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 |
|---|---|---|---|---|
330,488 | <p>On the complex plane <span class="math-container">$\mathbb C$</span> consider the half-open square <span class="math-container">$$\square=\{z\in\mathbb C:0\le\Re(z)<1,\;0\le\Im(z)<1\}.$$</span> </p>
<p>Observe that for every <span class="math-container">$z\in \mathbb C$</span> and <span class="math-container">$p\in\{0,1,2,3\}$</span> the set <span class="math-container">$(z+i^p\cdot\square)$</span> is the shifted and rotated square <span class="math-container">$\square$</span> with a vertex at <span class="math-container">$z$</span>.</p>
<blockquote>
<p><strong>Problem.</strong> Is it true that for any function <span class="math-container">$p:\mathbb C\to\{0,1,2,3\}$</span> there a subset <span class="math-container">$Z\subset\mathbb C$</span> such that the union of the squares
<span class="math-container">$$\bigcup_{z\in Z}(z+i^{p(z)}\cdot\square)$$</span>is not Borel in <span class="math-container">$\mathbb C$</span>?</p>
</blockquote>
<hr>
<p><strong>Added in Edit.</strong> As @YCor observed in his comment, the answer to this problem is affirmative under <span class="math-container">$\neg CH$</span>.</p>
<p>An affirmative answer to Problem would follow from an affirmative answer to another intriguing </p>
<blockquote>
<p><strong>Problem'.</strong> Is it true that for any partition <span class="math-container">$\mathbb C=A\cup B$</span> either <span class="math-container">$A$</span> contains an uncountable strictly increasing function or <span class="math-container">$B$</span> contains an uncountable strictly decreasing function? </p>
</blockquote>
<p>Here by a <em>function</em> I understand a subset <span class="math-container">$f\subset \mathbb C$</span> such that for any <span class="math-container">$x\in\mathbb R$</span> the set <span class="math-container">$f(x)=\{y\in\mathbb R:x+iy\in f\}$</span> contains at most one element.</p>
<hr>
<p><strong>Added in the Next Edit.</strong> In the discussion with @YCor we came to the conclusion that under CH the answer to both problems is negative. Therefore, both problems are independent of ZFC. Very strange.</p>
| Nate Eldredge | 4,832 | <p>(This addresses a misinterpretation of the question, where <span class="math-container">$p$</span> can be chosen. I'll try to fix it.)</p>
<p>This seems too easy, so maybe I've misunderstood the question, but: let <span class="math-container">$L$</span> be the diagonal line <span class="math-container">$\{z : \Re(z) = - \Im(z)\}$</span> and let <span class="math-container">$Z$</span> be a non-Borel subset of <span class="math-container">$L$</span>. Take <span class="math-container">$p \equiv 0$</span>. Then the set in question is <span class="math-container">$E = \bigcup_{z \in Z} (z + \Box)$</span>, but we have <span class="math-container">$E \cap L = Z$</span> so that <span class="math-container">$E$</span> is not Borel.</p>
|
2,149,006 | <p>While learning the power rule, one thing popped up in my mind which is confusing me. We know what the power rule states :</p>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}$$ where $n$ is a real number.</p>
<blockquote>
<p>But instead of $n$, if we have a trig function like $\sin(x)$, <strong>will the power rule still apply?</strong></p>
</blockquote>
<p>Eg. We have a function $y = x^{\sin(x)}$, and thus by the power rule;</p>
<p>$$\frac{dy}{dx} = sin(x)x^{sin(x)-1}$$. </p>
<p>Is this possible? Please tell me if even the function I wrote above really does exist or not.</p>
<p>I know this may seem a stupid question to many, but please help because I cannot find any explanation to this. </p>
| Dr. Sonnhard Graubner | 175,066 | <p>one possibility is taking the logarithm on both sides
$$\ln(y)=\sin(x)\ln(x)$$ and by the chain rule we get
$$\frac{y'}{y}=\cos(x)\ln(x)+\sin(x)\cdot \frac{1}{x}$$
you must multiply this equation by $$y(x)$$</p>
|
213,513 | <p>I need help to extrapolate these data:</p>
<pre><code>θ = {20.7, 28.62, 32.04};
ω = {5, 6, 7};
</code></pre>
<p>using this equation:</p>
<p>θ(ω)=θ(ω⟶∞)+ c /ω^n to know what the values of c and n</p>
<p>I found a result using the below plot:</p>
<pre><code>ListLinePlot[
Transpose[{ω^-4,#}]&/@{θ},
FrameLabel->{"1/\!\(\*SuperscriptBox[\(ω\), \(4\)]\)","θ"},
Axes->True,
Frame->True,
AxesStyle->Directive[{Bold,25},{Black,25}],
BaseStyle->{FontWeight->Bold,FontSize->14},
PlotRange->All,
PlotMarkers->{"◆",18}
]
</code></pre>
<p>But my problem how to obtain the fits using the extrapolation for the above equation using Mathematica.</p>
<p>Thanks in advance!</p>
| Henrik Schumacher | 38,178 | <p>Assuming that you meant the third key of your input to be <code>{{a1, a2}, {a1, a1}}</code>, this might do want you ask for.</p>
<pre><code>a = <|{{a1, a1}, {a2, a2}} -> 1, {{a1, a1}, {a1, a2}} -> 10, {{a1, a2}, {a1, a1}} -> 3|>;
f = Sort;
Merge[Thread[Rule[f /@ Keys[a], Values[a]]], Total]
</code></pre>
<p>You may replace <code>f</code> by any mapping that maps each key to a canonical representative of its equivalence class.</p>
|
213,513 | <p>I need help to extrapolate these data:</p>
<pre><code>θ = {20.7, 28.62, 32.04};
ω = {5, 6, 7};
</code></pre>
<p>using this equation:</p>
<p>θ(ω)=θ(ω⟶∞)+ c /ω^n to know what the values of c and n</p>
<p>I found a result using the below plot:</p>
<pre><code>ListLinePlot[
Transpose[{ω^-4,#}]&/@{θ},
FrameLabel->{"1/\!\(\*SuperscriptBox[\(ω\), \(4\)]\)","θ"},
Axes->True,
Frame->True,
AxesStyle->Directive[{Bold,25},{Black,25}],
BaseStyle->{FontWeight->Bold,FontSize->14},
PlotRange->All,
PlotMarkers->{"◆",18}
]
</code></pre>
<p>But my problem how to obtain the fits using the extrapolation for the above equation using Mathematica.</p>
<p>Thanks in advance!</p>
| Carl Woll | 45,431 | <p>You can use the <a href="http://reference.wolfram.com/language/ref/ResourceFunction" rel="nofollow noreferrer"><code>ResourceFunction</code></a> <a href="https://resources.wolframcloud.com/FunctionRepository/resources/GroupByList" rel="nofollow noreferrer">"GroupByList"</a> to do this:</p>
<pre><code>ResourceFunction["GroupByList"][Values[a], Sort /@ Keys[a], Total]
</code></pre>
<blockquote>
<p><|{{a1, a1}, {a2, a2}} -> 1, {{a1, a1}, {a1, a2}} -> 13|></p>
</blockquote>
|
2,063,038 | <p>Let <span class="math-container">$S$</span> be the region in the plane that is inside the circle <span class="math-container">$(x-1)^2 + y^2 = 1$</span> and outside the circle <span class="math-container">$x^2 + y^2 = 1 $</span>. I want to calculate the area of <span class="math-container">$S$</span>.</p>
<h3>Try:</h3>
<p>first, the circles intersect when <span class="math-container">$x^2 = (x-1)^2 $</span> that is when <span class="math-container">$x = 1/2$</span> and so <span class="math-container">$y =\pm \frac{ \sqrt{3} }{2} $</span>. So, using washer method, we have</p>
<p><span class="math-container">$$Area(S) = \pi \int\limits_{- \sqrt{3}/2}^{ \sqrt{3}/2} [ (1+ \sqrt{1-y^2})^2 - (1-y^2) ] dy $$</span></p>
<p>is this the correct setting for the area im looking for?</p>
| Omar Eafoe | 227,292 | <p>It will be complicated if you tried to solve such problems with cartesian coordinates it's much easier to solve by polar coordinates , and in that type of problems try to draw them if you draw you will find that the area of the upper half is same of the area of lower half so you can calculate 1 of them and multiply it by 2 . You brought the point of iterse ction bring it in terms of r and theta then integrate </p>
|
1,734,819 | <p>I think I'm on the right track with constructing this proof. Please let me know.</p>
<p>Claim: Prove that there exists a unique real number $x$ between $0$ and $1$ such that
$x^{3}+x^{2} -1=0$</p>
<p>Using the intermediate value theorem we get
$$r^{3}+r^{2}-1=c^{3}+c^{2}-1$$
......
$$r^{3}+r^{2}-c^{3}-c^{2}=0$$</p>
<p>Factoring gives us</p>
<p>$$(r-c)[(r^{2}+rc+c^{2})+(r+c)]=0$$
I'm lost now. How do I prove that there exists a number between $0$ and $1$.</p>
| Hagen von Eitzen | 39,174 | <p>You might rewrite the factor $(r^2+rc+c^2+r+c)$ as
$$r^2+rc+c^2+r+c = \underbrace{(r-c)^2}_{\ge0}+\underbrace{3rc}_{\ge0}+\underbrace r_{\ge0}+\underbrace c_{\ge0}\ge0 $$
with equality only if $r=c=0$.</p>
|
174,075 | <p>What is the difference when a line is said to be normal to another and a line is said to be perpendicular to other?</p>
| Prajjawal | 59,632 | <p>Normal is 90 degree to a surface but perpendicular is 90 degree to a line</p>
|
1,966,128 | <p>I know that by De Morgan's law that it is false. But how to disprove it?</p>
| RJM | 376,273 | <p>Suppose x $\in$ $A^c \cup B^c$ assume wlog that x $\in$ $A^c$. Then x $\in$ $A^c \cup B$, which implies x $\notin$ $(A \cup B)^c$</p>
|
4,490,778 | <p>Let <span class="math-container">$\mu$</span> be a sigma finite positive measure on <span class="math-container">$(X,\mathcal{A})$</span>. then exists <span class="math-container">$w\in L^1(\mu)$</span> such that <span class="math-container">$0< w(x) < 1$</span> for all <span class="math-container">$x\in X$</span>.</p>
<p>Since <span class="math-container">$\mu$</span> is a sigma finite measure we have that <span class="math-container">$$X=\bigcup_{n=1}^\infty E_n\quad \mu(E_n)<\infty.$$</span> We define <span class="math-container">$$w_n(x)=\frac{1}{2^n(1+\mu(E_n))}\quad\text{if}\;x\in E_n$$</span> zero otherwise.</p>
<p>Define <span class="math-container">$$w(x):=\sum_{n=1}^\infty w_n(x).$$</span></p>
<blockquote>
<p>I can't find a way to show that <span class="math-container">$$\int_X w\;d\mu <\infty$$</span> could someone give me a suggestion?</p>
</blockquote>
<blockquote>
<p>Why <span class="math-container">$0<w<1$</span>?</p>
</blockquote>
| Joe | 524,659 | <p>Since our measure is <span class="math-container">$\sigma$</span>-finite, we may write <span class="math-container">$X = \bigcup_{n=1}^\infty E_n$</span>, where <span class="math-container">$\mu(E_n) < \infty$</span> And <span class="math-container">$E_n \subseteq E_{n+1}$</span> for all <span class="math-container">$n$</span>. For a set <span class="math-container">$A$</span>, let <span class="math-container">$I_A(x)$</span> denote the indicator function (also called characteristic function) for the set <span class="math-container">$A$</span>. Now set:
<span class="math-container">$$\begin{align*} w_n(x) & := \frac{I_{E_n}(x)}{2^{n}(1 + \mu(E_n))} \quad n \in \mathbb{N} \\ S_n(x) & := \sum_{j=1}^n w_j(x) \\ w(x) & := \sum_{j=1}^\infty w_j(x) \end{align*} $$</span>
We may compute:
<span class="math-container">$$\begin{align*} \int_X S_n \: d\mu & = \sum_{j=1}^n \int_{X} \omega_n \: d\mu \\
& = \sum_{j=1}^n \frac{1}{2^{n}(1 + \mu(E_n))} \int_X I_{E_n} \: d \mu \\
& = \sum_{j=1}^n \frac{\mu(E)}{2^{n}(1 + \mu(E_n))} \\
& < \sum_{j=1}^n \frac{1}{2^n} \\
& \leq 1 \end{align*} $$</span>
Further, we see that <span class="math-container">$S_n \nearrow \omega$</span>. hence by the monotone convergence theorem, we have that <span class="math-container">$\int_X \omega \: d \mu \leq 1 < \infty$</span>.</p>
<p>We also see that for any <span class="math-container">$x \in X$</span>, there is some <span class="math-container">$E_k$</span> such that <span class="math-container">$x \in E_k$</span>. So we compute:
<span class="math-container">$$ \begin{align*} \omega(x) & \geq \omega_k(x) \\
& = \frac{1}{2^k(1 + \mu(E_k))} \\
& > 0 \end{align*}$$</span></p>
<p>We also have that:
<span class="math-container">$$ \begin{align*} \omega(x) & = \sum_{n=1}^\infty \frac{I_{E_n}(x)}{2^n(1 + \mu(E_n))} \\
& \leq \sum_{n=1}^\infty \frac{1}{2^n(1 + \mu(E_n))} \\
& < \sum_{n=1}^\infty \frac{1}{2^n} \\
& = 1 \end{align*}$$</span>
So <span class="math-container">$0 < w < 1$</span>.</p>
|
4,490,778 | <p>Let <span class="math-container">$\mu$</span> be a sigma finite positive measure on <span class="math-container">$(X,\mathcal{A})$</span>. then exists <span class="math-container">$w\in L^1(\mu)$</span> such that <span class="math-container">$0< w(x) < 1$</span> for all <span class="math-container">$x\in X$</span>.</p>
<p>Since <span class="math-container">$\mu$</span> is a sigma finite measure we have that <span class="math-container">$$X=\bigcup_{n=1}^\infty E_n\quad \mu(E_n)<\infty.$$</span> We define <span class="math-container">$$w_n(x)=\frac{1}{2^n(1+\mu(E_n))}\quad\text{if}\;x\in E_n$$</span> zero otherwise.</p>
<p>Define <span class="math-container">$$w(x):=\sum_{n=1}^\infty w_n(x).$$</span></p>
<blockquote>
<p>I can't find a way to show that <span class="math-container">$$\int_X w\;d\mu <\infty$$</span> could someone give me a suggestion?</p>
</blockquote>
<blockquote>
<p>Why <span class="math-container">$0<w<1$</span>?</p>
</blockquote>
| Davide Giraudo | 9,849 | <p>Actually, you can modify your construction by choosing the sequence <span class="math-container">$(E_n)$</span> to be pairwise disjoint instead of increasing. In this way,
<span class="math-container">$$
\int_X w(x)d\mu(x)=\sum_{n\geqslant 1}\int_X w_n(x)d\mu(x)\leqslant \sum_{n\geqslant 1}2^{-n}
$$</span>
and since for each <span class="math-container">$x$</span>, there exists exactly one <span class="math-container">$n(x)$</span> such that <span class="math-container">$x\in E_{n(x)}$</span>, we have <span class="math-container">$0<w(x)\leqslant 2^{-n(x)}<1$</span>.</p>
|
1,650,277 | <p>Does the category of partial orders have a subobject classifier? (Edit: No, see Eric's answer.)</p>
<p>If not, what is a category which is "close" to the category of partial orders (e.g. it should consists of special order-theoretic constructs) and has a subobject classifier? Bonus question: Is there also such an elementary topos? Notice that the category of partial orders has all limits, colimits and it is cartesian closed.</p>
| drhab | 75,923 | <p>Normally the disjoint union of topological spaces <span class="math-container">$(X_\alpha)_{\alpha\in A}$</span> is defined as: <span class="math-container">$$X:=\bigcup_{\alpha\in A}\left(X_{\alpha}\times\{\alpha\}\right)$$</span>and is accompanied by injections <span class="math-container">$\iota_{\alpha}:X_{\alpha}\to X$</span> prescribed by <span class="math-container">$x\mapsto\langle x,\alpha\rangle$</span>.</p>
<p>Preassuming that this is the case collection <span class="math-container">$\tau$</span> must actually be defined as:<span class="math-container">$$\tau:=\{U\in\wp(X)\mid\iota_{\alpha}^{-1}(U)\in\tau_{\alpha}\text{ for all }\alpha\in A\}$$</span></p>
<p>Evidently <span class="math-container">$\iota_{\alpha}^{-1}(X)=X_{\alpha}\in\tau_{\alpha}$</span> for all <span class="math-container">$\alpha\in A$</span>, so that we can conclude that <span class="math-container">$X\in\tau$</span>.</p>
|
1,307,280 | <p>I m in the point X. I m 2 blocks up from a point A and 3 blocks down from my home H. Every time I walk one block i drop a coin.</p>
<p>H
.
.
.
X
.
.
A</p>
<p>If the coin is face I go one block up and if it is not face I go one block down.</p>
<p>Which is the probability of arriving home before the point A?</p>
<hr>
<p>What I really want to do is to solve that problem in a recursive way. Maybe it can be solved with a binomial distribution... But is it also recursive?</p>
| Ittay Weiss | 30,953 | <p>You are correct. You present a nuance of the 'working definition', a special case where "getting closer" is misleading. You should interpret "getting closer" as "getting as close as you like". This is a more accurate 'working definition' in any case. Then the constant function scenario works just fine. You can get as close as you like to the constant, quite trivially. </p>
<p>The "getting closer" definition makes it sounds as if a limit is somehow an indefinite thing, moving around, getting closer to things. This is misleading. A better intuition is "the limit at $x_0$ of $f$ is $L$" is "the value $f(x)$ can be made as close as you like to $L$ for all $x$ that is sufficiently close to but not equal to $x_0$". </p>
|
165,487 | <p>After I use Simplify on an expression I get$\dfrac{1}{2}\sqrt{-\dfrac{\sqrt{(-b^2+16|c|^2)(4|c|^2+b\Im(c))^2}}{4a(4|c|^2+b\Im(c)])}}$. This expression can clearly be simplified further by noticing that the square bracket term in the numerator cancels the other bracket term in the denominator so $\dfrac{1}{2}\sqrt{-\dfrac{\sqrt{(-b^2+16|c|^2)}}{4a}}$. This is clearly a much simpler form since it includes less terms, so my question is why does not mathematica do this?</p>
<p>Edit: here is my code</p>
<pre><code>real[x_, y_] := -2 a x^3 - 2 a y^2 x + 2 Re[c] x + 2 Im[c] y + b/2 y;
imaginary[x_, y_] := -2 a x^2 y - 2 a y^3 + 2 Im[c] x - 2 Re[c] y - b/2 x;
sol = Solve[{real[x, y] == 0, imaginary[x, y] == 0},{x,y}];
FullSimplify[Sqrt[(x /. sol[[2, 1]])^2 + (y /. sol[[2, 2]])^2],
Assumptions -> {(a | b) ∈ Reals && c ∈ Complexes && (a | b | c) > 0}]
</code></pre>
| Ulrich Neumann | 53,677 | <p>The "bracket" you want to simplify is complex! </p>
<pre><code>bracket = r Exp[I φ];(* stands for (4 c Conjugate[c] + b Im[c])*)
expr = Sqrt[bracket^2]/bracket ;
FullSimplify[ expr, {Element[{r, φ}, Reals], r > 0 }]
(* E^(-I φ) Sqrt[E^(2 I φ)] *)
</code></pre>
<p>Further simplification needs information about <code>φ</code></p>
|
1,906,013 | <blockquote>
<p>Let $f$ be a smooth function such that $f'(0) = f''(0) = 1$. Let $g(x) = f(x^{10})$. Find $g^{(10)}(x)$ and $g^{(11)}(x)$ when $x=0$.</p>
</blockquote>
<p>I tried applying chain rule multiple times:</p>
<p>$$g'(x) = f'(x^{10})(10x^9)$$</p>
<p>$$g''(x) = \color{red}{f'(x^{10})(90x^8)}+\color{blue}{(10x^9)f''(x^{10})}$$</p>
<p>$$g^{(3)}(x)=\color{red}{f'(x^{10})(720x^7) + (90x^8)f''(x^{10})(10x^9)}+\color{blue}{(10x^9)(10x^9)f^{(3)}(x^{10})+f''(x^{10})(90x^8)}$$</p>
<p>The observation here is that, each time we take derivative, one "term" becomes two terms $A$ and $B$, where $A$ has power of $x$ decreases and $B$ has power of $x$ increases. $A$ parts will become zero when evaluated at zero, but what about $B$ parts?</p>
| Robert Z | 299,698 | <p>We have that $f(x)=f(0)+x+x^2/2+o(x^2)$. Therefore the expansion of $g$ at $0$ is
$$g(x)=f(x^{10})=f(0)+x^{10}+x^{20}/2+o(x^{20}).$$
Hence $g^{(10)}(0)/10!$, which is the coefficient of $x^{10}$, is equal to $1$, and we conclude that $g^{(10)}(0)=10!$. Are you able now to find $g^{(11)}(0)$?</p>
|
27,951 | <p>Something I notice is when there's an advanced/specialized question, it often receives very few upvotes. Even if it is seemingly well written. I try to upvote advanced questions <strong>that I might not even understand</strong>, if they appear well written. </p>
<p>Is this good behaviour? Should we encourage upvoting seemingly well-written questions <strong>even if you don't understand it</strong>? </p>
| Aloizio Macedo | 59,234 | <p>First off, you explicitly <a href="https://math.meta.stackexchange.com/a/25118/59234">mentioned this suggestion a while ago</a>, and the community seemed to think it is valuable.</p>
<p>That said, as the comments imply, it can be a little risky... but I don't think it is a significant risk. From my personal experience, any "nonsense" is swiftly dealt with by the community, regardless of how advanced it is. I don't know if this should be intuitive for some psychological reason, but people seem to be more active towards things which are significantly wrong than giving praise for something (for better or for worse). Furthermore, advanced topics will naturally have a lower rate of nonsense (or simply be <strong>completely</strong> nonsense, spotted even by people who don't know about said advanced topic), merely because of the level of intimacy with mathematics as a whole that it requires.</p>
<p>So, I don't think the risks are significant at all. However, one pertinent question remains: "what is the goal on doing so?"</p>
<p>It is a fact, no matter how we try to mask it, that reputation points are valuable. Be it because of the points numerically, be it simply because it is a ratification of the fact that your spent time was recognized by someone else, or some other reason which is specific to each one. If you spend time somewhere and you get no feedback on it, this will eventually wear you out. And this may eventually decline the number of answers/questions on specific topics. Likewise, if questions/answers on a topic are upvoted, there is an underlying motivation for them and also a feeling that said topic "is alive". This is less abstract than it seems.</p>
<p>Having made explicit my opinion of the <em>little risks</em> and of the <em>goal</em>, I say that this can be a good behaviour for the site... but I don't think this should be encouraged to the community as a whole, since an <em>en masse</em> upvoting of this sort could be potentially disruptive (this would never happen, but...). It also has "ethical" problems which were mentioned in the comment: upvoting something can be interpreted as you signing your name (anonimously, as oxymoronic as that might sound) under the post saying it is good quality. If you don't understand what you are signing under, this is bad. If you don't think of upvotes that way, I think you are fine.</p>
|
2,891,444 | <p>For the intersection of two line segments, how was it know to use the determinants shown <a href="http://mathworld.wolfram.com/Line-LineIntersection.html" rel="nofollow noreferrer">here</a>? </p>
<p>I'm trying to determine how it was shown that they could be used to compute the intersection point.</p>
| V. Vancak | 230,329 | <p>The probability that the first proofreader will miss the error is $0.02$. Same probability has the second proofreader. The complementary event of the event of interest is "both will miss the error". Due to independence, the probability of such event is multiplication of the two probabilities of missing the error, i.e., $0.02^2$, hence your answer is
$$
1-0.02^2=0.9996 .
