qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,671,933 | <p>In the question <a href="https://math.stackexchange.com/questions/1502484/the-application-of-nimbers-to-nim-strategy">on nimbers</a>, the original poster asks for the meaning of Nimber <a href="https://en.wikipedia.org/wiki/Nimber#Multiplication" rel="nofollow noreferrer">multiplication</a> in the context of impartial games.</p>
<hr>
<p><strong>Edit: As noted by Mark Fischler in the comments below, the following is wrong</strong></p>
<p>My gut instinct is $*a \times *b$ means that if $*a$ is a game equivalent to $a$ stones, and $*b$ is a game equivalent to $b$ stones, then if you replace every stone in the $*a$ game with a copy of the $*b$ game, you get a game with the Nimber $*a \times *b$, but I haven't been able prove it.</p>
| Théophile | 26,091 | <p>I don't think there's an intuitive way to understand nimber multiplication. Also note that it is distinct from repeated addition, so it isn't as simple as replacing stones with copies of piles. Multiplication is defined recursively as
$$ab = \text{mex}\left(\{a'b+ab'+a'b': a'<a, b'<b\}\right)$$
where $\text{mex}$ is the minimum excluded element.</p>
<p>Although it's hard to see what this means in terms of a concrete game, it's possible to understand why it's defined this way. The point is that we want to create an algebraic system without zero divisors, so $(a-a')(b-b') \neq 0$ whenever $a\neq a'$ and $b\neq b'$. In other words,
$$ab - ab' - a'b + a'b' \neq 0,$$
so
$$ab \neq a'b + ab' - a'b'.$$
Since subtraction and addition are the same operation, we have
$$ab \neq a'b + ab' + a'b'.$$
The definition above takes the first nimber that meets this criterion.</p>
|
979,656 | <p>Finding prime solutions to $100q+80 = p^3 + q^2$</p>
<p>Does them being prime imply some patterns on division modulo 3 or some other integer? How is this done?</p>
| André Nicolas | 6,312 | <p><strong>Outline:</strong> One can do it crudely. Rewrite the equation as $-p^3=q^2-100q-80$, and note that the smallest value of $q^2-100q-80$ is $-2580$. </p>
|
979,656 | <p>Finding prime solutions to $100q+80 = p^3 + q^2$</p>
<p>Does them being prime imply some patterns on division modulo 3 or some other integer? How is this done?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>solving your equation for $q$ we obtain
$q=50\pm\sqrt{2580-p^3}$
we get
$p\le 13$
therefore $p\in \{2,3,5,7,11,13\}$</p>
|
106,853 | <p>I have taken a lot of math in university, but chose to omit differential equations. Unfortunately, now I have to read computer science proofs that use them, mostly ODEs, and this is always a struggle. What textbook(s) should I read to take me from the basics to practical use of the theory?</p>
| Artes | 21,946 | <p>If you are looking for practical solving of differential equations I recommend :</p>
<ul>
<li><p>Advanced Mathematical Methods for Scientists and Engineers by Carl M. Bender and Steven A.Orszag.</p></li>
<li><p>Methods of Mathematical Physics R.Courant and D. Hilbert two volumes.</p></li>
</ul>
<p>The second positions is absolutely the best ever written textbook on differential equations.</p>
<p>Maybe not so practical for solving special types of ODE's but still close to applications is </p>
<ul>
<li>Ordinary Differential Equations by V.I.Arnold</li>
</ul>
<p>Another quite well written textbook dealing with differential equations is </p>
<ul>
<li>Dynamical Systems with Applications using Mathematica by Stephen Lynch</li>
</ul>
<p>This book however mainly describes qualitavie methods for dynamical systems, but there are many good examples. </p>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| SaeidAli | 371,048 | <p>Note that the corresponding space created by a vector(non-zero) is equivalent to a line passing through zero. Now since your assumption does not involve being closed under addition you only need to consider at least two lines which pass through zero.</p>
|
1,726,540 | <p>How should $x^{\frac{1}{x}}$ be differentiated? I know the answer is
$$\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$$</p>
<p>but I do not understand how to get there.</p>
<h2>Attempt at solution.</h2>
<p>I believe the following is true:</p>
<p>$$
\begin{aligned}\frac{d}{dx}x^u&=ux^{u-1}\cdot u^\prime\\
\frac{d}{dx}a^x&=a^x\cdot\ln(a)
\end{aligned}$$
but I don't know what to do when both the base and the exponent are functions of $x$.</p>
| Bernard | 202,857 | <p>Use the general definition of a <em>real</em> exponent:
$$x^{\tfrac1x}\overset{\text{def}}{=}\mathrm e^{\tfrac{\ln x}{x}}.$$
Hence
$$\Bigl(x^{\tfrac1x}\Bigr)'=\mathrm e^{\tfrac{\ln x}{x}}\cdot\frac{\cfrac1x\cdot x-\ln x}{x^2}=x^{\tfrac1x}\cdot\frac{1-\ln x}{x^2}.$$</p>
|
332,130 | <p>Here is my proof, I would appreciate it if someone could critique it for me:</p>
<p>To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.</p>
<hr>
<p>Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$</p>
<p>If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.</p>
<hr>
<p>To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement?</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>I know you were asking for a critique of your original proof. But since I find text-based proofs (like yours, and the one from another answer) more difficult to read than symbolic ones, let me answer by presenting the way in which I would write a proof for this.</p>
<p><em>Proof.</em> For all sets A and B,</p>
<p>$$
\begin{array}{ll}
& \mathcal{P}(A) ⊆ \mathcal{P}(B) \\
\equiv & \;\;\;\text{"definition of ⊆"} \\
& \langle \forall V : V \in \mathcal{P}(A) : V \in \mathcal{P}(B) \rangle \\
\equiv & \;\;\;\text{"definition of $\mathcal{P}$, twice"} \\
& \langle \forall V : V ⊆ A : V ⊆ B \rangle \;\;\;\;\;\;\;\;\;\;\;\; (*) \\
\equiv & \;\;\;\text{"see below"} \\
& A ⊆ B \\
\end{array}
$$</p>
<p>which proves this theorem. The forward direction in the last step is easily proven:</p>
<p>$$
\begin{array}{ll}
& \langle \forall V : V ⊆ A : V ⊆ B \rangle \\
\Rightarrow & \;\;\;\text{"choose $V := A$; $A ⊆ A$"} \\
& A ⊆ B \\
\end{array}
$$</p>
<p>For the other direction, assuming $A ⊆ B$ we prove $(*)$ as follows: for every $V$,</p>
<p>$$
\begin{array}{ll}
& V ⊆ A \\
\Rightarrow & \;\;\;\text{"A ⊆ B, and ⊆ is transitive"} \\
& V ⊆ B \\
\end{array}
$$</p>
<p><em>Note.</em> This is the style of proof writing used in, e.g., <a href="https://www.springer.com/computer/theoretical+computer+science/book/978-0-387-94115-8" rel="noreferrer">A Logical Approach to Discrete Math </a> by Gries and Schneider; it was originally designed by Dijkstra and Feijen, and is discussed in chapter "On our proof format" of Dijkstra and Scholten, <a href="https://www.springer.com/computer/theoretical+computer+science/book/978-0-387-96957-2" rel="noreferrer">Predicate calculus and program semantics</a>, and (to provide a more accessible source) near the end of <a href="http://www.cs.utexas.edu/users/EWD/transcriptions/EWD13xx/EWD1300.html" rel="noreferrer">EWD1300</a>.</p>
<p>The nice thing about this style is that the $\text{"hints"}$ show very clearly which properties are used: apart from predicate logic and the definitions of $\mathcal{P}$ and $⊆$, the above proof only uses the fact that $⊆$ is reflexive and transitive.</p>
|
165,888 | <p>I want to return the final value from the Table.For example something like Table[n-1] value.</p>
| J. M.'s persistent exhaustion | 50 | <p>This is slightly faster and certainly more compact than the uncompiled implementation in Henrik's answer:</p>
<pre><code>makeKMat[k1_?SquareMatrixQ, k2_?SquareMatrixQ] := Module[{m = Length[k1], n = Length[k2]},
KroneckerProduct[k1, IdentityMatrix[n, SparseArray]] +
KroneckerProduct[IdentityMatrix[m, SparseArray], k2]]
</code></pre>
<p>and as b3m2a1 notes, the extension can be done as</p>
<pre><code>makeKMatND[ks : {__?SquareMatrixQ}] := Fold[makeKMat[#2, #1] &, ks]
</code></pre>
<p>Using the examples in the OP:</p>
<pre><code>MatrixPlot[makeKMat[ConstantArray[1, {9, 9}], ConstantArray[2, {6, 6}]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/uyWar.png" alt="matrix plot of 2D case"></p>
<pre><code>AbsoluteTiming[tst =
makeKMatND[{ConstantArray[4, {5, 5}], ConstantArray[3, {4, 4}],
ConstantArray[2, {3, 3}], ConstantArray[1, {2, 2}]}];][[1]]
0.000406914
MatrixPlot[tst, MaxPlotPoints -> Infinity]
</code></pre>
<p><img src="https://i.stack.imgur.com/wCijV.png" alt="matrix plot of 4D case"></p>
|
3,025,375 | <p>What is<span class="math-container">$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$</span>So it is<span class="math-container">$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$</span>
I do not know what to do next, because my resuts is <span class="math-container">$∞$</span> but the answer from book is <span class="math-container">$\dfrac{1}{4\sqrt{2}}$</span>.</p>
| egreg | 62,967 | <p>Formally substitute <span class="math-container">$n=1/t$</span>; if the function you get has a limit for <span class="math-container">$t\to0^+$</span>, then it is the same as the limit you are looking for. So consider
<span class="math-container">$$
\lim_{t\to0^+}\frac{1}{t^3}\left(
\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\frac{\sqrt{2}}{t}
\right)=
\lim_{t\to0^+}\frac{\sqrt{1+\sqrt{1+t^4}}-\sqrt{2}}{t^4}
$$</span>
Now the dependency is only on <span class="math-container">$t^4$</span>, so the limit is the same as
<span class="math-container">$$
\lim_{u\to0^+}\frac{\sqrt{1+\sqrt{1+u}}-\sqrt{2}}{u}
$$</span>
which is the derivative at <span class="math-container">$0$</span> of <span class="math-container">$f(u)=\sqrt{1+\sqrt{1+u}}$</span>.</p>
<blockquote class="spoiler">
<p> Since <span class="math-container">$$f'(u)=\frac{1}{2\sqrt{1+\sqrt{1+u}}}\frac{1}{2\sqrt{1+u}}$$</span> we have <span class="math-container">$$f'(0)=\frac{1}{4\sqrt{2}}$$</span></p>
</blockquote>
|
2,786,610 | <p>I have to calculate $\int\operatorname{arccot}(\cot(x))\ dx.$ If I had to find the derivative it would be easy with the chain rule. How can I do this?</p>
| Will Jagy | 10,400 | <p>If you write down $(A-I)^2$ you see that any vector with the final entry nonzero is outside the kernel, so you can pick
$$
p_4 =
\left(
\begin{array}{c}
0 \\
0 \\
0 \\
1
\end{array}
\right)
$$
followed by $p_3 = (A-I) p_4 \;, \;$ $p_2 = (A-I) p_3 \;, \;$ which is automatically an eigenvector because $(A-I)p_2 = (A-I)^3 p_4 = 0.$
Then let the first column $p_1$ be a different eigenvector from $p_2.$ Notice that, with two eigenvectors available, I deliberately chose a linear combination so the entires of the first column would be small. I hoped it would reduce the determinant of $P,$ and this worked out nicely, the determinant came out to $2.$ You can see this is the factor of $\frac{1}{2}$ required to write all the matrices with integers. </p>
<p>$$
\frac{1}{2}
\left(
\begin{array}{cccc}
1&5&3&0 \\
-9&-35&-35&0 \\
2&8&8&0 \\
0&0&0&2
\end{array}
\right)
\left(
\begin{array}{cccc}
0&-8&-35&0 \\
1&1&4&0 \\
-1&1&5&0 \\
0&0&0&1
\end{array}
\right) =
\left(
\begin{array}{cccc}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
0&0&0&1
\end{array}
\right)
$$</p>
<p>$$
\frac{1}{2}
\left(
\begin{array}{cccc}
1&5&3&0 \\
-9&-35&-35&0 \\
2&8&8&0 \\
0&0&0&2
\end{array}
\right)
\left(
\begin{array}{cccc}
-7&-32&-32&-35 \\
1&5&4&4 \\
1&4&5&5 \\
0&0&0&1
\end{array}
\right)
\left(
\begin{array}{cccc}
0&-8&-35&0 \\
1&1&4&0 \\
-1&1&5&0 \\
0&0&0&1
\end{array}
\right) =
\left(
\begin{array}{cccc}
1&0&0&0 \\
0&1&1&0 \\
0&0&1&1 \\
0&0&0&1
\end{array}
\right)
$$
and
$$
\frac{1}{2}
\left(
\begin{array}{cccc}
0&-8&-35&0 \\
1&1&4&0 \\
-1&1&5&0 \\
0&0&0&1
\end{array}
\right)
\left(
\begin{array}{cccc}
1&0&0&0 \\
0&1&1&0 \\
0&0&1&1 \\
0&0&0&1
\end{array}
\right)
\left(
\begin{array}{cccc}
1&5&3&0 \\
-9&-35&-35&0 \\
2&8&8&0 \\
0&0&0&2
\end{array}
\right) =
\left(
\begin{array}{cccc}
-7&-32&-32&-35 \\
1&5&4&4 \\
1&4&5&5 \\
0&0&0&1
\end{array}
\right)
$$</p>
|
1,633,846 | <p>I'm currently a high school Pre-Calculus student and my textbook presents the following theorem without proof:</p>
<blockquote>
<p>Let <span class="math-container">$f(x)$</span> be a polynomial with real coefficients and a positive leading coefficient.</p>
<p>Let <span class="math-container">$a \geq 0$</span>. Then, if when <span class="math-container">$f(x)$</span> is divided by <span class="math-container">$x-a$</span>, all of the coefficients of the quotient and the remainder are non-negative, <span class="math-container">$a$</span> is an upper bound on the real zeroes of <span class="math-container">$f(x)$</span>.</p>
<p>Now, let <span class="math-container">$a \leq 0$</span>. Then, if when <span class="math-container">$f(x)$</span> is divided by <span class="math-container">$x-a$</span>, all of the coefficients of the quotient and the remainder alternate between non-positive and non-negative, <span class="math-container">$a$</span> is a lower bound on the real zeroes of <span class="math-container">$f(x)$</span>.</p>
</blockquote>
<p>I am trying to find a proof for this theorem and I think I found a proof for it <a href="http://mathweb.scranton.edu/monks/courses/ProblemSolving/POLYTHEOREMS.pdf" rel="nofollow noreferrer">pages 6 and 7 of this PDF file</a>. However, the proof does not seem correct:</p>
<blockquote>
<p>[...] since all of <span class="math-container">$q(x)$</span>'s remaining coefficients are positive [...]</p>
</blockquote>
<p>This is a quote from the proof of the first part of this theorem. Here, <span class="math-container">$q(x)$</span> is the resulting polynomial from dividing <span class="math-container">$f(x)$</span> by <span class="math-container">$x-b$</span> for some root <span class="math-container">$b$</span> for <span class="math-container">$f(x)$</span>. However, there is a clear counter example to this: If <span class="math-container">$a=5$</span> and <span class="math-container">$f(x)=x^2-5x+6$</span>, then <span class="math-container">$f(x)$</span> and <span class="math-container">$a$</span> meet the hypothesis since <span class="math-container">$\frac{f(x)}{x-a}=x+0+\frac{6}{x-a}$</span>, so there are all non-negative coefficients in the quotient and the remainder. Then, <span class="math-container">$b=3$</span> is a root of <span class="math-container">$f(x)$</span>, but, <span class="math-container">$q(x)=\frac{f(x)}{x-b}=x-2$</span> does not have all positive coefficients, contradicting the above.</p>
<p>The proof of the second part of the theorem is also wrong:</p>
<blockquote>
<p>Because <span class="math-container">$a < 0$</span> and the leading term in <span class="math-container">$q(x)$</span> has a positive coefficient, the constant term in <span class="math-container">$q(x)$</span> has the same sign as <span class="math-container">$q(a)$</span>.</p>
</blockquote>
<p>However, if we let <span class="math-container">$a=-4$</span> and <span class="math-container">$f(x)=x^2+3x+2$</span>, then <span class="math-container">$f(x)$</span> and <span class="math-container">$a$</span> meet the hypothesis since <span class="math-container">$\frac{f(x)}{x-a}=x-1+\frac{6}{x-a}$</span>, which alternatives between non-negative and non-positive coefficients in the quotient and remainder. Then, <span class="math-container">$b=2$</span> is a root of <span class="math-container">$f(x)$</span>, but <span class="math-container">$q(x)=\frac{f(x)}{x-b}=x+1$</span> and thus <span class="math-container">$q(a)=-3$</span> while the constant term of <span class="math-container">$q(x)$</span> is <span class="math-container">$1$</span>, which also clearly contradicts the above.</p>
<p>Thus, while I have found proofs for these theorems, I do not think they are valid. Could someone show me valid proofs for these theorems in my Pre-Calc textbook? Thank you!</p>
| Aops Vol. 2 | 706,407 | <p>A more elegant proof for the upper bound (the same logic can follow with the lower) is to notice that if the quotient <span class="math-container">$q(x)$</span> has only positive coefficients all numbers greater than <span class="math-container">$a$</span> will have a positive output (and thus never cross the x-axis).</p>
<p><span class="math-container">$$f(x)=(x-a)q(x)+r(x)$$</span>
For some any <span class="math-container">$c>a$</span>
<span class="math-container">$$f(c)=(c-a)q(c)+r(c)$$</span>
<span class="math-container">$$c>a>0$$</span>
<span class="math-container">$$q(c)>0$$</span>
<span class="math-container">$$c-a>0$$</span>
<span class="math-container">$$r(c)>0$$</span>
So...<span class="math-container">$$f(c)=(c-a)q(c)+r(c)>0$$</span>
Thus any value <span class="math-container">$>a$</span> cannot be a root and is thus an upper bound.</p>
|
3,699,640 | <p>How to show <span class="math-container">$x^2=20$</span> has no solution in <span class="math-container">$2$</span>-adic ring of integers <span class="math-container">$\mathbb{Z}_2$</span> ?</p>
<p>What is the general criterion fora solution of <span class="math-container">$x^2=a$</span> in <span class="math-container">$\mathbb{Z}_2$</span> ?</p>
<p>I know for odd prime <span class="math-container">$x^2=a$</span> has solution in <span class="math-container">$\mathbb{Z}_p$</span> or in other word <span class="math-container">$a$</span> is quadratic residue modulo <span class="math-container">$p$</span> if <span class="math-container">$a_0$</span> is quadratic residue modulo <span class="math-container">$p$</span>, where <span class="math-container">$a=a_0+a_1p+a_2p^2+\cdots \in \mathbb{Z}_p$</span>.</p>
<p>But what about for <span class="math-container">$p$</span> even i.e., <span class="math-container">$p=2$</span> ?</p>
<p>We have two following results as well.</p>
<blockquote>
<p>Result <span class="math-container">$1$</span>: For <span class="math-container">$p \neq 2$</span>, an <span class="math-container">$\epsilon \in \mathbb{Z}_p^{\times}$</span> is square in <span class="math-container">$\mathbb{Z}_p$</span> iff it is square in the residue field of <span class="math-container">$\mathbb{Z}_p$</span>.</p>
<p>Result <span class="math-container">$2$</span>: An unit <span class="math-container">$\epsilon \in \mathbb{Z}_2^{\times}$</span> is square iff <span class="math-container">$\epsilon \equiv 1 (\mod 8)$</span>.</p>
</blockquote>
<p>But as <span class="math-container">$20$</span> is not a unit in <span class="math-container">$\mathbb{Z}_2$</span>, the above Result <span class="math-container">$2$</span> is not applicable here.</p>
<p>So we have to use other way.</p>
<p>I am trying as follows:</p>
<p>For a general <span class="math-container">$2$</span>-adic integer <span class="math-container">$a$</span>, we have the following form <span class="math-container">$$ a=2^r(1+a_1 \cdot 2+a_2 \cdot 2^2+\cdots).$$</span>
Thus one necessary criterion for <span class="math-container">$a \in \mathbb{Q}_2$</span> to be square in <span class="math-container">$\mathbb{Q}_2$</span> is that <span class="math-container">$r$</span> must be <strong>even integer</strong>.</p>
<p>Are there any condition on <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span> ?</p>
<p>For, let <span class="math-container">$\sqrt{20}=a_0+a_1 2+a_22^2+a_32^3+\cdots$</span>.</p>
<p>Then squaring and taking modulo <span class="math-container">$2$</span>, we get <span class="math-container">$$ a_0^2 \equiv 0 (\mod 2) \Rightarrow a_0 \equiv 0 (\mod 2).$$</span> Thus, <span class="math-container">$\sqrt {20}=a_12+a_22^2+a_32^3+\cdots$</span>.</p>
<p>Again squaring and taking modulo (<span class="math-container">$2^2)$</span>, we get
<span class="math-container">$$a_1 \equiv 1 (\mod 2^2).$$</span> Thus we have <span class="math-container">$\sqrt{20}=2+a_22^2+a_32^3+\cdots$</span>.</p>
<p>Again squaring and taking modulo <span class="math-container">$(2^3)$</span>, we get no solution for <span class="math-container">$a_i,\ i \geq 2$</span>.</p>
<p>What do I conclude here from ?</p>
<p>How do I conclude that <span class="math-container">$x^2=20$</span> has no solution in <span class="math-container">$\mathbb{Z}_2$</span> ?</p>
<p>Help me</p>
| Merosity | 741,168 | <p>I want to show a more elementary approach by directly grinding out algebra steps just for the fun of showing how Hensel's lemma and the ultrametric inequality can be used. Let's begin by supposing there is an <span class="math-container">$x \in \mathbb{Z}_2$</span> such that <span class="math-container">$x^2=20$</span>.</p>
<p><span class="math-container">$$x^2=20$$</span>
<span class="math-container">$$|x^2|_2 = |20|_2$$</span>
<span class="math-container">$$|x|_2^2 = 1/4$$</span>
<span class="math-container">$$|x|_2 = 1/2$$</span></p>
<p>So we know <span class="math-container">$x = 2u$</span> for <span class="math-container">$|u|_2=1$</span>.</p>
<p><span class="math-container">$$(2u)^2 = 20$$</span>
<span class="math-container">$$u^2 = 5$$</span></p>
<p>Now let's use Hensel's lemma in its more general form with <span class="math-container">$f(u) = u^2-5=0$</span>. Simply, for there to be a <span class="math-container">$u \in \mathbb{Z}_2$</span> we need
<span class="math-container">$$|f(u)|_2 < |f'(u)|_2^2$$</span>
<span class="math-container">$$|u^2-5|_2 < |2u|_2^2$$</span>
<span class="math-container">$$|u^2-5|_2 < 1/4$$</span></p>
<p>At this point we can use with <span class="math-container">$|k|_2\le 1$</span>,
<span class="math-container">$$u = 1+2k$$</span>
<span class="math-container">$$u^2 = 1 + 4k + 4 k^2 $$</span>
<span class="math-container">$$u^2 = 1 + 4k(k+1) $$</span>
At this point we notice no matter if k is "even or odd" (in <span class="math-container">$2\mathbb{Z}_2$</span> or <span class="math-container">$1+2\mathbb{Z}_2$</span>), we must have <span class="math-container">$|k(k+1)|_2<1$</span>.</p>
<p><span class="math-container">$$u^2 = 1 + 4k(k+1) $$</span>
<span class="math-container">$$u^2 - 1 = 4k(k+1) $$</span>
<span class="math-container">$$|u^2 - 1|_2 = |4k(k+1)|_2 $$</span>
<span class="math-container">$$|u^2 - 1|_2 < 1/4 $$</span></p>
<p>But wait, earlier we showed <span class="math-container">$|u^2-5|_2 < 1/4$</span> is required for there to be a solution in <span class="math-container">$\mathbb{Z}_2$</span>. Because <span class="math-container">$|5-1|_2 = 1/4$</span> it's impossible for <span class="math-container">$u^2$</span> to be that close to both! Let's directly show the contradiction now, hinging upon the ultrametric inequality:</p>
<p><span class="math-container">$$|u^2-1|_2 = |u^2-5 + 4|_2 \le \max(|u^2-5|_2 , |4|_2)$$</span>
The ultrametric inequality's "greatest wins" property gives us the maximum whenever there's a competition. Since here <span class="math-container">$|u^2-5|_2 < 1/4$</span> and <span class="math-container">$|4|_2 = 1/4$</span> we have,</p>
<p><span class="math-container">$$|u^2-1|_2 = 1/4$$</span></p>
<p>This clearly contradicts what we showed a second ago with <span class="math-container">$|u^2-1|_2 <1/4$</span>, so there are no solutions.</p>
|
1,944,630 | <p>whether every positive definite matrix has to be symmetric? If not, what will be the example?</p>
| Martin Argerami | 22,857 | <p>If the definition is simply $x^TAx\geq0$ for all $x $, this can fail over the reals:
$$
A=\begin{bmatrix}1&1\\0&1\end {bmatrix} $$ satisfies $x^TAx\geq0$ for all $x\in\mathbb R^2$.</p>
<p>For complex matrices, the condition $x^*Ax\geq0$ for all $x\in\mathbb C^n $ implies that $A $ is selfadjoint. </p>
|
2,558,619 | <p>Compute
$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt
$$
In closed form, where $\delta(t)$ is the <a href="http://mathworld.wolfram.com/DeltaFunction.html" rel="nofollow noreferrer">Dirac Delta function </a>.</p>
<p>My attempt: </p>
<p>$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt = \int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|}\mathrm dt
$$</p>
<p>Then noting that $\sin(t)$ is zero whenever $t=n\pi$. By formula (2) and (7) in the above link,</p>
<p>\begin{align}
\int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|} \mathrm dt& = \sum_{n=-\infty}^{\infty} \frac{(n\pi)^2e^{-|n\pi|}}{|\cos(n\pi)|}
\\&
=2\pi^2\sum_{n=0}^{\infty} \frac{(n)^2e^{-n\pi}}{1}
\end{align}</p>
<p>However , I am stuck here, i do not know how to procede, Wolfram Alpha tells me that this sum doesn't converge so how can i compute it in closed form? I can only assume I have gone about this the wrong way or made a mistake. Any help would be great.</p>
| robjohn | 13,854 | <p>Note that
$$
\begin{align}
\sum_{n=0}^\infty n^2x^n
&=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}\\
&=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}\\[6pt]
&=\frac{x+x^2}{(1-x)^3}
\end{align}
$$
Therefore,
$$
\begin{align}
2\pi^2\sum_{n=0}^\infty n^2e^{-\pi n}
&=2\pi^2\frac{e^{-\pi}+e^{-2\pi}}{\left(1-e^{-\pi}\right)^3}\\
&=\frac{\pi^2}2\frac{\cosh\left(\frac\pi2\right)}{\sinh^3\left(\frac\pi2\right)}
\end{align}
$$</p>
|
3,707,132 | <p>I believe my proof of this simple fact is fine, but after a few false starts, I was hoping that someone could look this over. In particular, I am interested in whether there is an alternate proof.</p>
<blockquote>
<p>For a real number <span class="math-container">$a$</span> and non-empty subset of reals <span class="math-container">$B$</span>, define <span class="math-container">$a + B = \{a + b : b \in B\}$</span>. Show that if <span class="math-container">$B$</span> is bounded above, then <span class="math-container">$\sup(a + B) = a + \sup B$</span>.</p>
</blockquote>
<p>My attempt: </p>
<blockquote>
<p>Fix <span class="math-container">$a \in \mathbb{R}$</span>, take <span class="math-container">$B \subset \mathbb{R}$</span> to be nonempty and bounded above, and define
<span class="math-container">$$a + B = \{a + b : b \in B\}.$$</span>
Since <span class="math-container">$B$</span> is nonempty and bounded above, the least-upper-bound axiom guarantees the existence of <span class="math-container">$\sup B$</span>. For any <span class="math-container">$b \in B$</span>, we have
<span class="math-container">$$b \leq \sup B,$$</span>
which implies
<span class="math-container">$$a + b \leq a + \sup B.$$</span>
As this is true for any <span class="math-container">$b \in B$</span>, it follows that <span class="math-container">$a + \sup B$</span> is an upper bound of <span class="math-container">$a + B$</span>, and hence <span class="math-container">$\sup(a + B)$</span> exists, by the completeness axiom, since <span class="math-container">$B \neq \emptyset$</span> implies immediately that <span class="math-container">$a + B \neq \emptyset$</span>. I claim that <span class="math-container">$a + \sup B$</span> is in fact the least upper bound of <span class="math-container">$a + B$</span>. As we have already shown it to be an upper bound, it suffices to demonstrate that <span class="math-container">$a + \sup B$</span> is the least of the upper bounds. Let <span class="math-container">$\gamma$</span> be an upper bound of <span class="math-container">$a + B$</span>. Hence, for any <span class="math-container">$b \in B$</span>,
<span class="math-container">$$a + b \leq \gamma,$$</span>
which implies that
<span class="math-container">$$b \leq \gamma - a.$$</span>
As this holds for all <span class="math-container">$b \in B$</span>, <span class="math-container">$\gamma - a$</span> is an upper bound of <span class="math-container">$B$</span>. Hence, by the definition of supremum,
<span class="math-container">$$\gamma - a \geq \sup B,$$</span>
which implies that
<span class="math-container">$$\gamma \geq a + \sup B,$$</span>
as desired. </p>
</blockquote>
<p>I tried to write the proof initially be showing that <span class="math-container">$\sup(a + B) \leq a + \sup B$</span> and <span class="math-container">$\sup(a + B) \geq a + \sup B$</span>, but didn't have any luck. If there is a trick to it, I would be interested in hearing it.</p>
| Sebastian Monnet | 572,592 | <p>What you've done looks correct to me, but I think we can rework it more concisely using exactly the strategy that you suggest at the end of the question. Note that both suprema exist since the sets are non-empty.</p>
<p><strong>First direction:</strong> Let <span class="math-container">$\lambda \in a + B$</span>. Then <span class="math-container">$\lambda = a + b$</span> for some <span class="math-container">$b \in B$</span>. Since a supremum is an upper bound, <span class="math-container">$b \leq \sup B$</span>, so <span class="math-container">$\lambda \leq a + \sup B$</span>. Since <span class="math-container">$\lambda \in a + B$</span> was arbitrary, <span class="math-container">$a + \sup B$</span> is an upper bound for <span class="math-container">$a + B$</span>, hence <span class="math-container">$\sup(a + B) \leq a + \sup B$</span>.</p>
<p>At this point it might be worth pausing to try the other direction yourself - the idea is similar, so it would be a good test of understanding.</p>
<p><strong>Other direction:</strong> Let <span class="math-container">$b \in B$</span>. Then <span class="math-container">$a + b \in a + B$</span>, and since a supremum is an upper bound, <span class="math-container">$a + b \leq \sup(a + B)$</span>. Rearranging, <span class="math-container">$b \leq \sup(a + B) - a$</span>, so <span class="math-container">$\sup(a + B) - a$</span> is an upper bound on <span class="math-container">$B$</span>, and hence <span class="math-container">$\sup B \leq \sup(a + B) - a$</span>, and it follows that <span class="math-container">$\sup(a + B) \geq a + \sup B$</span>.</p>
<p><strong>Conclusion:</strong> It follows immediately that <span class="math-container">$\sup(a +B) = a + \sup B$</span>.</p>
|
4,183,734 | <p>I just started trigonometry, and I came across this problem. I know that this problem would be very simple with a calculator, but without one, I'm lost. How would you determine if <span class="math-container">$\sin 4$</span> is bigger than <span class="math-container">$\sin 3$</span> in negativity or positivity? How would you even determine if they are positive or negative in the first place, and how would you know which one is bigger?</p>
| Lalit Tolani | 913,165 | <p>@Ritam_Dasgupta perfectly answered your question.</p>
<p>I will help you with other question i.e. which out of <span class="math-container">$\sin 3$</span> and <span class="math-container">$\sin 4$</span> is greater in magnitude</p>
<p>For this I will assume <span class="math-container">$\pi\approx3.14$</span></p>
<p>Now <span class="math-container">$\sin 3= \sin(3.14-3)\approx\sin(\pi-0.14)\approx \sin(0.14)$</span></p>
<p>Now <span class="math-container">$\sin 4= \sin(3.14+0.86)\approx\sin(\pi+0.86)\approx -\sin(0.86)$</span></p>
<p>Clearly both <span class="math-container">$0.86,0.14<1.57\approx0.5\pi$</span></p>
<p>Therefore their <span class="math-container">$\sin()$</span> is positive</p>
<p>Now <span class="math-container">$\sin x$</span> is increasing function on the interval <span class="math-container">$(0,\frac{\pi}{2})$</span></p>
<p>Therefore <span class="math-container">$\sin(0.86)>\sin(0.14)$</span></p>
<p><span class="math-container">$|\sin 4| > \sin 3$</span></p>
<p>And since <span class="math-container">$|\sin 4|=-\sin 4$</span> as <span class="math-container">$\sin 4$</span> is negative</p>
<p>therefore <span class="math-container">$-\sin 4> \sin 3\implies \sin 3+\sin 4<0$</span></p>
|
39,184 | <p>Hi everybody,</p>
<p>Reading about Geometry Processing, I have realized that people in this area are very interested in regular vertices(degree=6) rather than irregular ones.</p>
<p>Can anybody give me reasons why this is an important property?
