, 2) .$$</p>
|
470,739 | <p>Assume $S$ and $T$ are diagonalizable maps on $\mathbb{R}^n$ such that $S\circ T$=$T \circ S$. Then $S$ and $T$ have a common eigenvector.</p>
<p>I already have proof, but I just need validation in one part.
My proof:
Let $F$ be an eigenvector of $T$. This means $\exists \; \lambda \in R$ such that $T(v)=\lambda v$. Then, using the fact that $S\circ T$=$T \circ S$, we have</p>
<p>$$ S(T(v)) = (S\circ T)(v)=(T \circ S)(v)=T(S(v)) \Longrightarrow T(S(v))=\lambda S(v)$$</p>
<p>Thus, $S(v)$ is also an eigenvector of $T$. So, $S$ maps eigenvectors of $T$ to eigenvevtors of $T$. Thus, $S$ must have an eigenvector of $T$.</p>
<p>How would one rigorously prove that if $S$ maps eigenvectors of $T$ to eigenvectors of $T$, then $S$ also has an eigenvector of $T$?</p>
<p>Thanks.</p>
| Community | -1 | <p>Since $T$ is diagonalizable then its minimal polynomial factors into distinct linear factors and the restriction of $T$ to any invariant subspace annihilates the minimal polynomial so this restriction is also diagonalizable since its minimal polynomial divides the minimal polynomial of $T$ and then also it factors into distinct linear factors.</p>
<p>We take $v$ in the eigenspace $E_\lambda(T)$ and we prove (as you did) that
$$T(S(v))=\lambda S(v)$$
so $S(v)\in E_\lambda(T)$ and this means that $E_\lambda(T)$ is invariant by $S$ hence the restriction of $S$ to $E_\lambda(T)$ is an endomorphism and has an eigenvector which's a common eigenvector of $S$ and $T$.</p>
|
2,684,805 | <p>This question is asked by my 12 yr old cousin and I seem to be failing to give him a convincing explanation. Here is the summary of our discussion so far - </p>
<p>Case1 : $a>0, b>0$<br>
<a href="https://i.stack.imgur.com/fuoZS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fuoZS.png" alt="enter image description here"></a></p>
<p>I asked him to put another block of length $a$ adjacent to $b$ and stare at the symmetry. He quickly told me that the mid point of $a, b$ equals half the length of $a+b$ :
<a href="https://i.stack.imgur.com/ShBQb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ShBQb.png" alt="enter image description here"></a></p>
<p>So far we're good. But when either one of $a,b$ is negative, I feel stuck. I fail to give him a similar explanation using symmetry. Greatly appreciate any help. Thanks! </p>
| Christian Blatter | 1,303 | <p>Let ${\mathbb Z}_3$ be the set of colors. There are six ways of assigning $\pm1$ to the edges of the square such that the sum is $=0$ mod $3$, namely four of cyclic type $(1,1,-1,-1)$ and two of type $(1,-1,1,-1)$. Given such an assignment $s$ there are three colorings $f$ of the vertices such that (in the obvious interpretation) $\partial f= s$. It follows that there are $18$ admissible colorings in all. </p>
|
3,775,749 | <p>So far I know that it’s possible to draw angles which are multiples of <strong>15°</strong> (ex. <em><strong>15°</strong></em>, <em><strong>30°</strong></em>, <em><strong>45°</strong></em> etc.).</p>
<p>Could anybody please tell me if it's possible to draw other angles which are not multiples of 15° using only a compass and a ruler.</p>
| Ethan Bolker | 72,858 | <p>You can construct a regular <span class="math-container">$n$</span>-gon with straightedge and compass if and only if <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> times a product of <a href="https://en.wikipedia.org/wiki/Fermat_number" rel="nofollow noreferrer">Fermat primes</a> - primes of the form <span class="math-container">$2^{2^j} +1$</span>.</p>
<p>That tells you what fractional angles you can construct. For example, the <span class="math-container">$17$</span>-gon is constructible, so you can construct an angle of <span class="math-container">$360/17$</span> degrees.</p>
|
158,896 | <p>Being interested in the very foundations of mathematics, I'm trying to build a rigorous proof on my own that $a + b = b + a$ for all $\left[a, b\in\mathbb{R}\right] $. Inspired by interesting properties of the complex plane and some researches, I realized that defining multiplication as repeated addition will lead me nowhere (at least, I could not work with it). So, my ideas:</p>
