qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,113,934 | <p>How to prove that $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ ?</p>
<p>(We treat it as field homomorphisms. )</p>
<p>I know that $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$ and $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.</p>
| Georges Elencwajg | 3,217 | <p>Let us show that every morphism of fields $f:\mathbb R\to \mathbb R$ is the identity, namely $f(r)=r$ for all real $r$. </p>
<p>0) Trivially $f(q)=q$ for all $q\in \mathbb Q$ </p>
<p>1) Notice that $f$ preserves the order relation in $\mathbb R$ : $x\leq y \implies f(x)\leq f(y)$.<br>
Indeed if $r\geq 0$ we can write $r=\rho^2$ for some $\rho\in \mathbb R$ and then $f(r)=f(\rho^2)=(f(\rho))^2$ .<br>
Since the square of a real number is nonnegative , we see that $f(r)\geq 0$.<br>
But then $x\leq y \implies y-x\geq 0 \implies f(y-x)\geq 0\implies f(y)-f(x)\geq 0\implies f(x)\leq f(y)$<br>
and the order relation in $\mathbb R$ is preserved, as claimed. </p>
<p>2) Given $r\in \mathbb R$ we have for any rational $q,q' \in \mathbb Q$ :<br>
$$q\leq r\implies f(q)=q\leq f(r) \quad\operatorname {and} \quad r\leq q'\implies f(r)\leq f(q')$$
Since $r$ and $f(r)$ have the same position relative to rational numbers they are equal: $f(r)=r$ and we have proved that, as announced, $f$ is the identity. </p>
<p>[By the way the definition of real numbers by Dedekind cuts, a definition not very popular nowadays, is based on the very idea that a real number $r$ is characterized by the rational numbers smaller (resp. larger) than $r$ ] </p>
|
267,963 | <p>Why <code>HoldForm</code> do this?</p>
<p><a href="https://i.stack.imgur.com/DDFDT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DDFDT.png" alt="enter image description here" /></a></p>
<pre><code>HoldForm[a/b \[Integral]f[x] \[DifferentialD]x]
</code></pre>
<p>Output is this ugly looking formula:
<span class="math-container">$$\frac{a \int f(x) \, dx}{b}$$</span></p>
<p>I want the output to look exactly same as I typed it:
<span class="math-container">$$\frac{a}{b} \int f(x) \, dx$$</span></p>
<p>What is the purpose of <code>HoldForm</code> when it does not hold the form of expression as it has been typed?</p>
<p>It always rearranges fractions from <span class="math-container">$\frac{a}{b} c$</span> to <span class="math-container">$\frac{a c}{b}$</span>.</p>
| Alexei Boulbitch | 788 | <p>It is just enough to wrap a/b. Try this:</p>
<p><a href="https://i.stack.imgur.com/cCmGW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cCmGW.png" alt="enter image description here" /></a></p>
<p>Have fun!</p>
|
267,963 | <p>Why <code>HoldForm</code> do this?</p>
<p><a href="https://i.stack.imgur.com/DDFDT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DDFDT.png" alt="enter image description here" /></a></p>
<pre><code>HoldForm[a/b \[Integral]f[x] \[DifferentialD]x]
</code></pre>
<p>Output is this ugly looking formula:
<span class="math-container">$$\frac{a \int f(x) \, dx}{b}$$</span></p>
<p>I want the output to look exactly same as I typed it:
<span class="math-container">$$\frac{a}{b} \int f(x) \, dx$$</span></p>
<p>What is the purpose of <code>HoldForm</code> when it does not hold the form of expression as it has been typed?</p>
<p>It always rearranges fractions from <span class="math-container">$\frac{a}{b} c$</span> to <span class="math-container">$\frac{a c}{b}$</span>.</p>
| Lukas Lang | 36,508 | <h3>Why it happens</h3>
<p>When you enter an expression and evaluate it in Mathematica, <code>MakeExpression</code> is used to convert the box expression (i.e. the description of the exact input you entererd) into an expression. For example, when you enter <code>f[a]</code> into a Notebook, what Mathematica sees is <code>RowBox[{"f","[","a","]"}]</code> (you can use Cell > Show expression to see this). Only after <code>MakeExpression</code> is applied, is it actually converted to <code>f[a]</code>.</p>
<p>Similar conversions happen for all other types of valid input. In particular, this process removes certain kinds of information from the input. For example, <code>a+b</code>, <code>a+(b)</code>, <code>Plus[a,b]</code> are all converted to <code>a+b</code>. At this point, it is too late to tell how <code>a+b</code> was entered. Something similar is happening in your case, only that <code>Times</code> and <code>Plus</code> have some additional simplification rules that are applied during these early phases. To prevent this from happening, we have to interfere with this process at or before the <code>MakeExpression</code> step, see below for an example of how to do that.</p>
<h3>Workaround</h3>
<p>Here's a solution similar to what the <code>Notation` </code> paclet does:</p>
<pre><code>HoldBoxesTemplate[t_] :=
TemplateBox[{t}, "HoldBoxes", DisplayFunction -> (# &)]
MakeExpression[TemplateBox[{b_}, "HoldBoxes", ___], frm_] :=
HoldComplete@HoldBoxes[b, frm]
MakeBoxes[HoldBoxes[b_, _], _] ^:=
HoldBoxesTemplate[b]
Normal@HoldBoxes[b_, frm_] ^:=
ReleaseHold@MakeExpression[b, frm]
AppendTo[CurrentValue[EvaluationNotebook[], InputAliases],
"hb" -> HoldBoxesTemplate["\[Placeholder]"]];
</code></pre>
<p>You can now press <kbd>Esc</kbd>hb<kbd>Esc</kbd> to insert a placeholder that will keep its contents in exactly the same shape:</p>
<p><a href="https://i.stack.imgur.com/fEse3.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/fEse3.gif" alt="enter image description here" /></a></p>
<p>The trick is to have an invisible <code>TemplateBox</code> around your expression that ensures the box structure is preserved through <code>MakeExpression</code>, rather than being converted into a normal expression (at which point there expression has already been "canonicalized", so that it's no longer possible to reconstruct the original shape). The expression returned is of the form <code>HoldBoxes[...]</code>:</p>
<p><a href="https://i.stack.imgur.com/injM6.png" rel="noreferrer"><img src="https://i.stack.imgur.com/injM6.png" alt="enter image description here" /></a></p>
|
2,785,449 | <p>So, it is obvious that the empty set is a subset of every set, such that <strong>∀A:∅⊆A</strong>.</p>
<p>Therefore, if the null set is an element of some set A, such that <strong>A = {∅}</strong>, is the empty set considered a singleton element of A?</p>
<p>Thank you.</p>
| Anuska | 542,827 | <p>Yes $$\emptyset$$ is the one and only element of the set $A$. </p>
|
2,785,449 | <p>So, it is obvious that the empty set is a subset of every set, such that <strong>∀A:∅⊆A</strong>.</p>
<p>Therefore, if the null set is an element of some set A, such that <strong>A = {∅}</strong>, is the empty set considered a singleton element of A?</p>
<p>Thank you.</p>
| fleablood | 280,126 | <p>Divorce your mind from any thought that of an element being a set in its own right, from the role of the element as an element in another set. If $x$ is the only element in $A$ then $x$ is a singleton element of $A$. It doesn't matter <em>what</em> $x$ is and if $x$ just happens to be a set, its elements have nothing whatsoever to do with what kind of element $x$ is to $A$.</p>
<p>So $A=\{\emptyset\}$. The $A$ has a single element. That singleton element is $\emptyset$. </p>
<p>That's all that needs to be said.</p>
<p>.....</p>
<p>Um... I'm assuming you are taking "singleton" to mean the only element of a set?</p>
|
4,516,006 | <p>Let <span class="math-container">$a,~b\in\Bbb Q$</span> and suppose <span class="math-container">$\sqrt{a},~\sqrt{b}$</span> is irrational and <span class="math-container">$\sqrt{a}-\sqrt{b}\in\Bbb Q$</span>. I want to prove that <span class="math-container">$\sqrt{a}-\sqrt{b}=0$</span>; that is, <span class="math-container">$\sqrt{a}=\sqrt{b}$</span>. It seems straightforward by observing some practical numbers. However I found it hard to write a formal proof.</p>
<p>I have tried squaring them, but gained nothing. And I have also searched on this site but nothing was found.</p>
| David Lui | 445,002 | <p><span class="math-container">$\sqrt{a} - \sqrt{b} = r$</span></p>
<p><span class="math-container">$\sqrt{a} = r + \sqrt{b}$</span></p>
<p><span class="math-container">$a = (r + \sqrt{b})^2 = r^2 + 2 r \sqrt{b} +b$</span></p>
<p>If <span class="math-container">$r \neq 0$</span>, then :</p>
<p><span class="math-container">$\frac{a - b - r^2}{2r} = \sqrt{b}$</span>.</p>
<p>The left side is rational.</p>
|
3,190,781 | <blockquote>
<p>Suppose the polynomial <span class="math-container">$P(x)$</span> with integer coefficients satisfies the following conditions:<br>
(A) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 − 4x + 3$</span>, the remainder is <span class="math-container">$65x − 68$</span>.<br>
(B) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 + 6x − 7$</span>, the remainder is <span class="math-container">$−5x + a$</span>.<br>
Then we know that <span class="math-container">$a =$</span>?</p>
</blockquote>
<p>I am struggling with this first question from the <a href="https://www.maa.org/sites/default/files/pdf/programs/JUEEDocument.pdf" rel="noreferrer">1990 Japanese University Entrance Examination</a>. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the <a href="https://en.wikipedia.org/wiki/Polynomial_remainder_theorem" rel="noreferrer">polynomial remainder theorem</a> but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?</p>
<p>So for some <span class="math-container">$g(x)$</span> and <span class="math-container">$h(x)$</span> we have:
<span class="math-container">$$P(x) = g(x)(x^2-4x+3) + (65x-68),\\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$</span>
which looks to have more unknowns than equations. How should I proceed from here?</p>
| ajotatxe | 132,456 | <p>Hint:</p>
<p>If you know that the remainder of the division of some polynomial <span class="math-container">$Q$</span> by, say, <span class="math-container">$x^2-5x+4$</span> is <span class="math-container">$7x-8$</span> you can find some values of <span class="math-container">$Q$</span> by substituting <span class="math-container">$x$</span> by the zeros of the divisor.</p>
<p>Indeed, the zeros of <span class="math-container">$x^2-5x+4$</span> are <span class="math-container">$x=1$</span> and <span class="math-container">$x=4$</span>. So you can find what is <span class="math-container">$Q(4)$</span>.</p>
<p><span class="math-container">$$Q(x)=h(x)(x^2-5x+4)+7x-8$$</span>
<span class="math-container">$$Q(4)=h(4)(4^2-5\cdot 4+4)+7\cdot 4-8=h(4)\cdot 0+28-8=20$$</span></p>
|
3,190,781 | <blockquote>
<p>Suppose the polynomial <span class="math-container">$P(x)$</span> with integer coefficients satisfies the following conditions:<br>
(A) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 − 4x + 3$</span>, the remainder is <span class="math-container">$65x − 68$</span>.<br>
(B) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 + 6x − 7$</span>, the remainder is <span class="math-container">$−5x + a$</span>.<br>
Then we know that <span class="math-container">$a =$</span>?</p>
</blockquote>
<p>I am struggling with this first question from the <a href="https://www.maa.org/sites/default/files/pdf/programs/JUEEDocument.pdf" rel="noreferrer">1990 Japanese University Entrance Examination</a>. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the <a href="https://en.wikipedia.org/wiki/Polynomial_remainder_theorem" rel="noreferrer">polynomial remainder theorem</a> but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?</p>
<p>So for some <span class="math-container">$g(x)$</span> and <span class="math-container">$h(x)$</span> we have:
<span class="math-container">$$P(x) = g(x)(x^2-4x+3) + (65x-68),\\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$</span>
which looks to have more unknowns than equations. How should I proceed from here?</p>
| Community | -1 | <p>You could use Polynomial Remainder Theorem here, it's just impractical. It states P(x) mod (x-b) is congruent to P(b). It never says: let b, be a number. <span class="math-container">$x^2-4x+3=x-(-x^2+5x-3)$</span>.</p>
<p>Easier to note the second divisor is <span class="math-container">$10x-10$</span> more than the first so <span class="math-container">$(10x-10)y-5x+a= 65x-68$</span> so y = 7 produces <span class="math-container">$70x-70-5x+a=65x-(70-a)$</span> so <span class="math-container">$a=2$</span> works. </p>
|
2,876,040 | <p>So i have the following problem: I need to cover the whole interval from $1$ to $10^{12}$ on a computer program. But i can´t do it on a for loop because it would be too slow. So can i iterate over a smaller amount ($\sqrt{n}$ for instance), and still cover the whole interval? </p>
<p>What i've done so far:</p>
<p>Iterate from $1$ to $\sqrt{n}$, then cover $(n-i)$, $\frac{n}{i}$,$\frac{n}{n-i}$.</p>
<p>But i'm still missing some numbers.</p>
<p><em>Thanks in advance.</em> </p>
| Ethan Bolker | 72,858 | <p>If you have to examine every one of those numbers you will have to execute some code $10^{12}$ times. It doesn't matter how cleverly you organize the loop.</p>
<p>Some languages (for example, Python) can optimize looping for you by clever indexing ("vectorizing") that saves a little of the looping overhead - but not the essential calculations.</p>
<p><strong>Edit</strong> in response to comment. If you want to look only at primes in that range, consider reading this file:</p>
<p><a href="http://compoasso.free.fr/primelistweb/page/prime/liste_online_en.php" rel="nofollow noreferrer">http://compoasso.free.fr/primelistweb/page/prime/liste_online_en.php</a></p>
|
2,487,709 | <p>I have to proof that if for any $z$:
$$\cos(z+\omega) = \cos(z)$$</p>
<p>then $\omega = 2\pi k \quad (k = 0, \pm 1, \pm 2 \ldots)$</p>
<p>I have no idea how to prrof it strictly</p>
| Mark Viola | 218,419 | <p>If $f(t)=a+(b-a)H(t)$ and $x<0$, then $\int_{-\infty}^x \delta'(t)f(t)\,dt =\int_{-\infty}^x \delta'(t)a\,dt=0$ since $\delta'$ is of compact support on $\{0\}$.</p>
<p>Next note that there is no meaning to the square of the Dirac Delta as a distribution (<a href="https://math.stackexchange.com/questions/2221429/why-is-the-square-of-dirac-delta-function-not-a-distribution">See Here</a>). Consequenlty, if $f(t)=a+(b-a)H(t)$, then the expressions of this pseudo equalities, $\int_{-\infty}^x \delta'(t)f(t)\,dt=-\int_{-\infty}^x \delta(t)f'(t)\,dt=-\int_{-\infty}^x\delta(t)\,\left((b-a)\delta(t)\right)\,dt$, are meaningless for all $x\ge 0$.</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>
| Raghav | 164,237 | <p>When You write 12 you actually mean 10+2. Or in general when you represent a number in decimal by writing $a_na_{n-1}a_{n-2} . . . a_2a_1a_0$ we actually mean $ \Sigma_{i=0}^n a_i10^i$.
Now, Perhaps It is easy for you to see that you can add term-wise.
For example
You can verify $a_i 10^i + b_i 10^i = d_i 10^{i+1} + c_i 10^i$ where $c_i = (a_i + b_i)mod 10 ; d_i = Quotient((a_i+b_i)/10)$.
Since, above is true for each $i$, your term-wise addition holds and $d_i$ is carried over. </p>
|
3,264,655 | <p>I've found myself where I've got a bidirectional graph of roughly 5500 nodes, 14000 edges.</p>
<p>I'd like to visit ~85 of these nodes, each <em>at least</em> once, starting and ending at the same node. I'd like to get a short route (don't need explicitly shortest, though that would be nice!). What graph processing algorithm should I use in this situation?</p>
<p>I feel like this isn't that unusual of a problem, but I think I'm searching for the wrong terms, as searching around has turned up fruitless. Or maybe I just haven't practiced math in too long!</p>
<p>The cost for traveling between two connected nodes is all the same.</p>
| Sudix | 470,072 | <p>Create a graph with only the nodes of interest, and calculate for each pair the shortest distance (there are polynomial procedures for calculating the transitive closure of a graph).</p>
<p>Now you're looking for a relaxed version of the Traveling Salesman Problem.</p>
<p>Therefore, in this modified version, you can at least find an approximation algorithm using the MST of the modified graph with a factor of 2x (i.e. it is, at worst, twice as bad as the optimal route).</p>
<p>I'm not 100% sure, but I think the relaxed version of TSP was NP-complete as well, so searching for anything better than an approximation algorithm would be out of reach.</p>
<p>Either way, with this modification done, you can use everything we know about the TSP and apply it onto it.</p>
|
1,591,990 | <p>A while ago I saw this question <a href="https://math.stackexchange.com/questions/1561877/quartic-diophantine-equation-16r4112r3200r2-112r16-s2">Quartic diophantine equation: $16r^4+112r^3+200r^2-112r+16=s^2$</a> which was very relevant to a undergraduate research paper I am currently working on. The answer given for this problem describes a method for determining solutions to diophantine equations by finding a "birational equivalence" to the solution set of an elliptic curve. </p>
<p>The result I am seeking depends on the solution set to a quartic diophantine equation of two variables and I believe that there is only one solution and no more (which was indicated by a program which checked possible values up to a high number).</p>
<p>So my question is, is there a general method for determining such a birational equivalence between solutions sets of diophantine equations and elliptic curves? And where is a good source for an undergraduate to learn to use these tools?</p>
| Tito Piezas III | 4,781 | <p>(<em>This is a long comment re MacLeod's answer</em>.)</p>
<p>We can also combine the two cases together. Assume a quartic polynomial to be made a square,</p>
<p><span class="math-container">$$pu^4+qu^3+ru^2+su+t=z_1^2\tag1$$</span></p>
<p>has a <strong><em>known</em></strong> rational point, call it <span class="math-container">$w$</span>. We substitute <span class="math-container">$u=v+w$</span> and collect the new variable <span class="math-container">$v$</span>, </p>
<p><span class="math-container">$$c_4v^4+c_3v^3+c_2v^2+c_1v+\color{blue}{c_0^2}=z_2^2\tag2$$</span></p>
<p>where the <span class="math-container">$c_i$</span> are polynomials in <span class="math-container">$w$</span> and the coefficients of <span class="math-container">$(1)$</span>. The constant term of <span class="math-container">$(2)$</span> turns out to be a square, specifically, <span class="math-container">$c_0^2:=pw^4+qw^3+rw^2+sw+t$</span>.</p>
<p>Let <span class="math-container">$v=1/x,\,$</span> <span class="math-container">$z_2=c_0\, z_3/x^2$</span> and <span class="math-container">$(2)$</span> becomes,</p>
<p><span class="math-container">$$x^4+d_3x^3+d_2x^2+d_1x+d_0=z_3^2\tag3$$</span></p>
<p>which is the Case 1 and then be solved as explained by MacLeod.</p>
|
4,244,565 | <p>In resources like Wikipedia the following equation holds for <span class="math-container">$k>0$</span>.
