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Calculate the derivate $dx/dy$ using $\int_0^x \sqrt{6+5\cos t} \, dt + \int_0^y \sin t^2 \, dt = 0$ I want to calculate $\frac{dx}{dy}$ using the equation below.
$$\int_0^x \sqrt{6+5\cos t}\;dt + \int_0^y \sin t^2\;dt = 0$$
I don't even know from where to start. Well I think that I could first find the integrals and then try to find the derivative. The problem with this approach is that I cannot find the result of the first integral.
Can someone give me a hand here?
| HINT: You have $$f(x)=\int_0^x\sqrt{6+5\cos t}\,dt=-\int_0^y\sin t^2 \,dt=g(y)\;.$$ What are $\dfrac{df}{dx}$ and $\dfrac{dg}{dy}$ according to the fundamental theorem? And when you have $\dfrac{df}{dx}$, what can you multiply it by to get $\dfrac{df}{dy}$?
| {
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Conditional probability. Targeting events
Electric motors coming off two assembly lines are pooled for storage in a common stockroom, and the room contains an equal number of motors from each line. Motors are periodically sampled from that room and tested. It is known that 10% of the motors from line I are defective and 15% of the motors from line II are defective. If a motor is randomly selected from the stock-room and found to be defective, find the probability that it came from line I.
Here is my way to solve it. First it is a conditional probability. The formula is
$$P(A \mid B) = \frac{P (A\cap B) }{ P(B) }.$$
$P(B)$ = probability that it came from line 1 = $2 P_1$.
Now here is where it gets interesting. What would be $P(A\cap B)$ in that case?
Is $P(A \cap B)=P(\text{came from line 1 * defective})$?
| P(B) is not P(came from line 1) in this problem. You are being asked to calculate P(came from line 1 | is defective) so B is "is defective" and A is "came from line 1". You're right that P(AB) is "Came from line 1 and is defective", and if you know how to calculate P(B) correctly in this case then you're essentially doing the same thing as you would if you used Bayes' Theorem.
To calculate P(B) correctly: P(B) needs to be the probability that any given motor in the entire factory is defective, not just from line I. Use the Law of Total Probability.
For P(AB): This is itself a conditional probability problem. Consider P(is defective | came from line 1) = P(defective and from line 1)/P(came from line 1)
| {
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Convergence of $\sum_{n=1}^\infty\frac{1}{2\cdot n}$ It is possible to deduce the value of the following (in my opinion) converging infinite series? If yes, then what is it?
$$\sum_{n=1}^\infty\frac{1}{2\cdot n}$$
where n is an integer. Sorry if the notation is a bit off, I hope youse get the idea.
| The series is not convergent, since it is half of the harmonic series which is known to be divergent$^1$.
$$\sum_{n=1}^{\infty }\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^{\infty }\frac{1}{n}.$$
--
$^1$ The sum of the following $k$ terms is greater or equal to $\frac{1}{2}$
$$\frac{1}{k+1}+\frac{1}{k+2}+\ldots +\frac{1}{2k-1}+\frac{1}{2k}\geq
k\times \frac{1}{2k}=\frac{1}{2},$$
because each term is greater or equal to $\frac{1}{2k}$.
| {
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Equation of straight line I know,
$Ax + By = C$
is the equation of straight line but a different resource says that:
$y = mx + b$
is also an equation of straight line? Are they both same?
| Yes. That is, they both give the equation of a straight line and the equation of any non-vertical line can be written in either form.
If $B\ne 0$. Then you can write $Ax+By=C$ as
$$
By=-Ax+C
$$
and, since $B\ne0$, the above can be written
$$
y=-\textstyle{A\over B}x +{C\over B}.
$$
If $B=0$, the equation is $Ax=C$, which is a vertical line when $A\ne0$. In this case you can't write it in the form $y=mx+b$ (which defines a function).
On the other hand, given $y=mx+b$, you can rewrite it as $-mx+y=b$.
Note that for the equation $Ax+By=C$ with $A$ and $B$ both non-zero:
The $y$-intercept of its graph is $C/B$ and is found by taking $x=0$.
The $x$-intercept is of its graph is $C/A$ and is found by taking $y=0$.
The slope of the line is then $ {C/B-0\over 0-C/A } = -A/B$.
| {
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When is $a^k \pmod{m}$ a periodic sequence? Let $a$ and $m$ be a positive integers with $a < m$. Suppose that $p$ and $q$ are prime divisors of $m$. Suppose that $a$ is divisible by $p$ but not $q$.
Is there necessarily an integer $k>1$ such that $a^k \equiv a \pmod{m}$?
Or is it that the best we can do is say there are $n>0$ and $k>1$ such that $a^{n+k} \equiv a^n \pmod{m}$
What can be said about $n$ and $k$?
EDIT: Corrected to have $k>1$ rather than $k>0$.
EDIT: The following paper answers my questions about $n$ and $k$ very nicely.
A. E. Livingston and M. L. Livingston, The congruence $a^{r+s} \equiv a^r \pmod{m}$, Amer. Math. Monthly $\textbf{85}$ (1978), no.2, 97-100.
It is one of the references in the paper Math Gems cited. Arturo seems to say essentially the same thing in his answer.
| A nice presentation of such semigroup generalizations of the Euler-Fermat theorem and related number theory is the following freely available paper
S. Schwarz, The role of semigroups in the elementary theory of numbers, Math. Slovaca, Vol. 31 (1981) pp. 369–395.
| {
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Finding a third point I have learnt that if we are given 3 points in the extended complex plane and their corresponding image points, we have a unique Möbius map that can perform the mapping. Suppose I have 2 orthogonally intersecting circles and I want to map them (individually) to the real and imaginary axes respectively by some Möbius map, is there a systematic way to do so? I have figured that the intersection points will have to be sent to $0$ and $\infty$ respectively but how might I determine a third point and its image so as to define such a map?
| Let one of the intersections be $p$. The inversion $z \to 1/(z-p)$ takes $p$ to $\infty$ and takes your two circles to straight lines intersecting orthogonally. Now just translate and rotate.
| {
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Numerical analysis textbooks and floating point numbers What are some recommended numerical analysis books on floating point numbers? I'd like the book to have the following
*
*In depth coverage on the representation of floating point numbers on modern hardware (the IEEE standard).
*How to do arbitrary precision floating point calculations with a reasonably fast modern algorithm.
*How to compute the closest 32-bit floating point representation of a dot product and cross product. And do this fast, so no relying on generic arbitrary precision calculations to get the bits of the 32-bit floating point number right.
From what I can infer from doing some searches most books tend to focus on stuff like the runge kutta and not put much emphasis on how to make floating point calculations that are ultra precise.
| You could try the book written by J.M. Muller, N. Brisebarre:
*
*Handbook of Floating Point Arithmetic (amazon.com)
The literature of numerical mathematics concentrates on algorithms for mathematical problems, not on implementation issues of arithmetic operations.
How to compute the closest 32-bit floating point representation of a dot product and cross product.
Since these are concatenations of addition and multiplication, I expect that you won't find much about dot and cross products themselves.
| {
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Easiest way to perform Euclid's division algorithm for polynomials Let's say I have the two polynomials $f(x) = x^3 + x + 1$ and $g(x) = x^2 + x$ over $\operatorname{GF}(2)$ and want to perform a polynomial division in $\operatorname{GF}(2)$.
What's the easiest and most bullet proof way to find the quotient $q(x) = x + 1$ and the remainder $r(x)=1$ by hand?
The proposal by the german edition of Wikipedia is rather awkward.
| Polynomial long division is the way to go. Especially over a finite field where you don't have to worry about fractional coefficients (working over for instance the rational numbers these can get extremely unwieldy surprisingly soon). Over $\mathbb Z/2\mathbb Z$ you don't even have to worry about dividing coefficients at all, the only question to be answered is "to substract or not to subtract", where as a bonus subtraction is actually the same as addition.
Note that the wikipedia article you refer to does not assume such a simple context, and avoids division by coefficients by doing a pseudo-division instead (for which instead of explosion of fractions you can get enormous coefficients).
| {
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Formulas for counting pairs in a set I have a few questions regarding Cartesian products that will help me optimize a complicated SQL query I'm working on.
Suppose I have 52 playing cards, and I want to know how many combinations of pairs (first two cards) a dealer can draw at the beginning. Obviously, this would be less than $52*52$ since the dealer cannot draw the same card twice. So, to me it seems the answer is $(52*52) - 52$, since there's 52 "pairs" of the same card, in other words $52*51$.
However, I'd like to better understand the math behind this so I can apply it to any number of cards and any size sets:
*
*Given n cards, how many ordered sets of y cards can be created?
For example, if I had 100,000 cards, how many unique sets of 10 cards could I make?
*Given n cards, how many unordered sets of y cards can be created?
For example, if I had 100 cards, how many unique unordered sets of 3 could I make?
What's the mathematical formula that represents both these answers? Thanks!
| The concepts you are looking for are known as "permutations" and "combinations."
*
*If you have $n$ items, and you want to count how many ordered $r$-tuples you can make without repetitions, the answer is someimtes written $P^n_r$, and:
$$P^{n}_{r} = n(n-1)(n-2)\cdots (n-r+1).$$
This follows from the "multiplication rule": if event $A$ can occur in $p$ ways, and event $B$ can occur in $q$ ways, then the number of ways in which both events $A$ and $B$ can occur is $pq$.
Your answer of $52\times 51$ for ordered pairs of playing cards is correct if you care about which one is the first card and which one is the second. Another way to see this is that there are 52 possible ways in which the first card is dealt; and there are 51 ways for the second card to be dealt (as there are only 51 cards left).
*If you don't care about the order, then you have what are called "combinations" (without repeititons). The common symbol is
$$\binom{n}{k}$$
which is pronounces "$n$ choose $k$". The symbol represents the number of ways in which you can select $k$ elements from $n$ possibilities, without repetition. In other words, the number of ways to choose subsets with $k$ elements from a set with $n$ elements. The formula is
$$\binom{n}{k}=\frac{n!}{k!(n-k)!},\quad \text{if }0\leq k\leq n$$
where $n! = n\times (n-1)\times\cdots\times 2\times 1$.
To see this, note that there are $\frac{n!}{(n-k)!} = n(n-1)\cdots(n-k+1)$ ways of selecting $k$ items if you do care about the order. But since we don't care about the order, how many times did we pick each subset? For instance, a subset consisting of $1$, $2$, and $3$ is selected six times: once as 1-2-3, once as 1-3-2, once as 2-1-3, once as 2-3-1, once as 3-1-2, and once as 3-2-1.
Well, there are $k$ items, and so there are $P^k_k$ ways of ordering them; this is exactly $k!$ ways. So we counted each $k$-subset $k!$ ways. So the final answer is $\frac{n!}{(n-k)!}$ divided by $k!$, giving the formula above,
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$$
See also this previous question and answer for the general principles and formulas.
| {
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How to prove that for $A\cap B\neq\varnothing$, $(\bigcap A)\cap(\bigcap B)\subseteq\bigcap(A\cap B)$? $A$ and $B$ are non empty sets with non empty intersection.
