qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,762,040 | <p>I started to think about this problem and then factored $n^5 - n$ to $(n^2 - 1)(n^2 + 1)(n)$, and later to $(n-1)(n)(n+1)(n^2 + 1)$. I know that $(n-1)(n)(n+1)$ is divisible by $6$, but it is not that case $5$ divides $n^2 + 1$ for any integer $n$, so i can´t use the multiplication property. Can anyone help me finish this proof?</p>
| David C. Ullrich | 248,223 | <p>One of the other answers requires that you know something - the other is very clever. A hint for a proof that doesn't require either non-trivial theorems or cleverness: $n=5k+j$, where $j\in\{0,1,2,3,4\}$.</p>
|
16,754 | <p>Let $c$ be an integer, not necessarily positive and not a square. Let $R=\mathbb{Z}[\sqrt{c}]$
denote the set of numbers of the form $$a+b\sqrt{c}, a,b \in \mathbb{Z}.$$
Then $R$ is a subring of $\mathbb{C}$ under the usual addition and multiplication.</p>
<p>My question is: if $R$ is a UFD (unique factorization domain), does it follow that
it is also a PID (principal ideal domain)?</p>
| Matt E | 221 | <p>The answer is yes. The argument is as follows: if $R$ is a UFD, then it is necessarily integrally closed in its fraction field $K = \mathbb Q(\sqrt{c})$, and thus is
equal to the full ring of algebraic integers in $K$. A general fact about such full rings of algebraic integers is that if they are UFDs then they are PIDs, the reason being that in these rings, one always has the unique factorization of non-zero ideals into prime ideals, and it is not hard to see that the UFD property forces prime ideals (and hence any product of prime ideals) to be principal.</p>
|
4,171,907 | <blockquote>
<p>If <span class="math-container">$3\sin x +5\cos x=5$</span> then prove that <span class="math-container">$5\sin x-3\cos x=3$</span></p>
</blockquote>
<p>What my teacher did in solution was as follows</p>
<p><span class="math-container">$$3\sin x +5\cos x=5 \tag1$$</span></p>
<p><span class="math-container">$$3\sin x =5(1-\cos x) \tag2$$</span></p>
<p><span class="math-container">$$3=\frac{5(1-\cos x)}{\sin x} \tag3$$</span></p>
<p><span class="math-container">$$3=\frac{5\sin x}{(1+\cos x)} \tag4$$</span></p>
<p><span class="math-container">$$5\sin x-3\cos x=3 \tag5$$</span></p>
<p>However this should not be true when <span class="math-container">$\sin x=0$</span>, as division by zero is not defined; and also, if <span class="math-container">$\sin x=0$</span>, then the expression we have to prove evaluates to <span class="math-container">$-3$</span>. In other words question is incomplete but my teacher denied it.</p>
<p>Am I correct in my reasoning?</p>
| Umesh shankar | 80,910 | <p>Yes you are right,he/she should have broken the problem into two cases.</p>
<p>Case <span class="math-container">$1.$</span> When <span class="math-container">$\sin x\ne 0$</span>, then according to your teacher <span class="math-container">$5\sin x-3\cos x=3$</span>.</p>
<p>Case <span class="math-container">$2.$</span> If <span class="math-container">$\sin x=0$</span>, then from the given equation we get <span class="math-container">$\cos x=1$</span>. Plugging these in to <span class="math-container">$5\sin x-3\cos x$</span> we get the required value as <span class="math-container">$-3$</span>.</p>
<p>We can do the same problem in an alternative way:</p>
<p><span class="math-container">$$3\sin x+5\cos x=5$$</span>
Let <span class="math-container">$$5\sin x-3\cos x =k$$</span></p>
<p>Squaring and adding both the above equations we get</p>
<p><span class="math-container">$$k^2+25=34$$</span>
<span class="math-container">$\implies$</span>
<span class="math-container">$$k=\pm 3$$</span></p>
|
59,046 | <p>Let $A \in M_n(\mathbb R)$ and suppose its minimal polynomial is:
$$M_A(t)=\prod_{i=1}^{k}(t-\lambda_i)^{\textstyle s_i}.$$</p>
<p>When $\lambda _1,\lambda_2,\lambda _3,......,\lambda _k$ are distinct eigenvalues.</p>
<p>We define a new matrix: $B\in M_{2n}(\mathbb R)$ by:
$$\left(\begin{matrix}
A &I_n \\
0 & A
\end{matrix}\right)$$</p>
<p>I need to prove that the minimal polynomial of $B$ is
$$M_B(t)=\prod_{i=1}^k(t-\lambda_i)^{\textstyle s_{i}+1}.$$</p>
<hr/>
<p><strong>Edit:</strong> <em>What follows below refered to the original version of this question, in which the definition of $B$ was:</em>
$$B = \left(\begin{array}{cc}A&I_n\\I_n&A\end{array}\right).$$</p>
<p><hr/>
I tried to do that in induction. I don't understand this basic case: for $1$x$1$ matrix: $\begin{pmatrix}
5
\end{pmatrix}$ we get that $ \begin{pmatrix}
5 & 1\\
1& 5
\end{pmatrix}$ 's minimal polynomial is $(x-4)(x-6)$ and it doesn't answer the condition. so I have also a problem with understanding the question I guess. </p>
<p>Thanks!!</p>
| Yuval Filmus | 1,277 | <p>Prove by induction that for any polynomial $P$,
$$ P(B) = P\left( \begin{bmatrix} A & I \\ I & A \end{bmatrix} \right) =
\frac{1}{2} \begin{bmatrix} P(A+I)+P(A-I) & P(A+I)-P(A-I) \\ P(A+I)-P(A-I) & P(A+I)+P(A-I) \end{bmatrix}. $$</p>
<p>Hence $$P(B)=0 \Leftrightarrow P(A+I) = P(A-I) = 0.$$
You take it from here.</p>
|
3,362,916 | <p>I'm trying to graph <span class="math-container">$|x+y|+|x-y|=4$</span>. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at <span class="math-container">$\pi/4$</span> to the horizontal axis (take it to be <span class="math-container">$x$</span>)<span class="math-container">$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$</span></p>
<p>However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.</p>
| fGDu94 | 658,818 | <p>Once <span class="math-container">$x$</span> is fixed you can look for solutions in three distinct regions: <span class="math-container">$y>x$</span>, <span class="math-container">$-x<y<x$</span>, <span class="math-container">$y<-x$</span>. By doing this you can simplify the absolute values and find solutions.</p>
|
3,362,916 | <p>I'm trying to graph <span class="math-container">$|x+y|+|x-y|=4$</span>. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at <span class="math-container">$\pi/4$</span> to the horizontal axis (take it to be <span class="math-container">$x$</span>)<span class="math-container">$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$</span></p>
<p>However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.</p>
| ASTRONO | 706,272 | <p>The given graph forms a square with side 4.Simple logic is that when (X+y)+(x-y)=c
When (X+y) approaches c,(x-y) approaches 0.
From the above thing the upper boundary will be 2x=4.
You get X=2,y=2.
Because there is mod if you can solve one boundary it gets mirrored about both X and y axis</p>
|
3,362,916 | <p>I'm trying to graph <span class="math-container">$|x+y|+|x-y|=4$</span>. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at <span class="math-container">$\pi/4$</span> to the horizontal axis (take it to be <span class="math-container">$x$</span>)<span class="math-container">$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$</span></p>
<p>However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.</p>
| Doug M | 317,162 | <p>Looks like you have said</p>
<p><span class="math-container">$u = \frac {\sqrt {2}}{2} (x + y)\\
v = \frac {\sqrt {2}}{2} (x - y)$</span></p>
<p>Which rotates coordinate system 45 degrees.</p>
<p>And your expression becomes</p>
<p><span class="math-container">$|\sqrt 2 u| + |\sqrt 2 v| = 4$</span> or <span class="math-container">$|u| + |v| = \sqrt 2$</span></p>
<p>So what does this graph look like in an unrotated frame?</p>
<p>Do you know about the taxicab metric? </p>
<p><a href="https://en.wikipedia.org/wiki/Taxicab_geometry" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Taxicab_geometry</a></p>
<p>What we have here is the set of points that are <span class="math-container">$\sqrt 2$</span> units from the origin by the taxicab metric.</p>
<p><a href="https://i.stack.imgur.com/6e6xW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6e6xW.png" alt="enter image description here"></a></p>
<p>And rotate coordinates back to the original system.</p>
|
3,058,019 | <blockquote>
<p>Two numbers <span class="math-container">$297_B$</span> and <span class="math-container">$792_B$</span>, belong to base <span class="math-container">$B$</span> number system. If the first number is a factor of the second number, then what is the value of <span class="math-container">$B$</span>?</p>
</blockquote>
<p>Solution:
<a href="https://i.stack.imgur.com/6ScrF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ScrF.jpg" alt="enter image description here"></a> </p>
<p>But base cannot be negative. Could someone please explain where I am going wrong?</p>
| nonuser | 463,553 | <p>Since <span class="math-container">$b+1>0$</span> and <span class="math-container">$$(b+1)(2b+7)\mid (7b+2)(b+1)\implies 2b+7\mid 7b+2$$</span></p>
<p>we have <span class="math-container">$$2b+7\mid (7b+2)-3(2b+7) = b-19$$</span></p>
<p>so if <span class="math-container">$b-19> 0$</span> we have <span class="math-container">$$2b+7\mid b-19 \implies 2b+7\leq b-19 \implies b+26\leq 0$$</span></p>
<p>which is not true. So <span class="math-container">$b\leq 19$</span>. By trial and error we see that <span class="math-container">$b=4$</span> and <span class="math-container">$b=19$</span> works.</p>
|
2,109,347 | <p>My statistics note states that the variance of the empirical distribution is
$v= \sum_{i=1}^{n}(x_i-\bar x )^2\frac {1} {n}$ which the author then re-writes as
$v= \sum_{i=1}^{n}x_i^2 (\frac {1} {n}) - \bar x^2$. How is this achieved?</p>
| szw1710 | 130,298 | <p>$$\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n}\left(\sum_{i=1}^n x_i^2-2\sum_{i=1}^n x_i\cdot \bar{x}+\sum_{i=1}^n \bar{x}^2\right)=\frac{1}{n}\left(\sum_{i=1}^nx_i^2-2\bar{x}\sum_{i=1}^nx_i+\bar{x}^2\sum_{i=1}^n 1\right).$$</p>
<p>Hence</p>
<p>$$
\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n}\sum_{i=1}^n x_i^2-2\bar{x}\cdot\bar{x}+\frac{1}{n}\cdot n \bar{x}^2=\frac{1}{n}\sum_{i=1}^n x_i^2-\bar{x}^2.
$$</p>
|
2,869,442 | <blockquote>
<p>Check whether the series
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$$
is convergent.</p>
</blockquote>
<p>I tried to sandwich the function by $\dfrac{1}{1+x^2}$ and $\dfrac{x}{1+x^2}$ , but this did not help at all.
Any other way of approaching?</p>
| Chappers | 221,811 | <p>Since $\sqrt{x}$ is increasing and $1/(1+x^2)$ is decreasing, we have
$$ \int_0^{1/n} \frac{\sqrt{x}}{1+x^2} \, dx < \frac{1}{\sqrt{n}} \int_0^{1/n}\frac{dx}{1+x^2} < \frac{1}{\sqrt{n}} \int_0^{1/n} dx = \frac{1}{n^{3/2}}. $$
And $\sum_n 1/n^{3/2}$ converges.</p>
|
2,291,310 | <p>I'm seeking an alternative proof of this result:</p>
<blockquote>
<p>Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that
$$|CI|^2=\frac12\left(\left(\;|BC|-|AB|\;\right)^2+|AC|^2\right)$$</p>
</blockquote>
<p><strong>My Proof.</strong> Draw $ID \perp AB$, $IE\perp BC$, and $IF\perp CE$.</p>
<p>We have $|ID|=|IE|=|IF|=x$. Since $\triangle ADI$ is right isosceles triangle, we also have that $|AD|=|ID|=x$. Respectively, we have: $$|ID|=|IF|=|IE|=|AD|=|AF|=x$$ </p>
<p>$\triangle BDI=\triangle BEI \Rightarrow |BD|=|BE|=y$. And $|CE|=|CF|=z$</p>
<p>We have:<br>
$$|CI|^2=|CE|^2+|IE|^2=x^2+z^2 \tag{1}$$</p>
<p>And
$$\begin{align}\frac12\left(\left(|BC|-|AB|\right)^2+|AC|^2\right) &=\frac12\left(\left(\;\left(y+z\right)-\left(x+y\right)\;\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac12\left(\left(x-z\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac22\left(x^2+z^2\right) \\[4pt]
&=x^2+z^2
\tag{2}\end{align}$$</p>
<p>From $(1);(2)$ we are done. $\square$</p>
| Jack D'Aurizio | 44,121 | <p>Alternative proof: by Stewart's theorem the length $\ell_c$ of the angle bisector through $C$ is given by
$$ \ell_c^2 = \frac{ab}{(a+b)^2}\left[(a+b)^2-c^2\right] $$
and by Van Obel's theorem and the bisector theorem $\frac{CI}{\ell_c}=\frac{a+b}{a+b+c}$. It follows that, in general:
$$ CI^2 = \frac{ab}{(a+b+c)^2}\left[(a+b)^2-c^2\right]=\frac{ab(a+b-c)}{(a+b+c)} $$
and it is enough to prove that the given hypothesis ($a^2=b^2+c^2$) ensure
$$ \frac{ab(a+b-c)}{(a+b+c)} = \frac{(a-c)^2+b^2}{2} $$
or
$$ 2ab(a+b-c) = (a+b+c)(a^2+b^2+c^2-2ac). $$
That is trivial since the difference of the RHS and the LHS of the last expression is
$$ (b+c-a)(a^2-b^2-c^2). $$</p>
|
19,285 | <p>Is anyone aware of Mathematica use/implementation of <a href="http://en.wikipedia.org/wiki/Random_forest">Random Forest</a> algorithm?</p>
| Seth Chandler | 5,775 | <p>I'm going to be bold and attempt to edit the Ross code so that it is (a) a little easier to understand and (b) takes the same form of argument as LinearModelFit and other Mathematica prediction creators. I've also added some annotations to the critical code. My variable names are now far longer than the Ross names but perhaps for informative. So far in my testing this code works the same as Ross, but it is possible I have messed something up in my rewrite.</p>
<p>Part 1. Same as Ross but I add a function informationGain that determines the information gain one obtains about y from knowing the ith feature of x.</p>
<pre><code>condEnt = Statistics`Library`NConditionalEntropy;
ent = Statistics`Library`NEntropy;
maxI[v_] := Ordering[v, -1][[1]];
informationGain[x_, y_, i_] :=
Statistics`Library`NEntropy[x] -
Statistics`Library`NConditionalEntropy[x[[i]], y]
</code></pre>
<p>Part 2. Same idea as Ross but x now contains not just the features (independent variables in statistics lingo) but the entire data set. It thus has the same structure as the first argument to LinearModelFit and NonlinearModelFit. </p>
<pre><code> cart[x_?MatrixQ /; ent[Last /@ x] == 0, m_] := First[Last /@ x]
</code></pre>
<p>Part 3. The idea here is again to let the first argument just be the entire data set. What I have then done is to use a lot of local variables with hopefully evocative names and some annotations to better explain what is going on in the elegant but terse Ross code. The text below explains the annotations.</p>
<pre><code>cart[x_?MatrixQ, m_Integer] :=
Block[{dims = Dimensions[x], numberOfInstances, numberOfAttributes,
allAttributes, y, byFeature,
h, keep, bestVar, ub, allValuesOfBestVar, mask, best,
xBiggerThanBestSplitQ, bestSplittingValue,
iXBestGTESplit,
iXBestLTSplit},
numberOfInstances = First[dims];
numberOfAttributes = Last[dims] - 1;(*1*)
allAttributes = Range[numberOfAttributes];(*2*)
y = Part[x, All, -1];(*3*)
byFeature = Transpose[(Most /@ x)];(*4*)
h = ent[y];(*5*)
keep = RandomSample[allAttributes, m];(*6*)
bestVar =
maxI[Table[
If[MemberQ[keep, i], informationGain[byFeature, y, i], 0], {i,
allAttributes}]];(*7*)
allValuesOfBestVar = byFeature[[bestVar]];(*8*)
ub = Union[allValuesOfBestVar];(*9*)
mask = UnitStep[allValuesOfBestVar - #] & /@
ub;(*10*)
best = maxI[(h - condEnt[#, y] & /@ mask)];(*11*)
xBiggerThanBestSplitQ = mask[[best]];(*12*)
bestSplittingValue = ub[[best]];(*13*)
iXBestGTESplit=
Pick[x, xBiggerThanBestSplitQ, 1];(*14*)
iXBestLTSplit =
Pick[x, xBiggerThanBestSplitQ, 0];(*15*)
If[Min[xBiggerThanBestSplitQ] === Max[xBiggerThanBestSplitQ],
RandomChoice[y] ,
{bestVar, bestSplittingValue,
cart[iXBestGTESplit
m],
cart[
iXBestLTSplit, m]}](*16*)
]
</code></pre>
<p>Part 3 Annotations.</p>
<ol>
<li><p>The number of attributes is just the number of columns minus one, because the last column is the class value.</p></li>
<li><p>I like the idea of iterating over a set of values, so I create it here.</p></li>
<li><p>The class values y are the last column of the x matrix.</p></li>
<li><p>To get the features in the same form as Ross, Transpose x and then take Most of each row.</p></li>
<li><p>Calculating the entropy of the y values in this level of the decision tree.</p></li>
<li><p>Use an m-length subset of the attributes. Note that in my code, m is the number to keep, not the number to drop.</p></li>
<li><p>Find the information gain for each selected attribute. Find the selected attribute that produces the greatest information gain. This produces an index into the attributes. We call it <strong>bestVar</strong>.</p></li>
<li><p>Get the features in the bestVar th column of x. Call it <strong>allValuesOfBestVar</strong>.</p></li>
<li><p>Find the different values of the features in <strong>allValuesOfBestVar</strong>. Call it <strong>ub</strong>. (Union of Best values). These now represent possible splitting points in our best feature.</p></li>
<li><p>This is the clever part. For each possible splitting point, determine whether each value is above or equal to that splitting point (1) or below it (9). This produces a matrix that is Length[ub] x Length[y]. Call this matrix <strong>mask</strong>.</p></li>
<li><p>Rather than compute the information gain from knowing the feature values, compute the information gain from knowing whether the feature for a particular instance is above a splitting point for all the splitting points. Pick the splitting point that produces the most information gain. The index of the splitting point into the list of splitting points is named <strong>best</strong>. Note that this is the second use of the idea of conditional entropy.</p></li>
<li><p>Now select the <strong>best</strong> part of the <strong>mask</strong>. This gives us the information on >= or < using the best splitting point for the data for the best feature of each instance.</p></li>
<li><p>We'll need that best splitting value again, so we capture it here as <strong>bestSplittingValue</strong>.</p></li>
<li><p>Pick the instances for which the best feature has a value above or equal to the best splitting value. Call those instances <strong>iXBestGTESplit</strong></p></li>
<li><p>Pick the instances for which the best feature has a value less than the best splitting value. Call those instances <strong>iXBestLTSplit</strong></p></li>
<li><p>The recursion step. Run cart again first on the <strong>iXBestGTESplit</strong> data and then on the
<strong>iXBestLTSplit</strong> data. But before doing so, create a list that shows what feature you split on and what the best splitting value was. Stop when your data has no more information to yield.</p></li>
</ol>
<p>Part 4. This code to classify an instance works exactly the same as in Ross. Notice that here, and now consistent with the way it is done above, x is a vector of attributes. It will also work as a vector that contains a list of which most of the elements are attributes and the last element is the class value. Thus, one can now just map over the instances in some testing set to get the predictions.</p>
<pre><code>classify[x_?MatrixQ, Except[_List, d_]] := d
classify[x_?MatrixQ, {best_, value_, d1_, d2_}] :=
classify[x, If[x[[best]] < value, d2, d1]]
</code></pre>
<p>Part 5. The random forest creation function is similar to Ross but, again, x is now a unified matrix containing the instances. I have added an optional argument subsetFactor that can speed up your code (at the risk of losing accuracy) by using just a subset of the instances to create the trees each time.</p>
<pre><code>rf[x_?MatrixQ, m_Integer, ntree_Integer, subsetFactor_: 1] :=
Table[With[{boot =
RandomChoice[Range[Length[x]], Round[subsetFactor*Length[x]]]},
cart[x[[boot]], m]], {ntree}]
</code></pre>
|
519,764 | <p>Question: show that the following three points in 3D space A = <-2,4,0>, B = <1,2,-1> C = <-1,1,2> form the vertices of an equilateral triangle.</p>
<p>How do i approach this problem?</p>
| Community | -1 | <p><strong>Hint</strong>: A triangle is equilateral if and only if all its sidelengths are equal. We compute</p>
<p>$$\|A - B\| = \|\langle 3, 2, 1\rangle\| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}$$</p>
<p>$$\|A - C\| = \|\langle -1, 3, -2\rangle\| = ? $$</p>
<p>$$\|B - C\| = \|\langle 2, 1, -3\rangle\| = ? $$</p>
|
519,764 | <p>Question: show that the following three points in 3D space A = <-2,4,0>, B = <1,2,-1> C = <-1,1,2> form the vertices of an equilateral triangle.</p>
<p>How do i approach this problem?</p>
| rnjai | 42,043 | <p>Find the distance between all the pairs of points
$$|AB|,|BC|,|CA|$$
and check if
$$|AB|=|BC|=|CA|$$
For example:
$$|A B| = \sqrt{(-2-1)^2 + (4-2)^2 + (0-(-1))^2} = \sqrt{14}$$</p>
|
519,764 | <p>Question: show that the following three points in 3D space A = <-2,4,0>, B = <1,2,-1> C = <-1,1,2> form the vertices of an equilateral triangle.</p>
<p>How do i approach this problem?</p>
| David Park | 99,469 | <p>This would be the Grassmann algebra approach. This is a fairly simple problem and the main advantage of Grassmann algebra is that we can set up the objects and calculate with them in a natural manner.</p>
<pre><code><< GrassmannCalculus`
SetPreferences["Grassmann3Space", "Vector"]
</code></pre>
<p>The following defines the three vertices as points in 3-space:</p>
<pre><code>vertexA = Origin + FreeBasis.{-2, 4, 0}
vertexB = Origin + FreeBasis.{1, 2, -1}
vertexC = Origin + FreeBasis.{-1, 1, 2}
</code></pre>
<p>Giving:</p>
<p><img src="https://i.stack.imgur.com/VkpxU.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/dHQlw.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/mfctj.jpg" alt="enter image description here"></p>
<p>The Measure routine gives the distance between any pair of points. We check if they are all equal.</p>
<pre><code>Measure[vertexB - vertexA] == Measure[vertexC - vertexB] ==
Measure[vertexA - vertexC]
True
</code></pre>
<p>If we want the values:</p>
<pre><code>{Measure[vertexB - vertexA], Measure[vertexC - vertexB],
Measure[vertexA - vertexC]}
</code></pre>
<p>Giving:</p>
<p><img src="https://i.stack.imgur.com/LJJN6.jpg" alt="enter image description here"></p>
|
496,011 | <p>Suppose that $ p_n(t) $ is the probability of finding n particle at a time t. And the dynamics of the particle is described by this equation : </p>
<p>$$ \frac{d}{dt} p_n(t) = \lambda \Delta p_n(t) $$</p>
<p>Defining one - dimensional lattice translation operator $ E_m = e^{mk} $ with $ km - mk = 1 $ and $\Delta = E_1 + E_{-1} - 2 $ is a lattice laplacian.</p>
<p>So, here are my questions:</p>
<ol>
<li>What is the one-dimensional lattice translation operator ? I think this one is a term from statistical physics, can you give me a simple explanation or a reference ?</li>
