qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,995,471 | <p>I'm trying to find all of the the roots to the following polynomial with a variable second coefficient:
$$P(x)=4x^3-px^2+5x+6$$
All of the roots are rational, and $p$ is too. It is also given that the difference of 2 roots equals the third, e.g. $r-s=t$. I would like to solve for the roots using relationships between roots & the rational roots theorem.</p>
<p>I know from relationships between roots (Vieta's formula) that $p/4=r+s+t$, which can be reduced to $p/4=2r$ per the previous equation, and therefore $p/8$ is a root. However, I'm not sure where to go from here-- performing the substitution with the other coefficients does not seem to yield anything that lets me solve for a root or $p$. For example, we know from the coefficient of $x^0$ that
$$5/4=rs+rt+st=rs+(r+s)(r-s)$$
but there is no obvious substitution that can be made here that would put things in terms of one variable.</p>
<p>How do I solve for the roots and $p$ using relationships between roots and the rational roots theorem here? Thanks!</p>
| James Ko | 385,100 | <p>Turns out it actually was possible to use relationships between roots here to help factor the equation, I just wasn't looking hard enough.</p>
<p>Since $r - s = t$, then $r = s + t$. As mentioned, $\frac 54 = rs + st + rt = st + (s + t)^2$. Now, both $st$ and $s + t$ can be written in terms of $r$, since</p>
<p>$$
-\frac 32 = rst
$$</p>
<p>$$
-\frac 3{2r} = st
$$</p>
<p>and $s + t = r$. So substituting those values in, we get</p>
<p>$$
\frac 54 = -\frac 3{2r} + r^2
$$</p>
<p>$$
\frac 54 r = -\frac 32 + r^3
$$</p>
<p>$$
0 = r^3 - \frac 54 r - \frac 32
$$</p>
<p>$$
0 = 4r^3 - 5r - 6
$$</p>
<p>at which point we can use synthetic division to solve for $r$ (since all roots are rational). Then we get $r = \frac 32$ as a root, and factoring out $2x - 3$ and using the quadratic formula to solve for the remaining roots, $s = -\frac 12$ and $t = 2$.</p>
|
3,167,062 | <p>Recently friend of mine showed me a way to write <span class="math-container">$f(x)=0$</span>, however I could not find anything about that in the Internet. Is it correct?</p>
<p><span class="math-container">$f(x)=x^2-1\stackrel{set}{=}0 \\
x = 1 \lor x = -1$</span></p>
| Mark Bennet | 2,906 | <p>Geometrically it gives the (signed) area of the parallelogram defined by the two vectors. </p>
<p>If you multiply by the appropriate unit normal to the plane you get the normal three dimensional cross product. You don't get a vector in the plane though.</p>
<p>If you try to define a "cross product" in four dimensions, you might appreciate that the familiar situation in three dimensions is a happy coincidence which trips up people who try to generalise in the wrong way.</p>
|
2,159,136 | <p>I arrived to this question while solving a question paper. The question is as follows:</p>
<blockquote>
<p>If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$</p>
</blockquote>
<p>I started as </p>
<p>$$\begin{align}
f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt]
&=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt]
&=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt]
&\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt]
&=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt]
&=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt]
&=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt]
&\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt]
&=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt]
&=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt]
&\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)}
\end{align}$$</p>
<p>Hence the answer should be ...</p>
<p>$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$</p>
<p>... but the answer given was $\frac{1}{12}$.</p>
<p>I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.</p>
| Katherine Wu | 817,994 | <p>‘Tis I have a better solution to solve this math problem...</p>
<p>Let <span class="math-container">$a = \cos^2 x$</span> and <span class="math-container">$b = \sin^2 x,$</span> so <span class="math-container">$a + b = 1.$</span> Then
[(a + b)^2 = a^2 + 2ab + b^2 = 1,]so <span class="math-container">$a^2 + b^2 = 1 - 2ab.$</span> Also,
[(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 = 1,]so
<span class="math-container">\begin{align*}
a^3 + b^3 &= 1 - (3a^2 b + 3ab^2) \\
&= 1 - 3ab(a + b) \\
&= 1 - 3ab.
\end{align*}</span>Therefore,
<span class="math-container">\begin{align*}
f_4(x) - f_6(x) &= \frac{\sin^4 x + \cos^4 x}{4} - \frac{\sin^6 x + \cos^6 x}{6} \\
&= \frac{a^2 + b^2}{4} - \frac{a^3 + b^3}{6} \\
&= \frac{1 - 2ab}{4} - \frac{1 - 3ab}{6} \\
&= \boxed{\frac{1}{12}}.
\end{align*}</span></p>
<p>BTW, I solved this</p>
|
58,912 | <p>In the board game <a href="http://en.wikipedia.org/wiki/Hex_%28board_game%29" rel="nofollow">Hex</a>, players take turns coloring hexagons either red or blue. One player tries to connect the top and bottom edges of the board, colored red; the other tries to connect the left and right edges, colored blue. It is known that a game of Hex will never end in a tie: no matter how it is played, there will always be either a blue path connecting the blue edges, or a red path connecting the red edges.</p>
<p>My question is, if this fact always holds for a finite grid of hexagons, does it also hold on the plane? If the top and bottom edges of a square are colored red, the left and right edges are colored blue, and the interior of the square is colored arbitrarily, must there be either a red path connecting the red edges, or a blue path connecting the blue edges?</p>
<p>More formally, let $S$ be any subset of $[0, 1]^2$. $S$ will represent the points that are red. Must there be either a path within $S$ whose endpoints are of the form $(x, 0)$ and $(x, 1)$, or a path within $[0, 1]^2 - S$ whose endpoints are of the form $(0, y)$ and $(1, y)$?</p>
| t22 | 11,209 | <p>It is even true that $I^{2}$ may be represented as an union of two closed sets
$S_{1}$ and $S_{2}$ none of them containing a topological arc connecting two
opposite sides of the square. Here is a possible example.</p>
<p>Let $D$ be a closed disk in the interior of $I^{2}$ , take 4 disjoint
topological spirals $J_{i}$ lying in $I^{2}$ and homeomorphic to $[0,\infty)$,
each of them beginning from a different side of the square and spiralling
around $D$ in such a way that $\overline{J_{i}}=J_{i}\cup\partial D$. Now
thicken slightly each $J_{i}$ to get some $J_{i}^{\prime}\supset J_{i}$
homeomorphic to a closed half-plane so that the $J_{i}^{\prime}$ are still
disjoint and spiralling around $D$. [$\overline{J_{i}^{\prime}}=J_{i}^{\prime
}\cup\partial D$.] To get the claimed representation of the square, it suffices
to set
$S_{1}=(\cup_{i=1}^{4}J_{i}^{\prime})\cup D$ and $S_{2}=\overline
{I^{2}\backslash S_{1}}$. </p>
|
2,776,388 | <p>I tried this problem and I first found the lim of $x^x$ as $x$ approaches zero from right to be 1 (I did this by re-writing $x^x$ as an exponential) and when I repeated the same process to find lim of $x^{(x^x)}$, I found 1 again but the final answer should be ZERO. Could I have an explanation on why it's a zero?</p>
| Rhys Hughes | 487,658 | <p>As $x\to 0^+$, we see $x^x\to 0^0$, which evidence suggests is $1$ for $x \in \Bbb R$. </p>
<p>Thus for $x^{x^x}$, as $x\to 0^+$ we get $0^1$, which is certainly $0$.</p>
<p>Incidentally, this implies that the limit of $x^{x^{x^x}}=1$, and $x^{x^{x^{x^x}}}=0$, etc...</p>
<p>Graphed for your perusal <a href="https://www.desmos.com/calculator/nq2wqz2sya" rel="nofollow noreferrer">here</a> </p>
|
1,299,630 | <p>I've been wondering if there is any use to defining a set that is isomorphic to $\mathbb{Q}^2$ (in the same way that $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$).</p>
<p>I immediately see a problem with <em>e.g.</em> Cauchy's theorem (there's no $\pi$ available) and perhaps even defining a "analytical function" concept because there are series defined in $\mathbb{Q}$ that do not converge within $\mathbb{Q}$.</p>
<p>Still, I wonder if there's a way around these difficulties.</p>
| mp8394 | 243,601 | <p>Given a bijective linear isometry $T:X\rightarrow Y$, the dual map $T^*:Y^*\rightarrow X^*$ is also a bijective linear isometry. (This follows from the fact that $T^*$ has inverse $(T^{-1})^*$ and $\|T^*\|=\|T\|$.) From this, we have that $T^{**}:X^{**}\rightarrow Y^{**}$ is a bijective linear isometry as well.</p>
<p>Let $J_X:X\rightarrow X^{**}$ is the canonical embedding of $X$ into its bidual, which we're assuming is surjective. If $\phi\in Y^{**}$, then there exists a unique $\psi \in X^{**}$ such that $T^{**}(\psi)=\phi$ (since $T^{**}$ is a bijective isometry). As $J_X$ is surjective, there is an $x\in X$ such that $\psi=J_X(x)$, so $\phi=T^{**}(J_X(x))$.</p>
<p>Now choose any $f\in Y^*$. From the above, $\phi(f)=T^{**}(J_X(x))(f)$. By definition of $T^{**}$ and of $J_X$, this is equal to $(J_X(x))(T^*(f))=(T^*(f))(x)=f(T(x))$. Notice that this is $J_Y(T(x))(f)$. As this holds for every $f\in Y^*$, we have that $J_Y(T(x))=\phi$.</p>
<p>This means that, for any $\phi\in Y^{**}$, we can find a $y\in Y$ (the element $T(x)$ in the above paragraph) such that $\phi=J_Y(y)$, so $J_Y$ is surjective, which means precisely that $Y$ is reflexive.</p>
|
942,846 | <p>Given 5 different green dyes,4 different blue dyes and 3 different red dyes the number of combinations does that can be chosen by taking at least one green and blue dyes is?</p>
| Marc van Leeuwen | 18,880 | <p>You can use inclusion-exclusion here. Count the number of combinations without restrictions (you did not specify the number in a selection, so I can't do the computation for you). Subtract the number of selections without any green dye by redoing the compoutation after removing all gree dyes. Similarly subtract the number of selections without any blue dye. The selections with neither green nor blue dyes have been doubly subtracted, so need to be added back once to get the proper answer.</p>
|
942,846 | <p>Given 5 different green dyes,4 different blue dyes and 3 different red dyes the number of combinations does that can be chosen by taking at least one green and blue dyes is?</p>
| Sooraj S | 223,599 | <p>Number of green dyes =5</p>
<p>Number of blue dyes=4</p>
<p>Number of red dyes =3</p>
<p>The required number of ways of choosing the dye is =
$$
[(^2C_1)^5-1]*[(^2C_1)^4-1]*(^2C_1)^3=[2^5-1]*[2^4-1]*2^3\\=(32-1)*(16-1)*8=31*15*8=3720
$$
where each $^2C_1=2$ corresponds to selecting or not selecting a particular dye and subtracting with $1$ is to eliminate the case of not selecting all the particular dyes, i.e.$2^5-1$ which excludes the case of not selecting any green dye from all possible ways.</p>
|
2,753,504 | <p>Where $a,b$ and $c$ are positive real numbers.</p>
<p>So far I have shown that $$a^2+b^2+c^2 \ge ab+bc+ac$$ and that $$a^2+b^2+c^2 \ge a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}$$ but I am at a loss what to do next... I have tried adding various forms of the two inequalities but always end up with something extra on the side of $ab+bc+ac$. Any help appreciated!</p>
| hamam_Abdallah | 369,188 | <p>Since $(a-b)^2\ge 0,$</p>
<p>we have</p>
<p>$$a^2+b^2\ge 2ab$$
$$b^2+c^2\ge 2bc $$
$$c^2+a^2\ge 2ca .$$</p>
<p>the sum gives</p>
<p>$$2 (a^2+b^2+c^2)\ge 2 (ab+bc+ac) $$</p>
|
47,724 | <p>I am looking at a past exam written by a student. There was a question I believed he got correct but received only 1/4. The marker wrote down "4 more compositions, order matters".</p>
<p>This is the problem:</p>
<p>List all 3 part compositions of 5. (recall that compositions have no zeros)</p>
<p>$(1, 1, 3)
(1, 2, 2)
(1, 3, 1)
(3, 1, 1)
(2, 1, 2)
(2, 2, 1)$ (what the student has on paper) <strong>EDIT: I made some mistakes copying</strong></p>
<p>My guess is that the student wrote only (1,1,3) and (2,2,1) ,but corrected it after exam was returned. I just want to make sure that this is the case, so that I don't miss something.</p>
| mathmath8128 | 11,976 | <p>You're correct... the only possible ways to add 5 from 3 parts is (1,1,2) and (1,2,2)... each one you have 3 choices on where to place the 2 and 1 respectively, so 3*2=6 possible ways.</p>
|
871,730 | <p>Given x and y we define a function as follow : </p>
<pre><code>f(1)=x
f(2)=y
f(i)=f(i-1) + f(i+1) for i>2
</code></pre>
<p>Now given x and y, how to calculate f(n)</p>
<p>Example : If x=2 and y=3 and n=3 then answer is 1</p>
<p>as f(2) = f(1) + f(3), 3 = 2 + f(3), f(3) = 1.</p>
<p>Constraints are : x,y,n all can go upto 10^9.</p>
| Empy2 | 81,790 | <p>Calculate the first dozen numbers. You should find a simple result for this algorithm.</p>
|
3,426,550 | <p>Are there parts in functional analysis that would be easier to learn without real analysis or is this just not possible?</p>
<p>When I say functional analysis I mean</p>
<ol>
<li>Lebesgue measure and integral</li>
<li>spaces of Lebesgue integrable functions</li>
<li>Banach spaces</li>
<li>duality</li>
<li>bounded linear operator and linear operators in general</li>
<li>Hilbert space</li>
<li>reproducing kernel Hilbert space</li>
<li>non-linear analysis in Banach spaces</li>
</ol>
| P Vanchinathan | 28,915 | <p>Items 1 and 2 listed by you specifies Lebesgue measure and integrable functions. It will be unwise to try to understand them without having been exposed Riemann integral which is part of standard first course in real analysis. That Lebesgue theory is logically a generalisation and so usual Riemann integration is a special case is correct but only academically so.</p>
<p>Also textbooks of functional analysis are not written for people not exposed to real analysis. So it will be difficult to lern it.</p>
<p>It is possible to do mathematics by communicating with other mathematicians by writing letters (Euler, Newton, Fermat, Gauss did so). But you want a modern internet based forum now. And email and LaTeX all are needed though logically it is not a requirement. Now you decide how you should learn Functional Analysis.</p>
|
2,812,472 | <p>I am trying to show that</p>
<p>$$
\frac{d^n}{dx^n} (x^2-1)^n = 2^n \cdot n!,
$$ for $x = 1$. I tried to prove it by induction but I failed because I lack axioms and rules for this type of derivatives. </p>
<p>Can someone give me a hint?</p>
| W. mu | 369,495 | <p>$$\frac{d^n}{dx^n}(x^2-1)^n=2nx\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^{n-1}+2n\frac{d^{n-2}}{dx^{n-2}}(x^2-1)^{n-1}$$</p>
<p>The second term is $0$ at $x=1$.</p>
<p>So we have the recursion $I_n=2nI_{n-1}$.</p>
|
1,060,213 | <p>I have searched the site quickly and have not come across this exact problem. I have noticed that a Pythagorean triple <code>(a,b,c)</code> where <code>c</code> is the hypotenuse and <code>a</code> is prime, is always of the form <code>(a,b,b+1)</code>: The hypotenuse is one more than the non-prime side. Why is this so?</p>
| rogerl | 27,542 | <p>All Pythagorean triples are of the form $k(p^2-q^2), 2kpq, k(p^2+q^2)$ where $p$ and $q$ are positive, relatively prime and of opposite parity. Since $a$ is to be prime, it must be $k(p^2-q^2)$, not $2kpq$; this forces $p^2-q^2=1$, which is impossible, or $k=1$ and $p^2-q^2$ is a prime. So the triangle is of the form $$a=p^2-q^2,\quad b=2pq,\quad c=p^2+q^2.$$ But then
$$a = p^2-q^2 = (p-q)(p+q),$$
so that since $a$ is prime we must have $p-q=1$. But then
$$c-b = p^2 + q^2 - 2pq = (p-q)^2 = 1.$$</p>
|
4,021,297 | <p><span class="math-container">$\Sigma$</span> is a <span class="math-container">$\sigma$</span>-algebra if it satisfies the following three properties:</p>
<ul>
<li><span class="math-container">$X$</span> is in <span class="math-container">$\Sigma$</span>, and <span class="math-container">$X$</span> is considered to be the universal set in
the following context.</li>
<li><span class="math-container">$\Sigma$</span> is closed under complementation: If <span class="math-container">$A$</span> is in <span class="math-container">$\Sigma$</span>, then
so is its complement, <span class="math-container">$A^{c}=X-A$</span>.</li>
<li><span class="math-container">$\Sigma$</span> is closed under countable unions: If <span class="math-container">$A_{1}$</span>, <span class="math-container">$A_{2}$</span>,
<span class="math-container">$A_{3}$</span>, ... are in <span class="math-container">$\Sigma$</span>, then so is <span class="math-container">$A=A_{1}\cup A_{2}\cup
A_{3}\cup\ldots$</span>.</li>
</ul>
<p>If the above three are our definitions.</p>
<ul>
<li>Can we show that if <span class="math-container">$A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$</span> and <span class="math-container">$A$</span> is in <span class="math-container">$\Sigma$</span> then each of the <span class="math-container">$A_{i}$</span> are in <span class="math-container">$\Sigma$</span>?</li>
<li>Do we need any additional conditions to show this or would the three basic axioms suffice?</li>
</ul>
<p>Notes:</p>
<ol>
<li>I found some related discussions, but am unable to see if my doubts can be cleared using the pointers from these threads. Listing them below for completeness:</li>
</ol>
<p><a href="https://math.stackexchange.com/questions/468577/can-every-member-of-a-sigma-algebra-be-represented-by-a-countable-union-of-di?rq=1">Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?</a></p>
<p><a href="https://math.stackexchange.com/questions/1142962/show-that-any-element-of-a-sigma-algebra-is-the-union-of-disjoint-sets?rq=1">Show that any element of a sigma algebra is the union of disjoint sets</a></p>
<ol>
<li>I was trying to understand the concept in the question asked since the corresponding property seemed to be required to show that If <span class="math-container">$E$</span> has <span class="math-container">$\sigma$</span>-finite measure, then <span class="math-container">$E$</span> is inner regular (Rudin RCA, Definition 2.16, some related threads below):</li>
</ol>
<p><a href="https://math.stackexchange.com/questions/199938/if-e-has-sigma-finite-measure-then-e-is-inner-regular">If $E$ has $\sigma$-finite measure, then $E$ is inner regular</a></p>
<p><a href="https://math.stackexchange.com/questions/325009/rudin-real-and-complex-definition-2-16?noredirect=1&lq=1">Rudin Real And Complex Definition 2.16</a></p>
<p>This point is clarified: If <span class="math-container">$E$</span> is in a sigma algebra <span class="math-container">$\Sigma$</span> and <span class="math-container">$E$</span> has <span class="math-container">$\sigma$</span>-finite measure then it is a countable union of sets, <span class="math-container">$E_{i}$</span> with finite measure (<span class="math-container">$\mu\left(E_{i}\right)<\infty$</span>). Which means the sets that make up the union of <span class="math-container">$E$</span> are already in <span class="math-container">$\Sigma$</span> since the measure <span class="math-container">$\mu$</span> is defined only on the members of the sigma algebra <span class="math-container">$\Sigma$</span>. Hence the <span class="math-container">$E_{i}$</span> belong to the sigma algebra, <span class="math-container">$\Sigma$</span></p>
<ol>
<li>Though my original doubt is clarified, I am still curios about the conditions under which the reverse of the third axiom hold. It is quite possible that I am missing something very basic. Happy to delete this question, if that is the case once my doubt is clarified.</li>
</ol>
| Kevin Arlin | 31,228 | <p>This is a pretty confusing question to ask early in one’s logical education! What’s going on is that, if we want to think about the statement “<span class="math-container">$p$</span> implies <span class="math-container">$q$</span>” and we can prove that the converse “<span class="math-container">$q$</span> implies <span class="math-container">$p$</span>” is false, then in fact we know that <span class="math-container">$q$</span> is true and <span class="math-container">$p$</span> is false, which means <span class="math-container">$p$</span> does imply <span class="math-container">$q$</span>. That is, symbolically, <span class="math-container">$$\neg(q\to p)\to (p\to q).$$</span>
But this last implication is not an equivalence! That is, knowing <span class="math-container">$p\to q$</span> is true doesn’t mean <span class="math-container">$q\to p$</span> is false, or equivalently, knowing <span class="math-container">$q\to p$</span> is true doesn’t mean <span class="math-container">$p\to q$</span> is false.</p>
<p>Thus a statement is to some extent independent of its converse, but not perfectly so, as it’s impossible for both a statement and its converse to be false. This comes somehow from the asymmetry of the truth table for implication, which has three T’s out of four.</p>
|
4,021,297 | <p><span class="math-container">$\Sigma$</span> is a <span class="math-container">$\sigma$</span>-algebra if it satisfies the following three properties:</p>
<ul>
<li><span class="math-container">$X$</span> is in <span class="math-container">$\Sigma$</span>, and <span class="math-container">$X$</span> is considered to be the universal set in
the following context.</li>
<li><span class="math-container">$\Sigma$</span> is closed under complementation: If <span class="math-container">$A$</span> is in <span class="math-container">$\Sigma$</span>, then
so is its complement, <span class="math-container">$A^{c}=X-A$</span>.</li>
<li><span class="math-container">$\Sigma$</span> is closed under countable unions: If <span class="math-container">$A_{1}$</span>, <span class="math-container">$A_{2}$</span>,
<span class="math-container">$A_{3}$</span>, ... are in <span class="math-container">$\Sigma$</span>, then so is <span class="math-container">$A=A_{1}\cup A_{2}\cup
A_{3}\cup\ldots$</span>.</li>
</ul>
<p>If the above three are our definitions.</p>
<ul>
<li>Can we show that if <span class="math-container">$A=A_{1}\cup A_{2}\cup A_{3}\cup\ldots$</span> and <span class="math-container">$A$</span> is in <span class="math-container">$\Sigma$</span> then each of the <span class="math-container">$A_{i}$</span> are in <span class="math-container">$\Sigma$</span>?</li>
<li>Do we need any additional conditions to show this or would the three basic axioms suffice?</li>
</ul>
<p>Notes:</p>
<ol>
<li>I found some related discussions, but am unable to see if my doubts can be cleared using the pointers from these threads. Listing them below for completeness:</li>
</ol>
<p><a href="https://math.stackexchange.com/questions/468577/can-every-member-of-a-sigma-algebra-be-represented-by-a-countable-union-of-di?rq=1">Can every member of a $\sigma$-algebra be represented by a countable union of disjoint members?</a></p>
<p><a href="https://math.stackexchange.com/questions/1142962/show-that-any-element-of-a-sigma-algebra-is-the-union-of-disjoint-sets?rq=1">Show that any element of a sigma algebra is the union of disjoint sets</a></p>
<ol>
<li>I was trying to understand the concept in the question asked since the corresponding property seemed to be required to show that If <span class="math-container">$E$</span> has <span class="math-container">$\sigma$</span>-finite measure, then <span class="math-container">$E$</span> is inner regular (Rudin RCA, Definition 2.16, some related threads below):</li>
</ol>
<p><a href="https://math.stackexchange.com/questions/199938/if-e-has-sigma-finite-measure-then-e-is-inner-regular">If $E$ has $\sigma$-finite measure, then $E$ is inner regular</a></p>
<p><a href="https://math.stackexchange.com/questions/325009/rudin-real-and-complex-definition-2-16?noredirect=1&lq=1">Rudin Real And Complex Definition 2.16</a></p>
<p>This point is clarified: If <span class="math-container">$E$</span> is in a sigma algebra <span class="math-container">$\Sigma$</span> and <span class="math-container">$E$</span> has <span class="math-container">$\sigma$</span>-finite measure then it is a countable union of sets, <span class="math-container">$E_{i}$</span> with finite measure (<span class="math-container">$\mu\left(E_{i}\right)<\infty$</span>). Which means the sets that make up the union of <span class="math-container">$E$</span> are already in <span class="math-container">$\Sigma$</span> since the measure <span class="math-container">$\mu$</span> is defined only on the members of the sigma algebra <span class="math-container">$\Sigma$</span>. Hence the <span class="math-container">$E_{i}$</span> belong to the sigma algebra, <span class="math-container">$\Sigma$</span></p>
<ol>
<li>Though my original doubt is clarified, I am still curios about the conditions under which the reverse of the third axiom hold. It is quite possible that I am missing something very basic. Happy to delete this question, if that is the case once my doubt is clarified.</li>
</ol>
| user2661923 | 464,411 | <p>If you translate <span class="math-container">$[A \implies B]~~$</span> to <span class="math-container">$~~[(\neg A) \vee B]~~$</span> then the assertion that (in general) <span class="math-container">$~~(P \implies Q)~~$</span> must hold when <span class="math-container">$~~\neg(Q \implies P)~~$</span> ::: is a <strong>true assertion</strong>, demonstrated as follows:</p>
<p>The only way that the statement <span class="math-container">$(Q \implies P)$</span> can be false is if<br>
<span class="math-container">$$(\neg P) \wedge Q.\tag1$$</span></p>
<p>The only way that the statement <span class="math-container">$(P \implies Q)$</span> can be false is if<br>
<span class="math-container">$$(\neg Q) \wedge P.\tag2$$</span></p>
<p>If statement (1) above is true, then statement (2) above <strong>can not</strong> be true.</p>
|
2,279,392 | <p>I could use some help again.