$$ </p>
|
2,678,406 | <p>I'm trying to compute the distance between a point and a plane of the form
<span class="math-container">$$
ax+bx+cz = d
$$</span>
not using the standard formula for analytical geometry.</p>
<ul>
<li>I am trying to compute it by coming up with the projection matrix onto the normal of the plane passing through the origin and then projecting a a vector that is shifted the amount required to make the plane pass through the origin and then taking the length of that vector.</li>
<li>I upload pictures of my work for the first one I used this method and it worked it gave me the correct distance, but for any other equation it doesn't seem to work and I cant figure out why or what to change to make it work.</li>
</ul>
<p><a href="https://i.stack.imgur.com/7Ouuk.jpg" rel="nofollow noreferrer">Example of the method working</a>, <a href="https://i.stack.imgur.com/LEmJl.jpg" rel="nofollow noreferrer">Example of method not working</a></p>
| knight5478 | 1,129,053 | <p>you can calculate the distance between those points is by that formula:</p>
<p>1)find two linearly independent vectors from the plane (<span class="math-container">$\vec u$</span>, <span class="math-container">$\vec v$</span>)</p>
<p>2)find a vector from any point on the plane to the desired point on the line (<span class="math-container">$\vec w$</span>).</p>
<p>3)calculate the determinant of
|<span class="math-container">$\vec v$</span>
<span class="math-container">$\vec u$</span>
<span class="math-container">$\vec w$</span>|,
now you have the volume of the parallelepiped that those 3 vectors make.</p>
<p>4)divide the volume of the parallelepiped by |<span class="math-container">$\vec v $</span> x <span class="math-container">$ \vec u$</span>| which is the surface of one face of the parallelepiped.</p>
<p>this formula always gives the shortest distance from a point to a surface.</p>
|
392,020 | <p>It is well known that the minimum number of monochromatic triangles in a red/blue coloring of the edges of the complete graph <span class="math-container">$K_n$</span> is given by Goodman's formula
<span class="math-container">$$M(n)=\binom n3-\left\lfloor\frac n2\left\lfloor\left(\frac{n-1}2\right)^2\right\rfloor\right\rfloor;$$</span>
see OEIS sequence <a href="http://oeis.org/A014557" rel="nofollow noreferrer">A014557</a> or the original paper by A. W. Goodman, <a href="https://www.jstor.org/stable/2310464?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">On sets of acquaintances and strangers at any party</a>, Amer. Math. Monthly 66 (1959), 778–783, or my answer to <a href="https://math.stackexchange.com/questions/2216943/counting-triangles-in-a-graph-or-its-complement">this math.stackexchange question</a>.</p>
<blockquote>
<p>Is there any literature on the more general question, what is the minimum number of monochromatic triangles in a red/blue coloring of the edges of the complete graph <span class="math-container">$K_n$</span> <strong>with a prescribed number of edges of each color?</strong></p>
</blockquote>
<p>...</p>
<blockquote>
<p>Is there any literature on the related question, given natural numbers <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, what is the minimum value of the quantity <span class="math-container">$m+kb$</span> over all red/blue colorings of the edges of the complete graph <span class="math-container">$K_n$</span>, where <span class="math-container">$m$</span> is the number of monochromatic triangles (of either color) and <span class="math-container">$b$</span> is the number of blue edges?</p>
</blockquote>
| Thomas Bloom | 385 | <p>Suppose that the red degree of each vertex is denoted by <span class="math-container">$r_x$</span>, and the total number of red edges is <span class="math-container">$R=\frac{1}{2}\sum_x r_x$</span>. Then the number of monochromatic triangles is exactly
<span class="math-container">$$ \binom{n}{3}-(n-1)R+\frac{1}{2}\sum_xr_x^2.$$</span></p>
<p>The minimum value can be derived from this, but the answer will be a little messy according as <span class="math-container">$2R$</span> is divisible by <span class="math-container">$n$</span> or not. If <span class="math-container">$2R$</span> is divisible by <span class="math-container">$n$</span>, the minimum this can be is when all the <span class="math-container">$r_x$</span> are equal (so <span class="math-container">$=2R/n$</span>), and so for example in this case the minimum number of monochromatic triangles is</p>
<p><span class="math-container">$$ \binom{n}{3}-(n-1)R+\frac{2R^2}{n}.$$</span></p>
<p>(Note this is indeed attainable by taking red edges to be a regular graph. Also this is always a lower bound for the number of monochromatic triangles by Cauchy-Schwarz, but will not be achievable if <span class="math-container">$R$</span> is not divisible by <span class="math-container">$n$</span>.)</p>
<p>The first formula is just a rearrangement of Goodman's more general formula, which says that for all coloured <span class="math-container">$K_n$</span> the number of monochromatic triangles is equal to</p>
<p><span class="math-container">$$ \frac{1}{2}\left(\sum_x \binom{r_x}{2}+\sum_x\binom{n-1-r_x}{2}-\binom{n}{3}\right).$$</span></p>
<p>(The proof of which is just to note that if we count all monochromatic pairs of edges from each vertex then each monochromatic triangle is counted 3 times and non-monochromatic triangles are counted once.)</p>
|
70,500 | <p>I am trying to find $\cosh^{-1}1$ I end up with something that looks like $e^y+e^{-y}=2x$. I followed the formula correctly so I believe that is correct up to this point. I then plug in $1$ for $x$ and I get $e^y+e^{-y}=2$ which, according to my mathematical knowledge, is still correct. From here I have absolutely no idea what to do as anything I do gives me an incredibly complicated problem or the wrong answer.</p>
| APW | 490,629 | <p>To find the inverse for any $x$, we are looking for
$$ y = \cosh^{-1} x, $$
i.e. $$x = \cosh y = \frac{1}{2} (e^y + e^{-y}). $$
Multiplying through by $2e^y$ gives
$$ (e^y)^2 -2x\,e^y + 1 = 0, $$
which is a quadratic in $e^y$. You can then use the quadratic formula, or here completing the square
$$ (e^y-x)^2 - x^2 + 1 = 0, $$
$$ e^y -x = \sqrt{x^2 - 1}, $$
$$ y = \ln \left(x+\sqrt{x^2-1}\right). $$
At the start I said "for any $x$", but observe that the result is only valid for $x\ge 1$. This is the inverse for the right-hand side of the (even) function $y=\cosh x$, with $x\ge0$ and $y\ge1$.</p>
|
3,857,698 | <p>Let <span class="math-container">$D_1, ..., D_n$</span> be arbitrary <span class="math-container">$n$</span> sets where <span class="math-container">$D_i \cap D_j \neq \emptyset$</span>. In the simplified case where <span class="math-container">$n = 2$</span>, we have that
<span class="math-container">$$
\begin{split}
| X \cap D_1 | + | X \cap D_2 | = &| X \cap (D_1 \setminus D_2) | + | X \cap (D_1 \cap D_2) | \\
&+ | X \cap (D_2 \setminus D_1) | + | X \cap (D_1 \cap D_2) | \\
= & |X| + | X \cap (D_1 \cap D_2) | \\
\leq & |X| + | D_1 \cap D_2 |.
\end{split}
$$</span></p>
<p>My question is that, can we generalize the above upper bound to something like
<span class="math-container">$$
\sum_{i = 1}^n | X \cap D_i | \leq |X| + c,
$$</span>
where <span class="math-container">$c$</span> is dependent on <span class="math-container">$(D_1, D_2, ..., D_n)$</span>? It is self-evident that, if <span class="math-container">$D_1, ..., D_n$</span> is a disjoint partition of a universe, then we have <span class="math-container">$\sum_{i = 1}^n | X \cap D_i | = |X|$</span>. However, it seems difficult for me to bound <span class="math-container">$c$</span> when <span class="math-container">$D_1, ..., D_n$</span> are not disjoint.</p>
<p>It would be appreciated if you could give me any hint.</p>
| dafinguzman | 123,170 | <p>You can arrive at complex arithmetic from geometric intuition if you start with transformations of the plane.</p>
<p>It is well known that matrices which preserve angles (i.e. map shapes to similar shapes) and orientation are of the form <span class="math-container">$cR(\theta)$</span>, where <span class="math-container">$c$</span> is a positive number and <span class="math-container">$R(\theta)$</span> is a <a href="https://en.wikipedia.org/wiki/Rotation_matrix" rel="nofollow noreferrer">rotation matrix</a>. That is,
<span class="math-container">$$cR(\theta) = c\pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta}. $$</span></p>
<p>Since <span class="math-container">$c$</span> and <span class="math-container">$\theta$</span> are arbitrary, these are all matrices of the form <span class="math-container">$$\pmatrix{a & -b \\ b & a}$$</span> for <span class="math-container">$a, b \in \mathbb R$</span> (except for the null matrix).</p>
<p>Now, after adjoining the null matrix, this set becomes a vector space of dimension two, closed under matrix multiplication, and where all non-null elements have a multiplicative inverse (<span class="math-container">$c^{-1} R(-\theta)$</span>).</p>
<p>The interesting part is that we can choose a basis like this:
<span class="math-container">$$ \pmatrix{a & -b \\ b & a} = a \pmatrix{1 & 0 \\ 0 & 1} + b \pmatrix{0 & -1 \\ 1 & 0} = a I + b J,$$</span> where <span class="math-container">$I$</span> is the identity matrix and <span class="math-container">$J=\pmatrix{0 & -1 \\ 1 & 0}$</span> is a matrix which, under matrix multiplication, has the property <span class="math-container">$J^2=-I$</span>. That is, it is in some sense the "square root" of <span class="math-container">$-I$</span>. It also represents rotation in 90º (like the complex <span class="math-container">$i$</span> does). Indeed: <span class="math-container">$J = R(\pi/2)$</span> and, as expected, <span class="math-container">$J e_1 = e_2$</span> and <span class="math-container">$J e_2 = -e_1$</span>.</p>
<p>Moreover, if you work out the product rule, it is exactly the one that arises in complex numbers:</p>
<p><span class="math-container">$$ \pmatrix{a & -b \\ b & a} \cdot \pmatrix{c & -d \\ d & c} \\ = \pmatrix{ac-bd & -(ad+bc) \\ ad+bc & ac-bd} \\ = (ac-bd)I + (ad+bc)J.$$</span></p>
<p>Furthermore, we can define subtraction, division, and all arithmetic operations for them in a way parallel to how they are defined for complex numbers.</p>
<p>Finally, add to this that the subspace generated by <span class="math-container">$I$</span> is an algebraic copy of <span class="math-container">$\mathbb R$</span>, so you can view the full space as an extension of <span class="math-container">$\mathbb R$</span>.</p>
<h3>To sum up</h3>
<ul>
<li>Angle and orientation preserving linear transformations carry great geometric meaning (similarity).</li>
<li>They form a two-dimensional space, which you can think of as an algebraically compatible extension of <span class="math-container">$\mathbb R$</span>.</li>
<li>They have two components, one in the direction of the identity/unit and one in the direction of a <span class="math-container">$\pi/2$</span> rotation.</li>
<li>This space can be constructed from <span class="math-container">$\mathbb R$</span> just by adjoining an outside element <span class="math-container">$J$</span> such that <span class="math-container">$J^2$</span> is minus the identity (and extending the usual algebraic rules).</li>
<li>This is essentially the same recipe as the one for constructing the complex numbers.</li>
</ul>
|
3,645,263 | <p>I have recently started studying Set Theory in a self-thaught way, for that purpose I have been following Kunen's book: Set Theory: An Introduction to Independence Proofs. I'm in Chapter I section 7 and it has been defined the ordinals addition but I don't quite understand that definition. I have seen that in other books authors defines the addition using transfinite induction and it seems easier but now I want to understand Kunen's one.</p>
<p><span class="math-container">$$\alpha + \beta=type(\alpha \times \{ 0 \} \cup \beta \times \{1\}, R) \:\text{where } $$</span>
<span class="math-container">$$R=\{ \langle \langle \xi,0 \rangle, \langle \eta , 0\rangle \rangle : \xi<\eta<\alpha\} \; \cup \{\langle \langle \xi,1 \rangle, \langle \eta , 1\rangle \rangle : \xi<\eta<\beta\} \; \cup [(\alpha\times\{0\})\times(\beta\times\{1\})]. $$</span></p>
<p>With <span class="math-container">$type(A,R)$</span> is the unique ordinal <span class="math-container">$C$</span> such that <span class="math-container">$\langle A, R\rangle \cong C$</span> when <span class="math-container">$\langle A,R\rangle$</span> is a well-ordering set.</p>
<p>What I think I understood so far is that this definition tries to order two non-disjoint sets having that <span class="math-container">$\alpha<\beta$</span> and keeping the order inside <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span>. What I can't understand is how I can get that cardinal (in for example a finite case), maybe I'm being stubborn and I should "ignore" this definition and trying to understand the more simplified one given in the later results.</p>
<p>Thank you for your time. </p>
| Community | -1 | <p>As mentioned by Brian, it is essentially the lexicographic ordering. </p>
<p>For example, say <span class="math-container">$2=\{0_2,1_2\}$</span> and <span class="math-container">$3=\{0_3,1_3,2_3\}$</span>. </p>
<p>According to the definition, we first extend <span class="math-container">$2$</span> and <span class="math-container">$3$</span> to ordered pairs: </p>
<p><span class="math-container">$$2\times\{0\}=\{(0_2,0),(1_2,0)\}\quad{\rm and}\quad 3\times\{1\}=\{(0_3,1),(1_3,1),(2_3,1)\}.$$</span></p>
<p>Then what is <span class="math-container">$R$</span>? Although it is written as the union of sets, we can write it in a chain like this: </p>
<p><span class="math-container">$$(0_2,0)<(1_2,0)<(0_3,1)<(1_3,1)<(2_3,1).$$</span></p>
<p>The set <span class="math-container">$2\times\{0\}\cup 3\times\{1\}$</span> is linearly ordered and isomorphic to <span class="math-container">$5=\{0_5,1_5,2_5,3_5,4_5\}$</span> in which </p>
<p><span class="math-container">$$0_5<1_5<2_5<3_5<4_5.$$</span></p>
<p>Thus, <span class="math-container">$2+3=5$</span>. </p>
|
245,756 | <p>I'm supposed to show that the jordan content is $0$. The definition for a set $S$ having jordan content zero I have to work with is :$\forall\epsilon>0$ there is a finite collection of generalized rectangles in $\mathbb{R}^n$ that covers $S$ the sum of these rectangles volumes being less that $\epsilon$.</p>
<p>So I have the set $S = \{(x,y,z) \in \mathbb{R}^3\mid x^2 + y^2 + z^2 =1\}$. I can't just build $1 \cdot 1 \cdot \frac{1}{k}$ boxes (with $k > \frac{1}{\epsilon}$) and call it a day because the sum obviously isn't going to be less than epsilon for all epsilon.</p>
<p>I think another approach would be to try and build a sequence of generalized rectangles such that the first sequence has two boxes each covering half of the sphere and then sub divide and keep only the boxes that intersect the sphere. But I'm having trouble formalizing this idea.</p>
<p>Can somebody point me in the right direction / give any advice?</p>
<p>Thanks in advance.</p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> In the context of a calculus course, I think you are first expected to argue informally that such a maximal cylinder must have axis that goes through the center of the circle, and that without loss of generality that axis is the $z$-axis.</p>
<p>So now suppose that the cylinder meets the $x$-$y$ plane in a circle of radius $t$. Find the height of the cylinder in terms of $t$, and hence the volume. Now use the ordinary tools to maximize. </p>
|
245,756 | <p>I'm supposed to show that the jordan content is $0$. The definition for a set $S$ having jordan content zero I have to work with is :$\forall\epsilon>0$ there is a finite collection of generalized rectangles in $\mathbb{R}^n$ that covers $S$ the sum of these rectangles volumes being less that $\epsilon$.</p>
<p>So I have the set $S = \{(x,y,z) \in \mathbb{R}^3\mid x^2 + y^2 + z^2 =1\}$. I can't just build $1 \cdot 1 \cdot \frac{1}{k}$ boxes (with $k > \frac{1}{\epsilon}$) and call it a day because the sum obviously isn't going to be less than epsilon for all epsilon.</p>
<p>I think another approach would be to try and build a sequence of generalized rectangles such that the first sequence has two boxes each covering half of the sphere and then sub divide and keep only the boxes that intersect the sphere. But I'm having trouble formalizing this idea.</p>
<p>Can somebody point me in the right direction / give any advice?</p>
<p>Thanks in advance.</p>
| Till Hoffmann | 23,894 | <p>Let $R$ be the radius of the sphere and let $h$ be the height of the cylinder centered on the center of the sphere. By the Pythagorean theorem, the radius of the cylinder is given by
$$
r^2 = R^2 - \left(\frac{h}{2}\right)^2.
$$</p>
<p>The volume of the cylinder is hence
$$
\begin{align}
V &= \pi r^2 h\\
&= \pi \left(h R^2 - \frac{h^3}{4}\right).
\end{align}
$$</p>
<p>Differentiating with respect to $h$ and equating to $0$ to find extrema gives
$$
\frac{dV}{dh}=\pi \left(R^2 - \frac{3h^2}{4}\right) = 0\\
\therefore h_0 = \frac{2R}{\sqrt{3}}
$$</p>
<p>The second derivative of the volume with respect to $h$ is negative if $h>0$ such that the volume is maximal at $h = h_0$. Substituting gives
$$
V_{max}=\frac{4 \pi R^3}{3\sqrt{3}}.
$$</p>
|
1,090,658 | <p>I'm doing some previous exams sets whilst preparing for an exam in Algebra.</p>
<p>I'm stuck with doing the below question in a trial-and-error manner:</p>
<p>Find all $ x \in \mathbb{Z}$ where $ 0 \le x \lt 11$ that satisfy $2x^2 \equiv 7 \pmod{11}$</p>
<p>Since 11 is prime (and therefore not composite), the Chinese Remainder Theorem is of no use? I also thought about quadratic residues, but they don't seem to solve the question in this case.</p>
<p>Thanks in advance</p>
| HSN | 58,629 | <p>A first step in this could be to find the inverse of $2$, which turns out to be $6$, yielding $6\cdot2x^2\equiv x^2\equiv 6\cdot 7\mod 11\equiv 9\mod 11$. The obvious solutions to this are $3$ and $-3\equiv 8$. An important remark is that since $11$ is prime, there are at most two solutions to any given quadratic equation, so these are your desired solutions.</p>
<p>Notice however, that there can indeed be more solutions for non-primes. In these cases, the Chinese remainder theorem might be helpful. In a case like $x^2\equiv1\mod 12$, the solutions are $1, 5, 7$ and $11$. These correspond to the four combinations of $\pm1\mod3$ and $\pm1\mod4$. However, it isn't always possible to use this theorem. How would you, for instance, tackle $x^2=0$ working modulo $16$?</p>
|
3,021,631 | <p>I've been strongly drawn recently to the matter of the fundamental definition of the exponential function, & how it connects with its properties such as the exponential of a sum being the product of the exponentials, and it's being the eigenfunction of simple differentiation, etc. I've seen various posts inwhich clarification or demonstration or proof of such matters as how such & such a property <em>proceeds</em> from its definition as <span class="math-container">$$e^z\equiv\lim_{k\rightarrow\infty}\left(1+\frac{z}{k}\right)^k .$$</span> I'm looking at how there is a <em>web</em> of interconnected theorems about this; and I am trying to <em>spin</em> the web <em>as a whole</em>. This is not necessary for proving <em>some particular item to be proven</em> - a path along some one thread or sequence of threads is sufficient for that; but I think the matter becomes 'unified', and by reason of that actually <em>simplified</em>, when the web is perceived as an entirety. This is why I <em>bother</em> with things like combinatorial demonstrations of how the terms in a binomial expansion <em>evolve towards</em> the terms in a Taylor series as some parameter tends to ∞, when the matter at hand is actually susceptible of a simpler proof by taking the logarithm of both sides ... & other seeming redundancies.</p>
<p>To this end, another 'thread' I am looking at is that of showing that the coefficients in the Taylor series for <span class="math-container">$\ln(1+z)$</span> actually are a <em>consequence</em> of the requirement that the logarithm of a product be the sum of the logarithms, and ... if not quite a combinatorial <em>derivation</em> of them <em>from</em> that requirement, at least a <em>reverse- (or sideways-) engineering <strong>equivalent</em></strong> of it - the combinatorially showing that if the coefficients <em>be plugged into</em> the Taylor series, then the property follows</p>
<p>Taking the approach that <span class="math-container">$$(1+x)(1+y) = 1+x+y+xy ,$$</span> plugging <span class="math-container">$z=x+y+xy$</span> into the series for <span class="math-container">$\ln(1+z)$</span> and hoping that all non-fully-homogeneous terms cancel out, leaving sum of the series for <span class="math-container">$\ln(1+x)$</span> & that for <span class="math-container">$\ln(1+y)$</span>, we are left with proving that</p>
<p><span class="math-container">$$\sum_{k=1}^\infty\left((-1)^k(k-1)!×\sum_{p\inℕ_0,q\inℕ_0,r\inℕ_0,p+q+r=k}\frac{x^{p+r}y^{q+r}}{p!q!r!}\right)$$</span><span class="math-container">$$=$$</span><span class="math-container">$$\sum_{k=1}^\infty(-1)^k\frac{x^k+y^k}{k} .$$</span></p>
<p>The <em>inner</em> sum of the LHS of this quite appalling-looking theorem is the trinomial expansion of <span class="math-container">$(x+y+xy)^k$</span> for arbitrary <span class="math-container">$k$</span>, & the outer sum is simply the logarithm Taylor series expansion (with its <span class="math-container">$k$</span> in the denominator 'absorbed' into the combinatorial <span class="math-container">$k!$</span> in the numerator of the inner sum) . This theorem can be quite easily verified by 'brute force' - simply <em>doing</em> the expansions at the first (very!) few terms; but the labour of it escalates <em>extremely</em> rapidly. An algebraic manipulation package would no-doubt verify it at a good few more terms; but what I am looking-for is a showing of the fully general case: but I do not myself have the combinatorial toolage for accomplishing this. </p>
<p>So I am asking whether anyone can show me an outline of what to do ... or even actually <em>do</em> it for me, although that would probably take up a <em>very</em> great deal of space and be an <em>extremely</em> laborious task for the person doing it ... so I'm content to ask for just an outline of doing it, or for some 'signposts' as to how to do it - maybe someone knows some text on this kind of thing that they would recommend.</p>
| Ira Gessel | 437,380 | <p>Here's another proof that <span class="math-container">$\log\bigl((1+x)(1+y)\bigr) = \log(1+x) + \log(1+y)$</span> as formal power series. This assumes some facts about derivatives of formal power series that are not difficult to verify (e.g., a very simple special case of the chain rule). It follows directly from the power series for <span class="math-container">$\log(1+x)$</span> that
<span class="math-container">$$\frac{d\ }{dx} \log(1+x) = \frac{1}{1+x}.$$</span>
Thus by the chain rule,
<span class="math-container">$$\frac{\partial\ }{\partial x} \log\bigl((1+x)(1+y)\bigr) = \frac{(1+y)}{(1+x)(1+y)}=\frac{1}{1+x}
=\frac{\partial\ }{\partial x}\bigl(\log(1+x)+\log(1+y)\bigr).$$</span>
Similarly
<span class="math-container">$$\frac{\partial\ }{\partial y} \log\bigl((1+x)(1+y)\bigr) = \frac{1}{1+y}=\frac{\partial\ }{\partial y}\bigl(\log(1+x)+\log(1+y)\bigr).$$</span>
Thus <span class="math-container">$\log\bigl((1+x)(1+y)\bigr) - \bigl(\log(1+x)+\log(1+y)\bigr)$</span> must be a constant and that constant must be <span class="math-container">$0$</span>.</p>
|
197,393 | <p>Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?</p>
| lab bhattacharjee | 33,337 | <p>$$\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{2+3}{1-2\cdot 3}\right)}=\tan^{-1}(-1)=n\pi-\frac \pi 4,$$ where $n$ is any integer.</p>
<p>Now the principal value of $\tan^{-1}(x)$ lies in $[-\frac \pi 2, \frac \pi 2]$ precisely in $(0, \frac \pi 2)$ if finite $x>0$.
So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0, \pi) $.</p>
<p>So, the principal value of $\tan^{-1}(2)+\tan^{-1}(3)$ will be $\frac {3\pi} 4$.</p>
<p>Interestingly, the principal value of $\tan^{-1}(-1)$ is $-\frac {\pi} 4$.</p>
<p>But the general values of $\tan^{-1}(2)+\tan^{-1}(3)$ and $\tan^{-1}(-1)$ are same.</p>
<p>Alternatively, $$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\tan^{-1}{\left(\frac{1+2+3-1\cdot 2\cdot 3}{1-1\cdot 2- 2\cdot 3 -3\cdot 1}\right)}=\tan^{-1}(0)=m\pi$$, where $m$ is any integer.</p>
<p>Now the principal value of $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)$ will lie in $(0 ,\frac {3\pi} 2)$ which is $\pi$.</p>
<p>The principal value of $\tan^{-1}(0)$ is $0\neq \pi$.</p>
|
197,393 | <p>Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?</p>
| kennytm | 171 | <p><img src="https://i.stack.imgur.com/zyXUh.png" alt="enter image description here"> </p>
<p>Consider $O=(0,0)$, $A=(1,1)$, $B=(-1,3)$, $D=(1,-3)$, $E=(1,0)$.</p>
<p>\begin{align}
2 &= \frac{AB}{AO} = \tan \angle AOB \\
1 &= \frac{AE}{EO} = \tan \angle AOE \\
3 &= \frac{DE}{DO} = \tan \angle DOE
\end{align}</p>
<p>The points B, O and D are collinear, i.e. $\angle BOD = \tan^{-1}2+\tan^{-1}1+\tan^{-1}3 = \pi$.</p>
|
119,561 | <p>I am interested in determining the <strong>minimum</strong> and <strong>maximum</strong> values of the real roots of polynomials of form $P(x)=\sum_{k=0}^n a_{k} x^k$ where $n$ will have a defined value (say 3,4,5...) and $a_k$ are chosen from the set $\{-1,1\}$ with equal probability.</p>
<p>I have tried creating a table of the roots, and then using <code>MinMax</code>; here is my (bad) attempt (with $n=3$):</p>
<pre><code>T = Table[Roots[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x], 25]
MinMax[T]
</code></pre>
<p>Unfortunately, <code>Roots</code> gives both the real and imaginary roots, I would only like the real roots . Also, <code>MinMax</code> cannot work on the table $T$ when the roots are not presented as a list (they have $||$ in between each root). </p>
<p>Any suggestions/help with this issue is immensely appreciated.
