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 |
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2,601,851 | <p>I have a binary variable $y_{t}$ that is equal to $1$ iff the job is scheduled at slot $t$. I need to write constraints that guarantee that if the job is scheduled somewhere, then it must be scheduled for a period of $A$ consecutive slots. I tried to write it this way:</p>
<p>$\sum_{t'=t}^{t+A-1}y_{t'}\geqslant A y_t$ for all $t$.</p>
<p>Here I can see that if $y_t=1$ then I must have $y_{t+1}=\ldots=y_{t+A-1}=1$. The problem with this is that I will have $y_{t'}=1$ for all $t'\geqslant t$ because of the recurrence relation I have in my constraints.</p>
<p>After some effort, I found this way to do it: introduce binary variable $z_t$and add the constraints</p>
<p>$\sum_{t}z_t\leqslant 1$and $z_t\sum_{t'=t}^{t+A-1}y_{t'}\geqslant A y_t z_t$ for all $t$.</p>
<p>But, if I am correct, this is a non-linear constraint.</p>
| user160110 | 160,110 | <p>Imagine this is the real number line</p>
<p>$--------------$</p>
<p>Let's put $U=U(f,P)$ and$L=L(f,P)$ on there</p>
<p>$---L----------U---$</p>
<p>Now we know that the distance between these two is less than $\epsilon$ </p>
<p>$---L----------U---$
$---|<---(\epsilon< )--->|--$</p>
<p>Given that the smallest upper area achieved by partition $U^*$ is smaller than $U$ and the largest lower area achieved $L^*$ is larger than $L$ we see that</p>
<p>$---L-L^*----U^*-U---$
$---|<---(\epsilon< )--->|--$</p>
<p>Meaning that</p>
<p>$U^*-L^*< \epsilon$</p>
|
3,511,660 | <p>Can you help me please I could not figure this out.</p>
<p>Given: </p>
<p><span class="math-container">$f:\mathbb{R}\to\mathbb{R}$</span>, <span class="math-container">$f'(0)$</span> exists, <span class="math-container">$f(x)\neq0$</span> and for all <span class="math-container">$a, b\in\mathbb{R}$</span>, <span class="math-container">$f(a+b)=f(a)f(b)$</span></p>
<p>How to prove that <span class="math-container">$f(x)$</span> is differentiable in <span class="math-container">$\mathbb{R}$</span>? </p>
| Dominik Kutek | 601,852 | <p>Since <span class="math-container">$f'(0)$</span> exists, there is <span class="math-container">$\delta >0$</span> such that <span class="math-container">$f$</span> is continuous on <span class="math-container">$(-\delta,\delta)$</span>. Since <span class="math-container">$f > 0$</span> (why), we can take <span class="math-container">$\ln$</span> both sides getting: <span class="math-container">$\ln(f(a+b)) = \ln(f(a)) + \ln(f(b))$</span>. Take <span class="math-container">$ g = \ln \circ f $</span></p>
<p>We have <span class="math-container">$g(a+b) = g(a) + g(b)$</span> and we know that <span class="math-container">$g$</span> is continuous on <span class="math-container">$(-\delta,\delta)$</span>, and since <span class="math-container">$g$</span> is linear, it is continuous on <span class="math-container">$\mathbb R$</span>.</p>
<p>Now it is an exercise from analysis to check, that only continuous solutions to this are <span class="math-container">$g(x) = Cx$</span> where <span class="math-container">$C \in \mathbb R$</span> is a constant.</p>
<p>We do it this way:
<span class="math-container">$g(0+0) = 2g(0)$</span> so <span class="math-container">$g(0) = 0$</span>
<span class="math-container">$g(2a) = 2g(a)$</span>, and by induction one can take <span class="math-container">$g(na) = ng(a)$</span>, for any <span class="math-container">$a \in \mathbb R$</span>.</p>
<p>Then take <span class="math-container">$g(a-a) = g(a) + g(-a)$</span>, so <span class="math-container">$g(a) = -g(-a)$</span>, and we have: for any <span class="math-container">$a \in \mathbb R, k \in \mathbb Z$</span> <span class="math-container">$g(ka) = kg(a)$</span>.</p>
<p>Now we go to rationals: take any <span class="math-container">$p,q \in \mathbb Z, a \in \mathbb R$</span>.</p>
<p>We have <span class="math-container">$pg(a) = pg(\frac{q}{q}a) = pq g(\frac{a}{q}) = q g(\frac{ap}{q})$</span>, so <span class="math-container">$\frac{p}{q}g(a) = g(a\frac{p}{q})$</span>.</p>
<p>We showed that for any <span class="math-container">$a \in \mathbb R, r \in \mathbb Q$</span> we have <span class="math-container">$g(ra) = rg(a)$</span>. And now by continuity, one have <span class="math-container">$g(xa) = xg(a)$</span> for any <span class="math-container">$x,a \in \mathbb R$</span>.</p>
<p>Take <span class="math-container">$a=1$</span>, we have <span class="math-container">$g(x) = xg(1) = Cx$</span>, where <span class="math-container">$C=g(1) = \ln(f(1))$</span> is some constant.</p>
<p>And we have <span class="math-container">$g(x) = \ln(f(x))$</span>, so <span class="math-container">$f(x) = e^{Cx}$</span> and it clearly is differentiable.</p>
|
4,624,058 | <p>Godsil&Royle <a href="https://doi.org/10.1007/978-1-4613-0163-9" rel="nofollow noreferrer">Algebraic Graph Theory</a> section 2.5 states (slightly paraphrased):</p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a transitive group acting on a set <span class="math-container">$V$</span>. A nonempty subset <span class="math-container">$S$</span> of <span class="math-container">$V$</span> is a <em>block of imprimitivity</em> for G if for any element <span class="math-container">$g\in G$</span>, either <span class="math-container">$g(S) = S$</span> or <span class="math-container">$g(S)\cap S = \emptyset$</span>. Because <span class="math-container">$G$</span> is transitive, it is clear that the translates of <span class="math-container">$S$</span> form a partition on <span class="math-container">$V$</span>.</p>
</blockquote>
<p>Assuming "translates of <span class="math-container">$S$</span>" refers to all the <span class="math-container">$g(S)$</span>, I understand why those <em>cover</em> <span class="math-container">$V$</span>.</p>
<p><strong>Question:</strong> But how does it follow that any two <span class="math-container">$g(S)$</span> and <span class="math-container">$h(S)$</span>, if not identical, have an empty intersection?</p>
<p>Put another way: why is the situation in this diagram not possible?</p>
<p><a href="https://i.stack.imgur.com/mlqU8m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mlqU8m.jpg" alt="overlapping g(S) and h(S)" /></a></p>
<p>The diagram shall describe <span class="math-container">$x_1,x_2\in S$</span> both mapped to some <span class="math-container">$x\notin S$</span> by <span class="math-container">$g$</span> and <span class="math-container">$h$</span> respectively where otherwise <span class="math-container">$g(S)\neq h(S)$</span>.
I tried to derive a contradiction from this situation, but failed.</p>
| Devo | 1,092,170 | <p>If <span class="math-container">$g(S)=f(S)$</span>, you are done. So, let's assume <span class="math-container">$g(S)\ne f(S)$</span>.</p>
<ul>
<li>Case 1: <span class="math-container">$g(S)=S$</span> and <span class="math-container">$f(S)\ne S$</span>; then, <span class="math-container">$g(S)\cap f(S)=S\cap f(S)=\emptyset$</span>.</li>
<li>Case 2: <span class="math-container">$g(S)\ne S$</span> and <span class="math-container">$f(S)=S$</span>; then, <span class="math-container">$g(S)\cap f(S)=g(S)\cap S=\emptyset$</span>.</li>
<li>Case 3: <span class="math-container">$g(S)\ne S$</span> and <span class="math-container">$f(S)\ne S$</span>; if <span class="math-container">$g(S)\cap f(S)\ne\emptyset$</span>, then <span class="math-container">$\exists s',s''\in S$</span> such that <span class="math-container">$g(s')=f(s'')$</span>, and hence <span class="math-container">$(f^{-1}g)(s')=s''\in S$</span>; but then <span class="math-container">$(f^{-1}g)(S)=S$</span>, because <span class="math-container">$S$</span> in a block of imprimitivity. Therefore, <span class="math-container">$g(S)=f(S)$</span>: contradiction, and hence <span class="math-container">$g(S)\cap f(S)=\emptyset$</span>.</li>
</ul>
<p>So, <span class="math-container">$g(S)=f(S)$</span> or <span class="math-container">$g(S)\cap f(S)=\emptyset$</span>.</p>
|
864,212 | <p>While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the <a href="http://en.wikipedia.org/wiki/Euclidean_domain">Euclidean domain</a> page of Wikipedia that</p>
<p>$$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$</p>
<p>is such a ring. After a good deal of searching, I have not been able to find any other (online) reference to this ring.</p>
<p>Can anyone confirm this result? Is there a reference for it (paper, textbook or website)?</p>
| Jessica B | 81,247 | <p>Here are some more details for those who, like me, are not so familiar with this material.</p>
<p>$1$) Take a non-zero prime ideal. We wish to show that it is maximal. It contains a prime element (since we are in a UFD, choose an element with a shortest factorisation). Take this to be $ax+by+c$. We will show that the ideal is maximal by showing that the quotient ring by this ideal is a field.</p>
<p>Using the relation $ax+by+c=0$ we can express any element as a polynomial in $x$ with coefficients in $\mathbb{R}$. With the quadratic relation we can make this a linear expression. In searching for an inverse we get two equations in two unknowns, and the determinant of the corresponding matrix is non-zero.</p>
<hr>
<p>$2$) Non-zero, non-invertible elements exist because otherwise we would be in a field and wouldn't be trying to prove the lemma in the first place.</p>
<p>$3$) Suppose $p$ is not prime. Then there exist $a$ and $b$ such that $p\mid ab$ but $p\nmid a$ and $p\nmid b$.
Write $a=q_a p+r_a$ with $\delta(r_a)<\delta(p)$. Then $r_a$ is invertible, with inverse $s_a$. Similarly, we have $q_b$, $r_b$, $s_b$.</p>
<p>Then $ab=q_a (q_b p+r_b)p +q_b r_a p +r_b r_a$, so $p\mid r_a r_b$. Then $p\mid r_a s_a r_b s_b =1$, a contradiction.</p>
<hr>
<p>$4$) $A^{\times}=\mathbb{R}^{\times}$ follows because every element of $A$ can be uniquely represented as $p(y)+x q(y)$ for polynomials $p$ and $q$.</p>
<p>$5$) From $1$) we know that each element of $A/pA$ can be expressed as a linear expression in $x$, and we have one relation $x^2(a^2+b^2)+2acx+c^2+1=0$. The roots of this are non-real, so by sending $x$ to one of them we can define an isomorphism to $\mathbb{C}$.</p>
<p>$6$) The projection homomorphism is injective because having an invertible element in $pA$ would show that $p$ is also invertible.</p>
<p>$7$) $\mathbb{R}^{\times}$ and $\mathbb{C}^{\times}$ are not isomorphic because $\mathbb{C}^{\times}$ has order $4$ elements whereas $\mathbb{R}^{\times}$ does not (see linked question).</p>
|
3,858,517 | <p>Is it possible to count exactly the number of binary strings of length <span class="math-container">$n$</span> that contain no two adjacent blocks of 1s of the same length? More precisely, if we represent the string as <span class="math-container">$0^{x_1}1^{y_1}0^{x_2}1^{y_2}\cdots 0^{x_{k-1}}1^{y_{k-1}}0^{x_k}$</span> where all <span class="math-container">$x_i,y_i \geq 1$</span> (except perhaps <span class="math-container">$x_1$</span> and <span class="math-container">$x_k$</span> which might be zero if the string starts or ends with a block of 1's), we should count a string as valid if <span class="math-container">$y_i\neq y_{i+1}$</span> for every <span class="math-container">$1\leq i \leq k-2$</span>.</p>
<p>Positive examples : 1101011 (block sizes are 2-1-2), 00011001011 (block sizes are 2-1-2), 1001100011101 (block sizes are 1-2-3-1)</p>
<p>Negative examples : 1100011 (block sizes are <strong>2-2</strong>), 0001010011 (block sizes are <strong>1-1</strong>-2), 1101011011 (block sizes are 2-1-<strong>2-2</strong>)</p>
<p>The sequence for the first <span class="math-container">$16$</span> integers <span class="math-container">$n$</span> is: 2, 4, 7, 13, 24, 45, 83, 154, 285, 528, 979, 1815, 3364, 6235, 11555, 21414. For <span class="math-container">$n=3$</span>, only the string 101 is invalid, whereas for <span class="math-container">$n=4$</span>, the invalid strings are 1010, 0101 and 1001.</p>
| Phicar | 78,870 | <p>I am gonna attempt to complement <strong>RobPratt</strong>'s proposed approach involving inclusion exclusion and stars and bars and be that person who posts a horribly long formula.<br><br>
Consider <span class="math-container">$$A_{n,k,r}=\left |\left \{0^{l_1}1^{k_1}\cdots 0^{l_r}1^{k_r}0^{l_{r+1}}\in \{0,1\}^n: k_i>0,k_i\neq k_{i+1}, \sum _{i=1}^rk_i=k \text{ and for $i\neq 1,r+1,$ }l_i>0\right \}\right |.$$</span> Our desired result will be <span class="math-container">$$A_n=\sum _{k=0}^n\sum _{r=0}^nA_{n,k,r}.$$</span>
Notice that we can, by the multiplication principle, express <span class="math-container">$A_{n,k,r}$</span> as
<span class="math-container">$$A_{n,k,r}=|B_{n,k,r}|\times |C_{k,r}|,$$</span>
where <span class="math-container">$$B_{n,k,r}=\left \{(l_1,\cdots ,l_{r+1})\in \left (\mathbb{Z}^{\geq 0}\right )^{r+1}:\sum l_i=n-k, l_i>0 \text{ for $1<i<r+1$}\right \}$$</span>
represents the way to place the <span class="math-container">$0'$</span>s and <span class="math-container">$$C_{k,r}=\{(k_1,\cdots ,k_r)\in \left (\mathbb{Z}^{> 0}\right )^{r}:\sum k_i=k,k_i\neq k_{i+1}\}.$$</span>
represents the way to place the 1's.<br><br>
By stars and bars we get that <span class="math-container">$|B_{n,k,r}|=\binom{n-k-(r-1)+(r+1)-1}{(r+1)-1}=\binom{n-k+1}{r}.$</span> Now, consider the following set <span class="math-container">$C_{k,r,x}=\{(k_1,\cdots k_r)\in C_{k,r}:k_x=k_{x+1}\}$</span> which carries the words with at least one consecutive chunk of 1's of the same size(at index <span class="math-container">$x$</span>). To illustrate the next step, notice that <span class="math-container">$|C_{k,r,x}|=\sum _{t=1}^{\lfloor k/2\rfloor}\binom{k-2t-1}{r-2-1},$</span> for any <span class="math-container">$1\leq x<r$</span> by assuming the summands at position <span class="math-container">$x$</span> and <span class="math-container">$x+1$</span> are the same, this value is given by <span class="math-container">$t.$</span><br><br>
We can then express <span class="math-container">$$C_{k,r}=\binom{k-1}{r-1}-\sum _{\ell =1}^{r-1}(-1)^{\ell -1}\sum _{X\in \binom{[r-1]}{\ell}}\left | \bigcap_{x\in X} C_{k,r,x}\right |,$$</span>
but now the problem starts again because we need to know how many chunks of elements(consecutive) are in the set <span class="math-container">$X$</span> for us to be able to know how many summands are equal. When we know this, we can use stars and bars. Call thr number of chunks <span class="math-container">$s$</span> and call the size of <span class="math-container">$i-$</span>th chunk <span class="math-container">$\ell _i.$</span> Notice that we want <span class="math-container">$\ell _i>0$</span> and <span class="math-container">$\sum \ell _i=\ell.$</span> We then associate a number <span class="math-container">$t_i$</span> to be the number in the summand of all elements in the <span class="math-container">$i-$</span>th chunk (for a total contribution of <span class="math-container">$(\ell _i+1)t_i$</span> to the whole sum). We get then that
<span class="math-container">$$C_{k,r}=\sum _{\ell =0}^{r-1}(-1)^{\ell}\sum _{s = 0}^{\ell}\sum _{\substack{ \ell_1+\cdots +\ell_s=\ell \\ \ell _i,t_i>0}}\binom{r-1-\ell -(s-1)+(s+1)-1}{s+1-1}\binom{k-\left (\sum _{i=1}^s(l_i+1)t_i\right )-1}{r-(\ell+s)-1}.$$</span>
Putting all this together we get
<span class="math-container">$$A_{n}=\sum _{k=0}^n\sum _{r=0}^n\binom{n-k+1}{r}\sum _{\ell =0}^{r-1}(-1)^{\ell}\sum _{s = 0}^{\ell}\sum _{\substack{ \ell_1+\ell_2+\cdots +\ell_s=\ell \\ \ell _i,t_i>0}}\binom{r-\ell}{s}\binom{k-\left (\sum _{i=1}^s(l_i+1)t_i\right )-1}{r-(\ell+s)-1}.$$</span>
In this formula,by the combinatorial interpretation, treat <span class="math-container">$\binom{-1}{-1}=1.$</span><br><br>
Using <em>sage</em> I get the sequence <span class="math-container">$2,4,7,13,24,45,83,154,\dots.$</span><br>
For the moment I do not see any way to make this less painful.<br><br></p>
<hr />
<p>Another approach that might yield to something. Construct the DFA associated with that language (fix the maximal number of chunks to be <span class="math-container">$n$</span> and the maximum number of <span class="math-container">$1's$</span> in each chunk to be <span class="math-container">$k.$</span>) The DFA looks like an <span class="math-container">$k\times n$</span> array and consider the <a href="https://en.wikipedia.org/wiki/Chomsky%E2%80%93Sch%C3%BCtzenberger_enumeration_theorem" rel="nofollow noreferrer">Chomsky-Schutzenberger technique</a> One has to solve a system of <span class="math-container">$k(n-k)$</span> equations and then try to take the limit as <span class="math-container">$n,k$</span> go to <span class="math-container">$\infty.$</span> The system to solve in variables <span class="math-container">$R_{i,j}\in \mathbb{Q}[[x]]$</span> looks like
<span class="math-container">$$R_{i,j}=\begin{cases}xR_{i,j+1}+xR_{j,0}+[i\neq j] & \text{If }j>0\\
xR_{i,1}+xR_{i,0}+1 & \text{If }j=0.\end{cases}$$</span></p>
|
286,798 | <blockquote>
<p>Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$</p>
</blockquote>
<p>I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$</p>
| Did | 6,179 | <p>The identity $k^2-1=(k+1)(k-1)$ shows that
$$
\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2,
$$
and the value of limit should follow.</p>
|
660,899 | <p>find the unit normal $\bf \hat{N}$ of</p>
<p>$${\bf r}=6 \mathrm{e}^{-14 t}\cos(t){\bf i}+6 \mathrm{e}^{-14 t}\sin(t){\bf j}$$</p>
<p>The answer should be in vector form. Use t as parameter. Write $e^x$ for exponentials.</p>
<p>Have been working with this a long time now but cant get the right answer.
