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 |
|---|---|---|---|---|
3,930,466 | <p>I want to prove by induction that</p>
<p><span class="math-container">$$\left(\tfrac{1}{n}\sum_{k=1}^{n} x_k\right)^2\leq\tfrac{1}{n}\sum_{k=1}^{n} x_k^2$$</span></p>
<p>But I have no idea how. For n=1 it is trivial that
<span class="math-container">$$\left(\tfrac{1}{1}\sum_{k=1}^{1} x_k\right)^2\leq\tfrac{1}{1}\sum_{k=1}^{1} x_k^2=x_k^2$$</span></p>
<p>But if i want to conclude n+1 from n I get</p>
<p><span class="math-container">$$\left(\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k\right)^2\leq\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k^2$$</span></p>
<p>And now I cannot seperate <span class="math-container">$\tfrac{1}{n+1}$</span> in order to substitute the rest with the induction hypothesis.</p>
<p>Maybe one needs to use a different version of induction? Thanks for any help.</p>
| Macavity | 58,320 | <p>With a little rewriting, the Induction hypothesis is
<span class="math-container">$$\color{blue}{\left(\sum_{k=1}^n x_k\right)^2 \leqslant n \sum_{k=1}^n x_k^2}$$</span></p>
<p>For the inductive step, note
<span class="math-container">$$\left(\sum_{k=1}^n x_k +x_{n+1} \right)^2 = \color{blue}{\left(\sum_{k=1}^n x_k \right)^2} + 2x_{n+1}\left(\sum_{k=1}^n x_k \right)+x_{n+1}^2 \color{blue}{ \leqslant n \sum_{k=1}^n x_k^2} + x_{n+1}^2 + 2x_{n+1}\left(\sum_{k=1}^n x_k \right)$$</span></p>
<p>Hence it remains to show
<span class="math-container">$$2x_{n+1}\left(\sum_{k=1}^n x_k \right) \leqslant \sum_{k=1}^n x_k^2+nx_{n+1}^2 \iff \sum_{k=1}^n \left(x_{n+1}-x_k \right)^2 \geqslant 0$$</span></p>
|
3,740,501 | <p>This question builds upon a previous question I asked <a href="https://math.stackexchange.com/questions/3548807/is-at-least-1-of-4-non-concyclic-points-contained-in-the-circle-through-the-othe">here</a></p>
<p>The answers helped me understand that for <span class="math-container">$4$</span> points that are not concyclic, and for which no <span class="math-container">$3$</span> fall on a straight line, of the <span class="math-container">$4$</span> circles that can be drawn through triples of the points, exactly <span class="math-container">$1$</span> or <span class="math-container">$2$</span> of the circles will contain the point they don't pass through.</p>
<p>I'm looking for an elementary proof (i.e. one that a high school geometry student could understand) as to why a point must be contained in a circle through <span class="math-container">$3$</span> given points. Consider the diagram below:</p>
<p><a href="https://i.stack.imgur.com/3Rnhj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Rnhj.jpg" alt="enter image description here" /></a></p>
<p>DLeMeur's answer to my original question helped me understand that circle <span class="math-container">$ABD$</span> will contain <span class="math-container">$C$</span> if and only if <span class="math-container">$D$</span> is placed in one of the gray areas. The arguments that I can make for this are only sort of convincing, but not really "air tight".</p>
<p>Case 1: <span class="math-container">$D$</span> is in the circular segment cut off by chord <span class="math-container">$\overline{AB}$</span>. Then circle <span class="math-container">$ABD$</span> has a greater radius than circle <span class="math-container">$ABC$</span>, and since <span class="math-container">$D$</span> and <span class="math-container">$C$</span> are on opposite sides of <span class="math-container">$\overleftrightarrow{AB}$</span>, <span class="math-container">$C$</span> must be contained in circle <span class="math-container">$ABD$</span>.</p>
<p>Case 2: <span class="math-container">$D$</span> is outside circle <span class="math-container">$ABC$</span>, on the same side of <span class="math-container">$\overleftrightarrow{AB}$</span> as point <span class="math-container">$C$</span>. Again, circle <span class="math-container">$ABD$</span> has greater radius than circle <span class="math-container">$ABC$</span>, thus the entire portion of circle <span class="math-container">$ABC$</span> below <span class="math-container">$\overleftrightarrow{AB}$</span> is contained in circle <span class="math-container">$ABD$</span>.</p>
<p>These arguments seem like they are missing some details. For instance, if someone asked, "How do you know circle <span class="math-container">$ABD$</span> has greater radius than circle <span class="math-container">$ABC$</span>?" I do not have a good answer. I would appreciate any input!</p>
| Axion004 | 258,202 | <p>Your answer is incorrect because you wrote <span class="math-container">$f(t+a)=f(t+5)$</span> when the step function had a right shift of <span class="math-container">$4$</span>. Applying</p>
<p><span class="math-container">$$\mathcal{L} \left(u(t-a)f(t)\right) = e^{-as} \mathcal {L} [f(t+a)],$$</span></p>
<p>we find</p>
<p><span class="math-container">$$\mathcal{L}[f(t+4)]=\mathcal L \left[(t+4)^2-8(t+4)+23\right] = \mathcal {L}[t^2+7] = \frac2{s^3} + \frac{7}{s}.$$</span></p>
<p>This gives the final answer of</p>
<p><span class="math-container">$$\mathcal{L} \left(u(t-4)f(t)\right)=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$</span></p>
<hr />
<p>I took a slightly different approach. We are given:</p>
<p><span class="math-container">$$f(t)=\begin{cases}
0, & t < 4, \\
t^2 -8t + 23, & t \geq 4, \\
\end{cases}=u(t-4)(t^2-8t+23).$$</span></p>
<p>Lets first complete the square</p>
<p><span class="math-container">$$t^2-8t+23=(t-4)^2+7.$$</span></p>
<p>Then the problem becomes</p>
<p><span class="math-container">$$f(t)=u(t-4)(t-4)^2+7u(t-4).$$</span></p>
<p>Applying</p>
<p><span class="math-container">$$\mathcal{L}(t^2)=\frac{2}{s^{3}},\quad \mathcal{L}\big[u(t-c)\big]=\frac{e^{-cs}}{s},\quad \mathcal{L}\big[u(t-c)g(t-c)\big]=e^{-cs}\mathcal{L}[g(t)],$$</span></p>
<p>we find</p>
<p><span class="math-container">$$\mathcal{L}[f(t)]=\mathcal{L}\left[u(t-4)(t-4)^2\right]+\mathcal{L}[7u(t-4)]=\frac{2e^{-4s}}{s^3}+\frac{7e^{-4s}}{s}=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$</span></p>
|
63,796 | <p>I'm looking for a neater way to achieve this, which to me looks awkward and suggests that I am missing something...</p>
<p>Given a Dataset...</p>
<pre><code>ds = Dataset[{
<|"a"->1,"b"->1,"c"->3|>,
<|"a"->1,"b"->2,"c"->4|>,
<|"a"->2,"b"->3,"c"->5|>}];
</code></pre>
<p>Return another Dataset given by the grouping by one column and the maximal by another...</p>
<pre><code>ds[GroupBy[#a &], MaximalBy[#b &]] // Values // Flatten
</code></pre>
<p>To return...</p>
<pre><code>Dataset[{<|"a" -> 1, "b" -> 2, "c" -> 4|>, <|"a" -> 2, "b" -> 3, "c" -> 5|>}
</code></pre>
<p>The GroupBy and MaximalBy return a list of Associations of row number to a List of Associations (of which there is only one) which then needs (something like) //Values//Flatten to retrieve the Dataset I want. </p>
<p>Is there a more Dataset-eque way of doing this?</p>
| Ymareth | 880 | <p>Typically as soon as I post this occurs to me...</p>
<pre><code>ds[GroupBy[#a &], MaximalBy[#b &]][Values, Merge[First]]
</code></pre>
<p>Since Values is a descending operator.</p>
|
51,732 | <p>A <em>Perron number</em> is a real algebraic integer $\lambda$ that is larger than the absolute value of any of its Galois conjugates. The Perron-Frobenius theorem says that any
non-negative integer matrix $M$ such that some power of $M$ is strictly positive has a
unique positive eigenvector whose eigenvalue is a Perron number. Doug Lind proved the converse: given a Perron number $\lambda$, there exists such a matrix, perhaps in dimension
much higher than the degree of $\lambda$. Perron numbers come up frequently in many places, especially in dynamical systems.</p>
<p>My question:</p>
<blockquote>
<p>What is the limiting distribution of Galois conjugates of Perron numbers $\lambda$ in
some bounded interval, as the degree goes to infinity?</p>
</blockquote>
<p>I'm particularly interested in looking at the limit as the length of the interval goes to
0. One way to normalize this is to look at the ratio $\lambda^g/\lambda$, as $\lambda^g$
ranges over the Galois conjugates. Let's call these numbers \emph{Perron ratios}.</p>
<p>Note that for any fixed $C > 1$ and integer $d > 0$, there are only
finitely many Perron numbers $\lambda < C$ of degree $< d$, since there is obviously a bound on the discriminant of the minimal polynomial for $\lambda$, so the question is only interesting when a bound goes to infinity. </p>
<p>In any particular field, the set of algebraic
numbers that are Perron lie in a convex cone in the product of Archimedean places of the field. For any lattice, among lattice points with $x_1 < C$ that are within this cone, the projection along lines through the origin to the plane $x_1 = 1$ tends toward the uniform
distribution, so as $C \rightarrow \infty$, the distribution of Perron
ratios converges to a uniform distribution in the unit disk (with a contribution for each complex place of the field) plus a uniform distribution
in the interval $[-1,1]$ (with a contribution for each real place of the field).</p>
<p>But what happens when $C$ is held bounded and the degree goes to infinity? This question seems
related to the theory of random matrices, but I don't see any direct translation from
things I've heard. Choosing a random Perron number seems very different from choosing
a random nonnegative integer matrix.</p>
<p>I tried some crude experiments, by looking at randomly-chosen polynomials of a fixed degree whose coefficients are integers in some fixed range except for the coefficient of $x^d$
which is $1$, selecting from those the irreducible polynomials whose largest real root is Perron. This is not the same as selecting a random Perron number of the given degree
in an interval. I don't know any reasonable way to do the latter except for small enough $d$ and $C$ that one could presumably find them by exhaustive search.
Anyway, here are some samples from what I actually tried.
First, from among the 16,807 fifth degree polynomials with coefficients in the range -3 to 3, there are $3,361$ that define a Perron number. Here is the plot of the Perron ratios:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints5%2C3.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints5%2C3.jpg</a></p>
<p>Here are the results of a sample of 20,000 degree 21 polynomials with coefficients
between -5 and 5. Of this sample, 5,932 defined Perron numbers:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21.jpg</a></p>
<p>The distribution decidedly does not appear that it will converge toward a uniform distribution on the disk plus a uniform distribution on the interval. Maybe the artificial bounds on the coefficients cause the higher density ring.</p>
<blockquote>
<p>Are there good, natural distributions for selecting random integer polynomials? Is there a
way to do it without unduly prejudicing the distribution of roots?</p>
</blockquote>
<p>To see if it would help separate what's happening,
I tried plotting the Perron ratios restricted to $\lambda$ in subintervals. For
the degree 21 sample, here is the plot of $\lambda$ by rank order:</p>
<p><a href="http://dl.dropbox.com/u/5390048/CDF21.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/CDF21.jpg</a></p>
<p>(If you rescale the $x$ axis to range from $0$ to $1$ and interchange $x$ and $y$ axes,
this becomes the plot of the sample cumulative distribution function of $\lambda$.)
Here are the plots of the Perron ratios restricted to the intervals $1.5 < \lambda < 2$
and $3 < \lambda < 4$:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21%281.5%2C2%29.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21%281.5%2C2%29.jpg</a></p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21%283%2C4%29.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21%283%2C4%29.jpg</a></p>
<p>The restriction to an interval seems to concentrate the absolute values of Perron ratios even more. The angular distribution looks like it converges to the uniform
distribution on a circle plus point masses at $0$ and $\pi$. </p>
<p>Is there an explanation for the distribution of radii? Any guesses for what it is?</p>
| Misha Belolipetsky | 13,859 | <p>You can ask a similar question about Pisot and Salem numbers. Last year together with my project student Charlie Scarr we were looking, in particular, at a possible connection between distribution of the roots inside the unit circle and the Mahler measure of a polynomial. We did not progress too far but Charlie made some interesting observations which can be found in his report <a href="http://www.maths.dur.ac.uk/~dma0mb/projects/C_Scarr.pdf">http://www.maths.dur.ac.uk/~dma0mb/projects/C_Scarr.pdf</a></p>
<p>Closer to your question, I think it would be very interesting to check what kind of picture comes out if one restricts to the polynomials with small Mahler measure. A large list of such polynomials is available at <a href="http://www.cecm.sfu.ca/~mjm/Lehmer/lists/">Michael Mossinghoff's page</a> on Lehmer's problem. It is in no way a random sample but the ordering by Mahler measure looks quite natural in this context.</p>
|
1,353,922 | <p>A function $f$ is continuous for all $x \geq 0$ and $f(x) \neq 0$ for all $x >0$. </p>
<p>If $\{f(x)\}^2 = 2\int_0^xf(t)dt $ then $f(x) = x$ for all $x \geq 0$.</p>
<p>But I am stuck with the sum. </p>
| Brian Cheung | 248,555 | <p>Differentiate both sides w.r.t. $x$ </p>
<p>Simplifying yields $2(f'(x)-1)f(x)=0$</p>
<p>i.e. $f'(x)=1$ or $f(x)=0$ (rej.) </p>
<p>Then integrate it back w.r.t. $x$ to get $f(x)=x+C$, use $f(0)=0$ and get $f(x)=x$.</p>
|
3,406,817 | <p>What intrinsic relationship is there between (riemann) integration and Euclidean geometry that enables one to get area under the curve as a integral? This can only be related to the definition of riemann integral as a limit of sums of areas of rectangles; or is it something else.</p>
| José Carlos Santos | 446,262 | <p>The Riemann integral is defined in such a way (through a limit of sums of areas of rectangles, as you wrote) that the area under the curve of a non-negative function is equal to the integral of that function.</p>
|
1,626,721 | <blockquote>
<p>In the <span class="math-container">$n$</span>-cell below, we are to tile it completely with cells of the form <span class="math-container">$\boxdot$</span> and <span class="math-container">$\boxtimes \hspace{-0.45 mm}\boxtimes$</span>. How many tilings are possible for a <span class="math-container">$12$</span>-cell?</p>
<p><a href="https://i.stack.imgur.com/Xn9lh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xn9lh.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Let <span class="math-container">$H_n$</span> denote the number of such tilings for an <span class="math-container">$n$</span>-cell. It is easy to see that <span class="math-container">$H_0 = 1, H_1 = 1, H_2 = 2, H_3 = 3,$</span> and <span class="math-container">$H_4 = 5$</span>. These seem to correspond to the fibonacci recursion of <span class="math-container">$_{n+1} = S_{n}+S_{n-1}$</span>, but how do I prove that this recursion is indeed <span class="math-container">$H_{n+1} = H_{n}+H_{n-1}$</span>? This is equivalent to saying, "the number of tilings for an <span class="math-container">$n+1$</span>-cell is equal to the number of tilings for an <span class="math-container">$n$</span>-cell plus the number of tilings for an <span class="math-container">$n-1$</span>-cell." Why is this true?</p>
| N. S. | 9,176 | <p><strong>Hint</strong> Let us figure out $H_{n+1}$. If the last tile is $\boxdot$, by erasing this we get a tiling of $n$ by the same tilings. There are $H_n$ such tilings, and by adding $\boxdot$ at the end we get a tiling of $n+1$.</p>
<p>If the last tiling is $\boxtimes \hspace{-0.45 mm}\boxtimes$ by erasing it we get a tiling of $n-1$. Also, any of the $H_{n-1}$ such tiling creates by adding $\boxtimes \hspace{-0.45 mm}\boxtimes$ a tiling of $n+1$.</p>
<p>It is easy to check that this process creates all tilings and each tiling exactly once, therefore
$$H_{n+1}=H_n+H_{n-1}$$</p>
|
479,851 | <p>I toss a coin a many times, each time noting down the result of the toss. If at any time I have tossed more heads than tails I stop.
I.e. if I get heads on the first toss I stop.
Or if toss T-T-H-H-T-H-H I will stop.
If I decide to only toss the coin at most 2n+1 times, what is the probability that I will get more heads than tails before I have to give up?</p>
| Brian M. Scott | 12,042 | <p>There is no harm in imagining that we toss the coin $2n+1$ times and simply look to see whether at some point we got a success (meaning more heads than tails at that point).</p>
<p>Say that a finite string of H’s and T’s is <em>balanced</em> if it contains an equal number of H’s and T’s, and that it is <em>completely unbalanced</em> if it has no balanced initial segment. Let $U$ be the set of completely unbalanced strings of length $2n+2$; the lemma in <a href="https://math.stackexchange.com/a/72661/12042">this proof</a> and the remarks preceding it show that $|U|=\binom{2n+2}{n+1}$.</p>
<p>Let $U_T$ be the set of completely unbalanced strings of length $2n+2$ that start with T; these are the ones that have an excess of T’s over H’s in every non-empty initial segment. There is clearly a bijection between $U_T$ and $U_H$, the set of $\sigma\in U$ that start with H, and $U_T\cap U_H=\varnothing$, so $|U_T|=\frac12\binom{2n+2}{n+1}$.</p>
<p>Let $\sigma\in U_T$. If you delete the initial T from $\sigma$, you get a string of length $2n+1$ whose non-empty initial segments all have at least as many T’s as H’s, i.e., strings that are not successes. Conversely, prepending a T to any unsuccessful string of length $2n+1$ produces a string in $U_T$. Thus, exactly $\frac12\binom{2n+2}{n+1}$ of the $2^{n+1}$ strings of length $2n+1$ are failures, and there must therefore be</p>
<p>$$2^{n+1}-\frac12\binom{2n+2}{n+1}$$</p>
<p>successes.</p>
|
1,252,857 | <p>Let $f\colon \mathbb R \to \mathbb R$ be defined by $f(x)= 5x^3+3$. Is it onto?</p>
<p>According to me, if $y=5x^3+3$, then $x = \sqrt[3]{(y-3)/5}$ is not an element of $\mathbb R$ for all $y \in (-\infty,3)$ so all numbers in the codomain $(-\infty,3)$ wont have pre-images.</p>
<p>But many say $5x^3+7$ as an odd degree equation will have at least one real root. Is it onto?</p>
| Noiralef | 224,757 | <p>Why do you say that $x = \sqrt[3]{\frac{y-3}5}$ is not an element of $\mathbb R$ for all $y$?</p>
<p>For every $b \in \mathbb R$ there is exactly one $a \in \mathbb R$ such that $a^3 = b$, we call this $a = \sqrt[3]{b}$.
Maybe you are confused because for $b < 0$ there is no $a \in \mathbb R$ such that $a^2 = b$, in other words we can't define $\sqrt b$.
But the situation is different for the third root.</p>
|
2,751,300 | <blockquote>
<p>Figure out with distinct real numbers the system of equations.</p>
<p><span class="math-container">$$\frac{xy}{x-y} = \frac{1}{30}$$</span>
<span class="math-container">$$\frac{x^2y^2}{x^2+y^2} = \frac{1}{2018}$$</span></p>
</blockquote>
<p>I multiplied x-y both side on the first equation and square on both side, and I stucked.</p>
<p>Help me...</p>
| g.kov | 122,782 | <p>Since $x,y\ne0$,
we can use $u=1/x$, $v=1/y$:
\begin{align}
v-u &= 30
\tag{1}\label{1}
,\\
v^2+u^2 &= 2018
\tag{2}\label{2}
.
\end{align} </p>
<p>\begin{align}
\eqref{2}-\eqref{1}^2:\quad
2uv&=1118
,\\
v^2+u^2+2uv &= 2018+1118
,\\
(u+v)^2&=56^2
,\\
u+v&=\pm56
\tag{3}\label{3}
.
\end{align}</p>
<p>Combination of \eqref{3} with \eqref{1}
provides two cases:</p>
<p>Case 1.</p>
<p>\begin{align}
v-u &= 30
,\\
v+u&=56
,\\
u&=13
\quad v=43
,\\
x&=\frac1{13},\quad y=\frac1{43}
.
\end{align} </p>
<p>Case 2.</p>
<p>\begin{align}
v-u &= 30
,\\
v+u&=-56
,\\
u&=-43
\quad v=-13
,\\
x&=-\frac1{43},\quad y=-\frac1{13}
.
\end{align} </p>
|
18,136 | <p>I introduced the hypercube (to undergraduate students in the U.S.) in the
context of generalizations of the Platonic solids, explained its
structure, showed it rotating.
I mentioned <a href="https://matheducators.stackexchange.com/a/1824/511">Alicia Stott</a>, who discovered the <span class="math-container">$6$</span> regular
polytopes in <span class="math-container">$\mathbb{R}^4$</span> (discovered after Schläfli).
I sense they largely did not grasp what is the hypercube, let alone
the other regular polytopes.</p>
<p>I'd appreciate hearing of techniques for getting students to
"grok" the fourth dimension.</p>
| Kafein | 13,579 | <p>I would posit it's impossible for most humans to truly grok higher dimensions. I've heard of extremely rare exceptions but I'm going to ignore them here.</p>
<p>Having an intuitive understanding of some fact about geometry requires having a <em>mental image</em> of what is going on. You can't have 4D mental images. Can <em>you</em> mentally "see" a tesseract? Not the common 3D bastardization of it but the real thing?</p>
<p>If anything, trying to represent 4D objects for 3D brains on a 2D medium seems counter-productive. The classical 3D representation of a tesseract is actually misleading. A better way to put it would be to film a cube changing colors over time, or to show 2 cubes while saying that there is a unit distance in the 4th dimension between the "same" vertices of the two cubes.</p>
<p>To me, complete visual representations are just the wrong tool for this. You might sort of get something out of it for the 4th dimension, but it becomes hopeless beyond that.</p>
<p>For inspiration, here's a 3blue1brown video which did an OK job giving intuition about a higher dimension problem, though a completely different one: <a href="https://youtu.be/zwAD6dRSVyI" rel="nofollow noreferrer">https://youtu.be/zwAD6dRSVyI</a></p>
|
18,136 | <p>I introduced the hypercube (to undergraduate students in the U.S.) in the
context of generalizations of the Platonic solids, explained its
structure, showed it rotating.
I mentioned <a href="https://matheducators.stackexchange.com/a/1824/511">Alicia Stott</a>, who discovered the <span class="math-container">$6$</span> regular
polytopes in <span class="math-container">$\mathbb{R}^4$</span> (discovered after Schläfli).
