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 |
|---|---|---|---|---|
3,977,687 | <p>A coin of radius 1 cm is tossed onto a plane surface that has been tessellated by right triangles whose sides are 8 cm, 15 cm, and 17 cm long. Find the probability that the coin lands within a triangle.</p>
<p>I know that this has to do with similarity because the inner triangle that is formed by the area where the coin can land is similar to the outer triangle. Therefore, I know the angles of this triangle, but I am not sure how to find one side of this triangle.</p>
| heropup | 118,193 | <p>If <span class="math-container">$O$</span> is the center of the coin, the locus of <span class="math-container">$O$</span> inside a single <span class="math-container">$8$</span>-<span class="math-container">$15$</span>-<span class="math-container">$17$</span> right triangle <span class="math-container">$\triangle ABC$</span> such that the coin does not intersect any of the sides, is another triangle, say <span class="math-container">$\triangle A'B'C'$</span>, that is similar in shape. Moreover, <span class="math-container">$\triangle A'B'C'$</span> shares the same <em>incenter</em> as <span class="math-container">$\triangle ABC$</span> because corresponding pairs of vertices, e.g. <span class="math-container">$A$</span> and <span class="math-container">$A'$</span>, lie on the same angle bisectors.</p>
<p>Note that <span class="math-container">$$|\triangle ABC| = rs$$</span> where <span class="math-container">$r$</span> is the inradius and <span class="math-container">$s = (a+b+c)/2$</span> is the semiperimeter. Similarly, <span class="math-container">$$|\triangle A'B'C'| = r's'$$</span> where <span class="math-container">$r' = r-1$</span> is the inradius of the smaller triangle. And because of similarity, <span class="math-container">$s'/s = r'/r$</span>; consequently <span class="math-container">$$\frac{|\triangle A'B'C'|}{|\triangle ABC|} = \frac{r's'}{rs} = \left(\frac{r'}{r}\right)^2 = \left(1 - \frac{1}{r}\right)^2.$$</span> Since <span class="math-container">$|\triangle ABC| = (8)(15)/2 = 60$</span> and <span class="math-container">$s = (8+15+17)/2 = 20$</span>, we have <span class="math-container">$r = 3$</span> and the desired probability is <span class="math-container">$$(1 - 1/3)^2 = \frac{4}{9}.$$</span></p>
|
2,300,613 | <p>I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way? </p>
<p>$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$</p>
| Luke | 450,423 | <p>$$
g(z) = a_0+a_1z+a_2z^2+...,\\
\frac{g(z)}{1-z}=a_0\frac{1}{1-z}+a_1\frac{z}{1-z}+a_2\frac{z^2}{1-z}+...
$$
Using the power series for $\frac{1}{1-z}$ gives
$$
g(z)=a_0(1+z+z^2+...)+a_1z\cdot(1+z+z^2+...)+a_2z^2\cdot(1+z+z^2+...),\\
g(z)=a_0(1+z+z^2+...)+a_1\cdot(z+z^2+z^3+...)+a_2\cdot(z^2+z^3+z^4...),\\
g(z)=a_0+(a_0+a_1)z+(a_0+a_1+a_2)z^2+...
$$
For the specific case $g(z)=e^z$
$$
\frac{e^z}{1-z}=\frac{1}{0!}+\left(\frac{1}{0!}+\frac{1}{1!}\right)z+\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}\right)z^2+...+\left(\sum_{r=0}^n\frac{1}{r!}\right)z^n+...
$$
as required.</p>
|
3,395,098 | <p>I am trying to work out for what <span class="math-container">$\lambda_1, \lambda_2 > 0$</span> is it true that <span class="math-container">$f(y) = \lambda_1 e^{y-\lambda_1 e^y} + \lambda_2 e^{y-\lambda_2 e^y}$</span> is unimodal?</p>
<p>Experimentally it seems it is unimodal when <span class="math-container">$\lambda_1 < \lambda_2$</span> and <span class="math-container">$\frac{\lambda_2}{\lambda{1}} < 7.5$</span> .</p>
<p>To work this out I started with:</p>
<p><span class="math-container">$$\frac{d}{dy} \left(\lambda_1 e^{y-\lambda_1 e^y} + \lambda_2 e^{y-\lambda_2 e^y} \right) = \lambda_1 e^{y - \lambda_1 e^y} (1 - \lambda_1 e^y ) + \lambda_2 e^{y - \lambda_2 e^y} (1 - \lambda_2 e^y )$$</span></p>
<p>It seems we then need to check when</p>
<p><span class="math-container">$$\lambda_1 e^{y - \lambda_1 e^y} (1 - \lambda_1 e^y ) + \lambda_2 e^{y - \lambda_2 e^y} (1 - \lambda_2 e^y ) = 0$$</span></p>
<p>has more than one solution when solved for <span class="math-container">$y \in \mathbb{R}$</span>. How can we determine the conditions under which it has different numbers of solutions?</p>
<h1>Added:</h1>
<p>Substituting <span class="math-container">$z = e^y$</span> and dividing by <span class="math-container">$e^{y-1}$</span> we are trying to determine how many solutions</p>
<p><span class="math-container">$$
\lambda_1 e^{1-\lambda_1 z}(1-\lambda_1 z) +\lambda_2e^{1-\lambda_2 z}(1-\lambda_2 z) = 0
$$</span></p>
<p>has with <span class="math-container">$z > 0$</span>.</p>
<h1>Examples:</h1>
<p>Example <span class="math-container">$\lambda_1 = 1, \lambda_2 = 7$</span> with only one mode (code in python):</p>
<pre><code>import matplotlib.pyplot as plt
import numpy as np
def pdf_func(y, params):
return sum([lambd*np.exp(y - lambd * np.exp(y)) for lambd in params])
params = [1, 7]
xs = np.linspace(-10,10,1000)
plt.plot(xs, [pdf_func(y, params) for y in xs])
</code></pre>
<p><a href="https://i.stack.imgur.com/iN2Yd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iN2Yd.png" alt="enter image description here" /></a></p>
<p>Example <span class="math-container">$\lambda_1 = 1, \lambda_2 = 50$</span> with two modes:</p>
<p><a href="https://i.stack.imgur.com/HFAnp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HFAnp.png" alt="enter image description here" /></a></p>
<h1>Questions</h1>
<ul>
<li>How can one prove (assuming it is true) that that the number of local maxima that <span class="math-container">$f(y)$</span> has is either 1 or 2 and there are no other possibilities?</li>
<li>Is it true that for <span class="math-container">$\lambda_2 > \lambda_1 > 0$</span>, there exists a threshold <span class="math-container">$c$</span> so that if <span class="math-container">$\frac{\lambda_2}{\lambda_1} < c$</span> then <span class="math-container">$f(y)$</span> is unimodal and if not it has two local maxima? (My guess is that the answer is yes and this threshold is around <span class="math-container">$7.5$</span>.)</li>
</ul>
| Szeto | 512,032 | <p>Define <span class="math-container">$f(x)=ae^{-ax}(1-ax)+be^{-bx}(1-bx)$</span> for <span class="math-container">$x>0$</span>, where <span class="math-container">$b>a>0$</span>. </p>
<p>The OP is interested in the roots of <span class="math-container">$f$</span>.</p>
<p>Let <span class="math-container">$r=\frac ba$</span>.</p>
<blockquote>
<p><strong>Lemma 1</strong> <span class="math-container">$f$</span> has exactly one root in <span class="math-container">$(0,\frac2b)$</span>.</p>
</blockquote>
<p><strong>Sketch of proof:</strong></p>
<ul>
<li><span class="math-container">$f(0)=a+b >0$</span></li>
<li><span class="math-container">$\displaystyle{f\left(\frac2b\right)<0}$</span> </li>
<li>The derivative of <span class="math-container">$g_c(x):= ce^{-cx}(1-cx)$</span> is <span class="math-container">$g_c’(x)=-c^2e^{-cx}(2-cx)$</span>. Thus, <span class="math-container">$g_c$</span> is decreasing on <span class="math-container">$(0,\frac2c)$</span>, and <span class="math-container">$f=g_a+g_b$</span> is decreasing on <span class="math-container">$(0,\min(\frac2a,\frac2b))=(0,\frac2b)$</span>.</li>
</ul>
<p>By intermediate value theorem, there exists at least one zero in <span class="math-container">$(0,\frac2b)$</span>. Monotonicity implies uniqueness of the zero.</p>
<blockquote>
<p><strong>Lemma 2</strong>
<span class="math-container">$$\text{number of zeroes in $\left[\frac2b,\infty\right)$}=
\begin{cases}
0, & r<\rho \\
1, & r=\rho \\
2, & r>\rho \\
\end{cases}
$$</span>
where <span class="math-container">$\rho\approx 7.566$</span> and satisfies <span class="math-container">$$\exp\left[h(\rho)\left(1-\frac1\rho\right)\right]+\rho^2\cdot\frac{1-h(\rho)}{\rho-h(\rho)}=0,\quad h(t)=\frac12\left(1+t+\sqrt{t^2-6t+1}\right)$$</span></p>
</blockquote>
<p><strong>Ideas:</strong></p>
<p>I have not been able to come up with a formal proof. But intuition and numerical experiments have led me to believe that the critical case occurs when the zero of <span class="math-container">$f$</span> is also a stationary point.</p>
<p><a href="https://i.stack.imgur.com/D1NYc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D1NYc.jpg" alt=""></a>
<a href="https://i.stack.imgur.com/MrULw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MrULw.png" alt=""></a>
<a href="https://i.stack.imgur.com/igwzE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/igwzE.jpg" alt=""></a></p>
<p>Therefore, we want to solve for <span class="math-container">$\rho=\frac ab$</span>, given that</p>
<p><span class="math-container">$$\quad f(x)=ae^{-ax}(1-ax)+be^{-bx}(1-bx)=0$$</span>
<span class="math-container">$$\implies ae^{-ax}(1-ax)=-be^{-bx}(1-bx)\qquad{(*)}$$</span></p>
<p><span class="math-container">$$\quad f'(x)=-a^2e^{-ax}(2-ax)-b^2e^{-bx}(2-bx)=0$$</span>
<span class="math-container">$$\implies a^2e^{-ax}(2-ax)=-b^2e^{-bx}(2-bx)$$</span></p>
<p>Dividing the two equations give
<span class="math-container">$$a\cdot\frac{2-ax}{1-ax}=b\cdot\frac{2-bx}{1-bx}$$</span></p>
<p>With a little algebra we get
<span class="math-container">$$ab\cdot x^2-(a+b)x+2=0$$</span></p>
<p>By quadratic formula,
<span class="math-container">$$x=\frac12\left(\frac1a+\frac1b+\sqrt{\frac1{a^2}+\frac1{b^2}-\frac6{ab}}\right)
=\frac1{2b}\left(1+\rho+\sqrt{\rho^2-6\rho+1}\right)=\frac{h(\rho)}b$$</span></p>
<p>(Again, experiments tell that only the positive root should be considered.)</p>
<p>Hence, <span class="math-container">$ax=h(\rho)/\rho$</span> and <span class="math-container">$bx=h(\rho)$</span>. Substituting this back into <span class="math-container">$(*)$</span>,
<span class="math-container">$$ae^{-h(\rho)/\rho}\left(1-\frac{h(\rho)}{\rho}\right)=-be^{-h(\rho)}(1-h(\rho))$$</span></p>
<p>or</p>
<p><span class="math-container">$$\exp\left[h(\rho)\left(1-\frac1\rho\right)\right]+\rho^2\cdot\frac{1-h(\rho)}{\rho-h(\rho)}=0$$</span></p>
<p>This equation has two roots, <span class="math-container">$\rho\approx7.566$</span> or <span class="math-container">$\rho’=\frac1\rho\approx 0.132$</span>. The latter is rejected as <span class="math-container">$b>a\implies \rho>1$</span>.</p>
<p>Any ideas on how to turn these observations into a formal proof?</p>
<hr>
<p><strong>23rd October 2019 edit</strong></p>
<p>I'd like to view the whole problem from a slightly different perspective. Alternatively, consider the system
<span class="math-container">$$
\begin{cases}
ye^{-yx}(1-yx)+ae^{-ax}(1-ax)=0 \\
y=b \\
y\ge a
\end{cases}
\qquad{(S)}
$$</span></p>
<p>One advantage of this approach is that the increasing trend of the number of solution (<span class="math-container">$1\to 2\to 3$</span>) is more obviously illustrated by the implicit function:
<a href="https://i.stack.imgur.com/rSZc0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rSZc0.png" alt=""></a></p>
<p>The proof may be thus simplified.</p>
<p><em>Intuition</em>: What does this graph actually mean? Imagine that you are plotting <span class="math-container">$y=be^{-bx}(1-bx)+ae^{-ax}(1-ax)$</span> for a fixed <span class="math-container">$a$</span> and a varying <span class="math-container">$b$</span>. Then:</p>
<ol>
<li>Take a snapshot of the x-axis of the graph.</li>
<li>Make a small increment to <span class="math-container">$b$</span>.</li>
<li>Take another snapshot of the x-axis of the graph.</li>
<li>Stack/append this snapshot on top of the previous one.</li>
<li>Repeat.</li>
</ol>
<p>Then you will get the graph above.</p>
<p>Since we are interested only in the distribution of x-intercepts (i.e. roots) of <span class="math-container">$y=be^{-bx}(1-bx)+ae^{-ax}(1-ax)$</span> for different values of <span class="math-container">$b$</span>, the graph above essentially captures all the information that is of our interest.</p>
<p><em>A few more words:</em> The implicit function has two branches, and they are separated by <span class="math-container">$x=\frac2y$</span>.<a href="https://i.stack.imgur.com/7jAYq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7jAYq.png" alt=""></a> (This observation is indeed a direct consequence of Lemma 1.)</p>
<p>In other words,
<span class="math-container">$$
\begin{cases}
ye^{-yx}(1-yx)+ae^{-ax}(1-ax)=0 \\
y>\frac2x \\
y\ge a
\end{cases}
$$</span>
uniquely defines a 'function', which is the parabola-like branch in the graph. </p>
<p>We will focus on this branch only. Call this branch <span class="math-container">$Y_a(x)$</span>.</p>
<hr>
<p>To prove Lemma 2, it suffices to prove the equivalent version</p>
<blockquote>
<p><span class="math-container">$$\text{The number of solutions of $Y_a(x)=b$ is }
\begin{cases}
0, & r<\rho \\
1, & r=\rho \\
2, & r>\rho \\
\end{cases}
$$</span></p>
</blockquote>
<p>The basic steps of the proof will be:</p>
<ol>
<li>Prove <span class="math-container">$\lim_{x\to 0^+}Y_a(x)=+\infty$</span>.</li>
<li>Prove <span class="math-container">$\lim_{x\to (1/a)^-}Y_a(x)=+\infty$</span>.</li>
<li><span class="math-container">$Y_a(x)$</span> has only one stationary point in <span class="math-container">$(0,\frac1a)$</span>, and it is a minimum attained at <span class="math-container">$y=a\rho$</span>.</li>
</ol>
<p>When these three statements are proved, it can be shown that</p>
<ol>
<li><span class="math-container">$Y_a(x)=b$</span> has no solutions when <span class="math-container">$r<\rho$</span>, as <span class="math-container">$b$</span> is smaller than the minimum of <span class="math-container">$Y_a$</span>.</li>
<li><span class="math-container">$Y_a(x)=b$</span> has exactly one solution when <span class="math-container">$r=\rho$</span>, and the solution is at the minimum.</li>
<li><p><span class="math-container">$Y_a(x)=b$</span> has two solutions when <span class="math-container">$r>\rho$</span>. Let the coordinates of the minimum be <span class="math-container">$(k,a\rho)$</span>. Consider <span class="math-container">$Z(x)=Y_a(x)-b$</span></p>
<ul>
<li>Since <span class="math-container">$Z(0)=+\infty>0$</span> and <span class="math-container">$Z(k)=a\rho-b<0$</span>, by intermediate value theorem (IVM) <span class="math-container">$Z(x)$</span> has roots in <span class="math-container">$(0,k)$</span>. Moreover, absence of local minimum in <span class="math-container">$(0,k)$</span> implies monotonic decrease of <span class="math-container">$Z$</span>. Therefore, <span class="math-container">$Z(x)$</span> has one unqiue root in <span class="math-container">$(0,k)$</span>.</li>
<li>Similarly, <span class="math-container">$Z(k)<0$</span> and <span class="math-container">$Z(\frac1a)=+\infty>0$</span> together with absence of local maximum in <span class="math-container">$(k,\frac1a)$</span> give another unqiue root in <span class="math-container">$(k,\frac1a)$</span>.</li>
</ul></li>
</ol>
<p>This is the full outline of the proof. What remains is the proof of the first three statements.</p>
<hr>
<p><strong>Here is the proof of the first statement:</strong></p>
<p>From the definition of <span class="math-container">$Y$</span>, <span class="math-container">$$Y>\frac2x\implies\lim_{x\to 0^+}Y>\lim_{x\to 0^+}\frac2x=+\infty$$</span></p>
<p>Therefore, <span class="math-container">$Y$</span> diverges to <span class="math-container">$+\infty$</span> as <span class="math-container">$x\to0^+$</span>.</p>
<p><strong>Proof of statement 2:</strong></p>
<p>Let <span class="math-container">$\phi(z)=ze^{-z}(1-z)$</span>.</p>
<p>Then <span class="math-container">$\phi(xY)+\phi(ax)=0$</span>.</p>
<p>Assume <span class="math-container">$Y$</span> is bounded in a neighbourhood of <span class="math-container">$x=\frac1a^-$</span>.</p>
<p>Taking limits on both sides,
<span class="math-container">$$\lim_{x\to(1/a)^-}\phi(xY)=0$$</span>
<span class="math-container">$$\phi\left(\frac1a\lim_{x\to(1/a)^-}Y\right)=0$$</span>
<span class="math-container">$$\implies Y\to0\text{ or }Y\to a$$</span></p>
<p>Either value does not satisfy the definition of <span class="math-container">$Y$</span>: <span class="math-container">$Y>\frac2x$</span>.</p>
<p>Hence <span class="math-container">$Y$</span> is not bounded near <span class="math-container">$x=\frac1a^-$</span>. If <span class="math-container">$Y$</span> diverges to <span class="math-container">$-\infty$</span>, then <span class="math-container">$\lim_{x\to(1/a)^-}\phi(xY)=-\infty\ne 0$</span>. Hence, <span class="math-container">$Y$</span> diverges to <span class="math-container">$+\infty$</span>.</p>
<p>The proof of statement 3 will be added soon.</p>
|
945,104 | <p>7 people are attending a concert.</p>
<p>(a) In how many different ways can they be seated in a row?</p>
<p>(b) Two attendees are Alice and Bob. What is the probability that Alice sits next to Bob?</p>
<p>(c) Bob decides to make Alice a rainbow necklace with 7 beads, each painted a different
colour on one side (red, orange, yellow, blue, green, indigo, violet), placed on a chain
that is then closed to form a circle. How many different necklaces can he make? (Since
the beads can slide along the chain, the necklace with beads R O Y G B I V would be
considered the same as O Y G B I V R for example. The beads are plain on the back, so
the necklace cannot be turned over.)</p>
<p>How should i approach these questions? Are they correct?</p>
<p>for the first one i understand that it is a permutation.
Therefore would a) = 7! = 5040 possible different ways of sitting in a row </p>
<p>b) p(7,2)
= 7!/(7-2)! = 5040/120 = 42 therefore probabiltiy = 42/5040 = 0.0083%</p>
<p>c) =6!/2
because the first bead doesnt matter and over 2 as it can either go left or right.</p>
| Marc van Leeuwen | 18,880 | <p>Since André Nicolas only left (c) to be answered I'll just mention that one. Since no starting point for the beads is fixed, one can choose any one bead as reference point and slide it to the first position; I'll arbitrarily choose the orange bead for this (O looks a bit like $0$). Then each necklace is uniquele determined by "seating" the remaining beads Y G B I V R in the six remaining positions relative to O; this is like problem (a) but with $6$ in place of $7$.</p>
|
4,350,695 | <p>My book was introducing the concept of integrals and wrote this:</p>
<p><span class="math-container">$$\text{Area under the curve of $f(x)$}=\lim_{\Delta x\to0}\sum_{n=1}^{N}f(x)\Delta x\tag{1}$$</span></p>
<p>My problem with <span class="math-container">$(1)$</span> is that there is no <span class="math-container">$n$</span> in the expression <span class="math-container">$f(x)\Delta x$</span>. So, what does <span class="math-container">$n=1$</span> below the summation notation even mean then? To elucidate my point:</p>
<p>Let,</p>
<p><span class="math-container">$y=f(x)$</span></p>
<p>Now,</p>
<p><span class="math-container">$$\text{Area under the curve of $f(x)$ from $y_1$ to $y_N$}=\lim_{\Delta x\to0}\sum_{n=1}^{N}y_n\Delta x\tag{2}$$</span></p>
<p>Now, the usage of <span class="math-container">$n$</span> makes sense because there is an actual <span class="math-container">$n$</span> present in the expression <span class="math-container">$y_n\Delta x$</span>.</p>
<p>So, is my book wrong?</p>
| peek-a-boo | 568,204 | <p>That is indeed sloppy notation. The proper way of writing it is to say the following. Suppose <span class="math-container">$f:[a,b]\to\Bbb{R}$</span> is a given bounded function. Let <span class="math-container">$P=\{x_0,\dots, x_N\}$</span> be a partition of the interval <span class="math-container">$[a,b]$</span>, meaning that <span class="math-container">$a=x_0<x_1<\cdots< x_N=b$</span>. Suppose also that we're given a collection of points <span class="math-container">$\{\xi_1,\dots, \xi_N\}$</span>, where <span class="math-container">$\xi_1\in[x_0,x_1], \cdots, \xi_N\in [x_{N-1},x_N]$</span>. We call <span class="math-container">$(P,\{\xi_i\}_{i=1}^N)$</span> a <em>tagged partition</em> of <span class="math-container">$[a,b]$</span>.</p>
<p>Corresponding to this tagged partition, we consider the Riemann sum
<span class="math-container">\begin{align}
R(f,P,\{\xi_i\}_{i=1}^N):=\sum_{i=1}^Nf(\xi_i)(x_i-x_{i-1})\equiv\sum_{i=1}^Nf(\xi_i)\Delta x_i
\end{align}</span>
where we defined <span class="math-container">$\Delta x_i:=x_i-x_{i-1}$</span> for each <span class="math-container">$i\in\{1,\dots, N\}$</span>. Here, <span class="math-container">$\xi_i$</span> is just an intermediate point of the interval <span class="math-container">$[x_{i-1},x_i]$</span>. Also, let us define <span class="math-container">$\|P\|:=\max\limits_{1\leq i\leq N}\Delta x_i$</span>; this is called the <em>mesh</em> of the partition <span class="math-container">$P$</span>.</p>
<p>Finally, with all the notation above, we say that the function <span class="math-container">$f$</span> is Riemann integrable on <span class="math-container">$[a,b]$</span> if the following limit exists:
<span class="math-container">\begin{align}
\lim_{\|P\|\to 0}R(f,P,\{\xi_i\}_{i=1}^N)
\end{align}</span>
i.e we require the limit
<span class="math-container">\begin{align}
\lim_{\|P\|\to 0}\sum_{i=1}^Nf(\xi_i)\Delta x_i\tag{$*$}
\end{align}</span>
to exist. More explicitly, what this means is:</p>
<blockquote>
<p>There is a number <span class="math-container">$I\in\Bbb{R}$</span> such that for every <span class="math-container">$\epsilon>0$</span>, there is a <span class="math-container">$\delta>0$</span> such that for any tagged partition <span class="math-container">$(P=\{x_i\}_{i=0}^N,\{\xi_i\}_{i=1}^N)$</span> of <span class="math-container">$[a,b]$</span>, if <span class="math-container">$\|P\|<\delta$</span> then
<span class="math-container">\begin{align}
\left|R(f,P,\{\xi_i\}_{i=1}^N)-I\right|:=\left|\sum_{i=1}^Nf(\xi_i)\Delta x_i - I\right|<\epsilon.
