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 |
|---|---|---|---|---|
18,048 | <p>When taking a MOOC in calculus the exercises contain 5 options to select from. I then solve the question and select the option that matches my answer. Obviously only one of the options is correct. But there are (quite a few) times where my solution is wrong even though it is one of the available options. </p>
<p>My question is, how does a maths teacher know how to create wrong options that are as plausible as the correct answer?</p>
<p>I've found these two good questions but I am wondering how would an educator come up with wrong answers in general.</p>
<p><a href="https://matheducators.stackexchange.com/questions/16573/what-are-some-common-ways-students-get-confused-about-finding-an-inverse-of-a-fu/16576#16576">What are some common ways students get confused about finding an inverse of a function?</a></p>
<p><a href="https://matheducators.stackexchange.com/questions/12045/how-are-students-messing-up-in-this-khan-academy-surface-area-problem-right-tr">How are students messing up in this Khan Academy surface area problem? (right triangular prism: 3-4-5 triangular base, height of 11)</a></p>
| ytrkptl | 13,678 | <p>One thing common, especially at middle school/high school levels is to list the unknown (for example, x = 3) as one of the options, especially when the question actually asks you to solve for something else that requires knowing x. At the middle/high school levels, several students have a habit of thinking that "x is the answer, or you're done once your find x". In such questions, listing x = 3 (or whatever the unknown actually is) as one of the answer choices tends to throw students off. This is how I come up with wrong answers sometimes.</p>
|
2,919,266 | <p>Let $(x_n)$ be a sequence in $(-\infty, \infty]$. </p>
<p>Could we define the sequence $(x_n)$ so that limsup$(x_n) = -\infty$? </p>
<p>My intuitive thought is no, but I’m not 100% sure. </p>
| Donald Splutterwit | 404,247 | <p>Rewrite the sum
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{i}{(i+1)(i+2)} 2^{i}
\end{eqnarray*}
Now do partial fractions
\begin{eqnarray*}
\frac{i}{(i+1)(i+2)} =\frac{2}{n+2} - \frac{1}{n+1}.
\end{eqnarray*}
Now not that the powers of $2$ cause this sum to be telescoping, so
Rewrite the sum
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{i}{(i+1)(i+2)} 2^{i}=\frac{2^{n+1}}{n+2}-1.
\end{eqnarray*}</p>
|
3,745,159 | <p>Let <span class="math-container">$A,B,C$</span> be <span class="math-container">$n\times n$</span> matrices with real entries such that their product is pairwise commutative. Also <span class="math-container">$ABC=O_{n}$</span>. If
<span class="math-container">$$k=\det\left(A^3+B^3+C^3\right).\det\left(A+B+C\right)$$</span>
then find the value or the range of values that <span class="math-container">$k$</span> may take.</p>
<p><strong>My Attempt</strong></p>
<p>I tried <span class="math-container">$k=\left(\det(A+B+C)\right)^2\left(\det(A^2+B^2+C^2-AB-BC-CA)\right)$</span>. but couldn't go further than this</p>
| Yes it's me | 767,521 | <p><span class="math-container">$\textbf{Hint:}$</span>Use euclidean algorithm.Denote by <span class="math-container">$(a,b)$</span> as the gcd of <span class="math-container">$a,b$</span></p>
<p>Observe::<span class="math-container">$\quad\begin{align}(2^m-1,2^n-1)&=(2^m-1,2^n-1-2^m+1)\\&=(2^m-1,2^m*(2^{n-m}-1))\\&=(2^m-1,2^{(n-m)}-1)\end{align}$</span> (Assume WLOG <span class="math-container">$n>m$</span>).</p>
<p>This is similar to, <span class="math-container">$(m,n)=(m-n,n)$</span> in the method of finding gcd using the euclidean algorithm.</p>
|
1,926,423 | <p>My problem with this is the final step. Through iterative substitution, I come up with the following:
$$T(n) = T(n-4) + (n-3) + (n-2) + (n-1) + n$$</p>
<p>which leads to the general form:
$$T(n) = T(n-k) + kn - \frac{[k(k-1)]}{2}$$</p>
<p>The restrictions are $T(1)=1$ and $n=2^k-1$. What I am doing at this point is solving for $k$, when $T(1) = 1$ which gives me :
$$n = 2^k - 1$$
$$n + 1 = 2^k$$
$$2 = 2^k$$
($n=1$, since this is for $T(1)=1$)</p>
<p>$$1 = k$$</p>
<p>plugging back in I just come up with the original formula which is obviously not correct. </p>
<p>What am I doing wrong in the final step (if that really is my problem)? Based on other examples I have seen, this would work out nicely if I can put $k$ in terms of $n$ instead of a constant, but I am drawing a blank on how to do this and I cannot find it anywhere.</p>
| GoodDeeds | 307,825 | <p>$$\begin{align}T(n)&=T(n-1)+n\\T(n)&=T(n-2)+(n-1)+n\\T(n)&=T(n-3)+(n-2)+(n-1)+n\\ &\vdots\\T(n)&=T(n-k)+(n-k+1)+(n-k+2)+\cdots+n\\ &\vdots\\T(n)&=T(1)+2+3+\cdots+n\\T(n)&=1+2+3+\cdots+n\\T(n)&=\frac{n(n+1)}{2}\end{align}$$</p>
|
1,134,177 | <p>Consider this primality test: Fix an initial segment of primes (e.g. 2,3,5,7), and combine a $b$-pseudoprime test for each b in that list. For several such initial segments, find the first $n$ for which the test gives an incorrect answer.</p>
<p>Hey all! I'm not quite understanding what the aforementioned question is asking... So help would be handy. :)</p>
<p>My code:</p>
<pre><code>primetest1 := proc
(Numprimes, N) local i, j;
for i to Numprimes do
for j to N do
if `mod`(ithprime(i)^(j-1), j) = 1 then
if isprime(j) = false then print(j*"is a false answer")
end if
end if
end do
end do
end proc
</code></pre>
| Asinomás | 33,907 | <p>I'm assuming what it means is given an initial segment of primes (the first $n$ primes). for example the numbers $2,3,5$ Use this number to test if a number is prime or not. That is: check wheter a number is divisible by $2,3$ and $5$. Clearly if it is the number is not prime.</p>
<p>If the number is not divisible by any of them assume this test tells you the number is a prime. Of course this is not true. The question presumably asks you to find the firs non-prime number that passes the prime test.</p>
<p>For example, when you take $2,3,5$ the first non-prime number that passes the test is $49$ since it is not prime and not divisible by either of those primes. In fact the first counterexample will always be the square of the smallest prime yu didn't take (in this case 7).</p>
|
2,369,081 | <blockquote>
<p>Evaluate the integral $$\int_0^1\frac{x^7-1}{\log (x)}\,dx $$</p>
</blockquote>
<p>[1]: <a href="https://i.stack.imgur.com/lcK2p.jpgplz" rel="nofollow noreferrer">https://i.stack.imgur.com/lcK2p.jpgplz</a> I'm trying to solve this definite integral since 2 hours. Please, I need help on this.</p>
| Mark Viola | 218,419 | <p>Note that using $\int_0^1 x^t \,dt=\frac{x-1}{\log(x)}$ we can write</p>
<p>$$\begin{align}
\int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 (x^6+x^5+x^4+x^3+x^2+x+1)\left(\int_0^1 x^t\,dt\right)\,dx\tag 1\\\\
&=\int_0^1\int_0^1 (x^{t+6}+x^{t+5}+x^{t+4}+x^{t+3}+x^{t+2}+x^{t+1}+x^t)\,dx\,dt\tag2 \\\\
&=\int_0^1 \left(\frac{1}{t+7}+\frac{1}{t+6}+\frac{1}{t+5}+\frac{1}{t+4}+\frac{1}{t+3}+\frac{1}{t+2}+\frac{1}{t+1}\right)\,dt\\\\
&=\log(8)
\end{align}$$</p>
<p>where the Fubini-Tonelli Theorem guarantees the legitimacy of interchanging the order of integration in going from $(1)$ to $(2)$.</p>
<hr>
<p>Note that if we first enforce the substitution $x^7\to x$, we obtain</p>
<p>$$\begin{align}
\int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 x^{1/7-1}\left(\frac{(x-1)}{\log(x)}\right)\\\\
&=\int_0^1 x^{1/7-1}\left(\int_0^1 x^t\,dt\right)\,dx\\\\
&=\int_0^1\int_0^1 x^{t+1/7-1} \,dx\,dt\\\\
&=\int_0^1 \frac{1}{t+1/7}\,dt\\\\
&=\log(8)
\end{align}$$</p>
<p>as expected!</p>
|
1,849,608 | <p>Given the irreducible fraction $\frac a b$, with $a, b \in \mathbb N$, what is the expression that enumerates all the irreducible fractions of integers that add up to $\frac a b$? Namely, an expression (in terms of $a$ and $b$) for all the $\frac c d$ and $\frac e f$, with $c,d,e,f \in \mathbb N$, such that $\frac c d + \frac e f = \frac a b$.</p>
| naslundx | 130,817 | <p>The best you can do is simplify your expression:</p>
<p>$$\frac{c}{d} + \frac{e}{f} = \frac{cf + ed}{df} = \frac{a}{b}$$</p>
<p>The numerator and denominator will give you a system of two linear equations. With $a$ and $b$ fixed there are four unknowns $c,d,e,f$. As $4>2$ this means we have infinitely many solutions (as would be expected):</p>
<p>$$a = cf + ed$$
$$b = df$$</p>
<p>Now, with two parameters $T, S \in \mathbb R$ and substituting $c = T, d = S$ (for example):</p>
<p>$$a = Tf + Se$$
$$b = Sf$$</p>
<p>gives you</p>
<p>$$f = \frac{b}{S}$$
$$e = \frac{a - T\frac{b}{S}}{S}$$
$$d = S$$
$$c = T$$</p>
<p>for any real numbers $S, T$.</p>
|
277,135 | <p><code>Graphics[{{Blue, Line[{{-5, 3}, {5, 8}}]}, {Dashed, Arrow[{{0, 0}, P}]}, Red, Arrow[{{0, 0}, v}]}, Axes -> True, AxesLabel -> {x, y}]</code>
I've tried combining Graphics and Plot to manipulate my graphic but it says that I can't.</p>
<p><a href="https://i.stack.imgur.com/8slBl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8slBl.png" alt="enter image description here" /></a></p>
| user293787 | 85,954 | <p>Code:</p>
<pre><code>(* General *)
alldice[{nfaces_,min_,max_}]:=Select[Tuples[Range[min,max],nfaces],OrderedQ];
findBruteForce[admissibleDice_,ndice_,diceinfo:{nfaces_,min_,max_}]:=Catch[
Outer[If[admissibleDice[##],Throw[{##}]]&,
Sequence@@ConstantArray[alldice[diceinfo],ndice],1];{}];
(* Example *)
Pevent1[dice1_,dice2_,dice3_]:=1/Length[dice1]*Count[dice1,0];
Pevent2[dice1_,dice2_,dice3_]:=1/Length[dice1]*Count[dice1,_?Positive];
Pevent3[dice1_,dice2_,dice3_]:=1/Length[dice1]*1/Length[dice2]*Count[Outer[Plus,dice1,dice2],_?Positive,{2}];
Pevent4[dice1_,dice2_,dice3_]:=1/Length[dice1]*1/Length[dice2]*1/Length[dice3]*Count[Outer[Plus,dice1,dice2,dice3],_?Positive,{3}];
admissibleDiceExample[tol_]:=And[
Abs[Pevent1[##]-0]<tol,
Abs[Pevent2[##]-1/2]<tol,
Abs[Pevent3[##]-2/3]<tol,
Abs[Pevent4[##]-5/6]<tol]&;
</code></pre>
<p>Search among all dice with 6 faces and numbers from $-1$ to $2$, runs in less than a second:</p>
<pre><code>dice123 = findBruteForce[admissibleDiceExample[0.01],3,{6,-1,2}]
(* {{-1,-1,-1,1,1,1},{-1,-1,2,2,2,2},{1,1,1,1,1,1}} *)
</code></pre>
<p>I used a small tolerance, but this happens to be an exact solution:</p>
<pre><code>Pevent1@@dice123
(* 0 *)
Pevent2@@dice123
(* 1/2 *)
Pevent3@@dice123
(* 2/3 *)
Pevent4@@dice123
(* 5/6 *)
</code></pre>
<p>The <code>Pevent</code> functions look a little complicated. But they use <code>Outer[Plus,...]</code> which may be faster than some other approaches.</p>
|
3,336,311 | <p><strong>Help me factor these polynomials</strong> </p>
<ul>
<li><span class="math-container">$(x+\sqrt{2})^2$</span> - 8</li>
<li>14a - 49<span class="math-container">$a^2$</span> + 100<span class="math-container">$b^2$</span> - 1</li>
</ul>
| Matteo | 686,644 | <p>For the first, use <span class="math-container">$a^2-b^2=(a+b)(a-b)$</span>, so: <span class="math-container">$$(x+\sqrt{2})^2-8=(x+\sqrt2+2\sqrt2)(x+\sqrt2-2\sqrt2)=(x+3\sqrt2)(x-\sqrt2)$$</span>
For the second case pick a <span class="math-container">$a$</span> and use the same rule as the previous answer, so: <span class="math-container">$$14a(2-7a)+(10b+1)(10b-1)$$</span> or <span class="math-container">$$(10b)^2-(7a-1)^2=(10b-7a+1)(10b+7a-1)$$</span></p>
|
2,876,050 | <p>I have There was a cowboy and 3 donkeys (Donkey A, Donkey B, Donkey C). The cowboy wears an eye cover and shoots randomly. What is the probability of donkey A to still be there after the cowboy shot 2 bullets? There is equal chances that the cowboy hits Donkey A, Donkey B, Donkey C or a Miss.</p>
<p>I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that donkey A still alive. But I want to solve it straight forward and directly. Is it possible? If yes, how?</p>
| Christian Blatter | 1,303 | <p>As remarked by hardmath the protocol is somewhat ambiguously defined. Since the cowboy is blind he may shoot at a dead donkey the second time. I add a dummy donkey D to account for the misses. Then donkey A survives the first as well as the second shot with probability ${3\over4}$ each. Assuming these shots to be independent the probability that A is still alive at the end is ${9\over16}$.</p>
|
1,823,736 | <p><a href="http://www.math.drexel.edu/~dmitryk/Teaching/MATH221-SPRING'12/Sample_Exam_solutions.pdf" rel="nofollow">Problem 10c from here</a>.</p>
<blockquote>
<p>Thirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?</p>
</blockquote>
<p>The given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.</p>
<p>My question is, why can't we set this up as the sum $\binom{3}{1} + \binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is <em>close</em> to the answer given, but it's $62$ short. I get that it's the "wrong" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.</p>
| Noble Mushtak | 307,483 | <p>As you said, we have $(0, 25)$, $(3, 0)$, and $(4, 9)$ as our intersection points. Therefore, we need to look at the regions of area in between those intersections points.</p>
<p>Between $x=0$ and $x=3$, the area is between the blue curve, $y=25-x^2$, and the purple curve, $y=25-\frac{25x}{3}$. Thus, we have the following integral:
$$\int_0^3 \left(25-x^2-\left(25-\frac{25x}{3}\right)\right)dx$$</p>
<p>Between $x=3$ and $x=4$, the area is between the blue curve, $y=25-x^2$, and the red curve, $y=9(x-3)$. Thus, we have the following integral:
$$\int_3^4 (25-x^2-(9(x-3)))dx$$</p>
<p>Add the two integrals together to get the total area.</p>
|
71,166 | <p>This question have been driving me crazy for months now. This comes from work on multiple integrals and convolutions but is phrased in terms of formal power series.</p>
<p>We start with a formal power series</p>
<p>$P(C) = \sum_{n=0}^\infty a_n C^{n+1}$</p>
<p>where $a_n = (-1)^n n!$</p>
<p>With these coefficients the formal power series can be expressed as a hypergeometric function</p>
<p>$P(C) = C \, _2F_0(1,1;;-C)$</p>
<p>I'm then interested in the formal power series $P_T(C)=\frac{P}{1-P}$ as well as, if possible, the series $P^n$ for arbitrary positive integer n (where this is the power series P raised to the nth power).</p>
<p>Specifically if</p>
<p>$P_T(C) = \sum_{n=0}^\infty b_n C^{n}$ </p>
<p>then want to construct the function</p>
<p>$f(x) = \sum_{n=0}^\infty \frac{b_{n+1}}{(n!)^2} x^{n}$</p>
<p>which, from other results, should converge for all x. We can think of this as the doubly-exponential generating function for the $b_n$ sequence.</p>
<p>There are rules for multiplying and dividing formal power series (see <a href="http://en.wikipedia.org/wiki/Formal_power_series#Operations_on_formal_power_series" rel="nofollow">here</a>) and I've used these to get a recurrence relation for the coefficients in $P_T(C)$ (as well as P^n(C)) but I've been unable to solve these recurrence relations explicitly. They're in a form where each $b_n$ depends on all the previous $b_n$'s and I've not been able to make progress with them.</p>
<p>Explicitly the recurrence relation for the $b_n$ is $b_0 = 1$, $b_n = \sum_{k=1}^n b_{n-k} a_k$ (for n > 1). This looks simple enough but I don't think has a nice closed-form expression.</p>
<p>Nevertheless I do know what the $b_n$ are. They are the sequence <a href="http://oeis.org/A052186" rel="nofollow">A052186</a> (up to plus and minus signs). So </p>
<p>$P_T(C) = C+C^3-3 C^4+14 C^5-77 C^6+497 C^7-3676 C^8+\ldots$</p>
<p>and</p>
<p>$f(x) = 1 + \frac{1}{4}x^2 - \frac{1}{12}x^3 + \frac{7}{288}x^4 - \frac{77}{14400} x^5 +
\frac{497}{518400}x^6 +\ldots$</p>
<p>The question is, is it possible to figure out the function $f(x)$!? Does it have a nice closed form? Perhaps in the form of a hypergeometric function? If it does, great, if it doesn't then at least I can stop searching for it!</p>
| Gottfried Helms | 7,710 | <p>[update] <em>remark: I shifted the original answer to the bottom although it was missing the question to keep it as reference.</em><br>
I've done a nice table, which shows the coefficients of a powerseries with constant term <em>=1</em>; for any <em>p</em> 'th integer power of the function you insert p into the formula.<br>
Unfortunately the Latex-notation of the table was too messy, so I insert it as an image here: </p>
<p><a href="https://i.stack.imgur.com/bccTS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bccTS.png" alt="table"></a><br>
<sub>(source: <a href="http://go.helms-net.de/math/images/mo_110724_decofun.png" rel="nofollow noreferrer">helms-net.de</a>)</sub> </p>
<p>If you have Pari/GP you can easily manipulate power series; the formula <span class="math-container">$ \small P \cdot (1 - P)^{-1}$</span>, where <em>P</em> is defined to be a powerseries can directly be evaluated : </p>
<pre><code>P = Ser( sum(k=0,20, (-1)^k*k! * x * x^k ))
print ( P*(1 - P)^-1 + O(x^16) )
</code></pre>
<p>The matrix-operator is simply the coefficients of each power of the taylor series of a function in a column of an (ideally infinite sized) matrix.</p>
<p><hr>
[original answer]
Assumed that I got your function correct ( <span class="math-container">$ \small P(x) = \sum_{k=1}^{\infty} (-1)^{k-1} \cdot (k-1)! \cdot x^k $</span> ) I have polynomials for the coefficients of <span class="math-container">$ \small P^{\text{ o } h} (x) $</span> Here I assume, that your notation <span class="math-container">$ \small P^n$</span> means iteration and not power. (If that was wrong I can delete this answer) I constructed the matrix-logarithm of the matrix-operator <span class="math-container">$ \small M$</span> for the function <span class="math-container">$ \small P(x)$</span> and got by <span class="math-container">$ \small \exp(h \cdot \log(M)) $</span> the following powerseries, where the coefficients at <em>x</em> are polynomials in the iteration-parameter <em>h</em> : </p>
<p><span class="math-container">$ \qquad \small \begin{array} {l} P^{\text{ o } h} (x) = & 1 \cdot x \\\
& - h \cdot x^2 \\\
& + (h^2 + h) \cdot x^3 \\\
& + (-h^3 - 5/2 \cdot h^2 - 5/2 \cdot h) \cdot x^4 \\\
& + ( h^4 + 13/3 \cdot h^3 + 9 \cdot h^2 + 29/3 \cdot h) \cdot x^5 \\\
& + (-h^5 - 77/12 \cdot h^4 - 125/6 \cdot h^3 - 511/12 \cdot h^2 - 295/6 \cdot h) \cdot x^6 \\\
& + O(x^8) \end{array} $</span> </p>
<p>I was not yet able to construct the other function <span class="math-container">$ \small P_T(C)$</span> although I've tried with a certain vague idea; please show some of the coefficents <span class="math-container">$ \small b_n $</span> so that I can compare.</p>
|
3,109,482 | <p><a href="https://i.stack.imgur.com/awp2x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/awp2x.png" alt="enter image description here"></a></p>
<p>I'm stumped on determining P(ABC) of Part A. My understanding is:</p>
<ul>
<li><p>Calculate the total number of patients (100)</p></li>
<li><p>Calculate individual <span class="math-container">$P(A), P(B),$</span> and <span class="math-container">$P(C)$</span> <span class="math-container">$(0.4; 0.35; 0.24 $</span> respectively)</p></li>
<li><p>Multiply <span class="math-container">$P(A)P(B)P(C) = ANS (0.0336)$</span></p></li>
</ul>
<p>This answer should be the same as <span class="math-container">$8/20$</span> or the population of serious patients, under 40, whose parents had diabetes but this fraction comes out to 0.08--what am I missing here? I appreciate your help.</p>
| Robert Lewis | 67,071 | <p>Since</p>
<p><span class="math-container">$A \ne B, \tag 1$</span></p>
<p>the only way <span class="math-container">$A$</span> and <span class="math-container">$B$</span> can be collinear is if</p>
<p><span class="math-container">$A = -B; \tag 2$</span></p>
<p>in this case <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are antipodal points and any plane containing both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> will pass through <span class="math-container">$(0, 0,0)$</span>; such a plane will intersect the sphere in a great circle containing <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, and we are done.</p>
<p>In the event that</p>
<p><span class="math-container">$A \ne B \ne -A, \tag 3$</span></p>
<p>then the plane passing through <span class="math-container">$(0, 0, 0)$</span> and normal to <span class="math-container">$A \times B$</span> will again cut the sphere in a great circle containing <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, which then forms a differentiable curve in said sphere connecting <span class="math-container">$A$</span> and <span class="math-container">$B$</span>; this is explained in greater detail, including explicit formulas for said curve, in my answer to <a href="https://math.stackexchange.com/questions/3098166/show-that-sigmat-lies-on-a-sphere-of-radius-1-and-center-at-the-origin">this question</a>.</p>
|
491,926 | <p>Let I have $(S,\Sigma,\mu)$ be a probability space, then $X,Y \in \Sigma$. Define $\rho (X,Y)$ by $\rho (X,Y)$ = correlation between random variable $I_X$ and $I_Y$, where $I_X$ and $I_Y$ are the indicator function of $X$ and $Y$. Express $\rho (X,Y)$ in term of $\mu (X)$, $\mu (Y)$, $\mu(XY)$. Conclude that $\rho(X,Y)=0$, if and only if $X$ and $Y$ are independent. </p>
<blockquote>
<p>How to show that: $$\rho(X,Y)= \frac{\mu(XY)\,\mu(X^cY^c)-\mu(XY^c)\,\mu(X^cY)}{(\mu(X) \,\mu(X^c)-\mu(Y) \,\mu(Y^c))^{1/2}};0<\mu(X)<1;\,0<\mu(Y)<1$$ </p>
</blockquote>
| Jonathan Y. | 89,121 | <p>This is not an answer for the quote, but definition we have
$$\rho(X,Y) = \frac{\operatorname{var}(I_X,I_Y)}{\sigma_{I_X}\sigma_{I_Y}} = \frac{\mathbb{E}(I_XI_Y)-\mathbb{E}(I_X)\mathbb{E}(I_Y)}{\sqrt{\left(\mathbb{E}(I_X^2)-\mathbb{E}(I_X)^2\right)\left(\mathbb{E}(I_Y^2)-\mathbb{E}(I_Y)^2\right)}}=\\
= \frac{\mathbb{E}(I_{X\cap Y})-\mathbb{E}(I_X)\mathbb{E}(I_Y)}{\sqrt{\left(\mathbb{E}(I_X)-\mathbb{E}(I_X)^2\right)\left(\mathbb{E}(I_Y)-\mathbb{E}(I_Y)^2\right)}} =
\frac{\mu(X\cap Y)-\mu(X)\mu(Y)}{\sqrt{\left(\mu(X)-\mu(X)^2\right)\left(\mu(Y)-\mu(Y)^2\right)}}$$
and clearly $\rho(X,Y)=0$ iff $\mu(X\cap Y)=\mu(X)\mu(Y)$.</p>
|
3,553,681 | <p>consider the identity
<span class="math-container">$$\frac{e^{-x}}{1-x}=\sum_{n=0}^{\infty}c_nx^n$$</span></p>
<p>Show that for each <span class="math-container">$n\ge0$</span>
<span class="math-container">$$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=1$$</span></p>
<p>My trial :
By cauchy product, </p>
<p><span class="math-container">$$c_k=\sum_{i=0}^{k}\frac{(-1)^i}{i!}$$</span></p>
<p>Then
<span class="math-container">$$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=\sum_{k=0}^{n}\sum_{i=0}^{k}\frac{(-1)^i}{(n-k)!i!}$$</span>
<span class="math-container">$$=\sum_{i=0}^{n}\sum_{k=i}^{n} \frac{(-1)^i}{(n-k)!i!}$$</span>
<span class="math-container">$$=\sum_{i=0}^{n}\sum_{k=i}^{n}\binom nk\frac{k!}{n!}$$</span></p>
<p>I was stuck in here.