<li>What is the lattice laplacian ? Is it similar to discrete laplacian ? Could you give me a reference for this one ?</li>
<li>Is it possible to solve this equation analytically ?</li>
</ol>
<p>Oh, all of this equation is about random walk with a boundary (random walk of a particle)</p>
<p>Thanks</p>
| Cameron Buie | 28,900 | <p>I'm afraid not. You should not have an equation with a "differential factor" on one side and not on the other (that is, $dx=2x$ is nonsense). For more on how to deal with differential factors, you might find the second part of <a href="https://math.stackexchange.com/a/155769/28900">this answer</a> (from "Now, if I wrote..." through "...let's get back to your problem.") useful.</p>
<p>What you <em>can</em> say is that $$\frac{du}{dx}=2x,$$ so that $$du=2x\,dx,$$ so that $$x\,dx=\frac12\,du.$$ Then your substitution gives you $$\int x^3\sqrt{x^2+1}\,dx=\frac12\int x^2\sqrt{u}\,du.$$ We're not quite there, yet, though. Can you rewrite $x^2$ in terms of $u$?</p>
|
496,011 | <p>Suppose that $ p_n(t) $ is the probability of finding n particle at a time t. And the dynamics of the particle is described by this equation : </p>
<p>$$ \frac{d}{dt} p_n(t) = \lambda \Delta p_n(t) $$</p>
<p>Defining one - dimensional lattice translation operator $ E_m = e^{mk} $ with $ km - mk = 1 $ and $\Delta = E_1 + E_{-1} - 2 $ is a lattice laplacian.</p>
<p>So, here are my questions:</p>
<ol>
<li>What is the one-dimensional lattice translation operator ? I think this one is a term from statistical physics, can you give me a simple explanation or a reference ?</li>
<li>What is the lattice laplacian ? Is it similar to discrete laplacian ? Could you give me a reference for this one ?</li>
<li>Is it possible to solve this equation analytically ?</li>
</ol>
<p>Oh, all of this equation is about random walk with a boundary (random walk of a particle)</p>
<p>Thanks</p>
| abiessu | 86,846 | <p>HINT: when completing a $u$ substitution, the typical flow is as follows:</p>
<p>$$\int x^3\sqrt{x^2+1}dx$$</p>
<p>Substitute $u=x^2+1$, then $du=2xdx$ and we have $x^2=u-1$:</p>
<p>$$\int {(u-1)\sqrt u du\over 2}$$</p>
<p>From there, the remaining steps are to integrate, then reverse-substitute to obtain the function relative to $x$, and to add the unknown constant.</p>
|
14,448 | <p>Here's the most common way that I've seen letter grades assigned in undergrad math courses. At the end of the semester, the professor: 1) computes the raw score (based on homework, quizzes, and tests) for each student; 2) writes down all the raw scores in order; 3) somewhat arbitrarily clusters the scores into groups corresponding to grades of A, B, C, etc.</p>
<p>There are a few problems with this approach though. One might object that the assignment of letter grades is too arbitrary. Students also might object that they don't know in advance what grade they are likely to get in the course.</p>
<p>On the other hand, it's difficult to assign letter grades in a less arbitrary manner. For example, if we declare in advance that a score of 80-90 on an exam corresponds to a B, we might find that the exam was too difficult and that scores on the exam were lower than expected.</p>
<p><strong>What do you think is the best approach to assigning letter grades in an undergraduate math course?</strong></p>
| Matt Ollis | 10,159 | <p>As @shoover hints at in their comment, the question assumes that you've already answered the question of why you are assigning letter grades at all. Assuming the answer to that is something along the lines of "because I have no choice" or "that's just how it's done", here's how I approach it. (Perhaps useful info: my college has a brief description of what letter grades should mean across disciplines. For example, a B is an indicator that the student is ready---and encouraged---to pursue further studies in the discipline.)</p>
<p>I don't use numbers at all any more, and grade assignments like (I assume) my colleagues in the humanities do. Does the work demonstrate mastery of the material? Put an A on it. Is it littered with errors, both small and large, but with evidence of a reasonable amount of understanding of the basics? C. Of course, there's written feedback on individual errors/questions and the assignment/exam as a whole as well. This written feedback should, and usually does, correlate with the grade.</p>
<p>Course grades are then compiled from these individual grades in some somewhat arbitrary but pre-defined and known way. (Best 6 assignments from 8 count 60% and the final exam counts 40%, or whatever.) I also reserve the right to move a student's grade up or down by upto one letter based on promptness of submission and class participation and whatever other sundries pop up during the semester. (This is much more often a boon for doing something not explicitly assigned credit rather than a dock.)</p>
<p>In my experience, this bypasses a lot of the concerns that seem to have driven the question.</p>
<p>[Bonus anecdote; related but not strictly part of an answer: I studied and then taught a little in the UK in places where 70ish earned an A and a passing mark was around 40. I moved to the US and started teaching without knowing about the usual US system. In my first Calc 1 assignment my students did well with some tough (in context) questions and most scored between 60 and 80 (so, I thought, at least a B- and some really strong As). They were not happy. This is part of what prompted the switch to my current system, but I like it regardless of its origin.] </p>
|
2,121,583 | <p>Question: Let $f,g: X \rightarrow \mathbb{R}$ continous (over $X$, and $X$ is a metric space). If $\overline{Y}\subset X $, and $f(y)=g(y)$ for every $y\in Y $, prove that $\left.f\right|_\overline{Y}= \left.g\right|_\overline{Y}$.</p>
<p>Attempt:</p>
<p>Since $\overline{Y} \subset X$, it follows that $\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ are continuous functions. Given $\epsilon>0$, let $a\in \overline{Y}$. Since $\overline{Y}=Y\cup Y'$, assume that $a \in Y'$. </p>
<p>$\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ are continuous, so there exists $\delta_1,\delta_2>0$ such that for $x\in \overline{Y}$, $|x-a|<\delta_1$ and $|x-a|<\delta_2$ we have:</p>
<p>$|f(x)-f(a)|<\epsilon/2$ and $|g(x)-g(a)|<\epsilon/2$, respectively.</p>
<p>Since we are supposing that $a\in Y'$, the continuity of $\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ also gives us that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Using the hyphothesis, we conclude that, in particular, $f(a)=g(a)$.</p>
<p>Hence, taking $\delta_0$=min$\{\delta_1,\delta_2\}$, $x \in \overline{Y}$ and $x\in(a-\delta_0,a+\delta_0)$ implies that :</p>
<p>$|f(x)-g(x)|\leq |f(x)-f(a)|+|f(a)-g(x)|<\epsilon/2+\epsilon/2=\epsilon.$</p>
<p>Since $\epsilon$ is arbitrary, it proves that $f(x)=g(x)$ for every $x\in \overline{Y}$.</p>
<p>Doing the same, supposing that $a\in Y$, we still get that $f(x)=g(x)$ for every $x\in \overline{Y}$.</p>
<p>In either case, we coclude that $\left.f\right|_\overline{Y}= \left.g\right|_\overline{Y}$.</p>
<p>Is it correct?</p>
| Arthur | 15,500 | <p>Assume there is a point $y_0\in\overline Y$ such that $f(y_0)\neq g(y_0)$. Then, by continuity, there must be an $\epsilon>0$ such that for any $y\in B_{\epsilon}(y_0)$, we have $f(y)\neq g(y)$. But because some of those $y$ necessarily must be in $Y$, we have reached a contradiction.</p>
|
29,861 | <p>The meta question <em><a href="https://math.meta.stackexchange.com/q/29857/290189">Not actually a question, just a rant!</a></em> has inspired me to ask for <em>what</em> an answerer can do in case of self-deletion by the question asker while the answerer is typing the answer.</p>
<p>Since per se site is supposed to be a collection of Q&A, can one re-post the deleted question provided that it has good context?</p>
| daniel | 18,124 | <p>Would it not be possible to add a feature that gives notice of imminent self-closure? Like a 5-minute warning? Or just require notice in a comment? I have closed quite a few of my own questions but try to indicate imminent closure in a comment in case someone is working on an answer so they have a chance to object. I think most posters would refrain from closing if asked, and they would keep a question open for the same reason they would have closed it: regard for the opinions/time of other users. </p>
<p>The legalistic argument leaves me cold. Deleting doesn't nullify posting...? It's surely a reflection of the poster's intent, which should be honored unless there is some compelling reason to do otherwise. I can't imagine insisting on legal ownership of a MSE question over the OP's sense that the question has no merit. There is no shortage of interesting unanswered questions on the site. </p>
|
4,168,223 | <p>Let <span class="math-container">$R$</span> be the row reduced echelon form of a <span class="math-container">$4 \times 4$</span> real matrix <span class="math-container">$A$</span> and
let the
third column of <span class="math-container">$R$</span> be <span class="math-container">$\left[\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right]$</span>. Then which is true?</p>
<p>P): If <span class="math-container">$\left[\begin{array}{l}\alpha \\ \beta \\ \gamma \\ 0\end{array}\right]$</span> is a solution of <span class="math-container">$A \mathrm{x}=0$</span>, then <span class="math-container">$\boldsymbol{\gamma}=0$</span>.</p>
<p>Q): For all <span class="math-container">$\mathrm{b} \in \mathbb{R}^{4}, \operatorname{rank}[A \mid \mathrm{b}]=\operatorname{rank}[R \mid \mathbf{b}]$</span>.</p>
<p>For <span class="math-container">$P$</span> <span class="math-container">\begin{aligned}
&{\left[\begin{array}{llll}
a & b & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & \phi \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
\alpha \\
\beta \\
\gamma \\
0
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0\\
0
\end{array}\right]} \\
&a \alpha+b \beta+0+0=0 \\
&0+0+\gamma+0=0 \\
&\Rightarrow \gamma=0
\end{aligned}</span></p>
<p>For <span class="math-container">$Q$</span></p>
<p>Rank of a matrix and rank of Row echelon matrix is same . So Q is also correct.
So both the statements are true. Is my approach correct?
Thanks in advance.</p>
| Anurag A | 68,092 | <p>Assuming that you are using the <strong>same</strong> <span class="math-container">$\bf{b}$</span> in <span class="math-container">$[A | \bf{b}]$</span> and <span class="math-container">$[R | \bf{b}]$</span> (i.e. you have not row reduced the right hand side column), then <span class="math-container">$Q$</span> need NOT be true. For example, we can have the last entry of <span class="math-container">$\bf{b}$</span> as (say) <span class="math-container">$1$</span>, then the rank of <span class="math-container">$[R | \bf{b}]$</span> is more than that of <span class="math-container">$[A | \bf{b}]$</span>.</p>
<p>Here is a simple example that can explain it better:
<span class="math-container">$$[A | \bf{b}] \rightarrow [R | \bf{c}] \implies \left[\begin{array}{lll|l}1&1&1&1\\1&1&1&1\end{array}\right] \rightarrow \left[\begin{array}{lll|l}1&1&1&1\\0&0&0&0\end{array}\right].$$</span>
Clearly <span class="math-container">$[A|\bf{b}]$</span> has rank <span class="math-container">$1$</span>. But if we look at <span class="math-container">$[R | \color{red}{\bf{b}}]$</span>, then we have
<span class="math-container">$$\left[\begin{array}{lll|l}1&1&1&\color{red}{1}\\0&0&0&\color{red}{1}\end{array}\right],$$</span>
which has rank <span class="math-container">$2$</span>.</p>
|
2,394,716 | <p>Let $A = \{\frac{x}{2} - \lfloor\frac{x+1}{2}\rfloor : x \in \mathbb{R} \}$ </p>
<p>Does <strong>supremum</strong> and <strong>infimum</strong> of $A$ exist ? If the answer is yes then find them .</p>
<p>My try : I rewrite the expression $\frac{x}{2} - (\lfloor 2x \rfloor - \lfloor x \rfloor)$ but it doesn't help really . Also I substituted $x$ with $n + p $ which $n$ is an integer and $0\le p \lt 1$ and wasn't helpful again .</p>
| Dietrich Burde | 83,966 | <p>It really depends on the context. For solving linear equations by substitution, the standard form of $-4x+3y=5$ would be multiplying by $\frac{1}{4}$ to obtain
$x=\frac{3y-5}{4}$ for substitution. So the normal form could be
$$
-4x+3y-5=0,
$$</p>
<p>or
$$
\; x-\frac{3}{4}y+\frac{5}{4}=0.
$$
The last normal form automatically determines the sign of the constant term.</p>
|
4,250,292 | <p>I am stuck trying to solve the following integral:</p>
<p><span class="math-container">$$\int_{0}^{3}\frac{1}{3}e^{3t-2}dt$$</span></p>
<p>I understand that I can take out <span class="math-container">$\frac{1}{3}$</span> of the integral, and that the integral of <span class="math-container">$e^{3t-2}$</span> is <span class="math-container">$\frac{1}{3t-2}e^{3t-2}$</span></p>
<p>However, when I insert the boundaries, I always receive <span class="math-container">$\frac{1}{7}e^7+\frac{1}{2}e^{-2}$</span> which yields <span class="math-container">$\frac{1}{21}e^7+\frac{1}{6}e^{-2}$</span> when multiplied with <span class="math-container">$\frac{1}{3}$</span>.</p>
<p>In the solution book, however, it is said to be <span class="math-container">$\frac{1}{9}(e^7-e^{-2})$</span>.</p>
<p>Does somebody see where my mistake is and know how to get to the correct solution?</p>
| Alessio K | 702,692 | <p>For <span class="math-container">$a\neq 0$</span> we have</p>
<p><span class="math-container">$$\int e^{at}dt=\frac{1}{a}e^{at}+C$$</span></p>
<p>Thus we have</p>
<p><span class="math-container">$$\int \frac{1}{3} e^{3t-2}dt=\frac{1}{3}\int e^{3t}e^{-2}dt$$</span>
<span class="math-container">$$=\frac{1}{3}e^{-2}\int e^{3t}dt=\frac{1}{3}e^{-2}\left[\frac{1}{3}e^{3t}\right]+C=\frac{1}{9}e^{3t-2}+C$$</span></p>
|
468 | <p>Textbook writers are blessed with only solving problems with neat answers. Numerical coefficients are small integers, many terms cancel, polynomials split into simple factors, angles have trigonometric functions with known values. Pure bliss.</p>
<p>The "real life" is different (as any of us knows).</p>
<p>Giving such questions for homework runs the risk that the student knows she is wrong when some crooked formula/value shows up. On the other hand, checking messy results (unless perhaps directly numerical values) is harder.</p>
<p>What do you think about posing questions which don't have neat derivations/results? I presume the answer could depend on the subject matter, the level of the students, and perhaps on exactly what the question should teach.</p>
| Markus Klein | 114 | <p><strong>I would encourage to have a significant amount of non-"round" numbers in your homework and also in exams.</strong> Some reasons:</p>
<ul>
<li>Except when you would teach calculations with numbers, normally the way you do things is important, not actual calculations. Even if the students feel that something is wrong with the numbers: As long as his/her way was correct, it is only a very small part of points not gained due to miscalculation.</li>
<li>If your students know that the only possibility is that number results are "nice", you add some sort of bias to the result: If the calculation is not too complex, students could try to do basic calculations with the given numbers in order to arrive at a "nice" result and can so perform "reverse engineering" in order to get the right idea only by guessing and trying.</li>
<li>Even when you construct a "nice" example with "round" numbers, there could be a chance that there are two possible correct ways: One with your constructed nice numbers and one with not so nice ones. (For example, if you perform a <a href="http://en.wikipedia.org/wiki/QR_decomposition#Using_Householder_reflections" rel="nofollow noreferrer">QR decomposition via Householder reflection</a>, there a freedom in the sign of one real number (on the wikipedia-link this is called $\alpha$). Depending on the choice of the sign, you get different values. There is a recommendation how to choose the sign, but in most examples this does not play a role and you are really free in the choice of the sign.).</li>
<li>You can train other skills in finding out if a calculation is completely wrong like perfoming rough calculation and estimating lower and upper bounds for some result. In real life this should be also very useful.</li>
<li>If you allow a calculator in the exam and there is nothing where you really need it, this could be misleading to the students.</li>
<li>As it was mentioned by the OP and in other answers, it is misleading that there will be only nice values as a result of a calculation since real world values are not nice.</li>
</ul>
<p><strong>However, it is nice to have "round" results on many examples (in particular those you perform on the board) because you can save time and not distract students from the actual topic.</strong></p>
|
468 | <p>Textbook writers are blessed with only solving problems with neat answers. Numerical coefficients are small integers, many terms cancel, polynomials split into simple factors, angles have trigonometric functions with known values. Pure bliss.</p>
<p>The "real life" is different (as any of us knows).</p>
<p>Giving such questions for homework runs the risk that the student knows she is wrong when some crooked formula/value shows up. On the other hand, checking messy results (unless perhaps directly numerical values) is harder.</p>
<p>What do you think about posing questions which don't have neat derivations/results? I presume the answer could depend on the subject matter, the level of the students, and perhaps on exactly what the question should teach.</p>
| Jyrki Lahtonen | 282 | <p>Like others I more often than not craft the problems to have nice solutions. Just to keep the students on their toes I occasionally insert something not so nice. As we are largely discussing eigenvalue problems, let me propose the following trick I picked up from a senior colleague. </p>
<p>Use a problem, where the precise eigenvalues are not needed, only their signs. For example you can ask whether a given 3-variable quadratic form is positive definite or not. You probably should warn students that they may need to estimate the zeros of the characteristic polynomial as opposed to find them. Also the students need to recall how to do that - always nice to use stuff from courses with very different topics. Your task as the problem designer is then to make sure that the zeros have distinct integral parts, so that it won't be too arduous a task to locate intervals with integral endpoints each containing a single eigenvalue.</p>
|
1,377,412 | <p>I am brand new to ODE's, and have been having difficulties with this practice problem. Find a 1-parameter solution to the homogenous ODE:$$2xy \, dx+(x^2+y^2) \, dy = 0$$assuming the coefficient of $dy \ne 0$
The textbook would like me to use the subsitution $x = yu$ and $dx=y \, du + u \, dy,\ y \ne 0$
Rewriting the equation with the subsititution:
$$2uy^2(y \, du + u \, dy)+(x^2+y^2) \, dy = 0$$
divide by $y^2$
$$2u(y \, du + u \, dy) + (u^2+1 ) \, dy=0$$
but after further simplification I end up getting ${dy \over y}$ which would mean I would get a logarithm after integrating, and the answer is given as $$3x^2y+y^3 = c$$
Could I get some help/hints as to how this answer was obtained? </p>
| Chiranjeev_Kumar | 171,345 | <p>$$2xy\mathrm dx+(x^2+y^2)\mathrm dy = 0$$</p>
<p>$$2xy \mathrm dx=-(x^2+y^2)\mathrm dy$$</p>
<p>$$\frac{dx}{dy}=-\frac{(x^2+y^2)}{2xy}$$</p>
<p>Now put, $x=yu \Rightarrow \mathrm dx=u\mathrm dy+y\mathrm du$, i.e.
$\frac {dx}{dy}=u+y\frac{du}{dy}$</p>
<p>Using these substitution and after simplification a you will get that</p>
<p>$$\frac{2u}{3u^2+1} \mathrm du=-\frac{1}{y}\mathrm dy$$</p>
<p>On integration you will get that,</p>
<p>$$\frac{1}{3}\ln(u^2+\frac{1}{3})+\ln y=\ln c$$</p>
<p>$$\ln {(u^2+\frac{1}{3})y^3=\ln c^3}$$</p>
<p>$$(u^2+\frac{1}{3})y^3=C$$ Now replace back $u=x/y$, you will get your answer.</p>
|
1,851,209 | <p>Let $L:X\to Y$ an linear operator. I saw that $L$ is bounded if $$\|Lu\|_Y\leq C\|u\|_X$$
for a suitable $C>0$. This definition looks really weird to me since such application is in fact not necessary bounded as $f:\mathbb R\to \mathbb R$ defined by $f(x)=x$. So, is there an error in <a href="https://en.wikipedia.org/wiki/Bounded_operator" rel="nofollow">wikipedia definition</a> ? And if the are true (what I supposed, since it's wikipedia), what is the interest to defined boundness as Lipschitz condition ? (it make no sense in fact...) </p>
| Surb | 154,545 | <p><strong>Answer 1</strong></p>
<p>Let $A\in \Bbb R^{n\times n}$ and $v\in\Bbb R^n$ such that $Av\neq 0$.</p>
<p>Then it is true that $$\lim_{s\to \infty} \|A(sv)\|=\|Av\|\lim_{s\to \infty} s=\infty.$$
However, there exists a $C>0$ such that
$$\frac{\|A(sv)\|}{\|sv\|}=\frac{\|Av\|}{\|v\|}\leq C \qquad \forall s>0.$$
Now, if you set
$$ \|A\| = \sup_{w\neq 0}\frac{\|Aw\|}{\|w\|}= \sup_{\|w\|=1}\|Aw\|$$
Then you can choose $C=\|A\|$ above. In this finite dimensional setting, we always have $\|A\|<\infty$ and $\|A\|$ is called an induced matrix norm. In <em>this sense</em>, $A$ is bounded because $\|A\|<\infty$.
Note that
$$\|A\|=\inf\{C>0 \mid \|Aw\|\leq C\|w\|, \forall w\}.$$
Now, in infinite dimensional space, things are less nice. In particular, there exist some linear operators such that $\|A\|$ defined as above equals $\infty$. In which case $A$ is unbounded.</p>
<p><strong>Answer 2</strong></p>
<p>This way to define the boundness has sense. Indeed, let $\mathcal L(X,Y)$ the set of the linear application from $X\longrightarrow Y$. Then, $$|||L|||:=\sup_{x}\frac{\|Lx\|_Y}{\|x\|_X}$$ is a norm over $\mathcal L(X,Y)$, and thus, $$\{f\in \mathcal L(X,Y)\mid |||f|||<\infty \}$$
is a normed vector space. A good example to me would be to take an integrale operator,</p>
<p>\begin{align*}
T:L^p(\mathbb R^n)&\longrightarrow L^q(\mathbb R^n)\\
f&\longmapsto Tf= \int_{\mathbb R^n}K(t,\cdot )f(t)\mathrm d t.
