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 |
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1,269,738 | <p>I'm looking for problems that due to modern developments in mathematics would nowadays be reduced to a rote computation or at least an exercise in a textbook, but that past mathematicians (even famous and great ones such as Gauss or Riemann) would've had a difficult time with. </p>
<p>Some examples that come to mind are <em><a href="http://en.wikipedia.org/wiki/Group_testing">group testing problems</a></em>, which would be difficult to solve without a notion of error-correcting codes, and -- for even earlier mathematicians -- calculus questions such as calculating the area of some $n$-dimensional body.</p>
<p>The questions have to be understandable to older mathematicians and elementary in some sense. That is, past mathematicians should be able to appreciate them just as well as we can. </p>
| Community | -1 | <p>In the nineteenth century expressing the antiderivative of an elementary function as an elementary function was an open problem.</p>
<p>Nowadays, <a href="https://en.wikipedia.org/wiki/Risch_algorithm" rel="nofollow">Risch algorithm</a>, which can be run on machines, decides whether such operation can be done and, if so, yields a version of the correct result.</p>
<p>I cannot speak for past mathematicians, but I think this is a useful tool.</p>
<p><strong>Added</strong>: @columbus8myhw made a very important technical remark in the comments, which is also explained in the last part of the wikipedia article.</p>
|
4,504,080 | <p>If it is given that <span class="math-container">$$\displaystyle \frac{1}{(20-x)(40-x)}+\displaystyle \frac{1}{(40-x)(60-x)}+....+\displaystyle \frac{1}{(180-x)(200-x)}= \frac{1}{256}$$</span> then how to find the maximum value of <span class="math-container">$x$</span> ? I tried solving it with <span class="math-container">$V_n$</span> method but it is getting more tedious.</p>
| Thomas Preu | 1,011,882 | <p>As <a href="https://math.stackexchange.com/users/16497/alex-youcis">Alex Youcis</a> encouraged me to do so, I try to answer myself.</p>
<p><strong>The mistake</strong> was that <span class="math-container">$A:=\text{Aut}_{\text{var}}(\mathbb{G}_m\times_kK)\cong\left\{a\cdot t^n:a\in K^{\times},n\in\mathbb{Z}^{\times}=\{\pm 1\}\right\}\not\cong K^{\times}\times\mathbb{Z}/2\mathbb{Z}$</span> (an abelian group) but rather <span class="math-container">$\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$</span> (a non-abelian group). Explicitly the group composition is given by <span class="math-container">$(a_1,b_1)\cdot(a_2,b_2)=(a_1\cdot a_2^{b_1},b_1\cdot b_2)$</span>.</p>
<p>Since the <span class="math-container">$N$</span>-<span class="math-container">$\Delta$</span>-approach only computes group cohomology when the <span class="math-container">$G$</span>-set in question is an abelian group where the abelian group structure is compatible with the <span class="math-container">$G$</span>-action (i.e. a <span class="math-container">$G$</span>-module) we cannot compute <span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},A)$</span> via <span class="math-container">$\text{ker}(N)/\text{im}(\Delta)\cong\mathbb{Z}/2\mathbb{Z}$</span>.</p>
<p><strong>The correct <span class="math-container">$\text{Tw}(\mathbb{G}_m)$</span></strong> should be computed via non-abelian group cohomology as presented e.g. in <a href="https://www.jmilne.org/math/CourseNotes/iAG200.pdf" rel="nofollow noreferrer">J.S. Milne's lecture notes "Algebraic groups"</a>, sec. 27.a, pp. 469ff. We use 1-cocycles and equivalence <span class="math-container">$\sim$</span> of 1-cocycles as described there.</p>
<p>A 1-cocycle in this context is given by a pair of elements of <span class="math-container">$A\cong K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}$</span> of the form <span class="math-container">$c=(c_{\text{id}},c_{\sigma})$</span>. The cocycle condition immediately yields <span class="math-container">$c_{\text{id}}=1_A=(1,1)$</span> and we will discard it from the notation and identify <span class="math-container">$c=c_{\sigma}=(a,b)=(s+ti,b)$</span>. Exploiting the cocycle condition further we get the set of 1-cocycles to be <span class="math-container">$$\text{Cocyc}^1=\{(s+ti,1):(s,t)\in k^2\setminus\{(0,0)\} \wedge s^2+t^2-1=0\} \cup \{(s,-1):s\in k^{\times}\}$$</span> <span class="math-container">$$= \{(a,1):a\in\text{U
}(1)\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^+\}\quad\dot{\cup}\quad\{(a,-1):a\in\mathbb{R}^-\}.$$</span>
Straight forward computation yields that these three disjoint sets are exactly the 3 equivalence classes of 1-cocycles.</p>
<p>In summary <span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},A)=\left(\text{Cocyc}^1/\sim\right)=\{\mathbb{G}_m,\text{S}^1,Y\}=\text{Tw}(\mathbb{G}_m)$</span></p>
<p><strong>Alternatively Alex Youcis</strong> suggested to use the short exact sequence of non-abelian groups (recall that abelian groups are also non-abelian groups) <span class="math-container">$$1\rightarrow K^{\times} \rightarrow K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow 1,$$</span>
which yields a long exact sequence of pointed sets, whose relevant part is given by
<span class="math-container">$$\ldots\rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \rightarrow \text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z}).$$</span>
By the classic Hilbert's Theorem 90 (cohomological version) we have <span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},K^{\times})\cong 1$</span> and by remark 1 from the question <span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},\mathbb{Z}/2\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$</span>, which makes the previous sequence more explicit
<span class="math-container">$$\ldots\rightarrow 1 \rightarrow \text{H}^1_{\text{grp}}(\text{G},K^{\times}\rtimes\mathbb{Z}/2\mathbb{Z}) \xrightarrow{\alpha} \mathbb{Z}/2\mathbb{Z}.$$</span>
<span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},A)$</span> consists of the fibers above <span class="math-container">$\mathbb{Z}/2\mathbb{Z}\cong\{\mathbb{G}_m, \text{S}^1\}$</span> of which <span class="math-container">$\alpha^{-1}\left(\mathbb{G}_m\right)=\{\mathbb{G}_m\}$</span> must be a singleton, because of the properties of exact sequences of pointed sets. This leaves us with computing the other fiber <span class="math-container">$\alpha^{-1}\left(\text{S}^1\right)$</span>.</p>
<p>Computing <span class="math-container">$\alpha^{-1}\left(\text{S}^1\right)$</span> explicitly seems to me not much simpler than computing <span class="math-container">$\text{H}^1_{\text{grp}}(\text{G},A)$</span> directly using 1-cocycles and equivalences as done above. At least I do not see how to do it differently, which gives this approach the feeling of a dead end. But more apt people might have better ideas than me.</p>
<p><strong>Edit note:</strong> removed the trailing dots in the long exact sequences as suggested by Alex Youcis.</p>
|
25,917 | <p>$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ </p>
<p>$\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$</p>
<p>Why they are different?</p>
| Bill Dubuque | 242 | <p>This follows from a more general result that's both simpler to prove and more insightful, viz. the result follows immediately by this frequently applicable <strong>multiplicative form of induction</strong>, which shows that for multiplicative sets we need only check the generators (here <span class="math-container">$1$</span> and primes).</p>
<p><strong>Lemma</strong> <span class="math-container">$\rm\, \mathbb N$</span> is the only set of naturals containing <span class="math-container">$\color{#c00}1$</span> and <span class="math-container">$\rm\color{#c00}{all\ primes}$</span> and <span class="math-container">$\rm\color{#0a0}{\text {closed under product}}$</span>.</p>
<p><strong>Proof</strong> <span class="math-container">$\ $</span> Suppose <span class="math-container">$\rm\!\ S\subset \mathbb N\,$</span> has said properties. <span class="math-container">$ $</span> We prove by strong induction every natural <span class="math-container">$\rm\!\ n\in S.\, $</span> If <span class="math-container">$\rm\,n\,$</span> is <span class="math-container">$\,\color{#c00}1\,$</span> or <span class="math-container">$\color{#c00}{\rm prime}$</span> then by hypothesis <span class="math-container">$\rm\,n\in S.\,$</span> Else <span class="math-container">$\rm\,n\,$</span> is composite, <span class="math-container">$ $</span> hence <span class="math-container">$\rm\ n = j k\ $</span> for <span class="math-container">$\rm\, j,k < n.\,$</span> By strong induction the smaller <span class="math-container">$\rm\ j,k\in S,\,$</span> <span class="math-container">$\rm\color{#0a0}{thus}$</span> <span class="math-container">$\rm\, n = jk\in S.\ \ \ $</span> <strong>QED</strong></p>
<p>This yields the sought result. Let <span class="math-container">$\rm\!\ S\!\ $</span> be the set of naturals that have the form <span class="math-container">$\rm\,2^{\,j}\!\ n\,$</span> for odd <span class="math-container">$\rm\,n\in \mathbb N.\ $</span> <span class="math-container">$\, 1\!\ $</span> and all primes <span class="math-container">$\rm\,p\,$</span> are in <span class="math-container">$\rm\!\ N,\, $</span> by <span class="math-container">$\, 2 = 2^{\!\ 1}\!\cdot\! 1\,$</span> and <span class="math-container">$\rm\, p = 2^{\!\ 0}\!\cdot\! p\,$</span> for odd <span class="math-container">$\rm\,p\,$</span> (and <span class="math-container">$1).$</span> <span class="math-container">$\rm\,S\!\ $</span> is closed under multiplication by <span class="math-container">$\rm\, (2^{\,j}\!\ m) (2^{\,k}\!\ n) = 2^{\,j+k}\!\ m\!\ n,\, $</span> with <span class="math-container">$\rm\!\ m\!\ n\!\ $</span> odd by <span class="math-container">$\rm\!\ m,n\!\ $</span> odd. <span class="math-container">$\!\ $</span> So <span class="math-container">$\rm\ S = \mathbb N\ $</span> by Lemma.</p>
<hr />
<p><strong>Corollary</strong> <span class="math-container">$\ $</span> Every natural <span class="math-container">$> 0\,$</span> is a product of primes (i.e. irreducibles).</p>
<p><strong>Proof</strong> <span class="math-container">$\, $</span> It is clear for <span class="math-container">$1$</span> and primes, and products of primes are closed under multiplication.</p>
<p><strong>Remark</strong> <span class="math-container">$ $</span> We could also deduce the sought result from the prior Corollary. Then it reduces to (inductively) proving that a product of <span class="math-container">$n$</span> odds remains odd. But this way is not an instructive example of strong induction, since the strong induction is hidden in the proof of the corollary.</p>
|
2,374,282 | <p>I am trying to find all connected sets containing $z=i$ on which $f(z)=e^{2z}$ is one to one.
I have no idea how to approach.
Can someone give me some hints?
Thank you</p>
| Grey Matters | 176,262 | <p>If you want to be able to do that in your head, then you'd start by working out the number of the day of both dates (as in, January 2nd would be day #2, December 30th in a non-leap year would be day #364, etc.).</p>
<p>You can do this fairly quickly by memorizing the number of days that occur prior to each month in a non-leap year.</p>
<p>Before January, there are 0 days.<br>
Before Februrary, there are 31 days.<br>
Before March, there are 59 days.<br>
Before April, there are 90 days.<br>
Before May, there are 120 days.<br>
Before June, there are 151 days.<br>
Before July, there are 181 days.<br>
Before August, there are 212 days.<br>
Before September, there are 243 days.<br>
Before October, there are 273 days.<br>
Before November, there are 304 days.<br>
Before December, there are 334 days.</p>
<p>These amounts of days are easy to memorize. All of them, not surprisingly, are close to their month number (January = 1, February = 2, etc.) minus 1 times 30. For leap years, of course, just add 1 to every month from March to December.</p>
<p>Using your example of August 17th, 1996, you'd first need to realize that 1996 is a leap year. Before August, there are 212 days before it, plus 1 for the leap day, making 213 days. Naturally, this makes August 17th, 1996 the 230th (213 + 17 = 230) day of the year.</p>
<p>As I write this, it's July 27, 2017, so let's figure out when the next July 27th would be after your start date. That would be July 27th, 1997. July 27th is the 208th day of the year (181 days before July, plus 27 days in July).</p>
<p>From the 230th day of a leap year to the end of the year (the 366th day) is 136 days. Adding another 208 days to that gives us 344 days between the 2 dates.</p>
<p>So, from August 17th, 1996 to July 27, 1997 is 344 days. July 27th, 2017 is 20 years after that, so if you desired, you could simply state that it's been exactly 20 years and 344 days since August 27, 1996.</p>
<p>To count the total number of days, you'd have to multiply 365 by 20 (which is 730 × 10 or 7300 days). Finally, you'd have to add 1 for each leap year between the 2 dates: 2000, 2004, 2008, 2012, and 2016. You get 7300 + 5 + 344 = 7649 days between the 2 dates. (<a href="http://www.wolframalpha.com/input/?i=days%20from%20August%2017,%201996%20to%20July%2027,%202017" rel="nofollow noreferrer">Wolfram|Alpha calculation</a>)</p>
<p>Of course, you'd have to practice enough to become comfortable with multiplying 365 times various integers to use this latter approach.</p>
|
604,836 | <p>Prove if <span class="math-container">$a \equiv c \pmod{n}$</span> and <span class="math-container">$b \equiv d \pmod n$</span> then <span class="math-container">$ab \equiv cd \pmod{n}$</span>.</p>
<p>I tried to use <span class="math-container">$(a-c)(b-d) = ab-ad-cb+cd$</span>, but it seem doesn't work.</p>
| Stefan4024 | 67,746 | <p>One good way to prove it is to use this fact:</p>
<p>$$a \equiv c \pmod n \implies n|a-c$$
$$b \equiv d \pmod n \implies n|b-d$$</p>
<p>Obviously then $n|b(a-c) + c(b-d)$</p>
<p>So we have: $n|ba - bc + bc - cd$</p>
<p>$$\implies n|ab - cd$$</p>
<p>Q.E.D.</p>
|
2,585,466 | <p>I have two growth curve data sets, A (Martians) and B (Venusians). Data point sets of age (0 (birth) - 250 months, X axis) against height (0 - 200 centimeters, Y axis). The first set (A) contains 67 X Y point pairs, the second set (B) contains 27 point pairs. I have fit both data sets to my favorite version of the Logistic Equation using NonlinearModelFit. NonlinearModelFit returns estimates for my two independent variables: Increment, (N0), and Time Coefficient (k). Then following I invoked "ParameterTable" calculating: (1) Standard Errors (2) t-Statistics and (3) P-Values for both of the curve fitting exercises, Martians and Venusians. Of the three Parameters: Standard Errors, t-Statistics, and P-Values, which parameter indicates a better fit to an energy conservative logistic equilibrium? Standard Errors on the calculated Time Coefficients (k)? t-Statistics on the calcuated Time Coefficents (k)? Is growth on Mars more of an energy conservative mechanical process than growth on Venus? Are data sets with different numbers of point pairs directly comparable on Standard Errors, t-Statistics and P-Values? </p>
| ratalan | 506,825 | <p>One possible way to do that is to compute the gradients at a point $(x,y)$ of the functions which define the two curves $f\left(x,y\right)=y^{2}-4ax$ and $g\left(x,y\right)=xy-c^{2}$.
The respective gradients are $\left(\begin{array}{c}
-4a\\
2y
\end{array}\right)$ and $\left(\begin{array}{c}
y\\
x
\end{array}\right)$, so their scalar product is $y\left(2x-4a\right)$. If the two curves
intersect at $\left(x,y\right)$, you can compute the coordinates $x$ and $y$ and then deduce the scalar product as a function of $a$ and $c$, and check that the latter vanishes when $c^{4}=32a^{4}$. </p>
|
3,281,503 | <blockquote>
<p>For natural numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, what is the greatest value of <span class="math-container">$b$</span> so that <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits?</p>
</blockquote>
<p>I knew that the greatest value of <span class="math-container">$b$</span> is <span class="math-container">$21$</span>, where <span class="math-container">$9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$</span>, but I am not sure how to prove that there is no greater value of <span class="math-container">$b$</span>.</p>
| Arthur | 15,500 | <p>Proof sketch: First show that <span class="math-container">$a=9$</span> is the best possible value for <span class="math-container">$a$</span> (you don't have to show that it is better than any other value, just that no other value can possibly be better). Then find the maximal <span class="math-container">$b$</span> that works for <span class="math-container">$a=9$</span> through brute force.</p>
<p>Alternately: If <span class="math-container">$a^b$</span> has <span class="math-container">$b$</span> digits, what can be said about <span class="math-container">$\frac{a^b}{10^b}=\left(\frac{a}{10}\right)^b$</span>? Now in a similar manner show that <span class="math-container">$a=9$</span> is optimal and then afterwards, find the best possible <span class="math-container">$b$</span>. This way it's a bit easier to show that nothing untoward happens for larger <span class="math-container">$b$</span>.</p>
|
900,884 | <p>If $ A = \{ m^n| \text{ } m, n \in Z \text { and } m, n \ge 2 \} $, then how find $\sum_{k \in A} \frac{1}{k-1} $?</p>
| gtrrebel | 169,563 | <p>The answer is 1.</p>
<p>The result is known as Goldbach-Euler theorem.</p>
<p>See <a href="http://en.wikipedia.org/wiki/Goldbach-Euler_theorem" rel="noreferrer">Wikipedia entry</a> for "proof".</p>
<p>For rigorous proof, you could consider sum of reciprocals of all perfect powers, $S$. Note that sum equals
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\ldots+\frac{1}{25}+\frac{1}{125}+\ldots
$$
where $B$ is set of all integers not perfect power. Why? All reciprocals of perfect powers are clearly in the sum, and terms in sum are distinct: if
$$
\frac{1}{x^{n}} = \frac{1}{y^{m}}
$$
and $x$ and $y$ are distinct, also $m$ and $n$ are. But since $x$ and $y$ aren't perfect powers, $x^{n}$ and $y^{m}$ are exactly $n$:th and $m$:th powers, respectively, for distinct $m$ and $n$ which is impossible.</p>
<p>But now
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \sum_{x = 2}^{\infty}\sum_{n = 2}^{\infty}\frac{1}{x^{n}}-\sum_{x \in A}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} \\
= \sum_{x = 2}^{\infty}\frac{1}{x(x-1)}-\sum_{x \in A}\frac{1}{x(x-1)} \\
= 1- \sum_{x \in A}\left(\frac{1}{x-1}-\frac{1}{x}\right) = 1- \sum_{x \in A}\frac{1}{x-1}+\sum_{x \in A}\frac{1}{x} \\
= 1- \sum_{x \in A}\frac{1}{x-1}+S
$$
so
$$
\sum_{x \in A}\frac{1}{x-1} = 1
$$</p>
|
2,072,347 | <p>I was trying to solve this problem, but couldn't figure it out. The solution goes like this:</p>
<p><a href="https://i.stack.imgur.com/1KSWH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1KSWH.png" alt="http://www.tkiryl.com/Calculus/Problems/Section%201.4/Calculating%20Limits/Solutions/Calc_S_59.png (I don't have the reputation to post the image)"></a></p>
<p>I don't understand the first step. Why is the limit multiplied by $\frac{4x}{5x}$? and $\frac{5}{4}$ ? </p>
| projectilemotion | 323,432 | <p>Alternatively, you could use <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hopital's rule</a>:</p>
<p>$\lim \limits_{x \to 0} \frac{\sin(5x)}{\sin(4x)}$=$\lim \limits_{x \to 0} \frac{\frac{d}{dx}\sin(5x)}{\frac{d}{dx}\sin(4x)}$=$\lim \limits_{x \to 0} \frac{5\cos(5x)}{4\cos(4x)}=\frac{5}{4}$</p>
|
2,786,656 | <p>I know that if a group is generated by a single element then the group is abelian but does this mean that if a group is abelian then its conjugacy class is composed of a single element?</p>
| Tsemo Aristide | 280,301 | <p>Since $T$ is an isomorphism and an isometry, $T^{-1}$ is also an isometry. We deduce that $T(\{x\in X:\|x\|=1\})=\{y\in Y:\|y\|=1\}$ This implies that $\|T^*(f)\|=sup_{\|x\|=1}||f(T(x))|=sup_{\|y\|=1}|f(y)|=\|f\|$.</p>
|
912,002 | <p>I'm solving some programming puzzle and it has come down to this:</p>
<p>I've a fraction, say 12/13, and I need to multiply it with a smallest possible natural number (say x) to get a whole number. How do I solve for x?</p>
<p>I intuitively feel I need to use LCM to solve this but haven't been able to pin down on a method.</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $\,\ c \dfrac{a}b\in\Bbb Z\iff b\mid ac\iff b\mid ac,bc\iff b\mid(ac,bc)= (a,b)c\iff b/(a,b)\mid c$</p>
|
871,744 | <p>When A and B are square matrices of the same order, and O is the zero square matrix of the same order, prove or disprove:-
$$AB=0 \implies A=0 \text{ or } \ B=0$$</p>
<p>I proved it as follows:-</p>
<p>Assume $A \neq O$ and $ B \neq O$:
then, $$ |A||B| \neq 0 $$
$$ |AB| \neq 0 $$
$$ AB \neq O $$
$$ \therefore A \neq O\ and\ B \neq O \implies AB \neq O $$
$$ \neg[ AB \neq O] \implies \neg [ A \neq O\ and\ B \neq O ] $$
$$AB=O \implies A=O \text{ or } \ B=O$$</p>
<p>But when considering,
A := \begin{pmatrix} 1&1 \\1&1
\end{pmatrix} and B:= \begin{pmatrix} -1& 1\\ 1 &-1
\end{pmatrix}then, AB=O and A$\neq $O and B $\neq$ O</p>
<p>I can't figure out which one and where I went wrong.</p>
| Barath | 316,775 | <p>You aren't wrong! AB = O does not imply that either A or B is zero. In matrices there is no such case. So even when A is not equal to O and B is not equal to zero, you can get AB =O. Voila!</p>
|
3,389,542 | <blockquote>
<p><strong>Proposition.</strong> If <span class="math-container">$\text{Ran}(R) \subseteq \text{Dom}(S)$</span>, then <span class="math-container">$\text{Dom}(S \circ R) = \text{Dom}(R)$</span></p>
</blockquote>
<p>My attempt:</p>
<p>Suppose <span class="math-container">$\text{Ran}(R) \subseteq \text{Dom}(S)$</span></p>
<p>We need to show:</p>
<p><span class="math-container">$(\rightarrow) $</span> <span class="math-container">$\text{Dom}(S \circ R) \subseteq \text{Dom}(R)$</span></p>
<p><span class="math-container">$(\leftarrow)$</span> <span class="math-container">$\text{Dom}(R) \subseteq \text{Dom}(S \circ R) $</span></p>
<hr>
<p><span class="math-container">$(\rightarrow)$</span></p>
<p>Consider arbitrary element <span class="math-container">$a$</span>, where <span class="math-container">$a \in \text{Dom}(S \circ R)$</span>. Then there must be some element <span class="math-container">$p$</span>, where <span class="math-container">$p = (a,x)$</span> and <span class="math-container">$p \in R$</span>. It implies that <span class="math-container">$a \in Dom(R)$</span>. Since <span class="math-container">$a$</span> was arbitrary, we have <span class="math-container">$\text{Dom}(S \circ R) \subseteq \text{Dom}(R)$</span></p>
<p><span class="math-container">$(\leftarrow$</span>)</p>
<p>Consider element <span class="math-container">$(x,y)$</span> where <span class="math-container">$(x,y) \in R$</span>. Since <span class="math-container">$y \in Ran(R)$</span> and <span class="math-container">$\text{Ran}(R) \subseteq \text{Dom}(S)$</span>, it follows that <span class="math-container">$y \in Dom(S)$</span>, which means that there must be some element <span class="math-container">$(y,p)$</span> such that <span class="math-container">$(y,p) \in S$</span>. </p>
<p>Since <span class="math-container">$(y,p) \in S$</span> and <span class="math-container">$(x,y) \in R$</span>, we have <span class="math-container">$(x,p) \in S \circ R$</span>, and it means that <span class="math-container">$x \in Dom(S \circ R)$</span>.</p>
<p>Since we've considered arbitrary element, we have <span class="math-container">$\text{Dom}(R) \subseteq \text{Dom}(S \circ R)$</span></p>
<p>We've shown both sides, hence <span class="math-container">$\text{Dom}(S \circ R) = \text{Dom}(R)$</span>. <span class="math-container">$\Box$</span></p>
<hr>
<p>Is it correct?</p>
<p>If it is, are there better ways to prove <span class="math-container">$(\rightarrow)$</span>?</p>
| Nicholas Viggiano | 709,871 | <p>I would argue that the best you can do is using a maximal triangle-free graph (per Turan's construction, this is a complete bipartite graph whose partitions are as balanced as possible). It at least serves as a lower bound: take <span class="math-container">$n$</span> even (for simplicity, I'll handle odd in a bit), and construct the complete bipartite graph with partitions each of size <span class="math-container">$n/2$</span>. This creates <span class="math-container">$n^2/4$</span> edges, and it's quick to note that no color can be shared among 3 or more vertices, because by pigeonhole, 2 would have to be on the same side (and thus are not adjacent). So every edge needs its own color, and thus <span class="math-container">$x(G) \geq n^2/4$</span> for <span class="math-container">$n$</span> even. (With <span class="math-container">$n$</span> odd, instead you have <span class="math-container">$\frac{n - 1}{2}\frac{n + 1}{2} = \frac{n^2 - 1}{4}$</span> edges and thus colors.)</p>
<p>As for why this is probably best possible: the proof of Turan's theorem is the justification behind this construction being maximally triangle-free. In this scenario, each edge has its own color, but if we have a triangle, suddenly all three of those edges can share a color, that is, if we add an edge to our current construction (say our current edge goes between vertices who share color 1, and we add an edge between the vertex in the "top" partition and a new vertex up top), then suddenly the new top vertex and the original bottom vertex are allowed to share color 1 instead of having to use whatever color their edge originally induced.</p>
<p>Like I said, this is wishy-washy, so I would love for someone to tighten it up, but I'm fairly confident triangle-free is the way to go.</p>
|
2,475,938 | <blockquote>
<p>How can I factor the polynomial <span class="math-container">$x^4-2x^3+x^2-1$</span>?</p>
</blockquote>
<p>This is an exercise in algebra. I have the solution showing that
<span class="math-container">$$
x^4-2x^3+x^2-1=(x^2-x-1)(x^2-x+1).