Let $f,g$ be functions from $\mathbb{N}\to \mathbb{N}$.
Also known is that $f(n) = g(2n)$ for every $n\in\mathbb{N}$.
Assuming that $f$ is a surjective function, how do you prove that $g$ is not a one-to-one function?</p>
<p>Cheers</p>
| CY Aries | 268,334 | <p>Let $m$ be a positive odd integer. Then $g(m)\in\mathbb{N}$.</p>
<p>As $f$ is surjective, there exists $n\in \mathbb{N}$ such that $f(n)=g(m)$.</p>
<p>But this implies that $g(m)=g(2n)$, with $m\ne 2n$.</p>
<p>So $g$ is not one-to-one.</p>
|
57,928 | <p>A function that satisfies both $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$ for all real $x$ is known to be the identity over $\mathbb Q$, but is it also the identity over $\mathbb R$? If not, can you provide me an example of such a function? Thanks.</p>
| Steven Stadnicki | 785 | <p>I was just about to supply a counter-example (e.g., set $f(\pi)=e$ and then assign the values at all of the points 'algebraically related' to $\pi$ according to the value of the same algebraic relation applied to $e$) when I realized that this has a clear flaw - for $f$ to be well-defined over $\mathbb{R}$, it must satisfy $f(x)\gt 0$ for all $x\gt 0$ (otherwise $f(\sqrt{x})$ is ill-defined). I suspect this is enough to prove that $f$ must be the identity because you can use the maps $x\rightarrow x+1$, $x\rightarrow x^2$ and their inverses to map each $x$ down into some arbitrarily small neighborhood of $0$ and bound its values there (the positivity condition bounds the value from below and I suspect that clever manipulations will let you use it to bound the value from above), forcing f to be continuous in a neighborhood of $0$ and thus (by 'blowing up' the neighborhood to $[0,1)$ using $f(\sqrt{x}) = \sqrt{f(x)}$ and then translating up and down the line) to be the identity.</p>
<p><strong>EDIT:</strong> shurtados's version of this argument does the trick. As noted there, if $f(x) \lt x$ for some $x\gt 0$, we can derive a contradiction because we can find integers $m$ and $n$ such that $f(x^{2^n}) \lt m \lt x^{2^n}$ and thus find a value $y\gt 0$ with $f(y)\lt 0$, giving us the contradiction above.</p>
<p>Now, we can finish with this argument: suppose $f(x)\gt x$ for some $x\gt 0$; choose an integer $m$ with $f(x)\lt m$. We get $(x-m)\lt f(x-m) \lt 0$, so $0\lt f((x-m)^2) = (f(x-m))^2 \lt (x-m)^2$, and with $y = (x-m)^2$ we have $f(y) \lt y$, giving the needed contradiction; ergo, $f(x)=x$ for all $x\in\mathbb{R}$.</p>
|
57,928 | <p>A function that satisfies both $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$ for all real $x$ is known to be the identity over $\mathbb Q$, but is it also the identity over $\mathbb R$? If not, can you provide me an example of such a function? Thanks.</p>
| shurtados | 14,682 | <p>if <span class="math-container">$f(x)<x$</span> for some <span class="math-container">$x$</span>, you can suppose <span class="math-container">$x$</span> and <span class="math-container">$f(x)$</span> being greater than <span class="math-container">$1$</span>. Then <span class="math-container">$f\left(x^{2^n}\right)< m < x^{2^n}$</span> for some integers <span class="math-container">$m,n$</span>, then <span class="math-container">$f\left(x^{2^n} - m\right) < 0$</span> being <span class="math-container">$ x^{2^n} - m > 0$</span> and that's a contradiction. You can do a similar argument with the other inequality , the argument is very close to the one in the previous answer.</p>
|
4,412,700 | <p>I've been working on a problem but have been stuck for several hours finishing it.</p>
<p>The problem is to show that
<span class="math-container">$$
\frac{x}{(e+x) \ln(e+x)} \leq \ln(\ln(e+x)) \leq \frac{x}{e}
$$</span>
for all <span class="math-container">$x > 0$</span>.</p>
<p>I proceeded by first integrating each component as the problem suggested, which changed it to proving that
<span class="math-container">$$
\int_0^x \frac{1}{(e+x) \ln(e+x)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{(e+t) \ln(e+t)} \,\mathrm{d}t
\leq
\int_0^x \frac{1}{e} \,\mathrm{d}t \,.
$$</span></p>
<p>From here I was able to show quite easily that when <span class="math-container">$x>0$</span>, the second inequality holds. However, I'm unsure of how to now prove the first inequality: is there a result which could help with this or something that I'm missing?</p>
<p>Thanks very much for any help, and sorry for the poor formatting, it's my first time using stack exchange.</p>
<p>Mark</p>
| RRL | 148,510 | <p>Using just the integral definition of <span class="math-container">$\ln$</span>, we have</p>
<p><span class="math-container">$$\ln (\ln(e+x)) = \int_1^{\ln(e+x)} \frac{dt}{t} \geqslant \frac{\ln(e+x) -1}{\ln(e+x)} = \frac{\ln (e \cdot (1+x/e))-1}{\ln(e+x)} = \frac{\ln(1+x/e)}{\ln(e+x)} \\= \frac{1}{\ln(e+x)}\int_1^{1+ x/e} \frac{dt}{t}\geqslant \frac{1}{\ln(e+x)}\frac{x/e}{1+x/e}= \frac{x}{(e+x)\ln(e+x)}$$</span></p>
|
2,149,006 | <p>While learning the power rule, one thing popped up in my mind which is confusing me. We know what the power rule states :</p>
<p>$$\frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1}$$ where $n$ is a real number.</p>
<blockquote>
<p>But instead of $n$, if we have a trig function like $\sin(x)$, <strong>will the power rule still apply?</strong></p>
</blockquote>
<p>Eg. We have a function $y = x^{\sin(x)}$, and thus by the power rule;</p>
<p>$$\frac{dy}{dx} = sin(x)x^{sin(x)-1}$$. </p>
<p>Is this possible? Please tell me if even the function I wrote above really does exist or not.</p>
<p>I know this may seem a stupid question to many, but please help because I cannot find any explanation to this. </p>
| Paramanand Singh | 72,031 | <p>It is important to understand the power rule of differentiation $$\frac{d}{dx}x^{n} = nx^{n - 1}\tag{1}$$ The $n$ in exponent is independent of $x$. There is another power rule where $n$ is base namely $$\frac{d}{dx}n^{x} = n^{x}\log n\tag{2}$$ Here also the base $n$ is independent of $x$. Note that there is no power rule to deal with $$\frac{d}{dx}u^{v}$$ where both $u, v$ are non-constant functions of $x$. But that does not mean that we can't differentiate $u^{v}$. The right approach is to use the definition $u^{v} = \exp(v\log u)$ and then differentiate via chain rule (and product rule) to get $$\frac{d}{dx}u^{v} = u^{v}\left(\frac{v}{u}\frac{du}{dx} + \log u\cdot\frac{dv}{dx}\right)\tag{3}$$ Note that if $u$ (or $v$) is constant then one of the terms in parenthesis vanishes and we essentially get one of the formulas $(1)$ and $(2)$.</p>
|
654,263 | <p>My intention is neither to learn basic probability concepts, nor to learn applications of the theory. My background is at the graduate level of having completed all engineering courses in probability/statistics -- mostly oriented toward the applications without much emphasis on mathematical rigor.</p>
<p>Now I am very interested in learning the core logic and mathematical framework of probability theory, as a math branch. More specifically, I would like to learn answers to the following questions:</p>
<blockquote>
<p>(1) What are the necessary axioms from which we can build probability theory?</p>
<p>(2) What are the core theorems and results in the mathematical theory of probability?</p>
<p>(3) What are the derived rules for reasoning/inference, based on the theorems/results in probability theory?</p>
</blockquote>
<p>So I am seeking a book that covers the "heart" of mathematical probability theory -- not needing much on applications, or discussion on extended topics.</p>
<p>I would like to appreciate your patience for reading my post and any informative responses.</p>
<p>Regards,
user36125</p>
| broccoli | 50,577 | <p>A good collection of books on probability theory and statistics is listed <a href="http://bayesianthink.blogspot.com/2012/12/the-best-books-to-learn-probability.html" rel="nofollow">here</a>. It is a good collection in the sense it has all ranges of books, from core concepts to lighter reading. Additionally, there is no better way to learn it than to learn a programming language like R and a scripting language like Perl/Python. Hope that helps.</p>
|
213,513 | <p>I need help to extrapolate these data:</p>
<pre><code>θ = {20.7, 28.62, 32.04};
ω = {5, 6, 7};
</code></pre>
<p>using this equation:</p>
<p>θ(ω)=θ(ω⟶∞)+ c /ω^n to know what the values of c and n</p>
<p>I found a result using the below plot:</p>
<pre><code>ListLinePlot[
Transpose[{ω^-4,#}]&/@{θ},
FrameLabel->{"1/\!\(\*SuperscriptBox[\(ω\), \(4\)]\)","θ"},
Axes->True,
Frame->True,
AxesStyle->Directive[{Bold,25},{Black,25}],
BaseStyle->{FontWeight->Bold,FontSize->14},
PlotRange->All,
PlotMarkers->{"◆",18}
]
</code></pre>
<p>But my problem how to obtain the fits using the extrapolation for the above equation using Mathematica.</p>
<p>Thanks in advance!</p>
| kglr | 125 | <pre><code>ClearAll[mergE]
mergE[f_, red_] := Merge[red]@*KeyValueMap[f[#] -> #2 &];
</code></pre>
<p><strong><em>Examples:</em></strong></p>
<pre><code>a = <|{{a1, a1}, {a2, a2}} -> 1, {{a1, a1}, {a1, a2}} -> 10, {{a1, a2}, {a1, a1}} -> 3|>;
mergE[Sort, Total]@a
</code></pre>
<blockquote>
<p><|{{a1, a1}, {a2, a2}} -> 1, {{a1, a1}, {a1, a2}} -> 13|></p>
</blockquote>
<pre><code>mergE[Sort, foo]@a
</code></pre>
<blockquote>
<p><|{{a1, a1}, {a2, a2}} -> foo[{1}], {{a1, a1}, {a1, a2}} -> foo[{10, 3}]|></p>
</blockquote>
|
4,219,492 | <p>This is a long question/explanation, where I more or less have a method to the madness; however, I've stumbled upon how I perform the calculation and I feel it is quite crude. This question is based off <a href="https://math.stackexchange.com/questions/2631501/finding-the-distribution-of-the-sum-of-three-independent-uniform-random-variable#">another post</a> about the sum of three variables. Since I'm am unsure of my method, I didn't want to make this an answer there but rather my own question for critiquing. Reproduced here, consider three random variables <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span>, all drawn from Uniform<span class="math-container">$(0,1)$</span>. What is the distribution of the sum of the three random variables, <span class="math-container">$W = X+Y+Z$</span>?</p>
<h2>Part 1: Convolution of two variables</h2>
<p>First define the random variable of the sum of two variables, <span class="math-container">$S = X+Y$</span>. We find the pdf of <span class="math-container">$S$</span> by convolution of the two distributions</p>
<p><span class="math-container">$$\begin{align}
f_S(s) &= \int_{-\infty}^\infty\! f_X(s-t) f_Y(t)\ \textrm{d}t,\\
&= \int_0^1\! f_X(s-t)\ \textrm{d}t,\\
&= \int_{s-1}^s f_X(u)\ \textrm{d}u.
\end{align}$$</span></p>
<p>Here I used the fact that <span class="math-container">$f_Y(t) = 1$</span> when <span class="math-container">$0 \leq t \leq 1$</span>, and is otherwise <span class="math-container">$0$</span>. Likewise, we know that <span class="math-container">$f_X(s-t)$</span> has the same properties, such that another bound is <span class="math-container">$0 \leq s-t \leq 1$</span>. This may be transform into <span class="math-container">$s-1 \leq t \leq s$</span>, which is "coincidently" the bounds over the variable <span class="math-container">$u$</span> (perhaps this "coincident" is where my enlightenment awaits).</p>
<p><span class="math-container">$\textbf{Here}$</span> is where I make a leap of faith of sorts. We now have two bounds for for our integral over <span class="math-container">$t$</span></p>
<ol>
<li><span class="math-container">$0 \leq t \leq 1$</span> from the pdf of <span class="math-container">$Y$</span>, and</li>
<li><span class="math-container">$s-1 \leq t \leq s$</span> from the shifted pdf of <span class="math-container">$X$</span>.</li>
</ol>
<p>I therefore, "mix-and-match" my bounds to arrive at</p>
<p><span class="math-container">$$\begin{align}
f_S(s) =
\cases{
\displaystyle\int_0^s \mathrm{d}t = s, & $0 \leq s < 1$\\
\displaystyle\int_{s-1}^1 \mathrm{d}t = 2-s, & $1 \leq s \leq 2$
}.
\end{align}$$</span></p>
<p>I can see the mixing of bounds either from naively swapping them (say the upper bounds). I can also appreciate from visualizing the convolution that there are two "behaviors" and therefore a piecewise function is appropriate. However, this feeling is not systematized and is just an intuition at best. I've seen these bounds explained via proof by "what else could it be", i.e., what makes sense to keep the integral within the domain <span class="math-container">$[0,1]$</span>, where we take that <span class="math-container">$s \in [0,2]$</span> since <span class="math-container">$S = X+Y$</span>... After arriving at the answer that seems fine, but not very satisfactory.</p>
<h3>Aside about how I visualize the convolution</h3>
<p>Imagining the convolution of the two unit boxes, fixing one distribution and sliding the other, there are three "moments"/instances of interest</p>
<ol>
<li>When the distributions begin to overlap, the leading edge of the sliding distribution is at <span class="math-container">$s=0$</span>.</li>
<li>When the distributions completely overlap, <span class="math-container">$s=1$</span>. This moment also coincides with when the distributions begin to diverge.</li>
<li>When the distributions last touch <span class="math-container">$s=2$</span>.</li>
</ol>
<p>Since "moment" two is instantaneous there are really only two "event," when the distributions are increasing their overlapping region and when the are decreasing it. From this, I argue there are two cases: <span class="math-container">$0 \leq s < 1$</span> and <span class="math-container">$1 \leq s \leq 2$</span>.</p>
<h2>Part 2: Convolution of three variables</h2>
<p>We now want <span class="math-container">$W = X+Y+Z = S+Z$</span>, meaning we will use our results from the convolution of two variables to do another convolution. Similar to before</p>
<p><span class="math-container">$$\begin{align}
f_W(w) &= \int_{-\infty}^\infty\! f_S(w-t) f_Z(t)\ \textrm{d}t,\\
&= \int_0^1\! f_S(w-t)\ \textrm{d}t,\\
&= \int_{w-1}^w f_S(u)\ \textrm{d}u.
\end{align}$$</span></p>
<p>Here again I have restricted <span class="math-container">$0 \leq t \leq 1$</span> from the distribution of <span class="math-container">$Z$</span>. Now this time <span class="math-container">$f_S(w-t)$</span> is a little more complicated since it is not a uniform distribution</p>
<p><span class="math-container">$$\begin{align}
f_S(w-t) =
\cases{
w-t, & $0 \leq w-t < 1\ \longrightarrow\ w-1 \leq t \leq w$\\
2-w+t, & $1 \leq w-t \leq 2\ \longrightarrow\ w-2 \leq t \leq w-1$
}.
\end{align}$$</span></p>
<p>Now I can repeat the "mixing-and-matching" of <span class="math-container">$0 \leq t \leq 1$</span> with the bounds of <span class="math-container">$f_S$</span> to arrive at a list of</p>
<ol>
<li><span class="math-container">$0 \leq t \leq w$</span></li>
<li><span class="math-container">$w-1 \leq t \leq 1$</span></li>
<li><span class="math-container">$0 \leq t \leq w-1$</span></li>
<li><span class="math-container">$w-2 \leq t \leq 1$</span>.</li>
</ol>
<p>Now I have to lean on my visualization of the convolution to say there are three cases of interest. These are, with what bounds I associated with them, when the distribution begin to overlap (1), while the overlap (2 and 3), and when the begin to stop overlapping (4). Thus, the first half of <span class="math-container">$f_S$</span> shows up by itself while they begin to overlap, both part of <span class="math-container">$f_S$</span> contribute while the distributions are completely overlapping, and only the later part of <span class="math-container">$f_S$</span> contributes when they begin to diverge. This leads me to the result that</p>
<p><span class="math-container">$$\begin{align}
f_W(w) =
\cases{
\displaystyle\int_0^w (w-t)\ \mathrm{d}t = \frac{w^2}{2}, & $0 \leq w < 1$\\
\displaystyle\int_{w-1}^1 (w-t)\ \mathrm{d}t + \int_0^{w-1} (2-w+t)\ \mathrm{d}t= -w^2 + 3w - \frac{3}{2}, & $1 \leq w \leq 2$\\
\displaystyle\int_{w-2}^1 (2-w+t)\ \mathrm{d}t= \frac{(w-3)^2}{2}, & $2 \leq w \leq 3$
}.
\end{align}$$</span></p>
<p>This is the correct answer from the original post. My intuition hasn't gotten me the wrong answer at the very least.</p>
<h2>Part 3: Convolution of four variables</h2>
<p>I'll skip the majority of the explanation, and summarize we now want <span class="math-container">$K = X+Y+Z+J$</span>, where <span class="math-container">$J$</span> is another random variable drawn from Uniform<span class="math-container">$(0,1)$</span>.</p>
<p><span class="math-container">$$\begin{align}
f_K(k) &= \int_{-\infty}^\infty\! f_W(k-t) f_J(t)\ \textrm{d}t,\\
&= \int_0^1\! f_W(k-t)\ \textrm{d}t,\\
&= \int_{k-1}^k f_W(u)\ \textrm{d}u.
\end{align}$$</span></p>
<p>Writing the shifted convolved three variable distribution and solving for its bounds</p>
<p><span class="math-container">$$\begin{align}
f_W(k-t) =
\cases{
\frac{(k-t)^2}{2}, & $0 \leq k-t < 1\ \longrightarrow k-1 \leq t < k$\\
-(k-t)^2 + 3(k-t) - \frac{3}{2}, & $1 \leq k-t \leq 2\ \longrightarrow k-2 \leq t < k-1$\\
\frac{(k-t-3)^2}{2}, & $2 \leq k-t \leq 3\ \longrightarrow k-3 \leq t < k-2$
}.
\end{align}$$</span></p>
<p>Then setting up the integrals and solving I arrive at</p>
<p><span class="math-container">$$\begin{align}
f_K(k) =
\cases{
\displaystyle\int_0^k \frac{(k-t)^2}{2}\ \mathrm{d}t = \frac{k^3}{6}, & $0 \leq k < 1$\\
\displaystyle\int_{k-1}^1 \frac{(k-t)^2}{2}\ \mathrm{d}t + \int_0^{k-1} \left(-(k-t)^2+3(k-t)-\frac{3}{2}\right)\ \mathrm{d}t= -\frac{k^3}{3} + 2 k^2 - 2k + \frac{2}{3}, & $1 \leq k < 2$\\
\displaystyle\int_{k-2}^1 \left(-(k-t)^2+3(k-t)-\frac{3}{2}\right)\ \mathrm{d}t + \int_0^{k-2} \frac{(k-t-3)^2}{2}\ \mathrm{d}t = \frac{k^3}{2} -4k^2 +10k - \frac{22}{3}, & $2 \leq k < 3$\\
\displaystyle\int_{k-3}^1 \frac{(k-t-3)^2}{2}\ \mathrm{d}t = -\frac{(k-4)^3}{6}, & $3 \leq k \leq 4$\\
}.