Suppose that we can have a mesh with all regular vertices (genus 1) but with a very poor geometry positioning for the vertices.</p>
<p>So, is there any specific reason why connectivity is this much important regardless of the geometry?</p>
<p>Thanks</p>
| sleepless in beantown | 8,676 | <p>Nima, geometry processing is used in meshes for 3-dimensional modeling of real world structures or physical items.</p>
<p>If you try to create a regularly spaced lattice in 3-dimensions such as $\mathbb{R}^3$, a starting point would be the regions of $\mathbb{Z}^3$ which are enclosed by the object in question. That would be the lattice which contains a point at every integer value of $x, y, $ and $z$ within the object's boundaries. If you connect an edge between every two lattice points that are Euclidean distance $1$ apart, $(\Delta x)^2+(\Delta y)^2+(\Delta z)^2 = 1$, then each vertex will have six edges from it leading to the six lattice points at the relative positions $\Delta x=-1$, $\Delta x=+1$, $\Delta y=-1$, $\Delta y=+1$, $\Delta z=-1$, $\Delta z=+1$.</p>
<p>Notice that the boundaries of such a mesh would be <em>clipped</em> and would not match the boundaries of the object. Such a mesh could be extended by having the outermost lattice-points moved from co-inciding with $(x,y,z) \in \mathbb{Z}^3$ to a position in $\mathbb{R}^3$ which is on the boundary of the object. Now, the lattice is no longer regular, but the vertex connectivity is still regular. (Alternatively, additional lattice points could be added on the object boundary in $\mathbb{R}^3$ and connected with the underlying regular lattice, leading to irregular vertex connectivity.)</p>
<p>This sort of 6-connectivity at vertices in 3-dimensions for lattice simulations allows for simulations such as 3-d lattice Boltzman numerical simulations. This type of 6-connectivity is also known as the Von Neumann neighborhood in cellular automata simulations, and in the 2-dimensional case the analogous 4-connectivity Von Neumann neighborhood for a point at $(a,b)$ are the lattice points $(a-1,b), (a+1,b), (a,b-1),$ and $(a,b+1)$.</p>
<p>It is possible to associate a value at every point of the lattice and allow that value to represent a particular material property.</p>
<p>If you take a solid object represented by a 3-d lattice and model applying a physical pressure to it, then the regular object modeled by a regular lattice with regular geometry will become deformed. The lattice points, originally defined as being located at integer values, can be allowed to move in $\mathbb{R}^3$, associating a real value for each of its 3-dimensional coordinates. Thus, deformation modeling can be carried out, such as when an automobile's fender changes its shape in response to a collision with another object.</p>
<p>In the realm of hydrodynamics and computational fluid dynamics, there are two ways to use lattice models with 6-connectivity at each vertex.</p>
<p>In Eulerian models, the lattice points stay fixed while the values associated with the lattice points are recalculated and changed during each step of the numerical simulation.</p>
<p>In Lagrangian models, the lattice points are defined at particular positions (not necessarily coinciding with integer lattice points in $\mathbb{Z}^3$), and the vertices in the mesh are moved in space during steps of the simulation.</p>
<p>If you give me more information about what specific topic you are studying, or what you wish to use "geometry processing" for (mesh creation?, mesh assessment?, mesh refinement?, mesh validation?), perhaps I could provide more detailed answers.</p>
<p>In response to your comment about which is better,
what is <strong>better</strong> has to be defined in terms of the end goal to be reached: numerical precision, model fidelity to the underlying physical objects, speed of computation.</p>
<p>If the numerical precision of the simulator will blow up with slim triangles (for example, triangles with one angle of less than 5 degrees), then it's better to accept irregular vertices. If the simulation would perform better (faster, fewer errors, etc.) with regular and similar vertex connectivity, then allow the irregular triangles and banish irregular nonconforming vertices. There is always a <em>trade-off</em> to make: precision vs. computational time; computational complexity vs.fidelity to the physical object, etc.</p>
<p>It is easier to calculate "subdivision surfaces" for regular meshes. It is easier to speed up calculations at homogeneous regions by using larger triangulations and to have more accuracy at inhomogeneous regions and irregular boundaries by using smaller triangulations.</p>
<p>Tetrahedral meshing <em>has</em> to use non-identical tetraheda, because it is impossible to completely fill space with identical regular tetrahedra. It is possible to fill space with regular cubes or regular rectangles. That is one reason that 6-connectivity at vertices is frequently used in 3-d meshes.</p>
|
2,738,023 | <p>We know we can check if two vectors are 'orthogonal' by doing an inner product.</p>
<p>$a*b=0$</p>
<p>tells us that these two vectors are orthogonal</p>
<p>here comes the question:</p>
<p>if there a way to compute if they are 'parallel'? i.e., they are pointing at the same direction. </p>
| Michael Hoppe | 93,935 | <p>Two vectors are parallel iff the dimension of their span is less than $2$.</p>
|
3,579,739 | <p>Its also mentioned that this question is an example of <strong>Sampling without replacement</strong></p>
<p>My question is , by the general method of how I do these kind of problems , I would assume that there are two possibilities</p>
<ol>
<li><p>You chose a defective bottle and then a non defective which will have the probability of </p>
<pre><code> 1/5 X 4/4 = 4/20
OR
</code></pre></li>
<li><p>Choosing a non defective one first and then a defective one which will have the probability of</p>
<pre><code>4/5 x 1/4 =4/20
</code></pre></li>
</ol>
<p>Total probability is 8/20= 2/5 which doesn't make sense at all, since there is just one bottle to begin with
But the problem is I cant find a fault in my logic. Though I feel both the options are redundant , isn't both two different ways of selecting the pair?. Or is the fact that the phrase "one after the other" is absent a reason on why this might be a wrong approach?</p>
<pre><code> Thank you in advance
</code></pre>
| JMP | 210,189 | <p>In the general case, say there are <span class="math-container">$n$</span> bottles, and you choose <span class="math-container">$k$</span> of them. There is only one flawed bottle.</p>
<p>The number of ways to pick <span class="math-container">$k$</span> bottles from <span class="math-container">$n$</span> is the binomial coefficient <span class="math-container">$\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}$</span>.</p>
<p>Now we count the number of these <span class="math-container">$\dbinom{n}{k}$</span> arrangements that contain the flawed bottle.</p>
<p>By hand is OK for small <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, for example with <span class="math-container">$n=4,k=2$</span> we have <span class="math-container">$\dbinom{4}{2}=6$</span> cases:</p>
<pre>
AB..
A.C.
A..D
BC..
.B.D
..CD
</pre>
<p>and if the flawed bottle is A (it doesn't actually matter) then there are three cases with an A in them, and so the probability is <span class="math-container">$\dfrac{3}{6}=0.5$</span>.</p>
<p>More generally, we need to know the number of arrangements that have the flawed bottle.</p>
<p>If we remove the flawed bottle, then we have <span class="math-container">$n-1$</span> bottles, and we want to know the number of arrangements of any <span class="math-container">$k-1$</span> of them, as we then add the flawed bottle back to give us the total number of arrangements we require. This is <span class="math-container">$\dbinom{n-1}{k-1}$</span>.</p>
<p>So the general probability is given by:</p>
<p><span class="math-container">$$\frac{\dbinom{n-1}{k-1}}{\dbinom{n}{k}}=\frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{k!(n-k)!}{n!}=\frac{k}{n}$$</span></p>
<p>A perhaps more intuitive method is to pick the <span class="math-container">$k$</span> bottles first, and then someone assigns a bottle as flawed from the <span class="math-container">$n$</span> bottles. You have a <span class="math-container">$k$</span> in <span class="math-container">$n$</span> chance of getting the flawed bottle.</p>
|
4,207,823 | <p>I came across the following question in a set of lecture notes:</p>
<p>If <span class="math-container">$(G, *)$</span> is a group, when, in general, is it true that there exists <span class="math-container">$X \subset \mathbb{R}^n$</span> such that <span class="math-container">$G \cong \text{Sym}(X)$</span>? Or, less formally, when does an abstract group have a geometric realization?</p>
<p>In this context, <span class="math-container">$\text{Sym}(X)$</span> is the group of isometries of <span class="math-container">$\mathbb{R}^n$</span> that permute <span class="math-container">$X$</span>.</p>
<p>I'm quite confused about this. As far as I understand, the elements of the group act on the vertices by some group action. I suppose in order for <span class="math-container">$G$</span> to be identified with the group of symmetries, the action should be faithful, and transitive. But I know far too little about groups in order to be able to say anything more general or detailed about what we can deduce about the group structure. I also don't understand the importance of the <span class="math-container">$X$</span> being elements of <span class="math-container">$\mathbb{R}^n$</span>, although this is a key part of the question.</p>
<p>So, summing up: what conditions does such a group need to satisfy, and how would the answer differ if we drop the condition that <span class="math-container">$X \subset \mathbb{R}^n$</span>?</p>
| Jacob Manaker | 330,413 | <p>The group of isometries of <span class="math-container">$\mathbb{R}^n$</span> is a <a href="https://en.wikipedia.org/wiki/Semidirect_product" rel="nofollow noreferrer">semidirect product</a> <span class="math-container">$\mathbb{R}^n\rtimes O(\mathbb{R}^n)$</span>, where <span class="math-container">$O(\mathbb{R}^n)$</span> are the orthogonal matrices and <span class="math-container">$\mathbb{R}^n$</span> acts by translation. (If you haven't seen semidirect products, don't worry! It's just a slightly-less-commutative generalization of the direct product.) To see the semidirect decomposition, first note that, by the polarization identity, any isometry must preserve dot products. Thus a isometry fixing the origin is orthogonal. Now choose any specific isometry <span class="math-container">$g$</span>; <span class="math-container">$g(0)$</span> is some point, so we can find a translation <span class="math-container">$t$</span> such that <span class="math-container">$(tg)(0)=0$</span>. Thus <span class="math-container">$tg\in O(\mathbb{R}^n)$</span> and the claim follows.</p>
<p>Now <span class="math-container">$O(\mathbb{R}^n)$</span> is itself the semidirect product <span class="math-container">$\{\pm1\}\rtimes\text{SO}(\mathbb{R}^n)$</span>. Thus <strong>your group must be a subgroup of <span class="math-container">$\mathbb{R}^n\rtimes\{\pm1\}\rtimes\text{SO}(\mathbb{R}^n)$</span>.</strong></p>
<p><strong>If <span class="math-container">$n$</span> is fixed and small, then this imposes a rather strict restriction on the sort of groups that one can encounter.</strong> A very accessible reference (for undergrads!) is Chapters 4 & 5 of Artin's <em>Algebra</em>, which essentially classify all <em>discrete</em> subgroups in 2 and 3 dimensions. The upshot is that the discrete subgroups of <span class="math-container">$\text{SO}(\mathbb{R}^n)$</span> are cyclic groups, dihedral groups, and a few sporadic examples corresponding to the platonic solids; if we allow translations, then one obtains the classical crystallographic <a href="https://en.wikipedia.org/wiki/Space_group" rel="nofollow noreferrer">space groups</a> as well.</p>
<p>As <span class="math-container">$n$</span> increases, more and more sporadic examples appear, so that one might hope that every group can be a geometric symmetry in high enough dimensions. Certainly, if a group can be the symmetries of an <span class="math-container">$n$</span>-dimensional set, then it is certainly the symmetries of an <span class="math-container">$m>n$</span>-dimensional set, just by adding randomly chosen asymmetric points to cut off the extra dimensions. I claim that <strong>if <span class="math-container">$n$</span> is allowed to grow without bound, then every <em>finite</em> group can be constructed</strong> via a similar process. (Infinite groups that are symmetries of points in <span class="math-container">$\mathbb{R}^n$</span> are much harder to classify; my guess is that without further structure they are "undecidable" in the sense that no countable set of invariants can do the trick. This is also why my answer is consistent with Moishe Kohan's: Moishe Kohan's counterexamples are all infinite.)</p>
<p>Before I begin, note that the isometries of <span class="math-container">$\mathbb{R}^n$</span> are generically <span class="math-container">$(n+1)$</span>-faithful: if we know where <span class="math-container">$(n+1)$</span>-many affinely-independent points <span class="math-container">$C$</span> are mapped, then we can recover the image of each element of <span class="math-container">$\mathbb{R}^n$</span> by measuring its distances to <span class="math-container">$C$</span> and applying triangulation. (Conversely, as long as we preserve distances and angles, then the isometries can take any <span class="math-container">$(n+1)$</span>-dimensional set to another; thus <span class="math-container">$\text{Isom}(\mathbb{R}^n)$</span> is <em>almost</em> sharply <span class="math-container">$(n+1)$</span>-transitive.)</p>
<p>Suppose <span class="math-container">$G$</span> is a finite group. Then <span class="math-container">$G$</span> acts on itself via left-multiplication; this gives an isomorphism to a permutation group. We can consider those permutations as acting on basis vectors of <span class="math-container">$\mathbb{R}^{|G|}$</span> and extended by linearity; such actions are clearly isometries. So we can construct <span class="math-container">$X\subseteq\mathbb{R}^{|G|}$</span> such that <span class="math-container">$G\leq\text{Sym}{(X)}$</span>; can we ensure the reverse inequality?</p>
<p>If <span class="math-container">$G$</span> is the full symmetric group <span class="math-container">$S_k$</span>, then the answer is clearly yes: let <span class="math-container">$X=\{\vec{0},\vec{e_1},\vec{e_2},\dots,\vec{e_k}\}\subseteq\mathbb{R}^k$</span>. Any symmetry of <span class="math-container">$X$</span> must fix <span class="math-container">$0$</span> (what else is distance <span class="math-container">$1$</span> away from all the other points?), and so acts as a permutation on the <span class="math-container">$k$</span>-element remainder. This gives a homomorphism <span class="math-container">$\text{Sym}(X)\to S_k$</span>; since the isometries of <span class="math-container">$\mathbb{R}^k$</span> are <span class="math-container">$(k+1)$</span>-faithful, said homomorphism is injective. We can clearly make all transpositions (rotate the corresponding axes by <span class="math-container">$90^{\circ}$</span>) to generate <span class="math-container">$S_k$</span>, so the homomorphism is surjective too.</p>
<p>Likewise if <span class="math-container">$G=\{1\}$</span>, then it is clear that <span class="math-container">$G$</span> is the symmetries of some set. We now assume that <span class="math-container">$G>\{1\}$</span>.</p>
<p>Suppose we can construct a permutation group <span class="math-container">$H>G$</span> as the symmetries of some finite set <span class="math-container">$Y\subseteq\mathbb{R}^n$</span> containing an affinely-independent set of size <span class="math-container">$n+1$</span> and such that all <span class="math-container">$H$</span> fixes some <span class="math-container">$y_0\in Y$</span>. Our first step is to show that we can add points to <span class="math-container">$Y$</span> without adding more symmetries.</p>
<p>Let <span class="math-container">$D(A)$</span> denote the set of distances between distinct points in <span class="math-container">$A$</span>, i.e. <span class="math-container">$D(A)=\{|a-b|:a\neq b\in A\}$</span>. (For any point <span class="math-container">$x$</span>, the sets <span class="math-container">$D(Y)$</span> and <span class="math-container">$D(Gx)$</span> are nonempty since <span class="math-container">$|Y|\geq n+1$</span> and <span class="math-container">$G>\{1\}$</span>.) Since <span class="math-container">$Y$</span> is finite, so is <span class="math-container">$D(Y)$</span>; by choosing <span class="math-container">$x$</span> sufficiently far from <span class="math-container">$y_0$</span>, we may ensure that the elements of <span class="math-container">$D(Gx)$</span> are sufficiently large, and thus ensure <span class="math-container">$D(Y)$</span> and <span class="math-container">$D(Gx)$</span> are disjoint. But then any <span class="math-container">$r\in\text{Sym}(Y\cup Gx)$</span> must preserve the partition <span class="math-container">$Y\sqcup Gx$</span>. Thus there is an element <span class="math-container">$h\in H$</span> whose action on <span class="math-container">$Y$</span> is the same as <span class="math-container">$r$</span>. But <span class="math-container">$Y$</span> contains <span class="math-container">$(n+1)$</span>-many affinely-independent points and the isometries of <span class="math-container">$\mathbb{R}^n$</span> are <span class="math-container">$(n+1)$</span>-faithful. So we must have <span class="math-container">$h=r$</span>.</p>
<p>Thus <span class="math-container">$G\leq\text{Sym}(Y\sqcup Gx)\leq H$</span> for any <span class="math-container">$x$</span> so chosen. So far, we <em>only</em> require <span class="math-container">$x$</span> far from <span class="math-container">$y_0$</span>; in particular, we do not care about its "angle". Now fix <span class="math-container">$h\in H\setminus G$</span>: it is time to pin down that angle. For each <span class="math-container">$g\in G$</span>, since <span class="math-container">$h\neq g$</span>, the vector space <span class="math-container">$\ker{(g-h)}\neq\mathbb{R}^n$</span>. Thus <span class="math-container">$\bigcup_{g\in G}{\ker{(g-h)}}\neq\mathbb{R}^n$</span> too. So there is some <span class="math-container">$x$</span> far from <span class="math-container">$y_0$</span> that does not lie in that union. Choose such an <span class="math-container">$x$</span>; then we never have <span class="math-container">$hx=gx$</span> for any <span class="math-container">$g\in G$</span>. Equivalently, <span class="math-container">$hx\notin Gx$</span>, so that <span class="math-container">$h\notin\text{Sym}(Y\sqcup Gx)$</span>.</p>
<p>Thus we can choose <span class="math-container">$x$</span> so that <span class="math-container">$\text{Sym}(Y\sqcup Gx)<H$</span>. If <span class="math-container">$G<\text{Sym}(Y\sqcup Gx)$</span>, then we can replace <span class="math-container">$H$</span> with <span class="math-container">$Y\sqcup Gx$</span> and induct on <span class="math-container">$|H|$</span>, so eventually <span class="math-container">$G=\text{Sym}(Y\sqcup Gx)$</span> as desired.</p>
|
621,438 | <p>Is there any Characterization for all measurable sets in $\mathbb{R}$?
Can I say that a set is measurable if an only if it has the property of Baire? (differs from an open set by a first category set). If the answer is no, Is there an example of a measurable set that does not have the property of Baire?</p>
<p>Thank you!
Shir </p>
| tomasz | 30,222 | <p>A useful characterization of measurable sets is that a set is measurable if and only if it is of the form $B\triangle N$ where $B$ is Borel (even $G_\delta$) and $N$ is null (measure zero) (this is easy to show using regularity of the measure, and is actually true for any regular Borel measure, on any space).</p>
<p>From that it's not hard to see that a measurable set can be rather pathological: $N$ can be any subset of the Cantor set, for instance. It can also be any subset of a comeager set which is of zero measure (like the set of Liouville numbers, for example), which can fail to have Baire property. It also makes it easy to see that there are $2^{\mathfrak c}$ measurable sets.</p>
|
451,001 | <p>To define the function $f(x)=|[x]|$ where $|[x]|$ is the greatest integer that is less or equal to $x$, we need to prove that indeed such an integer exists. In other words,</p>
<blockquote>
<p>$$\forall x\in \mathbb{R} \;\;\exists !\, n\in\mathbb{Z}\;\;:\;\;n\leq x<n+1.$$</p>
</blockquote>
<p>My first attempt is by defining the set $A=\{y\in\mathbb{Z}| y\leq x\}$. Since $\mathbb{N}$ is not bounded above then $\exists n\in \mathbb{N}(n>-x)$. Then $\exists n'\in \mathbb{Z}(n'<x)$, namely $n'=-n$. Therefore $A\neq\emptyset$. Also $A$ is bounded above by $x$. Then $\sup(A)$ exists. </p>
<p>**Here my problem is how to prove that $\sup(A)\in A$. </p>
<p>Also my question is that I'm not sure if this is enough because $sup(A)$ is unique or do I still need to prove uniqueness. </p>
<hr>
<h2><strong>EDIT:</strong></h2>
<p>After all the hints in the answers, chich I appreciate so much, I have arrived to this attempts of solution, please feel free to comment and help me saying if something is wrong: </p>
<p><strong>First approach:</strong> considering tree cases ($x<0, x=0, x>0$). </p>
<p><strong>1)</strong> If $x=0$ then I can think of $n=0$. </p>
<p><strong>2)</strong> If $x>0$ then I can think of the set $B=\{y\in \mathbb{N}|x\leq y\}$. By using the well-ordering principle then $\min(B)-1<x\leq \min(B)$ and then if $x=\min(B) $ I can make $n=\min(B)$; if $\min(B)-1<x<\min(B)$ then $n=\min(B)-1$. </p>
<p><strong>3)</strong> if $x<0$ then $-x>0$ and therefore $\exists m \in \mathbb{N}(m\leq -x < m+1)$. Then $\exists n'\in \mathbb{Z}(-n'<x\leq-n'-1)$, namely $n'=m+1$. Again, if $x=-n'-1$ then I can make $n=-n'-1$. If $-n'<x\leq-n'-1$ then $n=-n'$.</p>
<p><strong>Second approach</strong>. Since $\mathbb{N}$ is not bounded above then let $x_{0}\in \mathbb{N}$ be such that $x_{0}>\sup(A)$. Let's consider the set $B=\{x_{0}-y: y\in A\}$. Since $B\subseteq \mathbb{N}$ then, by the well ordering principle, $\min(B)$ exists. Moreover $\exists y_{0}\in A\forall y\in A(\min(B)=x_{0}-y_{0}\leq x_{0}-y)$. This means that $\exists y_{0}\in A\forall y\in A (y\leq y_{0})$. Therefore $\max(A)=y_{0}$. In this case I can make $n=\max(A)$. </p>
| Hagen von Eitzen | 39,174 | <p>By symmetry with respect to the mid-perpendicular of the side the motion is parallel to, the ball moves by an even number <span class="math-container">$2k$</span> of sides (which may be more than <span class="math-container">$\frac n2$</span>, as in the case <span class="math-container">$n=3$</span>) and the (polygonal or star) shape of the path is uniquely determined by the number <span class="math-container">$2k$</span>, except that <span class="math-container">$n-2k$</span> and <span class="math-container">$2k$</span> generate the same. So if <span class="math-container">$n=2m$</span> is even, we have to check <span class="math-container">$k$</span> ranging from <span class="math-container">$1$</span> to <span class="math-container">$\lfloor \frac m2\rfloor =\lfloor \frac n4\rfloor$</span> inclusive, if <span class="math-container">$n$</span> is odd <span class="math-container">$k$</span> ranging from <span class="math-container">$1$</span> to <span class="math-container">$\frac{n-1}2$</span> inclusive. For each such <span class="math-container">$k$</span>, one closed cycle involves every <span class="math-container">$d$</span>th edge where <span class="math-container">$d=\gcd(n,2k)$</span>, so we get <span class="math-container">$d$</span> distinct rotational copies of this path and the total number of closed cycles is
<span class="math-container">$$\tag1 \sum_{k=1}^{\lfloor\frac n4\rfloor\text{ or }\frac{n-1}2}\gcd(n,2k).$$</span>
It seems that this sequence
<span class="math-container">$$ 0, 0, 1, 2, 2, 2, 3, 6, 6, 4, 5, 12, 6, 6, 15, 16, 8, 12, 9, 22, 22, 10, 11, 34, 20, 12, 27, 32, 14, 30$$</span>
is not in <a href="http://oeis.org/search?q=1,%202,%202,%202,%203,%206,%206,%204,%205,%2012,%206,%206,%2015,%2016,%208,%2012,%209,%2022&language=english&go=Search" rel="nofollow noreferrer">OEIS</a> (yet).
