qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,927,502 | <p><span class="math-container">$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$</span>.</p>
<p>With epsilon definition I get my answer as <span class="math-container">$N=\left[ \dfrac{5}{9\varepsilon }\right] +1$</span>. But then I thought that how can I evaluate this sequence, in functions <span class="math-container">$\delta$</span>-<span class="math-container">$\varepsilon$</span> definition by substituting <span class="math-container">$n$</span> by <span class="math-container">$x$</span>. But I can't make it.</p>
<p>So here's my question:</p>
<p>What can I do is I have function like this one?
<span class="math-container">$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$</span>.</p>
<p>I'm thinking about that function. But how to find.</p>
<p><span class="math-container">$\forall \varepsilon>0$</span> be given, then <span class="math-container">$\exists M>0$</span> such that <span class="math-container">$x>M$</span> implies <span class="math-container">$| f\left( x\right) -l|<\varepsilon$</span></p>
<p>For function:
<span class="math-container">$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$</span></p>
<p>And this is where I stuck. Because <span class="math-container">$x\in \mathbb{R}$</span>, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!</p>
| Claude Leibovici | 82,404 | <p>If you use the long division, you have
<span class="math-container">$$\frac{n^{2}-n+2}{3n^{2}+2n-4}=\frac{1}{3}-\frac{5}{9 n}+\frac{40}{27
n^2}+O\left(\frac{1}{n^3}\right)$$</span> Therefore
<span class="math-container">$$ \frac{1}{3}-\frac{5}{9 n}<\frac{n^{2}-n+2}{3n^{2}+2n-4}<\frac{1}{3}-\frac{5}{9 n}+\frac{40}{27
n^2}$$</span></p>
|
3,927,502 | <p><span class="math-container">$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$</span>.</p>
<p>With epsilon definition I get my answer as <span class="math-container">$N=\left[ \dfrac{5}{9\varepsilon }\right] +1$</span>. But then I thought that how can I evaluate this sequence, in functions <span class="math-container">$\delta$</span>-<span class="math-container">$\varepsilon$</span> definition by substituting <span class="math-container">$n$</span> by <span class="math-container">$x$</span>. But I can't make it.</p>
<p>So here's my question:</p>
<p>What can I do is I have function like this one?
<span class="math-container">$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$</span>.</p>
<p>I'm thinking about that function. But how to find.</p>
<p><span class="math-container">$\forall \varepsilon>0$</span> be given, then <span class="math-container">$\exists M>0$</span> such that <span class="math-container">$x>M$</span> implies <span class="math-container">$| f\left( x\right) -l|<\varepsilon$</span></p>
<p>For function:
<span class="math-container">$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$</span></p>
<p>And this is where I stuck. Because <span class="math-container">$x\in \mathbb{R}$</span>, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!</p>
| Angelo | 771,461 | <p>We have to prove that</p>
<p><span class="math-container">$\forall\;\varepsilon>0\;\;\exists\;M>0\;$</span> such that <span class="math-container">$\;x>M\;$</span> implies <span class="math-container">$\;\left|f\left(x\right)-l\right|<\varepsilon\;.$</span></p>
<p>For any <span class="math-container">$\;\varepsilon>0\;,\;$</span> let <span class="math-container">$\;M=\max\left\{2,\dfrac{5}{9\varepsilon}\right\}>0\;.$</span></p>
<p>Moreover,</p>
<p><span class="math-container">$x>M\ge2\implies|-5x+10|=5x-10<5x\;,$</span></p>
<p><span class="math-container">$x>M\ge2\implies|9x^2+6x-12|=9x^2+6x-12>9x^2\;.$</span></p>
<p>Hence,</p>
<p><span class="math-container">$x>M\implies\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5x}{9x^2}=\dfrac{5}{9x}\;.$</span></p>
<p>Consequently,</p>
<p><span class="math-container">$x>M\implies|f(x)-l|=\left|\dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\dfrac{5}{9x}<\dfrac{5}{9M}\le\dfrac{5}{9\frac{5}{9\varepsilon}}=\varepsilon\;.$</span></p>
|
73,277 | <p>Let $\boldsymbol{\theta}=(\theta_1,\ldots,\theta_m)$ be a vector of real numbers in $[-\pi,\pi]$. For $t\ge 0$, define
$$ f(t,\boldsymbol{\theta}) = \binom{m+t-1}{t}^{-1}
\sum_{j_1+\cdots+j_m=t} \exp(ij_1\theta_1+\cdots+ij_m\theta_m),$$
where the sum is over non-negative integers $j_1,\ldots,j_m$ with sum $t$.
Note that the number of terms in the sum is $\binom{m+t-1}{t}$, so
$|f(t,\boldsymbol{\theta})|\le 1$ with equality occurring when all the $\theta_j$s
are equal.</p>
<p>For a problem in asymptotic combinatorics, we need a bound on
$|f(t,\boldsymbol{\theta})|$ that decreases rapidly as the $\theta_j$s move apart and
is valid for all $\boldsymbol{\theta}$.
Surely this problem has been studied before?</p>
<p>Note that $\binom{m+t-1}{t}f(t,\boldsymbol{\theta})$ is the coefficient of $x^t$ in
$$\prod_{j=1}^m (1-xe^{i\theta_j})^{-1},$$
which suggests some sort of contour integral approach.</p>
| Noam D. Elkies | 14,830 | <p>Another approach is to write ${m+t-1 \choose t} f(t,\theta)$ as a Schur function of the $z_j := \exp i \theta_j$, and thus as a quotient $\Delta' / \Delta$ of $m\times m$ determinants with unit-norm entries. Then $|\Delta'| \leq m^{m/2}$ by Hadamard, and $\Delta$ is the Vandermonde determinant of the $z_j$ so
$$
|\Delta| = \biggl| \prod_{1 \leq j < k \leq m} (z_j - z_k) \biggr| \phantom{+} = \prod_{1 \leq j < k < m} 2 \left| \sin (\theta_j^{\phantom{Y}} - \theta_k^{\phantom{Y}})/2 \right|.
$$
Hence
$$
|f(t,\theta)| \leq \frac{m^{m/2}\strut} {{m+t-1 \choose t} \prod_{1 \leq j < k \leq m} 2 \left| \sin (\theta_j^{\phantom{Y}} - \theta_k^{\phantom{Y}})/2 \right|}.
$$
This bound has the advantage of satisfying the desideratum of "decreasing rapidly as the $\theta_j$s move apart and [being] valid for all $\theta$", and of being sharp in some cases where the $\theta_j$ are equally spaced. It has the disadvantage of being larger than the trivial upper bound $|f(t,\theta)| \leq 1$ when some $\theta_j$ are very close, and indeed infinite when two or more $\theta_j$ coincide.</p>
<p><strong>EDIT</strong> Expanding $\Delta'$ by the $z_j^{t+m-1}$ row yields the formula
$$
{m+t-1 \choose t} f(t,\theta) = \sum_{j=1}^m \frac{z_j^{t+m-1}}{\prod_{k\neq j} (z_j-z_k)}.
$$
Hence
$$
|f(t,\theta)| \leq {m+t-1 \choose t}^{-1} \sum_{j=1}^m \phantom{Y} \left[ 1 \left/ \prod_{k\neq j} \phantom{Y} \left| 2 \sin \frac12(\theta_j^{\phantom{Y}} - \theta_k^{\phantom{Y}}) \right| \right. \right]
$$
which has the same overall advantages and disadvantages as before but is better when the $\theta_j$ are neither bunched together nor spaced exactly evenly.</p>
<p>The determinant formula also gives yet another interpretation of ${m+t-1 \choose t} f(t,\theta)$, in terms of polynomial interpolation: it is the $z^{m-1}$ coefficient of the unique polynomial $P$ with $\deg P < m$ such that $f(z_j) = z^{m+t-1}$ for each $j$.</p>
|
3,910,345 | <p>Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:</p>
<p><span class="math-container">$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$</span></p>
<p>The logic here was that this could be used as:</p>
<p><span class="math-container">$$\begin{align}
-x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\
&\equiv y\mathrm{d}x - x\mathrm{d}y
\end{align}
$$</span></p>
<p>Why is this legal?</p>
<p>I can see some kind of differentiation going on with the second term in the above equivalence, producing the <span class="math-container">$\frac{1}{x^2}$</span>, but having the single <span class="math-container">$\mathrm{d}$</span> seems like a really weird abuse of notation, and I don't quite follow why it splits the single <span class="math-container">$\frac{y}{x}$</span> fraction into two parts.</p>
| Bernard | 202,857 | <p>You should know that the <em>differential</em> at a point <span class="math-container">$\mathbf x_0$</span> of a function <span class="math-container">$\;\mathbf R^m\longrightarrow \mathbf R^n$</span> is the <em>linear map</em> <span class="math-container">$\:\ell:\mathbf R^m\longrightarrow \mathbf R^n$</span>, that yields the best linear approximation of <span class="math-container">$f(\mathbf x_0)$</span> in a neighbourhood of <span class="math-container">$\mathbf x_0$</span>, in the sense that we have
<span class="math-container">$$f(\mathbf x_0+\mathbf h)=f(\mathbf x_0)+\ell(\mathbf h)+o\bigl(\|\mathbf h\|\bigr).$$</span>
This differential is denoted <span class="math-container">$\:\mathrm d f_{\mathbf x_0}$</span> (or simply <span class="math-container">$\mathrm df$</span> for the differential at a generic point). A linear function is of course its own differential.</p>
<p>With the usual abuse of language that denotes a function <span class="math-container">$f$</span> by its value at a given point, we obtain that the differential of the <span class="math-container">$i$</span>-th projection <span class="math-container">$\:p_i:\mathbf x=(x_1,x_2,\dots,x_m)\longmapsto x_i$</span> is denoted <span class="math-container">$\mathrm dx_i$</span>.</p>
<p>As an example, in the case of a function of a single variable <span class="math-container">$x$</span>, the linear map defining the differential simply corresponds to the equation of the tangent to the graph of <span class="math-container">$f$</span> with abscissa <span class="math-container">$x_0$</span>:
<span class="math-container">$$f(x_0+h)=f(x_0)+f'(x_0)h,\enspace\text{i.e. }\quad \mathrm df_{x_0}:h\longmapsto f'(x_0)h,$$</span>
that we may write as <span class="math-container">$\enspace\mathrm df=f'(x)\,\mathrm dx$</span>. This notation is generalised to functions of <span class="math-container">$m$</span> variables under the form
<span class="math-container">$$\mathrm df=\frac{\partial f}{\partial x_1}\,\mathrm dx_1+\frac{\partial f}{\partial x_2}\,\mathrm dx_2+\dots+\frac{\partial f}{\partial x_m}\,\mathrm dx_m. $$</span>
The usual formulæ for the derivatives have a ‘differential version’:</p>
<ul>
<li><span class="math-container">$\mathrm d(f+g)=\mathrm df+\mathrm dg$</span>,</li>
<li><span class="math-container">$\mathrm d(fg)=f\,\mathrm dg+g\,\mathrm df$</span>,</li>
<li><span class="math-container">$\mathrm d\Bigl(\dfrac fg\Bigr)=\dfrac1{g^2}\bigl(g\,\mathrm df-f\,\mathrm dg\big),$</span></li>
<li><span class="math-container">$\mathrm d(g\circ f)=\mathrm dg_{f(x)}\circ\mathrm df_x$</span>.</li>
</ul>
|
351,030 | <p>for positive integer $n$, how can we show</p>
<p>$$ \sum_{d | n} \mu(d) d(d) = (-1)^{\omega(n)} $$</p>
<p>where $d(n)$ is number of positive divisors of $n$ and $mu(n)$ is $(-1)^{\omega(n)} $ if $n$ is square free, and $0$ otherwise. Also, what is</p>
<p>$$ \sum_{d | n} \mu(d) \sigma (d) $$ where $\sigma(n)$ is the sum of positive divisors of $n$</p>
| Harald Hanche-Olsen | 23,290 | <p><strong>Hint:</strong> It is enough to compute $$\lim_{n\to\infty}\int_0^1 A_n(x)f(x)\,dx$$ for continuous functions $f\colon[0,1]\to\mathbb{R}$, since these are dense in $L^1(\Omega)$. You will find that the limit is $$\frac{\alpha+\beta}{2}\int_0^1 f(x)\,dx\tag{1}$$ – just take the difference between the two integrals, consider the difference of integrals over each subinterval $[k\epsilon,(k+1)\epsilon]$, and use uniform continuity. (Perhaps you wish to insert a Riemann sum in the estimate.)</p>
<p><strong>Edit:</strong> For more detail, note $$\int_0^1 A_n(x)f(x)\,dx=\sum_{k=0}^{n-1}\Bigl(\alpha\int_{k\epsilon}^{(k+1/2)\epsilon}f(x)\,dx+\beta\int_{(k+1/2)\epsilon}^{(k+1)\epsilon}f(x)\,dx\Bigr)$$ in which you replace $f(x)$ in the integrals on the right by $f(k\epsilon)$, carefully estimating the error you introduce by doing so. (For any $\eta>0$ you can pick $\epsilon>0$ so that $\lvert f(x)-f(k\epsilon)\rvert<\eta$ for $x\in[k\epsilon,(k+1)\epsilon]$ – this is uniform continuity.) Now do the same with the integral in (1) and subtract.</p>
|
3,546,773 | <p>what are the real/complex zeros for:</p>
<p><span class="math-container">$t^9 - 1$</span></p>
<p>I also need to use the exponential form of complex numbers</p>
| Andrew Chin | 693,161 | <p><span class="math-container">\begin{align}
t^9-1&=(t^3-1)(t^6+t^3+1)\\
&=(t-1)(t^2+t+1)(t^6+t^3+1)
\end{align}</span></p>
<p>The complex zeroes can be solved by means of quadratics.</p>
|
3,546,773 | <p>what are the real/complex zeros for:</p>
<p><span class="math-container">$t^9 - 1$</span></p>
<p>I also need to use the exponential form of complex numbers</p>
| fleablood | 280,126 | <p>The truly <em>WONDERFUL</em> thing about exponential form of complex numbers is that if</p>
<p><span class="math-container">$z = re^{i\theta}$</span> and <span class="math-container">$w = se^{i\phi}$</span> then <span class="math-container">$z\cdot w = (rs)e^{i(\theta + \phi \pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle})}$</span></p>
<p>This means if <span class="math-container">$z^m = r^me^{i(m\theta\pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle})}$</span></p>
<p>And it means we can solve: if <span class="math-container">$z^n = re^{i\theta}$</span> then the <span class="math-container">$n$</span> roots are:</p>
<p><span class="math-container">$\sqrt[n]{r} e^{i(\frac \theta n + \text{some multiples }\frac {2\pi}k\text{ to take care of cases where angles swung around a full circle})}$</span></p>
<p>or in other words, the possible values for <span class="math-container">$z$</span> are </p>
<p><span class="math-container">$\sqrt[n]{r} e^{i(\frac \theta n +\frac {2\pi}k)}$</span> for <span class="math-container">$k=0....., n-1$</span>.</p>
<p>So since <span class="math-container">$1 = 1*e^{i*0}$</span> then if <span class="math-container">$z^9 = 1$</span> then the nine possible values are:</p>
<p><span class="math-container">$\sqrt[9]{1}*e^{i(\frac 09 + \frac {2\pi}k} = e^{i(0\frac {i2\pi}k}=$</span></p>
<p>Or the <span class="math-container">$9$</span> roots are <span class="math-container">$e^{i0} =1; e^{i\frac 29\pi}; e^{i\frac 39\pi};.... e^{i\frac 89\pi}$</span>.</p>
<p>=====</p>
<p>Now if you are like nearly every person first seeing this you are probably confuse be the <span class="math-container">$\frac {2\pi}k$</span> and the "<span class="math-container">$\pm\text{some multiples of }2\pi\text{ to take care of cases where angles swing around a full circle}$</span>" business.</p>
<p>The this is the if we have a number such as <span class="math-container">$z = e^{i\theta}$</span> where <span class="math-container">$\pi < \theta < 2\pi$</span> then <span class="math-container">$z^2 = e^{i\theta}\cdot e^{i\theta}= e^{i2\theta}$</span>. But <span class="math-container">$2\pi < 2\theta$</span> so the angle has swung "full circle". And <span class="math-container">$e^{i2\theta} = e^{i(2\theta - \2pi)}$</span>. And if we let <span class="math-container">$\eta = 2\theta - 2\pi$</span> then <span class="math-container">$z^2 = e^{i\eta}$</span>.</p>
<p>SO if we were given <span class="math-container">$w= e^{i\eta}$</span> at the beginning, and asked: find <span class="math-container">$z$</span> so that <span class="math-container">$z^2 = e^{i\eta}$</span> we would figure <span class="math-container">$z = e^{i\frac {\eta}2}$</span>. That's fine and that is <em>one</em> of the roots (Notice that <span class="math-container">$0< \eta<\pi$</span>) but we need to take into account what happens when the angle in <span class="math-container">$z$</span> is large enough that doubling it will "swing it past a full circle".</p>
<p>In other words: <span class="math-container">$w = e^{i\eta} = e^{i(\eta + 2\pi)}$</span> so when we have <span class="math-container">$z^2 = w$</span> we have that <span class="math-container">$z$</span> can be <span class="math-container">$e^{i\frac {\eta} 2}$</span> or <span class="math-container">$z$</span> can be <span class="math-container">$e^{i\frac {\eta+2\pi}2}=e^{i(\frac \eta 2 + \pi)}$</span>. Notice <span class="math-container">$\frac \eta 2 + \pi = \theta$</span>, our original big angle. And Notice that <span class="math-container">$\frac \eta 2 =\theta-\pi$</span>. </p>
|
519,325 | <p>Evaluate $\displaystyle\int \dfrac{1}{x^2+9} \, dx$.
I've only learned the normal way of solving integrals but it does not work.
I haven't learned how to use trigonometry to solve these problem.</p>
<p>I know you have to rearrange it into the form ${[f(x)]² + 1}$ and then integrate.</p>
<p>Can someone point me some rules to solve these kinds of question?</p>
<p>My teacher expected that the prerequisite course taught this but I have not learned it yet.</p>
| Mhenni Benghorbal | 35,472 | <p>Make the substitution</p>
<p>$$ x=3\tan t \implies dx = 3\sec^2 t \,dt .$$</p>
<p>Subs back in the integral and you need to use the identity</p>
<p>$$ 1+\tan^2 t = \sec^2 t. $$</p>
|
3,695,439 | <p>So I know that we can find <span class="math-container">$dy/dx$</span> of a curve in polar coordinates by leveraging the fact that <span class="math-container">$x=rcos\theta$</span> and <span class="math-container">$y=rsin\theta$</span>, and since <span class="math-container">$r$</span> is a function of <span class="math-container">$\theta$</span> we can take <span class="math-container">$dy/d\theta$</span> and <span class="math-container">$dx/d\theta$</span> and divide them. But my question is, staying in polar coordinates can we glean anything about the tangent slopes from just <span class="math-container">$dr/d\theta$</span>? For instance if we have <span class="math-container">$r=sin\theta$</span>, then <span class="math-container">$dr/d\theta=cos\theta$</span>. So when <span class="math-container">$\theta$</span> is equal to <span class="math-container">$0$</span>, <span class="math-container">$r$</span> is changing with respect to <span class="math-container">$\theta$</span> at a rate of <span class="math-container">$1$</span>? I can't seem to wrap my head around what this means. Obviously in cartesian coords. the slope of the tangent when <span class="math-container">$x=0$</span> is <span class="math-container">$0$</span> and infinite when <span class="math-container">$y=1/2$</span>. Is there no way to intuitively see this from just looking at <span class="math-container">$dr/d\theta$</span>?</p>
| SarGe | 782,505 | <p><span class="math-container">$\frac{dy}{dx}$</span> is, by definition, the limit of a secant line as the distance between two points approaches zero - it simply is the slope, nothing more to prove really (other than that the derivative actually exists, which is beyond the scope of this question). </p>
<p>Also, <span class="math-container">$tan(θ)=\frac{opposite}{adjacent}$</span> for right triangles; since we define slope as <span class="math-container">$\frac{Δy}{Δx}$</span> and this is basically what the derivative gives you: a small change in y over a small change in x. We can find the inclination <span class="math-container">$θ$</span> by taking the arctangent of both sides.</p>
<p>However, in polar coordinate system, there is a pole (analogous to origin in Cartesian coordinate system) and a polar axis as reference direction (analogous to <span class="math-container">$+ve$</span> <span class="math-container">$x$</span>-axis in Cartesian coordinate system) but there is no analogy for <span class="math-container">$y$</span>-axis. However, to define the slope of tangent, as mentioned above, we need the distances of the point from two fixed perpendicular oriented lines. </p>
<p>So, in order to tackle this we convert the polar coordinates <span class="math-container">$(r, \theta) $</span> to Cartesian ones <span class="math-container">$(x, y) $</span> as <span class="math-container">$$x=rcos\;\theta \text{and} y=rsin\;\theta$$</span>. We can get the slope as <span class="math-container">$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$</span>.</p>
<p>Now, your doubt regarding <span class="math-container">$\frac{dr}{d\theta}$</span>. Well, it just implies the rate of change of distance of the point from the pole w.r.t. to the polar angle. It has nothing to do with the slope of tangent. The Cartesian coordinate system is defined in such a way that the slope of tangent directly comes out to be <span class="math-container">$\frac{dy}{dx}$</span> (as <span class="math-container">$dy$</span> and <span class="math-container">$dx$</span> itself represent the opposite and adjacent side required for slope) but not <span class="math-container">$\frac{dr}{d\theta}$</span> in polar coordinate system. </p>
|
29,255 | <p>sorry! am not clear with these questions</p>
<ol>
<li><p>why an empty set is open as well as closed?</p></li>
<li><p>why the set of all real numbers is open as well as closed?</p></li>
</ol>
| Andrea Mori | 688 | <p>By definition, a set $A$ of real numbers is <em>open</em> when the following condition is met:
$$
\hbox{$\forall x\in A, \exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset A$,}
$$
where $(a,b)$ denotes the open interval $\{x\in{\Bbb R}\,|\,a<x<b\}$. It should be not hard to convince yourself that the subsets $A=\emptyset$ and $A={\Bbb R}$ satisfy this condition.</p>
<p>Then, remember that
$$
\hbox{$A$ is open $\iff {\Bbb R}\setminus A$ is closed}
$$
again by definition. You conclude since $\emptyset={\Bbb R}\setminus{\Bbb R}$ and
${\Bbb R}={\Bbb R}\setminus\emptyset$.</p>
|
29,255 | <p>sorry! am not clear with these questions</p>
<ol>
<li><p>why an empty set is open as well as closed?</p></li>
<li><p>why the set of all real numbers is open as well as closed?</p></li>
</ol>
| Ari Royce Hidayat | 435,467 | <p>By definition, a set <span class="math-container">$A$</span> of real numbers is open when the following condition is met:</p>
<p>(<em>Note that this applies equally well to the set of real numbers, just substitute <span class="math-container">$A = R$</span>.</em>)</p>
<p><span class="math-container">$$
\hbox{$\forall x\in A, \exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset A$.}
$$</span></p>
<p>By the definition above, the set of all <span class="math-container">$x$</span> is no other than the set <span class="math-container">$A$</span> itself, and <span class="math-container">$A$</span> is already defined above to be open.</p>
<p>As for why an empty set is also open, is a rather tricky business AND DEFINITELY NOT IMMEDIATELY CLEAR why it is so as many people seem to say. First of all, it is defined rather than concluded from some logical reasoning. The reason we want an empty set to be made open is:</p>
<ol>
<li><p>We want to maintain that empty set is a subset of any sets.</p>
</li>
<li><p>We want to maintain a theorem that says the union of open sets is an open set.</p>
</li>
</ol>
<p>Now watch that the definition of an open set is the same as saying it is the union of all sets of all open balls in the set. As we want to maintain that empty set is a subset of any sets (<em>reason number 1</em>), we have that empty set is a subset of all sets of those open balls, thus empty set is a subset of an open set. As we also want to maintain the theorem mentioned in reason number 2 to be true, there is no other way than to define an empty set as an open set.</p>
<p>For the last part of the question of why the set <span class="math-container">$A$</span> and empty set are both also close is to remember that the definition of a close set is:</p>
<p><span class="math-container">$$
\hbox{If $B \subset A$, then $A \setminus B$ is close $\iff B$ is open.}
$$</span></p>
<p>If we set <span class="math-container">$B = A$</span>, then we have <span class="math-container">$\emptyset = A \setminus A$</span> in which by the definition above is close because <span class="math-container">$B = A$</span> is open. If we set <span class="math-container">$B = \emptyset$</span>, then we have <span class="math-container">$A = A \setminus \emptyset$</span> which is close because <span class="math-container">$B = \emptyset$</span> is open.</p>
|
29,255 | <p>sorry! am not clear with these questions</p>
<ol>
<li><p>why an empty set is open as well as closed?</p></li>
<li><p>why the set of all real numbers is open as well as closed?</p></li>
</ol>
| Joe | 623,665 | <p>Of course, the answer depends on how you define "open" and "closed" sets of <span class="math-container">$\mathbb R$</span>. There are many equivalent definitions.</p>
<p>Here is a common one when we are considering <span class="math-container">$\mathbb R$</span> as a metric space with the absolute value norm: we say that <span class="math-container">$E\subset\mathbb R$</span> is open if every point of <span class="math-container">$E$</span> is an interior point of <span class="math-container">$E$</span>; we say that <span class="math-container">$E$</span> is closed if every limit point of <span class="math-container">$E$</span> is a point of <span class="math-container">$E$</span>.</p>
<p>According to these definitions, the assertion that <span class="math-container">$\varnothing$</span> is open is vacuously true. The assertion that <span class="math-container">$\mathbb R$</span> is open is true because <em>every</em> neighbourhood of every point of <span class="math-container">$\mathbb R$</span> is a subset of <span class="math-container">$\mathbb R$</span>, and so it follows <em>a fortiori</em> that each point of <span class="math-container">$\mathbb R$</span> has a neighbourhood which is included in <span class="math-container">$\mathbb R$</span>. Notice that neighbourhoods themselves are defined as certain subsets of <span class="math-container">$\mathbb R$</span>, making this whole business rather trivial.</p>
<p>The assertion that <span class="math-container">$\varnothing$</span> is closed is also fairly easy to verify. If <span class="math-container">$x\in\mathbb R$</span> were a limit point of <span class="math-container">$\varnothing$</span>, then every neighbourhood of <span class="math-container">$x$</span> would contain a point in <span class="math-container">$\varnothing$</span>. Since the empty set is, well, empty, this is impossible. So the empty set has no limit points, and the implication is vacuously true. Since <span class="math-container">$\mathbb R$</span> is the metric space under consideration, for <span class="math-container">$x$</span> to be considered to be a limit point of <span class="math-container">$\mathbb R$</span>, it has to member of <span class="math-container">$\mathbb R$</span>, and so <span class="math-container">$\mathbb R$</span> is closed as well.</p>
<p>This proof can be shortened if we have previously demonstrated that a set is open if and only if its complement is closed.</p>
|
2,310,441 | <p>I consider the sequence of composite odd integers: 9, 15, 21, 25, 27, 33, 35, 41, ...</p>
<p>I observe that there are certain large gaps between the composite odd integers and this may contribute towards the solution.</p>
<p>So I start by considering some sums first:</p>
<p>9 + 9 + 9 = 27, 9 + 9 + 15 = 33. So this means that 31 is potentially such a prime number.</p>
<p>Then I consider other sums and manage to obtain 39, 43 and 45. So now 41 becomes the potential contender.</p>
<p>But this method is clearly just trial and error. Is there a more elegant method?</p>
| lokodiz | 70,984 | <p>The largest such number is $47$.</p>
<p>Let $C$ be the set of positive odd composite numbers, so $C = \{9,15,21,25, \dots \}$. First check that $47$ can't be written as a sum of three elements of $C$. Now observe that $C$ contains $\{ 6k+3 \mid k \geqslant 1 \}$, so if we can write a prime $p$ as a sum of three elements of $C$, then the same is true of every larger prime $q$ with $p \equiv q \mod 6$. Moreover, every prime $ \geqslant 5$ is congruent to $1$ or $5$ modulo $6$. Finally, the smallest primes greater than $47$ which are congruent to $1$ and $5$ are, respectively, $61$ and $53$, and $53 = 9 + 9 + 35$ and $61 = 9+25+27$ are both sums of three elements of $C$.</p>
|
480,727 | <p>If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$</p>
<p>I have tried starting with taking logartithms, but that gives just some more equations.</p>
<p>Any specific way to solve these type of problems?</p>
<p>Any help will be appreciated.</p>
| Balbichi | 24,690 | <p>$2^x=3^y=6^{-z}=k $ say, then $2= k^{1\over x},3=k^{1\over y},6=k^{-1\over z}$ now can you go on?</p>
<p>then $k^{-1\over z}=6=2\times 3 = k^{1\over x}\times k^{1\over y}=k^{{1\over x}+{1\over y}}$</p>
|
480,727 | <p>If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$</p>
<p>I have tried starting with taking logartithms, but that gives just some more equations.</p>
<p>Any specific way to solve these type of problems?</p>
<p>Any help will be appreciated.</p>
| Harish Kayarohanam | 30,423 | <p>$$2^x = 3^y = 6^{-z} = k $$</p>
<p>so$$x = \log_2k$$
$$ y = \log_3k$$
$$z= -\log_6k$$</p>
<p>so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k2 + log_k3 -\log_k6$$
$$=\log_k{\frac{2\times3}{6}}$$
$$=0$$</p>
|
3,057,517 | <p>Hy everybody ! </p>
<p>I'm studying population dynamics for my calculus exam, and I don't understand something that seems really easy, so I thought you might be able to help me out ;)</p>
<p>Here's the thing. I have this differential equation <span class="math-container">$\frac{dN}{dt} = \sqrt{N}$</span>.</p>
<p>Our book makes us realize that both <span class="math-container">$N(t) = 0$</span> and <span class="math-container">$N(t) = \frac{t^2}{4}$</span> are solution, which makes sense so far, simply by replacing in the original equation.</p>
<p>Now suppose we start at <span class="math-container">$N(0) = 0$</span>. How can <span class="math-container">$N$</span> start growing like <span class="math-container">$\frac{t^2}{4}$</span> if it's derivative is <span class="math-container">$0$</span> at <span class="math-container">$t = 0$</span> ? Because zero derivative should mean no growth, so <span class="math-container">$\sqrt{N}$</span> should remain zero, which means still no growth, and so on. My brain is melting right now.</p>
| user619894 | 617,446 | <p>You are correct. There are indeed two solutions. If the population is zero it will stay zero. The only way it will grow is if the initial conditions are positive. However, the population zero solution is "unstable" in the sense that even if the initial condition is slightly positive, the <span class="math-container">$t^2$</span> solution will kick in and push the population away from zero.</p>
|
1,386,261 | <p>They start by choosing $m$ s.t. $|s_n - s| \lt \frac{1}{2} |s|$ if $n > m$.