So, I tried to solve it by taking <span class="math-container">$c_k=\frac{f^{k}(0)}{k!}$</span>. But I couldn't solve as well.. Could you please give me a few hint.. it will help me a lots. Thanks!</p>
| Ninad Munshi | 698,724 | <p>Just apply the integral twice and show the double integral reduces back down to the single integral via Fubini's theorem like so:</p>
<p><span class="math-container">$$(P^2f)(x) = \frac{1}{4\pi^2}\int_{-\pi}^\pi \int_{-\pi}^\pi [4\cos^2(x-y)-1][4\cos^2(y-z)-1]f(z)\:dz\:dy$$</span></p>
<p><span class="math-container">$$\frac{1}{4\pi^2}\int_{-\pi}^\pi f(z)\:dz \int_{-\pi}^\pi [4\cos^2(x-y)-1][4\cos^2(y-z)-1]\:dy$$</span></p>
<p><span class="math-container">$$\frac{1}{4\pi^2}\int_{-\pi}^\pi f(z)\:dz \int_{-\pi}^\pi [2\cos(2y-2x)+1][2\cos(2y-2z)+1]\:dy$$</span></p>
<p>(for formatting purposes only pay attention to the <span class="math-container">$y$</span> integral for now)</p>
<p><span class="math-container">$$\int_{-\pi}^\pi 2\cos(2z-2x)+2\cos(4y-2x-2z)+2\cos(2y-2x)+2\cos(2y-2z)+1\:dy$$</span></p>
<p><span class="math-container">$$\left[y(2\cos(2z-2x)+1)+\frac{1}{2}\sin(4y-2x-2z)+\sin(2y-2x)+\sin(2y-2z)\right]_{-\pi}^\pi$$</span></p>
<p>(now bring the <span class="math-container">$z$</span> integral back into play)</p>
<p><span class="math-container">$$\implies \frac{1}{2\pi}\int_{-\pi}^\pi [2\cos(2x-2z)+1]f(z)\:dz=\frac{1}{2\pi}\int_{-\pi}^\pi [4\cos^2(x-z)-1]f(z)\:dz \equiv (Pf)(x)$$</span></p>
<p>where we freely used the evenness of cosine and the <span class="math-container">$2\pi$</span>-periodicity of sine to make the other terms vanish.</p>
|
3,553,681 | <p>consider the identity
<span class="math-container">$$\frac{e^{-x}}{1-x}=\sum_{n=0}^{\infty}c_nx^n$$</span></p>
<p>Show that for each <span class="math-container">$n\ge0$</span>
<span class="math-container">$$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=1$$</span></p>
<p>My trial :
By cauchy product, </p>
<p><span class="math-container">$$c_k=\sum_{i=0}^{k}\frac{(-1)^i}{i!}$$</span></p>
<p>Then
<span class="math-container">$$\sum_{k=0}^{n}\frac{c_k}{(n-k)!}=\sum_{k=0}^{n}\sum_{i=0}^{k}\frac{(-1)^i}{(n-k)!i!}$$</span>
<span class="math-container">$$=\sum_{i=0}^{n}\sum_{k=i}^{n} \frac{(-1)^i}{(n-k)!i!}$$</span>
<span class="math-container">$$=\sum_{i=0}^{n}\sum_{k=i}^{n}\binom nk\frac{k!}{n!}$$</span></p>
<p>I was stuck in here.
So, I tried to solve it by taking <span class="math-container">$c_k=\frac{f^{k}(0)}{k!}$</span>. But I couldn't solve as well.. Could you please give me a few hint.. it will help me a lots. Thanks!</p>
| Jean-Claude Colette | 742,526 | <p>Before you show that P is a projector, you may notice that</p>
<p><span class="math-container">$2\cos^2(2x-2y)-1=2\cos(2x-2y)+1$</span></p>
<p><span class="math-container">$=2\cos(2x)\cos(2y)+2\sin(2x)\sin(2y)+1$</span></p>
<p>Hence</p>
<p><span class="math-container">$P(f)(x)=\frac{1}{\pi}\cos(2x)\int_{-\pi}^{\pi}\cos(2y)f(y)dy+\frac{1}{\pi}\sin(2x)\int_{-\pi}^{\pi}\sin(2y)f(y)dy+\frac{1}{\pi}\int_{-\pi}^{\pi}\frac12f(y)dy$</span></p>
<p>We can consider the basis: <span class="math-container">$\{x\mapsto\cos(nx),x\mapsto\sin(nx)|n\in\mathbb{N}^*\}\cup\left\{x\mapsto\frac{1}{\sqrt{2}}\right\}$</span></p>
<p>and the scalar product <span class="math-container">$<f,g>=\frac{1}{\pi}\int_{-\pi}^{\pi} f(y)g(y)dy$</span></p>
<p>The linerity of P is clear.</p>
<p>Let's denote <span class="math-container">$e_0:x\mapsto 1/\sqrt{2}$</span> and for <span class="math-container">$n\geq 1$</span> <span class="math-container">$e_n:x\mapsto\cos(nx)$</span>, <span class="math-container">$e'_n:x\mapsto\sin(nx)$</span>.</p>
<p>We have for <span class="math-container">$n\geq 0$</span>, <span class="math-container">$<e_n,e_n>=1$</span> and for <span class="math-container">$n\geq 1$</span>, <span class="math-container">$<e'_n,e'_n>=1$</span></p>
<p>Moreover for <span class="math-container">$p\neq q$</span>, <span class="math-container">$<e_p,e_q>=0$</span>, <span class="math-container">$<e'_p,e'_q>=0$</span> </p>
<p>and for all <span class="math-container">$p,q$</span>, <span class="math-container">$<e_p, e'_q>=0$</span></p>
<p>We remark that</p>
<p><span class="math-container">$P(f)=<f,e_0>.e_0+<f,e_2>e_2+<f,e'_2>e'_2$</span></p>
<p>To prove that <span class="math-container">$P^2=P$</span></p>
<p><span class="math-container">$P^2(f)=<f,e_0><e_0,e_0>e_0+<f,e_2><e_2,e_2>e_2+<f,e'_2><e'_2,e'_2>e'_2$</span></p>
<p><span class="math-container">$=<f,e_0>e_0+<f,e_2>e_2+<f,e'_2>e'_2$</span></p>
<p><span class="math-container">$=P(f)$</span></p>
<p><span class="math-container">$P$</span> is the orthogonal projection on <span class="math-container">$\text{Span}\{e_0, e_2, e'_2\}$</span></p>
|
3,353,826 | <p>All the vertices of quadrilateral <span class="math-container">$ABCD$</span> are at the circumference of a circle and its diagonals intersect at point <span class="math-container">$O$</span>. If <span class="math-container">$∠CAB = 40°$</span> and <span class="math-container">$∠DBC = 70°$</span>, <span class="math-container">$AB = BC$</span>, then find <span class="math-container">$∠DCO$</span>.</p>
| B. Goddard | 362,009 | <p>I have seen <span class="math-container">$(\!(x)\!)$</span> for "nearest integer." My memory is dim, but maybe it was Emil Grosswald's elementary number theory text. I like it because it's easy to type and it's not likely to be confused with another function.</p>
|
119,904 | <p>It is possible to do simple math between TemporalSeries objects. For example</p>
<pre><code>es=EventSeries[{{{2016, 1, 1}, 2}, {{2016, 1, 3}, 2.1}}];
td=TemporalData[{{{2016, 1, 1}, 2}, {{2016, 1, 3}, 3.1}}];
es*td (* works fine *)
</code></pre>
<p>This returns a TemporalData object with the path <code>{{3660595200, 4}, {3660768000, 6.51}}</code>.</p>
<p>This method does not work however if the objects being manipulated have differing dimensions.</p>
<pre><code>es=EventSeries[{{{2016, 1, 1}, 2}, {{2012, 1, 3}, 1.9}, {{2016, 1, 3}, 2.1}}];
td=TemporalData[{{{2016, 1, 1}, 2}, {{2016, 1, 3}, 3.1}}];
es*td (* ka-boom *)
</code></pre>
<p>With the data I'm using, one series or the other may be missing data at various time stamps, and it would be acceptable to either interpolate for those missing points or to simply omit them from the computation and the output.</p>
<p>I've been dealing with this problem in several steps,</p>
<ol>
<li>Make a list of the dates that all the series have in common with
<code>intersectionDates=Apply[Intersection, Map[#["Path"][[All, 1]] &,
listOfTD]</code></li>
<li>Extract the points for only those dates with <code>alignedData=Map[Select[#, MemberQ[intersectiondates, #[[1]]] &]&, listOfTD]</code></li>
<li>And finally turn the output back into TemporalData with <code>Map[TimeSeries[#] &, alignedData]</code></li>
</ol>
<p>In practice I have additional code to remove duplicate dates, if any exist, to convert EventSeries back to EventSeries, and some other bells and whistles, not shown above.</p>
<p>The problem is, this method is unusably slow for large datasets. With a few hundred points, it works fine, but with 5000+, it becomes intolerable.</p>
<p>I welcome any suggestions for approaches that would allow me to complete simple math between TemporalData objects of differing dimensions much faster.</p>
| Gosia | 5,372 | <p>If you want to interpolate then use TimeSeriesThread to combine TimeSeries (by default TimeSeries uses linear interpolation):</p>
<pre><code>In[96]:= ts1 = TimeSeries[{{{2016, 1, 1}, 2}, {{2012, 1, 3}, 1.9}, {{2016, 1, 3},
2.1}}];
ts2 = TimeSeries[{{{2016, 1, 1}, 2}, {{2016, 1, 3}, 3.1}}];
In[100]:= res = TimeSeriesThread[Times @@ # &, {ts1, ts2}];
During evaluation of In[100]:= InterpolatingFunction::dmval: Input value {3534537600} lies outside the range of data in the interpolating function. Extrapolation will be used.
In[101]:= Normal[res]
Out[101]= {{DateObject[{2012, 1, 3}], -1520.86}, {DateObject[{2016, 1,
1}], 4.}, {DateObject[{2016, 1, 3}], 6.51}}
</code></pre>
<p>If you use TemporalData, then default is the interpolation with order 0.</p>
<p>If you use EventSeries, no interpolation is applied and Missing[] is created:</p>
<pre><code> In[87]:=
es1 = EventSeries[{{{2016, 1, 1}, 2}, {{2012, 1, 3}, 1.9}, {{2016, 1, 3},
2.1}}];
es2 = EventSeries[{{{2016, 1, 1}, 2}, {{2016, 1, 3}, 3.1}}];
In[94]:= res = TimeSeriesThread[Times @@ # &, {es1, es2}];
In[95]:= Normal[res]
Out[95]= {{DateObject[{2012, 1, 3}],
1.9 Missing[]}, {DateObject[{2016, 1, 1}],
4}, {DateObject[{2016, 1, 3}], 6.51}}
</code></pre>
|
2,756,139 | <p>Let $I=[0,1]$ and $$X = \prod_{i \in I}^{} \mathbb{R}$$
That is, an element of $X$ is a function $f:I→\mathbb{R}$.</p>
<p>Prove that a sequence $\{f_n\}_n ⊆ X$ of real functions converges to some $f ∈ X$ in the product topology on $X$, if and only if it converges pointwise, i.e. for every $x ∈ I$, $f_n(x) → f(x)$ in the usual sense of convergence of sequences.</p>
<hr>
<p>I don't understand how a product indexed by an uncountably infinite set is like. I'm guessing a single element $f$ of $X$ is an uncountable set $(y_i)_{i \in I}$ in itself. But then how is $f(x)$ different from any other $f$? Is it the cartesian product indexed by $[0,x]$?</p>
<p>Note that I included the "prove" question just to show the context, but it isn't what I find problematic. It's probably easy enough if I get how the product works.</p>
| Community | -1 | <p>To make things easier to understand, let $I = {0, 1, 2}$. To say
$$X = \prod_{i \in I} \mathbb{R}$$
means that an element of $X$ is an technically an element of $\mathbb{R^3}$. But this can be interpreted as a function from $I \to \mathbb{R}$, in the sense that</p>
<p>$$(a, b, c) \cong f \text{ such that} f(0) = a, f(1) = b, f(2)= c$$</p>
<p>To be accurate the function $f$ would be $\{(0, a),(1, b), (2, c)\}$. So, you could say</p>
<p>$$X = \prod_{i \in I} (\{i\} \times \mathbb{R})$$</p>
<p>where we consider the <em>unordered</em> Cartesian product. Now in your case, $I = [0,1]$, but the concept is the same. </p>
|
1,480,179 | <p>Let $n ≥ 2$, and $A ∈ M_n$ be Hermitian, and let ${\rm{B }} \in {\rm{ }}{{\rm{M}}_{n - 1}}$ be a leading principal
submatrix of A.</p>
<p>If $B$ is positive semidefinite and $rank B = rank A$, why does $A$ is positive
semidefinite?</p>
| Bento | 46,364 | <p>The hermitian form $a$ associated with $A$ has a diagonalizing basis $\mathbf{b}$, so the matrix which rapresents the form can be considered diagonal. The rank is $n-1$ so we can assume (if needed, we can interchange the basis vectors order $\mathbf{b}_{i}$ to achieve this) the last row of the matrix as zero vector $\mathbf{0}$ (one row must be of zeros coordinates becouse there are $n-1$ independent rows in the matrix by hypothesis, and the matrix is diagonal, so one line must be linear combination of the remaining $n-1$).</p>
<p>$\qquad \qquad \qquad \qquad \qquad A=$$\begin{bmatrix} b_{11} & 0 & \cdots & 0 & 0 \\0 & b_{22} & \cdots & 0 & 0\\ \vdots & \vdots & \vdots &\vdots & \vdots \\ 0 & 0 & \cdots & b_{n-1 n-1} &0 \\ 0 & 0 & \cdots & 0 & 0 \end{bmatrix}$</p>
<p>The matrix $B$ corresponding to the form $b$ is the diagonal submatrix of $A$ with the $b_{ii}$ as diagonal elements.
The hermitian form $a(\mathbf{v},\mathbf{w})$ can be expressed as</p>
<p>$\qquad \qquad \sum_{i,j} x_i \bar y_ja(\mathbf{b}_i,\mathbf{b}_j) = \ ^\text{t}\mathbf{x}A \mathbf {\bar y}$ ($\mathbf{x}$ and $\mathbf{y}$ are respectively the $\mathbf{v}$ and $\mathbf{w}$ coordinate vectors) </p>
<p>So we have to test that $\sum_{i,j} x_i \bar x_ja(\mathbf{b}_i,\mathbf{b}_j) = \ ^\text{t}\mathbf{x}A \mathbf {\bar x}\ge 0$. This is an immediate consequence of the hypothesis that $B$ is positive semidefinite.</p>
<p>(To fix ideas I assumed rank $= n-1$, but the matter obviously remains substantially unchanged by assuming any rank - some $b_{ii}$ will be now $0$.)</p>
|
3,973,611 | <p>Let <span class="math-container">$$F(x)=\int_{-\infty}^x f(t)dt,$$</span>
where <span class="math-container">$x\in\mathcal{R}$</span>, <span class="math-container">$f\geq 0$</span> is complicated (it cannot be integrated analytically).</p>
<p>Can I used the Simpson's rule to approximate this integral, knowing that <span class="math-container">$f(-\infty)=0$</span>?</p>
| Robert Z | 299,698 | <p>There is no contradiction, just note that <span class="math-container">$C\cap D=\emptyset$</span> and therefore all intersection sets including <span class="math-container">$C\cap D$</span> have zero cardinality. Moreover, by a similar reason, also <span class="math-container">$A\cap B\cap C=\emptyset$</span> and <span class="math-container">$A\cap B\cap D=\emptyset$</span>. Thus it remains to compute
<span class="math-container">$$\begin{align}
&|U|-(|A|+|B|+|C|+|D|)+(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|)
\end{align}
$$</span>
P.S. One more thing, your number for <span class="math-container">$|D|$</span> should be multiplied by <span class="math-container">$8$</span>, i.e. the possible positions of the capital letter.</p>
|
1,493,965 | <p>We're given the power series $$ \sum_1^\inf \frac{j!}{j^j}z^j$$</p>
<p>and are asked to find radius of convergence R. I know the formula $R=1/\limsup(a_n ^{1/n})$, which leads me to compute $\lim \frac{j!^{1/j}}{j}$, and then I'm stuck.</p>
<p>The solution manual calculates R by $1/\lim|\frac{a_{j+1}}{a_j}|$, but I can't figure out the motivation for that formula.</p>
| Tsemo Aristide | 280,301 | <p>We suppose $t_0\in [a,b]$ and $f,f'_x$ continuous on $[a,b]$</p>
<p>$\|T(x(s)-T(y(s))\| =\|\int_{t_0-\delta}^{t_0+\delta}(f(x(s),s)-f((y(s),s))ds \|$.</p>
<p>Since $f'_x$ is continuous, $\| f(x(s),s)-f((y(s),s)\|\leq$ $Sup \|f'_x\|_{x\in [a,b]}\|x(s)-y(s)\|$.</p>
<p>$|T(x(s)-T(y(s))\|\leq 2\delta Sup \|f'_x\|_{x\in [a,b]}\|x(s)-y(s)\|$. We can choose $\delta$ such that $2\delta Sup \|f'_x\|_{x\in [a,b]}=q<1$.