\end{align*}</p>
<p>Then, if indeed $\|Tf\|_{L^q}\leq C \|f\|_{L^p}$ for a suitable $C$, then $T$ transform $L^p$ in $L^q$. So you can see the "boundness" as the fact that the operator is well defined. In the example I gave you, you can see it like the fact that $\|f\|<\infty $, implies that $Tf$ is well defined. </p>
|
113,843 | <p>Here comes some sample data</p>
<pre><code>data = {{0, 0.7, 0.4}, {1, 0.831177, 0.51854}, {2, 1.11106, 0.463533},
{3, 1.84226, -0.642571}, {4, 0.677049, -0.327877},
{5, 0.77886, -0.451322}, {6, 0.965874, -0.508772},
{7, 1.34397, -0.202473}, {8, 1.01761, -0.717013},
{9, -0.0507992, -1.3864}, {10, -0.102145, -0.957957},
{11, -0.00228078, -0.861489}}
</code></pre>
<p>The first integer is a counter, while the other two reals are the $(x,y)$ coordinates. </p>
<p>Let's plot these points</p>
<pre><code>d0 = data[[All, {2, 3}]];
L0 = ListPlot[d0, Joined -> True, Mesh -> All,
PlotStyle -> {Black, Thickness[0.002], PointSize[0.012]},
Frame -> True, Axes -> False, PlotRange -> All, ImageSize -> 500]
</code></pre>
<p>Now I want the customize the plot as follows:</p>
<p>(a). Add arrows showing the evolution of the points, like</p>
<p><a href="https://i.stack.imgur.com/RX1Kp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RX1Kp.jpg" alt="enter image description here"></a> </p>
<p>(b). The first point with counter 0 should be plotted in red, while the last point should be plotted in blue.</p>
<p>(c). Add labels near to the points indicating the corresponding counter (e.g, 2, 3, etc). The label of the first point should be $P_0$, instead of 0, while for the last point the label should be $P_f$.</p>
<p>Any suggestions? </p>
| Hubble07 | 7,009 | <pre><code>r3 = AppendTo[Table[{Graphics[{Text[
Which[i == 1, Subscript[P, 0], i == Length[d0], Subscript[P, f],
True, ToString[i - 1]], Offset[{0, 10}, d0[[i]]]]}],
Graphics[{PointSize[Large], Which[i == 1, Red],
Which[i == Length[d0] - 1, {Point[d0[[i]]], Blue,
Point[d0[[i + 1]]]}, True, Point[d0[[i]]]]}],
Graphics[{Arrow[d0[[i ;; i + 1]]]}],
Graphics[{White, Point[{0.5, 0.6}]}]}, {i, 1, Length[d0] - 1}],
Graphics[{Text[Subscript[P, f], Offset[{0, 10}, Last@d0]]}]]
Show[r3, Frame -> Automatic]
</code></pre>
<p><a href="https://i.stack.imgur.com/BDEmG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BDEmG.jpg" alt="enter image description here"></a></p>
|
3,406,056 | <p>By the definition of matrix exponentiation,</p>
<p><span class="math-container">$$A^k = \begin{cases}
I_n, & \text{if } k=0 \\[1ex]
A^{k-1}A, & \text{if } k\in \mathbb {N}_0 \\
\end{cases}$$</span></p>
<p>In my book, there's an exercise to do <span class="math-container">$D^k$</span>, where <span class="math-container">$D$</span> is a diagonal matrix.
In the solutions, though, they wrote</p>
<p><span class="math-container">$$D^k=\operatorname{diag}(d_1^k, \dots , d_n^k)$$</span></p>
<p>Which one is correct?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Both are correct but they are two different topics.</p>
<p><span class="math-container">$$A^k=\left\{
\begin{array}{c}
I_n, if\, k=0 \\
A^{k-1}A, if\, k\in {\displaystyle \mathbb {N} }_0 \\
\end{array}
\right.$$</span>
is the definition of the matrix power <span class="math-container">$A^k$</span> in general.</p>
<p>On the other hand <span class="math-container">$$D^k=diag(d_1\,^k, ... , d_n\,^k)$$</span> is a special case where your matrix is a diagonal matrix.</p>
<p>There is no conflict because the second one is just the application of the first one to the diagonal matrix <span class="math-container">$$D=diag(d_1,d_2, ... , d_n)$$</span> </p>
|
2,614,127 | <p>I have been trying to show the statement below using the $AC$ but I am starting to think that it is not strong enough to do it. </p>
<p><strong>Context:</strong> Let $\Gamma$ be an uncountable linearly ordered set with a smallest element (not necessarily well-ordered). </p>
<p>For each $\alpha\in\Gamma$, let $C_\alpha$ be a non-empty set such that $C_\alpha\supsetneqq C_\beta$ whenever $\alpha<\beta$.
For each $\alpha\in\Gamma$, let $P(C_\alpha)$ be a partition of $C_\alpha$ such that: whenever $\alpha<\beta$, for all $B\in P(C_\beta)$, there exists $A\in P(C_\alpha)$ such that $B\subsetneqq A$.</p>
<blockquote>
<p><strong>Statement:</strong> There exists $\{A_\alpha\}_{\alpha\in\Gamma}$ such that $A_\alpha\in P(C_\alpha)$ and $A_\alpha\supsetneqq A_\beta$ whenever
$\alpha<\beta$.</p>
</blockquote>
<p>By using the AC we can see that there exists $\{A_\alpha\}_{\alpha\in\Gamma}$ such that $A_\alpha\in P(C_\alpha)$, but (I think) there is nothing to ensure the monotonicity condition: $A_\alpha\supsetneqq A_\beta$ whenever $\alpha<\beta$.</p>
<p>Maybe the statement is a well-known result or conjecture that I am not aware of, I would appreciate some answer or reference.</p>
| Eric Wofsey | 86,856 | <p>Let me first ignore your requirements that $\Gamma$ is uncountable and that the containments are strict, since these requirements are wholly artificial and have nothing to do with what's going on, and just make counterexamples a bit messier. Here, then, is the prototypical counterexample. Let $\Gamma=\mathbb{N}$, let $C_n=\{m\in\mathbb{N}:m\geq n\}$, and let $P(C_n)$ be the partition of $C_n$ into singleton sets. There is then no sequence of the sort you ask for since each singleton set $\{k\}$ in any of the partitions stops existing once you reach $C_{k+1}$.</p>
<p>OK, now if you insist, we can bulk this example up to meet all your requirements. Let's take $\Gamma=[0,\infty)\subset\mathbb{R}$, and let $C_x=\{(y,z)\in\mathbb{R}^2:y,z\geq x\}$, and let $P(C_x)$ be the partition into sets of the form $\{y\}\times[x,\infty)$. Now $\Gamma$ is uncountable and all the inclusions are strict, but the sequence you ask for fails to exist for the same reason as in the first example.</p>
|
17,436 | <p>I have this list:</p>
<pre><code>a = {{{0, 0}, {1, 7}, {2, 0}, {3, 2}, {4, 7}}, {{0, 0}, {1, 0}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>and I'd like to transform it to this:</p>
<pre><code>a = {{{0, 0}, {1, 7}, {2, Na}, {3, 2}, {4, 7}}, {{0, 0}, {1, Na}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>ie I'd like to replace every 0 by Na except in the first sub-sub-list of each sub-list.</p>
<p>All I can come up with is:</p>
<pre><code>Part[#, 2 ;; All, All] & /@ a /. 0 -> Na
</code></pre>
<p>and it works I get </p>
<pre><code>{{{1, 7}, {2, Na}, {3, 2}, {4, 7}}, {{1, Na}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>but the modification is not attributed to <code>a</code>.</p>
<p>How can I do that?</p>
| image_doctor | 776 | <p>This will work if all your first elements of the sub-lists are of the form {0,0}</p>
<pre><code>a = a /. {{x_?Positive, 0} :> {x, Na}};
a
</code></pre>
<blockquote>
<p>{{{0, 0}, {1, 7}, {2, Na}, {3, 2}, {4, 7}}, {{0, 0}, {1, Na}, {2,
1}, {3, 2}, {4, 7}}}</p>
</blockquote>
|
17,436 | <p>I have this list:</p>
<pre><code>a = {{{0, 0}, {1, 7}, {2, 0}, {3, 2}, {4, 7}}, {{0, 0}, {1, 0}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>and I'd like to transform it to this:</p>
<pre><code>a = {{{0, 0}, {1, 7}, {2, Na}, {3, 2}, {4, 7}}, {{0, 0}, {1, Na}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>ie I'd like to replace every 0 by Na except in the first sub-sub-list of each sub-list.</p>
<p>All I can come up with is:</p>
<pre><code>Part[#, 2 ;; All, All] & /@ a /. 0 -> Na
</code></pre>
<p>and it works I get </p>
<pre><code>{{{1, 7}, {2, Na}, {3, 2}, {4, 7}}, {{1, Na}, {2, 1}, {3, 2}, {4, 7}}}
</code></pre>
<p>but the modification is not attributed to <code>a</code>.</p>
<p>How can I do that?</p>
| Rolf Mertig | 29 | <p>Another quick possibility</p>
<pre><code>a=a/. {n_, r:{_Integer,_Integer}...} :> Join[{n},{ r} /. 0->Na]
</code></pre>
|
747,519 | <p><img src="https://i.stack.imgur.com/jYzfz.png" alt="enter image description here"></p>
<p>I tried this problem on my own, but got 1 out of 5. Now we are supposed to find someone to help us. Here is what I did:</p>
<p>Let $f:[a,b] \rightarrow \mathbb{R}$ be continuous on a closed interval $I$ with $a,b \in I$, $a \leq b$</p>
<p>If $f(a), f(b) \in f(I)$ let $f(a)\leq y \leq f(b)$. Then by IVT there exists $x$, $a\leq x \leq b$ where $f(x)=y$ $Rightarrow$ The image is also an interval. </p>
<p>Show closed: Let m be the lowest upper bound and M the greatest lower bound of the image interval. $I=[a,b]$ must be a subset of $[M,m]$ and the function attains its bounds,
$m,M\in f(I)$. so $f(I)$ is a subset of $[M,m]$, thus is closed. </p>
<p>Can anyone provide a proof of this statement? Thanks!</p>
| DanielWainfleet | 254,665 | <p>(1).... Let $J=[a,b]$ with $a\leq b.$ Any sequence $(x_n)_{n\in N}$ of members of $J$ has a convergent subsequence. That is, there is a strictly increasing $g:N\to N$ such that $(x_{g(n)})_{n\in N}$ is a convergent sequence.</p>
<p>Proof: Let $J_1=[a,(a+b)/2]$ if $\{n: x_n\in [a,(a+b)/2]\}$ is an infinite set, otherwise let $J_1=[(a+b)/2,b].$ Recursively,when $J_n=[a_n,b_n],$ let $J_{n+1}=[a_n,(a_n+b_n)/2]$ if $\{n:x_n\in [a_n,(a_n+b_n)/2\}$ is infinite, otherwise let $J_{n+1}=[(a_n+b_n)/2,b_n].$</p>
<p>Now for every $n\in N$ the set $\{n:x_n\in J_n\}$ is infinite, and $J_{n+1}\subset J_n,$ and $b_n-a_n=2^{1-n}(b-a).$</p>
<p>Let $g(1)$ be the least (or any) $m$ such that $x_m\in J_1.$ Recursively, let $g(n+1)$ be the least (or any) $m$ such that $m>g(n)$ and $x_m\in J_{n+1}.$</p>
<p>Then $(x_{g(n)})_{n\in N}$ is a Cauchy sequence because $J_{n+1}\subset J_n\implies \{x_{g(n+1)},x_{g(n)}\}\subset J_n$ $\implies |x_{g(n+1)}-x_{g(n)}|\leq b_n-a_n=2^{1-n}(b-a)$ (which is sufficient to imply the Cauchy condition ). Of course a Cauchy sequence in $[a,b]$ converges to a member of $[a,b].$</p>
<p>(2).... A continuous $f:[a,b]\to R$ is bounded. Proof: (By contradiction).If not, let $x_n\in [a,b]$ with $|f(x_n)|>n$. By (1),let $(x_{g(n)})_{n\in N}$ be a convergent subsequence of $(x_n)_{n\in N}$ with limit point $x\in [a,b].$ Continuity of $f$ requires that $f(x_{g(n)})\to f(x)$ as $n\to \infty.$ But $|f(x_{g(n})|>|g(n)|=g(n)\geq n,$ so $(f(x_{g(n)})_{n\in N}$ is not a convergent sequence, a contradiction.</p>
<p>(3).... A continuous $f :[a,b]\to R$ attains it maximum and minimum. Proof: By (2), $M=\sup \{f(x):x\in [a,b]\}<\infty.$ Let $x_n\in [a,b]$ with $M\geq f(x_n)\geq M-2^{-n}.$ By (1) let $(x_{g(n)})_{n\in N}$ be convergent to $x^*\in [a,b].$ Continuity of $f$ implies $f(x^*)=\lim_{n\to \infty}f(x_{g(n)})=M.$ Obtaining $y^*\in [a,b]$ with $f(y^*)=m=\inf \{f(x):x\in [a,b]\}$ is done similarly.</p>
<p>(4)...By IVT, with $x^*$ and $y^*$ as in (3) we have $$\{f(x):x\in [a,b]\}\supset [f(y^*),f(x^*)]=[m,M].$$ By def'n of $ m$ and $M$ we have $$\{f(x):x\in [a,b]\}\subset [m,M].$$</p>
|
613,105 | <p>I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$.<br> I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them!<br><br> What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$.<br> Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$.<br> If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$.<br><br> I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.</p>
| N. S. | 9,176 | <p>Without loss of generality $a \leq b \leq c$. Then $a+b+c \leq 3c$ and hence</p>
<p>$$abc=a+b+c \leq 3c$$</p>
<p>Thus, either $c =0$, in which case $a=b=c=0$, or </p>
<p>$$ab \leq 3 \,.$$</p>
<p>This leads to only four possibilities to check: $a=0$ or $(a,b)=(1,1)$ or $(a,b)=(1,2)$ or $(a,b)=(1,3)$.</p>
|
794,389 | <p>Q1. I was fiddling around with squaring-the-square type algebraic maths, and came up with a family of arrangements of $n^2$ squares, with sides $1, 2, 3\ldots n^2$ ($n$ odd). Which seems like it would work with ANY odd $n$. It's so simple, surely it's well-known, but I haven't seen it in my (brief) web travels. I have a page with pics <a href="http://www.adamponting.com/squaring-the-plane/" rel="nofollow noreferrer">here</a> of the $n=7,9,11$ versions, and description of how to construct them; $n=11$ is:</p>
<p><img src="https://i.stack.imgur.com/jsBgg.png" alt="enter image description here"></p>
<p>It seems the same method could square the plane, well, fill greater than any specified area, no matter how huge, at least. Which is what 'infinite' means, practically, isn't it? Anyway, it seems, if somehow not known (which it must be, surely - if anyone has links etc to where it's discussed I would be very grateful) then it's another way of squaring the plane.</p>
<p>Q2. Each of these arrangements can be extended, but the ones I've tried (5 or 6) have a little gap to the south-east, (i.e. where the squares don't fit neatly together) but otherwise can be extended forever. Is there an $n$ for which there is no gap? There are more than 1 of each sized square in this arrangement, but still, it would be a nice tessellation with integer squares. <a href="http://www.adamponting.com/squaring-the-plane-ii/" rel="nofollow noreferrer">Here's</a> a picture of the 7x7 version extended, and detail of the centre.</p>
<p>Thanks for any answers or help.</p>
<hr>
<p>P.s. This was too long for the comments section:</p>
<p>I wrote to Jim Henle the other day asking about this method, he hadn't seen it, thought it was nice, but not plane-filling in the way his method is. He wrote in part:</p>
<blockquote>
<p>You are sort of "squaring the plane," but not in the sense that we did it. You are squaring larger and larger areas of the plane, but you don't square the whole thing. ... There are many meanings to "infinite". Aristotle distinguished between the "potential infinite" (more and more, without bound) and the "actual infinite" (all the numbers, all at once). Your procedure is the first sort, and ours is the second sort.</p>
</blockquote>
<p>Which is what I had thought. </p>
<p>But the more I think about it.. the less clear the difference seems. Well, e.g. 'there are an infinite number of primes' means: there is no highest one; any number you can say, there's a higher prime. That's how that is defined, spoken about, to my (non-mathematician's) understanding. Similarly, there are an infinite number of these $n\times n$ square groups, there's no biggest one, any area you name, there's a bigger one. The Henle's method consists in adding more squares, ideally forever, but practically, you stop at some point and say 'and so on forever'. The procedure requires an infinite number of steps. I can't quite see how this series of $n\times n$ squares is so different. You have to start drawing again with each new $n$, sure, but I can't see that matters so much - there are an infinite number of arrangements, and there's the same 'and so on forever'.. i.e. "there is, strictly speaking, no such thing as an infinite summation or process."</p>
<p>Nov 2015. [I can't add comments to questions below, not sure why.]
Ross M, sorry about the delay! It seems you might be confused between the 2 parts, my fault for combining them in one question. (I still haven't heard anything much about either 2 from anyone.)</p>
<p>The first part is the basic $n^2$ arrangements of squares. (The second part takes just one of these and tries to extend it outwards, wonders about the possibility and mathematics of there being no gaps, and has many more than 1 of each sized square)
I still don't quite see why 'having to rearrange each step' makes a huge difference to anything. Imagine I had a method of going from $n^2$ to $(n+1)^2$ squares by adding more around the edges. Then, according to what people seem to be saying, I 'could tile the whole thing'? The way I've done it has exactly the same area as that would be, just it has to be redrawn. I don't see how that affects whether it 'tiles the plane' or not, or anything else. If someone could explain that to me, I would be very grateful. </p>
| Ross Millikan | 1,827 | <p>It looks like you can tile an arbitrarily large section of the plane, but not the whole thing. As you say, it looks like you cannot extend it toward the southeast. If I challenge you to cover a $1,000,000 \times 1,000,000$ square you can do it, but you need to plan ahead by putting the correct number of small squares starting with $1$ on the first diagonal. You couldn't take a pattern that started out covering $1000 \times 1000$ and extend it to cover $1,000,000 \times 1,000,000$. This doesn't say there is anything wrong with what you have done-it is quite neat. You just need to be careful what you claim to be able to cover.</p>
|
102,721 | <p>This is probably a very simple question, but I couldn't find a duplicate.</p>
<p>As everybody knows, <code>{x, y} + v</code> gives <code>{x + v, y + v}</code>. But if I intend <code>v</code> to represent a vector, for example if I am going to substitute <code>v -> {vx, vy}</code> in the future, then the result <code>{x + v, y + v}</code> is meaningless.</p>
<p>How I can indicate to Mathematica that <code>v</code> is not a scalar and functions like <code>Plus</code> should not treat it as one? I tried setting <code>$Assumptions = v ∈ Vectors[2]</code> but that didn't help.</p>
| Ruslan | 5,208 | <p>Another option, which doesn't require activation/replacement-by-identity/manual-indexing/etc. is to create a custom Plus function, which only evaluates for lists:</p>
<pre><code>vectorPlus[a_?ListQ,b_?ListQ]=a+b;
</code></pre>
<p>With it you have:</p>
<pre><code>vectorPlus[x, {q,r,s}]
vectorPlus[{q,r,s}, y]
vectorPlus[{q,r,s}, y] /. y->{a,b,c}
vectorPlus[{q,r,s}, {x,y,z}]
</code></pre>
<blockquote>
<p>vectorPlus[x, {q, r, s}]</p>
<p>vectorPlus[{q, r, s}, y]</p>
<p>{a + q, b + r, c + s}</p>
<p>{q + x, r + y, s + z}</p>
</blockquote>
|
1,815,662 | <blockquote>
<p>Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.</p>
</blockquote>
<p>I've tried to use the cyclotomic polynomial as:
$$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$</p>
<p>So I have that my polynomial is
$$\frac{X^5-1}{X-1}$$ and now i have to prove that is irreducible. </p>
<p><strong>The lineal change of variables are ok*(I don't know why) so I substitute $X$ by $X+1$ then I have:
$$\frac{(X+1)^5-1}{X}=\frac{X^5+5X^4+10X^3+10X^2+5X}{X}=X^4+5X^3+10X^2+10X+5$$
And now we can apply the Eisenstein criterion with p=5. So my polynomial is irreducible in $\mathbb{Q}$</strong></p>
<p>Now let's prove that it has two different irreducible factors in $\mathbb{R}$</p>
<p>I've tryed this way: $X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D)$
and solve the system. But solve the system is quite difficult. Is there another way?</p>
| egreg | 62,967 | <p>Finding the complex roots of the polynomial is easy: if $\varphi=2\pi/5$, the roots are
$$
r_1=e^{i\varphi},\quad
r_2=e^{2i\varphi},\quad
r_3=e^{3i\varphi}=\bar{r}_2\quad
r_4=e^{4i\varphi}=\bar{r}_1
$$
and so the factorization over the reals is
$$
(X^2-(r_1+\bar{r}_1)X+1)((X^2-(r_2+\bar{r}_2)X+1).