Thank You!</p>
| Bob Hanlon | 9,362 | <p>To find the exact values for min and max roots</p>
<pre><code>roots = DeleteDuplicates@
Flatten[x /.
Solve[# == 0, x, Reals] & /@
(Tuples[{-1, 1}, {4}].{1, x, x^2, x^3})] //
SortBy[#, N] &;
{min, max} = roots[[{1, -1}]] // ToRadicals
</code></pre>
<p><a href="https://i.stack.imgur.com/nZX3o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nZX3o.png" alt="enter image description here"></a></p>
<p>The approximate numeric values are as shown by @MarcoB</p>
<pre><code>{min, max} // N
(* {-1.83929, 1.83929} *)
</code></pre>
|
1,041,134 | <p>I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$.
Can you provide me some hints please?</p>
| Henry | 6,460 | <p>Hint: If $\displaystyle a+\frac1a$ is an integer then $\displaystyle \left(a+\frac1a\right)^2,\left(a+\frac1a\right)^3, \ldots $ are integers.</p>
<p>Multiply the powers out and you should be able to see why $a^t+\dfrac1{a^t}$ is going to be an integer for positive integer $t$, using a combination of symmetry and induction.</p>
|
3,105,482 | <p>For the formula:</p>
<p><span class="math-container">$$1 = \sqrt{x^2 + y^2 + z^2 + w^2}$$</span></p>
<p>How to rewrite it to find <span class="math-container">$w$</span>?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>We have <span class="math-container">$$w^2=1-(x^2+y^2+z^2)=w^2$$</span> so we obtain <span class="math-container">$$w=\pm\sqrt{1-(x^2+y^2+z^2)}$$</span> for <span class="math-container">$$1\geq x^2+y^2+z^2$$</span></p>
|
2,761,658 | <blockquote>
<p>Two numbers $x$ and $y$ are chosen at random from the numbers $1,2,3,4,\ldots,2004$. The probability that $x^3+y^3$ is divisible by $3$ is?</p>
</blockquote>
<p>The correct answer is $\dfrac13$ while mine is $\dfrac{445}{2003}$</p>
<p>My attempt:</p>
<p>For $x^3+y^3$ to be divisible by $3$, EITHER both $x$ and $y$ should be a multiple of $3$ OR one of them should leave remainder $1$ when divided by $3$ and the other should leave remainder $2$.</p>
<p>Therefore, $$\text{no. of ways} = \frac{668 \times 667 + 668\times 668}{2004\times 2003} = \frac{445}{2003}$$ </p>
| Piquito | 219,998 | <p>It is immediate (by using arithmetic progression) to get that there are equally $668$ numbers of residues $0,1,2$ modulo $3$. We have
$$(3n)^3+(3n)^3\equiv 0\pmod3\\(3n)^3+(3n+1)^3\equiv 1\pmod3\\(3n)^3+(3n+2)^3\equiv 2\pmod3\\(3n+1)^3+(3n+1)^3\equiv 2\pmod3\\(3n+1)^3+(3n+2)^3\equiv 0\pmod3$$
It follows$$\binom{668}{2}+668^2=669002\space\space\text{favorable cases.}\\\binom{2004}{2}=2007006\space\space\text{possibilities}$$
Thus the probability is $$\frac{669002}{2007006}=\frac13$$</p>
|
108,110 | <p>How can I use <em>Mathematica</em> to expand such a product (only need a finite number of terms):</p>
<p>$$\prod^{\infty}_{n=1}\frac{({1-yq^{n+1}})({1-y^{-1}q^n})}{(1-q^n)^2}$$</p>
| Lukas | 21,606 | <p>Given that you only need a finite number of terms, I would do something like this:</p>
<pre><code>ClearAll[prod, expand];
prod[num_?NumericQ] :=Product[(1 - y*q^(n + 1))*(1 - q^n/y)/(1 -q^n)^2, {n, 1, num}];
expand[num_?NumericQ] := Total[Total[#] & /@ Numerator[#]/Flatten[DeleteDuplicates[#] & /@ Denominator[#]] &@GatherBy[List @@ Expand[prod[num]], Denominator]]
</code></pre>
<p>Of course, you can just run <code>Expand</code> on your product. But do it and observe the longish result. So my approach (which I bet is not the most beautiful one to accomplish this, btw) gathers by common denominators and sums their respective numerators. This, at least to me, makes the expanded result way more readable. Of course, that depends on what you want to do with your result.
Example:</p>
<p><a href="https://i.stack.imgur.com/wUbdg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wUbdg.jpg" alt="enter image description here"></a></p>
<p>Of course, you can also run <code>Simplify</code> on such constructs, but that will for sure need more time. However, there you may specify <code>TransformationFunctions</code> to end up with a form of your desire.</p>
<p><code>AbsoluteTiming</code> of <code>Simplify@Expand@prod[4]</code>: <code>0.134552</code></p>
<p><code>AbsoluteTiming</code> of <code>expand[4]</code>: <code>0.009244</code></p>
|
3,754,225 | <p>Let S be the circle with centre <span class="math-container">$(0,0)$</span>, radius <span class="math-container">$r$</span> units. The chord <span class="math-container">$C$</span> of the circle S subtends an angle of 2π/3 at its center. If R represents the region consisting of all points inside S which are closer to C than to circumference of S, then</p>
<p>(1) What is area of region <span class="math-container">$R$</span>, and,</p>
<p>(2) In what ratio does <span class="math-container">$C$</span> divide the region <span class="math-container">$R$</span> ?</p>
<p>I found the equation of a circle intersecting circle radius <span class="math-container">$r$</span> for the chord subtending angle <span class="math-container">$2\pi/3$</span> at center we should have</p>
<p><span class="math-container">$$ x=\dfrac{r}{2} $$</span></p>
<p>but don't know what to do next. Please help.</p>
| Math Lover | 801,574 | <p><a href="https://i.stack.imgur.com/J1wrS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J1wrS.jpg" alt="enter image description here" /></a></p>
<p>When you draw the circle and the chord, it should become simpler to understand. This is a bit difficult to visualize otherwise.</p>
<p>You could pick any chord that subtends an angle of <span class="math-container">$\frac{2\pi}{3}$</span> at the center.<br />
For simplicity, let's pick chord (C) with the equation <span class="math-container">$y = \frac{R}{2}$</span> where R is the radius of the circle. As you can see, it will subtend an angle of <span class="math-container">$120^0$</span> at the center.</p>
<p>Let's pick any point A <span class="math-container">$(r,\theta)$</span> inside the circle of radius R.</p>
<p>Now as per problem statement, distance from point A to the line C has to be smaller or equal to the distance from point A to the circumference of the circle.</p>
<p>d = distance from point <span class="math-container">$A (r, \theta)$</span> to the line C = <span class="math-container">$|\frac{R}{2}-rsin\theta|$</span><br />
r' = distance from point A to the circumference = R-r</p>
<p>As <span class="math-container">$d \le r', \space |\frac{R}{2}-rsin\theta| \le R-r$</span>.</p>
<p>A few tests for right half of the circle (<span class="math-container">$x \ge 0$</span>). We can double the area once we find for the right half.</p>
<p>For <span class="math-container">$r = R, \theta = \frac{\pi}{6}$</span><br />
For <span class="math-container">$\theta = -\frac{\pi}{2}, r \le \frac{R}{4}$</span><br />
For <span class="math-container">$\theta = \frac{\pi}{2}, r \le \frac{3R}{4}$</span></p>
<p>You can find out that there is discontinuity at <span class="math-container">$\frac{\pi}{6}$</span> and two different curves between<br />
<span class="math-container">$-\frac{\pi}{2} \le \theta \le \frac{\pi}{6}$</span> with Area <span class="math-container">$A_1$</span> (say) and <span class="math-container">$\frac{\pi}{6} \le \theta \le \frac{\pi}{2}$</span> with Area <span class="math-container">$A_2$</span> (say).</p>
<p><span class="math-container">$\begin{align*}
A_1 &= \int_{-\pi/2}^{\pi/6}\int_{0}^{\frac{R}{2(1-sin\theta)}}r dr d\theta\\\\
A_2 &= \int_{\pi/6}^{\pi/2}\int_{0}^{\frac{3R}{2(1+sin\theta)}}r dr d\theta
\end{align*}$</span></p>
<p>I hope you can take it from here.</p>
<p>Area of region R = 2(<span class="math-container">$A_1+A_2)$</span></p>
<p>Ratio in which chord C divides the region R = <span class="math-container">$\frac{A_1}{A_2}$</span>. It is basically the ratio of area of region R below the chord and above it.</p>
|
3,271,891 | <p>In <a href="https://en.wikipedia.org/wiki/Boolean_algebra_(structure)" rel="nofollow noreferrer">Wikipedia, the Boolean algebra</a> is defined as a 6-tuple
<span class="math-container">$(A,\wedge,\vee,\neg,0,1)$</span>. In Kuratowski1976, on the other side in the definition on page 34, there is no <span class="math-container">$1$</span>. Halmos1963 has the <span class="math-container">$1$</span>.</p>
<blockquote>
<p>Does the definition in Kuratowski1976 leads to something more general or somehow different theory? On page 37 he introduces the concept of unit <span class="math-container">$i$</span> that I assume is the <span class="math-container">$1$</span> of the other authors (also judging from the definition <span class="math-container">$a\wedge i=a$</span>).</p>
</blockquote>
<p><strong>Definition from Kuratowski1976:</strong></p>
<p><a href="https://i.stack.imgur.com/IRiIj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRiIj.png" alt="enter image description here"></a></p>
| Con | 682,304 | <p>I have no access to Kuratowski's definition currently, but I assume that he might want to define something less restrictive (depending on whether all the other axioms are the usual ones or equivalent) that we would usually not call boolean algebra and afterwards introduce the notion of <span class="math-container">$1$</span> and what we call boolean algebras. That is basically the same as the different notions for rings. There are associative rings, associative unitary rings, commutative rings and what I personally call a ring (commutative unitary). Depending on the context some notion might be better, appears more often or is of main interest such that it makes sense to choose a fitting definition.</p>
<p>Edit: As amrsa pointed out: In the case that really only the <span class="math-container">$1$</span> element is missing, then you should just be able to get it as the complement of <span class="math-container">$0$</span> with the other axioms.</p>
|
4,629,824 | <p>Let <span class="math-container">$k$</span> be a given positive integer. I want to solve the following system of Diophantine equations: <span class="math-container">$$\begin{cases} a^2 + b^2 + c^2 = k^2 \\ b^2 = ac \end{cases}$$</span>
where <span class="math-container">$a, b, c \in \mathbb{N}$</span> are non-zero.</p>
<p>There is an OEIS sequence for the numbers which are sums of three non-zero squares, but I don't know of any general expression (some squares, such as 25, are not there).</p>
<p>I know, by substituting the second equation into the first, that <span class="math-container">$k = m^2 + n^2 + mn$</span>, <span class="math-container">$a = m^2 - n^2$</span> and <span class="math-container">$c = 2mn + n^2$</span>, where <span class="math-container">$m, n$</span> are coprime integers, but I don't really know how to use this, since plugging this back into the second equation yields a very complicated looking one (namely, when is the product a square?)</p>
<p>I'm not used to solving Diophantine equations (much less systems of them), but I have tried direct computation and seriously believe that there is no solution.</p>
<p>Are there any hints or techniques for solving this type of problem?</p>
<p>Thanks in advance!</p>
| Will Jagy | 10,400 | <p>we may demand <span class="math-container">$ \gcd(a,b,c) = 1$</span> by dividing through by any common factor. This works because of homogeneity.</p>
<p>Next, <span class="math-container">$ac=b^2$</span> implies that
<span class="math-container">$$ a = x^2, \; \; b = xy, \; \; c = y^2 $$</span></p>
<p>At this point you have
<span class="math-container">$$ x^4 + x^2 y^2 + y^4 = k^2 $$</span>
with <span class="math-container">$\gcd(x,y) = 1.$</span> This is the result labelled (7') on page 19 of Mordell, <em>Diophantine Equations</em>. There are only trivial solutions. The proof is about a page, on pages 19 and 20. Worth looking up and going through it in detail. It is just taking the "smallest" positive solution and showing a contradiction. No elliptic curves.</p>
|
785,844 | <p>I've got the following limit to solve:</p>
<p>$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$</p>
<p>I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?</p>
| corporal | 148,632 | <p>You were on the right track using the <em>conjugate</em>.</p>
<p>$$\begin{align*}
\frac{\sqrt{s}-s^2}{1-\sqrt{s}}&=\frac{\sqrt{s}-s^2}{1-\sqrt{s}}\times\frac{1+\sqrt{s}}{1+\sqrt{s}}\\
&=\frac{\sqrt{s}-s^2+s-s^{5/2}}{1-s}\\
&=\frac{s-s^2+\sqrt{s}-s^{5/2}}{1-s}\\
&=\frac{s(1-s)+\sqrt{s}(1-s^{2})}{1-s}\\
&=\frac{s(1-s)+\sqrt{s}(1-s)(1+s)}{1-s}\\
&=s+\sqrt{s}(1+s)\\
\end{align*}$$</p>
<p>This is now ready for taking the limit as $s\to 1$.</p>
|
2,704,290 | <p>I'm prepared for the competitive exam like <a href="http://csirhrdg.res.in/mathCEN_June2015.pdf" rel="nofollow noreferrer">this</a> (a sample question).</p>
<p>In order to solve the problems, first to familiarize with the prerequisite for the each concept. It's ok! </p>
<p>My problem is: If I'm working certain problems on group theory, then it will take two days or a week. Ok! No problem. After the week, I can move on to the next concept like integration, so it will take some time. But in the third week, I lack some concepts bit about group theory (even sometimes, I'm forgetting what I'm doing in the first week). So I feel I waste lot of time. I really want to learn algebra, analysis, topology, and all that simultaneously (sorry if this word does not make sense) or to first learn some bit about algebra and then analysis? I don't no how to move on? </p>
<p>Suppose the question booklet start with group theory. That is, the first question involves group theory, the second one involves like analyticity and the 17th question (for example) involves again group theory etc.</p>
<p>My question is: Is the way to doing problems one by one (in the order)? or
By selecting the first one and do all the problems related to group theory and then move on the next concept (like analyticity)?
or
anything
else?</p>
<p>What is the best strategy to prepare this type of exam?</p>
| Joonas Ilmavirta | 166,535 | <p>Forgetting something is normal.
What you have to do then is to remind yourself of what you once remembered actively.
It is much easier the second time.
It is easy to forget things after focusing on them for a week.
A more permanent understanding of the topics needs a longer time frame than a couple of weeks.</p>
<p>This kind of thing is learned by repetition, and you have to <strong>keep repeating</strong>.
Even if you are satisfied with your algebra skills after one intensive week of algebra, do not give up doing exercises in algebra.
Otherwise you will forget easily.
Every day, do a little bit of algebra on the side while learning integration.</p>
<p>Ideally, you should <strong>study new things that build on old things</strong>.
That way you are constantly reminded of earlier material and have to keep rehearsing it.
For example, when you study proofs with $\epsilon$ and $\delta$ in analysis, you will be constantly reminded of how to deal with absolute values and basic algebra.</p>
<p>Designing the curriculum so that you build on previous knowledge is not trivial, especially if you don't already know what prerequisites each topic has.
Therefore I can only suggest to <strong>try to apply old knowledge when possible</strong>.</p>
<p>The most important thing is to <strong>keep all knowledge active</strong>.
The best way is to <strong>do exercises with mixed topics</strong> every once in a while, so that you have to keep using all the tools you have picked up over the last weeks.
You will sometimes forget something.
That's fine; then just go back and relearn it.</p>
<p>Do not try to learn the different topics in isolation.
(They are not isolated in the example exam, either.)
Many of the boundaries are artificial, which will come more and more evident as you progress in your studies.</p>
<p>In addition to learning how to mechanically execute tasks, learn <strong>why</strong> the methods work.
Insight and understanding is very valuable.</p>
<p>You might also be interested in the question <a href="https://matheducators.stackexchange.com/q/5586/2074">"How can you be perfect at maths (highschool)?"</a> and its answers.</p>
|
981,748 | <p>We know a closed-form of the first two powers of the integral of <a href="http://mathworld.wolfram.com/Trilogarithm.html" rel="nofollow noreferrer">trilogarithm function</a> between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>. From the result <a href="https://math.stackexchange.com/q/981650/153012">here</a> we know that</p>
<p><span class="math-container">$$I_1=\int_0^1 \operatorname{Li}_3(x)\,dx = \zeta(3)-\frac{\pi^2}{6}+1.$$</span></p>
<p>From <a href="https://math.stackexchange.com/q/980764/153012">here</a> we also know that</p>
<p><span class="math-container">$$I_2=\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)+\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$</span></p>
<p>Is there a closed-form of</p>
<p><span class="math-container">$$I_3=\int_0^1 \operatorname{Li}_3^3(x)\,dx$$</span>
and
<span class="math-container">$$I_4=\int_0^1 \operatorname{Li}_3^4(x)\,dx\,?$$</span></p>
<hr />
<p>Update (by editor after 6 years): By generalizing @Kirill's algorithm one may prove that:</p>
<blockquote>
<p><span class="math-container">$$\int_0^1 \text{Li}_3(x){}^4 \, dx=-51 \pi ^2 \zeta(6,2)+6480 \zeta(6,2)+\frac{4743}{8} \zeta(8,2)+\zeta (3)^4-\frac{2 \pi ^2 \zeta (3)^3}{3}-68 \zeta (3)^3-248 \pi ^2 \zeta (3)^2+16680 \zeta (3)^2+\frac{106 \pi ^6 \zeta (3)}{105}+126 \pi ^4 \zeta (3)+3920 \pi ^2 \zeta (3)+102 \pi ^2 \zeta (5) \zeta (3)-11160 \zeta (5) \zeta (3)-\frac{4743 \zeta (7) \zeta (3)}{4}-114240 \zeta (3)+42 \pi ^4 \zeta (5)+1260 \pi ^2 \zeta (5)-73080 \zeta (5)+\frac{819 \pi ^2 \zeta (7)}{2}-33030 \zeta (7)-9660 \zeta (9)-\frac{4023 \zeta (5)^2}{4}+\frac{563 \pi ^{10}}{39600}+\frac{83 \pi ^8}{42}-1064 \pi ^4-11200 \pi ^2-\frac{184 \pi ^6}{21}+369600$$</span></p>
</blockquote>
<p>Here <span class="math-container">$\zeta(6,2), \zeta(8,2)$</span> are irreducible. One may verify its correctness numerically.</p>
| Alexander Vlasev | 11,998 | <p>Here's my approach. To reduce clutter let</p>
<p>$$f_m(x) = Li_m(x)$$</p>
<p>and let</p>
<p>$$f_M(x) = f_{m_1}(x)\dots f_{m_n}(x)$$</p>
<p>where $M$ has $n$ entries. Then let</p>
<p>$$I_k(M) = \int f_M(x) dx$$</p>
<p>where we can interpret this as an indefinite or definite integral. Consider the following derivative</p>
<p>$$(x^{k+1}f_M)' = (k+1)x^k f_M + x^{k+1} \sum_{k=1}^n f_{m_k}'\prod_{i\neq k}f_{m_i} = (k+1)x^k f_M + x^{k} \sum_{k=1}^n f_{m_k-1}\prod_{i\neq k}f_{m_i}$$</p>
<p>Integrating with respect to $x$ (and ignoring the extra constant) yields</p>
<p>$$x^{k+1}f_M = (k+1)I_k(M) + \sum_{i=1}^n I_k(M)_i$$</p>
<p>where the notation $I_k(M)_i$ means we subtract $1$ from the $i$-th entry in $M$. This works for every $k+1 \neq 0$. For the case where $k+1 = 0$ we have</p>
<p>$$f_M' = \sum_{k=1}^n f_{m_k}'\prod_{i\neq k}f_{m_i} = \frac{1}{x}\sum_{k=1}^n f_{m_k-1}\prod_{i\neq k}f_{m_i}$$</p>
<p>Integrating yields the formula</p>
<p>$$f_M = \sum_{i=1}^n I_{-1}(M)_i$$</p>
<p>The integrals you seek are $I_0(3,3,3)$ and $I_0(3,3,3,3)$. The difficulty is in the number of terms and not necessarily the size of each term. We can write the recurrence relation</p>
<p>$$I_0(M) = x f_M - \sum_{i=1}^n I_0(M)_i$$</p>
<p>These identities should also hold for the definite version of the integrals.</p>
<p>Anyway, we can start with $I_0(3,3,3)$ and reduce it to $I_0(2,3,3)$. Then we reduce it further to $I_0(1,3,3)$ and $I_0(2,2,3)$ and so on. The question is where do we stop. One should ideally stop when one can compute all the quantities. If not, then one should go further. In the end I think stopping when the smallest entry is 1 might be a good strategy. We reduce</p>
<p>$$(3,3,3) \to (2,3,3) \to (1,3,3),(2,2,3)$$</p>
<p>and then</p>
<p>$$(2,2,3) \to (1,2,3),(2,2,2)$$</p>
<p>The definite version of $I_0(2,2,2)$ corresponds to $\mathcal{I}_3$ from Kirill's linked answer so we have</p>
<p>$$I_0(3,3,3) = 6xf_2^2 f_3- 3xf_2 f_3^2 + x f_3^3 - 6 I_0(2,2,2) - 12 I_0(1,2,3) + 3 I_0(1,3,3)$$</p>
<p>Letting $x \to 1$ and $x \to 0$ we find</p>
<p>$$I_0(3,3,3) = 6 \zeta(2)^2 \zeta(3) - 3 \zeta(2) \zeta(3)^2 + \zeta(3)^3 - 6 \mathcal{I_3} - 12 I_0(1,2,3) + 3 I_0(1,3,3)\\
= 540 - 108 \zeta(2) - 216 \zeta(3) + 72 \zeta(2) \zeta(3) + 6 \zeta(2)^2 \zeta(3) - 3 \zeta(3) \zeta(3)^2 + \zeta(3)^3 - 171 \zeta(4) - 36 \zeta(5) - \tfrac{105}{4} \zeta(6) - 12 I_0(1, 2, 3) + 3 I_0(1, 3, 3)$$</p>
<p>Therefore the problem reduces to calculating The quantities</p>
<p>$$I_0(1, 3, 3) = \int_0^1 Li_1(x)Li_3(x)^2 dx = - \int_0^1 \log(1-x)Li_3(x)^2 dx$$</p>
<p>$$I_0(1, 2, 3) = \int_0^1 Li_1(x)Li_2(x)Li_3(x) dx = -\int_0^1 \log(1-x)Li_2(x)Li_3(x) dx$$</p>
<p>I'm not sure how much of a reduction this is because the number of terms has not been decreased. However, from Kirill's answer, it seems that it'll make his approach easier since one would need fewer iterated integrals.</p>
<p>EDIT: As Kirill mentioned, the procedure reduces the overall weight $|M| = m_1+\dots+m_n$ which could be useful on its own. If we allowed the appearance of $0$'s in $M$ we get divergent integrals, so one would have to handle them in some more special way. If we just went ahead with it we'd get</p>
<p>$$I_0(3,3,3) = 90 f_1^3 x+18 \left(f_3-5 f_2\right) f_1^2 x+3 \left(12 f_2^2-6 f_3f_2+f_3^2\right) f_1 x-6f_2^3 x+f_3^3 x-3 f_2 f_3^2 x+6 f_2^2 f_3 x + R$$</p>
<p>where $R$ consists of</p>
<p>$$R = -270 I_0(0, 1, 1) + 180 I_0(0, 1, 2)- 36 I_0(0, 1, 3) - 36 I_0(0, 2, 2) + 18 I_0(0, 2,3) - 3 I_0(0, 3, 3)$$</p>
<p>We've reduced the weight from 9 to 6, 5, 4, 3 and 2. One could even go a little further and think about making the weight 0, although I'm not sure if that would be a good way to go about this. In any case, we can try this and we obtain a larger sum with $M$'s $(-6,3,3)$, $(-5,2,3)$, $(-4,1,3)$, $(-4,2,2)$, $ (-3,0,3)$, $(-3,1,2)$, $(-2,-1,3)$, $(-2,0,2)$, $(-2,1,1)$, $(-1,-1,2)$, $(-1,0,1)$ and $(0,0,0)$.</p>
|
3,956,112 | <p>I tried to evaluate
<span class="math-container">$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$</span>
in the following way:
<span class="math-container">$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$</span>
But when I typed the integral into a calculator I got a math error. Why did this happen?</p>
| Jan Eerland | 226,665 | <p>Well, what happends to:</p>
<p><span class="math-container">$$f(x)=\frac{1}{x^2}$$</span></p>
<p>When <span class="math-container">$x\in\left[-1,1\right]$</span>?</p>
|
3,956,112 | <p>I tried to evaluate
<span class="math-container">$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$</span>
in the following way:
<span class="math-container">$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$</span>
But when I typed the integral into a calculator I got a math error. Why did this happen?</p>
| Joe | 623,665 | <p>Say <span class="math-container">$F$</span> is an antiderivative of <span class="math-container">$f$</span>—that is, if you differentiate <span class="math-container">$F$</span>, then you get <span class="math-container">$f$</span>. What you're doing to compute the integral is the following:
<span class="math-container">$$
\int_{a}^{b} f(x) \, dx = [F(x)]_a^b=F(b)-F(a) \, .