$$</span></p>
<p>But the solution does not show any details. Using the distributive property I can check that this is indeed true:
<span class="math-container">$$
\begin{align}
&(x^2-x-1)(x^2-x+1)\\
&=x^2(x^2-x+1)-x(x^2-x+1)-(x^2-x+1)\\
&=x^4-x^3+x^2-x^3+x^2-x-x^2+x-1\\
&=x^4-2x^3+x^2-1,
\end{align}
$$</span></p>
<p>but I can't figure out the steps to get there. Can anyone help?</p>
| Clayton | 43,239 | <p>As @MathLover suggests, use the factorization $$x^4-2x^3+x^2=x^2(x-1)^2=(x(x-1))^2.$$ Now use the fact that it is of the form $y^2-1$ to factor it as the difference of squares.</p>
|
2,081,641 | <p>I figured out that $\lim_{n \to \infty}\frac{(3n^2−2n+1)\sqrt{5n−2}}{(\sqrt{n}−1)(1−n)(3n+2)} $ is $-\sqrt{5}$ but I don't know how to prove it.</p>
| Community | -1 | <p>The dominant terms in the numerator and denominator are both in $n^{5/2}$, hence the limit is finite. The corresponding coefficients are $3\sqrt5$ and $-3$. The ratio gives the answer.</p>
|
114,664 | <p>How would one evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$?</p>
<p>I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.</p>
| josh | 25,488 | <p>$$
I = \int_0^1 \frac{\ln(1+x)}{x}\,dx = \int_0^\infty \ln(1+e^{-t})\,dt\,,
$$
where $x = e^{-t}$. Then expand
$$
\ln(1 + e^{-t}) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,e^{-tn}\,,
$$
So we find
$$
I = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,\int_0^\infty e^{-tn}\,dt = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}\,.
$$
Is the last sum familiar to you? What about the closely related and easier sum
$$
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\,?
$$</p>
|
114,664 | <p>How would one evaluate $\int_0^1 {\ln(1+x)\over x}\,dx$?</p>
<p>I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.</p>
| Alex Becker | 8,173 | <p>The Taylor series for $\frac{\ln(1+x)}{x}$ is $\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}$, and this converges absolutely in $(0,1)$ thus we can use it for our integral. This means
$$\int_0^1\frac{\ln(1+x)}{x}=\int_0^1\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\int_0^1\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{1}{n^2}$$
and this series is equal to $\pi^2/12$ according to Wolfram|Alpha.</p>
|
2,117,054 | <p>Find all prime solutions of the equation $5x^2-7x+1=y^2.$</p>
<p>It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$ </p>
<p>In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$</p>
<p>How put together the two cases?</p>
<p>Computer find two prime solutions $(3,5)$ and $(11,23).$</p>
| Robert Israel | 8,508 | <p>Completing the square and dividing by $5$, we have </p>
<p>$$ (10 x - 7)^2 - 20 y^2 = 29$$</p>
<p>Thus $z = 10 x - 7$ and $w = 2 y$ are solutions of the Pell-type equation</p>
<p>$$ z^2 - 5 w^2 = 29$$</p>
<p>The positive integer solutions of this can be written as</p>
<p>$$\pmatrix{z\cr w\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{7\cr 2\cr} \ \text{or}\ \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{23\cr 10\cr}$$
for nonnegative integers $n$.</p>
<p>Now you want $w$ to be even and $z \equiv 3 \mod 10$. All the solutions will have $w$ even, and $z$ altermately $\equiv \pm 3 \mod 10$. Thus for $n$ even you get integers for $x,y$ with
$$ \pmatrix{z_n\cr w_n\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{23\cr 10\cr}$$
and for $n$ odd,
$$ \pmatrix{z_n\cr w_n\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{7\cr 2\cr}$$
You do get primes for $n=0$ ($z_0 = 23, w_0 = 10, x_0 = 3, y_0 = 5$)
and $n=1$ ($z_1 = 103, w_1 = 46, x_1 = 11, y_1 = 23$). In general,</p>
<ul>
<li>$x_n \equiv 0 \mod 3$ for $n \equiv 0$ or $3 \mod 4$. </li>
<li>$x_n \equiv 0 \mod 17$ for $n \equiv 2$ or $3 \mod 6$. </li>
<li>$x_n \equiv 0 \mod 5$ or $y_n \equiv 0 \mod 5$ for $n \equiv 0,3, 4, 5, 6, 9 \mod 10$. </li>
<li>$x_n \equiv 0 \mod 11$ for $n \equiv 1, 8 \mod 10$.</li>
<li>$x_n \equiv 0 \mod 13$ for $n \equiv 3, 5, 6, 7, 8, 10 \mod 14$.</li>
<li>$y_n \equiv 0 \mod 23$ for $n \equiv 1, 2, 5, 6 \mod 8$.</li>
</ul>
<p>And every integer $n$ is in at least one of these classes. We conclude there are no other prime solutions. </p>
|
3,225,553 | <p>Show that <span class="math-container">$4x^2+6x+3$</span> is a unit in <span class="math-container">$\mathbb{Z}_8[x]$</span>.</p>
<p>Once you have found the inverse like <a href="https://math.stackexchange.com/questions/3172556/show-that-4x26x3-is-a-unit-in-mathbbz-8x">here</a>, the verification is trivial. But how do you come up with such an inverse. Do I just try with general polynomials of all degrees and see what restrictions RHS = <span class="math-container">$1$</span> imposes on the coefficients until I get lucky? Also is there a general method to show an element in a ring is a unit?</p>
| Bill Dubuque | 242 | <p>By <a href="https://math.stackexchange.com/a/3224776/242">simpler multiples,</a> to invert <span class="math-container">$\, a - f\,$</span> where <span class="math-container">$\,a\,$</span> is invertible, say <span class="math-container">$\,\color{#0a0}{ab = 1},$</span> and <span class="math-container">$\,f\,$</span> is nilpotent <span class="math-container">$\color{#c00}{f^n = 0},\,$</span> we invert its <em>simpler multiple</em> <span class="math-container">$\, a^n-\color{#c00}{f^n} = \color{#0a0}{a^n},\,$</span> with obvious inverse <span class="math-container">$\,\color{#0a0}{b^n},\,$</span> explicitly</p>
<p><span class="math-container">$\ \ \ \ \ \ \ \ \ \color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \overbrace{\dfrac{1}{a-f} = \dfrac{a^{\large n-1}\!+\!\cdots\! +\! f^{\large n-1}}{\!\!\!\!\!\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}}}^{\large \text{check via cross multiply}} =\, \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})$</span>
<span class="math-container">$\!\begin{align}{\rm so}\ \ &\color{#0a0}{3(3)=1},\, \color{#c00}{f^{\large 3} = 0}\ \Rightarrow\ \dfrac{1}{3-f} = \dfrac{3^{\large 2}+\,3f\,+\, f^{\large 2^{\phantom{|^{|^.}}\!\!\!\!\!}}}{\color{#0a0}{3^{\large 3}-\color{#c00}{f^{\large 3}}}}\ \ \, =\, \ \ \color{#0a0}{3^{\large 3}}(3^{\large 2}\! +3f + f^{\large 2}) = \bbox[5px,border:1px solid #c00]{2x+3}\\[.1em]
&{\rm because}\ \ \ \color{#c00}{2^{\large 3}\mid f^{\large 3}}\ \ {\rm by}\ \ 2\mid f = -6x-4x^2,\,\ {\rm to\ invert}\ \ 3\!-\!f = 3\!+\!6x\!+\!4x^2 \in \smash[b]{\Bbb Z_{\large \color{#c00}{2^{\Large 3}}}}\end{align}$</span></p>
<p>Generally it is <a href="https://math.stackexchange.com/a/19145/242">easy to prove</a> that a polynomial is a unit iff its constant term is a unit and all other coefficients are nilpotent (the method of proof there can be made constructive - similar to above).</p>
<p>This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of <a href="https://math.stackexchange.com/a/26097/242">rationalizing denominators</a> and in <a href="https://math.stackexchange.com/a/174687/242">Gauss's algorithm</a> for computing modular inverses. <a href="https://math.stackexchange.com/a/3224776/242">Analogous methods</a> may be employed for for computing remainders via modular arithmetic.</p>
|
870,583 | <p><strong>Question:</strong>
Each user on a computer system has a password, which is six to eight characters long, where each character is an upper-case letter or a digit. Each password must contain at least one digit. How many possible passwords are there?</p>
<blockquote>
<p>I'm in the <strong>Basic of Counting</strong> section of my Discrete Mathematics book, and I have a problem with my reasoning with this. I will give you my reasoning and the books reasoning. Both give different answers, but I don't see a difference in the train of thought, so I need someone to point out the difference.</p>
</blockquote>
<p><strong>My Attempt:</strong>
Immediately I noticed 3 kinds of character length which allows me to break it down to three cases, respectively $P_6, P_7, P_8$, then add all of them. For $P_6$, $5$ will be made up of alpha numeric characters, and $1$ is made up of just digits due to the requirement of <em>"at least one digit"</em>, thus</p>
<p>$$P_6 = (36)^5*10$$</p>
<p>Should be enough to show my train of thought, now the books solution.</p>
<p><strong>Books Solution:</strong> The book did the same thing in dividing in 3 cases and adding them later so I'll go ahead and show you their train of thought for $P_6$. </p>
<p>$$P_6 = 36^6 - 26^6$$</p>
<p>Basically its the number of possible 6 alphanumeric minus just alpha numeric.</p>
<blockquote>
<p>I know that both give different answers, but I still can't tell why.</p>
</blockquote>
| David | 119,775 | <p>Your calculation for $P_6$ gives the number of words in which the first five characters can be letters or digits, and the last must be a digit. But there is nothing in the rules to say that the <strong>last</strong> must be a digit.</p>
<p>Alternatively, you could interpret your answer to say that, for example, symbols $1,2,4,5,6$ can be anything, and the $3$rd must be a digit. Same problem - there is no requirement that <em>any specific</em> symbol must be a digit.</p>
<p>The simplest solution to the problem is the one given in the book.</p>
|
1,255,368 | <p>How do I solve this? What steps? I have been beating my head into the wall all evening. </p>
<p>$$ x^2 + y^2 = \frac{x}{y} + 4 $$</p>
| N. F. Taussig | 173,070 | <p>We can multiply both sides of the equation
$$x^2 + y^2 = \frac{x}{y} + 4 \tag{1}$$
by $y$ to obtain
$$x^2y + y^3 = x + 4y \tag{2}$$
Differentiating equation 2 implicitly with respect to $x$ yields
\begin{align*}
2xy + x^2y' + 3y^2y' & = 1 + 4y'\\
(x^2 + 3y^2 - 4)y' & = 1 - 2xy\\
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4} \tag{3}
\end{align*}
Let's see why this answer is equivalent to that given by @CivilSigma. If we solve equation 1 for $x/y$, we obtain
$$\frac{x}{y} = x^2 + y^2 - 4$$
Hence,
\begin{align*}
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4}\\
& = \frac{1 - 2xy}{x^2 + y^2 - 4 + 2y^2}\\
& = \frac{1 - 2xy}{\dfrac{x}{y} + 2y^2}\\
& = \frac{y - 2xy^2}{x + 2xy^3}
\end{align*}</p>
|
1,557,015 | <p>This one looks simple, but apparently there is something more to it.
$$f{(x)=x^x}$$
I read somewhere that the domain is $\Bbb R_+$, a friend said that $x\lt-1, x\gt0$... </p>
<p>I'm really confused, because i don't understand why the domain isn't just all the real numbers.
According to any grapher online the domain is $\Bbb R_+$.
Any Thoughts on the matter?</p>
<p>Can someone explain what am I missing?</p>
| John Molokach | 90,422 | <p>I am not sure what your mathematics background is, but the function
$f(x)=x^x$ is defined for $\mathbb{R}_+$ as well as a countably infinite set of rational values in $\mathbb{Q}_-$. For example, we can find $f(-\frac n3)$ for all $n\in\mathbb{N}$. In fact, I cannot with confidence write down the entire set of negative values in the domain but any $x=\frac n {1-2n}$ will work for starters.</p>
<p>Some textbooks won't include the negative values, because they form a countable set of numbers that has Lebesgue measure zero.</p>
|
2,568,157 | <p>Consider the following:</p>
<p>$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$</p>
<p>$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$</p>
<p>$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$</p>
<p>In General is it true for further increase i.e.,</p>
<p>Is</p>
<p>$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$</p>
| DanielWainfleet | 254,665 | <p>Let the LHS be $A(n)$ and let the RHS be $2B(n)^4.$ </p>
<p>Then $A(n+1)-A(n)=(n+1)^7+(n+1)^5=(n+1)^5(n^2+2n+2).$ </p>
<p>$$\text {We have }\quad 2B(n+1)^4-2B(n)^4=$$ $$(*)\quad =2(B(n+1)^2+B(n)^2)\cdot (B(n+1)+B(n))\cdot (B(n+1)-B(n)).$$ Since $B(n)=n(n+1)/2$ we have $$B(n+1)^2+B(n)^2=(n+1)^2((n+2)^2+n^2)/4=(n+1)^2(n^2+2n+2)/2$$ $$\text { and}\quad B(n+1)+B(n)=(n+1)((n+2)+n)/2=(n+1)^2$$ $$\text { and }\quad B(n+1)-B(n)=n+1.$$ From these we compute $(*)$ (4th line from the top) and find it is equal to $(n+1)^5(n^2+2n+2)$, which is $A(n+1)-A(n).$ (2nd line from the top.). </p>
<p>So if $A(1)=B(1)$ then $A(n)=B(n)$ for all $n\in \Bbb N$ by induction on $n.$</p>
<p>Interesting identity. Another interesting one is $\sum_{x=1}^nx^3=(\sum_{x=1}^nx)^2=(n(n+1)/2)^2.$</p>
|
281,735 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/202452/why-is-predicate-all-as-in-allset-true-if-the-set-is-empty">Why is predicate “all” as in all(SET) true if the SET is empty?</a> </p>
</blockquote>
<p>In don't quite understand this quantification over the empty set:</p>
<p>$\forall y \in \emptyset: Q(y)$</p>
<p>The book says that this is always TRUE regardless of the value of the predicate $Q(y)$, and it explain that this is because this quantification adds no predicate at all, and therefore can be considered the weakest predicate possible, which is TRUE.</p>
<p>I know that TRUE is the weakest predicate because $ $P$ \Rightarrow$ TRUE is TRUE for every $P$.
I don't see what is the relationship between this weakest predicate and the quantification. </p>
| user58437 | 58,437 | <p>Because there are no members to the set, anything you say about a member can be considered <em>trivially</em> true. It's not really to say that there are actually members for which Q, but rather for all y, Q(y)--i.e. there are no circumstances where y and not Q(y). </p>
|
4,554,831 | <blockquote>
<p>Let <span class="math-container">$(X,d)$</span> be a metric space. Prove that if the point <span class="math-container">$x$</span> is on the boundary of the open ball <span class="math-container">$B(x_0,r)$</span> then <span class="math-container">$d(x_0,x)=r$</span>.</p>
</blockquote>
<p>I find this difficult because it seems intuitive yet not easy to prove. By definition, if <span class="math-container">$A\subset X$</span> then a point <span class="math-container">$x$</span> is on the boundary if for all <span class="math-container">$\epsilon>0$</span> we have <span class="math-container">$B(x,\epsilon)\cap A\ne\emptyset$</span> and also <span class="math-container">$(X\setminus A)\cap B(x,\epsilon)\ne\emptyset$</span>. However I don't know how to use this definition in any meaningful way.</p>
| geetha290krm | 1,064,504 | <p>If <span class="math-container">$d(x,x_0) <r$</span> then <span class="math-container">$x \in B(x_0,r)$</span> so <span class="math-container">$x$</span> is not a boundary point. (<span class="math-container">$B(x_0,r)$</span> is open). If <span class="math-container">$d(x,x_0) >r$</span> and <span class="math-container">$s=d(x,x_0) -r$</span> then there is no point of <span class="math-container">$B(x_0,r) $</span> in <span class="math-container">$B(x,s)$</span>, so <span class="math-container">$x$</span> is not a boundary point.</p>
<p>[If <span class="math-container">$y \in B(x_0,r) \cap B(x,s)$</span> then <span class="math-container">$s+r=d(x_0,x) \leq d(x_0,y)+d(y,x) <r+s$</span>, a contradiction].</p>
|
3,884,891 | <p>I am reading an article about "Longest Paths in Digraphs". In the proof there is a step that they considered as a trivial step, it seems easy but I am not being able to write or concieve an exact proof for it.</p>
<p>We have a strong digraph <span class="math-container">$D$</span>, that has minimum in degree <span class="math-container">$h$</span> and minimum out degree <span class="math-container">$k$</span>. Let <span class="math-container">$C$</span> be the longest circuit in <span class="math-container">$D$</span> of length <span class="math-container">$c$</span>; then <span class="math-container">$c\ge \max(h,k)+1$</span>.</p>
<p>Why is that true?</p>
<p>I think it is solved by contradiction and we can find a larger circuit.</p>
<p>Suppose to contradiction that <span class="math-container">$c< \max(h,k)+1$</span> and suppose without loss of generality that <span class="math-container">$\max(h,k)=k$</span>, then <span class="math-container">$c \le k$</span> which means that every vertex on the circuit has at least <span class="math-container">$1$</span> out neighbor outside <span class="math-container">$C$</span>.</p>
<p>But how can we continue to find a larger circuit?</p>
| Mike | 544,150 | <p>Let us assume here that <span class="math-container">$k \ge h$</span> as you did and let <span class="math-container">$P$</span> be a <em>path</em> of length <span class="math-container">$c \ge \max(h,k)+1 = k+1$</span>. Your line of reasoning shows that there is indeed such a path <span class="math-container">$P$</span>. We use this to show how to get a <em>circuit</em> <span class="math-container">$C$</span> with at least <span class="math-container">$k+1$</span> vertices as follows:</p>
<p>Now let us write (*) <span class="math-container">$P=y_1y_2\ldots y_c$</span> where <span class="math-container">$y_1$</span> is the first vertex in <span class="math-container">$P$</span> and where <span class="math-container">$y_c$</span> is the last vertex in <span class="math-container">$P$</span>.</p>
<ol>
<li><p>If <span class="math-container">$y$</span> has a neighbor <span class="math-container">$y'=y_{c+1}$</span> outside of <span class="math-container">$P$</span> then add the arc <span class="math-container">$y_cy' =y_cy_{c+1}$</span> to <span class="math-container">$P$</span>, now <span class="math-container">$P$</span>'s length is <span class="math-container">$c+1$</span> instead of <span class="math-container">$c$</span> as it was before.</p>
</li>
<li><p>If all of <span class="math-container">$y$</span>'s neighbours are in <span class="math-container">$P$</span>, then let <span class="math-container">$j$</span> be the smallest integer
such that <span class="math-container">$y_cy_j$</span> is an arc in <span class="math-container">$G$</span>, where <span class="math-container">$y_j$</span> is as in (*). Then <span class="math-container">$j \le c-k$</span> [why is this] so <span class="math-container">$C= y_jy_{j+1} \ldots y_cy_j$</span>
is a circuit of length at least <span class="math-container">$k+1$</span>.</p>
</li>
</ol>
<p>Now note that 1. can be true for at most <span class="math-container">$n-k$</span> times as the number of vertices in any path <span class="math-container">$P'$</span> in <span class="math-container">$G$</span> is of course upper-bounded by the number of vertices in the digraph <span class="math-container">$G$</span>. So you will indeed obtain a circuit of length at least <span class="math-container">$k+1$</span> after adding at most <span class="math-container">$n-k-1$</span> vertices to <span class="math-container">$P$</span> as in 1. above.</p>
<p>I trust that you can handle the case where <span class="math-container">$h > k$</span>.</p>
|
90,112 | <p>When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following result which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.</p>
<p>The result (slightly simplified) is</p>
<p>If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.</p>
<p>Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.</p>
<p>I'm sure this should be easy and I'm just missing an obvious trick, but oh well.</p>
| Bill Johnson | 2,554 | <p>Another book reference:</p>
<p>This is proved as Lemma 9.2 in ``Functional Analysis and Infinite-dimensional Geometry" by Fabian, Habala, Hajek, Montesinos Santalucia, Pelant, and Zizler.</p>
|
1,775,649 | <p>True or false and explain why?: a matrix with characteristic polynomial $\lambda^3 -3\lambda^2+2\lambda$ must be diagonalizable.</p>
<p>First I found the lambda's that make this zero (eigenvalues) and got $0, 1, 2$ but I don't know if having $0$ as an eigenvalue means that the matrix is not diagonalizable? I know that a matrix has $0$ as an eigenvalue if it is not invertible, but I don't know if a matrix needs to be invertible to be diagonalizable? Also if a matrix has complex eigenvalues does that also mean it cannot be diagonalizable?</p>
| marwalix | 441 | <p>The characteristic polynomial splits as follows $P_M(\lambda)=\lambda(\lambda-1)(\lambda-2)$. The matrix is $3\times 3$ (the degree of the characteristic polynomial) and has three distinct eigenvalues therefore it is diagonalisable.</p>
|
715,809 | <p>I have occasionally come across Leibniz's Law (left to right, ie the indiscernibility of identicals) written with schematic letters in the consequent, and occasionally with a bound predicate-variable taking the place of the schematic letters. What is the relevant difference between these formulations? If there is ongoing debate, any leads on relevant literature would be most helpful.</p>
| Peter Smith | 35,151 | <p>Suppose, to fix ideas, we are doing arithmetic, i.e. our first-order variables run over numbers, and suppose our language is countable.</p>
<p>The schematic version of Leibniz's Law, applied in this context, in effect tells us that if the numbers $m$ and $n$ are equal, then a first-order predicate $\varphi$ is true of $m$ if and only it is true of $n$.</p>
<p>The second-order version of Leibniz's Law tells us that if the numbers $m$ and $n$ are equal, then for any property at all $m$ has it if and only it is $n$ has it.</p>
<p>The second-order principle is stronger because it applies to all uncountably many properties of properties [or if you like to think extensionally, all uncountably many sets of numbers] not just to the countably many properties that can be expressed by predicates in our language. Moreover, the second-order version seems to be what we want to say when we claim that that identicals are indiscernible: i.e. if $m$ and $n$ are the same they share <em>all</em> properties.</p>
<p>Or at least the second-order version seems to be what we want, if we can indeed make sense of the idea of there being arbitrary properties of the numbers [arbitrary sets of numbers] which exist even when we have no way of specifying them. There are general issues about whether we can do this, according to various brands of constructivist or predicativist. Classical realists will say that there isn't a problem.</p>
<p>For an excellent treatment of issues surrounding the use of second-order logic (from a defender), and a discussion of the added richness we get from second-order formulations of various principles, see Stewart Shapiro's classic book <em>Foundations without Foundationalism: The Case for Second-Order Logic</em>.</p>
|
1,811,443 | <p>Let $(a_n)$ be a sequence of rational numbers, where <strong>all rational numbers are terms</strong>. (<em>i.e. enumeration of rational numbers</em>)</p>
<p>Then, is there any convergent sub-sequence of $(a_n)$?</p>
| Jason | 195,308 | <p>Even better - for every real $x$, there is a subsequence of $(a_n)$ converging to $x$.</p>
<p>Indeed, start by letting $a_{n_1}$ be any element of the sequence in $(x-1,x+1)$. Then, given $a_{n_1},\ldots,a_{n_k}$, since there are infinitely many rationals in each interval, there must exist $n_{k+1}>n_k$ such that $a_{n_{k+1}}\in(x-\frac1{k+1},x+\frac1{k+1})$. Inductively this creates a subsequence $(a_{n_k})$ which converges to $x$.</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| Fawad | 369,983 | <p>Hint:</p>
<p>$x-1={\left (\sqrt x \right )}^2 -1^2$</p>
<p>$x-1=(\sqrt x +1)(\sqrt x-1)$</p>
|
2,497,875 | <p>Define $\sigma: [0,1]\rightarrow [a,b]$ by $\sigma(t)=a+t(b-a)$ for $0\leq t \leq 1$. </p>
<p>Define a transformation $T_\sigma:C[a,b]\rightarrow C[0,1]$ by $(T_\sigma(f))(t)=f(\sigma(t))$ </p>
<p>Prove that $T_\sigma$ satisfies the following:</p>
<p>a) $T_\sigma(f+g)=T_\sigma(f)+T_\sigma(g)$</p>
<p>b) $T_\sigma(fg)=T_\sigma(f)*T_\sigma(g)$</p>
<p>c) $T_\sigma(f)\leq T_\sigma(g)$ iff $f\leq g$</p>
<p>d) $||T_\sigma(f)||=||f||$</p>
<p>e) $T_\sigma$ is both 1-1 and onto, moreover, $(T_\sigma)^{-1}=T_{\sigma^{-1}}$</p>
<p>This is a problem from "A Short Course on Approximation Theory."