\end{align}</span></p>
</blockquote>
<p>One can easily show that if the number <span class="math-container">$I$</span> exists, then it is unique.</p>
<p>In this case, we call the unique number <span class="math-container">$I$</span> the Riemann integral of <span class="math-container">$f$</span> on the interval <span class="math-container">$[a,b]$</span>, and write this as <span class="math-container">$\int_a^bf$</span>. We define this to be the (signed) area bounded by the graph of <span class="math-container">$f$</span> on the interval <span class="math-container">$[a,b]$</span> and the horizontal axis.</p>
<hr />
<p>So long story short, <span class="math-container">$(*)$</span> is what you should be writing. The choice of index <span class="math-container">$i$</span> is irrelevant. I can use <span class="math-container">$j$</span> or <span class="math-container">$\alpha$</span> or <span class="math-container">$n$</span> or any other letter I like, for example, <span class="math-container">$\sum_{\beta=1}^Nf(\xi_{\beta})\Delta x_{\beta}$</span>. Writing <span class="math-container">$\sum_{n=1}^Nf(x)\Delta x$</span> is incorrect for two reasons. First, it makes it seem like <span class="math-container">$\Delta x$</span> is a fixed number and <span class="math-container">$x$</span> is fixed, so the sum should just evaluate to <span class="math-container">$Nf(x)\Delta x$</span>; of course this isn't the intended meaning.</p>
<p>We have to allow several things to vary: the number of points in the partition (i.e <span class="math-container">$N$</span>), and also the spacing (i.e we're not assuming all the <span class="math-container">$\Delta x_i$</span>'s are equal), and also we have allow for arbitrary intermediate points (i.e arbitrary choice of <span class="math-container">$\xi_i$</span>'s).</p>
|
1,265,074 | <p>I simply don't know how to go about answering this question. I've done a good few other questions about point estimation, but I really don't know where I'm going with this one:</p>
<p><img src="https://i.stack.imgur.com/coWTD.png" alt="Unbiased estimator of 1 over lambda"></p>
<p>Thanks for the help!</p>
<p>EDIT: My question is regarding the unbiasedness section, but any input on the second part would also be great.</p>
| Michael Hardy | 11,667 | <p>What would go wrong if $1$ had not been added in the denominator is that the probability that the denominator is $0$ would be positive so the expected value would be $\infty\ne 1/\lambda$.</p>
<p>The problem is simply to find the expected value:
\begin{align}
& \operatorname{E}\left( \frac n {1+\sum_{i=1}^n X_i} \right) = \sum_{x=0}^\infty \frac n {1+x} \Pr\left( \sum_{i=1}^n X_i = x \right) = n\sum_{x=0}^\infty \frac 1 {1+x} \cdot \frac{e^{-n\lambda} (n\lambda)^x}{x!} \\[10pt]
= {} & n \sum_{x=0}^\infty \frac{e^{-n\lambda} (n\lambda)^x }{(x+1)!} = \frac n {n\lambda} \sum_{x=0}^\infty \frac{e^{-n\lambda} (n\lambda)^{x+1} }{(x+1)!} = \frac 1 \lambda e^{-n\lambda} \underbrace{\sum_{x=1}^\infty \frac{(n\lambda)^x}{x!}}_{\text{One term is missing.}}
\end{align}
In the last equality, a substitution was done: the new $x$ was the old $x+1$, and as the old $x$ goes from $0$ to $\infty$, the new $x$ goes from $1$ to $\infty$; thus the sum is missing the first term in a sum that evaluates to $e^{n\lambda}$. The last sum is therefore $e^{n\lambda}-1$.</p>
|
1,383,781 | <p>Given $\mathbb{X}$ = $\mathbb{R^2}$, consider $\| \cdot \|_2$ and $\| \cdot \|_\infty$ </p>
<p>We can show that </p>
<p>$\| x \|_\infty \leq \| x \|_2 \leq \sqrt2 \| x \|_\infty$ </p>
<p>Hence $\| \cdot \|_2$ and $\| \cdot \|_\infty$ are equivalent norms</p>
<p>Is there some deeper implication regarding this particular relationship? Why do we care if two norms are equivalent in this sense?</p>
| Stephan Kulla | 32,951 | <p>As pointed out by <a href="https://math.stackexchange.com/users/75808/clement-c">Clement C</a> in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent.</p>
<p>Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case when open sets for $\|\cdot\|_1$ are also open in $\|\cdot\|_2$ and vice versa. This is the case when in each open ball in $\|\cdot\|_1$ contains an open ball of $\|\cdot\|_2$ and the other way around. From this you can prove that there are constants $c$ and $C$ such that $c\|\cdot\|_1 \le \|\cdot\|_2 \le C \|\cdot\|_1$.</p>
<p>So the answer to your question is: Two norms are equivalent iff their induced topologies are the same.</p>
<p>What is the benefit if two topologies are the same? If a "topological property" is valid for one norm, then it is valid for the other norm. For example:</p>
<ul>
<li>Open, closed and compact sets are the same.</li>
<li>If a sequence converges in one norm, it converges also in the other norm.</li>
<li>If a sequence is a Cauchy sequence in one norm it is Cauchy in the other.</li>
<li>...</li>
</ul>
|
3,063,651 | <p>i am currently looking out for some possible topics i could study for my research project in high school. Algebra, trigonometry, Pythagoras’ theorem, geometry, circles and their properties, etc. and perhaps combined with a little knowledge from Physics i.e. Kinematics, Gravity, etc. could interest me. </p>
<p>Another particularly interesting one i stumbled upon was the Rubik’s cube since i myself do enjoy solving it. I am not very sure what type of research topics i could come up with that could be geared towards and tied with the math behind solving Rubik’s cube? Any suggestions or advices are greatly appreciated :)</p>
| J.G. | 56,861 | <p>What are the requirements of your research project? I'm guessing they can't expect a high schooler to discover something new; they probably just want you to work through a non-syllabus proof, possibly after you conjecture the result from special cases.</p>
<p>If you want to do something with a Rubik cube you could discuss its group theory, or number of configurations using combinatorics. Physics is harder because you might not have enough calculus, but <a href="https://youtu.be/xdIjYBtnvZU" rel="nofollow noreferrer">3blue1brown's video on why orbits are ellipses</a> might give you some ideas for a more geometry-based analysis of gravity.</p>
<p>I live in the UK, so what you're trying to do is probably similar to our GCSE maths coursework. Based on my memories of that, possible projects include:</p>
<ul>
<li>Given a 2D grid of numbers in which each row and column is an arithmetic progression, what is the difference between the products of endpoints of a rectangle's diagonals?</li>
<li>How many diagonals does a polygon have?</li>
<li>How is the sum of a positive integer's positive factors related to its prime factorisation? (Or you may find some other multiplicative function in number theory more interesting.)</li>
<li>What is the sum of the first <span class="math-container">$n$</span> positive cubes? Is there anything interesting about its square root?</li>
<li>In how many ways can you get from a point on a 2D grid to another point neither below not to the left of it, in unit steps that only go up or to the right?</li>
</ul>
|
646,109 | <p>For function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies $f\left(x+y\right)=f\left(x\right)f\left(y\right)$
and is not the zero-function I can prove that $f\left(1\right)>0$
and $f\left(x\right)=f\left(1\right)^{x}$ for each $x\in\mathbb{Q}$.
Is there a way to prove that for $x\in\mathbb{R}$?</p>
<p>This question has been marked to be a duplicate of the question whether $f(xy)=f(x)f(y)$ leads to $f(x)=x^p$ for some $p$. I disagree on that. Both questions are answered by means of construction of a function $g$ that suffices $g(x+y)=g(x)+g(y)$. In this question: $g(x)=\log f(x)$ and in the other $g(x)=\log f(e^x)$. So the answers are alike, but both questions definitely have another startpoint.</p>
| Hagen von Eitzen | 39,174 | <p>If $f(x_0)=0$ for some $x_0$ then $f(x)=f((x-x_0)+x_0)=f(x-x_0)f(x_0)=0$ for all $x$. As the zero function is excluded, $f(x)\ne 0$ for all $x$ and in fact $f(x)=f(\frac x2)^2>0$. Therefore we can define $g(x)=\ln f(x)$ and find the functional equation $$ g(x+y)=g(x)+g(y)$$ for $g$. As has been pointed out, this (i.e. Cauchy's) functional equation has precisely the affine linear functions as continuous solutions, but also many wild and non-continuous solutions. Each solution $f$ is of the form $f(x)=e^{g(x)}$ for such a $g$.</p>
|
3,861,324 | <p>Given a nonzero column vector <span class="math-container">$A$</span>=<span class="math-container">$[a_1 a_2.......a_n]^T$</span>. Find the non zero eigen values and eigen vectors for <span class="math-container">$A$$A^T$</span>.</p>
<p>I have no idea.what theorem should I apply or what I have to do to solve this. I know that fact <span class="math-container">$A$$A^T$</span> and <span class="math-container">$A^T$$A$</span> have the same eigen values and if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be square matrix of same order the <span class="math-container">$AB$</span> and <span class="math-container">$BA$</span> have the same eigen values. Is it related to this statements?</p>
| Zhanxiong | 192,408 | <p><span class="math-container">$\newcommand{\rank}{\mathrm{rank}}$</span></p>
<p>This is a well known exercise in eigenvalue-eigenvector, for which you actually do not need to start with solving the equation <span class="math-container">$\det(\lambda I - B) = 0$</span> to determine eigenvalues (as you usually do). The special structure of <span class="math-container">$AA^T$</span> as a product of two rank one vectors allows us to have the following shortcut.</p>
<p>Since <span class="math-container">$\rank(A) = 1$</span>, <span class="math-container">$\rank(AA^T) = \rank(A) = 1$</span>, hence <span class="math-container">$AA^T$</span> has one and only one non-zero eigenvalue, and all the remaining <span class="math-container">$n - 1$</span> eigenvalues are <span class="math-container">$0$</span>.</p>
<p>Now note <span class="math-container">$A^TA = a_1^2 + a_2^2 + \cdots + a_n^2 \neq 0$</span> is a scalar, denoted it by <span class="math-container">$\lambda$</span>, hence
<span class="math-container">\begin{align*}
(AA^T)A = AA^TA = (A^TA)A = \lambda A.
\end{align*}</span>
This shows <span class="math-container">$\lambda$</span> is an eigenvalue of <span class="math-container">$AA^T$</span>, and <span class="math-container">$A$</span> is the eigenvector associated with <span class="math-container">$\lambda$</span>.</p>
|
3,151,662 | <p>Consider <span class="math-container">$a_1,\dots,a_n\in\mathbb{R}^n$</span> and identify <span class="math-container">$a_j\in\mathcal{L}(\mathbb{R},\mathbb{R}^n)$</span> via <span class="math-container">$\varphi\mapsto \varphi1$</span>.</p>
<p>Also, consider <span class="math-container">$A\in\mathcal{L}(\mathbb{R}^n)$</span> given by
<span class="math-container">$$A\colon (x_1,\dots,x_n)\mapsto a_1x_1 + \dots + a_nx_n\tag{$\star$}$$</span></p>
<p>What's the name or symbol of the map
<span class="math-container">$$\mathcal{L}(\mathbb{R},\mathbb{R}^n)\times\dots\times\mathcal{L}(\mathbb{R},\mathbb{R}^n)\to\mathcal{L}(\mathbb{R}^n,\mathbb{R}^n),\quad(a_1,\dots,a_n)\mapsto A$$</span>
where <span class="math-container">$A$</span> and <span class="math-container">$a_1,\dots,a_n$</span> are related as in <span class="math-container">$(\star)$</span>?
I'd like to write e.g. <span class="math-container">$A=a_1\otimes\dots\otimes a_n$</span>.
Is there some higher level concept that induces that map? (E.g. some time ago I wondered if this map would be the tensor product). </p>
<p>The matrix equivalent would be saying that the columns of <span class="math-container">$A$</span> are the column vectors <span class="math-container">$a_1,\dots,a_n$</span> and writing <span class="math-container">$A=\begin{bmatrix}a_1& \dots &a_n\end{bmatrix}$</span>. However, I'd like to keep things matrix free.</p>
<p>Thanks in advance.</p>
| J.G. | 56,861 | <p>Here's an application in calculus. The multivariate generalisation of integration by substitution viz. <span class="math-container">$x=f(y)\implies dx=f^\prime(y)dy$</span> uses the determinant of a matrix called a Jacobian in place of the <span class="math-container">$f^\prime$</span> factor. In particular, the chain rule <span class="math-container">$dx_i=\sum_j J_{ij}dy_j,\,J_{ij}:=\frac{\partial x_i}{\partial y_j}$</span> for <span class="math-container">$n$</span>-dimensional vectors <span class="math-container">$\vec{x},\,\vec{y}$</span> can be summarised as <span class="math-container">$d\vec{x}=Jd\vec{y}$</span>. Then <span class="math-container">$d^n\vec{x}=|\det J|d^n\vec{y}$</span>.</p>
|
3,318,993 | <p>Consider the functions <span class="math-container">$x$</span> and <span class="math-container">$x^2$</span> on <span class="math-container">$\mathbb{R}$</span>. Clearly, they are linearly independent.<br>
But consider the following argument.</p>
<p>Consider the matrix
<span class="math-container">$$A = \begin{bmatrix}
x & x^2\\
0 & 0\\
\end{bmatrix}$$</span></p>
<p>Clearly, the determinant is zero. This implies the existence of a nonzero matrix, <span class="math-container">$$B =\begin{bmatrix} a\\ b\\ \end{bmatrix}$$</span> such that <span class="math-container">$$AB=0$$</span>.</p>
<p>This implies that <span class="math-container">$ax+bx^2=0$</span> for some nonzero <span class="math-container">$a$</span> or some nonzero <span class="math-container">$b$</span>. But this implies that <span class="math-container">$x$</span> and <span class="math-container">$x^2$</span> are linearly dependent.</p>
<p>Clearly, false.</p>
<hr>
<p>Where’s the flaw?</p>
| Mark | 470,733 | <p>They are linearly independent as functions of variable <span class="math-container">$x$</span>. Yes, you are right that given a specific value of <span class="math-container">$x$</span> you can find non-zero <span class="math-container">$a,b\in\mathbb{R}$</span> such that <span class="math-container">$ax+bx^2=0$</span>. But you will never find non-zero <span class="math-container">$a,b$</span> which will work <strong>for all</strong> <span class="math-container">$x\in\mathbb{R}$</span>. </p>
|
3,318,993 | <p>Consider the functions <span class="math-container">$x$</span> and <span class="math-container">$x^2$</span> on <span class="math-container">$\mathbb{R}$</span>. Clearly, they are linearly independent.<br>
But consider the following argument.</p>
<p>Consider the matrix
<span class="math-container">$$A = \begin{bmatrix}
x & x^2\\
0 & 0\\
\end{bmatrix}$$</span></p>
<p>Clearly, the determinant is zero. This implies the existence of a nonzero matrix, <span class="math-container">$$B =\begin{bmatrix} a\\ b\\ \end{bmatrix}$$</span> such that <span class="math-container">$$AB=0$$</span>.</p>
<p>This implies that <span class="math-container">$ax+bx^2=0$</span> for some nonzero <span class="math-container">$a$</span> or some nonzero <span class="math-container">$b$</span>. But this implies that <span class="math-container">$x$</span> and <span class="math-container">$x^2$</span> are linearly dependent.</p>
<p>Clearly, false.</p>
<hr>
<p>Where’s the flaw?</p>
| Arthur | 15,500 | <p>When discussing linear independence of the columns of a matrix, you must allow the coefficients in the linear combinations to come from the same space as the entries in the matrix (otherwise linear algebra doesn't in any way work the way you're used to). And clearly, if we allow <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to be polynomials, then <span class="math-container">$ax+bx^2=0$</span> has nontrivial solutions.</p>
|
821,875 | <p>A school director must randomly select 6 teachers to participate in a training session. There are 30 teachers at the school. In how many different ways can these teachers be selected, if the order of selection does not matter?</p>
| StumpyLeg | 143,237 | <p>This is just "30 choose 6." If you have a graphing calculator, under the Probability menu there should be something like nCr. But your textbook should contain the relevant formula--and explain the basis for it.</p>
|
684,892 | <p>My progress:</p>
<p>Let's take $a \in \mathbb{Z}\left[\frac{-1 + \sqrt{-3}}2\right]$ such that $a \mid 2$, and function $l(x) = x \bar x$.</p>
<p>$a \mid 2$ $\Rightarrow$ $2 = ab$ $\Rightarrow$ $l(ab) = l(a)l(b) = 4 = l(2)$</p>
<p>If $z \in \mathbb{Z}[\frac{-1 + \sqrt{-3}}2]$, then $z = x + y\frac{-1 + \sqrt{-3}}2$, $x, y \in Z$ and $l(z) = x^2 - xy + y^2 \in \mathbb{Z}$. </p>
<p>So $l(a), l(b) \in \mathbb{Z}$. </p>
<p>Thus there are three possible values for $l(a)$:</p>
<ol>
<li>$l(a) = 1$ and $l(b) = 4$ : $a \bar a = 1$, thus $a$ is a unit.</li>
<li>$l(a) = 4$ and $l(b) = 4$ : $b$ is a unit.</li>
<li>$l(a) = 2$ and $l(b) = 2$ : If $a = x + y\frac{-1 + \sqrt{-3}}2$ , $l(a) = x^2 - xy + y^2 = 2$. </li>
</ol>
<p>If the equation $x^2 - xy + y^2 = 2$ doesn't have a solution, then $2$ is a prime. How can I show that it doesn't have a solution?</p>
<p>Any other proof would be highly appreciated.</p>
| Daniel Fischer | 83,702 | <p>Multiply with $4$ and complete the square, you are looking for integer solutions to</p>
<p>$$(2x-y)^2 + 3y^2 = 8.$$</p>
<p>That constrains $\lvert y\rvert \leqslant 1$ already. $5$ isn't a square, $8$ isn't a square, so $y = \pm 1$ and $y = 0$ are also ruled out.</p>
<p>Another proof is: If $x^2 - xy + y^2 \equiv 0 \pmod{2}$, then $x\equiv y \equiv 0 \pmod{2}$. For if $x\equiv 1 \pmod{2}$ then $x^ - xy + y^2 \equiv 1 + y(y-1) \equiv 1 \pmod{2}$, and symmetrically for $y\equiv 1 \pmod{2}$. And thus, if the norm of $z = x + y \frac{-1+\sqrt{-3}}{2}$ is even, then $x$ and $y$ are even, and $z$ is a multiple of $2$.</p>
|
35,463 | <p>I was asked the following vector calculus problem:</p>
<blockquote>
<p>Let <span class="math-container">$D$</span> be the unit ball and let <span class="math-container">$S$</span> be the unit sphere in <span class="math-container">$\mathbb{R}^3$</span>. Suppose that <span class="math-container">$F:\mathbb{R}^3\rightarrow \mathbb{R}^3$</span> is a <span class="math-container">$C^1$</span> vector field on some open neighborhood of <span class="math-container">$D$</span> which satisfies:</p>
<p><span class="math-container">$(i) \nabla\times F=0$</span></p>
<p><span class="math-container">$(ii) \nabla\cdot F=0$</span></p>
<p><span class="math-container">$(iii)$</span> On <span class="math-container">$S$</span>, <span class="math-container">$F$</span> is orthogonal to the radial vector.</p>
<p>Prove that <span class="math-container">$F=0$</span> on all of <span class="math-container">$D$</span>.</p>
</blockquote>
<p>Conditions <span class="math-container">$(i)$</span> and <span class="math-container">$(ii)$</span> imply that <span class="math-container">$F=\nabla g$</span> for some <span class="math-container">$g:\mathbb{R}^3\rightarrow \mathbb{R}$</span> where <span class="math-container">$g$</span> must be harmonic as well.</p>
<p>I know one solution (see end), however my initial instinct was to try to use the max/min property of harmonic functions, and I couldn't get it to work. Since the gradient is always orthogonal to the sphere, there must be a point on the sphere where it is <span class="math-container">$0$</span>. (Hairy ball) If that was a local max or min in <span class="math-container">$\mathbb{R}^3$</span> we would be done, by taking a small neighborhood around it. If it is a saddle point this doesn't work. (We know that it must be a local max/min on <span class="math-container">$S$</span> since it is harmonic)</p>
<p><strong>My question is:</strong> Is there any way to modify this approach, and solve the problem?</p>
<p>Thanks!</p>
<p><strong>Other Solution:</strong> Here is one solution that first uses the fact that the radial vector is orthogonal, and then applies Gauss's Divergence theorem to the function <span class="math-container">$gF$</span>. (<span class="math-container">$\nabla g=F$</span>) That is <span class="math-container">$$0=\iint_S (gF\cdot n)dS=\iiint_D \nabla\cdot (gF)dV=\iiint_D \|F\|^2dV,$$</span> and since the integrand on the right hand side is non-negative, continuous and integrates to give zero, it must be zero.</p>
| Shuhao Cao | 7,200 | <p>Eric, I am simply rephrasing your second proof to give you a more PDE style approach to this problem, let $\vec{F} = \nabla u$, the problem you gave was equivalent to solve the following boundary value problem:
Find $u\in C^2(D)$ such that
$$
\left\{\begin{eqnarray}
-\Delta u &=& 0 \text{ in } D \\
\nabla u \cdot n &=& 0 \text{ on } \partial D
\end{eqnarray}
\right.$$
Now multiply both sides of the equation by a test function in some proper space(as large as possible), doing integration by parts to use the boundary condtion, and you will get the variational formulation of this problem:
Find $u\in V = W^{1,2}(D)$ such that
$$
\int_D \nabla u\cdot \nabla v = 0 \quad \forall v\in V
$$
Notice there is no Dirichlet type boundary condition enforced for the test function $v$, which makes this problem yield non-unique solutions; In order to make this problem well-posed, instead in the space $V$, consider a new space $V_0 = \{v\in V | \int_D v dx = 0\} = V/R$ this variational problem is equivalent to the following functional minimization problem:
If $u$ solves the original pde, then $u$ minimizes
$$
J(u) = \min_{v\in V_0}\int_D |\nabla v|^2\,dx
$$
Now assume $u$ is the solution, then
$$
J(u) = \int_D |\nabla u|^2\,dx \leq \sup_{v\in V_0} \int_D \nabla u\cdot \nabla v = 0
$$
which proves for the solution $u\in V_0$: $\nabla u = 0$(other solutions are different with some specific $u$ from a constant by the definition of that quotient space, hence the gradient does not change)</p>
|
2,088,346 | <p>I've got the domain of function and I've attempted to find the first derivative at zero but it results in a quartic equation that is too difficult for me to solve. </p>
<p>$f'(x) = \frac{4x-3}{\sqrt{2x^2-3x+4}} + \frac{2x-2}{\sqrt{x^2-2x}}$</p>
<p>For $f'(x) = 0$:</p>
<p>$(4x-3)^2(x^2-2x) = (2-2x)^2(2x^2-3x+4)$</p>
<p>Therefore $8x^4-28x^3+9x^2+26x-16 = 0$</p>
<p>I've probably made an error in my calculations but I'm sure that this is not how you approach the question and I'm not quite sure how to do it otherwise. According to the textbook the answer is 2.</p>
| Matt | 263,495 | <p>I would suggest sketching your first square root and your second square root separately on a graph. The solution then doesn't require any sort of complicated algebra.</p>
<p><a href="https://i.stack.imgur.com/pQDNg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pQDNg.png" alt="enter image description here"></a></p>
|
1,841,958 | <p>This is a claim on Wikipedia <a href="https://en.wikipedia.org/wiki/Partially_ordered_set">https://en.wikipedia.org/wiki/Partially_ordered_set</a></p>
<p>I am not sure how to make sense of the claim</p>
<p>What does it mean by ordered by inclusion? Inclusion as in $\subseteq$? </p>
<p>Can someone provide a small example of couple subspaces being "ordered" by inclusion?</p>
<p>Is this a linear order?</p>
| Eric Wofsey | 86,856 | <p>Most of your questions have been answered well by user247327, so let me just answer your last question. The poset of all subspaces of a vector space $V$ is not linearly ordered as long as $\dim V>1$. For instance, if $v$ and $w$ are two linearly independent vectors, then $\operatorname{span}(v)$ and $\operatorname{span}(w)$ are two subspaces of $V$, with neither contained in the other. (On the other hand, if $\dim V\leq 1$, the only subspaces are $V$ and $\{0\}$, with $\{0\}\subseteq V$, so then it is a linear order.)</p>
|
1,039,141 | <blockquote>
<p>Let <span class="math-container">$X = \mathbb{R}$</span> and <span class="math-container">$Y = \{x \in \mathbb{R} :x ≥ 1\}$</span>, and define <span class="math-container">$G : X → Y$</span> by <span class="math-container">$$G(x) = e^{x^2}.$$</span>
Prove that <span class="math-container">$G$</span> is onto.</p>
</blockquote>
<p>Is this going along the right path and if so how do get the function to equal <span class="math-container">$y$</span>?</p>
<blockquote>
<p><span class="math-container">$G: \mathbb{R} \to\mathbb{N}_1$</span>. Let <span class="math-container">$y$$\in $$\mathbb{N_1}$</span>.</p>
<p><em>claim:</em> <span class="math-container">$\sqrt{\ln y}$</span> maps to <span class="math-container">$y$</span>.</p>
<p>Does <span class="math-container">$\sqrt{\ln y}$</span> belong to <span class="math-container">$\mathbb{N_1}$</span>? Yes because <span class="math-container">$y \in \mathbb{N_1}$</span>, <span class="math-container">$G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}$</span>.</p>
</blockquote>
| Swapnil Tripathi | 117,387 | <p><strong>Hint:</strong> $g(x)=e^{x^2}$ is symmetric about the y-axis. Also $g$ is continuous and strictly increasing in $\mathbb{R}^+$ with $g(0)=1$</p>
|
3,201,797 | <p>I have three points <span class="math-container">$(x_1, y_1),~ (x_2, y_2),~ (x_3, y_3)$</span> that are on the same line. How to efficiently find which is the point in between.</p>
<p><a href="https://i.stack.imgur.com/e2wHq.png" rel="nofollow noreferrer">Example</a></p>
<p>Also, is there any efficient way to check if 3 random points are on the same line and then find the point in between?</p>
| Vasili | 469,083 | <p>Find three distances and use the fact that the distance between end points is the greatest. You have points <span class="math-container">$A(x_1,y_1), B(x_2, y_2), C(x_3, y_3)$</span>. The square of distance between <span class="math-container">$A$</span> and <span class="math-container">$B$</span> is <span class="math-container">$$(x_1-x_2)^2+(y_1-y_2)^2$$</span> You can use squared distance to simplify calculations.<p>To check if three points are on one line, there is a simple formula derived from area of a triangle formed by three points.: <span class="math-container">$$(y_3-y_1)(x_2-x_1)-(x_3-x_1)(y_2-y_1)$$</span> This is double of the area. If it's zero, the points are on one line.</p>
|
3,201,797 | <p>I have three points <span class="math-container">$(x_1, y_1),~ (x_2, y_2),~ (x_3, y_3)$</span> that are on the same line. How to efficiently find which is the point in between.</p>
<p><a href="https://i.stack.imgur.com/e2wHq.png" rel="nofollow noreferrer">Example</a></p>
<p>Also, is there any efficient way to check if 3 random points are on the same line and then find the point in between?</p>
| Intelligenti pauca | 255,730 | <p>Once you know that the three points are aligned,
compute
<span class="math-container">$$
t={x_3-x_2\over x_1-x_2}\quad
\text{(or}\quad
t={y_3-y_2\over y_1-y_2}\quad
\text{if $x_1-x_2\approx0$).}
$$</span>
Then:</p>
<ul>
<li>if <span class="math-container">$t>1$</span> then <span class="math-container">$P_2$</span> lies between <span class="math-container">$P_1$</span> and <span class="math-container">$P_3$</span>;</li>
<li>if <span class="math-container">$t<0$</span> then <span class="math-container">$P_1$</span> lies between <span class="math-container">$P_2$</span> and <span class="math-container">$P_3$</span>;</li>
<li>if <span class="math-container">$0<t<1$</span> then <span class="math-container">$P_3$</span> lies between <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>.</li>
</ul>
|
706,546 | <p>If we have a non-zero real $n$ by $m$ matrix $M$, then there may exist a non-zero unit vector $v$ of $m$ elements so that $Mv = 0$. I understand we can't call this an eigenvector with eigenvalue $0$. </p>
<blockquote>
<p>Why is this not a sensible definition of an eigenvector of a rectangular matrix?</p>
</blockquote>
<p>If $m > n$ then $M$ must have rank less than $m$. </p>
<blockquote>
<p>Can such a matrix still have no non-zero unit vectors $v$ of $m$
elements so that $Mv = 0$?</p>
</blockquote>
<p>Note that the $0$ term on the right hand side is $n$-dimensional and $v$ is $m$-dimensional.</p>
| Marc van Leeuwen | 18,880 | <p>The following observations suffice to prove the statement:</p>
<ol>
<li>A power $A^k$ is in the span of lower powers $A^0,\ldots,A^{k-1}$ if and only if there exists a (monic) polynomial$~P$ of degree$~k$ with $P[A]=0$.</li>
<li>If this happens for some $k=m$, it also happens for all $k>m$, so that by an immediate induction argument, $\operatorname{span}(A^0,\ldots,A^{m-1})$ contains <em>all</em> powers of$~A$ (and of course it has dimension${}\leq m$).</li>
<li>There exists a monic polynomial$~P$ of degree${}\leq n$ with $P[A]=0$.</li>
</ol>
<p>Point 1. is fairly obvious; it suffices to consider an equation expressing $A^k$ as linear combination of $A^0,\ldots,A^{k-1}$, and to bring all its terms to the same side of the equation as$~A^k$. Point 2. is also easy, since it suffices to multiply the polynomial $P$ by a power$~X^d$ to raise is degree; the result still annihilates$~A$, since $(X^dP)[A]=A^d\circ (P[A])=0$ by linearity of$~A$. (Alternatively one could argue for the remainder $R$ of $X^k$ after division by$~P$ that $A^k=R[A]\in\operatorname{span}(A^0,\ldots,A^{m-1})$, since $\deg R<m$.) For point 3. the Cayley-Hamilton theorem says the characteristic polynomial $P=\chi_A$, which has degree $n$, can be chosen.</p>
<p><em>Added</em>.