$$</span>
This method of evaluating integrals is an application of the fundamental theorem of calculus—which connects antiderivatives with integrals. To be clear,
<span class="math-container">$$
\int_{a}^{b} f(x) \, dx
$$</span>
represents the area under the graph of <span class="math-container">$f(x)$</span> between <span class="math-container">$x=a$</span> and <span class="math-container">$x=b$</span>. Meanwhile,
<span class="math-container">$$
\int f(x) \, dx
$$</span>
represents the family of functions which, when differentiated, gives you <span class="math-container">$f(x)$</span>. It should surprise you that these two ideas should should be so intimately connected. (This also explains why the integral and antiderivative notations are so similar.) However, the fundamental theorem of calculus only applies in certain cases. The example you gave,
<span class="math-container">$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx \, ,
$$</span>
is not one of those cases, since the function is undefined at <span class="math-container">$x=0$</span>. This is a problem many students have: they are so used to applying the fundamental theorem of calculus that they don't realise when it doesn't apply. In fact,
<span class="math-container">$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx
$$</span>
is not even an integral at all in the strict sense of the word. Just take a look at the graph of <span class="math-container">$y=x^2-\frac{1}{x^2}+3$</span>:</p>
<p><a href="https://i.stack.imgur.com/l5jZJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l5jZJ.png" alt="Graph of y=x^2-1/x^2+3" /></a></p>
<p>What would 'the area under the graph' mean to you in this case? What we are dealing with is called an 'improper integral'. The example you gave
<span class="math-container">$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx \, ,
$$</span>
is merely a shorthand for
<span class="math-container">$$
\lim_{a\to0^-}\int_{-1}^{a} x^2-\frac{1}{x^2}+3 \, dx + \lim_{b \to 0^+}\int_{b}^{1} x^2-\frac{1}{x^2}+3 \, dx \, .
$$</span>
Since neither of these limits exist, the 'integral' is said to be divergent. Here are some useful links:</p>
<ul>
<li>The Wikipedia articles on the <a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus" rel="nofollow noreferrer">fundamental theorem of calculus</a> and <a href="https://en.wikipedia.org/wiki/Improper_integral" rel="nofollow noreferrer">improper integrals</a>.</li>
<li>I also gave an introduction to improper integrals <a href="https://math.stackexchange.com/questions/3934619/what-does-infinity-as-an-upper-lower-limit-mean-in-a-definite-integral-eg-in/3934667#3934667">here</a>.</li>
</ul>
|
3,956,112 | <p>I tried to evaluate
<span class="math-container">$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$</span>
in the following way:
<span class="math-container">$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$</span>
But when I typed the integral into a calculator I got a math error. Why did this happen?</p>
| FFjet | 597,771 | <p>Recall when a function is Riemann integrable:</p>
<blockquote>
<p>A <strong>bounded function</strong> on a compact interval <span class="math-container">$[a, b]$</span> is Riemann integrable if and only if it is continuous almost everywhere.</p>
</blockquote>
<p>So, is <span class="math-container">$x^2-\dfrac{1}{x^2}+3$</span> bounded on <span class="math-container">$[-1,1]$</span>?</p>
|
1,270,584 | <p>I tried googling for simple proofs that some number is transcendental, sadly I couldn't find any I could understand.</p>
<p>Do any of you guys know a simple transcendentality (if that's a word) proof?</p>
<p>E: What I meant is that I wanted a rather simple proof that some particular number is transcendental ($e$ or $\pi$ would work), not a method to prove that any number is transcendental, sorry for the confusion.</p>
<p>Or even a proof about transcendental numbers being 'as common' as algebraic numbers?</p>
| Michael Hardy | 11,667 | <p>Cantor showed a simple way to list the real algebraic numbers in a sequence $a_1,a_2,a_3,\ldots$. Every term has only finitely many terms before it, and every algebraic number will be reached after only finitely many steps. Now let us seek a transcendental number in the interval $(0,1)$. Find the first number in the above sequence that is between $0$ and $1$; call it $b_1$, and narrow down the interval from $(0,1)$ to $(b_1,1)$. Find the first term in the sequence that comes after $b_1$ and falls in our now narrower interval; callit $c_1$, and narrow down the interval from $(b_1,1)$ to $(b_1,c_1)$. Then find the first number in the sequence that comes after $c_1$ and falls in our now yet narrower interval; call it $b_2$, and narrow down the interval to $(b_2,c_1)$. And so on: the next narrowing will be to $(b_2,c_2)$ then $(b_3,c_2)$, then $(b_3,c_3)$. We have $0<b_1<b_2<b_3<\cdots < c_3<c_2<c_1<1$. Some number must be between the $b$s and the $c$, and it can't be algebraic because our process has eliminated all of those.</p>
<p>This argument involving narrowing was in Cantor's very first paper on uncountability, published in 1874. His "diagonal argument" came later.</p>
<p><a href="http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof" rel="nofollow">http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof</a></p>
|
3,657,106 | <blockquote>
<p>Sam was adding the integers from <span class="math-container">$1$</span> to <span class="math-container">$20$</span>. In his rush, he skipped one of the numbers and forgot to add it. His final sum was a multiple of <span class="math-container">$20$</span>. What number did he forget to add?</p>
</blockquote>
<p>My idea was to use Gauss's trick to find this relatively simply so I proceeded as follows.</p>
<p>We have <span class="math-container">$S=1+2+3+ \dots+ 18+19+20$</span>. Using Gauss's trick we get <span class="math-container">$\frac{n(n+1)}{2} = \frac{20(21)}{2} = 210$</span>. Since we want this to equal some multiple of <span class="math-container">$20$</span> we have that <span class="math-container">$210 = 20n$</span>, but solving for <span class="math-container">$n$</span> results in <span class="math-container">$\frac{21}{2} = 10.5$</span>.</p>
<p>The correct answer for this was <span class="math-container">$10$</span>, but it seems that I'm missing something?</p>
| Toby Mak | 285,313 | <p>You don't actually need to calculate the sum to find the missing number.</p>
<p>Using Gauss's trick again, <span class="math-container">$1 + 19 = 20$</span>, <span class="math-container">$2 + 18 = 20$</span>, <span class="math-container">$3 + 17 = 20$</span> and so on, all the way up to <span class="math-container">$9 + 11 = 20$</span>. Observe that each pair is a multiple of <span class="math-container">$20$</span>.</p>
<p>The numbers which have not been paired up are <span class="math-container">$10$</span> and <span class="math-container">$20$</span>. Of these two numbers, <span class="math-container">$20$</span> is a multiple of <span class="math-container">$20$</span>, but <span class="math-container">$10$</span> is not a multiple of <span class="math-container">$20$</span>.</p>
<p>Therefore we must remove <span class="math-container">$10$</span> from the sum in order for all the other numbers form multiples of <span class="math-container">$20$</span>.</p>
|
3,163,355 | <p>I found that the Wieferich prime <span class="math-container">$1093$</span> divides <span class="math-container">$3^{1036}-1$</span>.
Does <span class="math-container">$1093$</span> divides infinitely many <span class="math-container">$3^{k}-1$</span>, with k a positive integer? And what features must <span class="math-container">$3^{k}-1$</span> have to be divisible by <span class="math-container">$1093$</span>?</p>
| Carl Schildkraut | 253,966 | <p>We see that <span class="math-container">$1093|3^7-1$</span>. Now, </p>
<p><span class="math-container">$$1093|3^k-1\Leftrightarrow 1093|(3^k-1)-(3^7-1) \Leftrightarrow 1093|3^k-3^7 \Leftrightarrow 1093|3^{k-7}-1.$$</span></p>
<p>We can see, by repeatedly subtracting <span class="math-container">$7$</span>, that <span class="math-container">$1093|3^{7m}-1$</span> for any <span class="math-container">$m$</span>. Now, if <span class="math-container">$1093|3^k-1$</span> for some other <span class="math-container">$k$</span>, then</p>
<p><span class="math-container">$$1093|3^{k-7m}-1$$</span></p>
<p>for any <span class="math-container">$m$</span>, and we can pick such an <span class="math-container">$m$</span> so <span class="math-container">$1\leq k-7m\leq 6$</span>. However,</p>
<p><span class="math-container">$$0<3^1-1<3^6-1=728<1093,$$</span></p>
<p>so <span class="math-container">$1093\nmid 3^k-1$</span> for any <span class="math-container">$7\nmid k$</span>. </p>
|
3,276,124 | <p>Consider the equation</p>
<p><span class="math-container">$$-242.0404+0.26639x-0.043941y+(5.9313\times10^{-5})\times xy-(3.9303\times{10^{-6}})\times y^2-7000=0$$</span></p>
<p>with <span class="math-container">$x,y>0$</span>. If you plot it, it'll look like below:</p>
<p><a href="https://i.stack.imgur.com/OqfFa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OqfFa.png" alt="enter image description here"></a></p>
<p>Now, I want to find the corner point/inflection point in this equation/graph, which roughly should be somewhere here. This is my manually pinpointed approximation, using my own eyes:</p>
<p><a href="https://i.stack.imgur.com/JawkV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JawkV.png" alt="enter image description here"></a></p>
<p>Any help on how to mathematically find this point would be really helpful.</p>
<p><strong>UPDATE</strong>
Based on Adrian's answer, I've got the following <span class="math-container">$(1.1842*10^{-4},0.6456*10^{-4})$</span> (wondering what can cause this slight error?):</p>
<p><a href="https://i.stack.imgur.com/oGXyw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGXyw.png" alt="enter image description here"></a></p>
<p>The actual corner point seems a little bit far from the one found by Adrian's approach (why?):</p>
<p><a href="https://i.stack.imgur.com/9j0tN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9j0tN.jpg" alt="enter image description here"></a></p>
<p><strong>Update 2</strong>
The problem was the aspect ratio of my drawing, I fixed the aspect ratio and Adrian's answer looks pretty accurate:</p>
<p><a href="https://i.stack.imgur.com/vzpoI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vzpoI.png" alt="enter image description here"></a></p>
| Adrian Keister | 30,813 | <p>Following Calvin Khor's line of reasoning, we will use the following algorithm:</p>
<ol>
<li>Find the center of the hyperbola, and translate the hyperbola so that the center coincides with the origin.</li>
<li>Find the angle of rotation necessary to put the hyperbola into the canonical form <span class="math-container">$x^2/a^2-y^2/b^2=1.$</span></li>
<li>The corner points are represented, at this point, by <span class="math-container">$x=\pm a.$</span></li>
<li>Rotate these two points back through the angle found in Step 2.</li>
<li>Translate these two points back through the translation performed in Step 1.</li>
</ol>
<p>The <a href="https://en.wikipedia.org/wiki/Hyperbola#Quadratic_equation" rel="nofollow noreferrer">wikipedia page on the general hyperbola</a> will be our guide, here.</p>
<p>Step 1. According to the wiki page, we must write the hyperbola in the form
<span class="math-container">$$A_{xx}x^2+2A_{xy}xy+A_{yy}y^2+2B_xx+2B_yy+C=0. $$</span>
We have
<span class="math-container">$$0x^2+\left(5.9313\times 10^{-5}\right)xy-\left(3.9303\times 10^{-6}\right)y^2 + 0.26639x-0.043941y-7242.0404=0, $$</span>
making
<span class="math-container">\begin{align*}
A_{xx}&=0 \\
A_{xy}&=\left(5.9313\times 10^{-5}\right)/2=2.96565\times 10^{-5} \\
A_{yy}&=-3.9303\times 10^{-6} \\
B_x&=0.26639/2=0.133195 \\
B_y&=-0.043941/2=-0.0219705 \\
C&= -7242.0404.
\end{align*}</span>
We check for the hyperbola nature, namely, that
<span class="math-container">$$D=\left|\begin{matrix}A_{xx}&A_{xy}\\ A_{xy} &A_{yy} \end{matrix}\right|<0,\quad\text{or}\quad D=\left|\begin{matrix}0&2.96565\times 10^{-5}\\ 2.96565\times 10^{-5} &-3.9303\times 10^{-6} \end{matrix}\right|=-8.79508\times 10^{-10}<0, $$</span>
which is clearly true. The center <span class="math-container">$(x_c,y_c)$</span> of the hyperbola is given by
<span class="math-container">\begin{align*}
x_c&=-\frac{1}{D}\left|\begin{matrix}B_x &A_{xy}\\ B_y &A_{yy}\end{matrix}\right| =\frac{1}{8.79508\times 10^{-10}}\left|\begin{matrix}0.133195 &2.96565\times 10^{-5}\\ -0.0219705 &-3.9303\times 10^{-6}\end{matrix}\right|=145.618\\
y_c&=-\frac{1}{D}\left|\begin{matrix}A_{xx} &B_x\\ A_{xy} &B_y\end{matrix}\right|
=\frac{1}{8.79508\times 10^{-10}}\left|\begin{matrix}0 &0.133195\\ 2.96565\times 10^{-5} &-0.0219705\end{matrix}\right|=-4491.26.\end{align*}</span></p>
<p>Step 2. The angle of rotation is given by
<span class="math-container">\begin{align*}
\tan(2\varphi)&=\frac{2A_{xy}}{A_{xx}-A_{yy}} \\
\varphi&=\frac12\,\arctan\left(\frac{2A_{xy}}{A_{xx}-A_{yy}}\right)=0.752315\,\text{rad}=43.1045^{\circ},
\end{align*}</span>
which definitely looks about right.</p>
<p>Step 3. The formula for <span class="math-container">$a^2$</span> is
<span class="math-container">$$a^2=-\frac{\Delta}{\lambda_1 D}, $$</span>
where
<span class="math-container">\begin{align*}
\Delta&=\left|\begin{matrix}A_{xx}&A_{xy}&B_x\\A_{xy}&A_{yy}&B_y\\B_x&B_y&C\end{matrix}\right|=6.26559\times 10^{-6} \\
0&=\lambda^2-(A_{xx}+A_{yy})\lambda+D.
\end{align*}</span>
Unfortunately, the wiki page fails to distinguish between <span class="math-container">$\lambda_1$</span> and <span class="math-container">$\lambda_2$</span>. If we examine the signs, we must have <span class="math-container">$a^2>0,$</span> which means, since <span class="math-container">$D<0$</span> and <span class="math-container">$\Delta>0,$</span> we must pick the positive root. We have
<span class="math-container">\begin{align*}
\lambda_2&=-3.16867\times 10^{-5}\\
\lambda_1&=2.77564\times 10^{-5},
\end{align*}</span>
so that
<span class="math-container">$$a=\pm 16020.6. $$</span>
Step 4. The point we need to rotate is <span class="math-container">$(16020.6, 0)$</span> counter-clockwise through an angle <span class="math-container">$\varphi=0.752315\,\text{rad}$</span>. The rotation matrix for doing that is given by
<span class="math-container">$$R=\left[\begin{matrix}\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi) &\cos(\varphi)\end{matrix}\right]=\left[\begin{matrix}0.730109&-0.683331\\ 0.683331 &0.730109\end{matrix}\right]. $$</span>
After rotating, the point is located at <span class="math-container">$(11696.8, 10947.4).$</span></p>
<p>Step 5. This is the moment of truth! We must translate back to the original coordinate system. The center of the original hyperbola was located at <span class="math-container">$(145.618, -4491.26).$</span> What we must do is add the coordinates together to get the un-translated version. The final point is located at <span class="math-container">$(11842.4, 6456.14).$</span></p>
<p>This is not too far off from my other answer! We check to make sure this point is on the curve</p>
<p><span class="math-container">$$x=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y}{0.26639+\left(5.9313\times10^{-5}\right)\!y}, $$</span>
and it is. So I say that this point is the "corner" of your graph.</p>
|
3,276,124 | <p>Consider the equation</p>
<p><span class="math-container">$$-242.0404+0.26639x-0.043941y+(5.9313\times10^{-5})\times xy-(3.9303\times{10^{-6}})\times y^2-7000=0$$</span></p>
<p>with <span class="math-container">$x,y>0$</span>. If you plot it, it'll look like below:</p>
<p><a href="https://i.stack.imgur.com/OqfFa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OqfFa.png" alt="enter image description here"></a></p>
<p>Now, I want to find the corner point/inflection point in this equation/graph, which roughly should be somewhere here. This is my manually pinpointed approximation, using my own eyes:</p>
<p><a href="https://i.stack.imgur.com/JawkV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JawkV.png" alt="enter image description here"></a></p>
<p>Any help on how to mathematically find this point would be really helpful.</p>
<p><strong>UPDATE</strong>
Based on Adrian's answer, I've got the following <span class="math-container">$(1.1842*10^{-4},0.6456*10^{-4})$</span> (wondering what can cause this slight error?):</p>
<p><a href="https://i.stack.imgur.com/oGXyw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGXyw.png" alt="enter image description here"></a></p>
<p>The actual corner point seems a little bit far from the one found by Adrian's approach (why?):</p>
<p><a href="https://i.stack.imgur.com/9j0tN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9j0tN.jpg" alt="enter image description here"></a></p>
<p><strong>Update 2</strong>
The problem was the aspect ratio of my drawing, I fixed the aspect ratio and Adrian's answer looks pretty accurate:</p>
<p><a href="https://i.stack.imgur.com/vzpoI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vzpoI.png" alt="enter image description here"></a></p>
| quarague | 169,704 | <p>You could also use more high powered maths as follows. </p>
<p>a) Find a parametrization <span class="math-container">$t \mapsto \gamma(t)=(x(t), y(t))$</span> of your curve.</p>
<p>b) Reparametrize to get a cuve parametrized by arc length, that is <span class="math-container">$x'(t)^2+y'(t)^2=1$</span> for all <span class="math-container">$t$</span>.</p>
<p>c) Compute the curvature <span class="math-container">$\left\|\gamma''(t)\right\|$</span></p>
<p>d) The point you are looking for is the point with maximum curvature</p>
<p>The wikipedia page <a href="https://en.wikipedia.org/wiki/Curvature" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Curvature</a> explains this in more detail. The answer should be exactly the same as the given by Adrian Keister.</p>
|
2,885,453 | <p>Evaluate $$\lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$$ where $[\cdot]$ denotes the greatest integer function.</p>
<p>Can anyone give me a hint to proceed?</p>
<p>I know that $$\frac {\sin x}{x} < 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$ and $$\frac {\tan x}{x} > 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$. But will these two inequalities be helpful here?</p>
| Kavi Rama Murthy | 142,385 | <p>Using the inequalities $\cos x < 1-\frac {x^{2}} {2} + \frac {x^{4}} {24}$ and $\sin x > x -\frac {x^{3}} {6}$ you can check that $0\leq \frac {x^{2}} {\sin x \tan x} <1$ for all $x>0$ sufficiently small. Hence the integer part of this fraction is $0$ for such $x$. The function is even so the limit from both sides are $0$. To obtain the stated inequality for $\cos x$ note that $\frac {x^{2n+2}} {(2n+2)!} <\frac {x^{2n}} {(2n)!}$ if $0<x<1$. Group the terms of the Taylor series two by two and use this inequality. A similar argument works for $\sin x$.</p>
|
3,107,858 | <p>I have to find the parametric equation for the line <span class="math-container">$M_1$</span> with the following info: </p>
<ul>
<li><span class="math-container">$M_1$</span> goes through the point <span class="math-container">$P(1,2,2)$</span></li>
<li>Is parallel with the plane <span class="math-container">$x + 3y + z = 1$</span></li>
<li>Has an intersection somewhere with the line <span class="math-container">$M_2 = 1+t, 2-2t, -1+2t$</span></li>
</ul>
<p>So if I understand it correctly, since <span class="math-container">$M_1$</span> is parallel with the plane <span class="math-container">$x + 3y + z = 1$</span>. I can do <span class="math-container">$x = 1-3y-z$</span> and replace <span class="math-container">$y$</span> and <span class="math-container">$z$</span> with <span class="math-container">$t$</span> and <span class="math-container">$s$</span> to get the direction of <span class="math-container">$M_1$</span>. The planes equation will then be:</p>
<p><span class="math-container">$$(x,y,z)=(1-3s-t , s , t)$$</span></p>
<p>So then the <span class="math-container">$M_1$</span> should be with the coordinates from <span class="math-container">$P$</span>:</p>
<p><span class="math-container">$$x = 1 + 1-3s-t$$</span></p>
<p><span class="math-container">$$y = 2 + s$$</span></p>
<p><span class="math-container">$$z = 2 + t$$</span></p>
<p>But it is incorrect because <span class="math-container">$M_1$</span> and <span class="math-container">$M_2$</span> have no intersections. Where did I go wrong here?
Thanks in advance for the answers!</p>
| lulu | 252,071 | <p>I expect there is a typo in the expression for <span class="math-container">$M_2$</span>. You've written <span class="math-container">$-1+2$</span> for the third coordinate which seems incorrect. I'll guess you meant <span class="math-container">$-1+2t$</span> and compute from there. If you meant something else, you should be able to adapt the calculation easily enough.</p>
<p>We note that the normal to the given plane is <span class="math-container">$$\vec n=(1,3,1)$$</span></p>
<p>We assume the intersection between the desired line and the given line is <span class="math-container">$P(t)=(1+t,2-2t,-1+2t)$</span>. Then the vector <span class="math-container">$\vec {v(t)}=P(t)-P=(t,-2t,-3+2t)$</span> should be perpendicular to <span class="math-container">$\vec n$</span> so we want to solve: <span class="math-container">$$t-6t-3+2t=0\implies 3t=-3\implies t=-1$$</span></p>
<p>Thus <span class="math-container">$M_1$</span> is given by <span class="math-container">$$P+t\vec {v(-1)}=(1,2,2)+t(-1,2,-5)$$</span></p>
<p>Check: Since <span class="math-container">$P$</span> is not in the given plane we should see that no point in <span class="math-container">$M_1$</span> is in the plane. We try to solve <span class="math-container">$$1-t+6+6t+2-5t=1$$</span> and confirm that there is no solution.</p>
|
181,839 | <p>Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.</p>
<p>Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.</p>
<p>Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)</p>
<p>Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.</p>
<p>I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)</p>
<p>But i couldn't. Is there any way to show this?</p>
<p>If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.</p>
<p>Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.</p>
<p>Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.</p>
<p>Please help.</p>
<p>If my argument is wrong, please give me a proper proof.</p>
| Asaf Karagila | 622 | <p>This is essentially the Baire category theorem. Indeed in ZF it holds for separable complete metric spaces.</p>
<p>The argument is as follows:</p>
<p>Suppose that $(X,d)$ is a separable complete metric space, and $\{a_k\mid k\in\omega\}=D\subseteq X$ is a countable dense subset.</p>
<p>By contradiction assume that we can write $X=\bigcup F_n$ where $F_n$ are closed and with empty interior, we can further assume that $F_n\subseteq F_{n+1}$.</p>
<p>Define by recursion the following sequence:</p>
<ol>
<li>$x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$; </li>
<li>$r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.</li>
<li>$x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;</li>
<li>$r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.</li>
</ol>
<p>Note that $x_n$ is a Cauchy sequence, therefore it converges to a point $x$. If $x\in F_n$ for some $n$, first note that $d(x_k,F_n)\leq d(x_k,x)$, by the definition of a distance from a closed set.</p>
<p>If so, for some $k$ we have that $d(x,x_k)<r_n$, in particular $d(F_n,x_k)<r_n$. First we conclude that $n<k$, otherwise $d(F_n,x_n)>r_n$. Now we note that:</p>
<p>$$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$$</p>
<p>It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.</p>
<hr>
<p><strong>Added:</strong> <em>Why is there $a_k$ in every step of the inductive definition?</em></p>
<p>Note that $D$ is dense therefore $\overline{D}=X$. In particular, if $F_n$ is closed and has a dense subset then $F=X$. Since we assume that the interior of $F_n$ is empty we have that $F_n\cap D$ is not dense in $X$, otherwise $F_n=X$ and has a non-empty interior.</p>
<p>We have that $D\setminus F_n$ is dense, since $D\cap F_n$ is not dense. In particular this means that in every open set there is some $a_k\in D\setminus F_n$, and thus the induction can be carried in full.</p>
|
3,714,418 | <p>I need to calculate the angle between two 3D vectors. There are plenty of examples available of how to do that but the result is always in the range <span class="math-container">$0-\pi$</span>. I need a result in the range <span class="math-container">$\pi-2\pi$</span>.</p>
<p>Let's say that <span class="math-container">$\vec x$</span> is a vector in the positive x-direction and <span class="math-container">$\vec y$</span> is a vector in the positive y-direction and <span class="math-container">$\vec z$</span> is a reference vector in the positive z-direction. <span class="math-container">$\vec z$</span> is perpendicular to both <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span>. Would it then be possible to calculate the angle between <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span> and get a result in the range <span class="math-container">$\pi-2\pi$</span>? </p>
<p>The angle value should be measured counter clockwise. I have not been able to figure out how to do that.
