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 |
|---|---|---|---|---|
2,820,796 | <p>In How many ways can a 25 Identical books can be placed in 5 identical boxes. </p>
<p>I know the process by counting but that is too lengthy .
I want different approach by which I can easily calculate required number in Exam hall in few minutes. </p>
<p>Process of Counting :
This problem can be taken partitions of 25 into 5 parts.</p>
<p>25 = 25+0+0+0+0</p>
<p>25 = 24 +1 + 0 + 0 +0</p>
<p>25 = 23+ 1 +1 +0 + 0
... ....
Like this way many combinations are made.: about 377 </p>
<p>How can we calculate it without this process of manual counting. </p>
| Sri-Amirthan Theivendran | 302,692 | <p>In general, you can use the following recursive formula. Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts. Let $p_{k}(n)$ be the number of partitions of $n$ into $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Then
$$
p_{k}(n)=p_{k-1}(n-1)+p_{k}(n-k)
$$
by classifying $\lambda=(\lambda_1,\dots,\lambda_k)\in P_k(n)$ (where the $\lambda_i$ are weakly decreasing and positive) based on whether $\lambda_k=1$ or $\lambda_k>1$. Let $p_{\leq k}(n)=\sum_{m=0}^kp_{m}(n)$ be the number of partitions of $n$ into at most $k$ parts.</p>
<p>In your case you want to compute
$$
p_{\leq 5}(25)=\sum_{m=0}^5p_{m}(25).
$$
But note that there are some special values of $k$ which make it easy to compute. Indeed $p_{0}(25)=0$, $p_{1}(25)=1$, $p_2(25)=\lfloor 25/2\rfloor=12$. </p>
<p>Alternatively, we can note that partitions of $n$ into at most $k$ parts correspond bijectively to partitions of $n$ where the largest part is at most $k$. Thus
$$
\begin{align}
\sum_{n\geq 0}p_{\leq k}(n)x^n&=(1+q+q^2+\dotsb)(1+q^2+q^4+\dotsb)\dotsb(1+q^k+q^{2k}+\dotsb)\\
&=\frac{1}{(1-q)(1-q^2)\dotsb(1-q^k)}
\end{align}
$$
In particular we find that in your case
$$
p_{\leq 5}(25)=[q^{25}]\left(\frac{1}{(1-q)(1-q^2)\dotsb(1-q^5)} \right)=377
$$
using a computer algebra system for example.</p>
|
11,266 | <p>I have a list of time durations, which are strings of the form: <code>"hh:mm:ss"</code>. Here's a sample for you to play with:</p>
<pre><code>durations = {"00:09:54", "00:31:24", "00:40:07", "00:11:58", "00:13:51", "01:02:32"}
</code></pre>
<p>I want to convert all of these into numbers in seconds, so that I can actually do useful things with this data!</p>
<p>How would I go about doing this? There must be some way to get <em>Mathematica</em> to extract the relevant things from each string.</p>
<p>Thanks.</p>
<p>Edit: I'm surprised that there doesn't seem to be a question related to this already - but it may be that I was using the wrong search terms.</p>
| Mr.Wizard | 121 | <p>You didn't specify that performance is important but I decided to examine that aspect.</p>
<p>First I generate a list of 10,000 durations:</p>
<pre><code>durations =
ToString@Row[#, ":"] & /@
Map[IntegerString[#, 10, 2] &,
RandomInteger[{0, 24}, {10000, 3}], {2}];
</code></pre>
<p>I then time each of the existing solutions (using <code>timeAvg</code>, posted many times before):</p>
<pre><code>AbsoluteTime /@ durations - AbsoluteTime["00:00:00"] // timeAvg
</code></pre>
<blockquote>
<p>7.722</p>
</blockquote>
<pre><code>3600 FromDMS[ToExpression[StringSplit[#, ":"]]] & /@ durations // timeAvg
</code></pre>
<blockquote>
<p>0.078</p>
</blockquote>
<p>We see that at least with my test set and on my machine the second method is more than two orders of magnitude faster than the first, but I think we can do better:</p>
<pre><code>FromDigits[#, 60] & /@ Map[FromDigits, StringSplit[durations, ":"], {2}] // timeAvg
</code></pre>
<blockquote>
<p>0.02868</p>
</blockquote>
<p>By using the undocumented function <a href="https://mathematica.stackexchange.com/a/1744/121"><code>StringToDouble</code></a> we can eek out a bit more:</p>
<pre><code>timeAvg[
FromDigits[#, 60] & /@ Map[Internal`StringToDouble, StringSplit[durations, ":"], {2}]
]
</code></pre>
<blockquote>
<p>0.02248</p>
</blockquote>
<p>This last method returns Reals instead of Integers, if that matters, but the time taken to <code>Round</code> is negligible:</p>
<pre><code>seconds =
FromDigits[#, 60] & /@ Map[Internal`StringToDouble, StringSplit[durations, ":"], {2}];
Round[seconds] // timeAvg
</code></pre>
<blockquote>
<p>0.00011488</p>
</blockquote>
|
1,136,192 | <p>I need to solve this integral but I have no idea about how to procede, this is the integral:</p>
<p>$$\int \frac{x-1}{x+4x^3}\mathrm dx$$</p>
<p>This is how I solve the first part:</p>
<p>$$\int \frac{x}{x+4x^3}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>$$\int \frac{1}{1+4x^2}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>So I solved the first integral:</p>
<p>$$\int \frac{1}{1 + (2x)^2}\mathrm dx = \frac{1}{2}\arctan(2x) + C$$</p>
<p>But how can I solve the second?</p>
<p>$$- \int \frac{1}{x(1+4x^2)}\mathrm dx$$</p>
| Thomas | 103,537 | <p>Not sure if the question is about $1^n$ or $(-1)^n$ but in the latter case :</p>
<p>Hint: you have to group terms by two : $\sum \frac{1}{\ln(2n)} - \frac{1}{\ln(2n+1)}$.</p>
<p>Then use the $O()$ notation and express $\ln(2n+1)$ in terms of $\ln(2n)$ and $O(\frac{1}{n})$</p>
<p>This should lead you to $\sum O(\frac{1}{n\ln(n)})$. If you are not familiar with this notation I will answer more</p>
|
1,136,192 | <p>I need to solve this integral but I have no idea about how to procede, this is the integral:</p>
<p>$$\int \frac{x-1}{x+4x^3}\mathrm dx$$</p>
<p>This is how I solve the first part:</p>
<p>$$\int \frac{x}{x+4x^3}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>$$\int \frac{1}{1+4x^2}\mathrm dx - \int \frac{1}{x+4x^3}\mathrm dx$$</p>
<p>So I solved the first integral:</p>
<p>$$\int \frac{1}{1 + (2x)^2}\mathrm dx = \frac{1}{2}\arctan(2x) + C$$</p>
<p>But how can I solve the second?</p>
<p>$$- \int \frac{1}{x(1+4x^2)}\mathrm dx$$</p>
| Marco Cantarini | 171,547 | <p>First of all the series must start from $n=1$. If you have $(-1)^{n}$ at numerator the series converges for the alternating series test <a href="http://en.wikipedia.org/wiki/Alternating_series_test" rel="nofollow">http://en.wikipedia.org/wiki/Alternating_series_test</a>, and its sum is $-0.92429...$ If you have $1^{n}=1$ at numerator, the series diverges because $$\underset{n\geq1}{\sum}\frac{1}{n}\leq\underset{n\geq1}{\sum}\frac{1}{\log\left(n+1\right)}.$$</p>
|
3,864,729 | <p>I am studying for an exam and I am almost grasping compactness. However, some examples are still unclear. E.g. for <span class="math-container">$A = (a, b)$</span> with <span class="math-container">$a < b$</span> and <span class="math-container">$a, b \in \mathbb{R}$</span>.
How can <span class="math-container">$(a, b) \subseteq \cup _{k \in \mathbb{N}} \in (a, b − \frac{1}{k})$</span>, making this an open covering. Even though <span class="math-container">$\frac{1}{k} \ge 0$</span> for <span class="math-container">$k\ge\infty$</span>, how is this set still covering <span class="math-container">$(a,b)$</span> when <span class="math-container">$b -\frac{1}{k}$</span> is smaller than <span class="math-container">$b$</span>? I hope you can make this clear for me, thanks!</p>
| fleablood | 280,126 | <p>"how is this set still covering (a,b) when b−1k is smaller than b"</p>
<p>Because <span class="math-container">$b$</span> is <em>not</em> in <span class="math-container">$(a,b)$</span>. so it never needs to be covered.</p>
<p><em>Every</em> point in <span class="math-container">$(a,b)$</span> is <em>also</em> smaller than <span class="math-container">$b$</span> and for every point <span class="math-container">$x \in (a,b)$</span> we have <span class="math-container">$x<b$</span> and there is some <span class="math-container">$b-\frac 1k$</span> so that <span class="math-container">$x < b-\frac 1k < b$</span>.</p>
<p>Notice: because <span class="math-container">$b$</span> is <em>not</em> in <span class="math-container">$(a,b)$</span> but <span class="math-container">$b$</span> <em>is</em> a limit point of <span class="math-container">$(a,b)$</span> that that is <em>why</em> <span class="math-container">$(a,b)$</span> is not closed. And therefore <span class="math-container">$(a,b)$</span> is not compact. If we consider <span class="math-container">$[a,b]$</span> so <span class="math-container">$b$</span> <em>was</em> in <span class="math-container">$[a,b]$</span> then those would not be an open cove because <span class="math-container">$b$</span> is <em>not</em> covered. To have an open cover we'd need an open set with <span class="math-container">$b$</span> and that set (because it is open) would have to have so points <em>beyond</em> <span class="math-container">$b$</span> as well. And that one open set because it goes "over" covers an infinite number of points. This is why (in <span class="math-container">$\mathbb R$</span>) compact sets must be closed.</p>
|
1,344,464 | <p>Consider <span class="math-container">$$f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^{3n+1}$$</span></p>
<p>It's a power series with a radius, <span class="math-container">$R=1$</span>. at <span class="math-container">$x=1$</span> it converges. Hence, by Abel's thorem:</p>
<p><span class="math-container">$$\lim_{x\to 1^-} f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$</span></p>
<p>Evaluating the derivative</p>
<p><span class="math-container">$$f'(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} (3n+1)x^{3n+1} = \ldots = \frac{1}{1+x^3}$$</span></p>
<p>Now, consider this claim: "Since <span class="math-container">$f(0) = 0$</span>: <span class="math-container">$$f(x) = \int_0^x \frac{1}{1+t^3} \ dt$$</span></p>
<p>I am familiar with the fundamental theorem of calculus, yet not sure why this claim true, More presicely; Why is the integral starts at <span class="math-container">$0$</span>?</p>
<p>Thanks.</p>
| Loreno Heer | 92,018 | <p>Note the following corollary:</p>
<blockquote>
<p>$$\int_a^b f(t)\, dt = F(b)-F(a)$$</p>
</blockquote>
<p>You are free to choose the points $a$ and $b$ as long the function $f$ is defined in $[a,b]$. But choosing $a$ such that $F(a)=0$ simplifies things.</p>
|
1,435,269 | <p>Let a sequence $x_n$ be defined inductively by $x_{n+1}=F(x_n)$. Suppose that $x_n\to x$ as $n\to \infty$ and $F'(x)=0$. Show that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.</p>
<p>I'm not sure how to do this. Any solutions are greatly appreciated. I think The Mean-Value Theorem will be useful and we can assume that $F$ is continuously differentiable.</p>
| Paramanand Singh | 72,031 | <p>This is a direct consequence of the mean value theorem and the assumption that $F$ is continuously differentiable (without this assumption the result may not hold).</p>
<p>We have via mean value theorem $$x_{n+2}-x_{n+1}=F(x_{n+1})-F(x_{n})=(x_{n+1}-x_{n})F'(c)$$ for some $c$ between $x_n$ and $x_{n+1}$. As $n\to\infty$ we can see that both $x_{n} $ and $x_{n+1}$ tend to $x$ and therefore so does $c$. By continuity of $F'$, the value $F'(c) \to F'(x) =0$ and thus we have the desired result. </p>
|
588,802 | <p>The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$</p>
<p>The first thing I did was use the divergence test which didn't help since the result of the limit was 0.</p>
<p>If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$</p>
<p>I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.</p>
| Hagen von Eitzen | 39,174 | <p>Note that $\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)$ so this is a telescoping sum
$$\sum_{n=1}^m \frac1{n(n+3)}=\frac13\left(1+\frac 12+\frac13-\frac1{m+1}-\frac1{m+2}-\frac1{m+3}\right)\to \frac{11}{18}.$$</p>
|
249,908 | <p>This seems like a very inefficient way of doing what I want. I generate all permutations of (for example) <code>{1, 1, 2, 2}</code>, then for each permutation I generate its rotations, select the first one in <code>Sort</code> order, and use that for <code>GatherBy</code>.</p>
<pre><code>list = {1, 1, 2, 2};
display@GatherBy[Permutations@list, Sort[Table[RotateLeft[#, i], {i, 0, Length@# - 1}]][[1]]& ]
(* {{"1122", "1221", "2112", "2211"}, {"1212", "2121"}} *)
</code></pre>
<p>How can I do this more efficiently?</p>
| kglr | 125 | <pre><code>GroupOrbits[CyclicGroup[Length @ #], Permutations @ #, Permute] & @ list
</code></pre>
<blockquote>
<pre><code>{{{1, 1, 2, 2}, {1, 2, 2, 1}, {2, 1, 1, 2}, {2, 2, 1, 1}},
{{1, 2, 1, 2}, {2, 1, 2, 1}}}
</code></pre>
</blockquote>
|
1,865,062 | <p>I have seen some examples of inner automorphisms of Lie algebras. Can anyone please give me an example of an automorphism of Lie algebras that is not inner (with proof).
Note - An automorphism is said to be inner if it is of the form $exp(adx)$ for $adx$ nilpotent where $adx(y)$=$[x,y]$.
Thanks for any help.</p>
| Hanno | 81,567 | <p>The simplest example I know is ${\mathfrak s}{\mathfrak l}_n({\mathbb C})$ for $n\geq 3$, which has the non-inner automorphism $A\mapsto -A^t$. This is not inner because is doesn't preserve conjugacy classes of matrices (look at the eigenvalues) which however any inner automorphism $\text{exp}(\text{ad}(X))=\text{conj}(e^X)$ does.</p>
<p>For a general semisimple complex Lie-algebra you get a non-inner automorphism for any non-trivial symmetry of the Dynkin diagram. In the example above you are using the horizontal symmetry of $A_{n-1}$ (note that for $n=2$ this is trivial, and $A\mapsto -A^t$ is indeed inner there).</p>
|
392,442 | <p>What would be the immediate implications for Math (or sciences as a general) if someone developed a formula capable of generating every prime number progressively and perfectly, also able to prove (or disprove) the primality of every N-th number. I know this is a very large and subjective answer, however, I would like to know some of these implications - like the breaking of many security systems based on the primes. Moreover, there are examples of practical Math's implication, not just theoretical, of a possible prime number formula discovery? There is for example in Physics, Chemistry, Geography or Astronomy any field which would be very improved with a so great and dreamed <em>Eureka</em>?</p>
| Community | -1 | <p>As others have mentioned there are many formulas for primes. </p>
<p>I can't pass up the opportunity to mention my favorite:</p>
<p>$$p_n=1+\sum^{2^n}_{m=1}\left\lfloor \sqrt[n]n \left( \sum^{m}_{x=1}\left\lfloor \cos^2\left( \pi \frac{(x-1)!+1}{x}\right) \right\rfloor \right)^{-1/n} \right\rfloor$$</p>
<p>Maybe it's just my favorite because it's so complicated and unwieldy! </p>
<p>I first found the formula in Hacker's Delight by Warren. The formula is cited as "Willan's Formula". The formula can be derived from the fact that $(x-1)!\equiv -1 (\mod x)$ iff x is 1 or prime. So if $((x-1)!+1)/x$ is an integer, $x$ is prime (or 1), so the cosine of this times $\pi$ will be -1 or 1 iff $x$ is prime or 1. $\cos^2$ gets rid of the negative, floor keeps only the values for which the $\cos^2$ is 1. </p>
<p>This formula appears to be described on <a href="http://mathworld.wolfram.com/PrimeFormulas.html" rel="nofollow">the mathworld page for prime formulas.</a></p>
|
2,802,959 | <p>If I write
$$
x\in [0,1] \tag 1
$$
does it mean $x$ could be ANY number between $0$ and $1$?</p>
<p>Is it correct to call $[0,1]$ a set? Or should I instead write $\{[0,1]\}$? </p>
<p>Q2:</p>
<p>If I instead have
$$
x\in \{0,1\} \tag 2
$$
does it mean $x$ could be only $0$ OR $1$?</p>
| user061703 | 515,578 | <p>$x\in[0;1]$ means $0\le x\le 1$, any real number $x$ that satisfies this is true. For example:</p>
<p>$x^2-x\le 0\Leftrightarrow x(x-1)\le 0\Leftrightarrow x-1\le 0\le x\Leftrightarrow 0\le x\le 1$</p>
<p>$x\in\{0;1\}$ means $x=0$ or $x=1$ is true. For example:</p>
<p>$x^2-x=0\Leftrightarrow x(x-1)=0 \Leftrightarrow x\in\{0;1\}.$ </p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Arturo Magidin | 3,959 | <p>The <a href="http://en.wikipedia.org/wiki/Abc_conjecture%20%22ABC%20conjecture%22">ABC conjecture</a> and Vojta's conjectures come to mind.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Roy Maclean | 3,537 | <p>Clicking on Toolbox-What Link's Here in the wikipedia article <a href="http://en.wikipedia.org/wiki/Conditional_result">Conditional proof</a> brings up <a href="http://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H">Schinzel's hypothesis H</a> which the article says is used to prove conditional results in diophantine geometry.</p>
|
28,811 | <p>There are lots of statements that have been conditionally proved on the assumption that the Riemann Hypothesis is true.</p>
<p>What other conjectures have a large number of proven consequences?</p>
| Gil Kalai | 1,532 | <p>In computational complexity there are several conjectures which are stronger than $NP \ne P$ which have important consequences. To mention a few</p>
<p>1) The conjecture that factoring is computationally hard is the basis to much theoretical and practical cryptography. </p>
<p>2) More broadly the conjecture that one-way functions exist has many consequences. </p>
<p>3) The conjecture that the polynomial hierarchy ($PH$) does not collapse has many consequences.</p>
<p>4) Khot's unique game conjecture has many important consequences for hardness of approximation.</p>
<p>5) The exponential time hypothesis ($ETH$)and strong exponential time hypothesis ($SETH$) are strong form of $NP \ne P$ with important consequences. </p>
<p>6) There are stronger and stronger versions of the "derandomization" conjecture with many consequences. </p>
|
889,719 | <p>Example $5.9$ on page $103$ of John Lee's Smooth Manifolds says the following:</p>
<p>The intersection of $S^n$ with the open subset $\{x:x^i>0\}$ is the graph of the smooth function
$$
x^i=f(x^1,\dots,x^{i-1},x^{i+1},\dots,x^{n+1})
$$
where $f\colon B^n\to\mathbb{R}$ is $f(u)=\sqrt{1-|u|^2}$. The intersection of $S^n$ with $\{x:x^i<0\}$ is the graph of $-f$. Since every point in $S^n$ is in one of these sets, $S^n$ satisfies the local $n$-slice condition, this is an embedded submanifold.</p>
<p>The terminology is that if $M$ is a smooth manifold, and $S\subset M$ a subset, then $S$ satisfies the local $k$-slice condition if each point of $S$ is contained in the domain of a smooth chart $(U,\varphi)$ for $M$ such that $S\cap U$ is a single $k$-slice in $U$. </p>
<p>I don't see how this makes $S^n$ satisfy the local $n$-slice condition. Presumably the chart on $\mathbb{R}^{m+1}$ is $(U=\{x:x^i>0\},\mathrm{id})$, so that $S^n\cap\{x:x^i>0\}$ is an $n$-slice of $U$? But this doesn't seem right since $\mathrm{id}(S^n\cap U)$ is a hemisphere of $S^n$, but that's not a $n$-slice in the corresponding half-place $\mathrm{id}(U)$?</p>
| Kelvin Lois | 322,139 | <p>I'm also confused by that. But after read Prof. Lee and koch comments i realized that maybe what it means is this :</p>
<p>$\mathbb{S}^n$ covered by the graphs of smooth function. The graph of smooth function is embedded submanifold. But an embedded submanifold satisfy local k-slice condition. So each graph that cover $\mathbb{S}^n$ satisfy local slice condition. Because each point in $\mathbb{S}^n$ is in these graphs, $\mathbb{S}^n$ satisfy local slice condition. Therefore $\mathbb{S}^n$ is embedded submanifold.</p>
|
1,032,714 | <p>'Let $X$ be a topological space and let $(U_i)_{i \in I}$ be a cover of $X$ by connected subspaces $U_i$. Supposed for all $i,j \in I$ there exists some $n \geq 0$ and $k_0,...,k_n \in I$ such that $k_0 = i, k_n = j$ and $$U_{k_0} \cap U_{k_1} \neq \emptyset, U_{k_1} \cap U_{k_2} \neq \emptyset, ..., U_{k_{n-1}} \cap U_{n} \neq \emptyset$$
Show that $X$ is connected.</p>
<p>My intuition tells me the proof is along the lines of since each $U_i$ is connected, there are no subsets $A,B \subseteq U_i$ for $i \in I$ such that $A$ and $B$ are disjoint (unless they are the empty set.) Also, since the intersection of $U_{k_i} \cap X/U_{k_i}$ is non empty (as there always exists at least some other subcover it intersects with) there are no two nonempty subsets $A,B \subset X$ such that $A \cap B = X$. Thus $X$ is connected.</p>
<p>I was looking for some help formulating this into a nice mathematical proof. Thanks.</p>
| Community | -1 | <p>The first one is a special case of the second definition.
