qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,032,950 | <p>I have the following Cauchy problem. I do not know where to start, so I would appreciate any help and tips.
<span class="math-container">$$\frac{\partial^2 Y(t, x)}{\partial t^2} = 9\frac{\partial^2 Y(t,x)}{\partial x^2} - 2Z(t,x)$$</span>
<span class="math-container">$$\frac{\partial^2 Z(t, x)}{\partial t^2} = 6\frac{\partial^2 Z(t,x)}{\partial x^2} - 2Y(t,x)$$</span>
<span class="math-container">$$Y(0, x) = \cos(x),\ Z(0, x) = 0$$</span>
<span class="math-container">$$\frac{\partial Y(t, 0)}{\partial t} = \frac{\partial Z(t, 0)}{\partial t} = 0$$</span></p>
| Frits Veerman | 273,748 | <p>Your equation is linear, so you could try expanding the solutions in a Fourier series. Moreover, as the initial condition is given in terms of <span class="math-container">$\cos(x)$</span>, it's a good idea to write
<span class="math-container">\begin{align}
Y(t,x) &= \sum_{k=0}^\infty a_Y(k,t) \cos(k x) + b_Y(k,t) \sin(k x)\\
Z(t,x) &= \sum_{k=0}^\infty a_Z(k,t) \cos(k x) + b_Z(k,t) \sin(k x)
\end{align}</span>
which yields a linear system of ordinary differential equations for the coefficients <span class="math-container">$a_{Y,Z}(k,t)$</span> and <span class="math-container">$b_{Y,Z}(k,t)$</span>. The initial conditions of the PDE system then translate in
<span class="math-container">\begin{equation}
a_Y(1,0) = 1,\quad a_Y(k,0) = 0 \; (k \neq 1),\quad b_Y(k,0) = 0 = b_Z(k,0),
\end{equation}</span>
which are initial conditions for the ODE system. The boundary condition(s) translate in
<span class="math-container">\begin{equation}
\frac{\partial Y}{\partial t}(t,0) = \sum_{k=0}^\infty \frac{\text{d} a_Y}{\text{d} t}(k,t) = 0 = \sum_{k=0}^\infty \frac{\text{d} a_Z}{\text{d} t}(k,t) = \frac{\partial Z}{\partial t}(t,0).
\end{equation}</span>
Note that, as it stands, the initial- and boundary conditions do not fully determine the solution. </p>
|
879,640 | <p>Does a matrix have only one inverse matrix (like the inverse of an element in a field)? If so, does this mean that</p>
<p>$A,B \text{ have the same inverse matrix} \iff A=B$?</p>
| vadim123 | 73,324 | <p>Specific counterexample to the non-square case: Let $A=\left(\begin{smallmatrix}1&0&0\\0&1&0\end{smallmatrix}\right)$. </p>
<p>Then there is a matrix $B$ such that $AB=I$, namely $B=\left(\begin{smallmatrix}1&0\\0&1\\x&y\end{smallmatrix}\right)$. Note that $x,y$ can each be anything, so this right inverse is not unique. Also note that $I$ is $2\times 2$.</p>
<p>On the other hand, there is <em>no</em> matrix $C$ such that $CA=I$, where now $I$ would have to be $3\times 3$. The reason is that the last column of $A$ is all zeroes, so the last column of $CA$ would be all zeroes as well. Thus $A$ has no left inverse at all.</p>
|
902,653 | <p>I have recently been learning some hyperbolic geometry and the professor briefly mentioned spherical geometry. From a modern, naive point of view, it seems quite easy to show that spherical geometry is an example of non euclidean geometry.</p>
<p>Clearly, this is not true since it took over a 1000 years to do this. Is the apparent difference in difficulty due to my modern viewpoint or were there technical difficulties like with hyperbolic geometry? </p>
| Mikhail Katz | 72,694 | <p>The key modern insight here is the distinction between axioms and models, in other words between syntax and semantics. Euclidean axioms were once thought to characterize a unique <em>space</em> whatever that may be. The idea that truth of a proposition may be relative to a model is a revolutionary idea that we take for granted today.</p>
|
1,598,545 | <p>Maybe I am not well versed with the actual definition of mean, but I have a doubt. On most resources, people say that arithmetic mean is the sum of $n$ observations divided by n. So my first question: </p>
<blockquote>
<p>How does this formula work? Is there any derivation to it? If not,
then while creating this definition, what was the creator thinking?</p>
</blockquote>
<p>Okay, so using my intuition, I thought that it is the value that lies in the centre. And it worked for some cases, like the mean of $1$ , $2$ and $3$ is $2$ , which is the central value. But, lets imagine a number line from numbers $0$ to $9$. Now, I choose $3$ numbers, say $1$, $8$ and $9$. By the formula, I get the mean is equal to $6$. But, if mean really is a central value, shouldn't it be $5$(I know we call $5$ the median in this case)? But it seems like the mean is getting closer to $8$ and $9$, which means it is not central? So my final question?</p>
<blockquote>
<p>Have I imagined mean incorrectly? What kind of central value really
mean is?</p>
</blockquote>
| Aloizio Macedo | 59,234 | <p>Let me give the bureaucratic answer first:</p>
<p>No, there is no "derivation" (supposing that you mean something like a "theorem") that concludes that $E(X)$ is what it is. $E(X)$ is <em>defined</em> that way.</p>
<p>Now to properly answer your question:</p>
<p>At a very naive level, you can think of the mean as "given a collection of values coming from $n$ persons, the value you should distribute to each person in order that all of them receive the same value". It is a nice intuition, but you may feel it is a bit circular... </p>
<p>Well, in my opinion, if you don't understand the intuition (or think it is not justified) of the definition, the proper way to get a feeling of what "mean" is is by taking a look at the <a href="https://en.wikipedia.org/wiki/Law_of_large_numbers" rel="nofollow">Law of Large Numbers</a>.</p>
|
1,726,187 | <p>Recently in class our teacher told us about the evaluating of the sum of reciprocals of squares, that is $\sum_{n=1}^{\infty}\frac{1}{n^2}$. We began with proving that $\sum_{n=1}^{\infty}\frac{1}{n^2}<2$ by induction. However, we actually proved a stronger result, namely that $\sum_{n=1}^{\infty}\frac{1}{n^2}<2-\frac{1}{n}$, because we were told that it is much easier to prove the stronger result than the weaker one. My question is: why exactly is it easier? Is it because our induction hypothesis is stronger? And how to prove specifically the weaker result using induction, without actually evaluating the sum to be $\frac{\pi^2}{6}$?</p>
| Mark Viola | 218,419 | <p>To arrive at the result without induction, we note that (<a href="https://en.wikipedia.org/wiki/Integral_test_for_convergence#Proof" rel="nofollow">See this for a proof</a>) an upper bound for the sum is given by</p>
<p>$$\begin{align}
\sum_{n=1}^N\frac{1}{n^2}&\le 1+\int_1^N\frac{1}{x^2}\,dx\\\\
&=2-\frac1N
\end{align}$$</p>
<hr>
<p>Now, if we proceed using induction, we first establish a base case. It is easy to see that for $N=1$</p>
<p>$$\sum_{n=1}^1 \frac1{n^2}=1\le 2-\frac11=1$$</p>
<p>Then, we assume that for some $N$ the inequality </p>
<p>$$\sum_{n=1}^N\frac1{n^2}\le 2-\frac1N$$</p>
<p>holds. Then, we see that</p>
<p>$$\begin{align}
\sum_{n=1}^{N+1}\frac1{n^2}&=\sum_{n=1}^N\frac1{n^2}+\frac{1}{(N+1)^2}\\\\
&\le 2-\frac{1}{N}+\frac{1}{(N+1)^2}\\\\
&= 2-\frac{1}{N+1}+\left(\frac{1}{N+1}-\frac{1}{N}+\frac{1}{(N+1)^2}\right)\\\\
&= 2-\frac{1}{N+1}-\frac{1}{N(N+1)^2}\\\\
&\le 2-\frac{1}{N+1}
\end{align}$$ </p>
<p>And we are done!</p>
|
2,721,836 | <p>I recently found a different method to compute prime number in $\mathcal O(\log(\log n))$ complexity. At present, that logic working fine for $300$ digits prime number, which I found on websites.I need to validate whether that logic will be working fine for a higher number of digits. At present, I have computed a prime number of $300\ 000$ digits(but I am not sure whether this would be valid),</p>
<p>My questions are:</p>
<ul>
<li>Where can I find a prime number of higher digits i.e., more than
$300\ 000$ digits?</li>
<li>Where can I validate $300\ 000$ digit prime number is valid one?</li>
</ul>
| Arnaud Mortier | 480,423 | <p>From what you write I understand that you want to prove that your method is fine. It probably is if you checked up to 300, digits. But the only way to validate a method is by analysing it step by step and actually <em>prove</em> that it works. No matter how many digits you try, that will not be a proof that your method has no flaws.</p>
|
18,960 | <p>I am making this post in regards to the ongoing delete/undelete skirmish (let's at least change the monotonicity of the use of "war"). The old version of the question is <a href="https://math.stackexchange.com/revisions/172652/3">here</a>, the current version (after edits today) <a href="https://math.stackexchange.com/q/172652/11619">here</a>, and the answer <a href="https://math.stackexchange.com/questions/172652/conics-passing-through-integer-lattice-points/172683#172683">here</a> </p>
<p>There are two facts:</p>
<ol>
<li>The original post is of poor quality: it is a problem statement of various simultaneous problems, which shows no effort from the OP.</li>
<li>Nonetheless, the post has received a very good answer, which is worth keeping (in my personal opinion) for the sake of future users.</li>
</ol>
<p>As a consequence, the moderator team decided to undelete the question some days ago. The question was again deleted by two of the three users that had deleted it before. Since the question is already closed, I find no reason to delete it, given it signifies the deletion of a useful answer. </p>
<p>The mod team (more precisely, Pedro and Jyrki) have undeleted, locked, edited, unlocked and reopened the question today.</p>
| Jyrki Lahtonen | 11,619 | <p>For my part I will add the following hopefully clarifying comments:</p>
<ul>
<li>The question showed no effort, and was a wrapper for five separate although closely related questions. I believe that the community at large disproves of questions of this type.</li>
<li>IMO the question was sufficiently non-trivial that many future readers will benefit from studying an answer.</li>
<li>I just edited the question in a way that hopefully makes it meet our quality standards by removing all but a single part, and adding details to that last part. I hope that the answerers will edit their posts to reflect this change.</li>
<li>The users who are worried about questions getting deleted CANNOT refer to this case as a precedent, and expect a moderator to step in on their behalf in the future. My personal advice (not speaking for the entire mod team in this bullet) is that the said users should act pre-emptively, and edit the contentious questions themselves. </li>
<li>The second bullet swayed me into action on this occasion. I also want to test, how the two factions receive this. Also I want to set a model example of what kind of edits will make a question palatable (assuming that this passes the test).</li>
</ul>
<p>Unless we get other tools to stop delete/undelete wars (bar on repeat votes or some such), the mods will act by locking posts. But you should not expect all the contentious questions to all be locked in either deleted or undeleted state. </p>
|
920,782 | <p>How do I find the number of integral solutions to the equation - </p>
<p>$$2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N$$</p>
<p>$$x_1,x_2,\ldots,x_7 \ge 1$$</p>
<p>I just thought that I should reduce this a bit more, so I replace $x_i$ with $(y_i+1)$, so we have:</p>
<p>$$y_1 + y_2 + \cdots + y_6 = \tfrac{1}{2}(N + 13 - y_7)$$</p>
<p>$$y_1,y_2,\ldots,y_7 \ge 0$$</p>
<p>I will be solving this as a programming problem by looping over $y_7$ from $[0, N+13]$. How do I find the number of solutions to this equation in each looping step?</p>
| Khosrotash | 104,171 | <p>$$x_{7}=2k+1 , x_{7}=2k \\ partition - by - x_{7}\\2x_{1}+2x_{2}+2x_{3}+...+x_{7}=N\\(1) x_{7}=2k+1 , 2x_{1}+2x_{2}+2x_{3}+...+2k+1=N\\2x_{1}+2x_{2}+2x_{3}+...+2K=N-1\\if -(N-1) - was- even -divide - by -2\\x_{1}+x_{2}+x_{3}+...+K=\frac{N-1}{2}\\\binom{\frac{N-1}{2}-1}{7-1}\\ \\(2) x_{7}=2k , 2x_{1}+2x_{2}+2x_{3}+...+2k=N\\if\\N \\was\\even\\divide -by -2 \\x_{1}+x_{2}+x_{3}+...+K=\frac{N}{2}\\\binom{\frac{N}{2}-1}{7-1} $$</p>
|
813,716 | <p>I am supposed to calculate the following as simple as possible.</p>
<p>Calcute:
$$1 + 101 + 101^2 + 101^3 + 101^4 + 101^5 + 101^6 + 101^7$$
Tip: $$ 101^8 = 10828567056280801$$</p>
<p>I have absolutely no idea how this tip is supposed to help me.<br>
Do I still have to calculate each potency?<br>
Can I somehow solve it with 101^7 * (1 + 101) = 10828567056280801?</p>
<p>As I am not allowed to use a calculator a simple technique for formulas like the one above would be welcome.</p>
| lhf | 589 | <p><em>Hint:</em> Let $S=1 + 101 + 101^2 + 101^3 + 101^4 + 101^5 + 101^6 + 101^7$ and consider $101S$.</p>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| riemann_lebesgue | 539,416 | <p>Here's a different way of doing it! Unfortunately I don't really know how to use latex, so here is the outline</p>
<p>Using the residue theorem, we know that <span class="math-container">${n \choose k}$</span> equals the contour integral of </p>
<p><span class="math-container">$(1+z)^N / z^{k+1}) {/}(2*pi*i)$</span></p>
<p>where <span class="math-container">$z$</span> is restricted to the unit circle, i.e. its magnitude is 1.</p>
<p>Now we write out the sum, but writing the <span class="math-container">${R \choose k}$</span> term as its complex integral representation.</p>
<p>Now bring the sum inside the integral. Rearrange, and use the fact that a contour integral of a function without a singularity equals 0 (so you can add terms without singularities at will). You will get the contour integral of </p>
<p><span class="math-container">$(1+z)^{R+M} / z^{N+1}) {/}(2*pi*i)$</span></p>
<p>as required.</p>
<p>(I will try and remember to update this when my latex skills improve!!)</p>
|
622,076 | <p>Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? </p>
<p>It seems to me like they are equal definitions in a way. </p>
<p>Can you give me a counter-example? </p>
<p>Thanks</p>
| Andrés E. Caicedo | 462 | <p><strong>I.</strong></p>
<p>Some of the answers reveal a confusion, so let me start with the definition. If $I$ is an interval, and $f:I\to\mathbb R$, we say that $f$ has the <em>intermediate value property</em> iff whenever $a<b$ are points of $I$, if $c$ is between $f(a)$ and $f(b)$, then there is a $d$ <em>between</em> $a$ and $b$ with $f(d)=c$. </p>
<p>If $I=[\alpha,\beta]$, this is significantly stronger than asking that $f$ take all values between $f(\alpha)$ and $f(\beta)$:</p>
<ul>
<li>For example, this implies that if $J\subseteq I$ is an interval, then $f(J)$ is also an interval (perhaps unbounded). </li>
<li>It also implies that $f$ cannot have <em>jump</em> discontinuities: For instance, if $\lim_{x\to t^-}f(x)$ exists and is strictly smaller than $f(t)$, then for $x$ sufficiently close to $t$ and smaller than $t$, and for $u$ sufficiently close to $f(t)$ and smaller than $f(t)$, $f$ does not take the value $u$ in $(x,t)$, in spite of the fact that $f(x)<u<f(t)$. This indicates that if $f$ is discontinuous, its discontinuities must be somewhat wild. </li>
<li>In particular, if $f$ is discontinuous at a point $a$, then there are $y$ such that the equation $f(x)=y$ has infinitely many solutions near $a$.</li>
</ul>
<p><strong>II.</strong></p>
<p>There is a nice survey containing detailed proofs of several examples of functions that both are discontinuous and have the intermediate value property: I. Halperin, <a href="http://cms.math.ca/10.4153/CMB-1959-016-1" rel="noreferrer"><em>Discontinuous functions with the Darboux property</em></a>, Can. Math. Bull., <strong>2 (2)</strong>, (May 1959), 111-118. It contains the amusing quote </p>
<blockquote>
<p>Until the work of Darboux in 1875 some mathematicians believed that [the intermediate value] property actually implied continuity of $f(x)$. </p>
</blockquote>
<p>This claim is repeated in other places. For example, <a href="https://www.encyclopediaofmath.org/index.php/Darboux_property" rel="noreferrer">here</a> one reads </p>
<blockquote>
<p>In the 19th century some mathematicians believed that [the intermediate value] property is equivalent to continuity. </p>
</blockquote>
<p>This is very similar to what we find in A. Bruckner, <strong>Differentiation of real functions</strong>, AMS, 1994. In page 5 we read</p>
<blockquote>
<p>This property was believed, by some 19th century mathematicians, to be equivalent to the property of continuity.</p>
</blockquote>
<p>Though I have been unable to find a source expressing this belief, that this was indeed the case is supported by the following two quotes from Gaston Darboux's <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1875_2_4_/ASENS_1875_2_4__57_0/ASENS_1875_2_4__57_0.pdf" rel="noreferrer"><em>Mémoire sur les fonctions discontinues</em></a>, Ann. Sci. Scuola Norm. Sup., <strong>4</strong>, (1875), 161–248. First, on pp. 58-59 we read:</p>
<blockquote>
<p>Au risque d'être trop long, j'ai tenu avant tout, sans y réussir peutêtre, à être rigoureux. Bien des points, qu'on regarderait à bon droit comme évidents ou que l'on accorderait dans les applications de la science aux fonctions usuelles, doivent être soumis à une critique rigoureuse dans l'exposé des propositions relatives aux fonctions les plus générales. Par exemple, on verra qu'il existe des fonctions continues qui ne sont ni croissantes ni décroissantes dans aucun intervalle, qu'il y a des fonctions discontinues qui ne peuvent varier d'une valeur à une autre sans passer par toutes les valeurs intermédiaires.</p>
</blockquote>
<p>The proof that derivatives have the intermediate value property comes later, starting on page 109, where we read:</p>
<blockquote>
<p>En partant de la remarque précédente, nous allons montrer qu'il existe des fonctions discontinues qui jouissent d'une propriété que l'on regarde quelquefois comme le caractère distinctif des fonctions continues, celle de ne pouvoir varier d'une valeur à une autre sans passer par toutes les valeurs intermediaires.</p>
</blockquote>
<p><strong>III.</strong></p>
<p>Additional natural assumptions on a function with the intermediate value property imply continuity. For example, <a href="https://math.stackexchange.com/q/7172/462">injectivity</a> or <a href="https://math.stackexchange.com/q/233054/462">monotonicity</a>. </p>
<p>Derivatives have the intermediate value property (see <a href="https://math.stackexchange.com/q/54843/462">here</a>), but there are discontinuous derivatives: Let $$f(x)=\left\{\begin{array}{cl}x^2\sin(1/x)&\mbox{ if }x\ne0,\\0&\mbox{ if }x=0.\end{array}\right.$$ (The example goes back to Darboux himself.) This function is <a href="https://math.stackexchange.com/q/232672/462">differentiable</a>, its derivative at $0$ is $0$, and $f'(x)=2x\sin(1/x)-\cos(1/x)$ if $x\ne0$, so $f'$ is discontinuous at $0$.</p>
<p>This example allows us to find functions with the intermediate value property that are not derivatives: Consider first $$g(x)=\left\{\begin{array}{cl}\cos(1/x)&\mbox{ if }x\ne0,\\ 0&\mbox{ if }x=0.\end{array}\right.$$ This function (clearly) has the intermediate value property and indeed it is a derivative, because, with the $f$ from the previous paragraph, if $$h(x)=\left\{\begin{array}{cl}2x\sin(1/x)&\mbox{ if }x\ne 0,\\ 0&\mbox{ if }x=0,\end{array}\right.$$ then $h$ is continuous, and $g(x)=h(x)-f'(x)$ for all $x$. But continuous functions are derivatives, so $g$ is also a derivative. Now take $$j(x)=\left\{\begin{array}{cl}\cos(1/x)&\mbox{ if }x\ne0,\\ 1&\mbox{ if }x=0.\end{array}\right.$$ This function still has the intermediate value property, but $j$ is <em>not</em> a derivative. Otherwise, $j-g$ would also be a derivative, but $j-g$ does not have the intermediate value property (it has a jump discontinuity at $0$). For an extension of this theme, see <a href="https://math.stackexchange.com/a/624109/462">here</a>.</p>
<p>In fact, a function with the intermediate value property can be extremely chaotic. Katznelson and Stromberg (<em>Everywhere differentiable, nowhere monotone, functions</em>, The American Mathematical Monthly, <strong>81</strong>, (1974), 349-353) give an example of a differentiable function $f:\mathbb R\to\mathbb R$ whose derivative satisfies that each of the three sets $\{x\mid f'(x)>0\}$, $\{x\mid f'(x)=0\}$, and $\{x\mid f'(x)<0\}$ is dense (they can even ensure that $\{x\mid f'(x)=0\}=\mathbb Q$); this implies that $f'$ is highly discontinuous. Even though their function satisfies $|f'(x)|\le 1$ for all $x$, $f'$ is not (Riemann) integrable over any interval. </p>
<p>On the other hand, derivatives must be continuous <em>somewhere</em> (in fact, on a dense set), see <a href="https://math.stackexchange.com/a/112133/462">this answer</a>. </p>
<p>Conway's <a href="https://en.wikipedia.org/wiki/Conway_base_13_function" rel="noreferrer">base 13 function</a> is even more dramatic: It has the property that $f(I)=\mathbb R$ for all intervals $I$. This implies that this function is discontinuous everywhere. Other examples are discussed in <a href="https://math.stackexchange.com/a/614435/462">this answer</a>.</p>
<p>Halperin's paper mentioned above includes examples with even stronger discontinuity properties. For instance, there is a function $f:\mathbb R\to\mathbb R$ that not only maps each interval onto $\mathbb R$ but, in fact, takes each value $|\mathbb R|$-many times on each uncountable closed set. To build this example, one needs a bit of set theory: Use transfinite recursion, starting with enumerations $(r_\alpha\mid\alpha<\mathfrak c)$ of $\mathbb R$ and $(P_\alpha\mid\alpha<\mathfrak c)$ of its perfect subsets, ensuring that each perfect set is listed $\mathfrak c$ many times. Now recursively select at stage $\alpha<\mathfrak c$, the first real according to the enumeration that belongs to $P_\alpha$ and has not been selected yet. After doing this, continuum many reals have been chosen from each perfect set $P$. List them in a double array, as $(s_{P,\alpha,\beta}\mid\alpha,\beta<\mathfrak c)$, and set $f(s_{P,\alpha,\beta})=r_\alpha$ (letting $f(x)$ be arbitrary for those $x$ not of the form $s_{P,\alpha,\beta}$). </p>
<p>To search for references: The intermediate value property is sometimes called the <a href="https://www.encyclopediaofmath.org/index.php/Darboux_property" rel="noreferrer"><em>Darboux property</em></a> or, even, one says that a function with this property is <em>Darboux continuous</em>.</p>
<p>An excellent book discussing these matters is A.C.M. van Rooij, and W.H. Schikhof, <strong>A second course on real functions</strong>, Cambridge University Press, 1982.</p>
|
3,009,345 | <p>I ran into this question which hints me to use Cauchy's Integral Theorem for Derivatives, however I don't seem to be able to fit this integral into the form of the Integral Formula</p>
<p><span class="math-container">$$\displaystyle \int_{|z|=2} \frac{\cos(z)}{z(z^2+8)}dz$$</span> I tried using the fact that <span class="math-container">$\displaystyle \int_\gamma f(z)dz=\int_a^b f(\gamma(t))\gamma'(t)dt$</span> for <span class="math-container">$\gamma(t)$</span> where <span class="math-container">$t \in [a, b]$</span>. But got nowhere. Is there a way I can transform the given integral into the form in which I can use the Integral Formula as stated below?</p>
<p><span class="math-container">$$f^{(k)}(w)=\frac{k!}{2\pi i}\int_\gamma \frac{f(z)}{(z-w)^{k+1}}dz$$</span> where <span class="math-container">$f^{(k)}(w)$</span> is the <span class="math-container">$k^{th}$</span> derivative of <span class="math-container">$f$</span></p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
\int_{0}^{2}{1 \over 4}
{\bracks{0 < z/x < x^{3}} \over \verts{x}}\,\dd x & =
{1 \over 4}\bracks{z > 0}
\int_{0}^{2}{\bracks{x > z^{1/4}} \over x}\,\dd x
\\[5mm] & =
{1 \over 4}\bracks{z > 0}\bracks{z^{1/4} < 2}\int_{z^{1/4}}^{2}
{\dd x\over x}
\\[5mm] & =
{1 \over 4}\bracks{0 < z < 16}\ln\pars{2 \over z^{1/4}}
\end{align}</span></p>
<p><a href="https://i.stack.imgur.com/jFMgo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jFMgo.png" alt="enter image description here" /></a></p>
|
871,542 | <p>I have the following theorem:</p>
<blockquote>
<p>Let <span class="math-container">$\rho$</span> be the traffic intensity.</p>
<p>a) If <span class="math-container">$\rho<1$</span>, then <span class="math-container">$X$</span> is positive recurrent.</p>
<p>b) If <span class="math-container">$\rho>1$</span>, then <span class="math-container">$X$</span> is transient.</p>
<p>c) If <span class="math-container">$\rho=1$</span>, then <span class="math-container">$X$</span> is null-reсurrent.</p>
</blockquote>
<p>It is not important what exactly <span class="math-container">$\rho$</span> and <span class="math-container">$X$</span> are. Let's assume I have proved a) and b). Then it remains to prove the case <span class="math-container">$\rho=1$</span>:</p>