By the way, if <span class="math-container">$n$</span> is an odd prime, then <span class="math-container">$(1)$</span> equals <span class="math-container">$\frac{n-1}{2}$</span>.</p>
|
451,001 | <p>To define the function $f(x)=|[x]|$ where $|[x]|$ is the greatest integer that is less or equal to $x$, we need to prove that indeed such an integer exists. In other words,</p>
<blockquote>
<p>$$\forall x\in \mathbb{R} \;\;\exists !\, n\in\mathbb{Z}\;\;:\;\;n\leq x<n+1.$$</p>
</blockquote>
<p>My first attempt is by defining the set $A=\{y\in\mathbb{Z}| y\leq x\}$. Since $\mathbb{N}$ is not bounded above then $\exists n\in \mathbb{N}(n>-x)$. Then $\exists n'\in \mathbb{Z}(n'<x)$, namely $n'=-n$. Therefore $A\neq\emptyset$. Also $A$ is bounded above by $x$. Then $\sup(A)$ exists. </p>
<p>**Here my problem is how to prove that $\sup(A)\in A$. </p>
<p>Also my question is that I'm not sure if this is enough because $sup(A)$ is unique or do I still need to prove uniqueness. </p>
<hr>
<h2><strong>EDIT:</strong></h2>
<p>After all the hints in the answers, chich I appreciate so much, I have arrived to this attempts of solution, please feel free to comment and help me saying if something is wrong: </p>
<p><strong>First approach:</strong> considering tree cases ($x<0, x=0, x>0$). </p>
<p><strong>1)</strong> If $x=0$ then I can think of $n=0$. </p>
<p><strong>2)</strong> If $x>0$ then I can think of the set $B=\{y\in \mathbb{N}|x\leq y\}$. By using the well-ordering principle then $\min(B)-1<x\leq \min(B)$ and then if $x=\min(B) $ I can make $n=\min(B)$; if $\min(B)-1<x<\min(B)$ then $n=\min(B)-1$. </p>
<p><strong>3)</strong> if $x<0$ then $-x>0$ and therefore $\exists m \in \mathbb{N}(m\leq -x < m+1)$. Then $\exists n'\in \mathbb{Z}(-n'<x\leq-n'-1)$, namely $n'=m+1$. Again, if $x=-n'-1$ then I can make $n=-n'-1$. If $-n'<x\leq-n'-1$ then $n=-n'$.</p>
<p><strong>Second approach</strong>. Since $\mathbb{N}$ is not bounded above then let $x_{0}\in \mathbb{N}$ be such that $x_{0}>\sup(A)$. Let's consider the set $B=\{x_{0}-y: y\in A\}$. Since $B\subseteq \mathbb{N}$ then, by the well ordering principle, $\min(B)$ exists. Moreover $\exists y_{0}\in A\forall y\in A(\min(B)=x_{0}-y_{0}\leq x_{0}-y)$. This means that $\exists y_{0}\in A\forall y\in A (y\leq y_{0})$. Therefore $\max(A)=y_{0}$. In this case I can make $n=\max(A)$. </p>
| Servaes | 30,382 | <p>Let <span class="math-container">$R$</span> be a regular <span class="math-container">$n$</span>-gon, and number the midpoints of its sides <span class="math-container">$0$</span> through <span class="math-container">$n-1$</span>.</p>
<p>Note that if a line passing through the midpoint of a side is parallel to another side, then it passes through the midpoint of another side. For odd <span class="math-container">$n$</span> the converse holds; a line passing through the midpoints of two distinct sides is parallel to another side. For even <span class="math-container">$n$</span> this only holds if the sides differ by an even number (assuming we have numbered the sides in a sensible way).</p>
<p>Hence we may describe the initial motion as starting from a point <span class="math-container">$p\in\{0,\ldots,n-1\}$</span>, towards another point <span class="math-container">$q\in\{0,\ldots,n-1\}$</span>, where <span class="math-container">$p\neq q$</span>. Denote the path starting from point <span class="math-container">$p$</span> moving towards <span class="math-container">$q$</span> by <span class="math-container">$P(p,q)$</span>.</p>
<p>Consider the path <span class="math-container">$P(0,m)$</span>. After meeting point <span class="math-container">$m$</span>, the ball is reflected and moves towards point <span class="math-container">$2m$</span>. By symmetry, the <span class="math-container">$k$</span>-th point to be met is the point <span class="math-container">$k\cdot m$</span>, so the points on the path <span class="math-container">$P(0,m)$</span> are the integer multiples of <span class="math-container">$m$</span>, modulo <span class="math-container">$n$</span> of course.</p>
<p>Counting orientation, this yields precisely <span class="math-container">$n-1$</span> distinct paths meeting <span class="math-container">$0$</span> if <span class="math-container">$n$</span> is odd, and precisely <span class="math-container">$\tfrac{n}{2}-1$</span> distinct paths meething <span class="math-container">$0$</span> if <span class="math-container">$n$</span> is even. Of course not every path meets the point <span class="math-container">$0$</span> necessarily. The length of <span class="math-container">$P(0,m)$</span>, i.e. the number of distinct points it meets, is <span class="math-container">$\operatorname{lcm}(m,n)/n$</span>, or equivalently <span class="math-container">$m/\gcd(m,n)$</span>, whichever you prefer, so there are precisely <span class="math-container">$\gcd(m,n)$</span> distinct 'rotations' of this path. The total number of distinct paths, counting orientation, is therefore</p>
<p><span class="math-container">$$T(n)=\left\{\begin{array}{cc}
\sum_{m=1}^{n-1}\gcd(m,n)&\text{ for } n \text{ odd,}\\
\sum_{m=1}^{\tfrac{n}{2}-1}\gcd(2m,n)&\text{ for } n \text{ even.}\end{array}\right.$$</span></p>
<p>This is not as bad as it looks; it turns out these sums are multiplicative. For details, see <a href="https://math.stackexchange.com/questions/1990320/how-do-i-simplify-sum-k-1n-gcdk-n?noredirect=1&lq=1">this question</a>. The accepted answer there shows that if <span class="math-container">$n=\prod_{p\mid n} p^{n_p}$</span> then
<span class="math-container">$$\sum_{m=1}^n\gcd(m,n)=n\prod_{p\mid n}\big(1+n_p(1-\tfrac1p)\big).$$</span></p>
<p>So indeed this shows that the number of paths on an <span class="math-container">$n$</span>-sided table is closely related to the factorization of <span class="math-container">$n$</span>.</p>
<p>Also, if we don't count orientation, we find the total numbers to be</p>
<p><span class="math-container">$$t(n)=\left\{\begin{array}{cc}
\frac{1}{2}(T(n)+n)&\text{ if } 4\mid n\\
\frac{1}{2}T(n)&\text{ otherwise}\end{array}\right..$$</span></p>
|
199,374 | <p>Consider the two table set below</p>
<pre><code>t1= {0.44, 0.62, 0.77, 0.87, 0.93, 0.96, 0.98, 1}
t2= {0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25}
</code></pre>
<p>how can I create a plot which wil plot a rectangle with a heat map normalized to the values of the table?</p>
<p>The result should look like</p>
<p><a href="https://i.stack.imgur.com/GgXBG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GgXBG.png" alt="h"></a></p>
| Vitaliy Kaurov | 13 | <p>Function <strong>Blend</strong> actually does a great color interpolation. For simplicity assume that all your values are always in <span class="math-container">$[0,1]$</span> interval. Choose a specific <a href="http://reference.wolfram.com/language/guide/ColorSchemes.html" rel="nofollow noreferrer">Color Scheme</a>, for example DeepSeaColors , and define:</p>
<pre><code>cf[v_][x_] := Blend[ColorData["DeepSeaColors"] /@ v, x]
</code></pre>
<p>You do not have to use built in color schemes, you could just have assigned some colors to your numbers. The point is in blending them. Then using detailed Raster make:</p>
<pre><code>Graphics[Raster[{Range[100]/100}, ColorFunction -> cf[t1]], AspectRatio -> .3]
</code></pre>
<p><a href="https://i.stack.imgur.com/ROP9f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ROP9f.png" alt="enter image description here"></a></p>
<pre><code>Graphics[Raster[{Range[100]/100}, ColorFunction -> cf[t2]], AspectRatio -> .3]
</code></pre>
<p><a href="https://i.stack.imgur.com/uSopr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uSopr.png" alt="enter image description here"></a></p>
<p>It is not clear from your question how exactly you want to rescale / normalize. But you can always use Rescale to adjust the above aproach.</p>
|
3,397,606 | <p>Find <span class="math-container">$[z^N]$</span> for <span class="math-container">$\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$</span>. Here <em>generalized harmonic numbers</em> should be used. </p>
<p><span class="math-container">$$H^{(2)}_n = 1^2 + \frac{1}{2}^2 + \dots + \frac{1}{N}^2.$$</span></p>
<p>For now, I was able to find <span class="math-container">$[z^N]$</span> for <span class="math-container">$\left(\ln \frac{1}{1-z}\right)^2$</span> which is the following convolution for <span class="math-container">$N \ge 2$</span>:</p>
<p><span class="math-container">$$\sum_{1\le k \le N-1} \frac{1}{N-k}\frac{1}{k}$$</span></p>
<p>Next step would be partial sum but I don't see how all this leads me to <em>generalized harmonic numbers</em>.</p>
| Jack D'Aurizio | 44,121 | <p>From
<span class="math-container">$$-\log(1-z)=\sum_{n\geq 1}\frac{z^n}{n}$$</span>
it follows that
<span class="math-container">$$\frac{-\log(1-z)}{1-z} = \sum_{n\geq 1} H_n z^n $$</span>
and by applying <span class="math-container">$\int_{0}^{x}\ldots dx$</span> to both sides
<span class="math-container">$$ \log^2(1-x) = \sum_{n\geq 1}\frac{2H_n}{n+1}x^{n+1} = \sum_{n\geq 2}\frac{2H_{n-1}}{n} x^n. $$</span>
It follows that for any <span class="math-container">$n\geq 2$</span> we have</p>
<p><span class="math-container">$$ [z^n]\frac{\log^2(1-z)}{1-z} = 2\sum_{k=1}^{n}\frac{H_{k}-\frac{1}{k}}{k}=2\sum_{k=1}^{n}\frac{H_k}{k}-2 H_{n}^{(2)}. $$</span>
On the other hand, by symmetry,
<span class="math-container">$$ \sum_{k=1}^{n}\frac{H_k}{k}=\sum_{1\leq j\leq k}\frac{1}{jk}=\frac{H_n^2+H_{n}^{(2)}}{2}. $$</span>
It follows that
<span class="math-container">$$ \frac{\log^2(1-z)}{1-z}=\sum_{n\geq 1}\color{red}{\left(H_n^2-H_n^{(2)}\right)}z^n. $$</span></p>
|
783 | <p>In a category I have two objects $a$ and $b$ and a morphism $m$ from $a$ to $b$ and one $n$ from $b$ to $a$. Is this always an isomorphism? Why is it emphasized that this has to be true, too: $m \circ n = \mathrm{id}_b$ and $n \circ m = \mathrm{id}_a$?</p>
<p>I am looking for an example in which the id-part is not true and therefore $m$ and $n$ are not isomorphic.</p>
| Dennis | 1,646 | <p>The important point here, I think, is about Hom(a,a) and Hom(b,b). nm is guaranteed to be an element of the first and mn an element of the second; to be an isomorphism, both of these maps must be the identity by definition. The definition of category requires that you have at least the identity in both endomorphism sets; if neither set has non-identity elements then you get an isomorphism, since the compositions don't have anything other than the identity to be, but this needn't hold in general.</p>
<p>Maximally toy example: take the full subcategory of sets given by the one element set A and the two element set B. Hom(A,A) is only the identity, but Hom(B,B) has four elements, two of which are not isomorphisms.</p>
|
61,406 | <p>Can I say this is a bad performance from the new V10 function <code>SubsetQ</code>?</p>
<p>Here are some tests comparing it to <code>Complement[l2, l1] === {}</code></p>
<pre><code>count1[data_,list_]:=Module[{r},
r=SubsetQ[#,list]&/@data;
Counts[r]
]
count2[data_,list_]:=Module[{r},
r=Complement[list,#]==={}&/@data;
Counts[r]
]
</code></pre>
<p>Small columns test:</p>
<pre><code>$HistoryLength = 0;
data = RandomInteger[100, {100000, 10}];
list = {4, 3, 2, 1};
count1[data, list] // AbsoluteTiming
count2[data, list] // AbsoluteTiming
</code></pre>
<blockquote>
<p>{2.760775, <|False -> 99995, True -> 5|>}
{0.450933, <|False -> 99995,True -> 5|>}</p>
</blockquote>
<p>Large columns test:</p>
<pre><code>$HistoryLength=0;
data=RandomInteger[100,{100000,100}];
list={4, 3, 2, 1};
count1[data,list]//AbsoluteTiming
count2[data,list]//AbsoluteTiming
</code></pre>
<blockquote>
<p>{3.345720, <|False -> 97745, True -> 2255|>} {0.910420, <|False ->
97745, True -> 2255|>}</p>
</blockquote>
<p><strong>Update:</strong></p>
<p>Still slow in V10.1</p>
| ubpdqn | 1,997 | <p><strong>EDIT</strong></p>
<p>After comments from RunnyKine...</p>
<p>Just another approach and timing:</p>
<pre><code>subs[u_, v_] := Length@Intersection[u, v] == Length@v
</code></pre>
<p>Performance:</p>
<pre><code> BenchmarkPlot[{count1[#, list] &, count2[#, list] &,
Tally@Map[Function[x, subs[x, list]], #] &},
RandomInteger[100, {100000, #}] &, PowerRange[1, 1000],
"IncludeFits" -> True, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/kSA76.png" alt="enter image description here"></p>
|
1,660,231 | <blockquote>
<p>Show that the function $g(x)=\sqrt{x^2+x+2}$
is defined and is continuous on $\mathbb{R}$.</p>
</blockquote>
<p>I have tried completion of square for $$x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}$$
This means that range, $r\ge 7/4$ in domain $\mathbb{R}$.</p>
<p>Cannot find any more logic in it, please help.</p>
| Community | -1 | <p>To prove that $g (x)$ is defined on $\mathbf {R}$, we need to prove that for <strong>any</strong> $x_0 \in \mathbf {R}$ the value $g (x_0)$ exists. Indeed, since $x^2 + x + 2 > 0$ for any $x \in \mathbf {R}$, the expression $$g (x) = \sqrt {x^2 + x + 2}$$ is meaningful. Additionally, $g (x)$ is always $> \sqrt {7}/2$.</p>
<p>We now prove the continuity. Let $c$ be an arbitrary real number, and consider the limit $$\lim_{x \to c} g (x) = \lim_{x \to c} \sqrt {x^2 + x + 2} = \sqrt {c^2 + c + 2} = g (c).$$ Since $g$ is defined on reals, $g (c)$ exists. The limit also exists, since left and right limits exist and are equal. Also, the value of the limit equals $g (c)$. These three prove that $g (x)$ is continuous at every $c \in \mathbf {R}$, that is, it is continuous on $\mathbf {R}$.</p>
|
51,732 | <p>A <em>Perron number</em> is a real algebraic integer $\lambda$ that is larger than the absolute value of any of its Galois conjugates. The Perron-Frobenius theorem says that any
non-negative integer matrix $M$ such that some power of $M$ is strictly positive has a
unique positive eigenvector whose eigenvalue is a Perron number. Doug Lind proved the converse: given a Perron number $\lambda$, there exists such a matrix, perhaps in dimension
much higher than the degree of $\lambda$. Perron numbers come up frequently in many places, especially in dynamical systems.</p>
<p>My question:</p>
<blockquote>
<p>What is the limiting distribution of Galois conjugates of Perron numbers $\lambda$ in
some bounded interval, as the degree goes to infinity?</p>
</blockquote>
<p>I'm particularly interested in looking at the limit as the length of the interval goes to
0. One way to normalize this is to look at the ratio $\lambda^g/\lambda$, as $\lambda^g$
ranges over the Galois conjugates. Let's call these numbers \emph{Perron ratios}.</p>
<p>Note that for any fixed $C > 1$ and integer $d > 0$, there are only
finitely many Perron numbers $\lambda < C$ of degree $< d$, since there is obviously a bound on the discriminant of the minimal polynomial for $\lambda$, so the question is only interesting when a bound goes to infinity. </p>
<p>In any particular field, the set of algebraic
numbers that are Perron lie in a convex cone in the product of Archimedean places of the field. For any lattice, among lattice points with $x_1 < C$ that are within this cone, the projection along lines through the origin to the plane $x_1 = 1$ tends toward the uniform
distribution, so as $C \rightarrow \infty$, the distribution of Perron
ratios converges to a uniform distribution in the unit disk (with a contribution for each complex place of the field) plus a uniform distribution
in the interval $[-1,1]$ (with a contribution for each real place of the field).</p>
<p>But what happens when $C$ is held bounded and the degree goes to infinity? This question seems
related to the theory of random matrices, but I don't see any direct translation from
things I've heard. Choosing a random Perron number seems very different from choosing
a random nonnegative integer matrix.</p>
<p>I tried some crude experiments, by looking at randomly-chosen polynomials of a fixed degree whose coefficients are integers in some fixed range except for the coefficient of $x^d$
which is $1$, selecting from those the irreducible polynomials whose largest real root is Perron. This is not the same as selecting a random Perron number of the given degree
in an interval. I don't know any reasonable way to do the latter except for small enough $d$ and $C$ that one could presumably find them by exhaustive search.
Anyway, here are some samples from what I actually tried.
First, from among the 16,807 fifth degree polynomials with coefficients in the range -3 to 3, there are $3,361$ that define a Perron number. Here is the plot of the Perron ratios:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints5%2C3.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints5%2C3.jpg</a></p>
<p>Here are the results of a sample of 20,000 degree 21 polynomials with coefficients
between -5 and 5. Of this sample, 5,932 defined Perron numbers:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21.jpg</a></p>
<p>The distribution decidedly does not appear that it will converge toward a uniform distribution on the disk plus a uniform distribution on the interval. Maybe the artificial bounds on the coefficients cause the higher density ring.</p>
<blockquote>
<p>Are there good, natural distributions for selecting random integer polynomials? Is there a
way to do it without unduly prejudicing the distribution of roots?</p>
</blockquote>
<p>To see if it would help separate what's happening,
I tried plotting the Perron ratios restricted to $\lambda$ in subintervals. For
the degree 21 sample, here is the plot of $\lambda$ by rank order:</p>
<p><a href="http://dl.dropbox.com/u/5390048/CDF21.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/CDF21.jpg</a></p>
<p>(If you rescale the $x$ axis to range from $0$ to $1$ and interchange $x$ and $y$ axes,
this becomes the plot of the sample cumulative distribution function of $\lambda$.)
Here are the plots of the Perron ratios restricted to the intervals $1.5 < \lambda < 2$
and $3 < \lambda < 4$:</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21%281.5%2C2%29.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21%281.5%2C2%29.jpg</a></p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints21%283%2C4%29.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints21%283%2C4%29.jpg</a></p>
<p>The restriction to an interval seems to concentrate the absolute values of Perron ratios even more. The angular distribution looks like it converges to the uniform
distribution on a circle plus point masses at $0$ and $\pi$. </p>
<p>Is there an explanation for the distribution of radii? Any guesses for what it is?</p>
| Bill Thurston | 9,062 | <p>I've gained some new perspective on this question, based partly on comments and on Hitachi Peach's answer. Instead of editing the original question, I'll write down some more thoughts as a (partial) answer in the hopes that it will inspire someone with different expertise to say more.</p>
<p>First, after Hitachi Peach's comment following his answer, I tried plotting a picture
of all the answers for a couple two of the simplest situations: quadratics and cubics
with a small value of $C$.</p>
<p>Below is a diagram in the coefficient space for quadratic polynomials. The horizontal axis is the coefficient of $x$ and the vertical axis is the constant.</p>
<p><a href="http://dl.dropbox.com/u/5390048/QuadraticSmallPerron.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/QuadraticSmallPerron.jpg</a></p>
<p>The unshaded area in the middle are polynomials whose roots are real with maximum absolute
value 5 and minimum absolute value 1; the left half of this area consists of Perron polynomials. The red lines are level curves of the maximum root.</p>
<p>Here is a similar plot for cubic polynomials, this time showing the region in coefficient space where all roots have absolute value less than 2. </p>
<p><a href="http://dl.dropbox.com/u/5390048/CubicRootsSmaller2.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/CubicRootsSmaller2.jpg</a></p>
<p>Among these are 31 Perron polynomials (where the maximum is attained for a positive real root. Here are their roots, and the normalized roots (divided by the Perron number):</p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints3%282%29.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints3%282%29.jpg</a></p>
<p><a href="http://dl.dropbox.com/u/5390048/PerronPoints3%282%29normalized.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/PerronPoints3%282%29normalized.jpg</a></p>
<p>After seeing and thinking about these pictures, it became clear that for polynomials
with roots bounded by $C > 1$, as the degree grows large, the volume in coefficient space
grows large quite quickly with degree, and appears to high volume/(area of boundary) ratio. The typical coefficients become large, and
most of the roots seem to change slowly as the coefficients change, so you don't
bump into the boundary too easily. If so, then
to get a random lattice point within this volume, it should work fairly well to first
find a random point chosen uniformly in coefficient space, and then move to the
nearest lattice point.</p>
<p>With that in mind, I tried to guess the distribution of roots (invariant by complex
conjugation), choose a random sample of $d$ elements chosen independently from this distribution, generate the polynomial with real coefficients having those roots, round
off the coefficients to the nearest integer, and looking at the resulting roots.
To my surprise, many of the roots were not very stable: the nearest integral polynomial
usually ended up with roots fairly far out of bounds, for any parameter values of
several distributions I tried. (Note: one constraint on the distribution is that the
geometric mean of absolute values must be an integer $\ge 1$. This rules out the
uniform distribution at least for small values of $C$).</p>
<p>After thinking harder about the stability question for roots (as the coefficients are
perturbed), I realized the importance of the interactions of nearby roots. Whenever
there is a double root, the roots move quickly when coefficients are changed --- i.e.,
the ratio of volume in coefficient space to volume in root space is relatively small.
It's as if nearby roots in effect have a repulsive force. The joint distribution of roots is important: you get wrong answers if you treat them as independent.</p>
<p>With this in mind, I tried an experiment where I clicked on a diagram to put in roots
for a controlling real polynomial by hand, while the computer found the roots of the nearest polynomial
with integer coefficients. With a little practice, this worked well. New roots "prefer" to
be inserted where the existing polynomial is high, so I shaded the diagram by absolute value of the polynomial, to indicate
good places for a new root. Sometimes, roots of the controlling polynomial become disassociated from roots of the nearest integer polynomial, and the result is often an out-of-bounds root not near any controlling root. In that case, deleting control roots that are disassociated brings it back into line. As the control roots are moved around, the algebraic integers jump in discrete steps, and these steps are much smaller when the control root distribution is in a good region of the parameter space.</p>
<p>Here's a screen shot from the experiment, (which is fun!):</p>
<p><a href="http://dl.dropbox.com/u/5390048/ControlRoots.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/ControlRoots.jpg</a></p>
<p>Here, the convention is that each control point above the real axis is duplicated with its complex conjugate.
Each control point below the real axis is projected to the real axis, and gives a real
root for the control polynomial. All the control roots are shown in black, and the roots of the nearest integer polynomial are shown in red. For these positions, the red roots are nicely associated with black roots. It is a Perron polynomial, because a
real root has been dragged so that it has maximum modulus.</p>
<p>In the next screenshot, I have dragged several control roots into a cluster around
11 o'clock. The red roots weren't happy there, so they disassociated from the control roots
and scattered off in different directions, one of them out to a much larger radius. This is a good indication that
the ratio of coefficient-space volume to root-space volume is small. This polynomial is not
Perron.</p>
<p><a href="http://dl.dropbox.com/u/5390048/RootPerturb-disassociated.jpg" rel="noreferrer">alt text http://dl.dropbox.com/u/5390048/RootPerturb-disassociated.jpg</a></p>
<p>This experiment is much trickier for $C$ close to $1$: the coefficients are much smaller for a given degree, which makes the roots much less stable. They become more stable when there
are lots of roots spread out fairly evenly, mostly near the outer boundary.</p>
<p>Here is one method that in principle should give a nearly uniformly-random choice of
a polynomial with roots bounded by $C$, and I think, by approximating with the nearest
polynomial having integer coefficients, give a nearly uniform choice of algebraic integer
whose conjugates are bounded by $C$: Start from any polynomial whose roots are bounded by
$C$, for instance, a cyclotomic polynomial. Choose a random vector in coefficient space,
and follow a $C^1$ curve whose tangent vector evolves by Brownian motion on the unit sphere.
Whenever a root hits the circle of radius $C$, choose a new random direction in which
the root decreases in absolute value (i.e, use diffuse scattering on the surfaces).
The distribution of positions should converge to the uniform distribution in the
given region in coefficient space.</p>
<p>This method should also probably work to find a random polynomial whose roots are inside
any connected open set, and subject to certain geometric limitations (for instance, it can't be
inside the unit circle) a nearly uniformly random algebraic integer of high degree
whose Galois conjugates are inside a given connected open set.</p>
<p>Of course still more interesting than an empirical simulation would be a
good theoretical analysis.</p>
|
553,040 | <p>The question is: A card is drawn from an ordinary pack(52 cards) and a gambler bets that either a spade or an ace is going to appear. The probability of his winning are?