My answer is
$$(-((e^{-14t})(\cos(t)-14\sin(t)))/((\sqrt{12})\sqrt{e^{-28t}}), (-((e^{-14t})(\sin(t)+14\cos(t)))/((\sqrt{12})\sqrt{(e^{-28t}})),0)$$
but it aint right. Thx for help!</p>
| Did | 6,179 | <p>Two different questions here:</p>
<blockquote>
<p>How do you derive that properly?</p>
</blockquote>
<p>For example, by noting that the definition of conditional distribution says exactly that $Y$ is distributed as $aX+b+\sigma_YZ$ where $Z$ is standard normal (and independent of $X$) hence $E(Y)=aE(X)+b+\sigma_Y E(Z)=a\mu_X+b$.</p>
<blockquote>
<p>This is what I have thought: $E(Y)=\displaystyle\int yp(y) dy$ (and) $p(y)=\displaystyle∫p(y|x)p(x) dx.$
How do I proceed from here?</p>
</blockquote>
<p>For example, by plugging the second identity into the first one, getting
$$E(Y)=\iint yp(y|x)p(x)\mathrm dx\mathrm dy,
$$
and inserting the values of $p(x)$ and $p(y|x)$ in the double integral on the RHS. The computations that follow are boring but they carry through. One can (prove and) use as an intermediate step the fact that
$$
\int yp(y|x)\mathrm dy=ax+b.
$$</p>
|
2,966,010 | <p>how to show that <span class="math-container">$$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$</span>? Can I use the Alternating Series test and how? </p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>We have that</p>
<p><span class="math-container">$$(-1)^{n}\dfrac{3n-1}{n^2 + n} =(-1)^{n}\frac1n\dfrac{3n+3-4}{n + 1} =3\frac{(-1)^{n}}n-4(-1)^{n}\dfrac{1}{n(n + 1)}=\ldots$$</span></p>
<p>and since by telescoping</p>
<p><span class="math-container">$$\dfrac{1}{n(n + 1)}=\frac1{n}-\frac1{n+1}$$</span></p>
<p>we obtain</p>
<p><span class="math-container">$$-\frac{(-1)^{n}}n+4\dfrac{(-1)^{-1}}{n + 1}$$</span></p>
<p>then recall that by <a href="https://en.wikipedia.org/wiki/Alternating_series#Alternating_series_test" rel="nofollow noreferrer"><strong>alternating harmonic series</strong></a></p>
<ul>
<li><span class="math-container">$\sum_{n=1}^\infty \frac{(-1)^{n}}n=\ln 2$</span></li>
</ul>
|
2,966,010 | <p>how to show that <span class="math-container">$$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$</span>? Can I use the Alternating Series test and how? </p>
| lab bhattacharjee | 33,337 | <p><span class="math-container">$$\dfrac{3n-1}{n(n+1)}=\dfrac{3(n+1)-4}{n(n+1)}=\dfrac3n-4\left(\dfrac1n-\dfrac1{n+1}\right)=\dfrac4{n+1}-\dfrac1n$$</span></p>
<p>This can also be achieved by <a href="http://mathworld.wolfram.com/PartialFractionDecomposition.html" rel="noreferrer">Partial Fraction Decomposition</a> <span class="math-container">$$\dfrac{3n-1}{n(n+1)}=\dfrac An+\dfrac B{n+1}$$</span></p>
<p>Now <span class="math-container">$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$</span> for <span class="math-container">$-1\le x<1$</span> by </p>
<p><a href="https://math.stackexchange.com/questions/1356517/what-is-the-correct-radius-of-convergence-for-ln1x">What is the correct radius of convergence for <span class="math-container">$\ln(1+x)$</span>?</a></p>
|
2,745,884 | <p>If random variables $X$ and $Y$ are independent and $X$ and $Z$ are independent, are $X$ and $Y \cup Z$ independent?</p>
| lulu | 252,071 | <p>No. </p>
<p>suppose you are tossing two coins, a penny and a dime. $X$ is the event that the penny comes up $H$, $Y$ is the event that the dime comes up $H$, and $Z$ is the event that the coins match. Then the events that comprise $Y\cup Z$ are $$HH,TH,TT\implies P(Y\cup Z)=\frac 34$$
Where $TH$, for example, denotes the event "the penny comes up $T$ and the dime comes up $H$".</p>
<p>But $$P\left(X\cap (Y\cup Z)\right)=\frac 14\neq P(X)\times P(Y\cup Z)$$</p>
|
919,699 | <p>I am very bad at problems involving expected return and was hoping some one could help me out.</p>
<p>You are offered a chance to play a game for $48 against 99 other players(100 including you) the game consists of 16 rounds and in each round you have 4 chances to win. The first winner picked in each round gets 30 dollars, the second winner gets 50 dollars, the third gets 30 dollars, and the fourth gets 100 dollars. It is also possible for one person to win more than one of the prizes in one round. The question is what is the expected payout of playing this game?</p>
<p><strong>Note</strong>: All rounds are independent of each other and each player has an equal chance to win each prize in each round.</p>
| Durin | 144,406 | <p>Lets consider $X_i$ to be the random variable which represents our earning in a given round.
We play 16 rounds so our total earning will be be represented by a random variable $Y=\sum_{n=1}^{16} X_i$.
The expected value of total earning is $E(Y) = E(X_1) + E(X_2) + ...E(X_{16})$
Since all random variables represent same event played again and again.
$E(Y) = 16.E(X_1)$</p>
<p>Now lets find $E(X_1)$.
You can win in 4 sub-rounds (the 4 chances) in 15 ways. You can win 1 subround, or 2 subround or 3 or even 4.
Total will be ${4 \choose 1} + {4 \choose 2} + {4 \choose 3} + {4 \choose 4} = 15$
However you need to write them down because each way yields different amount you earn. What I've come up with is
$30, 50, 30, 100, (30 + 50), (30 + 30), (30 + 100), (50 + 30), (50 + 100), (30 + 100), (30 + 50 + 30), (30 + 50 + 100), (30 + 30 + 100) , (50 + 30 + 100), (30 + 50 +30 + 100) $
So total 15 ways.
Now lets calculate probability of each of the 15 ways. It will be $\dfrac 1 {100} * {(\dfrac {99} {100})}^3$ where you win in just 1 subround, $\dfrac 1 {100^2}* {(\dfrac {99} {100})}^2$ where you win two, $\dfrac 1 {100^3}* {(\dfrac {99} {100})}$ where you win 3 and $\dfrac 1 {100^4}$ where you win all four.</p>
<p>The expected value $E(X) = \sum_{n=1}^{15} x_i . P(x_i)$
I won't do calculation here but it comes out to be
$E(X) = 2.0376279(for-winning-1-subround) + 0.0411642(for-winning-2-subround) + 4.455*10^{-4}(for-winning-3-subround) + 2.1*10^{-6}(for-winning-4-subround) = 2.0792397$</p>
<p>Now $E(Y) = 16.E(X_1) = 16 * 2.0792397 = 33.2678352$ </p>
<p>You had invested $48\$$ and you might get back only $33.2678352\$$, so don't play the game.</p>
|
34,049 | <p>A person has a sheets of metal of a fixed size.</p>
<p>They are required to cut parts from the sheets of metal. </p>
<p>It's desireable to waste as little metal as possible. </p>
<p>Assume they have sufficient requirements before making the first cut to more than use one sheet of metal</p>
<p>What is the name of the branch of math which is involved in optimizing the decision on how to do the cuts ?</p>
| phv3773 | 1,312 | <p>Depending on context and specifics, it could be considered an problem in analysis, optimization, combinatorics, or simply applied math. If I wanted a text to help me out, I'd look for terms like Operations Research or Management Science in the title. </p>
|
397,040 | <p>What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$</p>
<p>according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$</p>
<p>however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?</p>
| Mark Bennet | 2,906 | <p>When you convert $A \le B$ to $\cfrac 1 A \ge \cfrac 1 B$, you are in fact dividing by $AB$. This works if $AB$ is positive, but if $AB \lt 0$ you have to reverse the inequality. </p>
<p>The inequality in the question is true whenever $x \lt 0$ because the left hand side is negative and the right-hand side is positive.</p>
|
1,097,658 | <p>I read in a notes: A semi-function is a relation (not a function) with of the form $y^2=f(x)$. </p>
<p>It seems that we can get more that one values for $f(x)$ for a single value of $x$. </p>
<p>Could any-one please help me to understand this notion.</p>
<p>The link of the note is <a href="http://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=6&ved=0CDcQFjAF&url=http%3A%2F%2Fwww.milefoot.com%2Fabout%2Fpresentations%2Fpolypretzels.pdf&ei=7M6vVN2uL9CquQTkm4LYBg&usg=AFQjCNF97nqb0WA5hAcxlfF4MO-51yEKLA&bvm=bv.83339334,d.c2E" rel="nofollow">here</a>.</p>
| Fizz | 173,347 | <p>"Semi-function" is a rather seldom-encountered term, I think. <em><a href="https://en.wikipedia.org/wiki/Multivalued_function" rel="nofollow">Multivalued function</a></em> or <em>set-valued function</em> are the more common ones. If there are some specific issues you don't understand from those presentations, you should probably ask a more pointed question. </p>
<p>For the $ y^2 = f(x) $ example, it simply means that $ y = \pm \sqrt{f(x)} $, so there is a set of two values corresponding to every $y$... assuming $ f(x) $ is positive... which the paper doesn't even say. If $ f(x) $ is negative (for some $x$), then there will be no real value $y$ for that $x$, i.e. the empty set is associated with the corresponding $y$. I guess the "semi" term wants to suggest that there are only two (or perhaps zero-or-two) rather than more values for every <em>y</em>, but I'm not exactly sure what the author had in mind when he chose this seldom-used "semi-function" terminology instead of the more common one(s) that I mentioned in the previous paragraph. The paper you link to lacks a more general definition of "semi-function" beyond this $ y^2 = f(x) $ so we can't be sure what it would mean in general in this author's mind.</p>
|
160,818 | <p>Could someone help me with an simple example of a profinite group that is not the p-adics integers or a finite group? It's my first course on groups and the examples that I've found of profinite groups are very complex and to understand them requires advanced theory on groups, rings, field and Galois Theory. Know a simple example?</p>
<p>Last, how to prove that that $\mathbb{Z}$ not is a profinite group?</p>
| Akhil Mathew | 536 | <p>One way to get a profinite group is to start with any torsion abelian group $A$, and take $\hom(A, \mathbb{Q}/\mathbb{Z})$. This acquires a topology as the inverse limit topology: it is the inverse limit of $\hom(A_0, \mathbb{Q}/\mathbb{Z})$ for $A_0 \subset A$ a finitely generated (and necessarily torsion) subgroup. </p>
<p>In fact, this gives an anti-equivalence between profinite abelian groups and torsion abelian groups, which is sometimes useful (it means, for instance, that a filtered inverse limit of profinite abelian groups is always exact, since the corresponding fact for filtered direct limits is true in abelian groups). </p>
|
48,629 | <p>Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$.</p>
<p>On my desk you could find the following books: <strong>Algebraic Geometry</strong> by <em>Hartshorne</em>, <strong>Ideals, Varieties, and Algorithms</strong> by <em>Cox & Little & O'Shea</em>, <strong>Algorithms in Real Algebraic Geometry</strong> by <em>Basu & Pollack & Roy</em> and <strong>A SINGULAR Introduction to Commutative Algebra</strong> by <em>Greuel & Pfister</em>. Unfortunately, neither of them introduced notions and ideas I'm looking for.</p>
<p>If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: <em>is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space?</em></p>
<p>Further questions I have are, for example:</p>
<ol>
<li>Can I define <em>geodesics</em> (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities?</li>
<li>How about <em>curvature</em>? Is it defined for these objects?</li>
<li>Can we talk about <em>convexity</em> of subsets of the algebraic surface?</li>
<li>What other tools and term can be imported from differential/Riemannian geometry?</li>
</ol>
<p>I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help.</p>
| Ariyan Javanpeykar | 4,333 | <p>Let $X$ be a connected normal projective C-scheme of dimension 2, i.e., an algebraic surface. The topology on an algebraic surface is the Zariski topology. But you can associate to $X$ its analytification. (See Hartshorne's appendix B or the wonderful SGA1 Exposé XII available on Arxiv.) Let $X^a$ be the analytification of $X$. Hartshorne explains that $X^a$ is a complex analytic variety (by definition) of dimension 2. So in general it won't be a complex analytic surface. More precisely, a complex manifold to me is a <strong>nonsingular</strong> complex analytic variety and if you look in SGA1 exposé XII you will see that $X$ having property P is equivalent to $X^a$ having property P. Here you can take P to be the property of nonsingular. (In fact, in SGA1 Exposé XII P is a property of a morphism such as flat smooth etale , etc.) </p>
<p>In any case, being nonsingular is a <strong>local</strong> property, i.e., it is a condition imposed on the local rings. (A variety is nonsingular if and only if its local rings are (local noetherian) regular rings.) Therefore, if you take a nonsingular point in $X$ you can find a neighborhood $U$ of $x$ which is nonsingular. (So this means that the local rings of $U$ are all regular.) The analytification of $U$ is a manifold in the above sense.</p>
<p>So the point I'm trying to make is that if you start with a nonsingular connected projective $\mathbf{C}$-scheme $X$ and take its analytification $X^a$ you can use all your knowledge of manifolds to work with $X^a$. Vaguely speaking, sometimes you can use results in the complex analytic world to deduce results for $X$ itself. (Think about comparison theorems for fundamental groups and cohomology or the theory of characteristic classes.)</p>
<p>I hope this helps a bit! </p>
<p>Ow btw, the notion of a complex analytic variety is explained more detailed in some article by Grauert I think. It is also explained in Hartshorne's appendix very briefly. </p>
<p>EDIT: I took a <strong>normal</strong> scheme above. There is a good reason for this. Namely, the singularities of $X$ will be closed points. That is, the singular locus of $X$ is closed and of codimension 2. Moreover, there is a nice theory of resolving these singularities. See for example the book Complex complex surfaces by Barth-Hulek-Peters-Ven. If you take a resolution of singularities $Y\longrightarrow X$ and endow $Y^a$ with the analytic topology you get a complex manifold. Sometimes it's nice to be able to work with $Y^a$.</p>
|
536,187 | <p>Let $\mathbb{R}^{\omega}$ be the countable product of $\mathbb{R}$. Make t a topological space using the box topology. Let $\pi_{n}$ denote the usual projection maps. </p>
<p>Fix $N \in \mathbb{Z}_+$ and define $A_N = \{x \in \mathbb{R}^\omega$ $|$ $\pi_{k}(x) = 0$ $\forall k>N\}$. Show that $A_N$ is closed in the box topology. </p>
<p>We know that $A_N$ is closed in the box topology if it has a complement that is open in the box topology. </p>
| Brian M. Scott | 12,042 | <p>HINT: Just show that each $x\in\Bbb R^\omega\setminus A_N$ has an open nbhd that is disjoint from $A_N$. This is very straightforward: if $x\notin A_N$, there is a $k>N$ such that ... what?</p>
|
516,544 | <p>The following is an <a href="http://placement.freshersworld.com/placement-papers/IBM/Placement-Paper-Whole-Testpaper-37851" rel="nofollow">aptitude problem (question no: 29-32)</a>, I am trying to solve:- </p>
<blockquote>
<p>Questions 29 - 32:</p>
<p>A, B, C, D, E and F are six positive integers such that</p>
<p>B + C + D + E = 4A</p>
<p>C + F = 3A</p>
<p>C + D + E = 2F</p>
<p>F = 2D</p>
<p>E + F = 2C + 1</p>
<p>If A is a prime number between 12 and 20, then</p>
<ol>
<li>The value of F is</li>
</ol>
<p>(A) 14</p>
<p>(B) 16</p>
<p>(C) 20</p>
<p>(D) 24</p>
<p>(E) 28</p>
<ol>
<li>Which of the following must be true?</li>
</ol>
<p>(A) D is the lowest integer and D = 14</p>
<p>(B) C is the greatest integer and C = 23</p>
<p>(C) B is the lowest integer and B = 12</p>
<p>(D) F is the greatest integer and F = 24</p>
<p>(E) A is the lowest integer and A = 13</p>
</blockquote>
<p>Now there are 5 equations to solve 6 variables. So I am at a loss on how should I start solving the problem? Any help from anybody is appreciated.</p>
| Ramchandra Apte | 38,626 | <p>Keep substituting and work with the equations to end up having equations for the other variables in terms of <code>A</code>, i.e. something like <code>B = 10A+3</code>. There are only certain values of <code>A</code> that are a prime number between 12 and 20. For those values of <code>A</code>, using your equation for <code>F</code>, you can compute the possible values of <code>F</code>. Once the noninteger or nonpositive values are removed there should be only one value for <code>F</code>. Likewise the second question can be answered.</p>
|
2,540,992 | <blockquote>
<p>An infinite sequence of increasing positive integers is given with bounded first differences.</p>
<p>Prove that there are elements <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in the sequence such that <span class="math-container">$\dfrac{a}{b}$</span> is a positive integer.</p>
</blockquote>
<p>I think maybe computing the <a href="https://en.wikipedia.org/wiki/Natural_density" rel="noreferrer">Natural Density</a> of the sequence would lead to some contradiction. But don't know if it exists.</p>
<p>Any help will be appreciated.<br />
Thanks.</p>
| Abr001am | 223,829 | <p>I have doubts about the effeciency of this proof attempt, but it is just an extension of @MooS 's answer that is based upon the fact that <a href="https://oeis.org/A005250" rel="nofollow noreferrer">increasing prime gaps</a> are diverging to infinity, and <a href="https://primes.utm.edu/notes/gaps.html" rel="nofollow noreferrer">they are!</a>:</p>
<blockquote>
<p>The proof in the paper claims that for an arbitrary large $n$ there is a sequence $n!,n!+2,n!+3,n!+4....n!+n=n!,2(n!/2+1),3(n!/3+1)...$ of composite numbers intermediating the largest prime $p_0$ before $n$ and the nearest $p_1$ after $n!+n$ which converges respectively of larger $n$. </p>