<p>Can I say the following: <span class="math-container">$X$</span> is transient if and only if <span class="math-container">$\rho>1$</span>, and is positive recurrent if and only if <span class="math-container">$\rho<1$</span>. If <span class="math-container">$\rho=1$</span> then <span class="math-container">$X$</span> has no choice but null-recurrent.</p>
<p>Is that correct?</p>
| Community | -1 | <p>I disagree with <a href="https://math.stackexchange.com/users/88052/mark-fantini">@MarkFantini</a>'s <a href="https://math.stackexchange.com/a/871548">answer</a>.</p>
<p>In this section (<span class="math-container">$\S$</span>1.II.2 <em>Moving Particle Argument</em>), Needham is giving a heuristic argument in support of the fact that <span class="math-container">$e^{i \theta} = \cos \theta + i \sin \theta$</span>. He has <em>not</em> defined <span class="math-container">$e^{i \theta}$</span> to be <span class="math-container">$\cos \theta + i \sin \theta$</span>, and in fact he argues that doing so is "a low blow to Euler, reducing one of his greatest achievements to a mere tautology" (<span class="math-container">$\S$</span>1.II.1 <em>Introduction</em>, page 10).</p>
<p>Instead, what Needham is arguing in <span class="math-container">$\S$</span>1.II.2 is that if we suppose that it makes sense to talk about <span class="math-container">$e^{it}$</span>, then a natural property we would want it to have is <span class="math-container">$\frac{d}{dt}e^{it} = i e^{it}$</span> (eq. (11) on page 11), analogous to the known property of the real exponential function, namely <span class="math-container">$\frac{d}{dx} e^{kx} = k e^{kx}$</span> for any real <span class="math-container">$k$</span>.</p>
<p>Now, visualizing the function <span class="math-container">$t \mapsto e^{it}$</span> as describing the motion of a moving particle in the plane, the quantity <span class="math-container">$\frac{d}{dt} e^{it}$</span> is then the velocity of the particle. We know that multiplication by <span class="math-container">$i$</span> corresponds to a rotation of the plane by <span class="math-container">$\pi/2$</span> radians in the counter-clockwise direction. So, the position and velocity vectors are at right angles to each other at every point. From this, and the fact that at <span class="math-container">$t = 0$</span> the particle is at <span class="math-container">$e^0 = 1$</span>, one can deduce that the particle is just moving around the unit circle.</p>
<p>Hence, <span class="math-container">$\lvert e^{it} \rvert = 1$</span> for all <span class="math-container">$t$</span>.</p>
|
1,499,423 | <p>Are there a group of numbers whose squares are made up of squares? For example, $7$ would be one because $7^2$ is $49$ which has $2^2$ and $3^2$. $20$ would be another example.</p>
<p>What are these numbers called?</p>
<p>Please help me find good <strong>tags</strong> for this question.</p>
| Piquito | 219,998 | <p>I am not agree with taking $20$ as a number of that kind ( or trivial in whose case $10, 30,40,50,60,70,80,90$ are too). </p>
<p>We show here that the only two-digit number with four-digit square of this kind is $41$. Looking at the algorithm of extraction of the square root in the figure below we can write</p>
<p><a href="https://i.stack.imgur.com/NwKg6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NwKg6.png" alt="enter image description here"></a></p>
<p>$$ab=x^2\Rightarrow 4\le x\le 9$$
$$y(20x+y)=cd=z^2\Rightarrow y(20x+y)\le 81\Rightarrow x=4$$</p>
<p>Hence $$y(80+y)\le81\Rightarrow y=1$$
No other possibility. In fact $41^2=1681$ as we wished.</p>
|
1,262,305 | <p>$$f(x)=\int_0^x\left(\frac{t^3-2t^2-4}{t^2+1}\right)\ dt$$
I need to find the $x$ and $y$ intercepts, and the inflection points of the function $f(x)$ (with both $x$ and $y$ coordinates). I need to find it through the calculator and explain my answer. How do I find the antiderivative?</p>
| mathreadler | 213,607 | <p>So you already have how to calculate the integral explicitly from Olivier. However you can find the maximums and minimums with the fundamental theorem of calculus together with finding zeros to the polynomial $t^3 - 2t^2 - 4$. At least one $t$ root is easy to guess and then continue with polynomial division and completing the square.</p>
|
1,598,006 | <p>(Here, $B$ is relatively compact means the closure of $B$ is compact.)</p>
<blockquote>
<ol>
<li><p>$\hat A$ is compact.</p></li>
<li><p>$\hat A=\hat {\hat A}$.</p></li>
<li><p>$\hat A$ is connected.</p></li>
<li><p>$\hat A=X$.</p></li>
</ol>
</blockquote>
<p>I try to eliminate the options by using an example.</p>
<p>Consider $X=\Bbb R - \{1,2,3\}$ with metric topology and let $A=(-\infty,1)$.</p>
<p>Then $\hat A=(-\infty,1) \cup (1,2) \cup (2,3)$.</p>
<p>Hence options 1,3,4 are false.</p>
<p>So I select option 2 as an answer.</p>
<p>Is my method correct?</p>
| Forever Mozart | 21,137 | <p>I think $\hat A=A$ for your example. The closure in your $X$ of $(1,2)$ equals $(1,2)$, which is not compact, and similarly for $(2,3)$. But you can still mark off 1 and 4 because of this.</p>
<p>How about $X=\{1\}\cup \{2\}$ and $A=\{1\}$. Then $\hat A=X$, which is not connected. So you can rule out option 3.</p>
<p>Thus option 2 is correct (you may want to try a simple proof of this).</p>
|
1,341,486 | <p>Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:</p>
<p>\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.</p>
<p>Thanks Bob</p>
| Masacroso | 173,262 | <p>You can write it as $\sum_{k\ge0}\frac{2k+1}{(k+3)_3}=\sum_{k\ge0}(2k+1)(k)_{-3}$ and now it seems easy to solve by summation by parts:</p>
<p>$$\sum (2k+1)(k)_{-3}\delta k=(2k+1)\frac{(k)_{-2}}{-2}+\sum(k+1)_{-2}=\frac{2k+1}{-2(k+2)_2}-\frac{1}{k+2}=\frac{4k+3}{-2(k+2)_2}$$</p>
<p>And taking limits we have that the series converges to $\sum_{k\ge 0}\frac{2k+1}{(k+3)_3}=\frac{3}{4}$</p>
<p>The partial sum will be</p>
<p>$$\sum_{k=0}^{n}\frac{2k+1}{(k+3)_3}=\frac{4n+3}{-2(n+2)_2}+\frac{3}{4}$$</p>
|
872,657 | <p>For $1 \leq r < p < \infty$ prove the continuous injection of $L^p([0, 1])$ into $L^r([0, 1])$. </p>
<p>I am having a hard time starting. Any suggestions. I tried a straight forward approach. That is, given $\epsilon > 0$, I tried to find a $\delta >0$ such that $||f - g||_p < \delta$ implies that $||f - g||_r < \epsilon.$</p>
<p>Thanks for any help.</p>
| Vítězslav Štembera | 663,062 | <p>Interesting! The fulltext of the paper is here:</p>
<p><a href="https://cowles.yale.edu/sites/default/files/files/pub/cdp/m-0403.pdf" rel="nofollow noreferrer">https://cowles.yale.edu/sites/default/files/files/pub/cdp/m-0403.pdf</a></p>
|
1,876,287 | <p><strong>Question:</strong></p>
<p>Let P be a point where the normal (in the point where the x-coordinate is h) to the curve</p>
<p>$$y = e^{2x} - 2x$$</p>
<p>cuts the y-axis. Determine the y-coordinates of P when h goes to 0.</p>
<p><strong>Attempted solution:</strong></p>
<p>I first decided to draw the following image:</p>
<p><a href="https://i.stack.imgur.com/O1aQb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O1aQb.png" alt="enter image description here"></a></p>
<p>So basically, we have the function for the curve, but we need to get on the normal and then move along the normal until it reaches the y-axis.</p>
<p>Let us start by taking the derivative of the function:</p>
<p>$$y' = 2e^{2x} - 2$$</p>
<p>The point x = h is the relevant point here, so:</p>
<p>$$y'(h) = 2e^{2h} - 2$$</p>
<p>Finding the k value of the normal:</p>
<p>$$k_1 k_2 = -1 \rightarrow k_2 = \frac{-1}{2e^{2h} - 2}$$</p>
<p>The equation for the normal is then:</p>
<p>$$y - (2e^{2h} - 2) = \frac{-1}{2e^{2h} - 2}(x-h)$$</p>
<p>Simplification gives:</p>
<p>$$y = 2e^{2h} - 2 -\frac{x-h}{2e^{2h} - 2}$$</p>
<p>If h goes to 0, this goes towards infinity. </p>
<p>But that is hardly the case and cannot be true since it obviously has to cut the y-axis. Somewhere, something must have gone terribly, terribly wrong. The correct answer is $\frac{5}{4}$.</p>
| Jean Marie | 305,862 | <p>@Claude Leibovici @Ahmed Hussein </p>
<p>There is an alternative answer. </p>
<p>Let us first recall that the envelope of the normals to a curve (called its <a href="https://en.wikipedia.org/wiki/Evolute" rel="nofollow noreferrer">evolute</a>) is the locus of the centres of curvature for this curve.</p>
<p>Here, function $f$ is equivalent (see Remark 1) in the vicinity of $0$ to function $g$ defined by:</p>
<p>$$g(x)=1-\dfrac{x^2}{2}$$</p>
<p>(see graphics below where the curve of $f$ is black, and the curve of $g$ is red).</p>
<p>Let us switch then from the curve of $f$ to the equivalent curve $(C)$ of $g$, a parabola with summit $S=(0,1)$. (C) has clearly axis $Oy$ as its normal at point $S$.</p>
<p>The radius of curvature $R$ of curve (C) with equation $y=g(x)$ being the same as that of $y=f(x)$, the ordinate of the centre of curvature $P$ of (C) at point $S$ will be $R+1$. It suffices then to establish that $R=\dfrac{1}{4}$ for $x=0$.</p>
<p>This is immediate using <a href="http://mathworld.wolfram.com/RadiusofCurvature.html" rel="nofollow noreferrer">classical formula</a></p>
<p>$$R=\dfrac{(1+g'(0)^2)^{3/2}}{|g''(0)|}$$</p>
<p>Remark 1 : Adjective "equivalent" may be ambiguous : the important thing is that $f^{(k)}=g^{(k)}$ for $k=0,1,2$.</p>
<p>Remark 2: the evolute of a parabola is represented <a href="http://mathworld.wolfram.com/ParabolaEvolute.html" rel="nofollow noreferrer">here</a> has a cusp.</p>
<p>Remark 3 : have a look at this remarkable <a href="http://blogs.ams.org/visualinsight/2016/05/01/involutes-of-a-cubical-parabola/" rel="nofollow noreferrer">blog of the AMS</a>. Note that these curves are involutes ("building an/the involute" is the inverse operation of "building an/the evolute"). </p>
<p><a href="https://i.stack.imgur.com/An9Tw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/An9Tw.jpg" alt="enter image description here"></a></p>
|
2,481,046 | <p>I have a question that asks to show that $S^2 = \{(x,y,z) \in \mathbb{R}^3|x^2+y^2+z^2=1\}$ is a differentiable manifold. My professor says that one way to do this is to define the following 6 parametrizations of the sphere, which cover the entire sphere.</p>
<p>$\vec{\phi_{i}}:V \to \mathbb{R}^3$ where $V = \{(u,v) \in \mathbb{R}^2|u^2+v^2<1\}$</p>
<p>$\vec{\phi_{1}}(u,v) = (u,v,\sqrt{1-u^2-v^2}) \qquad (z>0)$</p>
<p>$\vec{\phi_{2}}(u,v) = (u,v,-\sqrt{1-u^2-v^2}) \qquad (z<0)$</p>
<p>$\vec{\phi_{3}}(u,v) = (u,\sqrt{1-u^2-v^2},v) \qquad (y>0)$</p>
<p>$\vec{\phi_{4}}(u,v) = (u,-\sqrt{1-u^2-v^2},v) \qquad (y<0)$</p>
<p>$\vec{\phi_{5}}(u,v) = (\sqrt{1-u^2-v^2},u,v) \qquad (x>0)$</p>
<p>$\vec{\phi_{6}}(u,v) = (-\sqrt{1-u^2-v^2},u,v) \qquad (x<0)$</p>
<p>I don't understand what these parameterizations mean at all and I don't understand what a parameterization is. From what I can read online, it's some function but I'm not sure why this specific function with $u$ and $v$ is what we're using to cover the entire sphere. Can someone explain this to me please?</p>
| Fred | 380,717 | <p>Convergence: Since $ \int_0^1 1 dx$ converges and $|\sin(x+1/x)| \le 1$, the integral $\int_{0}^{1} \sin(x+\frac{1}{x})dx$ converges absolutely by the comparison test.</p>
|
891,575 | <p>The circumference of a circle has length 90 centimeters, Three points on the circle divide the circle into three equal lengths. Three ants A, B, and C start to crawl clockwise on the circle, with starting from one of the three points. Initially A is ahead of B and B is ahead of C. Ant A crawls 3 centimeters per second, ant V 5 centimeters, and and C 10 centimeters. How long does it take for the three ants to arrive at the same spot for the first time?</p>
<p>I tried making a list and writing down the numbers, but they seem to never be the same. I know the distance formula is <em>d=rt</em>, but I don't know how to use it to solve this problem. Any help? Thanks!</p>
| paw88789 | 147,810 | <p>First focus on A and B. Since B goes 2 cm/sec faster than A, B first catches A after 15 seconds; and then every 45 seconds thereafter. </p>
<p>Now look at B and C. Since C goes 5 cm/sec faster than B, C first catches B after 6 seconds, and every 18 seconds there after.</p>
<p>So A and B coincide at times $15 + 45n$ seconds for $n = 0, 1, ...$; and B and C coincide at times $6 + 18k$ seconds for $k = 0, 1, 2,...$. </p>
<p>So you need $15 + 45n=6 + 18k$ for nonnegative integer $n,k$.</p>
<p>The smallest solution is $n=1, k=3$. Giving a time of $60$ seconds. </p>
|
2,694,740 | <p>$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = ? $$</p>
<p>These questions are making me confused because we're dealing with the terms like $10^x$. What are your professional tips? </p>
<p><strong>My attempt:</strong></p>
<p>$$\frac{2.10^{-7} - 4.10^{-7}}{10^{-8}} \tag{1} $$
$$\frac{ -8.10^{-7}}{10^{-8}} \tag{2} $$</p>
<p>And that's where I'm stuck. </p>
| CY Aries | 268,334 | <p>$(2)$ is incorrect.</p>
<p>From $(1)$, $\displaystyle \frac{2\cdot10^{-7} - 4\cdot10^{-7}}{10^{-8}}=\frac{(2-4)\cdot 10^{-7}}{10^{-8}}=\frac{-2\cdot 10^{-7}\cdot 10^8}{10^{-8}\cdot 10^8}=\frac{-2\cdot 10}{1}=-20$</p>
|
2,012,532 | <p>The following is all confirmed to be true:</p>
<p>Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$</p>
<p>U =
$
\begin{bmatrix}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
$</p>
<p>L =
$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-2 & 0 & 1 & 0\\
-1 & -1 & -1 & 0\\
\end{bmatrix}
$</p>
<p>Okay so using that I need to solve the following system:</p>
<p>$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$</p>
<p>So step one is solving $Ly = b$, where $y = Ux$</p>
<p>So that is:</p>
<p>$
y_1 = 0\\
y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$</p>
<p>How can we find $y_3$ in the last two equations? Because,</p>
<p>$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$</p>
<p>So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.</p>
| Siong Thye Goh | 306,553 | <p>Multiplying the first row of $L$ with the first column of $U$ gives us $-1$, which is not the $(1,1)$ entry of $A$. Hence, you have made a mistake in computation of LU decomposition. </p>
<p>There seems to be a missing permutation matrix being involved in your computation.</p>
<p>Your procedure to solve the linear system upon computing the LU decomposition is correct.</p>
|
87,466 | <p>I have some code which involves tiny numbers being put to the power of very large numbers. The function I'm looking at is</p>
<p>$\varphi = \omega(T) \left(1 - (1 - \epsilon)^{n_{e}(T)} \right)$</p>
<p>when $\epsilon $ is very small (~$10^{-16}$) and $n_{e}$ is large ($> 10^{10}$). Both $n_{e}$ and $\omega$ are known functions of $T$. At a known value ($T = 37$) the value of $\varphi$ is known, and we can in theory estimate $\epsilon$ by</p>
<p>$\epsilon = 1 - \left(1 - \frac{\varphi}{\omega} \right)^{\frac{1}{n_{e}}}$</p>
<p>However, when I put this into Mathematica I'm getting very strange behaviour; here's a MWE when $\varphi(37) = 0.2506$; </p>
<pre><code>gammaex = 0.2506;
omega[t_] := 2.43163218375*10^7*Exp[1700*(1/298.15 - 1/(273.15 + t))];
w[t_] := (3.414105049212413*10^12)/(omega[t]);
v[t_] := Sqrt[661.6469313477045*(t + 273.15)];
ne[t_] := (v[t]*5.104757516005496*(10^7));
epsilon = SetAccuracy[ 1 - ( 1 - gammaex/w[37])^(1/ne[37]), 30];
test = w[37]*(1 - (1 - epsilon)^(ne[37]))
</code></pre>
<p>Now when I evaluate $\varphi(37)$ via the last line I should get the value 0.2506 but I do not; instead I get $\varphi = 0.289104$, over 15% off the true value. I thought maybe this was a precision problem, so I tried some other commands (Surd, Power...) and got the same wrong value. The output value for the epsilon is $\epsilon \approx 1.11 \times 10^{-16} $ from mathematica.</p>
<p>However I've evaluated this with WolframAlpha and got a markedly different value of $\epsilon \approx 9.61 \times 10^{-17}$, and as <a href="http://www.wolframalpha.com/input/?i=%28112608%29*%281%20-%20%20%28%201%20-9.61369%20%C3%97%2010%5E-17%29%5E%282.31246*10%5E10%29%29" rel="nofollow">this link demonstrates</a> the calculation seems to work with WolframAlpha, returning close to the expected value around 0.25*. Any ideas why the calculations are different, and how I can make Mathematica behave with such extreme values? </p>
<p>Incidentally, if I dump the WolframAlpha equation ( (112608)*(1 - (1 - 9.61369*10^-17)^(2.31246*10^10)) ) directly into Mathematica, it still comes up with the wrong value, so I assume it's a radical issue?</p>
| Bob Hanlon | 9,362 | <p>You can use <code>Rationalize</code> to convert numbers to exact numbers</p>
<pre><code>gammaex = 0.2506 // Rationalize[#, 0] &;
omega[t_] =
2.43163218375*10^7*Exp[1700*(1/298.15 - 1/(273.15 + t))] //
Rationalize[#, 0] &;
w[t_] = (3.414105049212413*10^12)/(omega[t]) // Rationalize[#, 0] &;
v[t_] = Sqrt[661.6469313477045*(t + 273.15)] // Rationalize[#, 0] &;
ne[t_] = (v[t]*5.104757516005496*(10^7)) // Rationalize[#, 0] &;
epsilon = 1 - (1 - gammaex/w[37])^(1/ne[37]);
test = w[37]*(1 - (1 - epsilon)^(ne[37]))
</code></pre>
<blockquote>
<p>1253/5000</p>
</blockquote>
<pre><code>% // N
</code></pre>
<blockquote>
<p>0.2506</p>
</blockquote>
|
362,062 | <p>If you didn't know anything about stabilization phenomena in algebraic topology and were trying to discover/prove theorems about the homotopy theory of spaces, what clues would point you toward results such as Freudenthal suspension or the existence of stable homotopy groups of spheres? </p>
<p>References suggest that Freudenthal originally stated his result in <a href="http://www.numdam.org/item/?id=CM_1938__5__299_0" rel="noreferrer">this 1938 Paper</a>, although I'm unable to find an English translation. This was published only a few short years after <a href="https://eudml.org/doc/212801" rel="noreferrer">the discovery of the Hopf fibration</a>, so I find it pretty surprising that not only would there have been clear notions of <span class="math-container">$\pi_{\geq 2}$</span> at the time, but also enough evidence to suggest looking for things like the suspension map or stable homotopy groups.</p>
<p>Analogous stabilization phenomena do seem to occur elsewhere in mathematics: for instance, vector bundles that become isomorphic after taking Whitney sums with trivial bundles. From there, it may not be <em>that</em> much of a leap to suppose that something similar might work for fibrations. </p>
<p>However, it also seems that Freudenthal's paper predated results like this, and so historically, perhaps the flow of ideas was the other way around. What other results might have motivated his suspension theorem? Or in retrospect, what are some signs that such a thing might have worked and been useful?</p>
| Nicholas Kuhn | 102,519 | <p>For those whose German is shaky or non-existent, it is fun to copy and paste a couple of paragraphs of Freudenthal's paper into Google translate. The answer to your question emerges. His paper is concerned with the interplay of the then new Hopf invariant and "suspension" - "Einhängung" in German, and possibly named first in this paper. Another thing that seems to be named first here is the notion of "<span class="math-container">$k$</span>-stem" ("<span class="math-container">$k$</span>-Stamm").</p>
<p>In modern terms, he is exploring the exactness of the EHP sequence: his Satz I says that the kernel of <span class="math-container">$H$</span> (= Hopf) is the image of <span class="math-container">$E$</span> (= Einhängung), his Satz II is telling us that homotopy groups stabilize in the usual way, and his Satz III is showing that the first stable stem is <span class="math-container">$\mathbb Z/2$</span>, with a nonzero element represented by the suspension of any map with odd Hopf invariant.</p>
<p>His methods seem to consist of a careful analysis using simplicial approximation as a key tool. And this would be the answer to your question: anyone exploring such questions finds themselves thinking about general position, how we build up spaces, etc. To a modern student, I would observe that the stable range can be seen by considering the difference between the wedge of two <span class="math-container">$n$</span>-spheres and their product: one needs to attach a <span class="math-container">$2n$</span>-disk using a map from a <span class="math-container">$2n-1$</span>-sphere.</p>
<p>His paper is even more impressive when one remembers that it was written under the darkening cloud of Nazism.</p>
|
362,062 | <p>If you didn't know anything about stabilization phenomena in algebraic topology and were trying to discover/prove theorems about the homotopy theory of spaces, what clues would point you toward results such as Freudenthal suspension or the existence of stable homotopy groups of spheres? </p>
<p>References suggest that Freudenthal originally stated his result in <a href="http://www.numdam.org/item/?id=CM_1938__5__299_0" rel="noreferrer">this 1938 Paper</a>, although I'm unable to find an English translation. This was published only a few short years after <a href="https://eudml.org/doc/212801" rel="noreferrer">the discovery of the Hopf fibration</a>, so I find it pretty surprising that not only would there have been clear notions of <span class="math-container">$\pi_{\geq 2}$</span> at the time, but also enough evidence to suggest looking for things like the suspension map or stable homotopy groups.</p>
<p>Analogous stabilization phenomena do seem to occur elsewhere in mathematics: for instance, vector bundles that become isomorphic after taking Whitney sums with trivial bundles. From there, it may not be <em>that</em> much of a leap to suppose that something similar might work for fibrations. </p>
<p>However, it also seems that Freudenthal's paper predated results like this, and so historically, perhaps the flow of ideas was the other way around. What other results might have motivated his suspension theorem? Or in retrospect, what are some signs that such a thing might have worked and been useful?</p>
| Arun Debray | 97,265 | <p><span class="math-container">$\newcommand{\R}{\mathbb R}\newcommand{\inj}{\hookrightarrow}$</span>This will be an anachronistic answer, because it was discovered a bit later, but: Whitney proved that every
<span class="math-container">$n$</span>-manifold embeds in <span class="math-container">$\R^N$</span> for <span class="math-container">$N$</span> large enough, and Wu showed in 1958 that for <span class="math-container">$N\ge 2n+2$</span>, all such embeddings
are isotopic. This leads to a few interesting stability phenomena: most notably, that every manifold has
canonically the data of the isomorphism type of the normal bundle to the embedding <span class="math-container">$M\inj\R^N$</span>, up to direct sums
with trivial bundles. (And this leads to stable vector bundles, another stabilization in algebraic
topology…)</p>
<p>How might this have led to Freudenthal's theorem? One upshot is that is bordism groups of immersions stabilize, and
in high codimension are just abstract bordism groups. Thom's work on bordisms showed that bordism groups of
immersions are homotopy groups of certain spaces, called Thom spaces, and the Thom space for <span class="math-container">$n$</span>-manifolds in
<span class="math-container">$\R^{N+1}$</span> is the suspension of the Thom space for <span class="math-container">$n$</span>-manifolds in <span class="math-container">$\R^N$</span>. So there are two different reasons
these homotopy groups stabilize (the numbers don't quite match: Freudenthal's theorem is sharper). But in some
alternate history, where Whitney and Wu's work was earlier, one could imagine people asking, “so the homotopy
groups of Thom spaces stabilize, what about everything else?”</p>
<p>(If you modified this by asking for <span class="math-container">$M\inj\R^N$</span> to be equipped with a trivialization of its normal bundle, then the
Thom space is a sphere, so this provides another description/proof of the stable homotopy groups of the spheres.)</p>
|
345,735 | <p>If <strong>two planes</strong> are <strong>intersected</strong> <em>by making a straight line, like <span class="math-container">$AB$</span></em> then</p>
<blockquote>
<p>Does the angle between two planes (see figure) <strong>always</strong> given by the
angle between normal vectors (<span class="math-container">$n_1$</span> and <span class="math-container">$n_2$</span>) ?</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/Y054P.jpg" alt="enter image description here" /></p>
| Phil Wang | 67,822 | <p>In your figure, thing is right. If you reverse vector n1, the angle between two planes plus that between two normal vectors will be 2pi. There will be some difference.</p>
|
3,518,221 | <p>So I had this complex integral </p>
<blockquote>
<p>If <span class="math-container">$0 \leq y \leq 1$</span>, find the maximum value of the integral
<span class="math-container">$$
\int_0^y \left(x^4 + (y-y^2) \right)^{1/2}\, dx
$$</span></p>
</blockquote>
<p>I differentiated the integral using the leibniz rule and equated it to 0.