I think the answer is $\frac{16}{52} = \frac{4}{13}$. Did I go "probably" go wrong somewhere?</p>
| Ahaan S. Rungta | 85,039 | <p>Nope, you're correct! (Converting this to an answer.)</p>
|
1,252,857 | <p>Let $f\colon \mathbb R \to \mathbb R$ be defined by $f(x)= 5x^3+3$. Is it onto?</p>
<p>According to me, if $y=5x^3+3$, then $x = \sqrt[3]{(y-3)/5}$ is not an element of $\mathbb R$ for all $y \in (-\infty,3)$ so all numbers in the codomain $(-\infty,3)$ wont have pre-images.</p>
<p>But many say $5x^3+7$ as an odd degree equation will have at least one real root. Is it onto?</p>
| Hagen von Eitzen | 39,174 | <p>You overlook that real cube roots (in contrast to square roots) are defined for all reals For example $\sqrt[3]{-8}=-2$ because $(-2)^3=-8$.</p>
|
1,252,857 | <p>Let $f\colon \mathbb R \to \mathbb R$ be defined by $f(x)= 5x^3+3$. Is it onto?</p>
<p>According to me, if $y=5x^3+3$, then $x = \sqrt[3]{(y-3)/5}$ is not an element of $\mathbb R$ for all $y \in (-\infty,3)$ so all numbers in the codomain $(-\infty,3)$ wont have pre-images.</p>
<p>But many say $5x^3+7$ as an odd degree equation will have at least one real root. Is it onto?</p>
| Daniel W. Farlow | 191,378 | <p>Simply looking at Wolfram|Alpha shows that $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=5x^3+3$ is not just onto but also one-to-one. Let's prove that $f$ is onto though, using your choice of $x$ to do this. </p>
<p><strong>Claim:</strong> The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=5x^3+3$ is onto.</p>
<p><em>Proof</em>. Suppose $y\in\mathbb{R}$. Then let $x=\sqrt[3]{\frac{y-3}{5}}$. We have the following:
\begin{align}
f(x) &= f\left(\sqrt[3]{\frac{y-3}{5}}\right)\\[1em]
&= 5\left(\sqrt[3]{\frac{y-3}{5}}\right)^3+3\\[1em]
&= 5\cdot\frac{y-3}{5}+3\\[1em]
&=y-3+3\\[0.5em]
&= y.
\end{align}
Thus, $f$ is onto. $\blacksquare$</p>
|
2,932,139 | <blockquote>
<p>How would I show that the line <span class="math-container">$A=[(x,y,z)=(0,t,t)\mid t\in\mathbb{R}]$</span> is parallel to the plane <span class="math-container">$5x-3y+3z=1$</span>?</p>
</blockquote>
<p>I know the normal vector would be <span class="math-container">$(5,-3,3)$</span>, but how would I get the the directional of <span class="math-container">$A$</span>?</p>
| Community | -1 | <p>The direction of <span class="math-container">$A$</span> is <span class="math-container">$(0,1,1)$</span>. Check if <span class="math-container">$(0,1,1)$</span> and <span class="math-container">$(5,-3,3)$</span> are perpendicular. Since the dot product is zero, they are. </p>
|
2,932,139 | <blockquote>
<p>How would I show that the line <span class="math-container">$A=[(x,y,z)=(0,t,t)\mid t\in\mathbb{R}]$</span> is parallel to the plane <span class="math-container">$5x-3y+3z=1$</span>?</p>
</blockquote>
<p>I know the normal vector would be <span class="math-container">$(5,-3,3)$</span>, but how would I get the the directional of <span class="math-container">$A$</span>?</p>
| irchans | 372,582 | <p>In the equation <span class="math-container">$5x-3y+3z=1$</span>, replace the <span class="math-container">$x$</span> by 0, the <span class="math-container">$y$</span> by <span class="math-container">$t$</span>, and the <span class="math-container">$z$</span> by <span class="math-container">$t$</span>. After replacement, if the left hand side cannot be 1, then you know that the line never touched the plane and therefore the line and the plane are parallel. </p>
|
4,014,768 | <blockquote>
<p>Find the sationary points of the curve and their nature for the equation <span class="math-container">$y=e^x\cos x$</span> for <span class="math-container">$0\le x\le\pi/2$</span>.</p>
</blockquote>
<p>I derived it and got <span class="math-container">$e^x(-\sin x+\cos x)=0$</span>.</p>
<p><span class="math-container">$e^x$</span> has no solution but I don't know how to find the <span class="math-container">$x$</span> such that <span class="math-container">$-\sin x+\cos x=0$</span></p>
| Dr. Wolfgang Hintze | 198,592 | <p>Graphically we see immediately that <span class="math-container">$x=\frac {\pi}{4}$</span>. But how about deriving this analytically?</p>
<p>Consider <span class="math-container">$\cos(2x)$</span> and use the formula for the double argument <span class="math-container">$\cos(2x) = \cos(x)^2 - \sin(x)^2$</span></p>
<p>Hence if <span class="math-container">$\sin(x) = \cos(x)$</span> we find that</p>
<p><span class="math-container">$$\cos(2x)=0$$</span></p>
<p>Since <span class="math-container">$\cos(\alpha)$</span> vanishes if and only if the argument is of the form <span class="math-container">$\alpha=\pi(\frac{1}{2}+k), k=0,\pm1,\pm 2, ...$</span> $ we have</p>
<p><span class="math-container">$$x = \frac{\pi}{2}(\frac{1}{2}+k)|_{k\to 0}\to \frac{\pi}{4}$$</span></p>
<p>in the requested range.</p>
<p>We can also find the value of <span class="math-container">$a=\sin(x)=\cos(x)\gt0$</span>: from the relation <span class="math-container">$\sin(x)^2+\cos(x)^2=1$</span> we get <span class="math-container">$2 a^2 = 1$</span> and hence <span class="math-container">$a=\frac{1}{\sqrt{2}}$</span>.</p>
|
18,136 | <p>I introduced the hypercube (to undergraduate students in the U.S.) in the
context of generalizations of the Platonic solids, explained its
structure, showed it rotating.
I mentioned <a href="https://matheducators.stackexchange.com/a/1824/511">Alicia Stott</a>, who discovered the <span class="math-container">$6$</span> regular
polytopes in <span class="math-container">$\mathbb{R}^4$</span> (discovered after Schläfli).
I sense they largely did not grasp what is the hypercube, let alone
the other regular polytopes.</p>
<p>I'd appreciate hearing of techniques for getting students to
"grok" the fourth dimension.</p>
| guest | 13,762 | <p>I would just say you have a square in 2D, and a cube is the similar shape in 3D, then what is the next shape in 4D. Then show them the cube in cube view and cross like fold out. </p>
<p>Don't jump to Schlegel diagrams and the rotating pictures on Wiki so fast...they are confusing. Do like I said instead. </p>
<p>I think showing that transition from 2d to 3D, gets them thinking there must be some thing in 4D. Just doing that makes it simpler than talking about all the Platonic solids. Which themselves most people don't know.</p>
<p>Then next show them the Dali painting: <a href="https://en.wikipedia.org/wiki/Crucifixion_(Corpus_Hypercubus)" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Crucifixion_(Corpus_Hypercubus)</a> Mention that 4d has been intriguing to arts and writers (mention Flatland and how the sphere comes and explains 3D to the plane figure square. And then when the square says, fine, I bet there are higher dimensions than you, the sphere gets mad and runs off.</p>
<p>The point here is not to be having them list axes of symmetry or character groups or the like. It's to get them comfortable that there's some 4d stuff...and it's weird and hard to think about it. But at least they are in the mood to think about it!</p>
<p>I think you are a super nice guy and wicked smart. But too "hard".</p>
<p>Yes, if it becomes important to show all the polytopes than you will need to backtrack and discuss Platonic solids. Unless you played D&D, you don't know them as much. I mean how many non-chemists know there are C3 axes in the dodecahedron? But start with square-cube-tesseract before doing the Platonic solids. (They're harder than you think, even though all 3D.) Square-cube-tesseract is powerfully intuitive.</p>
<p>P.s. I am so tempted to make the Smith/Wellesly/Radcliff proverb a part of this...</p>
<p>P.s.s. <a href="https://en.wikipedia.org/wiki/%22%E2%80%94And_He_Built_a_Crooked_House%E2%80%94%22" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/%22%E2%80%94And_He_Built_a_Crooked_House%E2%80%94%22</a> "Stating that it "was, for many readers, the first introduction to four-dimensional geometry that held any promise of comprehensibility", Carl Sagan in 1978 listed "—And He Built a Crooked House—" as an example of how science fiction "can convey bits and pieces, hints and phrases, of knowledge unknown or inaccessible to the reader"."</p>
|
18,136 | <p>I introduced the hypercube (to undergraduate students in the U.S.) in the
context of generalizations of the Platonic solids, explained its
structure, showed it rotating.
I mentioned <a href="https://matheducators.stackexchange.com/a/1824/511">Alicia Stott</a>, who discovered the <span class="math-container">$6$</span> regular
polytopes in <span class="math-container">$\mathbb{R}^4$</span> (discovered after Schläfli).
I sense they largely did not grasp what is the hypercube, let alone
the other regular polytopes.</p>
<p>I'd appreciate hearing of techniques for getting students to
"grok" the fourth dimension.</p>
| Cees Timmerman | 13,523 | <p>Use arrays (aka <a href="https://www.doitpoms.ac.uk/tlplib/tensors/what_is_tensor.php" rel="nofollow noreferrer">tensors</a> in math lingo) to store the values of a place:</p>
<p>0 dimensions / zero rank (or order) tensor:</p>
<pre><code>a = 1
v = a
</code></pre>
<p>These simple values are also known as <em>scalars</em>.</p>
<p>1 dimension / first rank tensor:</p>
<pre><code>a = [1, 2, 3, ...]
v = a[x]
</code></pre>
<p>These lists of values are also known as <em>vectors</em>.</p>
<p>2 dimensions / second rank tensor:</p>
<pre><code>a =
[
[1, 2, 3, ...],
[1, 2, 3, ...],
[1, 2, 3, ...],
...
]
v = a[x][y]
</code></pre>
<p>These tables are also known as rows by columns <em>matrices</em>.</p>
<p>3 dimensions / third rank tensor:</p>
<pre><code>a =
[
[
[1, 2, 3, ...],
[1, 2, 3, ...],
[1, 2, 3, ...],
...
],
[
[1, 2, 3, ...],
[1, 2, 3, ...],
[1, 2, 3, ...],
...
],
[
[1, 2, 3, ...],
[1, 2, 3, ...],
[1, 2, 3, ...],
...
],
...
]
v = a[x][y][z]
</code></pre>
<p>4 dimensions / 4th rank tensor / 4D array/matrix/object with 2+ elements per dimension and numbered integer indices:</p>
<pre><code>a =
[
[
[
[1, 2, ...],
[1, 2, ...],
...
],
[
[1, 2, ...],
[1, 2, ...],
...
],
...
],
[
[
[1, 2, ...],
[1, 2, ...],
...
],
[
[1, 2, ...],
[1, 2, ...],
...
],
...
],
...
]
v = a[i1][i2][i3][i4]
</code></pre>
<p>Etc. Note the self-similarity.</p>
<p>When you scale the dimensional axes, the amount of data changes by the power of the number of axes. Doubling a 1D array doubles the data (2<sup>1</sup> = 2), doubling a 2D array squares the data (2<sup>2</sup> = 4), doubling a 3D array cubes the data (2<sup>3</sup> = 8), and so on.
To save data, one could simply store the coordinates on those axes, as a dimension is "a measurable extent of a particular kind, such as length, breadth, depth, or height" according to the dictionary.</p>
<p>Other self-similar things are fractals, so named because they have <a href="https://www.youtube.com/watch?v=gB9n2gHsHN4" rel="nofollow noreferrer">a fractional dimension</a>. Consider a Sierpiński triangle. Doubling its size leads to a data increase of 3, so 2<sup>D</sup> = 3, so its dimension is log<sub>2</sub>(3) = 1.58496250072.</p>
|
1,328,082 | <p>I've been trying to prove the following inequality, but until now I've had problems coming up with a solution:</p>
<p>$$
2^{mn} \ge m^n
$$</p>
<p>$m$ and $n$ can assume any natural number.</p>
<p>I wasn't able to find any counterexample that would invalidate this inequality, so I am assuming that this statement is generally true, but of course this still has to be proven.</p>
| Somabha Mukherjee | 21,617 | <p>Look at the function $f(x) = 2^x-x$. </p>
<p>$f'(x)=(2^x)(\log 2)- 1 > 0$ for $x \geq 1$. So $f$ is increasing. Now, $f(1) = 1$. Hence, $f(m) > 0$ for every natural number $m$. </p>
|
2,466,913 | <p>I am having trouble with a question on my homework and I have done some work but I'm not sure if I have completed all that is asked or how to take it further if needed.</p>
<p>Problem: </p>
<p>Let $\theta = \arccos(5x^3)$. Find $\frac{d\theta}{dx}$ in two ways:</p>
<p>(a) Use implicit differentiation starting from $\cos(\theta)= 5x^3$. Write your final answer in terms of $x$ by using SOH-CAH-TOA, a right triangle, and the Pythagorean Theorem.</p>
<p>(b) Use the Chain Rule extension of the formula $\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1-x^2}}$. Compare with your answer from part (a).</p>
<hr>
<p>Here is my work for part (a)</p>
<p>I drew a right triangle with an angle theta and labeled the hypotenuse as $1$ and the adjacent leg $5x^3$. Using the Pythagorean Theorem I found the opposite leg to be $\sqrt{1-25x^6}$. I'm not sure if this is all the question is asking or if I need to do something further. The wording of the problem is confusing me a little.</p>
<p>I also simply used this extra side to find $\sin(\theta) = \sqrt{1-25x^6}$. </p>
<p>Here is my work for part (b)</p>
<p>\begin{equation}
\begin{split}
\frac{d}{dx} \arccos 5x^3 &= \frac{-1}{\sqrt{1-(5x^3 )^2}} \frac{ d}{dx} (5x^3 )\\
&= \frac{-1}{\sqrt{1-25x^6}} 15x^2\\
&= \frac{-15x^2}{\sqrt{1-25x^6}}.
\end{split}
\end{equation}</p>
<p>If this is as far as I am supposed to take this problem are my answers supposed to be the same for part (a) and part (b)? If not then why not? I know I can simplify the answer for part (b) by rationalizing the denominator but it still wouldn't equal anything from part (a).</p>
<p>Any help would be greatly appreciated to help me finish this problem. I've looked at it several times for hours and am not sure where to go. Thanks!</p>
| quasi | 400,434 | <p>Since whenever a piece is cut into $7$ pieces, you gain $6$ pieces ($7$ new pieces minus the replaced old one), it follows that at any stage, the number of pieces remains the same mod $6$.
<p>
Since at the start, you have $7$ pieces, it follows that at any stage, the number of pieces is congruent to $1$ mod $6$.
<p>
But $2016$ is congruent to $0$ mod 6.</p>
|
3,742,707 | <p>How many different fractions can be made up of the numbers 3, 5, 7, 11, 13, 17 so that each fraction contains 2 different numbers? How many of them will be the proper fractions?</p>
| Mathematician | 804,713 | <p>The numbers of fraction which have two different numbers according to @Benjamin wang = 6C2×2!=30. Now you can find the solution of next question. To be proper fraction I can't take the value 2!
So, the answer of next question=6C2=15</p>
|
14,746 | <p>Is there a case to be made that the topic of line integrals should only involve vector fields?</p>
<p>My colleagues and our textbook take the position that line integrals should only be taught from a vector field perspective (specifically for calculating "work"). [In fact, our textbook <em>defines</em> a line integral as <span class="math-container">$\int _C \mathbf{F} \cdot d\mathbf{r}$</span>, where <span class="math-container">$\mathbf{F}$</span> is a vector field and <span class="math-container">$C$</span> is some parametrized curve.]</p>
<p>I think it makes more pedagogical sense to introduce line integrals as a way to generalize what students should have just done in their integral calculus class: integrate a function along some 1-dimensional direction. Now, that "direction" can be a path in 2-space or 3-space, so we can see an area as the result of the line integral. Then, after introducing vector fields, we can consider other, meaningful things to integrate along a path, such as a dot product of the field with the path.</p>
<p>My motivation here is that I would like new calculus topics to be easily connected to old topics, if possible. If we jump right into calculating work without any tie back to the "calculate the area" problem students are used to, I fear they may come away thinking that line integrals are just these weird things with their own rules.</p>
<p>In short: Is there a prevailing setting for introducing line integrals? If so, has there been a movement to minimize the teaching of line integrals over scalar fields, focusing primarily on work calculations?</p>
| Nick C | 470 | <blockquote>
<p>Please take 12-15 minutes to <em>introduce</em> your lesson</p>
</blockquote>
<p>Your intuition is correct that this is far too much for a 12-15 minute lesson. The committee knows that there is no way you could deliver a whole lesson, even an introductory one, on this topic.</p>
<p>Since you've only got 15 minutes, plan to use it to spark interest in the topic, knowing that you're only showing the committee the very beginning.</p>
<p>For example, why not show students the cosine function on a small interval around <span class="math-container">$x=0$</span>, and ask them what polynomial it reminds them of? "Does it look like some kind of <span class="math-container">$x^2$</span> or <span class="math-container">$x^3$</span>?" Let them decide which one, and then graph their guess on top of <span class="math-container">$y=\cos(x)$</span>. You will have the opportunity to discuss the vertical intercept and the width of the graph – that it can be adjusted (by multiplying a term by a constant). In this way, your lesson could start as a game about building the cosine function from simpler functions, and students will be involved in the guessing process. I see this occupying up to 5 minutes. Then do your steps 1 and 2, and you're probably out of time.</p>
<p>In my opinion, using your lesson time to draw on your students' previous experience with trig and power functions helps them see that they are already capable of doing much of the intuitive work leading to Taylor polynomials, and it validates that this <em>new</em> thing is something that doesn't have to seem like just applying another magic math formula.</p>
<p>You should definitely plan to tell the committee what else the lesson would include (if you had the time), but I also would tell them what they should already know. [e.g. "Hey, class...let's remember back when we studied graphs of power functions. Do you all remember how to draw the graph of <span class="math-container">$y=-x^2$</span>?"]</p>
<blockquote>
<p>Have a student centered formative assessment last 2-3 minutes and find the Taylor expansion for sin x</p>
</blockquote>
<p>You could considerably simplify this assessment (for an introduction) by asking them what <em>just the first term</em> of the series expansion should be. If you took the time to remind them of the "long-run behavior" of a polynomial when you discussed the cosine function, you could let them use their calculators to determine what the 2nd term of the series expansion of <span class="math-container">$\sin(x)$</span> is.</p>
|
2,346,921 | <blockquote>
<p>Evaluate the following integral using Laplace transform
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$</p>
</blockquote>
<p>I obtained this partial result
$$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$
and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer. </p>
| Siddhartha | 427,792 | <p>the standard form of LT is
$$\mathcal{L}f(t)=\int_{0}^{\infty}e^{-st}f(t)\,dt$$
So for the given integral you just have to calculate LT of $\frac{\sin bx}{x}$ then put $s=a$
these two properties of LT may be useful
$$\mathcal{L} \sin (bx)=\frac{b}{b^2+s^2}$$and $$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_{s}^{\infty}F(s)\,ds$$</p>
|
2,346,921 | <blockquote>
<p>Evaluate the following integral using Laplace transform
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$</p>
</blockquote>
<p>I obtained this partial result
$$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$
and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer. </p>
| robjohn | 13,854 | <p>This doesn't use the Laplace Transform, however, it is pretty simple.</p>
<p>$$
\begin{align}
\frac{\partial}{\partial a}\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x
&=-\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}x\tag{1}\\
&=-\frac1b+\frac ab\int_0^\infty e^{-ax}\cos(bx)\,\mathrm{d}x\tag{2}\\
&=-\frac1b+\frac{a^2}{b^2}\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}\tag{3}\\
&=-\frac{b}{a^2+b^2}\tag{4}
\end{align}
$$
Explanation:<br>
$(1)$: differentiate inside the integral with respect to $a$<br>
$(2)$: integrate by parts: $-e^{-ax}\sin(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\cos(bx)$<br>
$(3)$: integrate by parts: $e^{-ax}\cos(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\sin(bx)$<br>
$(4)$: $\frac{b^2}{a^2+b^2}$ times $(3)$ plus $\frac{a^2}{a^2+b^2}$ times $(1)$</p>
<p>Integrating $(4)$ with respect to $a$ gives
$$
\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x=\tan^{-1}\left(\frac ba\right)\tag{5}
$$</p>
|
1,422,666 | <p>I've just run into this problem, and was able to go as far, and understand the induction step up to the bolded section. The last part I found in the back of my book, italicized, I can't understand.</p>
<p>Use induction to prove: $n^2\geq 2n + 1$ for all $n\in \mathbb{N}$ and $n\geq 3$.</p>
<p>Base: $3^2\geq 6+1$.</p>
<p>Induction Hypothesis: Assume that $k^2\geq 2k+1$ for $k\geq 3$.</p>
<p>Induction Step: $(k + 1)^2=k^2+2k+1\geq 2k+1 + 2k + 1 = 2(k + 1) + 2k \geq 2(k + 1) + 1$</p>
<p>I went as far as $(k + 1)^2 \geq 2(k+1) + 1$?</p>
<p>Any help, much appreciated</p>
| Selene Routley | 10,549 | <p>By your expansion, </p>
<p>$$\left(k^2\geq 2\,k+1\right)\;\Rightarrow\; \left((k+1)^2 \geq 4\,k+2\right)\tag{1}$$</p>
<p>since, as you have found, $(k+1)^2 = k^2 + 2\,k+1 \geq 2\,(2\,k+1)$ by the predicate on the LHS of the $\Rightarrow$ in (1).</p>
<p>But, for $k\geq 1$, we also have $4\,k+2 > 2\, k+3 = 2(k+1)+1$, hence your induction step follows from (1) together with $4\,k+2 > 2 k+3$.</p>
<p>You've clearly got the right idea: you simply seem to be confusing yourself with symbols: this happens to most humans often so (as I have needed to do many times), if this happens to you, I susgest you write the above out in terms of a predicate with a formal definition $P:\mathbb{N}\to\{\text{true},\,\text{false}\}$ (in this case $P(n) = (k^2\geq 2\,k+1)$) and manipulate, with standard first order rules of inference, the truth functions beginning from $P(n_0)=\text{true}$ for the induction basis and $P(n)\Rightarrow P(n+1)$ for the step.</p>
|
643,237 | <p>This statement is suggested as a correction to <a href="https://math.stackexchange.com/questions/643122/splitting-field-containing-nth-root">this question</a>:</p>
<p>If $K$ is the splitting field of the polynomial $P(x)=x^n-a$ over $\mathbb{Q}$, prove that $K$ contains all the $n$th roots of unity.</p>
<p>How to prove it?</p>
| Robert Lewis | 67,071 | <p>Let $F$ be an <em>arbitrary</em> field, $P(x) = x^n - a \in F[x]$ with $p = \text{char} \; F \not\mid n$ and $a \ne 0$, and let $K$ be the splitting field of $P(x)$ over $F$. Then $P(x)$ splits into linear factors in $K$; that is, there exist $\beta_j \in K$, $1 \le j \le n$, with</p>
<p>$P(x) = x^n - a = \prod_1^n (x - \beta_j) \tag{1}$</p>
<p>in $K$. Furthermore, since</p>
<p>$P'(x) = nx^{n - 1}, \tag{2}$</p>
<p>we see that $P(x)$ and $P'(x)$ have no common root; since $p \not\mid n$, $n \ne 0$ in $F$, so the only zero of $P'(x)$ is $0$, which is not a root of $P(x)$. This implies the $\beta_j$ are in fact all distinct, each from the others. Now setting $\omega_i = \beta_1^{-1}\beta_i \in K$, we see that $\omega_i^n = \beta_1^{-n}\beta_i^n = a^{-n}a^n = 1$, since each $\beta_i$ satisfies $\beta_i^n = a$. Furthermore, the $\omega_i$ are distinct, since the $\beta_i$ are. Thus the $n$ $n$-th roots of unity $\omega_i$ are all contained in $K$. <strong><em>QED.</em></strong></p>
<p>Applying the above result to the case $F = \Bbb Q$, we have that the splitting field $K$ of
$x^n - a$ contains all the $n$-th roots of unity for <em>any</em> $n$, since $0 = \text{char} \; \Bbb Q \not\mid n$.</p>
<p>It should be observed that the above argument makes no assumptions about the roots of unity lying in $K$; rather, it deduces their existence from that of the distinct zeroes $\beta_i$ of $P(x)$, hence no explicit reference to the complex field $\Bbb C$ is required. Nor is the primitivity of any of the $\omega_i$ an issue, since we obtain $n$ distinct $\omega_i$ <em>via</em> the formula $\omega_i = \beta_1^{-1}\beta_i$. And apparently, the essential result is true for <em>many</em> fields other than $\Bbb Q$.</p>
<p>Hope this helps. Cheerio,</p>
<p>and as always,</p>
<p><em><strong>Fiat Lux!!!</em></strong></p>
|
444,865 | <p>Prove that any natural number n can be written as $$n=a^2+b^2-c^2$$ where $a,b,c$ are also natural.</p>
| individ | 128,505 | <p>I like this approach to the solution of this equation.</p>
<p>If we consider the Diophantine equation: $qX^2+Y^2=Z^2+j$ </p>
<p>If the root is a : $a=\sqrt{\frac{j}{q}}$ </p>
<p>We use the solutions of Pell's equation: $p^2-(q+1)s^2=1$ </p>
<p>Solutions can be written: </p>
<p>$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$ </p>
<p>$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$ </p>
<p>$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$ </p>
<p>$L$ - any integer number given by us </p>
<p>number: $b,c,f,t$ - are solutions of the following equations </p>
<p>$qb^2+c^2=f^2$ </p>
<p>$t^2-(q+1)f^2=\pm{q}$ </p>
<p>If we take the solutions of Pell's equation: $p^2-(q+1)s^2=k$ </p>
<p>number $b,c$ - are solutions of the equation: $qb^2+c^2=f^2$ </p>
<p>wherein: $c-b=k$ </p>
<p>number $t,f$ - solutions of the equation: $t^2-(q+1)f^2=\pm{qk^2}$ </p>
<p>These formulas allow us to find some solutions of Pell's equation using solutions of simpler equations. At least there will be another opportunity to find a solution to this equation. Later draw solutions with other factors.</p>
<p>All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$ </p>
<p>If the ratio is factored so: $a=(b-c)(b+c)$ </p>
<p>Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$ </p>
<p>where: $f=(q+1)k^2-2kt-(q-1)t^2$ </p>
<p>Then the solutions are of the form: </p>
<p>$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$ </p>
<p>$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$ </p>
<p>$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$ </p>
<p>All of numbers can be any character.</p>
|
3,176,593 | <p>I'm having a hard time understanding how to find all solutions of the form <span class="math-container">$a_n = a^{(h)}_n+a_n^{(p)}$</span></p>
<p>I show that <span class="math-container">$a_n=n2^n \to a_n=2(n-1)2^{n-1} +2^n=2^n(n-1+1)=n2^n$</span>.</p>
<p>I can show that <span class="math-container">$a_n^{(h)}$</span> characteristic equation <span class="math-container">$r-2=0 \to a_n^{(h)}=\alpha2^n$</span></p>
<p>But I'm stuck on <span class="math-container">$a_n^{(p)}$</span> characteristic equation <span class="math-container">$C2^n=2C\cdot2^{n-1}+2^n$</span></p>
<p>Simplifies to <span class="math-container">$C \neq C+1$</span>, Looking online I saw that the solution is <span class="math-container">$a_n=c\cdot2^n+n2^n$</span>, but I'm not sure how to get there. </p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$a_m=b_m+2^m(a_0+a_1m+a_2m^2+\cdots)$</span></p>
<p><span class="math-container">$$2^n=a_n-2a_{n-1}$$</span>
<span class="math-container">$$=ba_n-2b_{n-1}+2^n(a_0+a_1n+a_2n^2+\cdots)-2^n(a_0+a_1(n-1)+a_2(n-1)^2+\cdots)$$</span></p>
<p><span class="math-container">$$=ba_n-2b_{n-1}+2^n[a_1+a_2\{n^2-(n-1)^2\}+a_3\{n^3-(n-1)^3\}+\cdots]$$</span></p>
<p>For <span class="math-container">$a_r,$</span> the highest exponent of <span class="math-container">$n$</span> is <span class="math-container">$r-1$</span></p>
<p><span class="math-container">$a_r=0\ \forall r\ge2$</span> as the coefficient of <span class="math-container">$2^n$</span> in the left hand side is <span class="math-container">$1$</span></p>
<p>Set <span class="math-container">$a_1=1$</span> so that <span class="math-container">$b_n=2b_{n-1}=2^rb_{n-r}; 0\le r\le n$</span></p>
|
1,412,177 | <p>Let $W$ be the subset of $\mathbb{R}^5$ consisting of all vectors an odd number of the entries in which are equal to $0$. Is $W$ a subspace of $\mathbb{R}^5$?</p>
<p>I'm not sure how to do this. Any solutions or hints are greatly appreciated. I know that in order for anything to be a subspace of something the zero vector must be in it. How would I go about this? What exactly do we mean by subset here? Is it any $5$-tuple or could it be $1,2,3,4,5$-tuples?</p>
| Yes | 155,328 | <p>Since $W$ is a subspace of $\mathbb{R}^{5}$ if and only if (i) $0_{\mathbb{R}^{5}} \in W$, (ii) $x,y \in W$ implies $x+y \in W$, and (iii) $c \in \mathbb{R}$ and $x \in W$ imply $cx \in W$, and since $(1,2,0,3,4) \in W$ and $(4,0,3,2,1) \in W$ but their sum is a vector having no zero components, so $W$ is not a subspace of $\mathbb{R}^{5}$.</p>
<p>I misunderstood your question. :) </p>
|
1,412,177 | <p>Let $W$ be the subset of $\mathbb{R}^5$ consisting of all vectors an odd number of the entries in which are equal to $0$. Is $W$ a subspace of $\mathbb{R}^5$?</p>
<p>I'm not sure how to do this. Any solutions or hints are greatly appreciated. I know that in order for anything to be a subspace of something the zero vector must be in it. How would I go about this? What exactly do we mean by subset here? Is it any $5$-tuple or could it be $1,2,3,4,5$-tuples?</p>
| Jason | 195,308 | <p>No. For instance, let $v=(0,1,1,1,1)$ and $w=(1,0,1,1,1)$. Then $v,w\in W$ but $v+w=(1,1,2,2,2)\notin W$.</p>
|
1,344,001 | <p>I need to solve an expression of this kind (solve for $x$):</p>
<p>$e^{\pi i x} -e^{-\pi ix} = 2yi$</p>
<p>Both $x$ and $y$ are real numbers, $y$ is given. I have no clue on how to solve it analytically.</p>
<p>All I know is that I can rewrite this as:</p>
<p>$\sin(x\pi) = y$</p>
<p>so:</p>
<p>$x=\frac{\arcsin(y)}{\pi}$</p>
<p>But I don't know how to generate the complex solutions from this form (neither form actually).</p>
| Rory Daulton | 161,807 | <p>$$e^{\pi ix}-e^{-\pi ix}=y\cdot 2i$$
$$\left( e^{\pi ix}-e^{-\pi ix} \right)e^{\pi ix}=(2iy)e^{\pi ix}$$
$$\left(e^{\pi ix}\right)^2-1=2iye^{\pi ix}$$
$$\left(e^{\pi ix}\right)^2-2iy\left(e^{\pi ix}\right)-1=0$$
$$e^{\pi ix}=\frac{-(-2iy)\pm\sqrt{(-2iy)^2-4(1)(-1)}}{2(1)}$$
$$=\frac{2iy\pm\sqrt{-4y^2+4}}{2}$$
$$=iy\pm\sqrt{1-y^2}$$
$$\pi ix=\ln\left( iy\pm\sqrt{1-y^2} \right)$$
$$x=\frac{\ln\left( iy\pm\sqrt{1-y^2} \right)}{i\pi}$$</p>
<p>You can modify this, if you like, in several ways, especially to emphasize that $x$ is real. Note that the $\ln$ function is multi-valued, so you may get more than two values.</p>
|
1,481,106 | <p>I am having a difficult time solving this problem. I have tried this several different ways, and I get a different result, none of which is correct, every time. I've derived an answer geometrically and cannot replicate it with a double integral.</p>
<p>Here's the problem: Use a double integral to find the area between two circles $$x^2+y^2=4$$ and $$(x−1)^2+y^2=4.$$</p>
<p>Here is how I have tried to go about this problem:</p>
<p>First, I graphed it to get a good idea visually of what I was doing.