As far as I can see, the second definition <strong>does not</strong> follows from the first.</p>
<p>It is highly customary to accept the second as the definition.</p>
|
1,269,738 | <p>I'm looking for problems that due to modern developments in mathematics would nowadays be reduced to a rote computation or at least an exercise in a textbook, but that past mathematicians (even famous and great ones such as Gauss or Riemann) would've had a difficult time with. </p>
<p>Some examples that come to mind are <em><a href="http://en.wikipedia.org/wiki/Group_testing">group testing problems</a></em>, which would be difficult to solve without a notion of error-correcting codes, and -- for even earlier mathematicians -- calculus questions such as calculating the area of some $n$-dimensional body.</p>
<p>The questions have to be understandable to older mathematicians and elementary in some sense. That is, past mathematicians should be able to appreciate them just as well as we can. </p>
| zoli | 203,663 | <p>I would say that computing the Fourier coefficients of a tamed function is a triviality today even at an engineering math 101 level.</p>
<p><a href="http://www.amazon.co.uk/Mathematical-Experience-Penguin-Press-Science/dp/0140134743/ref=sr_1_1?ie=UTF8&qid=1430917805&sr=8-1&keywords=Reuben+Hersh">Ph. Davis and R. Hersh</a> tell the long and painful story of Fourier series.
I quote from their book:</p>
<blockquote>
<p>"Fourier didn't know Euler had already done this, so he did it over. And Fourier, like Bernoulli and Euler before him, overlooked the beautifully direct method of orthogonality [...]. Instead, he went through an incredible computation, that could serve as a classic example of physical insight leading to the right answer in spite of flagrantly wrong reasoning." </p>
</blockquote>
<p>(Fifth Ch. "Fourier Analysis".)</p>
|
25,917 | <p>$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ </p>
<p>$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$</p>
<p>Why they are different?</p>
| Kerry | 7,887 | <p>Proof:</p>
<p>By the fundamental theorem of algebra, every integer $N$ can be uniquely factored as $\prod^{n}_{i=1}p_{i}^{a_{i}}$. Now, mark $2=p_{1}$, note $a_{i}$ can take value of $0$. You got the theorem. </p>
<p>For the "inductive" proof, suppose for $n<k$ this is true. For $n+1$ its factors must be in previous $n$ numbers. Hence $n+1=\prod n_{i}$. Decompose $n_{i}$ by induction hypothesis you get the statement. </p>
|
2,223,163 | <p>I don't have any idea on how to prove it, and I need it for one of my questions which is still unanswered: <a href="https://math.stackexchange.com/questions/2192947/what-is-the-largest-number-smaller-than-100-such-that-the-sum-of-its-divisors-is?noredirect=1#comment4521040_2192947">What is the largest number smaller than 100 such that the sum of its divisors is larger than twice the number itself?</a>.</p>
| Misha Lavrov | 383,078 | <p>It's always even, and it has digits $999\dots996$ so it's always divisible by $3$: it's $3$ times $333\dots332$.</p>
|
3,281,540 | <p>I wrote an algorithm by combining Fermat's Little Theorem and Euler's Method. However, I am experiencing a problem in Euler's method. </p>
<p>For instance, If I take <span class="math-container">$(A, B, M)$</span> such that <span class="math-container">$A^B mod(M)$</span>.</p>
<p>When the initial values are <span class="math-container">$(12341, 123141, 12313)$</span> they reduce to <span class="math-container">$(12341, 7113, 12313)$</span> However after this the program won't go further. The reason I believe is that when <span class="math-container">$$B < \phi(M)$$</span> the algorithm stops working. So simply to algorithm work properly we need larger <span class="math-container">$B$</span> but small <span class="math-container">$M$</span>.</p>
<p><strong>At this point, my question is there a way to reduce M to small numbers by some mathematical operations ?</strong> </p>
<pre><code>def eulers_theorem(A, B, M):
totient = M
factors_of_M = [(i-1) / i for i in factorint(M).keys()]
for i in factors_of_M:
totient *= i
new_B = int(B % totient)
return (A, new_B, M)
</code></pre>
<p>So I am factorizing the M values and then calculating the totient of it( <span class="math-container">$\phi(M)$</span> ) and then My new B becomes </p>
<p><span class="math-container">$B_{new} = B~mod (\phi(M))$</span></p>
<p>For instance, if (A, B, M) = (123, 562, 100) </p>
<p><span class="math-container">$\phi(100) = 40$</span> so </p>
<p><span class="math-container">$$B_{new} = 562~mod (40) = 2$$</span></p>
<p>So (123, 562, 100) reduces to (123, 2, 100)</p>
<p>In the Above example when <span class="math-container">$(12341, 123141, 12313)$</span> it reduces to <span class="math-container">$(12341, 7113, 12313)$</span>. In this case the algorithm enters a loop since, </p>
<p><span class="math-container">$$B_{new} = 7113~mod (\phi(12313)) = 7113~mod(10548) = 7113$$</span> so <span class="math-container">$B_{new}$</span> is always the same.</p>
| Culver Kwan | 686,157 | <p>First, using modular arithmetic, we can reduce <span class="math-container">$12341$</span> to <span class="math-container">$12341-12313=28$</span>.</p>
<p>As <span class="math-container">$12313=7\times1759$</span>, we can first find the remainder of <span class="math-container">$28^{7113}$</span> divided by <span class="math-container">$7$</span> and <span class="math-container">$1759$</span>.
As <span class="math-container">$28$</span> is divisible by <span class="math-container">$7$</span>,<span class="math-container">$28^{7113}\equiv0$</span> (mod <span class="math-container">$7$</span>)</p>
<p>Then, we can use Euler's theorem and other theorems to find the remainder of <span class="math-container">$28^{7113}$</span> divided by 1759. Note that <span class="math-container">$\phi(1759)=1758$</span></p>
<p><span class="math-container">$28^{7113}\equiv28^{1758\times4+81}\equiv28^{81}\equiv267$</span> (mod <span class="math-container">$1759$</span>)</p>
<p>Then by Chinese Remainder theorem, the remainder of <span class="math-container">$12341^{123141}$</span> divided by <span class="math-container">$12313$</span> is <span class="math-container">$5544$</span>.</p>
<p>Hope that it helps!</p>
|
3,620,767 | <p><a href="https://imgur.com/a/i24lMmS" rel="nofollow noreferrer">https://imgur.com/a/i24lMmS</a></p>
<p>I tried solving this problem, but couldn't find an answer. Any suggestions? Thanks!</p>
| Quanto | 686,284 | <p>Let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> be the side segments of the largest square segmented by the middle-square vertexes. Then,</p>
<p><span class="math-container">$$ x^2+y^2=37^2,\>\>\>\>\>16(x+y) =xy$$</span></p>
<p>where the second equation matches the triangle area with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> as the sides. Together,</p>
<p><span class="math-container">$$(x+y)^2-32(x+y)-37^2=0$$</span></p>
<p>Thus the area is</p>
<p><span class="math-container">$$(x+y)^2= 1881+160\sqrt{65}$$</span></p>
|
2,435,505 | <p>It is a question from permutations and combination chapter and its ans is 48 as given in book! Please help me to do this. I am unable to figure out the solution! Please help!</p>
| Ross Millikan | 1,827 | <p>The answer is badly wrong. If you have to use all the digits you are asking for the number of permutations of your digits with the last one being even. You have $3$ choices for the units digit, then $5$ for the tens digit, $4$ for the hundreds and so on. You get $\frac {6!}2=360$ choices. If you don't have to use all the digits you have even more choices.</p>
|
723,707 | <p>I'm trying to understand what the relation is between the direct product and the quotient group. </p>
<p>If we let $H$ be a normal subgroup of a group $G$, then it is not too difficult to show that the set of all cosets of $H$ in $G$ forms a quotient group $G/H$:
\begin{equation}
G/H = \{ g H \mid g \in G \}
\end{equation}</p>
<p>On the other hand, the Cartesian product of two groups $G$ and $H$ is defined as:
\begin{equation}
G \times H = \{ (g,h) \mid g \in G \text{ and } h \in H \}
\end{equation}
where $(g,h)$ denotes the set of ordered pairs. The direct product operation on this set is defined as:
\begin{equation}
(g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2) \in G \times H
\end{equation}
and it is easy to see that the direct product forms a group.</p>
<p>Is the following statement true:
\begin{equation}
K = G \times H \implies G \simeq K / (\{e_G \} \times H)
\end{equation}
If so, under what conditions is it true? And how can we see it is true (or false)?</p>
<p><strong>Edit 26/03</strong>:</p>
<p>Up to this point, I believe I have found a method (see below) of showing the isomorphism relation. I would be really grateful if someone could tell me whether this proof is correct or not.</p>
<p>Let us identity the elements of $h \in H$ with element of $K$ by setting $h \equiv (e_G,h)$. The elements of $K/H$ are as usual defined by:
\begin{equation}
K/H = \{ (g,h) H \mid g \in G \text{ and } h \in H \}
\end{equation}
Since $h_1H=H$ for some $h_1 \in H$, we have:
\begin{equation}
(g,h)H = (g',h')H \iff g=g' \text{ and } h' = h h_1 \tag{1}
\end{equation}
and so without loss of generality we can write every element of $K/H$ in the form $(g,e_H)H$. Now, let the map:
\begin{equation}
f : G \to K/H
\end{equation}
be defined by:
\begin{equation}
f(g) = (g,e_H) H \tag{2}
\end{equation}
The map is one-to-one. This can be seen by equation $(1)$, because if:
\begin{equation}
f(g) = f(g')
\end{equation}
then:
\begin{equation}
(g,e_H) H = (g',e_H) H \implies g=g'
\end{equation}
Furthermore, the map is trivially onto:
\begin{equation}
\forall (g,h) H \in K/H \; \exists \; g \in G \; , \; f(g)=(g,h) H
\end{equation}
and thus the map is bijective. Finally, the map is also a homomorphism, because:
\begin{equation}
f(gg') = (gg',e_H) H = (g,e_H)(g',e_H) H = (g,e_H)(g',e_H) HH = (g,e_H) H (g',e_H) H = f(g) f(g')
\end{equation}
and so $f$ is a isomorphism. Thus, by definition of equation $(2)$, we have shown that $G \simeq K/H$.</p>
<p>Any input is much appreciated.</p>
| Manolito Pérez | 13,293 | <p>This statement is always true, since $H$ is a normal subgroup of $G \times H$ (i.e. $xH = Hx$ for all $x \in K$). To see why, consider two cases:</p>
<ol>
<li><p>The case $x \in H$</p></li>
<li><p>The case $x \notin H$</p></li>
</ol>
<p>Then, the statement follows from the fundamental theorem of homomorphisms. By the way, you should replace the second $=$ sign by the $\simeq$ sign, in your statement. </p>
|
3,281,503 | <blockquote>
<p>For natural numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, what is the greatest value of <span class="math-container">$b$</span> so that <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits?</p>
</blockquote>
<p>I knew that the greatest value of <span class="math-container">$b$</span> is <span class="math-container">$21$</span>, where <span class="math-container">$9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$</span>, but I am not sure how to prove that there is no greater value of <span class="math-container">$b$</span>.</p>
| Hagen von Eitzen | 39,174 | <p>A natural number <span class="math-container">$n$</span> has <span class="math-container">$b$</span> digits iff <span class="math-container">$10^{b-1}\le n<10^b$</span>. So for this to hold for <span class="math-container">$a^b$</span>, we clearly need <span class="math-container">$a<10$</span>, or: <span class="math-container">$a\le 9$</span>. On the other hand, we need <span class="math-container">$(\frac{10}{a})^{b-1}\le a$</span>. For given <span class="math-container">$a$</span>, this can be solved by taking logarithms. We obtain <span class="math-container">$b\le 21$</span> for <span class="math-container">$a=9$</span> and <span class="math-container">$b\le 10$</span> for <span class="math-container">$a\le8$</span>. So, as perhaps expected, we use <span class="math-container">$a=9$</span> and find
<span class="math-container">$$9^{21}=\underbrace{109,418,989,131,512,359,209}_{21}$$</span>
as largest case.</p>
|
813,715 | <p>Say I am asked to find, in expanded form without brackets, the equation of a circle with radius 6 and centre 2,3 - how would I go on about doing this?</p>
<p>I know the equation of a circle is $x^2 + y^2 = r^2$, but what do i do with this information?</p>
| Tunk-Fey | 123,277 | <p>The general solution for the circle with centre $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2.
$$
Now, we have the centre $(2,3)$ and the radius $6$, therefore the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=6^2\\
x^2-4x+4+y^2-6x+9&=36\\
x^2+y^2-4x-6y+4+9-36&=0\\
\large\color{blue}{x^2+y^2-4x-6y-23}&\large\color{blue}{=0}.
\end{align}</p>
|
2,704,102 | <p>Let $X,Y,Z$ be topological spaces. Is the following statement true?
$X \times Z \cong Y \times Z \implies X \cong Y$?
how would you prove it? </p>
<p>and I know that if $A \cong B$, and $a \in A$ that there is a $b \in B$, such that $A\setminus{\{a\}} \cong B\setminus{\{b\}}.$ How would you prove the same for removing lines from product topology, instead of point of normal topological spaces?</p>
| amir bahadory | 204,172 | <p>Let <span class="math-container">$ X=\mathbb{R^+} \cup \{0\} $</span> then <span class="math-container">$ \mathbb{R} \times X \cong X\times X $</span> with homeomorphism <span class="math-container">$(x,y) \to (x^2 -y^2,2xy)$</span> but <span class="math-container">$ \mathbb{R^+} \cup \{0\} \ncong \mathbb{R} $</span></p>
|
4,398,873 | <p>Given f: (0,1) <span class="math-container">$\rightarrow$</span> <span class="math-container">$\mathbb R$</span>. f is a continuous function and improper integrable.</p>
<p>If <span class="math-container">$\int_{0}^{x}f(t)dt = 0$</span> <span class="math-container">$\forall x \in [0,1] $</span>,</p>
<p>Does <span class="math-container">$f(x) = 0 $</span> <span class="math-container">$\forall x \in [0,1] $</span> ?</p>
| Sarah | 1,018,684 | <p>As pointed out in the comments, using the fundamental theorem of calculus we have that <span class="math-container">$F(x)-F(0)=0$</span> where <span class="math-container">$F'(x)=f(x)$</span>. Hence <span class="math-container">$F(x)=F(0)$</span> for all <span class="math-container">$x$</span>, which means <span class="math-container">$F(x)$</span> is a constant function, and hence <span class="math-container">$f(x)=F'(x)=0$</span> for all <span class="math-container">$x$</span>.</p>
|
4,398,873 | <p>Given f: (0,1) <span class="math-container">$\rightarrow$</span> <span class="math-container">$\mathbb R$</span>. f is a continuous function and improper integrable.</p>
<p>If <span class="math-container">$\int_{0}^{x}f(t)dt = 0$</span> <span class="math-container">$\forall x \in [0,1] $</span>,</p>
<p>Does <span class="math-container">$f(x) = 0 $</span> <span class="math-container">$\forall x \in [0,1] $</span> ?</p>
| B. S. Thomson | 281,004 | <p><strong>Can we appeal to the Fundamental Theorem of the Calculus?</strong></p>
<p>Certainly ... if you can find one in your textbook for <em>improper integrals</em>. Many textbooks make no such statement, assuming that you can use the regular version as needed.</p>
<p>If you cannot find one, then prove a simple one by basing the proof on the usual Fundamental theorem.</p>
<p><strong>THEOREM</strong>. Suppose that <span class="math-container">$f:(a,b)\to \mathbb R$</span> and that <span class="math-container">$f$</span> is integrable in the improper Riemann sense on <span class="math-container">$ (a,b)$</span>. Then,<br />
<span class="math-container">$$\frac{d}{dx} \int_{a}^x f(t)\,dt = f(x)$$</span>
at every point <span class="math-container">$x\in (a,b)$</span> at which <span class="math-container">$f$</span> is continous.</p>
<p>Proof. If <span class="math-container">$a<x_0<x<b$</span> then
<span class="math-container">$$ \int_{a}^x f(t)\,dt = \int_a^{x_0} f(t)\,dt + \int_{x_0}^x f(t)\,dt $$</span>
where the first integral on the right is improper Riemann and the second is ordinary Riemann. By the usual FTC, since <span class="math-container">$f$</span> is Riemann integrable here,
<span class="math-container">$$\frac{d}{dx} \int_{x_0}^x f(t)\,dt = f(x)$$</span>
at every point <span class="math-container">$x\in (a,b)$</span> at which <span class="math-container">$f$</span> is continous. Etc.</p>
<p><strong>Comment</strong>. Is this pedantic? Of course. But a theorem proved for the Riemann integral is not <em>automatically</em> true for the improper Riemann integral. One checks!</p>
|
587,878 | <p>Let $X$ be a reflexive Banach space of infinite dimension. </p>
<p>a) Prove that there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.</p>
<p>b) Let $x_n$ be a sequence such that $\forall f \in X' \quad \exists \lim\limits_{n\to\infty} f(x_n)<\infty$ .Prove that $x_n$ converges weakly.</p>
<p>c) Find a Banach space non riflexive where b is false.</p>
<p>My idea:
the weak closure of $S^1$ is the closed unit ball, thus $0\in$ the weak closure of $S^1$. But to prove that there exists a sequence in $S^1$ that converges weakly to 0 we must prove that the weak topology is metrizzable.Is it always true in riflexive spaces?</p>
<p>I would appreciate any idea on the solution.Thank you in advance.</p>
| gtoques | 638,316 | <p>What you said is not true in general, but there's a way around it. Find a closed subspace <span class="math-container">$S \subset X$</span> which is separable (and it is reflexive also by a common result). To do this, pick <span class="math-container">$\{x_{n}\}_{n \in \mathbb{N}}$</span> linearly independent vectors in <span class="math-container">$X$</span> and consider the subspace spanned by them. This should be separable (why? considering rational scalings of all <span class="math-container">$\{x_{n}\}_{n \in \mathbb{N}}$</span>). Now since <span class="math-container">$S$</span> is separable and reflexive, <span class="math-container">$S^{*}$</span> is separable (this is a theorem, not trivial), which shows that <span class="math-container">$B_{X} \cap S$</span> is metrizable. Apply your argument to <span class="math-container">$B_{X} \cap S$</span>.</p>
|
276,948 | <p>$a_n$ be a sequence of integers such that such that infinitely many terms are non zero, we need to show that either the power series $\sum a_n x^n$ converges for all $x$ or Radius of convergence is at most $1$. need some hint. thank you.</p>
| Davide Giraudo | 9,849 | <blockquote>
<p><strong>Claim:</strong> let <span class="math-container">$\{a_n\}$</span> a sequence of integers. The radius of convergence of <span class="math-container">$\sum_{n=1}^{+\infty}a_nx^n$</span> is</p>
<ul>
<li>infinite if <span class="math-container">$a_n=0$</span> when <span class="math-container">$n$</span> is large enough;</li>
<li>at most <span class="math-container">$1$</span> if infinitely many <span class="math-container">$a_n$</span> are different from <span class="math-container">$0$</span>.</li>
</ul>
</blockquote>
<p>The first case is trivial.</p>
<p>If <span class="math-container">$|z|>1$</span> and <span class="math-container">$n$</span> is such that <span class="math-container">$a_n\neq 0$</span>, as <span class="math-container">$a_n$</span> is integer we have that <span class="math-container">$|a_n|=1$</span> so <span class="math-container">$|a_nz^n|\geqslant |z|^n$</span>. As it occurs for infinitely many terms, this proves that the sequence <span class="math-container">$\{a_nz^n\}$</span> is not bounded whenever <span class="math-container">$|z|\geqslant 1$</span>, hence the radius of convergence is at most <span class="math-container">$1$</span>.</p>
|
2,475,938 | <blockquote>
<p>How can I factor the polynomial <span class="math-container">$x^4-2x^3+x^2-1$</span>?</p>
</blockquote>
<p>This is an exercise in algebra. I have the solution showing that
<span class="math-container">$$
x^4-2x^3+x^2-1=(x^2-x-1)(x^2-x+1).
