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 |
|---|---|---|---|---|
1,534,246 | <p>I'm trying to simplify this boolean expression:</p>
<p>$$(AB)+(A'C)+(BC)$$</p>
<p>I'm told by every calculator online that this would be logically equivalent:</p>
<p>$(AB)+(A'C)$</p>
<p>But so far, following the rules of boolean algebra, the best that I could get to was this: </p>
<p>$(B+A')(B+C)(A+C)$</p>
<p>All of the above are logically equivalent (I've made a truth table for each) but I don't understand what steps am I missing trying to simplify the expression.</p>
<p>I also couldn't find an "expression simplifier" tool online that could show me the steps that I'm missing.</p>
<p>Help / directions to go to would be much appreciated, thanks in advance.</p>
| Dhruv Joshi | 522,283 | <p>Though this has already some good answer, I know an interesting way to look at boolean expression with less than 4 elements(for greater than 3 it becomes <a href="https://math.stackexchange.com/questions/1475/why-can-a-venn-diagram-for-4-sets-not-be-constructed-using-circles">messy</a> ),it might be helpful for any future user struggling with a similar problem.</p>
<p>Imagine them as Venn diagrams, draw their intersection using the given expression, after this most of the time question becomes trivial.</p>
<p>You can easily see/confirm A'C +AB covers the colored region completely.</p>
<p>(ignore my drawing skills)
<a href="https://i.stack.imgur.com/obpl2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/obpl2.png" alt="ignore my drawing skills" /></a></p>
|
2,306,319 | <p>I have point $P_2(x_2, y_2, z_2)$, which is situated in euclidean coordinate system $xyz$. I want to transform this point to similar coordinate system $x'y'z'$ arbitrarily rotated (three directions are some known vectors) and centered at point $P_1(x_1, y_1, z_1)$ - center relative to system $xyz$. So new coordinates of point $P_2$ would be relative to $(0, 0, 0)$ at center of $x'y'z'$. How can I (easily) do that?
<a href="https://i.stack.imgur.com/uE2Kv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uE2Kv.png" alt="enter image description here"></a></p>
| JMJ | 295,405 | <p>If $x,y,z$ are the vectors in the first coordinate system and $x',y',z'$ the vectors in the second the rotation matrix is given by
$$
R =
\begin{bmatrix}
x'\cdot x & x'\cdot y & x'\cdot z\\
y'\cdot x & y'\cdot y & y'\cdot z\\
z'\cdot x & z'\cdot y & z'\cdot z\\
\end{bmatrix}
$$</p>
<p>If the coordinate systems have different origins you also need to add in the vector between the two origins. </p>
|
2,306,319 | <p>I have point $P_2(x_2, y_2, z_2)$, which is situated in euclidean coordinate system $xyz$. I want to transform this point to similar coordinate system $x'y'z'$ arbitrarily rotated (three directions are some known vectors) and centered at point $P_1(x_1, y_1, z_1)$ - center relative to system $xyz$. So new coordinates of point $P_2$ would be relative to $(0, 0, 0)$ at center of $x'y'z'$. How can I (easily) do that?
<a href="https://i.stack.imgur.com/uE2Kv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uE2Kv.png" alt="enter image description here"></a></p>
| amd | 265,466 | <p>This coordinate transformation consists of a translation followed by a rotation: $(x',y',z')=R(P_2-P_1)$. To construct the rotation matrix $R$, recall that the columns of a transformation matrix are images of the basis vectors. Let $\mathbf u_{x'}$, $\mathbf u_{y'}$ and $\mathbf u_{z'}$ be unit vectors in the original coordinate system that define the new coordinate axis directions. Then the rotation matrix which maps from the <em>rotated</em> coordinates to the original ones, $R^{-1}$ has these vectors as its columns. Since the inverse of a rotation matrix is its transpose, $R$ is the matrix that has these unit vectors for <em>rows</em>.</p>
<p>The transformation $R(P_2-P_1)$ therefore expands to $$\begin{align}x'&=(P_2-P_1)\cdot\mathbf u_{x'} \\ y'&=(P_2-P_1)\cdot\mathbf u_{y'} \\ z'&=(P_2-P_1)\cdot\mathbf u_{z'}\end{align}$$ which says that the new coordinates are the projections of $P_2$ onto the new coordinate axes. This is one of the definitions of coordinates, though, so we could’ve gotten this last set of equations directly from the definition. </p>
<p>Depending on your application, it might be more convenient to package this up into the homogeneous transformation matrix $$\begin{bmatrix}R&-RP_1\\\mathbf 0&1\end{bmatrix}$$ which includes the translation as part of the matrix multiplication.</p>
|
3,298,282 | <p>Find the number of terms in the expansion <span class="math-container">$(1+a^3+a^{-3})^{100}$</span></p>
<p>I used the concept <span class="math-container">$a^3+a^{-3}=T$</span>, while using this I have 101 terms, from <span class="math-container">$T^2$</span> to <span class="math-container">$T^{100}$</span> how do i find the number of terms that do not intersect</p>
| lab bhattacharjee | 33,337 | <p>Write <span class="math-container">$a^3=b$</span></p>
<p>Multinomial theorem says</p>
<p>The general term of <span class="math-container">$$(1+b+b^{-1})^{100}$$</span> is <span class="math-container">$$\dfrac{100!}{p!q!r!}b^{p-q}$$</span> with</p>
<p><span class="math-container">$p+q+r=100$</span> and <span class="math-container">$p,q,r\ge0$</span></p>
<p><span class="math-container">$\implies0\le p,q\le100$</span></p>
<p>So, <span class="math-container">$-100\le p-q\le100$</span></p>
<p>If <span class="math-container">$d=p-q,100=d+2q+r\iff d=100-2q-r$</span></p>
<p>If <span class="math-container">$r=0,p+q=100$</span></p>
<p><span class="math-container">$d=100-2q,d$</span> can attain all the even integer values in <span class="math-container">$[-100,100]$</span></p>
<p>If <span class="math-container">$r=1,p+q=99$</span></p>
<p><span class="math-container">$d=100-(2q+1)=99-2q$</span> can attain all the odd values in <span class="math-container">$[-99,+99]$</span></p>
<p>Clearly <span class="math-container">$p-q$</span> can attain all the possible <span class="math-container">$100-(-100)+1$</span> values</p>
|
1,930,933 | <blockquote>
<p>Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?</p>
</blockquote>
<p>The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?</p>
| ctst | 363,720 | <p>This is (for $n>2$) never an integer, since you always have a single prime number (meaning with exponent 1) and hence its $n$th root, namely the biggest prime smaller than $n$. Since the product of roots of different prime numbers are never integer, this number will never be an integer (for $n>2$).</p>
|
72,651 | <p>$G$ is a group and $H$ is a subgroup of $G$ such that $\forall a, b$ in $G, ab\in H\implies ba\in H$. Show that $H$ is normal in $G$</p>
| Bill Cook | 16,423 | <p>Your goal is to show $xH=Hx$ for all $x\in G$ so you should show set containment both ways.</p>
<p>Suppose $y \in xH$. This means that there exists some $h \in H$ such that $y=xh$. Notice that $x^{-1}y=h \in H$. Now invoke your hypothesis. I'll let you finish from there.</p>
<p>Do be careful though. If $ab=h_1 \in H$, then you know $ba \in H$. This does not mean $ba=h_1$ but just that $ba$ is some element of $H$ (it could be a different element).</p>
|
3,831,702 | <p>One can prove that for <span class="math-container">$x\in \mathbb{R}$</span>, the sequence
<span class="math-container">$$
u_0=x\text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
converges to <span class="math-container">$0$</span> if <span class="math-container">$x \in ]-\infty,\delta[$</span> and diverges to <span class="math-container">$+\infty$</span> if <span class="math-container">$x\in ]\delta,+\infty[$</span> for a fixed <span class="math-container">$\delta$</span>. I'm trying to find more information on the value <span class="math-container">$\delta$</span> (inequalities or expression) and on the specific sequence
<span class="math-container">$$
u_0=\delta \text{ and } \forall n\in \mathbb{N},\qquad u_{n+1}=\frac{e^{u_n}}{n+1}
$$</span>
Any reference or help are welcome. The only thing I can prove at the moment is <span class="math-container">$\ln \ln 2 \le \delta \le 1$</span>.</p>
| Simply Beautiful Art | 272,831 | <p>Numerical results:</p>
<p>It seems that <span class="math-container">$\delta\approx0.3132776395465557$</span>. This was computed using root-finding techniques to find when <span class="math-container">$u_n=y$</span> for <span class="math-container">$y=0.1,1,10$</span> and <span class="math-container">$n=10,100,1000$</span>. <a href="https://tio.run/##tZPNbqMwEIDvPMWIXpI0Yc22vVQlxz3BacXJQlXTmNQKMimYrCPEs2fHY2jdHFY5bCXGYob5@WY8NN3mdD5vZCtetawVJKs1lEswKUoOfQBQppBAyU1aWCV3Sm6VVu5cgIHewFOyBjag2brVDexBKoijKGYMLQAmw1AWPSxmJr01@ZyMZebyZQWpdMiSMvMyKyBJxvfUOWAai2My0kTVismce2a1DaYTn0borlEYGAxB0Ir3TqhXQdyKGnQNOByPW02ZMXH2ot8iYQ4zM/@x92p8JkdlsPmr@o9onjd1p7ZeCY05PmbMV2vTm5WfUw1LiJegbmNW@OPjCwSJVBE14iiasVk9AVX1bqYX@/lHsyOMtiDd4XAlSPy9KOdSqu3zVlT6hUBOBHLodAvh77o6SrWDjiu8a7jpTwNVxLWCmCEIcwcLbcgF@XSVHEm5KRBnBSektyHFv/3HgOv9XcCFP7UQ2mF/tshZhB89/ULDoQYu8Bdabe@0MUAX1QL@geQYBnYM0g3@zv5ES/iJcodyj/Lg5kIYjVQawpteDo/g5uQtIZdFYG@EiiLs19q0JP@3trd3Y@3zXw" rel="nofollow noreferrer">Code here</a>.</p>
<p>Proofs of bounds:</p>
<p>Disclaimer: Proof that the bounds are tight is not given, but is supported numerically.</p>
<p>Lower bounds can be proven by observing when <span class="math-container">$t=u_n=u_{n-1}$</span> occurs. If this occurs, then all future iterations are clearly decreasing. Solving for this gives</p>
<p><span class="math-container">$$t=\frac1ne^t\implies t=-W_{-1}\left(-\frac1n\right)$$</span></p>
<p>where <span class="math-container">$W_{-1}$</span> is the real branch of the Lambert W function which gives the largest value for <span class="math-container">$t$</span>. Working backwards from this point to find <span class="math-container">$x$</span> then gives</p>
<p><span class="math-container">$$x=\ln\left(1\cdot\ln\left(2\cdot\ln\left(\dots(n-1)\cdot\ln\left(-nW_{-1}\left(-\frac1n\right)\right)\dots\right)\right)\right)$$</span></p>
<p><span class="math-container">$\delta$</span> is then at least the supremum of this. Note that there are no solutions for <span class="math-container">$t<3$</span>. The first few values of this lower bound is given by</p>
<p><span class="math-container">$$\begin{array}{c|c}n&\delta\ge{}?\\\hline3&0.1013550034887759\\4&0.2751555022435671\\5&0.3044035425578071\\10&0.3132705224120361\\20&0.3132776395448800\\30&0.3132776395465558\end{array}$$</span></p>
<p>Note that <span class="math-container">$n=30$</span> corresponds to the estimated <span class="math-container">$\delta$</span> approximation. See also the code above.</p>
<p>Upper bounds can be shown in a similar manner by solving <span class="math-container">$u_n=u_{n-1}+1\ge3$</span>, which gives</p>
<p><span class="math-container">$$u_{n+1}=\frac1{n+1}e^{u_n}=\frac{en}{n+1}\cdot\frac1ne^{u_{n-1}}=\frac{en}{n+1}\cdot u_n\ge u_n+1$$</span></p>
<p>and is thus diverges to <span class="math-container">$\infty$</span>. (Again see code above).</p>
<p>It appears to be the case that these bounds are asymptotically tight. Proving so seems to be rather messy, however.</p>
|
84,312 | <p>In the topological sense, I understand that the unit circle $S^1$ is a retract of $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ where $\mathbb{0}$ is the origin. This is because a continuous map defined by $r(x)= x/|x|$ is a retraction of the punctured plane $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ onto the unit circle $S^1 \subset \mathbb{R}^2 \backslash \{\mathbb{0}\}$. Does this mean that $S^1$ is not a retract of $\mathbb{R}^2$? I would appreciate some clarification here.</p>
| Matthias Klupsch | 19,700 | <p>Indeed there is no retraction $r: \mathbb{R^2} \rightarrow S^1$
because if $\iota :S^1 \rightarrow \mathbb{R^2}$ is the inclusion
you would have a monomorphism $\iota^* : \pi(S^1) \rightarrow \pi(\mathbb{R^2})$ between the fundamental groups, i.e. a monomorphism $\mathbb{Z} \rightarrow \{0\}$ which would be absurd.</p>
|
84,312 | <p>In the topological sense, I understand that the unit circle $S^1$ is a retract of $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ where $\mathbb{0}$ is the origin. This is because a continuous map defined by $r(x)= x/|x|$ is a retraction of the punctured plane $\mathbb{R}^2 \backslash \{\mathbb{0}\}$ onto the unit circle $S^1 \subset \mathbb{R}^2 \backslash \{\mathbb{0}\}$. Does this mean that $S^1$ is not a retract of $\mathbb{R}^2$? I would appreciate some clarification here.</p>
| Thomas Andrews | 7,933 | <p>You can use the <a href="http://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem" rel="nofollow noreferrer">Brouwer Fixed Point Theorem</a> to show that $S^1$ is not a retraction of the unit disk, and hence not a retraction of the entire plane (since, if $X\subset Y\subset Z$, and $j:Z\rightarrow X$ is a retraction, then the restriction $j_{|Y}:Y\rightarrow X$ is a retraction.)</p>
<p>Assume there is a retraction $f:D^2\rightarrow S^1$. Define $g(x):D^2\rightarrow D^2$ as $g(x)=-f(x)$. Then if $x\in S^1$, $f(x)=x$, by the condition that $f$ is a retraction, so $g(x)=-x$, and hence $g(x)\neq x$. If $x\notin S^1$, then $g(x)\in S^1$, and so $g(x)\neq x$. So there are no fixed points for $g$, contradicting Brouwer.</p>
<p>Now, usually, Brouwer is proven the opposite way - using Algebraic Topology to show that there can be no retraction of $D^2\rightarrow S^1$, and then showing that if $g:D^2\rightarrow D^2$ has no fixed point, then you can get a retraction of $D^2$ to $S^1$.</p>
<p>However, Brouwer has other, non-Algebraic proofs. Even constructive proofs.</p>
|
1,946,637 | <p>I have come across this integral and have been unable to solve it so far.</p>
<p>$$I=\int_{-\infty}^{0.29881}e^{-0.5x^2}\,\mathrm dx$$</p>
| kobe | 190,421 | <p>If we express your series as $\sum\limits_{j = 0}^\infty a_j z^j$, then $$a_j = \begin{cases}\dfrac{2^{j/2}}{3^{j/2} + 4^{j/2}}& \text{if $j$ is even}\\0&\text{otherwise}\end{cases}.$$</p>
<p>Since $\lim\limits_{j\to \infty} \lvert a_{2j-1}\rvert^{1/(2j-1)} = 0$ and </p>
<p>$$\lim_{j\to \infty} \lvert a_{2j}\rvert^{1/{2j}} = \frac{2^{1/2}}{\lim\limits_{j\to \infty} (3^{j} + 4^{j})^{1/{2j}}} = \frac{2^{1/2}}{\max\{3,4\}^{1/2}} = \frac{1}{\sqrt{2}},$$</p>
<p>then $\limsup\limits_{j\to \infty}\, \lvert a_j\rvert^{1/j} = \dfrac{1}{\sqrt{2}}$. Hence, the radius of convergence is $\sqrt{2}$.</p>
|
866,885 | <p>I try to write math notes as clearly as possible. In practice, this means using letters and notation similar to what the reader is already familiar with.</p>
<p>A real function is often $f(x)$, an angle is often $\theta$, a matrix has size $m\times n$, and $i$ is often an index. The full theoretical list is long and complicated. For example, $\pi$ is very often a constant, but sometimes it's a variable for a permutation. Capital sigma $\Sigma$ can indicate summing a series, but it can also denote a matrix, as in the singular value decomposition. So things like context matter, and a great list would have to include more than just variable names. Another choice to make is how to write an inner product, for example.</p>
<p>Does such a list exist?</p>
| Fred Daniel Kline | 28,555 | <p>I just got: <a href="http://rads.stackoverflow.com/amzn/click/1466230525" rel="nofollow">Mathematical Notation</a> which is surprisingly comprehensive for such a thin book. It covers several mathmatical topics. It even has $\LaTeX$ examples. Edit: it even has conventions for notes and blackboards.</p>
|
2,584,405 | <p>During an excercise session in a basic course of probability it was shown that the secretary problem can be reduced to solving the following task: For a given natural $n$ optimize $2\leq k\leq n-1$ so that
$\frac{k-1}{n}\sum^n_{i=k}\frac 1{i-1} = \frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H}_{k-2})$ is the largest.</p>
<p>Solution that was presented to me:</p>
<p>$\operatorname{H}_l=\operatorname{ln}l+\lambda+\operatorname{o}(1)$</p>
<p>$\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\approx\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$</p>
<p>$f(k):=(k-1)\operatorname{ln} n-(k-1)\operatorname{ln}(k-1)$</p>
<p>$f^{\prime}(k)=\operatorname{ln}n-\operatorname{ln}(k-1)-1$</p>
<p>$f^{\prime}(k)=0\iff k=\frac{n}{\operatorname{e}}+1$</p>
<p>So the optimal $k$ is $\frac n {\operatorname{e}}+1$ and for this $k$ we have $\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))=\frac n {\operatorname{e}}$</p>
<p>It is likely because I've already forgotten quite a lot from my elementary courses on fields like mathematical analysis and discrete mathematics that I don't think I understand the above proof.</p>
<p>For me this proof leaves significant holes. First and foremost, the fact that</p>
<p>$\operatorname{H}_l=\operatorname{ln}l+\lambda+\operatorname{o}(1)$</p>
<p>does not yet prove that</p>
<p>$\frac{(k+1)-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{(k+1)-2}})>\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\iff \\\frac{(k+1)-1}n (\operatorname{ln}(n-1)-\operatorname{ln}((k+1)-2))>\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$</p>
<p>And this is a prerequisite of studying the monotonicity of $\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})$ by means of studying the monotonicity of $\frac{k-1}n (\operatorname{ln}(n-1)-\operatorname{ln}(k-2))$.</p>
<p>This $\operatorname{o}(1)$ tells us that any inaccuracies become insignificant <strong>for sufficiently large $k$</strong>; but we don't have "<em>sufficiently large $k$</em>" - on the contrary, we have $2\leq k\leq n-1$. So instead of saying anything about "<em>sufficiently large $k$</em>" we should prove that the inaccuracies are insignificant enough not to affect the result of our computations <strong>for $k$ as low as $3$!!</strong> and I don't know how to prove this.</p>
<p>We have:</p>
<p>$\frac{(k+1)-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{(k+1)-2}})>\frac{k-1}{n}(\operatorname{H}_{n-1}-\operatorname{H_{k-2}})\iff\operatorname{H}_{n-1}-\operatorname{H}_{k-2}>k(\operatorname{H}_{k-1}-\operatorname{H}_{k-2})\iff\\\operatorname{ln}(n-1)-\operatorname{ln}(k-2)+\operatorname{o}(1)>k(\operatorname{ln}(k-1)-\operatorname{ln}(k-2)+\operatorname{o}(1))$</p>
<p>And why can we simply discard these $\operatorname{o}(1)$s in the above expression?</p>
<p>What am I missing? What did I forget?</p>
| Dr. Wolfgang Hintze | 198,592 | <p>EDIT: graph and approximation improved</p>
<p>The following graph compares the exact optimal value of k with the approximations $k_{appr1} = n/e$ and $k_{appr2}=\left\lfloor (n+1)/e\right\rfloor$</p>
<p><a href="https://i.stack.imgur.com/5AYFe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5AYFe.jpg" alt="enter image description here"></a></p>
<p>We gather from the graph that $k_{appr2}$ is an almost pointwise correct approximation to $k_{opt}$ which fails only in very few points at the ends of the steps of the "staircase" (see $n=10$ and $n=29$).</p>
|
1,051,464 | <p>I was asked to find the truth value of the statement: </p>
<blockquote>
<p>$$
\pi + e \; \text{ is rational or } \pi - e\; \text{ is rational }
$$</p>
</blockquote>
<p>I am only allowed to use the fact that $\pi, e $ are irrational numbers and cannot use the theory of transcendental numbers.</p>
<p>Cannot proceed. any help would be appreciated. </p>
| Martin Brandenburg | 1,650 | <p>This seems to be an open problem. It is a conjecture that the statement is false, i.e. that $\pi + e$ and $\pi - e$ are irrational. According to <a href="http://en.wikipedia.org/wiki/Transcendental_number">Wikipedia</a> this remains unproven. (Just imagine the impact of the discovery of an equation such as $\pi=e+\frac{4233108252.........3123782}{31238295213.......0591231}$ ... unbelievable!)</p>
<p>Remark that at least one of those numbers is irrational, even transcendental (but this doesn't prove that both are irrational!). For if both would be algebraic, then their sum would be algebraic, which is $2 \pi$, a contradiction. Notice that this argument is not constructive at all, and again that it does not decide if "$\pi+e$ is rational or $\pi-e$ is rational" is false or not, it only proves that the stronger statement "$\pi+e$ is rational and $\pi-e$ is rational" is false.</p>
|
1,895,721 | <p>How to find $3\times3$ matrices that satisfy the matrix equation $A^2=I_3$? Can anyone please show me steps to do this question?</p>
| Asinomás | 33,907 | <p>All of the matrices you want can be obtained by taking $A$ of the following form:</p>
<p>$\begin{pmatrix}
a_1 & 0 & 0 \\
0 & a_2 & 0 \\
0 & 0 & a_3\\
\end{pmatrix}$</p>
<p>where $a_1,a_2$ and $a_3$ are $1$ or $-1$.</p>
<p>And then taking $B$ an invertible matrix (possibly with complex entries) and considering the matrix $BAB^{-1}$</p>
|
434,614 | <p>Is there a name for a type of grid you might find in Battleship? Where coordinates don't relate to points on a grid but rather the squares themselves?</p>
| amWhy | 9,003 | <p>"Grid" is as good a name as any: See <a href="http://en.wikipedia.org/wiki/Regular_grid" rel="nofollow">Regular Grid</a> in Wikipedia: In particular, see the "related" grid: the Cartesian Grid: </p>
<blockquote>
<p>"A Cartesian grid is a special case where the <strong>elements are unit squares or unit cubes</strong> [cubes in the case of a 3-D grid], and the vertices are integer points." [brackets, bold-face mine].</p>
</blockquote>
<p>You could also refer to this sort of "playing field" in a game like battleship as an <strong><a href="http://en.wikipedia.org/wiki/Incidence_matrix" rel="nofollow">incidence matrix</a></strong> of sorts: where a cell in the ith row and jth column might be occupied, using "$1$", or not occupied, using "$0$".</p>
|
4,297,177 | <p>I have a polynomial:</p>
<p><span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span></p>
<p>I am supposed to show that it is <span class="math-container">$\Omega(n^2)$</span> and <span class="math-container">$O(n^2)$</span>. Once I prove those, it should be easy to show that it is also <span class="math-container">$\Theta(n^2)$</span> using the theorems given, but I'm having difficulties with <span class="math-container">$\Omega$</span> and <span class="math-container">$O$</span>. Mainly the subtraction part is throwing me off, as it has been hard finding a constant and <span class="math-container">$k$</span> value that works.</p>
<p>For showing <span class="math-container">$\Omega(n^2)$</span>, I started with <span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span> and <span class="math-container">$g(n) = n^2$</span>, and have been trying to show that <span class="math-container">$f(n) \geq C\cdot g(n)$</span>, but the subtraction has made it hard for me to choose a value for <span class="math-container">$C$</span> and <span class="math-container">$k$</span>. Can I just chose any value that makes it work?</p>
<p>Pretty similar for showing <span class="math-container">$O(n^2)$</span>, I started with <span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span> and <span class="math-container">$g(n) = n^2$</span>, and have been trying to show that <span class="math-container">$f(n) \leq C \cdot g(n)$</span>, and again, I'm having troubles knowing where to go from here.</p>
| Keeley Hoek | 88,896 | <p>For checking that <span class="math-container">$f(n)$</span> is <span class="math-container">$O(n^2)$</span> I think you have just made a simple mistake and it seems like you understand what to do: if <span class="math-container">$C = \frac{1}{4}$</span> then <span class="math-container">$f(n) \leq C g(n)$</span> becomes
<span class="math-container">$$
\frac{1}{4} n^2 - 24 n - 16 \leq \frac{1}{4} n^2
$$</span>
which is automatically always true for <span class="math-container">$n \geq 0$</span>.</p>
<p>They key point for showing that <span class="math-container">$f(n)$</span> is <span class="math-container">$\Omega(n^2)$</span> is that you only need to show that there exists <span class="math-container">$C$</span> so that <span class="math-container">$f(n) \leq C g(n)$</span> <em>for all <span class="math-container">$n$</span> greater than some fixed <span class="math-container">$k$</span></em>. Here is how we use this fact via basically a simplifying trick: observe that for all <span class="math-container">$n \geq 1$</span> we have
<span class="math-container">$
6 < 24 n
$</span>.