\end{align}$$</span></p>
<p>Now this is where my earlier visualization intuition failed me, or at least evidently needed to be tweaked. Rather than the events: beginning to overlap, overlapping, and beginning to diverge; the real importance is the coupling of different regions. Meaning, the first piecewise in the 2 variable sum came from coupling the void (0) with the first half of the uniform distribution, and the void with the second half—this seems to be the biggest subtly in my altered thinking. However, it is then straightforward to see why we have 3 pieces for the 3 sum, and 4 for the 4 sum, and why each middle section is the sum of two integrals as we are coupling adjacent regions. In fact all pieces are the sum of two integrals, it is just that the first and last section are integrals over 0!</p>
<p>Now I have no reference for if these are correct, but it is what I get from following my method. Moreover, when I plot it in Mathematica, I got a reasonable looking distribution. By chance if this process is taken to infinitum, does this converge to a scaled Normal distribution (perhaps rather the of convolving the Uniform<span class="math-container">$(-1,1)$</span>)? I'd suppose not exactly, as the convolution of the normal with a uniform is not strictly the normal distribution from what I've seen.</p>
<p><a href="https://i.stack.imgur.com/voXsf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/voXsf.png" alt="Sum of 4 uniform random variables" /></a></p>
<h2>Question/Challenge</h2>
<p>With my thinking made clear, I am having difficulty convolving the triangle distribution with itself, i.e., it should be our 4 uniform variable sum. Now I want to see that <span class="math-container">$K = S+S$</span> such that</p>
<p><span class="math-container">$$\begin{align}
f_K(k) &= \int_{-\infty}^\infty\! f_S(k-t) f_S(t)\ \textrm{d}t,\\
&= \int_0^1\! t f_S(s-t)\ \textrm{d}t + \int_1^2 (2-t) f_S(s-t)\ \textrm{d}t.
\end{align}$$</span></p>
<p>How are the bound picked, and how do I systematically know which integrals combine with one another? I can get the limits <span class="math-container">$0 \leq t \leq 2$</span>, <span class="math-container">$z-1 \leq t \leq z$</span> and <span class="math-container">$z-2 \leq t \leq z-1$</span>. Naively, I can end up with 8 integrals, four for each term in <span class="math-container">$f_K(k)$</span> where I get four from mixing and matching bounds of the shifted triangle distribution. If there are really 8 integrals, which ones belong to which section of the piecewise function?</p>
| Tianyi Miao | 484,971 | <p>The core of this problem is how to systematically convolve 2 piecewise functions, which doesn't necessarily involve probability distributions per se.</p>
<p>This manuscript, which is also motivated by summing uniform random variables, provides such a method: <a href="http://www.mare.ee/indrek/misc/convolution.pdf" rel="nofollow noreferrer">http://www.mare.ee/indrek/misc/convolution.pdf</a></p>
<p>The essential ideas are copied below:</p>
<h2>Part 1: Each function is only defined on 1 interval</h2>
<p>Given 2 functions <span class="math-container">$F(x)$</span> and <span class="math-container">$G(x)$</span> such that
<span class="math-container">$$
F(x)=\begin{cases}
f(x), & a_f < x < b_f\\
0, & \text{otherwise}
\end{cases}
$$</span>
<span class="math-container">$$
G(x)=\begin{cases}
g(x), & a_g < x < b_g\\
0, & \text{otherwise}
\end{cases}
$$</span>
We can assume without the loss of generality that <span class="math-container">$b_f - a_f < b_g - a_g$</span>, because convolution is commutative and we can convert <span class="math-container">$F * G$</span> to <span class="math-container">$G * F$</span> when necessary.</p>
<p><span class="math-container">$$
\begin{align*}
(F*G)(t) &= \int_{-\infty}^\infty F(x)G(t-x)\ dx\\
&= \int_{-\infty}^\infty F(x)G(t-x)\ dx\\
&= \begin{cases}
\int_{a_f}^{t - a_g} f(x)g(t-x)\ dx, & a_f + a_g < t < b_f + a_g\\
\int_{a_f}^{b_f} f(x)g(t-x)\ dx, & b_f + a_g < t < a_f + b_g\\
\int_{t-b_g}^{b_f} f(x)g(t-x)\ dx, & a_f + b_g < t < b_f + b_g\\
0, & \text{otherwise}
\end{cases}
\end{align*}
$$</span></p>
<h2>Part 2: Multiple Intervals</h2>
<p>Given piecewise functions with non-overlapping intervals:
<span class="math-container">$$
F(x) = \begin{cases}
f_1(x), & a_f^1 < x < b_f^1\\
...\\
f_N(x), & a_f^N < x < b_f^N\\
0, &\text{otherwise}
\end{cases}\\
G(x) = \begin{cases}
g_1(x), & a_g^1 < x < b_g^1\\
...\\
g_N(x), & a_g^M < x < b_g^M\\
0, &\text{otherwise}
\end{cases}
$$</span></p>
<p>We can rewrite <span class="math-container">$F(x) = \sum_{i=1}^N F_i(x)$</span> where
<span class="math-container">$$F_i(x) = \begin{cases}
f_i(x), & a_f^i < x < b_f^i\\
0, & \text{otherwise}
\end{cases}
$$</span></p>
<p>Because convolution is distributive with respect to addition <span class="math-container">$f * (g + h) = f*g + f*h$</span>, we can calculate <span class="math-container">$F*G$</span> as follows
<span class="math-container">$$
F*G = \sum_{i=1}^N \sum_{j=1}^M F_i * G_j
$$</span></p>
<p>Going back to your original question. We can find the intervals for <span class="math-container">$t$</span> based on the intervals <span class="math-container">$(a_f, b_f)$</span> and <span class="math-container">$(a_g, b_g)$</span>. When there are multiple intervals, the problem reduces to the sum of piecewise functions, which is just finding overlaps of intervals.</p>
|
1,689,853 | <p>If I have that $B_t$ is a standard brownian motion process, is $B_t^2 - \frac{t}{2}$ a martingale w.r.t. brownian motion? I know that $B_t^2 - t$ is but can't see it for the latter. </p>
| Marcel | 257,844 | <p>No, it's not. Just calculate the conditional expectation. It is a submartingale.</p>
|
1,353,184 | <p>Let $a$, $b$, and $c$ be positive real numbers. Prove that</p>
<p>$$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$</p>
<p>Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?</p>
| Batominovski | 72,152 | <p><strong>An Algebraic Solution:</strong></p>
<p>Note that $$\sqrt{a^2-ab+b^2}=\sqrt{\left(a-\frac{b}{2}\right)^2+\left(\frac{\sqrt{3}b}{2}\right)^2}$$ and $$\sqrt{a^2-ac+c^2}=\sqrt{\left(\frac{c}{2}-a\right)^2+\left(\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Using the Triangle Inequality in the form $\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\geq \sqrt{\sum_{i=1}^n\left(x_i+y_i\right)^2}$ for real numbers $x_i$'s and $y_i$'s, we have $$\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq \sqrt{\Bigg(\left(a-\frac{b}{2}\right)+\left(\frac{c}{2}-a\right)\Bigg)^2+\left(\frac{\sqrt{3}b}{2}+\frac{\sqrt{3}c}{2}\right)^2}\,.$$ Expanding the right-hand side, we get $\sqrt{a^2-ab+b^2}+\sqrt{a^2-ac+c^2}\geq\sqrt{b^2+bc+c^2}$. The equality holds if and only if there exists $\lambda \geq 0$ such that $\lambda\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=\left(\frac{c}{2}-a,\frac{\sqrt{3}c}{2}\right)$ or $\left(a-\frac{b}{2},\frac{\sqrt{3}b}{2}\right)=(0,0)$ (equivalently, $(a,b,c)=(0,0,c)$ or $(a,b,c)=\left(\frac{\lambda b}{\lambda+1},b,\lambda b\right)$ for some $\lambda \geq 0$).</p>
<p>Although the geometric solution by thkim1011 is a more elegant solution, it is more complicated if $a$, $b$, or $c$ can be negative (although you can prove geometrically with essentially the same method, you have to deal with many cases). My algebraic solution is better in this sense.</p>
|
271,785 | <p>I have to prove that $f(x)=\log(1+x^2)$ is Uniform continuous in $[0,\infty)$ (with $\epsilon ,\delta$ formulas...)</p>
<p>I wrote the definition: (what I have to prove):</p>
<p>$\forall \epsilon>0 \quad \exists \delta>0 \quad s.t. \quad \forall x,y\in [0,\infty) : \quad |x-y|<\delta \quad \Rightarrow \quad |f(x)-f(y)|<\epsilon $</p>
<p>So I tried developing $|f(x)-f(y)| = |\log(1+x^2)-\log(1+y^2)| = |\log(\frac{1+x^2}{1+y^2})| $... </p>
<p>now this is where I try to make it bigger and simplify the expression so i can choose the right $\delta$ which depends on the $\epsilon$, and then say that if the simplified bigger expression is still smaller than $\epsilon$ then of course the original $|f(x)-f(y)|$ is smaller than $\epsilon$ </p>
<p>but what can I do with this expression? any algebraic tricks? </p>
| Fabian | 7,266 | <p><em>Hint:</em> You can use that $|\log (1+ z)| \leq z$ valid for $z\geq 1$. Thus, for $x>y$
$$\left|\log \left(\frac{1+x^2}{1+y^2}\right)\right| \leq \frac{x^2-y^2}{1+y^2}
\leq \frac{x+y}{1+y^2} (x-y)= \frac{2y +(x-y)}{1+y^2}(x-y) .$$</p>
<p>Moreover, for $x<y$
$$\left|\log \left(\frac{1+x^2}{1+y^2}\right) \right| =
\left|\log \left(\frac{1+y^2}{1+x^2}\right) \right| \leq \frac{y^2-x^2}{1+x^2}
\leq \frac{x+y}{1+x^2} (y-x) =\frac{2x+(y-x)}{1+x^2} (y-x).$$</p>
|
3,188,327 | <p>I derived the operator <span class="math-container">$\mathscr{L} = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}$</span> from the PDE <span class="math-container">$u_x + uu_y = 0$</span> in order to figure out whether it is linear. </p>
<p>The textbook solutions take the following steps in finding whether the operator is linear:</p>
<p><span class="math-container">$$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u + v)\dfrac{\partial{(u + v)}}{\partial{y}} = \dots = \mathscr{L}u + \mathscr{L}v + uv_y + vu_y,$$</span></p>
<p>which is obviously nonlinear. But I proceeded as follows:</p>
<p><span class="math-container">$$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}} = \dots = \mathscr{L}u + \mathscr{L}v,$$</span></p>
<p>which obviously <em>is</em> linear.</p>
<p>I don't understand why the author let <span class="math-container">$u$</span> in the operator become <span class="math-container">$u = u + v$</span>? It seemed to me that, since <span class="math-container">$u$</span> is part of the operator, it should remain as <span class="math-container">$u$</span> and <em>not</em> <span class="math-container">$u + v$</span>?</p>
<p>I would greatly appreciate it if people could please take the time to clarify this. <em>Why</em> is it wrong? My point is that <span class="math-container">$u$</span> is part of the operator, and the operator is acting upon <span class="math-container">$(u + v)$</span>, so why is it that <span class="math-container">$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u + v)\dfrac{\partial{(u + v)}}{\partial{y}}$</span> instead of <span class="math-container">$\mathscr{L}(u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + (u)\dfrac{\partial{(u + v)}}{\partial{y}}$</span>?</p>
<p>EDIT: For future reference, I would also like to present another interesting example.</p>
<p>Take the operator <span class="math-container">$\mathscr{L} u = u_x + u_y + 1$</span>. Therefore, <span class="math-container">$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u}$</span>. And so,</p>
<p><span class="math-container">$$\mathscr{L} (u + v) = \left( \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + \dfrac{1}{u + v} \right) (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1$$</span></p>
<p>EDIT2: </p>
<p>I think the derivation of <span class="math-container">$\mathscr{L}$</span> in the above edit is incorrect. Here is what I think the correct derivation is:</p>
<p>Take the operator <span class="math-container">$\mathscr{L} u = u_x + u_y + 1$</span>. Therefore, <span class="math-container">$\mathscr{L} = \dfrac{\partial}{\partial{x}} + \dfrac{\partial}{\partial{y}} + 1$</span>. And so,</p>
<p><span class="math-container">$$\mathscr{L} (u + v) = \dfrac{\partial{(u + v)}}{\partial{x}} + \dfrac{\partial{(u + v)}}{\partial{y}} + 1$$</span></p>
| Trebor | 584,396 | <p>Obviously the notation <span class="math-container">$$\mathscr{L} = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}$$</span> meant <span class="math-container">$$\mathscr{L}[u] = \dfrac{\partial}{\partial{x}} + u\dfrac{\partial}{\partial{y}}.$$</span> So the <span class="math-container">$u$</span> should be replaced too.</p>
|
1,170,708 | <p>What functions satisfy $f(x)+f(x+1)=x$?</p>
<p>I tried but I do not know if my answer is correct.
$f(x)=y$</p>
<p>$y+f(x+1)=x$</p>
<p>$f(x+1)=x-y$</p>
<p>$f(x)=x-1-y$</p>
<p>$2y=x-1$</p>
<p>$f(x)=(x-1)/2$</p>
| Christian Blatter | 1,303 | <p>This is a linear inhomogeneous problem. The general philosophy for such problems is the following:</p>
<p>(a) Find the general solution $f_h$ of the associated homogeneous problem
$$f(x)+f(x+1)=0\ .$$
To this end write $f_h(x):=e^{i\pi x} g(x)$ with a new unknown function $x\mapsto g(x)$. You will obtain a huge set of solutions.</p>
<p>(b) Find a single particular solution $x\mapsto f_p(x)$ of the given equation. To this end make the "Ansatz" $f_p(x):=a x+b$ with undetermined coefficients $a$ and $b$, and fix $a$ and $b$ such that $f_p$ fulfills the given functional equation.</p>
<p>(c) The general solution of the given functional equation is then
$$f(x)=f_h(x)+f_p(x)\ .$$</p>
|
3,860,199 | <p>What is the integration of <span class="math-container">$\int x^{|x|} dx$</span>?<br />
Actually through several google search finally I have found a solution for the problem <span class="math-container">$\int x^{x} dx$</span> using Gamma function function and I am hardly sure that the same method can be applied to solve the problem if <span class="math-container">$x^{x}$</span> is going to be replaced by <span class="math-container">$x^{|x|}$</span>.</p>
<p>So, what is the appropriate approach to solve the problem <span class="math-container">$\int x^{|x|} dx$</span></p>
| David Kipper | 764,256 | <p><span class="math-container">$x^x$</span> and <span class="math-container">$x^{-x}$</span> are not integrable in elementary terms. See <a href="https://en.wikipedia.org/wiki/Liouville%27s_theorem_%28differential_algebra%29" rel="nofollow noreferrer">This theorem by Liouville.</a> This theorem one way to approach that question.</p>
<p>This is obviously not something one would expect, and I too was surprised at this when I first asked myself your question.</p>
<p>It's possible (and a good exercise if you're interested) to find a closed form expression for the area under <span class="math-container">$x^x$</span> and <span class="math-container">$x^{-x}$</span> over the interval <span class="math-container">$[0,1]$</span>. Spoilers:</p>
<p><span class="math-container">$$\int_0^1x^xdx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^n}$$</span></p>
<p>Once you've done that, you can do essentially the same for <span class="math-container">$x^{-x}$</span>.
This, as far as I know, is the closest one could come to an answer to your question.</p>
|
174,075 | <p>What is the difference when a line is said to be normal to another and a line is said to be perpendicular to other?</p>
| Samuel Reid | 19,723 | <p>There is no difference between saying two lines are perpendicular and that one line is normal to another line. It is literally a synonymous term, like saying that you take the product of two numbers or you multiply two numbers.</p>
|
4,648,068 | <h1>Background</h1>
<p>There is a source code that generates surface triangles. The isosurface is generated for the iso-value of <code>0</code>. The source code uses a table for <code>2^8=256</code> possible <em>inside/outside</em>, i.e. <em>negative/positive</em>, combinations of <code>8</code> scalar values at <code>8</code> cube corners. The table returns an array. Every <code>3</code> consecutive array items would correspond to a triangle. The array items could be from <code>0</code> to <code>11</code>, pointing to the <code>12</code> edges a cube has. Probably this table comes from a published paper in the field of mathematics or computer science:</p>
<p><a href="https://github.com/deadsy/sdfx/blob/2d4e9502ec6fe898e8774020882cb8150f16a6a6/render/march3.go#L360" rel="nofollow noreferrer">https://github.com/deadsy/sdfx/blob/2d4e9502ec6fe898e8774020882cb8150f16a6a6/render/march3.go#L360</a></p>
<h1>Objective</h1>
<p>I'm trying to adapt the above <em>marching cubes</em> source code, and its <em>tables</em>, to generate tetrahedra throughout the volume of a 3D model. The code would extract tetrahedra elements with all the non-positive, i.e. <code><=0</code>, values. Non-positive means the 3D space <em>on</em> and <em>inside</em> the isosurface of the <code>0</code> value.</p>
<h1>Question</h1>
<p>For some reason, I cannot find any publication for extracting a tetrahedral mesh <em>on</em> and <em>inside</em> the isosurface from a three-dimensional discrete scalar field. Maybe I'm not looking at the right places. Am I missing something? Or do I have to come up with the <em>tables</em> myself? It looks like a daunting task to me.</p>
| kabenyuk | 528,593 | <p>This is possible only when <span class="math-container">$-1=1$</span>, otherwise <span class="math-container">$(-1)^4=1\neq-1$</span>.
For the example in the field <span class="math-container">$\mathbb{F}_2$</span> the identity <span class="math-container">$a^4=a$</span> holds.</p>
|
4,648,068 | <h1>Background</h1>
<p>There is a source code that generates surface triangles. The isosurface is generated for the iso-value of <code>0</code>. The source code uses a table for <code>2^8=256</code> possible <em>inside/outside</em>, i.e. <em>negative/positive</em>, combinations of <code>8</code> scalar values at <code>8</code> cube corners. The table returns an array. Every <code>3</code> consecutive array items would correspond to a triangle. The array items could be from <code>0</code> to <code>11</code>, pointing to the <code>12</code> edges a cube has. Probably this table comes from a published paper in the field of mathematics or computer science:</p>
<p><a href="https://github.com/deadsy/sdfx/blob/2d4e9502ec6fe898e8774020882cb8150f16a6a6/render/march3.go#L360" rel="nofollow noreferrer">https://github.com/deadsy/sdfx/blob/2d4e9502ec6fe898e8774020882cb8150f16a6a6/render/march3.go#L360</a></p>
<h1>Objective</h1>
<p>I'm trying to adapt the above <em>marching cubes</em> source code, and its <em>tables</em>, to generate tetrahedra throughout the volume of a 3D model. The code would extract tetrahedra elements with all the non-positive, i.e. <code><=0</code>, values. Non-positive means the 3D space <em>on</em> and <em>inside</em> the isosurface of the <code>0</code> value.</p>
<h1>Question</h1>
<p>For some reason, I cannot find any publication for extracting a tetrahedral mesh <em>on</em> and <em>inside</em> the isosurface from a three-dimensional discrete scalar field. Maybe I'm not looking at the right places. Am I missing something? Or do I have to come up with the <em>tables</em> myself? It looks like a daunting task to me.</p>
| Koro | 266,435 | <p>I think that I figured it out:</p>
<p>We have <span class="math-container">$(-a) ^4= a^4$</span> so <span class="math-container">$-a= a$</span> for every <span class="math-container">$a$</span>. It follows that <span class="math-container">$2a=0$</span>,hence the field is of char <span class="math-container">$2$</span>.</p>
|
4,566,926 | <p><span class="math-container">$a_n$</span> converges to <span class="math-container">$a \in \mathbb{R}$</span>, find <span class="math-container">$\lim\limits_{n \rightarrow \infty}\frac{1}{\sqrt{n}}(\frac{a_1}{\sqrt{1}} + \frac{a_2}{\sqrt{2}} +\frac{a_3}{\sqrt{3}} + \cdots + \frac{a_{n-1}}{\sqrt{n-1}} + \frac{a_n}{\sqrt{n}})$</span></p>
<p>I've tried to use Squeeze theorem. Nothing good. Maybe I've tried it in a wrong way. According to my assumptions, this sequence diverges (I mean "goes" to infinity).</p>
<p>Can we solve this using Stolz–Cesàro theorem? Is there easier/better/more correct approach? If there is, please, share your thoughts</p>
| Essaidi | 708,306 | <ol>
<li>If <span class="math-container">$a \neq 0$</span> then <span class="math-container">$a_n \sim a$</span> so <span class="math-container">$\dfrac{a_n}{\sqrt{n}} \sim \dfrac{a}{\sqrt{n}}$</span>.<br>
The series <span class="math-container">$\sum \dfrac{a}{\sqrt{n}}$</span> diverges then :
<span class="math-container">$$\sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \sim \sum_{k = 1}^n \dfrac{a}{\sqrt{k}}$$</span>
We deduce that :
<span class="math-container">$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \sim \dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a}{\sqrt{k}} = \dfrac{a}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}} \to a \int_0^1 \sqrt{x} \mathrm{d}x = 2 a$$</span>
Finally :
<span class="math-container">$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \to 2a$$</span></li>
<li>If <span class="math-container">$a = 0$</span> then <span class="math-container">$a_n = o(1)$</span> so <span class="math-container">$\dfrac{a_n}{\sqrt{n}} = o \left(\dfrac{1}{\sqrt{n}}\right)$</span>.<br>
The series <span class="math-container">$\sum \dfrac{1}{\sqrt{n}}$</span> diverges then :
<span class="math-container">$$\sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} = o\left(\sum_{k = 1}^n \dfrac{1}{\sqrt{k}}\right)$$</span>
We deduce that :
<span class="math-container">$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} = o \left(\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{1}{\sqrt{k}}\right) = o\left(\dfrac{1}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}}\right) = o(1)$$</span>
because :
<span class="math-container">$$\dfrac{1}{n} \sum_{k = 1}^n \dfrac{1}{\sqrt{\dfrac{k}{n}}} \to \int_0^1 \sqrt{x} \mathrm{d}x = 2$$</span>
Finally :
<span class="math-container">$$\dfrac{1}{\sqrt{n}} \sum_{k = 1}^n \dfrac{a_k}{\sqrt{k}} \to 0$$</span>
</li>
</ol>
|
90,130 | <p>So now that my term's over, I've been brushing up on my quantum field theory, and I came across the following line in my textbook without any justification:</p>
<p>$$\frac{1}{4\pi^2}\int_m^{\infty}\sqrt{E^2-m^2}e^{-iEt}dE \sim e^{-imt}\text{ as }t\to\infty$$</p>
<p>Well, I can see intuitively that <em>if</em> most of the integral cancels out, the main contribution will be from the region $E\approx m$, since (under a coordinate transformation) that's the region of stationary phase. But I'm a mathematician, dangit, not a physicist, and I want this to be rigourous.</p>
<p>The Riemann-Lebesgue lemma, if I'm not mistaken, doesn't apply since $\sqrt{E^2-m^2}$ is unbounded as $E\to\infty$, and it certainly isn't $L^1$. And I guess I could shift the path of the integral off the real axis in the complex plane, but I don't see why that would be the <em>right</em> way to take the integral. The whole thing is giving me the heebie-jeebies, and I was hoping one of you folks could assuage my fears.</p>
| Brightsun | 118,300 | <p>Considering the integrand as the Fourier transform of a tempered distribution, it makes sense then to write
$$
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE = -\frac{\partial^2}{\partial t^2}\int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho.