I sense they largely did not grasp what is the hypercube, let alone
the other regular polytopes.</p>
<p>I'd appreciate hearing of techniques for getting students to
"grok" the fourth dimension.</p>
| Aloizio Macedo | 4,817 | <p>My suggestion would be to demystify the concept and try to disassociate it from spatial interpretations at the first approach.</p>
<p>Having <span class="math-container">$n$</span> dimensions is just having <span class="math-container">$n$</span> variables. As one professor I had liked to repeat, a grocery store owner who is trying to maximize gains by selling oranges, apples, bananas and peaches with some constraint in his storage is attempting to solve a <span class="math-container">$4$</span>-dimensional problem without even thinking about it. This is much easier to use for motivation. Then, from there, <em>if it is useful/needed/appreciated</em>, one can introduce spatial visualizations like projections into space/plane etc and geometric structures like the inner product etc.</p>
<p>I've found (personally) that relying on "materializing" higher dimensions as an initial approach often backfires, both for the understanding of students (those who get interested frequently veer into philosophical aspects rather than mathematical ones) and their motivation (those who do not get interested will just act as if it is useless and/or nonsensical and/or obscure).</p>
|
3,886,643 | <p>I have doubts about how to solve this issue about bases in numbering systems:</p>
<p>"Determine the base b in <span class="math-container">$(104)_b = 8285$</span>"</p>
<p>can anybody help me?
Thanks.</p>
| Théophile | 26,091 | <p>Hint: In the number <span class="math-container">$104_b$</span>, the final digit represents the units, the middle digit represents the number of "<span class="math-container">$b$</span>"s, and the leftmost digit represents the number of "<span class="math-container">$b^2$</span>"s. (Compare to base <span class="math-container">$10$</span>, where these would be units, tens, and hundreds.) In other words, you have
<span class="math-container">$$b^2+4 = 8285.$$</span></p>
<p>Now solve for <span class="math-container">$b$</span>.</p>
|
3,140,052 | <p>I found a proof of this theorem in <a href="http://web.mit.edu/18.705/www/13Ed.pdf" rel="nofollow noreferrer">modern algebra</a> 2.30</p>
<blockquote>
<p>Proof: Set <span class="math-container">$S := \{ \text{ ideals } \mathfrak b \mid \mathfrak b ⊃ \mathfrak a \text{ and } \mathfrak b ∌ 1 \}$</span>. Then <span class="math-container">$\mathfrak a ∈ S$</span>, and <span class="math-container">$S$</span> is partially ordered by inclusion. Given a totally ordered subset <span class="math-container">$\{\mathfrak b_λ\}$</span> of <span class="math-container">$S$</span>, set <span class="math-container">$\mathfrak b := \bigcup_\lambda \mathfrak b_λ$</span>. Then <span class="math-container">$\mathfrak b$</span> is clearly an ideal, and <span class="math-container">$1 \notin \mathfrak b$</span>; so <span class="math-container">$\mathfrak b$</span> is an upper bound of <span class="math-container">$\{\mathfrak b_λ\}$</span> in <span class="math-container">$S$</span>. Hence by Zorn’s Lemma, <span class="math-container">$S$</span> has a maximal element, and it is the desired maximal ideal.</p>
</blockquote>
<p>I'm not satisfied with the proof, with follow questions:</p>
<ul>
<li>Why <span class="math-container">$S$</span> is a set, <span class="math-container">$S$</span> may not be a set as Russell's paradox.</li>
<li>How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number <span class="math-container">$(0,1)$</span>, and say that range <span class="math-container">$(0,1)$</span> has a maximal element ?</li>
</ul>
<p>I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.</p>
<h3>Edit 1: still confused about something here</h3>
<p>I'll start from an example:</p>
<blockquote>
<p>Proof: <span class="math-container">$S = \{ 1 - \frac 1 n \mid n \in \mathbb N \}$</span> has a maximal element</p>
<p>Obviously <span class="math-container">$S$</span> is partially ordered, let <span class="math-container">$x = 1 - \Pi_{e \in S} (1 - e) = 1 - \Pi_{n \in \mathbb N} \frac 1 n$</span>, then clearly <span class="math-container">$x \in S$</span>; so <span class="math-container">$x$</span> is an upper bound of <span class="math-container">$S$</span>. Hence by Zorn’s Lemma, <span class="math-container">$S$</span> has a maximal element.</p>
</blockquote>
<p>I just copied the proof word by word, then got an obviously wrong result. Where does my proof go wrong ?</p>
<p>If I'm not allowed to construct <span class="math-container">$x$</span> this way then it is also not allowed to construct <span class="math-container">$\mathfrak b$</span> in the origin proof.</p>
<p>I think the truth is <span class="math-container">$x=1$</span> and <span class="math-container">$x \notin S$</span>, and what about <span class="math-container">$\mathfrak
b = \bigcup_\lambda \mathfrak b_λ$</span> ? Things can go wild with infinite. I'm not satisfied with <span class="math-container">$1 \notin \mathfrak b$</span></p>
| Arthur | 15,500 | <p><span class="math-container">$S$</span> is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).</p>
<p>We cannot apply Zorn's lemma to <span class="math-container">$(0,1)$</span> with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says</p>
<blockquote>
<p>If <span class="math-container">$(P, \geq)$</span> is a partial order such that every totally ordered subset of <span class="math-container">$P$</span> has an upper bound in <span class="math-container">$P$</span>, then <span class="math-container">$P$</span> has a maximal element.</p>
</blockquote>
<p>The interval <span class="math-container">$(0,1)$</span> with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in <span class="math-container">$(0,1)$</span>. So Zorn's lemma cannot be applied.</p>
<p>For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in <span class="math-container">$S$</span> is still an ideal in <span class="math-container">$S$</span>. So any totally ordered subset of <span class="math-container">$S$</span> has an upper bound in <span class="math-container">$S$</span>, and therefore <span class="math-container">$S$</span> must have a maximal element.</p>
|
2,346,921 | <blockquote>
<p>Evaluate the following integral using Laplace transform
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$</p>
</blockquote>
<p>I obtained this partial result
$$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$
and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer. </p>
| Robert Z | 299,698 | <p>Your partial result does not hold. According to <a href="https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_improper_integrals" rel="noreferrer">Laplace transform properties</a>
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx=\cal{L}\left(\frac{\sin bx}{x}\right)(a)=\int_a^{+\infty}
\cal{L}\left(\sin bx\right)(p)dp=\int_a^{+\infty}
\frac{b}{p^2+b^2}\, dp.$$
Can you take it form here?</p>
|
3,138,450 | <p>Let <span class="math-container">$A \subseteq \mathbb{R}$</span> be bounded above and let <span class="math-container">$c \in \mathbb
{R}$</span>. Define the set <span class="math-container">$c + A = \{c + a : a \in A\} $</span></p>
<p>Now since <span class="math-container">$a \leq \sup A , \forall a\in A$</span>. Then <span class="math-container">$a + c \leq \sup A + c $</span>. So <span class="math-container">$A+c$</span> has an upper bound. Now the claim is that <span class="math-container">$\sup(c+A) = c + \sup A.$</span></p>
<p>So now I have to prove that <span class="math-container">$\forall \varepsilon> 0, c+a > c + \sup A - \varepsilon.$</span></p>
<p>How can I proceed? Thanks.</p>
| fleablood | 280,126 | <blockquote>
<p>So now I have to prove that ∀ε>0,c+a>c+supA−ε.</p>
</blockquote>
<p>Well if that <em>is</em> what you have to prove, then you'd have <span class="math-container">$\forall \epsilon > 0, a > \sup A - \epsilon$</span> so <span class="math-container">$c+a > c +\sup A - \epsilon$</span>.</p>
<p>But I think you meant <span class="math-container">$\forall \epsilon >0 \exists a \in A$</span> so <span class="math-container">$a > \sup A - \epsilon$</span>, and so for that <span class="math-container">$a$</span>, <span class="math-container">$c+a\in c+A$</span> and <span class="math-container">$c+a > c+\sup A - \epsilon$</span>.</p>
<p>Basically everything you can say about the set <span class="math-container">$A$</span>, its elements and any bounds, can be said about <span class="math-container">$c + A$</span> by adding <span class="math-container">$c$</span> to all pertainent parts. </p>
|
3,249,391 | <p>By Mathematica, we find <span class="math-container">$$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$</span></p>
<blockquote>
<p>How to find the closed form for general series:
<span class="math-container">$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$</span></p>
</blockquote>
| Robert Z | 299,698 | <p>Note that
<span class="math-container">$$\sum_{k=1}^\infty \frac{(4x)^n}{n^2{{2n}\choose n}}=2\arcsin^2(\sqrt{x}).$$</span>
Hence, for <span class="math-container">$p=3$</span> we have the integral form
<span class="math-container">$$\sum\limits\limits_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\int_{0}^1\frac{2\arcsin^2(\sqrt{x})}{x}\,dx.$$</span>
and you should be able to recover the result <span class="math-container">$\pi^2\ln(2)-\frac{7}{2}\zeta(3)$</span>.</p>
<p>As regards the case <span class="math-container">$p=4$</span>,
<span class="math-container">$$\sum_{n=1}^\infty \frac{4^n}{n^4\binom{2n}{n}}=\int_0^1\frac{1}{t}\int_{x=0}^t\frac{2\arcsin^2(\sqrt{x})}{x}\,dx\,dt$$</span>
which, according to ykcaZ's comment below, leads to
<span class="math-container">$$8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$</span>
that is equal to
<span class="math-container">$$8\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{3}\ln^4(2)+\frac{2\pi^2}{3}\ln^2(2)-\frac{19\pi^4}{360}$$</span>
(see <a href="https://math.stackexchange.com/questions/1640940/tough-definite-integral-int-0-frac-pi2x-ln2-sin-xdx">tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$</a> ).</p>
<p>More generally, for <span class="math-container">$p\geq 2$</span>,
<span class="math-container">$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}
=\frac{(-2)^p}{(p-2)!}\int_0^\frac{\pi}{2} x\ln^{p-2}(\sin x)\,dx.$$</span></p>
<p>Look through the paper <a href="https://www.emis.de/journals/INTEGERS/papers/g27/g27.pdf" rel="nofollow noreferrer">Sums of reciprocals of the central binomial coefficients</a> by R. Sprugnoli for more references. See also <a href="https://math.stackexchange.com/questions/3213038/on-binomial-sums-sum-n-1-infty-frac1nk-binom-2nn-and-log-sine-in">On binomial sums $\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$ and log sine integrals</a></p>
|
4,621,227 | <p>This is a very soft and potentially naive question, but I've always wondered about this seemingly common phenomenon where a theorem has some method of proof which makes the statement easy to prove, but where other methods of proof are incredibly difficult.</p>
<p>For example, proving that every vector space has a basis (this may be a bad example). This is almost always done via an existence proof with Zorn's lemma applied to the poset of linearly independent subsets ordered on set inclusion. However, if one were to suppose there exists a vector space <span class="math-container">$V$</span> with no basis, it seems (to me) that coming up with a contradiction given so few assumptions would be incredibly challenging.</p>
<p>With that said, I had a few questions:</p>
<ol>
<li>Are there any other examples of theorems like this?</li>
<li>Is this phenomenon simply due to the logical structure of the statements themselves, or is it something deeper? Is this something one can quantize in some way? That is, is there any formal way to study the structure of a statement, and determine which method of proof is ideal, and which is not ideal?</li>
<li>With (1) in mind, are there ever any efforts to come up with proofs of the same theorem using multiple methods for the sake of interest?</li>
</ol>
| Thomas | 128,832 | <p>Questions like this always remind me of Hamiltons search for a multiplication in <span class="math-container">$\mathbb{R}^3$</span> which somehow extends the multiplication of real and complex numbers (*). He searched, in vain, for years. And he was, at his time, a really famous and renowned mathematician.</p>
<p>Today it is easy to see that this is not possible. If a vector space <span class="math-container">$V$</span> with odd dimension is a division algebra, then for <span class="math-container">$0\neq a\in V$</span> the map <span class="math-container">$x\mapsto ax$</span> is a linear map which must have a real eigenvalue <span class="math-container">$\lambda$</span> (due to the mean value theorem). If <span class="math-container">$v\neq 0$</span> is an eigenvector we have <span class="math-container">$(a -\lambda) v =0$</span>, so <span class="math-container">$a= \lambda e$</span>. Since <span class="math-container">$a$</span> was chosen arbitrarily <span class="math-container">$V$</span> must be isomorphic to <span class="math-container">$\mathbb{R}$</span>.</p>
<p>What Hamilton was lacking were the concepts and definitions involved in this short proof. There is a lot of power and hidden knowledge and the result of decades of research effort in the definitions we are taught today when we learn mathematics.</p>
<p>((*) Summarized from the introduction to an article by Köcher and Remmert, from the book 'Numbers' by Ebbinghaus, Hermes, Hirzebruch et al., German version, Springer 1983)</p>
|
4,311,731 | <p>I was thinking about solutions of equation <span class="math-container">$x^2=i$</span> . First thought coming to my mind was <span class="math-container">$x=\pm \sqrt{i}$</span> . ( I know it's wrong ) .
Then I thought if we solve this equation like real problem then
<span class="math-container">$|x|=\sqrt i$</span> ( Again wrong ) .</p>
<p>But it got me thinking that we define |z| as distance of complex number z from origin . But what if this distance is not real so let consider complex numbers z whose <span class="math-container">$|z|=i$</span> .</p>
<blockquote>
<p>Does this question make sense ? Does such numbers exist on complex x - y plane ? Do we need to include another dimensions for such numbers ?</p>
</blockquote>
<p>Sorry if this question seems silly and thanks for help !</p>
| mathma | 270,091 | <p><span class="math-container">$\sqrt{i}$</span> can be defined, simply take and <span class="math-container">$8$</span>-th root of unity. You can also compute it explicitely in the following way, by Euler's formula <span class="math-container">$i=e^{i\pi/2}$</span> so a square root of <span class="math-container">$i$</span> is given by <span class="math-container">$\pm e^{i\pi/4}=\pm(\cos(\pi/4)+i\sin(\pi/4))=\pm\frac{\sqrt{2}}{2}(1+i)$</span>. You can check directly that these are the correct values.</p>
<p>Distance in the complex numbers is defined as a real number. I do not know of any generalisation of the distance where complex numbers are allowed, however I expect that such a thing would not match your intuition of distance.</p>
|
3,042,308 | <p>Suppose a matrix <span class="math-container">$A$</span> has eigenvalues 0, 3, 7 and eigenvectors <span class="math-container">$\mathbf{u, v, w,}$</span> respectively. Find the least square minimum length solution for <span class="math-container">$A\mathbf{x} = \mathbf{u+v+w}$</span>.</p>
<p>This was on our engineering math final exam last year and we've tried some techniques about Moore-Penrose pseudoinverse, which didn't seem to work. Can someone help?</p>
| AmbretteOrrisey | 613,228 | <p><span class="math-container">$${1\over2}(2\operatorname{heavi}(x)-\operatorname{kron\delta}(x))(2\operatorname{heavi}(y)+\operatorname{kron\delta}(y)-1)+\operatorname{kron\delta}(x) ,$$</span>where <span class="math-container">$$\operatorname{heavi}$$</span>is the <em>Heaviside</em> function, &<span class="math-container">$$\operatorname{kron\delta}$$</span>the <em>Kronecker</em> <span class="math-container">$\delta$</span> function.</p>
<p>This could be expressed as a <em>limit</em> of continuous functions: <span class="math-container">$$\lim_{a\to\infty}{1\over2}(\tanh(ax)-\exp(-(ax)^2)+1)(\tanh(ay)+\exp(-(ay)^2)+\exp(-(ax)^2) .$$</span>or<span class="math-container">$$\lim_{a\to\infty}{1\over2}(\tanh(ax)-\operatorname{sech}(ax)+1)(\tanh(ay)+\operatorname{sech}(ay))+\operatorname{sech}(ax) .$$</span></p>
|
2,699,536 | <blockquote>
<p>Given a continuous function $f:[0,1]\to\mathbb R$, prove that $$\forall t>0, \frac{1}{t}\cdot\ln\left(\int_{0}^{1}e^{-tf(x)}dx\right)\leq-\min f(x).$$</p>
</blockquote>
<p>I have no idea where to begin. Thought I could use the FTOC to come up with some form of antiderivative but I don't know if this is even the right intuition to approach this.</p>
| Michael Hardy | 11,667 | <p>The inequality says
$$ \int_0^1 e^{-tf(x)} \, dx \leq e^{-t\min_x f(x)}. $$</p>
<p>Since $t>0,$ the quantity $e^{-t\min_x f(x)}$ is the largest value of the function $e^{-tf(x)}$ within $0\le x\le 1.$ So this just says that</p>
<p>$$
\int_0^1 g(x) \,dx \le (1-0)\times \max\{ g(x) : 0\le x\le 1 \}.
$$</p>
|
3,091,162 | <p>Let <span class="math-container">$G_1, G_2$</span> be two groups with at least one nontrivial proper subgroup each.</p>
<p>Let <span class="math-container">$S_1, S_2$</span> be the sets of proper subgroups of, respectively <span class="math-container">$G_1, G_2$</span>.</p>
<p>Suppose there exists a bijective function <span class="math-container">$f: S_1 \rightarrow S_2$</span> such that <span class="math-container">$\forall A\in S_1, f(A)$</span> is isomorphic to <span class="math-container">$A$</span>.</p>
<p>When can I conclude that <span class="math-container">$G_1, G_2$</span> are isomorphic?</p>
<p>I think that, if <span class="math-container">$G_1$</span> and <span class="math-container">$G_2$</span> are finite and abelian we can conclude that they are isomorphic, but I can't prove It.
Moreover, I haven't found any counterexample for nonabelian finite groups.</p>
| verret | 191,246 | <p>There are two pairs of examples of order <span class="math-container">$16$</span>. These are the smallest examples. One of these two pairs is <span class="math-container">$C_4\times C_4$</span> and <span class="math-container">$C_4\rtimes C_4$</span>. For both of these, the complete list of proper subgroups is:</p>
<ul>
<li><p>1 trivial subgroup</p></li>
<li><p>3 subgroups isomorphic to <span class="math-container">$C_2$</span></p></li>
<li><p>6 subgroups isomorphic to <span class="math-container">$C_4$</span></p></li>
<li><p>1 subgroup isomorphic to <span class="math-container">$C_2\times C_2$</span></p></li>
<li><p>3 subgroups isomorphic to <span class="math-container">$C_4\times C_2$</span></p></li>
</ul>
<p>(See <a href="https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups" rel="nofollow noreferrer">https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups</a> for the subgroups of <span class="math-container">$C_4\rtimes C_4$</span>.)</p>
<p>Another easy pair of examples is <span class="math-container">$C_9\times C_3$</span> and <span class="math-container">$C_9\rtimes C_3$</span>.</p>
<p>(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)</p>
|
3,783,186 | <p>I am trying to prove that <span class="math-container">$$2≤\int_{-1}^1 \sqrt{1+x^6} \,dx ≤ 2\sqrt{2} $$</span> I learned that the equation <span class="math-container">$${d\over dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x) $$</span> is true due to Fundamental Theorem of Calculus and Chain Rule, and I was thinking about taking the derivative to all side of the inequality, but I am not sure that it is the correct way to prove this. Can I ask for a help to prove the inequality correctly? Any help would be appreciated! Thanks!</p>
| SarGe | 782,505 | <p><img src="https://i.stack.imgur.com/JhuEW.png" alt="enter image description here" /></p>
<p>[Edited after the <a href="https://math.meta.stackexchange.com/q/32326/782505">meta question</a> and <a href="https://chat.stackexchange.com/transcript/message/55188059#55188059">this</a> conversation.]</p>
<p>Let <span class="math-container">$f(x)=\sqrt{1+x^6}$</span>. It is evident that the argument of the square root <span class="math-container">$1+x^6\ge 1\ \forall\ x\in\mathbb R$</span> and there is only one point of global minima <span class="math-container">$(0, 1) $</span>. Hence, <span class="math-container">$f(x)$</span> is monotically decreasing and increasing for <span class="math-container">$x\le 0$</span> and <span class="math-container">$x\ge 0$</span> respectively.</p>
<blockquote class="spoiler">
<p> <span class="math-container">$$\text{Area}(\square CEFD)\le \text{Area under curve} \le \text{Area}(\square ABDC) \\ \implies 2\le \int_{-1}^{1} \sqrt{1+x^6}\ dx\le 2\sqrt 2$$</span></p>
</blockquote>
|
3,176,593 | <p>I'm having a hard time understanding how to find all solutions of the form <span class="math-container">$a_n = a^{(h)}_n+a_n^{(p)}$</span></p>
<p>I show that <span class="math-container">$a_n=n2^n \to a_n=2(n-1)2^{n-1} +2^n=2^n(n-1+1)=n2^n$</span>.</p>
<p>I can show that <span class="math-container">$a_n^{(h)}$</span> characteristic equation <span class="math-container">$r-2=0 \to a_n^{(h)}=\alpha2^n$</span></p>
<p>But I'm stuck on <span class="math-container">$a_n^{(p)}$</span> characteristic equation <span class="math-container">$C2^n=2C\cdot2^{n-1}+2^n$</span></p>
<p>Simplifies to <span class="math-container">$C \neq C+1$</span>, Looking online I saw that the solution is <span class="math-container">$a_n=c\cdot2^n+n2^n$</span>, but I'm not sure how to get there. </p>
| Minus One-Twelfth | 643,882 | <p>Your homogeneous solution has <span class="math-container">$2^n$</span> in it already. When this happens, for the particular solution part, we cannot just use <span class="math-container">$C2^n$</span> (you have seen what happens if we do). Instead, <em>the rule in this scenario is to modify the guess by <strong>multiplying by <span class="math-container">$\boldsymbol{n}$</span></strong>, i.e. try <span class="math-container">$a_n^{(p)}=C\color{red}{n}\cdot 2^n$</span></em>.</p>
|
632,850 | <blockquote>
<p>Suppose that $f(x)$ is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. It is also
known that $|f''(x)| \le A$ for every $x \in (0,1)$. Prove that
$|f'(x)| \le A/2$ for every $x \in [0,1]$.</p>
</blockquote>
<p>I'll explain what I did so far. First using Rolle's theorem, there is some point $c \in [0,1]$ so $f'(c) = 0$.</p>
<p>EDIT: My first preliminary solution was wrong so I tried something else.