Prove that
$(\bigcap A)\cap(\bigcap B) \subseteq \bigcap (A\cap B).$
The definition of intersection of a set is something like this, if $M$ is a nonempty set whose elements are themselves sets, then $x$ is an element of the intersection of $M$ if and only if for every element $A$ of $M$, $x$ is an element of $A$.
| For theorems like these, as Asaf wrote, expanding definitions and simplifying is the way to go. However, I do these kind of things more 'calculationally' using the rules of predicate logic.
In this case, we can easily calculate the elements $\;x\;$ of the left hand side:
\begin{align}
& x \in \bigcap A \;\cap\; \bigcap B \\
\equiv & \qquad\text{"definition of $\;\cap\;$; definition of $\;\bigcap\;$, twice"} \\
& \langle \forall V : V \in A : x \in V \rangle \;\land\; \langle \forall V : V \in B : x \in V \rangle \\
\equiv & \qquad\text{"logic: merge ranges of $\;\forall\;$ statements -- to simplify"} \\
& \langle \forall V : V \in A \lor V \in B : x \in V \rangle \\
\end{align}
And similarly for the right hand side:
\begin{align}
& x \in \bigcap (A \cap B) \\
\equiv & \qquad\text{"definition of $\;\bigcap\;$; definition of $\;\cap\;$"} \\
& \langle \forall V : V \in A \cap B : x \in V \rangle \\
\equiv & \qquad\text{"definition of $\;\cup\;$"} \\
& \langle \forall V : V \in A \land V \in B : x \in V \rangle \\
\end{align}
These two results look promisingly similar. We see that latter range implies the former, and predicate logic tells us that
$$
\langle \forall z : P(z) : R(z) \rangle \;\Rightarrow\; \langle \forall z : Q(z) : R(z) \rangle
$$
if $\;Q(z) \Rightarrow P(z)\;$ for all $\;z\;$. In our specific case, that means
\begin{align}
& \langle \forall V : V \in A \lor V \in B : x \in V \rangle \\
\Rightarrow & \qquad \text{"using the above rule, with $\;P \land Q \;\Rightarrow\; P \lor Q\;$"} \\
& \langle \forall V : V \in A \land V \in B : x \in V \rangle \\
\end{align}
Putting this all together, with the definition of $\;\subseteq\;$, tells us that
$$
\bigcap A \;\cap\; \bigcap B \;\subseteq\; \bigcap (A \cap B)
$$
which is what we set out to prove.
| {
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Tensoring with vector bundle is a dense endofunctor of $D^b(\text{coh }X) $? A functor $F:T\to R$ between triangulated categories is dense if every object of $R$ is isomorphic to a direct summand in the image of $F$.
Let $R=T=D^b(\text{coh }X)$ for a variety $X$ and consider the functor $-\otimes \mathcal{V}$, $\mathcal{V}$ a vector bundle.
I do not understand the following
claim: "$-\otimes\mathcal{V}$ is a is a dense functor, as any object $P\in D^b(\text{coh }X)$ is a summand of $(P\otimes V^\vee)\otimes V$."
Can anyone help?
| What part of the claim you don't understand?
For any vector bundle $V$ the bundle $V\otimes V^\vee$ contains trivial 1-dimensional vector bundle (spanned by the section "Id"$\in V\otimes V^\vee$; the map in the opposite direction is the evaluation map). So any object $P\in D^b$ is a summand of the image of the object $P\otimes V^\vee$.
| {
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How can one prove that the cube root of 9 is irrational? Of course, if you plug the cube root of 9 into a calculator, you get an endless stream of digits. However, how does one prove this on paper?
| This is essentially the same proof I gave in my answer here.
Suppose $9^{\frac{1}{3}}$ is rational. Then $3^2n^3 = m^3$ for some natural numbers $n$ and $m$. On left side of the equation, the power of $3$ is of the form $3k + 2$ and on the right side it is of the form $3l$. This is a contradiction, because each integer greater than one has a unique prime factorization by the fundamental theorem of arithmetic. Thus $9^{\frac{1}{3}}$ is not rational.
This same proof also works for a more general case. Let $p$ be prime and $n \geq 2$ an integer. Then $\sqrt[n]{p^k}$ is irrational when $n$ does not divide $k$. Just like before, assuming that $\sqrt[n]{p^k}$ is rational leads to a situation where we have a number with two different prime factorizations. One factorization has $p$ of power divisible by $n$, while the other has $p$ of power not divisible by $n$.
| {
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Find $DF$ in a triangle $DEF$ Consider we have a triangle $ABC$ where there are three points $D$, $E$ & $F$ such as point $D$ lies on the segment $AE$, point $E$ lies on $BF$, point $F$ lies on $CD$. We also know that center of a circle over ABC is also a center of a circle inside $DEF$. $DFE$ angle is $90^\circ$, $DE/EF = 5/3$, radius of circle around $ABC$ is $14$ and $S$ (area of $ABC$), K (area of DEF), $S/K=9.8$. I need to find $DF$. Help me please, I'd be very grateful if you could do it as fast as you can. Sorry for inconvenience.
| Here is a diagram. I may or may not post a solution later.
Edit I will not post a solution since it appears to be quite messy. Please direct votes towards an actual solution.
| {
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Simple expressions for $\sum_{k=0}^n\cos(k\theta)$ and $\sum_{k=1}^n\sin(k\theta)$?
Possible Duplicate:
How can we sum up $\sin$ and $\cos$ series when the angles are in A.P?
I'm curious if there is a simple expression for
$$
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta
$$
and
$$
\sin\theta+\sin 2\theta+\cdots+\sin n\theta.
$$
Using Euler's formula, I write $z=e^{i\theta}$, hence $z^k=e^{ik\theta}=\cos(k\theta)+i\sin(k\theta)$.
So it should be that
$$
\begin{align*}
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta &= \Re(1+z+\cdots+z^n)\\
&= \Re\left(\frac{1-z^{n+1}}{1-z}\right).
\end{align*}
$$
Similarly,
$$
\begin{align*}
\sin\theta+\sin 2\theta+\cdots+\sin n\theta &= \Im(z+\cdots+z^n)\\
&= \Im\left(\frac{z-z^{n+1}}{1-z}\right).
\end{align*}
$$
Can you pull out a simple expression from these, and if not, is there a better approach? Thanks!
| Take the expression you have and multiply the numerator and denominator by $1-\bar{z}$, and using $z\bar z=1$:
$$\frac{1-z^{n+1}}{1-z} = \frac{1-z^{n+1}-\bar{z}+z^n}{2-(z+\bar z)}$$
But $z+\bar{z}=2\cos \theta$, so the real part of this expression is the real part of the numerator divided by $2-2\cos \theta$. But the real part of the numerator is $1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}$, so the entire expression is:
$$\frac{1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}}{2-2\cos\theta}=\frac{1}{2} + \frac{\cos {n\theta} - \cos{(n+1)\theta}}{2-2\cos \theta}$$
for the cosine case. You can do much the same for the case of the sine function.
| {
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injection from double dual to finite-dimensional vector space (Note: I'm using the word "natural" to mean "without the need to choose a basis." I'm aware that there is a precise category-theoretic meaning of this word, but I don't have great intuition for it yet and am hoping, perhaps naively, it's not necessary to understand the following.)
There exists a natural injection $V\rightarrow (V^{*})^{*}$ defined by sending $v\in V$ to a functional $\mu_{v}$ on $V^{*}$ such that $\mu_{v}(f)=f(v)$ for all $f\in V^{*}$. When $V$ is finite dimensional, this map is an isomorphism by comparing dimensions, so there is also an injection $(V^{*})^{*}\rightarrow V$. Is there a "natural" (again, in this context I understand this to mean basis-free) way to write down this injection, other than as simply the reverse of the first one?
| For the sake of having an answer: no. Any good definition of "natural" would imply that this map also existed for infinite-dimensional vector spaces, which it doesn't. You shouldn't be able to do any better than "the inverse, when it exists, of the natural map $V \to (V^{\ast})^{\ast}$."
| {
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100 Soldiers riddle One of my friends found this riddle.
There are 100 soldiers. 85 lose a left leg, 80 lose a right leg, 75
lose a left arm, 70 lose a right arm. What is the minimum number of
soldiers losing all 4 limbs?
We can't seem to agree on a way to approach this.
Right off the bat I said that:
85 lost a left leg, 80 lost a right leg, 75 lost a left arm, 70 lost a right arm.
100 - 85 = 15
100 - 80 = 20
100 - 75 = 25
100 - 70 = 30
15 + 20 + 25 + 30 = 90
100 - 90 = 10
My friend doesn't agree with my answer as he says not all subsets were taken into consideration. I am unable to defend my answer as this was just the first, and most logical, answer that sprang to mind.
| You can easily do it visually with a Venn diagram with the four sets of soliders with each limb. For mimimum number of soliders losing all four limbs, none of the inner sets overlap. So $100 - (15+20+25+30) = 10$.
| {
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Help to find the domain to this function $$
\sqrt{\log_\frac{1}{2}\left(\arctan\left(\frac{x-\pi}{x-4}\right)\right)}
$$
Please, could someone show me the steps to find the domain of this function?
It's the sixth time that I try to solve it, and I'm going to burn everything...
| I assume that you are talking about the so-called "natural domain" of a real valued function of real variable (a common concept in Calculus, at least in the U.S.): given a formula, such as the above, and no words about its domain, we assume the domain is to be taken to be a subset of the real numbers, and that this subset should be "as large as possible". That is, we want the know all real numbers for which the expression makes sense and yields a real number.
So, let's analyze the expression step by step, just as you would if you were trying to evaluate it.
*
*First, given an $x$, you would compute both $x-\pi$ and $x-4$. No problems there, that can be done with any real number $x$.
*Then you would compute $\frac{x-\pi}{x-4}$. In order to be able to do this, you need $x-4\neq 0$. So we are going to have to exclude $x=4$. That is, the domain so far is "all $x\neq 4$".
*Then we would compute $\arctan\left(\frac{x-\pi}{x-4}\right)$. Since the domain of the arctangent is "all real numbers", this can be done with any $x$ for which the fraction makes sense. We don't need to exclude any new values of $x$.
*Then we would try to compute the logarithm (base $\frac{1}{2}$) of this number. In order to be able to compute the logarithm, we need the argument to be positive. So we are going to need
$$\arctan\left(\frac{x-\pi}{x-4}\right)\gt 0.$$
When is the arctangent positive? When the argument is positive. So we need
$$\frac{x-\pi}{x-4}\gt 0.$$
When is a fraction positive? When both numerator and denominator are positive, or when they are both negative.
So we need either $x-\pi\gt 0$ and $x-4\gt 0$ (this happens when $x\gt 4$); or $x-\pi\lt 0$ and $x-4\lt 0$ (this happens when $x\lt \pi$). So we now need to restrict our $x$s to $(-\infty,\pi)\cup(4,\infty)$. (Note that this also maintains the exclusion of $4$).