<span class="math-container">\begin{equation}
\sum_{n=0}^{k} [x^n]=\frac{1-x^{k+1}}{1-x}\hspace{1cm}x\neq1
\end{equation}</span>
If you put the right hand side in a limit it holds for <span class="math-container">$k\geq0$</span>. But why do we rarely show the bounds of the k value and why not put the k in an absolute value on the right hand side (Other than the aversion to absolute values)?</p>
| Robert Shore | 640,080 | <p>You're trying to say that there are <span class="math-container">$m!$</span> bijections <span class="math-container">$f:X \to Y$</span> if <span class="math-container">$\vert X \vert = \vert Y \vert = m$</span>. Prove this by induction on <span class="math-container">$\vert X \vert$</span>.</p>
<p>If <span class="math-container">$\vert X \vert = \vert Y \vert =1$</span>, then there is only <span class="math-container">$1$</span> bijective function <span class="math-container">$f:X \to Y$</span>.</p>
<p>Now assume we know the result for <span class="math-container">$\vert X \vert = \vert Y \vert = n$</span>. Assume <span class="math-container">$X = \{ x_1, x_2, \ldots, x_{n+1} \}, Y= \{ y_1, y_2, \ldots, y_{n+1} \}$</span> and let <span class="math-container">$f:X \to Y$</span> be a bijection. Then there are <span class="math-container">$n+1$</span> choices for <span class="math-container">$f(x_{n+1})$</span>, and because <span class="math-container">$f$</span> is a bijection from <span class="math-container">$X$</span> to <span class="math-container">$Y$</span>, <span class="math-container">$f$</span> restricted to <span class="math-container">$X \setminus \{x_1 \}$</span> is a bijection to <span class="math-container">$Y \setminus \{f(x_1) \}$</span>.</p>
<p>By our inductive hypothesis, there are <span class="math-container">$n!$</span> choices for the restricted map for each choice of <span class="math-container">$f(x_1)$</span>, so since there are <span class="math-container">$n+1$</span> choices for <span class="math-container">$f(x_1)$</span>, there are a total of <span class="math-container">$(n+1)!$</span> bijections from <span class="math-container">$X$</span> to <span class="math-container">$Y$</span>.</p>
|
3,249,926 | <p>I’ve tried applying Vandermonde’s identity, but got stuck. Any help would be appreciated!</p>
| PM. | 416,252 | <p><a href="https://i.stack.imgur.com/fuHBW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fuHBW.png" alt="enter image description here"></a></p>
<p>In this case you can consider the graph of <span class="math-container">$x^2$</span> to understand the relevant interval and the lower bound <span class="math-container">$0$</span>.</p>
|
403,977 | <blockquote>
<p>Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.</p>
</blockquote>
<p>Please provide me a hint or a full solution.</p>
| Tim | 74,128 | <p>Hint: think about the partial sums
$$ S_k = \sum_{n = 2^k+1}^{2^{k+1}} \frac{u_n}n$$</p>
<p>Notice that $$\sum_{n=1}^\infty \frac{u_n}n = u_1 + \frac{u_2}2 + \sum_{k=1}^\infty S_k$$
so it's enough to show that the sequence $\sum_{k=1}^\infty S_k$ diverges.</p>
<p>Notice that the terms in the partial sums $S_k$ are decreasing. Therefore the smallest term in the sum is the last one, $\frac{u_{2^{k+1}}}{2^{k+1}}$.</p>
<p>As there are $2^k$ terms in each sum we must have
$$\sum_{n = 2^k+1}^{2^{k+1}} \frac{u_n}n \geq 2^{k} \frac{u_{2^{k+1}}}{2^{k+1}}=\frac{u_{2^{k+1}}}{2} $$</p>
<p>So $$\sum_{n=1}^\infty \frac{u_n}n \geq u_1 + \frac 12\left(u_2 + u_4 + u_8 + \dots\right)$$which we are given as divergent.</p>
|
2,228,913 | <p>For some geometric problem that I am trying to solve, I am interested
in characterizing the cross product between a Bezier curve and its second order derivative.</p>
<p>So, let's $\mathbf{c}(t)$ be a Bezier curve, and $\ddot{\mathbf{c}}(t)$ it's derivative.</p>
<p>Is $\mathbf{w}(t) = \mathbf{c}(t) \times \ddot{\mathbf{c}}(t)$ a Bezier curve?</p>
<p>If it is the case, is it also the case in general ?</p>
<p>Thanks a lot for your time and answers. Sincerely,</p>
<p>Steve</p>
| Jean Marie | 305,862 | <p>@Henning Makholm One can say a little more.</p>
<p>Let us recall that a Bézier curve with degree <span class="math-container">$n$</span> (particular cases <span class="math-container">$n=2$</span>: quadratic, <span class="math-container">$n=3$</span> cubic) can be <strong>defined</strong> as any expression of the form:</p>
<p><span class="math-container">$$\tag{1}\vec{OM}=\vec{u_0}+\vec{u_1}t+\vec{u_2}t^2+\vec{u_3}t^3+\cdots+\vec{u_{n-1}}t^{n-1}+\vec{u_n}t^n \ \ \text{with} \ \ \vec{u_n}\neq 0$$</span></p>
<p>(see example at the bottom of this text.)</p>
<p>(this well known result is due to the fact that every polynomial with degree <span class="math-container">$\leq n$</span> can be decomposed onto the Bernstein polynomials of order <span class="math-container">$n$</span> that constitute a basis of the vector space of polynomials with degree <span class="math-container">$\leq n$</span>).</p>
<blockquote>
<p><strong>Result:</strong> If <span class="math-container">$\mathbf{c}(t)$</span> describes a Bézier curve with degree <span class="math-container">$n$</span>, the curve</p>
<p><span class="math-container">$$\mathbf{w}(t) = \mathbf{c}(t) \times \ddot{\mathbf{c}}(t)$$</span></p>
<p>still is a Bézier curve, with degree <span class="math-container">$\leq 2n-3.$</span></p>
</blockquote>
<p>Proof: The second derivative of (1) is</p>
<p><span class="math-container">$$\tag{2}\ddot{\vec{OM}}=2\vec{u_2}+6\vec{u_3}t+\cdots+(n-2)(n-1)\vec{u_{n-1}}t^{n-3}+n(n-1)\vec{u_n}t^{n-2}$$</span></p>
<p>Taking the cross product of (1) and (2), one gets, by distributivity:</p>
<p><span class="math-container">$$\vec{u_0} \times 2\vec{u_2}+ \cdots + \vec{v}t^{n-3}+n(n-1)\underbrace{\vec{u_n}\times \vec{u_n}}_{\vec{0}}t^{2n-2}$$</span></p>
<p>with <span class="math-container">$\vec{v}:=2(n-1) \vec{u_{n-1}} \times \vec{u_{n-2}}$</span>.</p>
<p><strong>Two particular cases:</strong></p>
<ul>
<li><p>(A) A quadratic Bézier curve is transformed into a straight line (<span class="math-container">$n=2 \to 2n-3=1$</span>).</p>
</li>
<li><p>(B) A cubic Bézier curve is transformed into a cubic Bézier curve (<span class="math-container">$n=3 \to 2n-3=3$</span>).</p>
</li>
</ul>
<hr />
<p>Example of a quadratic Bézier curve :</p>
<p><span class="math-container">$$\vec{OM}=(1-t)^2\vec{OA}+2t(1-t)\vec{OB}+t^2\vec{OC}=\vec{OA}+2t\vec{AB}+t^2(\vec{BA}+\vec{BC}).$$</span></p>
|
495,863 | <p>I am trying to show that this sequence $\{a_n\}$ = (2n+1)/(3n+5) does not converge to $42$ if it is not bounded above. I have already showed that it converges to $2/3$. For this I want to use a proof by contradiction,i.e, I assume that the sequence does converge to $42$, which will lead to a contradiction to the initial assumption that is the sequence will be bounded above. Any ideas or theorems that can help me solve this question? </p>
| Pete L. Clark | 299 | <p>Your approach seems distinctly strange. For one thing, if the sequence converged to $42$, then it <em>would</em> be bounded above!</p>
<p>On the other hand, you have a specific sequence that you already know is converging to $\frac{2}{3}$, so assuming that it converges to something else is simply contradictory (I assume you know that limits are unique). </p>
<p>Let's back up several steps. Try to show that a convergent sequence is bounded above: that's logically equivalent to your title question and less convoluted. Can you do that? </p>
|
3,926,580 | <p>I'm trying to prove with squeeze theorem that the limit of the following series equals 1:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the left side of the inequality I did:</p>
<p><span class="math-container">$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the right side, at first I did the following:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$</span></p>
<p>But then I realized it wasn't true and that the direction of this inequality is the opposite.</p>
<p>Do you have any idea which series with limit 1 is bigger from the original series?</p>
<p>Thanks!</p>
| Yiorgos S. Smyrlis | 57,021 | <p>One may combine the following two facts:</p>
<p><a href="https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation" rel="nofollow noreferrer">1.</a> If <span class="math-container">$a_n\to a,\,$</span> then <span class="math-container">$\,\frac{1}{n}(a_1+\cdots+a_n)\to a$</span>.</p>
<p><a href="https://math.stackexchange.com/questions/28348/proof-of-lim-n-to-infty-sqrtnn-1">2.</a> <span class="math-container">$\sqrt[n]{n}\to 1$</span>.</p>
<p>Another way to show it is the following:
<span class="math-container">$$
\sqrt[2k]{k}=1+a_k\Longrightarrow \sqrt{k}=(1+a_k)^{k}\ge 1+ka_k
\Longrightarrow 0\le a_k<\frac{1}{\sqrt{k}}
$$</span>
and hence
<span class="math-container">$$
1<\sqrt[n]{n}=(1+a_n)^2=1+2a_n+a_n^2<1+\frac{2}{\sqrt{n}}+\frac{1}{n}\le
1+\frac{3}{\sqrt{n}}
$$</span>
and thus
<span class="math-container">$$
1<\frac{1}{n}(1+\sqrt{2}+\cdots+\sqrt[n]{n})<1+\frac{3}{n}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) \\ <1+\frac{3}{n}\cdot (2\sqrt{n}+1)\to 1.
$$</span>
It remains to show that
<span class="math-container">$$
1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}+1
$$</span>
which can be easily done inductively.</p>
|
15,875 | <p>It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is <em>continuously</em> differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations" rel="noreferrer">Wikipedia</a>). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below. </p>
<p>Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then<br>
\begin{equation*}
u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|)
\end{equation*}
and
\begin{equation*}
v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|).
\end{equation*}
Multiplying the second equation by $i$ and adding the two together gives
\begin{align*}
(u+iv)(z+h)-(u+iv)(z) &= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\
&= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|).
\end{align*}
Now dividing by $h$ gives us the desired result. </p>
<p>Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? </p>
| lhf | 589 | <p>There's also the <a href="http://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem">Looman–Menchoff theorem</a>.</p>
|
15,875 | <p>It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is <em>continuously</em> differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations" rel="noreferrer">Wikipedia</a>). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below. </p>
<p>Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then<br>
\begin{equation*}
u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|)
\end{equation*}
and
\begin{equation*}
v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|).
\end{equation*}
Multiplying the second equation by $i$ and adding the two together gives
\begin{align*}
(u+iv)(z+h)-(u+iv)(z) &= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\
&= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|).
\end{align*}
Now dividing by $h$ gives us the desired result. </p>
<p>Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? </p>
| paul garrett | 12,291 | <p>Thinking of the Cauchy-Riemann operator as an elliptic partial differential operator, the basic elliptic regularity result implies that any <em>distribution</em> satisfying the C-R equation is a holomorphic function. For example, locally integrable suffices. This result was used in Gunning' "Riemann Surfaces", for example, in the discussion of Serre duality.</p>
|
1,826,094 | <p>$i\cdot\bar{z} = 2+2i$</p>
<p>I know that $\bar{z} = a-bi$ so then i get $i(a-bi)=2+2i$
Then $ai+b=2+2i$ (because $i^2=-1$)</p>
<p>When 2 complex numbers are equal you usually can equal their parts
Ex: $2+2i=a+bi$ so $a=2$ and $b=2$</p>
<p>But in this case I aint got $a+bi$ I have $ai+b$</p>
<p>The answer is $z=2-2i$ but i dont understand how to get there</p>
<p>Thanks</p>
| Thomas | 26,188 | <p>It looks like you are trying to solve
$$
i\bar{z} = 2 + 2i.
$$
You can start by multiplying both sides by $i$:
$$
i^2\bar{z} = 2i + 2i^2 = -2 + 2i
$$
This gives you
$$
-\bar{z} = -2 + 2i.
$$
or
$$
\bar{z} = 2 - 2i
$$
Now just take the complex conjugate on both sides:
$$
z = \overline{\bar{z}} = \overline{2 - 2i} = 2 + 2i.
$$</p>
|
47,177 | <p>Recently I've been reading "The Wild Book" which applies semigroup theory to, among other things, chemical reactions. If I google for mathematics and chemistry together, most of the results are to do with physical chemistry: cond-mat, fluids, QM of molecules, and analysis of spectra. I'm more interested in learning about biochemistry, molecular biology, and organic chemistry — and would prefer to learn from a mathematical perspective.</p>
<p>What other books aim to teach (bio- <code>||</code> organic) chemistry specifically to those with a mathematical background?</p>
| Fujita | 58,711 | <p>(1) Because my monograph:</p>
<p>S. Fujita, <a href="http://www.springer.com/chemistry/organic+chemistry/book/978-3-540-54126-4" rel="noreferrer">Symmetry and Combinatorial Enumeration in Chemistry</a> (Springer 1991)</p>
<p>has been referred to as an endeavor of dealing with topics of organic chemistry on the basis of mathematics, I would like to add two monographs aiming at organic reactions and stereochemistry coupled with mathematics (group theory): </p>
<p>S. Fujita,
<a href="http://xymtex.com/fujitas/rd/choshoe.html#FBOOK5d" rel="noreferrer">Computer-Oriented Representation of Organic Reactions</a>
(Yoshioka Shoten, Kyoto, 2001)</p>
<p>S. Fujita,
<a href="http://xymtex.com/fujitas/rd/choshoe.html#FBOOK5g" rel="noreferrer">Diagrammatical Approach to Molecular Symmetry and Enumeration of Stereoisomers</a>
(University of Kragujevac, Kragujevac, 2007)</p>
<p>which have appeared more recently. The first and the third books are concerned with Fujita's unit-subduced-cycle-index (USCI) method for symmetry-itemized enumeration of 3D structures as well as graphs. </p>
<p>(2) For a more recent publication (not a book) on interdisciplinary topics between mathematics and chemistry, I would like to introduce an account article, which is freely available: </p>
<p>S. Fujita, <a href="https://www.jstage.jst.go.jp/article/bcsj/83/1/83_20090008/_pdf" rel="noreferrer">
"Numbers of Alkanes and Monosubstituted Alkanes. A Long-Standing Interdisciplinary Problem over 130 Years"</a>
(Bull. Chem. Soc. Japan, 83, 1--18 (2010), access-free). </p>
<p>This account deals with combinatorical enumeration of three-dimensional trees,
where trees as graphs are extended to 3D trees having 3D structures. </p>
<p>(3) Another book on interdisciplinary topics between mathematics and chemistry:</p>
<p>S. Fujita, <a href="http://match.pmf.kg.ac.rs/mcm15.html" rel="noreferrer">
Combinatorial Enumeration of Graphs, Three-Dimensional Structures, and
Chemical Compounds</a>
(University of Kragujevac, Kragujevac, 2013), </p>
<p>has published to introduce Fujita's proligand method (Chapter 7), which provides us with a powerful tool for gross enumeration of 3D structures. This is a substantial extension of Polya's theorem, which aims at gross enumeration of graphs. </p>
<p>The landmark article on Polya's theorem appeared originally in 1937 and was translated into English in 1987 (after 50 years!): </p>
<p>G. Polya and R. C. Read, <a href="http://link.springer.com/book/10.1007/978-1-4612-4664-0/page/1" rel="noreferrer">
Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds</a>
(Springer, 1987). </p>
<p>As found in their book titles, chemical compounds are regarded as three-dimensional structures
in Fujita's book, while they are regarded as graphs in Polya-Read's book.
This difference is critical to discuss stereochemistry: </p>
<p>S. Fujita, <a href="http://hrcak.srce.hr/file/8789" rel="noreferrer">
Sphericities of Cycles. What Polya's Theorem is Deficient in for Stereoisomer Enumeration</a><br>
(Croat. Chem. Acta, 79, 411--427 (2006), access-free).</p>
<p>(4) A further book on interdisciplinary topics between mathematics and chemistry has appeared recently:</p>
<p>S. Fujita,
<a href="http://d-nb.info/1059131072" rel="noreferrer">Mathematical Stereochemistry</a> (De Gruyter, Berlin, 2015). xviii + 437pp</p>
<p>This book deals with Fujita's stereoisogram approach, where <em>RS</em>-stereoisomers as new concepts are represented by stereoisograms diagrammatically. </p>
|
1,482,564 | <p>In school I'm learning about concavity and finding points of inflection using the second-derivative test. A typical question will look like</p>
<blockquote>
<p><strong>Determine the open intervals on which the graph is concave upward or concave downward.</strong></p>
</blockquote>
<p>Here is how I would solve a problem like that.</p>
<ol>
<li><p>Find the derivative, and then second derivative of f.</p>
</li>
<li><p>Find the critical numbers of f'(x) by setting f''(x) = 0 and f''(x) is undefined, then simplify. Let's say I get the critical numbers <strong>a</strong> and <strong>b</strong> (a < b).</p>
</li>
<li><p>Take f'' of a number in the open interval <span class="math-container">$(-\infty, a)$</span> If it is negative, f is concave downward on <span class="math-container">$(-\infty, a)$</span> if it is positive, it is concave upward. Repeat this process for a number in <span class="math-container">$(a, b)$</span> and in <span class="math-container">$(b, \infty)$</span></p>
</li>
</ol>
<p>In general, I will get a result like</p>
<blockquote>
<p>f is concave upward on <span class="math-container">$(-\infty, a) \cup (b, \infty)$</span></p>
<p>and concave downward on <span class="math-container">$(a, b)$</span></p>
</blockquote>
<p>But now I'm wondering, is it possible for the concavity to stay the same even on an interval containing a critical number of f prime? For example, if f'(x) is increasing when x < a, f'(x) = 0 @ x = a, and f'(x) is still increasing when a < x < b? Does that mean that f is concave upward on <span class="math-container">$(-\infty, b)$</span> Can this even happen?</p>
| math_nix | 234,249 | <p>No.</p>
<p>Consider $f(x) = x^4$. </p>
<p>In order for there to be a change in concavity there needs to be a change of sign for the second derivative around our potential inflection point. However, $f''(x)=12x^2$ is positive to both sides of $x=0$ so it's not an inflection point.</p>
<p>You can visually tell this isn't an inflection point because the graph looks like a steeper parabola.</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>
| Mikhail Katz | 72,694 | <p>Since your curve is in the plane, you can find the normal vector without differentiating. If the components of the unit tangent vector are $(a,b)$ then the components of a unit normal are $(-b,a)$. "Outward" doesn't really make sense until you have a closed curve, and involves global considerations.</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>This problem [restated as : Given a point, $(x_0, y_0)$ on the given curve, compute the unit normal to the curve at $(x_0, y_0)$.] cannot be solved. Why? Because there might be two distinct normal lines to the curve at $(x_0, y_0)$. The following picture shows this:
<img src="https://i.stack.imgur.com/IP0lP.png" alt="enter image description here"></p>
<p>At points where the plot intersects itself, there are two distinct normals, so any unambiguous formula in terms of $(x_0, y_0)$ will give at most one...which means it's wrong, (esp. if the other one is the one you wanted). </p>
<p>Pretty plot, though. </p>
|
95,168 | <p>I try to prove the following</p>
<p>$$\binom{2\phi(r)}{\phi(r)+1} \geq 2^{\phi(r)}$$</p>
<p>with $r \geq 3$ and $r \in \mathbb{P}$. Do I have to make in induction over $r$ or any better ideas?</p>
<p>Any help is appreciated.</p>
| sdcvvc | 12,523 | <p>Combinatorial proof of ${2n \choose n+1} \geq 2^n$ where $n \geq 2$:</p>
<p>Let's take set $\{x_1,y_1,\dots,x_{n-2},y_{n-2},a,b,c,d\}$ which has $2n$ elements; select three elements out of $\{a,b,c,d\}$ and for all $i$, a single element of $\{x_i,y_i\}$, you'll select $n+1$ in total. So</p>
<p>${2n \choose n+1} \geq {4 \choose 3} 2^{n-2}=2^n$</p>
|
680,501 | <blockquote>
<p>Let $M$ be an $R$-module and let $F$ be a free $R$-module of finite rank. Let $\phi : M \to F$ be an epimorphism. Then show that $M$ has a submodule $F' \cong F $ such that $M=F' \oplus \ker\phi$.</p>
</blockquote>
<p>I am new to Module theory. If I apply fundamental theorem of homomorphism then $M/\ker\phi \cong F$, what to do next?</p>
| user149792 | 149,792 | <p>In the interests of completeness, we provide a complete solution.</p>
<p><strong>Lemma.</strong> Let $X_1, X_2, \dots$ be a sequence of Bernoulli random variables $($not necessarily independent, not necessarily with same $p$-value$)$. Define$$Y = \sum_{i=1}^\infty {{2X_i}\over{3^i}}.$$Then $\sigma(Y) = \sigma(X_1, X_2, \dots)$.</p>
<p><em>Proof.</em> Fix $n$, and consider the event $\{X_n \in I\}$, where $I \subset \mathbb{R}$ is an interval. If $I$ contains $0$ and $1$, then $\{X_n \in I\} = \Omega \in \sigma(Y)$. If $I$ contains neither $0$ nor $1$, then $\{X_n \in I\} = \emptyset \in \sigma(Y)$. If $I$ contains $1$, but not $0$, then $\{X_n \in I\} = \{X_n = 1\}$. This is the event that the $n$th digit in the ternary expansion of $Y$ is $2$, which occurs exactly when$$Y \in \bigcup_{k=1}^{3^{n-1}} \left[{k\over{3^{n-1}}} - {1\over{3^n}}, {k\over{3^{n-1}}}\right],$$which is a Borel set. Thus in this case, $\{X_n \in I\} \in \sigma(Y)$. If $I$ contains $0$, but not $1$, then $\{X_n \in I\} = \{X_n = 0\}$ and a similar argument to the preceding case shows $\{X_n \in I\} \in \sigma(Y)$. Since intervals generate the Borel $\sigma$-algebra on $\mathbb{R}$, it follows that $\sigma(X_1, X_2, \dots) \subset \sigma(Y)$.</p>
<p>Now, choose $x \in [0, 1]$ and write down a ternary expansion $x = 0.a_1a_2a_3\dots$. Then we have$$\{Y \ge x\} = \bigcup_{k=1}^\infty \left(\{2X_1 \le a_1\} \cap \{2x_2 \le a_2\} \cap \dots \cap \{2X_k > a_k\}\right) \in \sigma(X_1, X_2, \dots).$$Since the intervals $[x, \infty)$ generate the Borel $\sigma$-algebra on $\mathbb{R}$, it follows that $\sigma(Y) \subset \sigma(X_1, X_2, \dots)$.$\tag*{$\square$}$</p>
<p>Let $X: \Omega \to \mathbb{R}$ be a random variable. Recall that the Borel $\sigma$-algebra on $\mathbb{R}$ is countably generated; let $\{A_i\}$ be a countable generating collection for the Borel sets. It is an exercise left to the reader to show that every Borel set may be obtained by a sequence of countable unions and complementations of the $A_i$ $($alternatively, one can use the Dynkin $\pi$-$\lambda$ Theorem$)$. Since these operations are preserved under taking preimages, every event of the form $\{X \in B\}$, where $B$ is a Borel set, is included in $\sigma\left(X^{-1}(A_1), X^{-1}(A_2), \dots\right)$. Hence $\sigma(X)$ is countably generated, by the collection $\left\{X^{-1}(A_i)\right\}$.</p>