This implies that $\|T(x(s)-T(y(s))\|\leq q\| x(s)-y(s)\|$</p>
|
1,224,180 | <p>Q: evaluate $\lim_{x \to \infty}$ $ (x-1)\over \sqrt {2x^2-1}$</p>
<p>What I did:</p>
<p>when $\lim_ {x \to \infty}$ you must put the argument in the form of $1/x$ so in that way you know that is equal to $0$</p>
<p>but in this ex. the farest that I went was</p>
<p>$\lim_{x \to \infty}$ $x \over x \sqrt{2}$ $1-(1/x) \over (1/x) - ??$</p>
| Clement C. | 75,808 | <p><strong>Hint:</strong> write $$
\frac{(x-1)}{\sqrt {2x^2-1}} = \frac{x(1-\frac{1}{x})}{x\sqrt{2-\frac{1}{x^2}}}= \frac{1-\frac{1}{x}}{\sqrt{2-\frac{1}{x^2}}}
$$
for $x > 0$. What happens to numerator and denominator when $x\to\infty$?</p>
|
1,685,423 | <p>$$
x_n=\begin{cases}\frac{1}{n^2} & \text{if n is even} \\ \frac{1}{n} & \text{if n is odd}\end{cases}$$.</p>
<p>How can I show that $$ \sum x_n$$ is convergent?</p>
| DonAntonio | 31,254 | <p>I think you have</p>
<p>$$\sum_{n=1}^\infty x_n\;\;,\;\;\;x_n=\begin{cases}\frac1{n^2},\,&n\text{ is even}\\{}\\\frac1n,\,&n\text{ is odd}\end{cases}$$</p>
<p>If the series was convergent, supose to $\;S\;$ , then as it is a positive one, you can "split it":</p>
<p>$$S=\sum_{n=1}^\infty x_n=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{2n-1}$$</p>
<p>But the first series on the right is convergent, so we'd get the rightmost one (a harmonic series) is also convergent as a difference of two convergent ones, which is false.</p>
|
480,504 | <p>If <span class="math-container">$M$</span> is a symmetric positive-definite matrix, is it possible to get a <strong>positive</strong> lower bound on the smallest eigenvalue of <span class="math-container">$M$</span> in terms of a matrix norm of <span class="math-container">$M$</span> or elements of <span class="math-container">$M$</span>? E.g., I want <span class="math-container">$$\lambda_{\text{min}} \geq f(\lVert M \rVert)$$</span> or something like that. <span class="math-container">$M$</span> is a Gram matrix, if that helps.</p>
| Saeed Manaffam | 124,075 | <p>There is one lower bound on minimum eigenvalue of symmetric p.d. matrix given in [Applied Math. Sc., vol. 4, no. 64] which is based on Frobenius norm (F) and Euclidean norm (E)</p>
<p>$$ \lambda_{min} \gt \sqrt{\frac{||A||_F^2-n||A||_E^2}{n(1-||A||_E^2/|det(A)|^{2/n})}} $$</p>
<p>if it helps.</p>
<p>[reference]: K. H. Schindler, "A New Lower Bound for the Minimal Singular Value for Real Non-Singular Matrices by a Matrix Norm and Determinant", Journal of Applied Mathematical Sciences, Vol. 4, No. 64, 2010.</p>
|
1,074,360 | <p>I know that $\phi(n)$, Euler's totient function, defines the number of all integers less than or equal to $n$ that are relatively prime to $n$. </p>
<p>I know that there is a trick to finding this with the larger non-prime numbers, but now I cannot find it anywhere. Could someone please explain how to find $\phi(n)$ for some large, non prime $n$. </p>
<p>For example:
$\phi(352)$?</p>
| Casteels | 92,730 | <p>There are two basic facts that you should know that will help you remember the formula. </p>
<p>First, if $n$ is a power of a prime, say $n=p^k$, then $\phi(p^k)=p^k-p^{k-1}$. This is because the only numbers less than or equal to $p^k$ that are <em>not</em> relatively prime to $p$ are $p,p^2,\ldots,p^{k-1}$. Note that $\phi(n)=\phi(p^k)=p^k(1-\frac{1}{p})$. </p>
<p>The second is that if $m$ and $n$ are relatively prime, then $\phi(mn)=\phi(m)\phi(n)$. This can be shown with not too much trouble using the chinese remainder theorem. </p>
<p>Putting all this together, if you can find the prime factorization $n=p_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}$, where $p_1,\ldots,p_m$ are distinct primes, then $$\phi(n)=p_1^{k_1}(1-\frac{1}{p_1})p_2^{k_2}(1-\frac{1}{p_2})\cdots p_m^{k_m}(1-\frac{1}{p_m})\\ =n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_m}).$$</p>
<p>For example, if $n=140$ and you find that $n=(2^2)(5)(7)$, then $\phi(140)=140(1-\frac{1}{2})(1-\frac{1}{5})(1-\frac{1}{7})$.</p>
|
1,074,360 | <p>I know that $\phi(n)$, Euler's totient function, defines the number of all integers less than or equal to $n$ that are relatively prime to $n$. </p>
<p>I know that there is a trick to finding this with the larger non-prime numbers, but now I cannot find it anywhere. Could someone please explain how to find $\phi(n)$ for some large, non prime $n$. </p>
<p>For example:
$\phi(352)$?</p>
| lhf | 589 | <p>The main tool is multiplicativity: $\phi(ab)=\phi(a)\phi(b)$ if $\gcd(a,b)=1$.</p>
<p>Also, if $p$ is a prime, then $\phi(p^k)=p^{k-1}(p-1)$.</p>
<p>In particular, if $n=2^k m$, with $m$ odd, then $\phi(n)=2^{k-1} \phi(m)$.</p>
<p>For $n=352=2^5\cdot 11$, we get $\phi(352)=2^4 \phi(11)=2^4 \cdot 10 = 160$.</p>
|
4,558,387 | <p>So I need to determine the scalarform of a plane that contain <span class="math-container">$L:\begin{cases} 3x + 4y + z = 5 \\
x − y = −6 \end{cases}$</span> and goes through the midpoint of <span class="math-container">$(1, 1, 2)$</span> and <span class="math-container">$(3, 1, 4)$</span>.</p>
<p>My try solving this. Call the midpoint <span class="math-container">$P$</span> then <span class="math-container">$P:(2,1,3)$</span>. The direction vector <span class="math-container">$v$</span> of the line is obtained by the crossproduct of the normal of each plane, so <span class="math-container">$(3,4,1)\times (1,-1,0)=(1,1,-7)$</span>. Thus we have <span class="math-container">$x+y-7z=d$</span> and putting in <span class="math-container">$(2,1,3)$</span> gives <span class="math-container">$x+y-7z=20$</span>. But the answer is <span class="math-container">$13x + 36y + 7z = 83$</span>.</p>
| Robert Z | 299,698 | <p>Your approach is not correct. <span class="math-container">$x+y−7z=-18$</span> is the plane which is <em>orthogonal</em> to the line <span class="math-container">$L$</span> and passes through <span class="math-container">$(2,1,3)$</span>.</p>
<p>Notice that all the planes which <em>contain</em> the line <span class="math-container">$L$</span> (<a href="https://en.wikipedia.org/wiki/Sheaf_of_planes" rel="nofollow noreferrer">sheaf of planes</a>) are given by
<span class="math-container">$$a(3x+4y+z-5)+b(x-y+6)=0\quad \text{with $a,b\in\mathbb{R}$}.$$</span>
We have to find <span class="math-container">$a,b$</span> such that the above equation is satisfied for <span class="math-container">$(x,y,z)=(2,1,3)$</span>:
<span class="math-container">$$a(6+4+3-5)+b(2-1+6)=0\implies 8a+7b=0.$$</span>
Therefore may take <span class="math-container">$a=7$</span> and <span class="math-container">$b=-8$</span>:
<span class="math-container">$$7(3x+4y+z-5)-8(x-y+6)=13x + 36y + 7z - 83=0$$</span>
and your answer is confirmed.</p>
|
4,558,387 | <p>So I need to determine the scalarform of a plane that contain <span class="math-container">$L:\begin{cases} 3x + 4y + z = 5 \\
x − y = −6 \end{cases}$</span> and goes through the midpoint of <span class="math-container">$(1, 1, 2)$</span> and <span class="math-container">$(3, 1, 4)$</span>.</p>
<p>My try solving this. Call the midpoint <span class="math-container">$P$</span> then <span class="math-container">$P:(2,1,3)$</span>. The direction vector <span class="math-container">$v$</span> of the line is obtained by the crossproduct of the normal of each plane, so <span class="math-container">$(3,4,1)\times (1,-1,0)=(1,1,-7)$</span>. Thus we have <span class="math-container">$x+y-7z=d$</span> and putting in <span class="math-container">$(2,1,3)$</span> gives <span class="math-container">$x+y-7z=20$</span>. But the answer is <span class="math-container">$13x + 36y + 7z = 83$</span>.</p>
| Glorious Nathalie | 948,761 | <p><span class="math-container">$(1, 1, -7)$</span> is the direction vector of the line of intersection of the two given planes (line <span class="math-container">$L$</span>). It is not the normal vector of the required plane.</p>
<p>In addition to the direction vector of line <span class="math-container">$L$</span>, you need a point on that line. Such a point can be found by Gauss-Jordan elimination of the two linear equations of the two planes</p>
<p><span class="math-container">$ 3 x + 4 y + z = 5 \hspace{25pt}(1) $</span></p>
<p><span class="math-container">$ x - y = -6 \hspace{25pt}(2)$</span></p>
<p>Set <span class="math-container">$x = 0$</span>, and solve the resulting system for <span class="math-container">$y$</span> and <span class="math-container">$z$</span></p>
<p><span class="math-container">$ 4 y + z = 5 \hspace{25pt}(1')$</span></p>
<p><span class="math-container">$ - y = - 6 \hspace{25pt}(2')$</span></p>
<p>So <span class="math-container">$(2')$</span> implies <span class="math-container">$ y = 6 $</span> and then <span class="math-container">$(1')$</span> implies <span class="math-container">$z = -19 $</span></p>
<p>So a point on line <span class="math-container">$L$</span> is <span class="math-container">$Q=(0, 6, -19) $</span></p>
<p>The midpoint you found to be <span class="math-container">$P = (2,1,3)$</span></p>
<p>Calculate the vector</p>
<p><span class="math-container">$QP = P - Q = (2, -5, 22) $</span></p>
<p>And you have the direction vector of line <span class="math-container">$L$</span>, namely,</p>
<p><span class="math-container">$ V = (1, 1, -7)$</span></p>
<p>Hence, the normal to the plane is</p>
<p><span class="math-container">$ N = QP \times V = (2, -5, 22) \times (1, 1, -7) = (13, 36, 7 ) $</span></p>
<p>The equation of the plane is</p>
<p><span class="math-container">$ N \cdot (p - Q) = 0 $</span></p>
<p><span class="math-container">$ (13, 36, 7) \cdot ((x,y,z) - (0, 6, -19) ) = 0 $</span></p>
<p>which simplifies to</p>
<p><span class="math-container">$ 13 z + 36 y + 7 z - 83 = 0 $</span></p>
|
2,950,813 | <blockquote>
<p>Take <span class="math-container">$G$</span> to be a group of order <span class="math-container">$600$</span>. Prove that for any element <span class="math-container">$a$</span> <span class="math-container">$\in$</span> G there exist an element <span class="math-container">$b$</span> <span class="math-container">$\in$</span> G such that <span class="math-container">$a = b^7$</span>. </p>
</blockquote>
<p>My thought process: since <span class="math-container">$a = b^7$</span> <span class="math-container">$\implies$</span> <span class="math-container">$|a| = |b^7|$</span>. Consequently
<span class="math-container">$\operatorname{lcm}(1,|a|) = \dfrac{1}{7}\operatorname{lcm}(7,|b|)$</span> implies <span class="math-container">$7|a| = 7|b|$</span> so <span class="math-container">$|a| = |b|$</span>. I don't know where I would go from here or if this is even the right approach. </p>
| lhf | 589 | <p><em>Hint:</em> <span class="math-container">$\gcd(600,7)=1 \implies 1=600m+7n$</span>. Indeed, <span class="math-container">$ 1 = 3 \cdot 600 - 257 \cdot 7$</span>.</p>
|
1,817,367 | <blockquote>
<p>Prove that $\forall k = m^2 + 1. \space m \in \mathbb{Z}^+$, if $k$ is divisible by any prime then that prime is congruent to $1, 2 \pmod 4$.</p>
</blockquote>
<p>I am unable to realize why it can't have $2$ prime factors congruent to $3 \pmod 4$. Can anyone please help me proceed?</p>
<p>Thanks.</p>
| JMoravitz | 179,297 | <p><em>Hiding the major revelations in spoiler quotes.</em></p>
<hr>
<p>Suppose that $p\mid m^2+1$</p>
<p>Suppose for contradictory purposes that $p=4z+3$ for some $z\in\Bbb Z$.</p>
<p>Note that $p\not\mid m$ and $2\neq p$</p>
<blockquote class="spoiler">
<p> This is because $4z+3$ is always odd. Further $k=m^2+1$ so $k\equiv 1\pmod{m}$ implying that $p\nmid m$ for any $p\mid k$.</p>
</blockquote>
<p>Then $m^{p-1}\equiv 1\pmod{p}$ by fermat's little theorem.</p>
<blockquote class="spoiler">
<p> Further, we know that $m^2+1\equiv 0\pmod{p}$ so $m^2\equiv -1\pmod{p}$</p>
</blockquote>
<p>.</p>
<blockquote class="spoiler">
<p> But then, that implies $m^{p-1}=m^{4z+2}=(m^2)^{2z+1}\equiv (-1)^{2z+1}\equiv -1\pmod{p}$</p>
</blockquote>
<p>A contradiction.</p>
|
1,817,367 | <blockquote>
<p>Prove that $\forall k = m^2 + 1. \space m \in \mathbb{Z}^+$, if $k$ is divisible by any prime then that prime is congruent to $1, 2 \pmod 4$.</p>
</blockquote>
<p>I am unable to realize why it can't have $2$ prime factors congruent to $3 \pmod 4$. Can anyone please help me proceed?</p>
<p>Thanks.</p>
| lhf | 589 | <p>If $p$ is an odd prime that divides $m^2+1$, then $m^2 \equiv -1 \bmod p$ and $m^4 \equiv 1 \bmod p$. Therefore, $m$ has order $4$ mod $p$. By Lagrange's theorem, we must have that $4$ divides $p-1$. In other words, $p \equiv 1 \bmod 4$.</p>
|
1,305,935 | <p>Let $f(n)$ be non-negative real valued function defined for each natural number $n$.</p>
<p>If $f$ is convex and $lim_{n\to\infty}f(n)$ exists as a finite number, then can we conclude that $f$ is non-increasing?</p>
| Jacky1205 | 305,093 | <p><a href="https://i.stack.imgur.com/oDLS5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oDLS5.png" alt="SVM" /></a></p>
<p>Let <span class="math-container">$\textbf{x}_+$</span> be a positive example on one gutter, such that
<span class="math-container">$$\textbf{w} \cdot \textbf{x}_+ - b = 1$$</span></p>
<p>Let <span class="math-container">$\textbf{x}_-$</span> be a negative example on another gutter, such that <span class="math-container">$$\textbf{w} \cdot \textbf{x}_- - b = -1$$</span></p>
<p>The width of margin is the <a href="https://en.wikipedia.org/wiki/Scalar_projection" rel="nofollow noreferrer">scalar projection</a> of <span class="math-container">$\textbf{x}_+ - \textbf{x}_-$</span> on unit normal vector , that is the dot production of <span class="math-container">$\textbf{x}_+ - \textbf{x}_-$</span> and <span class="math-container">$\frac{\textbf{w}}{\|\textbf{w}\|}$</span></p>
<p><span class="math-container">\begin{align}
width & = (\textbf{x}_+ - \textbf{x}_-) \cdot \frac{\textbf{w}}{\|\textbf{w}\|} \\
& = \frac {(\textbf{x}_+ - \textbf{x}_-) \cdot {\textbf{w}}}{\|\textbf{w}\|} \\
& = \frac{\textbf{x}_+ \cdot \textbf{w} \,{\bf -}\, \textbf{x}_-\cdot \textbf{w}}{\|\textbf{w}\|} \\
& = \frac{1-b-(-1-b)}{\lVert \textbf{w} \rVert} \\
& = \frac{2}{\|\textbf{w}\|}
\end{align}</span></p>
<p>The above refers to <a href="https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-034-artificial-intelligence-fall-2010/lecture-videos/lecture-16-learning-support-vector-machines/" rel="nofollow noreferrer">MIT 6.034 Artificial Intelligence</a></p>
|
2,260,051 | <p>Let $M$ and $N$ be two square matrices of same order, and $M^2 = N^4$.</p>
<p>Can any such $M,N$ exist when the following relations do not hold?</p>
<ol>
<li><p>$M = N^2$, and </p></li>
<li><p>$M = -N^2$ ?</p></li>
</ol>
| Gerry Myerson | 8,269 | <p>$$M=N=\pmatrix{0&1\cr0&0\cr}$$</p>
|
198,116 | <p>A finite <a href="http://en.wikipedia.org/wiki/Lattice_(order)" rel="nofollow noreferrer">lattice</a> is planar if it admits a <a href="http://en.wikipedia.org/wiki/Hasse_diagram" rel="nofollow noreferrer">Hasse diagram</a> which is a <a href="http://en.wikipedia.org/wiki/Planar_graph" rel="nofollow noreferrer">planar graph</a> (we want the Hasse diagram to be represented in the plane so that the <span class="math-container">$y$</span>-coordinate of each element respects the order).</p>
<p><em>Remark</em>: See the paper <a href="http://www.sciencedirect.com/science/article/pii/0095895676900241" rel="nofollow noreferrer">Planar lattices and planar graphs</a> (1976) by C. R. Platt :<br />
It is shown that a finite lattice is planar if and only if the (undirected) graph obtained from its (Hasse) diagram by adding an edge between its least and greatest elements is a planar graph.</p>
<p>The <span class="math-container">$B_3$</span> lattice (below) is not planar.<br />
<img src="https://i.stack.imgur.com/Nxdxm.png" alt="enter image description here" /></p>
<p><em>Question</em>: Is a finite <a href="http://en.wikipedia.org/wiki/Distributive_lattice" rel="nofollow noreferrer">distributive lattice</a> planar iff it admits no sublattice isomorphic to <span class="math-container">$B_3$</span>?</p>
| Joseph O'Rourke | 6,094 | <p>If I may be permitted to extract Wille's theorem
("On modular lattices of order dimension two")
to which Professor Stanley refers
(remember, this is 1973!):</p>
<hr>
<blockquote>
<p><strong>Theorem:</strong> A modular lattice $\mathfrak{M}$ is planar iff $\mathfrak{M} - \{ d \in \mathfrak{M} \mid d \text{ is doubly irreducible} \}$ is a planar distributive lattice iff $\mathfrak{M}$ does not contain any of the posets of figures 1 and 2 as a subposet.</p>
</blockquote>
<hr>
<p>
<img src="https://i.stack.imgur.com/LvNDe.png" alt="Figs1&2"></p>
<hr>
|
1,055,832 | <p>How can one prove distributivity of a <a href="http://ncatlab.org/nlab/show/Heyting+algebra" rel="nofollow"> Heyting Algebra</a> via the <a href="http://ncatlab.org/nlab/show/Yoneda+lemma" rel="nofollow"> Yoneda lemma</a>?</p>
<p>I'm able to prove it using the Heyting algebra property $(x \wedge a) \leq b$ if and only if $x \leq (a \Rightarrow b)$. But using Yoneda Lemma, I'm unable to get that.
Help Greatly appreciated . Thank You Guys. </p>
| Hanno | 81,567 | <p>For a Heyting algebra $H$, the category ${\mathcal H}$ associated to the poset underlying $H$ is a small, skeletal, Cartesian closed category with finite coproducts in which any two parallel morphisms are equal - and in fact this assignment is a bijection. Now if $x\in H$ then $x\wedge -: {\mathcal H}\to{\mathcal H}$ is a left adjoint with right adjoint $x\Rightarrow -$, hence preserves colimits up to canonical isomorphism; in particular, for any $y_1,...,y_n\in{\mathcal H}$, the canonical morphism $\varphi: (x\wedge y_1)\vee ...\vee (x\wedge y_n)\to x\wedge (y_1\vee ...\vee y_n)$ is an isomorphism in ${\mathcal H}$. By the above-mentioned properties of ${\mathcal H}$, this means that $(x\wedge y_1)\vee ...\vee (x\wedge y_n)=x\wedge (y_1\vee ...\vee y_n)$ and that $\varphi$ is the identity.</p>
<p>The Yoneda lemma is hidden here in the proof of the fact that left adjoints preserve colimits.</p>
|
4,092,473 | <p>I am trying to prove if the following set is bounded <span class="math-container">$$S=\bigcup_{a\in(0,1)} M_{a},$$</span> where
<span class="math-container">$$M_{a}=\{(x,y) \in \mathbb{R}^2: ax+(1-a)y=b, x>0, y>0, b \text{ is a fixed positive real number}\}.$$</span>
I think this set is bounded since all lines in the set are bounded by the vertical line <span class="math-container">$x = b$</span> and the horizontal line <span class="math-container">$y=b$</span>. However, I am not sure how to prove it formally?
Thanks!</p>
<p><strong>Edited: the expression set is reformed and my attempt for this is as follows</strong>
<span class="math-container">$$\lim_{a \rightarrow 0} M_a = {(x,y)}\in \mathbb{R}^2: y=b, x\in (0,\infty)$$</span>
and
<span class="math-container">$$\lim_{a \rightarrow 1} M_a = {(x,y)}\in \mathbb{R}^2: x=b, y\in (0,\infty).$$</span>
Does this implies that <span class="math-container">$S$</span> is bounded?</p>
| 5xum | 112,884 | <p>According to your definition of <span class="math-container">$S$</span>, which is <span class="math-container">$$S=\cup_{a\in(0,1)} \{(x,y)\in \mathbb R_+^2| ax+(1-a)y = b\},$$</span></p>
<p>the set <span class="math-container">$S$</span> is <strong>not</strong> bounded.</p>
<p>For an intuitive idea, think of it this way. For any <span class="math-container">$M>0$</span>, let's try to find a point <span class="math-container">$(x, y)$</span> such that <span class="math-container">$\|(x,y)\|$</span> is at least <span class="math-container">$M$</span>. One way to achieve this is to have <span class="math-container">$x$</span> be at least <span class="math-container">$M$</span>. Well, if <span class="math-container">$x$</span> is <span class="math-container">$M$</span>, then <span class="math-container">$$ax+(1-a)y = aM + (1-a)y.$$</span></p>
<p>So we know that <span class="math-container">$aM$</span>, plus something <strong>positive</strong>, must equal <span class="math-container">$b$</span>. We can achieve this by making <span class="math-container">$aM$</span> smaller than <span class="math-container">$b$</span>. Can we do that? <strong>YES</strong>. If <span class="math-container">$a$</span> is small enough (and we still have the freedom to pick <span class="math-container">$a$</span>), then <span class="math-container">$aM$</span> can become arbitrarily small, and we will be able to then pick a <span class="math-container">$y$</span> such that the total sum will be <span class="math-container">$b$</span>. In particular, we can choose <span class="math-container">$a$</span> such that <span class="math-container">$aM = \frac b2$</span>, which means that <span class="math-container">$(1-a)y$</span> must also be <span class="math-container">$\frac b2$</span> - this is again something we <em>can</em> achieve by setting <span class="math-container">$y=\frac{b}{2(1-a)}$</span>.</p>
<hr />
<p>To put this all together, you can show that <span class="math-container">$S$</span> is not bounded. In particular, for any <span class="math-container">$M>b$</span>, the point</p>
<p><span class="math-container">$$\left(M, \frac{bM}{2M-b}\right)$$</span> is an element of the set. This is because, for <span class="math-container">$a=\frac{b}{2M}$</span>, the point <span class="math-container">$\left(M, \frac{bM}{2M-b}\right)$</span> is an element of the set <span class="math-container">$$\{(x,y)\in \mathbb R_+^2| ax+(1-a)y = b\}.$$</span></p>
|
3,540,613 | <p>The integral to solve:</p>
<p><span class="math-container">$$
\int{5^{sin(x)}cos(x)dx}
$$</span></p>
<p>I used long computations using integration by parts, but I don't could finalize:</p>
<p><span class="math-container">$$
\int{5^{sin(x)}cos(x)dx} = cos(x)\frac{5^{sin(x)}}{ln(5)}+\frac{1}{ln(5)}\Bigg[ \frac{5^{sin(x)}}{ln(5)}-\frac{1}{ln(5)}\int{5^{sin(x)}cos(x)dx} \Bigg]
$$</span></p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$5^{\sin x}=y,$</span></p>
<p>As <span class="math-container">$5=e^{\ln5},$</span></p>
<p><span class="math-container">$dy=5^{\sin x}\cos x\ln 5\ dx$</span> right?</p>
|
1,406,280 | <p>Why does taylor series have ample amount of importance in calculus? </p>
<p>I like to know some insights behind taylor series. </p>
| N. F. Taussig | 173,070 | <p>I will assume that by the natural numbers that you mean the positive integers (as opposed to the nonnegative integers).</p>
<p>Observe that $a + c = 100 - 2b$ is an even number. Thus, $a$ and $c$ must have the same parity.</p>
<p><strong>Case 1:</strong> The numbers $a$ and $c$ are both even. Let $a = 2u$; let $c = 2v$. Then $u, v \in \mathbb{N}$. Moreover,
\begin{align*}
a + 2b + c & = 100\\
2u + 2b + 2v & = 100\\
u + b + v & = 50 \tag{1}
\end{align*}
which is an equation in positive integers. The number of solutions of equation 1 is the number of ways we can place two addition signs in the $49$ spaces between consecutive ones in a row of $50$ ones, which is $\binom{49}{2}$. </p>
<p><strong>Case 2:</strong> The numbers $a$ and $c$ are both odd. Let $a = 2s - 1$; let $b = 2t - 1$. Then $s, t \in \mathbb{N}$. Moreover,
\begin{align*}
a + 2b + c & = 100\\
2s - 1 + 2b + 2t - 1 & = 100\\
2s + 2b + 2t & = 102\\
s + b + t & = 51 \tag{2}
\end{align*}
which is an equation in positive integers. The number of solutions of equation 2 is the number of ways two addition signs can be placed in the $50$ spaces that appear between consecutive ones in a row of $51$ ones, which is $\binom{50}{2}$.</p>
<p>Thus, the number of solutions of the equation $a + 2b + c = 100$ in the positive integers is
$$\binom{49}{2} + \binom{50}{2} = 1176 + 1225 = 2401$$</p>
|
317,601 | <blockquote>
<p>Let <span class="math-container">$F$</span> be a ring, let <span class="math-container">$f(x)=a_0+a_1x+\cdots+a_nx^n$</span> be in <span class="math-container">$F[x]$</span>, and <span class="math-container">$f'(x)$</span> be the regular derivative of <span class="math-container">$f(x)$</span>.</p>
<p>Prove that <span class="math-container">$(f+g)'(x)=f'(x)+g'(x)$</span>.</p>
<p>Conclude that we can define a homomorphism of abelian groups <span class="math-container">$D:F[x]\to F[x]$</span> by <span class="math-container">$D(f(x))=f'(x)$</span>.</p>
</blockquote>
<p>How to prove that <span class="math-container">$(fg)'=f'g+fg'$</span>?</p>
| Jesko Hüttenhain | 11,653 | <p>The condition that $f'$ does not vanish <a href="http://en.wikipedia.org/wiki/Inseparable_extension#Separable_and_inseparable_polynomials" rel="nofollow">is equivalent to $f$ being separable</a>.</p>
|
4,550,991 | <p>This is question is taken from an early round of a Norwegian national math competition where you have on average 5 minutes to solve each question.</p>
<p>I tried to solve the question by writing every number with four digits and with introductory zeros where it was needed. For example 0001 and 0101 would be the numbers 1 and 101. I then divided the different numbers into four groups based on how many times the digit 1 appeared in the number. I called these group 1,2,3 and 4. 0001 would then belong to group 1 and 0101 to group 2. I first found out in how many ways I could place the digit 1 in each group, then multiplied it by the number of combinations with the other possible eight digits (0,2,3,5,6,7,8,9). This would be the number of combinations for each group and I lastly multiplied it with the number of times 1 appeared in the number. This is done for all of the groups under:</p>
<p><span class="math-container">$\binom{4}{1}\cdot8^3\cdot1$</span> times in group 1</p>
<p><span class="math-container">$\binom{4}{2}\cdot8^2\cdot2$</span> times in group 2</p>
<p><span class="math-container">$\binom{4}{3}\cdot8^1\cdot3$</span> times in group 3</p>
<p><span class="math-container">$\binom{4}{4}\cdot8^0\cdot4$</span> times in group 4</p>
<p>The sum of all these calculations will be the 2916 and the correct number of times 1 appears (I think). Is this calculation/way of thinking correct? And is there a more efficient way to do it?</p>
| Henry | 6,460 | <p>One handwaving quick approach</p>
<ul>
<li>How many digits in total do you write for <span class="math-container">$0000$</span> to <span class="math-container">$9999$</span>?
<ul>
<li><span class="math-container">$4 \times 10^4$</span></li>
</ul>
</li>
<li>How many of those are <span class="math-container">$1$</span>s?