$$
What you want to prove is that this factorization is not over $\mathbb{Q}$, thereby deducing that the polynomial is irreducible over $\mathbb{Q}$.</p>
<p>The procedure is standard: let $r$ be any root of the polynomial; then
$$
r^2+r+1+\frac{1}{r}+\frac{1}{r^2}=0
$$
and so
$$
\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)-1=0
$$
Since the polynomial $X^2+X-1$ has no rational root, you have proved that
$$
r_1+\bar{r}_1=r_1+\frac{1}{r_1}
$$
is not rational.</p>
<hr>
<p>If $p$ is prime, then $X^{p-1}+X^{p-2}+\dots+X+1$ is irreducible over $\mathbb{Q}$. Write it as
$$
\frac{X^p-1}{X-1}
$$
and substitute $X=Y+1$. You'll see that Eisenstein applies.</p>
|
167,575 | <p>I have 6 sets of 4D points. Here is an example of one set :</p>
<pre><code>{{30., 5., 111.925, 113.569}, {30., 7.5, 114.7, 158.286}, {30., 10., 115.625, 206.023},
{30., 12.5, 115.625, 257.528}, {30., 15., 117.475, 294.663}, {30., 17.5, 119.325, 328.03},
{30., 20., 121.175, 357.982}, {30., 22.5, 122.1, 393.646}, {30., 25., 122.1, 437.384},
{30., 27.5, 122.1, 481.123}}
</code></pre>
<p>I want to plot the x,y coordinates of the points on the 2D plane and use the z coordinate to define the size of the symbol (bubble radius or area) and the last coordinate to define a color for that bubble. So the color will be different depending on the fourth coordinate. Any help would be appreciated !</p>
<p>I would like to have a 4D graphic like that :
<a href="https://i.stack.imgur.com/wk30Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wk30Z.png" alt="Bubble chart 4D"></a></p>
| Carl Woll | 45,431 | <p>Why not use <a href="http://reference.wolfram.com/language/ref/BubbleChart" rel="nofollow noreferrer"><code>BubbleChart</code></a> with style wrappers?:</p>
<pre><code>BubbleChart[
Replace[
{{3,4,3,5},{4,1,4,8}},
{a_, b_, c_, d_} :> Style[{a, b, c}, Lighter[Green, d/10]],
{1}
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/TWQBr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TWQBr.png" alt="enter image description here"></a></p>
|
65,059 | <p>I have two points ($P_1$ & $P_2$) with their coordinates given in two different frames of reference ($A$ & $B$). Given these, what I'd like to do is derive the transformation to be able to transform any point $P$ ssfrom one to the other.</p>
<p>There is no third point, but there <em>is</em> an extra constraint, which is that the y axis of Frame $B$ is parallel to the $X$-$Y$ plane of Frame $A$ (see sketch below). I <em>believe</em> that is enough information to be able to do the transformation.</p>
<p><img src="https://i.stack.imgur.com/2d6QH.png" alt="Two frames"></p>
<p>Also:</p>
<ul>
<li>The points are the same distance apart in both frames (no scaling).</li>
<li>The points don't coincide.</li>
<li>The origins don't necessarily coincide.</li>
</ul>
<p>As you may have gathered, I'm <em>not</em> a mathematician (ultimately this will end up as code), so please be gentle...</p>
<p><sub>I've seen this question (<a href="https://math.stackexchange.com/questions/23197/finding-a-rotation-transformation-from-two-coordinate-frames-in-3-space">Finding a Rotation Transformation from two Coordinate Frames in 3-Space</a>), but it's not <em>quite</em> the same as my problem, and unfortunately I'm not good enough at math to extrapolate from that case to mine.</sub></p>
<p><strong>EDIT</strong> I've updated the diagram, which makes it a bit cluttered, but (I hope) shows all the 'bits': $P3_B$ is what I'm trying to calculate...</p>
| Ben Blum-Smith | 13,120 | <p>I think that the two points should be enough information even without the additional information about the $y$-axis of the second frame. <strong>EDIT:</strong> joriki pointed out that this isn't true (see comments). The additional piece of information about the $y$-axis is needed, but then the problem should be solvable.</p>
<p>The transformation you want is <a href="http://en.wikipedia.org/wiki/Affine_transformation" rel="nofollow">affine</a> because it sends lines to lines and its linear part is <a href="http://en.wikipedia.org/wiki/Orthogonal_transformation" rel="nofollow">orthogonal</a> because it preserves distances. This means it will have the form</p>
<p>$$\begin{align}
x' &= a_{11}x+a_{12}y+a_{13}z +b_1\\
y' &= a_{21}x+a_{22}y+a_{23}z +b_2\\
z' &= a_{31}x+a_{32}y+a_{33}z +b_3\\
\end{align}$$</p>
<p>This form makes it affine. To make sure it is orthogonal the $a$'s are subject to the constraints</p>
<p>$$\begin{align}
&a_{11}^2+a_{21}^2+a_{31}^2=\\
&a_{12}^2+a_{22}^2+a_{32}^2=\\
&a_{13}^2+a_{23}^2+a_{33}^2=1
\end{align}$$</p>
<p>and</p>
<p>$$\begin{align}
&a_{11}a_{12}+a_{21}a_{22}+a_{31}a_{32}=\\
&a_{11}a_{13}+a_{21}a_{23}+a_{31}a_{33}=\\
&a_{12}a_{13}+a_{22}a_{23}+a_{32}a_{33}=0
\end{align}$$</p>
<p><strong>EDIT:</strong> The fact that the new $y$-axis is parallel to the old $xy$-plane means that $a_{23}=0$.</p>
<p>Right now there are 11 unknown quantities (the $a$'s and $b$'s).</p>
<p>If you plug in one of your points (with both old and new coordinates) to the transformation at the top, you will get 3 equations. If you plug in the other point (both old and new coordinates) you will get 3 more. The constraints on the $a$'s give 6 more equations; there are 12 in all. <strong>EDIT:</strong> Due to joriki's point, one of these equations will be redundant, but we should still have enough information to solve them. (There may be a small finite number of solutions because the equations in the $a$'s are quadratic.)</p>
<p>Of course, solving this system of equations is its own challenge. A general method is <a href="http://en.wikipedia.org/wiki/Elimination_theory" rel="nofollow">elimination</a> using <a href="http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis" rel="nofollow">Grobner bases</a>, which is too elaborate for me to explain quickly. But the problem has many special features that make it seem like a shortcut should be available. I'll look for a shortcut and add it to this answer if I find one.</p>
<p>(P.S. I am taking your question as primarily practical, so not justifying my claims above about the form of the transformation. If you would like some theoretical background let me know and I will either explain briefly if I can or point you to good sources.)</p>
|
4,639,966 | <p>I recently saw the expansion <span class="math-container">$(1+ \frac{1}{x})^n = 1 + \frac{n}{x} + \frac{n(n-1)}{2!x^2} + \frac{n(n-1)(n-2)(n-3)}{3!x^3}.... $</span> where <span class="math-container">$n \in \mathbb Q$</span></p>
<p>From what I understood, they have taken the Taylor series of <span class="math-container">$(1+x)^n$</span> and put <span class="math-container">$x=\frac{1}{t}$</span>. This doesn't make sense to me because the Taylor series used, uses successive derivatives at zero but derivatives at zero won't be defined for <span class="math-container">$f(x)=\frac{1}{x}$</span>.</p>
<p>How can I directly calculate the Taylor series for <span class="math-container">$(1+ \frac{1}{x})^n$</span>?</p>
| aschepler | 2,236 | <p>This equation is not a Taylor series, but it is correct, at least for <span class="math-container">$x \geq 1$</span>.</p>
<p>The usual Taylor series</p>
<p><span class="math-container">$$ (1+t)^n = 1 + nt + \frac{n(n-1)}{2!} t^2 + \frac{n(n-1)(n-2)}{3!} t^3 + \cdots $$</span></p>
<p>is a true equation for every rational <span class="math-container">$n$</span> and every real <span class="math-container">$t$</span> with <span class="math-container">$-1 < t \leq 1$</span>, in the sense that the infinite series converges to <span class="math-container">$(1+t)^n$</span>. (Also for some other domains, but in general we may need to worry about when <span class="math-container">$a^b$</span> even has a clear meaning.) This fact is related to repeated derivatives, but we don't need those derivatives to just say that the equation holds true.</p>
<p>If <span class="math-container">$x \geq 1$</span>, then <span class="math-container">$0 < \frac{1}{x} \leq 1$</span>. So just by substitution, it is true that</p>
<p><span class="math-container">$$ \left(1+\frac{1}{x}\right)^n = 1 + \frac{n}{x} + \frac{n(n-1)}{2!\, x^2} + \frac{n(n-1)(n-2)}{3!\, x^3} + \cdots $$</span></p>
|
2,124,068 | <p>I came across the following problem in a book I was reading on continuous probability distributions:-</p>
<p>$Q.$ Let $Y$ be uniformly distributed on $(0,1)$. Find a function $\phi$ such that $\phi(Y )$ has the gamma density $\Gamma(\frac12,\frac12)$.</p>
<p>I know that the probability density represented by $\Gamma(\frac12,\frac12)$ is the following:-</p>
<p>$$\Gamma\left(\frac12,\frac12\right)=\begin{cases}\frac1{\sqrt{2\pi x}}.e^{-\frac x2} &&&& x \ge 0 \\ 0 &&&& x <0\end{cases}$$</p>
<p>I don't have any idea what to do after this. I would also like to have insight on similar questions.</p>
| m_goldberg | 72,018 | <p>Consider that </p>
<pre><code>54 {Cos[t], Sin[t], t/10 + 12.5/54} /. t -> 0
</code></pre>
<p>gives</p>
<blockquote>
<p><code>{54, 0, 12.5}</code>.</p>
</blockquote>
<p>Since </p>
<pre><code>54 {Cos[t], Sin[t], t/10 + 12.5/54}
</code></pre>
<p>is the parametric expression of a helix with radius 54, it represents a helix that solves your problem. This is visualized by</p>
<pre><code>Show[
ParametricPlot3D[54 {Cos[t], Sin[t], t/10 + 12.5/54}, {t, -10, 10}],
Graphics3D[{Red, Sphere[{54, 0, 12.5}, Scaled[.015]]}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/jK1oo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jK1oo.png" alt="plot"></a></p>
|
2,017,993 | <p>Is $0$ an eigenvalue for a compact normal operator?</p>
<p>Many texts mention that compact normal operators have a complete orthonormal basis of eigenvectors. If they do, then what about the kernel of the operator? The elements in the kernel, may not be eigenvectors.</p>
<p>Where is the mistake in my understanding?</p>
| Martin Argerami | 22,857 | <p>The elements of the kernel are precisely the eigenvectors for zero.</p>
<p>So, when $T $ is normal and compact you can form an orthonormal basis by joining orthonormal bases for each eigenspace (including the kernel, if nonzero).</p>
|
422,948 | <p>How could I/is it possible to take a fourier transform of text? i.e. What domain would/does text exist in? Any help would be great.</p>
<p>NOTE: I do not mean text as an image. I understand it's value, but I'm wondering if it is possible to map text to some domain and transform text on the basis of letters. This is in hopes of performing frequency filtering on said text.</p>
| Bert Newton | 986,909 | <p>I had a similar idea last night when I was trying to explain the concept of FFTs for fundamental analysis and synthesis of sounds to someone, and the analogy that popped into my head was of the prevalence of lowercase letters, uppercase letters, and punctuation in a page of text corresponding to signals that occur with high, medium and low frequency.</p>
<p>I haven't tried this yet, but I was thinking of converting the symbols to numbers (using their ASCII values might be enough) and feeding the resulting sequence into an FFT analysis to see if a paragraph of text could be decomposed into the sum of a reasonably finite series of sine waves such that the list of coefficients would be smaller than the original text.</p>
<p>I don't think it would have any <em>meaning</em> as such; it certainly wouldn't be useful to count word frequencies or to synthesize texts, but it's a very interesting question!</p>
|
4,106,273 | <p>In how many ways can a committee of four be formed from 10 men (including Richard) <br>
and 12 women (including Isabel and Kathleen) if it is to have two men and two women <br></p>
<p>a) Isabel refuses to serve with Richard,</p>
<p>b) Isabel will serve only if Kathleen does, too</p>
<p>My Thoughts : <br>
a) Total number of ways to select 4 people = 22C4 now for part a) I just need to deduct the <br>
pairs where Isabel and Richard are both in the committe which = 20C2
<br>
I am not sure how to proceed with part (b) and if part (a) is entirely correct</p>
| herb steinberg | 501,262 | <p>Part (b). There are <span class="math-container">$^{20}C_3$</span> ways when Isabel is on and Kathleen is not.</p>
<p>Part (a) is correct.</p>
|
3,999,652 | <p>Let triangle <span class="math-container">$ABC$</span> is an equilateral triangle. Triangle <span class="math-container">$DEF$</span> is also an equilateral triangle and it is inscribed in triangle <span class="math-container">$ABC \left(D\in BC,E\in AC,F\in AB\right)$</span>. Find <span class="math-container">$\cos\measuredangle DEC$</span> if <span class="math-container">$AB:DF=8:5$</span>.</p>
<p><a href="https://i.stack.imgur.com/FZkBd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FZkBd.png" alt="enter image description here" /></a>
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.</p>
<p>Let <span class="math-container">$\measuredangle DEC=\alpha$</span>. We can note that <span class="math-container">$\triangle AEF \cong \triangle BFD \cong CDE$</span>. This is something we can always use in such configuration. So <span class="math-container">$$AE=BF=CD, $$</span> <span class="math-container">$$AF=BD=CE.$$</span> Let <span class="math-container">$AB=BC=AC=8x$</span> and <span class="math-container">$DF=DE=EF=5x$</span>. If we denote <span class="math-container">$CD=y,$</span> then <span class="math-container">$CE=AC-AE=AC-CD=8x-y$</span>. Cosine rule on <span class="math-container">$CED$</span> gives <span class="math-container">$$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$</span> which is a homogenous equation. I got that <span class="math-container">$\dfrac{y}{x}=4\pm\sqrt{3}.$</span> Now using the sine rule on <span class="math-container">$CED$</span> <span class="math-container">$$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$</span> Now we can use the trig identity <span class="math-container">$\sin^2x+\cos^2x=1$</span> but it doesn't seem very rational. Can you give me a hint? I was able to find <span class="math-container">$\sin\measuredangle DEC$</span> in acceptable way, but I can't find <span class="math-container">$\cos\measuredangle DEC$</span>...</p>
| Semiclassical | 137,524 | <p>As others have done, I'll assume <span class="math-container">$AB=8,DF=5$</span> without loss of generality. Then equilateral triangles <span class="math-container">$\triangle ABC$</span> and <span class="math-container">$\triangle DEF$</span> have area <span class="math-container">$(\sqrt{3}/4)AF^2 = 16\sqrt{3}$</span> and <span class="math-container">$(\sqrt{3}/4)DF^2=(25/4)\sqrt{3}$</span> respectively. Since the triangles <span class="math-container">$\triangle FAE,\triangle DBF, \triangle ECD$</span> are all congruent, they must each have area <span class="math-container">$$\frac{1}{3}\left(16\sqrt{3}-\frac{25}{4}\sqrt{3}\right)=\frac{13\sqrt{3}}{4}.$$</span> But each triangle has a 60-degree base angle, so we can also write the area of <span class="math-container">$\triangle FAE$</span> as</p>
<p><span class="math-container">$$\frac12 AF\cdot AE \cdot \sin 60^\circ$$</span></p>
<p>Since <span class="math-container">$AE=AC-EC=8-AF$</span>, this becomes <span class="math-container">$$\frac12 AF(8-AF)\frac{\sqrt{3}}{2}= \frac{13\sqrt{3}}{4}\implies AF(8-AF)=13$$</span>
with <span class="math-container">$x=4\pm \sqrt{3}$</span> as solutions. As a minor variation on the other approaches, we may then use the law of cosines on triangle <span class="math-container">$\triangle CED$</span> to write</p>
<p><span class="math-container">$$CD^2=CE^2+ED^2-2CE\cdot ED\cos\measuredangle DEC$$</span>
and therefore</p>
<p><span class="math-container">\begin{align}
\cos\measuredangle DEC
&=\frac{CE^2+ED^2-CD^2}{2CE\cdot ED}\\
&=\frac{(4\pm \sqrt{3})^2+5^2-(4\mp \sqrt{3})^2}{2(5)(4\mp \sqrt{3})}\cdot \frac{4\mp \sqrt{3}}{4\mp \sqrt{3}}\\
&=\frac{(25\pm 16\sqrt{3})(4\mp \sqrt{3})}{130}\\
&=\frac{52\pm 39\sqrt{3}}{130}\\
&=\frac{4\pm 3\sqrt{3}}{10}
\end{align}</span>
as others have obtained.</p>
|
420,294 | <p>While reading Bill Thurston's <a href="http://www.ams.org/publications/journals/notices/201601/rnoti-p31.pdf" rel="noreferrer">obituary</a> in the Notices of the AMS I came across the following fascinating anecdote (pg. 32):</p>
<blockquote>
<p>Bill’s enthusiasm during the early stages of mathematical discovery was infectious. Once, while sitting in his
living room, Bill said to me, “I can do this group with
grep,” which was sort of strange to hear at first. But being
his student I knew just enough computerese to have an
inkling of what he was saying: he was able to compute
in that group with the UNIX utility for processing regular
expressions using finite deterministic automata. From
there, it was exciting to observe the quick unfolding of
the theory of automatic groups.</p>
</blockquote>
<p>Looking through David Epstein's <em>Word Processing in Groups</em> I can see that there are indeed connections between <a href="https://en.wikipedia.org/wiki/Automatic_group" rel="noreferrer">automatic groups</a> and regular expressions that should allow faster algorithms for certain computations e.g. solving the word problem.</p>
<p>But is there a concrete example of a group-theoretic computation with <code>grep</code>?</p>
| Wahome | 122,319 | <p>As Derek Holt suggested in a comment, it seems Thurston was indeed thinking of word acceptors that returned normal forms for elements of automatic groups. From a 1989 research report of his titled <a href="http://timo.jolivet.free.fr/docs/ThurstonLectNotes.pdf" rel="nofollow noreferrer">Groups, tilings and finite state automata:</a> (pg. 41)</p>
<blockquote>
<p>A good example is the Unix utility <code>egrep</code>. The word acceptor for Z, for instance, could
be specified by the regular expression
<code>a*|A*</code>
where the symbol <code>*</code> denotes zero or more repetitions of the preceding object,
and the
symbol <code>|</code> means ‘or’. The command <code>egrep '^a*|A*$'</code>
prints out all lines of its inputs which are accepted by WA <span class="math-container">$\dots$</span></p>
</blockquote>
<p>He goes on in page 42:</p>
<blockquote>
<p>For instance, a word acceptor which accepts only words in reduced form for
the free group <span class="math-container">$\langle ab|\rangle$</span> is illustrated in 11.3. The corresponding <code>egrep</code> command is
<code>egrep '^(b+|B+)?((a+|A+)(b+|B+))*(a+|A+)?$'</code> <span class="math-container">$\dots$</span></p>
</blockquote>
<p>With a text file as input one can easily check that these commands are the correct word acceptors.</p>
|
1,382,479 | <p>I would like to know why $a^p \equiv a \pmod p$ is the same as $a^{p-1} \equiv 1 \pmod p$, and also how Fermat's little theorem can be used to derive Euler's theorem, or vice versa. </p>
<p>Please keep in mind that I have little background in math, and I am trying to understand these theorems to understand the math behind RSA encryption, so I would appreciate it if the explanation could be as simple as possible (i.e. not using too much mathematical notation). I have had trouble finding resources online to explain these theorems that explain it in simple enough terms that I can understand.</p>
| hmakholm left over Monica | 14,366 | <p>It is not obvious how to derive Euler's theorem in its full generality from Fermat's little theorem -- if the modulus has a non-trivial square factor, then Fermat's little theorem doesn't seem to provide enough.</p>
<p>Fortunately, for RSA you don't need Euler's theorem in its full generality; it is enough to know:</p>
<blockquote>
<p>If $p$ and $q$ are two <em>different</em> primes, then $a^{(p-1)(q-1)+1}\equiv a\pmod{pq}$ for all $a$.</p>
</blockquote>
<p>Here you can start by Fermat's little theorem and then prove
$$ a^{(p-1)k+1}\equiv a \pmod p $$
for all $k$, by induction on $k$. Then do then same for $q$, and we have that
$ a^{(p-1)(q-1)+1}\equiv a $ modulo both $p$ and $q$. The Chinese Remainder Theorem now concludes the claim above.</p>
|
1,382,479 | <p>I would like to know why $a^p \equiv a \pmod p$ is the same as $a^{p-1} \equiv 1 \pmod p$, and also how Fermat's little theorem can be used to derive Euler's theorem, or vice versa. </p>
<p>Please keep in mind that I have little background in math, and I am trying to understand these theorems to understand the math behind RSA encryption, so I would appreciate it if the explanation could be as simple as possible (i.e. not using too much mathematical notation). I have had trouble finding resources online to explain these theorems that explain it in simple enough terms that I can understand.</p>
| Thomas Andrews | 7,933 | <p>(For the first part.)</p>
<p>More generally, if $b\mid ac$ and $b$ and $a$ are relatively prime, then $b\mid c$. In this case, $b=p$ and $c=a^{p-1}-1$.</p>
<p>This more general theorem can be seen by "unique factorization," also known as the fundamental theorem of arithmetic. But it can also be proved first as a lemma for proving unique factorization.</p>
<p>It uses something called <a href="https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" rel="nofollow">Bézout's identity</a>, that says that if $a,b$ are relatively prime, then there is an integer solution to $ax+by=1$. Then multiplying by $c$ you get: $acx + bcy=c$. $acx$ is divisible by $b$, by assumption, and $bcy$ is obviously divisible by $b$, so $c$ is divisible by $b$.</p>
|
4,487,494 | <blockquote>
<p><strong>Problem:</strong> Let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> be non-zero vectors in <span class="math-container">$\mathbb{R}^n$</span>.<br>
(a) Suppose that <span class="math-container">$\|x+y\|=\|x−y\|$</span>. Show that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> must be perpendicular.<br>
(b) Suppose that <span class="math-container">$x+y$</span> and <span class="math-container">$x−y$</span> are non-zero and perpendicular. Show that <span class="math-container">$x$</span>
and <span class="math-container">$y$</span> must have the same length.</p>
</blockquote>
<p><strong>Attempt</strong>:</p>
<p>(a)<span class="math-container">\begin{align*}\|x+y\|^2 & =\|x-y\|^2 \\
(x+y)\cdot (x+y) & =(x-y)\cdot (x-y) \\
\|x\|^2+\|y\|^2+2x\cdot y & =\|x\|^2+\|y\|^2-2x\cdot y \\
2x\cdot y & =-2x\cdot y \\
x\cdot y & =0.
\end{align*}</span>(b)<span class="math-container">\begin{align*}(x+y)\cdot (x-y) & =0 \\
\|x\|^2-x\cdot y+x\cdot y-\|y\|^2 & =0 \\
\|x\|^2-\|y\|^2 & =0.
\end{align*}</span>Are these attempts correct?</p>
<p><strong>EDITED</strong></p>
| Sourav Ghosh | 977,780 | <p><span class="math-container">$a)$</span> Using <a href="https://en.m.wikipedia.org/wiki/Polarization_identity" rel="nofollow noreferrer">polarization identity</a></p>
<p><span class="math-container">$4\langle x, y\rangle =\|x+y\|^2-\|x-y\|^2=0$</span></p>
<p><span class="math-container">$\langle x, y\rangle =x\cdot y=0$</span></p>
<hr />
<p><span class="math-container">$b)$</span> Again using polarization identity</p>
<p><span class="math-container">$\begin{align}4\langle x+y, x-y\rangle &=\|(x+y)+(x-y)\|^2-\|(x+y)-(x-y)\|^2\\&=4(\|x\|^2 -\|y\|^2)\end{align}$</span></p>
<p>Given <span class="math-container">$\langle x+y, x-y\rangle=0$</span> implies <span class="math-container">$\|x\|=\|y\|$</span></p>
|
4,226,030 | <blockquote>
<p>I want to solve
<span class="math-container">$$C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$$</span>
The solution must be valid for all <span class="math-container">$\theta$</span> in <span class="math-container">$\mathbb{R}$</span> and all <span class="math-container">$m$</span> in <span class="math-container">$\mathbb{Z}$</span>, but <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, and <span class="math-container">$\lambda$</span> are to be determined and can be in <span class="math-container">$\mathbb{C}$</span>.</p>
</blockquote>
<p>The solutions I've found by guessing are <span class="math-container">$(C\ $</span>arbitrary<span class="math-container">$, D\ $</span>arbitrary<span class="math-container">$, \lambda = n^2)$</span>, where <span class="math-container">$n$</span> is any integer, and <span class="math-container">$(C = 0, D = 0, \lambda\ $</span>arbitrary<span class="math-container">$)$</span>.</p>
<p>Is there some algebra I can do to show that these are the only solutions, or find the rest of the solutions to this equation?</p>
| David K | 139,123 | <p>Let <span class="math-container">$f_{C,D,\lambda}(\theta)
= C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta).$</span>
Your want the values of the parameters <span class="math-container">$C,$</span> <span class="math-container">$D,$</span> and <span class="math-container">$\lambda$</span> such that</p>
<p><span class="math-container">$$ f_{C,D,\lambda}(\theta) = f_{C,D,\lambda}(\theta + 2m\pi) $$</span></p>
<p>for all <span class="math-container">$\theta \in \mathbb R$</span> and all <span class="math-container">$m \in \mathbb Z.$</span>
In other words, <span class="math-container">$f_{C,D,\lambda}$</span> must either be constant or must be non-constant with period <span class="math-container">$2\pi$</span> (implying a minimal period that divides <span class="math-container">$2\pi$</span>).</p>
<p>You found the constant versions of <span class="math-container">$f_{C,D,\lambda},$</span> which require either <span class="math-container">$C=D=0$</span>
or <span class="math-container">$\lambda = 0.$</span></p>
<p>For non-constant <span class="math-container">$f_{C,D,\lambda},$</span> note that <span class="math-container">$f_{C,D,\lambda}$</span> is sinusoidal
with period <span class="math-container">$2\pi/\sqrt\lambda.$</span>
Hence <span class="math-container">$2\pi/\sqrt\lambda$</span> must divide <span class="math-container">$2\pi,$</span>
hence <span class="math-container">$\sqrt\lambda$</span> must be a non-zero integer.