The problem is that I obtain a expression containing a biquadratic in a square root and an integral. The biquadratic inside the root is always positive when I checked it and the integral was easily evaluated. The complexity of the problem kept on increasing as I started getting fractional exponents.
Please help me on this.</p>
| jacky | 14,096 | <p>I am assuming that you mean <span class="math-container">$\max$</span> of</p>
<p><span class="math-container">$$\displaystyle \int^{y}_{0}\sqrt{x^4+(y-y^2)^2}dx$$</span> subjected to <span class="math-container">$0 \leq y\leq 1$</span></p>
<p><span class="math-container">$\bullet\; $</span> From <span class="math-container">$$\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$$</span> for all <span class="math-container">$a,b\geq 0$</span></p>
<p><span class="math-container">$$\displaystyle \int^{y}_{0}\sqrt{x^4+(y-y^2)^2}dx\leq \int^{y}_{0}x^2dx+\int^{y}_{0}y(1-y)dx$$</span></p>
<p><span class="math-container">$$= \frac{y^3}{3}+y^2(1-y)=\frac{1}{3}y\cdot y \cdot (3-2y)$$</span></p>
<p><span class="math-container">$$=\frac{1}{3}\bigg[y\cdot y \cdot (3-2y)\bigg]\leq \frac{1}{3}\bigg[\frac{y+y+(3-2y)}{3}\bigg]^3=\frac{1}{3}$$</span></p>
<p><span class="math-container">$$\Longrightarrow \bigg(\int^{y}_{0}\sqrt{x^4+(y-y^2)^2}dx\bigg)_{\max}=\frac{1}{3}$$</span></p>
|
3,537,654 | <p><span class="math-container">$$\lim_{x\to 0^{+}} (\tan x)^x$$</span></p>
<p><span class="math-container">$$\lim_{x\to 0^{+}} e^{\ln((\tan x)^x)}=\lim_{x\to 0^{+}} e^{x\ln(\tan x)}=\lim_{x\to 0^{+}} e^{x[\ln(\sin x)-\ln(\cos x)]}$$</span></p>
<p>We can continue to create an expression that may help us use L'Hospital but it does not seem to be correct </p>
<p>P.S can we write:</p>
<p><span class="math-container">$$1=(\frac{-1}{-1})^x\leq \lim_{x\to 0^+}\Bigl(\frac{\sin x}{\cos x}\Bigr)^x\leq \Bigl(\frac{1}{1}\Bigr)^x=1$$</span>?</p>
| Kavi Rama Murthy | 142,385 | <p><span class="math-container">$\lim_{x \to 0+} x \ln (\tan x)=\lim_{x \to 0+} \frac {\ln (\tan x)} {1/x}=-\lim_{x \to 0+} \frac {\sec^{2}x} {\tan x /x^{2}}$</span>. You can write this as <span class="math-container">$-\lim_{x \to 0+} \frac {x^{2}} { \sin x \cos x}$</span> and this limit is <span class="math-container">$0$</span>. (Why?). Hence the given limit is <span class="math-container">$e^{0}=1$</span>. </p>
|
2,453,126 | <p>I'm at a loss here. I tried everything I could think of, can't seem to get the correct answer after plugging in the new boundaries (the teacher wants us to use the new boundaries instead of the ones given). How do I find where the radical values of $\sin(\arctan(x/3))$? I am supposed to evaluate the integral using trig substitution for the following:</p>
<p>$$\int_\sqrt3 ^3 \frac{\sqrt{9+x^2}}{x^6}dx $$ </p>
<p>I inserted a picture of my work, sorry if you can't read my terrible writing, let me know if you have any questions regarding what I wrote.</p>
<p><img src="https://i.stack.imgur.com/d7ssh.jpg" alt="Here is what I have so far"></p>
| Michael Rozenberg | 190,319 | <p>Let $AB=1$.</p>
<p>Thus, by law if sines for $\Delta FEB$ we obtain:
$$\frac{\sin\measuredangle BFE}{BE}=\frac{\sin\measuredangle FBE}{FE}$$ or
$$\frac{\sin\measuredangle BFE}{2\sin54^{\circ}}=\frac{\sin18^{\circ}}{1}.$$
Id est,
$$\sin\measuredangle BFE=2\sin54^{\circ}\sin18^{\circ}=2\cos36^{\circ}\cos72^{\circ}=$$
$$=\frac{4\sin36^{\circ}\cos36^{\circ}\cos72^{\circ}}{2\sin36^{\circ}}=\frac{\sin144^{\circ}}{2\sin36^{\circ}}=\frac{1}{2}$$
and since $\angle BFE$ is an acute angle, we obtain $$\measuredangle BFE=30^{\circ}.$$
Done!</p>
|
2,453,126 | <p>I'm at a loss here. I tried everything I could think of, can't seem to get the correct answer after plugging in the new boundaries (the teacher wants us to use the new boundaries instead of the ones given). How do I find where the radical values of $\sin(\arctan(x/3))$? I am supposed to evaluate the integral using trig substitution for the following:</p>
<p>$$\int_\sqrt3 ^3 \frac{\sqrt{9+x^2}}{x^6}dx $$ </p>
<p>I inserted a picture of my work, sorry if you can't read my terrible writing, let me know if you have any questions regarding what I wrote.</p>
<p><img src="https://i.stack.imgur.com/d7ssh.jpg" alt="Here is what I have so far"></p>
| Edward Porcella | 403,946 | <p><a href="https://i.stack.imgur.com/GVdg7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GVdg7.jpg" alt="angle outside pentagon=30^o"></a>Using only elementary geometry, in the given figure, join $EB$. From the nature of the regular pentagon, $\angle ABE=36^o$. Through $E$ draw $EH$ perpendicular to $FB$, meeting $BA$ extended at $K$. Since $EK$ is parallel to $CB$, then $\angle EKA$, supplementary to $\angle ABC,=72^o$. But also, $\angle KAE$, supplementary to $\angle EAB$,$=72^o$. Therefore $\triangle EKA$ is isosceles, with $EK=EA$. Further, since triangles $EKA$ and $EKB$ are similar, then $BE=BK$. And since $BH$ is perpendicular to $EK$, then $EH=HK$, making $$EH=\frac12EK$$But $EK=EA=EF$. Therefore $$EH=\frac12EF$$And since $EF$ is the hypotenuse of right triangle $EHF$, then$$\angle EFH=30^o$$ </p>
|
556,054 | <p>Please help me to prove the inequality
$$
\sqrt{a^2 + b^2} \geq \frac{|a-b|}{\sqrt{2}}.
$$</p>
| Community | -1 | <p><strong>Hint</strong>: Square both sides and multiply by $2$, and you'll find that this is equivalent to proving that</p>
<p>$$2(a^2 + b^2) \ge |a - b|^2 = (a - b)^2$$</p>
|
2,208,755 | <p>I got stuck on this question: find all solutions $x$ for $a\in R$:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$</p>
<p>I see that if we simplify we get:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$</p>
<p>From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.</p>
<p>[EDIT]</p>
<p>Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?</p>
<p>From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.</p>
<p>Also, from here:</p>
<p>$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$</p>
<p>in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.</p>
<p>With these two rules we can now keep generating roots until we have 6 total.</p>
<p>If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.</p>
<p>If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.</p>
<p>Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$</p>
<p>The 6 obtained values are distinct, so they cover all the roots.</p>
<p>[EDIT2]</p>
<p>I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>factorizing the given equation and cancel the denominators we get
$$(ax-a-x)(ax-x+1)(x-1+a)(-x+a)(ax-1)(ax-a+1)=0$$</p>
|
4,080,385 | <p>Would you please compute the behavior of the following composed generalized function?</p>
<p><span class="math-container">$g(t) = $</span> <span class="math-container">$\delta(e^t)$</span></p>
<p><strong>Is it even a valid generalized function?</strong></p>
<p>Thank you very much for your time.</p>
| Mohammad Ali Dastgheib | 559,629 | <p>Thank you, everyone. I think I am ready now to write the answer to my question. Let me write a synopsis first:</p>
<p>In Distribution Theory (or theory of Generalized Functions), a generalized function is considered not a function in itself but only in relation to how it affects other functions when "integrated" against them. In keeping with this philosophy, to define the delta function properly, it is enough to say what the "integral" of the delta function is against a set of sufficiently "well-behaved" test functions. Hence: <strong>A different class of test (or testing) functions will define a different generalized function.</strong></p>
<p><em>Answer Begins:</em><br></p>
<ol>
<li><p><em>from <strong>engineering (and applied physics)</strong> point of view:</em><br />
Test functions are usually infinitely differentiable complex-valued (or sometimes real-valued) functions with compact support. Roughly speaking, having compact support means that these test functions vanish (produce only <strong>zero</strong> output) outside some fixed interval. Due to this property of test functions, the generalized function <span class="math-container">$g(t)$</span> = <span class="math-container">$\delta(e^t)$</span> maps any test function to the scalar <strong>zero</strong> (necessary calculations to verify this are straightforward). Indeed, the composed generalized function is mathematically valid (it is the identically zero distribution).</p>
</li>
<li><p><em>from <strong>mathematics</strong> point of view:</em><br />
If we would like to analyze any arbitrary vector space of test functions that are equipped with a notion of convergence, then we should refer to the precise definition of "<span class="math-container">$\delta \circ f$</span>" or "<span class="math-container">$\delta (f(t))$</span>" which can be found here: [1] (Chapter 1; §1.9 Change of variables in generalized functions; page 22). <em>Simply put, it says:</em></p>
<ul>
<li><p>Firstly, we must prepare a sequence of regular distributions converging to the singular delta distribution, e.g. by using "<a href="https://en.wikipedia.org/wiki/Bump_function" rel="nofollow noreferrer">bump functions (Ψ)</a>".</p>
</li>
<li><p>Then we must compose every element of the sequence with <span class="math-container">$f(t)$</span>, e.g. by computing <span class="math-container">$Ψ_n(f(t))$</span> where <span class="math-container">$n$</span> is the index of the sequence of distributions.</p>
</li>
<li><p>Finally, we must investigate the convergence of this sequence for our test functions; if it converges, then this converged value is the output of the "<span class="math-container">$\delta (f(t))$</span>" distribution. If it doesn't converge, then "<span class="math-container">$\delta (f(t))$</span>" is not a valid distribution over our test function vector space.</p>
</li>
</ul>
</li>
</ol>
<p>[1] <em>Vladimirov, V. S.</em> (2002), <strong>Methods of the theory of generalized functions, Analytical Methods and Special Functions</strong>, ISBN 0-415-27356-0</p>
|
78,641 | <p>I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance.</p>
| Nathan | 18,662 | <p>Lest there be any question on this point, ZFC alone implies the existence of some non-separable ccc spaces. The tables in the back of Counterexamples in Topology list four of them. Probably the most interesting is #63, the Countable Complement Extension Topology. This is the minimal extension of the Euclidean topology on the real line given by letting all countable sets be closed.</p>
|
3,964,429 | <p>Zeckendorf : <em>Every positive integer N can be expressed uniquely as a sum of distinct non-consecutive Fibonacci numbers</em></p>
<p>I was wondering if this theorem can be applied with the extended Fibonacci numbers, and especially I am looking for a way to <strong>find the Zeckendorf-like representation of <span class="math-container">$ N $</span>, with Fibonacci numbers <span class="math-container">$ F_n $</span> having only negative indexes</strong>. (First for <span class="math-container">$ N \in \mathbb{N} $</span> then maybe <span class="math-container">$ N \in \mathbb{Z} $</span>)</p>
<p><a href="https://mathworld.wolfram.com/ZeckendorfRepresentation.html" rel="nofollow noreferrer">Mathworld</a> states that this theorem only applies on positive numbers but do not says if Fibonacci numbers can be negative.
Online <a href="https://www.dcode.fr/zeckendorf-representation" rel="nofollow noreferrer">Zeckendorf Representation</a> pages show positive indexes only and mostly use Binet's formula.</p>
<p>I've read that Binet's formula can be applied to generalized Fibonacci numbers (<a href="https://proofwiki.org/wiki/Euler-Binet_Formula/Negative_Index" rel="nofollow noreferrer">Proof</a>), but it does not prove that Zeckendorf theorem can be used as well.</p>
<p><strong>EDIT :</strong> I missed a paragraph in <a href="https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem#Representation_with_negafibonacci_numbers" rel="nofollow noreferrer">this Wikipedia page</a> called Representation with Negafibonacci numbers that states this Zeckendorf-like representation exists and is unique. I probably can use <a href="https://en.wikipedia.org/wiki/NegaFibonacci_coding" rel="nofollow noreferrer">NegaFibonacci</a> coding method to find the indexes.</p>
| wendy.krieger | 78,024 | <p>The negative fibbonacci numbers run as F(-n)=F(n) for 2|n and F(-n)=-F(n) for n odd.</p>
<p>So we get for fibonacci numbers</p>
<pre><code> -21 13 -8 5 -3 2 -1 1
-1
-3 1
-3
-3 -1
-8 2 1
-8 2
-8 t0 12
-21 5 2 1
-21 -8 -3 -1
</code></pre>
<p>So yes, all of the natural numbers can be expressed as a sum of non-consecutive fibonacci numbers of negative index.</p>
<p>It ought be a straight-forward conversion from the positive to negative schema.</p>
<pre><code> eg -21 5 2 1
21 -5 -2 -1
13 8 -5 -2 -1
13 3 -2 -1
13
</code></pre>
<p>It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.</p>
|
4,206,039 | <p>Find the radius of convergence of the following power series <span class="math-container">$$\sum_{n=1}^\infty \frac{(-1)^n z^{n(n+1)}}{n}$$</span></p>
<p>Here's my working
<span class="math-container">$$\lim_{n\to \infty}| \frac{(-1)^{n+1} z^{(n+1)(n+2)}}{n+1} \frac{n}{(-1)^nz^{n(n+1)}}|$$</span>
<span class="math-container">$$=\lim_{n\to \infty}\big| \frac{(-1)^n(-1) z^{(n^2+3n+2)}}{n+1} \frac{n}{(-1)^nz^{n^2}z^n}\big|$$</span>
<span class="math-container">$$= \lim_{n\to \infty}\big| \frac{ -z^{3n}z^2}{n+1} \frac{n}{{}z^n}\big|$$</span>
<span class="math-container">$$=\lim_{n\to \infty}\big| \frac{ -nz^{2n}z^2}{n+1}\big|$$</span>
I am stuck after this. Is the limit greater than 1 since anything raised to the power infinity is very huge? How do I do this? Also what happens if z=i?</p>
| Oliver Díaz | 121,671 | <p>The series in the posting can be expressed as
<span class="math-container">$$
\sum^\infty_{n=1}\frac{(-1)^n}{n} z^{n(n+1)}=\sum^\infty_{m=0}a_mz^m$$</span></p>
<p>where <span class="math-container">$a_m=0$</span> unless <span class="math-container">$m\in\{n(n+1):n\in\mathbb{N}\}$</span>, in which case <span class="math-container">$a_m=\frac{(-1)^n}{n}$</span>, where <span class="math-container">$n=\sqrt{m+\frac14}-\frac12$</span>.</p>
<p>Then by the <a href="https://en.wikipedia.org/wiki/Root_test" rel="nofollow noreferrer">root test</a> criteria, the series converges for all <span class="math-container">$z$</span> with <span class="math-container">$|z|<R$</span> and diverges for all <span class="math-container">$|z|>R$</span>, where
<span class="math-container">$$\frac{1}{R}=\limsup_m\sqrt[m]{|a_m|}=\lim_m\frac{1}{\Big(\sqrt{m+\frac14}-\frac12\Big)^{1/m}}=1$$</span>
As can be seen from the fact that <span class="math-container">$\lim_{k\rightarrow\infty}\sqrt[k]{k}=1$</span>.</p>
|
2,624,498 | <p>Evaluate $$\lim_{n \rightarrow\infty} \sqrt[n]{3^{n} +5^{n}}$$</p>
<p>Attempt:</p>
<p>The only sort of manipulation that has come to mind is: $$e^{\frac{1}{n}ln(e^{n\ln(3)} + e^{n\ln(5)})}$$</p>
<p>So what is the trick to successfully evaluate this?</p>
| marty cohen | 13,079 | <p>By Bernoulli's inequality,
if $x > 0$
and $n \ge 1$,
$(1+x)^n
\ge 1+nx$.</p>
<p>Therefore
$(1+x/n)^n
\ge 1+x$
so that
$(1+x)^{1/n}
\le 1+x/n
$.</p>
<p>Therefore,
if $0 < a < b$
then
$\sqrt[n]{a^n+b^n}
=b\sqrt[n]{1+(a/b)^n}
\le b(1+\frac{(a/b)^n}{n})
= b+\frac{b(a/b)^n}{n}
\lt b+\frac{b}{n}
$
since
$a/b < 1$
and
$\sqrt[n]{a^n+b^n}
=b\sqrt[n]{1+(a/b)^n}
\gt b
$.</p>
<p>Thererfore
$b
\lt \sqrt[n]{a^n+b^n}
\lt b + \frac{b}{n}
$
so that
$\lim_{n \to \infty} \sqrt[n]{a^n+b^n}
= b
$.</p>
|
204,612 | <blockquote>
<p>Is it possible to verify the following <code>lhs,rhs</code> involving the sums
are equal, with Mathematica?</p>
</blockquote>
<p>I can verify it for individual values of <span class="math-container">$d$</span> variable:</p>
<pre><code>ClearAll[d, q, h, eq1, eq2, x, lhs, rhs];
eq1[d_: d, q_: q, h_: h] := Sum[
((-1)^(d + 1 - k))*(1/((d + 1) Factorial[(k - 1)] Factorial[(d + 1 - k)]))*
Product[(q/h + (k - i)), {i, 1, d + 1}] *
Product[(1 - j h), {j, k, d}] *
Product[(1 + l h), {l, d - k + 2, d}]
, {k, 1, d + 1}]
eq2[d_: d, q_: q, h_: h] := Sum[
((-1)^(d - k))*(1/((d) Factorial[(k - 1)] Factorial[(d - k)]))*
Product[(q/h + (k - i)), {i, 1, d}] *
Product[(1 + (k - j) h), {j, 1, d}]
, {k, 1, d}]
lhs[d_: d, x_: x, h_: h] := (x - 1)^d eq1[d, 1/(x - 1), h]
rhs[d_: d, x_: x, h_: h] := (x)^d eq2[d, 1/x, (((x - 1) h)/x)]
t = 0;
Do[r = PossibleZeroQ[lhs[d] - rhs[d]]; Print[d, " ", r], {d, 1, 100}];
</code></pre>
<p>And it is true for first <span class="math-container">$100$</span> values of <span class="math-container">$d$</span>, for example.</p>
<p>But <code>PossibleZeroQ[lhs[d] - rhs[d]]</code> returns <code>false</code> for general <span class="math-container">$d$</span>. </p>
<p>I've tried simplifying and expanding, with no success.</p>
<p>I've added method <code>Method -> "ParallelBestQuality"</code> to both sums to obtain expressions in terms of <code>Gamma</code> and <code>HypergeometricPFQRegularized</code>, as following:</p>
<p><br>
lhs: </p>
<pre><code>((-1)^d (-h)^d (-1 + x)^d Gamma[1 + d - 1/h] Gamma[1 + 1/(h (-1 + x))] HypergeometricPFQRegularized[{-d, -((1 + d h)/ h), 1 + 1/(h (-1 + x))},{(-1 + h)/h, -d + 1/(h (-1 + x))}, 1])/Gamma[2 + d]
</code></pre>
<p>rhs:</p>
<pre><code>((-1)^(1 + d) (h (-1 + 1/x))^d x^d Gamma[1 + 1/(h (-1 + x))] Gamma[(x + h (-1 + d + x - d x))/(h (-1 + x))] Gamma[d + x/(h - h x)] HypergeometricPFQRegularized[{1 - d, 1 + 1/(h (-1 + x)), 1 + x/(h (-1 + x))}, {1 - d + 1/(h (-1 + x)), (x + h (-1 + d + x - d x))/(h (-1 + x))},1])/(Gamma[1 + d] Gamma[x/(h - h x)])
</code></pre>
<blockquote>
<p>Is it possible for Mathematica to simplify one of these two into the
other? (Show difference is <span class="math-container">$0$</span>?)</p>
</blockquote>
<hr>
<p>Solving this will also answer the following question from MSE: <a href="https://math.stackexchange.com/q/3330637/318073">Equality like Pascal triangle</a>.</p>
<p>(The source of sums)</p>
| Chris K | 6,358 | <p>The function is <a href="https://reference.wolfram.com/language/ref/Expectation.html" rel="nofollow noreferrer"><code>Expectation</code></a> not <code>ExpectedValue</code>. Unfortunately,</p>
<pre><code>Expectation[b*x*(1 + ω*x^ρ)^κ, x \[Distributed] LogNormalDistribution[μ, σ]]
</code></pre>
<p>does not yield an answer.</p>
<p>If <code>κ</code> is an integer, it does appear to work:</p>
<pre><code>Expectation[b*x*(1 + ω*x^ρ)^3, x \[Distributed] LogNormalDistribution[μ, σ]]
(* b E^(μ + σ^2/2) (1 + 3 E^(μ ρ + 1/2 ρ (2 + ρ) σ^2) ω +
3 E^(2 ρ (μ + (1 + ρ) σ^2)) ω^2 + E^(3/2 ρ (2 μ + (2 + 3 ρ) σ^2)) ω^3) *)
</code></pre>
|
204,612 | <blockquote>
<p>Is it possible to verify the following <code>lhs,rhs</code> involving the sums
are equal, with Mathematica?</p>
</blockquote>
<p>I can verify it for individual values of <span class="math-container">$d$</span> variable:</p>
<pre><code>ClearAll[d, q, h, eq1, eq2, x, lhs, rhs];
eq1[d_: d, q_: q, h_: h] := Sum[
((-1)^(d + 1 - k))*(1/((d + 1) Factorial[(k - 1)] Factorial[(d + 1 - k)]))*
Product[(q/h + (k - i)), {i, 1, d + 1}] *
Product[(1 - j h), {j, k, d}] *
Product[(1 + l h), {l, d - k + 2, d}]
, {k, 1, d + 1}]
eq2[d_: d, q_: q, h_: h] := Sum[
((-1)^(d - k))*(1/((d) Factorial[(k - 1)] Factorial[(d - k)]))*
Product[(q/h + (k - i)), {i, 1, d}] *
Product[(1 + (k - j) h), {j, 1, d}]
, {k, 1, d}]
lhs[d_: d, x_: x, h_: h] := (x - 1)^d eq1[d, 1/(x - 1), h]
rhs[d_: d, x_: x, h_: h] := (x)^d eq2[d, 1/x, (((x - 1) h)/x)]
t = 0;