</blockquote>
<p>On this ground you should remark that picking up consecutive primes iterately will force us into a pitfall! what is it ? it's like a cul-de-sac after consecutive leaps over prime gaps it should exist some infinitely divergent one when you have to choose a composite number as forced to be bound to some limit of differences.</p>
<p>What if a range of primes are skipped ? The following numbers should be mutually indivisible written in the form of skipped prime numbers. $p_0^ip_1^j...$ in the forthcoming must have decreasing prime weights, which means descending sets of {i,j..} thus they wouldn't divide their previous. Which immediately indicates that picking up a composite $p_0^ip_1^jp_2^k$ for example must consequently lead to a following $p_0^{i'}p_1^{j'}p_2^{k'}\prod_l p_l^l$ where $i+j+k>i'+j'+k'$.</p>
<p>Let's denote a <em>prime quotient</em>: a ratio between two composite mutually indivisible numbers chosen consequently. The prime quotient between the two previous composites is $\frac{p_0^{i'}p_1^{j'}p_2^{k'}\prod_l p_l^l}{p_0^ip_1^jp_2^k}$ their values are continuously increasing. The only steady sequence that guarantees a smallest prime quotient is $p_jp_{j+1}$ the prime quotient is ultimately small which equals $\frac{p_{j+2}}{p_j}$, consequently the gap between them $p_{j+1}(p_{j+2}-p_{j})$ is divergent.</p>
<p>I was planning to check my claims with an exhaustive sweepment program but there is already a <a href="https://codegolf.stackexchange.com/questions/148999/different-ways-of-defining-primes">handful here</a> by which one can ensure that increasing n-prime gaps are also divergent.</p>
|
19,585 | <p>Since I'm fairly new to Mathematica, I'm trying to learn better ways to improve my coding skills so I've turned to Project Euler and this site to speed up my learning pace. Anyways, I was trying to solve problem 32 on the project Euler forum and came up with the following code</p>
<pre><code>PanDigital[n_, m_] := Sort[Flatten[IntegerDigits /@ {n, m, n m}]] == Range[9];
Module[{k}, k = Select[Flatten[Table[{i j, PanDigital[i, j]}, {i, 2, 9}, {j, 1234,
9876}] ~Join~ Table[{i j, PanDigital[i, j]}, {i, 12, 98}, {j, 123, 987}],
1], #[[2]] == True &]; k = Union@Select[Flatten[k], IntegerQ] // Total] // Timing
</code></pre>
<blockquote>
<p>{2.620817, 45228}</p>
</blockquote>
<p>By the way here is the question:</p>
<blockquote>
<p>We shall say that an n-digit number is pandigital if it makes use of
all the digits 1 to n exactly once; for example, the 5-digit number,
15234, is 1 through 5 pandigital.</p>
<p>The product 7254 is unusual, as the identity, 39 × 186 = 7254,
containing multiplicand, multiplier, and product is 1 through 9
pandigital.</p>
<p>Find the sum of all products whose multiplicand/multiplier/product
identity can be written as a 1 through 9 pandigital.</p>
<p>HINT: Some products can be obtained in more than one way so be sure to
only include it once in your sum.</p>
</blockquote>
<p>My question is, how can I make this code better, in terms of style and speed. I imagine using things like <code>Reap</code> and <code>Sow</code> could improve the readability and also speed. Also while my code run in under 3 seconds, I saw other people claiming less than 10 milliseconds time. Of course, using <code>ParallelTable</code> decreased the time to about 0.6 seconds on my PC but this is still not comparable to those times. Any advise is greatly appreciated.</p>
| Mr.Wizard | 121 | <p>I'm really uncomfortable with hosting solutions to Project Euler problems here, but <a href="https://mathematica.meta.stackexchange.com/questions/634/should-we-allow-project-euler-questions">apparently the community feels otherwise</a>.</p>
<p>I'll remark that it is often best to find the smallest set that encompasses the problem and test those cases. For example, you could consider only the numbers that might be products and then test those to see if they can be formed as the product of the remaining digits. This allows for about an order of magnitude improvement over your code I believe.</p>
<hr>
<p>In light of the argument presented in the comments I shall go against my usual policy and post my solution using the principle described above.</p>
<pre><code>test =
MemberQ[
Union @@@ Table[#[[{i, -i}]], {i, 2, Length@#/2}] &@IntegerDigits@Divisors@FromDigits@#,
Range@9 ~Complement~ #
] &;
FromDigits /@ Select[
TakeWhile[Reverse@Range@9 ~Permutations~ {4}, # =!= {3, 1, 8, 5} &],
test
] // Tr
</code></pre>
|
19,585 | <p>Since I'm fairly new to Mathematica, I'm trying to learn better ways to improve my coding skills so I've turned to Project Euler and this site to speed up my learning pace. Anyways, I was trying to solve problem 32 on the project Euler forum and came up with the following code</p>
<pre><code>PanDigital[n_, m_] := Sort[Flatten[IntegerDigits /@ {n, m, n m}]] == Range[9];
Module[{k}, k = Select[Flatten[Table[{i j, PanDigital[i, j]}, {i, 2, 9}, {j, 1234,
9876}] ~Join~ Table[{i j, PanDigital[i, j]}, {i, 12, 98}, {j, 123, 987}],
1], #[[2]] == True &]; k = Union@Select[Flatten[k], IntegerQ] // Total] // Timing
</code></pre>
<blockquote>
<p>{2.620817, 45228}</p>
</blockquote>
<p>By the way here is the question:</p>
<blockquote>
<p>We shall say that an n-digit number is pandigital if it makes use of
all the digits 1 to n exactly once; for example, the 5-digit number,
15234, is 1 through 5 pandigital.</p>
<p>The product 7254 is unusual, as the identity, 39 × 186 = 7254,
containing multiplicand, multiplier, and product is 1 through 9
pandigital.</p>
<p>Find the sum of all products whose multiplicand/multiplier/product
identity can be written as a 1 through 9 pandigital.</p>
<p>HINT: Some products can be obtained in more than one way so be sure to
only include it once in your sum.</p>
</blockquote>
<p>My question is, how can I make this code better, in terms of style and speed. I imagine using things like <code>Reap</code> and <code>Sow</code> could improve the readability and also speed. Also while my code run in under 3 seconds, I saw other people claiming less than 10 milliseconds time. Of course, using <code>ParallelTable</code> decreased the time to about 0.6 seconds on my PC but this is still not comparable to those times. Any advise is greatly appreciated.</p>
| chyanog | 2,090 | <pre><code>Union @@ Table[
If[a*b <= 9876 &&
Union[IntegerDigits[a], IntegerDigits[b], IntegerDigits[a*b]] ==
Range[9], a*b, 0], {a, 123, 1987}, {b, 2, 98}] // Tr // Timing
(*v7*)(*{0.686, 45228}*)
(*v8*)(*{0.078, 45228}*)
</code></pre>
|
631,163 | <p>As a student in high school, I never bothered to memorize equations or methods of solving, rather I would try to identify the logic behind the operations and apply them. However, now that I've begun to teach Algebra in high school, I find it rather frustrating when students either a) memorize methods of solving the textbook problems or b) look for a general formula/method to "just plug in to"</p>
<p>I've tried to throw them curveballs as my old Algebra teacher did, but usually they just dismiss it as "a weird problem" and continue using whatever method they have been.</p>
<p>My objection to A is that it often impedes actual learning. Upon seeing a chunk of 6 similar problems in the textbook, many students just apply the same steps to every problem in the section (and usually get quite a couple wrong).</p>
<p>My objection to B is that from my experience, students who flat out memorize equations (like $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ for quadratic equations) often fail to extend the same logic (completing the square/simplification) when faced with different but similar problems. They also frequently misapply the "magic formulas" they were taught before (i.e. solving simple quartics $ax^4+bx^2+c$ with the quadratic formula) and needing plenty of prompting after the suggestion of substituting $x^2$.</p>
<p>This is the problem identified in <a href="https://math.stackexchange.com/questions/85681/is-memorization-a-good-skill-to-learn-or-master-mathematics#comment202224_85689">this question</a> and in particular the issue raised in <a href="https://math.stackexchange.com/questions/85681/is-memorization-a-good-skill-to-learn-or-master-mathematics#comment202224_85689">this comment</a>.</p>
| David | 651,991 | <p>I think that "lazy memorization" will always be the default strategy taken by your students, since that is what works in most of the subjects they are tought. Their minds are wired that way and that is hard to change.</p>
<p>If you really want students to think by themselves, you will have no choice but to force them to. Present them problems where some reasoning must be done before jumping into the application of the method. Use that type of problems in tests so that there is no workaround! It will be tough at first, but the reward is high if this strategy is successful.</p>
<p>In the end, it is a matter of the teacher's choice. There is a limited amount of time for teaching, so you can't encompass everything! You have to make a compromise. What I mean is that maybe you can teach fewer "mechanichal methods" in exchange for dedicating time to "teach how to think"</p>
|
1,386,682 | <p>How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?</p>
<p>I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>And that I could not get out, can anyone help me?</p>
| mvw | 86,776 | <p>How about
$$
\lim_{z\to 0} \frac{\bar{z}^2}{z} =
\lim_{z\to 0} \frac{\bar{z}^2z^2}{z^3} =
\lim_{z\to 0} \frac{\lvert z\rvert^4}{z^3} =
\lim_{z\to 0} \frac{\lvert z\rvert^4}{\lvert z \rvert^3 e^{3i\phi(z)}}
= 0
$$</p>
|
2,317,625 | <p>How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)</p>
<p>Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).</p>
| TomGrubb | 223,701 | <p>We have $\sqrt{3}\leq 1.8$ so $6-2\sqrt{3}\geq 2.4$, whereas $\sqrt{2}\leq 1.42$ so $3\sqrt{2}-2\leq 2.26$.</p>
|
192,125 | <p>Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<</p>
| Zarrax | 3,035 | <p>Square both sides, and you get
$$x - 4 + 20\sqrt{x - 4} + 100 = x + 4$$
This simplifies to
$$20\sqrt{x - 4} = -92$$
or just
$$\sqrt{x - 4} = -\frac{92}{20}$$
Since square roots of numbers are always nonnegative, this cannot have a solution.</p>
|
192,125 | <p>Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<</p>
| Ross Millikan | 1,827 | <p>As others have said, there are no solutions within the usual rules. However, once we get to $\sqrt {x-4}=-4.6$ we can remember that square roots can be negative (despite the convention that $\sqrt x \ge 0$). Then we can square and add $4$to find $x=25.16$. Checking, we find $\sqrt {x+4}=5.4, \sqrt{x-4}=-4.6$ and the difference is truly $10$. You can decide if this is better than no answer at all.</p>
|
111,183 | <p><img src="https://i.stack.imgur.com/1MOuo.jpg" alt="Problem">
<img src="https://i.stack.imgur.com/bdRXi.png" alt="New Solution"></p>
<p>I believe I have gotten all of the ways now - thanks for the hints below Yun, Andre Nicolas, and Gerry Myerson. If anyone could confirm my answer (I feel there should be more possibilities, but for the numbers to be increasing left-right, I have found them all I believe). Sorry I don't have the headings on this chart, but from left to right it is: Numbers 1-6 (one for each column) with the last column being "Sum," if it was not already obvious. </p>
| davidlowryduda | 9,754 | <p>I can verify your solution. I have one more than you. I've also put up a list (everything is divided by 2). I also put it up in the order that mirrors the smaller example's list (at least, as I interpreted it - perhaps, putting the list in strictly number-increasing order would have been a good idea too). </p>
<p>1,1,1,1,1,8</p>
<p>1,1,1,1,2,7</p>
<p>1,1,1,1,3,6</p>
<p>1,1,1,2,2,6</p>
<p>1,1,1,1,4,5</p>
<p>1,1,1,2,3,5</p>
<p>1,1,2,2,2,5</p>
<p>1,1,1,2,4,4</p>
<p>1,1,1,3,3,4</p>
<p>1,1,2,2,3,4</p>
<p>1,2,2,2,2,4</p>
<p>1,1,2,3,3,3</p>
<p>1,2,2,2,3,3</p>
<p>2,2,2,2,2,3</p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| Aryabhata | 1,102 | <p>Let $a = \cos x$ and $b = \sin x$, and so $a,b \in [-1,1]$.</p>
<p>We have to solve $\sin a = \cos b$. </p>
<p>We can assume that $0 \le b \le 1$, because, if $x$ is a root, so is $-x$.</p>
<p>Since $\sin a = \cos b$ and $b \ge 0$, we must have that $a \ge 0$ (remember, $a,b \in [-1,1]$)</p>
<p>Thus if the equation has roots, at least one of those is such that $x \in [0,\pi/2]$.</p>
<p>It is clear that $0, \pi/2$ are not roots. So we can assume $x \in (0, \pi/2)$.</p>
<p>Now $ \sin y \lt y$ for all $y \in (0, \pi/2)$.</p>
<p>Thus $\cos (\sin x) \gt \cos x$ (as $\sin x \lt x$ and $\cos $ is decreasing )</p>
<p>We also have $\cos x \gt \sin (\cos x)$ (using $\cos x = y \gt \sin y = \sin (\cos x)$).</p>
<p>Thus we have that $$\cos (\sin x) \gt \sin (\cos x), \quad x \in (0, \pi/2)$$</p>
<p>Contradiction, and the equation has no real roots.</p>
|
3,077,629 | <p>Assume <span class="math-container">$(f_i)_{i\in I}$</span> is an orthonormal/orthogonal system in an (complex) inner product space. Does <span class="math-container">$$\sum_{i\in I}\langle f_i,f\rangle f_i$$</span> always converges for any <span class="math-container">$f$</span> (may not to <span class="math-container">$f$</span>)? Especially, when <span class="math-container">$I$</span> is uncountable.</p>
<p>And we define convergency of a sum <span class="math-container">$\sum_I x_i$</span> on an arbitrary set by: if there exists <span class="math-container">$g$</span> and for any <span class="math-container">$\epsilon>0$</span>, there exists a finite set <span class="math-container">$F$</span> and for all other finite set <span class="math-container">$H\supseteq F,|g-\sum_H x_i|<\epsilon$</span> as an analogue to the limit of series in analysis.</p>
| lonza leggiera | 632,373 | <p>The constraint <span class="math-container">$\ x^2 - y^2 - z^2 + 16 \le 0\ $</span> is not satisfied by <span class="math-container">$\ \left(\pm 4, 0, 0\right)\ $</span>, so it's not even a feasible point, let alone an extremum.</p>
<p>All points of the form <span class="math-container">$\ \left(\,0, y, z\,\right)\ $</span> with <span class="math-container">$\ y^2 + z^2 = 16\ $</span> satisfy the Lagrange extremality conditions, and are all maxima, since <span class="math-container">$\ f\left(\,0,y,z\,\right)\ = \frac{1}{16}\ $</span> for such points, and the constraint <span class="math-container">$\ x^2 - y^2 - z^2 \le 16\ $</span> implies that its value cannot be any larger than that.</p>
<p>There are also arbitrarily large values of <span class="math-container">$\ x, y,\ \mbox{and } z\ $</span> that can satisfy the constraint <span class="math-container">$\ x^2 - y^2 - z^2 \le 16\ $</span>. If <span class="math-container">$\ x=a^2 + 1, y=2a,\ $</span> and <span class="math-container">$ z=a^2 - 1\ $</span>, for instance, then <span class="math-container">$\ x^2 - y^2 - z^2 = 0 \le 16\ $</span> for any value of <span class="math-container">$\ a\ $</span>, and <span class="math-container">$\ a\ $</span> can be made aribitrarily large. Thus, although the value of the objective function is always <em>strictly</em> positive, it can nevertheless be made arbitrarily close to zero, which is therefore its infimum, and <span class="math-container">$\ f\left(R\right) = \left(0, \frac{1}{16}\,\right]\ $</span>.</p>
|
210,658 | <p>$$\sum_{i=1}^n t^{i-1}$$
I am stuck with the proof of this equality. </p>
| Alex | 38,873 | <p>Use perturbation method from Concrete Mathematics:
$$
S_n=\sum_{k=1}^{n}t^{i-1}\\
S_{n+1}=S_n + t^{n}=\sum_{k=1}^{n}t^{k} + 1=t\sum_{k=1}^{n}t^{k-1}+1=t S_n + 1
$$
After the algebra you get
$$
S_n=\frac{1-t^{n}}{1-t}
$$</p>
|
2,336,535 | <p>I have a limit:</p>
<p>$$\lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^4+y^2}$$</p>
<p>I need to show that it doesn't equal 0.</p>
<p>Since the power of $x$ is 3 and 4 down it seems like that part could go to $0$ but the power of $y$ is 3 and 2 down so that seems like it's going to $\infty$.</p>
<p>I wonder if that even makes sense.</p>
<p>So how can I solve this limit, with substitution or changing it into a polar representation?</p>
| Glorfindel | 228,959 | <p>The limit just <strong>doesn't exist</strong> ('is undefined'), for exactly the reasons you describe. Therefore, it is not equal to zero.</p>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| Nasser | 70 | <p>Another option is is to use basic Do loop or a Table.</p>
<pre><code>update[a_Integer, b_Integer] := (a + 2*b)/(a + b)
a = 1;
b = 2;
lis = Last@
Reap@Do[ t = update[a, b]; a = Numerator[t]; b = Denominator[t];
Sow[a/b],
{n, 1, 10}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/TL2KI.png" alt="Mathematica graphics" /></p>
<pre><code>N[lis]
</code></pre>
<p><img src="https://i.stack.imgur.com/YJUik.png" alt="Mathematica graphics" /></p>
<pre><code>Sqrt[2]//N
</code></pre>
<p><img src="https://i.stack.imgur.com/92DWE.png" alt="Mathematica graphics" /></p>
<p>There are at least 8 more ways to do this in Mathematica.</p>
|
279,808 | <p>I was working on a way of calculating the square root of a number by the method of x/y → (x+4y)/(x+y) as shown by bobbym at <a href="https://math.stackexchange.com/questions/861509/">https://math.stackexchange.com/questions/861509/</a></p>
<p>I tried to do it via functions on mathematica, everything seems correct. Why am I not seeing 2.5 as the answer? How can I fix it?
