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 |
|---|---|---|---|---|
3,165,494 | <blockquote>
<p>Consider the function <span class="math-container">$\phi(t,x)$</span> where <span class="math-container">$x=(x_1,x_2) \in \mathbb{R}^2$</span>. How can we check if <span class="math-container">$\phi(t,x)$</span> is a flow for a vector field? <span class="math-container">$$\phi(t,x)=(x_1\cos(r^2t)+x_2\sin(r^2t),-x_1\sin(r^2t)+x_2\cos(r^2t))$$</span> where <span class="math-container">$r^2=x_1^2+x_2^2$</span>.</p>
</blockquote>
<hr />
<p>My try:</p>
<p>We need to check if <span class="math-container">$\phi(0,x)=\phi(t,\phi(-t,x))$</span>. I am wondering if there is a shorter way?</p>
| Lutz Lehmann | 115,115 | <p>If you write <span class="math-container">$\zeta(t,z)=ϕ_1(t,x)+iϕ_2(t,x)$</span> where <span class="math-container">$z=x_1+ix_2$</span>, you will find that
<span class="math-container">$$
\zeta(t,z)=e^{-i|z|^2t}z
$$</span>
and as a consequence <span class="math-container">$|\zeta(t,z)|=|z|$</span> constant along the trajectories, so that the group conditions for the flow are easy to prove.</p>
|
2,702,329 | <blockquote>
<p>Let $A$ be an $m\times n$ matrix, let $v \in \mathbb{R}^m$, and let $S$ be the set of least squares solutions to the system $Ax=b$. Show that there exists a unique <em>minimal least square solution</em> to $Ax=b$; that is, there exists some $x\in S$ such that $||x|| \le||y||$ for all $y\in S$.</p>
</blockquote>
<p>I have two questions about this.</p>
<ol>
<li><p>How can there be a unique least square solution if the stipulation is $||x|| \le ||y||$, shouldn't it just be less than?</p></li>
<li><p>This seems tautological to me, because you're just trying to show that there is a least element in S, which should clearly exist because the norm gives you a scalar that you can compare. Is there any more to this proof?</p></li>
</ol>
| copper.hat | 27,978 | <p>Since $x \mapsto \|Ax-b\|^2$ is convex, the set of minimisers $S$ is a closed, convex set.</p>
<p>(In particular, if $x_0$ is a solution, then $S = \{x_0\} + \ker A$.)</p>
<p>A standard Hilbert space theorem states that a closed convex set has a unique point of minimum norm.</p>
<p>The $x$ is the question is the point of minimum norm in $S$.</p>
|
1,501,736 | <p>Let $x_1 = 1$ and for $n \ge2, x_n = \sqrt{3 + x_{n-1}}$, I just want to know how to determine the limit. </p>
| Jack D'Aurizio | 44,121 | <p><strong>Hint:</strong> prove that your sequence is increasing and bounded, hence converging to its supremum, then notice that there is only one real number $r$ such that $r=\sqrt{r+3}$.</p>
|
477,250 | <p>Let $u=\{u_\alpha\}$ be an open cover of $[a,b], a <b $. Let $S=\{r \in [a,b]$ such that $[a,r]$ is covered by some finite collection of open sets belonging to $u\}$. Is $S$ non-empty ? How ?</p>
| Gerry Myerson | 8,269 | <p>Let $v=(1,1,1,1,1)$. Note $Av=v$. Deduce $A^{-1}v=v$. What does this say about row sums of $A^{-1}$?</p>
|
1,607,108 | <p>Using the ratio test:
$$\frac{1}{3}\lim\limits_{n\to\infty}\left|\frac{(n+1)^3(\sqrt{2}+(-1)^{n+1})^{n+1}}{n^3(\sqrt{2}+(-1)^n)^n}\right|$$</p>
<p>Without evaluating the limit, the numerator is greater then denominator and the series is divergent. Is there an easier method for checking the convergence of this sequence?</p>
| Community | -1 | <p>To answer your question without resorting to any explanation beyond what you have provided:</p>
<p>No. This part is seriously off "Assume 2|x−1 and 4|x+1" without any explanation. When you say "assume this", "assume that", you have to cover all options.</p>
<p>2 is prime so it must divide either (x+1) or (x-1), that essential part is missing.</p>
<p>Now, if (x+1) divides 2, then either (x+1) divides 4, or (x-1) divides 4. If (x-1) divides 2, then either (x-1) divides 4, or (x+1) divides 4.</p>
<p>Now make all combinations</p>
<ol>
<li>(x+1) divides 4 (here you need to add that (x-1) divides 2)</li>
<li>(x+1) divides 2 and (x-1) divides 4</li>
<li>(x-1) divides 4 (here you need to add that (x+1) divides 2)</li>
<li>(x-1) divides 2 and (x+1) divides 4</li>
</ol>
<p>Now you can conclude.</p>
|
11,435 | <p><strong>Question</strong>: What is the correct notion of a <em>product</em> of integral (or rational) polytopes which induces a factorization of its Ehrhart (quasi-)polynomial into two primitive Ehrhart (quasi-)polynomials corresponding to its constituent polytopes, viz., $L_{P \times Q}(t) = L_{P}(t) L_{Q}(t)$? </p>
<p>(<em>Motivation</em>) Given two closed integral polytopes $P$ and $Q$ each with vertices at $\{ \mathbf{0} , b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n} \}$ and $\{ \mathbf{0} , d_{1} \mathbf{e}_{1}, \dots, d_{m} \mathbf{e}_{m} \}$, respectively, where $n, b_{i}, m, d_{j} \in \mathbb{N}$, define the integral polytope $R$ with vertices at $\{ \mathbf{0}, b_{1} \mathbf{e}_{1}, \dots, b_{n} \mathbf{e}_{n}, d_{1} \mathbf{e}_{n+1}, \dots, d_{m} \mathbf{e}_{n+m} \}$. </p>
<p>The above construction cannot be the sought after product $P \times Q$. Suppose $P$ and $Q$ are defined by $b_{1} = b_{2} = d_{1} = d_{2} = 2$. It is easy to show that $L_{P}(1) = L_{Q}(1) = 6$. Define $R$ as above with vertices of $P$ and $Q$. It is true that $L_{R}(1) = 15 \neq 6^{2}$.</p>
<p><strong>Question</strong>: What is $R$ in terms of $P$ and $Q$? Is it special in some way?</p>
<p>Thanks!</p>
| Dan Moore | 1,782 | <p>The (Motivation) description appears to be similar to a direct sum (a.k.a. tegum product) as defined in Section 1.2 of this paper:</p>
<p><a href="http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf" rel="nofollow">http://people.reed.edu/~davidp/homepage/seniors/mcmahan.pdf</a></p>
<p>The difference is that for a direct sum, 0 is in the relative interior of both of the factor polytopes, rather than a common vertex.</p>
<p>The same article answers your question about what the nm vertices of a Cartesian product are.</p>
|
2,118,230 | <blockquote>
<p>Let $G$ be a group and let $H,K \le G$ be subgroups of finite index, say $|G:H| = m$ and $|G:K|=n$. Prove that $\mathrm{lcm}(m,n) \le |G:H \cap K| \le mn$. </p>
</blockquote>
<p>I was able to establish the upper bound, but I am having difficulty establishing the lower bound, so I consulted <a href="https://crazyproject.wordpress.com/2010/03/25/bounds-on-the-index-of-an-intersection-of-two-subgroups/" rel="nofollow noreferrer">this</a>. However, I am having trouble following the author's reasoning. Here is the relevant part I am referring to:</p>
<blockquote>
<p>Now...we have $H \cap K \le H \le G$. Thus, $m$ divides $|G:H \cap K|$ and $n$ divides $|G:H \cap K|$, so that $\mathrm{lcm}(m,n)$ divides $|G:H \cap K|$.</p>
</blockquote>
<p>Exactly what theorem is being used to make this conclusion?</p>
| Andrew Chiriac | 172,569 | <p>The theorem you are looking for is the following: If $G_1\geq G_2\geq G_3$, then $[G_1:G_3]=[G_1:G_2][G_2:G_3]$. The proof of this theorem can be found in most introductory texts on abstract algebra. </p>
<p>Recall that the least common multiple of two integers $a$ and $b$ is the smallest positive integer that is divisible by both $a$ and $b$. Using the above theorem, one can see that both $m$ and $n$ divides $[G:H\cap K]$. Thus $lcm(m,n)\leq [G:H\cap K]$.</p>
|
1,129,175 | <p>In my assignment I have to find a Jordan normal form for this matrix:</p>
<p><img src="https://i.stack.imgur.com/8yVcy.jpg" alt="Original Matrix"></p>
<p><img src="https://i.stack.imgur.com/eor4m.jpg" alt="Jordan Normal Form">
<img src="https://i.stack.imgur.com/wEPtW.jpg" alt="Nilpotent">
<img src="https://i.stack.imgur.com/mvWt5.jpg" alt="Similair">
<img src="https://i.stack.imgur.com/gzK2j.jpg" alt="Don't understand"></p>
<p>Thank you for your help, and I'm sorry the question is pronunced with Latex.</p>
| Peter | 82,961 | <p>Hint : The Caley-hamilton-theorem states that a matrix with characteristic equation
$$a_n\lambda^n+...+a_1\lambda+a_0=0$$
satisfies the equation
$$a_nA^n+...+a_1A+a_0I=0$$</p>
<p>Also note, that A can be written as AI.</p>
<p>Now, do you see how $A^{-1}$ can be expressed by powers of $A$ , including I ?</p>
|
1,129,175 | <p>In my assignment I have to find a Jordan normal form for this matrix:</p>
<p><img src="https://i.stack.imgur.com/8yVcy.jpg" alt="Original Matrix"></p>
<p><img src="https://i.stack.imgur.com/eor4m.jpg" alt="Jordan Normal Form">
<img src="https://i.stack.imgur.com/wEPtW.jpg" alt="Nilpotent">
<img src="https://i.stack.imgur.com/mvWt5.jpg" alt="Similair">
<img src="https://i.stack.imgur.com/gzK2j.jpg" alt="Don't understand"></p>
<p>Thank you for your help, and I'm sorry the question is pronunced with Latex.</p>
| Karl | 203,893 | <p>I had a lot of trouble getting my head around this myself and apologise if anything in my answer is incorrect.</p>
<p>Consider the matrix expression
$$\begin{bmatrix}
3&6\\
1&2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}$$
I can take a factor of 3 out of the 1st row
$$
3\begin{bmatrix}
1&2\\
1&2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}$$
The matrix has a zero determinant by inspection or $2-2=0$</p>
<p>A matrix with zero determinant is singular and has no inverse.</p>
<p>Notice that the 1st row is obviously a linear combination of the second row and so they are linearly dependent. This was just an example to get a feeling for what is happening. It is more appropriate to think of the determinant as being designed to test for linear dependence. Herbert Gross has excellent lectures from MIT. </p>
|
2,617,555 | <p>So the cost per hour of running a cruiser is $\$ \left(\frac {V^2}{40} + 10\right)$, where $V $is the speed in knots. So I’ve answered the first question showing the cost would be $\$\frac DV \left(\frac {V^2}{40} + 10\right)$. And then they asked me to find the most economical speed for running the cruiser, and I have no idea how to get it</p>
| Redsbefall | 97,835 | <p>Let your cost function be $C(V)$, which is dependent on $V$, the velocity of the cruise.
$$ C(V) = \frac{D}{V}\left( \frac{V^{2}}{40} + 10 \right) $$
with $D$ as a constant. </p>
<p><em>The most economical velocity</em> is the velocity that makes the <em>smallest cost</em>.</p>
<p>Since we already have the cost function, we can find this <em>economical velocity</em> by using first derivative of $C(V)$. Since $C(V)$ is continuous for $V>0$, $C(V)$ will attain a minima or maxima value at a certain velocity when its first derivative equals 0 :
$$ C(V) = \frac{DV}{40} + \frac{10D}{V} $$
$$ C'(V) = \frac{D}{40} - 10DV^{-2} $$
now you may try to solve
$$ C'(V) = 0 $$
You could get more than 1 velocities,
choose the one that makes $C(V)$ as minimum as possible. Is this okay?</p>
|
3,612,316 | <p>I am trying to interpret an equation, but can't understand how the less-than signs work:</p>
<p><span class="math-container">$\sum _{n=1}^{\infty } \frac{1}{n^2}<1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}=1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)=1+1=2$</span></p>
<p>The first part returns False when I put it into Mathematica which I don't think is how it should be interpreted since this value approximates 1.644. </p>
<p><span class="math-container">$\sum _{n=1}^{\infty } \frac{1}{n^2}<1$</span></p>
| CHAMSI | 758,100 | <p>Since <span class="math-container">$ \left(\forall n\geq 2\right),\ n^{2}\geq n\left(n-1\right) $</span>, we get that <span class="math-container">$ \left(\forall n\geq 2\right),\ \frac{1}{n^{2}}\leq\frac{1}{n\left(n-1\right)} $</span>, thus : <span class="math-container">$$ \sum_{n=1}^{+\infty}{\frac{1}{n^{2}}}=1+\sum_{n=2}^{+\infty}{\frac{1}{n^{2}}}<1+\sum_{n=2}^{+\infty}{\frac{1}{n\left(n-1\right)}}=1+\sum_{n=1}^{+\infty}{\frac{1}{n\left(n+1\right)}} $$</span></p>
<p>I suppose you can now understand why the first inequality is true.</p>
|
1,679,383 | <p>Find numbers $\alpha,\beta, \gamma \in \mathbb{C}$ so that the integral $$\int_{-1}^{1}|x^3-\alpha-\beta x-\gamma x^2|^2dx$$ is minimal.
Here I want to apply Hilbert space and projection lemma by considering the integral as the distance square between a point and a closed linear subspace. So how can I start my work? </p>
| Disintegrating By Parts | 112,478 | <p>If $\mathcal{M}$ is the subspace of polynomials $p(x)=\alpha + \beta x + \gamma x^2$, then you are looking for $p\in \mathcal{M}$ closest to $x^3$. Equivalently, $(x^3-p)\perp\mathcal{M}$, which gives equations
$$
\int_{-1}^{1}(x^3-\alpha-\beta x-\gamma x^2)dx =0 \\
\int_{-1}^{1}(x^3-\alpha-\beta x-\gamma x^2)xdx = 0 \\
\int_{-1}^{1}(x^3-\alpha-\beta x-\gamma x^2)x^2dx = 0.
$$
That's 3 equations in the 3 unknowns $\alpha,\beta,\gamma$.</p>
|
1,600,266 | <p>I'm trying to work out sum of this series $$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots$$</p>
<p>I know one method is to do substitutions and getting the series into a form of a known series. So far I've converted the series into $$ 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \ldots $$ where $x=2$ and I'm trying to get it into the form of the $\ln(1+x)$ series somehow. I have tried differentiating, integrating and nothing is working out. The closest I got is by inverting which gave me $ 1 + \frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} \ldots $ </p>
<p>Now I'm just lost and have no idea what to do.</p>
<p>The other idea I had was converting it into $$ \Large{\sum_{n=1}^\infty{\frac{n}{2^{n-1}}}}, $$ but I have no idea how to do anything further to it.
How would you do this? Thnx.</p>
| fleablood | 280,126 | <p>Okay. I found the text online and this is my interpretation.</p>
<p>However the formatting was hard to read, a page or two seems to be missing, and it's a subject I have never studied not even an iota....</p>
<p>but... I think</p>
<p>He's talking about writing a string parser. You give it a binary string and the program is supposed to analyze whether the string is member of a described set. The set could be ... anything. Lp would be the set of prime numbers. But Lc would be the set of valid C programs. I imagine other sets could be capital cities transposed to binary strings but I could be misunderstanding.</p>
<p>So, I think he is saying sometimes it's easy for the program to recognize if a string is a member of a set. If the set is of strings that start with "010" it'd be trivial. Sometimes it is hard. My understanding is if the set is Lc terminating C programs the only way to tell is to actually run it. (Again, I have <em>never</em> studied this field so I might just have said something utterly idiotic and stupid.) So he's saying (I think) the set of Lp is sometimes easy: If the string begins with a 0 then the string isn't an integer at all and can not be a prime. But if the string begins with a 1 and <em>is</em> therefore an integer, there isn't necessarily any easy way to tell if a number is prime.</p>
<p>That's what I think he is saying. 0011101 is not a number therefore not prime. But 11101 is 29 which... we have to do some programming to determine if it is prime simply by <em>looking</em> at it.</p>
<p>I think.</p>
<p>Excerpt:</p>
<p>============</p>
<p>"32 CHAPTER 1. AUTOMATA: THE METHODS AND THE MADNESS </p>
<p>Set-Formers as a Way to Define Languages </p>
<p>It is common to describe a language using a "set-former" : </p>
<p>{w | something about w) </p>
<p>This expression is read "the set of words w such that (whatever is said
about w to the right of the vertical bar}." Examples are: </p>
<p>1 . {vj | w consists of an equal number of O's and 1 's } . </p>
<ol start="2">
<li><p>{w | w is a binary integer that is prime }. </p></li>
<li><p>{w J id is a syntactically correct C program }. </p></li>
</ol>
<p>It is also common to replace w by some expression with parameters and
describe the strings in the language by stating conditions on the parame-
ters. Here are some examples; the first with parameter n, the second with
parameters i and j: </p>
<ol>
<li><p>{0™1" | n > 1}. Read "the set of 0 to the n 1 to the n such that n
is greater than or equal to 1," this language consists of the strings
{01,0011,000111,...}. Notice that, as with alphabets, we can raise
a single symbol to a power n in order to represent n copies of that
symbol. </p></li>
<li><p>{0 l V | 0 < i < j}. This language consists of strings with some O's
(possibly none) followed by at least as many l's. </p></li>
</ol>
<p>...*page missing????....</p>
<p>decision is easy. For instance, 0011101 cannot be the representation of a prime,
for the simple reason that every integer except 0 has a binary representation
that begins with 1. However, it is less obvious whether the string 11101 belongs
to L p , so any solution to this problem will have to use significant computational
resources of some kind: time and/or space, for example. □ </p>
<p>One potentially unsatisfactory aspect of our definition of "problem' 7 is that
one commonly thinks of problems not as decision questions (is or is not the
following true?) but as requests to compute or transform some input (find the
best way to do this task). For instance, the task of the parser in a C compiler
can be thought of as a problem in our formal sense, where one is given an ASCII
string and asked to decide whether or not the string is a member of L c , the set
of valid C programs. However, the parser does more than decide. It produces a
parse tree, entries in a symbol table and perhaps more. Worse, the compiler as
a whole solves the problem of turning a C program into object code for some "</p>
|
835,454 | <p>I was struggling with the following problem (from linear algebra):</p>
<blockquote>
<p>Let $V$ be the vector space of the $2 \times 2$ matrices with real coefficients. Consider the action of the group $SL_2(\mathbb{R})$ on $V$, namely for any matrix $T \in SL_2(\mathbb{R})$ and for every matrx $A \in V$ we define $\Phi_T(A)=T\cdot A\cdot T^{-1}$. Prove that for every $T\in SL_2(\mathbb{R})$, $\Phi_T$ is an endomorphism of $V$. Then starting from a basis of $V$, write the corresponding homomorphism $SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$, then show that the image lies inside $SL_4(\mathbb{R})$</p>
</blockquote>
<p>First of all. Does anybody know the origin of this problem? Is there a book with standard solutions to this kind of problem? And is my solution any good? Apparently it is generalizable to many more dimensions, giving a nontrivial result on the determinant of such matrices.