<a href="https://i.stack.imgur.com/i7cNa.png" rel="nofollow noreferrer">Here's the graph I scribbled on.</a>
The region I'm interested is where these two circles overlap. This region can easily be divided into two separate areas. There are clearly a number of ways to go about solving this...but the one I opted for is to find the shaded region. The bounds for $x$ in this case are between $D$ and $C$. D can be found by setting $C_1=C_2$, and $x$ turns out to be $\frac{1}{2}$. On the right, $x$ is where $C_1(y)=0$, $x=\pm2$, so $x=2$ at point $C$. $y$ is greater than $B_y$ and less than $A_y$, which are also found where $C_1=C_2$, and $y$ turns out to be $\pm\sqrt{\frac{15}{4}}$. So far so good. Now I know my limits of integration. But here's what I don't understand. What am I actually integrating? $x$ has constant bounds, and $y$ does not, and looking at other double integral problems, that would lead me to believe that I should integrate $y$ first as a function of $x$, evaluate it at its bounds, and then integrate $x$ and evaluate it at its bounds giving me half the area I am looking for. However, when I try to do this, I get utter nonsense for an answer, or I get lost trying to set up the problem.</p>
<p>I could really use the help, I've spent entirely too much time trying to puzzle through this. Thank you in advance!</p>
<p>P.s. I determined the area geometrically using a CAD program to calculate the area, and it should be approximately $8.46$.</p>
| Michael Medvinsky | 269,041 | <p><a href="https://math.stackexchange.com/questions/670029/area-between-two-circles-as-a-double-integral-in-polar-coordinates">Here</a> is a similar problem solved.</p>
<p><a href="http://www.math.utah.edu/~mmedvin/Teaching/Math1321/LectureNotes/27_1321L_TripleIntegrals.pdf" rel="nofollow noreferrer">Here</a> is an example of less similar problem, it is for volume between two balls and therefore, of course, it is using a triple integral. </p>
<p>Hope this helps.</p>
<p>ps: note the links</p>
|
2,306,122 | <p>Show $X=\{n \in \mathbb{N}: \text{n is odd and} \ n = k(k+1) \text{for some} \ k \in \mathbb{N}\}=\emptyset$</p>
<p>My proof is as follow, please point if I have made any mistake. </p>
<p><strong>proof:</strong></p>
<p>we have $\emptyset \subseteq X$
suppose $X≠\emptyset$ pick $n \in X$</p>
<p>Then there are 2 cases
1st case: n is odd then n=(k+1)k</p>
<p>Then
suppose k is odd $\implies$ k+1 is even $\implies$ n is even</p>
<p>2nd case: consider k is even $\implies k+1$ is odd</p>
<p>then
n=(k+1)k for some $k \in \mathbb{N}=\emptyset \implies n$ is even</p>
<p>Therefore, n is neither even nor odd, so $k \in \mathbb{N} \implies n \not\in X$
and $\implies X= \emptyset$ </p>
<p>Q.E.D</p>
| Sri-Amirthan Theivendran | 302,692 | <p>Here is a combinatorial proof that $k(k+1)$ is always even for $k\in\mathbb{N}$. The number of two element subsets of a $k+1$ element set is
$$
\binom{k+1}{2}=\frac{k(k+1)}{2}
$$
as there are $k+1$ choices for the first element of the subset and $k$ choices for the second element of the subset, but as $\{a,b\}$ and $\{b,a\}$ represent the same set, we divide the product by $2$ to find the number of such subsets. The result follows.</p>
|
2,645,406 | <p>I am reading Rudin's real and complex analysis. On page 14 we see $\underset{n} {\text{sup }} f_n$, and on page 15 we see max$\{f,g\}$. These two are obviously different. I searched some answers and found an example: </p>
<p>sup $\{f,g\} = f(x)$ wehn $f(x)\geq g(x)$ and $g(x)$ when $f(x) < g(x)$ </p>
<p>But then I saw max$\{f,g\}$ which makes me confused. Are they the same? What are the differences?</p>
<p>Can someone also give some explanation of $(\underset{n\to \infty}{\text{lim sup }} f_n)(x)$? How to visualize/understand it?</p>
| saulspatz | 235,128 | <p>"max" stands for "maximum" the greatest value in a set. "sup" stands for "supremum" or "least upper bound" of the values in a set. When the set is finite, these mean the same thing, and which one you use is a matter of state. When the set is infinite, it may not have a maximum. For example, what is $\max \{ x | 0 < x < 1\}?$ Obviously, there is no maximum, but the supremum is $1$. So for infinite sets, you shouldn't use "max" unless somehow you know that the set has a maximum. For example, we know that a continuous function attains its maximum on a closed interval, so it would be okay to use "max" in that case, though "sup" would be just as good.</p>
|
2,414,721 | <p>For all perfect numbers $N$, $\sigma (N) = 2N$, where $\sigma$ is the divisor sigma function.</p>
<p>Let $s$ be a perfect number of the form $3^m 5^n 7^k$, where $m,n,k \geq 1$ are integers.</p>
<p>Then $\sigma (s)= \sigma (3^m 5^n 7^k)$</p>
<p>$ =\sigma (3^m) \sigma (5^n) \sigma (7^k)$ since $3, 5,$ and $7$ are coprime to each other.</p>
<p>$ =\left(\frac{3^{m+1}-1}{2}\right)\left(\frac{5^{m+1}-1}{4}\right)\left(\frac{7^{k+1}-1}{6}\right)$</p>
<p>$ =2(3^m 5^n 7^k)$ since $s$ is a perfect number.</p>
<p>$\implies 9 (3^m 5^n 7^k) = 3^{m+1} 5^{n+1}+3^{m+1} 7^{k+1} + 5^{n+1} 7^{k+1} - 3^{m+1}-5^{n+1} - 7^{k+1}-1$ after some algebra.</p>
<p>This is as far as I got using this method. Any and all help would be appreciated.</p>
| Joffan | 206,402 | <p>In order to get a perfect number, we need the ratio $\dfrac{\sigma(n)}{n}=2$, which also means that we need the prime powers in the denominator to appear in the sigma values on top. And we need exactly and only one power of $2$ to appear in that sigma value, because the denominator is odd.</p>
<p>Unrolling this into prime powers, </p>
<p>$$\frac{\sigma(n)}{n}=\frac{\sigma(3^m)}{3^m} \frac{\sigma(5^n)}{5^n} \frac{\sigma(7^k)}{7^k}$$</p>
<p>For primes in general, $p$ does not divide $\sigma(p^j)$. So we would need to match the powers of each prime from the $\sigma$ values of the other primes.</p>
<p>For $\sigma(3^m)$, any odd value of $m$ makes $\sigma(3^m)$ divisible by $4$, so we need $m$ even - but in that case, $\sigma(3^m)$ is not divisible by $5$ - we actually need $m{+}1$ divisible by $4$ to get any power of $5$, which can't happen.</p>
<p>Similarly, for $\sigma(7^k)$, any odd value of $k$ makes $\sigma(7^k)$ divisible by $8$, so we need $k$ even - but again in that case, $\sigma(7^k)$ is not divisible by $5$ which once again only happens if $k{+}1$ is divisible by $4$.</p>
<p>So a number of that form cannot have $\dfrac{\sigma(n)}{n}=2$ and thus cannot be perfect.</p>
|
1,136,486 | <p>Can anyone help me find the first digit of $2015^{2015}$?</p>
<p>It is easy to find the last digit but I have no idea for the first digit.</p>
| Luigi D. | 164,401 | <p>The units digit is of course a 5. </p>
<p>For the leading digit, see that $n^n = 10^{n\cdot log_{10}(n)} \approx 10^{2015\cdot3.304275} \approx 10^{6658.114} = 10^{6658}10^{0.114} \approx 1.3008\cdot 10^{6658}$. Therefore, the first digits of your number are 13008...</p>
<p>Check: <a href="http://www.wolframalpha.com/input/?i=2015%5E2015">http://www.wolframalpha.com/input/?i=2015^2015</a></p>
|
674,621 | <p>I am trying to figure out what the three possibilities of $z$ are such that </p>
<p>$$
z^3=i
$$</p>
<p>but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.</p>
| lab bhattacharjee | 33,337 | <p>As $\displaystyle i^2=-1,i=-i^3\implies z^3=i=-i^3=(-i)^3\iff z^3-(-i)^3=0$</p>
<p>Now, $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2),$</p>
<p>$\displaystyle\implies\{z-(-i)\}\{z^2+z(-i)+(-i)^2\}=\{z-(-i)\}(z^2-iz-1)$</p>
<p>If $z-(-i)=0, z=-i$</p>
<p>Else $z^2-iz-1=0\implies z=\dfrac{i\pm\sqrt{i^2-4(-1)}}2=\dfrac{i\pm\sqrt3}2$</p>
|
1,601,427 | <blockquote>
<p>Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$</p>
</blockquote>
<p>It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?</p>
| asomog | 183,714 | <p>Let $a=x^3$, $b=y^3$, $c=z^3$, then it can be rewritten as:
$$
x^6+y^6+z^6+3 x^2 y^2 z^2-2 \left(x^3 y^3+x^3 z^3+y^3 z^3\right)\geq 0
$$
Use the following notations:
$$S_{3}:=xyz\qquad S_2:=xy+yz+xz\qquad S_1=x+y+z$$
Then:
$$
x^6+y^6+z^6=S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-12 S_2 S_3 S_1-2 S_2^3+3 S_3^2
$$
$$
x^3 y^3+x^3 z^3+y^3 z^3=S_2^3-3 S_1 S_3 S_2+3 S_3^2
$$
$$
3x^2y^2z^2=3S_3^2
$$
Then we only have to prove:
$$
S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\geq 0
$$
Now put $S_2=S_1^2$, and notice that with this:
$$
\left.S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\right|_{S_2=S_1^2}=0
$$
Thus this can be factorised as:
$$
\left(S_1^2-S_2\right) \left(S_1^4-5 S_2 S_1^2+6 S_3 S_1+4 S_2^2\right)\geq0
$$
Since: $(x+y+z)^2\geq 3(xy+yz+xz)\Rightarrow S_1^2\geq 3S_2$ by rearrangement, it is enough to prove that the second factor is non-negative. Return to our previous notations, enough to show:
$$
x^4+y^4+z^4+(x+y+z)xyz-x^3y-y^3x-y^3z-z^3y-x^3z-xz^3=
$$
$$
=x^2(x-y)(x-z)+y^2(y-x)(y-z)+z^2(z-x)(z-y)\geq 0
$$
Which is trivially true by applying <a href="https://en.wikipedia.org/wiki/Schur's_inequality" rel="nofollow">Schur's inequality</a></p>
|
1,601,427 | <blockquote>
<p>Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$</p>
</blockquote>
<p>It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?</p>
| Michael Rozenberg | 190,319 | <p>Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.</p>
<p>Hence, our inequality it's $f(v^2)\geq0$, where $f(v^2)=3u^2-4v^2+w^2$.</p>
<p>Thus, $f$ is a linear function, which says that $f$ get's a minimal value for an extremal value of $v^2$,
which happens for equality case of two variables.</p>
<p>Let $b=a=x^3$ and $c=1$.</p>
<p>Hence, we need to prove that $x^6+2+3x^2\geq2(2x^3+1)$,</p>
<p>which is $(x-1)^2(x^2+2x+3)x^2\geq0$. Done!</p>
|
243,743 | <p>I have been studying this sequence (<a href="https://oeis.org/A266882" rel="nofollow">A266882</a> in the OEIS) and found the following pattern:</p>
<p>$13 + 17 + 19 + 23 + 29 = 101$ (101 is prime)</p>
<p>37 does not hold.</p>
<p>$223 + 227 + 229 + 233 + 239 = 1151$ (1151 is prime)</p>
<p>The same is true for 1087, 1423, 1483, and 2683.</p>
<p>So, is 37 the only prime of this sequence (A266882) for whom the sum is not also a prime?</p>
| aosjckajsd | 94,470 | <p>Here's a Haskell program to display the values for which the conjecture is false.</p>
<p>Within a minute, the largest value it found was the following, rather nice quintuple: $$[468883,468887,468889,468893,468899]$$</p>
<pre><code>import Math.NumberTheory.Primes
specialPrime n = ((nthPrime n) + (nthPrime (n + 3)) == (nthPrime (n + 1)) + (nthPrime (n + 2))) && ((nthPrime n) + (nthPrime (n + 4)) == (nthPrime (n + 2)) + (nthPrime (n + 3)))
specialPrimes = filter (specialPrime) [1..]
specialTuples n = let v = specialPrimes !! (n-1)
in fmap nthPrime [v..(v + 4)]
checkConjecture n = isPrime (foldr (+) 0 $ specialTuples n)
falseValues = fmap specialTuples $ filter (not . checkConjecture) [1..]
main = do
sequence_ $ fmap print falseValues
</code></pre>
|
729,441 | <p>This question comes up after going over Arzela-Ascoli theorems.</p>
<p>For a set of continuous functions $\mathbb F$ from $\mathbb R$ to $\mathbb R$ that is equicontinuous. How do I show that if sup{$|f(0)|:f \in F$} $< \infty$ , then $\mathbb F$ is pointwise bounded?</p>
<p>I know I need to show that since sup{$|f(0)|:f \in F$} = $M_0 < \infty$ then for each $x \in \mathbb R$ sup{$|f(x)|:f \in F$} = $M_x < \infty$.</p>
<p>What I have gotten so far is that I should fix $\epsilon$ and use the fact that the domain is $\mathbb R$.</p>
<p>I think I just need some help using equicontinuity and sup together.</p>
| Marc van Leeuwen | 18,880 | <p>If you know that change of basis is realised by conjugating by an appropriate invertible matrix, then you can reason in terms of matrices as follows. $E_{i,j}$ is the matrix with unique nonzero entry $1$ at position $i,j$.</p>
<ul>
<li><p>The (unique) matrix $M$ of $T$ can have no nonzero off-diagonal entries: if $a_{i,j}$ were such an entry, then conjugating by $I+E_{j,i}$ adds $a_{i,j}$ to the diagonal entries at $(j,j)$, and subtracts it from the entry at $(i,i)$, while it was supposed to leave all entries unchanged.</p></li>
<li><p>Being diagonal, $M$ must have all diagonal entries equal, since conjugating by a permutation matrix permutes the diagonal entries.</p></li>
</ul>
|
238,809 | <p><a href="https://i.stack.imgur.com/W5ILn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W5ILn.jpg" alt="enter image description here" /></a>
How to construct a tree like <a href="https://i.stack.imgur.com/XKYl9.png" rel="nofollow noreferrer">this</a>? I was looking <code>CompleteKaryTree</code> initially, there are some similarities overall, but it's still different.</p>
<pre><code>CompleteKaryTree[5, 2, GraphLayout -> "LayeredEmbedding", AspectRatio -> 1/4]
</code></pre>
<p><a href="https://i.stack.imgur.com/xrW63.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xrW63.png" alt="enter image description here" /></a></p>
<p>Another way, I've generated the coordinates of all the points, but I don't know how to connect them</p>
<pre><code>n=4;
pts=Join @@ Table[{1/2 (1+(n-j)!)+(i-1) (n-j)!,n-j-1},{j,0,n},{i,FactorialPower[n,j]}];
Graphics[{Point@pts}, ImageSize->Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/L3hGw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L3hGw.png" alt="enter image description here" /></a></p>
| Daniel Huber | 46,318 | <p>We create the points recursively. Given the numbers of siblings in every generation by e.g. <code>ngen=ngen = {4, 3, 2, 1}</code>, in a first step we create 4 descendants. Then for every sibling we create another 3 descendants, then 2, then 1. Finally we use <code>TreePlot</code>. You may play with labels, I simply number the edges here:</p>
<pre><code>ngen = {4, 3, 2, 1};
p = 0;
Clear[step];
step[n0_,
gen_] := (next =
Table[n0 \[UndirectedEdge] ++p, ngen[[gen]]]; {next,
If[gen == Length@ngen, Nothing[], step[#[[2]], gen + 1] & /@ next]})
tr = step[0, 1];
TreePlot[tr // Flatten(*,0,VertexLabels\[Rule]"Name"*),
EdgeLabels -> "Index"]
</code></pre>
<p><a href="https://i.stack.imgur.com/BvE1T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BvE1T.png" alt="enter image description here" /></a></p>
|
238,809 | <p><a href="https://i.stack.imgur.com/W5ILn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W5ILn.jpg" alt="enter image description here" /></a>
How to construct a tree like <a href="https://i.stack.imgur.com/XKYl9.png" rel="nofollow noreferrer">this</a>? I was looking <code>CompleteKaryTree</code> initially, there are some similarities overall, but it's still different.</p>
<pre><code>CompleteKaryTree[5, 2, GraphLayout -> "LayeredEmbedding", AspectRatio -> 1/4]
</code></pre>
<p><a href="https://i.stack.imgur.com/xrW63.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xrW63.png" alt="enter image description here" /></a></p>
<p>Another way, I've generated the coordinates of all the points, but I don't know how to connect them</p>
<pre><code>n=4;
pts=Join @@ Table[{1/2 (1+(n-j)!)+(i-1) (n-j)!,n-j-1},{j,0,n},{i,FactorialPower[n,j]}];
Graphics[{Point@pts}, ImageSize->Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/L3hGw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L3hGw.png" alt="enter image description here" /></a></p>
| Szabolcs | 12 | <p>Using my package <a href="http://szhorvat.net/mathematica/IGraphM/" rel="nofollow noreferrer">IGraph/M</a>,</p>
<pre><code>Needs["IGraphM`"]
IGSymmetricTree[{4, 3, 2, 1}, GraphLayout -> "LayeredEmbedding"]
</code></pre>
<p><a href="https://i.stack.imgur.com/qyxFc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qyxFc.png" alt="enter image description here" /></a></p>
<p>See <a href="http://szhorvat.net/mathematica/IGDocumentation/#igsymmetrictree" rel="nofollow noreferrer">its documentation</a>, which shows precisely the tree you are asking for.</p>
|
3,466,707 | <p>In that problem</p>
<p><span class="math-container">$$\lim\limits_{x \to \infty} \left(1-\frac{1}{x}\right)^\left(e^x\right)$$</span></p>
<p>I use <span class="math-container">$\ln$</span>, then it gave me <span class="math-container">$0\times\infty$</span> indeterminate, then I use L'Hospital Rule but I cannot reach the any answer.</p>
<p>Sorry for grammatical mistakes, my native language is not English. </p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>Let</p>
<p><span class="math-container">$$ \left(1-\dfrac{1}{x}\right)^{(e^x)}= \left[\left(1-\dfrac{1}{x}\right)^{x}\right]^{\dfrac{e^x}x}$$</span></p>
<p>and use standard limits.</p>
<p>As an alternative we can use that </p>
<p><span class="math-container">$$\left(1-\frac1x\right)^{e^x}=e^{e^x\log \left(1-\frac1x\right)}=e^{\frac{e^x}{x}\frac{\log \left(1-\frac1x\right)}{\frac1x}}$$</span> </p>
<p>and use l’Hospital separately for the two parts</p>
<ul>
<li><p><span class="math-container">$\frac{e^x}{x}$</span></p></li>
<li><p><span class="math-container">$\frac{\log \left(1-\frac1x\right)}{\frac1x}$</span></p></li>
</ul>
|
1,524,109 | <p>Could anyone help me with this proof without using determinant? I tried two ways. </p>
<blockquote>
<p>Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix. </p>
</blockquote>
<p>I then get stuck. The other way was to prove by contradiction, and I failed too. </p>
| Hippalectryon | 150,347 | <p>If the sum of the rows is zero, <a href="https://math.stackexchange.com/questions/347408/prove-that-if-the-sum-of-each-row-of-a-equals-s-then-s-is-an-eigenvalue-o?rq=1">then the matrix has the eigenvalue</a> $0$. As a result its $\ker$ is of dimension $\ge1$, i.e. there is a nonzero solution to $AX=0$, hence it's noninvertible</p>
|
1,524,109 | <p>Could anyone help me with this proof without using determinant? I tried two ways. </p>
<blockquote>
<p>Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix. </p>
</blockquote>
<p>I then get stuck. The other way was to prove by contradiction, and I failed too. </p>
| Nitin Uniyal | 246,221 | <p><strong>Hint:</strong> For a matrix $A$ having such a property has vector $(1,1,...,1)$ in its kernel thereby giving $\dim(\ker A)>0$.</p>
|
1,524,109 | <p>Could anyone help me with this proof without using determinant? I tried two ways. </p>
<blockquote>
<p>Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix. </p>
</blockquote>
<p>I then get stuck. The other way was to prove by contradiction, and I failed too. </p>
| user548010 | 548,010 | <p>I'm not too sure but my reasoning is that if the sum of a row is 0, then the rows of the matrix A are linearly dependent because they are a linear combination. If the rows of A are linearly dependent then the columns of A transpose is linearly dependent. Therefore matrix A is not invertible by the Invertible Matrix Theorem and the determinant is equal to 0. </p>
|
81,019 | <p>In my youthful naiveté (i.e. fifteen minutes ago), I was trying to prove that if $X$ was a random variable with zero mean, then $XY$ also had zero mean for any random variable $Y$. My proof used integration by parts:</p>
<p>$\int_{\Omega} x(\omega)y(\omega)f(\omega)d\omega = y(\omega) \int_{\Omega}x(\omega)f(\omega)d\omega - \int_{\Omega}[\int_{\Omega}x(\omega)f(\omega)d\omega] dy(\omega)$</p>
<p>Since $X$ has zero mean, $\int_{\Omega} x(\omega)f(\omega)d\omega$ should be $0$, so the entire expression should be $0$. However, I don't think this is true. In fact, if $X \sim N(0, 1)$, then $E(X) = 0$ but $E(X^2) = 1$. What am I doing wrong?</p>
<p>Thanks!</p>
| Ashok | 16,921 | <p>If $X$ and $Y$ have $0$ mean then $\mathbb{E}(XY)=0$ if $X$ and $Y$ are independent; otherwise one cannot conclude anything about $\mathbb{E}(XY)$.