$$</span></p>
<p>But the solution does not show any details. Using the distributive property I can check that this is indeed true:
<span class="math-container">$$
\begin{align}
&(x^2-x-1)(x^2-x+1)\\
&=x^2(x^2-x+1)-x(x^2-x+1)-(x^2-x+1)\\
&=x^4-x^3+x^2-x^3+x^2-x-x^2+x-1\\
&=x^4-2x^3+x^2-1,
\end{align}
$$</span></p>
<p>but I can't figure out the steps to get there. Can anyone help?</p>
| Mr Pie | 477,343 | <p>Remember that for all values <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, <span class="math-container">$$(a + b)^2 = (a + b)(a + b)$$</span> which we then <a href="http://www.mathwords.com/f/foil_method.htm" rel="nofollow noreferrer"><em>FOIL</em></a> to work out that <span class="math-container">$$(a + b)^2 = a^2 + b^2 + 2ab$$</span></p>
<p>Because the last term on the furthest <span class="math-container">$RHS$</span> (<em>Right Hand Side</em> of the equation) being <span class="math-container">$2ab$</span> is a multiple of <span class="math-container">$2$</span>, if you have an equation for example:</p>
<p><span class="math-container">$$x^2 + 6x = 7$$</span></p>
<p>What you do to solve for <span class="math-container">$x$</span> is <em>divide</em> <span class="math-container">$6$</span> by <span class="math-container">$2$</span> and then add the square of the result.</p>
<p><span class="math-container">$$x^2 + 6x + \bigg(\frac{6}{2}\bigg)^2 - 9 = 7$$</span></p>
<p>Here we also subtract <span class="math-container">$9$</span> since <span class="math-container">$\dfrac{6}{2} = 3$</span> and <span class="math-container">$3^2 = 9$</span>, because we do not want to change the <span class="math-container">$RHS$</span>. And now, we can factor the <span class="math-container">$LHS$</span> (<em>Left Hand Side</em> of the equation):</p>
<p><span class="math-container">$$\underbrace{x^2 + 6x + 9}_{(x + 3)^2} - 9 = 7 \Rightarrow (x + 3)^2 - 9 = 7 \Rightarrow (x + 3)^2 = 7 + 9 = 16 \Rightarrow x + 3 = \pm \sqrt{16} = \pm 4 \ \ \ \ \ \therefore \ \ x = 1 \quad \text{ or } \quad x = -7$$</span></p>
<p>You can also solve for <span class="math-container">$x$</span> using the <a href="https://en.m.wikipedia.org/wiki/Quadratic_formula" rel="nofollow noreferrer"><em>Quadratic Formula</em></a> as an alternative method by re-writing the equation as <span class="math-container">$x^2 + 6x - 7 = 0$</span>, but this does not help us know how to factor. The technique that I showed you instead, however, is known as <a href="http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/quadequationshirev2.shtml" rel="nofollow noreferrer"><em>Completing the Square</em></a>.</p>
<hr>
<p>So now let's see where your equation fits. We have <span class="math-container">$$x^4 - 2x^3 + x^2 - 1$$</span> Since <span class="math-container">$(a + b)^2 = a^2 + b^2 + 2ab \Rightarrow (a - b)^2 = a^2 + b^2 - 2ab$</span>, we can rearrange the terms. <span class="math-container">$$x^4 - 2x^3 + x^2 - 1 = \underbrace{(x^2)^2 + x^2 - 2\cdot x^2\cdot x}_{(x^2 - x)^2} - 1 = (x^2 - x)^2 - 1$$</span> Now we can <a href="https://mathspace.co/learn/world-of-maths/algebra/distributive-law-non-linear-terms-6702/expanding-brackets-295/" rel="nofollow noreferrer"><em>expand the bracket</em></a> of <span class="math-container">$x^2 - x$</span> (simply, <em>expand</em> <span class="math-container">$x^2 - x$</span>) so that <span class="math-container">$x^2 - x = x(x - 1)$</span>, thus we have <span class="math-container">$$[x(x - 1)]^2 - 1$$</span> And now we can actually use a seperate rule: <span class="math-container">$$a^2 - b^2 = (a + b)(a - b)$$</span> since <span class="math-container">$1 = 1^2$</span>. <span class="math-container">$$\begin{align} \therefore [x(x - 1)]^2 - 1 &= [x(x - 1)]^2 - 1^2 \\ &= \big[[x(x - 1)] + 1\big]\big[[x(x - 1) - 1\big] \\ &= (x^2 - x + 1)(x^2 - x - 1) \\ &= (x^2 - x - 1)(x^2 - x + 1) \qquad \color{green}{\checkmark} \end{align}$$</span>
Remember that these rules are called <em>Binomial Formulas</em> (or <em>Formulae</em>) derived from the <a href="http://www.purplemath.com/modules/binomial.htm" rel="nofollow noreferrer"><em>Binomial Theorem</em></a> (or go <a href="https://www.mathsisfun.com/algebra/binomial-theorem.html" rel="nofollow noreferrer">here</a> instead if you found the previous link too difficult to understand) and there you go! In fact, since the equation you want to factor has <em>all</em> of its terms being a product of <span class="math-container">$x$</span> (apart from <span class="math-container">$1$</span> of course), you can expand the entire equation and re-write it as <span class="math-container">$x\cdot c - 1$</span> for some value <span class="math-container">$c$</span>. Then use the binomial formulas to factor <span class="math-container">$c$</span> and go from there! So there are two ways to complete the problem! </p>
<p>After you have completed the problem yourself, try and factor <span class="math-container">$a^3 - b^3$</span> to improve your knowledge! If you don't know, tap/click the yellow box below for the answer or go <a href="https://www.tiger-algebra.com/drill/a~3-b~3=0/" rel="nofollow noreferrer">here</a>!</p>
<blockquote>
<blockquote class="spoiler">
<p> <span class="math-container">$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$</span></p>
</blockquote>
</blockquote>
<p>Another way to factor <span class="math-container">$a^3 - b^3$</span> that is seperate to the link provided is shown below if needed.</p>
<blockquote class="spoiler">
<p> <span class="math-container">$$\frac{a^3 - b^3}{a - b} = \frac{a^3 - ba^2 + ba^2 - b^3}{a - b} = a^2 + \frac{ba^2 - b^3}{a - b} = a^2 + ab + \frac{b^2a - b^3}{a - b} = a^2 + ab + b^2$$</span></p>
</blockquote>
<p>:)</p>
|
4,065,797 | <p>Just to give a simple numerical example but in general the variables <span class="math-container">$x,y,z,u,v$</span> are not equal.</p>
<p><span class="math-container">$113= 2*4^2 + 2*4^2 +2*4^2 + 4^2 +1^2$</span></p>
<p>I am looking for a general method to solve this type of equation or a piece of software to do the same. I already looked in this site for methods that could help but could not find anything dealing with this ind of case.</p>
<p><strong>Question 2</strong> It is also useful to know if there is a test that can tell if the equation does not have a solution.</p>
| giobrach | 332,594 | <p>As was hinted at in @Quanto's answer, we have the Wallis integral
<span class="math-container">$$W(\alpha) := \int_0^{\pi/2} \sin^\alpha(x)\ dx = \frac{\sqrt \pi}{2} \frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)}, $$</span>
for all <span class="math-container">$\alpha \geqslant 0$</span>. Differentiating <span class="math-container">$W$</span> with respect to <span class="math-container">$\alpha$</span>, we find (by the symmetry of the integrand)
<span class="math-container">$$\frac{dW}{d\alpha} (\alpha) = \int_0^{\pi/2} \sin^{\alpha}(x) \log(\sin(x))\ dx = \frac 1 2 I(\alpha), $$</span>
so that, setting <span class="math-container">$g(\alpha) := \Gamma\left(\dfrac{\alpha+1}{2}\right)\Big/\Gamma\left(\dfrac \alpha 2 + 1\right)$</span>,
<span class="math-container">$$\begin{split}
\frac{1}{\sqrt{\pi}}I(\alpha) &= \frac{dg}{d\alpha} (\alpha) = \frac{d}{d\alpha}e^{\log g(\alpha)} = e^{\log g(\alpha)} \frac{d}{d\alpha}\log(g(\alpha)) \\
&=g(\alpha) \frac{d}{d\alpha}\left\{ \log \Gamma\left(\frac{\alpha+1}{2}\right) -\log\Gamma\left(\frac \alpha 2 + 1\right)\right\} \\
&= \frac 1 2\frac{\Gamma\left(\frac{\alpha+1}{2}\right)}{\Gamma\left(\frac \alpha 2 + 1\right)} \left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\},
\end{split}$$</span>
where <span class="math-container">$\psi^{(0)}(z)=(\log \Gamma)'(z)$</span> is the <a href="https://mathworld.wolfram.com/PolygammaFunction.html" rel="nofollow noreferrer">digamma function</a>.</p>
<hr />
<p><strong>Bonus.</strong> We can go further. Using the identity <span class="math-container">$\psi^{(0)}(t+1) = \psi^{(0)}(t) + 1/t$</span> for all <span class="math-container">$t>0$</span>, and taking advantage of the fact that
<span class="math-container">$$\psi^{(0)}\left(\frac{s+1}{2s} \right) - \psi^{(0)}\left(\frac{1}{2s} \right) = \sum_{k=0}^\infty \frac{(-1)^k}{sk+1}$$</span>
for <span class="math-container">$s>0$</span>, choosing <span class="math-container">$s = (\alpha+2)^{-1}$</span> we find
<span class="math-container">$$\frac{\alpha+2}{2}\left\{\psi^{(0)}\left(\frac{\alpha+1}{2}\right) -\psi^{(0)}\left(\frac \alpha 2 + 1\right) \right\} = - \frac{\alpha+2}{\alpha+1} + \sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1}, $$</span>
whence
<span class="math-container">$$\begin{split}
I(\alpha) &= \sqrt\pi g(\alpha)\left\{-\frac1{\alpha+1} + \frac{1}{\alpha+2}\sum_{k=0}^\infty \frac{(-1)^k}{k/(\alpha+2)+1}\right\}\\
&=\sqrt\pi g(\alpha) \sum_{k=1}^\infty \frac{(-1)^k}{\alpha+k}.
\end{split}$$</span>
Now we restrict ourselves to nonnegative integers <span class="math-container">$\alpha = n \geqslant 1$</span>, and we observe that
<span class="math-container">$$\sum_{k=0}^\infty \frac{(-1)^k}{n+k} = \sum_{j=n}^\infty \frac{(-1)^{j-n}}{j} = (-1)^{n+1} \sum_{j=n}^\infty \frac{(-1)^{j+1}}{j} =(-1)^{n+1}\left[-\tilde H_{n} + \log(2) \right], $$</span>
where <span class="math-container">$\tilde H_{n} = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$</span> is the <span class="math-container">$n$</span>-th alternating <a href="https://mathworld.wolfram.com/HarmonicNumber.html" rel="nofollow noreferrer">harmonic number</a>. we obtain
<span class="math-container">$$\boxed{I(n) = (-1)^n\sqrt\pi \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac n 2 + 1\right)} \big( \tilde H_n - \log(2)\big).}$$</span>
One may then distinguish even-<span class="math-container">$n$</span> and odd-<span class="math-container">$n$</span> cases and employ the duplication formula for <span class="math-container">$\Gamma$</span> to express <span class="math-container">$I(n)$</span> in terms of factorials and double factorials.</p>
|
849,583 | <p>I have several derivatives to find:</p>
<blockquote>
<p>For <span class="math-container">$g(s)=3s^3-s+4$</span>, <span class="math-container">$g'(s)=$</span></p>
<p>For <span class="math-container">$p(t)=\frac1t+t^2$</span>, <span class="math-container">$\frac{\mathrm dp}{\mathrm dt}=$</span></p>
<p>For <span class="math-container">$w(u)=\sqrt u-2u^2-10$</span>, <span class="math-container">$\frac{\mathrm dw}{\mathrm du}=$</span></p>
</blockquote>
<p>Each has multiple terms, and as a result is difficult to determine how to solve.<br />
I need to find out how to go about solving these, for example I can't use the power rule because these functions have more than just one term.</p>
<p>I am looking for insight on what steps to take.</p>
| Hakim | 85,969 | <p><strong>Hint:</strong> <br> Derivatives are linear, i.e. $$\dfrac{\mathrm d}{\mathrm dx}\Big[f(x)+g(x)\Big]=\dfrac{\mathrm d}{\mathrm dx}f(x)+\dfrac{\mathrm d}{\mathrm dx}g(x).$$
As an example: let $h(x)=x^2-x$, we then have: $$h'(x)=\dfrac{\mathrm d}{\mathrm dx}\Big[x^2-x\Big]=\dfrac{\mathrm d}{\mathrm dx}x^2+\dfrac{\mathrm d}{\mathrm dx}(-x)=2x-1.$$</p>
|
1,687,868 | <p>This might be a really obvious question so I apologize in advance, but I'm having trouble seeing when matrices are commutative for general nxn matrices. For example, when proving tr(AB)=tr(BA), I can easily prove this in a 2x2 matrix but I'm getting confused for proving it in a nxn matrix.</p>
<p>I've searched online and the recurring solution that comes up is:</p>
<p><span class="math-container">$$ \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{nk}B_{kn} = \sum_{k=1}^{n} \sum_{i=1}^{n} B_{kn}A_{nk} = \sum_{i=1}^{n} (BA)_{ii} $$</span></p>
<p>How come we're able to switch the sums in line 3? I originally tried solving the question by trying to do TR(AB) and TR(BA) separately:</p>
<p><span class="math-container">$$ \mbox{tr}(AB) = \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{nk}B_{kn} $$</span></p>
<p><span class="math-container">$$ \mbox{tr}(BA) = \sum_{i=1}^{n} (BA)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} B_{nk}A_{kn} $$</span></p>
<p>I would greatly appreciate it if someone could perhaps point out where my reasoning went wrong.</p>
| Salcio | 821,280 | <p>Of course one can prove this directly as above.
Another proof is to assume first that <span class="math-container">$A$</span>, and <span class="math-container">$B$</span> are invertable (which is fine since in general case one can approximate any matrix by an invertable one) and check that <span class="math-container">$det(AB-\lambda *I) = det(BA-\lambda*I)$</span>. In words, that characteristic polynomials of <span class="math-container">$AB$</span> and <span class="math-container">$BA$</span> are the same (this is an interesting fact on its own).
Since trace is the second coefficient of characteristic polynomial we get what we want. Now, <span class="math-container">$det(AB-\lambda*I) = det(AB-\lambda*B^{-1}B) = det(A-\lambda* B^{-1})*det(B) = det(B^{-1}BA-\lambda*B^{-1})*det(B) = det(B^{-1})det(BA-\lambda*I)*det(B) = det(BA-\lambda*I)$</span></p>
|
3,377,353 | <p>Given a purely real, rational integer <span class="math-container">$p$</span> that is prime in <span class="math-container">$\mathbb{Z}$</span>, we know very well that it ramifies in <span class="math-container">$\mathbb{Q}(\sqrt{pm})$</span> (where <span class="math-container">$m$</span> is a nonzero integer coprime to <span class="math-container">$p$</span>), it is inert in some of the other quadratic rings and it splits in the others.</p>
<p>In a ring of degree <span class="math-container">$4$</span>, things are of course a bit more complicated than that. For example, in <span class="math-container">$\mathcal{O}_{\mathbb{Q}(\zeta_8)}$</span>, we see that <span class="math-container">$2$</span> is ramified, since it's ramified in each of the three intermediate fields (<span class="math-container">$\mathbb{Q}(i)$</span>, <span class="math-container">$\mathbb{Q}(\sqrt{-2})$</span> and <span class="math-container">$\mathbb{Q}(\sqrt{2})$</span>).</p>
<p>Furthermore, we see that <span class="math-container">$(1 - \zeta_8)(1 + \zeta_8) = 1 - i$</span> and <span class="math-container">$(1 - {\zeta_8}^3)(1 + {\zeta_8}^3) = 1 + i$</span>. Is this what they call "ramifies completely"?</p>
<p>Turning our attention to <span class="math-container">$3$</span>, we see that it is not prime in <span class="math-container">$\mathcal{O}_{\mathbb{Q}(\zeta_8)}$</span>, because, although it does not split in two of the intermediate fields, it does split in <span class="math-container">$\mathbb{Z}[\sqrt{-2}]$</span>. I may have overlooked something, but as far as I can tell, the equation <span class="math-container">$x^4 + b x^3 + c x^2 + d x \pm 3 = 0$</span> has no solutions in <span class="math-container">$\mathcal{O}_{\mathbb{Q}(\zeta_8)}$</span>.</p>
<p>If I'm right, this would mean that <span class="math-container">$3$</span> does not split as "completely" as <span class="math-container">$2$</span> ramifies. Assuming I'm correct in these assertions, am I using the correct terminology? And if not, what is the correct terminology?</p>
| Bob Happ | 215,060 | <p>The terminology of "ramify" and "split" carries over, but now needs the adjectives "completely" and [placeholder for others to add appropriate adjectives].</p>
<p>The ring of algebraic integers of <span class="math-container">$\mathbb{Q}(\sqrt{2} + \sqrt{3})$</span> provides richer examples, in my opinion. We see that <span class="math-container">$2$</span> ramifies, since it ramifies in each of the intermediate fields (<span class="math-container">$\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3}), \mathbb{Q}(\sqrt{6})$</span>; if the second one of these gives you any doubt, remember that, in <span class="math-container">$\mathbb{Z}[\sqrt{3}]$</span>, <span class="math-container">$\langle 1 - \sqrt{3} \rangle \subseteq \langle 1 + \sqrt{3} \rangle$</span> and vice-versa, so in fact <span class="math-container">$\langle 1 - \sqrt{3} \rangle = \langle 1 + \sqrt{3} \rangle$</span> and <span class="math-container">$\langle 2 \rangle$</span> is a square ideal).</p>
<p>Thus we say that <span class="math-container">$2$</span> "ramifies completely." Furthermore, the [quartic integer of norm <span class="math-container">$2$</span> I haven't found it yet]</p>
<p>The situation with <span class="math-container">$3$</span> is a bit more interesting than that. Obviously <span class="math-container">$3$</span> ramifies in <span class="math-container">$\mathbb{Z}[\sqrt{3}]$</span> and <span class="math-container">$\mathbb{Z}[\sqrt{6}]$</span>, but it's actually inert. Thus we say that <span class="math-container">$3$</span> ramifies in a [someone else please fill in right term]</p>
|
3,225,553 | <p>Show that <span class="math-container">$4x^2+6x+3$</span> is a unit in <span class="math-container">$\mathbb{Z}_8[x]$</span>.</p>
<p>Once you have found the inverse like <a href="https://math.stackexchange.com/questions/3172556/show-that-4x26x3-is-a-unit-in-mathbbz-8x">here</a>, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = <span class="math-container">$1$</span> imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?</p>
| Cornman | 439,383 | <p>To find an inverse polynomial for that holds <span class="math-container">$p(x)(4x^2+6x+3)=1$</span> so it has to be <span class="math-container">$3y=1\mod 8$</span> [edit: For more context on <span class="math-container">$y$</span>, see the comments below]. So <span class="math-container">$y=3$</span> and the polynomial might look like this:</p>
<p><span class="math-container">$p(x)=(ax+3)$</span></p>
<p>Then <span class="math-container">$(4x^2+6x+3)(ax+3)=4ax^3+(6a+12)x^2+(3a+18)x+9$</span>. Now it has to be <span class="math-container">$4a\equiv 0\mod 8$</span> and <span class="math-container">$6a+12\equiv 0\mod 8$</span> and <span class="math-container">$3a+18\equiv 0\mod 8$</span>.</p>
<p>Is there such an <span class="math-container">$a$</span>?.