This means that
<span class="math-container">$$
\frac{1}{4} n^2 - 24 n - 16
\geq \frac{1}{4} n^2 - 24 n - 24 n
= \frac{1}{4} n^2 - 48 n.
$$</span>
(In effect this means that it's enough to solve the problem for <span class="math-container">$\frac{1}{4} n^2 - 48 n$</span> instead.)</p>
<p>Moreover also note that whenever <span class="math-container">$n > 4 \times 48$</span> we have <span class="math-container">$\frac{1}{16} n^2 > 48 n$</span>. That means that
<span class="math-container">$$
\frac{1}{4} n^2 - 48 n \geq \frac{1}{4} n^2 - \frac{1}{16} n^2 = \frac{3}{16} n^2 \quad \text{whenever} \quad n > 4 \times 48.
$$</span>
Putting this all together, if we set <span class="math-container">$C = \frac{3}{16}$</span> then we have just shown that
<span class="math-container">$$
f(n) = \frac{1}{4} n^2 - 24 n - 16
\geq \frac{3}{16} n^2 = \frac{3}{16} g(n)
$$</span>
for all <span class="math-container">$n > k = 4 \times 48$</span>, which is enough to establish that <span class="math-container">$f(n)$</span> is <span class="math-container">$\Omega(n^2)$</span>.</p>
|
156,013 | <p>I would like to <code>FoldList</code> a simple function, with desired output:</p>
<pre><code>f[a,b,1]
f[b,c,1]
f[c,a,1]
f[f[a,b,1],f[b,c,1],2]
f[f[b,c,1],f[c,a,1],2]
f[f[c,a,1],f[a,b,1],2]
f[f[f[a,b,1],f[b,c,1],2],f[f[b,c,1],f[c,a,1],2],3]
f[f[f[b,c,1],f[c,a,1],2],f[f[c,a,1],f[a,b,1],2],3]
f[f[f[c,a,1],f[a,b,1],2],f[f[a,b,1],f[b,c,1],2],3]
</code></pre>
<p>which I can hack with</p>
<pre><code>t2[li_] := Take[#, 2] &@RotateLeft[li, #] & /@ Range[0, Length@li - 1];
sf[n_] := "" <> ToString[# - 1] <> "], " <> ToString[# - 2] <> "]" ->
"" <> ToString[# - 1] <> "], " <> ToString[#] <> "]" &@n;
nl = ToExpression@With[{string = ToString /@
Rest[FoldList[f[#[[1]], #[[2]]] & /@ t2@# &, {a, b, c},
Range@#]]}, Fold[StringReplace, string,
Flatten[{{"]" -> ", 1]", "1], 1]" -> "1], 2]"},
sf /@ Range[3, #]}, 1]]] &[5];
Column[Column /@ Take[nl, 3]]
</code></pre>
<p>with huge inefficiencies for large <code>n</code> (there is no problem with the <code>t2</code> part of the function). All I want to do is add in indexing into the last position of <code>f[_,_,_]</code>.</p>
<p>I'd really like to do it with something simple like</p>
<pre><code>FoldList[f[#[[1]], #[[2]],#2] & /@ t2@# &, {a, b, c}, Range@3]
</code></pre>
<p>but I can't use <code>#2</code> since <code>MapThread</code> gets in the way. I know I am over-complicating things - what is a more sensible way to achieve the above?</p>
| Carl Woll | 45,431 | <p>Here is one possibility. Create a helper function:</p>
<pre><code>g[l_List,i_] := f[##, i]& @@@ Partition[l, 2, 1, 1]
</code></pre>
<p>Then, fold this helper function:</p>
<pre><code>FoldList[g, {a, b, c}, {1, 2, 3}]
</code></pre>
<blockquote>
<p>{{a, b, c}, {f[a, b, 1], f[b, c, 1],
f[c, a, 1]}, {f[f[a, b, 1], f[b, c, 1], 2], f[f[b, c, 1], f[c, a, 1], 2],
f[f[c, a, 1], f[a, b, 1], 2]}, {f[f[f[a, b, 1], f[b, c, 1], 2],
f[f[b, c, 1], f[c, a, 1], 2], 3],
f[f[f[b, c, 1], f[c, a, 1], 2], f[f[c, a, 1], f[a, b, 1], 2], 3],
f[f[f[c, a, 1], f[a, b, 1], 2], f[f[a, b, 1], f[b, c, 1], 2], 3]}}</p>
</blockquote>
|
166,666 | <p>For which values of the coefficient $c$ does the quantity
$$
\cos\alpha\cos\beta- c\sin\alpha\sin\beta
$$
depend on $\alpha$ and $\beta$ only through their sum?</p>
<p>(I'll post a quick answer below. This will be the first time I've posted a question with intent to immediately post an answer.)</p>
| Michael Hardy | 11,667 | <p>Let $f(\alpha,\beta) = \cos\alpha\cos\beta- c\sin\alpha\sin\beta$. Since this depends on $\alpha$ and $\beta$ only through their sum we have</p>
<p>$$f(\alpha,\beta)=f(\alpha+\beta,0).$$</p>
<p>Then</p>
<p>$$
f(\alpha+\beta,0) = \cos(\alpha+\beta)\cos 0 - c\sin(\alpha+\beta)\sin 0 = \cos(\alpha+\beta).
$$</p>
<p>So by the usual identity, $c=1$.</p>
<p><b>Later edit:</b> Another way would be to write
$$
\begin{align}
\cos\alpha\cos\beta - c\sin\alpha\sin\beta & = \Big(\cos\alpha\cos\beta - \sin\alpha\sin\beta\Big) - (1-c)\sin\alpha\sin\beta \\[8pt]
& = \cos(\alpha+\beta) - (1-c)\sin\alpha\sin\beta
\end{align}
$$
and then observe that the last term doesn't depend on $\alpha$ and $\beta$ only through their sum. From one point of view, this seems like the obvious way to do it---far more so than what I did above, and yet what I did above seems simpler.</p>
|
166,666 | <p>For which values of the coefficient $c$ does the quantity
$$
\cos\alpha\cos\beta- c\sin\alpha\sin\beta
$$
depend on $\alpha$ and $\beta$ only through their sum?</p>
<p>(I'll post a quick answer below. This will be the first time I've posted a question with intent to immediately post an answer.)</p>
| Community | -1 | <p>We want $f(\alpha, \beta) = g(\alpha + \beta)$. This means that $f(\alpha,-\alpha) = g(0) = f(0,0)$, $\forall \alpha$.</p>
<p>$f(\alpha,-\alpha) = f(0,0)$ for all $\alpha$. Hence, we get that $$\cos^2(\alpha) + c \, \sin^2(\alpha) = 1, \, \, \forall \alpha$$
$$c \, \sin^2(\alpha) = \sin^2(\alpha), \, \, \forall \alpha$$
Hence, $c = 1$.</p>
|
913,998 | <p>What can we do with this function, so the function will be continuous in $(0,0)$?</p>
<p>$f:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto \frac{x^2+y^2-x^3y^3}{x^2+y^2}$</p>
<p>What I think we should do, is:</p>
<p>approximate $(0,0)$ via the line $y=x$, so substitute $y=x$ and take the limit of that function, i.e. $\lim_{x\rightarrow0}$:</p>
<p>$\lim_{x\rightarrow0} 1 - \frac{x^6}{2x^2} = \lim_{x\rightarrow0}1-\frac{x^4}{2} = 1$</p>
<p>So the new function, that is continuous in $(0,0)$ is defined by:</p>
<p>$F:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto F(x) = \begin{cases} 1 & \mbox{if } (x,y) = (0,0) \\ f(x) & \mbox{if } (x,y) \neq (0,0) \end{cases}$</p>
<p>I'm sorry for my english, but if you understand my question, could you say if I'm right? </p>
| lab bhattacharjee | 33,337 | <p>$$x^4+2x^2+2=(x^2+1)^2+1$$</p>
<p>Now for real $x,x^2\ge0$</p>
|
710,045 | <p><a href="https://i.stack.imgur.com/7rJAB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7rJAB.png" alt="question" /></a></p>
<p>For (a) I have said</p>
<p>By MVT:</p>
<p><span class="math-container">$|f(x)-f(y)|\le K|x-y|$</span></p>
<p>Choose <span class="math-container">$\delta=\epsilon/K$</span></p>
<p><span class="math-container">$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|\le K|x-y| \le K\delta=\epsilon$</span></p>
<p>For (b) I have said</p>
<p><span class="math-container">$f'(x)=-4x^3/(1+x^4)^2$</span> so <span class="math-container">$|f'(x)| \le B$</span> where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.</p>
<p>Any help would be appreciated, thank you.</p>
| Joshua Pepper | 43,511 | <p>Every finite-dimensional normed space of dimension $n$ over $\mathbb{F}$ is isomorphic to $\mathbb{F}^n$, and hence complete (assuming $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$).</p>
<p>Your student's argument works! The space that this norm defines (in $n$ dimensions) is somtimes called $l_1^n$ (which is complete). You can also apply the same norm (infinite sum) to the space of sequences $(a_n)$ such that $\sum |a_n| < \infty$, called $l_1$ (which is also complete, although you'll need a more sophisticated argument).</p>
<p>Here is the full argument written out for clarity:</p>
<p>Suppose $(z_n) \in l^2_1$ is Cauchy. Write $z_n=(x_n,y_n)$.
Note that $|x_n| \leq \|z_n\|= |x_n| + |y_n|$, so $(x_n)$ is a Cauchy sequence in $\mathbb{F}$. Similarly, $(y_n)$ is Cauchy in $\mathbb{F}$. Hence $x_n \to x$, $y_n \to y$ for some $x,y \in \mathbb{F}$, using completeness of $\mathbb{F}$. It's easy to see that this implies that $\|z_n - (x,y)\| \to 0$, and so $z_n \to (x,y) \in \mathbb{F}^2$.</p>
<p>Note that you <em>do</em> switch norms in the proof, and then switch back. However this is valid, as you are invoking the definition of the norm you put on the product space.</p>
|
2,767,392 | <p>I have the following curve:</p>
<p>$x^4=a^2(x^2-y^2)$</p>
<p>Prove that the area of its loop is $\frac{2a^2}{3}$.</p>
<p><strong>My approach</strong></p>
<p>This curve has four loops. So the required area should be:</p>
<p>$4\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx$</p>
<p>But, After solving, the area turned out to be $\frac{4a^2}{3}$. What am I doing wrong here?</p>
<p>Thanks</p>
| user | 293,846 | <p>For the proof it suffices to show that
$$
\sum_{i=0}^k (-1)^i\binom{n}{i}=(-1)^k\binom{n-1}{k}.
$$</p>
<p>The simplest way to demonstrate this is by induction. The statement is obviously true for $k=0$. Assumption that it is true for $k$ implies that it is true for $k+1$ as well:
$$
\sum_{i=0}^{k+1} (-1)^i\binom{n}{i}=\sum_{i=0}^{k} (-1)^i\binom{n}{i}+
(-1)^{k+1}\binom{n}{k+1}\stackrel{I.H.}{=}(-1)^k\binom{n-1}{k}+(-1)^{k+1}\binom{n}{k+1}\\
=(-1)^{k+1}\left[\binom{n}{k+1}-\binom{n-1}{k}\right]=(-1)^{k+1}\binom{n-1}{k+1}.
$$</p>
|
6,340 | <p>Just to clarify, I want to show that:</p>
<p>If $f$ is entire and $\int_{\mathbb{C}} |f|^p dxdy <\infty$, then $f=0$.</p>
<p>I think I can show that this is the case for $p=2$, but I'm not sure about other values of $p$...</p>
| Braindead | 2,499 | <p>Right! Thank you, I didn't think about using Hölder's.</p>
<p>In short, </p>
<p>\begin{equation}
|f(0)| = \left|\int_{|z|=R} \frac{f(z)dz}{z} \right| < \left(\int_{|z|=R} |f(z)|^p |dz|\right)^{1/p} \left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q}
\end{equation}</p>
<p>And $\left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$ </p>
<p>\begin{equation}
\int_0^\infty \left((2\pi)^{1/q} |f(0)| R^{1/p}\right)^p dR < || f ||_p^p <\infty
\end{equation}</p>
<p>And the only way for this to fly is for $f(0)=0$, for $p\le 1$.</p>
|
6,340 | <p>Just to clarify, I want to show that:</p>
<p>If $f$ is entire and $\int_{\mathbb{C}} |f|^p dxdy <\infty$, then $f=0$.</p>
<p>I think I can show that this is the case for $p=2$, but I'm not sure about other values of $p$...</p>
| kahen | 1,269 | <p>$\newcommand\RR{\mathbb{R}}$
$\newcommand\NN{\mathbb{N}}$
$\newcommand\e{\varepsilon}$
$\newcommand\lbg{\lambda}$</p>
<p>If $f: \RR \to \RR$ is an analytic unbounded Lebesgue integrable function, then it must in particular satisfy this property (because it is continuous):
\[
\forall \e,M > 0 \; \exists x_0 : |x| > |x_0| \Rightarrow
\lbg(\{x \in \RR : |f(x)| > M\}) < \e,
\]
where $\lbg$ is the usual Lebesgue measure. </p>
<p>Then you just use that $f$ is analytic iff $\forall \text{ compact } K \subset \RR \; \exists C > 0 : x \in K, n \in \NN \Rightarrow \left\vert\frac{\partial^n f}{\partial x^n}(x)\right\vert
\leq C^{n+1}n!$</p>
<p>But it is easy to see that this is impossible since $|f'|$ will have to be arbitrarily large - or in other words that you can go far enough out to the right on the x-axis that $|f|$ has to go from being larger than some big $M$ to smaller than some small $m$ on a very small interval.</p>
<p>This is missing quite a few details, but it should be pretty easy to fill out.</p>
|
3,065,331 | <p>In <a href="https://rads.stackoverflow.com/amzn/click/0199208255" rel="nofollow noreferrer">Nonlinear Ordinary Differential Equations: An Introduction for Scientists and Engineers</a> one of the very first stated equations are, as in the title of the question,</p>
<p><span class="math-container">$$
\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2} \dot{x}^2 \right).