Do[r = PossibleZeroQ[lhs[d] - rhs[d]]; Print[d, " ", r], {d, 1, 100}];
</code></pre>
<p>And it is true for first <span class="math-container">$100$</span> values of <span class="math-container">$d$</span>, for example.</p>
<p>But <code>PossibleZeroQ[lhs[d] - rhs[d]]</code> returns <code>false</code> for general <span class="math-container">$d$</span>. </p>
<p>I've tried simplifying and expanding, with no success.</p>
<p>I've added method <code>Method -> "ParallelBestQuality"</code> to both sums to obtain expressions in terms of <code>Gamma</code> and <code>HypergeometricPFQRegularized</code>, as following:</p>
<p><br>
lhs: </p>
<pre><code>((-1)^d (-h)^d (-1 + x)^d Gamma[1 + d - 1/h] Gamma[1 + 1/(h (-1 + x))] HypergeometricPFQRegularized[{-d, -((1 + d h)/ h), 1 + 1/(h (-1 + x))},{(-1 + h)/h, -d + 1/(h (-1 + x))}, 1])/Gamma[2 + d]
</code></pre>
<p>rhs:</p>
<pre><code>((-1)^(1 + d) (h (-1 + 1/x))^d x^d Gamma[1 + 1/(h (-1 + x))] Gamma[(x + h (-1 + d + x - d x))/(h (-1 + x))] Gamma[d + x/(h - h x)] HypergeometricPFQRegularized[{1 - d, 1 + 1/(h (-1 + x)), 1 + x/(h (-1 + x))}, {1 - d + 1/(h (-1 + x)), (x + h (-1 + d + x - d x))/(h (-1 + x))},1])/(Gamma[1 + d] Gamma[x/(h - h x)])
</code></pre>
<blockquote>
<p>Is it possible for Mathematica to simplify one of these two into the
other? (Show difference is <span class="math-container">$0$</span>?)</p>
</blockquote>
<hr>
<p>Solving this will also answer the following question from MSE: <a href="https://math.stackexchange.com/q/3330637/318073">Equality like Pascal triangle</a>.</p>
<p>(The source of sums)</p>
| mikado | 36,788 | <p>The expression whose expectation you seek can be expanded as a power series in <code>x</code>. (You might need to worry about whether this converges).</p>
<pre><code>expr = b*x*(1 + ω*x^ρ)^κ;
coeff = Assuming[κ > 0 && n >= 0,
SeriesCoefficient[expr /. x^ρ -> z, {z, 0, n}]];
</code></pre>
<p>We see that the power series converges to your original expression.</p>
<pre><code>Sum[coeff x^(ρ n), {n, 0, ∞}] == expr
(* True *)
</code></pre>
<p>We can take the expectation of a general term in this series</p>
<pre><code>expectation =
FullSimplify[
Expectation[coeff x^(ρ n),
x \[Distributed] LogNormalDistribution[μ, σ]]];
</code></pre>
<p>You can them sum the series to obtain the expectation.</p>
<pre><code>Sum[expectation, {n, 0, ∞}] // InputForm
(* Sum[b*E^(((1 + n*ρ)*(2*μ + (1 + n*ρ)*σ^2))/2)*ω^n*Binomial[κ, n],
{n, 0, Infinity}] *)
</code></pre>
<p>Unfortunately, Mathematica doesn't return a simple expression for this sum, but you might find for the parameter values of interest it converges quite quickly.</p>
|
3,442,862 | <blockquote>
<p>Let <span class="math-container">$F $</span> be a subset family of the set {<span class="math-container">$ 1, 2, ..., 2017 $</span>} such that
for any <span class="math-container">$ A, B \in F $</span>, worth that <span class="math-container">$A \cap B$</span> has exactly one
element. Determine as many as possible of <span class="math-container">$ F $</span> elements</p>
</blockquote>
<p><strong>Solution:</strong> Generalization: if the total number set is <span class="math-container">$\{1, 2, ..., n\}$</span>, then the maximum of <span class="math-container">$|F|$</span> is <span class="math-container">$n$</span>. In the original problem <span class="math-container">$n=2017$</span>, so <span class="math-container">$max|F|=\boxed{2017}$</span>.</p>
<ol>
<li><p>We claim that <span class="math-container">$|F| \leq n$</span>
Consider a map from one subset <span class="math-container">$A$</span> of <span class="math-container">$\{1, 2, ..., n\}$</span> to a <span class="math-container">$n$</span>-dimension vector <span class="math-container">$V=(v_1, v_2, ..., v_n)^T$</span>. For <span class="math-container">$1 \leq i \leq n$</span>, if <span class="math-container">$i \in A$</span> then <span class="math-container">$v_i=1$</span>, else <span class="math-container">$v_i=0$</span>.
Consider the vector set mapped from <span class="math-container">$F$</span>: <span class="math-container">$\{V_1, V_2, ..., V_m\}$</span>. We could prove the vectors are linearly independent.
For <span class="math-container">$i \ne j$</span>, <span class="math-container">$<V_i, V_j>=V_i^TV_j=1$</span>, since there is exactly one element included in any two elements of <span class="math-container">$F$</span>.
For <span class="math-container">$i = j$</span>, <span class="math-container">$<V_i, V_i>=|V_i| \geq 1$</span>, where <span class="math-container">$|V_i|$</span> counts the number of <span class="math-container">$1$</span> appeared in <span class="math-container">$V_i$</span>.
Consider <span class="math-container">$S=\sum_{k=1}^{m} a_kV_k$</span>, if <span class="math-container">$S=(0,0,...,0)^T$</span>, then <span class="math-container">$<S, S>=0$</span>.
However, <span class="math-container">$<S, S>=\sum_{i=1}^{m} a_i^2<V_i, V_i>+\sum_{i \ne j} 2a_ia_j<V_i, V_j>=(\sum_{i=1}^{m} a_i)^2+\sum_{i=1}^{m} a_i^2(|V_i|-1) \geq 0$</span>.
So the equality holds only when <span class="math-container">$a_i=0$</span>, that means <span class="math-container">$\{V_1, V_2, ..., V_m\}$</span> are linearly independent, so that <span class="math-container">$m \leq n$</span>, so that <span class="math-container">$|F| \leq n$</span>.</p></li>
<li><p>A construction of <span class="math-container">$|F|=n$</span>
Consider <span class="math-container">$F=\{\{a,n\} | 1 \leq a \leq n-1\} \cup\{n\}$</span>. <span class="math-container">$|F|=n$</span>, and any <span class="math-container">$ A, B \in F$</span>, <span class="math-container">$A \cap B=\{n\}$</span>.</p></li>
</ol>
<p>I didn't understand the logic of this solution</p>
| oshill | 651,041 | <p>How is your linear algebra? It is saying we are constructing vectors in a 2017-dimensional space. I will give the example in <span class="math-container">$3$</span>-D. Let <span class="math-container">$\{1,2,3\}$</span> be the base set, so the subset <span class="math-container">$\{1,3\}$</span> maps to the vector <span class="math-container">$\begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix}$</span>. Look at what happens if we have linear dependence (this means a "good" combination of the non-zero vectors becomes zero) for example:</p>
<p><span class="math-container">$\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}-\begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}-\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}=0$</span>. Now see that this is like saying in terms of our original sets <span class="math-container">$\{1,2\} -\{1\} - \{2\} = \emptyset$</span>. But notice that <span class="math-container">$|\{1,2\}\cap \{1\}|=1, |\{1,2\}\cap \{2\}|=1,$</span> BUT <span class="math-container">$|\{1\}\cap \{2\}|=0$</span>. </p>
<p>Try to extend this idea of linear dependence requiring the sets to over or under intersect to the full argument given.</p>
|
1,028,371 | <p>I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.</p>
<p>$$\phi:a+bi \rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$</p>
<p>This is what I have so far:</p>
<p>Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.</p>
<p>$$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$</p>
<p>So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.</p>
<p>Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$. </p>
<p>Surjective:</p>
<p>I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.</p>
<p>I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated! </p>
| AlexR | 86,940 | <p>You could prove injectivity simply by showing
$$z\ne w \Rightarrow \phi(z) \ne \phi(w)$$
wich is next to obvious.<br>
Regarding surjectivity: A function is always surjective on its range. All you need to show here is that any Matrix $\pmatrix{a&-b\\b&a}$ is in the range of $\phi$, explicitly $a+bi$ is the required argument (sounds like nothing to show, because it's simply the definition of $\phi$ wich guarantees this)</p>
|
448 | <p>Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?</p>
| bryn | 106 | <p>For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB. </p>
<p>The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.</p>
|
2,677,584 | <p>I have the following question:</p>
<blockquote>
<p>Find the real values of $a$ for which the equation
$$(1+\tan^2\theta)^2 + 4a\tan\theta(\tan^2\theta + 1) + 16\tan^2\theta = 0$$
has four distinct real roots in $\left(0, \dfrac{\pi}{2}\right)$.</p>
</blockquote>
<p>I tried to solve the above equation by dividing the entire equation by $\tan^2\theta$ and then substituting $\tan\theta + \dfrac{1}{\tan\theta}$ as $y$ and then solving for $y$. Then I tried to apply the inequality $\tan\theta + \dfrac{1}{\tan\theta} \geqslant 2$ but couldn't find a proper range of values of $a$.</p>
<p>Please help. Please point out if there is any mistake in my work. Thanks in advance.</p>
| giuseppe mancò | 191,154 | <p>The four roots are:</p>
<p><span class="math-container">$tan(\theta_{1})=-a+\sqrt{a^{2}-4}-\sqrt{-5+2a^{2}-2a\sqrt{a^2-4}}$</span></p>
<p><span class="math-container">$tan(\theta_{2})=-a+\sqrt{a^{2}-4}+\sqrt{-5+2a^{2}-2a\sqrt{a^2-4}}$</span></p>
<p><span class="math-container">$tan(\theta_{3})=-a-\sqrt{a^{2}-4}-\sqrt{-5+2a^{2}+2a\sqrt{a^2-4}}$</span></p>
<p><span class="math-container">$tan(\theta_{4})=-a-\sqrt{a^{2}-4}+\sqrt{-5+2a^{2}-2a\sqrt{a^2-4}}$</span>,</p>
<p>and become real when:</p>
<p><span class="math-container">$-\frac{5}{2}<=a<=-2$</span></p>
<p>and</p>
<p><span class="math-container">$2<=a<=\frac{5}{2}$</span>.</p>
|
1,984,178 | <p>I have a problem with the following exercise:</p>
<p>We have the operator $T: l^1 \to l^1$ given by</p>
<p>$$T(x_1,x_2,x_3,\dots)=\left(\left(1-\frac11\right)x_1, \left(1-\frac12\right)x_2, \dots\right)$$ for $(x_1,x_2,x_3,\dots)$ in $l^1$. Showing that this operator is bounded is easy, but I am really desperate with showing that the norm $\|T\| = 1$.</p>
<p>I know that for bounded operators the norm is defined as $\|T\|=\sup{\left\{\|T(x)\|: \|x\| \le 1\right\}}$.</p>
<p>I am also wondering if there exists a x in $ l^1$ such that $\|x\|=1 $</p>
<p>and $\|T(x)\|= \|T\|$</p>
<p>Thank you! :)</p>
| J.R. | 44,389 | <p>First observe that $\|T\|\le 1$.</p>
<p>Fix a large $n$ and let $x_n=(\dots,0,1,0,\dots)$ with $1$ exactly at the $n$th position and $0$ everywhere else. We have $Tx_n = (1-\frac1n) x_n$, $\|x_n\|=1$ and $\|Tx_n\| = 1-\frac1n.$
This shows $\|T\|\ge 1- \frac1n$. Now let $n\to\infty$ to conclude $\|T\|\ge 1$.</p>
|
1,662,226 | <p>Find a sufficient statistic for $σ^2$ with $μ$ known, where $X_i$ is a random sample from $N(μ,σ^2)$</p>
<p>I was able to find a sufficient statistic for $μ$ with $σ^2$ known, but I'm stuck on finding one for $σ^2$ when $μ$ is known. Can anyone give me some help? </p>
<p>I was using the factorization method before, is this the best way?</p>
| dbanet | 220,258 | <p>This absolutely is the worst way to solve this problem, but I though it might be useful to you, or at least interesting (I bet ability to evaluate things numerically on paper <em>is</em> something interesting).</p>
<p>From \begin{align}\sqrt{0.016}&=\sqrt{16\cdot10^{-3}}=4\sqrt{10^{-1}10^{-2}}=4\cdot10^{-1}\sqrt{10^{-1}}=\frac4{10}\sqrt{1\over10}=\frac25\frac1{\sqrt{10}}\frac{\sqrt{10}}{\sqrt{10}}=\\&=\frac25\frac{\sqrt{10}}{10}=\frac1{25}\sqrt{10}\end{align}
you can always evaluate $\sqrt{10}$ (or any square root, and many other different things) numerically to arbitrary precision with <em>numerical root finding algorithms</em>, which given some function $$f:D\subseteq\mathbb R\mapsto\mathbb R$$ and (for most of them) some initial guess $$x_0\in D$$ will provide $x_n\in D$ such that $$\lim_{n\to\infty}f(x_n)=0,$$ that is, the bigger the $n$ gets (more <em>iterations</em> of the algorithm are computed), the more accurate the result $x_n$ gets relative to the actual root $x:\,f(x)=0$. The simplest algorithm is via Newton and goes like this: you want $\sqrt{10}$, i.e. some $x$ such that $x^2=10$, that is, $x^2-10=0$; let $$f(x)=x^2-10.$$
Then by Newton's method you need to <em><a href="https://en.wikipedia.org/wiki/Derivative" rel="nofollow">differentiate</a></em> that to get $$f'(x)=2x,$$ and also set some initial guess $x_0$, for which $$x_0=3$$ seems reasonable. The first step goes like this:
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=3-\frac{9-10}6=3+\frac16=\frac{19}6;$$
the second one:
$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=\frac{19}6-\frac{361/36-10}{38/6}=\frac{19}6-\frac{1/36}{38/6}=\frac{19}6-\frac1{6\cdot38}=\frac{721}{228};$$
you get the idea. Each step produces bigger and bigger nominators and denominators, which ratio approaches $\sqrt{10}$. Long division yields $$x_2=\frac{721}{228}=3.16\overline{228070175438596491\,\,}\approx3.16228,$$ which with $$x=\sqrt{10}=3.16227766016837933199889\dotsc\approx3.16228$$ is accurate to <strong>six</strong> figures (and only via <em>two</em> iterations).</p>
<p>Now, dividing $3.16228$ by $25$ (remember to cut off at the sixth significant figure to avoid <a href="https://en.wikipedia.org/wiki/False_precision" rel="nofollow">overprecision</a> — you ain't have no more information!) yields $$0.12649,$$ which shows that $0.13$ is the closet number in the given set to $\sqrt{0.016}\,.$</p>
|
3,403,855 | <p>Construct an example of a set of real numbers E that has no points of accumulation and yet has the property
that for every ε > 0 there exist points x, y ∈ E so that 0 < |x − y| < ε.</p>
<p>so i know we need a convergent sequence to show that the difference between two elements can be as small as we like, also it should be a set not an interval to avoid accumulation points. but if a sequence converge to L, wouldnt L be an accumulation point too?</p>
| pancini | 252,495 | <p>Consider</p>
<p><span class="math-container">$$E=\left\{\sum_{k=1}^n k^{-1}:n\in \Bbb N\right\}.$$</span></p>
|
1,200,919 | <p>Let $x$ be the solution of the equation $x^x=2$. Is $x$ irrational? How to prove this?</p>
| Thomas Andrews | 7,933 | <p>If $x$ is rational, $p/q$, then $2^{q/p}$ is rational. That's only possible if $p=1$ and $p/q$ is an integer.</p>
<p>A quick way to write it, assuming $p,q$ are relatively prime:</p>
<p>$$\left(\frac pq\right)^{p/q}=2\implies p^p=2^qq^p$$</p>
<p>But $p$ and $q$ are relatively prime, using unique factorization, we see that $p^p,q^p$ are relatively prime, so $q^p=1$ which means either that $p=0$ (not possible) or $q=1$. If $q=1$, then $p^p=2$, so again by unique factorization, $p=2^k$ and $1=k2^k$, which can't be solved.</p>
|
1,384,053 | <p>Which number is bigger? $1.01^{101}$ or $2$? and how about $e^{\pi}$ or $\pi^e$?</p>
<p>Tried some algebraic manipulations to no end, so would love some suggestions or some different ways to approach those kind of problem</p>
| Khosrotash | 104,171 | <p>$$x \to 0 , (1+x)^n \approx 1+nx \\(1.01)^101=(1+\frac{1}{100})^101 \approx 1+ 101(\frac{1}{100}) >2$$ for the second one
Consider this function $$x^{\frac{1}{x}}$$.
$f'=x^{\frac{1}{x}}(\frac{1}{x^2})(1-\ln x)$,
function has global maximum at $x=e$.</p>
<p>so $e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}} \to $, and it is clear that the inequality is strict,$$(e^{\frac{1}{e}} \geq \pi^{\frac{1}{\pi}})^{e\pi} \to $$ so <strong>$$e^{\pi}>\pi^{e}$$</strong></p>
|
2,436,268 | <p>My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the expression. My "toolbox" for solving problems like these is very limited... If polar form doesn't work, then I usually have no clue on how to continue.</p>
<p><strong>Edit</strong>: I think this is the solution.
$$
\lim_{(x,y)\to(0,0)}\left|\frac{x^5+y^2}{x^4+|y|}\right| = \frac{|x^5+y^2|}{|x^4+|y||}
$$
Applying the triangle inequality gives
$$
\frac{|x^5+y^2|}{|x^4+|y||} \leq \left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right|
$$
Inspecting the denominators on the RHS gives:
$$
\left|\frac{x^5}{x^4+|y|}\right| \leq |x|, \quad\left|\frac{y^2}{x^4+|y|}\right| \leq |y|
$$
So
$$
\left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right| \leq |x| + |y|
$$
Since $|x| + |y| \to 0$ when $x,y\to 0$, the sandwich theorem states that $|\frac{x^5+y^2}{x^4+|y|}| \to 0$. And if $\lim |f(x)|=0$ then $\lim f(x)=0$ which solves the original problem.</p>
| Peter Szilas | 408,605 | <p>$0\le |\dfrac{x^5+y^2}{x^4 +|y|}| \le$</p>
<p>$\dfrac{|x^5| +|y^2|}{|x^4+|y||} \le$</p>
<p>$\dfrac{|x^5|}{x^4} + \dfrac{y^2}{|y|} =$</p>
<p>$|x| + |y|.$</p>
<p>And now?</p>
|
3,549,072 | <p>The following are given</p>
<p><span class="math-container">$$ \lim_{x \to \infty}{\log(x)} = \infty$$</span></p>
<p><span class="math-container">$$ \lim_{x \to \infty}{\cosh(x)} = \infty$$</span></p>
<p><span class="math-container">$$ \lim_{x \to \infty}{\sinh(x)} = \infty$$</span></p>
<p><span class="math-container">$$
\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \quad \text { for } x \in \mathbb{R}
$$</span></p>
<p><span class="math-container">$$ \lim_{x \to \infty}{x^2} = \infty$$</span></p>
<p><span class="math-container">$$ \lim_{x \to \infty}{\exp(x)} = \infty$$</span></p>
<p>Both the denominator and the numerator goes to infinity. I do not know how to handle this. A hint or a bit more will be helpful. </p>
<p>Differentiation are not allowed.</p>
| Axion004 | 258,202 | <p><span class="math-container">\begin{align}\frac{\tanh (x)-1}{e^{-2 x}}&=e^{2x}\big(\tanh (x)-1\big)\\&=
e^{2x}\left( \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}-1\right)\\&=
e^{2x}\left(\frac{-2e^{-x}}{e^{x}+e^{-x}}\right)\\&=
\frac{-2e^{x}}{e^{x}+e^{-x}}\\&=\frac{2}{e^{2x}+1}-2
\end{align}</span></p>
|
4,347,308 | <p><em><strong>Definition:</strong></em></p>
<p>Let <span class="math-container">$(X,\mathscr{A},\mu)$</span> be a measurable space, an atom of the measure <span class="math-container">$\mu$</span> is a set <span class="math-container">$A \in\mathscr{A}$</span> with the property that
<span class="math-container">$\mu(A) > 0$</span> and for any <span class="math-container">$B\in \sigma (A)$</span> either <span class="math-container">$\mu(B) = 0$</span>, or <span class="math-container">$\mu(A \setminus B) = 0$</span>. If a measure has
atoms it is called atomic; in the opposite case, the measure is called non-atomic (or
atomeless). A measure is called purely atomic if <span class="math-container">$X$</span> can be written as the union of a
finite or countable number of atoms.</p>
<p>From the definition of atoms, we get the following corollary:</p>
<p><em><strong>Corollary:</strong></em></p>
<p><em>Every purely atomic measure is an atomic measure.</em></p>
<p>I am trying to find an example of an atomic measure that is not purely atomic, can anyone help me?</p>
| Bonnaduck | 92,329 | <p>The answer you are conjecturing can be written as</p>
<p><span class="math-container">$$\frac{1+x^3}{1+x^2}=1-x^2+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})$$</span></p>
<p>Multiplying both sides by <span class="math-container">$1+x^2$</span> gives us</p>
<p><span class="math-container">\begin{align*}
1+x^3&=(1+x^2)(1-x^2)+(1+x^2)\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(1+x^2)(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})
\end{align*}</span></p>
<p>The sum on the RHS is telescoping; that is, <span class="math-container">$\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})$</span> is equal to</p>
<p><span class="math-container">\begin{align*}
(x^3+x^4+x^5+x^6)-(x^5+x^6+x^7+x^8)+(x^7+x^8+x^9+x^{10})-\dots=x^3+x^4.