And what does it happen when the trace is $0$? Can we show an homomorphism $SL_2(\mathbb{R}) \to SL_3(\mathbb{R})$ in the same way?</p>
| Olivier Bégassat | 11,258 | <p>Let $\Bbb k$ be a field, and let $V$ be a finite dimensional $\Bbbk$-vector space (of dimension $n$). The special linear group $SL(V)$ is generated <a href="http://en.wikipedia.org/wiki/Shear_mapping" rel="nofollow">shear mappings</a>. All shear mappings are conjugate in $GL(V)$, so that the linear maps "conjugation by a shear map" are conjugate as endomorphisms of $\mathrm{End}(V)$, and thus have the same determinant. If we prove tha conjugation by a shear mapping induces a linear map of determinant one on $\mathrm{End}(V)$, then from what precedes we see that conjugation by a map of determinant one has determinant one aswell.</p>
<p>I'll describe two paths to the result. The first is a direct computation in an appropriately chosen basis of $\mathrm{End}(V)$. The second approach is simpler and more formal : it uses the determinant and the fact that the square of a shear mapping is a shear mapping (or the identity in characteristic $2$).</p>
<h2>First proof</h2>
<p>Let $u$ be a shear mapping. By definition there exists nonzero linear form $f:V\to \Bbbk$ and a nonzero vector $h\in H=\ker(f)$ such that for all $v\in V$
$$u(v)=v+f(v)h$$
Consider a basis $\mathcal{B}$ of $V$ of the form $(e_1,\dots,e_n)=(h_1,\dots,h_{n-2},h,x)$ where $h$ is as above, $f(x)=1$ and $(h_1,\dots,h_{n-2},h)$ is a basis of $H$. Then
$$\mathrm{Mat}(u;\mathcal{B})=\begin{pmatrix}1\\&1\\&&\ddots\\&&&1\\&&&&1&1\\&&&&&1\end{pmatrix}$$</p>
<p>Consider the basis $(E_{i,j})_{1\leq i,j\leq n}$ of $\mathrm{End}(V)$ associated to the basis $\mathcal{B}$ defined on basis elements by
$$E_{i,j}(e_k)=\begin{cases}k=j:& e_i\\ k\neq j:&0\end{cases}$$
Then $u=\mathrm{id}+E_{n-1,n}$ and $u^{-1}=\mathrm{id}-E_{n-1,n}$; for any $i,j$
$$\begin{array}{rcl}
u\circ E_{i,j}\circ u^{-1} &=&(\mathrm{id}+E_{n-1,n})E_{i,j}(\mathrm{id}-E_{n-1,n})\\
&=&E_{i,j}+E_{n-1,n}E_{i,j}-E_{i,j}E_{n-1,n}-E_{n-1,n}E_{i,j}E_{n-1,n}\\
&=&
\begin{cases}
i\neq n,j\neq n-1:&E_{i,j}\\
i= n,j\neq n-1:&E_{n,j}+E_{n-1,j}\\
i\neq n,j= n-1:&E_{i,n-1}-E_{i,n}\\
i=n,j=n-1:&E_{n,n-1}+E_{n-1,n-1}-E_{n-1,n}-E_{n,n}
\end{cases}\\
&=&
\begin{cases}
i\neq n,j\neq n-1:&E_{i,j}\\
i= n,j\neq n-1:&E_{i,j}+\cdots\in X\\
i\neq n,j= n-1:&E_{i,j}+\cdots\in X\\
i=n,j=n-1:&E_{i,j}+\cdots\in Y
\end{cases}
\end{array}$$
where $X=\mathrm{Vect}(E_{i,j})_{i\neq n,j\neq n-1}$ and $Y=\mathrm{Vect}(E_{i,j})_{(i,j)\neq(n,n-1)}$, so that the matrix of conjugation by $u$, when represented in some basis obtained by ordering the vectors $(E_{i,j})_{i\neq n,j\neq n-1}\cup(E_{n,j})_{j\neq n-1}\cup(E_{i,n-1})_{i\neq n}\cup(E_{n,n-1})$ inside each set of parantheses is upper triangular with a diagonal made of ones.</p>
<h2>Second proof</h2>
<p>Another line of reasoning goes as follows: we have a linear group action $\rho$ of $SL(V)$ on $\mathrm{End}(V)$
$$\rho:SL(V)\to GL(\mathrm{End}(V)),\quad u\mapsto u\circ-\circ u^{-1}$$
Which we may compose with the determinant
$$\det:GL(\mathrm{End}(V))\to\Bbbk^{\times}$$
Notice that the square of a shear mapping is either a shear mapping (if $\mathrm{char}(\Bbbk)\neq 2$) or the identity (if $\mathrm{char}(\Bbbk)=2$) so that for any shear mapping $u$
$$\det(\rho(u))^2=
\begin{cases}
\det(\rho(u))&\text{if }\mathrm{char}(\Bbbk)\neq 2\\
1 &\text{if }\mathrm{char}(\Bbbk)=2
\end{cases}$$
So that, in both cases (since $1=-1$ in characteristic $2$) $\det(\rho(u))=1$. The generation and conjugation argument from above then tells us that $\rho$ maps $SL(V)$ into
$SL(\mathrm{End}(V))$.</p>
|
2,598,848 | <p>I have a project I'm doing in desmos and want to know if this is possible.</p>
<p>I want to be able to set some parameter, a, equal to the solution of f(x) = 0. I plugged my specific equation into Maple and it could not arrive at an explicit solution for x.</p>
<p>However, when I type in the equation f(x) = 0 into desmos, it does a fine job of plotting the x values for which this equation is true.</p>
<p>My question is this: Is it possible to take this x value(s) and set my parameter (AKA constant or 'slider') equal to it? Does desmos provide this functionality?</p>
<p>Thanks!</p>
| Doug M | 317,162 | <p>The figure will be symmetric across the line $x = y$ (along with $x = 0$ and $y = 0$)</p>
<p>We can analyze $\frac 18$ the figure, and then use this symmetry to our advantage.</p>
<p>Suppose $a<b$</p>
<p>$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ will be on the "inside" from $0$ to the point of intersection.</p>
<p>Parameterize the area inside the curve:</p>
<p>$x = ar\cos\theta\\
y = br\sin\theta\\
dx\ dy = ab r\ dr\ d\theta$ </p>
<p>The last line is the Jacobean for this coordinate system.</p>
<p>And the point of intersection is where $x = y$</p>
<p>$a\cos\theta = b \sin \theta\\
\tan \theta = \frac {a}{b}$</p>
<p>$8\int_0^{\arctan \frac {a}{b}}\int_0^{1} abr\ dr\ d\theta$</p>
<p>$4 ab \arctan\frac {a}{b}$</p>
|
2,278,120 | <p>In one of my algorithm courses, there is this: </p>
<blockquote>
<p>A subset $S$ of vertices in a directed graph $G$ is
strongly connected if for each pair of distinct
vertices ($v_i$, $v_j$) in $S$, $v_i$ is connected to $v_j$ and
$v_j$ is connected to $v_i$.</p>
</blockquote>
<p>And then the following example graph is given for this proposition: </p>
<p><a href="https://i.stack.imgur.com/m6PLe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6PLe.jpg" alt="enter image description here"></a></p>
<p>Maybe i do not understand what that phrase means. What i think it means is this: A node, say $E$ can be a member of an ordered pair $(v_i, v_j)$ only once, ie. if $(E, A)$, then $\lnot (E, [someOtherNode]) $. But we clearly see here, that we have $(E,A)$ and $(E,D)$. </p>
<p>How should i correctly interpret this phrase. What does it mean exactly? Thanks.</p>
| Matthew Leingang | 2,785 | <p>From your subset $S$ of vertices you can construct a set of pairs $P$, where $(v,w) \in P$ if there is a directed path from $v$ to $w$ (not necessarily confined to $S$). The set is strongly connected if
$$
\forall v,w\in S, ( v\neq w \Rightarrow (v,w) \in P \wedge (w,v) \in P)
$$
Let $S=G$ in the example graph. Notice that $(A,B)$, $(B,C)$, $(C,D)$, $(D,E)$, $(E,A)$ are all in $P$. So by transitivity, every pair of distinct vertices is in $P$.</p>
<p>The word <em>distinct</em> in the definition rules out the requirement that $(v,v) \in P$. However, it seems that $(v,w) \in P \wedge (w,v) \in P$ would imply that $(v,v)\in P$ anyway.</p>
<p>I'm not sure what you mean by “valid” pair. Any pair <em>may</em> be in $P$; the condition is only that certain pairs <em>must</em> be in $P$.</p>
|
195,432 | <p>Mathematica 12 has a new function <code>FunctionCompileExportLibrary</code>, which can export a pure function as a <code>.dll</code> file like this:</p>
<pre><code>FunctionCompileExportLibrary["function.dll", Function[
Typed[arg, "MachineInteger"], arg + 1]]
</code></pre>
<blockquote>
<p><code>"...\\function.dll"</code></p>
</blockquote>
<p>But how to call it in an external program (e.g., visual studio)? I have no <code>.h</code> file, I have no <code>.lib</code> file. I even don't know the function name that I want to call... Can anyone tell me how to use it?</p>
<p>The documentation of <a href="https://reference.wolfram.com/language/ref/FunctionCompileExportLibrary.html" rel="noreferrer"><code>FunctionCompileExportLibrary</code></a> states that</p>
<blockquote>
<p>The library generated by FunctionCompileExportLibrary is suitable for linking into external programs. It can also be loaded into the Wolfram System using LibraryFunctionLoad.</p>
</blockquote>
| sunt05 | 260 | <p>This is a follow-up of <a href="https://mathematica.stackexchange.com/a/195557/260">@ilian's answer</a>, which doesn't work under macOS but provides a promising basis. </p>
<p>The error reported by <a href="https://mathematica.stackexchange.com/questions/195432/how-to-use-the-library-exported-from-functioncompileexportlibrary-outside-of-mat#comment508843_195557">@happy fish</a> that <code>dyld: Library not loaded: @rpath/function.dylib Referenced from: ... Reason: image not found Abort trap: 6</code> suggests the linking was not correct. Then I tried to correct the @rpath issue (which seems to be a bug but I'm not quite sure; this needs further investigation) and still got other linking issues which clearly indicated <code>libWolframRTL_Static_Minimal.a</code> was not statically linked (which is <a href="https://developer.apple.com/library/archive/qa/qa1118/_index.html" rel="nofollow noreferrer">a well-known pain on macOS</a>).</p>
<p>So below I provide a workaround for macOS, which used the <code>FunctionCompileExport</code> instead of <code>FunctionCompileExportLibrary</code> due to possible linking bug in it:</p>
<pre><code>pathStaticLib="/Applications/Mathematica.app/Contents/SystemFiles/Libraries/MacOSX-x86-64/libWolframRTL_Static_Minimal.a";
lib=FunctionCompileExport["function.o",Function[Typed[arg,"MachineInteger"],arg+1];
Needs["CCompilerDriver`"]
test=CreateExecutable["
#include <stdio.h>
#include <inttypes.h>
int64_t Main(int64_t);
int main() {
printf(\"%\" PRId64, Main(10));
}","test","CompileOptions"->pathStaticLib,"ExtraObjectFiles"->{lib}];
RunProcess[test]
(*<|"ExitCode"->0,"StandardOutput"->"11","StandardError"->""|>*)
</code></pre>
|
244,048 | <p>A couple days ago I posted this on MSE (<a href="https://math.stackexchange.com/questions/1853774/determining-a-function-is-harmonic-from-mean-value-property-for-just-three-ra">here</a>) but in retrospect it might be more appropriate for this site.</p>
<p>This theorem is well-known (maybe it can be called Morera's theorem):</p>
<blockquote>
<p>A continuous function satisfying the mean value property on balls is harmonic.</p>
</blockquote>
<p>I was recently surprised to hear in a talk that the conclusion still holds if you only check the mean value property on three (I think) radii. Does anyone have a reference or name for this result? I would enjoy seeing the details and a proof.</p>
| user111 | 89,429 | <p>I guess this is the theorem :</p>
<p>Let $f$ be an infinitely differentiable function defined in $\mathbb R^{n}$, and $u(x,r)$ the mean value of $f$ taken over the sphere with center at $x$ and radius $r$, and let $a$ and $b$ denote two fixed positive numbers. If $u(x,a)=u(x,b)=f(x)$ in $\mathbb R^{n}$, then $f$ is harmonic. When $n>3$, exception must be made of a finite number of ratios $a/b$ which are independent of the function $f$.</p>
<p>The theorem is proved in :</p>
<p>J. Delsarte, J.-L. Lions, Moyennes généralisées, Comment. Math. Helv. 33 1959 59–69. </p>
<p>also in the lecture notes (see in particular Chapter III) :</p>
<p>J. Delsarte, Lectures on topics in mean periodic functions and the two-radius theo-
rem, Tata Institute, Bombay, 1961. <a href="http://www.math.tifr.res.in/~publ/ln/tifr22.pdf" rel="nofollow">http://www.math.tifr.res.in/~publ/ln/tifr22.pdf</a></p>
|
116,494 | <p>I am trying to plot contour plots and some of them show a white patch because of a change in sign of the points plotted. How can I make the contour plot show this change in sign without having to take the absolute of the values?</p>
<p>The data is the following:</p>
<pre><code>axialP1plot= {{30, 0, 0.509185}, {60, 0, 0.474159}, {90, 0, 0.452413}, {120, 0,
0.450468}, {0, 0.6, 0.422016}, {30, 0.6, 0.365962}, {60, 0.6,
0.263496}, {90, 0.6, 0.200892}, {120, 0.6, 0.188312}, {0, 1.2,
0.316512}, {30, 1.2, 0.140834}, {60, 1.2, -0.129596}, {90,
1.2, -0.248149}, {120, 1.2, -0.246671}, {0, 1.8, 0.211008}, {30,
1.8, -0.454357}, {60, 1.8, -1.08961}, {90, 1.8, -1.13311}, {120,
1.8, -0.979351}, {0, 2.4, 0.105504}, {30, 2.4, -3.50926}, {60,
2.4, -3.56817}, {90, 2.4, -2.62564}, {120, 2.4, -1.96434}, {0, 2.9,
0.017584}, {30, 2.9, -13.3785}, {60, 2.9, -6.04938}, {90,
2.9, -3.59638}, {120, 2.9, -2.49518}, {0, 0, 0.52752}}
</code></pre>
<p>And the contour plot:</p>
<pre><code>ListContourPlot[axialP1plot,
ColorFunction -> "Rainbow",
Contours -> 10,
PlotLegends -> Automatic,
PlotLabel -> Style["Axial Force on Plate 1", FontSize -> 14]]
</code></pre>
<p>Which yields:</p>
<p><a href="https://i.stack.imgur.com/PxgKQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PxgKQ.jpg" alt="enter image description here"></a></p>
| m_goldberg | 3,066 | <p>As george2079 say in his comment to your question, add the option <code>PlotRange -> All</code></p>
<pre><code>ListContourPlot[axialP1plot,
PlotRange -> All,
ColorFunction -> "Rainbow",
Contours -> 10,
PlotLegends -> Automatic,
PlotLabel -> "Axial Force on Plate 1"]
</code></pre>
<p><a href="https://i.stack.imgur.com/egqCW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/egqCW.png" alt="plot"></a></p>
|
3,827,302 | <p>In part of Kenneth Falconer's proof of the countable stability of Hausdorff dimension on p. 49, sect 3.2 of <a href="https://www.wiley.com/en-us/Fractal+Geometry%3A+Mathematical+Foundations+and+Applications%2C+3rd+Edition-p-9781119942399" rel="nofollow noreferrer">Fractal Geometry</a>, I understand him to say that
<span class="math-container">$$\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}\;,$$</span>
because when <span class="math-container">$s>\dim_H F_i$</span> for all <span class="math-container">$i$</span>, <span class="math-container">${\cal H}^s(F_i)=0$</span>, and thus
<span class="math-container">$${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0\;.$$</span>
Here <span class="math-container">$\dim_H$</span> is Hausdoff dimension, and <span class="math-container">${\cal H}^s$</span> is <span class="math-container">$s$</span>-dimensional Hausdorff measure.</p>
<p>I understand everything after <span class="math-container">$s>\dim_H F_i$</span> above, but I'm confused about why <span class="math-container">${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0$</span> implies <span class="math-container">$\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}$</span>.</p>
<p>I understand that Hausdorff dimension of a set <span class="math-container">$G$</span> is the Hausdorff measure for which <span class="math-container">${\cal H}^s(G)$</span> is finite, such that for <span class="math-container">$s>\dim_H G$</span>, the Hausdorff measure is <span class="math-container">$0$</span>, and for <span class="math-container">$s<\dim_H G$</span>, the Hausdorff measure is infinite. I don't see why the fact that <span class="math-container">${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq 0$</span> for an <span class="math-container">$s$</span> that is larger than the dimension implies the first inequality above, which concerns a <span class="math-container">$\sup$</span> for Hausdorff dimensions that are possibly greater than <span class="math-container">$0$</span>. I'm sure there must be something obvious that I'm not seeing. I've been thinking about it for a week and I'm still confused.</p>
<p><a href="https://math.stackexchange.com/a/772605/73467">This answer</a> gives a detailed proof of the part I already understand, but leaves my question unanswered.</p>
| Kevin Arlin | 31,228 | <p>"Composability" and "invertibility" are not, as you've noted, really the relevant primitive notions in a quasicategory. But horn-filling accounts for all the possibilities you want. The way to make this all make sense is to consider your quasicategory as generalizing the nerve of a 2-category. Given a 2-category <span class="math-container">$\mathcal K$</span>, its nerve has <span class="math-container">$0$</span>-simplices the objects of <span class="math-container">$\mathcal K$</span> and 1-simplices the 1-morphisms; a 2-simplex with boundary
<span class="math-container">\begin{array}{ccc}
x&\xrightarrow{f}&y\\&\searrow \scriptsize{h}&\downarrow \scriptsize g\\&&z
\end{array}</span>
is a 2-morphism <span class="math-container">$\alpha:g\circ f\to h$</span>. Higher simplices then arise from pasting diagrams in <span class="math-container">$\mathcal K$</span>, much as for the nerve of an ordinary category. Thus the 2-simplices in a quasicategory aren't quite what you think of when you picture a 2-morphism; if <span class="math-container">$f$</span> is an identity, though, then such a 2-simplex corresponds precisely to a 2-morphism <span class="math-container">$g\to h$</span>.</p>
<p>With this perspective, the construction you suggest does indeed capture the notion of composition of <span class="math-container">$\sigma_1$</span> and <span class="math-container">$\sigma_2$</span>. Specifically, if the edges <span class="math-container">$0\to 1$</span> and <span class="math-container">$1\to 2$</span> are degenerate, then choosing the doubly degenerate 2-simplex for the <span class="math-container">$0\to 1\to 2$</span> face defines a composite <span class="math-container">$\sigma_1\circ \sigma_2$</span> that agrees with the composite in the 2-category <span class="math-container">$\mathcal K$</span> in case your quasicategory is the nerve of <span class="math-container">$\mathcal K$</span>.</p>
<p>As for invertibility, we can tell a similar story. Given <span class="math-container">$\sigma_1$</span> with, again, <span class="math-container">$0\to 1$</span> degenerate, one can construct an "inverse" by filling a horn with <span class="math-container">$\sigma_1$</span> as the <span class="math-container">$0\to 1\to 3$</span> face, the <span class="math-container">$0\to 1\to 2$</span> face double degenerate, and the <span class="math-container">$0\to 2\to 3$</span> face degenerate on the nondegenerate edge of <span class="math-container">$\sigma_1$</span>. Again, in case your quasicategory is the nerve of the 2-category <span class="math-container">$\mathcal K$</span>, this reconstructs the inverse of the 2-morphism represented by <span class="math-container">$\sigma_1$</span>.</p>
<p>Your construction gives a good generalization of composition to 2-morphisms, but in fact the most natural notion of composition of 2-morphisms in a quasicategory is to compose together any <em>three</em> 2-morphisms that fit together into an outer horn. That is, there's no good reason, from the perspective of the quasicategory, to focus on filling horns where the <span class="math-container">$0\to 1\to 2$</span> face is degenerate.</p>
<p>On the other hand, to talk about invertibility in a quasicategory it really helps to make some edges degenerate. If we picture a 2-simplex as a 2-morphism <span class="math-container">$(g,f)\to h$</span>, then it doesn't make sense to ask for an inverse <span class="math-container">$h\to (g,f)$</span>. A quasicategorical way of stating formally that a quasicategory "is" an <span class="math-container">$(\infty,1)$</span>-category is, then, that "every special outer horn has a filler", where an outer horn is special if its <span class="math-container">$0\to 1$</span> edge (in the case of a 0-horn) or its <span class="math-container">$n-1\to n$</span> edge (in the case of an <span class="math-container">$n$</span>-horn) is an equivalence (which means it might as well be degenerate.)</p>
|
3,411,982 | <p>I know the answer using
Total-none of it is <span class="math-container">$6 = 900−648=252$</span></p>
<p>My doubt is that if we you it like</p>
<p>when unit digit is <span class="math-container">$6 = 9\cdot10\cdot1 = 90$</span></p>
<p>when tens place is <span class="math-container">$6 = 9\cdot1\cdot10 = 90$</span></p>
<p>when hundredth place is <span class="math-container">$6 = 1\cdot10\cdot10 = 100$</span></p>
<p>So total <span class="math-container">$= 90 + 90 + 100 = 280$</span></p>
<p>why my answer is not the same </p>
| Community | -1 | <p>If exactly one digit is a <span class="math-container">$6,$</span> there are <span class="math-container">$1\cdot 9\cdot 9+8\cdot1 \cdot 9+8\cdot 9 \cdot 1=225$</span> possibilities.</p>
<p>If exactly two digits are a <span class="math-container">$6,$</span> there are <span class="math-container">$9+9+8=26$</span> possibilities.</p>
<p>If exactly three digits are <span class="math-container">$6,$</span> there is only one possibility.</p>
<p>Thus there are <span class="math-container">$252$</span> possibilities in total.</p>
<p>Just to clear things up,</p>
<p>If the units digit is <span class="math-container">$6,$</span> then you should have <span class="math-container">$8\cdot9\cdot1$</span> possibilities, not what you got. </p>
<p>If the tens digit is <span class="math-container">$6,$</span> then you should have <span class="math-container">$8\cdot1\cdot9=81$</span> possibilities. </p>
<p>If the hundreds digit is <span class="math-container">$6,$</span> then there are <span class="math-container">$1\cdot9\cdot9$</span> possibilities. This is because you cannot include <span class="math-container">$0$</span> as the hundreds digit and you must subtract <span class="math-container">$1$</span> from <span class="math-container">$10$</span> since the other digits cannot be <span class="math-container">$6.$</span> Also you need to add the possibilities where there is more than one <span class="math-container">$6.$</span> </p>
|
1,007,529 | <p>Suppose that the function $f:[-1,1]\rightarrow[-1,1]$ is continuous. Use the intermediate value theorem to prove that there exists a $c \in [-1,1]$ such that $f(c)=c^5$. You should carefully justify each of the hypothesis of the theorem. Question is to be done very formally.</p>
<p>First can I just say, shouldn't $c$ be $c \in (-1,1)$ because that is what the theorem has. So I am just going to assume that it is meant to say $c \in (-1,1)$. If this is not the case, then please let me know. Also this is what my thoughts are:</p>
<p>$f$ is continuous and is defined by $f(x)=x^5$ for $x\in[-1,1]$ so $f(-1)=-1<0$ and $f(1)=1$. </p>
<p>So $f(-1)$ is not equal to $f(1)$. And $f(-1) < 0 < f(1)$. By the intermediate value theorem there exists a point $c\in(-1,1)$ such that $f(c)=0$ that is $c^5=0$.</p>
<p>Is this correct?</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Take $g(x)=f(x)-x^5$. </p>
<p>Then $g(-1)=f(-1)+1\ge 0$, while $g(1)=f(1)-1\le 0$. Using the Intermediate Value Theorem for $g$, we obtain that there exists a $c\in[-1,1]$, such that
$$
f(c)=c^5.
$$</p>
|
1,557,206 | <p>Given the series $\sum\limits_{n=1}^\infty a_n^2$ and $\sum\limits_{n=1}^\infty b_n^2$ converge. Show that the series $\sum\limits_{n=1}^\infty a_n b_n$ converges absolutely.</p>
<p>My idea so far:</p>
<ul>
<li>It's quite quite obvious that both given series converge absolutely</li>
<li>So the Cauchy-Produc tells me that $\sum\limits_{n=1}^\infty a_n^2 b_n^2 = \sum\limits_{n=1}^\infty (a_n b_n)^2$ converges absolutely</li>
</ul>
<p>I got stuck at that point. Can somehow give me a hint how to solve this ?</p>
<p>Thanks in advance!</p>
| Michael Hardy | 11,667 | <p>\begin{align}
0 & \le (a+b)^2 = a^2 + b^2 + 2ab \\
0 & \le (a-b)^2 = a^2 + b^2 - 2ab \\[10pt]
\text{Therefore} \\
-2ab & \le a^2+b^2, \\
2ab & \le a^2 + b^2, \\[10pt]
\text{and consequently} \\
2|ab| & \le a^2 + b^2.