When using Cayley-Hamilton, I always wonder if one could also do it by more elementary means (because the proof of C-H is somewhat subtle). For point 3. this is is in fact the case.</p>
<p>One can use strong induction on the dimension. The inductive hypothesis takes the following concrete form: for any $A$-stable subspace $V$, there exists a monic polynomial $P\in K[X]$ with $\deg P\leq\dim V$ such that $V\subseteq\ker(P[A])$. By expressing the restriction of $A$ to$~V$ on a basis, this is just point 3. for that restriction. So assume this result holds for any <em>proper</em> subspace. If $n=0$ one can take $P=1$, so also assume $n>0$ and choose a nonzero vector $v\in K^n$. The $n+1$ vectors $v,Av,A^2v,\ldots,A^nv$ are certainly linearly dependent, so one can take $d$ minimal such that $v,Av,A^2v,\ldots,A^dv$ are linearly dependent. Then $A^dv$ is a linear combination of the preceding vectors, and this gives a polynomial $R$ of degree$~d$ such that $R[A]v=0$. But $\ker(R[A])$ is $A$-stable so $v,Av,A^2v,\ldots,A^{d-1}v\in\ker(R[A])$, and since these vectors are linearly independent, one obtains $\dim\ker(R[A])\geq d$. Putting $V=\operatorname{Im}(R[A])$ the rank-nuullity theorem gives $\dim V\leq n-d$, and since $V$ is $A$-stable, the induction hypothesis gives a polynomial $Q$ with $\deg Q\leq n-d$ such that $V\subseteq\ker(Q[A])$. The latter means that $0=Q[A]\circ R[A]=(QR)[A]$; then taking $P=QR$ works, since one gets $\deg P=\deg Q+\deg R\leq(n-d)+d=n$.</p>
|
3,364,316 | <p>While I'm reading E. Landau's <em>Grundlagen der Analysis</em> (tr. <em>Foundations of Analysis</em>, 1966), I couldn't understand the proof of <em>Theorem 3</em> at the segment of <em>Natural Numbers</em> which I've quoted below.</p>
<blockquote>
<p><strong>Theorem 3:</strong> <em>If</em><br>
<span class="math-container">$$x \neq 1$$</span>
<em>then there exists one</em> (hence, by Axiom 4, exactly one) <span class="math-container">$u$</span> <em>such that</em><br>
<span class="math-container">$$x = u'$$</span>
<strong>Proof:</strong> Let <span class="math-container">$\mathbb{S}$</span> be the set consisting of the number <span class="math-container">$1$</span> and of all those <span class="math-container">$x$</span> for which there exists such a <span class="math-container">$u$</span>. (For any such <span class="math-container">$x$</span>, we have of necessity that<br>
<span class="math-container">$$x \neq 1$$</span>
by Axiom 3.)<br>
I) <span class="math-container">$1$</span> belongs to <span class="math-container">$\mathbb{S}$</span>.<br>
II) If <span class="math-container">$x$</span> belongs to <span class="math-container">$\mathbb{S}$</span>, then, with <span class="math-container">$u$</span> denoting the number <span class="math-container">$x$</span>, we have<br>
<span class="math-container">$$x'=u'$$</span><br>
so that <span class="math-container">$x'$</span> belongs to <span class="math-container">$\mathbb{S}$</span>.<br>
By Axiom 5, <span class="math-container">$\mathbb{S}$</span> therefore contains all the natural numbers. <span class="math-container">$\square$</span> </p>
</blockquote>
<p>Sir Landau refers to the Axioms of Peano on proof text. Can someone explain what's going on?</p>
| Keith Backman | 29,783 | <p>Let <span class="math-container">$g_n=p_{n+1}-p_n$</span>. Then <span class="math-container">$\frac{p_n}{p_n+p_{n+1}}=\frac{p_n}{2p_n+g_n}=\frac{1}{2+\frac{g_n}{p_n}}$</span>. The lim inf for increasing <span class="math-container">$n$</span> will occur when <span class="math-container">$g_n$</span> is as large as possible with respect to <span class="math-container">$p_n$</span>. However, improvements to <a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate" rel="nofollow noreferrer">Bertrand's postulate</a> suggest that as <span class="math-container">$n$</span> becomes large, <span class="math-container">$\frac{g_n}{p_n}$</span> becomes very small, on the order of <span class="math-container">$\frac{1}{\ln^3 p_n}$</span>. This is much smaller than Bertrand's postulate per se, which requires only that <span class="math-container">$\frac{g_n}{p_n}<1$</span></p>
<p>Plainly, <span class="math-container">$\lim_{n \to \infty}\frac{1}{\ln^3 p_n} \to 0$</span>, so <span class="math-container">$\frac{1}{2+\frac{g_n}{p_n}}\approx \frac{1}{2+\frac{1}{\ln^3 p_n}} \to \frac{1}{2}$</span></p>
<p>Since the absolute limit is <span class="math-container">$\frac{1}{2}$</span>, the lim inf is the same (as noted in a previous answer).</p>
|
692,998 | <p>The inner product in a $L^2$ space can be defined as:</p>
<p>$$\langle f,g\rangle =\int_a^b \bar{f}(x)g(x)w(x)dx$$</p>
<p>For Legendre polynomials, we define it as:</p>
<p>$$\langle P_m,P_n\rangle =\int_0^1 \bar{P}_m(x)P_n(x)dx$$
so $w(x)=1$.</p>
<p>But there are case in which $w(x)\neq 1$. For example, Laguerre $w(x)=e^{-x}$ and Hermite polynomials $w(x)=e^{-x^2}$.</p>
<p><strong>Is there any intuition/motivation behind different weight functions of orthogonal polynomials</strong>? I think it might be related to measure theory and Sturm-Liouville problems.</p>
| typedrums | 596,646 | <p>I think you mean this question in one of two ways.</p>
<p>Either you mean why do e.g. Laguerre polynomials have that specific weight? </p>
<p>The answer to which is just that you start with the weight and the polynomials follow from it. You can often in principle retrieve the weight from the polynomials though. The zeros of polynomials will be in the support of the weight, so you need a weight that satisfies
<span class="math-container">$$
\int_{support(w)} x^k w(x) dx < \infty, \qquad \forall k \geq 0.
$$</span>
If the zeros are not bounded by a compact interval than a weight like <span class="math-container">$w=1$</span> won't cut it.</p>
<p>The other way you might have meant this question is whether the polynomials and the weight have any connecting properties that can be understood intuitively. Below I present an example of this that I came across in M. E. H. Ismail's book "Classical and Quantum Orthogonal Polynomials in One Variable".</p>
<p>There is some electrostatic interpretation (might help right?) behind <em>some</em> weights for orthogonal polynomials.
Weights are often of the form
<span class="math-container">$$
w(x) = e^{-v(x)},
$$</span>
where <span class="math-container">$v$</span> is analytic on the interior of the support of the weight. </p>
<p>For example</p>
<ul>
<li><span class="math-container">$v(x)=x^2$</span> on <span class="math-container">$\mathbb{R}$</span> for Hermite</li>
<li><span class="math-container">$v(x)= a\log(x)x$</span> on <span class="math-container">$(0,\infty)$</span> for Laguerre</li>
<li><span class="math-container">$v(x)= a\log(x-1)+b\log(x+1)$</span> on <span class="math-container">$(-1,1)$</span> for Jacobi</li>
</ul>
<p>This <span class="math-container">$v$</span> can the <strong>roughly</strong> be interpreted as an electrostatic potential. Then if you put <span class="math-container">$n$</span> charged particles in this potential, their equilibrium positions will be the zeros of the orthogonal polynomial of degree <span class="math-container">$n$</span> (with respect to <span class="math-container">$w$</span>). </p>
<p><strong>This isn't the whole story!</strong> What mass and charges do the particles get for example? And in the case of Legendre one needs to add some extra static charges at the endpoints of the support for there to even be an equilibrium position.</p>
<p>For more details see for example section 6.7 of "Orthogonal Polynomials" by G. Szegö or maybe a bit lighter the introduction in "Electrostatic interpretation for the zeros of certain polynomials and the Darboux process" by A Grünbaum (2001).</p>
|
4,274,314 | <blockquote>
<p>Find all <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> such that
<span class="math-container">$$f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $$</span>
for all <span class="math-container">$x,y \in \mathbb{R}$</span>.</p>
</blockquote>
<p>Help me solving this. My expectation of the answer is <span class="math-container">$f(x) = x+1$</span>.</p>
<p>My try:
<span class="math-container">$$
P(x, y): f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y \text. \\
P(x, 0): f\bigl(xf(0)\bigr)+f\bigl(-f(x)\bigr) = f(0) \text. \\
P(0, y): f(y)+f\bigl(-f(0)\bigr) = f\Bigl(y\bigl(f(0)-1\bigr)\Bigr)+y \text. \\
P(0, 0): f(0)+f\bigl(-f(0)\bigr) = f(0) \implies f\bigl(-f(0)\bigr)=0 \text. \\
\text {Assume) } f(a)=1 \text. \\
P(x, a): f(x+a) + f\bigl(-f(x)\bigr)=f\bigl(af(x)-a\bigr)+a \text. \\
x = 0 \text; \ 1 = f\Bigl(a\bigl(f(0)-1\bigr)\Bigr)+a \text. \\
x = a \text; \ f(2a)+f(-1) = f(0)+a \text. \\
$$</span></p>
| Sathvik | 516,604 | <p><span class="math-container">$$f(xf(y)+y)+f(−f(x))=f(yf(x)−y)+y$$</span>
Define <span class="math-container">$f(0)=c$</span> for some <span class="math-container">$c\in \mathbb{R}$</span>.</p>
<hr />
<ul>
<li><span class="math-container">$(x,y)\equiv (0,0)$</span>
<span class="math-container">$$f(0)+f(-f(0))=f(0)\implies f(-c)=0$$</span></li>
<li><span class="math-container">$(x,y)\equiv (-c,-c)$</span>
<span class="math-container">$$f(-c)+c=f(c)-c\implies f(c)=2c$$</span></li>
<li><span class="math-container">$(x,y)\equiv (0,c)$</span>
<span class="math-container">$$f(c)+f(-c)=f(c^2-c)+c\implies c=f(c^2-c)$$</span></li>
<li><span class="math-container">$(x,y)\equiv(0,-c)$</span>
<span class="math-container">$$f(-c)+f(-c)=f(-c^2+c)-c\implies c=f(-(c^2-c))$$</span></li>
<li><span class="math-container">$(x,y)\equiv(-c,c^2-c)$</span>
<span class="math-container">$$f(-c)+c=f(-c^2+c)+c^2-c \implies c^2=c\implies c\in\{0,1\}$$</span></li>
</ul>
<hr />
<p>If <span class="math-container">$c=f(0)=0$</span>,</p>
<ul>
<li><span class="math-container">$(x,y)\equiv(x,0)$</span>
<span class="math-container">$$f(-f(x))=0\;, \;\;\; \forall x\in \mathbb{R}$$</span></li>
<li><span class="math-container">$(x,y)\equiv(-x/f(x),x)$</span>
<span class="math-container">$$-f(xf(-x/f(x))-x)=x \\ f(-f(xf(-x/f(x))-x))=f(x) \\
\;\;\;\therefore\; f(x)=0\;, \;\;\; \forall x\in \mathbb{R}\;\;$$</span>
But <span class="math-container">$f(x)=0$</span> for real <span class="math-container">$x$</span> is not a solution of the original F.E.</li>
</ul>
<hr />
<p>If <span class="math-container">$c=f(0)=1$</span>,</p>
<ul>
<li><span class="math-container">$(x,y)\equiv(0,x)$</span>
<span class="math-container">$$f(x)=f(cx-x)+x\implies f(x)=x+1.$$</span>
Plugging this back in the original F.E, we verify that <span class="math-container">$\boxed{f(x)=x+1}\; \forall x\in \mathbb{R}$</span> is the only solution.</li>
</ul>
<hr />
|
207,515 | <p>Suppose I have the following list, </p>
<pre><code>l = {{"b", "c", "d"}, {"e", "b"}, {"a", "b", "d", "e"}}
</code></pre>
<p>and further suppose I have the following association, </p>
<pre><code>l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
</code></pre>
<p>I wonder how can I replace the keys into my list such that I get, </p>
<pre><code>{{2, 3, 4}, {5, 2}, {1, 2, 4, 5}}
</code></pre>
| Lukas Lang | 36,508 | <p>The following gives you a "reversed" version of your association, with keys and values flipped:</p>
<pre><code>l1 = <|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>;
lookup = First /@ PositionIndex@l1
(* <|"a" -> 1, "b" -> 2, "c" -> 3, "d" -> 4, "e" -> 5|> *)
</code></pre>
<p>You can then use <a href="https://reference.wolfram.com/language/ref/ReplaceAll.html" rel="noreferrer"><code>ReplaceAll</code></a> (<code>/.</code>) to do the replacement:</p>
<pre><code>l = {{"b", "c", "d"}, {"e", "b"}, {"a", "b", "d", "e"}};
l /. lookup
(* {{2, 3, 4}, {5, 2}, {1, 2, 4, 5}} *)
</code></pre>
<h3>Alternatives</h3>
<p>Other possible solutions for creating <code>lookup</code>:</p>
<pre><code>lookup = <|Reverse /@ Normal@l1|>
lookup = Reverse /@ Normal@l1 (* doesn't need to be an association *)
</code></pre>
|
207,515 | <p>Suppose I have the following list, </p>
<pre><code>l = {{"b", "c", "d"}, {"e", "b"}, {"a", "b", "d", "e"}}
</code></pre>
<p>and further suppose I have the following association, </p>
<pre><code>l1=<|1 -> "a", 2 -> "b", 3 -> "c", 4 -> "d", 5 -> "e"|>
</code></pre>
<p>I wonder how can I replace the keys into my list such that I get, </p>
<pre><code>{{2, 3, 4}, {5, 2}, {1, 2, 4, 5}}
</code></pre>
| Suba Thomas | 5,998 | <pre><code>l /. Reverse /@ Normal[l1]
(* {{2, 3, 4}, {5, 2}, {1, 2, 4, 5}} *)
</code></pre>
<p>or</p>
<pre><code>l /. AssociationMap[Reverse, l1]
(* {{2, 3, 4}, {5, 2}, {1, 2, 4, 5}} *)
</code></pre>
|
167,326 | <p>Let $ S = \sum_{i=1}^n X_i$ where:</p>
<ul>
<li>Each $X_i$ is independently 3 or 9 (with equal probability), and</li>
<li>The sample size $n$ is itself an independent random variable where $N \sim \text{NegativeBinomial}(r,p)$ e.g. $r = 5$ and $p = \frac34$</li>
</ul>
<p>Let $W = \begin{cases}S-10 & S > 10 \\ 0 & S \leq 10 \end{cases} . \quad$ </p>
<p>Find: (i) the pmf of $S\quad$ (ii) the pmf of $W \quad$ (iii) If not the former, can one at least find $\mathbb{E}[W]$ ?</p>
<hr>
<p>How can I obtain the distribution of $W$ using mathematica? Although it may not be able to give me a reasonable PDF, <strong>can I at least find the expected value of $W$ with mathematica? Also, can I also draw random samples from the distribution of $W$?</strong> If so, how would I do this? Thank you.</p>
| JimB | 19,758 | <p>Here is a basic principles approach to obtaining the probability mass function for both $S$ and $W$ (which uses the clarification asked for by @wolfies):</p>
<pre><code>(* Set parameters *)
r = 5;
p = 3/4;
pBin = 1/2;
(* Function to generate the combinations of n (sample size) and
k (number of successes) that result in S: {n,k} as S = 9n - 6k *)
f[s_] := Table[{(s - 2 i)/3, s/3 - i}, {i, 0, s/3, 3}]
(* Individual probability for a combination of n and k *)
pr[n_, k_] := PDF[BinomialDistribution[n, pBin], k]*
PDF[NegativeBinomialDistribution[r, p], n]
(* Probability mass function for S *)
(* Note that only non-negative multiples of 3 have a positive probability *)
prS[s_] := If[IntegerQ[s/3], Total[pr[#[[1]], #[[2]]] & /@ f[s]], 0]
(* Probability mass function for W *)
prW[w_] := If[w == 0, prS[0] + prS[3] + prS[6] + prS[9],
If[IntegerQ[(w + 1)/3], prS[w + 10], 0]]
(* Approximate expectation of W *)
N[Sum[w prW[w], {w, 0, 500}]]
(* 3.79827 *)
</code></pre>
<p>Plots of the probability mass functions are as follows:</p>
<pre><code>ListPlot[Table[prS[s], {s, 0, 50}], PlotRange -> All,
Frame -> True, FrameLabel -> {"S", "Probability"}, Filling -> 0]
ListPlot[Table[prW[w], {w, 0, 50}], PlotRange -> All,
Frame -> True, FrameLabel -> {"W", "Probability"}, Filling -> 0]
</code></pre>
<p><a href="https://i.stack.imgur.com/2ycPo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2ycPo.png" alt="pmf of S"></a>
<a href="https://i.stack.imgur.com/ydleU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ydleU.png" alt="pmf of W"></a></p>
|
112,651 | <p>What is known about the set of well orderings of $\aleph_0$ in set theory without choice? I do not mean the set of countable well-order types, but the set of all subsets of $\aleph_0$ which (relative to a pairing function) code well orderings. And I would be interested in an answer in, say, ZF without choice. My actual concern is higher order arithmetic.</p>
<p>I would not be surprised if ZF proves there are continuum many. But I don't know.</p>
<p>At the opposite extreme, is it provable in ZF that there are not more well orderings of $\aleph_0$ than there are countable well-order types?</p>
| Stefan Geschke | 7,743 | <p>There are two answers already, but I think this argument is simpler than both of the previous answers.</p>
<p>Any two permutations of $\omega$ give two different wellorderings of order type $\omega$.
We show that there are $2^{\aleph_0}$ permutations of $\omega$.
Given a function $f:\omega\to 2$ let $\sigma_f$ be the permutation that for all $n\in\omega$
exchanges $2n$ and $2n+1$ iff $f(n)=1$.