<p>Conversely, suppose $\mathcal{G}$ is countably generated, say by the collection $\{A_i\}$. For each $i \in \mathbb{N}$, define the random variable $X_i = \mathbb{1}_{A_i}$. Evidently, $\sigma(X_1, X_2, \dots) = \mathcal{G}$, since the event $\{X_i \in A_i\}$ is just $A_i$ for each $i \in \mathbb{N}$. Note that each $X_i$ is Bernoulli distributed, taking value $1$ with probability $P(A_i)$ and value $0$ with probability $1 - P(A_i)$. Thus, setting $Y = \sum(2X_i)/3^i$, the lemma implies that $\mathcal{G} = \sigma(Y)$.</p>
|
4,426,699 | <p>I want to solve this functional</p>
<p><span class="math-container">\begin{equation}
L(y)=\int_{-1}^7\sqrt{1+y'^2} \ dx
\end{equation}</span></p>
<p>with IC: <span class="math-container">$y(0)=1,\ y(1)=2$</span></p>
<p>I start using the Euler Lagrange equation</p>
<p><span class="math-container">\begin{equation}
\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0
\end{equation}</span></p>
<p>But I encounter immediately a problem, namely that</p>
<p><span class="math-container">$\frac{d}{dx}\frac{\partial F}{\partial y'}=0$</span>, since <span class="math-container">$\frac{\partial F}{\partial y'}=\frac{y'}{\sqrt{1+y'^2}}$</span> which has no x-variable. Also <span class="math-container">$\frac{\partial F}{\partial y}=0$</span>, hence it does not make sense.</p>
<p>Alterantively, I set this equal to a constant instead, and get:</p>
<p><span class="math-container">\begin{equation}
\begin{array}
f\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=C \\
\frac{y'}{\sqrt{1+y'^2}}=C_1x + C_3\\
\frac{y'}{\sqrt{1+y'^2}}=C_2x+C_4
\end{array}
\end{equation}</span></p>
<p>How do I solve this?</p>
<p>Thanks</p>
| Luthier415Hz | 924,287 | <p>I found a second method:</p>
<p>Using the formula</p>
<p><span class="math-container">\begin{equation}
F-y'F_{y'}=C
\end{equation}</span></p>
<p>I get:</p>
<p><span class="math-container">\begin{equation}
\begin{array}
a\sqrt{1+y'^2}-y'\frac{y'}{\sqrt{1+y'^2}}=C \\
-y'^2+1+y'^2=C\sqrt{1+y'^2}\\
1=C(1+y'^2)\\
Cy'^2+C=1\\
C(y'^2+1)=1\\
y'=\sqrt{C-1} \ set\ \sqrt{C-1}=A\\
\int y'=\int A \\
y(x)=Ax+B
\end{array}
\end{equation}</span></p>
<p>With initial conditions, this gives <span class="math-container">$y(x)=x+1$</span></p>
|
3,984,652 | <p>Let <span class="math-container">$(X_n,d_n) $</span> a family of metric spaces and <span class="math-container">$(X=\prod_{n \in \mathbb N} X_n,\mathcal d ) $</span> a metric space with <span class="math-container">$\mathcal T_p$</span> the product topology is equal to the topology defined by the distance <span class="math-container">$d$</span>, such that for all <span class="math-container">$x=(x_n)_n, y=(y_n)_n\in X$</span> we have <span class="math-container">$d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}d_n(x_n,y_n) $</span> and <span class="math-container">$d_n(x,y) \leq M$</span> for all <span class="math-container">$x, y\in X_n$</span>.</p>
<p>Show that <span class="math-container">$X$</span> is compact iff <span class="math-container">$X_n$</span> compact for all <span class="math-container">$n\in \mathbb N$</span>.</p>
<hr />
<p>For the first implication, I have used the the fact that <span class="math-container">$X$</span> is compact and the projection maps are continuous. For the other one, I get stuck with it. any help is appreciated.</p>
| Prime Mover | 466,895 | <p>Obviously an <span class="math-container">$n$</span> digit number is represented as <span class="math-container">$a_{n-1} a_{n-2} \ldots a_0$</span> not <span class="math-container">$a_n a_{n-1} \ldots a_1$</span>.</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>
| B. Goddard | 362,009 | <p>The sphere is considered to be a 2-dimensional surface embedded in 3-dimensional space. But the embedding is sort of extraneous. It's not as handy, but the same map embeds in 2-dimensional space.</p>
|
812,143 | <p>City a is separated by a 2km wide river and are located as shown in Figure 1 (not drawn to scale). A road is to be built between city A to B that crosses a bridge straight across the river. </p>
<p>Use the Sign Test to determine the type of stationary point </p>
<p>Test to see if you have a minimum</p>
| David | 119,775 | <p>Move $B$ $2$km closer to the river, in a direction perpendicular to the river, and ignore the river. Then you won't even need to use calculus.</p>
<p>Specifically, the shortest distance between $A$ and$B$ is now the straight line distance
$$\sqrt{17^2+12^2}=\sqrt{433}$$
and the value of $x$ will be given by similar triangles:
$$x=17\times\frac{9}{12}=\frac{51}{4}\ .$$
Moving $B$ back where it was adds the $2$ km length of the bridge for a total of $2+\sqrt{433}$ km.</p>
|
2,866,768 | <p>Question: Use the trigonometric identity $\cos(2A)=1-2\sin^2(A)$ to show that $$\sin\left(\frac{\pi}{12}\right)=\sqrt{\frac{1}{2} - \frac{\sqrt{3}}{4}}$$</p>
<p>What are good strategies to figure out this question in Particular?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$\cos(2A)=1-2\sin^2(A)$$</p>
<p>$$\cos(\pi /6)=1-2\sin^2(\pi /{12})$$</p>
<p>$$1-2\sin^2(\pi /{12})=\frac {\sqrt 3}{2}$$</p>
<p>$$2\sin^2(\pi /{12})=1-\frac {\sqrt 3}{2}$$
$$
\sin^2(\pi /{12})=\frac {1}{2}-\frac {\sqrt 3}{4}$$
$$\sin(\pi /{12})=\sqrt {\frac {1}{2}-\frac {\sqrt 3}{4}}$$</p>
|
3,669,266 | <p>The equation <span class="math-container">$(1-x)^n = x$</span> has a solution in <span class="math-container">$x' \in (0,1)$</span> and indeed the solution <span class="math-container">$x' \to 0$</span> as <span class="math-container">$n \to \infty$</span>. (Consider <span class="math-container">$f(x) = (1-x)^n$</span> noting that <span class="math-container">$f(0) = 1$</span> and <span class="math-container">$f(1) = 0$</span>. As <span class="math-container">$n$</span> increases, <span class="math-container">$f(x)$</span> is 'flat' in an increasingly-large neighbourhood around <span class="math-container">$x=1$</span>. Hence <span class="math-container">$f(x)$</span> crosses the line <span class="math-container">$y=x$</span> at a point that gets closer to <span class="math-container">$x=0$</span> as <span class="math-container">$n \to \infty$</span>.) </p>
<p><strong>What is the rate of convergence of <span class="math-container">$x' \to 0$</span> as <span class="math-container">$n \to \infty$</span>?</strong></p>
<p>What I have tried:</p>
<ul>
<li><p>I have obtained solutions for <span class="math-container">$x_n$</span> for <span class="math-container">$n = 1 \dots 17$</span> as shown below. (The ratio <span class="math-container">$x_{x+1}/x_n$</span> is increasing so it's not exponential decay (?))</p></li>
<li><p>I did a quick literature look for the bounds on the real-valued roots of a polynomial <a href="https://link.springer.com/chapter/10.1007/978-3-319-18275-9_10" rel="nofollow noreferrer">On Geometry of the Zeros of a Polynomial</a>. I was seeking an upper bound that decreases with <span class="math-container">$n$</span>. Unfortunately, the bounds in the literature (that I found) were all <span class="math-container">$1 + something$</span> which is unhelpful.</p></li>
</ul>
<p>Any suggestions most appreciated. Many thanks in advance.</p>
<p><a href="https://i.stack.imgur.com/nyuXD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nyuXD.png" alt="Solutions to <span class="math-container">$(1-x)^n = x$</span> obtained numerically"></a></p>
| Aphelli | 556,825 | <p>Then <span class="math-container">$\ln{x_n}=n\ln(1-x_n)\sim -nx_n$</span>. </p>
<p>Write <span class="math-container">$x_n=e^{-y_n}$</span> with <span class="math-container">$y_n \rightarrow \infty$</span>. Then, taking logarithms again, <span class="math-container">$y_n+\ln{y_n}=\ln{n}+o(1)$</span>. Thus <span class="math-container">$y_n=(\ln{n})(1+z_n)$</span> with <span class="math-container">$z_n \rightarrow 0$</span>. </p>
<p>It follows then that <span class="math-container">$\ln{\ln{n}}+z_n\ln{n}=o(1)$</span>, hence <span class="math-container">$z_n=-\frac{\ln{\ln{n}}}{\ln{n}}+o((\ln{n})^{-1})$</span>, ie <span class="math-container">$y_n=\ln{n}-\ln{\ln{n}}+o(1)$</span>. Therefore <span class="math-container">$x_n=e^{-\ln{n}+\ln{\ln{n}}+o(1)} \sim \frac{\ln{n}}{n}$</span>. </p>
|
4,143,759 | <p><strong>What I understand so far :</strong></p>
<p>Intuitively this question gives me a yuuuuuge hint, it tells me at t=0, t = -4. So that let's me know that I have a phase shift of -4 immediately.</p>
<p>I understand what a period is, I understand what a phase shift is, I understand what amplitude and vertical displacement are and how they work.</p>
<p>I'm just not sure how to build a function even with that info at my disposal.</p>
<p><em><strong>My ultimate question here isn't to find an answer, but to be given a hint or a pointing in the right direction. Is there anything that you, my acquaintances at stackexchange have to offer?</strong></em></p>
| David G. Stork | 210,401 | <p><span class="math-container">$$V = \frac{4 \pi}{3} r^3$$</span></p>
<p>so</p>
<p><span class="math-container">$$dV = \frac{4 \pi}{3} 3 r^2 dr$$</span></p>
<p>You know <span class="math-container">$dV/dt$</span>. Solve for <span class="math-container">$dr/dt$</span>.</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>
| David Steinberg | 1,231 | <p>One can use the machinery of the fundamental group and of covering spaces to easily prove that any subgroup of a free group must be free.</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>
| Douglas Zare | 2,954 | <p><strong>Problem:</strong> Suppose you care about the real world and objects you can hold in your hands. Show that any <a href="http://mathworld.wolfram.com/FlexiblePolyhedron.html" rel="noreferrer">flexible polyhedron</a> maintains a constant volume while it is flexed. This was known as the Bellows Conjecture.</p>
<p><strong><a href="http://www.math.ucdavis.edu/~deloera/MISC/MATHRESEARCH/TEMPO/Connelly3.pdf" rel="noreferrer">Solution</a>:</strong> With a little commutative algebra, you can prove that 12*volume is an algebraic integer in $\mathbb Q$ adjoin the lengths of the sides. Any continuous function from $\mathbb R$ to a countable set is constant. In fact, the volume is a root of a single polynomial. </p>
|
2,159,092 | <p>$\sum _{k=1}^{n}\mathrm{sin}kz=\frac{\mathrm{sin}\frac{n+1}{{2}}\cdot \mathrm{sin}\frac{nz}{2}}{\mathrm{sin}\frac{z}{2}}$
Proof for $z\neq 0$</p>
| Derek Elkins left SE | 305,738 | <p>Before focusing on the specific question, I'd like to provide some context. First, every (non-trivial) topos has a Boolean subtopos. This is essentially what you say, and it roughly corresponds to the double negation interpretation of classical logic into intuitionistic logic. However, one of the things that makes toposes interesting is that many things are toposes, but relatively few things are <em>Boolean</em> toposes. Similarly, (part of) what makes intuitionistic type theory interesting is that it's the internal language of a topos. Restricting to a classical theory discards many relevant examples. For example, as the name "topos" suggests, a significant application is to topology with a topological space giving rise to a topos. Restricting to Boolean toposes is like only considering <a href="https://en.wikipedia.org/wiki/Stone_space" rel="noreferrer">Stone spaces</a> which largely defeats the purpose of topology. (One could say this paragraph is an [partial] argument for why the law of excluded middle is <em>not</em> a "good principle", though it is certainly convenient when it holds.)</p>
<p>The way predicates are interpreted in categorical logic for an arbitrary category is as subobjects. A predicate $P$ on "individuals" of sort $A$ is viewed as a subobject of the object $A$ in a category, i.e. as a(n equivalence class of) monomorphism(s) $P \hookrightarrow A$. A subobject classifier allows us to reify these subobjects as <em>terms</em>. That is, the <em>formula</em> $a:A\vdash P(a)$ becomes the <em>term</em> $a:A \vdash \chi_P(a):\Omega$. We can then recover the formula as $a:A\vdash P(a)\Leftrightarrow \chi_P(a)=_\Omega\top$. In fact, this equivalence is the heart of what a subobject classifier is. Characteristic functions $\chi_P$ that factor through $\mathbf{2}$ correspond to <em>decidable</em> predicates, i.e. predicates for which the law of excluded middle holds. As a formula $a:A\vdash P(a)\lor\neg P(a)$ or via a characteristic function $a:A\vdash (\chi_P(a)\lor\neg\chi_P(a))=_\Omega\top$. The former states that $P$ is a <a href="https://ncatlab.org/nlab/show/complemented+subobject" rel="noreferrer">complemented or decidable subobject</a> of $A$. In the latter, $\lor$ and $\neg$ are operations on $\Omega$ rather than logical connectives.</p>
<p>A topos where $\Omega \ncong \mathbf{2}$ is one whose internal logic has some propositions that are not decidable in the above sense. If you restrict yourself to Boolean-valued characteristic functions, then you are restricting yourself to only the decidable propositions in your internal logic. This is a completely coherent thing to do, but it means there are formulas you can state which will have no Boolean-valued characteristic function. They will, however, always have an $\Omega$-valued characteristic function. If you <em>identify</em> predicates with Boolean-valued characteristic functions, then what you are doing is equivalent to interpreting into a Boolean subtopos of whatever topos you started with.</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>
| user376343 | 376,343 | <p><strong>Geometric solution</strong></p>
<p>Locus of points equidistant to a circle and a point (outside the circle) is a hyperbola.<br />
These equidistant points are centers of circles through the given point, and touching the given circle.</p>
<p><a href="https://i.stack.imgur.com/wPGR8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wPGR8.png" alt="enter image description here" /></a></p>
<p>In the present case, the given blue circle is denoted <span class="math-container">$\gamma$</span> and is centered at <span class="math-container">$A.$</span> The hyperbola has foci <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and passes through the middle point of the segment <span class="math-container">$BG,$</span> where <span class="math-container">$G$</span> is intersection of <span class="math-container">$AB$</span> and <span class="math-container">$\gamma.$</span><br />
Since the centers of circles touching <span class="math-container">$\gamma$</span> must lie at <span class="math-container">$BC,$</span> they are at the intersection of <span class="math-container">$BC$</span> and the hyperbola.</p>
<p>The picture gives two possible configurations assuming that <span class="math-container">$BC$</span> has empty intersection with the circle:</p>
<ol>
<li>the line cuts each branch of the hyperbola at a single point</li>
<li>the line cuts one branch of hyperbola at two points</li>
</ol>
<p>(analytic solution can be added à la commande)</p>
|
3,718,255 | <p>So i solved the first part of this question, that there's 588 ways for choosing 5 digit integers, but i can't solve "Find the number of integers in (i) which satisfies "if there's 8, it's followed by 6"
i tried to solve it using different cases:</p>
<p>Case 1) there's two 8s and two 6s => (3 * 3!)/2! - 1</p>
<p>Case 2) there's two 6s and 8 => 4 * 4!-3*3!</p>
<p>Case 3) 8 and 6 => 4!-3!</p>
<p>the sum was 104 but i don't think that's right, should i make another case for no 8s? or is there an easier way.</p>
| fleablood | 280,126 | <p>Add up the ways to do it with no <span class="math-container">$8$</span>, one <span class="math-container">$8$</span> or two <span class="math-container">$8$</span>s.</p>
<p>If there are no <span class="math-container">$8$</span>s you must choose <span class="math-container">$5$</span> components (with order) from <span class="math-container">$\{5,2,6,6,0\}$</span> There are <span class="math-container">$\frac {5!}2$</span> ways to do this but as your number can't start with <span class="math-container">$0$</span> we must subtract the <span class="math-container">$\frac {4!}2$</span> ways to start with <span class="math-container">$0$</span>.</p>
<p>If there is one <span class="math-container">$8$</span> then you must place an <span class="math-container">$86$</span> in four possible locations. the <span class="math-container">$3$</span> remaining slots must be filled with <span class="math-container">$\{5,2,6,0\}$</span> and there are <span class="math-container">$3!$</span> ways to do that. but we must subtract the ways that start with <span class="math-container">$0$</span>. If it starts with <span class="math-container">$0$</span> there are <span class="math-container">$3$</span> spots for the <span class="math-container">$86$</span> and <span class="math-container">$2!$</span> ways to do the rest.</p>
<p>If there are two <span class="math-container">$86$</span>s then you must place <span class="math-container">$86,86, x$</span> into <span class="math-container">$3$</span> slots. There are <span class="math-container">$3$</span> places to put the one digit that <em>isn't</em> one of the <span class="math-container">$86$</span> and that <span class="math-container">$1$</span> slot can be filled with <span class="math-container">$\{5,2,0\}$</span> and there are <span class="math-container">$3$</span> ways to do that. But there is <span class="math-container">$1$</span> way the first slot is a <span class="math-container">$0$</span> and we must omit that.</p>
<p>So putting it together:</p>
<p><span class="math-container">$(\frac {5!}2 - \frac {4!}2) + (4*3!- 3*2!) + (3*3 - 1)$</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>
| zhw. | 228,045 | <p>Another approach: Note the set <span class="math-container">$\{x_1,x_2,\dots\}\cup \{x\}$</span> is bounded, hence is contained in <span class="math-container">$[-M,M]$</span> for some positive <span class="math-container">$M.$</span> Since <span class="math-container">$[-M,M]\times [0,1]$</span> is compact, <span class="math-container">$f$</span> is uniformly continuous there.</p>
<p>Let <span class="math-container">$\epsilon >0.$</span> Then there exists <span class="math-container">$\delta>0$</span> such that <span class="math-container">$z,w\in[-M,M]\times [0,1],$</span> <span class="math-container">$ |z-w|< \delta,$</span> implies <span class="math-container">$|f(z)-f(w)|<\epsilon.$</span></p>
<p>Since <span class="math-container">$x_n\to x,$</span> we can choose <span class="math-container">$N$</span> such that <span class="math-container">$|x_n-x|<\delta$</span> for <span class="math-container">$n>N.$</span> For such <span class="math-container">$n,$</span> <span class="math-container">$|(x_n,y)-(x,y)| = |x_n-x|<\delta$</span> for all <span class="math-container">$y\in [0,1].$</span> Hence for <span class="math-container">$n>N,$</span> <span class="math-container">$|g(x_n,y)-g(x,y)|<\epsilon$</span> for all <span class="math-container">$y\in [0,1]$</span> as desired.</p>
|
3,631,714 | <p>If <span class="math-container">$f: \mathbb{R}^n \to \mathbb{R} $</span> is a function on the Schwartz space <span class="math-container">$\mathcal{S} (\mathbb{R}^n) $</span> we define the Fourier transform of <span class="math-container">$f$</span> as the function
<span class="math-container">$$ \hat f(\xi)= \int_{\mathbb{R}^n} f(x)e^{2 \pi ix\xi} dx $$</span>
How can I prove that if <span class="math-container">$f \in L^1(\mathbb{R}^n)$</span>, then the same formula for <span class="math-container">$\hat f$</span> holds? I'd like to prove that that formula extends the Fourier transform on <span class="math-container">$L^1$</span>.</p>
| Mathsmerizing | 757,478 | <p>Well, going by the video you posted, I will assume that you wish to explore more competition oriented resources. The one resource for IIT-JEE exam is <a href="https://www.iitmind.com/2018/12/tmh-iitjee-mathematics-pdf.html" rel="nofollow noreferrer">TMH Jee Advanced</a> and this is the book I recommend the most for IIT JEE aspirants. It primarily is a problem book and that too unstructured but it captures the spirit and essence of IIT JEE.</p>
|
3,631,714 | <p>If <span class="math-container">$f: \mathbb{R}^n \to \mathbb{R} $</span> is a function on the Schwartz space <span class="math-container">$\mathcal{S} (\mathbb{R}^n) $</span> we define the Fourier transform of <span class="math-container">$f$</span> as the function
<span class="math-container">$$ \hat f(\xi)= \int_{\mathbb{R}^n} f(x)e^{2 \pi ix\xi} dx $$</span>
How can I prove that if <span class="math-container">$f \in L^1(\mathbb{R}^n)$</span>, then the same formula for <span class="math-container">$\hat f$</span> holds? I'd like to prove that that formula extends the Fourier transform on <span class="math-container">$L^1$</span>.</p>
| Confused Simpleton | 722,188 | <p>These books covers the essence of what you are looking for. Almost everybody in 11th and 12th study from these books. </p>
<p><a href="https://jeemain.guru/best-books-for-mathematics-part-1/" rel="nofollow noreferrer">https://jeemain.guru/best-books-for-mathematics-part-1/</a></p>
<p>PS- The books written by NCERT are basic books suitable for self study. I may add, NCERT also publishes Exemplar problems books (For Class 11th , and 12th) which is tougher version of those basic books for 11th and 12th.</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>
| mmfrgmpds | 284,638 | <p>To find the distributional equivalent of <span class="math-container">$f''$</span>, one calculates, say <span class="math-container">$\varphi\,' \in \mathcal{C^\infty_c([a-1,1])}$</span>, and calculates the Schwartz bracket quantity,
<span class="math-container">$$
\begin{align}
\langle f'',\,\varphi'\rangle &= - \langle f',\,\varphi''\rangle \\
&= -\int\limits_{a-1}^1 f'(x)\,\varphi''(x)\;{\rm d} x \\
&= -2\int\limits_{a-1}^{a+1} (a-x)\,\varphi''(x)\;{\rm d} x -\int\limits_{a+1}^1 0\cdot\,\varphi''(x)\;{\rm d} x \\
&= -2\int\limits_{a-1}^{a+1} (a-x)\,\varphi''(x)\;{\rm d} x \\
&= -2a\int\limits_{a-1}^{a+1}\varphi''(x)\;{\rm d} x + 2\int\limits_{a-1}^{a+1} x\,\varphi''(x)\;{\rm d} x \\
&= 2\varphi'(a+1) - 2\varphi(a+1) \\
&= 2\langle(\delta_{a+1},\,\varphi'\rangle + 2\langle(h_{a+1},\,\varphi'\rangle \\
&= \langle f'',\,\varphi'\rangle.