From here, it looks like they use the triangular inequality $|s_n - s| + |s_n| < |s|$ to come up with this statement but I'm not sure as they introduce some variable m.</p>
<p>Next, given $\epsilon > 0$, there is a $N > m$ s.t $n \geq N$ (why not just state this in the first place?) implies $|s_n - s| < \frac{1}{2}|s^2|\epsilon$.</p>
<p>Hence for $n \geq N$ we have:</p>
<p>$|\frac{1}{s_n} - \frac{1}{s}| < |\frac{s_n-s}{s_ns}|<\frac{2}{|s^2|}|s_n-s|<\epsilon$.</p>
<p>I can follow the inequalities, but I'm not sure what techniques they used to come to this conclusion. I would appreciate it if someone could shed some light on this, thanks!</p>
| Sinister Cutlass | 235,860 | <p>Yes, the space of all n-by-n matrices with entries in field F itself admits the structure of an affine space and hence also admits the Zariski topology. The general linear group is, as the previous poster said, a Zariski-open subset of the affine space. The general linear group may <em>itself</em> be given the subspace topology induced from the Zariski topology on the affine space, and in this way, the general linear group may be said to "have" the Zariski topology.</p>
|
3,350,251 | <p>The integral of velocity plots position and not change in position. But the definition of the integral is the area under the velocity curve and the area under the velocity curve is change in position. So why doesn't the integral of velocity plot change in position?</p>
| Saketh Malyala | 250,220 | <p>The information presented in the question seems a bit misguided. The plot of the integral of velocity, that is <span class="math-container">$\displaystyle f_1(t) = \int_{a}^{t}v(x)\,dx$</span> does show the change in position. In this case, we will have <span class="math-container">$f(a)=0$</span>. In fact, if we change <span class="math-container">$f$</span> so that <span class="math-container">$\displaystyle f_2(t)=\int_a^tv(x)\,dx+p(a)$</span>, then <span class="math-container">$f(t)=p(t)$</span>. So while <span class="math-container">$f_1(t)$</span> may not show the position itself, it DOES show the change in position from some reference point. </p>
|
565,046 | <blockquote>
<p>The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.</p>
</blockquote>
<p>I have that
$$D_6=\left< a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right>$$
$$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$
My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that
$$a^3b=ba^3$$
and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?</p>
<p>For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.</p>
<p>Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible. </p>
| Jay | 608,546 | <p>Check that <span class="math-container">$a^{k}b=(a^{k}b)^{-1},k=1,2,3,4,5$</span>. Then since <span class="math-container">$ba^k=(a^{6-k}b)^{-1}$</span> (multiply them), then since <span class="math-container">$(a^{6-k}b)^{-1}=a^{6-k}b, a^{6-k}b=ba^{k}\Rightarrow $</span>
every element of <span class="math-container">$D_{6}$</span> may be written as as <span class="math-container">$a^{k}b^{l}$</span>.Then two things: </p>
<p>(1) If <span class="math-container">$a^kb=ba^k\Rightarrow a^kb=a^{6-k}b\Rightarrow a^{k}=a^{6-k}\Rightarrow k\equiv 6-k(\text{mod }6)\Rightarrow k=3\Rightarrow a^3$</span> is the only element not
equal to <span class="math-container">$b$</span> that commutes with <span class="math-container">$b$</span>. Since <span class="math-container">$ab = ba^5, b$</span> is not in the center. </p>
<p>(2) <span class="math-container">$a^{3}
(a^{k}b)=a^{3}(ba^{6-k})=(a^{3}b)a^{6-k}=(ba^3)a^{6-k}=b(aa^{6-k})=b(a^{6-k}a)=
(ba^{6-k})a^3=(a^{k}b)a^3\Rightarrow a^{3}$</span> commutes with all <span class="math-container">$a^{k}b$</span>.</p>
<p>We already know that <span class="math-container">$a^3$</span> commutes with every other <span class="math-container">$a^{k}$</span>. By (2), <span class="math-container">$a^3$</span> is in the center. By (1), <span class="math-container">$a^3$</span> is the only nontrivial element in the center. </p>
|
2,840,333 | <p>I know that the easy way to evaluate the mean and variance of the Binomial distribution is by considering it as a sum of Bernoulli distributions.</p>
<p>However, I was wondering just for fun if there is a way to evaluate them directly. I got the mean easily: it only involves some fiddling around with the binomial coefficient to absorb the 'extra' $k$ in the summation, followed by a direct application of the binomial theorem. However, in the process of evaluating the variance I need to compute a sum of the form:</p>
<p>$$
\sum\limits_{k=0}^{n}k^2 \binom{n}{k} r^k
$$</p>
<p>The extra $k$ in the sum now doesn't let me apply my previous trick. Wolfram Alpha has no problem evaluating this sum, but it won't give me a step-by-step solution. Any leads would be appreciated.</p>
| Robert Israel | 8,508 | <p>More generally, $$\sum_{k=0}^n k a_k r^k= r \dfrac{d}{dr} \sum_{k=0}^n a_k r^k$$
and so $$\sum_{k=0}^n k^2 a_k r^k = r \dfrac{d}{dr} \left( r \dfrac{d}{dr} \sum_{k=0}^n a_k r^k \right)$$
Here
$$ \sum_{k=0}^n {n \choose k} r^k = (1+r)^n $$
so
$$ \sum_{k=0}^n k^2 {n \choose k} r^k = r \dfrac{d}{dr} \left( r \dfrac{d}{dr} (1+r)^n \right) = r^2 (n^2-n) (1+r)^{n-2} + r n (1+r)^{n-1}$$</p>
|
1,571,099 | <blockquote>
<p>Consider the rectangle formed by the points $(2,7),(2,6),(4,7)$ and
$(4,6)$. Is it still a rectangle after transformation by $\underline
A$= $ \left( \begin{matrix} 3&1 \\ 2&\frac {1}{2} \\ \end{matrix}
\right) $ ?By what factor has its area changed ?</p>
</blockquote>
<p>I've defined the point $(2,6)$ as the origin of my vectors $\vec v $ and $\vec u$ with
$\vec v = \left(\begin{matrix}0 \\1 \\\end{matrix} \right)$ and $\vec u = \left(\begin{matrix}2 \\0 \\\end{matrix} \right)$ which get transformed to $\underline A \vec v=$$\left(\begin{matrix}1 \\\frac{1}{2} \\\end{matrix} \right)$ and $\underline A \vec u=$$\left(\begin{matrix}6 \\4 \\\end{matrix} \right)$.</p>
<p>So my new figure(which is not a rectangle anymore,but is now a parallelogram) has vertices $(2,6)(3,6 \frac{1}{2}),(8,10)$ and $(9,10 \frac{1}{2})$</p>
<p>Now the rectangle has area equal to $2 \cdot 1=2$, and after the transformation I have that the area of the resulting parallelogram is $\underline A \vec v \times \underline A \vec u =|1\cdot 4 -\cfrac{1}{2}\cdot 6|=1$</p>
<p>Now my problem is that when I calculate the area by geometric methods I have:</p>
<p><a href="https://i.stack.imgur.com/JtCyj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JtCyj.jpg" alt="enter image description here"></a></p>
<p>You see I get a different answer,so it's clear that I've had it all wrong since the beginning but I don't see where.</p>
<p>I upload now the image of the parallelogram where I've applied law of cosines in the last step of the above image.</p>
<p><a href="https://i.stack.imgur.com/8N3ug.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8N3ug.png" alt="enter image description here"></a></p>
<p>I've tried to be as specific as possible about my steps.Can someone help me ?</p>
<p>Thanks in advance.</p>
| Charles Bronson | 161,483 | <p>Just to expand what Rob Arthan said above.</p>
<p>So, you have got the following Kripke model $M= (S, R, V)$:</p>
<h2><ul></h2>
<ul>
<li>$S = \{u_1, u_2, u_3, u_4\}$, </li>
<li>$R (a) = \{u_1 \xrightarrow {a} u_1, u_2 \xrightarrow {a} u_1, u_4 \xrightarrow {a} u_1\}$,
<li>$V(p) = \{u_1, u_2\}, V(q) = \{u_3, u_4\}$. (or $\pi$)</li>
</ul></li>
</ul>
<p>Semantics for the diamond operator is as follows:</p>
<p>$(M,u) \models [a]\varphi \text{ iff for all } v:(u,v) \in R(a) \text{ implies } (M,v) \models \varphi$.</p>
<p>In your case we have:</p>
<p>$(M, u_3) \models [a](p \wedge q) \text{ iff }$ </p>
<p>$(M, u_3) \models [a]p \text{ and } (M, u_3) \models [a]q \text{ iff }$ </p>
<p>$\text {for all } v:(u_3,v) \in R(a) \text{ implies } (M,v) \models p $, which is true (since $u_3$ is a dead-end world). The same holds for $q$. </p>
|
11,457 | <p>In their paper <em><a href="http://arxiv.org/abs/0904.3908">Computing Systems of Hecke Eigenvalues Associated to Hilbert Modular Forms</a></em>, Greenberg and Voight remark that</p>
<p>...it is a folklore conjecture that if one orders totally real fields by their discriminant, then a (substantial) positive proportion of fields will have strict class number 1.</p>
<p>I've tried searching for more details about this, but haven't found anything. </p>
<p>Is this conjecture based solely on calculations, or are there heuristics which explain why this should be true? </p>
| Jonah Sinick | 683 | <p>For ordinary class number 1 in the real quadratic case, see Cohen and Lenstra's <em>Heuristics on Class Groups of Number Fields</em>
<a href="https://openaccess.leidenuniv.nl/retrieve/2845/346_069.pdf" rel="nofollow">https://openaccess.leidenuniv.nl/retrieve/2845/346_069.pdf</a></p>
<p>Maybe it's not so much of a jump from their to see a heuristic arguing that a substantial positive portion have strict class number 1.</p>
|
3,436,515 | <p>Please help!</p>
<p>How to show that <span class="math-container">$ \lim _{n→∞} \frac{x_{(n+1)}}{x_n} =\frac{1+\sqrt 5}{2}$</span> for a dynamical system
<span class="math-container">$$x_{(n+1)}=x_n + y_n\\
y_{(n+1)}=x_n$$</span></p>
<p>Thank you!</p>
| Brian S. | 1,065,386 | <p>Let <span class="math-container">$ \lim _{n\to\infty} x(n+1)/x(n) = a$</span></p>
<p><span class="math-container">$\lim_{n\to\infty} \ { x(n+1)/x(n)= \lim_{n\to\infty} \ x(n)/x(n-1) = a} $</span></p>
<p>Given <span class="math-container">$y(n)=x(n-1)$</span></p>
<p><span class="math-container">$x(n+1)=x(n)+x(n-1)$</span> (Eq. 1)</p>
<p><span class="math-container">$x(n+1)/x(n)=1+x(n-1)/x(n)$</span></p>
<p><span class="math-container">$a=1+1/a$</span></p>
<p><span class="math-container">$a^2-a-1=0$</span> (Eq. 2)</p>
<p><span class="math-container">$a=\frac{1+\sqrt 5}{2}$</span> is a solution to (Eq. 2)</p>
|
442,759 | <p>I was reading a book on groups, it points out about the uniqueness of the neutral element and the inverse element. I got curious, are there algebraic structures with more than one neutral element and/or more than one inverse element?</p>
| Hagen von Eitzen | 39,174 | <p>A structure can have more than one left neutral element ($e$ with $e\circ x=x$ for all $x$) or more than one right neutral element ($e$ with $x\circ e=x$ for all $x$).
For example consider the set of functions $f\colon \mathbb N_0\to\mathbb N$ under composition (using $\mathbb N\subset \mathbb N_0$ of course).
Then any $f$ with $f(x)=x$ for all $x\ge1$ (but $f(0)$ arbitrary!) is left-neutral. You can do something similar with right-neutral.
On the other hand, if an element $e$ is left-neutral and $e'$ is right-neutral, then necessarily $e=e\circ e'=e'$, so if 'neutral' is defined as 'left and right neutral' then uniqueness follows.</p>
<p>A structure with only one neutral can have several left inverses. For example consider the set of functions $f\colon X\to X$ under composition with identity as neutral (obviously). Then a left inverse of $f$ exists if and only if $f$ is injective, a right inverse exists if and only if $f$ is surjective. Unless $f$ is also bijective (which need not be the case as soon as $X$ is infinite), these inverses are not uniquely determined. Admittedly, this is a structure where not all elements have at least one inverse in the first place, but you get the feeling.</p>
|
1,798,855 | <p>I'm trying to understand this proof that:</p>
<p>$M$ connected $\iff$ $M$ and $\emptyset$ are the only subsets of $M$ open and closed at the same time</p>
<p>Which is:</p>
<p>If $M=A\cup B$ is a separation, then $A$ and $B$ are open and closed. Recriprocally, if $A\subset M$ is open and closed, then $M = A\cup(M-A)$. What? I know that if $M=A\cup B$ is a separation, $A$ and $B$ are both open. But why closed? Also, the 'recriprocally' part is totally nonsense to me. Anybody could help?</p>
<p>Also, there's another proof, which states: $M$ and $\emptyset$ are the only subsets of $M$ at the same time closed and open $\iff$ if $X\subset M$ has empty boundary, theb $X=M$ or $X=\emptyset$</p>
<p>which is proved as the following:</p>
<p>given $X\subset M$, we know the condition $X\cap \partial X = \emptyset$ implies $X$ is open, while the condition $\partial X \subset X$ implies $X$ is closed. Then, $X$ is open and closed $\iff$ $\partial X = X\cap \partial X = \emptyset$, this show $\iff$ for the theorem above.</p>
<p>First of all, I think that the condition $X$ has boundary empty implies that $X\cap \partial X$ is empty, but who said anything about $\partial X\subset X$? Also, where's the $\rightarrow$ of this proof? I can only see $\leftarrow$</p>
| user332239 | 332,239 | <p>For the first proof, if you have a separation, $M = A \cup B$, then $A$ and $B$ are both open, and $A \cap B = \emptyset$. But, $A$ is also closed since $B$ is open, and $A = M \setminus B$. Same goes for $B$. So this is the contrapositive of the reverse direction of the statement. </p>
<p>When they say reciprocally, they are referring to the forward implication of the statement. Namely, proving that if $M$ is connected, then $M$ and $\emptyset$ are the only sets that are both open and closed. Again, they prove the contrapositive. Assume there exists some set $A \subset M$ that is both open and closed. Then $M \setminus A$ is open. Since $A \cap (M \setminus A) = \emptyset$, and $A \cup (M \setminus A) = M$, then $M$ is not connected.</p>
<p>Now for the second proof, they are assuming that $\partial X = \emptyset$. Thus $\partial X = \emptyset \subset X$. To see the forward implication, assume that $M$ and $\emptyset$ are the only subsets of $M$ that are both open and closed. Then $M$ and $\emptyset$ are the only sets with empty boundary. Therefore, if $X \subset M$ has empty boundary, then $X$ is either $M$ or $\emptyset$.</p>
|
2,835,175 | <p>I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$</p>
<p>here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}$</p>
<p>now observe that for $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$,</p>
<p>$2.5-(1-\cos^5(\frac{\pi}{4}- x))\leq\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5-(\sin^2 (\frac{\pi}{4}- x))$ </p>
<p>then
$\lim\limits_{x \to \pi/4}(2.5-(1-\cos^5(\frac{\pi}{4}- x)))\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq \lim\limits_{x \to \pi/4} (2.5-(\sin^2 (\frac{\pi}{4}- x)))$</p>
<p>$2.5\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5$ </p>
<p>this implies $4\lim\limits_{x \to \pi/4}\frac{(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}=4\cdot 2.5=10$</p>
<p>is my answer correct? wolfram gave the same answer. is there any other simpler method? </p>
| BruceET | 221,800 | <p><strong>Extended Comment:</strong> As indicated in the Comment by @HagenvonEitzen, one way to work the initial problem (on the probability D6 shows a larger value than D10) is to enumerate
cases. In particular, you might make a $10 \times 6$ array of possible pairs
of outcomes and highlight the pairs that satisfy your condition.
When I did that, it was pretty clear that there are 15 favorable outcomes out of 60, so the probability $P(\text{D6 > D10}) = 1/4.$ </p>
<pre><code>D10\D6 1 2 3 4 5 6
1 11 12* 13* 14* 15* 16*
2 21 22 23* 24* 25* 26*
3 31 32 33 34* 35* 36*
4 41 42 43 44 45* 45*
5 51 52 53 54 55 56*
6 61 62 63 64 65 66
...
</code></pre>
<p>A brief simulation can sometimes help to provide insurance against miscounting outcomes.
In a simulation of a million pairs of dice rolls (D6 and D10), results ought
to give two place accuracy, and that was the result. (The code is for R
statistical software: <code>event</code> is a logical vector with a million <code>TRUE</code>s and
<code>FALSE</code>s, and its <code>mean</code> is its proportion of <code>TRUE</code>s.)</p>
<pre><code>m = 10^6
event = replicate(m, sample(1:6, 1) > sample(1:10, 1))
mean(event)
[1] 0.250383 # aprx 16/50 = 1/4
</code></pre>
<p>Using the array method might
suggest a way to generalize to cases in which the two dice have <em>consecutively</em> numbered faces, such as in the 'generalization' of your first paragraph. I will
leave it to you to figure that out. Or maybe you can see how to generalize
Hagen von Eitzen's computation.</p>
<p>My guess is that it will be considerably messier to solve the 'bonus' problem.</p>
|
2,835,175 | <p>I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$</p>
<p>here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}$</p>
<p>now observe that for $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$,</p>
<p>$2.5-(1-\cos^5(\frac{\pi}{4}- x))\leq\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5-(\sin^2 (\frac{\pi}{4}- x))$ </p>
<p>then
$\lim\limits_{x \to \pi/4}(2.5-(1-\cos^5(\frac{\pi}{4}- x)))\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq \lim\limits_{x \to \pi/4} (2.5-(\sin^2 (\frac{\pi}{4}- x)))$</p>
<p>$2.5\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5$ </p>
<p>this implies $4\lim\limits_{x \to \pi/4}\frac{(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}=4\cdot 2.5=10$</p>
<p>is my answer correct? wolfram gave the same answer. is there any other simpler method? </p>
| Graham Kemp | 135,106 | <p>Use the Law of Total Probability: for example $d6, d10$ the results of independen six and ten sided dice.</p>
<p>$$\begin{align}\mathsf P(d6>d10) &= \mathsf P(d10>6)\mathsf P(d6>d10\mid d10>6)+\mathsf P(d10\leq 6)\mathsf P(d6>d10\mid d10\leq 6) \\ &=\tfrac 4{10}\cdot 1+\tfrac 6{10}\cdot\mathsf P(d6>d10\mid d10\leq 6)\\ &=\tfrac 2{5}+\tfrac 3 5\cdot\mathsf P(d6>d10\mid d10\leq 6)\end{align}$$</p>
<p>All that is left is to evaluate that last term</p>
<p>Hints: $1=\mathsf P(d6>d10\mid d10\leq 6)+\mathsf P(d6=d10\mid d10<6)+\mathsf P(d6<d10\mid d10\leq 6)\\\mathsf P(d6>d10\mid d10\leq 6)=\mathsf P(d6<d10\mid d10\leq 6)$</p>
<p>Also, when given the condition that it is at most 6, the distribution of a 10 sided die is identical to the distribution of a six sided die .</p>
<p>Extend this principle to account for selections from any two independent uniform discrete distributions from non-identical supports.</p>
|
694,279 | <p>I am learning convex analysis by myself and I need help.</p>
<p>How to show that if $X=U=\mathbb{R}$
and $f\left(x\right)=\frac{|x|^{p}}{p}$
then the convex conjugate $f^{*}\left(u\right)=\frac{|u|^{q}}{q}$
when $\frac{1}{p}+\frac{1}{q}=1$?