\end{align*}</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$1+x^3=1-x^4+x^3+x^4=1+x^3.$$</span></p>
<p>You should be able to work backwards to fomulate a proof.</p>
|
1,533,745 | <p>$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$</p>
<p>I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?</p>
| lisyarus | 135,314 | <p>Since $x \approx \sin x$ when $x \rightarrow 0$, the denominator is of order $x^4$.</p>
<p>The numerator needs more careful analysis of $\sin x$. Using $\sin x = x - \frac{x^3}{6} + o(x^3)$, we get $\sin^2 x = x^2 - \frac{x^4}{3} + o(x^4)$, so the numerator is $\sin^2 x - x^2 \approx -\frac{x^4}{3}$.</p>
<p>Dividing and ignoring terms of lower order, we get the limit: $-\frac{1}{3}$.</p>
|
1,533,745 | <p>$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$</p>
<p>I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?</p>
| Idris Addou | 192,045 | <p>Hint:
\begin{equation*}
\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\left( \frac{\sin x-x}{x^{3}}%
\right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2}
\end{equation*}
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\sin x+x}{x}\right) &=&1+1=2 \\
\lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2} &=&1^{2}=1.
\end{eqnarray*}</p>
|
1,087,874 | <p>I want to understand how I can count the terms of the expression $x^{m-1} + x^{m-2} +\ldots+ x^0$ when $x=1$.</p>
<p>The result is $m$, I dont know how to count them formally, any advice would be helpful. I'm desperated, not because it is required to do the above, but how can be done, I need to understand the subject. Sorry for my bad english.</p>
<p>PS: It is related to this limit:
$$\lim_{x \to 1} \frac{x^m-1}{x-1} = m$$</p>
<p>I dont want to use L'Hôpital's rule, I just use a simple factorization and a change of a variable.</p>
| hmakholm left over Monica | 14,366 | <p>When you set $x=1$, then the value of each of the terms is $1$! (Note that there are no explicit coefficients). Adding the ones then simply tells you how many ones there are.</p>
|
1,087,874 | <p>I want to understand how I can count the terms of the expression $x^{m-1} + x^{m-2} +\ldots+ x^0$ when $x=1$.</p>
<p>The result is $m$, I dont know how to count them formally, any advice would be helpful. I'm desperated, not because it is required to do the above, but how can be done, I need to understand the subject. Sorry for my bad english.</p>
<p>PS: It is related to this limit:
$$\lim_{x \to 1} \frac{x^m-1}{x-1} = m$$</p>
<p>I dont want to use L'Hôpital's rule, I just use a simple factorization and a change of a variable.</p>
| Mark Bennet | 2,906 | <p>The terms of the polynomial can be indexed by the exponents of $x$ which are $m-1, m-2, \dots 0$ and these are simply $m$ consecutive integers in reverse order. Each term has a different exponent and no exponent is omitted.</p>
|
795 | <p>Please observe the following thread <a href="https://math.stackexchange.com/questions/4489/proving-that-the-given-diophantine-equation-has-a-solution">Proving that the given Diophantine equation has a solution</a>.</p>
<p>There is a long boring argument/discussion about whether it should be posted, who should post it - how to get more people involved in the site... There is also a trend of converting mathematical problems into google search problems - I don't think much insight or understanding comes from this.</p>
<p>Is this necessary?</p>
<hr>
<p>It is frustrating to be downvoted when I post questions like this. <em>Please give me some idea as to what the problem is if you would be so kind</em>.</p>
| Larry Wang | 73 | <p>I removed those comments that were not related to <a href="https://math.stackexchange.com/questions/4489/proving-that-the-given-diophantine-equation-has-a-solution">the question Chandru1 posted</a>, and reproduce them here so that they will not be lost. Both on-topic and off-topic comments have been provided for context.</p>
<hr />
<blockquote>
<p>Chandru1, may I suggest in the future
that you ask your friends to post
their interesting problems
themselves... – Qiaochu Yuan 13 hours
ago</p>
<p><span class="math-container">$\mathbb{Q}(\sqrt{-11})$</span> has class
number one. – Robin Chapman 12 hours
ago</p>
<p>@Qioachu Yuan: Whether "my friend" posts or i post whats the
difference? – Chandru1 12 hours ago</p>
<p>@Chandru1: It is fine to post your
friend's question (and I'm not sure
why you put that in quotes), but I
agree with Qiaochu. If your friends
post their own questions, they can
interact directly with the answerers.
Also, you have posted 98 questions in
40 days; it might be nice to take a
breather and at the same time
encourage more people to get involved
with this website. – Jonas Meyer 12
hours ago</p>
<p>deleted comment from Chandru1</p>
<p>@Chandru1: No, I don't
mean to say that there is a
restriction on the number of questions
you can post (nor am I in any way an
authority), just that it would be
beneficial to the website to have more
people involved, including your
friends, and hopefully beneficial to
your friends to use the website
directly. There is also some
discussion on meta about how rapidly
posting questions makes it more
difficult to follow-up appropriately
on each one. – Jonas Meyer 12 hours
ago</p>
<p>Here's the source: Mathematical
Olympiad in China: problems and
solutions by Bin Xiong, Peng Yee Lee -
Page 115 – baudrillard 12 hours ago</p>
<p>@Jonas Meyer: Agreed! – Chandru1 12 hours ago</p>
<p>deleted comment from Chandru1</p>
<p>@Chandru1 I've made a
simple Google Books search, it took me
less than a minute. The exact search
line was: diophantine "with * being
odd numbers". By the way, there's a
complete solution in the book. –
baudrillard 12 hours ago</p>
<p>@baudrillard: Great thinking! I was trying with Diophantine equation
has a solution :x)! – Chandru1 12
hours ago</p>
<p>@Chandru: Maybe you should edit the
source back in quickly, before we got
another complain ;) – KennyTM♦ 12
hours ago</p>
<p>Chandru: why exactly don't you want your friends to come here? This
site could use more people, you know.
It would also be better that they can
directly ask people who "know" instead
of having to use an intermediary. – J.
M. 10 hours ago</p>
<p>@Robin Chapman, I noticed that
<span class="math-container">$z^2+z+3$</span> has discriminant <span class="math-container">$-11$</span> but
how does the class number come into
play? – muad 5 hours ago</p>
</blockquote>
|
169,097 | <p>In the beginning of chapter two in The HoTT Book there is a discussion about synthetic vs. analytic geometry:</p>
<blockquote>
<p>An important difference between homotopy type theory and classical homotopy theory is that homotopy type theory provides a <em>synthetic</em> description of spaces, in the following sense. Synthetic geometry is geometry in the style of Euclid [EucBC]: one starts from some basic notions (points and lines), constructions (a line connecting any two points), and axioms (all right angles are equal), and deduces consequences logically. This is in contrast with analytic geometry, where notions such as points and lines are represented concretely using cartesian coordinates in $\mathbb{R}^n$ — lines are sets of points — and the basic constructions and axioms are derived from this representation. While classical homotopy theory is analytic (spaces and paths are made of points), homotopy type theory is synthetic: points, paths, and paths between paths are basic, indivisible, primitive notions.</p>
</blockquote>
<p>If a circle is a cohesive closed curve independent of points, is the circle necessarily inhabited by points? Or can we have a "pointless circle", i.e. a type $C$ representing a circle without any objects? Is there a circle without any subspaces such as points and paths? If so, is a circle inhabited by a point or a path something else than a plain circle? Or do we automatically have an infinite set of points as inhabitants when we consider a cohesive geometric object in HoTT, as in Euclidean geometry?</p>
| Mike Shulman | 49 | <p>Francois' answer is good; let me add a bit more. Homotopy type theory is synthetic <em>homotopy theory</em>, which means that the "spaces" in question are not the same sort of "spaces" that you find in point-set topology: it's better to think of them as "homotopy types" or "$\infty$-groupoids". They are not "geometric" in the sense of circles as "closed curves" in a plane; you should think instead of the "circle" as the fundamental $\infty$-groupoid of a topological circle.</p>
<p>(There are refinements of homotopy type theory in which there also exist "cohesive" or "geometric" objects in addition to "homotopy types", but that's an extra layer of confusion that's best ignored when learning the basic subject.)</p>
<p>Another point to emphasize is that in homotopy type theory, a type does not have a "set of points" at all. The analogy to Euclidean geometry may seem a bit misleading because in the latter one can talk about "the set of points that lie on some line" whether or not that line is identified with its set of points, but in homotopy type theory there is no such thing. A type does <em>have</em> points, but we don't have a "set of" such points, at least not for the meaning of "set" internal to the theory. (So actually, the analogy to Euclidean geometry is okay, because <em>internal</em> to Euclidean geometry we don't even have a meaning of "set"; thus "the set of points on a line" must also there be external to the theory. But it's confusing because in homotopy type theory we <em>do</em> have an internal meaning of "set".)</p>
<p>Finally, there was also a recent <a href="https://groups.google.com/d/msg/homotopytypetheory/hE1eY-v_Kes/FEbadDz1hREJ">discussion</a> on the homotopy type theory mailing list about "circles without points".</p>
|
271,824 | <p>I have a list= {4, 8, 10, 11, 12, 14, 16, 7, 9}</p>
<p>How can i partition the list by group of Arithmetic Progression with common difference 1 :</p>
<p>{{4}, {8}, { 10, 11, 12}, {14}, {16}, {7}, {9}}</p>
| lericr | 84,894 | <p>Something like this, maybe?</p>
<pre><code>Map[If[MemberQ[li2, #], #, x] &, li1]
</code></pre>
|
4,547,918 | <p>Given the torus and given the point p <span class="math-container">$\in$</span> M corresponding to the parameters <span class="math-container">$s=\frac{\pi }{4}$</span> and <span class="math-container">$t=\frac{\pi }{3}$</span>.
Determine the cartesian equation of the tangent plane to M in p.</p>
<p><span class="math-container">$\begin{cases} x=\left(3+\sqrt{2}cos\left(s\right)\right)cos\left(t\right) \\ y=\left(3+\sqrt{2}cos\left(s\right)\right)sin\left(t\right) \\ z=\sqrt{2}sin\left(s\right) \end{cases}$</span></p>
<p>Could someone give me a hint or help me? I'm not sure if I firstly should go from the given parametric equation to a cartesian equation.</p>
| P. Lawrence | 545,558 | <p>Use the "tan half-ange" substitution, <span class="math-container">$viz$</span> <span class="math-container">$$u=\tan (x/2)$$</span> to convert the integral to the integral of a rational function.</p>
|
204,560 | <p>I am a computer programmer interested in prime numbers. I have implementations of several algorithms related to prime numbers at my <a href="http://programmingpraxis.com" rel="nofollow">blog</a>. I want to add an implementation of the AKS primality prover to my collection, but I am having trouble, and my knowledge of math is insufficient to make sense of some of the things I read, so I ask for help here. I have two questions: how to compute the exponentiation of a polynomial mod another polynomial and an integer, and how to compute the r in the AKS algorithm. I begin with polynomial exponentiation, using an example.</p>
<p>Consider the problem of squaring polynomial x^3 + 4 x^2 + 12 x + 3 modulo (x^5 - 1, 17). Polynomial multiplication is exactly the same as grade-school multiplication, except there are no carries, so the process looks like this:</p>
<pre><code> 1 4 12 3
1 4 12 3
--- --- --- ---
3 12 36 9
12 48 144 36
4 16 48 12
1 4 12 3
--- --- --- --- --- --- ---
1 8 40 102 168 72 9
</code></pre>
<p>Thus, 1 x^3 + 4 x^2 + 12 x + 3 squared is 1 x^6 + 8 x^5 + 40 x^4 + 102 x^3 + 168 x^2 + 72 x + 9. To reduce the result modulo 1 x^5 - 1 we divide by the grade-school long-division algorithm and take the remainder, which gives 1 x + 8 with a remainder of 40 x^4 + 102 x^3 + 168 x^2 + 73 x + 17:</p>
<pre><code> 1 8
+ --- --- --- --- --- --- ---
1 0 0 0 0 -1 | 1 8 40 102 168 72 9
1 0 0 0 0 -1
--- --- --- --- --- ---
8 40 102 168 73 9
8 0 0 0 0 -8
--- --- --- --- --- ---
40 102 168 73 17
</code></pre>
<p>We can confirm the calculation by multiplying and adding the remainder:</p>
<pre><code> 1 0 0 0 0 -1
1 8
--- --- --- --- --- ---
8 0 0 0 0 -8
1 0 0 0 0 -1
--- --- --- --- --- --- ---
1 8 0 0 0 -1 -8
40 102 168 73 17
--- --- --- --- --- --- ---
1 8 40 102 168 72 9
</code></pre>
<p>Then we simply reduce each of the coefficients of the remainder modulo 17, giving the result 6 x^4 + 0 x^3 + 15 x^2 + 5 x + 0. The whole computation can be organized as shown below. Note how the division and reduction modulo x^5 - 1 is accomplished, eliminating the high-order coefficients and adding them back to the low-order coefficients; we are relying here on the simple form of the polynomial modulus, and would have to do more work if it was more complicated:</p>
<pre><code> 1 4 12 3 multiplicand
1 4 12 3 multiplier
--- --- --- ---
3 12 36 9 3 * 1 4 12 3 * x^0
12 48 144 36 12 * 1 4 12 3 * x^1
4 16 48 12 4 * 1 4 12 3 * x^2
1 4 12 3 1 * 1 4 12 3 * x^3
--- --- --- --- --- --- ---
1 8 40 102 168 72 9 1 4 12 3 * 1 4 12 3
-1 -8 1 8 reduce modulo x^5 - 1
--- --- --- --- --- --- ---
40 102 168 73 17 reduce modulo 17
6 0 15 5 0 final result
</code></pre>
<p>Now that we can perform modular multiplication of a polynomial, we return to the modular exponentiation of a polynomial, which is done using the same square-and-multiply algorithm as modular exponentiation of integers, except that we use polynomial modular multiplication instead of integer modular multiplication. Here is the algorithm on integers; to adapt it to polynomials, just replace the integer modular multiplications with polynomial modular multiplications.</p>
<pre><code>func powermod(b, e, m) # b^e (mod m)
r := 1
while e > 0
if e is odd
r := r * b (mod m)
e := floor(e/2)
b := b * b (mod m)
return r
</code></pre>
<p>So, the first question: Have I correctly stated the algorithm for modular exponentiation of polynomials?</p>
<p>For the second question, I start with the statement of the AKS algorithm, as given at the <a href="http://primes.utm.edu/prove/prove4_3.html" rel="nofollow">Prime Pages</a>:</p>
<pre><code>Input: Integer n > 1
Output: either PRIME or COMPOSITE
if (n has the form a^b with b > 1)
then output COMPOSITE
r := 2
while (r < n) {
if (gcd(n,r) is not 1)
then output COMPOSITE
if (r is prime greater than 2)
then {
let q be the largest factor of r-1
if (q > 4 * sqrt(r) * log2 n)
and ( n^{(r-1)/q} is not 1 (mod r) )
then break
}
r := r+1
}
for a = 1 to 2 * sqrt(r) * log2 n {
if ( (x-a)^n is not (x^n-a) (mod x^r-1,n) )
then output COMPOSITE
}
output PRIME;
</code></pre>
<p>Again I will work with a specific example, trying to prove that n = 89 is prime. We first consider if n is a perfect power of the form a^b with b > 1. When a = 2, the powers of 2 are 4, 8, 16, 32, 64 and 128, so 2 fails. When a = 3, the powers of 3 are 9, 27, 81 and 243, so 3 fails. When a = 5, the powers of 5 are 25 and 125, so 5 fails. When a = 7, the powers of 7 are 49 and 343, so 7 fails. When a = 11, or any higher prime, a^2 is greater than 89, so we are finished with the perfect power tests.</p>
<p>Second we compute the value of r. Since r and q = (r-1)/2 must both be prime, and q must be greater than 4 * log n = 26, the only possibility is r = 59, but 4 * sqrt(59) * log2(89) = 199, so there is no early break from the loop, and r must be 89.</p>
<p>So, the second question: Have I correctly computed r = 89?</p>
<p>I think that must be incorrect. With r = 89, a will range from 1 to 122. Let's take a = 17 as an example. Modular exponentiation of the polynomial (x - 17) ^ 89 (mod x^89 - 1, 89) is 73; that is, the number 73, with all coefficients of powers of x equal to 0. But that doesn't equal x^89 - 17, suggesting that 89 is composite. Of course, 89 is prime, so something is wrong.</p>
<p>Code is available at <a href="http://codepad.org/4jwgScdX" rel="nofollow">http://codepad.org/4jwgScdX</a>. Please let me know what I have done wrong. Thank you for reading this far.</p>
| user448810 | 20,808 | <p>Ok, I've got it. The calculation of r was incorrect; it should be 191. Working code, including the full AKS prover, is at <a href="http://programmingpraxis.codepad.org/6ZHrsEmx" rel="nofollow">http://programmingpraxis.codepad.org/6ZHrsEmx</a>. I'll have a full write-up at my blog later this week: <a href="http://programmingpraxis.com/2012/10/02/aks-primality-prover-part-1/" rel="nofollow">Part 1</a> and <a href="http://programmingpraxis.com/2012/10/05/aks-primality-prover-part-2/" rel="nofollow">Part 2</a>.</p>
|
204,560 | <p>I am a computer programmer interested in prime numbers. I have implementations of several algorithms related to prime numbers at my <a href="http://programmingpraxis.com" rel="nofollow">blog</a>. I want to add an implementation of the AKS primality prover to my collection, but I am having trouble, and my knowledge of math is insufficient to make sense of some of the things I read, so I ask for help here. I have two questions: how to compute the exponentiation of a polynomial mod another polynomial and an integer, and how to compute the r in the AKS algorithm. I begin with polynomial exponentiation, using an example.</p>
<p>Consider the problem of squaring polynomial x^3 + 4 x^2 + 12 x + 3 modulo (x^5 - 1, 17). Polynomial multiplication is exactly the same as grade-school multiplication, except there are no carries, so the process looks like this:</p>
<pre><code> 1 4 12 3
1 4 12 3
--- --- --- ---
3 12 36 9
12 48 144 36
4 16 48 12
1 4 12 3
--- --- --- --- --- --- ---
1 8 40 102 168 72 9
</code></pre>
<p>Thus, 1 x^3 + 4 x^2 + 12 x + 3 squared is 1 x^6 + 8 x^5 + 40 x^4 + 102 x^3 + 168 x^2 + 72 x + 9. To reduce the result modulo 1 x^5 - 1 we divide by the grade-school long-division algorithm and take the remainder, which gives 1 x + 8 with a remainder of 40 x^4 + 102 x^3 + 168 x^2 + 73 x + 17:</p>
<pre><code> 1 8
+ --- --- --- --- --- --- ---
1 0 0 0 0 -1 | 1 8 40 102 168 72 9
1 0 0 0 0 -1
--- --- --- --- --- ---
8 40 102 168 73 9
8 0 0 0 0 -8
--- --- --- --- --- ---
40 102 168 73 17
</code></pre>
<p>We can confirm the calculation by multiplying and adding the remainder:</p>
<pre><code> 1 0 0 0 0 -1
1 8
--- --- --- --- --- ---
8 0 0 0 0 -8
1 0 0 0 0 -1
--- --- --- --- --- --- ---
1 8 0 0 0 -1 -8
40 102 168 73 17
--- --- --- --- --- --- ---
1 8 40 102 168 72 9
</code></pre>
<p>Then we simply reduce each of the coefficients of the remainder modulo 17, giving the result 6 x^4 + 0 x^3 + 15 x^2 + 5 x + 0. The whole computation can be organized as shown below. Note how the division and reduction modulo x^5 - 1 is accomplished, eliminating the high-order coefficients and adding them back to the low-order coefficients; we are relying here on the simple form of the polynomial modulus, and would have to do more work if it was more complicated:</p>
<pre><code> 1 4 12 3 multiplicand
1 4 12 3 multiplier
--- --- --- ---
3 12 36 9 3 * 1 4 12 3 * x^0
12 48 144 36 12 * 1 4 12 3 * x^1
4 16 48 12 4 * 1 4 12 3 * x^2
1 4 12 3 1 * 1 4 12 3 * x^3
--- --- --- --- --- --- ---
1 8 40 102 168 72 9 1 4 12 3 * 1 4 12 3
-1 -8 1 8 reduce modulo x^5 - 1
--- --- --- --- --- --- ---
40 102 168 73 17 reduce modulo 17
6 0 15 5 0 final result
</code></pre>
<p>Now that we can perform modular multiplication of a polynomial, we return to the modular exponentiation of a polynomial, which is done using the same square-and-multiply algorithm as modular exponentiation of integers, except that we use polynomial modular multiplication instead of integer modular multiplication. Here is the algorithm on integers; to adapt it to polynomials, just replace the integer modular multiplications with polynomial modular multiplications.</p>
<pre><code>func powermod(b, e, m) # b^e (mod m)
r := 1
while e > 0
if e is odd
r := r * b (mod m)
e := floor(e/2)
b := b * b (mod m)
return r
</code></pre>
<p>So, the first question: Have I correctly stated the algorithm for modular exponentiation of polynomials?</p>