\end{align}
$$</span>
The above is true for all <span class="math-container">$\varphi'$</span> in the test function space, and so therefore one must conclude from the last two lines that,
<span class="math-container">$$
f'' \doteq 2\delta_{a+1} + 2h_{a+1},
$$</span>
where the symbol <span class="math-container">$\doteq$</span> means "is equal in the sense of distributions", and <span class="math-container">$h(x)$</span> is the Heaviside function which is <span class="math-container">$1$</span> for <span class="math-container">$x > 0$</span> and <span class="math-container">$0$</span> for <span class="math-container">$x<0$</span>. <span class="math-container">$\qquad\blacksquare$</span></p>
|
200,414 | <p>Some people may carelessly say that you need calculus to find such a thing as a local maximum of $f(x) = x^3 - 20x^2 + 96x$. Certainly calculus is <em>sufficient</em>, but whether it's <em>necessary</em> is another question.</p>
<p>There's a global maximum if you restrict the domain to $[0,8]$, and $f$ is $0$ at the endpoints and positive between them. Say the maximum is at $x_0$. One would have
$$
\frac{f(x)-f(x_0)}{x-x_0}\begin{cases} >0 & \text{if }x<x_0, \\ <0 & \text{if }x>x_0. \end{cases}
$$
This difference quotient is undefined when $x=x_0$, but mere algebra tells us that the numerator factors and we get
$$
\frac{(x-x_0)g(x)}{x-x_0} = g(x)
$$
where $g(x)$ is a polynomial whose coefficients depend on $x_0$. Then of course one seeks its zeros since it should change signs at $x_0$.</p>
<p>Have we tacitly used the intermediate value theorem, or the extreme value theorem? To what extent can those be avoided? Must one say that <em>if</em> there is a maximum point, <em>then</em> it is at a zero of $g(x)$? And can we say that without the intermediate value theorem? (At least in the case of this function, I think we stop short of needing the so-called fundamental theorem of algebra to tell us some zeros of $g$ exist!)</p>
| Mathemagician1234 | 7,012 | <p>There's actually a famous book by Ivan Niven on precisely this subject i.e. finding extrema by purely algebraic or geometric methods. It's called, simply enough, <em>Maxima and Minima without Calculus</em>. It's currently published by the Mathematicial Association of America. I think you'll find it fascinating and full of problems very much like this. </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>
| yomplex | 12,996 | <p>A <em>good</em> answer to this question is one that (is correct, and) <em>she</em> finds convincing.</p>
<p>As kids are growing up and making sense of the world around them, <em>experimentation</em> is often one of their key sources of truth. Perhaps, experimentation can also serve as a reliable and effective gateway to the abstract world of mathematical truths!</p>
<p>Here's one possible roadmap that leads convincingly to arithmetical truths.</p>
<ul>
<li><p><strong>Experiment with an Object</strong>: Take <span class="math-container">$2$</span> blocks, add <span class="math-container">$6$</span> blocks. Count the total number of blocks to discover that the result is <span class="math-container">$8$</span> blocks. Therefore, <span class="math-container">$(2 \mbox{ blocks}) + (6 \mbox{ blocks}) = (8 \mbox{ blocks})$</span>.</p></li>
<li><p><strong>Discover that these results apply to all Other Objects</strong>: Experiment to discover that this rule applies to other objects such as apples, almonds, chairs, tables, toys, candy and anything else you can think of. Perhaps she would be willing to take the leap with you to agree that this would apply to other things that she has not yet experimented with, such as ships, planets and whales. Therefore, <span class="math-container">$(2 \mbox{ things}) + (6 \mbox{ things}) = (8 \mbox{ things})$</span>.</p></li>
<li><p><strong>General result</strong>: Since these results are valid irrespective of the object or thing being added, it is therefore acceptable (and more efficient) to write our experimental result concisely as: <span class="math-container">$2+6 = 8$</span> without reference to any object in particular.</p></li>
</ul>
<p>Soon we point out that, in particular, these rules also apply to fingers (just like any other object), which are always at hand(!) so we can use them no matter what it is that we are in fact trying to add.</p>
<p>The key feature of the approach just outlined is that the act of writing <span class="math-container">$2+6 = 8$</span> is a symbolic recording (by pencil on paper, or chalk on board) of an experimental truth that we have discovered by exploring the world around us. The question of logically or abstractly proving it, therefore, does not arise.</p>
<p>When we arrive at the question, "what is <span class="math-container">$4+4$</span>?", we must now resort to experimenting with our fingers (or any other object) to discover the result of this addition. Having discovered the result, we can record it symbolically as <span class="math-container">$4+4 = 8$</span>.</p>
<p>Of course, we can get creative with our experiments. For example, to count the number of dots below, </p>
<p>: : : : </p>
<p>we can count them as </p>
<p>: <span class="math-container">$+$</span> : : : </p>
<p>or as </p>
<p>: : <span class="math-container">$+$</span> : :</p>
<p>This provides an intuitive justification that <span class="math-container">$2+6$</span> and <span class="math-container">$4+4$</span> are really just two different ways of visualizing <span class="math-container">$8$</span>.</p>
<p>Also, I can't help pointing out that the following is a staggering gap in her present education:</p>
<blockquote>
<p>But she has not learnt the Peano Axioms at this age.</p>
</blockquote>
<p>I hope you will remedy it at the earliest! ;)</p>
|
3,131,791 | <p>I read this:</p>
<blockquote>
<p><strong>Definition 1.1</strong>. <em>The</em> complex projective line <span class="math-container">$\mathbb{CP}^1$</span> <em>(or just</em> <span class="math-container">$\mathbb{P}^1$</span><em>) is the set of all ordered pairs of complex numbers</em> <span class="math-container">$\{(x, y)\in\mathbb{C}^2\mid(x, y)\neq(0, 0)\}$</span> <em>where we identify pairs</em> <span class="math-container">$(x, y)$</span> <em>and</em> <span class="math-container">$(x', y')$</span> <em>if one is a scalar multiple of the other</em>: <span class="math-container">$(x, y)=\lambda(x', y')$</span> <em>for some</em> <span class="math-container">$\lambda\in\mathbb{C}^*$</span><em>, where</em> <span class="math-container">$\mathbb{C}^*$</span> <em>is the set of nonzero complex numbers</em>.</p>
<p><span class="math-container">$\quad$</span> So for example <span class="math-container">$(1, 2)$</span>, <span class="math-container">$(3, 6)$</span>, and <span class="math-container">$(2+3i, 4+6i)$</span> all represent the same point of <span class="math-container">$\mathbb{P}^1$</span>.</p>
<p><span class="math-container">$\quad$</span> This construction is an example of the quotient of a set by an equivalence relation. See Exercise 2.</p>
<p><span class="math-container">$\quad$</span> The idea is that <span class="math-container">$\mathbb{P}^1$</span> can be thought of as the union of the set of complex numbers <span class="math-container">$\mathbb{C}$</span> and a single point "at infinity". To see this, consider the following subset <span class="math-container">$U_0 \subset \mathbb{P}^1$</span>:</p>
<p><span class="math-container">$$ U_0 = \{(x_0, x_1) \in\mathbb{P}^1\mid x_0\neq 0 \}.$$</span></p>
<p>Then <span class="math-container">$U_0$</span> is in one-to-one correspondence with <span class="math-container">$\mathbb{C}$</span> via the map</p>
<p><span class="math-container">$$\tag{1} \phi_0 : U_0 \to \mathbb{C} : \qquad (x_0, x_1) \mapsto \frac{x_1}{x_0}. $$</span></p>
<p>Note that <span class="math-container">$\phi_0$</span> is well defined on <span class="math-container">$U_0$</span>. First of all, <span class="math-container">$x_0$</span> is not <span class="math-container">$0$</span> so the division makes sense. Secondly, if <span class="math-container">$(x_0, x_1)$</span> represent the same point of <span class="math-container">$\mathbb{P}^1$</span> as <span class="math-container">$(x_0', x_1')$</span> then <span class="math-container">$x_0 = \lambda x_0'$</span> and <span class="math-container">$x_1 = \lambda x_1'$</span> for some nonzero <span class="math-container">$\lambda \in \mathbb{C}$</span>. Thus, <span class="math-container">$\phi_0((x_0, x_1)) = x_1 / x_0 = (\lambda x_1')/(\lambda x_0')$</span> <span class="math-container">$= x_1' / x_0' = \phi_0((x_0', x_1')) $</span> and <span class="math-container">$\phi_0$</span> is well defined as claimed. The inverse map is given by</p>
<p><span class="math-container">$$ \psi_0 : \mathbb{C}\to U_0 : \qquad z \mapsto (1, z). $$</span></p>
<p><span class="math-container">$\quad$</span> The complement of <span class="math-container">$U_0$</span> is the set of all points of <span class="math-container">$\mathbb{P}^1$</span> of the form <span class="math-container">$(0, x_1)$</span>. But since <span class="math-container">$(0, x_1) = x_1(0, 1)$</span>, all of these points coincide with <span class="math-container">$(0,1)$</span> as a point of <span class="math-container">$\mathbb{P}^1$</span>. So <span class="math-container">$\mathbb{P}^1$</span> is obtained from a copy of <span class="math-container">$\mathbb{C}$</span> by adding a single point.</p>
<p><span class="math-container">$\quad$</span> This point can be thought of as the point of infinity. To see this, consider a complex number <span class="math-container">$t$</span>, and identify it with a point of <span class="math-container">$U_0$</span> using <span class="math-container">$\psi_0$</span>; i.e., we identify it with <span class="math-container">$\psi_0(t) = (1, t)$</span>. Now let <span class="math-container">$t \to \infty$</span>. The beautiful feature is that the limit now exists in <span class="math-container">$\mathbb{P}^1$</span>! To see this, rewrite <span class="math-container">$(1, t)$</span> as <span class="math-container">$(1/t, 1)$</span> using scalar multiplication by <span class="math-container">$1/t$</span>. This clearly approaches <span class="math-container">$(0,1)$</span> as <span class="math-container">$t \to \infty$</span>, so <span class="math-container">$(0, 1)$</span> really should be thought of as the point at infinity!</p>
<p><span class="math-container">$\quad$</span> We have been deliberately vague about the precise meaning of limits in <span class="math-container">$\mathbb{P}^1$</span>. This is a notion from topology, which we will deal with later in Chapter 4. The property that limits exist in a topological space is a consequence of the <em>compactness</em> of the space, and the process of enlarging <span class="math-container">$\mathbb{C}$</span> to the compact space <span class="math-container">$\mathbb{P}^1$</span> is our first example of the important process of <em>compactification</em>. This makes the solutions to enumerative problems well-defined, by preventing solutions from going off to infinity. A precise definition of compactness will be given in Chapter 4. </p>
<p><span class="math-container">$\quad$</span> We now have to modify our description of complex polynomials by associating to them polynomials <span class="math-container">$F(x_0, x_1)$</span> on <span class="math-container">$\mathbb{P}^1$</span>. Before turning to their definition, note that the equation <span class="math-container">$F(x_0, x_1) = 0$</span> need not make sense as a well-defined equation of <span class="math-container">$\mathbb{P}^1$</span>, since it is conceivable that a point could have different representations <span class="math-container">$(x_0, x_1)$</span> and <span class="math-container">$(x_0', x_1')$</span> such that <span class="math-container">$F(x_0, x_1) = 0$</span> while <span class="math-container">$F(x_0', x_1') \neq 0$</span>. We avoid this problem by requiring that <span class="math-container">$F(x_0, x_1)$</span> be a <em>homogenous polynomial</em>; i.e., all terms in <span class="math-container">$F$</span> have the same total degree, which is called the degree of <span class="math-container">$F$</span>. So</p>
<p><span class="math-container">$$\tag{2} F(x_0, x_1) = \sum_{i=0}^d a_i x_0^i x_1^{d-i} $$</span></p>
<hr>
</blockquote>
<p>In the last paragraph, I tried to take one non-homogeneous polynomial and checked that indeed, that problem happens but when I took an homogeneous polynomial, the problem vanishes! I am baffled: it looks like sorcery! I tried to explain myself why that happens but couldn't so, why does that happen?</p>
| Travis Willse | 155,629 | <p>Suppose <span class="math-container">$F$</span> is homogeneous of degree <span class="math-container">$d$</span>, so that
<span class="math-container">$$F(x_0, x_1) = \sum_{i = 0}^d a_i x_0^i x_1^{d - i} .$$</span>
Any other nonzero point <span class="math-container">$(x_0', x_1')$</span> on the line spanned by <span class="math-container">$(x_0, x_1) \neq (0, 0)$</span> can be written as <span class="math-container">$(\lambda x_0, \lambda x_1)$</span> for some <span class="math-container">$\lambda \neq 0$</span>, and substituting into the formula for <span class="math-container">$F$</span> gives
<span class="math-container">$$F(\lambda x_0, \lambda x_1) = \sum_{i = 0}^d a_i (\lambda x_0)^i (\lambda x_1)^{d - i} = \sum_{i = 0}^d a_i \lambda^i x_0^i \lambda^{d - i} x_1^{d - i} \stackrel{(\ast)}{=} \lambda^d \sum_{i = 0}^d a_i x_0^i x_1^{d - i} = \lambda^d F(x_0, x_1) .$$</span>
In particular, <span class="math-container">$F(x_0, x_1) = 0$</span> if and only if <span class="math-container">$F(\lambda x_0, \lambda x_1) = \lambda^d (0) = 0$</span>. Put another way, for homogeneous <span class="math-container">$F$</span> the equation <span class="math-container">$F(x_0, x_1) = 0$</span> is well-defined on <span class="math-container">$\Bbb P^1$</span>.</p>
<p>On the other hand, if <span class="math-container">$F$</span> is not homogeneous, we cannot factor all of the occurrences of <span class="math-container">$\lambda$</span> out of the sum as we did in the equality marked <span class="math-container">$(\ast)$</span> above, and so the conclusion does not follow.</p>
|
2,501,450 | <p>I was trying for a while to prove non-existence of theu following limit:</p>
<p>$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$$</p>
<p>Unfortunately, with no results.</p>
<p>My hope was to show, that:</p>
<p>$$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}\not=0$$</p>
<p>But showing that was harder than I thought.</p>
<p>Can anyone show me how to solve this problem?</p>
| DRPR | 347,579 | <p>Hint:Divide the numerator and denominator by $2^n$. You dont need Lhopital rule .</p>
|
1,216,576 | <p><img src="https://i.stack.imgur.com/p4xbH.png" alt="enter image description here"></p>
<blockquote>
<p>$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the midpoint of
$BC$. If $AE$ intersects $BC$ at $F$, show that $Area(\triangle BFE)=2Area(\triangle FED)$</p>
</blockquote>
<p><strong>Solution given</strong><img src="https://i.stack.imgur.com/fBsFE.png" alt="enter image description here"> :</p>
<p><img src="https://i.stack.imgur.com/wsWIe.png" alt="enter image description here"></p>
<p>I didn't understand in the solution that $FB=2FD$ .</p>
<p>Any hint is appreciated.</p>
| Salomo | 226,957 | <p>Imagine there are two group of people, with each group $n$ people. Now, we want to choose $n$ people from these $2n$ people. On one hand, the number of combinations we can have is $\sum^n_{i=0} \binom{n}{i} \binom{n}{n-i}=\sum^n_{i=0} \binom{n}{i}^2$, since we can choose $i$ from one group and $n-i$ from another group. On the other hand, we can neglect the groups, and it is just $\binom{2n}{n}=\frac{2n!}{n!n!}$.</p>
|
2,804,086 | <p>A distribution is an element of the continuous dual space of some function space. Let us take the Schwartz space $\mathcal{S} := \mathcal{S}(\mathbb{R}^n)$ just as an example. A distribution $\phi \in \mathcal{S}'$ is then a map
$$ \phi: \mathcal{S} \rightarrow \mathbb{C}.$$</p>
<p>My question is this: how do I interpret $\phi(x)$? I see this written a lot, but I don't understand how to work with it. What does for example $\phi(x) = \phi(-x)$ mean? The only thing I can think of is that $\phi(f) = \phi(\hat{f})$, where $\hat{f}(x) = f(-x)$.</p>
<p>And more specifically for the problem I'm working on: I have a distribution
$$ \mathcal{W} : \mathcal{S}(\underbrace{\mathbb{R}^4 \times \dots \times \mathbb{R}^4}_{n \text{ times}}) \rightarrow \mathbb{C} $$</p>
<p>and then they say that $\mathcal{W}$ is translation invariant, i.e.
$$\mathcal{W}(x_1 +a,\dots, x_n + a) = \mathcal{W}(x_1,\dots,x_n)$$
so it can be writthen as a distribution $\mathfrak{W}$ that only depends on the differences $x_1-x_2,\dots,x_{n-1} - x_n$:
$$\mathcal{W}(x_1,\dots,x_n) = \mathfrak{W}(x_1-x_2,\dots,x_{n-1}-x_n).$$</p>
<p>How do I interpret this last line?</p>
| Abdelmalek Abdesselam | 244,562 | <p>If the motivation is Wightman functions, it's better to use the differences $x_1-x_n,\ldots,x_{n-1}-x_n$ with a single point of reference $x_n$ rather than the consecutive differences $x_1-x_2,
\ldots,x_{n-1}-x_n$.
So let me consider the statement that as elements in $S'(\mathbb{R}^{4n})$
$$
\mathcal{W}(x_1,\ldots,x_n)=V(x_1-x_n,\ldots,x_{n-1}-x_n)
$$
for some $V$ in $S'(\mathbb{R}^{4(n-1)})$.</p>
<p>First do some nonrigorous calculations as follows. Let $f(x_1,\ldots,x_n)$ be a test function in $S'(\mathbb{R}^{4n})$.</p>
<p>$$
\int_{\mathbb{R}^{4n}} V(x_1-x_n,\ldots,x_{n-1}-x_n)f(x_1,\ldots,x_n)
\ dx_1\cdots dx_n
$$
$$
=\int_{\mathbb{R}^{4n}} V(z_1,\ldots,z_{n-1})f(z_1+x_n,\ldots,z_{n-1}+x_n,x_n)
\ dz_1\cdots dz_{n-1}dx_n
$$
by the change of variables $z_i=x_i-x_n$ for $1\le i\le n-1$.
By Fubini the last integral becomes
$$
\int_{\mathbb{R}^{4(n-1)}} V(z_1,\ldots,z_{n-1})g(z_1,\ldots,z_{n-1})
\ dz_1\cdots dz_{n-1}
$$
where
$g=\Gamma (f)$ is defined by
$$
g(z_1,\ldots,z_{n-1})=\int_{\mathbb{R}^4}
f(z_1+x_n,\ldots,z_{n-1}+x_n,x_n)
\ dx_n\ .
$$
Now you can make rigorous sense of your RHS as the composition $V\circ \Gamma$
of the distribution $V:S(\mathbb{R}^{4(n-1)})\longrightarrow\mathbb{C}$ with the continuous linear map
$$
\Gamma:S(\mathbb{R}^{4n})\longrightarrow S(\mathbb{R}^{4(n-1)})\ .
$$</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>
| colormegone | 71,645 | <p>To stay consistent with your labelling, write the position of the Vancouver cyclist as $ \ x_A = 25 \ (t+2) \ $ and the position of the Seattle cyclist as $ \ x_B = 300 - 32t \ . $ We are calling the location of Vancouver $ \ x = 0 \ $ by doing so. Remember that the Vancouver cyclist is the one who has been pedalling two hours <em>longer</em>. </p>
<p>You want to solve for the time when $ \ x_A = x_B \ . $ That time $ \ T \ $ can then be entered into either position function to find the distance from Vancouver where they meet.</p>
<p>Note: the answer <em>rounds off</em> to 160 km., in case you are wondering about the "ugly" fractions.</p>
|
4,083,588 | <p>Can somebody help me how to solve</p>
<p><span class="math-container">$x* 23 \equiv_{60} 1 $</span> with <span class="math-container">$ x \in \mathbb{N} $</span> and <span class="math-container">$x > 100$</span></p>
<p>What would be a good approach?</p>
<p>I know that x = 107 would be a solution. However how can I find solutions and how to prove that there are no solutions, if there aren't any?</p>
| Doug M | 317,176 | <p>Use the Euclidean algorithm.</p>
<p>Since <span class="math-container">$\gcd(23,60) = 1$</span> there exists integers <span class="math-container">$m,n$</span> such that <span class="math-container">$23m + 60n = 1.$</span></p>
<p><span class="math-container">$m$</span> will be our inverse.</p>
<p>The Euclidean algorithm:</p>
<p><span class="math-container">$\begin {array}{rrc}
n&m&60n+23m\\ \hline
1&&60\\
&1&23\\
1&-2&14\\
-1&3&9\\
2&-5&5\\
5&-13&1\\
\end{array}$</span></p>
<p>What are we doing? The first column are candidates for <span class="math-container">$n.$</span> The second column are candidates for <span class="math-container">$m.$</span> The third column is <span class="math-container">$60n + 23m.$</span></p>
<p>We then subtract the largest multiple in the lines below from lines above to generate smaller numbers for <span class="math-container">$60n + 23m.$</span></p>
<p>We get down to <span class="math-container">$-13$</span> as a multiplicative inverse. And if <span class="math-container">$-13$</span> is an inverse, so is <span class="math-container">$47.$</span></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>
| Gabriel Romon | 66,096 | <p>Assuming $X$ is real-valued, $$\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\} = ((X-\limsup_n X_n)=0)\cap ((X-\liminf_n X_n)=0) $$</p>
<p>Since $\limsup_n X_n$, $\liminf_n X_n$ and $X$ are measurable, $((X-\limsup_n X_n)=0)\cap ((X-\liminf_n X_n)=0)$ is measurable.</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>
| Théophile | 26,091 | <p>Your claim is wrong. You can choose any nonnegative number as the limit:
<span class="math-container">$$\lim_{x\to0^+}\left(e^{-1/x}\right)^{ax} = e^{-a}.$$</span></p>
<p>This example is from the <a href="https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero" rel="noreferrer">Wikipedia article</a> on <span class="math-container">$0^0$</span>, as @PrasunBiswas suggested. There are other examples there showing a limit of infinity, etc.</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>
| Robert Lewis | 67,071 | <p>To see that the given expression is equal to $1$, set</p>
<p>$\omega = e^{\frac{ik\pi}{n + 1}}, \tag{1}$</p>
<p>then the expression becomes </p>
<p>$\dfrac{1}{1 - \omega} + \dfrac{1}{1 - \bar{\omega}} = \dfrac{1 - \bar{\omega} + 1 - \omega}{(1 - \omega)(1 - \omega)} = \dfrac{2 - (\omega + \bar{\omega})}{(1 - \omega)(1 - \bar{\omega})}$