It is clear that $f\mapsto\sigma_f$ is 1-1. Hence there are at least $2^{\aleph_0}$
wellorderings of order type $\omega$ on $\omega$.</p>
<p>As Andres pointed out, this transfers to every countable order type $\alpha$.</p>
|
2,435,596 | <p>Suppose we have an unsigned $8$ bit number (min=$0$, max=$255$).</p>
<p>the result of "$200 + 200$" overflows to $144$</p>
<p>the result of "$100 - 200$" (under?)overflows to $156$</p>
<p>Is there are mathematical symbol to represent this?</p>
| Siong Thye Goh | 306,553 | <p>You can denote it using $x \pmod {y+1}$ as long as your minimum is $0$ and the maximum value is $y$.</p>
<p>Example $1$:
$$200 + 200 = 400 = 256 + 144$$</p>
<p>$$200 + 200 \equiv 144 \pmod {256}$$</p>
<p>Example $2$:
$$100 - 200 = - 100 = -256 + 156$$</p>
<p>$$100 - 200 \equiv -100 \pmod {256}$$</p>
|
3,608,441 | <p>It is well known that we can define <span class="math-container">$e^x$</span> by the following limit</p>
<p><span class="math-container">$$e^{x}=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$</span></p>
<p>I would like to show that the RHS sequence is always less than or equal to <span class="math-container">$e^x$</span> for all <span class="math-container">$-n\le x\le0$</span> and <span class="math-container">$n>1$</span>, and what I had currently done is to study the property of this sequence (which is defined as <span class="math-container">$f(x)$</span>)</p>
<p><span class="math-container">$$f(x)=\left(1+{x\over n}\right)^n$$</span></p>
<p>I also find its first and second derivative <span class="math-container">$f'(x)=\left(1+{x\over n}\right)^{n-1}$</span> and <span class="math-container">$f''(x)={n-1\over n}\left(1+{x\over n}\right)^n$</span> and show that they are strictly positive for all <span class="math-container">$x\in[-n,0]$</span>. As a result <span class="math-container">$f(x)$</span> is monotonically increasing and concave up. By plugging in the end points I found that <span class="math-container">$f(0)=1\le e^0$</span> and <span class="math-container">$f(-n)=0<e^{-n}$</span>, I wonder if these conditions allow me conclude that <span class="math-container">$f(x)\le e^x$</span> for all <span class="math-container">$x\in[-n,0]$</span>.</p>
| Gabriel Romon | 66,096 | <p>Here's an elementary proof that only requires Bernoulli's inequality. </p>
<p>Let <span class="math-container">$\displaystyle u_n:=\left(1+\frac xn\right)^n$</span>
and let <span class="math-container">$N$</span> be the smallest integer strictly greater than <span class="math-container">$|x|$</span>. Consider <span class="math-container">$n\geq N$</span>. </p>
<p>Then <span class="math-container">$n>|x|$</span>, hence <span class="math-container">$0<1+\frac xn$</span>.
Besides, <span class="math-container">$$u_{n+1}=\left(1+\frac xn \right)^{n+1}\left(1-\frac{x}{n(n+1)\left(1+\frac xn\right)} \right)^{n+1}$$</span>
Note that for <span class="math-container">$n\geq N$</span>, <span class="math-container">$$-\frac{x}{n(n+1)\left(1+\frac xn\right)}\geq -1 \iff \frac{x}{n(n+1)} \leq 1+\frac xn$$</span>
and <span class="math-container">$$\frac{x}{n(n+1)} \leq 1+\frac xn \iff -x\leq n+1$$</span></p>
<p>Bernoulli's inequality applies and yields <span class="math-container">$\forall n\geq N$</span>, <span class="math-container">$$\left(1-\frac{x}{n(n+1)\left(1+\frac xn\right)} \right)^{n+1}\geq 1-\frac{x}{n\left(1+\frac xn\right)}$$</span>
Since <span class="math-container">$\displaystyle 1-\frac{x}{n\left(1+\frac xn\right)} = \frac{1}{1+\frac xn}$</span>, we get that
for <span class="math-container">$n\geq N$</span>, <span class="math-container">$$u_{n+1}\geq u_n$$</span></p>
|
3,608,441 | <p>It is well known that we can define <span class="math-container">$e^x$</span> by the following limit</p>
<p><span class="math-container">$$e^{x}=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$</span></p>
<p>I would like to show that the RHS sequence is always less than or equal to <span class="math-container">$e^x$</span> for all <span class="math-container">$-n\le x\le0$</span> and <span class="math-container">$n>1$</span>, and what I had currently done is to study the property of this sequence (which is defined as <span class="math-container">$f(x)$</span>)</p>
<p><span class="math-container">$$f(x)=\left(1+{x\over n}\right)^n$$</span></p>
<p>I also find its first and second derivative <span class="math-container">$f'(x)=\left(1+{x\over n}\right)^{n-1}$</span> and <span class="math-container">$f''(x)={n-1\over n}\left(1+{x\over n}\right)^n$</span> and show that they are strictly positive for all <span class="math-container">$x\in[-n,0]$</span>. As a result <span class="math-container">$f(x)$</span> is monotonically increasing and concave up. By plugging in the end points I found that <span class="math-container">$f(0)=1\le e^0$</span> and <span class="math-container">$f(-n)=0<e^{-n}$</span>, I wonder if these conditions allow me conclude that <span class="math-container">$f(x)\le e^x$</span> for all <span class="math-container">$x\in[-n,0]$</span>.</p>
| pre-kidney | 34,662 | <p>To show that <span class="math-container">$(1+x/n)^n\leq e^x$</span> for all <span class="math-container">$x\geq -n$</span> is equivalent to showing that
<span class="math-container">$$
1+\frac{x}{n}\leq e^{x/n},\qquad x\geq -n.
$$</span>
This is a special case of the following inequality
<span class="math-container">$$
1+y\leq e^y,\qquad (\star)
$$</span>
which holds for all <span class="math-container">$y\in\mathbb R$</span>, when we substitute <span class="math-container">$y=x/n$</span>.</p>
<p>To prove <span class="math-container">$(\star)$</span>, you can observe that it is an equality when <span class="math-container">$y=0$</span>, and the derivative of the left side is less than the derivative of the right side for all <span class="math-container">$y>0$</span>, and it is larger for <span class="math-container">$y<0$</span>.</p>
|
1,283,325 | <p>I got two questions about $p$-adic numbers:</p>
<blockquote>
<ol>
<li>I often read that the field $\mathbb Q_p$ is much different than the field $\mathbb R$.</li>
</ol>
</blockquote>
<p>An element of $\mathbb Q_p$ is of the form $\sum_{i=-k}^{\infty}a_ip^i$ where $a_i\in \{0,...,p-1\}$.</p>
<p>But isn't this just a real number? So at least the elements of $\mathbb Q_p$ are a subset of $\mathbb R$? That would mean that these fields are especially different in terms of their operation?</p>
<blockquote>
<ol start="2">
<li>Let $x\in \mathbb Q_p^*$. Why $x$ can be written uniquely like this: $x=p^na$ where $a$ is an element of the $p$-adic integers?</li>
</ol>
</blockquote>
<p>Thanks in advance!</p>
| Crostul | 160,300 | <p>As for the first question, the answer is no. For example, what real number should represent this
$$\sum_{n=0}^{\infty} 5^{n!}$$
$5$-adic number? Note that in real numbers this series is obviously divergent.</p>
<p>As for the second question: you should know that $\Bbb{Z}_p$ is a local ring whose unique maximal ideal is $p\Bbb{Z}_p$. This means that every $p$-adic integer which is not divisible by $p$ is invertible.
Moreover it is a UFD, and every element can be factorized as $up^k$ for some unit $u$, some $k \ge 0$.</p>
<p>So every element of $\Bbb{Q}_p$ has the form
$$\frac{up^k}{vp^h} = (uv^{-1})p^{k-h}$$</p>
<p>EDIT: The confusion comes to your mind, since you are thinking these numbers as they were real numbers: but they are not! Let's consider for example the sequence of integers (actual integers in $\Bbb{Z}$)
$$1, \ \ 1+5, \ \ 1+5+5^2, \ \ 1+5+5^2+5^3, \dots$$
in $\Bbb{R}$ these sequence diverges. However, if we think it inside $\Bbb{Q}_5$, this sequence converges to the $5$-adic number
$$A=\sum_{n=0}^{\infty} 5^n$$
actually, this is the inverse of $-4$ in $\Bbb{Q}_5$ since
$$-4A=A-5A = (1+5+5^2+5^3+5^4+ \dots)-(5+5^2+5^3+\dots) = 1$$
(all of this is not true in $\Bbb{Q}_p$ for $p \neq 5$, where the sequence diverges). So you have
$$\sum_{n=0}^{\infty} 5^n = -\frac{1}{4} \ \ \ \ \mbox{ in } \Bbb{Q}_5$$
This is possible because of the strange topological structure of $p$-adic integers.</p>
|
1,858,297 | <p>Suppose the diameter of a nonempty set $A$ is defined as </p>
<p>$$\sigma(A) := \sup_{x,y \in A} d(x,y)$$</p>
<p>where $d(x,y)$ is a metric.</p>
<p>Is $\sigma(.)$ a 'measurement'? I.e., how do I prove the countable additivity for this particular case?</p>
| Ethan Bolker | 72,858 | <p>It's not even finitely additive. If $X$ and $Y$ are two disjoint closed intervals on the real line then the diameter of their union is not the sum of their diameters.</p>
|
1,070,870 | <p>"Write down (say, as a power series) a holomorphic function $f(z)$ on $D(1, 1)$ which satisfies $f(z)^5 = z$ and $f(1) = 1$. What is the result of analytically continuing $f$ along a path which travels once counterclockwise around the origin, returning to the point $1$? What about if you go $N$ times counterclockwise around the origin, where $N$ is an integer?"</p>
<p>For the analytic continuation, I understand how to do it in the way where I explicitly write down square root functions in successive disks around the origin in terms of polar coordinates. Is there a more general/"abstract" way to do it though?</p>
<p>Thanks in advance!</p>
| JohnD | 52,893 | <p>Unsure of your notation/assumptions, but here's a hint: </p>
<ol>
<li>For real matrices, $$\text{rank}(A^*A)=\text{rank}(AA^*)=\text{rank}(A)=\text{rank}(A^*)$$</li>
<li>For complex matrices, $$\text{rank}(A^*A)=\text{rank}(A)=\text{rank}(A^*)$$</li>
</ol>
<p>Mouse over for more after you've pondered it a bit:</p>
<blockquote class="spoiler">
<p> $\text{rank}(A)=n\iff \text{rank}(A^*A)=n\iff A^*A\text{ has full rank}\iff A^*A\text{ invertible}$</p>
</blockquote>
|
3,450,598 | <blockquote>
<p>Prove that <span class="math-container">$\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = \sum_{i = m}^p a_i$</span>, where <span class="math-container">$m ≤ n<p$</span> are integers, and <span class="math-container">$a_i$</span> is a real number assigned to each integer <span class="math-container">$m ≤ i ≤ p$</span>. (Hint: you might want to use induction)</p>
</blockquote>
<p>Let's follow the hint and use induction on <span class="math-container">$p-m = k$</span><br>
Base case: <span class="math-container">$k = 1$</span>, then <span class="math-container">$p = m + 1$</span> and <span class="math-container">$n = m$</span>. <span class="math-container">$\sum_{i = m}^n a_i + \sum_{i = n + 1}^p a_i = a_m + a_{m+1}$</span> and <span class="math-container">$\sum_{i = m}^p a_i = a_m+a_{m+1}$</span>. Therefore, the right-hand side is equal to the left-hand side.<br>
Inductive step: Assume for <span class="math-container">$p-m=k$</span> the statement holds, show for <span class="math-container">$p-m = k + 1$</span>. We know that <span class="math-container">$\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k} a_i = \sum_{i = m}^{m+k} a_i$</span>. Now, <span class="math-container">$\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k+1} a_i = \sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k} a_i + a_{m+k+1} = \sum_{i = m}^{m+k} a_i + a_{m+k+1}$</span> by inductive hypothesis. Therefore, we get <span class="math-container">$$\sum_{i = m}^n a_i + \sum_{i = n + 1}^{m+k+1} a_i =\sum_{i = m}^{m+k+1} a_i$$</span></p>
<p>Is this prove plausible. At this point about the finite sum, I can use the following facts: </p>
<p>if <span class="math-container">$ m < n \sum_n^m a_i= 0$</span><br>
if <span class="math-container">$n \ge m - 1 \sum_{m}^{n+1}a_i = \sum_{m}^{n}a_i + a_{n+1}$</span> </p>
| Community | -1 | <p><span class="math-container">$$\sum_{i=m}^n a_i+\sum_{i=n+1}^{p}a_i=\sum_{i=m}^n a_i+a_{n+1}-a_{n+1}+\sum_{i=n+1}^{p}a_i
=\sum_{i=m}^{n+1} a_i+\sum_{i=n+2}^{p}a_i$$</span></p>
<p>is a base case and by induction,<span class="math-container">$$\sum_{i=m}^n a_i+\sum_{i=n+1}^{p}a_i
=\sum_{i=m}^{n+k} a_i+\sum_{i=n+k+1}^{p}a_i.$$</span></p>
<p>Form this the initial claim follows by setting <span class="math-container">$n+k=p$</span>.</p>
|
1,626,821 | <p>Is there a name for a vector with all equal elements? If so, what is it?</p>
<p>For example,</p>
<p>$$ (7, 7, 7, 7, 7) $$</p>
| Zubin Mukerjee | 111,946 | <p>This is <a href="https://mathoverflow.net/questions/9898/notation-for-the-all-ones-vector">relevant</a>. Given how much of a hullabaloo there is over just the all-ones vector, I'm guessing there is no standard name for the more general vectors with all equal elements. </p>
<p>If we use $\vec{1}$ for the all-ones vector, in $\mathbb{R}^5$ we could write</p>
<p>$$(7,7,7,7,7) = 7 \cdot \vec{1}$$</p>
|
40,348 | <p>I'm trying to prove the following statement (an exercise in Bourbaki's <em>Set Theory</em>): </p>
<p><em>If $E$ is an infinite set, the set of subsets of $E$ which are equipotent to $E$ is equipotent to $\mathfrak{P}(E)$.</em> </p>
<p>As a hint, there is a reference to a proposition of the book, which reads: </p>
<p><em>Every infinite set $X$ has a partition $(X_\iota)_{\iota\in I}$ formed of countably infinite sets, the index set $I$ being equipotent to $X$.</em> </p>
<p>I don't have any idea how that proposition might help. </p>
<p>If $E$ is countable, then a subset of $E$ is equipotent to $E$ iff it is infinite. But the set of all finite subsets of $E$ is equipotent to $E$. So its complement in $\mathfrak{P}(E)$ has to be equipotent to $\mathfrak{P}(E)$ by Cantor's theorem. Hence the statement is true if $E$ is countable. Unfortunately, I don't see a way to generalize this argument to uncountable $E$.</p>
<p>I'd be glad for a small hint to get me going. </p>
| Mike F | 6,608 | <p>Another approach would make use of the fact that $\kappa_1 + \kappa_2 = \max(\kappa_1,\kappa_2)$ when $\kappa_1,\kappa_2$ are cardinals at least one of which is infinite. From this it follows that, for any subset of $S \subset X$, either $S$ or its complement has cardinality $|X|$. Since more straightforward cardinal arithmetic shows there are as many complementary pairs of subsets as there are subsets of $X$, we get the result.</p>
<p>By the way, JDH's approach could allow you to show the (slightly) stronger statement that there are $2^X$ sets $S \subset X$ with <em>both</em> of $S$ and $X - S$ equipotent to $X$. Simply write $X$ as a disjoint union $X = X_0 \cup X_1 \cup X_2$ with all parts equipotent to $X$ and consider subsets of the form $X_0 \cup A$ where $A \subset X_1$. </p>
<p>An interesting follow-up question might be to see if you can produce $2^X$ sets $S \subset X$ so that $|S|=|X|$, but $|X-S| < |X|$. </p>
|
1,613,645 | <p>Let's get started:</p>
<p>$$\hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} |x|e^{-inx} dx$$</p>
<p>since $|x|$ is an even function:</p>
<p>$$= \frac{1}{\pi}\int_0^{\pi} xe^{-inx} dx$$</p>
<p>Integration by parts yields:</p>
<p>$$e^{-inx}\Big|_0^{\pi} + \frac{1}{in} \int_0^\pi e^{-inx} dx = (-1)^n - 1 + \frac{1}{in} \left( \frac{(-1)^n}{-in} + \frac{1}{in} \right) \\ = (-1)^n - 1 + \frac{(-1)^n - 1}{n^2}$$</p>
<p>So if $n$ is even then $\hat f(n) = 0$. Otherwise:</p>
<p>$$\hat f(n) = \frac{1}{\pi} \left( -2 -\frac{2}{n^2} \right)$$</p>
<p>but that doesn't make sense since we know that $\hat f(n) \to 0$.</p>
<p>Where is my mistake? </p>
<p><strong>EDIT</strong>
it should be </p>
<p>$$x e^{-inx}\Big|_0^{\pi} + \frac{1}{in} \int_0^\pi e^{-inx} dx = \frac{\pi e^{-in\pi}}{-in} + \frac{(-1)^n - 1}{n^2}$$</p>
<p>So $$\hat f(n) = \frac{1}{\pi} \left( \frac{(-1)^n}{-in} + \frac{(-1)^n - 1}{n^2} \right)$$</p>
| Igor | 165,838 | <p>First you can set up a proportion to find the central angle it has moved,</p>
<p>35/60=x/360→x=210</p>
<p>Then you can use the arc length formula
arc length = 2πr(x/360) = 2π(15)(210/360)=35π/2.</p>
|
3,673,014 | <p>If you take an <span class="math-container">$2r\times 2r\times 2r$</span> cube, and divide it to 27 equal cubes, and then remove all the "axis" cubes (all the cubes which are straight left, straight right, straight up, etc. from the middle cube) then divide each cube into 125 equal cubes and remove all the axis cubes, and repeat this process for every odd number to the power of three, you will get a nice fractal cube.</p>
<p>If we try to work out the area of this cube we get (after some short algebra):<span class="math-container">$$A=8r^3\prod_{n=1}^{\infty}\frac{(2n)^2(2n+3)}{(2n+1)^3}=8r^3\frac{2\times2\times\not5\times4\times4\times\not7\times6\times6\times\dots}{3\times3\times3\times\not5\times5\times5\times\not7\times7\times7\times\dots}$$</span>
Now, we extract the Wallis product and get<span class="math-container">$$A=8r^3\frac{\pi}6$$</span>Which is the area of a sphere with radius <span class="math-container">$r$</span>(!!!)</p>
<p>Now I want to know what is the fractal dimension of this sphere cube.</p>
| amd | 265,466 | <p>In the method that you propose to use, the expression that defines the set is a convex combination of the vertices of the polyhedron. It’s a bit unfortunate that it was presented to you in simplified form since that obscures the simple construction: It’s just a linear combination <span class="math-container">$\sum_i \lambda_iv_i$</span> of the vertices with the constraints <span class="math-container">$\lambda_i\ge0$</span> and <span class="math-container">$\sum_i\lambda_i=1$</span>. So, for this tetrahedron, we would have the combination <span class="math-container">$$\lambda_1(0,0,0)+\lambda_2(1,0,0)+\lambda_3(0,1,0)+\lambda_4(0,1,1) = \lambda_2e_1+(\lambda_3+\lambda_4)e_2+\lambda_4e_3.$$</span> Even though it doesn’t appear in the final expression, you still need to account for <span class="math-container">$\lambda_1$</span> in the constraint <span class="math-container">$\sum_i\lambda_i=0$</span>. Leaving <span class="math-container">$\lambda_1$</span> out of that constraint amounts to fixing it at <span class="math-container">$0$</span>, in which case all of the points lie on the plane defined by the second through last vertices: the tetrahedron collapses into the triangle <span class="math-container">$\triangle{v_2v_3v_4}$</span>. </p>
<p>For a small problem like this one, you could take a more direct approach instead. Take the vertices three at a time and find the equation of the corresponding plane, choosing the direction of the normal <span class="math-container">$n$</span> so that if the equation of the plane is <span class="math-container">$n\cdot x=b$</span> the remaining vertex <span class="math-container">$v$</span> satisfies <span class="math-container">$n\cdot v\le b$</span>. You then stack the coefficients of the resulting equations into the augmented matrix <span class="math-container">$[A\mid b]$</span>. For example, taking <span class="math-container">$v_2$</span>, <span class="math-container">$v_3$</span> and <span class="math-container">$v_4$</span>, we can obtain the normal <span class="math-container">$(1,1,0)$</span>, so an equation of the plane is <span class="math-container">$x+y=1$</span>. Comparing this to the remaining vertex, we see that the appropriate half-space is <span class="math-container">$x+y\le1$</span>, so the first row of <span class="math-container">$A$</span> is <span class="math-container">$(1,1,0)$</span> and the first element of <span class="math-container">$b$</span> is <span class="math-container">$1$</span>. Continuing with the other faces will give you the other three rows.</p>
|
2,238,734 | <p>Let G be a group of rationals under addition, if $G_1$ and $G_2$ are two non empty subgroups of G, then prove that $G_1 \cap G_2 \neq${0}</p>
| Community | -1 | <p>It is the first definition. Because you know that $\mathbb{R^n}$ is a vector space, the second thing you wrote is trivial.</p>
<p>To show that $W$ is a subspace of $V$, you have to show it is closed under scalar multiplication and vector addition, so you have to show that:</p>
<p>$\forall x,y \in W: x+y \in W$</p>
<p>$\forall x \in W, \forall \alpha \in \mathbb{R}: \alpha x \in W$</p>
<p>or, equivalently:</p>
<p>$\forall x,y \in W, \forall \alpha, \beta \in \mathbb{R}: \alpha x + \beta y \in W$</p>
|
1,983,745 | <p>I need to choose weather this is a product notation or a summation. I can figure out which one it is.</p>
<p>I have this expression:</p>
<p>$$2 \times 4 \times 6 \times 8 \times 10 \ldots \times 40$$</p>
<p>The answer is either:</p>
<blockquote>
<p>$$\sum_{m=2}^{40} m$$</p>
</blockquote>
<p><strong>or</strong> </p>
<blockquote>
<p>$$\prod_{m=2}^{40} m$$</p>
</blockquote>
| Siong Thye Goh | 306,553 | <p>Neither of your proposal is correct.</p>
<p>For your first guess, it means $$2+3+4+\ldots+ 40$$</p>
<p>For your second guess, it means $$2(3)(4) \ldots (40)$$</p>
<p>You are multiplying even numbers, it should be</p>
<p>$$\prod_{i=1}^{20} (2i)$$</p>
|
1,983,745 | <p>I need to choose weather this is a product notation or a summation. I can figure out which one it is.</p>
<p>I have this expression:</p>
<p>$$2 \times 4 \times 6 \times 8 \times 10 \ldots \times 40$$</p>
<p>The answer is either:</p>
<blockquote>
<p>$$\sum_{m=2}^{40} m$$</p>
</blockquote>
<p><strong>or</strong> </p>
<blockquote>
<p>$$\prod_{m=2}^{40} m$$</p>
</blockquote>
| Nambiar M. | 374,565 | <p>Neither is correct. It is in fact the product notation, as summation notation would be $2+4+6+8+...+40$, however, the expression is incorrect. The correct expression is $\prod_{m=1}^{20}2m$. There is a way to do it with summation notation, but I don't think that's what you're looking for.</p>
|
3,197,540 | <p>Let a function be defined as:</p>
<p><span class="math-container">$ f(x)=x^2\sin{\left(\frac 1x\right)}$</span> for <span class="math-container">$x \neq 0$</span> and
<span class="math-container">$ f(x)=0$</span> for <span class="math-container">$x=0$</span></p>
<p>I'm trying to prove that f is differentiable at 0 using the definition of derivative. However in the process of doing this I was stopped by this limit:</p>
<p><span class="math-container">$$ \lim_{h \to 0} \frac{\sin\left({\frac{1}{x+h}}\right)-\sin\left({\frac{1}{x}}\right)}{h}
$$</span></p>
<p>Is it possible to solve this limit question without using l'Hopital's rule?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Use that <span class="math-container">$$\sin(x)-\sin(y)=2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \cos
\left(\frac{x}{2}+\frac{y}{2}\right)$$</span></p>
|
2,317,391 | <p>This question popped up somewhere on the internet and I thought it was interesting. I attempted to solve it but I don't know if it is correct.</p>
<blockquote>
<p>Find the derivative of $$F(x)=\int_{\cos{x^3}}^{\int_{1}^{x} {1/(1+t^2)dt}} {\sin{w} dw}$$</p>
</blockquote>
<p>$$\begin{align}
\implies F(x) & =-\cos{w}]_{\cos{x^3}}^{\arctan{x}-\frac{\pi}{4}}\\
& = -\cos{(\arctan{x}-\frac{\pi}{4})}+\cos{(\cos{x^3})}\\
\implies F'(x) & = \frac{1}{1+x^2} \sin{(\arctan{x}-\frac{\pi}{4}})+3x^2\sin{(x^3)} \sin{(\cos{(x^3)})}\\
\end{align}$$</p>
<p>Is it this simple? Or is there something I should know before solving this that changes the normal differentiation and integration techniques?</p>
| Yes | 155,328 | <p>Yes, your concept is correct. But note that we have
$$
F(x) = \int_{g(x)}^{f(x)}\sin w dw = \int_{a}^{f(x)} \sin w dw - \int_{a}^{g(x)} \sin w dw
$$
where $f,g$ are the given functions appearing in the original upper and lower integral limits, respectively; note that $f,g,F$ are "nice" enough to allow us to write
$$
DF(x) := F'(x) = D\int_{a}^{f(x)}\sin w dw - D\int_{a}^{g(x)} \sin w dw = \sin \circ f(x)f'(x) - \sin \circ g(x)g'(x)
$$
for all suitable $x$ by the chain rule.</p>
|
1,322,016 | <p>Find the vertical asymptotes (if any) of the graph of the function. (Use $n$ as an arbitrary integer if necessary.)</p>
<p>$$s(t)= \frac{8t}{\sin{t}}$$</p>
<p>$t= ?$, where n cannot $=?$</p>
<p>I need a general rule for the asymptotes with where the exception of $n$ is. </p>
| RowanS | 246,774 | <p>We are not allowed to divide by $0$ so $sin(t) = 0$ is an asymptote. So $t = n\pi \quad \forall n \in \mathbb Z - \{ 0 \} $ </p>
|
1,322,016 | <p>Find the vertical asymptotes (if any) of the graph of the function. (Use $n$ as an arbitrary integer if necessary.)</p>
<p>$$s(t)= \frac{8t}{\sin{t}}$$</p>
<p>$t= ?$, where n cannot $=?$</p>
<p>I need a general rule for the asymptotes with where the exception of $n$ is. </p>
| Irddo | 71,356 | <p>So, you have a vertical asymtote, very times that $f(x)\to+\infty$ or $f(x)\to-\infty$, when $x\to a\in \mathbb{R}$, so $x=a$ will be a vertical asymtote.</p>
<p><strong>Hint:</strong> For which $x\in \mathbb{R}$, you have the $f(x)\to+\infty$ or $f(x)\to-\infty$? </p>
<p>(Try to annul the denominator of the function for "blow up" the function.)</p>
|
2,439,340 | <p>How would one proceed to prove this statement?</p>
<blockquote>
<p>The set of the strictly increasing sequences of natural numbers is not enumerable.</p>
</blockquote>
<p>I've been trying to solve this for quite a while, however I don't even know where to start.</p>
| David K | 139,123 | <p>For any infinite subset of the natural numbers, you can list its members in increasing order, and then you have a sequence that is strictly increasing.</p>
<p>Moreover, if you take two distinct infinite subsets of the natural numbers,
you get two different sequences.