<a href="https://i.stack.imgur.com/TXA8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXA8S.png" alt="code" /></a></p>
| E. Chan-López | 53,427 | <p>Following Syed's idea, but using <code>FoldList</code>:</p>
<pre><code>FoldList[(Numerator@# + 2 Denominator@#)/(Numerator@# + Denominator@#) &, 1/2, Range[5]]
(*{1/2, 5/3, 11/8, 27/19, 65/46, 157/111}*)
</code></pre>
|
2,558,267 | <p>Let $M$ be a finite dimensional von-Neumann algebra. We know this algebra is generated by its projections. My question maybe simple. Can one computing these projections? What about its minimal or central projections? </p>
<p>If possible please give me a reference for this. </p>
<p>Thanks </p>
| Martin Argerami | 22,857 | <p>You would have to define "compute". To "compute" an element of an algebra (or any other mathematical object, for that matter) you need to have some kind of presentation of the object. </p>
<p>In the case of a finite-dimensional von Neumann/C$^*$-algebra, one can prove that they are always (isomorphic to) a direct sum of full matrix algebras. Are you starting with the former or the latter? </p>
<p>And again with the word "compute". Can you compute all rank-2 orthogonal projections in $M_3(\mathbb C)$? It depends on what you mean. They are all of the form $vv^*+ww^*$ with $v,w$ unit mutually orthogonal vectors in $\mathbb C^3$, so in a sense they are very explicit. On the other hand, I'm not saying what $v$ and $w$ are. If you try to somehow parametrize them, it can probably be done, but it is not pretty and the general expression for $vv^*+ww^*$ will be headache that likely contributes nothing. </p>
<p>Anyway, I haven't really answered your question. It depends on how you express your algebra, and what you mean by "compute". </p>
|
3,274,172 | <p>Let <span class="math-container">$X$</span> a compact set. Prove that if every connected component is open then the number of components is finite.</p>
<p>Ok, <span class="math-container">$X = \bigcup C(x)$</span> where <span class="math-container">$C(x)$</span> is the connected component of <span class="math-container">$x \in X.$</span> I know that <span class="math-container">$X \subset UA_\lambda$</span>, where <span class="math-container">$A_\lambda$</span> is a finite family of opens set but how can i conclude that the components are finite??</p>
| Henno Brandsma | 4,280 | <p>Under the assumption that components are open, they form an <em>open</em> cover of the compact space <span class="math-container">$X$</span>. They form a partition of non-empty sets (always), so we cannot omit any one of them and still have a cover. </p>
<p>So <span class="math-container">$X$</span> has a finite subcover of components which is not really a <strong>proper</strong> subcover, but just the <em>original</em> cover of components. Which is thus finite.</p>
|
2,601,088 | <p>I'm new to the group theory and want to get familar with the theorems in it, so I choose a number $52$
to try making some obseveration on all group that has this rank. Below are my thoughts. I don't know if there is any better way to think of these (i.e., an experienced group theorist would think), and I still have some questions not being solved. So I post this.</p>
<p>Let $G$ be a group and $|G|=52$. Since $52$ in not prime, then there is no theorem guarantee that $G$ must be cyclic. On the other hand, since $\Bbb Z_{52}$ is cyclic, so what we can know is only that "there exists a group $G$ with rank $52$ that is cyclic." Besides, $52$ is not of the form $pq$, where $p,~q$ are prime numbers, so again we can not use such theorem to say that $G$ must be cyclic. (Also, $52$ is not of the form $p^2$, so we're unable to conclude that $G$ is abelian anyway.)</p>
<p>So there remains a question: can $G$ be non-cyclic? I don't know how to answer it.</p>
<p>Next, move on to the obseverations on its subgroup. If $G$ is cyclic (i.e., isomorphic to $\Bbb Z_{52}$), then by a theorem that I don't know its name, $G$ has <em>exactly</em> one nontrival proper subgroup of rank $2$ (also $4,~13,~26$, respectively). By Lagrange Theorem we know that these $4$ subgroup, plus the $\{e\}$ and $G$, are the all $6$ subgroups of $G$. However, since I don't know whether $G$ can be non-cyclic, so I discuss such case below. The prime divisors of $52$ are $2,~13$. By Cauchy theorem, there exists at least one element that has order $2$ and $13$ respectively. So we know that there are two subgroups of $G$ with rank $2$ and $13$ respectively (but maybe $G$ has more subgroup with these ranks). However, how about subgroup with rank $26$ and $4$, etc.? Is their a way to tell whether such subgroup exists?</p>
| ajotatxe | 132,456 | <p>Consider the group $D_{26}$. That is, take a regular polygon with $26$ sides. Let $V_1,\ldots,V_{26}$ be its vertices (in order). Define these elements:</p>
<ul>
<li><p>$\tau$ is the rotation that maps each vertex to the next one. That is, $\tau(V_n)=V_{n+1}$ if $1\le n\le 25$ and $\tau(V_{26})=V_1$. The order of this element is $26$.</p></li>
<li><p>$\rho$ is a symmetry with axis $V_1V_{14}$, and it has order two, like any other symmetry.</p></li>
</ul>
<p>The group $\langle\tau,\rho\rangle$ is often called $D_{26}$, it has $52$ elements and $\rho\tau\neq\tau\rho$ (for example, $\tau\rho(V_1)=V_2$ and $\rho\tau(V_1)=V_{26}$), so this group is not abelian, and hence, is not cyclic either.</p>
|
2,393,872 | <p>I was doing the following problem</p>
<blockquote>
<p>An isoceles triangle is a triangle in which two sides are equal. Prove that the angles opposite to the equal sides are equal.</p>
</blockquote>
<p>I drew this diagram (sorry for the large picture):</p>
<p><a href="https://i.stack.imgur.com/uxLlm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uxLlm.jpg" alt=""></a></p>
<p>Name the triangle $ABC$ such that $AB=BC$. The angle bisector of $\angle B$ intersects line $AC$ at the point $D$. Now we have the two triangles $ABD$ and $CBD$. They both share side $BD$, $m\angle ABD = m\angle CBD$, and by hypothesis $AB=BC$. so the two triangles are congruent by $SAS$.</p>
<p>So what I know so far is that there exists a way to place triangle $ABD$ onto triangle $BDC$ so that they overlap perfectly. From the picture it is clear that this could be achieved by reflecting triangle $BDC$ over the line $BD$, which would imply the desired conclusion. But is this sort of "from the picture" argument really valid/rigorous? What if the actual way to make the triangles overlap is to put side $BD$ of triangle $BDC$ onto side $AB$ of triangle $ABD$? </p>
| Jay Zha | 379,853 | <p>Yes, you are right that "there exists a way to place triangle $ABD$ onto triangle $BDC$ so that they overlap perfectly", and you might argue: "well there are more than one way to place, how do I know which way they overlap perfectly" - and that's exactly your question. </p>
<p>However, do not forget that you get the congruence by $SAS$, which means that you know which two sides are equal respectively, and you also know which angle (i.e. the angle of those two equal sides) are the same for the two triangles.</p>
<p>Then, it is obvious which way you should place the triangles.</p>
|
2,393,872 | <p>I was doing the following problem</p>
<blockquote>
<p>An isoceles triangle is a triangle in which two sides are equal. Prove that the angles opposite to the equal sides are equal.</p>
</blockquote>
<p>I drew this diagram (sorry for the large picture):</p>
<p><a href="https://i.stack.imgur.com/uxLlm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uxLlm.jpg" alt=""></a></p>
<p>Name the triangle $ABC$ such that $AB=BC$. The angle bisector of $\angle B$ intersects line $AC$ at the point $D$. Now we have the two triangles $ABD$ and $CBD$. They both share side $BD$, $m\angle ABD = m\angle CBD$, and by hypothesis $AB=BC$. so the two triangles are congruent by $SAS$.</p>
<p>So what I know so far is that there exists a way to place triangle $ABD$ onto triangle $BDC$ so that they overlap perfectly. From the picture it is clear that this could be achieved by reflecting triangle $BDC$ over the line $BD$, which would imply the desired conclusion. But is this sort of "from the picture" argument really valid/rigorous? What if the actual way to make the triangles overlap is to put side $BD$ of triangle $BDC$ onto side $AB$ of triangle $ABD$? </p>
| Intelligenti pauca | 255,730 | <p>Even if your proof is the preferred one in high-school textbooks, I think a simpler proof is worth mentioning.</p>
<p>Compare triangles $ABC$ and $CBA$: they are congruent by $SAS$ and thus $\angle A\cong\angle C$.</p>
|
2,736,426 | <p>Let's imagine a point in 3D coordinate such that its distance to the origin is <span class="math-container">$1 \text{ unit}$</span>.</p>
<p>The coordinates of that point have been given as <span class="math-container">$x = a$</span>, <span class="math-container">$y = b$</span>, and <span class="math-container">$z = c$</span>.</p>
<p>How can we calculate the angles made by the vector with each of the axes?</p>
| The Integrator | 538,397 | <p>Suppose you have a vector $\vec v = xi+yj+zk$ where $i,j,k $ are the basis unit vectors then the angles $\alpha,\beta, \gamma$ of the vector to the $x,y,z $ axes respectively is given by ;</p>
<p>$\alpha = \frac{x}{\sqrt{x^2+y^2+z^2}} = \cos(a)\\\beta = \frac{y}{\sqrt{x^2+y^2+z^2}}=\cos(b)\\\gamma = \frac{z}{\sqrt{x^2+y^2+z^2}}=\cos(c)$ </p>
<p>It follows that squaring the 3 equations and adding them results in </p>
<p>$\cos^2(a)+\cos^2(b)+\cos^2(c) =\alpha^2+\beta^2+\gamma^2 = 1 $ </p>
<p><a href="https://i.stack.imgur.com/oQP9C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oQP9C.png" alt="enter image description here"></a></p>
|
3,757,213 | <blockquote>
<p>Prove that the maximum area of a rectangle inscribed in an ellipse <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span> is <span class="math-container">$2ab$</span>.</p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>Equation of ellipse: <span class="math-container">$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$</span>.</p>
<p>Assume that <span class="math-container">$a>b$</span> & let <span class="math-container">$l$</span> & <span class="math-container">$w$</span> are length and width of rectangle then area will be <span class="math-container">$$A=l\times w \tag{1}$$</span></p>
<p>Now, I substituted the point <span class="math-container">$\left(\dfrac {l}{2}, \dfrac {w}{2}\right)$</span> in the equation of ellipse
<span class="math-container">$$\frac{(l/2)^2}{a^2}+\frac{(w/2)^2}{b^2}=1,$$</span>
<span class="math-container">$$b^2l^2+a^2w^2=4a^2b^2.\tag{2}$$</span></p>
<p>I am not sure how to proceed from here.</p>
| Harish Chandra Rajpoot | 210,295 | <p>Here is an easier method to prove (without using Calculus).</p>
<p>The sides (i.e. length and width) of inscribed rectangle have to be parallel to the axes of ellipse to make all four vertices of inscribed rectangle lie on the ellipse.</p>
<p>Consider <span class="math-container">$(\pm a\cos\theta, \pm b\sin\theta)$</span> as four vertices of inscribed rectangle lying on the ellipse: <span class="math-container">$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$</span></p>
<p>Then the area of inscribed rectangle will be <span class="math-container">$A=(2a\cos\theta)(2b\sin\theta)=2ab\sin2\theta$</span></p>
<p>Now, the area <span class="math-container">$2ab\sin2\theta$</span> will be maximum when <span class="math-container">$\color{red}{\sin2\theta}$</span> is maximum i.e. <span class="math-container">$\color{red}{1}$</span></p>
<p>Therefore the maximum area of rectangle inscribed in an ellipse is <span class="math-container">$\color{blue}{2ab}$</span></p>
|
1,040,932 | <p>I have a system of congruence equations</p>
<p>$$
\begin{cases}
x \equiv 17 \pmod{15} \\
x \equiv 14 \pmod{33}
\end{cases}
$$</p>
<p>I need to investigate the system and see if they've got any solutions.</p>
<p>I know that I should use the Chinese remainder theorem "in a reverse order" so I think I should split each congruence equation in two new systems of two congruence equations.</p>
<p>From the CRT two congruence equations can be joined in a single congruence equation by</p>
<p>$$
x \equiv b_1 + c n_1 (b_2 - b_1) \pmod{n_1 n_2}
$$</p>
<p>From the first congruence equation I can get these two
$$
b_1 + c n_1 (b_2 - b_1) = 17 \\
n_1 n_2 = 15
$$</p>
<p>and from the second I can get
$$
b_1 + cn_1 (b_2 - b_1) = 14 \\
n_1 n_2 = 33
$$</p>
<p>but the unknown variables are not combined so I cannot just solve the system of four equations.</p>
<p>I need a hint :-)</p>
| lab bhattacharjee | 33,337 | <p>$$x\equiv17\pmod{15}\equiv2$$</p>
<p>$$\implies x\equiv2\pmod3\ \ \ \ (1),$$</p>
<p>$$x\equiv2\pmod5\ \ \ \ (2)$$</p>
<p>$$x\equiv14\pmod{33}\implies x\equiv14\pmod3\equiv2,$$</p>
<p>$$x\equiv14\pmod{11}\equiv3\ \ \ \ (3)$$</p>
<p>Now apply <a href="http://mathworld.wolfram.com/ChineseRemainderTheorem.html" rel="nofollow">CRT</a> on $(1),(2),(3)$ as $3,5,11$ are pairwise prime</p>
|
1,146,824 | <p>The Russel's Paradox, showing $X=\{x|x\notin x\}$ can't exist is not very hard.
If $X \in X$, then $X \notin X$ by definiition, in the other case, $X \notin X$, then $X \in X$ by definition. Both cases are impossible.</p>
<p>But how about whole things $X=\{x|x=x\}$? $X \in X$ probably cause the problem, but I don't know why violation of axiom of foundation in proper class is problem.</p>
| Mike Earnest | 177,399 | <p>Cantor's Theorem says that for any set $Y$, the power set of $Y$ is strictly bigger than $Y$. But $X$ contains all the elements of any set, and is therefore at least as big as any set, including its own power set, giving the contradiction
$$
\mathcal P(X)\le X<\mathcal P(X)
$$</p>
|
3,013,355 | <p>I have been asked to prove that </p>
<p>(<span class="math-container">$a \to $</span>b) <span class="math-container">$\vee$</span> (<span class="math-container">$a \to $</span>c) = <span class="math-container">$a \to ($</span>b <span class="math-container">$\vee$</span> c).</p>
<p>I believe it is just the simple case of using the distributive law:</p>
<p><span class="math-container">$a \wedge ($</span>b <span class="math-container">$\vee$</span> c)= (a <span class="math-container">$\wedge c) \vee ($</span>a <span class="math-container">$\wedge$</span> b).</p>
<p>But I am not sure.</p>
| Trevor Gunn | 437,127 | <p>I'm not sure if this is exactly what you're looking for, but the equivalence of <span class="math-container">$y^2$</span> and <span class="math-container">$(1 - x)(1 + x)$</span> for <span class="math-container">$\mathbf{S}^1$</span> says that you can look at the divisor <span class="math-container">$D = 2((1,0) + (-1,0))$</span> in two ways:</p>
<ol>
<li><p>By intersecting the lines <span class="math-container">$x = \pm 1$</span> with the circle (the lines are tangent so the intersection has multiplicity <span class="math-container">$2$</span>)</p></li>
<li><p>By intersecting the line <span class="math-container">$y = 0$</span> with the circle. The points above are doubled so really we want the doubled line <span class="math-container">$y^2 = 0$</span>.</p></li>
</ol>
<p>So we have <span class="math-container">$D = \mathcal{V}(y^2) = \mathcal{V}((1-x)(1+x))$</span>.</p>
<p>Also notice that every principal divisor <span class="math-container">$\mathcal{V}(f)$</span> on <span class="math-container">$\mathbf{S}^1$</span> has even degree. For example, every line intersects the circle in two distinct points or one doubled point (a tangential intersection). This classifies the principal divisors (i.e. every even degree divisor is principal).</p>
<p>So <span class="math-container">${\rm Cl}(\mathbf S^1) = \mathbf Z/2$</span>. The problem here, is somehow we want to be able to break our divisor <span class="math-container">$D$</span> up into 4 parts, not just two. That is, if <span class="math-container">$(1,0) = \mathcal{V}(f)$</span> and <span class="math-container">$(-1,0) = \mathcal{V}(g)$</span> then we could say</p>
<p><span class="math-container">$$ \mathcal{V}(y) = \mathcal{V}(fg), \mathcal{V}(1-x) = \mathcal{V}(f^2), \mathcal{V}(1 + x) = \mathcal{V}(g^2) $$</span></p>
<p>Which would give us <span class="math-container">$y^2 = (fg)^2$</span> and <span class="math-container">$(1 - x)(1 + x) = (f^2)(g^2)$</span> (up to a constant) so that <span class="math-container">$y^2$</span> and <span class="math-container">$(1 - x)(1 + x)$</span> represent the same factorization. But <span class="math-container">$(1,0)$</span> and <span class="math-container">$(-1,0)$</span> are not principal divisors so we don't get this.</p>
|
3,382,464 | <p>Let <span class="math-container">$g$</span> be a <strong>smooth</strong> Riemannian metric on the closed <span class="math-container">$n$</span>-dimensional unit disk <span class="math-container">$\mathbb{D}^n$</span>. Let <span class="math-container">$f$</span> be a harmonic function w.r.t <span class="math-container">$g$</span>.</p>
<blockquote>
<p>Is it true that <span class="math-container">$f$</span> must be real-analytic?</p>
</blockquote>
<p>I <em>think</em> that this is true if we assume that <span class="math-container">$g$</span> is real-analytic, but I am not sure. Is it true in that case? I would like to find a reference.</p>
<p>This should be related to whether or not the Riemannian laplacian <span class="math-container">$\Delta_g$</span> is "analytically hypoelliptic".</p>
| Allawonder | 145,126 | <p><em>Hint.</em> No, multiply through by <span class="math-container">$a$</span> instead, and with a little rearrangement, you now have <span class="math-container">$$a^6=a^4-a^2+2a,$$</span> whose right hand side is a biquadratic in <span class="math-container">$a.$</span> You may still attempt to complete squares on RHS.</p>
|
1,939,382 | <p>I've read about integration, and i believe i understood concept correctly. But, unfortunately, the simplest exercise already got my stumbled. I need to find an integral of $x{\sqrt {x+x^2}}$. So i proceed as follows,</p>
<p>By the fundamental theorem of calculus:</p>
<p>$f(x)=\int[f'(x)]=\int[x\sqrt{x+x^2}]$,</p>
<p>First I've tried to apply chain rule and i end up with: </p>
<p>$f'(x)=xu^\frac{1}{2}\frac{du}{2x+1}$ , not sure how i can proceed in this case.</p>
<p>Next I've tried to apply product rule:</p>
<p>If $f(x)=i(x)j(x)$, then $x\sqrt{x+x^2}=i'(x)j(x)+j'(x)i(x)$,</p>
<p>Using sum rule i could assume that, $f(x)=i(x)j(x)-\int[j'(x)i(x)]$,</p>
<p>Now, finding that $i'(x)=x, i(x)=\frac{x^2}{2}, j(x)=\sqrt{x+x^2}$ and $j'(x)=\frac{2x+1}{2\sqrt{x+x^2}}$, </p>
<p>$f(x)$ should be of form $f(x)=\frac{x^2\sqrt{x+x^2}}{2}-\int\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}$, so now i should find integral of this fraction,</p>
<p>If i can assume, that $p(x)=\frac{a(x)}{b(x)}$, then $\frac{x^2(2x+1)}{ 4\sqrt{x+x^2}}=\frac{a'(x)b(x)-a(x)b'(x)}{(b(x))^2}$, hence:</p>
<p>$b(x)=2(x+x^2)^\frac{1}{4}, b'(x)=\frac{1}{2(x+x^2)^\frac{3}{4}}$ and as a result $a(x)$, should be $a(x)=\frac{(x+x^2)^\frac{3}{4}(4a'(x)(x+x^2)^\frac{1}{4}-4x^3-2x^2)}{2x+1}$, but now i don't now how substitute $a'(x)$, if i differentiate this expression i will get $a''(x)$. </p>
<p>So my question is, what substitution i shall perform to obtain a(x), a'(x)?</p>
<p>Thank you! And forgive me my ignorance. </p>
| Jack D'Aurizio | 44,121 | <p>I hope you do not mind if I prefer to start from scratch. We have
$$ \int (2x+1)\sqrt{x^2+x}\,dx = C+\frac{2}{3}(x^2+x)^{3/2} \tag{1}$$
and the problem boils down to computing $\int\sqrt{x^2+x}\,dx$. Integration by parts gives
$$ \int \sqrt{x^2+x}\,dx = x\sqrt{x^2+x}-\int\frac{x+2x^2}{2\sqrt{x+x^2}}\,dx \tag{2}$$
hence
$$ 2\int\sqrt{x^2+x}\,dx = x\sqrt{x^2+x}+\frac{1}{2}\int \frac{x}{\sqrt{x^2+x}}\,dx \tag{3}$$
and the problem boils down to computing $\int\frac{x}{\sqrt{x^2+x}}$. With an argument similar to $(1)$, we have:
$$ \int\frac{2x+1}{\sqrt{x^2+x}}\,dx = C+2\sqrt{x^2+x}\tag{4} $$
so it is enough to compute $\int\frac{dx}{\sqrt{x^2+x}}$ and here it comes an interesting trick. Since
$$ \frac{1}{\sqrt{x^2+x}}=2\cdot\frac{\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}}{\sqrt{x}+\sqrt{x+1}}\tag{5}$$
we have
$$ \int\frac{dx}{\sqrt{x^2+x}}=2\log\left(\sqrt{x}+\sqrt{x+1}\right)\tag{6}$$
and by putting everything together
$$ \int x\sqrt{x^2+x}\,dx = \color{red}{C+\frac{x\left(8x^3+10x^2-x-3\right)}{24\sqrt{x+x^2}}+\frac{\log(\sqrt{x}+\sqrt{x+1})}{8}}.\tag{7}$$
It is not that simple, indeed!</p>
|
256,666 | <p>Let $X$ be a set. Suppose $\beta$ is a basis for the topology $\tau_\beta$ of $X$. Since each base element is open (with respect to $\tau_\beta$) we have that $$B\in \beta\Rightarrow B\in \tau_\beta.$$ Thus, $\beta\subset \tau_\beta$. </p>
<p>However, since $\beta$ is a union of base elements (I assume a set can always be written as a union of its elements) and topologies are closed under arbitrary unions, we have $\beta \in \tau_\beta.$ Is it possible for a set to be a subset an an element of another set?</p>
| Brian M. Scott | 12,042 | <p>Let $S$ be any set. Then $$S=\bigcup_{x\in S}\{x\}\;,$$ but in general $$S\ne\bigcup_{x\in S}x\;.$$ That is, $S$ is the union of the <em>singletons</em> of its elements, but it is not in general the union of its elements. It can only be the union of subsets, and in general $x\in S$ does not imply that $x\subseteq S$. (Sets $S$ with the property that $x\subseteq S$ whenever $x\in S$ are called <a href="http://en.wikipedia.org/wiki/Transitive_set" rel="nofollow"><em>transitive</em></a> and are very important in set theory, but they’re the exception, not the rule.)</p>
<p>In particular, your $\beta=\bigcup_{B\in\beta}\{B\}=\{B:B\in\beta\}$; $\bigcup_{B\in\beta}B$ is a completely different thing. In fact we know that $\bigcup_{B\in\beta}B=X$, because one of the conditions defining a base for a topology on $X$ is that every point of $x$ belong to some member of the base.</p>
|
355,740 | <p>Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$</p>
<p>$$\begin{array}{c | c | c | c}
x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline
-1 & \frac{1}{2} & 0.346 & 0.693 \\
0 & 1 & 0.693 & 0.693 \\
1 & 2 & 1.38 & 0.693\\
2 & 4 & 2.77 &0.693&
\end{array}$$ </p>
<p>So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is <em>just like that</em>? Furthermore, what's the use of knowing this?</p>
<p>EDIT:</p>
<p>It seems that I have missed the fairly simple </p>
<p>$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{b^x}}}=\ln\,b$$</p>
<p>But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily? </p>
| spin | 12,623 | <p>In the ring $\mathbb{Z}_4 = \mathbb{Z}/4\mathbb{Z}$ the characteristic is $4$, but $2 + 2 = 0$. You could call this the order of an element in the additive group, perhaps "additive order" would be a good term.</p>
|
1,431,464 | <p>Does anyone know a good reference where it is shown that the Schwartz class $\mathcal{S}(\mathbb R)$ is a dense subset of $L^2(\mathbb R)$?</p>
<p>Many thanks</p>
| Luigi Nocera | 687,083 | <p>A proof that <span class="math-container">$C_{0}$</span> is dense in <span class="math-container">$L^{p}$</span> can be found in Naylor and Sell's "Linear Operator Theory in Engineering and Science", Appendix D, paragraph 12 "Dense Subspaces in <span class="math-container">$L^{p}$</span>, <span class="math-container">$1\le p<\infty$</span>".</p>
<p>Another proof is in Brezis' "Functional analysis, Sobolev spaces and partial differential equations", Theorem 4.12.</p>
|
183,077 | <p>A complex Lie group may have several real forms.
Are there any duality/trinity... between them?