\end{align}</p>
<p>So
$$
\sum_n |a_n b_n| \le \frac 1 2 \left( \sum_n a_n^2 + \sum_n b_n^2 \right) < \infty.
$$</p>
|
2,014,806 | <p>I was hoping to get some help on a review problem from my text book. The problem reads: </p>
<p>Let $A$ be $m\times{n}$ matrix. Prove that</p>
<p>a) $A^*A$ is self-adjoint.</p>
<p>b) All eigenvalues of $A^*A$ are non-negative.</p>
<p>c) $A^*A + I$ is invertible.</p>
<p>For part a), it seems to follow very clearly that $(A^* A)^* =(A)^* (A^*)^* =A^*A$. So this first part is clear.</p>
<p>With parts b) and c) I am unsure how to begin. For b), would one want to take the inner product of $(A^*{A}x,x)$ and then say some $Ax=qx$ where $q$ is an eigenvalue and then get $(Ax,Ax)=(qx,qx)= |q|^2\|x\|^2$? So the eigenvalue is positive? That is what I am thinking but I am unsure if it shows what I am wanting to show/ does not have a ton of holes. With part c) I am just lost so would appreciate any advice possible on that one. Thank you!</p>
<p>Upon more thought, I have a new idea for b). So we have some A* Ax =qx. So (Ax,Ax)=(A* Ax,x)=q(x,x). So q=((Ax,Ax)/(x,x))=||Ax||/||x||, which is greater than or equal to zero, so non-negative. would that work?</p>
| Robert Lewis | 67,071 | <p>For part (b), if</p>
<p><span class="math-container">$A^\dagger A \vec v = \mu \vec v, \; \vec v \ne 0, \tag 1$</span></p>
<p>then</p>
<p><span class="math-container">$\mu \langle \vec v, \vec v \rangle = \langle \vec v, \mu \vec v \rangle = \langle \vec v, A^\dagger A \vec v \rangle = \langle A\vec v, A \vec v \rangle \ge 0; \tag 2$</span></p>
<p>since</p>
<p><span class="math-container">$\langle \vec v, \vec v \rangle > 0, \tag 3$</span></p>
<p>(2) yields</p>
<p><span class="math-container">$\mu = \dfrac{\langle A\vec v, A \vec v \rangle}{\langle \vec v, \vec v \rangle} \ge 0. \tag 4$</span></p>
<p>For (c), if <span class="math-container">$A^\dagger A +I$</span> is <em>not</em> invertible, then</p>
<p><span class="math-container">$\exists \vec v \ne 0, \; A^\dagger A \vec v + \vec v = (A^\dagger A + I) \vec v = 0; \tag 5$</span></p>
<p>then</p>
<p><span class="math-container">$A^\dagger A \vec v = -\vec v, \tag 6$</span></p>
<p>whence</p>
<p><span class="math-container">$\langle \vec v, A^\dagger A \vec v \rangle = -\langle \vec v, \vec v \rangle; \tag 7$</span></p>
<p>but</p>
<p><span class="math-container">$\langle \vec v, A^\dagger A, \vec v \rangle = \langle A\vec v, A\vec v \rangle \ge 0, \tag 8$</span></p>
<p>and</p>
<p><span class="math-container">$-\langle \vec v, \vec v \rangle < 0; \tag 9$</span></p>
<p>combining (7), (8), and (9) yields</p>
<p><span class="math-container">$0 \le \langle \vec v, A^\dagger A \vec v \rangle = -\langle \vec v, \vec v \rangle < 0, \tag{10}$</span></p>
<p>or</p>
<p><span class="math-container">$0 < 0; \tag{11}$</span></p>
<p>this contradictory statement shows that there can be no such <span class="math-container">$\vec v \ne 0$</span>; hence</p>
<p><span class="math-container">$\vec v = 0, \tag{12}$</span></p>
<p>and thus <span class="math-container">$A^\dagger A + I$</span> is invertible.</p>
|
2,278,436 | <p>Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer? I have no clue how to begin. I've tried to simplify this somehow, but with no effect.</p>
| Robert Israel | 8,508 | <p>If there is a prime (or, for that matter a prime power) dividing only one of the denominators, then the sum is not an integer. Now by the prime number theorem in arithmetic progressions, for any fixed $a$ this holds for sufficiently large $n$, as there will be a prime $p = 1 + k a$ for some $k$ with $n/2 < k \le n$. I don't know if there is a good enough result to
work for every $a$ and $n$.</p>
<p>EDIT: I tried all $a$ from $2$ to $100$ and $n$ from $2$ to $10000/a$. In all these cases there was at least one prime that divided only one of the denominators. Of course that is not a proof that it works for all $n$ and $a$.</p>
|
868,663 | <blockquote>
<p>Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.</p>
</blockquote>
<p>I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.</p>
| Adriano | 76,987 | <p>Suppose we add $372$ and $594$. Using "the addition method that we know", we would first add $2$ and $4$ to get $6$. Then we would add $7$ and $9$ to get $16$, so we write down a $6$ then carry the $1$. Then we add the carried $1$ with the $3$ and $5$ to get $9$, for a final answer of $966$. Why does this work?</p>
<p>Well what we're really doing is taking advantage of how numbers are written in base $10$:
\begin{align*}
372 + 594
&= (300 + 70 + 2) + (500 + 90 + 4) \\
&= (300 + 500) + (70 + 90) + (2 + 4) \\
&= (300 + 500) + (70 + 90) + (6) \\
&= (300 + 500) + (160) + (6) \\
&= (300 + 500) + (100 + 60) + (6) \\
&= (300 + 500 + 100) + (60) + (6) \\
&= (900) + (60) + (6) \\
&= 966
\end{align*}
This idea indeed generalizes, even to numbers written in other bases.</p>
|
3,772 | <p>In <a href="https://math.stackexchange.com/questions/102764/non-polynomial-functions-on-fields-of-finite-characteristic">this</a> question, there are two answers, both of which I up-voted, but I didn't "accept" either since there didn't appear to be a reason why one stands out as clearly preferable to the other. Apparently, my lack of "acceptance" offended someone who's taking me to task for it in comments on one of my questions. Some people (or at least one person) seem to get rather passionate about acceptances. Has anyone defined codified some criteria that say when one should "accept" an answer and what that is supposed to mean?</p>
| Willie Wong | 1,543 | <p>Firstly, I fully endorse Rahul's comment. <code>:-)</code></p>
<p>Secondly, let me explain, or at least give my point of view, on why some users are getting worked up about the acceptance rate. There are two main ways of user-interaction which can be used to express gratitude for providing an answer beyond a simple "thanks." They are "up votes" and "answer acceptances". (There's also "bounties", which we'll ignore for now.) Many people have their own ideas about what each of the two means, so I won't try to attach meaning to them. But I will point out that there are two main technical differences:</p>
<ol>
<li>Up votes are worth less reputation than acceptances (+10 vs +15)</li>
<li>Acceptances can be more easily rescinded than votes (a vote can only be changed within a certain time frame, or after an edit was made to the corresponding answer; an answer acceptance can be removed any time). </li>
</ol>
<p>Because of the difference, people regard the two with different values. Now, in spite of the actual intentions behind a questioner deciding not to accept an answer, some users may interpret a low acceptance rate as (the list is not intend to be complete, but just a description of some of the comments I've seen)</p>
<ol>
<li>the questioner being "stingy" with an intangible commodity with no actual worth and which violates the "have a cake and eat it too" rule. </li>
<li>the questioner is "leeching" from the community; while answers here are given voluntarily and time spent here are of the users' own wills, some people still think that people who ask questions should "give something back" to the community, be it providing answers or giving people reputation points. </li>
</ol>
<p>I will not debate whether either of the views are reasonable. This is just to give you some ideas on what may be behind comments calling you on your accept rate. </p>
<p>Thirdly, for (not-so-)new users with low reputation and a habit of asking poorly motivated questions, I do sometimes decide not to answer a question because of extremely low acceptance rates. This is due to mainly lack of feedback. If the user consistently refuse to provide the context in which the questions are asked, nor any details about his/her personal mathematical background, <em>and</em> if the user does not indicate whether previous answers to his or her questions are actually useful, how am I to know whether the answer I have in mind will be useful to the person who asked? This is especially frustrating in the case when I know I can explain the same thing in several different levels, but the user gives no indication which one level is most suitable. </p>
|
44,310 | <p>I wanted to draw (lattice-aligned) unit cubes inside a sphere with a given radius centered on the origin, so I wrote this program:</p>
<pre><code>Cube[x_, y_, z_] := Cuboid[{x - 1, y - 1, z - 1}, {x, y, z}];
Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@ # & /@
With[{t = Sqrt[r^2 - 1]},
Select[Flatten[Table[{x, y}, {x, 1, t}, {y, 1, x}], 1],
Norm[#] <= t &]];
Cubes[r_] :=
Cube @@ # & /@
Union[Flatten[Permute[#, SymmetricGroup[3]] & /@ Coords[r], 1]];
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250, 500, 1000, 2500, 6054}
</code></pre>
<p>(Note that it shows only the cubes on the outside -- drawing the inner ones takes too long and usually Mathematica runs out of memory.) It gets the job done, but it's pretty ugly. I'm not a regular Mathematica user, and tips for improving my code?</p>
<p>That is, how can I write idiomatically-better Mathematica code?</p>
<hr>
<p>Improved code:</p>
<pre><code>Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@@
With[{t = Sqrt[r^2 - 1]},
Select[Join @@ Table[{x, y}, {x, t}, {y, 1, x}], Norm[#] <= t &]];
Cubes[r_] := (Cuboid /@ (Union @@ (Permute[#, SymmetricGroup[3]] & /@
Coords[r]) - 1));
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250}
</code></pre>
| Michael E2 | 4,999 | <p><a href="http://reference.wolfram.com/mathematica/ref/Translate.html" rel="nofollow noreferrer"><code>Translate</code></a> is probably the most efficient way to represent and display such a figure.</p>
<pre><code>With[{r = 30},
Graphics3D[{
Translate[Cuboid[],
Union @@ (Permute[#, SymmetricGroup[3]] & /@ Coords[r])
],
{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}
}
]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/P9GFJ.png" alt="Mathematica graphics"></p>
|
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$</p>
<p>But apparently not. </p>
| k170 | 161,538 | <p>$$ \int \frac{1}{\sqrt{x^2-c}}dx = \int \frac{1}{\sqrt{c\left(\frac{x^2}{c}-1\right)}}dx=\frac{1}{\sqrt{c}} \int \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2-1}}dx $$
Let $u=\frac{x}{\sqrt{c}}$, then $du=\frac{1}{\sqrt{c}} dx$. So now we have
$$ \int \frac{1}{\sqrt{u^2-1}}du =\mathrm{arcosh}(u)+C = \mathrm{arcosh}\left( \frac{x}{\sqrt{c}}\right)+C $$</p>
|
4,365,859 | <p>My professor had given us a problem. It goes like -</p>
<blockquote>
<p>Find equation of circle passing through intersection of circle <span class="math-container">$S=x^2+y^2-12x-4y-10=0$</span> and <span class="math-container">$L:3x+y=10$</span> and having radius equal to that of circle <span class="math-container">$S$</span>.</p>
</blockquote>
<p>So when he was telling the solution, he setup an equation like -</p>
<p><span class="math-container">$x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$</span> where <span class="math-container">$t \in \mathbb{R}$</span></p>
<p><span class="math-container">$= x^2+y^2+2(3t-6)x+2(-2+t)y-(10+2t)=0$</span></p>
<p>So now we can find out that radius of <span class="math-container">$S=\sqrt{50}$</span></p>
<p>Now we can use for any general circle <span class="math-container">$r=\sqrt{g^2+f^2-c}$</span> to get ,</p>
<p><span class="math-container">$(6-3t)^2+(t-2)^2+10+20t=50\implies t=0,2$</span></p>
<p>Now we can remove <span class="math-container">$t=0$</span> as we will get <span class="math-container">$S$</span></p>
<p>So putting <span class="math-container">$t=2 \implies \boxed{x^2+y^2=50}$</span></p>
<p>Now this solution is ok, but I didn't get why putting <span class="math-container">$t$</span> worked? I mean why is <span class="math-container">$x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$</span> working? Can someone explain this?</p>
| Cye Waldman | 424,641 | <p>In spite of your obfuscating figure, you are asking for the surface area of a torus whose inner radius, <span class="math-container">$R$</span> (to the center of the cross-section) and outer radius, <span class="math-container">$r$</span> (that <em>of</em> the cross-section) are the same. This is well known to be <span class="math-container">$S=4\pi^2Rr$</span> (see, for example the CRC Mathematical Tables). So in your case, <span class="math-container">$S=4\pi^2a^2$</span></p>
<p>We can derive this result with Pappus's (<span class="math-container">$1^{st}$</span>) Centroid Theorem, which states that the surface area <span class="math-container">$S$</span> of a surface of revolution generated by rotating a plane curve <span class="math-container">$C$</span> about an axis external to <span class="math-container">$C$</span> and on the same plane is equal to the product of the arc length <span class="math-container">$s$</span> of <span class="math-container">$C$</span> and the distance <span class="math-container">$d$</span> traveled by its geometric centroid. Simply put, <span class="math-container">$S=2πRL$</span>, where <span class="math-container">$R$</span> is the normal distance of the centroid to the axis of revolution and <span class="math-container">$L$</span> is the curve length. In your case, <span class="math-container">$R=a$</span> and <span class="math-container">$L$</span> is the circumference of the circle, i.e., <span class="math-container">$=2\pi a$</span>, so that <span class="math-container">$S=4\pi^2a^2.$</span></p>
|
4,365,859 | <p>My professor had given us a problem. It goes like -</p>
<blockquote>
<p>Find equation of circle passing through intersection of circle <span class="math-container">$S=x^2+y^2-12x-4y-10=0$</span> and <span class="math-container">$L:3x+y=10$</span> and having radius equal to that of circle <span class="math-container">$S$</span>.</p>
</blockquote>
<p>So when he was telling the solution, he setup an equation like -</p>
<p><span class="math-container">$x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$</span> where <span class="math-container">$t \in \mathbb{R}$</span></p>
<p><span class="math-container">$= x^2+y^2+2(3t-6)x+2(-2+t)y-(10+2t)=0$</span></p>
<p>So now we can find out that radius of <span class="math-container">$S=\sqrt{50}$</span></p>
<p>Now we can use for any general circle <span class="math-container">$r=\sqrt{g^2+f^2-c}$</span> to get ,</p>
<p><span class="math-container">$(6-3t)^2+(t-2)^2+10+20t=50\implies t=0,2$</span></p>
<p>Now we can remove <span class="math-container">$t=0$</span> as we will get <span class="math-container">$S$</span></p>
<p>So putting <span class="math-container">$t=2 \implies \boxed{x^2+y^2=50}$</span></p>
<p>Now this solution is ok, but I didn't get why putting <span class="math-container">$t$</span> worked? I mean why is <span class="math-container">$x^2+y^2-12x-4y-10=0+2t(3x+y-10)=0$</span> working? Can someone explain this?</p>
| robjohn | 13,854 | <h3>Comment on the Question</h3>
<p>I believe that this is the standard setup for surface of revolution about the <span class="math-container">$x$</span>-axis
<span class="math-container">$$
\int_{-a}^a\overbrace{\quad2\pi y\quad\vphantom{\frac{a}{\sqrt{a^2}}}}^{\substack{\text{account for}\\\text{revolution}}}\overbrace{\frac{a}{\sqrt{a^2-x^2}}}^{\mathrm{d}s/\mathrm{d}x}\,\mathrm{d}x
$$</span>
However, the part for the upper arc of the circle does not give the same area as that for the lower arc of the circle, so we need to compute both separately:
<span class="math-container">$$
\begin{align}
&\int_{-a}^a2\pi\left(a+\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{upper}\\
&+\int_{-a}^a2\pi\left(a-\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{lower}\\
&=2\pi\int_{-a}^a2a\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\[6pt]
&=4\pi^2a^2
\end{align}
$$</span></p>
<hr />
<h3>Other Approaches</h3>
<p><strong>Revolution Around the <span class="math-container">$\boldsymbol{x}$</span>-axis</strong></p>
<p>As I had posted before I realized the question was revolving around the <span class="math-container">$y$</span>-axis, if we revolve around the <span class="math-container">$x$</span>-axis, the upper and lower parts of the surface have the same area, so we can just multiply the upper integral by <span class="math-container">$2$</span> in this case. Thus, the formula is
<span class="math-container">$$
\begin{align}
2\int_0^{2a}2\pi x\,\frac{a}{\sqrt{a^2-(x-a)^2}}\,\mathrm{d}x
&=2\int_{-a}^a2\pi(x+a)\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{1a}\\
&=4\pi a^2\int_{-1}^1(x+1)\frac1{\sqrt{1-x^2}}\,\mathrm{d}x\tag{1b}\\
&=4\pi a^2\int_{-\pi/2}^{\pi/2}(\sin(x)+1)\,\mathrm{d}x\tag{1c}\\[9pt]
&=4\pi^2a^2\tag{1d}
\end{align}
$$</span>
Explanation:<br />
<span class="math-container">$\text{(1a)}$</span>: substitute <span class="math-container">$x\mapsto x+a$</span><br />
<span class="math-container">$\text{(1b)}$</span>: substitute <span class="math-container">$x\mapsto ax$</span><br />
<span class="math-container">$\text{(1c)}$</span>: substitute <span class="math-container">$x\mapsto\sin(x)$</span><br />
<span class="math-container">$\text{(1d)}$</span>: integrate</p>
<p><strong>Parametrization</strong></p>
<p>Parametrize the torus as follows: at each point of the circle around the <span class="math-container">$z$</span>-axis, <span class="math-container">$a(\cos(\phi),\sin(\phi),0)$</span> put a circle perpendicular to this circle:
<span class="math-container">$$
\begin{align}
p(\phi,\theta)
&=\overbrace{a(\cos(\phi),\sin(\phi),0)}^\text{primary circle}+\overbrace{a(\cos(\phi)\cos(\theta),\sin(\phi)\cos(\theta),\sin(\theta))}^\text{secondary circle around the primary circle}\\
&=a(\cos(\phi)(1+\cos(\theta)),\sin(\phi)(1+\cos(\theta)),\sin(\theta))\tag{2a}]\\[6pt]
p_1(\phi,\theta)
&=a(-\sin(\phi)(1+\cos(\theta)),\cos(\phi)(1+\cos(\theta)),0)\\
&=a(1+\cos(\theta))(-\sin(\phi),\cos(\phi),0)\tag{2b}\\[6pt]
p_2(\phi,\theta)
&=a(-\cos(\phi)\sin(\theta),-\sin(\phi)\sin(\theta),\cos(\theta))\tag{2c}
\end{align}
$$</span>
Thus, we get
<span class="math-container">$$
\begin{align}
|p_1(\phi,\theta)\times p_2(\phi,\theta)|
&=a^2(1+\cos(\theta))\,|(\cos(\theta)\cos(\phi),\cos(\theta)\sin(\phi),\sin(\theta))|\\
&=a^2(1+\cos(\theta))\tag3
\end{align}
$$</span>
and we can compute
<span class="math-container">$$
\int_0^{2\pi}\int_0^{2\pi}a^2(1+\cos(\theta))\,\mathrm{d}\phi\,\mathrm{d}\theta=4\pi^2a^2\tag4
$$</span>
<strong>Theorem of Pappus</strong></p>
<p>As I mentioned in a comment, we can apply the <a href="https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem" rel="nofollow noreferrer">Theorem of Pappus</a>: the primary circle has circumference <span class="math-container">$2\pi a$</span> and the secondary circle has circumference <span class="math-container">$2\pi a$</span>, so the area is
<span class="math-container">$$
(2\pi a)(2\pi a)=4\pi a^2\tag5
$$</span></p>
|
3,949,872 | <p>I am to expand <span class="math-container">$\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$</span> using the quotient rule of logs. I arrived at:</p>
<p><span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>Whereas my textbook solution is:</p>
<p><span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>It looks like I got it right for the denominator part but not the numerator. I don't see how to arrive at <span class="math-container">$\log_3(x+3)$</span> from <span class="math-container">$7x^2+21x$</span>?</p>
<p>My train of thought was to divide <span class="math-container">$7x^2$</span> in the numerator by the <span class="math-container">$7x$</span> in the denominator leaving me with <span class="math-container">$7x+21x = 28x$</span> in the numerator and just <span class="math-container">$(x-1)(x-2)$</span> in the denominator.</p>
<p>With <span class="math-container">$28x$</span> in the numerator I expanded out to be <span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x).$$</span></p>
<p>Where did I go wrong, and how can I arrive at <span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
| 19aksh | 668,124 | <p>First isolate <span class="math-container">$7x$</span> from the numerator. You then have <span class="math-container">$\log_3\left[\frac{7x(x+3)}{7x(x-1)(x-2)}\right] = \log_3\left[\frac{(x+3)}{(x-1)(x-2)}\right] _{(x\ne0)} = \log_3(x+3)-\log_3(x-1)-\log_3(x-2)$</span></p>
|
3,949,872 | <p>I am to expand <span class="math-container">$\log_3\frac{7x^2+21x}{7x(x-1)(x-2)}$</span> using the quotient rule of logs. I arrived at:</p>
<p><span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>Whereas my textbook solution is:</p>
<p><span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
<p>It looks like I got it right for the denominator part but not the numerator. I don't see how to arrive at <span class="math-container">$\log_3(x+3)$</span> from <span class="math-container">$7x^2+21x$</span>?</p>
<p>My train of thought was to divide <span class="math-container">$7x^2$</span> in the numerator by the <span class="math-container">$7x$</span> in the denominator leaving me with <span class="math-container">$7x+21x = 28x$</span> in the numerator and just <span class="math-container">$(x-1)(x-2)$</span> in the denominator.</p>
<p>With <span class="math-container">$28x$</span> in the numerator I expanded out to be <span class="math-container">$$\log_3(7)+\log_3(4)+\log_3(x).$$</span></p>
<p>Where did I go wrong, and how can I arrive at <span class="math-container">$$\log_3(x+3)-\log_3(x-1)-\log_3(x-2)$$</span></p>
| CACM6 | 814,401 | <p>You can factor the numerator and denominator, both have a common factor <span class="math-container">$7x$</span>, so you have
<span class="math-container">$$\frac{7x^2+21x}{7x(x-1)(x-2)}=\frac{x+3}{(x-1)(x-2)},$$</span> now you can use the <span class="math-container">$\log$</span> rules and get the answer that the book gives</p>
|
1,531,525 | <p>I would like to compute the following sum:</p>
<p>$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$</p>
<p>I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated. </p>
| 2'5 9'2 | 11,123 | <p>If you are only looking for a hint, write your sum as $$\sum_{n=0}^\infty\frac{e^{in\theta}+e^{-in\theta}}{2\cdot2^n}$$ Break it up as two sums, each of which are geometric, so you can use the geometric series formula.</p>
|
209,259 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/16374/universal-chord-theorem">Universal Chord Theorem</a> </p>
</blockquote>
<p>I am having a problem with this exercise. Could someone help?</p>
<p>Suppose $a \in (0,1)$ is a real number which is not of the form $\frac{1}{n}$ for any natural number n
n. Find a function f which is continuous on $[0, 1]$ and such that $f (0) = f (1)$ but which does not satisfy $f (x) = f (x + a)$ for any x with $x$, $x + a \in [0, 1]$.</p>
<p>I noticed that this condition is satisfied if and only if $f(x) \geq f(0)$</p>
<p>Thank you in advance</p>
| fgp | 42,986 | <p>Look at $f(x) = \sin(2\pi x)$. For which values of $a$ can you find an $x \in [0,1]$ with $x+a \in [0,1]$ and $f(x) = f(x+a)$? In particular, if you additionally require $a > \frac{1}{2}$, can such an $a$ exist at all?</p>
<p>Once you've answered that you've solved your problem for some values of $a$. Which are those?</p>
<p>A general solution can be found in the answers to <a href="https://math.stackexchange.com/questions/16374">Universal Chord Theorem</a> (Link found by the user who asked the question). To quote $$
f(x) = \sin^2\left(\frac{\pi x}{a}\right) - x \ \sin^2\left(\frac{\pi}{a}\right)
$$</p>
<p>is a solution. This works because $f(x) = f(x+a)$ implies $a \sin^2\left(\frac{\pi}{a}\right) = 0$ and thus $a=\frac{1}{n}$ for some $n \in \mathbb{N}$. The answers to the linked questions also prove that $a \neq \frac{1}{n}$ for every $n \in \mathbb{N}$ is a necessary condition for a solution to exist.</p>
|
224,474 | <p>I want to generate the following matrix for any n:</p>
<pre><code> Table[A[i, j], {i, 0, n}, {j, 0, n}];
</code></pre>
<p>Where,</p>
<pre><code> A ={{0,1,0},{0,0,2},{0,0,0}} when n=2;
A={{0,1,0,0},{0,0,2,0},{0,0,0,3},{0,0,0,0}} when n=3;
A={{0,1,0,0,0},{0,0,2,0,0},{0,0,0,3,0},{0,0,0,0,4},{0,0,0,0,0}} when n=4, and so on for any n.