(There is some number $k$ that is in one of the subsets and not the other,
and this number occurs in one sequence and not the other.)</p>
<p>So the number of strictly increasing sequences of natural numbers is at least as great as the number of infinite subsets of natural numbers.</p>
|
231,036 | <p>I wonder if there is an example of rational homology sphere that is not a Seifert manifold. If there is, how can one construct such a rational homology sphere from a surgery of a knot in $S^3$?</p>
| Daniel Valenzuela | 52,936 | <p>By Thurston, all but finitely many $(p,q)$-surgeries on a hyperbolic knot in $S^3$ result in hyperbolic rational homology spheres for $p\neq 0$. In particular there are infinitely many integral homology spheres among them.</p>
|
3,950,098 | <p>I can evaluate the limit with L'Hospital's rule:</p>
<p><span class="math-container">$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$</span></p>
<p>But is there any way to do it without using L'Hospital's rule?</p>
| Sewer Keeper | 213,667 | <p>Another approach using known limit</p>
<blockquote>
<p><span class="math-container">$$ \lim_{n \to +\infty} \frac{\mathrm{e}^{a_n}-1}{a_n} = 1 $$</span>
where <span class="math-container">$a_n$</span> is a sequence such that <span class="math-container">$ \lim_{n \to +\infty} a_n = 0$</span>.</p>
</blockquote>
<p><span class="math-container">$$
\begin{split}
\lim_{n \to +\infty} \frac{4^{\frac{1}{n}}-1}{\frac{1}{n}}&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&=\lim_{n \to +\infty} \frac{\mathrm{e}^{\frac{\ln 4}{n}}-1}{\frac{\ln 4}{n}}\cdot \lim_{n \to +\infty}\frac{\frac{\ln 4}{n}}{\frac{1}{n}}\\
&= 1 \cdot \ln 4 = \ln 4
\end{split}
$$</span></p>
|
3,380,081 | <p>Question: Suppose <span class="math-container">$n(S)$</span> is the number of subset of <span class="math-container">$S$</span> and <span class="math-container">$|S|$</span> be the number of elements of <span class="math-container">$S$</span>. If <span class="math-container">$n(A)+n(B)+n(C)=n(A\cup B\cup C)$</span> and <span class="math-container">$|A|=|B|=100$</span>, Find the minimum value of <span class="math-container">$|A\cap B\cap C|$</span>.</p>
<p>Now, I realise PIE is the only way to go, but I don't know how to handle the intersections of the sets taken two at a time.
Also, I know that this is a duplicate, but the original on wasn't answered fully. I'd request you o kindly provide a solution before mercilessly closing it off.</p>
<p>Plus, I'm not good at even very basic set theory. If you could recommend a short but good book for set-theory, I'd be much obliged.</p>
<p>Thank you all!</p>
| Jack D'Aurizio | 44,121 | <p><a href="https://i.stack.imgur.com/vpzAQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vpzAQ.png" alt="enter image description here"></a></p>
<p>The area of the <span class="math-container">$A$</span>-region can be computed as the difference between the area of a <span class="math-container">$4\times 2$</span> rectangle and the area of the <span class="math-container">$B$</span>-region. Since the inverse function of <span class="math-container">$y=\sqrt{x}$</span> is <span class="math-container">$x=y^2$</span> the area of the <span class="math-container">$B$</span>-region is given by</p>
<p><span class="math-container">$$ \lim_{n\to +\infty}\frac{2}{n}\sum_{k=1}^{n}\left(\frac{2k}{n}\right)^2 = 8 \lim_{n\to +\infty}\frac{n(n+1)(2n+1)}{6n^3}=\frac{8}{3}$$</span>
and the area of the <span class="math-container">$A$</span>-region equals <span class="math-container">$8-\frac{8}{3}=\frac{16}{3}$</span>. Similarly the area you want to compute is given by
<span class="math-container">$$ 3+\left[3-\left(\int_{1}^{2}y^2\,dy-1\right)\right] = 7-\frac{8-1}{3}=\color{red}{\frac{14}{3}}.$$</span></p>
|
621,461 | <p>I'm having trouble understanding division when the divisor is greater than the dividend, for ex 1/4.</p>
<p>I think of division as "how many times can the divisor fit into the dividend evenly". </p>
<p>Intuitively, when I see 1/4 in the context of slices of pizza, I think of it as 1 "out of" 4, but I can't seem to grasp it in terms of "how many times does 4 fit into 1" if that makes any sense.</p>
<p>In other words my question could be why do we use division to represent "one out of 4"?.</p>
<p>If it helps you guys understand, this came about as I was trying to find the percentage representation of two populations, as in "there are 1253 A's and 747 B's, what is the proportion of each in % ?".</p>
<p>Conceptually I understand that I need to add up those two populations and then find the proportion they represent of that total. However, when I got to that second part, I couldn't reason through whether I needed to divide the total by a population, or a population by the total in order to find the desired result. </p>
<p>Obviously I eventually found the right way to do it, but it still doesn't make sense to me.</p>
<p>Sorry if it's very vague, this is really bothering me; I can't seem to reconcile those two ways of thinking about division.</p>
| user21820 | 21,820 | <p>I'm not sure of your level, but it might not be helpful to think of $\frac{a}{b}$ as "the number of times" $b$ fits into $a$ simply because that is not so intuitive when it comes to non-integer number of times. Instead, $\frac{a}{b}$ (where $b \ne 0$) is the amount such that $\frac{a}{b} \times b = a$. In other words, dividing by $b$ is the inverse (opposite) of multiplying by $b$. This also means that $\frac{a \times b}{b} = a$. For example, $\frac{2}{5}$ is the amount such that $5$ times of it is $2$, and $\frac{2 \times 5}{5}$ is the amount such that 5 times of it is $2 \times 5$. This definition remains valid for irrational amounts (and later complex numbers and more generally mathematical structures called fields). For example, a square with length $2r$ has greater area than a circle with radius $r$, and the ratio of the square's area to the circle's area is $\frac{4r^2}{\pi r^2} = \frac{4}{\pi} \approx 1.27324$. Intuitively, this means that if you divide up both shapes into tiny squares of equal size using a grid (and ignore the pieces that aren't square), the ratio of the number of pieces will get closer to $\frac{4}{\pi}$ as the size of the grid square decreases.</p>
|
1,984,076 | <p>How to prove that
$$
\sum_{k=1}^\infty\frac{k^k}{k!}x^k=\frac{1}{2}, ~\text{where}~~ x=\frac{1}{3}e^{-1/3}~?
$$
I found this sum in my notes, but I don't remember where I got it. Any hints or references would be nice.</p>
| Andreas | 317,854 | <p>Let's consider $\sum_{k=1}^\infty\frac{k^k}{k!}\big(\frac{1}{ae^{1/a}}\big)^k=\frac{1}{a-1}$ for $a > 1$ (as in the question). </p>
<p>Then $\frac{1}{a-1} = - 1+ \frac{1}{1-1/a} = \sum_{k=1}^\infty (\frac1a)^k$ by the geometric series. So one has to show </p>
<p>$\sum_{k=1}^\infty\Big[-1 + \frac{k^k}{k!}{e^{-k/a}}\Big](\frac{1}{a}\big)^k=0$</p>
<p>Writing some terms</p>
<p>$0 = \Big[-1 + {e^{-1/a}}\Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}{e^{-2/a}}\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}{e^{-3/a}}\Big](\frac{1}{a}\big)^3 + \cdots$</p>
<p>Expanding the exponentials and writing the terms up to $(1/a)^3$, say, requires to take the terms </p>
<p>$0 = \Big[-\frac{1}{a} + \frac12 (\frac{1}{a})^2 + \cdots \Big](\frac{1}{a}\big) + \Big[-1 + \frac{2^2}{2!}(1-\frac{2}{a} + \cdots)\Big](\frac{1}{a}\big)^2 + \Big[-1 + \frac{3^3}{3!}(1 + \cdots)\Big](\frac{1}{a}\big)^3 + \cdots$</p>
<p>Sorting for powers of $1/a$ gives</p>
<p>$0 = \Big[0 \Big](\frac{1}{a}\big) + \Big[-1 -1 + \frac{2^2}{2!}\Big](\frac{1}{a}\big)^2 + \Big[\frac12 - 2 \frac{2^2}{2!} -1 + \frac{3^3}{3!}\Big](\frac{1}{a}\big)^3 + \cdots$</p>
<p>and all the prefactors are zero, as required. The full solution then comes upon expanding all exponentials. Clearly this requires some more formal work.</p>
|
3,936,187 | <blockquote>
<p>Consider the differential equation <span class="math-container">$$(1+t)y''+2y=0$$</span>
with the variabel coefficient <span class="math-container">$(1+t)$</span>, with <span class="math-container">$t\in \mathbb{R}$</span>.</p>
<p>Set <span class="math-container">$y(t)=\sum_{n=0}^{\infty}a_nt^n$</span>. What are the first 4 terms in the associated power series?</p>
<p><a href="https://i.stack.imgur.com/yYwaI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yYwaI.png" alt="enter image description here" /></a></p>
</blockquote>
<h2>An Attempt</h2>
<p>We write
<span class="math-container">$$(1+t)y''=\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}$$</span>
<span class="math-container">$$2y=\sum_{n=0}^{\infty}2a_nt^{n} $$</span></p>
<p>Combining these three gives us</p>
<p><span class="math-container">$$\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}+ \sum_{n=0}^{\infty}2a_nt^{n}$$</span></p>
<p>We now want all the powers to be <span class="math-container">$t^n$</span>.
<span class="math-container">$$\sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=0}^{\infty}2a_nt^{n} $$</span></p>
<p>We finally want the index of summation to all be the same.
<span class="math-container">$$2a_0+2a_2+ \sum_{n=1}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=1}^{\infty}2a_nt^{n}$$</span></p>
<p>Which simplifies to
<span class="math-container">$$2a_0+2a_2+ \sum_{n=1}^{\infty}\big[a_{n+2}(n+2)(n+1)+a_{n+1}n(n+1)+2a_n \big]t^n$$</span></p>
<p>Clearly, I'm not getting anywhere close to any of the answers given. I just can't find what I'm doing wrong. I have also tried with Maple, but that didn't give me any real result either.</p>
<p>I hope someone can help me out.</p>
| mathcounterexamples.net | 187,663 | <p>Take <span class="math-container">$t=0$</span>. You get from the ODE <span class="math-container">$y^{\prime \prime}(0)+y(0)=0$</span>.</p>
<p>And answer b) is the only valid option <em>providing that at least one is valid</em>.</p>
|
670,292 | <p>Could someone assist with the following three surface integrals? </p>
<p><strong>Q1</strong> The portion of the cone $z=\sqrt{x^2+y^2}$ that lies inside the cylinder $x^2+y^2 =2x$. </p>
<p><strong>Q2</strong> The portion of the paraboloid $z=1-x^2-y^2$ that lies above the $xy$-plane.</p>
<p><strong>Q3</strong> The portion of the paraboloid $2z = x^2+y^2$ that is inside the cylinder $x^2+y^2=8$.</p>
<p>Any assistance will be greatly appreciated.</p>
| Ben Grossmann | 81,360 | <p><strong>Hint:</strong> Define
$$
f(x) = \frac{x}{1-x}
$$
Noting that $f(x)$ is continuous except at $x = 1$. Suppose that $b_n = \frac{a_n}{1+a_n} \to l \neq 1$, what can we say about the sequence $f(b_n)$?</p>
|
1,762,001 | <p>I recently watched a <a href="https://www.youtube.com/watch?v=SrU9YDoXE88" rel="noreferrer">video about different infinities</a>. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \ldots, \omega_\omega$, etc..</p>
<p>I can't find myself in all of this. Why there are so many infinities, and why even bother to classify infinity, when infinity is just... infinity? <strong>Why do we use all of these symbols? What does even each type of infinity mean?</strong></p>
| PMar | 335,420 | <p>One issue not yet addressed is why we have both $\aleph$s and $\omega$s. These both exist because Cantor introduced two separate concepts about infinite sets - their <em>sizes</em> ($\aleph$) and their <em>order-types</em> ($\omega$). Everyone else explained sizes, so I won't go over that.</p>
<p>For order-types, begin by considering the following set:</p>
<pre><code>{ 1/2, 2/3, 3/4, ..., N/N+1, ... }
</code></pre>
<p>This is of course an infinite set of size $\aleph_0$. But the elements can also be ordered by <, and so this is also a set of a particular infinite order-type, which Cantor chose to name $\omega$. Now consider a variant of this set:</p>
<pre><code>{ 1/2, 2/3, 3/4, ..., N/N+1, ..., 1 }
</code></pre>
<p>This is also of size $\aleph_0$ (I leave proof to you). But it is a different order-type, because its internal ordering is structurally different - i.e it is NOT possible to biject the two sets <em>preserving order</em>. This can be seen by noting that the element 1 in the second set has infinitely many predecessors, which no element in the first set has. Cantor names this set's order-type $\omega+1$. If I added 2 into the second set, the resutling order-type would then be $\omega+2$; and so on. Actually, that doesn't do it justice - there are an infinity of 'and so on's coming - compounded 'to infinity and beyond'.</p>
<p>So what Cantor discovered is a very sophisticated notation for describing highly complicated sets of real numbers (or as he thought of it, points on the number line), a notation which he realized could be abstracted to <em>a system of actually infinite numbers</em> having its own rules of arithmetic (inferred from the point sets resulting from catenating other point sets).</p>
|
1,762,001 | <p>I recently watched a <a href="https://www.youtube.com/watch?v=SrU9YDoXE88" rel="noreferrer">video about different infinities</a>. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \ldots, \omega_\omega$, etc..</p>
<p>I can't find myself in all of this. Why there are so many infinities, and why even bother to classify infinity, when infinity is just... infinity? <strong>Why do we use all of these symbols? What does even each type of infinity mean?</strong></p>
| Mikhail Katz | 72,694 | <p>The preoccupation expressed in the OP's question goes back (at least) to the 17th century, when Nieuwentijt objected to Leibnizian hierarchy of infinite numbers by claiming that there should be only one level of infinity, and reciprocally, only one level of infinitesimal, say $\epsilon$ (though Nieuwentijt didn't denote it that way), so that the square of $\epsilon$ would be zero: $\epsilon^2=0$. Leibniz and his school objected to this by saying essentially that this would involve a violation of the <em>law of continuity</em>. A modern formalisation of the Leibnizian law of continuity is the transfer principle of Robinson's framework. For a discussion of these issues in a historical context see <a href="http://u.cs.biu.ac.il/~katzmik/guillaume14.pdf" rel="nofollow">Guillaume's review</a>.</p>
|
3,568,230 | <p>My question is: why, in general we cannot write down an formula for the <span class="math-container">$n-$</span>th term, <span class="math-container">$S_{n}$</span>, of the sequence of partial sums?</p>
<p>I will explain better in the following but the question is basically that one above.</p>
<p>Suppose then you have an <em>infinite sequence</em> in your pocket, <span class="math-container">$\{a_{1},a_{2},a_{3},...\}$</span>, or,</p>
<p><span class="math-container">$$\{a_{1},a_{2},a_{3},...\} \equiv \{a_{n}\}_{n=0}^{\infty} \tag{1}$$</span></p>
<p><span class="math-container">$(1)$</span> then is a fundamental object because then you can "sum up" all the terms of this particular sequence, just like: <span class="math-container">$a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot$</span> to define another object. Well, doing that procedure you construct that object, called <em>infinite series of the infinite sequence <span class="math-container">$\{a_{n}\}_{n=0}^{\infty}$</span></em> </p>
<p><span class="math-container">$$a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot \equiv \sum^{\infty}_{n=0}a_{n} \tag{2}$$</span></p>
<p>The next procedure you might like to do is then question yourself if a infinite series have some value <span class="math-container">$s \in \mathbb{K}$</span> (<span class="math-container">$\mathbb{K}$</span> a field) indeed. The procedure to answer that question is then firstly construct another infinite sequence called <em>the sequence of partial sums of the series</em>:</p>
<p><span class="math-container">$$\{S_{0},S_{1},S_{2},S_{3},...,S_{k},...\} \equiv \{S_{n}\}_{n=0}^{\infty} \tag{3} $$</span></p>
<p>Which is:</p>
<p><span class="math-container">$$\begin{cases} S_{0} = \sum^{0}_{n=0}a_{n} = a_{0}\\S_{1} = \sum^{1}_{n=0}a_{n} = a_{0} + a_{1} \\ S_{1} = \sum^{2}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} \\ S_{3} = \sum^{3}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3} \\\vdots\\ S_{k} = \sum^{k}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3}+\cdot \cdot \cdot+a_{k}\\ \vdots \end{cases} $$</span></p>
<p>and then calculate the limit of this sequence <span class="math-container">$(3)$</span>, like:</p>
<p><span class="math-container">$$ \lim_{n\to \infty} \sum^{n}_{j=0}a_{j} \equiv \lim_{n\to \infty} S_{n} \tag{4} $$</span></p>
<p>Now, if the limit <span class="math-container">$(4)$</span> has a value <span class="math-container">$s = L$</span> then the can say that the <em>Sum of the Series</em> is that limit:</p>
<p><span class="math-container">$$ \sum^{\infty}_{n=0}a_{n} = s \tag{5}$$</span></p>
<p><span class="math-container">$$ * * * $$</span></p>
<p>Now, if we do not have a proper expression for <span class="math-container">$S_{n} = \sum^{k}_{n=0}a_{n}$</span>, then the whole "direct limit calculus" do not work and then we need other methods for search the value (more generally the convergence) of a series (e.g. integral test). The thing is, I do not see (understand) why we cannot in general write down a formula for <span class="math-container">$S_{n}$</span> and some times we can. For instance, I do not see why in one hand we can write down a formula for geometric series but on the other hand we cannot for harmonic series, for me the <span class="math-container">$S_{n}$</span> term, of the harmonic series, to plug up in the limit is given by:</p>
<p><span class="math-container">$$ S_{n} = \sum^{n}_{k=0}\frac{1}{n} = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \equiv \Big( 1+ \frac{1}{2} + \frac{1}{3} + ... \Big) + \frac{1}{n} = $$</span></p>
<p><span class="math-container">$$= C + \frac{1}{n} $$</span></p>
<p><span class="math-container">$C$</span> a constant since it's a finite sum. Then,</p>
<p><span class="math-container">$$\lim_{n\to \infty} C + \frac{1}{n} = C$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\sum^{\infty}_{n=0}\frac{1}{n} = C$$</span></p>
<p>I know that what I wrote above isn't write, but I simply do not understand why. There's a subtle thing that I do not understand. Anyway, the question is posted above.</p>
<p>Thank You. </p>
| G Cab | 317,234 | <p>Actually you can express the partial sums of the Harmonic sequence
<span class="math-container">$$
\sum\limits_{k = 1}^n {{1 \over k}} = \psi (n + 1) - \psi (1) = \gamma + \psi (n + 1)
$$</span>
through the <a href="https://en.wikipedia.org/wiki/Digamma_function" rel="nofollow noreferrer">Digamma Function</a>.</p>
<p>For the general question <em>Josè</em> answered already.</p>
|
180,169 | <p>Can anyone give me suggestions for new books about Besicovitch's almost periodic functions? Thanks a lot. </p>
| Robert Bryant | 13,972 | <p>The questions you are asking have been carefully studied in the literature, but, usually, with carefully chosen global hypotheses so that reasonable results can be achieved. You should look at the works of William Goldman, Mischa Gromov, and Scott Adams, just to name a few, as well as more recent authors studying parabolic geometries (e.g. Andreas Cap, Jan Slovak, Mike Eastwood, etc.) for more information about this. Here, I'll just make a few remarks.</p>