$$</span></p>
<p>However, I'm having trouble seeing why this should be true. Could anyone clarify this? Thank you for your time in advance.</p>
| Robert Lewis | 67,071 | <p><span class="math-container">$\dfrac{d}{dx} \left (\dfrac{1}{2} \dot x^2 \right ) = \dfrac{d \dot x}{dx}\dot x = \dfrac{d\dot x}{dx} \dfrac{dx}{dt} = \dfrac{d \dot x(x(t))}{dt} = \dfrac{d \dot x}{dt} = \dfrac{d^2 x}{dt^2} = \ddot x \tag 1$</span></p>
<p>by the chain rule.</p>
<p>The key here is the observation that, for one-dimensional motion (as (1) appears to describe), the variables <span class="math-container">$x$</span> and <span class="math-container">$t$</span> may <em>both</em> be taken to be parameters along the curve <span class="math-container">$x(t)$</span>.</p>
|
1,423,728 | <p>The definition of a limit in <a href="http://rads.stackoverflow.com/amzn/click/0321888545" rel="nofollow noreferrer">this book</a> stated like this </p>
<p><a href="https://i.stack.imgur.com/tKjaa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tKjaa.png" alt="enter image description here"></a></p>
<p>1) why we require ƒ(x) be defined on an open interval about $x_0$ ?</p>
<p>2) does the definition mean it is impossible to talk about limit of function of this sort?</p>
<p>$f(x) = \begin{cases} undefined & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$</p>
<p><strong>Edit</strong>: Since some similar answers to my first question is that we want to talk about two-sided limit, thus require ƒ(x) be defined on an open interval about $x_0$. However, these answers doesn't clear my intended question. Now if I restrict $x>=1$, then can we talk about the limit of that function , especially for $\displaystyle \lim_{x \to 1^{+}}f(x)$?</p>
<p>As for my second question, I found a more precise definition of limit in Courant's book Introduction_to_Calculus_and_Analysis stated like this
<a href="https://i.stack.imgur.com/8J8Wh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8J8Wh.png" alt="enter image description here"></a>
which doesn't require ƒ(x) to be defined on every point of the an open interval about $x_0$, so I think it is possible to talk about limit of that function.</p>
| Steven Alexis Gregory | 75,410 | <p>(1) Because they want to approach $x_0$ from both sides.</p>
<p>(2) If you define $f:\mathbb Q \to \mathbb R$ by $f(x) = 1$ for all $x \in \mathbb Q$, then your definition of limit will not allow for $\displaystyle \lim_{x \to x_0}f(x)$ to exist. In topology, for example, the domain of the function is taken into account and, in that case $\displaystyle \lim_{x \to x_0}f(x)$ will be equal to $1$ for all rational $x_0$.</p>
|
1,423,728 | <p>The definition of a limit in <a href="http://rads.stackoverflow.com/amzn/click/0321888545" rel="nofollow noreferrer">this book</a> stated like this </p>
<p><a href="https://i.stack.imgur.com/tKjaa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tKjaa.png" alt="enter image description here"></a></p>
<p>1) why we require ƒ(x) be defined on an open interval about $x_0$ ?</p>
<p>2) does the definition mean it is impossible to talk about limit of function of this sort?</p>
<p>$f(x) = \begin{cases} undefined & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$</p>
<p><strong>Edit</strong>: Since some similar answers to my first question is that we want to talk about two-sided limit, thus require ƒ(x) be defined on an open interval about $x_0$. However, these answers doesn't clear my intended question. Now if I restrict $x>=1$, then can we talk about the limit of that function , especially for $\displaystyle \lim_{x \to 1^{+}}f(x)$?</p>
<p>As for my second question, I found a more precise definition of limit in Courant's book Introduction_to_Calculus_and_Analysis stated like this
<a href="https://i.stack.imgur.com/8J8Wh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8J8Wh.png" alt="enter image description here"></a>
which doesn't require ƒ(x) to be defined on every point of the an open interval about $x_0$, so I think it is possible to talk about limit of that function.</p>
| Paramanand Singh | 72,031 | <p>The idea of a <em>limit of a function $f$ at a point $x_{0}$</em> is invented to give answer to the following question:</p>
<p><em>How does function $f$ behave near the point $x_{0}$?</em></p>
<p>Note that the question does not ask as to what happens at $x_{0}$ but rather what happens near $x_{0}$. Hence it does not matter whether $f$ is defined at $x_{0}$, but <strong>it does matter that $f$ is defined near $x_{0}$</strong>. Moreover we would like the function $f$ to be defined at points which are as close to $x_{0}$ as we want. The central idea is that the domain $D$ of $f$ need not contain point $x_{0}$, but it should contain points which are as near to $x_{0}$ as we please.</p>
<p>More formally, <em>a basic prerequisite for defining limit of a function $f$ at a point $x_{0}$ is that for every real number $\delta > 0$ there must exist at least one point $x_{\delta}$ in the domain $D$ of $f$ such that $0 < |x_{\delta} - x_{0}| < \delta$</em>.</p>
<p>When the condition mentioned in above paragraph holds we say that <em>$x_{0}$ is a limit point (accumulation point, cluster point) of $D$</em>.</p>
<p>In general setting the distance $|x_{\delta} - x_{0}|$ is replaced by a metric $d(x_{\delta}, x_{0})$ (which satisfies properties similar to the absolute value function).</p>
<p>For a beginner learning calculus it is better not to indulge in such generalities (of metric / topological spaces), but rather deal with the concrete spaces like set of real numbers. In that case we just say that $f$ needs to be defined in an interval which contains $x_{0}$ in the interior with the possibility that $f$ may not be defined at $x_{0}$.</p>
<p>Note that some authors prefer the general notion and define limits in the following manner.</p>
<p><em>Let $f: D\to \mathbb{R}$ be a function with $D\subseteq\mathbb{R}$ and let $x_{0}$ be any real number. A real number $L$ is said to be the limit of function $f$ at point $x_{0}$ (denoted by $L = \lim_{x \to x_{0}}f(x)$) if for any given number $\epsilon > 0$ it is possible to find a number $\delta > 0$ such that $|f(x) - L| < \epsilon$ for all points $x \in D$ with $0 < |x - x_{0}| < \delta$.</em></p>
<p>Based on this definition the limit of the function $f$ in your question is $1$.</p>
<p>The advantage of this definition over the usual one is that it deals with the left-hand and right-hand limits without any special treatment. However it is preferable to stick to the usual definition and define left-hand and right-hand limits separately (especially for a beginner).</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Dustin G. Mixon | 29,873 | <p>I don't have the time to regularly edit Wikipedia. I've made time for MO lately because asking and answering these questions positively affects my research. Based on my experience writing lecture notes, I find that organizing and presenting math that I already understand really well takes a lot of time compared to the benefit I get from the clearer understanding of the material. For the more specialized math I personally care about, I take the time to write notes on my research blog (like MO, this positively affects my research).</p>
<p>Having said that, if I find an error in the math posted on Wikipedia, I will certainly make an edit for the sake of the community. But this hasn't happened yet. (!)</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| ex0du5 | 24,133 | <p>There are also mathematical models of why the number of editors is decreasing that have nothing to do with a decreasing opinion of Wikipedia or a migration elsewhere. In the beginning, Wikipedia was tabula rasa, and editors could add sections on anything. Addition, multiplication, trigonometry, basic set theory, etc. all needed <em>something</em>, even if it was a very trivial entry. The level of expertise required to contribute was very low. Over time, the articles have indeed improved, and the information content has clearly grown. Those areas with little barrier to entry have been filled out quite extensively.</p>
<p>So, I expect that the data presented in the question is not something to be overly alarmed about. There is much of the trend that should be expected. As the requirements for constructively contributing increase, fewer individuals will be able to add useful information or correct improper or awkwardly worded entries.</p>
<p>I think this is true throughout Wikipedia. The type of editing that is easy these days are more focused on current events. Other data will natural decrease to some convergence and only be edited as cultural changes make different expositions better, a very slow process. Different data would be needed to separate this phenomenon from the other causes for edit rate changes.</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Salix alba | 34,268 | <p>Replying to Dick Palais: Any edit to Wikipedia, no matter how small, is good even if it's just correcting a typo. One thing I've done is to make a small edit to every page I visit. There is always something to be fixed. Once you've done a number of small edits, you can do something more substantial. The mathematics reference desk is quite similar to MathOverflow and could be a good place to get started: <a href="http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics</a>.</p>
<p>Replying to Marc Palm: It is quite easy to see who made edits to a page, just clicking on the history tab. A user's contributions can be seen by going to a user's page and selecting user contributions on the left.</p>
|
361,060 | <blockquote>
<p>Consider the ring of Gaussian integers $D=\lbrace a+bi\mid a,b \in \mathbb{Z \rbrace}$, where $i \in \mathbb{C}$ such that $i^2=-1$. Consider the map $f$ from $D$ to $\mathbb{Z}[x]/(x^2+1)$ sending $i$ to the class of $x$ modulo $x^2+1$. Show that $f$ is a ring isomorphism.</p>
</blockquote>
<p>I got a confusion in this question. I don't understand the map sending $i$ to the class of $x$ modulo $x^2+1$. Can anyone help me to clear my confusion?</p>
<p>EDIT: I am having trouble to show the map is injective and surjective. Can anyone guide me?</p>
| baharampuri | 50,080 | <p>The map in general format will be $f(a+ib)=a+bx + \langle x^2 +1 \rangle$
it is a ring homomorphism since $f(a+ib +c +id )= (a+bx +c+dx) + \langle x^2 +1 \rangle = a+bx +\langle x^2 +1 \rangle c+dx +\langle x^2 +1 \rangle =f(a+ib)+f(c+id)$</p>
|
57,642 | <p>I'm looking for the shortest and the clearest proof for this following theorem:</p>
<p>For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , I need to show $V= \ker T^n \oplus $ Im $T^n$.</p>
<p>Any hints? I don't know where to start from.</p>
<p>Thank you.</p>
| Robert Israel | 8,508 | <p>The main point is that ${\rm Ker}\ T^n \cap {\rm Im}\ T^n = \{0\}$. Show that if $T^k x \ne 0$ and $T^{k+1} x = 0$, then $x, Tx, \ldots T^k x$ are linearly independent, and therefore $k \le n-1$. No need here for Jordan Canonical Form - in particular this works over any field.</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| mau | 89 | <p>unknown source:</p>
<p>Could the plane be colored with two different colors (say, red and blue) so that there is no equilateral triangle whose vertices are all of the same color?</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| ThudanBlunder | 1,770 | <p>Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother and the monkey's father. </p>
<p>If in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, and assuming no <em>a priori</em> knowledge of the monkeys' longevity, find their respective ages. :-D</p>
|
566 | <h3>We all love a good puzzle</h3>
<p>To a certain extent, any piece of mathematics is a puzzle in some sense: whether we are classifying the homological intersection forms of four manifolds or calculating the optimum dimensions of a cylinder, it is an element of investigation and inherently puzzlish intrigue that drives us. Indeed most puzzles (cryptic crosswords aside) are somewhat mathematical (the mathematics of Sudoku for example is hidden in <a href="http://en.wikipedia.org/wiki/Latin_squares" rel="nofollow noreferrer">latin squares</a>). Mathematicians and puzzles get on, it seems, rather well.</p>
<h3>But what is a good puzzle?</h3>
<p>Okay, so in order to make this question worthwhile (and not a ten-page wade-athon through 57 varieties of the men with red and blue hats puzzle), we are going to have to impose some limitations. Not every puzzle-based answer that pops into your head will qualify as a good puzzle—to do so it must</p>
<ul>
<li><strong>Not be widely known:</strong> If you have a terribly interesting puzzle that motivates something in cryptography; well done you, but chances are we've seen it. If you saw that <em>hilarious</em> scene in the film 21, where Kevin Spacey explains the Monty hall paradox badly and want to share, don't do so here. Anyone found posting the liar/truth teller riddle will be immediately disembowelled.</li>
<li><strong>Be mathematical (as much as possible):</strong> It's true that logic <em>is</em> mathematics, but puzzles beginning '<em>There is a street where everyone has a different coloured house …</em>' are tedious as hell. Note: there is a happy medium between this and trig substitutions.</li>
<li><strong>Not be too hard:</strong> Any level of difficulty is cool, but if coming up with an answer requires more than two sublemmas, you are misreading your audience.</li>
<li><strong>Actually have an answer:</strong> Crank questions will not be appreciated! You can post the answers/hints in <a href="http://en.wikipedia.org/wiki/Rot_13" rel="nofollow noreferrer">Rot-13</a> underneath as comments as on MO if you fancy.</li>
<li><strong>Have that indefinable spark that makes a puzzle awesome:</strong> Like a situation that seems familiar, requiring unfamiliar thought …</li>
</ul>
<p>Ideally include where you found the puzzle so we can find more cool stuff like it. For ease of voting, one puzzle per post is best.</p>
<hr />
<h1>Some examples to set the ball rolling</h1>
<blockquote>
<p>Simplify <span class="math-container">$\sqrt{2+\sqrt{3}}$</span></p>
</blockquote>
<p><strong>From:</strong> problem solving magazine</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Try a two term solution</p>
</blockquote>
<hr />
<blockquote>
<p>Can one make an equilateral triangle with all vertices at integer coordinates?</p>
</blockquote>
<p><strong>From:</strong> Durham distance maths challenge 2010</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> This is equivalent to the rational case</p>
</blockquote>
<hr />
<blockquote>
<p>The collection of <span class="math-container">$n \times n$</span> <a href="http://en.wikipedia.org/wiki/Magic_squares" rel="nofollow noreferrer">Magic squares</a> form a vector space over <span class="math-container">$\mathbb{R}$</span> prove this, and by way of a linear transformation, derive the dimension of this vector space.</p>
</blockquote>
<p><strong>From:</strong> Me, I made this up (you can tell, can't you!)</p>
<p><strong>Hint:</strong></p>
<blockquote class="spoiler">
<p> Apply the rank nullity theorem</p>
</blockquote>
| Steven Stadnicki | 785 | <p>A personal favorite (albeit not ideally-worded): Jamie has a windowbox where he plants a row of iris flowers. He plants in just two colors - blue and yellow - but never plants two yellow irises next to each other (the result is just too garish). Assuming that he wants to keep the same number of flowers in his box every day, how many flowers will he need if he wants a different arrangement of blue and yellow every day for a year?</p>
|
844,887 | <p>I am trying to prove by induction that every non-zero natural number has at least one predecessor. However, I don't know what to use as a base case, since 0 is not non-zero and I haven't yet established that 1 is the number following zero.</p>
<p>My axioms are: </p>
<ol>
<li>$0$ is a natural number. </li>
<li>if $b$ is a natural number then $S(b)$ is also a natural number. </li>
<li>$0$ is not a successor of any natural number. </li>
<li>different numbers have different successors.</li>
</ol>
<p>Any advice?</p>
| martini | 15,379 | <p><strong>Hint</strong>: Let $P(n) ={}$ "all numbers $<n$ have a predecessor". Then $P(0)$ is true (as there are no numbers $< 0$. I'm sure you can do the induction step.</p>
|
4,561,863 | <p>This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$</span>
where <span class="math-container">$a>0$</span>, <span class="math-container">$b>0$</span> are some constants.
I know how to calculate this limit using the L'Hopital's rule:
<span class="math-container">$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$</span>
<span class="math-container">$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$</span>
<span class="math-container">$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$</span>
<span class="math-container">$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$</span>
I'm allowed to use the limits <span class="math-container">$\lim_{x\to0}\frac{\sin x}{x}=1$</span>,
<span class="math-container">$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$</span>,
<span class="math-container">$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$</span> and <span class="math-container">$\lim_{x\to0} (1+x)^{1/x}=e$</span>.</p>
| Feng | 624,428 | <p>Taking logarithm, then using <span class="math-container">$\sin x\sim x$</span> and <span class="math-container">$\ln(1+x)\sim x$</span> when <span class="math-container">$x\to0$</span>, we have
<span class="math-container">\begin{align*}
\frac1x\cdot\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)&\sim\frac1{\sin x}\cdot\left(\frac{a^{\sin x}+b^{\sin x} }{2}-1\right)\\
&=\frac12\frac{a^{\sin x}-1}{\sin x}+\frac12\frac{b^{\sin x}-1}{\sin x}\\
&\rightarrow\frac12(\ln a+\ln b),\qquad x\to0,
\end{align*}</span>
where in the last line we've used
<span class="math-container">$$\lim_{x\to0}\frac{a^{\sin x}-1}{\sin x}=\lim_{t\to0}\frac{a^{t}-1}{t}=\ln a. $$</span>
Therefore,
<span class="math-container">$$\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.$$</span></p>
|
506,767 | <p>What is the largest number on multiplying with itself gives the same number as last digits of the product?</p>
<p>i.e., $(376 \times 376) = 141376$</p>
<p>i.e., $(25\times 25) = 625$</p>
<p>If the largest number cant be found out can you prove that there is always a number greater than any given number? (only in base $10$)</p>
| mercio | 17,445 | <p>For any integer $n$,there are $2$ solutions to $x^2 = x \pmod {2^n}$ and $x^2 = x \pmod {5^n}$, which are $0$ and $1$.</p>
<p>Hence by the chinese remainder theorem, there are $4$ solutions to $x^2 = x \pmod {10^n}$. Two of those are the obvious $0$ and $1$, there is one solution which is $0$ mod $5^n$ and $1$ mod $2^n$, and the last one is $0$ mod $2^n$ and $1$ mod $5^n$.</p>
<p>So there is no largest such number.</p>
<p>You can summarize as saying that in the $10$-adic numbers (numbers with an infinite decimal expansion on the left), there are $4$ numbers satisfying $x^2 = x$ : </p>
<p>$...00000000\\
...00000001\\
...87109376\\
...12890625$</p>
|
2,084,671 | <p>I've tried simplifying the term by <strong>substituting</strong> $y := x^2$ and also to use a certain <strong>algorithm</strong> that outputs the f(a), but it gets too big for the input r, that is a variable and not a number to begin with (added little example):</p>
<p><a href="https://i.stack.imgur.com/8C9sV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8C9sV.png" alt="enter image description here"></a></p>
<p>If you have a good idea how to do this, please let me know</p>
| Maverick | 171,392 | <p>Let $y=x^2$ which reduces the equation to $y^2-6qy+q^2=0$</p>
<p>$y=\frac{6q\pm\sqrt{36q^2-4q^2}}{2}$</p>
<p>$x^2=\left(3 \pm 2\sqrt{2}\right)q$</p>
<p>$x^2=\left(\sqrt{2}\pm1\right)^2 q$</p>
<p>$x=\pm\left(\sqrt{2}\pm1\right)\sqrt{q}$</p>
<p>$x=\pm\left(\sqrt{2q}\pm\sqrt{q}\right)$</p>
<p>Q.E.D</p>
|
2,703,323 | <p>How can one show that the limit of the following is $1$?</p>
<p>$$\lim_{x\to 0}\frac{\frac{1}{1-x}-1}{x}=1$$</p>
| user | 505,767 | <p>By definition of derivative with </p>
<p>$$f(x)=\frac{1}{1-x}\implies f'(x)=\frac{1}{(1-x)^2}$$</p>
<p>we have</p>
<p>$$\lim_{x\to 0}\frac{\frac{1}{1-x}-1}{x-0}=f'(0)=1$$</p>
|
4,196,125 | <p>Can someone tell me where this calculation goes wrong?
I get (2 3 4)(1 2 3 4 5 6)^-1 = (1 6 5 2).
My book and Mathematica get (1 6 5 4).
I have read several explanations of how to multiply permutations in cycle notation and have
worked dozens of examples successfully, but I always get this one wrong.</p>
<p>(2 3 4)(1 2 3 4 5 6)^-1 = (2 3 4)(6 5 4 3 2 1) = (2 3 4)(1 6 5 4 3 2)</p>
<p>1 -> 6 then 6 is unchanged giving (1 6</p>
<p>6 -> 5 then 5 in unchanged giving (1 6 5</p>
<p>5 -> 4 then 4 -> 2 giving (1 6 5 2</p>
<p>2 -> 1 then 1 is unchanged completing the cycle giving (1 6 5 2).</p>
| Math Lover | 801,574 | <p>Please note that you have a circle</p>
<p><span class="math-container">$x^2 + (y-2)^2 \leq 1$</span></p>
<p>As the circle forms for <span class="math-container">$-1 \leq x \leq 1$</span>, your bounds for <span class="math-container">$x$</span> is incorrect.</p>
<p>Also for limits of <span class="math-container">$y$</span>, <span class="math-container">$ \ 2 - \sqrt{1-x^2} \leq y \leq 2 + \sqrt{1-x^2}$</span></p>
<p>And that gives height of each shell as <span class="math-container">$2 \sqrt{1-x^2}$</span>.</p>
<p>As the circle is being rotated around <span class="math-container">$x = 2$</span>, radial distance of shell from the axis of rotation is <span class="math-container">$(2-x)$</span>.</p>
<p>So the integral to find volume using shell method is,</p>
<p><span class="math-container">$\displaystyle \int_{-1}^1 2 \pi (2-x) \cdot 2 \sqrt{1-x^2} \ dx = 4 \pi \int_{-1}^1 (2-x) \sqrt{1-x^2} \ dx$</span></p>
<p>Can you take it from here?</p>
|
114,898 | <p>A) Let $f:F\rightarrow S$ be a flat proper morphism of schemes with geometrically normal fibers. Then supposedly the number of $\textbf{connected}$ components of the geometric fibers is constant. Why is this? Without some kind of vanishing of cohomology or information on the base, I don't see why this is true. </p>
<p>B) Furthermore, supposedly if $F$ is now a flat proper morphism with reduced, connected, nodal curves as geometric fibers, then there is a Zariski open subset of $S$ on which the fibers all have the same number of $\textbf{irreducible}$ components. Why is this?</p>
<p>Finally, how far can these results be generalized? For example, is B) true for any flat proper morphism?</p>
| Qfwfq | 4,721 | <p>Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$</p>
<p>for $|z|<1$, where $\varphi: S^1\to\mathbb{C}$ is a Peano curve (i.e. its image has nonempty interior), and $\phi(z):=\varphi(z)$ for $z\in S^1$. [<strong>Edit</strong>: this construction doesn't work because $\phi$, as I defined it, may not be continuous up to the boundary - see the comments]</p>
|
416,387 | <p>In the process of solving a DE and imposing the initial condition I came up with the following question.</p>
<p>I've reached the stage that</p>
<p>$$\ln y + C = \int\left(\frac{2}{x+2}-\frac{1}{x+1}\right)dx$$
$$\Rightarrow \ln y +C=2\ln|x+2|-\ln|x+1|$$
$$\Rightarrow y=A\frac{(x+2)^2}{|x+1|}.$$</p>
<p>Now I had also found that the curve passes through $(-4,-3)$. Susbstituting in the expression above, we find $A=-9/4$. However, the solution in the markscheme (the problem was from a past exam) drops the modulus sign and so it gives $A=9/4$.</p>
<p>So my question is, why do they drop the modulus sign and when is one allowed to do so in dealing with integrals of the form $\int 1/t ~dt$?</p>
<p>Thanks in advance.</p>
| Ted Shifrin | 71,348 | <p>First of all, you should presumably have had
$$\ln |y| + C = 2\ln |x+2| - \ln |x+1|\,, \quad\text{leading to}\quad |y| = A\frac{(x+2)^2}{|x+1|}\,.$$
Since a solution curve through $(-4,-3)$ cannot cross the lines $x=-1$ and $y=0$, we infer that throughout $y<0$ and $x<-1$, so $|y|=-y$ and $|x+1|=-(x+1)$. Thus, the solution is
$$y=\frac94\frac{(x+2)^2}{x+1}\,, \quad x<-1,\ y<0\,.$$
Check the markscheme carefully! :)</p>
|
4,353,891 | <p>How to prove that <span class="math-container">$A=\{(x,y)\in \mathbb{R}^2: |x|+|y|^{1/2}<1\}$</span> is convex? I tried using the definition but couldn’t go far, since the second component involves square root(tried squaring, that made it complicated).</p>
<p>I plotted it in mathematica and the graph comes out to be convex, but how to prove it mathematically?</p>
<p>I started with two points (x,y) and (a,b) in A, and tried to prove (ta+(1-t)x, tb +(1-t)y) is in A, for which I need to show, <span class="math-container">$| ta+(1-t)x|+|t b +(1-t)y|^{1/2}<1$</span>, but since the square root term is there, I couldn’t simplify this.</p>
| Zim | 793,815 | <p>It's easily seen <a href="https://www.desmos.com/calculator/hvwbfhhlav" rel="nofollow noreferrer">graphically</a> that this set is not convex.</p>
|
3,170,629 | <p>I am trying to define a sequence.
The first few terms of the sequence are:</p>
<p><span class="math-container">$2,5,13,43,61$</span></p>
<p>Not yet found other terms because I am working with paper and pen, no software.</p>
<p>Why the first term is <span class="math-container">$5$</span>?</p>
<p>Let be <span class="math-container">$\pi(x)$</span> the celebrated prime counting function.
Well 5-<span class="math-container">$\pi(5)$</span>=<span class="math-container">$5-3$</span>=2 which is a prime.
If we repeat the same thing with the new prime <span class="math-container">$2$</span>, we have 2-<span class="math-container">$\pi(2)=1$</span>, which is not a prime. So starting the sequence from prime <span class="math-container">$5$</span>, we have the cycle <span class="math-container">$5\rightarrow 2\rightarrow 1$</span>. The arrows stop when a not prime is reached. No prime below <span class="math-container">$5$</span> has a longer cycle. Infact starting for example from <span class="math-container">$3$</span> you get <span class="math-container">$3-\pi(3)=1$</span>, which is not prime so the cycle is simply <span class="math-container">$3\rightarrow 1$</span>.
The second term of the sequence is <span class="math-container">$13$</span> because below <span class="math-container">$13$</span> no other prime has a larger cycle. Infact <span class="math-container">$13-\pi(13)=7$</span>, which is prime. Then <span class="math-container">$7-\pi(7)=3$</span>, which is prime and eventually <span class="math-container">$3-\pi(3)=1$</span>, which is not prime. So the cycle is <span class="math-container">$13\rightarrow 7\rightarrow 3\rightarrow 1$</span></p>
<p>The cycle for 43 is longer so it is the third term of the above sequence.
Could you suggest to me a nice and elegant definition for this sequence: <span class="math-container">$5,13,43,61...$</span> (I don't know if it is infinite)
Could you find other terms with Pari if you want?</p>
| Community | -1 | <pre><code>`my(a=0,b=0);forprime(x=1,50000,y=x;while(isprime(y-primepi(y)),y-=primepi(y);b++);if(b>a,a=b;print(x));b=0)`
</code></pre>
<p>produces 14897 as the next one. Then no more below 500000. There's not too much to say except primes in the sequence will be primes at even indices after the first, simply because most primes are more than 2 away from their indices. </p>
|
3,170,629 | <p>I am trying to define a sequence.
The first few terms of the sequence are:</p>
<p><span class="math-container">$2,5,13,43,61$</span></p>
<p>Not yet found other terms because I am working with paper and pen, no software.</p>
<p>Why the first term is <span class="math-container">$5$</span>?</p>
<p>Let be <span class="math-container">$\pi(x)$</span> the celebrated prime counting function.
Well 5-<span class="math-container">$\pi(5)$</span>=<span class="math-container">$5-3$</span>=2 which is a prime.
If we repeat the same thing with the new prime <span class="math-container">$2$</span>, we have 2-<span class="math-container">$\pi(2)=1$</span>, which is not a prime. So starting the sequence from prime <span class="math-container">$5$</span>, we have the cycle <span class="math-container">$5\rightarrow 2\rightarrow 1$</span>. The arrows stop when a not prime is reached. No prime below <span class="math-container">$5$</span> has a longer cycle. Infact starting for example from <span class="math-container">$3$</span> you get <span class="math-container">$3-\pi(3)=1$</span>, which is not prime so the cycle is simply <span class="math-container">$3\rightarrow 1$</span>.
The second term of the sequence is <span class="math-container">$13$</span> because below <span class="math-container">$13$</span> no other prime has a larger cycle. Infact <span class="math-container">$13-\pi(13)=7$</span>, which is prime. Then <span class="math-container">$7-\pi(7)=3$</span>, which is prime and eventually <span class="math-container">$3-\pi(3)=1$</span>, which is not prime. So the cycle is <span class="math-container">$13\rightarrow 7\rightarrow 3\rightarrow 1$</span></p>
<p>The cycle for 43 is longer so it is the third term of the above sequence.