I am no math guru but I have basic understanding of vectors at least.
Thank you very much for the help! </p>
| fleablood | 280,126 | <p>That's straightforward element chasing and in my opinion the way you did it <em>IS</em> the best proof.</p>
<p>We can go to "concepts" but this assume the reader has a strong intuition which... I'm learning not all do ... and what good is a proof if it's hard to follow:</p>
<p>For any sets <span class="math-container">$W,K, M$</span> where <span class="math-container">$K\subset M$</span> then <span class="math-container">$W-M \subset W-K$</span>. Is that clear?</p>
<p>If not. <span class="math-container">$W-M = $</span> all elements of <span class="math-container">$W$</span> minus set of <em>all</em> elements of <span class="math-container">$M$</span> which "clearly" a subset of all elements of <span class="math-container">$W$</span> minus only <em>some</em> of the elements of <span class="math-container">$M$</span> which is what <span class="math-container">$W-K$</span> is as <span class="math-container">$K\subset M$</span> so <span class="math-container">$K$</span> is only <em>some</em> of the elements of <span class="math-container">$K$</span>.</p>
<p>(Of course, the <em>emotional</em> appeal to "clear" obviousness is precisely what we invented cold hard <em>mathematics</em> to <em>avoid</em>....)</p>
<p>So as <span class="math-container">$B-A\subset B$</span> then <span class="math-container">$A- B \subset A- (B-A)$</span>.</p>
<p>=======</p>
<p>Actually let's make that better:</p>
<p>Lemma: If <span class="math-container">$M \subset K$</span> then <span class="math-container">$K^c = \{x\not \in K$</span> where it is understood there is some universal collection of all valid elements<span class="math-container">$\} \subset M^c =\{x\not \in M\}$</span>.</p>
<p>Pf: If <span class="math-container">$x \not \in K$</span> then <span class="math-container">$x \not \in M$</span> as <span class="math-container">$M \subset K$</span>. So <span class="math-container">$K^c \subset M^c$</span>.</p>
<p>Lemma 2: If <span class="math-container">$M \subset K$</span> then <span class="math-container">$W- K\subset W-M$</span>.</p>
<p>Pf: If <span class="math-container">$x \in W-K$</span> then <span class="math-container">$x\in W$</span> and <span class="math-container">$x \not \in K$</span>. But then <span class="math-container">$x \not \in M$</span> because <span class="math-container">$M \subset K$</span>. And <span class="math-container">$x\in W$</span>. So <span class="math-container">$x\in W-M$</span> so <span class="math-container">$W-K \subset W-K$</span>.</p>
<p>Lemma 3 (just to be thourough): <span class="math-container">$B-A \subset B$</span>.</p>
<p>Pf: Well, doi! if <span class="math-container">$x\in B-A$</span> then <span class="math-container">$x \in B$</span> and <span class="math-container">$x\not \in A$</span>. SO <span class="math-container">$x \in B$</span>.</p>
<p>and that's it.</p>
<p><span class="math-container">$B-A \subset B$</span> so</p>
<p><span class="math-container">$A-B \subset A-(B-A)$</span>.</p>
<p>....</p>
<p>Hmmm, I changed my mind. That <em>is</em> more elegant and more basic.... albeit it much longer....</p>
<p>That's my proof not and I'm sticking with it....until I change my mind again.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Henry Segerman | 6,015 | <p>(I learned this puzzle from Ravi Vakil.) Suppose you have an infinite grid of squares, and in each square there is an arrow, pointing in one of the 8 cardinal directions, with the condition that any two orthogonally adjacent arrows can differ by at most 45 degrees.</p>
<p>Can there be a closed cycle? (i.e. start at some arrow, move to the square that arrow points to, follow where the arrow there points and so on, and come back to the square you started at).</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p><strong>Fork in the road 1</strong></p>
<p>You're on a path on an island, come to a fork in the road. Both paths lead to villages of natives; the entire village either always tells the truth or always lies <em>(both villages could be truth-telling or lying villages, or one of each)</em>. There are two natives at the fork - they could both be from the same village, or from different villages <em>(so both could be truth-tellers, both liars, or one of each)</em>.</p>
<p>One path leads to safety, the other to doom. You're allowed to ask only one question to each native to figure out which path is which.</p>
<p>What do you ask?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| BlueRaja | 2,883 | <p><strong>Fork in the road 2</strong></p>
<p>You're once again at a fork in the road, and again, one path leads to safety, the other to doom.</p>
<p>There are three natives at the fork. One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers. Of course you don't know which is which.</p>
<p>Moreover, the natives answer "pish" and "posh" for yes and no, but you don't know which means "yes" and which means "no."</p>
<p>You're allowed to ask only two yes-or-no questions, each question being directed at one native.</p>
<p>What do you ask?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| reb | 50 | <p>What is the resistance between 2 adjacent vertices of an infinite checkerboard if every edge is a 1 ohm resistor? </p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Tom Boardman | 5,869 | <p>Okay, so it's somewhat more numeric than the others, but I quite enjoy the simplicity of:</p>
<blockquote>
<h3>Simplify:</h3>
<p>$$\sqrt{2+\sqrt3}$$</p>
</blockquote>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Ken Fan | 7,434 | <p>Here's a balance scale problem that I decided to post because a little bit of googling around for it came up negative. It differs from most balance scale puzzles I've seen because it doesn't involve "bad weights". I learned of it from a friend of mine who is an engineer.</p>
<p>There are 10 balls which come in two possible weights. Using a balance scale at most 3 times, determine whether all the balls are the same weight or not.</p>
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29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| schnitzi | 7,740 | <p>Okay, I've got one, and as far as I know it hasn't been analyzed before.</p>
<p>I was watching a travel show the other night -- they were in Korea, and a group of people were playing a drinking game. It works like this:</p>
<p><em>One person is "it". This person says something like, "ready, set..." then points at one other player and calls out a number between 2 and n (where n is the number of people playing). At the same time, everyone else also points at one other player. Then, for whatever number got call out, you jump that many steps from the "it" person, and that person has to drink. So if I call out "two" and point at Joe, and Joe points at Bob, then Bob has to drink.</em></p>
<p>I think the game is pretty interesting, mathematically, especially when you allow numbers greater than n to be called. One interesting thing I found: with n=3, if you call 7 (or 7+6x, where x is a non-negative integer), you are guaranteed to stick the player you
initially point at, no matter who points to whom.</p>
<p>I think an interesting question is, given n players, what is the smallest number the 'it' person can call that guarantees he will not stick himself? (I have an answer, but I want to see if you all come up with the same thing. :-) And what's the best strategy for the caller if you enforce the rule that you must call out a number between 2 and n? What's the best strategy for the other players, if they're allowed to collude on who they're going to point to? Etc.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Tracy Hall | 7,936 | <p>You are the captain of a team of <i>N</i> players, in charge of choosing a strategy that your adversary will overhear (and therefore rig the game for you to lose unless the strategy is perfect). To play the game, the adversary writes a distinct name on each player's forehead and you are brought into a situation where each of you can learn the name given to every other player, but not your own. Naturally you cannot communicate once the game has started. Each of you is blindfolded and given a single invertible glove. On a signal, each of you silently places your glove on one hand or the other. You are then lined up in alphabetical order by the names on your foreheads, all facing the same direction, and you join hands in one long chain. If any of you touches another player's glove with your bare hand the team loses, but if it is always hand-to-hand and glove-to-glove, you are victorious.</p>
<p>For what values of <i>N</i> can you give your team a winning strategy, and what is it? </p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| klaraspina | 6,882 | <p>There are some dwarves approaching to a bridge. They have to cross it to come back home from the cave where they work. Unfortunately, a dragon has just decided to reside under that bridge, and it's hungry. But it's also bored, so it doesn't want just to eat the dwarves, but proposes them a game: it will put on each of them one hat, either black or white, in no specific proportion (for example, it can happen all hats to be white). Of course they can't see their own hat, but they can see the others'. They will be then queueing at the beginning of the bridge, and each of them can just say one word. If this word matches with the colour of the hat that dwarf is wearing, then he's allowed to pass and to come back home. Otherwise, he'll be eaten by the dragon. Of course, they can decide for a strategy before the game begins.</p>
<p>What's the best strategy, and how many dwarves die on average?</p>
<p>EDIT: (deleted the previous PS, modified into this one)
PS: Since I didn't solve all previous puzzles posted, I would be glad if someone could point me at equivalent puzzles, if any!</p>
|
557,963 | <p>Suppose that $V = X ⊕Y$, and let $P$ be the projector onto $X$ along
$Y$. Prove that
$R(P) = N (I − P) = X$ and $R(I − P) = N (P) = Y$.</p>
<p>I know that from $V = X ⊕Y$ I got $v=x+y$ for $v,x,y$ are element of $V,X,Y$ and the intersect of $X$ and $Y$ is zero. But I don't know what to do next.</p>
| Cameron Buie | 28,900 | <p><strong>Hint</strong>: For any $v\in V$, we have $v=P(v)+y,$ where $y=v-P(v)=(I-P)(v).$</p>
<p>Recall that for any linear transformation $T:V\to W,$ we have $$N(T)=\{v\in V:T(v)=0_W\}$$ and $$R(T)=\{w\in W:T(v)=w\text{ for some }v\in V\}.$$</p>
<p>In particular, then, recalling that all $v\in V$ can be uniquely written in the form $x+y$ where $x\in X,y\in Y,$ and so $P(v)=x,$ then $$N(P)=\{v\in V:x=0\}\\R(I-P)=\{w\in V:w=(I-P)(v)=v-P(v)=v-x=y\text{ for some }v\in V\}.$$ It shouldn't be too difficult to show that these two sets are equal, and in particular are equal to $Y$. Similarly, we can show that $N(I-P)=R(P)=X$.</p>
|
557,963 | <p>Suppose that $V = X ⊕Y$, and let $P$ be the projector onto $X$ along
$Y$. Prove that
$R(P) = N (I − P) = X$ and $R(I − P) = N (P) = Y$.</p>
<p>I know that from $V = X ⊕Y$ I got $v=x+y$ for $v,x,y$ are element of $V,X,Y$ and the intersect of $X$ and $Y$ is zero. But I don't know what to do next.</p>
| Tom | 103,715 | <p>Hint: For a projection $P$, $P^2 = P$. So, for example, $P(I-P)x = Px - P^2x = Px - Px = 0$.</p>
|
4,344,139 | <p>I have broken down this equation into factorials, but I'm unsure of where to go from here. This may not even be the right approach to solve this binomial transform. Any help would be appreciated.</p>
<p><a href="https://i.stack.imgur.com/oqNqm.png" rel="noreferrer">Binomial transform identity</a>:</p>
<p><span class="math-container">\begin{align*}
&\sum_{k=0}^n(-1)^{n-k}{n\choose k}{n+jk\choose jk}=j^n\\
&\sum_{k=0}^n(-1)^{n-k}\frac{(jk+1)(jk+2)\dots(jk+n)}{(n-k)!\,k!}=j^n
\end{align*}</span></p>
| Carl Schildkraut | 253,966 | <p>Recall that, if <span class="math-container">$p(x)$</span> is a polynomial of degree <span class="math-container">$n$</span>,
<span class="math-container">$$\sum_{k=0}^n(-1)^{n-k}\binom nkp(x+k)$$</span>
is <span class="math-container">$n!$</span> times the leading coefficient of <span class="math-container">$p(x)$</span> (this may be proven by induction). Now, select
<span class="math-container">$$p(x)=\frac{(jx+n)(jx+n-1)\cdots (jx+1)}{n!},$$</span>
which is a polynomial of degree <span class="math-container">$n$</span> with leading coefficient <span class="math-container">$j^n/n!$</span>. For integer <span class="math-container">$k$</span>,
<span class="math-container">$$p(k)=\frac{(jk+n)(jk+n-1)\cdots(jk+1)}{n!}=\binom{jk+n}{n}=\binom{n+jk}{jk}.$$</span>
So,
<span class="math-container">$$\sum_{k=0}^n(-1)^{n-k}\binom nk\binom{n+jk}{jk}$$</span>
is <span class="math-container">$n!$</span> times the leading coefficient of <span class="math-container">$p(x)$</span>, i.e. <span class="math-container">$j^n$</span>.</p>
|
4,344,139 | <p>I have broken down this equation into factorials, but I'm unsure of where to go from here. This may not even be the right approach to solve this binomial transform. Any help would be appreciated.</p>
<p><a href="https://i.stack.imgur.com/oqNqm.png" rel="noreferrer">Binomial transform identity</a>:</p>
<p><span class="math-container">\begin{align*}
&\sum_{k=0}^n(-1)^{n-k}{n\choose k}{n+jk\choose jk}=j^n\\
&\sum_{k=0}^n(-1)^{n-k}\frac{(jk+1)(jk+2)\dots(jk+n)}{(n-k)!\,k!}=j^n
\end{align*}</span></p>
| Marko Riedel | 44,883 | <p>In trying to prove</p>
<p><span class="math-container">$$\sum_{k=0}^n (-1)^{n-k} {n\choose k} {n+jk\choose jk} = j^n$$</span></p>
<p>we start with</p>
<p><span class="math-container">$$\sum_{k=0}^n (-1)^{n-k} {n\choose k} {n+jk\choose n}
\\ = [z^n] (1+z)^n
\sum_{k=0}^n (-1)^{n-k} {n\choose k} (1+z)^{jk}
\\ = [z^n] (1+z)^n ((1+z)^j-1)^n.$$</span></p>
<p>Now <span class="math-container">$((1+z)^j - 1)^n = (jz + \cdots)^n$</span> so the only term that contributes
to the coefficient extractor in <span class="math-container">$z$</span> is <span class="math-container">$j^n z^n.$</span> (The <span class="math-container">$n$</span> factors
contribute at least <span class="math-container">$z$</span> so for the power to be less than or equal to <span class="math-container">$n$</span>
we have to choose <span class="math-container">$z$</span> from each factor. As soon as we choose just one
factor to a power at least two the term produces a power that is larger
than <span class="math-container">$n$</span> and does not contribute to the coefficient extractor.) We get</p>
<p><span class="math-container">$$[z^n] (1+z)^n j^n z^n = j^n$$</span></p>
<p>as claimed.</p>
|
4,245,205 | <p>Suppose <span class="math-container">$X$</span> is a CW-complex of dimension <span class="math-container">$n$</span>. If <span class="math-container">$e_i$</span> is an <span class="math-container">$n$</span>-cell then is <span class="math-container">$H_n(X) \to H_n(X, X \setminus e_i)$</span> the zero map? If not, then is <span class="math-container">$H_n(X, X\setminus e_i) \to H_n(X, X \setminus \{x\})$</span> the zero map where <span class="math-container">$x$</span> is a point in the interior of <span class="math-container">$e_i$</span>?</p>
<p>I cannot prove either, I suspect the proof would boil down to some fact about the homology of CW complexes which I do not know.</p>
<p>For what its worth, these statements are relevant for a proof I am reading where <span class="math-container">$X$</span> is a compact manifold of dimension <span class="math-container">$n$</span> admitting a CW structure and homology with <span class="math-container">$\mathbb{Z}_2$</span> coefficients are taken.</p>
| user965894 | 965,894 | <p>No, take <span class="math-container">$X = S^n = e_0 \cup e_n$</span> as a counterexample. The map <span class="math-container">$H_n(X) \to H_n(X, X \setminus \{x\})$</span> is an isomorphism.</p>
<p>More generally your map <span class="math-container">$X \to X/(X \setminus e_n) \cong S^n$</span> is a cellular map. It induces a chain map <span class="math-container">$f: C_*^{CW}(X, *) \to C_*^{CW}(S^n, *) = \Bbb Z[n]$</span>, where the latter notation means we have a copy of the integers in degrees 0 and n and no differential. The map <span class="math-container">$f$</span> collapses every generator except <span class="math-container">$e_n$</span>.</p>
<p>If <span class="math-container">$c = \sum c(j) e_n^j$</span> is a sum of n-cells with let's say <span class="math-container">$e_n^0 = e_n$</span> is the fixed one in the discussed above, then <span class="math-container">$f(c) = c(0)$</span>. If <span class="math-container">$f$</span> is zero on homology then every cycle has <span class="math-container">$c(0) = 0$</span>; if <span class="math-container">$f$</span> is nonzero on homology there is some cycle with <span class="math-container">$c(0) \neq 0$</span>.</p>
<p>Putting this all together your map is zero on homology if and only if <span class="math-container">$\partial e_n$</span> is not a sum of boundaries of other cells. I'm not sure if this is what you're looking for, it's more or less just from definition.</p>
|
3,530,285 | <blockquote>
<p>Let <span class="math-container">$\text{char}(\mathbb{K}) = 0$</span>. It then follows that <span class="math-container">$AB-BA \ne 1 \, (A, B \in \Bbb K^{n \times n})$</span>.</p>
</blockquote>
<p>I first showed that <span class="math-container">$\text{trace}(AB) = \text{trace}(BA)$</span> for every <span class="math-container">$A, B \in \Bbb K^{n \times n}$</span>:</p>
<p>Let <span class="math-container">$A=(a_{ij}), B=(b_{ij}) \in \Bbb K^{n \times n}$</span>, so <span class="math-container">$AB=(c_{ij}) \in \Bbb K^{n \times n}$</span> where <span class="math-container">$c_{ij}= \sum_{k=1}^na_{ik}b_{kj}$</span>. One now gets: </p>
<p><span class="math-container">$$\text{trace}(AB) = \sum_{i=1}^n \sum_{k=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n b_{ki}a_{ik} = \text{trace}(BA)\tag{*}.$$</span></p>
<p>Now, assume <span class="math-container">$AB-BA=1$</span>. Consider</p>
<p><span class="math-container">$$AB-BA=1 \Leftrightarrow \text{trace}(AB)-\text{trace}(BA)=\text{trace}(1) \xrightarrow{(*)} \\ 0 = n,$$</span></p>
<p>which is a contradiction, so the assumption <span class="math-container">$AB-BA=1$</span> was wrong.</p>
<p>My question is: Where exactly do you need <span class="math-container">$\text{char}(\Bbb K)=0$</span> in this proof?</p>
<p>Thanks in advance!</p>
| G. Chiusole | 436,096 | <p><span class="math-container">$S$</span> is not closed in <span class="math-container">$\mathbb{Q}$</span>: </p>
<p>Assume it were, then for any sequence <span class="math-container">$(x_n)_{n \in \mathbb{N}} \subseteq S$</span> which converges to <span class="math-container">$a \in \mathbb{Q}$</span> we would have <span class="math-container">$a \in S$</span>. However, take the sequence <span class="math-container">$(\frac{1}{n})_{n \in \mathbb{N}} \subseteq S$</span>. As a sequence in <span class="math-container">$\mathbb{Q}$</span>, it converges to <span class="math-container">$0$</span>. However, <span class="math-container">$0 \not\in S$</span>. Hence <span class="math-container">$S$</span> is not closed in <span class="math-container">$\mathbb{Q}$</span>. </p>
<p>In more detail: Why does <span class="math-container">$(\frac{1}{n})_{n \in \mathbb{N}} \subseteq \mathbb{Q}$</span> converge to <span class="math-container">$0 \in \mathbb{Q}$</span>? </p>
<p>By definition, a sequence converges to <span class="math-container">$0$</span> iff for any open neighborhood <span class="math-container">$N$</span> of <span class="math-container">$0$</span> there exists an <span class="math-container">$n_0 \in \mathbb{N}$</span> s.t. <span class="math-container">$\forall n \geq n_0: x_n \in N$</span>. Here, since <span class="math-container">$\mathbb{Q}$</span> has the subspace topology w.r.t. <span class="math-container">$\mathbb{R}$</span>, the neighborhoods of <span class="math-container">$0$</span> are completely characterized as the open neighborhoods of <span class="math-container">$0$</span> in <span class="math-container">$\mathbb{R}$</span>, intersected with <span class="math-container">$\mathbb{Q}$</span>. Furthermore, since in <span class="math-container">$\mathbb{R}$</span>, every open neighborhood of <span class="math-container">$0$</span> contains an open ball, we only need to consider those. So, since we know what the open balls of <span class="math-container">$\mathbb{R}$</span> are, we know that <span class="math-container">$N = \mathbb{Q} \cap B_{\varepsilon}(0)$</span> for some <span class="math-container">$\varepsilon > 0$</span>. </p>
<p>And indeed, for every <span class="math-container">$\varepsilon > 0$</span> there exists a <span class="math-container">$n_0 \in \mathbb{N}$</span> s.t. <span class="math-container">$\frac{1}{n} \in \mathbb{Q} \cap B_{\varepsilon}(0)$</span> i.e. <span class="math-container">$\vert \frac{1}{n} \vert < \varepsilon$</span>.</p>
|
4,242,093 | <p><em><strong>Question:</strong></em></p>
<blockquote>
<p>Let <span class="math-container">$G=(V_n,E_n)$</span> such that:</p>
<ul>
<li>G's vertices are words over <span class="math-container">$\sigma=\{a,b,c,d\}$</span> with length of <span class="math-container">$n$</span>, such that there aren't two adjacent equal chars.</li>
<li>An edge is defined to be between two vertices that are different by only one char.</li>
</ul>
<p>A. Does the graph contain an Euler cycle?</p>
<ul>
<li>Find a pattern.</li>
</ul>
<p>B. Does the graph contain a Hamiltonian cycle</p>
<ul>
<li>This can be proven by induction.</li>
</ul>
</blockquote>
<hr />
<p><span class="math-container">$Solution.A.$</span></p>
<p>Now, when <span class="math-container">$n=1$</span>, we have 4 vertices: <span class="math-container">$$v_1= \ 'a'$$</span> <span class="math-container">$$v_2= \ 'b'$$</span> <span class="math-container">$$v_3= \ 'c'$$</span> <span class="math-container">$$v_4= \ 'd'$$</span></p>
<p>Therefore, for each <span class="math-container">$v\in \{v_1,v_2,v_3,v_4\}$</span>, <span class="math-container">$N(v)=\{v_1,v_2,v_3,v_4\}/ \{v\}$</span> so we get that their degree is 3, so by a theorem we get that there isn't an Euler cycle.</p>
<p>In addition, when <span class="math-container">$n=4$</span>, considering the string <span class="math-container">$"abad"$</span> we have 2 options to replace the edges of the string. In order to replace the second char we have 2 options, replacing it by <span class="math-container">$'c'$</span> and <span class="math-container">$'d'$</span>. For the third char, we can replace it only by <span class="math-container">$'c'$</span>. In total, we got 7 edges with this vertex, so by a theorem, we get that there isn't an Euler cycle.</p>
<p>I cannot find here a pattern, because if we take a look at <span class="math-container">$n=2$</span> we get an Euler cycle.</p>
<p><span class="math-container">$Solution.B.$</span></p>
<p>First, we examine whether each vertex has at least <span class="math-container">$\frac{n}{2}$</span> neighbors. Hence, we should take the vertex to have the least number of neighbors. This vertex should be the string with disjoint chars. i.e. the string "abcd" when <span class="math-container">$n=4$</span>. The first and last chars has always 2 neighbors, so we get that the least degree is: <span class="math-container">$$2+2+\binom{n-2}{n-3} \cdotp 1=4+n-2=n+2\geq \frac{n}{2}$$</span></p>
<p>Thus, we get that the graph always has a Hamiltonian cycle.</p>
<hr />
<p>I don't get why I didn't get a pattern in <span class="math-container">$A$</span>, and how <span class="math-container">$B$</span> can be proven by induction. In addition, is my answer correct?</p>
| ploosu2 | 111,594 | <p>For part A use the words</p>
<p><span class="math-container">$$w = a(bc)^k$$</span></p>
<p>and</p>
<p><span class="math-container">$$w = a(bc)^kb$$</span></p>
<p>for <span class="math-container">$n>2$</span> odd and even respectively.</p>
<p>These have odd degrees since the first <span class="math-container">$b$</span> is the only letter that can be changed to only <span class="math-container">$1$</span> other. All others can be changed to <span class="math-container">$2$</span>. By symmetry you get too many odd degree vertices for the graph to be Eulerian.</p>
<p>(This fails in the case <span class="math-container">$n=2$</span> when that <span class="math-container">$b$</span> becomes an "edge letter".)</p>
<p>EDIT: I'll put my code here for safe keeping:</p>
<pre><code>import itertools
def makeGraph(n):
letters = ['a', 'b', 'c', 'd']
forbids = [c*2 for c in letters]
g = Graph()
for t in itertools.product(*[letters]*n):
v = "".join(t)
if all(f not in v for f in forbids):
g.add_vertex(v)
V = list(g.vertices())
for v1 in V:
for v2 in V:
if sum(1 if c1!=c2 else 0 for c1,c2 in zip(v1,v2))==1:
g.add_edge(v1, v2)
return g
g = makeGraph(3)
g.show()
#for hc in g.subgraph_search_iterator(graphs.CycleGraph(g.order())):
# print (hc)
found, hCyc = g.hamiltonian_cycle(algorithm='backtrack')
print(hCyc)
</code></pre>
|
78,946 | <p>I want to find min of the function
$$\frac{1}{\sqrt{2
x^2+\left(3+\sqrt{3}\right)
x+3}}+\frac{1}{\sqrt{2
x^2+\left(3-\sqrt{3}\right)
x+3}}+\sqrt{\frac{1}{3}
\left(2 x^2+2 x+1\right)}.$$
I know, the exact value minimum is $\sqrt{3}$ at $x = 0$. With <em>Mathematica</em>, I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
NMinimize[A, {x}]
</code></pre>
<p>And I got </p>
<blockquote>
<p>{1.73205, {x -> -2.57345*10^-16}}</p>
</blockquote>
<p>When I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
Minimize[A, {x}]
</code></pre>
<p>my computer ran about 20 minutes and I did not got the result. How can I get the exact value minimum of the given function?</p>
| k_v | 24,727 | <pre><code>a[x_] := Sqrt[1 + 2 x + 2 x^2]/Sqrt[3] + 1/Sqrt[
3 + (3 - Sqrt[3]) x + 2 x^2] + 1/Sqrt[3 + (3 + Sqrt[3]) x + 2 x^2]
Plot[a[x], {x, -5, 5}]
</code></pre>
<p><img src="https://i.stack.imgur.com/LBzs5.jpg" alt="enter image description here"></p>
<pre><code>extr = NSolve[a'[x] == 0, x, Reals]
</code></pre>
<blockquote>
<p>{{x -> -2.14393}, {x -> -1.20438}, {x -> 0.}}</p>
</blockquote>
<pre><code>{a[x] /. #, #} & /@ extr
</code></pre>
<blockquote>
<p>{{2.42668, {x -> -2.14393}}, {3.4091, {x -> -1.20438}}, \ {1.73205, {x
-> 0.}}}</p>
</blockquote>
|
3,298,282 | <p>Find the number of terms in the expansion <span class="math-container">$(1+a^3+a^{-3})^{100}$</span></p>
<p>I used the concept <span class="math-container">$a^3+a^{-3}=T$</span>, while using this I have 101 terms, from <span class="math-container">$T^2$</span> to <span class="math-container">$T^{100}$</span> how do i find the number of terms that do not intersect</p>
| Mike | 544,150 | <p>Well, note that </p>
<p><span class="math-container">$$(a^{-3} + 1+a^3)^{100} = (a^{-3}+a^0+a^{3})^{100} = \sum_{i=-100}^{100} c_ia^{3i},$$</span></p>
<p>for some <span class="math-container">$c_i$</span> and each <span class="math-container">$c_i$</span> is strictly positive. In fact, each <span class="math-container">$c_i$</span>
is precisely the number of ordered multisets <span class="math-container">$\{b_1,\ldots, b_{100}\}$</span>;
<span class="math-container">$b_l \in \{-1,0,1\}$</span> such that <span class="math-container">$i=\sum_{l=1}^{100} b_l$</span>.