There exists a particular technique that I have to apply in order to compute the convex conjugate? </p>
| Falcon | 766,785 | <p>Another nice way to show it is by using Young's inequality: For <span class="math-container">$x, \xi \in \mathbb R^n$</span>, we have
<span class="math-container">$$\langle\xi, x \rangle \le \frac{|\xi|^p}{p} + \frac{|x|^q}{q},$$</span>
with equality if <span class="math-container">$|\xi|^p = |x|^q$</span>. Therefore,
<span class="math-container">$$f^*(\xi) = \sup_{x \in \mathbb R^n}\left(\langle\xi, x \rangle - \frac{|x|^q}{q}\right) = \frac{|\xi|^p}{p}.$$</span></p>
|
7,761 | <p>Our undergraduate university department is looking to spruce up our rooms and hallways a bit and has been thinking about finding mathematical posters to put in various spots; hoping possibly to entice students to take more math classes. We've had decent success in finding "How is Math Used in the Real World"-type posters (mostly through AMS), but we've been unable to find what I would call interesting/informative math posters. </p>
<p>For example, I remember seeing a poster once (put out by Mathematica) that basically laid out how to solve general quadratics, cubics, and quartics. Then it had a good overview of proving that no formula existed for quintics. So not only was it pretty to look at, but if you stopped to read it you actually learned something.</p>
<p>Does anyone know of a company or distributor that carries a variety of posters like this? </p>
<p>I've tried searching online but all that comes up is a plethora of posters of math jokes. And even though the application/career-based posters are nice and serve a purpose, I don't feel like you actually gain mathematical knowledge by reading them.</p>
| Gerhard Paseman | 3,468 | <p>Various conference attendees sometimes have informative posters as part of their advertising campaigns. I have an old poster (somewhere!) of the graph of the real and imaginary parts of the zeta function on the critical line, produced I think by Wolfram Research. You might ask colleagues about promotional materials they have seen, and then check out the vendors and see what they have in their libraries/collections.</p>
<p>Gerhard "And There's 'Men Of Mathematics'" Paseman, 2015.04.01</p>
|
2,629,744 | <p>I have done the sum by first plotting the graph of the function in the Left Hand Side of the equation and then plotted the line $y=k$. For the equation to have $4$ solutions, both these two curves must intersect at $4$ different points, and from the two graphs, I could see that for the above to occur, the value of $k$ must lie between $\cfrac{1}{4}$ and $6$, that is k belongs to $( 0.25,6)$. Thus , the integral values of $k$ would be $1,2,3,4,5$; and so the number of integral values of $k = 5$. This was the answer given in the book. But as I was looking at the graph, I realized that $k=0$ also gives $4$ solutions namely $x=-2,-3,2,3$. So, should the number of integral values of $k$ be $6$ $(0,1,2,3,4,5)$? </p>
| Renji Rodrigo | 522,531 | <p>I will show you a more general way to find solutions, you can apply that method later to your problem ( and others of the same type)</p>
<blockquote>
<p><strong>Theorem(Solution of the recurrence)</strong>
Given sequences $g(n) \neq 0$ and $b(n)$, we have that $f(n)$
the solution of the recurrence
$$f(n+1)=g(n).f(n)+b(n)$$
is given by
$$f(n)= \bigg(\sum^{n-1}_{p=1}\frac{b(p)}{\prod\limits^{p}_{k=1}g(k)}+f(1) \bigg)\prod^{n-1}_{k=1}g(k). $$</p>
</blockquote>
<p><strong>Moreover,:</strong> in the same way we can show
$$ f(n)= h(c)\prod^{n-1}_{k=a}g(k) + \prod^{n-1}_{k=a}g(k)\bigg(\sum^{n-1}_{p=c}\frac{b(p)}{\prod\limits^{p}_{k=a}g(k)} \bigg)$$
<strong>Notation:</strong> $\Delta f(x)=f(x+1)-f(x)$</p>
<p><strong>Proof\deduction</strong></p>
<p>Let $h(n)$ be define by $$h(n)=\frac{f(n)}{\prod\limits^{n-1}_{k=a}g(k)},$$ so that,
$$f(n)=h(n)\prod^{n-1}_{k=a}g(k) ~~~~and~~~~f(n+1)=h(n+1)\prod^{n}_{k=a}g(k) $$
substitute tha term on the recurrence,
$$h(n+1)\prod^{n}_{k=1}g(k)=h(n)g(n)\prod^{n-1}_{k=1}g(k)+b(n)=h(n)\prod^{n}_{k=1}g(k)+b(n) $$
then
$$ h(n+1)\prod^{n}_{k=1}g(k)-h(n)\prod^{n}_{k=1}g(k)= b(n)$$
$$\Delta h(n)\prod^{n}_{k=1}g(k)=b(n)$$
$$\Delta h(n)=\frac{b(n)}{\prod\limits^{n}_{k=1}g(k)} $$
apply the sum $\sum\limits^{n-1}_{p=1}$ on both sides, it's telescopic
$$\sum^{n-1}_{p=1}\Delta h(p)=h(n)-h(1)=\sum^{n-1}_{p=1}\frac{b(p)}{\prod\limits^{p}_{k=1}g(k)}, $$
so
$$h(n)=\sum^{n-1}_{p=1}\frac{b(p)}{\prod\limits^{p}_{k=1}g(k)}+h(1) $$
then
$$f(n)= \prod^{n-1}_{k=1}g(k)\bigg(\sum^{n-1}_{p=1}\frac{b(p)}{\prod\limits^{p}_{k=1}g(k)}+h(1) \bigg). $$</p>
<p>Obs: $h(1)=f(1)$</p>
|
246,862 | <p>I have stumbled upon this problem which keeps me from finishing a proof:</p>
<p>$(\sum_{n} {|X_n|})^a \leq \sum_{n} {|X_n|}^a$,
where $n \in \mathbb{N}$ and $ 0 \leq a \leq 1 $</p>
<p>I have no idea how to prove this. It is something like the Cauchy-Schwarz inequality which applies in case $0 \leq a \leq 1$?</p>
<p>Any tip is welcome.
Thanks!</p>
| Robert Israel | 8,508 | <p>I'll leave the case $a=0$ to you. Otherwise, let $a = 1/b$, $b \ge 1$. If $y_n = |X_n|^a$, we have
$|X_n| = y_n^b$, and your inequality says
$$ \left(\sum_n y_n^b\right)^{1/b} \le \sum_n y_n$$
which is essentially Minkowski's inequality for counting measure: if
$v(n)$ is the vector with $v(n)_n = y_n$, $v(n)_j = 0$ otherwise, then $\|v(n)\|_p = y_n$, and your inequality becomes
$$ \| \sum_n v(n) \|_p \le \sum_n \|v(n)\|_p $$</p>
|
2,185,585 | <p>Triangular numbers (See <a href="https://en.wikipedia.org/wiki/Triangular_number" rel="noreferrer">https://en.wikipedia.org/wiki/Triangular_number</a> )</p>
<p>are numbers of the form $$\frac{n(n+1)}{2}$$</p>
<p>In ProofWiki I found three claims about triangular numbers. The three claims are that a triangular number cannot be a cube, not a fourth power and not a fifth power. Unfortunately, neither was a proof given nor did I manage to do it myself. Therefore my qeustions :</p>
<blockquote>
<p>Does someone know a proof that a triangular number cannot be a cube, a fourth power or a fifth power ?</p>
</blockquote>
| Pete Caradonna | 164,325 | <p>Here is a somewhat lower-brow 'proof by picture' of how such a process of gluing might go. To save you from having to read my poor handwriting, the steps are:</p>
<p>1) Embed $S^1$ into the $(x,y)$-plane in $\mathbb{R}^3$.</p>
<p>2) Pick an antipodal pair and glue (together) to the $z$-axis. </p>
<p>3) Pick another pair of non-glued antipodal points, and snip the two loops at this pair.</p>
<p>4) Glue the antipodal pairs of line segments (together) along the $z$-axis. You'll have but one pair left unglued.</p>
<p>5) Glue final pair.</p>
<p><a href="https://i.stack.imgur.com/nb3Z2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nb3Z2.jpg" alt="enter image description here"></a></p>
|
255,252 | <p>Let $\mathfrak{S}_n$ be the permutation group on an $n$-element set. For each fixed $k\in\mathbb{N}$, consider the two sets
$$A_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i)-i=k$}\}$$
and
$$B_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i+1)-\sigma(i)=k$}\}.$$ </p>
<blockquote>
<p><strong>QUESTION.</strong> I believe the following is true, for each $n$ and $k$:
$$\# A_n(k)=\# B_n(k).$$
Is there a <strong>combinatorial</strong> proof of this? If it is known, then can you provide references?</p>
</blockquote>
| Richard Stanley | 2,807 | <p>A simple variant of the "transformation fondamentale" of Rényi and
of Foata-Schützenberger does the trick. Write a permutation $\sigma$ in
disjoint cycle form, with the smallest element of each cycle first,
and the cycles arranged in decreasing order of the smallest element,
e.g., $(7,8)(5,6,9)(3)(1,4,2)$. Erase the parentheses to get another
permutation $\hat{\sigma}$, written as a word, e.g., $785693142$. This gives
a bijection $\mathfrak{S}_n\to\mathfrak{S}_n$ with the property that
$\sigma(i)-i=k>0$ if and only if
$\hat{\sigma}(j+1)-\hat{\sigma}(j)=k$, where $\hat{\sigma}(j)=i$. </p>
|
287,947 | <p>For example, $\sqrt 2 = 2 \cos (\pi/4)$, $\sqrt 3 = 2 \cos(\pi/6)$, and $\sqrt 5 = 4 \cos(\pi/5) + 1$. Is it true that any integer's square root can be expressed as a (rational) linear combinations of the cosines of rational multiples of $\pi$?</p>
<p>Products of linear combinations of cosines of rational multiples of $\pi$ are themselves such linear combinations, so it only needs to be true of primes. But I do not know, for example, a representation of $\sqrt 7$ in this form.</p>
| David E Speyer | 297 | <p>Someone should actually record the formula. If $p$ is a prime $\equiv 1 \bmod 4$, then
$$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \cos \frac{2 k \pi}{p}$$
where $\left( \tfrac{k}{p} \right)$ is the quadratic residue symbol. Note that $\left( \tfrac{k}{p} \right) = \left( \tfrac{p-k}{p} \right)$, so every term appears twice. </p>
<p>Similarly, if $p \equiv 3 \bmod 4$, then
$$\sqrt{p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \sin \frac{2 k \pi}{p}.$$
Again, $k$ and $p-k$ make the same contribution. </p>
<p>These are usually both written together as
$$\sqrt{(-1)^{(p-1)/2} p} = \sum_{k=1}^{p-1} \left( \frac{k}{p} \right) \exp \frac{2 k \pi i}{p}.$$
This is a formula of <a href="https://en.wikipedia.org/wiki/Quadratic_Gauss_sum" rel="noreferrer">Gauss</a>.</p>
|
50,521 | <p>I would like to know if there are some open mathematical problems in General Relativity, that are important from the point of view of Physics. </p>
<p>Is there something that still needs to be justified mathematically in order to have solid foundations? </p>
| timur | 2,473 | <p>There are some problems related to initial data, such as if gravity can be "insulated", or if there exist generic set of spacetimes which do not admit CMC spacelike hypersurface.</p>
|
158,810 | <p>Let $ (\hat i, \hat j, \hat k) $ be unit vectors in Cartesian coordinate and $ (\hat e_\rho, \hat e_\theta, \hat e_z)$ be on spherical coordinate.
Using the relation, $$ \hat e_\rho = \frac{\frac{\partial \vec r}{\partial \rho}}{ \left | \frac{\partial \vec r}{\partial \rho} \right |}, \hat e_\theta = \frac{\frac{\partial \vec r}{\partial \theta}}{ \left | \frac{\partial \vec r}{\partial \theta} \right |}, \;\; \hat e_z = \frac{\frac{\partial \vec r}{\partial z}}{ \left | \frac{\partial \vec r}{\partial z} \right |} $$
We have the relation $$\begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix}$$
$$\text { Let } A = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}$$
To express unit vectors of Cartesian coordinate in Spherical coordinates, the author uses,
$$\hat i = \frac { \begin{vmatrix} \hat e_{\rho} & \sin \phi & 0\\ \hat e_{\theta} & \cos \phi & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\ \hat j = \frac { \begin{vmatrix} \cos \phi & \hat e_{\rho} & 0\\ -\sin \phi & \hat e_{\theta} & 0\\ 0 & 0 & 1 \end{vmatrix} }{|A|} \\\hat k = \frac { \begin{vmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & \hat e_z \end{vmatrix} }{|A|}
$$
Which I cannot understand! Can anyone help me to understand it?
$$ \begin{bmatrix} \hat i\\ \hat j\\ \hat k \end{bmatrix} = \begin{bmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0\\ 0 & 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} \hat e_{\rho}\\ \hat e_{\theta}\\ \hat e_{z} \end{bmatrix} $$
Looks intuitive but certainly the previous way is faster. I would like to know above relation works if it works.
Thank you!!</p>
| Valentin | 31,877 | <p>In fact unit vectors are components of the <em>determinant</em>, not the <em>matrix</em> $A$. There is nothing wrong with it. Determinant is really an antisymmetric linear form, so you still have vector quantities on both sides of the relation.
<strong>EDIT</strong>
After a closer look the formulae do not seem entirely correct. According to Cramer's rule the numerator must be look like the system determinant with one column replaced by the RHS:</p>
<p>$$\hat i = \frac { \begin{vmatrix} \hat e_{\rho} & \sin \theta & 0\\ \hat e_{\theta} & \cos \theta & 0\\ \hat e_z & 0 & 1 \end{vmatrix} }{|A|} \\ \hat j = \frac { \begin{vmatrix} \cos \theta & \hat e_{\rho} & 0\\ -\sin \theta & \hat e_{\theta} & 0\\ 0 & \hat e_z & 1 \end{vmatrix} }{|A|} \\\hat k = \frac { \begin{vmatrix} \cos \theta & \sin \theta & \hat e_{\rho}\\ -\sin \theta & \cos \theta & \hat e_{\theta}\\ 0 & 0 & \hat e_z \end{vmatrix} }{|A|}
$$
Which computationally lead to the same result. The symbol for polar angle should be consistent throughout.</p>
|
66,951 | <p>I am asked to find all rows in a matrix in reduced row echelon form which contain nothing but pivots (pivot is $1$, all other entries are $0$).</p>
<p>For example, in this matrix:</p>
<p>$$
\begin{bmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\sim
\begin{bmatrix}
\color{red}1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & \color{red}1
\end{bmatrix}
$$
the rows whose pivots are marked in red are such rows.</p>
<p>I wrote the following code:</p>
<pre><code>matrix = {{1, 1, 1, 1}, {0, 1, 1, 0}, {0, 0, 0, 1}}; (* Same example. *)
reduced = RowReduce[matrix]
For[i = 1, i <= Length[reduced], ++i,
row = reduced[[i]];
onlyPivot = False;
Clear[pivot];
For[j = 1, j <= Length[row], ++j,
If[And[row[[j]] != 0, Not[ValueQ[pivot]]],
pivot = row[[j]];
onlyPivot = True,
If[And[row[[j]] != 0, ValueQ[pivot]],
onlyPivot = False
]
];
];
Print[onlyPivot];
];
</code></pre>
<p>As you can notice, this is very... C-like (at least it works), and probably very inefficient. Is there a better way to do this in Mathematica? What should I be looking into?</p>
| Silvia | 17 | <h1>Update</h1>
<p>According to <a href="https://mathematica.stackexchange.com/users/3246/kennycolnago">KennyColnago</a>'s advice, post-processing is not needed, as <code>StreamColorFunction</code> can handle it essentially by using <code>VertexColors</code> on <code>Line</code>-s:</p>
<pre><code>ListStreamPlot[
testdata,
StreamPoints -> {samplePoints, Automatic, 10},
StreamStyle -> "Line",
Background -> Black,
StreamColorFunction -> (GrayLevel[1, #5] &)
]
</code></pre>
<h1>Original</h1>
<p>With <a href="https://mathematica.stackexchange.com/a/15911/17">Simon Woods' neat trick</a> and some post-processing:</p>
<pre><code>testdata = Table[{{x, y}, RandomReal[{-1, 1}, 2]}, {x, -3, 3, 1}, {y, -3, 3, 1}] // Flatten[#,1]&;
samplePoints = Tuples[Range[-3, 3, 1/5], 2];
ListStreamPlot[
testdata,
StreamPoints -> {samplePoints, Automatic, 10},
StreamStyle -> "Line",
Background -> Black
] //
# /. ln : Line[idxes_] :>
MapIndexed[
{
GrayLevel[1, 1 - Rescale[#2[[1]], {1, Length[idxes] - 1}]],
Line[#1]
} &,
Partition[idxes, 2, 1]
] &
</code></pre>
<p><img src="https://i.stack.imgur.com/GUhAw.png" alt="random wind map"></p>
<p>Note for a more meaningful graphics, one might want to determine the <code>GrayLevel</code> according to the local wind scale.</p>
|
46,631 | <p>I'm writing a program to play a game of <a href="http://en.wikipedia.org/wiki/Pente" rel="noreferrer">Pente</a>, and I'm struggling with the following question:</p>
<blockquote>
<p>What's the best way to detect patterns on a two-dimensional board?</p>
</blockquote>
<p>For example, in Pente a pair of neighboring stones of the same color can be captured when they are flanked from both sides by an opponent; how can we find all the stones that can be captured with the next move for the following board?</p>
<p><img src="https://i.stack.imgur.com/cV3Gb.png" alt="sample board"></p>
<p>Below I show one possible straightforward solution, but with a defect: it's hard to extend it for other interesting patterns, i.e. three stones of the same color in a row surrounded by empty spaces, or four stones of the same color in a row which are flanked from one side but open from another, etc.</p>
<blockquote>
<p>I'm wondering whether there is a way to define a DSL for detecting 2-dimensional structures like that on a board - sort of a <em>2D pattern matching</em>.</p>
</blockquote>
<p>P.S. I would also appreciate any advice on how to simplify the code below and make it more idiomatic - for example, I don't really like the way how <code>sortStones</code> is defined.</p>
<h2>Straightforward solution</h2>
<p>Here is one way to solve this problem (see below for graphics primitives to generate and display random boards):</p>
<ul>
<li>Enumerate all subsets of 3 stones from the board above</li>
<li>Select those that form an <em>AABE</em> or <em>ABBE</em> pattern, where E denotes an unoccupied space</li>
</ul>
<p>Lets store the board as a list of black and white stones,</p>
<pre><code>a = {black[2, 1], black[4, 3], black[2, 5], black[4, 2], black[5, 3],
black[1, 2], black[1, 3], black[5, 4], black[1, 5], white[3, 1],
white[4, 1], white[4, 4], white[3, 5], white[3, 4], white[5, 1],
white[5, 2], white[3, 3], white[1, 1]}
</code></pre>
<p>First, we define <code>isTriple</code> which checks whether three stones sorted by their x and y coordinates are in the same row next to each other and follow an ABB or AAB pattern:</p>
<pre><code>isTriple[{a_, b_, c_}] := And[
(* A A B or A B B *)
Head[a] != Head[c] /. {black -> 1, white -> 0},
(* x and y coordinates are equally spaced *)
a[[1]] - b[[1]] == b[[1]] - c[[1]],
a[[2]] - b[[2]] == b[[2]] - c[[2]],
(* and are next to each other *)
Abs[a[[1]] - b[[1]]] <= 1,
Abs[a[[2]] - b[[2]]] <= 1]
</code></pre>
<p>Next, we determine the coordinates and the color of the stone that will kill the pair:</p>
<pre><code>killerStone[{a_, b_, c_}] :=
If[Head[a] == Head[b] /. {black -> 1, white -> 0},
Head[c][2 a[[1]] - b[[1]], 2 a[[2]] - b[[2]]],
Head[a][2 c[[1]] - b[[1]], 2 c[[2]] - b[[2]]]]
</code></pre>
<p>Finally, we only select those triples where killer stone's space is not already occupied:</p>
<pre><code>sortStones[l_] :=
Sort[l, OrderedQ[{#1, #2} /. {black -> List, white -> List}] &]
triplesToKill[board_] := Module[
{triples = Select[sortStones /@ Subsets[board, {3}], isTriple]},
Select[triples,
Block[
{ks = killerStone[#]},
FreeQ[board, _[ks[[1]], ks[[2]]]]] &]]
displayBoard[a, #] & /@ triplesToKill[a] //
Partition[#, 3, 3, {1, 1}, {}] & // GraphicsGrid
</code></pre>
<p><img src="https://i.stack.imgur.com/QHj2c.png" alt="straightforward solution"></p>
<h2>Graphics primitives</h2>
<pre><code>randomPoints[n_] := RandomSample[Block[{nn = Ceiling[Sqrt[n]]},
Flatten[Table[{i, j}, {i, 1, nn}, {j, 1, nn}], 1]], n];
(* n is number of moves = 2 * number of points *)
randomBoard[n_] := Module[
{points = randomPoints[2 n]},
Join[
Take[points, n] /. {x_, y_} -> black[x, y],
Take[points, -n] /. {x_, y_} -> white[x, y]
]]
grid[minX_, minY_, maxX_, maxY_] :=
Line[Join[
Table[{{minX - 1.5, y}, {maxX + 1.5, y}}, {y, minY - 1.5, maxY + 1.5,
1}],
Table[{{x, minY - 1.5}, {x, maxY + 1.5}}, {x, minX - 1.5, maxX + 1.5,
1}]]];
displayBoard[board_] := Module[
{minX = Min[First /@ board], maxX = Max[First /@ board],
minY = Min[#[[2]] & /@ board], maxY = Max[#[[2]] & /@ board], n},
Graphics[{
grid[minX, minY, maxX, maxY],
board /. {
black[n__] -> {Black, Disk[{n}, .4]},
white[n__] -> {Thick, Circle[{n}, .4], White, Disk[{n}, .4]}
}}, ImageSize -> Small, Frame -> True]];
displayBoard[board_, points_] := Show[
displayBoard[board],
Graphics[
Map[{Red, Disk[{#[[1]], #[[2]]}, .2]} &, points]]]
</code></pre>
| george2079 | 2,079 | <p>This may be a bit un-mathematicaesque, but it turns out to be convenient to store the board as a flat vector:</p>
<p>(larger board for illustration)</p>
<pre><code> n = 12;
board0 = Flatten[ Table[0, {n^2}], 1];
v[icol_, jrow_] = icol + n (jrow - 1);
</code></pre>
<p>Now we can create lists of indices representing structures such as rows,columns, and diagonals. Here the function <code>diag</code> returns a list of the indices in the flat vector along each of the 8 directions in order away from a given row,column position:</p>
<pre><code> diag[icol_, jrow_, p_, q_] :=
Table[ (icol + p (k - 1) + n (jrow + q (k - 1) - 1)),
{k, Min[
((1 - n (p - 2)) (p + 1))/2 - p icol,
((1 - n (q - 2)) (q + 1))/2 - q jrow]}];
diag[ipos_, p_, q_] :=
diag[Mod[ipos - 1, n] + 1 , Floor[(ipos - 1)/n] + 1, p, q];
alldir = Cases[Tuples[{-1, 0, 1}, 2], Except[{0, 0}]];
</code></pre>
<p>manipulator illustrating how <code>diag</code> works</p>
<pre><code> Manipulate[
board = board0;
MapIndexed[ ((board[[#[[1]]]] =
Table[#[[2]], {Length[#[[1]]]}]) &@
{diag[col, row, Sequence @@ #], First@#2}) & , alldir ];
board[[v[col, row]]] = "X";
Partition[ board , n] // MatrixForm,
{{col, 3}, 1, n, 1}, {{row, 3}, 1, n, 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/rHAeL.png" alt="enter image description here"></p>
<p>now a random board, with 0-> empty, 1-> Red , -1->Black</p>
<pre><code> n = 6
board1 =Table[ RandomChoice[{-1, 0, 0, 1}], {n^2}];
GraphicsGrid[
Partition[
Graphics[{Switch[#, 1, Red, -1, Black, 0, White], Disk[{0, 0}],
Black, Circle[{0, 0}]}] & /@ board1 , n]]
</code></pre>
<p><img src="https://i.stack.imgur.com/Yjahq.png" alt="enter image description here"></p>
<p>now find all empty positions and search over all adjacent rows,columns,diagonals for the desired pattern:</p>
<pre><code> open = Flatten[Position[board1, 0]];
hits = Last@
Reap[ Function[{dir},
If[ MatchQ[board1[[d = diag[#, Sequence @@ dir]]] ,
{0, x_ /; x != 0, x_, y_ /; y != 0, ___} /; x != y],
Sow[d[[;; 4]]]]] /@ alldir & /@ open ];
GraphicsGrid[
Partition[
Graphics[{Switch[#, 1, Red, -1, Black, 0, White, 2, Green],
Disk[{0, 0}], Black, Circle[{0, 0}]}] & /@
MapIndexed[
If[Count[ (First@hits)[[;; , 1]] , First@#2] == 1, 2, #] &, board1] , n]]
</code></pre>
<p><img src="https://i.stack.imgur.com/qBQWW.png" alt="enter image description here"></p>
<p>just for fun a <a href="http://en.wikipedia.org/wiki/Reversi" rel="noreferrer">reversi</a> simulation (pattern is different from Pente)</p>
<pre><code> h = 5; n = 2 h; board1 = Table[0, {n^2}];
board1[[{(h - 1) n + h, (h - 1) n + h + 1, h n + h, h n + h + 1}]] = {1, -1, -1, 1};
pb = GraphicsGrid[Partition[ Graphics[
{Switch[#, 1, Red, -1, Black, 0, White, 2, LightRed, -2 , Gray],
Disk[{0, 0}], Black, Circle[{0, 0}]}] & /@ # , n]] &;
up = 1; down = -1;
First@Last@Reap[
Sow[pb@board1 ];
While[0 < Length[
{up, down} = {down, up};
hits = Select[ Union@Flatten[Last@Reap[Function[{dir},
If[ MatchQ[
bb = board1[[d = diag[#, Sequence @@ dir]]] ,
{0, down .., up, ___}],
Sow[d[[;; First@First@Position[bb, up]]]]]] /@
alldir ]] &
/@ Flatten[Position[board1, 0]] , # != {} &] ],
board1[[choice = RandomChoice[(Length /@ hits) -> hits]]] = 2 up;
Sow[gg = pb@board1 ];
board1[[choice]] = up]]
</code></pre>
<p><img src="https://i.stack.imgur.com/q1fX4.gif" alt="enter image description here"></p>
|
394,085 | <p>How is it possible to establish proof for the following statement?</p>
<p>$$n = \frac{1}{2}(5x+4),\;2<x,\;\text{isPrime}(n)\;\Rightarrow\;n=10k+7$$</p>
<p>Where $n,x,k$ are $\text{integers}$.</p>
<hr>
<p>To be more verbose:</p>
<p>I conjecture that;</p>
<p>If $\frac{1}{2}(5x+4),\;2<x$ is a prime number, then $\frac{1}{2}(5x+4)=10k+7$</p>
<p>How could one prove this?</p>
| Alex R. | 22,064 | <p>If $n$ is prime , clearly $x$ must be even and moreover $x$ has only one power of 2, eg $x=2y$ where $y$ is odd, $y=2z+1$. Thus we have </p>