Could you suggest to me a nice and elegant definition for this sequence: <span class="math-container">$5,13,43,61...$</span> (I don't know if it is infinite)
Could you find other terms with Pari if you want?</p>
| lucasb | 654,079 | <p>Using <span class="math-container">$S$</span> to denote the sequence you're trying to define, one may do so in
terms of two auxiliary functions <span class="math-container">$N$</span> and <span class="math-container">$L$</span>, where <span class="math-container">$N$</span> assigns to every prime number <span class="math-container">$x$</span> a sequence whose first term, denoted by <span class="math-container">$(N(x))(0)$$^*$</span>, is <span class="math-container">$x$</span> itself, and each next term, denoted by <span class="math-container">$(N(x))(n + 1)$</span>, is given by <span class="math-container">$(N(x))(n) - \pi((N(x))(n))$</span>, and <span class="math-container">$L$</span> is the function which gives the number of terms of a sequence returned by <span class="math-container">$N$</span> up to when the first non-prime term is reached. <span class="math-container">$S$</span> is then defined to be such that the first term equals <span class="math-container">$5$</span>, and given any term <span class="math-container">$S(n)$</span>, the next term in the sequence is then the smallest prime number <span class="math-container">$p$</span> such that <span class="math-container">$L(N(p)) > L(N(S(n)))$</span>.</p>
<p>In formula:</p>
<p><span class="math-container">$S(0) = 5$</span><br>
<span class="math-container">$S(n + 1) = \langle\downarrow p : p \in \mathbb{P} : L(N(p)) > L(N(S(n)))\rangle$</span></p>
<p><span class="math-container">$(N(x))(0) = x$</span><br>
<span class="math-container">$(N(x))(n + 1) = (N(x))(n) - \pi((N(x))(n))$</span></p>
<p><span class="math-container">$L(N(x)) = \langle\downarrow n : n \in \mathbb{N} : (N(x))(n) \notin
\mathbb{P}\rangle$</span></p>
<p>The notation <span class="math-container">$\langle\downarrow x : R(x) : T(x)\rangle$</span> here denotes the minimum element <span class="math-container">$x$</span> that satisfies <span class="math-container">$T(x)$</span> from the set of all elements satisfying <span class="math-container">$R(x)$</span>. <span class="math-container">$R(x)$</span> and <span class="math-container">$T(x)$</span> denote arbitrary predicates (i. e. boolean-valued functions) which generally depend on <span class="math-container">$x$</span>.</p>
<hr>
<p><span class="math-container">$^*$</span><strong>Note:</strong> Here we use the definition that a <em>sequence</em> is any function whose domain consists of either all natural numbers or all natural numbers <span class="math-container">$n$</span> such that <span class="math-container">$0 \le n \lt m$</span> for arbitrary natural constant <span class="math-container">$m$</span>. We admit <span class="math-container">$0$</span> as the smallest natural number.</p>
|
2,976,613 | <p>I want to show, that <span class="math-container">$a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$</span> is not converging, because <span class="math-container">$\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$</span>. Therefore, the series can't be absolute converge too.</p>
<p>Firstly, I try to simplify the term. After that I want to find the limit.</p>
<p>Unfortunately, I can't seem to find any good equation with that I can clearly show <span class="math-container">$(*)$</span>.
<span class="math-container">\begin{align}
\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\
&=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\
&= \cdots
\end{align}</span></p>
<p>How to go on?</p>
| user | 505,767 | <p>Recall that </p>
<p><span class="math-container">$$\sum_0^\infty a_n <\infty \implies a_n \to 0$$</span></p>
<p>therefore if <span class="math-container">$a_n \not \to 0$</span> the series can’t converge.</p>
<p>In that case for <span class="math-container">$n\ge 3$</span> we have</p>
<p><span class="math-container">$$\left|\dfrac{2n+n^3}{3-4n}\right|=\dfrac{2n+n^3}{4n-3}>\dfrac{n^3}{4n}=\frac{n^2}4\ge2$$</span></p>
<p>and then</p>
<p><span class="math-container">$$|a_n|=\left|\dfrac{2n+n^3}{3-4n}\right|^n\ge 2^n$$</span></p>
<p>Refer also to the related:</p>
<ul>
<li><a href="https://math.stackexchange.com/q/2701344/505767">Does the series <span class="math-container">$\sum_{n=1}^\infty (-1)^n\ln(n)$</span> converge or diverge?</a></li>
</ul>
|
2,976,613 | <p>I want to show, that <span class="math-container">$a:=\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n$</span> is not converging, because <span class="math-container">$\lim \limits_{n \to \infty}(a)\neq 0 \; (*)$</span>. Therefore, the series can't be absolute converge too.</p>
<p>Firstly, I try to simplify the term. After that I want to find the limit.</p>
<p>Unfortunately, I can't seem to find any good equation with that I can clearly show <span class="math-container">$(*)$</span>.
<span class="math-container">\begin{align}
\sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\
&=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\
&= \cdots
\end{align}</span></p>
<p>How to go on?</p>
| Peter Szilas | 408,605 | <p>Let <span class="math-container">$n \ge 4$</span>.</p>
<p><span class="math-container">$|a_n|:=\left |\dfrac{n^3+2n}{4n-3}\right |^n \gt$</span></p>
<p><span class="math-container">$\left (\dfrac{n^3}{4n}\right )^n \ge n^n.$</span></p>
<p><span class="math-container">$\lim_{n \rightarrow \infty} |a_n| \not = 0$</span>.</p>
<p>Hence the series does not converge.</p>
|
1,143,889 | <p>I am interested in counting the number of hyperedges in the complete, $t$-uniform hypergraph on $n$ vertices which intersect a specified set of $k$ vertices. This is trivial, the answer is:</p>
<p>$$\sum_{i=1}^t {k \choose i}{n-k \choose t-i}.$$</p>
<p>My questions is whether there is a nice simplification of this expression; I'd like to get rid of the sum if possible. Anyone know?</p>
<p>Thanks a lot for the help!</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
We'll use the integral representation
$\ds{\pars{a \atop b}=\oint_{\verts{z} = R}\frac{\pars{1 + z}^{a}}{z^{b + 1}}\,
\frac{\dd z}{2\pi\ic}\,,\quad R > 0}$.
<hr>
\begin{align}&\color{#66f}{%
\sum_{j = 1}^{t}\pars{k \atop j}\pars{n - k \atop t - j}}
=-\pars{n - k \atop t}
+\sum_{j = 0}^{\infty}\pars{k \atop j}\pars{n - k \atop t - j}
\\[5mm]&=-\pars{n - k \atop t} + \sum_{j = 0}^{\infty}\pars{k \atop j}\
\overbrace{\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n - k}}{z^{t - j + 1}}
\,\frac{\dd z}{2\pi\ic}}^{\dsc{\pars{n - k \atop t - j}}}
\\[5mm]&=-\pars{n - k \atop t}
+\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n - k}}{z^{t+ 1}}\ \overbrace{%
\sum_{j = 0}^{\infty}\pars{k \atop j}z^{j}}^{\dsc{\pars{1 + z}^{k}}}\
\,\frac{\dd z}{2\pi\ic}
=-\pars{n - k \atop t}\ +\ \overbrace{%
\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n}}{z^{t+ 1}}\,\frac{\dd z}{2\pi\ic}}
^{\dsc{\pars{n \atop t}}}
\end{align}
<hr>
Then,
$$
\color{#66f}{%
\sum_{j = 1}^{t}\pars{k \atop j}\pars{n - k \atop t - j}}
=\color{#66f}{\pars{n \atop t} - \pars{n - k \atop t}}
$$</p>
|
813,121 | <p>I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?</p>
| Michael Hardy | 11,667 | <p>Letting $w=1/x$, we have
$$
\lim_{x\downarrow 0}\frac{2-x}{x^3}e^{(x-1)/x^2} = \lim_{w\to+\infty} \left(2 - \frac 1 w \right) w^3 e^{w^2\left(\frac 1 w - 1\right)} = \lim_{w\to+\infty} (2w^3 - w^2) e^{w-w^2}
$$
$$
= \lim_{w\to+\infty} \frac{2w^3-w^2}{e^{w^2-w}}.
$$
L'Hopital should handle that.</p>
<p>Maybe I'll post something on $x\uparrow 0$ later . . .</p>
|
173,745 | <p>I am trying to prove the following: 'If $(X_1,d_1)$ and $(X_2,d_2)$ are separable metric spaces (that is, they have a countable dense subset), then the product metric space $X_1 \times X_2$ is separable.' It seems pretty straightforward, but I would really appreciate it if someone could verify that my proof works.</p>
<p>Since $(X_1,d_1)$ and $(X_2,d_2)$ are separable, they each contain a countable dense subspace, say $D_1 \subset X_1$ and $D_2 \subset X_2$. We will show that $D_1 \times D_2 \subset X_1 \times X_2$ is dense and countable. First, $D_1 \times D_2$ is countable since both $D_1$ and $D_2$ are.</p>
<p>Now let $x=(x_1,x_2) \in X_1 \times X_2$ and let $d$ be the product metric on $X_1 \times X_2$ (given by $d(x,y)=(\displaystyle\sum_{i=1}^2 d_i(x_i,y_i)^2)^{1/2}$). We will show that every open ball $B_d(x,\varepsilon)$ containing $x=(x_1,x_2)$ also contains a distinct point of $D_1 \times D_2$. Let $a_1 \in B_{d_1}(x_1,\frac{\sqrt{2}}{2}\varepsilon)\cap D_1$ and let $a_2 \in B_{d_2}(x_2,\frac{\sqrt{2}}{2}\varepsilon)\cap D_2$ (such points exist because $D_1$ and $D_2$ are dense in $X_1$ and $X_2$, respectively.) Letting $a=(a_1,a_2)$, we then have $d(x,a)=(\displaystyle\sum_{i=1}^2 d_i(x_i,a_i)^2)^{1/2})=(d_1(x_1,a_1)^2 + d_2(x_2,a_2)^2)^{1/2} < ((\frac{\sqrt{2}}{2}\varepsilon)^2 + (\frac{\sqrt{2}}{2}\varepsilon)^2)^{1/2}=\varepsilon$, so we have that $a \in B_d(x,\varepsilon)$, so $D_1 \times D_2$ is dense in $X_1 \times X_2$. Then since $D_1 \times D_2 \subset X_1 \times X_2$ is a countable dense subspace of $X_1 \times X_2$, we have that $X_1 \times X_2$ is separable.</p>
<p>I can see how this would easily generalize to finite products, but does it also extend to countable products?</p>
| Paul | 17,980 | <p>I think it is not the answer for your question. It's better to see it as a comment after answering of Kevin Carlson. However I know there is a <a href="http://planetmath.org/HewittMarczewskiPondiczeryTheorem.html" rel="nofollow">result from Pondiczery, Hewitt and Marczewski</a> that (recently when I read a textbook I found):</p>
<blockquote>
<p>If there are not more than $\mathfrak{c}$ ( which is the continuum), separable topological spaces, then their product is still separable.</p>
</blockquote>
|
2,237,963 | <p>One-point compactification of $S_{\Omega}$ is homeomorphic with $\bar S_{\Omega}$.</p>
<p>Let $X$ be a topological space. Then the One-point compactification of $X$ is a certain compact space $X^*$ together with an open embedding $c : X \to X^*$ such that the complement of $X$ in $X^*$ consists of a single point, typically denoted $\infty$. </p>
<p>Let $X$ be a well-ordered set. Given $\alpha \in X$, let $S_{\alpha}$ denote the set $S_{\alpha} = \{x \mid x \in X \text{ and }x < \alpha \}$.
It is called the section of $X$ by $\alpha$. </p>
<p>I am finding difficulty to do the problem!</p>
| Calvin Lin | 54,563 | <p>Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of <span class="math-container">$\sin$</span> on <span class="math-container">$[0, \pi]$</span> quickly reduces the problem to showing the inequality
<span class="math-container">$$2 \sin 1 > \frac{8}{5} .$$</span></p>
<p>Then, from Maclaurin expansion, we have</p>
<p><span class="math-container">$$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$</span></p>
<p>Observe that the absolute value of each of these terms is decreasing, hence in this alternating sum, we may conclude that</p>
<p><span class="math-container">$$ \sin 1 = 1 - \frac{1}{3!} + \left( \frac{1}{5!} - \frac{1}{7!} \right) + \left( \ldots \right) + \ldots > 1 - \frac{1}{3!} = \frac{5}{6} > \frac{4}{5}. $$</span></p>
<hr>
<p>In particular, we could have strengthened the original inequality to</p>
<p><span class="math-container">$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{5}{3}.$$</span></p>
|
175,576 | <p><a href="https://i.stack.imgur.com/wtjxT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wtjxT.png" alt="enter image description here"></a></p>
<p>This should work? But it is giving me the wrong t values</p>
| kglr | 125 | <pre><code>r[t_] := {6 Cos[ 2.9 Pi t], Cos[3.5 Pi t] + 5 t};
pp = ParametricPlot[r[t], {t, 0, 1} ];
pt = First @ Graphics`Mesh`FindIntersections[pp];
Show[pp, Graphics @ {Red, PointSize[Large], Point @ pt},
PlotLabel -> Style[Row[{"intersection = ", pt}], 16]]
</code></pre>
<p><a href="https://i.stack.imgur.com/xwZ8o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xwZ8o.png" alt="enter image description here"></a></p>
<p>To get the same result using <code>FindRoot</code>, we can use the approach from <a href="https://mathematica.stackexchange.com/a/10288/125">this answer</a> to (1) add the constraints <code>t2 > t1</code> and (2) randomly change the starting values until no error message is issued by <code>FindRoot</code> :</p>
<pre><code>constraints = 0 < t1 < 1 && t2 > t1;
startingvals = {RandomReal[], RandomReal[]};
While[err == Quiet@Check[
sol = FindRoot[1 - Boole[constraints] + Boole[constraints] (r[t2] - r[t1]),
Transpose[{{t1, t2}, startingvals}]], err],
startingvals = {RandomReal[], RandomReal[]};];
r[t1] /. sol
</code></pre>
<blockquote>
<p>{-0.320402, 3.3467}</p>
</blockquote>
<p>To get multiple intersections remove <code>First</code> in definition of <code>pt</code> above:</p>
<pre><code>pp = ParametricPlot[r[t], {t, 0, 10}];
pt = Graphics`Mesh`FindIntersections[pp];
Labeled[Show[pp, Graphics@{Red, PointSize[Large], Point@pt}, AspectRatio -> 1],
Style[Column[{"intersections", MatrixForm@pt}], 16], Left]
</code></pre>
<p><a href="https://i.stack.imgur.com/8LHld.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8LHld.png" alt="enter image description here"></a></p>
|
175,576 | <p><a href="https://i.stack.imgur.com/wtjxT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wtjxT.png" alt="enter image description here"></a></p>
<p>This should work? But it is giving me the wrong t values</p>
| Akku14 | 34,287 | <p>NSolve can do the job, if you use rationalized parameters. Even Solve can do it, gives Root expression.</p>
<pre><code>r[t_] = {6 Cos[29/10 Pi t], Cos[35/10 Pi t] + 5 t};
eq1 = And @@ Thread[r[t1] == r[t2]] && 0 < t1 < t2 < 1
eq2 = eq1 // TrigToExp // Simplify
(* E^(-(29/10) I \[Pi] t2) + E^((29 I \[Pi] t2)/10) ==
E^(-(29/10) I \[Pi] t1) + E^((29 I \[Pi] t1)/10) &&
E^(-(7/2) I \[Pi] t2) + E^((7 I \[Pi] t2)/2) + 10 t2 ==
E^(-(7/2) I \[Pi] t1) + E^((7 I \[Pi] t1)/2) + 10 t1 &&
0 < t1 < t2 < 1 *)
nsol = NSolve[eq2, {t1, t2}, WorkingPrecision -> 30] // Chop //
Simplify[#, t1 \[Element] Reals && t2 \[Element] Reals] &
(* {{t1 -> 0.511377264145361554521949670476,
t2 -> 0.867933080682224652374602053662}} *)
sol = Solve[eq2, {t1, t2}] // Chop //
Simplify[#, t1 \[Element] Reals && t2 \[Element] Reals] &
(* {{t1 -> Root[{-29 (-1 + (-1)^(1/29)) (1 + (-1)^(1/29)) (1 + (-1)^(
2/29)) (1 - (-1)^(1/29) + (-1)^(2/29)) (1 + (-1)^(
1/29) + (-1)^(2/29)) (1 + (-1)^(4/29)) (1 - (-1)^(
2/29) + (-1)^(4/29)) (1 - (-1)^(4/29) + (-1)^(8/29)) -
400 E^((7 I \[Pi] #1)/2) + 29 E^(7 I \[Pi] #1) -
29 E^(7/29 I \[Pi] (-20 + 29 #1)) +
580 E^((7 I \[Pi] #1)/2) #1 &,
0.511377264145361554521949670476156189687360774384 +
0.*10^-49 I}],
t2 -> 40/29 -
Root[{-29 (-1 + (-1)^(1/29)) (1 + (-1)^(1/29)) (1 + (-1)^(
2/29)) (1 - (-1)^(1/29) + (-1)^(2/29)) (1 + (-1)^(
1/29) + (-1)^(2/29)) (1 + (-1)^(4/29)) (1 - (-1)^(
2/29) + (-1)^(4/29)) (1 - (-1)^(4/29) + (-1)^(8/29)) -
400 E^((7 I \[Pi] #1)/2) + 29 E^(7 I \[Pi] #1) -
29 E^(7/29 I \[Pi] (-20 + 29 #1)) +
580 E^((7 I \[Pi] #1)/2) #1 &,
0.511377264145361554521949670476156189687360774384 +
0.*10^-49 I}]}} *)
</code></pre>
|
175,576 | <p><a href="https://i.stack.imgur.com/wtjxT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wtjxT.png" alt="enter image description here"></a></p>
<p>This should work? But it is giving me the wrong t values</p>
| Bob Hanlon | 9,362 | <pre><code>r[t_] := {6 Cos[29 Pi t/10], Cos[7 Pi t/2] + 5 t};
ParametricPlot[r[t], {t, 0, 1},
ColorFunction -> Function[{x, y, t}, ColorData["Rainbow"][t]],
PlotLegends -> BarLegend["Rainbow"]]
</code></pre>
<p><a href="https://i.stack.imgur.com/hRME1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hRME1.png" alt="enter image description here"></a></p>
<p>Using the colors to determine the initial estimates,</p>
<pre><code>sol = FindRoot[{(r[t0][[1]]) == (r[t1][[1]]), (r[t0][[2]]) == (r[
t1][[2]])}, {{t0, 1/2}, {t1, 9/10}}, WorkingPrecision -> 15]
(* {t0 -> 0.511377264145361, t1 -> 0.867933080682225} *)
</code></pre>
<p>Verifying,</p>
<pre><code>r[t0] == r[t1] /. sol
(* True *)
</code></pre>
|
4,320,849 | <p>I had this problem in an exam I recently appeared for:</p>
<blockquote>
<p>Find the range of
<span class="math-container">$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$</span></p>
</blockquote>
<p>By randomly assuming the value of <span class="math-container">$x$</span>, I got the lower range of this expression as <span class="math-container">$3/7$</span>. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.</p>
<p>Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.</p>
<p>Another guy I met outside the exam center, told me he used an approach of <span class="math-container">$x$</span> tending to infinity in both cases and got the maximum value of this expression as <span class="math-container">$1/2$</span>. But before I could ask him to explain more on this method, he had to leave for his work.</p>
<p>So, will someone please throw some light on this method of <span class="math-container">$x$</span> tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).</p>
| Darshan P. | 992,496 | <p><span class="math-container">$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$</span></p>
|
2,441,488 | <p>We are not unfamiliar to imaginary angles, so what can be $\tan(a+ib)$</p>
<blockquote>
<p>If$$\tan(a+ib)=x+iy$$then find $$x,y$$</p>
</blockquote>
<p>My attempt,</p>
<p>Let $x$ and $z$ be two complex bombers such that,
$$\tan(z)=x$$
$$z=\tan^{-1} (x)$$
$$z=\tan^{-1}(re^{i\theta}).$$
But this doesn't seem to get me anywhere, how do we proceed?</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>From the complex definitions of the trigonometric functions,</p>
<p>$$\tan z=\frac2{2i}\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}.$$</p>
<p>When you substitute $z=a+ib$,</p>
<p>$$\tan(a+ib)=\frac1i\frac{e^{-b+ia}-e^{b-ia}}{e^{-b+ia}+e^{b-ia}}=-i\frac{e^{-b+ia}-e^{b-ia}}{e^{-b+ia}+e^{b-ia}}\,\frac{e^{-b-ia}-e^{b+ia}}{e^{-b-ia}+e^{b+ia}}.$$</p>
<p>The denominator is now a real function,</p>
<p>$$e^{-2b}+e^{2ia}+e^{-2ia}+e^{2b}=e^{2b}+e^{-2b}+2\cos(2a).$$ (You can also use the hyperbolic functions and write $2\cosh(2b)+2\cos(2a)$.)</p>
<p>The real and imaginary parts of the numerator easily follow. </p>
|
1,206,528 | <p>Find the matrix $A^{50}$ given</p>
<p>$$A = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}$$ as well as for $$A=\begin{bmatrix} 2 & 0 \\ 2 & 1\end{bmatrix}$$</p>
<p>I was practicing some questions for my exam and I found questions of this form in a previous year's paper.</p>
<p>I don't know how to do such questions.</p>
<p>Please assist over this question.</p>
<p>Thank You</p>
| Community | -1 | <p>An unorthodoxical but quick solution.</p>
<p>Let us compute the powers by recurrence:</p>
<p>$$
\begin{bmatrix} a_{n+1} & b_{n+1} \\ c_{n+1} & d_{n+1} \end{bmatrix}=
\begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} a_n & b_n \\ c_n & d_n \end{bmatrix}
=\begin{bmatrix} 2a_n-c_n & 2b_n-d_n \\ c_n & d_n \end{bmatrix}
,$$
with the initial values
$$\begin{bmatrix} a_0 & b_0 \\ c_0 & d_0 \end{bmatrix}=
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
.$$
Thanks to invariance, we immediately have $c_n=0$ and $d_n=1$. Then solving simple recurrences
$$a_{n+1}=2a_{n}=2^{n+1},$$
$$b_{n+1}=2b_n-1=1-2^{n+1},$$
and
$$\begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix}^{50}=
\begin{bmatrix} 2^{50} & 1-2^{50} \\ 0 & 1 \end{bmatrix}
.$$</p>
|
1,206,980 | <p>Here is the problem: if $v^1$, $v^2$, ..., $v^d$ is an orthonormal basis in $\mathbb{R}^d$, then show that
$$
||A - A\sum_{i = 1}^k v^i(v^i)^T ||^2_F = \sum_{i = k+1}^d||Av^i ||_2^2.