$= \dfrac{2 - (\omega + \bar{\omega})}{(1 + \omega \bar{\omega}) - (\omega + \bar{\omega})} = \dfrac{2 - (\omega + \bar{\omega})}{2 - (\omega + \bar{\omega})} = 1, \tag{2}$</p>
<p>since $\omega \bar{\omega} = 1$.</p>
<p>Hope this helps. Cheerio,</p>
<p>and as always,</p>
<p><em><strong>Fiat Lux!!!</em></strong></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>
| Andreas Blass | 48,510 | <p>Since you wrote "short1 $\circ$ left1", you must be using $\circ$ to denote composition in the usual order for functions, where $f\circ g$" means first apply $g$ and then apply $f$. But in your "tests for composition", you seem to be using $\circ$ to mean composition in the opposite order, following the order of the arrows instead. I conjecture that, if you used one convention consistently, your problem(s) would disappear. </p>
|
2,904,432 | <blockquote>
<p>Let the subspace <span class="math-container">$S=\{(x, y, z) \in R^3; x = -2z \}$</span></p>
<p>a) Determine its dimension and a basis B of S;</p>
<p>b) Complete B such that it becomes a basis M of <span class="math-container">$R^3$</span>.</p>
</blockquote>
<p><span class="math-container">$\operatorname{dim}(S) = 2$</span> because there are two free variables, but how to determine a basis <span class="math-container">$B$</span> for <span class="math-container">$S$</span>? Does the vector have to belong to <span class="math-container">$S$</span>? I think so. What about <span class="math-container">$M$</span>? Does the third vector have to belong to <span class="math-container">$S$</span>? I don't think so, otherwise <span class="math-container">$B$</span> would not generate <span class="math-container">$R^3$</span>.</p>
| CyclotomicField | 464,974 | <p>So a typical vector would normally be written as $\begin{bmatrix}x\\y\\z\end{bmatrix}$ but we can make a substitution here using $x=-2z$ so that our subspace vectors are written as $\begin{bmatrix}-2z\\y\\z\end{bmatrix}$. From here we separate the variables using addition then factor out the variables using scalar multiplication as follows. $$\begin{bmatrix}-2z\\y\\z\end{bmatrix}=\begin{bmatrix}0\\y\\0\end{bmatrix}+\begin{bmatrix}-2z\\0\\z\end{bmatrix}=y\begin{bmatrix}0\\1\\0\end{bmatrix}+z\begin{bmatrix}-2\\0\\1\end{bmatrix}$$</p>
<p>Which gives us the basis vectors $\{\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}-2\\0\\1\end{bmatrix}\}$. Since there are two of them the subspace is two dimensional as you deduced from the number of free variables.</p>
<p>To give a complete basis for $R^3$ you'll need an additional vector not in the subspace spanned by these two. I would suggest picking a vector at random because the probability that it will be in the subspace is zero so you'll find one quickly. You just have to show it's not in the subspace to make it linearly independent of the other two vectors.</p>
|
2,330,496 | <p>Players Ruby and Bob are given an undirected graph and a number $N$. First Ruby colors $N$ vertices red, then Bob colors $N$ vertices blue (they must be distinct from Ruby's choices). Afterward, all other points on the graph are given the color of whichever color they are closest to (shortest path) with ties left blank. The player with more of their color on the resulting graph wins.</p>
<p>Can the first player always win (or tie)?</p>
<p>Some context: This problem arose for me out of a mobile puzzle game. My knowledge in graph theory is pretty minimal, so I don't have much machinery to solve it (but I have spent a while with small graphs, always able to find an unbeatable strategy for the first player). My thoughts so far are to mark points as having an advantage in comparison to their neighbors, the maximum of which I'd guess provides a win with N=1. For higher N though the dynamics get much more challenging for me to express. There seems to be an aspect of even-spacing which I'm not sure how to formalize (perhaps it's picking vertices which minimize their shortest distance to any point).</p>
<p>Also if anyone has heard of a similar problem before (specifically related to coloring a graph based on the shortest path to a colored vertex) or has references I'd be happy to read them, but was unable to find much since I'm not certain what to search for.</p>
| Alex Ravsky | 71,850 | <p>I can prove that for the first player (Ruby) is possible not to lose in some simple cases for the given graph $G$ with $n$ vertices. </p>
<ol>
<li><p><a href="https://en.wikipedia.org/wiki/Dominating_set" rel="nofollow noreferrer">Domination number</a> $\gamma(G)\le N$. Indeed, in this case at her move the first player can assure that the red vertices constitute a dominating set. Then after the second player move only $N$ vertices colored by him will become blue, because any not red vertex has a red neighbor, so if it was not colored by the second player then it becomes red if it has no blue neighbor or remains blank otherwise. </p></li>
<li><p>The graph $G$ is a tree and $N=1$. By <a href="https://pdfs.semanticscholar.org/96cb/20f1d3b622709366280d3cd439b4f900ee69.pdf" rel="nofollow noreferrer">Jordan Theorem from 1869</a> (which I proved independently few hours ago :-) ), $G$ has a vertex $v$ whose removal disconnects the tree into components of size at most $n/2$. So if the first player color the vertex $v$, the second player after his move will make blue only vertices from one of the connected components, whose size is at most $n/2$. The remaining vertices will become red. </p></li>
<li><p>The graph $G$ is a cycle. Let the first player color clockwise vertices $v_1,\dots, v_N,v_{N+1}=v_1$ assuring that their removal disconnects the cycle into components whose sizes differs by at most $1$. Let these sizes be $l$ and $l+1$. For any $i<N$ denote the clockwise path (without its second endpoint) from a vertex $v_i$ to a vertex $v_{i+1}$ by $[v_i,v_{i+1})$. Consider a component $C$ bounded by vertices $v_i$ and $v_{i+1}$. If the second player color one vertex $v$ in $C$, then $C$ with its two red boundary vertices will be split into two paths $v_i,\dots, v$ and $v,\dots, v_{i+1}$ with the endpoints of different colors. By symmetry, after the final coloring each of them will have equal number of red and black vertices. But here the blue vertex $v$ was counted two times, as a endpoint of each of the paths. Taking this into account we see that the second player obtain no gain at the clockwise path $[v_i,v_{i+1})$. If the second player color more than one vertex $v$ in $C$ then he can assure all of vertices $[v_i,v_{i+1})$ but $v_i$ be blue, but at the cost that all vertices a some clockwise path $[v_j,v_{j+1})$ will be red. So he wins at most $l+1$ vertices at $[v_i,v_{i+1})$ but loses at least $l+1$ vertices at $[v_j,v_{j+1})$. Thus he obtains no gain in this case too.</p></li>
</ol>
|
4,537,390 | <p>Let <span class="math-container">$C$</span> be a circle centred at <span class="math-container">$O$</span> and <span class="math-container">$P$</span> be a point inside the circle.</p>
<p>The line <span class="math-container">$OP$</span> intersects the circle <span class="math-container">$C$</span> at <span class="math-container">$Q_1$</span> and <span class="math-container">$Q_2$</span>.</p>
<p>Let <span class="math-container">$Q$</span> be a point on the circle <span class="math-container">$C$</span>.</p>
<p>Then the length of <span class="math-container">$PQ$</span> is in between the lengths of <span class="math-container">$PQ_1$</span> and <span class="math-container">$PQ_2$</span>.</p>
<p><a href="https://i.stack.imgur.com/bVoba.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bVoba.png" alt="enter image description here" /></a></p>
<p>I have observed that if <span class="math-container">$P = O$</span>, then we have <span class="math-container">$PQ_1 = PQ_2 = PQ$</span>.</p>
<p>How to prove the general case. Any hints will be useful.</p>
<hr />
<p>Source: Homework Assignment.</p>
| Glorious Nathalie | 948,761 | <p>Let <span class="math-container">$Q_1 = (r, 0), Q_2 = (-r, 0), Q = r (\cos \theta, \sin \theta), P = (a, 0)$</span> where <span class="math-container">$ 0 \le a \le r $</span></p>
<p><span class="math-container">$PQ_1 = r - a$</span></p>
<p><span class="math-container">$PQ_2 = a + r$</span></p>
<p><span class="math-container">$PQ = \sqrt{ (r \cos \theta - a)^2 + r^2 \sin^2 \theta } = \sqrt{ r^2 + a^2 - 2 a r \cos \theta } $</span></p>
<p>Since</p>
<p><span class="math-container">$$ -1 \le \cos \theta \le 1 $$</span></p>
<p>then by multiplying through by <span class="math-container">$(-2 a r) $</span> we get</p>
<p><span class="math-container">$$ -2 a r \le - 2 a r \cos \theta \le 2 a r $$</span></p>
<p>and further</p>
<p><span class="math-container">$$ a^2 + r^2 - 2 a r \le a^2 + r^2 - 2 a r \cos \theta \le a^2 + r^2 + 2 a r $$</span></p>
<p>Applying the square root</p>
<p><span class="math-container">$$ r - a \le PQ \le a + r $$</span></p>
<p>Thus</p>
<p><span class="math-container">$$ P Q_1 \le PQ \le PQ_2 $$</span></p>
|
2,604,660 | <p>In a box there are $16$ ice-creams: $6$ lemon flavor,$4$ mint flavor and $6$ strawberry flavor.When we extract two ice-creams,what's the probability of getting two different flavors,given that at least one is strawberry flavor.</p>
<p>That's my solution : </p>
<p>$P(A)$ = "Different flavors" = $\frac{\binom{16}{2}}{16!}$</p>
<p>$P(B^c)$ = "At least one is strawberry flavor" = $1 - \frac{\binom{10}{2}}{16!}$</p>
<p>So...We want $P(A|B^c)$ using conditional probability, where I go wrong?</p>
| Masacroso | 173,262 | <p>I dont see exactly what you did (edit: I will provide a complete anwer to clarify things). </p>
<p>There is a way to do it using the law of total probability: assuming that each ice-cream have the same probability to be taken then it is the probability that the first one will be some flavor and that the second will be different, that is:</p>
<p>$$P[\text{ two distinct flavor }]=\\=\sum_{x\in\{\text{ flavors }\}}P[\text{ first extraction is }x\text{ and second extraction is not } x]$$</p>
<p>Using formal notation: let $X_1$ the random variable that represent choose the first ice-cream, and $X_2$ the random variable that choose the second. The sample space is</p>
<p>$$\Omega:=\{\text{ lemon , strawberry, mint }\}$$</p>
<p>Now we assign $X_j(\text{ lemon }):=1$, $X_j(\text{ strawberry }):=2$, $X_j(\text{ mint }):=3$, for $j=1,2$. Thus the above can be rewritten formally as</p>
<p>$$P[X_1\neq X_2]=\sum_{x=1}^3P[X_1=x,X_2\neq x]=\sum_{x=1}^3P[X_2\neq x|X_1=x]P[X_1=x]\\=\frac{4+6}{15}\cdot\frac6{16}+\frac{4+6}{15}\cdot\frac6{16}+\frac{6+6}{15}\cdot\frac4{16}\\=\frac1{15\cdot 16}(2\cdot 6\cdot10+4\cdot 12)=\frac7{10}$$</p>
<hr>
<p>For the second probability you just need to consider two cases and add up: if the first extracted ice-cream is of strawberry the probability that the second is of different flavor, and the probability that the first extracted ice-cream is not strawberry, that is</p>
<p>$$P[\text{ second is not strawberry considering the first one is }]+\\+P[\text{ first is not strawberry considering that the second will be }]$$</p>
<p>or alternatively using conditional probability notation</p>
<p>$$P[\text{ two distinct flavors }|\text{ first one is strawberry }]+\\+P[\text{ two distinct flavors }|\text{ second one is strawberry }]$$</p>
<p>Using the above formal notation this can be written as</p>
<p>$$P[X_1\neq X_2|X_1=2]+P[X_1\neq X_2|X_2=2]\\=P[X_1=2|X_2\neq 2]+P[X_1\neq 2|X_2=2]\\=\frac{P[X_1=2,X_2\neq 2]}{P[X_2\neq 2]}+\frac{P[X_1\neq 2,X_2=2]}{P[X_2=2]}\\=P[X_2\neq 2|X_1=2]\frac{P[X_1=1]}{P[X_2\neq 2]}+P[X_2= 2|X_1\neq2]\frac{P[X_1\neq1]}{P[X_2= 2]}$$</p>
<p>where</p>
<p>$$P[X_2\neq 2]=P[X_2\neq 2|X_1=2]P[X_1=2]+P[X_2\neq 2|X_1\neq 2]P[X_1\neq 2]\\=\frac{6+4}{15}\cdot\frac6{16}+\frac{6+4-1}{15}\frac{4+6}{16}=\frac1{15\cdot 16}(60+90)=\frac58$$</p>
<p>I leave the other calculations to you.</p>
|
26,736 | <p>I am still a newbie to <em>Mathematica</em>, so I am still getting used to the syntax and logic of the program - but I've been trying to do the following 3D plot:</p>
<pre><code>g[x_, t_] := y /. FindRoot[y == Tanh[(2*y + x)/t], {y, x, t}];
p1 = ContourPlot3D[{z ==
Sech[(2 g[x, t] + x)/t]^2/(t - 2 Sech[(2 g[x, t] + x)/t]^2)}, {t,
0.001, 3}, {x, -1.5, 1.5}, {z, 0, 2.5}, Mesh -> None,
ContourStyle -> Opacity[0.25],
ColorFunction -> ColorData["ThermometerColors"]];
Rasterize[
Show[p1, PlotRange -> {{0.001, 3}, {-1.5, 1.5}, {0, 2.5}},
ViewPoint -> {0.9*Pi, -0.45*Pi, 1.05*Pi}], ImageResolution -> 1200]
</code></pre>
<p>The <code>ContourPlot3D</code> of the transcendental equation <code>y == Tanh[(2*y + x)/t]</code> itself comes out perfect - but after some 10-15 mins of evaluating z, I seem to be getting the right 3D surface from the 2D solutions I've done (cuts through the 3D surface I am trying to plot), but the plot is filled with artifacts. I am the tinkering type, so I tend to need to see things work right a few times, before things like the syntax and logic of <em>Mathematica</em> start to stick in my memory - and I can't figure out this one. What am I doing wrong?</p>
| Silvia | 17 | <p>Michael's fantastic answer is a general way to accomplish implicit plots. But for the very problem in OP, the method I used in <a href="https://mathematica.stackexchange.com/a/23565/17">this</a> and <a href="https://mathematica.stackexchange.com/a/27235/17">this</a> answers is applicable, too.</p>
<p>First we define the functions we want to plot:</p>
<pre><code>Clear[mainFunc]
constraintEq = y == Tanh[(2 y + x)/t];
mainFunc[x_, y_, t_] := Sech[(2 y + x)/t]^2/(t - 2 Sech[(2 y + x)/t]^2)
</code></pre>
<p>As we can see from a sketch plot, there is a discontinuity in <code>mainFunc</code>, so we need to exclude the area near it when generating the feasible set. Unfortunately, <code>ContourPlot3D</code> does not support <code>Exclusions</code> option, so we have to simulate it with <code>RegionFunction</code>. Also we should note that $y$ dimension will not present in the final plot, which is in the $txz$ space, so it would be informative to represent it by surface color.</p>
<pre><code>paraRegion =
ContourPlot3D[Evaluate@constraintEq,
{x, -1.5, 1.5}, {y, -1, 1}, {t, .01, 3},
RegionFunction -> Function[{x, y, t},
Abs[mainFunc[x, y, t]^-1] > .2], (* Excluding the discontinuity area *)
MeshFunctions -> {#3 &, mainFunc[#1, #2, #3] &}, (* Meshes in t and z directions *)
Mesh -> {10, Range[-6, 6, .5]},
ColorFunction -> Function[{x, y, t}, ColorData["Rainbow"][y]],
]
</code></pre>
<p><img src="https://i.stack.imgur.com/KYZs8.png" alt="plot in xyt"></p>
<p>Now we define the transformation function which will transform the feasible set from $xyt$ space to $txz$ space:</p>
<pre><code>Clear[plotTransformation]
plotTransformation[paraRegion_, plotTransOpts__] :=
Graphics3D[
Cases[paraRegion,
GraphicsComplex[pts_, others_, opts1___, VertexNormals -> vn_, opts2___] :>
GraphicsComplex[
Function[{x, y, t}, {t, x, mainFunc[x, y, t]} ] @@@ pts,
others, opts1, opts2],
∞][[1]],
Axes -> True, plotTransOpts]
</code></pre>
<p>Transform the plot as our wish:</p>
<pre><code>plotTransformation[paraRegion,
AxesLabel -> (Style[#, 20, Bold] & /@ {t, x, z}),
BoxRatios -> {3, 2, 2}, PlotRange -> {-3, 3},
Lighting -> {{"Ambient", White}}]
</code></pre>
<p><img src="https://i.stack.imgur.com/HpvHJ.png" alt="plot in txz"></p>
|
1,369,428 | <p>I am trying to solve the following equation;
$$\int_{-1}^{1}e^{i(x+a\cos x)} \, \mathrm{d}(\cos x)$$ or $$\int_{0}^{\pi}e^{i(x+a\cos x)} \sin x \, \mathrm{d}x$$</p>
<p>I tried this in <em>Wolfram Alpha</em>, but it says that answer cannot be obtained.</p>
| Mark Viola | 218,419 | <p>Let $F(a)$ be defined as</p>
<p>$$F(a)=\int_0^{\pi}e^{i(x+a\cos x)}\sin x\,dx \tag 1$$</p>
<p>Now, it is easy to show that the real part of $F$ is identically $0$. To do this, we write the real part of $(1)$ as </p>
<p>$$\text{Re}\lbrace F(a)\rbrace=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{\pi/2}^{\pi}\cos(x+a\cos x)\sin x\,dx \tag 2$$</p>
<p>Then, enforcing the substitution $x\to \pi -x$ in the second integral on the right-hand side of $(2)$ reveals</p>
<p>$$\begin{align}
\text{Re}\lbrace F(a)\rbrace&=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{0}^{\pi/2}\cos(\pi-x+a\cos (\pi-x))\sin (\pi-x)\,dx\\\\
& =\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx-\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx\\\\
&=0 \tag 3
\end{align}$$</p>
<p>as was to be shown. We can write, therefore, $F(a)$ as </p>
<p>$$\begin{align}
F(a)&=i\int_0^{\pi}\sin(x+a\cos x)\sin x\,dx \\\\
&=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)+\sin(x)\cos(x)\sin(a\cos x)\,\right)dx\tag 4
\end{align}$$</p>
<p>where we arrived at the right-hand side of $(4)$ using a parallel development that led to $(3)$.</p>
<hr>
<p>The second integral on the right-hand side of $(4)$ is easily evaluated. Substituting $x\to \cos(x)$ and integrating by parts reveals that </p>
<p>$$\begin{align}
i\int_0^{\pi}\sin(x)\cos(x)\sin(a\cos x)\,dx&=i2\int_{0}^1 x\sin(ax)\,dx\\\\
&=i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 5
\end{align}$$</p>
<p>Using $(5)$ in $(4)$ reduces $F(a)$ to </p>
<p>$$\begin{align}
F(a)&=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)\,dx\\\\
&+i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 6
\end{align}$$</p>
<hr>
<p>In order to evaluate the integral in $(6)$ we will enforce two successive substitutions. First, we let $a\cos x\to x$. This yields</p>
<p>$$i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)=\frac{i2}{a}\int_0^a
\sqrt{1-\left(\frac{u}{a}\right)^2} \cos (x)\,dx$$</p>
<p>where we exploited the fact that the integrand is an even function. Next, we enforce the substitution $x\to a\sin x$ which reveals </p>
<p>$$\begin{align}
i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)&=i\int_0^{\pi}\cos^2(x)\cos(a\sin x)\,dx \tag 7\\\\
&=i\int_0^{\pi}\cos(a\sin x)\,dx-i\int_0^{\pi}\sin^2(x)\cos(a\sin x)\,dx\\\\
&=i\pi J_0(a)+i\pi J_1'(a)\\\\
&=i\pi\frac{J_1(a)}{a} \tag 8
\end{align}$$</p>
<p>where $J_0(a$) and $J_1(a)$ are the <a href="https://en.wikipedia.org/wiki/Bessel_function">Bessel Function</a> of the first kind and order zero and one, respectively, while $J_1'(a)$ is the first derivative of $J_1$. In deriving $(8)$ we relied on Equations $(B5)$, $(B6)$, and $(B8)$ in the section on Bessel Functions.</p>
<p>Aside, we exploited the symmetry of $\cos^2(x)\cos(a\sin x)$ around $\pi/2$ to arrive at $(7)$.</p>
<hr>
<p><strong>FINAL RESULT</strong></p>
<p>Putting $(6)$ and $(8)$ together finally yields</p>
<p>$$\bbox[5px,border:2px solid #C0A000]{F(a)=i2\frac{-a\cos(a)+\sin(a)}{a^2}+i\pi\frac{J_1(a)}{a}} $$</p>
<p>which agrees with the empirically obtained hypothesis of @ClaudeLeibovici and @JoshBroadhurst!</p>
<hr>
<p><strong>BESSEL FUNCTION ASIDE</strong></p>
<p>We note an integral representation for the first kind Bessel Function of order $n$</p>
<p>$$\begin{align}
J_n(a)&=\frac{1}{\pi}\int_0^{\pi}\cos(nx-a\sin x)dx \\\\
&=\frac{1}{\pi}\int_0^{\pi} \left(\cos (nx)\cos(a\sin x)+\sin(nx)\sin(a\sin x)\right) \,dx \tag{B1}
\end{align}$$</p>
<p>along with recurrence relationships</p>
<p>$$J_n(a)=\frac{a}{2n}\left(J_{n-1}(a)+J_{n+1}(a)\right) \tag{B2}$$</p>
<p>and</p>
<p>$$J_n'(a)=\frac12
\begin{cases}
J_{n-1}(a)-J_{n+1}(a),&n\ne 0\\\\
-J_1(a),&n = 0 \tag{B4}
\end{cases}$$</p>
<p>Setting $n=0$ in $(B1)$ yields</p>
<p>$$\bbox[5px,border:2px solid #C0A000]{J_0(a)=\frac{1}{\pi}\int_0^{\pi} \left(\cos(a\sin x)\right) \,dx} \tag{B5}$$</p>
<p>Taking the two derivatives of $J_0(a)$ and multiplying by $-1$ and using $(B4)$ yields</p>
<p>$$\bbox[5px,border:2px solid #C0A000]{J_1'(a)=-\frac{1}{\pi}\int_0^{\pi} \sin^2(x)\cos(a\sin x)\,dx}\tag{B6}$$</p>
<p>Note that using $(B3)$ and $(B4)$, we can eliminate $J_{n+1}$ and write $J_n$ as</p>
<p>$$J_n(a)=\frac{a}{n}\left(J_{n-1}(a)-J_n'(a)\right) \tag{B7}$$</p>
<p>Setting $n=1$ in $(B7)$ and rearranging we obtain</p>
<p>$$\bbox[5px,border:2px solid #C0A000]{\frac{J_1(a)}{a}=J_0(a)+J_1'(a)} \tag{B8}$$</p>
<hr>
|
176,240 | <p>I <a href="https://math.stackexchange.com/questions/2832912/sum-infty-infty-frac-exp-n21-4n2-in-closed-form/2832965#2832965">came across</a> a possible bug in Mathematica's evaluation of integrals that I thought I'd share. Consider the code:</p>
<pre><code>zmax = 4;
a = Integrate[Exp[-x^2] Abs[Sin[x]], {x, -z , z }] /. z -> zmax
b = NIntegrate[Exp[-x^2] Abs[Sin[x]], {x, -zmax, zmax}]
a - b
</code></pre>
<p>If <code>zmax</code> is smaller than $\pi$ it evaluates close to $0$ (to the precision limit) while for <code>zmax</code> greater than $\pi$ it evaluates to $-4.50908\cdot 10^{-6}$ (which happens to be roughly $4\cdot \int_{\pi}^{\infty}e^{-x^2}\sin(x){\rm d}x$). The condition $z< \pi$ is not listed in Matematica's (11.2.0.0) analytical expression for the integral <code>a</code> above:</p>
<pre><code>ConditionalExpression[( Sqrt[\[Pi]] (2 Erfi[1/2] - Abs[Sin[z]] Csc[z] (Erfi[1/2 - I z] + Erfi[1/2 + I z])))/(2 E^(1/4)), Re[z] >= 0 && Im[z] == 0]
</code></pre>
<p>Anybody knows why this happens? If I were to guess, based on the result above, it looks like Mathematica uses the symmetry to rewrite the integral as $2$ times the integral over $[0,z]$ and then for some reason removes the absolute value over $\sin(x)$ (which is allowed only if $z<\pi$), but forgets to mention it.</p>
| b3m2a1 | 38,205 | <p>I know halirutan has already provided an answer, but here's a nice way to deal with GIF import in a frame-wise manner (or really in chunks of frames).</p>
<p>Mathematica does a very dumb thing when it reads GIFs. It calls the function <code>System`Convert`CommonGraphicsDump`IEImageRead</code> which, if intelligent, would merely process the requested frames. Instead it always takes <em>every single frame</em>, which is returned as an <code>Image3D</code> calls <code>Image3DSlices</code>.</p>
<p>Why it does this is truly beyond me, but you can check it like so:</p>
<pre><code>Block[{Image3DSlices = Echo[Head[#]] &},
Import[gif, {"ImageList", 5}]
]
Image3D
Import::fmterr: Cannot import data as GIF format.