*Finally, we need to take the square root of the answer. That means that the logarithm must be nonnegative. When is $\log_{\frac{1}{2}}(a)\geq 0$? When $0\lt a \leq 1$ (taking exponentials with base $\frac{1}{2}$ flips the inequality, because $(\frac{1}{2})^x$ is decreasing). So we actually need
$$0\lt \arctan\left(\frac{x-\pi}{x-4}\right)\leq 1.$$
When is $0\lt \arctan(a)\leq 1$? When $0\lt a \lt \frac{\pi}{4}$ (thanks to Jonas Meyer for the heads up!). When $\tan(0)\lt a \leq \tan(1)$. So we need
$$0 \lt \frac{x-\pi}{x-4}\leq \tan(1).$$
Since $\tan(1)\gt 0$ This happens if either
$$0\lt (x-\pi) \lt \tan(1)(x-4)$$
or
$$\tan(1)(x-4)\lt (x-\pi)\lt 0.$$
So check the inequalities; then remember that $x$ must be greater than $4$ for both $x-\pi$ and $x-4$ to be positive; or less than $\pi$ for both to be negative.
| {
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Why do all circles passing through $a$ and $1/\bar{a}$ meet $|z|=1$ are right angles? In the complex plane, I write the equation for a circle centered at $z$ by $|z-x|=r$, so $(z-x)(\bar{z}-\bar{x})=r^2$. I suppose that both $a$ and $1/\bar{a}$ lie on this circle, so I get the equation
$$
(z-a)(\bar{z}-\bar{a})=(z-1/\bar{a})(\bar{z}-1/a).
$$
My idea to show that the circles intersect at right angles is to show that the radii at the point of intersection are at right angles, which is the case when the sum of the squares of the lengths of the radii of the circles is the square of the distance to the center of the circle passing through $a$ and $1/\bar{a}$. However, I'm having trouble finding a workable situation, since I don't think there is not a unique circle passing through $a$ and $1/\bar{a}$ to give a center to work with. What's the right way to do this?
| I have a solution that relies on converting the complex numbers into ordered pairs although I believe there must be a solution with just the help of complex numbers.
Two circles intersect orthogonally, if their radii are perpendicular at the point of intersection. So, using this we can have a condition for orthogonality.
$\hskip 2.5in$
Here's a trick of how you will get a condition. Let us consider two circles,
$$C_1,A:x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$C_2,B:x^2+y^2+2g_2x+2f_2y+c_2=0$$
From your high school course in analytical geometry in high school, it must be clear that the centres $A$ and $B$ are $A(-g_1,-f_1)$ and $B(-g_2,-f_2)$. And the radii of such a circle is $r_1=\sqrt{g_1^2+f_1^2-c_1}$ and similarly $r_2=\sqrt{g_2^2+f_2^2-c_2}$.
Now invoke Pythagoras here, I'll leave the actual computation to you, the condition would turn out to be, $$2g_1g_2+2f_1f_2=c_1+c_2$$
Now, find a parametric equation for a circle passing through the complex numbers $a$ and $\dfrac{1}{\bar a}$. How do you do this?
Since the circle always passes through, $a\cong(l,m)$ and $\dfrac{1}{\bar a}=\dfrac{a}{|a|^2}\cong\left(\dfrac{l}{l^2+m^2},\dfrac{m}{l^2+m^2}\right)$, you have the following will be the equation of the circle:
$$(x-l)\left(x-\dfrac{l}{l^2+m^2}\right)+(y-m)\left(y-\dfrac{m}{l^2+m^2}\right)+\lambda(ly-mx)=0$$
The second circle is, $$x^2+y^2-1=0$$
So, you should now see that $g_2=f_2=0$ and $c_2=-1$. Also, after a little inspection, note that we need not care for what those $g_1$ and $f_1$ are. And, thankfully, $c_1=1$. So, you have the required condition for orthogonality.
I know this is lengthy and not instructive, but this is all I can recollect from high school geometry. So, I only hope this is of some help!
| {
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How can you find the number of sides on this polygon? I'm currently studying for the SAT. I've found a question that I can't seem to figure out.
I'm sure there is some logical postulate or assumption that is supposed to be made here. Here is the exact problem:
I don't really care for an answer, I would rather know steps on how to solve this. I'm trying to be prepared for all types of questions on the SAT.
Thanks!
EDIT:
Thanks so much for the help guys! I've figured it out and wanted to explain it in detail for anybody who wanted to know:
SOLUTION SPOILER:
1. The figure displayed is a non-regular quadrilateral.
2. Because we know that, we know that the interior angles of the shape are 360 because of the formula (n-2)180.
3. I then created two statements. x+y=80 and x+y+z where z is the top two full angles (above both x and z).
4. I then simply solved for z (the total of the two angles above x and y).
5. I found z to equal 280. When split between the two angles it represented, I determined that each angle in the shape was equal to 140 degrees.
6. Because each angle is congruent (the shape is regular) we now know the measurement of every angle.
7. I then plugged this into the interior angle formula: (n-2)180=n140.
8. After solving for n, you learn that the number of sides is 9 :)
Hope this helps!
| Hints:
The sum of the measures of the interior angles of an $n$-sided convex polygon is $(n-2)*180^\circ$. So, if the polygon is regular, each interior angle has measure $180^\circ-{360^\circ\over n}$. (You could also use the fact that the sum of the exterior angles of a convex polygon is $360^\circ$. An "exterior angle" is the angle formed by a line coinciding with a side and the "next" side.)
| {
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Is this group a semidirect product? $G=\langle x,y,z:xy=yx,zxz^{-1}=x^{-1},zyz^{-1}=y^{-1}\rangle$, could you help me to understand if this group is a semidirect product of the type $\langle x,y\rangle\rtimes_\varphi\langle z\rangle$.
I was trying to prove that $\langle x,y\rangle\triangleleft G$ and $\langle x,y\rangle\cap\langle z\rangle=\{1\}$, but I'm having trouble with the second, and actually I even don't know if this is true, it's possible that this group is not a semidirect product. Could you help me?
| Showing that the intersection of two subgroups is trivial in a group described by generators and relations is a little tricky.
Clearly, it is enough to show that if $i,j,k$ are integers and $x^i y^j z^k = 1$, then $i=j=k=0$. This is of course equivalent to showing that if $i,j,k$ are not all zero, then $x^i y^j z^k \ne 1$ in $G$.
In a group described by generators and relations, in order to show that some word $w \ne 1$, you need to show that there is some group $H$ such that: (1) $H$ contains elements that satisfy the relations; (2) $w \ne 1$ in $H$. (This proves that the relations together with the group axioms do not force $w=1$; hence $w \ne 1$ in $G$.)
So we have to show, for each $(i,j,k)$, there is some group $H$ such that (1) $H$ contains three elements $x,y,z$ satisfying the given relations; (2) in $H$, we have $x^i y^j z^k \ne 1$.
You need such an $H$ for each nonzero triple $(i,j,k)$, so you'll need to find lots of groups containing 3 elements satisfying the given relations. A good source of such groups are the dihedral groups: $x,y$ can be any 2 rotations, and $z$ any reflection. It is easy to check that $x,y,z$ satisfy the given relations.
The dihedral groups should give you enough $H$'s to rule out $x^i y^j z^k = 1$ for any $(i,j,k) \ne (0,0,0)$.
| {
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Is there a first-order-logic for calculus? I just finished a course in mathematical logic where the main theme was first-order-logic and little bit of second-order-logic. Now my question is, if we define calculus as the theory of the field of the real numbers (is it?) is there a (second- or) first-order-logic for calculus? In essence I ask if there is a countable model of calculus.
I hope my question is clear, english is my third language.
| I take the view that the proper logical framework in which to do model theory for structures in analysis is continuous logic. For more information on the subject, look up the webpage of Ward Henson.
| {
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Perfect squares always one root? I had an exam today and I was thinking about this task now, after the exam of course.
$f(x)=a(x-b)^2 +c$
Now, the point was to find C so that the function only has one root. Easy enough, I played with the calculator and found this. But I hate explanations like that, yes. You get a few points but far from full score. But overall I should still get an A, I hope. If $C=0$ then the expression is a perfect square and they only have one root? Is that far of?
$a(x-b)^2= - c$
$\frac{a(x-b)^2}{a}= - \frac{c}{a}$
$(x-b)^2= - \frac{c}{a}$
This also argues that c should be 0 for it to only be one root?
| An alternative way to think about it is geometrically. The graph of $y=x^2$ is a parabola that opens up with vertex at the origin. The graph of
$$y = (x-b)^2$$
is then a horizontal shift by $b$ units (so $b$ units to the right if $b\geq 0$, and $|b|$ units to the left if $b\lt 0$) of the same graph. There is still only one root: the vertex.
If $a\neq 0$, then
$$y = a(x-b)^2$$
is a vertical stretch of this graph, possibly with a flip (if $a\lt 0$); it does not change the number of intersections with the $x$-axis.
Finally, $$y=a(x-b)^2 + c$$
is a vertical shift by $c$ units (up if $c\gt 0$, down if $c\lt 0$).
If $y=a(x-b)^2$ is a parabola that opens "up" (if $a\gt 0$), then shifting it up ($c\gt 0$) will remove all intersections with the $x$-axis; and shifting it down ($c\lt 0$) will create two intersections with the $x$-axis as the vertex moves down.
If $y=a(x-b)^2$ is a parabola that opens "down" (if $a\lt 0$), then the situation is reversed: $c\gt 0$ will create two intersections with the $x$-axis, and $c\lt 0$ will remove all intersections with the $x$-axis.
Either way, in order to maintain one and only one intersection, you need the vertex of the parabola to stay on the $x$-axis, so you need $c=0$. Conversely, if $c=0$, you have a parabola with vertex on the $x$-axis, hence with a single intersection.
| {
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How to test any 2 line segments (3D) are collinear or not? if we have two line segments in 3D, what would be the way to test whether these two lines are collinear or not? (I fogot to mentioned that my line segments are 3D. So, I edited the original post. Sorry for the inconveniences)
I wish to check the direction of the lines and the perpendicular distance between them.
Does these two factors are enough to decide whether 2 line segments are collinear or not.
Thank you in advance.
| If the two line segments $AB$ and $CD$ are given by 4 distinct points A, B, C and D, it is also sufficient that both $AB \parallel CD$, $AC \parallel BD$ and $AD\parallel BC$.
To see if $A(a_1,a_2)B(b_1,b_2) \parallel C(c_1,c_2)D(d_1,d_2)$, you test whether or not $\vec{BA} = B-A $ and $\vec{DC} = C-D$ are linearly dependent vectors.