EDIT2: Another revision :\</p>
<p>I define a Taylor series of a second order around the point $1$:
$$ f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2 $$
$$ f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2 $$
$$ |f'(1)| = \frac12 |f''(d_1)| <= \frac12 A $$</p>
<p>Now I develop a Taylor series of a first order for $f'(x)$ around $1$:
$$ f'(x) = f'(1) + f''(d_2)(x-1) $$
$$ |f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2} $$</p>
<p>It looks correct to me, what do you guys think?</p>
<p><strong>Note: I cannot use integrals, because we have not covered them yet.</strong></p>
| Disintegrating By Parts | 112,478 | <p>If $f''$ exists on $[0,1]$ and $|f''(x)| \le A$, then $f'$ is Lipschitz continuous, which implies that $f'$ is absolutely continuous. So $f'(a)-f'(b)=\int_{a}^{b}f''(t)\,dt$ for any $0 \le a, b \le 1$. And, of course, the same is true of $f$ because $f$ is continuously differentiable.</p>
<p>Because $f(0)=0$, one has the following for $0 \le x \le 1$:
$$
f(x) = \int_{0}^{x}f'(t)\,dt = tf'(t)|_{0}^{x}-\int_{0}^{x}tf''(t)\,dt=xf'(x)-\int_{0}^{x}tf''(t)\,dt.
$$
Similarly, because $f(1)=0$, one has the following for $0 \le x \le 1$:
$$
f(x)=\int_{1}^{x}f'(t)\,dt = (t-1)f'(t)|_{1}^{x}-\int_{1}^{x}(t-1)f''(t)\,dt=(x-1)f'(x)-\int_{1}^{x}(t-1)f''(t)\,dt.
$$
Subtracting the second equation from the first gives
$$
0=f'(x)-\int_{0}^{x}tf''(t)\,dt+\int_{1}^{x}(t-1)f''(t)\,dt.
$$
Therefore,
$$
|f'(x)| \le A\left(\int_{0}^{x}t\,dt + \int_{x}^{1}(1-t)\,dt\right)=\frac{A}{2}(x^{2}+(1-x)^{2}).
$$
The maximum of the expression on the right occurs at $x=0$ and $x=1$, with a value of $A/2$. Therefore, $|f'(x)| \le A/2$.</p>
|
4,401,028 | <p>Let <span class="math-container">$p$</span> be an odd prime and <span class="math-container">$ω=e^{2\pi i /p}$</span>. Determine if <span class="math-container">$1-ω$</span> is prime in <span class="math-container">$\mathbb{Z}[ω]=\mathbb{Z}+\mathbb{Z}ω+\mathbb{Z}ω^2+...+\mathbb{Z}ω^{p-2}$</span></p>
<p><strong>My attempt</strong></p>
<p>I have tried using the definition of prime and also tried to show <span class="math-container">$\langle 1-\omega \rangle$</span> is maximal but I end up with more unknowns than equations.</p>
<p>I know the norm of the ideal is <span class="math-container">$N(\langle 1-\omega \rangle)=|1-\omega|^{p-2}$</span>.</p>
| B. Goddard | 362,009 | <p>People sometimes work with "pure functions." So if you are considering <span class="math-container">$0$</span> to be a function, then you could write <span class="math-container">$\sin =0$</span>. It would be a false statement, because the two functions are not the same. The CAS Maple, for instance, will return "<span class="math-container">$\cos$</span>" if you ask for the derivative of "<span class="math-container">$\sin$</span>".</p>
<p>But the equation <span class="math-container">$\sin x = 0$</span> is a conditional equation. It's asking for which <span class="math-container">$x$</span> is the equation true, in which case you need the <span class="math-container">$x$</span>.</p>
|
123,712 | <p>How can I prove that the function
$$ f(x) = \left\{\begin{array}{l l} x &\text{if }x \in \mathbb{Q} \\ -x & \text{if } x \notin \mathbb{Q}
\end{array} \right. $$
is discontinuous for $x \neq 0$, using $\epsilon$'s and $\delta$'s?</p>
<p>I see that is truth. But I cannot prove using only $\epsilon$'s and $\delta$'s.</p>
| David Mitra | 18,986 | <p>Let $x_0\ne 0$ be fixed. Informally, $f$ is continuous at $x_0$ would mean that $f(x)$ can be made as close to $f(x_0)$ as desired by taking $x$ sufficiently close to $x_0$. With the epsilon business, this means that $\color{maroon}{ \text{for every}}$ $\epsilon>0$, there is a $\delta>0$ such that $$\tag{1}
\underbrace{|f(x_0)-f(x)|<\epsilon}_{f(x)\text{ is close to }f(x_0)}
\quad\text{ whenever }\quad
\underbrace{|x-x_0|<\delta\phantom{f}}_{x\text{ is close to }x_0}.$$</p>
<p>If $f$ were not continuous at $x_0$, this would mean that there $\color{maroon}{ \text{is some}}$ $\epsilon>0$ such that no matter what $\delta$ you choose, equation $(1)$ is not true. Informally, this means that no matter how close you take $x$ to $x_0$, you are not guaranteed that $f(x)$ is close to $f(x_0)$ and, in fact, you can find an $x$ as close to $x_0$ as you wish with $f(x)$ "far" away from $f(x_0)$. </p>
<p>So you need to show the following is true: </p>
<p><b>S:</b> There is some $\epsilon>0$ such that for every $\delta>0$, there is a $x_1$ such that both $|x_1- x_0|<\delta$ and $|f(x_1)-f(x_0)|\ge \epsilon$.</p>
<p>Well, how to accomplish this? Let's consider the graph of our function (as far as we can; of course, it's impossible to actually draw the graph of $f$ but we'll use our imaginations) and try to get some intuition as to what the required $\epsilon$ should be.
The graph of your function resembles the graphs of the lines $y=\pm x$; or, if you will, an X shape. For rational values of $x$, the point $\bigl(x,f(x)\bigr)$ is on the graph of $y=x$; and for irrational values of $x$, the point $\bigl(x,f(x)\bigr)$ is on the graph of $y=-x$.</p>
<p>I suggest you draw the picture now (when I have time, I'll provide one though).</p>
<p>Now if you take any open interval $I$ containing $x_0$, you can find both a rational number and an irrational number in $I$. (Why? This is the gist of the whole argument, so it's worthwhile that you convince yourself that you can.) Thus, you can find a point $x_1$ in $I$ that is irrational if $x_0$ is rational and which is rational if $x_0$ is irrational. It will follow then that $f(x_0)$ and $f(x_1)$ are far apart.
In fact, they are approximately $2|x_0|$ away from each other when $I$ has small length, since, if you'll allow a sloppy and vague statement, $f(x_0)$ and $f(x_1)$ are on different arms of the X.</p>
<p>This leads us to suspect that $|x_0|$ would serve for the $\epsilon$ needed in statement <b>S</b>.
And
now it's a matter to formalize things. We need to show that statement <b>S</b> holds.</p>
<p>Let $0 < \epsilon < |x_0|$. </p>
<p>Let $\delta>0$ be arbitrary.</p>
<p>We need to show that there is an $x_1$ with $|x_0-x_1|<\delta$ such that
$|f(x_0)-f(x_1)|\ge\epsilon $.</p>
<p>The preceding discussion points to a way towards finding $x_1$. Choose $x_1$ to be an irrational number with $|x_0-x_1|<\delta$ if $x_0$ is rational. If $x_0$ is irrational, choose $x_1$ to be an rational number with $|x_0-x_1|<\delta$.</p>
<p>Now we show this choice of $x_1$ "works":</p>
<p>Either </p>
<p>$\ \ \ \ \ f(x_0)=x_0$ and $f(x_1)=-x_1$ </p>
<p>or</p>
<p>$\ \ \ \ \ f(x_0)=-x_0$ and $f(x_1)= x_1$. </p>
<p>In either case $|f(x_0)-f(x_1)| = |x_0|+|x_1|\ge |x_0|>\epsilon. $</p>
<p>And that's it...</p>
|
1,591,197 | <p>Let $ A \trianglelefteq G $ and $ B \trianglelefteq A $ a Sylow normal subgroup of $ A $. My textbook says then that $ B \trianglelefteq G $.</p>
<p>I don’t understand why that is.</p>
| Tsemo Aristide | 280,301 | <p>Since $B$ is a normal sylow $p$-subgroup there is one $p$-sylow subgroup in $A$ since all the sylow $p$-subgroups are conjugated. Let $g\in G, gBg^{-1}\subset A$ since $A$ is normal in $G$ and is a $p$-sylow subgroup since it has the same cardinal than $B$ thus is $B$ since $B$ is the unique $p$-sylow subgroup of $A$,</p>
|
416,407 | <blockquote>
<p>What examples are there of habitual but unnecessary uses of the axiom of
choice, in any area of mathematics except topology?</p>
</blockquote>
<p>I'm interested in standard proofs that use the axiom of choice, but where
choice can be eliminated via some judicious and maybe not quite obvious
rephrasing. I'm less interested in proofs that were originally proved
using choice and where it took some significant new idea to remove the
dependence on choice.</p>
<p>I exclude topology because I already know lots of topological examples. For
instance, Andrej Bauer's <a href="https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/" rel="noreferrer">Five stages of accepting constructive
mathematics</a>
gives choicey and choice-free proofs of a standard result (Theorem 1.4):
every open cover of a compact metric space has a Lebesgue number. Todd
Trimble told me about some other topological examples, e.g. a compact
subspace of a Hausdorff space is closed, or the product of two compact
spaces is compact. There are more besides.</p>
<p>One example per answer, please. And please sketch both the habitual proof
using choice and the alternative proof that doesn't use choice.</p>
<p>To show what I'm looking for, here's an example taken from that paper of Andrej Bauer. It would qualify as an answer except that it comes from
topology.</p>
<p><strong>Statement</strong> Every open cover <span class="math-container">$\mathcal{U}$</span> of a compact metric space
<span class="math-container">$X$</span> has a Lebesgue number <span class="math-container">$\varepsilon$</span> (meaning that for all <span class="math-container">$x \in X$</span>, the
ball <span class="math-container">$B(x, \varepsilon)$</span> is contained in some member of <span class="math-container">$\mathcal{U}$</span>).</p>
<p><strong>Habitual proof using choice</strong> For each <span class="math-container">$x \in X$</span>, choose some
<span class="math-container">$\varepsilon_x > 0$</span> such that <span class="math-container">$B(x, 2\varepsilon_x)$</span> is contained in some
member of <span class="math-container">$\mathcal{U}$</span>. Then <span class="math-container">$\{B(x, \varepsilon_x): x \in X\}$</span> is a
cover of <span class="math-container">$X$</span>, so it has a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_{x_1}),
\ldots, B(x_n, \varepsilon_{x_n})\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_{x_i}$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
<p><strong>Proof without choice</strong> Consider the set of balls <span class="math-container">$B(x, \varepsilon)$</span>
such that <span class="math-container">$x \in X$</span>, <span class="math-container">$\varepsilon > 0$</span> and <span class="math-container">$B(x, 2\varepsilon)$</span> is
contained in some member of <span class="math-container">$\mathcal{U}$</span>. This set covers <span class="math-container">$X$</span>, so it has
a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_1), \ldots, B(x_n,
\varepsilon_n)\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_i$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
| Tom Leinster | 586 | <p>Turning ZM's <a href="https://mathoverflow.net/questions/416407/unnecessary-uses-of-the-axiom-of-choice#comment1068341_416407">comment</a> into an answer: Zorn's lemma is sometimes invoked to show that the maximal atlas (in the definition of differentiable manifolds) exists, but it is <a href="https://math.stackexchange.com/questions/66554">unnecessary</a>.</p>
<p>(This question is community wiki.)</p>
|
416,407 | <blockquote>
<p>What examples are there of habitual but unnecessary uses of the axiom of
choice, in any area of mathematics except topology?</p>
</blockquote>
<p>I'm interested in standard proofs that use the axiom of choice, but where
choice can be eliminated via some judicious and maybe not quite obvious
rephrasing. I'm less interested in proofs that were originally proved
using choice and where it took some significant new idea to remove the
dependence on choice.</p>
<p>I exclude topology because I already know lots of topological examples. For
instance, Andrej Bauer's <a href="https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/" rel="noreferrer">Five stages of accepting constructive
mathematics</a>
gives choicey and choice-free proofs of a standard result (Theorem 1.4):
every open cover of a compact metric space has a Lebesgue number. Todd
Trimble told me about some other topological examples, e.g. a compact
subspace of a Hausdorff space is closed, or the product of two compact
spaces is compact. There are more besides.</p>
<p>One example per answer, please. And please sketch both the habitual proof
using choice and the alternative proof that doesn't use choice.</p>
<p>To show what I'm looking for, here's an example taken from that paper of Andrej Bauer. It would qualify as an answer except that it comes from
topology.</p>
<p><strong>Statement</strong> Every open cover <span class="math-container">$\mathcal{U}$</span> of a compact metric space
<span class="math-container">$X$</span> has a Lebesgue number <span class="math-container">$\varepsilon$</span> (meaning that for all <span class="math-container">$x \in X$</span>, the
ball <span class="math-container">$B(x, \varepsilon)$</span> is contained in some member of <span class="math-container">$\mathcal{U}$</span>).</p>
<p><strong>Habitual proof using choice</strong> For each <span class="math-container">$x \in X$</span>, choose some
<span class="math-container">$\varepsilon_x > 0$</span> such that <span class="math-container">$B(x, 2\varepsilon_x)$</span> is contained in some
member of <span class="math-container">$\mathcal{U}$</span>. Then <span class="math-container">$\{B(x, \varepsilon_x): x \in X\}$</span> is a
cover of <span class="math-container">$X$</span>, so it has a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_{x_1}),
\ldots, B(x_n, \varepsilon_{x_n})\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_{x_i}$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
<p><strong>Proof without choice</strong> Consider the set of balls <span class="math-container">$B(x, \varepsilon)$</span>
such that <span class="math-container">$x \in X$</span>, <span class="math-container">$\varepsilon > 0$</span> and <span class="math-container">$B(x, 2\varepsilon)$</span> is
contained in some member of <span class="math-container">$\mathcal{U}$</span>. This set covers <span class="math-container">$X$</span>, so it has
a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_1), \ldots, B(x_n,
\varepsilon_n)\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_i$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
| Tim Campion | 2,362 | <p>The existence of a Haar measure on any locally compact group was first proven by Weil using the axiom of choice. Cartan later supplied a choice-free proof.</p>
<p>Because the Haar measure is unique up to a scalar factor, this is an example where it seems "obvious" that choice really shouldn't be necessary.</p>
<p>If anybody wants to edit to sketch one or both of the proofs, that would be most welcome!</p>
|
416,407 | <blockquote>
<p>What examples are there of habitual but unnecessary uses of the axiom of
choice, in any area of mathematics except topology?</p>
</blockquote>
<p>I'm interested in standard proofs that use the axiom of choice, but where
choice can be eliminated via some judicious and maybe not quite obvious
rephrasing. I'm less interested in proofs that were originally proved
using choice and where it took some significant new idea to remove the
dependence on choice.</p>
<p>I exclude topology because I already know lots of topological examples. For
instance, Andrej Bauer's <a href="https://www.ams.org/journals/bull/2017-54-03/S0273-0979-2016-01556-4/" rel="noreferrer">Five stages of accepting constructive
mathematics</a>
gives choicey and choice-free proofs of a standard result (Theorem 1.4):
every open cover of a compact metric space has a Lebesgue number. Todd
Trimble told me about some other topological examples, e.g. a compact
subspace of a Hausdorff space is closed, or the product of two compact
spaces is compact. There are more besides.</p>
<p>One example per answer, please. And please sketch both the habitual proof
using choice and the alternative proof that doesn't use choice.</p>
<p>To show what I'm looking for, here's an example taken from that paper of Andrej Bauer. It would qualify as an answer except that it comes from
topology.</p>
<p><strong>Statement</strong> Every open cover <span class="math-container">$\mathcal{U}$</span> of a compact metric space
<span class="math-container">$X$</span> has a Lebesgue number <span class="math-container">$\varepsilon$</span> (meaning that for all <span class="math-container">$x \in X$</span>, the
ball <span class="math-container">$B(x, \varepsilon)$</span> is contained in some member of <span class="math-container">$\mathcal{U}$</span>).</p>
<p><strong>Habitual proof using choice</strong> For each <span class="math-container">$x \in X$</span>, choose some
<span class="math-container">$\varepsilon_x > 0$</span> such that <span class="math-container">$B(x, 2\varepsilon_x)$</span> is contained in some
member of <span class="math-container">$\mathcal{U}$</span>. Then <span class="math-container">$\{B(x, \varepsilon_x): x \in X\}$</span> is a
cover of <span class="math-container">$X$</span>, so it has a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_{x_1}),
\ldots, B(x_n, \varepsilon_{x_n})\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_{x_i}$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
<p><strong>Proof without choice</strong> Consider the set of balls <span class="math-container">$B(x, \varepsilon)$</span>
such that <span class="math-container">$x \in X$</span>, <span class="math-container">$\varepsilon > 0$</span> and <span class="math-container">$B(x, 2\varepsilon)$</span> is
contained in some member of <span class="math-container">$\mathcal{U}$</span>. This set covers <span class="math-container">$X$</span>, so it has
a finite subcover <span class="math-container">$\{B(x_1, \varepsilon_1), \ldots, B(x_n,
\varepsilon_n)\}$</span>. Put <span class="math-container">$\varepsilon = \min_i
\varepsilon_i$</span> and check that <span class="math-container">$\varepsilon$</span> is a Lebesgue number.</p>
| Kameryn Williams | 64,676 | <p>A good number of theorems in Ramsey theory and related areas are what logicians call <span class="math-container">$\Pi^1_2$</span> statements—those of the form "for every set of integers <span class="math-container">$X$</span> there is a set of integers <span class="math-container">$Y$</span> satisfying some property which only quantifies over integers". Often, the easiest proofs of these results use AC, e.g. in the guise of using a nonprincipal ultrafilter or using nonstandard methods. But a consequence of Shoenfield's absoluteness theorem is that no theorem of this form can require choice for its proof.</p>
<p>A good example of this is Hindman's theorem (any finite coloring of <span class="math-container">$\mathbb N$</span> admits an infinite set whose set of finite sums is monochromatic). There's a very nice, quick proof through idempotent ultrafilters, which of course need (a fragment of) AC. There is an elementary proof, but it is much more involved and intricate, requiring you to do all the bookkeeping details by hand.</p>
|
1,481,106 | <p>I am having a difficult time solving this problem. I have tried this several different ways, and I get a different result, none of which is correct, every time. I've derived an answer geometrically and cannot replicate it with a double integral.</p>
<p>Here's the problem: Use a double integral to find the area between two circles $$x^2+y^2=4$$ and $$(x−1)^2+y^2=4.$$</p>
<p>Here is how I have tried to go about this problem:</p>
<p>First, I graphed it to get a good idea visually of what I was doing.
<a href="https://i.stack.imgur.com/i7cNa.png" rel="nofollow noreferrer">Here's the graph I scribbled on.</a>
The region I'm interested is where these two circles overlap. This region can easily be divided into two separate areas. There are clearly a number of ways to go about solving this...but the one I opted for is to find the shaded region. The bounds for $x$ in this case are between $D$ and $C$. D can be found by setting $C_1=C_2$, and $x$ turns out to be $\frac{1}{2}$. On the right, $x$ is where $C_1(y)=0$, $x=\pm2$, so $x=2$ at point $C$. $y$ is greater than $B_y$ and less than $A_y$, which are also found where $C_1=C_2$, and $y$ turns out to be $\pm\sqrt{\frac{15}{4}}$. So far so good. Now I know my limits of integration. But here's what I don't understand. What am I actually integrating? $x$ has constant bounds, and $y$ does not, and looking at other double integral problems, that would lead me to believe that I should integrate $y$ first as a function of $x$, evaluate it at its bounds, and then integrate $x$ and evaluate it at its bounds giving me half the area I am looking for. However, when I try to do this, I get utter nonsense for an answer, or I get lost trying to set up the problem.</p>
<p>I could really use the help, I've spent entirely too much time trying to puzzle through this. Thank you in advance!</p>
<p>P.s. I determined the area geometrically using a CAD program to calculate the area, and it should be approximately $8.46$.</p>
| SchrodingersCat | 278,967 | <p>Surface area
$$=\int \int_S dS$$
$$=\int \int_S \frac{dx \,\ dy}{\hat n\cdot\hat k}$$
since the surface considered here is $z=0$
$$=\int^{y=2}_{y=-2} \int^{x=\sqrt{4-y^2}}_{x=1-\sqrt{4-y^2}} dx \,\ dy$$
$$=\int^{y=2}_{y=-2} (2\sqrt{4-y^2}-1)dy$$
$$= 4(\pi - 1)$$</p>
|
1,481,106 | <p>I am having a difficult time solving this problem. I have tried this several different ways, and I get a different result, none of which is correct, every time. I've derived an answer geometrically and cannot replicate it with a double integral.</p>
<p>Here's the problem: Use a double integral to find the area between two circles $$x^2+y^2=4$$ and $$(x−1)^2+y^2=4.$$</p>
<p>Here is how I have tried to go about this problem:</p>
<p>First, I graphed it to get a good idea visually of what I was doing.
<a href="https://i.stack.imgur.com/i7cNa.png" rel="nofollow noreferrer">Here's the graph I scribbled on.</a>
The region I'm interested is where these two circles overlap. This region can easily be divided into two separate areas. There are clearly a number of ways to go about solving this...but the one I opted for is to find the shaded region. The bounds for $x$ in this case are between $D$ and $C$. D can be found by setting $C_1=C_2$, and $x$ turns out to be $\frac{1}{2}$. On the right, $x$ is where $C_1(y)=0$, $x=\pm2$, so $x=2$ at point $C$. $y$ is greater than $B_y$ and less than $A_y$, which are also found where $C_1=C_2$, and $y$ turns out to be $\pm\sqrt{\frac{15}{4}}$. So far so good. Now I know my limits of integration. But here's what I don't understand. What am I actually integrating? $x$ has constant bounds, and $y$ does not, and looking at other double integral problems, that would lead me to believe that I should integrate $y$ first as a function of $x$, evaluate it at its bounds, and then integrate $x$ and evaluate it at its bounds giving me half the area I am looking for. However, when I try to do this, I get utter nonsense for an answer, or I get lost trying to set up the problem.</p>
<p>I could really use the help, I've spent entirely too much time trying to puzzle through this. Thank you in advance!</p>
<p>P.s. I determined the area geometrically using a CAD program to calculate the area, and it should be approximately $8.46$.</p>
| amd | 265,466 | <p>As a double integral, this is $I=\int_R dA$, where $R$ is the region given by the intersection of the two disks. This integral can be evaluated as an iterated integral in several ways. Picking up where you left off, we can integrate with respect to $y$ first, then with respect to $x$, i.e., $$I=2\int_\frac12^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\;dx.$$ You could take further advantage of symmetry by changing the lower bound of $y$ to $0$ and multiplying by $2$, but that doesn’t really make things any simpler. </p>
<p>You could instead integrate with respect to $x$ first:$$I=4\int_0^{\frac{\sqrt{15}}2}\int_\frac12^{\sqrt{4-y^2}} dx\;dy.$$ The key thing in both cases is that the bounds of the inner integral will be variable. I often find it helpful to draw a typical line segment for the inner integral to make sure that I get the bounds right. For example, in this second integral, for each value of $y$, the integral runs over a line segment parallel to the $x$-axis that runs from $x=\frac12$ out to the edge of the left-hand disk. </p>
<p>If you don’t want to deal with iterated integrals at all, this problem is a good candidate for Green’s theorem: $I=\int_R dA=\int_{\partial R}y\;dx$. You can often make the line integral simpler without changing its value by adding the differential of some function to the integrand. If you parametrize the bounding arcs as $\langle 2\cos t,2\sin t\rangle$ and $\langle 1-2\cos t,-2\sin t\rangle$, respectively, and use the differential form $\frac12(x\;dy-y\;dx)$ for the line integral, the integrals end up being very simple to evaluate:$$I=\int_{-\arccos\frac14}^{\arccos\frac14}2\;dt+\int_{-\arccos\frac14}^{\arccos\frac14}2-\cos t\;dt=\int_{-\arccos\frac14}^{\arccos\frac14}4-\cos t\;dt.$$</p>
|
1,136,486 | <p>Can anyone help me find the first digit of $2015^{2015}$?</p>
<p>It is easy to find the last digit but I have no idea for the first digit.</p>
| Empy2 | 81,790 | <p>Take the base-ten logarithm.<br>
By the log laws, that will equal $2015\log2015$. Suppose that is $x+y$, where $x$ is a whole number and $0<y<1$.