$$
<a href="https://i.stack.imgur.com/wovat.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wovat.png" alt="enter image description here"></a>
Now, in the complex plane, let us consider the contour in the figure: a quarter of circle centered at $1$ of radius $R$ with arc going from $R$ to $1-iR$ and a small indent around $1$. Integrating
$$
f(z)=\frac{\sqrt{z^2-1}}{z^2}e^{-imtz}
$$
along such a contour and choosing the branch cut from $-1$ to $+1$, we get a vanishing contribution (by Jordan's lemma) from the arc and hence
$$
\int_{1}^{+\infty}\frac{\sqrt{\rho^2-1}}{\rho^2}e^{-imt\rho}d\rho =
\int_0^{+\infty}\frac{\sqrt{y^2+i2y}}{(1-iy)^2}e^{-imt(1-iy)}dy.
$$
Differentiating twice, by the above consideration,
\begin{align}
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE
&=m^2\int_0^{+\infty}\sqrt{y^2+i2y} \, e^{-mt(y+i)}dy
\end{align}
and rescaling $s=mty$
\begin{align}
\int_{m}^{+\infty}\sqrt{E^2-m^2}e^{-iEt}dE
&= e^{-imt}\sqrt{m}t^{-3/2} \int_0^{+\infty} \sqrt{\frac{s^2}{mt}+i2s}\,e^{-s}ds\\
&= e^{-imt}\sqrt{m}t^{-3/2} \left( \sqrt\frac{i\pi}{2} + O\left(t^{-1}\right) \right),
\end{align}
asymptotically as $t\to\infty$.</p>
|
158,451 | <p>Suppose that the contents of an urn are $w$ red balls, $x$ yellow balls, $y$ green balls, and $z$ blue balls collectively, where $w \geq 3$, $x\geq 1$, $y\geq 1$, and $z\geq 1$. We draw balls randomly from this urn without replacement.</p>
<p>What is the probability of our having drawn at least 1 yellow ball by (and including) the seventh draw, at least 1 green ball by (and including) the eight draw, and at least 3 red balls and 1 blue ball by (and including) the ninth draw? </p>
<p>Note that this is one single event and not four separate events.</p>
| Alex Becker | 8,173 | <p>Let $x=(x_1,\ldots,x_n), y=(y_1,\ldots,y_n), x' = (\bar x,\ldots,\bar x)$ and $y' = (\bar y,\ldots,\bar y)$. Using this, we can simplify notation by writing
$$r_{xy}=\frac{\sum\limits_{i=1}^n(x_i-\bar x)(y_i-\bar y)}{\sqrt{\sum\limits_{i=1}^n(x_i-\bar x)^2\sum\limits_{i=1}^n(y_i-\bar y)^2}}\leq\frac{\sum\limits_{i=1}^n|x_i-\bar x||y_i-\bar y|}{\sqrt{\sum\limits_{i=1}^n(x_i-\bar x)^2\sum\limits_{i=1}^n(y_i-\bar y)^2}}=\frac{\|(x-x')(y-y')\|_1}{\|x-x'\|_2\|y-y'\|_2}$$
where $\|\cdot \|_1$ denotes the $L^1$ norm and $\|\cdot \|_2$ denotes the $L^2$ norm. We then only need that
$$\|(x-x')(y-y')\|_1\leq\|x-x'\|_2\|y-y'\|_2$$
which follows from the <a href="http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality" rel="nofollow">Cauchy-Schwarz inequality</a>.</p>
|
1,494,167 | <p>Using only addition, subtraction, multiplication, division, and "remainder" (modulo), can the absolute value of any integer be calculated?</p>
<p>To be explicit, I am hoping to find a method that does not involve a piecewise function (i.e. branching, <code>if</code>, if you will.)</p>
| Empy2 | 81,790 | <p>EDIT:<br>
$$m=n\%(n^2-n+2)\\
p=m\%(n^2+2)\\
|n|=2p-n$$</p>
<p>If $n\ge0$ then $m=n$ and $p=n$.<br>
If $n<0$ then $m=n^2+2$ and $p=0$. </p>
|
524,073 | <p>Hey can some help me with this textbook question</p>
<p>Let $R^{2×2}$ denote the vector space of 2×2 matrices, and let</p>
<p>$S =\left\{
\left[\begin{matrix}
a \space b \\
b \space c \\
\end{matrix}\right]\mid a,b,c \in \mathbb{R}\right\}$</p>
<p>Find (with justication) a basis for $S$ and determine the dimension of $S$.</p>
| rst | 73,496 | <p>$S =\left\{
\left[\begin{matrix}
a \space b \\
b \space c \\
\end{matrix}\right]\mid a,b,c \in \mathbb{R}\right\}$</p>
<p>or [a b b c]---------------(1)</p>
<p>Put values {[1 0 0 0 ],[0 1 0 0],[0 0 1 0],[0 0 0 1]} in (1)
we get</p>
<p>{[1 0 0 0 ],[0 1 1 0],[0 1 1 0],[0 0 0 1]}</p>
<p>$
\left[\begin{matrix}
1 \space 0 \space 0 \space 0 \\
0 \space1 \space 1 \space 0 \\
0 \space1 \space 1 \space 0 \\
0 \space 0 \space 0 \space 1 \\
\end{matrix}\right]$</p>
<p>I will reduce to</p>
<p>$
\left[\begin{matrix}
1 \space 0 \space 0 \space 0 \\
0 \space1 \space 1 \space 0 \\
0 \space0 \space0 \space 0 \\
0 \space 0 \space 0 \space 1 \\
\end{matrix}\right]$</p>
<p>its rank is 3,So dimension of basis of S is 3</p>
<p>Basis ={[1 0 0 0 ],[0 1 1 0],[0 0 0 1]} (see the reduced matrix)</p>
<p>Basis ={
$
\left[\begin{matrix}
1 \space 0 \\
0 \space0 \\
\end{matrix}\right]$,$
\left[\begin{matrix}
0 \space 1 \\
1 \space0 \\
\end{matrix}\right]$,$
\left[\begin{matrix}
0 \space 0 \\
0 \space1 \\
\end{matrix}\right]$}</p>
|
2,678,406 | <p>I'm trying to compute the distance between a point and a plane of the form
<span class="math-container">$$
ax+bx+cz = d
$$</span>
not using the standard formula for analytical geometry.</p>
<ul>
<li>I am trying to compute it by coming up with the projection matrix onto the normal of the plane passing through the origin and then projecting a a vector that is shifted the amount required to make the plane pass through the origin and then taking the length of that vector.</li>
<li>I upload pictures of my work for the first one I used this method and it worked it gave me the correct distance, but for any other equation it doesn't seem to work and I cant figure out why or what to change to make it work.</li>
</ul>
<p><a href="https://i.stack.imgur.com/7Ouuk.jpg" rel="nofollow noreferrer">Example of the method working</a>, <a href="https://i.stack.imgur.com/LEmJl.jpg" rel="nofollow noreferrer">Example of method not working</a></p>
| Community | -1 | <p>A line perpendicular to the plane, through the point $(u,v,w)$ has the parametric equation</p>
<p>$$x=u+ta,\\y=v+tb,\\z=w+tc.$$</p>
<p>It intersects the plane when</p>
<p>$$a(u+ta)+b(v+tb)+c(w+tc)=d$$ or</p>
<p>$$t=\frac{d-au-bv-cw}{a^2+b^2+c^2}$$</p>
<p>and the distance between the two points is given by</p>
<p>$$\sqrt{(ta)^2+(tb)^2+(tc)^2}=|t|\sqrt{a^2+b^2+c^2}=\frac{|d-au-bv-cw|}{\sqrt{a^2+b^2+c^2}}.$$</p>
|
1,090,658 | <p>I'm doing some previous exams sets whilst preparing for an exam in Algebra.</p>
<p>I'm stuck with doing the below question in a trial-and-error manner:</p>
<p>Find all $ x \in \mathbb{Z}$ where $ 0 \le x \lt 11$ that satisfy $2x^2 \equiv 7 \pmod{11}$</p>
<p>Since 11 is prime (and therefore not composite), the Chinese Remainder Theorem is of no use? I also thought about quadratic residues, but they don't seem to solve the question in this case.</p>
<p>Thanks in advance</p>
| Did | 6,179 | <p>Hints: </p>
<ul>
<li>$6\cdot2=1\pmod{11}$</li>
<li>$6\cdot7=9\pmod{11}$</li>
<li>$ab=0\pmod{11}\iff (a=0\pmod{11}\vee b=0\pmod{11})$</li>
</ul>
|
88,880 | <p>In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$:</p>
<ul>
<li>Using equivalent classes of smooth curves through $x$</li>
<li>Using derivations near $x$</li>
<li>Using cotangent vectors at $x$</li>
</ul>
<p>Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same.</p>
| painday | 109,535 | <p>Here is an explanation based on the category-theoretical formulation of tangent spaces.</p>
<p><strong>In short:</strong> To give a definition of tangent spaces, one actually define a functor from the category of (pointed)manifolds to the category of vector spaces that is a <strong>natural extension</strong> of
the tangent space of Euclidean spaces satisfying the principle of localization(see below). It can be shown that the tangent space functor is uniquely determined by these two conditions, up to a natural isomorphism.</p>
<p>Hence to check that a concrete definition of tangent space is reasonable, one only need to show that this definition generalizes the concept of tangent spaces of Euclidean spaces, and satisfies the principle of localization.
If you have two such definitions, then they are equivalent in the sense that there exists a natural isomorphism between the functors representing the two definitions.</p>
<p><strong>In detail:</strong> The category of pointed manifold <span class="math-container">$\mathcal M$</span> has objects in the form <span class="math-container">$(x,M),$</span> where <span class="math-container">$M$</span> is any smooth manifold and <span class="math-container">$x$</span> an element of <span class="math-container">$M.$</span>
A morphism from <span class="math-container">$(x,M)$</span> to <span class="math-container">$(y,N)$</span> is an ordered quadruple <span class="math-container">$(x,f,M,N),$</span> in which <span class="math-container">$f$</span> is a <strong>functional relation</strong> on <span class="math-container">$M\times N$</span> with <span class="math-container">$\mathrm{dom}(f)$</span> a (not necessarily open)neighborhood of <span class="math-container">$x,$</span> <span class="math-container">$f$</span> differentiable at <span class="math-container">$x$</span> and <span class="math-container">$f(x)=y.$</span></p>
<p>More exactly, <span class="math-container">$f$</span> is a subset of <span class="math-container">$M\times N$</span> such that if <span class="math-container">$(x,y),(x,y')\in f$</span> then <span class="math-container">$y=y'.$</span>
By definition, if <span class="math-container">$O$</span> is an open neighborhood of <span class="math-container">$x$</span> in <span class="math-container">$M$</span> such that <span class="math-container">$\mathrm{dom}(f)\subset O,$</span> and <span class="math-container">$O'$</span> is an open subset of <span class="math-container">$N$</span> with <span class="math-container">$\mathrm{Im}(f)\subset O',$</span> then <span class="math-container">$(x,f,O,O')$</span> is a morphism from <span class="math-container">$(x,O)$</span> to <span class="math-container">$(y,O').$</span></p>
<p>The composition of morphisms is given by
<span class="math-container">\begin{equation}
\boxed{(y,g,N,Q)\circ (x,f,M,N):=(x,g\circ f,M,Q)}
\end{equation}</span>
where <span class="math-container">$g\circ f:=\{(x_1,x_3)|\ \exists \ x_2\in N,\ s.t. \ (x_1,x_2)\in f,\ (x_2,x_3)\in g \}$</span> is the composition of relations.</p>
<p>If we consider Euclidean spaces only, then we get a full subcategory <span class="math-container">$\mathcal M(\mathbb E)$</span> of <span class="math-container">$\mathcal M.$</span> Now we define the <strong>canonical tangent space</strong> of <span class="math-container">$(x,\mathbb R^n)$</span> to be simply <span class="math-container">$\mathbb R^n,$</span> and the <strong>canonical tangent map</strong> <span class="math-container">$T_{\rm canonical}(x,f,\mathbb R^n,\mathbb R^m):=D_xf,$</span>
then we get a functor <span class="math-container">$T_{\rm canonical}:\mathcal M(\mathbb E)\to \mathsf{Vec}(\mathbb R).$</span>
<span class="math-container">\begin{equation}
\boxed{T_{\rm canonical}\big[(x,\mathbb R^n)\stackrel{(x,f,\mathbb R^n,\mathbb R^m)}{\longrightarrow} (y,\mathbb R^m)\big]:=\mathbb R^n \stackrel{D_xf}{\longrightarrow}\mathbb R^m}
\end{equation}</span></p>
<p>A tangent space functor defined on the full category <span class="math-container">$\mathcal M$</span> must generalize the concept of tangent spaces of Euclidean spaces, that is, it must be an extension of <span class="math-container">$T_{\rm canonical}.$</span> In fact this <strong>extension should be natural</strong>, that is to say,
the restriction of <span class="math-container">$T$</span> to <span class="math-container">$\mathcal M(\mathbb E)$</span> is nutural isomorphic to <span class="math-container">$T_{\rm canonical}.$</span> In other words,
for any object <span class="math-container">$(x,\mathbb R^n)$</span> there is an isomorphism <span class="math-container">$\alpha_{(x,\mathbb R^m)}:T(x,\mathbb R^n)\to T_{\rm canonical}(x,\mathbb R^n)=\mathbb R^n$</span> such that
if <span class="math-container">$(x,f,\mathbb R^n,\mathbb R^m):(x,\mathbb R^n)\to (y,\mathbb R^m)$</span> is a morphism, then the following diagram commutes:
<a href="https://i.stack.imgur.com/AcCIg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AcCIg.png" alt="enter image description here" /></a>
We say that a functor <span class="math-container">$T:\mathcal M\to\mathsf{Vec}(\mathbb R)$</span> gives a definition of tangent spaces, if it satisfies the followng two conditions:</p>
<ol>
<li><span class="math-container">$T$</span> is a natural extension of <span class="math-container">$T_{\rm canonical}.$</span></li>
<li>(<strong>Principle of localization</strong>)If <span class="math-container">$(x,f,M,N),(x,g,M,N)$</span> are morphisms and <span class="math-container">$f,g$</span> coincide on a neighborhood of <span class="math-container">$x$</span> then <span class="math-container">$T(x,f,M,N)=T(x,g,M,N).$</span></li>
</ol>
<p>Of course, if <span class="math-container">$T$</span> and <span class="math-container">$\widetilde T$</span> are two such functors, then their restriction to <span class="math-container">$\mathcal M(\mathbb E)$</span> are naturally isomorphic through the transformation <span class="math-container">$\beta:=\widetilde\alpha^{-1}\circ\alpha,$</span> since we have the commutative diagram
<a href="https://i.stack.imgur.com/5ckXm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5ckXm.png" alt="enter image description here" /></a></p>
<p>To show the uniqueness of tangent space functor, we need to find a natural isomorphism between <span class="math-container">$T$</span> and <span class="math-container">$\widetilde T.$</span></p>
<p>For any object <span class="math-container">$(x,M)$</span> of <span class="math-container">$\mathcal M,$</span> we define a vector space isomorphism
<span class="math-container">$\gamma_{(x,M)}:T(x,M)\to \widetilde T(x,M)$</span> by
<span class="math-container">\begin{equation}\boxed{\gamma_{(x,M)}:=\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)}\end{equation}</span>
in which <span class="math-container">$(U,\varphi)$</span> is a coordinate chart for <span class="math-container">$x$</span> in <span class="math-container">$M.$</span> To show that <span class="math-container">$\gamma_{(x,M)}$</span> is well-defined, we have to verify that if
<span class="math-container">$(V,\psi)$</span> is another coordinate chart for <span class="math-container">$x$</span> in <span class="math-container">$M$</span> then
<span class="math-container">\begin{equation}\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(x),\psi^{-1},\mathbb R^n,M)\circ \beta_{(\psi(x),\mathbb R^n)}\circ T(x,\psi,M,\mathbb R^n)\end{equation}</span></p>
<p>In fact, principle of localization yields the commutative diagram
<a href="https://i.stack.imgur.com/XbIeX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XbIeX.png" alt="enter image description here" /></a></p>
<p>A similar digram commutes for the functor <span class="math-container">$\widetilde T.$</span> On the other hand, since <span class="math-container">$\psi\circ\varphi^{-1}$</span> is a functional relation on <span class="math-container">$\mathbb R^n\times\mathbb R^n,$</span> it follows that the following diagram commutes:
<a href="https://i.stack.imgur.com/W2qXq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W2qXq.png" alt="enter image description here" /></a></p>
<p>Thus we have a combined commutative diagram</p>
<p><a href="https://i.stack.imgur.com/R7H18.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R7H18.png" alt="enter image description here" /></a></p>
<p>which is desired.</p>
<p>It remains to show that <span class="math-container">$\gamma$</span> is a natural isomorphism between <span class="math-container">$T$</span> and <span class="math-container">$\widetilde T.$</span>
Let <span class="math-container">$(x,f,M,N)$</span> be a morphism from <span class="math-container">$(x,M)$</span> to <span class="math-container">$(y,N)$</span> and let <span class="math-container">$(U,\varphi)$</span> and <span class="math-container">$(V,\psi)$</span> be a coordinate chart for <span class="math-container">$x$</span> and <span class="math-container">$y,$</span> respectively, then <span class="math-container">$(\varphi(x),\psi\circ f\circ\varphi,\mathbb R^n,\mathbb R^n)$</span> is a morphism from <span class="math-container">$(\varphi(x),\mathbb R^n)$</span> to <span class="math-container">$(\psi(y),\mathbb R^n),$</span> hence
<span class="math-container">\begin{equation}\label{key}\widetilde T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\circ\beta_{\varphi(x),\mathbb R^n}=\beta_{\psi(y),\mathbb R^n}\circ T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\end{equation}</span>
Plugging
<span class="math-container">\begin{equation}(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)=(y,\psi,N,\mathbb R^n)\circ (x,f,M,N)\circ(\varphi(x),\varphi^{-1},\mathbb R^n,M)\end{equation}</span>
into the former equation gives
<span class="math-container">\begin{equation}\widetilde T(x,f,M,N)\circ \widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(y),\psi^{-1},\mathbb R^n,N)\circ\beta_{(\psi(y),\mathbb R^n)}\circ T(y,\psi,N,\mathbb R^n)\circ T(x,f,M,N)\end{equation}</span>
hence we have the commutative diagram
<a href="https://i.stack.imgur.com/TTgFA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TTgFA.png" alt="enter image description here" /></a>
as expected.</p>
<hr />
<p>The principle of localization is equivalent to the following statement: for any two open subsets <span class="math-container">$U,U'$</span> of any manifold <span class="math-container">$M,$</span> if <span class="math-container">$x\in U\cap U'$</span> then there is an isomorphism <span class="math-container">$\Delta^{(x,M)}_{U,U'}$</span>(abbreviated as <span class="math-container">$\Delta_{U,U'}$</span>) satisfying the cocycle condition <span class="math-container">$\Delta_{U',U''}\circ \Delta_{U,U'}=\Delta_{U,U''},\ \Delta_{U,U}=\mathrm{id}$</span> and
if <span class="math-container">$(x,f,M,N)$</span> is a morphism from <span class="math-container">$(x,M)$</span> to <span class="math-container">$(y,N),$</span> then the following diagram commutes:
<a href="https://i.stack.imgur.com/1Flgp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Flgp.png" alt="enter image description here" /></a>
Here <span class="math-container">$f|_{U}^V:=f\cap (U\times V)$</span> and it's easy to see that <span class="math-container">$(x,f|_U^V,U,V)$</span> is a morphism from <span class="math-container">$(x,U)$</span> to <span class="math-container">$(y,V).$</span></p>
<p>In fact, assume that the principle of localization hold, then defining
<span class="math-container">\begin{equation}\Delta_{U,U'}^{(x,M)}:=T(x,\mathrm{id}_{U\cap U'},U,U')\end{equation}</span>
will make the cocycle diagram commute.</p>
<p>Conversely,assume that the cocycle diagram commute, and assume <span class="math-container">$f$</span> coincide <span class="math-container">$g$</span> on a neighborhood <span class="math-container">$U$</span> of <span class="math-container">$x,$</span> then we have the commutative diagram
<a href="https://i.stack.imgur.com/pC8qk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pC8qk.png" alt="enter image description here" /></a>
Since <span class="math-container">$\Delta_{U,M}\circ\Delta_{M,U}=\mathrm{id},\ \Delta_{N,N}=\mathrm{id},$</span> it follows that <span class="math-container">$T(x,f,M,N)=T(x,g,M,N),$</span> which establishes the equivalence of the two statements.</p>
|
119,561 | <p>I am interested in determining the <strong>minimum</strong> and <strong>maximum</strong> values of the real roots of polynomials of form $P(x)=\sum_{k=0}^n a_{k} x^k$ where $n$ will have a defined value (say 3,4,5...) and $a_k$ are chosen from the set $\{-1,1\}$ with equal probability.</p>
<p>I have tried creating a table of the roots, and then using <code>MinMax</code>; here is my (bad) attempt (with $n=3$):</p>
<pre><code>T = Table[Roots[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x], 25]
MinMax[T]
</code></pre>
<p>Unfortunately, <code>Roots</code> gives both the real and imaginary roots, I would only like the real roots . Also, <code>MinMax</code> cannot work on the table $T$ when the roots are not presented as a list (they have $||$ in between each root). </p>
<p>Any suggestions/help with this issue is immensely appreciated.