</code></pre>
<p>Thanks</p>
| Pillsy | 531 | <p>I sometimes like using <code>Module</code> to define local functions (more precisely, local symbols with <code>DownValues</code>) to use the ridiculous generality Mathematica provides for such definitions:</p>
<pre><code>ClearAll[specialMatrix];
specialMatrix[n_Integer?Positive] := specialMatrix[n] =
Module[{a, between = Between[{0, n}]},
a[i_Integer?between, j_Integer?between] /; j == i + 1 = i + 1;
a[_Integer, _Integer] = 0;
Table[a[i, j], {i, 0, n}, {j, 0, n}]]
specialMatrix[3]
(* {{0, 1, 0, 0}, {0, 0, 2, 0}, {0, 0, 0, 3}, {0, 0, 0, 0}} *)
specialMatrix[4]
(* {{0, 1, 0, 0, 0}, {0, 0, 2, 0, 0}, {0, 0, 0, 3, 0}, {0, 0, 0, 0, 4}, {0, 0, 0, 0, 0}} *)
</code></pre>
<p>The memoization in the definition of <code>specialMatrix</code> is optional but may be worthwhile if you repeatedly use <code>specialMatrix</code> to generate large matrices.</p>
|
199,946 | <p>I came across this step in an inductive proof, but my algebra skills seem a bit rusty..</p>
<p>$7(7^{k+2}) + 64(8^{2k+1}) = 7(7^{k+2}+8^{2k+1})+57(8^{2k+1})$</p>
<p>How did they do this?</p>
<p>Note: The point was to show that the expressions are divisible by 57.</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\ $ With $\rm\: A = 7^{\,k+2}\:$ and $\rm\: B = 8^{\,2k+1}\:$ it is more easy to comprehend</p>
<p>$$\rm 7\, A + 64\ B\ =\ 7\,(A + B) + 57\ B $$</p>
|
1,221,206 | <p>In this computer, numbers are stored in <span class="math-container">$12$</span>-bits. We will also assume
that for a floating point (real) number, <span class="math-container">$6$</span> bits of these bits are reserved for
the mantissa (or significand) with <span class="math-container">$2^{k-1}-1$</span> as the exponent bias (where
<span class="math-container">$k$</span> is the number of bits for the characteristic).</p>
<p><span class="math-container">$011100100110010111110011$</span></p>
<p>What pair of floating point numbers could be represented by these
<span class="math-container">$24$</span>-bits?</p>
<p>I have gone this far:</p>
<p>As described above that each number is of <span class="math-container">$12$</span> bit so we get each number </p>
<p><span class="math-container">$011100100110$</span></p>
<p>First one is <span class="math-container">$0$</span> bit so it is positive and </p>
<p>Mantissa will be <span class="math-container">$100110$</span></p>
<p>Exponent will be <span class="math-container">$11100b=28$</span></p>
<p>my unbiased exponent will be <span class="math-container">$2^{28-15}=2^{13}$</span></p>
<p>How to find the floating point number from here?</p>
| Rellek | 228,621 | <p>In a summation, any constant can be pulled out (your constant is 3). If you do this, you will see that this is a geometric series. This means that each term only differs by a constant from the preceding term. In this case it is easy to see that this constant will be 1/4. Note that $$\frac{1}{4^n} = \frac{1^n}{4^n} = (\frac{1}{4})^n$$</p>
<p>Perhaps rewriting your summation in this way will help you see what is happening in this series a little more easily.</p>
<p>Also, a little hint: If you were to denote your summation as $3S$, where $S$ is only the series consisting of your $(\frac{1}{4})^n$ term, then consider what $$3(S-\frac{1}{4}S)$$ might simplify to.</p>
|
388,875 | <p>For $V$ a vector space.</p>
<p>Let $T\colon V\rightarrow V$ be a linear transformation of a finite dimensional inner product space.</p>
<p>How can I prove that $\ker(T^* T + TT^*) = $\ker(T)$\cap$$\ker(T^*)$?</p>
<p>And what is the meaning of $\ker(T^*T + TT^*)$ ?</p>
<p>Is $\ker(T^*T + TT^*)=\ker(T^*T)+\ker(TT^*)$ ?</p>
| Hagen von Eitzen | 39,174 | <p>$\ker(T^*T+TT^*)$ is the set of all $x\in V$ such that $$(T^*T+TT^*)x=0, $$
that is
$$ T^*Tx+TT^*x=0.$$
This is not the same as the set of all vectors $x$ that can writtten as $x=y+z$ with $T^*Ty=0$ and $TT^*z=0$. I.e., in general $\ker(T^*T+TT^*)\ne\ker(T^*T)+\ker(TT^*)$.</p>
|
497,609 | <p>There are 30 balls in a bag. There are 6 balls each of 5 different colors. What is the minimum number of times I have to pick a ball to pick 2 balls of each color?</p>
<ol>
<li>24</li>
<li>25</li>
<li>26</li>
<li>27</li>
<li>28</li>
</ol>
<p>Is the question correct? Someone asks me this and I'm confused.</p>
| obataku | 54,050 | <p>What we're counting is the minimum number of balls we could blindly pick yet be entirely sure we have two of each color. For example, liken it to picking socks blindly from a drawer -- we want to determine how many socks we have to pull out blindly to ensure we have a pair. </p>
<p>Think of the worst case scenario, where you go through all 6 of the first 4 colors before you get a ball of the last color. This means you would need to pick $4\times6$ balls before getting to your last color and to pick 2 of the last color we end with $4\times6+2=26$ balls. </p>
|
1,256,688 | <p>Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of 2015 distinct real numbers such that after one initial move is applied to the sequence — no matter what move — there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.</p>
| futurebird | 2,563 | <p><strong>(1) Suppose that I want to prove that if the statements A, B,C hold true, then Z holds.</strong> </p>
<p>So you want to prove this:</p>
<p>(A and B and C) ==> z</p>
<p>Here is the truth table: </p>
<pre><code>(A and B and C) | Z | (A and B and C) ==> Z
----------------------------------------------------
True | True | True
True | False | False
False | True | True
False | False | True
</code></pre>
<p>Notice that when (A and B and C) is False the implication is true either way. </p>
<p><strong>To prove this, I would assume A,B,C, which then implies D,E,F,G,H, which in turn implies I,J,K which in turn implies …, and so on. And I eventually get Z.</strong> </p>
<p>So your proof shows that whenever (A and B and C) is true Z is true. It does not show "Z is true."</p>
<p><strong>But then, how do I know that Z is actually true?</strong> </p>
<p>You don't. Z could be false if (A and B and C) is false. But you still proved that (A and B and C) ==> Z.</p>
<p><strong>That is, how do I know that some combinations of the statements say J,K,F which are obtained along the way, do not contradict Z?</strong></p>
<p>This is a different question. Since you are using those statements to show that Z is true, provided that your axioms or "first principles" are consistent you will not have a contradiction. </p>
<p><strong>(2) I know that if assuming A yields a contradiction, then A is false. But, what if assuming A does not contradict anything?</strong></p>
<p>A must be either true or false. But, there is no guarantee that we can prove that A is true or false if our axioms are consistent. The ambiguity in this question comes from the word "anything" -do you mean anything we can write with the given axioms and symbols? </p>
<p>If that is the case I don't know how you could prove that "A does not contradict anything" except by proving that it is true since (unless we are in a very boring simple system) we know there are some true statements with no proof within the system. In most example we don't know what these statements are, so how do we check if there is a contradiction.</p>
<p>If we instead show that A contradicts none of the axioms of the system that shows A is true, so ... </p>
<p><strong>Can we conclude that A is true? or what can we say about A?</strong></p>
<p>I will say yes though this 2nd question was a bit deeper than I thought at first.</p>
|
27,896 | <p>How would I solve these differential equations? Thanks so much for the help!</p>
<p>$$P'_0(t) = \alpha P_1(t) - \beta P_0(t)$$
$$P'_1(t) = \beta P_0(t) - \alpha P_1(t)$$</p>
<p>We also know $P_0(t)+P_1(t)=1$</p>
| Ross Millikan | 1,827 | <p>Use $P_0(t)+P_1(t)=1$ to turn it into $P_0'(t)=\alpha-(\alpha+\beta)P_0(t)$, which you should be able to solve.</p>
|
27,896 | <p>How would I solve these differential equations? Thanks so much for the help!</p>
<p>$$P'_0(t) = \alpha P_1(t) - \beta P_0(t)$$
$$P'_1(t) = \beta P_0(t) - \alpha P_1(t)$$</p>
<p>We also know $P_0(t)+P_1(t)=1$</p>
| Aryabhata | 1,102 | <p>There is a general method to solve such equations, if we view them as a linear system of equation</p>
<p>$$y'(x) = A y(x)$$</p>
<p>When $A$ is a matrix with constants, the solution can be written in terms of the <a href="http://en.wikipedia.org/wiki/Matrix_exponential" rel="nofollow">exponent matrix</a> $e^{Ax}$.</p>
<p>More details can be found <a href="http://en.wikipedia.org/wiki/Ordinary_differential_equation#Linear_ordinary_differential_equations" rel="nofollow">here</a>.</p>
|
1,060,933 | <p>It is necessary for me to find unit outward normal vector for the curve:</p>
<p>$$\gamma=(x(t),y(t))$$</p>
<p>where
$$x(t)=(0.6)\cos(t)-(0.3)\cos(3t)$$
and
$$y(t)=(0.7)\sin(t)+(0.07)\sin(7t)+(0.1)\sin(3t)$$
I know how to find unit outward normal vector for this: using</p>
<p>$$T=\frac{\gamma'(t)}{||\gamma(t)||},\;\text{ so }\,N=\frac{T'(t)}{||T(t)||}$$</p>
<p>but my problem is that I do not have $t$. Just I have $x(t)$ and $y(t)$.
How could I find $t$ or $N$ without need to $t$.</p>
<p>Is there any command in MATLAB or MAPLE to this?</p>
| John Hughes | 114,036 | <p>$$
\gamma'(t) = (-\sin(t) + 1.5 \sin(3t), \cos(t)+7\cos(7t)+3\cos(3t))\\
\| \gamma'(t) \| = \sqrt{(-\sin(t) + 1.5 \sin(3t))^2, (\cos(t)+7\cos(7t)+3\cos(3t))^2}
$$
In general, for a vector $(a, b)$ in the plane, $(-b, a)$ is perpendicular to it. </p>
<p>So your normal vector is
$$
N(t) = \frac{\pm (-\cos(t)-7\cos(7t)-3\cos(3t), -\sin(t) + 1.5 \sin(3t))}{\| \gamma'(t) \|}
$$</p>
<p>To choose the sign, you may want to make it point in the direction that has a positive dot product with $T'$; to do so by differentiating the quotient that defines $T$ would be a pain in the neck, but fortunately, you can merely make your nromal vector point in the direction that has positive dot product with $\gamma''$, since $T'$ ends up being a linear combination of this and a vector in the $T$ direction, which will not affect the dot product. So:</p>
<p>Compute
$$
s = (-\cos(t)-7\cos(7t)-3\cos(3t), -\sin(t) + 1.5 \sin(3t)) \cdot
(-\cos(t)+4.5\cos(3t), -\sin(t)-49\sin(7t)-9\sin(3t))
$$
and if $s$ is positive, select the "+" choice in the $\pm$ formula I gave you above. </p>
<p>If $s$ changes sign as a function of $t$, then the means that the curve has an inflection, and there's no way to continuously choose the normal vector to be on the "concave" side of the curve. </p>
|
2,193,306 | <p>I have to evaluate the integral of the $\sec x$ function and I do it as follows $$\int\sec xdx=\int\frac{dx}{\cos x}=\int\frac{\cos^2 x+\sin^2 x}{\cos x}dx=\sin x+\int\frac{\sin^2x}{\cos x}dx$$ Now we make a change of variables $\cos x=t$ so our integral becomes $$\sin x-\int\frac{\sqrt{1-t^2}}{t}dt$$ Now we make another change of variable $\sqrt{1-t^2}=z$ so our integral becomes $$\sin x-\int\frac{z^2-1+1}{z^2-1}dz=\sin x-\int\left(1+\frac{1}{z^2-1}\right)dz=$$ $$=\sin x-\sqrt{1-\cos^2x}+\frac12\ln\left|\frac{1+\sqrt{1-\cos^2x}{}}{1-\sqrt{1-\cos^2x}}\right|+C=\frac12\ln|\tan^2x+\sec^2x|+C$$ The <a href="http://www.integral-calculator.com/" rel="nofollow noreferrer">calculator</a> however evaluates this integral as $$\ln|\tan x+\sec x|+C$$
but I can't figure out where I made a mistake in my calculations.</p>
| Jacky Chong | 369,395 | <p>Hint: Use the trick
\begin{align}
\sec x= \frac{\sec x+\tan x}{\sec x+\tan x} \sec x.
\end{align}
then $u$-sub.</p>
|
3,787,576 | <p><span class="math-container">$$
\mbox{Prove}\quad
\int_{0}^{1}{\mathrm{d}x \over
\left(\,{x - 2}\,\right)\,
\sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,}
=
-\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}
$$</span></p>
<ul>
<li>Being honest I havent got a clue where to start. I dont think any obvious substitutions will help (<span class="math-container">$x \to 1-x, \frac{1}{x}, \sqrt{x},$</span> more).</li>
<li>The indefinite integral involves hypergeometric function so some miracle substitution has to work with the bounds I suspect.</li>
<li>Maybe gamma function is involved some how ??.</li>
</ul>
<p>If anyone has an idea and can provide help I would appreciate it.</p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
\text { Let } x&=\sin ^2 \theta, \quad \textrm{ then } d x=2 \sin \theta \cos \theta d \theta, \textrm{ and }\\
I&=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\left(\sin ^2 \theta-2\right) \sqrt[5]{\sin ^4 \theta \cos ^6 \theta}}\\
&=2 \int_0^{\frac{\pi}{2}} \frac{\tan ^{\frac{1}{5}} \theta d \theta}{\cos ^2 \theta\left(\tan ^2 \theta-2 \sec ^2 \theta\right)}\\
&=-2 \int_0^{\infty} \frac{t^{\frac{1}{5}}}{t^2+2} d t ,\quad \textrm{ where }t=\tan \theta\\
&=-2\left(\frac{2^{\frac{1}{10}} \pi}{\sqrt{5+\sqrt{5}}}\right) \quad \textrm{(via beta function)} \\
&=-\frac{2^{\frac{11}{10}} \pi}{\sqrt{5+\sqrt{5}}}
\end{aligned}
$$</span></p>
|
4,205,179 | <p>Solve the equation</p>
<blockquote>
<blockquote>
<p><span class="math-container">$\sin x-\cos x-4\cos^2 x \sin x=4\sin^2 x$</span></p>
</blockquote>
</blockquote>
<p>My attempt:</p>
<p>I have rewritten the equation as:
<span class="math-container">$$\sin x=\frac{4\cos ^2 x-\cos x-4}{4\cos^2 x-1}$$</span></p>
<p>I tried drawing graphs of LHS and RHS. But is there any analytical way?</p>
| Theo Bendit | 248,286 | <p>Try using the half-angle substitution: <span class="math-container">$t = \tan(x/2)$</span>. Then the following relations hold:
<span class="math-container">\begin{align*}
\cos(x) &= \frac{1 - t^2}{1 + t^2} \\
\sin(x) &= \frac{2t}{1 + t^2}.
\end{align*}</span>
Then the equation becomes:
<span class="math-container">$$\frac{2t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} - 8 \frac{t(1 - t^2)^2}{(1 + t^2)^3} = 16\frac{t^2}{(1 + t^2)^2}.$$</span>
Multiplying both sides by <span class="math-container">$(1 + t^2)^3$</span>, we get
<span class="math-container">$$(t^2 - 2t - 1)(1 + t^2)^2 - 8t(1 - t^2)^2 = 16t^2(1 + t^2).$$</span>
When we expand and simplify, we get
<span class="math-container">$$t^6 - 10 t^5 - 15 t^4 + 12 t^3 - 17 t^2 - 10 t - 1 = 0.$$</span>
Wolfram Alpha gives no elementary factors or roots, with two being non-real complex conjugates. But, if <span class="math-container">$t$</span> is a root of the above polynomial, then <span class="math-container">$x = 2\tan^{-1}(t)$</span> will be one such solution.</p>
|
4,205,179 | <p>Solve the equation</p>
<blockquote>
<blockquote>
<p><span class="math-container">$\sin x-\cos x-4\cos^2 x \sin x=4\sin^2 x$</span></p>
</blockquote>
</blockquote>
<p>My attempt:</p>
<p>I have rewritten the equation as:
<span class="math-container">$$\sin x=\frac{4\cos ^2 x-\cos x-4}{4\cos^2 x-1}$$</span></p>
<p>I tried drawing graphs of LHS and RHS. But is there any analytical way?</p>
| Aaron | 9,863 | <p>Let <span class="math-container">$\omega=\cos(x)+i\sin(x)$</span>, so that <span class="math-container">$\cos(x)=(\omega+\omega^{-1})/2$</span> and <span class="math-container">$\sin(x)=(\omega-\omega^{-1})/2i$</span>. Then the equation we are trying to solve, when expressed in terms of <span class="math-container">$\omega$</span> and simplified is:
<span class="math-container">$$\frac{i\omega^6+2\omega^5-\omega^4-4\omega^3-\omega^2+2\omega-i}{2\omega^3}=0.$$</span></p>
<p>Unfortunately, there are no obvious factors to this equation, although wolfram alpha suggests that 3 of the 6 complex roots are of length 1, which means they correspond to actual solutions to the original problem.</p>
|
114,831 | <p>$$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$</p>
<p>I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?</p>
| Did | 6,179 | <p>Note that
$$
\log2-A_n=\int_n^{2n}\frac{\mathrm dx}x-\sum_{k=1}^n\frac1{n+k}=\sum_{k=1}^n\int_0^1\frac{1-x}{(n+k-1+x)(n+k)}\mathrm dx.
$$
To get an upper bound, use $n+k-1+x\geqslant n+k-1$, hence
$$
\log2-A_n\leqslant\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}\int_0^1(1-x)\mathrm dx.
$$
The integral is $\frac12$ and the sum is
$$
\sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac1{n+k}\right)=\frac1n-\frac1{2n}=\frac1{2n},
$$
hence
$$
\log2-A_n\leqslant\frac1{4n}.
$$
To get a lower bound, use $n+k-1+x\leqslant n+k\leqslant n+k+1$, hence
$$
\log2-A_n\geqslant\sum_{k=1}^n\frac{1}{(n+k)(n+k+1)}\int_0^1(1-x)\mathrm dx.
$$
The integral is still $\frac12$ and the sum is
$$
\sum_{k=1}^n\left(\frac{1}{n+k}-\frac1{n+k+1}\right)=\frac1{n+1}-\frac1{2n+1}=\frac{n}{(n+1)(2n+1)},
$$
hence
$$
\log2-A_n\geqslant\frac{n}{2(n+1)(2n+1)}=\frac{u_n}{4n},
$$
with
$$
u_n=\frac1{(1+1/n)(1+1/(2n))}.
$$
Finally,
$\frac14u_n\leqslant n(\log2-A_n)\leqslant\frac14$ and $u_n\to1$ hence
$n(\log2-A_n)\to\tfrac14$.</p>
|
217,711 | <p>Set
$$
g(x)=\sum_{k=0}^{\infty}\frac{1}{x^{2k+1}+1} \quad \text{for} \quad x>1.
$$</p>
<p>Is it true that
$$
\frac{x^{2}+1}{x(x^{2}-1)}+\frac{g'(x)}{g(x)}>0 \quad \text{for}\quad x>1?
$$
The answer seems to be positive. I spent several hours in proving this statement but I did not come up with anything reasonable. Maybe somebody else has (or will have) any bright idea? </p>
<p><strong>Motivation?</strong> Nothing important, I was just playing around this question:
<a href="https://mathoverflow.net/questions/217530/a-problem-of-potential-theory-arising-in-biology">A problem of potential theory arising in biology</a></p>
| Iosif Pinelis | 36,721 | <p>Writing $k=\sum_{j=1}^k 1$ and then summing by parts, one has
$$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2}
=\sum_{j\ge 1}(-1)^{j-1}s_j
=\sum_{k\ge0}[s_{2k+1}-s_{2k+2}],
$$
where
$$s_j:=\sum_{k=j}^\infty(-1)^{k-j} a(k)=\sum_{m\ge0}(-1)^m a(m+j)
=\sum_{k\ge0}[a(2k+j)-a(2k+1+j)]
$$
and
$a(q):=\dfrac1{(x^q+1)^2}$.