The number has $x+1$ digits.<br>
$y$ tells you the first digit. Or rather, $10^y$ starts with the same digits $2015^{2015}$ does.</p>
|
1,136,486 | <p>Can anyone help me find the first digit of $2015^{2015}$?</p>
<p>It is easy to find the last digit but I have no idea for the first digit.</p>
| Mario Carneiro | 50,776 | <p>Now, you didn't say "by hand" as these calculations often do, but I'll assume that at least you're from before 1950 so that solutions like just <a href="http://www.wolframalpha.com/input/?i=2015%5E2015">asking Wolfram Alpha</a> about a mere 8000-digit number are beyond you.</p>
<p>We can calculate $\log_{10}(2015^{2015})=2015(\log_{10}(2.015)+3)$, and consulting my trusty <a href="http://www.sosmath.com/tables/logtable/logtable.html">tables</a> I find that $$\log_{10}(2.015)\approx\frac{\log_{10}(2.01)+\log_{10}(2.02)}2\approx\frac{0.30320+0.30535}2\approx0.30428,$$ so $\log_{10}(2015^{2015})\approx6658.1$. (I need to do this calculation well enough to show it is between $6658$ and $6658+\log_{10} 2\approx6658.3$ to prove the claim.)</p>
<p>Then, $6658\le\log_{10}(2015^{2015})<6658+\log_{10} 2$ gives $10^{6658}\le2015^{2015}<2\cdot 10^{6658}$, so the first digit is $1$.</p>
|
674,621 | <p>I am trying to figure out what the three possibilities of $z$ are such that </p>
<p>$$
z^3=i
$$</p>
<p>but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.</p>
| Dan | 1,374 | <p>Taking the absolute value of both sides: $|z^3| = |i|$, gives $|z| = 1$. So, $z = \cos (\theta) + i \sin (\theta)$ for some real $\theta$.</p>
<p>Using De Moivre's formula gives $z^3 = \cos(3\theta) + i \sin(3\theta)$. Given that $z^3 = i = 0 + 1i$, this means that $\cos(3\theta) = 0$ and $\sin(3\theta) = 1$. Solving this system gives $3\theta = \frac{\pi}{2} + 2\pi n$, or $\theta = \frac{\pi}{6} + \frac{2 \pi n}{3}$, for any $n \in \mathbb{Z}$.</p>
<p>Plugging in a few values for $n$ gives:</p>
<ul>
<li>$n = 0$ → $\theta = \frac{\pi}{6}$ → $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i$</li>
<li>$n = 1$ → $\theta = \frac{5\pi}{6}$ → $z = \frac{-\sqrt{3}}{2} + \frac{1}{2} i$ </li>
<li>$n = 2$ → $\theta = \frac{3\pi}{2}$ → $z = -i$</li>
</ul>
<p>And we can stop there because this is a polynomial equation of degree 3, and the Fundamental Theorem of Algebra guarantees that it has at most 3 distinct roots. The solution set is thus $z \in \{ \frac{\sqrt{3}}{2} + \frac{1}{2} i, \frac{-\sqrt{3}}{2} + \frac{1}{2} i, -i \}$.</p>
|
674,621 | <p>I am trying to figure out what the three possibilities of $z$ are such that </p>
<p>$$
z^3=i
$$</p>
<p>but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.</p>
| IQ WANTER | 511,144 | <p>
$z^3=i$ $⟹ z=\sqrt [3] {i}$<br />
</p>
<p>We know,
$\sqrt [3] {i} =a+bi$</p>
<p>$⟹ i=(a+bi)^3 = a^3-3ab^2+3a^2bi-b^3i= a^3-3ab^2+(3a^2b-b^3)i$</p>
<p>From this we can say, </p>
<p>$a^3-3ab^2=0$________________(i)</p>
<p>$3a^2b-b^3=1$_______________(ii)</p>
<p>From (i) we get,</p>
<p>$a(a^2-3b^2)=0$<br /> </p>
<p>So, <strong>$a=0$</strong></p>
<p>Or,
$a^2-3b^2=0$ $⟹ a=±b\sqrt {3}$<br /><br>
</p>
By setting the value of $a$ in (ii) we get,</p>
<p>
If $a=0$, $ b=-1$<br />
</p>
<p></p>
<p>
If $a=±b\sqrt {3}$, $ b=1/2$<br />
</p>
<p></p>
<p>Then, $a=±b\sqrt {3}=±(1/2)\sqrt {3}=±\sqrt {3}/2$</p>
<p>So, $(a,b)=(0,-1),(±\sqrt {3}/2,1/2)$</p>
<p>Now, there are 3 values of $z$. </p>
<p>(1) $\sqrt [3] {i} =a+bi= 0+(-1)i= -i$</p>
<p>(2) $\sqrt [3] {i} =a+bi= \dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$</p>
<p>(3) $\sqrt [3] {i} =a+bi= -\dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$</p>
|
1,601,427 | <blockquote>
<p>Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$</p>
</blockquote>
<p>It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?</p>
| Mufasa | 49,003 | <p>I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...</p>
<p>Using AM-GM we can show that:
$$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$
$$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$
We can also show that:
$$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$
We can therefore infer that:
$$a^2+b^2+c^2\ge ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{4}$$
Using these facts we can say that:
$$a^2+b^2+c^2=3\sqrt[3]{a^2b^2c^2}+\delta_1\text{ where }\delta_1\ge0\tag{5}$$
$$ab+bc+ca=3\sqrt[3]{a^2b^2c^2}+\delta_2\text{ where }\delta_2\ge0\tag{6}$$
We can then use (4) to further infer that:
$$\delta_1\ge\delta_2\ge0$$
Finaly we can state that:
$$\begin{align}
a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&=6\sqrt[3]{a^2b^2c^2}+\delta_1\\
&\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\
&\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\
&\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\
\end{align}$$
I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.</p>
|
615,396 | <p>What is considered a good format for writing problem sets in mathematics? Are there any good examples of problem sets that are well-written and formatted that you can show me?</p>
| Tyler Clark | 3,791 | <p>I like to write my proofs in LaTeX. I have gotten some nice templates offline for this and modified them to my preference. I have also created a commands file so I am not constantly writing out long commands (i.e. \mathbb{R} vs. \bb{R}). I can send you the .tex files for these if you are interested. </p>
<p>I also like to include any definitions or theorem I will be using (in boxes after the question and before the solution). That way, if I look back at the problems and I don't remember what some definition is or what theorem makes a certain fact true in the proof, I have it right there and I don't have to hunt for it.</p>
|
767,474 | <blockquote>
<p>If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$</p>
</blockquote>
| evil999man | 102,285 | <p>By Cauchy Schwarz Lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geqslant \frac{(a+b+c)^2}{2(a+b+c)}$$</p>
<p>Also By $HM\leqslant AM$</p>
<p>$$\frac 3 {\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}}\leqslant \frac{2(a+b+c)}{3}$$</p>
<p>$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\geqslant\frac9 {2(a+b+c)}$$</p>
<p>Add both of them. Then apply $AM\geqslant GM$ on RHS</p>
|
189,218 | <p>I have raw images that I can import and view with ImageJ. They are the output of a program which I don't have the source code for, so I'm stuck with the output format.</p>
<p>They are imported as follows with ImageJ:</p>
<blockquote>
<p>Image type: 16 bit Unsigned</p>
<p>Width: 320</p>
<p>Height: 25600</p>
<p>Offset to fist image: 0 bytes</p>
<p>Number of images: 1</p>
<p>Gap Between images: 0 bytes</p>
<p>White is zero unchecked</p>
<p>Little-endian byte order unchecked</p>
<p>Open all files in folder unchecked</p>
<p>use virtual stack unchecked.</p>
</blockquote>
<p>Essentially each RAW file is a stack of 100 320 x 256 images.</p>
<p>When I try importing via <code>Import[]</code> in <em>Mathematica</em>, I get</p>
<blockquote>
<pre><code>LibraryFunction::rterr: An error with return code -2 was encountered evaluating the function ReadImageRAW.
</code></pre>
</blockquote>
<blockquote>
<pre><code>Import::fmterr: Cannot import data as Raw format.
</code></pre>
</blockquote>
<p>I can't seem to find any info on the first error message.</p>
| yode | 21,532 | <p>After some dig in version 13, I found a package that can do this indeed. And I have tried Canon and Sony, so far so good.</p>
<pre><code><< ImageFileTools`
</code></pre>
<p>Here is a function <code>ImageFileTools`Raw`RawGet</code>, and its usage is:</p>
<pre><code>? ImageFileTools`Raw`RawGet
</code></pre>
<p><a href="https://i.stack.imgur.com/F4F8N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F4F8N.png" alt="enter image description here" /></a></p>
<h1>Usage</h1>
<p><a href="https://i.stack.imgur.com/6SZ3U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6SZ3U.png" alt="enter image description here" /></a></p>
|
308,436 | <p>$P$ is probability. We have: $P(A) \ge \frac{2}{3}$, $P(B) \ge \frac{2}{3}$, $P(C) \ge \frac{2}{3}$ and $P(A \cap B \cap C)=0$. We have to find $P(A)$.
How to do it? Of course we have $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B)- P(A \cap C)- P(B \cap C)$, but what next? Please, help me.</p>
| dtldarek | 26,306 | <p>First, observe that </p>
<p>$$P(X \cap Y) = 0 \text{ implies } P(X \cap Y^c) = P(X). \tag{1}$$</p>
<p>Then, bound $P(B \cap C)$ by
$$ P(B \cap C) = P(B) + P(C) - P(B \cup C) \geq 2\frac{2}{3}-1 = \frac{1}{3}.$$</p>
<p>Finally, apply $(1)$ to $P(A\cap (B \cap C)) = 0$ to get
$$P(A) = P(A \cap (B \cap C)^c) \leq \frac{2}{3}.$$</p>
<p>Concluding, $P(A) = \frac{2}{3}$.</p>
|
3,466,707 | <p>In that problem</p>
<p><span class="math-container">$$\lim\limits_{x \to \infty} \left(1-\frac{1}{x}\right)^\left(e^x\right)$$</span></p>
<p>I use <span class="math-container">$\ln$</span>, then it gave me <span class="math-container">$0\times\infty$</span> indeterminate, then I use L'Hospital Rule but I cannot reach the any answer.</p>
<p>Sorry for grammatical mistakes, my native language is not English. </p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>Determine first the limit of the log with an <em>asymptotic equivalent</em>:
<span class="math-container">$$\mathrm e^x\ln\Bigl(1-\frac 1x\Bigr)\sim_{x\to\infty} \mathrm e^x\cdot\Bigl(-\frac1x\Bigr).$$</span></p>
|
878,816 | <p>Solve inequality: $-5 < \frac{1}{x} < 0$ </p>
<p>I thought about how I can solve this. If I multiply all sides by $x$ I'm afraid I'm removing the answer, cause $\frac{x}{x}=1$. And when $x$ 'leaves' the inequality I'm left with no letter. </p>
<p>How do I get just $x$ in the middle without adding $x$ to other sides or removing $x$?<br>
I then saw that: $\frac{1}{x} = -x$. So can I multiply all sides with $-1$. This also changes the signs. So I'm left with: $5> x > 0$. </p>
<p>Is this correct? If not what did I do wrong?</p>
| J126 | 2,838 | <p>Since you know that $x\neq 0$ you can multiply by $x$. Just remember that $x<0$. So, multiplying by $x$ reverses the inequalities.</p>
<p>Then you get $-5x>1>0$. This gives $-5x>1$. Now divide by $-5$.</p>
|
1,360,658 | <p>prove: $$\frac{1-\cos 2\theta}{1-\cos\theta}=2\cos\theta-2$$</p>
<p>I'm thinking that there will be something to square in this? Because I notice that the $LHS$ looks like the half-angle identity....</p>
<p>Edit: I am so sorry guys, my grave mistake, the expression should have been equal to 2 instead like,</p>
<p>$$\frac{1-\cos 2\theta}{1-\cos\theta}-2\cos\theta=2$$</p>
<p>BUT THANKS A LOT!</p>
| Socre | 223,538 | <p>$$\frac{(1-(cos^{2}(\theta)-sin^2(\theta))}{(1-cos(\theta)}=\frac{1-cos(2\theta)}{(1-cos(\theta)}$$
Found By using property $cos(2\theta)=cos^{2}(\theta)-sin^{2}(\theta)$
$$\frac{2sin^{2}(\theta)}{1-cos(\theta)}=\frac{2(1-cos^{2}(\theta))}{(1-cos(\theta)}$$
Found By using the fact that $ 1- cos^{2}(\theta)=sin^{2}(\theta)$
$$\frac{2(1-cos(\theta)(1+cos(\theta))}{1-cos(\theta)}$$
Found by the property $a^2-b^2=(a+b)(a-b)$
Which results in$$2(1+cos(\theta))$$
Not $2(1-\cos(\theta))$</p>
|
1,330,362 | <p>I need to define in ZFC the following things:</p>
<ul>
<li>image and domain of a binary relation ($\{ x \mid (x,y)\in f \}$ would be a definition of domain, but it is a class for which is for me is not quite clear why it is a set)</li>
<li>f[X] for a binary relation $f$ and a set $X$</li>
</ul>
<p>Please describe a more or less formal way to describe these things in ZFC.</p>
| Ben Grossmann | 81,360 | <p>The best we can do (in terms of algebraic solutions) is as follows: rewrite the equation as
$$
a = \left(e^x\right)^{\ln(b)} + \left(e^x\right)^{\ln(c)}
$$
setting $y = e^x$, this becomes an equation on $y$:
$$
a = y^{\ln(b)} + y^{\ln(c)}
$$
Or, rearranging,
$$
y^{\ln(b)} + y^{\ln(c)} - a = 0
$$
In other words, we're looking for $x > 0$ that satisfy
$$
x^\beta + x^\gamma - a = 0
$$
There is no general way to solve for $x$ if $\beta,\gamma$ are integers, let alone if $\beta$ and $\gamma$ can be arbitrary real numbers.</p>
|
1,524,109 | <p>Could anyone help me with this proof without using determinant? I tried two ways. </p>
<blockquote>
<p>Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix. </p>
</blockquote>
<p>I then get stuck. The other way was to prove by contradiction, and I failed too. </p>
| Jimmy R. | 128,037 | <p><strong>Hint:</strong> You can sum the elements of a row by multiplying this row with a vector of $1$'s. Can you find now a matrix $X$ (with appropriate columns) such that $AX=Ο$?</p>
|
1,524,109 | <p>Could anyone help me with this proof without using determinant? I tried two ways. </p>
<blockquote>
<p>Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix. </p>
</blockquote>
<p>I then get stuck. The other way was to prove by contradiction, and I failed too. </p>
| Pietro Paparella | 414,530 | <blockquote>
<p><strong>Theorem.</strong> If $A$ is an $n$-by-$n$ matrix, then $A$ is not invertible if and only if zero is an eigenvalue of $A$.</p>
</blockquote>
<p>If $e$ denotes the all-ones vector of appropriate size, then, by hypothesis, $Ae = 0 = 0e$, i.e., zero is an eigenvalue of $A$.</p>
|
2,053,017 | <p>The first thing I was trying to do is to substitute $a$ with something like $1 \over b$ or $1 + \frac{1}{b}$, but I got no results. Then I tried to use Squeeze theorem here, but I also got confused because, for example, $(a + 1)^{1/n}$ is not a decreasing function. </p>
| Mark Viola | 218,419 | <p>Without Loss of Generality, we assume $0<a<1$. Let $x_n=(1/a)^{1/n}-1$ so that $0\le x_n$. Then, we see using Bernoulli's Inequality that</p>
<p>$$a=\frac{1}{(1+x_n)^n}\le \frac{1}{1+nx_n}$$</p>
<p>whereupon isolating $x_n$ reveals</p>
<p>$$0 \le x_n \le \frac{1-a}{a\,n}\tag 1$$</p>
<p>Applying the squeeze theorem to $(1)$ yields the coveted limit </p>
<p>$$\lim_{n\to \infty}a^{1/n}=\lim_{n\to \infty}\frac{1}{1+x_n}=1$$</p>
<p>Note for $a>1$, simply analyze the limit of $(1/a)^{1/n}$ using the same way forward.</p>
|
194,312 | <p>I'm preparing myself to a combinatorics test. A part of it will concentrate on the pigeonhole principle. Thus, I need some hard to very hard problems in the subject to solve.
I would be thankful if you can send me links\books\or just a lone problem.</p>
| Sungjin Kim | 67,070 | <p>IF $\alpha>0$ is irrational, then the fractional parts of $n\alpha$ forms a dense subset of $[0,1]$. </p>
<p>Given a positive integer $n>0$, Fibonacci numbers modulo $n$ is periodic. </p>
<p>These are very famous ones from number theory. There are proofs of these using pigeonhole principle. </p>
|
1,260,722 | <blockquote>
<p>Prove that <span class="math-container">$f=x^4-4x^2+16\in\mathbb{Q}[x]$</span> is irreducible.</p>
</blockquote>
<p>I am trying to prove it with Eisenstein's criterion but without success: for <strong>p=2</strong>, it divides <strong>-4</strong> and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find <span class="math-container">$f(x\pm c)$</span> which is irreducible:</p>
<blockquote>
<p><span class="math-container">$f(x+1)=x^4+4x^3+2x^2-4x+13$</span>, but 13 has the divisors: <strong>1 and 13</strong>, so don't exist a prime number <strong>p</strong> such that to apply the first condition: <span class="math-container">$p|a_i, i\ne n$</span>; the same problem for <span class="math-container">$f(x-1)=x^4+...+13$</span></p>
<p>For <span class="math-container">$f(x+2)=x^4+8x^3+20x^2+16x+16$</span> is the same problem from where we go, if we set <strong>p=2</strong>, that means <span class="math-container">$2|8, 2|20, 2|16$</span>, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for <strong>x-2</strong></p>
</blockquote>
<p>Now I'll verify for <span class="math-container">$f(x\pm3)$</span>, but I think it will be fall... I think if I verify all constant <span class="math-container">$f(x\pm c)$</span> it doesn't work with this method... so have any idea how we can prove that <span class="math-container">$f$</span> is irreducible?</p>
| Jyrki Lahtonen | 11,619 | <p>The polynomial has no real roots, because it is equal to $(x^2-2)^2+12$. The remaining possibility is thus that it is a product of two quadratic factors. By Gauss' Lemma these need to have integer coefficients, so we are looking for a possibile factorization like
$$
p(x)=x^4-4x^2+16=(x^2+ax+b)(x^2+cx+d)
$$
with some integers $a,b,c,d$. Modulo $3$ we have the factorization
$$p(x)=(x^2-2)^2+12\equiv(x^2+1)^2.$$ Therefore $a$ and $c$ must both be divisible by three, and $b\equiv d\equiv 1\pmod3$. Modulo $5$ we have
$$
p(x)\equiv x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1).