<p>The cases $n\ge 3$ are very different from the $n=2$ case and so have to be treated differently. </p>
<p>In general, when $n\ge 3$, and one is looking at pseudo-Riemannian metrics of type $(p,q)$ (where $q = n{-}p$), there is the 'standard conformally flat model' $Q_{p,q}=G_{p,q}/P_{p,q}$, where $G_{p,q} = \mathrm{O}(p{+}1,q{+}1)$, and $P_{p,q}$ is a maximal parabolic in $G_{p,q}$. </p>
<p>When $\bigl(M^n,[g]\bigr)$ is conformally flat, Cartan showed that one can canonically construct a $P_{p,q}$-bundle $B_{p,q}\to M$ with a flat Cartan connection $\phi$. If $M$ is connected and $\tilde M\to M$ is the universal cover, then there is a mapping $g:\tilde B_{p,q}\to G_{p,q}$ unique up to left translation, that pulls back the canonical left-invariant form on $G_{p,q}$ to be $\phi$. This mapping is equivariant with respect to the natural right actions of $P_{p,q}$ on $\tilde B_{p,q}$ and $G_{p,q}$. This induces a <em>monodromy homomorphism</em> $\mu:\pi_1(M)\to G_{p,q}$ that is well-defined up to conjugation in $G_{p,q}$. Then the existence of a flat metric in the conformal class is equivalent to having the image of this monodromy homomorphism be conjugate to a subgroup of $H_{p,q}\subset G_{p,q}$, the subgroup that fixes a metric on an open subset of $Q_{p,q}$ that contains the induced image of $\tilde M$ in $Q_{p,q}$. (This almost never happens without special hypotheses on $\pi_1(M)$ or curvature restrictions on some metric on the conformal class, etc.) In any case, the methods involved are ODE methods and information about the geometry of discrete subgroups of the conformal group. </p>
<p>When $n=2$, you are asking about a global PDE problem. Of course, there are answers in simple cases, such as when $M$ is compact, or complete with positive curvature, or some such. For example, if you assume that the metric is positive definite and that the conformal type is parabolic, then, yes, there is such a flat metric in the class, but this is a very restrictive assumption. Generally, conformal structures (even in the definite case) can be very complicated (even when the topology is 'finite'), and there will be no simple universal criterion for the existence of a flat metric in the given conformal class. </p>
|
234,348 | <p>There is a nice <a href="https://www.wolfram.com/mathematica/new-in-10/basic-and-formula-regions/compute-region-distance.html.en" rel="nofollow noreferrer">example</a> of how to generate an isometric visualisation of surfaces which are at a constant distance from a given region.</p>
<p>All these examples are just working on one single region. If one tries to apply this functionality to RegionUnions then the evaluation does not yield any results. E.g. a simple union of two cuboids with subsequent plotting of the surface which is 0.25 away from their surfaces yields no result:</p>
<pre><code>ContourPlot3D[Evaluate@RegionDistance[RegionUnion[
Cuboid[{-5, -5, 0}, {5, 5, 1}],
Cuboid[{-10, -10, -10}, {10, 10, 0}]], {x, y, z}],
{x, -7, 7}, {y, -7, 7}, {z, 0, 2},
Mesh -> None, Contours -> {0.25},
ContourStyle -> ColorData[94, "ColorList"], Lighting -> "Neutral",
BoxRatios -> Automatic]
</code></pre>
<p>whereas the simpler command with just one cuboid generates the expected result:</p>
<pre><code>ContourPlot3D[Evaluate@RegionDistance[
Cuboid[{-5, -5, 0}, {5, 5, 1}], {x, y, z}],
{x, -7, 7}, {y, -7, 7}, {z, 0, 2},
Mesh -> None, Contours -> {0.25},
ContourStyle -> ColorData[94, "ColorList"], Lighting -> "Neutral",
BoxRatios -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/BFFGB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BFFGB.png" alt="enter image description here" /></a></p>
<p>Potentially the functionality is just limited to single entities and not unions? Any ideas how to achieve this for Region Unions or Intersections?</p>
| cvgmt | 72,111 | <p><code>Min</code> work or not?</p>
<pre><code>reg1 = Cuboid[{-5, -5, -12}, {5, 5, 12}];
reg2 = Cuboid[{-10, -10, -10}, {10, 10, 10}];
ContourPlot3D[
Min[RegionDistance[reg1, {x, y, z}],
RegionDistance[reg2, {x, y, z}]] // Evaluate, {x, -15,
15}, {y, -15, 15}, {z, -15, 15}, Mesh -> None, Contours -> {0.25},
ContourStyle -> ColorData[94, "ColorList"], Lighting -> "Neutral",
BoxRatios -> Automatic, PlotRange -> All]
</code></pre>
|
234,348 | <p>There is a nice <a href="https://www.wolfram.com/mathematica/new-in-10/basic-and-formula-regions/compute-region-distance.html.en" rel="nofollow noreferrer">example</a> of how to generate an isometric visualisation of surfaces which are at a constant distance from a given region.</p>
<p>All these examples are just working on one single region. If one tries to apply this functionality to RegionUnions then the evaluation does not yield any results. E.g. a simple union of two cuboids with subsequent plotting of the surface which is 0.25 away from their surfaces yields no result:</p>
<pre><code>ContourPlot3D[Evaluate@RegionDistance[RegionUnion[
Cuboid[{-5, -5, 0}, {5, 5, 1}],
Cuboid[{-10, -10, -10}, {10, 10, 0}]], {x, y, z}],
{x, -7, 7}, {y, -7, 7}, {z, 0, 2},
Mesh -> None, Contours -> {0.25},
ContourStyle -> ColorData[94, "ColorList"], Lighting -> "Neutral",
BoxRatios -> Automatic]
</code></pre>
<p>whereas the simpler command with just one cuboid generates the expected result:</p>
<pre><code>ContourPlot3D[Evaluate@RegionDistance[
Cuboid[{-5, -5, 0}, {5, 5, 1}], {x, y, z}],
{x, -7, 7}, {y, -7, 7}, {z, 0, 2},
Mesh -> None, Contours -> {0.25},
ContourStyle -> ColorData[94, "ColorList"], Lighting -> "Neutral",
BoxRatios -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/BFFGB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BFFGB.png" alt="enter image description here" /></a></p>
<p>Potentially the functionality is just limited to single entities and not unions? Any ideas how to achieve this for Region Unions or Intersections?</p>
| Carl Woll | 45,431 | <p>If you try evaluating your <a href="http://reference.wolfram.com/language/ref/RegionDistance" rel="nofollow noreferrer"><code>RegionDistance</code></a> object for a point you will see that it doesn't work:</p>
<pre><code>RegionDistance[
RegionUnion[Cuboid[{-5,-5,0},{5,5,1}],Cuboid[{-10,-10,-10},{10,10,0}]],
{1,-1,2}
] //InputForm
</code></pre>
<blockquote>
<p>RegionDistance[Polyhedron[{{-5., -5., 0.}, {5., -5., 0.}, {5., -5., 1.}, {-5., -5., 1.},
{5., 5., 0.}, {-5., 5., 0.}, {-5., 5., 1.}, {5., 5., 1.}, {5., -5., -10.}, {-5., -5.,
-10.}, {-5., 5., -10.}, {5., 5., -10.}, {10., -5., -10.}, {10., 5., -10.}, {10., 5.,
0.}, {10., -5., 0.}, {-10., 5., -10.}, {-10., -5., -10.}, {-10., -5., 0.}, {-10., 5.,
0.}, {10., 10., -10.}, {-10., 10., -10.}, {-10., 10., 0.}, {10., 10., 0.}, {-10.,
-10., -10.}, {10., -10., -10.}, {10., -10., 0.}, {-10., -10., 0.}},
{{1, 2, 3, 4}, {5, 6, 7, 8}, {6, 1, 4, 7}, {2, 5, 8, 3}, {4, 3, 8, 7}, {9, 10, 11,
12}, {13, 14, 15, 16}, {14, 13, 9, 12}, {16, 15, 5, 2}, {17, 18, 19, 20}, {10, 18,
17, 11}, {6, 20, 19, 1}, {21, 22, 23, 24}, {22, 17, 20, 23}, {14, 21, 24, 15}, {25,
26, 27, 28}, {18, 25, 28, 19}, {26, 13, 16, 27}, {22, 21, 14, 12, 11, 17}, {15, 24,
23, 20, 6, 5}, {13, 26, 25, 18, 10, 9}, {28, 27, 16, 2, 1, 19}}], {1, -1, 2}]</p>
</blockquote>
<p>This is why you don't get a plot. One possible workaround is to discretize your cuboids. Another optimization is to create the <a href="http://reference.wolfram.com/language/ref/RegionDistanceFunction" rel="nofollow noreferrer"><code>RegionDistanceFunction</code></a> and then to use it. So:</p>
<pre><code>rdf = RegionDistance @ RegionUnion[
DiscretizeRegion @ Cuboid[{-5,-5,0},{5,5,1}],
DiscretizeRegion @ Cuboid[{-10,-10,-10},{10,10,0}]
];
ContourPlot3D[
rdf[{x,y,z}],
{x,-7,7}, {y,-7,7}, {z,0,2},
Mesh->None,
Contours->{0.25},
ContourStyle->ColorData[94,"ColorList"],
Lighting->"Neutral",
BoxRatios->Automatic
]
</code></pre>
<p><a href="https://i.stack.imgur.com/4A0iI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4A0iI.png" alt="enter image description here" /></a></p>
|
236,927 | <p>I thought this would be a hard problem but I found a link that seems to ask the answer to this question as a homework problem? Can somone help me out here, are there an infinite number of prime powers that differ by 1? or are there a finite number of them? If so which are they?</p>
| xtimz | 164,004 | <p>Finally, I know why this problem is elementary.</p>
<p>$p^r-q^s=1$ where $p$, $q$ are primes, and $r$, $s$ are integers greater than $1$.</p>
<p>If $p$ and $q$ are both odd primes, then $p^r$ and $q^s$ are both odd, then $p^r-q^s$ can only be even. So, either $p$ or $q$ is a even prime, that is $2$.</p>
<p>It seems the problem becomes elementary when $p=2$ or $q=2$. In the following, we will prove the claim when $p=2$ or $q=2$. It is not important whether the other is prime or not.</p>
<p>Case 1. $p=2$, that is $q^s=2^r-1$.</p>
<p>$2^r-1$ is a $4k-1$ type number, so $q^s$ has to be $4k-1$ type.</p>
<p>For an odd number $q$, there are two choices: $4k+1$ type or $4k-1$ type.</p>
<p>For $4k+1$ type numbers, any power is $4k+1$ type.</p>
<p>For $4k-1$ type numbers, odd power is $4k-1$ type, even power is $4k+1$ type.</p>
<p>So, $q$ has to be $4k-1$ type, and $s$ has to be odd.</p>
<p>$2^r=q^s+1=(q+1)(q^{s-1}-q^{s-2}+\cdots-q+1)$</p>
<p>Notice that $q^{s-1}-q^{s-2}+\cdots-q+1$ has odd number of terms and each term is odd, so it is an odd number, which contradicts that $2^r$ only has factor $2$.</p>
<p>Case 2. $q=2$, that is $p^r=2^s+1$.</p>
<p>$2^s+1$ is a $4k+1$ type number, so:</p>
<p>If $p$ is $4k-1$ type, $r$ has to be even.</p>
<p>If $p$ is $4k+1$ type, $r$ can be even or odd.</p>
<p>Case 2.1. $p$ is $4k-1$ type, and $r=2m$ is even.</p>
<p>$2^s=p^{2m}-1=(p^m+1)(p^m-1)$</p>
<p>$p^m-1$ is $4k-2$ type. If it only has factor $2$, it has to be $p^m-1=2$, and then we obtain the only solution in this case: $3^2-2^3=1$.</p>
<p>Case 2.2. $p$ is $4k+1$ type, and $r=2m$ is even.</p>
<p>$2^s=p^{2m}-1=(p^m+1)(p^m-1)$</p>
<p>$p^m+1$ is $4k+2$ type. If it only has factor $2$, it has to be $p^m+1=2$, which has no solution.</p>
<p>Case 2.3. $p$ is $4k+1$ type, and $r$ is odd.</p>
<p>$2^s=p^r-1=(p-1)(p^{r-1}+p^{r-2}+\cdots+p+1)$</p>
<p>Notice that $p^{r-1}+p^{r-2}+\cdots+p+1$ has odd number of terms and each term is odd, so it is an odd number, which contradicts that $2^s$ only has factor $2$.</p>
|
945,736 | <p>:)</p>
<p>I have this matrix:</p>
<p>B = \begin{bmatrix}
0.626 & 2.56 & 2.15 & \\
0.835 & 6.66 & 5.16 & \\
0 & 0 & -1.65 &
\end{bmatrix}</p>
<p>I was wondering how to find a givens matrix such that I could apply it from the right side of the matrix and eliminate B[2][1] (0.835). </p>
<p>B*g = \begin{bmatrix}
* & * & * & \\
0 & * & * & \\
0 & 0 & * &
\end{bmatrix}
Best regards,
rox</p>
| John Alexiou | 3,301 | <p>See <a href="https://stackoverflow.com/a/4361442/380384">https://stackoverflow.com/a/4361442/380384</a></p>
<blockquote>
<pre><code> | a b tx |
A = | c d ty |
| 0 0 1 |
</code></pre>
<p>which transforms the coordinates [x,y,1] into:</p>
<pre><code>[x',y',1] = A * [x,y,1]
</code></pre>
<p>Thus set the traslation into <code>[dx,dy]=[tx,ty]</code></p>
<p>The scale is <code>sx=sqrt(a^2+b^2)</code> and <code>sy=sqrt(c^2+d^2)</code></p>
<p>The rotation angle is <code>t=atan(c/d)</code> or <code>t=atan(-b/a)</code> as also they should be the same.</p>
</blockquote>
<p>The inverse matrix is $$ A^{-1} = \frac{1}{a d-b c} \begin{bmatrix}
d & -b & b t_y-d t_x \\
-c & a & c t_x-a t_y \\
0 & 0 & a d - b c \end{bmatrix} $$</p>
<p>Or you can try</p>
<p>$$ \begin{bmatrix} a & b & t_x \\ c & d & t_y \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ -\frac{c}{d} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =
\begin{bmatrix} a-\frac{b c}{d} & b & t_x \\ 0 & d & t_y \\ 0 & 0 & 1 \end{bmatrix} $$</p>
|
1,278,848 | <p>Based on <a href="https://math.stackexchange.com/questions/1267021/let-m-subseteq-mathbbrk-manifold-topology-vs-trace-topology/1267760?noredirect=1#comment2573732_1267760">this</a> question I'd like to know: Are there compact (sub)manifolds without boundary in $\mathbb{R}^n$ ? Because, as that question shows, the topology of the manifolds has to be the trace topology; thus compact subspace (in particular, <em>compact</em> manifolds) are characterized by the Heine-Borel theorem: They are precisely those sets in $\mathbb{R}^n$ that are closed and bounded.</p>
<p>But, as far as I know (and I'm just starting to read about manifolds and haven't got a good grasp on the formal definitions yet), manifolds without boundary aren't closed, so they can't be compact ?</p>
| Autolatry | 25,097 | <p>Yes; there are some. If $M$ is a compact, oriented $n$-manifold without boundary, then there must be some $n$-forms which do not arise from taking the exterior derivatives of $n-1$-forms. </p>
<p>If $M$ is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on $M$ which are not the divergence of any vector field on $M$.</p>
|
2,948,327 | <p>Being an undergraduate student I find difficult to understand the perfect differences between normal and partial differential equations. Elaborate the answer </p>
| Community | -1 | <p>An <em>ordinary</em> differential equation involves a derivative over a single variable, usually in an univariate context, whereas a <em>partial</em> differential equation involves several (partial) derivatives over several variables, in a multivariate context.</p>
<p>E.g.
<span class="math-container">$$\frac{dz(x)}{dx}=z(x)$$</span></p>
<p>vs.</p>
<p><span class="math-container">$$\frac{\partial z(x,y)}{\partial x}+\frac{\partial z(x,y)}{\partial y}=z(x,y).$$</span></p>
<p>PDEs are notably more difficult to solve than ODEs.</p>
<hr>
<p><strong>Remark:</strong></p>
<p>Even though several variables appear,</p>
<p><span class="math-container">$$\frac{\partial z(x,y)}{\partial x}=z(x,y)$$</span></p>
<p>can very well be considered as an ODE, where <span class="math-container">$y$</span> is just an independent parameter.</p>
|
869,341 | <p>What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$</p>
| Joel | 85,072 | <p>A generating function approach would make this straightforward:</p>
<p>$$G(x) = \sum_{n=0}^\infty a_n x^n = 1+2x+\sum_{n=2}^\infty a_n x^n$$
$$=1+2x+\sum_{n=2}^\infty (a_{n-1} + 2a_{n-2} + 2^n)x^n$$
$$=1+2x+\sum_{n=1}^\infty a_n x^{n+1} + 2 \sum_{n=0}^\infty a_n x^{n+2} + \sum_{n=2}^\infty 2^n x^n$$
$$=1+2x+x(G(x)-1)+2x^2G(x)+\frac{2^2x^2}{1-2x}$$</p>
<p>Thus $$G(x)(1-x-2x^2) = 1+x+\frac{2^2x^2}{1-2x}$$ and $$G(x) = \frac{1+x}{1-x-2x^2} + \frac{2^2x^2}{(1-2x)(1-x-2x^2)} = \frac{1+x}{(1+x)(1-2x)} + \frac{2^2x^2}{(1-2x)(1+x)(1-2x)}$$
$$= \frac{1}{(1-2x)} + \frac{2^2x^2}{(1-2x)^2(1+x)}$$</p>
<p>Now you can use partial fractions, and then expand in terms of geometric series to find the taylor coefficients of $G(x)$ which are the terms $a_n$.</p>
|
869,341 | <p>What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$</p>
| Deathkamp Drone | 56,720 | <p>The best course of action for simple inhomogeneous recurrences is to make use of a smart "change of variables" (read: substitute with another recurrence relation) to turn it into a homogeneous recurrence. A nice observation here is to notice that <strong>$2^n$ is itself a recurrence relation</strong>, namely: </p>
<p>$$y_n=2y_{n-1},\,\,(y_0=1)$$
Let's use this to our advantage:</p>
<p>$$\begin{align}
a_n=a_{n-1}+2a_{n-2}+2^n &\Leftrightarrow 2^n=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow 2\cdot2^{n-1}=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow 2\cdot (a_{n-1}-a_{n-2}-2a_{n-3})=a_n-a_{n-1}-2a_{n-2} \\
&\Leftrightarrow a_n-3a_{n-1}+4a_{n-3}=0
\end{align}$$</p>
<p>Bam, homogeneous recurrence. Can apply the characteristic equation and finish it by yourself?</p>
<p><strong>Edit:</strong> The characteristic equation for the last recurrence is $x^3-3x+4=0$, whose roots are $2,2$ and $-1$. It is well-known then, that given these roots, the solution is of the form:
$$a_n=(An+B)2^n+C(-1)^n$$</p>
<p>For some constants $A,B,C$. To find out what the constants are, we plug in values of $n$ for which the value of $a_n$ is known. The values of $a_0,a_1$ were given, but we need three to solve a linear system with three variables. We can easily calculate $a_2$: $$a_2=a_1+2a_0+2^2=8$$</p>
<p>Now we solve the system:</p>
<p>$$\begin{cases}
(A\cdot0+B)2^0+C(-1)^0=1 \\
(A\cdot1+B)2^1+C(-1)^1=2 \Rightarrow \\
(A\cdot2+B)2^2+C(-1)^2=8
\end{cases}$$</p>
<p>$$\Rightarrow \begin{cases}
B+C=1 \\
2(A+B)-C=2 \\
4(2A+B)+C=8
\end{cases}$$</p>
<p>Solving this will yield $A=\frac{2}{3}$,$B=\frac{5}{9}$ and $C=\frac{4}{9}$. Therefore, the closed form is:
$$a_n=\frac{(6n+5)2^n+4(-1)^n}{9}$$</p>
|
869,341 | <p>What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$</p>
| Mathsource | 12,624 | <p>This is nonhomogeneneous difference equations. First solve the homogeneous equation
$$
a_n - a_{n-1} - 2a_{n-2} = 0 \quad (1)
$$
Let $a_n = r^n$, so that $a_{n-1} = r^{n-1}$ and $r_{n-2} = r^{n-2} = 0$. Replacing in (1), we have
$$
r^n - r^{n-1} - 2r^{n-2} = 0 \quad r^2 - r - 2 = 0 \quad \Rightarrow \ r_1 = -1, \ r_2 = 2
$$
$$
a_{nh} = C_1(-1)^n + C_22^n
$$
Particular solution: Let $a_{np} = a_n = An2^n$. The presence of n is due to the fact that $2$ is a root of the equation (1). So, $a_{n-1} = A(n-1)2^{n-1}$ and $a_{n-2} = A(n-2)2^{n-2}$. Substituting in the given equation, we have
$$
An2^n + (An - A)2^{n-1} - 2A(n-2)2^{n-2} = 2^n \quad \Rightarrow \quad A = \frac{2}{3}
$$
Thus, $a_n = a_{nh} + a_{np} = C_1(-1)^n + C_22^n + \frac{2n2^n}{3}$. But, $a_0 = a(0) = 1$ and $a_1 = a(1) = 2$. We have the system
$$
\begin{cases}
1 = a_0 = C_1 + C_2\\
2 = a_1 = -C_1 + C_2 + \frac{8}{3}
\end{cases}
\quad \Rightarrow \quad C_1 = 8/9, \quad C_2 = 1/9
$$
Thus,
$$
a_n = \frac{8(-1)^n}{9} + \frac{1}{9}2^n + \frac{2n2^n}{3}
$$</p>
|
1,846,592 | <p>I know that a discrete topological space is where all singletons are open.</p>
<p>For example, $\mathbb{N}$ with the subspace topology inherited from $(\mathbb{R}, \mathfrak{T}_{usual})$. This is the case because we can find $\{n\} = (a,b) \cap \mathbb{N}$ which is open. Hence all singletons are open.</p>
<p>But are all sets are clopen? Closed?</p>
<p>My thoughts: Suppose we take a singleton $\{x\}$ in a discrete space $X$, we know singleton is open, hence $\{x\}^c$ is closed. But it is the arbitrary union of singletons, so it is also open, so all sets are clopen. </p>
| drhab | 75,923 | <p>Let $F$ be an arbitrary subset of $X$ where $X$ is equipped with discrete topology. </p>
<p>As you said: in a discrete topological space all singletons are open.</p>
<p>As you said: arbitrary unions of singletons are open so $F^c=\bigcup_{x\in F^c}\{x\}$ is an open set. </p>
<p>(You don't even need this subroute: in a discrete space all sets are open by definition)</p>
<p>Then its complement $F$ is a closed set.</p>
|
1,322,076 | <p>Hey can anybody help me with the following proof? I am trying to solve the following limit using epsilon delta and I have found the limit to be 1/3 using the squeeze theorem and have got to this thus far but am a bit confused where I go now as I have both a 3x and a sinx when trying to find an epsilon??