Yes indeed. For <span class="math-container">$a=2$</span> we have <span class="math-container">$8\equiv 0\mod 8$</span></p>
<p><span class="math-container">$24\equiv 0\mod 8$</span> and <span class="math-container">$24\equiv 0\mod 8$</span>.</p>
<p>If we would fail to find this <span class="math-container">$a$</span> in this step, we would have to try with <span class="math-container">$p(x)=(ax^2+bx+3)$</span> and proceed as above, which gets more and more complicated.</p>
<p>So it is <span class="math-container">$(4x^2+6x+3)(2x+3)\equiv 1\mod 8$</span></p>
|
3,225,553 | <p>Show that <span class="math-container">$4x^2+6x+3$</span> is a unit in <span class="math-container">$\mathbb{Z}_8[x]$</span>.</p>
<p>Once you have found the inverse like <a href="https://math.stackexchange.com/questions/3172556/show-that-4x26x3-is-a-unit-in-mathbbz-8x">here</a>, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = <span class="math-container">$1$</span> imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?</p>
| lhf | 589 | <p>Write
<span class="math-container">$$
4x^2+6x+3 = 3(4x^2+2x+1) = 3((2x)^2+(2x)+1) = 3 \frac{(2x)^3-1}{2x-1} = \frac{-3}{2x-1}
$$</span>
Therefore,
<span class="math-container">$$
\frac{1}{4x^2+6x+3} = \frac{2x-1}{-3} = 3(1-2x) = 3-6x = 2x+3
$$</span></p>
|
3,577,249 | <p>Let be <span class="math-container">$X$</span> a topological space and let be <span class="math-container">$Y\subseteq X$</span> such that <span class="math-container">$\mathscr{der}(Y)=\varnothing$</span>: so for any neighborhood <span class="math-container">$I_y$</span> of <span class="math-container">$y\in Y $</span> it result that <span class="math-container">$I_y\cap Y={y}$</span> and so Y is a set of isolated point. Now of <span class="math-container">$Y$</span> is a set of isolated point clearly none of its point could be a accumulation point: anyway if <span class="math-container">$x\notin Y$</span> how prove that for any its neighborhood <span class="math-container">$I_x$</span> it result that <span class="math-container">$I_x\cap Y=\varnothing$</span>?</p>
<p>Could someone help me, please?</p>
| Soma | 91,108 | <p>It is not possible. Let us consider <span class="math-container">$\mathbb{N}$</span> endowed with the discrete topology and let <span class="math-container">$X=\mathbb{N}\cup \{\infty\}$</span> as the one point compactification of such a space (<span class="math-container">$X$</span> is homeomorphic to a convergent sequence). Let <span class="math-container">$Y=\mathbb{N}\subset X$</span>. Since the subspace topology on <span class="math-container">$Y$</span> is equivalent to the discrete topology, all the points of <span class="math-container">$Y$</span> are isolated points. On the other hand the limit point <span class="math-container">$\{\infty\}$</span> is an accumulation point for <span class="math-container">$Y$</span>, therefore each neighborhood of <span class="math-container">$\{\infty\}$</span> intersects <span class="math-container">$Y$</span>.</p>
|
2,378,508 | <p>I am reading about Arithmetic mean and Harmonic mean. From <a href="https://en.wikipedia.org/wiki/Harmonic_mean#In_physics" rel="nofollow noreferrer">wikipedia</a>
I got this comparision about them:</p>
<blockquote>
<p>In certain situations, especially many situations involving rates and ratios, the harmonic mean provides the truest average. For instance, if a vehicle travels a certain distance at a speed x (e.g., 60 kilometres per hour -
km
/
h
) and then the same distance again at a speed y (e.g., 40
km
/
h
), then its average speed is the harmonic mean of x and y (48
km
/
h
), and its total travel time is the same as if it had traveled the whole distance at that average speed. However, if the vehicle travels for a certain amount of time at a speed x and then the same amount of time at a speed y, then its average speed is the arithmetic mean of x and y, which in the above example is 50 kilometres per hour. The same principle applies to more than two segments: given a series of sub-trips at different speeds, if each sub-trip covers the same distance, then the average speed is the harmonic mean of all the sub-trip speeds; and if each sub-trip takes the same amount of time, then the average speed is the arithmetic mean of all the sub-trip speeds. </p>
</blockquote>
<pre><code> distance time velocity remark
1st section d/2 t1 60
2nd section d/2 t2 40
1st + 2nd section d (t1+t2) v use harmonic mean to calculate v
1st section d1 t/2 60
2nd section d2 t/2 40
1st + 2nd section d1+d2 t v use arithmetic mean to calculate v
</code></pre>
<p>How <code>distance</code> and <code>time</code> are pushing us to compute harmonic mean and arithmetic mean respectively for computing "average <code>v</code>" in this case?</p>
| Vaas | 291,134 | <blockquote>
<p>Polynomials: These are functions which can be written in the form <span class="math-container">$f(x) = a_0 +a_1x+···+a_dx^d$</span> for some numbers <span class="math-container">$a_j, 0 ≤ j ≤ d, d ∈ N$</span>. If <span class="math-container">$a_d \neq 0$</span>, we say that f is a polynomial of degree d with leading coefficient <span class="math-container">$a_d$</span>. The maximal domain of any polynomial is R. We say that r is a root or a zero of the polynomial f if <span class="math-container">$f(r) = 0$</span>.</p>
</blockquote>
<p>Where as</p>
<blockquote>
<p>A differential equation is an equation of the form <span class="math-container">$F(x,y,y',...,y^{n})= 0, x ∈ I$</span>, with the following properties:</p>
<p>(1) n ∈ N is the order of the equation.</p>
<p>(2) F is a given function of n+2 variables.</p>
<p>(3) y is the unknown function of the independent variable x.</p>
<p>(4) I is the interval in which the equation is to hold.</p>
</blockquote>
<p>Therefore <span class="math-container">$\sum^n_{j=1} (y'')^{j}=0$</span> is still of degree/order 2</p>
<p>(Both extracts are lifted from lecture notes)</p>
<p>Further if were talking about degrees of a differential equation we talk about the highest derivative. If we talk about degrees of a polynomial we talk about the exponent.</p>
<p>Btw for the equation <span class="math-container">$x^{4}=x^{9}$</span> where <span class="math-container">$x=y''$</span>it is not in its simplist form</p>
<p>Thus <span class="math-container">$1 = x^{5} \Rightarrow x=1$</span></p>
|
3,536,135 | <blockquote>
<p>There are five multiple choice questions on a test, with four choices per question. A student was given 10 questions to study for the test and the teacher picked 5 out of 10 questions to put on the test. The student memorizes 7 of the 10 answers of the questions given. If the student encounters the three questions the student does not know the answer to, the four choices will be equally likely to be guessed upon.
<p>a. What is the probability the student gets the first question right?
<p>b. What is the probability the student gets a 100 on the test?</p>
</blockquote>
<p><p> I think I got the first part. Let A be the event the student gets the first question correct. For all questions to be rewarded a point, the probability is <span class="math-container">$\frac{7}{10}$</span> since she knows 7 of the the 10 questions. So we will call question one X1 and question two X2 and so on..
<p>So we have <span class="math-container">$P(A)=P(A|X1=1)P(X1=1)+P(A|X1=0)P(X1=0)$</span>=<span class="math-container">$(1)$</span>(<span class="math-container">$\frac{7}{10}$</span>)+(<span class="math-container">$\frac{1}{4}$</span>)(<span class="math-container">$\frac{3}{10}$</span>)<span class="math-container">$=.7+.075=.775$</span>
<p> The second question I am more confused on. Let C1 mean question one is correct C2 question 2 is correct etc.
<p>P(C1)=P(C2)=...=P(C5)=<span class="math-container">$\frac{31}{40}$</span>
<p> We will assume the C's are independent and do <span class="math-container">$(\frac{31}{40})^5$</span>. After here I am not sure what to do and I am not sure if I am going on the right track. If somebody can guide me on the right track if the work looks good so far that be appreciated!</p>
| kccu | 255,727 | <p>Your answer to the first question is correct, but your notation is poor. You say <span class="math-container">$X1$</span> denotes question one, so what does it mean to write <span class="math-container">$X1=1$</span>? Instead I would define <span class="math-container">$B$</span> to be the event that question 1 is one of the 7 the student studied, and <span class="math-container">$B^C$</span> to be the event that question 1 is one of the 3 unknown questions. Then
<span class="math-container">$$P(A)=P(A|B)P(B)+P(A|B^C)P(B^C) = 1 \cdot \frac{7}{10} + \frac{1}{4}\cdot \frac{3}{10} = 0.775.$$</span></p>
<p>For the second question, you're off to a good start, but we cannot assume the questions are independent. In fact, we know they are NOT independent. For instance, if the student gets questions 1, 2, and 3 all wrong, then those must have been the unknown questions and so they have to get questions 4 and 5 right.</p>
<p>Instead, let <span class="math-container">$E$</span> be the event that the student gets 100%. We can condition on the number of unkown questions <span class="math-container">$N$</span> the student sees. They either see <span class="math-container">$0$</span>, <span class="math-container">$1$</span>, <span class="math-container">$2$</span>, or <span class="math-container">$3$</span> unknown questions, so:
<span class="math-container">$$P(E)=P(E|N=0)P(N=0)+P(E|N=1)P(N=1)+P(E|N=2)P(N=2)+P(E|N=3)P(N=3).$$</span>
So now you need to calculate <span class="math-container">$P(E|N=k)$</span> and <span class="math-container">$P(N=k)$</span> for <span class="math-container">$k=0,1,2,3$</span>. As an example, I'll do <span class="math-container">$k=3$</span>. Given that the student encounters <span class="math-container">$N=3$</span> unknown questions, the probability they get 100% is <span class="math-container">$1^2\left(\frac{1}{4}\right)^3 = \frac{1}{64}$</span>. The probability they see <span class="math-container">$N=3$</span> unknown questions is <span class="math-container">$\frac{\binom{3}{3}\binom{7}{2} }{\binom{10}{5}}=\frac{21}{252}=\frac{1}{12}$</span> since there are <span class="math-container">$\binom{3}{3}$</span> ways for the teacher to choose the unknown questions, <span class="math-container">$\binom{7}{2}$</span> ways to choose the known questions, and <span class="math-container">$\binom{10}{5}$</span> total ways to choose 5 questions for the test. Can you do the rest now?</p>
|
75,005 | <p>Let's imagine a guy who claims to possess a machine that can each time produce a completely random series of 0/1 digits (e.g. $1,0,0,1,1,0,1,1,1,...$). And each time after he generates one, you can keep asking him for the $n$-th digit and he will tell you accordingly.</p>
<p>Then how do you check if his series is <em>really completely random</em>?</p>
<p>If we only check whether the $n$-th digit is evenly distributed, then he can cheat using:</p>
<blockquote>
<p>$0,0,0,0,...$<br>
$1,1,1,1,...$<br>
$0,0,0,0,...$<br>
$1,1,1,1,...$<br>
$...$</p>
</blockquote>
<p>If we check whether any given sequence is distributed evenly, then he can cheat using:</p>
<blockquote>
<p>$(0,)(1,)(0,0,)(0,1,)(1,0,)(1,1,)(0,0,0,)(0,0,1,)...$<br>
$(1,)(0,)(1,1,)(1,0,)(0,1,)(0,0,)(1,1,1,)(1,1,0,)...$<br>
$...$</p>
</blockquote>
<p>I may give other possible checking processes but as far as I can list, each of them has flaws that can be cheated with a prepared regular series.</p>
<p>How do we check if a series is really random? Or is randomness a philosophical concept that can not be easily defined in Mathematics?</p>
| cyborg | 17,917 | <p>Leaving aside the theoretical aspect of your question, there are also pragmatic answers to it because there are real world uses for high-quality <a href="http://en.wikipedia.org/wiki/Random_number_generation" rel="nofollow">random generators</a> (whether hardware or algorithmic). For statistical uses, <a href="http://en.wikipedia.org/wiki/Statistical_randomness" rel="nofollow">"statistical randomness"</a> is a used. For example you can use these <a href="http://en.wikipedia.org/wiki/Diehard_tests" rel="nofollow">"diehard tests"</a> or <a href="http://en.wikipedia.org/wiki/TestU01" rel="nofollow">TestU01</a>.</p>
|
281,735 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/202452/why-is-predicate-all-as-in-allset-true-if-the-set-is-empty">Why is predicate “all” as in all(SET) true if the SET is empty?</a> </p>
</blockquote>
<p>In don't quite understand this quantification over the empty set:</p>
<p>$\forall y \in \emptyset: Q(y)$</p>
<p>The book says that this is always TRUE regardless of the value of the predicate $Q(y)$, and it explain that this is because this quantification adds no predicate at all, and therefore can be considered the weakest predicate possible, which is TRUE.</p>
<p>I know that TRUE is the weakest predicate because $ $P$ \Rightarrow$ TRUE is TRUE for every $P$.
I don't see what is the relationship between this weakest predicate and the quantification. </p>
| zrbecker | 19,536 | <p>Well if $\forall y \in \emptyset : Q(y)$ were false, we would be able to find some $y \in \emptyset$ such that $Q(y)$ were false. However, there are no $y \in \emptyset$. So $\forall y \in \emptyset : Q(y)$ should be true.</p>
|
281,735 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/202452/why-is-predicate-all-as-in-allset-true-if-the-set-is-empty">Why is predicate “all” as in all(SET) true if the SET is empty?</a> </p>
</blockquote>
<p>In don't quite understand this quantification over the empty set:</p>
<p>$\forall y \in \emptyset: Q(y)$</p>
<p>The book says that this is always TRUE regardless of the value of the predicate $Q(y)$, and it explain that this is because this quantification adds no predicate at all, and therefore can be considered the weakest predicate possible, which is TRUE.</p>
<p>I know that TRUE is the weakest predicate because $ $P$ \Rightarrow$ TRUE is TRUE for every $P$.
I don't see what is the relationship between this weakest predicate and the quantification. </p>
| Victor Jose Arana Rodriguez | 54,341 | <p>I think I got the idea, but it confuses me when I compare this reasoning with the existencial quantifier over the empty set
$ \exists x \in \emptyset : P(x) $</p>
<p>I think this is TRUE because I'm quantifying over the empty set too.</p>
|
438,490 | <p>I'm trying to find the eigenvalues and eigenvectors of the following <span class="math-container">$n\times n$</span> matrix, with <span class="math-container">$k$</span> blocks.
<span class="math-container">\begin{gather*}
X = \left( \begin{array}{cc}
A & B & \cdots & \\
B & A & B & \cdots \\
B & B & A & B & \cdots \\
\cdots\end{array} \right) \\
A = \left( \begin{array}{cc}
a & \cdots \\
a & \cdots \\
\cdots\end{array} \right) \in R^{\frac{n}{k} \times \frac{n}{k}} \\
B = \left( \begin{array}{cc}
b & \cdots \\
b & \cdots \\
\cdots\end{array} \right) \in R^{\frac{n}{k} \times \frac{n}{k}}.
\end{gather*}</span></p>
<p>I know that the first eigenvector is the constant <span class="math-container">$1$</span> vector. <br>
So the first eigenvalue is: <span class="math-container">$X1 = \left(\frac{n}{k}a+(k-1)\frac{n}{k} b\right)1$</span>.</p>
<p>How do I find all of the rest of the eigenvalues and eigenvectors?</p>
| Toni Mhax | 121,643 | <p>The block matrix <span class="math-container">$X$</span> is a particular one. Set <span class="math-container">$\frac{n}{k}=m$</span>; you may diagonalize <span class="math-container">$A$</span> or <span class="math-container">$B$</span> by the same unitary <span class="math-container">$U$</span>, thus taking the diagonal block matrix <span class="math-container">$V$</span> with diagonal blocks <span class="math-container">$U$</span>, the matrix <span class="math-container">$D=VXV^*$</span> will have its <span class="math-container">$m\times m$</span> blocks, diagonals with one non zero entry as the leading entry <span class="math-container">$ma$</span> or <span class="math-container">$mb$</span>. Apply a perfect shuffle (that is a permutation <span class="math-container">$P$</span>) to <span class="math-container">$D$</span> so <span class="math-container">$PDP^*=M\oplus 0$</span>, where <span class="math-container">$M=mbJ_k+m(a-b)I_k$</span>; (<span class="math-container">$J_k $</span> is the all one matrix of dimension <span class="math-container">$k$</span>, <span class="math-container">$I_k$</span> the identity matrix).</p>
|
271,489 | <p>I have a function for an ellipse that depends on time:</p>
<pre><code>elip = 1/2 (-((-0.476431 t + x) ((0.551344 (-0.476431 t + x))/t - (0.692939 (0.00786131 t + y))/t))
- (0.00786131 t + y) (-((0.692939 (-0.476431 t + x))/t) + (50.1052 (0.00786131 t + y))/t))
</code></pre>
<p>I want to plot this ellipse at a few different times and so use the following code:</p>
<pre><code>ContourPlot[Log[.5] == (elip /. {t -> time}), {y, -150, 100}, {x, 0, 4500}, ContourStyle -> Black]
</code></pre>
<p>For <code>time = 1796.66</code>, <code>time = 2032.74</code>, and <code>time = 2299.85</code>. For the first and last time this plots fine but for the middle time ContourPlot does not output anything. If I change the bounds of x to be <code>{x,0,4000}</code> then the middle time plots fine but the last time does not plot. I have the bounds so large to match the plot area of some other functions and because I would like to keep increasing time without having to dynamically update the plot range as the ellipse grows. Is there some reason for this strange behavior?</p>
| Bob Hanlon | 9,362 | <pre><code>Clear["Global`*"]
elip = 1/
2 (-((-0.476431 t +
x) ((0.551344 (-0.476431 t + x))/
t - (0.692939 (0.00786131 t + y))/t)) - (0.00786131 t +
y) (-((0.692939 (-0.476431 t + x))/
t) + (50.1052 (0.00786131 t + y))/t)) //
Rationalize[#, 0] & // Simplify;
times = {2299.85, 2032.74, 1796.66};
ContourPlot[
Evaluate[
(Log[1/2] == (elip /. t -> #)) & /@
times],
{y, -150, 100}, {x, 0, 4500},
PlotPoints -> 50,
MaxRecursion -> 5,
PlotLegends ->
LineLegend[times,
LegendLabel -> Style["time", 14]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/tvcBO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tvcBO.png" alt="enter image description here" /></a></p>
|
271,489 | <p>I have a function for an ellipse that depends on time:</p>
<pre><code>elip = 1/2 (-((-0.476431 t + x) ((0.551344 (-0.476431 t + x))/t - (0.692939 (0.00786131 t + y))/t))
- (0.00786131 t + y) (-((0.692939 (-0.476431 t + x))/t) + (50.1052 (0.00786131 t + y))/t))
</code></pre>
<p>I want to plot this ellipse at a few different times and so use the following code:</p>
<pre><code>ContourPlot[Log[.5] == (elip /. {t -> time}), {y, -150, 100}, {x, 0, 4500}, ContourStyle -> Black]
</code></pre>
<p>For <code>time = 1796.66</code>, <code>time = 2032.74</code>, and <code>time = 2299.85</code>. For the first and last time this plots fine but for the middle time ContourPlot does not output anything. If I change the bounds of x to be <code>{x,0,4000}</code> then the middle time plots fine but the last time does not plot. I have the bounds so large to match the plot area of some other functions and because I would like to keep increasing time without having to dynamically update the plot range as the ellipse grows. Is there some reason for this strange behavior?</p>
| Michael E2 | 4,999 | <p>Algebraically, <code>elip</code> is easy to analyze and much easier if the coefficients are exact than floating-point numbers (easier for Mma, that is):</p>
<pre><code>elipRat = Rationalize[Rationalize@elip, 0];
reg[t_] = ImplicitRegion[Log[1/2] == elipRat, {y, x}];
Show[
Module[{color = 0},
HighlightMesh[
DiscretizeRegion@reg[#],
{Style[0, None], Style[1, ColorData[97][++color]]}
] & /@ Rationalize@{1796.66, 2032.74, 2299.85}
],
Frame -> True, FrameLabel -> Last@reg[t],
AspectRatio -> 1
]
</code></pre>
<img src="https://i.stack.imgur.com/8XZ8s.png" width="360">
<p>Switch <code>{y, x}</code> to <code>{x, y}</code> if you want the orientation to be in school-traditional order.</p>
<p>If you want to use features of <code>ContourPlot</code>, then the following shows where the plot should be focused:</p>
<pre><code>N@RegionBounds@reg[#] & /@ Rationalize@{1796.66, 2032.74, 2299.85}
(*
{{{-21.2367, -7.01153}, {788.18, 923.789}},
{{-23.5454, -8.41455}, {896.339, 1040.58}},
{{-26.127, -10.0327}, {1019.01, 1172.43}}}
*)
</code></pre>
<p>If you want to programmatically determine the range for <code>ContourPlot</code> with approximately optimal (minimal) sampling:</p>
<pre><code>bnds = N@RegionBounds@reg[#] & /@ (* use reg[t] above *)
Rationalize@{1796.66, 2032.74, 2299.85};
ctr = Mean@*Flatten /@ Transpose@bnds; (* center of sampling *)
{dy, dx} = (* min. diam. scaled by number < 1/Sqrt[2] *)
Min /@ Map[Differences, Transpose@bnds, {2}]/GoldenRatio;
{yrange, xrange} = MinMax /@ Transpose@bnds; (* bounding ranges *)
yrange = {-150, 100}; xrange = {0, 4500}; (* desired ranges if different *)
{ny, nx} = Ceiling[Abs@({yrange, xrange} - ctrs[[All, 2]])/{dy, dx}];
ContourPlot[
Evaluate@ (* makes three equations *)