$$
This means that $b\equiv d\equiv1\pmod 5$ as well. The Chinese Remainder Theorem (or case-by-case check) then shows that $b\equiv d\equiv 1\pmod{15}$.</p>
<p>Because $bd=16$ the only remaining possibility is that they are $1$ and $16$ in some order. But this is impossible because modulo $2$ we have
$$p(x)\equiv x^4,$$ so all of $a,b,c,d$ must be even.</p>
<p>The conclusion is that $p(x)$ is irreducible.</p>
|
1,260,722 | <blockquote>
<p>Prove that <span class="math-container">$f=x^4-4x^2+16\in\mathbb{Q}[x]$</span> is irreducible.</p>
</blockquote>
<p>I am trying to prove it with Eisenstein's criterion but without success: for <strong>p=2</strong>, it divides <strong>-4</strong> and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find <span class="math-container">$f(x\pm c)$</span> which is irreducible:</p>
<blockquote>
<p><span class="math-container">$f(x+1)=x^4+4x^3+2x^2-4x+13$</span>, but 13 has the divisors: <strong>1 and 13</strong>, so don't exist a prime number <strong>p</strong> such that to apply the first condition: <span class="math-container">$p|a_i, i\ne n$</span>; the same problem for <span class="math-container">$f(x-1)=x^4+...+13$</span></p>
<p>For <span class="math-container">$f(x+2)=x^4+8x^3+20x^2+16x+16$</span> is the same problem from where we go, if we set <strong>p=2</strong>, that means <span class="math-container">$2|8, 2|20, 2|16$</span>, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for <strong>x-2</strong></p>
</blockquote>
<p>Now I'll verify for <span class="math-container">$f(x\pm3)$</span>, but I think it will be fall... I think if I verify all constant <span class="math-container">$f(x\pm c)$</span> it doesn't work with this method... so have any idea how we can prove that <span class="math-container">$f$</span> is irreducible?</p>
| Sil | 290,240 | <p>After substitution
$$
g(x)=f(2x-1)=16x^4-32x^3+8x^2+8x+13
$$
it becomes irreducible by <a href="http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf" rel="nofollow noreferrer">Murty's criterion</a> (see Theorem 1) for $n=4$, since $g(4)=2221$ is a prime. The criterion is not so well spread, so here it is:</p>
<blockquote>
<p>Let $f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0$ be a polynomial of degree $m$ in $\mathbb{Z}[x]$ and set $$H=\max_{0\leq i\leq m-1} |a_i/a_m|.$$
If $f(n)$ is prime for some integer $n\geq H+2$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$. </p>
</blockquote>
<p>Just a note, other substitutions are possible as well. For example, reverting coefficients - basically a substitution $x^4f(1/x)$ - yields again irreducibility by the criterion above, this time with $g(5)=9901$ a prime.</p>
|
223,718 | <p>Suppose I have two lists to which I apply the <code>Line</code> command for example</p>
<pre><code>A = {{-4,0},{4,0}}
B = {{0,4},{0,-4}}
</code></pre>
<p>and I take <code>Line[A]</code> and <code>Line[B]</code>. Is there a way to get Mathematica to tell me the intersection points of the line? Of course this is a very simple example, in practice the lines would have many defining points to approximate a curve.</p>
<p>further questions: How about if I had <span class="math-container">$n$</span> lists? Could I ask to find the intersection points in some bounded region only?</p>
| kglr | 125 | <h3><code>Graphics`Mesh`FindIntersections</code></h3>
<pre><code>Graphics`Mesh`MeshInit[];
findIntersections = Complement[Graphics`Mesh`FindIntersections[Join[##]],
Join @@ Graphics`Mesh`FindIntersections /@ {##}] &;
</code></pre>
<p>Using <code>redLines</code> and <code>blueLines</code> from flinty's answer:</p>
<pre><code>intersections = findIntersections[redLines, blueLines]
</code></pre>
<blockquote>
<pre><code>{{-1.73595,3.07582}, {-0.648352,0.472527}, {0.385965,0.192982}, {-1.14815,1.37037},
{1.34783,1.86957}, {2.12405,-0.255696}}
</code></pre>
</blockquote>
<pre><code>Graphics[{Red, redLines, Blue, blueLines, Black, PointSize[Large], Point@intersections},
PlotRange -> {{-3, 3}, {-2, 5}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/mymhH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mymhH.png" alt="![enter image description here"></a></p>
<pre><code>SeedRandom[1]
greenLines = {Line[RandomReal[{-3, 3}, {8, 2}]],
BezierCurve[RandomReal[{-2, 2}, {7, 2}]]};
intersections = findIntersections[redLines, blueLines, greenLines]
</code></pre>
<blockquote>
<pre><code> {{-1.73595,3.07582}, {-1.63671,1.27266}, {-1.43717,0.798769},
{-1.214,1.3572}, {-1.14815,1.37037}, {-1.13457,1.33642},
{-0.648352,0.472527},{-0.601914,0.459977}, {-0.541359,0.312038},
{-0.487964,0.42918},{-0.444653,0.222327}, {-0.373226,0.398169},
{-0.295139,0.147569},{0.385965,0.192982}, {1.01337,0.0234133},
{1.16581,-0.0177874}, {1.34783,1.86957}, {2.12405,-0.255696},
{2.41381,1.48283}}
</code></pre>
</blockquote>
<pre><code>Graphics[{Red, redLines, Blue, blueLines, Green, greenLines,
Black, PointSize[Medium], Point@intersections},
PlotRange -> {{-3, 3}, {-3, 5}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/Ykkhp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ykkhp.png" alt="enter image description here"></a></p>
|
2,598 | <p>Should Wolfram Alpha Notebook questions be considered on-topic?</p>
<p>Here's an example: <a href="https://mathematica.stackexchange.com/questions/240780/calculating-double-integral-bounded-by-domain-in-wolfram-alpha-notebook">Calculating double integral bounded by domain in Wolfram Alpha Notebook</a></p>
<p>Here are some related meta Q&A:</p>
<ul>
<li><p><a href="https://mathematica.meta.stackexchange.com/questions/68/other-wri-product-discussion">Other WRI product discussion?</a></p>
</li>
<li><p><a href="https://mathematica.meta.stackexchange.com/questions/265/are-questions-about-doing-symbolic-math-in-wolfram-alpha-on-topic-here">Are questions about doing symbolic math in Wolfram Alpha on topic here?</a></p>
</li>
</ul>
<p><a href="https://www.wolfram.com/wolfram-alpha-notebook-edition/" rel="nofollow noreferrer">Wolfram Alpha Notebooks</a> are a new WRI product that hybridizes W|A and Mathematica. I looked only briefly, but it resembles a <em>Mathematica</em> notebook in which the only valid input starts with single-equals (probably without having to type <code>=</code>), though the sample inputs are sometimes interpreted differently in the examples shown than in my <em>Mathematica</em>.</p>
<p>(For those who may not know, it was decided to consider questions about <a href="https://mathematica.stackexchange.com/help/on-topic">W|A off-topic</a>.)</p>
| Penelope Benenati | 74,045 | <p><strong>Yes, most of the time (but not always). It seems to me that many users do not know well Notebook Edition and are not aware that Notebook Edition accepts almost all code syntax of Mathematica.</strong></p>
<p>To answer this question, one should first define "Wolfram Alpha Notebook questions." I use Notebook Edition, and during the last nine months, I verified that for all the problems I had to solve, all the answers to my questions that I received here were beneficial (you can take a look at my questions and discussions here). The only difference I had to implement to use the suggestions I obtained was changing the variable name using only one character. About all the rest, I never found any difference.</p>
<p><strong>As I said, Notebook Edition accepts almost all code syntax of Mathematica.</strong></p>
<p><strong>A question cannot be on-topic or off-topic based only on the software used by the user who asks it.</strong> Indeed, in principle, the user could also ask an interesting question about the Mathematica language by using his/her smartphone without any computer or math software (at all).</p>
<hr />
<p>Hence, <strong>the answer is simple: questions about the language of Mathematica are on topic independently of the software used by the user who asks them;</strong> questions that are specifically related to the syntax of Notebook Edition (that cannot be used with Mathematica) or that are about the software Notebook Edition itself are off-topic.</p>
<hr />
<p>Questions are on-topic or off-topic, not the software used by people who ask them. Otherwise, you should close <strong>all my questions of the last nine months</strong>, which were considered useful by several people here, just because I am "revealing" that I always used Notebook Edition while implementing the suggestions obtained (just using variables of one character solely).</p>
<p>Finally, since you mentioned above <a href="https://mathematica.stackexchange.com/questions/240780/calculating-double-integral-bounded-by-domain-in-wolfram-alpha-notebook">Calculating double integral bounded by domain in Wolfram Alpha Notebook</a> as an example to clarify the above question, I am glad to show the screenshot I obtained by using Notebook Edition. The question is now: <em><strong>Do you solve that problem differently with Mathematica, in a way that I cannot implement with Notebook Edition</strong></em>? Thank you.
<a href="https://i.stack.imgur.com/7td8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7td8m.png" alt="enter image description here" /></a></p>
<hr />
<hr />
<p>I also add a screenshot of the result I get attempting to use the answer I received here: <a href="https://mathematica.stackexchange.com/q/246519/74045">Plotting a complete graph with a given image as vertices</a> :</p>
<p><a href="https://i.stack.imgur.com/QQUmO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QQUmO.png" alt="enter image description here" /></a></p>
|
1,575,532 | <p>I'm having some difficulty with this problem out of my Calculus Book: </p>
<blockquote>
<p>Find the total area between region and the $x$-axis: $y=x^3-x^2-6x$, $-2 \le x \le 3.$</p>
</blockquote>
<p>I know I start with setting the function equal to $0$:</p>
<p>$$0=x^3-x^2-6x$$
$$0=x(x-3)(x-2)$$
Hence $x=3$ or $x=2$.</p>
<p>I then need to take the integral of the Top Function minus the bottom function of each respective area:</p>
<p>$$\text{Area} = \int\limits_{a}^{b} (\text{Top}_f - \text{Btm}_f)\, dx$$</p>
<p>The problem I'm having is getting the initial function split in half? I need the function for the portion above the $x$-axis to the $x$-axis, PLUS the portion below the $x$-axis to the $x$-axis added together to get the entire area...?</p>
<p>Can someone walk me through what I am missing here?</p>
| Vincenzo Zaccaro | 269,380 | <p>You use thr Rienmann integral. Then $A=\int_{-2 }^3(x^3-x^2-6x) dx=[x^4/4-x^3/3-3x^2]_{x-2}^{x=3}$. This is the area with sign. If you want the area without sign: $A= \int_{-2} ^0(x^3-x^2-6x) dx- \int_{0} ^3(x^3-x^2-6x) dx $. The polynomial is negative in $[0,3]$.</p>
|
1,575,532 | <p>I'm having some difficulty with this problem out of my Calculus Book: </p>
<blockquote>
<p>Find the total area between region and the $x$-axis: $y=x^3-x^2-6x$, $-2 \le x \le 3.$</p>
</blockquote>
<p>I know I start with setting the function equal to $0$:</p>
<p>$$0=x^3-x^2-6x$$
$$0=x(x-3)(x-2)$$
Hence $x=3$ or $x=2$.</p>
<p>I then need to take the integral of the Top Function minus the bottom function of each respective area:</p>
<p>$$\text{Area} = \int\limits_{a}^{b} (\text{Top}_f - \text{Btm}_f)\, dx$$</p>
<p>The problem I'm having is getting the initial function split in half? I need the function for the portion above the $x$-axis to the $x$-axis, PLUS the portion below the $x$-axis to the $x$-axis added together to get the entire area...?</p>
<p>Can someone walk me through what I am missing here?</p>
| gt6989b | 16,192 | <p>You factored incorrectly. Note that $$f(x) = x^3 - x^2 - 6x = x(x-3)(x+2)$$ so the roots lie at $-2,0,3$. The plot looks like</p>
<p><a href="https://i.stack.imgur.com/WoR2l.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WoR2l.gif" alt="enter image description here"></a></p>
<p>If you want the real area between the curve and the $x$-axis, the region to the right of $0$ must be negated, since it is under the $x$-axis and you get $$\int_{-2}^0 f(x) dx - \int_0^3 f(x) dx$$</p>
|
742,160 | <p>Answer true or false to each of the following questions. If a statement is true, prove it. If a statement is false, give a counterexample.</p>
<ol>
<li>For all sets $A$,$B$ and $C$: IF $A ⊆ B$ and $A ⊆ C$, Then $A ⊆ (B ∩ C)$</li>
<li>For all sets $A$ and $B$, if $|A| \le |B|$, then $A ⊆ B$</li>
</ol>
| Michal Boska | 140,880 | <p>Take a look at this picture (from <a href="https://en.wikipedia.org/wiki/File:Surjection.svg" rel="noreferrer">Wikipedia</a>):</p>
<p><a href="https://i.stack.imgur.com/XVMeJm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XVMeJm.png" alt="][2]][2]"></a></p>
<p>This function is NOT injection, because two arrows point into single point in that picture.</p>
<p>Now imagine injections at the doctor. Injections usually hurt and you, sure as hell, woudln't want anyone to stick that injection into the same point on your body multiple times.</p>
<p>So that's why injective functions cannot have multiple arrows pointing into the same point (value)</p>
<p>:)</p>
|
6,090 | <p>I have been reading about Riemann Zeta function and have been thinking about it for some time.</p>
<p>Has anything been published regarding upper bound for the real part of zeta function zeros as the imaginary part of the zeros tend to infinity?</p>
<p>Thanks</p>
| timur | 2,473 | <p>De La Vallee-Pousin's theorem has been improved by Korobov and Vinogradov in the 50's and I believe their result is the strongest known asymptotic zero free region, cf. <em>The Riemann zeta-function: Theory and applications</em> by Alexandar Ivic. One can find more recent papers but my impression is that they don't touch the main exponents and so relatively tractable problems seem to be improving the constants, or giving explicit bounds on the constants etc.</p>
|
4,506,196 | <p>If <span class="math-container">$R$</span> is a commutative Hilbert ring then each of its prime ideals is an intersection of maximal ideals. Is there a similar class of commutative rings for which every ideal is an intersection of prime ideals (that is, in which all ideals are radical)? Does this class have a name? Have the rings in it been characterized?</p>
| Eric Wofsey | 86,856 | <p>As Qiaochu Yuan's answer observed, such a ring must have the property that every finitely generated ideal is generated by an idempotent and therefore every ideal is generated by idempotents. Conversely, I claim that if every principal ideal in a commutative ring <span class="math-container">$R$</span> is generated by an idempotent then every ideal in <span class="math-container">$R$</span> is radical. To prove this, suppose <span class="math-container">$I\subseteq R$</span> is an ideal and <span class="math-container">$r^n\in I$</span>. Let <span class="math-container">$e$</span> be an idempotent that generates the principal ideal <span class="math-container">$(r)$</span>. Then since <span class="math-container">$e$</span> is a multiple of <span class="math-container">$r$</span>, <span class="math-container">$e^n=e$</span> is a multiple of <span class="math-container">$r^n$</span> so <span class="math-container">$e\in I$</span>. But since <span class="math-container">$(e)=(r)$</span>, this means <span class="math-container">$r\in I$</span>.</p>
<p>So, all ideals are radical in a commutative ring iff every principal ideal is generated by an idempotent. Such a ring is called a <a href="https://en.wikipedia.org/wiki/Von_Neumann_regular_ring" rel="noreferrer">von Neumann regular ring</a> and much is known about them. For instance, one alternative characterization is that a commutative ring is von Neumann regular iff it is <span class="math-container">$0$</span>-dimensional (i.e., every prime is maximal) and reduced.</p>
|
1,027,707 | <p>In how many ways can 100 identical chairs be divided among 4 different rooms so that each room will have 10,20,30,40 or 50 chairs?</p>
<p>I'm having problems coming up with the generating function for this question. </p>
<p>The answer given is 68. </p>
| Community | -1 | <p>The question is incorrect, you need a tensor product in place of the product.
With that modification you can use/prove that</p>
<p>$\mathrm{Hom}_\mathrm{FDVect}(V,W) \cong W \otimes V^*$</p>
|
2,095,349 | <p>I test my skills in statistics and probabilities and I decided to work with distributions. So, I tried to solve the below problem </p>
<p><strong>Problem</strong></p>
<p>Suppose that a hospital serves in average $80$ citizens daily from a city with $11000$ citizens.
In a random day, what is the probability that the hospital serves at most $8$ citizens?</p>
<p><strong>My solution</strong></p>
<p>I defined a random variable $X$ {number of citizens who will be served in one day }. </p>
<p>$X \sim b(x;n=11000,p)$, where
\begin{align}
p &= \frac{E(X)}{n} = \frac{80}{11000} = 0.07
\end{align}</p>
<p>Provided that $npq = 76.4 > 10$:</p>
<p>$b(x;n=11000,p) \sim N(pq,npq)$ </p>
<p>According to the central limit theorem,
\begin{align}
Z = \frac{X - np}{\sqrt{npq}} = \frac{8-80}{8.74} = -8.23
\end{align}
So $P(Z\le -8.23) = 0$. </p>
<p>Where is my fault? I think my reasoning is not correct. </p>
| Sean Roberson | 171,839 | <p>Using the appropriate commands in R and Excel, the actual probability is $9.360672 \cdot 10^{-17}$. However, this is extremely small, so an answer of zero would also be acceptable.</p>
<p>When it comes to statistics and the normal distribution, don't expect "exact" answers. Why? Well, the probabilities are generated by this integral:</p>
<p>$$ P(Z < z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty} ^z e^{-\frac{x^2}{2}} \ dx $$</p>
<p>which has no elementary antiderivative. Hence, only estimates can be given.</p>
<p>TL;DR, you're fine.</p>
|
2,008,653 | <p>According to Riemann (I think) the (exact) prime counting function is given by:
<span class="math-container">$$
\pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1}
$$</span>
with <span class="math-container">$ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$</span> and <span class="math-container">$\rho$</span> running over <strong>all</strong> the zeros of the <span class="math-container">$\zeta$</span> function.</p>
<p>Why isn't this function used in general to calculate <span class="math-container">$\pi(x)$</span>? Shouldn't it be a great approximation if many zeros of the <span class="math-container">$\zeta$</span> function are used? Probably nowadays there are many zeros known? Until now I thought <span class="math-container">$\pi(x)$</span> is a very "hard" function, because the distribution of the primes is quite hard, but this explicit formula does not look that hard.</p>
<p>Thank you.</p>
| Greg Hurst | 93,700 | <p>Borwein, Bradley, and Crandall state on page 249 <a href="https://cr.yp.to/bib/2000/borwein.pdf" rel="noreferrer">here</a> that</p>
<blockquote>
<p>... it should be the case that, in some appropriate sense
$$ \pi(x) \sim \text{Ri}(x) - \sum_{\rho} \text{Ri}(x^\rho) \tag{4} $$
with $\text{Ri}$ denoting the Riemann function defined:
$$ \text{Ri}(x) = \sum_{m=1}^\infty \frac{\mu(m)}{m}\text{li}(x^{1/m}). \tag{5} $$
This relation (4) has been called “exact” [94], yet we could not locate a proof in the literature; such
a proof should be nontrivial, as the conditionally convergent series involved are problematic. In any case relation (4) is quite accurate ...</p>
</blockquote>
<p>So it appears we don't really know if this relation is indeed true or not.</p>
<p>But as the quote says, it does <em>appear</em> this relation at least provides a good estimate of $\pi$. In fact $\text{Ri}(x)$ alone provides a good estimate of $\pi(x)$. For example, $\pi(10^{20}) = 2220819602560918840 $, and here's $\text{Ri}(10^{20})$ evaluated in Mathematica:</p>
<pre><code>In[186]:= Floor[RiemannR[10^20]]
Out[186]= 2220819602556027015
</code></pre>
<p>This gives a relative error of about $2.2 \cdot 10^{-12}$, meaning the first 11 digits are correct!</p>
<p>Now how about incorporating the zeros $\rho$? Well they actually seem to make things worse (at least for a 'small' number of zeros). I took the first $14400$ zeros of $\zeta$, to a precision of 30 digits, and got an answer with relative error $3.1 \cdot 10^{-7}$. In fact the more zeros I chose, the worse the relative error became.</p>
<p><a href="https://i.stack.imgur.com/L8aeX.png" rel="noreferrer"><img src="https://i.stack.imgur.com/L8aeX.png" alt="enter image description here"></a></p>
<p>So to answer your question, I think this formula seems to provide an excellent approximation for $\pi(x)$. However, at the end of the day we'll only be able to get an approximation, not an exact answer.</p>
|
3,465,607 | <p>If I have a cubic where I know the turning points, can I find what its equation is?</p>
<p>I already know that the derivative is 0 at the turning points. Other than that, I'm not too sure how I can continue.</p>
<p>Suppose I have the turning points (-2,5) and (4,0). I have started doing the following:
<span class="math-container">$$
\frac{dy}{dx} = 0 \text{ at turning points}\\
\text{So, } 0 = (x+2)(x-4)\\
$$</span>
<span class="math-container">$$
\int (x+2)(x-4) dx\\
= \int x^2-2x-8 dx\\
= \frac{x^3}{3} - x^2 - 8x + C
$$</span></p>
<p>Graphing this, you get correct <span class="math-container">$x$</span> coordinates at the turning points, but not correct <span class="math-container">$y$</span>.</p>
<p>I'm aware that only with that information you can't tell how steep the cubic will be, but you should at least be able to find some sort of equation.</p>
<p>Finally, would a <span class="math-container">$y$</span>-intercept be helpful? If so, then suppose for the above example that the <span class="math-container">$y$</span>-intercept is 4.</p>
| Bernard | 202,857 | <p>Welcome! You simply forgot that having the turning points provides the derivative up to a nonzero constant factor, i.e.