For example, take $X$ to be a random variable with mean $0$ and variance $1$ and $Y=X$.</p>
|
81,019 | <p>In my youthful naiveté (i.e. fifteen minutes ago), I was trying to prove that if $X$ was a random variable with zero mean, then $XY$ also had zero mean for any random variable $Y$. My proof used integration by parts:</p>
<p>$\int_{\Omega} x(\omega)y(\omega)f(\omega)d\omega = y(\omega) \int_{\Omega}x(\omega)f(\omega)d\omega - \int_{\Omega}[\int_{\Omega}x(\omega)f(\omega)d\omega] dy(\omega)$</p>
<p>Since $X$ has zero mean, $\int_{\Omega} x(\omega)f(\omega)d\omega$ should be $0$, so the entire expression should be $0$. However, I don't think this is true. In fact, if $X \sim N(0, 1)$, then $E(X) = 0$ but $E(X^2) = 1$. What am I doing wrong?</p>
<p>Thanks!</p>
| Ilya | 5,887 | <p>The formula you used in your answer is incorrect and not only because of using integration by parts. It seems that you're mixing definitions of an expectation because sometimes one think of three definitions while there is only one. Just for the case, let us recall some theory. Maybe it seems boring - but I hope if you put a bit of attention to it, you will benefit.</p>
<p>Let us talk only about real-valued random variables. Given <em>some</em> probability space $(\Omega,\mathscr F,\mathsf P)$ the random variable $X$ is a <em>measurable</em> function
$$
X:(\Omega,\mathscr F)\to(\mathbb R,\mathscr B)
$$
where $\mathscr B$ is the Borel $\sigma$-algebra of reals. That means that for any Borel set (i.e. any open or closed set) $B$ the probability $\mathsf P\{X\in B\}$ is well-defined because measurability of $X$ implies
$$
\{\omega:X(\omega)\in B\}\in\mathscr F
$$ and the probability measure $\mathsf P$ is defined for any element of $\mathscr F$. Due to this reason, we can define the expectation of $X$ as just its <a href="http://en.wikipedia.org/wiki/Lebesgue_integration" rel="nofollow">Lebesgue integral</a>:
$$
\mathsf E[X] = \int\limits_\Omega X(\omega)\mathsf P(d\omega).
$$
In such construction you cannot apply any kind of intergartion by parts in general, since there is no a right analogue for the derivative. On the other hand, you can recall that if we define
$$
\mathsf Q_X(B) = \mathsf P\{X\in B\}
$$
for any $B\in\mathscr B$ then $\mathsf Q_X$ will be a probability measure on $(\mathbb R,\mathscr B)$ called <em>the distribution</em> of $X$. Note that $\mathsf Q_X$ is not the same as $\mathsf P$: first of all, they are defined on different spaces; second, $\mathsf Q_X$ certainly depends on $X$ so probability measure $\mathsf P$ can produce a lot of distributions if we vary $X$.</p>
<p>The benefit from using distributions is that calculation of expectations becomes easier:
$$
\mathsf E[X] = \int\limits_\Omega X(\omega)\mathsf P(d\omega) = \int\limits_\mathbb R x\,\mathsf Q_X(dx)\quad (1)
$$
where $x$ and $\omega$ are just integration variables so that we can call them $y$ and $\zeta$ as an example - it does not matter. Note that going from $\omega$ to $x$ is just a sort of change of variables.</p>
<p>As we already discussed, there are lots of distributions so we should find a kind of common denominator for most of them to be able to calculate the last integral in $(1)$. Such a denominator was chosen to be Lebesgue measure (which is the length of an interval). Further, if $\mathsf Q_X$ has a density w.r.t. to Lebesgue measure, i.e.
$$
\mathsf Q_X ((a,b]) = \int\limits_a^b f_X(x)\,dx
$$
for some function $f$ then
$$
\mathsf E[X] = \int\limits_\mathbb R x\,\mathsf Q_X(dx) = \int\limits_{-\infty}^\infty x\cdot f_X(x)dx\quad (2)
$$
which is maybe the most popular formula to calculate expectations.</p>
<p>Now, what is $\Omega$ if we one to describe only a single random variable? Ok, it appears that in this case that set of reals has enough randomness to model <em>single</em> real-valued random variable: no matter of the original construction we can put $\Omega = \mathbb R, \mathscr F = \mathscr B$ and $\mathsf P = \mathsf Q_X,X = \operatorname{Id}$, i.e. for any $\omega\in \mathbb R$ it holds that $X(\omega) = \omega$. This construction immediately implies $(1)$ since spaces, measures and functions are the same. Bur it does not imply your formula in which you have <em>multiple</em> random variables which cannot be always modeled by taking $\Omega = \mathbb R$.</p>
<p>When you're dealing with two real random variables $X,Y$, you should think not of them as a joint random variable $Z = (X,Y)$ with values in $\mathbb R^2$ because of possible dependence between $X$ and $Y$. That's why if $\mathsf Q_{XY}$ is their joint distribution and $f_{XY}$ is a density of this distribution then you have
$$
\mathsf E[XY] = \int\limits_{\mathbb R^2}xy\,\mathsf Q_{XY}(dxdy) = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty xy \cdot f_{XY}(x,y)\,dx\,dy
$$
and so if you want to apply integration by parts in this integral you should do it in a different way, not like you've done in your question.</p>
|
2,053,017 | <p>The first thing I was trying to do is to substitute $a$ with something like $1 \over b$ or $1 + \frac{1}{b}$, but I got no results. Then I tried to use Squeeze theorem here, but I also got confused because, for example, $(a + 1)^{1/n}$ is not a decreasing function. </p>
| Thomas Andrews | 7,933 | <p>Let $b_n = a^{1/n}$. </p>
<p>Assume $a\neq 1$ ($a=1$ is easy.) </p>
<p>Use that $$\frac{1-b_n}{1-a} =\frac{1-b_n}{1-b_n^n}= \frac{1}{1+b_n+b_n^2+\cdots + b_n^{n-1}}$$
Use this to get a good lower and upper bound on $1-b_n$ for use with the squeeze theorem.</p>
<p>--</p>
<p>The Mean Value Theorem version is that $\frac{b_n^n-1}{b_n-1} = nc^{n-1}$ for some $c$ in $[b_n,1]$, and again using this to get upper and lower bounds on $1-b_n$.</p>
|
194,312 | <p>I'm preparing myself to a combinatorics test. A part of it will concentrate on the pigeonhole principle. Thus, I need some hard to very hard problems in the subject to solve.
I would be thankful if you can send me links\books\or just a lone problem.</p>
| Austin Mohr | 11,245 | <p>There are many such problems (with solutions!) in the <a href="https://math.stackexchange.com/search?q=pigeonhole">Math.SE archives</a>. You can try other keywords in the search box at the top right of the page to look for more problems.</p>
|
194,312 | <p>I'm preparing myself to a combinatorics test. A part of it will concentrate on the pigeonhole principle. Thus, I need some hard to very hard problems in the subject to solve.
I would be thankful if you can send me links\books\or just a lone problem.</p>
| A.Schulz | 35,875 | <p>The famous <a href="http://en.wikipedia.org/wiki/Proofs_from_THE_BOOK" rel="nofollow">Proofs from the book</a> contains a chapter on <strong>Pigeon-hole and double counting</strong>. You can find there several cute applications of the pigeon-hole principle. </p>
|
194,312 | <p>I'm preparing myself to a combinatorics test. A part of it will concentrate on the pigeonhole principle. Thus, I need some hard to very hard problems in the subject to solve.
I would be thankful if you can send me links\books\or just a lone problem.</p>
| VividD | 121,158 | <hr>
<p><em>I will divide my answer into two parts: resources from internet, and resources from this very site.</em></p>
<hr>
<h2>Some resources on the internet</h2>
<p><a href="http://www.math.lsa.umich.edu/~hderksen/ProblemSolving/PS7.pdf" rel="noreferrer">This short paper</a> contains a lot of pigeonhole principle-related problems, both easy and hard ones, and both with and without solution.</p>
<p><a href="http://artofproblemsolving.com/wiki/index.php?title=Pigeonhole_Principle" rel="noreferrer">This web page</a> contains also a number of pigeonhole problems, from basic to very complex, with all solutions.</p>
<p>Take a look also at <a href="http://mindyourdecisions.com/blog/2008/11/25/16-fun-applications-of-the-pigeonhole-principle/" rel="noreferrer">these fun applications of the pigeonhole principle</a></p>
<hr>
<h2>Some problems from this site</h2>
<p>You can find a lot of interesting problems that are solved with pigeonhole principle on this site.</p>
<p><strong><em>Numbers</em></strong></p>
<p><a href="https://math.stackexchange.com/questions/282589/a-pigeonhole-principle-problem">101 positive integers placed on a circle</a></p>
<p><em>101 positive integers whose sum is 300 are placed on a circle. Prove that it is possible to choose some consecutive numbers from these numbers whose sum is equal to 200.</em></p>
<p><a href="https://math.stackexchange.com/questions/181609/choose-100-numbers-from-1200-one-less-than-16-prove-one-is-divisible-by-ano">Choose 100 numbers from 1~200 (one less than 16) - prove one is divisible by another!</a></p>
<p><em>Prove that if 100 numbers are chosen from the first 200 natural numbers and include a number less than 16, then one of them is divisible by another.</em></p>
<p><a href="https://math.stackexchange.com/questions/604635/of-any-52-integers-two-can-be-found-whose-difference-is-divisible-by-100">Of any 52 integers, two can be found whose difference of squares is divisible by 100</a></p>
<p><em>Prove that for any 52 integers two can always be found such that the difference of their squares is divisible by 100.</em></p>
<p><a href="https://math.stackexchange.com/questions/421057/given-n-numbers-prove-that-difference-of-at-least-one-of-the-pair-of-these-numb">Given n numbers, prove that difference of at least one pair of these numbers is divisible by n-1</a></p>
<p><em>Suppose you have a list of n numbers, n≥2. Let A be the set of differences of pairs of these n numbers. Prove or disprove that at least one element of A must be divisible by n−1.</em></p>
<p><a href="https://math.stackexchange.com/questions/155500/the-pigeonhole-principle-question">Proving an interesting feature of any $1000$ different numbers chosen from $\{1, 2, \dots,1997\}$</a></p>
<p><em>Assume you choose 1000 different numbers from the group {1,2,…,1997}. Prove that within the 1000 chosen numbers, there is a couple which sum is 1998.</em></p>
<p><strong><em>People and objects</em></strong></p>
<p><a href="https://math.stackexchange.com/questions/252427/discrete-mathematics-ice-cream-random-samples">Discrete Mathematics - Ice Cream random samples</a></p>
<p><em>Suppose there are 5 different types of ice cream you like. How many random samples ice cream must be eaten to guarantee that you have had at least 7 samples of one type?</em></p>
<p><a href="https://math.stackexchange.com/questions/320555/a-question-related-to-pigeonhole-principle">A question related to Pigeonhole Principle</a></p>
<p><em>In a room there are 10 people, and none of them is older than 60, and each is at least 1 year old. Prove that one can always find two groups of people (with no common person) such that the sum of ages is the same in both groups.</em></p>
<p><a href="https://math.stackexchange.com/questions/155200/combinatorics-the-pigeonhole-principle">combinatorics: The pigeonhole principle</a></p>
<p><em>Assume that in every group of 9 people, there are 3 in the same height. Prove that in a group of 25 people there are 7 in the same height.</em></p>
<p><a href="https://math.stackexchange.com/questions/121561/pigeonhole-principle-question">Pigeonhole Principle question</a></p>
<p><em>There is a row of 35 chairs. Find the minimum number of chairs that must be occupied such that there is a consecutive set of 4 or more occupied chairs.</em></p>
<p><a href="https://math.stackexchange.com/questions/40212/another-pigeonhole-principle-question">Another pigeonhole principle question</a></p>
<p><em>A course has seven elective topics, and students must complete exactly three of them in order to pass the course. If 200 students passed the course, show that at least 6 of them must have completed the same electives as each other.</em></p>
<p><strong><em>Geometry</em></strong></p>
<p><a href="https://math.stackexchange.com/questions/415679/pigeonhole-principle-for-a-triangle">Pigeonhole principle for a triangle</a></p>
<p><em>Consider a equilateral triangle of total area 1. Suppose 7 points are chosen inside. Show that some 3 points form a triangle of area $\leq\frac 14$.</em></p>
<p><a href="https://math.stackexchange.com/questions/479909/pigeonhole-principle-100-points-in-13-times18-rectangle">Arrangement of $100$ points inside $13\times18$ rectangle</a></p>
<p><em>Prove that you can't arrange 100 points inside a 13×18 rectangle so that the distance between any two points is at least 2.</em></p>
<p><a href="https://math.stackexchange.com/questions/156154/the-pigeonhole-principle-question-diamond">Pigeon principle question: Nine points in a diamond</a></p>
<p><em>A diamond (a parallelogram with equal sides) is given, and its sides are 2 cm long. The sharp angels are 60 degrees. If there are nine points inside the diamond, prove that there must be two of them so that the distance between them is at most 1 cm.</em></p>
<p><a href="https://math.stackexchange.com/questions/436630/pigeonhole-principle-and-a-decagon">Pigeonhole principle and a decagon</a></p>
<p><em>The numbers ${0,1,2,.....9}$ are randomly assigned to the vertices ${x_0,x_1,...x_9}$ of a decagon. Show that there are 3 consecutive vertices whose sum is at least 14.</em> </p>
<p><a href="https://math.stackexchange.com/questions/629969/polygon-and-pigeon-hole-principle-question">Polygon and Pigeon Hole Principle Question</a></p>
<p><em>Seven vertices are chosen in each of two congruent regular 16-gons. Prove that these polygons can be placed one atop another in such a way that at least four chosen vertices of one polygon coincide with some of the chosen vertices of the other one.</em></p>
<p><a href="https://math.stackexchange.com/questions/1223486/smallest-number-of-points-on-plane-that-guarantees-an-angle-of-at-most-18-circ/1243763#1243763">Smallest number of points on plane that guarantees existence of a small angle</a></p>
<p><em>What is the smallest number n, that in any arrangement of n points on the plane, there are three of them making an angle of at most 18∘?</em></p>
|
1,260,722 | <blockquote>
<p>Prove that <span class="math-container">$f=x^4-4x^2+16\in\mathbb{Q}[x]$</span> is irreducible.</p>
</blockquote>
<p>I am trying to prove it with Eisenstein's criterion but without success: for <strong>p=2</strong>, it divides <strong>-4</strong> and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find <span class="math-container">$f(x\pm c)$</span> which is irreducible:</p>
<blockquote>
<p><span class="math-container">$f(x+1)=x^4+4x^3+2x^2-4x+13$</span>, but 13 has the divisors: <strong>1 and 13</strong>, so don't exist a prime number <strong>p</strong> such that to apply the first condition: <span class="math-container">$p|a_i, i\ne n$</span>; the same problem for <span class="math-container">$f(x-1)=x^4+...+13$</span></p>
<p>For <span class="math-container">$f(x+2)=x^4+8x^3+20x^2+16x+16$</span> is the same problem from where we go, if we set <strong>p=2</strong>, that means <span class="math-container">$2|8, 2|20, 2|16$</span>, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for <strong>x-2</strong></p>
</blockquote>
<p>Now I'll verify for <span class="math-container">$f(x\pm3)$</span>, but I think it will be fall... I think if I verify all constant <span class="math-container">$f(x\pm c)$</span> it doesn't work with this method... so have any idea how we can prove that <span class="math-container">$f$</span> is irreducible?</p>
| k1.M | 132,351 | <p>Suppose that $f$ is reducible, observe that $f$ is monic, hence $f$ can be written as the product of two polynomials $g$ and $h$ of degree at least one and with integer coefficients.<br>
Now observe that $f$ has no integer root, because with rational root theorem an integer root to $f$ must be a divisor of $f(0)=16$ and you can check it's cases.<br>
So $f$ has no linear factor, hence $g$ and $h$ are two monic polynomials with degree two, with letting
$$f(x)=g(x)h(x)=(x^2+ax+b)(x^2+cx+d)=x^4-4x^2+16$$
We get
$$
a=-c,b+d+ac=-4,ad+bc=0,bd=16
$$
Now if $a\neq0$, then $b=d$,$b+d+ac=-4$, so $b^2=16$, $2b-a^2=\pm8-a^2=-4$, which contradicts.<br>
Now if $a=-c=0$, then $bd=16$, and $b+d=-4$ which again contradicts.</p>
|
1,260,722 | <blockquote>
<p>Prove that <span class="math-container">$f=x^4-4x^2+16\in\mathbb{Q}[x]$</span> is irreducible.</p>
</blockquote>
<p>I am trying to prove it with Eisenstein's criterion but without success: for <strong>p=2</strong>, it divides <strong>-4</strong> and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find <span class="math-container">$f(x\pm c)$</span> which is irreducible:</p>
<blockquote>
<p><span class="math-container">$f(x+1)=x^4+4x^3+2x^2-4x+13$</span>, but 13 has the divisors: <strong>1 and 13</strong>, so don't exist a prime number <strong>p</strong> such that to apply the first condition: <span class="math-container">$p|a_i, i\ne n$</span>; the same problem for <span class="math-container">$f(x-1)=x^4+...+13$</span></p>
<p>For <span class="math-container">$f(x+2)=x^4+8x^3+20x^2+16x+16$</span> is the same problem from where we go, if we set <strong>p=2</strong>, that means <span class="math-container">$2|8, 2|20, 2|16$</span>, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for <strong>x-2</strong></p>
</blockquote>
<p>Now I'll verify for <span class="math-container">$f(x\pm3)$</span>, but I think it will be fall... I think if I verify all constant <span class="math-container">$f(x\pm c)$</span> it doesn't work with this method... so have any idea how we can prove that <span class="math-container">$f$</span> is irreducible?</p>
| Adhvaitha | 228,265 | <p>Below is an explicit proof. Note that $x^4-4x^2+16 = (x^2-2)^2 + 12$, which clearly has no real root. Hence, the only possible way to reduce $x^4-4x^2+16$ over $\mathbb{Q}$ is $(x^2+ax+b)(x^2+cx+d)$. However, the roots of $x^4-4x^2+16$ are $x = \pm \sqrt{2 \pm i\sqrt{12}}$, which are all complex numbers. Since complex roots occur in conjugate pairs, $\sqrt{2 \pm i\sqrt{12}}$ must be the roots of one of the factored quadratic. Hence, the factored quadratic must be
$$x^2-(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2-2\sqrt3 x+4$$
The other factored quadratic must be
$$x^2+(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2+2\sqrt3 x+4$$
Hence, $x^4-4x^2+16$ is irreducible over $\mathbb{Q}$.</p>
|
1,260,722 | <blockquote>
<p>Prove that <span class="math-container">$f=x^4-4x^2+16\in\mathbb{Q}[x]$</span> is irreducible.</p>
</blockquote>
<p>I am trying to prove it with Eisenstein's criterion but without success: for <strong>p=2</strong>, it divides <strong>-4</strong> and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find <span class="math-container">$f(x\pm c)$</span> which is irreducible:</p>
<blockquote>
<p><span class="math-container">$f(x+1)=x^4+4x^3+2x^2-4x+13$</span>, but 13 has the divisors: <strong>1 and 13</strong>, so don't exist a prime number <strong>p</strong> such that to apply the first condition: <span class="math-container">$p|a_i, i\ne n$</span>; the same problem for <span class="math-container">$f(x-1)=x^4+...+13$</span></p>
<p>For <span class="math-container">$f(x+2)=x^4+8x^3+20x^2+16x+16$</span> is the same problem from where we go, if we set <strong>p=2</strong>, that means <span class="math-container">$2|8, 2|20, 2|16$</span>, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for <strong>x-2</strong></p>
</blockquote>
<p>Now I'll verify for <span class="math-container">$f(x\pm3)$</span>, but I think it will be fall... I think if I verify all constant <span class="math-container">$f(x\pm c)$</span> it doesn't work with this method... so have any idea how we can prove that <span class="math-container">$f$</span> is irreducible?</p>
| egreg | 62,967 | <p>The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square:
\begin{align}
x^4-4x^2+16
&=x^4+8x^2+16-12x^2\\
&=(x^2+4)^2-(\sqrt{12}\,x)^2\\
&=(x^2-\sqrt{12}\,x+4)(x^2+\sqrt{12}\,x+4)
\end{align}
These two polynomials have negative discriminant (no need to verify it) and so they're irreducible in $\mathbb{R}[x]$. If the given polynomial were reducible over the rationals, the two factorizations in $\mathbb{Q}[x]$ and $\mathbb{R}[x]$ would coincide.</p>
<p>Therefore the given polynomial is irreducible over the rationals.</p>
<hr>
<p>What's the general rule? Suppose you have $x^4+px^2+q$, with $p,q$ integers and $p^2-4q<0$ (so $q>0$). Write $q=r^2$, with $r>0$ (it need not be integer), and
$$
x^4+px^2+q=x^4+2rx^2+r^2-(2r-p)x^2
$$
Note that $2r-p>0$: it's obvious if $p<0$; if $p\ge0$ it's the same as $4q>p^2$, which is true by hypothesis. Then
$$
x^4+px^2+q=(x^2-\sqrt{2r-p}\,x+r)(x^2+\sqrt{2r-p}\,x+r)
$$
is the decomposition of the polynomial in $\mathbb{R}[x]$. It is in $\mathbb{Q}[x]$ if and only if $\sqrt{q}$ and $\sqrt{2\sqrt{q}-p}$ are integers.</p>
<p>For example, $q=4$ and $p=0$ is a case. For $q=16$ we need $8-p$ to be a square, so $q=16$ and $p=4$ is another case.</p>
|
2,598 | <p>Should Wolfram Alpha Notebook questions be considered on-topic?</p>
<p>Here's an example: <a href="https://mathematica.stackexchange.com/questions/240780/calculating-double-integral-bounded-by-domain-in-wolfram-alpha-notebook">Calculating double integral bounded by domain in Wolfram Alpha Notebook</a></p>
<p>Here are some related meta Q&A:</p>
<ul>
<li><p><a href="https://mathematica.meta.stackexchange.com/questions/68/other-wri-product-discussion">Other WRI product discussion?</a></p>
</li>
<li><p><a href="https://mathematica.meta.stackexchange.com/questions/265/are-questions-about-doing-symbolic-math-in-wolfram-alpha-on-topic-here">Are questions about doing symbolic math in Wolfram Alpha on topic here?</a></p>
</li>
</ul>
<p><a href="https://www.wolfram.com/wolfram-alpha-notebook-edition/" rel="nofollow noreferrer">Wolfram Alpha Notebooks</a> are a new WRI product that hybridizes W|A and Mathematica. I looked only briefly, but it resembles a <em>Mathematica</em> notebook in which the only valid input starts with single-equals (probably without having to type <code>=</code>), though the sample inputs are sometimes interpreted differently in the examples shown than in my <em>Mathematica</em>.</p>
<p>(For those who may not know, it was decided to consider questions about <a href="https://mathematica.stackexchange.com/help/on-topic">W|A off-topic</a>.)</p>
| Szabolcs | 12 | <p>Thanks to Penelope Benenati, it came to light that Wolfram|Alpha Notebook Edition does not behave like Wolfram|Alpha. Unlike Wolfram|Alpha (which frequently interprets input differently from Mathematica), it appears to be able to understand Wolfram Language syntax.</p>
<p>Based on her screenshots, it appears to me that W|A Notebook Edition might work as follows (please correct me if I got any of this wrong):</p>
<ul>
<li>It contains a Mathematica kernel which can evaluate Wolfram Language code locally.</li>
<li>It does not allow inputting Wolfram Language code directly. Instead, it interprets free-form input using Wolfram Alpha, in the same way as Mathematica does when we start an input using "=". Then it shows its interpretation as the actual Wolfram Language code that will be evaluated.</li>
<li>In most cases (in all examples shown so far), if the input happens to be valid Wolfram Language code, it interprets it literally. It is unclear to me if this is <em>always</em> the case of only sometimes.</li>
</ul>
<p>Here is my updated proposal on what questions should be on- or off-topic:</p>
<ul>
<li><p>Naturally, any question about the Wolfram Language should be on-topic regardless of software used.</p>
</li>
<li><p>However, questions about how to coerce W|A Notebook Edition to interpret free-form input in a certain way should stay off-topic. This is because there is no clear specification of what free-form input is allowed to look like. It's also unclear how frequently its interpretation changes. For the sake of example, if W|A Notebook Edition happens to interpret "Sqrt[x]" not as <code>Sqrt[x]</code> but as <code>Surd[x,2]</code>, questions about how to get it to understand it as <code>Sqrt[x]</code> should continue to be off-topic.</p>
</li>
<li><p>If it is indeed the case that W|A Notebook Edition understands <em>most</em> Wolfram Language input with the exception of some specific functionality, it should be fine to ask how to solve a problem without using that functionality. For example, if some standard packages such as TriangleLink are not available, it is fine to ask how to solve a problem without TriangleLink. This is contingent on the asker being able to clearly explain what is and isn't available. In my view, this is no different from asking questions about the cloud which also does not support all functionality (but note that it does support <em>most</em> functionality and it is quite clear what isn't available and why).</p>
</li>
</ul>
<p>It is crucial that W|A Notebook Edition explicitly shows the Wolfram Language code that it is evaluating. This ensures that there won't be any debates about differences in how W|A Notebook Edition and Mathematica interprets input. Without this certainly, I would not be in favour of allow W|A Notebook Edition questions. But thanks to this feature, it is easy to separate the interpretation of free-form input (questions about this should be off-topic) from the usage of Wolfram Language (which has always been on-topic, even for subsets of the Wolfram Language).</p>
<hr />
<p>The question that triggered this debate is the following (now closed):</p>
<p><a href="https://mathematica.stackexchange.com/q/246519/12">Plotting a complete graph with a given image as vertices</a></p>
<p>According to what I wrote above, I still consider this question to be off-topic. Notice that the asker mentions that the following code did not work:</p>
<pre><code>g = CompleteGraph[8, VertexShape-> Import["~/google-chrome-yellow-png-image-69499.png"]]
</code></pre>
<p>However, this is perfectly valid Wolfram Language code that works in Mathematica. Neither the asker, nor any of the commenters seem to be certain <em>why</em> W|A Notebook Edition does not accept this input, other than the usual explanation that it accept unspecified free-form input, and it doesn't happen to consider this one valid.</p>
<p>Also notice this screenshot from the question:</p>
<p><img src="https://i.stack.imgur.com/3U9wX.png" alt="" /></p>
<p>Here we see valid Wolfram Language code, given as free-form input. W|A Notebook Edition fails to understand it as Wolfram Language code for unknown reasons. This is precisely the kind of problem which I believe should be off-topic here.</p>
<p>Finally, notice that it is completely unclear in what way the code in the accepted answer differs from the code that OP claims did not work. It probably has to do with the idiosyncrasies of how W|A interprets free-form input.</p>
<p>For these reasons, I did not vote to reopen the question.</p>
|
3,366,442 | <p>I am asked to figure out when this limit exists for polynomials <span class="math-container">$f$</span> and <span class="math-container">$g$</span>, if so what the limit is, and then to prove my findings. So far I have gathered that the limit exists if and only if <span class="math-container">$\deg f\leq \deg g$</span>, in which case the limit is the division of the leading coefficients (if <span class="math-container">$\deg f = \deg g$</span>) and otherwise <span class="math-container">$0$</span> (if <span class="math-container">$\deg g > \deg f$</span>).</p>
<p>I am stuck on proving this rigorously with the definition of a limit. This is my first course in real analysis and I find that most of our problems have quite intuitive results but the proofs are super involved. Hints are appreciated.</p>
| Kavi Rama Murthy | 142,385 | <p>Hint: <span class="math-container">$\frac { \sum\limits_{k=0}^{p} a_kx^{k}} {\sum\limits_{k=0}^{m} b_kx^{k}}= x^{p-m} \frac { \sum\limits_{k=0}^{p} a_kx^{k-p}} {\sum\limits_{k=0}^{m} b_kx^{k-m}}$</span> and <span class="math-container">$\frac { \sum\limits_{k=0}^{p} a_kx^{k-p}} {\sum\limits_{k=0}^{m} b_kx^{k-m}} \to \frac {a_p} {b_m}$</span> as <span class="math-container">$x \to \infty$</span>. </p>
<p>So you only have to see what happens to <span class="math-container">$x^{p-m}$</span> as <span class="math-container">$x \to \infty$</span> and this is very easy. </p>
|
742,160 | <p>Answer true or false to each of the following questions. If a statement is true, prove it. If a statement is false, give a counterexample.</p>
<ol>
<li>For all sets $A$,$B$ and $C$: IF $A ⊆ B$ and $A ⊆ C$, Then $A ⊆ (B ∩ C)$</li>
<li>For all sets $A$ and $B$, if $|A| \le |B|$, then $A ⊆ B$</li>
</ol>
| Ellya | 135,305 | <p>The best way to remember is to only remember one, then by elimination you know the other.</p>
<p>I choose to remember injective as follows:</p>
<p>Injections cure things, and you have one injection for one cure. I.e. one to one.</p>
|
742,160 | <p>Answer true or false to each of the following questions. If a statement is true, prove it. If a statement is false, give a counterexample.</p>
<ol>
<li>For all sets $A$,$B$ and $C$: IF $A ⊆ B$ and $A ⊆ C$, Then $A ⊆ (B ∩ C)$</li>
<li>For all sets $A$ and $B$, if $|A| \le |B|$, then $A ⊆ B$</li>
</ol>
| Hovercouch | 36,369 | <p>The way I remember it is that when you get a flu shot your entire body doesn't turn into a giant flu virus, because the needle is smaller than your arm is. Then you can easily remember surjection as "the other one".</p>
<p>Another one is that <strong>in</strong>-jections are <strong>in</strong>-ferior and <strong>su</strong>-rjections are <strong>su</strong>-perior. </p>
|
4,532,322 | <p>I was reading <em>"Ricci flow, an introduction"</em> by B. Chow and D. Knopf and I was stuck by the equation on Page 13. As below, <span class="math-container">$x,y,z>0$</span> satisfying
<span class="math-container">\begin{equation}
\begin{array}{ll}
\dfrac{\mathrm{d}x}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{y^2+z^2-x^2}{yz},\\
\dfrac{\mathrm{d}y}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+z^2-y^2}{xz},\\
\dfrac{\mathrm{d}z}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+y^2-z^2}{xy}.