[Make sure you see why]. This is at least 1 for each <span class="math-container">$i \in \{-100, -99,\ldots, -1, 0,1, \ldots, 100\}$</span>; indeed if <span class="math-container">$i$</span> is positive let <span class="math-container">$b_l=1$</span> for each <span class="math-container">$l=1,2,\ldots, i$</span> and <span class="math-container">$b_l=0$</span> for each other <span class="math-container">$i$</span>,
and if <span class="math-container">$i$</span> is negative let <span class="math-container">$b_l=-1$</span> for each <span class="math-container">$i=1,2,\ldots, |i|$</span>, and <span class="math-container">$b_l=0$</span> for each other <span class="math-container">$l$</span>. If <span class="math-container">$i$</span> is 0 then let all the <span class="math-container">$b_l$</span>s are 0.</p>
<p>So, as each <span class="math-container">$c_i$</span> is positive this would be 201 terms.</p>
<p>Now with that said <span class="math-container">$(1+a^3)^{100}$</span> would have 101 terms. Do you see why?</p>
|
1,930,933 | <blockquote>
<p>Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?</p>
</blockquote>
<p>The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?</p>
| robjohn | 13,854 | <p>In <a href="https://math.stackexchange.com/a/590901">this answer</a>, it is shown that the number of factors of $p$ that divide $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$.</p>
<p>For $n!$ be an $n^{\text{th}}$ power, $(1)$ must be a multiple of $n$ for any prime $p$.</p>
<p>For any $n\ge1$, we have $\sigma_p(n)\ge1$ for any prime $p$. Thus, $(1)$ is less than $n$, and since it must be a multiple of $n$, it must be $0$.</p>
<p>Thus, either $n=0$ or the number of factors of any prime $p$ that divides $n!$ must be $0$. Therefore, we either have $n=0$ or $n=1$.</p>
|
121,784 | <p>I hope this is not too trivial for this forum. I was wondering if someone has come across this polytope.</p>
<p>You start with the <a href="http://mathworld.wolfram.com/RhombicDodecahedron.html" rel="nofollow noreferrer">rhombic dodecahedron</a>, subdivide it into four parallellepipeds, <a href="https://i.stack.imgur.com/639Xy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/639Xy.jpg" alt="enter image description here"></a></p>
<p>and then fill the space between the four parallellepipeds with a tetrahedron, six parallellepipeds and four prisms (hopefully I counted correctly), so as to obtain a convex polytope. </p>
<p>Does this have a name? could someone provide a link to a picture?</p>
| Joseph O'Rourke | 6,094 | <p>Is this it?
<br />
<img src="https://i.stack.imgur.com/lTqnQ.jpg" alt="Rhombic Dodecahedron + Tetrahedron" />
<br /></p>
|
121,784 | <p>I hope this is not too trivial for this forum. I was wondering if someone has come across this polytope.</p>
<p>You start with the <a href="http://mathworld.wolfram.com/RhombicDodecahedron.html" rel="nofollow noreferrer">rhombic dodecahedron</a>, subdivide it into four parallellepipeds, <a href="https://i.stack.imgur.com/639Xy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/639Xy.jpg" alt="enter image description here"></a></p>
<p>and then fill the space between the four parallellepipeds with a tetrahedron, six parallellepipeds and four prisms (hopefully I counted correctly), so as to obtain a convex polytope. </p>
<p>Does this have a name? could someone provide a link to a picture?</p>
| F. C. | 10,881 | <p>It seems to ressemble the "Self-Dual Icosioctahedron #4" :</p>
<p><a href="http://dmccooey.com/polyhedra/SelfDualIcosioctahedron4.html" rel="nofollow noreferrer">http://dmccooey.com/polyhedra/SelfDualIcosioctahedron4.html</a></p>
<p>Some code:</p>
<pre><code>sage: P = polytopes.rhombic_dodecahedron()
sage: Q = polytopes.tetrahedron()
sage: R = P.minkowski_sum(Q)
sage: R.f_vector()
(1, 28, 54, 28, 1)
sage: R.plot()
</code></pre>
|
72,651 | <p>$G$ is a group and $H$ is a subgroup of $G$ such that $\forall a, b$ in $G, ab\in H\implies ba\in H$. Show that $H$ is normal in $G$</p>
| Arturo Magidin | 742 | <p>Let $G$ be a group, and let $H$ be a subgroup. In analogy to the definition of congruence for integers, $a\equiv b\pmod{m}$ if and only if $m|a-b$, if and only if $a-b\in\langle m\rangle$, define:</p>
<ul>
<li>$a$ is congruent on the left modulo $H$ to $b$, $a\mathrel{{}_H{\equiv}} b$, if and only if $a^{-1}b\in H$.</li>
<li>$a$ is congruent on the right modulo $H$ to $b$, $a\mathrel{\equiv_H} b$, if and only if $ab^{-1}\in H$.</li>
</ul>
<p><strong>Proposition.</strong> Both $\mathrel{\equiv_H}$ and $\mathrel{{}_H{\equiv}}$ are equivalence relations on $G$.</p>
<p><strong>Theorem.</strong> Let $G$ be a group, and let $H$ be a subgroup of $G$. The following are equivalent:</p>
<ol>
<li>For all $a\in G$ there exists $b\in G$ such that $aH=Hb$; that is, every left coset of $H$ is also a right coset of $H$.</li>
<li>For all $a\in G$, $aH = Ha$.</li>
<li>For all $a\in G$, $a^{-1}Ha=H$.</li>
<li>For all $a\in G$, $a^{-1}Ha\subseteq H$.</li>
<li>For all $a,b\in G$, $a\mathrel{\equiv_H}b$ if and only if $a\mathrel{{}_H{\equiv}} b$; that is, congruence on the left modulo $H$ and congruence on the right modulo $H$ are the same equivalence relation on $G$.</li>
<li>If $aH=bH$ and $xH=yH$, then $axH=byH$ (multiplication of left cosets is well-defined).</li>
<li>If $Ha = Hb$ and $Hx = Hy$, then $Hab = Hxy$ (multiplication of right cosets is well-defined).</li>
</ol>
<p>If you prove this theorem (you may want to; it's a good exercise and it gives you other ways of checking that a group is normal) then condition 5 gives you the result:
$$\begin{align*}
a\mathrel{{}_H{\equiv}} b &\iff a^{-1}b\in H\\
&\iff ba^{-1}\in H\\
&\iff b\mathrel{\equiv_H} a\\
&\iff a\mathrel{\equiv_H} b
\end{align*}
$$</p>
|
3,831,702 | <p>One can prove that for <span class="math-container">$x\in \mathbb{R}$</span>, the sequence
<span class="math-container">$$
u_0=x\text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
converges to <span class="math-container">$0$</span> if <span class="math-container">$x \in ]-\infty,\delta[$</span> and diverges to <span class="math-container">$+\infty$</span> if <span class="math-container">$x\in ]\delta,+\infty[$</span> for a fixed <span class="math-container">$\delta$</span>. I'm trying to find more information on the value <span class="math-container">$\delta$</span> (inequalities or expression) and on the specific sequence
<span class="math-container">$$
u_0=\delta \text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
Any reference or help are welcome. The only thing I can prove at the moment is <span class="math-container">$\ln \ln 2 \le \delta \le 1$</span>.</p>
| dan_fulea | 550,003 | <p>I am proving something about <span class="math-container">$\delta$</span>, namely the <em>divergence</em> of the sequence starting from <span class="math-container">$\delta$</span>, see the <em>Result</em> below. This was the question in the OP. Some numerical aid is added to show how the "sequence works", and to give some support to the arguments involved.</p>
<hr />
<p>To have a precise notation, i will fix an <span class="math-container">$x>0$</span> and define for it recursively <span class="math-container">$u_0(x)=x$</span> and <span class="math-container">$u_n(x)=\frac 1 n\exp u_{n-1}(x)$</span> for an integer <span class="math-container">$n>0$</span>.
A first useful observation is the following one.</p>
<p><em>Lemma 1:</em> Assume <span class="math-container">$u_n(x)\ge u_{n+1}(x)$</span>. Then
<span class="math-container">$$
u_n(x)
\ge u_{n+1}(x)
> u_{n+2}(x)
> u_{n+3}(x)
> u_{n+4}(x)
> \dots
$$</span>
so <span class="math-container">$(u_N(x))_{N\ge n}$</span> is a strictly decreasing sequence of positive numbers, and thus a convergent sequence.</p>
<p><em>Proof:</em> Let <span class="math-container">$s$</span> be <span class="math-container">$u_n(x)$</span>. We know <span class="math-container">$s\ge \frac{e^s}{n+1}$</span>. Then
<span class="math-container">$$
\begin{aligned}
u_{n+1}(x)=\frac{\exp s}{n+1}
&> \frac{\exp\frac{e^s}{n+1}}{n+2}=u_{n+2}(x)
&&\text{ is equivalent to}\\
\underbrace{
\frac {n+2}{n+1}}_{>1}
&>\exp\underbrace{\left(\frac{e^s}{n+1}-s\right)}_{\le 0}\ ,
\end{aligned}
$$</span>
and the last relation is clear, since the <span class="math-container">$1$</span> separates the two expressions.
Inductively, all other comparison signs between two consecutive terms remain <span class="math-container">$>$</span> and we obtain convergence.</p>
<hr />
<p><em>Lemma 2:</em>
Fix <span class="math-container">$x>0$</span>.
Assume that the sequence <span class="math-container">$(u_n(x))_{n\ge 0}$</span> is monotone and increasing. Then it is unbounded.</p>
<p><em>Proof:</em> Assume there is a constant <span class="math-container">$M$</span> (upper bound) such that
<span class="math-container">$$
x=u_0(x)\le u_1(x)\le u_2(x)\le \dots \le M\ .
$$</span>
We adjust <span class="math-container">$M$</span> to be minimal with this property. For any <span class="math-container">$\epsilon>0$</span> we have infinitely many terms of the sequence in the interval <span class="math-container">$(M-\epsilon,\epsilon]$</span>, all terms with index <span class="math-container">$n\ge n_0$</span> for some suitable <span class="math-container">$n_0=n_0(\epsilon)$</span>, so that for any <span class="math-container">$n\ge n_0$</span>
<span class="math-container">$$
M-\epsilon\le u_{n+1}(x)=\frac 1{n+1}\exp u_n(x)\le\frac 1{n+1}\exp M\ .
$$</span>
We let <span class="math-container">$n$</span> not go to infinity, obtaining a contradiction for <span class="math-container">$\epsilon=M/2$</span> from <span class="math-container">$M/2\le\frac 1{n+1}e^M$</span>.</p>
<p><span class="math-container">$\square$</span></p>
<hr />
<p>Consider now the set <span class="math-container">$C$</span> of all <span class="math-container">$x$</span> so that <span class="math-container">$(u_n(x))$</span> is convergent. This set is an interval (is convex) since from <span class="math-container">$x\in C$</span> we obtain <span class="math-container">$y\in C$</span> for all positive <span class="math-container">$y\le x$</span>. The OP denotes by <span class="math-container">$\delta$</span> the number <span class="math-container">$\delta=\sup C$</span>. It is clear that for any <span class="math-container">$x>\delta$</span> we have <span class="math-container">$u_n(x)\nearrow\infty$</span>.</p>
<p>The question in the OP asks explicitly for the nature of the series <span class="math-container">$x_n(\delta)$</span>.</p>
<hr />
<p>To investigate what happens "in <span class="math-container">$\delta$</span>" we use the analysis in the part of <span class="math-container">$(0,\infty)$</span> that "we better know", which is the convergence domain <span class="math-container">$C$</span>. It is maybe useful to introduce some numerical discussion, just to show how the "sequence works". We have <span class="math-container">$u_0(x)=x<x+1\le e^x=u_1(x)$</span>. Assume now <span class="math-container">$x\in C$</span>. By Lemma 2, the sequence <span class="math-container">$u_n(x)$</span> is not strictly increasing. So at some point it decreases. By Lemma 1 the sequence keeps the decreasing character after this point. So it makes sense to speak of the <em>first point</em> <span class="math-container">$N=N(x)$</span> where the sequence becomes a decreasing sequence, i.e.
<span class="math-container">$$
u_0(x)<u_1(x)\le \dots\le u_{N-1}(x)\le \boxed{u_N(x)}>n_{N+1}(x)>u_{N+2}(x)>\dots\ .
$$</span>
For instance, for <span class="math-container">$x=0.313$</span> we have numerically</p>
<pre><code>? {a=0.313;
for(n=1, 20, aa=exp(a)/n;
print1("u_", n, " = ", aa);
if(a < aa, print(" BIGGER"), print(" smaller"));
a = aa;)}
u_1 = 1.3675215310276052580156244485205873317 BIGGER
u_2 = 1.9628045639746968825527107313466628204 BIGGER
u_3 = 2.3730885094556717601582056878568478087 BIGGER
u_4 = 2.6826205886314247378465092031314923147 BIGGER
u_5 = 2.9246729901579251786653681066911054540 BIGGER
u_6 = 3.1046889118003658683779460180009380929 BIGGER
u_7 = 3.1860399885563785616430186856926073497 BIGGER
u_8 = 3.0240543978924587401724975895522475423 smaller
u_9 = 2.2860600201376727553284809922492204883 smaller
u_10 = 0.98361071715125928138755201449968838114 smaller
u_11 = 0.24309947539391373305256660173652401198 smaller
u_12 = 0.10626628903174564052002058114668744999 smaller
u_13 = 0.085547537069273768253858963967751083770 smaller
u_14 = 0.077808095919794710713595034674642307547 smaller
u_15 = 0.072061013781783719666740623208530533727 smaller
u_16 = 0.067170057178175596313653825639495171561 smaller
u_17 = 0.062910431471108603831434935127199016802 smaller
u_18 = 0.059162858367558800252445283831682027556 smaller
u_19 = 0.055839368996440348276611168751081353904 smaller
u_20 = 0.052871390720140815464124735441909317362 smaller
</code></pre>
<p>So <span class="math-container">$N(0.313)=7$</span>, and the same code starting with <code>a=0.3132</code> gives <span class="math-container">$N(0.3132)=8$</span>:</p>
<pre><code>u_1 = 1.3677950626860648528324517882175013160 BIGGER
u_2 = 1.9633415265971033194641148584052431230 BIGGER
u_3 = 2.3743631114611039626890639195230221387 BIGGER
u_4 = 2.6860420422461167457807641681804189877 BIGGER
u_5 = 2.9346967612874001121755801606161241533 BIGGER
u_6 = 3.1359660986784763145732917033109680528 BIGGER
u_7 = 3.2872651234829478358588813251548230858 BIGGER
u_8 = 3.3461940041560052698109444803350566130 BIGGER
u_9 = 3.1549398373315806806605530408402914704 smaller
u_10 = 2.3451626136580891634913767329764798075 smaller
u_11 = 0.94863358710409358130678000785727277093 smaller
u_12 = 0.21518157708850363361007658962635940641 smaller
u_13 = 0.095391311378459099032647088551719786951 smaller
u_14 = 0.078577803453709167722261045262294803777 smaller
u_15 = 0.072116501038733476706519017436400138429 smaller
u_16 = 0.067173784363802268615874765719256293850 smaller
u_17 = 0.062910665950401524353995891488337534298 smaller
u_18 = 0.059162872240025623837696802595148633706 smaller
u_19 = 0.055839369771070147502677863568128329238 smaller
u_20 = 0.052871390761096570258333115181710141051 smaller
</code></pre>
<p>We need a lemma cementing this observation.</p>
<hr />
<p><em>Lemma 3:</em>
If <span class="math-container">$x\le x'$</span> then <span class="math-container">$N(x)\le N(x')$</span>.</p>
<p>In other words, if we define for <span class="math-container">$N\ge 1$</span> the set <span class="math-container">$C_N$</span> to be the set of all <span class="math-container">$x$</span> so that <span class="math-container">$N(x)=N$</span>, then <span class="math-container">$C_1 <C_2<C_3<C_4<\dots$</span> .</p>
<p><em>Proof:</em></p>
<p>Starting from <span class="math-container">$x=0$</span> we obtain <span class="math-container">$u_1(0)=1$</span>, <span class="math-container">$u_2(0)\approx 1.359\dots$</span>, <span class="math-container">$u_3(0)\approx 1.29$</span>, so <span class="math-container">$0\in C_2$</span>.</p>
<ul>
<li><p>The set <span class="math-container">$C_1$</span> is thus empty. (It is the set of all <span class="math-container">$x$</span> so that
<span class="math-container">$$ x=u_0(x)< \boxed{u_1(x)}\ge u_2(x)\ . $$</span>
The inequality <span class="math-container">$\le $</span> is valid on the interval with <span class="math-container">$u_1(x)\ge \frac 12\exp u_1(x)$</span>. This happens, if it happens, for <span class="math-container">$x$</span> on an interval <span class="math-container">$C_1=(0,c_1]$</span>,
so that for <span class="math-container">$y=u_1(x)=e^x>0$</span> we have <span class="math-container">$y\ge \frac 12\exp y$</span>. There is no such <span class="math-container">$y$</span>, so the set <span class="math-container">$C_1$</span> is empty.)</p>
</li>
<li><p>The set <span class="math-container">$C_2$</span> is the set of all <span class="math-container">$x$</span> so that
<span class="math-container">$$ x=u_0(x) < u_1(x)< \boxed{u_2(x)}\ge u_3(x)\ . $$</span>
The inequality <span class="math-container">$\le $</span> is valid on the interval with <span class="math-container">$u_2(x)\ge \frac 13\exp u_2(x)$</span>. This happens, and it happens, for <span class="math-container">$x$</span> on an interval <span class="math-container">$C_2=(0,c_2]$</span>,
so that for <span class="math-container">$y=u_2(x)>0$</span> we have <span class="math-container">$y\ge \frac 13\exp y$</span>.
The maximal value is <span class="math-container">$y=u_2(c_2)$</span>, and this is the solution <span class="math-container">$w_2>1$</span> of the transcendental equation
<span class="math-container">$$
y=\frac 13 e^y
\ .
$$</span>
We have solutions, and <span class="math-container">$w_2\approx 1.5121345516\dots$</span>, so
<span class="math-container">$c_2=u_2^{-1}(w_2)\approx 0.101355\dots$</span> .
For all other values <span class="math-container">$x>c_2$</span> we have
<span class="math-container">$u_0(x) < u_1(x)< u_2(x) <u_3(x)$</span>.</p>
</li>
<li><p>The set <span class="math-container">$C_3$</span> is the set of all <span class="math-container">$x>0$</span>, <span class="math-container">$x\not\in C_2$</span>, so <span class="math-container">$x>c_2$</span>, so that
<span class="math-container">$$ x=u_0(x) < u_1(x)< u_2(x) < \boxed{u_3(x)}\ge u_4(x)\ . $$</span>
The inequality <span class="math-container">$\le $</span> is valid on the interval with <span class="math-container">$u_3(x)\ge \frac 14\exp u_3(x)$</span>. This happens, if it happens, for <span class="math-container">$x$</span> on an interval <span class="math-container">$C_3=(c_2,c_3]$</span>,
so that for <span class="math-container">$y=u_3(x)> u_3(c_2)=u_2(c_2)$</span> we have <span class="math-container">$y\ge \frac 14\exp y$</span>.
The maximal value is <span class="math-container">$y=u_3(c_3)$</span>, and this is the solution <span class="math-container">$w_3>1$</span> of the transcendental equation
<span class="math-container">$$
y=\frac 14 e^y
\ .
$$</span>
We have solutions, and <span class="math-container">$w_3\approx 2.15329236411\dots$</span>, so
<span class="math-container">$c_3=u_3^{-1}(w_3)\approx 0.27515550\dots$</span> .
For all other values <span class="math-container">$x>c_3$</span> we have
<span class="math-container">$u_0(x) < u_1(x)< u_2(x) <u_3(x) < u_4(x)$</span>.</p>
</li>
<li><p>And the procedure goes on. For a general <span class="math-container">$n$</span> we define <span class="math-container">$w_n>1$</span> to be the solution of the equation
<span class="math-container">$$
y = \frac 1{n+1}e^y\ ,
$$</span>
then we set <span class="math-container">$c_n=u_n^{-1}(w_n)$</span>. Then <span class="math-container">$C_n=(c_{n-1},c_n]$</span> defined to be the set where we have
<span class="math-container">$$ x=u_0(x) < u_1(x)< u_2(x) < \dots <\boxed{u_n(x)}\ge u_{n+1}(x)\ . $$</span></p>
</li>
</ul>
<p>For all other values (not considered yet, i.e.) <span class="math-container">$x>c_n$</span> we have
<span class="math-container">$u_0(x) < u_1(x)< u_2(x) <\dots <u_n(x) < u_{n+1}(x)$</span>.</p>
<p>(At the last place we have a strict inequality making the last term strictly bigger than the previous one(s).)</p>
<ul>
<li>The only thing we have to show is the following one: <em>For all <span class="math-container">$n>2$</span> we have the inequality marked with an exclamation mark:</em>
<span class="math-container">$$
c_n\overset !< c_{n+1}\ .
$$</span>
Equivalently, after applying <span class="math-container">$u_{n+1}$</span> first, we have successively:
<span class="math-container">$$
\begin{aligned}
c_n& \overset !< c_{n+1}\ , \\
u_{n+1}(c_n)& \overset !< u_{n+1}(c_{n+1})\ , \\
\underbrace{\frac 1{n+1}(\exp u_n(c_n)}_{=w_n}& \overset !< w_{n+1}\ , \\
w_n&\overset !< w_{n+1}\ .\\
n+1&\overset !< n+2
\end{aligned}
$$</span>
The last step is clear because the function <span class="math-container">$\frac 1ye^y$</span> is strictly monotone on the interval of interest <span class="math-container">$[3,\infty)$</span>.</li>
</ul>
<hr />
<p><em>Result:</em> The sequence <span class="math-container">$(c_n)$</span> is bounded. (For instance <span class="math-container">$c_n\le 1$</span>.)</p>
<p>Let <span class="math-container">$\delta$</span> be its limit, <span class="math-container">$\delta=\lim c_n=\sup c_n$</span>.</p>
<p><strong>Then <span class="math-container">$(u_n(\delta))$</span> diverges.</strong></p>
<p><em>Proof:</em> Because <span class="math-container">$\delta>c_n$</span> for all <span class="math-container">$n$</span> we have
<span class="math-container">$$u_0(\delta)<u_1(\delta)<u_2(\delta)<u_3(\delta)<\dots<u_n(\delta)<\dots\ .
$$</span>
By Lemma 2 we get an unbounded sequence.</p>
<p><span class="math-container">$\square$</span></p>
<hr />
<p><em>Numeric evidence:</em></p>
<p>I will user again pari/gp . Then the first values for <span class="math-container">$w_=w(n)n$</span> are:</p>
<pre><code>? \p 60
realprecision = 77 significant digits (60 digits displayed)
? for(n=2, 10, wn = solve(y=1, 10, exp(y)-(n+1)*y); print("w(", n, ") ~ ", wn);)
w(2) ~ 1.51213455165784247389673967807203870460365038513535945425929
w(3) ~ 2.15329236411034964916909915009298137553620648531947769588451
w(4) ~ 2.54264135777352642429380615666184829016147490752943176711693
w(5) ~ 2.83314789204934214261167464234313256401468427714756512323373
w(6) ~ 3.06642134506926941241072450697636490912388231023308379327649
w(7) ~ 3.26168568457648877690566236430873973172114539334780952204022
w(8) ~ 3.42969628915899382743138865331419512387682622454177359601026
w(9) ~ 3.57715206395729721840939196351199488040179625779307592368353
w(10) ~ 3.70853115627103351156855721025086632923025483851609038044510
</code></pre>
<p>(Pari/GP has only the pricipal brance of the <code>lambertw</code> function.)