Thank You!</p>
| MarcoB | 27,951 | <p>Try something like </p>
<pre><code>x /. Solve[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x, Reals]
</code></pre>
<p>instead of your <code>Roots</code> expression. Use <code>NSolve</code> if you want the numerical value of the roots, rather than a symbolic representation.</p>
<p>For instance:</p>
<pre><code>t = Table[x /. NSolve[Sum[RandomChoice[{-1, 1}] x^k, {k, 0, 3}] == 0, x, Reals], 4500];
realroots = DeleteDuplicates@Flatten@t
(* Out: {-1., 1., 1.83929, -1.83929, 0.543689, -0.543689} *)
</code></pre>
<p>As you can see, these seem to be the only admissible real root values. By inspection, or using <code>MinMax</code>, you obtain the largest and smallest one:</p>
<pre><code>MinMax@realroots
(* Out: {-1.83929, 1.83929} *)
</code></pre>
<hr>
<p>Consider also that you can pretty simply <em>enumerate</em> all possible polynomials of your desired form:</p>
<pre><code>Tuples[{-1, 1}, {4}].{1, x, x^2, x^3}
(* Out:
{-1 - x - x^2 - x^3, -1 - x - x^2 + x^3, -1 - x + x^2 - x^3,
-1 - x + x^2 + x^3, -1 + x - x^2 - x^3, -1 + x - x^2 + x^3,
-1 + x + x^2 - x^3, -1 + x + x^2 + x^3, 1 - x - x^2 - x^3,
1 - x - x^2 + x^3, 1 - x + x^2 - x^3, 1 - x + x^2 + x^3,
1 + x - x^2 - x^3, 1 + x - x^2 + x^3, 1 + x + x^2 - x^3,
1 + x + x^2 + x^3}
*)
Flatten[x /. NSolve[# == 0, x, Reals] & /@ %] // MinMax
(* Out: {-1.83929, 1.83929} *)
</code></pre>
|
2,576,466 | <p>One says a bounded $f$ is Riemann integrable on [a,b] if the Upper and lower Riemann integrals are equal. Another sufficient condition for Riemann integrablity is that the set of discontinuity of $f$ must be countable set. The following function is continuous only at $x=1/2$ and so the set of discontinuity of $f$ is uncountable set. This shows that $f$ is not Riemann integrable. But I want to justify this by computing the upper and lower Riemann integrals. Any one who can help me how to get the upper and lower Riemann integrals.
Let $f\colon[0,1]\rightarrow\mathbb{R}$ such that
$$f(x)=\left\{\begin{array}{c}
x, x\in\mathbb{Q}\\ 1-x, x\notin\mathbb{Q}
\end{array}\right.$$ Show that $f$ is not Riemann integrable on $[0,1].$</p>
| JH vd Walt | 514,810 | <p>Let </p>
<p>$$g(x) = \left\{\begin{array}{lll}
1-x & if & 0\leq x\leq 1/2 \\
x & if & 1/2<x\leq 1 \\
\end{array}\right.$$</p>
<p>and</p>
<p>$$h(x) = \left\{\begin{array}{lll}
x & if & 0\leq x\leq 1/2 \\
1-x & if & 1/2<x\leq 1 \\
\end{array}\right.$$</p>
<p>Both $g$ and $h$ are continuous, and so Riemann integrable. Now show that the upper integral for $f$ is equal to the integral of $g$ and the lower integral of $f$ is equal to the integral of $h$. Here you need to use the following facts.
1. $h\leq f\leq g$\
2. Both $\mathbb{Q}\cap[0,1]$ and $[0,1]\setminus \mathbb{Q}$ are dense in $[0,1]$. Therefore every interval $[a,b]\subseteq [0,1]$ with $a<b$ will intersect both $\mathbb{Q}\cap[0,1]$ and $[0,1]\setminus \mathbb{Q}$, so </p>
<p>$$
\sup_{x\in [a,b]} f(x) = \sup_{x\in [a,b]} g(x)~~{\rm and}~~ \inf_{x\in [a,b]} f(x) = \inf_{x\in [a,b]} h(x)
$$</p>
|
771,997 | <p>I'm working on a matrix extension of Fermat's Little Theorem, but I'm stuck on trying to show that if $A^p \equiv A$ mod $p$, then $A$ does not have to be diagonalizable.</p>
<p>Any help would be appreciated!</p>
<p><strong>Edit::</strong> I would like to either find a reason why $A$ does not have to be diagonalizable, or somehow be able to categorize the matrices that are / are not diagonalizable in a way that would suggest a pattern. An example would be that if $A^p \equiv A$ mod $p$, then $A$ is diagonalizable when xxx or not diagonalizable when xxx.</p>
| Stella Biderman | 123,230 | <p>What people generally think of as the generalization of Fermat's Little Theorem is as follows: </p>
<p>Let $p$ be prime and $A\in GL_n(\mathbb{Z})$. Then $tr(A^p)=tr(A)$ mod $p$, where $tr(A)$ denotes the trace of the matrix $A$. In fact, this holds even more generally: $tr(A^{p^k})=tr(A^{p^{k-1}})$ mod $p^k$ where the first statement is the case for $k=1$</p>
<p>The first statement was proven by V.I. Arnold (who also conjectured the second). The second was proven by <a href="http://link.springer.com/article/10.1134/S008154380804007X" rel="nofollow">Alexander Zarelua in 2008</a>.</p>
|
3,105,482 | <p>For the formula:</p>
<p><span class="math-container">$$1 = \sqrt{x^2 + y^2 + z^2 + w^2}$$</span></p>
<p>How to rewrite it to find <span class="math-container">$w$</span>?</p>
| NicNic8 | 24,205 | <p><span class="math-container">\begin{align}
1 &= \sqrt{x^2 + y^2 + z^2 + w^2} \\
1 &= x^2 + y^2 + z^2 + w^2 \\
w^2 &= 1 - x^2 - y^2 - z^2 \\
w &= \pm\sqrt{1 - x^2 - y^2 - z^2}
\end{align}</span></p>
|
1,173,643 | <p>I saw the following statement written, but I can't understand why it is true.</p>
<p>$$
\dfrac {P(A \text{ and } B)}{P(B)} = \dfrac{P(A)-P(A \text{ and }B^c)}{ 1-P(B^c)}
$$</p>
<p>Any help understanding why these are equivalent would be appreciated.</p>
| Siméon | 51,594 | <p>Since $B$ and $B^c$ satisfy $B \wedge B^c = \emptyset$ and $B \vee B^c = \Omega$, we have$$
P(A \wedge B) + P(A \wedge B^c) = P(A \wedge \Omega) = P(A).
$$
As a consequence, $$\tag{1}P(A \wedge B) = P(A) - P(A \wedge B^c).$$</p>
<p>In addition, the special case $A = \Omega$ gives
$$
\tag{2}P(B) = 1 - P(B^c).
$$
Finally, $\dfrac{(1)}{(2)}$ yields what you want.</p>
|
963,125 | <p>I'm not exactly sure where to start on this one. Any help would be greatly appreciated.</p>
<p>Show that $4$ does not divide $12x+3$ for any $x$ in the integers.</p>
<p>Here's what I have so far:</p>
<p>There exist c in the integers such that 4 | 12x+3. Then 12x+3 = 4y for some y in the integers. This is a contradiction since y = 3x + 3/4 and y is in the integers.</p>
<p>Is that correct?</p>
| gamma | 88,524 | <p>Note that $12x + 3$ is always odd.</p>
<p>To be divisible by $4$ means to contain at least two factors of $2$.</p>
<p>Odd numbers contain no factors of $2$ so $12x + 3$ can never be divisible by $4$ </p>
|
330,065 | <p>What is an example of a manifold <span class="math-container">$M$</span> with <span class="math-container">$\dim(M)>1$</span> whose Lie algebra <span class="math-container">$\chi^{\infty}(M)$</span> of smooth vector fields admit an elliptic operator <span class="math-container">$D:\chi^{\infty}(M)\to \chi^{\infty}(M)$</span> such that <span class="math-container">$D$</span> is a Lie algebra derivation on <span class="math-container">$\chi^{\infty}(M)$</span>?</p>
<p>Does every manifold admit such an operator?</p>
<p>Is there a Riemannian manifold for which the Laplace operator <span class="math-container">$D=\Delta$</span>, naturally defined on <span class="math-container">$\chi^{\infty}(M)\simeq \Omega^1(M)$</span>, would be a derivation of <span class="math-container">$\chi^{\infty}(M)$</span>?</p>
| Bazin | 21,907 | <p>First a classical result about elliptic operators: in dimension <span class="math-container">$\ge 3$</span> the order of elliptic operators is even. In dimension 2, say in <span class="math-container">$\mathbb R^2$</span> you have elliptic vector fields such as
<span class="math-container">$$
\bar \partial=\frac12(\partial_x+i\partial_y),\quad \partial=\frac12(\partial_x-i\partial_y).
$$</span>
Of course vector fields are derivations and I suspect that, since a derivation on a manifold must be a differential operator, that differential operator must be first-order from Leibniz' formula and thus should be a vector field since no mutiplication by a (non-zero) function is a derivation.</p>
|
3,754,225 | <p>Let S be the circle with centre <span class="math-container">$(0,0)$</span>, radius <span class="math-container">$r$</span> units. The chord <span class="math-container">$C$</span> of the circle S subtends an angle of 2π/3 at its center. If R represents the region consisting of all points inside S which are closer to C than to circumference of S, then</p>
<p>(1) What is area of region <span class="math-container">$R$</span>, and,</p>
<p>(2) In what ratio does <span class="math-container">$C$</span> divide the region <span class="math-container">$R$</span> ?</p>
<p>I found the equation of a circle intersecting circle radius <span class="math-container">$r$</span> for the chord subtending angle <span class="math-container">$2\pi/3$</span> at center we should have</p>
<p><span class="math-container">$$ x=\dfrac{r}{2} $$</span></p>
<p>but don't know what to do next. Please help.</p>
| g.kov | 122,782 | <p>Using the symmetry, we can consider a upper semicircle
where <span class="math-container">$DE$</span> is a half of the chord.</p>
<p><a href="https://i.stack.imgur.com/XQbzd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XQbzd.png" alt="enter image description here" /></a></p>
<p>The chord splits the area in question
in two parts.
Let the points <span class="math-container">$V_{1,2}$</span> are the points on the boundary,
for which
the closest points on a circumference are <span class="math-container">$U_{1,2}$</span>,
and the closest points on a chord are <span class="math-container">$W_{1,2}$</span>,
and the value of the closed distances are <span class="math-container">$x_{1,2}$</span>.</p>
<p>Then we can find <span class="math-container">$x_{1,2}$</span> in terms of the angles <span class="math-container">$\phi_{1,2}$</span> as</p>
<p><span class="math-container">\begin{align}
x_1&=
\tfrac12\,\frac{R(2\cos\phi_1-1)}{\cos\phi_1-1}\tag{1}\label{1}
,\\
x_2&=
\tfrac12\,\frac{R(2\cos\phi_2-1)}{\cos\phi_2+1}\tag{2}\label{2}
.
\end{align}</span></p>
<p>The coordinates of the point
on the boundary of the first region in therms of <span class="math-container">$\phi$</span> are:
<span class="math-container">\begin{align}
\frac{\tfrac12\,R}{1-\cos\phi}
\cdot(\cos\phi,\, \sin\phi)\tag{3}\label{3}
.
\end{align}</span></p>
<p>Similarly, the coordinates of the point
on the boundary of the second region in therms of <span class="math-container">$\phi$</span> are:
<span class="math-container">\begin{align}
\frac{\tfrac32\,R}{\cos\phi+1}
\cdot(\cos\phi,\, \sin\phi)\tag{4}\label{4}
.
\end{align}</span></p>
<p>Now we can find the functions of the boundary curves
<span class="math-container">\begin{align}
y_1(x)&=
\tfrac12\,\sqrt{R^2+4R\,x}\tag{5}\label{5}
,\\
y_2(x)&=
\tfrac12\,\sqrt{9R^2-12R\,x}\tag{6}\label{6}
.
\end{align}</span></p>
<p>So the areas can be found as</p>
<p><span class="math-container">\begin{align}
S_1&=\int_{-\tfrac14\,R}^{\tfrac12\,R}
y_1(x)\, dx
=\tfrac{\sqrt3}4\,R^2
\tag{7}\label{7}
,\\
S_2&=\int_{\tfrac12\,R}^{\tfrac34\,R}
y_2(x)\, dx
=\tfrac{\sqrt3}{12}\,R^2
\tag{8}\label{8}
.
\end{align}</span></p>
<p>The total area of the region</p>
<p><span class="math-container">\begin{align}
S_{\textsf{tot}}
&=
2S_1+2S_2=
\tfrac23\,\sqrt3\,R^2
\tag{9}\label{9}
,
\end{align}</span></p>
<p>and the chord splits the area as
<span class="math-container">\begin{align}
S_1:S_2&=3:1
\tag{10}\label{10}
.
\end{align}</span></p>
<p>Edit</p>
<p>As @Narasimham pointed out,
both boundary lines are parabolas,
with the focus at the origin
and lines
<span class="math-container">$x=-\tfrac12\,R$</span>
and
<span class="math-container">$x=\tfrac32\,R$</span>
as directrices.</p>
|
139,135 | <p>Suppose $R = \mathbb{Q}[x_1, ..., x_n]/I$, and $J \subset R$ is a given height one ideal. Is there a quick algorithm one could write to determine if $J$ is a principal ideal or necessarily not principal? Is it not possible to do this with Groebner bases?</p>
| Thomas Kahle | 5,495 | <p>In the graded situation the concept of "minimal generators" is well-defined. Just think about the minimal generators as part of the minimal free resolution. Their number is the first total Betti number and those can be computed with Gröbner bases. Macaulay2 has the command <a href="http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.6/share/doc/Macaulay2/Macaulay2Doc/html/_mingens_lp__Groebner__Basis_rp.html%20mingens" rel="nofollow">mingens</a> which does it directly.</p>
<p>I don't know how to do it in the non-graded situation since I think the concept of minimal generators of an ideal is not really well-defined. Maybe principal vs. non-principal can be decided, though.</p>
|
3,271,891 | <p>In <a href="https://en.wikipedia.org/wiki/Boolean_algebra_(structure)" rel="nofollow noreferrer">Wikipedia, the Boolean algebra</a> is defined as a 6-tuple
<span class="math-container">$(A,\wedge,\vee,\neg,0,1)$</span>. In Kuratowski1976, on the other side in the definition on page 34, there is no <span class="math-container">$1$</span>. Halmos1963 has the <span class="math-container">$1$</span>.</p>
<blockquote>
<p>Does the definition in Kuratowski1976 leads to something more general or somehow different theory? On page 37 he introduces the concept of unit <span class="math-container">$i$</span> that I assume is the <span class="math-container">$1$</span> of the other authors (also judging from the definition <span class="math-container">$a\wedge i=a$</span>).</p>
</blockquote>
<p><strong>Definition from Kuratowski1976:</strong></p>
<p><a href="https://i.stack.imgur.com/IRiIj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRiIj.png" alt="enter image description here"></a></p>
| amrsa | 303,170 | <p>All Boolean algebras have a top element.</p>
<p>I don't have access to that Kuratowski definition, but tipically, what happens is these alternative definitions have axioms enough to define a partial order relation relative to which the complement operation is order-reversinng. </p>
<p>Given that <span class="math-container">$0$</span> is the bottom element, it will follow that <span class="math-container">$0'$</span> is the top one.</p>
|
235,940 | <p>Consider a sphere $S$ of radius $a$ centered at the origin. Find the average distance between a point in the sphere to the origin.</p>
<p>We know that the distance $d = \sqrt{x^2+y^2+z^2}$. </p>
<p>If we consider the problem in spherical coordinates, we have a 'formula' which states that the average distance $d_{avg} = \frac{1}{V(S)}\iiint \rho dV$</p>
<p>I think that this is reminiscent of an average density function which I've seen in physics courses, and it is clear that the $\iiint \rho dV$ is equal to the volume of the sphere, but I'm not sure as to why we must integrate over the distance and then divide by the actual volume to calculate the average distance.</p>
<p>I am looking for a way to explain this to my students without presenting the solution as a formula, any insights would be appreciated.</p>
| anon | 190,539 | <p>All points are equidistant from the origin ... that is why it is a sphere. Therefore, the average distance of a point on the sphere from its center is the radius. </p>
|
24,912 | <p>I have the category-theoretic background of the occasional stroll through MacLane's text, so excuse my ignorance in this regard. I was trying to learn all that I could on the subject of tensor algebras, and higher exterior forms, and I ran into the notion of cohomological determinants. Along this line of inquiry, I ran into the general use of the notion of a Picard category, and kept running into frustration in trying to find some sort of exposition of what these structures are. So where can I find out more about these structures, and about (cohomological) determinants in K-theory, which seem to be a hot topic among AG, AT, and RT researchers alike at the moment.</p>
| Timo Schürg | 473 | <p>I ran across Picard categories in a totally different area of mathematics, but maybe it helps.</p>
<p>In short, a Picard category is a group object in the category of groupoids. </p>
<p>Picard categories come up when you study Picard stacks. Roughly, a Picard stack is a sheaf of Picard categories. The classical example is taking a two-term (perfect??) complex of sheaves, and associating to such a complex the groupoid quotient of one term by the other. This is important when you want to produce a geometric object from such a complex. This is an important tool in defining virtual fundamental classes as in <a href="http://arxiv.org/abs/alg-geom/9601010" rel="nofollow noreferrer">http://arxiv.org/abs/alg-geom/9601010</a>. </p>
<p>Before I tell you too many things that are not true, here are the references I know of: </p>
<p><a href="http://math.berkeley.edu/~molsson/MSRISummer07.pdf" rel="nofollow noreferrer">Lecture notes of Martin Olsson</a>, Lecture 5. You can even <a href="https://www.msri.org/summer_schools/419/schedules/3015" rel="nofollow noreferrer">watch it on video</a>.</p>
<p>The definitive reference is <a href="http://www.cmls.polytechnique.fr/perso/laszlo/sga4/SGA4-3/sga43.pdf" rel="nofollow noreferrer">Exposé XVIII of SGA 4</a>.</p>
<p>And finally there are very friendly and down-to-earth lectures of Barbara Fantechi at <a href="http://www.openeya.org/sissa/" rel="nofollow noreferrer">http://www.openeya.org/sissa/</a> [link dead]. I think lecture 3 or lecture 4 is about Picard categories.</p>
|
2,067,553 | <p>I'm currently working on another problem: let $x_1,x_2,x_3$ be the roots of the polynomial: $x^3+3x^2-7x+1$, calculate $x_1^2+x_2^2+x_3^2$. Here is what i did: $x^3+3x^2-7x+1=0$ imply $x^2=(7x-x^3-1)/3$. And so $x_1^2+x_2^2+x_3^2= (7x_1-x_1^3-1)+7x_2-x_2^3-1+7x_3-x_3^3-1)/3= 7(x_1+x_2+x_3)/3+(x_1^3+x_2^3+x_3^3)-1$. Then I don't know what to do anymore.</p>
| Semiclassical | 137,524 | <p>Hint: Writing your polynomial as $p(x)$, observe that
\begin{align}
p(x)p(-x)
&=(x-x_1)(x-x_2)(x-x_3)\times -(x+x_1)(x+x_2)(x+x_3)\\
&=-(x^2-x_1^2)(x^2-x_2^2)(x^2-x_3^2).