You found those solutions too.</p>
<p>Even if we allow <span class="math-container">$C,$</span> <span class="math-container">$D,$</span> <span class="math-container">$\lambda,$</span> and/or <span class="math-container">$\theta$</span> to be complex, <span class="math-container">$f_{C,D,\lambda}$</span> is still a single-valued function that is either constant or has period <span class="math-container">$2\pi/\sqrt\lambda.$</span>
We can write
<span class="math-container">$$
f_{C,D,\lambda}(\theta) =
\frac12(C-iD)e^{i\sqrt\lambda \theta}
+ \frac12(C+iD)e^{-i\sqrt\lambda \theta}.
$$</span>
For non-constant <span class="math-container">$f_{C,D,\lambda}$</span>, therefore,
<span class="math-container">$2\pi/\sqrt\lambda$</span> must still divide <span class="math-container">$2\pi$</span> evenly,
hence <span class="math-container">$\sqrt\lambda$</span> must still be a non-zero integer,
and <span class="math-container">$\lambda$</span> must still be the square of a non-zero integer,
which gives a set of solutions that you already found.</p>
<p>As far as I can see there are no other solutions.</p>
|
634,929 | <p>How can I evaluate this integral?</p>
<p>$$
\int{x^{3}\,{\rm d}x \over \left(x - 1\right)^{2}\sqrt{x^{2} + 2x + 4}}$$</p>
<p>I would be grateful for any tips.</p>
| lab bhattacharjee | 33,337 | <p>As $x^2+2x+4=(x+1)^2+(\sqrt3)^2,$ using <a href="http://en.wikipedia.org/wiki/Trigonometric_substitution" rel="nofollow">Trigonometric substitution</a></p>
<p>let us set $x+1=\sqrt3\tan\psi$ and assuming $0<\psi<\frac\pi2$ so that $\sec\psi,\tan\psi>0$</p>
<p>$$I=\int\frac{x^3}{(x-1)^2\sqrt{x^2+2x+4}}dx=\int\frac{(\sqrt3\tan\psi-1)^3}{(\sqrt3\tan\psi-2)^2\sqrt3\sec\psi}\sqrt3\sec^2\psi d\psi$$</p>
<p>$$I=\int\frac{(\sqrt3\tan\psi-2+1)^3}{(\sqrt3\tan\psi-2)^2}\sec\psi d\psi$$</p>
<p>$$=\int\left((\sqrt3\tan\psi-2)+3+\frac3{\sqrt3\tan\psi-2}+\frac1{(\sqrt3\tan\psi-2)^2}\right)\sec\psi d\psi$$</p>
<p>$$=\sqrt3\int\tan\psi\sec\psi d\psi+\int\sec\psi d\psi+3\int\frac{\sec\psi}{\sqrt3\tan\psi-2}d\psi+\int\frac{\sec\psi}{(\sqrt3\tan\psi-2)^2}d\psi$$</p>
<p>The first two integral are too easy to be described</p>
<p>For the third, $\displaystyle\int\frac{\sec\psi}{\sqrt3\tan\psi-2}d\psi=\int\frac1{\sqrt3\sin\psi-2\cos\psi}d\psi$ asking for <a href="http://en.wikipedia.org/wiki/Tangent_half-angle_substitution" rel="nofollow">Weierstrass substitution</a></p>
<p>For the fourth $\displaystyle I_4= \int\frac{\sec\psi}{(\sqrt3\tan\psi-2)^2}d\psi=\int\frac{\cos\psi}{(\sqrt3\sin\psi-2\cos\psi)^2}d\psi$</p>
<p>We can write $\displaystyle\cos\psi=A\frac{d(\sqrt3\sin\psi-2\cos\psi)}{d\psi}+B(\sqrt3\sin\psi-2\cos\psi)$ so that</p>
<p>$\displaystyle I_4$ becomes $\displaystyle A\int\frac{d(\sqrt3\sin\psi-2\cos\psi)}{(\sqrt3\sin\psi-2\cos\psi)^2}+B\int\frac1{\sqrt3\sin\psi-2\cos\psi}d\psi $</p>
<p>Now solve for $A,B$</p>
|
1,842,365 | <p>I recently got acquainted with a theorem:</p>
<p>If $f(x)$ is a periodic function with period $P$, then $f(ax+b)$ is periodic with period $\dfrac{P}{a}$ , $a>0$.</p>
<p>I am having a difficulty in understanding this theorem. Does this theorem mean that $f(ax+b)=f(ax+b+ \dfrac{P}{a})$? </p>
<p>If the above meaning is true, then how can a function, initially having a single period, acquire another period by just changing the arguments?</p>
| Gordon | 169,372 | <p>We need only show that, for any Borel set $A \in \mathbb{R}$,
\begin{align*}
\int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP.
\end{align*}
We denote by $F$ the common cumulative distribution function of $Z_1$ and $Z_2$. Then, from the independence assumption,
\begin{align*}
\int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \right)\\
&=\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x) dF(y).
\end{align*}
Analogously,
\begin{align*}
\int_{Z_1+Z_2 \in A} Z_2 dP &= E\left(Z_2 \pmb{1}_{Z_1+Z_2 \in A} \right)\\
&=\iint_{\mathbb{R}^2} u \pmb{1}_{v+u \in A} dF(u) dF(v)\\
&=\iint_{\mathbb{R}^2} u \pmb{1}_{u+v \in A} dF(u) dF(v).
\end{align*}
That is,
\begin{align*}
\int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP.
\end{align*}
In other words,
\begin{align*}
E\left(Z_1 \mid Z_1+Z_2 \right) = E\left(Z_2 \mid Z_1+Z_2 \right).
\end{align*}</p>
|
1,504,214 | <p>When Mr. and Mrs. Smith took the airplane, they had together 94 pounds of baggage.
He paid 1.50 and she paid 2.00 for excess weight. If Mr. Smith made the trip by himself with the combined baggage of both of them, he would have to pay $13.50. How many pounds of baggage can one person take along without being charged.</p>
<p>The variable in the problem is not explicitly stated and I have tried multiple ways of starting it. I realized i understand what the question is asking and can figure out how to answer it, but not algebraically. How can I answer it algebraically? Any help would be appreciated. Thanks!</p>
| Tavasanis | 39,690 | <p>An exclusively algebraic solution follows. Generally, the cost a passenger should pay if he/she has excess baggage, is $$c=(w_t-w_f)\; q,$$where $w_t$ is the weight the passanger shows over the counter, $w_f$ is the maximum free weight, and $q$ is the amount per pound in excess.</p>
<p>The couple showed 94 pounds. Husband paid 1.50 and wife paid 2.00, individually. If Mr Smith made the trip by himself with the combined baggage of both of them, he would paid 13.50.</p>
<p>Let $w_w$ be the wife's weight, and $h_w$ the husband's weight. Then
\begin{align*}
2&=(w_w-w_f)\;q\\
1.5&=(w_h-w_f)\;q
\end{align*}</p>
<p>There is a relation between his weight and her weight: $w_w=94-w_h$. Then we can write\begin{align*}
2&=(94-w_h-w_f)\;q\\
1.5&=(w_h-w_f)\;q
\end{align*}</p>
<p>Now we add both equations and simplify. We have $$3.5=q(94-2w_f).$$</p>
<p>A second equation comes from the chance they both travelled but as a single passanger: $(94-w_f)\;q=13.50$. Now we have two equations in $q$ and $w_f$. Solving the system, we get $w_f=40$ pounds, and $q=0.25$ money units.</p>
|
79,270 | <p>For integer $n\ge 3$, consider the graph on the set of all even vertices of the $n$-dimensional hypercube $\{0,1\}^n$ in which two vertices are adjacent whenever they differ in exactly two coordinates. This is an $(n(n-1)/2)$-regular graph on $2^{n-1}$ vertices. Is there any standard name / notation for this graph? Is there a way to construct it from some "basic" graphs using standard graph operations (like products of graphs)? Has anybody ever studied the isoperimetric problem for this graph?</p>
<p>Thanks!</p>
| Zack Wolske | 18,086 | <p>Conway & Sloane's "Sphere Packings, Lattices and Groups" references Coxeter's "Regular Polytopes" for the phrase "halfcube", but Coxeter only uses the notation $h\Pi_n$, saying $h$ can be taken to stand for half- or hemi-, for an arbitrary polytope $\Pi_n$ {$p, q, \ldots, w$} with even $p$ (in your case, {$4,3,3,\ldots, 3$}) This construction is section 8.6 in Coxeter. Since then, halfcube seems to have lost favour, and hemi-cube has become the name for a construction of quotienting out vertices, while the term demicube (or demihypercube if you want to be explicit about using hypercubes and not cubes) is reserved for the construction of deleting vertices of a hypercube. See Conway, Burgiel and Goodman-Strass's "Symmetries of Things." Chapter 26 covers this, where they call them hemicubes, and draw some lovely pictures.</p>
<p>Specific dimensional cases have different names. Your $n=3$ case is the complete $K_4$. $n=4$ is the 16-cell, also called a hexadecachoron in older books, and happens to be a cross-polytope (this does not continue in higher dimensions). By $n=5$, the polytopes begin to take shape as their own specific family and no longer have multiple names. See <a href="http://en.wikipedia.org/wiki/Demihypercube" rel="noreferrer">http://en.wikipedia.org/wiki/Demihypercube</a>, and various dimension specific pages there.</p>
<p>I do not know anything about the isoperimetric problem for these graphs, but there has likely been work done on the $n \leq 4$ cases, since those graphs also show up as other constructions.</p>
|
1,255,803 | <p>My understanding is that the thesis is essentially a <em>definition</em> of the term "computable" to mean something that is computable on a Turing Machine.</p>
<p>Is this really all there is to it? If so, what makes this definition so important? What makes this definition so significant as to warrant having it's own name? </p>
<p>In most other branches of mathematics, a definition is an important part of the scaffolding, but not a result onto itself. In the case of the Church Turing Thesis, it seems like there must be more, but all I can see is the definition.</p>
<p>So, what is the significance of the Church-Turing Thesis?</p>
| Christopher | 73,985 | <p>The idea is that we have an informal notion of "computable" - that is, "something that can be computed". (This is explicitly not a precise definition). We also have a formal definition of "computable", that is, "computable by a Turing machine". The Church-Turing thesis is that these two notions coincide, that is, anything that "should" be computable is in fact computable by a Turing machine. (It's pretty clear that anything that is computable by a Turing machine is computable in the more informal sense).</p>
<p>Put another way, the Church-Turing thesis says that "computable by a Turing machine" is a correct definition of "computable". </p>
|
1,189,814 | <p>Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?</p>
<p>It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.</p>
| user26857 | 121,097 | <p>Let $R$ be a UFD which is not a PID, e.g. $R=\mathbb Z[X]$, and $M=\bigoplus_{p\text{ prime}} R/(p)$. Note that every non-invertible element of $R$ is a zero-divisor on $M$. Let $I=(p_1,p_2)$ with $p_1,p_2$ primes such that $I\ne R$. Since $I$ does not contain invertible elements, every element of $I$ is a zero-divisor on $M$. Moreover $(0:_MI)=0$.</p>
<p>Now consider the <em><a href="https://math.stackexchange.com/a/207468/121097">idealization</a></em> $A=R(+)M$ of the $R$-module $M$. Let $J=IA$. We have that $J$ is finitely generated ideal and consists of zero-divisors, but no non-zero element of $A$ annihilates $J$.</p>
|
1,477,871 | <p>If the space $X$ is banach , then I want to show that any linear map $T:X \to X$ is continuous iff the null space is closed. I could show that if $T$ is continuous then the null space is closed. But I am unable to prove the converse. Any hints are appreciated. Thanks</p>
| gerw | 58,577 | <p>This is not true. See the $T$ in <a href="https://math.stackexchange.com/a/426494/58577">this answer</a>. Since this $T$ is injective, its kernel is $\{0\}$.</p>
<p>Note that a similar assertion is true for linear functionals $T : X \to \mathbb{R}$.</p>
|
1,516,450 | <p>When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$</p>
<p>When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$</p>
<p>How does wolfram arrive at that inequality:</p>
<p>$a< -500\left(\sqrt{2} - 2\right)$</p>
<p>Here is the link: <a href="http://www.wolframalpha.com/input/?i=a%2Bb%2Bc%3D1000%2Ca%5E2%2Bb%5E2%3Dc%5E2%2C0%3Ca%3Cb%3Cc" rel="nofollow">Wolfram</a></p>
| Brian Tung | 224,454 | <p>For the inequality, we observe that since $b > a$, we must have</p>
<p>$$
c^2 = a^2+b^2 > 2a^2
$$</p>
<p>or</p>
<p>$$
c > a\sqrt{2}
$$</p>
<p>Thus</p>
<p>$$
1000 = a+b+c > a+a+a\sqrt{2} = a(2+\sqrt{2})
$$</p>
<p>or</p>
<p>$$
a < \frac{1000}{2+\sqrt{2}} = \frac{1000(2-\sqrt{2})}{2^2-(\sqrt{2})^2}
= 500(2-\sqrt{2})
$$</p>
<p>As regards the original problem: Primitive Pythagorean triples can be obtained using the well-known schema</p>
<p>$$
a, b = u^2-v^2, 2uv
$$</p>
<p>in some order, with $u > v$ both integers, and</p>
<p>$$
c = u^2+v^2
$$</p>
<p>Note that in this case,</p>
<p>$$
a+b+c = 2u^2+2uv = 2u(u+v)
$$</p>
<p>In order for $ka+kb+kc = 1000$ for some integer $k$, we need $u(u+v) \mid 500$. For example, with $u = 20, v = 5$, we obtain $a = 2uv = 200, b = u^2-v^2 = 375, c = u^2+v^2 = 425$. Note that $a+b+c = 200+375+425 = 1000$, and $a^2+b^2 = 40000+140625 = 180625 = c^2$.</p>
<p>That this is the only solution can be demonstrated by noting first that $500 = 2^2 \times 5^3$, and observing that for $u$ and $u+v$, we need two disjoint subsets of factors whose separate products differ by less than a factor of $2$.* This only happens for the above case with $20$ and $25$ (yielding $u = 20, v = 5$), and also with $4$ and $5$ (yielding $u = 4, v = 1$, and requiring us to scale the resulting triple by a factor of $25$). Since these two pairs are in direct proportion to each other, they produce the same unique solution.</p>
<p>*ETA: Only if we require $a > 0$. If $a$ can be less than $0$, then we only need disjoint factor subsets $u$ and $u+v$.</p>
|
1,516,450 | <p>When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$</p>
<p>When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$</p>
<p>How does wolfram arrive at that inequality:</p>
<p>$a< -500\left(\sqrt{2} - 2\right)$</p>
<p>Here is the link: <a href="http://www.wolframalpha.com/input/?i=a%2Bb%2Bc%3D1000%2Ca%5E2%2Bb%5E2%3Dc%5E2%2C0%3Ca%3Cb%3Cc" rel="nofollow">Wolfram</a></p>
| poetasis | 546,655 | <p>Let <span class="math-container">$(A+B+C)=P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn$</span></p>
<p><span class="math-container">$$\text{We solve for }\\ n=\frac{P-2m^2}{2m}\text{ where }\biggl\lceil\frac{\sqrt{P}}{2}\space\biggr\rceil\le m\le\biggl\lfloor\sqrt\frac{P}{2}\biggr\rfloor$$</span></p>
<p>Given <span class="math-container">$$P=1000\implies m_{min}=\biggl\lceil\frac{\sqrt{1000}}{2}\space\biggr\rceil=16, \quad m_{max}= \biggl\lfloor\sqrt{\frac{1000}{2}} \biggr\rfloor =22\\
\text{ and
we test }16\le m\le 22 \quad \text{to see which yield }\quad
n \in\mathbb{N}$$</span></p>
<p>We find only one where <span class="math-container">$f(20,5)=(375,200,425)$</span>. Note that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are reversible if <span class="math-container">$a<b<c$</span> is a concern. In any case it's a multiple: <span class="math-container">$25(15,8,17)$</span></p>
|
2,859,463 | <blockquote>
<p>Prove or disprove. All four vertices of every regular tetrahedron in $ \mathbb{R}^3$ have at least two irrational coordinates.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/hYrWv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYrWv.png" alt="enter image description here"></a></p>
<p>This question arose from my inability to construct a tetrahedron in $\mathbb{R}^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y \;\mathrm
{and}\; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers. </p>
| Ross Millikan | 1,827 | <p>You can place one of the vertices at the origin. A second one can be at $(1,0,0)$</p>
|
3,831,198 | <p>Suppose that <span class="math-container">$100$</span>kg of a radioactive substance decays to <span class="math-container">$80$</span>kg in <span class="math-container">$20$</span> years.</p>
<p>a) Find the half-life of the substance (round to the nearest year).</p>
<p>b)Write down a function <span class="math-container">$y(t)$</span> (<span class="math-container">$t$</span> in years) modeling the amount (in kg) of the radioactive substance at time <span class="math-container">$t$</span>.</p>
| user577215664 | 475,762 | <p><strong>Hint:</strong>
You have the radioactive decay law:
<span class="math-container">$$N(t)=N_0e^{-\lambda t}$$</span>
You have <span class="math-container">$N=80$</span> the time <span class="math-container">$t$</span> and <span class="math-container">$N_0=100$</span> deduce <span class="math-container">$\lambda$</span>.
Then half life <span class="math-container">$T$</span> is:
<span class="math-container">$$N=\dfrac {N_0}2$$</span>
<span class="math-container">$$\dfrac {N_0}2=N_0e^{-\lambda T}$$</span>
<span class="math-container">$$\implies \dfrac 12=e^{-\lambda T}$$</span>
<span class="math-container">$$ T=\dfrac {\ln 2}{\lambda}$$</span></p>
|
3,757,222 | <p>Let <span class="math-container">$n_{1}, n_{2}, ... n_{k} $</span> be a sequence of k consecutive odd integers. If <span class="math-container">$n_{1} + n_{2} + n_{3} = p^3$</span> and <span class="math-container">$n_{k} + n_{k-1} + n_{k-2} + n_{k-3} + n_{k-4} = q^4$</span> where both p and q are prime, what is k?</p>
<p>I am struggling with this question. I know the first sum can be written as <span class="math-container">$3n_{1} + 6 = p^3$</span> and the second sum can be written as <span class="math-container">$5n_{k} - 20 = q^4$</span>. I believe the second sum is also <span class="math-container">$5n_{1} +10k - 30 = q^4$</span>. However rearranging these I get no workable equations.</p>
| fisura filozofica | 356,159 | <p>You got it to <span class="math-container">$3n_1+6=p^3$</span> but how about <span class="math-container">$3(n_1+2)=p p^2$</span> and <span class="math-container">$p$</span> is prime so <span class="math-container">$p=3$</span>. Using this you can find <span class="math-container">$n_1$</span> then <span class="math-container">$q$</span> and after all this <span class="math-container">$k$</span>.</p>
|
265,635 | <p>I'm trying to create a user-defined function that computes the equivalent resistance of <span class="math-container">$n$</span> resistors in parallel.</p>
<p>As we know, such formula is:</p>
<p><span class="math-container">$R_\text{eq.p} = \dfrac{1}{\displaystyle\sum_{k=1}^{n} \dfrac{1}{R_k}} = \left( \displaystyle\sum_{k=1}^{n} R_k^{-1} \right)^{-1} \tag*{}$</span></p>
<p>The code would seem straight forward. I tried:</p>
<pre><code>Rp[list_] := 1/Sum[1/list[[k]], {k, Length[list]}];
Attributes[Rp] = {Listable};
</code></pre>
<p>where I'm using <a href="https://reference.wolfram.com/language/ref/Listable.html" rel="noreferrer"><code>Listable</code></a> because the input of the function is a list/vector. To test it, I created the list <code>test = {1, 2, 3}</code>, yet when I enter <code>Rp[test]</code> I get the error <em>Power: Infinite expression 1/0 encountered</em>. Why isn't this working?</p>
| Syed | 81,355 | <pre><code>Clear[Rp]
Rp[rin_List] := Module[{r},
r = DeleteCases[rin, \[Infinity]];
If[Total[r] === 0
, \[Infinity]
, Times @@ r/Total[Times @@@ Subsets[r, {Length@r - 1}]]
]
]
testCases = {{4 k, 4 k}, {Quantity[6, "KiloOhms"],
Quantity[4, "KiloOhms"]}, {1, 2, 3}, {r1, r2}, {r1, r2, r3}, {1,
2, 0}, {1, -1}, {1, \[Infinity]}
};
Rp /@ testCases
</code></pre>
<p><a href="https://i.stack.imgur.com/EJDBB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EJDBB.png" alt="enter image description here" /></a></p>
|
4,020,464 | <p>So <span class="math-container">$Ax = λx$</span>;</p>
<p><span class="math-container">$A(Ax) = λ(Ax) \to (A^2)x = (λ^2)x$</span></p>
<p>I kind of dont know how to get the <span class="math-container">$1$</span> and <span class="math-container">$I$</span> here...</p>
<p>Any help is appreciated</p>
| Vons | 274,987 | <p>Let's look at the pieces</p>
<p><span class="math-container">$$\begin{split}X&\sim N(\mu_1,\sigma_1^2)\\
Y&\sim N(\mu_2,\sigma_2^2)\end{split}$$</span></p>
<p>Then what the book means by <span class="math-container">$\bar X-\bar Y$</span> is a sum of m+n independent normal random variables is that <span class="math-container">$\bar X=\frac{\sum X}{m}=\sum {\frac{X}{m}}$</span> and likewise for <span class="math-container">$\bar Y$</span> with</p>
<p><span class="math-container">$$\begin{split}\frac {X}{m}&\sim N\left(\frac{\mu_1}{m},\frac{\sigma_1^2}{m^2}\right)\\
\frac{Y}{n}&\sim N\left(\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)\end{split}$$</span></p>
<p>As you know the sum of k independent normal random variables <span class="math-container">$X_1,...,X_k$</span> with means <span class="math-container">$\mu_1,...,\mu_k$</span> and variances <span class="math-container">$\sigma_1^2,...,\sigma_k^2$</span> is normally distributed with mean <span class="math-container">$\sum \mu$</span> and variance <span class="math-container">$\sum \sigma^2$</span></p>
<p>Also note that</p>
<p><span class="math-container">$$-\frac{Y}{n}\sim N\left(-\frac{\mu_2}{n},\frac{\sigma_2^2}{n^2}\right)$$</span></p>
<p>Hence</p>
<p><span class="math-container">$$\begin{split}\bar X-\bar Y&=\frac{X_1}{m}+...+\frac{X_m}{m}+\frac{Y_1}{n}+...+\frac{Y_n}{n}\\
&\sim N\left(m\cdot \frac{\mu_1}{m}-n\cdot \frac{\mu_2}{n}, m\cdot \frac{\sigma_1^2}{m^2}+n\cdot \frac{\sigma_2^2}{n^2}\right)\\
&= N\left(\mu_1-\mu_2, \frac{\sigma_1^2}{m}+\frac{\sigma_2^2}{n}\right)\end{split}$$</span></p>
<p>We seek</p>
<p><span class="math-container">$$P(|\bar X-\bar Y-(\mu_1-\mu_2)|\le 1)\ge.95$$</span></p>
<p>Standardize via division by the standard error with m=n here</p>
<p><span class="math-container">$$\begin{split}P\left(\left|\frac{\bar X-\bar Y-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right|\le \frac 1{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{n}}}\right)&\ge.95\\
P(|Z|\le\sqrt{\frac{n}{\sigma_1^2+\sigma_2^2}})&\ge.95\end{split}$$</span></p>
<p>Using R we find that the critical value is</p>
<pre><code>> qnorm(.975)
[1] 1.959964
</code></pre>
<p>Thus</p>
<p><span class="math-container">$$\begin{split}\sqrt{\frac{n}{4.5}}&\ge1.96\\
n&\ge17.29\end{split}$$</span></p>
<p>Conservatively we pick n=18</p>
|
1,691,605 | <p>I've been given the following definition:</p>
<blockquote>
<p>For a THMC with one step transition matrix $\mathbf{P}$, the row vector $\mathbf{\pi}$ with elements $(\pi_{i})_{i \in S}$ (where $S$ is the state space) is a stationary distribution iff $\mathbf{\pi \; P} = \mathbf{\pi}$</p>
</blockquote>
<p>However, I also know that many THMCs will have multiple stationary distributions. </p>
<p>This leaves me with the following questions:</p>
<ol>
<li>How can you tell how many stationary distributions a THMC has?</li>
<li>How can I show that a THMC has only one stationary distribution?</li>
<li>The equation $\mathbf{\pi \; P} = \mathbf{\pi}$ looks like it should only have one solution, so how is it possible to have multiple stationary distributions?</li>
</ol>
| frog | 84,997 | <p>Using l'Hospitals rule you get
$$
\lim_{x\to a}\frac{a^{a^x}a^x(\ln a)^2-a^{x^a}ax^{a-1}\ln a}{a^x\ln a - ax^{a-1}}=a^{a^a}\ln a.