So,
$$s_j-s_{j+1}=\sum_{k\ge0}[a(2k+j)-2a(2k+1+j)+a(2k+2+j)]>0,
$$
since $a''>0$ and hence $a$ is strictly convex.
So, $\sum_{k\geq 1} \dfrac{(-1)^{k-1}k}{(x^{k}+1)^2}>0$, and the result follows by <strong>Max Alekseyev</strong>'s answer. </p>
|
1,933,410 | <p>I'm reading a book about differential geometry and there's a part where he talks about the standar spherical coordinates on $S^2$, which are given by:</p>
<p>$$x:(0,2\pi)\times (0,\pi)\to S^2, x(\theta, \phi) = (\cos\theta\sin\phi,\sin\theta\sin\phi, \cos\phi)$$</p>
<p>Shouldn't it be in $S^3$?</p>
| Emilio Novati | 187,568 | <p>No. It is a subset of $\mathbb{R}^3$, so we have three coordinates $ (x_1,x_2,x_3)$ but it is a surface because, as you can see: $x_1^2+x_2^2+x_3^2=1$ so it is the surface of a sphere in the $3D$ space. </p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| lhf | 532 | <p>Also, differential Galois theory for differential equations.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Mariano Suárez-Álvarez | 1,409 | <p>One set of problems to which homological algebra applies surprisingly well is those posed by topology, as evinced by algebraic topology :P</p>
<p>This is of course quite anhistorical...</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Qiaochu Yuan | 290 | <p>This example is more closely related to <a href="https://mathoverflow.net/questions/5449/combinatorial-results-without-known-combinatorial-proofs">a question of mine</a>, but I'll give it here anyway.</p>
<p>A graded poset $P$ is <em>Sperner</em> if no antichain is larger than the largest level $P_i$ of $P$. This property is named after Sperner, who proved that the Boolean posets $B_n$ of subsets of $\{ 1, 2, ... n \}$ are Sperner. the Boolean posets $B_n$ are also <em>rank-symmetric</em> because they satisfy $(B_n)_i = (B_n)_{n-i}$, i.e. ${n \choose i} = {n \choose n-i}$ and <em>unimodal</em> because the sequence $(B_n)_i$ is at first increasing and then decreasing. </p>
<p>Let $G$ be a group acting on $\{ 1, 2, ... n \}$. Then $G$ acts on $B_n$ in the obvious way by order- and grade-preserving automorphisms. Define the quotient poset $B_n/G$ in the obvious way, which inherits the grading on $B_n$. It's not hard to see that $B_n/G$ is also rank-symmetric.</p>
<p><strong>Theorem:</strong> $B_n/G$ is unimodal and Sperner.</p>
<p>The only known proofs of this theorem are algebraic (the one I know uses linear algebra), and even for special cases a purely combinatorial proof is not known. For example, in the case that $n = ab$ is a product of two positive integers and $G = S_b \wr S_a$ we recover the poset $L(a, b)$ of Young diagrams that fit into an $a \times b$ box; then the above theorem implies that the coefficients of the q-binomial coefficient ${a + b \choose b}_q$ are unimodal. This was first proven by Sylvester in 1878, and a combinatorial proof was not found until 1989. A combinatorial proof of the Sperner property is still not known. This example is all the more remarkable because the statement that $L(m, n)$ is Sperner requires almost no mathematics to understand.</p>
<p><strong>Edit:</strong> I should probably provide a reference. This material is from some <a href="http://math.mit.edu/~musiker/rstan5-6.pdf" rel="nofollow noreferrer">notes on algebraic combinatorics</a> by Stanley.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Martin Brandenburg | 2,841 | <p>As for <b>category theory</b>, I don't think that there is a motivating example which has not already a category theoretic flavor. The leading theme is to unify and then generalize constructions resp. arguments, which come up in all areas of mathematics. Historically, natural transformations were introduced for the foundations of homology theory of topological spaces. But to start with an easy example, you may observe that for abelian groups $A,B,C$ there is a canonical isomorphism $(A \oplus B) \oplus C \cong A \oplus (B \oplus C)$, which reminds you of other <i>associativity</i> results such as $(X \cup Y) \cup Z \cong X \cup (Y \cup Z)$ for sets (here $\cup$ means disjoint union). Within category theory, you can see what's the real content of this: direct sum and disjoint unions are examples of coproducts, and coproducts are always associative. Even more striking, Yoneda's Lemma, which lies at the heart of foundations of category theory, tells you that the case of sets already settles the general case!</p>
<p>But category theory is more than just a language, it also provides general constructions: Assume you want to <i>approximate</i> a theory with another theory. This may be formalized by finding an adjunction between two categories. Freyds/Special Adjoint Functor Theorem tell you when this is possible. Although in some situations you can't write down the adjunction, the only thing you need is to know that it exists. For example, what is the categorical coproduct of an infinite family of compact hausdorff spaces? Can you write it down without using Stone-Cech?</p>
<p>There are also somewhat <i>global</i> motivations: Some Theories behave like other theories, and thus you may develope a theory for a large class of categories at once: monodial categories, topological categories, algebraic categories, locally presentable categories, etc. Of course, the same is true for other notions of category theory (functors, natural transformations, types of morphisms, etc.).</p>
<p>But if one has not heard of category theory before, the first motiviation should be to <i>think in categories</i> (in the colloquial sense). For example the set, which underlies a group, <i>really differs</i> from the group. In almost every book and lecture, this is absorbed by abuse of notation. The existence of bases in vector spaces is no reason at all to restrict linear algebra to vector spaces of the form $K^{(B)}$. Similarily, vector bundles should <i>not</i> be defined as bundles which are locally isomorphic to some $\mathbb{R}^n \times X$, which most topologists still ignore! Rather, it is first of all a vector space object in the category of bundles over $X$.</p>
<p>Let's conclude with an example which both introduces functors and <b>algebraic geometry</b>: Assume you have a system of polynomial equations $f_1(x)=...=f_n(x)=0$ in $m$ variables defined over $\mathbb{Z}$ and you want to study the solutions in arbitrary rings <i>at once</i>, using a <i>single mathematical object</i>, e.g. having in mind some local-global results of algebraic number theory. So for every ring $R$, we put $F(R) = \{x \in R^m : f_1(x)=...=f_n(x)=0\}$. Observe that for every ring homomorphism $R \to S$ there is a set map $F(R) \to F(S)$ and that this is compatible with composition of homomorphisms. This exactly means that $F$ is a functor from the category of rings to the category of sets. Algebraic geometry studies functors which locally look like the functor above.</p>
|
3,104,058 | <p>I need to make a part of a program in java that calculates a circle center. It has to be a circle through a given point that touches another circle, and the variable circle center has the possibility to move over a given line.</p>
<p><a href="https://i.stack.imgur.com/RQgBC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RQgBC.png" alt="Example of the problem"></a></p>
<p>Here the coordinates of A, B and C and the radius of the circle around A are given. I need to know how to get the coordinates of P and P' when they touch the blue circle around A.</p>
| John Hughes | 114,036 | <p><strong>A strong hint, but not a complete solution</strong></p>
<p>Let
<span class="math-container">$$
\newcommand{\bu} {{\mathbf u}}
\newcommand{\bv} {{\mathbf v}}
\begin{align}
s &= \|B - C\|\\
\bu &= \frac{B - C}{s}
\end{align}
$$</span>
The point <span class="math-container">$P$</span> is then <span class="math-container">$C + t\bu$</span> for some <span class="math-container">$t \in \Bbb R$</span>, and I'll follow the picture and consider the case <span class="math-container">$t < s$</span> so that we find <span class="math-container">$P$</span> instead of <span class="math-container">$P'$</span>. We'll work out some constraints on <span class="math-container">$t$</span>. </p>
<p>The first constraint is that the distance from <span class="math-container">$P$</span> to <span class="math-container">$B$</span> (namely <span class="math-container">$s - t$</span>), which is the radius of the circle around <span class="math-container">$P$</span>, must, when added to <span class="math-container">$r$</span>, the radius of the blue circle, give the distance from <span class="math-container">$P$</span> to <span class="math-container">$A$</span>. Thus:</p>
<p><span class="math-container">$$
(s - t) + r = \| A - (C + t \bu) \|.
$$</span>
Squaring both sides, and letting <span class="math-container">$\bv = A - C$</span> and <span class="math-container">$e = \|A - C \| = \|\bv\|$</span>, we get
<span class="math-container">\begin{align}
(s - t)^2 + 2r(s-t) + r^2 &= \| (A - C) - t \bu) \|^2\\
(s - t)^2 + 2r(s-t) + r^2 &= [ (A - C) - t \bu) ] \cdot [ (A - C) - t \bu) ] \\
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= (A - C)\cdot(A-C) - 2t \bu \cdot (A - C) + t^2 \bu \cdot bu \\
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= (A - C)\cdot(A-C) - 2t \bu \cdot (A - C) + t^2 & \text{, because $\bu$ is a unit vector}\\
s^2 - 2st + t^2 + 2rs-2rt + r^2 &= e^2 - 2t \bu \cdot \bv + t^2 & \text{, defn's of $\bv$ and $e$}\\
s^2 - 2st + 2rs-2rt + r^2 &= e^2 - 2t \bu \cdot \bv & \text{algebra}\\
2t \bu \cdot \bv - 2st -2rt &= e^2 -s^2 -2rs - r^2& \text{algebra}\\
t( -2 \bu \cdot \bv + 2s + 2r) &= (s+r)^2 - e^2& \text{algebra}\\
t &= \frac{(s+r)^2 - e^2}{-2 \bu \cdot \bv + 2s + 2r }& \text{algebra}\\
\end{align}</span>
...so that gives you the point <span class="math-container">$P$</span> (you just compute <span class="math-container">$P + t\bu$</span>). Now you have to do the same thing, but starting with <span class="math-container">$(t - s) + r = ...$</span> to find the point on the other side of <span class="math-container">$B$</span>. </p>
<p>Here is (not pretty) Matlab code to implement this, and a plot of the result of </p>
<pre><code>circles([0 3], [1, 1], [-3, 0], 1)
</code></pre>
<p>being run in the Command window. </p>
<pre><code>function circles(a, b, c, r)
clf;
a = a(:); b = b(:); c = c(:);
vv = b - c;
s = sqrt(dot(vv, vv));
u = (b-c)/s;
v = a - c;
e = sqrt(dot(v, v));
numerator = (s+r)^2 - e^2;
denominator = -2 * dot (u, v)+ 2*s + 2*r;
t = numerator/denominator;
P = c + t*u
% draw line from C to B
point(c);
point(b);
plot([c(1) b(1)], [c(2) b(2)], 'k');
circle(a, r);
circle(P, (s-t));
axis equal
figure(gcf);
function point(pt)
hold on;
plot(pt(1), pt(2), 'ro');
hold off;
function circle(ctr, radius)
t = linspace(0, 2*pi, 100);
x = ctr(1) + radius * cos(t);
y = ctr(2) + radius * sin(t);
hold on;
plot(x, y);
point(ctr);
hold off;
</code></pre>
<p><a href="https://i.stack.imgur.com/7uMX9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7uMX9.jpg" alt="enter image description here"></a></p>
<p>The blue circle is the one around point <span class="math-container">$A$</span>; the red-orange circle is the computed on. The black line segment goes from <span class="math-container">$C$</span> to <span class="math-container">$B$</span>. </p>
<p>Of course, you still have to work through the case where <span class="math-container">$t > s$</span> to find the coordinates of the center <span class="math-container">$P'$</span> and the radius of the second circle. </p>
|
443,013 | <p>If a linear map $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is injective, then there exists $c>0$ such that $|Ax|\geq c|x|$ for all $x\in\mathbb{R}^m$</p>
<p>Could someone give any solution or hint?</p>
<p>Thanks.</p>
| Nick Peterson | 81,839 | <p><strong>Hint:</strong> Consider
$$
\alpha:=\inf_{\|x\|=1}\|Ax\|.
$$
This must exist and be non-negative, since $\|Ax\|\geq 0$ for all $x\in\mathbb{R}^m$. If $\alpha>0$, then you can use linearity to prove the desired result. What happens if $\alpha=0$? </p>
|
443,013 | <p>If a linear map $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is injective, then there exists $c>0$ such that $|Ax|\geq c|x|$ for all $x\in\mathbb{R}^m$</p>
<p>Could someone give any solution or hint?</p>
<p>Thanks.</p>
| user71352 | 71,352 | <p>First note that if such an inequality holds on the unit sphere in the Euclidean norm then it is true for all $x\in\mathbb{R}^{n}$. This is because if we have</p>
<p>$\vert\vert Ax\vert\vert\ge C\vert\vert x\vert\vert=C$ for some constant $C>0$ on the unit sphere then we simply note that for any $x\in\mathbb{R}^{n}-\{0\}$ we have that $\frac{x}{\vert\vert x\vert\vert}$ has unit norm. So $\vert\vert A(\frac{x}{\vert\vert x\vert\vert})\vert\vert\ge \vert\vert \frac{x}{\vert\vert x\vert\vert}\vert$. Multiplying by $\vert\vert x\vert\vert$ which is simply a positive constant gives the inequality we desire.The inequallity holds trivially for $x=0$ so it holds for all of $\mathbb{R}^{n}$.</p>
<p>Note that $A$ is injective implies that $Ax=0$ only if $x=0$. Now we simply look at the function $f:S^{n-1}\to\mathbb{R}$ defined as $f(x)=\frac{\vert\vert Ax\vert\vert}{\vert\vert x\vert\vert}=\vert\vert Ax\vert\vert$. This is a continuous function on the compact surface $S^{n-1}$ and so attains a minimum on the unit sphere. Since A is injecive the lower bound is some $C>0$. This gives the bound we desire.</p>
|
4,009,117 | <p>Suppose <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> are exhaustive events with <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> being mutually exclusive. Can <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> be treated as a partition? Also, as a follow-up, can the law of total probability be applied to <span class="math-container">$X$</span> this way: <span class="math-container">$P(^cX) = P(^cX \cap Y) + P(^cX\cap Z)$</span>, where <span class="math-container">$^cX$</span> is the complement of <span class="math-container">$X$</span>?</p>
| Arturo Magidin | 742 | <p>It depends on exactly what your definition of limit is.</p>
<p>In calculus, limits of real-valued functions are often defined in such a way that in order for the limit as <span class="math-container">$x\to a$</span> to exist, the function must be defined on some interval of the form <span class="math-container">$(a-\delta,a+\delta)$</span>, except perhaps at <span class="math-container">$a$</span>. For limits as <span class="math-container">$x\to\infty$</span>, you require the function to be defined on some interval of the form <span class="math-container">$(M,\infty)$</span>. The definition using <span class="math-container">$\epsilon$</span>s would look something like:</p>
<blockquote>
<p><span class="math-container">$\lim_{x\to\infty} f(x)=L$</span> if and only if for all <span class="math-container">$\epsilon\gt 0$</span> there exists <span class="math-container">$M$</span> such that if <span class="math-container">$x\gt M$</span>, then <span class="math-container">$|f(x)-L|\lt\epsilon$</span>.</p>
</blockquote>
<p>However, sometimes limits are defined differently (slightly more generally, applicable to more functions), and the prerequisite for the limit as <span class="math-container">$x\to a$</span> to exist is that <span class="math-container">$a$</span> be an “accumulation point” of the domain of <span class="math-container">$f$</span>. That is, that for all <span class="math-container">$\delta\gt 0$</span>, there exist points <span class="math-container">$b$</span> in the domain of <span class="math-container">$f$</span> such that <span class="math-container">$0\lt |b-a|\lt \delta$</span>. For limits as <span class="math-container">$x\to\infty$</span>, one requires that for all <span class="math-container">$M$</span> there exist <span class="math-container">$b$</span> in the domain of <span class="math-container">$f$</span> such that <span class="math-container">$b\gt M$</span>.</p>
<p>Here, the definition using <span class="math-container">$\epsilon$</span>s would be:</p>
<blockquote>
<p><span class="math-container">$\lim_{x\to\infty} f(x)=L$</span> if and only if (i) for every <span class="math-container">$M$</span> there exists <span class="math-container">$x\gt M$</span> such that <span class="math-container">$x\in\mathrm{dom}(f)$</span>; and (ii) for every <span class="math-container">$\epsilon\gt 0$</span> there exists <span class="math-container">$N$</span> such that if <span class="math-container">$x\gt N$</span> <strong>and</strong> <span class="math-container">$x\in\mathrm{dom}(f)$</span>, then <span class="math-container">$|f(x)-L|\lt\epsilon$</span>.</p>
</blockquote>
<p>If your definition of limits is of the first type, then the limit you write does not exist, because the function does not meet the requirement to be able to even talk about a potential limit as <span class="math-container">$x\to\infty$</span>: it is not defined on any interval of the form <span class="math-container">$(M,\infty)$</span>, so you can always find an <span class="math-container">$x$</span> with <span class="math-container">$x\gt M$</span>, and where <span class="math-container">$|f(x)-1|\lt\epsilon$</span> fails because the expression doesn’t even make sense. On the other hand, if your definition of limit is of the second type, then the limit is <span class="math-container">$1$</span>, since the satisfies both conditions of that definition.</p>
<p>In short, neither opinion is right and neither is wrong: it depends on exactly what the symbol <span class="math-container">$\lim_{x\to\infty}f(x)=L$</span> means.</p>
|
3,069,369 | <p>I know how to perform polynomial regression. But is there any method to use for estimating the degree of the polynomial that is best suited? Some kind of meta-regression.</p>
<p>With best suited I mean the grade that has the highest probability of being the true degree of the source for the data.</p>
<p>For example, if we look at this picture we can easily "see" that a polynomial of degree 4 would fit nicely:</p>
<p><a href="https://i.stack.imgur.com/ZOIbY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZOIbY.png" alt="enter image description here"></a></p>
<p>A more generalized question is if there is any method to determine if the source is polynomial at all or if it is exponential or something else.</p>
| David G. Stork | 210,401 | <p>Starting with <span class="math-container">$2 x^4+x^3-7 x^2-x+5$</span> and using the insight that <span class="math-container">$x=1$</span> must be a factor:</p>
<p>Divide out:</p>
<p><span class="math-container">$${2 x^4+x^3-7 x^2-x+5 \over x-1} = 2 x^3+3 x^2-4 x-5$$</span></p>
<p>Notice again, from the sum of coefficients that <span class="math-container">$x = -1$</span> is a root, so divide out again:</p>
<p><span class="math-container">$${2 x^3+3 x^2-4 x-5 \over x+1} = 2 x^2+x-5.$$</span></p>
<p>So:</p>
<p><span class="math-container">$$(x-1)(x+1)(2 x^2+x-5)$$</span></p>
<p>Use the quadratic equation:</p>
<p><span class="math-container">$$(x-1)(x+1)(x - 1/4 (-1 - \sqrt{41}))(x+1/4 (-1 - \sqrt{41}))$$</span></p>
|
853,735 | <p>Can anyone show, step-by-step, how the expression on the LHS can be turned into the expression on the RHS?</p>
<p>$x^ay^b=a^ab^b(a+b)^{-(a+b)}(x+y)^{a+b}$</p>
| Juanito | 153,015 | <p>Take x=0, y=1, a=1, b=1. The LHS is not equal to RHS.
So, the identity is False.</p>
|
4,581,481 | <p>How would one go about ALGEBRAICALLY finding the range of a semicircle?</p>
<p>eg. <span class="math-container">$y = \sqrt{4-x^2}$</span></p>
<p>Since there is no plus-minus sign, we know that <span class="math-container">$y$</span> must at least have a range of <span class="math-container">$[0, \infty)$</span>. Yet this must not be the complete case because we know a semicircle has restrictions on how small AND how big the <span class="math-container">$y$</span> value can be.</p>
| N. F. Taussig | 173,070 | <p>Since <span class="math-container">$x, y$</span> are real numbers, <span class="math-container">$y = \sqrt{4 - x^2}$</span> is only defined if <span class="math-container">$4 - x^2 \geq 0$</span>.