$$</p>
<p>I am having trouble understanding this step of a longer proof. The text that I got this from just says that this holds because $v^1$, $v^2$, ..., $v^d$ is an orthonormal basis of $\mathbb{R}^d$. </p>
| Chappers | 221,811 | <p>Interchanging the order of integration allows us to write $T$ as a single integral:
$$ T(p(x))(t) = \int_0^t \int_0^y p(x) \, dx \, dy = \int_0^t p(x) \int_x^t \, dy \, dx = \int_0^t (t-x)p(x) \, dx $$
(the region of integration is $0<x<y<t$, which tells you what the new limits should be)
Now, given $p(x)=x^n$, we find
$$ T(x^n)(t) = \int_0^t (t-x)x^n \, dx = \left[ t\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^t = t^{n+2} \left( \frac{1}{n+1}-\frac{1}{n+2} \right) = \frac{t^{n+2}}{(n+1)(n+2)}, $$
so you get
$$
\left(
\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & 0 \\
0 & \frac{1}{6} & 0 & 0 \\
0 & 0 & \frac{1}{12} & 0 \\
0 & 0 & 0 & \frac{1}{20} \\
\end{array}
\right),
$$
I suppose.</p>
|
2,815,900 | <p>Consider the proposition $∀x (P (x) → Q (x))$, where $P (x)$ and $Q (x)$ are predicates on a domain $U$. Prove or disprove the following statement, justifying the answer.</p>
<p>If the statement $∀x (P (x) → Q (x))$ is FALSE then it's FALSE also $∀x P (x)$</p>
<p>How can I solve this exercise?</p>
| Prajwal Kansakar | 49,781 | <p>Consider the statement, </p>
<p>For all $x$, if $x^2\geq 0$, then $x\geq 0$.</p>
<p>with the domain consisting of real numbers. This is false. What about </p>
<p>For all $x$, $x^2\geq 0$?</p>
|
1,721,055 | <p>This afternoon I've been studying the pythagorean identities & compound angles. I've got a problem with a question working with 2 sets of compound angles:</p>
<blockquote>
<p>Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$ </p>
</blockquote>
<p>I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding? </p>
<p>$$\cos\theta\cos 25^\circ-\sin\theta \sin 25^\circ+\sin\theta\cos 65^\circ+\sin 65^\circ\cos\theta=1$$</p>
<p>$$\cos \theta\;\left(\cos 25^\circ+\sin 65^\circ\right) +\sin\theta\;\left(\cos 65^\circ-\sin 25^\circ\right)=1$$</p>
<p>Could you tell me if I've made a mistake or how I could continue;
thanks</p>
<p><em>coffee is wearing out</em></p>
| Doug M | 317,162 | <p>$\cos(\theta + 25) + \sin(\theta+65) = 1$</p>
<p>$\cos(\theta + 25) + \cos(25-\theta) = 1$</p>
<p>$\cos(\theta + 25) + \cos(\theta-25) = 1$</p>
<p>$2\cos\theta \cos(25) = 1$</p>
<p>$\cos \theta = (1/2) \sec 25$</p>
<p>$\theta$ ~ 57 degrees, 303 degrees</p>
|
1,571,083 | <p>Given the 2 terms
$$ \frac{k + a}{k + b}$$
and $$\frac{a}{b}$$
with $a, b, k \in \mathbb{R^+}$ and $a > b$</p>
<p>I want to show, that the first term is always bigger than the second one.</p>
<p><strong>My try</strong>
$$
\frac{k + a}{k + b} > \frac{a}{b} \\
\frac{k + b + (a - b)}{k + b} > \frac{b + (a - b)}{b} \\
1 + \frac{a - b}{k + b}> 1 + \frac{a - b}{b} \\
\frac{1}{k + b} > \frac{1}{b} \\
k + b > b
$$
the problem is, that I think the inequality sign has to change to $<$ in the last step because of calculating the inverse on both sides. That would mean
$$ k + b < b$$
which is nonsense.</p>
| fredq | 297,080 | <h1>What my answer is about</h1>
<ul>
<li>I give the correct calculation to answer the question.</li>
<li>I point out that maybe the reason the event seems unlikely is because of the way we look at it.</li>
<li>I do some more calculation to show that similar events may be in fact common.</li>
</ul>
<h2>The correct calculation is the following :</h2>
<p>For the first drawing, the probability of winning is the probability of winning the first draw, plus the probability of losing the first draw and winning the second: </p>
<p>$$p_1 = \frac{1}{31} + \frac{30}{31}\times\frac{1}{30} \approx 0.0645$$</p>
<p>For the second drawing, the probability is strightfoward to compute : </p>
<p>$$p_2 = \frac{1}{25} = 0.04$$</p>
<p>The total probability is the product of individual probabilities : </p>
<p>$$P \approx 0.06*0.04 \approx 0.026 = 2.6\%$$</p>
<p>This seems low. </p>
<h2>However</h2>
<p>When doing probabilities, you must be careful not to introduce bias! In your example, you include only the two drawings he won, but have not considered all the drawings that take place! </p>
<p>I will now show you that the event you consider suspicious might not be. </p>
<p>From your question, I will consider the following statement to be what you find suspicious :</p>
<p><strong>"Mr. x won twice on 3 consecutive draws"</strong></p>
<p>Now, say you have a drawing each week, with on average 30 people in each. Mr x participates every time.</p>
<p>The probability of him winning 2 times in 3 draws are : </p>
<p>$$p = 3\times (\frac{1}{30})^2 = \frac{1}{300}$$</p>
<p>The probability of this event happening in 52 weeks is the following : </p>
<p>$$P = 1- (1- \frac{1}{300})^{50} \approx 0.17 = 17\%$$</p>
<p>The term in parenthesis is the probability of not winning twice in any 3-week period, and there is 50 3-week periods in a year.</p>
<p>Those odds are a little bit less suspicious.</p>
<h2>Furthermore</h2>
<p>Now, if you are interested in the probability that <em>anyone</em> at some point in the year will win 2 times in 3 draws, the calculation is : </p>
<p>$$ P = 1- (1-0.17)^{30} \approx 0.99 = 99\%$$</p>
<p>(we consider 30 persons playing each week for a year)</p>
<p>It is very likely that someone, at some point, will win twice out of 3 consecutive draws.</p>
<p>Here, I made some assumptions, but the idea is right.</p>
|
1,887,395 | <p>Let $D=\{Z\in\mathbb{C}:|z|<1\}$. Suppose $f:\overline D\to\mathbb{C}$ is continuous, holomorphic in $D$, and $f(\partial D)\subset \mathbb{R}$. Define $h:\mathbb{C}\to\mathbb{C}$ by
$$
h(z)=\begin{cases}
f(z) &\text{if}\ z\in \overline D\\
\overline{f(1/\overline z)} &\text{if}\ z\in\mathbb{C}\setminus \overline{D}
\end{cases}$$
Show that $h$ is entire and $f$ is a constant function.</p>
<p>My work: I can show that $h$ is holomorphic in $\mathbb{C}\setminus \overline{D}$. </p>
<p>To show $h$ is entire, by using Morera's theorem, considering arbitrary triangular path, the integral of $h$ is zero if it does not cross the boundary of the unit disc, however, what if the triangular path cross the unit disc? How to show in this case the integral is also zero.</p>
<p>Could anyone kindly help? Thanks!</p>
| Zubzub | 349,735 | <p>Some ideas here :
Write it as
$$
\sum_{n=0}^\infty \sum_{i=1}^n \frac{1}{i\ n} \ \ - \sum_{n=0}^\infty \sum_{i=1}^n \frac{1}{i(n+k)}
$$
Now let's explicitly write some terms of the positive sum and of the negative sum in a triangle-shape:
$$
\frac{1}{1}\frac{1}{1} + \\
\frac{1}{1}\frac{1}{2} + \frac{1}{2} \frac{1}{2} + \\
\frac{1}{1} \frac{1}{3} + \frac{1}{2} \frac{1}{3} + \frac{1}{3} \frac{1}{3} + \\
...
$$
$$
- ( \\
\frac{1}{1}\frac{1}{1+k} + \\
\frac{1}{1}\frac{1}{2+k} + \frac{1}{2} \frac{1}{2+k} + \\
\frac{1}{1} \frac{1}{3+k} + \frac{1}{2} \frac{1}{3+k} + \frac{1}{3+k} \frac{1}{3} + \\
... )
$$</p>
<p>Let's call the $j$th diagonal of such a triangle the "line" that starts from the $j$'th row at the most left term and goes down "diagonally" to the bottom right corner. For example the $1$-diagonal starts at the top of the triangle and goes through all the terms on the right side.</p>
<p>We see that the sum of these two triangles will cancel leaving only the $k$'th first diagonals (that is all the term in diagonal $1,2,...,k$) of the "positive triangle" :
$$
\sum_{i=1}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)}
$$
Where the outer sum selects the diagonal and the inner sum goes through all the elements. In particular we can extract the first sum where $i=1$ like so :
$$
\sum_{j=1}^\infty \frac{1}{j^2} + \sum_{i=2}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)} = \frac{\pi^2}{6} + \sum_{i=2}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)}
$$
My mathematical knowledge stops me here because Mathematica yields a too complicated expression for the rest of the sum. Hope it helps a bit :)</p>
|
4,295,652 | <p>In the integration formula <span class="math-container">$\int dx/x = log x + c$</span>, Is the log natural or log base 10? The formula appears in many problems and i just got a problem wrong for apparently using the wrong log. Could you please enlighten me about the right log to be used in integration.</p>
| Angel | 109,318 | <p>Consider the differential equation <span class="math-container">$y'=y$</span> with <span class="math-container">$y(0)=1$</span> on <span class="math-container">$\mathbb{R}$</span>. It can be proven that this has a unique solution. We <em>define</em> the exponential function on <span class="math-container">$\mathbb{R}$</span> to be equal to this solution. We denote it as <span class="math-container">$\exp$</span>, but many people abuse notation and denote it <span class="math-container">$e^x$</span>, which in this context, is fine, as long as everyone knows what is being meant. Now, it can be proven that <span class="math-container">$\exp$</span> is function whose range is <span class="math-container">$\mathbb{R}^+$</span>. In other words, <span class="math-container">$\exp[\mathbb{R}]=\mathbb{R}^+$</span>. It can also be proven that <span class="math-container">$\exp$</span> is a function that has a compositional inverse. This compositional inverse has a name: the natural logarithm, and we denote it <span class="math-container">$\ln$</span>, or <span class="math-container">$\log$</span>. Of course, one must remember that <span class="math-container">$\ln[\mathbb{R}^+]=\mathbb{R}$</span>. The relationship that this function has with other logarithmic functions is complicated, but in some sense, it is <strong>the</strong> logarithmic function, due to the fact that it can be defined independently of any notion of powers. Hence the name "natural" logarithm. In terms of other logarithmic functions, the natural logarithm can be interpreted as being base <span class="math-container">$e$</span>, where <span class="math-container">$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$</span>.</p>
<p>Now, given the equation <span class="math-container">$\exp'=\exp$</span>, it can be proven via the chain rule that <span class="math-container">$\ln'(x)=\frac{1}{x}$</span> for every <span class="math-container">$x\in\mathbb{R}^+$</span>. This is because <span class="math-container">$(\exp\circ\ln)'(x)=1$</span>, yet by the chain rule, <span class="math-container">$(\exp\circ\ln)'(x)=(\exp'\circ\ln)(x)\ln'(x)=(\exp\circ\ln)(x)\ln'(x)=x\ln'(x)$</span>, so <span class="math-container">$x\ln'(x)=1$</span>, implying <span class="math-container">$\ln'(x)=\frac{1}{x}$</span>. Because of this, <span class="math-container">$\ln(x)$</span> is an antiderivative of <span class="math-container">$\frac{1}{x}$</span> on <span class="math-container">$\mathbb{R}^+$</span>.</p>
<p>However, this is not the complete story. Consider the function <span class="math-container">$\ln^-$</span> defined by <span class="math-container">$(\ln^-)(x)=\ln(-x)$</span> for every <span class="math-container">$\mathbb{R}^-$</span>. It turns out that <span class="math-container">$(\ln^-)'(x)=\frac{1}{x}$</span> is true on <span class="math-container">$\mathbb{R}^-$</span> as well! So to talk about the antiderivatives of <span class="math-container">$\frac{1}{x}$</span> properly, we need to consider functions <span class="math-container">$f_{A,B}:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$</span> such that <span class="math-container">$$f_{A,B}(x)=\begin{cases} \ln(-x)+A & x\lt0 \\ \ln(x)+B & x\gt0 \end{cases}.$$</span> These functions satisfy the property that <span class="math-container">$f_{A,B}'(x)=\frac{1}{x}$</span> for every <span class="math-container">$x\in\mathbb{R}\setminus\{0\}$</span>. And in fact, this is every function that satisfies the equation on that domain, there are no other functions. So you can say <span class="math-container">$$\int\,\frac{\mathrm{d}x}{x}=f_{A,B}(x).$$</span> If you want to limit yourself to <span class="math-container">$x\gt0$</span>, though, then <span class="math-container">$$\int\,\frac{\mathrm{d}x}{x}=\ln(x)+B.$$</span> Hopefully, this covers everything.</p>
|
1,429,000 | <p>Prove that for any integer $n$, the integer $n^2 + 7n + 1$ is odd.</p>
<p>I have $n=2k+1$ for some $k\in Z$</p>
<p>I really do not how to do this problem. any help in understanding would be greatly appreciated.</p>
| Robert Soupe | 149,436 | <p>There are only two scenarios here: $n$ is odd or $n$ is even. That alone is enough to determine whether $n^2 + 7n + 1$ is odd or even.</p>
<ul>
<li>If $n$ is odd, then so is $n^2$, as well as $7n$. Then $n^2 + 7n$ is even, because two odd numbers add up to an even number. But then there's the $+ 1$, which makes the number odd again.</li>
<li>If $n$ is even, then so is $n^2$, as well as $7n$. Two even numbers also add up to an even number. But again we have the $+ 1$, which then makes the number odd.</li>
</ul>
<p>Try it with a few specific odd and even values of $n$. I'll do two for you:</p>
<ul>
<li>$n = 1$, then we have $1 + 7 + 1 = 9$, which is odd.</li>
<li>$n = 2$, then we have $4 + 14 + 1 = 19$, which is also odd.</li>
</ul>
|
2,349,312 | <blockquote>
<p>Let $F$ be a subfield of the complex numbers. Let $A,B\in F^{n\times n}$ and $X\in F^{n\times 1}.$ If $A\ne B,$ then $AX\ne BX$ for every non-zero $X.$</p>
</blockquote>
<p>I am NOT sure if the above statement is right. I did some computation with random matrices and it seems true to me. This is the argument I tried :</p>
<p>Let $A=[A_1,A_2,\dots,A_n]$ and $B=[B_1,B_2,\dots,B_n]$ where $A_i,B_i \in F^{n\times 1}$ are columns of $A$ and $B$ respectively. Let $X=\begin{bmatrix}x_1\\\vdots \\x_n\end{bmatrix}.$</p>
<p>Since $A\ne B,$ choose $j$ such that $A_j\ne B_j.$ Let $J_1$ be the set of all such $j$'s and $J_2=I$ \ $ J_1$ where $I=\{1,2,\dots ,n\}.$ Then, $AX=\sum_{i\in J_1} x_iA_i +\sum_{i\in J_2} x_iA_i \ne \sum_{i\in J_1} x_iB_i + \sum_{i\in J_2} x_iB_i= BX.$</p>
<p><strong>(1) Is my argument correct?<br>(2) Also, can $F$ be generalized?</strong> </p>
| Wuestenfux | 417,848 | <p>In an integral domain $R$, we have $ax=bx\Rightarrow a=b$ for all $a,b,x\in R$ with $x\ne 0$. To see this, write $(a-b)x=0$. Since $R$ has no zero divisors and $x\ne 0$ we must have $a-b=0$, i.e., $a=b$. But an $n\times n$ matrix ring is not an integral domain for $n\geq 2$.</p>
|
762,964 | <blockquote>
<p>Let <span class="math-container">$X$</span> be from a normal distribution <span class="math-container">$N(\theta,1)$</span>.</p>
<p>a) Find a sufficient statistic for <span class="math-container">$\theta$</span>.<br />
b) Is <span class="math-container">$S_n^2$</span> a
sufficient statistic for <span class="math-container">$\theta$</span>?</p>
</blockquote>
<p><strong>My answers</strong></p>
<p>For part a)</p>
<p>Since the joint p.d.f is <span class="math-container">$1 \over (2\pi)^{n/2}$$e^{{-1 \over 2}\sum(x_i-\theta)^2}$</span> I can say that <span class="math-container">$\sum X_i$</span> is a sufficient statistic for <span class="math-container">$\theta$</span> because <span class="math-container">$e^{{-1 \over 2}\sum(x_i-\theta)^2}$</span> depends on X only through the values of <span class="math-container">$\sum X_i$</span> right? Because if I know the value of <span class="math-container">$\sum X_i$</span> then I know <span class="math-container">$\sum X_i^2$</span> as well.</p>
<p>For part b)</p>
<p>Expanding the joint p.d.f as <span class="math-container">$$\frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i- \bar x + \bar x-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[\sum(x_i- \bar x)^2+n(\bar x-\theta)^2\Big]} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[{\sum(x_i- \bar x)^2 \over n-1}n-1+n(\bar x-\theta)^2\Big]}$$</span></p>
<p>Now can I say <span class="math-container">$S_n^2$</span> is a sufficient statistic for <span class="math-container">$\theta$</span> . Is it a problem that I have <span class="math-container">$\bar x$</span> in the function <span class="math-container">$g(S_n^2,\theta)$</span>?.<br />
Because <span class="math-container">$\bar x$</span> is a particular value I thought <span class="math-container">$g(S_n^2,\theta)$</span> depends on <span class="math-container">$\theta $</span> only through the values of <span class="math-container">$S_n^2$</span>.</p>
| Michael Hardy | 11,667 | <p>$$
\sum_{i=1}^n (x_i - \theta)^2 = \left( \sum_{i=1}^n x_i^2 \right) -2\theta \left( \sum_{i=1}^n x_i \right) + n\theta^2
$$
Therefore
$$
\exp \left( \frac {-1} 2 \sum_{i=1}^n (x_i-\theta)^2 \right) = \underbrace{ e^{-n\theta^2/2}\cdot \exp\left( \theta\sum_{i=1}^n x_i \right)}_{\large\text{first factor}} \cdot \underbrace{ \exp\left( \frac{-1} 2\sum_{i=1}^n x_i^2 \right) }_{ \large \text{second factor}}
$$</p>
<p>The first factor depends on $(x_1,\ldots,x_n)$ only through $\displaystyle\sum_{i=1}^n x_i.$ The second factor does not depend on $\theta.$</p>
<p>Therefore by Fisher's factorization theorem, $\displaystyle\sum_{i=1}^n x_i$ is sufficient for $\theta.$</p>
<p>(As your question now stands, it says "known mean", but "$N(\theta,1)$" means the mean is unknown and the variance is known.)</p>
|
531,895 | <p>Assume K is a field that is not algebraically closed, then how to prove that for any m>0, there is a polynomial with m variables over K that possess a unique zero point?</p>
| Slade | 33,433 | <p>Take any $\alpha \in \overline{K}\setminus K$, let $m_\alpha$ be its minimal polynomial, and let $n$ be its degree.</p>
<p>For $m=2$, let $f(X,Y) = Y^n m_\alpha (X/Y)$. Then $f(X,Y)\in K[X,Y]$ and $f$ has only $(0,0)$ as a root in $K^2$.</p>
<p>For $m=3$, write $g(X,Y,Z) = f(f(X,Y),Z)$. If $g(a,b,c)=0$ for $a,b,c\in K$, then $c=0$ and $f(a,b)=0$, so $a=b=0$.</p>
<p>And so on.</p>
|
1,283,541 | <p>Find the least value of $f(x)=3^{-x+1} + e^{-x-1}$.</p>
<p>I tried to use the maxima/minima concept but it was of no use. Please help.</p>
| Giuseppe Negro | 8,157 | <p>I don't know if it is "algebraic", but you might start from the observation that the function
$$
\theta\in \mathbb{R}\mapsto e^{i\theta}\in \mathbb{C}
$$
is Lipschitz with constant $1$. This can be proved via the fundamental theorem of calculus:
$$
\left\lvert e^{i\theta+h}-e^{i\theta}\right\rvert=\left\lvert e^{i\theta}\int_0^h ie^{i\, h'}\, dh'\right\rvert\le\lvert h\rvert.$$
So
$$
\left\lvert\frac{e^{i\theta}-1}{\theta}\right\rvert^2\le 1, $$
and applying Euler's formula one obtains
$$
\frac{(\cos \theta -1)^2}{\theta^2}+\frac{\sin^2\theta}{\theta^2}\le 1.
$$
In particular,
$$
\left\lvert \frac{\sin \theta}{\theta}\right\rvert\le 1,\qquad\left\lvert\frac{1-\cos\theta}{\theta}\right\rvert\le 1$$ for all $\theta\in \mathbb{R}$. </p>
|
1,080,572 | <p>I am having problem with this question , kindly please help me with this ,</p>
<p>Let $$S = \{x : x^{6} -x^{5} \leq 100\}$$</p>
<p>And $$T =\{x^{2} - 2x : x \in ( 0, \infty)\}$$</p>
<p>Then I have to show that set $S \cap T$ is closed and bounded. $S$ appears to be closed and bounded as it less than or equal to $100$.</p>
| drhab | 75,923 | <p>Hint on being closed:</p>
<p>$T$ can also be written as: $\{x^2-2x:x\in[0,\infty)\}$ since $0=2^2-2.2\in T$</p>
<p>So both sets $S$ and $T$ are preimages of closed sets under continuous functions.</p>
<p>Hint on being bounded.</p>
<p>If a set is bounded then so are its subsets.</p>
|
2,345,142 | <p>Can you use Skolemization to reduce a formula to the variables you want it to be about? I was trying to think of a nice algorithmic way to do it but only ended up having problems. </p>
<p>Say you have a formula $G$ in prenex normal form, i.e. something looking like this:</p>
<p>$$G \equiv Q_1x_1...Q_nx_n\;F$$</p>
<p>where $Q_i$ are quantifiers binding variables $x_i$ and occuring exclusively on the left of formula $F$ (which, consequently does not contain quantifiers itself). It is known that $G$ can be transformed into Skolem normal form, i. e. all existential quantifiers can be eliminated by substituting the variables they are binding by functions. But can <i>any</i> variable be substituted so that in the end the formula expresses something about the variables you wish to keep?</p>
<p>The possibly cheapest way to do it involves bracketing $F$ with its nearest quantifier. </p>
<p><em>[EDIT: as you can read in the comments, this manoeuvre involves a major flaw, which shall stay undeleted for the record.]</em></p>
<p>If the quantifier nearest to $F$ is $\exists$, it would go like this:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;\exists x_n\;F$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\exists x_n\;F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(F[x_n/c_1])$$</p>
<p>...where $F[x_n/c_1]$ expresses that $x_n$ has been substituted by constant $c_1$ in $F$.</p>
<p>If the quantifier nearest to $F$ is $\forall$, the procedure takes some extra legwork and two negations:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall x_n F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\neg \exists x_n\;\neg F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;\neg (\exists x_n\;\neg F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\neg F[x_n/c_1])$$</p>
<p>Iterating the process for all bound variables would leave you with a formula with no quantifiers but many constants you cannot assume much about.</p>
<p>To sustain "expessive power" regarding the variables designated to stay, there ought to be a way to substitute the rest of the variables by functions with arities higher than 0. I was thinking of (if required, repeatedly) putting the formula in the following form:</p>
<p>$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall y_{k} ... \forall y_n\;\exists x_n F)$$
$$Q_1x_1...Q_{n-1}x_{n-1}\;(\forall y_{k} ... \forall y_n\;F[x_n/f(y_{k},..., y_n)])$$</p>
<p>...where $y_i$ shall be the variables preserved in the final form and $x_i$ the variables to be eliminated.</p>
<p>Two problems arise:</p>
<p>1) As any quantifiers can bind $y_i$ in the original prenex normal form, the existential quantifiers binding $y_i$ must first be rephrased into universal ones for the Skolemization to yield functions of $y_i$. This does not always lead to a chain of universal quantifiers with no negations in between. It seems like it's not always possible to get rid of the negation in $\forall y_i \; \neg \forall y_{i-1}$.</p>
<p>2) Consequently, opportunities to rephrase $Q_ix_i\; Q_{i+1}x_{i+1}$ as $Q_{i+1}x_{i+1}\;Q_ix_i$ are limited.</p>
<p>So far, I managed to fabricate formulas containing only the variables I picked as survivors in the initial stage. Anyway, I could not avoid constants in every case. Can you think of a procedure with rules that would prevent constants appearing in any resulting formula?</p>
| Bram28 | 256,001 | <p>Unless I misunderstand you, this doesn't work.</p>
<p>Say you have $\forall x \exists y Loves(x,y)$ : Everyone loves someone.</p>
<p>By your method you get (again, if I understand you correctly):</p>
<p>$\forall x Loves(x,c_1)$</p>
<p>But that can't be right, for that would imply that there is something $c_1$ that everyone loves. But if I have $a$ and $b$ (and not other objects in my domain) and $a$ loves $b$, $b$ loves $a$, but $a$ and $b$ do not love themselves, then $\forall x \exists y Loves(x,y)$ is true but $\forall x Loves(x,c_1)$ is false: $c_1$ cannot be $a$, and also not be $b$, and there are no other objects.</p>
<p>This is exactly why you can only eliminate existentials by using a function that shows that the 'something' is dependent on the choice of objects for the universal quantifiers in front of it: if I pick $a$, I can point to $b$ as the something that $a$ loves, but if I pick $b$, I point to $a$. So you have to think about this process 'outside-in', rather than 'inside-out'.</p>
|
121,631 | <p>How to prove most simply that if a polyonmial $f$, has only real coefficients and $f(c)=0$, and $k$ is the complex conjugate of $c$, then $f(k)=0$?</p>
| Jacob Salomon | 71,482 | <p>I've long forgotten the details so maybe someone else can complete this. I also don't see how folks have formatted the conjugate with the overhead bar so I will use ~ for conjugate.</p>
<p>When I took complex variable, the proof was along the lines of:</p>
<p>Suppose z1 is indeed a zero of f(z).