$$
EDIT: If you're looking for method that does not involve taking derivatives, I suggest the following:
$$\lim_{x\to a}\frac{a^{a^x}-a^{x^a}}{a^x-x^a}=\lim_{x\to a}a^{a^x}\frac{1-a^{x^a-a^x}}{a^x-x^a}=a^{a^a}\underbrace{\lim_{h\to 0}\frac{a^h-1}{h}}_{=\ln a}.$$
In this way you only need $\displaystyle \lim_{h\to 0}\frac{a^h-1}{h}=\ln a$, $\displaystyle\lim_{x\to a}a^x-x^a=0$ and to set $h:=a^x-x^a$.</p>
|
1,691,605 | <p>I've been given the following definition:</p>
<blockquote>
<p>For a THMC with one step transition matrix $\mathbf{P}$, the row vector $\mathbf{\pi}$ with elements $(\pi_{i})_{i \in S}$ (where $S$ is the state space) is a stationary distribution iff $\mathbf{\pi \; P} = \mathbf{\pi}$</p>
</blockquote>
<p>However, I also know that many THMCs will have multiple stationary distributions. </p>
<p>This leaves me with the following questions:</p>
<ol>
<li>How can you tell how many stationary distributions a THMC has?</li>
<li>How can I show that a THMC has only one stationary distribution?</li>
<li>The equation $\mathbf{\pi \; P} = \mathbf{\pi}$ looks like it should only have one solution, so how is it possible to have multiple stationary distributions?</li>
</ol>
| Fnacool | 318,321 | <p>Let $f(t) = a^t=e^{(\ln a) t}$. Now $f'(t) = (\ln a) (a^t)$. </p>
<p>By the mean value theorem, </p>
<p>$$\frac{f(t)-f(s) }{ t-s} = f'(c)$$</p>
<p>for some intermediate $c$.</p>
<p>In our case $t$ and $s$ are functions of $x$, satisfying $\lim_{x\to a} t(x) =\lim_{x\to a} s(x) = a^a$. Therefore, the intermediate point $c=c(x)$ tends to $a^a$. Since the derivative of $f$ is continuous, the answer is $f'(a^a)= (\ln a) a^{a^a} $. </p>
|
546,123 | <p>Let <span class="math-container">$X$</span> be any uncountable set with the cofinite topology. Is this space first countable?</p>
<p>I don't think so because it seems that there must be an uncountable number of neighborhoods for each <span class="math-container">$ x \in X$</span>. But I am not sure if this is true.</p>
| Ari Royce Hidayat | 435,467 | <p><em>(Just a rewrite of an already excellence answer above and its comment.)</em></p>
<p>Cofinite topology is not first countable. To prove it, we show that some point of <span class="math-container">$X$</span> does not have a countable local base.</p>
<p>Let <span class="math-container">$x \in X$</span>, and suppose that <span class="math-container">$\mathscr{B_x}=\{B_n:n\in\Bbb N\}$</span> is a countable local base of open sets at the point <span class="math-container">$x$</span>.</p>
<p>For each <span class="math-container">$n\in\Bbb N$</span>, let <span class="math-container">$F_n = X \setminus B_n$</span>. <span class="math-container">$B_n$</span> is open, so by definition <span class="math-container">$F_n$</span> is finite. Let <span class="math-container">$F = \{x\} \cup \bigcup_{n \in \Bbb N}F_n$</span>; <span class="math-container">$F$</span> is the union of countably many finite sets, so <span class="math-container">$F$</span> is countable. <span class="math-container">$X$</span> is uncountable, so there is some <span class="math-container">$b\in X\setminus F$</span> <em>(and <span class="math-container">$b \neq x$</span>)</em>.</p>
<blockquote>
<p>Note that if <span class="math-container">$X$</span> is not a cofinite topology, then <span class="math-container">$F$</span> is not necessarily countable, then <span class="math-container">$b$</span> does not necessarily exist.</p>
</blockquote>
<p>Let <span class="math-container">$O = X \setminus \{b\}$</span>. <span class="math-container">$\{b\}$</span> is finite, so by definition <span class="math-container">$O$</span> is an open set, and it contains <span class="math-container">$x$</span> <em>(because of how <span class="math-container">$b$</span> defined)</em>. For every <span class="math-container">$n \in \Bbb N$</span>, we know that <span class="math-container">$b \notin F_n$</span>, so <span class="math-container">$b \in B_n$</span>, but <span class="math-container">$b \notin O$</span>, so <span class="math-container">$b \in B_n \setminus O$</span>, and therefore <span class="math-container">$B_n \notin O$</span>.</p>
<p>In summary, there is an open set <span class="math-container">$O$</span> containing <span class="math-container">$x$</span> but not containing <span class="math-container">$B_n$</span>. This is a contradiction.</p>
<blockquote>
<p>We may wonder though what happens if we define <span class="math-container">$O = X \setminus \{b\}$</span> in the usual topology. If the local base is an open ball <span class="math-container">$B(x, \epsilon)$</span>, there would always be <span class="math-container">$\epsilon < d(x, b)$</span>.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/9A8RG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9A8RG.jpg" alt="Picture to see things clearer" /></a></p>
|
4,562,451 | <p>I had this maths question:</p>
<blockquote>
<p>Given that <span class="math-container">$$8\sqrt{p} = q\sqrt{80}$$</span> where <span class="math-container">$p$</span> is prime, find the value of <span class="math-container">$p$</span> and the value of <span class="math-container">$q$</span></p>
</blockquote>
<p>I did this by simplifying the RHS to <span class="math-container">$4q\sqrt{5}$</span> and comparing clearly gives <span class="math-container">$p=5$</span> and <span class="math-container">$q=2$</span></p>
<p>However, I also thought why not do this by getting unitary surds on either side, eg <span class="math-container">$$8\sqrt{p} = q\sqrt{80} \Rightarrow \sqrt{64p} = \sqrt{80q^2}$$</span> This tells me that <span class="math-container">$64p=80q^2$</span> or equivalently <span class="math-container">$4p = 5q^2$</span>.</p>
<p>How would I be able to get <span class="math-container">$p$</span> and <span class="math-container">$q$</span> from this method? How do I know that the solution is unique?</p>
<p>If so, is it fortuitous that we get a unique solution with these particular numbers or will it always be unique - I think I just need to see a proof to convince myself!</p>
| Oscar Lanzi | 248,217 | <p>Clearly <span class="math-container">$4p$</span> has to have a prime factor of <span class="math-container">$5$</span> but <span class="math-container">$4$</span> does not have that factor. So the prime <span class="math-container">$p$</span> has to have a factor of <span class="math-container">$5$</span>, therefore ... .</p>
|
4,562,451 | <p>I had this maths question:</p>
<blockquote>
<p>Given that <span class="math-container">$$8\sqrt{p} = q\sqrt{80}$$</span> where <span class="math-container">$p$</span> is prime, find the value of <span class="math-container">$p$</span> and the value of <span class="math-container">$q$</span></p>
</blockquote>
<p>I did this by simplifying the RHS to <span class="math-container">$4q\sqrt{5}$</span> and comparing clearly gives <span class="math-container">$p=5$</span> and <span class="math-container">$q=2$</span></p>
<p>However, I also thought why not do this by getting unitary surds on either side, eg <span class="math-container">$$8\sqrt{p} = q\sqrt{80} \Rightarrow \sqrt{64p} = \sqrt{80q^2}$$</span> This tells me that <span class="math-container">$64p=80q^2$</span> or equivalently <span class="math-container">$4p = 5q^2$</span>.</p>
<p>How would I be able to get <span class="math-container">$p$</span> and <span class="math-container">$q$</span> from this method? How do I know that the solution is unique?</p>
<p>If so, is it fortuitous that we get a unique solution with these particular numbers or will it always be unique - I think I just need to see a proof to convince myself!</p>
| Nicolino | 415,177 | <p>The answer is very different depending on whether q is required to be an integer or not. Let's look at both cases, starting from the equation <span class="math-container">$4p = 5q^2$</span> that you derived.</p>
<p>If q is restricted to the integers, then that means <span class="math-container">$5q^2$</span> is divisible by 5, and because of equality, <span class="math-container">$4p$</span> is as well. Since 4 does not divide 5, <span class="math-container">$p$</span> must. But <span class="math-container">$p$</span> also has to be prime, and the only prime number divisible by <span class="math-container">$5$</span> is <span class="math-container">$5$</span> itself. Therefore, <span class="math-container">$\boxed{\textbf{p = 5}}$</span>.</p>
<p>If q is unrestricted, then it's a whole different story. You can plug in any prime number for <span class="math-container">$p$</span>, and <span class="math-container">$4p = 5q^2$</span> will always have a real number solution since both sides will always be positive. In this situation, there are <strong>infinitely many solutions</strong>, because there are infinitely many prime numbers that <span class="math-container">$p$</span> could equal.</p>
|
2,417,542 | <p>$$\sum_{i,j}{n\brack i+j}\binom{i+j}i$$
Does this have a combinatorial interpretation? I don't see how to use Stirling numbers of the first kind in interpretations. I know that the answer is $(n+1)!$ , but the original question didn't provide it.</p>
| Behnam Esmayli | 283,487 | <p>To find the effect of this matrix on unit vector $\vec{I}$ multiply the matrix into $(1,0)^T$. So, $\vec{i}$ goes to $(o,-1)=-\vec{j}$. Thus a rotation of $-\pi /2$ or $3\pi /2$. Which one? Look at effect on $\vec{j}$. Multiply into $(0,1)$... You'll see that $-\pi/2$ is the right rotation whose matrix is the one given.</p>
|
4,035,300 | <p>If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!</p>
<p>'''</p>
<p>When a company bids for contracts it estimates the probability of winning each contract is <span class="math-container">$0.18$</span>, independent on whether other contracts have been won or lost.</p>
<p>(a) If the company bids for 5 contracts, what is the probability it wins:</p>
<p>i) at least one contract</p>
<p>ii) at least two contracts</p>
<p>'''</p>
<p>i)I used the OR rule for this question so if there's a <span class="math-container">$0.18$</span> chance then surely it's <span class="math-container">$0.18$</span> OR <span class="math-container">$0.18$</span>, etc which is just <span class="math-container">$0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$</span></p>
<p>ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did <span class="math-container">$0.18 * 0.18 = 0.0324$</span></p>
<p>If someone could just correct me if I took my stupid pills this morning I would be very grateful!</p>
| Neat Math | 843,178 | <p>Let <span class="math-container">$f(x)=x^3-x^2+2x-1, g(x)=x^3+x^2-1$</span>. We have</p>
<p><span class="math-container">$$-g(x)\cdot g(-x)=-(x^3+x^2-1)((-x)^3+(-x)^2-1)\\
=(x^3+x^2-1)(x^3-x^2+1) = x^6 - (x^2-1)^2\\
=x^6-x^4+2x^2-1=f(x^2)$$</span></p>
<p>Now if <span class="math-container">$g(x)=(x-r)(x-s)(x-t)$</span>, then</p>
<p><span class="math-container">$$f(x^2)= - (x-r)(x-s)(x-t) (-x-r)(-x-s)(-x-t)\\=(x^2-r^2)(x^2-s^2)(x^2-t^2)
\\\implies f(x)=(x-r^2)(x-s^2)(x-t^2)$$</span>
and we are done.</p>
|
1,724,419 | <p>I can create a large collection of normalized real valued $n$-dimensional vectors from some random process which I hypothesis should be equidistributed on the unit sphere. I would like to test this hypothesis.</p>
<ul>
<li>What is a good way numerically to test if vectors are equidistributed on the unit sphere? I am writing computer code so I will be testing that way</li>
<li>Is there some way to visualise the distribution given that my vectors are in $n$ dimensions?</li>
</ul>
| G Cab | 317,234 | <p>I would proceed on the basis that a (hollow) sphere with $N$ mass = $1$ points uniformly distributed shall have mass-centre (1st moment) =$0$, moment of inertia (2nd moment) = $\rho (n)N$, around any ax.
Where $\rho (n) = 2/3$ in the case $n=3$, while for the n-dimensional sphere in general it shall be calculated .</p>
<p>Therefore the $n$-vector of the mass-centre and that of the inertia could be a start to statistically evaluate the hypothesis of uniform distribution.</p>
|
2,373,175 | <p>It is probably an easy solution to this problem but I am either too overwhelmed already or not smart enough. Please help me out with this problem from the last year test</p>
<p><a href="https://i.stack.imgur.com/rzQWv.jpg" rel="nofollow noreferrer">Here is the problem</a></p>
| farruhota | 425,072 | <p>Note:
$$
\begin{array}{c|l}
n & \text{# of perfect squares} \\
\hline
1 & 1 \\
2 & 1 \\
3 & 1 \\
4 & 2 \\
5 & 2 \\
\cdot & \cdot \\
8 & 2 \\
9 & 3 \\
10 & 3 \\
\cdot & \cdot \\
15 & 3 \\
16 & 4 \\
\cdot & \cdot \\
n & [\sqrt{n}]
\end{array}
$$
where $[n]$ is the largest integer less than or equal to $n$.</p>
<p>Hence:
$$P(perfect \ square)=\frac{\# \ of \ perfect \ squares}{n}=\frac{[\sqrt{n}]}{n}.$$
For example, if $n=8:$
$$P(perfect \ square) = \frac{[\sqrt{8}]}{8}=\frac{2}{8}=\frac{1}{4}.$$</p>
|
6,831 | <p>I would like for the autocomplete feature to search through contexts, for example if I have a symbol named A`B`C`MyFunction, when I type A` and press "cmd + shift + k" it will complete it.</p>
<p><em>Edit</em></p>
<p>To be clear, I don't want to have to type the path because it's usually very long, and I don't want to have to type the function name again, even if the path itself gets auto completed. I want the following:</p>
<p>If I have these functions:</p>
<pre><code>Very`Long`Context`For`My`Function1
Very`Long`Context`For`My`Function2
...
</code></pre>
<p>I want to be able to type Very` and then press CMD+Shift+k, to get a dropdown menu saying exactly</p>
<pre><code>Very`Long`Context`For`My`Function1
Very`Long`Context`For`My`Function2
...
</code></pre>
| Brett Champion | 69 | <blockquote>
<p><em><strong>This is obsolete in Mathematica 9, which automatically includes contexts in completions.</strong></em></p>
<p><em><strong>Undocumented function: use at your own risk, subject to change in future versions, etc....</strong></em></p>
</blockquote>
<hr />
<p>The function you're interested in is <code>FE`FC</code>.</p>
<p>It's been around for a while (here's a <a href="http://www.mathematica-journal.com/issue/v6i3/columns/wagner/contents/63wagner.pdf" rel="noreferrer">Mathematica Journal article</a> that references it, near the end) although it has changed argument structure at least once that I'm aware of.</p>
<p>Anyway, here's the code I currently use to a similar end as what Mike would like. (Most of this is boilerplate from the original definition; the main difference is the use of a new function <code>FE`names</code>.)</p>
<pre><code>(* Nice little hack to have command completion (cmd-k) include contexts *)
Unprotect[FE`FC];
ClearAll[FE`FC]
FE`FC[FE`nameString_, FE`ignoreCase_:False] /; $Notebooks:=
MathLink`CallFrontEnd[FrontEnd`CompletionsListPacket[
FE`names[FE`nameString<>"*"], FE`ignoreCase],
FE`NoResult]
FE`names[FE`str_, FE`ignoreCase_:False] :=
Join[FE`shortContexts[FE`str], Names[FE`str, IgnoreCase -> FE`ignoreCase]];
FE`shortContexts[FE`patt_]:=
With[{FE`brettclen = Length[StringSplit[FE`patt, "`"]]},
Union[StringJoin[
Riffle[Take[#, Min[FE`brettclen, Length[#]]], "`", {2, -1, 2}]] & /@
StringSplit[Contexts[FE`patt], "`"]]
]
Protect[FE`FC];
</code></pre>
<p>The end result is that when I use command completion, I get contexts that match in addition to symbols. This isn't quite the same as Mike's request, since it gives the contexts one at a time:</p>
<p><img src="https://i.stack.imgur.com/abIqb.png" alt="enter image description here" /></p>
<p><img src="https://i.stack.imgur.com/EhDFZ.png" alt="enter image description here" /></p>
<p>since otherwise the list can get a bit overwhelming. For example, if you typed <code>Int</code> and then tried to complete to <code>IntegerPart</code>, there's a factor of ten difference:</p>
<pre><code>In[5]:= {Length[Names["Int*"]] + Length[Contexts["Int*"]],
Length[Names["Int*`*"]]}
Out[5]= {41, 419}
</code></pre>
|
1,859,178 | <p>Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,</p>
<p><a href="https://i.stack.imgur.com/wCPAN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wCPAN.png" alt="enter image description here"></a>
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure</p>
<p>Thanks help</p>
| Peter G. Chang | 339,525 | <p>The confusion arises from ignoring an implicit condition of the first equation, $y^2 = 4x$. Since $y^2 \geq 0$, we have the condition $x \geq 0$.</p>
<p>Thus of the two solutions for $x$ you have, the positive $x = 2 + \sqrt{5}$ is the only solution.</p>
<p>We see that there are two values of $y$ that satisfy the equations with the $x$ value above, and this agrees with the graphical representation.</p>
|
1,859,178 | <p>Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,</p>
<p><a href="https://i.stack.imgur.com/wCPAN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wCPAN.png" alt="enter image description here"></a>
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure</p>
<p>Thanks help</p>
| John Molokach | 90,422 | <p>$x_B$ is an extraneous solution. Try finding the value of $y$ in $y^2=4x_B$ and you'll find nonreal answers. </p>
|
1,859,178 | <p>Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,</p>
<p><a href="https://i.stack.imgur.com/wCPAN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wCPAN.png" alt="enter image description here"></a>
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1
\end{cases}$$
we have
$$x^2-4x-1=0\Longrightarrow x=2+\sqrt{5},2-\sqrt{5}$$ so the two point $\color{red}{x_{A}=2+\sqrt{5},x_{B}=2-\sqrt{5}}?$,then
contradiction it is well known$x_{A}=x_{B}$,where is mistake?and the second root $\color{blue}{x=2-\sqrt{5}}$ where is from?or where is point to this figure</p>
<p>Thanks help</p>
| DRF | 176,997 | <p>Ahh your edit clarified your mistake. While $x_A=x_B$ if you are looking at the $x$-coordinates of the two points you have in your picture these $x_A$ and $x_B$ don't correspond to the two solutions of the quadratic you found.</p>
<p>One of those solutions $x=2-\sqrt{5}$ does not solve the pair of equations since assuming we are working over the reals, $x$ must be positive. The two solutions you see in your picture both come from the $x=2+\sqrt{5}$ solution for which you get two different $y$ values when you plug back in to $y^2=4x$ to solve for $y$.</p>
|
1,025,321 | <p>$\ln(1+xy) = xy$</p>
<p>When I try to implicitly differentiate this I get</p>
<p>$\frac{1}{1+xy}(y + xy')$ = (y + xy')</p>
<p>At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.</p>
<p>However, the answer to this is $-\frac{y}{x}$... How do you get this?</p>
| C-star-W-star | 79,762 | <p>Not every smooth function induces a smooth map:
$$\Phi:\mathbb{B}\to\mathbb{R}:\quad\varphi(|x|):=\frac{1}{1-|x|}$$
Just have a careful look at its diagram:</p>
<p><img src="https://i.stack.imgur.com/p4Ck8.gif" alt="Diffeomorphic Ball"></p>
<p><em>(Note that it is not even differentiable at zero!)</em></p>
<p>The problem is that the norm is not smooth in general:
$$\|\cdot\|:V\to[0,\infty)$$</p>
<p>One can master this almost only by patching a bump on top:
$$\varphi(r):=e^{-\frac{1}{r^2}}\cdot\frac{1}{1-r}$$</p>
<p>For a mere diffeomorphism still the whole story highly depends on the chosen norm!</p>
|
1,025,321 | <p>$\ln(1+xy) = xy$</p>
<p>When I try to implicitly differentiate this I get</p>
<p>$\frac{1}{1+xy}(y + xy')$ = (y + xy')</p>
<p>At which point the $(y + xy')$ terms cancel out, leaving no $y'$ to solve for.</p>
<p>However, the answer to this is $-\frac{y}{x}$... How do you get this?</p>
| C-star-W-star | 79,762 | <p><strong>Standard Approach</strong></p>
<p>Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_E^2}}$$
Its inverse is given explicitely by:
$$\Psi:\mathbb{R}^n\to\mathbb{B}^n:\Psi(y):=\frac{y}{\sqrt{1+\|y\|_E^2}}$$
The argument of the roots never vanish:
$$1-\|x\|_E^2\neq0\quad1+\|y\|_E^2\neq0$$
So they're both differentiable.</p>
<p><strong>Alternative Approach</strong></p>
<p>Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{1-\|x\|_E^2}$$
By the inverse function theorem:
$$\mathcal{N}d\Phi(x)\equiv(0)\implies\mathrm{d}\Psi(y)=\mathrm{d}\Phi(\Psi(y))\in\mathcal{L}(\mathbb{R}^n)$$
But this holds globally as the identification is a bijection.</p>
<p><strong>Problematic Approach</strong></p>
<p>Consider the identification:
$$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_\infty^2}}$$
This is most likely not differentiable.</p>
|
355,489 | <p>What are suggestions for reducing the transmission rate of the current epidemics?</p>
<p>In summary, my best one so far is (once we are down to the stay home rule) to discretize time, i.e., to introduce the following rule for the general populace not directly involved in necessary services:</p>
<p><em>If members of your household go to public services on a certain day, the whole household should not use any public service for 2 weeks after. That way you still can get infected but cannot infect without knowing it.</em></p>
<p>Do you have some suggestions? Good models to look at? No predictions please, just advices what to do.</p>
<p>Edit: In more detail:</p>
<p>Model (the simplest version to make things as clear as possible): There are several categories of people <span class="math-container">$C_i$</span> that constitute portion <span class="math-container">$p_i$</span> of the population and have a certain matrix <span class="math-container">$A$</span> of interactions per day. Then, if <span class="math-container">$x_i(t)$</span> is the number of ever infected people in category <span class="math-container">$C_i$</span> by the time <span class="math-container">$t$</span>, the driving ODE is
<span class="math-container">$$
\dot x(t)=\alpha A[x(t)-x(t-\tau)]
$$</span>
where <span class="math-container">$\tau$</span> is the ("typical") time after which the sick person is removed from the population and <span class="math-container">$\alpha$</span> is the transmission probability. In this model the exponential growth is unsustainable if <span class="math-container">$\alpha\lambda(A)\tau<1$</span> where <span class="math-container">$\lambda$</span> is the largest eigenvalue of <span class="math-container">$A$</span>. We do not know <span class="math-container">$\alpha$</span> (though we can try to make suggestions how to reduce it, most such suggestions are already made by the government). The government can modify <span class="math-container">$A$</span> by issuing orders. Some orders merely reduce <span class="math-container">$a_{ij}$</span> to <span class="math-container">$0$</span>, but the government cannot shut essential public services completely this way.</p>
<p>Questions: What is <span class="math-container">$A$</span>, which entries <span class="math-container">$a_{ij}$</span> are most important to reduce, how to issue a sensible order that will modify them, and by how much they will reduce the eigenvalue? </p>
<p>First suggestion for these 4 answers: There are two categories of people: ordinary population that only goes to public services and<br>
public servants that both provide services and go to them. There is only one ("averaged") type of service involving a dangerous client-server interaction and all infection goes there. The portion of public servants in the population is <span class="math-container">$p$</span>. The server sees <span class="math-container">$M$</span> clients a day. Then the current social interaction matrix (say, for the grocery store I've seen yesterday) is <span class="math-container">$A=\begin{bmatrix}0 & M(1-p)\\ Mp & 2Mp\end{bmatrix}$</span> (ordinary population does not transmit to ordinary population, servers transmit to clients who can be both servers and clients, clients transmits to servers. The largest eigenvalue is <span class="math-container">$M(p+\sqrt p)$</span>. The lion's share comes from <span class="math-container">$\sqrt p$</span>, which is driven by the off-diagonal entries, The order should be issued as above, the effect that ordinary people never come to the service infected, which will remove the left bottom corner and drop the largest eigenvalue to <span class="math-container">$2Mp$</span>. Assuming <span class="math-container">$p=1/9$</span> (not too unrealistic), the drop will be two-fold even if you leave the service organization as it is. </p>
<p>That ends the solution I propose in mathematical language.