Thread[Log[0.5] == (elip /. t -> {1796.66, 2032.74, 2299.85})],
(* align initial sampling grid on ctr *)
{y, ctr[[1]] - First@ny*dy, ctr[[1]] + Last@ny*dy},
{x, ctr[[2]] - First@nx*dx, ctr[[2]] + Last@nx*dx},
(* adjust PlotPoints to {dy, dx} via {ny, nx} *)
PlotPoints -> (Max[# + 1, 15] & /@ Total /@ {ny, nx})]
</code></pre>
<img src="https://i.stack.imgur.com/SRY1S.png" width="360">
<p>The following shows the initial sample mesh in the area around the ellipses. The important thing is to get at least one point inside each ellipse. I picked <code>GoldenRatio</code> for a safety factor (exceeds <code>Sqrt[2]</code>). Dividing by <code>Sqrt[2]</code> is on the boundary of optimal <em><strong>if</strong></em> the axes of the ellipses are parallel to the coordinate axes, which is not quite the case here. I didn't worry about the general case. The requirement is that any <code>dy</code> by <code>dx</code> rectangle should have at least one vertex in each ellipse. It works here. Clearly, extreme cases such as an ellipse that is nearly an oblique line segment may be extremely hard to deal with using <code>ContourPlot</code>.</p>
<p>I bumped up <code>MaxRecursion</code> to make the ellipses look pretty. The small ones above are jagged. In this case, the large <code>MaxRecursion</code> leads to refinement only in the neighborhood of the ellipses. It does not lead to the explosion of meshing one often gets and is quite quick in this case.</p>
<pre><code>Show[
ContourPlot[1, (* initial sample mesh *)
{y, ctr[[1]] - First@ny*dy, ctr[[1]] + Last@ny*dy}, {x,
ctr[[2]] - First@nx*dx, ctr[[2]] + Last@nx*dx},
PlotPoints -> (Max[# + 1, 15] & /@ Total /@ {ny, nx}),
ColorFunction -> (LightBlue &),
Mesh -> All, MaxRecursion -> 0],
ContourPlot[ (* ellipses from above *)
Evaluate@
Thread[Log[0.5] == (elip /. t -> {1796.66, 2032.74, 2299.85})],
{y, ctr[[1]] - First@ny*dy, ctr[[1]] + Last@ny*dy}, {x,
ctr[[2]] - First@nx*dx, ctr[[2]] + Last@nx*dx},
PlotPoints -> (Max[# + 1, 15] & /@ Total /@ {ny, nx}),
MaxRecursion -> 5],
Graphics[{Red, PointSize@Medium, Point@ctr}],
PlotRange -> MinMax /@ Transpose@bnds
]
</code></pre>
<img src="https://i.stack.imgur.com/OZZkx.png" width="360">
|
3,080,124 | <p>Let <span class="math-container">$X$</span> be a topological space. Let <span class="math-container">$a\in X$</span>. Is it always true that <span class="math-container">$a$</span> is contained in a proper open set of <span class="math-container">$X$</span>? I don't know how to derive it directly by the axioms of a topological space.</p>
| jmerry | 619,637 | <p>No. Indeed, the "trivial topology" in which only the empty set and the full space are open, is a valid topology. It's not very interesting, but it does satisfy the axioms.</p>
|
3,881,029 | <p><span class="math-container">$R(A)$</span> is range of <span class="math-container">$A$</span> <br>
<span class="math-container">$N(A)$</span> is nullspace of <span class="math-container">$A$</span> <br>
<span class="math-container">$R(A^T)$</span> is range of <span class="math-container">$A^T$</span> <br>
<span class="math-container">$N(A^T)$</span> is nullspace of <span class="math-container">$A^T$</span> <br></p>
<p>Suppose <span class="math-container">$y \in R(A)$</span> and <span class="math-container">$x \in N(A^T)$</span>.</p>
<p>How would one go about showing that <span class="math-container">$x^Ty=0$</span> (aka <span class="math-container">$x$</span> is perpendicular to <span class="math-container">$y$</span>)?</p>
| Hagen von Eitzen | 39,174 | <p>With <span class="math-container">$y=Az$</span>,
<span class="math-container">$$ \langle x, y\rangle =\langle x, Az\rangle =\langle A^Tx,z\rangle=\langle 0,z\rangle =0.$$</span></p>
|
117,933 | <p>I couldn't find similar question being asked here. The closest one I can find is <a href="https://mathoverflow.net/questions/11366/when-to-split-merge-papers">When to split/merge papers?</a>. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is: Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part? </p>
| Rodrigo A. Pérez | 13,923 | <p>It is probably too much to ask from one editor to find space for two 50 page papers on the same topic, so it is natural to send the second part elsewhere. The exceptions I know involve solutions to big problems and/or big names.</p>
<p>About your second problem... include paper one as a helpful reference for the referee.</p>
|
117,933 | <p>I couldn't find similar question being asked here. The closest one I can find is <a href="https://mathoverflow.net/questions/11366/when-to-split-merge-papers">When to split/merge papers?</a>. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is: Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part? </p>
| Noah Snyder | 22 | <p>One option to consider, which we did with the small index subfactor papers, is to submit to the same editor but at <em>different</em> journals. This retains the logistical advantage of having the same referees, but doesn't require as much space in one journal. Furthermore you can put the better half in a better journal.</p>
|
117,933 | <p>I couldn't find similar question being asked here. The closest one I can find is <a href="https://mathoverflow.net/questions/11366/when-to-split-merge-papers">When to split/merge papers?</a>. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is: Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part? </p>
| Stefan Kohl | 28,104 | <p>First of all, I wonder how you can be sure that a 100-pages-proof is correct before you even have typed it in completely -- at least unless it consists mostly of routine computations!</p>
<p>As to your question, I think one proof should ideally be given in one paper, or a least in two papers in the same journal. -- Well, of course there are some proofs distributed over MANY papers, like the one of the Classification of Finite Simple Groups, but that's probably something different ... .</p>
|
2,930,003 | <p>Can there be a relation which is reflexive, symmetric, transitive, and antisymmetric at the same time? I tried to find so.</p>
<p>If <span class="math-container">$A = \{ a,b,c \}$</span>. Let <span class="math-container">$R$</span> be a relation which is reflexive, symmetric, transitive, and antisymmetric.</p>
<p><span class="math-container">$R = \{ (a,a), (b,b), (c,c) \}$</span></p>
<p>Is this correct? If I'm wrong, can you help me understand it?</p>
<p>Since if <span class="math-container">$(a, b)$</span> and <span class="math-container">$(b, c)$</span> are elements of <span class="math-container">$R$</span> by transitive there would be <span class="math-container">$(a, c)$</span>, but then there should be <span class="math-container">$(b, a)$</span>, <span class="math-container">$(c, b)$</span> and <span class="math-container">$(c, a)$</span> by symmetry, but then it would not be antisymmetric. If I'm not mistaken.</p>
| 5xum | 112,884 | <p>For any set <span class="math-container">$A$</span>, there exists only one relation which is both reflexive, symmetric and assymetric, and that is the relation <span class="math-container">$R=\{(a,a)| a\in A\}$</span>.</p>
<p>You can easily see that any reflexive relation must include all elements of <span class="math-container">$R$</span>, and that any relation that is symmetric <strong>and</strong> antisymmetric cannot include any pair <span class="math-container">$(a,b)$</span> where <span class="math-container">$a\neq b$</span>. So already, <span class="math-container">$R$</span> is your only candidate for a reflexive, symmetric, transitive and antisymmetric relation.</p>
<p>Since <span class="math-container">$R$</span> is also transitive, we conclude that <span class="math-container">$R$</span> is the only reflexive, symmetric, transitive and antisymmetric relation.</p>
|
2,137,530 | <blockquote>
<p>Given $N$ trials of a die roll, where we have defined $D$ as the number of distinct outcomes, what would be the mean and standard deviation of $D$?</p>
</blockquote>
<p>If we have defined $I(k)$ as an indicator random variable which equals 1 if outcome $k$ (such as 6) appears at least once, and 0 otherwise, for $k\in\{ 1,\dots,6\}$, then by definition
$$D = \sum\limits_{k=1}^6 I(k)$$
How do the dependencies between the $I(k)$ play into the solution? (Which is the part that is tripping me up the most.)</p>
| Graham Kemp | 135,106 | <p>Good tactic. The distributions of the indicators are identical, though <em>not</em> independent, as you noticed. Fortunately that is not a critical issue.</p>
<p>A really useful though counterintuitive, property of Expectation is that it is Linear and this holds whether the random variables are independent or not. So the expectation of the count of distinct results is the sum of the expectation of these indicator random variables.</p>
<p>$$\begin{align}\mathsf E(D) ~&=~ \sum_{k=1}^6 \mathsf E(I(k))\\[1ex] & =~ 6~\mathsf E(I(1))\end{align}$$</p>
<p>Unfortunately this does <em>not</em> quite apply to variance; however, Covariance is Bilinear, which is almost as useful.</p>
<p>$$\begin{align}\mathsf {Var}(D) ~&=~ \sum_{k=1}^6 \mathsf {Var}(I(k))~+ \mathop{2~~\sum}\limits_{k,j~:~1\leq k < j\leq 6}\mathsf {Cov}(I(k), I(j))\\[1ex] &=~ 6~\mathsf {Var}(I(1)) + 30~\mathsf{Cov}(I(1),I(2)) \end{align}$$</p>
<p>You can also use the definition:</p>
<p>$$\begin{align}\mathsf{Var}(D) ~&=~ \mathsf E\left(\left(\sum_{k=1}^6 I(k)\right)^2\right)-\left(\mathsf E\left(\sum_{k=1}^6 I(k)\right)\right)^2
\\[1ex] &=~ 6~\mathsf E\left(I(1)^2\right) + 30~\mathsf E(I(1)I(2))- 36~\mathsf E(I(1))^2
\end{align}$$</p>
<p>So find $\mathsf E(I(1)), \mathsf {Var}(I(1)), \text{ and }\mathsf {Cov}(I(1),I(2))$ and apply them.</p>
<p>Shall we leave that up to you?</p>
<hr>
<blockquote class="spoiler">
<p> You should have for all supported $k$ that $\mathsf E(I(k)^2)=\mathsf E(I(k)) = 1-(\tfrac 56)^N$, and for all supported $j\neq k$ that $\mathsf E(I(j)\cdot I(k))=1-2(\tfrac 56)^N+(\tfrac 46)^N$. </p>
</blockquote>
|
2,137,530 | <blockquote>
<p>Given $N$ trials of a die roll, where we have defined $D$ as the number of distinct outcomes, what would be the mean and standard deviation of $D$?</p>
</blockquote>
<p>If we have defined $I(k)$ as an indicator random variable which equals 1 if outcome $k$ (such as 6) appears at least once, and 0 otherwise, for $k\in\{ 1,\dots,6\}$, then by definition
$$D = \sum\limits_{k=1}^6 I(k)$$
How do the dependencies between the $I(k)$ play into the solution? (Which is the part that is tripping me up the most.)</p>
| Parcly Taxel | 357,390 | <p>We approach this problem from a combinatorial perspective. The number of $n$-roll sequences with $k$ distinct values ($1\le k\le6$), out of $6^n$ sequences total, is
$$D(n,k)=\binom6kk!\left\{n\atop k\right\}=\binom6k\sum_{j=0}^k(-1)^{k-j}\binom kjj^n$$
where $\left\{n\atop k\right\}$ is the <a href="https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind" rel="nofollow noreferrer">Stirling number of the second kind</a> and counts the number of ways to partition the rolls into homogeneous subsets, and $\binom6kk!$ is the number of ways to fill those subsets with dice rolls. Letting $n$ vary across the positive integers we get
$$D(n,1)=6$$
$$D(n,2)=15\cdot2^n-30$$
$$D(n,3)=-60\cdot2^n+20\cdot3^n+60$$
$$D(n,4)=90\cdot2^n-60\cdot3^n+15\cdot4^n-60$$
$$D(n,5)=-60\cdot2^n+60\cdot3^n-30\cdot4^n+6\cdot5^n+30$$
$$D(n,6)=15\cdot2^n-20\cdot3^n+15\cdot4^n-6\cdot5^n+6^n-6$$
$\frac{D(n,k)}{6^n}$ then gives the probability an $n$-roll sequence will have $k$ distinct values. The expected value of $D$ for a given $n$ is then
$$\mu_n=\sum_{k=1}^6k\cdot\frac{D(n,k)}{6^n}=6\left(1-\left(\frac56\right)^n\right)$$
and the standard deviation is
$$\sigma_n=\sqrt{\sum_{k=1}^6\frac{D(n,k)}{6^n}(k-\mu_n)^2}=\sqrt{\frac{5\cdot144^n-6\cdot150^n+180^n}{6^{3n-1}}}$$
Note that
$$\lim_{n\to\infty}\mu_n=6\text{ and }\lim_{n\to\infty}\sigma_n=0$$
which match our intuition, since for large $n$ almost all roll sequences should contain all six outcomes.</p>
<p>The SymPy code that generated these results can be found <a href="https://gist.github.com/Parclytaxel/b0fd7b3386e5ccee9d8250c3fc3db634" rel="nofollow noreferrer">here</a>.</p>
|
88,511 | <p>In version <a href="http://reference.wolfram.com/language/guide/SummaryOfNewFeaturesIn102.html">10.2</a> there is a new experimental function: <a href="http://reference.wolfram.com/language/ref/FindFormula.html"><code>FindFormula[]</code></a>.</p>
<p>I suspect that a <a href="https://en.wikipedia.org/wiki/Symbolic_regression">genetic programming algorithm (symbolic regression)</a> is behind this new feature, but I can't find any references.</p>
<p><strong>Question</strong></p>
<ul>
<li>What is behind this new function?</li>
</ul>
| TheDoctor | 2,263 | <p>I doubt that this is very robust. Consider a simple change in the DE example in the Documentation:</p>
<pre><code>sol = y /. NDSolve[{y'[x] == y[x] Cos[x], y[0] == 2}, y, {x, -5, 300}][[1]];
times = N[Range[-5, 600]/9];
data = Transpose[{times, sol[times] + RandomReal[0.05, Length[times]]}];
lp = ListPlot[data, PlotRange -> All]
</code></pre>
<p>Now</p>
<pre><code>FindFormula[data, x, 1, TargetFunctions -> {Exp, Sin, Cos}]
</code></pre>
<p>thinks the best solution is <code>2.27414 Sin[x] + 2.5479</code>. Whereas a <em>much better</em> solution, obviously compatible with the selected <strong>TargetFunctions</strong>, is <code>2 Exp[Sin[x]]</code>.</p>
|
1,991,950 | <p>Let $m$ and $n$ be relatively prime integers, with $n \ne 0,1$, so that $a=m/n$ is a non-integer rational fraction. Let $p$ be an odd prime.</p>
<blockquote>
<p><strong>QUESTION 1</strong>: Can Fermat's Little Theorem be applied, <em>i.e.</em>, can one say $a^p \equiv a\!\pmod{p}$, without any further consideration/explanation?</p>
</blockquote>
<p>To that end,</p>
<blockquote>
<p><strong>QUESTION 2</strong>: Are there any proofs (preferably well-known results) that do this?</p>
</blockquote>
| Jyrki Lahtonen | 11,619 | <p>It can be applied in the following sense.</p>
<ul>
<li>You need to assume that <span class="math-container">$p\nmid n$</span> (division by <span class="math-container">$p$</span> amounts to division by zero when doing arithmetic <span class="math-container">$\pmod p$</span>, so is a no-no).</li>
<li>We can work in the ring <span class="math-container">$$R_p=\{\frac mn\in\Bbb{Q}\mid \gcd(n,p)=1\}.$$</span></li>
<li>This ring has an ideal <span class="math-container">$$P_p=\{\frac mn\in\Bbb{Q}\mid \gcd(n,p)=1,\ \text{$m$ is divisible by $p$}\}.$$</span> Basically the ideal consists of the numbers <span class="math-container">$px, x\in R_p$</span>, so we can say that the ideal <span class="math-container">$P_p$</span> consists of the elements of <span class="math-container">$R_p$</span> that are divisible by <span class="math-container">$p$</span>.</li>
<li>We have <span class="math-container">$$x^p\equiv x\pmod{P_p}$$</span> for all <span class="math-container">$x\in R_p$</span>. In other words, the difference <span class="math-container">$x^p-x$</span> is an element of <span class="math-container">$P_p$</span> (divisible by <span class="math-container">$p$</span> if you wish).</li>
</ul>
<hr />
<p>Alternatively (when <span class="math-container">$\gcd(n,p)=1$</span>) the residue class of <span class="math-container">$n$</span> has an inverse modulo <span class="math-container">$p$</span>. IOW there exists an integer <span class="math-container">$n'$</span> such that <span class="math-container">$nn'\equiv1\pmod p$</span>. We can then define <span class="math-container">$m/n=mn'$</span> (think: <span class="math-container">$m/n=m\cdot n^{-1}$</span>). Of course, the Little Fermat relation <span class="math-container">$a^p\equiv a\pmod p$</span> holds for <span class="math-container">$a=mn'$</span>, and in this sense the result also makes sense.</p>
|
400,838 | <p>I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$</p>
<p>I tried and tried... friends of mine tried as well and we don't know how to get out of:</p>
<p>$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$</p>
<p>(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)</p>
<p>How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.</p>
| amWhy | 9,003 | <p>You had the right idea: the issue is in your simplification of the numerator:</p>
<p>$$\begin{align}
(2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) & = 2^2 - \left(\sqrt{(3 + x)}\right)^2 \\ \\
& = 4 - (3 + x) \\ \\
& = 4 - 3 - x \\ \\
& = 1 - x = -(x - 1)
\end{align}$$</p>
<p>That gives you $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} & = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}}
& = -\frac 14
\end{align}$$</p>
|
1,811,443 | <p>Let $(a_n)$ be a sequence of rational numbers, where <strong>all rational numbers are terms</strong>. (<em>i.e. enumeration of rational numbers</em>)</p>
<p>Then, is there any convergent sub-sequence of $(a_n)$?</p>
| Aloizio Macedo | 59,234 | <p>The argument given by Elio Joseph can be refined in order to prove that for any given real number, there always exists a subsequence which converges to it.</p>
<p>For that, let $x \in \mathbb{R}$ be a real number. Consider the interval $[x-1,x+1]$. It contains infinitely many rationals, but in particular it contains one rational, which is some $a_{n_1}$ of our enumeration.</p>
<p>Now, having chosen $a_{n_i}$, we consider the interval $[x-\frac{1}{i+1}, x+\frac{1}{i+1}]$. This contains infinitely many rationals. In particular, we have that it contains one rational which is not one of the first $n_i$ of our enumeration. We take such rational to be our $a_{n_{i+1}}$.</p>
<p>It is clear that $a_{n_i} \to x$.</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| James Garrett | 457,432 | <p>Noticing that $x-1=-(1-\sqrt{x})(1+\sqrt{x})$ then $$\lim_{x \to 1} \frac{\sin(1-\sqrt{x})}{-(1-\sqrt{x})(1+\sqrt{x})}=\lim_{x \to 1} \frac{\sin(1-\sqrt{x})}{1-\sqrt{x}}\cdot \lim_{x \to 1} \frac{-1}{1+\sqrt{x}}=\lim_{x\to 1} \frac{-1}{1+\sqrt{x}}=\frac{-1}{2},$$ using the fact that $$\lim_{u \to 0} \frac{\sin(u)}{u}=1.$$</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| Michael Rybkin | 350,247 | <p>In order to find this limit, you're going to have to employ the following well-known fact from elementary limit theory:
$$
\lim \limits_{x \to 0} \frac{\sin{x}}{x}=1
$$</p>
<p>What it says is that as $x$ approaches zero, sine of $x$ over $x$ approaches $1$. Here's <a href="https://www.youtube.com/watch?v=7Dy8ASSE8ok&index=8&list=PLF5E22224459D23D9#t=12m29s" rel="nofollow noreferrer">the proof</a> of that (YouTube video). Also notice that as $x$ approaches $1$, the expression $1-\sqrt{x}$ approaches 0. Lastly, recall your conjugate multiplication facts from the high-school days and you're all set to take this limit.</p>
<p>$$
\lim \limits_{x \to 1} \frac{\sin\left(1-\sqrt{x}\right)}{x-1}=
\lim \limits_{x \to 1} \left(\frac{\sin\left(1-\sqrt{x}\right)}{x-1}\cdot\frac{1-\sqrt{x}}{1-\sqrt{x}}\right)=\\
\lim \limits_{x \to 1} \left(\frac{\sin\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1-\sqrt{x}}{x-1}\right)=
\lim \limits_{1-\sqrt{x} \to 0} \frac{\sin\left(1-\sqrt{x}\right)}{1-\sqrt{x}} \cdot \lim \limits_{x \to 1} \left(\frac{1-\sqrt{x}}{x-1}\cdot\frac{1+\sqrt{x}}{1+\sqrt{x}}\right)=\\
1 \cdot \lim \limits_{x \to 1} \frac{1-x}{(x-1)(1+\sqrt{x})}=
-\lim \limits_{x \to 1} \frac{x-1}{(x-1)(1+\sqrt{x})}=
-\lim \limits_{x \to 1} \frac{1}{1+\sqrt{x}}=\\
-\frac{1}{1+1}=-\frac{1}{2}
$$</p>
<p>This is just one of the several possible ways to find this limit. Siong Thye Goh, however, has offered you a more elegant way to do that (elegance in the fact that it's just a tad shorter).</p>
|
3,657,026 | <p>I need to prove expression using mathematical induction <span class="math-container">$P(1)$</span> and <span class="math-container">$P(k+1)$</span>, that:</p>
<p><span class="math-container">$$
1^2 + 2^2 + \dots + n^2 = \frac{1}{6}n(n + 1)(2n + 1)
$$</span></p>
<p>Proving <span class="math-container">$P(1)$</span> gave no difficulties, however I was stuck with <span class="math-container">$P(k+1)$</span>, I've reached this point:</p>
<p><span class="math-container">$$
1^2 + \dots + (k+1)^2 = 1^2 + \dots + k^2 + (k+1)^2 = \\ \frac{1}{6}k(k+1)(2k+1) + (k+1)^2
$$</span></p>
<p>I've checked answer from the exercise book, the next step would be:</p>
<p><span class="math-container">$$
= \frac{1}{6}(k+1)(k(2k+1)+6(k+1))
$$</span></p>
<p>How it was converted like that? Could you provide some explanation?</p>
<p>Thank you in advance</p>
| fleablood | 280,126 | <p><span class="math-container">$\frac{1}{6}k(k+1)(2k+1) + (k+1)^2=$</span></p>
<p><span class="math-container">$\frac{1}{6}k(k+1)(2k+1) + \frac 16\cdot 6(k+1)^2=$</span> </p>
<p><span class="math-container">$\color{green}{\frac{1}{6}}k\color{blue}{(k+1)}(2k+1) + \color{green}{\frac 16}\cdot 6\color{blue}{(k+1)}^2=$</span></p>
<p>.....Now just factor out <span class="math-container">$\color{green}{\frac{1}{6}}\color{blue}{(k+1)}$</span> and you get....</p>
<p><span class="math-container">$\color{green}{\frac{1}{6}}k\color{blue}{(k+1)}(2k+1) + \color{green}{\frac 16}\cdot 6\color{blue}{(k+1)}^2=$</span></p>