<p>$$n\equiv 5y+2 \pmod{10}\equiv 5+2\equiv 7$$ </p>
|
425,969 | <p>It seems striking that the cardinalities of <span class="math-container">$\aleph_0$</span> and <span class="math-container">$\mathfrak c = 2^{\aleph_0}$</span> each admit what I will call a "homogeneous cyclic order", via the examples of <span class="math-container">$ℚ/ℤ$</span> and <span class="math-container">$ℝ/ℤ$</span>. By which I mean a cyclic order (as defined in <a href="https://ncatlab.org/nlab/show/cyclic+order" rel="nofollow noreferrer">https://ncatlab.org/nlab/show/cyclic+order</a>) such that for any two elements <span class="math-container">$x, y$</span> of the cardinal, there is a bijection of the cardinal to itself taking <span class="math-container">$x$</span> to <span class="math-container">$y$</span> and preserving the cyclic order.</p>
<p>In ZFC, is there any reason to believe that either a) all infinite cardinals admit a homogeneous cyclic order, or b) there exists an infinite cardinal admitting no homogeneous cyclic order?</p>
| Alessandro Codenotti | 49,381 | <p>While Andreas's answer is a very simple construction let me point out that one can get even more.</p>
<p><strong>Theorem:</strong> for every cardinal <span class="math-container">$\lambda$</span> there exists a ultra-homogeneous circularly ordered set <span class="math-container">$(X,R)$</span> with <span class="math-container">$|X|=\lambda$</span> (ultra-homogeneous here means that any isomorphism between finite substructures of <span class="math-container">$X$</span> extends to an automorphism of <span class="math-container">$X$</span>).</p>
<p>This has been noted independentently by Pestov and Truss, but neither of them published it. It was published, with Pestov's proof, by Glasner and Megrelishvili in <em>Circular orders, ultra-homogeneous order structures and
their automorphism groups</em>, the construction below is due to Pestov and taken from the paper just mentioned.</p>
<p><strong>Proof:</strong> Let <span class="math-container">$k$</span> be a linearly ordered field of size <span class="math-container">$\lambda$</span> (to construct one start with <span class="math-container">$\Bbb Q$</span> and do a tower of transcendental extensions, for <span class="math-container">$\alpha<\lambda$</span>. At successor stages <span class="math-container">$\Bbb Q_{\alpha+1}=\Bbb Q_\alpha(x_\alpha)$</span> ordered in the usual way that makes <span class="math-container">$x_\alpha$</span> a positive infinitesimal over <span class="math-container">$\Bbb Q_\alpha$</span>, while at limit stages take the union of the previous extensions). Now for <span class="math-container">$x\in k$</span> let <span class="math-container">$|x|=\max\{-x,x\}$</span> and let <span class="math-container">$\mathrm{fin}(k)=\{x\in k\mid \exists n\in\Bbb Z(|x|<n)\}$</span> (note that <span class="math-container">$k$</span> has characteristic zero being ordered). Now we mimic the <span class="math-container">$\Bbb R/\Bbb Z$</span> construction by considering <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span>. We set <span class="math-container">$(a,b,c)\in R$</span> if and only if there are <span class="math-container">$a'\leq b'\leq c'$</span> with <span class="math-container">$a=a'+\Bbb Z$</span>, <span class="math-container">$b=b'+\Bbb Z$</span> and <span class="math-container">$c=c'+\Bbb Z$</span>.<br />
It remains to check that <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span> is ultra-homogeneous. Let <span class="math-container">$a_1,\ldots,a_n$</span> and <span class="math-container">$b_1,\ldots,b_n$</span> be two positively circularly ordered subsets of <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span>, meaning that whenever <span class="math-container">$[i,j,k]$</span> in the circularly ordered finite group <span class="math-container">$\Bbb Z_k$</span>, one has <span class="math-container">$[a_i,a_j,a_k]$</span> and <span class="math-container">$[b_i,b_j,b_k]$</span> in <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span>. Wlog assume <span class="math-container">$a_1=b_1=0$</span>. Identifying <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span> with the interval <span class="math-container">$[0,1)$</span> in <span class="math-container">$k$</span> we get representatives <span class="math-container">$0=a_1'<a_2'<\ldots<a_n'<1$</span> and <span class="math-container">$0=b_1'<b_2'<\ldots<b_n'<1$</span>. Now apply a piecewise linear transformation <span class="math-container">$[0,1)_k\to[0,1)_k$</span> mapping <span class="math-container">$a_i'$</span> to <span class="math-container">$b_i'$</span> for every <span class="math-container">$i$</span>, and lift it to an automorphism of <span class="math-container">$\mathrm{fin}(k)/\Bbb Z$</span> mapping <span class="math-container">$a_i$</span> to <span class="math-container">$b_i$</span> for every <span class="math-container">$i$</span>.</p>
|
1,181,631 | <p>Let $f : \mathbb R \to \mathbb R$ continuous. Prove that graph $G = \{(x, f(x)) \mid x \in \mathbb R\}$ is closed.</p>
<p>I'm a little confused on how to prove $G$ is closed. I get the general strategy is to show that every arbitrary convergent sequence in $G$ converges to a point in $G$.</p>
<p>Here is what I tried so far:</p>
<ol>
<li>Let $x_k$ be a sequence which converges to $x$.</li>
<li>Since $f$ is continuous, this implies that $f(x_k)$ converges to $f(x)$.</li>
<li>At this point, can you say every $(x_k, f(x_k))$ converges to a $(x, f(x))$, so $G$ is closed?</li>
</ol>
| Gregory Grant | 217,398 | <p>Almost, but better to start the argument with a general convergent sequence $(x_n,y_n)\rightarrow (x,y)$ and then write $y_n$ as $f(x_n)$ to show $y_n$ converges to $f(x)$. Then conclude $(x_n,y_n)$ converges to $(x,f(x))$ which is in $G$ and therefore an arbitrary sequence in $G$ that converges, converges in $G$. Therefore $G$ is closed.</p>
|
1,529,324 | <p>I've read that if $\Phi$ is a Poisson point process (on $\mathbb{R}^d$, say), then conditional on there being $k$ points in some $A \subseteq \mathbb{R}^d$, the positions $X_1,\ldots,X_k$ of these points are uniformly distributed in $A$.</p>
<p>I'm having trouble making sense of what this means. "Conditional on $\Phi(A)=k$ I guess means consider the process $\Phi 1_{\Phi(A)=k}$ and then divide probabilities by $P(\Phi(A)=k)$. But, probabilities of what exactly? How am I labeling the points $X_1,\ldots,X_k$? In $\mathbb{R}$ If I did so by $X_1< X_2 < \cdots < X_k$ then clearly they are not uniformly distributed, so clearly the way that I label them matters. Hence my question, what is meant by saying $X_1,\ldots,X_k$ are uniformly distributed? </p>
| MaxW | 23,782 | <blockquote>
<blockquote>
<p>I've read that if $\Phi$ is a Poisson point process (on $\mathbb{R}^d$, say), then conditional on there being $k$ points in some $A \subseteq \mathbb{R}^d$, the positions $X_1,\ldots,X_k$ of these points are uniformly distributed in $A$.</p>
</blockquote>
</blockquote>
<p>$\Phi$ is a process which generates iid distributed points in $\mathbb{R}^d$, then if $A \subseteq \mathbb{R}^d$ with $k$ points, then the $k$ points within $A$ are uniformly distributed in $A$.</p>
|
76,753 | <p>I am having a hard time getting to factor this binomial: I have tried other methods but they do not seem to work... ah well.
$$4m^2-\frac{9}{25}.$$</p>
<p>Thanks.</p>
| picakhu | 4,728 | <p>I assume this is homework, so only ideas from me here. </p>
<p>Note that the formula is of the form $A^2-B^2$.</p>
<p>and we know that $A^2-B^2$ can be factored easily. </p>
|
918,788 | <p>How to do this integral</p>
<p>$$\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x$$</p>
<p>for any $k > 0$ ?.</p>
<p>I tried to use gamma function, but sometimes the series doesn't converge.</p>
| gaoxinge | 94,751 | <p>We assume
$$F(k)=\int_{-\infty}^{\infty}e^{-x^2}\cos kxdx$$
Consider $F'(k)$, we have
$$F'(k)=\int_{-\infty}^{\infty}-xe^{-x^2}\sin kxdx$$
$$=\frac{1}{2}(e^{-x^2}\sin kx|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}ke^{-x^2}\cos kxdx)$$
$$=-\frac{1}{2}kF(k)$$
Then we solve the ordinary differential equation with $F(0)=\sqrt\pi$, and we get
$$F(k)=\sqrt\pi e^{\frac{-k^2}{4}}$$</p>
|
3,511,118 | <p>I can't see how <span class="math-container">$$e^\left(2i\pi\right) = 1$$</span>
will result in:
<span class="math-container">$$e^\left(i\pi\right) +1 = 0$$</span>
thanks</p>
| Fred | 380,717 | <p>If <span class="math-container">$f$</span> is continuous, then</p>
<p><span class="math-container">$$\frac{d}{d x} \int_{3}^{x} f(t)^{2} d x= f(x)^2,$$</span></p>
<p>by the Fundamental Theorem of Calculus. </p>
|
2,772,895 | <p>I've noticed one classical way of defining certain topologies is to define them as the "weakest" (or coarsest) topology such that a certain set of functions is continuous. For example,</p>
<blockquote>
<p>The <strong>product topology</strong> on <span class="math-container">$X=\prod X_i$</span> is the weakest topology such that the canonical projections <span class="math-container">$p_i : X\to X_i$</span> are continuous.</p>
<p>The <strong>weak*-topology</strong> on a Banach space <span class="math-container">$X$</span> is the weakest topology such that the <em>evaluation map</em> (the natural isomorphism from <span class="math-container">$X$</span> to <span class="math-container">$X^{**}$</span>, <span class="math-container">$J(x)(\phi) = \phi(x)$</span>) is continuous.</p>
</blockquote>
<p>I have a very poor intuition behind what these "mean". When I look at the definition of a topology, the ones that make the most sense to me are the ones in which the open sets (or at least a base of open sets) are explicitly constructed, as in the Euclidean topology (or a general metric space topology, or a norm induced topology).</p>
<p>I get a little stuck when the definition of the topology is given in some "abstract" sense, where the open sets are "chosen" to satisfy a certain other property. How am I supposed to visualize the open sets in these spaces, or work with them?</p>
<p>If <span class="math-container">$\tau$</span> is the weakest topology on <span class="math-container">$X$</span> such that <span class="math-container">$f : X\to Y$</span> is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets in <span class="math-container">$Y$</span> under <span class="math-container">$f$</span>? This follows directly from the definition of a continuous function on topological spaces. Is this always the coarsest topology?</p>
<p>Furthermore, what benefit do these topologies provide? What interesting, and potentially theoretically useful, properties do they possess? Why should I care about them?</p>
| Martin Argerami | 22,857 | <blockquote>
<p>If $τ$ is the weakest topology on $X$ such that $f:X→Y$ is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets under $Y$? This follows directly from the definition of a "continuous" function. Is this always the coarsest topology?</p>
</blockquote>
<p>Yes. For $f$ to be continuous, you need the topology on $X$ to contain all preimages of open sets through $f$. The topology induced by a family $\mathcal T$ of functions is generated by
$$
\{f^{-1}(E):\ f\in\mathcal T,\ E\subset Y\ \text{ open }\}.
$$
Some reasons why one cares about these topologies are</p>
<ul>
<li><p>They often appear naturally, as when considering duals and preduals of normed spaces;</p></li>
<li><p>In several cases the topology is coarse enough that some interesting sets become compact (for instance the unit ball in a Banach space, see the <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem" rel="noreferrer">Banach-Alaoglu Theorem</a>). </p></li>
</ul>
|
2,111,833 | <p>Let $A=(a_{ij})\in M_n$ be an arbitrary matrix and let
$A_1=\begin{pmatrix}
a_{11}\\
a_{21}\\
\vdots\\
a_{n1}\\
\end{pmatrix}$
$A_2=\begin{pmatrix}
a_{12}\\
a_{22}\\
\vdots\\
a_{n2}\\
\end{pmatrix}\ldots$
$A_n=\begin{pmatrix}
a_{1n}\\
a_{2n}\\
\vdots\\
a_{nn}\\
\end{pmatrix}\in M_{n1}$ be columns of $A$. Prove that if the set$\{A_1,A_2,...,A_n\} $ is linearly dependent in vector space $M_{n1}$, then $\det A=0$.</p>
<p>I know this already has an answer <a href="https://math.stackexchange.com/questions/1387075/proving-that-deta-0-when-the-columns-are-linearly-dependent">here</a> but I don't understand OP's solution. </p>
<p>$\lambda_1 A_1 + \ldots + \lambda_n A_n = 0$ where not all $\lambda_i$ are zero.
Suppose that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = - \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n. \end{align*}</p>
<p>Now what happens after that, with the determinant? </p>
| egreg | 62,967 | <p>Since exchanging two columns only switches sign to the determinant, it is not restrictive to assume that the last column is a linear combination of the previous $n-1$ columns:
$$
A_n=\alpha_1A_1+\dots+\alpha_{n-1}A_{n-1}
$$
By multilinearity of the determinant, you have
$$
\det A=
\det\begin{bmatrix} A_1 & \dots & A_{n-1} &
\sum\limits_{i=1}^{n-1}\alpha_iA_i\end{bmatrix}=
\sum_{i=1}^{n-1}\alpha_i
\det\begin{bmatrix} A_1 & \dots & A_{n-1} & A_i\end{bmatrix}=0
$$
because a matrix with two equal columns has zero determinant again by the above mentioned property above that exchanging two columns changes the sign of the determinant.</p>
<p>With $\begin{bmatrix} v_1 & \dots & v_{n-1} & v_n\end{bmatrix}$ I denote the matrix whose columns are the column vectors $v_1,\dots,v_{n-1},v_n$.</p>
|
2,111,833 | <p>Let $A=(a_{ij})\in M_n$ be an arbitrary matrix and let
$A_1=\begin{pmatrix}
a_{11}\\
a_{21}\\
\vdots\\
a_{n1}\\
\end{pmatrix}$
$A_2=\begin{pmatrix}
a_{12}\\
a_{22}\\
\vdots\\
a_{n2}\\
\end{pmatrix}\ldots$
$A_n=\begin{pmatrix}
a_{1n}\\
a_{2n}\\
\vdots\\
a_{nn}\\
\end{pmatrix}\in M_{n1}$ be columns of $A$. Prove that if the set$\{A_1,A_2,...,A_n\} $ is linearly dependent in vector space $M_{n1}$, then $\det A=0$.</p>
<p>I know this already has an answer <a href="https://math.stackexchange.com/questions/1387075/proving-that-deta-0-when-the-columns-are-linearly-dependent">here</a> but I don't understand OP's solution. </p>
<p>$\lambda_1 A_1 + \ldots + \lambda_n A_n = 0$ where not all $\lambda_i$ are zero.
Suppose that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = - \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n. \end{align*}</p>
<p>Now what happens after that, with the determinant? </p>
| Bernard | 202,857 | <p>I'll give a variant of the proof, hoping you'll understand better.</p>
<p>Suppose there's a non-trivial linear relation between the columns:
$$\lambda_1A_1+\lambda_2A_2+\dots+\lambda_nA_n=0.$$
Say $\lambda_1\ne 0$. By linearity w.r.t. the 1st column, $\;\det(
\lambda_1A_1,A_2,\dots,A_n)=\lambda_1\det(A_1,A_2,\dots,A_n)$. Also, the determinant is <code>alternating</code> and <code>linear</code> in each column, so
$$ \lambda_1\det A=\det(\lambda_1A_+\lambda_2A_2+\dots+\lambda_nA_n,A_2,\dots,A_n)= \det(0,A_2,\dots,A_n)=0,$$
whence $\det A=0$.</p>
|
2,111,833 | <p>Let $A=(a_{ij})\in M_n$ be an arbitrary matrix and let
$A_1=\begin{pmatrix}
a_{11}\\
a_{21}\\
\vdots\\
a_{n1}\\
\end{pmatrix}$
$A_2=\begin{pmatrix}
a_{12}\\
a_{22}\\
\vdots\\
a_{n2}\\
\end{pmatrix}\ldots$
$A_n=\begin{pmatrix}
a_{1n}\\
a_{2n}\\
\vdots\\
a_{nn}\\
\end{pmatrix}\in M_{n1}$ be columns of $A$. Prove that if the set$\{A_1,A_2,...,A_n\} $ is linearly dependent in vector space $M_{n1}$, then $\det A=0$.</p>
<p>I know this already has an answer <a href="https://math.stackexchange.com/questions/1387075/proving-that-deta-0-when-the-columns-are-linearly-dependent">here</a> but I don't understand OP's solution. </p>
<p>$\lambda_1 A_1 + \ldots + \lambda_n A_n = 0$ where not all $\lambda_i$ are zero.
Suppose that $\lambda_1 \neq 0$. Then we get \begin{align*} A_1 = - \frac{\lambda_2}{\lambda_1} A_2 - \ldots - \frac{\lambda_n}{\lambda_1} A_n. \end{align*}</p>
<p>Now what happens after that, with the determinant? </p>
| user646399 | 646,399 | <p>The determinant of <span class="math-container">$A$</span> is the product of the elements of the main diagonal when <span class="math-container">$A$</span> is converted to row echelon form. For a linearly dependent set of columns, when <span class="math-container">$A$</span> is converted to row echelon form, there will be a <span class="math-container">$0$</span> in the main diagonal of the matrix corresponding to the column of the free variable and hence, the determinant is <span class="math-container">$0$</span>.</p>
|
1,734,102 | <blockquote>
<p>Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Prove that there is a constant $c$ for which $f(x) = cx$ for all $x \in \mathbb{Q}$.</p>
</blockquote>
<p>We know that $f(0) = 0$. Now set $x = n$ and $y=n$ to get $f(2n) = 2f(n)$ where $n$ is a rational number. But doesn't this condition imply all constant functions work? So we must get another condition.</p>
| Derek Allums | 17,736 | <p>Notice:
$$
f\left(\frac{m}{n}\right) = m\cdot f\left(\frac{1}{n}\right) = m\cdot \left( \frac{f(1)}{n}\right) = f(1)\cdot \frac{m}{n}
$$
where $f\left(\frac{1}{n}\right) = \frac{f(1)}{n}$ because
$$
f\left(\frac{1}{n}\right) = f\left( \frac{n}{n} - \frac{n-1}{n}\right) = f(1) - (n-1)f\left(\frac{1}{n}\right)
$$
by the linearity assumption. </p>
|
1,734,102 | <blockquote>
<p>Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Prove that there is a constant $c$ for which $f(x) = cx$ for all $x \in \mathbb{Q}$.</p>
</blockquote>
<p>We know that $f(0) = 0$. Now set $x = n$ and $y=n$ to get $f(2n) = 2f(n)$ where $n$ is a rational number. But doesn't this condition imply all constant functions work? So we must get another condition.</p>
| bgins | 20,321 | <p>You can demonstrate that for $n\in\mathbb{Z}$, $f(n)=nf(1)$ and $f(1)=nf(\frac1n)$ for $n\ne0$, so that for $x=\frac{p}{q}\in\mathbb{Q}$, $f(x)=\frac{p}{q}f(1)$, so $c=f(1)$.</p>
|
68,428 | <p>I am looking at the description of LTI systems in the time domain.</p>
<p>Intuitively, I'd have guessed it would be the composition of the input function and some "system function".
$$ y(t) = f(x(t)) = (f\circ x)(t)$$
Where $x(t)$ is the input, $y(t)$ output and $f(x)$ a "system function".</p>
<p>Why is it not that way? Could such a "system function" be found for, say, an R-C-Circuit?</p>
<p>The actual output function y(t), is defined as
$$ y(t) = (h * x)(t) $$
Where $h(t)$ is the response to a dirac impulse.
This is hard to grasp for me. Why is it so? I have looked at various explanations, drawings of rectangles becoming infinitely narrow, which I sort of understood, but it is still "hard to grasp"! I am looking for a simple explanation in one or two sentences here.</p>
<p><a href="http://en.wikipedia.org/wiki/LTI_system_theory" rel="nofollow">http://en.wikipedia.org/wiki/LTI_system_theory</a></p>
| Mike Stay | 756 | <p>Simon Willerton explains it all very well here: <a href="http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf" rel="nofollow">http://www.simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf</a></p>
|
3,207,767 | <p>What is the general solution of differential equation <span class="math-container">$y\frac{d^{2}y}{dx^2} - (\frac{dy}{dx})^2 = y^2 log(y)$</span>.</p>
<p>The answer to this DE is <span class="math-container">$log(y) = c_1 e^x + c_2 e^{-x}$</span></p>
<p>I don't know the method to solve differential equation with degree more than 1. Please tell me how to solve these types of equations.</p>
| Michael Rozenberg | 190,319 | <p>We'll prove that
<span class="math-container">$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq\frac{3\sqrt3}{4},$$</span> where the equality occurs for <span class="math-container">$a=b=c=d=\frac{1}{\sqrt3}$</span> only.</p>
<p>Indeed, let <span class="math-container">$a=\frac{x}{\sqrt3},$</span> <span class="math-container">$b=\frac{y}{\sqrt3},$</span> <span class="math-container">$c=\frac{z}{\sqrt3}$</span> and <span class="math-container">$d=\frac{t}{\sqrt3}.$</span></p>
<p>Thus, we need to prove that
<span class="math-container">$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$$</span>
Consider five cases.</p>
<ol>
<li><span class="math-container">$x\geq1\geq y\geq z\geq t$</span>.</li>
</ol>
<p>Thus, by C-S <span class="math-container">$$\prod_{cyc}(x^2+3)=(x^2+3)\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1+4)=$$</span>
<span class="math-container">$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)(z^2-1)+\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=$$</span>
<span class="math-container">$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-\prod_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)\right)\geq$$</span>
<span class="math-container">$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-(1-y^2)(1-z^2)\right)\geq$$</span>
<span class="math-container">$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=16(x^2+1+1+1)(1+y^2+z^2+t^2)\geq$$</span>
<span class="math-container">$$\geq16(x+y+z+t)^2;$$</span>
2. <span class="math-container">$x\geq y\geq1\geq z\geq t.$</span></p>
<p>Thus, by C-S again we obtain:
<span class="math-container">$$\prod_{cyc}(x^2+3)=\prod_{cyc}(4+(x^2-1))\geq(16+4(x^2+y^2-2))(16+4(z^2+t^2-2))=$$</span>
<span class="math-container">$$=16(x^2+y^2+1+1)(1+1+z^2+t^2)\geq16(x+y+z+t)^2;$$</span>
The cases </p>
<ol start="3">
<li><p><span class="math-container">$x\geq y\geq z\geq1\geq t$</span>,</p></li>
<li><p><span class="math-container">$1\geq x\geq y\geq z\geq t$</span> and</p></li>
<li><p><span class="math-container">$x\geq y\geq z\geq t\geq1$</span> for you.</p></li>
</ol>
<p>Can you end it now?</p>
|
1,684,095 | <p>How can I evaluate the following series.</p>
<p>$$\sum_{k=1}^{\infty}\frac{1}{(k+1)(k-1)!}.$$</p>
| Stefan Mesken | 217,623 | <p>A trivial example is $\operatorname{On} \cup \{*\}$ for some $* \not \in \operatorname{On}$ that is put "on top", i.e. we extend the order on $\operatorname{On}$ by letting $\alpha < *$ for all $\alpha \in \operatorname{On}$. Clearly $<$ remains to be a strict well order and its order type is "$\operatorname{On} + 1$", since $*$ has class many predecessors.</p>
|
3,747,453 | <p>Isn't it wrong to write the following with only the percent sign? Instead of <span class="math-container">$100 \%$</span>?</p>
<blockquote>
<p>The change in height as a percentage is
<span class="math-container">$$
\frac{a - b}{a} \% \tag 1
$$</span>
where <span class="math-container">$a$</span> is the initial height and <span class="math-container">$b$</span> is the final height.</p>
</blockquote>
<p>Because if <span class="math-container">$a=10$</span>, <span class="math-container">$b=5$</span> we have
<span class="math-container">$$
\frac{10-5}{10}\%=\frac{1}{2} \% = 0.5 \frac{1}{100} = 0.005
\quad \text{what?!}
\tag 2
$$</span></p>
<p>If we convert a decimal number to percent we multiply it by <span class="math-container">$100$</span> and add the percent sign. We have <span class="math-container">$1\%=\frac{1}{100}$</span>, so with <span class="math-container">$100 \%$</span> we multiply the number by <span class="math-container">$1$</span>, i.e.