It looks like an easy problem and $\sigma$ is clearly an affine transformation, but I cannot figure out how to work with the transformation within another function. The results from this are used to extend the Weierstrass theorem from $C[0,1]$ to $C[a,b]$</p>
| Gono | 384,471 | <p>A different solution without needing $\frac{sin(x)}{x}$:
$$\frac{\sin(1-\sqrt{x})}{x-1} = \frac{\sin(1-\sqrt{x}) - \sin(1-\sqrt{1})}{x-1} = -\frac{\cos(1-\sqrt{\xi})}{2\sqrt{\xi}}$$ for a $\xi \in (1,x)$ by the mean value theorem.
And because $\xi \to 1$ if $x\to 1$ we get:</p>
<p>$$\lim_{x\to 1}\frac{\sin(1-\sqrt{x})}{x-1} = \lim_{\xi\to 1} -\frac{\cos(1-\sqrt{\xi})}{2\sqrt{\xi}}$$</p>
|
13,889 | <p><strong>Question:</strong> Are there intuitive ways to introduce cohomology? Pretend you're talking to a high school student; how could we use pictures and easy (even trivial!) examples to illustrate cohomology?</p>
<p><strong>Why do I care:</strong> For a number of math kids I know, doing algebraic topology is fine until we get to homology, and then it begins to get a bit hazy: why does all this quotienting out work, why do we make spaces up from other spaces, how do we define attaching maps, etc, etc. I try to help my peers do basic homological calculations through a sequence of easy examples (much like the ones Hatcher begins with: taking a circle and showing how "filling it in" with a disk will make the "hole" disappear --- ) and then begin talking about what kinds of axioms would be nice to have in a theory like this. I have attempted to begin studying co-homology through "From Calculus to Cohomology" and Hatcher's text, but I cannot see the "picture" or imagine easy examples of cohomology to start with. </p>
| Jjm | 196,095 | <p>There is a nice and, to my opinion, more natural way to motivate cohomology - a geometric one, rather than an analytical one. Please read carefully the following question and answer in math.stackexchange:</p>
<p><a href="https://math.stackexchange.com/questions/1112419/intuitive-approach-to-de-rham-cohomology">Intuitive Approach to de Rham Cohomology</a></p>
|
2,430,690 | <p>(<a href="https://i.stack.imgur.com/NlAPf.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/NlAPf.jpg</a>)</p>
<p>This is a question in my school's exercise book
I couldn't think of any equation to be formed to solve it
Please teach me</p>
| George Coote | 445,167 | <p>Remember, </p>
<p>$$\sin^2\theta + \cos^2\theta \equiv 1$$ </p>
<p>For all $\theta$. Given this, and that $\sin\theta = k$, how can we find $\cos\theta$ in terms of $k$? How can we then find $\tan\theta$? Can you think of where to go from there seeing that $w = 360 - \theta$?</p>
|
2,430,690 | <p>(<a href="https://i.stack.imgur.com/NlAPf.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/NlAPf.jpg</a>)</p>
<p>This is a question in my school's exercise book
I couldn't think of any equation to be formed to solve it
Please teach me</p>
| gen-ℤ ready to perish | 347,062 | <p>First, the diagram makes it clear that $w^\circ = \theta$.</p>
<p>Identically, it is true that $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. From the Pythagorean theorem, we know that $\cos^2\theta = 1 - \sin^2\theta$.</p>
<p>$$\begin{align}
\tan w^\circ &= \tan\theta \\
&= \frac{\sin\theta}{\cos\theta} \\
&= \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} \\
&= \frac{k}{\sqrt{1 - k^2}}
\end{align}$$</p>
<p>However, notice that $w^\circ$ is in quadrant III, where $\tan$ is negative. Since we did some work with squaring and principle roots, we need to correct our signs:</p>
<p>$$\bbox[yellow,5px]{\tan w^\circ = -\frac{k}{\sqrt{1 - k^2}}}$$</p>
<p>You should find this <a href="https://en.wikipedia.org/wiki/List_of_trigonometric_identities" rel="nofollow noreferrer">list of trigonometric identities</a> <em>very</em> useful in the future.</p>
|
4,444,504 | <p>We have measure theory in this semester.I found the statement of Lusin's theorem on the internet to be:</p>
<blockquote>
<p>Let <span class="math-container">$f:\mathbb{R\to R}$</span> be a Lebesgue measurable function.Then for each <span class="math-container">$\epsilon>0$</span> there exists a closed set <span class="math-container">$F
_\epsilon\subset \mathbb R$</span> such that <span class="math-container">$f|_{F_{\epsilon}}$</span> is continuous and <span class="math-container">$|\mathbb R\setminus F_{\epsilon}|<\epsilon$</span>.</p>
</blockquote>
<p>But in another book I saw the following version:</p>
<blockquote>
<p>Let <span class="math-container">$f:\mathbb{R\to R}$</span> be a Lebesgue measurable function .Then for each <span class="math-container">$\epsilon>0$</span> there exists a compact set <span class="math-container">$K_{\epsilon}\subset \mathbb R$</span> such that <span class="math-container">$f|_{K_{\epsilon}}$</span> is continuous and <span class="math-container">$|\mathbb R-K_\epsilon|<\epsilon$</span>.</p>
</blockquote>
<p>Things turned out getting worse when our instructor told us the following version of Lusin's theorem:</p>
<blockquote>
<p>Any continuous function on <span class="math-container">$\mathbb R$</span> with compact support is Lebesgue integrable.</p>
</blockquote>
<p>Now I am really confused.I cannot understand why these all are equivalent.I also tried to prove these results but couldn't.In the book by Sheldon Axler I found a proof but that proof is given for Borel measurable functions not Lebesgue measurable functions.How can I prove these results and how to show they are indeed same?</p>
| coudy | 716,791 | <p>The second statement is wrong. There is no compact subset <span class="math-container">$K$</span> in <span class="math-container">${\bf R}$</span> such that <span class="math-container">$|{\bf R} \setminus K| < \varepsilon$</span>. A compact subset of the real line is bounded and his complement is of infinite Lebesgue measure.</p>
<p>The first statement is correct. The last statement is correct but is unrelated to Lusin's theorem. These three statements are definitely not the same and they can't be deduced from each others.</p>
|
118,742 | <p>The PDF for $Y$ is
$$f_Y(y) = \begin{cases}
0 & |y|> 1 \\
1-|y| & |y|\leq 1
\end{cases}$$</p>
<p>How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get
$$F_Y(y)=\begin{cases}
1/2 -y/2-y^2/2 & [-1,0] \\
1/2-y/2+y^2/2 & [0,1]
\end{cases}
$$
by using the fact that $F_Y(y)=\int _{-\infty}^{y}{f_Y(y)}\,dy$, however my text claims the answer is
$$F_Y(y)=\begin{cases}
1/2 +y+y^2/2 & [-1,0] \\
1/2+y-y^2/2 & [0,1]
\end{cases}
$$
I am struggling with pdf and cdfs, so I asssume I did something wrong other than the simple integration. Who's correct? Me or the Text!?? $:)$ </p>
| anon | 11,763 | <p>First work with $y\le 0$ to obtain</p>
<p>$$F_Y(y)=\int_{-1}^y 1-|u|du=\int_{-1}^y 1+u du=y-(-1)+\frac{y^2-(-1)^2}{2}=\frac{1}{2}+y+\frac{1}{2}y^2 $$</p>
<p>Now work with $1\ge y\ge0$ by splitting (using the fundamental theorem of calculus)</p>
<p>$$F_Y(y)=\int_{-1}^y f_Y(u)du=F_Y(0)+\int_0^y 1-u du $$</p>
<p>Now figure out what $F_Y$ must be for $y\le-1$ and $y\ge+1$...</p>
|
4,028,534 | <p>I have three points with coordinates: <span class="math-container">$A (5,-1,0),B(2,4,10)$</span>, and <span class="math-container">$C(6,-1,4)$</span>.</p>
<p>I have the following vectors <span class="math-container">$\overrightarrow {CA} = (-1, 0, -4)$</span> and <span class="math-container">$\overrightarrow{CB} = (-4, 5, 6)$</span>.</p>
<p>To find the area of the triangle I used the dot product between these vectors to get the angle and then applied the formula <span class="math-container">$A=0.5ab\sin{C}$</span> to find the area of the triangle which gave me <span class="math-container">$15.07(2dp)$</span>.</p>
<p>However in the given solutions the answer is given as <span class="math-container">$(3*\sqrt(102))/2$</span></p>
<p>I think they have used the trig identity <span class="math-container">$\cos^2(\theta) + \sin^2(\theta) = 1$</span> to find the value of <span class="math-container">$\sin(\theta)$</span> rather than <span class="math-container">$\arccos(\theta)$</span> to find the angle ACB. However I don't understand why there would be such a discrepancy between the two answers; one using <span class="math-container">$\arccos$</span> and the other using the trig identity.</p>
| Quanto | 686,284 | <p>Note that <span class="math-container">$a=\sqrt{17}$</span>, <span class="math-container">$b=\sqrt{77}$</span> and <span class="math-container">$\cos C= -\frac{20}{ab}$</span>, which yields the area</p>
<p><span class="math-container">$$A= \frac12 ab \sqrt{1-\cos^2C}=\frac12 \sqrt{17\cdot77-400}=\frac{3\sqrt{101}}2$$</span></p>
|
2,352,527 | <p><em>This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.</em></p>
<p>What is the value of </p>
<p>$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$ </p>
<p><em>Do I just use</em> </p>
<p>$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$</p>
| tattwamasi amrutam | 90,328 | <p>Hint: Take $$g(z)=\frac{f(z)}{z}, z \ne 0$$
and $$g(0)=f'(0)$$.</p>
<p>Show that $g$ is an entire function which is bounded. </p>
|
2,352,527 | <p><em>This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.</em></p>
<p>What is the value of </p>
<p>$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$ </p>
<p><em>Do I just use</em> </p>
<p>$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$</p>
| zhw. | 228,045 | <p>The integral approach you tried is indeed problematic, but let me show you one that works. The key is to look at the integrals of $|f|^2.$ From the given growth estimate we have</p>
<p>$$\tag 1 \int_{-\pi} ^\pi |f(re^{it})|^2\, dt \le 2\pi M^2r^2\,\,\text { for } r\ge 0.$$</p>
<p>On the other hand, we have $f(z) = \sum_{n=0}^{\infty}a_n z^n$ just as you wrote. Note then that $f(re^{it}) = \sum_{n=0}^{\infty}a_n r^ne^{int}.$ So by Parseval's identity,</p>
<p>$$\tag 2 \int_{-\pi} ^\pi |f(re^{it})|^2\,dt = 2\pi \sum_{n=0}^{\infty}|a_n|^2 r^{2n}.$$</p>
<p>Now the right side of $(2)$ grows faster than $r^2$ if $a_n\ne 0$ for some $n>1.$ By $(1),$ that can't happen. Therefore $a_n = 0$ for $n>1$ and we have $f(z) = a_0 +a_1z.$ Because $f(0)=0,a_0=0.$ We thus conclude $f(z) = a_1z$ for all $z.$</p>
|
505,848 | <p>It is known that <a href="http://mathworld.wolfram.com/SomosSequence.html" rel="nofollow noreferrer">the $k$-Somos sequences</a> always give integers for $2\le k\le 7$.</p>
<p>For example, the $6$-Somos sequence is defined as the following : </p>
<p>$$a_{n+6}=\frac{a_{n+5}\cdot a_{n+1}+a_{n+4}\cdot a_{n+2}+{a_{n+3}}^2}{a_n}\ \ (n\ge0)$$</p>
<p>where $a_0=a_1=a_2=a_3=a_4=a_5=1$.</p>
<p>Then, here is my question.</p>
<p><strong>Question</strong> : If a sequence $\{b_n\}$ is difined as
$$b_{n+6}=\frac{b_{n+5}\cdot b_{n+1}+b_{n+4}\cdot b_{n+2}+{b_{n+3}}^2}{b_n}\ \ (n\ge0)$$$$b_0=b_1=b_2=b_3=1, b_4=b_5=2,$$
then is $b_n$ an integer for any $n$?</p>
<p>We can see
$$b_6=5,b_7=11,b_8=25,b_9=97,b_{10}=220,b_{11}=1396,b_{12}=6053,b_{13}=30467$$
$$b_{14}=249431,b_{15}=1381913,b_{16}=19850884,b_{17}=160799404$$
$$b_{18}=1942868797,b_{19}=36133524445, \cdots.$$</p>
<p><strong>Motivation</strong> : I've been interested in seeing what happens when we change the first few terms. It seems true, but I can neither find any counterexample nor prove that the sequence always gives an integer. As far as I know, it seems that this question cannot be solved in the way which proved that the $6$-Somos sequence always gives an integer.</p>
<p><em>Update</em> : I crossposted to <a href="https://mathoverflow.net/questions/143609/does-this-sequence-always-give-an-integer">MO</a>.</p>
| njguliyev | 90,209 | <p>Hints:
1. For any $\varepsilon>0$ let $\delta := \frac{a}{N}$ and apply (b).<br>
2. It is obvious that $f(0) = 0$. Since $f$ has a limit at $0$, we can put $\varepsilon = 1$ and write: $\exists \delta > 0,\ \forall x,\ |x|<\delta\colon\ |f(x)|<1$.</p>
|
505,848 | <p>It is known that <a href="http://mathworld.wolfram.com/SomosSequence.html" rel="nofollow noreferrer">the $k$-Somos sequences</a> always give integers for $2\le k\le 7$.</p>
<p>For example, the $6$-Somos sequence is defined as the following : </p>
<p>$$a_{n+6}=\frac{a_{n+5}\cdot a_{n+1}+a_{n+4}\cdot a_{n+2}+{a_{n+3}}^2}{a_n}\ \ (n\ge0)$$</p>
<p>where $a_0=a_1=a_2=a_3=a_4=a_5=1$.</p>
<p>Then, here is my question.</p>
<p><strong>Question</strong> : If a sequence $\{b_n\}$ is difined as
$$b_{n+6}=\frac{b_{n+5}\cdot b_{n+1}+b_{n+4}\cdot b_{n+2}+{b_{n+3}}^2}{b_n}\ \ (n\ge0)$$$$b_0=b_1=b_2=b_3=1, b_4=b_5=2,$$
then is $b_n$ an integer for any $n$?</p>
<p>We can see
$$b_6=5,b_7=11,b_8=25,b_9=97,b_{10}=220,b_{11}=1396,b_{12}=6053,b_{13}=30467$$
$$b_{14}=249431,b_{15}=1381913,b_{16}=19850884,b_{17}=160799404$$
$$b_{18}=1942868797,b_{19}=36133524445, \cdots.$$</p>
<p><strong>Motivation</strong> : I've been interested in seeing what happens when we change the first few terms. It seems true, but I can neither find any counterexample nor prove that the sequence always gives an integer. As far as I know, it seems that this question cannot be solved in the way which proved that the $6$-Somos sequence always gives an integer.</p>
<p><em>Update</em> : I crossposted to <a href="https://mathoverflow.net/questions/143609/does-this-sequence-always-give-an-integer">MO</a>.</p>
| Community | -1 | <p>Your Question is completely answered in the following book (Theorem 1.2, page 3). Also you can find some further related topics.</p>
<p>Pl. Kannappan, Functional Equations and Inequalities with Applications, Springer, 2009.</p>
<p>Theorem 1.2. Suppose $A : R\rightarrow R$ satisfies $A(x+y)=A(x)+A(y)$ with $c = A(1) > 0$. Then the following conditions are equivalent:</p>
<p>(i) $A$ is continuous at a point $x_{0}$.</p>
<p>(ii) $A$ is monotonically increasing.</p>
<p>(iii) $A$ is nonnegative for nonnegative $x$.</p>
<p>(iv) $A$ is bounded above on a finite interval.</p>
<p>(v) $A$ is bounded below on a finite interval.</p>
<p>(vi) $A$ is bounded above (below) on a bounded set of positive Lebesgue measure.</p>
<p>(vii) $A$ is bounded on a bounded set of positive measure (Lebesgue).</p>
<p>(viii) $A$ is bounded on a finite interval.</p>
<p>(ix) $A(x) = cx$.</p>
<p>(x) $A$ is locally Lebesgue integrable.</p>
<p>(xi) $A$ is differentiable.</p>
<p>(xii) A is Lebesgue measurable.</p>
|
2,333,847 | <p>A function $f(x) = k$ and the domain is $\{-2,-1,\dotsc,3\}$. Would I say
$$x = \{-2,-1,\dotsc,3\}\quad\text{or}\quad x \in \{-2,-1,\dotsc,3\} \ ?$$
Thanks. </p>
| Community | -1 | <p>Use $\in$. And be careful with your commas: $\{-2, -1 \color{red}, \dots \color{red}, 3\}$</p>
<p>In general:</p>
<p>$x \in S$ means "$x$ is an element of $S$" or "$x$ is in $S$." (The $\LaTeX$ code to produce "$\in$" is <code>$\in$</code>.)</p>
<p>$x = s$ means "$x$ is $S$" or "$x$ is equal to $S$." In this case you would be saying $x$ <strong>is</strong> the domain, but this is not what you want to say.</p>
|
174,528 | <p>I am editing my original question, as I have figured out a method of doing what I want.</p>
<p>Now my question is if there is a more elegant, efficient way to do the following:</p>
<pre><code>Options[f] = {
Energy -> energy,
Temperature -> Func[Energy*Frequency, {Energy, Frequency}],
Frequency -> freq
};
f[OptionsPattern[]] := Module[{energy, temperature, frequency},
energy = OptionValue[Energy];
frequency = OptionValue[Frequency];
temperature = If[
Head[OptionValue[Temperature]] === Func,
OptionValue[Temperature][[1]] /. Map[# -> OptionValue[#] &,
Flatten[{OptionValue[Temperature][[2]]}]],
OptionValue[Temperature]
];
{
Energy -> energy,
Frequency -> frequency,
Temperature -> temperature
}]
</code></pre>
<p>Please note that this has the needed generalization to change what options are used as arguments to the passed function or if desired a value (or variable) can be directly inserted as an option.</p>
<p>For this application I have >70 Options (parameters) and I would like to be able to insert (if desired) a function with arguments based on any of the other options, or directly insert a value (if a function "Func" isn't desired.)</p>
<p>Is this more clear?</p>
<p>Thanks again!</p>
| Henrik Schumacher | 38,178 | <p>Apparently, the only problem was that you scoped <code>f</code> by <code>Module</code> so that <code>OptionValue</code> had a hard time to find the default values for <code>f</code>. (<code>Option</code> stores <code>Option[f]</code> within <code>f</code>, not as <code>DownValue</code> of <code>Option</code>.) </p>
<p>The following should work without any problems.</p>
<pre><code>Options[f] = {
Energy -> 25,
Frequency -> 60,
TemperatureFunction ->
Function[{Energy, Frequency}, Energy*Frequency^2]
};
f[x_, OptionsPattern[]] := Module[{en, freq, tempFunction},
en = OptionValue[Energy];
freq = OptionValue[Frequency];
tempFunction = OptionValue[TemperatureFunction];
(* **Missing Necessary Code***)
x*tempFunction[en, freq]
]
</code></pre>
|
660,315 | <p>Let $A,B,C$ be sets. Identify a condition such that $A \cap C = B \cap C$ together with your condition implies $A=B$. Prove this implication. Show that your condition is necessary by finding an example where $A \cap C = B \cap C$, but $ A \neq B$</p>
<p>Edit: I've read the wrong proposition/definition. UGH! The question was probably about having sets being equal, not empty. </p>
<p>Now, suppose that $ A \neq B$... that would mean that $A$ isn't equal to $B$. So they are different sets, but $A \cap C = B \cap C$ are equal sets. </p>
<hr>
<p>I'm lost on this. I need to find a condition, but where do I even start? These are my thoughts about the question so far. </p>
<p>We need to find a condition for $A \cap C = B \cap C$. </p>
<p>Proposition 3.1.12 states that if $A$ and $B$ are both empty sets, then $A =B$.</p>
<p>So $A \cap C$ and $B \cap C$ are both empty sets which is why $A \cap C = B \cap C$.</p>
<p>It seems that there are empty sets everywhere because proposition 3.1.12 claims that if $A$ and $B$ are empty set, then $A =B$. It's like there aren't any elements at all. There's nothing. </p>