Thanks in advance!!<img src="https://i.stack.imgur.com/eUP2B.jpg" alt="enter image description here"></p>
| RowanS | 246,774 | <p>You can easily show that $$\frac{\sin(x)+1}{3x+1}<\frac{2}{3x}$$
and then it is easy to show that $$\frac{2}{3x} < \epsilon $$</p>
|
1,516,450 | <p>When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$</p>
<p>When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$</p>
<p>How does wolfram arrive at that inequality:</p>
<p>$a< -500\left(\sqrt{2} - 2\right)$</p>
<p>Here is the link: <a href="http://www.wolframalpha.com/input/?i=a%2Bb%2Bc%3D1000%2Ca%5E2%2Bb%5E2%3Dc%5E2%2C0%3Ca%3Cb%3Cc" rel="nofollow">Wolfram</a></p>
| lulu | 252,071 | <p>Note: seeing the edited form of the question, the following isn't really on point. It addresses an efficient algorithm for finding the triple, it does not speak to whatever Wolfram Alpha might be doing.</p>
<p>Playing with the code, the following emerges: Suppose $$a^2+b^2=c^2\;\;\;\&\;\;\;a+b+c=S$$
We remark that $S^2-2Sb=2a^2+2ac=2a(a+c)=2a(S-b)$ Whence: $$a=\frac {S^2-2Sb}{2(S-b)}$$</p>
<p>Inelegant, perhaps, but it makes the computer search very efficient.</p>
|
3,831,198 | <p>Suppose that <span class="math-container">$100$</span>kg of a radioactive substance decays to <span class="math-container">$80$</span>kg in <span class="math-container">$20$</span> years.</p>
<p>a) Find the half-life of the substance (round to the nearest year).</p>
<p>b)Write down a function <span class="math-container">$y(t)$</span> (<span class="math-container">$t$</span> in years) modeling the amount (in kg) of the radioactive substance at time <span class="math-container">$t$</span>.</p>
| Math Lover | 801,574 | <p>If <span class="math-container">$A_t$</span> is the amount left at time <span class="math-container">$t$</span>, <span class="math-container">$A_0$</span> is the initial amount and <span class="math-container">$k$</span> is the rate constant for the first order reaction, we know for the radioactive decay (which is first order reaction) -</p>
<p><span class="math-container">$\displaystyle A_t = A_0e^{-kt} \,$</span> ...(i)</p>
<p>or <span class="math-container">$ \, \displaystyle \ln (\frac{A_t}{A_0}) = -kt$</span></p>
<p>As <span class="math-container">$A_0 = 100, A_t = 80, t = 20, -> \, \displaystyle \ln (\frac{80}{100}) = -20k$</span></p>
<p><span class="math-container">$\displaystyle k = -\frac{\ln 0.8}{20}$</span> ...(ii)</p>
<p>At half life, <span class="math-container">$\displaystyle \ln (\frac{50}{100}) = -kt_{1/2}$</span> or <span class="math-container">$\displaystyle t_{1/2} = - \frac {\ln (0.5)}{k} = 20 \times \frac {ln (0.5)} {ln(0.8)} \approx 62.1$</span> years.</p>
<p>Now, find <span class="math-container">$k$</span> using equation <span class="math-container">$(ii)$</span> and plug <span class="math-container">$k$</span> and <span class="math-container">$A_0 = 100$</span> into equation <span class="math-container">$(i)$</span> and that is your answer to <span class="math-container">$(b)$</span>.</p>
|
3,015,149 | <p>I’m using the formula that the number of conjugacy class is given to be <span class="math-container">$\frac{1}{|G|}\sum|C_{G}(g)|$</span>, where <span class="math-container">$C_{G}(g)=\{h \in G ; gh=hg\}$</span>, which is a special result by Burnside’s theorem.</p>
<p>I found that the number of conjugacy class in <span class="math-container">$D_8$</span> is 5, so to double check, I listed down all <span class="math-container">$C_{G}(g)=\{h \in G ; gh=hg\}$</span>.</p>
<p>Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.</p>
<p>Well, <span class="math-container">$|C_G(1)|=|D_8|=16$</span>,
<span class="math-container">$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7\}|=8$</span>, </p>
<p><span class="math-container">$|C_G(r^4)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m\}|=9$</span> since <span class="math-container">$r^4m=mr^4$</span>
Similarly, we can count for the reflections;
<span class="math-container">$|C_G(m)|=|C_G(r^4m)|=|\{1,m\}|=2$</span>, while the rest of the elements with any reflections only has one element, i.e <span class="math-container">$|C_G(r^nm)|=|\{1\}|=1,n \neq 0,4$</span></p>
<p>So the problem comes that my summation is 83, while my <span class="math-container">$|G|=16$</span>. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...</p>
| Doug M | 317,162 | <p><span class="math-container">$G = \{r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}\}$</span></p>
<p><span class="math-container">$e$</span> commutes with all elements and is in a conjugacy class all by itself.</p>
<p><span class="math-container">$\{e\}$</span></p>
<p>Rotations fall into conjugacy classes that include their inverses.</p>
<p><span class="math-container">$(r^n)(r^m)(r^{-n}) = r^m$</span></p>
<p><span class="math-container">$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$</span></p>
<p><span class="math-container">$\{r, r^7\},\{r^2, r^6\},\{r^3, r^5\}, \{r^4\}$</span></p>
<p>Reflections:</p>
<p><span class="math-container">$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$</span></p>
<p>and <span class="math-container">$(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$</span></p>
<p>creating congugacy classes of <span class="math-container">$\{ar, ar^3, ar^5, ar^7\}$</span> and <span class="math-container">$\{a, ar^2, ar^4, ar^6\}$</span></p>
<p>That gives 7 conjugacy classes.</p>
<p>Counting the centralizers.</p>
<p>The identity commutes with everything.</p>
<p>Rotations commute with rotations.</p>
<p><span class="math-container">$r^4$</span> commutes with every reflection, not just one reflection (as suggested above). Which means that <span class="math-container">$r^4$</span> commutes with everything.</p>
<p>Every reflection commutes with the identity, <span class="math-container">$r^4$</span>, itself, and one other reflection.</p>
<p><span class="math-container">$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$</span></p>
<p><span class="math-container">$|C_G(e)| = 16\\
|C_G(r)| = 8 \text { times 6}\\
|C_G(r^4)| = 16\\
|C_G(a)| = 4 \text { times 8}$</span></p>
<p><span class="math-container">$\frac {16\times 2 + 8\times 6 + 4\times 8}{16} = 7$</span></p>
|
120,992 | <p>An algorithm book <a href="http://rads.stackoverflow.com/amzn/click/1849967202" rel="nofollow">Algorithm Design Manual</a> has given an description:</p>
<blockquote>
<p>Consider a graph that represents the street map of Manhattan in New York City. Every junction of two streets will be a vertex of the graph. Neighboring junctions are connected by edges. <strong>How big is this graph? Manhattan is basically a grid of 15 avenues each crossing roughly 200 streets. This gives us about 3,000 vertices and 6,000 edges</strong>, since each vertex neighbors four other vertices and each edge is shared between two vertices.</p>
</blockquote>
<p>If it says "The graph is a grid of 15 avenues each crossing roughly 200 streets", how can I calculate the number of vertices and edges? Although the description above has given the answers, but I just can't understand.</p>
<p>Can anyone explain the calculation more easily?</p>
<p>Thanks</p>
| Brian M. Scott | 12,042 | <p>Each street crossing is a vertex of the graph. An avenue crosses about $200$ streets, and each of these crossings is a vertex, so each avenue contains about $200$ vertices. There are $15$ avenues, each of which contains about $200$ vertices, for a total of $15\cdot 200=3000$ vertices.</p>
<p>To make the description easier, imagine that the avenues all run north-south and the other streets east-west. Then each intersection of an avenue and a street has another intersection due north along the avenue. It also has one due south along the avenue, one due east along the cross street, and one due west along the cross street. That’s a total of four neighboring vertices. There must be an edge from the given vertex to each of those four, so we count $4\cdot 3000=12,000$ edges. But that’s counting each edge twice, once at each end, so there are really only half that many edges, or $6000$.</p>
<p>Now you might object that the vertex (i.e., intersection) in the northwest corner, say, has only two neighboring vertices, one to the east and one to the south, and similarly for the other three corners. You might also worry about the non-corner vertices along the edges, since they seem to have only three neighboring vertices each. But remember, the original figure of $200$ cross streets was only an approximation in the first place, so we might as well ignore these relatively minor edge effects: they probably don’t affect the result much more than the approximation in the $200$ figure already does.</p>
|
2,724,686 | <blockquote>
<p>Set $B = \{1,2,3,4,5\}$, $S$ - equivalence relation. It is given that for all $x,y \in B$ if $(x,y)\in S$ and if $x+y$ is an even number then $x = y$. In such case is it true that:</p>
<ol>
<li>the number of elements in each equivalence class of $S$ is at most $2$</li>
<li>any relation $S$ would have an equivalence class made up of just one even number.</li>
</ol>
</blockquote>
<p>As far as I understand $S$ could be only of such form:
$$
S = \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5 \end{pmatrix}
$$
because all pairs of $(x,y), x \neq y$ which are both odd numbers can't be in $S$ as well as all pairs $(x,y), x \neq y$ which are both even numbers for example:
$$
\begin{pmatrix}1\\3 \end{pmatrix}, \begin{pmatrix}2\\4 \end{pmatrix}
$$
because their sum will be even but $x \neq y$.</p>
<p>In addition $(x,y), x\neq y$ where one of them is odd and one is even also can't be in $S$ because then the relation will not be transitive and hence will not be an equivalence relation. </p>
<p>In such case I think the statement 1 is false because all equivalence classes are exactly of size $1$ and statement 2 is true because we have for example the equivalence class $\{2\}$ which is one even number.</p>
<p>I'm not sure about my logic because the question is quite tricky.</p>
| Mark Bennet | 2,906 | <p>The answer to the first part is that if an equivalence class has three elements in it then two are odd or two are even, and the pair of common parity gives a counterexample to the condition.</p>
<p>For the second part, any equivalence class of size $2$ must consist of an odd number and an even number. Can you create a set of equivalence classes without an even singleton set?</p>
<p>I've left you some work to do, because only by working this through for yourself will you really get to understand. But you should appreciate that specifying an equivalence relation as a set of pairs is itself equivalent to expressing it by specifying the associated equivalence classes.</p>
|
4,562,451 | <p>I had this maths question:</p>
<blockquote>
<p>Given that <span class="math-container">$$8\sqrt{p} = q\sqrt{80}$$</span> where <span class="math-container">$p$</span> is prime, find the value of <span class="math-container">$p$</span> and the value of <span class="math-container">$q$</span></p>
</blockquote>
<p>I did this by simplifying the RHS to <span class="math-container">$4q\sqrt{5}$</span> and comparing clearly gives <span class="math-container">$p=5$</span> and <span class="math-container">$q=2$</span></p>
<p>However, I also thought why not do this by getting unitary surds on either side, eg <span class="math-container">$$8\sqrt{p} = q\sqrt{80} \Rightarrow \sqrt{64p} = \sqrt{80q^2}$$</span> This tells me that <span class="math-container">$64p=80q^2$</span> or equivalently <span class="math-container">$4p = 5q^2$</span>.</p>
<p>How would I be able to get <span class="math-container">$p$</span> and <span class="math-container">$q$</span> from this method? How do I know that the solution is unique?</p>
<p>If so, is it fortuitous that we get a unique solution with these particular numbers or will it always be unique - I think I just need to see a proof to convince myself!</p>
| lone student | 460,967 | <p>The answer is "no". Because, we don't have an important restriction <span class="math-container">$q\in\mathbb Z^{+}$</span>. Otherwise, we have infinitely many solutions:</p>
<p><span class="math-container">$$q=2\sqrt {\frac p5}$$</span>
where, <span class="math-container">$p$</span> is an any prime number.</p>
|
1,218,354 | <p>I've read in some textbooks that $\vdash$ and $\vDash$ are symbols present only in metalanguage. From this, I infer that their use in object language is unacceptable.</p>
<p>I would like to know why. Can't we define them as relation symbols in a structure? Or introduce them in statements for the sake of formal proofs?</p>
| goblin GONE | 42,339 | <p>I can explain it like this.</p>
<p>Given a set $X$, write $X^*$ for the collection of all finite sequences in $X$, including the empty sequence. Now let $L$ denote an arbitrary set, which we think of as a "language"; so you should be thinking of the elements of $L$ as "formulae". Then:</p>
<blockquote>
<p><strong>Definition.</strong> An <em>inference relation</em> over $L$ is subset $\vdash$ of $L^* \times L$ subject to certain axioms, like:</p>
<ol>
<li><p>For all $\Gamma \in L^*$ and all $\varphi \in L$, if $\varphi$ occurs somewhere in $\Gamma$, then $\Gamma \vdash \varphi$.</p></li>
<li><p>etc.</p></li>
</ol>
<p>Note that we write $\Gamma \vdash \varphi$ as a more readable alternative to the more correct $(\Gamma,\varphi) \in \;\vdash$. This notation can be seen in axiom 1, for example.</p>
</blockquote>
<p>Now here's where the ambiguity creeps in. Suppose $L$ is the set of all strings featuring the symbols $0,1$, the comma symbol, and the symbol $\vdash$. So a generic element of $L$ looks like:</p>
<p>$$0\vdash 1,,0\vdash,1$$</p>
<p>You can see the issue, right? <strong>If we suppose furthermore that $\vdash$ is an inference relation on $L$,</strong> then we cannot tell what "$01,1 \vdash 1$" means. It could be an element of $L$. Or, it could be the writer attempting to claim that $(\langle 01,1\rangle,1) \in \;\vdash$. Without further information, we cannot know.</p>
<p>To avoid this kind of ambiguity, we would choose a different symbol for the inference relation, as in:</p>
<blockquote>
<p>Suppose furthermore that $\vdash'$ is an inference relation on $L$.</p>
</blockquote>
<p>It now becomes clear that $01,1 \vdash 1$ is intended to denote an element of $L$, whereas $01,1 \vdash' 1$ is expressing the proposition that $(\langle 01,1\rangle,1) \in \;\vdash'$.</p>
<p>Make sense?</p>
|
4,035,300 | <p>If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!</p>
<p>'''</p>
<p>When a company bids for contracts it estimates the probability of winning each contract is <span class="math-container">$0.18$</span>, independent on whether other contracts have been won or lost.</p>
<p>(a) If the company bids for 5 contracts, what is the probability it wins:</p>
<p>i) at least one contract</p>
<p>ii) at least two contracts</p>
<p>'''</p>
<p>i)I used the OR rule for this question so if there's a <span class="math-container">$0.18$</span> chance then surely it's <span class="math-container">$0.18$</span> OR <span class="math-container">$0.18$</span>, etc which is just <span class="math-container">$0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$</span></p>
<p>ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did <span class="math-container">$0.18 * 0.18 = 0.0324$</span></p>
<p>If someone could just correct me if I took my stupid pills this morning I would be very grateful!</p>
| Darth Geek | 163,930 | <p>We will prove the last statement.</p>
<p>Let <span class="math-container">$\alpha$</span> be a root of <span class="math-container">$p_1:= X^3 + X^2-1$</span>, i.e., <span class="math-container">$\alpha^3 +\alpha^2-1 = 0$</span>. We will prove that <span class="math-container">$\beta := \alpha^2$</span> is a root of <span class="math-container">$p_2:= X^3-X^2+2X-1$</span>.</p>
<p>We have the following identities:</p>
<p><span class="math-container">$$\beta = \alpha^2,$$</span>
<span class="math-container">$$\beta^2 = \alpha^4 = \alpha\cdot \alpha^3 = \alpha (-\alpha^2 +1) = -\alpha^3 + \alpha,$$</span>
<span class="math-container">$$\beta^3 = \alpha^6 = (\alpha^3)^2 = (-\alpha^2+1)^2 = \alpha^4 -2\alpha^2 +1 = -\alpha^3 -2\alpha^2 + \alpha +1.$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\beta^3 - \beta^2 +2\beta -1 = (-\alpha^3 -2\alpha^2 + \alpha +1) - (-\alpha^3 + \alpha) +2(\alpha^2) -1 = 0.$$</span></p>
<p>Thus <span class="math-container">$\beta$</span> is a root of <span class="math-container">$p_2$</span>. Note that we did not require <span class="math-container">$\alpha$</span> to be real, this means that the roots of <span class="math-container">$p_2$</span> are precisely the squares of the roots of <span class="math-container">$p_1$</span>. Particularly, the square of the real root of <span class="math-container">$p_1$</span> is a (the) real root of <span class="math-container">$p_2$</span>.</p>
|
1,658,846 | <ul>
<li><p>How can one prove the existence of an order preserving bijection from $\mathbb{Q}$ to $\mathbb{Q}\backslash\lbrace{0}\rbrace$?</p></li>
<li><p>Can you give an example of such a bijection? </p></li>
</ul>
| hmakholm left over Monica | 14,366 | <p>Choose an irrational number $\alpha$.</p>
<p>Let $x_1, x_2, \ldots$ be a strictly <em>increasing</em> sequence of rational numbers that converge towards $\alpha$.</p>
<p>Let $y_1, y_2, \ldots$ be a strictly <em>decreasing</em> sequence of rational numbers that converge towards $\alpha$.</p>
<p>Then define $f:\mathbb Q\to\mathbb Q\setminus\{0\}$ as:</p>
<ul>
<li>$f$ maps $(-\infty,x_1]$ to $(-\infty,-1]$ by subtracting $x_1+1$ from everything.</li>
<li>For every $n$, $f$ maps $[x_n,x_{n+1}]$ to $[-\frac1n,-\frac1{n+1}]$, by linear interpolation between the endpoints.</li>
<li>For every $n$, $f$ maps $[y_{n+1},y_n]$ to $[\frac1{n+1},\frac1n]$, by linear interpolation between the endpoints.</li>
<li>$f$ maps $[y_1,\infty)$ to $[1,\infty)$ by subtracting $y_1-1$ from everything.</li>
</ul>
|
1,658,846 | <ul>
<li><p>How can one prove the existence of an order preserving bijection from $\mathbb{Q}$ to $\mathbb{Q}\backslash\lbrace{0}\rbrace$?</p></li>
<li><p>Can you give an example of such a bijection? </p></li>
</ul>
| Igor Rivin | 109,865 | <p>See the proof of theorem 3.7 here: <a href="http://www.math.wustl.edu/~freiwald/ch8.pdf" rel="nofollow">http://www.math.wustl.edu/~freiwald/ch8.pdf</a></p>
|
1,557,353 | <p>Context: I'm taking calc based physics, and we are supposed to be able to integrate moment of inertia for a cylinder. I referenced a mit vid, and though I have no education on multiple integrals, I got all but one thing. <a href="https://www.youtube.com/watch?v=iYFogDTPlRo" rel="nofollow">https://www.youtube.com/watch?v=iYFogDTPlRo</a></p>
<p>My progress is:</p>
<p>1) $\int r^2 dm$</p>
<p>2) $ \int r^2 \delta dV $ because $ \delta = \frac{dm}{dV} $ -> $dm = \delta dV$ where $\delta$ is density</p>
<p>3) $\delta \int_0^{2\pi} \int_0^r \int_0^h r^2 * r dzdrd\theta$ <- I dont fully understand this step.</p>
<p>Can someone explain why $dV = r dz dr d\theta$. Can it be justified in a similar manner as step 2, perhaps in multiple parts even? I get the idea I believe, but can't justify it.</p>
<p>EDIT:
Thank you everyone. I don't have enough rep to upvote, but the answers given by David, Hamed, B. Pasternak, and root were all helpful. I have a pretty good understanding now.</p>
| Hamed | 191,425 | <p>Other than David's geometrical method which is nicer, here is an algebraic method. (Remember $r$ is the radial length in the <strong>x-y</strong> plane)
$$
(x,y,z)=(r\cos\theta, r\sin\theta, z)
$$
Now $dx dy dz = Jdrd\theta dz$, where $J$ is the Jacobian and is defined as (this works for any change of coordinates)
$$
J=\begin{vmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z}\\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z}
\\
\frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z}
\end{vmatrix}=
\begin{vmatrix}
\cos\theta & -r\sin\theta & 0\\
\sin\theta & r\cos\theta & 0
\\
0 & 0 & 1
\end{vmatrix}=r(\cos^2\theta + \sin^2 \theta) = r
$$
Hence $dV=rdrd\theta dz$ (by the way $|\bullet|$ above means the determinant).</p>
|
1,557,353 | <p>Context: I'm taking calc based physics, and we are supposed to be able to integrate moment of inertia for a cylinder. I referenced a mit vid, and though I have no education on multiple integrals, I got all but one thing. <a href="https://www.youtube.com/watch?v=iYFogDTPlRo" rel="nofollow">https://www.youtube.com/watch?v=iYFogDTPlRo</a></p>
<p>My progress is:</p>
<p>1) $\int r^2 dm$</p>
<p>2) $ \int r^2 \delta dV $ because $ \delta = \frac{dm}{dV} $ -> $dm = \delta dV$ where $\delta$ is density</p>
<p>3) $\delta \int_0^{2\pi} \int_0^r \int_0^h r^2 * r dzdrd\theta$ <- I dont fully understand this step.</p>
<p>Can someone explain why $dV = r dz dr d\theta$. Can it be justified in a similar manner as step 2, perhaps in multiple parts even? I get the idea I believe, but can't justify it.</p>
<p>EDIT:
Thank you everyone. I don't have enough rep to upvote, but the answers given by David, Hamed, B. Pasternak, and root were all helpful. I have a pretty good understanding now.</p>
| root | 264,845 | <p>For more-dimensional change of coordinates (in this case: cartesian $\rightarrow$ cylindric) one uses, that (under certain circumstances you can find in every textbook) substitution is given by</p>
<p>$$ \int_{g(D)} f(\mathbf{y}) d\mathbf{y} = \int_D (f \circ g)(x) \left| \det \left( \frac{dg}{d\mathbf{x}}(\mathbf{x}) \right) \right| d\mathbf{x}$$</p>
<p>This is the substitution rule in $\Bbb R^n$. $ \frac{dg}{d\mathbf{x}}(\mathbf{x})$ is the Jacobi matrix, given by</p>
<p>$$J = \frac{dg}{d\mathbf{x}}(\mathbf{x}) = \begin{pmatrix} \nabla g^1(\mathbf{x}) \\ \vdots \\ \nabla g^m (\mathbf{x}) \end{pmatrix}$$</p>
<p>So the entry $(J_{kj})$ is given by $\frac{\partial g^k(\mathbf{x})}{\partial x_j}$, where $g: \Bbb R^n \to \Bbb R^m$ is in your case the mapping cartesian $\rightarrow$ cylindric coordinates and $g^k(\mathbf{x})$ is the $k^{\text{th}}$ coordinate of $g$, which is in your case</p>
<p>$$g: (x,y,z) \rightarrow (r \cos(\varphi), r \sin(\varphi), z) $$</p>
<p>E. g. the first entry of $J$ would be $ J_{11}= \frac{\partial g^1(\mathbf{x})}{\partial x_1} = \cos(\varphi)$, since $(x_1,x_2,x_3) = (r,\varphi,z)$. Summarizing:</p>
<p>$$d\mathbf{x} \equiv dxdydz = \left| \det \left( \frac{dg}{d(r,\varphi,z)}(r,\varphi,z) \right) \right| d(r,\varphi,z) = r drd\varphi dz$$</p>
|
3,772,534 | <p>Tangents to a circumference of center O, drawn by an outer point C, touch the circle at points A and B. Let S be any point on the circle. The lines SA, SB and SC cut the diameter perpendicular to OS at points A ', B' and C ', respectively. Prove that C 'is the midpoint of A'B'.</p>
<p><a href="https://i.stack.imgur.com/Ggwjj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ggwjj.jpg" alt="My draw (geogebra)" /></a></p>
<p>I saw a solution by Projective Geometry. I want to know if there is a solution by euclidean geometry. I think that is possible to do with Menelaus Theorem, but I don't know wich triangles I have to use. Thanks for attention.</p>
| Riemann'sPointyNose | 794,524 | <p>To expand a little on @Angina Seng's point, essentially the thing to notice is that if we can find the fourth root of <span class="math-container">${-1}$</span>, and replace <span class="math-container">$x$</span> with <span class="math-container">$x$</span> multiplied by this fourth root you get</p>
<p><span class="math-container">$${\frac{((-1)^{\frac{1}{4}}x)^{4n}}{(4n)!}=\frac{((-1)^{\frac{1}{4}})^{4n}x^{4n}}{(4n)!}=\frac{(-1)^nx^{4n}}{(4n)!}}$$</span></p>
<p>There are actually <strong>four different fourth roots of <span class="math-container">${-1}$</span></strong>:</p>
<p><span class="math-container">$${e^{\frac{i\pi}{4}},e^{\frac{-i\pi}{4}},e^{\frac{3i\pi}{4}},e^{\frac{-3i\pi}{4}}}$$</span></p>
<p>So any one of these will suffice. If we take the principal root (<span class="math-container">${e^{\frac{i\pi}{4}}}$</span>) though, this would make your answer</p>
<p><span class="math-container">$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{i\pi}{4}}x\right)+\cosh\left(e^{\frac{i\pi}{4}}x\right)}{2}}$$</span></p>
<p>Just as others have said. But we don't have to pick that root,</p>
<p><span class="math-container">$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{3i\pi}{4}}x\right)+\cosh\left(e^{\frac{3i\pi}{4}}x\right)}{2}}$$</span></p>
<p>is also just as valid</p>
<p><strong>Edit</strong>: I won't write the full working (I will leave it as an exercise to you) - but we can actually further then simplify the result to</p>
<p><span class="math-container">$${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right)}$$</span></p>
|
1,690,346 | <p>How can we solve this integral?