<p><span class="math-container">$\color{green}{\frac{1}{6}}\color{blue}{(k+1)}[k(2k+1)+6(k+1)]$</span></p>
<p>... Now from here we need to convert <span class="math-container">$[k(2k+1)+6(k+1)]\implies ((k+1)+1)(2(k+1)+1)=(k + 2)(2k+3)=2k^2 + 7k + 6$</span>...</p>
<p><span class="math-container">$=\frac 16(k+1)[k(2k+1)+6(k+1)]=$</span></p>
<p><span class="math-container">$\frac 16(k+1)(2k^2 + 7k +6)=$</span></p>
<p><span class="math-container">$\frac 16(k+1)[(2k^2 + 4k)+(3k+6)]=$</span></p>
<p><span class="math-container">$\frac 16(k+1)[2k\color{red}{(k+2)} + 3\color{red}{(k+2)}]=$</span></p>
<p><span class="math-container">$\frac 16(k+1)\color{red}{(k+2)}[2k+3]=$</span></p>
<p><span class="math-container">$\frac 16(k+1)(k+2)(2k+3)=$</span></p>
<p><span class="math-container">$\frac 16(\color{purple}{k+1})(\color{purple}{k+1} + 1)(2(\color{purple}{k+1})+1)$</span>.</p>
<p>ANd we are done</p>
|
3,718,347 | <p>If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are real numbers and <span class="math-container">$$x^2+y^2-4x-6y-1=0$$</span>what is the minimum value of <span class="math-container">$x + y$</span>?</p>
<p>I tried to factor it but I couldn't. Then I tried to make squares so all the squares could equal to 0. But I couldn't find a way. So please kindly give me the solution for this question.</p>
| lab bhattacharjee | 33,337 | <p>Hint</p>
<p>Set <span class="math-container">$x+y=c$</span></p>
<p>Replace <span class="math-container">$y$</span> with <span class="math-container">$c-x$</span> to form a quadratic equation in <span class="math-container">$x$</span></p>
<p>As <span class="math-container">$x$</span> is real, the discriminant must be <span class="math-container">$$\ge0$$</span></p>
|
3,718,347 | <p>If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are real numbers and <span class="math-container">$$x^2+y^2-4x-6y-1=0$$</span>what is the minimum value of <span class="math-container">$x + y$</span>?</p>
<p>I tried to factor it but I couldn't. Then I tried to make squares so all the squares could equal to 0. But I couldn't find a way. So please kindly give me the solution for this question.</p>
| Angelo | 771,461 | <p><span class="math-container">$x^2+y^2-4x-6y-1=$</span></p>
<p><span class="math-container">$=x^2-4x+y^2-6y-1=$</span></p>
<p><span class="math-container">$=(x^2-4x+4)-4+(y^2-6y+9)-9-1=$</span></p>
<p><span class="math-container">$=(x-2)^2-4+(y-3)^2-9-1=$</span></p>
<p><span class="math-container">$=(x-2)^2+(y-3)^2-14$</span></p>
<p>The problem of finding the minimum value of <span class="math-container">$x+y$</span> can be solved by using Lagrange multipliers method.</p>
<p>Nevertheless, it is possible to solve this problem in the following way.</p>
<p>Let <span class="math-container">$X=x-2$</span> and <span class="math-container">$Y=y-3$</span>.</p>
<p>The minimum value of <span class="math-container">$X+Y$</span> is a number <span class="math-container">$k$</span> such that the straight line <span class="math-container">$X+Y=k$</span> is tangent to the circle <span class="math-container">$X^2+Y^2-14=0$</span>.</p>
<p>Now, we are going to find the two straight lines parallel to <span class="math-container">$X+Y=0$</span> and tangent to the circle <span class="math-container">$X^2+Y^2-14=0$</span></p>
<p><span class="math-container">$\begin{cases}
X+Y=k \\
X^2+Y^2-14=0
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
Y=k-X \\
X^2+(k-X)^2-14=0
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
Y=k-X \\
X^2+k^2-2kX+X^2-14=0
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
Y=k-X \\
2X^2-2kX+k^2-14=0
\end{cases}$</span></p>
<p><span class="math-container">$\begin{cases}
Y=k-X \\
X^2-kX+\frac{1}{2}k^2-7=0
\end{cases}$</span></p>
<p><span class="math-container">$\Delta=k^2-2k^2+28=28-k^2$</span></p>
<p>The straight line <span class="math-container">$X+Y=k$</span> is tangent to the circle <span class="math-container">$X^2+Y^2-14=0$</span> if and only if <span class="math-container">$\Delta=0$</span>.</p>
<p><span class="math-container">$28-k^2=0$</span></p>
<p><span class="math-container">$k^2=28$</span></p>
<p><span class="math-container">$k=\pm\sqrt{28}$</span></p>
<p><span class="math-container">$k=\pm 2\sqrt{7}$</span></p>
<p>The two straight lines parallel to <span class="math-container">$X+Y=0$</span> and tangent to the circle <span class="math-container">$X^2+Y^2-14=0$</span> are:</p>
<p><span class="math-container">$X+Y=2\sqrt{7}$</span> and</p>
<p><span class="math-container">$X+Y=-2\sqrt{7}$</span></p>
<p>It means that the minimum value of <span class="math-container">$X+Y$</span> is <span class="math-container">$-2\sqrt{7}$</span>, but <span class="math-container">$X=x-2$</span> and <span class="math-container">$Y=y-3$</span>, so the minimum value of <span class="math-container">$x+y-5$</span> is <span class="math-container">$-2\sqrt{7}$</span>.</p>
<p>Hence the minimum value of <span class="math-container">$x+y$</span> is <span class="math-container">$5-2\sqrt{7}$</span>.</p>
|
2,482,669 | <p>Find sum of the expression,
$$x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$$
where $x,y$ are real numbers and $n$ is a natural number.</p>
| Macavity | 58,320 | <p><strong>Hint:</strong> It’s a geometric progression with common ratio $\dfrac{y}x$.</p>
|
211,623 | <p>Let $G$ be a (finite) group, and $a, b \in G$ be any two elements. Consider the sequence defined by
\begin{eqnarray*}
s_0 &=& a, \\
s_1 &=& b, \text{and} \\
s_{n+2} &=& s_{n+1} s_{n} \text{ for $n \geq 0$}.
\end{eqnarray*}
This sequence is what we get when replace the letters in the algae L-system (cf: <a href="https://en.wikipedia.org/wiki/L-system#Example_1:_Algae" rel="nofollow">https://en.wikipedia.org/wiki/L-system#Example_1:_Algae</a>) with the two elements of our group. As $G$ is finite, there are finitely many pairs $(s_{n+1}, s_{n})$ and hence that the sequence will eventually start repeating. Do we know more about this sequence $(s_n)$? Like, when is it periodic?</p>
| Alireza Abdollahi | 19,075 | <p>You may start with the PhD thesis of P. P. Campbell </p>
<p><a href="http://www-circa.mcs.st-and.ac.uk/Theses/ppcphd.pdf" rel="nofollow">http://www-circa.mcs.st-and.ac.uk/Theses/ppcphd.pdf</a></p>
|
118,742 | <p>The PDF for $Y$ is
$$f_Y(y) = \begin{cases}
0 & |y|> 1 \\
1-|y| & |y|\leq 1
\end{cases}$$</p>
<p>How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get
$$F_Y(y)=\begin{cases}
1/2 -y/2-y^2/2 & [-1,0] \\
1/2-y/2+y^2/2 & [0,1]
\end{cases}
$$
by using the fact that $F_Y(y)=\int _{-\infty}^{y}{f_Y(y)}\,dy$, however my text claims the answer is
$$F_Y(y)=\begin{cases}
1/2 +y+y^2/2 & [-1,0] \\
1/2+y-y^2/2 & [0,1]
\end{cases}
$$
I am struggling with pdf and cdfs, so I asssume I did something wrong other than the simple integration. Who's correct? Me or the Text!?? $:)$ </p>
| Community | -1 | <p>We have that $F(y) = \displaystyle \int_{-\infty}^y f(x) dx$. In your case, we are given that $$f(x) = \begin{cases} 0 & x <-1\\ 1 + x & x \in[-1,0]\\ 1-x & x \in [0,1]\\ 0 & x > 1\end{cases}$$</p>
<ul>
<li>If $y < -1$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^y 0 dx =0 $. We have the integrand $f(x) = 0$ since $x \leq y < -1$.</li>
<li>If $y \in [-1,0]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-1}^{y} f(x) dx = \displaystyle \int_{-1}^{y} \left( 1+x \right) dx = \left( x + \frac{x^2}{2} \right)_{-1}^{y} $$
$$F(y) = \left(y + \frac{y^2}{2} \right( - \left( -1 + \frac12 \right) = \frac12 + y + \frac{y^2}{2}.$$</li>
<li>If $y \in [0,1]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx = \displaystyle \int_{-1}^{0} \left( 1+x \right) dx + \displaystyle \int_{0}^{y} (1-x) dx$$
Hence, $$F(y) = \frac12 + \left( x - \frac{x^2}{2}\right)_0^{y} = \frac12 + y - \frac{y^2}{2}$$</li>
<li>For $y > 1$, since $f(x) = 0$ for all $x>1$, we have that $F(y) = F(1)$ for all $y > 1$. Hence, $F(y) = F(1) = 1$.</li>
</ul>
<p>Hence, $$F(y) = \begin{cases} 0 & y <-1\\ \frac12 + y + \frac{y^2}{2} & y \in[-1,0]\\ \frac12 + y - \frac{y^2}{2} & y \in[0,1]\\ 1 & y > 1\end{cases}$$</p>
|
118,742 | <p>The PDF for $Y$ is
$$f_Y(y) = \begin{cases}
0 & |y|> 1 \\
1-|y| & |y|\leq 1
\end{cases}$$</p>
<p>How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get
$$F_Y(y)=\begin{cases}
1/2 -y/2-y^2/2 & [-1,0] \\
1/2-y/2+y^2/2 & [0,1]
\end{cases}
$$
by using the fact that $F_Y(y)=\int _{-\infty}^{y}{f_Y(y)}\,dy$, however my text claims the answer is
$$F_Y(y)=\begin{cases}
1/2 +y+y^2/2 & [-1,0] \\
1/2+y-y^2/2 & [0,1]
\end{cases}
$$
I am struggling with pdf and cdfs, so I asssume I did something wrong other than the simple integration. Who's correct? Me or the Text!?? $:)$ </p>
| Dilip Sarwate | 15,941 | <p>This is the kind of problem that gives integration a bad name among students.</p>
<ul>
<li><p><strong>Draw</strong> a graph of the density function. It looks like an isoceles
right triangle with hypotenuse $2$ and apex at $(0,1)$ and very obviously has area $1$ (useful as a check on one's work.)</p></li>
<li><p>For any $x_0$, $F(x_0)$ is the area under the density function to the
left of $x_0$. It should be obvious that $F(x_0) = 0$ if $x_0 \leq -1$ and
$F(x_0) = 1$ if $x_0 > 1$.</p></li>
<li><p>Pick an $x_0$ in $[-1,0]$. The area to the left of $x_0$ is a right
triangle with altitude $1+x_0$ and base $1+x_0$ (or $x_0 - (-1)$ if
you like, and so $F(x_0) = \frac{1}{2}(1+x_0)^2$. Quick check: value
is $\frac{1}{2}$ at $x_0 = 0$ and $0$ at $-1$.</p></li>
<li><p>Pick an $x_0$ in $[0,1]$. By symmetry, the area to the <em>right</em>
of $x_0$ is $\frac{1}{2}(1-x_0)^2$. Quick check: value
is $\frac{1}{2}$ at $x_0 = 0$ and $0$ at $1$. Hence,
$F(x_0) = 1 - \frac{1}{2}(1-x_0)^2$.</p></li>
</ul>
<p>Putting it all together, we get the same answer as Sivaram Ambikasaran.
It takes longer to write out the instructions than to just work off the
diagram.</p>
|
2,352,527 | <p><em>This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.</em></p>
<p>What is the value of </p>
<p>$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$ </p>
<p><em>Do I just use</em> </p>
<p>$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$</p>
| Fred | 380,717 | <p>For $z \ne 0$ let $g(z)=\frac{f(z)}{z}$. Then $g$ is holomorphic on $ \mathbb C \setminus \{0\}$ and in $0$ the function $g$ has an isolated singularity.</p>
<p>From $|f(z)|\leq M|z|$, we see that $|g(z)| \le M$ for $ z \in\mathbb C \setminus \{0\}$.</p>
<p>Riemann says: $0$ is a removable singularity of $g$. Hence there is an entire function $h$ such that $h=g$ on $ \mathbb C \setminus \{0\}$. By continuity we have $|h(z)| \le M$ for all $ z \in\mathbb C $.</p>
<p>Now Liouville says: there is $c \in \mathbb C$ with</p>
<p>$$h(z)=c$$</p>
<p>for all $z \in \mathbb C$ . This gives</p>
<p>$$f(z)=cz$$</p>
<p>for all $z \in \mathbb C$, as desired. </p>
|
756,662 | <p>I am trying to figure out what random variables are measurable with respect to sigma algebra given by $[1-4^{-n}, 1]$ where $n= 0, 1, 2, ....$ if $[0,1]$ is the sample space. I believe I can do with with indicator functions but I'm not sure how to write this.
Thanks!</p>
| user139388 | 139,388 | <p>Well, <em>any</em> indicator function won't do. You can see that there is no way to write $(0,25,0.5)$ as an element of the $\sigma$-algebra generated by those sets, and so the indicator of that set won't be measurable. But the set of simple functions which are measurable in that space is most likely dense with respect to any measure that is absolutely continuous with respect Lebesgue measure.</p>
|
272,468 | <p>I wrote the <a href="https://www.researchgate.net/publication/334884635_On_a_Bivariate_Frechet_Distribution" rel="nofollow noreferrer">Frechet Distribution</a> as follows:</p>
<pre><code>dist1 = ProbabilityDistribution[{"PDF", \[Lambda]1/\[Alpha]1 (x/\[Alpha]1)^(-\[Lambda]1 - 1)E^-((x/\[Alpha]1)^-\[Lambda]1)}, {x, 0, \[Infinity]}, Assumptions -> {\[Lambda]1 > 0, \[Alpha]1 > 0}]
dist2 = ProbabilityDistribution[{"PDF", \[Lambda]2/\[Alpha]2 (x/\[Alpha]2)^(-\[Lambda]2 - 1)E^-((x/\[Alpha]2)^-\[Lambda]2)}, {x, 0, \[Infinity]}, Assumptions -> {\[Lambda]2 > 0, \[Alpha]2 > 0}]
bidist = CopulaDistribution[{"FGM", \[Gamma]}, {dist1, dist2}]
</code></pre>
<p>when i write cdf of coupla dist1,dist2 as follows</p>
<pre><code>CDF[bidist, {x, y}]
</code></pre>
<p>The output is as follows:</p>
<p><a href="https://i.stack.imgur.com/NUKOb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NUKOb.png" alt="11" /></a></p>
<p>That is not the output from the linked publication.</p>
<p><a href="https://i.stack.imgur.com/tXowD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tXowD.jpg" alt="d3" /></a></p>
<p>I want to find maximum likelihood
of FGM copula, if it is possible.</p>
| Michael Seifert | 27,813 | <p>The two problems seem to be related, or at least can be cured together:</p>
<pre><code>obj2[a_?NumericQ] :=
Chop[NIntegrate[(E^(1/2 (-x[1]^2 - x[2]^2)) Log[
Abs[1 - a (x[1]^2 + x[2]^2)]] x[1]^2)/(\[Pi] (x[1]^2 +
x[2]^2)), {x[1], -\[Infinity], \[Infinity]}, {x[
2], -\[Infinity], \[Infinity]}] + Log[2]/2];
obj2[.915] (* -0.00319086 *)
FindRoot[obj2[a], {a, 0.92}] (* {a -> 0.918611} *)
</code></pre>
<p>The two modifications are:</p>
<ul>
<li><p>Using a test to sanitize the inputs of <code>obj2</code>. Using the pattern <code>a_?NumericQ</code> inside a function definition means that the function will only be evaluated if <code>NumericQ</code> evaluates to <code>True</code> on an argument, and will be return unevaluated if it does not. In particular, this prevents <code>FindRoot</code> from trying to take symbolic derivatives of <code>obj2</code> in its internal algorithms, which it will do by default.</p>
</li>
<li><p><code>Chop</code> removes "small" complex parts from numerical outputs. Like you, I'm unsure why <code>NIntegrate</code> is returning a complex part for this integrand at all; but it apparently causes <code>FindRoot</code> to invoke some kind of algorithm that does not converge. Making sure that <code>obj2</code> only ever returns a real output makes <code>FindRoot</code> much happier.</p>
</li>
</ul>
|
80,966 | <p>I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism
$\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism).</p>
<p>For compact Lie groups, using the exponential map, the answers is no, but in general I don't know. </p>
| Andreas Thom | 8,176 | <p>The answer is no in general, but this is a rather deep fact.</p>
<blockquote>
<p><strong>Theorem:</strong> (Nikolov, Segal) If $G$ is any compact Hausdorff topological group, then every finitely generated (abstract) quotient of $G$ is finite.</p>
</blockquote>
<p>N. Nikolov and D. Segal, <em>Generators and commutators in finite groups; abstract quotients of compact groups</em>, arXiv, <a href="http://arxiv.org/abs/1102.3037">http://arxiv.org/abs/1102.3037</a></p>
|
80,966 | <p>I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism
$\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism).</p>
<p>For compact Lie groups, using the exponential map, the answers is no, but in general I don't know. </p>
| Alain Valette | 14,497 | <p>Andreas shot first, but I still encourage everybody to have a look at the lemma on p.263 of R. Alperin,Locally compact groups acting on trees and property $T$. Monatsh. Math. 93 (1982), no. 4, 261–265: any homomorphism from a locally compact group to $\mathbb{Z}$, is continuous. This answers Florent's question.</p>
|
456,581 | <p>If one requires simply the existence of partial derivatives of first order rather than all orders, then a standard example is the function</p>
<p>$$ f(x,y) = \left\{\begin{array}{l l}
\frac{2xy}{x^2+y^2} & \quad \text{if $(x,y)\neq(0,0),$}\\
0 & \quad \text{if $(x,y)=(0,0).$}
\end{array} \right.$$</p>
<p>However, this does not constitute an answer to my question since the partial derivative of $\frac{\partial f}{\partial x}$ with respect to $y$ does not exist at the origin.</p>
<p>PS: This question rose out of my wonder as to whether, in the definition of a smooth function, continuity of partials is an essential requirement or not.</p>
| user | 89,499 | <p>For ease of reading, I TeXified a part of <a href="http://mathforum.org/kb/message.jspa?messageID=5449291">sci.math post</a> by Dave L. Renfro (this post is a CW). </p>
<hr>
<p>What follows is from The American Mathematical Monthly 67 #8 (October 1960), 813-814.</p>
<p><strong>Discontinuous Function with Partial Derivatives Everywhere</strong> 4876 [1959, 921]. Proposed by Naoki Kimura, University of Washington</p>
<p>If a real valued function $f(x,y)$ of two real variables
possesses all of its partial derivatives
$$ \frac{\partial^{m+n} f(x,y)}{\partial x^m \partial y^n}$$
at every point, is it necessarily continuous?</p>
<p><em>Solution by John Burr, University of New England, Australia.</em></p>
<p>The following example shows that the function need not be continuous. The function $$f(x,y) = \begin{cases} \exp\left(-\frac{x^2}{y^2} - \frac{y^2}{x^2}\right) \quad & xy \ne 0 \\ 0 & \text{otherwise}\end{cases}$$
is discontinuous at $(0,0)$, since $\lim_{t\to 0}f(t,t) = e^{-2}\ne f(0,0)$. Suppose
that it has been proved that a particular partial derivative $\phi(x,y)$ has the properties </p>
<ol>
<li>$\phi(0,y) = \phi(x,0) = 0$ </li>
<li>if $xy \ne 0$ then $\phi(x,y) = R(x,y) f(x,y)$ where $R(x,y)$ is a rational function with denominator of the form $ x^p y^q$.</li>
</ol>
<p>Then, by <code>1</code>, $ \phi_y (0,y) = \phi_x (x,0) = 0$;
by <code>2</code>, $\phi_x(0,y) = 0$ when $y \ne 0$, since when $x\to 0$,
$f(x,y) \to 0$ more rapidly than any power of $x$. Similarly $\phi_y (x,0) = 0$. Hence $\phi_x (x,y)$ and $\phi_y (x,y)$ both have the property <code>1</code>, and it is clear from <code>2</code> that
they also have the property <code>2</code>. Since $f(x,y)$ has these properties, it follows by induction that all the partial derivatives have them, and that these derivatives exist at every point.</p>
<p>It may be noted that this function satisfies conditions more stringent than those prescribed in the problem; namely the additional conditions that the mixed partial derivatives all exist, and are independent of the order in which the differentiations are performed.</p>
<hr>
<p>Now, to the question</p>
<blockquote>
<p>whether, in the definition of a smooth function, continuity of partials is an essential requirement or not.</p>
</blockquote>
<p>I'll say that defining smoothness via partials is a wrong approach to begin with. The definition should be in terms of the total derivative. Then one can inquire into its relation with partial derivatives and obtain results such as: existence and continuity of partials implies the existence of the total derivative of that order. </p>
|
2,747,509 | <p>How would you show that if <span class="math-container">$d\mid n$</span> then <span class="math-container">$x^d-1\mid x^n-1$</span> ?</p>
<p>My attempt :</p>
<blockquote>
<p><span class="math-container">$dq=n$</span> for some <span class="math-container">$q$</span>. <span class="math-container">$$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$</span> in fact, <span class="math-container">$$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$</span></p>
<p>By multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$(x-1)$</span> we get that <span class="math-container">$1-x^d\mid 1-x^n$</span> which is the final result</p>
</blockquote>
<p>Is this an ok proof?</p>
| DonAntonio | 31,254 | <p>An idea for you:</p>
<p>$$d\,\mid\,n\implies n=qd\;,\;\;q\in\Bbb Z\;,\;\;\text{and from here}: $$</p>
<p>$$x^n-1=\left(x^d\right)^q-1=\left(x^d-1\right)\left(\left(x^d\right)^{q-1}+\left(x^d\right)^{q-2}+\ldots+x^d+1\right)$$</p>
<p>The above uses the basic relation from geometric series:</p>
<p>$$x^a-1=(x-1)(x^{a-1}+x^{a-2}+\ldots+x+1)\;,\;\;a\in\Bbb N$$</p>
|
2,747,509 | <p>How would you show that if <span class="math-container">$d\mid n$</span> then <span class="math-container">$x^d-1\mid x^n-1$</span> ?</p>
<p>My attempt :</p>
<blockquote>
<p><span class="math-container">$dq=n$</span> for some <span class="math-container">$q$</span>. <span class="math-container">$$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$</span> in fact, <span class="math-container">$$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$</span></p>
<p>By multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$(x-1)$</span> we get that <span class="math-container">$1-x^d\mid 1-x^n$</span> which is the final result</p>
</blockquote>
<p>Is this an ok proof?</p>
| Xiao | 131,137 | <p>Hint from Dummit and Foote: </p>
<p>Suppose $n= kd+r$, where $0 \leq r < d$ we have
$$x^n -1 = x^{kd+r} - x^r + x^r -1 = x^r (x^{kd}-1) + (x^r -1)$$
then
can you conclude $d|n$ if and only if $x^d-1 | x^n-1$.</p>
|
2,532,156 | <p>As the title suggests I am confused between what arguments will qualify a explanation as a proof and when does the intuition betrays us. Here is the question that made me think about this:</p>
<p>On a certain planet Tau Cetus more than half of its land is dry. Prove that a tunnel can be dug straight through center of the planet (assuming their technology is sufficiently advanced) starting and ending at dry lands.</p>
<p>Isn't it obvious?