In sage we can use also other branches of this function. We rewrite the transcendental equation
<span class="math-container">$$
y =\frac 1{n+1}e^y
$$</span>
in the form
<span class="math-container">$$
(-y)e^{-y}=-\frac 1{n+1} \ ,
$$</span>
which gives <span class="math-container">$w_n=W_{-1}(-1/(n+1)$</span> . Sage gives then the values:</p>
<pre><code>for n in [2..10
wn = -lambert_w(-1, -1/(n+1)).n(digits=60
print(f"w({n}) ~ {wn}")
w(2) ~ 1.51213455165784247389673967807203870460365038513535945425929
w(3) ~ 2.15329236411034964916909915009298137553620648531947769588451
w(4) ~ 2.54264135777352642429380615666184829016147490752943176711693
w(5) ~ 2.83314789204934214261167464234313256401468427714756512323373
w(6) ~ 3.06642134506926941241072450697636490912388231023308379327649
w(7) ~ 3.26168568457648877690566236430873973172114539334780952204022
w(8) ~ 3.42969628915899382743138865331419512387682622454177359601026
w(9) ~ 3.57715206395729721840939196351199488040179625779307592368353
w(10) ~ 3.70853115627103351156855721025086632923025483851609038044510
</code></pre>
<p>Since i rather trust the precision in pari/gp, i will compute some <span class="math-container">$c_n$</span>-values in pari/gp:</p>
<pre><code>? {c(n) = wn = solve(y=1, n, exp(y)-(n+1)*y);
a = wn;
for( kk=1, n, k=(n+1)-kk; a = log(k*a) );
a; }
? \p 50
realprecision = 57 significant digits (50 digits displayed)
? for(n=2, 60, print("c(", n, ") &\\sim", c(n), "\\dots\\\\"));
</code></pre>
<p>The results are suited to a copy+paste action into an aligned block:</p>
<p><span class="math-container">$$
\begin{aligned}
c(2) &\sim0.10135500348877619648835046893307802134647843854859\dots\\
c(3) &\sim0.27515550224356722642700029292669340425790411988586\dots\\
c(4) &\sim0.30440354255780715384928950088935566512074662284688\dots\\
c(5) &\sim0.31111934109691133707464026297298705606522340312816\dots\\
c(6) &\sim0.31275095307818668153209240588697305474570060282017\dots\\
c(7) &\sim0.31315034197961514476112541848049188150250410074878\dots\\
c(8) &\sim0.31324730472844731441045222549671358612256305068358\dots\\
c(9) &\sim0.31327052241203616368893190165080929026601998047499\dots\\
c(10) &\sim0.31327599571763732522624466142920222534452300093615\dots\\
c(11) &\sim0.31327726562968841826282081670459211837048707198094\dots\\
c(12) &\sim0.31327755572894389701305007274609305684358656586721\dots\\
c(13) &\sim0.31327762101787184872963296682754422073736624513741\dots\\
c(14) &\sim0.31327763550439768085446246483894501667765926971232\dots\\
c(15) &\sim0.31327763867571006703555595324736062516361585176649\dots\\
c(16) &\sim0.31327763936115504573935501667287678447210707448810\dots\\
c(17) &\sim0.31327763950752672055982251214461261883812504160744\dots\\
c(18) &\sim0.31327763953842719229730321612969220129090235414359\dots\\
c(19) &\sim0.31327763954488001621055885793746377925086787133883\dots\\
c(20) &\sim0.31327763954621367609922944945670217886704627412091\dots\\
c(21) &\sim0.31327763954648661595904784671869579247553003455047\dots\\
c(22) &\sim0.31327763954654195287253067161058269140404032890270\dots\\
c(23) &\sim0.31327763954655307206776515886125074035389761846850\dots\\
c(24) &\sim0.31327763954655528726715108843199348672484120918294\dots\\
c(25) &\sim0.31327763954655572498212771123141999427090570670146\dots\\
c(26) &\sim0.31327763954655581079567023002593270768368799867357\dots\\
c(27) &\sim0.31327763954655582749282669405759711979793525735967\dots\\
c(28) &\sim0.31327763954655583071818919874759817293462259261239\dots\\
c(29) &\sim0.31327763954655583133689251460813922039357123411747\dots\\
c(30) &\sim0.31327763954655583145477928547388335070258877443942\dots\\
c(31) &\sim0.31327763954655583147709599602419284510657691971650\dots\\
c(32) &\sim0.31327763954655583148129431164142194407675946396456\dots\\
c(33) &\sim0.31327763954655583148207935057505826128744893134496\dots\\
c(34) &\sim0.31327763954655583148222528727352500245553858308174\dots\\
c(35) &\sim0.31327763954655583148225226318204998596936649150874\dots\\
c(36) &\sim0.31327763954655583148225722227611323237580761358174\dots\\
c(37) &\sim0.31327763954655583148225812908456600452851730734627\dots\\
c(38) &\sim0.31327763954655583148225829404662758602316288107775\dots\\
c(39) &\sim0.31327763954655583148225832390544808433831911964086\dots\\
c(40) &\sim0.31327763954655583148225832928371927863800658191742\dots\\
c(41) &\sim0.31327763954655583148225833024788691497864879533556\dots\\
c(42) &\sim0.31327763954655583148225833041993809498564822525567\dots\\
c(43) &\sim0.31327763954655583148225833045050213708406158795592\dots\\
c(44) &\sim0.31327763954655583148225833045590796516903409149109\dots\\
c(45) &\sim0.31327763954655583148225833045686001554718024594137\dots\\
c(46) &\sim0.31327763954655583148225833045702698985929467652069\dots\\
c(47) &\sim0.31327763954655583148225833045705615574487227907134\dots\\
c(48) &\sim0.31327763954655583148225833045706123006980155619276\dots\\
c(49) &\sim0.31327763954655583148225833045706210949522978129814\dots\\
c(50) &\sim0.31327763954655583148225833045706226133158177531860\dots\\
c(51) &\sim0.31327763954655583148225833045706228744990391540019\dots\\
c(52) &\sim0.31327763954655583148225833045706229192644777682540\dots\\
c(53) &\sim0.31327763954655583148225833045706229269099154514550\dots\\
c(54) &\sim0.31327763954655583148225833045706229282111523143202\dots\\
c(55) &\sim0.31327763954655583148225833045706229284318695085698\dots\\
c(56) &\sim0.31327763954655583148225833045706229284691835169076\dots\\
c(57) &\sim0.31327763954655583148225833045706229284754712279749\dots\\
c(58) &\sim0.31327763954655583148225833045706229284765273792594\dots\\
c(59) &\sim0.31327763954655583148225833045706229284767042269982\dots\\
c(60) &\sim0.31327763954655583148225833045706229284767337485177\dots\\
\end{aligned}
$$</span></p>
|
220,618 | <p>The cyclic group of $\mathbb{C}- \{ 0\}$ of complex numbers under multiplication generated by $(1+i)/\sqrt{2}$</p>
<p>I just wrote that this is $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ making a polar angle of $\pi/4$. I am not sure what to do next. My book say there are 8 elements. </p>
<p>Working backwards, maybe the angle divides the the four quadrant into 8 areas? I have no idea</p>
| Community | -1 | <p><strong>HINT</strong> Find the minimum $k$ such that $\left(\dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}} \right)^k = 1$. It will be of help to write $\dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}}$ as $\exp(i \pi/4)$.</p>
<p>The element in the group generated by $r = \dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}}$ will then be $$\{r^m : m = 0,1,2,\ldots,k-1\}$$</p>
|
3,353,483 | <p>I have just begun studying finite fields today, and it is clear in GF(2) why 1+1=0. (I just show that 1+1 can't equal 1, or 1=0, which contradicts an axiom that states that 1 is not 0).</p>
<p>If we interpreted these symbols "1", "+", "1", "0" as we would in primary school, clearly this breaks arithmetic rules in Real numbers.</p>
<p>Given that, I have lost all confidence in how arithmetic can be applied in a finite field. <strong>How do I even know how to do basic arithmetic on GF(n) where n is prime?</strong>
For example, for GF(7), how do I even know that 4+1=5?
Can anyone show with just the 9 axioms of finite fields that 4+1=5?</p>
<p>Axioms: associativity of addition, additive identity, additive inverse, commutatitivity of addition, associativity of multiplication, multiplicative inverse, commutatitivity of mulitplication, distributive law</p>
| Robert Shore | 640,080 | <p>This actually brings up a subtle point. What do we mean by <span class="math-container">$5$</span> in a finite field? Or if you choose to define <span class="math-container">$5$</span> in terms of <span class="math-container">$1 ~(5=1+1+1+1+1)$</span>, then what do we mean by <span class="math-container">$1$</span>?</p>
<p>One answer is to define <span class="math-container">$5$</span> in terms of equivalence classes. Say that two integers <span class="math-container">$m$</span> and <span class="math-container">$n$</span> are equivalent if <span class="math-container">$p \vert (m-n).$</span> First, you prove this really is an equivalence relation on the integers. Then you define <span class="math-container">$[m]+[n]=[m+n]$</span> and <span class="math-container">$[m][n]= [mn]$</span>. So by <span class="math-container">$5$</span> we actually mean the equivalence class <span class="math-container">$[5]$</span>.</p>
<p>You need to prove that your field operations are well-defined (you get the same answer no matter which representative of an equivalence class you choose) and that <span class="math-container">$[0]$</span> and <span class="math-container">$[1]$</span> really are the additive and multiplicative identities, as you'd expect. But once you've done that, you can see that <span class="math-container">$[4]+[1]=[5]$</span> (and usually we abuse notation by dropping the brackets) <em>because we've defined it that way</em>.</p>
|
2,046,957 | <p>The random variable $X$ is $N(5,2)$ and $y=2X+4$.
Find:</p>
<p>a) $\eta_y$</p>
<p>b) $\sigma_y$</p>
<p>c) $f_Y(y)$</p>
<p>My attempt:</p>
<p>I have solved a and b as follow:</p>
<p>a) $\eta_y = 2\eta_X+4 = 14$</p>
<p>b) $\sigma_y^2 = 4\sigma_{x}^2 = 16, \sigma_y = 4$</p>
<p>c) how can I solve $f_Y(y)$?</p>
| caverac | 384,830 | <p>You need to guarantee that </p>
<p>$$
f_X(x)dx = f_Y(y)dy ~~~\Rightarrow~~~ f_Y(y) = f_X(x)\left| \frac{dx}{dy} \right|
$$</p>
<p>where the absolute value is used here to ensure that $f_Y(y) \ge 0$. In this case $dx/dy = 1/2$ therefore</p>
<p>$$
f_Y(y) = \frac{1}{2}\frac{1}{\sqrt{2\pi \sigma_x^2}} e^{-\left(\frac{y-4}{2} - \eta_x\right)^2/2\sigma_x^2} = \frac{1}{\sqrt{2\pi(4\sigma_x^2)}}e^{(y - (2\eta_x + 4))/2(2\sigma_x)^2} = \frac{1}{\sqrt{2\pi\sigma_y^2}}e^{(y-\eta_y)^2/2\sigma_y^2} \tag{1}
$$</p>
<p>This last step just shows that $Y\sim \mathcal{N}(\eta_y = 14, \sigma_y = 4)$. Below there's a small simulation, the blue histogram is built by generating $10^5$ points $ X ~\mathcal{N}(5,2)$ and then transforming them according to $Y = 2 X + 4$. The red line is Eq. (1)</p>
<p><a href="https://i.stack.imgur.com/FDcee.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FDcee.png" alt="enter image description here"></a></p>
|
3,361,489 | <p><strong>Question:</strong></p>
<p>Do there exist functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> such that
<span class="math-container">$$\lim_{x \to c} f(x) = 1 \text{ and } \lim_{x \to c} f(x) g(x) - g(x) \neq 0 \, ?$$</span>
(Allowing, of course, for <span class="math-container">$\lim_{x \to c}$</span> g(x) to not exist.)</p>
<p><strong>Context:</strong></p>
<p>I am thinking about the limit property that <span class="math-container">$\lim_{x \to c} f(x) \cdot g(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)$</span>.</p>
<p>My understanding is that for this to be guaranteed to hold, we need both limits on the RHS to exist. Indeed, I am familiar with examples for which the two limits on the RHS diverge while the limit on the LHS exists (like in <a href="https://math.stackexchange.com/questions/513822/can-the-limit-of-a-product-exist-if-neither-of-its-factors-exist">this post</a>), as well as examples like <span class="math-container">$f(x) = 1/x, g(x) = x$</span> for which the LHS exists but one of the limits on the RHS is zero and the other diverges.</p>
<p>If, however, only one of the limits on the RHS diverges but the other exists <em>and is nonzero</em>, will we ever run into trouble by applying this limit property? In some sense, can we modify the requirement that BOTH limits on the RHS exist to the requirement that AT LEAST ONE of the limits on the RHS exists and is nonzero?</p>
| David C. Ullrich | 248,223 | <p><span class="math-container">$c=0$</span>, <span class="math-container">$f(x)=1+x$</span>, <span class="math-container">$g(x)=1/x$</span>.</p>
|
434,614 | <p>Is there a name for a type of grid you might find in Battleship? Where coordinates don't relate to points on a grid but rather the squares themselves?</p>
| Will Orrick | 3,736 | <p>How about <a href="http://en.wikipedia.org/wiki/Alpha-numeric_grid" rel="nofollow">alpha-numeric grid</a>? That the coordinates refer to squares rather than vertices is no big deal. Squares can be identified with vertices in the <a href="http://en.wikipedia.org/wiki/Dual_graph" rel="nofollow">dual</a> of the square lattice (which is also a square lattice).</p>
|
3,532,173 | <p>I have seen this problem somewhere on the internet but I could not prove it.</p>
<p>Let <span class="math-container">$$I_{0}=\int^{\infty}_{0}\frac{\sin x}{x}dx$$</span> and then define
<span class="math-container">$$I_{n+1}=\int^{I_{n}}_{0}\frac{\sin x}{x}dx.$$</span></p>
<p>Show that
<span class="math-container">$$\lim_{n\rightarrow\infty}\sqrt{n}\ I_{n}=3.$$</span></p>
| xpaul | 66,420 | <p>It is easy to see that <span class="math-container">$I_n\to0$</span> as <span class="math-container">$n\to\infty$</span>. So by Stolze's Theorem, one has
<span class="math-container">\begin{eqnarray}
\lim_{n\rightarrow\infty}n I^2_{n}&=&\lim_{n\rightarrow\infty}\frac{n}{I^{-2}_{n}}\\
&=&\lim_{n\rightarrow\infty}\frac{1}{I^{-2}_{n+1}-I^{-2}_{n}}\\
&=&\lim_{n\rightarrow\infty}-\frac{I^{2}_{n+1}I^{2}_{n}}{I^{2}_{n+1}-I^{2}_{n}}\\
&=&\lim_{x\to0}-\frac{x^2\left(\int_0^x\frac{\sin t}{t}dt\right)^2}{\left(\int_0^x\frac{\sin t}{t}dt\right)^2-x^2}\\
&=&\lim_{x\to0}-\frac{x^2\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2}{\left(\int_0^x(1-\frac16t^2+O(t)^5)dt\right)^2-x^2}\\&=&9.
\end{eqnarray}</span></p>
|
2,767,392 | <p>I have the following curve:</p>
<p>$x^4=a^2(x^2-y^2)$</p>
<p>Prove that the area of its loop is $\frac{2a^2}{3}$.</p>
<p><strong>My approach</strong></p>
<p>This curve has four loops. So the required area should be:</p>
<p>$4\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx$</p>
<p>But, After solving, the area turned out to be $\frac{4a^2}{3}$. What am I doing wrong here?</p>
<p>Thanks</p>
| Sri-Amirthan Theivendran | 302,692 | <p>We use the generalized hockey-stick identity
$$
\sum_{i=0}^k\binom{m+i}{i}=\binom{m+k+1}{k}\quad(m\in\mathbb{C})\tag{H.S.}
$$
which follows from the regular hockey-stick identity via the polynomial method to prove that
$$
\sum_{i=0}^k(-1)^{i}\binom{n}{i}
=(-1)^k\binom{n-1}{k}.$$
Indeed, since $(-1)^{i}\binom{n}{i}=(-1)^{i}(n)_{i}/i!=(-n)^{(i)}/i!=\binom{i-n-1}{i}$ (where $(n)_i$ is the falling factorial and $n^{(i)}$ is the rising factorial), we have that
$$
\sum_{i=0}^k(-1)^{i}\binom{n}{i}=\sum_{i=0}^k\binom{i-n-1}{i}\stackrel{\text{H.S.}}{=}\binom{k-n}{k}=(-1)^k\binom{n-1}{k}
$$
as desired.</p>
|
2,438,795 | <blockquote>
<p>Why are any two initial objects of a category equivalent?</p>
</blockquote>
<p>By definition:</p>
<p>If $A,B$ are initial objects of a category $C$, then for each $X \in \text{obj } C$, there exists a unique morphisms $f : A \rightarrow X, g : B \rightarrow X$.</p>
<p>How can these be equivalent? This seems to just indicate that each $A$ and $B$ have their own set of maps.</p>
| Community | -1 | <ul>
<li>Since $A$ is initial, there is a unique map $A \to B$. </li>
<li>Since $B$ is initial, there is a unique map $B \to A$. </li>
</ul>
<p>Furthermore, they are inverses; the products have to be identities because</p>
<ul>
<li>Since $A$ is initial, there is a unique map $A \to A$. </li>
<li>Since $B$ is initial, there is a unique map $B \to B$. </li>
</ul>
|
158,483 | <p>When I started mathematica, this message popped up.</p>
<pre><code>Part::partw: Part 5 of PacletManager`Utils`Private`$taskData[2] does not exist.