Assume ~z1 is <strong>not</strong> a zero of f(z).
Then 1/(fz) does not have a singularity at ~z1 i.e. 1/f(z1) is a permitted operation.</p>
<p>And that's where my memory fails me: Somehow, the professor (and the book) show that this leads to a contradiction.</p>
<p>Just a starter..</p>
<p>-- JS</p>
|
282,780 | <p>Let $R = k[x_1, \ldots, x_n]$ for $k$ a field of characteristic zero and let $S \subset R$ be a graded sub-$k$-algebra (for the standard grading: $\deg x_i = 1$) such that $R$ is a free $S$-module of finite rank. Does this imply $S \cong k[y_1,\ldots,y_n]$?</p>
| Neil Strickland | 10,366 | <p>Note that $R$ and $S$ each have only one graded maximal ideal, so they are local in the graded sense, so most of the standard results for ungraded local rings are applicable. There is an obvious finite resolution of $k$ by modules that are finitely generated and free over $R$ and thus also over $S$. This implies that $S$ has finite global dimension, and so is a regular local ring by a theorem of Serre (Theorem 19.2 in Matsumura). Together with the grading assumptions this forces $S$ to be polynomial.</p>
|
292,780 | <p>I have to realize an NFA that recognizes the language of strings on the alphabet {a, b} ending with: bb, ba, baa. I thought that there must be the following states:</p>
<p>$q_0$: the string ends with bb.</p>
<p>$q_1$: the string ends with ba.</p>
<p>$q_2$: the string ends with baa.</p>
<p>Is right the definition of the states? Or missing some state?</p>
<p>Thanks.</p>
| Rick Decker | 36,993 | <p>As you seem to understand, the states in a FA can be designed to correspond to having seen a particular form of input. One solution (not with the minimal number of states, but easier to understand) uses five states:</p>
<ul>
<li>$q_0$: Processes any input and takes special action if a $b$ is seen. (start state)</li>
<li>$q_1$: Just saw input $\dots b$</li>
<li>$q_2$: Just saw input $\dots ba$ (final state)</li>
<li>$q_3$: just saw input $\dots bb$ (final state)</li>
<li>$q_4$: Just saw input $\dots baa$ (final state)</li>
</ul>
<p>Since we're looking for $ba, bb, baa$ at the end of the input, we'll have no moves from $q_3$ and $q_4$ and only an $a$ transition from $q_2$, to handle the possibility that the input ends after having seen $\dots baa$.</p>
<p>The transition table for this NFA is
$$
\begin{array}{ccc}
& \mathbf{a} & \mathbf{b}\\
q_0 & \{q_0\} & \{q_0, q_1\}\\
q_1 & \{q_2\} & \{q_3\}\\
q_2 & \{q_4\} & \varnothing\\
q_3 & \varnothing & \varnothing\\
q_4 & \varnothing & \varnothing
\end{array}
$$</p>
<p>You might want to try designing a NFA for this language with four, rather than five states.</p>
|
136,570 | <p>Good day everyone. I was reading the more advanced lectures on complex analysis and encountered a lot of questions, concerning the determination of complex logarithm. As far I don't even understand the concept of it, but I'll do provide you with several practical questions concerning the topic. </p>
<p>First of all, the determination of complex logarithm on open $\Omega \subset \mathbb{C}$ is continuous $f(w)$ such that:</p>
<p>$$\forall w \in \Omega \text{ }\exp(f(w))=w$$</p>
<p>So then it starts. Maybe someone could explain what presumptions does the following statement contradicts.</p>
<p>There is no continuous determination for complex logarithm in $\mathbb{C}$ \ ${0}$. </p>
<p>Second part is more practical. </p>
<p>We say that determination of complex logarithm is called principal if it given as a complement of $\mathbb{C}$ and semi axis of negative or zero reals $\Omega_{\pi}=\mathbb{C}$ \ $\{z\in \mathbb{C} : \Re(z) \leq 0\}$. Such as
$$ f(z) = \log(|z|)+ i\begin{cases} \arcsin{(y/|z|)} & x \geq 0, \\ \pi - \arcsin{(y/|z|)} & x \leq 0,\, y \geq 0, \\ -\pi - \arcsin{(y/|z|)} & x \leq 0,\, y \leq 0. \end{cases}$$</p>
<p>We can see that the argument belongs to $(-\pi,\pi]$, but I don't understand neither why a set without a negative reals define such argument neither how does it happens. After this there is an example saying that if we take $\Omega_0=\mathbb{C}$ \ $\{z \in \mathbb{C} : \Re \geq 0\}$ then the argument will be $(0,2\pi]$ but I also didn't get how this happens. What will happen if we take out non positive imaginary part? What kind of argument we will have then?</p>
<p>There is no such topic explained in wiki so maybe the deep answer will help others who encountered the same problem.</p>
| Dan | 1,374 | <p>By inverting <a href="http://en.wikipedia.org/wiki/Euler%27s_formula" rel="nofollow">Euler's formula</a> $e^{ix} = \cos{x} + i \sin{x}$, you get</p>
<p>$\log(r(\cos{\theta} + i \sin{\theta}))$</p>
<p>$ = \log{r} + \log(\cos{\theta} + i \sin{\theta})$</p>
<p>$ = \log{r} + i \theta$</p>
<p>The problem here is that $\theta$, the argument or phase angle of a complex number, is not unique, due to sin and cos being periodic. For example, the following are all valid polar representations of $-1-i$:</p>
<ul>
<li>$\sqrt{2} \text{ cis } \frac{-3\pi}{4}$</li>
<li>$\sqrt{2} \text{ cis } \frac{5\pi}{4}$</li>
<li>$\sqrt{2} \text{ cis } \frac{-11\pi}{4}$</li>
</ul>
<p>So, how do you decide whether $\log (-1-i)$ is $\frac{\log{2}}{2} - \frac{3\pi}{4} i$, $\frac{\log{2}}{2} + \frac{5\pi}{4} i$, or $\frac{\log{2}}{2} - \frac{11\pi}{4} i$? By arbitrary picking a $2\pi$-wide interval to which $\theta$ is restricted.</p>
<ul>
<li>If you choose $\theta \in (-\pi, \pi]$, then $\log (-1-i) = \frac{\log{2}}{2} - \frac{3\pi}{4} i$.</li>
<li>If you choose $\theta \in [0, 2\pi)$, then $\log (-1-i) = \frac{\log{2}}{2} + \frac{5\pi}{4} i$.</li>
</ul>
<p>Whichever "branch cut" you choose, there will be a discontinuity across it. For example, using $\theta \in (-\pi, \pi]$, then if $x$ is a negative real number,</p>
<ul>
<li>$\lim_{y \rightarrow 0^+} (x + yi) = \log{|x|} + \pi i$</li>
<li>$\lim_{y \rightarrow 0^-} (x + yi) = \log{|x|} - \pi i$</li>
</ul>
|
1,610,800 | <p>When dividing $f(x)$ by $g(x)$: $f(x)=g(x)Q(x)+R(x)$.
How to find the quotient $Q(x)$ and the remainder $R(x)$?
For example: $f(x)=\ 2x^4+13x^3+18x^2+x-4 \ $
, $g(x)=\ x^2+5x+2 \ $</p>
<blockquote>
<p>At first $g(x)= (x+4.56)(x+0.44)$ then we use synthetic division : $Q(x)=
\ 2x^2+3x-1 \ $ , $R(x)= \ 0.04x-1.98 \ $</p>
</blockquote>
<p>how to show that $f(x)$ and $g(x)$ have no common roots?</p>
<blockquote>
<p><strong>I was wrong</strong> when I used synthetic division rather than long division
so that, $Q(x)= \ 2x^2+3x-1 \ $ , $R(x)=-2 \ $</p>
</blockquote>
<p>$f(x)= (2x^2+3x-1) g(x)\ $ $- 2 \ $ </p>
<p>let $x=α\ $ is a root for $f(x)=0\ $ and $g(x)=0\ $
and substitute in the equation which proves α does not exist so they have no common roots.</p>
<p>Is that true? </p>
| Alex | 12,133 | <p>Well
$$(x + 4.56)(x + 0.44) = x^2 + 5x + 2.0064 \neq x^2 + 5x + 2 = g(x), $$
so no, anything using that factorization couldn't tell you anything about $g(x)$. Your $Q(x)$ looks OK, but your $R(x)$ certainly doesn't satisfy $f(x) = Q(x)g(x) + R(x)$. Now, once you use $g(x)$ to get the correct $R(x)$ you will have an expression of the form
$$ R(x) = f(x) - Q(x)g(x) $$
and so if some factor divides both $f(x)$ and $g(x)$, what can you say about $R(x)$? Can you see why this shows $f$ and $g$ have no common roots?</p>
|
2,994,440 | <p><strong>Caution</strong>: Axiom of Completeness is not assumed here.</p>
<p><strong>Before reading my attempt, I want you to think up a proof of your own.</strong></p>
<p>Here is my attempt:</p>
<blockquote class="spoiler">
<p> Let <span class="math-container">$D$</span> be the set of all upper bounds of <span class="math-container">$A$</span> as well as <span class="math-container">$B$</span>. Since <span class="math-container">$\sup A$</span> exists, <span class="math-container">$\sup A \le d$</span> for all <span class="math-container">$d \in D$</span>. It also means <span class="math-container">$D$</span> is bounded below.
Now I have to show that <span class="math-container">$\inf D$</span> exists and is equal to <span class="math-container">$\sup A$</span>. (This is the part that I have no idea how to approach.)
Then since <span class="math-container">$\inf D$</span> exists, <span class="math-container">$\sup B$</span> exists and is equal to <span class="math-container">$\inf D$</span>. Therefore, <span class="math-container">$\sup A = \sup B$</span>.</p>
</blockquote>
<p>This concludes my proof. Two questions:</p>
<ol>
<li>How do I show the part in my proof where I was stuck?</li>
<li>Is there any other way to prove it? Preferably much simpler. I want to see some other ways.</li>
</ol>
| Misha Lavrov | 383,078 | <p>You need to at some point invoke the definition of <span class="math-container">$\sup$</span> (or something that follows from it), because otherwise you're just talking about a symbol you know nothing about.</p>
<p>One definition of <span class="math-container">$\sup A$</span> is a real number <span class="math-container">$M$</span> such that:</p>
<ul>
<li>For every <span class="math-container">$a \in A$</span>, <span class="math-container">$a \le M$</span>.</li>
<li>For every <span class="math-container">$\epsilon>0$</span>, there is an <span class="math-container">$a \in A$</span> such that <span class="math-container">$a > M-\epsilon$</span>.</li>
</ul>
<p>So you want to prove:</p>
<ol>
<li>For every <span class="math-container">$b \in B$</span>, <span class="math-container">$b \le M$</span>. This is true because <span class="math-container">$M$</span> is an upper bound on <span class="math-container">$A$</span>, therefore it is an upper bound on <span class="math-container">$B$</span>.</li>
<li>For every <span class="math-container">$\epsilon>0$</span>, there is a <span class="math-container">$b \in B$</span> such that <span class="math-container">$b > M - \epsilon$</span>. This is easiest to prove by contradiction. Suppose that there is some <span class="math-container">$\epsilon>0$</span>, such that for all <span class="math-container">$b \in B$</span>, <span class="math-container">$b \le M-\epsilon$</span>. Then <span class="math-container">$M-\epsilon$</span> is an upper bound on <span class="math-container">$B$</span>, therefore it is an upper bound on <span class="math-container">$A$</span>. This means that for all <span class="math-container">$a \in A$</span>, <span class="math-container">$a \le M-\epsilon$</span>, which contradicts the definition above. Therefore no such <span class="math-container">$\epsilon$</span> can exist.</li>
</ol>
<p>Obviously this is a bit specific to the definion you're using. Another definiton of <span class="math-container">$\sup A$</span>, instead of the statement with <span class="math-container">$\epsilon$</span>, specifies that if <span class="math-container">$M'$</span> is another upper bound on <span class="math-container">$A$</span>, then <span class="math-container">$M \le M'$</span>. This would lead to a different proof, but the idea is the same.</p>
|
2,994,440 | <p><strong>Caution</strong>: Axiom of Completeness is not assumed here.</p>
<p><strong>Before reading my attempt, I want you to think up a proof of your own.</strong></p>
<p>Here is my attempt:</p>
<blockquote class="spoiler">
<p> Let <span class="math-container">$D$</span> be the set of all upper bounds of <span class="math-container">$A$</span> as well as <span class="math-container">$B$</span>. Since <span class="math-container">$\sup A$</span> exists, <span class="math-container">$\sup A \le d$</span> for all <span class="math-container">$d \in D$</span>. It also means <span class="math-container">$D$</span> is bounded below.
Now I have to show that <span class="math-container">$\inf D$</span> exists and is equal to <span class="math-container">$\sup A$</span>. (This is the part that I have no idea how to approach.)
Then since <span class="math-container">$\inf D$</span> exists, <span class="math-container">$\sup B$</span> exists and is equal to <span class="math-container">$\inf D$</span>. Therefore, <span class="math-container">$\sup A = \sup B$</span>.</p>
</blockquote>
<p>This concludes my proof. Two questions:</p>
<ol>
<li>How do I show the part in my proof where I was stuck?</li>
<li>Is there any other way to prove it? Preferably much simpler. I want to see some other ways.</li>
</ol>
| DanielWainfleet | 254,665 | <p>Let <span class="math-container">$U(A)$</span> be the set of upper bounds for <span class="math-container">$A.$</span> The definition of <span class="math-container">$\sup A$</span> is <span class="math-container">$\min U(A),$</span> which exists iff <span class="math-container">$U(A)$</span> has a least member.</p>
<p>Suppose <span class="math-container">$U(A)=U(B)$</span> and that <span class="math-container">$\sup A$</span> exists. Then <span class="math-container">$\min U(A)$</span> exists. So <span class="math-container">$\min U(B)=\min U(A)$</span> exists because <span class="math-container">$U(A)$</span> and <span class="math-container">$U(B)$</span> are the same thing.</p>
<p>Hence <span class="math-container">$\sup A=\min U(A)=\min U(B)=\sup B.$</span></p>
|
225,730 | <p>I am reading a paper "2-vector spaces and groupoid" by <a href="http://arxiv.org/abs/0810.2361" rel="noreferrer">Jeffrey Morton </a> and I need a help to understand the following.</p>
<p>Let $X$ and $Y$ be finite groupoids. Let $[X, \mathbb{Vect}]$ be a functor category from $X$ to $\mathbb{Vect}$. Suppose we have a functor $f: Y \rightarrow X$. The author defines a pushforward $f_{*} :[Y, \mathbb{Vect}] \rightarrow [X, \mathbb{Vect}]$ as follows.</p>
<p>For each object $x \in X$, the comma category $(f \downarrow x)$ has object which are objects $y \in Y$ with maps $f(y) \rightarrow x$ in $X$, and morphisms which are morphisms $a: y \rightarrow y'$ whose images make the evident triangle in $X$ commute.</p>
<p>Then the author defines for each $F\in [Y, \mathbb{Vect}]$, $f_{*}(F)(x):=colim F(f \downarrow x)$.</p>
<p>He also shows that $f_{*}(F)(y)=\bigoplus_{f(x)\cong y} \mathbb{C}[Aut(y)]\otimes_{\mathbb{C}[Aut(x)]}F(x)$ and fimilarly calculate $f^*f_*$ and $f_*f^*$.</p>
<p>Later he says that this description accords with the susal description of these functors in the left adjunction. Then he mentions that the right adjoint is given as</p>
<p>$f_*F(x)= \bigoplus_{[y], f(y)\cong x}hom_{\mathbb{C}[Aut(x)]}(\mathbb{C}[Aut(y), F(y)).$</p>
<p>Then we says:</p>
<blockquote>
<p>The Nalayama isomorphism gives the duality between the two descriptions of $f_*$, in terms of $hom_{\mathbb{C}[Aut(x)]}$ and $\otimes_{\mathbb{C}[Aut(x)]}$ by means of the exterior trace map.</p>
</blockquote>
<p>I understood the calculation of $f_*F(x)$ using a colimit defined above. But I don't know what it means by saying left or right adjoint. Also I don't know how to get the formula for the right adjoint. I even don't know what is Nakayama isomorphism. I searched for it but I couldn't find a good resource.</p>
<p>So I would like to know what is going on here. Especially, I'd like to know;</p>
<ol>
<li><p>Why is the first construction using a colimit called "left adjoint"?</p></li>
<li><p>What is the right adjoint and how is it defined and how to calculate it to get the formula above.</p></li>
<li><p>What is Nakayama isomorphism?</p></li>
</ol>
<p>I have never studied these things so I don't know where to look up. I also want good references.</p>
<p>I appreciate any help. Thank you in advance.</p>
| Bart Michels | 43,288 | <p>To expand on Qiaochu Yuan's answer: The Nakayama isomorphism is what you get when you take the "natural" isomorphism of <span class="math-container">$\mathbb C[G]$</span> with its left <span class="math-container">$\mathbb C[H]$</span>-dual in the sequence of isomorphisms
<span class="math-container">$$\begin{align*}
\operatorname{Coind}_K^G V
& = \operatorname{Hom}_{\mathbb C[H]}(\mathbb C[G], V) \\
&\cong \mathbb C[G]^* \otimes_{\mathbb C[H]} V \\
& \cong \mathbb C[G] \otimes_{\mathbb C[H]} V \\
&= \operatorname{Ind}_H^G V \,.
\end{align*}$$</span>
What is this natural isomorphism of <span class="math-container">$(G, H)$</span>-bimodules? There is a natural homomorphism of left <span class="math-container">$G$</span>-modules
<span class="math-container">$$\begin{align*}
\mathbb C[G] &\to \operatorname{Hom}_{\mathbb C}(\mathbb C[G], \mathbb C) \subset \operatorname{Hom}_{\mathbb C}(\mathbb C[G], \mathbb C[H]) \\
g & \mapsto (g^{-1})^* \,.