<span class="math-container">\begin{align}
\frac{10-5}{10} \cdot \color{blue}{1}
&=
\frac{10-5}{10} \cdot \color{blue}{100 \%}
=
\frac{5}{10} \cdot
\color{blue}{100 \frac{1}{100}}
=\frac{1}{2} \cdot
\color{blue}{100 \frac{1}{100}} \tag 3
\\
&=0.5\cdot \color{blue}{100 \frac{1}{100}}
=50 \color{blue}{\frac{1}{100}} = 50 \color{blue}{\%} \tag 4
\end{align}</span>
So, shouldn't we instead write <span class="math-container">$(1)$</span> as
<span class="math-container">$$
\frac{a - b}{a} 100 \% \quad ? \tag 5
$$</span></p>
| Brian M. Scott | 12,042 | <p>Let <span class="math-container">$S=\wp(\Bbb N)$</span>, with symmetric difference as addition and intersection as multiplication. Let <span class="math-container">$R=T=\wp(2\Bbb N)$</span> with the same operations. Let <span class="math-container">$f:R\to S:x\mapsto x$</span>, and let <span class="math-container">$g:S\to T:x\mapsto x\cap 2\Bbb N$</span>. Then <span class="math-container">$g\circ f$</span> is just the identity map, but <span class="math-container">$g$</span> is clearly not injective.</p>
|
3,188,298 | <blockquote>
<p>Let <span class="math-container">$f: X\to Y$</span> be bijective, and <span class="math-container">$f^{-1}: Y\to X$</span> be it's inverse. If
<span class="math-container">$V\subseteq Y$</span>, show that the forward image of <span class="math-container">$V$</span> under <span class="math-container">$f^{-1}$</span> is
the same set as the inverse image of <span class="math-container">$V$</span> under <span class="math-container">$f$</span>.</p>
</blockquote>
<p>I have interpreted this as: show that <span class="math-container">$f(f^{-1}(V))=f^{-1}(f(V))$</span></p>
<p>I really do not know what to do from here.</p>
| David | 119,775 | <p>Actually, what you have to prove is that
<span class="math-container">$$f^{-1}(V)=f^{-1}(V)\ .$$</span>
To see why this is <strong>not obvious</strong>, you have to carefully unpack the meanings of all terms, noting in particular that both <span class="math-container">$f$</span> and <span class="math-container">$f^{-1}$</span> actually have multiple meanings.</p>
<p>First, the inverse image of <span class="math-container">$V$</span> under <span class="math-container">$f$</span> is by definition
<span class="math-container">$$A=\{\,x\in X\ \mid\ f(x)\in V\,\}\ .$$</span></p>
<p>Second, what is the forward image of <span class="math-container">$V$</span> under <span class="math-container">$f^{-1}$</span>? We note that since <span class="math-container">$f$</span> is a bijection, <span class="math-container">$f^{-1}$</span> is a function from <span class="math-container">$Y$</span> to <span class="math-container">$X$</span> defined by
<span class="math-container">$$f^{-1}(y)=x\quad\hbox{if and only if}\quad f(x)=y\ .$$</span>
The forward image of <span class="math-container">$V$</span> under this function is by definition
<span class="math-container">$$B=\{\,f^{-1}(v)\ \mid\ v\in V\,\}\ .$$</span></p>
<p>So, you have to prove that <span class="math-container">$A=B$</span>, and now I hope you understand why putting it as "<span class="math-container">$f^{-1}(V)=f^{-1}(V)$</span>" is so confusing!!</p>
<p>Suppose that <span class="math-container">$x\in A$</span>. Then
<span class="math-container">$$\eqalign{f(x)=v\quad \hbox{for some $v\in V$}\quad
&\Rightarrow\quad x=f^{-1}(v)\quad \hbox{for some $v\in V$}\cr
&\Rightarrow\quad x\in B\ ,\cr}$$</span>
so <span class="math-container">$A\subseteq B$</span>. You also need to show <span class="math-container">$B\subseteq A$</span>, now that you understand the problem better please try this for yourself.</p>
|
78,443 | <p>Let $F$ be a number field and $A$ an abelian variety over $F$. It is known that if $A$ has complex multiplication, then it has potentially good reduction everywhere, namely there exists a finite extension $L$ of $F$ such that $A_L$ has good reduction over every prime of $L$.</p>
<p>And what about the inverse: if $A$ is known to be of potential good reduction everywhere, how far is it from having complex multiplication?</p>
<p>As the reduction behavior is determined by the Galois representations of the decompositon groups, one can reformulate the problem as follows: let $A$ be an abelian variety over $F$, $p$ a fixed rational prime, $V$ the p-adic Tate module of $A$; and for $\lambda$ primes of $F$, $\rho_\lambda$ is the $p$-adic representation on $V$ of the decomposition group $G_\lambda$ at $\lambda$. If $\rho_\lambda$ is potentially unramified for $\lambda$ not dividing $p$, and potentially cristalline for $\lambda$ dividing $p$, do we know that the global Galois representation $\rho$ on $V$ is potentially abelian, i.e. when shifting to some open subgroup, the image of $\rho$ is contained in a torus of $GL_{\mathbb{Q}_p}(V)$? What do we know about the Fontaine-Mazur conjecture in this case?</p>
<p>Thanks!</p>
| David Roberts | 4,177 | <p>In his <a href="http://www.math.harvard.edu/~lurie/papers/SAG-rootfile.pdf" rel="nofollow">recently posted draft of <em>Spectral Algebraic Geometry</em></a>, Lurie defines Nisnevich covers for quasi-compact, quasi-separated spectral algebraic spaces as follows. </p>
<p>Consider a family of étale morphisms $\{X_\alpha \to X\}$. They generate a Nisnevich covering if there is a sequence of open immersions
$$
\emptyset = U_0 \hookrightarrow \ldots \hookrightarrow U_n \simeq X
$$
and for $i=1,\ldots,n$, the composite $K_i \hookrightarrow U_i \hookrightarrow X$ factors through some $X_\alpha \to X$, where $K_i \subset U_i$ is the reduced closed substack [DR: why substack and not subspace? I'm not sure] of $U_i$ complementary to $U_{i-1}$.</p>
<p>One can see fairly immediately that the definition reduces to that given for affine ordinary schemes in my question. Again, one can interpret this as saying that the surjective étale map $\coprod_\alpha X_\alpha \to X$ is Nisnevich is there are local sections over some locally closed subspaces $K_i$ arising from a filtration $\emptyset = U_0 \hookrightarrow \ldots \hookrightarrow U_n \simeq X$.</p>
|
3,984,930 | <p>I am studying maths purely out of interest and have come across this question in my text book:</p>
<p>A rectangular piece of paper ABCD is folded about the line joining points P on AB and Q on AD so that the new position of A is on CD. If AB = a and AD = b, where <span class="math-container">$a \ge\frac{2b}{\sqrt3}$</span>, show that the least possible area of the triangle APQ is obtained when the angle AQP is equal to <span class="math-container">$\frac{\pi}{3}$</span>. What is the significance of the condition <span class="math-container">$a \ge\frac{2b}{\sqrt3}$</span>?</p>
<p>I realise I need to get an equation for the area of the fold, which I can then differentiate. Even then, I am not sure how to relate this to angle AQP.I have looked at solutions on the internet but they tend to look at the length of the crease, not the area of the fold.</p>
<p>I have taken A in my diagram to represent its position <em>before</em> the fold. I could be mistaken.</p>
<p>This is how I visualise it:</p>
<p><a href="https://i.stack.imgur.com/y1v2x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y1v2x.png" alt="enter image description here" /></a></p>
<p>I have said:</p>
<p><span class="math-container">$x^2 = m^2 + (b - x)^2 = m^2 + b^2 - 2bx +x^2 \implies m^2 = 2b(x - \frac{b}{2})$</span></p>
<p><span class="math-container">$L^2 = (L - m)^2 + b^2 \implies L = \frac{m^2 + b^2}{2m}$</span></p>
<p><span class="math-container">$y^2 = x^2 + L^2 = x^2 + \frac{(m^2 + b^2)^2}{4m^2} = x^2 + \frac{(2bx - b^2 + b^2)^2}{4m^2} = x^2 + \frac{4b^2x^2}{8b(x - \frac{b}{2})} = \frac{bx^2}{2x - b}$</span></p>
<p>After this I am not sure how to proceed.</p>
| Will Orrick | 3,736 | <p>I think you were almost there. You don't need <span class="math-container">$y$</span> to compute the area, so you can omit that part of your calculation. So you have two variables, <span class="math-container">$x$</span> and <span class="math-container">$m$</span>, but these are related. You've solved for <span class="math-container">$m^2$</span> in terms of <span class="math-container">$x$</span>, but you can instead solve for <span class="math-container">$x$</span> in terms of <span class="math-container">$m^2$</span>. Then you can minimize <span class="math-container">$A=\frac{1}{2}xL$</span>, which is a function of the single variable <span class="math-container">$m$</span>.</p>
<p>Notice that there are natural endpoints for this minimization. Start by assuming that the piece of paper is semi-infinite. That is, let <span class="math-container">$a$</span> go to infinity so you don't have to worry about not getting a triangle when you make the fold. You should be able to see that <span class="math-container">$m$</span> must be greater than <span class="math-container">$0$</span>. (As <span class="math-container">$m$</span> approaches <span class="math-container">$0$</span>, the area of the triangle approaches infinity.) On the other hand, <span class="math-container">$m$</span> can be at most <span class="math-container">$b$</span> (since <span class="math-container">$m\le x\le b$</span>), at which point the triangle area is <span class="math-container">$\frac{1}{2}b^2$</span>. So you are minimizing on the interval <span class="math-container">$(0,b]$</span>.</p>
|
4,246,048 | <p>As I understand it, Cantor defined two sets as having the same cardinality iff their members can be paired 1-to-1. He applied this to infinite sets, so ostensibly the integers (Z) and the even integers (E) have the same cardinality because we can pair each element of Z with exactly one element of E.</p>
<p>For infinite sets, this definition seems problematic no matter which direction we come at it from: We don't know up front that two infinite sets have the same cardinality, so we cannot conclude that their elements can be <em>exactly</em> paired. And we do not know up front that two infinite sets' elements can be paired up exactly (because we don't know with certainty what happens beyond the finite cases we can verify). So we cannot conclude that their cardinalities are the same. The definition above therefore seems useless, since we cannot start from either side of the "iff".</p>
<p>It might be argued that if we state it as follows: "For each element e of E, pair it with element e/2 of Z," then we have expressed the general case symbolically, and it works. But we can only verify that for finite values of E and Z. We can't know what happens beyond finite elements of those sets. So expressing it symbolically does not seem to help.</p>
<p>Why is Cantor's definition not circular and therefore useless for deciding the question of infinite set cardinalities?</p>
| Bertrand Einstein IV | 824,733 | <p>You are approaching cardinality and infinity in the wrong way. For two finite sets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> the sets are of the same size if and only if there exists a bijection between the two sets. This is somewhat obvious, and the proof is done in the first few weeks of any discrete math or set theory course.</p>
<p>Of course we cannot think about infinite sets in the exact same way, but we use the same idea of a bijection to identify "families" or "classes" of infinite sets, and oftentimes we say that these equivalent classes of infinities have the same size; although that is not entirely accurate it gets the point across that these infinities exist within the same group, and share attributes.</p>
<p>As someone commented there is a clear bijection between integers and even numbers, you can try find a bijection between integers and natural numbers, or natural numbers and fractions on your own time to indeed convince yourself that these infinities are in the same "family" as we defined above. We call this "family" of infinities Countably Infinite.</p>
<p>You may be wondering what the point of this is...if we are just mapping primes, to natural numbers, to fractions, to even numbers, etc. that doesn't give us any extra information or really say anything about the sets, that is until you realise there are infinite sets that cannot be bijected to the natural numbers, a classic example is the Real Numbers, this is proven in Cantor's Argument of Diagonalization (One of the most elegant mathematical arguments ever in my opinion.)</p>
|
3,579,065 | <p>It is known that the quantity <span class="math-container">$\cos \frac{2π}{17}$</span> is a root of the <span class="math-container">$8$</span>'th degree equation,
<span class="math-container">$$x^8 + \frac{1}{2} x^7 - \frac{7}{4} x^6 - \frac{3}{4} x^5 + \frac{15}{16} x^4 + \frac{5}{16} x^3 - \frac{5}{32} x^2 - \frac{x}{32} + \frac{1}{256} = 0$$</span>
It is known that the regular <span class="math-container">$17$</span> sided polygon can be constructed from <span class="math-container">$cos \frac{2π}{17}$</span> , if this can be expressed in square roots.
Can you show that this equation can be derived from the relation <span class="math-container">$9\theta=-\sin 8\theta$</span>, where <span class="math-container">$\theta = \frac{2π}{17}$</span>?</p>
<p>Can you demonstrate that it is possible to solve this particular <span class="math-container">$8$</span>'th degree equation, even though there is no <span class="math-container">$8$</span>'th degree formula? ( there is more than one possible form of solution; it is known that there is often more than one expression in radicals for the same quantity)
Also, can you find a square root form for <span class="math-container">$\cos \frac{2\pi}{17}$</span> which has minimum number of terms?</p>
| J. W. Tanner | 615,567 | <p>You know <span class="math-container">$\dfrac1{(1-x)^2}=1+2x+3x^2+4x^3+\cdots.$</span></p>
<p>Therefore, <span class="math-container">$\dfrac x{(1-x)^2}=x(1+2x+3x^2+4x^3+\cdots)=x+2x^2+3x^3+\cdots$</span></p>
<p>and <span class="math-container">$\dfrac {x^2}{(1-x)^2}=x^2(1+2x+3x^2+4x^3+\cdots)=x^2+2x^3+\cdots.$</span></p>
<p>Therefore, adding these up, </p>
<p><span class="math-container">$\dfrac {1+x+x^2}{(1-x)^2}=1+(2+1)x+(3+2+1)x^2+(4+3+2)x^3+\cdots$</span></p>
<p><span class="math-container">$=1+3x+6x^2+9x^3+\cdots$</span>.</p>
|
2,962,203 | <p>I got stuck at : <span class="math-container">$a^2/b^2 = 12+2 \sqrt 35$</span></p>
<p>I understand that <span class="math-container">$12$</span> is rational and now I need to prove that <span class="math-container">$\sqrt{35}$</span> is irrational.</p>
<p>so I defined <span class="math-container">$∀c,d∈R$</span> while <span class="math-container">$d$</span> isn't <span class="math-container">$0$</span> that:
<span class="math-container">$c^2/d^2 = \sqrt 35$</span>
so <span class="math-container">$- c^2=(d^2)\sqrt{35}$</span>
It means that <span class="math-container">$c$</span> divide with <span class="math-container">$5$</span> and <span class="math-container">$7$</span>?
Also, how do I prove that if for example <span class="math-container">$X^2/4$</span> then <span class="math-container">$X/4$</span>? </p>
| Connor Harris | 102,456 | <p>If <span class="math-container">$a = \sqrt{5} + \sqrt{7}$</span> is rational, then <span class="math-container">$b = \sqrt{5} - \sqrt{7} = a - 2 \sqrt{7}$</span> is irrational. But <span class="math-container">$ab = -2$</span>, so <span class="math-container">$a$</span> and <span class="math-container">$b$</span> must be both rational or irrational.</p>
|
42,787 | <p>I am using <code>ListPlot</code> to display from 5 to 12 lines of busy data. The individual time series in my data are not easy to distinguish visually, as may be evident below, because the colors are not sufficiently different.</p>
<p><img src="https://i.stack.imgur.com/PiMMh.png" alt="enter image description here"></p>
<p>I have been trying to use <code>PlotStyle</code>, <code>ColorData</code>and related functions to get better colors. I would rather not have to specify a specific list of colors because the number of plot items varies from test to test. I created a toy plot to experiment with - the problem is illustrated by lines "F" and "G", which seem to be almost the same color. <code>PlotStyle</code> -> <code>ColorData</code> doesn't seem to work. Is there a simple way to do this?</p>
<pre><code>ListPlot[Table[i*Range[0, 10], {i, 1, 5, 0.5}]
, Frame -> True, Joined -> True, PlotRange -> All
, PlotLegends -> SwatchLegend[Characters["ABCDEFGHIGKLMNOP"]]
, PlotStyle -> ColorData["TemperatureMap"] ]
</code></pre>
<p><img src="https://i.stack.imgur.com/Wdi0O.png" alt="enter image description here"></p>
<p>It looks like </p>
<pre><code>ListLinePlot[Table[data2*i, {i, k}], PlotStyle -> Thick,
ColorFunction -> Function[{x1, x2}, ColorData[c1][x2]]]
</code></pre>
<p>from another <a href="https://mathematica.stackexchange.com/questions/27131/plotstyle-in-listplot-change-color-scheme-manually-choose-color-of-first-plot?rq=1">question</a> may be the answer. I didn't see that before. I'll try it out. I don't think I really understand <code>ColorData</code>. Meanwhile, if anyone has generally enlightening comments, I would appreciate them.</p>
| Dr. belisarius | 193 | <p>In case all else fails :) ...</p>
<pre><code>n = 8;
Manipulate[
ListPlot[d,
Frame -> True,
Joined -> True,
PlotRange -> All,
PlotLegends -> SwatchLegend[Characters["ABCDEFGHIGKLMNOP"]],
PlotStyle -> cs],
{{cs, Array[RGBColor[0.`, 0.`, 0.`] &, n]}, ControlType -> None},
Column[Outer[Legended[ColorSlider[Dynamic[cs[[#1]]], AppearanceElements -> "Swatch"], #1]&,
Range[n]]],
Initialization -> (d = Table[i*Range[0, 10], {i, 1, n}]),
ControlPlacement -> Right]
</code></pre>
<p><img src="https://i.stack.imgur.com/NUTIe.png" alt="Mathematica graphics"></p>
|
227,797 | <p>I have this function and I want to see where it is zero.
<span class="math-container">$$\frac{1}{16} \left(\sinh (\pi x) \left(64 \left(x^2-4\right) \cosh \left(\frac{2 \pi x}{3}\right) \cos (y)+\left(x^2+4\right)^2+256 x \sinh \left(\frac{2 \pi x}{3}\right) \sin (y)\right)+\left(x^2-12\right)^2 \sinh \left(\frac{7 \pi x}{3}\right)-2 \left(x^2+4\right)^2 \sinh \left(\frac{5 \pi x}{3}\right)\right)+2 \left(x^2-4\right) \sinh \left(\frac{\pi x}{3}\right)$$</span>
I use ContourPlot</p>
<pre><code>f[x_, y_] :=
2 (-4 + x^2) Sinh[(π x)/3] +
1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] +
256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] -
2 (4 + x^2)^2 Sinh[(5 π x)/3] + (-12 + x^2)^2 Sinh[(
7 π x)/3]);
ContourPlot[
f[x, y] == 0, {x, 3.465728, 3.465729}, {y, 1.046786, 1.046795},
PlotPoints -> 500]
</code></pre>
<p>and I obtain this plot</p>
<p>Now, my question is that can I trust this plot and conclude that the curves do not cross?</p>
<p>Or, I should increase the precision of the plot? And if so, how can I ask Mathematica to give higher precision for the axis in ContourPlot?</p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
f[x_, y_] :=
2 (-4 + x^2) Sinh[(π x)/3] +
1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] +
256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] -
2 (4 + x^2)^2 Sinh[(5 π x)/
3] + (-12 + x^2)^2 Sinh[(7 π x)/3]);
ContourPlot[f[x, y] == 0,
{x, 3464/1000, 3468/1000}, {y, 103/100, 106/100},
PlotPoints -> 100,
MaxRecursion -> 5,
WorkingPrecision -> 15,
FrameLabel -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/ZkJpd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZkJpd.png" alt="enter image description here" /></a></p>
<pre><code>({min, arg} = NMinimize[{f[x, y]^2,
3464/1000 < x < 3468/1000, 1030/1000 < y < 1060/1000},
{x, y}, WorkingPrecision -> 30]) /. r_Real :> N[r]
(* {0, {x -> 3.46572, y -> 1.04691}} *)
</code></pre>
<p><strong>EDIT:</strong> Verifying the solution,</p>
<pre><code>f[x, y] /. arg
(* 0.*10^-20 *)
</code></pre>
<p><strong>EDIT 2:</strong> Using <a href="https://reference.wolfram.com/language/ref/Plot3D.html" rel="nofollow noreferrer"><code>Plot3D</code></a></p>
<pre><code>Plot3D[f[x, y],
{x, 3464/1000, 3468/1000}, {y, 103/100, 106/100},
MeshFunctions -> {#3 &},
Mesh -> {{0}},
MeshStyle -> Directive[Red, Thick],
PlotPoints -> 100,
MaxRecursion -> 5,
WorkingPrecision -> 15,
ClippingStyle -> None]
</code></pre>
<p><a href="https://i.stack.imgur.com/UTiyr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UTiyr.png" alt="enter image description here" /></a></p>
|
227,797 | <p>I have this function and I want to see where it is zero.
<span class="math-container">$$\frac{1}{16} \left(\sinh (\pi x) \left(64 \left(x^2-4\right) \cosh \left(\frac{2 \pi x}{3}\right) \cos (y)+\left(x^2+4\right)^2+256 x \sinh \left(\frac{2 \pi x}{3}\right) \sin (y)\right)+\left(x^2-12\right)^2 \sinh \left(\frac{7 \pi x}{3}\right)-2 \left(x^2+4\right)^2 \sinh \left(\frac{5 \pi x}{3}\right)\right)+2 \left(x^2-4\right) \sinh \left(\frac{\pi x}{3}\right)$$</span>
I use ContourPlot</p>
<pre><code>f[x_, y_] :=
2 (-4 + x^2) Sinh[(π x)/3] +
1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] +
256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] -
2 (4 + x^2)^2 Sinh[(5 π x)/3] + (-12 + x^2)^2 Sinh[(
7 π x)/3]);
ContourPlot[
f[x, y] == 0, {x, 3.465728, 3.465729}, {y, 1.046786, 1.046795},
PlotPoints -> 500]
</code></pre>
<p>and I obtain this plot</p>
<p>Now, my question is that can I trust this plot and conclude that the curves do not cross?</p>
<p>Or, I should increase the precision of the plot? And if so, how can I ask Mathematica to give higher precision for the axis in ContourPlot?</p>
| Akku14 | 34,287 | <p><code>f[x, y] == 0</code> can be separated in lhs[y] == rhs[x]. This shows, rhs gets immaginary in a small range of x. So curves do not intersect.</p>
<pre><code>f[x_, y_] =
2 (-4 + x^2) Sinh[(\[Pi] x)/3] +
1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 \[Pi] x)/3] +
256 x Sin[y] Sinh[(2 \[Pi] x)/3]) Sinh[\[Pi] x] -
2 (4 + x^2)^2 Sinh[(5 \[Pi] x)/
3] + (-12 + x^2)^2 Sinh[(7 \[Pi] x)/3]);
eq = Equal @@@ (First@
Solve[f[x, y] == 0 && Sin[y]^2 + Cos[y]^2 == 1, Cos[y],
Sin[y]]) // First
(* Cos[y] == (-65536 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/
3] Sinh[\[Pi] x] +
32768 x^2 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] -
4096 x^4 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] +
8192 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 +
2048 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 -
512 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 -
128 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 -
16384 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] -
4096 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] +
1024 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] +
256 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] +
73728 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] -
30720 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] +
3584 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] -
128 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/
3] - \[Sqrt]((65536 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/
3] Sinh[\[Pi] x] -
32768 x^2 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/
3] Sinh[\[Pi] x] +
4096 x^4 Cosh[(2 \[Pi] x)/3] Sinh[(\[Pi] x)/
3] Sinh[\[Pi] x] -
8192 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 -
2048 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 +
512 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 +
128 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x]^2 +
16384 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/
3] + 4096 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(
5 \[Pi] x)/3] -
1024 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/
3] - 256 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(
5 \[Pi] x)/3] -
73728 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/
3] + 30720 x^2 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(
7 \[Pi] x)/3] -
3584 x^4 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/
3] + 128 x^6 Cosh[(2 \[Pi] x)/3] Sinh[\[Pi] x] Sinh[(
7 \[Pi] x)/3])^2 -
4 (65536 Cosh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 -
32768 x^2 Cosh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 +
4096 x^4 Cosh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 +
65536 x^2 Sinh[(2 \[Pi] x)/
3]^2 Sinh[\[Pi] x]^2) (16384 Sinh[(\[Pi] x)/3]^2 -
8192 x^2 Sinh[(\[Pi] x)/3]^2 +
1024 x^4 Sinh[(\[Pi] x)/3]^2 -
4096 Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] -
1024 x^2 Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] +
256 x^4 Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] +
64 x^6 Sinh[(\[Pi] x)/3] Sinh[\[Pi] x] +
256 Sinh[\[Pi] x]^2 + 256 x^2 Sinh[\[Pi] x]^2 +
96 x^4 Sinh[\[Pi] x]^2 + 16 x^6 Sinh[\[Pi] x]^2 +
x^8 Sinh[\[Pi] x]^2 -
65536 x^2 Sinh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 +
8192 Sinh[(\[Pi] x)/3] Sinh[(5 \[Pi] x)/3] +
2048 x^2 Sinh[(\[Pi] x)/3] Sinh[(5 \[Pi] x)/3] -
512 x^4 Sinh[(\[Pi] x)/3] Sinh[(5 \[Pi] x)/3] -
128 x^6 Sinh[(\[Pi] x)/3] Sinh[(5 \[Pi] x)/3] -
1024 Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] -
1024 x^2 Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] -
384 x^4 Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] -
64 x^6 Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] -
4 x^8 Sinh[\[Pi] x] Sinh[(5 \[Pi] x)/3] +
1024 Sinh[(5 \[Pi] x)/3]^2 +
1024 x^2 Sinh[(5 \[Pi] x)/3]^2 +
384 x^4 Sinh[(5 \[Pi] x)/3]^2 +
64 x^6 Sinh[(5 \[Pi] x)/3]^2 +
4 x^8 Sinh[(5 \[Pi] x)/3]^2 -
36864 Sinh[(\[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
15360 x^2 Sinh[(\[Pi] x)/3] Sinh[(7 \[Pi] x)/3] -
1792 x^4 Sinh[(\[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
64 x^6 Sinh[(\[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
4608 Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] +
1536 x^2 Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] -
64 x^4 Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] -
32 x^6 Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] +
2 x^8 Sinh[\[Pi] x] Sinh[(7 \[Pi] x)/3] -
9216 Sinh[(5 \[Pi] x)/3] Sinh[(7 \[Pi] x)/3] -
3072 x^2 Sinh[(5 \[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
128 x^4 Sinh[(5 \[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
64 x^6 Sinh[(5 \[Pi] x)/3] Sinh[(7 \[Pi] x)/3] -
4 x^8 Sinh[(5 \[Pi] x)/3] Sinh[(7 \[Pi] x)/3] +
20736 Sinh[(7 \[Pi] x)/3]^2 -
6912 x^2 Sinh[(7 \[Pi] x)/3]^2 +
864 x^4 Sinh[(7 \[Pi] x)/3]^2 -
48 x^6 Sinh[(7 \[Pi] x)/3]^2 +
x^8 Sinh[(7 \[Pi] x)/3]^2)))/(2 (65536 Cosh[(2 \[Pi] x)/
3]^2 Sinh[\[Pi] x]^2 -
32768 x^2 Cosh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 +
4096 x^4 Cosh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2 +
65536 x^2 Sinh[(2 \[Pi] x)/3]^2 Sinh[\[Pi] x]^2)) *)
Plot[eq[[2]], {x, 3.465728, 3.465729}, PlotPoints -> 100,
WorkingPrecision -> 30]
Plot[Im[eq[[2]]], {x, 3.465728, 3.465729}, PlotPoints -> 200,
WorkingPrecision -> 60, PlotRange -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/DggUd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DggUd.jpg" alt="enter image description here" /></a></p>
<pre><code>eq /. x -> Rationalize[3.46572839, 0] // N[#, 20] &
(* Cos[y] == 0.50035241364172394089 - 9.794261144042*10^-8 I *)
</code></pre>
<p><strong>Edit</strong></p>
<p>f does not reach zero as it should, in the imaginary range.</p>
<pre><code>Plot[f[Rationalize[3.46572839, 0], y], {y, 1030/1000, 1060/1000},
PlotRange -> 10, PlotPoints -> 200]
Maximize[{f[Rationalize[3.46572839, 0], y],
1030/1000 < y < 1060/1000}, y] // N[#, 10] &
(* {-7.787746817*10^-6, {y -> 1.046790571}} *)
ContourPlot[
f[x, y] == 0, {x, 34657283/10000000, 6931457/2000000}, {y,
1046786/1000000, 1046795/1000000}, PlotPoints -> 200,
MaxRecursion -> 5, WorkingPrecision -> 30, FrameLabel -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/9Z9IB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Z9IB.jpg" alt="enter image description here" /></a></p>
|
2,642,144 | <p>How would I prove or disprove the following statement?