<p>We need to find a condition that demonstrates that $A \cap C = B \cap C$ which are empty sets, but $A \neq B$ means that there are elements in $A$ and $B$. </p>
<p>How is this even possible? </p>
<p>$A \cap C$ by definition 3.2.1 is $[x: x: \in A \land x \in C]$</p>
<p>x belongs in A and x belongs in C. </p>
<p>$B \cap C$ by definition 3.2.1 is $[x: x: \in B \land x \in C]$</p>
<p>x belongs in B and x belongs in C</p>
<p>The only way I could think of is a contradiction to this... but that would mean that $A = B$ ... there are no elements in A and B, but $A \cap C \neq B \cap C$ means that there are elements. There may not be elements in A and B, but there are elements in C. </p>
| Asaf Karagila | 622 | <p><strong>HINT:</strong> What happens if both $A=A\cap C$ and $B=B\cap C$? (Note that this generalizes the case of both $A$ and $B$ being empty.)</p>
|
2,747,509 | <p>How would you show that if <span class="math-container">$d\mid n$</span> then <span class="math-container">$x^d-1\mid x^n-1$</span> ?</p>
<p>My attempt :</p>
<blockquote>
<p><span class="math-container">$dq=n$</span> for some <span class="math-container">$q$</span>. <span class="math-container">$$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$</span> in fact, <span class="math-container">$$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$</span></p>
<p>By multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$(x-1)$</span> we get that <span class="math-container">$1-x^d\mid 1-x^n$</span> which is the final result</p>
</blockquote>
<p>Is this an ok proof?</p>
| fleablood | 280,126 | <p>Notice that for any $x$ and and natural $n$ that $$(x-1)(x^{n-1} + ..... + x + 1) = (x^n + x^{n-1} +....... +x) - (x^{n-1} + x^{n-2} +.... +1) = x^n -1$$ so that $x-1|x^n - 1$ always.</p>
<p><strong>Lemma:</strong> $x-1|x^n-1$ for natural $n$.</p>
<p>Now $d|n$ so let $m = \frac nd$ and let $y= x^d$.</p>
<p>Then $y-1|y^m -1$.</p>
<p>But $y-1 = x^d -1$ and $y^m -1 = x^n - 1$.</p>
<p>In particular: $(x^d -1)(x^{n-d} + x^{n-2d} + ..... + x^d + 1) = x^n - 1$</p>
|
2,747,509 | <p>How would you show that if <span class="math-container">$d\mid n$</span> then <span class="math-container">$x^d-1\mid x^n-1$</span> ?</p>
<p>My attempt :</p>
<blockquote>
<p><span class="math-container">$dq=n$</span> for some <span class="math-container">$q$</span>. <span class="math-container">$$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$</span> in fact, <span class="math-container">$$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$</span></p>
<p>By multiplying both sides of <span class="math-container">$(1)$</span> by <span class="math-container">$(x-1)$</span> we get that <span class="math-container">$1-x^d\mid 1-x^n$</span> which is the final result</p>
</blockquote>
<p>Is this an ok proof?</p>
| Anastassis Kapetanakis | 342,024 | <p>You have:
$$d\mid n \Rightarrow n=ad$$
Then:
$$x^n-1=x^{ad}-1=(x^d)^a-1$$
Setting $x^d=y$ (just for simplifying the process) we have:
$$y^a-1=(y-1)(y^{a-1}+\dots+y+1)=(x^d-1)((x^d)^{a-1}+\dots + x^d+1)$$
In other words we showed that:
$$x^n-1= (x^d-1)((x^d)^{a-1}+\dots + x^d+1) $$
Which obviously implies that:
$$x^d-1 \mid x^n-1$$</p>
|
4,144,083 | <p>I am currently doing an internship in a research laboratory ( I am in my third year of Bachelor ) and I'm really struggling with the things I have to do.</p>
<p>For instance, here's something I'm having trouble with.</p>
<p>Let <span class="math-container">$L$</span> be a finite Galois extension of <span class="math-container">$\mathbb{Q}$</span>, <span class="math-container">$O_L$</span> its ring of integers, and <span class="math-container">$I$</span> an ideal of <span class="math-container">$O_L$</span>. Let <span class="math-container">$K$</span> be the decomposition field of <span class="math-container">$I$</span>. Let <span class="math-container">$R=I\cap O_K$</span>. Suppose <span class="math-container">$n=[L : \mathbb{Q} ]$</span> and <span class="math-container">$g=[ K : \mathbb{Q} ]$</span>. Suppose also that we have a basis <span class="math-container">$(b_i)_{1 \leq i \leq n/g}$</span> of <span class="math-container">$O_L$</span> over <span class="math-container">$O_K$</span>.</p>
<p>In order to find an isomorphism between <span class="math-container">$(O_K/R)^{n/g}$</span> and <span class="math-container">$O_L/I$</span>, I wanted to proceed like that :</p>
<p><span class="math-container">$(O_K)^{\frac{n}{g}} \overset{f_1}{\longrightarrow} O_L \overset{f_2}{\longrightarrow} O_L /I$</span></p>
<p><span class="math-container">$f_1$</span> being : <span class="math-container">$ (x_1,\cdots , x_{n/g}) \longrightarrow \sum\limits_{i=1}^{n/g} x_i b_i$</span></p>
<p>and <span class="math-container">$f_2$</span> being the canonical surjection.</p>
<p>In order to find my isomorphism, I need to prove that the kernel of the composition of <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> is <span class="math-container">$R^{\frac{n}{g}}$</span>.</p>
<p>The Kernel of <span class="math-container">$f_2$</span> being <span class="math-container">$I$</span>, what's left to prove is that :</p>
<p><span class="math-container">$R^{\frac{n}{g}}=\{ x\in (O_K)^{\frac{n}{g}} ~:~ f_1(x)\in I \}$</span></p>
<p>The <span class="math-container">$\subset$</span> of the equality is easy, but I can't prove the <span class="math-container">$\supset$</span>.</p>
<p>Please forgive my mistakes, I'm still learning, and English isn't my first language.</p>
<p>Thank you for your help !</p>
| reuns | 276,986 | <p>If <span class="math-container">$K$</span> is the subfield of <span class="math-container">$L$</span> fixed by <span class="math-container">$\{ \sigma\in Gal(L/\Bbb{Q}), \sigma(I)\subset I\}$</span>
then try with <span class="math-container">$O_L=\Bbb{Z}[i], I=(1+i)^3,O_K=\Bbb{Z},I\cap O_K=(4)$</span>,</p>
<p>As a group <span class="math-container">$O_L/I\cong C_4\times C_2$</span></p>
<p>(<span class="math-container">$C_4$</span> is the subgroup generated by <span class="math-container">$1$</span> and <span class="math-container">$C_2$</span> the subgroup generated by <span class="math-container">$1+i$</span>)</p>
<p>If <span class="math-container">$I=P^d$</span> and <span class="math-container">$P$</span> is <strong>unramified</strong> then yes your statement holds, this follows from <span class="math-container">$|O_L/I|=|O_K/(P^d\cap O_K)|^{f(P)}$</span> where <span class="math-container">$n=efg,e=1$</span> so the kernel of your surjective map <span class="math-container">$O_K^{n/g}\to O_L/I$</span> can't be larger than <span class="math-container">$(O_K\cap I)^{n/g}$</span>.</p>
<p>When <span class="math-container">$I$</span> is a product of distinct unramified primes powers it should often fail, say when the decomposition group is not the same for each prime.</p>
|
657,047 | <p>So I have $a^n = b$. When I know $a$ and $b$, how can I find $n$?</p>
<p>Thanks in advance!</p>
| lsp | 64,509 | <p>Apply Logarithm on both sides:</p>
<p>$$ \log a^n = \log b$$
$$n \log a = \log b$$
$$n = \frac{\log b}{\log a}$$</p>
<p>If you are a starter in Logarithms, you can refer <a href="http://en.wikipedia.org/wiki/Logarithm" rel="nofollow">here</a>.</p>
|
1,563,044 | <blockquote>
<p>Prove by transfinite induction that there is a function $f:\mathbb R \to\mathbb R$ such that $|f^{-1}(r) \bigcap (a,b)| = 2^{\omega}$ for every $a, b, r \in\mathbb R$ and $a < b$.</p>
</blockquote>
<p>I have:</p>
<p>Let $F = \{(a,b) \times \{r\}\, |\, a, b, r \in\mathbb R\text{ and }a < b\}$ and $\{F_\xi\, |\, \xi < 2^{\omega}\}$ be an enumeration of $F$.</p>
<p>Also, I have that $(x_\xi, y_\xi) \in F_\xi \setminus \bigcup_{\zeta < \xi} (\{x_\zeta\} \times\mathbb R)$ where $f(x_\xi) = y_\xi$ for all $\xi < 2^{\omega}$.</p>
<p>I'm not sure where to go from here. </p>
| Tormod Haugland | 240,556 | <p>Sadly I don't have enough reputation to comment, so I guess I will post an answer instead. I hope its not inappropriate.</p>
<p>As indicated by the comments, the answer varies on your definition of $e^x$. In fact one definition is implicitly by:</p>
<p>$$ \lim_{h \to 0} \frac{e^h - 1}{h} = 1$$</p>
<p>Defining $e^x$ this way, you are finished. If you define it as </p>
<p>$$ e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$$</p>
<p>you are left with a bit more work.</p>
<p>Regardless, you will probably find all the answers you're looking for in the following thread: <a href="https://math.stackexchange.com/questions/199447/proof-of-derivative-of-ex-is-ex-without-using-chain-rule">Proof of derivative of $e^x$ is $e^x$ without using chain rule</a></p>
|
1,563,044 | <blockquote>
<p>Prove by transfinite induction that there is a function $f:\mathbb R \to\mathbb R$ such that $|f^{-1}(r) \bigcap (a,b)| = 2^{\omega}$ for every $a, b, r \in\mathbb R$ and $a < b$.</p>
</blockquote>
<p>I have:</p>
<p>Let $F = \{(a,b) \times \{r\}\, |\, a, b, r \in\mathbb R\text{ and }a < b\}$ and $\{F_\xi\, |\, \xi < 2^{\omega}\}$ be an enumeration of $F$.</p>
<p>Also, I have that $(x_\xi, y_\xi) \in F_\xi \setminus \bigcup_{\zeta < \xi} (\{x_\zeta\} \times\mathbb R)$ where $f(x_\xi) = y_\xi$ for all $\xi < 2^{\omega}$.</p>
<p>I'm not sure where to go from here. </p>
| Akiva Weinberger | 166,353 | <p>As I've said fairly often in the last few days (for some reason), one of my favorite equations is:
$$e^x\ge x+1$$
The reason, partly, is that it uniquely defines $e$ without calculus. Hint for a proof: use <a href="https://en.m.wikipedia.org/wiki/Bernoulli%27s_inequality" rel="noreferrer">this</a>. (By the way, do equations need equals signs? Or is it <em>equalities</em> that need equals signs?)</p>
<p>Now, replacing $x$ by $-x$, we get $e^{-x}\ge1-x$, so:
$$e^x\le\frac1{1-x}$$
(The inequality gets reverse for $x>1$, as the right-hand side is negative there. But we only care about when $x$ is near zero.)</p>
<p>Thus:
\begin{align}
x+1\le{}&e^x\le\frac1{1-x}\\
x\le{}&e^x-1\le\frac x{1-x}\\
1\le^*{}&\frac{e^x-1}x\le^*\frac1{1-x}
\end{align}
*Since we just divided by $x$, the inequalities get reversed if $x$ is negative. It doesn't affect the argument.</p>
<p>Let $x$ tend to zero. By the squeeze theorem:
$$1=\lim_{x\to0}\frac{e^x-1}x$$</p>
|
118,311 | <p>Let $A$ be an $n\times n$ matrix with entries in an arbitrary field $k$. </p>
<p>Is the characteristic polynomial $\det(tI_n-A)$ dependent only on the trace and determinant of $A$?</p>
| Community | -1 | <p>Consider the representation $\tau_3$ (or $\tau_4$) and an eigenvalue $w$ of the the of $x$. Then $w$ is a $7^{th}$ root of unity. Since $x^2$ and $x^4$ are conjugate to $x$, $w^2$ and $w^4$ are also eigenvalues of $x$. All eigenvalues of $x$ cannot be $1$ by column orthonormality. So, $w$, $w^2$ and $w^4$ are distinct.
Filling the rest of the table should not be hard now.</p>
|
336,943 | <p>Given $P(A) = 0.5$ and $P(A \cup (B^c \cap C^c)^c)=0.8$.</p>
<p>Determine $P(A^c \cap (B \cup C))$.</p>
<p>I know from DeMorgans law that: $(B^c \cap C^c)^c = (B \cup C)$.</p>
<p>Edit:</p>
<p>Also how can I "prove" that P(X)=P(Y) if and only if $P(X \cap Y^c) = P(X^c \cap Y)$? </p>
| Cameron Buie | 28,900 | <p>Note that $$\begin{align}P\bigl(A^c\cap(B\cup C)\bigr) &= P(A^c\cap A)+P\bigl(A^c\cap(B\cup C)\bigr)\\ &= P\bigl(A^c\cap (A\cup B\cup C)\bigr)\\ &= P\bigl((A\cup B\cup C)-A\bigr),\end{align}$$ and that if $A$ occurs, then $A\cup B\cup C$ occurs. Thus, $$P\bigl(A^c\cap(B\cup C)\bigr)=P(A\cup B\cup C)-P(A).$$ This is even easier to see with a Venn diagram.</p>
<p>For the second bit, use the fact that $P(X)=P(X\cap Y)+P(X\cap Y^c)$ and $P(Y)=P(X\cap Y)+P(X^c\cap Y).$</p>
|
106,396 | <p>An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)</p>
<p>$$\sin x \approx \frac{{16x\left( {\pi - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi - x} \right)}}$$</p>
<p>for $(0,\pi)$</p>
<p>Here's an image. Cyan for the sine and blue for the approximation.
<img src="https://i.stack.imgur.com/mgmoA.jpg" alt="enter image description here"></p>
<p>¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations? </p>
| Community | -1 | <ul>
<li><p>Here is an <a href="http://www.jstor.org/pss/10.4169/math.mag.84.2.098" rel="noreferrer">article</a> which is written by Shailesh Shirali. Unfotunately my university doesn't have access to it.</p></li>
<li><p>Here is one more <a href="http://www.dli.gov.in/rawdataupload/upload/insa/INSA_1/20005af0_121.pdf" rel="noreferrer">article.</a></p></li>
</ul>
|
3,632,431 | <blockquote>
<p>Consider the function <span class="math-container">$f: \mathbb{N} \to \mathbb{N}$</span> defined by <span class="math-container">$f(x)=\frac{x(x+1)}{2}$</span>. Show that <span class="math-container">$f$</span> is injective but not surjective.</p>
</blockquote>
<p>So I started by assuming that <span class="math-container">$f(a)=f(b)$</span> for some <span class="math-container">$a,b \in \mathbb{N}$</span>.
I want to show that <span class="math-container">$a=b$</span>.</p>
<p><span class="math-container">$$\Rightarrow \frac{a(a+1)}{2} = \frac{b(b+1)}{2}\\
\Rightarrow a(a+1)=b(b+1) \\
\Rightarrow a^2+a=b^2+b$$</span></p>
<p>I don't know where to go from here.</p>
| Shaun | 104,041 | <p>To show <span class="math-container">$f(x)=f(y)$</span> implies <span class="math-container">$x=y$</span>, show the contrapositive, namely, </p>
<p><span class="math-container">$$a\neq b\implies f(a)\neq f(b).$$</span></p>
<p>So suppose WLOG that <span class="math-container">$a<b$</span>. Then <span class="math-container">$a+1<b+1$</span>, so that <span class="math-container">$a(a+1)<a(b+1)<b(b+1)$</span>, but then <span class="math-container">$$\frac{a(a+1)}{2}\neq\frac{b(b+1)}{2}.$$</span></p>
<p>Thus <span class="math-container">$f$</span> is injective from <span class="math-container">$\Bbb N$</span> to <span class="math-container">$\Bbb N$</span>.</p>
<hr>
<p>To show <span class="math-container">$f$</span> is not surjective, consider <span class="math-container">$n$</span> such that <span class="math-container">$f(n)=4$</span>.</p>
|
448,694 | <p>Let $k$ be a field, and $A$ be a finitely generated $k$-algebra. Then does $k$ also have to be a finitely generated field?</p>
<p>Motivation: Let $A$ be generated by $\{a_1,a_2,\dots,a_n\}$, and $k$ be generated by $\{k_1,k_2,k_3,\dots\}$. Then the element $k_1k_2k_3\dots a_1\in A$ is not generated by a finite number of elements. </p>
<p>I suppose the case becomes even clearer when $A$ is a unital $k$-algebra. Then the set of generators of $A$ becomes $\{a_1,a_2,a_3,\dots,a_n,k_1,k_2,\dots\}$</p>
<p>Thanks in advance!</p>
| celtschk | 34,930 | <p>A very simple counter example:</p>
<p>Be $A=\{0\}$ the trivial algebra over $K$: $A$ clearly is finitely generated (it contains just one element!), even if $K$ is not.</p>
|
1,324,062 | <p>Evaluate: </p>
<blockquote>
<p>$$\lim_{h \rightarrow 0} \frac{e^{2h}-1}{h}$$</p>
</blockquote>
<p>Now one way would be using the Maclaurin expansion for $e^{2x}$</p>
<p>However, can we solve it using the definition of the derivative (perhaps considering $f(x)=e^x$)? Many thanks for your help! $$$$
EDIT: I forgot to mention to please not use L'Hopital's Rule. Using it, the problem becomes trivial and loses all chances of getting a beautiful solution.</p>
| wythagoras | 236,048 | <p>Using $e^{2h}-1=(e^h-1)(e^h+1)$, we get </p>
<p>\begin{align*}
\lim_{h \rightarrow 0} \frac{e^{2h}-1}{h} &= 2 \lim_{h \to 0}\frac{
(e^h-1)(e^h+1)}{h} \\&= \lim_{h \to 0}(e^h+1)\lim_{h \to 0}\frac{e^h-1}{h} \\&= 2\lim_{h \to 0}\frac{e^h-1}{h} \\&=2\lim_{h \to 0}\frac{e^{0+h}-e^0}{h} \\&=2f'(0)=2e^0=2\cdot1=2\end{align*}</p>
<p>with $f(x)=e^x$ of course. </p>
|
3,853,980 | <p>Let <span class="math-container">$A$</span> be an <span class="math-container">$n×n $</span> complex matrix such that the three matrices <span class="math-container">$A+I$</span> , <span class="math-container">$A^2+I $</span> , <span class="math-container">$ A^3+I$</span> are all unitary .Prove that<span class="math-container">$ A$</span> is the zero matrix</p>
<p>I try to show that</p>
<p><span class="math-container">$Trace( A^{\theta}A) =0$</span> where <span class="math-container">$A^{\theta }$</span> is conjugate transpose of matrix <span class="math-container">$A$</span></p>
<p><span class="math-container">$\because $</span>
<span class="math-container">$Trace( A^{\theta}A)$</span> = <span class="math-container">$|a_{11}|^2 + |a_{12}|^2....|a_{nn}|^2$</span></p>
<p><span class="math-container">$A+I$</span> is unitary ,so</p>
<p><span class="math-container">$(A+I)^{\theta}(A+I)= I $</span></p>
<p><span class="math-container">$\implies (A^ {\theta}+I)(A +I) =I $</span></p>
<p><span class="math-container">$A^ {\theta}A+ A^ {\theta}+A = 0$</span></p>
<p><span class="math-container">$ Trace( A^{\theta}A)= -( Trace( A^{\theta}+A))$</span>
<span class="math-container">$ \implies Trace( A^{\theta}A)=-2$</span>( sum of real parts of each diagonal entry of A</p>
<p>I don't know how to proceed further
Please help</p>
| Michael Hardy | 11,667 | <p>In fact <span class="math-container">$\Pr(\max\{X_1,X_2\}=x) = 0.$</span> I assume you must have meant that the value of the probability <b>density</b> function of <span class="math-container">$\max\{X_1,X_2\}$</span> at <span class="math-container">$x$</span> is <span class="math-container">$2x.$</span></p>
<p><span class="math-container">$$
\operatorname E(\max\{t,X_2\}) = \operatorname E(\operatorname E(\max\{t,X_2\} \mid \mathbf 1[X_2>t]))
$$</span>
where <span class="math-container">$\mathbf 1[X_2>t] = 1$</span> or <span class="math-container">$0$</span> according as <span class="math-container">$X_2>t$</span> or not.
<span class="math-container">$$
\operatorname E(\max\{t,X_2\} \mid \mathbf 1[X_2>t]) = \begin{cases} t & \text{if } X_2\le t, \\ (1+t)/2 & \text{if } X_2 > t. \end{cases}
$$</span>
And the expected value of that is
<span class="math-container">\begin{align}
& t\cdot\Pr(X_2\le t) + \frac{1+t} 2\cdot\Pr(X_2>t) \\[8pt]
= {} & t^2 + \frac{1+t} 2\cdot(1-t) = \frac{1+t^2} 2 .