$Failed
</code></pre>
<p>Instead of using that we'll use the built in <code>"GIFTools`"</code> library which is intelligent about this. The core functions we'll want are these:</p>
<pre><code>{
GIFTools`Private`$ReadOneFrame,
GIFTools`Private`$ReadPalettes,
GIFTools`Private`$ReadAllFrames,
GIFTools`Private`$ReadRasterBits,
GIFTools`Private`$ReadFileMetadata,
GIFTools`Private`$ReadFrameMetadata,
GIFTools`Private`$ReadGlobalPalette
}
</code></pre>
<p>The most useful parts here are <code>$ReadFileMetadata</code> and <code>$ReadOneFrame</code>, but all of these can be used to dramatically outperform the default <code>Import</code> implementation.</p>
<p>Now we can effectively map a function over blocks of frames extracted using <code>Range</code>.</p>
<p>Here's the full code. Most of it is just making the interface nice, implementing protections and validations, etc.</p>
<pre><code>gifBlockMapLoadLib[] :=
(
Replace[
If[DownValues@
System`Convert`CommonGraphicsDump`ImageReadMetadataGIF == {},
Quiet@Import[
"https://i.imgur.com/7Wkns4d.gif", {"GIF", "Elements"}];
];
DownValues@System`Convert`CommonGraphicsDump`ImageReadMetadataGIF,
{} :>
Throw[
Failure["BadGIFInit", <|
"MessageTemplate" -> "GIF tools not loaded"|>], "GIFBlockMap"]
];
);
gifBlockMapGetFile[gif_] :=
With[{base = ExpandFileName@Replace[gif, File[f_] :> f]},
If[! FileExistsQ@base,
Throw[
Failure["BadGIF",
<|"MessageTemplate" -> "GIF file `` does not exist",
"MessageParameters" -> {gif}|>
],
"GIFBlockMap"
],
base
]
];
gifBlockMapGetImageCount[gif_] :=
Replace[
Quiet@
System`Convert`CommonGraphicsDump`ImageReadMetadataGIF[gif,
"ImageCount"],
{
{i_Integer} :> i,
_ :>
Throw[
Failure["BadGIF",
<|"MessageTemplate" -> "`` is not a valid GIF file",
"MessageParameters" -> {gif}|>
],
"GIFBlockMap"
]
}
];
gifBlockMapGetBlockSpec[frameCount_, block_] :=
Module[
{
start, stop, step
},
If[! IntegerQ@frameCount,
Throw[
Failure["BadGIF",
<|"MessageTemplate" -> "GIF frame count `` is not an integer",
"MessageParameters" -> {frameCount}|>
],
"GIFBlockMap"
]
];
If[block[[3]] <= 0,
Throw[
Failure["BadBlock",
<|"MessageTemplate" -> "Block step `` must be greater than 0",
"MessageParameters" -> {block[[3]]}|>
],
"GIFBlockMap"
]
];
{start, stop, step} =
Replace[List @@ block,
{
All -> frameCount,
i_Integer?Negative :>
frameCount - i,
i_Integer?(# > frameCount &) :>
Throw[
Failure["BadBlock",
<|
"MessageTemplate" ->
"Block parameter `` is larger than frame count ``",
"MessageParameters" -> {i, frameCount}|>
],
"GIFBlockMap"
],
Scaled[s_] :>
Max@{Ceiling[s*frameCount], 1}
},
1
];
If[start >= stop,
Throw[
Failure["BadBlock",
<|"MessageTemplate" ->
"Block start `` must be smaller than block end ``",
"MessageParameters" -> {start, stop}|>
],
"GIFBlockMap"
]
];
{start, stop, step}
];
gifBlockMapRead["GIF", frame_, ___][gif_, ___] :=
Replace[
Switch[
frame,
_Integer,
{GIFTools`Private`$ReadOneFrame[gif, frame]},
{__Integer},
GIFTools`Private`$ReadOneFrame[gif, #] & /@ frame,
All,
GIFTools`Private`$ReadAllFrames[gif],
_,
$Failed
],
Except[{__Image}] :>
Throw[
Failure["BadFrames",
<|
"MessageTemplate" ->
"Can't read frames `` from ``",
"MessageParameters" -> {frame, gif}
|>
],
"GIFBlockMap"
]
];
gifBlockMap[
func_,
gifFile : (_String | _File),
mapFunc : Map | Scan : Map,
blockSpec :
Span[
_Integer | Scaled[_?(0 <= # < 1 &)],
_Integer | Scaled[_?(0 < # <= 1 &)] | All,
_Integer?Positive | Scaled[_?(0 <= # <= 1 &)] | All
],
sampling : _Integer?Positive : 1
] :=
Function[Null, Catch[#, "GIFBlockMap"], HoldFirst ]@
Block[
{
System`Convert`CommonGraphicsDump`IEImageRead = gifBlockMapRead
},
Module[
{
gif,
frameCount,
start,
end,
step,
curFrame,
oldFrame,
frameRange
},
gifBlockMapLoadLib[];
gif = gifBlockMapGetFile[gifFile];
frameCount = gifBlockMapGetImageCount[gif ];
{start, end, step} =
gifBlockMapGetBlockSpec[frameCount, blockSpec];
curFrame = start;
Switch[mapFunc, Map, Table, Scan, Do][
oldFrame = curFrame;
curFrame = 1 + Min@{curFrame + step, end};
frameRange = Range[oldFrame, curFrame - 1, sampling];
If[Length@frameRange > 0,
func@Import[gif, {"GIF", "ImageList", frameRange}],
Nothing
],
Ceiling[frameCount/step]
]
]
];
gifBlockMap[
func_,
gifFile : (_String | _File),
mapFunc : Map | Scan : Map,
blockSpec : _Integer?Positive | Scaled[_?(0 < # < 1 &)] | All : 10,
sampling : _Integer?Positive : 1
] :=
gifBlockMap[
func, gifFile, mapFunc,
1 ;; All ;; blockSpec, sampling
]
</code></pre>
<p>Then we can see this in action:</p>
<pre><code>gif = URLDownload["https://i.imgur.com/7Wkns4d.gif"];
gifBlockMap[Blend, gif, Map, 10] // GraphicsRow[#, ImageSize -> 800] &
</code></pre>
<p><a href="https://i.stack.imgur.com/yi7Y8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/yi7Y8.png" alt="raccoon"></a></p>
<p>It performs okay (at minimum it cuts down our memory consumption):</p>
<pre><code>Blend /@ Partition[Import[gif, "ImageList"], 10] //
MaxMemoryUsed // AbsoluteTiming
{0.253601, 41266472}
gifBlockMap[Blend, gif, 10] // MaxMemoryUsed // AbsoluteTiming
{0.807644, 11808856}
</code></pre>
<p>We're just under four times slower and use bit under four times less memory.</p>
<p>Amusingly, this outperforms the direct <code>"ImageList</code> import, since we're being less general and don't use <code>Image3D</code> and things:</p>
<pre><code>Import[gif, "ImageList"] // MaxMemoryUsed // AbsoluteTiming
{0.137272, 41266400}
gifBlockMap[Identity, gif, Map, All][[1]] //
MaxMemoryUsed // AbsoluteTiming
{0.118931, 20667552}
</code></pre>
<p>Even better though is to just directly call <code>$ReadAllFrames</code>:</p>
<pre><code>Image3DSlices@GIFTools`Private`$ReadAllFrames[First@gif] //
MaxMemoryUsed // AbsoluteTiming
{0.034736, 41164192}
</code></pre>
<p>Or if you can work with the <code>Image3D</code> object instead:</p>
<pre><code>GIFTools`Private`$ReadAllFrames[First@gif] //
MaxMemoryUsed // AbsoluteTiming
{0.017178, 20520712}
</code></pre>
<p>On the other hand for a big GIF this method makes things more tractable:</p>
<pre><code>bigGIF =
URLDownload["https://www.wolframcloud.com/objects/b3m2a1/chem/gradient_descent_walkers.gif"];
tm1 =
gifBlockMap[Head, bigGIF, 35] // MaxMemoryUsed // AbsoluteTiming
{8.12303, 42231864}
tm2 =
Head /@ Partition[Import[bigGIF, "ImageList"], UpTo[35]] //
MaxMemoryUsed // AbsoluteTiming
{1.83082, 470784832}
MapThread[
((# - #2)/Max@{##}) &,
{tm1, tm2}
] // N
{0.774613, -0.910295}
</code></pre>
<p>It takes 77% longer but use 91% less memory</p>
|
1,085,620 | <p>Which number between the interval 7902 & 7918 can be divided without remainder only to itself and to the number 1?</p>
| egreg | 62,967 | <p>The requirement is that the number is prime, so this excludes the even numbers, leaving
$$
7903;7905;7907;7909;7911;7913;7915;7917.
$$
We can remove $7905$ and $7915$ because they're divisible by $5$; $7911$ and $7917$ are divisible by $3$. Moreover $7903$ is divisible by $7$ and $7909$ is divisible by $11$; note that the next odd multiples of $7$ and $11$ are larger than $7917$. So we're left with
$$
7907; 7913.
$$
At this point there's nothing else to do than divisions. Since we have already tried $2$, $3$, $5$, $7$ and $11$, the list of primes to check is
$$
13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 53; 59; 61; 67; 71; 73; 79; 83.
$$
The next one is $89$, but since $89^2=7921>7913$, if a prime bigger than $83$ divides one of the numbers, also one not larger than $83$ would.</p>
<p>You can save some work by noticing that when you divide $7907$ by $13$ is $3$, so $7913=7907+6$ isn't divisible by $13$ either. Similarly, $7907\bmod 17=2$ (this denotes the remainder), so $7913$ isn't divisible by $17$ either. Go on until you find a prime divisor for one of the numbers and continue with the other alone.</p>
<p>Note that there's no guarantee that in a given interval of numbers you'll find a prime.</p>
<p>This is what Pari-GP says:</p>
<pre><code>? isprime(7907)
%1 = 1
? isprime(7913)
%2 = 0
? factor(7913)
%3 =
[ 41 1]
[193 1]
</code></pre>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| Atmos | 516,446 | <p>Consider the set $S$ formed of $x=(a,c) \in \mathbb{R}^2$ such as $\left|a\right|\geq \left|c\right|$.
Then for $b \in \mathbb{R}$
$$
bx=\left(ba,bc\right)
$$
and $\left|ba\right| \geq \left|bc\right|$.</p>
<p>However, with $(4,1)$ and $(-3,2)$ which are both in the set
$$
(4,1)+(-3,2)=(1,3)
$$
which is not in the set so it is not a vector space.</p>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| M. Winter | 415,941 | <p>Choose any two linearly independent vectors $v,w$. Then the set</p>
<p>$$\{bv\mid b\in\Bbb R\}\cup\{bw\mid b\in\Bbb R\}.$$</p>
<p>satisfies your condition, but is not a vector space since $v+w$ is not contained (vector spaces must contain all sums of their vectors).</p>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| mathmandan | 198,422 | <p>Let $S$ be the union of the first and third quadrants (along with the $x$- and $y$-axes, let's say). Scaling by a real number $r$ will either keep each vector in its same quadrant (if $r>0$), or send the first and third quadrants to each other (if $r<0$), or send all vectors to the origin (if $r=0$). So $S$ is closed under scalar multiplication.</p>
<p>However, notice that
$$
(-2, -1) + (1, 2) = (-1, 1)
$$
which is in the second quadrant, so $S$ is not closed under vector addition.</p>
<p>Extra note: for an example of something closed under addition but not under scalar multiplication, take just the first quadrant.</p>
|
1,726,540 | <p>How should $x^{\frac{1}{x}}$ be differentiated? I know the answer is
$$\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$$</p>
<p>but I do not understand how to get there.</p>
<h2>Attempt at solution.</h2>
<p>I believe the following is true:</p>
<p>$$
\begin{aligned}\frac{d}{dx}x^u&=ux^{u-1}\cdot u^\prime\\
\frac{d}{dx}a^x&=a^x\cdot\ln(a)
\end{aligned}$$
but I don't know what to do when both the base and the exponent are functions of $x$.</p>
| C Marius | 327,443 | <p>$x^{\frac{1}{x}} = e^{\frac{1}{x} ln(x)}$ therefore $\left(x^{\frac{1}{x}}\right)' = \left(e^{\frac{1}{x} ln(x)}\right)' = e^{\frac{1}{x} ln(x)}\left(\frac{1}{x} ln(x)\right)' = e^{\frac{1}{x} ln(x)} \left(\frac{-1}{x^2}ln(x) + \frac{1}{x}\frac{1}{x}\right) = e^{\frac{1}{x} ln(x)}\left(\frac{1-ln(x)}{x^2}\right) = x^{\frac{1}{x}}x^{-2}\left(1-ln(x)\right) = \frac{\left(1-ln(x)\right)}{x^{2 - \frac{1}{x}}}$ </p>
|
1,726,540 | <p>How should $x^{\frac{1}{x}}$ be differentiated? I know the answer is
$$\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$$</p>
<p>but I do not understand how to get there.</p>
<h2>Attempt at solution.</h2>
<p>I believe the following is true:</p>
<p>$$
\begin{aligned}\frac{d}{dx}x^u&=ux^{u-1}\cdot u^\prime\\
\frac{d}{dx}a^x&=a^x\cdot\ln(a)
\end{aligned}$$
but I don't know what to do when both the base and the exponent are functions of $x$.</p>
| rahul joshi | 351,574 | <p>its very simple to get answer </p>
<p>Let $u=x^{\frac{1}{x}}$</p>
<p>take logarithm of both sides</p>
<p>$log\ u=log(x^{\frac{1}{x}})$</p>
<p>$log\ u=\frac{1}{x}log(x)$</p>
<p>differentiation wrt 'x'
$$\frac{d}{dx}log\ u=\frac{d}{dx}\frac{1}{x}log(x)$$
$$to\ differentiate\ this,\ use\ chain\ rule\ in\ LHS\ and\ product\ rule\ in\ RHS$$
$$\Rightarrow \frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{x} \cdot \frac{d}{dx}(log\ x)+(log\ x) \cdot \frac{d}{dx}\Big(\frac{1}{x}\Big)$$</p>
<p>$$\Rightarrow \frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{x} \cdot \frac{1}{x}+log\ x\Big(\frac{-1}{x^2}\Big)$$
$$\Rightarrow \frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{x^2}-\frac{log\ x}{x^2}$$
$$\Rightarrow \frac{1}{u}\cdot \frac{du}{dx}=\frac{1}{x^2}\Big[1-log\ x\Big]$$
$$\Rightarrow \frac{du}{dx}=u \cdot \frac{1}{x^2}\Big[1-log\ x\Big]$$
Put $$u=x^{\frac{1}{x}}$$
$$\Rightarrow \frac{d}{dx}\big(x^{\frac{1}{x}}\big)=x^{\frac{1}{x}} \cdot \frac{1}{x^2}\Big[1-log\ x\Big]$$
$$\Rightarrow \frac{d}{dx}\big(x^{\frac{1}{x}}\big)=x^{\frac{1}{x}-2}\Big[1-log\ x\Big]$$</p>
|
1,726,540 | <p>How should $x^{\frac{1}{x}}$ be differentiated? I know the answer is
$$\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$$</p>
<p>but I do not understand how to get there.</p>
<h2>Attempt at solution.</h2>
<p>I believe the following is true:</p>
<p>$$
\begin{aligned}\frac{d}{dx}x^u&=ux^{u-1}\cdot u^\prime\\
\frac{d}{dx}a^x&=a^x\cdot\ln(a)
\end{aligned}$$
but I don't know what to do when both the base and the exponent are functions of $x$.</p>
| Community | -1 | <p>There is a nice trick that can rescue you: differentiate all the functions obtained by considering all instances of $x$ but one as constants, and take the sum.</p>
<p>This gives ($X$ is a constant)</p>
<p>$$\left(x^{1/X}\right)'=\frac1Xx^{1/X-1}$$
and
$$\left(X^{1/x}\right)'=\ln X\left(\frac1x\right)'X^{1/x}=-\ln X\frac{X^{1/x}}{x^2}.$$</p>
<p>Then restoring the variable and regrouping,</p>
<p>$$\left(x^{1/x}\right)'=\frac1xx^{1/x-1}-\ln x\frac{x^{1/x}}{x^2}=(1-\ln x)x^{1/x-2}.$$</p>
<hr>
<p>Alternatively, if you are familiar with functions of several variables, you can replace the occurrences of $x$ by independent variables and work with total differentials.</p>
<p>$$x^{1/x}\to z^{1/y},$$ then</p>
<p>$$d\left(z^{1/y}\right)=\frac1yz^{1/y-1}\,dz-\ln z\frac{z^{1/y}}{y^2}\,dy.$$</p>
<p>Now you can evaluate the differential at $x=z=y$, and</p>
<p>$$d\left(x^{1/x}\right)=(1-\ln x)\frac{x^{1/x-2}}{x^2}\,dx.$$</p>
|
3,998,484 | <p>For the purpose of clarity, I do not want anybody to prove this statement, I am just looking to get some help translating it into the contrapositive using <em>symbolic logic</em>. Right now I have the following translation:</p>
<hr />
<p>Original Statement:</p>
<p><span class="math-container">$$(\exists k \in \mathbb{Z} \quad n = 2k) \quad \land \quad (\exists m \in \mathbb{Z} \quad n+1 =m^2) \quad \Rightarrow \quad \text{n is divisible by 8}$$</span></p>
<p>Contrapositive Statement:</p>
<p><span class="math-container">$$ \text{n is not divisible by 8} \quad \Rightarrow \quad (\exists k \in \mathbb{Z} \quad n = 2k+1) \quad \lor \quad (\forall m \in \mathbb{Z} \quad n+1 \neq m^2) $$</span></p>
<hr />
<p>A couple of points of clarification:</p>
<ul>
<li>I wasn't sure how to write <span class="math-container">$n$</span> is divisible by 8. Perhaps it could have been stated as <span class="math-container">$\exists p \in \mathbb{Z} \quad n = 8p$</span>?</li>
<li>For the negation of <span class="math-container">$\exists k \in \mathbb{Z} \quad n = 2k$</span>, I know that typically you might want to reverse the <span class="math-container">$\exists$</span> to an <span class="math-container">$\forall$</span>, but I think what I wrote makes sense since the negation of even numbers is odd numbers.</li>
</ul>
<p>Eventually I will try to prove this statement by contrapositive, but I want to get some practice in with translating mathematical statements into a more formal structure.</p>
<p>My question(s) are following:</p>
<ol>
<li>Is the original translation correct? How might it be improved?</li>
<li>How could I more formally translate '<span class="math-container">$n$</span> is divisible by 8'?</li>
<li>Is the contrapositive stated correctly?</li>
</ol>
<p>I also am not able to use <span class="math-container">$mod$</span> notation, <span class="math-container">$a|b$</span> notation, or anything that is unique to Abstract Algebra.</p>
| open problem | 876,065 | <p>You have fixed the homomorphism proof.</p>
<p>Continuing from there:</p>
<p>α((a,b))=e</p>
<p>Whenever 3b = e</p>
<p>This is true for b=0,9,18.</p>
<p>So <span class="math-container">$ker(\alpha) = Z_{9}\times (0,9,18)$</span></p>
|
332,130 | <p>Here is my proof, I would appreciate it if someone could critique it for me:</p>
<p>To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.</p>
<hr>
<p>Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$</p>
<p>If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.</p>
<hr>
<p>To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement?</p>
| Brian M. Scott | 12,042 | <p>Your argument is fine, if a little more roundabout than necessary: both directions can be done easily with direct proofs. </p>
<p>Suppose first that $\wp(A)\subseteq\wp(B)$. $A\subseteq A$, so $A\in\wp(A)\subseteq\wp(B)$, so $A\in\wp(B)$, and hence $A\subseteq B$.</p>
<p>Now suppose that $A\subseteq B$. Then for any $X\in\wp(A)$ we have $X\subseteq A\subseteq B$, so $X\subseteq B$, and therefore $X\in\wp(B)$. Thus, $\wp(A)\subseteq\wp(B)$.</p>
|
750,906 | <p>I would like to show my Calc I class that $f(x)=f(a)+f'(a)(x-a)+(f''(c)/2)(x-a)^2$ for some $c$ in $(a,x)$ (for $f$ smooth). Just from the form of the statement, it seems as though this should be possible using only the ordinary MVT, with no integrals. However, I can't seem to get the symbols to behave properly; all my attempts have ended with remainder term $f''(c)(d-a)(x-a)$ for some $d$ in $(a,x)$, which is also correct but not the formula that I want. Help me out or let me know if what I seek is unreasonable. Thanks.</p>
| solstafir | 142,542 | <p>Let $$g(x)=f(b)-f(x)-(b-x)f'(x)-\left(\frac{b-x}{b-a}\right)^2[f(b)-f(a)-(b-a)f'(a)]$$</p>
<p>Then $$g'(x)=\frac{2(b-x)}{(b-a)^2}[f(b)-f(a)-(b-a)f'(a)-\frac{1}{2}(b-a)^2f''(x)]$$</p>
<p>(Differentiate, then factor out $\frac{2(b-x)}{(b-a)^2}$.)</p>
<p>Applying the mean value theorem, $g(b)-g(a)=g'(c)(b-a)$ for some $a<c<b$. $g(a)=g(b)=0$, so $g'(c)$ must equal zero for some $a<c<b$. Therefore, $$f(b)=f(a)+(b-a)f'(a)+\frac{1}{2}(b-a)^2f''(c)$$ for some $a<c<b$.</p>
<p>This proof is from Hardy's "A Course of Pure Mathematics".</p>
|
750,906 | <p>I would like to show my Calc I class that $f(x)=f(a)+f'(a)(x-a)+(f''(c)/2)(x-a)^2$ for some $c$ in $(a,x)$ (for $f$ smooth). Just from the form of the statement, it seems as though this should be possible using only the ordinary MVT, with no integrals. However, I can't seem to get the symbols to behave properly; all my attempts have ended with remainder term $f''(c)(d-a)(x-a)$ for some $d$ in $(a,x)$, which is also correct but not the formula that I want. Help me out or let me know if what I seek is unreasonable. Thanks.</p>
| xpaul | 66,420 | <p>Another way to solve the problem: Let
$$F(x)=f(x)-f(a)-f'(a)(x-a)-k(x-a)^2$$
where $k$ is a constant determined later. Then $F(a)=0$. Let $F(x)=0$; then we have
$$ k=\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}. $$
Since $F(a)=F(x)=0$ and $F$ is differentiable in $(a,x)$, by Rolle's Theorem, there is $c\in(a,x)$ such that $F'(c)=0$, namely
$$ f'(c)-f'(a)-2(c-a)k=0. $$
By Langrange's MVT, there is $\bar{c}\in(a,c)$ such that
$$ f'(c)-f'(a)=f''(\bar{c})(c-a). $$
Thus we have
$$ f''(\bar{c})(c-a)-2(c-a)k=0 $$
or $k=\frac{1}{2}f''(\bar{c})$. This completes the proof.</p>
|
4,365,005 | <p>The set of all homomorphisms between two spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> is denoted as <span class="math-container">$\text{Hom}(X,Y)$</span>, the set of endomorphisms of a space <span class="math-container">$X$</span> is denoted as <span class="math-container">$\text{End}(X)$</span>,... these are standard notations, but what about the set of all maps between two spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> (without any extra structure) Are <span class="math-container">$\text{Map}(X,Y)$</span> or <span class="math-container">$\text{Fun}(X,Y)$</span> standard notations?</p>
<p>This might be an stupid question, but I want to make sure I am using the right notation, or at least the most usual notation for this spaces. Thanks for your help.</p>
| John Bentin | 875 | <p>As mentioned in the comments, the notation <span class="math-container">$Y^X$</span> is standard for the set of maps from <span class="math-container">$X$</span> to <span class="math-container">$Y$</span>. Moreover, there is no other common notation for this.</p>
|
1,234,320 | <p>It is given that $4 x^4 + 9 y^4 = 64$.</p>
<p>Then what will be the maximum value of $x^2 + y^2$?</p>
<p>I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .</p>
| Rellek | 228,621 | <p>Since you tagged Lagrange Multipliers, I will use that method.</p>
<p>Allow that $F(x,y) = 4x^4+9y^4-64$ and $G(x,y)=x^2+y^2-C$, where $C$ is treated as a constant and is the maximum you wish to find.</p>
<p>We allow that $\nabla F(x,y) = \lambda \nabla G(x,y)$ and find two new equations:</p>
<p>$16x^3 = \lambda 2x$ and
$36y^3 = \lambda 2y$</p>
<p>$\implies x^4 = \frac{\lambda ^2}{64}$ and $y^4 = \frac{\lambda ^2}{324}$.</p>
<p>Plugging those into $F(x,y)$ we find that:</p>
<p>$$\lambda = 26.6256...$$</p>
<p>We now go back to our equations after applying the gradient and solve for $x_{max}$ and $y_{max}$. We get:
$x_{max}=1.8245...$ and $y_{max}=1.21622...$</p>
<p>Plugging this into $x^2+y^2$ we obtain the maximum, $4.80727$.</p>
|
1,234,320 | <p>It is given that $4 x^4 + 9 y^4 = 64$.</p>
<p>Then what will be the maximum value of $x^2 + y^2$?</p>
<p>I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .</p>
| HK Lee | 37,116 | <p>$x^2=s,\ y^2=t \geq 0$ then $$ (2s)^2+ (3t)^2=8^2 \Rightarrow
2s=4\cos\ \theta,\ 3t=4\sin\ \theta,\ 0\leq\theta \leq
\frac{\pi}{2} $$</p>
<p>Then $$ s+t=2\cos\ \theta + \frac{4}{3}\sin\ \theta=\sqrt{2^2+
\bigg(\frac{4}{3}\bigg)^2} \sin\ (\theta+\alpha),\ 0< \alpha <
\frac{\pi}{2} $$</p>
<p>Let $\theta =\frac{\pi}{2}-\alpha$ </p>
|
2,558,619 | <p>Compute
$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt
$$
In closed form, where $\delta(t)$ is the <a href="http://mathworld.wolfram.com/DeltaFunction.html" rel="nofollow noreferrer">Dirac Delta function </a>.</p>
<p>My attempt: </p>
<p>$$
\int_{-\infty}^{\infty} t^2 \delta(\sin(t)) e^{-|t|} \mathrm dt = \int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|}\mathrm dt
$$</p>
<p>Then noting that $\sin(t)$ is zero whenever $t=n\pi$. By formula (2) and (7) in the above link,</p>
<p>\begin{align}
\int_{-\infty}^{\infty} \delta(\sin(t))t^2 e^{-|t|} \mathrm dt& = \sum_{n=-\infty}^{\infty} \frac{(n\pi)^2e^{-|n\pi|}}{|\cos(n\pi)|}
\\&
=2\pi^2\sum_{n=0}^{\infty} \frac{(n)^2e^{-n\pi}}{1}
\end{align}</p>
<p>However , I am stuck here, i do not know how to procede, Wolfram Alpha tells me that this sum doesn't converge so how can i compute it in closed form? I can only assume I have gone about this the wrong way or made a mistake. Any help would be great.</p>
| Tom | 103,715 | <p>I'll comment on calculating the sum. First, the sum $\sum\limits_{n=0}^{\infty}n^2 e^{-\pi n}$ certainly does converge, just as $\int_0^{\infty} x^2 e^{-\pi x} \, dx$ converges. Here's how I would go about finding a closed expression which is equal that sum:</p>
<p>Note that $n^2 e^{-\pi n} = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}e^{-\lambda n}$, and since we will be evaluating $\lambda$ at $\pi$, we can always assume $\lambda > 1$. Now:</p>
<p>$$
\sum_{n=0}^{\infty} n^2 e^{-\pi n} = \sum\limits_{n=0}^{\infty}\frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}e^{-\lambda n} = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi}\sum\limits_{n=0}^{\infty}e^{-\lambda n} \\
=\frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi} \sum\limits_{n=0}^{\infty} \left(e^{-\lambda}\right)^n = \frac{d^2}{d\lambda^2}\Big|_{\lambda = \pi} \frac{1}{1 - e^{-\lambda}} = \cdots
$$
where the last evaluated equality comes from the geometric series formula. From here you only need to evaluate the differentiation.</p>
|
457,654 | <p>This question is a continuation of: <a href="https://math.stackexchange.com/questions/456118/finding-a-set-of-values-with-the-binomial-theorem">Finding a set of values with the binomial theorem</a></p>
<p>For $n \in \mathbb N$ and the function $p(x) = \left(x + \frac 1x\right)^{n}$.</p>
<p>By the binomial theorem:<br />
$$\left(x + \frac 1x\right)^{n} = \sum_{k=0}^n {n \choose k} x^{n-2k}$$</p>
<p>1) For $p(x) = a_ox^{b_o}+a_1x^{b_1}+...+a_nx^{b_n}$ and $D=$ {$b_o, b_1, ..., b_n$}.<br />
Determine D and its maximum and minimum values.<br />
<strong>Answer:</strong> $D = {n, n-2, n-4, ..., -n}$ with a minimum of $-n$ and a maximum of $n$.</p>
<p>2) Determine the coefficient $a_k$ of $x^{b_k}$ where $k=0,1,...,n$ and give a formula.</p>
<p><strong>Can someone give me a hint of what to do? I don't know where to start for the second part...</strong></p>
| Jyrki Lahtonen | 11,619 | <p>Obviously the polynomial $a(x)$ has to be coprime to $m(x)$ for otherwise the common factor will always prevent this. If the factorization of $m(x)$ to powers of irreducible polynomials is
$$
m(x)=\prod_i p_i(x)^{a_i},
$$
then by the Chinese remainder theorem
$$
\mathbb{F_p}[x]/(m(x))\simeq\bigoplus_i\mathbb{F_p}[x]/(p_i(x)^{a_i}).