<span class="math-container">$$y'(x)=K(x+2)(x-4),\quad K\in \mathbf R^*, \quad
\text{ whence }\;y(x)=K\biggl(\frac{x^3}3 -x^2-8x\biggr)+C.$$</span>
You now have two constants to adjust the ordinates.</p>
|
3,465,607 | <p>If I have a cubic where I know the turning points, can I find what its equation is?</p>
<p>I already know that the derivative is 0 at the turning points. Other than that, I'm not too sure how I can continue.</p>
<p>Suppose I have the turning points (-2,5) and (4,0). I have started doing the following:
<span class="math-container">$$
\frac{dy}{dx} = 0 \text{ at turning points}\\
\text{So, } 0 = (x+2)(x-4)\\
$$</span>
<span class="math-container">$$
\int (x+2)(x-4) dx\\
= \int x^2-2x-8 dx\\
= \frac{x^3}{3} - x^2 - 8x + C
$$</span></p>
<p>Graphing this, you get correct <span class="math-container">$x$</span> coordinates at the turning points, but not correct <span class="math-container">$y$</span>.</p>
<p>I'm aware that only with that information you can't tell how steep the cubic will be, but you should at least be able to find some sort of equation.</p>
<p>Finally, would a <span class="math-container">$y$</span>-intercept be helpful? If so, then suppose for the above example that the <span class="math-container">$y$</span>-intercept is 4.</p>
| marty cohen | 13,079 | <p>If the turning points
of a cubic polynomial
<span class="math-container">$f(x)$</span> are
<span class="math-container">$(a, b)$</span> and <span class="math-container">$(c, d)$</span>
then
<span class="math-container">$f(x)
=k(\dfrac{x^3}{3}-\dfrac{(a+c)x^2}{2}+acx)+h
$</span>
where
<span class="math-container">$k
=-\dfrac{6(b-d)}{(a-c)^3}
$</span>
and
<span class="math-container">$h
=\dfrac{b+d}{2}-\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{2(a-c)^3}
$</span>.</p>
<p>Here's the details.</p>
<p>If the turning points are
<span class="math-container">$(a, b)$</span> and <span class="math-container">$(c, d)$</span>
then
<span class="math-container">$f'(x)
=k(x-a)(x-c)
=k(x^2-(a+c)x+ac)
$</span>.</p>
<p>Then
<span class="math-container">$f(x)
=k(\dfrac{x^3}{3}-\dfrac{(a+c)x^2}{2}+acx)+h
$</span>.</p>
<p><span class="math-container">$f(a)
=k(\dfrac{a^3}{3}-\dfrac{(a+c)a^2}{2}+a^2c)+h
=b
$</span>
and
<span class="math-container">$f(c)
=k(\dfrac{c^3}{3}-\dfrac{(a+c)c^2}{2}+ac^2)+h
=d
$</span>.</p>
<p>Subtracting,</p>
<p><span class="math-container">$\begin{array}\\
b-d
&=k(\dfrac{a^3-c^3}{3}-\dfrac{(a+c)(a^2-c^2)}{2}+ac(a-c))\\
&=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{(a+c)(a+c)}{2}+ac)\\
&=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{(a+c)(a+c)}{2}+ac)\\
&=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{a^2+2ac+c^2}{2}+ac)\\
&=k(a-c)(\dfrac{2(a^2+ac+c^2)-3(a^2+2ac+c^2)+6ac}{6})\\
&=k(a-c)(\dfrac{-a^2-c^2+2ac}{6})\\
&=-k(a-c)(\dfrac{(a-c)^2}{6})\\
&=-k\dfrac{(a-c)^3}{6}\\
&=k\dfrac{(c-a)^3}{6}\\
\end{array}
$</span></p>
<p>so
<span class="math-container">$k
=-\dfrac{6(b-d)}{(a-c)^3}
$</span>.</p>
<p>To maintain symmetry,
I'll add the two equations.</p>
<p><span class="math-container">$\begin{array}\\
b+d
&=k(\dfrac{a^3+c^3}{3}-\dfrac{(a+c)(a^2+c^2)}{2}+ac(a+c))+2h\\
&=k\dfrac{2(a^3+c^3)-3(a+c)(a^2+c^2)+6ac(a+c)}{6}+2h\\
&=k\dfrac{2(a^3+c^3)-3(a^3+ac^2+a^2c+c^3)+6a^2c+6ac^2}{6}+2h\\
&=k\dfrac{-a^3-c^3+3a^2c+3ac^2}{6}+2h\\
&=k\dfrac{-(a^3+c^3)+3ac(a+c)}{6}+2h\\
&=k\dfrac{-(a+c)(a^2-ac+c^2)+3ac(a+c)}{6}+2h\\
&=k\dfrac{-(a+c)(a^2+c^2-4ac)}{6}+2h\\
&=-\dfrac{6(b-d)}{(a-c)^3}\dfrac{-(a+c)(a^2+c^2-4ac)}{6}+2h\\
&=\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{(a-c)^3}+2h\\
\end{array}
$</span></p>
<p>so
<span class="math-container">$2h
=(b+d)-\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{(a-c)^3}
$</span>
and
<span class="math-container">$k
=-\dfrac{6(b-d)}{(a-c)^3}
$</span>.</p>
|
3,630,482 | <p>Show that, if 13 divides <span class="math-container">$n^2$</span> + <span class="math-container">$3n$</span> + <span class="math-container">$51$</span> then 169 divides <span class="math-container">$21n^2$</span> + <span class="math-container">$89n$</span> + <span class="math-container">$44$</span></p>
<p>We have 13 <span class="math-container">$|$</span> <span class="math-container">$n^2$</span> + 3n + 51</p>
<p>Using some congruency rules, this becomes:</p>
<p>13 <span class="math-container">$|$</span> <span class="math-container">$n^2$</span> + <span class="math-container">$3n$</span> - <span class="math-container">$1$</span></p>
<p>Or 13 divides <span class="math-container">$n(n+3)$</span> + 1</p>
<p>At this point, I was feeling kinda lazy, so I just listed the factors of 13, added 1 to each and saw which one can be broken down into two numbers such that one is 3 less than the other instead of trying to look for a more elegant solution</p>
<p>I quite quickly arrived at n = 5 (5 × 5 + 3 = 40 = 39 + 1)</p>
<p>I plugged n = 5 into the other equation, and got something that's divisible by 169</p>
<p>Now, how do I do the final thing, which is to prove either that no other such value of n can be found, or if it can be found, it would satisfy the other condition as well?</p>
<p>Never mind, found it</p>
| Joshua Wang | 773,061 | <p>First, turn the inequalities into lines by using an equals sign in place of the inequality sign. Then take every possible pair and find their point of intersection. Then, substitute each intersection point into the original system of inequalities to see if it lies on the solution region. Of all of the points that work, take the rightmost one (the point with the highest <span class="math-container">$x$</span>-coordinate), and that is your answer. If two points have the highest <span class="math-container">$x$</span>-coordinate, they are both solutions, along with the entire line in between them. </p>
|
4,260,645 | <p>I can't find any insights online on how useful the graph of function <span class="math-container">$f(x)$</span> (on the <span class="math-container">$y$</span> axis) versus it's derivative <span class="math-container">$f'(x)$</span>? (on the <span class="math-container">$x$</span> axis) does it provide some useful informations if any?</p>
<p>For example, I see that when I plot <span class="math-container">$\sin(x)$</span> against <span class="math-container">$\cos(x)$</span> the plot is a circle which is reminiscent of parametric equations.</p>
<p>My question is: Assuming the function is nice (nice in the common sense that it is continuous) does that kind of graph provide any useful information? What about this graph where <span class="math-container">$x(t) = f'(t)$</span>:</p>
<p><a href="https://i.stack.imgur.com/DIYa9.png" rel="nofollow noreferrer">f(t) versus x(t) where x(t) = f'(t)</a></p>
| Narasimham | 95,860 | <p>Such a graph treats <span class="math-container">$x$</span> as a <em>parameter</em>.</p>
<p>In dynamic systems ( vibrations/ oscillations etc. ) plots <span class="math-container">$ (f(t), f'(t))$</span> with time <span class="math-container">$t$</span> as parameter are called <em>phase portraits</em>; they are very useful, as they help to represent non-linearities of oscillatory phenomena through derivatives of motion e.g., in Van der Pol equation and Limit cycles.. to include effects of damping. Another example id population growth dynamics.</p>
<p>In coupled differential equations with several state variables phase portrait for each state variable <span class="math-container">$\{(x(t),x'(t)),(y(t),y'(t))\}$</span> can be plotted separately to study.</p>
|
2,518,213 | <p>We work in the vector space $\mathbb{R}^k$ over $\mathbb{R}$. We define the (usual) infinite norm on $\mathbb{R}$:</p>
<p>$$\|x\|_{\infty} = \max\{|x_1|,|x_2|,\dots,|x_n|\}$$ where $x_1, x_2, \dots, x_n $ are the coordinates of of $x$ in $\mathbb{R}^k$. </p>
<p>We would like to prove the following fact:
$$\exists c> 0, \forall x\in \mathbb{R}^k, \|x\|_{\infty} \le c \|x\|$$</p>
<p>where $\|\cdot \|$ is any norm. I am also told that if this isn't true, then I can construct a sequence $(x_n)$ such that $\|x_n\|$ is bounded but $\|x\|_{\infty}$ diverges to infinity.</p>
<p>I am slightly puzzled by that hint, and I'm stuck trying to prove the existence of such a sequence. Here is what I've done: first negate the statement:</p>
<p>$$\forall c> 0, \exists x\in \mathbb{R}^k, \|x\|_{\infty} > c \|x\|$$</p>
<p>Then I thought, whynot construct the obvious sequence and see where that takes us? So we define $(x_n)$ such that $$\|x_n\|_{\infty} > n \|x_n\|$$ This at least tells us that $\frac{\|x_n\|}{\|x_n\|_{\infty}}$ tends to $0$, but I haven't gotten much further. What bugs me the most is trying to somehow bound $\|x_n\|$. </p>
<p>Could someone suggest an approach to tackle that hint?</p>
| daw | 136,544 | <p>You can choose the sequence $(x_n)$ such that $\|x_n\|_\infty=1$. Then $\|x_n\|\to0$.</p>
<p>To arrive at a contradiction, I will use the Weierstrass theorem. First, we show that $x\mapsto \|x\|$ is continuous wrt the $\infty$-norm:
Write $x=\sum x_ie_i$ with the unit vectors $e_i$, then:
$$
\|x\|\le \sum_{i=1}^n |x_i| \|e_i\| \le \|x\|_\infty \sum_{i=1}^n \|e_i\|.
$$
Then for arbitrary $x,y$, we get
$$
|\|x\|-\|y\||\le \|x-y\|\le \|x-y\|_\infty \sum_{i=1}^n \|e_i\|.
$$
Then the continuous function $x\mapsto \|x\|$ attains its minimum on $\{x:\|x\|_\infty=1\}$. Since $\|x\|\ne0$ on this set, there is $c>0$ such that $\|x\|\ge c$ for all $\|x\|_\infty=1$. </p>
<p>Now, going back to the original sequence, we arrive at a contradiction.</p>
|
97,329 | <p>This is inspired by <a href="https://mathoverflow.net/questions/97307/polynomials-all-of-whose-roots-are-rational">this </a> question. Let $f(x)=a_nx^n+...+a_0$ be a polynomial with rational coefficients. The sandard procedure of finding a rational root $p/q$ involves checking all $p$ that divide $a_0$ and all $q$ that divide $a_0$. This is not very complicated but involves factoring $a_0$ and $a_n$. The factoring problem is not known to be in P. If $n\le 4$, then the fact that the group $S_4$ is solvable and the well known formulas for roots of polynomials of degree $\le 4$ give easy polynomial time algorithm of finding rational roots. </p>
<p><b> Question</b> Is the problem of finding a rational root of $f(x)$ in P for every $n$ (say, for $n=5$)?</p>
<p><b> Update 1</b> After I posted the question, I noticed an answer by Robert Israel to the previous <a href="https://mathoverflow.net/questions/97307/polynomials-all-of-whose-roots-are-rational"> question</a> (of Joseph O'Rourke). That could give an answer to my question but I am still not sure how one can avoid factoring numbers $a_0,a_n$. </p>
<p><b> Update 2 </b> Robert Israel's explanations (see his comment <a href="https://mathoverflow.net/questions/97307/polynomials-all-of-whose-roots-are-rational">here </a>) convince me that his algorithm of checking whether a polynomial has a rational root (all roots rational) runs in polynomial time. </p>
<p>I removed Question 2 so that I can accept Michael Stoll's answer. I will post Question 2 as a separate question. </p>
| Bruno | 16,178 | <p>You do not need the full power of LLL algorithm for finding rational roots. In particular, <a href="https://en.wikipedia.org/wiki/Berlekamp%E2%80%93Zassenhaus_algorithm" rel="nofollow noreferrer">Berlekamp-Zassenhaus algorithm</a> exponential-time algorithm to compute the irreducible factorization can be used to get a polynomial-time algorithm for finding rational roots: The exponential part of this algorithm is the recombination of different modular factors to build <em>true</em> factors, but no such recombination is needed to build the linear factors. </p>
<p>A complete description and analysis of such an algorithm has been given by Loos in 1983: </p>
<p>Loos, R. <em>Computing Rational Zeros of Integral Polynomials by P-Adic Expansion.</em> SIAM Journal on Computing 12, no. 2 (May 1, 1983): 286–93. doi:<a href="http://epubs.siam.org/doi/abs/10.1137/0212017" rel="nofollow noreferrer">10.1137/0212017</a>.</p>
|
3,502,507 | <p>This is very similar to the <a href="https://math.stackexchange.com/questions/3501693/given-that-you-started-with-one-chip-what-is-the-probability-that-you-will-win">question</a> I've just asked, except now the requirement is to gain <span class="math-container">$4$</span> chips to win (instead of <span class="math-container">$3$</span>) </p>
<p>The game is:</p>
<blockquote>
<p>You start with one chip. You flip a fair coin. If it throws heads, you gain
one chip. If it throws tails, you lose one chip. If you have zero
chips, you lose the game. If you have <strong>four</strong> chips, you win. What is the
probability that you will win this game?</p>
</blockquote>
<p>I've tried to use the identical reasoning used to solve the problem with three chips, but seems like in this case, it doesn't work.</p>
<p>So the attempt is:</p>
<p>We will denote <span class="math-container">$H$</span> as heads and <span class="math-container">$T$</span> as tails (i.e <span class="math-container">$HHH$</span> means three heads in a row, <span class="math-container">$HT$</span> means heads and tails etc)</p>
<p>Let <span class="math-container">$p$</span> be the probability that you win the game. If you throw <span class="math-container">$HHH$</span> (<span class="math-container">$\frac{1}{8}$</span> probability), then you win. If you throw <span class="math-container">$HT$</span> (<span class="math-container">$\frac{1}{4}$</span> probability), then your probability of winning is <span class="math-container">$p$</span> at this stage. <strong>If you throw heads <span class="math-container">$HHT$</span> (<span class="math-container">$\frac{1}{8}$</span> probability), then your probability of winning <span class="math-container">$\frac{1}{2}p$</span></strong></p>
<p>Hence the recursion formula is </p>
<p><span class="math-container">$$\begin{align}p & = \frac{1}{8} + \frac{1}{4}p+ \frac{1}{8}\frac{1}{2}p \\
&= \frac{1}{8} + \frac{1}{4}p +\frac{1}{16}p \\
&= \frac{1}{8} + \frac{5}{16}p
\end{align}$$</span></p>
<p>Solving for <span class="math-container">$p$</span> gives</p>
<p><span class="math-container">$$\frac{11}{16}p = \frac{1}{8} \implies p = \frac{16}{88}$$</span></p>
<p>Now, to verify the accuracy of the solution above, I've tried to calculate the probability of losing using the same logic, namely:</p>
<p>Let <span class="math-container">$p$</span> denote the probability of losing. If you throw <span class="math-container">$T$</span> (<span class="math-container">$\frac{1}{2}$</span> probability), you lose. If you throw <span class="math-container">$H$</span> (<span class="math-container">$\frac{1}{2}$</span> probability), the probaility of losing at this stage is <span class="math-container">$\frac{1}{2}p$</span>. If you throw <span class="math-container">$HH$</span>(<span class="math-container">$\frac{1}{4}$</span> probability), the probability of losing is <span class="math-container">$\frac{1}{4}p$</span>. Setting up the recursion gives</p>
<p><span class="math-container">$$\begin{align}p & = \frac{1}{2} + \frac{1}{4}p+ \frac{1}{8}\frac{1}{2}p \\
&= \frac{1}{2} + \frac{1}{4}p +\frac{1}{16}p \\
&= \frac{1}{2} + \frac{5}{16}p
\end{align}$$</span></p>
<p>Which implies that </p>
<p><span class="math-container">$$\frac{11}{16}p = \frac{1}{2} \implies p = \frac{16}{22} = \frac{64}{88}$$</span></p>
<p>Which means that probabilities of winning and losing the game do not add up to <span class="math-container">$1$</span>.</p>
<p>So the <em>main</em> question is: Where is the mistake? How to solve it using recursion? (Note that for now, I'm mainly interested in the recursive solution)</p>
<p>And the bonus question: Is there a possibility to generalize? I.e to find the formula that will give us the probability of winning the game, given that we need to gain <span class="math-container">$n$</span> chips to win? </p>
| antkam | 546,005 | <p>This answer only addresses <strong>what's wrong with your recursion</strong>, since the other answers (both in this question and your earlier question) already gave many different ways to set up the right recursions (or use other methods).</p>
<p>The key mistake is what you highlighted. When you throw <span class="math-container">$HHT$</span>, you now have <span class="math-container">$2$</span> chips. <em>For the special case of this problem</em>, <span class="math-container">$2$</span> chips is right in the middle between <span class="math-container">$0$</span> and <span class="math-container">$4$</span> chips, so the winning prob is obviously <span class="math-container">$\color{red}{\frac12}$</span> by symmetry. But you had it as <span class="math-container">$\color{red}{\frac12 p}$</span> which is wrong. Thus the correct equation is:</p>
<p><span class="math-container">$$p = P(HHH) + P(HT) p + P(HHT) \color{red}{\frac12}= \frac18 + \frac14 p + \frac18 \color{red}{\frac12}$$</span></p>
|
1,251,334 | <blockquote>
<p>$||x-2|-3| >1$, then $x$ belongs to:<BR> (a) $(-\infty, -2) \cup (0,4) \cup(6,\infty)$<br> (b) $(-1,1)$<br> (c) $(-\infty, 1)\cup (1,
> \infty)$ <br> (d) $(-2,2)$</p>
</blockquote>
<p><strong><em>Answer:(a)</em></strong></p>
<p>My solution:</p>
<p>let $|x-2| = p,$<br>
$|p-3|>1$ and finally i got four inequalities after more calculations :<br></p>
<p>$x>6, x<-2,x>0, x<4$</p>
<p>I think I am doing this question wrong but i don't know how to do this question either way. How do I do this?</p>
| layman | 131,740 | <p>Ok, so $|x - 2| = p$, and so $|p - 3| > 1$.</p>
<p>Then that means $p - 3 > 1$ or $p - 3 < -1$.</p>
<p>Then that means $p > 4$ or $p < 2$.</p>
<p>But $p = |x - 2|$, so $|x - 2| > 4$ or $|x - 2| < 2$.</p>
<p>That means either $x - 2 > 4$ or $x - 2 < -4$ or $-2 < x - 2 < 2$.</p>
<p>That means either $x > 6$ or $x < -2$ or $0 < x < 4$.</p>
<p>Notice that the last set of inequalities (the $0 < x < 4$ one) is an <em>and</em>. Your answer isn't represented exactly like this. You wrote "$x > 0$ or $x < 4$" but that's different than writing $0 < x < 4$.</p>
|
188,503 | <p>A Cayley table of an finite group has to have every element exactly once in every row and exactly once in every column.</p>
<p><strong>Proof</strong> that every element of a group has to be at most once in every row and at most once in every column:</p>
<p>Let $(G, \circ)$ be a group and $a, b, c, d \in G$ with:</p>
<p>(I) $a \circ b = d$</p>
<p>(II) $a \circ c = d \Leftrightarrow a = d \circ c^{-1}$</p>
<p>Then:</p>
<p>$\begin{align}
(a \circ c) \circ (a \circ b) &= d \circ d \\
\Leftrightarrow d \circ (d \circ c^{-1} \circ b) &= d \circ d \\
\Leftrightarrow d \circ c^{-1} \circ b &= d\\
\Leftrightarrow c^{-1} \circ b &= e\\
\Leftrightarrow b &= c
\end{align}$</p>
<p>As the group is finite, this also means it is exactly once in every row/column ($\forall a,b \in G: a \circ b = x$ with $x \in G$).</p>
<p>Now my question is:</p>
<p><strong>Does a group with an infinite number of elements exist, that has not every element in every row/column of its Cayley table?</strong></p>
<p>(I know that Cayley tables usually get used only for finite groups. But if set of the group has a countable number of elements, you can imagine a Cayley table. For example, $(\mathbb{Z}, +)$ has obviously every element in every row/column).</p>
| William | 13,579 | <p>In the general setting, having at least one element in every row and column just means that given any $g$ and any $a$, there exists a $b$ and $b'$ such that $ab = g$ and $b'a = g$. This is true because you can let $b = a^{-1}g$ and $b' = ga^{-1}$. </p>
<p>Having exactly one element in each row or column is equivalent to $ac = g$ and $ac' = g$ implying that $c = c'$. This is because $ac = g = ac'$. Multiplying $a^{-1}$ to both side gives $c = c'$. Do the same thing for $ca = g$ and $c'a = g$. </p>
|
2,458,613 | <p>Say that we have $k$ red balls and $n-k$ blacks balls for $n$ balls total. Then, say we partition the balls into equal sized groups of size $m$. What is the expected number of groups with a red ball?</p>
<p>It seems clear that I should use linearity of expectation of some sort. I tried calculating the probability that any one group has at least one red ball, but I can't seem to get my equation to match my code simulation result.</p>
<p>Any help would be appreciated</p>
| true blue anil | 22,388 | <p>Let $X_i$ be an indicator random variable $=1$ if the $i^{th}$ group has a red ball, and $=0$ otherwise.</p>
<p>Then $P[i^{th}$ group has a red ball] $= \left[1 - \dfrac{\binom{n-k}{m}}{\binom{n}{m}}\right]$<br>
Now the expectation of an indicator r.v. is just the probability of the event it indicates, so $E[X_i] = \left[1 - \dfrac{\binom{n-k}{m}}{\binom{n}{m}}\right]$ </p>
<p>By linearity of expectation we have <em>expectation of sum = sum of expectations</em>,<br>
$E[\sum{(X_i)}] = \sum{E(X_i)} = \dfrac{n}{m}\left[1 - \dfrac{\binom{n-k}{m}}{\binom{n}{m}}\right]$ </p>
|
3,267,110 | <p>This integral
<span class="math-container">$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$</span>
looks suspiciously interesting as it is independent of the parameter <span class="math-container">$b$</span>. The question is: What is the best way of proving or disproving this?</p>
| Z Ahmed | 671,540 | <p><strong>One way:</strong></p>
<p>Denote the integral as <span class="math-container">$$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$</span>
<span class="math-container">$$(x^2-b^2)^2+a^2x^2=(x^2+p^2)(x^2+q^2) \Rightarrow p=(a+c)/2,q=(a-c)/2, c=\sqrt{a^2-4b^2}.$$</span>
Then <span class="math-container">$$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx=
\frac{2a^2}{p^2-q^2}[p\tan^{-1}(x/p)-q \tan^{-1}(x/q)]_{0}^{\infty}=\frac{a^2\pi}{p+q}=a\pi.$$</span></p>
|
3,267,110 | <p>This integral
<span class="math-container">$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$</span>
looks suspiciously interesting as it is independent of the parameter <span class="math-container">$b$</span>. The question is: What is the best way of proving or disproving this?</p>
| Z Ahmed | 671,540 | <p><strong>How about this, please check it critically</strong></p>
<p>Denote the integral as <span class="math-container">$$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$</span>
<span class="math-container">$$(x^2-b^2)^2+a^2x^2=(x^2-r^2)(x^2-s^2) \Rightarrow r=(d+ia)/2,s=(d-ia)/2, d=\sqrt{4b^2-a^2}.$$</span>