So the two line segments are contained in the same line if
$$ \begin{cases} (a_1-b_1)(c_2-d_2) - (c_1-d_1)(a_2-b_2) = 0 \\ (a_1-c_1)(b_2-d_2) - (b_1-d_1)(a_2-c_2) = 0 \\ (a_1-d_1)(c_2-b_2) - (c_1-b_1)(a_2-d_2) = 0 \end{cases}$$
| {
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is there a connection between the following? Assume $A$ is $m \times n$ and $B$ is $m \times n$.
Is there a connection between the eigenvalues of $AB'$ and the eigenvalues of $B'A$?
One is an $m \times m$ and the other is $n \times n$.
($B'$ stands for the transpose of $B$)
| It seems easier for me to assume that $B$ is an $n \times m$ matrix. In that case, a classical argument shows that $AB$ and $BA$ have the same nonzero eigenvalues, not counting multiplicity. The case that these eigenvalues are distinct is dense in the general case, so $AB$ and $BA$ have the same nonzero eigenvalues counting multiplicity. Of course one of them has $|n-m|$ more zero eigenvalues than the other.
| {
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Nomenclature of random variables $\{X=0, Y=0\}$ same as $\{X=0\}\cap \{Y=0\}$? just a small doubt.
My exercises keep oscillating their nomenclature on this small detail and I always have the other version.
Let $X,Y$ be random variables.
Is $\{X=0, Y=0\}$ the same as $\{X=0\}\cap \{Y=0\}$?
Another example.
Let $N$ be the number of Users on a webpage. Two files are available for download, one with 200 kb and another with 400 kb size.
$$
\begin{align}
X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}
\end{align}
$$
I want to express, at least one user downloaded the 200 kb file. Here's how I expressed it
$\{X_1 + X_2 + \cdots + X_n \geq 1\}$. Would this be ok? The book expressed it as $\{X_1=1\}\cup\{X_1=3\}\cup \cdots \cup\{X_n=1\}\cup\{X_n=3\}$.
Another thing to express: no user downloaded the 200 kb file. I expressed it as
$|\{X_k=1, 1 \leq k \leq N\}|=0$. The book as $\{X_1 \neq 1\}\cap \cdots \cap \{X_n \neq 1\}$. Would my solution be ok?
I'm always in doubt when I'm allowed to use symbols like $+$ and $|\mathrm{modulo}|$ (to get the number of elements). Is this generally always allowed? Many thanks in advance!
Thanks in advance guys!
| $\{X=0,Y=0\}$ and $\{X=0\}\cap\{Y=0\}$ are the same thing. Both notations refer to
$$
\{\omega\in\Omega : X(\omega)=0\ \ \&\ \ Y(\omega)=0\} = \{\omega\in\Omega : X(\omega)=0\}\cap\{\omega\in\Omega : Y(\omega)=0\}.
$$
Your notation saying
$$
\begin{align}
X_n(w) := w_n = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}
\end{align}
$$
seems confused. I suspect maybe you meant
$$
\begin{align}
\Omega = \{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\},
\end{align}
$$
although even that may differ from what's appropriate if you're bringing in $n$ different random variables. Your later notation makes it look as if what the author of the book had in mind is that $X_k$ is the number of kb downloaded by the $k$th user, for $k=1,\ldots,n$. Just what $w$ is, you're not clear about, and at this point I'm wondering if you're confusing $w$ with $\omega$. Probably what is needed is this:
$$
\begin{align}
\{ & 0:=\text{user downloads no file}, \\
& 1:=\text{user downloads the first file (200 kb)}, \\
& 2 :=\text{user downloads the second file (400 kb)}, \\
& 3:=\text{user downloads both files (600 kb)}\}^n
\end{align}
$$
i.e. then $n$th power of that set of four elements. This is the set of all $n$-tuples where each component of an $n$-tuple is one of these four elements. Then, when $\omega$ is any such $n$-tuple, $X_k(\omega)$ is its $k$th component, which is one of those four elements.
For example, if $n=3$, so there are three users, then
$$
\begin{align}
\Omega = \{ & (0,0,0), (0,0,1), (0,0,2), (0,0,3), (0,1,0), (0,1,1), (0,1,2), (0,1,3),\ldots\ldots\ldots \\ \\
& \ldots\ldots\ldots, (3,3,3) \},
\end{align}
$$
with $64$ elements. If, for example, $\omega=(2,3,0,1)$, then $X_2(\omega)=3$.
| {
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Does $\int_0^1 \sum_{n=0}^{\infty}x e^{-nx}\;dx = \sum_{n=0}^{\infty}\int_0^1 x e^{-nx}\;dx$? Does $$\int_0^1 \sum_{n=0}^\infty x e^{-nx}\;dx = \sum_{n=0}^\infty \int_0^1 x e^{-nx}dx$$ ?
This exercise leaves me stumped. On the one hand, it seems the series $\sum_{n=0}^\infty xe^{-nx}$ is not uniformly convergent in $[0,1]$ (it equals $\frac{xe^x}{(e^x-1)}$ in $(0,1]$ and 0 in $x_0=0$, so it cannot be uniformly convergent since it is a series of continuous functions that converges to a non-continuous function). On the other hand, if this is the case, how do I deal with that... thing?
Perhaps the series is uniformly convergent and I made a mistake?
Thanks!
| You can use Fubini's theorem, but it seems overkill. Note that for all integer $N$ we have
$$\sum_{n=0}^N\int_0^1xe^{-nx}dx=\int_0^1\sum_{n=0}^Nxe^{-nx}dx\leq \int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx,$$
so
$$\sum_{n=0}^{+\infty}\int_0^1xe^{-nx}dx\leq \int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx.$$
For the reversed inequality, fix $\varepsilon>0$. Since $\sum_{n=0}^{+\infty}xe^{-nx}$ is integrable, we can find a $\delta>0$ such that $\int_0^{\delta}\sum_{n=0}^{+\infty}xe^{-nx}\leq \varepsilon$. And the series $\sum_{n=0}^{+\infty}xe^{-nx}$ is normally convergent on $[\delta,1]$. So we have
\begin{align*}
\int_0^1\sum_{n=0}^{+\infty}xe^{-nx}dx&=\int_0^\delta\sum_{n=0}^{+\infty}xe^{-nx}dx+\int_\delta^1\sum_{n=0}^{+\infty}xe^{-nx}dx\\
&\leq\varepsilon +\int_\delta^1\sum_{n=0}^{+\infty}xe^{-nx}dx\\
&=\varepsilon +\sum_{n=0}^{+\infty}\int_\delta^1xe^{-nx}dx\\
&\leq \varepsilon +\sum_{n=0}^{+\infty}\int_0^1xe^{-nx}dx,
\end{align*}
and since $\varepsilon$ is arbitrary we can conclude the equality.
| {
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Conditional Probability Question Bowl A contains 6 red chips and 4 blue chips. Five chips are randomly chosen and transferred without replacement to Bowl B. One chip is drawn at random from Bowl B. Given that this chip is blue, find the conditional probability that 2 red chips and 3 blue chips are transferred from bowl A to bowl B.
Attempt:
$$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$
Let $B$ = chip is blue and $A$ = 2 red and 3 blue are chosen.
$$\begin{align}
&P(A) = \frac {\binom 6 2 \cdot \binom 4 3}{\binom {10} 5}\\
&P(B|A) = \frac 3 5
\end{align}$$
By Bayes Rule, $P(A|B) = \left(\dfrac 3 5\right)\cdot \dfrac{ \binom 6 2 \binom 4 3}{\binom {10} 5\cdot \dfrac{4}{10}}$.
Is this correct?
| There are $\frac{10!}{6!4!}$ (= 210) possible arrangements for the chips, and $\frac{5!}{2!3!}$ arrangements for the chips desired in bowl B. Any given arrangement of bowl B can occur for every corresponding arrangement in bowl A (also $\frac{5!}{2!3!}$ combinations)
The total number of possiblilities with the correct bowl B is therefore $\frac{5!}{2!3!}\dot{}\frac{5!}{2!3!}=100$
Substitute P(A) = 100/210 to get P(A|B) = (3/5)(100/210)/(4/10) = 5/7, or about 71%
| {
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Is the following derivative defined? I am new to this site so I am not sure if this is the right place to be asking this question but I will try anyway. I am reading an economics paper for my research and the author does the following:
$$\frac{\partial}{\partial C_t(j)} \int_0^1 P_t(j) C_t(j) dj = P_t(j)$$
I feel that this derivative is not properly defined but I am probably missing something obvious because the author knew what he was doing. Could someone please tell me if this is a legitimate derivative?
Thanks,
| That derivative can be properly defined if and only if
there exists an appropriate pair of values $t$ and $j$ such that $\: C_t(j) = 0 \:$
and
all appropriate values of $t$ such that
[there exists an appropriate value of $j$ such that $\: C_t(j) = 0 \:$]
give the same value of the integral
and
$0$ is a limit point of the set of values taken by $C_t(j)$ for appropriate values of $t$ and $j$
("appropriate values" are those associated with the definition of $C_t(j)$, and even if the derivative "can be properly defined", it may still fail to exist, as is the case with non-differentiable functions).
If either of the first two fail there is not a unique "base value" to plug into the difference quotient.
If the last one fails, it is not clear what the definition of "limit" would be.
| {
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Circle and Line segment intersection I have a line segment (begin $(x_1,y_1)$, end $(x_2,y_2)$, with $D=5$, let’s say) and a circle (radius $R$, center $(x_3,y_3)$)
How can I check that if my line segment intersects my circle?
picture
http://kepfeltoltes.hu/120129/inter_www.kepfeltoltes.hu_.png
| The points $(x,y)$ on the line segment that joins $(x_1,y_1)$ and $(x_2,y_2)$ can be represented parametrically by
$$x=tx_1+(1-t)x_2, \qquad y=ty_1+(1-t)y_2,$$
where $0\le t\le 1$.
Substitute in the equation of the circle, solve the resulting quadratic for $t$. If $0\le t\le 1$ we have an intersection point, otherwise we don't. The value(s) of $t$ between $0$ and $1$ (if any) determine the intersection point(s).
If we want a simple yes/no answer, we can use the coefficients of the quadratic in $t$ to determine the answer without taking any square roots.
| {
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Counterexample of G-Set
If every element of a $G$-set is left fixed by the same element $g$ of $G$,
then $g$ must be the identity $e$.
I believe this to be true, but the answers say that it's false. Can anyone provide a counter-example? Thanks!
| For concreteness, let $G$ be the group of isometries of the plane, and let $g$ be reflection in the $x$-axis. Let $S$ be the $x$-axis. Then $g(v)=v$ for every point $v$ in $S$.
| {
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Every manifold is locally compact?
Theorem. Every Manifold is locally compact.
This is a problem in Spivak's Differential Geometry.
However, don't know how to prove it. It gives no hints and I don't know if there is so stupidly easy way or it's really complex.
I good example is the fact that Heine Borel Theorem, I would have no clue on how to prove it if I didn't see the proof.
So can someone give me hints. I suppose if it's local, then does this imply that it's homeomorphic to some bounded subset of a Euclidean Space?
| I do not think the above answers are completely right, since the "Hausdorff" condition in the definition of topological manifolds must be needed.