<p>For the second question, I start with the statement of the AKS algorithm, as given at the <a href="http://primes.utm.edu/prove/prove4_3.html" rel="nofollow">Prime Pages</a>:</p>
<pre><code>Input: Integer n > 1
Output: either PRIME or COMPOSITE
if (n has the form a^b with b > 1)
then output COMPOSITE
r := 2
while (r < n) {
if (gcd(n,r) is not 1)
then output COMPOSITE
if (r is prime greater than 2)
then {
let q be the largest factor of r-1
if (q > 4 * sqrt(r) * log2 n)
and ( n^{(r-1)/q} is not 1 (mod r) )
then break
}
r := r+1
}
for a = 1 to 2 * sqrt(r) * log2 n {
if ( (x-a)^n is not (x^n-a) (mod x^r-1,n) )
then output COMPOSITE
}
output PRIME;
</code></pre>
<p>Again I will work with a specific example, trying to prove that n = 89 is prime. We first consider if n is a perfect power of the form a^b with b > 1. When a = 2, the powers of 2 are 4, 8, 16, 32, 64 and 128, so 2 fails. When a = 3, the powers of 3 are 9, 27, 81 and 243, so 3 fails. When a = 5, the powers of 5 are 25 and 125, so 5 fails. When a = 7, the powers of 7 are 49 and 343, so 7 fails. When a = 11, or any higher prime, a^2 is greater than 89, so we are finished with the perfect power tests.</p>
<p>Second we compute the value of r. Since r and q = (r-1)/2 must both be prime, and q must be greater than 4 * log n = 26, the only possibility is r = 59, but 4 * sqrt(59) * log2(89) = 199, so there is no early break from the loop, and r must be 89.</p>
<p>So, the second question: Have I correctly computed r = 89?</p>
<p>I think that must be incorrect. With r = 89, a will range from 1 to 122. Let's take a = 17 as an example. Modular exponentiation of the polynomial (x - 17) ^ 89 (mod x^89 - 1, 89) is 73; that is, the number 73, with all coefficients of powers of x equal to 0. But that doesn't equal x^89 - 17, suggesting that 89 is composite. Of course, 89 is prime, so something is wrong.</p>
<p>Code is available at <a href="http://codepad.org/4jwgScdX" rel="nofollow">http://codepad.org/4jwgScdX</a>. Please let me know what I have done wrong. Thank you for reading this far.</p>
| Copperfield | 326,292 | <blockquote>
<p>The calculation of r was incorrect; it should be 191.</p>
</blockquote>
<p>wait what?, that is wrong the r for 89 is not 191, is 43</p>
<p>first star with the definition of r in the context of the AKS test:</p>
<blockquote>
<p>find the smallest r such that such that: ord<sub>r</sub>(n) > (log<sub>2</sub>(n))<sup>2</sup></p>
</blockquote>
<p>where ord<sub>r</sub>(n) is the multiplicative order of n modulo r, and that is: if gcd(n,r)=1, then is smallest k>=1 such that n<sup>k</sup>=1 (mod r)</p>
<p>programing that is very easy, here are the first ord<sub>r</sub>(89) for r>=2: </p>
<p>ord<sub>2</sub>(89)=1, ord<sub>3</sub>(89)=2, ord<sub>4</sub>(89)=1, ord<sub>5</sub>(89)=2, ord<sub>6</sub>(89)=2, ord<sub>7</sub>(89)=6, ord<sub>8</sub>(89)=1, ord<sub>9</sub>(89)=2, ord<sub>10</sub>(89)=2, ord<sub>11</sub>(89)=1, ord<sub>12</sub>(89)=2, ord<sub>13</sub>(89)=12, ord<sub>14</sub>(89)=6, ord<sub>15</sub>(89)=2, ord<sub>16</sub>(89)=2, ord<sub>17</sub>(89)=4, ord<sub>18</sub>(89)=2, ord<sub>19</sub>(89)=18, ord<sub>20</sub>(89)=2, ord<sub>21</sub>(89)=6, ord<sub>22</sub>(89)=1, ord<sub>23</sub>(89)=22, ord<sub>24</sub>(89)=2, ord<sub>25</sub>(89)=10, ord<sub>26</sub>(89)=12, ord<sub>27</sub>(89)=6, ord<sub>28</sub>(89)=6, ord<sub>29</sub>(89)=28, ord<sub>30</sub>(89)=2, ord<sub>31</sub>(89)=10, ord<sub>32</sub>(89)=4, ord<sub>33</sub>(89)=2, ord<sub>34</sub>(89)=4, ord<sub>35</sub>(89)=6, ord<sub>36</sub>(89)=2, ord<sub>37</sub>(89)=36, ord<sub>38</sub>(89)=18, ord<sub>39</sub>(89)=12, ord<sub>40</sub>(89)=2, ord<sub>41</sub>(89)=40, ord<sub>42</sub>(89)=6, ord<sub>43</sub>(89)=42, ord<sub>44</sub>(89)=1, ord<sub>45</sub>(89)=2, ord<sub>46</sub>(89)=22, ord<sub>47</sub>(89)=23, ord<sub>48</sub>(89)=2, ord<sub>49</sub>(89)=42, ord<sub>50</sub>(89)=10, ord<sub>51</sub>(89)=4, ord<sub>52</sub>(89)=12, ord<sub>53</sub>(89)=13, ord<sub>54</sub>(89)=6, ord<sub>55</sub>(89)=2, ord<sub>56</sub>(89)=6, ord<sub>57</sub>(89)=18, ord<sub>58</sub>(89)=28, ord<sub>59</sub>(89)=58, ord<sub>60</sub>(89)=2</p>
<p>now (log<sub>2</sub>(89))<sup>2</sup> = ( 6.4757 )<sup>2</sup> = 41.9351</p>
<p>then r=43 because ord<sub>43</sub>(89)=42</p>
<p>Now, the other problem I see is when you explain how you check for n=a<sup>b</sup> for b>1, you said that you check only primes power, so unless that you do something else behind the scene, that is wrong too.</p>
<p>Take for example this number 8281, there is no prime such that p<sup>b</sup>=8291 for b>=1, but 8281 is a perfect square, it is 91<sup>2</sup> (91=7*13) </p>
<p>for a more accurate check you can calculate the b-th root of n, n<sup>1/b</sup>, round that and ask if that to the power of n is equal to n, that is: round(n<sup>1/b</sup>)<sup>b</sup>=n</p>
<p>About the rest I don't know, my knowledge about polynomials operations is not enough because I am also a programmer that want to add the AKS test to my collection...</p>
|
1,817,035 | <p>Gradient descent reduces the value of the objective function in each iteration. This is repeated until convergence happens.</p>
<p>The question is if the norm of gradient has to decrease as well in every iteration of gradient descent?</p>
<hr>
<p><strong>Edit:</strong> How about when the objective is a convex function?</p>
| Rodrigo de Azevedo | 339,790 | <p>Suppose we have the following objective function</p>
<p>$$f (x) := \frac{1}{2} x^T Q x - r^T x$$</p>
<p>where $Q \in \mathbb R^{n \times n}$ is symmetric and positive definite and $r \in \mathbb R^n$. From the symmetry of $Q$, we conclude that its eigenvalues are real. From the positive definiteness of $Q$, we conclude that its eigenvalues are positive and that $f$ is a <strong>convex</strong> function.</p>
<p>Taking the gradient of $f$,</p>
<p>$$\nabla f (x) = Q x - r$$</p>
<p>The gradient vanishes at $\bar{x} := Q^{-1} r$, which is the unique solution to the linear system $Q x = r$.</p>
<p>Doing gradient descent,</p>
<p>$$\begin{array}{rl} x_{k+1} &= x_k - \alpha \nabla f (x_k)\\\\ &= x_k - \alpha Q x_k + \alpha r\\\\ &= (I_n - \alpha Q) x_k + \alpha r\end{array}$$</p>
<p>Hence,</p>
<p>$$x_k = \bar{x} + (I_n - \alpha Q)^k (x_0 - \bar{x})$$</p>
<p>Convergence to $\bar{x}$ happens when</p>
<p>$$|1 - \alpha \lambda_i (Q)| < 1$$</p>
<p>for all $i \in \{1,2,\dots,n\}$, i.e., when</p>
<p>$$0 < \alpha < \frac{2}{\lambda_{\max} (Q)} =: \alpha_{\max}$$</p>
<p>Evaluating the gradient,</p>
<p>$$\begin{array}{rl} \nabla f (x_{k+1}) &= Q x_{k+1} - r\\\\ &= Q (I_n - \alpha Q) x_k + \alpha Q r - r\\\\ &= (I_n - \alpha Q) (Q x_k - r)\\\\ &= (I_n - \alpha Q) \nabla f (x_k)\end{array}$$</p>
<p>Thus,</p>
<p>$$\|\nabla f (x_{k+1})\|_2 \leq \|I_n - \alpha Q\|_2 \|\nabla f (x_k)\|_2$$</p>
<p>If $0 < \alpha < \alpha_{\max}$, then $|1 - \alpha \lambda_i (Q)| < 1$ and $\|I_n - \alpha Q\|_2 < 1$. Thus, if we have convergence, then the $2$-norm of the gradient contracts with each iteration</p>
<p>$$\|\nabla f (x_{k+1})\|_2 < \|\nabla f (x_k)\|_2$$</p>
|
289,864 | <p>Let $C_{1}$ be a circle of unit radius. Let A and B be two points inside $C_{1}$. Now I want to construct another circle $C_{2}$ such that A and B lie on $C_{2}$ and $C_{2}$ is orthogonal to $C_{1}$ at their point of intersection(I want $C_{2}$ in such a way that it intersects $C_{1}$). I tried and failed to find a way to construct such a $C_{2}$.</p>
<p>Any help is appreciated.</p>
| Neal | 20,569 | <p>If I understand your question correctly, it is answered in Construction 2.1 of this paper: <a href="http://comp.uark.edu/~strauss/papers/hypcomp.pdf" rel="nofollow">http://comp.uark.edu/~strauss/papers/hypcomp.pdf</a></p>
<p>Edit upon request: Given two points $A$ and $B$ in the interior of a circle $C_1$, invert $A$ through $C_1$ to $A'$. Now construct the circle through $A$, $B$, and $A'$. Sigur's answer contains a picture of this construction.</p>
<p>The paper discusses general compass-and-straightedge constructions in hyperbolic geometry, where, in particular, geodesics can be modeled as circles perpendicular to the unit circle in $\mathbb{C}$.</p>
|
2,166,075 | <blockquote>
<p>Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively.</p>
</blockquote>
<p>Where $a_i=a_{i+1}-r$.</p>
<p>I tried to start proving it inductively, but any try lead to a bad conclusion, so I ended up proving it by making $a_n$ depend on $a_i$.</p>
<p>But I didn't know how to prove it inductively, so there is the problem.</p>
<p><strong>EDIT:</strong></p>
<p>I'm looking for a valid induction steps to reach the conclusion.</p>
| Bram28 | 256,001 | <p>No induction needed ... just use a simple trick famously used by Gauss when he was 10 years old:</p>
<p>Take two of these series, one going from $a_1$ to $a_n$, and the other one going back from $a_n$ to $a_1$, put them under each other, and add them up by entry (that is, add the first entries of the two series, then add the second entries, etc):</p>
<p>$a_1 + a_2 + ... + a_{n-1} + a_n$</p>
<p>$a_n + a_{n-1} + ... + a_2 + a_1$</p>
<p>Added together gives:</p>
<p>$(a_1 + a_n) + (a_2 + a_{n-1}) + ... + (a_{n-1} + a_2) + (a_n+a_1)$</p>
<p>Now note that $a_{1+i} + a_{n-i} = a_1 + i*r + a_n - i*r = a_1 + a_n$ </p>
<p>(put differently: each time you move an entry to the right, the first number of the pair increases by $r$, while the second of the pair decreases by $r$, so the sum stays the same)</p>
<p>So, each pair adds up to $a_1 + a_n$, and since you have n pairs, you get a total of $n*(a_1 + a_n)$.</p>
<p>Since that is the sum of two series, one series has a sum of half of that, i.e.:</p>
<p>$$a_1 + ... + a_n = \frac{n*(a_1 + a_n)}{2}$$</p>
|
2,166,075 | <blockquote>
<p>Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively.</p>
</blockquote>
<p>Where $a_i=a_{i+1}-r$.</p>
<p>I tried to start proving it inductively, but any try lead to a bad conclusion, so I ended up proving it by making $a_n$ depend on $a_i$.</p>
<p>But I didn't know how to prove it inductively, so there is the problem.</p>
<p><strong>EDIT:</strong></p>
<p>I'm looking for a valid induction steps to reach the conclusion.</p>
| Stefano | 387,021 | <p>Statement true for $n=2$. Assume</p>
<p>$$ a_1 + \dots + a_n = \frac{(a_1+a_n)n}{2}.$$</p>
<p>Then, using the inductive hypothesis and adding and subtracting terms, we get</p>
<p>$$a_1 + \dots+ a_n + a_{n+1} = \frac{(a_1+a_n)n}{2}+ a_{n+1}$$</p>
<p>$$= \frac{(a_1+ a_{n+1})(n+1) }{2} + \frac{a_nn-a_1}{2}+ a_{n+1} - \frac{a_{n+1}(n+1)}{2}. $$</p>
<p>Finally,</p>
<p>$$ \frac{a_nn-a_1}{2}+ a_{n+1} - \frac{a_{n+1}(n+1)}{2} = \frac{na_n-a_1+a_{n+1}-na_{n+1}}{2}$$</p>
<p>$$ = \frac{a_{n+1}-a_1-nr}{2}=0$$</p>
<p>since $a_{n+1} = a_1 + nr$.</p>
|
4,047,601 | <p>I did a question <span class="math-container">$\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$</span>, and evaluating this is divergent integral yes? Then as a general form <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx$</span>, <span class="math-container">$p \in \mathbb{R}$</span>, what values of <span class="math-container">$p$</span> can give me <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$</span>? This is a easy integral to calculate, make it <span class="math-container">$\int_{0}^{1}x^{-p}dx$</span> and calculate, etc. Then how do I get this to solve <span class="math-container">$p$</span>? I am using the fundamental theorem of calculus and confused here.</p>
| José Carlos Santos | 446,262 | <p>Actually,<span class="math-container">$$\int_0^1x^{-1/2}\,\mathrm dx=\left[2x^{1/2}\right]_{x=0}^{x=1}=2.$$</span></p>
<p>On the other hand,<span class="math-container">$$\int_0^1x^{-p}\,\mathrm dx=\left[\frac{x^{1-p}}{1-p}\right]_{x=0}^{x=1}=\frac1{1-p}.$$</span>So, take <span class="math-container">$p=\frac14$</span>.</p>
|
4,047,601 | <p>I did a question <span class="math-container">$\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$</span>, and evaluating this is divergent integral yes? Then as a general form <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx$</span>, <span class="math-container">$p \in \mathbb{R}$</span>, what values of <span class="math-container">$p$</span> can give me <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$</span>? This is a easy integral to calculate, make it <span class="math-container">$\int_{0}^{1}x^{-p}dx$</span> and calculate, etc. Then how do I get this to solve <span class="math-container">$p$</span>? I am using the fundamental theorem of calculus and confused here.</p>
| Obsessive Integer | 864,339 | <p><span class="math-container">$\int_0^1{\frac{1}{x^\frac{1}{2}}}dx$</span> is not divergent.</p>
<p><span class="math-container">$\int_0^1{\frac{1}{x^\frac{1}{2}}}dx=\int_0^1 x^{-\frac{1}{2}}dx=2x^\frac{1}{2}|_0^1=2$</span></p>
<p>Similarly solving: <span class="math-container">$\int_0^1 \frac{1}{x^p}dx=\frac{4}{3}$</span></p>
<p><span class="math-container">$\int_0^1 \frac{1}{x^p}dx=\frac{x^{p-1}}{1-p}|_0^1=
\frac{1}{1-p}=\frac{4}{3}$</span></p>
<p>So, <span class="math-container">$p=\frac{1}{4}$</span></p>
|
3,005,100 | <p>Given the following formula
<span class="math-container">$$
\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,.
$$</span>
How can I show that this is equal to
<span class="math-container">$$
\frac{n!}{x(x+1)\cdots(x+n)}\,?
$$</span></p>
| ajotatxe | 132,456 | <p>Induction step:</p>
<p><span class="math-container">$$\begin{align}
\sum_{k=0}^{n+1}&\frac{(-1)^k}{x+k}\binom{n+1}k=\frac1x+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\left[\binom nk+\binom n{k-1}\right]
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\binom{n}{k-1}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\sum_{k=0}^{n-1}\frac{(-1)^k}{(x+1)+k}\binom{n}{k}
\\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}+\frac{(-1)^n}{x+n+1}
\\&=\frac{n!(x+n+1)-n!x}{x(x+1)\cdots(x+n+1)}=\frac{(n+1)!}{x(x+1)\cdots(x+n+1)}
\end{align}$$</span></p>
|
2,965,989 | <p>Why <span class="math-container">$$p(y)=\int_0^\infty x\delta (y-x)dx=y\ \ ?$$</span></p>
<p>For me, <span class="math-container">$$p(y)=\int_0^\infty x\delta (y-x)dx=\int_{\{y\}}xdx=0.$$</span></p>
<p>If it would be written <span class="math-container">$\int_0^\infty xd\delta _y$</span>, then I would be agree with the answer. But here it's written <span class="math-container">$$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$$</span>
what I interpret as <span class="math-container">$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$</span>.</p>
| Masacroso | 173,262 | <p>The <span class="math-container">$\delta_y$</span> of Dirac is a measure that is defined as</p>
<p><span class="math-container">$$
\delta_y(A)=\begin{cases}1,& y\in A\\ 0,&y\notin A\end{cases}
$$</span></p>
<p>for any subset <span class="math-container">$A$</span> of the measure space (in your case the measure space seems to be <span class="math-container">$(0,\infty)$</span> or <span class="math-container">$\Bbb R$</span>). In your case you have</p>
<p><span class="math-container">$$
\int_0^\infty x\delta_y(dx)=\int_{\{y\}\cap (0,\infty)}x\delta_y(dx)+\underbrace{\int_{(0,\infty)\setminus\{y\}}x\delta_y(dx)}_{=0}\\=\delta_y\big(\{y\}\cap(0,\infty)\big)\,x|_y=\begin{cases}y,& y\in(0,\infty)\\0,&y\notin (0,\infty)\end{cases}
$$</span> </p>
<p>The reasons why this is the value of the integral can be understood knowing the theory of the Lebesgue integral (or Bochner integral) for arbitrary measures.</p>
<hr>
<p>The integral <span class="math-container">$\int_X f(x)\mu(dx)$</span> for some arbitrary measure space <span class="math-container">$(X,\mu,\Bbb R)$</span> is defined as the limit of a sequence of integrals of simple functions that approximate the value of the integral of <span class="math-container">$f$</span> in <span class="math-container">$X$</span> in some sense.</p>
<p>A simple function is a finite sum of weighted characteristic functions of measurable subsets of <span class="math-container">$X$</span> with finite measure (similarly to Riemann sums), a simple function have the form</p>
<p><span class="math-container">$$
s(x):=\sum_{k=0}^m \chi_{A_k}(x)\,c_k\tag1
$$</span></p>
<p>where <span class="math-container">$\mu(A_k)<\infty$</span> for each <span class="math-container">$k=1,\ldots,m$</span> and the <span class="math-container">$c_k$</span> are constants. </p>
<p>Let <span class="math-container">$(f_j)$</span> a sequence of simple functions. If <span class="math-container">$(f_j)\to f$</span> point-wise almost everywhere then</p>
<p><span class="math-container">$$\int_X f(x)\mu(dx):=\lim_{j\to\infty}\int_X f_j(x)\mu(dx)\tag2$$</span></p>
<p>And for a simple function the Lebesgue integral is defined by</p>
<p><span class="math-container">$$\int_X f_j(x)\mu(dx):=\sum_{k=0}^{n_j}\mu(A_{k,n_j})c_{k,n_j}\tag3$$</span></p>
<p>In your case we have the sequence of simple functions <span class="math-container">$f_j(x):=\sum_{k=0}^{n_j}\chi_{A_{k,j}}(x)c_{k,j}$</span> defined by</p>
<p><span class="math-container">$$
A_{k,j}:=[jk2^{-j},j(k+1)2^{-j}),\quad\text{for }n_j:=2^j,\quad\text{and }c_{k,j}=jk2^{-j}\tag4
$$</span></p>
<p>Then we find that <span class="math-container">$\lim_{x\to\infty}f_j(x)= x$</span> for each <span class="math-container">$x\in(0,\infty)$</span> and
<span class="math-container">$$
\int_0^\infty f_j(x)\delta_y(dx)=\sum_{k=0}^{2^j}\delta_y(A_{k,2^j})\, jk2^{-j}=\begin{cases}jm2^{-j}, &y\in A_{m,2^j}\text{ for some }m\\0,&\text{otherwise}\end{cases}\tag5
$$</span></p>
<p>With a bit of work it can be shown that <span class="math-container">$\int_0^\infty x\delta_y(dx)=\lim_{j\to\infty}\int_0^\infty f_j(x)\delta_y(dx)=y$</span> whenever <span class="math-container">$y\in(0,\infty)$</span>.</p>
|
947,770 | <p>Here I have a diophantine equation featuring a homogeneous polynomial:</p>
<p>$$x^2+5y^2+34z^2+2xy-10xz-22yz=0; x, y, z\in\mathbb{Z}$$</p>
<p>I have no idea how to approach this, I've tried various substitutions like $x=py, x=qz$ but then I get a non-homogeneous polynomial of 2 variables which is no better than this. Wolfram|Alpha finds a solution easily, but I still don't understand how.</p>
<p>For reference, W|A claims that the solution is $x=7n, y=3n, z=2n; n\in\mathbb{Z}$</p>
| Mahmoud. A .Solomon | 361,251 | <p>assuming </p>
<p>$x^2+5y^2+2xy=c$</p>
<p>and solving the equation,</p>
<p>$34z^2-(10x+22y)z+c=0$</p>
<p>one can find that </p>
<p>$z=(10x+22y∓(√(-4(3x-7y)^2 ))/(2*34)$</p>
<p>which mean that the discriminant D is negative and z is complex
but z∈Z then $d=0$</p>
<p>or ,$3x=7y$</p>
<p>and $34z=5x+11y$</p>
<p>suppose $x=7n$ ∀ n ∈ Z</p>
<p>then $y=3n$
and $34z=35n+33n$ or $z=2n$</p>
|
4,069,499 | <p>If we let <span class="math-container">$x = 0$</span>.</p>
<p><span class="math-container">\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}</span></p>
<p>Then <span class="math-container">$9y$</span> should always equal <span class="math-container">$21$</span>?
Solving for <span class="math-container">$y$</span> finds <span class="math-container">$\frac{7}{3}$</span>.</p>
<p>But <span class="math-container">$3(x+7)-\frac{7}{3}(2(x)+9)$</span> does not have the same result for diffrent values of <span class="math-container">$x$</span>.</p>
<p>Where am I going wrong?</p>
| Royi | 33 | <p>You have the function:</p>
<p><span class="math-container">$$ f \left( x, y \right) = 3 \left( x + 7 \right) - y \left( 2 x + 9 \right) $$</span></p>
<p>Since the function is constant for <span class="math-container">$ x $</span> then it means:</p>
<p><span class="math-container">$$ \nabla_{x} f \left( x, y \right) = 0 = 3 - 2 y \Rightarrow y = \frac{3}{2} $$</span></p>
<p>Another way thinking about it, is by definition the function with respect to <span class="math-container">$ x $</span> must be constant (Specific case of Linear).<br />
Hence its coefficients (Which are given by the derivative since it's a specific case of Linear Function) must vanish (As seen in @Deepak answer).</p>
|
7,130 | <p>I'm looking for an explanation on how reducing the Hamiltonian cycle problem to the Hamiltonian path's one (to proof that also the latter is NP-complete). I couldn't find any on the web, can someone help me here? (linking a source is also good).</p>
<p>Thank you.</p>
| Prashant | 96,758 | <p>The answer mentioned is correct for the purpose of showing reduction, but I have this idea which is faster than the idea presented. Please correct me if I am wrong.
If ham path says no for even a single vertex then ham cycle says no.