<ul>
<li><p>Defining addition $a+b$ as a kind of "walking" to right $\left(b>0\right)$ or to the left $\left(b<0\right)$ a space $b$ from $a$. Adding a number $b$ to a number $a$ (denoted by $a+b$) involves doing the following operation:</p>
<ol>
<li>Consider the real line $\lambda$ and its origin at $0$. Mark a point $a$, draw another real line $\omega$ above $\lambda$ such what $\omega \parallel \lambda$ and mark a point $b$ on $\omega$. Now, draw a line $\sigma$ such that $\sigma \perp \omega$ and the only point in commom between $\sigma$ and $\omega$ is $b$. Consider the point that $\lambda$ and $\sigma$ have in commom; this point is nicely denoted as $a + b$.</li>
</ol></li>
</ul>
<p>(Note that all my work is based here. Any problems, and my proof goes to trash)</p>
<ul>
<li><p>This definition can be used to see the properties of adding two numbers $a$ and $b$, for all $a, b \in\mathbb{R}$.</p></li>
<li><p>Using geometric properties may lead us to a rigorous proof (if not, I would like to know the problems of using it).</p></li>
</ul>
<p>So, I started:</p>
<ul>
<li>$a, b \in\mathbb{N}$:</li>
</ul>
<p>$a+b = \overbrace{\left(1+1+1+\cdots+1\right)}^a + \overbrace{\left(1+1+1+\cdots+1\right)}^b = \overbrace{1+1+1+1+\cdots+1}^{a+b} = \overbrace{\left(1+1+1+\cdots+1\right)}^b + \overbrace{\left(1+1+1+\cdots+1\right)}^a = b + a$</p>
<p>(Implicity, I'm using the fact that $\left(1+1\right)+1 = 1+\left(1+1\right)$, which I do not know how to prove and interpret it as cutting a segment $c$ in two parts -- $a$ and $b$. However, this result can be extended to $\mathbb{Z}$ in the sense that $-a$ $(a > 0)$ is a change; from right to left).</p>
<ul>
<li>$a, b \in\mathbb{R}$:</li>
</ul>
<p>Here, we have basically two cases:</p>
<ul>
<li>$a$ and $b$ are either positive or negative;</li>
<li>$a$ and $b$, where one of them is negative.</li>
</ul>
<p>Since in my definition $-b, b>0$ means drawing a point $b$ to the left of the real line, there's no big deal interpretating it; <em>subtracting</em> can be interpreted now. So, it starts:</p>
<p>$a + b = c$. However, $c$ can be cut in two parts: $b$ and $a$. Naturally, if $a>c$, then $b<0$ -- many cases can be listed. So, $c = b + a$. But $c = a + b$; it follows that $a + b = b + a$. My questions:</p>
<p><em>Is there any problem in using my definition of adding two numbers $a$ and $b$, which uses many geometric properties? Is there any way to solve it from informality? Is there anything right here?</em></p>
<p>Thanks in advance.</p>
| mboratko | 17,349 | <p>Please don't let the comments discourage you - it is good that you are trying to understand the fundamentals of Mathematics, and your effort is to be applauded.</p>
<p>In order to prove a theorem, one must have in place a set of axioms and definitions (as well as acceptable rules of logic) from which the theorem can be deduced. It is quite helpful to see some specific examples of how this is achieved before striking out on your own. In addition (pun intended), it is helpful to have a book (or better yet, a teacher) which can guide your proofs along.</p>
<p>The answer to your question is that your proof is missing some key parts, most importantly the axiomatic framework. Your proof seems to boil down to a proof by intuition, where you define addition in a geometric way and rely on the geometric intuition that no matter which way you perform your construction, the total length would be the same. This is not enough, and to understand why you need to see what <em>is</em> enough.</p>
<p>I would recommend starting a book on set theory, which often is the fist place one encounters proofs of this nature. This free online book looks accessible to someone with your background:
<a href="http://math.boisestate.edu/~holmes/holmes/head.pdf" rel="nofollow">http://math.boisestate.edu/~holmes/holmes/head.pdf</a></p>
|