Maybe a trivial question to ask, is $SL(3,\mathbb{C})$ a real form of $SL(3,\mathbb{C})\times SL(3,\mathbb{C})$ ?</p>
| Geoff Robinson | 14,450 | <p>As Amritanshu Prasad points out, the $6$-dimensional irreducible complex representation of $S_{5}$ is indeed monomial (with respect to a suitable basis), and thinking about how to prove this directly led me to a general observation: let $G$ be a finite group, and $\chi$ be a non-linear complex irreducible character of $G$ of minimal degree. Let $H$ be a proper subgroup of $G$ of index less than $2\chi(1).$ Let $\lambda$ be a linear character of $H$ of order $m$ ( that is, $\lambda^{m}$ is the trivial character, but no smaller positive integer power of $\lambda$ is trivial). Then if $m$ does not divide $[G:G^{\prime}],$ the induced character $\theta = {\rm Ind}_{H}^{G}(\lambda)$ is irreducible. For otherwise, (by Frobenius reciprocity), there must be a linear constituent $\mu$ of $\theta$ such that $\langle {\rm Res}^{G}_{H}(\mu),\lambda \rangle \neq 0.$ But then the order of $\mu$ must be divisible by $m,$ whereas the order of $\mu$ divides $[G:G^{\prime}],$ a contradiction. </p>
<p>To apply this result to $G = S_{5},$ take $H$ to be the normalizer of a Sylow $5$-subgroup. Then $[G:G^{\prime}] = 2,$ and $H$ has a linear character $\lambda$ of order $4,$ so the above result can be used to see that ${\rm Ind}_{H}^{G}(\lambda)$ is irreducible (for note that $S_{5}$ has no non-linear irreducible character of degree less than $4$).</p>
<p>Incidentally, taking $G = A_{5},$ and $K$ to be a Sylow $5$-normalizer of $G$, we see that $[G:G^{\prime}] = 1,$ that $[G:K] = 6$ and that $G$ has no non-linear irreducible character of degree less than $3.$ However, $K$ has a linear character of order $2$ such that ${\rm Ind}_{K}^{G}(\lambda)$ is not irreducible ( it is a sum of two irreducible characters of degree $3$). Hence the bound $[G:H] < 2 \chi(1)$ in the result above can't be sharpened in general.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Harun Šiljak | 4,925 | <p>Just my 2c: Being a student with a limited mathematical education, I used V.S. Vladimirov's <em>Generalized Functions in Mathematical Physics</em> (Mir Moscow 1979) and it was not as hard as I expected it to be - Vladimirov was rigorous and pedantic, as a book in mathematics should be, but not too complicated in explaining the concepts.</p>
|
4,291,864 | <p>I have the following equation:</p>
<p><span class="math-container">$y=\frac{3x}{x^{2}+1}$</span></p>
<p>and I want to obtain x in terms of y, so far what I have done is the following:</p>
<p><span class="math-container">$3x=y(x^{2}+1)$</span></p>
<p><span class="math-container">$3x=x^{2}y+y$</span></p>
<p><span class="math-container">$3x-x^{2}y=y$</span></p>
<p>and at this point I got stucked, because that cuadratic term would not allow me to just get one x in one side of the equation. Any advice? Thanks</p>
| José Carlos Santos | 446,262 | <p>Yes, <span class="math-container">$\Bbb R$</span> is an open set. Nonetheless, it is also a closed set (the whole space is <em>always</em> a closed set for <em>any</em> topology) and, with respect to the co-finite topology, it is the smallest closed set that contains <span class="math-container">$(a,b)$</span>. Therefore <span class="math-container">$\overline{(a,b)}=\Bbb R$</span>.</p>
|
4,411,096 | <p>I know closure of connected set in a topological space must be connected as well. However, I can't understand why this counterexample fails.
Take <span class="math-container">$X=[0,2)\cup\{3\}, B_2(1)=(0,2)$</span> which is connected. Now take the closed ball <span class="math-container">$C_2(1)=[0,2)\cup \{3\}$</span> which is clearly not connected. I appreciate your help</p>
| Ken Hung | 626,360 | <p>In this case, the closure of the open ball <span class="math-container">$ B_2(1) $</span> is not the closed ball <span class="math-container">$ C_2(1)$</span>. This is because we have <span class="math-container">$ (B_{1/2}(3) \cap X) \cap B_2(1) = \varnothing $</span> and this shows that <span class="math-container">$ 3 $</span> is not an adherent point of <span class="math-container">$ B_{2}(1) $</span> in <span class="math-container">$ X $</span>.</p>
|
3,369,069 | <p>Let <span class="math-container">$l_1$</span> and <span class="math-container">$l_2$</span> be two distributions in disjoint variables <span class="math-container">$x_1, ..., x_n$</span> and <span class="math-container">$y_1, ..., y_m$</span>. Then it is said to be possible to define a product distribution.</p>
<p>However, I am fundamentally confused. Distributions are in fact linear functionals on the space of smooth and compactly supported functions. Then, how does the 'product' of linear functionals again become a linear functional?</p>
<p>Especially, what can be a definition of <span class="math-container">$\delta(x_1)\delta(x_2)$</span> such that it is equal to <span class="math-container">$\delta(x_1, x_2)$</span>? I am just stuck......</p>
| quarague | 169,704 | <p>If you want to multiply <span class="math-container">$\delta(x_1)$</span> and <span class="math-container">$\delta(x_2)$</span> you first need to make them into functions acting on the space space, so you multiply <span class="math-container">$\delta(x_1)Id(x_2)$</span> and <span class="math-container">$Id(x_1)\delta(x_2)$</span>, where <span class="math-container">$Id$</span> is just the identity map. With that interpretation you get <span class="math-container">$\delta(x_1)\delta(x_1)=\delta(x_1, x_2)$</span>.</p>
|
1,125,842 | <p>In $\sf ZFC$ we have the axiom of infinity and thus can define the natural numbers $$\mathbb N \equiv \bigcap\{X:\emptyset\in X\land \forall n(n\in X\implies n\cup\{n\}\in X)\}.$$ From this it's not particularly hard (exercises 1.6 and 1.7 in Jech - <em>Set Theory</em>) to prove that, firstly, every $n\in\mathbb N$ obeys foundation and secondly, $\mathbb N$ itself obeys it, all without explicitly using the axiom of foundation. My question is: does this extend to arbitrary finite sets or even other countable sets? If not, are there any other specific important examples for which we can verify foundation?</p>
| Rene Schipperus | 149,912 | <p>The von Neumann herirarchy is defined as
$$V_0=\emptyset$$
$$V_{\alpha+1}=P(V_{\alpha})$$
$$V_{\lambda}=\cup_{\alpha < \lambda} V_{\alpha}$$</p>
<p>Now all elements of $V$ satisfy foundation. What foundation really means is that all sets belong to $V$. Since in practice all the sets we naturally deal with are already in $V$ foundation is not so essential to the development of set theory or mathematics. </p>
|
3,752,455 | <blockquote>
<p><strong>Problem.</strong> Show that for <span class="math-container">$n\ge 2$</span> there are no solution <span class="math-container">$$x^n+y^n=z^n$$</span> such that <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> are prime numbers.</p>
</blockquote>
<p>Personally I'd consider this a relatively cute problem which can be given to students when talking about <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="noreferrer">Fermat's Last Theorem</a> - and which should be relatively easily solvable.
(I can post my solution - but I suppose that the solutions which will be given here are very likely to be cleverer than mine.)</p>
<p>I will stress that we're looking that the solutions where <em>simultaneously</em> all three numbers are primes - unlike a more difficult problem posted here: <a href="https://math.stackexchange.com/q/1317346">Diophantine Equation <span class="math-container">$x^n + y^n =z^n (x<y, n>2)$</span></a>.</p>
<p>I have searched on the site a bit to see whether this problem has been posted here before. I only found this deleted question: <a href="https://math.stackexchange.com/q/1311295">How we can deal with this equation <span class="math-container">$a^n+b^n=c^n$</span> if it was given to have solutions in primes numbers not integers numbers?</a> (Of course, it is quite possible that I might have missed something. After all, searching on this site is not easy.)</p>
| Devansh Kamra | 625,028 | <p>It can be observed that if <span class="math-container">$k$</span> is prime, then either <span class="math-container">$k\equiv 1 \space(\text {mod 6})$</span> or <span class="math-container">$k\equiv 5 \space(\text {mod 6})$</span>.</p>
<p>It is also observable that if <span class="math-container">$k\equiv 1 \space(\text {mod 6})$</span> or <span class="math-container">$k\equiv 5 \space(\text {mod 6})$</span>, then <span class="math-container">$k^n\equiv 1 \space(\text {mod 6})$</span> or <span class="math-container">$k^n\equiv 5 \space(\text {mod 6})$</span>.</p>
<p>Let us analyze <span class="math-container">$x$</span> and <span class="math-container">$y$</span> modulo <span class="math-container">$6$</span></p>
<p>Now as <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are primes, both of them can either be congruent to <span class="math-container">$1$</span> modulo <span class="math-container">$6$</span> or congruent to <span class="math-container">$5$</span> modulo 6.</p>
<p><span class="math-container">$\therefore x^n\equiv 1 \space(\text {mod 6})$</span> or <span class="math-container">$x^n\equiv 5 \space(\text {mod 6})$</span></p>
<p>Similarily <span class="math-container">$y^n\equiv 1 \space(\text {mod 6})$</span> or <span class="math-container">$y^n\equiv 5 \space(\text {mod 6})$</span></p>
<p>If both <span class="math-container">$x^n\equiv 1\space \text{(mod 6)}$</span> and <span class="math-container">$y^n\equiv 1\space \text{(mod 6)}$</span>, then <span class="math-container">$(x^n+y^n=z^n)\equiv 2\space \text{(mod 6)}$</span> which means that <span class="math-container">$z$</span> is not prime.</p>
<p>Similarly if both <span class="math-container">$x^n\equiv 5\space \text{(mod 6)}$</span> and <span class="math-container">$y^n\equiv 5\space \text{(mod 6)}$</span>, then <span class="math-container">$(x^n+y^n=z^n)\equiv 4\space \text{(mod 6)}$</span> which means that <span class="math-container">$z$</span> is not prime.</p>
<p>Finally if <span class="math-container">$x^n\equiv 1\space \text{(mod 6)}$</span> and <span class="math-container">$y^n\equiv 5\space \text{(mod 6)}$</span>, then <span class="math-container">$(x^n+y^n=z^n)\equiv 0\space \text{(mod 6)}$</span>, which means that <span class="math-container">$z$</span> again is not a prime.</p>
<p>We have proved that in any of the cases where <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are primes, it is impossible for <span class="math-container">$z$</span> to be prime.</p>
<p><span class="math-container">$\therefore$</span> There are no solutions for <span class="math-container">$x^n+y^n=z^n$</span> for <span class="math-container">$n\geq 2$</span> where <span class="math-container">$x, y$</span> and <span class="math-container">$z$</span> are primes</p>
|
2,647,000 | <p>Consider a function $ϕ$ such that $$\lim_{h→0} ϕ(h) = L$$ and $$L − ϕ(h) ≈ ce^{−1/h}$$ for some constant $c$. By combining $ϕ(h)$, $ϕ(h/2)$, and $ϕ(h/3)$, find an accurate estimate of $L$.</p>
<p>Isn't $ϕ(h)=-ce^{−1/h}+L$? I think I am over-simplfying this...</p>
| videlity | 70,729 | <p>It would probably be best to talk to someone at a university who researches in number theory. There's many different area and aspects of research which would largely depend on a possible supervisor that you would have. They would also have possible projects that you could look into.</p>
|
753,881 | <p>I want to know some typical forms of system of equations generating from practical problems in engineering/economics/physics,etc.</p>
<p>Some examples or research articles would be good.</p>
<p>Specifically, I am looking for some examples of nonlinear system of equations generated from practical problems.</p>
<p>Thanks.</p>
| celtschk | 34,930 | <p>If no odd digits can be placed on odd places, you must fill all odd places with even digits. There are four odd places, and you've got four even digits ($2$, $2$, $4$, $4$), so they are just enough to fill the four odd slots. The number of ways to do so is
$$n_{\text{odd places}} = \frac{4!}{2!\cdot 2!} = 6.$$
Now you have to fill the even places. Since you've used up all even digits for the odd places, you have to fill the even places with the odd digits, $1$, $1$, $1$, $3$. It is quite obvious that there are four possibilities to do so, since all you can decide is where to put the $3$, and then fill the rest with $1$. But you can also calculate this mechanically:
$$n_{\text{even places}} = \frac{4!}{3!\cdot 1!} = 4.$$
Now it turns out that the filling of odd and even places is independent of each other (this would have been different if, e.g., you would have had three each of $2$ and of $4$, and only two odd digits). For this reason you find the total number of possibilities by just multiplying both numbers:
$$n_{\text{total}} = n_{\text{odd places}}\cdot n_{\text{even places}} = 6\cdot 4 = 24.$$</p>
|
3,837,548 | <p>The triple integral
<span class="math-container">$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}=\zeta(3) \dots (1)$$</span>
is not separable in <span class="math-container">$x,y,z$</span> and the integral representation of reciprocal: <span class="math-container">$\frac{1}{1-xyz}=\int_{0}^{1} t^{-xyz} dt$</span> also doesn't help separation of the integrand. I want help to prove (1).</p>
| FDP | 186,817 | <p><span class="math-container">\begin{align}\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{1-xyz}&=-\int_0^1\int_0^1 \frac{\ln u}{1-uz}dudz\\
&=-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln (uz)}{1-uz}dudz\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 v}{1-v}dv\\
&=\frac{1}{2}\times 2\zeta(3)\\
&=\boxed{\zeta(3)}
\end{align}</span></p>
<p>NB: i have used the following results: If <span class="math-container">$f$</span> is a continuous function on <span class="math-container">$[0;1]$</span> then <span class="math-container">$\displaystyle \int_0^1 \int_0^1 f(xy)dxdy=-\int_0^1 f(z)\ln zdz$</span></p>
<p><span class="math-container">$\displaystyle \int_0^1 \frac{\ln^2 x}{1-x}dx=2\zeta(3)$</span></p>
|
1,762,268 | <p>Let $X$ be a Hausdorff space and let $f:X\to \mathbb{R}$. If grapph of $f$ is compact we have to show that $f$ is continuous. </p>
<p>Since every closed subset of a Hausdorff space is closed, therefore grapph of $f$ is closed. WE know that if $f:X\to Y$ and $Y$ is compact, then graph of $f$ is clsed implies $f$ is continuous. But here $\mathbb R$ is not compact. Please help!</p>
| Rick Sanchez | 332,412 | <p>Fix $x_0 \in X$ and let $x_a$ be a net converging to $x_0$ in $X$. You need to show $f(x_a)\rightarrow f(x_0)$ in $\mathbb{R}$. Let $G_f$ denote the graph of $f$, which is a subset of $X \times \mathbb{R}$. Then $(x_a,f(x_a))$ is a net in $G_f$, which is compact, so we can extract a convergent subnet $(x_b,f(x_b)) \rightarrow (x,y) \in G_f$. But then $x_b \rightarrow x$ and since $x_a \rightarrow x_0$, we have $x=x_0$ by uniqueness of limits (Hausdorff). Then $y=f(x_0)$ since $(x,y) \in G_f$. This means $f(x_b) \rightarrow f(x_0)$. </p>
<p>Now, do this argument for ANY subnet of $x_a$ to conclude that the whole net $f(x_a)$ converges to $f(x_0)$.</p>
|
1,724,554 | <p>Say, $A$ is an $ n\times n $ matrix over $\Bbb R$, with</p>
<p>$$ A_{ij} = \begin{cases} a \qquad \text{if } i=j\\
b \qquad \text{otherwise.}
\end{cases} $$</p>
<p>How do we compute the determinant of this symmetrix matrix $A$?</p>
| mathreadler | 213,607 | <p>If the first row is the vector $\bf v$ and it's DFT (discrete fourier-transform) is $\bf w$, then the determinant is $\prod_i {\bf w}_i$. This is true for any <a href="https://en.wikipedia.org/wiki/Circulant_matrix" rel="nofollow">circulant</a> matrix as it's eigenvalues coincide with those fourier coefficients $$\lambda_i = {\bf w}_i$$ together with the fact that the determinant equals the product of eigenvalues $$\det({\bf A}) = \prod_i \lambda_i({\bf A})$$So this can be used for a wider class of matrices as long as we can compute the Fourier transform.</p>
|
3,077,084 | <blockquote>
<p>If <span class="math-container">$a,b,c>0.$</span> Then minimum value of</p>
<p><span class="math-container">$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$</span></p>
</blockquote>
<p>Try: Arithmetic geometric inequality</p>
<p><span class="math-container">$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$</span></p>
<p>and <span class="math-container">$(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$</span></p>
<p>so <span class="math-container">$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2\geq 18\sqrt{2}(abc)^{-1/3}$</span></p>
<p>could some help me to solve it. answer is <span class="math-container">$64$</span></p>
| Martin R | 42,969 | <p><em>Hint:</em> Apply <span class="math-container">$AM \ge GM$</span> not to <span class="math-container">$8a^2 + b^2 + c^2$</span>, but to
<span class="math-container">$$
(2a)^2 + (2a)^2 + b^2 + c^2
$$</span>
and <span class="math-container">$HM \le GM$</span> not to <span class="math-container">$a^{-1}+b^{-1}+c^{-1}$</span>, but to
<span class="math-container">$$
\frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c}
$$</span></p>
<p>The “partitions” are chosen in such a way that equality can hold <em>simultaneously</em> in both estimates, in this case when <span class="math-container">$2a=b=c$</span>.</p>
<p>One can also use the relationships between <a href="https://en.wikipedia.org/wiki/Generalized_mean" rel="nofollow noreferrer">generalized means</a>, here “harmonic mean <span class="math-container">$\le$</span> quadratic mean”:
<span class="math-container">$$
\frac{4}{\frac{1}{2a} + \frac{1}{2a} + \frac{1}{b} + \frac{1}{c}}
\le \sqrt{\frac{(2a)^2 + (2a)^2 + b^2 + c^2}{4}}
$$</span> </p>
|
44,562 | <p>The question is motivated from the definition of $C^r(\Omega)$ I learned from S.S.Chern's <em>Lectures on Differential Geometry</em>:</p>
<p>Suppose $f$ is a real-valued function defined on an open set $\Omega\subset{\bf R}^m$. If all the $k$-th order partial derivatives of $f$ exist and are continuous for $k\leq r$, then we say $f\in C^r(\Omega)$. Here $r$ is some positive integer.</p>
<p>While in Folland's <em>Introduction to Partial Differential Equations</em>, $C^r(\Omega)$ denotes the space of functions possessing continuous derivatives up to order $r$ on $\Omega$, where $\Omega$ is an open subset of ${\bf R}^m$ and $k$ is a positive integer.</p>
<p>It's trivial to show that these two definitions are equivalent when $m=1$. So here is my question:</p>
<blockquote>
<p>Are these two definitions equivalent in the higher dimensions? How to prove it?</p>
</blockquote>
<hr>
<p><strong>Edit</strong>: The question was partially answered by Didier <a href="https://math.stackexchange.com/questions/44355/can-being-differentiable-imply-having-continuous-partial-derivatives/44364#44364">here</a>. I do not think it is trivial for me. It boils down to the following one:</p>
<blockquote>
<p>In the higher dimension case, say,
$f:{\bf R}^n\to {\bf R}$($n\geq 2$), what's
the relationship between the higher
order partial derivative of $f$ and
the high order derivative $f^{(k)}$?</p>
</blockquote>
<p>When $1\leq k\leq 2$, $f^{(k)}$ is the <a href="http://en.wikipedia.org/wiki/Gradient" rel="nofollow noreferrer">gradient</a> and the <a href="http://en.wikipedia.org/wiki/Hessian_matrix" rel="nofollow noreferrer">Hessian matrix</a> respectively. The answer to the above questions in these two cases is clear. I have no idea for $k>2$. I don't know much about tensor, I'm not sure if the question is related to the topic of multilinear algebra.</p>
| Condor | 11,915 | <p>A sketch proof...</p>
<p>Consider $\mathbb{R}$ as the set of equivalence classes of Cauchy sequences of rationals. We define "the normal" partial order on $\mathbb{R}$ by $x \leq y$ iff $(x = y)$ OR $(\forall \langle x_{i} \rangle)(\forall \langle y_{i} \rangle)(\langle x_{i} \rangle \in x$ AND $\langle y_{i} \rangle \in y \rightarrow (\exists n)(n \in \mathbb{N} \rightarrow (\forall m)(m > n \rightarrow x_{m} < y_{m}))))$. For $A$ to be well-ordered under this partial order there can only be at most $\aleph_{0}$ distinct $n$ in the last part of the definition as they are drawn from $\mathbb{N}$.</p>
|
3,491,816 | <p>Find the min and max values of the function <span class="math-container">$$f(x,y)=10y^2-4x^2$$</span> with the constraint <span class="math-container">$$g(x,y)=x^4+y^4=1$$</span>
I have done the following working;
<span class="math-container">$$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$</span><span class="math-container">$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$</span><span class="math-container">$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$</span><span class="math-container">$$\lambda x^3=2\\\lambda y^3=5$$</span> My question is how can I find the value of lambda for the constraint to hold.</p>
| José Carlos Santos | 446,262 | <p>Your system of equations is<span class="math-container">$$\left\{\begin{array}{l}-8x=4\lambda x^3\\20y=4\lambda y^3\\x^4+y^4=1.\end{array}\right.$$</span>If <span class="math-container">$x=0$</span>, then you can obviously take <span class="math-container">$y=\pm1$</span> and if <span class="math-container">$y=0$</span>, you can obviously take <span class="math-container">$x=\pm1$</span>; these are the only solutions in which one of the numbers <span class="math-container">$x$</span> or <span class="math-container">$y$</span> is <span class="math-container">$0$</span>.</p>