In layman terms, the public will completely do this part (you cannot ask for more) and still have some life, and we can concentrate on the models of how servers should be organized.</p>
<p>Edit:</p>
<p>Time to remove the non-relevant part and add some relevant thoughts about what else we can help with plus the response to JCK.</p>
<p>First of all, <em>It is very hard to formulate the orders correctly</em>. The stay home rule really means "avoid all close contacts outside your household except the necessary interactions with public servants providing vital services to you" (and even that version is, probably flawed). It is not about dogs, etc., as the Ohio version reads now. If everybody understood and implemented that meaning, my suggestion could be formulated as I said. However the intended meaning really is</p>
<p><em>When going to public services, minimize the probability that you can infect others as much as feasible and consider it to be <span class="math-container">$0$</span> if in the last two weeks nobody from your household had a contact with a stranger and nobody in the household had any symptoms</em>. </p>
<p>Now it is more to the point, but also more complicated. And if a professional mathematician like myself is so inept, imagine the difficulties of other people. </p>
<p>So, within that model, what would be the best formulation of the order to give?</p>
<p>Second, the set of questions I asked is clearly incomplete.
One has to add for instance "What assumption can be wrong and what effect that will have on the outcome <em>under the condition that the order is given in the currently stated form</em>. I have never seen a book that teaches the influence of the order formulation on the possible model behavior and that may be a crucial thing now. The interaction between the formal logic and differential equations within a given scenario is a non-existing science (or am I just ignorant of something? <em>That</em> reference would really be useful).</p>
<p>Third, if we have a particular question (say, how much to reduce and how organize the public transportation, which is NY and Tel Aviv headache now), what would be a good mathematical model for just that and what would be the corresponding order statement under this model? </p>
<p>The questions like that are endless and if there were ready answers in textbooks, the governments would just implement them already instead of having 7-hour meetings. So I can fairly safely conclude that they are not there.</p>
<p>What I tried with my model example was, in particular, to show that there may be some non-trivial moves in even seemingly optimal situations (strict stay home order and running only the absolutely vital services at the minimal rate that still allows to serve the population) that also make common sense and can be used by everyone right now and right here. Finding such moves can really help now. The main real life question now is "What can I (as government, business, or individual) do to reduce the largest eigenvalue of the social interaction matrix?" Now show me the textbook that teaches that and I'll stop the "ballspitting" and apologize for the wasted time of the people reading all this. </p>
| Gil Kalai | 1,532 | <p>Mathematically speaking it looks that if we split the categories <span class="math-container">$C_i$</span> into subcategories with zero (or small) cross transmission edges then this suppress the exponential growth at some scale. (And this is also the case for other geometric limitation on the graph of possible transmission, e.g if there are no edges for people who arge geographically remote). In technical terms one need to make the transmission graph (and every large subgraph) a very "bad" expander. </p>
|
808,144 | <p>Here is a fun looking one some may enjoy. </p>
<p>Show that:</p>
<p>$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$</p>
| Tolaso | 203,035 | <p>I understand that this an old question but I would like to share my 2 cents. First of all we recall the following Fourier identities:</p>
<p><strong>Lemma 1:</strong> Let <span class="math-container">$x \in [-\pi, \pi]$</span> then </p>
<p><span class="math-container">$$\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{\pi^2}{6} -\frac{\pi x}{2} + \frac{x^2}{4} \tag{1}$$</span></p>
<p><strong>Lemma 2:</strong> Let <span class="math-container">$x \in [-\pi, \pi]$</span> then</p>
<p><span class="math-container">$$\sum_{n=1}^{\infty} \frac{(-1)^n \sin nx}{n} = - \frac{x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^n \cos nx}{n^2} = \frac{x^2}{4} - \frac{\pi^2}{12} \tag{2}$$</span></p>
<p><strong>Lemma 3:</strong> It holds that:</p>
<p><span class="math-container">$$\ln \left ( 1 - 2x \cos a + x^2 \right ) = -2\sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \tag{3}$$</span></p>
<p>Hence for <span class="math-container">$a \in [0, \pi]$</span></p>
<p><span class="math-container">\begin{align*}
\ln \left ( \frac{1+2x \cos a + x^2}{1-2x \cos a + x^2} \right ) &= \ln \left ( 1 + 2x \cos a + x^2 \right ) - \ln \left ( 1 - 2x \cos a + x^2 \right ) \\
&=-2 \sum_{n=1}^{\infty} \frac{(-x)^n \cos na}{n} + 2 \sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \\
&= -2 \sum_{n=1}^{\infty} \frac{\left ( (-x)^n - x^n \right ) \cos na}{n} \\
&= 4 \sum_{n=1}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1}
\end{align*}</span></p>
<p>since <span class="math-container">$(-x)^n - x^n$</span> is <span class="math-container">$0$</span> whenever <span class="math-container">$n$</span> is even. Hence,</p>
<p><span class="math-container">\begin{align*}
\int_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x &= 4\int_{0}^{1} \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1} \, \mathrm{d}x \\
&=4 \sum_{n=0}^{\infty} \frac{\cos(2n+1)a}{2n+1} \int_{0}^{1} x^{2n} \, \mathrm{d}x \\
&=4 \sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2}
\end{align*}</span></p>
<p>This is nothing else than <span class="math-container">$\chi_2$</span> <a href="https://en.wikipedia.org/wiki/Legendre_chi_function" rel="nofollow noreferrer">Legendre function</a> directly associated with the series in <span class="math-container">$(1)$</span>, <span class="math-container">$(2)$</span>. Hence,</p>
<p><span class="math-container">\begin{align*}
\sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2} & = \mathfrak{Re} \left ( \sum_{n=0}^{\infty} \frac{e^{i(2n+1)a}}{(2n+1)^2} \right ) \\
&=\frac{1}{2}\mathfrak{Re} \left ( \mathrm{Li}_2 \left ( e^{ia} \right ) - \mathrm{Li}_2 \left ( - e^{ia} \right ) \right )
\end{align*}</span></p>
<p>The real part of <span class="math-container">$\mathrm{Li}_2(e^{ix})$</span> is equation <span class="math-container">$(1)$</span> whereas the real part of <span class="math-container">$\mathrm{Li}_2(-e^{ix})$</span> is equation <span class="math-container">$(2)$</span>. Thus, </p>
<p><span class="math-container">\begin{align*}
4\sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{\left ( 2n+1 \right )^2} &= 2 \left ( \frac{\pi^2}{6} - \frac{\pi a}{2} + \frac{a^2}{4} - \frac{a^2}{4} + \frac{\pi^2}{12} \right ) \\
&=\frac{\pi^2}{2} - \pi a
\end{align*}</span></p>
<p>Because the LHS is even so must be the RHS. So, the result can be extended for <span class="math-container">$a \in [-\pi, 0]$</span>. Therefore, </p>
<p><span class="math-container">$$\int_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x = \frac{\pi^2}{2} - \pi \left| a \right|$$</span></p>
|
2,477,107 | <p>Okay so I have to prove this. I can write that if 2 divides n and 7 divides n, then there must be integers k and m such that
$2*k=n$
and
$7*m=n$</p>
<p>So $14*k*m=n^2$</p>
<p>But what to do after that?</p>
<p>If I say that then 14 divides $n^2$, I get bit of a circular argument, but if I write that n divides $14*k*m$, then I don't know what to do next.</p>
<p>Any help/suggestions?</p>
| Math Lover | 348,257 | <p>Following from what you have written, $$n = 2k=7m \implies k=\frac{7m}{2}.$$
Since $k$ is an integer and $\gcd(2,7)=1$, $m/2$ must be an integer; i.e., $m/2=r \implies m=2r$, where $r$ is an integer. Therefore,
$$n=7m=7\times 2r = 14 r.$$
Q.E.D.</p>
|
2,875 | <p>I've heard that irreducible unitary representations of noncompact forms of simple Lie groups, the first example of such a group <code>G</code> being <code>SL(2, R)</code>, can be completely described and that there is a discrete and continuous part of the spectrum of <code>L^2(G)</code>.</p>
<ol>
<li>How are those representations described?</li>
<li>Do all unitary representations come from <code>L^2(G)</code>?</li>
<li>How are those related to representation of compact <code>SO(3, R)</code>? </li>
<li>What happens in the flat limit between <code>SL(2, R)</code> and <code>SO(3, R)</code>?</li>
</ol>
<p>Also, is it possible to answer the questions above simultaneously for all Lie groups, not just <code>SL(2, R)</code>?</p>
| David Bar Moshe | 1,059 | <p>I just want to elaborate more on questions 3. and 4. I'll consider the locally isomorphic groups SU(1,1) of SL2(R) and SU(2) of SO(3)</p>
<p>There is an analogy between the discrete series of SU(1,1) and the unitary irreps of SO(3). Both have holomorphic representations on the group's orbit on the flag manifold S^2 = SL(2,C)/B (B is a Borel subgroup). In the case of SU(2), the orbit is the whole of SU(2) while for SU(1,1) its is a noncomapct supspace: The Poicare disc. In both cases the representation space is a reproducing kernel Hilbert space and the group action is throug a Mobius transformation. This analogy generalizes to other non-compact groups having a holomorphic discrete series and it can be considered as a generalization of the Borel-Weil construction for compact groups.</p>
<p>Concerning question 4. I think that you are talking about Wigner's theory of Lie group contraction, in which a Lie group with the same dimension and with more "flat" directions is associated to the original Lie group. For example there is a contraction of SU(2) to Eucledian group in two dimensions and SU(1,1) to the Poncare group in two dimensions. There are interseting connections to the group representations of the contracted versions, and also of the Casimirs.</p>
|
228,224 | <p>I am faced with the following expression</p>
<p><span class="math-container">$$
-\frac{(1 - a x^{2})^{b/2}}{b} {{}_2F_1} (1, \frac{b}{2}; \frac{c}{2}; 1 - a x^{2}) = - p t
$$</span></p>
<p>where <span class="math-container">$ a, b, c, p $</span> are constant values. Also, <span class="math-container">$ {{}_2F_1} $</span> is a hypergeometric function. Now, I am trying to plot <span class="math-container">$ x $</span> versus <span class="math-container">$ t $</span>. I am thankful for any suggestions.</p>
<hr />
<p>I don't know why the code doesn't work in my machine. I am trying a different way by this</p>
<p><code>f[a_, b_, c_, p_] := -((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2] == -p t; Plot[Evaluate@Table[f[1, 2, 3, p], {p, 1, 2, 3}], {t, -1, 2}, {x, -2, 2}]</code></p>
<p>is that correct?</p>
| cvgmt | 72,111 | <p>Since we can solve t ,so we can also use <code>ParametricPlot</code></p>
<pre><code>Clear["`*"];
a = 1;
b = 2;
c = 3;
t = ((1 - a x^2)^(b/2)/b) Hypergeometric2F1[1, b/2, c/2, 1 - a x^2]/p;
ParametricPlot[
Table[{t, x}, {p, {1, 2, 3}}] // Evaluate, {x, -2, 2}, {t, -1, 2},
Axes -> False, FrameLabel -> {"t", "x"}]
</code></pre>
<p><a href="https://i.stack.imgur.com/0ueGT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/0ueGT.png" alt="enter image description here" /></a></p>
|
3,700,814 | <p>Consider the vector space <span class="math-container">$V=C(\mathbb{R},\mathbb{R})$</span> and <span class="math-container">$V\ni U=\{f\in C(\mathbb{R},\mathbb{R}) |f(0)=0\}$</span>. I want to find a complement of <span class="math-container">$U$</span>, such that <span class="math-container">$V=U\oplus W$</span>. This condition is the same as finding a set <span class="math-container">$W$</span> that satisfies <span class="math-container">$V=U+W$</span> and <span class="math-container">$U\cap W={f_0}$</span> where <span class="math-container">$f_0$</span> is the null element. At a lecture, the professor defined <span class="math-container">$W=Span({\mathbb{\mathbf{1}}})$</span>, where <span class="math-container">$\mathbf{1}$</span> is the function defined by <span class="math-container">$\mathbf{1}(x)=1\ \forall x\in \mathbb{R}$</span> and proved that this is a complement. He did not, however, give an explanation of how intuitively to find this complement without knowing beforehand that it indeed does satisfy the conditions. </p>
<p>Looking at the definition of <span class="math-container">$U$</span>, I would rather define <span class="math-container">$W$</span> as:
<span class="math-container">$$
W=\{f \in C(\mathbb{R}, \mathbb{R}): f(0) \neq 0\} \cup\left\{f_{0}\right\}
$$</span>
But I would imagine that if it were that simple, the professor would have done the same. Is there something wrong with my definition?</p>
| Digitallis | 741,526 | <p>As others have pointed out the <span class="math-container">$W$</span> you defined is not a subspace. It also looks like you think the choice of <span class="math-container">$W$</span> should always be <span class="math-container">$W = U^c \cup \{ f_0\}$</span> that is not the case, precisely because this will not always be a subspace.</p>
<p>Now how do you derive that <span class="math-container">$W$</span> should be the constant functions ?</p>
<p>The idea behind the decomposition <span class="math-container">$V = U + W$</span> and <span class="math-container">$W \cap U = \{ f_0 \}$</span> is that every element <span class="math-container">$f$</span> of <span class="math-container">$V$</span> can be written as a unique <span class="math-container">$f = g + h$</span> with <span class="math-container">$g \in U$</span> and <span class="math-container">$h \in W$</span>. </p>
<p>Your looking for <span class="math-container">$g,h$</span> such <span class="math-container">$f = g + h$</span> where <span class="math-container">$g$</span> is such that <span class="math-container">$g(0) = 0$</span> so define <span class="math-container">$g(x) = f(x)- f(0).$</span> </p>
<p><span class="math-container">$$ f(x) = f(x) - f(0) + f(0)$$</span> </p>
<p>with <span class="math-container">$h(x) = f(0)$</span> a constant function therefore define
<span class="math-container">$$W = \{h \in C(\mathbb R , \mathbb R) \; \vert \; h \text{ is a constant function } \}$$</span> </p>
<p>Since the only constant function that vanishes at <span class="math-container">$0$</span> is <span class="math-container">$f_0$</span> we have <span class="math-container">$U \cap W = \{ f_0\}$</span> and we also have <span class="math-container">$V = U + W.$</span></p>
|
1,607,190 | <p>Prove by induction that $8^{n} − 1$ for any positive integer $n$ is divisible by $7$. </p>
<p>Hint: It is easy to represent divisibility by $7$ in the following way: $8^{n} − 1 = 7 \cdot k$ where k is a positive integer.</p>
<p>This question confused me because I think the hint isn't true. If $n = 1$ and $k = 2$ for example, then we end up with $7 = 14$ which is obviously invalid. Does this mean the $n \leq k$ in order for the hint to be true.</p>
| Harish Chandra Rajpoot | 210,295 | <p>Notice the following steps of mathematical induction, </p>
<ol>
<li><p>Setting $n=1$, $$8^1-1=7$$ above number is divisible by $7$ for $n=1$</p></li>
<li><p>assume that $8^n-1$ is divisible by $7$ for $n=k$ then $$8^k-1=7m$$
or $$8^k=7m+1\tag 1$$
where, $m$ is some integer </p></li>
<li><p>setting $n=k+1$, $$8^{k+1}-1$$
$$=8\cdot 8^k-1$$
setting the value of $8^k$ from (1), $$=8(7m+1)-1$$
$$=7(8m+1)$$</p></li>
</ol>
<p>since, $(8m+1)$ is an integer hence the number $7(8m+1)$ is divisible by $7$ thus $8^n-1$ is divisible by $7$ for $n=k+1$</p>
<p>hence, $8^n-1$ is divisible by $7$ for all integers $n\ge 1$</p>
|
1,016,884 | <p>Four friends, Andrew, Bob, Chris and David, all have different heights. The sum of their heights is 670 cm.
Andrew is 8cm taller than Chris and Bob is 4cm shorter than David.
The sum of the heights of the tallest and shortest of the friends is 2cm more than the sum of the heights of the other two.
Find the height of each friend.</p>
| advocateofnone | 77,146 | <p>Let the heights of andrew bob chris and david be $a,b,c,d$. Then
$$a+b+c+d=670$$
$$a=c+8$$
$$d=b+4$$</p>
<p>This tells us that either $a$ or $d$ is the tallest height and $c$ or $b$ the shortest. So there are in total $4$ options try each of them and see if all the constraints are satisfied. For example if andrew is tallest and chris is shortest.