$\int_0^1\:\frac{\ln(x)\:\Big[1+x^{-\frac{1}{3}}\Big]}{(1-x)\sqrt[3]{x}}\:dx$</p>
| Yuriy S | 269,624 | <p>We have</p>
<p>$$\int_0^1 x^a dx = \frac{1}{a+1}$$</p>
<p>$$\int_0^1 x^a \log x ~dx = -\frac{1}{(a+1)^2}$$</p>
<p>$$\int_0^1 x^b x^q \log x ~dx = -\frac{1}{(b+q+1)^2}$$</p>
<p>$$\int_0^1 \sum_{b=0}^{\infty} x^b x^q \log x ~dx = -\sum_{b=0}^{\infty} \frac{1}{(b+q+1)^2}$$</p>
<p>$$\int_0^1 \frac{x^q \log x}{1-x} ~dx = -\sum_{b=0}^{\infty} \frac{1}{(b+q+1)^2}=-\zeta(2,q+1)=-\psi^{(1)}(q+1) $$</p>
<p>With the last two expressions being Hurwitz zeta and polygamma functions respectively.</p>
<p>So, in your case:</p>
<p>$$\int_0^1 \frac{(x^{-1/3}+x^{-2/3}) \log x}{1-x} ~dx =-\zeta \left(2,\frac{2}{3} \right)-\zeta \left(2,\frac{1}{3} \right) =-\psi^{(1)} \left(\frac{2}{3} \right)-\psi^{(1)} \left(\frac{1}{3} \right)$$</p>
<p>Using the reference provided by @tired</p>
<p>$$\psi^{(1)} \left(z \right)+\psi^{(1)} \left(1-z \right)=\frac{\pi^2}{\sin^2 (\pi z)}$$</p>
<p>$$\int_0^1 \frac{(x^{-1/3}+x^{-2/3}) \log x}{1-x} ~dx =-\frac{4\pi^2}{3}$$</p>
|
1,859,178 | <p>Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,</p>
<p><a href="https://i.stack.imgur.com/wCPAN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wCPAN.png" alt="enter image description here"></a>
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure</p>
<p>Thanks help</p>
| Giacomo Ascione | 354,026 | <p>I know it's rude to give complex solutions to real problems, but I think we can easily see where the second solution $x=2-\sqrt{5}$ came from.</p>
<p>We want to find solutions for: $$\begin{cases} y^2=4x \\ x^2-y^2=1 \end{cases}$$ Let's try find them in $\mathbb{C}$.</p>
<p>We got, as you said before, the equation $x^2-4x-1=0$ that gives us the solutions you found. We will call them $x_1=2+\sqrt{5}$ and $x_2=2-\sqrt{5}$.</p>
<p>But now we have to find $y$. Let's use $x_1$ in the first equation to find $y^2=8+4\sqrt{5}$ which will give us $y_1=\sqrt{8+4\sqrt{5}}$ and $y_2=-\sqrt{8+4\sqrt{5}}$.</p>
<p>You can see that the points you have in the picture are $A=(2+\sqrt{5},\sqrt{8+4\sqrt{5}})$ and $B=(2+\sqrt{5},-\sqrt{8+4\sqrt{5}})$, and that solves part of the contraddiction.</p>
<p>But where does $x_2$ come from? Let us see that $x_2=2-\sqrt{5}<0$ (in fact you noticed $x_1x_2=-1<0$) so the equation $y^2=8-4\sqrt{5}$ has no real solution! Instead in $\mathbb{C}$ we could get $y_3=i\sqrt{8+4\sqrt{5}}$ and $y_4=-i\sqrt{8+4\sqrt{5}}$ which define other two points $C=(2-\sqrt{5},+i\sqrt{8+4\sqrt{5}})$ and $D=(2-\sqrt{5},-i\sqrt{8+4\sqrt{5}})$. These two points are not seen in the picture because they are complex points (they've got imaginary $y$ coordinate) so thay cannot be represented in a picture on the $\mathbb{R}^2$ plane. </p>
<p>In your problem you want to discuss only real intersections of the curves, so you don't notice there are two complex intersections (and you don't want to because you're working with real numbers, so the second equation we solved in $\mathbb{C}$ doesn't give any solution in $\mathbb{R}$). Then the second solution for the equation in $x$ is to be rejected because cannot provide a real solution for the system.</p>
<p>Hope I helped</p>
|
2,522,342 | <p>So far I have only got 9 from just guess and check. I am thinking of using Vieta's Formula, but I am struggling over the algebra. Can someone give me the first few steps?</p>
| user502959 | 502,959 | <p>Assuming that $x$ is a natural number, it's quite trivial, since $2^x-1,\ 2^x,\ 2^x+1$ are three consecutive numbers, so one of them is divisible by 3, and it obviously isn't $2^x$.</p>
<p>3 is the only prime divisible by 3, so:</p>
<p>$2^x-1=3$ leads to $x=2$ - good ($2^x+1=5$ - prime).</p>
<p>$2^x+1=3$ leads to $x=1$ - contradiction ($2^x-1=1$ - not prime).</p>
<p>EDIT:
Since the Mersenne prime problem is an open conjecture,
I assumed the author means both numbers togoether prime.</p>
|
1,182,432 | <p>Is it possible, that everyone is a pseudo-winner in a tournament with 25 people?
(pseudo-winner means that either he won against everyone, or if he lost against someone, then he beated someone else, who beated the one who he lost to).</p>
<p>In the "language" of graph theory: Is it possible in the directed K(25) (all edges drawn with 25 vertices), that from all vertices, every vertices can be reached with a maximum of 2-long route.</p>
<p>I thought about this, and my answer is no. I tried to do the proving, starting with K(3), which is a triangle. Triangle works, but I don't know how to continue this until 25 vertices.</p>
<p>Maybe any easier solutions?</p>
<p>Thanks!</p>
| Misha Lavrov | 383,078 | <p>The answer is a <a href="https://www.umop.com/rps25.htm" rel="nofollow noreferrer">well-known graph</a> generalizing a classical 3-vertex construction:</p>
<p><a href="https://i.stack.imgur.com/sZt9F.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sZt9F.jpg" alt="enter image description here"></a></p>
<p>To see that RPS-25 satisfies the conditions, it's enough to check that each shape beats exactly $12$ others. As a result, if shape $A$ loses to shape $B$, then among the $12$ shapes that shape $A$ beats, at most $11$ can lose to shape $B$: so there must be a shape $C$ that beats $B$ and loses to $A$.</p>
<p>(And so it's easy to find other examples, for any odd $n$: just take any regular tournament.)</p>
|
2,935,693 | <p>I am trying to prove that </p>
<p><span class="math-container">$(a\to(b\to c))\to((a\to b)\to(a\to c))$</span></p>
<p>holds in natural deduction, in particular when I work backwards from a Fitch style proof I can only get so far:</p>
<p><a href="https://i.stack.imgur.com/w2BYf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w2BYf.png" alt="enter image description here"></a></p>
<p>How can I prove it?</p>
| Graham Kemp | 135,106 | <p><span class="math-container">$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$</span>
How exactly does one go about proving a conditional statement? A Conditional Proof is usually used. Step one: assume the antecedant. Step two: derive the consequent. Step three: introduce the conditional.</p>
<p>Here we have a nesting of conditionals, so we need to nest Conditional Proofs.
<span class="math-container">$$\fitch{}{\fitch{a\to(b\to c)}{\fitch{a\to b}{\fitch{a}{\vdots\\c}\\a\to c}\\(a\to b)\to(a\to c)}\\(a\to(b\to c))\to((a\to b)\to(a\to c))}$$</span></p>
<p>Well, you understand that, I see, but your trouble is how do we get from <span class="math-container">$a$</span> to <span class="math-container">$c$</span>? Now, the assumptions we have made are <span class="math-container">$a$</span> and <em>two</em> condtionals with <span class="math-container">$a$</span> as an antecedant. This indicates that <em>both</em> those conditionals may be eliminated, and doing so gives <span class="math-container">$b$</span> and <span class="math-container">$b\to c$</span>. </p>
<p>One more elimination and we are done.</p>
<p><span class="math-container">$$\fitch{}{\fitch{~~1.~a\to(b\to c)~;\textsf{assumption}}{\fitch{~~2.~a\to b~;\textsf{assumption}}{\fitch{~~3.~a~;\textsf{assumption}}{~~4.~b~: \textsf{conditional elimination (2,3)}\\~~5.~b\to c~: \textsf{conditional elimination (1,3)}\\~~6.~c~: \textsf{conditional elimination (4,5)}}\\~~7.~a\to c~: \textsf{conditional introduction (3-6)}}\\~~8.~(a\to b)\to(a\to c)~: \textsf{conditional introduction (2-7)}}\\~~9.~(a\to(b\to c))\to((a\to b)\to(a\to c))~: \textsf{cond. intro. (1-8)}}$$</span></p>
|
1,025,321 | <p>$\ln(1+xy) = xy$</p>
<p>When I try to implicitly differentiate this I get</p>
<p>$\frac{1}{1+xy}(y + xy')$ = (y + xy')</p>
<p>At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.</p>
<p>However, the answer to this is $-\frac{y}{x}$... How do you get this?</p>
| PtF | 49,572 | <p>As to the book recommendations I suggest the following:</p>
<p><strong>(i)</strong> A course in Differential Geometry (W. Klingenberg)</p>
<p><strong>(ii)</strong> Elementary Differential Geometry (A. Pressley)</p>
<p><strong>(iii)</strong> Differential-Geometrie und Minimal-Flächen (J. Jost) (in german but it's a wonderful book).</p>
<p>Some more advanced books dealing with differential geometry in the context of differentiable manifolds I'd suggest:</p>
<p><strong>(v)</strong> Differential Geometry: Manifolds, Curves and Surfaces (M. Berger, B. Gostiaux)</p>
<p><strong>(vii)</strong> A Comprehensive Introduction to Differential Geometry (M. Spivak) (all the volumes)</p>
<p>I hope it helps.</p>
<p>As to your question about the diffeomorphism between the open unit ball and $\mathbb R^n$ the idea is defining a function which will stretch the open ball in all directions.</p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| fedja | 1,131 | <p>Response to Steven Landsburg:</p>
<p>The main point Steven made was that M is at least as important as <span class="math-container">$A$</span>.
How to reduce M? (the interaction between public servants and ordinary people)? That, as Steven correctly pointed out, is in the hands of people. The servants just serve what you need. If you reduce what you need, you can fill your supplies to the household capacity less frequently and thus reduce <span class="math-container">$M$</span> locally in the matrix <span class="math-container">$A$</span> (at last I have enough preliminary explanations to speak in the language in which I think). Example: I eat mainly grain now (bought a few bags at Costco some time ago). That produces zero garbage emission, so my garbage bin fills slowly and I can put it out, probably, 5 times less frequently, so most of the time the truck bypasses my house and I am not afraid of contaminated garbage been 5 times longer than an average person, so I have an egoistic incentive to do it as well. Besides, I just have no supply of required garbage bags :-) ). That potentially reduces M five times for the garbage pickup part of <span class="math-container">$A$</span>. And that is where the virus teleports: the garbage trucks go everywhere. But it teleports at the speed of car and we can teleport ideas at the speed of light (Internet). The proposed modus operandi for propagation of thought is exactly what I posted in the beginning: share your thoughts on what to do in the widest network available to you where you have enough reputation to be listened to. We'll fight the car teleportaion of virus with the light speed teleportation of ideas. That is not even restricted to country: explain to your friends everywhere what is going on, what's the global strategy, why you are doing what you are doing to support it, and talk to them in the common language between you and them trying to keep it as simple as possible, but not simpler than that. Everyone is free to implement the rules that are not weaker than the current regulations in his household (I follow max(Israel,Ohio)) and think how to advance them more. That's why the government does not impose the draconian regulations itself. They hope that graduate reduction will be met by gradual self-restriction and we'll bargain not at the strict line but on a curve allowing freedoms where they are needed and using freedoms only if they are needed. That's the advantage of democracy, used properly. The advantage of dictatorship is the speed of bargaining. What we'll get depends on how we act. </p>
<p>If you get lost, please don't downwote, just ask questions :-) </p>
|
669,207 | <p>For any function $f$ and any $x∈Dom(f)$, if for any neighbourhood $S$ of $x$,</p>
<p>$\qquad f(t)=0$ for some $t∈S$ </p>
<p>$\qquad f(u)=1$ for some $u∈S$ </p>
<p>then $ f$ is discontinuous at $x$.</p>
<p>Why is this true? I find it very hard to understand it :(.</p>
| Patrick Da Silva | 10,704 | <p>Inverse image applied to elements only makes sense if the function is injective, in which case we can restrict the range to the image and get a bijective map for which $f^{-1}$ is a well-defined map. Otherwise we reserve the notation $f^{-1}$ for the map given by $f^{-1} : \mathcal P(Y) \to \mathcal P(X)$ which sends a subset of $Y$ to a subset of $X$.</p>
<p>However, it is common to write $f^{-1}(a)$ for $f^{-1}(\{a\})$, for purposes of brevity. It is up to the context to determine if the function was bijective or not to begin with. One can write $f^{-1}(a) = \{b\}$ to make this clear, if it would so happen that the inverse image was a singleton.</p>
<p>Hope that helps,</p>
|
2,875 | <p>I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group <code>G</code> being <code>SL(2, R)</code>, can be completely described and that there is a discrete and continuous part of the spectrum of <code>L^2(G)</code>.</p>
<ol>
<li>How are those representations described?</li>
<li>Do all unitary representations come from <code>L^2(G)</code>?</li>
<li>How are those related to representation of compact <code>SO(3, R)</code>? </li>
<li>What happens in the flat limit between <code>SL(2, R)</code> and <code>SO(3, R)</code>?</li>
</ol>
<p>Also, is it possible to answer the questions above simultaneously for all Lie groups, not just <code>SL(2, R)</code>?</p>
| Hadi | 9,962 | <p>Regarding classification of irreducible unitary representations of <em>arbitrary</em> Lie groups, Michel Duflo showed (I guess in the early 80's) that at least for algebraic Lie groups the classification can be reduced to the case of reductive Lie groups. See Vogan's <em>Unitary representations of reductive Lie groups</em> for more information.</p>
|
692,085 | <p>It appears that taking the cube root of a negative number will yield a negative number, which when squared, will yield a positive number. But all the calculators and books I have seen show this particular problem yields a negative number. Any help here?</p>
| Thomas | 26,188 | <p>Indeed
$$
(-64)^{2/3} = 16.
$$
This can be seen two ways:
$$\begin{align}
(-64)^{2/3} &= ((-64)^{1/3})^2 = (-4)^2 = 16 \\
(-64)^{2/3} &= ((-64)^2)^{1/3} =(4096)^{1/3} = 16.
\end{align}
$$
Here remember that
$$
a^{1/3} = \sqrt[3]{a}.
$$</p>
|
692,085 | <p>It appears that taking the cube root of a negative number will yield a negative number, which when squared, will yield a positive number. But all the calculators and books I have seen show this particular problem yields a negative number. Any help here?</p>
| kitkat | 131,788 | <p>Well let's take this step by step though you do seem to have an understanding of it.</p>
<p>The cube root of -64 is -4. So yes It does yield a negative number. Then the square of that is 16, so like the previous person said perhaps you have not entered it into the calculator properly. </p>
<p>Ah technology, a blessing but yet a curse.</p>
|
2,155,740 | <p>This is a homework problem for which I think I've missed the point or have incorrectly done the proof (or both). There are two parts to the problem: Let $(a_n)$ be a sequence with $a_n \ge 0,$ for all $n$.</p>
<p>Part 1:<br>
Suppose that $a_n \rightarrow 0$. Show that $\sqrt{a_n}\rightarrow 0$. I recognized that these are actually two different sequences. Here is my proof.<br>
<strong>Proof</strong><br>
Let $(b_n)=\sqrt{a_n}$ and $\forall a\in (a_n), a\ge 0$.<br>
By assumption, $(a_n) \rightarrow 0$.<br>
By definition of <em>Square Root</em>, $\forall b_n, b_n < a_n$.<br>
So, since $b_n < a_n$ and $(a_n) \rightarrow 0, (b_n) \rightarrow 0 \square$.</p>
<p>Part 2:<br>
Suppose that $a_n \rightarrow L$. Show that $\sqrt{a_n}\rightarrow \sqrt{L}$.</p>
<p>Assuming I did things correctly in part 1, shouldn't this be a virtual <em>ditto</em>? My teacher gave this <em>hint</em> which makes me think I'm way off base:<br>
<em>you can assume that $L\ne 0$. Use that $\sqrt{x} - \sqrt{y} = \frac{x-y}{\sqrt{x}+\sqrt{y}}$</em>.</p>
<p>Since my proof in part 1 didn't use anything like this, I'm assuming I'm off in the weeds. Since this homework hasn't yet been graded, I'll need hints rather than solutions. Thanks.</p>
| RRL | 148,510 | <p>Hint: </p>
<p>For an all-encompassing proof with $L \geqslant 0$ note that</p>
<p>$$|\sqrt{a_n} - \sqrt{L}|^2 \leqslant |\sqrt{a_n} - \sqrt{L}||\sqrt{a_n} + \sqrt{L}| = |a_n - L|$$</p>
|
1,607,190 | <p>Prove by induction that $8^{n} − 1$ for any positive integer $n$ is divisible by $7$. </p>
<p>Hint: It is easy to represent divisibility by $7$ in the following way: $8^{n} − 1 = 7 \cdot k$ where k is a positive integer.</p>
<p>This question confused me because I think the hint isn't true. If $n = 1$ and $k = 2$ for example, then we end up with $7 = 14$ which is obviously invalid. Does this mean the $n \leq k$ in order for the hint to be true.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>let $$T_k=8^k-1$$ then $$T_{k+1}=8(8^k-1)+7$$</p>
|
2,784,784 | <p>Let $f(x)=$$x-1 |x \in \mathbb{Q} \brace 5-x| x \in \mathbb{Q}^c$</p>
<p>Show that $\lim_{x \to a}f(x)$ does not exists for any $a \not= 3$</p>
<p>I first showed that $lim_{x \to 3}f(x)=2$. </p>
<p>I don't know how to approach this part. Can anyone please guide? I was thinking of using density theorem at first to show that there will exist sequences of rationals and irrationals that will approach a but their limits would not equal but that is problematic as I don't really know the limit.</p>
<p>thank you</p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>Of course, you must use the fact that for any point, there is a sequence of rationals and another sequence of irrationals which converge to that point.</p>
<p>Here is the conventional limit definition :</p>
<blockquote>
<p>Given $f : D \to \mathbb R$ and $x \in D$, suppose there is a number $L$ such that for every $\epsilon > 0$ there is a $\delta > 0$ such that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. Then, $\lim_{t \to x}f(t)$ is said to exist and equal $L$.</p>
</blockquote>
<p>Using a usual trick, we bring this down to convergence of sequences, and then get a sequential definition :</p>
<blockquote>
<p>Given $f : D \to \mathbb R$ and $x \in D$, if there exists a number $L$ such that for every sequence $x_n$ converging to $x$, the sequence $f(x_n)$ converges to $L$. Then $\lim_{t \to x}f(t)$ is said to exist and equal $L$.</p>
</blockquote>
<p>Now, if the limit <em>does not</em> exist, then we must negate the above statement:</p>
<blockquote>
<p>Given $f : D \to \mathbb R$ and $x \in D$, if <em>for all</em> numbers $L$, <em>there exists a</em> sequence $x_n$ converging to $x$ such that $f(x_n)$ <em>does not</em> converge to $L$, then the limit $\lim_{x \to a} f(x)$ is said to <em>not</em> exist.</p>
</blockquote>
<p>Now, we are in shape to attack the problem : all we need to do, is given any $a \neq 3$ and a candidate $L$ for the limit, find a sequence of points which converge to $a$ such that the function values don't go to $L$. To do this, we use the piecewise definition of our function.</p>
<p>Let $a \neq 3$. Suppose that $L$ is a candidate limit. </p>
<p>EDITED : At this stage, we want a sequence of numbers which does converge to $a$, but whose function values don't converge to $L$.</p>
<p>The answer to your question below is this : The pieces of $f$, <em>when treated as functions from the real line to itself</em>, are continuous, and this fact can be used to prove that $f$ is not continuous at any point, other than $3$.</p>
<p>First, let $p_n$ is a sequence of rationals converging to $a$. We claim that $f(p_n) \to a-1$. </p>
<p>This follows from the fact that $p_n$ lies inside one piece of $f$, so we may use continuity of the function which defines $f$ on that piece. But if you want to argue by basics, then : $f(p_n) = p_n - 1$ from the fact that $p_n$ is rational. Now, since $p_n \to a$ and (the constant sequence)$-1 \to -1$, we can add limits to get $p_n -1 \to a-1$, and therefore $f(p_n) \to a-1$.</p>
<p>By uniqueness of limits, if $L \neq a-1$, then $p_n$ serves as a candidate for the sequence of values not converging to $L$, since $p_n \to a$ but $f(p_n) \not\to L$.</p>
<p>On the other hand, if $q_n$ is a sequence of irrationals converging to $a$, then we may repeat the above argument for the irrational piece to get $f(q_n) \to 5-a$. So $q_n$ now serves as a candidate for the sequence of values not converging to $L$, whenever $5-a \neq L$.</p>
<p>Now, if $a-1 \neq 5-a$, <em>which happens precisely when</em> $a \neq 3$, we see that for any $L$, of course $L$ can't be equal to both $a-1$ and $5-a$, so take the case which it does not equal, and that sequence works.</p>
<p>Hence, the limit of the given function does not exist, for any $a \neq 3$. Of course, if $f$ has to be continuous at $a$, then $\lim_{x\to a} f(x)$ has to <em>exist</em> and equal $f(a)$, but the limit doesn't exist if $a \neq 3$, so $f$ is not continuous at any such point.</p>
|
4,150,776 | <p>Let <span class="math-container">$(a_n)_{n=1}^\infty$</span> Let be a positive, increasing, and unbounded sequence. Prove that the series:</p>
<p><span class="math-container">$$\sum_{n=1}^\infty\left(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}\right)$$</span></p>
<p>convergent.</p>
<hr />
<p>We know that since <span class="math-container">$a_n$</span> is increasing and unbounded, than <span class="math-container">$\lim_{n \to \infty}a_n=\infty$</span>, so I want to apply that to say that, <span class="math-container">$\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$</span> is decreasing and its limit will be <span class="math-container">$0$</span>.</p>
<p>My problem is that every time I get confused while it says <span class="math-container">$a_{2n}$</span> or <span class="math-container">$a_{2n-1}$</span>, it is less intuitive for me than just "normal" <span class="math-container">$a_n$</span>...</p>
<p>Appreciate your help!</p>
<p>Thanks a lot!</p>
| Mark Viola | 218,419 | <p><strong>HINT:</strong></p>
<p>Since <span class="math-container">$a_n$</span> is positive and increasing , then <span class="math-container">$\frac1{a_n}$</span> is positive and decreasing. Moreover, we have</p>
<p><span class="math-container">$$\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{a_n}=\sum_{n=1}^N \left(\frac1{a_{2n-1}}-\frac1{a_{2n}}\right)$$</span></p>
<p>Now use what you know about <span class="math-container">$\lim_{n\to \infty}a_n$</span>. Can you finish now?</p>
|
172,124 | <p>$$\int \frac{1}{x^{10} + x}dx$$</p>
<p>My solution :</p>
<p>$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$</p>
<p>Is there completely different way to solve it ?</p>
| Venus | 146,687 | <p>Let we generalise the problem with a slightly different way. Consider
$$\int\frac{\mathrm dx}{x^n+x}$$
Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get
\begin{align}
\int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt]
&=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[9pt]
&=-\frac{\ln |1+z|}{n-1}+C\\[9pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln |x|-\frac{\ln \left|1+x^{n-1}\right|}{n-1}+C}}
\end{align}
In your case
$$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$</p>
|
185,549 | <blockquote>
<p>Suppose $F$ is a nondecreasing and right continuous function, and the sequence $\{x_n\}_{n\geq1}$ converges to $x$. Then $\liminf\limits_{n\to\infty}F(x_n)\leq F(x)$. </p>
</blockquote>
<p>How can I prove this?</p>
| Did | 6,179 | <p><strong>Hints:</strong></p>
<ul>
<li>For every $z\gt F(x)$, there exists $y\gt x$ such that $F(u)\leqslant z$ for every $u\leqslant y$. </li>
<li>Since $x_n\to x$, $x_n\leqslant y$ for every $n$ large enough.</li>
<li>Hence...</li>
</ul>
<p><strong>Edit:</strong></p>
<p>... $F(x_n)\leqslant F(y)\leqslant z$ for every $n$ large enough. In particular, $\limsup\limits_{n\to\infty}F(x_n)\leqslant z$. This is valid for every $z\gt F(x)$, hence...</p>
|
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