$$
Denote $R_i=\mathbb{F_p}[x]/(p_i(x)^{a_i}).$ The coset of $a(x)$ is a unit of $R_i$ as $p_i(x)\nmid a(x)$. Can you count the size of the unit group of $R_i$?
If you know that $|R_i^*|=n_i$, then Lagrange's theorem tells you that
$$
a(x)^{n_i}\equiv 1\pmod{p_i(x)^{a_i}}.
$$
For the rest you are expected to mimic the integer case and apply Chinese remainder theorem backwards.</p>
<hr>
<p>In some cases it is possible to replace the number $n_i$ by its proper divisor.
This is because the group $R_i^*$ is not always cyclic, and thus its exponent is strictly smaller than its cardinality. Consider the following. As explained in Alex Youcis' answer, when $a_i=1$, $R_i$ is a finite field, and thus $R_i^*$ is cyclic of order $q-1, q=|\mathbb{F_p}[x]/(p_i(x)|=p^{\deg p_i(x)}$. So we always have the congruence
$$
a(x)^{q-1}\equiv1 \pmod {p_i(x)},
$$
and thus $a(x)^{q-1}\equiv 1+p_i(x) b(x)\pmod {p_i(x)^2}$ for some polynomial $b(x)$. Raising this congruence to $p^{th}$ power gives then that
$$
a(x)^{p(q-1)}\equiv1\pmod {p_i(x)^2}
$$
even though $|\mathbb{F_p}[x]/(p_i(x)^2)^*|=q(q-1)>p(q-1)$ whenever $\deg p_i>1$.
This shows that for $R_i^*$ to be cyclic when $a_i>1$, it is necessary that $\deg p_i=1$. It would not be too difficult to analyze the structure of $R_i^*$ further, but at this point the reader can hopefully already appreciate that the exponent of $R_i$ depends on the parameters $\deg p_i$, $p$, and $a_i$ in a non-trivial way.</p>
|
2,572,428 | <p>$$z^2=3+4i ,(z=x+iy)$$
Seems easy? That's what I thought!</p>
<p>I get a system of equations that I can't solve:
$x^2-y^2=3$ and $2xiy=4i$ which I then get $x=\frac{2}{y}$</p>
<p>I can't solve the system of equations?</p>
| WaveX | 323,744 | <p>Hint</p>
<p>$$3+4i = 4+4i-1$$ $$ = 4+4i+i^2 $$</p>
|
4,192,879 | <p>This question is about the <em>provided solution</em> to Question 11d from Chapter 8 of Spivak's Calculus. The relevant background material (i.e. 11a-11c) is as follows:</p>
<blockquote>
<p><span class="math-container">$(a)$</span> If <span class="math-container">$\{a_n\}$</span> is a sequence of positive terms such that <span class="math-container">$$a_{n+1}\leq a_n/2$$</span> then for every <span class="math-container">$\epsilon>0$</span> there exists an <span class="math-container">$n$</span> with <span class="math-container">$a_n<\epsilon$</span>.</p>
<p><span class="math-container">$(b)$</span> Let <span class="math-container">$P$</span> be a regular polygon inscribed in a circle. Let <span class="math-container">$P'$</span> be the regular polygon with twice the sides of <span class="math-container">$P$</span>. Let <span class="math-container">$A$</span> be the area of <span class="math-container">$P$</span>, <span class="math-container">$A'$</span> be the area of <span class="math-container">$P'$</span> and <span class="math-container">$C$</span> that of the circle. Prove</p>
<p><span class="math-container">$$(C-P')\leq \frac 1 2 (C-P)$$</span></p>
<p><span class="math-container">$(c)$</span> Prove there is a regular inscribed polygon with area as close as that to the circle.</p>
</blockquote>
<p>Part d reads as follows:</p>
<blockquote>
<p>Using the fact that the areas of two regular polygons with the same number of sides are in the same relation as the squares of their sides, prove the areas of two circles are in the same relation as the squares of their radii.</p>
</blockquote>
<p>Here is Spivak's solution (<strong>I will include an indexed dagger symbol <span class="math-container">$\dagger_i$</span> in places that I will subsequently ask questions about</strong>):</p>
<hr />
<p>Let <span class="math-container">$r_1$</span> and <span class="math-container">$r_2$</span> be the radii of the two circles <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span>, and let <span class="math-container">$A_i$</span> be the area of the region bounded by <span class="math-container">$C_i$</span>. We know that there are numbers <span class="math-container">$\delta_1, \delta_2 \gt 0$</span> such that</p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{B_1}{B_2} \Big\rvert \lt \epsilon \quad \dagger_1$$</span></p>
<p>for any numbers <span class="math-container">$B_1, B_2$</span> with <span class="math-container">$|A_i-B_i| \lt \delta_i \quad \dagger_2$</span>. By part (c), there are numbers <span class="math-container">$n_i$</span> such that the area of a regular polygon, with <span class="math-container">$n_i$</span> sides, inscribed in <span class="math-container">$C_i$</span> differs from <span class="math-container">$A_i$</span> by less than <span class="math-container">$\delta_i$</span>. Let <span class="math-container">$P_i$</span> be the area of a regular polygon inscribed in <span class="math-container">$C_i$</span> with <span class="math-container">$\max(n_1,n_2)$</span> sides. Then</p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{P_1}{P_2} \Big\rvert \lt \epsilon$$</span></p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{r_1^{\ 2}}{r_2^{\ 2}} \Big\rvert \lt \epsilon$$</span></p>
<p>Since this is true for each <span class="math-container">$\epsilon \gt 0$</span>, it follows that <span class="math-container">$A_1/A_2=r_1^{\ 2}/r_2^{\ 2}$</span></p>
<hr />
<p>Referencing <span class="math-container">$\dagger_1$</span> and <span class="math-container">$\dagger_2$</span>, which I view as being directly connected, how is the equation <span class="math-container">$\Big\lvert \frac{A_1}{A_2}-\frac{B_1}{B_2} \Big\rvert \lt \epsilon$</span>, where <span class="math-container">$\epsilon$</span> is <em>arbitrary</em>, derived from the two equations: (1) <span class="math-container">$|A_1-B_1| \lt \delta_1$</span> and <span class="math-container">$|A_2-B_2| \lt \delta_2$</span>. Further, where are the statements <span class="math-container">$|A_1-B_1| \lt \delta_1$</span> and <span class="math-container">$|A_2-B_2| \lt \delta_2$</span> coming from? (If <span class="math-container">$B_1$</span> and <span class="math-container">$B_2$</span> are arbitrary numbers, and <span class="math-container">$\delta_1$</span> and <span class="math-container">$\delta_2$</span> are constructed numbers, I am having difficulties seeing how the <span class="math-container">$\delta$</span>'s can be related to an arbitrary <span class="math-container">$\epsilon$</span>)</p>
| Hagen von Eitzen | 39,174 | <p>The map <span class="math-container">$(0,\infty)\times (0,\infty)\to (0,\infty)$</span>, <span class="math-container">$(x,y)\mapsto \frac xy$</span> is continuos. Use the <span class="math-container">$\epsilon\delta$</span>-definition of continuity (together with a simple lower bound on distances in <span class="math-container">$(0,\infty)\times (0,\infty)$</span>).</p>
<hr />
<p>Or calculate directly: If <span class="math-container">$0<\delta_i<A_i$</span>, then for <span class="math-container">$|B_i-A_i|<\delta_i$</span>, we have
<span class="math-container">$$\frac{A_1-\delta_1}{A_2+\delta_2}<\frac{B_1}{B_2}<\frac{A_1+\delta_1}{A_2-\delta_2}.$$</span>
Now show that you can make the <span class="math-container">$\delta_i$</span> small enough to ensure
<span class="math-container">$$ \frac{A_1+\delta_1}{A_2-\delta_2}-\frac{A_1-\delta_1}{A_2+\delta_2}<\epsilon$$</span></p>
|
2,738,023 | <p>We know we can check if two vectors are 'orthogonal' by doing an inner product.</p>
<p>$a*b=0$</p>
<p>tells us that these two vectors are orthogonal</p>
<p>here comes the question:</p>
<p>if there a way to compute if they are 'parallel'? i.e., they are pointing at the same direction. </p>
| user1228123 | 148,520 | <p>Two parallel vectors $a$ and $b$ will have a null cross product:
$$a\times b=0\iff a\parallel b.$$</p>
|
633,482 | <p>Consider a cube that exactly fills a certain cubical box. As in Examples 8.7 and 8.10, the ways in which the cube
can be placed into the box corresponds to a certain group of permutations of the vertices of the cube. This is the group
of <strong>group of rigid motions (or rotations) of the cube</strong>. (It should not be confused with th e <em>group of symmetries
of the figure</em>, which will be discussed in the exercises of Section 12.) How many elements does this group have? Argue
geometrically that this group has at least three different subgroups of order 4 and at least four different subgroups of
order 3.</p>
<p><strong>Fraleigh Solution</strong> The group has $24$ elements. </p>
<p>The first subgroup of order $4.$ For any one of the $6$ faces can be on top, and for each such face on top, the cube can be rotated in $4$ different positions leaving that face on top. The $4$ such rotations, leaving the top face on top and the bottom face on the bottom, form a cyclic subgroup of order $4$. </p>
<p>The second rotation group of order $4$ is formed by the rotations leaving the front and
back faces in those positions.</p>
<p>The third rotation groups of order $4$ is formed by the rotations leaving the side faces in those positions. </p>
<p>One exhibits a subgroup of order $3$ by $\color{red}{\text{taking hold of a pair of diagonally opposite vertices and
rotating through the three possible positions}}$, corresponding to the three edges emanating from each
vertex.<br>
There are $4$ such diagonally opposite pairs of vertices, giving the desired $4$ groups of order $3$.</p>
<p>Question 1. I feel this is easier than <a href="https://math.stackexchange.com/a/632954/54547">my other post</a>. But I can't see how 'taking hold a pair of diagonally opposite vertices' causes 'three possible positions'? I tried the animation at that other post but no luck. </p>
<p>Question 2. How do you decide on the classifications of the rotations of a shape? These two solutions just say what they are. They never revealed how they prefigured this group has at least three different subgroups of order 4 and at least four different subgroups of
order 3. I don't mean just playing around with the shape. I tried that and it didn't help me here. </p>
<p>Question 3. Does this solution break down the type of rotations differently than the other post? That solution talks about " a line joining the centers of opposite faces" and "a line joining diagonally opposite vertices ". This one doesn't. </p>
<p>This is from John B. Fraleigh page 86 exercise 8.45 A First Course in Abstract Algebra</p>
| Yiorgos S. Smyrlis | 57,021 | <p>The function $f(t)=\mathrm{e}^{at}$ can be viewed as the solution of the initial value problem:
$$
x'=ax, \quad x(0)=1.
$$
More generally, if $A\in\mathbb R^{n\times n}$ and $u_0\in\mathbb R^n$, then the solution of
the
initial value problem:
$$
x'=Ax, \quad x(0)=u_0,
$$
is $u(t)=\mathrm{e}^{tA}u_0$, where $\mathrm{e}^{tA}$ is the exponential of the matrix $A$. It $A$ possessed imaginary eigenvalues, then $u(t)$ returns to a multiple of itself, for suitable initial $u_0$.</p>
|
673,229 | <blockquote>
<p>The medians to the two legs of a right angled triangle are $10$ and $4\sqrt{10}$. Find the hypotenuse of the right angled triangle.</p>
</blockquote>
<p>I'm confused. Can somebody please explain to me how to do this step by step? Not just the answer I want to know what you did to get the answer and why you did it please. Thank you!</p>
| Kim Fierens | 172,717 | <p>Some additional examples that may be of interest: </p>
<ol>
<li>Spherical trigonometry.</li>
<li>Quaternion analysis (in the 19th century a rather hot and active area of research).</li>
<li>Synthethic projective geometry (in the tradition of Steiner and Poncelet).</li>
<li>Analytical theory of generalized continued fractions.</li>
<li>Geometry of straight-line linkages (popular from the 1870s to the 1890s); mathematical kinematics in general.</li>
</ol>
<p>Personally, I'd say quaternion analysis is the most spectacular example of an area that was once very fashionable and now all but forgotten.</p>
<p>All these things are relative, of course. It would be very hard, I think, to find mathematical disciplines in which the amount of active research is exactly zero. </p>
|
2,064,403 | <p>Example. Let $V$ be a finite dim vector space with two different bases</p>
<p>$S = \{ u_1,u_2 \} = \{ (1,2),(3,5) \}$ and $S' = \{ v_1, v_2 \} = \{ (1,-1), (1,-2) \} $</p>
<p>You can check that $v_1 = -8u_1 + 3u_2$ and $v_2 = -11u_1+4u_2$ and $P = \begin{bmatrix}
-8 &-11 \\
3& 4
\end{bmatrix}$ is the change of basis where the columns are the coords.</p>
<p>But why can't we define change of basis by their rows? Since it works nicely that</p>
<p>$\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix} = \begin{bmatrix}
-8 &3 \\
-11& 4
\end{bmatrix}\begin{bmatrix}
u_1 \\
u_2
\end{bmatrix}$</p>
<p>So to move from the old basis $S$, you apply the matrix $\begin{bmatrix}
-8 &3 \\
-11& 4
\end{bmatrix}$ to get a new basis. Why do we have to transpose? If you transpose, how do you even use this change of basis? Why can't I use this definition of change of basis.</p>
| Christian Blatter | 1,303 | <p>When a basis, or coordinate, transform is in action then we are not so much interested in how the new basis vectors are defined (this is dealt with at the start), but we need to transform the coordinates of vectors a thousand times. Therefore things are set up in such a way that the the formulas for the transformation of coordinates are as simple as possible.</p>
<p>A second thing: Per definition of matrix multiplication matrices should operate on, resp., are multiplied with, tuples of numbers, and not tuples of vectors. But in the setup you propose the latter is the case.</p>
|
3,579,739 | <p>Its also mentioned that this question is an example of <strong>Sampling without replacement</strong></p>
<p>My question is , by the general method of how I do these kind of problems , I would assume that there are two possibilities</p>
<ol>
<li><p>You chose a defective bottle and then a non defective which will have the probability of </p>
<pre><code> 1/5 X 4/4 = 4/20
OR
</code></pre></li>
<li><p>Choosing a non defective one first and then a defective one which will have the probability of</p>
<pre><code>4/5 x 1/4 =4/20
</code></pre></li>
</ol>
<p>Total probability is 8/20= 2/5 which doesn't make sense at all, since there is just one bottle to begin with
But the problem is I cant find a fault in my logic. Though I feel both the options are redundant , isn't both two different ways of selecting the pair?. Or is the fact that the phrase "one after the other" is absent a reason on why this might be a wrong approach?</p>
<pre><code> Thank you in advance
</code></pre>
| Jamāl | 506,409 | <p>You can think of this problem in two ways. One way is the one you did which is perfectly fine.