<span class="math-container">\begin{align*}
4 - x^2 & \geq 0\\
4 & \geq x^2\\
2 & \geq |x| && \text{(take the square root of each side of the equation)}
\end{align*}</span>
Since <span class="math-container">$|x| \leq 2$</span>, <span class="math-container">$-2 \leq x \leq 2$</span>. Thus, as you noted in the comments, the domain of the function is <span class="math-container">$[-2, 2]$</span>.</p>
<p>Since <span class="math-container">$\sqrt{t}$</span> denotes the nonnegative square root of <span class="math-container">$t$</span>, <span class="math-container">$y = \sqrt{4 - x^2} \geq 0$</span>.</p>
<p><span class="math-container">\begin{align*}
y & = \sqrt{4 - x^2}\\
y^2 & = 4 - x^2 && \text{(square each side of the equation)}
\end{align*}</span>
subject to the restriction that <span class="math-container">$y \geq 0$</span>. Since <span class="math-container">$x^2 \geq 0$</span>, with equality holding if and only if <span class="math-container">$x = 0$</span>, <span class="math-container">$y^2 \leq 4$</span> and <span class="math-container">$y \geq 0$</span>.</p>
<p><span class="math-container">\begin{align*}
y^2 & \leq 4\\
|y| & \leq 2 && \text{(take the square root of each side of the inequality)}
\end{align*}</span>
Since <span class="math-container">$|y| \leq 2$</span>, <span class="math-container">$-2 \leq y \leq 2$</span>. Since <span class="math-container">$-2 \leq y \leq 2$</span> and <span class="math-container">$y \geq 0$</span>, <span class="math-container">$0 \leq y \leq 2$</span>. Thus, the range of the function is <span class="math-container">$[0, 2]$</span>.</p>
|
3,737,831 | <p>The problem is</p>
Let <span class="math-container">$\phi$</span> be an endomorphism of a <span class="math-container">$n$</span>- dimensional vector space V. Suppose that <span class="math-container">$\phi$</span> has n different eigenvalues. Prove that exists a vector <span class="math-container">$v$</span> of <span class="math-container">$V$</span> such <span class="math-container">$\{v,\phi(v), . . . , \phi^{n-1}(v)\}$</span> is a basis of <span class="math-container">$V$</span>.
<p>I tried to prove it by induction. But after proving it to be true for 1 vector, I can't figure out the way to proceed.
How would you prove this?</p>
| alphaomega | 775,794 | <p>I know it's not exactly related but it's interesting to observe that the compact condition cannot be dropped. Consider for example,</p>
<ul>
<li><span class="math-container">$f(x,y)=(x+y)^2$</span>, <span class="math-container">$(x,y)\in [0, +\infty)^2$</span></li>
<li><span class="math-container">$x_n=\frac{1}{n}$</span>, <span class="math-container">$x=0$</span></li>
</ul>
<p>Then, <span class="math-container">$$|g_n(y)-g(y)|=|\big( \frac{1}{n}+y \big)^2 - y^2| = |2y+\frac{1}{n^2}|.$$</span></p>
<p>Thus, <span class="math-container">$\sup\{ |g_n(y)-g(y)| : \quad y\in[0, +\infty) \} = \infty$</span>.</p>
|
1,117,924 | <p>How can I show with linear approximation that $y \approx x$ for small x? I know the rule $$f(x) \approx f(a) + f^{\prime}(a) (x-a),$$ but I don't know how to put it to use in this case.</p>
| A. Donda | 49,995 | <p>The function you are trying to approximate is
$$
f(x) = \ln(x)
$$
and you need an approximation around $a = 1$.</p>
<p>For the rule we need the derivative of the function, and we <a href="http://en.wikipedia.org/wiki/Logarithm#Derivative_and_antiderivative" rel="noreferrer">know</a> that the derivative of the natural logarithm is the inverse:
$$
f'(x) = \frac1x.
$$</p>
<p>Let's evaluate both the function and the derivative at $a = 1$,
$$
f(a) = \ln(1) = 0
\quad \text{and} \quad
f'(a) = \frac11 = 1,
$$
and apply the rule.</p>
<p>$$
f(x) \approx f(a) + f'(a) (x-a)
\quad \text{for} ~ x \approx a
$$
means
$$
\ln(x) \approx 0 + 1 (x - 1) = x - 1
\quad \text{for} ~ x \approx 1
$$
or
$$
\ln(x + 1) \approx x
\quad \text{for} ~ x \approx 0.
$$</p>
<p>We could also haven directly chosen $f(x) = \ln(1 + x)$ and $a = 0$, at the price of a slightly harder computation of the derivative, but of course with the same result.</p>
|
22,662 | <p>Suppose $V_1$ and $V_2$ are two $(g,K)$ modules of some reductive group $G$ with maximal compact $K$. Let $P$ be the minimal parabolic of $G$, $U$ its unipotent part, and $u$ its Lie algebra. Suppose the quotients $V_1/uV_1$ and $V_2/uV_2$ are isomorphic as modules for the Levi component of $P$, then what else do we need to know to conclude that $V_1$ and $V_2$ are isomorphic as $G$ modules?</p>
<p>edit:Thanks for Kevin and Emerton's comments and sorry for the confusion about the base field. Here I'm assuming REAL reductive group. </p>
| Victor Protsak | 5,740 | <p>This is a comment, mostly on terminology that I think caused some confusion, not an answer, but I don't have enough "influence" to post this as a comment.</p>
<p>For real reductive groups, the Jacquet functor $V\mapsto J(V)$ (Jacquet-Casselman functor, Jacquet module, etc) is defined <em>differently</em> from the nonarchimedean case: it is $\varinjlim (V/n^iV)^*$, which is dual to the n-adic completion of V. Thus J(V) is a p-finite g-module (n is the nilradical, p is the parabolic Lie subalgebra) and the target category of the functor J is parabolic category O for g. On the one hand, this provides more structure: we get a g-module and, for example, the infinitesimal character of V can be read off the infinitesimal character of J(V) (Casselman-Osborne); more generally, V and J(V) have the same annihilator in U(g). On the other hand, J(V) is morally a highest weight module, not a Harish-Chandra module, so some information is lost. The 0th n-homology that this problem is asking about, V/nV, is just the the top layer of J(V), so even more information is lost. By the way, unlike the p-adic Jacquet functor, n-homology is not exact (there may be higher homology), but V/nV is always non-zero (short proof was given by Beilinson and Bernstein).</p>
<p>I assume that the modules V1 and V2 in the formulation were presumed to be simple? </p>
|
215,077 | <p>This question is partly connected with the following <a href="https://mathoverflow.net/questions/214931/connection-between-stallings-end-theorem-and-seifert-van-kampen-theorem">Connection between Stalling's end theorem and Seifert-van Kampen Theorem</a>. </p>
<p>By Stalling's Theorem a group with more than one end splits over a finite subgroup, i.e. can be written as an HNN-Extension or a free product with amalgamation (over a finite subgroup).</p>
<p>However when I was working through the proofs of Dunwoody, Dunwoody & Krön and Krön, which all use Bass-Serre Theory, I was wondering if there is any way to detect if we are dealing with an HNN-Extension or a free product with amalgamation. Primarily I was thinking of some property of the Cayley graph or the action on it by the group in question. For example one could consider the induced action on the end space of the Cayley graph.</p>
<p>In my opinion this is a natural question and I would be grateful for any comment or references regarding this.</p>
<p>EDIT: I am looking for properties of the group, its Cayley graph, the action of the group on its Cayley graph and/or its end space etc., which help to distinguish if the group, supposed to split by Stalling's Theorem, is indeed an HNN-Extension <em>or</em> a free product with amalgamation.</p>
<p>For some statements one probably has to make additional assumptions such as the group is finitely generated or finitely presented etc. </p>
| NWMT | 38,698 | <p>There is nothing <em>intrinsic</em> about whether a group splits as a free product with amalgamation or an HNN extension. For example free groups split both ways $$
\langle a,b \rangle = \langle a \rangle * \langle b \rangle = \langle a,b ; t \mid t^{-1} a t = b \rangle $$ as do many other groups. If you're cutting up a Cayley graph like in the papers you cite, one cut may give you an HNN extension, while the other can give you a free product with amalgamation.</p>
<p>Actually free products with amalgamation and HNN extensions aren't that different. It's just that because of the <em>celebrated</em> Seifert Van-Kampen theorem, HNN extensions don't get no respect! Bass-Serre theory unifies and generalizes these notions.</p>
<p>If your group acts on a simplicial tree with one edge orbit and two vertex orbits, then it splits as an amalgamated free product. If it acts with one edge orbit and one vertex orbit then it splits as an HNN extension. Any characterization must follow somehow from this definition. For example an HNN extension is dsitinguished from a FPA if the group cannot be generated by elements that act elliptically (i.e. fix a vertex) on the the Bass-Serre tree.</p>
<p>With respect to ends, the most canonical cut set may give you neither a free product with amalgamation nor an HNN extension, but a more general graph of groups decomposition, namely a Dunwoody decomposition. Such cut sets would consist of disjoint orbits of finite cut sets.</p>
<p>Edit: By the way it is "obvious" that the ends of the Bass-Serre tree dual to the Dunwoody decomposition are in natural bijective corresponsdence with the set of ends of your group; at least in the f.p. (or accessible) case.</p>
|
1,923,226 | <p>I have been trying to prove this with some continuity theorems but haven't put together a good proof yet. </p>
| kobe | 190,421 | <p>Let $m\in \Bbb Z$ and $\epsilon > 0$. Set $\delta = 1$. Show that for all $n$, $\lvert n - m\rvert < \delta$ implies $\lvert f(n) - f(m) \rvert < \epsilon$.</p>
|
1,923,226 | <p>I have been trying to prove this with some continuity theorems but haven't put together a good proof yet. </p>
| TheGeekGreek | 359,887 | <p>Let us take the Weierstrass' $\varepsilon$-$\delta$ criterion. If $f:E\subseteq \mathbb{R} \to \mathbb{R}$ is continuous at $x_0 \in E$, then $$\forall \varepsilon > 0 \exists \delta > 0 \forall x \in E \left( \left| x - x_0\right| < \delta \to \left| f(x) - f(x_0)\right| < \varepsilon \right)$$ Now Let $E := \mathbb{Z}$ and $\varepsilon > 0$, $x_0 \in E$ be fixed. Taking $\delta := 1$, we get, that the only point which fulfills $\left| x - x_0\right| < \delta$ is $x_0$ itself, but then trivially $$\left| f(x_0) - f(x_0)\right| = 0 < \varepsilon$$ holds. Since $x_0 \in E$ was arbitrary, $f$ is continuous on $E$.</p>
|
4,153,364 | <blockquote>
<p>If <span class="math-container">$f(x) = x^2$</span> and <span class="math-container">$g(x) = \sqrt{x-1}$</span>, find <span class="math-container">$f(g(x))$</span> and specify the domain.</p>
</blockquote>
<p><strong>My solution</strong></p>
<p><span class="math-container">$$
f(g(x)) = \left(\sqrt{x-1}\right)^2 = x-1,
$$</span>
Domain: <span class="math-container">$x$</span> such that <span class="math-container">$x$</span> is any real number.</p>
<p>Solution at the back of the book states that the domain is, <span class="math-container">$x$</span> such that <span class="math-container">$x \ge 1$</span>.</p>
<p>Now I understand why that is the domain for <span class="math-container">$g(x)$</span>, for any number less than 1 would make <span class="math-container">$g(x)$</span> negative and the square root of a negative number is an imaginary number.</p>
<p>But, that is not the cases with <span class="math-container">$f(g(x))$</span>. When is <span class="math-container">$(\sqrt{x-1})^2$</span> not a real number? isn't any input valid?</p>
| gt6989b | 16,192 | <p>I think you have to differentiate between
<span class="math-container">$$
f(g(x)) = \left(\sqrt{x-1}\right)^2
$$</span>
and <span class="math-container">$h(x) = x-1$</span>. For <span class="math-container">$h(x)$</span>, indeed the domain are all reals, but <span class="math-container">$f(g(x))$</span> is defined to only take the <em>output</em> of <span class="math-container">$g(x)$</span>, so you have to go through <span class="math-container">$g(x)$</span> first, resulting in the domain being <span class="math-container">$x \ge 1$</span>.</p>
|
4,153,364 | <blockquote>
<p>If <span class="math-container">$f(x) = x^2$</span> and <span class="math-container">$g(x) = \sqrt{x-1}$</span>, find <span class="math-container">$f(g(x))$</span> and specify the domain.</p>
</blockquote>
<p><strong>My solution</strong></p>
<p><span class="math-container">$$
f(g(x)) = \left(\sqrt{x-1}\right)^2 = x-1,
$$</span>
Domain: <span class="math-container">$x$</span> such that <span class="math-container">$x$</span> is any real number.</p>
<p>Solution at the back of the book states that the domain is, <span class="math-container">$x$</span> such that <span class="math-container">$x \ge 1$</span>.</p>
<p>Now I understand why that is the domain for <span class="math-container">$g(x)$</span>, for any number less than 1 would make <span class="math-container">$g(x)$</span> negative and the square root of a negative number is an imaginary number.</p>
<p>But, that is not the cases with <span class="math-container">$f(g(x))$</span>. When is <span class="math-container">$(\sqrt{x-1})^2$</span> not a real number? isn't any input valid?</p>
| Kevin Arlin | 31,228 | <p>Another way to put the answer, which has already been given, is that it’s <em>not</em> always the case that <span class="math-container">$\left(\sqrt{x-1}\right)^2=x-1$</span>! This equation holds if and only if <span class="math-container">$x\ge 1$</span>. If <span class="math-container">$x=0$</span>, for instance, then the left hand is undefined while the right is <span class="math-container">$-1$</span>. So it’s incorrect to simplify this composite to <span class="math-container">$x-1$</span> without explicitly restricting the domain.</p>
|
3,839,153 | <p>Let <span class="math-container">$f:[a-1,1] \to \mathbb R_{+}$</span> such that <span class="math-container">$f(x)=\begin{cases} 1-(a-x)^2 &a-1 \lt x \leq a+1 \\ 0 & a+1 < x \leq 1 \end{cases}$</span></p>
<p>for some real <span class="math-container">$a<0$</span>.</p>
<p>If I compute the first order derivative of <span class="math-container">$f\;$</span>, I obtain <span class="math-container">$f^{\prime}(x)=\begin{cases} 2(a-x) &a-1 \lt x \leq a+1 \\ 0 & a+1 < x \leq 1 \end{cases} \quad$</span> which has a discontinuity of first kind at <span class="math-container">$x=a+1$</span>.</p>
<p>I am interested in estimating <strong>the sign of the second order derivative at <span class="math-container">$x=a+1$</span>.</strong> To this end, I tried to compute this derivative in the sense of distributions. If <span class="math-container">$\phi \in \mathcal C^1_c([a-1,1])$</span> then</p>
<p><span class="math-container">$\begin{align} \frac{d}{dx} f^{\prime}(\phi)=-\int_{a-1}^{a+1} 2(a-x)\phi^{\prime}(x)\;dx&=2\phi(a+1)+2\int_{a-1}^{a+1} \phi(x)\;dx\\&=2\delta_{a+1}(\phi)+2\int_{a-1}^{a+1} \phi(x)\;dx \end{align}$</span></p>
<p>At this point, I got stuck! Are my computations correct? If yes, then what am I missing? I don't see which will the sign of the second derivative of <span class="math-container">$f$</span> be at <span class="math-container">$x=a+1$</span>... Any help is much appreciated.</p>
<p>Thanks in advance!</p>
| md2perpe | 168,433 | <p>You have <span class="math-container">$f'((a+1)^-) = -2$</span> and <span class="math-container">$f'((a+1)^+) = 0$</span> so <span class="math-container">$f'$</span> increases at <span class="math-container">$x=a+1$</span>. Therefore the sign of <span class="math-container">$f''$</span> at <span class="math-container">$x=a+1$</span> is positive.</p>
|
1,401,661 | <blockquote>
<p>Dice are cubes with pips (small dots) on their sides, representing
numbers 1 through 6. Two dice are considered the same if they can be
rotated and placed in such a way that they present matching numbers on
the top, bottom, left, right, front, and back sides.</p>
<p>Below is an example of two dice that can be rotated to show that they
are the same if the 2-pip and 4-pip sides are opposite and the 3-pip
and 5-pip sides are also opposite.
<a href="https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0" rel="nofollow">https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0</a></p>
<p>How many different dice exist? That is, how many ways can you make
distinct dice that cannot be rotated to show they are the same? Note:
This problem does not involve rolling the dice or the probability of
roll outcomes.</p>
</blockquote>
<p>I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.</p>
<p>Could somebody make an attempt at rewording this? Or walking me through how to solve this?</p>
| Christian Blatter | 1,303 | <p>Put your dice on a table such that the $1$ is on the bottom for all of them. You still can rotate them freely around the vertical axis passing through the single dot on the bottom. Now begins the counting: You have five choices for the top face. The remaining four numbers can be paired in three ways to form pairs of opposite numbers on the vertical faces. For each chosen pairing you can realize two different orientations leading to two cubes which are mirror copies of each other. It follows that the total number of cubes is $5\cdot 3\cdot 2=30$.</p>
<p>By the way: There is a famous puzzle (<a href="http://www.mathematische-basteleien.de/macmahon.htm" rel="nofollow">Macmahon's colored cubes</a>) using $30$ colored cubes instead of numbered ones. Take any of the $30$ cubes as model and put together a $2\times2\times2$ cube looking like the model, with the extra condition that in the interior only faces with equal colors may touch.</p>
|
100,801 | <p>When doing calculations in notebook, I modify the code a lot, and there are several versions of similar codes in the same notebook. But I found myself easily forgot which version is the newest, and I have to confirm which cell is the newest.</p>
<p>So if mathematica can provide a menu options like "cell latest modification time" will be useful. But now, I can't find this option. </p>
<p>However, mathematica provide a menu option "Cell-> Notebook History" can do this job.</p>
<p>The problem is that on my computer, mathematica ver 10.3, if the notebook has a long and rich history, then the Notebook History makes mma froze again and again, keep popping dialog of "Formatting Notebook Contents" as follows</p>
<p><a href="https://i.stack.imgur.com/ufdhX.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ufdhX.png" alt="enter image description here"></a></p>
<p><strong>How can I start the notebook history for example only recent 3 days or more generally between specific date 10 Nov to 13 Nov? And also how to start notebook history with <code>selected cell</code> as default?</strong></p>
| matheorem | 4,742 | <p>Albert Retey just created a great button that can give a cell's latest modification time directly.</p>
<p>Here is a another workaround. It is small trick I just found, but works perfectly : )</p>
<p>The frozen of mma when open Notebook History windows is due to the default setting is checked for "All cell" instead of "selected cell". So mma collecting too many data to analysis dynamically.</p>
<p>If we could make "selected cell" the default option, then we solve the problem. But I don't know how to do it.</p>
<p><strong>Anyway, here is the "bug" trick.</strong></p>
<p>First, you have to open a fresh new notebook. </p>
<p>second, in this new notebook, you open Notebook History window, and check the "selected cell" option. like this</p>
<p><a href="https://i.stack.imgur.com/aDRHl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aDRHl.png" alt="enter image description here"></a></p>
<p>Third, switch back to the large notebook, now since "Notebook History window" is already set to analyze only selected cell, frozen problem is solved and you can select any cell to see its history smoothly without lag.</p>
|
553,103 | <p>consider the series of function $\sum f_n$ with $f_n(x)=\frac{x}{x^2+n^2}$.
It is easy to see that there is pointwise convergence on $\mathbb{R}$ (to a function that we'll call $f$) but not normal convergence. I want to know whether there is uniform convergence towards $f$ or not. </p>
<p>I tried to disprove uniform convergence by looking at $|\sum_{k=0}^n f_k(n)-f(n)|$ : the sum is unbounded and equivalent to $\ln n$. The problem is that I don't know how to estimate $f(n)$. </p>
<p>I'm sure I'm missing something not too difficult here, so help would be appreciated.</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Without using Complex Calculus, divide the numerator & the denominator by $\cos^2x$ and substitute $\tan x$ with $u$</p>
|
553,103 | <p>consider the series of function $\sum f_n$ with $f_n(x)=\frac{x}{x^2+n^2}$.