</code></pre>
<p>Does anyone know what this is? My version is mma 11.2.</p>
| Stefan R | 5,678 | <p>We believe this should be fixed in <em>Mathematica</em> 12.0. We were never able to reproduce this internally so we're not completely sure about this, though. Please let us know if this continues to happen in 12.0 for you.</p>
|
651,731 | <blockquote>
<p>Let $\{f_n\}$be sequence of bounded real valued functions on $[0,1]$ converging at all points of this interval. Then
If $\int^1_0 |f_n(t)-f(t)|dt\, \to 0$ as $n \to \infty$,does $\lim_{n \to \infty} \int^1_0 f_n(t) dt\,=\int^1_0 f(t)dt\,$</p>
</blockquote>
<p>I just know that if somehow we can show uniform convergence then we can say it would be true, but in other case what happens I don't know!</p>
| Did | 6,179 | <p>$$\left|\int^1_0 f_n(t)\,\mathrm dt-\int^1_0 f(t)\,\mathrm dt\right|\leqslant\int^1_0 \left|f_n(t) - f(t)\right|\,\mathrm dt$$</p>
|
2,555,200 | <p>Let $l$ be the smallest positive linear combination of $a,b\in \mathbb{Z}^+$ i.e.,$$l := \min\{ax+by >0 : x,y\in\mathbb{Z}\}.$$
Now, according to @Brahadeesh's answer here, <a href="https://math.stackexchange.com/questions/2553324/proof-for-gcd-being-the-smallest-linear-combination-of-a-b-in-mathbb-z">Proof for $\gcd$ being the smallest linear combination of $a,b \in \mathbb {Z}$.</a>, we can show that $l\mid a$ and $l\mid b$ (simultaneously).</p>
<p>But, consider $k>l$ such that $k\in \{ax+by >0 : x,y\in\mathbb{Z}\}$. Then, is it possible to show that $k\not\mid a $ and $k\not\mid b$ (simultaneously)? This indirectly gives us that there are no common divisors of $a$ and $b$ that are greater than $l$ which then proves that $l = \gcd(a,b)$.</p>
<p>The direct proof of the fact that $l=\gcd(a,b)$ would be to show that $c\mid a $ and $c\mid b \; \Rightarrow \; c\mid l$, which is quite trivial from the definition of $l$. However, I don't want to make use of this fact and instead want to show that </p>
<blockquote>
<p>Claim: If $k>l$ such that $k\in \{ax+by >0 : x,y\in\mathbb{Z}\}$ then $k\not\mid a $ and $k\not\mid b$ (simultaneously).</p>
</blockquote>
<p>Thanks in advance.</p>
| user | 505,767 | <p>Since the g.c.d. is in the form:</p>
<p>$$l=ax+by$$</p>
<p>Every other $k>l$ in the form:</p>
<p>$$k=ax'+by'$$</p>
<p>is a multiple of $l$,</p>
<p>thus it can't divide a and b simultaneously.</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Dick Palais | 7,311 | <p>Here is my rather personal answer. I like Wikipedia---both in general and the mathematically oriented part, and I would like to collaborate in its improvement and contribute in any way I can. But I don't know where to begin. With MO, when I have free time I can start "at the top" so to speak, and continue on down until I start seeing familiar entries---but in Wikipedia how does one decide what to look for that might benefit from one's inspection and possible editing. Getting over that intimidating problem is what prevents me from even getting started. I think it would help if "we" (i.e., the math community) could set up a website where there was listed mathematical subjects that either someone had spotted as being ignored by Wikipedia, or whose Wikipedia article had been recently edited, or which someone had read and felt would benefit from expansion or improvement. It would be good if such a site was maintained in chronological order by date of entry so one could view it from the top for new stuff as one had time to do so, and it should allow comments and "answers" to each entry to show any progress in fixing the problem. I guess some sort of "stack" is what I have in mind. I would certainly be willing to visit such a site frequently myself and contribute to it as I could. </p>
|
361,060 | <blockquote>
<p>Consider the ring of Gaussian integers $D=\lbrace a+bi\mid a,b \in \mathbb{Z \rbrace}$, where $i \in \mathbb{C}$ such that $i^2=-1$. Consider the map $f$ from $D$ to $\mathbb{Z}[x]/(x^2+1)$ sending $i$ to the class of $x$ modulo $x^2+1$. Show that $f$ is a ring isomorphism.</p>
</blockquote>
<p>I got a confusion in this question. I don't understand the map sending $i$ to the class of $x$ modulo $x^2+1$. Can anyone help me to clear my confusion?</p>
<p>EDIT: I am having trouble to show the map is injective and surjective. Can anyone guide me?</p>
| Andreas Caranti | 58,401 | <p>As the others have said, you have also to specify what happens to the integers.</p>
<p>And then, it's the <em>inverse</em> map that occurs more naturally in the context of the structure of simple extensions. </p>
<p>Consider the ring homomorphism
$$
\begin{align}
\varphi : & \Bbb{Z}[x] \to \Bbb{Z}[i]\\
&a \mapsto a, \quad \text{for $a \in \Bbb{Z}$}\\
&x \mapsto i,\end{align}$$
which is uniquely determined by the <a href="http://en.wikipedia.org/wiki/Polynomial_ring#Quotient_ring_of_K.5BX.5D" rel="nofollow">universal property of polynomial rings</a>.</p>
<p>This is clearly surjective, and its kernel is $(x^{2} + 1)$, since $x^{2} + 1$ is the <a href="http://en.wikipedia.org/wiki/Minimal_polynomial_%28field_theory%29" rel="nofollow">minimal polynomial</a> of $i$ over $\Bbb{Z}$. Now apply the <a href="http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem_2" rel="nofollow">first isomorphism theorem</a>.</p>
|
597,949 | <p>$\begin{cases}x\equiv 1 \pmod{3}\\
x\equiv 2 \pmod{5}\\
x\equiv 3 \pmod{7}\\
x\equiv 4 \pmod{9}\\
x\equiv 5 \pmod{11}\end{cases}$ </p>
<p>I am supposed to solve the system using the Chinese remainder theorem but $(3,5,7,9,11)\neq 1$
How can I transform the system so that I will be able to use the theorem?</p>
| John Butnor | 185,327 | <p>Solving the first two equations simultaneously you get X = 7(mod 15).</p>
<p>Solving the third and fourth simultaneously you get X = 31(mod 63).</p>
<p>Solving these two results simultaneously you get X = 157(mod 315).</p>
<p>Solving this result with the fifth equation simultaneously,</p>
<p>you get the final answer X = 1732(mod 3465).</p>
|
538,870 | <p>Evaluate</p>
<p>$$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$</p>
<p>This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?</p>
| Start wearing purple | 73,025 | <p>Use <a href="http://en.wikipedia.org/wiki/Stirling_formula">Stirling's approximation</a>:
$$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}=\left(\frac{n}{e}\right)^{n}e^{\frac12 \ln 2\pi n}$$
It transforms your limit into
$$\lim_{n\rightarrow\infty}\frac{n}{\ln n}\frac{e^{\frac{\ln n}{2n}}-1}{e}=\lim_{x\rightarrow 0}\frac{e^{\frac{x}{2}}-1}{ex}=\frac{1}{2e}.$$
When obtaining the first expression, we neglect $\frac{\ln 2\pi}{2n}$ in the exponential (it is $o\left(\frac{\ln n}{2n}\right)$), and then we make the change of variables $x=\frac{\ln n}{n}$.</p>
|
538,870 | <p>Evaluate</p>
<p>$$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$</p>
<p>This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?</p>
| Mikhail Katz | 72,694 | <p>There is a gap in @O.L.'s solution as I explained in a comment below that solution. I don't have time now to provide a complete solution but the OP should be aware of this. I see that the wiki page provides a formula with an error term, but the $O(1/n)$ occurs inside an argument and is not stated as a theorem, nor is there a source for this. It is probably true but note also that we need the error estimate for $\frac{\sqrt[n]{n!}}{n}$ rather than for $n!$ itself.</p>
|
354,885 | <p>Let <span class="math-container">$X_1,...,X_n$</span> be iid normal random variables. </p>
<p>I am looking for a strategy to establish the following limit for fraction of expectation values</p>
<p><span class="math-container">$$\lim_{N \rightarrow \infty} \frac{E(\prod_{1\le i < j\le n} \vert X_i-X_j \vert^{1/n})}{E(\prod_{1\le i < j\le n-1} \vert X_i-X_j \vert^{1/n})}=1.$$</span></p>
<p>Does anybody have any ideas what to use for this limit?</p>
| Matt F. | 44,143 | <p>The numerator and denominator can also be written as
<span class="math-container">$$E\left[\exp\left(\frac1n\sum\ln|X_i-X_j|\right)\right]$$</span>
where the numerator has <span class="math-container">$n(n-1)/2$</span> summands and the denominator has <span class="math-container">$(n-1)(n-2)/2$</span> summands.</p>
<p>Let <span class="math-container">$\mu$</span> and <span class="math-container">$\sigma$</span> be the mean and standard deviation of <span class="math-container">$\ln|X_i-X_j|$</span>. Since <span class="math-container">$X_i-X_j$</span> is a normal distribution with mean <span class="math-container">$0$</span> and standard deviation <span class="math-container">$\sqrt{2}$</span>,
<span class="math-container">$$\mu=E\left[\ln\Big|N(0,\sqrt{2})\Big|\right]=\frac{-\gamma}{2}$$</span>
where <span class="math-container">$\gamma$</span> is the <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant#Integrals" rel="nofollow noreferrer">Euler-Mascheroni constant</a>.</p>
<p><em>If the summands were independent</em>, then the fraction in the question would be approximated by a ratio of means of lognormal distributions:</p>
<p><span class="math-container">$$
\frac
{E\left[LN\left(\frac{n(n-1)}{2n}\mu,\sqrt{\frac{n(n-1)}{2n^2}}\sigma\right)\right]}
{E\left[LN\left(\frac{(n-1)(n-2)}{2n}\mu,\sqrt{\frac{(n-1)(n-2)}{2n^2}}\sigma\right)\right]}
=
\frac
{\exp\left(\frac{n(n-1)}{2n}\mu+\frac{n(n-1)}{2n^2}\frac{\sigma^2}{2}\right)}
{\exp\left(\frac{(n-1)(n-2)}{2n}\mu+\frac{(n-1)(n-2)}{2n^2}\frac{\sigma^2}{2}\right)}
$$</span></p>
<p>In the limit this is
<span class="math-container">$$
\exp\left(\frac{2(n-1)}{2n}\mu+\frac{2(n-1)}{2n^2}\frac{\sigma^2}{2}\right)
\rightarrow
\exp(\mu)=\exp(-\gamma/2) \sim 0.749.$$</span> </p>
<p>Empirically the number seems to be between that and <span class="math-container">$1$</span>.</p>
<p>With some numerical integration we could compute <span class="math-container">$\sigma$</span> and the correlation <span class="math-container">$\rho$</span> between <span class="math-container">$\ln|X_i-X_j|$</span> and <span class="math-container">$\ln|X_i-X_k|$</span>; with some combinatorics we could compute how often those <span class="math-container">$\rho$</span>'s show up in the standard deviations of the two lognormals; combining those results we could get a more accurate value for the overall expectation.</p>
|
1,299,552 | <p>Can you tell me why my answer is wrong?</p>
<p>$$\frac {x+y} {x-y} + \frac 1 {x+y} - \frac {x^2+y^2} {y^2-x^2} = \frac {x^2 + y^2} {x^2-y^2} + \frac {x-y} {x^2-y^2} + \frac {x^2+y^2} {x^2-y^2} = 2x^2 + 2y^2 + x-y$$</p>
| abel | 9,252 | <p>on the second line, the first expression you have $$\frac{x+y}{x-y} = \frac{x^2+y^2}{x^2-y^2} $$ it should be $$\frac{x+y}{x-y} = \frac{(x+y)^2}{x^2-y^2} = \frac{x^2+2xy + y^2}{x^2-y^2}$$ instead. the rest of them looks good. </p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| ThudanBlunder | 1,770 | <p>A regular tetrahedron and a regular square pyramid both have unit length. If a triangular face of the tetrahedron is glued to a triangular face of the square pyramid, the resulting shape has how many edges? </p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Oscar Lanzi | 248,217 | <p>A time bomb has 25 switches engaged, all in a row and numbered from 1 to 25. A spy has told you that you could defuse the bomb by flipping every switch, then flipping every multpile of 2, then flipping every multiple of 3, etc through multiples of 25 (then you are done, there are no numbers greater than equal to 26). But the bomb will go off shortly. Which switches do you flip to get the correct combination as quickly as possible?</p>
|
3,255,174 | <p>How can I find the volume bounded between <span class="math-container">$z=(x^2+y^2)^2$</span> and <span class="math-container">$z=x$</span>? </p>
<p>My idea so far is to use cylindrical polar coordinates and <span class="math-container">$z$</span> limit is from <span class="math-container">$(x^2+y^2)^2$</span> to <span class="math-container">$x$</span> quite clearly but I am struggling to parametrize the surface <span class="math-container">$(x^2+y^2)^2<x$</span> for the other integration limits, could someone please help?</p>
| user10354138 | 592,552 | <p>Clearly <span class="math-container">$x>0$</span>, or equivalently <span class="math-container">$\theta\in(-\frac\pi2,\frac\pi2)$</span>. Then
<span class="math-container">$$
(x^2+y^2)^2<x\iff r^4<r\cos\theta\iff 0<r^3<\cos\theta.
$$</span></p>
|
387,519 | <p>The domain of the following function $$y=2$$ is just 2? And the image of it?</p>
<p>I don't think I quiet understand what the image of a function means. The domain is all values that it can assume, correct?</p>
<p>Could you please try to define the image of this equation too: $$y = 2x - 6$$ so I can try to understand it?</p>
| rurouniwallace | 35,878 | <p>An image is a subset of the co-domain with respect to a certain pre-image, which is a subset of the domain. For the function $y=2x-6$, for example, given a pre-image of $[2,10]$, the image is $[-2,14]$. For the function $y=2$, since any input value in the domain will result in $y=2$, besides the null set, the only possible pre-image is ${2}$, and the only possible image is $2$.</p>
|
387,519 | <p>The domain of the following function $$y=2$$ is just 2? And the image of it?</p>
<p>I don't think I quiet understand what the image of a function means. The domain is all values that it can assume, correct?</p>
<p>Could you please try to define the image of this equation too: $$y = 2x - 6$$ so I can try to understand it?</p>
| Wishwas | 19,992 | <p>The value of f when applied to x is the image of x under f. y is alternatively known as the output of f for argument x.
Now Y=2 is a constant function. The image of Y is always 2 for any value of x.
So the Image of function Y=2 is the set containing element 2.
Image of 2, under Y=2x−6, is the set containing the element 2(2)-6=4-6=-2. </p>
|
387,519 | <p>The domain of the following function $$y=2$$ is just 2? And the image of it?</p>
<p>I don't think I quiet understand what the image of a function means. The domain is all values that it can assume, correct?</p>
<p>Could you please try to define the image of this equation too: $$y = 2x - 6$$ so I can try to understand it?</p>
| Alistair Savage | 74,366 | <p>This really depends on the context of the question. I'm going to make the assumption here that you're discussing functions from the real numbers to the real numbers. Strictly speaking, the domain of the function should be given explicitly when you define the function. However, often one defines a function and then says that the domain is the set of all real numbers for which the function is defined. In this interpretation, the domain of your function $y=f(x)=2$ is all of $\mathbb{R}$ and its image is $\{2\}$. Similarly, the domain of $y=f(x)=2x-6$ is all of $\mathbb{R}$ and its image is $\mathbb{R}$.</p>
<p>To give you a different example, the domain of the function $y=f(x)=\frac{1}{x-1}$ is $\mathbb{R} \setminus \{1\}$ and the image is $\mathbb{R} \setminus \{0\}$.</p>
|
506,767 | <p>What is the largest number on multiplying with itself gives the same number as last digits of the product?</p>
<p>i.e., $(376 \times 376) = 141376$</p>
<p>i.e., $(25\times 25) = 625$</p>
<p>If the largest number cant be found out can you prove that there is always a number greater than any given number? (only in base $10$)</p>
| Leen Droogendijk | 95,972 | <p>The numbers can be arbitrarily large.
Take an arbitrary number $k$, we will produce a number with the desired property that is
at least as big as $5^k$.</p>
<p>$2^k$ and $5^k$ are coprime, so we can find integers $p,q$ such that $p2^k+q5^k=1$, where we
may choose $q$ to be a positive number less than $2^k$.
Now $q5^k$ is our desired number, by our choices it is at least as big as $5^k$ and smaller than $10^k$.
Furthermore $q5^k.q5^k=q5^k(1-p2^k)=q5^k-pq10^k$, so $q5^k.q5^k\equiv q5^k\pmod{10^k}$ as desired.</p>
|
4,196,125 | <p>Can someone tell me where this calculation goes wrong?
I get (2 3 4)(1 2 3 4 5 6)^-1 = (1 6 5 2).
My book and Mathematica get (1 6 5 4).
I have read several explanations of how to multiply permutations in cycle notation and have
worked dozens of examples successfully, but I always get this one wrong.</p>
<p>(2 3 4)(1 2 3 4 5 6)^-1 = (2 3 4)(6 5 4 3 2 1) = (2 3 4)(1 6 5 4 3 2)</p>
<p>1 -> 6 then 6 is unchanged giving (1 6</p>
<p>6 -> 5 then 5 in unchanged giving (1 6 5</p>
<p>5 -> 4 then 4 -> 2 giving (1 6 5 2</p>
<p>2 -> 1 then 1 is unchanged completing the cycle giving (1 6 5 2).</p>
| Henry Lee | 541,220 | <p>yep so what you have is the circle:
<span class="math-container">$$x^2+(y-2)^2=1$$</span>
rotated about the <span class="math-container">$y$</span>-axis. Rearranging you have:
<span class="math-container">$$y=\pm\sqrt{1-x^2}+2$$</span>
just remember to account for above and below the axis. I am not sure where the <span class="math-container">$5$</span> and <span class="math-container">$1$</span> have come from, I think it should be <span class="math-container">$[0,1]$</span></p>
|
222,105 | <p>I want to use geometric shapes in Mathematica to build complex shapes and use my raytracing algorithm on it. I have a working example where we can get the intersections from a combination of a <code>Cone[]</code> and <code>Cuboid[]</code>, e.g </p>
<pre><code>shape1 = Cone[];
shape2 = Cuboid[];
(* add shapes in this list to make a more complicated shape *)
shapes = {shape1, shape2};
(* this constains the shapes so the shape is considered as a whole *)
constraints[shapes__] :=
And[## & @@ (Not /@
Through[(RegionMember[RegionIntersection@##] & @@@
Subsets[{shapes}, {2}])@#]),
RegionMember[RegionUnion @@ (RegionBoundary /@ {shapes})]@#] &
direction = {-0.2, -0.2, -1};
point = {0.5, 0.5, 1.5};
line = HalfLine[{point, point + direction}];
intersections[l_, s__] :=
NSolve[# ∈ l && constraints[s][#], #] &@({x, y, z}[[;; RegionEmbeddingDimension[l]]])
(* find intersection *)
intersection = intersections[line, ##] & @@ shapes;
points = Point[{x, y, z}] /. intersection;
Graphics3D[{{Opacity[0.2], shapes}, line, {Red, points}},
PlotRange -> {{-1, 1}, {-1, 1}, {-2, 2}}, Axes -> True]
</code></pre>
<p>This works well, and we get the external intersections as we expect. </p>
<p><a href="https://i.stack.imgur.com/3SCo1m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3SCo1m.png" alt="enter image description here"></a></p>
<p>Now, let us try to take the difference between two shapes, modelling something like </p>
<pre><code>square = Cuboid[];
ball = Ball[{0, 0, 1}, 1];
Region[RegionDifference[square, ball]]
</code></pre>
<p><a href="https://i.stack.imgur.com/tz64fm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tz64fm.png" alt="enter image description here"></a></p>
<pre><code>shapes = {RegionDifference[square, ball]};
direction = {0, 0, -1};
point = {0.5, 0.5, 5};
line = HalfLine[{point, point + direction}];
intersection = intersections[line, ##] & @@ shapes
</code></pre>
<p>Does not work, with an error that the constraints are "<em>not a quantified system of equations and inequalities</em>"...even though the constraints look fine </p>
<pre><code>constraints[shapes]
(* (##1 &) @@
Not /@ Through[
Apply[RegionMember[RegionIntersection[##1]] &,
Subsets[{{BooleanRegion[#1 && ! #2 &, {Cuboid[{0, 0, 0}],
Ball[{0, 0, 1}, 1]}]}}, {2}], {1}][#1]] &&
RegionMember[
RegionUnion @@
RegionBoundary /@ {{BooleanRegion[#1 && ! #2 &, {Cuboid[{0, 0,
0}], Ball[{0, 0, 1}, 1]}]}}][#1] & *)
</code></pre>
| Tim Laska | 61,809 | <p>This is not a direct answer to your question, but an alternate approach. You could create a list of primitives and a build function that contains the Computational Solid Geometry (CSG).</p>
<pre><code>square = Cuboid[];
ball = Ball[{0, 0, 1}, 1];
buildList = {square, ball};
(* Constraints *)
buildFn = ¬ #2 ∧ #1 &;
reg = Region[
Style[BooleanRegion[buildFn, buildList], Opacity[0.5], Green]];
direction = {0, 0, -1};
point = {0.5, 0.5, 5};
line = HalfLine[{point, point + direction}];
rint = Region[RegionIntersection[reg, line],
BaseStyle -> {Blue, Thick}];
intpoints = Point[Transpose@RegionBounds@rint];
Show[reg, rint, Graphics3D[{PointSize[Large], Red, intpoints}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/O0Hrr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/O0Hrr.png" alt="Difference"></a></p>
<p>Here is how it would look for the initial case:</p>
<pre><code>shape1 = Cone[];
shape2 = Cuboid[];
buildList = {shape1, shape2};
(* Constraints *)
buildFn = #2 || #1 &;
reg = Region[
Style[BooleanRegion[buildFn, buildList], Opacity[0.5], Green]];
direction = {-0.2, -0.2, -1};
point = {0.5, 0.5, 1.5};
line = HalfLine[{point, point + direction}];
rint = Region[RegionIntersection[reg, line],
BaseStyle -> {Blue, Thick}]; intpoints =
Point[Transpose@RegionBounds@rint];
Show[reg, rint, Graphics3D[{PointSize[Large], Red, intpoints}],
PlotRange -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/39Ah2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/39Ah2.png" alt="Initial Case"></a></p>
<h1>Update to Increase Speed</h1>
<p>@Tomi mentioned in the comments that speed is a concern. As addressed in my answer to the MSE question <a href="https://mathematica.stackexchange.com/a/219645/61809">Why is Ray Tracing Slow?</a> I created a solver that that used the fast region functions <a href="https://reference.wolfram.com/language/ref/RegionDistance.html" rel="noreferrer"><code>RegionDistance</code></a>and <a href="https://reference.wolfram.com/language/ref/RegionNormal.html" rel="noreferrer"><code>RegionNormal</code></a> to solve a 1000 multiple bounce ray traces in 3D geometry including geometry produced by a commercial CAD package. I will adapt that approach to look at the bouncing of single ray.</p>
<h2>Set up the Geometry</h2>
<p>The <a href="https://reference.wolfram.com/language/OpenCascadeLink/tutorial/UsingOpenCascadeLink.html" rel="noreferrer">OpenCascadeLink</a> does a pretty good job at constructing geometry that snaps to features while keeping the triangle count down. The following workflow will create the the initial Box-Cone geometry.</p>
<pre><code>Needs["OpenCascadeLink`"]
Needs["NDSolve`FEM`"]
pp = Polygon[{{0, 0, 0}, {0, 0, 1}, {1, 0, 1}}];
shape = OpenCascadeShape[pp];
axis = {{0, 0, 0}, {0, 0, 1}};
sweep = OpenCascadeShapeRotationalSweep[shape, axis];
bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[sweep];
Show[Graphics3D[{{Red, pp}, {Blue, Thick, Arrow[axis]}}],
bmesh["Wireframe"], Boxed -> False]
cu = OpenCascadeShape[Cuboid[{0, 0, 0}, {1, 1, 1}]];
union = OpenCascadeShapeUnion[cu, sweep];
bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[union];
groups = bmesh["BoundaryElementMarkerUnion"];
temp = Most[Range[0, 1, 1/(Length[groups])]];
colors = ColorData["BrightBands"][#] & /@ temp;
bmesh["Wireframe"["MeshElementStyle" -> FaceForm /@ colors]]
mrd = MeshRegion[bmesh, PlotTheme -> "Lines"]
</code></pre>
<h2>Solve a Single Ray Trace</h2>
<p>The following workflow solves for a single ray trace. Each bounce will cause the ray to attenuate the representative sphere size by 10%. This solves and plots quickly.</p>
<pre><code>(* Set up Region Operators on Differenced Geometry *)
rdf = RegionDistance[mrd];
rnf = RegionNearest[mrd];
(* Setup and run simulation *)
(* Time Increment *)
dt = 0.01;
(* Collision Margin *)
margin = (1 + dt) dt;
(* Conditional Particle Advancer *)
advance[r_, x_, v_, c_] :=
Block[{xnew = x + dt v}, {rdf[xnew], xnew, v, c}] /; r > margin
advance[r_, x_, v_, c_] :=
Block[{xnew = x , vnew = v, normal = Normalize[x - rnf[x]]},
vnew = Normalize[v - 2 v.normal normal];
xnew += dt vnew;
{rdf[xnew], xnew, vnew, c + 1}] /; r <= margin
(* Starting Point for Emission *)
sp = {0, 0, 0.25};
nparticles = 1;
ntimesteps = 800;
tabres = Table[
NestList[
advance @@ # &, {rdf[sp],
sp, { Cos[2 Pi #[[1]]] Sin[Pi #[[2]]],
Sin[ Pi #[[2]]] Sin[2 Pi #[[1]]], Cos[ Pi #[[2]]]} &@
First@RandomReal[1, {1, 2}], 0}, ntimesteps], {i, 1,
nparticles}];
epilog[i_] := {ColorData["Rainbow", (#4 - 1)/10],
Sphere[#2, 0.04 0.9^#4]} & @@@ tabres[[i]]
Graphics3D[{White, EdgeForm[Thin], Opacity[0.25], mrd, Opacity[1]}~
Join~epilog[1], Boxed -> False, PlotRange -> RegionBounds[mrd],
ViewPoint -> {-1.7742436871276688`, 1.5459832360779067`,
2.431459473742817`},
ViewVertical -> {0.052110700162003136`, -0.06948693625348555`,
0.9962208794332359`}]
</code></pre>
<p><a href="https://i.stack.imgur.com/BLKji.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BLKji.png" alt="Cone Box Single Ray Trace"></a></p>
<h1>A More Complex Case</h1>
<p>The following produces a shape with concavity that could find rays that intersect but would be blocked by an intervening surface. Because the solver use a fine time increment, these intersections are not found because the collision of the intervening surface is detected.</p>
<pre><code>pp = Polygon[{{0, 0, 0}, {0, 0, 1}, {1, 0, 1}}];
shape = OpenCascadeShape[pp];
axis = {{0, 0, 0}, {0, 0, 1}};
sweep = OpenCascadeShapeRotationalSweep[shape, axis];
bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[sweep];
Show[Graphics3D[{{Red, pp}, {Blue, Thick, Arrow[axis]}}],
bmesh["Wireframe"], Boxed -> False]
cu = OpenCascadeShape[Cuboid[{0, 0, 0}, {1, 1, 1}]];
ball = OpenCascadeShape[Ball[{1/2, 1/2, 2.4}, 1.5]];
union = OpenCascadeShapeUnion[cu, sweep, ball];
bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[union];
groups = bmesh["BoundaryElementMarkerUnion"];
temp = Most[Range[0, 1, 1/(Length[groups])]];
colors = ColorData["BrightBands"][#] & /@ temp;
bmesh["Wireframe"["MeshElementStyle" -> FaceForm /@ colors]]
mrd = MeshRegion[bmesh, PlotTheme -> "Lines"]
(* Set up Region Operators on Differenced Geometry *)
rdf = RegionDistance[mrd];
rnf = RegionNearest[mrd];
(* Setup and run simulation *)
(* Time Increment *)
dt = 0.01;
(* Collision Margin *)
margin = (1 + dt) dt;
(* Conditional Particle Advancer *)
advance[r_, x_, v_, c_] :=
Block[{xnew = x + dt v}, {rdf[xnew], xnew, v, c}] /; r > margin
advance[r_, x_, v_, c_] :=
Block[{xnew = x , vnew = v, normal = Normalize[x - rnf[x]]},
vnew = Normalize[v - 2 v.normal normal];
xnew += dt vnew;
{rdf[xnew], xnew, vnew, c + 1}] /; r <= margin
(* Starting Point for Emission *)
sp = {0, 0, 0.5};
nparticles = 1;
ntimesteps = 1600;
(*tabres= Table[NestList[advance@@#&,{rdf[sp],sp,{ Cos[2 Pi #[[1]]] \
Sin[Pi #[[2]]],Sin[ Pi #[[2]]] Sin[2 Pi #[[1]]], Cos[ Pi \
#[[2]]]}&@First@RandomReal[1,{1,2}],0},ntimesteps],{i,1,nparticles}];*)
tabres = Table[
NestList[
advance @@ # &, {rdf[sp],
sp, { Cos[2 Pi #[[1]]] Sin[Pi #[[2]]],
Sin[ Pi #[[2]]] Sin[2 Pi #[[1]]], Cos[ Pi #[[2]]]} &@
First@{{0.3788624698388783`, 0.8749177935911279`}}, 0},
ntimesteps], {i, 1, nparticles}];
epilog[i_] := {ColorData["Rainbow", (#4 - 1)/12],
Sphere[#2, 0.04 0.9^#4]} & @@@ tabres[[i]]
Graphics3D[{White, EdgeForm[Thin], Opacity[0.25], mrd, Opacity[1]}~
Join~epilog[1], Boxed -> False, PlotRange -> RegionBounds[mrd],
ViewPoint -> {-3.102894731729034`, -1.0062787100553268`,
0.8996929706836663`},
ViewVertical -> {-0.34334064946409365`, -0.07403103185215265`,
0.93628874005217`}]
</code></pre>
<p><a href="https://i.stack.imgur.com/4lz9b.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4lz9b.png" alt="Concave Surface"></a></p>
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