\end{align*}$$</span>
If you force those <span class="math-container">$(g^{-1})^*$</span> to become <span class="math-container">$H$</span>-equivariant by averaging over <span class="math-container">$H$</span>, you get a left <span class="math-container">$\mathbb C[G]$</span>-linear map
<span class="math-container">$$\begin{align*}
\mathbb C[G] &\to \mathbb C[G]^* \\
g & \mapsto \sum_{h \in H} h \cdot (hg^{-1})^* \,,
\end{align*}$$</span>
which turns out to be right <span class="math-container">$H$</span>-equivariant. This is the isomorphism of <span class="math-container">$(G, H)$</span>-bimodules <span class="math-container">$\mathbb C[G] \cong \mathbb C[G]^*$</span> that gives rise to the Nakayama isomorphism.</p>
<p>One other comment: this works over any field, provided that you write the end result as
<span class="math-container">$$ \operatorname{Nak}(\phi) = \sum_{g \in (H \backslash G)} g^{-1} \otimes \phi(g) \,. $$</span></p>
|
2,777,051 | <p>When talking about operator semigroups, some books use the notation $(e^{tA})_{t \geq 0}.$ Doesn't this create an ambiguity? Why are results proved for this semigroup instead of taking a general semigroup, say $(T(t))_{t \geq 0}$? </p>
| Pedro | 70,305 | <p><em>When talking about operator semigroups, some books use the notation <span class="math-container">$(e^{tA})_{t \geq 0}.$</span> Doesn't this create an ambiguity?</em></p>
<p>No, because for these authors the notation <span class="math-container">$(e^{tA})_{t \geq 0}$</span> means the semigroup <span class="math-container">$(T(t))_{t\geq 0}$</span> whose generator is <span class="math-container">$A$</span>.</p>
<ul>
<li><p>This notation makes sense because each semigroup has exactly one generator and no operator generates more than one semigroup.</p>
</li>
<li><p>This notation is natural because for the particular case in which <span class="math-container">$A$</span> is bounded, the semigroup generated by <span class="math-container">$A$</span> is exactly <span class="math-container">$\sum_{n=0}^\infty\frac{(tA)^n}{n!}$</span>. In other words, "the semigroup generated by <span class="math-container">$A$</span>" is the generalization of "exponential of <span class="math-container">$A$</span>" and thus it is natural to use the same notation for both.</p>
</li>
</ul>
<p><em>Why are results proved for this semigroup instead of taking a general semigroup, say <span class="math-container">$(T(t))_{t \geq 0}$</span>?</em></p>
<p>The results are proved for a general semigroup. But, in order to be concise, this general semigroup is called <span class="math-container">$(e^{tA})_{t \geq 0}$</span>. Then we mention the semigroup and the generator in the same notation and can write things like</p>
<blockquote>
<p>As <span class="math-container">$\mathbf i\mathbb R\subset \rho(A)$</span>, we conclude that <span class="math-container">$(e^{tA})_{t \geq 0}$</span> is strongly stable.</p>
<p>Consider two semigroups <span class="math-container">$(e^{tA})_{t \geq 0}$</span> and <span class="math-container">$(e^{tB})_{t \geq 0}$</span>.</p>
<p><span class="math-container">$e^{t\sigma(A)}\subset \sigma(e^{tA})$</span>.</p>
</blockquote>
<p>instead of</p>
<blockquote>
<p>As <span class="math-container">$\mathbf i\mathbb R\subset \rho(A)$</span>, we conclude that the semigroup <span class="math-container">$(T(t))_{t\geq 0}$</span> generated by <span class="math-container">$A$</span> is strongly stable.</p>
<p>Consider two semigroups <span class="math-container">$(T(t))_{t \geq 0}$</span> and <span class="math-container">$(S(t))_{t \geq 0}$</span> generated by <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, respectively.</p>
<p><span class="math-container">$e^{t\sigma(A)}\subset \sigma(T(t))$</span> where <span class="math-container">$(T(t))_{t\geq 0}$</span> is the semigroup generated by <span class="math-container">$A$</span>.</p>
</blockquote>
<p>In fact, many authors do not use this notation. But some authors (of books and papers) do.</p>
|
542,397 | <p>Consider the sequence $\{a_n\}_1^\infty$ such that $ a_n = $ ${n^m + 1}\over {n^{m+1} + 1}$ and $m \in \mathbb{R}$</p>
<p>EDIT: This is incorrect for $m < -1$, Then add the condition $m \geq -1$</p>
<p>I want to show this sequence is monotonically decreasing.<br>
This is not homework, I've seen many specific examples of this however and graphed many cases, and I'd like to show this for the general case.</p>
<p>I've tried considering the continuous case and taking the derivative, using the quotient rule I get a square in the denominator, and since I'm only interested in the sign of the derivative I can throw away the denominator in the quotient rule, I then have:</p>
<p>$mx^{m-1}(x^{m+1} + 1) - (m+1)x^m(x^m + 1)$</p>
<p>Plugging in $x = 1$ gives a negative result for the derivative showing the continuous function is decreasing at that point, so I get a feeling my "conjecture" is correct. <br>
Not sure where I can go from here, and if there's a simpler way without going to the continuous case.
Expanding the brackets didn't help much either</p>
| N. S. | 9,176 | <p>$$\frac{n^m + 1}{n^{m+1} + 1} \geq \frac{(n+1)^m + 1}{(n+1)^{m+1} + 1} \Leftrightarrow \\
(n^m + 1)((n+1)^{m+1} + 1) \geq ((n+1)^m + 1)(n^{m+1} + 1)\Leftrightarrow \\
n^m (n+1)^{m+1} + n^m +(n+1)^{m+1} + 1\geq (n+1)^m n^{m+1} + (n+1)^m + n^{m+1} + 1 \Leftrightarrow \\
n^m (n+1)^{m} +n(n+1)^{m} \geq n^m (n-1) \Leftrightarrow \\
(n+1)^{m} +n(1+\frac{1}{n})^{m} \geq (n-1)$$</p>
<p>As $m \geq -1$ and the exponentials $(n+1)^x$ and $(1+\frac{1}{n})^{x}$ are increasing, we have</p>
<p>$$(n+1)^{m} +n(1+\frac{1}{n})^{m} \geq (n+1)^{-1} +n(1+\frac{1}{n})^{-1} = \frac{1}{n+1}+\frac{n^2}{n+1}> \frac{n^2-1}{n+1}=n-1 $$</p>
|
1,598,335 | <p>Find the transformation that takes $y=3^x$ to $y=\textit{e}^x$.
I have tried: </p>
<p>Let
$y=3^x$ to $y=e^{x'}$ </p>
<p>$$\log_{3}(y)=x\quad\text{hence}\quad\log_{3}(y)=\frac{\log_{e}(y)}{\log_{e}(3)}$$</p>
<p>$$x\log_{e}(3)=x'$$</p>
<p>Gives the transformation as dilate by $\log_e(3)$
And also this:</p>
<p>$$3^{\log_{3}e}=e$$</p>
<p>$$e^x=3^{(\log_{3}e)x}$$</p>
<p>And the transformation is dilate by $\frac{1}{\log_3e}$</p>
<p>Could I please get an explanation which is right?</p>
| Narasimham | 95,860 | <p>Using short-hand $s,c$ for $\sin \theta, \cos \theta$
$$x= s/ \sqrt 2 + c/2; \, y = s/ \sqrt 2 - c/2 $$</p>
<p>Remember as formulas the results/identities:</p>
<p>$$ ( x+y)^2 +( x+y)^2 = 2(x^2 +y^2);\, ( x+y)^2-( x+y)^2 = 4 x y ; $$</p>
<p>$$ 3 (x^2 +y^2) = 3 s^2 + 3 c^2/2$$</p>
<p>$$ -2 x y = c^2/2 - s^2 $$</p>
<p>Add, RHS</p>
<p>$$\rightarrow 2$$</p>
<p>It represents the equation of an ellipse originally without $xy$ term rotated by $45^0$ thereby bringing in the $xy$ term.</p>
|
382,935 | <p>I reading my textbook and I don't understand the concept of distributions or number of solutions to an equation. It's explained that this problem is 1/4 types of sampling/distributions problems. An example is provided to illustrate:</p>
<blockquote>
<p>In how many ways can 4 identical jobs (indistinguishable balls) be distributed among 26 members (urns) without exclusion (since one member can do multiple jobs)?</p>
<p>A sample outcome might be:<br>
$\text{_____________}$<br>
| A | B | C |...| Z | <br>
$\text{--------------------}$<br></p>
<p>$\text{_____________}$<br>
| o | oo| | | ||| o | <br>
$\text{_____________}$<br>
| A | B | C |... | Z | <br>
$\text{--------------------}$<br></p>
<p>Thus, the question is reduced to, "How many $(26-1+4)$ letter words are there consisting of four circles and $(26-1)$ vertical lines?" Therefore, the solution is: $\binom{26-1+4}{4}$</p>
</blockquote>
<p>I really don't understand why its $(26-1+4)$. There's only 26 different spots or "urns" to place the 4 jobs. Can someone please explain? </p>
<p>I looked through another text to try and understand and I found it explained as such: </p>
<blockquote>
<p>There are $\binom{n+r-1}{r-1}$ distinct nonnegative integer valued vectors $(x_1,...,x_r$ satisfying the equation $(x_1 + ... + x_r = x_n$ for $x\ge0$. $\spadesuit$</p>
</blockquote>
<p>How in the world are they deriving this? For <em>distinct positive integers</em> I understand:</p>
<p>Assume I have 8 balls (n=8) and I have 3 urns (r=3):
o^o^o^o^o^o^o^o, where o represents a ball and ^ represents a place holder where an urn could be placed. For this scenario:</p>
<blockquote>
<p>There are $\binom{n-1}{r-1}$ distinct positive integer valued vectors $(x_1,...,x_n)$ satisfying the equation: $x_1 + ... + x_n = n, x_i>0, i=1,..,r$ $\clubsuit$</p>
</blockquote>
<p>It's clear that I could have this specific case ooo|ooo|oo. Here the bar represents a divide for the urn and you see I have 3 sections. So that case is clear. Can anyone please explain this problem to me? I don't understand the nonnegative integer case. </p>
<p>Also, people who post tend to be crazy smart and explain things very in a complicated manner. I'd appreciate it if it could be explained in layman's terms as much as possible. </p>
<p>Thank you!!!</p>
| user1527227 | 73,163 | <p>Let's say we have the following problem: Distribution of n identical objects into k boxes. </p>
<p>Let's consider a specific problem: How many ways can you put 4 identical balls into 2 boxes? We will draw a picture to understand the solution to this problem. Imagine 4 identical items are {<em>*</em>*}. If we draw a vertical bar somewhere among these 4 stars, this can represent a unique assignment of the balls to the boxes. For example:</p>
<p>$|****$, Box 1=0, Box2=4 <br>
$*|***$, Box 1=1, Box2=3 <br>
$**|**$, Box 1=2, Box2=2 <br>
$***|*$, Box 1=3, Box2=1 <br>
$****|$, Box 1=4, Box2=0 <br></p>
<p>Since there is a correspondence between a star/bar picture and assignments of balls to boxes, we can count the star/bar pictures and get an answer to our balls-in-boxes problem. So the question of finding the number of ways to put four identical balls into 2 boxes is the same as the number of ways of arranging four identical stars and one vertical bar which is $\binom{5}{4}=\binom{5}{1}=5$. In this case, we had n indistinguishable objects and k indistinguishable boxes:</p>
<blockquote>
<p>The number of ways n identical (indistinguishable) objects can be distributed into k boxes is: $\binom{n+k-1}{k-1}=\binom{n+k-1}{n}$</p>
</blockquote>
<p>Now consider the above case where you have 26 distinguishable people and 4 identical jobs which are to be distributed. As in the previous case, we can also represent this with stars and bars. However, in this case we will have 25 bars representing the dividers for the 26 people and 4 stars {<em>*</em>*} for the 4 jobs. The total characters in a linear sequence will be 25+4 or (n-1+k) and we will select either 4 jobs or the 25 dividers. So $\binom{29}{4}=\binom{29}{25}$. Thus:</p>
<blockquote>
<p>The number of unordered samples of r objects, with replacement, from n distinguishable objects is $\binom{n+r-1}{r}=\binom{n+r-1}{n-1}$. This is equivalent to the number of ways to distribute r indistinguishable balls into n distinguishable urns without exclusion.</p>
</blockquote>
|
1,654,354 | <p>Why is $A=\{(x_1,x_2,...,x_n)|\exists_{i\ne j}: x_i=x_j\}$ a null set?</p>
<p>This claim was shown in a solution I ran into, and I don't see how it holds. I try to follow the formal definition of nullity, which is having the possibility to be covered by a collection of open cubes, whose volume is as small as desired. I can't, however, tell how it is achieved here. There seem to be too many options and combinations here. I could use some help here.</p>
| orangeskid | 168,051 | <p>HINT:</p>
<p>For $n\ge 2$ denote by $S_n$ the subset of $\mathbb{R}^n $ consisting of points with the first two coordinates equal. Then with the indentification $\mathbb{R}^n = \mathbb{ R}^2 \times \mathbb{R}^{n-2} $ we have
$$S_n = S_2 \times \mathbb{R}^{n-2} $$</p>
<p>Hence it is enough to show that $S_2 \subset \mathbb{R}^2$ is a subset of measure $0$. In fact, it is enough to find a countable cover of $S_2$ by rectangles with finite total area $A$. Indeed, since $S_2$ is invariant under dilations, for any $\delta > 0$ we can find a cover of total area $\delta^2 \cdot A$. Now, consider a sequence of positive numbers $d_n$ so that $\sum d_m = \infty$ while $\sum d_m^2 < \infty$ (for instance $d_m = \frac{1}{m}$). Arrange along the diagonal starting from the origin squares with sides $d_1$, $d_2$, $\ldots$, $d_m$, $\ldots$ going one sequence $NE$, another $SW$. Since $\sum d_m = \infty$ the full line $x_1 = x_2$ gets covered, and the total area of the squares will be $2 \sum d_m^2 < \infty$. </p>
|
1,529,731 | <p>I'm currently studying for the GRE and a specific type of question has me stumped. I have a two part question, but first here is one problem and my work:</p>
<p><strong>"If x is the remainder when a multiple of 4 is divided by 6, and y is the remainder when a multiple of 2 is divided by 3, what is the greatest possible value of x+y"</strong></p>
<p>Since it says greatest I thought GCM and wrote(with arbitrary variables)</p>
<p>$$N = 24a + x$$</p>
<p>$$M = 6b + y$$
I thought that since 2 & 3 are factors of 4 & 6, respectively, maybe I could equate them:
$$24a+x=6b+y$$
Or maybe N is twice a large as M:
$$24a+x=2(6b+y)$$
<strong>1) How do I solve this? Am I even on the right track?</strong></p>
<p>This kind of question has had me stumped for a while but the books explanation has me even more lost. </p>
<p><strong>"The greatest possible remainder for a multiple of 4 being divided by 6, happens when 4 is divided by 6. When 4 is divided by 6 the result is 0 with a remainder of 6."</strong></p>
<p>.....Don't remainders by nature have to be smaller than the divisor......</p>
<p><strong>2)How is that possible? Shouldn't the result be 0 with a remainder of 4?</strong> </p>
<p>How can there be a remainder of 6 when the divisor is 6? I found a similar problem and solution here: <a href="https://math.stackexchange.com/questions/346172/finding-a-number-given-its-remainder-when-divided-by-other-numbers?newreg=e42984c5474e4b2dbb127333cfb3656d">Finding a number given its remainder when divided by other numbers</a> and here: <a href="https://math.stackexchange.com/questions/1401850/how-to-find-the-greatest-remainder-of-a-number-that-is-a-multiple-of-another-num">How to find the greatest remainder of a number that is a multiple of another number</a> But the more I read the more lost I was. </p>
<p>At this point I feel like a child throwing spaghetti about trying to make art. I have no idea how to even start on this problem any help is appreciated.</p>
| Edem | 868,915 | <p>Multiples of 4 are: 4,8,12,16,20,24 etc.. and multiples of 2: 2,4,6,8,10,12,14 etc..
Divide the listed multiples of 4 by 6 to check the remainder by finding the remainder in each multiple. It is obvious that 6 goes into 16 with a remainder of 4 giving the value for X. Remember the question said the greatest possible value and in this case is remainder 4. Apply the same method to multiples of 2 divided by 3. 3 goes into 8, 2 times with a remainder of 2 giving the value for Y. Therefore X+Y=6</p>
|
30,789 | <p>I'm trying to identify the frequencies in my time history samples, and I can see a frequency in the time history, but can't see it in its Fourier transform.
Here it is :</p>
<p>the sample data:</p>
<pre><code>dt = 0.01;(*0.01 second per sample*)
ls={7.18384,9.08503,7.13301,9.03243,7.23692,8.82911,7.48153,8.50053,7.8291,8.09453,8.22514,7.67123,8.60473,7.29656,8.90489,7.02926,9.07228,6.91356,9.07469,6.96968,8.90404,7.19156,8.58122,7.54573,8.15062,7.97689,7.67665,8.41549,7.23096,8.78898,6.88426,9.03305,6.69172,9.10231,6.68698,8.97821,6.87272,8.6724,7.22283,8.22643,7.68225,7.70488,8.17857,7.18711,8.62957,6.75303,8.9597,6.47266,9.10889,6.39174,9.04594,6.52704,8.77139,6.85901,8.32088,7.33824,7.7584,7.88776,7.16829,8.41967,6.64141,8.84366,6.2609,9.0868,6.08919,9.10045,6.15615,8.87392,6.45525,8.43236,6.94108,7.83828,7.53853,7.17843,8.15053,6.55427,8.67686,6.06288,9.02603,5.78403,9.13458,5.76455,8.97352,6.01177,8.55858,6.48966,7.94428,7.12458,7.22101,7.8152,6.49678,8.4485,5.88528,8.91739,5.48272,9.13786,5.35709,9.06375,5.53107,8.69409,5.98259,8.07629,6.64157,7.29757,7.40537,6.47512,8.14882,5.73434,8.74894,5.19349,9.09985,4.93916,9.13459,5.01757,8.83338,5.41929,8.23096,6.08644,7.41062,6.91285,6.49317,7.76769,5.61863,8.50777,4.92325,9.00806,4.51966,9.1747,4.47503,8.96814,4.80297,8.40434,5.455,7.55984,6.33198,6.55581,7.29271,5.54458,8.18165,4.6821,8.847,4.10633,9.17184,3.91187,9.08704,4.13559,8.59066,4.74753,7.74309,5.65489,6.66645,6.71462,5.51928,7.75486,4.47976,8.6024,3.70977,9.10911,3.33653,9.17844,3.42289,8.77952,3.96388,7.95795,4.87716,6.82561,6.02223,5.55073,7.21349,4.32513,8.25592,3.34263,8.96977,2.75835,9.22521,2.67336,8.96055,3.10556,8.19668,3.99626,7.03496,5.20525,5.6427,6.54315,4.23034,7.78851,3.01522,8.73391,2.19205,9.20894,1.89461,9.11841,2.17992,8.45091,3.00859,7.29079,4.25733,5.80068,5.727,4.20335,7.18224,2.74288,8.37806,1.64996,9.10926,1.10137,9.23383,1.19232,8.70843,1.91736,7.5871,3.1692,6.02681,4.75307,4.25335,6.41448,2.53923,7.8804,1.14864,8.89922,0.307694,9.28655,0.154088,8.95041,0.725381,7.91675,1.93765,6.31868,3.60713,4.38985,5.46625,2.41628,7.21338,0.707136,8.55287,-0.470752,9.2482,-0.919297,9.15831,-0.561298,8.26386,0.562301,6.67532,2.27715,4.61492,4.31759,2.39053,6.34975,0.340327,8.03907,-1.21045,9.08947,-2.01384,9.30485,-1.9283,8.61208,-0.958302,7.08603,0.756819,4.93354,2.94639,2.47094,5.26353,0.0706556,7.32281,-1.89254,8.77714,-3.10504,9.358,-3.36416,8.93794,-2.61487,7.53742,-0.963082,5.34406,1.33854,2.66827,3.9233,-0.0836731,6.37173,-2.49189,8.2693,-4.17005,9.28401,-4.84777,9.20861,-4.3986,8.01348,-2.88195,5.83756,-0.523299,2.99222,2.304,-0.10749,5.14496,-2.98081,7.52648,-5.18357,9.04878,-6.32959,9.4304,-6.18952,8.56459,-4.73628,6.56344,-2.13332,3.70414,1.31944,0.401559,5.19872,-2.85418,9.0291,-5.5357,12.3241,-7.17454,14.6869,-7.4465,15.8834,-6.29832,15.9227,-3.97558,15.044,-0.986409,13.6463,2.07958,12.1451,4.72194,10.8576,6.6742,9.93404,7.90659,9.37837,8.56294,9.10124,8.84598,8.99267,8.93815,8.96116,8.95636,8.95544,8.95677,8.95391,8.95455,8.95365,8.95341,8.95371,8.95173,8.95356,8.95088,8.95381,8.9499,8.95321,8.94933,8.95282,8.94887,8.9513,8.94838,8.95018,8.94857,8.94853};
</code></pre>
<p>Fourier transform function:</p>
<pre><code>DFT[A_, ht_] := RotateRight[ht/Sqrt[2 \[Pi]]*Fourier[RotateLeft[A, Length[A]/2 - 1],
FourierParameters -> {1, 1}], Length[A]/2 - 1];
(*shift the zero frequency to the center*)
</code></pre>
<p>plot the time history:</p>
<pre><code>ListPlot[ls, PlotRange -> All, Joined -> True, DataRange -> {0, dt*Length[ls]}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/4hnHe.png" alt="enter image description here"></p>
<p>We can see there are a fast frequency with period about 0.02s and a slow frequency with period about 1.5s, but in the Fourier transform we only see the fast frequency(except the zero frequency)</p>
<pre><code>ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {- 1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/3qthH.png" alt="enter image description here"></p>
<p>So where is the low frequency?</p>
<hr>
<p><strong>Update</strong></p>
<p>As Simon and bill suggest, the slow oscillation is the beating of two close high frequencies.
Since the Fourier transform resolution is 2Pi/(N*dt), where N is the number of sample points, so if I increase the resolution by increasing the number of sample points I should see two separated peaks.
So I tried to increase the number of sample points, but I can only see one peak all the time.
Here is how I did it:</p>
<pre><code>w = 5.0; dt = 0.66125;
f[x_] := Sin[w x]
ls = Table[f[x], {x, dt, 200 dt, dt}];
</code></pre>
<p>we see beating in the plot</p>
<pre><code>ListPlot[ls, PlotRange -> All, Joined -> True, DataRange -> {dt, 200 dt}]
</code></pre>
<p><img src="https://i.stack.imgur.com/h30HL.png" alt="enter image description here"></p>
<p>but one on peak in the Fourier transform:</p>
<pre><code>ListPlot[Abs[DFT[ls, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/U8r7E.png" alt="enter image description here"></p>
<p>If we increase the number of sample points, we still only see one peak:</p>
<pre><code>ls2 = Table[f[x], {x, dt, 800 dt, dt}];
ListPlot[Abs[DFT[ls2, dt]]^2, PlotRange -> All, Joined -> True, DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/ynyfB.png" alt="enter image description here"></p>
<p>So where is the problem?</p>
| bill s | 1,783 | <p>Simon has hit the problem on the head (those are not real frequency components you see). But there is a way to discover what these frequencies are, by taking the envelope of the signal and transforming (DFTing) the envelope. </p>
<pre><code>data = ls - MeanFilter[ls, 5];
decay = 0.06; rise = 0.2;
filt[z_, u_] := Max[decay z + (1 - decay) u, rise z + (1 - rise) u];
env = Drop[FoldList[filt, 0, Abs[data]], 1];
</code></pre>
<p>The first line removes the "DC" components from the data and the following lines are a method of extracting the envelope, taken from my answer to <a href="https://mathematica.stackexchange.com/questions/27748/elegant-way-of-obtaining-the-envelope-of-oscillating-function">this question</a>. (Detail: the <code>Drop</code> is needed because your DFT function doesn't work for odd-length sequences).</p>
<p><img src="https://i.stack.imgur.com/evd2I.png" alt="enter image description here"> </p>
<p>Taking the DFT (using your function) of this signal <code>env</code>, and plotting</p>
<pre><code>ListPlot[Abs[DFT[env, dt]]^2, PlotRange -> All, Joined -> True,
DataRange -> {-1/dt/2, 1/dt/2}, Axes -> False, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/zWVAy.png" alt="enter image description here"></p>
<p>shows the low frequency part of the signal that you are looking for.</p>
|
1,232,439 | <p><strong>EDIT:</strong> This is for a production scheduling problem with quadratic production and linear inventory costs. </p>
<p>The goal is to \begin{equation*}
\max_{u} \int_{0}^{T} -(c_{1}u^{2} + c_{2}y) dt
\end{equation*}
subject to
\begin{align*}
y' &= u,\\
y(0) &= 0,\\
y(T) &=B.\\
\end{align*}</p>
<p>So the Bellman equation is: \begin{equation*}
\max_{u} {-c_{1}u^{2} + c_{2}y + \frac{\partial V}{\partial y}u + \frac{\partial V}{\partial y}} =0.