\end{align}</span></p>
|
3,532,033 | <p>Let <span class="math-container">$(M_1,+,\times)$</span> be an algebraic structure, lets say, for example, a ring. If we have another structure <span class="math-container">$(M_2,+,\times)$</span> isomorphic to the first one does that mean that <span class="math-container">$(M_2,+,\times)$</span> is also a ring ?</p>
<p>Edit : To give more context, in 2 different exercises I had to prove that the following are rings :</p>
<p>1) <span class="math-container">$(\mathbb{C},+,•)$</span> where "<span class="math-container">$+$</span>" is simple addition and <span class="math-container">$z_1•z_2 = z_1z_2 + \operatorname{Im}(z_1) \operatorname{Im}(z_2)$</span>, where <span class="math-container">$\operatorname{Im}(z) = $</span> the imaginary component of z.</p>
<p>2) <span class="math-container">$(\mathbb{R} \times \mathbb{R},+,•)$</span> where <span class="math-container">$(a,b)+(x,y) = (a+x,b+y)$</span> and <span class="math-container">$(a,b)•(x,y) = (ax,ay+bx)$</span> </p>
<p>In fact, these are the same with different notations: <span class="math-container">$x+yi = (x,y)$</span></p>
| 9sven6 | 669,847 | <p>What you say is technically correct, but isomorphic doesn't just mean <span class="math-container">$(M_2,+,\times)$</span> is a ring, but with the exact same structures. <span class="math-container">$M_1$</span> and <span class="math-container">$M_2$</span> can be mapped one-to-one on eachother. </p>
|
35,220 | <p>It is a basic result of group cohomology that the extensions with a given abelian normal subgroup <em>A</em> and a given quotient <em>G</em> acting on it via an action $\varphi$ are given by the second cohomology group $H^2_\varphi(G,A)$. In particular, when the action is trivial (so the extension is a central extension), this is the second cohomology group $H^2(G,A)$ for the trivial action. In the special case where <em>G</em> is also abelian, we classify all the class two groups with <em>A</em> inside the center and <em>G</em> as the quotient group.</p>
<p>I am interested in the following: given a sequence of abelian groups $A_1, A_2, \dots, A_n$, what would classify (up to the usual notion of equivalence via commutative diagrams) the following: a group <em>E</em> with an ascending chain of subgroups:</p>
<p>$$1 = K_0 \le K_1 \le K_2 \le \dots \le K_n = E$$</p>
<p>such that the $K_i$s form a central series (i.e., $[E,K_i] \subseteq K_{i-1}$ for all <em>i</em>) and $K_i/K_{i-1} \cong A_i$?</p>
<p>The case $n = 2$ reduces to the second cohomology group as detailed in the first paragraph, so I am hoping that some suitable generalization involving cohomology would help describe these extensions.</p>
<p>Note: As is the case with the second cohomology group, I expect the object to classify, not isomorphism classes of possibilities of the big group, but a notion of equivalence class under a congruence notion that generalizes the notion of congruence of extensions. Then, using the actions of various automorphism groups, we can use orbits under the action to classify extensions under more generous notion of equivalence.</p>
<p>Note 2: The crude approach that I am aware of involves building the extension step by step, giving something like a group of groups of groups of groups of ... For intsance, in the case $n = 3$:</p>
<p>$$1 = K_0 \le K_1 \le K_2 \le K_3 = G$$</p>
<p>with quotients $A_i \cong K_i/K_{i-1}$, I can first consider $H^2(A_3,A_2)$ as the set of possibilities for $K_3/K_1$ (up to congruence). For each of these possibilities <em>P</em>, there is a group $H^2(P,A_1)$ and the total set of possibilities seems to be:</p>
<p>$$\bigsqcup_{P \in H^2(A_3,A_2)} H^2(P,A_1)$$</p>
<p>Here the $\in$ notation is being abused somewhat by identifying an element of a cohomology group with the corresponding extension's middle group.</p>
<p>What I really want is some algebraic way of thinking of this unwieldy disjoint union as a single object, or some information or ideas about its properties or behavior.</p>
| David Corwin | 1,355 | <p>You might find the material on the interpretation of $\mathrm{Ext}$ in terms of extensions at <a href="http://en.wikipedia.org/wiki/Ext_functor#Ext_and_extensions" rel="nofollow">Ext and Extensions</a> to be useful. You probably know that $H^n(G,M) = \mathrm{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z},M)$ (which is not exactly the same as the relation between $H^2$ and extensions, but it is similar). You might be able to construct a kind of Baer sum on these central series by taking pullbacks and pushouts, which makes the set of central series into a group. This makes sense, given that the method of adding group extensions which puts it in isomorphism with $H^2$ is known as <em>Baer multiplication</em> (c.f. Weiss, Cohomology of Groups).</p>
|
4,348,455 | <p>Each digits of the decimal expansion of the integer <span class="math-container">$2022$</span> (this year) consists of <span class="math-container">$0$</span> or <span class="math-container">$2$</span> and also, each digits of the ternary expansion of the same integer <span class="math-container">$2022$</span> (which is <span class="math-container">$2202220_3$</span>) consists of <span class="math-container">$0$</span> or <span class="math-container">$2$</span>. I wonder if there are infinitely many such integers.</p>
| paw88789 | 147,810 | <p>I am writing this as an answer rather than as a comment because the formatting is better. In a computer search I found the following thirteen values that satisfy the conditions (all values are given in decimal):</p>
<p><span class="math-container">$0\\
2\\
20\\
222\\
2000\\
2022\\
22220202002202\\
22220202002220\\
22220202002222\\
22220202020022\\
22220202020202\\
2200222000020222022200\\
2200222000020222022202$</span></p>
<p>I extended the search up to (approximately) <span class="math-container">$2.02\times 10^{27}$</span> and found nothing further.</p>
|
105,071 | <p>As one may know, a <b>dynamical system</b> can be defined with a monoid or a group action on a set, usually a manifold or similar kind of space with extra structure, which is called the <i>phase space</i> or <i>state space</i> of the dynamical system. The monoid or group doing the acting is what I call the <i>time space</i> of the dynamical system, and is usually the naturals, integers, or reals. Often, one may require the evolution map to be continuous, differentiable, etc.</p>
<p>But has anyone studied a generalization in which we allow the <i>time space</i> to be something more general & exotic, a multidimensional space like $\mathbb{R}^n$, $\mathbb{C}$ (viewed as "2-dimensional" by considering it to be like $\mathbb{R}^2$), etc.? I'm especially curious about the case where the time space is $\mathbb{C}$ and the phase space is $\mathbb{C}^n$ or another complex manifold and the map is required to be holomorphic in both its arguments, as that holomorphism provides a natural linkage between the two dimensions that lets us think of the complex time as a single 2-dimensional time as opposed to two real times (any dynamical system with a timespace of $\mathbb{R}^n$ can be decomposed into a bunch of mutually-commutative evolution maps with timespaces of $\mathbb{R}$). My questions are:</p>
<ol>
<li><p>Is it true that the only dynamical system with phase-space $\mathbb{C}$ and time-space $\mathbb{C}$ where the evolution map is required to be holomorphic in both its arguments (the time and point to evolve) is the linear one given by
$$\phi^t(z) = e^{ut} z + K \frac{1 - e^{ut}}{1 - e^u}, K, u \in \mathbb{C}$$
? I suspect so, because an injective entire function is linear (in the sense of a “linear equation” not <em>necessarily</em> a “linear map”), and $\phi^{-t}$ must be the inverse of $\phi^t$. Thus, $\phi^t$ must have the form $a(t) z + b(t)$ with $a: \mathbb{C} \rightarrow \mathbb{C}$, $b: \mathbb{C} \rightarrow \mathbb{C}$. Am I right?</p></li>
<li><p>Are there any interesting (e.g. with complicated, even chaotic behavior) dynamical systems of this kind on $\mathbb{C}^n$? On more sophisticated complex manifolds? (for the phase space, that is) If so, can you provide an example? Or does the holomorphism requirement essentially rule this out? EDIT: I provide one below.</p></li>
<li><p>There is something else here, an interesting observation I made. Consider the above complex-time, holomorphic dynamical system. We can investigate the two prime behaviors represented by the real and imaginary times. We'll just set u = 1, K = 0 for here.</p></li>
</ol>
<p>In “real time”, the dynamics looks like an “explosion” in the plane: all points “blast” away from z = 0 at exponentially increasing velocity.</p>
<p>In “imaginary time”, the result is cyclic motion, swirling around $z = 0$ with constant angular velocity that depends only on the distance of the point from $z = 0$.</p>
<p>But if we trace the contours of these two evolutions, formed from different points on the plane, and then superimpose them, we have contours intersecting in what looks like contours from a contour graph of the image of the complex plane under the function $\exp(z)$! Conversely, we could say it looks like a countour plot of $\log(z)$ with a cut along a ray from $0$. So, somehow “naturally” related to the dynamical system $\phi^t(z) = \exp(t) z$ is the function $\exp(z)$ (or, perhaps, $\log(z)$).</p>
<p><img src="https://i.stack.imgur.com/kPq32.png" alt="plot of evoluton countours for first CTHDS"></p>
<p>Note that my plotting facilities are unfortunately pretty limited, so I can't give really nice graphs with lots of contours, just a few selected ones taken from evolutions of various points in both real and imag times.</p>
<p>But we have another case. To see this, we must turn our attention away from a phase space given by the complex plane to one given by the Riemann sphere, $\hat{\mathbb{C}}$. In this case, we still have the dynamical system as above, but we have an additional class of dynamical systems given by the “Moebius transformations”, which include the above linear-function dynamical systems as a special case. One example is
$$\phi^t(z) = \frac{(1 + e^{i\pi t}) z + (1 – e^{i\pi t})}{(1 – e^{i\pi t})z + (1 + e^{i\pi t})}$$.
It is easy to check that this is indeed a Moebius transformation of the Riemann sphere. This map is holomorphic everywhere on the Riemann sphere. Note that for integer step t, the unit-step map is the reciprocal map.</p>
<p>Now we consider the contour lines of the real and imaginary evolution, as before. They look like this:</p>
<p><img src="https://i.stack.imgur.com/MECR7.png" alt="plot of evolution contours for second CTHDS"></p>
<p>(Physics buffs may notice that the real evolution (concentric circles) reminds one of the lines of a <i>magnetic</i> field of a <i>magnetic</i> dipole (like a bar magnet), while the imaginary evolution (arcs joining at points) looks like the lines of an <i>electric</i> field of an <i>electric</i> dipole.)</p>
<p>Again, notice how the lines meet at right angles. It looks again like the image of the plane (or the Riemann sphere, perhaps?) under some function which may be holomorphic, though I'm not sure what that function is in this case. Is this the case? Is there such a function, with a special relation to this CTHDS in the same way as $\exp$ (or $\log$) is to the other?</p>
<p>But in any case, it appears that for a complex-time holomorphic dynamical system, or CTHDS, there exists an associated natural function. What is the significance of this function/map? How does it relate to the CTHDS? If you give me a CTHDS that I don't have a closed form for, can I find its natural map?</p>
| Community | -1 | <p>I have seen talks about this, though somewhat over my head. But, yes absolutely, people do research such things. A. Katok is one name that comes to mind here. See this paper: <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.33.2082" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.33.2082</a></p>
<p>They consider actions of $\mathbb{Z}^k$ and $\mathbb{R}^k$ that generalize the usual time-like case of $\mathbb{Z}$ and $\mathbb{R}$.</p>
<p>B. Kalinin and V. Sadovskaya also work in this area, seeing their talks was how I was exposed to this field: <a href="http://www.southalabama.edu/mathstat/personal_pages/sadovska/Research/tns.pdf" rel="nofollow">http://www.southalabama.edu/mathstat/personal_pages/sadovska/Research/tns.pdf</a></p>
<p>A lot of the work I've been exposed to is about classifying such systems up to a conjugacy, and in particular they consider Anosov systems and see if they can find conjugacies in the general case to canonical examples. </p>
<p>This is not my expertise, though, so I can't say what this says about the $\mathbb{C}$ case you're interested in. But at least I can say, yes, indeed there is work in the higher-dimensional case.</p>
|
148,624 | <p>What does it mean to say that , A bounded linear operator is not "generally" bounded function.
Can anybody explain ? </p>
| Robert Israel | 8,508 | <p>A nonzero linear operator on any vector space (over $\mathbb R$ or $\mathbb C$) is never a bounded function: $\|A(tx)\| = |t| \|Ax\| \to \infty$ as $t \to \infty$ whenever $Ax \ne 0$.</p>
|
148,624 | <p>What does it mean to say that , A bounded linear operator is not "generally" bounded function.
Can anybody explain ? </p>
| Alex Becker | 8,173 | <p>A linear operator $T:V\to W$ is said to be "bounded" if there exists some $M\in\mathbb R$ such that $\|Tx\|\leq M\|x\|$ for all $x\in V$. A function $f:V\to W$ is said to be "bounded" if there exists some $M\in\mathbb R$ such that $\|f(x)\|\leq M$ for all $x\in V$. Note that this is a stronger condition than the previous one, as the upper bound for $\|T(x)\|$ varies with $x$ while the upper bound for $\|f(x)\|$ does not. In fact, any linear operator which is "bounded" <em>as a function</em> is simply $0$.</p>
|
623,709 | <p>I make the following conjecture: the function
$$
d(x, y):=\frac{||x-y||}{\max(||x||, ||y||)}
$$
is a distance on $H$, where $H$ is a normed vector space or a Hilbert space, and $x, y \in H$ (the function $d$ is defined to be $0$ in the case $x=y=0$). Note that $d$ is scale invariant, i.e., $d(\lambda x, \lambda y)=d(x, y)$ for $0 \neq \lambda \in \mathbb{R}$. The property of $d$ which needs to be explicitly proved or disproved is the triangle inequality
$$
d(x, y) \leq d(x,z)+d(z,y).
$$
The triangle inequality (TI) can be easily proved for $H=\mathbb{R}$; moreover, due to scale invariance, it is sufficient to prove it for $||x||,||y||, ||z|| \leq 1$. The TI has been numerically tested by a program which has generated $10^8$ triples of random points uniformly distributed in $[-1, 1]^3$, and the same number in $[-1,1]^6$: all the generated triples satisfied the TI. This test supports therefore the conjecture for $H=\mathbb{R}^3$ and $H=\mathbb{R}^6$. Since the subspace generated by three linearly independent vectors of a real (complex) Hilbert space is isometrically isomorphic to $\mathbb{R}^3$ ($\mathbb{R}^6$), the numerical test supports the conjecture also for a generic Hilbert space*. </p>
<p>Does somebody know if this conjecture has been already proved, or is able to prove (or disprove) it?</p>
<ul>
<li>Before posting this question, I exchanged some e-mail with prof. Egor Makimenko, of the Instituto Politécnico Nacional, México. I did by myself a program for the numerical test of the TI, but the test cited above has been performed by a program that prof. Maximenko sent me. Moreover, the generalization from $\mathbb{R}^3$-$\mathbb{R}^6$ to a generic Hilbert space is due to prof. Maximenko.</li>
</ul>
| Community | -1 | <p>We know that <span class="math-container">$p(x,y):= \|x-y\|$</span> is a metric. Hence we get for free</p>
<p><span class="math-container">$$\|x-z\|\leq \|x-y\|+\|y-z\|.$$</span></p>
<p>Remark: The case where exactly one of the vectors is the zero vector is trivial because we get an inequality of the form:</p>
<p><span class="math-container">$$1\leq 1+\varepsilon.$$</span></p>
<p>If exactly two are zero, we get either:
<span class="math-container">$$1\leq 1+0,$$</span>
OR
<span class="math-container">$$0\leq 1+1.$$</span></p>
<p>If all three are zero:</p>
<p><span class="math-container">$$0\leq 0+0.$$</span></p>
<p>Now consider the case <span class="math-container">$0<\|x\|\leq \|y\| \leq \|z\|$</span>:</p>
<p><span class="math-container">\begin{align}
\max(\|x\|,\|z\|)&=\|z\|, \\
\max(\|x\|,\|y\|)&=\|y\|, \\
\max(\|y\|,\|z\|)&=\|z\|.
\end{align}</span></p>
<p>So, dividing through the first inequality by <span class="math-container">$\max(\|x\|,\|z\|)$</span> we get</p>
<p><span class="math-container">$$\frac{\|x-z\|}{\max(\|x\|,\|z\|)}\leq \frac{\|x-y\|}{\max(\|x\|,\|z\|)} + \frac{\|y-z\|}{\max(\|x\|,\|z\|)},$$</span></p>
<p>but <span class="math-container">$$d(x,y)\geq \frac{\|x-y\|}{\max(\|x\|,\|z\|)} \text{ and } d(y,z)=\frac{\|y-z\|}{\max(\|x\|,\|z\|)}.$$</span></p>
<p>Hence, the triangle inequality is satisfied.</p>
<p>I am optimistic that the other 5 cases can be proven similarly. Don’t hold me to it! I hope you can make some progress from here.</p>
|
623,709 | <p>I make the following conjecture: the function
$$
d(x, y):=\frac{||x-y||}{\max(||x||, ||y||)}
$$
is a distance on $H$, where $H$ is a normed vector space or a Hilbert space, and $x, y \in H$ (the function $d$ is defined to be $0$ in the case $x=y=0$). Note that $d$ is scale invariant, i.e., $d(\lambda x, \lambda y)=d(x, y)$ for $0 \neq \lambda \in \mathbb{R}$. The property of $d$ which needs to be explicitly proved or disproved is the triangle inequality
$$
d(x, y) \leq d(x,z)+d(z,y).
$$
The triangle inequality (TI) can be easily proved for $H=\mathbb{R}$; moreover, due to scale invariance, it is sufficient to prove it for $||x||,||y||, ||z|| \leq 1$. The TI has been numerically tested by a program which has generated $10^8$ triples of random points uniformly distributed in $[-1, 1]^3$, and the same number in $[-1,1]^6$: all the generated triples satisfied the TI. This test supports therefore the conjecture for $H=\mathbb{R}^3$ and $H=\mathbb{R}^6$. Since the subspace generated by three linearly independent vectors of a real (complex) Hilbert space is isometrically isomorphic to $\mathbb{R}^3$ ($\mathbb{R}^6$), the numerical test supports the conjecture also for a generic Hilbert space*. </p>
<p>Does somebody know if this conjecture has been already proved, or is able to prove (or disprove) it?</p>
<ul>
<li>Before posting this question, I exchanged some e-mail with prof. Egor Makimenko, of the Instituto Politécnico Nacional, México. I did by myself a program for the numerical test of the TI, but the test cited above has been performed by a program that prof. Maximenko sent me. Moreover, the generalization from $\mathbb{R}^3$-$\mathbb{R}^6$ to a generic Hilbert space is due to prof. Maximenko.</li>
</ul>
| coffeemath | 30,316 | <p>Let the norm on the plane be the taxicab norm, norm of $(x,y)$ being $|x|+|y|.$ This induces a metric (satisfying triangle inequality etc.) on the metric space $\mathbb{R}^2$, making it a normed linear space.</p>
<p>But with your metric obtained by division of max norms, take the points $x,y,z$ as $(1,2),(2,1),(2,2)$ respectively. Then the usual distances (before division) are $2$ for $x,y$ and each usual distance for both $x,z$ and $y,z$ is $1.$ The norms of $x,y,z$ are respectively $3,3,4$, so that after division by max norms we get
$$d(x,y)=2/3, \ d(x,z)=d(z,y)=1/4.$$
This is against the triangle inequality since $2/3>1/4+1/4.$</p>
<p>So in conclusion it looks to me like the "metric" you propose does not in general satisfy the triangle inequality on an arbitrary normed linear space. I would still like to see if it holds in Euclidean space, since some features of taxicab metric are peculiar.</p>
|
1,770,804 | <p>I am a high school student my maths teacher said that if $\,ax+b=cx+d,\,$ then is $\,a=c\,$ and $\,b=d.\,$ Can someone give me a prove of this?</p>
| Community | -1 | <p>By the fundamental theorem of algebra (FTA), a first degree polynomial has exactly $1$ root. Thus if there are $2$ or more values of $x$ for which the equation holds, then $(a-c)x+(b-d)$ is the zero polynomial. So $a=c$ and $b=d$.</p>
|
9,168 | <p>I'm having a doubt about how should we users encourage the participation of new members. So far I have only presented MSE to three of my fellow colleagues in grad school. In an overall way I feel like if MSE becomes too open and wide known, some of the high-rank researchers and top-class grad and undergrads users will be frustrated. Maybe I get this impression for seeing how some really complicated questions are well-received.</p>
<p>But today I saw someone asking some really simple algebra questions like "solve this equation for x" and started thinking what is the general feel of MSE when this kind of question arises. I'm not judging anything and no one. I just would like to get some opinions on this.</p>
| Thomas Klimpel | 12,490 | <p>Community growth in the form of people willing to answer question should be encouraged. If these simple questions give some people the opportunity to write nice and clear answers, great. The answers from 10K+ users to these trivial questions might indeed be something worrying. I think the trivial questions themselves should be neither encouraged nor discouraged, but users well experienced in writing answers should perhaps try to refrain from answering such questions.</p>
|
4,026,149 | <p>If f is continuous on <span class="math-container">$[a,b]$</span> and <span class="math-container">$f(a)=f(b)$</span> then show that there exists <span class="math-container">$x,y \in (a,b)$</span> such that <span class="math-container">$f(x)=f(y)$</span></p>
<p>It looks obvious if I imagine the graph. But I am not able to prove it.
I am trying to employ intermediate value property, but not able to reach to the conclusion.</p>
| user10354138 | 592,552 | <p><strong>Hint</strong>: Either <span class="math-container">$f$</span> is a constant (so you are done), or there is some <span class="math-container">$t\in(a,b)$</span> such that <span class="math-container">$f(t)\neq f(a)$</span>, in which case employ IVT for <span class="math-container">$f$</span> on <span class="math-container">$[a,t]$</span> and <span class="math-container">$[t,b]$</span>.</p>
|
863,846 | <p>Steven Strogatz has a great informal textbook on Nonlinear Dynamics and Chaos. I have found it to be incredibly helpful to get an intuitive sense of what is going on and has been a great supplement with my much more formal text from Perko.</p>
<p>Anyways I was wondering if anyone knew of any similar informal, intuitive textbooks covering Numerical Analysis? I currently study out of Atkinson's Intro to Numerical Analysis. I am looking for a more informal numerical text aimed for upper level undergrads and first/second year grad students. I am currently a first year grad student studying for my qualifying exams. Any recommendations would be appreciated. Thank you</p>
| BeaumontTaz | 147,480 | <p>It can be a little dense sometimes, but <em>Numerical Analysis</em> by Burden and Faires is a classic book on the subject.</p>
|
1,823,556 | <p>Let be $X \subset F_1 \cup F_2$, where $F_1$ and $F_2$ are closed. If the function $f\colon X \longrightarrow \mathbb{R}$ is such that $f|_{X \cap F_1}$ and $f|_{X \cap F_2}$ are continuous, so prove that $f$ is continuous. </p>
<p>My attempt:</p>
<p>Suppose that $f$ is discontinuously, so exists $x \in X$ such that $f$ is discontinuously in $x$, but $X \subset F_1 \cup F_2$, therefore, if $x \in X \cap F_1$, so $f|_{X \cap F_1}$ is discontinuously in $x$ which is absurd because contradicts the hypothesis. Analogously, if $x \in X \cap F_2$, we have a contradiction.</p>
<p>That's my answer, but I don't sure if it's correct, because I didn't use the hypothesis that $F_1$ and $F_2$ are closeds. I would like to know if my attempt is correct. Thanks in advance!</p>
<p>EDIT: $X \subset \mathbb{R}$</p>
| Brian M. Scott | 12,042 | <p>HINT: Something similar to your approach is workable, but you’ve omitted most of the crucial details. In particular, you’ve not justified the assertion that $f\upharpoonright X\cap F_1$ is discontinuous at $x$. </p>
<p>Suppose that $f$ is not continuous at $x$; then there are an $\epsilon>0$ and a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converging to $x$ such that $|f(x_n)-f(x)|\ge\epsilon$ for each $n\in\Bbb N$. (Why?) Let </p>
<p>$$N_1=\{n\in\Bbb N:x_n\in F_1\}$$</p>
<p>and</p>
<p>$$N_2=\{n\in\Bbb N:x_n\in F_2\}\;;$$</p>
<p>clearly $N_1\cup N_2=\Bbb N$, so at least one of the sets $N_1$ and $N_2$ must be infinite. Without loss of generality suppose that $N_1$ is infinite. </p>
<ul>
<li>Explain why $x\in X\cap F_1$. </li>
<li>Use $N_1$ to show that $f\upharpoonright X\cap F_1$ cannot be continuous at $x$.</li>
</ul>
|
1,823,556 | <p>Let be $X \subset F_1 \cup F_2$, where $F_1$ and $F_2$ are closed. If the function $f\colon X \longrightarrow \mathbb{R}$ is such that $f|_{X \cap F_1}$ and $f|_{X \cap F_2}$ are continuous, so prove that $f$ is continuous. </p>
<p>My attempt:</p>
<p>Suppose that $f$ is discontinuously, so exists $x \in X$ such that $f$ is discontinuously in $x$, but $X \subset F_1 \cup F_2$, therefore, if $x \in X \cap F_1$, so $f|_{X \cap F_1}$ is discontinuously in $x$ which is absurd because contradicts the hypothesis. Analogously, if $x \in X \cap F_2$, we have a contradiction.</p>
<p>That's my answer, but I don't sure if it's correct, because I didn't use the hypothesis that $F_1$ and $F_2$ are closeds. I would like to know if my attempt is correct. Thanks in advance!</p>
<p>EDIT: $X \subset \mathbb{R}$</p>
| Qiyu Wen | 310,935 | <p>We can be a bit more general. Let $X = A\cup B$ be such that $A-B$ and $B-A$ are separated. For any subset $E$ of $X$, its $X$-closure is the union of the $A$-closure of $E\cap A$ and $B$-closure of $E\cap B$. That is,
$$
\bar{E} = \left(\overline{E\cap A} \cap A\right) \cup \left(\overline{E\cap B} \cap B\right).