<ul>
<li><span class="math-container">$\frac{1}{10} \times 4 \times 10^4 = 4 \times 10^3$</span></li>
</ul>
</li>
</ul>
<p>But you are missing anything with a particular digit, which is like working in base <span class="math-container">$9$</span> rather than base <span class="math-container">$10$</span>, so the answer is presumably <span class="math-container">$4 \times 9^3 = 2916$</span></p>
|
302 | <p>I know that the Fibonacci numbers converge to a ratio of .618, and that this ratio is found all throughout nature, etc. I suppose the best way to ask my question is: where was this .618 value first found? And what is the...significance?</p>
| Tracy Hall | 7,936 | <p>What is the significance? Most of the nice properties of the golden mean can be attributed to the fact that its continued fraction coefficients are uniformly bounded, as will be true in particular for any periodic continued fraction, which is to say any quadratic irrational, such as arises as the spectral radius of an indecomposable two-term linear recurrence relation. Among such continued fractions, the unique one with the minimum possible upper bound of 1 naturally exhibits these effects most prominantly, and it arises from (arguably) the simplest such recurrence.</p>
|
255,773 | <p>As it is known that
Integrate[A+B]= Integrate[A] + Integrate[B]</p>
<p>I am facing problem with the following integral, when I integrate</p>
<pre><code>Integrate[(-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) -
2 (EulerGamma + Log[(b x)/2])) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2) , {x,
1, \[Infinity]}]
</code></pre>
<p>I obtain an answer but when I wish to solve a part</p>
<pre><code>Integrate[-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2) , {x,
1, \[Infinity]}]
</code></pre>
<p>Mathematica 11 do not returns any value and processing goes on.</p>
| Akku14 | 34,287 | <p>Don't know why the definite integral of the part integral does not work, but with indefinite integration you get the right result (in MMA version 8.0)</p>
<pre><code>f[x_, b_] =
(-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) -
2 (EulerGamma + Log[(b x)/2])) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2);
g[x_, b_] =
-(1/2) b^2 x^2 (-1 + EulerGamma + Log[(b x)/2]) 1/
x ((-1 + x) Log[-1 + x]^2 - 2 (-1 + x) Log[-1 + x] (1 + Log[x]) +
2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2);
h[x_, b_] =
1/x (-2 (EulerGamma + Log[(b x)/2])) ((-1 + x) Log[-1 + x]^2 -
2 (-1 + x) Log[-1 + x] (1 + Log[x]) + 2 x Log[x] (2 + Log[x]) -
2 (1 + x) (1 + Log[x]) Log[1 + x] + (1 + x) Log[1 + x]^2);
f[x, b] - g[x, b] - h[x, b] // Simplify
(* 0 *)
</code></pre>
<p>.</p>
<pre><code>intf[b_] = Integrate[f[x, b], {x, 1, \[Infinity]}]
(* (17 \[Pi]^4)/180 - 8 EulerGamma Log[2] + 4 Log[2]^2 +
4 EulerGamma Log[2]^2 + 1/3 \[Pi]^2 Log[2]^2 - 4 Log[2]^3 -
Log[2]^4/3 + Log[256] - 8 PolyLog[4, 1/2] - (7 Zeta[3])/2 +
7/2 EulerGamma Zeta[3] - 21/2 Log[2] Zeta[3] +
1/144 b^2 (\[Pi]^2 (-5 + 6 EulerGamma - 6 Log[2]) - 8 Log[2] +
48 EulerGamma (-1 + Log[2]) Log[2] - 16 Log[2]^2 - 48 Log[2]^3 +
63 Zeta[3]) +
1/24 Log[b] (b^2 (\[Pi]^2 + 8 (-1 + Log[2]) Log[2]) +
12 (-16 Log[2] + 8 Log[2]^2 + 7 Zeta[3])) *)
</code></pre>
<p>.</p>
<pre><code>iintg = Integrate[g[x, b], x]
(* -(1/1296)b^2 (169 - 216 EulerGamma - 27 Log[4] + 396 Log[1 - x] -
396 EulerGamma Log[1 - x] + 216 Log[-1 + x] +
180 EulerGamma Log[-1 + x] - 612 x Log[-1 + x] +
216 EulerGamma x Log[-1 + x] - 576 x^2 Log[-1 + x] +
432 EulerGamma x^2 Log[-1 + x] + 576 x^3 Log[-1 + x] -
432 EulerGamma x^3 Log[-1 + x] - 48 Log[8] Log[-1 + x] -
198 Log[-1 + x]^2 + 108 EulerGamma Log[-1 + x]^2 +
486 x^2 Log[-1 + x]^2 - 324 EulerGamma x^2 Log[-1 + x]^2 -
288 x^3 Log[-1 + x]^2 + 216 EulerGamma x^3 Log[-1 + x]^2 -
216 Log[x] + 1224 x Log[x] - 432 EulerGamma x Log[x] -
1152 x^3 Log[x] + 864 EulerGamma x^3 Log[x] +
36 Log[1 - x] Log[x] - 216 EulerGamma Log[1 - x] Log[x] +
360 Log[-1 + x] Log[x] - 972 x^2 Log[-1 + x] Log[x] +
648 EulerGamma x^2 Log[-1 + x] Log[x] +
576 x^3 Log[-1 + x] Log[x] -
432 EulerGamma x^3 Log[-1 + x] Log[x] -
108 Log[-1 + x]^2 Log[x] - 576 x^3 Log[x]^2 +
432 EulerGamma x^3 Log[x]^2 + 216 Log[(b x)/2] +
144 Log[1 - x] Log[(b x)/2] - 504 Log[-1 + x] Log[(b x)/2] +
216 x Log[-1 + x] Log[(b x)/2] +
432 x^2 Log[-1 + x] Log[(b x)/2] -
432 x^3 Log[-1 + x] Log[(b x)/2] +
108 Log[-1 + x]^2 Log[(b x)/2] -
324 x^2 Log[-1 + x]^2 Log[(b x)/2] +
216 x^3 Log[-1 + x]^2 Log[(b x)/2] - 432 x Log[x] Log[(b x)/2] +
864 x^3 Log[x] Log[(b x)/2] -
216 Log[1 - x] Log[x] Log[(b x)/2] +
648 x^2 Log[-1 + x] Log[x] Log[(b x)/2] -
432 x^3 Log[-1 + x] Log[x] Log[(b x)/2] +
432 x^3 Log[x]^2 Log[(b x)/2] + 144 Log[-1 + x] Log[b x] -
888 Log[1 + x] + 396 EulerGamma Log[1 + x] - 612 x Log[1 + x] +
216 EulerGamma x Log[1 + x] + 576 x^2 Log[1 + x] -
432 EulerGamma x^2 Log[1 + x] + 576 x^3 Log[1 + x] -
432 EulerGamma x^3 Log[1 + x] - 216 Log[x] Log[1 + x] +
216 EulerGamma Log[x] Log[1 + x] + 972 x^2 Log[x] Log[1 + x] -
648 EulerGamma x^2 Log[x] Log[1 + x] +
576 x^3 Log[x] Log[1 + x] -
432 EulerGamma x^3 Log[x] Log[1 + x] +
36 Log[(b x)/2] Log[1 + x] + 216 x Log[(b x)/2] Log[1 + x] -
432 x^2 Log[(b x)/2] Log[1 + x] -
432 x^3 Log[(b x)/2] Log[1 + x] +
216 Log[x] Log[(b x)/2] Log[1 + x] -
648 x^2 Log[x] Log[(b x)/2] Log[1 + x] -
432 x^3 Log[x] Log[(b x)/2] Log[1 + x] + 198 Log[1 + x]^2 -
108 EulerGamma Log[1 + x]^2 - 486 x^2 Log[1 + x]^2 +
324 EulerGamma x^2 Log[1 + x]^2 - 288 x^3 Log[1 + x]^2 +
216 EulerGamma x^3 Log[1 + x]^2 + 108 Log[-x] Log[1 + x]^2 -
108 Log[(b x)/2] Log[1 + x]^2 +
324 x^2 Log[(b x)/2] Log[1 + x]^2 +
216 x^3 Log[(b x)/2] Log[1 + x]^2 + 324 Log[2 (1 + x)] -
324 EulerGamma Log[2 (1 + x)] - 324 Log[x] Log[2 (1 + x)] +
324 Log[(b x)/2] Log[2 (1 + x)] - 48 Log[3 (1 + x)] +
144 EulerGamma Log[3 (1 + x)] + 144 Log[x] Log[3 (1 + x)] -
144 Log[(b x)/2] Log[3 (1 + x)] -
216 Log[-1 + x] PolyLog[2, 1 - x] +
36 (-5 + 6 EulerGamma + 6 Log[x] + 6 Log[(b x)/2]) PolyLog[
2, -x] + 180 PolyLog[2, x] - 216 EulerGamma PolyLog[2, x] -
216 Log[x] PolyLog[2, x] - 216 Log[(b x)/2] PolyLog[2, x] +
216 Log[1 + x] PolyLog[2, 1 + x] + 216 PolyLog[3, 1 - x] -
432 PolyLog[3, -x] + 432 PolyLog[3, x] - 216 PolyLog[3, 1 + x]) *)
limo = Limit[iintg, x -> \[Infinity]]
(* 1/1296 b^2 (-169 - 396 I \[Pi] - 90 \[Pi]^2 - 54 Log[2] -
108 \[Pi]^2 Log[2] + 324 Log[2]^2 + 48 Log[3] -
144 Log[2] Log[3] +
36 EulerGamma (6 + 11 I \[Pi] + 3 \[Pi]^2 + Log[512/81]) +
36 I \[Pi] Log[16] +
36 (-6 - 4 I \[Pi] + 3 \[Pi]^2 - 9 Log[2] + Log[81]) Log[b]) *)
limu = Limit[iintg, x -> 1, Direction -> -1]
(* 1/1296 b^2 (-169 - 396 I \[Pi] - 36 Log[2] - 36 I \[Pi] Log[2] +
468 Log[2]^2 - 324 I \[Pi] Log[2]^2 + 432 Log[2]^3 + 48 Log[3] -
144 Log[2] Log[3] + 27 Log[4] - 180 PolyLog[2, 2] -
432 Log[2] PolyLog[2, 2] +
36 Log[b] (-6 - 12 Log[2]^2 + I \[Pi] (-4 + Log[64]) + Log[648] +
6 PolyLog[2, 2]) +
36 EulerGamma (6 - 12 Log[2]^2 + I \[Pi] (11 + Log[64]) +
Log[2097152/81] + 6 PolyLog[2, 2]) + 216 PolyLog[3, 2] -
756 Zeta[3]) *)
intg[b_] = limo - limu // FullSimplify
(* -(1/144) b^2 (16 Log[
2] (3 EulerGamma + Log[2] - EulerGamma Log[8] +
Log[2] Log[8]) + \[Pi]^2 (5 - 6 EulerGamma + Log[64]) +
Log[256] - 6 (\[Pi]^2 + 8 (-1 + Log[2]) Log[2]) Log[b] -
63 Zeta[3]) *)
</code></pre>
<p>.</p>
<pre><code>inth[b_] = Integrate[h[x, b], {x, 1, \[Infinity]}]
(* -2 (-((17 \[Pi]^4)/360) - 4 Log[2] - 2 Log[2]^2 -
2 EulerGamma Log[2]^2 - 1/6 \[Pi]^2 Log[2]^2 + 2 Log[2]^3 +
Log[2]^4/6 + EulerGamma Log[16] - 2 (-2 + Log[2]) Log[2] Log[b] +
4 PolyLog[4, 1/2] - 7/4 (-1 + EulerGamma + Log[b/8]) Zeta[3]) *)
intf[b] - intg[b] - inth[b] // FullSimplify
(* 0 *)
</code></pre>
|
1,853,846 | <p>Prove that the equation <span class="math-container">$$x^2 - x + 1 = p(x+y)$$</span> has integral solutions for infinitely many primes <span class="math-container">$p$</span>.</p>
<p>First, we prove that there is a solution for at least one prime, <span class="math-container">$p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is always odd so there is no solution for <span class="math-container">$p=2$</span>. We prove there is a solution for <span class="math-container">$p=3$</span>. If <span class="math-container">$p=3$</span>, <span class="math-container">$y = (x-2)^2/3-1$</span>. We get integral solutions whenever we get <span class="math-container">$x = 3m +2$</span>, where <span class="math-container">$m$</span> is any integer.<span class="math-container">$\\$</span>
We provide a proof by contradiction which is similar to Euclid's proof of there being infinitely many primes. \Let us assume that it is is true for only finitely many primes, and name the largest prime for which the equation is true <span class="math-container">$P$</span>.\\We set <span class="math-container">$$x = 2\cdot3\cdot5\dots P$$</span> <span class="math-container">$x$</span> is the product of all primes upto <span class="math-container">$P.\\$</span>
Then, the term <span class="math-container">$x(x-1) + 1$</span> is either prime or composite. If it is prime, then we set <span class="math-container">$p = x(x-1) + 1, y = 1 - x$</span> and get a solution. If it is composite, we write <span class="math-container">$x(x-1) + 1 = p\times q$</span>, where <span class="math-container">$p$</span> is any prime factor of <span class="math-container">$x(x-1)+1$</span>, and <span class="math-container">$q$</span> is an integer, <span class="math-container">$q = (x(x-1)+1)/p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is not divisible by any prime upto <span class="math-container">$P$</span> since it leaves a remainder of <span class="math-container">$1$</span> with all of them. So, <span class="math-container">$p > P$</span>. We set <span class="math-container">$y=q-x$</span> for a solution.<span class="math-container">$\\$</span>In either case, we get a solution for a prime <span class="math-container">$p > P$</span>, which means there's no largest prime for which this equation has solutions. This contradicts the assumption that there are solutions for only finitely many primes.</p>
<p>I feel like I'm missing some step. Is this correct ?</p>
| Majid | 254,604 | <p>Replacing $cos\theta=\pm\sqrt{1-sin^2\theta}$ in your equation, $8\sin\theta = 4 + \cos\theta$, and considering $u=sin\theta$, we have</p>
<p>$8u=4\pm\sqrt{1-u^2}$ which leads to $64u^2 +16 - 64u=1-u^2\to 65u^2-64u+15=0$.</p>
<p>Now, you only need to solve this equation to find $u$ that is $sin\theta$.</p>
|
1,853,846 | <p>Prove that the equation <span class="math-container">$$x^2 - x + 1 = p(x+y)$$</span> has integral solutions for infinitely many primes <span class="math-container">$p$</span>.</p>
<p>First, we prove that there is a solution for at least one prime, <span class="math-container">$p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is always odd so there is no solution for <span class="math-container">$p=2$</span>. We prove there is a solution for <span class="math-container">$p=3$</span>. If <span class="math-container">$p=3$</span>, <span class="math-container">$y = (x-2)^2/3-1$</span>. We get integral solutions whenever we get <span class="math-container">$x = 3m +2$</span>, where <span class="math-container">$m$</span> is any integer.<span class="math-container">$\\$</span>
We provide a proof by contradiction which is similar to Euclid's proof of there being infinitely many primes. \Let us assume that it is is true for only finitely many primes, and name the largest prime for which the equation is true <span class="math-container">$P$</span>.\\We set <span class="math-container">$$x = 2\cdot3\cdot5\dots P$$</span> <span class="math-container">$x$</span> is the product of all primes upto <span class="math-container">$P.\\$</span>
Then, the term <span class="math-container">$x(x-1) + 1$</span> is either prime or composite. If it is prime, then we set <span class="math-container">$p = x(x-1) + 1, y = 1 - x$</span> and get a solution. If it is composite, we write <span class="math-container">$x(x-1) + 1 = p\times q$</span>, where <span class="math-container">$p$</span> is any prime factor of <span class="math-container">$x(x-1)+1$</span>, and <span class="math-container">$q$</span> is an integer, <span class="math-container">$q = (x(x-1)+1)/p$</span>. Now, <span class="math-container">$x(x-1) + 1$</span> is not divisible by any prime upto <span class="math-container">$P$</span> since it leaves a remainder of <span class="math-container">$1$</span> with all of them. So, <span class="math-container">$p > P$</span>. We set <span class="math-container">$y=q-x$</span> for a solution.<span class="math-container">$\\$</span>In either case, we get a solution for a prime <span class="math-container">$p > P$</span>, which means there's no largest prime for which this equation has solutions. This contradicts the assumption that there are solutions for only finitely many primes.</p>
<p>I feel like I'm missing some step. Is this correct ?</p>
| Burrrrb | 322,248 | <p>$$8\sin(\theta)- \cos(\theta) =4$$
Dividing by $\sqrt{1^2 +8^2}$
$$\frac{8\sin(\theta)}{\sqrt{65}}- \frac{\cos(\theta)}{\sqrt{65}} =\frac{4}{\sqrt{65}}$$
Let $\sin(\alpha) = \frac{1}{\sqrt{65}}$, we $\cos(\alpha) = \frac{8}{\sqrt{65}}$.</p>
<p>Thus
$$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$
$$\sin(\theta -\alpha) = \frac{4}{\sqrt{65}}$$
$$\theta = \sin^{-1}(\frac{4}{\sqrt{65}}) + \alpha $$
$$= \sin^{-1}(\frac{4}{\sqrt{65}}) + \sin^{-1}(\frac{1}{\sqrt{65}})$$</p>
|
542,148 | <p>Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)</p>
<p>\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}\\
0 & 0 & 0 & x^{12} & x^{13}\\
0 & 0 & 0 & x^{14} & x^{15}
\end{bmatrix}</p>
<p>Can someone help me with this question?</p>
| nbubis | 28,743 | <p><strong>Hint</strong>:
$$\det\begin{pmatrix}A & 0\\ C & D\end{pmatrix} = \det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det(A) \det(D)$$
Where $A,B,C,D$ are block matrices. </p>
|
4,122,732 | <p>I was solving exercise 3.125 of Wackerly's Probability book and i did not understand the solution given in the solutions manual.</p>
<p>The problem says:</p>
<p>Customers arrive at a shop following a poisson distribution for an average of 7 customers per hour. What is the probability that exactly two clients arrive in 2 hours between :</p>
<p>(a). 2:00pm and 4:00pm (a continous period of two hours)?
(b). between 1:00pm and 2:00pm or between 3:00pm and 4:00pm (two periods of 1 hour that in total add to two hours)?</p>
<p>For part (a) i responded correctly. That is <span class="math-container">$P(Y=2) = \frac{14^2 e^{-14} }{2!}$</span> since initially <span class="math-container">$\lambda$</span>=7 (per hour), but because we are considering a continous period of 2 hours, our random variable Y has <span class="math-container">$\lambda_1$</span> = 14.</p>
<p>However for part (b) the solutions manual says the result is the same as (a), even though we are considering separate 1 hour periods. I would have said that the answer was <span class="math-container">$2P(Y=2) = 2 \frac{7^2 e^{-7} }{2!}$</span>. Why is this not the answer? Does it have something to do with the "Poisson process"? The book does not dive much into this, and wikipedia's article was overwhelming.</p>
<p>Thank you</p>
| angryavian | 43,949 | <p>The question is asking that the probability that exactly two people arrive in the combined two-hour period. In particular, it is ok for one person to arrive at 1:30 and one other to arrive at 2:30.</p>
<p>You computed the probability that exactly two people arrived between 1:00 and 2:00 and added it to the probability that exactly two people arrived between 2:00 and 3:00, which is not what the question is asking.</p>
|
8,567 | <p>When highlighting text using <code>Style</code> and <code>Background</code>, as in <code>Style["Test ", White, Background -> Lighter@Blue]</code> is there a way to pad (ie, enlarge) the bounding box? </p>
<p>The bottom of the background seems coincident with the base of the text: <img src="https://i.stack.imgur.com/SLimA.jpg" alt="here"></p>
| Rojo | 109 | <p>Does this work as you want to?</p>
<pre><code>SetAttributes[f, HoldAllComplete];
{first, rest___} ^:= HoldComplete[rest]
f[args___] := {first, args}
f[own, down[1], sub[1][2], N[n], up]
</code></pre>
<blockquote>
<p>HoldComplete["OwnValue", "DownValue", "SubValue", 3.14, up]</p>
</blockquote>
|
161,678 | <p>Assume a process with Itô dynamics of the generic form
$$dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$$</p>
<p>and let $f:\mathbb{R}\to\mathbb{R}$ be borel-measurable. Is the following function smooth ?
$$g(t,x)=\mathbb{E}[f(X_T)|\mathcal{F}_t]$$</p>
<p>I remember comming upon the proof of above once but I cannot find it any longer.</p>
| Dan | 44,169 | <p>This depends on what you assume of $\mu$, $\sigma$, and $f$. I'll provide a few examples. Let's assume $f$ is bounded throughout, since what I'll say below is still true in the unbounded case under suitable growth assumptions.</p>
<p>First, as long the SDE is well-posed (in the sense of weak existence and uniqueness in law), then the solution is a Feller process (this is somewhere in Stroock and Varadhan's book), and your function $g$ is continuous in $x$ whenever $f$ is continuous. If $f \in C^2$, then in fact $g$ is in $C^{1,2}$ (see Theorem 24.1 of Kallenberg's "Foundations of modern probability").</p>
<p>If you're interested in higher order derivatives, consult the vast literature on stochastic flows of diffeomorphisms. For example, if $\mu = \mu(x)$ and $\sigma = \sigma(x)$ are time-homogeneous and $C^\infty$ with bounded first derivative, then $g(t,\cdot)$inherits derivatives from $f$; i.e. whenever $f \in C^k$ then $g(t,\cdot) \in C^k$ for each $t$. (See Theorem V.13.8 of Rogers & Williams, "Diffusions, Markov processes, and martingales" vol 2.)</p>
<p>Finally, if you are really only willing to assume $f$ is measurable, the story is a bit more complicated. You'll want uniformly nondegenerate volatility, i.e. $\sigma\sigma^\top \ge \delta I$ for some $\delta > 0$, so that the noise helps you recover some smoothness. In this case, at least if $\mu = \mu(x)$ and $\sigma = \sigma(x)$ are time-homogeneous, then the map from the initial condition $x$ to $\text{Law}(X^x_t)$ (the distribution of the SDE solution at some time $t > 0$) is typically continuous in total variation. See sections 11.3 and 11.4 of Stroock and Varadhan's book.</p>
|
3,009,112 | <p>I am a geographer/ecologist and I want to know how to accurately calculate volume of a lake or a reservoir? I am not looking for a vague estimate which is generally calculated using surface area and mean height parameters assuming the body is of a certain shape (truncated cone/triangle or circular). Since reservoirs are completely irregular in shape I am having difficulties in using the traditional volume formulae. Any help or suggestion would be greatly appreciated. Thank you. </p>
| Derek Holt | 2,820 | <p>I think the following is essentially the same argument as in the proof of the Brauer-Fowler result cited by the_fox.</p>
<p>Let <span class="math-container">$I$</span> be the set of involutions in <span class="math-container">$G$</span> and <span class="math-container">$m=|I|$</span>, <span class="math-container">$n=|G|$</span>. </p>
<p>For <span class="math-container">$x \in G$</span>, define <span class="math-container">$C^*_G(x) = \{ g \in G \mid g^{-1}xg = x^{\pm 1} \}$</span>. Then either <span class="math-container">$C^*_G(x) = C_G(x)$</span>, or <span class="math-container">$|C^*_G(x):C_G(x)|=2$</span>.
Notice that, if <span class="math-container">$u,v \in I$</span> and <span class="math-container">$x=uv \not\in I$</span>, then <span class="math-container">$u,v \in C^*_G(x) \setminus C_G(x)$</span>.</p>
<p>Now, for <span class="math-container">$x \in G$</span>, let <span class="math-container">$\beta(x)$</span> be the number of ordered pairs <span class="math-container">$(u,v) \in I \times I$</span> with <span class="math-container">$uv=x$</span>. Then <span class="math-container">$m^2 = \sum_{x \in G}\beta(x)$</span>.</p>
<p>To estimate <span class="math-container">$\beta(x)$</span>, we consider three cases.</p>
<p>If <span class="math-container">$x = 1$</span>, then clearly <span class="math-container">$\beta(x)=m$</span>.</p>
<p>If <span class="math-container">$x \in I$</span> and <span class="math-container">$uv = x$</span> with <span class="math-container">$u,v \in I$</span>, then <span class="math-container">$u,v \in C_G(x)$</span>, so <span class="math-container">$\beta(x)$</span> is the number of involutions in <span class="math-container">$C_G(x) \setminus \{x\}$</span>, which is at most <span class="math-container">$|C_G(x)|-2$</span>.</p>
<p>Otherwise, if <span class="math-container">$x \ne 1$</span> and <span class="math-container">$x \not\in I$</span>, then either <span class="math-container">$\beta(x)=0$</span>, or <span class="math-container">$\beta(x) \le |C^*_G(x)| - |C_G(x)| = |C_G(x)|$</span>.</p>
<p>We will use the well-known result that the only finite nonabelian simple group with a proper subgroup of index at most 5 is <span class="math-container">$A_5$</span>, and that the result is true in <span class="math-container">$A_5$</span>.
(In fact <span class="math-container">$|I|=|G|/4$</span> in <span class="math-container">$A_5$</span>.)</p>
<p>So, for <span class="math-container">$m \in I$</span>, we have <span class="math-container">$|C_G(x)| \le n/6$</span>, and for <span class="math-container">$m \in G \setminus (I \cup \{1\})$</span>, <span class="math-container">$|C^*_G(x)| \le n/6$</span> and hence <span class="math-container">$|C_G(x)| \le n/12$</span>. So we get</p>
<p><span class="math-container">$$m^2 \le m + m(n/6-2) + (n-m-1)n/12,$$</span>
and then, putting <span class="math-container">$k=n/m$</span>, we have
<span class="math-container">$$n^2/k^2 \le n/k + n(n/6-2)/k + (n-n/k-1)n/12= n^2/(12k) - n/k +n^2/12- n/12,$$</span>
so <span class="math-container">$$n\left(\frac{1}{k^2} -\frac{1}{12k}-\frac{1}{12}\right) \le -\frac{1}{k} - \frac{1}{12},$$</span>
but the left hand side is positive for <span class="math-container">$k \le 3$</span>, so <span class="math-container">$|I| < |G|/3$</span>, QED.</p>
|
3,335,060 | <blockquote>
<p>The numbers of possible continuous <span class="math-container">$f(x)$</span> defiend on <span class="math-container">$[0,1]$</span> for which <span class="math-container">$I_1=\int_0^1 f(x)dx = 1,~I_2=\int_0^1 xf(x)dx = a,~I_3=\int_0^1 x^2f(x)dx = a^2 $</span> is/are</p>
<p><span class="math-container">$(\text{A})~~1~~~(\text{B})~~2~~(\text{C})~~\infty~~(\text{D})~~0$</span></p>
<p>I have tried the following:
Applying ILATE (the multiplication rule for integration) - nothing useful comes up, only further complications like the primitive of the primitive of f(x). No use of the given information either.