It is easy to see that there is pointwise convergence on $\mathbb{R}$ (to a function that we'll call $f$) but not normal convergence. I want to know whether there is uniform convergence towards $f$ or not. </p>
<p>I tried to disprove uniform convergence by looking at $|\sum_{k=0}^n f_k(n)-f(n)|$ : the sum is unbounded and equivalent to $\ln n$. The problem is that I don't know how to estimate $f(n)$. </p>
<p>I'm sure I'm missing something not too difficult here, so help would be appreciated.</p>
| Ron Gordon | 53,268 | <p>Hint: Let $z=e^{i x}$; $dx=-i dz/z$; $\cos{x}=(z+z^{-1})/2$. The integral is then equal to</p>
<p>$$-i \oint_{|z|=1} \frac{dz}{z} \frac{1}{1+\frac{3}{4} (z+z^{-1})^2} $$</p>
<p>Multiply out, determine the poles, figure out which poles, if any, lie within the unit circle, find the residues of those poles, multiply the sum of those residues (there may only be one, or none) by $i 2 \pi$, and you are done.</p>
|
17,398 | <p>I am teaching my daughter, who is currently about <span class="math-container">$46$</span> months old, additions. She is very curious and asks a lot of <em>good</em> questions. For example, when I told her that <span class="math-container">$2+6=8$</span> and <span class="math-container">$4+4=8$</span>, she asked me the following question:</p>
<blockquote>
<p>Why are they the same?</p>
</blockquote>
<p>Surely I know the logical answer this question using <a href="https://en.wikipedia.org/wiki/Peano_axioms" rel="noreferrer">Peano Axioms</a> and the definition of natural numbers. But she has not learnt the Peano Axiom at this age.</p>
<p>So how to answer my three-year-old daughter's question that </p>
<blockquote>
<p>Why <span class="math-container">$2+6$</span> is the same as <span class="math-container">$4+4$</span>?</p>
</blockquote>
| Nick C | 470 | <p>I'm nearly <em>sure</em> I did this with my child when she was young.</p>
<p>First, establish that she understands that a number, like three, is equal to <span class="math-container">$1+1+1$</span>. Hold three fingers up and ask her "how many is this"?</p>
<p><a href="https://i.stack.imgur.com/HthuC.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/HthuC.jpg" alt="enter image description here"></a></p>
<p>Then spread them out and ask the same question. Are we adding <span class="math-container">$1+1+1$</span>?</p>
<p><a href="https://i.stack.imgur.com/3kLwo.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/3kLwo.jpg" alt="enter image description here"></a></p>
<p>Try holding up eight fingers (keep your thumbs down, for example) and ask her to count them.</p>
<p><a href="https://i.stack.imgur.com/1xGsj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/1xGsj.jpg" alt="enter image description here"></a></p>
<p>Then hold them together, but move your hands apart. What numbers do you see added? What is the total?</p>
<p><a href="https://i.stack.imgur.com/Ubowj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Ubowj.jpg" alt="enter image description here"></a></p>
<p>Then make a V with one hand, holding two fingers apart from the other six. What numbers are being added? What is the total?</p>
<p><a href="https://i.stack.imgur.com/II6F2.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/II6F2.jpg" alt="enter image description here"></a></p>
<p>Then hold three fingers apart from the other five. What numbers are being added? What is the total?</p>
<p><a href="https://i.stack.imgur.com/NZJdA.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/NZJdA.jpg" alt="enter image description here"></a></p>
<p>Finally, hold one finger apart from the other seven. What numbers are being added? What is the total?</p>
<p><a href="https://i.stack.imgur.com/Ug4zc.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Ug4zc.jpg" alt="enter image description here"></a></p>
<p>Then you can talk about different ways to add numbers to make <span class="math-container">$10$</span>. This time, have her act it out with her hands.</p>
|
17,398 | <p>I am teaching my daughter, who is currently about <span class="math-container">$46$</span> months old, additions. She is very curious and asks a lot of <em>good</em> questions. For example, when I told her that <span class="math-container">$2+6=8$</span> and <span class="math-container">$4+4=8$</span>, she asked me the following question:</p>
<blockquote>
<p>Why are they the same?</p>
</blockquote>
<p>Surely I know the logical answer this question using <a href="https://en.wikipedia.org/wiki/Peano_axioms" rel="noreferrer">Peano Axioms</a> and the definition of natural numbers. But she has not learnt the Peano Axiom at this age.</p>
<p>So how to answer my three-year-old daughter's question that </p>
<blockquote>
<p>Why <span class="math-container">$2+6$</span> is the same as <span class="math-container">$4+4$</span>?</p>
</blockquote>
| user21820 | 1,550 | <p>Here is a written explanation to accompany <a href="https://matheducators.stackexchange.com/a/17401">Nick C's excellent 'hand-waving' answer</a>. =)</p>
<p><span class="math-container">$2+6 \quad = \quad 1+1 \quad + \quad 1+1+1+1+1+1$</span></p>
<p>        <span class="math-container">$ = \quad 1+1+1+1+1+1+1+1$</span></p>
<p>      <span class="math-container">$ = \quad 1+1+1+1 \quad + \quad 1+1+1+1 \quad = \quad 4+4$</span></p>
<p>Explain that adding the same numbers always gives the same result no matter which order you add them, so we can add the ones up in any way we like. Also explain that we define <span class="math-container">$2 = 1+1$</span> and <span class="math-container">$4 = 1+1+1+1$</span> and <span class="math-container">$6 = 1+1+1+1+1+1$</span>, so if we choose the right order to add the ones up, we can get either <span class="math-container">$2+6$</span> or <span class="math-container">$4+4$</span>.</p>
|
933,515 | <p>Let $T$ be a bounded operator acting on a Banach space $X$. The point spectrum $\sigma_p(T)$ is of $T$ is defined to be $$\sigma_p(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has nonempty kernel}\}$$ and the residual spectrum $\sigma_r(T)$ is defined to be $$\sigma_r(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is not even dense in }X\}.$$
For the continuous spectrum there seem to exist two non-equivalent definitions. For example, Dunford-Schwartz defines
$$\sigma_c(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is a dense proper subset of } X\}$$ whereas for example Kato or Schmüdgen define
\begin{align*}\sigma_c'(T)&:=\{\lambda\in\mathbb C~|~\text{Ran}(T-\lambda)\text{ is not a closed subset of } X\}\\
\sigma_r'(T)&:=\{\lambda\in\mathbb C~|~T-\lambda\text{ has a bounded inverse defined only on a proper subset of } X\}.
\end{align*}</p>
<p>Does anyone have insights on the difference of these definitions? What are the advantages? The former one gives a partition of the spectrum, so that's kind of nice, but why is the second definition (more?) useful?</p>
<hr>
<p><strong>Edit:</strong> I just created a table that categorizes parts of the spectrum in terms of the questions:</p>
<ul>
<li>Is $K_\lambda:=$Ker$(T-\lambda)$ zero or nonzero?</li>
<li>Is $R_\lambda:=$Ran$(T-\lambda)$ closed or not?</li>
<li>Is $R_\lambda:=$Ran$(T-\lambda)$ dense or not?</li>
</ul>
<p><img src="https://i.stack.imgur.com/PeFKz.png" alt="Table of spectral Decomposition"></p>
<p>According to this table the point spectrum can be subdivided into four subtypes and the residual spectrum can be subdivided into two subtypes. The alternative definitions would correspond to \begin{align*}\sigma_c'(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\\
\sigma_r'(T)&=\sigma_{r,1}(T)
\end{align*} </p>
<p>The approximate point spectrum and compression spectrum on the other hand are given by
\begin{align*}
\sigma_{ap}(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T)\\
\sigma_{com}(T)&=\sigma_{r,1}(T)\cup\sigma_{r,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T).
\end{align*} </p>
| Disintegrating By Parts | 112,478 | <p>The common classification of spectrum is the first one you gave. I'm not sure where $\sigma_{c}'(T)$ would have come from, or why it was defined that way, but it doesn't strike me as a standard or common definition, even though I can see a reason for the alternative definition.</p>
<p>If $T$ is closed, then this alternative definition would probably be better described as "approximate point spectrum," which consists of $\lambda\in\mathbb{C}$ for which there is a sequence of unit vectors $\{ v_{n}\}$, none of which is in $\mathcal{N}(T-\lambda I)$, for which $\lim_{n}\|Tv_{n}-\lambda v_{n}\|=0$. This alternative definition may help explain the authors' motives.</p>
<p>The first classification of spectrum is somewhat lacking when it comes to approximate spectrum because the presence of an eigenvector with eigenvalue $\lambda$ automatically puts $\lambda$ in the point spectrum, even though there may be more to the story for $\lambda$ because it is still possible for there to be approximate eigenvectors with approximate eigenvalue $\lambda$. For example, let $Tf=xf$ on $L^{2}_{\mu}[0,1]$. If $\mu$ has an atom at $\lambda \in [0,1]$, but $\mu([\lambda-\epsilon,\lambda+\epsilon]\setminus\{\lambda\})=0$ for some $\epsilon > 0$, then there is no approximate eigenvector sequence for $T$ with approximate eigenvalue $\lambda$; but if this is not the case, then there is an approximate point spectrum component. Using the first classification, both cases of $T$ are classified with $\lambda$ in their point spectrum ... end of story. But there really is more to the story because $(T-\lambda I)$ may or not may not have a closed range.</p>
|
13,553 | <p>(This issue has been discussed before, specifically <a href="https://math.meta.stackexchange.com/questions/8485/how-civil-should-we-be/8493">here</a>. A different kind of censorship has been discussed <a href="https://math.meta.stackexchange.com/questions/8306/defacement-censorship-of-peoples-answers">here</a>.)</p>
<p>There are two reasons I am bringing this up again. Firstly, the previous discussion on this issue did not seem conclusive. In particular, the <a href="https://math.meta.stackexchange.com/a/8487/68107">top answer</a> from Mad Scientist states that this is "a matter of professionalism" and that "if the swear words are not necessary for the post, I see no problem in removing them." Conversely, the <a href="https://math.meta.stackexchange.com/a/8520/68107">second-to-top answer</a> (which I agree with much more) states that "I know this is supposed to be a professional environment, and that is precisely why we need to allow every form of expression possible."</p>
<p>Secondly, a moderator has recently edited <a href="https://math.stackexchange.com/a/775883/68107">this post</a> to remove a swear word. Perhaps this was a response to a flag. Instead of rolling back and starting a potential edit war, I am bringing it up here to solicit the broader community opinion.</p>
<p>With regards to this specific post, I am of the strong opinion that the swear word should not have been replaced, because the way it was used was effective and clear. The replacement--"horrible open access journals"--simply doesn't get the point across. I think this censorship should be reversed.</p>
<p>More generally, I disagree with removing swear words merely because they aren't necessary, and think they should only be removed if they <em>take away</em> from a post. I believe it is unprofessional and pointless to censor any form of effective communication. I feel we should tolerate swearing to the extent that it is used well, and I see no need to favor the small percentage of us who are offended by it.</p>
<p>What do others think?</p>
| Thomas | 26,188 | <p>In my little narrow minded world I can't comprehend why someone would suggest that we should allow swear words (or "bad" words in general). But then I am reminded that the world actually is bigger than the one I live in. I realize that my view of right and wrong doesn't necessarily overlap with other peoples views. I often find myself thinking that the Stack Exchange system is American and as such should follow American standards on everything. But, again, thinking about it I realize that this is a site for everyone in the world.</p>
<p>Because of this I believe that we need to be as respectful of each others cultures. We have to acknowledge that some people find swearing offensive. So why not be respectful?</p>
<p>My opinion is that swearing is never needed. If you have to resort to that type of language it must be because (in my opinion) because you can't find any other way of saying it. So I would suggest that when you feel the urge to call a journal "shitty" maybe stop and think if you could find a more clever way to articulate that.</p>
<p>I appreciate that the word was removed because it was <em>flagged</em>. When a comment is flagged it is because someone found it offensive. Again, in your culture that might not make much sense, but maybe we can agree that while we don't understand we constantly strive to make as many people welcome as possible.</p>
<p>About the discussion in the <a href="http://meta.math.stackexchange.com/questions/8485/how-civil-should-we-be">other meta-thread</a> you write that you agree much more with <a href="http://meta.math.stackexchange.com/questions/8485/how-civil-should-we-be/8520#8520">other answer</a>. You don't say that you agree completely with the other answer, but it does state that:</p>
<blockquote>
<p>...we need to allow every form of expression possible.</p>
</blockquote>
<p>I am wondering what the author means by <em>possible</em>. Hopefully he doesn't mean that every form of expression that is physically possibly should be allowed. We would hopefully not allow expressing the same views as <a href="http://www.nambla.org/" rel="nofollow noreferrer">NAMBLA</a> on this site (even if we do believe that we can make a mathematical point more forceful by doing so). So not every form of expression should be allowed. The author of the answer makes the argument that we should allow a greater form of expression because it helps people express their ideas (about mathematics) and that limiting ways of expression (communication) is unprofessional. I would disagree with this. Professionalism isn't about allowing all forms of expression. When I teach in the classroom I have to behave professionally. That means that I have to limit myself. I might feel like expressing that a student is unintelligent or a lazyass because they failed a test, but I refrain from doing that because I want to behave professionally. </p>
<p><strong>The key question is: What should be allowed?</strong></p>
<p>We could start by reading <a href="https://math.stackexchange.com/help/behavior">https://math.stackexchange.com/help/behavior</a> about what behavior is allowed by the Stack Exchange. It clearly says:</p>
<blockquote>
<p>Please note that expletives are not allowed. If you use expletives on this site, you may be issued a warning or a suspension.</p>
</blockquote>
<p>Granted, using expletives is not the same as swearing, but it still sets a tone. It points out that the Stack Exchange network isn't built on the idea that "anything goes". We, for example, close questions that are off-topic. If you are familiar with CB radios, it is a commonly known problem in the US that the channels are filled with all kinds of profanity. Here you have a system where you can say what ever you want under relative anonymity, but it also has its consequences.</p>
<p>Unless we resort to arguing from a specific religious point of view, I don't know that we can appeal to any object morals that can guide us in answering the above key question. So what do we do? Maybe we simply let the community decide. If a comment is flagged and it is "commonly known" that the comment could cause someone to be offended, and if removing the comment (or the "bad" word) doesn't radically change the content, why not remove it?</p>
<p>Last: I am not saying that you are saying this, but I can sometimes "feel" that discussions can turn into a question about <em>rights</em>. What are my <em>rights</em> as a user on this site? I have a right to express myself. You should limit my right to be who I am. Maybe it is better to humbly focus on how we can treat each other with respect. And that might mean that you have to give up some of your "rights" to express yourself.</p>
|
3,459,178 | <p>Is there a general way to factor <span class="math-container">$ax^2 + bx$</span> into <span class="math-container">$($</span>some expression<span class="math-container">$)^2$</span>, where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are constants?</p>
<p>The reason I am asking is to understand how the integral <span class="math-container">$\int_{-\infty}^{\infty} e^{ax^2+bx}\ dx$</span> is turned into the form <span class="math-container">$\int_{-\infty}^{\infty} e^{-u^2}\ du$</span> which is an important step in solving it.</p>
| Donald Splutterwit | 404,247 | <p><span class="math-container">\begin{eqnarray*}
ax^2+bx= a \left( x+ \frac{b}{2a} \right)^2 - \frac{b^2}{4a}.
\end{eqnarray*}</span></p>
|
2,327,181 | <p>I'm a bit confused about the definition of finite sets/intervals. I know that a set S is called finite when it has a finite number of elements, or formally, when there exists a bijection $f:S\to\{1,...,n\}$ for some natural number n. </p>
<p>However, the interval $(1,2)$ is called finite. I don't understand why; $(1,2)$ is not even countable, and there definitely does not exist a bijection $f:(1,2)\to\{1,...,n\}$. </p>
<p>Why do we call $(a,b)$ with $a,b\in\mathbb{R}$, finite? Did we just agree to do so, or is it incorrect to use the interval $(a,b)$ as a set like I did in the definition of finiteness above?</p>
<p>Thanks!</p>
| Asaf Karagila | 622 | <p>There are different ways to think about the size of a set. In the case of the real numbers, and specifically intervals, we can talk about their length (and generally, their Lebesgue measure in the case of measurable sets).</p>
<p>If you think about the real numbers as a model of time or space, then the distance between you and the screen through which you are reading this is a finite interval. But in this model, based on the real numbers, it is an uncountable interval, not a finite set.</p>
<hr>
<p>One thing to remember about terminology, is that it should highlight to the reader or listener something about a certain relevant property. In the case of intervals, we already know they all have the same cardinality (in the case of non-degenerate intervals). So we can use "finite" or "infinite" to talk about their length (and formally, their measure). Thus setting the importance on that aspect, rather than their cardinality.</p>
|
118,985 | <p>Here's a question I came across:</p>
<blockquote>
<p>count the number of permutations of 10 men, 10 women and one child,
with the limitation that a man will not sit next to another man, and a
woman will not sit next to another woman</p>
</blockquote>
<p>What I tried was using the inclusion-exclusion principle but it got way to complicated. I don't have the right answer, but I know it should be simple (it was a 5 points question in an exam I was going through).</p>
<p>Any ideas? Thanks!</p>
| Robert Israel | 8,508 | <p>Try comparing to $e^{-tx}$ where $0 < t < 1$.</p>
|
118,985 | <p>Here's a question I came across:</p>
<blockquote>
<p>count the number of permutations of 10 men, 10 women and one child,
with the limitation that a man will not sit next to another man, and a
woman will not sit next to another woman</p>
</blockquote>
<p>What I tried was using the inclusion-exclusion principle but it got way to complicated. I don't have the right answer, but I know it should be simple (it was a 5 points question in an exam I was going through).</p>
<p>Any ideas? Thanks!</p>
| Did | 6,179 | <p>For every $x\geqslant2$, $x^2\mathrm e^{-x}\leqslant16\mathrm e^{-2}\mathrm e^{-x/2}$ hence
$$\int\limits_2^{+\infty}x^2\mathrm e^{-x}\mathrm dx\leqslant16\mathrm e^{-2}\int\limits_2^{+\infty}\mathrm e^{-x/2}\mathrm dx=16\mathrm e^{-2}\cdot2\mathrm e^{-1}\lt2.
$$</p>
|
690,729 | <p>Vancouver is 300 km away from Seattle. Two friends, one leaving from each city on a bike want to find a campsite between the two cities. One cyclist starts from vancouver at 25 km/h. His friend will start 2.0h later from seattle at 32 km/h. How far from Vancouver do the friends meet?</p>
<p>Attempt:</p>
<p>dA + dB = 300 </p>
<p>vA(t) + vB(t+2) = 300</p>
<p>25t + 32t + 2 = 300 </p>
<p>t = 4.14 s</p>
<p>now how do i find the distance travelled by each cyclist?</p>
<p>not sure if this is even the correct direction, but any help would be greatly appreciated! Thank you!!</p>
<p>btw- correct answer is 160 km</p>
| MCT | 92,774 | <p><strong>Hint</strong>: Suppose that the friends begin cycling at the same time, but instead of starting from $300$ kilometers away from each other, they start $250$ kilometers away from each other. If $x$ is the answer to this equation, the answer to your original equation will be $x + 50$.</p>
|
2,365,120 | <p>I'm doing some revision on indices and surds. How do you simplify </p>
<blockquote>
<p>$(xy^2)^p\sqrt{x^q} $</p>
</blockquote>
<p>Bit confused because my textbook says the answer is </p>
<blockquote>
<p>$x^{p+q/2}y^{2p}$</p>
</blockquote>
<p>I can understand simplifying but only when it's the same base. I'm confused with this specific question - step-by-step help would be much appreciated!</p>
<p>[[Edit: changed $x^{p+1/2q}$ to $x^{p + q/2}$.]]</p>
| user247327 | 247,327 | <p>You need parentheses to clarify your expression, and you seem to have dropped a "^"- you mean (xy^2)^p sqrt(x^q). Apply the "rules of exponents": (xy^2)^p= x^p (y^2)^p= x^p y^{2p}. Also sqrt(x^q)= (x^q)^(1/2)= x^(q/2). </p>
<p>So (xy^2)^p sqrt(x^q)= (x^p y^(2p))(x^(q/2))= x^(p+ q/2) y^{2p} </p>
<p>Also "q/2" is NOT the same as "1/2q"!</p>
|
2,501,235 | <p>Many books define almost sure convergence as follows:</p>
<blockquote>
<p>The sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$ defined on the probability space $(\Omega, \mathcal{F}, P)$ converges almost surely to a random variable $X$ defined on the same probability space, if
$$
P(\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}) = 1.
$$</p>
</blockquote>
<p>In connection to <a href="https://math.stackexchange.com/questions/2498575/equivalent-conditions-for-almost-sure-convergence">this question</a>, I wonder if the set $A := \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$ is implicitly assumed to be measurable or whether it is actually a priori measurable. If the latter is true, how can one show this?</p>
| Michael Greinecker | 21,674 | <p>$$A=\{\omega\in\Omega\mid\text{ for all }\epsilon>0\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<\epsilon\text{ for all }n\geq N\}$$
$$=\{\omega\in\Omega\mid\text{ for all }m\text{ there is } N\text{ such that } |X_n(\omega)-X(\omega)|<1/m\text{ for all }n\geq N\}$$
$$=\bigcap_{m=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n=N}^\infty\{\omega\in\Omega\mid|X_n(\omega)-X(\omega)|<1/m\}.$$</p>
|
3,936,694 | <p>I would like to construct a deterministic TM that decides <span class="math-container">$L=$</span>{<span class="math-container">$<G,s,t,k>$</span> | <span class="math-container">$G$</span> is a directed graph that has a path of length at most <span class="math-container">$k$</span> from vertex <span class="math-container">$s$</span> to vertex <span class="math-container">$t$</span>} in polynomial time.</p>
<p>So far, my construction is as follows.</p>
<p><span class="math-container">$M=$</span>"On input <span class="math-container">$<G,s,t,k>:$</span></p>
<ol>
<li>For each path <span class="math-container">$p$</span> from <span class="math-container">$s$</span> to a vertex <span class="math-container">$v \in G$</span> of <span class="math-container">$k$</span> length:</li>
</ol>
<p>------1.0 Clear all marks from vertices in <span class="math-container">$G$</span>.</p>
<p>------1.1 Mark the vertices in <span class="math-container">$p$</span>.</p>
<p>------1.2 Scan input. If <span class="math-container">$t$</span> is marked, accept.</p>
<ol start="2">
<li>Reject, since <span class="math-container">$t$</span> is not within <span class="math-container">$k$</span> steps from <span class="math-container">$s$</span>."</li>
</ol>
<p>(1.0), (1.1), and (1.2) are each <span class="math-container">$O(n)$</span>, and <span class="math-container">$(2)$</span> is <span class="math-container">$O(1)$</span>.</p>
<p>However, I think (1) is <span class="math-container">$O(n!)$</span>.</p>
<p>What is a better approach that would make this <span class="math-container">$O(n^m)$</span> (i.e., polynomial)?</p>
<p>Edit:</p>
<p>The TM does not have to be "rigorously made".</p>
<p>For example, to find determine whether PATH={<G,s,t>| G is a directed graph that has a path from s to t} is in <span class="math-container">$P$</span>, the following TM is acceptable:</p>
<p>M="On input <G,s,t>"</p>
<ol>
<li><p>Mark vertex s.</p>
</li>
<li><p>Repeat until no additional vertex is marked:</p>
</li>
</ol>
<p>------Scan all edges of G. If there is an edge from a marked vertex u to an unmarked vertex v, mark v.</p>
<ol start="3">
<li>If t is marked, accept. Otherwise, reject."</li>
</ol>
<p>((1) is <span class="math-container">$O(1)$</span>, (2) is <span class="math-container">$O(\#$</span>of edges <span class="math-container">$*\#$</span>of vertices<span class="math-container">$)=O(n^2*n=n^3)$</span>, and (3) is <span class="math-container">$O(1)$</span>).</p>
| Mike | 544,150 | <p>That's not true in general. Take <span class="math-container">$$y(t)=e^{-1/t}$$</span> and <span class="math-container">$$x(t)=-5t$$</span> say.