\end{align}</p>
|
785,844 | <p>I've got the following limit to solve:</p>
<p>$$\lim_{s\to 1} \frac{\sqrt{s}-s^2}{1-\sqrt{s}}$$</p>
<p>I was taught to multiply by the conjugate to get rid of roots, but that doesn't help, or at least I don't know what to do once I do it. I can't find a way to make the denominator not be zero when replacing $s$ for $1$. Help?</p>
| Community | -1 | <p>$$\lim_{s\to 1}\dfrac{\sqrt{s}-s^2}{1-\sqrt{s}}=\lim_{s\to 1}\dfrac{\dfrac{1}{2\sqrt{s}}-2s}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}\dfrac{\dfrac{1-4s\sqrt{s}}{2\sqrt{s}}}{-\dfrac{1}{2\sqrt{s}}}=\lim_{s\to 1}(-1+4s\sqrt{s})$$
By L'Hopital's rule.</p>
|
2,704,290 | <p>I'm prepared for the competitive exam like <a href="http://csirhrdg.res.in/mathCEN_June2015.pdf" rel="nofollow noreferrer">this</a> (a sample question).</p>
<p>In order to solve the problems, first to familiarize with the prerequisite for the each concept. It's ok! </p>
<p>My problem is: If I'm working certain problems on group theory, then it will take two days or a week. Ok! No problem. After the week, I can move on to the next concept like integration, so it will take some time. But in the third week, I lack some concepts bit about group theory (even sometimes, I'm forgetting what I'm doing in the first week). So I feel I waste lot of time. I really want to learn algebra, analysis, topology, and all that simultaneously (sorry if this word does not make sense) or to first learn some bit about algebra and then analysis? I don't no how to move on? </p>
<p>Suppose the question booklet start with group theory. That is, the first question involves group theory, the second one involves like analyticity and the 17th question (for example) involves again group theory etc.</p>
<p>My question is: Is the way to doing problems one by one (in the order)? or
By selecting the first one and do all the problems related to group theory and then move on the next concept (like analyticity)?
or
anything
else?</p>
<p>What is the best strategy to prepare this type of exam?</p>
| polfosol | 301,977 | <p>I think the one good advice that you can get is the fact that the best learning strategy is totally dependent on the learner himself. Joonas has already mentioned some good points, but at the end of the day, it is only you who can evaluate which is the best path to take. In short:</p>
<ul>
<li>Trust yourself and your judgement about which learning method is more effective.</li>
<li>Have self esteem, and don't lose confidence in yourself if you forgot something. Forgetting stuff is totally natural and happens all the time for all of us.</li>
<li>Usually a lot of courage and perseverance is needed for these kinda exams. So you need a good motivation and strong desire to nail it.</li>
<li>One common misconception that many people have is that mathematics is all about being smart and fast in problem solving. On the contrary the most important thing that learning math needs is hard work and perseverance. So keep it up and try again.</li>
<li>You also need a good hobby for your spare times. Preferably a physical one.</li>
<li>Gamification of the learning process is a great help and boosts your progress.</li>
</ul>
<p>I might add some other stuff later if I remember them!</p>
|
1,270,584 | <p>I tried googling for simple proofs that some number is transcendental, sadly I couldn't find any I could understand.</p>
<p>Do any of you guys know a simple transcendentality (if that's a word) proof?</p>
<p>E: What I meant is that I wanted a rather simple proof that some particular number is transcendental ($e$ or $\pi$ would work), not a method to prove that any number is transcendental, sorry for the confusion.</p>
<p>Or even a proof about transcendental numbers being 'as common' as algebraic numbers?</p>
| Yuval Filmus | 1,277 | <p>The easiest proof is via Liouville's criterion.</p>
<p><strong>Lemma.</strong> Suppose $\alpha$ is an irrational algebraic number. There exist integers $C,n$ such that for all integers $p/q$,
$$
\left| \alpha - \frac{p}{q} \right| \geq \frac{C}{q^n}.
$$</p>
<p>You can find a proof of the lemma on <a href="http://en.wikipedia.org/wiki/Liouville_number#Liouville_numbers_and_transcendence" rel="noreferrer">Wikipedia</a>, and in many other sources. Now consider the number
$$
\alpha = \sum_{m=0}^\infty \frac{1}{10^{m!}},
$$
sometimes known as Liouville's number. Since the decimal expansion of $\alpha$ is aperiodic, it is irrational. On the other hand, for all $r$ we have
$$
\left| \alpha - \frac{\sum_{m=0}^r 10^{r!-m!}}{10^{r!}} \right| \leq \sum_{m=n+1}^\infty \frac{1}{10^{m!}} \leq \sum_{m=r+1}^\infty \frac{1}{10^{(r+1)!+m-(r+1)}} \leq \frac{2}{10^{(r+1)!}}.
$$
Since $2/10^{(r+1)!}$ is so much smaller than $10^{r!}$ for large $r$, it is not hard to check that $\alpha$ doesn't satisfy the conclusion of the lemma. Indeed, suppose that it did, for some $C,n$. Then for all $r$ we must have
$$
\frac{2}{10^{(r+1)!}} \geq \frac{C}{10^{r!n}} \Longleftrightarrow 10^{r!n} \geq (C/2) 10^{(r+1)!} \Longleftrightarrow 1 \geq (C/2) 10^{r! (r+1-n)},
$$
which fails for large enough $r$.</p>
<p>Since the conclusion of the lemma doesn't hold, one of its premises must be false. Since $\alpha$ is definitely irrational, it must be not algebraic. In other words, $\alpha$ is transcendental.</p>
|
3,291,303 | <p>this very same problem appeared in a different thread but the questions was slightly different. In my case, I'm looking precisely for the answer.</p>
<p>This is how I solved it, I only need confirmation of whether this actually is correct:</p>
<p>So I assumed that 1111... (100 ones) is going to be exactly divided by 1111111 (7 ones) 14 times (100/7 = 14). Hence, if 14 times 7 equals 98 '1's, then the remainder is 2 '1's.</p>
<p>Thanks.</p>
| hmakholm left over Monica | 14,366 | <p>Yes, this is a perfectly sound reasoning.</p>
<p>You can also express it as</p>
<p><span class="math-container">$$ 100000010000001\cdots1000000100\cdot 1111111 + 11 = \underbrace{111\cdots111}_{100\text{ ones}} $$</span></p>
|
2,025,069 | <p>$$\sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n}}{(2n)!}) * \sum_{n=0}^{\infty} C_{n}x^{n} = \sum_{n=0}^{\infty} (-1)^{n}(\frac{x^{2n+1}}{(2n+1)!})$$</p>
<p>How to find coefficients until the $x^{7} $ ? It is basically sin and written in form of sums but i'm not sure how should i use this form to calculate coefficients from this factor sum. How should i precede? </p>
| hamam_Abdallah | 369,188 | <p>Observe that </p>
<p>$$\sum_{n=0}^{+\infty}C_nx^n=\tan(x)$$</p>
<p>and $C_0=C_2=C_4=C_6=0$.</p>
|
3,163,355 | <p>I found that the Wieferich prime <span class="math-container">$1093$</span> divides <span class="math-container">$3^{1036}-1$</span>.
Does <span class="math-container">$1093$</span> divides infinitely many <span class="math-container">$3^{k}-1$</span>, with k a positive integer? And what features must <span class="math-container">$3^{k}-1$</span> have to be divisible by <span class="math-container">$1093$</span>?</p>
| fleablood | 280,126 | <p>Notice <span class="math-container">$a^m -1 = (a-1)(a^{m-1} + a^{m-2}+... + a + 1)$</span></p>
<p>So if <span class="math-container">$1093|a^{1036}-1$</span> then <span class="math-container">$1093|a^{1036m} - 1$</span> so there are infinite <span class="math-container">$k = 1036m$</span> where this is true.</p>
<p>Also remember Fermat's little Theorem. If <span class="math-container">$1093$</span> is prime than <span class="math-container">$3^{1092}\equiv 1 \pmod{1093}$</span> (and so <span class="math-container">$1093|3^{1092}-1$</span>). As a consequence. If <span class="math-container">$3^a \equiv 1\pmod {1093}$</span> and <span class="math-container">$a$</span> is the smallest such number (other than zero) then <span class="math-container">$m|1092$</span>.</p>
<p>Also if <span class="math-container">$3^{m}\equiv 1 \pmod {1093}$</span> and <span class="math-container">$3^k \equiv 1 \pmod {1093}$</span> then <span class="math-container">$3^{\gcd(m,n)}\equiv 1 \pmod {1093}$</span>. (and if <span class="math-container">$a$</span> is the smallest power so that <span class="math-container">$3^a \equiv 1 \pmod{1093}$</span> then <span class="math-container">$a|\gcd(m,n)$</span>.</p>
<p><span class="math-container">$1092 = 2^2*3*7*3*13$</span>. And <span class="math-container">$1036 = 2^2*7*37$</span> and <span class="math-container">$\gcd(1092,1036)=28$</span>. So <span class="math-container">$3^{28} \equiv 1 \pmod {1093}$</span>. </p>
<p>Its possible that some factors of <span class="math-container">$28$</span> are such powers and simple testing finds <span class="math-container">$3^7 = 2187 = 2*1093 + 1$</span> so <span class="math-container">$1093|3^7 -1$</span>.</p>
<p>So <span class="math-container">$1093|3^{7m}-1$</span>.</p>
<p>If <span class="math-container">$3^k\equiv 1 \pmod {1093}$</span> then <span class="math-container">$3^{\gcd(7,k)}\equiv 1 \pmod {1093}$</span> but if <span class="math-container">$k $</span> is not a multiple of <span class="math-container">$7$</span> then <span class="math-container">$\gcd(7,k) = 1$</span> but <span class="math-container">$3^1 \not \equiv 1 \pmod{1093}$</span>.</p>
<p>So <span class="math-container">$k = 7m$</span> are all and the only solutions to <span class="math-container">$1093|3^k -1$</span>.</p>
|
3,714,418 | <p>I need to calculate the angle between two 3D vectors. There are plenty of examples available of how to do that but the result is always in the range <span class="math-container">$0-\pi$</span>. I need a result in the range <span class="math-container">$\pi-2\pi$</span>.</p>
<p>Let's say that <span class="math-container">$\vec x$</span> is a vector in the positive x-direction and <span class="math-container">$\vec y$</span> is a vector in the positive y-direction and <span class="math-container">$\vec z$</span> is a reference vector in the positive z-direction. <span class="math-container">$\vec z$</span> is perpendicular to both <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span>. Would it then be possible to calculate the angle between <span class="math-container">$\vec x$</span> and <span class="math-container">$\vec y$</span> and get a result in the range <span class="math-container">$\pi-2\pi$</span>? </p>
<p>The angle value should be measured counter clockwise. I have not been able to figure out how to do that.
I am no math guru but I have basic understanding of vectors at least.
Thank you very much for the help! </p>
| zwim | 399,263 | <p><span class="math-container">$\begin{align}A-(B-A)&=A\cap \overline{(B-A)}=A\cap(\overline{B\cap\overline{A}})=A\cap(A\cup \overline{B})=(A\cap A)\cup(A\cap \overline{B})\\&=A\cup(A-B)\\&=A\end{align}$</span></p>
<p>Thus it is clear from second line that <span class="math-container">$(A-B)$</span> is a subset of this set.</p>
<p>Note that in the end, the conclusion is no more than <span class="math-container">$A-B\subset A$</span> since the above set is just <span class="math-container">$A$</span>.</p>
|
689,591 | <p>Today at class, my teacher stated the following proposition saying it is obvious:</p>
<p>Let $x_0 \in U \subset \mathbb{R}^d$, $U$ open, and $f: U \to \mathbb{R}^m$ differentiable at $x_0$, then for any $v \in \mathbb{R}^d$ we have</p>
<p>$$ D_v f(x_0) = D_f (x_0)(v) $$</p>
<p>It is not obvious to me. Can someone explain to me why this is true? Thanks a lot.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Let us first remember what is the meaning of each symbol:</p>
<ul>
<li><p>$D_\nu f(x_0)$ is the derivative of $f$ at $x=x_0$ in the $\nu-$direction, i.e.,
$$
D_\nu f(x_0)=\lim_{h\to 0}\frac{f(x_0+hv)-f(x_0)}{h}. \tag*{(*)}
$$</p></li>
<li><p>On the other hand, we have the following the definition:</p></li>
</ul>
<p><em>The function $f:U\to\mathbb R^m$, where $U\subset\mathbb R^d$ open and $x_0\in U$, is said to be differentiable at $x=x_0$ if there exists a linear function $A: \mathbb R^d\to\mathbb R^m$, such that
$$
\lim_{\|w\|\to 0}\frac{\|f(x_0+w)-f(x_0)-Aw\|}{\|w\|}=0. \tag{1}
$$
This unique linear functional $A$ in $(1)$ is called the differential of $f$ at $x_0$, and is denoted by $Df(x_0)$.</em></p>
<p>If $\nu\in \mathbb R^d\smallsetminus\{0\}$ and $w=h\nu$, and $f$ is differentiable at $x=x_0$, then $(1)$ implies that
$$
\lim_{\|h\|\to 0}\frac{\|f(x_0+h\nu)-f(x_0)-hDf(x_0)\nu\|}{\lvert h\rvert}=0,
$$
or, dividing numerator and denominator by $h$
$$
\lim_{\|h\|\to 0}\,\Big\|\,\frac{f(x_0+h\nu)-f(x_0)}{h}-Df(x_0)\nu\,\Big\|. \tag{**}
$$
Now $(*)$ and $(**)$ implies that
$$
D_\nu f(x_0)=Df(x_0)\nu.
$$</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Douglas S. Stones | 2,264 | <p>Simplify (x-a)(x-b)...(x-z).</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| matthias beck | 3,193 | <p>(I learned this problem from Persi Diaconis.)
A deck of $n$ different cards is shuffled and laid on
the table by your left hand, face down. An identical deck of cards,
independently shuffled, is laid at your right hand, also face down. You
start turning up cards at the same rate with both hands, first the top
card from both decks, then the next-to-top cards from both decks, and
so on. What is the probability that you will simultaneously turn up identical
cards from the two decks? What happens as $n \to \infty$? And does the answer for small $n$ (say, $n=7$) differ greatly from $n=52$?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Andrea Ferretti | 828 | <p>There is a square with seven monkeys on the floor and seven bananas on the top. Seven ladders go up the square, from one monkey to the banana over it, and the monkeys can climb them. Moreover there are some ropes which connect the ladders.</p>
<p>A monkey will go up towards the bananas, but whenever it meets a rope it cannot resist the temptation to stray and hang on it. Prove that every monkey will reach a banana, no matter the configuration of ropes.</p>
<p>There are at least two different solutions to this.</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Malik Younsi | 1,162 | <p>I really like the following puzzle, called the blue-eyed islanders problem, taken from Professor Tao's blog :</p>
<p>"There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).</p>
<p>Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).</p>
<p>One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.</p>
<p>One evening, he addresses the entire tribe to thank them for their hospitality.</p>
<p>However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.</p>
<p>What effect, if anything, does this faux pas have on the tribe?"</p>
<p>For those of you interested, there is a huge discussion of the problem at <a href="http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/">http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/</a></p>
<p>Malik</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Richard Dore | 27 | <p>A certain rectangle can be covered by 25 coins of diameter 2. Can it always be covered with 100 coins of diameter 1?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Tony Huynh | 2,233 | <p>You have 1000 bottles of wine. Exactly one of the bottles contains a deadly poison, but you don't know which one. The killing time of the poison varies from person to person, but death is imminent in at most $t$ hours after ingestion. You are allowed to use 10 notorious criminals as poison fodder (they are on death row). How much time do you need to correctly determine the poisoned bottle?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| Gjergji Zaimi | 2,384 | <p>Here is another of my favorites: Player 1 thinks of a polynomial P with coefficients that are natural numbers. Player 2 has to guess this polynomial by asking only evaluations at natural numbers (so one can not ask for $P(\pi)$). How many questions does the second player need to ask to determine P?</p>
|
29,323 | <p>You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.</p>
<p>I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.</p>
<p>So: What are your favorite dinner conversation math puzzles?</p>
<p>I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.</p>
<p>One problem per answer.</p>
<p>If you post the answer, please obfuscate it with something like <a href="http://www.rot13.com/">rot13</a>. Don't spoil the fun for everyone else.</p>
| ACL | 10,696 | <p>When you watch yourself in a mirror, left and right are exchanged. But why aren't top and bottom?</p>
|
4,159,227 | <p>Given the problem of finding the Laplace transform of the function <span class="math-container">$$f(t)=t^ne^{at}$$</span> with <span class="math-container">$n\in\mathbb{N}$</span> and <span class="math-container">$a\in\mathbb{R}$</span>, I realize it can be shown the transform is <span class="math-container">$$\frac{n!}{(s-a)^{n+1}}$$</span> by more than one direct method. However, I'd like to show this using strong induction. I began the problem by showing that
<span class="math-container">\begin{align*}
\mathcal{L}\{te^{at}\}=\int_{0}^{\infty}te^{-(s-a)t}\,dt&=\frac{1}{(s-a)^2},\\
\mathcal{L}\{t^2e^{at}\}=\int_{0}^{\infty}t^2e^{-(s-a)t}\,dt&=\frac{2}{(s-a)^3},\text{ and}\\
\mathcal{L}\{t^3e^{at}\}=\int_{0}^{\infty}t^3e^{-(s-a)t}\,dt&=\frac{6}{(s-a)^4}.
\end{align*}</span>
I originally did this so I could personally see the pattern. Then, I began the proof as follows.</p>
<p><span class="math-container">$\textbf{Proof:}$</span> Let <span class="math-container">$P(n)$</span> be the statement
<span class="math-container">$$P(n): \mathcal{L}\{t^ne^{at}\}=\frac{n!}{(s-a)^{n+1}},$$</span>
for all <span class="math-container">$n\in\mathbb{N}.$</span> We have already shown <span class="math-container">$P(1), P(2), P(3)$</span> are true, so the base case has been proven.</p>
<p><em>Inductive Step:</em> Assume <span class="math-container">$P(k)$</span> is true for all <span class="math-container">$k$</span>. We must then show that <span class="math-container">$P(k)\implies P(k+1).$</span>
We see
<span class="math-container">\begin{align*}
\mathcal{L}\{t^{k+1}e^{at}\}&=\frac{(k+1)!}{(s-a)^{k+2}}\\
&=\frac{k+1}{s-a}\frac{k!}{(s-a)^{k+1}}\\
&=\frac{k+1}{s-a}\mathcal{L}\{t^{k}e^{at}\} \quad\text{(by the inductive step)}
\end{align*}</span>
At this point, I'm not really sure where to go. I'm pretty awful at proofs, so I assume I'm probably not going in the right direction after the inductive hypothesis.</p>
| user577215664 | 475,762 | <p>You can usethe integral definition of the Laplace transform:
<span class="math-container">$$\mathcal{L}\{t^{n+1}e^{at}\}=\int_0^{\infty}t^{n+1}e^{-t(s-a)} dt$$</span>
Integrate by part and use the reccurence for the second integral.
<span class="math-container">$$\mathcal{L}\{t^{n+1}e^{at}\}= \lim_{N\to\infty} \left[t^{n+1}\dfrac {e^{-t(s-a)} }{a-s}\right]_0^{N}- \dfrac {n+1}{a-s}\mathcal{L}\{t^{n}e^{at} \}$$</span></p>
|
1,194,440 | <p>How do I prove if the following formulas are consistent?</p>
<p>∀x$\neg$S(x,x)</p>
<p>∃x P(x)</p>
<p>∀x∃y S(x,y)</p>
<p>∀x(P(x)$\to$∃y S(y,x))</p>
<p>I think I proved part of it...</p>
<p>There is at least one value for x such that P(x) is <em>True</em></p>
<p>This means that for ∀x(P(x)$\to$∃y S(y,x)) ... the P(x) part is <em>False</em> because we do not know if it is <em>True</em> for all x.</p>
<p>I am not even sure if my approach is correct, can someone please explain to me how to prove if they are consistent? Unfortunately my teacher does not provide good resources for me to figure this out on my own, Thanks.</p>
| Graham Kemp | 135,106 | <blockquote>
<p>This means that for $∀x(P(x)→ ∃y S(y,x))$ ... the $P(x)$ part is False because we do not know if it is True for all x.</p>
</blockquote>
<p>No. This statement does not require $P(x)$ to be universally true, only that <em>whenever</em> it is true, <em>then</em> so too is $\exists y\;S(y,x)$.</p>
<p>This is called "restricted universal quantification". </p>
<p>$∀x(P(x)→ ∃y S(y,x))\;$ reads: "For every $x$ which satisfies $P(x)$, there is some $y$ which satisfies $S(y,x)$."</p>
|
2,623,621 | <p>I'm struggling to find the derivative of this function :</p>
<p>$$y=\bigg\lfloor{\arccos\left(\frac{1}{\tan\left(\sqrt{\arcsin x}\right)}\right)}\bigg\rfloor$$</p>
<p>I've been told it should be 0 but how can I find that ?!</p>
| Arnaud Mortier | 480,423 | <p><em>This is a partial solution that works in spaces where closure coincides with sequential closure, such as Frechet spaces.</em> </p>
<p>Pick $$x\in \overline{\bigcup_{i\in I} S_i}$$</p>
<p>Then because $\mathcal{S}$ is locally finite there is a neighbourhood $x\in U$ such that $U$ has a non-empty intersection with only finitely many $S_i$'s. </p>
<p>Now pick a sequence $x_k\in \bigcup_{i\in I} S_i$ that converges to $x$. You can choose the $x_k$ to lie in $U$ because it is a neighbourhood of $x$ (or in any case they will all lie in $U$ except finitely many of them). </p>
<p>Since there are infinitely many $x_k$ but finitely many $S_i$ where they can possibly lie, by the pigeonhole principle, there is a subsequence that lies entirely in one and the same $S_i$. It follows that $x$ lies in the closure of that particular $S_i$.</p>
<p>Therefore we have proved $$\overline{\bigcup_{i\in I} S_i}\subset \bigcup_{i\in I} \overline{S_i}$$</p>
<p>The other direction is trivial.</p>
|
1,534,246 | <p>I'm trying to simplify this boolean expression:</p>
<p>$$(AB)+(A'C)+(BC)$$</p>
<p>I'm told by every calculator online that this would be logically equivalent:</p>
<p>$(AB)+(A'C)$</p>
<p>But so far, following the rules of boolean algebra, the best that I could get to was this: </p>
<p>$(B+A')(B+C)(A+C)$</p>
<p>All of the above are logically equivalent (I've made a truth table for each) but I don't understand what steps am I missing trying to simplify the expression.</p>
<p>I also couldn't find an "expression simplifier" tool online that could show me the steps that I'm missing.</p>
<p>Help / directions to go to would be much appreciated, thanks in advance.</p>
| Kevin Zakka | 235,367 | <p>This is known as the consensus rule.</p>
<p>$BC$ is the <em>"consensus"</em> term of $AB$ and $A'C$ and can be removed or added to the boolean expression.</p>
<p><strong>Derivation</strong>
$$(X.Y)+(X'.Z)+(Y.Z)$$
$$=(X.Y)+(X'.Z)+(X+X')(Y.Z)$$
$$=(X.Y)+(X'.Z)+(X.Y.Z)+ (X'.Y.Z)$$
$$=(X.Y)+(X.Y.Z)+(X'.Z)+(X'.Y.Z)$$
$$=X.Y(1+Z) + X'.Z(1+Y)$$
$$=X.Y + X'.Z$$</p>
<p>Thus $$(X.Y)+(X'.Z)+(Y.Z)= X.Y + X'.Z$$</p>
|
1,534,246 | <p>I'm trying to simplify this boolean expression:</p>
<p>$$(AB)+(A'C)+(BC)$$</p>
<p>I'm told by every calculator online that this would be logically equivalent:</p>
<p>$(AB)+(A'C)$</p>
<p>But so far, following the rules of boolean algebra, the best that I could get to was this: </p>
<p>$(B+A')(B+C)(A+C)$</p>
<p>All of the above are logically equivalent (I've made a truth table for each) but I don't understand what steps am I missing trying to simplify the expression.</p>
<p>I also couldn't find an "expression simplifier" tool online that could show me the steps that I'm missing.</p>
<p>Help / directions to go to would be much appreciated, thanks in advance.</p>
| Brian M. Scott | 12,042 | <p>$$\begin{align*}
AB+A'C+BC&=AB(C+C')+A'C(B+B')+BC\\
&=ABC+ABC'+A'BC+A'B'C+BC\\
&=(A+A')BC+BC+ABC'+A'B'C\\
&=BC+BC+ABC'+A'B'C\\
&=BC+ABC'+A'B'C\\
&=(A+A')BC+ABC'+A'B'C\\
&=ABC+A'BC+ABC'+A'B'C\\
&=AB(C+C')+A'(B+B')C\\
&=AB+A'C
\end{align*}$$</p>
|
3,221,547 | <p>I'm trying to prove that there are no simple groups of order <span class="math-container">$240$</span>. So let <span class="math-container">$G$</span> be a simple group such that <span class="math-container">$|G|=240=2^4\cdot3\cdot5$</span>. Then
<span class="math-container">$$n_2\in\{1,3,5,15\}\quad n_3\in\{1,4,10,40\}\quad n_5\in\{1,6,16\}$$</span>
After some reasoning, we get <span class="math-container">$n_2=15$</span> and two mutually exclusive cases:
<li> If for every two distinct <span class="math-container">$P,Q\in\text{Sylow}_2(G)$</span> we have <span class="math-container">$P\cap Q=1$</span>, then, by counting elements of order a power of <span class="math-container">$2$</span> we get only one <span class="math-container">$3$</span>-Sylow and only one <span class="math-container">$5$</span>-Sylow (so <span class="math-container">$G$</span> is not simple)
<li> If there are tow distinct <span class="math-container">$P,Q\in\text{Sylow}_2(G)$</span> such that <span class="math-container">$P\cap Q$</span> is not trivial then <span class="math-container">$|P\cap Q|\in\{2,4,8\}$</span> and I don't know how to proceed from here. Trying to prove <span class="math-container">$P\cap Q\unlhd G$</span> fails</p>
<p></p>
| David Leep | 1,025,208 | <p>There is a mistake, or at least a gap, in the provided answer. The case
<span class="math-container">$n_3 = 16$</span> is not considered. In addition, the case <span class="math-container">$n_3 = 40$</span> doesn't seem to be dealt with in the given link. Here is a proof that there are no simple groups of order <span class="math-container">$240$</span>.</p>
<p>Let <span class="math-container">$G$</span> be a simple group of order <span class="math-container">$240$</span>.