$$
To prove it, note that (since closure of union of two sets is the union of their closures)
$$
\bar{E} = \overline{E\cap A} \cup \overline{E\cap(B-A)}.
$$
Consequently
$$
\bar{E}\cap A = \left(\overline{E\cap A} \cap A\right)\cup \left(\overline{E\cap(B-A)}\cap A\right).
$$
Since $X = A\cup B$ and $\overline{B-A} \cap A-B = \varnothing$, $\overline{B-A} \subset B$. Hence $\overline{E\cap(B-A)} \subset B$. Note that we also have $\overline{E\cap(B-A)}\subset \overline{E\cap B}$. Therefore, $\overline{E\cap(B-A)} \subset \overline{E\cap B} \cap B$. To conclude,
$$
\bar{E}\cap A \subset \left(\overline{E\cap A} \cap A\right)\cup \left(\overline{E\cap B} \cap B\right).
$$
The same holds for $\bar{E}\cap B$. Hence
$$
\bar{E} \subset \left(\overline{E\cap A} \cap A\right)\cup \left(\overline{E\cap B} \cap B\right).
$$
The other direction is rather obvious.</p>
<p>With the same $X$, suppose that $f:X\to Y$ is such that both $f|_A$ and $f|_B$ are continuous. Then $f$ is continuous on $X$. To prove it, let $F$ be a closed set in $Y$, and put $E = f^{-1}(F)$. Then ${f|_A}^{-1}(F) = E\cap A$, and ${f|_B}^{-1}(F) = E\cap B$. Since $f$ is continuous on $A$ and $B$, $E\cap A$ and $E\cap B$ are closed in $A$ and $B$ respectively. By our previous formula,
$$
\bar{E} = (E\cap A) \cup (E\cap B) = E,
$$
so $E$ is closed. Hence $f$ is continuous on $X$.</p>
<p>To apply the result to your case, note that when $A$ and $B$ are both closed, $A-B$ and $B-A$ are separated. The reason is that both are open in the subspace $A-B \cup B-A$.</p>
|
109,213 | <p>In classical complex analysis it is easy to prove that a meromorphic function has at most one analytic continuation (on an open connected subset of $\mathbb C$, say).</p>
<p>The problem of non-uniqueness of analytic continuation is one of the reasons why it is not possible (if one wants a good theory) to translate the complex theory to the $p$-adic case without some modifications, and so it is one of the motivation for introducing rigid analytic variety. However, I am not able to find a precise statement that explains under which hypothesis the uniqueness of analytic continuation holds for rigid analytic varieties. So the question is the following.</p>
<p>Let $k$ be a non-archimedean field and let $X$ be a connected rigid analytic space over $k$. Let $f \colon X \to k$ be a rigid analytic function that vanishes on $Y$, that is an admissible subdomain of $X$. It is always true that $f$ vanishes on $X$? If this is not the case, under which assumptions it is true?</p>
<p>Any references will be very appreciated!</p>
| user27056 | 27,056 | <p>No: let $X$ be the union of the coordinate axes in the affine plane. As over $\mathbf{C}$, the answer is affirmative on a connected <em>normal</em> analytic space. Hint: prove in any rigid-analytic space that connected components are witnessed via finite linked chains of connected affinoid opens (and after thereby reducing to the affinoid case, assume smoothness if you are unfamiliar with the excellence properties of affinoid algebras). </p>
<p>The answer is applicable to meromorphic functions as well, but proving that requires more care (e.g., one has to first figure out how to appropriately define the concept of meromorphicity and prove some basic features of it).</p>
|
3,106,574 | <p>Let <span class="math-container">$(a_n) _{n\ge 0}$</span> <span class="math-container">$a_{n+2}^3+a_{n+2}=a_{n+1}+a_n$</span>,<span class="math-container">$\forall n\ge 1$</span>, <span class="math-container">$a_0,a_1 \ge 1$</span>. Prove that <span class="math-container">$(a_n) _{n\ge 0}$</span> is convergent.<br>
I could prove that <span class="math-container">$a_n \ge 1$</span> by mathematical induction, but here I am stuck. </p>
| Barry Cipra | 86,747 | <p>This is a variant on maxmilgram's answer. If we write <span class="math-container">$a_n=1+u_n$</span> with <span class="math-container">$u_n\ge0$</span> (as the OP observes), the recursion becomes</p>
<p><span class="math-container">$$u_{n+2}^3+3u_{n+2}^2+4u_{n+2}=u_{n+1}+u_n$$</span></p>
<p>This implies <span class="math-container">$4u_{n+2}\le u_{n+1}+u_n$</span>, at which point we have <span class="math-container">$u_n\le v_n$</span> with <span class="math-container">$v_0=u_0$</span>, <span class="math-container">$v_1=u_1$</span>, and</p>
<p><span class="math-container">$$v_{n+2}={v_{n+1}+v_n\over4}$$</span></p>
<p>so that</p>
<p><span class="math-container">$$0\le u_n\le v_n=C_1\left(1-\sqrt{17}\over8\right)^n+C_2\left(1+\sqrt{17}\over8\right)^n\to0$$</span></p>
<p>as in maxmilgram's answer.</p>
|
3,927,502 | <p><span class="math-container">$\lim\limits_{n\to\infty}\dfrac{n^2-n+2}{3n^2+2n-4}=\dfrac{1}{3}$</span>.</p>
<p>With epsilon definition I get my answer as <span class="math-container">$N=\left[ \dfrac{5}{9\varepsilon }\right] +1$</span>. But then I thought that how can I evaluate this sequence, in functions <span class="math-container">$\delta$</span>-<span class="math-container">$\varepsilon$</span> definition by substituting <span class="math-container">$n$</span> by <span class="math-container">$x$</span>. But I can't make it.</p>
<p>So here's my question:</p>
<p>What can I do is I have function like this one?
<span class="math-container">$\lim\limits_{x\to+\infty}\dfrac{x^2-x+2}{3x^2+2x-4}=\dfrac{1}{3}$</span>.</p>
<p>I'm thinking about that function. But how to find.</p>
<p><span class="math-container">$\forall \varepsilon>0$</span> be given, then <span class="math-container">$\exists M>0$</span> such that <span class="math-container">$x>M$</span> implies <span class="math-container">$| f\left( x\right) -l|<\varepsilon$</span></p>
<p>For function:
<span class="math-container">$\left| \dfrac{x^{2}-x+2}{3x^{2}+2x-4}-\dfrac{1}{3}\right|=
\left| \dfrac{-5x+10}{9x^{2}+6x-12}\right|<\left| \dfrac{-5x+10}{9x^{2}}\right|$</span></p>
<p>And this is where I stuck. Because <span class="math-container">$x\in \mathbb{R}$</span>, so I think we can't remove the brackets. Isn't it?
For y'all's helps I solved it thank you all!</p>
| zipirovich | 127,842 | <p>How is it any different from what you've already done? Renaming <span class="math-container">$n$</span> into <span class="math-container">$x$</span> and <span class="math-container">$N$</span> into <span class="math-container">$M$</span> doesn't really change anything. The same answer still works. Or you can simplify it a little by dropping the rounding, since <span class="math-container">$M$</span> doesn't have to be an integer anymore.</p>
|
133,418 | <p>Let $\langle R,0,1,+,\cdot,<\rangle$ be the standard model for R, and let S be a countable model of R (satisfying all true first-order statements in R). Is it true that the set 1,1+1,1+1+1,… is bounded in S? My intuition says "no", but I am yet to find a counter example. I read something about rational functions, but I cannot verify it is, indeed, a non-standard model of R.</p>
| hmakholm left over Monica | 14,366 | <p>Let $T$ be the set of all true sentences about $\mathbb R$ and construct $T'$ by adding to $T$ a new constant $c$ together with the axioms $c>1$, $c>1+1$, $c>1+1+1$, ...</p>
<p>Every finite subset of $T$ has $\mathbb R$ as a model, so $T'$ is consistent by the compactness theorem, and has a countable model because $T'$ contains only countably many symbols. This shows that a countable $S$ <em>can</em> be non-Archimedean.</p>
<p>On the other hand, there must also be an Archimedean countable model. This follows directly from the downward Löwenheim-Skolem theorem, which produces a subset of $\mathbb R$ that is closed under the operations and is elementarily equivalent to $\mathbb R$ itself.</p>
|
480,727 | <p>If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$</p>
<p>I have tried starting with taking logartithms, but that gives just some more equations.</p>
<p>Any specific way to solve these type of problems?</p>
<p>Any help will be appreciated.</p>
| Mark Bennet | 2,906 | <p>Take logarithms (any base) to obtain $$x\log 2= y\log 3=-z\log 6=-z\log 3-z\log 2$$</p>
<p>Then note that $(y+z)\log 3=-z\log 2$ and $x\log 2 = y\log 3$</p>
<p>Multiply left- and right- hand sides to obtain $$(xy+yz)\log 3 \log 2=-yz\log 2\log 3$$</p>
<p>Whence $xy+yz+zx=0$</p>
<p>Note that we have done no division so far, except by the non-zero term $\log 2\log 3$, and also that $x=y=z=0$ is a solution. If $xyz\neq 0$ we can divide by $xyz$ to obtain the required equation.</p>
|
3,057,517 | <p>Hy everybody ! </p>
<p>I'm studying population dynamics for my calculus exam, and I don't understand something that seems really easy, so I thought you might be able to help me out ;)</p>
<p>Here's the thing. I have this differential equation <span class="math-container">$\frac{dN}{dt} = \sqrt{N}$</span>.</p>
<p>Our book makes us realize that both <span class="math-container">$N(t) = 0$</span> and <span class="math-container">$N(t) = \frac{t^2}{4}$</span> are solution, which makes sense so far, simply by replacing in the original equation.</p>
<p>Now suppose we start at <span class="math-container">$N(0) = 0$</span>. How can <span class="math-container">$N$</span> start growing like <span class="math-container">$\frac{t^2}{4}$</span> if it's derivative is <span class="math-container">$0$</span> at <span class="math-container">$t = 0$</span> ? Because zero derivative should mean no growth, so <span class="math-container">$\sqrt{N}$</span> should remain zero, which means still no growth, and so on. My brain is melting right now.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Zero derivative at one point does not mean no growth at other points. </p>
<p>For example the derivative of function <span class="math-container">$$N(t)=\frac {t^2}{4}$$</span> is <span class="math-container">$$N'(t)=t/2$$</span> That is the derivative is zero at <span class="math-container">$t=0$</span> but it is not zero at other points. The point of the problem is that in this case the solution is not unique. </p>
|
1,032,535 | <p>I know $n \in \mathbb{N}$ and...</p>
<p>$$
a_n = \begin{cases}
0 & \text{ if } n = 0 \\
a_{n-1}^{2} + \frac{1}{4} & \text{ if } n > 0
\end{cases}
$$</p>
<ol>
<li><strong>Base Case:</strong></li>
</ol>
<p>$$a_1 = a^2_0 + \frac{1}{4}$$</p>
<p>$$a_1 = 0^2 + \frac{1}{4} = \frac{1}{4}$$</p>
<p>Thus, we have that $0 < a_1 < 1$. So our base case is ok.</p>
<ol start="2">
<li><strong>Inductive hypothesis</strong>:</li>
</ol>
<p>Assume $n$ is arbitrary. Suppose
$$0 < a_{n} < 1$$
$$0 < a_{n-1}^{2} + \frac{1}{4} < 1$$
is true, when $n > 1$.</p>
<ol start="3">
<li><strong>Inductive step</strong>:</li>
</ol>
<p>Let's prove
$$0 < a_{n+1} < 1$$
$$0 < a_{n}^{2} + \frac{1}{4} < 1$$</p>
<p>is also true when $n > 1$.</p>
<p>My guess is that we have to prove that $a^2_{n}$ has to be less than $\frac{3}{4}$, which otherwise would make $a_{n+1}$ equal or greater than $1$.</p>
<p>So we have $(a_{n-1}^{2} + \frac{1}{4})^2 < \frac{3}{4}$... I don't know if this is correct, and how to continue...</p>
| Adhvaitha | 191,728 | <p>Starting with $a_0 = 0$, it is easier to show the stronger inequality, $0 < a_n < 1/2$ for $n \in \mathbb{Z}^+$. This conclusion immediately falls out from the recurrence relation.</p>
|
199,148 | <p><a href="https://i.stack.imgur.com/9BuHp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9BuHp.png" alt="enter image description here"></a> </p>
<pre><code>pbdomains = <|
"Overall " -> Around[2.6, 0.04],
"PB" -> Around[4.25, 0.06]
|>;
BarChart[pbdomains, ChartStyle -> "BrightBands",
LabelStyle -> {FontFamily -> "Times New Roman", 28, Bold,
GrayLevel[0]}, Frame -> True, FrameLabel -> {"", " Count"},
BarSpacing -> Tiny,
ChartLabels -> Callout[Automatic, Above, Appearance -> "Balloon"]]
</code></pre>
| SelfHorizons Work | 65,424 | <p><a href="https://i.stack.imgur.com/lsGrC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lsGrC.png" alt="enter image description here"></a>Based on Rohit Namjoshi solution:</p>
<pre><code>pbdomains = <|"Overall " -> Around[2.6, 0.04],
"PB" -> Around[4.25, 0.06]|>;
BarChart[pbdomains, ChartStyle -> "BrightBands",
LabelStyle -> {FontFamily -> "Times New Roman", 28, Bold,
GrayLevel[0]}, Frame -> True, FrameLabel -> {"", " Count"},
BarSpacing -> Tiny,
ChartLabels -> Callout[Automatic, Above, Appearance -> "Balloon"],
PlotRange -> {{0.5, 2.5}, All}]
</code></pre>
|
565,046 | <blockquote>
<p>The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.</p>
</blockquote>
<p>I have that
$$D_6=\left< a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right>$$
$$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$
My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that
$$a^3b=ba^3$$
and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?</p>
<p>For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.</p>
<p>Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible. </p>
| jasmine | 557,708 | <p>General method to find the centre of <span class="math-container">$D_{2n}$</span></p>
<p>if <span class="math-container">$n=1$</span> or <span class="math-container">$n=2$</span> ,then <span class="math-container">$D_{2n}$</span> is abelian and hence <span class="math-container">$Z(D_{2n})= D_{2n}.$</span></p>
<p>Now suppose <span class="math-container">$ n \ge 3$</span> .By definition we have</p>
<p><span class="math-container">$D_{2n} = \{ a^ib^j :i=0,1 , j=0,1,....., n-1\}$</span> where <span class="math-container">$a$</span> is an element of order <span class="math-container">$2$</span>, <span class="math-container">$b$</span> is an element of order <span class="math-container">$n$</span> and <span class="math-container">$a, b$</span> are given by relation <span class="math-container">$ ba=ab^{-1}$</span></p>
<p>Then its implies <span class="math-container">$bba=b^2a = bab^{-1}=ab^{-1}b^{-1}=ab^{-2}.$</span></p>
<p>In general form we can write</p>
<p><span class="math-container">$b^ra=ab^{-r}\tag1$</span></p>
<p>For all integer <span class="math-container">$r \ge 0$</span>.Now , since <span class="math-container">$a$</span> and <span class="math-container">$b$</span> together generate <span class="math-container">$D_{2n}$</span>, an element of <span class="math-container">$D_{2n}$</span> is in center if and only if its commutes with both <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.</p>
<p>Therefore <span class="math-container">$ x= a^ib^j \in Z(D_{2n})$</span> if and only if <span class="math-container">$xa=ax$</span> and <span class="math-container">$xb=bx$</span>.</p>
<p><span class="math-container">$x b= a^ib^j b=a^ib^{j+1}= bx= ba^ib^j$</span> where <span class="math-container">$x= a^ib^j$</span></p>
<p>Therefore <span class="math-container">$a^ib=ba^i \tag 2$</span></p>
<p>we can see clearly <span class="math-container">$i=0$</span> in <span class="math-container">$(2)$</span> because <span class="math-container">$a^0b=ba^0$</span> implies <span class="math-container">$ b=b$</span></p>
<p>we can see clearly <span class="math-container">$i \neq 1$</span> because if <span class="math-container">$i=1$</span> then <span class="math-container">$ab=ba$</span> but from <span class="math-container">$(1)$</span> , for <span class="math-container">$n \ge 3$</span> we have <span class="math-container">$ab=ba^{-1} \neq ba$</span>
so we are getting contradiction if <span class="math-container">$i=1$</span></p>
<p>we have have <span class="math-container">$x=a^ib^j$</span> . If <span class="math-container">$i=0$</span> then <span class="math-container">$x= a^0b^j=b^j$</span></p>
<p>Now the condition <span class="math-container">$xa=ax$</span> t becomes <span class="math-container">$b^ja=ab^j$</span></p>
<p>from <span class="math-container">$(1)$</span> relation we have <span class="math-container">$ab^{-j}=ab^j$</span></p>
<p>There <span class="math-container">$ b^{-j}=b^j$</span> implies <span class="math-container">$b^{2j}=1 \tag 3$</span></p>
<p>Since order<span class="math-container">$(b)=n$</span> (given), we get from <span class="math-container">$(3)$</span> that <span class="math-container">$$n |2j$$</span>
Therefore if <span class="math-container">$n$</span> divide <span class="math-container">$2j$</span> then <span class="math-container">$j=0$</span> or <span class="math-container">$2j=n$</span></p>
<p>Now if <span class="math-container">$j=0$</span> then <span class="math-container">$x=b^j=b^0=1$</span></p>
<p>If <span class="math-container">$2j=n$</span> , then <span class="math-container">$n$</span> is even , that is <span class="math-container">$j= \frac{n}{2}$</span></p>
<p>so <span class="math-container">$x=b^j=b^{n/2}$</span></p>
<p>so we have proved that <span class="math-container">$$Z(D_{2n})= \begin{cases} D_{2n} \ \text{if n=1 ,2} \\ \{1\}\ \text{if n > 2 is odd} \\ \{1,b^{n/2}\} \ \text{if n >2 is even } \end{cases}$$</span></p>
|
2,172,399 | <p>Equation of the segment : $2x + 4y-3 = 0$
Equation of the hyperbola : $7x^2 - 4y^2 =14$</p>
<p>How do you find the equation of the two linear functions that are both perpendicular to the segment and tangent to the hyperbola?</p>
<p>Thanks</p>
| Mengchun Zhang | 420,459 | <p>The segment has an equation of $\,2x+4y-3=0\,$, hence the equation you are finding should be </p>
<p>$$4x-2y+c=0$$</p>
<p>where $\,c\,$ is a real constant</p>
<p>Now differentiate the hyperbola equation w.r.t. $\,x\,$ </p>
<p>$$14x-8y\frac{dy}{dx}=0$$
$$\Rightarrow\quad\frac{dy}{dx}=\frac{7x}{4y}\qquad$$</p>
<p>Since the slope of $\,4x-2y+c=0\,$ is $\,2$, we have</p>
<p>$$\frac{dy}{dx}=\frac{7x}{4y}=2$$
$$\Rightarrow\quad\frac78x=y\qquad$$</p>
<p>Substitute $\,x\ \,\text{for}\,\ y\,$ in the hyperbola equation</p>
<p>$$7x^2-4\left(\frac78x\right)^2=14$$</p>
<p>$$\Rightarrow\quad x=\pm\frac43\sqrt2,\ \ \ y=\pm\frac76\sqrt2\quad$$</p>
<p>Use the values of $\,x\,$ and $\,y\,$ to calculate $\,c\,$, then we get</p>
<p>$$c=2y-4x=\pm3\sqrt2$$</p>
<p>Thus, the straight line that is perpendicular to the segment and tangent to the hyperbola has an equation of:</p>
<p>$$4x-2y+3\sqrt2=0$$
$$\text{or}\quad 4x-2y-3\sqrt2=0\qquad$$</p>
|
411,717 | <p>Let $G$ be a group. By an automorphism of $G$ we mean an isomorphism $f: G\to G$
By an inner automorphism of $G$ we mean any function $\Phi_a$ of the following form:
For every $x\in G$, $\Phi_a(x)=a x a^{-1}$.
Prove that every inner automorphism of $G$ is an automorphism of $G$
which means I should prove $\Phi_a$ is isomorphism? any suggestion? thanks</p>
| jim | 44,551 | <p>$\phi_a(xy)=a(xy)a^{-1}=axa^{-1}aya^{-1}=\phi_a(x)\phi_a(y)$</p>
<p>$\phi_a(x)=\phi_a(y)\implies axa^{-1}=aya^{-1}\implies x=y$</p>
<p>$\phi_a$ is also surjective since for each $y \in G $ there exists $x=a^{-1}ya$ s.t. $\phi_a(x)=y$ </p>
|
3,005,842 | <blockquote>
<p>Let <span class="math-container">$(X,Y)$</span> be the coordinates of a point uniformly chosen from a
quadrilateral with vertices <span class="math-container">$(0,0)$</span>, <span class="math-container">$(1,0)$</span>, <span class="math-container">$(1,1)$</span>, <span class="math-container">$(0,2)$</span>. Find
the marginal probability density functions of <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>.</p>
</blockquote>
<h2>Try</h2>
<p>Well, these points are vertices of a quadriteral whose area is <span class="math-container">$\frac{2}{3}$</span>. Since it is uniformly chosen the points, then</p>
<p><span class="math-container">$$ f_{X,Y}(x,y) = \frac{1}{\text{Area(S)}} = \frac{1}{2/3} = \frac{3}{2} $$</span></p>
<p>Where <span class="math-container">$(x,y) \in S $</span>, the quadrilateral. Hence,</p>
<p><span class="math-container">$$ f_X(x) = \int\limits_0^2 \frac{2}{3} dy = \frac{4}{3} $$</span></p>
<p>and</p>
<p><span class="math-container">$$ f_Y(y) = \int\limits_0^1 \frac{2}{3} = \frac{2}{3} $$</span></p>
<p>Is this correct?</p>
| Siong Thye Goh | 306,553 | <p>This is the region:</p>
<p><a href="https://i.stack.imgur.com/kpASM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kpASM.png" alt="enter image description here"></a></p>
<p>The area should be <span class="math-container">$\frac32$</span>.