The key is to prove the following: If $V\subset U\subset X$, and X is Hausdorff, $\bar{V}_{U}$ is compact. Then $\bar{V}_{U}=\bar{V}$.
Proof: By definition, we only need to show that $\bar{V}\subset \bar{V}_U$. Suppose $x\in\bar{V}$, if $x\notin \bar{V}_U$. Since X is Hausdorff and $\bar{V}_{U}$ is compact, $\bar{V}_{U}$ is closed in X. So $ (\bar{V}_{U})^{c}$ is open and $x\in (\bar{V}_{U})^{c}$. Since $x\in \bar{V}$, we must have $V\cap (\bar{V}_{U})^{c}$ is not empty. It is impossible since $\bar{V}_{U}$ contains $V$.
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Proof that $\mathbb{Q}$ is dense in $\mathbb{R}$ I'm looking at a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$, using only the Archimedean Property of $\mathbb{R}$ and basic properties of ordered fields.
One step asserts that for any $n \in \mathbb{N}$, $x \in \mathbb{R}$, there is an integer $m$ such that $m - 1 \leq nx < m$. Why is this true? (Ideally, this fact can be shown using only the Archimedean property of $\mathbb{R}$ and basic properties of ordered fields...)
| Assume first that $x>0$, so that $nx>0$. By the Archimedean property there is a $k\in\mathbb{N}$ such that $k>nx$; let $m$ be the least such $k$. Clearly $m-1\le nx<m$. If $x=0$, just take $m=1$. Finally, if $x<0$, then $-nx>0$, so by the first part of the argument there is an integer $k$ such that $k-1\le -nx<k$, and hence $-k<nx\le 1-k$. If $nx\ne 1-k$, you’re done: just take $m=1-k$. If $nx=1-k$, take $m=2-k$.
| {
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If a coin is flipped 10 times, what is the probability that it will land heads-up at least 8 times? I absolutely remember learning this is middle school, yet I cannot remember how to solve it for the life of me. Something to do with nCr, maybe? ...
Thanks for any help.
| What we'd like to do is find a way to set the problem up in some way that we know how to solve it.
$P($At least $8$ heads) = $P(X \geq 8)$ where $X$ is the Random Variable associated with the number of heads attained.
Well, since $X$ can only have the values $0$ through $10$, perhaps we should split $P$ up:
$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$
We can split them up like this because there is no "overlap" between the events (You can't get 8 heads and then get either 9 or 10 heads too.)
Now we just need to apply the definition of probability:
$P(S) = n(E)/n(S)$
where $n(E)$ is the number of items in our event set, and $n(S)$ is the number of items in our sample space.
Well, for each of the probabilities, $n(S)$ = $2^{10}$ by the multiplication principle. Now, what are each of the $n(E)$?
You thought it would have to do with Combinations (nCr), and you were right. We use combinations instead of permutations because we really don't care which order we get the heads in, right?
So, for $X = 8$: $n(E) = $${10}\choose{8}$
and so on. Can you take it from here?
| {
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Finding points in a grid with exactly k paths to them? Suppose that we begin at (0, 0) and are allowed to take two types of steps - steps one unit up and steps one unit to the right. For example, a legal path might be (0, 0) → (1, 0) → (2, 0) → (2, 1) → (3, 1).
Now, suppose that you are given a number k. Is there an efficient algorithm to list of all points (x, y) with exactly k paths to them? For example, given the number 6, we would list (1, 5), (5, 1) and (2, 2), since these points have exactly six paths to them.
Thanks!
| This sounds to me like a combinatorial problem.
Say you start in (x, y) and want to go to (x+3, y+3). If we represent all "up" movements by 'U' and all "right" movements by 'R', such a path could be UUURRR. The total number of possible paths would be all possible permutations of UUURRR, namely 6!/(3!3!) = 20.
An algorithm finding all these paths could be to put all 'U's and 'R's in a pool and select one from the pool. This will be your first move. Then find all permutations involving the rest of the pool. Finally swap your first choice (i.e. an 'U') for the opposite choice (this time an 'R') and do it again. Recursively you'll now have found all possible paths between the two points.
Updating the answer reflecting templatetypedef's comments below:
If you want the number of paths reachable within k number of steps, the solution is still feasible and similar. Perhaps even simpler. Choose a 'U' then calculate the number of paths using (k-1) steps from there to the destination. After this is complete, choose an 'R' and calculate the number of paths using (k-1) steps from there to the destination. These two numbers added together will be your answer. Use recursion on the (k-1) subpath-steps.
If you want the points with exactly n subpaths leading to them, it gets trickier. One way could be to go by binomial numbers. Find all i and j such as i!/((i-j)!j!)=n. This will take O(n) time since i+j<=n. Then you can use my proposition above for finding the number of paths reachable within k number of steps. OleGG's solution below might be cleaner though. I leave it to you to benchmark :)
| {
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Pythagorean Theorem Proof Without Words 6 Your punishment for awarding me a "Nice Question" badge for my last question is that I'm going to post another one from Proofs without Words.
How does the attached figure prove the Pythagorean theorem?
P.S. No, I will not go through the entire book page-by-page asking for help.
P.P.S. No, I am not a shill for the book. Just a curious math student.
| If you have a diameter of a circle and a point on the circle, the length of the altitude from that point to the diameter is the geometric mean of the lengths of the two parts of the diameter. Perhaps drawing in the two chords from the top point on the circle to the endpoints of the diameter and looking for similar triangles would help.
| {
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Groups of symmetries What are the symmetries of a solid rectangular box whose length, width and height are all different?
I get a group of order 4 by rotation 180, flipping along a vertical and horizontal axis and itself.
| Your contention that you'll have cyclic subgroup of order $4$ is quite natural but needs a little rethinking. Since, you consider the symmetries of a mattress like object (where symmetry is to mean any rigid motion in 3-space which will move a copy of the mattress in any fashion and place the copy back on the original), a rotation of a "scalene" object by $\frac{\pi}{2}$ will not keep this object in the same place. So, rotation by $\pi$ is a valid symmetry.
Technically, the mattress group is called the Klein $4$-group.
$\hskip 2.5in$
$H$ stands for a horizontal flip about the axis parallel to $13$ through the midpoint of $12$; $V$ for a vertical flip about the axis parallel to $12$ and through the mid point of $13$ and $R$ for rotation through the mid-point of the mattress.
| {
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prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
| Here is a solution using induction:
Let $f(x)=x^3+6x^2+11x+6$
Since we want to see if it is divisible by 3 let us assume that $f(x)=3m$.
For the case where $x=0$, $f(0)=6$ which is divisible by 3.
Now that we have proved for one case let us prove for the case of $f(x+1)$
$$f(x+1)=(x+1)^3+6(x+1)^2+11(x+1)+6$$
$$= x^3+3x^2+3x+1+6x^2+12x+6+11x+11+6$$
$$=(x^3+6x^2+11x+6)+3x^2+15x+18$$
And since $x^3+6x^2+11x+6=3m$
$$f(x+1)=3m+3x^2+15x+18=3(m+x^2+5x+6)$$
Which is divisible by 3.
| {
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Rotations and the parallel postulate. If we take a full rotation to be $360^\circ$, then it seems that we can prove the following
Starting from the red point, we walk clockwise along the triangle. At each vertex, we must turn through the green angles marked to proceed down the adjacent sides of the triangle. When we return to the red point, we will have turned through one full rotation. This means that the sum of the exterior angles is given as $360^\circ$, implying the interior angles of the triangle sums of $180^\circ$.
The fact that the angles of a triangle sum to $180^\circ$ is well known to be equivalent to the parallel postulate and this made me wonder whether if the fact that a full rotation being $360^\circ$ is also equivalent to the parallel postulate?
I avoided stating the question using "exterior angles of a triangle sums to $360^\circ$" and instead used the more ambiguous term "rotations" to emphasize the fact that rotations seem to be more general. We can for example show that the interior angles of a heptagram sum to $180^\circ$ by noting that three full rotations are made while "walking" thge heptagram. This should generalize to arbitrary closed polygons and seems stronger than the fact that the exterior angles sum to $180^\circ$.
In summary, I would be interested in knowing the connections that this technique has to the parallel postulate as well as if this technique is a "rigorous" way of finding the internal angles of more complex shapes such as the heptagram.
| Your picture, and perhaps your assumptions, are lying in the Euclidean plane. Take the same idea and put it on the sphere, where the parallel postulate is false, and we get something like the following:
Notice that, in this case, the sum of the exterior angles is $270^\circ$, not $360^\circ$.
However, in answer to your question about the sum of the interior angles of a polygon, since an external and the corresponding interior angle sum to $180^\circ$, the sum of the exterior angles and the interior angles is $180^\circ\times$ the number of sides. Since, as you have noted, in the Euclidean plane, the sum of the exterior angles is $360^\circ$, we get that the sum of the interior angles of a polygon with $n$ sides is $(n-2)180^\circ$.
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Hofstadter's TNT: b is a power of 2 - is my formula doing what it is supposed to? If you've read Hofstadter's Gödel, Escher, Bach, you must have come across the problem of expressing 'b is a power of 2' in Typographical Number Theory. An alternative way to say this is that every divisor of b is a multiple of 2 or equal to 1. Here's my solution:
b:~Ea:Ea':Ea'':( ((a.a')=b) AND ~(a=(a''.SS0) OR a=S0) )
It is intended to mean: no divisor of b is odd or not equal to 1. E, AND and OR are to be replaced by the appropriate signs.
Is my formula OK? If not, could you tell me my mistake?
| Your idea is sound, but the particular formula you propose
$$\neg\exists a:\exists a':\exists a'':( ((a\cdot a')=b) \land \neg (a=(a''\cdot SS0) \lor a=S0) )$$
does not quite express it. The problem is that the quantifier for $a''$ has too large scope -- what your formula says is that it will prevent $b$ from being a power of two if there is some even number that is different from some factor of $b$. For example, your formula claims that $2$ itself is not a power of two, because you can make $((a\cdot a')=2) \land \neg (a=(a''\cdot SS0) \lor a=S0)$ true by setting $a=2$, $a'=1$, $a''=42$. The first part is true because $2\cdot 1$ is indeed $2$, and the second (negated) part is true because it is neither the case that $2=42\cdot SS0$ nor $2=S0$.
What you want is
$$\neg\exists a:\exists a':( ((a\cdot a')=b) \land \neg (\exists a'':(a=(a''\cdot SS0)) \lor a=S0) )$$
Moving the quantifier inside one negation switches the "burden of proof" -- now it says that there isn't any number that is half of $a$, rather than there is some number that isn't half of $a$.