158,896 | <p>Being interested in the very foundations of mathematics, I'm trying to build a rigorous proof on my own that $a + b = b + a$ for all $\left[a, b\in\mathbb{R}\right] $. Inspired by interesting properties of the complex plane and some researches, I realized that defining multiplication as repeated addition will lead me nowhere (at least, I could not work with it). So, my ideas:</p>
<ul>
<li><p>Defining addition $a+b$ as a kind of "walking" to right $\left(b>0\right)$ or to the left $\left(b<0\right)$ a space $b$ from $a$. Adding a number $b$ to a number $a$ (denoted by $a+b$) involves doing the following operation:</p>
<ol>
<li>Consider the real line $\lambda$ and its origin at $0$. Mark a point $a$, draw another real line $\omega$ above $\lambda$ such what $\omega \parallel \lambda$ and mark a point $b$ on $\omega$. Now, draw a line $\sigma$ such that $\sigma \perp \omega$ and the only point in commom between $\sigma$ and $\omega$ is $b$. Consider the point that $\lambda$ and $\sigma$ have in commom; this point is nicely denoted as $a + b$.</li>
</ol></li>
</ul>
<p>(Note that all my work is based here. Any problems, and my proof goes to trash)</p>
<ul>
<li><p>This definition can be used to see the properties of adding two numbers $a$ and $b$, for all $a, b \in\mathbb{R}$.</p></li>
<li><p>Using geometric properties may lead us to a rigorous proof (if not, I would like to know the problems of using it).</p></li>
</ul>
<p>So, I started:</p>
<ul>
<li>$a, b \in\mathbb{N}$:</li>
</ul>
<p>$a+b = \overbrace{\left(1+1+1+\cdots+1\right)}^a + \overbrace{\left(1+1+1+\cdots+1\right)}^b = \overbrace{1+1+1+1+\cdots+1}^{a+b} = \overbrace{\left(1+1+1+\cdots+1\right)}^b + \overbrace{\left(1+1+1+\cdots+1\right)}^a = b + a$</p>
<p>(Implicity, I'm using the fact that $\left(1+1\right)+1 = 1+\left(1+1\right)$, which I do not know how to prove and interpret it as cutting a segment $c$ in two parts -- $a$ and $b$. However, this result can be extended to $\mathbb{Z}$ in the sense that $-a$ $(a > 0)$ is a change; from right to left).</p>
<ul>
<li>$a, b \in\mathbb{R}$:</li>
</ul>
<p>Here, we have basically two cases:</p>
<ul>
<li>$a$ and $b$ are either positive or negative;</li>
<li>$a$ and $b$, where one of them is negative.</li>
</ul>
<p>Since in my definition $-b, b>0$ means drawing a point $b$ to the left of the real line, there's no big deal interpretating it; <em>subtracting</em> can be interpreted now. So, it starts:</p>
<p>$a + b = c$. However, $c$ can be cut in two parts: $b$ and $a$. Naturally, if $a>c$, then $b<0$ -- many cases can be listed. So, $c = b + a$. But $c = a + b$; it follows that $a + b = b + a$. My questions:</p>
<p><em>Is there any problem in using my definition of adding two numbers $a$ and $b$, which uses many geometric properties? Is there any way to solve it from informality? Is there anything right here?</em></p>
<p>Thanks in advance.</p>
| André Nicolas | 6,312 | <p>Something of this kind was done by Hilbert a bit over a century ago, in his <a href="http://en.wikipedia.org/wiki/Hilbert%27s_axioms" rel="nofollow">Foundations of Geometry.</a> The axioms of the geometric substrate were laid out in <em>great detail</em>. Then an arithmetic was defined on the points of a particular line $\ell$, with addition defined in a way reminiscent of what you did, and then multiplication and division. For addition one picks an arbitrary point $O$ on $\ell$ to serve as $0$. For multiplication one needs another arbitrary point $P\ne O$, to serve as $1$. </p>
<p>Hilbert then showed that the points on the line, under these geometric operations, form what we now call a complete ordered field. This is of some interest, in that it shows that classical plane geometry, done correctly (that is, with the axioms missed by Euclid added), yields a structure isomorphic to the familiar coordinate plane $\mathbb{R}^2$. </p>
<p>However, in my opinion, this work of Hilbert had little long-term significance. Although it shows that the real numbers can be developed within a classical geometric framework, the standard approach continues to be the arithmetical one pioneered by Weierstrass, Dedekind, and Cantor. The natural numbers are either taken as fundamental or (later) defined set-theoretically. Then one uses set-theoretic tools to build in succession the integers, the rationals, and the reals. </p>