<p>And there are no more solutions. In fact, if <span class="math-container">$x,y\neq0$</span>, then it follows from the first two equations that <span class="math-container">$x^2=-\frac2\lambda$</span> and that <span class="math-container">$y^2=\frac5\lambda$</span>. This is impossible, of course: if <span class="math-container">$\lambda>0$</span> there is no such <span class="math-container">$x$</span>, and if <span class="math-container">$\lambda<0$</span> there is no such <span class="math-container">$y$</span>.</p>
|
3,491,816 | <p>Find the min and max values of the function <span class="math-container">$$f(x,y)=10y^2-4x^2$$</span> with the constraint <span class="math-container">$$g(x,y)=x^4+y^4=1$$</span>
I have done the following working;
<span class="math-container">$$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$</span><span class="math-container">$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$</span><span class="math-container">$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$</span><span class="math-container">$$\lambda x^3=2\\\lambda y^3=5$$</span> My question is how can I find the value of lambda for the constraint to hold.</p>
| Michael Rozenberg | 190,319 | <p><span class="math-container">$$-4=-4\sqrt{x^4+y^4}\leq10y^2-4x^2\leq10\sqrt{x^4+y^4}=10.$$</span>
The equality occurs for <span class="math-container">$(x,y)=(1,0)$</span> for the left inequality and for <span class="math-container">$(x,y)=(1,0)$</span> for the right inequality, which says that we got the minimal value and the maximal value. </p>
|
3,491,816 | <p>Find the min and max values of the function <span class="math-container">$$f(x,y)=10y^2-4x^2$$</span> with the constraint <span class="math-container">$$g(x,y)=x^4+y^4=1$$</span>
I have done the following working;
<span class="math-container">$$\frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\\frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y}$$</span><span class="math-container">$$-8x= \lambda 4x^3\\20y= \lambda 4y^3$$</span><span class="math-container">$$x(2- \lambda x^3)=0\\y(5- \lambda y^3)=0$$</span><span class="math-container">$$\lambda x^3=2\\\lambda y^3=5$$</span> My question is how can I find the value of lambda for the constraint to hold.</p>
| DeepSea | 101,504 | <p>The solutions by Lagrange Multipliers are shown above. But if you like a traditional method, here is one. One way is to lower the "power" of the constraint, namely: <span class="math-container">$x^4+y^4 =1$</span>. So let's put <span class="math-container">$a = x^2, b = y^2$</span>, thus <span class="math-container">$a, b \ge 0$</span> and <span class="math-container">$a^2+b^2=1$</span>.Thus we have: <span class="math-container">$f(a,b) = 10b-4a$</span> with <span class="math-container">$a^2+b^2=1$</span> and <span class="math-container">$a,b \ge 0$</span>.From this we can use a trig substitution: <span class="math-container">$a = \cos \theta, b = \sin \theta, \theta\in [0,\frac{\pi}{2}]$</span>, and we have: <span class="math-container">$f(\theta) = 10\sin \theta - 4\cos \theta, \theta \in [0,\frac{\pi}{2}]$</span>. Taking derivative of <span class="math-container">$f$</span> we have: <span class="math-container">$f'(\theta) = 10\cos \theta + 4\sin \theta > 0 \implies f_{\text{min}} = f(0) = -4, f_{\text{max}} = f(\frac{\pi}{2}) = 10$</span> as claimed. </p>
|
1,392,209 | <blockquote>
<p>Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$</p>
</blockquote>
<p>My attempt </p>
<p>So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$</p>
<p>$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$</p>
<p>Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$</p>
<p>$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$</p>
<p>Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$</p>
<p>Thanks for any help. </p>
| CivilSigma | 229,877 | <p>Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following:</p>
<p>$$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$</p>
<p>We take the derivative of the numerator and denominator again:</p>
<p>$$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$</p>
|
1,392,209 | <blockquote>
<p>Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$</p>
</blockquote>
<p>My attempt </p>
<p>So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$</p>
<p>$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$</p>
<p>Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$</p>
<p>$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$</p>
<p>Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$</p>
<p>Thanks for any help. </p>
| Rio Alvarado | 253,991 | <p>For your final problem I'd use L'Hospital's Rule to obtain:</p>
<p>$$\lim_{x \to 0} \frac{x \sin(x)}{3x^2} \implies \lim_{x \to 0} \frac{\sin(x)}{3x} \\ \hspace{.1cm} \text{using L'Hospital's again}, \hspace{.1cm} \\ \lim_{x \to 0}\frac{\cos(x)}{3} = \frac{1}{3}.$$</p>
|
1,680,862 | <p>I am attempting to find the expected value and variance of the random variable $X$ analytically (in addition to a decimal answer). $X$ is the random variable <code>expression(100)[-1]</code> where <code>expression</code> is defined by:</p>
<pre><code>def meander(n):
x = [0]
for t in range(n):
x.append(x[-1] + 3*random.random())
return x
</code></pre>
<p>For those that do not understand the Python, $X$ is essentially the sum of a sequence of $100$ values with value of <code>3*random.random()</code>, where <code>random.random()</code> is uniformly distributed on $[0,1)$.</p>
<p>I am almost certain that I will need to apply the concepts of:
$$\text{mean}\left(\bar{X}\right)=E\left(\frac{1}{n}\left(X_1+X_2+...+X_n\right)\right)=E\left(X\right)$$
$$\text{and}$$
$$\text{var}\left(\bar{X}\right)=var\left(\frac{1}{n}\left(X_1+X_2+...+X_n\right)\right)=\frac{1}{n}\text{var}\left(X\right)$$
$$\text{where, }\bar{X}=\frac{1}{n}\left(X_1+X_2+...+X_n\right)$$</p>
<p>I am having difficulty understand how I should be plugging in this equation and representing it symbolically, let alone calculating it. I created a simulation in order to better understand the distribution of the data (in addition to getting an estimate of the expected value) and it seems to be a Gaussian distribution (<a href="https://i.stack.imgur.com/1NmfF.png" rel="nofollow noreferrer">histogram of distribution after 100,000 trials</a>). The simulation suggests an estimated expected value of $150.038527551$.</p>
<p>These solutions will culminate in the usage of the Central Limit Theorem in finding an analytical expression that approximates the pdf of $X$.</p>
<p>Any guidance or help to point me in the right direction would be very much appreciated!</p>
| Clement C. | 75,808 | <p>So, your random variable is $$
X = 3X_1+\dots+3X_{100} = \sum_{k=1}^n 3X_k
$$
with $n=100$, where $X_1,\dots, X_n$ are independent, identically distributed random variables that are uniform in $[0,1)$. <a href="https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)" rel="nofollow">In particular</a>, $\mathbb{E}\left[ X_k \right] = \frac{1}{2}$ and $\operatorname{var} X_k = \frac{1}{12}$ for every $1\leq k\leq n$.</p>
<p>By <a href="https://en.wikipedia.org/wiki/Expected_value#Linearity" rel="nofollow">linearity of expectation</a>, you get
$$
\mathbb{E}[X] = \mathbb{E}\left[ \sum_{k=1}^n 3X_k \right]
= \sum_{k=1}^n 3\mathbb{E}\left[ X_k \right]
=\sum_{k=1}^n 3\cdot \frac{1}{2} = n\cdot \frac{3}{2} = 150.
$$
(this does not rely on the fact that the $X_k$'s are independent, only on the fact that they all have a well-defined expectation).</p>
<p>By <a href="https://en.wikipedia.org/wiki/Variance#Basic_properties" rel="nofollow">properties of variance</a> (detailed below), crucially relying on the fact that the $X_k$'s are independent, you obtain
$$
\operatorname{var}(X) = \operatorname{var}\left( \sum_{k=1}^n 3X_k \right)
= \sum_{k=1}^n \operatorname{var}(3 X_k)
= \sum_{k=1}^n 9\operatorname{var} X_k
=\sum_{k=1}^n 9\cdot \frac{1}{12} = n\cdot \frac{3}{4} = 75
$$
where we used first the fact that <em>"the variance of the sum of (pairwise) independent random variables is the sum of their variances"</em>,* and then that $ \operatorname{var}(aY) = a^2 \operatorname{var}(Y)$ for any real number $a$.</p>
<p>(*) Provided the variances are well-defined, i.e. the random variables are in $L^2$.</p>
|
685,681 | <p>I want to prove that $\dim V/(X \cap Y)$ in finite, if $V$ be a vector space and $X$, $Y$ two sub spaces of $V$ such that $\dim V/Y$ and $\dim V/X$ are finite.</p>
| ajd | 90,897 | <p>Let $\{a_1 + Y,\ldots,a_r +Y\}$ be a basis of $V/Y$, where the $a_i$s are elements of $V$. Since $V/X$ is finite-dimensional, we have that $Y/(Y\cap X)$ is finite-dimensional (since we have maps $Y\to V\to V/X$ and the composition has kernel $Y\cap X$, so we have an injection $Y/(Y\cap X)\to V/X$, so $\dim Y/(Y\cap X)\le \dim V/X<\infty$), so let $\{b_1 + Y\cap X,\ldots,b_s + Y\cap X\}$ be a basis of $Y/(Y\cap X)$, where the $b_i$s are all elements of $Y$. Let $v+(X\cap Y)\in V/(X\cap Y)$. Now $v+Y\in V/Y$, so we can write $v + Y = \sum \lambda_i(a_i+Y)$, which means that $v = y+\sum \lambda_ia_i$ for some $y\in Y$. Then $y+Y\cap X\in Y/(Y\cap X)$, so we can write $y = \sum\nu_ib_i +z$ with $z\in Y\cap X$. Combining these together, we get that $v = \sum \lambda_ia_i + \sum \nu_ib_i + z$, where $z\in X\cap Y$. Therefore $v + X\cap Y = \sum \lambda_i(a_i+X\cap Y) + \sum \nu_i(b_i + X\cap Y$.</p>
<p>Therefore, $\{a_1 + X\cap Y,\ldots,a_r+X\cap Y,b_1+X\cap Y,\ldots,b_s+X\cap Y\}$ span $V/(X\cap Y)$, so $V/(X\cap Y)$ is finite-dimensional. (Note that the spanning set is likely not a basis.)</p>
|
2,544,261 | <p>The question:</p>
<blockquote>
<p>Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.</p>
</blockquote>
<p>Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.</p>
<p>\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\
& = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\
& = a+b-a^2+ab-b^2 \\
0 & = a^2-ab+b^2-a-b+1
\end{align}</p>
<p>\begin{align}
0 & = a^2-ab+b^2-a-b+1 \\
& = 2a^2-2ab+2b^2-2a-2b+2 \\
& = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\
& = (a-b)^2 + (a-1)^2 + (b-1)^2
\end{align}</p>
<p>This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?</p>
| stressed out | 436,477 | <p>If the sum of a finite number of non-negative expressions is $0$, each of them has to be zero. </p>
<p>In other words, when $a,b,c\geq0$</p>
<p>$a+b+c=0 \implies a=b=c=0$</p>
<p>You have done the hard part by showing that it can be written as the sum of three squares.
This means that $a=b$, $a=1$ and $b=1$. What does that tell you about $x$?</p>
|
2,073,410 | <p>If $$ a-(a \bmod x)<b$$ how do I prove that $$c-(c\bmod x)<b \;\forall c<a?$$ </p>
| Sergei Golovan | 400,926 | <p>The expression $c - (c\mathrel{\mathrm{mod}} x)$ represents the greatest integer $kx$ which is not greater than $c$. So, the larger $c$, the larger $c - (c\mathrel{\mathrm{mod}} x)$.</p>
|
1,240,212 | <blockquote>
<p>How to find the degree of an extension field ?</p>
</blockquote>
<p>Let $f:=T^3-T^2+2T+8\in\mathbb Z[T]$ and $\alpha$ be the real root of $f$. Why is then $\mathbb Q(\alpha)$ is a number field of degree $3$ ?</p>
<p>I've seen somewhere that $[\mathbb Q(r):\mathbb Q]\le n$ if $r$ is a root of an irreducible polynomial with coefficients in $\mathbb Q$ of degree $n$. What does it change in my case, if the extension field would contain also the other roots, they're also roots of the polynomial $f$, how does the degree increase ?</p>
<p>Obviously, by finding an element in $\mathbb Q(\alpha)$, which is not in $\mathbb Q$, the degree cannot be $1$, so it remains to show that, it is also not $2$. Or is there a better way, can we find $3$ field embeddings ?</p>
| A.P. | 65,389 | <blockquote>
<p><strong>Fact</strong>: Consider two polynomials $f$ and $p$ over $\Bbb{Q}$, with $p$ irreducible. It can be proved that if $f$ and $p$ share a root, then $p$ divides $f$.</p>
</blockquote>
<p>How does this help? Suppose that $\alpha$ is a root of an irreducible polynomial $f \in \Bbb{Q}[X]$ of degree $n$ and that $[\Bbb{Q}(\alpha):\Bbb{Q}] = m < n$. Since $[\Bbb{Q}(\alpha):\Bbb{Q}]$ is the dimension of $\Bbb{Q}(\alpha)$ over $\Bbb{Q}$ as a <em>vector space</em>, this means that there are $q_0,\dotsc,q_{m-1} \in \Bbb{Q}$ such that
$$
\alpha^m = q_0 + q_1 \alpha + \dotsb + q_{m-1} \alpha^{m-1}
$$
hence $\alpha$ is a root of $g(X) = X^m - q_{m-1} X^{m-1} - \dotsb - q_0$. By the aforementioned fact it follows that $g$ divides $f$ in $\Bbb{Q}(X)$, but this is absurd because $f$ is irreducible.</p>
<hr>
<p>As for what happens when adjoining other roots, that depends on how much your field differs from being Galois.</p>
<p>For example, consider $f(X) = X^3 - d$, with $d$ a cube-free integer. Then its roots are $\sqrt[3]{d},\zeta_3\sqrt[3]{d},\zeta_3\sqrt[3]{d}$, where $\zeta_3$ is a primitive root of unity, i.e. a root of $X^2 + X + 1$. Then $\Bbb{Q}(\alpha_1, \alpha_2) = \Bbb{Q}(\sqrt[3]{d},\zeta_3)$ has degree $6$ for any two distinct roots $\alpha_1,\alpha_2$ of $f$.</p>
<p>On the other hand, say, adding any root of $X^4 + X^3 + X^2 + X + 1$ to $\Bbb{Q}(\zeta_5)$ won't change the degree of the extension (which is $5$).</p>
<hr>
<p><strong>Regarding your specific problem</strong>: The splitting field of an irreducible cubic polynomial $f$ with rational coefficients can have either degree $3$ or degree $6$. In particular if, like in your case, $f$ has only one real root, then its splitting field has degree $6$. For example, you can find the full treatment of this at <a href="http://planetmath.org/galoisgroupofacubicpolynomial" rel="nofollow">PlanetMath</a>, where there is a proof that the splitting field of $f$ is
$$
\Bbb{Q}(\alpha, \sqrt{D})
$$
where $\alpha$ and $D$ are, respectively, a root and the discriminant of $f$.</p>
<p><em>Note</em>: $f$ has one real and two complex roots <a href="https://en.wikipedia.org/wiki/Discriminant#Cubic" rel="nofollow">iff</a> its discriminant is negative, thus in this case it is always true that $\sqrt{D} \notin \Bbb{Q}$.</p>
|
17,134 | <p>On a very regular basis we see new users that are not accustomed with the use of MathJaX on MSE. Sometimes even some users that aren't that new to the site. Most of us, when this happens, kindly bring to this users attention that there is a <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">MathJax basic tutorial and quick reference</a> and give a proper link. </p>
<p>Now this is not a bad habit at all, because you can't expect new users to magically know this. My idea however was that we could promote this a bit more actively. </p>
<p>For instance: a new user is taken by the hand when he/she is taking the <a href="https://math.stackexchange.com/tour">tour</a>. Would it not be to this new user's benefit to point out in the <a href="https://math.stackexchange.com/tour">tour</a>, that we write our maths in MathJaX. For example place a link to <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">-MathJax basic tutorial and quick reference-</a> or even a link to a -MathJaX tour page- (which would probably have the same structure as the normal <a href="https://math.stackexchange.com/tour">tour</a> page). And perhaps even -but this is really going ahead of things- award a badge for reading this page too.</p>
<p>This was a thought of mine that I wanted to share. I would like to know how the community feels about this. Or perhaps someone could make clear why this would not be helpfull.</p>
| Paradox 101 | 177,844 | <p>That's a really good suggestion. I'm relatively new here and had no idea that MathJax even existed. Your suggestion made me aware of it so thank you. The link should be given with the email.</p>
|
1,178,361 | <p>The surface with equation $z = x^{3} + xy^{2} $ intersects the plane with equation $2x-2y = 1$ in a curve. What is the slope of that curve at $x=1$ and $ y = \frac{1}{2} $</p>
<p>So I put $ x^{3} + xy^{2} = 2x - 2y - 1 $</p>
<p>We have $ x^{3} + xy^{2} - 2x + 2y + 1 $</p>
<p>Do I then differentiate wrt x and y simultaneously?</p>
<p>I know how to differentiate at a point with directional derivatives. But how do I go about the above question considering the fact that direction isn't mentioned.</p>
<p>Maybe I'm going completely down the wrong route... any help is hugely appreciated !</p>
| John Brevik | 210,492 | <p>Sorry to add yet another answer, but for a learner I think that a straightforward approach is best. How do you prove two sets equal? Prove that each is contained in the other. So let $y\in C$. Since $f$ is onto, there exists $x\in X$ such that $f(x)=y.$ Now $x\in f^{-1}(C)$. But then $x\in f^{-1}(D)$, so $f(x)\in D$. Maybe I'll let the OP finish...</p>
|
4,058,884 | <p>I have an orthonormal basis <span class="math-container">${\bf{b}}_1$</span> and <span class="math-container">${\bf{b}}_2$</span> in <span class="math-container">$\mathbb{R}^2$</span>. I want to find out the angle of rotation. I added a little picture here. I essentially want to find <span class="math-container">$\theta$</span></p>
<p><a href="https://i.stack.imgur.com/YH2Vt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YH2Vt.png" alt="enter image description here" /></a></p>
<p>I know that I can compute the angle between two vectors but then there are <span class="math-container">$4$</span> combinations here</p>
<ul>
<li><span class="math-container">$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$</span></li>
<li><span class="math-container">$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$</span></li>
</ul>
<p>How would one know which angle is correct? Importantly, here I used the labels <span class="math-container">$b_1$</span>, <span class="math-container">$b_2$</span> in the same order as <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> but that's not necessarily the same order geometically!</p>
| Glärbo | 892,839 | <p>Let <span class="math-container">$\mathbf{b}_1 = (x_1, y_1)$</span>. Then, <span class="math-container">$\theta = \operatorname{atan2}(y_1, x_1)$</span>.</p>
<p><span class="math-container">$\operatorname{atan2}(y_1, x_1)$</span> is the two-argument form of arcus tangent, equivalent to <span class="math-container">$\arctan(y_1 / x_1)$</span> except that the two-argument form takes the quadrant (signs of both <span class="math-container">$x_1$</span> and <span class="math-container">$y_1$</span>) into account.</p>
<hr />
<p>If you do not know which of <span class="math-container">$\mathbf{b}_1 = (x_1, y_1)$</span> and <span class="math-container">$\mathbf{b}_2 = (x_2, y_2)$</span> to use, then
<span class="math-container">$$\begin{aligned}
d &= x_1 y_2 - x_2 y_1 \\
\theta &= \begin{cases}
\operatorname{atan2}(y_1, x_1), & d \ge 0 \\
\operatorname{atan2}(y_2, x_2), & d \lt 0 \\
\end{cases} \end{aligned}$$</span>
Here, <span class="math-container">$d \gt 0$</span> if <span class="math-container">$\mathbf{b}_2$</span> is counterclockwise from <span class="math-container">$\mathbf{b}_1$</span>, and <span class="math-container">$d \lt 0$</span> if clockwise. This uses the vector which is the clockwise one of the pair.</p>
<hr />
<p>Note that for arbitrary coordinate system <span class="math-container">$\mathbf{e}_1$</span>, <span class="math-container">$\mathbf{e}_2$</span>, you can use
<span class="math-container">$$\begin{aligned}
x_1 &= \mathbf{b}_1^T \mathbf{e}_1 \\
y_1 &= \mathbf{b}_1^T \mathbf{e}_2 \\
x_2 &= \mathbf{b}_2^T \mathbf{e}_1 \\
y_2 &= \mathbf{b}_2^T \mathbf{e}_2 \\
\end{aligned}$$</span>
above.</p>
|
4,058,884 | <p>I have an orthonormal basis <span class="math-container">${\bf{b}}_1$</span> and <span class="math-container">${\bf{b}}_2$</span> in <span class="math-container">$\mathbb{R}^2$</span>. I want to find out the angle of rotation. I added a little picture here. I essentially want to find <span class="math-container">$\theta$</span></p>
<p><a href="https://i.stack.imgur.com/YH2Vt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YH2Vt.png" alt="enter image description here" /></a></p>
<p>I know that I can compute the angle between two vectors but then there are <span class="math-container">$4$</span> combinations here</p>
<ul>
<li><span class="math-container">$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$</span></li>
<li><span class="math-container">$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$</span></li>
<li><span class="math-container">$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$</span></li>
</ul>
<p>How would one know which angle is correct? Importantly, here I used the labels <span class="math-container">$b_1$</span>, <span class="math-container">$b_2$</span> in the same order as <span class="math-container">$e_1$</span> and <span class="math-container">$e_2$</span> but that's not necessarily the same order geometically!</p>
| Widawensen | 334,463 | <p>The simplest way is to construct orthogonal matrix with column vectors <span class="math-container">$B=[b_1 \ \ b_2 \ \ b_1 \times b_2]$</span>.<br />
(I assume here that <span class="math-container">$b_1$</span> and <span class="math-container">$b_2$</span> are normalized to unit length)
and to use trace of such matrix <span class="math-container">$\text{tr}(B)$</span>.</p>
<p>Then we have immediately <span class="math-container">$\text{tr}(B)=1+2\cos(\theta)$</span>.</p>
<p>See also <a href="https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle</a></p>
|
2,646,363 | <p>Let $A_1, A_2, \ldots , A_{63}$ be the 63 nonempty subsets of $\{ 1,2,3,4,5,6 \}$. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_i$. Then what is the value of $\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})$?</p>
<p>Here is the solution </p>
<p>For size 1: sum of the elements, which is 21
For size 2: $ 1 \cdot (2 + 3 + 4 + 5 + 6) = 20 $, $ 2 \cdot (3 + 4 + 5 + 6) = 36 $, $ 3 \cdot (4 + 5 + 6) = 45 $, $ 4 \cdot (5 + 6) = 44 $, $ 5 \cdot 6 = 30 $. Sum is 175.