Or perhaps more directly expressed:
$$\forall c:\Big(\exists d:( c\cdot d = b )\to \big(c=S0 \lor \exists a:(c=SS0\cdot a)\big)\Big)$$
| {
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Equivalence Class for Abstract Algebra Class Let
$$R_3= \{(a,b)\mid a,b \in \mathbb{Z}\text{ and there exists }k \in \mathbb{Z} \text{ such that }a-b=3k\}.$$
I know there is an equivalence relation but I'm not 100% on what it means to be an equivalence class for this problem. In class we got 3: $\{0,3,6,9,\ldots\}$ and $\{1,4,7,10,-2,-5,\ldots\}$ and $\{2, 5, 8, 11, -1, -4,\ldots\}$.
I don't understand where these cells came from. Help?
| I'll try to put it this way:
Define a relation $\sim$ on $\mathbb Z$, such that $a \sim b \iff \exists k \in \mathbb Z ~~ \text{such that}~~~~a-b=3k$
What does this say?
Integers $a$ and $b$ are related if and only if on their difference is a multiple of $3$. Since, the remainder when $a-b$ is divided by $3$ is the difference of the remainders when $a$ and $b$ are divided by $3$, taken(all taken$\mod 3$).
So, integers $a$ and $b$ are related if and only if they leave the same remainder when divided by $3$.
Now try to put all those numbers that are related to each other in the same "cell" and those that are not related in different "cells".
But, now notice that the number of distinct cells you'll need for the purpose is no more than $3$ and no less! (Why?)
Construct these "cells" to see how they coincide with what you have written down in your class.
And, now call these cells "equivalence classes".
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Soccer and Probability MOTIVATION: I will quote Wikipedia's article on a soccer goalkeeper for the motivation:
Some goalkeepers have even scored goals. This most commonly occurs where a goalkeeper has rushed up to the opposite end of the pitch to give his team an attacking advantage in numbers. This rush is risky, as it leaves the goalkeeper's goal undefended. As such, it is normally only done late in a game at set-pieces where the consequences of scoring far outweigh those of conceding a further goal, such as for a team trailing in a knock-out tournament.
The mathematical question:
Consider the following game (simplified soccer):
A single player starts with a score of 0.5 and plays N turns.
In each turn, the player has to choose one of 2 strategies: $(p_{-1},p_0,p_1)$ or
$(q_{-1},q_0,q_1)$ (these are probability vectors) and then her score is increased by -1, 0 or 1 according to the probabilites dictated by the chosen strategy. The player wins if at the end of the game she has a positive score, and loses if she has a negative score (the player's objective is to win, the only thing that matters is whether the final score is positive or negative).
What is the optimal global strategy given $N$, $(p_{-1},p_0,p_1)$ and
$(q_{-1},q_0,q_1)$?
A global strategy is a function of the number of turns left, the current score and the 2 probability vectors (which are constant for all turns).
If this question is hard, it may still may interesting to approximate an optimal global strategy (in what sense?).
| In the case where you know the number of turns in advance, you can construct an optimal strategy in time $O(N^2)$ by reasoning backwards from the last round.
If, before the last turn, you find yourself with score $-0.5$, choose your strategy by comparing $p_1$ to $q_1$. On the other hand, if the score is $0.5$, compare $p_{-1}$ to $q_{-1}$. If the score is anything else, it's too late to make a difference: either you have already won, or already lost.
Now you know the "value" of the game (that is, the probability of eventually winning, given optimal play) after $N-1$ turns, as a function of your score at that time.
Then look at your options before the second-to-last turn. For each possible score you have the option of playing $p$s or $q$s, and each of this will give you a certain probability of eventually winning, which you can easily compute because you already have a table of the probabilities after the turn. The optimal play at each score is, of course, the one that will yield you the best probability of winning.
Continue this backwards until the first turn. What you end up with is a two-dimensional table that tells you, given the number of turns left and your instant score, what your chance of winning is, and whether you should play $p$ or $q$. This table constitutes an optimal strategy.
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Rudin's assertion that if $t = x/(1 + x)$ then $0 \leq t < 1$ I'm having trouble understanding one step in the proof of Theorem 1.21 in Rudin's Principles of Mathematical Analysis.
Theorem 1.21 For every real $x > 0$ and every integer $n > 0$ there is one and only one positive real $y$ such that $y^{n} = x$.
In the proof he makes the following claim: Let $E$ be the set consisting of all positive real numbers $t$ such that $t^{n} < x$. If $t = \frac{x}{1 + x}$ then $0 \leq t < 1$.
I don't understand how he got that inequality. If $t = 0$ that implies that $x = 0$ which is a contradiction since every $x > 0$. And if $x \rightarrow \infty$, then $t = 1$.
| Notice that in this proof $x$ is a fixed positive real number, and that we are assuming $t=\frac{x}{x+1}$.
Since $x>0$, we have $x+1>0$ so $t=\frac{x}{x+1}>0$ hence $t>0$ thus $t\geq 0$. Furthermore, since $x<x+1$ and neither of these are $0$ we have $t=\frac{x}{x+1}<1$. Putting these together gives $0\leq t< 1$.
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Number of critical points in a period of a periodic function I am interested in a relationship (if any) between the number of critical points of a periodic function $f$ of class $C^3([0,T])$ and the number of critical points of $f''$ in $[0,T]$.
| Consider a $C^2$ function $F\colon\mathbb{R}\to\mathbb{R}$ periodic of period $T>0$ and assume that $F$ has $N$ distinct zeroes $\{x_1,\dots,x_N\}\subset[0,T]$. By Rolle's theorem, $F'$ has at least $N-1$ zeroes in $(0,T)$, one in each interval $(x_i,x_{i+1})$, $1\le i\le N$.
*
*If $x_1=0$ (and hence $x_N=T$ ), then $F'$ may have exactly $N-1$ zeroes.
*If $x_1>0$ (and hence $x_N<T$ ), then $F'$ has at least one zero between $x_N$ and $x_1+T$; call it $\xi$. Then either $\xi\in(x_N,T]$ or $\xi-T\in(0,x_1)$. Conclude that $F'$ has at least $N$ zeroes. If $\xi=T$, then $F'$ has at least $N+1$ zeroes.
Applying the above argument to $F'$ shows that $F''$ has at least $N-1$ zeroes. For $F''$ to have exactly $N-1$ zeroes it must be that $F(0)=F(T)=0$.
Returning to your original question, $f''$ has at least as many critical points as $f$, except when $0$ and $T$ are critical points, in which case $f''$ may have one less critical point than $f$.
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Solve $\theta''+g\sin(\theta)=0$ I encountered the following differential equation when I tried to derive the equation of motion of a simple pendulum:
$\frac{\mathrm d^2 \theta}{\mathrm dt^2}+g\sin\theta=0$
How can I solve the above equation?
| replacing $\sin\theta$ by $\theta$ (physically assuming small angle deflection) gives you a homogeneous second order linear differential equation with constant coefficients, whose general solution can be found in most introductory diff eq texts (or a google search). this new equation represents a simple harmonic oscillator (acceleration proportional to displacement, like a spring force).
$$
\theta''+g\theta=0
$$
has solutions $A\cos(\sqrt{g}t)+B\sin(\sqrt{g}t)$.
so, for example, if the initial displacement is $\theta_0$ and initial angular velocity is $0$ then the solution is
$$
\theta_0\cos(\sqrt{g}t)
$$
| {
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basic calculus proof - using theorems to prove stuff A function $f(x)$ is defined and continuous on the interval $[0,2]$ and $f(0)=f(2)$.
Prove that the numbers $x,y$ on $[0,2]$ exist such that $y-x=1$ and $f(x) = f(y)$.
I can already guess this is going to involve the intermediate value theorem.
So far I've defined things as such:
I'm looking to satisfy the following conditions for values x, y:
*
*$f(x) = f(x+1)$
*$f(x) = f(y)$
I've defined another function, $g(x)$ such that $g(x) = f(x+1) - f(x)$
If I can show that there exists an $x$ such that $g(x) = 0$ then I've also proven that $f(x) = f(x+1)$.
since I'm given the interval [0,2], I can show that:
$g(1) = f(2) - f(1)$,
$g(0) = f(1) - f(0)$
I'm told that $f(2) = f(0)$ so I can rearrange things to show that $g(1) = f(0) - f(1) = -g(0)$.
Ok, So i've shown that $g(0) = -g(1)$
How do I tie this up? I'm not able to close this proof. I know I need to incorporate the intermediate value theorem which states that if there's a point c in $(a,b)$ then there must be a value $a<k<b$ such that $f(k) = c $ because there's nothing else.
I thought maybe to use Rolle's theorem to state that since $f(0) = f(2)$ I know this function isn't monotonic. And if it's not monotonic it must have a "turning point" where $f'(x) = 0$ but it's not working out.
Anyway I need help with this proof in particular and perhaps some advice on solving proofs in general since this type of thing takes me hours.
Thanks.
| if $g(0)$ is positive, $g(1)$ will be negative and vice versa, so the IVT provides a root. if both are zero, $g(0)=g(1)=0=f(1)-f(0)=f(2)-f(1)$ and you're done as well.
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Question regarding infinite Blaschke product According to Gamelin's $\textit{Complex Analysis}$, a finite Blaschke product is a rational function of the form $B(z)= e^{i \varphi} (\frac{z-a_1}{1-\bar{a_1} z} \cdots \frac{z-a_n}{1-\bar{a_n} z})$ where $a_1, ..., a_n \in \mathbb{D}$ and $0 \leq \varphi \leq 2\pi$. Similarly, I would guess that an infinite Blaschke product would be of the form $e^{i \varphi} \prod_{n=1}^\infty\frac{z-a_n}{1-\bar{a_n} z}$. I believe this is supposed to satisfy what is known as the Blaschke condition, i.e. $\sum_{n=1}^\infty (1-|a_n|) < \infty$, but how is that so? Can this be verified using the log function on the infinite product?
| Actually, the infinite Blaschke product, for $|a_n|\le1$ and $|z|<1$, is defined as
$$
e^{i\varphi}\prod_{n=1}^\infty\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}\tag{1}
$$
The factor of $\;{-}\dfrac{|a_n|}{a_n}$ simply rotates $\dfrac{z-a_n}{1-\overline{a}_n z}$, which, for finite products, is incorporated into $e^{i\varphi}$. However, for infinite products, it is needed for convergence.
First, note that
$$
\begin{align}
\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}
&=|a_n|\frac{z-a_n}{|a_n|^2 z-a_n}\\
&=(1-(1-|a_n|))\left(1+\frac{z(1-|a_n|^2)}{|a_n|^2 z-a_n}\right)\\
&=(1-(1-|a_n|))\left(1+\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}(1-|a_n|)\right)\tag{2}
\end{align}
$$
where
$$
\begin{align}
\left|\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}\right|
&\le\frac{1+|a_n|}{|a_n|^2}\frac{|z|}{1-|z|}\\
&\le6\frac{|z|}{1-|z|}\tag{3}
\end{align}
$$
when $|a_n|\ge\frac12$.