$ \forall a \in \mathbb{Z} \forall b \in \mathbb{N}$ , if $a < b$ then $a^2 < b^2$</p>
| AbstractNonsense | 429,931 | <p>take $a=-1$ and $b=1$, then $a^2=1 \nless1=b^2$</p>
|
1,716,656 | <p>I am having trouble solving this problem</p>
<blockquote>
<p>Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?</p>
</blockquote>
<p>My attempt:</p>
<p>I first want to find the deposit per month.</p>
<p>I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,</p>
<p>$D*100(Ia_{30|0.08}) = 100,000$</p>
<p>However, the $D$ I got was 8.12, which is clearly not right.</p>
<p>Can someone help?</p>
| Em. | 290,196 | <p>To get yourself started, you could draw a table. The rows could be one roll, and the columns could be the other roll. Then the checkmark shows where the rolls are "two away" from each other.</p>
<p>\begin{array}{r|c|c|c|c|c|c}
&1&2&3&4&5&6\\\hline
1&&&\checkmark&&&\\\hline
2&&&&\checkmark&&\\\hline
3&\checkmark&&&&\checkmark&\\\hline
4&&\checkmark&&&&\checkmark\\\hline
5&&&\checkmark&&&\\\hline
6&&&&\checkmark&&
\end{array}
Notice that, since all pairs are equally likely, we have a $8/36 = 2/9$ chance of being "two away".</p>
|
1,716,656 | <p>I am having trouble solving this problem</p>
<blockquote>
<p>Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?</p>
</blockquote>
<p>My attempt:</p>
<p>I first want to find the deposit per month.</p>
<p>I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,</p>
<p>$D*100(Ia_{30|0.08}) = 100,000$</p>
<p>However, the $D$ I got was 8.12, which is clearly not right.</p>
<p>Can someone help?</p>
| user | 79,303 | <p>If the first die is 1, the other can only be 3, probability = 1/6</p>
<p>If the first die is 2, the other can only be 4, probability = 1/6</p>
<p>If the first die is 5, the other can only be 3, probability = 1/6</p>
<p>If the first die is 6, the other can only be 4, probability = 1/6</p>
<p>If the first die is 3, the other can only be 1 or 5, probability = 2/6</p>
<p>If the first die is 4, the other can only be 2 or 6, probability = 2/6</p>
<p>Total probability is (1+1+1+1+2+2)/(6+6+6+6+6+6) = 8/36 = 2/9</p>
|
435,298 | <p>Define
$$\langle X,Y \rangle := \operatorname{tr}XY^t,$$
where $X,Y$ are square matrices with real entries and $t$ denotes transpose.</p>
<p>I have some troubles in proving that
$$ \langle [X,Y],Z \rangle = - \langle Y,[X,Z] \rangle,$$
where square brackets denote commutator.</p>
<p>Let me update my <strong>questin</strong> to <strong>part ii</strong>. You have proven that my commutation relation without a transpose is wrong, while it is correct if we put a $t.$</p>
<p>Then I'd say I'm in trouble, because the next step would be to define
$$\operatorname{ad}_XY:=[X,Y]$$ and claim that by the above (false) property we have that $\operatorname{ad}$ is antisymmetric, i.e.
$$\langle \operatorname{ad}_XY,Z\rangle =- \langle Y,\operatorname{ad}_XZ\rangle:$$
Do you know of a way to recover such a nice property or something similar?</p>
| Avitus | 80,800 | <p>The key properties to use are </p>
<p>$$\langle A,B\rangle=\langle B, A\rangle,$$</p>
<p>i.e. with $tr((AB)^t)=tr(AB),$ and</p>
<p>$$tr(ABC^t)=tr(BC^tA),$$</p>
<p>for all $A,B\in M_n(\mathbb R)$.</p>
|
2,565,204 | <p>I have following problem that I cannot solve... I have a triangle with sides <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> which is split into two smaller triangles, <span class="math-container">$E$</span> and <span class="math-container">$F$</span>, like <a href="https://i.stack.imgur.com/rJ1hs.jpg" rel="nofollow noreferrer">this</a>.</p>
<p>I need to find the area of <span class="math-container">$F$</span>.</p>
<p>I also know that the perimeter of <span class="math-container">$F$</span> equals the perimeter of <span class="math-container">$E$</span>.</p>
<p>I tried to equate the perimeters of <span class="math-container">$E$</span> and <span class="math-container">$F$</span>, but I could not figure it out.</p>
<p>Thank you so much for any hint!</p>
| induction601 | 444,022 | <p>For any $j$, there is $N_j$ such that for all $n > N_j$,
$$ |g_n^j(x) - g(x)| \le \epsilon/2$$
Then we have
\begin{align}
|g^i_n(x) - g(x)| &\leq |g^i_n(x) - g^j_n(x)| + |g^j_n(x) - g(x)|\\
&\leq |g^i_n(x) - g^j_n(x)| + \epsilon/2.
\end{align}
By taking sup only w.r.t. $i$, we have
\begin{align}
\sup_{i\in I}|g^i_n(x) - g(x)| \leq \sup_{i\in I} |g^i_n(x) - g^j_n(x)| + \epsilon/2.
\end{align}
Since $\sup_{i\in I} |g^i_n(x) - g^j_n(x)| \le \sup_{i,j\in I}|g^i_n(x) - g^j_n(x)|$, we obtain
\begin{align}
\sup_{i\in I}|g^i_n(x) - g(x)| \leq \sup_{i,j \in I} |g^i_n(x) - g^j_n(x)| + \epsilon/2.
\end{align}
Since $\sup_{i,j \in I} |g^i_n(x) - g^j_n(x)| \to 0$ as $n \to \infty$, by choose $N$ appropriately, we conclude that
$$\sup_{i\in I}|g^i_n(x) - g(x)| \le \epsilon$$</p>
|
2,565,204 | <p>I have following problem that I cannot solve... I have a triangle with sides <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> which is split into two smaller triangles, <span class="math-container">$E$</span> and <span class="math-container">$F$</span>, like <a href="https://i.stack.imgur.com/rJ1hs.jpg" rel="nofollow noreferrer">this</a>.</p>
<p>I need to find the area of <span class="math-container">$F$</span>.</p>
<p>I also know that the perimeter of <span class="math-container">$F$</span> equals the perimeter of <span class="math-container">$E$</span>.</p>
<p>I tried to equate the perimeters of <span class="math-container">$E$</span> and <span class="math-container">$F$</span>, but I could not figure it out.</p>
<p>Thank you so much for any hint!</p>
| grand_chat | 215,011 | <p>The claim you're trying to prove is intended to hold for each $x$, so you can suppress $x$. Pick <em>any</em> $j$, say $j=J$. Then for each $i$ and $n$ we have
$$
|g_n^i-g|\le |g_n^i-g_n^J| + |g_n^J-g|\le \sup_{i,j}|g_n^i-g_n^j|+|g_n^J-g|\tag1
$$
The RHS of (1) depends on $n$ only, which implies the sup over all $i$ of the LHS is bounded by the same thing:
$$
\sup_i |g_n^i-g|\le \sup_{i,j}|g_n^i-g_n^j|+|g_n^J-g|
$$
Now let $n\to\infty$, and you're done.</p>
|
474,587 | <p>Does $\|Tv\|\leq\|v\|$ (for all $v \in V$) leads to $T$ is normal?</p>
<p>If not, when I add the additional information that every e.e of $T$ is of the absolute value 1, can I prove $T$ is unitary? </p>
<p>Thanks!</p>
| M Turgeon | 19,379 | <p>As Daniel Fischer has noted in the comments, the answer is no. One way to see this is as follows: on the vector space of linear operators on $V$, you can define a <em>norm</em>. The norm of $T$ is the smallest non-negative real number $c$ such that
$$\|Tv\|\leq c\|v\|,$$
for all $v\in V$. Since this is a norm, for all scalar $\lambda$, you have $\|\lambda T\|=\vert\lambda\vert\|T\|$ (where $\|T\|$ is the norm defined above). Now, going back to your problem, you are assuming that
$$\|T\|\leq 1.$$
But for any operator $T$, by picking $\lambda$ suitably, you can make $\|\lambda T\|\leq 1$. Therefore (and this is what Daniel Fischer mentioned above), if you take a non-normal $T$, you can make $\lambda T$ into a counterexample.</p>
|
122,471 | <p>Can anyone explain how I can prove that either $\phi(t) = \left|\cos (t)\right|$ is characteristic function or not? And which random variable has this characteristic function? Thanks in advance.</p>
| GEdgar | 442 | <p>W. Feller, <em>An Introduction to Probability Theory and Applications</em>, Volume I, XIX.4, Theorem 1.<br>
A continuous function $\phi$ with period $2\pi$ is a characteristic function iff its Fourier coefficients (4.2) satisfy $\phi_k \ge 0$ and $\phi(0) = 1$. </p>
<p>$$
\phi_k = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i k \zeta} \phi(\zeta)\,d\zeta
\tag{4.2}
$$</p>
|
12,717 | <p>In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.</p>
<p>What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points? Is there a characterization of the DVR's that aren't induced by closed points?</p>
<p>And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?</p>
<p>How about complete integral schemes that are regular in codimension 1?</p>
| Emerton | 2,874 | <p>Suppose that $X$ is a projective variety, and that $v$ is a discete valuation on $K(X)$
(trivial on $k$) whose corresponding valuation ring we will denote by $R$. The valuative criterion shows that
the map Spec $K(X) \rightarrow X$ extends to a map Spec $R \rightarrow X$. If I have the terminology correct, the image of the closed point of Spec $R$ is called the centre of
the valuation $v$ on $X$. It has codimension anywhere between $1$ and dim $X$.
Note that if $x \in X$ is the centre, then $R$ dominates $\mathcal O_x$ in $K(X)$
(i.e. we have a local inclusion of local rings $\mathcal O_x \subset R$).</p>
<p>Let's suppose for a moment that $X$ is a smooth surface. If the centre $x$ is codimension 1,
then both $\mathcal O_x$ and $R$ are (discrete) valuation rings. Since valuation rings
are (characterized by being) maximal for the partial order of dominance, $R$ and $\mathcal O_x$ coincide, and so the discrete valuation $v$ is just that given by the divisor of which
$x$ is the generic point.</p>
<p>Suppose instead that $x$ is a closed point.
Now we can blow up $x$ in $X$, to get a projective variety $X_1$, and the centre of $v$ in $X_1$ will now be contained in the exceptional divisor of $X_1$ (i.e. the preimage of $x$). If it coincides with the exceptional divisor,
then we have found a curve on $X_1$ giving rise to $v$; otherwise it is a point
$x_1$, which we can blow up again.</p>
<p>Either we eventually obtain a divisor on some iterated blow-up of $X$, or we
obtain a sequence of points $x \in X, x_1 \in X_1, \ldots,$ with each $X_n$ a blow-up
of the previous. In this case one sees that $R = \bigcup \mathcal O_{x_n}.$</p>
<p>There are a couple of exercises related to this issue in Hartshorne, namely II.4.5, II.4.12, and V.5.6. If I understand them correctly, any such sequence of $x\_n$ gives
a valuation ring $R$ in this way, and $R$ is a discrete valuation ring <I>unless</I> one constructs the sequene $x_n$ in the following manner: choose an irreducible curve $C$ in $X$ and define $x_n$ to
be the intersection of the proper transform of $C$ in $X_n$ with the exceptional
divisor. For a sequence $x_n$ constructed in this latter manner, one obtains not
a discrete valuation ring, but rather a rank 2 valuation ring: the valuation is determined
by first taking the valuation at the generic point of $C$, and then (for those functions
which are defined and non-zero at this generic point) restricting to $C$ and computing
the order of zero or pole at $x$. </p>
<p>What is the geometric intuition for the discrete valuation rings that correspond
to an infinite sequence $x_n$ rather than to some curve on $X$? One can think of
them as a transcendental curve on $X$, passing through $x$.<br>
Indeed, imagine you had such a curve.
Then you could restrict a rational function to it; since it is transcendental,
a non-zero rational function would not have a zero or pole along this curve, and so
would restrict to give a non-zero meromorphic function on the curve. We could then
compute the order of the zero or pole of this meromorphic function at $x$. In other
words, because the curve is transcendental, we get a rank one valuation, in contrast to the rank two valuations that arise when we apply this process with an algebraic curve
$C$ passing through $x$.</p>
<p>I'm not sure about the details of the higher dimensional case. (Among other things, I am worried about the possibility of the center being codim > 1, but singular, which seems like it could complicate the analysis.) Does anyone here know how it goes?</p>
|
187,959 | <p>This is the question:</p>
<p>Use the integral test to determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{1+2n}$.</p>
<p>I started by writing:</p>
<p>$$\int_1^\infty\frac{1}{1+2x}dx=\lim_{a \rightarrow \infty}\left(\int_1^a\frac{1}{1+2x}dx\right)$$</p>
<p>I then decided to use u-substitution with $u=1+2n$ to solve the improper integral. I got the answer wrong and resorted to my answer book and this is where they went after setting $u=1+2n$:</p>
<p>$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$</p>
<p>And the answer goes on...</p>
<p>What I can't figure out is where the $\frac{1}{2}$ came from when the u-substitution began and also, why the lower bound of the integral was changed to 3.</p>
<p>Can someone tell me? </p>
| DonAntonio | 31,254 | <p>$$u=1+2x\Longrightarrow du=2dx\Longrightarrow dx=\frac{1}{2}du$$</p>
<p>Remember, not only you substitute the variable and nothing more: you also have to change the $\,dx\,$ and the integral's limits:
$$u=1+2x\,\,,\,\text{so}\,\, x=1\Longrightarrow u=1+2\cdot 1 =3$$</p>
|
187,959 | <p>This is the question:</p>
<p>Use the integral test to determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{1+2n}$.</p>
<p>I started by writing:</p>
<p>$$\int_1^\infty\frac{1}{1+2x}dx=\lim_{a \rightarrow \infty}\left(\int_1^a\frac{1}{1+2x}dx\right)$$</p>
<p>I then decided to use u-substitution with $u=1+2n$ to solve the improper integral. I got the answer wrong and resorted to my answer book and this is where they went after setting $u=1+2n$:</p>
<p>$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$</p>
<p>And the answer goes on...</p>
<p>What I can't figure out is where the $\frac{1}{2}$ came from when the u-substitution began and also, why the lower bound of the integral was changed to 3.</p>
<p>Can someone tell me? </p>
| Santosh Linkha | 2,199 | <p>Let $1 + 2x = u$ then $du = 2 dx \implies du = {1 \over 2} dx$</p>
<p>$x $ goes from $1$ to $\infty$, since we are using new variable $u$ here, it would have different bound. When $x \rightarrow \infty$, $ u \rightarrow \infty$ and when $x = 1$, $ u = 1 + 2 \cdot ( 1) = 3 $, so $u$ goes from $3$ to $\infty$</p>
|
187,959 | <p>This is the question:</p>
<p>Use the integral test to determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{1+2n}$.</p>
<p>I started by writing:</p>
<p>$$\int_1^\infty\frac{1}{1+2x}dx=\lim_{a \rightarrow \infty}\left(\int_1^a\frac{1}{1+2x}dx\right)$$</p>
<p>I then decided to use u-substitution with $u=1+2n$ to solve the improper integral. I got the answer wrong and resorted to my answer book and this is where they went after setting $u=1+2n$:</p>
<p>$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$</p>
<p>And the answer goes on...</p>
<p>What I can't figure out is where the $\frac{1}{2}$ came from when the u-substitution began and also, why the lower bound of the integral was changed to 3.</p>
<p>Can someone tell me? </p>
| Thomas Russell | 32,374 | <p>You have $\int_{1}^{\infty}{\frac{1}{1+2n}\:dn}=\lim_{a\to\infty}\left(\int_{1}^{a}{\frac{1}{1+2n}\:dn}\right)$. Using the substitution you mentioned $u=1+2n$, we have our lower bound as $u(1)=1+2(1)=3$ and our upper bound as $u(a)=1+2(a)=1+2a$.</p>
<p>We also have $\frac{du}{dn}=2\implies dn=\frac{du}{2}$, therefore we have:</p>
<p>$$\int_{3}^{1+2a}{\frac{1}{u}\frac{du}{2}}=\frac{1}{2}\int_{3}^{1+2a}{\frac{1}{u}\:du}$$</p>
<p>Taking our previous limit, we have:</p>
<p>$$\lim_{a\to\infty}{\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}\:du\right)}$$</p>
<p>Which is what you have in your answer book.</p>
|
202,034 | <p>Is finding the largest prime factor of a number computationally easier than factoring the number into powers of primes? </p>
| gnasher729 | 137,175 | <p>Of course it is easier, since factoring a number into prime powers gives you the largest prime factor for free. In general, it's only a <em>tiny</em> bit easier. There will be extreme examples, like $4^k-1 = (2^k-1)(2^k+1)$, where it might be much easier to prove that one of the numbers is a prime and therefore the largest prime factor, than to factor the other number further. </p>
|
121,245 | <p><em>What is the difference between how Matlab and Mathematica solve <strong>State-Space</strong> and <strong>Transfer Function</strong> models</em>?</p>
<p>I have a $16 \times 16$ state space system for which I am calculating transfer function. Mathematica and Matlab give me completely different answers. I can imagine truncation may lead to slightly different answers but what I get is a <strong>huge difference</strong>. I have checked the difference in numerical values to check the conditioning. There is no significant difference. I give the matrix as seen in Mathematica to you to check.</p>
<p>Please find the code attached!</p>
<pre><code>a={{-(350103/48500), -14.9811, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {1.75898, -6.08528, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0, 0, 0, 0,
0}, {2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0}, {1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0, 0,
0, 0, 0, 0}, {0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0}, {0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757}, {0, 0, 0, 0, 0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0,
0, 0, 0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0}};
b={{116319/1940}, {10.695}, {0}, {0}, {0}, {0}, {0}, {0}, {0}, {0},
{0}, {0}, {0}, {0}, {0}, {0}};
c={{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1}};
sys = StateSpaceModel[{a,b,c}]
BodePlot[sys, {2 \[Pi] 0.01, 2 \[Pi] 100},
ScalingFunctions -> {{Automatic, Automatic}, {Automatic, "Degree"}},
PhaseRange -> {-\[Pi], \[Pi]}, GridLines -> Automatic, ImageSize ->Large]
</code></pre>
<p>Corresponding Matlab code with the same set of matrices:</p>
<pre><code>[num,den] = ss2tf(a,b,c,[0;0;0]);
P1 = tf(num(1,:),den);
P2 = tf(num(2,:),den);
P3 = tf(num(3,:),den);
figure; bode(P1,P2,P3);
</code></pre>
<p>The transfer function that I can from Matlab and Mathematica are completely different in nature. I am pretty sure that <strong><em>according to theory Matlab gives me the right answer</em></strong>. But why is Mathematica so off the mark?</p>
<p>Could it be any of the following?</p>
<ol>
<li>Precision/truncation</li>
<li>Size of the system</li>
<li>Conditioning</li>
</ol>
<p>Edit:
I have included the plots for both. Left is Mathematica and right is Matlab. The colors correspond to each other. Please take note of the Magnitude scale for both!
<a href="https://i.stack.imgur.com/E2go8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E2go8.png" alt="enter image description here"></a></p>
<p>Any insight would be appreciated!</p>
| JimB | 19,758 | <p>Using <em>Mathematica</em> 10.4.1 (Windows 7) the following code finds a pretty good match with the Matlab results. (And I'd argue that any differences between the two are due to differences in how the precision of the input numbers are handled.)</p>
<pre><code>a = Rationalize[a, 0.00001];
b = Rationalize[b];
sys = StateSpaceModel[{a, b, c, d}];
BodePlot[sys, {2 π 0.01, 2 π 100},
ScalingFunctions -> {{Automatic, Automatic}, {Automatic, "Degree"}},
PhaseRange -> {-π, π}, GridLines -> Automatic,
ImageSize -> Large,
PlotRange -> {{Full, {0, -400}}, {Full, {-180, 180}}},
PlotStyle -> {Blue, Red, Green}]
</code></pre>
<p><a href="https://i.stack.imgur.com/Dgh6f.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Dgh6f.png" alt="First BodePlot"></a></p>
<p><a href="https://i.stack.imgur.com/WKxm0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/WKxm0.png" alt="Second BodePlot"></a></p>
|
121,245 | <p><em>What is the difference between how Matlab and Mathematica solve <strong>State-Space</strong> and <strong>Transfer Function</strong> models</em>?</p>
<p>I have a $16 \times 16$ state space system for which I am calculating transfer function. Mathematica and Matlab give me completely different answers. I can imagine truncation may lead to slightly different answers but what I get is a <strong>huge difference</strong>. I have checked the difference in numerical values to check the conditioning. There is no significant difference. I give the matrix as seen in Mathematica to you to check.</p>
<p>Please find the code attached!</p>
<pre><code>a={{-(350103/48500), -14.9811, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {1.75898, -6.08528, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0, 0, 0, 0,
0}, {2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0}, {1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0, 0,
0, 0, 0, 0}, {0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0}, {0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757}, {0, 0, 0, 0, 0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0,
0, 0, 0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0}};
b={{116319/1940}, {10.695}, {0}, {0}, {0}, {0}, {0}, {0}, {0}, {0},
{0}, {0}, {0}, {0}, {0}, {0}};
c={{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1}};
sys = StateSpaceModel[{a,b,c}]
BodePlot[sys, {2 \[Pi] 0.01, 2 \[Pi] 100},
ScalingFunctions -> {{Automatic, Automatic}, {Automatic, "Degree"}},
PhaseRange -> {-\[Pi], \[Pi]}, GridLines -> Automatic, ImageSize ->Large]
</code></pre>
<p>Corresponding Matlab code with the same set of matrices:</p>
<pre><code>[num,den] = ss2tf(a,b,c,[0;0;0]);
P1 = tf(num(1,:),den);
P2 = tf(num(2,:),den);
P3 = tf(num(3,:),den);
figure; bode(P1,P2,P3);
</code></pre>
<p>The transfer function that I can from Matlab and Mathematica are completely different in nature. I am pretty sure that <strong><em>according to theory Matlab gives me the right answer</em></strong>. But why is Mathematica so off the mark?</p>
<p>Could it be any of the following?</p>
<ol>
<li>Precision/truncation</li>
<li>Size of the system</li>
<li>Conditioning</li>
</ol>
<p>Edit:
I have included the plots for both. Left is Mathematica and right is Matlab. The colors correspond to each other. Please take note of the Magnitude scale for both!