|
623,190 | <p>What would be the formula, to determine a rectagles edges, when given the perimeter and space? for example, the rectagles space is 80, and the perimeter is 36, and the edge would be 8 and 10, but how do I find them.</p>
<p>I know that the formula for the perimeter would be
2x+2y=per, or 2(space/y)+2y=per
However I'm trying to figure out how to find x and y, when I only know space and perimeter.</p>
| lsp | 64,509 | <p>length = $l$, breadth = $b$</p>
<p>space, $lb = 80$</p>
<p>perimeter, $2(l+b) = 36$</p>
<p>$$l + \frac{80}{l} - 18 = 0$$
$$l^2-18.l+80=0$$
$$(l-10)(l-8)$$
So, $l=10$ or $l=8$.</p>
<p>But since length > breadth:
$$length=10$$
$$breadth=8$$</p>
|
3,267,883 | <p>I apologize in advanced as my literacy in this subject is not too great and this question may either be trivial or impossible as of yet. </p>
<p>I have seen many questions on stack exchange utilizing the Chinese Remainder Theorem to find solutions of <span class="math-container">$a^2\equiv 1\mod (p*q)$</span>, where <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are distinct primes. </p>
<p>My question is whether we can find nontrivial (besides <span class="math-container">$a\equiv \pm1\mod 2^k$</span>) solutions of <span class="math-container">$a^2\equiv 1\mod 2^k$</span> by any method other than brute force, or whether solutions exist at all. </p>
<p>(Off the top of my head, <span class="math-container">$3^2\equiv 1\mod 2^2$</span>, and furthermore, <span class="math-container">$3^2=2^2+1$</span>, but I do not think that <span class="math-container">$a^2=2^k+1$</span> for any other <span class="math-container">$k$</span>.) </p>
| J. W. Tanner | 615,567 | <p><strong>Hint:</strong> You're asking for <span class="math-container">$2^k|(a+1)(a-1).$</span> <span class="math-container">$\gcd(a+1,a-1)$</span> is at most <span class="math-container">$2$</span>.</p>
|
1,182,523 | <p>I have stumbled across this question:
Let $a,b,c$ be integers, not all $0$ such that $\max(|a|,|b|,|c|)<10^6$. Prove that $|a+b \sqrt{2} + c \sqrt{3}| > 10^{-21}$. </p>
<p>Could anybody help by solving this? Elementary solution is preferred.</p>
| math110 | 58,742 | <p>Let $$\begin{cases}f_{1}=a+b\sqrt{2}+c\sqrt{3}\\
f_{2}=a-b\sqrt{2}+c\sqrt{3}\\
f_{3}=a-b\sqrt{2}-c\sqrt{3}\\
f_{4}=a+b\sqrt{2}-c\sqrt{3}\end{cases}$$
It is clear $$f_{1}f_{2}f_{3}f_{4}\in Z,a,b,c\in Z$$
since $a,b,c$ are integer,and not all 0 ,so $f_{k}\neq 0,k=1,2,3,4$.and Note
$$\max\{|a|,|b|,|c|\}<10^6\Longrightarrow |f_{k}|<10^7,k=1,2,3,4$$ so we have
$$|f_{1}f_{2}f_{3}f_{4}|\ge 1\Longrightarrow |f_{1}|\ge\dfrac{1}{|f_{2}f_{3}f_{4}|}>10^{-21}$$</p>
|
2,646,890 | <blockquote>
<p>If <span class="math-container">$p(x)$</span> is a polynomial of degree <span class="math-container">$n$</span> such that <span class="math-container">$$p(-2)=-15,\ p(-1)=1,\ p(0)=7,\ p(1)=9,\ p(2)=13,\ p(3)=25.$$</span>
Then smalest possible value of <span class="math-container">$n$</span> is</p>
<p>Options <span class="math-container">$(a)\; 2\;\;(b)\; 3\;\; (c)\;\; 4\;\; (d)\; 5$</span></p>
</blockquote>
<p>Try: Tracing curve on coordinate axis, it gave one point of intersection Further <span class="math-container">$p(x)$</span> must be an odd degree polynomial. And slope of function is not same in each interval. So it is not linear. So it must have least degree <span class="math-container">$3$</span>.</p>
<p>Can someone explain me if I am doing right? Thanks.</p>
<p>Otherwise please provide solution.</p>
| user5713492 | 316,404 | <p>Construct a difference table.
$$\begin{array}{rrrrr}-15&&&&\\
&16&&&\\
1&&-10&&\\
&6&&6&\\
7&&-4&&0\\
&2&&6&\\
9&&2&&0\\
&4&&6&\\
13&&8&&\\
&12&&&\\
25&&&&\end{array}$$
Since the fourth differences are all $0$ and the third differences are not, $p(x)$ can be fitted with a third degree polynomial but not a second degree one.</p>
|
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