\end{equation*}</p>
<p>By first-order condition, we obtain $u^{*} = \frac{1}{2c_{1}}\left ( \frac{\partial V}{\partial y} \right )$.</p>
<p>We insert this $u^{*}$ into the Bellman equation to obtain:</p>
<p>$$0 = -c_{1}\left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial y} \left (\frac{1}{2c_{1}} \frac{\partial V}{\partial y} \right ) + \frac{\partial V}{\partial t}$$ </p>
<p>And then I obtain the following PDE: $$0 = \frac{1}{4c_{1}}\left (\frac{\partial V}{\partial y} \right )^{2} + \frac{\partial V}{\partial t}$$</p>
<p>I have to solve for $V(y,t)$.</p>
<p>My work: $V = e^{ay + bt}$. </p>
<p>Then I find $V_{y} = ae^{ay+bt}$, and $V_{t} = be^{ay+bt}$.</p>
<p>Then I put it values of $V_{y}$ and $V_{t}$ in the PDE to solve for $b$: </p>
<p>$$\frac{1}{4c_{1}}(ae^{ay+bt})^{2} + be^{ay+bt} = 0$$ </p>
<p>$$\Rightarrow a^{2}e^{2(ay+bt)} + be^{ay+bt} = 0$$</p>
<p>$$\Rightarrow a^{2}e^{2} e^{(ay+bt)} = -be^{ay+bt}$$</p>
<p>Assuming that $e^{ay+bt} \neq 0$, we obtain $b = -a^{2}e^{2}$.</p>
<p>Now, substitute the value of $b$ back in the equation for $V$, to obtain $$V(y,t) = e^{ay - a^{2} e^{2}t} = e^{a(y-aAt)},$$ where $A = e^{2}$.</p>
<p>Does this look correct?</p>
<p>Also, the boundary conditions are given with $y(0)=0$, and $y(T)=B$. </p>
| Archaick | 191,173 | <p>First, we need to find the number of people in the room before Mr. L arrived, call it $N$. $N$ is the greatest integer such that ${N \choose 2} \leq 100$. This is because ${{N} \choose 2}+(N-1) < {{N+1} \choose 2}$. We find then that $N=14$ since ${14 \choose 2}+9=100$. That is, 5 people left the room, since Mr. L only shook hands with 9.</p>
|
84,977 | <p>In how many ways 3 flags of colors black, purple & yellow can be arranged at the corners of an equilateral triangle?</p>
| Quixotic | 2,109 | <p>This appears to me as a problem of <a href="http://mathworld.wolfram.com/CircularPermutation.html" rel="nofollow">circular permutation</a>. </p>
<p>Fix one flag of any color (say black) at a particular corner of the specified equilateral triangle, then the other two flags (purple and yellow) can be arranged in $2!$ ways at the other two corners. Hence, the answer should be $4$.</p>
|
2,226,150 | <blockquote>
<p>The following diophantine equation came up in the past paper of a Mathematics competition that I am doing soon: $$ 2(x+y)=xy+9.$$ </p>
</blockquote>
<p>Although I know that the solution is $(1,7)$, I am unsure as of how to reach this result. Clearly, the product $xy$ must be odd since $2(x+y)$ must be even, however beyond that, I am unable to see anything else that I can do to solve the problem. I have also tried using the AM-GM inequality, however, it did not simplify the problem much:$$(x+y)+(x-xy+y)\le(\frac{(x+y)+(x+y-xy)}{2})^2.$$
Any help would be greatly be appreciated.</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
& 2\pars{x + y} = xy + 9 \implies y = {2x - 9 \over x - 2} = 2 + {5 \over 2 - x}
\implies
\left\{\begin{array}{ll}
{\large\bullet} & \pars{2 - x} \mid 5
\\
{\large\bullet} & \pars{x \leq 2}\quad\mbox{or}\quad\pars{x \geq 5}
\\
{\large\bullet} & x\ odd
\end{array}\right.
\end{align}</span>
If <span class="math-container">$\ds{x \geq 0}$</span> the only possibility is <span class="math-container">$\ds{\pars{x,y}
= \pars{\color{red}{\large 1},\color{red}{\large 7}}}$</span> or <span class="math-container">$\ds{\pars{x,y}
= \pars{\color{red}{\large 7},\color{red}{\large 1}}}$</span> because
<span class="math-container">$\ds{\verts{2 - x} \leq 5}$</span>.</p>
|
4,373,262 | <p>Let G be a graph with order 9 so that the degree of each vertex is either 5 or 6. Prove
that there are either at least 5 vertices of degree 6 or at least 6 vertices of degree 5</p>
| Acccumulation | 476,070 | <p>By the Pigeonhole Principle, one of the degree classes has at least 5 members. If that degree class is 6, we're done. So assume there are at least 5 vertices with degree 5. We can't have an odd number of odd degrees, so there must be at least 6 vertices with degree 5.</p>
<p>If you're confused by the last statement, the statement "The number of vertices with odd degree is even" is true for any graph. For instance, given any group of people, the number of people who have shaken hands with an odd number of other people in that group is even. This is because each handshake involves two people, so the sum over all people of how many people they've shaken hands with is even. See <a href="https://math.stackexchange.com/questions/606899/number-of-people-having-shaken-hands-an-odd-number-of-times">Number of people having shaken hands an odd number of times</a></p>
|
504,997 | <p>I have a dynamic equation,
$$ \frac{\dot{k}}{k} = s k^{\alpha - 1} + \delta + n$$
Where $\dot{k}/k$ is the capital growth rate as a function of savings $s$, capital $k$, capital depreciation rate $\delta$, and population growth rate $n$.</p>
<p>I have been asked to find the change in the growth rate as $k$ increases. This is of course
$$\frac{\partial \dot{k}/k}{\partial k} = (\alpha - 1) s k^{\alpha -2}$$
But what I want to find now is the change in growth rate as $k$ increases <em>proportionately</em>. This should be
$$\frac{\partial \dot{k}/k}{\partial \ln(k)} = ?$$
How do you calculate the partial derivative with respect to the logarithm of a variable? I'm sure the answer is simple, but my analytical calculus is pretty rusty.</p>
| baudolino | 67,409 | <p>The simplest way is via chain rule:
$$
\dfrac{\partial \dot{k}/k}{\partial k} = \dfrac{\partial \dot{k}/k}{\partial \ln{k}}\dfrac{\partial \ln{k}}{\partial k}
$$
...from which you can move things around to get your quantity of interest, e.g.
$$\dfrac{\partial \dot{k}/k}{\partial \ln{k}} = \dfrac{\partial \dot{k}/k}{\partial k}k = (\alpha-1)sk^{\alpha-1}$$
(since $(\ln{x})' = 1/x$).</p>
|
3,469,582 | <p>I have a feeling that <span class="math-container">$\lim_{x\rightarrow\infty}\frac{\log f(x^2)}{f(x)}=0$</span> for any positive monotone increasing function such that <span class="math-container">$\lim_{x\rightarrow\infty}f(x)=\infty$</span>, (in reality I am thinking mostly about super-exponential growth here but I do not think it might make a difference). The idea, intuitively, is that the logarithm takes away more growth than the square in the argument boosts. Though, I am not sure if this is true (maybe adding some more hypotheses) because I find it hard to prove neatly, having <span class="math-container">$f$</span> in between mixing these two contributions up. Tried L'Hospital but did not work and standard limit stuff neither, unsuccessfully. Anybody sees which way of approaching this could be productive? Don't need detailed answers, if you have a hunch that'll be enough.</p>
| Maximilian Janisch | 631,742 | <p>This is not always the case because it would imply that <span class="math-container">$$\exp(f(x))>f(x^2)$$</span> for all <span class="math-container">$x$</span> large enough, which can't be the case for all <span class="math-container">$f$</span> that satisfy your conditions. More precisely, it is not true if <span class="math-container">$f$</span> grows very fast.</p>
<p>For example, take any monotonic (for example piecewise linear) continuation of the <a href="https://en.m.wikipedia.org/wiki/Tetration" rel="nofollow noreferrer">tetration</a> <span class="math-container">$$f:\mathbb N\to\mathbb N, n\mapsto 3\uparrow\uparrow n.$$</span></p>
<p>Then we have <span class="math-container">$f(n^2)\gg\exp(f(n))$</span> for all <span class="math-container">$n\in\mathbb N_{>2}$</span> and I think that your limit actually equals <span class="math-container">$\infty$</span>.</p>
|
8,756 | <p>I'd like to suggest that after a Question receives an Accepted Answer, some consideration be given to revising the title (if appropriate) to reflect what the real issue turned out to be.</p>
<p>It seems to me users often pick titles when first posting a question that are uninformative and are worth revisiting once the actually point is identified. Improved subject lines will help in searches, and by extension help with identifying duplicates.</p>
<p>High rep users, it seems to me, are not excessively shy about making changes to tags, which helps with later searches somewhat. But I think a good title trumps all that, and the SE search engine can use all the help we can give it.</p>
<p>Discussion?</p>
| hardmath | 3,111 | <p>Searching meta.SO for "retitled questions" gives only eight hits (but some relevant ideas), while "editing title" returns more than 1,000.</p>
<p>One thread discusses <a href="https://meta.stackexchange.com/q/139970/155839">What do you look for in a question title?</a> and broaches the appropriateness of making changes.</p>
<p>Another thread, <a href="https://meta.stackexchange.com/q/101923/155839">I've found a particular set of “question”s with bad titles; how should I proceed?</a> sparked not only a discussion but a clean-up "posse". One target was (as in Michael Greinecker's example) titles containing the word "Question". Some work was broken out by tags, so one feels at least a tiny corner of the realm has been cleaned up.</p>
<p>I think priority should be given to reworking titles to improve not only the chance of a question appearing in search results, but the recognizability of the specific topic to someone glancing quickly down a search results list, where the title is prominent and accompanied by about four lines of text from the question or answer.</p>
|
8,756 | <p>I'd like to suggest that after a Question receives an Accepted Answer, some consideration be given to revising the title (if appropriate) to reflect what the real issue turned out to be.</p>
<p>It seems to me users often pick titles when first posting a question that are uninformative and are worth revisiting once the actually point is identified. Improved subject lines will help in searches, and by extension help with identifying duplicates.</p>
<p>High rep users, it seems to me, are not excessively shy about making changes to tags, which helps with later searches somewhat. But I think a good title trumps all that, and the SE search engine can use all the help we can give it.</p>
<p>Discussion?</p>
| Gerry Myerson | 8,269 | <p>There are many many questions with titles such as "combinatorics problem", "about a combinatorics problem", "doubt in a combinatorics problem", and so on. I have just retitled three of them. I'll stop there (for the time being). </p>
|
2,607,668 | <p>I am trying to prove/disprove $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)$. Apparently $\operatorname{Arg}(zw)=\operatorname{Arg}(z)+\operatorname{Arg}(w)+2k\pi$ where $k=0,1,\text{ or }-1$, but I have no idea why. I keep on finding that answer online. I am very lost on how to prove this statement, any help would be great. Thank you. </p>
| Alvaro | 518,181 | <p>"but I have no idea why" ... Because exponential is periodic with period $2 \pi i$ over the complex plane. This is a picture from a Mathcad worksheet, but try this with your calculator:</p>
<p><a href="https://i.stack.imgur.com/7rHpu.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7rHpu.gif" alt="enter image description here"></a></p>
|
3,771,972 | <blockquote>
<p>If gcd<span class="math-container">$(m,n)=1$</span>, then <span class="math-container">$\phi(mn)=\phi(m)\phi(n)$</span>.</p>
</blockquote>
<p>My text book write that <span class="math-container">$f:\Bbb Z/_{(mn)}\Bbb Z\to\Bbb Z/_m\Bbb Z \times \Bbb Z/_n\Bbb Z$</span> by <span class="math-container">$f([a]_{mn})=([a]_m,[a]_n)$</span> is an isomorphism.</p>
<p>So far, I think he discusses the additive group.</p>
<p>But when he claims that <span class="math-container">$f(U(_{mn}\Bbb Z))=U(_m\Bbb Z)\times U(_n\Bbb Z)$</span>, he using the multiplication on <span class="math-container">$\Bbb Z/_{(mn)}\Bbb Z$</span> and I'm confused that whether changing operations is alright or not. If he want to use the multiplication on <span class="math-container">$\Bbb Z/_{(mn)}\Bbb Z$</span>, why don't he just write that <span class="math-container">$f:(\Bbb Z/_{(mn)}\Bbb Z)^{\times}\to(\Bbb Z/_m\Bbb Z)^\times \times (\Bbb Z/_n\Bbb Z)^\times$</span> although <span class="math-container">$(\Bbb Z/_n\Bbb Z)^\times$</span> and <span class="math-container">$(\Bbb Z/_m\Bbb Z)^\times$</span> may not be group.
Hope someone can help me to understand the notation here.</p>
| Wuestenfux | 417,848 | <p>The first isomorphism refers to a ring isomorphism.</p>
<p>The second assertion refers to the mapping of the multiplicative groups of the involved rings. They are mapped accordingly.</p>
|
50,023 | <p>two problems from Dugundji's book page $156$. (I don't know why the system deletes the word hi in this sentence)</p>
<p>$1$. Let $X$ be a Hausdorff space. Show that:</p>
<p>a) $\bigcap \{F: x \in F , F \ \textrm{closed}\} = \{x\}$.</p>
<p>b) $\bigcap \{U : x \in U, U \ \textrm{open}\} = \{x\}$.</p>
<p>c) The above properties are not equivalent to being Hausdorff.</p>
<p>$2$. Prove every infinite Hausdorff space contains a countably infinite discrete subspace.</p>
<p>My work:</p>
<ol>
<li>(a) So let $y \not \in \{x\}$ then $x \neq y$. Since $X$ is Hausdorff we can find disjoint open sets $U_{x}$, $V_{y}$. Define $F = X \setminus V_{y}$ then $x \in F$, $F$ is closed and $y \not \in F$, so $y \not \in \bigcap \{F: x \in F , F \ \textrm{closed}\}$.</li>
</ol>
<p>(b) Same as above, just note that $y \not \in U_{x}$, hence not in the intersection.</p>
<p>(c) Not sure here, can we take $X$ any infinite set endowed with the cofinite topology?</p>
<p>Claim: $\bigcap \{F: x \in F , F \ \textrm{closed}\} = \{x\}$. Suppose $y \not \in \{x\}$ then $y \neq x$, that is: $y \not \in \{x\}$. Note that since $X$ is a $T_{1}$ space then $F=\{x\}$ is closed and contains $x$, so $y$ is not in the intersection.</p>
<p>Similarly for $(b)$. But $X$ is not Hausdorff since any two non-empty open sets intersect.</p>
<p>2) How do you prove this one?</p>
| Brian M. Scott | 12,042 | <p>Some style comments on 1(a): As user6312 says, you should just begin by supposing that $y \ne x$; starting with $y \notin \{x\}$ is an unnecessary complication. More important, what Hausdorffness gives you is not just disjoint open sets $U_x$ and $V_y$, but disjoint open sets $U_x$ and $V_y$ <strong>such that $x \in U_x$ and $y \in V_y$</strong>, and for clarity you should say so. (Minor point: if you use different letters for the sets, there's no reason to use subscripts: just call them $U$ and $V$. If you use the subscripts, you might as well use a single basic name, e.g., $U_x$ and $U_y$.)</p>
<p>1(c): Just a brief expansion of what user 6312 said: $\{x\}$ is finite, so by definition its complement is open, and $\{x\}$ <strong>is</strong> therefore one of the closed sets containing $x$, and you don't need to look further to see that $\{x\}$ is the intersection of the closed sets containing $x$. Alternatively, if you've already proved that (a) is true in any $T_1$ space, you merely observe that if $x \ne y$, then $X \setminus \{y\}$ is an open set containing $x$ but not $y$, so $X$ is $T_1$. (But in that case you could have done (a) simply by noting that every Hausdorff space is $T_1$.) For the second part of (c) you do need the argument that $x$ is the intersection of the open sets $X \setminus \{y\}$ for $y \ne x$.</p>
<p>(2) If $x \in X$ has a finite open nbhd, then $\{x\}$ is open (i.e., $x$ is an isolated point). (Why?) Let $I$ be the set of isolated points of $X$; clearly $I$ is discrete, so if $I$ is infinite, just pick any countably infinite subset of it. That's the easy case; you have to work harder when $I$ is finite. </p>
<p>In that case let $Y = X \setminus I$. Note that if $y \in Y$, every open nbhd $V$ of $y$ is infinite. Now build the desired discrete set recursively. Start by choosing distinct points $y_0,y_1 \in Y$, and let $W_0,V_1$ be disjoint open sets with $y_0 \in W_0$ and $y_1 \in V_1$. (Here there's a reason for using different letters: the $V_n$'s are needed temporarily for the construction, but it's the $W_n$'s that we really want.) $V_1 \cap Y$ is infinite (why?), so we can pick a point $y_2 \in V_1 \setminus \{y_1\}$. Now let $W_1$ and $V_2$ be disjoint open subsets of $V_1$ such that $y_1 \in W_1$ and $y_2 \in V_2$. Now $V_2 \cap Y$ is infinite, so we can pick a point $y_3 \in V_2 \setminus \{y_2\}$ and let $W_2$ and $V_3$ be disjoint open subsets of $V_2$ such that $y_2 \in W_2$ and $y_3 \in V_3$. Continue in this fashion to get points $y_n$ and open sets $W_n$ for each $n \in \omega$. Clearly $y_n \in W_n$ for each $n$, and with a little thought you should be able to see that if $m \ne n$, $y_m \notin W_n$. (It's probably best to consider the cases $m<n$ and $m>n$ separately. It may also help to make a sketch of the first few steps of the construction.)</p>
|
1,461,484 | <p>For example, look at this sentence from Perko's text on dynamical system</p>
<p>"It follows from Cauchy Schwarz inequality that if $T \in L(R^n)$ is represented by the matrix $A$ with respect to the standard basis for $R^n$ $_\cdots$" pg 11</p>
<p>What does it mean for a $T: R^n \to R^n$ to be represented by a matrix? Isn't it by definition that $T: R^n \to R^n$ is equivalent to an n by n matrix? Can someone translate exactly what it means by "represented" versus "not represented"?</p>
| skyking | 265,767 | <p>The concept of matrix can be useful in other situation than for linear operators, and there exists linear operators that can not be represented by matrices.</p>
<p>The matrix is just an array of elements and are added and multiplied according to certain rules. For this it is only required that there is a meaningful definition for addition and multiplication for the elements. This means that the elements does not need to be real numbers, or even complex numbers. They can for example be integers (which may not seem that alarming, but it makes a difference). There's even more "odd" things one could put in a matrix as for example integers using modulo-n operations. These later cases means that the matrix do not represent a linear operator.</p>
<p>Then there are linear operators that does not operate on vector spaces of finite dimension or produce values from vector spaces of finite dimensions. For example a linear operator could map from or to a vector space of sequences or even functions. These can obviously not be represented as a matrix.</p>
<p>As for the normal case from linear algebra 101 one should remember that a linear operator is still the same no matter which basis for the vector space(s) you use, but the matrix representing the operator will be different depending on the choice of base. Even though you may think of a vector space as having a canonical base, this is not necessarily the case (it's better to think of them as being a space in it's own right and the base you select be arbitrary and no base is "better" than another).</p>
|
2,282,359 | <p>I am trying to calculate this limit:
$$\lim_{x \to \infty} x^2(\ln x-\ln (x-1))-x$$
The answer is $1/2$ but I am trying to verify this through proper means. I have tried L'Hospital's Rule by factoring out an $x$ and putting that as $\frac{1}{x}$ in the denominator (indeterminate form) but it becomes hopeless afterwards. Also I am a little hesitant about series involving the natural log because of restricted interval of convergence as $x$ is going to infinity. Is there a different approach how to do evaluate this limit? Thanks. </p>
| marty cohen | 13,079 | <p>$\begin{array}\\
\lim_{x \to \infty}x^2(\ln x-\ln (x-1))-x
&= \lim_{x \to \infty} (-x^2(\ln (1-1/x))-x)\\
&= \lim_{y \to 0} (-\frac1{y^2}(\ln (1-y))-\frac1{y})\\
&= \lim_{y \to 0} (-\frac1{y^2}((y+\frac{y^2}{2}+O(\frac1{y^3}))-\frac1{y})\\
&= \lim_{y \to 0} ((\frac1{y}+\frac{1}{2}+O(\frac1{y}))-\frac1{y})\\
&= \lim_{y \to 0} (\frac{1}{2}+O(\frac1{y}))\\
&=\frac12\\
\end{array}
$</p>
<p>You could also write
$(-\frac1{y^2}(\ln (1-y))-\frac1{y})
=-\frac{\ln (1-y)+y}{y^2}
$
and use
L'Hopital.</p>
|
2,282,359 | <p>I am trying to calculate this limit:
$$\lim_{x \to \infty} x^2(\ln x-\ln (x-1))-x$$
The answer is $1/2$ but I am trying to verify this through proper means. I have tried L'Hospital's Rule by factoring out an $x$ and putting that as $\frac{1}{x}$ in the denominator (indeterminate form) but it becomes hopeless afterwards. Also I am a little hesitant about series involving the natural log because of restricted interval of convergence as $x$ is going to infinity. Is there a different approach how to do evaluate this limit? Thanks. </p>
| hamam_Abdallah | 369,188 | <p>Using fact that
$$\ln (1+u)=u-\frac {u^2}{2}(1+\epsilon (u)) $$ with $\lim_0 \epsilon (u)=0$,
we get
$$x^2 \ln (\frac{x}{x-1})-x=$$</p>
<p>$$x^2\ln(1+\frac {1}{x-1})-x =$$
$$=x^2 (\frac {1}{x-1}-\frac{1}{2 (x-1)^2} (1+\epsilon (x)) -x$$
with $\lim_{+\infty}\epsilon (x)=0$,
$$=\frac {x}{x-1}-\frac {x^2}{2 (x-1)^2}\left(1+\epsilon (x)\right) $$</p>
<p>the limit is
$$1-\frac {1}{2}=\frac 12$$</p>
|
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