<a href="https://i.stack.imgur.com/E2go8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E2go8.png" alt="enter image description here"></a></p>
<p>Any insight would be appreciated!</p>
| Steffen Jaeschke | 61,643 | <p>In M11.3 this is much much easier to solve. Simply: use instead of d Automatic:</p>
<pre><code>a = {{-(350103/48500), -14.9811, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0}, {1.75898, -6.08528, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0, 0, 0, 0,
0}, {2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0}, {1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0,
0, 0, 0, 0, 0}, {0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757, 0, 0, 0, 0}, {0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0}, {0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 12.4114, -32.0709, 310.284,
0, -19.63, -102.059, -310.284, 77.9757}, {0, 0, 0, 0, 0, 0, 0, 0,
2.21387, -5.72064, 55.3468, 0, -0.454893, -21.6178, -55.3468,
13.9089}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0,
0, 0, 0, 0, 0, 0, 1, -2.584, 25, 0, -1, -7.016, -25, 0}};
b = {{116319/
1940}, {10.695}, {0}, {0}, {0}, {0}, {0}, {0}, {0}, {0}, {0}, \
{0}, {0}, {0}, {0}, {0}};
c = {{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1}};
a = Rationalize[a, 0.00001];
b = Rationalize[b];
sys = StateSpaceModel[{a, b, c, Automatic}]
BodePlot[sys, {2 \[Pi] 0.01, 2 \[Pi] 100},
ScalingFunctions -> {{Automatic, Automatic}, {Automatic, "Degree"}},
PhaseRange -> {-\[Pi], \[Pi]}, GridLines -> Automatic,
ImageSize -> Large,
PlotRange -> {{Full, {0, -400}}, {Full, {-180, 180}}},
PlotStyle -> {Blue, Red, Green}]
</code></pre>
<p>Both MATLAB and Mathematica 11.3 provide the very same results.<a href="https://i.stack.imgur.com/i1RQq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i1RQq.png" alt="enter image description here"></a></p>
|
674,259 | <p>I am trying to understand why the derivative of $f(x)=x^\frac{1}{2}$ is $\frac{1}{2\sqrt{x}}$ using the limit theorem. I know $f'(x) = \frac{1}{2\sqrt{x}}$, but what I want to understand is how to manipulate the following limit so that it gives this result as h tends to zero:</p>
<p>$$f'(x)=\lim_{h\to 0} \frac{(x+h)^\frac{1}{2}-x^\frac{1}{2}}{h} = \frac{1}{2\sqrt{x}}$$</p>
<p>I have tried writing this as :</p>
<p>$$\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$</p>
<p>But I can't see how to get to the limit of $\frac{1}{2\sqrt{x}}$.
Whilst this is not homework I have actually been set, I would like to understand how to evaluate the limit.</p>
| Ben Grossmann | 81,360 | <p><strong>Alternative approach:</strong></p>
<p>Noting that $h$ and $x+h$ need to be positive in order for this to make sense, we have
$$
\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{h} =
\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{(\sqrt{x+h})^2 - (\sqrt{x})^2} =
\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}
{(\sqrt{x+h}+ \sqrt x)(\sqrt{x+h} - \sqrt{x})}
$$</p>
|
3,316,730 | <p>I'm taking a course in Linear Algebra right now, and am having a hard time wrapping my head around bases, especially since my prof didn't really explain them fully. I would really appreciate any insight you could give me as to what bases are! Also, can there can be multiple different bases for a single subspace?</p>
<p>Thanks in advance. </p>
| jack.f | 269,635 | <p>While a bit late to the game, I thought another perspective might help.</p>
<hr>
<p>Consider the following physical example. Now, without being too pedantic about definition, a basis for a vector space is much like a building block of a biological system. We can <i>build</i> a human body from a set of cells. That is, we can construct all aspects of our anatomy beginning with a certain set of cells (e.g. nerve cells, blood cells, germ cells, epithelial cells, etc). Thus, if we take our various tissue as <i>vectors</i>, then we have as a <i>basis</i> our cells.</p>
<p>But we could certainly have another <i>biological basis</i> from which to build our <i>biological vectors</i>. Namely, biomolecules. Indeed, we could express our other defined basis using this basis. Thus, our <i>biological vector space</i> has more than one <i>biological basis</i>.</p>
<p>Some might argue that there's not a "full correspondence" here with the mathematical notion of a basis for a vector space because, for example, how could one exhibit a change of basis from biomolecules into cells (i.e. how does one express a biomolecule as "linear combination" of cells)? But I argue the idea of <i>building blocks</i> captures the underlying spirit of a basis for a first pass.</p>
|
909,741 | <blockquote>
<p><strong>ALREADY ANSWERED</strong></p>
</blockquote>
<p>I was trying to prove the result that the OP of <a href="https://math.stackexchange.com/questions/909712/evaluate-int-0-frac-pi2-ln1-cos-x-dx"><strong><em>this</em></strong></a> question is given as a hint.</p>
<p>That is to say: <em>imagine that you are not given the hint and you need to evaluate</em>:</p>
<blockquote>
<p>$$I = \int^{\pi/2}_0 \log{\cos{x}} \, \mathrm{d}x \color{red}{\overset{?}{=} }\frac{\pi}{2} \log{\frac{1}{2}} \tag{1}$$ </p>
</blockquote>
<p><em>How would you proceed?</em></p>
<hr>
<p>Well, I tried the following steps and, despite it seems that I am almost there, I have found some troubles:</p>
<ul>
<li>Taking advantage of the fact: $$\cos{x} = \frac{e^{ix}+e^{-ix}}{2}, \quad \forall x \in \mathbb{R}$$</li>
<li>Plugging this into the integral and performing the change of variable $z = e^{ix}$, so the line integral becomes a contour integral over <em>a quarter of circumference of unity radius centered at $z=0$</em>, i.e.:
$$ I = \frac{1}{4i} \oint_{|z|=1}\left[ \log{ \left(z+\frac{1}{z}\right)} - \log{2} \right] \, \frac{\mathrm{d}z }{z}$$</li>
</ul>
<blockquote>
<p>$\color{red}{\text{We cannot do this because the integrand is not holomorphic on } |z| = 1 }$</p>
</blockquote>
<ul>
<li>Note that the integrand has only one pole lying in the region enclosed by the curve $\gamma : |z|=1$ and it is holomorphic (is it?) almost everywhere (except in $z =0$), so the residue theorem tells us that:</li>
</ul>
<p>$$I = \frac{1}{4i} \times 2\pi i \times \lim_{z\to0} \color{red}{z} \frac{1}{\color{red}{z}} \left[ \underbrace{ \log{ \left(z+\frac{1}{z}\right)} }_{L} - \log{2} \right] $$</p>
<ul>
<li>As I said before, it seems that I am almost there, since the result given by eq. (1) follows iff $L = 0$, which is not true (I have tried L'Hôpital and some algebraic manipulations).</li>
</ul>
<p>Where did my reasoning fail? Any helping hand?</p>
<p>Thank you in advance, cheers!</p>
<hr>
<p>Please note that I'm not much of an expert in either complex analysis or complex integration so please forgive me if this is trivial.</p>
<hr>
<p>Notation: $\log{x}$ means $\ln{x}$.</p>
<hr>
<p>A graph of the function $f(z) = \log{(z+1/z)}$ helps to understand the difficulties:</p>
<p><img src="https://i.stack.imgur.com/jAjTP.png" alt="enter image description here"></p>
<p>where $|f(z)|$, $z = x+i y$ is plotted and the white path shows where $f$ is not holomorphic.</p>
| Matthias | 164,923 | <p>Hint: with $\cos x=u$
$$\int_0^{\pi/2}\log\cos x\mathrm{d}x=-\int_0^1\frac{\log u}{\sqrt{1-u^2}}\mathrm{d}u$$</p>
|
792,924 | <p>If a quantity can be either a scalar or a vector, how would one call that property? I could think of scalarity but I don't think such a term exists.</p>
| Gabriel Romon | 66,096 | <p>Suppose for contradiction that such a function exists. </p>
<p>Without resorting to Cauchy-Schwarz, since $f$ is continuous on a closed/compact interval, it is bounded by some positive $M$. </p>
<p>Now $$1=\left|\int_{0}^1f(t)t^n dt \right| \leq \frac{M} {n+1}$$</p>
<p>The RHS goes to $0$ whereas the LHS is $1$. </p>
|
140,754 | <p>Pleas tell me that what a "Kink" is and what this sentence means: </p>
<blockquote>
<p>Distance functions have a kink at the interface where $d = 0$ is a local minimum.</p>
</blockquote>
| Alex Becker | 8,173 | <p>In this case, I believe a "kink" in the function refers to a point at which the function fails to be differentiable. For example, the function $f(x)=|x|$ (which gives the distance between $x$ and $0$) is not differentiable at $x=0$, where the function is $0$ as well.</p>
|
152,880 | <p>I know that for every $n\in\mathbb{N}$, $n\ge 1$, there exists $p(x)\in\mathbb{F}_p[x]$ s.t. $\deg p(x)=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.</p>
<blockquote>
<p>I am interested in counting how many such $p(x)$ there exist (that is, given $n\in\mathbb{N}$, $n\ge 1$, how many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$).</p>
</blockquote>
<p>I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist $X$ irreducible polynomials of degree $n$ where $X$ is the number of...".</p>
<p>What are your thoughts ?</p>
| Qiaochu Yuan | 232 | <p><strong>Theorem:</strong> Let $\mu(n)$ denote the Möbius function. The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ is the <a href="http://en.wikipedia.org/wiki/Necklace_polynomial" rel="noreferrer">necklace polynomial</a>
$$M_n(q) = \frac{1}{n} \sum_{d | n} \mu(d) q^{n/d}.$$</p>
<p>(To get the number of irreducible polynomials just multiply by $q - 1$.)</p>
<p><em>Proof.</em> Let $M_n(q)$ denote the number in question. Recall that $x^{q^n} - x$ is the product of all the monic irreducible polynomials of degree dividing $n$. By counting degrees, it follows that
$$q^n = \sum_{d | n} d M_d(q)$$</p>
<p>(since each polynomial of degree $d$ contributes $d$ to the total degree). By <a href="http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula" rel="noreferrer">Möbius inversion</a>, the result follows. </p>
<p>As it turns out, $M_n(q)$ has a combinatorial interpretation for all values of $q$: it counts the number of <em>aperiodic necklaces</em> of length $n$ on $q$ letters, where a necklace is a word considered up to cyclic permutation and an aperiodic necklace of length $n$ is a word which is not invariant under a cyclic permutation by $d$ for any $d < n$. More precisely, the cyclic group $\mathbb{Z}/n\mathbb{Z}$ acts by cyclic permutation on the set of functions $[n] \to [q]$, and $M_n(q)$ counts the number of orbits of size $n$ of this group action. This result also follows from Möbius inversion.</p>
<p>One might therefore ask for an explicit bijection between aperiodic necklaces of length $n$ on $q$ letters and monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ when $q$ is a prime power, or at least <a href="https://mathoverflow.net/questions/769/exhibit-an-explicit-bijection-between-irreducible-polynomials-over-finite-fields">I did a few years ago</a> and it turns out to be quite elegant. </p>
<p>Let me also mention that the above closed form immediately leads to the "function field prime number theorem." Let the absolute value of a polynomial of degree $d$ over $\mathbb{F}_q$ be $q^d$. (You can think of this as the size of the quotient $\mathbb{F}_q[x]/f(x)$, so in that sense it is analogous to the norm of an element of the ring of integers of a number field.) Then the above formula shows that the number of monic irreducible polynomials $\pi(n)$ of absolute value less than or equal to $n$ satisfies
$$\pi(n) \sim \frac{n}{\log_q n}.$$</p>
|
4,344,571 | <p>In a previous exam assignment, there is a problem that asks for a proof that <span class="math-container">$\mathbb{Z}_{24}$</span> and <span class="math-container">$\mathbb{Z}_{4}\times\mathbb{Z}_6$</span> are <strong>not</strong> isomorphic.</p>
<p>We have <span class="math-container">$\mathbb{Z}_{24}$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span> if there exists a bijective function <span class="math-container">$f∶ \mathbb{Z}_{24}\rightarrow\mathbb{Z}_{4}\times\mathbb{Z}_6$</span> such that <span class="math-container">$f(a+b)=f(a)+f(b)$</span> and <span class="math-container">$f(ab)=f(a)f(b) \forall a,b\in R$</span>. Since there are exactly <span class="math-container">$24$</span> unique elements in both <span class="math-container">$\mathbb{Z}_{24}$</span> and <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span>, we can construct a bijective function <span class="math-container">$f∶ \mathbb{Z}_{24}\rightarrow\mathbb{Z}_{4}\times\mathbb{Z}_6$</span>. Consider then
<span class="math-container">$$\begin{aligned}
&f\left([a]_{24}+[b]_{24}\right) \\
&=f\left([a+b]_{24}\right) \\
&=\left([a+b]_{4},[a+b]_{6}\right) \\
&=\left([a]_{4}+[b]_{4},[a]_{6}+[b]_{6}\right) \\
&=\left([a]_{4},[a]_{6}\right)+\left([b]_{4},[b]_{6}\right) \\
&=f\left([a]_{24}\right)+f\left([b]_{24}\right)
\end{aligned}$$</span>
and
<span class="math-container">$$\begin{aligned}
&f\left([a]_{24}[b]_{24}\right) \\
&=f\left([a b]_{24}\right) \\
&=\left([a b]_{4},[a b]_{6}\right) \\
&=\left([a]_{4}[b]_{4},[a]_{6}[b]_{6}\right) \\
&=\left([a]_{4},[a]_{6}\right)\left([b]_{4},[b]_{6}\right) \\
&=f\left([a]_{24}\right) f\left([b]_{24}\right).
\end{aligned}$$</span>
It therefore seems to me that this function shows that <span class="math-container">$\mathbb{Z}_{24}$</span> <em>is</em> isomorphic to <span class="math-container">$\mathbb{Z}_4\times\mathbb{Z}_6$</span>.</p>
<p>Can someone tell me where I go wrong with this "proof", and tell me how I can show that the rings are <em>not</em> isomorphic?</p>
| Shaun | 104,041 | <p>Since <span class="math-container">$(\Bbb Z_{24},+)$</span> has an element of order <span class="math-container">$24$</span> and <span class="math-container">$(\Bbb Z_4\times \Bbb Z_6, +)$</span> does not, the two groups cannot be isomorphic as groups; hence the respective rings cannot be isomorphic as rings.</p>
|
544,464 | <p>Show that any subset of $\{1, 2, 3, ..., 200\}$ having more than $100$ members must contain at least one pair of integers which add to $201$.</p>
<p>I think it is doable using the Pigeonhole Principle.</p>
| Ross Millikan | 1,827 | <p>Hint: think about the pairs that add to $201$. How many such pairs are there?</p>
|
1,436,867 | <p>I don´t know an example wich $ \rho (Ax,Ay)< \rho (x,y) $ $ \forall x\neq y $ is not sufficient for the existence of a fixed point .
can anybody help me? please</p>
| Hamou | 165,000 | <p>$f:(0,1)\to (0,1)$ with $f(x)=x/2$.</p>
|
71,636 | <p>For a self-map $\varphi:X\longrightarrow X$ of a space $X$, many important notions of entropy are defined through a limit of the form $$\lim_{n\rightarrow\infty}\frac{1}{n}\log a_n,$$ where in each case $a_n$ represents some appropriate quantity (see, for example, <a href="https://mathoverflow.net/questions/69218/if-you-were-to-axiomatize-the-notion-of-entropy/69231#69231">this answer</a> to one of my previous questions.) Let $h(\varphi)$ denote a typical entropy that is defined by a limit as above and <strong>after</strong> Ian's example, assume that $h(\varphi)>0$. Does anyone know if limits of the form
\begin{equation}
\lim_{n\rightarrow\infty}\ \ \frac{a_n}{\exp(n\cdot h(\varphi))}
\end{equation} have been studied anywhere? I will appreciate any possible information about such limits. For example, is there a known case where the limit exists? If so, what is the limit called? etc.</p>
<p><strong>EDIT</strong>: As pointed out later by Ian, even if we assume $h(\varphi)>0$ this limit may not exist. I was curios to know if there were instances where the limit is known to exist. Or even better, can one characterize self-maps $\varphi$ for which the limit exists? </p>
| Ian Morris | 1,840 | <p>For the topological entropy of a subshift on finitely many symbols, I think that this limit will typically be infinite. Here is an example where this is the case.</p>
<p>Let $\Sigma_2= \{0,1\}^{\mathbb{N}}$ with the infinite product topology and let $T \colon \Sigma_2 \to \Sigma_2$ denote the shift transformation given by $T[(x_i)_{i=1}^\infty]:=(x_{i+1})_{i=1}^\infty$. The map $T$ is a continuous surjective transformation of the compact metrisable space $\Sigma_2$. Let us define a <em>cylinder set</em> of depth $n$ to be a set $Z \subseteq \Sigma_2$ having the form
$$Z=\{(x_i) \in \Sigma_2 \colon x_j=z_j \text{ for all }j\text{ such that }1 \leq j \leq n\}$$
for some finite sequence of symbols $z_1,\ldots,z_n \in \{0,1\}$. If $K$ is a nonempty compact subset of $\Sigma_2$ such that $TK \subseteq K$, then the topological entropy of $T$ restricted to $K$ admits the following description: if for each $n \geq 1$ we let $a_n$ be the number of distinct cylinder sets of depth $n$ which have nonempty intersection with $K$, then $h_{top}(K) = \lim_{n \to \infty} \frac{1}{n} \log a_n$. This holds because the cylinder sets form the smallest-growing family of open covers in the sense of the usual definition of topological entropy on a compact space.</p>
<p>Now, let $K \subset \Sigma_2$ be a compact $T$-invariant set of Sturmian words with some specified irrational slope (for the definition and fundamental properties of Sturmian words see e.g. Wikipedia). Such sets exist and satisfy $a_n=n+1$ for all $n \geq 1$. In particular the restriction of the shift map $T$ to $K$ has topological entropy zero and $a_ne^{-nh}=n+1 \to \infty$.</p>
<p>More generally, a little further thought shows that $a_ne^{-nh} \to \infty$ for every nonempty compact minimal invariant subset of $\Sigma_2$ which has zero topological entropy and is not equal to a periodic orbit.</p>
|
4,157,841 | <p>Q is to prove that integer just above(<span class="math-container">$\sqrt{3} + 1)^{2n}$</span> is divisible by <span class="math-container">$2^{n+1}$</span> for all n belongs to natural numbers.</p>
<p>In Q , by integer just above means that:
For an example , which is the integer just above 7.3 . It is 8. Then , Q wants you to prove that 8 is divisible by <span class="math-container">$2^{n+1}$</span>.</p>
<p><span class="math-container">\begin{equation}
\begin{array}{l}
(\sqrt{3}+1)^{2 n}=(4+2 \sqrt{3})^{n}=2^{n}(2+\sqrt{3})^{n}\\
=2^{n}(2+\sqrt{3})^{n}=2^{n}\left[^n C _{0}2^{n}+^n C_{1} 2^{n-1} \sqrt{3}+^n C_{{2}} 2^{n-2} \sqrt{3}^{2}+\right.\\
\begin{array}{l}
(\sqrt{3}-1)^{2 n}=(4-2 \sqrt{3})^{n}=2^{n}(2-\sqrt{3})^{n} \\
=2^{n}(2-\sqrt{3})^{n}=2^{n}\left[{ }^{n} c_{0} 2^{n}-n_{C}, 2^{n-1} \sqrt{3}+{ }^{n} c_{2} 2^{n-1} \sqrt{1}^{2} \ldots\right.
\end{array}\\
I+f+f)=2^{n}\left[2(\text { Integer) }]=2^{n+1}\right. \text { . Integer }\\
I+1=2^{n+1} \text { . Integer }
\end{array}
\end{equation}</span></p>
<p>In the image is the way this question is solved.</p>
<p>My Q from this method of solving is that if we notice at the end , we somehow got <span class="math-container">$ 2^{n+1}$</span>. If the Q has taken some other value like <span class="math-container">$3^{n+3}$</span> or Something else. Then , it was not possible to prove this question.</p>
<p>What is another method to prove this Q or can you help me justify that the above method can be used for all kinds of Q.</p>
<p>Thank you.</p>
| robjohn | 13,854 | <p>Part of the problem is that <a href="https://i.stack.imgur.com/9nPlD.jpg" rel="nofollow noreferrer">the image</a> was a bit obscured and so the transcription was a bit off. Here is a more accurate transcription with tags added for reference:</p>
<blockquote>
<p><span class="math-container">$$\require{cancel}
\left(\sqrt3+1\right)^{2n}=\left(4+2\sqrt3\right)^n=2^n\left(2+\sqrt3\right)^n\tag1
$$</span>
<span class="math-container">$$
\!\!\!\!I+f=2^n\left(2+\sqrt3\right)^n=2^n\left[{}^nC_02^n+\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2+\dots\right]\tag2
$$</span>
<span class="math-container">$$
\left(\sqrt3-1\right)^{2n}=\left(4-2\sqrt3\right)^n=2^n\left(2-\sqrt3\right)^n\tag3
$$</span>
<span class="math-container">$$
f'=2^n\left(2-\sqrt3\right)^n=2^n\left[{}^nC_02^n-\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2-\dots\right]\tag4
$$</span></p>
<hr />
<p><span class="math-container">$$
I+\bbox[3px,border:1px solid black]{f+f'}=2^n[2(\text{Integer})]=2^{n+1}\cdot\text{Integer}\tag5
$$</span>
<span class="math-container">$$
I+1=2^{n+1}\cdot\text{Integer}\quad\checkmark\tag6
$$</span></p>
</blockquote>
<p><span class="math-container">$(1)$</span> is just simple algebraic manipulation<br />
<span class="math-container">$(2)$</span> is expanding via the Binomial Theorem (<span class="math-container">$I$</span> and <span class="math-container">$f$</span> are the integer and fraction parts)<br />
<span class="math-container">$(3)$</span> is <span class="math-container">$(1)$</span> with the substitution <span class="math-container">$\sqrt3\mapsto-\sqrt3$</span>; note that <span class="math-container">$\left(\sqrt3-1\right)^{2n}=\left(-\sqrt3+1\right)^{2n}$</span><br />
<span class="math-container">$(4)$</span> is expanding via the Binomial Theorem<br />
<span class="math-container">$(5)$</span> adding <span class="math-container">$(2)$</span> and <span class="math-container">$(4)$</span> cancels all of the terms with <span class="math-container">$\sqrt3$</span> to an odd power<br />
<span class="math-container">$\phantom{\text{(5)}}$</span> and doubles all of the terms with <span class="math-container">$\sqrt3$</span> to an even power (these terms are integers)<br />
<span class="math-container">$(6)$</span> since <span class="math-container">$0\lt\left(4-2\sqrt3\right)\lt1$</span>, <span class="math-container">$f'\in(0,1)$</span> and by definition, <span class="math-container">$f\in[0,1)$</span><br />
<span class="math-container">$\phantom{\text{(6)}}$</span> since <span class="math-container">$I+f+f'\in\mathbb{Z}$</span>, we must have <span class="math-container">$f+f'=1$</span></p>
<hr />
<p>Here is, in my experience, a more usual proof of this.</p>
<p><span class="math-container">$\left(1\pm\sqrt3\right)^2=4\pm2\sqrt3$</span> are roots of <span class="math-container">$x^2-8x+4$</span>. Therefore, the solution to the <a href="https://en.wikipedia.org/wiki/Recurrence_relation#Solving_homogeneous_linear_recurrence_relations_with_constant_coefficients" rel="nofollow noreferrer">linear recurrence equation</a>
<span class="math-container">$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[3pt]
&=4(2u_{n-1}-u_{n-2})\tag7
\end{align}
$$</span>
is
<span class="math-container">$$
\begin{align}
u_n
&=a\left(1+\sqrt3\right)^{2n}+b\left(1-\sqrt3\right)^{2n}\\[3pt]
&=a\left(4+2\sqrt3\right)^n+b\left(4-2\sqrt3\right)^n\tag8
\end{align}
$$</span>
In particular, the sequence with <span class="math-container">$a=b=1$</span> starts out with
<span class="math-container">$$
u_0=2,u_1=8,u_2=56\tag9
$$</span>
Note that <span class="math-container">$\left.2^{n+1}\,\middle|\,u_n\right.$</span> for <span class="math-container">$n=1$</span> and <span class="math-container">$n=2$</span>. Then <span class="math-container">$(7)$</span> guarantees that
<span class="math-container">$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[6pt]
&=8\cdot2^n\frac{u_{n-1}}{2^n}-4\cdot2^{n-1}\frac{u_{n-2}}{2^{n-1}}\\
&=2^{n+1}\left(4\frac{u_{n-1}}{2^n}-\frac{u_{n-2}}{2^{n-1}}\right)\tag{10}
\end{align}
$$</span>
By induction, <span class="math-container">$(9)$</span> and <span class="math-container">$(10)$</span> show that <span class="math-container">$\left.2^{n+1}\,\middle|\,u_n\right.$</span> for all <span class="math-container">$n\ge1$</span>.</p>
<p>Since <span class="math-container">$u_n\in\mathbb{Z}$</span>,
<span class="math-container">$$
\begin{align}
u_n
&=\left(1+\sqrt3\right)^{2n}+\overbrace{\left(1-\sqrt3\right)^{2n}}^{\text{in }(0,1)}\\[3pt]
&=\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\tag{11}
\end{align}
$$</span>
Therefore, <span class="math-container">$(10)$</span> and <span class="math-container">$(11)$</span> show that
<span class="math-container">$$
\left.2^{n+1}\,\middle|\,\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\right.\tag{12}
$$</span></p>
|
126,120 | <p>Python has generators which save memory, is there a technique for generating in memory examples for your training set "on the fly".</p>
<p>For example purposes, I constructed here a regressor for blur:</p>
<pre><code>randomMask[img_] :=
Module[{t, h, g, d = ImageDimensions[img]},
t = Table[{PointSize@RandomReal[{0, .1}],
RandomChoice[{Point,
Rectangle[#, # + RandomReal[{-200, 200}, {2}]] &}]@
RandomPoint[Rectangle[{0, 0}, d]]}, {RandomChoice[{0, 1, 2, 3,
4, 8, 14, 20, 50, 200}]}];
g = Graphics[t, PlotRange -> Transpose[{{0, 0}, d}], ImageSize -> d];
{g, Area@DiscretizeGraphics@g/Times @@ d}]
makeExample[img_] := Module[{g, v},
{g, v} = randomMask[img];
ImageCompose[img, SetAlphaChannel[Blur[img, 15], ColorNegate@g]] ->
v
];
imgs = ConformImages[ExampleData /@ ExampleData["TestImage"], {100, 100}];
(* this is a large set that I don't want to precompute !!! *)
train = Table[makeExample@RandomChoice[imgs], {3000}]
test = Table[makeExample@RandomChoice[imgs], {500}];
convnet=NetChain[{
ConvolutionLayer[20,{5,5}],
ElementwiseLayer[Ramp],
PoolingLayer[{2,2},{2,2}],
ConvolutionLayer[50,{5,5}],
ElementwiseLayer[Ramp],
PoolingLayer[{2,2},{2,2}],
FlattenLayer[],
DotPlusLayer[500],
ElementwiseLayer[Ramp],
DotPlusLayer[50],
ElementwiseLayer[Ramp],
DotPlusLayer[1]
},
"Input"->NetEncoder[{"Image",{100,100}}],
"Output"->NetDecoder["Scalar"]
]
trainedConvnet = NetTrain[convnet, train, TargetDevice -> "GPU"]
output = trainedConvnet /@ Keys[test];
target = test // Values;
meanSquareLoss = Mean@Flatten[(#Output - #Target)^2, Infinity] &;
data = <|"Output" -> {{output}}, "Target" -> {{target}}|>;
N@meanSquareLoss@data
</code></pre>
| Taliesin Beynon | 7,140 | <p>It's not very well tested but you can supply a <code>"$AugmentationFunction" -> f</code> option to the Image <code>NetEncoder</code> in which you can put a <code>Blur</code> or whatever (anything that takes an image and produces an image). This option is not officially supported and it'll probably be replaced with something superior in future.</p>
<p>EDIT: we plan on supporting the ability to generate batches via a callback function in 11.1. The training data spec will look something like <code>{f, n}</code>, where f is your callback function (which should take a batchsize and produce an association mapping port to input data) and <code>n</code> is the number of examples that should count as a training round. </p>
<p>SECOND EDIT: this functionality has landed in 11.1 builds, so you can now supply an arbitrary function to produce batches of data on demand. You might want to sign up to be a beta tester.</p>
|
3,031,460 | <blockquote>
<p>Give an example of an assertion which is not true for any positive
integer, yet for which the induction step holds.</p>
</blockquote>
<p>First of all, definition.</p>
<blockquote>
<p>In <strong>inductive step</strong>, we suppose that <span class="math-container">$P(k)$</span> is true for some positive
integer <span class="math-container">$k$</span> and then we prove that <span class="math-container">$P(k + 1)$</span> is true.</p>
</blockquote>
<p><strong>My thoughts</strong> : Since the assertion has to satisfy the induction step, it must be false for <span class="math-container">$n=1$</span>. </p>
<p>But I cannot think of any such assertion.