Equations $(2)$ and $(3)$ say that the infinite product in $(1)$ converges absolutely when $|z|<1$ and
$$
\sum_{n=1}^\infty(1-|a_n|)\tag{4}
$$
converges. That is, the infinite product $\prod\limits_{n=1}^\infty(1+z_n)$ converges absolutely when $\sum\limits_{n=1}^\infty|z_n|$ converges.
| {
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"source": "stackexchange",
"question_score": "7",
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The differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$
The cosets of $\mathbb{Z}$ in $\mathbb{R}$ are all sets of the form $a+\mathbb{Z}$, with $0 ≤ a < 1$ a real number. Adding such cosets is done by adding the corresponding real numbers, and subtracting 1 if the result is greater than or equal to 1. -- Examples of Quotient Group, Wiki
I cannot figure out the differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$. Besides, "subtracting 1 if the result is greater than or equal to 1", what does "the result" mean here? Why do we need to subtract 1? I was wondering what is the background of $\mathbb{R}/ \mathbb{Z}$.
| $(a+\mathbb Z)+(b+\mathbb Z)$ is found by adding $a$ and $b$, the result of which is $a+b$. If $a+b<1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b)+\mathbb Z$. If $a+b\geq 1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b-1)+\mathbb Z$.
But this is only if you follow the stated convention of only listing representatives from $[0,1)$. The fact is, $(a+b)+\mathbb Z$ and $(a+b-1)+\mathbb Z$ are different names for the exact same set, so you don't really need to subtract $1$.
| {
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Existence of universal enveloping inverse semigroup (similar to "Grothendieck group") Context
In its simplest form, the Grothendieck group construction associates an abelian group to a commutative semigroup in a "universal way".
Now I'm interested in the following nilpotent commutative semigroup $N$ consisting of two elements $a$ and $b$ such that $a^2=b^2=ab=ba=a$. The corresponding Grothendieck group is the trivial group with just one element, which is a bit boring. So I asked myself whether it would be possible to construct a "universal enveloping inverse semigroup"(*) in a similar way as the Grothendieck group, and whether it would be more interesting.
I tried to compute the "universal enveloping inverse semigroup" $N_I$ for the semigroup $N=\{a,b\}$. I got $N_I = \{ a, b, b^{-1}, bb^{-1}, b^{-1}b \}$ with $(b^{-1})^2=ab^{-1}=b^{-1}a=a$. What I find surprising is that $N_I$ is not commutative, even so $N$ is commutative. So I tried to compute the "universal enveloping commutative inverse semigroup" $N_C$ instead and got $N_C = \{ a \}$.
(*)Note: In a semigroup $S$, we say that $y\in S$ is an inverse element of $x\in S$, if $xyx=x$ and $yxy=y$. A semigroup $S$ is called an inverse semigroup, if each $x\in S$ has a unique inverse element $x^{-1}\in S$. It's easy to see that $xx^{-1}$ and $x^{-1}x$ are idempotent, that all idempotent elements in an inverse semigroup commute, and that $(xy)^{-1}=y^{-1}x^{-1}$. So at least superficially, inverse semigroups seem to be nice generalization of groups and can have a zero element without being trivial.
Question
Does the "universal enveloping inverse semigroup" (and the "universal enveloping commutative inverse semigroup") of a semigroup $S$ always exist? I guess the answer is yes and this probably follows from some theorem of universal-algebra. Similarly, I guess that the "universal enveloping regular semigroup" doesn't always exist and wonder whether this also follows from some theorem of universal-algebra.
| I now found out how to prove that no "universal enveloping regular semigroup" exists for the example given in the question. (The existence of the other two cases has already been proved in the answer by Martin Wanvik.)
Let
$a=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$,
$b=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,
$b'=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$,
$c=\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$,
$c'=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$,
$d=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$,
$d'=\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$,
$e=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, and
$f=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. Let $N := \{ a,b \}$ and $N_I := \{ a,b,b',e,f \}$ (as in the question with $b'=b^{-1}$, $e=bb^{-1}$ $f=b^{-1}b$). It is easy to see that $S := \{ a,b,b',c,c',d,d',e,f \}$ is a regular group, and that $N_I$ and $R := \{ a,b,c,d,e \}$ are regular sub-semigroups of $S$. Now $b^{-1}=b'$ in $N_I$ and $c$ is the unique inverse element of $b$ in $R$. So if there would exist a regular semigroup $N_R$ containing $N$ as sub-semigroup, for which it is possible to uniquely extend any homomorphism from $N$ to $N_I$ or $R$, then the extension to $N_R$ of a homomorphism $h$ from $N$ to $S$ (with $h(a)=a$ and $h(b)=b$) won't be unique.
| {
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"source": "stackexchange",
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Number of solutions of $x^2=1$ in $\mathbb{Z}/n\mathbb{Z}$ Next is what I have worked out to the moment.
$1$ and $-1$ are roots for all $n$.
$x \in \mathbb{Z}/n\mathbb{Z},\ $ $x^2\equiv1 \Leftrightarrow (x-1)(x+1)\equiv0 \Leftrightarrow \exists k \in \mathbb{Z}/n\mathbb{Z}: k(k+2)\equiv0 $.
But how can it be applied to find other roots?
| (I'd have to check the details in the following but it provides some rough ideas.)
Write $n = \prod_i p_i^{\nu_i}$ and use the Chinese remainder theorem to obtain a system of equations
$$ x^2 \equiv 1 \pmod{p_i^{\nu_i}}$$
Of course, for every $p_i$, $x=\pm 1$ provides a solution. Based on some quick calculations, I think that:
*
*If $p_i>2$, these are the only solutions.
*If $p_i=2$ and $\nu_2 \geq 3$, then there are 4 solutions.
*If $p_i=2$ and $\nu_2 = 2$, then $x=\pm 1$ are solutions.
*If $p_i=2$ and $\nu_2 = 1$, there is $1$ solution, because $+1 = -1$.
To confirm this: investigate when a prime power can divide two numbers that differ by $2$, i.e when $p_i^{\nu_i} \mid (x+1)(x-1)$.
We may then use the Chinese remainder theorem to reassemble the solutions to find solutions mod $n$: each combination of solutions modulo the prime-powers will uniquely determine a solution mod $n$.
So the number of solutions is (where $\omega(n)$ is the number of
distinct prime factors of $n$):
*
*$2^{\omega(n)}$ if $n$ is odd or $\nu_2 = 2$.
*$2^{\omega(n)+1}$ if $\nu_2 \geq 3$
*$2^{\omega(n)-1}$ if $\nu_2 = 1$
A proof of the above can be based on theorem 4.19 and 4.20 in Basic Algebra vol 1, by N. Jacobson, discussing the structure of $U_{p^\nu}$, this is the group of units in $\mathbb Z/p^\nu \mathbb Z$: it is cyclic if $p>3$, $p^\nu=2$ or $p^\nu=4$ and isomorphic to $C_2\times C_{2^{\nu-2}}$ if $p=2$ and $\nu\geq 3$.
| {
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show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ $n = 4k + 3 $
We start by letting $a \not\equiv 0\pmod n$ $\Rightarrow$ $a \equiv k\pmod n$ .
$\Rightarrow$ $a^{4k+2} \equiv 1\pmod n$
Now, I know that the contradiction will arrive from the fact that if we can show $a^2 \equiv 1 \pmod n $ then we can say $b^2 \equiv -1 \pmod n $ then it is not possible since solution exists only for $n=4k_2+1 $ so $a \equiv b\equiv 0 \pmod n $
So from the fact that $a^{2^{2k+1}} \equiv 1 \pmod n$ I have to conclude something.
| This is solution from Prasolov V.V. Zadachi po algebre, arifmetike i analizu (В. В. Прасолов. Задачи по алгебре, арифметике и анализу.) This book (in Russian) is freely available at http://www.mccme.ru/free-books/
The problem appears there as Problem 31.2.
We want to show that if $p=4k+3$ is a prime number and $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$.
My translation of the solution given there:
Suppose that one of the numbers $a$, $b$ is not divisible by $p$. Then the other one is not divisible by $p$, either. Thus from Fermat's little theorem we get $a^{p-1} \equiv 1 \pmod p$ and $b^{p-1} \equiv 1 \pmod p$. This implies $a^{p-1}+b^{p-1} \equiv 2 \pmod p$.
On the other hand, the number $a^{p-1}+b^{p-1}=a^{4k+2}+b^{4k+2}=(a^2)^{2k+1}+(b^2)^{2k+1}$ is divisible by $a^2+b^2$, and thus it is divisible by $p$.
| {
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An ideal and all elements of the quotient ring
Defintion 1 An ideal I of R is an additive subgroup so that $$a\in R , s\in I \Rightarrow as,sa \in I$$
The ring R/I is called the quotient ring.
Example 1 : $R=\mathbb{Z}[x], I=nR (n\in \mathbb{Z})$. Then $R/I "=" (\mathbb{Z}/n\mathbb{Z})[x]$. Or $I=xR$. What are the $A+I (A\in \mathbb{Z}[x])?$
A=a+bx+... , so they are equal to the $a+I (a\in \mathbb{Z})$ because if $A=a+bx+...$, then $A-a = bx+... \in I$. So $R/I "=" \mathbb{Z}$ or : $I=2R+xR = \{ 2A+xB; A,B \in R\}$. So we can say that there are at most $2$ different side classes $0+I, 1+I$ and they can not be equal because otherwise $1\in I$, so $1=2A+xB; A,B\in \mathbb{Z}[x]$. It is true that $R/I$ is identical to $\mathbb{F}_{2}=\{0,1\}$
Example 2: R=$\mathbb{F}_{2}[x]$, I=xR, $R/I = \{0+I, 1+I\} "=" \mathbb{F}_{2}$. Or $I=x^{2}R, A+I = (a+bx+...)+I = (a+bx)+I (a,b \in \mathbb{F}_{2})$. So there are at most 4 side classes : $0+I,1+I,x+I,x+1+I$ and they are all different from eachother.
This is an outtake from my script. I do not understand how in example 1 it is concluded that there are at most 2 different side classes and 4 side classes with $(\mathbb{F}_{2}/x^{2}\mathbb{F}_{2})[x]$ in the example 2. How does one find the side classes? Are they elements of the ring? It seems to be easy to find them for simple rings (the elements of $\mathbb{F}_{5}$ are the equivalence classes 0,1,2,3,4) , but how to find them (the elements) for quotient rings?
| "Example 1" seems to be three examples: (i) $R/I$ with $R=\mathbb{Z}[x]$, $I=nR$; (ii) $R/I$ with $I=xR$; and (iii) $R/I$ with $I=2R+xR$.
In the third case, $I$ consists of all polynomials with integer coefficients that have even constant term: all such polynomials can be written as a multiple of $2$ plus a multiple of $x$; and conversely, if a polynomial is a multiple of $2$ plus a multiple of $x$, then its constant term must be even.