Using the rule <span class="math-container">$$ \int_a^b g(x)dx = \int_a^b g(a+b-x)dx$$</span>
I solved all three constraints to get
<span class="math-container">$$ \int_0^1 x^2f(1-x)dx = (a-1)^2 \\
\text{or} \int_0^1 x^2[f(1-x)+f(x)]dx = (a-1)^2 +a^2 \\$$</span>
Then I did the following - if f(x) + f(1-x) is constant, solve with the constraints to find possible solutions. Basically I was looking for any solutions where the function also follows the rule that f(x) + f(1-x) is constant.
Solving with the other constraints, I obtained that f(x) will only follow all four constraints if the constant [= f(x) + f(1-x)] is 2, and a is <span class="math-container">$\frac{√3\pm1}{2}$</span>.</p>
</blockquote>
| rrainbow | 642,066 | <p>Similar to your last proof, I found a positive function whose integral is <span class="math-container">$e^3-20$</span>.</p>
<p>For <span class="math-container">$$f(x)=\frac{1}{186}(x-1)^2(x-2)^4e^x\ge0$$</span></p>
<p>we have <span class="math-container">$$\int_{0}^{3}f(x)dx=e^3-20$$</span></p>
|
2,777,555 | <p>Prove that if $f(0)=0$ and $f'(0)=0$, then $f(x)=0$ for all $x$. </p>
<p>Hint: The idea
is to multiply both sides of the equation $f''(x)+ f(x) = 0$ by something that
makes the left-hand side of the equation into the derivative of something.</p>
<p>I'm not sure how to proceed and don't really understand the hint.</p>
| marty cohen | 13,079 | <p>Multiply by f'.
It becomes
$((f^2)'+(f'^2)')/2$.</p>
|
264,587 | <p><strong>NOTE</strong></p>
<p>I'm sorry, my question was not clear. I want to know all the ways to split a list with a given length simply, <strong>rather than split a cyclic substitution</strong>.
If a given list has length <span class="math-container">$N$</span> and the rule is <span class="math-container">${m, n, p, ...}$</span>, we should get a list of length <span class="math-container">${}_{N} C_{m} {}_{N-m} C_{n} {}_{N-m-n} C_{p} \dots = \frac{N!}{m! n! p! \cdots}$</span> if all elements of <span class="math-container">$N$</span> are independent.</p>
<p>Other examples:
<code>Length@partitionList[{a, b, c, d, e, f, g, h}, {2, 2, 4}]</code>
returns <span class="math-container">$420 = \frac{8!}{2! 2! 4!}$</span></p>
<p><code>Length@partitionList[{a, b, b, c, d, e, e, f}, {2, 2, 4}]</code>
returns 173</p>
<hr>
<p><strong>Question</strong></p>
<p>I want to split a given list into sets of lists, whose lengths are given.
For example, this means if we split a list <code>{a, b, c, d}</code> (length 4) to two lists with length <code>{1, 3}</code> (the sum of lengths should be 4), we obtain</p>
<pre><code>{{{a}, {b, c, d}}, {{b}, {c, d, a}}, {{c}, {d, e, a}}, {{d}, {e, a, b}}
</code></pre>
<p>(here, we don't care about ordering for elements in each sublist).</p>
<p>To achieve this, I prepared the following function:</p>
<pre class="lang-Mathematica prettyprint-override"><code>partitionList[l_List, p_List] :=
DeleteDuplicates@(Module[{$tmp, $deleteList, $lastchoose,
l2 = Range[Length@l]},
$tmp = Subsets[l2, {p[[1]]}];
$deleteList = Flatten /@ $tmp;
If[Length@p > 1,
Do[
$lastchoose =
Table[Subsets[
Delete[l2, {#} & /@ $deleteList[[$j]]], {p[[$i]]}], {$j,
Length@$deleteList}];
$tmp =
Replace[Flatten[
Tuples /@ Transpose[{{#} & /@ $tmp, $lastchoose}], 1],
x_ /; Depth@x > 2 :> Sequence @@ x, {2} ];
$deleteList = Flatten /@ $tmp;
, {$i, 2, Length@p}]
];
Map[l[[#]] &, $tmp, {2}]
]
)
</code></pre>
<p>Here, the argument <span class="math-container">$l$</span> is a list which we want to split, and <span class="math-container">$p$</span> is a list of lengths of sublists.
In the previous example, <span class="math-container">$l$</span> is <code>{a, b, c, d}</code> and <span class="math-container">$p$</span> is <code>{1, 3}</code>.</p>
<p>However, since it is based on procedural programming, I believe there are more efficient ways.
Could you please suggest such a method?</p>
| Simon Woods | 862 | <p>This isn't any faster than the original code, just a different approach.</p>
<pre><code>partitionList2[list_, pat_] := Module[{rn, p},
rn = Range@Length@pat;
p = Flatten[MapThread[ConstantArray, {rn, pat}]];
DeleteDuplicates@
Table[Flatten /@ Reap[MapThread[Sow, {list, q}], rn][[2]], {q, Permutations[p]}]]
AbsoluteTiming[
a = partitionList[Range[12], {3, 1, 4, 2, 2}];]
(* {27.6628, Null} *)
AbsoluteTiming[
b = partitionList2[Range[12], {3, 1, 4, 2, 2}];]
(* {27.5801, Null} *)
Sort[a] == Sort[b]
(* True *)
</code></pre>
|
838,400 | <p>One question asking if $\mathbb{Z}^*_{21}$ is cyclic.</p>
<p>I know that the cyclic group must have a generator which can generate all of the elements within the group.</p>
<p>But does this kind of question requires me to exhaustively find out a generator? Or is there any more efficient method to quickly determine if a group is a cyclic group?</p>
| Ittay Weiss | 30,953 | <p>A finite group is cyclic if, and only if, it has precisely one subgroup of each divisor of its order. So if you find two subgroups of the same order, then the group is not cyclic, and that can help sometimes. </p>
<p>However, $Z^*_{21}$ is a rather small group, so you can easily check all elements for generators.</p>
|
2,269,042 | <p>Consider $f(x) = \lfloor x \rfloor + \lfloor -x \rfloor $ . Now find value of $\lim_{x \to \infty} f(x) $ . I know that if $x_0 \in \mathbb{R}$ then $\lim_{x \to x_0} f(x) = -1$ but I don't know whether it is true or not in the infinity . </p>
| G Cab | 317,234 | <p>Defining with $ \left\{ x \right\}$ the fractional part of $x$, i.e.
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}
$$
and denoting by $[P]$ the <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow noreferrer"><em>Iverson bracket</em></a>
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
then we have that
$$
\begin{array}{l}
f(x) = \left\lfloor x \right\rfloor + \left\lfloor { - x} \right\rfloor = \left\lfloor x \right\rfloor - \left\lceil x \right\rceil = - \left( {\left\lceil x \right\rceil - \left\lfloor x \right\rfloor } \right) = \\
= - \left\lceil {\left\{ x \right\}} \right\rceil = - 1 + \left[ {x \in Z} \right] = \left\{ {\begin{array}{*{20}c}
{ - 1} & {\left| {\;x \notin Z} \right.} \\
0 & {\left| {\;x \in Z} \right.} \\
\end{array}} \right. \\
\end{array}
$$</p>
<p>and the limit for $x \to \infty$ does not exist.</p>
|
404,574 | <p>Suppose that:</p>
<p>$Y \pmod B = 0$</p>
<p>$Y \pmod C = X$</p>
<p>I know $B$ and $C$. $Y$ is unknown, it might be an extremely large number, and it does not interest me. </p>
<p>The question is: Is it possible to find $X$, and if so, how?</p>
| Key Ideas | 78,535 | <p>$\bigg\lbrace\begin{eqnarray}y\equiv 0\pmod b\\ y\equiv x\pmod c\end{eqnarray}\bigg\rbrace $ $\!\iff\! \bigg\lbrace\begin{eqnarray} y &=& j b,\ \ &{\rm some}\ \ j\in \Bbb Z\\ y &=& x\! +\! k c,\ \ &{\rm some}\ \ k\in \Bbb Z\end{eqnarray}\bigg\rbrace\!\iff\! x = jb\!-\!kc,\ \ {\rm some}\ \ j,k\in\Bbb Z$</p>
<p>Thus, by Bezout, the possible values of $\,x\,$ are precisely the multiples of $\,\gcd(b,c).$</p>
|
363,767 | <p>An ellipse is specified $ x^2 + 4y^2 = 4$, and a line is specified $x + y = 4$. I need to find the max/min distances from the ellipse to the line.</p>
<p>My idea is to find two points $(x_1, y_1)$ and $(x_2,y_2)$ such that the first point is on the ellipse and the second point is on the line. Furthermore, the line segment formed by these two points should be perpendicular to the line (slope = 1). This gives 3 constraints $g_i$ and an objective $f$. </p>
<p>$$ g_1: x_1 ^2 + 4y_1 ^2 - 4 = 0$$
$$ g_2: x_2 + y_2 - 4= 0 $$
$$ g_3:\frac{(y_2 - y_1)}{(x_2 - x_1)} - 1 = 0 $$
$$ f: (y_2 - y_1)^2 + (x_2- x_1)^2 $$ </p>
<p>Then I compute $\nabla g_i$ and $\nabla f$:</p>
<p>$$ \nabla g_1 = (2x_1, 8y_1, 0, 0) $$
$$ \nabla g_2 = (0, 0, 1, 1)$$
$$ \nabla g_3 = ( (y_2 - y_1)(x_2 - x_1)^{-2}, -(x_2 - x_1)^{-1}, -(y_2 - y_1)(x_2 - x_1)^{-2}, (x_2 - x_1)^{-1} )$$
$$ \nabla f = (-2(x_2-x_1), -2(y_2-y_1), 2(x_2-x_1), 2(y_2 - y_1))$$ </p>
<p>At this point I try to solve $\nabla f = \sum\lambda_i\nabla g_i$, which, together with the constraints, gives me 7 equations with 7 variables. I'm not sure how to solve this system. </p>
<p>$$\lambda_1 2 x_1 + \lambda_3(y_2-y_1)(x_2-x_1)^{-2} = -2(x_2-x_1)$$
$$\lambda_1 8 y_1 - \lambda_3 (x_2 - x_1)^{-1} = -2 (y_2 - y_1)$$
$$\lambda_2 - \lambda_3 (y_2 - y_1)(x_2 - x_1)^{-2} = 2(x_2 - x1)$$
$$\lambda_2 + \lambda_3 (x_2 - x_1)^{-1} = 2(y_2 -y_1)$$ </p>
<p>Is there an easy way to see the solutions of this system in $x_1, y_1, x_2, y_2$? If not, is there an easier formulation of this optimization problem?</p>
| Will Jagy | 10,400 | <p>much easier. if your line and ellipse intersect, the minimum distance is zero. if not, it occurs on a line segment that is orthogonal to both your line and to the ellipse. which is not hard to find. there should be two such line segments, one gives the actual min, one gives a local max...draw some graphs! also, no max, there are points on the line arbitrarily far away from your ellipse. </p>
|
1,278,329 | <p>Solve the recurrence $a_n = 4a_{n−1} − 2 a_{n−2}$</p>
<p>Not sure how to solve this recurrence as I don't know which numbers to input to recursively solve?</p>
| Scott | 89,128 | <p>We can represent this recursion with matrices:</p>
<p>$$
\begin{bmatrix}a_n \\ a_{n-1} \end{bmatrix}
=
\begin{bmatrix}4 & -2 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix}a_{n-1} \\ a_{n-2} \end{bmatrix}
$$</p>
<p>Or $x_n = A\cdot x_{n-1}$ where $x_n$ is the vector specified above. This is a simple difference equation and we can see that</p>
<p>$$
x_n = A^n \cdot x_0
$$</p>
<p>by expanding $x_n = A x_{n-1} = A \cdot A \cdot x_{n-2} = \cdots = A^n \cdot x_0 $</p>
<p>Computing this matrix power is difficult and hard numerically; there's no easy solution that I know of. To get around this, let's use eigenvectors $V$ and the matrix of eigenvalues $\Lambda$.</p>
<p>$$
x_n = (V\Lambda V^{-1})^n x_0 = V \Lambda^n V^{-1}x_0
$$ </p>
<p>The computation of $\Lambda^n$ is straightforward. $\Lambda$ is diagonal so the computation of this matrix power is just each term raised to a power.</p>
<p>For this $A$,</p>
<p>$$
\begin{aligned}
V &=
\begin{bmatrix}
2+\sqrt{2} & 2-\sqrt{2} \\ 1 & 1
\end{bmatrix}
\\
\Lambda &=
\begin{bmatrix} 2+\sqrt{2} & 0 \\ 0 & 2-\sqrt{2} \end{bmatrix}
=
\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}
\end{aligned}
$$</p>
<p>Carrying out this computation, we see that</p>
<p>$$
x_n = \frac{1}{\sqrt{8}}
\begin{bmatrix}
\lambda_1^{n+1} - \lambda_2^{n+1} & \lambda_1^{n+1}\lambda_2 + \lambda_1\lambda_2^{n+1} \\
\lambda_1^n - \lambda_2^n & \lambda_1^n\lambda_2 + \lambda_1 \lambda_2^n
\end{bmatrix}
\cdot x_0
$$</p>
|
4,121,607 | <p>I want to find a function which satisfies certain following limits.</p>
<p>The question is:
Find a function which satisfies</p>
<p><span class="math-container">$$
\lim_{x\to5} f(x)=3, \text{ and } f(5) \text{ does not exist}
$$</span></p>
<p>I would think that because it says <span class="math-container">$f(5)$</span> doesn't exist, there must be a fraction with <span class="math-container">$(x-5)$</span> on the bottom. I would think <span class="math-container">$f(x) = \frac{15}{x-5}$</span> but that tends to infinity as <span class="math-container">$x\to5$</span></p>
| hamam_Abdallah | 369,188 | <p><span class="math-container">$$f(x)=\frac{\sin(3(x-5))}{x-5}$$</span>
or
<span class="math-container">$$f(x)=\frac{e^{3(x-5)}-1}{x-5}$$</span>
or
<span class="math-container">$$f(x)=\frac{\ln(3x-14)}{x-5}$$</span>
satifies the wanted conditions.</p>
|
4,408,507 | <p>We study the definition of Lebesgue measurable set to be the following:</p>
<p>Let <span class="math-container">$A\subset \mathbb R$</span> be called Lebesgue measurable if <span class="math-container">$\exists$</span> a Borel set <span class="math-container">$B\subset A$</span> such that <span class="math-container">$|A-B|=0$</span>,where <span class="math-container">$|.|$</span> denotes the Lebesgue outer measure of a set.</p>
<p>Then we have theorems like:</p>
<p><span class="math-container">$A\subset \mathbb R$</span> is Lebesgue measurable iff</p>
<p><span class="math-container">$(1)$</span> Given any <span class="math-container">$\epsilon>0$</span> there exists <span class="math-container">$F\subset A$</span> closed such that <span class="math-container">$|A-F|<\epsilon$</span>.</p>
<p><span class="math-container">$(2)$</span> Given any <span class="math-container">$\epsilon>0$</span> there exists <span class="math-container">$G\supset A$</span> open such that <span class="math-container">$|G-A|<\epsilon$</span>.</p>
<p>I have two questions here.</p>
<p>First is that what motivates the definition of Lebesgue measurable sets and second is that why we are approximating Lebesgue measurable sets from below by closed sets and from above by open sets.I am studying the topic measure theory from Sheldon Axler's book that does not give motivation behind these definitions and theorems.Can someone give me motivation behind these things?</p>
| L. F. | 620,160 | <p>Here is an alternative trigonometric proof of the inequality for fun. Denote <span class="math-container">$$
x = \frac{A}{2}, \qquad y = \frac{B}{2}, \qquad \{x, y, x + y\} \subset [0, 90^\circ].
$$</span>
Then, using the <a href="https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities" rel="nofollow noreferrer">product-to-sum formula</a>
<span class="math-container">$$
\sin x \sin y = \frac{\cos (x - y) - \cos (x + y)}{2}
$$</span>
as well as the <a href="https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means" rel="nofollow noreferrer">AM-GM inequality</a>,
we have
<span class="math-container">$$ \begin{split}
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2}
&= \sin x \sin y \sin (90^\circ - x - y) \\
&= \sin x \sin y \cos (x + y) \\
&= \frac{\cos (x - y) - \cos (x + y)}{2} \cos(x + y) \\
&\le \frac{1 - \cos(x + y)}{2} \cos(x + y) \\
&\le \frac{1}{2} \Bigl( \frac{1}{2} \Bigr)^2 = \frac{1}{8},
\end{split} $$</span>
where equality holds for the first <span class="math-container">$\le$</span> sign if and only if <span class="math-container">$x = y$</span>, and equality holds for the second <span class="math-container">$\le$</span> sign if and only if <span class="math-container">$\cos(x + y) = \frac{1}{2}$</span>. Therefore, we have
<span class="math-container">$$
\sin \frac{A}{2} \, \sin \frac{B}{2} \, \sin \frac{C}{2} \le \frac{1}{8}
$$</span>
where equality holds if and only if <span class="math-container">$A = B = C = 60^\circ$</span>.</p>
|
4,526,852 | <p>I need some help understanding the fallacy in the following reasoning (thank you in advance!). It is essentially implying that for a single sample from a population, you can know a population parameter (like the EV) precisely with a few seemingly reasonable assumptions. I’ll frame it as an example.</p>
<p>Let’s say that we are trying to infer something about the return distribution (assume normal) of a financial instrument (population distribution). We make an assumption about the variance of this distribution. We then create samples and calculate their averages, and each sample consists of 100 draws from the population distribution (which again we don’t know).</p>
<p>Using our assumption about the variance of the population distribution, we can calculate the variance of the sampling distribution, which is the distribution which results after sampling (again, each sample being of 100 draws from the population distribution and the resultant average) infinitely many times. We can compute a 95% confidence or Bayesian credible interval using the variance of the sampling distribution, which implies that if we take 1 incremental sample and calculate its average, the probability that the population mean falls within said interval around the sample average is 95% (Yes I realize frequentists will challenge this for confidence intervals, but the credible interval unequivocally states this). So, if we take an incremental sample, it would follow that the probability that the population mean falls within the band described by the confidence interval around the estimated mean of the sample is 95%.</p>
<p>We’ve now basically created a probability distribution for the population mean itself, and the midpoint of the distribution is the estimated sample mean. If we assume that this distribution is normally distributed or even just that the likelihood that the population mean/EV falls within the 95% confidence band around the estimated sample mean or the in the 2.5% tails on either side of the band is uniform, then that would imply that in expectation, the expected value of the population mean/EV is the estimated sample mean.</p>
<p>It would seem obvious that that cannot be true given it would imply that that you could then know your true EV for any population simply by referencing a sample of any size. You can’t say Steph Curry’s free throw make probability is 50% after watching him shoot 2 free throws and miss 1.</p>
<p>Where exactly does this go wrong?</p>
| Lelouch | 991,491 | <p><span class="math-container">$\frac{1}{\sin(x)}$</span> is surely not bounded between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>.</p>
<p>You are falling into a pitfall since the inverse function is undefined in <span class="math-container">$0$</span>, you actually have to apply it twice to both inequalities <span class="math-container">$-1 \leq \sin(x) \leq 0$</span> and <span class="math-container">$0 \leq \sin(x) \leq 1$</span> to obtain <span class="math-container">$-1 \geq \frac{1}{\sin(x)} \geq -\infty$</span> and <span class="math-container">$1 \geq \frac{1}{\sin(x)} \geq +\infty$</span> and thus <span class="math-container">$ \frac{1}{\sin(x)} \in (-\infty, -1] \cup [1, +\infty)$</span> .</p>
<p>You need to be careful when manipulating inequalities, you can only apply a function on the inequalities if it's well defined on the whole interval (also note that the inverse function is not a decreasing function on <span class="math-container">$\mathbb{R}^*$</span> which is basically what you tried to use).</p>
|
4,526,852 | <p>I need some help understanding the fallacy in the following reasoning (thank you in advance!). It is essentially implying that for a single sample from a population, you can know a population parameter (like the EV) precisely with a few seemingly reasonable assumptions. I’ll frame it as an example.</p>
<p>Let’s say that we are trying to infer something about the return distribution (assume normal) of a financial instrument (population distribution). We make an assumption about the variance of this distribution. We then create samples and calculate their averages, and each sample consists of 100 draws from the population distribution (which again we don’t know).</p>
<p>Using our assumption about the variance of the population distribution, we can calculate the variance of the sampling distribution, which is the distribution which results after sampling (again, each sample being of 100 draws from the population distribution and the resultant average) infinitely many times. We can compute a 95% confidence or Bayesian credible interval using the variance of the sampling distribution, which implies that if we take 1 incremental sample and calculate its average, the probability that the population mean falls within said interval around the sample average is 95% (Yes I realize frequentists will challenge this for confidence intervals, but the credible interval unequivocally states this). So, if we take an incremental sample, it would follow that the probability that the population mean falls within the band described by the confidence interval around the estimated mean of the sample is 95%.</p>
<p>We’ve now basically created a probability distribution for the population mean itself, and the midpoint of the distribution is the estimated sample mean. If we assume that this distribution is normally distributed or even just that the likelihood that the population mean/EV falls within the 95% confidence band around the estimated sample mean or the in the 2.5% tails on either side of the band is uniform, then that would imply that in expectation, the expected value of the population mean/EV is the estimated sample mean.</p>
<p>It would seem obvious that that cannot be true given it would imply that that you could then know your true EV for any population simply by referencing a sample of any size. You can’t say Steph Curry’s free throw make probability is 50% after watching him shoot 2 free throws and miss 1.</p>
<p>Where exactly does this go wrong?</p>
| Bernkastel | 551,169 | <p>The reciprocal function <span class="math-container">$t \mapsto 1/t$</span> is a decreasing functions on the two separate intervals <span class="math-container">$(-\infty,0)$</span> and <span class="math-container">$(0,\infty$</span>). So, you cannot take reciprocals and reverse inequality sign if your functions not entirely belongs to <span class="math-container">$(-\infty,0)$</span> or <span class="math-container">$(0,\infty)$</span>. For example, it is true that <span class="math-container">$-2 < 3$</span> but it is false that <span class="math-container">$1/3<-1/2$</span>.</p>
<p>So you must distinguish two cases: <span class="math-container">$0< \sin x \le 1$</span> or <span class="math-container">$-1 \le \sin x < 0$</span>. In both cases, we can take reciprocals and get <span class="math-container">$\frac{1}{\sin x} \le -1$</span> or <span class="math-container">$\frac{1}{\sin x} \ge 1$</span>. This is where the "or" come from.</p>
|
2,352,313 | <p>If $f_n$ is the number of permutations of numbers $1$ to $n$ that no number is in it's place(I think same as $D_n$)and $g_n$ is the number of the same permutations with exactly one number in it's place Prove that $\mid f_n-g_n \mid =1$.</p>
<p>I need a proof using mosly combinatorics not mostly algebra.I think we should find sth like below:</p>
<p>$f_n-g_n=g_{n-1}-f_{n-1}$</p>
<p>But I can't do that.</p>
| drhab | 75,923 | <p>Concerning <a href="https://en.wikipedia.org/wiki/Derangement#Counting_derangements" rel="nofollow noreferrer">derangements</a> we can prove with inclusion/exclusion that:
$$!n=n!\sum_{i=0}^n\frac{(-1)^i}{i!}$$</p>
<p>The RHS can also be written as:$$n\left[(n-1)!\sum_{i=0}^{n-1}\frac{(-1)^i}{i!}\right]+(-1)^n=n\left[!(n-1)\right]+(-1)^n$$</p>
<p>Here $f_n=!n$ and as suggested in the comments $g_n=nf_{n-1}=n[!(n-1)]$.</p>
<p>This proves that $f_n-g_n=(-1)^n$ and consequently $|f_n-g_n|=1$.</p>
|
1,928,439 | <p>Is there a space whose dual is $F^m$? ($F$ is the field w.r.t. which the original set is a vector space)</p>
<p>I'm trying to do the following exercise:</p>
<p><a href="https://i.stack.imgur.com/j9WMP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j9WMP.png" alt="enter image description here"></a></p>
<p>I only need help with the direction from $\Gamma$ is injective.</p>
<p>I was thinking of using the theorem that states that if a dual map is injective then the linear map must be surjective. And if a linear map $T$ from $\mathcal{L}(F^m,V)$ with $T(e_i)=v_i$ is surjective, then the $v_i$ span $V$.</p>
<p>Let's assume that $\Gamma^*(\phi)=(\phi(v_1),...,\phi(v_m))$, where $v_i \in V$, and $\phi \in \mathcal{L}(V,F)$. Is there a way to think of $\Gamma^*$ as the dual map of a linear map $\Gamma$?</p>
| quid | 85,306 | <p>Let $e_1, \dots, e_m$ a basis of $F^m$. </p>
<p>Let $f:F^m \to F$ be a linear form, that is an element of the dual space. </p>
<p>Then for any $v= a_1e_1 + \dots + a_me_m$ we have $f(v)= f(a_1e_1 +\dots + a_me_m) = a_1f(e_1) +\dots + a_mf(e_m)$. Thus $f$ is uniquely determined by
$f(e_1), \dots, f(e_m)$ that is an $m$ elements from $F$, viz. an element of $F^m$. </p>
<p>Conversely for every $(b_1, \dots, b_m) \in F^m$ we can define an associated linear form via defining $g(e_i)=b_i$ and thus $g(a_1e_1 +\dots + a_me_m) = a_1b_1 +\dots + a_mb_m$.</p>
<p>It follows the elements of the dual space are in bijection with the elements of $F^m$. </p>
<p>It is not hard to show that this bijection is linear, and thus an isomorphism.</p>
<p>Note: the bijection depends on the basis. This is why one says there is no canonical isomorphism, but there still is an isomorphism. </p>
<p>However the spaces are not literally equal. The one is made up of $m$-tuples of elements of $F$ the other of linear maps from $F^m$ to $F$, that is certain subsets of the Cartesian product of $F^m \times F$. </p>
<p>However, when somebody says "the dual of $V$ is {something}", then usually this {something} is just isomorphic to the dual not literally the sets of linear maps from $V$ to $F$. </p>
<hr>
<p>For the supplementary question that motivated the question: </p>
<p>Suppose the function $\Gamma$ is injective. This means that the only form that vanishes on all of $v_1, \dots , v_m$ is the form that maps the whole space to $0$. But this means that $v_1, \dots, v_m$ must span $V$, for if not then we could define a form by saying it is $0$ on the span of $v_1, \dots, v_m$ and $1$ for $w_1, \dots, w_n$ where $w_1, \dots, w_n$ is an independent set that completes $v_1, \dots, v_m$ to a generating set of the space. </p>
|
2,386,602 | <p>This is a question from an exam I recently failed. </p>
<p>What is the radius of convergence of the following power series? $$(a) \sum_{n=1}^\infty(n!)^2x^{n^2}$$ and $$(b) \sum_{n=1}^\infty \frac {x^{n^2}}{n!}$$</p>
<p>Edit: Here's my attempt at the first one, if someone could tell me if it's any good...</p>
<p>$\sum_{n=1}^\infty(n!)^2x^{n^2}$=$\sum_{n=1}^\infty a_nx^n$ where $a_n= 0$ for any $n\notin \{k\in N| \exists t\in N: k=t^2\}$ and $a_n = (n!)^2$ else. So the radius of convergence would be the inverse of $\lim_{n\rightarrow \infty}{(n!)^{2/n}}=\lim e^{2/n\cdot log(n!) }$. The exponent with log of factorial becomes a series, $\sum_{n=1}^{\infty} \frac{logn}{n}$ which diverges by comparison test with $\frac{1}{n}$, so the radius of convergence would be equal to $0$.</p>
<p>Second edit: This is wrong. Correct answer below.</p>
| J.G. | 56,861 | <p>Use the ratio test. The ratio of consecutive terms is of the form $n^p x^{2n-1}$. For $|x|<1$, the $n\to\infty$ limit is $0$; for $|x|>1$, it is $\infty$. Thus the radius of convergence is $1$.</p>
|
883,620 | <p>$a,b,c \geq 0$ and $a^2+b^2+c^2+abc=4$ prove that $ab+bc+ac-abc \leq 2$
can any one help me with this problem,I believe Dirichlet's theorem is the key for this
sorry for making mistake over and over again,but i'm certain that the inequality is true now.</p>
| Macavity | 58,320 | <p>Note $a^2+b^2+c^2+abc=1 \implies 0\le a,b,c \le 1$. Hence
$$(1-a)(1-b)c\ge 0 \implies c\ge bc+ca-abc$$ </p>
<p>So now it is enough to prove $a+b+c \le 2$, which is obvious from AM-QM
$$\frac{a+b+c}3 \le \sqrt{\frac{a^2+b^2+c^2}3}\le \sqrt{\frac13}$$ </p>
|
883,620 | <p>$a,b,c \geq 0$ and $a^2+b^2+c^2+abc=4$ prove that $ab+bc+ac-abc \leq 2$
can any one help me with this problem,I believe Dirichlet's theorem is the key for this
sorry for making mistake over and over again,but i'm certain that the inequality is true now.</p>
| Macavity | 58,320 | <p>For your last revision of the question (you'll need to change the title too)...</p>
<p>Note among any three numbers $a,b,c$, some two will be either $\le 1$ or $\ge 1$. WLOG let these be $a,b$. So $(1-a)(1-b)c \ge 0 \implies bc + ca \le c+abc \implies ab+bc+ca -abc\le ab+c$. Now you need to show $ab \le 2-c$. For that, note:</p>
<p>$$4-c^2 = a^2+b^2+abc \ge 2ab+abc = ab(2+c)$$ </p>
<p>P.S. You may want to check <a href="http://www.artofproblemsolving.com/Forum/blog.php?b=78664" rel="nofollow">http://www.artofproblemsolving.com/Forum/blog.php?b=78664</a> for more variants, but don't keep changing the question!</p>
|
1,647,327 | <p>suppose S is a metric space and $B(S)$ is the set of bounded functions and $C_b(S)$ is the set consisting of bounded continuous functions.</p>
<p>Prove that $C_b(S)$ is a closed subspace of $B(S)$.</p>
<p>I thought of looking at the complement $B(S) \backslash C_b(S) = \{f| \text{ f is bounded and not continuous}\}$ and proving this is open. </p>
<p>But how does one check that for each element $f \in B(S) \backslash C_b(S)$, there exists an $\epsilon$ such that $B_f(\epsilon) \subset B(S) \backslash C_b(S)$?</p>
| noctusraid | 185,359 | <p>An alternative answer (based more on OP's knowledge):</p>
<p>In a metric space $(X, \Vert \cdot \Vert_X)$ and $A \subset X$ we have the following assertion:</p>
<p>$$ x \in \bar{A} \iff \exists (x_n)_{n \in \mathbb N} \subset A: x_n\to x \in X $$</p>
<p>This means in words: an element $x$ of a subspace $A \subset X$ is in the closure of $A$ (the smalles closed set containing $A$) if and only if there is a sequence <strong>in A</strong> that <strong>converges in X</strong>. To see that this is equivalent to the definition of closed I leave for you to check with 2 hints:</p>
<p>First: Use the toplogical definition of a closure.