We have <span class="math-container">$n_5 \in \{1,6,16\}$</span>. Since <span class="math-container">$G$</span> is simple, we have <span class="math-container">$n_5 \ne 1$</span>, and
<span class="math-container">$n_5 \ne 6$</span> because otherwise <span class="math-container">$G$</span> would be isomorphic to a subgroup of
<span class="math-container">$A_6$</span> but <span class="math-container">$240 \nmid 360$</span>. Thus <span class="math-container">$n_5 = 16$</span> and so <span class="math-container">$G$</span> is isomorphic to a
subgroup of <span class="math-container">$A_{16}$</span>.</p>
<p>Let <span class="math-container">$P_1, \ldots, P_{16}$</span> be the Sylow <span class="math-container">$5$</span>-subgroups of <span class="math-container">$G$</span>. Then
<span class="math-container">$|N_G(P_i)| = 15$</span>, <span class="math-container">$1 \le i \le 16$</span>.
Every group of order <span class="math-container">$15$</span> is a cyclic group and thus each
<span class="math-container">$N_G(P_i)$</span> contains a unique subgroup of order <span class="math-container">$5$</span>, namely <span class="math-container">$P_i$</span>.
It follows that each subgroup in <span class="math-container">$G$</span> of order <span class="math-container">$5$</span> lies in exactly one subgroup
<span class="math-container">$N_G(P_i)$</span>.<br />
Let <span class="math-container">$N_G(P_i) = \langle g_i \rangle$</span>, where <span class="math-container">$g_i$</span> has order <span class="math-container">$15$</span>.
Then <span class="math-container">$g_i^3$</span> has order <span class="math-container">$5$</span> and fixes <span class="math-container">$P_i$</span>, but fixes no other Sylow <span class="math-container">$5$</span>-subgroup.
Thus <span class="math-container">$g_i$</span> fixes <span class="math-container">$P_i$</span> but fixes no other Sylow <span class="math-container">$5$</span>-subgroup <span class="math-container">$P_j$</span>, <span class="math-container">$i \ne j$</span>.
Therefore, <span class="math-container">$g_i$</span> is a <span class="math-container">$15$</span>-cycle because the only element of
<span class="math-container">$A_{16}$</span> that fixes exactly one element and has order <span class="math-container">$15$</span> is a <span class="math-container">$15$</span>-cycle.</p>
<p>It follows that <span class="math-container">$|N_G(P_i) \cap N_G(P_j)| = 1$</span> for all <span class="math-container">$i \ne j$</span> because the non-identity elements of <span class="math-container">$N_G(P_i)$</span> fix <span class="math-container">$P_i$</span> and move every other <span class="math-container">$P_k$</span>.
This gives <span class="math-container">$16 \times 14 = 224$</span> distinct elements of orders <span class="math-container">$3, 5, 15$</span> in <span class="math-container">$G$</span>.
This leaves only <span class="math-container">$16$</span> elements not accounted for, and thus there must be a
unique Sylow <span class="math-container">$2$</span>-subgroup of <span class="math-container">$G$</span>, a contradiction because <span class="math-container">$G$</span> is simple.</p>
|
78,946 | <p>I want to find min of the function
$$\frac{1}{\sqrt{2
x^2+\left(3+\sqrt{3}\right)
x+3}}+\frac{1}{\sqrt{2
x^2+\left(3-\sqrt{3}\right)
x+3}}+\sqrt{\frac{1}{3}
\left(2 x^2+2 x+1\right)}.$$
I know, the exact value minimum is $\sqrt{3}$ at $x = 0$. With <em>Mathematica</em>, I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
NMinimize[A, {x}]
</code></pre>
<p>And I got </p>
<blockquote>
<p>{1.73205, {x -> -2.57345*10^-16}}</p>
</blockquote>
<p>When I tried </p>
<pre><code>A = 1/Sqrt[2 x^2 + (3 + Sqrt[3]) x + 3] +
1/Sqrt[2 x^2 + (3 - Sqrt[3]) x + 3] + Sqrt[(2 x^2 + 2 x + 1)/3]
Minimize[A, {x}]
</code></pre>
<p>my computer ran about 20 minutes and I did not got the result. How can I get the exact value minimum of the given function?</p>
| kirma | 3,056 | <p>In this case you can find minimum by comparing values at zero derivates:</p>
<pre><code>TakeSmallestBy[{A /. #, #} & /@ Solve[D[A, x] == 0, x, Reals], First, 1]
</code></pre>
<blockquote>
<p>{{Sqrt[3], {x -> 0}}}</p>
</blockquote>
<p><code>TakeSmallestBy</code> is a v10.1 function similar to <code>MinimalBy</code>, but performs numerical comparisons.</p>
|
3,298,282 | <p>Find the number of terms in the expansion <span class="math-container">$(1+a^3+a^{-3})^{100}$</span></p>
<p>I used the concept <span class="math-container">$a^3+a^{-3}=T$</span>, while using this I have 101 terms, from <span class="math-container">$T^2$</span> to <span class="math-container">$T^{100}$</span> how do i find the number of terms that do not intersect</p>
| farruhota | 425,072 | <p>It is equivalent to finding the number of terms in:
<span class="math-container">$$(1+a^3+a^{-3})^{100}=a^{-300}(a^6+a^3+1)^{100} \Rightarrow (a^6+a^3+1)^{100},$$</span>
whose terms will have powers (since the signs are positive and exponents are multiples of <span class="math-container">$3$</span>):
<span class="math-container">$$0,3,6,...,597,600$$</span>
The number of terms of the arithmetic progression is:
<span class="math-container">$$n=\frac{a_n-a_1}{d}+1=\frac{600-0}{3}+1=201.$$</span></p>
|
2,742,497 | <p>When two events $A$ and $B$ are independent, $P(A)\cdot P(B)$ equals the probability of both events occurring together.</p>
<p>$$P(A|B) = \frac{P(A\cap B)}{P(B)}.$$</p>
<p>If $A$ and $B$ are not independent, $P(A\cap B)$ is either greater than or less than $P(A)\cdot P(B)$.</p>
<ol>
<li><p>I want to understand the intuitive meaning of $P(A)\cdot P(B)$ for non independent event. What does it signify?</p></li>
<li><p>Also, what is intuition behind </p>
<p>If $P(A|B) > P(A)$, then $P(B|A) > P(B)$.</p>
<p>I understand it is direct consequence of $P(A\cap B) > P(A)\cdot P(B)$, but it is not very intuitive to me.</p></li>
</ol>
| farruhota | 425,072 | <p>Consider the following events $M$={pass midterm exam}, $F$={pass final exam} and the probabilities:
$$P(M)=0.8, P(F)=0.7,P(M\cap F)=0.6.$$
Then:
$$P(M|F)=\frac{P(M\cap F)}{P(F)}=\frac{0.6}{0.7}=\frac67>0.8=P(M); \\
P(F|M)=\frac{P(F\cap M)}{P(M)}=\frac{0.6}{0.8}=\frac34>0.7=P(F).$$
Hence, passing one exam increases the probability of passing the other exam, which is a consequence of a positive correlation. </p>
|
1,930,933 | <blockquote>
<p>Does there exist an $n \in \mathbb{N}$ greater than $1$ such that $\sqrt[n]{n!}$ is an integer?</p>
</blockquote>
<p>The expression seems to be increasing, so I was wondering if it is ever an integer. How could we prove that or what is the smallest value where it is an integer?</p>
| Winther | 147,873 | <p>If $\sqrt[n]{n!} = k \in \mathbb{N}$ then $n! = k^n$. When $n\geq 2$ we have $2\mid n!$ so we must also have $2\mid k$ which means that we can write $k = 2^{m} \ell$ for some integers $m$ and $\ell$. This again means that</p>
<p>$$n! = 2^{mn}\ell^n \implies 2^{mn} \mid n!$$</p>
<p>so the power of two that divides $n!$ is $mn$ which is greater or equal to $n$. On the other hand the power of two that divides $n!$ can be computed as</p>
<p>$$\left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{4}\right\rfloor + \left\lfloor\frac{n}{8}\right\rfloor + \ldots$$</p>
<p>This expression is less than $\frac{n}{2} + \frac{n}{4} + \frac{n}{8} +\ldots = n$ which gives us a contradiction.</p>
|
482,010 | <p>Sorry for my bad english And also my bad math literature in asking my first question.</p>
<p>Assume we have 15 glass full of water.</p>
<p>I need to count the number of possible ways which I can empty 5 glass in a way that no 2 empty glass remain in sequence.</p>
| Accidental Statistician | 88,206 | <p>If the glasses are in a line, say we draw a full glass as a star $*$, and an empty glass as a vertical bar $|$. Then a possible way to empty five glasses is drawn as a line of five stars and ten bars, where the bars are not next to each other. One such way could be drawn as</p>
<p>$|**|**|**|**|**$</p>
<p>This is similar to the problem discussed as Theorem One in <a href="http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29" rel="nofollow">Stars and bars notation</a>, except the bars can be placed on the ends as well. We can solve the problem by considering four cases:</p>
<ol>
<li><p>There is no bar on either the left end or the right end. This is then the same as Theorem One in the link, with $k=10$ and $n=6$, so there are $\binom{9}{5}$ of these.</p></li>
<li><p>There is a bar on the left end. Then ignore this bar, and the rest of the line follows Theorem One with $k=10$ and $n=5$, so there are $\binom{9}{4}$ of these.</p></li>
<li><p>There is a bar on the right end. Same as case 2, $\binom{9}{4}$ of these.</p></li>
<li><p>There is a bar on both ends. Ignore them both, and now $n=4$, so $\binom{9}{3}$ of these.</p></li>
</ol>
<p>The total number of arrangements is then $\binom{9}{5} + 2\binom{9}{4} + \binom{9}{3} = \binom{10}{5} + \binom{10}{4} = \binom{11}{5}.$</p>
|
1,030,050 | <p>Assignment:</p>
<blockquote>
<p>Let $f$ be Lebesgue - measurable and $a,b \in \mathbb{R}$ with the property:
$$\frac{1}{\lambda(M)} \cdot \int_Mf\ d\lambda \in [a,b]$$
for all Lebesgue - measurable sets $M \subset \mathbb{R}^n$ with $0 < \lambda(M) < \infty$.</p>
<p>Show that: $f(x) \in [a,b]$ almost everywhere.</p>
</blockquote>
<p>So I need to show that the set $N$ defined as $N:= \{x\in\mathbb{R}^n; \ f(x) \notin [a,b]\}$ has a measure of zero. However, I cannot really see how to go from there and how the property given could help me.</p>
<p>I'd appreciate any help.</p>
| RHP | 13,463 | <p>Suppose $N = \{x:f(x)\notin[a,b]\}$ is not a null set.</p>
<p>Then at least one of $N_1 = \{x:f(x)<a\}$ or $N_2 = \{x:f(x)>b\}$ is not a null set since $N = N_1\cup N_2$</p>
<p>If $N_1$ is not a null set, then (since $f(x)<a$ on $N_1$)</p>
<p>$$\frac{1}{\lambda(N_1)}\int_{N_1}f~d\lambda<\frac{1}{\lambda(N_1)}\int_{N_1}a~d\lambda=a$$</p>
<p>If $N_2$ is not a null set, then(since $f(x)>b$ on $N_2$)</p>
<p>$$\frac{1}{\lambda(N_2)}\int_{N_2}f~d\lambda>\frac{1}{\lambda(N_2)}\int_{N_2}b~d\lambda=b$$</p>
<p>Either way you contradict your assumption.</p>
|
3,353,483 | <p>I have just begun studying finite fields today, and it is clear in GF(2) why 1+1=0. (I just show that 1+1 can't equal 1, or 1=0, which contradicts an axiom that states that 1 is not 0).</p>
<p>If we interpreted these symbols "1", "+", "1", "0" as we would in primary school, clearly this breaks arithmetic rules in Real numbers.</p>
<p>Given that, I have lost all confidence in how arithmetic can be applied in a finite field. <strong>How do I even know how to do basic arithmetic on GF(n) where n is prime?</strong>
For example, for GF(7), how do I even know that 4+1=5?
Can anyone show with just the 9 axioms of finite fields that 4+1=5?</p>
<p>Axioms: associativity of addition, additive identity, additive inverse, commutatitivity of addition, associativity of multiplication, multiplicative inverse, commutatitivity of mulitplication, distributive law</p>
| Lubin | 17,760 | <p>There’s really no big deal. You are really working in integers (<em>never</em> in the reals), and whenever you get an answer that’s too big or too negative, subtract or add a multiple of your prime number <span class="math-container">$n$</span> (conventionally, we call the modulus <span class="math-container">$p$</span> in these cases). So, if you’re working modulo <span class="math-container">$7$</span>, to add <span class="math-container">$4+1$</span>, you do it in the integers <span class="math-container">$\Bbb Z$</span>. Answer <span class="math-container">$5$</span>. Is it at least <span class="math-container">$7$</span>? No, so just leave it be. But to add <span class="math-container">$4+5$</span>, the integer sum is <span class="math-container">$9$</span>, so you subtract <span class="math-container">$7$</span> to get <span class="math-container">$2$</span>, and in the system of integers modulo <span class="math-container">$7$</span>, you have <span class="math-container">$4+5=2$</span>. A standard way of writing this is <span class="math-container">$4+5\equiv2\pmod7$</span>, which you read, “four plus five is congruent to two modulo seven”. This notation and terminology goes back to Gauss (1801), maybe even farther.</p>
|
3,353,483 | <p>I have just begun studying finite fields today, and it is clear in GF(2) why 1+1=0. (I just show that 1+1 can't equal 1, or 1=0, which contradicts an axiom that states that 1 is not 0).</p>
<p>If we interpreted these symbols "1", "+", "1", "0" as we would in primary school, clearly this breaks arithmetic rules in Real numbers.</p>
<p>Given that, I have lost all confidence in how arithmetic can be applied in a finite field. <strong>How do I even know how to do basic arithmetic on GF(n) where n is prime?</strong>
For example, for GF(7), how do I even know that 4+1=5?
Can anyone show with just the 9 axioms of finite fields that 4+1=5?</p>
<p>Axioms: associativity of addition, additive identity, additive inverse, commutatitivity of addition, associativity of multiplication, multiplicative inverse, commutatitivity of mulitplication, distributive law</p>
| fleablood | 280,126 | <p>Beneath it all... we count.</p>
<p>And it doesn't matter what we count; we just count.</p>
<p>So we define <span class="math-container">$4$</span> as what we get if we add <span class="math-container">$1$</span> a "tick-tick-tick-tick" number of times. And we define <span class="math-container">$5$</span> as what we get if we add <span class="math-container">$1$</span> a "tick-tick-tick-tick-tick" numbers of times.</p>
<p>Well every time we add <span class="math-container">$4$</span> to <span class="math-container">$5$</span> we <em>always</em> count <span class="math-container">$9$</span> times. THat's because if you put "tick-tick-tick-tick-tick" and then put another "tick-tick-tick-tick" you get "tick-tick-tick-tick-tick-tick-tick-tick-tick" which we define as what <span class="math-container">$9$</span> is.</p>
<p>It's just that in the finite field <span class="math-container">$GF(2)$</span> you know that <span class="math-container">$1+1 = 0$</span>.</p>
<p>So you count</p>
<p>1 tick <span class="math-container">$\to 1$</span></p>
<p>2 tick <span class="math-container">$\to 1+1 = 0$</span></p>
<p>3 tick <span class="math-container">$\to 0+1 = 1$</span></p>
<p>4 tick <span class="math-container">$\to 1+1 = 0$</span>.</p>
<p>5 tick <span class="math-container">$\to 0+1 = 1$</span>.</p>
<p>Okay that was "tick-tick-tick-tick-tick". Now to add "tick-tick-tick-tick</p>
<p>1 tick <span class="math-container">$\to 1+1 =0$</span></p>
<p>2 tick <span class="math-container">$\to 0+1 = 1$</span></p>
<p>3 tick <span class="math-container">$\to 1+1 = 0$</span></p>
<p>4 tick <span class="math-container">$\to 0+1 = 1$</span>.</p>
<p>So <span class="math-container">$5 +4$</span> <em>still</em> equals <span class="math-container">$9$</span>. We just have <span class="math-container">$5 = 1$</span> and <span class="math-container">$4 = 0$</span> and <span class="math-container">$9=1$</span>.</p>
<p>The only rule we've changed is "adding one gives us a new number we never had before". That rule tells us <span class="math-container">$9\ne 1$</span> but without it <span class="math-container">$9=1$</span> is perfectly acceptable.</p>
|
2,779,841 | <p>I'm wondering how the following recursion for <span class="math-container">$D(n)$</span>, the derangement of <span class="math-container">$[n]$</span> distinct elements, can be proved</p>
<p><span class="math-container">$$D(n) = nD(n-1) + (-1)^n$$</span></p>
<hr>
<p>I see the argument for </p>
<p><span class="math-container">$$D(n) = (n-1)(D(n-1) + D(n-2))$$</span></p>
<p>where we consider the two cases for when an element, say <span class="math-container">$a$</span>, is sent to the "position" of any of the other <span class="math-container">$n-1$</span> element, say <span class="math-container">$b$</span>. We then have two cases, first when <span class="math-container">$b$</span> is sent to <span class="math-container">$a$</span>, hence the <span class="math-container">$D(n-2)$</span>. Second when <span class="math-container">$b$</span> is sent to somewhere else, hence <span class="math-container">$D(n-1)$</span>.</p>
<p>But this logic doesn't really work for <span class="math-container">$D(n) = nD(n-1) + (-1)^n$</span>, can someone provide a proof?</p>
| sku | 341,324 | <p>Algebraic proof:</p>
<p>$$D_{n+1} = nD_n + nD_{n-1} = (n+1)D_n - D_n + nD_{n-1}$$</p>
<p>We want to prove that $nD_{n-1} - D_n = (-1)^n$</p>
<p>$$nD_{n-1} - D_n = (n-1)D_{n-1} + D_{n-1} - ((n-1)D_{n-1} + (n-1)D_{n-2}) = D_{n-1} - (n-1)D_{n-2}$$</p>
<p>$$D_{n-1} - (n-1)D_{n-2} = D_{n-1} - (n-2)D_{n-2} - D_{n-2}$$
$$=(n-2)D_{n-2} + (n-2)D_{n-3} -(n-2)D_{n-2} - D_{n-2}$$
$$ = (n-2)D_{n-3} - D_{n-2}$$</p>
<p>So we have: $$nD_{n-1} - D_n = D_{n-1} - (n-1)D_{n-2} = (n-2)D_{n-3} - D_{n-2}$$</p>
<p>If we start with $n$ even and using the second equality, we will end up with</p>
<p>$$4D_3 - D_4 = 2D_1 - D_2 = -1$$</p>
<p>If we start with $n$ odd and using the first equality, we will end up with</p>
<p>$$3D_2 - D_3 = D_2 - 2D_1 = +1$$</p>
|
3,361,489 | <p><strong>Question:</strong></p>
<p>Do there exist functions <span class="math-container">$f$</span> and <span class="math-container">$g$</span> such that
<span class="math-container">$$\lim_{x \to c} f(x) = 1 \text{ and } \lim_{x \to c} f(x) g(x) - g(x) \neq 0 \, ?$$</span>
(Allowing, of course, for <span class="math-container">$\lim_{x \to c}$</span> g(x) to not exist.)</p>
<p><strong>Context:</strong></p>
<p>I am thinking about the limit property that <span class="math-container">$\lim_{x \to c} f(x) \cdot g(x) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)$</span>.</p>
<p>My understanding is that for this to be guaranteed to hold, we need both limits on the RHS to exist. Indeed, I am familiar with examples for which the two limits on the RHS diverge while the limit on the LHS exists (like in <a href="https://math.stackexchange.com/questions/513822/can-the-limit-of-a-product-exist-if-neither-of-its-factors-exist">this post</a>), as well as examples like <span class="math-container">$f(x) = 1/x, g(x) = x$</span> for which the LHS exists but one of the limits on the RHS is zero and the other diverges.</p>
<p>If, however, only one of the limits on the RHS diverges but the other exists <em>and is nonzero</em>, will we ever run into trouble by applying this limit property? In some sense, can we modify the requirement that BOTH limits on the RHS exist to the requirement that AT LEAST ONE of the limits on the RHS exists and is nonzero?</p>
| Bernard | 202,857 | <p><em>Counter-example</em>:</p>
<p><span class="math-container">$\lim_{x\to 0}\cos x=1$</span>, and
<span class="math-container">$$\lim_{x\to0}\cos x\,\frac1{x^2}-\frac1{x^2}=\lim_{x\to 0}\frac{\cos x-1}{x^2}=-\frac12.$$</span></p>
|
434,614 | <p>Is there a name for a type of grid you might find in Battleship? Where coordinates don't relate to points on a grid but rather the squares themselves?</p>
| Patrick Hew | 190,548 | <p><em>Cell-centered grid</em> may be a slightly better name, for when the coordinates refer to the cell. By contrast in a <em>node-centered grid</em>, the coordinates refer to the points on the grid. </p>
<p>The distinction is important in applications such as <a href="https://au.mathworks.com/help/images/image-coordinate-systems.html" rel="nofollow noreferrer">digital image processing</a>, <a href="https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20110002899.pdf" rel="nofollow noreferrer">computational fluid dynamics</a> and data sets for <a href="https://ngdc.noaa.gov/mgg/global/gridregistration.html" rel="nofollow noreferrer">topography and bathymetry</a>.</p>
|
1,103,478 | <p>$ r = 2\cos(\theta)$ has the graph<img src="https://i.stack.imgur.com/yvLb1.png" alt="enter image description here"></p>
<p>I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>whereas this one does:</p>
<p>$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$</p>
<p>Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?</p>
| abel | 9,252 | <p>the limit is from $0$ to $\pi$ not $0$ to $2\pi$ if you do the latter, you will go around the circle twice.</p>
<p>at $\theta = 0$ you are $(2,0)$ at $\theta = \pi/3$ you are at $x = 1, y = 1$ at $\theta = \pi/2$ you are at $(0,0)$ and then you move in the lower semicircle for the next $\pi/2$ to $\pi$ if you go farther in $\theta$ you will go back to where you started.</p>
<p>one has to be careful when setting up polar integral about the limits and how the curve is traced. you will lots of one inside other in cardiods and more complicated figures like $r = \cos(5 \theta),$ etc</p>
|
4,297,177 | <p>I have a polynomial:</p>
<p><span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span></p>
<p>I am supposed to show that it is <span class="math-container">$\Omega(n^2)$</span> and <span class="math-container">$O(n^2)$</span>. Once I prove those, it should be easy to show that it is also <span class="math-container">$\Theta(n^2)$</span> using the theorems given, but I'm having difficulties with <span class="math-container">$\Omega$</span> and <span class="math-container">$O$</span>. Mainly the subtraction part is throwing me off, as it has been hard finding a constant and <span class="math-container">$k$</span> value that works.</p>
<p>For showing <span class="math-container">$\Omega(n^2)$</span>, I started with <span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span> and <span class="math-container">$g(n) = n^2$</span>, and have been trying to show that <span class="math-container">$f(n) \geq C\cdot g(n)$</span>, but the subtraction has made it hard for me to choose a value for <span class="math-container">$C$</span> and <span class="math-container">$k$</span>. Can I just chose any value that makes it work?</p>
<p>Pretty similar for showing <span class="math-container">$O(n^2)$</span>, I started with <span class="math-container">$f(n)=\frac{1}{4}n^2-24n-16$</span> and <span class="math-container">$g(n) = n^2$</span>, and have been trying to show that <span class="math-container">$f(n) \leq C \cdot g(n)$</span>, and again, I'm having troubles knowing where to go from here.</p>
| Gary | 83,800 | <p>It is easy to show that
<span class="math-container">$$
a_n: n \mapsto \frac{{f(n)}}{{n^2 }}
$$</span>
is an increasing sequence and <span class="math-container">$\lim_{n \to +\infty}a_n=1/4$</span>. Thus, <span class="math-container">$a_n <1/4$</span> for all <span class="math-container">$n\geq 1$</span>. Inparticular <span class="math-container">$$f(n)<\frac{1}{4}n^2$$</span> for all <span class="math-container">$n\geq 1$</span>. Beacuse of the properties of <span class="math-container">$a_n$</span>, there has to be a <span class="math-container">$k$</span> such that <span class="math-container">$a_n>1/10$</span> say, for all <span class="math-container">$n\geq k$</span>. You can check that <span class="math-container">$k=161$</span> will do. Thus
<span class="math-container">$$f(n)>\frac{1}{10}n^2$$</span> for all <span class="math-container">$n\geq 161$</span>.</p>
|
3,443,226 | <blockquote>
<p>Prove that
<span class="math-container">$$
\int_{0}^{2\pi}\frac{d\theta}{(a+\cos\theta)^2}=\frac{2\pi a}{(a^2-1)^{\frac{3}{2}}}.