For <span class="math-container">$x \in (0,1)$</span>,
<span class="math-container">$$f_X(x) = \int_0^{2-x} \frac23 \, dy$$</span>
For <span class="math-container">$y \in (0,2)$</span>,
<span class="math-container">$$f_Y(y) = \int_0^{\min(1,2-y)} \frac23 \, dx$$</span></p>
|
7,761 | <p>Our undergraduate university department is looking to spruce up our rooms and hallways a bit and has been thinking about finding mathematical posters to put in various spots; hoping possibly to entice students to take more math classes. We've had decent success in finding "How is Math Used in the Real World"-type posters (mostly through AMS), but we've been unable to find what I would call interesting/informative math posters. </p>
<p>For example, I remember seeing a poster once (put out by Mathematica) that basically laid out how to solve general quadratics, cubics, and quartics. Then it had a good overview of proving that no formula existed for quintics. So not only was it pretty to look at, but if you stopped to read it you actually learned something.</p>
<p>Does anyone know of a company or distributor that carries a variety of posters like this? </p>
<p>I've tried searching online but all that comes up is a plethora of posters of math jokes. And even though the application/career-based posters are nice and serve a purpose, I don't feel like you actually gain mathematical knowledge by reading them.</p>
| Marius Kempe | 369 | <p>The German organization <em>Imaginary</em> made a lovely set of posters showing algebraic surfaces. You can <a href="http://www.imaginary-exhibition.com/poster.php" rel="noreferrer">buy a set</a>, though they're in German. Rice University maths department has them on hanging in the corridors. Here's an example:</p>
<p><img src="https://i.stack.imgur.com/N28sv.png" alt="enter image description here"></p>
<p>There's an English translation available on their website, though it's not very idiomatic; I've sometimes thought of retranslating it, and I would if there was demand.</p>
|
7,761 | <p>Our undergraduate university department is looking to spruce up our rooms and hallways a bit and has been thinking about finding mathematical posters to put in various spots; hoping possibly to entice students to take more math classes. We've had decent success in finding "How is Math Used in the Real World"-type posters (mostly through AMS), but we've been unable to find what I would call interesting/informative math posters. </p>
<p>For example, I remember seeing a poster once (put out by Mathematica) that basically laid out how to solve general quadratics, cubics, and quartics. Then it had a good overview of proving that no formula existed for quintics. So not only was it pretty to look at, but if you stopped to read it you actually learned something.</p>
<p>Does anyone know of a company or distributor that carries a variety of posters like this? </p>
<p>I've tried searching online but all that comes up is a plethora of posters of math jokes. And even though the application/career-based posters are nice and serve a purpose, I don't feel like you actually gain mathematical knowledge by reading them.</p>
| Karl | 4,668 | <p>This is a slightly left of field answer but have you thought of no posters but something interactive instead? I'd go for installing whiteboards with a hanging pen and stick a problem of the week on it. The questions could be made to be colourful, inspiring and relevant to what you're teaching. The most creative solutions to be discussed in class. Better still have the students create the problem! Or question for the professor? In my classroom all the walls have magnetic boards so that the students can do puzzles on them. I also have magnetic matches for sequences and flip over answers to simple questions.</p>
|
2,629,744 | <p>I have done the sum by first plotting the graph of the function in the Left Hand Side of the equation and then plotted the line $y=k$. For the equation to have $4$ solutions, both these two curves must intersect at $4$ different points, and from the two graphs, I could see that for the above to occur, the value of $k$ must lie between $\cfrac{1}{4}$ and $6$, that is k belongs to $( 0.25,6)$. Thus , the integral values of $k$ would be $1,2,3,4,5$; and so the number of integral values of $k = 5$. This was the answer given in the book. But as I was looking at the graph, I realized that $k=0$ also gives $4$ solutions namely $x=-2,-3,2,3$. So, should the number of integral values of $k$ be $6$ $(0,1,2,3,4,5)$? </p>
| csar | 446,038 | <p>The closed form is $a_n=\begin{cases}a\in\mathbb{R},&n=1\\1,&n\geq2\end{cases}$</p>
<p>It can be verified by substituting the first few $n$'s in the formula (also very easy to show by induction).</p>
|
4,375,994 | <blockquote>
<p>Question:</p>
<p>Show that, <span class="math-container">$$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$</span></p>
</blockquote>
<p><em>My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.</em></p>
<p>Proof: <span class="math-container">$$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})
$$</span><span class="math-container">$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$</span></p>
<p>As <span class="math-container">$\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$</span></p>
<p>The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.</p>
<p>I.e. <span class="math-container">$$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$</span></p>
<p><span class="math-container">$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$</span></p>
<p>and the RHS will reduce down to <span class="math-container">$\sqrt{3}$</span>. Hence LHS=RHS.</p>
<p>Some things that I've noticed about this method of proof:</p>
<ul>
<li>It could be used to (incorrectly) prove that <span class="math-container">$$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$</span></li>
</ul>
<p>So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?</p>
<ul>
<li>Instead of proving (*), wouldn't this method of proof actually prove that? <span class="math-container">$$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$</span></li>
</ul>
<p>for some <span class="math-container">$k\in \mathbb{Z}$</span> which we must find. In this case being when <span class="math-container">$k=0$</span>.</p>
| José Carlos Santos | 446,262 | <p>You have<span class="math-container">$$3\arccos\left(\frac5{\sqrt{28}}\right)\in[0,3],$$</span>since <span class="math-container">$\frac{\sqrt3}2<\frac5{\sqrt{28}}<1$</span>, and therefore<span class="math-container">$$3>3\frac\pi6>3\arccos\left(\frac5{\sqrt{28}}\right)>0.$$</span>You also have<span class="math-container">$$0\leqslant3\arctan\left(\frac{\sqrt3}2\right)<3\arctan\left(\sqrt3\right)=\pi,$$</span>and therefore<span class="math-container">$$\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\in\left[0,\frac\pi3+1\right].$$</span>What you did shows that<span class="math-container">$$\tan\left(\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\right)=\tan\left(\frac\pi3\right).$$</span>But the only number in <span class="math-container">$\left[0,\frac\pi3+1\right]$</span> whose tangent is <span class="math-container">$\tan\left(\frac\pi3\right)$</span> is <span class="math-container">$\frac\pi3$</span>. So, your proof is indeed correct.</p>
|
4,601,727 | <p>I'm aware that the title might be a bit off, I am unsure on how to describe this.</p>
<p>For <span class="math-container">$n\in \mathbb{N}$</span>, define <span class="math-container">$n+1$</span> independent random variables <span class="math-container">$X_0, \ldots , X_n$</span> which are uniformly distributed over the interval <span class="math-container">$[0,1]$</span>, We focus on this group:<span class="math-container">$$S=\{X_i|i\geq 1 , X_i<X_0\}$$</span>
For all <span class="math-container">$0\leq k\leq n$</span>, show that: <span class="math-container">$$P(|S|=k)=\int_0^1 {n\choose k}\cdot x^k\cdot (1-x)^{n-k}$$</span></p>
<p>I've reduced this to the following:
<span class="math-container">$$P(|S|=k)=P(exactly \ k \ elements\ are\ bigger\ than\ X_0)$$</span>
Due to independence, we can write:
<span class="math-container">$$=P(X_0 <X_i)\cdot \ \ldots \ \cdot P(X_0<X_{i+k}) , i\in\{1, \ldots ,n-k\}$$</span></p>
<p>I'm stuck here, cant find how to calculate <span class="math-container">$P(X_0<X_i)$</span>, which prevents me from proving the statement.</p>
<p>I'm aware that <span class="math-container">$n\choose k$</span> is because we're 'checking' every k-sized group out of the n available R.Vs,</p>
<p><span class="math-container">$x^k%$</span> which translates to <span class="math-container">$P(X_0<X_{i\rightarrow (i+k)})$</span> and <span class="math-container">$(1-x)^{n-k}$</span> to 'disable' the other R.Vs from being bigger than <span class="math-container">$X_0$</span></p>
| Snoop | 915,356 | <p>The random variables <span class="math-container">$Y_\ell=\mathbf{1}_{[0,X_0)}(X_\ell),\,1\leq \ell \leq n$</span> indicate if <span class="math-container">$X_\ell<X_0$</span>. Then by total expectation:
<span class="math-container">$$P(|S|=k)=E\bigg[P\bigg(\sum_{1\leq \ell \leq n}Y_\ell=k\bigg|X_0\bigg)\bigg]=\int_{[0,1]}\underbrace{\binom{n}{k}x^{k}(1-x)^{n-k}}_{P(\sum_{1\leq \ell \leq n}Y_\ell=k|X_0=x)}dx$$</span></p>
<hr />
<p>To see that <span class="math-container">$P(\sum_{1\leq \ell \leq n}Y_\ell=k|X_0=x)=\binom{n}{k}x^{k}(1-x)^{n-k}$</span>, recall that if <span class="math-container">$W,Z$</span> are independent, then <span class="math-container">$E[f(W,Z)|W]=E[f(w,Z)]|_{w=W}$</span> for admissible <span class="math-container">$f$</span> by <a href="https://math.stackexchange.com/questions/768648/a-question-about-conditional-expectations?noredirect=1">this result</a>. In our case, <span class="math-container">$Z=(X_1,...,X_n)$</span>, <span class="math-container">$W=X_0$</span> and <span class="math-container">$f(w,z)=\mathbf{1}_{\{k\}}(\sum_{1\leq \ell \leq n}\mathbf{1}_{[0,w]}(z_\ell))$</span>. So
<span class="math-container">$$E\bigg[\mathbf{1}_{\{k\}}\bigg(\sum_{1\leq \ell \leq n}\mathbf{1}_{[0,x]}(X_\ell)\bigg)\bigg]=P\bigg(\sum_{1\leq \ell \leq n}\mathbf{1}_{[0,x]}(X_\ell)=k\bigg)=\binom{n}{k}x^k(1-x)^{n-k}$$</span>
Because <span class="math-container">$\mathbf{1}_{[0,x]}(X_\ell)$</span> are independent Bernoulli rvs with probability of success <span class="math-container">$x$</span>.</p>
|
3,137,599 | <p>I actually have a doubt about the solution of this question given in my book. It uses the equations tan 2A = - tan C (from A=B, A+B+C = 180 degrees) and 2 tan A + tan C = 100, thereby formulating the cubic equation <span class="math-container">$x^3 - 50x^2 + 50=0$</span>. The discriminant is <span class="math-container">$3x^2- 100x$</span>. Since <span class="math-container">$f(0) f(100/3) <0 $</span>, it is found there are three distinct real roots. After this my book simply states that one of these roots is obtuse so the other two values are taken. Why is one of the values of A obtuse when we have used 2A + C = 180 degrees and an obtuse value of A would not satisfy this equation? Would someone please help?</p>
| J.G. | 56,861 | <p>Since <span class="math-container">$A$</span> is the repeated angle, <span class="math-container">$A$</span> is acute so <span class="math-container">$x:=\tan A>0$</span>. We must thus count the positive roots of <span class="math-container">$x^3-50x^2+50=0$</span>. The roots have a negative product, <span class="math-container">$-50$</span>, so there's one negative root plus two positive ones**. (They can't all be negative because their sum is <span class="math-container">$50$</span>).</p>
<p>** Or a conjugate pair of non-real complex roots, which would result in no solutions to the original problem. But we can verify from a sign change on <span class="math-container">$[1, 2]$</span> that <span class="math-container">$x^3-50x^2+50$</span> has at least one positive root, so in fact it has two.</p>
|
3,378,004 | <p>If <span class="math-container">$H$</span> and <span class="math-container">$K$</span> are abelian subgroups of a group <span class="math-container">$G$</span>, then <span class="math-container">$H\cap K$</span> is a normal subgroup of <span class="math-container">$\left\langle H\cup K\right\rangle$</span>.</p>
<p>I proved <span class="math-container">$H\cap K$</span> is a subgroup and need to prove it is normal subgroup of <span class="math-container">$\left\langle H\cup K\right\rangle$</span>. But isn't it obvious that <span class="math-container">$H\cap K$</span> is normal? All the elements in <span class="math-container">$H\cap K$</span> communicates with all the elements in <span class="math-container">$H$</span> and <span class="math-container">$K$</span>, hence <span class="math-container">$H\cap K$</span> is normal. Am I missing something?</p>
| egreg | 62,967 | <p>Suppose <span class="math-container">$L$</span> is a subgroup of <span class="math-container">$\langle X\rangle$</span>, where <span class="math-container">$X$</span> is a subset of a group <span class="math-container">$G$</span>. In order to show that <span class="math-container">$L$</span> is normal in <span class="math-container">$\langle X\rangle$</span>, it is sufficient to prove that, for every <span class="math-container">$z\in L$</span> and <span class="math-container">$x\in X$</span>, <span class="math-container">$xzx^{-1}\in L$</span>.</p>
<p>This is because, defining <span class="math-container">$\varphi_x\colon G\to G$</span> by <span class="math-container">$\varphi_x(g)=xgx^{-1}$</span>, we have <span class="math-container">$\varphi_{xy}=\varphi_x\circ\varphi_y$</span> and <span class="math-container">$\varphi_x^{-1}=\varphi_{x^{-1}}$</span>.</p>
<p>In your case <span class="math-container">$X=H\cup K$</span>; if <span class="math-container">$x\in H$</span>, then <span class="math-container">$xzx^{-1}=z$</span>, as this operation is inside <span class="math-container">$H$</span>. Similarly if <span class="math-container">$x\in K$</span>.</p>
|
153,902 | <p>Let $A_i$ be open subsets of $\Omega$. Then $A_0 \cap A_1$ and $A_0 \cup A_1$ are open sets as well.</p>
<p>Thereby follows, that also $\bigcap_{i=1}^N A_i$ and $\bigcup_{i=1}^N A_i$ are open sets.</p>
<p>My question is, does thereby follow that $\bigcap_{i \in \mathbb{N}} A_i$ and $\bigcup_{i \in \mathbb{N}} A_i$ are open sets as well?</p>
<p>And what about $\bigcap_{i \in I} A_i$ and $\bigcup_{i \in I} A_i$ for uncountabe $I$?</p>
| Cameron Buie | 28,900 | <p>Any union of a set of open sets is again open. However, infinite intersections of open sets need not be open. For example, the intersection of intervals $(-1/n,1/n)$ on the real line (for positive integers $n$) is precisely the singleton $\{0\}$, which is not open.</p>
|
46,631 | <p>I'm writing a program to play a game of <a href="http://en.wikipedia.org/wiki/Pente" rel="noreferrer">Pente</a>, and I'm struggling with the following question:</p>
<blockquote>
<p>What's the best way to detect patterns on a two-dimensional board?</p>
</blockquote>
<p>For example, in Pente a pair of neighboring stones of the same color can be captured when they are flanked from both sides by an opponent; how can we find all the stones that can be captured with the next move for the following board?</p>
<p><img src="https://i.stack.imgur.com/cV3Gb.png" alt="sample board"></p>
<p>Below I show one possible straightforward solution, but with a defect: it's hard to extend it for other interesting patterns, i.e. three stones of the same color in a row surrounded by empty spaces, or four stones of the same color in a row which are flanked from one side but open from another, etc.</p>
<blockquote>
<p>I'm wondering whether there is a way to define a DSL for detecting 2-dimensional structures like that on a board - sort of a <em>2D pattern matching</em>.</p>
</blockquote>
<p>P.S. I would also appreciate any advice on how to simplify the code below and make it more idiomatic - for example, I don't really like the way how <code>sortStones</code> is defined.</p>
<h2>Straightforward solution</h2>
<p>Here is one way to solve this problem (see below for graphics primitives to generate and display random boards):</p>
<ul>
<li>Enumerate all subsets of 3 stones from the board above</li>
<li>Select those that form an <em>AABE</em> or <em>ABBE</em> pattern, where E denotes an unoccupied space</li>
</ul>
<p>Lets store the board as a list of black and white stones,</p>
<pre><code>a = {black[2, 1], black[4, 3], black[2, 5], black[4, 2], black[5, 3],
black[1, 2], black[1, 3], black[5, 4], black[1, 5], white[3, 1],
white[4, 1], white[4, 4], white[3, 5], white[3, 4], white[5, 1],
white[5, 2], white[3, 3], white[1, 1]}
</code></pre>
<p>First, we define <code>isTriple</code> which checks whether three stones sorted by their x and y coordinates are in the same row next to each other and follow an ABB or AAB pattern:</p>
<pre><code>isTriple[{a_, b_, c_}] := And[
(* A A B or A B B *)
Head[a] != Head[c] /. {black -> 1, white -> 0},
(* x and y coordinates are equally spaced *)
a[[1]] - b[[1]] == b[[1]] - c[[1]],
a[[2]] - b[[2]] == b[[2]] - c[[2]],
(* and are next to each other *)
Abs[a[[1]] - b[[1]]] <= 1,
Abs[a[[2]] - b[[2]]] <= 1]
</code></pre>
<p>Next, we determine the coordinates and the color of the stone that will kill the pair:</p>
<pre><code>killerStone[{a_, b_, c_}] :=
If[Head[a] == Head[b] /. {black -> 1, white -> 0},
Head[c][2 a[[1]] - b[[1]], 2 a[[2]] - b[[2]]],
Head[a][2 c[[1]] - b[[1]], 2 c[[2]] - b[[2]]]]
</code></pre>
<p>Finally, we only select those triples where killer stone's space is not already occupied:</p>
<pre><code>sortStones[l_] :=
Sort[l, OrderedQ[{#1, #2} /. {black -> List, white -> List}] &]
triplesToKill[board_] := Module[
{triples = Select[sortStones /@ Subsets[board, {3}], isTriple]},
Select[triples,
Block[
{ks = killerStone[#]},
FreeQ[board, _[ks[[1]], ks[[2]]]]] &]]
displayBoard[a, #] & /@ triplesToKill[a] //
Partition[#, 3, 3, {1, 1}, {}] & // GraphicsGrid
</code></pre>
<p><img src="https://i.stack.imgur.com/QHj2c.png" alt="straightforward solution"></p>
<h2>Graphics primitives</h2>
<pre><code>randomPoints[n_] := RandomSample[Block[{nn = Ceiling[Sqrt[n]]},
Flatten[Table[{i, j}, {i, 1, nn}, {j, 1, nn}], 1]], n];
(* n is number of moves = 2 * number of points *)
randomBoard[n_] := Module[
{points = randomPoints[2 n]},
Join[
Take[points, n] /. {x_, y_} -> black[x, y],
Take[points, -n] /. {x_, y_} -> white[x, y]
]]
grid[minX_, minY_, maxX_, maxY_] :=
Line[Join[
Table[{{minX - 1.5, y}, {maxX + 1.5, y}}, {y, minY - 1.5, maxY + 1.5,
1}],
Table[{{x, minY - 1.5}, {x, maxY + 1.5}}, {x, minX - 1.5, maxX + 1.5,
1}]]];
displayBoard[board_] := Module[
{minX = Min[First /@ board], maxX = Max[First /@ board],
minY = Min[#[[2]] & /@ board], maxY = Max[#[[2]] & /@ board], n},
Graphics[{
grid[minX, minY, maxX, maxY],
board /. {
black[n__] -> {Black, Disk[{n}, .4]},
white[n__] -> {Thick, Circle[{n}, .4], White, Disk[{n}, .4]}
}}, ImageSize -> Small, Frame -> True]];
displayBoard[board_, points_] := Show[
displayBoard[board],
Graphics[
Map[{Red, Disk[{#[[1]], #[[2]]}, .2]} &, points]]]
</code></pre>
| Victor K. | 1,351 | <p>Here is my own <strong>rough</strong> answer - it turns out that asking a question on SE helps clarifying one's thinking! I would still appreciate if some of the experts can weigh in.</p>
<p>First, we'll store the board as a square matrix of symbols <code>B</code>, <code>W</code> and <code>"."</code>:</p>
<pre><code>m = Partition[RandomChoice[{B, W, "."}, 25], 5] // MatrixForm
</code></pre>
<p>$\left(
\begin{array}{ccccc}
W & . & B & B & W \\
W & . & B & . & . \\
W & B & W & B & W \\
W & B & . & W & . \\
W & . & . & . & W \\
\end{array}
\right)$</p>
<p>Next, we'll generate a list of all possible <em>segments</em>, that is, horizontal, vertical or diagonal subsets of the matrix of length $k$. For example, the above matrix has 12 segments of length 5 - all rows, all columns and two big diagonals, and $10+10+4+4=28$ segments of length 4.</p>
<pre><code>flatten1 := Flatten[#, 1] &
(* Give all segments of length k - horizontal, vertical and diagonal
- of a square matrix. Each segment is represented by a pair:
the elements themselves and their staring position and orientation in the matrix*)
segments[mat_, k_] := Module[{n = Length[mat]},
flatten1@Join[
(* vertical *)
Table[
{
mat[[i ;; i + k - 1, j]],
{i, j, vertical}
},
{i, n - k + 1}, {j, n}],
(* horizontal *)
Table[
{
mat[[i, j ;; j + k - 1]],
{i, j, horizontal}
},
{i, n}, {j, n - k + 1}],
(* diagonal SW *)
Table[
{
Table[mat[[i + x, j + x]], {x, 0, k - 1}],
{i, j, diagSW}
},
{i, n - k + 1}, {j, n - k + 1}],
(* diagonal NW *)
Table[
{
Table[mat[[i - x, j + x]], {x, 0, k - 1}], {
i, j, diagNW}},
{i, k, n}, {j, n - k + 1}]]]
</code></pre>
<p>For example,</p>
<pre><code>segments[m[[1 ;; 3, 1 ;; 3]], 2] // Grid
</code></pre>
<p>returns</p>
<p>$\left(
\begin{array}{cc}
\{W,W\} & \{1,1,\text{vertical}\} \\
\{.,.\} & \{1,2,\text{vertical}\} \\
\{B,B\} & \{1,3,\text{vertical}\} \\
\{W,W\} & \{2,1,\text{vertical}\} \\
\{.,B\} & \{2,2,\text{vertical}\} \\
\{B,W\} & \{2,3,\text{vertical}\} \\
\{W,.\} & \{1,1,\text{horizontal}\} \\
\{.,B\} & \{1,2,\text{horizontal}\} \\
\{W,.\} & \{2,1,\text{horizontal}\} \\
\{.,B\} & \{2,2,\text{horizontal}\} \\
\{W,B\} & \{3,1,\text{horizontal}\} \\
\{B,W\} & \{3,2,\text{horizontal}\} \\
\{W,.\} & \{1,1,\text{diagSW}\} \\
\{.,B\} & \{1,2,\text{diagSW}\} \\
\{W,B\} & \{2,1,\text{diagSW}\} \\
\{.,W\} & \{2,2,\text{diagSW}\} \\
\{W,.\} & \{2,1,\text{diagNW}\} \\
\{.,B\} & \{2,2,\text{diagNW}\} \\
\{W,.\} & \{3,1,\text{diagNW}\} \\
\{B,B\} & \{3,2,\text{diagNW}\} \\
\end{array}
\right)$</p>
<p>Finally, once we have all the segments, comparison to a pattern is easy - notice how in <code>matchPattern</code>, we generate all 4 patterns <code>{B,W,W,"."}</code>, <code>{W,B,B,"."}</code>, <code>{".",W,W,B}</code> and <code>{".",B,B,W}</code> from the pattern <code>{B,W,W,"."}</code> since our comparison is literal:</p>
<pre><code>(* match a single pattern *)
matchPattern1[p_] :=
Function[mat, Select[segments[mat, Length[p]], #[[1]] == p &]];
(* match multiple patterns *)
matchPattern2[p_] := Function[mat, matchPattern1[#][mat] & /@ p];
(* match all variations of a pattern *)
matchPattern[p_] :=
Function[mat,
flatten1[matchPattern2[{p, Reverse[p], p /. {W -> B, B -> W},
Reverse[p /. {W -> B, B -> W}]}][mat]]]
</code></pre>
<p>Now we can easily define a function to select all killable pairs:</p>
<pre><code>killablePair = matchPattern[{B, W, W, "."}];
</code></pre>
<p>and apply it to the above matrix</p>
<pre><code>killablePair[m]
</code></pre>
<blockquote>
<p>{{{".", B, B, W}, {1, 2, horizontal}}}</p>
</blockquote>
|
4,288,188 | <p>I am trying to obtain a formulae for a summation problem under section (d) given in a solutions manual for "Data Structures and Algorithm Analysis in C - Mark Allen Weiss", here's the screen shot</p>
<p><a href="https://i.stack.imgur.com/fMie6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fMie6.png" alt="enter image description here" /></a></p>
<p>As the pdf is protected i could not download it.