This means that all vertices should now say yes for Ham path(hence one method id to ask each egde).A faster method would be: Pick any vertex. For each of its edge remove the edge and add 2 vertices and join in same fashion as mentioned in the answer. If for any edge removal and addition of 2 new vertices ham path says yes, then output Yes.</p>
<p>Please correct me if I am wrong. </p>
|
3,248,123 | <p>I have this statement:</p>
<blockquote>
<p>If <span class="math-container">$a$</span> belongs to the interval <span class="math-container">$[- 4, - 1]$</span> and <span class="math-container">$b$</span> belongs to the
interval <span class="math-container">$[- 2, 3]$</span>, what interval does it contain? all possible
values of <span class="math-container">$(2a - b)$</span>?</p>
</blockquote>
<p>I have developed it, but according to the guide my answer is incorrect and I would like to know why.</p>
<p><span class="math-container">$-4 \leq a \leq -1$</span>, multiply by <span class="math-container">$2$</span></p>
<p><span class="math-container">$-8 \leq 2a \leq -2$</span></p>
<p>Now, the interval of <span class="math-container">$b$</span> is:</p>
<p><span class="math-container">$-2 \leq b \leq 3$</span></p>
<p>I know the extreme values of each interval, I will subtract the interval of <span class="math-container">$ b$</span> :</p>
<p><span class="math-container">$-8 - -2 \leq 2a - b\leq -2 -3$</span></p>
<p><span class="math-container">$-6 \leq 2a - b \leq -5$</span>, so my answer is: <span class="math-container">$[-6,-5]$</span></p>
<p>I would like to know why, my development is incorrect. Thanks in advance.</p>
| herb steinberg | 501,262 | <p>Max of <span class="math-container">$2a=-2$</span>, min of <span class="math-container">$b=-2$</span>, max of <span class="math-container">$2a-b=0$</span>.</p>
<p>Min of <span class="math-container">$2a=-8$</span>, max of <span class="math-container">$b=3$</span>, min of <span class="math-container">$2a-b=-11$</span>.</p>
|
3,248,123 | <p>I have this statement:</p>
<blockquote>
<p>If <span class="math-container">$a$</span> belongs to the interval <span class="math-container">$[- 4, - 1]$</span> and <span class="math-container">$b$</span> belongs to the
interval <span class="math-container">$[- 2, 3]$</span>, what interval does it contain? all possible
values of <span class="math-container">$(2a - b)$</span>?</p>
</blockquote>
<p>I have developed it, but according to the guide my answer is incorrect and I would like to know why.</p>
<p><span class="math-container">$-4 \leq a \leq -1$</span>, multiply by <span class="math-container">$2$</span></p>
<p><span class="math-container">$-8 \leq 2a \leq -2$</span></p>
<p>Now, the interval of <span class="math-container">$b$</span> is:</p>
<p><span class="math-container">$-2 \leq b \leq 3$</span></p>
<p>I know the extreme values of each interval, I will subtract the interval of <span class="math-container">$ b$</span> :</p>
<p><span class="math-container">$-8 - -2 \leq 2a - b\leq -2 -3$</span></p>
<p><span class="math-container">$-6 \leq 2a - b \leq -5$</span>, so my answer is: <span class="math-container">$[-6,-5]$</span></p>
<p>I would like to know why, my development is incorrect. Thanks in advance.</p>
| DanielWainfleet | 254,665 | <p>The error is assuming that <span class="math-container">$A\le B$</span> and <span class="math-container">$C\le D$</span> imply <span class="math-container">$A-B\le C-D.$</span> E.g. <span class="math-container">$4\le 6$</span> and <span class="math-container">$2\le 5$</span> but <span class="math-container">$\neg (4-2\le 6-5).$</span></p>
<p>On the other hand if <span class="math-container">$E\le F$</span> and <span class="math-container">$G\le H$</span> then <span class="math-container">$\{x+y:x\in [E,F]\land y\in [G,H]\}=[E+G,F+H].$</span></p>
<p><span class="math-container">$\{2a:a\in [-4,-1]\}=[-8,-2].$</span></p>
<p><span class="math-container">$\{-b:b\in [-2,3]\}=[-3,2].$</span></p>
<p>So <span class="math-container">$\{2a-b: a\in [-4,-1]\land b\in [-3,2]\}=\{x+y:x\in [-8,-2]\land y\in [-3,2]\}=[-11,0].$</span></p>
|
2,189,029 | <p>Can all numbers $a, b, c$ that $a^2+b^2=c^2$ be sides of a right triangle? ($c$ is the hypotenuse)
I saw this problem:</p>
<p>Find the number of right triangles that the hypotenuse and one side are both prime numbers. </p>
<p>I said that we don't have any property about the other side, so we can assume it is a real number and not only natural. Thus the sides would be $$p, \sqrt{q^2-p^2}, q $$
Where p and q are primes and q is the hypotenuse, and as result there exists infinte number of triangles of this type. </p>
<p>But what I am not sure about is that all these triples could be sides of a right triangle or not (being correct in triangle inequality).
$$p<\sqrt{q^2-p^2}+q$$
And for other sides. Can all numbers all numbers in this form be sides of a right triangle or not (why)?</p>
| Claude Leibovici | 82,404 | <p><em>As Friedrich Philipp commented.</em></p>
<p>If you apply the fundamental theorem of calculus $$\begin{equation}
g(z-z_0)=\int_{z_0}^{z}~f(t-z_0)\,g(z-t)\,dt
\end{equation}$$
$$\begin{equation}
\frac{dg(z-z_0)}{dz}= f(z-z_0)\,g(0)+\int_a^z f(t-z_0)\, g'(z-t) \, dt
\end{equation}$$</p>
|
2,189,029 | <p>Can all numbers $a, b, c$ that $a^2+b^2=c^2$ be sides of a right triangle? ($c$ is the hypotenuse)
I saw this problem:</p>
<p>Find the number of right triangles that the hypotenuse and one side are both prime numbers. </p>
<p>I said that we don't have any property about the other side, so we can assume it is a real number and not only natural. Thus the sides would be $$p, \sqrt{q^2-p^2}, q $$
Where p and q are primes and q is the hypotenuse, and as result there exists infinte number of triangles of this type. </p>
<p>But what I am not sure about is that all these triples could be sides of a right triangle or not (being correct in triangle inequality).
$$p<\sqrt{q^2-p^2}+q$$
And for other sides. Can all numbers all numbers in this form be sides of a right triangle or not (why)?</p>
| Zaid Alyafeai | 87,813 | <p>$$g(z-z_0)=\int_{z_0}^{z}f(t-z_0)g(z-t)dt
$$</p>
<p>Let $z-t = y$
$$g(z-z_0)=\int_{0}^{z-z_0} f(z-z_0 -y)g(y)dy $$</p>
<p>This implies that for $z-z_0 = t \geq 0$ </p>
<p>$$g(t) = (g*f)(t)$$</p>
<p>Then apply Laplace to both sides </p>
<p>$$G(s) = G(s) F(s)$$</p>
<p>I think you can go from here. </p>
|
154,173 | <p>The cross product $a \times b$ can be represented by the determinant</p>
<p>$$\mathbf{a}\times\mathbf{b}= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
\end{vmatrix}.$$</p>
<p>Does the <em>matrix</em> whose determinant is this have any significance?</p>
| Phira | 9,325 | <p>You can let the matrix act by ordinary matrix multiplication on ordinary vectors in three-dimensional space.</p>
<p>This will transform a vector in a triple containing the original vector and the lengths of the two projections on $a$ and $b$.</p>
<p>While I feel that this counts as "any significance", it isn't very satisfactory, because the matrix, as you presented it, does not allow for matrix multiplication.</p>
<p>I think it is a much more useful point of view, to first view $i$,$j$,$k$ as three scalar variables (better denoted by $x$, $y$, $z$), then take the determinant of your matrix and then regard the cross product as the gradient vector this determinant.</p>
|
154,173 | <p>The cross product $a \times b$ can be represented by the determinant</p>
<p>$$\mathbf{a}\times\mathbf{b}= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
\end{vmatrix}.$$</p>
<p>Does the <em>matrix</em> whose determinant is this have any significance?</p>
| TheEmptyFunction | 609,020 | <p>You could think of the entries as all being Quaternions, i.e. the matrix is an element of <span class="math-container">$M_3(\mathbb{H})$</span>. This gives you a genuine way to act on vectors and do matrix multiplication. But you need to be careful about converting between the two different interpretations <span class="math-container">$i=(1,0,0)\in\mathbb{R}^3$</span> versus <span class="math-container">$i\in\mathbb{H}$</span></p>
<p><a href="https://en.m.wikipedia.org/wiki/Quaternion" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Quaternion</a></p>
|
4,618,433 | <p>Just a heads up: "<span class="math-container">$a$</span>" and "<span class="math-container">$α$</span>" are different</p>
<p>Let <span class="math-container">$a,b \in \Bbb R$</span> and suppose <span class="math-container">$a^2 − 4b \neq 0$</span>. Let <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> be the (distinct) roots of the polynomial <span class="math-container">$x^2 + ax + b$</span>. Prove that there is a real number <span class="math-container">$c$</span> such that either <span class="math-container">$\alpha − \beta = c$</span> or
<span class="math-container">$\alpha − \beta = ci$</span>.</p>
<p>I have no idea how to prove this mathematically. Can someone explain how they would this, including how they would implement this using a proof tree?</p>
<p>This is what I was trying to do.</p>
<p><span class="math-container">$$(x - \alpha)(x - \beta) = x^2 + ax + b$$</span></p>
<p><span class="math-container">$$x^2 - \alpha x - \beta x + \alpha \beta = x^2 + ax + b$$</span></p>
<p><span class="math-container">$$-\alpha x - \beta x = ax$$</span></p>
<p><span class="math-container">$$-x(\alpha + \beta) = ax$$</span></p>
<p><span class="math-container">$$(\alpha + \beta) = -a$$</span></p>
<p><span class="math-container">$$\alpha \beta = b$$</span></p>
<p>However, I'm confused where to go from here and wondering if what I'm doing is wrong.</p>
| user2661923 | 464,411 | <p>Instead of factoring, you can apply the formula for the quadratic equation directly.</p>
<p>If <span class="math-container">$Ax^2 + Bx + C = 0$</span>, then the two roots are given by</p>
<p><span class="math-container">$$\frac{1}{2A} \left[-B \pm \sqrt{B^2 - 4AC}\right].$$</span></p>
<p>This means that (when the roots are distinct), the difference in the two roots will be</p>
<p><span class="math-container">$$\frac{1}{2A} \times 2 \times \pm \sqrt{B^2 - 4AC} ~: ~B^2 - 4AC \neq 0.$$</span></p>
<p>In your problem, you have :</p>
<ul>
<li><p><span class="math-container">$A = 1.$</span></p>
</li>
<li><p><span class="math-container">$B = a.$</span></p>
</li>
<li><p><span class="math-container">$C = b.$</span></p>
</li>
</ul>
<p>So, the difference in the two roots is</p>
<p><span class="math-container">$$\pm \sqrt{a^2 - 4b}. \tag1 $$</span></p>
<p>The expression in (1) above will either be real or imaginary, depending on whether <span class="math-container">$(a^2 - 4b)$</span> is positive or negative.</p>
|
4,618,433 | <p>Just a heads up: "<span class="math-container">$a$</span>" and "<span class="math-container">$α$</span>" are different</p>
<p>Let <span class="math-container">$a,b \in \Bbb R$</span> and suppose <span class="math-container">$a^2 − 4b \neq 0$</span>. Let <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> be the (distinct) roots of the polynomial <span class="math-container">$x^2 + ax + b$</span>. Prove that there is a real number <span class="math-container">$c$</span> such that either <span class="math-container">$\alpha − \beta = c$</span> or
<span class="math-container">$\alpha − \beta = ci$</span>.</p>
<p>I have no idea how to prove this mathematically. Can someone explain how they would this, including how they would implement this using a proof tree?</p>
<p>This is what I was trying to do.</p>
<p><span class="math-container">$$(x - \alpha)(x - \beta) = x^2 + ax + b$$</span></p>
<p><span class="math-container">$$x^2 - \alpha x - \beta x + \alpha \beta = x^2 + ax + b$$</span></p>
<p><span class="math-container">$$-\alpha x - \beta x = ax$$</span></p>
<p><span class="math-container">$$-x(\alpha + \beta) = ax$$</span></p>
<p><span class="math-container">$$(\alpha + \beta) = -a$$</span></p>
<p><span class="math-container">$$\alpha \beta = b$$</span></p>
<p>However, I'm confused where to go from here and wondering if what I'm doing is wrong.</p>
| Andrew Chin | 693,161 | <p>Solving the equation <span class="math-container">$x^2+ax+b=0$</span>, we have <span class="math-container">$$x=\frac{-a\pm\sqrt{a^2-4b}}{2},$$</span>
or more specifically, wlog, we can write <span class="math-container">$$
\alpha=\frac{-a+\sqrt{a^2-4b}}{2},\quad \beta=\frac{-a-\sqrt{a^2-4b}}{2}.$$</span>
The difference, <span class="math-container">$\alpha-\beta$</span>, can then be simplified: <span class="math-container">$$\alpha-\beta=\sqrt{a^2-4b}.$$</span></p>
<p>We are given <span class="math-container">$a^2-4b\neq0$</span>, and this leads into two cases:<br />
If <span class="math-container">$a^2-4b>0$</span>, then <span class="math-container">$\alpha-\beta=c$</span> (the square root of a positive number is a real number).<br />
If <span class="math-container">$a^2-4b<0$</span>, then <span class="math-container">$\alpha-\beta=ci$</span> (the square root of a negative number is an imaginary number).</p>
|
2,022,423 | <p>You are asked to <strong>permute the neighboring sub-sequence</strong> of the sequence $n,n-1,n-2,\cdots,1$ until the sequence is brought to the increasing order. </p>
<p>By <em>permute the neighboring sub-sequence</em> I mean for example:
$5,4,3,2,1 \to 5,3,4,2,1 $ or $5,4,3,2,1\to 5,2,4,3,1$ or $5,4,3,2,1\to5,2,1,4,3$. </p>
<p><strong>What is the least number of permutations needed?</strong></p>
<h2>Edit</h2>
<p>A first nontrivial case I can come up with is:</p>
<p>$$
54321\to52143\to14523\to12345
$$</p>
<h2>Edit 2</h2>
<p>A second nontrivial case I come up with to show $T(n)<f(n)$ is possible regarding the answer by @Brian M. Scott:
$$
7654321\to7632154\to7215634\to1567234\to1234567
$$
maybe $\frac{n+1}{2}$?</p>
| Dap | 467,147 | <p>You need at least <span class="math-container">$\lceil (n+1)/2\rceil.$</span> Consider the number <span class="math-container">$G$</span> of terms whose right neighbor is greater. <span class="math-container">$G$</span> starts at zero and ends as <span class="math-container">$n-1.$</span></p>
<p>For example
<span class="math-container">$$G(5,4,3,2,1)=0$$</span>
<span class="math-container">$$G(5,2,1\color{red},4,3)=1$$</span>
<span class="math-container">$$G(1\color{red},4\color{red},5,2\color{red},3)=3$$</span>
<span class="math-container">$$G(1\color{red},2\color{red},3\color{red},4\color{red},5)=4$$</span>
I have colored in red the comma between each term and its right neighbour whenever that neighbour is greater.</p>
<p>By Brian M. Scott's definition given in the comments, each swap is of the form <span class="math-container">$WXYZ\to WYXZ$</span> where <span class="math-container">$W,X,Y,Z$</span> are possibly empty subwords. Let <span class="math-container">$q,r,s,t$</span> be the lengths of <span class="math-container">$W,X,Y,Z$</span> respectively. Let</p>
<p><span class="math-container">$$w_1,\dots,w_q,x_1,\dots,x_r,y_1,\dots,y_s,z_1,\dots,z_t=WXYZ$$</span>
so
<span class="math-container">$$w_1,\dots,w_q,y_1,\dots,y_s,x_1,\dots,x_r,z_1,\dots,z_t=WYXZ.$$</span></p>
<p>For example for <span class="math-container">$5,4,3,2,1\to 5,2,1,4,3$</span> we have <span class="math-container">$(q,r,s,t)=(1,2,2,0).$</span>
Note
<span class="math-container">\begin{align*}
G(WYXZ)-G(WXYZ)=[w_q<y_1]+[y_s<x_1]+[x_r<z_1]\\-[w_q<x_1]-[x_r<y_1]-[y_s<z_1]\tag{1}
\end{align*}</span>
where <span class="math-container">$[a<b]=1$</span> if <span class="math-container">$a<b$</span> and <span class="math-container">$[a<b]=0$</span> otherwise (Iverson bracket notation). If <span class="math-container">$q=0$</span> ignore the terms using <span class="math-container">$w_q,$</span> and if <span class="math-container">$t=0$</span> ignore the terms using <span class="math-container">$z_1.$</span>
We can write the entire sequence of swaps as <span class="math-container">$P_0\to P_1\to \dots\to P_m$</span> where <span class="math-container">$m$</span> is the number of swaps, and <span class="math-container">$P_0=(n,n-1,\dots,1),$</span> and <span class="math-container">$P_m=(1,2,\dots,n).$</span></p>
<p>In the first swap <span class="math-container">$WXYZ=P_0\to P_1=WYXZ,$</span> we always have <span class="math-container">$w_q>y_1$</span> and <span class="math-container">$x_r>z_1.$</span> So by (1) we must have <span class="math-container">$G(P_1)-G(P_0)\leq 1.$</span>
In the final swap <span class="math-container">$WXYZ=P_{m-1}\to P_m=WYXZ$</span> we always have <span class="math-container">$w_q<x_1$</span> and <span class="math-container">$y_s<z_1,$</span> so <span class="math-container">$G(P_m)-G(P_{m-1})\leq 1.$</span>
Now consider an arbitrary swap <span class="math-container">$WXYZ=P_{i}\to P_{i+1}=WYXZ.$</span> From (1) we know <span class="math-container">$G(P_{i+1})-G(P_i)\leq 3,$</span> and it cannot be exactly <span class="math-container">$3$</span> because that would give <span class="math-container">$w_q<y_1<x_r<z_1<y_s<x_1<w_q.$</span> So <span class="math-container">$G(P_{i+1})-G(P_i)\leq 2.$</span> To summarize:</p>
<p><span class="math-container">\begin{align*}
&G(P_{1})&-&&&G(P_0)&\leq 1\\
&G(P_{i+1})&-&&&G(P_i)&\leq 2&\qquad\text{ for }1\leq i\leq m-2\\
&G(P_{m})&-&&&G(P_{m-1})&\leq 1
\end{align*}</span></p>
<p>which sum to <span class="math-container">$n-1=G(P_m)-G(P_0)\leq 2m-2.$</span> So <span class="math-container">$m\geq\lceil (n+1)/2\rceil.$</span></p>
|
1,942,854 | <blockquote>
<p>Does there exist an $n \not \equiv 3 \pmod{4}$ where $n \in \mathbb{N}$ and is greater than $1$ such that there exists a prime $p >5$ such that \begin{cases}3^{n^2-1} &\equiv 1 \pmod{p}\\2^{n^2-1} &\equiv 1 \pmod{p}?\end{cases}</p>
</blockquote>
<p>I tried finding such a prime but I couldn't. How should we think about it?</p>
| DonAntonio | 31,254 | <p>$$(-5)^{-\frac43}=\left[(-5)^{-4}\right]^{1/3}=\frac1{625^{1/3}}=\frac1{\underbrace{5\sqrt[3]5}_{=5^{4/3}}}>0$$</p>
<p>Or also</p>
<p>$$(-5)^{-\frac43}=\left[(-5)^{1/3}\right]^{-4}=\frac1{(\sqrt[3]{-5})^4}=\frac1{(\sqrt[3]{-5})^3(\sqrt[3]{-5})}=\frac1{(-5)\cdot(-5^{1/3})}=\frac1{5^{4/3}}>0$$</p>
|
1,611,052 | <p>Let H an infinite-dimensional Hilbert space in $\mathbb{R}$</p>
<p>If $x_1, x_2, \ldots x_n \in H$, how to prove: </p>
<p>$\sum_{1\leq i,j\leq n} {\lvert\lvert x_i - x_j \rvert\rvert}^2 \leq \sum_{1\leq i,j\leq n} ({\lvert\lvert x_i \rvert\rvert}^2 + {\lvert\lvert x_j \rvert\rvert}^2)$</p>
| mkk030572 | 210,906 | <p>The triangle inequality yields only $\|x_i - x_j\|^2 \le (\|x_i\| + \|x_j\|)^2 \le \|x_i\|^2 + 2 \|x_i\| \|x_j\| + \|x_j\|^2 \le 2 (\|x_i\|^2 + \|x_j\|^2)$...</p>
|
142,105 | <p>In trying to deduce the lower bound of the ramsey number R(4,4) I am following my book's hint and considering the graph with vertex set $\mathbb{Z}_{17}$ in which $\{i,j\}$ is colored red if and only if $i-j\equiv\pm2^i,i=0,1,2,3$; the set of non-zero quadratic (mod 17) and blue otherwise. This graph shows that $R(4,4)\ge 18$. That's all fine but how am I expected to convince myself of this without drawing a 17-vertex graph.</p>
<p>Is there some way to mathematically justify that such a graph will not contain a monochromatic $K_4$ without drawing this graph?</p>
<p>Thanks.</p>
| Han Altae-Tran | 28,011 | <p>We know that $N(3,4;2)=9$. By symmetry, we have that $N(4,3;2)=9$.
Thus we have the following: $
\lbrack
N(4,4;2)\leq N(3,4;2)+N(4,3;2)=18
\rbrack
$</p>
<p>Thus we create a counterexample, $K_{17}$ that does not contain any
red $K_{4}$ or green $K_{4}$. To do this, we label the vertices
$v_{1}=1,\dots,v_{17}=17$. Then we take all edges, $(v_{i},v_{j})$
with $i-j\equiv\pm1,2,4\text{or}8$ to be red. First off, it is easy
to show that the only 3 node closed paths that can be formed are by
the vertices $T_{1}=\{v_{i},v_{i+1},v_{i+2}\}$ or $T_{2}=\{v_{i},v_{i+4},v_{i+8}\}$,
or $T_{3}=\{v_{i},v_{i+2},v_{i+4}\}$ (i.e. sequences $i_{1},i_{2},i_{3}$
whose elements differ by $1,2,4,\text{or}8$), or some $T$ containing
permutations of these nodes. Thus in order to form $K_{4}$ from one
of these triangles, we must be able to find a vertex $v_{i_{4}}$
such that $v_{i_{4}}$ is connected to each $v_{i_{1}},\dots,v_{i_{3}}$(i.e.