\end{array}
\end{equation}</span>
We assume that <span class="math-container">$x(0)\geqslant y(0)\geqslant z(0)$</span>, then we can calculate that
<span class="math-container">\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=4(x-z)\frac{y^2-(x+z)^2}{xyz}.
\tag{$\ast$}
\end{equation}</span>
Then the author conclude that <span class="math-container">$x(t)\geqslant y(t)\geqslant z(t)$</span> persist for as long as the solution exists.</p>
<p>Here is my <strong>question</strong>: how can we judge <span class="math-container">$x(t)\geqslant y(t)\geqslant z(t)$</span> from <span class="math-container">$(\ast)$</span>? My attempt is
<span class="math-container">\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{ln}|x-z|)=4\frac{y^2-(x+z)^2}{xyz}.
\end{equation}</span>
But then what should I do?</p>
| K.defaoite | 553,081 | <p>It is shown <a href="https://math.stackexchange.com/questions/318169/determinant-of-matrix-exponential">here</a> that
<span class="math-container">$$\det(\exp\mathbf A)=\exp(\operatorname{tr}\mathbf A)$$</span>
But <span class="math-container">$\mathbf A\in \mathfrak{su}(2)\implies \operatorname{tr}\mathbf A=0$</span> which shows that
<span class="math-container">$$\det(\exp \mathbf A)=1$$</span>
As expected. There is no problem.</p>
|
4,532,322 | <p>I was reading <em>"Ricci flow, an introduction"</em> by B. Chow and D. Knopf and I was stuck by the equation on Page 13. As below, <span class="math-container">$x,y,z>0$</span> satisfying
<span class="math-container">\begin{equation}
\begin{array}{ll}
\dfrac{\mathrm{d}x}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{y^2+z^2-x^2}{yz},\\
\dfrac{\mathrm{d}y}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+z^2-y^2}{xz},\\
\dfrac{\mathrm{d}z}{\mathrm{d}t} \!\!\!\!\!&=-8+4\dfrac{x^2+y^2-z^2}{xy}.
\end{array}
\end{equation}</span>
We assume that <span class="math-container">$x(0)\geqslant y(0)\geqslant z(0)$</span>, then we can calculate that
<span class="math-container">\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(x-z)=4(x-z)\frac{y^2-(x+z)^2}{xyz}.
\tag{$\ast$}
\end{equation}</span>
Then the author conclude that <span class="math-container">$x(t)\geqslant y(t)\geqslant z(t)$</span> persist for as long as the solution exists.</p>
<p>Here is my <strong>question</strong>: how can we judge <span class="math-container">$x(t)\geqslant y(t)\geqslant z(t)$</span> from <span class="math-container">$(\ast)$</span>? My attempt is
<span class="math-container">\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}(\mathrm{ln}|x-z|)=4\frac{y^2-(x+z)^2}{xyz}.
\end{equation}</span>
But then what should I do?</p>
| orangeskid | 168,051 | <p><a href="https://en.wikipedia.org/wiki/Lie_product_formula" rel="nofollow noreferrer">Lie product formula</a> implies that if <span class="math-container">$\exp t A$</span>, <span class="math-container">$\exp t B$</span> are in a (closed) matrix group for all <span class="math-container">$t$</span>, then so is <span class="math-container">$\exp(A+B)$</span>. That should ease some of your concerns.</p>
|
4,202,451 | <p>I have been reading about the Miller-Rabin primality test. So far I think I got it except the part where the accuracy is stated.<br />
E.g from <a href="https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test#Accuracy" rel="nofollow noreferrer">wiki</a></p>
<blockquote>
<p>The error made by the primality test is measured by the probability
for a composite number to be declared probably prime. The more bases a
are tried, the better the accuracy of the test. It can be shown that
if n is composite, then at most 1⁄4 of the bases a are strong liars
for n. As a consequence, if n is composite then running k
iterations of the Miller–Rabin test will declare n probably prime with
a probability at most <span class="math-container">$4^{−k}$</span>.</p>
</blockquote>
<p>So if I understand correctly if we have a large number <span class="math-container">$N$</span> and if we have <span class="math-container">$k$</span> random witnesses then if none of them observes the non-primality of <span class="math-container">$N$</span>, then the probability that <span class="math-container">$N$</span> is <strong>not</strong> a prime is <span class="math-container">$1$</span> in <span class="math-container">$4^k$</span></p>
<p>What I am not clear is where does this <span class="math-container">$\frac{1}{4}$</span> come from.<br />
I understand we have <span class="math-container">$4$</span> conditions to be met (in order) i.e.:</p>
<ol>
<li><span class="math-container">$a \not\equiv 0 \mod N$</span></li>
<li><span class="math-container">$a^{N-1} \not\equiv 1 \mod N$</span></li>
<li><span class="math-container">$x^2 \equiv 1 \mod N$</span></li>
<li><span class="math-container">$x \equiv \pm 1 \mod N$</span></li>
</ol>
<p>The process is the following:<br />
In the above <span class="math-container">$a$</span> is the witness. We first check condition (1).<br />
If that passes we check condition (2).<br />
Do do that we start multiplying <span class="math-container">$a, a \cdot a, a\cdot a\cdot a ....$</span> until we calculate <span class="math-container">$a^{N-1}$</span>.<br />
Do do that efficiently we can use the squaring method. If in the process of the multiplication during squaring we encounter a number e.g. <span class="math-container">$x$</span> such that the <span class="math-container">$x^2 \equiv 1$</span> but <span class="math-container">$x \not\equiv 1$</span> and <span class="math-container">$x \not\equiv -1$</span>. (E.g <span class="math-container">$19^2 \equiv 1 \pmod {40}$</span> but <span class="math-container">$19 \not \equiv 1 \pmod {40}$</span> and <span class="math-container">$19 \not \equiv -1 \pmod {40}$</span>) then the conditions (3) and (4) fails otherwise we proceed multiplying.<br />
We check the final product for condition (2)</p>
<p>Does the <span class="math-container">$1/4$</span> mean that at most <span class="math-container">$1$</span> of these can be indicate a prime? If so, how is that validated?</p>
| egglog | 626,167 | <p>Since induction is required, consider this:
<span class="math-container">$\displaystyle\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=\prod_{i=1}^{n}\frac{n+i}{2i-3}\Big(\frac{(2n+1)(2n+2)}{(2n-1)(n+1)}\Big) \implies$</span><br />
<span class="math-container">$2^{n+1}(2n+1) = 2^n(2n-1) \frac{(2n+1)(2n+2)}{(2n-1)(n+1)}$</span> is true under assumption, which means assuming the first case is true, induction holds.</p>
|
2,008,653 | <p>According to Riemann (I think) the (exact) prime counting function is given by:
<span class="math-container">$$
\pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1}
$$</span>
with <span class="math-container">$ \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$</span> and <span class="math-container">$\rho$</span> running over <strong>all</strong> the zeros of the <span class="math-container">$\zeta$</span> function.</p>
<p>Why isn't this function used in general to calculate <span class="math-container">$\pi(x)$</span>? Shouldn't it be a great approximation if many zeros of the <span class="math-container">$\zeta$</span> function are used? Probably nowadays there are many zeros known? Until now I thought <span class="math-container">$\pi(x)$</span> is a very "hard" function, because the distribution of the primes is quite hard, but this explicit formula does not look that hard.</p>
<p>Thank you.</p>
| Raymond Manzoni | 21,783 | <p>I don't know a formal proof of the identity
$$\pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1}$$
either but the way this identity was obtained is pretty clear and in fact "hinted" in <a href="http://www.claymath.org/publications/riemanns-1859-manuscript" rel="nofollow noreferrer">Riemann's famous paper</a> (cf the german original and english translation linked at the right).<br>
The missing part appears to be proof of the convergence of the series over $\rho$ (supposing the non-trivial zeros sorted by increasing distance to the real axis). </p>
<p>The idea for the derivation (not proof !) is to start with von Mangoldt's proof that (cf <a href="https://books.google.com/books?id=ruVmGFPwNhQC&pg=PA48" rel="nofollow noreferrer">Edwards' book p.$48$</a>) :
$$\tag{2}f^*(x)=\operatorname{li}(x)-\sum_{\rho} \operatorname{li}(x^{\rho}),\quad(x>1)$$
(notation: $f$ may be replaced by $J$ or $\Pi$ or (confusingly) $\pi$ and $f^*$ appear as $f_0$)<br>
with $\ \displaystyle\operatorname{li}(x):=\int_2^x \frac{dt}{\log\,t}\,$ (Riemann's variant of the <a href="https://en.wikipedia.org/wiki/Logarithmic_integral_function" rel="nofollow noreferrer">logarithmic integral</a>)<br>
and with $\ \displaystyle f^*(x):=\sum_{p^k\le x}^{*}\frac 1k$ (the $^*$ means that the last term $\dfrac 1k$ must be replaced by $\dfrac 12\dfrac 1k\;$ if $p^k=x$)</p>
<p>linked to the <a href="https://en.wikipedia.org/wiki/Prime-counting_function" rel="nofollow noreferrer">prime-counting function</a> $\ \displaystyle\pi^*(x):=\sum_{p\le x}^{*}1\ $ by
$\ \displaystyle f^*(x)=\sum_{k>0} \frac{\pi^{*}\bigl(x^{1/k}\bigr)}k\qquad (3)$</p>
<p>In fact we need only to apply the <a href="https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula" rel="nofollow noreferrer">Möbius inversion formula</a> $\ \displaystyle\pi^{*}(x):=\sum_{n=1}^{\infty} \frac{\mu(k)}k f^*\bigl(x^{1/k}\bigr)\ $ to $(2)$ to get (with questionable convergence...) :
$$\tag{4}\boxed{\displaystyle\pi^*(x)=R(x)-\sum_{\rho} R(x^{\rho})},\quad(x>1)$$
Where Riemann's $\,\displaystyle R(x):=\sum_{n=1}^{\infty} \frac{\mu(k)}k \operatorname{li}\bigl(x^{1/k}\bigr)\,$ may be written as a <a href="http://mathworld.wolfram.com/GramSeries.html" rel="nofollow noreferrer">Gram series</a> (this part is an edit of my <a href="https://math.stackexchange.com/questions/269997/two-representations-of-the-prime-counting-function/282848#282848">more complete derivation</a>).</p>
<p>$$-$$</p>
<p>Back to your question "Why isn't this function used in general to calculate $\pi(x)$?". In fact it was used and in a rather successful ways as showed <a href="http://www.primefan.ru/stuff/primes/table.html#theory" rel="nofollow noreferrer">here</a> with some references and the <a href="http://www.math.uni-bonn.de/people/jbuethe/topics/AnalyticPiX.html" rel="nofollow noreferrer">history by Büthe here</a>. Note that $(1)$ had sometimes to be replaced by more effective variants :</p>
<ul>
<li>Lagarias and Odlyzko (1987) <a href="http://www.dtc.umn.edu/~odlyzko/doc/arch/analytic.pi.of.x.pdf" rel="nofollow noreferrer">"Computing $\pi(x)$: An Analytic Method"</a></li>
<li>Galway thesis (2004) <a href="https://books.google.fr/books/about/Analytic_Computation_of_the_Prime_counti.html?id=e92aHAAACAAJ" rel="nofollow noreferrer">"Analytic Computation of the Prime-Counting Function"</a>
(see too his <a href="http://www.math.uiuc.edu/~galway/PhD_Thesis/index.html" rel="nofollow noreferrer">homepage</a>)</li>
<li>Kotnik (2008) <a href="http://lbk.fe.uni-lj.si/pdfs/aicm2008.pdf" rel="nofollow noreferrer">"The prime-counting function and its analytic approximations"</a></li>
<li>Platt (2013) <a href="https://arxiv.org/abs/1203.5712" rel="nofollow noreferrer">"Computing π(x) Analytically"</a></li>
<li>Büthe (2015) <a href="https://arxiv.org/pdf/1410.7008.pdf" rel="nofollow noreferrer">"An improved analytic Method for calculating $\pi(x)$"</a></li>
</ul>
<p>Following animation used $(1)$ with increasing number of zeros to generate $\pi(x)$ :
<img src="https://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/picomp.gif" alt="animation"></p>
<p>(I created it for Matthew Watkins' neat page <a href="https://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/encoding1.htm" rel="nofollow noreferrer">" 'encoding' of the distribution of prime numbers by the nontrivial zeros of the Riemann zeta function"</a>)</p>
<p>It is important to understand that $\;\displaystyle R(x)-\sum_{\rho\ \text{real zero}}R(x^{\rho})$ is the initial smooth part. Without the non trivial zeros you would observe no steps at all! But for the steps to be visible you need enough zeros (see for example p.$12$ of Platt who used nearly $70$ billions zeros to compute $\pi(10^{24})$ and a comparison with the combinatorial methods used by Oliveira e Silva).</p>
|
3,630,482 | <p>Show that, if 13 divides <span class="math-container">$n^2$</span> + <span class="math-container">$3n$</span> + <span class="math-container">$51$</span> then 169 divides <span class="math-container">$21n^2$</span> + <span class="math-container">$89n$</span> + <span class="math-container">$44$</span></p>
<p>We have 13 <span class="math-container">$|$</span> <span class="math-container">$n^2$</span> + 3n + 51</p>
<p>Using some congruency rules, this becomes:</p>
<p>13 <span class="math-container">$|$</span> <span class="math-container">$n^2$</span> + <span class="math-container">$3n$</span> - <span class="math-container">$1$</span></p>
<p>Or 13 divides <span class="math-container">$n(n+3)$</span> + 1</p>
<p>At this point, I was feeling kinda lazy, so I just listed the factors of 13, added 1 to each and saw which one can be broken down into two numbers such that one is 3 less than the other instead of trying to look for a more elegant solution</p>
<p>I quite quickly arrived at n = 5 (5 × 5 + 3 = 40 = 39 + 1)</p>
<p>I plugged n = 5 into the other equation, and got something that's divisible by 169</p>
<p>Now, how do I do the final thing, which is to prove either that no other such value of n can be found, or if it can be found, it would satisfy the other condition as well?</p>
<p>Never mind, found it</p>
| gt6989b | 16,192 | <p>You region consists of some linear constraints, and you would like to solve the linear optimization problem to maximize <span class="math-container">$x$</span> subject to your constraints.</p>
<p>You can use, for example, the <a href="https://en.wikipedia.org/wiki/Simplex_algorithm" rel="nofollow noreferrer">Simplex Method</a> to do that.</p>
<p>There are also useful implementations in all conceivable programming languages I know, please specify if interested in anything specific. Here is one implementation for Python to get you started <a href="https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html" rel="nofollow noreferrer">scipy.optimize.linprog</a></p>
|
1,277,827 | <p>I came across a question from Munkres:
Problem 24.11: If A is a connected subspace of X, does it follow that IntA and BdA are connected? Does the converse hold?</p>
<p>Thinking about the converse, I know that in general, if Bd A is connected, it doesn't necessarily mean that A is connected (e.g. the rationals). But what if X is connected, A is closed and the Bd A is connected? Would that then imply A is also connected?</p>
| Matt Samuel | 187,867 | <p>A discrete space with more than one point is a trivial counterexample. The boundary of any subset is empty, hence connected, yet there exist disconnected closed subsets of the space (e.g. any subset with more than one point).</p>
<p>If the space is connected however, we may prove that the result is true as follows. Suppose $C$ is a closed subset such that $\mathrm{Bd}\ C$ is connected and suppose $C$ is disconnected. Then there exist disjoint, nonempty open sets $U,V$ separating $C$. Being connected, the boundary of $C$ lies entirely within exactly one of $U$ and $V$, say it is $V$. Then it follows that $C-V=(\mathrm{Int}\ C)\cap U$. The left hand side is clearly closed while the right hand side is clearly open, thus there is a proper nonempty subset of $X$ that is both open and closed, contradicting connectedness.</p>
|
263,718 | <p>I have 2d data in the form <code>{x,y,f(x,y)}</code> which is randomly stored. the random <code>{x,y}</code> for specific shape (square for simplification here) can be created as</p>
<pre><code>Regn = {{-\[Pi], -\[Pi]}, {-\[Pi], \[Pi]}, {\[Pi], \[Pi]}, {\[Pi], -\
\[Pi]}, {-\[Pi], -\[Pi]}};
xylist=Select[RegionMember[Polygon@(Regn)]][
Join @@ CoordinateBoundsArray[CoordinateBounds@(Regn), Into[50]]]
</code></pre>
<p>then these points are passed to a function <code>f</code> here it is just <code>sin</code> and we get</p>
<pre><code>datxy = Table[{xylist[[i, 1]], xylist[[i, 2]],
Sin[xylist[[i, 1]] xylist[[i, 2]]]}, {i, 1, Length[xylist]}];
ListDensityPlot[datxy, ColorFunction -> "TemperatureMap"]
</code></pre>
<p><a href="https://i.stack.imgur.com/jR7l8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jR7l8.png" alt="enter image description here" /></a></p>
<p>my question is how can I plot the x or y derivative of such data?</p>
<p><strong>Update</strong></p>
<p>here I will compare the solution provided by the answers which indicate that the built-in interpolation gives bad results compared to the method provided by @Shin Kim</p>
<p><a href="https://i.stack.imgur.com/k8xlG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k8xlG.png" alt="enter image description here" /></a></p>
<p>I used <code>PlotPoints->50</code> in <code>DensityPlot</code> for the interpolated data</p>
| Daniel Huber | 46,318 | <p>You can first calculate an interpolation of your data. As your arguments are not on a grid, only Interpolation order of 1 can be used:</p>
<pre><code>fun = Interpolation[datxy, InterpolationOrder -> 1];
</code></pre>
<p><a href="https://i.stack.imgur.com/urQCf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/urQCf.png" alt="enter image description here" /></a></p>
<p>With a function we may now calculate the derivatives like:</p>
<pre><code>derx[x_, y_] = D[fun[x, y], x]
</code></pre>
<p><a href="https://i.stack.imgur.com/xHkGD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xHkGD.png" alt="enter image description here" /></a></p>
<pre><code>dery[x_, y_] = D[fun[x, y], y]
</code></pre>
<p><a href="https://i.stack.imgur.com/bgwoq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bgwoq.png" alt="enter image description here" /></a></p>
|
263,718 | <p>I have 2d data in the form <code>{x,y,f(x,y)}</code> which is randomly stored. the random <code>{x,y}</code> for specific shape (square for simplification here) can be created as</p>
<pre><code>Regn = {{-\[Pi], -\[Pi]}, {-\[Pi], \[Pi]}, {\[Pi], \[Pi]}, {\[Pi], -\
\[Pi]}, {-\[Pi], -\[Pi]}};
xylist=Select[RegionMember[Polygon@(Regn)]][
Join @@ CoordinateBoundsArray[CoordinateBounds@(Regn), Into[50]]]
</code></pre>
<p>then these points are passed to a function <code>f</code> here it is just <code>sin</code> and we get</p>
<pre><code>datxy = Table[{xylist[[i, 1]], xylist[[i, 2]],
Sin[xylist[[i, 1]] xylist[[i, 2]]]}, {i, 1, Length[xylist]}];
ListDensityPlot[datxy, ColorFunction -> "TemperatureMap"]
</code></pre>
<p><a href="https://i.stack.imgur.com/jR7l8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jR7l8.png" alt="enter image description here" /></a></p>
<p>my question is how can I plot the x or y derivative of such data?</p>
<p><strong>Update</strong></p>
<p>here I will compare the solution provided by the answers which indicate that the built-in interpolation gives bad results compared to the method provided by @Shin Kim</p>
<p><a href="https://i.stack.imgur.com/k8xlG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k8xlG.png" alt="enter image description here" /></a></p>
<p>I used <code>PlotPoints->50</code> in <code>DensityPlot</code> for the interpolated data</p>
| Hugh | 12,558 | <p>Here is my suggestion for interpolation. I start by making your data.</p>
<pre><code>Regn = {{-π, -π}, {-π, π}, {π, π}, {π, -\
π}, {-π, -π}};
xylist = Select[RegionMember[Polygon@(Regn)]][
Join @@ CoordinateBoundsArray[CoordinateBounds@(Regn), Into[50]]];
datxy = Table[{xylist[[i, 1]], xylist[[i, 2]],
Sin[xylist[[i, 1]] xylist[[i, 2]]]}, {i, 1, Length[xylist]}];
ListDensityPlot[datxy, ColorFunction -> "TemperatureMap"]
</code></pre>
<p><a href="https://i.stack.imgur.com/KQaur.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KQaur.png" alt="enter image description here" /></a></p>
<p>I extract the coordinate points from your data and make a mesh.</p>
<pre><code>pts = datxy /. {x_, y_, z_} :> {x, y};
Needs["NDSolve`FEM`"]
mesh = ToElementMesh[pts];
</code></pre>
<p>Now I interpolate the data and we can use the interpolated data as if it was an equation or function.</p>
<pre><code>int = ElementMeshInterpolation[{mesh}, datxy[[All, 3]]];
Plot3D[int[x, y], {x, y} ∈ mesh]
</code></pre>
<p><a href="https://i.stack.imgur.com/91Hbm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/91Hbm.png" alt="enter image description here" /></a></p>
<p>To take the derivative we use the usual derivative operation.</p>
<pre><code>dx = Head[D[int[x, y], x]];
Plot3D[dx[x, y], {x, y} ∈ mesh]
</code></pre>
<p><a href="https://i.stack.imgur.com/IWW8E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IWW8E.png" alt="enter image description here" /></a></p>
<p>Possibly we need some smoothing after taking the derivative but that's another question.
Hope that helps.</p>
|
401,967 | <p>This question is about logical complexity of sentences in third order arithmetic. See <a href="https://en.wikipedia.org/wiki/Arithmetical_hierarchy" rel="nofollow noreferrer">Wikipedia</a> for the basic concepts.</p>
<p>Recall that the Continuum Hypothesis is a <span class="math-container">$\Sigma^2_1$</span> sentence. Furthermore (loosely speaking) it can't be reduced to a <span class="math-container">$\Pi^2_1$</span> sentence, as stated in <a href="https://mathoverflow.net/a/218649/170446">Emil Jeřábek's answer to <em>Can we find CH in the analytical hierarchy?</em></a>.</p>
<p>Is there an example of a <span class="math-container">$\Sigma^2_2$</span> sentence with no known reduction to a <span class="math-container">$\Pi^2_2$</span> sentence? (Equivalently, a <span class="math-container">$\Pi^2_2$</span> sentence with no known reduction to a <span class="math-container">$\Sigma^2_2$</span> sentence.) I mean that there should be no known reduction even under large cardinal assumptions.</p>
<p>I'd prefer an example that's either famous or easy to state. But to begin, any example will do.</p>
<p><em>Update:</em> Sentences such as "<span class="math-container">$\mathfrak{c} \leqslant \aleph_2$</span>" and "<span class="math-container">$\mathfrak{c}$</span> is a successor cardinal" are <span class="math-container">$\Delta^2_2$</span>, meaning that they're simultaneously <span class="math-container">$\Sigma^2_2$</span> and <span class="math-container">$\Pi^2_2$</span>. The reason is that each such sentence (and also its negation) can be expressed in the form "<span class="math-container">$\mathbb{R}$</span> has a well-ordering <span class="math-container">$W$</span> such that <span class="math-container">$\phi(W)$</span>" where <span class="math-container">$\phi$</span> is <span class="math-container">$\Sigma^2_2$</span>.</p>
| Noah Schweber | 8,133 | <p>Here are two <em>non</em>-examples, one erring in each direction:</p>
<ul>
<li><p><strong>Too simple</strong>: "The continuum is an <span class="math-container">$\aleph$</span>-fixed point," that is <span class="math-container">$\mathfrak{c}=\aleph_\mathfrak{c}$</span>. This is equivalent of course to <span class="math-container">$\mathfrak{c}\ge\aleph_\mathfrak{c}$</span>, which means it can be expressed as "There is an <span class="math-container">$\mathbb{R}$</span>-indexed family of sets of reals of pairwise distinct cardinalities," which is <span class="math-container">$\Sigma^2_2$</span>. Contra my original guess, however, this <em>does</em> have a <span class="math-container">$\Pi^2_2$</span> equivalent observed by Farmer S in the comments below <em>(and I'll add his argument here later when I have more time)</em>.</p>
</li>
<li><p><strong>Too complicated (so far!)</strong>: "The continuum is a limit cardinal." This can be expressed in a <span class="math-container">$\Pi^2_3$</span> way <em>(which I originally miscounted - thanks to Andreas Blass for bringing this to my attention)</em> as "For every set of reals <span class="math-container">$X$</span>, either there is a surjection <span class="math-container">$X\rightarrow\mathbb{R}$</span> or there is a set of reals <span class="math-container">$Y$</span> such that there is no surjection <span class="math-container">$X\rightarrow Y$</span> or <span class="math-container">$Y\rightarrow\mathbb{R}$</span>," and I don't see how to get a <span class="math-container">$\Sigma^2_3$</span> equivalent even granting large cardinals. <em>(In particular, note that "The continuum is <span class="math-container">$\ge$</span> some uncountable limit cardinal" is easy to express in a <span class="math-container">$\Sigma^2_2$</span> way as "There exists an <span class="math-container">$\omega$</span>-sequence of sets of reals of strictly increasing cardinality," but this doesn't seem to be useful here.)</em></p>
</li>
</ul>
<p>Of course, the first example doesn't work, and the second example almost certainly doesn't work. That said, I think this is a good indication that general continuum combinatorics is a good place to look for high-complexity third-order sentences.</p>
|
4,337,320 | <p>Let <span class="math-container">$G$</span> be a group and <span class="math-container">$F:G^n \to G$</span> with the following property: If <span class="math-container">$x_1,…,x_n,h \in G$</span>, then <span class="math-container">$F(hx_1,…,hx_n)=hF(x_1,…,x_n)$</span>. Is there a name for this type of function property? It is something I’ve been investigating lately. For instance, if <span class="math-container">$G$</span> is a vector space and <span class="math-container">$F$</span> outputs the average vector, then <span class="math-container">$F$</span> has this property.</p>
| Thomas Andrews | 7,933 | <p>The set of such functions are in one-to-one correspondence with the set of all functions <span class="math-container">$G^{n-1}\to G.$</span></p>
<p>For example, if <span class="math-container">$H:G^{n-1}\to G$</span> we can define:</p>
<p><span class="math-container">$$F(g_1,\dots,g_n)=g_nH(g_n^{-1}g_1,g_n^{-1}g_2,\dots,g_n^{-1}g_{n-1})$$</span></p>
<p>On the other hand, given an <span class="math-container">$F,$</span> we can get back <span class="math-container">$H$</span> by: <span class="math-container">$$H(g_1,\dots,g_{n-1})=F(g_1,\dots,g_{n-1},1).$$</span></p>
<p>So such functions <span class="math-container">$F$</span> don’t seem too interesting.</p>
<hr />
<p>One might ask, more generally, if <span class="math-container">$G$</span> acts on a set <span class="math-container">$X,$</span> then <span class="math-container">$G$</span> acts on <span class="math-container">$X^n,$</span> and what can we say about functions <span class="math-container">$F:X^n\to X$</span> which is a map in the category of sets acted on by <span class="math-container">$G?$</span> In your case, <span class="math-container">$X=G.$</span></p>
<p>This might be more complicated. For example, if <span class="math-container">$G$</span> acts on <span class="math-container">$X$</span> <span class="math-container">$2$</span>-transitively, and <span class="math-container">$n=2,$</span> then <span class="math-container">$F$</span> is entirely determined by one value of the form <span class="math-container">$F(x,x)$</span> and one value of the form <span class="math-container">$F(x,y), x\neq y.$</span> So there are at most <span class="math-container">$|X|^2$</span> such functions in that case.</p>
<p>In general, <span class="math-container">$X^n/G$</span> is very complicated, but when <span class="math-container">$X=G$</span> with the simple action, it is fairly simple.</p>
|
440,534 | <p>Given two <em>origin symmetric convex</em> polytopes <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> (that is <span class="math-container">$P_i=-P_i$</span>) with the same edge-graph, but potentially of different dimensions and combinatorial types.