Then <span class="math-container">$$\lim_{t \rightarrow 0^+} y(t) = \lim_{t \rightarrow 0^+} x(t)=0,$$</span> but <span class="math-container">$$\lim_{t \rightarrow 0^+} (e^{-1/t})^{-5t}= e^5 >1.$$</span></p>
<p>For an even more dramatic example take <span class="math-container">$y(t)=e^{-1/t^2}$</span> and <span class="math-container">$x(t)=-t$</span>.</p>
<p>HOWEVER, if <span class="math-container">$y(t)$</span> and <span class="math-container">$x(t)$</span> as above are both positive for then infact the limit will indeed fall between 0 and 1, as any positive number less than 1 raised to a positive power is in [0,1].</p>
|
2,661,718 | <p>My basic understanding is that each time
I have 30% chance of winning the prize
so between 3 and 4 tries I should win it</p>
<p>cause .3 +.3 +.3 = 90% as I need to win only once with 3 try
.3+.3+.3+.3 = 120% with 4 try</p>
<p>I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly</p>
<p>thanks for answering this noob question my math are way behind me now</p>
| alepopoulo110 | 351,240 | <p>the derivative is $f'(x)=2x-2\log(1+x)-\log(1+x)^2$. By setting $t=\log(x+1)$ one gets $h(t)=2e^t-2t-t^2-2$. We get $h'(t)=2e^t-2t-2\geq 0$, by the Taylor expansion of $e^t$. But $h(0)=0$, so for $t<0$, which corresponds to $x<0$, we have $f'(x)<0$ and for $t>0$ we have $f'(x)>0$. Also $f(0)=0$, so $f\geq 0$.</p>
|
1,825,392 | <p>Let $f(z)$ be an entire function so that,</p>
<p>$$ \int \frac{|f(z)|}{1 + |z|^3} dA(z) < \infty$$</p>
<p>where the integral is taken over the entire complex plane. Show that $f$ is a constant.</p>
<p>I believe that the idea is to use the mean value property; that is:</p>
<p>$$f(z) = \frac{1}{\pi\delta^2}\int_{D(z, \delta)}f(w)dA(w)$$</p>
<p>and then do some manipulation to relate the two integrals. But I'm not sure otherwise how to proceed. Can anyone help? </p>
| Hmm. | 227,501 | <p>Fix any $z_0\in \Bbb C$. $\forall r > 0$, we have by Mean Value Theorem, $$~~~~~~~~~~f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta})\, d\theta$$ </p>
<p>$$\implies |f(z_0)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})|\, d\theta$$ </p>
<p>For every $R>0$, we have,
$$\int_0^{R} \lvert f(z_0)\rvert~ \dfrac{r}{1+r^3}\, dr \le \frac{1}{2\pi}\int_0^R \int_0^{2\pi} \lvert f(z_0 + re^{i\theta})| ~\dfrac{r}{1+r^3}\, d\theta\, dr $$ </p>
<p>which means, for $$A_R=\int_0^R \dfrac{rdr}{1+r^3}$$</p>
<p>we obtain,
$$\lvert f(z_0)\rvert A_R \le \frac{1}{2\pi}\iint_{D(0;R)} \dfrac{|f(z)|}{1+|z|^3}\,dA$$ </p>
<p>Therefore, letting $R\to\infty$, for some constants $C$ and $M$, </p>
<p>$$C |f(z_0)|\le \iint_{\mathbb{C}} \dfrac{|f(z)|}{1+|z|^3}\, dx\, dy=M$$</p>
<p>So $f(z_0) \leq M/C$. Since $z_0$ was arbitrary, $f$ is constant.</p>
|
294,383 | <p>Evaluate :
$$\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}$$</p>
| xpaul | 66,420 | <p>We also can use this way to calculate instead of using complex analysis or special functions. Noting that
$$ \int_0^\infty te^{-xt}dt=\frac{1}{x^2} $$
we have
\begin{eqnarray}
&&\int_{0}^{\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}\\
&=&\int_{0}^{\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\left(\int_0^\infty te^{-xt}\text{d}t\right)\text{d}x}\\
&=&\int_{0}^{\infty}t\left(\int_0^\infty \left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)e^{-xt}\text{d}t\right)\\
&=&\frac12\int_{0}^{\infty}t\left(\int_0^\infty \frac{2xe^{-x}-1+e^{-2x}}{1-e^{-2x}}e^{-xt}\text{d}x\right)\text{d}dt\\
&=&\frac12\int_{0}^{\infty}t\left(\int_0^\infty(2xe^{-x}-1+e^{-2x})e^{-xt}\sum_{n=0}^\infty e^{-2nx}\text{d}x\right)\text{d}dt\\
&=&\frac12\int_{0}^{\infty}t\sum_{n=0}^\infty\left(-\frac1{2n+t}+\frac2{(2n+t+1)^2}+\frac1{2n+t+2}\right)\text{d}dt\\
&=&-\sum_{n=0}^\infty \left(1+n\ln n+\ln(n+\frac{1}{2})-(n+1)\ln(n+1)\right)\\
&=&-\lim_{n\to 0^+}\left(1+n\ln n+\ln(n+\frac{1}{2})-(n+1)\ln(n+1)\right)-\sum_{n=1}^\infty \left(1+n\ln n+\ln(n+\frac{1}{2})-(n+1)\ln(n+1)\right)\\
&=&-(1-\ln 2)+1-\frac32\ln 2\\
&=&-\frac12\ln 2.
\end{eqnarray}</p>
|
686,167 | <p>I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where</p>
<blockquote>
<p>$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$</p>
<p>Here $x_j=j/(n+1)$.</p>
</blockquote>
<p>So, working through the algebra:</p>
<blockquote>
<p>$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots
=\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}
$$</p>
</blockquote>
<p>Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$
which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.</p>
<p>My question: How does one see that it equals 1?</p>
| Michael Hardy | 11,667 | <p>$$
\frac{1}{1-e^{-a}} = \frac{1}{1-e^{-a}}\cdot\frac{e^a}{e^a} = \frac{e^a}{e^a-1} = \frac{-e^a}{1-e^a}.
$$</p>
<p>Hence
$$
\frac{1}{1-e^a} + \frac{1}{1-e^{-a}} = 1.
$$</p>
|
658,807 | <p>i am having difficulty understanding why the diagram on page 8 of this <a href="http://www.cs.toronto.edu/~sme/presentations/cat101.pdf" rel="nofollow">presentation</a> is a category. the author claims that it is on page 43.</p>
<p>it looks like the two smaller arrows on the left must be idempotent or we would have an infinite number of arrows there. it would also seem that composing one of the smaller arrows on the left with one of the longer (left pointing) arrows must result in the small arrow. composing either left with right must be the identity on the set on the right</p>
<p>i have some code and tests run fine with the categories with 0,1,2,3, and 4 elements as well as another simple example. in these examples the objects are just things.</p>
<p>one of the users here (mike stay) came up with a concrete example using 2 sets for the things. the thing on the left is a set with two elements where each of the small arrows pick out one element. and the thing on the right is a set with a single element. where the long right arrow maps everything to that single element and the two long left arrows map to one of the elements of the set on the right.</p>
<pre><code>A={a1,a2}
B={b1}
short1(x)=a1, short2(x)=a2
right(x)=b1
left1(x)=a1
left2(x)=a2.
</code></pre>
<p>i define short1 o left1 = short1 etc. and right(left?) = identity.</p>
<p>but this causes my tests for composition to fail since the source(short1 o left1)!=source(short1).</p>
<p>associativity seems to fails also in a manner similar to the the thing on the right <a href="http://yogsototh.github.io/Category-Theory-Presentation/#slide-20" rel="nofollow">here</a></p>
<p>the tests fails if i omit left2 from the category.</p>
<p>edit - @MartianInvader 's idea seems to work. the composition table for:</p>
<pre><code>r:A->B, s1:A->A, s2:A->A, l1:B->A, l2:B->A
</code></pre>
<p>is</p>
<pre><code>iA·iA=iA
iA·s1=s1
iA·s2=s2
iA·l1=l1
iA·l2=l2
iB·iB=iB
iB·r=r
r·iA=r
r·s1=r
r·s2=r
r·l1=iB
r·l2=iB
s1·iA=s1
s1·s1=s1
s1·s2=s1
s1·l1=l1
s1·l2=l1
s2·iA=s2
s2·s1=s2
s2·s2=s2
s2·l1=l2
s2·l2=l2
l1·iB=l1
l1·r=s1
l2·iB=l2
l2·r=s2
</code></pre>
<p>edit: seems like there are two ways to make this thing a category:</p>
<pre><code>one way:
iA iB r s1 s2 l1 l2
iA iA · · s1 s2 l1 l2
iB · iB r · · · ·
r r · · r r iB iB
s1 s1 · · s1 s1 l1 l1
s2 s2 · · s2 s2 l2 l2
l1 · l1 s1 · · · ·
l2 · l2 s2 · · · ·
another way:
iA iB r s1 s2 l1 l2
iA iA · · s1 s2 l1 l2
iB · iB r · · · ·
r r · · r r iB iB
s1 s1 · · s1 s1 l2 l2
s2 s2 · · s2 s2 l1 l1
l1 · l1 s2 · · · ·
l2 · l2 s1 · · · ·
</code></pre>
| Andrej Bauer | 30,711 | <p>There is a problem when you say "it would also seem that composing one of the smaller arrows on the left with one of the longer (left pointing) arrows must result in the small arrow."</p>
<p>The category you've given is ok, but you are confused about the compositions. Given morphisms $f : A \to B$ and $g : B \to C$, the composition $g \circ f$ is a morphism $A \to C$. That is, the source of $g \circ f$ is the same as the source of $f$. If you apply this to $\text{short1} \circ \text{left1}$ you will see that all is well.</p>
|
2,975,647 | <p>Is this following statement valid (where both <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are positive)?</p>
<blockquote>
<p>If <span class="math-container">$x>y$</span>, then <span class="math-container">$\dfrac{1}{x} < \dfrac{1}{y}$</span>. </p>
</blockquote>
| user10354138 | 592,552 | <p>In fact, on <span class="math-container">$\{\lvert z\rvert=1,z\neq 1\}$</span>, <span class="math-container">$\sideset{_2}{_1}F(a,b;c;z)$</span> converges only conditionally if <span class="math-container">$0\geq\operatorname{Re}(c-a-b)>-1$</span> and diverges if <span class="math-container">$\operatorname{Re}(c-a-b)\leq -1$</span>.</p>
|
2,061,655 | <p>Could we compute the limits
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$
without using the l'Hospital rule and the Taylor expansion?</p>
| Vincenzo Oliva | 170,489 | <p>The first one is equal to $1+\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}.$ We can apply to this limit the technique in <a href="https://math.stackexchange.com/a/2028545/170489">this answer of mine.</a> </p>
<p>Let $L=\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}$. This is not infinite if it exists because near $0$ one has $$-\frac12\leftarrow\frac{\cos x-1}{x^2}<\frac{\frac{\sin x}{x}-1}{x^2}<\frac{1-1}{x^2}=0.$$ Then, considering a similar limit, $$\lim_{x\to0}\frac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\frac{\sin(2x)-2x}{2x^3}=4L$$ we can deduce$$\lim_{x\to0}\frac{\sin x\cos x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12=3L,$$that is $L=-\frac16.$ So your limit equals $\frac56$.</p>
<p>Getting back to this; due to the finiteness of the first limit, the second one is the same as $$\lim_{x\to0}\frac{e^x-1-x}{x^2}+\lim_{x\to0}\frac{x-\sin x}{x^2}=\lim_{x\to0}\frac{\frac{e^x-1}{x}-1}{x},$$<em>i.e.</em> $f'(0)$ where $f(x)=\frac{e^x-1}{x}, f(0)=1.$ One has $$f'(0)=\lim_{x\to0}\frac{e^x(x-1)+1}{x^2}=\lim_{x\to0}\frac{e^x-1}{x}-\lim_{x\to0}\frac{e^x-1-x}{x^2}=1-f'(0),$$ so it equals $\frac12$.</p>
<p>Alternatively, let $L_2$ be this limit. It is largely known that $e^x>1+x$; this also means $e^{-x}>1-x$, or equivalently for $x<1$, $e^x<\frac1{1-x},$ and it's easy to prove $\frac1{1-x}<1+x+2x^2 $ for $0\ne x<\frac12.$ Thus, $0\le L_2\le2.$ Assuming it exists, we have $$L_2=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{2x}-(1+x)^2}{x^2(e^x+1+x)}=\frac12\lim_{x\to0} \frac{e^{2x}-1-2x-x^2}{x^2}=2L_2-\frac12,$$whence $L_2=\frac12$.</p>
|
2,176,503 | <p>A topological space $X$ is called quasi-compact if whenever ${(U_i)}_{i∈S}$ are a family of
open subsets such that $∪_{i∈S}(U_i) = X$ then there are a finite number of $(U_i)$’s which actually
cover X.</p>
<p>I'm thinking first we show that if $(U_{fi})$, $i ∈ S$ (where ${f_i}$, $i ∈ S$ a collection of elements of $R[X]$) is a family of principal open subsets which cover an affine
variety $X$, then there is a finite number which cover $X$.<br>
Then we use the fact that principal open subsets are a basis for the Zariski topology and above statement to show
that a finite number of the $(U_j)$ are sufficient to cover $X$. </p>
| Alex Mathers | 227,652 | <p>Affine schemes are Noetherian as topological spaces, meaning every descending chain of closed subspaces $Z_1\supseteq Z_2\supseteq\cdots$ stabilizes. This is the same as saying that each increasing chain of open subspaces $U_1\subseteq U_2\subseteq\cdots$ stabilizes.</p>
<p>Now try to show that if $X$ is not quasicompact, you can get an infinite strictly increasing chain of open subsets, giving a contradiction.</p>
|
2,176,503 | <p>A topological space $X$ is called quasi-compact if whenever ${(U_i)}_{i∈S}$ are a family of
open subsets such that $∪_{i∈S}(U_i) = X$ then there are a finite number of $(U_i)$’s which actually
cover X.</p>
<p>I'm thinking first we show that if $(U_{fi})$, $i ∈ S$ (where ${f_i}$, $i ∈ S$ a collection of elements of $R[X]$) is a family of principal open subsets which cover an affine
variety $X$, then there is a finite number which cover $X$.<br>
Then we use the fact that principal open subsets are a basis for the Zariski topology and above statement to show
that a finite number of the $(U_j)$ are sufficient to cover $X$. </p>
| Andres Mejia | 297,998 | <p>Every descending chain of Zariski closed subsets in $X$,
$$X_1 \supseteq X_2,...$$</p>
<p>corresponds by inclusion reversal to
$$I(X_1) \subseteq I(X_2)...$$</p>
<p>However, we know that Polynomial rings are Noetherian, so the ascending chain of ideals stabilizes.</p>
<p>Hence, every descending chain of closed sets does as well. By taking complements, we see that every ascending chain of open subsets stabilizes.</p>
<p>Let $\{B_i\}$ be some open covering of $X$. Consider the open sets that can be written as the finite union of elements in our open cover:$$\mathcal{A}:=\{ U \mid U=\bigcup_{i=1}^{k}B_i\}.$$</p>
<p>By the <a href="https://en.wikipedia.org/wiki/Axiom_of_dependent_choice" rel="nofollow noreferrer">axiom of dependent choice</a>, we know that if $\mathcal{A}$ does not contain $X$, then there exists some infinite ascending chain of open sets, which is a contradiction.</p>
|
1,098,028 | <p>Let $T: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}_{\leq 2}[X]$ be a linear map defined as $T(f(x)) = f'(x)$, and let $\beta$ and $\gamma$ be the standard ordered bases for resp. $\mathbb{R}_{\leq 3}[X]$ and $\mathbb{R}_{\leq 2}[X]$. Then the matrixrepresentation of the linear map is given as \begin{align*} A = [T]_{\beta}^{\gamma} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}. \end{align*} Let $\phi_{\beta}: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}^4$ and $\phi_{\gamma}: \mathbb{R}_{\leq 2}[X] \rightarrow \mathbb{R}^3$ be linear maps that represent de coordinates of the respective polynomials as columnvectors. We search a relationship between $L_A, \phi_{\beta}, \phi_{\gamma}$ and $T$, where $L_A$ is the left-multiplication transformation. Choose the polynomial $g(x)$, defined as $g(x) = 2 + x - 3x^2 + 5x^3 $. Then $\phi_{\beta} = \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix}$. Hence we have \begin{align*} L_A \circ \phi_{\beta} (g(x)) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} Because $T(g(x)) = g'(x) = 1 -6x + 15x^2$, we have also \begin{align*} \phi_{\gamma}\circ (T(g(x))) = \phi_{\gamma} (g'(x)) = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} So we can conclude that \begin{align*} L_A \circ \phi_{\beta} = \phi_{\gamma} \circ T. \end{align*} How should I understand this now? And how can I prove it in a rigorous manner? Some explanation would be helpful.</p>
| hmakholm left over Monica | 14,366 | <p>Do it one step at a time.</p>
<p>$f(-2)=0$ gives you one equation.</p>
<p>If $f$ has $y=6-6x$ as a tangent <em>at one of its zero crossings</em>, then the zero crossings in question must be at the point where $y=6-6x$ crosses the $x$ axis. Setting $f(x)=0$ there gives you another equation. Finally, the slope of $f$ at that point must be $-6$, so setting $f'(x)=-6$ at that point gives you a third equation. Solve. Now you know $f$.</p>
<blockquote>
<p>Alternatively, you know one zero of $f$; and another zero where the $6-6x=0$. Since $f$ is even, you can reflect those two zeroes in the $y$-axis to get two more, and then you have all the four possible zeroes of a fourth-degree polynomial. So $f(x)=c(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ for some $c$ -- but it turns out that setting $c=1$ already gives you the right slope at the point of tangency. How nice!</p>
</blockquote>
<p>Now that you know $f$ you can sketch it and find that it has exactly one local maximum, which must be where it crosses $g$. That gives you one equation for the coefficients of $g$; knowing that $g(0.5)=5.25$ and $g'(0.5)=0$ gives you two more. Solve again!</p>
<blockquote>
<p>Alternatively since you know the <em>apex</em> of $g$, you can be sure that $g(x)=d(x-0.5)^2+5.25$ for some factor $d$; then you just need to find the right $d$ such that $g$ intesects $f$ at the maximum, and multiply out.</p>
</blockquote>
<hr>
<p>(The alternatives at each step may seem more involved, but at least for me they're actually the simpler way. This way around I can keep everything in my head, up to and including finding $g(x)=-5x^2+5x+4$ -- whereas I wouldn't be able to do the three-equations-in-three-unknowns route without taking it to pencil and paper).</p>
|
1,576,007 | <p>I just completed an exam and was curious as to the following question:</p>
<blockquote>
<p>How many surjections are there from $\{1, 2, 3, ..., n\}$ to $\{1, 2\}$ if $n \geq 3$?</p>
</blockquote>
<p>My assumption was the following:</p>
<p>The first element of the domain has the possibility to map to either 1 or 2.</p>
<p>The second element then has the possibility to map to the opposite element of the codomain, in which the first element hasn't mapped to.</p>
<p>Now the functions are surjective, and then any remaining elements of the domain ($n-2$ of them) can then map to either of the 2 elements of the codomain.</p>
<p>My answer was $2^{n-1}$. Is this answer correct? Is there a fallacy in my thought process?</p>
| sbares | 198,998 | <p>You had some of the right ideas, but you are not quite right. Let's start with finding the number of functions from ${1,2,...,n}$ to ${1,2}$ without any restrictions on the function. Then each of the numbers $1,2,...,n$ can be mapped to either $1$ or $2$, and there are hence $2^n$ such functions. How many of these are not surjective? For the function not to be surjective, it would have to map every number to $1$, or to map every number to $2$. Hence there are exactly $2$ non-surjective functions, and the answer to the question is $2^n-2$. </p>
|
3,208,003 | <p>How do you calculate the sum of this series?</p>
<p><span class="math-container">$$ \sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n $$</span></p>
| Kandinskij | 657,309 | <p><strong>A solution withouth calculus</strong></p>
<p>We want to calculate the sum:</p>
<p><span class="math-container">$$\sum_{n=1}^{m} n a^n $$</span></p>
<p>Notice that:</p>
<p><span class="math-container">$$\sum_{n=1}^{m} n a^n=1a^1+2a^2+3a^3+...+m a^m =$$</span> <span class="math-container">$$=(a^1+a^2+a^3+...+a^m)+(a^2+a^3+...+a^m)+...+(a^{m-1}+a^m)+(a^m)=
a^1(1+a^1+...+a^{m-1})+a^2(1+a^1+..+a^{m-2})+...+a^m(1)$$</span></p>
<p>By geometric series formula this becomes:</p>
<p><span class="math-container">$$a^1 \frac{a^{m}-1}{a-1}+a^2\frac{a^{m-1}-1}{a-1}+...+a^m\frac{a^{1}-1}{a-1}=\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}$$</span></p>
<p>Now we're almost done:</p>
<p><span class="math-container">$$\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}=\frac{1}{a-1}\sum_{n=1}^{m} (a^{m+1}-a^{n})=\frac{1}{a-1}\left[\sum_{n=1}^{m} a^{m+1}-\sum_{n=1}^{m}a^{n}\right]=\left[ma^{m+1}-\frac{a^{m+1}-1}{a-1}+1\right]=\left[\frac{ma^{m+2}-ma^{m+1}-a^{m+1}+1+a-1}{(a-1)^2}\right]$$</span></p>
<p>As <span class="math-container">$m$</span> tend to infinity all the exponential terms tend to <span class="math-container">$0$</span> since <span class="math-container">$|a|<1$</span>, so it becomes:</p>
<p><span class="math-container">$$\frac{a}{(a-1)^2}$$</span></p>
<p>If <span class="math-container">$a=\frac 23$</span> this is equal to:</p>
<p><span class="math-container">$$6$$</span></p>
<p>Your sum presents also a factor <span class="math-container">$0.5$</span> so the final answer is .</p>
<p><span class="math-container">$$3$$</span></p>
<p>:)</p>
|
3,415,211 | <p>I was solving my Rubik's cube when I though that if i toss my cube what is the probability of getting red colour twice in a row. So I calculate that it is 1/6 for first time, 1/6 possibility for second time getting a red and 1/6 * 1/6 = 1/36 is the answer.</p>
<p>But then I thought if the colour is unspecified i.e the colour that come when i toss my cube one time is taken to be the input colour to come twice in a row. </p>
<p>If i toss my cube then a random colour, it can be red, white etc. come then for the second time it has to be same, would it change the probability. </p>
<p>In short, will the probability of getting a random colour twice on tossing a cube be same a getting as specified colour?</p>
<p>(English is not my native language so it might not make sense)</p>
| José Carlos Santos | 446,262 | <p>Both are wrong. The first one is wrong because from the fact that <span class="math-container">$\frac1{x\log x}<\frac1x$</span> and from the fact that <span class="math-container">$\int_1^\infty\frac{\mathrm dx}x$</span> diverges you deduce nothing. And the second one is wrong because you have to deal not only with the limit at <span class="math-container">$1$</span> but also with the limit at <span class="math-container">$0$</span>.</p>
|
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