<span class="math-container">$$I=\frac{a^2}{r^2-s^2} \int_{-\infty}^{\infty} \left( \frac{r^2}{x^2-r^2}-\frac{s^2}{x^2-s^2} \right)dx.$$</span>
By considering a semi-circle contour in the upper half plane and applying the residue theorem, we get
<span class="math-container">$$I=\frac{a^2}{r^2-s^2} 2i\pi \left( r^2 Res \left (\frac{1}{x^2-r^2} \right)_{x=r}-s^2 Res \left ( \frac{1}{x^2-s^2} \right)_{x=-s}\right)=\frac{i\pi a^2}{r-s}=a\pi.$$</span></p>
|
132,957 | <p>Could you show me that how can I solve this as fast as possible, please? <br /></p>
<p>$$ ABCDEF \, \land \, DEFABC \; \large{ \in \, \mathbb{N^{+}} } $$ $$ ABCDEF \, = \, 6 \times (DEFABC) \\ $$ $$ (A+B+C+D+E+F)=\; ? $$</p>
<p>Thank you very much... :)</p>
| Wonder | 27,958 | <p>DEFABC = 142857 obviously here.</p>
<p>More systematically though, let the sum be X. The equation shows that X is divisible 3, so applying it again shows that X is divisible by 9. Assuming that ABCDEF are all distinct, we have to choose 4 digits to reject. Overall sum of 0-9 is 45, so it must be less than that. Sum of rejects being 9 would make it hard to construct DEFABC as 0,1 would have to be rejected. Sum of rejects being 27 would have required us to reject 9, making ABCDEF hard to construct. So sum of rejects must be 18, thus X is 27.</p>
|
195,759 | <p>I have three wavelengths. Each of them has a different colour. I want to plot the resulting wave with colour that corresponds to the mixture of these colour intensities at every point. I tried this code:</p>
<pre><code>Plot[I1[y]+I2[y]+I3[y],{y,-0.0000001,0.0000001},ColorFunction ->colf]
colf[y_] := Blend[{{I1[y],Blue},{I2[y],Yellow},{I3[y],Red}}]
</code></pre>
<p>But it didn't work out. What's wrong?</p>
<p><a href="https://i.stack.imgur.com/JzC4v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JzC4v.png" alt="enter image description here"></a></p>
| Roman | 26,598 | <p>This works:</p>
<pre><code>Plot[I1[y] + I2[y] + I3[y], {y, -0.0000001, 0.0000001},
ColorFunctionScaling -> False,
ColorFunction -> (Blend[{Blue, Yellow, Red}, {I1[#], I2[#], I3[#]}] &)]
</code></pre>
|
30,191 | <p>Does anybody know about software that exactly calculates the tree-width of a given graph and outputs a tree-decomposition? I am only interested in tree-decompositions of reasonbly small graphs, but need the exact solution and a tree-decomposition. Any comments would be great. Thanks!</p>
| supercooldave | 2,620 | <p>Some tree decomposition software is given here: <a href="http://hein.roehrig.name/dipl/" rel="nofollow">http://hein.roehrig.name/dipl/</a>. I haven't used it so can say nothing about it's quality.</p>
|
30,191 | <p>Does anybody know about software that exactly calculates the tree-width of a given graph and outputs a tree-decomposition? I am only interested in tree-decompositions of reasonbly small graphs, but need the exact solution and a tree-decomposition. Any comments would be great. Thanks!</p>
| Bart Jansen | 5,200 | <p>You could try the LibTW software, which is freely available from <a href="http://www.treewidth.com/" rel="nofollow">http://www.treewidth.com/</a> (also read their FAQ linked at the bottom of the page). It's written in Java so you can easily extend it with your own functionality.</p>
|
3,350,091 | <p>In 2d space I have two cameras with different field of view (<span class="math-container">$40$</span> and <span class="math-container">$46$</span> degrees in my example) and have some distance between them.<br>
Is would possible to describe this distance using just this two numbers? And using degrees as dimension of result?<br>
<img src="https://i.stack.imgur.com/wWV5n.png" alt="fig 1"></p>
<p>It's theoretical question. If it impossible it is okay.</p>
| Matematleta | 138,929 | <p>I think what you want follows from the definition of the "partials" in an arbitrary manifold. If I misunderstood your question, please advise, and I will delete my answer. </p>
<p>Let <span class="math-container">$x\in \mathbb R^n$</span>, and <span class="math-container">$r_i$</span> the projections from <span class="math-container">$\mathbb R^n$</span> to the <span class="math-container">$i^{th}$</span> coordinate. As you point out, Tu proves that the derivations <span class="math-container">$\{\partial/\partial r_i\}_x$</span> form a basis for <span class="math-container">$T_x\mathbb R^n$</span>.</p>
<p>Now let <span class="math-container">$p\in M$</span>, an <span class="math-container">$n$</span>-dimensional manifold and consider a chart <span class="math-container">$(\phi,U)$</span> about <span class="math-container">$p$</span>. Then, since <span class="math-container">$\phi$</span> is a diffeomorphism onto <span class="math-container">$\phi(U),\ \phi_*^{-1}: T_{\phi(p)}R^n\to T_pU\cong T_p M$</span> is an isomorphism. </p>
<p>Now, if we <span class="math-container">$\textit{define}\ \left(\partial/\partial x_i\right)_p$</span> to be <span class="math-container">$\phi_*^{-1}\left(\partial/\partial r_i\right)_p,$</span> we get a basis for <span class="math-container">$T_pM$</span>.</p>
<p>From this definition, the rest follows: the covectors <span class="math-container">$dx^i$</span> are simply the (dual) basis for <span class="math-container">$T_p^*M.$</span></p>
|
3,350,091 | <p>In 2d space I have two cameras with different field of view (<span class="math-container">$40$</span> and <span class="math-container">$46$</span> degrees in my example) and have some distance between them.<br>
Is would possible to describe this distance using just this two numbers? And using degrees as dimension of result?<br>
<img src="https://i.stack.imgur.com/wWV5n.png" alt="fig 1"></p>
<p>It's theoretical question. If it impossible it is okay.</p>
| hal4math | 699,910 | <p><span class="math-container">$\newcommand{\R}{\mathbb{R}}$</span>
So because we are in <span class="math-container">$\mathbb{R}^n$</span> we can all imagine (or at least were told) how we should visualize what a tangent space is at a certain point <span class="math-container">$p$</span> in <span class="math-container">$\mathbb{R}^n$</span>, i.e. some arrows that are tangent. The key point here is that this concept only holds because one can think of the <span class="math-container">$\mathbb{R}^n$</span> as vector space (and usually does it). Then a point in <span class="math-container">$\R^n$</span> is really nothing else then a vector in <span class="math-container">$\R^n$</span>. But this is something that will not be true for manifolds. So when you think of <span class="math-container">$\R^n$</span> as being a manifold you should not think of any point in <span class="math-container">$\mathbb{\R}^n$</span> as vector but really just as point. Now, you can attach a vector space, that is the tangent space, at any point in <span class="math-container">$\mathbb{R}^n$</span>, and then from there you have a vector space <span class="math-container">$T_p(\mathbb{\R}^n)$</span> associated with that point <span class="math-container">$p$</span> of the manifold <span class="math-container">$\R^n$</span>. </p>
<p>Of course now that seems like just a lot of words and really for the <span class="math-container">$\R^n$</span> it probably is, but as we get more and more abstract those difference are key in the theory of manifolds.</p>
<p>Now, think of a donut or a sphere in <span class="math-container">$\R^n$</span>, then again you can attach to any point in that <em>manifold</em> a tangent space. Intuitively you will know how to do it, because that object is imbedded into a euclidian space. But later on in your studies of manifolds this will not be the case! (At least not in any practical way). </p>
<p>So, we need to find a concept of tangent vectors that only rely on the local properties of a point of a manifold. And this is were the (partial) derivative notation comes into play. Because these are concepts that can be also defined for manifolds in general.</p>
<p>To the question at hand: As it was pointed out, isomorphic means it is the same thing! Both are <span class="math-container">$n$</span> dimensional linear vector space which are isomorphic to each other and now the only question is which kind of "names" you want to give these vectors. But that is all that is different: the symbols. </p>
<p>Edit: Maybe to make it more precise: Take the basis <span class="math-container">$e_1,\dots,e_n$</span> of <span class="math-container">$T_p(\R^n)$</span> and let <span class="math-container">$\varphi : T_p(\R^n) \to D_p(\R^n)$</span> be the isomorphism. Then <span class="math-container">$\varphi(e_i) = \partial/\partial_{x^i}|_p$</span> for all <span class="math-container">$i\in\{1,\dots,n\}$</span>. Now, you do the calculations for <em>this basis</em> in <span class="math-container">$D_p(\R^n)$</span>. So, whenever you now have a element <span class="math-container">$v \in D_p(\R^n)$</span> and you had enough of this derivative notation you simply do <span class="math-container">$\varphi^{-1}(v)$</span> or even more concrete: you have <span class="math-container">$v = \sum_{i=1}^n v^i \partial/\partial_{x^i}|_p$</span>, so <span class="math-container">$\varphi^{-1}(v) = \sum_{i=1}^n v^i e_i$</span>.</p>
|
2,243,900 | <p>I've researched this topic a lot, but couldn't find a proper answer to this, and I can't wait a year to learn it at school, so my question is:</p>
<blockquote>
<p>What exactly is calculus? </p>
</blockquote>
<p>I know who invented it, the Leibniz controversy, etc., but I'm not exactly sure what it is. I think I heard it was used to calculate the area under a curve on a graph. If anyone can help me with this, I'd much appreciate it.</p>
| erfink | 376,021 | <p>As other answers have broken down some of the applications and the categories of calculus, I'll try to give a more intuitive and motivating explanation.</p>
<p>At its core, calculus is about trying to do meaningful computations with quantities that are infinitely large or infinitely small ("infinitesimals"). What math student, upon being introduced to the idea of infinity, hasn't wondered about <span class="math-container">$\infty - \infty$</span> or <span class="math-container">$0 \cdot \infty$</span>? Or wondered what <span class="math-container">$\frac{0}{0}$</span> should be? As Siri will tell you:</p>
<blockquote>
<p>Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn’t make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends.</p>
</blockquote>
<p>Furthermore, there are similar ancient questions such as <a href="https://en.wikipedia.org/wiki/Zeno%27s_paradoxes" rel="noreferrer">Zeno's Paradox</a>---what happens if you take infinitely many steps that are infinitely small? Even the great Archimedes had grappled with such questions and worked with a "proto-calculus" of sorts, almost two millennia before Newton and Leibniz.</p>
<p>In it's original formulations by Newton and Leibniz, calculus was about trying to perform these sorts of computations and get meaningful answers. In Leibniz's version, <span class="math-container">$\mathrm{d}x$</span> and <span class="math-container">$\mathrm{d}y$</span> were literally meant to be infinitesimal quantities in the <span class="math-container">$x$</span>- and <span class="math-container">$y$</span>-directions that were smaller than any real number, but still non-zero (Newton had a similar sort of notation).
Consider the slope of a curve at a single point, which would ordinarily be <span class="math-container">$\frac{0}{0} = \frac{f(a) - f(a)}{a-a}$</span>. In Leibniz's calculus, we could compute it as <span class="math-container">$\frac{\mathrm{d} y}{\mathrm{d}x}$</span>---the infinitesimal change in the <span class="math-container">$y$</span>-direction divided by the infinitesimal change in the <span class="math-container">$x$</span>-direction.</p>
<p>While this "infinitesimal" approach worked for nearly a hundred years, the logical inconsistencies grew more problematic. Thus Cauchy and other luminaries introduced the modern limit definitions of derivatives and integrals to give calculus a rigorous foundation.
While modern calculus is phrased in terms <em>limits</em>, generally of a sequence of approximations that become arbitrarily precise, the entire field is fundamentally about trying to make sense of calculations using infinitely small and infinitely large quantities.</p>
<h3>Examples</h3>
<ul>
<li>What happens if you add up <span class="math-container">$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$</span>? Intuitively, it feels like the answer should be <span class="math-container">$1$</span>, as you cover half the remaining distance with each additional term. However, you never actually get to <span class="math-container">$1$</span>, no matter how many terms you add up. Furthermore, how can adding up infinitely many non-zero numbers give a finite result?</li>
<li>How does a car speedometer work? What does it mean to be moving at <span class="math-container">$60mph$</span> when you haven't actually traveled <span class="math-container">$60$</span> miles or driven for an hour in the instant that you glance at the speedometer? It certainly makes sense to calculate an average speed over a finite distance, but that requires some finite amount of time as well. What does it mean to be traveling a certain speed at one instant? We can try to think of this as moving an "infinitesimal distance in an infinitesimal amount of time," but how can we actually calculate anything with that?</li>
<li>Given an arbitrary geometric shape, we can approximate its length/area/volume with nice shapes, such as rectangles. If we allow ourselves to approximate a shape with lots and lots of rectangles, we seem to get a better approximation. What if we use infinitely many rectangles that are infinitely small? How can we use this to find the true area?</li>
</ul>
|
292,639 | <p>Traditional Fourier analysis picks a period and then describes a function as: $$f(x) = \frac{1}{2} a_0 + \sum_{k=1}^\infty\, (a_k \cos{(\omega \cdot kx)} + b_n \sin{(\omega \cdot kx)})$$</p>
<p>I am wondering whether there is a way to Fourier-analyze a function in a way that the period is dependent on $x$. Let $g$ be a continuous (or differentiable, if necessary) function that is positive for all arguments $x$. Is there a representation of $f$ in a form that looks something like this? $$f(x) = \frac{1}{2} a'_0 + \sum_{k=1}^\infty\, (a'_k \cos{(g(x) \cdot kx)} + b'_n \sin{(g(x) \cdot kx)})$$</p>
<p>Perhaps one can first make $f(x)$ periodic in the traditional sense, then apply Fourier analysis, and then backtransform the result. This is just an idea.</p>
<p>By the way, $g$ needs to have certain properties for this question to make sense. If $g(x)$ falls so steeply that no period is ever completed, Fourier analysis might not be sensible. If someone would like to work out the details, he may feel free to do so here.</p>
| copper.hat | 27,978 | <p>Note that you are solving $\int_{-a}^a f(x) dx$ where $f$ is odd. What does that say about $\int_{-a}^0 f(x) dx$ and $\int_{0}^a f(x) dx$?</p>
|
162,147 | <p>I have</p>
<pre><code>J = Table[{x10, y10, x10*y10}, {x10, 0, 1, 0.5}, {y10, 0, 1, 0.5}]
L = Table[{x10, y10, 2.0*x10*y10}, {x10, 0, 1, 0.5}, {y10, 0, 1, 0.5}]
</code></pre>
<p>I want the third elements of J and L to be added and the first and second elements are as they are (as they are the same in both cases) for an example.</p>
| kglr | 125 | <pre><code>tJ = Table[{x10, y10, x10*y10}, {x10, 0, 1, 0.5}, {y10, 0, 1, 0.5}];
tL = Table[{x10, y10, 2.0*x10*y10}, {x10, 0, 1, 0.5}, {y10, 0, 1, 0.5}];
</code></pre>
<p>Several alternatives:</p>
<pre><code>tK1 = Map[{0, 0, 1} # &, tJ, {-2}] + tL;
tK2 = MapAt[2 # &, Mean[{tJ, tL}], {{All, All, -1}}];
tK3 = Mean[{tJ, tL}]; tK2[[All, All, -1]] = 2 tK2[[All, All, -1]];
Grid[tK1, Dividers -> All] // TeXForm
</code></pre>
<blockquote>
<p>$\begin{array}{|c|c|c|}
\hline
\{0.,0.,0.\} & \{0.,0.5,0.\} & \{0.,1.,0.\} \\
\hline
\{0.5,0.,0.\} & \{0.5,0.5,0.75\} & \{0.5,1.,1.5\} \\
\hline
\{1.,0.,0.\} & \{1.,0.5,1.5\} & \{1.,1.,3.\} \\
\hline
\end{array}$</p>
</blockquote>
<pre><code>tK1 == tK2 == tK3
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
177,091 | <p>What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral? </p>
<p>I saw this in the Fourier transform, and am unsure how to solve this.</p>
| copper.hat | 27,978 | <p>Formally it is solved exactly as in the real case. The antiderivative is $\frac{1}{i k} e^{ikx}$. So, $\int e^{ikx} dx = \frac{1}{i k} e^{ikx} + C$.</p>
|
177,091 | <p>What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral? </p>
<p>I saw this in the Fourier transform, and am unsure how to solve this.</p>
| jrand | 8,140 | <p>The definition of the Fourier inverse:</p>
<p>$$ f(t) = \mathfrak{F}^{-1}\{F(jw)\} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(jw) e^{jwt} dw $$</p>
<p>The Fourier pair (in the angular frequency domain):</p>
<p>$$ \delta(t) \leftrightarrow 1 $$</p>
<p>The integral in the question: </p>
<p>$$ 2\pi \times \frac{1}{2\pi} \int_{-\infty}^{+\infty} 1 \times e^{jxk} dx = 2\pi \times \delta(k) = 2\pi \delta(k)$$</p>
<p>The variable substitution $k=t$ was made and the u-substitution $w=x$ was made, for clarity.</p>
|
3,986,785 | <p>On page 153 of <em>Linear Algebra Done Right</em> the second edition, it says:</p>
<blockquote>
<p>Define a linear map <span class="math-container">$S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$</span> by:</p>
</blockquote>
<blockquote>
<p><strong>7.43:</strong> <span class="math-container">$S_1 (\sqrt{T^* T}v)=Tv$</span></p>
</blockquote>
<blockquote>
<p>First we must check that <span class="math-container">$S_1$</span> is <strong>well defined</strong>. To do this, suppose <span class="math-container">$v_1, v_2 \in V$</span> are such that <span class="math-container">$\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$</span>. For the definition given by 7.43 to make sense, we must show that <span class="math-container">$Tv_1=T v_2$</span>.</p>
</blockquote>
<p>It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?</p>
<p>Thanks</p>
| Will Harris | 851,396 | <p>A function <span class="math-container">$f: A \to B$</span> is well defined if for every <span class="math-container">$x \in A$</span>, <span class="math-container">$f(x)$</span> is equal to a single value in <span class="math-container">$B$</span>. So if <span class="math-container">$f(x)$</span> could equal two different values in <span class="math-container">$B$</span>, then it is not well defined. Or if <span class="math-container">$f(x)$</span> is not equal to a value in the set <span class="math-container">$B$</span>, then it is not well defined.</p>
|
3,986,785 | <p>On page 153 of <em>Linear Algebra Done Right</em> the second edition, it says:</p>
<blockquote>
<p>Define a linear map <span class="math-container">$S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$</span> by:</p>
</blockquote>
<blockquote>
<p><strong>7.43:</strong> <span class="math-container">$S_1 (\sqrt{T^* T}v)=Tv$</span></p>
</blockquote>
<blockquote>
<p>First we must check that <span class="math-container">$S_1$</span> is <strong>well defined</strong>. To do this, suppose <span class="math-container">$v_1, v_2 \in V$</span> are such that <span class="math-container">$\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$</span>. For the definition given by 7.43 to make sense, we must show that <span class="math-container">$Tv_1=T v_2$</span>.</p>
</blockquote>
<p>It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?</p>
<p>Thanks</p>
| Ruy | 728,080 | <p>When authors claim that something needs to be proven to be well defined that is because the definition given is under suspicion of having some flaw: it might not define anything at all. For example:</p>
<p>(Fake) Definition. Let <span class="math-container">$x$</span> be the smallest strictly positive real number.</p>
<p>Since there is no smallest positive real number, the definition above doesn't define anything.</p>
<p>Most of the time, this occurs in the literature when one is attempting to define a function, as lavishly explained in several other answers, but whenever there is reason for suspicion, a definition must be proven to be ok.</p>
|
3,960,689 | <p>I study maths as a hobby and have come across this problem.
Two circles with centres A and B intersect at points P and Q, such that <span class="math-container">$\angle APB$</span> is a right angle. If AB = xcm and <span class="math-container">$\angle PAQ = \frac{1}{3}\pi$</span> radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.</p>
<p>I calculate the area of the segment APQ to be <span class="math-container">$\frac{\pi}{6}r^2$</span>, where r = radius.
The area of <span class="math-container">$\triangle APQ$</span> I calculate as <span class="math-container">$\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$</span></p>
<p>I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.</p>
<p>But I cannot proceed any further and certainly not in terms of the length x.</p>
<p>This is the diagram as I visualise it:</p>
<p><a href="https://i.stack.imgur.com/CaRcC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CaRcC.png" alt="enter image description here" /></a></p>
| A-Level Student | 764,058 | <p><strong>HINT</strong></p>
<p>We know that <span class="math-container">$\angle PAB=\frac{\pi}{6}$</span>, hence the radius of the larger circle is <span class="math-container">$x\cos{\frac{\pi}{6}}=\frac{x\sqrt3}{2}$</span>. Similarly, the radius of the smaller circle is
<span class="math-container">$x\sin{\frac{\pi}{6}}=\frac{x}{2}$</span>.</p>
<p>Does that help? If you need any more help, please don't hesitate to ask. I have a full solution ready to post if you need it.</p>
|
3,960,689 | <p>I study maths as a hobby and have come across this problem.
Two circles with centres A and B intersect at points P and Q, such that <span class="math-container">$\angle APB$</span> is a right angle. If AB = xcm and <span class="math-container">$\angle PAQ = \frac{1}{3}\pi$</span> radians, find in terms of x the length of the perimeter and the area of the region common to the two circles.</p>
<p>I calculate the area of the segment APQ to be <span class="math-container">$\frac{\pi}{6}r^2$</span>, where r = radius.