For size 3: Those with least element 1: $ 6, 8, 10, 12, 12, 15, 18, 20, 24, 30 = 155 $. Those with least element 2: $ 24, 30, 36, 40, 48, 60 = 238 $. Those with least element 3: $ 60 + 72 + 90 = 222 $. Those with least element 4: only one possible subset, which is $ \{4, 5, 6\} $, the $ \pi $ of which is 120. The total sum here is 735.
For size 4: Least element 1: $ 24 + 30 + 36 + 40 + 48 + 60 + 60 + 72 + 90 + 120 = 580 $; least element 2: $ 120 + 144 + 180 + 240 + 360 = 1044 $; least element 3: only one, which is $ 3 \cdot 4 \cdot 5 \cdot 6 = 360 $. The total sum here is 1984.
For size 5: Exclude each one individually to get $ 720 + 360 + 240 + 180 + 144 + 120 = 1764 $
For size 6: $ 6! = 720 $</p>
<p>The final answer is $ 21 + 175 + 735 + 1984 + 1764 + 720 = \boxed{5399} $</p>
<p>Is there any shorter way for doing this ?</p>
<p>Thank a lot </p>
| Clive Newstead | 19,542 | <p>Let $X$ be a set containing $6$ geese a-laying, $5$ gold rings, $4$ calling birds, $3$ French hens, $2$ turtle doves and $1$ partidge in a pear tree.</p>
<p>For a fixed $A \subseteq \{ 1, 2, 3, 4, 5, 6 \}$, the value $\pi(A)$ is the number of ways of picking one of each of the animals (or rings, I guess) from $X$ as indicated by $A$. For example, $\pi(\{2,5\})$ is the number of ways of picking one turtle dove and one gold ring from $X$.</p>
<p>So $\sum_{A \subseteq X} \pi(A)$ is the number of ways of (i) selecting a subset of $[6]$ and (ii) selecting one of each animal/ring as indicated by that subset.</p>
<p>This value can also be calculated by, for each $k \in [6]$, deciding if you will choose an animal/ring as indicated by $k$ and, if so, selecting one. If an animal/ring is chosen, there are $k$ choices for which one to pick; if not, there is only $1$ choice (pick nothing). Thus
$$\sum_{A \subseteq X} \pi(A) = \prod_{k=1}^6 (k+1) = 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$$</p>
<p>This takes into account the possibility of selecting nothing, so we must subtract one, to obtain
$$\sum_{i=1}^{63} \pi(A_i) = \sum_{\varnothing \ne A \subseteq X} \pi(A) = 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 - 1 = \boxed{5039}$$</p>
|
3,085,842 | <p>What can be said about the uniform Convergence of <span class="math-container">$\sum_{n=1}^{\infty}\frac{x}{[(n-1)x+1][nx+1]}$</span> in the interval <span class="math-container">$[0,1]$</span>?</p>
<p>The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.</p>
| vidyarthi | 349,094 | <p>The series can be computed by using telescoping. We have <span class="math-container">$\frac{x}{[(n-1)x+1][nx+1]}=\frac1{(n-1)x+1}-\frac1{nx+1}$</span>. Thus, the sequence of partial sums would be <span class="math-container">$s_n=\frac{nx}{nx+1}$</span>. Hence, sum is equal to <span class="math-container">$$\begin{cases}0\ \text{when}\ x=0\\1\ \text{when}\ x\neq0\end{cases}$$</span>. This easily shows non-uniform convergence, as the limit is discontinuous at <span class="math-container">$0$</span>.</p>
|
3,101,098 | <p>From 11, 12 in the book Logic in Computer Science by M. Ryan and M. Huth:</p>
<p>**</p>
<blockquote>
<p>"What we are saying is: let’s make the assumption of ¬q. To do this,
we open a box and put ¬q at the top. Then we continue applying other
rules as normal, for example to obtain ¬p. But this still depends on
the assumption of ¬q, so it goes inside the box. Finally, we are ready
to apply →i. It allows us to conclude ¬q → ¬p, but that conclusion no
longer depends on the assumption ¬q. Compare this with saying that ‘If
you are French, then you are European.’ The truth of this sentence
does not depend on whether anybody is French or not. Therefore, we
write the conclusion ¬q → ¬p outside the box."</p>
</blockquote>
<p>**</p>
<p>My question is about the scope of assumptions in propositional logic and proving techniques. I am not sure I fully understand what this text is trying to say. </p>
<p>How can an assumption only have scope inside the box, but once you finish what you want to prove it is no more part of the assumption box and is accessible universally in the proof? WHY is this possible? Why does it not break things in the proof? This looks too convenient and random.</p>
<p>Secondly, I do not understand the French and European example connection to what is written in this text. If somebody could please connect this example to what the author is actually trying to explain through this. </p>
| Mees de Vries | 75,429 | <blockquote>
<p>once you finish what you want to prove it is no more part of the assumption box and is accessible universally in the proof?</p>
</blockquote>
<p>This is not what happens. You open a proof box with <span class="math-container">$\neg q$</span>, and within the proof box <span class="math-container">$\neg q$</span> holds. Then you do some reasoning and conclude <span class="math-container">$\neg p$</span>. Within the proof box, starting at that line, <span class="math-container">$\neg p$</span> holds. Once you close the box, neither <span class="math-container">$\neg q$</span> nor <span class="math-container">$\neg p$</span> is directly accessible: what you are allowed to conclude <em>from the box as a whole</em> is <span class="math-container">$\neg q \to \neg p$</span>. Because you were able to derive <span class="math-container">$\neg p$</span> from the assumption <span class="math-container">$\neg q$</span>, the implication <span class="math-container">$\neg q \to \neg p$</span> holds.</p>
<blockquote>
<p>I do not understand the French and European example connection to what is written in this text.</p>
</blockquote>
<p>Suppose you start a proof box with the assumption "Let <span class="math-container">$x$</span> be French." Then the statement "<span class="math-container">$x$</span> is French" holds inside the proof box. By doing some reasoning, you are able to conclude "<span class="math-container">$x$</span> is European"; this statement still goes inside the proof box. Then you can draw <em>as a conclusion of the whole proof box</em> that "If <span class="math-container">$x$</span> is French, then <span class="math-container">$x$</span> is European." However, this does not say that <span class="math-container">$x$</span> -- whatever it refers to -- is French! It just says that <em>if</em> <span class="math-container">$x$</span> is French, then <span class="math-container">$x$</span> is European. In particular, you do not "have access" to the "<span class="math-container">$x$</span> is French" that was true inside the proof box.</p>
|
4,247,888 | <p>I'm having a lot of trouble about an apparently simple task. I have the following trigonometric equation:</p>
<p><span class="math-container">$A\cos(\omega_1t+\phi_1)=B\cos(\omega_2t+\phi_2)$</span></p>
<p>which holds for every <span class="math-container">$t \in [0,+\infty)$</span>, where <span class="math-container">$\omega_1,\omega_2,\phi_1,\phi_2 \in \mathbb{R}$</span> are fixed, with <span class="math-container">$\omega_1 \neq 0$</span> and <span class="math-container">$\omega_2 \neq 0$</span>.</p>
<p>With <span class="math-container">$A$</span> and <span class="math-container">$B$</span> positive, I need to show that <span class="math-container">$A=B$</span>, but I'm really stuck. I tried to find a value of <span class="math-container">$t$</span> such that <span class="math-container">$\omega_1t+\phi_1=\pi$</span> and so on, but after a lot of calculation I can't conclude anything. Any help or hint would be really appreciated!</p>
| vonbrand | 43,946 | <p>You know that <span class="math-container">$\cos \theta = 0$</span> if and only if <span class="math-container">$\theta = (2 k + 1) \pi / 2$</span>, use that to find a relation between <span class="math-container">$\omega_1 t + \phi_1$</span> and <span class="math-container">$\omega_2 t + \phi_2$</span>. Then the value of <span class="math-container">$A \cos(\omega_1 t + \phi_1)$</span> and <span class="math-container">$B \cos(\omega_2 t + \phi_2)$</span> must agree when <span class="math-container">$\omega_1 t + \phi_1 = 0$</span>. You'll find two solutions, <span class="math-container">$A = B$</span> and <span class="math-container">$A = -B$</span>.</p>
<p>Alternatively, <span class="math-container">$\cos \theta$</span> has maxima exactly at <span class="math-container">$\theta = 2 k \pi$</span>, use that to match up the functions like above.</p>
|
1,222,064 | <p>Given is an ellipse with $x=a\cos(t),~~y=b\sin(t)$</p>
<p>I do this by using $S=|\int_c^d x(t)y'(t) dt|$, so calculating the area regarding the vertical axis.
Since $t$ runs from $0$ to $2\pi$ I figured I only had to calculate it from $c=\pi/2$ to $d=3\pi/2$ and then this times $2$. But when I integrate over those I get zero...</p>
<p>My steps:
\begin{align*}
0.5S & = |\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d a\cos(t)*b\cos(t) dt|\\
S & = 2|\int_c^d ab\cos^2(t) dt|\\
S & = 2ab|\int_c^d 2\cos(t)*(-\sin(t)) dt|\\
S & = 4ab|\int_c^d -\cos(t)\sin(t) dt|
\end{align*}</p>
<p>On those bounds $\cos(t)$ is zero, so how will this work?!</p>
<p>I notice that to integrate $\cos^2(x)$, most use $\cos^2x = 1/2\cos(2x) + 1/2 $ and thus find $1/4 sin(2x) + 1/2 x$. </p>
<p>This works for those boundaries (except for a factor 2 somewhere ???), but how do I know NOT to use $\cos(x)\sin(x)$ ?!</p>
| MvG | 35,416 | <h1>Projective transformations in general</h1>
<p>Projective transformation matrices work on <a href="http://en.wikipedia.org/wiki/Homogeneous_coordinates" rel="nofollow noreferrer">homogeneous coordinates</a>. So the transformation</p>
<p>$$\begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix}\cdot\begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = \lambda\begin{bmatrix} x' \\ y' \\ z' \\ 1 \end{bmatrix}$$</p>
<p>actually means</p>
<p>$$\begin{bmatrix}x\\y\\z\end{bmatrix}\mapsto\frac1{mx+ny+oz+p}\begin{bmatrix} ax+by+cz+d \\ ex+fy+gz+h \\ ix+jy+kz+l \end{bmatrix}$$</p>
<p>so compared to linear transformations you gain the power to express translation and division.</p>
<p>Just like with linear transformations you get the inverse operation by computing the inverse matrix. Unlike affine transformations, a multiple of the matrix will describe the same operation, so you can in fact compute the <a href="http://en.wikipedia.org/wiki/Adjugate_matrix#3_.C3.97_3_generic_matrix" rel="nofollow noreferrer">adjunct</a> to describe the inverse transformation. Then you plug in the points $(±1, ±1, ±1, 1)$ and apply that inverse transformation in order to obtain the corners of your frustrum in view coordinates.</p>
<p><a href="https://math.stackexchange.com/a/339033/35416">This post</a> describes how you can find a projective transformation if you know the images of five points in space, no four of which lie on a common plane.</p>
<h1>Your perspective transformations</h1>
<p>There are two ways to understand your perspective transformations. One is by computing the coordinates of the corners of the frustrum, the other by investigating the structure of the matrix. I'll leave the former as an excercise, and follow the latter.</p>
<p>The first example you quote has its geometrix effect already denoted in the upper left corner. You can think about this as two distinct steps. In the first step, the vector $(x,y,z)$ is divided by $z$. This means you're projecting your line of sight onto the plane $z=1$. Afterwards, everything is scaled by $D$ in the $x,y,z$ directions. So you have a plane of fixed depth somewhere in your image space. Which is of little use in case you want to do something like depth comparisons.</p>
<p>The second example appears to be the standard OpenGL perspective projection, as implemented by <a href="https://www.opengl.org/sdk/docs/man2/xhtml/gluPerspective.xml" rel="nofollow noreferrer"><code>gluPerspective</code></a>, except for some sign changes. This doesn't map everything to the same depth, but instead maps the range between <code>NearZ</code> and <code>FarZ</code> to the interval $[-1,1]$. The new $(x',y')$ coordinates, on the other hand, are essentially the old $(x,y)$ coordinates, divided by $z$ to effect a projection onto the image plane, followed by a scaling which depends on an angle $\alpha$ denoting the field of view. The scaling also takes a factor <code>ar</code> into account, which denotes the aspect ratio.</p>
|
2,776,089 | <p>Let $T = \mathbb{S}^1 \times \mathbb{S}^1$ be a torus and $x \in T$. Prove or disprove: There exists a continuous
surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H_1(T,x) \rightarrow
H_1(T,x)$is the zero-map. </p>
<p>I have no idea how to solve this kind of problems. All I know is that since the fundamental group of the torus is abelian we may think of $f^*$ as a map of fundamental groups instead. We can also say by the lifting lemma that our maps lifts to its universal cover $\mathbb{R}^2$ which is contractible, but I don't know if this is relevant. </p>
<p>Any help will be much appreciated</p>
| Jason DeVito | 331 | <p>Here is an outline on constructing such a map.</p>
<p>I'm thinking of $S^1 = \{z\in \mathbb{C}: |z|=1\}$. Then define $g:S^1\rightarrow S^1$ by $g(z) = \begin{cases} z^2 & \operatorname{Im}(z)\geq 0\\ \overline{z}^2 & \operatorname{Im}(z)\leq 0\end{cases}$.</p>
<p>Note that if $\operatorname{Im}(z) = 0$, then $z = \pm 1$, so $z^2 = \overline{z}^2 =1$ in this case. Thus, $g$ really is continuous.</p>
<p>Intuitively, $g$ wraps the top half of $S^1$ fully around $S^1$ one way, and wraps the bottom half of $S^1$ fully around $S^1$ the other way.</p>
<p>Can you prove $g$ is surjective? Can you prove that $g$ is the $0$ map on $H_1$? Can you use $g$ to construct $f:T\rightarrow T$ with the properties you're looking for?</p>
|
2,776,089 | <p>Let $T = \mathbb{S}^1 \times \mathbb{S}^1$ be a torus and $x \in T$. Prove or disprove: There exists a continuous
surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H_1(T,x) \rightarrow
H_1(T,x)$is the zero-map. </p>
<p>I have no idea how to solve this kind of problems. All I know is that since the fundamental group of the torus is abelian we may think of $f^*$ as a map of fundamental groups instead. We can also say by the lifting lemma that our maps lifts to its universal cover $\mathbb{R}^2$ which is contractible, but I don't know if this is relevant. </p>
<p>Any help will be much appreciated</p>
| Igor Sikora | 464,503 | <p>You can try also the following solution: Take any surjective continuous map from $T$ to $I$ - unit interval, for example a height function. Then using Peano curve you can find surjective continuous map from $I$ to $I^2$, and then the quotient map from $I^2$ to $T$. This map will be continuous and surjective, and induces $0$ in $H_1(X)$, since it factors through contractible space. </p>
<p>Actually, this map is null-homotopic.</p>
<p>To be totally fair, my friend gave me this solution, so I am not claiming I had this idea myself. Thanks Andrzej :)</p>
|
2,414,965 |
<p>I am following along and reading this notes: <a href="https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf" rel="nofollow noreferrer">https://www.maths.tcd.ie/~levene/221/pdf/cantor.pdf</a></p>
<p>I am having trouble understanding why we necessarily have $e_n=d_n+1$,
$d_{n+1}= d_{n+2} =···= 2$
and $e_{n+1} = e_{n+2} = ··· = 0$
when $d_n > e_n$. It would much appreciated if someone can guide me through this. </p>
| kishlaya | 369,027 | <p>You'll have to play around with the inequalities a little bit to establish that. </p>
<p>To make life easy, assume without loss of generality $d_1 \neq e_1$. Next, notice,</p>
<p>$$x = \sum_{n \geq 1} \frac{e_n}{3^n} = \frac{e_1}{3} + \sum_{n \geq 2} \frac{e_n}{3^n} \geq \frac{e_1}{3}$$</p>
<p>And also, </p>
<p>$$x = \sum_{n \geq 1} \frac{d_n}{3^n} = \frac{d_1}{3} + \sum_{n \geq 2} \frac{d_n}{3^n} \leq \frac{d_1}{3} + \sum_{n \geq 2} \frac{2}{3^n} = \frac{d_1+1}{3}$$</p>
<p>So, we have, $d_1 < e_1 \leq d_1 + 1 \implies e_1 = d_1+1$</p>
|
172,131 | <p>Given <span class="math-container">$P$</span>, a polynomial of degree <span class="math-container">$n$</span>, such that <span class="math-container">$P(x) = r^x$</span> for <span class="math-container">$x = 0,1, \ldots, n$</span> and some real number <span class="math-container">$r$</span>, I need to calculate <span class="math-container">$P(n+1)$</span>.</p>
<p>Can this be done without Lagrange interpolation?</p>
| Théophile | 26,091 | <p>$P(n+1) = r^{n+1}-(r-1)^{n+1}$.</p>
<p>Construct the successive differences between terms, thus:</p>
<pre><code>1 r r^2 r^3 ... r^n
r-1 r(r-1) r^2(r-1) ... r^(n-1)*(r-1)
(r-1)^2 r(r-1)^2 ...