$$</span></p>
</blockquote>
<p>This is an exercise in Stein's <em>Complex Analysis</em>.</p>
<p>By letting <span class="math-container">$z=e^{i\theta}$</span>, we have
<span class="math-container">$$
\int_{0}^{2\pi}\frac{d\theta}{(a+\cos\theta)^2}=4\int_C\frac{dz}{iz(z+\bar z+2a)^2},
$$</span>
where <span class="math-container">$C$</span> is the unit circle.</p>
<p>I find that <span class="math-container">$z+\bar z+2a$</span> has no zero in the closure of the unit disc. So by the residual formula, I get
<span class="math-container">$$
\int_C\frac{dz}{iz(z+\bar z+2a)^2}=\frac{2\pi}{4a^{2}},
$$</span>
which is false.
Note that <span class="math-container">$\bar z$</span> is not a holomorphic function, so how to use the residual formula?</p>
| Jack D'Aurizio | 44,121 | <p>Without complex analysis, for any <span class="math-container">$a>1$</span> we have
<span class="math-container">$$ J(a)=\int_{0}^{2\pi}\frac{d\theta}{a+\cos\theta}=2\int_{0}^{\pi}\frac{d\theta}{a+\cos\theta}=2\int_{0}^{\pi/2}\left(\frac{1}{a+\cos\theta}+\frac{1}{a-\cos\theta}\right)\,d\theta $$</span>
or
<span class="math-container">$$ J(a) = 2\int_{0}^{\pi/2}\frac{2a d\theta}{a^2-\cos^2\theta}\stackrel{\theta\mapsto\arctan t}{=}2\int_{0}^{+\infty}\frac{2a\,dt}{a^2(1+t^2)-1}\stackrel{t\mapsto s/a}{=}4\int_{0}^{+\infty}\frac{ds}{s^2+(a^2-1)}=\frac{2\pi}{\sqrt{a^2-1}}. $$</span>
The given integral just equals <span class="math-container">$-J'(a)=\frac{2\pi a}{(a^2-1)^{3/2}}$</span>.</p>
|
710,045 | <p><a href="https://i.stack.imgur.com/7rJAB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7rJAB.png" alt="question" /></a></p>
<p>For (a) I have said</p>
<p>By MVT:</p>
<p><span class="math-container">$|f(x)-f(y)|\le K|x-y|$</span></p>
<p>Choose <span class="math-container">$\delta=\epsilon/K$</span></p>
<p><span class="math-container">$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|\le K|x-y| \le K\delta=\epsilon$</span></p>
<p>For (b) I have said</p>
<p><span class="math-container">$f'(x)=-4x^3/(1+x^4)^2$</span> so <span class="math-container">$|f'(x)| \le B$</span> where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.</p>
<p>Any help would be appreciated, thank you.</p>
| Tom Cooney | 8,580 | <p>In finite dimensions, all norms are equivalent.
(For example, see <a href="https://math.stackexchange.com/q/599824/8580">Any two norms on finite dimensional space are equivalent</a>.)</p>
<p>This means the following: if $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are the two norms on the space $X$, there are constants $a>0$ and $b>0$ such that
$$
a \Vert x \Vert_1 \leq \Vert x \Vert_2 \leq b \Vert x \Vert_1,
$$
for every $x \in X$. If you have this, then it is easy to show that $\Vert x_n -y \Vert_1 \to 0$ if and only if $\Vert x_n -y \Vert_2 \to 0$. So changing the norms will not change whether the space is complete (in the finite dimensional case).
So if you add this fact to your proof, then this proof works in the finite dimensional case. In particular, it works for $\mathbb F^2$. Without using this fact (or a similar one), this proof is incomplete.</p>
<p>If the space is infinite-dimensional, then you are right to say that changing the norm can cause problems. For example, consider $C[0,1]$ (the continuous functions on the interval $[0,1]$) with the following two norms:
$$
\Vert f \Vert_\infty = \sup_{x \in [0,1]} |f(x)|;
$$
$$
\Vert f \Vert_1=\int_0^1 |f(x)| \ \mathrm dx.
$$</p>
<p>The space $(C[0,1],\Vert \cdot \Vert_\infty)$ is a Banach space. (For example, see the answers to <a href="https://math.stackexchange.com/q/83830/8580">How to show that $C=C[0,1]$ is a Banach space</a>.) However the space $(C[0,1],\Vert \cdot \Vert_1)$ is not a Banach space (for example, <a href="https://math.stackexchange.com/q/152233/8580">Showing $(C[0,1], d_1)$ is not a complete metric space</a>). So here the proof would not work. Changing the norms does affect whether or not the space is complete.</p>
|
301,724 | <p>There's an example in my textbook about cancellation error that I'm not totally getting. It says that with a $5$ digit decimal arithmetic, $100001$ cannot be represented.</p>
<p>I think that's because when you try to represent it you get $1*10^5$, which is $100000$. However it goes on to say that when $100001$ is represented in this floating point system (when it's either chopped or rounded) it comes to $100000$.</p>
<p>If what I said above is correct, does $100001$ go to $100000$ because of the fact that it can only be represented like $1*10^5$? </p>
<p>If I'm completely off the mark, clarification would be great.</p>
| ncmathsadist | 4,154 | <p>That is correct. You have what is called type overflow. Want to be amused? If you have a C compiler run this program</p>
<pre><code>#include<stdio.h>
int main(void)
{
int x = 1;
while (x > 0)
{
x = x + 1;
}
printf("x = %d\n", x);
}
</code></pre>
<p>It's educational. This is in integer-world so it's simpler than floating point numbers, but these have an analogous problem: loss of precision. I recommend Forman Acton's <code>Real Computing Made Real</code> for a discussion of this issue.</p>
|
811,535 | <p>I want to show that every prime power $p^k$ that divides $\binom{2m}{m}$ is smaller than or equal to $2m$.</p>
<p>As a first step, I looked at
$$\binom{2m}{m}
= \frac{(2m)!}{(m!)^2}
= \frac{2m(2m-1) \ldots (m+2)(m+1)}{m!} \, .$$
Here I'm essentially stuck. I can apply the prime factorization to numerator and denominator, then I can cancel and I know that $p^k$ is left over in the numerator. But I cannot conclude $p^k \leq 2m$.</p>
<p>I feel that some vital ingredient is missing here, but I don't know what it is.</p>
<p>(Post edited with respect to the helpful comment.)</p>
| Chen Yihan | 763,624 | <p>Let <span class="math-container">$c_k$</span> be the integer such that <span class="math-container">$p^{c_k}\leq k\leq p^{c_k+1}$</span>. For integer <span class="math-container">$n$</span>, the number of elements in <span class="math-container">$[n]=\{1,2,\dots,n\}$</span> divided by <span class="math-container">$p^j$</span> is <span class="math-container">$\lfloor \frac{n}{p^j} \rfloor$</span> then <span class="math-container">$c_{n!}=\sum\limits_{i=1}^{+\infty}\lfloor \frac{n}{p^i} \rfloor$</span> thus <span class="math-container">$c_{\binom{2m}{m}}=c_{\frac{(2m)!}{(m!)^2}}=\sum\limits_{i=1}^{c_{2m}}\lfloor \frac{2m}{p^i} \rfloor-2(\sum\limits_{i=1}^{c_{2m}}\lfloor \frac{m}{p^i} \rfloor)=\sum\limits_{i=1}^{c_{2m}}(\lfloor \frac{2m}{p^i} \rfloor-2\lfloor \frac{m}{p^i} \rfloor)$</span> but <span class="math-container">$\lfloor \frac{2m}{p^i} \rfloor-2\lfloor \frac{m}{p^i} \rfloor=0$</span> or <span class="math-container">$1$</span> by Euclidean division, thus <span class="math-container">$c_{\binom{2m}{m}}\leq c_{2m}$</span>, <span class="math-container">$p^k\leq p^{c_{\binom{2m}{m}}}\leq p^{c_{2m}}\leq 2m$</span>.</p>
|
458,472 | <p>Just doing some revision for ODEs and came across this problem. Find the general solution to $$u''+4u=0.$$</p>
<p>So far I've applied the characteristic polynomial:
$$\begin{array}{r c l}
\lambda^2 +4 & = & 0 \\
\lambda^2 & = & -4 \\
\lambda & = & i\sqrt{4} \\
\lambda & = & 2i, -2i. \\
\end{array}$$</p>
<p>So the general solution should be:
$$\begin{array}{l c l}
u_H & = & Ae^{2ix}+Be^{-2ix} \\
& = & A(\cos{2x}+i\sin{2x})+B(\cos{(-2x)}+i\sin{(-2x)}) \\
& = & A\cos{2x}+iA\sin{2x}+B\cos{2x}-iB\sin{2x} \\
& = & (A+B)\cos{2x}+i(A-B)\sin{2x} \\
& = & C_1\cos{2x}+iC_2\sin{2x}. \\
\end{array}$$</p>
<p>The answers have $u=C_1\cos{2x}+C_2\sin{2x}$, and my question is "what happened to the $i$?" Does it drop out somewhere or is there an error in the answers? </p>
<p>Many thanks for a quick explanation/link to the appropriate website explaining this. :)</p>
| Amzoti | 38,839 | <p>If we wrote:</p>
<p>$$\tag 1 e^{a+ 2i} = e^{a t}(\cos 2t + i \sin 2t)v_1$$</p>
<p>where $v_1$ is the eigenvector, and for your problem $a = 0$.</p>
<p>When we expand $(\cos 2t + i \sin 2t)v_1$, we get an expression of the form:</p>
<p>$$(\alpha) + i(\beta)$$</p>
<p>Because we know that the real imaginary parts are both solutions, we have:</p>
<p>$$c_1(\alpha) + c_2(\beta).$$</p>
|
866,144 | <p>Prove that
$$1-2^{-x}\geq \frac{\sqrt 2}{2}\sin\left(\frac{\pi}{4} x\right)$$
for $x\in[0,1]$.
Any suggestions please?</p>
| Robert Israel | 8,508 | <p>Let $f(x) = 1 - 2^{-x} - \frac{\sqrt{2}}{2} \sin(\frac{\pi}{4} x)$. Note that
$f(0) = f(1) = 0$ and $f'(0) > 0$ while $f'(1) < 0$. </p>
<ol>
<li>Show that $f'' < 0$ for
$0 \le x \le 0.8$, and note that $f'(0.8) < 0$ and $f(0.8) > 0$.</li>
<li>Show that $f' < 0$ for $0.8 \le x \le 0.9$.</li>
<li>Show that $f' < 0$ for $0.9 \le x \le 1$.</li>
</ol>
|
158,483 | <p>When I started mathematica, this message popped up.</p>
<pre><code>Part::partw: Part 5 of PacletManager`Utils`Private`$taskData[2] does not exist.
</code></pre>
<p>Does anyone know what this is? My version is mma 11.2.</p>
| Dmitriy Sh. | 79,885 | <p>If you encounter this message at the MMA's startup, please check init.m located in %AppData%\Mathematica\FrontEnd. I solved the same problem (I use MMA 11.3) just by deleting entire text/commands in the %AppData%\Mathematica\FrontEnd\init.m file. This init.m is automatically initialized at every system startup and automatically updated by the MMA. Backup the file before deleting any text/commands in this initialization file.</p>
|
3,065,331 | <p>In <a href="https://rads.stackoverflow.com/amzn/click/0199208255" rel="nofollow noreferrer">Nonlinear Ordinary Differential Equations: An Introduction for Scientists and Engineers</a> one of the very first stated equations are, as in the title of the question,</p>
<p><span class="math-container">$$
\ddot{x} = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{1}{2} \dot{x}^2 \right).
$$</span></p>
<p>However, I'm having trouble seeing why this should be true. Could anyone clarify this? Thank you for your time in advance.</p>
| WarreG | 519,466 | <p>Try putting <span class="math-container">$$ \frac{dx}{dt} = y(x,t).$$</span>
Notice that <span class="math-container">$y$</span> is a function of <span class="math-container">$x$</span> <em>and</em> <span class="math-container">$t$</span>. You can then use the chain rule for <span class="math-container">$$\frac{dy(x,t)}{dt}.$$</span>
From this you should be able to find the expression you want.</p>
|
1,423,728 | <p>The definition of a limit in <a href="http://rads.stackoverflow.com/amzn/click/0321888545" rel="nofollow noreferrer">this book</a> stated like this </p>
<p><a href="https://i.stack.imgur.com/tKjaa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tKjaa.png" alt="enter image description here"></a></p>
<p>1) why we require ƒ(x) be defined on an open interval about $x_0$ ?</p>
<p>2) does the definition mean it is impossible to talk about limit of function of this sort?</p>
<p>$f(x) = \begin{cases} undefined & x \not\in \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases}$</p>
<p><strong>Edit</strong>: Since some similar answers to my first question is that we want to talk about two-sided limit, thus require ƒ(x) be defined on an open interval about $x_0$. However, these answers doesn't clear my intended question. Now if I restrict $x>=1$, then can we talk about the limit of that function , especially for $\displaystyle \lim_{x \to 1^{+}}f(x)$?</p>
<p>As for my second question, I found a more precise definition of limit in Courant's book Introduction_to_Calculus_and_Analysis stated like this
<a href="https://i.stack.imgur.com/8J8Wh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8J8Wh.png" alt="enter image description here"></a>
which doesn't require ƒ(x) to be defined on every point of the an open interval about $x_0$, so I think it is possible to talk about limit of that function.</p>
| Eric M. Schmidt | 48,235 | <p>It certainly would be possible to define limits in a more general situation, provided that you're careful enough. The given definition has the advantage that you don't have to worry about the function being undefined, provided $\delta$ is small enough. If you wanted to be more general, you could drop the open interval requirement and stipulate that the $\epsilon$-$\delta$ condition is only required to hold at points where the function is defined. This would allow talking about limits in the function you specified. However, it is essential that limits cannot be defined at "isolated" points. For instance, if a function $f$ is defined only for $|x| \ge 1$ and $x=0$, then $\displaystyle\lim_{x\to0} f(x)$ makes no sense, but our modified definition would allow the limit to be anything! So we have to exclude isolated points, which would require defining precisely what an isolated point is. At the level of elementary calculus, this is starting to become overkill.</p>
<p>Thomas' definition of limit, together with "one-sided limits", covers all cases that actually occur in elementary one-variable calculus. Thus, there is not much reason to make things more complicated.</p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Waldemar | 34,050 | <p>"(...)most academics – despite goodwill to contribute – still perceive major barriers to participation, which typically include a general lack of time to contribute, but also barriers of a technical, social and cultural nature. These encompass the lack of incentives from the perspective of a professional career, the poor recognition of one’s expertise within Wikipedia, the widespread perception of Wikipedia as a non-authoritative source. In combination with the apparent anomaly of collaborative – and often anonymous – authorship and the resulting fluidity of Wikipedia articles, these factors create an environment that significantly differs from the ones experts are accustomed to"
Quote from <a href="http://blog.wellcome.ac.uk/2011/02/25/wikipedia/">http://blog.wellcome.ac.uk/2011/02/25/wikipedia/</a></p>
|
131,435 | <p>Wikipedia is a widely used resource for mathematics. For example, there are hundreds of mathematics articles that average over 1000 page views per day. <a href="http://en.wikipedia.org/wiki/Wikipedia:WikiProject_Mathematics/Popular_pages" rel="noreferrer">Here is a list of the 500 most popular math articles</a>. The number of regular Wikipedia readers is increasing, while the number of editors is decreasing (<a href="http://stats.wikimedia.org/EN/SummaryEN.htm" rel="noreferrer">graphs</a>), which is causing growing concern within the Wikipedia community. </p>
<p><a href="http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics" rel="noreferrer">WikiProject Mathematics</a> is relatively active (compared to other WikiProjects, but not compared to MathOverflow!), and there is always the need for more experts who really understand the material. An editor continually faces the tension between (1) providing a lot of advanced material and (2) explaining things well, which generates many productive discussions about how mathematics articles should be written, and which topics should have their own article. </p>
<p>Regardless of the long term concerns raised about whether Wikipedia is capable of being a resource for advanced mathematics (see <a href="https://mathoverflow.net/questions/19631/can-wikipedia-be-a-reliable-and-sustainable-resource-for-advanced-mathematics">this closed question</a>), the fact is, people are attempting to learn from Wikipedia's mathematics articles right now. So improvements made to articles today will benefit the readers of tomorrow. </p>
<hr>
<p>Wikipedia is a very satisfying venue for summarizing topics you know well, and explaining things to other people, due to its large readership. Based on the number of mathematicians at MathOverflow, who are willing to spend time (for free!) answering questions and clarifying subjects for other people, it seems like there is a lot of untapped volunteer potential here. So in the interests of exposing the possible obstacles to joining Wikipedia, I would like to know:</p>
<blockquote>
<p>Why don't mathematicians spend more time improving Wikipedia articles?</p>
</blockquote>
<p>Recent efforts intended to attract new participants and keep existing ones include the friendly atmosphere of the <a href="http://en.wikipedia.org/wiki/Wikipedia:Teahouse" rel="noreferrer">Teahouse</a>, as well as WikiProject Editor Retention. </p>
<p>In case anyone is interested, my Wikipedia username is User:Mark L MacDonald (which is my real name). </p>
| Federico Poloni | 1,898 | <p>My feeling towards Wikipedia is that it is great as a quick reference to check facts and results, but it doesn't work as well for explaining well introductory material. The reasons are various.</p>
<p>It is difficult to provide a good explanation for <em>everyone</em>. Often, different intended targets conflict: you can't explain "kernel" to a layman, to an engineer, and to a mathematician at the same time, while making the exposition clear for all of them.</p>
<p>Moreover, often one would like to start from scratch and lay down the whole article from a different point of view; this is diplomatically difficult because it would trash the contributions of the previous editors (good or bad as they are). Yet, for some quickly-evolving topics it might be necessary to rewrite the page from scratch every few years, to reflect the recent advances and the new understanding.</p>
<p>The notation of an article might need changing; however, it's a difficult job to get to agree on "the best notation" for a given topic, even among three co-authors. Edit wars are a concern, and democracy often isn't the best way to solve them.</p>
<p>Pages in other languages are another concern. There is nothing wrong with having a good English page translated in several languages; on the other hand, local writers will tend to start their own new pages independently, even if their quality is worse. Even if someone translates an English page at a certain moment, future changes will get de-synchronized among the different languages.</p>
<p>I feel that many mathematicians would rather write a (part of) a textbook or a review article, and publish it on the internet. Often, too many cook spoil the broth. A good writer working alone can trump many writers working collectively. So, maybe we'd rather need to focus efforts on making good textbooks and lecture notes easy to find on the internet.</p>
|
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