The area of <span class="math-container">$\triangle APQ$</span> I calculate as <span class="math-container">$\frac{1}{2}r^2\sin \frac{\pi}{3} = \frac{1}{2}r^2\sqrt3$</span></p>
<p>I also know that to find the right hand side of the central segment I need to subtract the area of the triangle APQ from the area of the segment APQ.</p>
<p>But I cannot proceed any further and certainly not in terms of the length x.</p>
<p>This is the diagram as I visualise it:</p>
<p><a href="https://i.stack.imgur.com/CaRcC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CaRcC.png" alt="enter image description here" /></a></p>
| Shubham Johri | 551,962 | <p><strong>Hint.</strong> For the common area,
<span class="math-container">$$\text{area of sector PBQ}+\text{area of sector PAQ}=\text{area of }\square\text{PBQA}+\text{common area of circles}$$</span></p>
<p>where <span class="math-container">$$\begin{align*}&\text{area of sector PAQ}=\frac16\pi r_1^2\\
&\text{area of sector PBQ}=\frac13\pi r_2^2\\
&\text{area of }\square\text{PBQA}=AB\times PD=xr_1\sin\angle PAD\end{align*}$$</span></p>
<p>Can you find <span class="math-container">$r_1,r_2$</span> in terms of <span class="math-container">$x$</span>? Use the fact that <span class="math-container">$\triangle APB$</span> is right-angled.</p>
|
2,957,481 | <p>How many ways are there to roll a die six times such that there are more ones than twos?</p>
<p>I broke this up into six cases: </p>
<p><span class="math-container">$\textbf{EDITED!!!!!}$</span></p>
<p><span class="math-container">$\textbf{Case 1:}$</span> One 1 and NO 2s --> 1x4x4x4x4x4 = <span class="math-container">$4^5$</span>. This can be arranged in six ways: <span class="math-container">$\dfrac{6!}{5!}$</span>. So there are <span class="math-container">$\dfrac{6!}{5!}$$4^5$</span> ways for this case. </p>
<p><span class="math-container">$\textbf{Case 2:}$</span> Two 1s and One 2 OR NO 2s --> 1x1x5x4x4x4 = <span class="math-container">$5x4^3$</span>. This can be arranged in <span class="math-container">$\dfrac{6!}{2!3!}$</span> ways. So there are <span class="math-container">$(6x4^3)$$\dfrac{6!}{2!3!}$</span> ways for this case. </p>
<p><span class="math-container">$\textbf{Case 3:}$</span> Three 1s and Two, One or NO 2s --> 1x1x1x5x5x4 = 4x<span class="math-container">$5^2$</span>. This can be arranged in <span class="math-container">$\dfrac{6!}{2!3!}$</span> ways. So there are <span class="math-container">$(4x5^2)$$\dfrac{6!}{2!3!}$</span> ways for this case. </p>
<p><span class="math-container">$\textbf{Case 4:}$</span> Four 1s and Two, One or NO 2s --> 1x1x1x1x5x5 = <span class="math-container">$5^2$</span>. This can be arranged in <span class="math-container">$\dfrac{6!}{4!2!}$</span> ways. So there are <span class="math-container">$(5^2)$$\dfrac{6!}{4!2!}$</span> ways for this case. </p>
<p><span class="math-container">$\textbf{Case 5:}$</span> Five 1s and One or NO 2s --> 1x1x1x1x1x5 = 5. This can be arranged in <span class="math-container">$\dfrac{6!}{5!}$</span> ways = 6 ways. So there are <span class="math-container">$5^3$</span> ways for this case. </p>
<p><span class="math-container">$\textbf{Case 6:}$</span> Six 1s and NO 2s --> 1x1x1x1x1x1 = 1. There is only one way to arrange this so there is only 1 way for this case. </p>
<p>With this logic...I would add the number of ways from each case to get my answer. </p>
| Ross Millikan | 1,827 | <p>Case 1 is wrong because there are <span class="math-container">$4$</span> numbers that are neither <span class="math-container">$1$</span> nor <span class="math-container">$2$</span>, so there are <span class="math-container">$6 \cdot 4^5$</span> ways to roll one <span class="math-container">$1$</span> and no <span class="math-container">$2$</span>s. That error repeats. </p>
<p>Added: Case 2 is wrong because the number of ways to arrange the numbers is dependent on whether there is a <span class="math-container">$2$</span> or not. The <span class="math-container">$1 \cdot 1 \cdot 5 \cdot 4^3$</span> is really <span class="math-container">$1 \cdot 1 \cdot (4+1) \cdot 4^3$</span> where the <span class="math-container">$4+1$</span> is four ways to not get a <span class="math-container">$2$</span> and one way to get one. If you don't get one there are <span class="math-container">$\frac {6!}{2!4!}={6 \choose 2}=15$</span> ways to arrange them.</p>
|
4,305,593 | <blockquote>
<p>Prove that
<span class="math-container">$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$</span></p>
</blockquote>
<p>Integration by parts gives</p>
<p><span class="math-container">\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl( \int_{0}^{R} \frac{\sin^2x}{x^3} \, dx - \int_{0}^{R} \frac{\sin x}{x^2} \, dx \biggr)\\
&= \lim_{R\to\infty} \biggl( \frac{\sin^2 x}{-2x^2}\Biggr\rvert_{0}^{R} - \int_{0}^{R} \frac{\sin (2x)}{-2x^2} \, dx - \biggl(-\frac{\sin x}{x} \Biggr\rvert_{0}^{R} + \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr) \biggr) \\
&= \lim_{R\to \infty} \biggl(\frac{1}{2} + \int_{0}^{2R} \frac{\sin u}{u^2/2} \, \Bigl(\frac{1}2 \, du\Bigr) - \biggl( 1 + \int_{0}^{R} \frac{\cos x}{x} \, dx \biggr)\biggr) \\
&\hspace{22em}\text{(using the substitution $u\mapsto 2x$)}\\
&= -\frac{1}{2} + \lim_{R\to \infty} \biggl(\int_{0}^{2R} \frac{\sin u}{u^2} \, du - \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr)\\
&= \frac{1}{2} + \lim_{R\to\infty} \biggl(\int_{R}^{2R} \frac{\cos x}{x} \, dx \biggr)
\end{align*}</span></p>
<p>Thus it suffices to show that <span class="math-container">$\lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, dx = \ln 2$</span>. The Taylor series expansion of <span class="math-container">$\cos x$</span> is given by <span class="math-container">$\cos x = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{(2i)!}$</span>.</p>
<blockquote>
<p>(If the step below (the one involving the interchanging of an infinite sum and integral) is valid, why exactly is it valid? For instance, does it use uniform convergence?)</p>
</blockquote>
<p>The limit equals</p>
<p><span class="math-container">$$
\lim_{R\to\infty} \int_{R}^{2R} \biggl( \frac{1}{x} + \sum_{i=1}^{\infty} \frac{(-1)^i x^{2i-1}}{(2i)!} \biggr) \, dx
= \ln 2 + \lim_{R\to\infty} \sum_{i=1}^{\infty} \biggl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\biggr]_{R}^{2R} . $$</span></p>
<p>But I don't know how to show <span class="math-container">$\lim_{R\to\infty} \lim_{R\to\infty} \sum_{i=1}^{\infty} \Bigl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\Bigr]_{R}^{2R} = -2 \ln 2$</span>.</p>
| Random Variable | 16,033 | <p>Alternatively, we could use the fact that <span class="math-container">$$\int_{0}^{\infty} t^{2} e^{-xt} \, \, \mathrm dt = \frac{2}{x^{3}}, \quad x >0,$$</span> and then apply Fubini's theorem.<span class="math-container">$$\begin{align} \int_{0}^{\infty} \frac{\sin^{2}x - x \sin x}{x^{3}} \, \mathrm dx &= \frac{1}{2}\int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) \int_{0}^{\infty} t^{2}e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2} - \frac{\cos 2x}{2} - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}+ \frac{\mathrm d}{\mathrm d t}\frac{1}{t^{2}+1} \right) \, \mathrm dt \tag{1}\\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}- \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} t^{2}\left(\frac{2}{t(t^{2}+4)} - \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} \left(\frac{2t}{t^{2}+4} - \frac{2t}{t^{2}+1} + \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \left(\ln \left(\frac{t^{2}+4}{t^{2}+1} \right) - \frac{1}{t^{2}+1} \right)\Bigg|_{0}^{\infty} \\ &= \frac{1}{2} \left(-\ln 4 +1 \right) \\ &= \frac{1}{2} - \ln 2. \end{align} $$</span></p>
<hr />
<p><span class="math-container">$(1)$</span> <span class="math-container">$\int_{0}^{\infty} x \sin(x) e^{-xt} \, \mathrm dx = -\frac{\mathrm d}{\mathrm dt} \int_{0}^{\infty} \sin(x) e^{-xt} \, \mathrm dx$</span></p>
|
52,364 | <p>A given user can interact in multiple ways with a website. Let's simplify a bit and say say a user can:</p>
<ul>
<li>Post a message</li>
<li>Comment a message</li>
<li>"like" something on the website via Facebook</li>
</ul>
<p><em>(after that we could add, following the site on twitter, buying something on the site & so on, but for readability's sake let's stick to these 3 cases)</em></p>
<p>I'm trying to find a formula that could give me a number between 0 and 100 that reflects accurately the user interaction with the given website.</p>
<p><strong>It has to take the following into account:</strong></p>
<ul>
<li><p>A user with 300 posts and a one with 400 should have almost the same score, very close to the maximum</p></li>
<li><p>A user should see his number increase faster at the beginning. For instance a user with 1 post would have 5/100, a user with 2 would have 9/100, one with 3 would have 12/100 and so on.</p></li>
<li><p>Each of these interactions have a different weight because they do not imply the same level of involvement. It would go this way: <code>Post > Comment > Like</code></p></li>
<li><p>In the end, the repartition of data should be a bit like the following, meaning a lot of user around 0-50, and then users really interacting with the website.</p></li>
</ul>
<p><img src="https://i.stack.imgur.com/65JVm.jpg" alt="enter image description here"></p>
<hr>
<p>This is quite specific and data-dependent, but I'm not looking for the perfect formula but more for how to approach this problem.</p>
| lucasdicioccio | 13,732 | <p>Just use fixed scores which adds up for each action. The data distribution will likely behave like a power-low.
If you want to prevent people from spamming the system, reward less and less as the number of actions grow (like XP in video games). You can use any $o(n)$ function (e.g. $\log$, $\mathrm{sqrt}$) with $n$ the number of actions so far.</p>
|
619,607 | <p>$\mathbf{(1)}$ Find $y^{\prime}$ of $y=8^{\sqrt x}$ </p>
<p>My try: </p>
<p>$\ln y=\ln(8)^{\sqrt x}$<br>
$\dfrac{1}{y}y^{\prime}=\sqrt{x}\ln8$<br>
I don't know how to proceed with right side. </p>
<p>$\mathbf{(2)}$ Find $y^{\prime}$ of $y=(t+4)(t+6)(t+7).$</p>
<p>This one I have no idea what to do so I don't have any work to show. My text says to use logarithmic differentiation, but still I don't how to solve this. </p>
<p>Thank you. </p>
| Brian | 114,928 | <p><strong>(1)</strong> Note that $\ln\left(a^{b}\right) = b\ln\left(a\right)$ so, for $y = 8^{\sqrt{x}}$, $\ln y = \sqrt{x}\ln\left(8\right)$. Differentiating, we get $\frac{y'}{y} = \frac{\ln\left(8\right)}{2\sqrt{x}}$ and so $y'=\ln\left(8\right)\frac{8^{\sqrt{x}}}{2\sqrt{x}}$.</p>
<p><strong>(2)</strong> I assume you're required to use logarithmic differentiation for this.</p>
<p>Remember that $\ln\left(ab\right)=\ln a + \ln b$. Therefore, for $y = \left(t+4\right)\left(t+6\right)\left(t+7\right)$, we have $\ln y = \ln\left(t+4\right) + \ln\left(t+6\right) + \ln\left(t+7\right)$. From there, differentiating, we have $\frac{y'}{y} = \frac{1}{t+4}+\frac{1}{t+6}+\frac{1}{t+7}$. Therefore, $y' = \left(t+4\right)\left(t+6\right)\left(t+7\right)\left(\frac{1}{t+4}+\frac{1}{t+6}+\frac{1}{t+7}\right)$.</p>
|
4,218,513 | <p>This is an old Schwarz lemma problem from the August 2020 UMD qualifying exam for analysis, which is posted <a href="https://www-math.umd.edu/images/pdfs/quals/Analysis/Analysis-August-2020.pdf" rel="nofollow noreferrer">here</a>. The precise wording from the test is:</p>
<p>Suppose <span class="math-container">$f(z)$</span> is a holomorphic function on the unit disk with <span class="math-container">$|f(z)| \le 3$</span> for all <span class="math-container">$|z| < 1$</span>, and
<span class="math-container">$f(1/2) = 2$</span>. Show that <span class="math-container">$f(z)$</span> has no zeros in the disk <span class="math-container">$|z| < 1/8$</span>. (Hint: first show <span class="math-container">$f(0) \neq 0$</span>).</p>
<p>There is another kind of standard problem that assumes <span class="math-container">$|f(0)| \ge r$</span> and asks to prove <span class="math-container">$|f(z)| \ge \frac{r - |z|}{1 - r|z|}$</span> on the disk <span class="math-container">$|z| < r$</span>; see problem 5 from chapter IX.1 of Gamelin. This problem instead gives you a value for <span class="math-container">$f(a)$</span>, <span class="math-container">$a \neq 0$</span>.</p>
| Geoffrey Sangston | 444,923 | <p>We know that the hyperbolic unit disk metric, <span class="math-container">$d(z, w) := |\frac{z - w}{1 - \bar{w}z}|$</span>, satisfies <span class="math-container">$d(g(z), g(w)) \le d(z, w)$</span> for analytic maps <span class="math-container">$g : \mathbb{D} \to \mathbb{D}$</span>. Note that <span class="math-container">$|\frac{f(z)}{3}| \le 1$</span> when <span class="math-container">$|z| < 1$</span>, and the maximum principle implies we can assume <span class="math-container">$|\frac{f(z)}{3}| < 1$</span>, otherwise <span class="math-container">$|\frac{f(z)}{3}|$</span> is constantly <span class="math-container">$1$</span>, contradicting <span class="math-container">$f(\frac{1}{2}) = 2$</span>. Hence</p>
<p><span class="math-container">\begin{align*}
\left|\frac{\frac{2}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}}\cdot \frac{2}{3}}\right| = d(\frac{2}{3}, \frac{f(0)}{3}) = d(\frac{f(\frac{1}{2})}{3}, \frac{f(0)}{3}) \le d(\frac{1}{2}, 0) = \frac{1}{2}.
\end{align*}</span>
If we assume that <span class="math-container">$|\frac{f(0)}{3}| \le \frac{2}{3}$</span>, then the triangle inequality applied twice implies</p>
<p><span class="math-container">\begin{align*}
\frac{2}{3} - \frac{|f(0)|}{3} \le \frac{1}{2} + \frac{|f(0)|}{9}.
\end{align*}</span>
Simplifying this implies <span class="math-container">$|\frac{f(0)}{3}| \ge \frac{1}{8}$</span>. This inequality also holds with the negation of our assumption, and so it holds generally.</p>
<p>Recall that <span class="math-container">$\phi_a (z) := \frac{z - a}{1 - \overline{a}z}$</span> is a conformal map <span class="math-container">$\mathbb{D} \to \mathbb{D}$</span> which swaps <span class="math-container">$a$</span> and <span class="math-container">$0$</span>. So <span class="math-container">$\phi_{\frac{f(0)}{3}} \circ (\frac{f}{3})$</span> maps <span class="math-container">$0$</span> to <span class="math-container">$0$</span>. The Schwarz lemma implies
<span class="math-container">\begin{align*}
\frac{|\frac{f(0)}{3}| - |\frac{f(z)}{3}|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \frac{\left||\frac{f(0)}{3}| - |\frac{f(z)}{3}|\right|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \left| \frac{\frac{f(z)}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}} \cdot \frac{f(z)}{3}} \right|= |\phi_{\frac{f(0)}{3}}(\frac{f(z)}{3})| \le |z|.
\end{align*}</span></p>
<p>Rearranging this inequality implies</p>
<p><span class="math-container">\begin{align*}
\frac{|\frac{f(0)}{3}| - |z|}{1 + |\frac{f(0)}{3}||z|} \le |\frac{f(z)}{3}|.
\end{align*}</span>
If <span class="math-container">$|z| < \frac{1}{8}$</span>, then <span class="math-container">$|\frac{f(0)}{3}| \ge \frac{1}{8}$</span> implies the first term in the above inequality is nonzero. And so we finally conclude that <span class="math-container">$f(z) \neq 0$</span> if <span class="math-container">$|z| < \frac{1}{8}$</span>.</p>
|
81,887 | <p>The problem: Prove that for $n \in \mathbb N$:</p>
<p>$$ \left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right). $$</p>
<p>The hint is to use the binomial theorem. So the left side can become:</p>
<p>$$ \sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$</p>
<p>I don't really know where to go from here, I've tried manipulating the expressions to make them look similar but I'm not really getting anywhere.</p>
| Arturo Magidin | 742 | <p><strong>Hint.</strong> $\displaystyle 1 - \frac{k}{n} = \frac{n-k}{n}$. So
$$\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right) = \frac{(n-1)(n-2)\cdots(n-(m-1))}{(n)(n)\cdots(n)}.$$</p>
|
1,794,673 | <p>Prove that there exist two real Gaussian distribution $X$ and $Y$ such that the vector $(X,Y)$ is not Gaussian. </p>
<p>The way that i tried to solve this problem was choosing $X=X$, and take before the linear combination $X-X$ or stuff like that; but in the definition of Gaussian distribution that i have the constants are also Gaussians distribution. The statement of this problem (and its generalization) is true if we assume that the constants are Gaussians? </p>
| Sergiy Maksymenko | 809,333 | <p>You can find the detailed proof in my joint paper with Olexandra Khokhliyuk for the case of weak Whitney topology:</p>
<p>[1] O. Khokhliuk, S. Maksymenko,
Smooth approximations and their applications to homotopy types.
Proceedings of the International Geometry Center, vol. 13, no. 2 (2020) 68-108
<a href="http://doi.org/10.15673/tmgc.v13i2.1781" rel="nofollow noreferrer">http://doi.org/10.15673/tmgc.v13i2.1781</a>
or
<a href="https://arxiv.org/pdf/2008.11991.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/2008.11991.pdf</a></p>
<p>I needed this kind of results two years ago and searched for the proof but failed to find it.
Finally we decided to write the detailed proof.
The paper also contains variants for manifolds with corners, inclusions of open sets (as mentioned by Ryan Budney <a href="https://mathoverflow.net/questions/35180/topology-of-function-spaces">here</a>), and relative variant for maps with fixed restriction to some subset.
It also contains references to partial results and some posts in the internet including the present <a href="https://math.stackexchange.com/questions/1794666/reference-request-inclusion-of-smooth-maps-into-continuous-maps-between-smooth">post</a> by Adrian Clough.</p>
<p>When we put the preliminary version of the paper to arXiv, Helge Glockner informed us that such a statement follows from his (currently yet unpublished) paper</p>
<p>[2] Helge Glockner.
Homotopy groups of ascending unions of infinite-dimensional manifolds (2008)
<a href="https://arxiv.org/pdf/0812.4713.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/0812.4713.pdf</a></p>
<p>Then we also included (in Remark 7.5 of [1]) explanations of how to get the mentioned result from Helge Glockner's paper [2].</p>
|
3,464,383 | <p>I couldn't find any substantial list of 'strange infinite convergent series' so I wanted to ask the MSE community for some. By <em>strange</em>, I mean infinite series/limits that <strong>converge when you would not expect them to and/or converge to something you would not expect</strong>.</p>
<p>My favorite converges to Khinchin's (sometimes Khintchine's) constant, <span class="math-container">$K$</span>. For almost all <span class="math-container">$x \in
\mathbb{R}$</span> (those for which this does not hold making up a measure zero subset) with infinite c.f. representation:
<span class="math-container">$$x = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac1{\ddots}}}$$</span>
We have:
<span class="math-container">$$\lim_{n \to \infty} =\root n \of{\prod_{i=1}^na_i} = \lim_{n \to \infty}\root n \of {a_1a_2\dots a_n} = K$$</span>
Which is...wow! That it converges independent of <span class="math-container">$x$</span> really gets me.</p>
| user97357329 | 630,243 | <p>You might find some interesting examples in the book, <strong>(Almost) Impossible Integrals, Sums, and Series</strong>. Here you have two examples:</p>
<p><strong>First example:</strong></p>
<blockquote>
<p><span class="math-container">$$\small\zeta(4)=\frac{4}{45}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty} \frac{(i-1)!(j-1)!(k-1)!}{(i+j+k-1)!}\left((H_{i+j+k-1}-H_{k-1})^2+H_{i+j+k-1}^{(2)}-H_{k-1}^{(2)}\right),$$</span>
where <span class="math-container">$H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1,$</span> denotes the <span class="math-container">$n$</span>th generalized harmonic number of order <span class="math-container">$m$</span>.</p>
</blockquote>
<p><strong>Second example:</strong></p>
<blockquote>
<p>Let <span class="math-container">$n\ge2$</span> be a natural number. Prove that
<span class="math-container">$$\sum_{k_1=1}^{\infty}\left(\sum_{k_2=1}^{\infty}\left(\cdots \sum_{k_n=1}^{\infty} (-1)^{\sum_{i=1}^n k_i} \left(\log(2)-\sum_{k=1}^{\sum_{i=1}^n k_i} \frac{1}{\sum_{i=1}^n k_i +k}\right)\right)\cdots\right)$$</span>
<span class="math-container">$$=(-1)^n\biggr(\frac{1}{2}\log(2)+\frac{1}{2^{n+1}}\log(2)+\frac{H_n}{2^{n+1}}-\sum_{i=1}^n\frac{1}{i2^{i+1}} -\frac{\pi}{2^{n+2}}\sum_{j=0}^{n-1} \frac{1}{2^j}
\binom{2j}{j}$$</span>
<span class="math-container">$$+\frac{1}{2^{n+1}}\sum_{j=1}^{n-1}\frac{1}{2^j}\binom{2j}{j}\sum_{i=1}^{j}\frac{2^i}{\displaystyle i \binom{2i}{i}}\biggr),$$</span>
where <span class="math-container">$H_n=\sum_{k=1}^n\frac{1}{k}$</span> denotes the <span class="math-container">$n$</span>th harmonic number.</p>
</blockquote>
|
3,464,383 | <p>I couldn't find any substantial list of 'strange infinite convergent series' so I wanted to ask the MSE community for some. By <em>strange</em>, I mean infinite series/limits that <strong>converge when you would not expect them to and/or converge to something you would not expect</strong>.</p>
<p>My favorite converges to Khinchin's (sometimes Khintchine's) constant, <span class="math-container">$K$</span>. For almost all <span class="math-container">$x \in
\mathbb{R}$</span> (those for which this does not hold making up a measure zero subset) with infinite c.f. representation:
<span class="math-container">$$x = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac1{\ddots}}}$$</span>
We have:
<span class="math-container">$$\lim_{n \to \infty} =\root n \of{\prod_{i=1}^na_i} = \lim_{n \to \infty}\root n \of {a_1a_2\dots a_n} = K$$</span>
Which is...wow! That it converges independent of <span class="math-container">$x$</span> really gets me.</p>
| Oscar Lanzi | 248,217 | <p>I would like to nominate an infinite product:</p>
<p><span class="math-container">$\prod_{n=2}^{\infty}\dfrac{n^3-1}{n^3+1}=\dfrac{2}{3}$</span></p>
<p>Proof: Factor thusly:</p>
<p><span class="math-container">$n^3-1=(n-1)(n^2+n+1)=((n-2)+1)(n^2+n+1)$</span></p>
<p><span class="math-container">$n^3+1=(n+1)(n^2-n+1)=(n+1)((n-1)^2+(n-1)+1)$</span></p>
<p>and the product then telescopes.</p>
|
294,972 | <p>I am seeking a particular integral of the differential equation:</p>
<p>$u^{(4)}(t) - 5u''(t) + 4u(t) - t^3 = 0$</p>
<p>I simply interested in the technique, not just an answer (Mathematica suffices for that). How would one apply undetermined coefficients in this case?</p>
| Mikasa | 8,581 | <p>We have $(D^4-5D+4)u=t^3$ where in $Du=u'$. So $$(D^2-1)(D^2-4)u=t^3$$ so $$D^4(D^2-1)(D^2-4)=0$$ So the general solution of the OE has a form of $$u=(At^3+Bt^2+Ct+E)+y_c$$ where in $y_c=C_1e^t+C_2e^{-t}+C_3e^{-2t}+C_4e^{2t}$.</p>
|
2,623,521 | <p>Prove that: $\tan (2\tan^{-1} (x))=2\tan (\tan^{-1} (x)+\tan^{-1} (x^3))$</p>
<p>My Attempt'
Let $\tan^{-1} (x)=A$
$$x=\tan (A)$$
Now,
L.H.S.$=\tan (2\tan^{-1} (x))$
$$=\tan (2A)$$
$$=\dfrac {2\tan (A)}{1-\tan^2 (A)}$$
$$=\dfrac {2x}{1-x^2}$$</p>
| mvw | 86,776 | <p>I still do no not understand the problem. But let us try to start anyway and discuss further.</p>
<p>With $m$ at rest at $A$ we have a force equilibrium
$$
0 = F_G + T_1 + T_{AC}
$$
In $x$-direction it is
$$
0 = 0 + \lVert T_1 \rVert \sin(30^\circ) - \lVert T_{AC} \rVert
$$
and in $y$-direction
$$
0 = -mg+ \lVert T_1 \rVert \cos(30^\circ) + 0
$$
so
$$
\lVert T_1 \rVert = \frac{mg}{\cos(30^\circ)}
$$
and
$$
T_1 =
\frac{mg}{\cos(30^\circ)}
\begin{pmatrix}
\sin(30^\circ) \\
\cos(30^\circ)
\end{pmatrix}
=
mg
\begin{pmatrix}
\tan(30^\circ) \\
\cos^2(30^\circ)
\end{pmatrix}
$$</p>
<p><strong>Update:</strong> (After the problem was clarified)</p>
<p>So we have a pendulum. First fixed at $CA$ and then released.</p>
<p>At point $B$ we have the force equilibrium
\begin{align}
m \ddot{r} &= F_G + T_2 \iff \\
m \lVert \ddot{r} \rVert
\begin{pmatrix}
-\cos(30^\circ) \\
-\sin(30^\circ)
\end{pmatrix}
&=
m
\begin{pmatrix}
0 \\
-g
\end{pmatrix}
+
\lVert T_2 \rVert
\begin{pmatrix}
-\sin(30^\circ) \\
\cos(30^\circ)
\end{pmatrix}
\end{align}
This gives
$$
\lVert \ddot{r} \rVert = \lVert T_2 \rVert \frac{\tan(30^\circ)}{m} \\
$$
and
$$
-\lVert T_2 \rVert \tan(30^\circ) \sin(30^\circ) =
-m g + \lVert T_2 \rVert \cos(30^\circ) \iff \\
\lVert T_2 \rVert = \frac{mg}{\cos(30^\circ) + \tan(30^\circ)\sin(30^\circ)}
= mg\cos(30^\circ)
$$
which finally gives
$$
T_2 =
mg \cos(30^\circ)
\begin{pmatrix}
-\sin(30^\circ) \\
\cos(30^\circ)
\end{pmatrix}
=
mg
\begin{pmatrix}
-\cos(30^\circ) \sin(30^\circ) \\
\cos^2(30^\circ)
\end{pmatrix}
$$
So $T_1$ and $T_2$ are equal in the $y$-component, but not in the $x$-component. Aside from the direction the $T_1$ $x$-component is larger.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.