The other is to think about the possible outcomes you can have at the end of the two selections. There are <span class="math-container">$5\choose 2$</span> total number of selections you can make. Now assume that you have already selected the fractured bottle. There are <span class="math-container">${4\choose 1}=4$</span> different ways of selecting the second bottle. Therefore probability that the fractured bottle is chosen is
<span class="math-container">$$
\frac{4\choose 1}{5\choose 2}=\frac{4}{10}=\frac{2}{5}
$$</span></p>
<p>PS: If order of selection is taken into consideration then the total number of outcomes will simply be doubled (since the permutation of the selected pair also matters). But this doesn't affect the final answer as <span class="math-container">$\frac{{4\choose 1}\times 2}{{5\choose 2}\times 2}=\frac{4\choose 1}{5\choose 2}$</span></p>
|
159,948 | <p>If I have this function (just and example) </p>
<pre><code> Plot3D[-Abs[Cos[ y] Sqrt[x] Sqrt[ y] Cos[ x]], {x, 0, 20}, {y, 0, 20}]
</code></pre>
<p>I am interested in finding the locations of the smallest (minimum) 6 values of this function? How can I do that. Apparently, Findminimum and its likes give me one value.</p>
<p>Thanks for help.</p>
| J. M.'s persistent exhaustion | 50 | <p>Another idea is to use the <code>MeshFunctions</code> option of <code>ContourPlot[]</code> (as previously shown <a href="https://mathematica.stackexchange.com/a/46569">here</a>) to extract putative positions for the minima, before polishing them with <code>FindMinimum[]</code>:</p>
<pre><code>(* pretend that the function is a black box *)
f[x_?NumericQ, y_?NumericQ] := -Abs[Cos[y] Sqrt[x] Sqrt[y] Cos[x]]
res = Cases[Normal[ContourPlot[Derivative[1, 0][f][x, y] == 0, {x, 0, 20}, {y, 0, 20},
ContourStyle -> None, Mesh -> {{0}},
MeshFunctions -> Function[{x, y, z},
Derivative[0, 1][f][x, y]],
PlotPoints -> 55]],
Point[{x0_, y0_}] :> (FindMinimum[{f[x, y], 0 <= x <= 20 && 0 <= y <= 20},
{{x, x0}, {y, y0}}]), ∞];
MinimalBy[DeleteDuplicates[res, First[#1] == First[#2] &], First, 6]
{{-18.8628, {x -> 18.876, y -> 18.876}}, {-17.222, {x -> 18.876, y -> 15.7397}},
{-15.7239, {x -> 15.7397, y -> 15.7397}}, {-15.4082, {x -> 18.876, y -> 12.606}},
{-14.0678, {x -> 12.606, y -> 15.7397}}, {-13.352, {x -> 18.876, y -> 9.47749}}}
</code></pre>
<p>(Replace <code>FindMinimum[]</code> with <code>FindMinValue[]</code> or <code>FindArgMin[]</code> if needed.)</p>
|
199,374 | <p>Consider the two table set below</p>
<pre><code>t1= {0.44, 0.62, 0.77, 0.87, 0.93, 0.96, 0.98, 1}
t2= {0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25}
</code></pre>
<p>how can I create a plot which wil plot a rectangle with a heat map normalized to the values of the table?</p>
<p>The result should look like</p>
<p><a href="https://i.stack.imgur.com/GgXBG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GgXBG.png" alt="h"></a></p>
| Carl Woll | 45,431 | <p>Another idea is to use <a href="http://reference.wolfram.com/language/ref/VertexColors" rel="nofollow noreferrer"><code>VertexColors</code></a> option of <a href="http://reference.wolfram.com/language/ref/Line" rel="nofollow noreferrer"><code>Line</code></a> (or <a href="http://reference.wolfram.com/language/ref/Rectangle" rel="nofollow noreferrer"><code>Rectangle</code></a>):</p>
<pre><code>Graphics[
{
Thickness[.4],
Line[Thread[{t1,0}], VertexColors->ColorData["DeepSeaColors"]/@t1]
},
AspectRatio->1/4
]
</code></pre>
<p><a href="https://i.stack.imgur.com/Tdy68.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tdy68.png" alt="enter image description here"></a></p>
|
4,390,127 | <p>I ma trying to prove that if <span class="math-container">$A$</span> is an <span class="math-container">$n\times n$</span> matrix, then <span class="math-container">$f:\mathbb{R}^n\to\mathbb{R},\ f(\vec{x})=A\vec{x}\cdot\vec{x}=\vec{x}^T A\vec{x}$</span> is differentiable and <span class="math-container">$Df(\vec{a})\vec{h}=A\vec{a}\cdot\vec{h}+A\vec{h}\cdot\vec{a}.$</span></p>
<p>(NOTE: computation edited according to Snaw's comment below)</p>
<p>Now, since a function <span class="math-container">$f:\mathbb{R}^n\to\mathbb{R}^m$</span> is differentiable if <span class="math-container">$$
\lim\limits_{\vec{h}\to\vec{0}}\frac{\vec{f}(\vec{x}+\vec{h})-\vec{f}(\vec{x})-Df(\vec{a})\vec{h}}{||\vec{h}||}=\vec{0}
$$</span>
we have that <span class="math-container">$f$</span> is differentiable at <span class="math-container">$\vec{a}\in\mathbb{R}^n$</span> if <span class="math-container">$$\lim\limits_{\vec{h}\to\vec{0}}\frac{(\vec{a}+\vec{h})^TA(\vec{a}+\vec{h})-\vec{a}^TA\vec{a}-A\vec{a}\cdot\vec{h}-A\vec{h}\cdot\vec{a}}{||\vec{h}||}\\
=\lim\limits_{\vec{h}\to\vec{0}}\frac{\vec{a}^TA\vec{h}+\vec{h}^TA\vec{a}+\vec{h}^TA\vec{h}-\vec{h}^TA\vec{a}-\vec{a}^TA\vec{h}}{||\vec{h}||}=\lim\limits_{\vec{h}\to\vec{0}}\frac{\vec{h}^TA\vec{h}}{||\vec{h}||}
$$</span></p>
<p>but at this point I am not sure if I can claim that, as <span class="math-container">$\vec{h}\to\vec{0}$</span>, the limit is <span class="math-container">$\vec{0}$</span>, since it seems to me that I would encounter an indeterminate form of the type <span class="math-container">$\vec{0}/0$</span>.</p>
<p>For example, if <span class="math-container">$A\in\mathbb{M}_{2\times 2}$</span> we would have <span class="math-container">$$\lim\limits_{(h_1,h_2)\to (0,0)}\frac{h_1^2 A_{11}+(A_{12}+A_{21})h_1h_2+h_2^2 A_{22}}{\sqrt{h_1^2+h_2^2}}.$$</span></p>
<p>How could I justify this claim?</p>
<p>Thanks</p>
| Henno Brandsma | 4,280 | <p>Let <span class="math-container">$U$</span> be an open neighbourhood of <span class="math-container">$a$</span>. By local convexity (and regularity (we don't need Hausdorff, all uniform spaces are completely regular and regular, whether separated (<span class="math-container">$T_2$</span>) or nor) there is an open convex neighbourhood <span class="math-container">$W$</span> of <span class="math-container">$a$</span> so that <span class="math-container">$\overline{W} \subseteq U$</span>.</p>
<p>As <span class="math-container">$x_d \to a$</span> we get that for some <span class="math-container">$d_0$</span> we have that for all <span class="math-container">$d \ge d_0$</span>: <span class="math-container">$X_d \subseteq \overline{W}$</span> (as in Kavi Rama's proof, using that <span class="math-container">$W$</span> is convex) and so also <span class="math-container">$(y_d \in) \overline{X_d} \subseteq \overline{W} \subseteq U$</span>, and so <span class="math-container">$y_d \to a$</span>, as witnessed by <span class="math-container">$d_0$</span> for the <span class="math-container">$U$</span> etc.</p>
|
2,751,300 | <blockquote>
<p>Figure out with distinct real numbers the system of equations.</p>
<p><span class="math-container">$$\frac{xy}{x-y} = \frac{1}{30}$$</span>
<span class="math-container">$$\frac{x^2y^2}{x^2+y^2} = \frac{1}{2018}$$</span></p>
</blockquote>
<p>I multiplied x-y both side on the first equation and square on both side, and I stucked.</p>
<p>Help me...</p>
| Martin R | 42,969 | <p>Inverting the equations gives
$$
30 = \frac{x-y}{xy} = \frac 1y - \frac 1x \\
2018 = \frac{x^2+x^2}{x^2y^2} = \frac 1{y^2} + \frac 1{x^2}
$$
This suggests to substitute $a= \frac 1x, b= \frac 1y$:
$$
30 = b - a \\
2018 = b^2 + a^2 \,
$$
This leads to a simple quadratic equation for $a$ (or $b$) which has two different real (actually
integral) solutions.</p>
|
4,014,768 | <blockquote>
<p>Find the sationary points of the curve and their nature for the equation <span class="math-container">$y=e^x\cos x$</span> for <span class="math-container">$0\le x\le\pi/2$</span>.</p>
</blockquote>
<p>I derived it and got <span class="math-container">$e^x(-\sin x+\cos x)=0$</span>.</p>
<p><span class="math-container">$e^x$</span> has no solution but I don't know how to find the <span class="math-container">$x$</span> such that <span class="math-container">$-\sin x+\cos x=0$</span></p>
| InsideOut | 235,392 | <p>Hint:</p>
<p><span class="math-container">$$-\sin x+\cos x=0$$</span>
If and only if</p>
<p><span class="math-container">$$\tan x=1$$</span></p>
<p>Can you get it from here?</p>
|
4,014,768 | <blockquote>
<p>Find the sationary points of the curve and their nature for the equation <span class="math-container">$y=e^x\cos x$</span> for <span class="math-container">$0\le x\le\pi/2$</span>.</p>
</blockquote>
<p>I derived it and got <span class="math-container">$e^x(-\sin x+\cos x)=0$</span>.</p>
<p><span class="math-container">$e^x$</span> has no solution but I don't know how to find the <span class="math-container">$x$</span> such that <span class="math-container">$-\sin x+\cos x=0$</span></p>
| Jean-Claude Arbaut | 43,608 | <p>For all <span class="math-container">$x$</span>, <span class="math-container">$\cos(x+\pi/2)=-\sin(x)$</span> hence the equation is equivalent to
<span class="math-container">$$\cos(x)+\cos(x+\pi/2)=0$$</span></p>
<p>But we also have the identity</p>
<p><span class="math-container">$$\cos p+\cos q=2\cos\frac{p+q}{2}\cos\frac{p-q}2$$</span></p>
<p>So your equation is also equivalent to</p>
<p><span class="math-container">$$2\cos(x+\frac{\pi}4)\cos\frac{\pi}4=0$$</span></p>
<p>That is, <span class="math-container">$\cos(x+\frac\pi4)=0$</span>.</p>
<p>But <span class="math-container">$\cos t=0$</span> iff <span class="math-container">$t=\frac{\pi}{2}+k\pi$</span>, <span class="math-container">$k\in\Bbb Z$</span>. Therefore, your equation is also equivalent to</p>
<p><span class="math-container">$$x=\frac\pi4+k\pi$$</span></p>
<p>And the only value of <span class="math-container">$x$</span> in <span class="math-container">$[0,\pi/2]$</span> occurs for <span class="math-container">$k=0$</span>.</p>
|
4,038,685 | <p>Suppose that <span class="math-container">$u(x)$</span> is continuous and satisfies the integral equation</p>
<p><span class="math-container">\begin{equation}\label{1.4.1}
u(x) = \int_0^x \sin(u(t))u(t)^p dt
\end{equation}</span></p>
<p>on the interval <span class="math-container">$0 \leq x \leq 1$</span>. show that <span class="math-container">$u(x) = 0$</span> on this interval if <span class="math-container">$p \geq 0$</span>.</p>
<p>This is what I have:</p>
<p>Since <span class="math-container">$\sin(u(t))u(t)^p$</span> is continuous, it follows from the integral definition that <span class="math-container">$u(x)$</span> is differentiable. Let us differentiate both sides of equation above with respect to <span class="math-container">$x$</span>. This yields:</p>
<p><span class="math-container">\begin{equation}
u'(x) = \sin(u(x))u(x)^p
\end{equation}</span></p>
<p>This ODE is separabale and becomes:</p>
<p><span class="math-container">\begin{equation}
\frac{1}{\sin(u(x))u(x)^p}du(x) = dx
\end{equation}</span></p>
<p>However, this doesn't seem easily solvable so I'm not sure how to show that <span class="math-container">$u(x) = 0$</span> from this.</p>
| Matthew H. | 801,306 | <p>Assume there exists <span class="math-container">$t\in(0,1)$</span> such that <span class="math-container">$u(t)>0$</span>. Because <span class="math-container">$u$</span> is continuous and <span class="math-container">$u(0)=0$</span>, we can identify <span class="math-container">$a,b\in [0,1]$</span> such that <span class="math-container">$u(a)=0,a<t<b,$</span> and <span class="math-container">$u>0$</span> on <span class="math-container">$(a,b)$</span>. Since <span class="math-container">$\sin\big(u(t)\big)$</span> is also continuous, we can find <span class="math-container">$c\in (a,b)$</span> such that <span class="math-container">$\sin \big(u(t)\big)>0$</span> on <span class="math-container">$(a,c)$</span>. Now consider the IVP: <span class="math-container">$$\frac{dy}{dx}=y^p\sin(y):x\in(a,c)$$</span> <span class="math-container">$$y(a)=0$$</span> We know <span class="math-container">$u$</span> is a solution to this IVP, so for <span class="math-container">$x_1,x_2\in (a,c)$</span> we have <span class="math-container">$$\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}=\int_{x_1}^{x_2}dx=x_2-x_1$$</span> Take <span class="math-container">$x_1 \longrightarrow a^{+}$</span> to get <span class="math-container">$$\lim_{x_1 \longrightarrow a^{+}}\Bigg[\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$</span> This reduces to <span class="math-container">$$\lim_{h \longrightarrow 0^{+}}\Bigg[\int_{h}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$</span> However, if <span class="math-container">$p\geq 0$</span> the limit on the left side doesn't exist. This is by a limit comparison test to <span class="math-container">$\lim_{h\longrightarrow 0^{+}}\Big[\int_h^{u(x_2)}\frac{dy}{y^{p+1}}\Big]$</span>. You will arrive at a similar contradiction if we assume <span class="math-container">$u(t)<0$</span>.</p>
|
1,519,070 | <p>How can I show that the sequence is defined recursively?</p>
<blockquote>
<p>Show that the recursively defined sequence $(x_n)_{n\in\mathbb{N}}$ with
$$x_1=1, \qquad\qquad x_{n+1}=\sqrt{6+x_n}$$
converges and determine its limit </p>
</blockquote>
<p><a href="https://i.stack.imgur.com/cL7BR.png" rel="nofollow noreferrer">Image</a></p>
| Abr001am | 223,829 | <p>$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{7}}}}<x$</p>
<p>If we go by induction :</p>
<p>$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{7}}}}=U_n<x$</p>
<p>$(U_{n+1})²-6=U_n<x$</p>
<p>$U_{n+1}<\sqrt{x+6}$</p>
<p>$\sqrt{x+6}=x$</p>
<hr>
<ul>
<li>Now it is left to you calculating $x$ and $U_0$ testing part.</li>
</ul>
|
1,328,082 | <p>I've been trying to prove the following inequality, but until now I've had problems coming up with a solution:</p>
<p>$$
2^{mn} \ge m^n
$$</p>
<p>$m$ and $n$ can assume any natural number.</p>
<p>I wasn't able to find any counterexample that would invalidate this inequality, so I am assuming that this statement is generally true, but of course this still has to be proven.</p>
| coffeemath | 30,316 | <p>Let $A=\{1,2,...,n\}$ and $B=\{1,2,...,m\}.$ Then $m^n$ is the number of functions from $A$ to $B,$ while $2^{m \cdot n}$ is the number of relations from $A$ to $B$ (functions or not). And there's an injection from the set of functions to the set of relations.</p>
|
2,466,913 | <p>I am having trouble with a question on my homework and I have done some work but I'm not sure if I have completed all that is asked or how to take it further if needed.</p>
<p>Problem: </p>
<p>Let $\theta = \arccos(5x^3)$. Find $\frac{d\theta}{dx}$ in two ways:</p>
<p>(a) Use implicit differentiation starting from $\cos(\theta)= 5x^3$. Write your final answer in terms of $x$ by using SOH-CAH-TOA, a right triangle, and the Pythagorean Theorem.</p>
<p>(b) Use the Chain Rule extension of the formula $\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1-x^2}}$. Compare with your answer from part (a).</p>
<hr>
<p>Here is my work for part (a)</p>
<p>I drew a right triangle with an angle theta and labeled the hypotenuse as $1$ and the adjacent leg $5x^3$. Using the Pythagorean Theorem I found the opposite leg to be $\sqrt{1-25x^6}$. I'm not sure if this is all the question is asking or if I need to do something further. The wording of the problem is confusing me a little.</p>
<p>I also simply used this extra side to find $\sin(\theta) = \sqrt{1-25x^6}$. </p>
<p>Here is my work for part (b)</p>
<p>\begin{equation}
\begin{split}
\frac{d}{dx} \arccos 5x^3 &= \frac{-1}{\sqrt{1-(5x^3 )^2}} \frac{ d}{dx} (5x^3 )\\
&= \frac{-1}{\sqrt{1-25x^6}} 15x^2\\
&= \frac{-15x^2}{\sqrt{1-25x^6}}.
\end{split}
\end{equation}</p>
<p>If this is as far as I am supposed to take this problem are my answers supposed to be the same for part (a) and part (b)? If not then why not? I know I can simplify the answer for part (b) by rationalizing the denominator but it still wouldn't equal anything from part (a).</p>
<p>Any help would be greatly appreciated to help me finish this problem. I've looked at it several times for hours and am not sure where to go. Thanks!</p>
| Temidire Adesiji | 490,025 | <p>Note that every time you cut one piece of paper into $7$, the total number of pieces of paper reduces by $1$ $piece$ $of$ $paper$ increases by $7$. Thus, the student will have $7, 13 ,19, ...$ Which will give rise to a sequence of odd integers, Hence, Impossible to get $2016$ pieces of paper.</p>
|
3,886,643 | <p>I have doubts about how to solve this issue about bases in numbering systems:</p>
<p>"Determine the base b in <span class="math-container">$(104)_b = 8285$</span>"</p>
<p>can anybody help me?
Thanks.</p>
| md2perpe | 168,433 | <p>Since <span class="math-container">$(104)_b = 1\cdot b^2 + 0 \cdot b + 4$</span> you get the equation
<span class="math-container">$$b^2+4 = 8285,$$</span>
which you should be able to solve.</p>
|
2,585,828 | <p>I saw in an article that for every real number, there exists a Cauchy sequence of rational numbers that converges to that real number. This was stated without proof, so I'm guessing it is a well-known theorem in analysis, but I have never seen this proof. So could someone give a proof (preferably with a source) for this fact? It can use other theorems from analysis, as long as they aren't too obscure.</p>
| Paramanand Singh | 72,031 | <p>Based on the context gathered from your question and your comments it is clear that you are concerned about the equivalence of construction of reals as Dedekind cuts of rationals and as equivalence classes of Cauchy sequences of rationals.</p>
<p>We can do this in the following manner. First we show that corresponding to a given Dedekind cut of rationals there exists a Cauchy sequence of rationals. To do so in proper manner it is best to explicitly state the definition of a Dedekind cut of rationals.</p>
<p>A Dedekind cut of rationals is a set $A\subseteq\mathbb{Q} $ with the following properties:</p>
<ul>
<li>$\emptyset\neq A\neq \mathbb{Q} $. </li>
<li>If $p\in A$ and $q\in\mathbb{Q} $ with $q<p$ then $q\in A$. </li>
<li>If $p\in A$ then there exists a $q\in A$ with $p<q$.</li>
</ul>
<p>The above definition is one provided by Rudin in his <em>Principles of Mathematical Analysis</em>. But the true picture of a Dedekind cut is revealed when one considers not only the set $A$ but also its complement $B=\mathbb{Q} - A$ as done by Dedekind himself.</p>
<p>It is easy to prove that given any positive rational number $\epsilon$ one can find rationals $p, q$ such that $p\in A, q\in B$ such that $0<q-p<\epsilon $. Thus for every positive integer $n$ we can find rationals $a_n, b_n$ such that $a_n\in A, b_n\in B, 0<b_n-a_n<1/n$ and further we can ensure that $b_{n} \geq b_{n+1},a_{n}<a_{n+1}$. Now we can show that $\{a_n\} $ is a Cauchy sequence of rationals. This is how one can find a Cauchy sequence of rationals corresponding to a given Dedekind cut of rationals. Note that the sequence $b_n$ can also be considered here instead of $a_n$. </p>
<p>To do the reverse let us consider a Cauchy sequence of rationals $\{a_n\} $ and we then need to find a Dedekind cut corresponding to it. Consider a rational number $r$. If there is a positive integer $m$ such that $a_n>r$ for all $n\geq m$ then we say that that $r$ is <em>inferior</em> to the sequence $\{a_n\} $. Consider the set $$A=\{p\mid p\in\mathbb {Q}, \exists q\in\mathbb{Q}, q>p, q\text{ is inferior to sequence }\{a_n\} \} $$ Since a Cauchy sequence is bounded it is clear that the set $A$ and its complement $B=\mathbb{Q} - A$ are both non-empty. Also if $p\in A$ then there is some $r\in\mathbb{Q} $ such that $r>p$ and $r$ is inferior to $\{a_n\} $. And if we take a rational $q$ with $q<p$ then $q<r$ and hence $q\in A$. Moreover we also have $(p+r) /2<r$ so that $p<(p+r) /2\in A$. Thus $A$ satisfies all the properties of a Dedekind cut. This is how one finds a Dedekind cut of rationals corresponding to a given Cauchy sequence of rationals.</p>
<p>We still need to show that if $\{b_n\} $ is another Cauchy sequence of rationals which is equivalent to $\{a_n\} $ (ie $a_n-b_n\to 0$) and if we apply the above mentioned procedure to find a Dedekind cut $A'$ corresponding to the sequence $\{b_n\} $ then $A'=A$. This is not difficult to prove and you should try it yourself.</p>
<hr>
<p>The full equivalence of the construction of reals via Dedekind cuts and via Cauchy sequences of rationals can be done by proving that the correspondence between these approaches described above is an isomorphism which preserves arithmetic as well as order relationship. But carrying out such a proof is boring and lengthy. </p>
|
3,140,052 | <p>I found a proof of this theorem in <a href="http://web.mit.edu/18.705/www/13Ed.pdf" rel="nofollow noreferrer">modern algebra</a> 2.30</p>
<blockquote>
<p>Proof: Set <span class="math-container">$S := \{ \text{ ideals } \mathfrak b \mid \mathfrak b ⊃ \mathfrak a \text{ and } \mathfrak b ∌ 1 \}$</span>. Then <span class="math-container">$\mathfrak a ∈ S$</span>, and <span class="math-container">$S$</span> is partially ordered by inclusion. Given a totally ordered subset <span class="math-container">$\{\mathfrak b_λ\}$</span> of <span class="math-container">$S$</span>, set <span class="math-container">$\mathfrak b := \bigcup_\lambda \mathfrak b_λ$</span>. Then <span class="math-container">$\mathfrak b$</span> is clearly an ideal, and <span class="math-container">$1 \notin \mathfrak b$</span>; so <span class="math-container">$\mathfrak b$</span> is an upper bound of <span class="math-container">$\{\mathfrak b_λ\}$</span> in <span class="math-container">$S$</span>. Hence by Zorn’s Lemma, <span class="math-container">$S$</span> has a maximal element, and it is the desired maximal ideal.</p>
</blockquote>
<p>I'm not satisfied with the proof, with follow questions:</p>
<ul>
<li>Why <span class="math-container">$S$</span> is a set, <span class="math-container">$S$</span> may not be a set as Russell's paradox.</li>
<li>How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number <span class="math-container">$(0,1)$</span>, and say that range <span class="math-container">$(0,1)$</span> has a maximal element ?</li>
</ul>
<p>I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.</p>
<h3>Edit 1: still confused about something here</h3>
<p>I'll start from an example:</p>
<blockquote>
<p>Proof: <span class="math-container">$S = \{ 1 - \frac 1 n \mid n \in \mathbb N \}$</span> has a maximal element</p>
<p>Obviously <span class="math-container">$S$</span> is partially ordered, let <span class="math-container">$x = 1 - \Pi_{e \in S} (1 - e) = 1 - \Pi_{n \in \mathbb N} \frac 1 n$</span>, then clearly <span class="math-container">$x \in S$</span>; so <span class="math-container">$x$</span> is an upper bound of <span class="math-container">$S$</span>. Hence by Zorn’s Lemma, <span class="math-container">$S$</span> has a maximal element.</p>
</blockquote>
<p>I just copied the proof word by word, then got an obviously wrong result. Where does my proof go wrong ?</p>
<p>If I'm not allowed to construct <span class="math-container">$x$</span> this way then it is also not allowed to construct <span class="math-container">$\mathfrak b$</span> in the origin proof.</p>
<p>I think the truth is <span class="math-container">$x=1$</span> and <span class="math-container">$x \notin S$</span>, and what about <span class="math-container">$\mathfrak
b = \bigcup_\lambda \mathfrak b_λ$</span> ? Things can go wild with infinite. I'm not satisfied with <span class="math-container">$1 \notin \mathfrak b$</span></p>
| Bernard | 202,857 | <ul>
<li><span class="math-container">$S$</span> is a set by the <em>Axiom schema of specification</em>.</li>
<li>You cannot apply <em>Zorn's lemma</em> to the interval <span class="math-container">$(0,1)$</span> for the usual order because not every chain in <span class="math-container">$(0,1)$</span> has an upperbound <code>in (0,1)</code>.</li>
</ul>
|
22,269 | <p>How to verify, which version of MathJax is used here? It seems that v. 2.5 is not implemented yet. It has very useful features, as <code>\delimiterfactor</code>, which is needed here: <a href="https://tex.stackexchange.com/questions/284809/getting-large-braces-in-mathjax">https://tex.stackexchange.com/questions/284809/getting-large-braces-in-mathjax</a></p>
| Davide Cervone | 7,798 | <p>MathJax does not implement the <code>\delimiterfactor</code> control sequence, but does have an internal value for <code>delimiterfactor</code> (the default value of 901 from TeX) that it uses in its computations for stretchy delimiters. That value can be set from Javascript, but not from within a TeX expression. One reason for this is that MathJax doesn't implement the localization of changes within braces, so changes to <code>\delimiterfactor</code> would be global.</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.