Here's my attempt at it.
Let <span class="math-container">$S_{N} = \sum_{i=0}^\infty \frac{i^N}{4^i}$</span>, then starting from <span class="math-container">$N = 4$</span> we have
<span class="math-container">$S_{0} = \frac{4}{3}, S_{1} = \frac{4}{9} , S_{2} = \sum_{i=0}^\infty \frac{2i + 1}{3*4^i}, S_{3} = \sum_{i=0}^\infty \frac{3i^2 + 3i + 1}{3*4^i},S_{4} = \sum_{i=0}^\infty \frac{4i^3 + 6i^2 + 4i + 1}{3*4^i}$</span>.</p>
<p>Using recursion, we have
<span class="math-container">$S_{0} = \frac{4}{3}, S_{1} = \frac{1}{3}S_{0} , S_{2} = \frac{2S_{1} + S_{0}}{3} = \frac{5}{9}S_{0} ,
S_{3} = \frac{3S_{2} + 3S_{1} + S_{0}}{3} = \frac{11}{9}S_{0}, S_{4} = \frac{4S_{3} + 6S_{2} + 4S_{1} + S_{0}}{3} = \frac{95}{27}S_{0}$</span>.</p>
<p>After cleaning up further we have
<span class="math-container">$$S_{0} = \frac{4}{3}, S_{1} = \frac{1}{3}S_{0} , S_{2} = \frac{5}{9}S_{0} ,
S_{3} = \frac{11}{9}S_{0}, S_{4} = \frac{95}{27}S_{0}$$</span>
I dont see a pattern emerging to make a formulae.I may be doing something wrong, any help is greatly appreciated.</p>
| Thomas Andrews | 7,933 | <p>The technique given in the question gives the recurrence:</p>
<p><span class="math-container">$$3S_N=\sum_{i}\frac{(i+1)^N-i^N}{4^i}=\sum_{j=0}^{N-1} \binom Nj S_j$$</span></p>
<p>This follows from the binomial theorem: <span class="math-container">$$(i+1)^N-i^n=\sum_{j=0}^{N}\binom Nj i^j$$</span></p>
<p>But “solving” this recurrence is very painful. I don’t know how to solve it. It’s not even clear the book is asking you to solve the recurrence.</p>
|
3,521,534 | <p>I tried solving a calculus problem and I got the right result, but I don't understand the solution provided at the end of the exercise. Even though I got the same answer, I would like to understand what's happening in the given solution aswell.</p>
<blockquote>
<p>Consider the function: <span class="math-container">$$\ f(x) = \begin{cases}
x^2+ax+b & x\leq 0 \\
x-1 & x>0 \\ \end{cases} \ $$</span> Find the antiderivatives of the function <span class="math-container">$f$</span> if they exist.</p>
</blockquote>
<p>The solution provided goes something like this:</p>
<blockquote>
<p>For <span class="math-container">$f$</span> to have antiderivatives the function <span class="math-container">$f$</span> must have the Darboux
property. (...Some calculations...), therefore <span class="math-container">$f$</span> has the Darboux
property if and only if <span class="math-container">$b = -1$</span> (I understood that now the function
is continuous, therefore it has an anitederivative). Using the
consequences of Lagrange's theorem on the intervals <span class="math-container">$(-\infty, 0)$</span> and
<span class="math-container">$(0, \infty)$</span> any antiderivative <span class="math-container">$F : \mathbb{R} \rightarrow
\mathbb{R}$</span> of <span class="math-container">$f$</span> has the form:</p>
<p><span class="math-container">$$ F(x) = \ \begin{cases}
\dfrac{x^3}{3} + a \dfrac{x^2}{2} - x + c_1 & x < 0 \\
\ c_2 & x=0 \\
\dfrac{x^2}{2} - x + c_3 & x>0 \end{cases} \ $$</span></p>
<p><span class="math-container">$F$</span> being differentiable, it is also continous, so <span class="math-container">$F(0) = c_2 = c_1 =
c_3 $</span>.</p>
<p>Therefore the antiderivatives of <span class="math-container">$f$</span> have the form:</p>
<p><span class="math-container">$$ F(x) = c + \ \begin{cases}
\dfrac{x^3}{3} + a \dfrac{x^2}{2} - x & x\leq 0 \\
\dfrac{x^2}{2} - x & x>0 \end{cases} \ $$</span></p>
</blockquote>
<p>Again, I got the same result, but I don't understand a lot of the work done above. </p>
<p>The first thing I didn't understand is the part where they say that <span class="math-container">$f$</span> has an antiderivative iff it has the Darboux property. I searched a bit online and I found that a function accepts antiderivatives only if it has the Darboux property. So I guess I have to accept that as a fact.</p>
<p>The second (and more important thing) that I didn't understand was the part where they said that they used the consequences of Lagrange's Theorem on the intervals <span class="math-container">$(-\infty, 0)$</span> and <span class="math-container">$(0, \infty)$</span> to find that first form of the antiderivative. What theorem are they refering to? How did they use it on those intervals? Why is there a separate case for <span class="math-container">$x = 0$</span> with an aditional constant, <span class="math-container">$c_2$</span>. I only used <span class="math-container">$2$</span> constants, why were there needed <span class="math-container">$3$</span>? Long story short, I just don't understand at all how they arrived at that first form of the antiderivative and how they used these "consequences of Langrange's theorem". I understood the second form of the antiderivative, that's what I also got, but the first form put me in the dark.</p>
<p>I know these are all just details, but I really want to understand what was used here, why was it used and how was it used.</p>
| Martin Argerami | 22,857 | <p>At zero, the function <span class="math-container">$f$</span> is either continuous or has a jump; by considering a small enough interval around <span class="math-container">$0$</span>, you can conclude that <span class="math-container">$f$</span> has to be continuous. That gives you <span class="math-container">$b=-1$</span>. </p>
<p>I assume that by "Lagrange's Theorem" they mean the Mean Value Theorem. And what they are likely using it ("the consequences of Lagrange's theorem") is to say that any two antiderivatives of <span class="math-container">$f$</span> differ by a constant (that's what lets you find the antiderivative as your are used to, and by adding a constant be sure that you have all possible antiderivatives). Because if a function <span class="math-container">$g$</span> has zero derivative on an interval <span class="math-container">$(a,b)$</span>, then for any <span class="math-container">$x,y\in (a,b)$</span> we have <span class="math-container">$$ g(x)-g(y)=g'(c)(x-y)=0,$$</span>so <span class="math-container">$g$</span> is constant. </p>
<p>Because the above reasoning requires open intervals, they cannot apply it at <span class="math-container">$0$</span>; that's why there are three cases initially. From there they work with continuity to reduce the constants. </p>
|
3,624,662 | <p><a href="https://i.stack.imgur.com/kwAMn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kwAMn.png" alt=" c"></a></p>
<p>In my mind, I can think of below example which seems to work.</p>
<p>If <span class="math-container">$(X,T) = \mathbb{R}$</span>, and <span class="math-container">$A = (0,\infty)$</span>, then as far as I know it comes in the standard (order) topology of <span class="math-container">$\mathbb{R}$</span>, but what I dont know is whether it will be a proper closed subset of <span class="math-container">$\mathbb{R}$</span>. If it works, then it will be perfect b/c for all positive integer <span class="math-container">$i$</span>, if I let <span class="math-container">$P_i$</span> be the open interval <span class="math-container">$(0,i)$</span>, then clearly <span class="math-container">$A \subset \cup$</span> <span class="math-container">$O_i$</span>, where <span class="math-container">$i = 1$</span> to <span class="math-container">$\infty$</span>, </p>
<p>however there doesnt exist <span class="math-container">$i_1, i_2,......, i_n$</span>, such that <span class="math-container">$A \subset (0,i_1),(0,i_2),....,(0,i_n)$</span>, therefore by the definition of compactness we can see that we dont have a finite sub-cover, so its not compact.</p>
<p>Kindly check my proof, let me know if there is anything wrong. Also, give it better style and notation if required. </p>
| Qurultay | 338,156 | <p>Noting that any bounded closed subset of <span class="math-container">$\mathbb{R}$</span> is compact, so we need to think of unbounded closed subsets like: <span class="math-container">$$[a,\infty), \mathbb{N}, \ldots$$</span></p>
|
1,356,545 | <p>Given a fair 6-sided die, how can we simulate a biased coin with P(H)= 1/$\pi$ and P(T) = 1 - 1/$\pi$ ?</p>
| Asinomás | 33,907 | <p>Throwing a dice $n$ times gives you a space of $6^n$ outcomes, so take an $n$ so that you can approximate $\frac{1}{\pi}$ by $\frac{m}{6^{n}}$ as precisely as required.</p>
<p>After this just pick $m$ of the $6^m$ results to represent heads and the other results represent tails.</p>
<p>The probability of getting tails is very close to $\frac{1}{\pi}$ (you can make it as close as you want).</p>
|
365,631 | <p>Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.</p>
<blockquote>
<p><strong>What is your favourite example of a proof of this type?</strong></p>
</blockquote>
<p>Here are some examples:</p>
<ul>
<li><p><strong>The probabilistic method in combinatorics</strong> As I understand it, a
typical pattern of argument is as follows. We have a set <span class="math-container">$X$</span> and want to
show that at least one element of <span class="math-container">$X$</span> has property <span class="math-container">$P$</span>. We choose some
function <span class="math-container">$f: X \to \{0, 1, \ldots\}$</span> such that <span class="math-container">$f(x) = 0$</span> iff <span class="math-container">$x$</span> satisfies
<span class="math-container">$P$</span>, and we choose a probability measure on <span class="math-container">$X$</span>. Then we show that
with respect to that measure, <span class="math-container">$\mathbb{E}(f) < 1$</span>. It follows that
<span class="math-container">$f^{-1}\{0\}$</span> has positive measure, and is therefore nonempty.</p>
</li>
<li><p><strong>Real analysis</strong> One example is <a href="http://www.artsci.kyushu-u.ac.jp/%7Essaito/eng/maths/Cauchy.pdf" rel="noreferrer">Banach's
proof</a>
that any measurable function <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> satisfying
Cauchy's functional equation <span class="math-container">$f(x + y) = f(x) + f(y)$</span> is linear.
Sketch: it's enough to show that <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span>, since then
it follows from additivity that <span class="math-container">$f$</span> is continuous everywhere, which makes
it easy. To show continuity at <span class="math-container">$0$</span>, let <span class="math-container">$\varepsilon > 0$</span>. An
argument using Lusin's theorem shows that for all sufficiently small
<span class="math-container">$x$</span>, the set <span class="math-container">$\{y: |f(x + y) - f(y)| < \varepsilon\}$</span> has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives <span class="math-container">$|f(x)| < \varepsilon$</span>.</p>
<p>Another example is the existence of real numbers that are
<a href="https://en.wikipedia.org/wiki/Normal_number" rel="noreferrer">normal</a> (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.</p>
</li>
<li><p><strong>Set theory</strong> Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set <span class="math-container">$X$</span> being a finitely additive <span class="math-container">$\{0, 1\}$</span>-valued
probability measure defined on the full <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$P(X)$</span>. Some
existence proofs work by proving that the subset of elements with the
desired property has measure <span class="math-container">$1$</span> in the ultrafilter, and is therefore nonempty.</p>
<p>One example is a proof that for every measurable cardinal
<span class="math-container">$\kappa$</span>, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a <span class="math-container">$\kappa$</span>-complete ultrafilter on <span class="math-container">$\kappa$</span>. Make an inspired choice of function <span class="math-container">$\kappa \to \{\text{cardinals } <
\kappa \}$</span>. Push the ultrafilter forwards along this function to give
an ultrafilter on <span class="math-container">$\{\text{cardinals } < \kappa\}$</span>. Then prove that the set
of inaccessible cardinals <span class="math-container">$< \kappa$</span> belongs to that ultrafilter ("has
measure <span class="math-container">$1$</span>") and conclude that, in particular, it's nonempty.</p>
<p>(Although it has a similar flavour, I would <em>not</em> include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an <em>explicit construction</em>
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)</p>
</li>
</ul>
<p>(Mathematicians being mathematicians, someone will probably observe that
<em>any</em> existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)</p>
<p><strong>PS</strong> I'm aware of the earlier question <a href="https://mathoverflow.net/questions/34390">On proving that a certain set is
not empty by proving that it is actually
large</a>. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on <em>positive measure</em>, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.</p>
| R W | 8,588 | <ol>
<li>The proofs of existence of expanders by Barzdin - Kolmogorov and Pinsker,</li>
</ol>
<p>and (somewhat related)</p>
<ol start="2">
<li>Gromov's proof of the existence of groups with no coarse embedding into a Hilbert space.</li>
</ol>
|
365,631 | <p>Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.</p>
<blockquote>
<p><strong>What is your favourite example of a proof of this type?</strong></p>
</blockquote>
<p>Here are some examples:</p>
<ul>
<li><p><strong>The probabilistic method in combinatorics</strong> As I understand it, a
typical pattern of argument is as follows. We have a set <span class="math-container">$X$</span> and want to
show that at least one element of <span class="math-container">$X$</span> has property <span class="math-container">$P$</span>. We choose some
function <span class="math-container">$f: X \to \{0, 1, \ldots\}$</span> such that <span class="math-container">$f(x) = 0$</span> iff <span class="math-container">$x$</span> satisfies
<span class="math-container">$P$</span>, and we choose a probability measure on <span class="math-container">$X$</span>. Then we show that
with respect to that measure, <span class="math-container">$\mathbb{E}(f) < 1$</span>. It follows that
<span class="math-container">$f^{-1}\{0\}$</span> has positive measure, and is therefore nonempty.</p>
</li>
<li><p><strong>Real analysis</strong> One example is <a href="http://www.artsci.kyushu-u.ac.jp/%7Essaito/eng/maths/Cauchy.pdf" rel="noreferrer">Banach's
proof</a>
that any measurable function <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> satisfying
Cauchy's functional equation <span class="math-container">$f(x + y) = f(x) + f(y)$</span> is linear.
Sketch: it's enough to show that <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span>, since then
it follows from additivity that <span class="math-container">$f$</span> is continuous everywhere, which makes
it easy. To show continuity at <span class="math-container">$0$</span>, let <span class="math-container">$\varepsilon > 0$</span>. An
argument using Lusin's theorem shows that for all sufficiently small
<span class="math-container">$x$</span>, the set <span class="math-container">$\{y: |f(x + y) - f(y)| < \varepsilon\}$</span> has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives <span class="math-container">$|f(x)| < \varepsilon$</span>.</p>
<p>Another example is the existence of real numbers that are
<a href="https://en.wikipedia.org/wiki/Normal_number" rel="noreferrer">normal</a> (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.</p>
</li>
<li><p><strong>Set theory</strong> Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set <span class="math-container">$X$</span> being a finitely additive <span class="math-container">$\{0, 1\}$</span>-valued
probability measure defined on the full <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$P(X)$</span>. Some
existence proofs work by proving that the subset of elements with the
desired property has measure <span class="math-container">$1$</span> in the ultrafilter, and is therefore nonempty.</p>
<p>One example is a proof that for every measurable cardinal
<span class="math-container">$\kappa$</span>, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a <span class="math-container">$\kappa$</span>-complete ultrafilter on <span class="math-container">$\kappa$</span>. Make an inspired choice of function <span class="math-container">$\kappa \to \{\text{cardinals } <
\kappa \}$</span>. Push the ultrafilter forwards along this function to give
an ultrafilter on <span class="math-container">$\{\text{cardinals } < \kappa\}$</span>. Then prove that the set
of inaccessible cardinals <span class="math-container">$< \kappa$</span> belongs to that ultrafilter ("has
measure <span class="math-container">$1$</span>") and conclude that, in particular, it's nonempty.</p>
<p>(Although it has a similar flavour, I would <em>not</em> include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an <em>explicit construction</em>
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)</p>
</li>
</ul>
<p>(Mathematicians being mathematicians, someone will probably observe that
<em>any</em> existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)</p>
<p><strong>PS</strong> I'm aware of the earlier question <a href="https://mathoverflow.net/questions/34390">On proving that a certain set is
not empty by proving that it is actually
large</a>. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on <em>positive measure</em>, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.</p>
| zeraoulia rafik | 51,189 | <p><a href="https://en.wikipedia.org/wiki/Zeta_function_universality" rel="nofollow noreferrer">Universality of Riemann zeta function</a> , Which related to the approximation of every Holomorphic function <span class="math-container">$f(z)$</span> by Riemann zeta function in the strip .</p>
<blockquote>
<p><strong>Corollary</strong>: Let <span class="math-container">$K_0$</span> be a compact set in the right half of the critical stripe <span class="math-container">$1/2< \Re z<1$</span>.
Let <span class="math-container">$f$</span> be a continuous function on <span class="math-container">$K_0$</span>, which is holomorphic on an open set containing <span class="math-container">$K_0$</span> and does not have any zeros in <span class="math-container">$K_0$</span> . For every <span class="math-container">$\epsilon_0>0$</span>, we have that the limit (lower density )
<span class="math-container">$$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big (\{ t\in[0,T]: \max\limits_{z \in K_0} \left| {\zeta(z+it) -f(z) )}\right| < \epsilon_0\Big\}) $$</span>
is positive for <span class="math-container">$\lambda$</span> being the Lebesgue measure.</p>
</blockquote>
|
365,631 | <p>Suppose we want to prove that among some collection of things, at least one
of them has some desirable property. Sometimes the easiest strategy is to
equip the collection of all things with a measure, then show that the set
of things with the desired property has positive measure. Examples of this strategy
appear in many parts of mathematics.</p>
<blockquote>
<p><strong>What is your favourite example of a proof of this type?</strong></p>
</blockquote>
<p>Here are some examples:</p>
<ul>
<li><p><strong>The probabilistic method in combinatorics</strong> As I understand it, a
typical pattern of argument is as follows. We have a set <span class="math-container">$X$</span> and want to
show that at least one element of <span class="math-container">$X$</span> has property <span class="math-container">$P$</span>. We choose some
function <span class="math-container">$f: X \to \{0, 1, \ldots\}$</span> such that <span class="math-container">$f(x) = 0$</span> iff <span class="math-container">$x$</span> satisfies
<span class="math-container">$P$</span>, and we choose a probability measure on <span class="math-container">$X$</span>. Then we show that
with respect to that measure, <span class="math-container">$\mathbb{E}(f) < 1$</span>. It follows that
<span class="math-container">$f^{-1}\{0\}$</span> has positive measure, and is therefore nonempty.</p>
</li>
<li><p><strong>Real analysis</strong> One example is <a href="http://www.artsci.kyushu-u.ac.jp/%7Essaito/eng/maths/Cauchy.pdf" rel="noreferrer">Banach's
proof</a>
that any measurable function <span class="math-container">$f: \mathbb{R} \to \mathbb{R}$</span> satisfying
Cauchy's functional equation <span class="math-container">$f(x + y) = f(x) + f(y)$</span> is linear.
Sketch: it's enough to show that <span class="math-container">$f$</span> is continuous at <span class="math-container">$0$</span>, since then
it follows from additivity that <span class="math-container">$f$</span> is continuous everywhere, which makes
it easy. To show continuity at <span class="math-container">$0$</span>, let <span class="math-container">$\varepsilon > 0$</span>. An
argument using Lusin's theorem shows that for all sufficiently small
<span class="math-container">$x$</span>, the set <span class="math-container">$\{y: |f(x + y) - f(y)| < \varepsilon\}$</span> has positive
Lebesgue measure. In particular, it's nonempty, and additivity then
gives <span class="math-container">$|f(x)| < \varepsilon$</span>.</p>
<p>Another example is the existence of real numbers that are
<a href="https://en.wikipedia.org/wiki/Normal_number" rel="noreferrer">normal</a> (i.e. normal to every base).
It was shown that almost all real numbers have this property
well before any specific number was shown to be normal.</p>
</li>
<li><p><strong>Set theory</strong> Here I take ultrafilters to be the notion of measure, an
ultrafilter on a set <span class="math-container">$X$</span> being a finitely additive <span class="math-container">$\{0, 1\}$</span>-valued
probability measure defined on the full <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$P(X)$</span>. Some
existence proofs work by proving that the subset of elements with the
desired property has measure <span class="math-container">$1$</span> in the ultrafilter, and is therefore nonempty.</p>
<p>One example is a proof that for every measurable cardinal
<span class="math-container">$\kappa$</span>, there exists some inaccessible cardinal strictly smaller than
it. Sketch: take a <span class="math-container">$\kappa$</span>-complete ultrafilter on <span class="math-container">$\kappa$</span>. Make an inspired choice of function <span class="math-container">$\kappa \to \{\text{cardinals } <
\kappa \}$</span>. Push the ultrafilter forwards along this function to give
an ultrafilter on <span class="math-container">$\{\text{cardinals } < \kappa\}$</span>. Then prove that the set
of inaccessible cardinals <span class="math-container">$< \kappa$</span> belongs to that ultrafilter ("has
measure <span class="math-container">$1$</span>") and conclude that, in particular, it's nonempty.</p>
<p>(Although it has a similar flavour, I would <em>not</em> include in this list the cardinal arithmetic proof of the
existence of transcendental real numbers, for two reasons. First,
there's no measure in sight. Second -- contrary to
popular belief -- this argument leads to an <em>explicit construction</em>
of a transcendental number, whereas the other arguments on this list
do not explicitly construct a thing with the desired properties.)</p>
</li>
</ul>
<p>(Mathematicians being mathematicians, someone will probably observe that
<em>any</em> existence proof can be presented as a proof in which the set of things
with the required property has positive measure. Once you've got a thing
with the property, just take the Dirac delta on it. But obviously I'm
after less trivial examples.)</p>
<p><strong>PS</strong> I'm aware of the earlier question <a href="https://mathoverflow.net/questions/34390">On proving that a certain set is
not empty by proving that it is actually
large</a>. That has some good
answers, a couple of which could also be answers to my question. But my
question is specifically focused on <em>positive measure</em>, and excludes
things like the transcendental number argument or the Baire category
theorem discussed there.</p>
| Ian Agol | 1,345 | <p><a href="https://arxiv.org/abs/1405.6410" rel="nofollow noreferrer">Lubotzky, Maher, and Wu showed</a> for any <span class="math-container">$n\in \mathbb{Z}, g\in \mathbb{N}$</span> the existence of homology 3-spheres of <a href="https://en.wikipedia.org/wiki/Heegaard_splitting" rel="nofollow noreferrer">Heegaard genus</a> <span class="math-container">$g$</span> and <a href="https://en.wikipedia.org/wiki/Casson_invariant" rel="nofollow noreferrer">Casson invariant</a> <span class="math-container">$n$</span> via a probabilistic argument. The idea is to take an appropriate subgroup of the Torelli subgroup of the mapping class group of a genus <span class="math-container">$g$</span> surface, and modify a Heegaard splitting of <span class="math-container">$S^3$</span> of genus <span class="math-container">$g$</span> by a random walk on this subgroup. On this subgroup, the Casson invariant is realized by a homomorphism to <span class="math-container">$\mathbb{Z}$</span>. Since random walks are recurrent, each integer is realized as a Casson invariant infinitely often. And they show that with probability tending to 1, the Heegaard genus is <span class="math-container">$g$</span>. Hence there exists manifolds with the desired invariants.</p>
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