My main motive is too understand proofs and methods of arguments better.</p>
| neptun | 192,385 | <p>You are asking one of the hardest questions of mathematics. Much about mathematics is seeing the reason for something to be true and then carefully writing it out, because very often the sheer complexity of a problem makes you overlook certain things that could go wrong. There are many statements that are obvious in two dimensional euclidean geometry that just fail to hold in three or four dimensions. A proof should be like a mechanical recipe to verify that the intuition is right. However, a proof need not be a mechanical string of formal symbols: If the explanation is worded carefully with unambiguous words, it can often suffice, because it can easily be translated to formal symbols by anyone who would want that.</p>
<p>With that said, the point of a proof is to make <em>other people agree with you that your reasoning is correct</em>. How precise and detailed you have to be depends a lot on who the reader is! Researchers sometimes state that things are obvious that would take weeks for a student to prove, because they have seen the problem before and/or just feel very confident that they can write it out. Conversely, when you prove something in a course, it is important to ground every line of reasoning in something related to the course or its prerequisites.</p>
|
733,908 | <p>How do i start off with integrating the below function? i tried applying trig substitution and U substitution. how do i go about solving this function? should i split them up further into 2 separate functions ? need some help in this as i can't seem to figure out how to continue on with it </p>
<p>$$\int\frac{x^{3}}{({x^{2}-1})^{0.5}}dx$$</p>
| André Nicolas | 6,312 | <p>Rewrite the integrand as $\frac{x(x^2-1)}{(x^2-1)^{1/2}}+\frac{x}{(x^2-1)}$, which is $x(x^2-1)^{1/2}+\frac{x}{(x^2-1)^{1/2}}$. Then $u=x^2-1$ finishes things. </p>
|
163,816 | <p>I'm looking for a good exercise book for probability theory, preferably at least partially with solutions to it. I want it to be detailed, not trivial, providing me solid fundamentals in the topic to be developed in the future. I'd wish it to be more of a "applied" technical university approach than the highly "abstract" one tailored for pure mathematics student at a university, however it need not to be so.</p>
<p>The topics I would like it to cover are more or less like in the part one of the book "Probability, Random Variables and Stochastic Processes" by Papoulis and Pillai ( <a href="http://www.mhhe.com/engcs/electrical/papoulis/graphics/toc.pdf" rel="nofollow">table of contents</a> ). Thank you in advance.</p>
| Dilip Sarwate | 15,941 | <p><em>One Thousand Exercises in Probability</em> by Grimmett and Stirzaker, Oxford Press, 2001 might suit your needs.</p>
|
3,717,243 | <p>I am reading Introduction to automata theory, languages, and computation 3ed, by John E. Hopcroft, et al. The wikipedia article (<a href="https://en.wikipedia.org/wiki/Turing_machine#Formal_definition" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Turing_machine#Formal_definition</a>) for Turing machine also cites the TM definition in the 1ed of the book. (I assume the 3ed still use the same TM definition as 1ed)</p>
<p>When the tape head of a TM is scanning a blank symbol, what will happen?</p>
<ul>
<li><p>Is the transition function undefined on a blank tape symbol?</p>
</li>
<li><p>Does the TM halt?</p>
</li>
</ul>
<p>Thanks.</p>
| Mostafa Ayaz | 518,023 | <p>I don't think the equality generally holds, specially by differentiating both sides we must have <span class="math-container">$$\sum_{i=1}^{\infty} {1\over (i+x)\cdot i}={1\over x}$$</span>which for <span class="math-container">$x\in\Bbb N-\{1\}$</span> means that <span class="math-container">$$\sum_{i=1}^{x}{1\over i}={1\over x}$$</span>which doesn't hold.</p>
|
2,816,965 | <p>Hey so I was wondering how to differentiate $(e^2)^x$ without using the chain rule.</p>
<p>I tried but I always end up using the chain rule in this case.</p>
<p>Would appreciate some help! (No hints please).</p>
| Rhys Hughes | 487,658 | <p>Indeed, using the chain rule is certainly the easiest: $$f(x)=e^{2x}\to f'(x)=2e^{2x}$$</p>
<hr>
<p>You could however, use the product rule:
$$f(x)=(e^2)^x=e^xe^x$$
Then $p=q=e^x$, and $p'=q'=e^x$</p>
<p>Product rule is $f'(x)=p'q+q'p\to e^xe^x+e^xe^x=2e^{2x}$</p>
<p>Or the quotient rule can also be used:
$$f(x)=(e^2)^x=\frac{e^x}{e^{-x}}$$
Then $p=p'=e^x$, $q=e^{-x}\to q'=-e^{-x}$</p>
<p>Quotient rule is $f'(x)=\frac{p'q-q'p}{q^2}\to\frac{e^xe^{-x}-e^x(-e^{-x})}{e^{-2x}}=\frac{2}{e^{-2x}}=2e^{2x}$</p>
|
118,311 | <p>Let $A$ be an $n\times n$ matrix with entries in an arbitrary field $k$. </p>
<p>Is the characteristic polynomial $\det(tI_n-A)$ dependent only on the trace and determinant of $A$?</p>
| Noah Snyder | 8 | <p>By column orthogonality you can immediately work out the entries in the last two columns. For the first two missing columns, the key observation is that x and $x^{-1}$ are not conjugate. This means that the first two missing columns are complex conjugates of each other. You then have to split into cases based on whether the entries are real or not, but column orthogonality (both the column with itself and with the identity column) gets you down to only one possibility.</p>
|
26,062 | <p>Given a <code>TabView</code> panel like this one</p>
<pre><code>TabView[{
DynamicModule[{x = False}, {Checkbox[Dynamic[x]], Dynamic[x]}],
"foo"}]
</code></pre>
<p>I would like to reset the value of <code>x</code> to its initial value (<code>False</code>) every time the first tab gets selected - that is, I basically want to "reset"/"reload" the tab on each selection.</p>
<p>Also, the variable(s) in question should remain <em>local</em> variables on a per-tab basis, if possible. This is, I hope to avoid a solution declaring <code>x</code> globally.</p>
<p>Any help very much appreciated (either on reloading the whole tab or on setting specific dynamic variables).</p>
| Albert Retey | 169 | <p>Here is a version which fully localizes the variables. Other than that it does the same as Kubas answer so I'd suggest to accept his answer:</p>
<pre><code>With[{xDefault = False},
DynamicModule[{x = xDefault, visibleTab = 1},
TabView[{
Row[{Checkbox[Dynamic[x]], " ", Dynamic[x]}],
Dynamic[x]
}, Dynamic[visibleTab, (visibleTab = #; If[visibleTab == 1, x = xDefault]) &]
]
]
]
</code></pre>
<p>Here is another solution which will regenerate the content completely, including initialization. It needs some hacky tricks to persuade Mathematica to really update the outermost <code>Dynamic</code> in the first tab, Mathematica seems to put quite some effort into not destroying the state of that <code>Checkbox</code>. Anyway, this seems to work for me:</p>
<pre><code>DynamicModule[{visibleTab = 1, trigger = AbsoluteTime[]},
TabView[{
Dynamic[
DynamicModule[{x = False},
Row[{Pane[trigger, {0, 1}], Checkbox[Dynamic[x]], Dynamic[x]}]]
],
"foo"
}, Dynamic[
visibleTab,
(visibleTab = #; trigger = AbsoluteTime[]) &
]
]
]
</code></pre>
<p>You should note that in many cases it is good enough to e.g. do <code>Row[{trigger;Checkbox[Dynamic[x]],Dynamic[x]}]</code>, but here it seems that you really have to trick Mathematica into believing that the value of dummy-variable <code>trigger</code>is actually seen on screen (which it actually is, but at a size of 0x1 pixels that shouldn't be too irritating :-)). The above technique can be used to create a wrapper which makes it somewhat simpler to use:</p>
<pre><code>reinitializingTabView[{tabs___}, opts : OptionsPattern[TabView]] :=
DynamicModule[{tabidx, trigger},
TabView[
Dynamic[Row[{Pane[trigger, {0, 1}], #}]] & /@ {tabs},
Dynamic[tabidx, (tabidx = #; trigger = AbsoluteTime[]) &],
opts
]
]
</code></pre>
<p>and then can be used e.g. like this:</p>
<pre><code>reinitializingTabView[{
DynamicModule[{x = False}, {Checkbox[Dynamic[x]], Dynamic[x]}],
DynamicModule[{x = False}, {Checkbox[Dynamic[x]], Dynamic[x]}],
"foo"
}]
</code></pre>
<p>handling of the second argument and options could probably be improved, but whether that's worth the effort depends on how often you'll use it...</p>
<p>Here is yet another solution which will localize per tab but doesn't regenerate, also quite complicated and somewhat hacky:</p>
<pre><code>DynamicModule[{visibleTab = 1, needsupdate = False},
TabView[{
DynamicModule[{x = False},
DynamicWrapper[
Row[{Checkbox[Dynamic[x]], " ", Dynamic[x]}]
,
Hold@Print["called"];
If[TrueQ[needsupdate],
Hold@Print["doing something"];
x = False;
needsupdate = False
],
UpdateInterval -> Dynamic[If[TrueQ[needsupdate], 1, Infinity]],
TrackedSymbols -> {needsupdate}
]
],
"foo"
}, Dynamic[visibleTab, (visibleTab = #; needsupdate = True) &]
]
]
</code></pre>
<p>If you delete the <code>Hold</code> wrapper for the <code>Print</code>s you'll note that it's now updated 3 times but it only actually does something one time.</p>
<p>I'd really wish this would be possible somewhat simpler but couldn't find something. I consider solutions like that to just be workarounds and hope there is something cleaner to achieve the same thing.</p>
|
1,774,084 | <p>I think it is convergent to $1$ because as $n$ tends to $\infty$ , $1/\sqrt(n)$ tends to $0$. Is it true?</p>
<p>Thanks!</p>
| Piquito | 219,998 | <p>Let $y_n=n^{\frac{1}{\sqrt n}}$ so we have $\log y_n=\frac{\log n}{\sqrt n}$.</p>
<p>Applying now the Hospital rule you have $$\lim_{n\to \infty}y_n=\lim_{n\to \infty}\frac{\log n}{\sqrt n}=\lim_{n\to \infty}\frac{\frac 1n}{-2\sqrt{n^3}}=0$$</p>
<p>Thus your limit is equal to $1$ so your sequence is convergent to $1$ as you think.</p>
|
448,694 | <p>Let $k$ be a field, and $A$ be a finitely generated $k$-algebra. Then does $k$ also have to be a finitely generated field?</p>
<p>Motivation: Let $A$ be generated by $\{a_1,a_2,\dots,a_n\}$, and $k$ be generated by $\{k_1,k_2,k_3,\dots\}$. Then the element $k_1k_2k_3\dots a_1\in A$ is not generated by a finite number of elements. </p>
<p>I suppose the case becomes even clearer when $A$ is a unital $k$-algebra. Then the set of generators of $A$ becomes $\{a_1,a_2,a_3,\dots,a_n,k_1,k_2,\dots\}$</p>
<p>Thanks in advance!</p>
| Chris Eagle | 5,203 | <p>No. For example, $\Bbb{C}$ is a finitely generated $\Bbb{C}$-algebra but not a finitely generated field (since finitely generated fields are countable).</p>
|
3,546,661 | <p>The following is from <em>Elementary Differential Geometry</em> by A.N. Pressley, page 102.</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/KkYsL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KkYsL.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/t2fa6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t2fa6.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/IjJC2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IjJC2.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/s4sY1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s4sY1.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Why is the author using the different choices of curves to prove <span class="math-container">$E_1=E_2$</span>, <span class="math-container">$F_1=F_2$</span> and <span class="math-container">$G_1=G_2$</span>, when it can be obtained by comparing the coefficients Eq(3)?</p>
| Emilio Novati | 187,568 | <p>Because the first curve, with <span class="math-container">$v=v_0$</span> has <span class="math-container">$\dot v =0$</span>,
the second, with <span class="math-container">$u=u_0$</span> has <span class="math-container">$\dot u=0$</span> so are used to compare only the coefficients <span class="math-container">$E_i$</span> and <span class="math-container">$G_i$</span> . Than we need another curve to compare the other coefficients.</p>
|
2,231,391 | <p>I'd like a low-discrepancy sequence of points over a 3D-hypercube <span class="math-container">$[-1,1]^3$</span>, but don't want to have to commit to a fixed number <span class="math-container">$n$</span> of points beforehand, that is just see how the numerical integration estimates develop with increasing numbers of low-discrepancy points.</p>
<p>I'd like to avoid have to start all over again, if the results with a fixed <span class="math-container">$n$</span> are unsatisfactory. Of course, one could just employ random numbers, but then the convergence behavior would be poorer.</p>
<p>"A sequence of n-tuples that fills n-space more uniformly than uncorrelated random points, sometimes also called a low-discrepancy sequence. Although the ordinary uniform random numbers and quasirandom sequences both produce uniformly distributed sequences, there is a big difference between the two." (mathworld.wolfram.com/QuasirandomSequence.html)</p>
<p>This question has also just been put on the mathematica.stack.exchange
(<a href="https://mathematica.stackexchange.com/questions/143457/how-can-one-generate-an-open-ended-sequence-of-low-discrepancy-points-in-3d">https://mathematica.stackexchange.com/questions/143457/how-can-one-generate-an-open-ended-sequence-of-low-discrepancy-points-in-3d</a>)</p>
<p>Since in his answer below, Martin Roberts advances a very interesting, appealing approach to the open-ended low-discrepancy problem, I’d like to indicate an (ongoing) implementation of his approach I’ve just reported in <a href="https://arxiv.org/abs/1809.09040" rel="nofollow noreferrer">https://arxiv.org/abs/1809.09040</a> . In sec. XI (p. 19) and Figs. 5 and 6 there, I analyze two problems—one with sampling dimension <span class="math-container">$d=36$</span> and one with <span class="math-container">$d=64$</span>—both using the parameter <span class="math-container">$\bf{\alpha}_0$</span> set to 0 and also to <span class="math-container">$\frac{1}{2}$</span>. To convert the quasi-uniformly distributed points yielded by the Roberts’ algorithm to quasi-uniformly distributed normal variates, I use the code developed by Henrik Schumacher in his answer to <a href="https://mathematica.stackexchange.com/questions/181099/can-i-use-compile-to-speed-up-inversecdf">https://mathematica.stackexchange.com/questions/181099/can-i-use-compile-to-speed-up-inversecdf</a></p>
| user32038 | 438,631 | <p>another good solution to get an open-ended sequence is using the Halton method. It is also very easy to implement, even for any dimension! For d<8 it has usually good properties, beyond this more difficult will typically outperform Halton.</p>
|
2,323,223 | <p>I am having a hard time with this question for some reason. </p>
<p>You and a friend play a game where you each toss a balanced coin. If the upper faces on
the coins are both tails, you win \$1; if the faces are both heads, you win \$2; if the coins
do not match (one shows head and the other tail), you lose \$1.
Calculate the expected value and standard deviation for your total winnings from this
game if you play 50 times.</p>
<p>PMF Values:
\begin{array}{c|c}
$& p\\\hline
+$1 & .25\\
+$2 & .25\\
-$1 & .50
\end{array}</p>
<p>I have calculated the expectation as $$1(.25)+2(.25)+(-1)(.5) = .25,$$ so $$E(50X) = 50\cdot.25 = \$12.5,$$ which I have confirmed is correct.</p>
<p>I know I need to get $\operatorname{Var}(50X)$, but doing a standard variance calculation and then using the formula $a^2\operatorname{Var}(X)$ is not giving me the correct value.</p>
<p>What step am I missing?</p>
| robjohn | 13,854 | <p>Variance is the mean of the squares minus the square of the mean.
$$
0.25\cdot1^2+0.25\cdot2^2+0.5\cdot(-1)^2-0.25^2=1.6875
$$
For independent events, the variance of a sum is the sum of the variances, so the variance for $50$ events is
$$
50\cdot1.6875=84.375
$$</p>
|
2,323,223 | <p>I am having a hard time with this question for some reason. </p>
<p>You and a friend play a game where you each toss a balanced coin. If the upper faces on
the coins are both tails, you win \$1; if the faces are both heads, you win \$2; if the coins
do not match (one shows head and the other tail), you lose \$1.