Does anyone have any hints? Thank you.</p>
<p><strong>Edit</strong>: Just realised that my thoughts don't make much sense, as the question demands the assertion to be false for every positive integer. </p>
| Bram28 | 256,001 | <p><span class="math-container">$n < n$</span> does not hold for any <span class="math-container">$n$</span> but the inductive step holds: </p>
<p>If <span class="math-container">$k <k$</span> then <span class="math-container">$k+1<k+1$</span></p>
|
22,753 | <p>I've learned the process of orthogonal diagonalisation in an algebra course I'm taking...but I just realised I have no idea what the point of it is.</p>
<p>The definition is basically this: "A matrix <span class="math-container">$A$</span> is orthogonally diagonalisable if there exists a matrix <span class="math-container">$P$</span> which is orthogonal and <span class="math-container">$D = P^tAP$</span> where <span class="math-container">$D$</span> is diagonal". I don't understand the significance of this though...what is special/important about this relationship?</p>
| Community | -1 | <p>Arturo has given a nice answer. Here are couple of my additions to Arturo's answer.</p>
<p>Any symmetric matrix which can be diagonalized can be re-written in an orthogonal diagonalized form. Another aspects of this orthogonal diagonalization is that this essentially means that the left and right singular vectors are the same (except probably for a change of sign). The singular values for such a matrix are nothing but the magnitude of the eigen values. So in essence, the orthogonal diagonalization gives the Singular Value Decomposition as well and knowing the SVD is all you need to know about any matrix.</p>
|
164,896 | <p>I want to create a list length <code>l</code> with the function $f(x_n)=x_{n-1}+r$ where $r$ is a random real number between -1 and 1 and $x_0=1$. It got it working like this:</p>
<pre><code>l = 50; a = Range[l]; a[[1]] = 0;
For[i = 2, i <= l, i++, a[[i]] = a[[i - 1]] + RandomReal[{-1, 1}]];
a
</code></pre>
<p>But I have a feeling I'm doing it "wrong" and this can be heavily optimized and written shorter.</p>
<p>Any ideas ?</p>
<p>For example the fact that I use <code>Range</code> to create a list with elements that I never use seems wrong...</p>
| user42582 | 42,582 | <p>It's not clear if the first entry should be zero or one (the body of the question uses $x_0=1$ but in the code there is <code>a[[1]]=0</code>) and if it's <code>RandomReal</code> or <code>RandomInteger</code> the relevant function to use; here, the first entry in the list is set to <em>zero</em> (see <code>xo</code> below) and <code>RandomReal</code> is used to generate the random numbers.</p>
<pre><code>(* init *)
With[{l = 50, xo = 0},
(* make everything reproducible *)
BlockRandom[
(* localize the generation of random numbers *)
Block[{r = RandomReal[{-1, 1}, l - 1]},
(* perform the actual iteration *)
FoldList[#1 + #2 &, xo, r]
], RandomSeeding -> 123]
]
</code></pre>
<hr>
<p><strong>edit #1</strong></p>
<p>Let $X_l=\{x_t\}_{t=0}^{l-1}$ denote a series of $l$ numbers, namely $X_l=\{x_0,x_1,...,x_{l-1}\}$. Please note that when using zero-based indexing for a list of $l$ numbers, the indexes themselves run from $0$ to $l-1$. </p>
<p>Consider the following example: for $l=3$, the list of numbers is going to be $X_3=\{x_0,x_1,x_2\}$. </p>
<p>Writing something like $x_s = x_{s-1} + c$, where $c$ is a <em>scalar</em> and $s$ is an index, running from $1$ to $l-1$, allows one to impose structure on list $X_l$. </p>
<p>A recursion like this, in the case of our example with $X_3$ above, would imply that, instead of $\{x_0,x_1,x_2\}$, list $X_3$ should-effectively-be transformed into $\{x_0,x_0+c,x_0+2c\}$ (example 1). </p>
<p>A further extension of the recursion model would allow $c$ to be dependent on the <em>recursion step</em>, $s$, too. In that case, the recursion would be written as $x_s = x_{s-1} + c_s$; in which case, the $X_3$ example should be modified to read $\{x_0,x_0+c_1,x_0+c_1+c_2\}$ (example 2).</p>
<p>Now, in the body of the question, function $f$, defines a recursion that is <em>seemingly</em> (that's <em>my</em> interpretation of the question in conjunction with the code provided) similar to the first example above ($x_s = x_{s-1} + c$, where $c$ a <em>scalar</em>); furthermore, it is explicitly stated in the question that $x_0=1$ (the first number in the sequence should be equal to one).</p>
<p>When ones inspects the code segment, however, one can't help but notice that the sequence <code>a=Range[l]; a[[1]]=0;</code> effectively imposes $x_0=0$ (the <code>a[[1]]=0</code> part), <em>contrary</em> to what was originally stated. </p>
<p>Additionally, calling <code>RandomReal</code> in <code>a[[i]] = a[[i - 1]] + RandomReal[{-1, 1}]</code> will produce a <em>new</em> random real number for <em>every</em> iteration step; this behavior implies that the relevant recursion model is that of example 2 (see above).
This, however, <em>contrasts</em> with what was initially adhered to (see quote "[...] $f(x_n)=x_{n-1}+r$ where r is a random real number between -1 and 1" from the question; here, $r$ is not dependent on the recursion step).</p>
<p>Having said that, if the relevant recursion model for the original question is the one exemplified in example 1, then a straightforward way to generate a list $X_l$ is</p>
<pre><code>xo + Range[0, l-1] RandomReal[{-1, 1}]
</code></pre>
<p>On the other hand, if what's relevant for the question, is the recursion model of example 2, then a straightforward way to generate list $X_l$ is</p>
<pre><code>FoldList[#1 + #2 &, xo, RandomReal[{-1, 1}, l - 1]]
</code></pre>
<p><strong>edit #2</strong></p>
<p>The use of <code>BlockRandom</code> is intended to assist in producing <em>reproducible</em> results. </p>
<p>For the function definition <code>f[x___] := RandomReal[]</code>, evaluating <code>f[]</code> two consecutive times will produce two different random numbers.
On the contrary, evaluating <code>BlockRandom[f[]]</code> two or more consecutive times will <em>always</em> produce the <em>same</em> random number. </p>
<p>The same effect can be gotten with <code>BlockRandom[f[],RandomSeeding->123456789]</code> where <code>RandomSeeding</code> provides the seed for the random number generators (see the documentation for <a href="http://reference.wolfram.com/language/ref/BlockRandom.html" rel="nofollow noreferrer">BlockRandom</a>).</p>
<p>The same effect as <code>BlockRandom</code> can be practically achieved by, using instead <a href="http://reference.wolfram.com/language/ref/SeedRandom.html" rel="nofollow noreferrer"><code>SeedRandom</code></a> before the evaluation of <code>f</code> in the example above (although there are caveats, please refer to the documentation).</p>
<p>What this means for this answer is that you can practically have a functioning solution without using <code>BlockRandom</code> if you choose to not use it.</p>
<p>According to the documentation, <code>BlockRandom</code> was introduced in 2007 (v.6.0) and was updated in 2017 (v.11.2)</p>
|
1,702,616 | <p>I was working on a programming problem to find all 10-digit perfect squares when I started wondering if I could figure out how many perfects squares have exactly N-digits. I believe that I am close to finding a formula, but I am still off by one in some cases.</p>
<p>Current formula where $n$ is the number of digits:</p>
<p>$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}}\rfloor$ </p>
<p>How I got here: </p>
<ol>
<li>Range of possible 10-digit numbers is from $10^9$ to $10^{10}-1$</li>
<li>10-digit perfect squares should fall into the range of $\sqrt{10^9}$ to $\sqrt{10^{10}-1}$</li>
</ol>
<p>Results of program for $n = 1,2,3,4,5$:</p>
<pre><code>DIGITS=1, ACTUAL_COUNT=3, COMPUTED_COUNT=2
DIGITS=2, ACTUAL_COUNT=6, COMPUTED_COUNT=6
DIGITS=3, ACTUAL_COUNT=22, COMPUTED_COUNT=21
DIGITS=4, ACTUAL_COUNT=68, COMPUTED_COUNT=68
DIGITS=5, ACTUAL_COUNT=217, COMPUTED_COUNT=216
</code></pre>
<p>Program: </p>
<pre><code>#!/usr/bin/perl
use strict;
use warnings;
sub all_n_digit_perfect_squares {
my ($n) = @_;
my $count = 0;
my $MIN = int( sqrt( 10**($n-1) ) );
my $MAX = int( sqrt( (10**$n)-1 ) );
foreach my $i ( $MIN .. $MAX ) {
if ( ($i * $i) >= 10**($n-1) ) {
$count++;
}
}
print "DIGITS=$n"
. ", ACTUAL_COUNT=$count"
. ", COMPUTED_COUNT="
. ($MAX-$MIN),
"\n";
return;
}
all_n_digit_perfect_squares($_) for (1..5);
</code></pre>
<p>Any advice on where I went wrong would be helpful.</p>
| ALam | 685,033 | <p>I'm interested in the question that you're asking. I've been thinking about this for the last few days and couldn't find anything online about this subject. If you have anything references, I would appreciate it. Here is what I have so far and sorry if the formatting is off, this is my first post.</p>
<p><span class="math-container">$$ \text{Let}\; \mathbb{N} \;\text{be the set of Natural numbers} \; = \{1,2,3,4,..,N\},$$</span>
Then there exists a set <span class="math-container">$\mathbb{S}$</span>, such that for
<span class="math-container">$$ s\in\mathbb{S}\; ;\; s_i = \{10^{i-1},...,(10^i-1)\} \;, \text{for} \; i = 1,2,3,...,M. $$</span>
Example : <span class="math-container">$$s_1 = \{1,2,3,...,9\}\; ; \;\;\; i=1 = 1\;\text{digit}$$</span>
<span class="math-container">$$ s_2 = \{10,11,12,...,99\}\; ; \;\; i=2 = 2\;\text{digits} $$</span>
<span class="math-container">$$ s_3 = \{100,101,102,...,999\}\; ; \;\; i=3 = 3\;\text{digits} $$</span>
<span class="math-container">$$ ... $$</span>
<span class="math-container">$$ s_i = \{10^{i-1},...,(10^i - 1)\} ; \;\; i\; \text{digits} $$</span></p>
<p>Within each set <span class="math-container">$s_i$</span>, there exists a set <span class="math-container">$\mathbb{A} $</span> such that for all members of <span class="math-container">$\mathbb{A}$</span>, call it <span class="math-container">$a_p$</span>, when squared are still a member of <span class="math-container">$s_i$</span>: <span class="math-container">$$a\in\mathbb{A}\; ; \;\; a = \{ a_1,a_2,...,a_k \} $$</span>
<span class="math-container">$$(a_p)^2 \in s_i ,\; \text{for}\; p = 1\; \text{to}\; k$$</span> </p>
<p>Example :
<span class="math-container">$$s_1 = \{1,2,3,...,9\}\; ; \;\;\; a_1 = \{1,2,3\}$$</span>
<span class="math-container">$$ s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{ a_1 , 4, 5,6,...,9\} = \{1,2,3 , 4, 5,6,...,9\} $$</span></p>
<p>I'm going to start omit the previous sets before the current <span class="math-container">$i^{th}$</span> set we're looking at. So it'll look like this: </p>
<p><span class="math-container">$$s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{4, 5,6,...,9\}, $$</span></p>
<p>but also know that all previous sets before the current <span class="math-container">$i^{th}$</span> can included, but is not unique to the <span class="math-container">$i^{th}$</span> digits.</p>
<p>I'm going to list a few in the long version, just as an example.
<span class="math-container">$$ s_3 = \{100,101,102,...,999\}\; ; \;\; a_3 = \{10,11,12,...,31 \}$$</span>
<span class="math-container">$$ s_4 = \{1000,1002,1003,...,9999\}\; ; \;\; a_4 = \{32,33,34,...,99 \}$$</span>
<span class="math-container">$$ s_5 = \{10000,...,99999\}\; ; \;\; a_5 = \{100,...,316 \} $$</span>
<span class="math-container">$$ s_6 = \{100000,...,999999\}\; ; \;\; a_6 = \{317,...,999 \} $$</span>
<span class="math-container">$$ s_7 = \{1000000,...,9999999\}\; ; \;\; a_7 = \{1000,...,3162 \} $$</span></p>
<p>Now we can write these compactly as, the first i=1,2,3 are written in the long way and i > 4 is written in the short hand version as given by <span class="math-container">$s_i$</span>:
<span class="math-container">$$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$</span>
<span class="math-container">$$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6 $$</span><br>
<span class="math-container">$$s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22 $$</span><br>
<span class="math-container">$$s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$</span>
<span class="math-container">$$s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$</span></p>
<p>I also introduced <span class="math-container">$t_i$</span>, which is the count of the number of members of <span class="math-container">$a_i$</span>, (this is exactly what you're solving for, I believe). Notice, we can get an exact solution for <span class="math-container">$t_i$</span> given by:
<span class="math-container">$$ t_i = (\{a_i\}_{max} - \{a_i\}_{min}) + 1 $$</span>
(I think this could be calculated using a norm)</p>
<p>I'm going to list out a few sequences and point out an interesting patterns.</p>
<p>Starting with <span class="math-container">$i=3$</span>, (you could extend this to <span class="math-container">$i=1$</span> if you want),</p>
<p>if <span class="math-container">$i = odd$</span> :
The lower bound of <span class="math-container">$a_i$</span>, call this <span class="math-container">$\{a_i\}_{min}$</span> is exactly <span class="math-container">$10^l$</span>, where <span class="math-container">$l = (i-1)/2$</span> ;</p>
<p>if <span class="math-container">$i = even$</span> :
The upper bound of <span class="math-container">$a_i$</span>, call this <span class="math-container">$\{a_i\}_{max}$</span> is exactly <span class="math-container">$10^z - 1$</span>, where <span class="math-container">$z = (i)/2$</span> ;</p>
<p>I claim that for <span class="math-container">$i=even$</span>, you have</p>
<p><span class="math-container">$$ s_{i-1} = \{10^{i-2},...,(10^{i-1}-1)\}\; ; \;\; a_{i-1} =\{10^l,...,upper\;bound_{(i-1)}\} \; ; \; t_{i-1} = ? $$</span>
<span class="math-container">$$ s_{i} = \{10^{i-1},...,(10^{i}-1)\}\; ; \;\; a_{i} =\{lower\; bound_{(i)},...,(10^z-1)\} \; ; \; t_{i} = ?.$$</span></p>
<p>I'm still working/thinking about I can get the upper bounds, lower bounds, and <span class="math-container">$t_i$</span>. </p>
<p>Here are the first 16 digits, and their lists. I also have done this out until 34 digits, took my computer a good few hours to compute by brute force. I am almost done with an algorithm that can do it much quicker. </p>
<p>Also note, that if you know the lower bounds and upper bounds, you can easily obtain <span class="math-container">$t$</span>, which is the number of elements that are contained in <span class="math-container">$\{a_i\}$</span>, since <span class="math-container">$\{a_i\}$</span> is ordered.</p>
<p><span class="math-container">$$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$</span>
<span class="math-container">$$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6 $$</span><br>
<span class="math-container">$$ s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22 $$</span><br>
<span class="math-container">$$ s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$</span>
<span class="math-container">$$ s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$</span>
<span class="math-container">$$ s_6= \{10^5,...,(10^6 -1)\}\; ; \; a_6 = \{317,...,999\}\; ; \; t_6 = 683$$</span>
<span class="math-container">$$ s_7= \{10^6,...,(10^7 -1)\}\; ; \; a_7 = \{10^3,...,3162\}\; ; \; t_7 = 2163$$</span>
<span class="math-container">$$ s_8= \{10^7,...,(10^8 -1)\}\; ; \; a_8 = \{3163,...,9999\}\; ; \; t_8 = 6837$$</span>
<span class="math-container">$$ s_9= \{10^8,...,(10^9 -1)\}\; ; \; a_9 = \{10^4,...,31622\}\; ; \; t_9 =21623$$</span>
<span class="math-container">$$ s_{10}= \{10^9,...,(10^{10} -1)\}\; ; \;\; a_{10} = \{31623,...,(10^5-1)\}\; ; \; t_{10} = 68377$$</span>
<span class="math-container">$$ s_{11}= \{10^{10},...,(10^{11} -1)\}\; ; \;\; a_{11} = \{10^5,...,316227\} \; ; \; t_{11} = 216228$$</span>
<span class="math-container">$$ s_{12}= \{10^{11},...,(10^{12} -1)\}\; ; \;\; a_{12} = \{316228,...,(10^6-1)\}\; ; \; t_{12} = 683772$$</span>
<span class="math-container">$$ s_{13}= \{10^{12},...,(10^{13} -1)\}\; ; \;\; a_{13} = \{10^6,...,3162277\}\; ; \; t_{13} = 2162278$$</span>
<span class="math-container">$$ s_{14}= \{10^{13},...,(10^{14} -1)\}\; ; \;\; a_{14} = \{3162278,...,(10^7-1)\} \; ; \; t_{14} = 6837722$$</span>.
<span class="math-container">$$ s_{15}= \{10^{14},...,(10^{15} -1)\}\; ; \;\; a_{15} = \{10^7,...,31622776\} \; ; \; t_{15} = 21622777$$</span>
<span class="math-container">$$ s_{16}= \{10^{15},...,(10^{16} -1)\}\; ; \;\; a_{16} = \{31622777,...,(10^8-1)\} \; ; \; t_{16} = 68377223$$</span></p>
|
1,491,484 | <p>Let $a,b,x \in Z^+$. Prove that $\operatorname{lcm}(ax,bx) = \operatorname{lcm}(a,b)\cdot x$.</p>
<p>Here are my thoughts: </p>
<p>Let $d = \operatorname{lcm}(ax, bx)$. By definition $ax|d$ and $bx|d$. Now it can be seen that $a|d$ and $b|d$. So, let e = lcm(a,b). e is merely the lcm(ax, bx) (which equals d) multiplied by x. So, ex = d, which means that x $\cdot$ lcm(a,b) = d." </p>
<p>I am not certain of my proof's validity and I feel it is too informal to be considered valid. Any thoughts?</p>
| 5xum | 112,884 | <p>Your proof, as written now, is incorrect:</p>
<blockquote>
<p>Now it can be seen that $d|a$ and $d|b$.</p>
</blockquote>
<p>This is false, since if $a=6$, $b=4$, then $d=12$, and it is not true that $d|a$.</p>
|
1,942,364 | <p>How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.</p>
<p>How can I go about counting the number of squares of each size?</p>
| Jean-Claude Arbaut | 43,608 | <p>Another way to view this: for each size, there is one square that you move inside the larger one.</p>
<p>How much can it move? A square of size $k\times k$ inside a square of size $n\times n$ has $n-k+1$ possible positions in each direction (up-down and left-right): consider the position of the top left corner for instance. Hence there are $(n-k+1)^2$ such squares.</p>
<p>Can you finish from here?</p>
|
1,942,364 | <p>How many squares exist in an $n \times n$ grid? There are obviously $n^2$ small squares, and $4$ squares of size $(n-1) \times (n-1)$.</p>
<p>How can I go about counting the number of squares of each size?</p>
| Sahil Kumar | 363,778 | <p><strong>NOTE:</strong></p>
<p>For $n\times n$ grid the answer is $n^2+(n-1)^2+(n-2)^2+.....+1^2$</p>
<p>So, the answer is
$$
\frac{n(n+1)(2n+1)}{6}
$$</p>
|
210,655 | <p>The lower density of $A\subseteq\mathbb{N}$ is defined to be $\lambda(A)=\lim\text{inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$. We set $${\cal C} = \{A\subseteq \mathbb{N}: \lambda(\mathbb{N}\setminus A) = 1 - \lambda(A)\}.$$</p>
<p>Do both ${\cal C}$ and ${\cal P}(\mathbb{N})\setminus {\cal C}$ have cardinality $2^{\aleph_0}$?</p>
| Stefan Kohl | 28,104 | <p>The answer is <em>yes</em>: Given a set $A$ in $\mathcal{C}$ or in
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$, you can take
any subset $B \subset \mathbb{N} \setminus A$ of lower density $0$,
and $A \cup B$ is still in $\mathcal{C}$ or in
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$, depending on
which of the two classes $A$ itself belongs to.
Now, unless $A$ is cofinite, the cardinality of the set of subsets
of $\mathbb{N} \setminus A$ with lower density $0$ is $2^{\aleph_0}$,
so you have $2^{\aleph_0}$ choices for $B$. The result follows since
each of the classes $\mathcal{C}$ and
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$ has at least one
member which is not cofinite, and since the set of cofinite subsets of $\mathbb{N}$
is countable.</p>
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3,706,332 | <p>Let <span class="math-container">$f: \mathbb{R} \to \mathbb{R} $</span> be a differentiable function. Is it true that <span class="math-container">$f$</span> is strictly increasing on <span class="math-container">$\mathbb R$</span> if and only if <span class="math-container">$f'(x) \geq 0$</span> on <span class="math-container">$\mathbb R$</span> and the set <span class="math-container">$X = \{x \in \mathbb{R} \,| \, f'(x) = 0\}$</span> is countable?</p>
| jacob bradley | 658,280 | <p>Edit: the op has edited the question to strictly increasing from simply increasing thus my answer is no longer valid. I think this isn't true and here is why, Take <span class="math-container">$f(x) = -x^2$</span> when <span class="math-container">$x<0$</span>, <span class="math-container">$f(x) = 0$</span> when <span class="math-container">$x \in [0,1]$</span> and <span class="math-container">$f(x) = x^2-1$</span> when <span class="math-container">$x>1$</span>, this function is clearly continuous, differentiable and increasing (not strictly). Now for the set of points between [0,1] f'(x) = 0, but x is an element of the reals so any closed interval such as [0,1] is uncountable.</p>
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