Second: How does a topology induces by a metric look like? How is convergence defined?</p>
<p>Now with this at hands we pick an arbitrary sequence $f_n$ in $C_b(S)$ which <strong>does convergence in $B(S)$ !</strong> If the limit, say $f$, now lies in $C_b(S)$ we have shown that $C_b(S)=cl(C_b(S))$. To conclude see post above.</p>
|
1,386,367 | <p>I'm interested in the definite integral</p>
<p>\begin{align}
I\equiv\int_{-\infty}^{\infty} \frac{1}{x^2-b^2}=\int_{-\infty}^{\infty} \frac{1}{(x+b) (x-b)}.\tag{1}
\end{align}</p>
<p>Obviously, it has two poles ($x=b, x=-b$) on the real axes and is thus singular. I tried to apply the contour integration methods mentioned <a href="https://math.stackexchange.com/questions/564952/complex-integration-poles-real-axis">here</a>, where they discuss the integral
\begin{align}
\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx = -\frac{\pi}{b}\sin(ab),\tag{2}
\end{align}
where the r.h.s. is the solution derivable in multiple ways as shown in the above thread (e.g. circumventing the poles with infinitesimal arcs).</p>
<p>However, since in the seemingly more simple case (1) the nominator is symmetric in constrast to the situation in (2), I obtain
$$I=0,$$
as the residues equal up to different signs. E.g. consider the limit $a\rightarrow 0$ in (2) which gives $\sin(ab)\rightarrow 0$.</p>
<p>Based on some literature (in the context in which the integral is appearing) it seems that one <em>should</em> obtain</p>
<p>\begin{align}
I=-\frac{i\pi}{b}.
\end{align}</p>
<p>Of course, this can be realized by considering the modified integral
\begin{align}
I_{mod}\equiv\lim_{\eta\rightarrow 0^+} \int_{-\infty}^{\infty} dx \frac{1}{x^2-b^2+i\eta},
\end{align}
and closing the contour (e.g. a box closed at infinity) in the lower half plane.</p>
<p>However, in this approach one seems to have some freedom (sign of the infinitesimal contribution, why shift one pole upwards and another pole downwards and not e.g. both upwards?)</p>
<p>So let me explicitly phrase my questions:</p>
<ol>
<li>Is the value of the definite integral in (1) well-defined?</li>
<li>Is it equal to zero?</li>
<li>In any case, why would I include an infinitesimal shift as in (2) and not in another way?</li>
</ol>
<p>Thank you very much in advance!</p>
| sebigu | 32,185 | <p>If $0 \neq 2$ in the field and $P^2=P$, then the minimal Polynomial of $P$ divides $f := x^2-x$, which means it is $f$, $x$, or $x-1$. If it is $x$, $P=0$, and if it is $x-1$, $P=1$. Those cases are clear.</p>
<p>So suppose it is $x^2-x$. Then $I+P$ has minimal polynomial $(x-1)(x-2)=x^2-3x+2$. This means that $I$ is $((I+P)^2-3(I+P))/(-2)$ and so
$(I+P)^{-1}$ is $(I+P-3I)/(-2)=(P-2I)/(-2)$</p>
|
3,944,628 | <p>I'm reading a book and, in its section on the definition of a stopping time(continuous), the author declares at the start that for the whole section every filtration will be complete and right-continuous.</p>
<p>So, in the definition of a Stopping Time, how important are these conditions? Why would they matter?</p>
| J.G. | 56,861 | <p>Since <span class="math-container">$\sin x$</span> is <a href="https://en.wikipedia.org/wiki/Concave_function" rel="nofollow noreferrer">concave</a> on acute <span class="math-container">$x$</span>, by <a href="https://en.wikipedia.org/wiki/Jensen%27s_inequality" rel="nofollow noreferrer">Jensen's inequality</a> the maximum is found at <span class="math-container">$A/2=B/2=C/2=\pi/6$</span>, as <span class="math-container">$3\sin\pi/6=3/2$</span>.</p>
<p>Edit: since the OP mentioned in a comment on @B.Goddard's answer that they know differentiation, here's another proof the equilateral case achieves a maximum:</p>
<p>Keep using <span class="math-container">$\frac{C}{2}=\frac{\pi}{2}-\frac{A+B}{2}$</span>. To extemize <span class="math-container">$\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}$</span> simultaneously solve<span class="math-container">$$\tfrac12\cos\tfrac{A}{2}-\tfrac12\cos\tfrac{C}{2}=0,\,\tfrac12\cos\tfrac{B}{2}-\tfrac12\cos\tfrac{C}{2}=0$$</span>viz. <span class="math-container">$A=B=C$</span>. I'll leave the reader to check it's a maximum by considering second derivatives.</p>
|
4,090,408 | <p>Show that <span class="math-container">$A$</span> is a whole number: <span class="math-container">$$A=\sqrt{\left|40\sqrt2-57\right|}-\sqrt{\left|40\sqrt2+57\right|}.$$</span>
I don't know if this is necessary, but we can compare <span class="math-container">$40\sqrt{2}$</span> and <span class="math-container">$57$</span>: <span class="math-container">$$40\sqrt{2}\Diamond57,\\1600\times2\Diamond 3249,\\3200\Diamond3249,\\3200<3249\Rightarrow 40\sqrt{2}<57.$$</span> Is this actually needed for the solution? So <span class="math-container">$$A=\sqrt{57-40\sqrt2}-\sqrt{40\sqrt2+57}.$$</span> What should I do next?</p>
| fleablood | 280,126 | <p>First of all is <span class="math-container">$40\sqrt 2$</span> greater or less than <span class="math-container">$57$</span>? One assumes it's a little less.</p>
<p><span class="math-container">$(\frac {57}{40})^2 = \frac {(60-3)^2}{1600} = \frac {3600 +9 -2\cdot 3\cdot 60}{1600}= \frac {3600-360+ 9}{1600}= \frac {3249}{1600} > 2$</span> so <span class="math-container">$40\sqrt 2 < 57$</span>.</p>
<p>Well if you square <span class="math-container">$A$</span> you get</p>
<p><span class="math-container">$|40\sqrt 2 + 57| + |40\sqrt 2 - 57|-2\sqrt(|40\sqrt 2+57||40\sqrt 2-57|)=$</span></p>
<p><span class="math-container">$(57+ 40\sqrt 2) + (57 - 40 \sqrt 2) - 2\sqrt{(57+40\sqrt 2)(57-40\sqrt 2)}=$</span></p>
<p><span class="math-container">$57 + 57 - 2\sqrt{57^2-(40\sqrt 2)^2}=$</span></p>
<p><span class="math-container">$114 - 2\sqrt{57^2 - 1600\cdot 2}=$</span></p>
<p><span class="math-container">$114 -2 \sqrt {3249 - 3200} =$</span></p>
<p><span class="math-container">$114 - 2\sqrt{49} = $</span></p>
<p><span class="math-container">$114 - 2\cdot 7= $</span></p>
<p><span class="math-container">$114 - 14= 100$</span>.</p>
<p>So <span class="math-container">$A^2 = 100$</span> and <span class="math-container">$A = \pm 10$</span>.</p>
<p>How cute.</p>
<p>.....</p>
<p>Oh, how clever....<span class="math-container">$57 \pm 40\sqrt 2 = 57 \pm 2\times 20\sqrt 2= 57 \pm 2\times a(b\sqrt 2)$</span> where <span class="math-container">$ab = 20$</span>.</p>
<p>To solve <span class="math-container">$a^2 + 2b^2 = 57$</span> we have <span class="math-container">$57 = 25 + 2*16$</span> and so <span class="math-container">$57\pm 2\times 40 \sqrt 2 = 5^2 \pm 2\times 5\cdot 4\sqrt 2 + (4\sqrt 2)^2 = (5 \pm 4\sqrt 2)^2$</span>.</p>
<p>So the who thing works out.</p>
<p><span class="math-container">$A =(4\sqrt 2 -5) -(4\sqrt 2 + 5) = -10$</span>.</p>
<p>Cute.</p>
|
165,900 | <p>Let $R=k[u,v,w]$ and $p\in R$ be a cubic form. Let $G$ be the group of graded automorphisms of $R$ which preserve $p$, i.e., $G$ is the subgroup
of $GL_3(k)$ consisting of elements $g$ such that $g(p) \in k p$. My question: is $G$ some well known algebraic group? </p>
| Steven Sam | 321 | <p>To elaborate on abx's comment: modding out by scalars, i.e., working in $PGL_3$ instead of $GL_3$, by definition the stabilizer of $p$ is the group of projective automorphisms of the curve $p=0$ which preserves the embedding of the curve.</p>
<p>If we assume that the curve is smooth and that there is a rational point, for example, if the field is algebraically closed, then this is an elliptic curve embedded by a degree 3 line bundle. For simplicity, assume $3 \ne 0$ in the field. </p>
<p>EDIT: (taking into account abx's comments) This stabilizer is the subgroup of automorphisms of the elliptic curve which preserves its 3-torsion. This includes translations by 3-torsion elements (which is a finite group abstractly isomorphic to $Z/3 \times Z/3$) but also automorphisms of the curve which preserve the identity This is generically $Z/2$ given by the inversion map but can be bigger for special elliptic curves. So we get a semidirect product of this automorphism group with the 3-torsion.</p>
<p>If we do not assume there is a rational point (but still assume the curve $C$ is smooth), then we can say that $C$ is a torsor for an elliptic curve $E$ and identify the stabilizer with the semidirect product of the automorphism group of $E$ preserving the identity with the 3-torsion $E[3]$ of $E$. The latter is a group scheme which is isomorphic to $Z/3 \times Z/3$ after passing to an algebraically closed field.</p>
<p>The extension of $E[3]$ by the scalars $G_m$ is an example of a Heisenberg group scheme, and one can use the Weil pairing to reconstruct it.</p>
<p>I'm not exactly sure what to say when $p$ is not smooth -- in the case of mild singularities probably one gets a similar description.</p>
|
4,436,210 | <p>I have been given this exercise: Calculate the double integral:</p>
<blockquote>
<p><span class="math-container">$$\iint_D\frac{\sin(y)}{y}dxdy$$</span>
Where <span class="math-container">$D$</span> is the area enclosed by the lines: <span class="math-container">$y=2$</span>, <span class="math-container">$y=1$</span>, <span class="math-container">$y=x$</span>, <span class="math-container">$2y=x$</span> (not <span class="math-container">$y = 2x$</span>).</p>
</blockquote>
<p>Visualising <span class="math-container">$D$</span> is easy. You can split D in two sub areas and get the bounds for the integrals. The problem I face is:</p>
<p>Let's split D in two sub areas, <span class="math-container">$D_1$</span> and <span class="math-container">$D_2$</span>. <span class="math-container">$D_1$</span> is the left, upright triangle of <span class="math-container">$D$</span> and <span class="math-container">$D_2$</span> is the right, upside down one.</p>
<p>Then <span class="math-container">$D_1$</span> is defined by the lines <span class="math-container">$y=1$</span>, <span class="math-container">$y=x$</span>, and <span class="math-container">$x=2$</span>.</p>
<p>You can express the area in a <span class="math-container">$y$</span>-normal form as:
<span class="math-container">$$\begin{align}
1 \le y \le 2\\
y \le x \le 2
\end{align}$$</span>
then the integral can be written as
<span class="math-container">$$ \begin{align}
&\int_1^2\int_y^2\frac{\sin(y)}{y}dxdy \\
&=\int_1^2\frac{\sin(y)}{y}[x]^2_y \space dxdy \\
&=\int_1^2\frac{2\sin(y)}{y} - \sin(y)dy \\
&=2\int_1^2\frac{\sin(y)}{y}dy -\int_1^2 \sin(y)dy \\
\end{align}$$</span></p>
<p>The second integral is trivial, but in the first one is not. I have tried substituting, integrating by parts but to no avail. What am I doing wrong?</p>
<p>Any answer is really appreciated.</p>
| Doug M | 317,176 | <p>Rather than proceeding with Gauss-Jordan elimination....</p>
<p>For what values of <span class="math-container">$a$</span> are the vectors <span class="math-container">$(1,2,-3),(3,-1,5),(4,1,a^2-14)$</span> linearly independent or dependent?</p>
<p>If they are independent we have a unique solution.</p>
<p>If they are dependent, either the system of equations is consistent and we have infinitely many solutions, or the system is inconsistent and we no solution.</p>
<p>Adding the first two vectors, <span class="math-container">$(1,2,-3)+(3,-1,5) = (4,1,2).$</span> We have dependency if <span class="math-container">$a^2-14 = 2$</span> or <span class="math-container">$a = \pm 4$</span></p>
<p>Now to check for consistency... From the sum of the first two equations.</p>
<p><span class="math-container">$4x + y + 2z = 6$</span><br />
and from the third</p>
<p><span class="math-container">$4x + y+ 4z = \pm 4 + 2$</span></p>
<p>If we choose positive 4 our equations are consistent, and if we choose <span class="math-container">$-4$</span> they are not.</p>
|
4,436,210 | <p>I have been given this exercise: Calculate the double integral:</p>
<blockquote>
<p><span class="math-container">$$\iint_D\frac{\sin(y)}{y}dxdy$$</span>
Where <span class="math-container">$D$</span> is the area enclosed by the lines: <span class="math-container">$y=2$</span>, <span class="math-container">$y=1$</span>, <span class="math-container">$y=x$</span>, <span class="math-container">$2y=x$</span> (not <span class="math-container">$y = 2x$</span>).</p>
</blockquote>
<p>Visualising <span class="math-container">$D$</span> is easy. You can split D in two sub areas and get the bounds for the integrals. The problem I face is:</p>
<p>Let's split D in two sub areas, <span class="math-container">$D_1$</span> and <span class="math-container">$D_2$</span>. <span class="math-container">$D_1$</span> is the left, upright triangle of <span class="math-container">$D$</span> and <span class="math-container">$D_2$</span> is the right, upside down one.</p>
<p>Then <span class="math-container">$D_1$</span> is defined by the lines <span class="math-container">$y=1$</span>, <span class="math-container">$y=x$</span>, and <span class="math-container">$x=2$</span>.</p>
<p>You can express the area in a <span class="math-container">$y$</span>-normal form as:
<span class="math-container">$$\begin{align}
1 \le y \le 2\\
y \le x \le 2
\end{align}$$</span>
then the integral can be written as
<span class="math-container">$$ \begin{align}
&\int_1^2\int_y^2\frac{\sin(y)}{y}dxdy \\
&=\int_1^2\frac{\sin(y)}{y}[x]^2_y \space dxdy \\
&=\int_1^2\frac{2\sin(y)}{y} - \sin(y)dy \\
&=2\int_1^2\frac{\sin(y)}{y}dy -\int_1^2 \sin(y)dy \\
\end{align}$$</span></p>
<p>The second integral is trivial, but in the first one is not. I have tried substituting, integrating by parts but to no avail. What am I doing wrong?</p>
<p>Any answer is really appreciated.</p>
| user2661923 | 464,411 | <p>Alternative approach:</p>
<p>Compute the determinant of the matrix.</p>
<p>If the determinant is non-zero, then there will automatically be exactly <span class="math-container">$1$</span> solution.</p>
<p>If the determinant is <span class="math-container">$0$</span>, then there will either be <span class="math-container">$0$</span> solutions or an infinite number of solutions, depending on whether there is an <strong>inconsistency</strong> (explained at the end of this answer) among the values to the right of the equal signs.</p>
<hr />
<p>The determinant is</p>
<p><span class="math-container">$(14 - a^2 - 5) + [(-2)(3a^2 - 62)] + [(-3)(7)] = 112 - 7a^2 = 7(16 - a^2).$</span></p>
<p>So, if <span class="math-container">$(16 - a^2) \neq 0$</span>, then you know immediately that there is exactly one solution.</p>
<p>The problem then reduces to a consideration of <span class="math-container">$a^2 = 16.$</span></p>
<hr />
<p>Here, I will explain the idea of inconsistency among linear equations as follows:</p>
<p>Consider the following two pairs of linear equations:</p>
<ul>
<li><span class="math-container">$x + 2y = 6, ~2x + 4y = 12.$</span></li>
<li><span class="math-container">$x + 2y = 6, ~2x + 4y = 13.$</span></li>
</ul>
<p>Both pairs of equations will have a determinant of <span class="math-container">$(1 \times 4) - (2 \times 2) = 0.$</span></p>
<p>However, the first pair of equations above is consistent, and therefore permits an infinite number of solutions. The second pair of equations above is inconsistent, so there will be <span class="math-container">$0$</span> solutions.</p>
<hr />
<p>The first thing to notice is that the superficial judgement that the evaluation of <span class="math-container">$a=4$</span> and <span class="math-container">$a=-4$</span> will yield identical results is wrong. It is true that the LHS makes no distinction between <span class="math-container">$a=4,$</span> and <span class="math-container">$a=-4$</span>, since the LHS only features an <span class="math-container">$a^2$</span> item.</p>
<p>However the RHS features an <span class="math-container">$(a+2)$</span> value which requires that <span class="math-container">$a=4$</span> and <span class="math-container">$a=-4$</span> be evaluated separately, to determine which values of <span class="math-container">$a$</span> (if any) results in consistent values that yield an infinite number of solutions, and which values of <span class="math-container">$a$</span> (if any) result in inconsistent values that yield <span class="math-container">$0$</span> solutions.</p>
<hr />
<p>With <span class="math-container">$a$</span> equal to either <span class="math-container">$+4$</span> or <span class="math-container">$-4$</span>, the equations become</p>
<p><span class="math-container">$1x + 2y - 3z = 4$</span> <br>
<span class="math-container">$3x - 1y + 5z = 2$</span> <br>
<span class="math-container">$4x + 1y + 2z = 6 ~\text{or}~ -2.$</span></p>
<p>Adding the 2nd and 3rd equations above together yields:</p>
<p><span class="math-container">$7x + 7z = 8 ~\text{or}~ 0.$</span></p>
<p>Multiplying the 2nd equation above by <span class="math-container">$(2)$</span> and adding it to the 1st equation yields:</p>
<p><span class="math-container">$7x + 7z = 8.$</span></p>
<p>At this point, you know immediately that <span class="math-container">$a=-4$</span> causes an inconsistency, and that therefore, if <span class="math-container">$a=-4$</span>, there are no solutions.</p>
<p>However, the value of <span class="math-container">$a=4$</span> must be explored further, to determine whether it yields consistent results.</p>
<p>Treating <span class="math-container">$z$</span> as a fixed (unknown) value, results in :</p>
<p><span class="math-container">$1x + 2y = 4 + 3z$</span> <br>
<span class="math-container">$3x - 1y = 2 - 5z$</span> <br>
<span class="math-container">$4x + 1y= 6 - 2z$</span>.</p>
<p>Adding the 2nd and 3rd of these revised equations together yields</p>
<p><span class="math-container">$\displaystyle 7x = 8 - 7z \implies x = \frac{8}{7} - z.$</span></p>
<p>Using the 3rd equation above, this implies that</p>
<p><span class="math-container">$\displaystyle y = (6 - 2z) - 4\left(\frac{8}{7} - z\right) = \frac{10}{7} + 2z.$</span></p>
<p>It only remains to verify that the following values also satisfy the 1st and 2nd equations above, which they do:</p>
<ul>
<li><span class="math-container">$\displaystyle x = \frac{8}{7} - z, ~y = \frac{10}{7} + 2z.$</span></li>
</ul>
<p>Therefore, <span class="math-container">$a=4$</span> yields consistent results.</p>
<p>Therefore, <span class="math-container">$a=4$</span> yields an infinite number of solutions.</p>
|
152,405 | <p>This question complement a previous MO question: <a href="https://mathoverflow.net/questions/95837/examples-of-theorems-with-proofs-that-have-dramatically-improved-over-time">Examples of theorems with proofs that have dramatically improved over time</a>.</p>
<p>I am looking for a list of</p>
<h3>Major theorems in mathematics whose proofs are very hard but was not dramatically improved over the years.</h3>
<p>(So a new dramatically simpler proof may represent a much hoped for breakthrough.) Cases where the original proof was very hard, dramatical improvments were found, but the proof remained very hard may also be included.</p>
<p>To limit the scope of the question:</p>
<p>a) Let us consider only theorems proved at least 25 years ago. (So if you have a good example from 1995 you may add a comment but for an answer wait please to 2020.)</p>
<p>b) Major results only.</p>
<p>c) Results with very hard proofs.</p>
<p>As usual, one example (or a few related ones) per post.</p>
<p>A similar question was asked before <a href="https://mathoverflow.net/questions/44593/still-difficult-after-all-these-years">Still Difficult After All These Years</a>. (That question referred to 100-years old or so theorems.)</p>
<h2>Answers</h2>
<p>(Updated Oct 3 '15)</p>
<p><strong>1)</strong> <a href="https://mathoverflow.net/a/152457">Uniformization theorem for Riemann surfaces</a>( Koebe and Poincare, 19th century)</p>
<p><strong>2)</strong> <a href="https://mathoverflow.net/a/152546">Thue-Siegel-Roth theorem</a> (Thue 1909; Siegel 1921; Roth 1955)</p>
<p><strong>3)</strong> <a href="https://mathoverflow.net/a/152412">Feit-Thompson theorem</a> (1963);</p>
<p><strong>4)</strong> <a href="https://mathoverflow.net/a/152419">Kolmogorov-Arnold-Moser (or KAM) theorem</a> (1954, Kolgomorov; 1962, Moser; 1963)</p>
<p><strong>5)</strong> <a href="https://mathoverflow.net/a/218043/1532">The construction of the <span class="math-container">$\Phi^4_3$</span> quantum field theory model.