Let <span class="math-container">$\phi: G_{P_1}\to G_{P_2}$</span> be an isomorphism between their edge-graphs.</p>
<blockquote>
<p><strong>Question:</strong> Does for each vertex <span class="math-container">$v\in P_1$</span> hold <span class="math-container">$\phi(-v)=-\phi(v)$</span>?</p>
</blockquote>
<p>Inuitively, I am asking whether the edge-graph already determines which vertices form an antipodal pair.</p>
<p>The following example shows that for a vertex (black) its antipodal vertex (white) is not necessarily a vertex of maximal graph-distance (gray).</p>
<img src="https://i.stack.imgur.com/Dm541.png" width="250"/>
<hr />
<p>This question is a more precise formulation of <a href="https://mathoverflow.net/q/439787/108884">this older question</a>.</p>
| Dima Pasechnik | 11,100 | <p>An extended comment:</p>
<p>A graph-theoretic reformulation might be as follows. Can a graph <span class="math-container">$\Gamma=(V,E)$</span>,
<span class="math-container">$2n=|V|$</span>, with two different antipodal structures of class size 2, be the edge graph of a polytope?</p>
<p>Here an <em>antipodal structure of class size 2</em> is an equivalence relation on <span class="math-container">$V$</span> with all classes <span class="math-container">$\{v_k,v'_k\}$</span> of size 2, such that none of the classes <span class="math-container">$\{v_k,v'_k\}$</span> is an edge, and the involutory permutation <span class="math-container">$\phi:=(v_1,v'_1)(v_2,v'_2)...(v_n,v'_n)$</span> is an automorphism of <span class="math-container">$\Gamma$</span>.</p>
<p>If you have two such antipodal structures, <span class="math-container">$\phi_1$</span> and <span class="math-container">$\phi_2$</span>, they together generate a dihedral subgroup <span class="math-container">$H$</span> in the automorphism group of <span class="math-container">$\Gamma$</span>.
(That's what two involutions generate.) Then notice that <span class="math-container">$H$</span> itself acts on these antipodal structures. It seems to lead somewhere...</p>
|
1,740,535 | <p>If <span class="math-container">$X\sim U(0,1)$</span> and <span class="math-container">$Y\sim U(0,X)$</span> what is the density (distribution) function <span class="math-container">$f_Y(y)$</span>?</p>
<p>I know the answer and I also found it on this site (link bellow). However, I just can't get the intuition why the last integral boundaries become from <span class="math-container">$y$</span> to <span class="math-container">$1$</span>?</p>
<p>Step by step solution attempt:</p>
<p><span class="math-container">$f_Y(y)=\displaystyle\int_\mathbb{R} f_{Y,X}(y,x)dx=\int_\mathbb{R} f_{Y|X=x}(y)f_{X}(x)dx=\displaystyle\int_\mathbb{R}\frac{1}{X}dx=^{?}\displaystyle\int_y^1\frac{1}{X}dx=-\ln(y)$</span></p>
<p><a href="https://math.stackexchange.com/questions/1738765/let-x-sim-mathcalu0-1-given-x-x-let-y-sim-mathcalu0-x-ho">Let. X∼U(0,1). Given X=x, let Y∼U(0,x). How can I calculate E(X|Y=y)?</a></p>
| Arnaud Mégret | 322,412 | <p>The equality :</p>
<p>$\int_\mathbb{R} f_{Y|X=x}(y)f_{X}(x)dx=\displaystyle\int_\mathbb{R}\frac{1}{X}dx$
is incorrect</p>
<p>it should be :</p>
<p>$\int_\mathbb{R} f_{Y|X=x}(y)f_{X}(x)dx=\displaystyle\int_{-\infty}^y0dx+\displaystyle\int_y^1 \frac{1}{x}dx+\displaystyle\int_1^{+\infty} 0dx$ </p>
<p>as for $x \leq y$ , $f_{Y|X=x}(y)=0$
and for $x \geq 1$ $f_X(x)=0$</p>
|
195,759 | <p>I have three wavelengths. Each of them has a different colour. I want to plot the resulting wave with colour that corresponds to the mixture of these colour intensities at every point. I tried this code:</p>
<pre><code>Plot[I1[y]+I2[y]+I3[y],{y,-0.0000001,0.0000001},ColorFunction ->colf]
colf[y_] := Blend[{{I1[y],Blue},{I2[y],Yellow},{I3[y],Red}}]
</code></pre>
<p>But it didn't work out. What's wrong?</p>
<p><a href="https://i.stack.imgur.com/JzC4v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JzC4v.png" alt="enter image description here"></a></p>
| Kevin Ausman | 54,593 | <p>This was harder than it should have been. But I learned some stuff along the way.</p>
<p>Let's first set up the functions:</p>
<pre><code>f[l_, y_] = 1/2 + Cos[ Pi y/(l 10^-9)]/2;
Plot[{f[450, y], f[550, y], f[650, y]}, {y, -1 10^-6, 1 10^-6},
PlotStyle -> {Blue, Yellow, Red}]
</code></pre>
<p><a href="https://i.stack.imgur.com/7Y08p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Y08p.png" alt="enter image description here"></a></p>
<p>Now let's make the mixture.</p>
<pre><code>mix[y_] = f[450, y] + f[550, y] + f[650, y];
Plot[mix[y], {y, -1 10^-6, 1 10^-6}]
</code></pre>
<p><a href="https://i.stack.imgur.com/Wnuis.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wnuis.png" alt="enter image description here"></a></p>
<p>Now we will plot it using the appropriate color function, turning off ColorFunctionScaling.</p>
<pre><code>Plot[mix[y], {y, -1 10^-6, 1 10^-6}, PlotStyle -> Thickness[0.02],
ColorFunction -> (Blend[{Blue, Yellow, Red}, {f[450, #1], f[550, #1],
f[650, #1]}] &), ColorFunctionScaling -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/hsXvu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hsXvu.png" alt="enter image description here"></a></p>
<p>Or perhaps when all three are low, you might want it to be closer to black (indicating that there isn't as much intensity), in which case:</p>
<pre><code>Plot[mix[y], {y, -1 10^-6, 1 10^-6}, PlotStyle -> Thickness[0.02],
ColorFunction -> (Blend[{Blue, Yellow, Red, Black}, {f[450, #1],
f[550, #1], f[650, #1], 3 - mix[#1]}] &), ColorFunctionScaling -> False]
</code></pre>
<p><a href="https://i.stack.imgur.com/kKYLk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kKYLk.png" alt="enter image description here"></a></p>
<p>In debugging some of this, I set up a manipulate (shown below as an animate) to explore the mixing. It is kind of instructive.</p>
<pre><code>Animate[Show[
Plot[{f[450, y 10^-6], f[550, y 10^-6], f[650, y 10^-6]},
{y, -1 , 1}, PlotStyle -> {Blue, Yellow, Red},
PlotRange -> {{-1, 1}, {0, 3.2}}, ImageSize -> Large],
Plot[mix[y 10^-6], {y, -1, 1}, PlotStyle -> Thickness[0.02],
ColorFunction -> (Blend[{Blue, Yellow, Red,
Black}, {f[450, #1 10^-6], f[550, #1 10^-6], f[650, #1 10^-6],
3 - mix[#1 10^-6]}] &), ColorFunctionScaling -> False],
Graphics[{Black, Thick, Circle[{-x , mix[-x 10^-6]}, {rx, ry}],
Circle[{x , mix[x 10^-6]}, {rx, ry}],
Line[{{-x - rx, mix[-x 10^-6]}, {-x + rx, mix[-x 10^-6]}}],
Line[{{-x, 0}, {-x , mix[-x 10^-6] + ry}}],
Line[{{x - rx, mix[x 10^-6]}, {x + rx, mix[x 10^-6]}}],
Line[{{x, 0}, {x , mix[x 10^-6] + ry}}],
Blend[{Blue, Yellow, Red, Black}, {f[450, -x 10^-6],
f[550, -x 10^-6], f[650, -x 10^-6], 3 - mix[x 10^-6]}],
Disk[{-0.5, 2.8}, {2 rx, 2 ry}], ,
Blend[{Blue, Yellow, Red, Black}, {f[450, x 10^-6],
f[550, x 10^-6], f[650, x 10^-6], 3 - mix[x 10^-6]}],
Disk[{0.5, 2.8}, {2 rx, 2 ry}],
Blend[{Black, Blue}, f[450, -x 10^-6]],
Disk[{-0.5, 1}, {rx, ry}],
Blend[{Black, Blue}, f[450, x 10^-6]],
Disk[{0.5, 1}, {rx, ry}],
Blend[{Black, Yellow}, f[550, -x 10^-6]],
Disk[{-0.5, 1.5}, {rx, ry}],
Blend[{Black, Yellow}, f[550, x 10^-6]],
Disk[{0.5, 1.5}, {rx, ry}],
Blend[{Black, Red}, f[650, -x 10^-6]],
Disk[{-0.5, 2}, {rx, ry}],
Blend[{Black, Red}, f[650, x 10^-6]],
Disk[{0.5, 2}, {rx, ry}]}]], {x, 0, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/h5aKB.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h5aKB.gif" alt="enter image description here"></a></p>
<p>Or the version without Black (code modifications for this left as an exercise to the reader, unless requested):</p>
<p><a href="https://i.stack.imgur.com/u9ZkU.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u9ZkU.gif" alt="enter image description here"></a></p>
|
6,825 | <p>Consider four <span class="math-container">$n$</span>-sided polygons arranged so that their centers are the vertices of a square. The square is exactly as large as the diameter of the smallest circle enclosing each polygon. The polygons do not overlap. Polygons are oriented so that each one has either exactly two sides in common or exactly one side and one vertex in common with the other shapes.</p>
<p>Trying this with small numbers gives me <span class="math-container">$f: 4 \to 0, \ 6 \to 4,\ 8 \to 4,\ 10 \to 8,\ 12 \to 8.$</span> This suggests that <span class="math-container">$$f(2n) = 4 \times \left( \left\lceil \frac{2n}{4} \right\rceil - 1 \right), 2n > 4.$$</span></p>
<p>Can this result be extended for <span class="math-container">$n \to \infty$</span>? Bonus: What about odd values of <span class="math-container">$n$</span>?</p>
| Ross Millikan | 1,827 | <p>Yes, and your answer is correct. Following your figure, assume the polygon has a side at the top and bottom. If n is even, so the polygons have a multiple of 4 sides, there will be a pair of sides at the left and right. There will be $n/2-1$ sides between them and four sets of that bounding the central region. If n is odd, there will be points at the left and right. From the point to the flat on top is $(n-1)/2$ sides, again multiplied by four to make the boundary of the central region. This supports your formula, using the ceiling function to combine the cases.</p>
|
292,176 | <p>I'm having a difficult time finding any theory on an inverse problem I've come up against. Let's say I have an unknown function $f:[0,1] \rightarrow \mathbb{R}$, and I know $\int_{a}^{b} f$ for some collection $A$ of pairs $(a,b)\in[0,1]^2$. I'm looking for pointers to any material that discusses condtions on $f$ and $A$ that are sufficient recover (all of? some of?) the values of $f$. Google searches just keep turning up elementary-calculus-help-type pages. I'm a beginning graduate student, if it matters. Thanks.</p>
<p><strong>Edit:</strong> I'm actually looking for something broader than I asked for. I know that recovering the values of $f$ is a lot to ask and is very unlikely unless $A=[0,1]^2$; I'm also looking for approximations to $f$, anything that can be said about its properties/behaviour, etc. when $A\subsetneqq [0,1]^2$.</p>
| Dirk | 9,652 | <p>Here is how you make an inverse problem of this problem: Choose a space <span class="math-container">$X$</span> for the function <span class="math-container">$f$</span> you are looking for (e.g. <span class="math-container">$L^2(0,1)$</span> to work in Hilbert spaces, but other spaces may be more suitable, depending on your needs).</p>
<p>I assume that you only have finitely many definite integrals (since I assume that this is a practical problem where the definite integrals come from measurements).
Now let us denote your tuples as <span class="math-container">$(a_1,b_1),\dots (a_N,b_N)$</span>. You forward operator is
<span class="math-container">$$\newcommand{\RR}{\mathbb{R}}
K:X\to\RR^N
$$</span>
mapping <span class="math-container">$f$</span> to the <span class="math-container">$N$</span>-vector with components <span class="math-container">$\int_{a_i}^{b_i}f(x)\,dx$</span>. So you are given a vector <span class="math-container">$g\in\RR^N$</span> and want some solution to
<span class="math-container">$$
Kf = g.
$$</span>
Now you are in business with the standard theory for linear inverse problems.</p>
<p>You have some of the usual problems coming with an inverse problem: Non-uniqueness (the operator is not injective) and probably instability in some sense (depending on you data and values <span class="math-container">$(a_i,b_i)$</span>). (As far as I see, non-solvability should not be an issue as <span class="math-container">$K$</span> should be surjective for meaningful tupels <span class="math-container">$(a_i,b_i)$</span>).</p>
<p>To deal with non-uniqueness: You may view this as an advantage as you can choose among all solutions of <span class="math-container">$Kf=g$</span>. To pick one, you can choose regularization functional <span class="math-container">$R:X\to [0,\infty]$</span> and define a minimum-<span class="math-container">$R$</span>-solution as solution of
<span class="math-container">$$
\min\{R(f)\mid f\in X,\ Kf=g\}.
$$</span>
From a computational point of view, convex functional <span class="math-container">$R$</span> are beneficial and you can choose <span class="math-container">$R$</span> to impose some structure on your solution, e.g. <span class="math-container">$R(f) = \int_0^1 |f'(x)|^2\, dx$</span> imposes some smoothness (effectively this means that you constrain your solutions to the Sobolev space <span class="math-container">$H^1$</span>). The most straight-forward choice would be <span class="math-container">$R(f) = \int_0^1 |f(x)|^2\, dx$</span> which should produce a linear equality as optimality condition (and you are effectively computing the Moore-Penrose pseudo-inverse). I could say more about regularizing functionals if needed.</p>
<p>If your data vector <span class="math-container">$g$</span> is also uncertain, i.e. it may be given by measurement data with an error, you may want to relax your problem and look for solutions of
<span class="math-container">$$
\min\{R(f)\mid d(Kf,g)\leq\delta\}
$$</span>
for some discrepancy functional <span class="math-container">$d$</span> and some value <span class="math-container">$\delta>0$</span>. Both should be related to the error in your data. Note that this is in some way equivalent to (generalized) Tikhonov regularization which would be solving
<span class="math-container">$$
\min_f d(Kf,g) + \lambda R(f)
$$</span>
for some regularization parameter <span class="math-container">$\lambda>0$</span>. The most simple case of this would be standard Tikhonov regularization in Hilbert spaces:
<span class="math-container">$$
\min_f \|Kf-g\|_{2}^2 + \lambda\|f\|_{L^2(0,1)}^2
$$</span>
leading to the linear optimality condition
<span class="math-container">$$
K^*(Kf-g) + \lambda f = 0.
$$</span>
The adjoint operator <span class="math-container">$K^*:\RR^N\to L^2(0,1)$</span> is given by
<span class="math-container">$$
K^*g = \sum_{i=1}^N g_i\chi_{[a_i,b_i]}
$$</span>
(where <span class="math-container">$\chi_{[a_i,b_i]}$</span> is the characteristic function of <span class="math-container">$[a_i,b_i]$</span>).
So the optimality condition is actually
<span class="math-container">$$
\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} + \lambda f = 0.
$$</span>
This shows that the regularized solution is also a linear combination of the characteristic functions <span class="math-container">$\chi_{[a_i,b_i]}$</span> and thus, we still get a finite dimensional linear problem for the coefficients.</p>
<p>If you want some smoothness, try <span class="math-container">$R(f) = \int_0^1 |f'(x)|^2\, dx$</span>. This would give an optimality conditions like
<span class="math-container">$$
\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} - \lambda f'' = 0
$$</span>
and thus the solution is piecewise quadratic.</p>
|
803,687 | <p>How to prove : </p>
<blockquote>
<p>$A/m^n$ is Artinian for all $n\geq 0$ if $A$ is a Noetherian ring and $m$ maximal ideal.</p>
</blockquote>
<p>Any suggestions ?</p>
| rschwieb | 29,335 | <p>Suggestion/hint: <a href="https://en.wikipedia.org/wiki/Hopkins%E2%80%93Levitzki_theorem" rel="nofollow">Hopkins-Levitzki</a></p>
|
2,243,900 | <p>I've researched this topic a lot, but couldn't find a proper answer to this, and I can't wait a year to learn it at school, so my question is:</p>
<blockquote>
<p>What exactly is calculus? </p>
</blockquote>
<p>I know who invented it, the Leibniz controversy, etc., but I'm not exactly sure what it is. I think I heard it was used to calculate the area under a curve on a graph. If anyone can help me with this, I'd much appreciate it.</p>
| J.G. | 56,861 | <p>There are several subdisciplines in calculus, which I'll summarise simply below. But what they have in common is the study of how one quantity changes when another changes by a small amount.</p>
<p>Differential calculus computes gradients of tangents to curves or surfaces, integral calculus computes the sizes of regions (including the areas of 2D regions), and it turns out differentiation reverses integration. (To see why, imagine expanding an area slightly with a narrow strip you can approximate as a rectangle, and changing a function slightly along the direction of its input increasing. That function might as well be the cumulative area left of a moving cutoff.) Newton co-invented calculus because he wanted to study motion this way. The distance traveled over time is the area under a graph of velocity.</p>
<p>Intrinsic calculus measures curve lengths. It's like applying Pythagoras to lots of small right-angled triangles, then adding their hypotenuses to make an arc. Relativity can be restated as a claim that a certain measure of paths through space and time don't change when you do the equivalent of rotating axes.</p>
|
2,243,900 | <p>I've researched this topic a lot, but couldn't find a proper answer to this, and I can't wait a year to learn it at school, so my question is:</p>
<blockquote>
<p>What exactly is calculus? </p>
</blockquote>
<p>I know who invented it, the Leibniz controversy, etc., but I'm not exactly sure what it is. I think I heard it was used to calculate the area under a curve on a graph. If anyone can help me with this, I'd much appreciate it.</p>
| MattPutnam | 170,998 | <p>For people familiar with the basic concept of limits, I like to describe it as the geometric application of indeterminate forms.</p>
<p>A derivative is the slope of a function computed at every point. For a line, we can take any two points and calculate the slope as $\frac{\Delta y}{\Delta x}$. For a curve, we do exactly the same, but we have to take the limit as the two points converge, leaving us with the indeterminate form $\frac{0}{0}$. Differential calculus is effectively the study of this limit.</p>
<p>An intergral is the area under a curve. For a constant function $f(x)=c$ from $0$ to $a$, this is just $c \times a$. For a curve, we can approximate it with a bunch of skinny rectangles with height $f(x)$ and width $\Delta x$. It's an approximation as long as the rectangles have some nonzero width, so we make it exact by taking the limit as $\Delta x \to 0$, and equivalently the number of rectangles goes to infinity. This leaves us with the indeterminate form $0 \times \infty$. Integral calculus is effectively the study of this limit.</p>
|
2,243,900 | <p>I've researched this topic a lot, but couldn't find a proper answer to this, and I can't wait a year to learn it at school, so my question is:</p>
<blockquote>
<p>What exactly is calculus? </p>
</blockquote>
<p>I know who invented it, the Leibniz controversy, etc., but I'm not exactly sure what it is. I think I heard it was used to calculate the area under a curve on a graph. If anyone can help me with this, I'd much appreciate it.</p>
| George Frank | 30,674 | <p>Differential calculus is not about infinitesimals or limits. if we think of a smooth function as a curve, that curve has a steepness for every value of the independent variable. (For roads the corresponding term is "grade.") For good reason the value of that steepness is taken to be the slope of the line tangent to that point. The new function that gives the steepness shouldn't be called the "derivative" of the original but we're stuck with the name. The limit is only relevant when we want to actually derive that function from the original function. Similarly, over any interval of the independent variable of a function, there is an area between the function and the axis of the independent variable. The new function that gives that area is called the "integral" of the original function. Once again this is not a descriptive name for what the new function IS and once again the limit is used to derive the integral of a function. We should never confuse a device used to obtain something (a limit in this case) with the thing obtained, in this case two functions (the "derivative" and the "integral") that describe properties of the original function. Abstraction and formality can be egregiously confusing.
bs</p>
|
3,856,715 | <p>Let <span class="math-container">$k$</span> be a field and <span class="math-container">$A,B$</span> be two (finite-dimensional) central simple <span class="math-container">$k$</span>-algebras. We usually say that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are <em>Brauer equivalent</em> (or <em>similar</em>) if their underlying division algebras (given by Wedderburn's theorem) are isomorphic or, equivalently, if <span class="math-container">$M_n(A)\cong M_m(B)$</span> for some integers <span class="math-container">$n,m$</span>.</p>
<p>The Wikipedia article about the Brauer group says that this is equivalent to demanding that the categories <span class="math-container">$A$</span>-Mod and <span class="math-container">$B$</span>-Mod of left modules are equivalent. However, I couldn't prove this fact neither did I found it in any books.</p>
<p>(PS: I know absolutely nothing about Morita equivalence, so I would prefer a direct answer to an answer that uses a basic fact from the theory of Morita equivalence.)</p>
<p>Why is it true?</p>
| rschwieb | 29,335 | <p>If <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are Morita equivalent, then <a href="https://en.wikipedia.org/wiki/Morita_equivalence#Criteria_for_equivalence" rel="nofollow noreferrer">there exists</a> an <span class="math-container">$n$</span> and a full idempotent <span class="math-container">$e\in M_n(A)$</span> such that <span class="math-container">$B\cong eM_n(A)e$</span>.</p>
<p>Now if <span class="math-container">$A\cong M_{n'}(D)$</span> using the Artin-Wedderburn theorem, rewrite the right hand side via an isomorphism to be <span class="math-container">$fM_{n'}(D)f$</span>.</p>
<p>There must exist some unit <span class="math-container">$u$</span> diagonalizing <span class="math-container">$f$</span> to a matrix <span class="math-container">$\hat f$</span> which is some number of <span class="math-container">$1$</span>'s on the diagonal followed by zeros on the diagonal (this just amounts to selecting an eigenbasis for the transformation, putting the nonzero eigenvectors up front.) Conjugation by <span class="math-container">$u$</span> makes the right hand side isomorphic to <span class="math-container">$ufu^{-1}M_{n'}(D)u^{-1}ufu^{-1}=\hat fM_{n'}(D)\hat f$</span>, but as you can see the diagonalized idempotent <span class="math-container">$\hat f$</span> simply selects out some upper left corner of <span class="math-container">$M_{n'}(D)$</span>, which is apparently of the form <span class="math-container">$M_m(D)$</span> for some <span class="math-container">$m < n'$</span>. Going back through the isomorphisms, you have <span class="math-container">$B\cong M_m(D)$</span>. From there it is easy that <span class="math-container">$M_{m}(A)\cong M_{n'}(B)$</span>.</p>
<p>The other direction has already been discussed above: if <span class="math-container">$M_n(A)\cong M_m(B)$</span>, then they are <em>a priori</em> Morita equivalent, and it is well known that <span class="math-container">$R$</span> is equivalent to <span class="math-container">$M_n(R)$</span> for any ring, so by transitivity of equivalence you get that <span class="math-container">$A$</span> is Morita equivalant to <span class="math-container">$B$</span> if <span class="math-container">$A$</span> is Brauer equivalent to <span class="math-container">$B$</span>.</p>
|
3,856,715 | <p>Let <span class="math-container">$k$</span> be a field and <span class="math-container">$A,B$</span> be two (finite-dimensional) central simple <span class="math-container">$k$</span>-algebras. We usually say that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are <em>Brauer equivalent</em> (or <em>similar</em>) if their underlying division algebras (given by Wedderburn's theorem) are isomorphic or, equivalently, if <span class="math-container">$M_n(A)\cong M_m(B)$</span> for some integers <span class="math-container">$n,m$</span>.</p>
<p>The Wikipedia article about the Brauer group says that this is equivalent to demanding that the categories <span class="math-container">$A$</span>-Mod and <span class="math-container">$B$</span>-Mod of left modules are equivalent. However, I couldn't prove this fact neither did I found it in any books.</p>
<p>(PS: I know absolutely nothing about Morita equivalence, so I would prefer a direct answer to an answer that uses a basic fact from the theory of Morita equivalence.)</p>
<p>Why is it true?</p>
| Qiaochu Yuan | 232 | <p>All you need to know is that <span class="math-container">$M_n(R)$</span> and <span class="math-container">$R$</span> are Morita equivalent. What this says explicitly is that the category of modules over <span class="math-container">$M_n(R)$</span> and over <span class="math-container">$R$</span> are equivalent, and this equivalence can be written down explicitly: it sends</p>
<p><span class="math-container">$$\text{Mod}(R) \ni M \mapsto M \otimes_R R^n \cong M^n \in \text{Mod}(M_n(R)).$$</span></p>
<p>Actually we only need that this is an equivalence for <span class="math-container">$R$</span> a central simple algebra, but it's true in general. Now:</p>
<p><span class="math-container">$\Rightarrow$</span>: if two central simple algebras <span class="math-container">$A, B$</span> are Brauer equivalent then <span class="math-container">$A \cong M_n(D)$</span> and <span class="math-container">$B \cong M_m(D)$</span> for some central division algebra <span class="math-container">$D$</span>, and <span class="math-container">$\text{Mod}(A) \cong \text{Mod}(D) \cong \text{Mod}(B)$</span>.</p>
<p><span class="math-container">$\Leftarrow$</span>: if <span class="math-container">$A \cong M_n(D)$</span> and <span class="math-container">$B \cong M_n(E)$</span> where <span class="math-container">$D, E$</span> are central division algebras, then <span class="math-container">$\text{Mod}(A) \cong \text{Mod}(D)$</span> and <span class="math-container">$\text{Mod}(B) \cong \text{Mod}(E)$</span>. A division algebra can be recovered from its category of modules: it's the algebra of endomorphisms of the unique simple module. So <span class="math-container">$\text{Mod}(D) \cong \text{Mod}(E)$</span> implies <span class="math-container">$D \cong E$</span>.</p>
<p>(We need the isomorphism to be <span class="math-container">$k$</span>-linear to get that <span class="math-container">$D \cong E$</span> over <span class="math-container">$k$</span>.)</p>
<p>The last observation can be extended to give a module-theoretic characterization of module categories over central simple algebras, or equivalently of module categories over division rings: they are precisely the semisimple module categories with a unique simple object.</p>
|
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