We have
$$a+c=2+b+d$$
This gives c=164,a=172,b=165,d=169. This fits all the constraints. This is one possible solution. Try similar other 3 possibilities.</p>
|
4,150,776 | <p>Let <span class="math-container">$(a_n)_{n=1}^\infty$</span> Let be a positive, increasing, and unbounded sequence. Prove that the series:</p>
<p><span class="math-container">$$\sum_{n=1}^\infty\left(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}\right)$$</span></p>
<p>convergent.</p>
<hr />
<p>We know that since <span class="math-container">$a_n$</span> is increasing and unbounded, than <span class="math-container">$\lim_{n \to \infty}a_n=\infty$</span>, so I want to apply that to say that, <span class="math-container">$\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$</span> is decreasing and its limit will be <span class="math-container">$0$</span>.</p>
<p>My problem is that every time I get confused while it says <span class="math-container">$a_{2n}$</span> or <span class="math-container">$a_{2n-1}$</span>, it is less intuitive for me than just "normal" <span class="math-container">$a_n$</span>...</p>
<p>Appreciate your help!</p>
<p>Thanks a lot!</p>
| Thomas Andrews | 7,933 | <p>Hint: Let</p>
<p><span class="math-container">$$b_n=\frac1{a_{n-1}}-\frac1{a_{n}}$$</span></p>
<p>Each <span class="math-container">$b_n$</span> is non-negative, since <span class="math-container">$a_n$</span> is increasing.</p>
<p>Then your series is: <span class="math-container">$$\sum_{n=1}^\infty b_{2n}$$</span> But:</p>
<p><span class="math-container">$$\sum_{n=2}^{2N} b_n\geq \sum_{n=1}^{N} b_{2n}\geq 0$$</span></p>
<p>And <span class="math-container">$$\sum_{n=2}^{2N} b_n=\frac1{a_{1}}-\frac1{a_{2N}}.$$</span></p>
|
4,118,161 | <p>Hello I am solving the following problem and could use some help.</p>
<p>Let (C[0,1],<span class="math-container">$d_\infty$</span>) be the metric space of continuous functions on [0,1] where the distance function is defined by</p>
<p>Let <span class="math-container">$d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $</span></p>
<p>Consider the function <span class="math-container">$T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$</span>) defined by</p>
<p>Let <span class="math-container">$(Tf)(x)=\int_0^xf(t)dt$</span></p>
<p>Prove:</p>
<p>(a) <span class="math-container">$T$</span> has a unique fixed point, i.e. there is a unique <span class="math-container">$f ∈ C[0,1]$</span> satisfies <span class="math-container">$Tf = f$</span> .</p>
<p>(b) <span class="math-container">$T^2$</span> is a contraction.</p>
<p>I know I'd like to use the function <span class="math-container">$f(x)=c*e^x$</span>, but I don't know how ton put the proof together.</p>
| mathcounterexamples.net | 187,663 | <p><strong>For (a)</strong></p>
<p><span class="math-container">$f \in \mathcal C^1([0,1], \mathbb R) =V$</span> is a fixed point if and only if</p>
<p><span class="math-container">$$f(x) = \int_0^x f(t) \ dt$$</span> for all <span class="math-container">$x \in [0,1]$</span>. As <span class="math-container">$f$</span> is supposed to be continuous, it is also differentiable as <span class="math-container">$\int_0^x f(t) \ dt$</span> is differentiable. Therefore, you get the differential equation <span class="math-container">$$f^\prime(x) = f(x)$$</span> whose solutions are indeed <span class="math-container">$f(x) =ce^x$</span>. You also have <span class="math-container">$f(0)=0$</span>. Hence <span class="math-container">$c=0$</span> and <span class="math-container">$f(x) =0$</span> is the unique fixed point.</p>
<p><strong>For (b)</strong></p>
<p>For <span class="math-container">$f,g \in V$</span> and <span class="math-container">$x \in [0,1]$</span></p>
<p><span class="math-container">$$\begin{aligned}
\vert (T^2(f) - T^2(g))(x) \vert &= \left\vert \int_0^x \left(\int_0^t (f(t) - g(t)) \ dt\right) dx\right\vert\\
&\le \int_0^x \left(\int_0^t \Vert f - g\Vert \ dt\right) dx\\
&\le \left\Vert f - g \right\Vert\int_0^x \left(\int_0^t \ dt\right) dx\\
&= \left\Vert f - g \right\Vert\int_0^x t \ dt\\
&=\frac{1}{2} \left\Vert f-g \right\Vert
\end{aligned}$$</span> This implies that <span class="math-container">$$\Vert T^2(f) - T^2(g) \Vert \le \frac{1}{2} \Vert f - g \Vert$$</span>proving that <span class="math-container">$T^2$</span> is indeed a contraction.</p>
|
4,118,161 | <p>Hello I am solving the following problem and could use some help.</p>
<p>Let (C[0,1],<span class="math-container">$d_\infty$</span>) be the metric space of continuous functions on [0,1] where the distance function is defined by</p>
<p>Let <span class="math-container">$d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $</span></p>
<p>Consider the function <span class="math-container">$T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$</span>) defined by</p>
<p>Let <span class="math-container">$(Tf)(x)=\int_0^xf(t)dt$</span></p>
<p>Prove:</p>
<p>(a) <span class="math-container">$T$</span> has a unique fixed point, i.e. there is a unique <span class="math-container">$f ∈ C[0,1]$</span> satisfies <span class="math-container">$Tf = f$</span> .</p>
<p>(b) <span class="math-container">$T^2$</span> is a contraction.</p>
<p>I know I'd like to use the function <span class="math-container">$f(x)=c*e^x$</span>, but I don't know how ton put the proof together.</p>
| Kavi Rama Murthy | 142,385 | <p>Answer for the second part: <span class="math-container">$|T^{2}f(x)|=|\int_0^{x}\int_u^{x}dt du|=|\int_0^{x}(x-u)f(u)du|\leq \|f\|\int_0^{x}f(u)du=\frac 1 2 \|f\|$</span>.</p>
|
405,648 | <p>Is there a sensible characterization of groups <span class="math-container">$G$</span> with the following property?</p>
<blockquote>
<p>Every extension of groups <span class="math-container">$1\to G\to H\to K\to 1$</span> is split.</p>
</blockquote>
<p>A complete group <span class="math-container">$G$</span> has that property and in fact such a group has a <em>normal</em> complement in every group that contains it as a normal subgroup (moreover, completeness is characterized by this)</p>
<p>For comparison, a group <span class="math-container">$K$</span> has the property that every extension <span class="math-container">$1\to G\to H\to K\to 1$</span> is split iff it is free (there is such an extension with <span class="math-container">$H$</span> free, and the splitting map <span class="math-container">$K\to H$</span> is injective, so gives an isomorphism of <span class="math-container">$K$</span> with a subgroup of <span class="math-container">$H$</span>, which is free by the Nielsen–Schreier theorem)</p>
<p><strong>NB:</strong> the title does use the old meaning of «extension of a group»...</p>
| YCor | 14,094 | <blockquote>
<p><strong>Proposition.</strong> Given a group <span class="math-container">$G$</span>, this happens (every exact sequence <span class="math-container">$1\to G\to H\to H/G\to 1$</span> splits) iff <span class="math-container">$G$</span> has a trivial center and <span class="math-container">$1\to G\to \mathrm{Aut}(G)\to\mathrm{Out}(G)\to 1$</span> splits.</p>
</blockquote>
<p><strong>Lemma:</strong> Let <span class="math-container">$Z$</span> be embedded as central subgroup in two groups <span class="math-container">$G$</span>, <span class="math-container">$S$</span>, and let <span class="math-container">$H$</span> be the quotient of <span class="math-container">$G\times S$</span> by the diagonal <span class="math-container">$D_Z$</span> of <span class="math-container">$Z$</span>. Then the extension <span class="math-container">$1\to G\to H\to H/G\to 1$</span> splits if and only if there exists a homomorphism <span class="math-container">$S\to G$</span> extending the identity of <span class="math-container">$S$</span>.</p>
<p><em>Proof.</em> Interpreting the existence of a splitting by pulling back in <span class="math-container">$G\times S$</span>, we see that the exact sequence splits if and only if there exists a subgroup <span class="math-container">$L$</span> of <span class="math-container">$G\times S$</span> such that (a) the projection of <span class="math-container">$L$</span> on <span class="math-container">$S/Z$</span> is surjective, and such that (b) <span class="math-container">$L\cap (G\times Z)=D_Z$</span>. Note that given (b), the condition (a) means that the projection of <span class="math-container">$L$</span> on <span class="math-container">$S$</span> is surjective. Thus a subgroup of <span class="math-container">$G\times S$</span> has this form if and only if it is the graph of a homomorphism <span class="math-container">$S\to G$</span> that is identity on <span class="math-container">$Z$</span>. <span class="math-container">$\Box$</span></p>
<p><strong>Lemma:</strong> Let <span class="math-container">$G$</span> be a group and <span class="math-container">$A$</span> an abelian group. Then there exists a group <span class="math-container">$S$</span> such that <span class="math-container">$S$</span> has no nontrivial quotient of cardinal <span class="math-container">$\le|G|$</span> and such that the center of <span class="math-container">$S$</span> contains an isomorphic copy of <span class="math-container">$A$</span>.</p>
<p><strong>Proof:</strong> for a field of characteristic zero <span class="math-container">$K$</span>, consider the semidirect product <span class="math-container">$S_K=\mathrm{SL}_2(K)\ltimes\mathrm{Hei}_3(K)$</span> (this is a central extension by <span class="math-container">$K$</span> of the standard semidirect product <span class="math-container">$\mathrm{SL}_2(K)\ltimes K^2$</span>; <span class="math-container">$\mathrm{Hei}$</span> refers to Heisenberg). This group has the property that every nontrivial quotient admits the simple group <span class="math-container">$\mathrm{PSL}_2(K)$</span> as quotient (<span class="math-container">$*$</span>). Hence is has no proper quotient of cardinal <span class="math-container">$\le |K|$</span>. Now the center of <span class="math-container">$S$</span> is isomorphic to <span class="math-container">$K$</span> as additive group, which is a vector space over <span class="math-container">$\mathbf{Q}$</span> which can be prescribed to be of arbitrary nonzero dimension. Since every abelian group <span class="math-container">$A$</span> is subquotient of a group of this form (<span class="math-container">$A$</span> is quotient of <span class="math-container">$\mathbf{Z}^{(A)}$</span> which is subgroup of <span class="math-container">$\mathbf{Q}^{(A)}$</span>), it therefore embeds into some quotient of <span class="math-container">$S_K$</span> provided <span class="math-container">$K$</span> is large enough.</p>
<p>(<span class="math-container">$*$</span>) hint: use that the normal subgroups of <span class="math-container">$\mathrm{SL}_2(K)\rtimes K^2$</span> are <span class="math-container">$\{0\}$</span>, <span class="math-container">$K^2$</span>, <span class="math-container">$\{\pm 1\}\ltimes K^2$</span> and the whole group. To conclude, use that a normal subgroup whose projection modulo the center is the whole group <span class="math-container">$\mathrm{SL}_2(K)\rtimes K^2$</span> has to contain the center.<span class="math-container">$\Box$</span></p>
<p><strong>Lemma:</strong> let <span class="math-container">$G$</span> be a group with nontrivial center <span class="math-container">$Z$</span>. Then there exists a nonsplit extension <span class="math-container">$1\to G\to H\to H/G\to 1$</span>.</p>
<p><strong>Proof:</strong> choose for <span class="math-container">$A=Z$</span> a group <span class="math-container">$S$</span> as in the previous lemma. By the first lemma, the resulting central extension is not split, since a splitting would imply the existence of a quotient of <span class="math-container">$S$</span> of size between <span class="math-container">$|Z|\ge 2$</span> and <span class="math-container">$|G|$</span>. <span class="math-container">$\Box$</span></p>
<p><strong>Proof of the proposition:</strong> suppose that <span class="math-container">$G$</span> satisfies the splitting property. By the previous lemma, <span class="math-container">$G$</span> has trivial center. Hence the splitting of the exact sequence <span class="math-container">$1\to G\to \mathrm{Aut}(G)\to \mathrm{Out}(G)\to 1$</span> follows by assumption.</p>
<p>Conversely, suppose that <span class="math-container">$G$</span> has trivial center and this exact sequence splits. Let <span class="math-container">$1\to G\to H\to K\to 1$</span> be an exact sequence. Let <span class="math-container">$Z$</span> be the centralizer of <span class="math-container">$G$</span> in <span class="math-container">$H$</span>. Then <span class="math-container">$N\cap G=1$</span>. So this induces an exact sequence <span class="math-container">$1\to G\to H/Z\to K/Z\to 1$</span>, which is split, so <span class="math-container">$H/Z=G\rtimes L/Z$</span> for some subgroup <span class="math-container">$L$</span> of <span class="math-container">$H$</span> containing <span class="math-container">$Z$</span>. Hence <span class="math-container">$H:G\rtimes L$</span>. <span class="math-container">$\Box$</span></p>
<hr>
<p>Note that the construction of a "big" central extension as in the second lemma is disappointing if one wishes to somewhat preserve the cardinalities (which is not required by OP). It might be possible to improve this, but with some technical cost.</p>
<hr>
<p>Note: (Oct 8 '21) I rewrote the proof since the initial one from Oct 7 '21 (when dealing with groups with nontrivial center) was fatally flawed; the issue was pointed out in comments by Mariano Suarez-Alvarez.</p>
|
2,889,929 | <p>I have following function:</p>
<p>$$f(x)=x^2\cdot\left({\sin{\frac 1 x}}\right)^2$$</p>
<p>I want to find the limit of the function for $x\rightarrow0^\pm$. First I analyze $\frac 1 x$:</p>
<ul>
<li>$\frac {1}{x}\rightarrow +\infty$ for $x\rightarrow0^+$</li>
</ul>
<p>but the $\sin$ of infinity does not exist. Then I use the comparison theorem (I don't know how it's called in English) and conclude that, because</p>
<p>$$\left|{x^2\left({\sin{\frac 1 x}}\right)}^2 \right| \le \frac{1}{x^2}\rightarrow0^+$$</p>
<p>therefore the initial function tends to $0$. Is this reasoning correct? Are there better ways? </p>
| Jan | 254,447 | <p>I am not going to be giving you the answer, but I will hopefully give you the means to find it yourself.</p>
<p>So we have three shooters who can all hit or miss the target. So there are many possible outcomes of the shooting. For example, A can miss, B can hit and C can miss. Clearly this is different from A hitting, B missing and C missing. Even though the amount of them hitting the target in total is the same, the events are not the same. In fact, they don't even have the same probability!</p>
<p>The first event, A miss, B hit, C miss has a certain probability. Do you know how to compute this? Assuming that they shoot independently (A hitting or missing does not impact the chances of B or C hitting or missing) we can multiply the probabilities to get the probability of this event.
$$P(\mathrm{A \ miss, \ B \ miss, \ C \ miss})=P(\mathrm{A\ miss}) \cdot P(\mathrm{B \ hit}) \cdot P(\mathrm{C\ miss}) = (1-\frac{1}{2}) \cdot \frac{1}{3} \cdot (1-\frac{1}{4})$$</p>
<p>Now see if you can compute this for the other event. </p>
<p>Now that you understand how to calculate the probability for an event, let's look at the question. </p>
<p>We are looking for the probability of 2 or more shooters hitting the target. So this is the cases where 2 of the 3 shooters hit and the case where they all hit the target.
If you list all these events, you see that there are 4 events in total for which you will have to compute the probabilities. Adding them together will give you the required answer!</p>
<p>Good luck :)</p>
|
1,550,923 | <p>I know of two applications of vector spaces over $\mathbb Q$ to problems posed by people not specifically interested in vector spaces over $\mathbb Q$:</p>
<ul>
<li>Hilbert's third problem; and</li>
<li>The Buckingham pi theorem.</li>
</ul>
<p>What others are there?</p>
| user293794 | 293,794 | <p>One nice application is proving the following theorem:</p>
<blockquote>
<p>A rectangle
R
with side lengths
$1$
and
$x$
, where
$x$
is irrational, cannot be “tiled” by finitely many squares (so that the squares
have disjoint interiors and cover all of
R
).</p>
</blockquote>
<p>The proof can be found here: <a href="http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf" rel="nofollow">http://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf</a>, as Miniature 12 (pg. 39).</p>
|
4,576,868 | <p>I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself.
The generating function is
<span class="math-container">$F(z)^8$</span>
Where
<span class="math-container">$$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$</span></p>
<p>Any help appreciated</p>
| Leucippus | 148,155 | <p>A method that will take a while but gives the first few terms for the equation
<span class="math-container">$$ F(x) = x + x^6 F(x)^5 + x^{11} F(x)^{10} + x^{16} F(x)^{15} + x^{21} F(x)^{20} $$</span>
is to let <span class="math-container">$F(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots$</span>.</p>
<p>First notice that the right-hand side does not have a constant term. This leads to <span class="math-container">$F(x)$</span> being <span class="math-container">$F(x) = x + a_{2} x^2 + \cdots$</span>. Also notice that <span class="math-container">$a_2 = a_{3} = \cdots = a_{10} = 0$</span> and further reduces <span class="math-container">$F(x)$</span> to <span class="math-container">$F(x) = x + x^{11} + \cdots$</span></p>
<p>This leads to
<span class="math-container">\begin{align}
x + x^{11} + a_{21} \, x^{21} + \cdots &= x + x^{11} (1 + 5 \, x^{10} + \cdots) + x^{21} (1 + 10 x^{10} + \cdots) + \cdots \\
&= x + x^{11} + (5 + 1) \, x^{21} + \cdots.
\end{align}</span>
Continuing gives the result mentioned by Gary in the comments.</p>
|
3,003,672 | <p>Say I have an infinte 2D grid (ex. a procedurally generated world) and I want to get a unique number for each integer coordinate pair. How would I accomplish this?</p>
<p>My idea is to use a square spiral, but I cant find a way to make a formula for the unique number other than an algorythm that just goes in a square spiral and stops at the wanted coords.</p>
<p>The application for this converstion could be for example a way to save an n dimensional shape to a file where each line represents a chunk of the shape (by using <span class="math-container">$u(x, y, z) = u(u(x, y), u(y, z))$</span> ), or have a very unique random seed for each integer point (ex. a way to hash an integer vector to a data point in an n dimensional array)</p>
| Kamil Maciorowski | 331,040 | <p>Other answers state how to convert integers to naturals, I won't repeat this step. Let's suppose you have two naturals, e.g.:</p>
<p><span class="math-container">$$ 123 $$</span>
<span class="math-container">$$ 98765 $$</span></p>
<p>Add leading zeros to obtain equal number of digits:</p>
<p><span class="math-container">$$ 00123 $$</span>
<span class="math-container">$$ 98765 $$</span></p>
<p>And "interleave":</p>
<p><span class="math-container">$$ 0908172635 $$</span></p>
<p>Reverting is trivial: you pick digits from either odd or even positions.</p>
<p>Notes:</p>
<ul>
<li>the representation depends on the base of the numeral system you use;</li>
<li>you can expand the method to non-negative reals in a quite obvious way;</li>
<li>similarly you can create a method that takes any fixed number of numbers and yields one number.</li>
</ul>
|
3,807,665 | <p>Let <span class="math-container">$ V $</span> a finite dimensional vector space and <span class="math-container">$ V^* $</span> the dual space. Further let be <span class="math-container">$ B=(v_1,...,v_n) $</span> a base of <span class="math-container">$ V $</span> and <span class="math-container">$ B^*=(v_1^*,...,v_n^*) $</span> a dual base of <span class="math-container">$ V^*$</span>. Show:</p>
<p>For all <span class="math-container">$ v,w\in V $</span> is <span class="math-container">$ v=w $</span> if and only if <span class="math-container">$ f(v)=f(w) $</span> for all <span class="math-container">$ f\in V^* $</span>.</p>
<p>My idea:</p>
<p>At first we have <span class="math-container">$ f=f(v_1)\cdot v_1^*+...+f(v_n)\cdot v_n^* $</span>.
Then we get with <span class="math-container">$ v=\sum\limits_{i=1}^n a_i\cdot v_i $</span> and <span class="math-container">$ w=\sum\limits_{i=1}^n b_i\cdot v_i $</span> for all <span class="math-container">$ v,w\in V $</span></p>
<p><span class="math-container">$ f(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n a_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot a_j\\=\sum\limits_{j=1}^n f(v_j)\cdot b_j\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n b_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(w)\\=f(w) $</span></p>
<p>Now I want to show that <span class="math-container">$ a_j=b_j $</span> for all <span class="math-container">$ j=1,...,n $</span> but I don't know how.</p>
| Peter Franek | 62,009 | <p>This seems too complicated. Just note that <span class="math-container">$v_i^*(\sum_j a_j v_j) = v_i^*(\sum_j b_j v_j)$</span> implies <span class="math-container">$a_i=b_i$</span>.</p>
|
3,807,665 | <p>Let <span class="math-container">$ V $</span> a finite dimensional vector space and <span class="math-container">$ V^* $</span> the dual space. Further let be <span class="math-container">$ B=(v_1,...,v_n) $</span> a base of <span class="math-container">$ V $</span> and <span class="math-container">$ B^*=(v_1^*,...,v_n^*) $</span> a dual base of <span class="math-container">$ V^*$</span>. Show:</p>
<p>For all <span class="math-container">$ v,w\in V $</span> is <span class="math-container">$ v=w $</span> if and only if <span class="math-container">$ f(v)=f(w) $</span> for all <span class="math-container">$ f\in V^* $</span>.</p>
<p>My idea:</p>
<p>At first we have <span class="math-container">$ f=f(v_1)\cdot v_1^*+...+f(v_n)\cdot v_n^* $</span>.
Then we get with <span class="math-container">$ v=\sum\limits_{i=1}^n a_i\cdot v_i $</span> and <span class="math-container">$ w=\sum\limits_{i=1}^n b_i\cdot v_i $</span> for all <span class="math-container">$ v,w\in V $</span></p>
<p><span class="math-container">$ f(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n a_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot a_j\\=\sum\limits_{j=1}^n f(v_j)\cdot b_j\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n b_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(w)\\=f(w) $</span></p>
<p>Now I want to show that <span class="math-container">$ a_j=b_j $</span> for all <span class="math-container">$ j=1,...,n $</span> but I don't know how.</p>
| Kevin Aquino | 804,948 | <p>It can be useful to note that if <span class="math-container">$\mathcal{B}= (v_{1}, \ldots, v_{n})$</span> is a basis for <span class="math-container">$V$</span> and <span class="math-container">$\mathcal{B}^{*}= (\varphi_{1}, \ldots, \varphi_{n})$</span> is the corresponding dual basis, then for any <span class="math-container">$v \in V$</span>,</p>
<p><span class="math-container">$$ v = \varphi_{1}(v) v_{1} + \ldots + \varphi_{n}(v)v_{n}, $$</span></p>
<p>so <span class="math-container">$\mathcal{B}^{*}$</span> gives us the coordinates of any vector <span class="math-container">$v \in V$</span> with respect to the basis <span class="math-container">$\mathcal{B}$</span>.</p>
|
3,807,665 | <p>Let <span class="math-container">$ V $</span> a finite dimensional vector space and <span class="math-container">$ V^* $</span> the dual space. Further let be <span class="math-container">$ B=(v_1,...,v_n) $</span> a base of <span class="math-container">$ V $</span> and <span class="math-container">$ B^*=(v_1^*,...,v_n^*) $</span> a dual base of <span class="math-container">$ V^*$</span>. Show:</p>
<p>For all <span class="math-container">$ v,w\in V $</span> is <span class="math-container">$ v=w $</span> if and only if <span class="math-container">$ f(v)=f(w) $</span> for all <span class="math-container">$ f\in V^* $</span>.</p>
<p>My idea:</p>
<p>At first we have <span class="math-container">$ f=f(v_1)\cdot v_1^*+...+f(v_n)\cdot v_n^* $</span>.
Then we get with <span class="math-container">$ v=\sum\limits_{i=1}^n a_i\cdot v_i $</span> and <span class="math-container">$ w=\sum\limits_{i=1}^n b_i\cdot v_i $</span> for all <span class="math-container">$ v,w\in V $</span></p>
<p><span class="math-container">$ f(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(v)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n a_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot a_j\\=\sum\limits_{j=1}^n f(v_j)\cdot b_j\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*\left(\sum\limits_{i=1}^n b_i\cdot v_i\right)\\=\sum\limits_{j=1}^n f(v_j)\cdot v_j^*(w)\\=f(w) $</span></p>
<p>Now I want to show that <span class="math-container">$ a_j=b_j $</span> for all <span class="math-container">$ j=1,...,n $</span> but I don't know how.</p>
| janmarqz | 74,166 | <p>Choose each <span class="math-container">$v^*_i$</span> and apply <span class="math-container">$v^*_i(v)=v^*_i(w)$</span> then <span class="math-container">$a_i=b_i$</span>, so <span class="math-container">$v=w$</span>.</p>
|
96,437 | <p>In Mathematica 9.0, the documentation for the Curl function states that in n-dimensions "the resulting curl is an array with depth n-k-1 of dimensions". Accordingly, if a 2-dimensional array is feeded in the Curl function in 3-D space, it returns a scalar value. </p>
<p>However, it does not agree with the definition I met in other sources!
$$
\mathbf{\nabla}\times\mathbf{S}=e_{ijk}S_{mj,i}\mathbf{e}_k\otimes\mathbf{e}_m
$$
where the curl of a second-order tensor is also a second order tensor. Is it possible to calculate in Mathematica the curl according to the above equation?</p>
| Kagaratsch | 5,517 | <p>Assuming that by $e_{ijk}$ you mean the totally anti-symmetric tensor $\epsilon_{ijk}$, the expression you cite only is valid in three dimensions (since only in three dimensions $\epsilon_{ijk}$ has three indices). With the above assumption, the equation you provide can be implemented as follows</p>
<pre><code>twotensorCurl3D[S_List] := Module[{},
Id = IdentityMatrix[3];
eps = LeviCivitaTensor[3];
var = Table[x[i], {i, 1, 3}];
Sum[eps[[i, j, k]] D[S[[m, j]], var[[i]]] TensorProduct[Id[[All, k]],Id[[All, m]]], {i, 1, 3}, {j, 1, 3}, {k, 1, 3}, {m, 1, 3}]
]
</code></pre>
<p>The bits of code mean the following. <code>twotensorCurl3D[S_List]</code> defines a function name and makes sure that only a list can be passed to the function. <code>:=</code> as opposed to simply <code>=</code> makes sure that the right hand side is only evaluated once the function is actually called with a specific input. <code>Module[{},...]</code> is simply a wrapper that allows several computational steps to be done within the function before the results are put out. You could define local variables in the <code>{}</code> like <code>{a,b,c}</code>. We write the three dimensional identity matrix into the variable <code>Id</code>. Then the unit vectors $e_i$ can be accessed through the columns of this matrix: <code>Id[[All,i]]</code>. The position variables, in respect to which the nabla is taking derivatives, are defined as <code>x[i]</code> with <code>i=1,2,3</code>. The remaining code is self explanatory. The result of the last line in the <code>Module</code> is given back by the function after evaluation because it is not suppressed by use of <code>;</code> at the end of the line.</p>
|
4,380,274 | <p>Indefinite integral is pretty easy to solve, I did it by substitution and I'm pretty sure it can be done relatively easy via integration by parts.
The problem are boundaries.</p>
<p>After substitution <span class="math-container">$arcsin x=t$</span> we get</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{(\sin t)^3}dt$$</span>
so we have</p>
<p><span class="math-container">$$\int_0^\frac{\pi}{2} \frac{\cos t}{(\sin t)^2} dt+ \int_0^\frac{\pi}{2} \frac{t \cos t}{(\sin t)^3}dt$$</span></p>
<p>Now for example first integral is easy to compute and I get <span class="math-container">$\frac{-1}{\sin t}$</span> from 0 to <span class="math-container">$\frac{\pi}{2}$</span>.But in 0 the value is <span class="math-container">$\infty$</span>.
The second integral can also be solved using partial integration with <span class="math-container">$t$</span> and <span class="math-container">$\frac{cost}{(sint)^3}$</span>.</p>
<p>The only method to bypass this that I know of is linear substitution (I'm not sure how you call it) but that doesn't give anything useful.</p>
<p>Any hints?</p>
| Quanto | 686,284 | <p>You may not split the integral into two divergent integrals. Instead, integrate as a whole as shown below</p>
<p><span class="math-container">\begin{align}\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{\sin^3t}dt
&=\frac12\int_0^\frac{\pi}{2} (t-\sin t)d\left(\frac{1}{\sin^2t}\right)\\
&= \frac12\frac{t-\sin t}{\sin^2t}\bigg|_0^{\frac\pi2}-\frac12
\int_0^\frac{\pi}{2} \frac{1-\cos t}{\sin^2t} dt\\
&= \frac12(\frac\pi2-1)-\frac12\lim_{t\to0}\frac{t-\sin t}{\sin^2t}-\frac14
\int_0^\frac{\pi}{2} \sec^2\frac t2dt\\
&= \frac12(\frac\pi2-1)-\frac12\cdot 0-
\frac12=\frac\pi4-1
\end{align}</span>
where the lower limit term evaluates to zero per L’Hopital rule.</p>
|
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