This was done in the early seventies by Glimm, Jaffe, Feldman, Osterwalder, Magnen and Seneor.</a> (NEW)</p>
<p><strong>6)</strong> <a href="https://mathoverflow.net/a/152752">Weil conjectures</a> (1974, Deligne)</p>
<p><strong>7)</strong> <a href="https://mathoverflow.net/a/152418">The four color theorem</a> (1976, Appel and Haken);</p>
<p><strong>8)</strong> <a href="https://mathoverflow.net/a/152613">The decomposition theorem for intersection homology</a> (1982, Beilinson-Bernstein-Deligne-Gabber); (<strong>Update:</strong> A new simpler proof by de Cataldo and Migliorini is now available)</p>
<p><strong>9)</strong> <a href="https://mathoverflow.net/a/218061/1532">Poincare conjecture for dimension four</a>, 1982 Freedman (NEW)</p>
<p><strong>10)</strong> <a href="https://mathoverflow.net/a/152416">The Smale conjecture</a> 1983, Hatcher;</p>
<p><strong>11)</strong> <a href="https://mathoverflow.net/a/152741">The classification of finite simple groups</a> (1983, with some completions later)</p>
<p><strong>12)</strong> <a href="https://mathoverflow.net/a/152537">The graph-minor theorem</a> (1984, Robertson and Seymour)</p>
<p><strong>13)</strong> <a href="https://mathoverflow.net/a/152424">Gross-Zagier formula</a> (1986)</p>
<p><strong>14)</strong> <a href="https://mathoverflow.net/a/218066/1532">Restricted Burnside conjecture</a>, Zelmanov, 1990. (NEW)</p>
<p><strong>15)</strong> <a href="https://mathoverflow.net/a/219182/1532">The Benedicks-Carleson theorem</a> (1991)</p>
<p><strong>16)</strong> <a href="https://mathoverflow.net/a/152454">Sphere packing problem in R 3 , a.k.a. the Kepler Conjecture</a>(1999, Hales)</p>
<p><strong>For the following answers some issues were raised in the comments.</strong></p>
<p><a href="https://mathoverflow.net/a/152598">The Selberg Trace Formula- general case</a> (Reference to a 1983 proof by Hejhal)</p>
<p><a href="https://mathoverflow.net/a/152423">Oppenheim conjecture</a> (1986, Margulis)</p>
<p><a href="https://mathoverflow.net/a/152459">Quillen equivalence</a> (late 60s)</p>
<p><a href="https://mathoverflow.net/a/152434">Carleson's theorem</a> (1966) (Major simplification: 2000 proof by Lacey and Thiele.)</p>
<p><a href="https://mathoverflow.net/a/152432">Szemerédi’s theorem</a> (1974) (Major simplifications: ergodic theoretic proofs; modern proofs based on hypergraph regularity, and polymath1 proof for density Hales Jewett.)</p>
<p><strong>Additional answer:</strong></p>
<p><a href="https://mathoverflow.net/a/152658">Answer about fully formalized proofs for 4CT and FT theorem</a>.</p>
| Vincent | 41,139 | <p>I'm a bit surprised not to see the <a href="http://en.wikipedia.org/wiki/Weil_conjectures">Weil conjectures</a> here since their proof by Deligne is so often mentioned as a primary example of something Very Hard. Is there a more simple recent proof that I haven't heared of?</p>
|
27,490 | <h2>Motivation</h2>
<p>The common functors from topological spaces to other categories have geometric interpretations. For example, the fundamental group is how loops behave in the space, and higher homotopy groups are how higher dimensional spheres behave (up to homotopy in both cases, of course). Even better, for nice enough spaces the (integral) homology groups count <span class="math-container">$n$</span>-dimensional holes.</p>
<hr />
<p>A groupoid is a category where all morphisms are invertible. Given a space <span class="math-container">$X$</span>, the fundamental groupoid of <span class="math-container">$X$</span>, <span class="math-container">$\Pi_1(X)$</span>, is the category whose objects are the points of <span class="math-container">$X$</span> and the morphisms are homotopy classes of maps rel end points. It's clear that <span class="math-container">$\Pi_1(X)$</span> is a groupoid and the group object at <span class="math-container">$x \in X$</span> is simply the fundamental group <span class="math-container">$\pi_1(X,x)$</span>. My question is:</p>
<blockquote>
<p>Is there a geometric interpretation <span class="math-container">$\Pi_1(X)$</span> analogous to the geometric interpretation of homotopy groups and homology groups explained above?</p>
</blockquote>
| Donu Arapura | 4,144 | <p>I'm not sure how to answer this, because it already seems pretty geometric to me. So let me
answer a slightly different question: what is the fundamental groupoid good for?
Since one knows that the fundamental group and groupoid are equivalent as categories for
path connected spaces, it's tempting to view the groupoid as giving nothing new.
But in fact, there are situations when it seems more natural. For example, if a group
$G$ acts continuously on a space $X$, then unless one knows something more, we only get an outer action of $G$ on $\pi_1(X,x)$ i.e. it's only well defined up to inner automorphisms.
However, $G$ will act on the fundamental groupoid $\Pi_1(X)$ on the nose, and in fact,
the previous statement becomes easier to see from this point of view.</p>
|
27,490 | <h2>Motivation</h2>
<p>The common functors from topological spaces to other categories have geometric interpretations. For example, the fundamental group is how loops behave in the space, and higher homotopy groups are how higher dimensional spheres behave (up to homotopy in both cases, of course). Even better, for nice enough spaces the (integral) homology groups count <span class="math-container">$n$</span>-dimensional holes.</p>
<hr />
<p>A groupoid is a category where all morphisms are invertible. Given a space <span class="math-container">$X$</span>, the fundamental groupoid of <span class="math-container">$X$</span>, <span class="math-container">$\Pi_1(X)$</span>, is the category whose objects are the points of <span class="math-container">$X$</span> and the morphisms are homotopy classes of maps rel end points. It's clear that <span class="math-container">$\Pi_1(X)$</span> is a groupoid and the group object at <span class="math-container">$x \in X$</span> is simply the fundamental group <span class="math-container">$\pi_1(X,x)$</span>. My question is:</p>
<blockquote>
<p>Is there a geometric interpretation <span class="math-container">$\Pi_1(X)$</span> analogous to the geometric interpretation of homotopy groups and homology groups explained above?</p>
</blockquote>
| The Mathemagician | 3,546 | <p>A very geometric approach to the fundamental groupoid can be found in Ronald Brown's <em>Topology and Groupoids</em>. Since EVERYTHING is expressed from the beginning in terms of the category of equivelence classes of paths,the formulation is very straightforward and simple. I highly recommend the book to all mathematicians:I have seen the future of point-set topology courses and Brown's text is the crystal ball. </p>
|
28,347 | <p>I often end up with a function that contains the term $1/(1 + x^2/y^2)$, and I need to evaluate this in the limit $y\rightarrow 0$. By hand, I can rewrite this as $y^2/(y^2 + x^2)$, but how can I tell <em>Mathematica</em> to make such a simplification?</p>
<p>I have tried using <code>1/(1 + x^2/y^2) // Simplify</code> and <code>Expand</code>, but neither work as intend (actually, they doesn't change anything).</p>
| bill s | 1,783 | <p>Like many limits involving two variables, this one is not as straightforward as it might seem. Consider</p>
<pre><code>Limit[1/(1 + x^2/y^2), x -> 0]
1
</code></pre>
<p>while</p>
<pre><code>Limit[1/(1 + x^2/y^2), y -> 0]
0
</code></pre>
<p>Coupling the two:</p>
<pre><code>Limit[1/(1 + x^2/y^2) /. y -> x, x -> 0]
1/2
</code></pre>
<p>In other words, which direction you approach in x,y space is important to the limit. You can make this take almost any value between zero and one by choosing how the x,y are coupled. For example:</p>
<pre><code>Limit[1/(1 + x^2/y^2) /. y -> 10 x, x -> 0]
Limit[1/(1 + x^2/y^2) /. y -> x/10, x -> 0]
</code></pre>
<p>give limits of 100/101 and 1/101. The same answers occur if the original form is replaced by the OPs simpler form, y^2/(x^2 + y^2). </p>
<p>There is a discussion of the use of <a href="http://reference.wolfram.com/mathematica/tutorial/GenericAndNonGenericSolutions.html" rel="nofollow">generic solutions</a>, which appears to be the reason that the first two give different answers: effectively the second is assuming that "generically" x is not zero. The tutorial doesn't explicitly mention <code>Limit</code> though presumably <code>Solve</code> is being used rather than <code>Reduce</code> (or at least, the methodology of <code>Solve</code> is being used rather than the methodology of <code>Reduce</code>) in terms of generic solutions.</p>
|
2,069,507 | <p><a href="https://i.stack.imgur.com/B4b88.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B4b88.png" alt="The image of parallelogram for help"></a></p>
<p>Let's say we have a parallelogram $\text{ABCD}$.</p>
<p>$\triangle \text{ADC}$ and $\triangle \text{BCD}$ are on the same base and between two parallel lines $\text{AB}$ and $\text{CD}$, So, $$ar\triangle \text{ADC}=ar\triangle \text{BCD}$$
Now the things those should be noticed are that:</p>
<p>In $\triangle \text{ADC}$ and $\triangle \text{BCD}$:</p>
<p>$$\text{AD}=\text{BC}$$ $$\text{DC}=\text{DC}$$ $$ar\triangle \text{ADC}=ar\triangle \text{BCD}.$$</p>
<p>Now in two different triangles, two sides are equal and their areas are also equal, so the third side is also equal or $\text{AC}=\text{BD}$. Which make this parallelogram a rectangle.</p>
<p>Isn't it a claim that every parallelogram is a rectangle or a parallelogram does not exist?</p>
| Narasimham | 95,860 | <p>Area of a triangle is</p>
<p>$$ Area = \frac12 b\, c \sin A $$</p>
<p>Among the three, two parameter values excepting $A$ are given, then the remaining angle $A$ and <em>its supplement</em> can satisfy this relation.</p>
<p>If you take base $b$ and $ h= c \sin A $ as height all the following shaded areas have same magnitude:</p>
<p>When area and height remain same (red line is height) $\sin A $ is same. Taking inverse of sine we can have two solutions $\alpha,\beta$ which are supplementary :</p>
<p>$$\alpha+\beta = \pi . $$</p>
<p>The above is <em>general</em> solution of two values. </p>
<p>$$ A= \sin^{-1}{ \frac{2 Area}{bc} } , \quad \pi - \sin^{-1}{ \frac{2 Area}{bc} }$$</p>
<p>This is <em>general</em> situation for a parallelogram.</p>
<p>However, only when </p>
<p>$$\alpha= \beta $$</p>
<p>do we have </p>
<p>$$\alpha= \beta = \pi/2 $$ </p>
<p>which is situation of right angles for <em>particular</em> case of rectangle.</p>
<p><a href="https://i.stack.imgur.com/v1iNE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v1iNE.png" alt="ParalloArea"></a>
<a href="https://i.stack.imgur.com/pknd2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pknd2.png" alt="Same Area"></a></p>
|
11,698 | <p><strong>Bug introduced in 8.0 or earlier and fixed in 13.2.0 or earlier</strong></p>
<hr />
<p>So I have been fighting with this for a while. I'm trying to get custom frame ticks on both the left and right side of a <code>DistributionChart</code>. It's not going very well. It just keeps throwing errors saying tick position needs to be a number. Which in my code it is. Here is an example:</p>
<pre><code>With[
{fps = {120, 60, 50, 40, 30, 25, 20, 15, 10}},
DistributionChart[
RandomVariate[SkewNormalDistribution[##], 100] & @@@ {{20, 13, 5}, {30, 12, 10}},
ChartLabels -> {1, 2},
PlotRange -> {0, 70},
GridLines -> {None, N@Table[1/i*1000, {i, fps}]},
FrameTicks -> {
{
Table[{N[1/i*1000], NumberForm[N[1/i*1000], 3]}, {i, fps}],(*Left*)
Table[{1/i*1000, i}, {i, fps}] (*Right*)
},
{
None,(*Bottom*)
None (*Top*)
}
}
]
]
</code></pre>
<p>Any help would be much appreciated.</p>
| Sjoerd C. de Vries | 57 | <p>There seems to be a bug here as even </p>
<pre><code>DistributionChart[{{1, 2, 3, 4, 5}, {1, 1, 2, 3}}, FrameTicks -> {{None, None}, {None, None}}]
</code></pre>
<p>yields an error, but you can work around it as follows:</p>
<pre><code>With[{fps = {120, 60, 50, 40, 30, 25, 20, 15, 10}},
DistributionChart[
RandomVariate[SkewNormalDistribution[##], 100] & @@@ {{20, 13, 5}, {30, 12, 10}},
ChartLabels -> {1, 2}, PlotRange -> {0, 70},
GridLines -> {None, N@Table[1/i*1000, {i, fps}]}
] /.
Rule[FrameTicks, __] ->
Rule[FrameTicks, {{Table[{N[1/i*1000],NumberForm[N[1/i*1000], 3]}, {i, fps}],(*Left*)
Table[{1/i*1000, i}, {i, fps}] (*Right*)}, {None,(*Bottom*) None (*Top*)}}]
]
</code></pre>
<p><img src="https://i.stack.imgur.com/zV9jr.png" alt="Mathematica graphics"></p>
|
1,674,676 | <p>Let $p$ be an odd prime number. I want to show that $\mathbb{F}_{p^2}$ has a primitive 8th root of unity $\zeta$. </p>
<ul>
<li>I know that $\zeta^8 = 1$. So my idea is to define $f = X^8 - 1$ such that $\zeta$ is a root of $f$. But this is for a field extension of degree 8 and $p^2$ is at least 9.</li>
</ul>
<p>any hints?</p>
| DonAntonio | 31,254 | <p>We have the cyclic group $\;\Bbb F_{p^2}^*\;$ of order $\;p^2-1=0\pmod8\;$ and thus there exists a (unique) subgroup of order eight there. Any generator of this subgroup will do it.</p>
|
4,440,233 | <blockquote>
<p>Find all the functions <span class="math-container">$f:\mathbb{Z}^+ \to \mathbb{Z}^+$</span> such that <span class="math-container">$f(f(x)) = 15x-2f(x)+48$</span>.</p>
</blockquote>
<p>If <span class="math-container">$f$</span> is a polynomial of degree <span class="math-container">$n$</span>, we have that <span class="math-container">$\deg(f(f(x))) = n^2$</span> and <span class="math-container">$\deg(15x-2f(x)+48)=n$</span>. Therefore, the only possible polynomials that satisfy the condition have degree <span class="math-container">$0$</span> or <span class="math-container">$1$</span>.</p>
<p>Let <span class="math-container">$f:\mathbb{Z}^+ \to \mathbb{Z}^+$</span> be a function that holds the condition of the problem given by <span class="math-container">$f(x)=ax+b$</span> for some constants <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. Since <span class="math-container">$$f(f(x)) = f(ax+b) = a(ax+b)+b = a^2x + (a+1)b$$</span> and <span class="math-container">$$15x-2f(x)+48 = 15x-2(ax+b)+48 = (15-2a)x+(48-2b),$$</span> it follows that
<span class="math-container">$$a^2+2a-15=0 \quad\text{and}\quad (a+1)b=48-2b.$$</span> From the first equation, we get that <span class="math-container">$a=-5$</span> or <span class="math-container">$a=3$</span>. If <span class="math-container">$a=-5$</span>, from the second equation we get that <span class="math-container">$b=-24$</span>, and it contradicts that <span class="math-container">$f[\mathbb{Z}^+]\subseteq \mathbb{Z}^+$</span>. If <span class="math-container">$a=3$</span>, then <span class="math-container">$b=8$</span>. Therefore, <span class="math-container">$f(x)=3x+8$</span> is the only polynomial that satisfies the condition of the problem. I guess that this is the only solution, but I do not know how to prove it.</p>
<p><strong>Edit:</strong> I was trying to prove that the iterations of any function <span class="math-container">$f$</span> that satisfies the problem have the same behaviour. For instance, by iterating <span class="math-container">$f$</span> we have that <span class="math-container">$2f^3(x)+f^4(x)-15f^2(x)=48$</span>, so this functions are almost the same except for constant terms. Is this usefull this idea to complete the problem?</p>
| Jyrki Lahtonen | 11,619 | <p>Let us study the two variable affine transformation <span class="math-container">$g:\Bbb{R}^2\to\Bbb{R}^2$</span>
<span class="math-container">$$
g(x,y)=(y,15x-2y+48).
$$</span>
The connection to the problem is, of course, that <span class="math-container">$g(x,f(x))=(f(x),f(f(x)))$</span>. I am using the idea from Tob's answer that unless we are on the straight and narrow path of <span class="math-container">$f(x)=3x+8$</span> we will diverge to a point where negative values will appear.</p>
<p>The linear part of the transformation <span class="math-container">$g$</span> comes from the matrix
<span class="math-container">$$
A=\left(\begin{array}{rr}0&1\\ 15&-2\end{array}\right).
$$</span>
Suggestively, the eigenvalues of <span class="math-container">$A$</span> are <span class="math-container">$\lambda_1=3$</span> with eigenvector <span class="math-container">$(1,3)^T$</span> and <span class="math-container">$\lambda_2=-5$</span> with eigenvector <span class="math-container">$(1,-5)^T$</span>. The idea is that if we iterate <span class="math-container">$g$</span> from a starting point that has a non-zero component belonging to that large negative eigenvalue, then eventually <span class="math-container">$(-5)^m$</span> dominates over <span class="math-container">$3^m$</span>, kicking us out of the positive zone.</p>
<p>Let's first linearize. We easily find that <span class="math-container">$P=(-4,-4)$</span> is a fixed point of <span class="math-container">$g$</span> (as <span class="math-container">$P$</span> is on the expected line <span class="math-container">$y=3x+8$</span> this, again, boosts our optimism). So if we move the origin to <span class="math-container">$P$</span> we need to write <span class="math-container">$u=x+4, v=y+4$</span>, and replace <span class="math-container">$g(x,y)$</span> with
<span class="math-container">$$h(u,v)=(v,15u-2v).$$</span>
The connection to the functional equation now reads:
<span class="math-container">$$
h(x+4,f(x)+4)=(f(x)+4,f(f(x))+4).
$$</span></p>
<p>We can now prove that we run into a contradiction, if <span class="math-container">$f(a)=b\neq 3a+8$</span> for some integer <span class="math-container">$a>0$</span>. We write <span class="math-container">$(a+4,b+4)$</span> using the eigenbasis above
<span class="math-container">$$
(a+4,b+4)=c_1(1,3)+c_2(1,-5).