Calculate the expected value and standard deviation for your total winnings from this
game if you play 50 times.</p>
<p>PMF Values:
\begin{array}{c|c}
$& p\\\hline
+$1 & .25\\
+$2 & .25\\
-$1 & .50
\end{array}</p>
<p>I have calculated the expectation as $$1(.25)+2(.25)+(-1)(.5) = .25,$$ so $$E(50X) = 50\cdot.25 = \$12.5,$$ which I have confirmed is correct.</p>
<p>I know I need to get $\operatorname{Var}(50X)$, but doing a standard variance calculation and then using the formula $a^2\operatorname{Var}(X)$ is not giving me the correct value.</p>
<p>What step am I missing?</p>
| Graham Kemp | 135,106 | <p>You are confusing the distribution of $50X_1$ and $\sum_{k=1}^{50}X_k$ when ${(X_k)}_{k=1}^n$ is a sequence of independent and identically distributed random variables.</p>
<p>It is true that $\mathsf E(50X_1)=50\mathsf E(X_1)$ and $\mathsf{Var}(50X_1)=2500\mathsf {Var}(X_1)$. However, that is not what you are dealing with.</p>
<hr>
<p>Due to the Linearity of Expectations, the expectation of the series is the series of expectations. Its because the distributions are identical that this series is equal to $50$ times an individual expectation .</p>
<p>$$\begin{align}\mathsf E(\sum_{k=1}^{50}X_k) ~&=~ \sum_{k=1}^{50}\mathsf E(X_k) & \text{Linearity of Expectation} \\[1ex] & =~ 50\,\mathsf E(X_1) & \text{Indentical Distributions}\end{align}$$</p>
<p>Similar result, different reasoning.</p>
<p>(Note: We have not use independence at thi point.)</p>
<hr>
<p>The distinction becomes apparent in dealing with the variance.</p>
<p>When it comes the the variance of the series, we have to employ the Bilinearity of Covariance.</p>
<p>$$\begin{align}\mathsf {Var}(\sum_{k=1}^{50} X_k) ~&=~ \mathsf {Cov}(\sum_{k=1}^{50}X_k,\sum_{j=1}^{50}X_j) \\ &=~ \sum_{k=1}^{50}\sum_{j=1}^{50}\mathsf{Cov}(X_k,X_j) &&\text{Bilinearity of Covariance} \\ &=~ \sum_{k=1}^{50}\mathsf {Cov}(X_k,X_k) ~+~ 0 &&\text{Independence: }\mathsf{Cov}(X_j,X_k)=0\text{ when }j\neq k \\[1ex] &=~ 50\mathsf {Var}(X_1) && \text{Identical distributions}
\end{align}$$</p>
|
35,220 | <p>It is a basic result of group cohomology that the extensions with a given abelian normal subgroup <em>A</em> and a given quotient <em>G</em> acting on it via an action $\varphi$ are given by the second cohomology group $H^2_\varphi(G,A)$. In particular, when the action is trivial (so the extension is a central extension), this is the second cohomology group $H^2(G,A)$ for the trivial action. In the special case where <em>G</em> is also abelian, we classify all the class two groups with <em>A</em> inside the center and <em>G</em> as the quotient group.</p>
<p>I am interested in the following: given a sequence of abelian groups $A_1, A_2, \dots, A_n$, what would classify (up to the usual notion of equivalence via commutative diagrams) the following: a group <em>E</em> with an ascending chain of subgroups:</p>
<p>$$1 = K_0 \le K_1 \le K_2 \le \dots \le K_n = E$$</p>
<p>such that the $K_i$s form a central series (i.e., $[E,K_i] \subseteq K_{i-1}$ for all <em>i</em>) and $K_i/K_{i-1} \cong A_i$?</p>
<p>The case $n = 2$ reduces to the second cohomology group as detailed in the first paragraph, so I am hoping that some suitable generalization involving cohomology would help describe these extensions.</p>
<p>Note: As is the case with the second cohomology group, I expect the object to classify, not isomorphism classes of possibilities of the big group, but a notion of equivalence class under a congruence notion that generalizes the notion of congruence of extensions. Then, using the actions of various automorphism groups, we can use orbits under the action to classify extensions under more generous notion of equivalence.</p>
<p>Note 2: The crude approach that I am aware of involves building the extension step by step, giving something like a group of groups of groups of groups of ... For intsance, in the case $n = 3$:</p>
<p>$$1 = K_0 \le K_1 \le K_2 \le K_3 = G$$</p>
<p>with quotients $A_i \cong K_i/K_{i-1}$, I can first consider $H^2(A_3,A_2)$ as the set of possibilities for $K_3/K_1$ (up to congruence). For each of these possibilities <em>P</em>, there is a group $H^2(P,A_1)$ and the total set of possibilities seems to be:</p>
<p>$$\bigsqcup_{P \in H^2(A_3,A_2)} H^2(P,A_1)$$</p>
<p>Here the $\in$ notation is being abused somewhat by identifying an element of a cohomology group with the corresponding extension's middle group.</p>
<p>What I really want is some algebraic way of thinking of this unwieldy disjoint union as a single object, or some information or ideas about its properties or behavior.</p>
| Richard Borcherds | 51 | <p>There is probably no easy answer to this question. Even the special case when all groups are of order 2 seems hopeless. While it is not the same as the problem of classifying all finite groups of order a power of 2, it is similar, and probably of about the same difficulty. Classifying groups of order 2<sup>n</sup> is known (or at least thought) to be a complete mess. The number of such groups grows rapidly with n:</p>
<p>1, 1, 2, 5, 14, 51, 267, 2328, 56092, 10494213, 49487365422<br>
<a href="http://oeis.org/A000679" rel="nofollow">http://oeis.org/A000679</a></p>
<p>Moreover if one actually looks at the collection of groups one gets, there does not seem to be much obvious nice structure. And if you change the prime 2 to some other prime such as 3, the answer you gets seems to change qualitatively: there are 2-groups that do not seem to be analogous to any 3-groups, and vice versa. Looking at the smallish groups of this form does not seem to give any hints of any usable structure on the set of groups with such an ascending chain of subgroups. </p>
<p>The people who classify 2-groups have presumably thought quite hard about this problem, and as far as I know have not come up with any easy solution: the groups are classified with a lot of hard work and computer time.</p>
|
160,165 | <p>For an elliptic curve $E$ over $\Bbb{Q}$, we know from the proof of the Mordell-Weil theorem that the weak Mordell-Weil group of $E$ is $E(\Bbb{Q})/2E(\Bbb{Q})$. It is well known that
$$
0 \rightarrow E(\Bbb{Q})/2E(\Bbb{Q}) \rightarrow S^{(2)}(E/\Bbb{Q}) \rightarrow Ш(E/\Bbb{Q})[2] \rightarrow 0
$$
is an exact sequence which gives us a procedure to compute the generators for $E(\Bbb{Q})/2E(\Bbb{Q})$. </p>
<p>(Relatively) recently I found out that there is another way to compute the rank of $E$ using $3$-descent. I was wondering, since the natural structure of the weak Mordell-Weil group is $E(\Bbb{Q})/2E(\Bbb{Q})$, what is the motivation behind using $3$-descent? Also does $3$-descent similarly produce the generators of $E(\Bbb{Q})/2E(\Bbb{Q})$ or does it simply tell us the structure of $E(\Bbb{Q})$ via the Mordell-Weil theorem by giving us only the rank of $E$? Finally does it help us get around the issue of $Ш(E/\Bbb{Q})$ containing an element that is infinitely $2$-divisible?</p>
| Alex B. | 3,212 | <p>There is no "natural" structure. The point is that if $E(\mathbb{Q})\cong \Delta\oplus \mathbb{Z}^r$, where $\Delta$ is a finite group, then for any positive integer $n$,
$$
E(\mathbb{Q})/nE(\mathbb{Q}) \cong \Delta/n\Delta\oplus (\mathbb{Z}/n\mathbb{Z})^r.$$
Note that this is a finite group. Since we can basically assume that we know everything about $\Delta$, computing $|E(\mathbb{Q})/nE(\mathbb{Q})|$ is equivalent to computing the rank $r$.</p>
<p>Moreover, for any $n$, we have the short exact sequence$^1$
$$
0 \rightarrow E(\Bbb{Q})/nE(\Bbb{Q}) \rightarrow S^{(n)}(E/\Bbb{Q}) \rightarrow Ш(E/\Bbb{Q})[n] \rightarrow 0.$$</p>
<p>The point is now that if you can compute the middle term, you get an upper bound on the rank. Depending on the precise shape of your elliptic curve, this term can be easier of harder to compute for various $n$. Moreover, depending on the size of the Tate-Shafarevich group, the upper bound you get might be tighter or less tight for different $n$. That's the reason people try to do $n$-descent for as many different $n$ as possible. The number 2 plays no special role here. You should read up on all this in Silverman.</p>
<p>To recapitulate, 3-descent produces (under ideal circumstances!) the generators of $E(\Bbb{Q})/3E(\Bbb{Q})$. But it might fail to do so, if the Tate-Shafarevich group has non-trivial 3-torsion. Then you might need to try a different $n$.</p>
<p>$^1$ In fact, such a sequence exists for any isogeny between two elliptic curves, but let's not go there for now. This is all nicely explained in Silverman.</p>
|
2,856,180 | <p>I've been learning some about counting and basic combinatorics. But some scenarios were not explained in my class...</p>
<p><strong>Example problem:</strong> You are given 6 tiles. 1 is labeled "1", 2 are labeled "2", and 3 are labeled "3".</p>
<p><strong>Problem 1:</strong> How many different ways can you arrange groups of three tiles (order matters)?</p>
<p><strong>Answer 1: 19</strong>, {1,2,2} {1,2,3} {1,3,2} {1,3,3} {2,1,2} {2,1,3} {2,2,1} {2,2,3} {2,3,1} {2,3,2} {2,3,3} {3,1,2} {3,1,3} {3,2,1} {3,2,2} {3,2,3} {3,3,1} {3,3,2} {3,3,3}.</p>
<p><strong>Question 1:</strong> You can't use the normal $\frac{5!}{\left(5-3\right)!}$ because you have repeated symbols. And you can't divide by the factorials of repeated symbols $\frac{6!}{\left(6-3\right)!\cdot3!\cdot2!}$ like you can when n = k and you make different arrangements of all n of symbols $\frac{n!}{\prod_{i=1}^mn_i}$. What formula would be used to find the answer to a more complex version of this problem?</p>
<p><strong>Problem 2:</strong> How many different ways can you make groups of three tiles (order doesn't matter)?</p>
<p><strong>Answer 2: 6</strong>, {1,2,2} {1,2,3} {1,3,3} {2,2,3} {2,3,3} {3,3,3}.</p>
<p><strong>Question 2:</strong> You can't use the normal $\frac{n!}{\left(n-k\right)!\cdot k!}$ because you have repeated symbols. What formula would be used to find the answer to a more complex version of this problem?</p>
| Shrey Joshi | 572,044 | <p>Question 1:
This is the same as finding the number of permutations of the 'word' $ABBCCC$. Let us distinguish the same letters by subscripts so that we now have $A_1B_1B_2C_1C_2C_3$. The number of of permutations of this is just $6!$. But for each valid permutation, there are $2!3!$ repeats due to permutations of the $B_i$ and $C_i$. Therefore our answer is $\frac{6!}{2!3!}=60$.</p>
<p>Question 2:
My best guess is organized casework.</p>
|
2,856,180 | <p>I've been learning some about counting and basic combinatorics. But some scenarios were not explained in my class...</p>
<p><strong>Example problem:</strong> You are given 6 tiles. 1 is labeled "1", 2 are labeled "2", and 3 are labeled "3".</p>
<p><strong>Problem 1:</strong> How many different ways can you arrange groups of three tiles (order matters)?</p>
<p><strong>Answer 1: 19</strong>, {1,2,2} {1,2,3} {1,3,2} {1,3,3} {2,1,2} {2,1,3} {2,2,1} {2,2,3} {2,3,1} {2,3,2} {2,3,3} {3,1,2} {3,1,3} {3,2,1} {3,2,2} {3,2,3} {3,3,1} {3,3,2} {3,3,3}.</p>
<p><strong>Question 1:</strong> You can't use the normal $\frac{5!}{\left(5-3\right)!}$ because you have repeated symbols. And you can't divide by the factorials of repeated symbols $\frac{6!}{\left(6-3\right)!\cdot3!\cdot2!}$ like you can when n = k and you make different arrangements of all n of symbols $\frac{n!}{\prod_{i=1}^mn_i}$. What formula would be used to find the answer to a more complex version of this problem?</p>
<p><strong>Problem 2:</strong> How many different ways can you make groups of three tiles (order doesn't matter)?</p>
<p><strong>Answer 2: 6</strong>, {1,2,2} {1,2,3} {1,3,3} {2,2,3} {2,3,3} {3,3,3}.</p>
<p><strong>Question 2:</strong> You can't use the normal $\frac{n!}{\left(n-k\right)!\cdot k!}$ because you have repeated symbols. What formula would be used to find the answer to a more complex version of this problem?</p>
| Karn Watcharasupat | 501,685 | <p><strong>For the case where order does not matter.</strong></p>
<p>Suppose you have $x$, $y$, and $z$ objects A, B and C respectively. You are choosing $k$ tiles. </p>
<p>If $k-(y+z)>0$, then you need to pick at least $(k-(y+z))$ of object A. Otherwise, at least zero.</p>
<p>For $i$ of object A chosen, you must choose $(k-i)$ of objects B and C combined.</p>
<p>If $(k-i)-z>0$, then you need to pick at least $((k-i)-z)$ of object B. Otherwise, at least zero.</p>
<p>For $j$ of object B chosen, you must choose $(k-(i+j))$ of object C.</p>
<p>In total, you have $i$ of object A, $j$ of object B, and $(k-(i+j))$ of object C.</p>
<p>If order does not matter, you permute the chosen $k$ objects and account for the identical objects.</p>
<p>$$\sum_{i=\max\{0,k-(y+z)\}}^{i=\min\{k,x\}}\sum_{j=\max\{0,(k-i)-z\}}^{i=\min\{k-i,y\}}\dfrac{k!}{i!j!(k-i-j)!}$$</p>
<hr>
<p><strong>For the case where order matters</strong></p>
<p>We don't really have a formula for this. But an organized method is more crucial.</p>
<p>Check if any of $x,y,z$ is less than $k$. Those that are equal or more than $k$ is in <em>excess supply</em> and we don't worry about them. Those that is less than $k$ introduces cases.</p>
<p>For your case, let me rename some stuff. You have 1 A, 2 B's and 3 C's. The <em>limiting agents</em> are $A$ and $B$.</p>
<p>So we consider all pairs of $(a,b)$ such that $0\le a+b \le k$. For your case, we have
$$(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$$</p>
<p>\begin{array}{c|c}
(a,b) & \text{number of ways} \\
\hline
(0,0) & \frac{3!}{3!}=1\\
(0,1) & \frac{3!}{2!}=3\\
(0,2) & \frac{3!}{2!}=3\\
(1,0) & \frac{3!}{2!}=3 \\
(1,1) & 3!=6\\
(1,2) & \frac{3!}{2!}=3\\
\end{array}</p>
<p>Total: 19 ways.</p>
<p>Of course you can derive a summation from this, <strong>but</strong> a generalized summation is not very worth deriving unless you have a specific context.</p>
|
35,585 | <p>Let H be an infinite dimensional and separable Hilbert space. Let C be a closed and
connected subset of H containing more than one point. Can C ever be the countable union
of closed and totally disconnected subsets of H?</p>
| BS. | 6,451 | <p>Let $E_c$ be the <em>complete Erdös space</em> (Erdös, Annals of Math vol 41 1940), defined as the subspace of $\ell^2(\mathbb{N})$ where all coordinates are irrationals. It is <a href="http://en.wikipedia.org/wiki/Polish_space" rel="nofollow">polish</a> (separable and completely metrizable) and totally disconnected, but admits a "connectification" namely a (still polish) topology on $E_c\cup\{p\}$ that makes it connected (and of course induces the one on $E_c$). The crucial point is the fact that any nonempty closed and open subset of $E_c$ is unbounded. Then, as in Bill Johnson's answer $E_c$ is the union of the closed and totally disconnected subspaces $\overline{B}(0,n)\cup\{p\}$, $n\geq 1$. It remains to remark that, like any polish space, $E_c\cup\{p\}$ embeds as a closed subset of $H$ (as remarked in Gerald Edgar's answer).</p>
|
3,710,804 | <p><span class="math-container">$$f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$$</span></p>
<p>I have been up on this problem for an hour, but without any clues. </p>
<p>Can someone please help me solving this?</p>
| Vishu | 751,311 | <p>The integral is <span class="math-container">$$ \int \frac{\cos x(5-8\sin^2x)}{\sin x(5-4\sin^2x)} dx$$</span>
Substitute <span class="math-container">$t=\sin x \implies dt =\cos x dx$</span> to get <span class="math-container">$$\int \frac{5-8t^2}{t(5-4t^2)} dt$$</span> Now, use partial fraction decomposition: <span class="math-container">$$ =\int\frac 1t dt+\int\frac{4t}{4t^2-5}dt $$</span> Substitute <span class="math-container">$u=4t^2-5 \implies du=8tdt$</span> in the second integral:</p>
<p><span class="math-container">$$= \ln|t| + \frac 12 \int \frac{du}{u} \\ =\ln|\sin x| +\frac 12 \ln|4\sin^2-5| +C$$</span></p>
|
2,466,947 | <p>In class we are beginning discrete probability and we are being introduced to counting. He said in an example today that the word "anagram" (which has $7$ letters) can be rearranged to have $7!/3!$ possible words when we don't count the order of the a's. To me this seemed like a combination but this result reduces to $7 \cdot 6 \cdot 5 \cdot 4$ which means we are choosing $7$ letters on $4$ spots. So a bit confused if anyone could help?</p>
| JMoravitz | 179,297 | <p>This can be explained as the following using multiplication principle:</p>
<ul>
<li>Pick which of the seven available spaces is occupied by the <code>n</code></li>
<li>Pick which of the six remaining available spaces is occupied by the <code>g</code></li>
<li>$\vdots$</li>
<li>Pick which of the four remaining spaces is occupied by the <code>m</code></li>
<li>All remaining three spaces will be occupied by the <code>a</code>'s.</li>
</ul>
<p>Applying multiplication principle, there are $7\cdot6\cdot5\cdot4$ ways to do this.</p>
|
1,003,096 | <p>Let $G=(\mathbb{Q}-\{0\},*)$ and $H=\{\frac{a}{b}\mid a,b\text{ are odd integers}\}$.</p>
<ol>
<li>Show $H$ is a normal subgroup of $G$.</li>
<li>Show that $G/H \cong (\mathbb{Z},+)$</li>
</ol>
<p>I know that there are multiple definitions for normal subgroup and I am having a hard time to develop the proof for these particular sets. </p>
<p>For part 2. I need help developing a function from $G/H \to (\mathbb{Z},+)$.</p>
| Mike Earnest | 177,399 | <p>Subgroups of abelian groups are always normal. One way to show normality is to show that for any $g\in G$, and $h\in H$, that $ghg^{-1}\in H$. But this follows immediately , since $ghg^{-1}=gg^{-1}h=h\in H$, since multiplication is commutative.</p>
|
57,131 | <p>Evaluating the following lines in my computer takes near six seconds in Mathematica 10 and near 5 in Mathematica 9. I consider this very slow as only quantiles are being calculated from data to create this simple set of charts. I think that Mathematica should do much better than this. Do you agree? Thoughts?</p>
<pre><code>data = Table[RandomVariate[NormalDistribution[RandomInteger[5], 1], 100000], {10}];
Timing@BoxWhiskerChart[data, ChartLabels -> {"a", "b", "c"}, PerformanceGoal -> "Speed"]
</code></pre>
| Verbeia | 8 | <p>It is calculating the quantiles that takes the time. Your code takes a bit under 4 seconds on my machine in version 9, but if I cut the sample sizes to 10,000 instead of 100,000, it's just under 0.4 seconds. Clearly calculating quantiles is roughly linear in performance with respect to the sample size. </p>
<pre><code>Table[With[{data = Table[RandomVariate[
NormalDistribution[RandomInteger[5], 1], i], {10}]},
{i, First@Timing@ BoxWhiskerChart[data, PerformanceGoal -> "Speed"]}],
{i, 20000, 120000, 20000}]
20000 0.748805
40000 1.497610
60000 2.230814
80000 2.979619
100000 3.728424
120000 4.524029
</code></pre>
<p><img src="https://i.stack.imgur.com/IAPyy.png" alt="enter image description here"></p>
<p>But note that calculating quantiles separately from the plotting function, using <code>Quantile</code>, is much faster. For the 100,000 size samples, I obtained a timing of about 0.12 seconds for the following:</p>
<pre><code>Timing[Quantile[#, {0.1, 0.25, 0.5, 0.75, 0.9}] & /@ data]
</code></pre>
<p>So, yes, calculating quantiles in the context of <code>BoxWhiskerChart</code> does seem to be less efficient than calculating the quantiles directly. It might be something to do with the automatic outliers functionality in that charting function.</p>
|
9,168 | <p>I'm having a doubt about how should we users encourage the participation of new members. So far I have only presented MSE to three of my fellow colleagues in grad school. In an overall way I feel like if MSE becomes too open and wide known, some of the high-rank researchers and top-class grad and undergrads users will be frustrated. Maybe I get this impression for seeing how some really complicated questions are well-received.</p>
<p>But today I saw someone asking some really simple algebra questions like "solve this equation for x" and started thinking what is the general feel of MSE when this kind of question arises. I'm not judging anything and no one. I just would like to get some opinions on this.</p>
| Community | -1 | <p>As the faq has,</p>
<p><code>Mathematics - Stack Exchange is for people studying mathematics at any level and professionals in related fields.</code></p>
<p>So I would suggest and strongly enourage in spreading the word around (I certainly spread the word among my co-grad students and Profs).</p>
<p>That said, I also strongly believe that the following three questions, which I assume are the ones you are also referring to in your question, should be closed as too localized. </p>
<ul>
<li><a href="https://math.stackexchange.com/questions/369866/">How to solve the equation $-5(y+7)=25$?</a></li>
<li><a href="https://math.stackexchange.com/questions/369817/">How can I solve the equation $8=5x+2$?</a></li>
<li><a href="https://math.stackexchange.com/questions/369783/">Solve the equation $x-7=28$</a></li>
</ul>
<p>All three have been asked by the <a href="https://math.stackexchange.com/users/71668/eli-arriaga">same person</a> in a short span of time. It makes sense to answer one of these in a general setting and close the rest as duplicate.</p>
|
65,582 | <p>Let $G$ be a locally compact group on which there exists a Haar measure, etc..</p>
<p>Now I am supposed to take such a metrisable $G$, and given the existence of some metric on $G$, prove that there exists a translation-invariant metric, i.e., a metric $d$ such that $d(x,y) = d(gx,gy)$ for all $x,y,g \in G$. How to go about this?</p>
| Lucas Kaufmann | 12,501 | <p>What if you take the original metric $d_0$ and define the new metric by<br>
$d(x,y) = \int_G d_0(gx,gy) d\mu$</p>
<p>where $d\mu$ is the Haar measure?<br></p>
<p>The invariance of the measure implies the invariance of the integral, and hence of $d$.</p>
<p>EDIT: As Theo pointed out, this works only when $G$ is compact.</p>
|
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