each 3-shuffle of any of the 4 vertices: $(v_{i_{j_{1}}},\dots v_{i_{j_{3}}})$
should contain a triangle, and thus $v_{i_{4}}$ such form a triangle
with each of the other $v_{i_{j}}$. However, looking at the available
triangles, we find that this is impossible. E.g. for $T_{1}$, adding
a $v_{\ell}$ would not produce a triangle between $\{v_{i},v_{i+1},v_{\ell}\}$
for $\ell=i+4$, $i-4$, $i+8$, $i-8$. The same holds for $T_{2}$,
$T_{3}$ for similar reasons.</p>
<p>Now we look at the set of green edges (the edges $(v_{i},v_{j})$
with $i-j\equiv3,5,6,7$). (note that we do not include $9$ because
in this situation, a difference of $8$ is equivalent to a difference
of $9$ by symmetry of the graph. We can only have the following triangles
$T_{1}=\{v_{i},v_{i+3},v_{i+6}\}$, $T_{2}=\{v_{i-5},v_{i},v_{i+7}\}$
since $-12\bmod17=5$, and $T_{3}=\{v_{i-5},v_{i},v_{i+6}\}$
for similar reasons. However, by the same reasoning above, we cannot
form $K_{4}$ from these triangles using the available vertices and
edges. E.g. For $T_{1}$, we cannot add $v_{i-5}$ (because $\{v_{i-5},v_{i},v_{i+3}\}$
is not an available triangle) or $v_{i+5}$ or $v_{i-7}$ or $v_{i+7}$
for similar reasons. Thus we see that we cannot form a red or green
$K_{4}$ in this counter example, so we have shown that
\begin{eqnarray*}
N(4,4;2) & \nless & 18\implies\\
N(4,4;2) & = & 18
\end{eqnarray*}</p>
|
761,616 | <p>How do you integrate $\sqrt{(x^4 + x^2)}$? </p>
| user103816 | 103,816 | <p>Hint: Put $x^2+1=t$ and and notice $2xdx=dt.$</p>
|
761,616 | <p>How do you integrate $\sqrt{(x^4 + x^2)}$? </p>
| Anastasiya-Romanova 秀 | 133,248 | <p>\begin{align}
\int\sqrt{x^4+x^2}\,dx&=\int\sqrt{x^2(x^2+1)}\,dx\\
&=\int x\sqrt{x^2+1}\,dx\\
&=\int \sqrt{x^2+1}\,xdx\\
\end{align}
Let $u=x^2+1$, then $du=2x\,dx$ or $x\,dx=\frac{1}{2}du$. Hence
\begin{align}
\int\sqrt{x^4+x^2}\,dx
&=\int \sqrt{x^2+1}\,xdx\\
&=\int \sqrt{u}\frac{1}{2}du\\
&=\frac{1}{2}\int u^{\frac{1}{2}}\,du\\
&=\frac{1}{2}\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}+C\\
&=\frac{1}{3}u^{\frac{3}{2}}+C\\
&=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C\\
&=\frac{1}{3}\sqrt{(x^2+1)^3}+C
\end{align}</p>
|
761,616 | <p>How do you integrate $\sqrt{(x^4 + x^2)}$? </p>
| Andrew Thompson | 92,231 | <p>Note that you may write, as David pointed out, the simplification $$\int \sqrt{x^4 + x^2}\space dx = \int x\sqrt{x^2+1}\space dx$$</p>
<p>Now, notice the substitution $u = x^2+1$, $du = 2x\space dx$. So we have $$\frac{1}{2}\int\sqrt{u}\space du = \frac{1}{2}\int u^{1/2}\space du = \frac{1}{2}\cdot\frac{2}{3}u^{3/2}+C = \frac{1}{3}\sqrt{(x^2+1)^3}+C$$</p>
|
3,939,620 | <p>Given a polynomial of the form <span class="math-container">$R(z):=\frac{P(z)}{Q(z)}$</span> such that <span class="math-container">$R(z)$</span> has no real roots and <span class="math-container">$deg(Q) \geq deg(P) + 2$</span>, then the integral can be expressed as</p>
<p><span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$</span>
where <span class="math-container">$\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$</span>.</p>
<p>Now for proving this statement, we have first to show that the following limit exists:
<span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$</span>
The argument they used there, is that <span class="math-container">$\exists M >0$</span> s.d. <span class="math-container">$ \vert R(z)\vert \leq \frac{M}{x^2}$</span>. But I don't see, why it exists such an <span class="math-container">$M$</span>. Many thanks for some help!</p>
| Tryst with Freedom | 688,539 | <p>There is perhaps a better way to approach, consider:</p>
<p><span class="math-container">$$ f= \sin^2 x + \sin x$$</span></p>
<p>Now, we can see there are zero set for this function: <span class="math-container">$ \sin x = \{0,-1 \}$</span>, in the interval <span class="math-container">$[0, 2 \pi]$</span> the zero set in terms of 'x' : <span class="math-container">$\{0, \pi, \frac{ 3 \pi}{2} \}$</span>. Hence, the domain of <span class="math-container">$x$</span> in which the function is greater than zero and also belongining the original domain constrained is given by:</p>
<p><span class="math-container">$$ \left[ 0 , \pi \right] \cup \left[ \frac{3 \pi}{2} , 2 \pi\right]$$</span></p>
<p><strong>Explanation:</strong></p>
<p>The function 'sign flips' whenever it hits a root, so we can partition the given domain based on root as as:</p>
<p><span class="math-container">$$ \left[ 0, \pi \right] \cup \left[ \pi, \frac{ 3 \pi}{2} \right] \cup \left[ \frac{3 \pi}{2} , 2 \pi \right]$$</span></p>
<p>Now since in the first domain the function is positive, the second it will be negative and third it will be positive again. Hence, the final answer is given by:</p>
<p><span class="math-container">$$ \left[ 0, \pi \right] \cup \left[ \frac{ 3\pi}{2} , 2\pi \right]$$</span></p>
<hr />
<h3>Edit</h3>
<p>Oops! The function above is slightly more complicated than a regular quadratic in the sense that a zero doesn't correspond exactly as a sign flip!</p>
<p>As correctly pointed out by the @The 2nd, the function retain's it's negative sign even after the zero <span class="math-container">$ \frac{3 \pi}{2}$</span> , hence the whole domain of <span class="math-container">$x$</span> values greater than <span class="math-container">$ \frac{3 \pi}{2}$</span> is neglected. But, there is still one more root to consider, that is the one at <span class="math-container">$ x = 2 \pi$</span></p>
<p>Hence, the final answer is given by:</p>
<p><span class="math-container">$$ \left[ 0 , \pi \right] \cup \left[ \frac{ 3 \pi}{2} \right] \cup \left[ 2 \pi \right]$$</span></p>
|
3,939,620 | <p>Given a polynomial of the form <span class="math-container">$R(z):=\frac{P(z)}{Q(z)}$</span> such that <span class="math-container">$R(z)$</span> has no real roots and <span class="math-container">$deg(Q) \geq deg(P) + 2$</span>, then the integral can be expressed as</p>
<p><span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$</span>
where <span class="math-container">$\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$</span>.</p>
<p>Now for proving this statement, we have first to show that the following limit exists:
<span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$</span>
The argument they used there, is that <span class="math-container">$\exists M >0$</span> s.d. <span class="math-container">$ \vert R(z)\vert \leq \frac{M}{x^2}$</span>. But I don't see, why it exists such an <span class="math-container">$M$</span>. Many thanks for some help!</p>
| The 2nd | 751,538 | <p>Let <span class="math-container">$t = \sin x \,\, (-1 \leq t \leq 1)$</span></p>
<p>The domain of the function will become:</p>
<p><span class="math-container">$$t^2 + t \geq 0$$</span></p>
<p><span class="math-container">$$\implies t \in (-\infty, -1] \cup [0, +\infty)$$</span></p>
<p>But <span class="math-container">$-1 \leq t \leq 1 \implies t = -1$</span> or <span class="math-container">$0 \leq t \leq 1$</span></p>
<ul>
<li><p><span class="math-container">$t = -1 \implies \sin x = -1 \implies x = -\dfrac{\pi}{2} + k2\pi \implies x = \dfrac{3\pi}{2}$</span> (since <span class="math-container">$x \in [0, 2\pi]$</span>)</p>
</li>
<li><p><span class="math-container">$0 \leq t \leq 1 \implies 0 \leq \sin x \leq 1 \implies x \in [0, \pi] \cup \{2\pi\}$</span></p>
</li>
</ul>
<p>Therefore, the domain is:</p>
<p><span class="math-container">$$D = [0, \pi] \cup \left\{\dfrac{3\pi}{2}, 2\pi \right\}$$</span></p>
|
574,614 | <p>if $\gamma:[0,2\pi]\mapsto\Bbb C,\quad \gamma(t)=1+e^{it}$ then show that $|\int_\gamma\frac{dz}{z-\frac{3}{2}}|\le4\pi$ (without computing)</p>
<p>I tried :
$ |\int_\gamma\frac{dz}{z-\frac{3}{2}}| \le \int_\gamma|\frac{1}{z-\frac{3}{2}}|dz$ and $ |z-\frac{3}{2}|\ge||z|-\frac{3}{2}|$ we should find its max value$ .\quad$
$z=\gamma(t)=1+e^{it} $ then we can say $x(t)=1+cost\quad and\quad y(t)=isint$ so we have $
(x-1)^2+y^2=1 $ circle. how can we continue? $max{|z|}=1$ so $||z|-\frac{3}{2}|\le |1-\frac{3}{2}| \quad =\frac{1}{2}? $</p>
<p>edit :
$||z|-\frac{3}{2}|\le |1-\frac{3}{2}| =\frac{1}{2} \Rightarrow \frac{1}{|z-\frac{3}{2}|}\le2\Rightarrow \int_c\frac{1}{|z-\frac{3}{2}|}dz\le\int_c2dz=4\pi \Rightarrow |\int_\gamma\frac{dz}{z-\frac{3}{2}}|\le4\pi $</p>
| MichalisN | 8,432 | <p>Without replacing $z$ with $|z|$ you need to find the minimal value of $|e^{it}-1/2|$ which is $1/2$ (on the unit circle $1$ is closest to 1/2). The length of your path is $2\pi$ so the integral is smaller than $\pi$.</p>
|
3,059,571 | <p><span class="math-container">$$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$</span></p>
<p>I only know of L'hopital method but that is very long. Is there a shorter method to solve this?</p>
| A.Γ. | 253,273 | <p>Another trick is to multiply both the numerator and denominator by
<span class="math-container">$(1+\tan(x/2))(1+\sin x)$</span> and use that
<span class="math-container">\begin{align}
1-\sin^2x&=\cos^2x,\\
1-\tan^2(x/2)&=\frac{\cos^2(x/2)-\sin^2(x/2)}{\cos^2(x/2)}=\frac{\cos x}{\cos^2(x/2)}.
\end{align}</span>
Then you get after simplification
<span class="math-container">$$
\frac{1}{(1+\tan(x/2))^2(1+\sin x)\cos^2(x/2)}\frac{1}{2^3}\color{red}{\left(\frac{\cos x}{\pi/2-x}\right)^3}
$$</span>
where the only uncertainty when <span class="math-container">$x\to\pi/2$</span> is in the red part. The limit of <span class="math-container">$\frac{\cos x}{\pi/2-x}$</span> can be calculated by L'Hopital.</p>
|
376,517 | <p>Let <span class="math-container">$U$</span> be a smooth variety, and <span class="math-container">$U\hookrightarrow X$</span> an smooth compactification with snc boundary <span class="math-container">$D=X\setminus U$</span>. Suppose that <span class="math-container">$\omega\in H^0(U,\Omega^n_U)$</span> is global algebraic <span class="math-container">$n$</span>-form on <span class="math-container">$U$</span>. It defines a class in <span class="math-container">$H^n(U,\mathbb{C})=\mathbb{H}^n(X,\Omega_X^\bullet(\log D))$</span>.</p>
<p>The form <span class="math-container">$\omega$</span> extends to a meromorphic form on <span class="math-container">$X$</span>, denote it by <span class="math-container">$\tilde{\omega}$</span>. This is not necessarily an element of <span class="math-container">$H^0(X,\Omega_X^n(\log D))$</span>, since <span class="math-container">$\tilde{\omega}$</span> can have poles of higher order. Is there an element <span class="math-container">$\omega'\in H^0(X,\Omega_X^n(\log D))$</span> such that <span class="math-container">$\omega'|_U$</span> defines the same cohomology class as <span class="math-container">$\omega$</span> in <span class="math-container">$H^n(U,\mathbb{C})$</span>?</p>
<p>Here are my thoughts: the Hodge spectral sequence degenerates, and so we have
<span class="math-container">$$Gr_F^i(H^n(U,\mathbb{C}))=H^{n-i}(X,\Omega^i(\log D)),$$</span>
and so non-canonically (I believe)
<span class="math-container">$$H^n(U,\mathbb{C})=\bigoplus_i H^{n-i}(X,\Omega^i(\log D)).$$</span>
Now it seems that the class defined by <span class="math-container">$\omega$</span> should be contributed by the summand <span class="math-container">$H^{0}(X,\Omega^n(\log D))$</span>, which would imply the claim. However, to prove this I think one would need something analogous to what Peters and Steenbrink call a "Hodge decomposition in the strong sense" (page 45). However, I do not know if this kind of result exists for non-compact <span class="math-container">$U$</span>?</p>
| AG learner | 74,322 | <p>This rarely happens. For example, when <span class="math-container">$U$</span> is an affine smooth variety, then by Grothendieck's algebraic de Rham theorem (or degeneration of Hodge spectral sequence at <span class="math-container">$E_2$</span>),</p>
<p><span class="math-container">$$H^n(U,\mathbb C)\cong\{\alpha\in H^0(U,\Omega^n)|d\alpha=0\}/\{d\beta|\beta\in H^0(U,\Omega^{n-1})\}.$$</span></p>
<p>In other words, <span class="math-container">$H^n(U,\mathbb C)$</span> are represented by closed algebraic <span class="math-container">$n$</span>-forms on <span class="math-container">$U$</span>. So if any closed algebraic <span class="math-container">$n$</span>-form is cohomologous to some logarithmic form, then it would imply <span class="math-container">$$H^n(U,\mathbb C)=H^0(U,\Omega_U(\log D))=F^nH^n(U,\mathbb C),$$</span></p>
<p>which is a pretty strong condition.</p>
|
1,203,269 | <p>I am trying to compute the hitting time of a linear Brownian motion on a two-sided boundary. More specifically, let $W_t$ be a (one-dimensional) Wiener process. Let $T = \inf \{t: |W_t| = a \}$ for some $ a > 0$. I want to find $\mathbb{P}\{ T > t\}$. </p>
<p>I know that probability distribution hitting time of a positive level, $\inf \, \{t: W_t = b\,, \ b > 0 \}$ can be computed quite easily, but I am not sure how to deal with it when dealing with the two-sided hitting time, i.e. with the absolute value. I am thinking of the minimum of hitting times of level $a$ and $-a$, but I can't get a promising conclusion.</p>
| C.Koca | 426,569 | <p>The distribution of hitting times for a Brownian Motion, <span class="math-container">$S$</span> starting at <span class="math-container">$0$</span> with barriers at <span class="math-container">$c$</span> and <span class="math-container">$-c$</span> and step size <span class="math-container">$l=1$</span> is given by</p>
<p><span class="math-container">$$
P(T>x)=\left(1-\sum_{i=-\infty}^\infty(-1)^{i+1}\left[\Phi\left(\frac{(2i+1)c}{\sqrt{x}}\right)-\Phi\left(\frac{(2i-1)c}{\sqrt{x}}\right)\right]\right),
$$</span>
where <span class="math-container">$\Phi()$</span> is the CDF of the standard normal distribution.</p>
<p>I am looking for a tidier representation myself <a href="https://math.stackexchange.com/questions/4309598/what-is-this-lognormal-like-distribution">in this question</a>.</p>
<p>The proof comes from the reflection principle. The probability of the <span class="math-container">$S$</span> hitting the lower bound <span class="math-container">$-c$</span> (<span class="math-container">$c$</span>) and returning back to the region <span class="math-container">$[-c,c]$</span> before time <span class="math-container">$x$</span> is equal to <span class="math-container">$S$</span> being in <span class="math-container">$[-3c,-c]$</span> (<span class="math-container">$[c,3c]$</span>) region at time step <span class="math-container">$x$</span>. For one sided boundary, this would be it. However, it is also possible for <span class="math-container">$S$</span> to hit <span class="math-container">$-c$</span>, return back to <span class="math-container">$[-c,c]$</span> and then hit <span class="math-container">$c$</span>. <span class="math-container">$S$</span> being in <span class="math-container">$[-3c,-c]\cup [c,3c]$</span> counts <span class="math-container">$S$</span> hitting both barriers twice. So, we need to take <span class="math-container">$S$</span> hitting both barriers separately.</p>
<p>The probability of <span class="math-container">$S$</span> hitting first <span class="math-container">$-c$</span> and hitting <span class="math-container">$c$</span> and then returning back to <span class="math-container">$[-c,c]$</span> is equal to <span class="math-container">$S$</span> being in <span class="math-container">$[-5c,-3c]$</span>.</p>
<p>You can extend this approach to infinity to reach the correct distribution.</p>
|
945,395 | <p>Let $a_1$ be real, and define $$a_{n+1}=\frac{2a_n^3}{1+a_n^4}$$ How can I prove that this $\{a_n\}$ to have limit. </p>
<p>I find it is hard to track. What I can do is just when $a_1=1$ then $a_n=1$; when $a_1=-1$, then $a_n=-1$; when $|a_1|<1$, $a_n\to 0$. When $|a_1|>1$, I have not find any idea.</p>
| Macavity | 58,320 | <p>It is sufficient to consider $a_n > 0$, as otherwise you could do the same analysis with $b_n = -a_n$ and $a_1=0 \iff a_n=0$, so that's trivial.</p>
<p>Note that $1+x^4 \ge 2x^2$ by AM-GM with equality iff $x=1$, so we have $a_{n+1} < a_n$, unless $a_k=1 \implies a_{n> k}=1$. Thus we have a decreasing, positive (i.e. bounded) sequence ...</p>
|
2,884,785 | <p>Find all integral pairs (x,y) such that - $$( xy - 1)^2 = (x +1)^2 + ( y+1)^2$$</p>
<hr>
<p><strong>My Approach :</strong></p>
<p>I just expanded this equation and wrote it in another form - $$\frac{(xy+1)(xy-1)}{(x+y)}-2=x+y$$ and from this we can say that $(x+y)|(xy+1) \ \mathrm{or}\ (x+y)|(xy-1) $ . But i don't know how to solve it further. Please help me with this. </p>
| nonuser | 463,553 | <p>Or $$x^2y^2 = (x+y)^2+2(x+y)+1 \implies x^2y^2 = (x+y+1)^2$$</p>
<p><strong>1. case:</strong> $$xy =x+y+1\implies (x-1)(y-1)=2\implies ....$$</p>
<p><strong>2. case:</strong> $$xy =-x-y-1\implies (x+1)(y+1)=0\implies ....$$</p>
|
257,623 | <p>Consider the following ellipse, generated by the bounding region of the following points</p>
<pre><code>ps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};
rec = N@BoundingRegion[ps, "FastEllipse"];
Graphics[{rec, Red, Point@ps}]
</code></pre>
<p><a href="https://i.stack.imgur.com/gvtUB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gvtUB.png" alt="enter image description here" /></a></p>
<p>We have that the ellipse 'rec' is given in the form</p>
<pre><code>Ellipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]
</code></pre>
<p>How can I retrieve the lengths of the two main axes of such ellipsoid? Following <a href="https://en.wikipedia.org/wiki/Ellipsoid#As_a_quadric" rel="nofollow noreferrer">this representation</a> and Mathematica's general definition of <code>Ellipsoid</code></p>
<p><a href="https://i.stack.imgur.com/vaCtl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vaCtl.png" alt="enter image description here" /></a></p>
<p>I tried the following, using the eigenvalues of <code>rec[[2]]</code></p>
<pre><code>eigs = Eigenvectors[Inverse[rec[[2]]]]
eigv = Eigenvalues[Inverse[rec[[2]]]];
lens = 2/Sqrt[eigv]
Out[]= {2.06559, 3.57771}
</code></pre>
<p>where the <code>2</code> factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get</p>
<pre><code>Graphics[{rec, Red, Point@ps,
Blue, Line[
RegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/
2, (lens[[1]] eigs[[1]])/2}],
Line[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/
2, (lens[[2]] eigs[[2]])/2}]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/1N5gL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1N5gL.png" alt="enter image description here" /></a></p>
<p>Is this correct? Is there a quicker way of doing this?</p>
| halmir | 590 | <p>Here's the alternate way using SingularValueDecomposition:</p>
<pre><code>pt = rec[[1]]; mat = rec[[2]];
{u, s, v} = SingularValueDecomposition[(mat + Transpose[mat])/2];
func = Composition[AffineTransform[{u, pt}], ScalingTransform[Sqrt[Diagonal[s]]]];
</code></pre>
<p>and the length:</p>
<pre><code>EuclideanDistance @@@ {func@{{0, -1}, {0, 1}}, func@{{-1, 0}, {1, 0}}}
</code></pre>
<blockquote>
<p>{2.06559, 3.57771}</p>
</blockquote>
<pre><code>Graphics[{rec, Blue, Line[func@{{-1, 0}, {1, 0}}],
Line[ func@{{0, -1}, {0, 1}}]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/T9g1r.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/T9g1r.jpg" alt="enter image description here" /></a></p>
|
7,981 | <p>I've read so much about it but none of it makes a lot of sense. Also, what's so unsolvable about it?</p>
| Sniper Clown | 21,855 | <p>You have to...have to....read <a href="http://www.math.jhu.edu/~wright/RH2.pdf" rel="noreferrer">this friendly introduction</a>. <em>(Making it CW since it's just a link.)</em></p>
|
188,102 | <p>I have the following list: </p>
<pre><code>m={{14, "extinguisher"}, {54, "virgule"}, {55, "turnoff"}, {51,
"sofa"}, {77, "beachcomber"}, {61, "stoic"}, {6,
"isomorphism"}, {34, "leftist"}, {84, "spline"}, {42,
"heartiness"}, {35, "postnatal"}, {41, "stratified"}, {66,
"silkworm"}, {95, "conformance"}, {38, "hemophiliac"}, {19,
"abdication"}, {13, "reimpose"}, {82, "cowhide"}, {78,
"banteringly"}, {26, "contention"}};
</code></pre>
<p>I wonder if it is possible to make a spiral bubble chart of this on Mathematica, where the number is represented by how the bubble should be big and each bubble would be labeled by the corresponding words. </p>
<p>In fact I am expecting to make something as follow:
<a href="https://i.stack.imgur.com/nErYz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nErYz.png" alt="enter image description here"></a></p>
| OkkesDulgerci | 23,291 | <p><strong>Edit:</strong></p>
<pre><code>m = Sort[{{14, "extinguisher"}, {54, "virgule"}, {55, "turnoff"}, {51,
"sofa"}, {77, "beachcomber"}, {61, "stoic"}, {6,
"isomorphism"}, {34, "leftist"}, {84, "spline"}, {42,
"heartiness"}, {35, "postnatal"}, {41, "stratified"}, {66,
"silkworm"}, {95, "conformance"}, {38, "hemophiliac"}, {19,
"abdication"}, {13, "reimpose"}, {82, "cowhide"}, {78,
"banteringly"}, {26, "contention"}}];
rad = m[[All, 1]];
center = CirclePoints[50, 20];
Show[Table[
Graphics[{ColorData["ThermometerColors"][
Rescale[rad[[i]], MinMax@rad]],
Disk[center[[i]], (rad[[i]])^(1/2.1)],
Text[Style[m[[i, 2]], Black], center[[i]]]}], {i, Length@rad}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/gdO86.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gdO86.png" alt="enter image description here"></a>
<strong>Original answer:</strong></p>
<p>Here is my attempt. It is not elegant. Note this is misleading!!</p>
<pre><code> m = Sort[{{14, "extinguisher"}, {54, "virgule"}, {55, "turnoff"}, {51,
"sofa"}, {77, "beachcomber"}, {61, "stoic"}, {6,
"isomorphism"}, {34, "leftist"}, {84, "spline"}, {42,
"heartiness"}, {35, "postnatal"}, {41, "stratified"}, {66,
"silkworm"}, {95, "conformance"}, {38, "hemophiliac"}, {19,
"abdication"}, {13, "reimpose"}, {82, "cowhide"}, {78,
"banteringly"}, {26, "contention"}}];
rad = m[[;; -2, 1]];
center = CirclePoints[40, 19];
Show[Graphics[{RGBColor[0.6, 0, 0],
Disk[{0, 0}, (m[[-1, 1]])^(1/1.3)],
Text[Style[m[[-1, 2]], Black], {0, 0}]}],
Table[Graphics[{ColorData["ThermometerColors"][
Rescale[rad[[i]], MinMax@rad]],
Disk[center[[i]], (rad[[i]])^(1/2.1)],
Text[Style[m[[i, 2]], Black], center[[i]]]}, Frame -> True], {i,
Length@rad}]]
</code></pre>
<p><a href="https://i.stack.imgur.com/7tR16.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7tR16.png" alt="enter image description here"></a></p>
|
2,267,005 | <p>I have been asked to evaluate the $\int{Fdr}$ over a curve $C$ where $F = yz\mathbf{i} + 2xz\mathbf{j} + e^{xy}\mathbf{k}$ and $C$ is the curve $x^2 + y^2 = 16, z =5$ with downward orientation</p>
<p>I want to use Stokes theorem, so I am thinking of parametrizing this surface as $(x, y, z)=(4 \cos t, 4\sin t,z)$ but I am confused as to what the bounds for $z$ would be since I have only been given $z =5$. I know $t \in (0, 2 \pi)$.</p>
<p>Any help would be greatly appreciated. Thanks</p>
| marshal craft | 167,793 | <p>Ok if I interpret the question so that it is interesting (as currently phrased it is trivial that answer is, it doesnt matter the max distance is 450) then we ask if the front wears through 100% of the tire in 450 km and back in 600 km then what is the maximum distance one can travel on good tires provided they swap the tires exactly one time?</p>
<p>We now consider the front and back separately for swap $x\le 450$,</p>
<p>$$D_{max \ f}(x)= 450\frac{600-x}{600}+x$$</p>
<p>$$D_{max \ r}(x)=600\frac{450-x}{450}+x$$</p>
<p>Now for final max we take $$D_{max}(x)=MIN(D_{max \ f}(x), D_{max \ r}(x))$$</p>
<p>The question now is to find the maximum of $D_{max}(x)$</p>
<p>For $x \le 300$, $D_{max}(x)=D_{max \ f}(x)$ and for $x \ge 300$, $D_{max}(x)=D_{max \ r}(x)$.</p>
<p>We can see a clear maximum at $x=300$ which would be the best place to swap yielding $500 \ km$ total. <img src="https://i.stack.imgur.com/6pFqd.gif" alt="enter image description here"></p>
<p><strong>Edit</strong> I've made a bit of a clerical error in the algebra but the procedure is sound.</p>
|
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