$$</span>
The contrapositive assumption is equivalent to <span class="math-container">$c_2\neq0$</span>.</p>
<p>When we iterate <span class="math-container">$m$</span> times we arrive at
<span class="math-container">$$
(a_m+4,b_m+4)=3^mc_1(1,3)+(-5)^mc_2(1,-5),
$$</span>
where, <span class="math-container">$a_0=a$</span>, <span class="math-container">$b_0=b$</span> and for all <span class="math-container">$m>0$</span>, <span class="math-container">$b_m=f(a_m)$</span> and <span class="math-container">$a_{m+1}=b_m$</span>.</p>
<p>It is then clear that the assumption <span class="math-container">$c_2\neq0$</span> forces <span class="math-container">$b_m<0$</span> for some large enough value of <span class="math-container">$m$</span>, which is a contradiction.</p>
|
3,618,791 | <p>Given <span class="math-container">$p\in[0,1]$</span>, prove or disprove that the sum
<span class="math-container">$$\sum_{n=k}^\infty\sum_{j=0}^k\left(\matrix{n\\j}\right)p^j(1-p)^{n-j}$$</span>
is bounded by a constant that does not depend on <span class="math-container">$k$</span>.</p>
<p>The terms <span class="math-container">$\left(\matrix{n\\j}\right)p^j(1-p)^{n-j}$</span> do remind me of the binomial expansion. But using that, the most naive estimate is
<span class="math-container">$$\sum_{j=0}^k\left(\matrix{n\\j}\right)p^j(1-p)^{n-j}\leq\sum_{j=0}^n\left(\matrix{n\\j}\right)p^j(1-p)^{n-j}=1$$</span>
And then summing over <span class="math-container">$n$</span> still gives <span class="math-container">$\infty$</span>. How should I make the correct estimate?</p>
<p>p.s. <strong>I am not absolutely sure that this holds</strong>. There is a proof that I was reading where such an estimate could yield the final result. I am a bit confused now because one comment says this is wrong while an answer seems to have proved this.</p>
| lonza leggiera | 632,373 | <p><strong>Hint:</strong> The sum in part (b) is not a geometric series. But we do know that <span class="math-container">$\ \ln\left(a^n\right)=n\ln(a)\ $</span> for any positive <span class="math-container">$\ a\ $</span>, so <span class="math-container">$\ \sum_\limits{n=0}^{10}\ln\left(0.5^n\right)= \sum_\limits{n=0}^{10}n\ln(0.5)\ $</span>. Can you now work out the sum of this series?</p>
|
192,072 | <p>Bonjour!<br>
I'm trying this number-theory problem, but i don't have any idea how to solve it.<br>
Can you give me some hints ?</p>
<p>We have got any $\mathbb{Z_+}$ number. Let it be $n$.<br>
Then we must proof that $2 \nmid \sigma(n) \implies n = k^2 \vee n = 2k^2$.<br>
Thanks for any help </p>
| André Nicolas | 6,312 | <p>Let $n=2^e m$ where $m$ is odd. Note that $\sigma(2^e)$ is odd. So by the multiplicativity of $\sigma$, $\sigma(n)$ is odd iff $\sigma(m)$ is odd. Any power of $2$ is a square or twice a square. So we need only show that if $\sigma(m)$ is odd for the odd number $m$, then $m$ is a perfect square. </p>
<p>If $m$ is <strong>not</strong> a perfect square, there is a prime $p$, necessarily odd, such that the highest power of $p$ that divides $m$ is $p^t$, where $t$ is odd. </p>
<p>But by multiplicativity, $1+p+\cdots+p^t$ divides $\sigma(m)$. And $1+p+\cdots+p^t$ is even, since it is the sum of an even number of odd numbers. This contradicts the fact that $\sigma(m)$ is odd. </p>
<p><strong>Remark:</strong> The converse is straightforward: If $n$ has shape $w^2$ or $2w^2$, then $\sigma(n)$ is odd. </p>
|
1,079,995 | <p>I can't understand how: $$ \frac {2\times{^nC_2}}{5} $$</p>
<p>Equals:</p>
<p>$$ 2\times \frac {^nC_2}{5} $$</p>
<p>If we forget the combination and replace it with a $10$, the result is clearly different. $1$ in the first example and and $0.5$ in the second.</p>
| DeepSea | 101,504 | <p>Remember some fact about fractions: $\dfrac{a}{b}\times \dfrac{c}{d} = \dfrac{ac}{bd} \Rightarrow \dfrac{2\times \binom{n}{2}}{5} = \dfrac{2\times \binom{n}{2}}{1\times 5} = \dfrac{2}{1}\times \dfrac{\binom{n}{2}}{5} = 2\times \dfrac{\binom{n}{2}}{5}$</p>
|
1,133,817 | <p>Using Legendre polynomial generating function
\begin{equation}
\sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}}
\end{equation}
Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$ </p>
<p>Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $$
And $$ P_{2n+1}(0)=0$$</p>
<p>I expressed $$(x^2 -1)^{2n}= \sum_{k=0}^{2n} {2n \choose k}x^{4n-2k}(-1)^k$$
And using second formula given, the only term that remains after differentiating 2n times and substituting x=0 is where $$4n-2k=2n$$ so $$2n=2k$$ k=n i.e $$(-1)^n \frac{(2n)!}{(n!)^2} $$ but multiplying this by$$ \frac{1}{2^{2n} (2n)!} $$ doesn't give desired solution. Where have I gone wrong?</p>
| Ivo Terek | 118,056 | <p>If $x \to 0$, then $x$ is close to zero, but it is not zero. So we can do: $$\lim_{x \to x_0} f(x) = \lim_{x \to x_0}\frac{f(x)}{x}x = \lim_{x \to x_0}\frac{f(x)}{x}\cdot \lim_{x \to x_0}x = 0 \cdot x_0 = 0.$$</p>
|
2,296,256 | <p>I need help how to mathematically interpret an ODE (Newton's second law). I used to the ODE in this form:
$$
m\ddot x(t)=F(t)\tag{1}
$$</p>
<p>However, in another book they wrote:
$$
m\ddot x=F(x,\dot x) \tag{2}
$$
where $F: \mathbb{R}^n \times \mathbb{R}^n\rightarrow \mathbb{R}^n$.</p>
<p><strong>Questions:</strong></p>
<ol>
<li><p>I guess $F(x,\dot x)$ is an abbreviation for $F(x(t),\dot x(t))$, is it correct?</p></li>
<li><p>What it the difference between writing $F(t)$ or $F(x,\dot x)$?</p></li>
<li><p>What is the meaning of the notation $F: \mathbb{R}^n \times \mathbb{R}^n\rightarrow \mathbb{R}^n$?</p></li>
</ol>
<p>Thanks!</p>
| 5xum | 112,884 | <p>First of all, your two conditions are redundant, because the second follows from the first:</p>
<ul>
<li>Just to make it more readable, the first condition is equivalent to $$\forall x,y: x>y \implies f(x)>f(y)$$</li>
<li>If $i < j$, then $j>i$.</li>
<li>Set $x=j$ and $y=i$.</li>
<li>Therefore, $x>y$.</li>
<li>Therefore, from the first condition, we know that $f(x)>f(y)$</li>
<li>Therefore, since $x=j$ and $y=i$, we know that $f(j)>f(i)$</li>
<li>$f(j)>f(i)$ is equivalent to $f(i)<f(j)$</li>
</ul>
<hr>
<p>Second of all, yes, the answer is you <em>can</em> conclude that if $f(i)=f(j)$, then $i=j$. This is because $i\neq j$ leads to a contradiction:</p>
<ul>
<li>Assume $i\neq j$, </li>
<li>then $i<j$ or $i>j$. Without loss of generality, we can say $i<j$ (otherwise, set $i'=j$ and $j'=i$ and we have $i'<j'$</li>
<li>Then, from condition $1$, you know that $f(i)<f(j)$</li>
<li>$f(i)=f(j)$ cannot be true since $f(i)=f(j)$.</li>
</ul>
|
2,510,723 | <p>Does the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{19+(n+5)!}{(n+7)!}$ Converge or diverge</p>
<blockquote>
<p>I tried using the ratio test, but it gave me this which I don't see how you'd be able to solve.
$$\lim_{n\to\infty} |\frac{a_{n+1}}{a_{n}}| = \lim_{n\to\infty}\frac{19+(n+6)!}{19+(n+5)!*(n+8)}$$</p>
</blockquote>
<p>Wolfram alpha told me to use the comparison test, but I can't for the love of god see which series I'd compare it to.</p>
| ajotatxe | 132,456 | <p>$$\sum_{n=1}^\infty\frac{19+(n+5)!}{(n+7)!}=\sum_{n=1}^\infty\frac{1+\frac{19}{(n+5)!}}{(n+6)(n+7)}<\sum_{n=1}^\infty\frac2{n^2}$$</p>
|
2,125,018 | <blockquote>
<p>You toss a fair coin 3x, events:</p>
<p>A = "first flip H"</p>
<p>B = "second flip T"</p>
<p>C = "all flips H"</p>
<p>D = "at least 2 flips T"</p>
<p><strong>Q:</strong> Which events are independent?</p>
</blockquote>
<p>From the informal def. it is where one doesnt affect the other.</p>
<p>So in this case, $AB$ seem independent? Any others?</p>
| Mick | 42,351 | <p>The picture clearly show $\dfrac {\beta}{\alpha}$ is not constant.</p>
<p><a href="https://i.stack.imgur.com/Y0ikm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y0ikm.png" alt="enter image description here"></a></p>
|
2,125,018 | <blockquote>
<p>You toss a fair coin 3x, events:</p>
<p>A = "first flip H"</p>
<p>B = "second flip T"</p>
<p>C = "all flips H"</p>
<p>D = "at least 2 flips T"</p>
<p><strong>Q:</strong> Which events are independent?</p>
</blockquote>
<p>From the informal def. it is where one doesnt affect the other.</p>
<p>So in this case, $AB$ seem independent? Any others?</p>
| Mick | 42,351 | <p>Since A is fixed, I assume that it is located at (h, k) where h and k are known quantities.</p>
<p><a href="https://i.stack.imgur.com/CiY2R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CiY2R.png" alt="enter image description here"></a></p>
<p>Further assumptions:- OY = m, OA = n, $\angle XOY = \alpha$. Let the horizontal line through Y cut OA at Z and Z divide OA in the ratio $\rho : 1$. The target is to find $\beta’$ (instead of $\beta$) in term of other known quantities.</p>
<p>$Y = (m \cos \alpha, m \sin \alpha)$.</p>
<p>$Z = (?,m \sin \alpha)$; where $m \sin \alpha = \dfrac {0 \times 1 + k \times \rho}{\rho + 1}$.</p>
<p>From which, we get $\rho = \dfrac {m \sin \alpha}{k – m \sin \alpha}$, which is then known.</p>
<p>Since Z divides OA (whose length is n) in the ratio $\rho : 1$, $OZ = … = \dfrac {\rho \times n}{\rho + 1}$ and $ZA = … = \dfrac {n}{\rho + 1}$. Also $\theta = \sin^{-1} (\dfrac { m \sin \alpha}{\rho}).$</p>
<p>$\beta’$ can then be found by applying sine/cosine laws to the $\triangle AZY$.</p>
|
154,893 | <p>I am having trouble figuring this out.</p>
<p>$$\sqrt {1+\left(\frac{x}{2}- \frac{1}{2x}\right)^2}$$</p>
<p>I know that $$\left(\frac{x}{2} - \frac{1}{2x}\right)^2=\frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}$$ but I have no idea how to factor this since I have two x terms with vastly different degrees, 2 and -2.</p>
| DonAntonio | 31,254 | <p>Hint: for any (real, complex) numbers $\,a,b,\,$: $$4ab+(a-b)^2=(a+b)^2$$</p>
|
42,881 | <p>In short, my question is the same as my previous one except that everything is now wrapped up in a module.</p>
<p>The relevant code I'm working with is:</p>
<pre><code>getinter[a_, b_, u0_, k_, m_, hbar_, Nu_, Np_, up_] :=
Module[{ekp, ms, LUs, env, eenv, envpart, f, kppart, g, approx,
approx1, papprox, approx2, hard, ereal, psiparts, real, real1,
realp, real2, er, inter},
ekp = energies[a, b, u0, k, 1, m, hbar, 0.001, 10^-15][[1]];
ms = effmass[a, b, u0, k, 0.2, 1, m, hbar, 0.001, 10^-15];
LUs = BuildLUs[a, b, Nu, Np, u0, up];
env[x_] := Abs[Det[envfunc[ms, hbar, x, up, Nu, Np, a, b]]]^2;
eenv = zeros[env, up + 0.000001, 0.01, 0, 1, 10^-15, 0.02][[1]];
envpart = getpsipieces[LUs[[1]], LUs[[2]], eenv, ms, hbar];
f[x_] := Evaluate@Piecewise[envpart];
Return[f[0.231]];
]
</code></pre>
<p>envpart is a properly formatted object for piecewise and the variable it uses is named x.
When I evaluate this, it's returning the structure I put into piecewise but now formatted as a case structure as opposed to an array.</p>
<p>What I'd like it to do is to return f[0.231]='some number' like you would expect. Also, f[x_]:= Evaluate@Piecewise[envpart]; works fine if I take it out of the module.</p>
<p>Thanks for the help. Also, I get the sense that the solution to this problem, if I fully understand it, will allow me to work out a lot of other problems I encounter, so I'd really appreciate any suggestions on where I can learn about the issues at play.</p>
<p>Edited for slightly less awful formatting.</p>
| Dr. belisarius | 193 | <p>Use pure functions as a return value :</p>
<pre><code>fOne[x_] :=(*just return a piecewise*)
Piecewise[{{#^2, # < x}, {#^2 + (x - #)^3 Sin[3 #], # > x}}] &
fTwo[y_, z_, w_] := Module[{vars},
Plot[fOne[y][x], {x, z, w}]]
fTwo[3, -3, 8]
</code></pre>
<p><img src="https://i.stack.imgur.com/oxTtp.png" alt="Mathematica graphics"></p>
|
2,268,299 | <p>Let's say I have a series of real values $y_0,y_1,y_2\cdots$.
My question is if it's always possible to find (at least one) $C^\infty$ real function such that
\begin{equation}
f^{(n)}(0)=y_n
\end{equation}
and in the affirmative case, how.
It is a kind of "reverse taylor" problem... any hints?</p>
| Jean Marie | 305,862 | <p>I am going to address a complementary issue: the fact that if a solution to this problem exists, it is not unique.</p>
<p>This is due to the existence of (Laurent) Schwartz function defined by </p>
<p>$\varphi(x)=\exp(-\tfrac{1}{x^2})$ for $x \neq 0$ and $\varphi(0) = 0$, which is such that $\forall n\in\mathbb{N}, \ \varphi^{(n)}(0)=0$.</p>
<p>Thus, if one has a function $f$ such that $f^{(n)}(0)=y_n,$ then $f+k\varphi$ is also a solution.</p>
<p>Remark: Using this function, one can define more general so-called "bump functions" (<a href="https://en.wikipedia.org/wiki/Bump_function" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Bump_function</a>).</p>
<p><a href="https://i.stack.imgur.com/3Tfl2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Tfl2.jpg" alt="enter image description here"></a></p>
<p>Graphical representation of function $\varphi$: the very flat behavior in $0$ accounts for the fact that $0$ has an infinite order of multiplicity. One notes also a horizontal asymptote with equation $y=e^0=1$</p>
|
474,048 | <p>I am stuck with the following problem from a book.</p>
<p>It asks whether or not $f_n \rightarrow f$ converges uniformly on $A$ if for every $[a,b], f_n\rightarrow f$ uniformly on $A\cap [a,b]$.</p>
<p>The statement seems false to me (i.e. not necessarily true) because of this intuition I had:</p>
<p>If $A$ is not compact then it does not have a finite subcover. I can then construct open intervals which cover $A$. In turn, these open intervals contain closed intervals on which $f_n \rightarrow f$ uniformly on $ A\cap [x,y]$. Consequently because obtaining the maximum of an infinite number of elements is tricky (i.e. no finite subcover), we cannot conclude uniform convergence. [The maxima I am pertaining to is the maxima of $N$ such that $n>N$].</p>
<p>How should I proceed? Suggestions very welcome</p>
| Colliot | 80,507 | <p>Write $A$ (say non-negative real numbers) as a infinite union of closed sets (say intervals) $A_j$, denote the (maximal) distance from limit on $A_j$ down to $f_i$ as $D(i,j)$. We know already $\forall\epsilon>0,\forall j, \exists N(j,\epsilon):i>N(j,\epsilon)\Rightarrow D(i,j)<\epsilon$. We don't whant $\forall\epsilon>0,\exists N(\epsilon):i>N(\epsilon)\Rightarrow D(i,j)<\epsilon$.</p>
<p>There are two viewpoints.</p>
<ol>
<li>Take $N(j,\epsilon)$ minimal. Fix an $\epsilon$, let $N(j,\epsilon)\rightarrow \infty$ as $j\rightarrow\infty$. This is your intuition. And an example is in the above answer.</li>
<li>Suppose there is an $N(\epsilon)$, let $D(i,j)\rightarrow\infty$ when $i>N$ and $j\rightarrow \infty$, thus getting a contradition. This produces something like on $(n,n+1)$, $f_i=n/i$.</li>
</ol>
|
208,744 | <p>I was asked to show that $\frac{d}{dx}\arccos(\cos{x}), x \in R$ is equal to $\frac{\sin{x}}{|\sin{x}|}$. </p>
<p>What I was able to show is the following:</p>
<p>$\frac{d}{dx}\arccos(\cos(x)) = \frac{\sin(x)}{\sqrt{1 - \cos^2{x}}}$</p>
<p>What justifies equating $\sqrt{1 - \cos^2{x}}$ to $|\sin{x}|$?</p>
<p>I am aware of the identity $ \sin{x} = \pm\sqrt{1 - \cos^2{x}}$, but I still do not see how that leads to that conclusion.</p>
| bradhd | 5,116 | <p>In general, $\sqrt{a^2} = |a|$, as $\sqrt y$ refers to the <em>non-negative</em> number whose square is $y$. Then since $1-\cos^2x = \sin^2x$, we have $\sqrt{1-\cos^2 x} = |\sin x|$.</p>
|
10,600 | <p>As mentioned in <a href="https://matheducators.stackexchange.com/questions/1538/counterintuitive-consequences-of-standard-definitions">this question</a> students sometimes struggle with the fact that continuity is only defined at points of the function's domain. For example the function $f:\mathbb R\setminus\{0\} \to \mathbb R: x \mapsto \tfrac 1x$ is continuous although it has a "jump" at $x=0$ (<a href="https://matheducators.stackexchange.com/a/1686/5097">cf. this answer with more details</a>). So:</p>
<p><em>Why is continuity only defined on the function's domain? What's the benefit?</em> How should a lecturer answer to such a question of a student?</p>
<hr>
<p><strong>My attempt to answer the question:</strong> I would give two arguments:</p>
<ul>
<li>When we take the <a href="https://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_sequences" rel="nofollow noreferrer">sequence limit definition of continuity</a> $\lim_{n\to\infty} f(x_n) = f\left(\lim_{n\to\infty} x_n\right) = f(x_0)$, then this definition makes only sense when $x_0 = \lim_{n\to\infty} x_n$ is in the domain of $f$.</li>
<li>The concept students have in mind is "continuous continuation" and not "continuity". Thus, one have to distinguish between both concepts.</li>
</ul>
<p>What do think about my answer? Have I missed something or are there other good arguments?</p>
<hr>
<p><strong>Note:</strong> This is another follow up question of <a href="https://matheducators.stackexchange.com/questions/10597/how-can-i-motivate-the-formal-definition-of-continuity">How can I motivate the formal definition of continuity?</a> I hope that's okay since I ask here for another aspect of continuity. I want to write an introductory article for continuity. That's the reason why I ask all these questions here...</p>
| Amir Asghari | 1,217 | <p>Have a look at the paper written by Nunez et all: </p>
<p><a href="http://link.springer.com/article/10.1023%2FA%3A1003759711966">EMBODIED COGNITION AS GROUNDING FOR SITUATEDNESS AND CONTEXT IN MATHEMATICS EDUCATION</a>. </p>
<p>In essence, they argue that it is better to be causious if you want to "motivate the formal definition of continuity starting from the intuition" you have suggested in your question. In the following passage, "natural continuity" refers to drawing without lifting the pen. </p>
<blockquote>
<p>For the purposes of this article, the pedagogical problem can be
summarized as follows: students are introduced to natural continuity
using concepts, ideas, and examples which draw on inferential patterns
sustained by the natural human conceptual system. Then, they are
introduced to another concept – Cauchy-Weierstrass continuity – that
rests upon radically different cognitive contents (although not
necessarily more complex). These contents draw on different
inferential structures and different entailments that conflict with
those from the previous idea. The problem is that students are never
told that the new definition is actually a completely different
human-embodied idea. <strong>Worse, they are told that the new definition
captures the essence of the old idea</strong>, which, by virtue of being
‘intuitive’ and vague, is to be avoided. </p>
</blockquote>
<p>If I had more time, I would summarize their argument. For the time being, I make this a community wiki to invite other people to complete this answer. </p>
|
1,354,953 | <p>Solve for the function f(x):</p>
<p>$$f(x)=\frac{x}{x+f\left(\frac{x}{x+f(x)}\right)}$$
I'm not able to solve this. </p>
<p>[For instance, I tried solving for $f(\frac{x}{x+f(x)})$, but this doesn't lead me anywhere as the value obtained, when substituted into the original equation, just yields $f(x)=f(x)$]</p>
<p>Question: <strong>What is f(x) = ?</strong></p>
<p><em>I suppose I need to mention that f(x) should be continuous, and that it's a subset of $R$.</em></p>
<p>(Related : <a href="https://math.stackexchange.com/questions/1280806/functional-equation-fx-frac11f-frac11fx">Functional equation $f(x)=\frac{1}{1+f(\frac{1}{1+f(x)})}$</a> , whose confirmed solutions are $-\phi$ and $\frac{1}{\phi}$)</p>
| Servaes | 30,382 | <p>If such an $f$ exists, plugging in $x=0$ we find that
$$f(0)=\frac{0}{0+f\left(\tfrac{0}{0+f(0)}\right)}=0,$$
and to avoid having to divide by $0$ this requires that
$$0+f(0)\neq0\qquad\text{ and }\qquad0+f\left(\frac{0}{0+f(0)}\right)\neq0.$$
But then $f(0)=0$ so $0+f(0)=0$, a contradiction. Hence no such $f$ exists.</p>
|
3,671,223 | <p>First and foremost, I have already gone through the following posts:</p>
<p><a href="https://math.stackexchange.com/questions/2463561/prove-that-for-all-positive-integers-x-and-y-sqrt-xy-leq-fracx-y">Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$</a></p>
<p><a href="https://math.stackexchange.com/questions/64881/proving-the-am-gm-inequality-for-2-numbers-sqrtxy-le-fracxy2">Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$</a></p>
<p>The reason why I open a new question is because I do not understand after reading the two posts.</p>
<p>Question: Prove that for any two positive numbers x and y, <span class="math-container">$\sqrt{ xy} \leq \frac{x + y}{2}$</span></p>
<p>According to my lecturer, he said that the question should begin with <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span>. Lecturer also said that this is from a "well-known" fact. Now, both posts also mentioned this exact same thing in the helpful answers.</p>
<p>My question is this - <strong>how</strong> and <strong>why</strong> do I know that I need to use <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span>? What "well-known" fact is this? Can't I simply just subtract <span class="math-container">$\sqrt{xy}$</span> to both side and conclude at <span class="math-container">$0 \leq {(x-y)}^2$</span>? I do not know how this <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span> come back and why it even appear.</p>
<p>Thanks in advance.</p>
<p>Edit: <strong>I am not looking for the direct answer to this question.</strong> I am looking for an answer on <strong>why</strong> <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span> is even considered in the first place as the first step to this question. Is this from a mathematical theorem or axiom etc?</p>
| Narasimham | 95,860 | <p>Let <span class="math-container">$ x=u^2, y=v^2$</span> since <span class="math-container">$ (x,y)$</span> are given positive.</p>
<p>You have to prove that</p>
<p><span class="math-container">$$ u^2+v^2-2 u v \ge 0$$</span></p>
<p>or that</p>
<p><span class="math-container">$$(u-v)^2\ge 0 $$</span></p>
<p>which is true for all real numbers.</p>
|
3,671,223 | <p>First and foremost, I have already gone through the following posts:</p>
<p><a href="https://math.stackexchange.com/questions/2463561/prove-that-for-all-positive-integers-x-and-y-sqrt-xy-leq-fracx-y">Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$</a></p>
<p><a href="https://math.stackexchange.com/questions/64881/proving-the-am-gm-inequality-for-2-numbers-sqrtxy-le-fracxy2">Proving the AM-GM inequality for 2 numbers $\sqrt{xy}\le\frac{x+y}2$</a></p>
<p>The reason why I open a new question is because I do not understand after reading the two posts.</p>
<p>Question: Prove that for any two positive numbers x and y, <span class="math-container">$\sqrt{ xy} \leq \frac{x + y}{2}$</span></p>
<p>According to my lecturer, he said that the question should begin with <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span>. Lecturer also said that this is from a "well-known" fact. Now, both posts also mentioned this exact same thing in the helpful answers.</p>
<p>My question is this - <strong>how</strong> and <strong>why</strong> do I know that I need to use <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span>? What "well-known" fact is this? Can't I simply just subtract <span class="math-container">$\sqrt{xy}$</span> to both side and conclude at <span class="math-container">$0 \leq {(x-y)}^2$</span>? I do not know how this <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span> come back and why it even appear.</p>
<p>Thanks in advance.</p>
<p>Edit: <strong>I am not looking for the direct answer to this question.</strong> I am looking for an answer on <strong>why</strong> <span class="math-container">$(\sqrt{x}- \sqrt{y})^2 \geq 0$</span> is even considered in the first place as the first step to this question. Is this from a mathematical theorem or axiom etc?</p>
| Aderinsola Joshua | 395,530 | <p><span class="math-container">$$\sqrt(xy) \leq \frac{x+y}{2}$$</span>
<span class="math-container">$$xy \leq \frac{(x+y)^2}{4}$$</span>
<span class="math-container">$$4xy \leq x^2+2xy+y^2$$</span>
<span class="math-container">$$2xy \leq x^2+y^2$$</span>
<span class="math-container">$$2 \leq \frac{x^2}{xy}+\frac{y^2}{xy}$$</span>
<span class="math-container">$$2 \leq \frac{x}{y}+\frac{y}{x}$$</span>
Since <span class="math-container">$\frac{x}{y}$</span> is a fraction, assume <span class="math-container">$x$</span> is greater than <span class="math-container">$y$</span>, say
<span class="math-container">$$z = \frac{x}{y}$$</span>
This means that the lowest value of <span class="math-container">$z$</span> is <span class="math-container">$1$</span>, if <span class="math-container">$x > y$</span>
<span class="math-container">$$2 \leq z + \frac{1}{z}$$</span>
If <span class="math-container">$z$</span> is lowest, <span class="math-container">$\frac{1}{z}$</span> would be highest, but the minimum value of their sum comes when they're minimum.....equal to <span class="math-container">$1$</span>
<span class="math-container">$$z + \frac{1}{z} = ≥ 2$$</span></p>
|
2,885,754 | <p>I just sat a real analysis exam and this was a question in it that I couldn't answer...</p>
<p>Prove that $\left|e^\frac{-x^2}{2t}-e^\frac{-y^2}{2t}\right| \leq \frac{|x-y|}{t}$ for $x,y \in [-1,1] ,t>0$</p>
<p>I ended up trying to set $f(x,y)=e^\frac{-x^2}{2t}-e^\frac{-y^2}{2t}$, then attempted trying $f(-1,-1) =f(1,1)$ but never ended up getting anywhere.</p>
<p>Any tips on how this is actually solved? I've never seen an inequality problem like this before.</p>
| xbh | 514,490 | <p>Hint. consider $F(z)= \exp(-z^2/2t)\;[|z|\leqslant 1]$, $t$ is fixed. Then MVT. </p>
|
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