...
(r-1)^n
</code></pre>
<p>These are all known. Consider the top line to be the $0$th line, so that the last one is the $n$th line. Now interpolate the polynomial: the next term on the last line will also be $(r-1)^n$. You can then work your way back up to the top, and it is straightforward to show that the last term on the $k$th line will be $(r-1)^k\big((r-1)^{n+1-k}-r^{n+1-k}\big)$.</p>
<p>Example with $r=10$, $n=3$ (the interpolation is on the right):</p>
<pre><code> 1 10 100 1000 | 3439 = 9^0*(10^4 - 9^4)
9 90 900 | 2439 = 9^1*(10^3 - 9^3)
81 810 | 1539 = 9^2*(10^2 - 9^2)
729 | 729 = 9^3*(10^1 - 9^1)
</code></pre>
|
172,131 | <p>Given <span class="math-container">$P$</span>, a polynomial of degree <span class="math-container">$n$</span>, such that <span class="math-container">$P(x) = r^x$</span> for <span class="math-container">$x = 0,1, \ldots, n$</span> and some real number <span class="math-container">$r$</span>, I need to calculate <span class="math-container">$P(n+1)$</span>.</p>
<p>Can this be done without Lagrange interpolation?</p>
| David E Speyer | 448 | <p>By the binomial theorem,
$$r^x = \sum_{k=0}^{\infty} \binom{x}{k} (r-1)^k$$
for any $x$. Now, if $x$ is a integer from $0$ to $n$, then $\binom{x}{k}=0$ for $k>n$. So
$$r^x = \sum_{k=0}^n \binom{x}{k} (r-1)^k \quad \mbox{for} \ x \in \{ 0,1,2,\ldots, n \}.$$</p>
<p>Notice that the right hand side is a degree $n$ polynomial in $x$. So
$$P(x) = \sum_{k=0}^n \binom{x}{k} (r-1)^k \ \textrm{and}$$
$$P(n+1) = \sum_{k=0}^n \binom{n+1}{k} (r-1)^k.$$
Using the binomial theorem one more time
$$P(n+1) = r^{n+1}-(r-1)^{n+1}$$
as in Theophile's answer.</p>
|
3,225,784 | <p>Solve for x:</p>
<blockquote>
<p><span class="math-container">$$2\sin(x) + 3\sin(2x) = 0 $$</span></p>
<p><span class="math-container">$$2\sin(x)(1 + 3\cos(x)) = 0$$</span></p>
</blockquote>
<p>Stuck here. The solution mentions some arccos function, but I need a detailed explanation on this one.</p>
| Peter Foreman | 631,494 | <p>If you have that
<span class="math-container">$$2\sin{(x)}(1+3\cos{(x)})=0$$</span>
then one or both of the factors must be equal to zero, hence either
<span class="math-container">$$2\sin{(x)}=0$$</span>
<span class="math-container">$$\sin{(x)}=0$$</span>
<span class="math-container">$$x=\pi k $$</span>
or
<span class="math-container">$$1+3\cos{(x)}=0$$</span>
<span class="math-container">$$\cos{(x)}=-\frac13$$</span>
<span class="math-container">$$x=\pm\arccos{\left(-\frac13\right)}+2\pi k$$</span>
where <span class="math-container">$k$</span> is an arbitrary integer. The function <span class="math-container">$\arccos{(x)}$</span> gives the value of <span class="math-container">$y\in\left[0,\pi\right]$</span> such that <span class="math-container">$\cos{(y)}=x$</span>.</p>
|
573,964 | <blockquote>
<p>Let set $S$ be the set of all functions $f:\mathbb{Z_+} \rightarrow \mathbb{Z_+}$. Define a realtion $R$ on $S$ by $(f,g)\in R$ iff there is a constant $M$ such that $\forall n (\frac{1}{M} < \frac{f(n)}{g(n)}<M). $ Prove that $R$ is an equivalence relation and that there are infinitely mane equivalence classes.</p>
</blockquote>
<p><strong>Attempt</strong>: Would it work if I define $f$ as $f_k(n)=kn$ and g as $g_k(n)=(M-k)n$ where $M>k$. Then, $$\forall n ((\frac{1}{M} < \frac{f(n)}{g(n)}<M)=(\frac{1}{M} < \frac{k}{M-k} <M))$$ is true as long as $M>k$. </p>
<p>So, $R$ is <strong>reflexive</strong>: $(f,f) \in R$
$$\forall n ((\frac{1}{M} < \frac{f(n)}{f(n)}<M)=(\frac{1}{M} < 1 <M)), \space M>1$$</p>
<p>$R$ is <strong>symmetric</strong>: $(f,g)\in R \Rightarrow (g,f)\in R$
$$\forall n ((\frac{1}{M} < \frac{f(n)}{g(n)}<M)=(\frac{1}{M} < \frac{k}{M-k} <M))$$
$$\forall n ((\frac{1}{M} < \frac{g(n)}{f(n)}<M)=(\frac{1}{M} < \frac{M-k}{k} <M))$$</p>
<p>for $M>k$.</p>
<p>$R$ is <strong>transitive</strong>: $(f,g)\in R \wedge (g,h) \in R \Rightarrow (f,h)\in R$
$$\frac{f(n)}{g(n)} \in R, \space \frac{g(n)}{h(n)} \in R \Rightarrow \frac{f(n)g(n)}{h(n)g(n)}=\frac{f(g)}{h(n)}\Rightarrow \forall n ((\frac{1}{M} < \frac{f(n)}{h(n)}<M) $$</p>
<p>Then, $R$ is an equivalence relation.</p>
| user107952 | 107,952 | <p>It is not a subspace of <span class="math-container">$\mathbb{ R}^2$</span> because <span class="math-container">$kx$</span> has to be in it for every real <span class="math-container">$k$</span>, not just integer <span class="math-container">$k$</span>.</p>
|
783,502 | <p>Here in my exercise I have to study the function and draw its graph. Can you please tell me what's the best method to do this, because I don't think that's reasonable to use the input output method, it's quite imprecise.
$$f(x)={|x+1|\over x}$$</p>
<p>Thank you!!!</p>
| evil999man | 102,285 | <p>First you have to break the mod into cases when $x>-1$ and...</p>
<p>Let me talk of case : $\frac{x+1}{x}=1+\frac 1 x$</p>
<p>$1/x $ is odd function. It tends to infinity at $0^+$ and tends to $0$ at infinity.</p>
<p>Make graph of $\frac 1 x $ and shift it one unit upward. Erase all part left of $x=-1$. Can you do the same for other part?</p>
<p><strong>PS:</strong> : The best method is to plot using one of many graphical calculators available and then copy them in you notebook of course, <strong>by hand</strong>.</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| Caleb Stanford | 68,107 | <p>You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.</p>
<p>You should not believe everything you read in a photo on the internet.</p>
|
251,705 | <p>I would like to find the residue of $$f(z)=\frac{e^{iz}}{z\,(z^2+1)^2}$$ at $z=i$. One way to do it is simply to take the derivative of $\frac{e^{iz}}{z\,(z^2+1)^2}$. Another is to find the Laurent expansion of the function.</p>
<p>I managed to do it using the first way, and the answer is $-3/(4e)$. However, I'm out of ideas as to how to find the expansion.</p>
<p>Any help is greatly appreciated.</p>
| Community | -1 | <p>$$e^{iz} = e^{i(z-i) + i^2} = \dfrac{e^{i(z-i)}}{e} = \dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}$$
$$\dfrac1z = \dfrac1{z-i+i} = \dfrac1i \dfrac1{1 + \dfrac{z-i}i} = \dfrac1i \sum_{k=0}^{\infty} (-1)^k\left(\dfrac{z-i}i \right)^k = \sum_{k=0}^{\infty} i^{k-1} (z-i)^k$$
$$\dfrac1{(z+i)^2} = \dfrac1{(z-i+2i)^2} = -\dfrac14 \dfrac1{\left(1 + \dfrac{z-i}{2i} \right)^2} = -\dfrac14 \sum_{k=0}^{\infty} (-1)^k (k+1) \left(\dfrac{z-i}{2i}\right)^k\\ = -\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k$$
Hence, we have
$$\dfrac1{(z-i)^2} \left(\dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}\right) \left(\sum_{k=0}^{\infty} i^{k-1} (z-i)^k \right) \left(-\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k\right)$$
Hence, the coefficient of $\dfrac1{z-i}$ is nothing but the coefficient of $(z-i)$ in $$\left(\dfrac1e \sum_{k=0}^{\infty} \dfrac{i^k(z-i)^k}{k!}\right) \left(\sum_{k=0}^{\infty} i^{k-1} (z-i)^k \right) \left(-\sum_{k=0}^{\infty} \dfrac{i^k (k+1)}{2^{k+2}} \left(z-i\right)^k\right)$$
Hence, the answer is
$$- \dfrac1e \times \left(\dfrac{i}{1!} \times \dfrac1i \times \dfrac14 + 1 \times 1 \times \dfrac14 + 1 \times \dfrac1i \times \dfrac{2i}{8} \right) = -\dfrac3{4e}$$</p>
|
2,741,686 | <p>If I have the following vector space $ V, \text{{$e_0, e_1, e_2$}} \text{ where } e_0(x) = 1, e_1(x) = x \text{ and } e_2(x) = x^2$.I want to know the linear dependency of it how can I proceed? I thought of following the definition of linearly independent $$c_0e_0 + c_1e_1 + c_2e_2 = c_0+ c_1x + c_2x^2=0\iff c_0 = c_1 = c_2 = 0$$
but I can not mount a system because of the $x^2$</p>
<p>I know that $c_0 = c_1 = c_2 = 0$ is solution, but i want to know if there is another solution for the equation $c_0+ c_1x + c_2x^2=0$ with $c_0 \neq 0, c_1 \neq 0 \text{ and }c_2 \neq 0 $ </p>
| Theo Bendit | 248,286 | <p>One way to show this is repeated differentiation. If
$$c_0 + c_1 x + c_2 x^2 \equiv 0,$$
then
\begin{align*}
c_1 + 2c_2 x &\equiv 0, \\
2c_2 &\equiv 0.
\end{align*}
From evaluating all these polynomials at $x = 0$, we obtain $c_0 = c_1 = c_2 = 0$.</p>
|
1,112,081 | <p>Does $\int_0^\infty e^{-x}\sqrt{x}dx$ converge? Thanks in advance.</p>
| Angelo | 208,573 | <p>yes, $\sqrt{\pi}/2$ should be your answer. let $\sqrt{x}=t$ make the substitution, then integrate by parts. you'll get an integral involving $\int_{0}^\infty e^{-t^2}\,dt$ which is equal to the answer stated above.</p>
<p>you should integrate the resulting integral, i.e., $2\int_{0}^\infty t^2\,e^{-t^2}\,dt$ by parts by breaking it up as $t$ $\cdot$ $2te^{-t^2}$ (I didn't want to give away all the tricks). </p>
|
1,921,302 | <p>I can't believe I am asking such a silly question. So I have the function
$$\ln\tan^{-1}x$$
I am asked to find the range of this function. I know that the range of $\ln x$ is all real numbers and that the range of $\tan^{-1}(x)$ is $(-\frac\pi2$, $\frac\pi2)$. Wouldn't the range of $\ln\tan^{-1}x$ also be $(-\frac\pi2$, $\frac\pi2)$, or am I doing something really stupid?</p>
| Hagen von Eitzen | 39,174 | <p>We are given that $\beta I\subseteq I$.
If $I$ is principal, say $I=(\iota)$ with $\iota\ne 0$, then $\beta \iota= r\iota $ for some $r\in R$ and hence $\beta=r\in R$.</p>
<p>For any other non-zero ideal $J$, we can replace $I$ with $IJ$: If $\beta I \subseteq I$ then also $\beta IJ\subseteq IJ$.
Using Theorem 8.19 we can pick $J$ so that $IJ$ is a non-zero principal ideal and are done.</p>
|
992,068 | <p>I am having a little trouble understanding this question.</p>
<p>For a DFA M = (Q, Σ, δ, q0, F), we say that a state q ∈ Q is reachable if there
exists some string w ∈ Σ∗ such that q = δ∗(q0, w).</p>
<p>Give an algorithm that, given as input a DFA expressed as a five-tuple M =
(Q, Σ, δ, q0, F), returns the set of all of M’s reachable states. </p>
<p>Can anyone help me understand and answer this question?</p>
| Compiii | 633,845 | <p>I hope that this algorithm will be useful for you:</p>
<pre><code>algorithm: determine the set if reachable states of an DFA;
intput: an DFA M = (Q, Σ, δ, q0, F);
output: an list of reachable state;
begin
create a list result and insert q0;
create a stack s and push q0;
while s is not empty
do
q = pop(s);
if q != q0 then
insert q in result;
for each symbol a of Σ
do
q' = δ(q, a);
if q' does not belongs to result then
push(s,q');
done
done
return result;
end.
</code></pre>
|
7,237 | <p>this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?</p>
| Gerhard Paseman | 3,468 | <p>I think it is important to emphasize Mark Fantini's remark: factoring is different from dividing, and factoring is the way to get a complete solution. I would also suggest that the
problem be rearranged and then factored to give 6x(x-2)=0, or even 2*3*x*(x-2)=0. Then
one can see that one of 2 , 3, x, or x-2 must be zero.</p>
<p>When one is comfortable with using the factoring method, then one can consider "shortcuts" by jumping from the given equation to saying "Suppose 6x is not 0: then divide..."</p>
<p>Gerhard "Shortcuts Can Cut One Short" Paseman, 2015.01.16</p>
|
32,137 | <p>I have an equation that I evaluate at some point (let's say $x=1$) that have terms of the form</p>
<pre><code>f[1,y] D[g[1,y],{y,2}]
</code></pre>
<p>Is there an easy way to replace [1,y] by [x,y] with a simple replacement rule? The thing is that </p>
<pre><code>g[1,y] /. g[1,y] -> g[x,y]
</code></pre>
<p>will indeed replace g, but it won't work for any of its derivatives, and there are far too many functions and derivatives for me to make a replacement rule for all of them. </p>
<p>Would there be a clever way to do this? So far, my only solution has been copy-pasting my equations in Word and using the text replacement tool in there. It would be neat not having to go outside of Mathematica to do this though.</p>
| Szabolcs | 12 | <p>Taking the question as "how to replace the first argument in $g(a,b)$ as well as all of its derivatives", you can do this:</p>
<p>Check the InputForm of the derivative:</p>
<pre><code>D[g[1, y], {y, 2}] // InputForm
(* ==> Derivative[0, 2][g][1, y] *)
</code></pre>
<p>Add a corresponding pattern to the replacement rules:</p>
<pre><code>f[1, y] D[g[1, y], {y, 2}] /.
{g[1, y] -> g[x, y], Derivative[d___][g][1, y] :> Derivative[d][g][x, y]}
</code></pre>
|
1,323,845 | <p>For a nonnegative integer $n$, a composition of $n$ means a partition in which the order of the parts matters.</p>
<p>Consider the generating function
$$C(x) = \sum_{n=0}^{\infty} c_nx^n,$$
where $c_n$ is the number of distinct compositions of $n$ (note that $c_0=1$ by convention).</p>
<p>What is the value of $C\left(\tfrac 15\right)$?</p>
<hr>
<p>How can I start this?</p>
| Lee Mosher | 26,501 | <p>Let $X$ be a path connected space with more than one point, and let $B \subset X$ be an open ball. If $B$ is just a single point then by an easy argument $B=X$, a contradiction.</p>
<p>So there are two points $x \ne y \in B$, and using them I'll prove that $B$ has contains a subset of the cardinality of the reals.</p>
<p>Applying path connectedness of $X$, let $\gamma : [0,1] \to X$ be a continuous path with $\gamma(0)=x$ and $\gamma(1)=y$. By continuity of $\gamma$ there exists $T \in (0,1]$ such that $\gamma[0,T] \subset B$. It cannot happen that $\gamma(T)=x$ for all such $T$, because if so then taking $T_0$ to be the supremum of such $T$ we would have $\gamma(T_0)=x \ne y$, and so $T_0 \ne 1$, but then by taking small enough $\delta>0$ we would have $\gamma[0,T_0+\delta] \subset B$ contradicting that $T_0$ is the supremum.</p>
<p>So we may therefore take $T \in (0,1]$ so that $\gamma[0,T] \subset B$ and $\gamma(T) \ne x$. Now consider the function
$$f : [0,T] \to \mathbb{R}, \quad f(t) = d(x,\gamma(t))
$$
This is a continuous function. By the intermediate value theorem its image $f[0,T]$ contains the entire nontrivial closed interval $[0,d(x,f(T))]$. So for each $D \in [0,d(x,f(T))]$ there exists a point $z \in B$ such that $d(x,z)=D$. It follows that the image $\gamma[0,T] \subset B$ contains a subset of cardinality equal to the reals.</p>
|
4,128,046 | <p>I am working through a pure maths book as a hobby. This question puzzles me.</p>
<p>The line y=mx intersects the curve <span class="math-container">$y=x^2-1$</span> at the points A and B. Find the equation of the locus of the mid point of AB as m varies.</p>
<p>I have said at intersection:</p>
<p><span class="math-container">$mx = x^2-1 \implies x^2 - mx - 1=0$</span></p>
<p>Completing the square:</p>
<p><span class="math-container">$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$</span></p>
<p><span class="math-container">$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$</span></p>
<p><span class="math-container">$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$</span></p>
<p>x-coordinates for points of intersection are</p>
<p><span class="math-container">$x= \frac{-\sqrt(4+m^2) +m}{2}$</span> and <span class="math-container">$x= \frac{\sqrt(4+m^2)+m}{2}$</span></p>
<p><span class="math-container">$\implies$</span> x-co-ordinate of P is mid-way between the two above points, namely <span class="math-container">$\frac{2m}{2} = m$</span></p>
<p>So <span class="math-container">$x=m, y = mx = m^2\implies y = x^2$</span></p>
<p>But my book says <span class="math-container">$y=2x^2$</span></p>
| Math Lover | 801,574 | <p>If <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> are x-coordinates of intersection points, x-coordinate of midpoint,</p>
<p><span class="math-container">$x_m = \frac{x_1 + x_2}{2} = \frac{m}{2}$</span></p>
<p><span class="math-container">$y = mx \implies y_m = 2 x_m^2$</span> so locus of midpoint is <span class="math-container">$y = 2 x^2$</span>.</p>
<p>Your approach is correct but note that you can also get to it quickly using Vieta's formula for quadratic equation.</p>
<p>If <span class="math-container">$ax^2+bx+c = 0, x_1 + x_2 = - \frac{b}{a}, \ x_1 \ x_2 = \frac{c}{a}$</span></p>
<p>(where <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> are roots of the quadratic)</p>
<p>Here we have, <span class="math-container">$x^2 - mx - 1=0 \implies x_1 + x_2 = m \ $</span> so <span class="math-container">$x_m = \frac{m}{2}$</span>.</p>
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2,638,679 | <p><a href="https://i.stack.imgur.com/S4p0Y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S4p0Y.jpg" alt="enter image description here"></a></p>
<p>Due apologies for this rustic image. But while drawing this lattice arrangement about the "square numbers" , I discovered a pattern here wherein if I add the alternate red dots (as depicted in the image above) to the square number, I get the next square number. For instance, $4 + 5(red\ dot) = 9$ , $9+7(red\ dot)=16$, $16+9(red\ dot)=25$, $25+11(red\ dot)=36$, $36+13 (red\ dot)=49$.</p>
<p>The red dotted numbers themselves have a pattern as is obvious from the image. Is there any mathematical explanation to this pattern.</p>
| Michael Hardy | 11,667 | <p>\begin{align}
\lambda_1 + \lambda_2 = 3 \\
\lambda_1\lambda_2 = 3
\end{align}
So you get $\lambda_2 = \dfrac 3 {\lambda_1},$ so the first equation above becomes
$$
\lambda_1 + \frac 3 \lambda_1 = 3.
$$
Multiply both sides by $\lambda_1$ and you have an ordinary quadratic equation.</p>
<p>The answer posted by "N.S." also leads to the same quadratic equation.</p>
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