When are $a_0 + a_1x+\cdots+a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ in the same lateral class modulo this $I$? If and only if
$$(a_0+\cdots + a_nx^n) - (b_0 + \cdots + b_mx^m) = (a_0-b_0) + (a_1-b_1)x + \cdots \in I.$$
In order for the difference to be in $I$ what do we need? We need $a_0-b_0$ to be even; so the two polynomials are in the same coset if $a_0$ and $b_0$ have the same parity. So there are at most two lateral classes: one for polynomials with even constant term, one for polynomials with odd constant term.
(Conversely, polynomials with constant terms of opposite parity are not in the same lateral class, so there are in fact exactly two lateral classes).
In Example 2, you have $R=F_2[x]$ and $I=x^2R$. A polynomial is in $I$ if and only if it is a multiple of $x^2$. If $a_0+a_1x+\cdots + a_nx^n$ and $b_0+b_1x+\cdots+b_mx^m$ have $a_0=b_0$ and $a_1=b_1$, then their difference is a multiple of $x^2$; since there are two possible choices for $a_0$ and two possible choices for $a_1$, there are at most four possible lateral classes: one for each choice combination. Because every polynomial will have constant and linear terms equal to $0$, both equal to $1$, constant term $0$ and linear term $1$, or constant term $1$ and linear term $0$. So there are at most 4 possibilities that yield different lateral classes.
(Conversely, those four possibilities are pairwise distinct, since the difference of, say, a poynomial with constant term 1 and linear term 0, and one of linear and constant terms equal to $1$, would not be a multiple of $x^2$).
| {
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roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to
$$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$$
So now we need to get the values from t for $k = 0,...5$.
How to get the values of t from the following identity then?
$$
\begin{align}
\frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
(t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\
\end{align}
$$
And now?
$$
t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1}
$$
So I've got six roots for $k = 0,...5$ as follows
$$
t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1}
$$
Is this right? But how can it be that the bottom equals $0$ for $k=0$?
I don't exactly know how to simplify this:
$$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$$
| Notice that $t=1$ is not a root. Divide by $(t-1)^6$.
If $\omega$ is a root of $z^6 - 1$, then a root of the original equation is given by $\frac{t+1}{t-1} = \omega$.
| {
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A better approximation of $H_n $ I'm convinced that
$$H_n \approx\log(n+\gamma) +\gamma$$ is a better approximation of the $n$-th harmonic number than the classical $$H_n \approx \log(n) +\gamma$$
Specially for small values of $n$. I leave some values and the error:
Just to make things clearer, I calculate the value between two numbers as follows.
Say $n$ is the optimal and $a$ is the apporximation, then $E = \frac{n-a}{n}$. $L_1$ stands for my approximation and $L_2$ for the classical one, and the errors $E_2$ and $E_1$ correspond to each of those (I mixed up the numbers).
It is clear that this gives an over estimate but tends to the real value for larger $n$.
So, is there a way to prove that the approximation is better?
NOTE: I tried using the \begin{tabular} environment but nothing seemed to work. Any links on table making in this site?
| The asymptotic expansion of the Harmonic numbers $H_n$ is given by
$$\log n+\gamma+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right).$$
The Maclaurin series expansion of the natural logarithm tells us $\log(1+x)=x+\mathcal{O}(x^2)$, and we can use this in your formula by writing $\log(n+\epsilon)=\log n+\log(1+\epsilon/n)$ and expanding:
$$\log(n+\gamma)+\gamma=\log n+\gamma\;\;\;+\frac{\gamma}{n}+\mathcal{O}\left(\frac{1}{n^2}\right).$$
Your approximation is asymptotically better than the generic one because the $\gamma=0.577\dots$ in its expansion is closer to the true coefficient $\frac{1}{2}$ than the illicit coefficient $0$ in the generic formula given by the usual $H_n\sim \log n +\gamma+0/n$. This also explains why it is asymptotically an over estimation.
As marty said in his answer, the expansion comes from the Euler-Maclaurin formula:
$$\sum_{n=a}^b f(n)=\int_a^b f(x)dx+\frac{f(a)+f(b)}{2}+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right).$$
Here we let $a=1,b=n$ (rewrite the index to a different letter) and $f(x)=1/x$.
| {
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Union of uncountably many subspaces Supposing, $\{V_t\}, t > 0$ are an uncountable number of linear subspaces of $\mathbb{R}^n$. If $\bigcup_{t>0} V_t = \mathbb{R}^n$ is it true that $V_T = \mathbb{R}^n$ for some $T>0$?
Any help is appreciated. Thanks.
EDIT: I have forgot to add the condition that $V_t$ are increasing.
| In general the answer is no. Consider family of subspaces of the form
$$
V_t=\{x\in\mathbb{R}^n:x_1\cos\frac{2\pi}{t+1}+x_2\sin\frac{2\pi}{t+1}=0\}
$$
| {
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chromatic number and subgraph
Prove that any graph $G$ with $n$ vertices and $ \chi(G)=k$ has a subgraph $H$ such that $ H \simeq \overline{K_p}$ where $p=n/k$ and $K_p$ is the complete graph with $n/k$ vertices.
My attempt: Because $ \chi(G)=k$ it must be $G \subseteq K_{p_1 p_2 \cdots p_k} $ where $\displaystyle{\sum_{j=1}^{k} p_j =n}$.
Can I consider now that $ p_j =n/k$ for all j?
If no then the other cases is to have $ p_j >n/k$ for some $j \in \{1, \cdots ,k\}$.
But now how can I continue?
| If $\chi(G) = k$, it means we can color the graph with $k$ colors, $c_1, \ldots, c_k$. Each color class, $c_i$, consists of some vertices $V_i$. Necessarily, the vertices in $V_i$ are independent, or we could not color them all the same color, $c_i$.
Now, assume that every color class contains less than $n / k$ vertices. Then the total number of vertices in the graph is $|G| < k \cdot (n/k) = n$. This isn't possible since we assume $|G| = n$. Therefore, some color class contains at least $n / k$ vertices. Since the vertices in a color class are independent, we have an independent set of size at least $n / k$.
| {
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Integrable function Please help me prove this.
Suppose $f$ is defined in$\left [ a,b \right ]$ ,$f\geq 0$, $f$ is integrable in$ [a,b]$ and $\displaystyle\int_{a}^{b}fdx=0$
prove:
$\displaystyle\int_{a}^{b}f(x)^2dx=0$
Thanks a lot!
| My first inclination would be to say that since $f\geq0$ and $\int_a^bfdx=0$, $f(x)=0$ for $x$ such that $a\leq x\leq b$. Because of this, $\int_b^af(x)^2dx=0$. You will probably need to fill in a fair amount of reasoning to make this hold up, though.
Also, suggestion - accept some answers to your questions. It makes people more likely to answer in the future.
| {
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Proving that the "real part" function is continuous I want to use the definition of a limit,
$|f(z) - w_0| < \varepsilon$ whenever $0 < |z - z_0| < \delta$
to prove
$$\lim_{z \to z_0} \mathop{\rm Re}(z) = \mathop{\rm Re}(z_0)$$
By intuition this is obvious but I dont know how to show it using the defn. of a limit. This is question 1(a) from the book Complex Variables and Applications.
Here's the basic manipulation I have made going by an example in the book, I dont know where to go from here...
$$|\mathop{\rm Re}(z)-\mathop{\rm Re}(z_0)| = |x - x_0| = |x| - |x_0| = x - x_0$$
| We have
$|z_0-z|^2=(\Re(z_0-z))^2+(\Im(z_0-z))^2\geq (\Re(z_0-z))^2$ so $|z_0-z|\geq |z_0-z|$. Now, check that $\delta=\varepsilon$ in the definition of the limit works.
| {
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eigenvalues and eigenvectors of $vv^T$ Given a column vector $v$ in $\mathbb{R}^n$, what are the eigenvalues of matrix $vv^T$ and associated eigenvectors?
PS: not homework even though it may look like so.
| The columns of the matrix are $v_1v,\ldots,v_nv$ so if we take two column these one are linearly dependent, and so $vv^T$ has a rank of at most $1$. It's $0$ if $v=0$, and if $v\neq 0$, we have $\mathrm{Tr}A=|v|^2$ so the eignevalues are $0$ with multiplicity $n-1$ and $|v|^2$ with multiplicity $1$.
| {
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proof that if $AB=BA$ matrix $A$ must be $\lambda E$ Let $A \in Mat(2\times 2, \mathbb{Q})$ be a matrix with $AB = BA$ for all matrices $B \in Mat(2\times 2, \mathbb{Q})$.
Show that there exists a $\lambda \in \mathbb{Q}$ so that $A = \lambda E_2$.
Let $E_{ij}$ be the matrix with all entries $0$ except $e_{ij} = 1$.
$$
\begin{align}
AE_{11} &= E_{11}A \\
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\\
\left( \begin{array}{cc}
a_{11} & 0 \\
a_{21} & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
a_{11} & a_{12} \\
0 & 0 \\
\end{array} \right) \\
\end{align}
$$
$\implies a_{12} = a_{21} = 0$
And then the same for the other three matrices $E_{12}, E_{21}, E_{22}$ …
I guess it's not the most efficient way of argueing or writing it down … ? Where's the trick to simplify this thingy?
| You wish to show that the center of the ring of matrices is the set of scalar matrices. Observe that the matrices commuting with diagonal matrices must be diagonal matrices. Hence the center must be contained in the set of diagonal matrices. Now a diagonal matrix that commutes with an arbitrary matrix must be a scalar matrix.
| {
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Ring theory notation question I was wondering what the notation $R[a]$ really stands for, if $a\in K$, where $K$ is a ring and $R$ is a subring of $K$.
In my book they define $\mathbb{Z}[\sqrt2]=\{a+b\sqrt2|a,b \in \mathbb{Z}\}$.
So, my guess is that $R[a] = \{P(a)\mid P \in R[X]\}$. Since for $\mathbb{Z}[\sqrt2]$ this is the case, or is this just a coincidence?
| By definition, $R[a]$ is the smallest subring of $K$ that contains both $R$ and $a$.
As you note, if $p(x)\in R[x]$, then $p(a)\in R[a]$ necessarily. Therefore,
$$\{p(a)\mid p(x)\in R[x]\} \subseteq R[a].$$
Conversely, note that
$$\{p(a)\mid p(x)\in R[x]\}$$
contains $a$ (as $p(a)$ where $p(x)=x$), contains $R$ (the constant polynomials); and is a subring of $K$: every element like is $K$; it is nonempty; it is closed under differences (since $p(a)-q(a) = (p-q)(a)$); and it is closed under products (since $p(a)q(a) = (pq)(a)$). Thus, $R[a]\subseteq \{p(a)\mid p(x)\in R[x]\}$. Hence we have equality.
More generally, if $a$ is integral over $R$ (satisfies a monic polynomial with coefficients in $R$, then letting $n$ be the smallest degree of a monic polynomial that is satisfied by $a$), then $R[a] = \{r_0 + r_1+\cdots + r_{n-1}a^{n-1}\mid r_i\in R\}$.
To see this, note that clearly the right hand side is contained on the left hand side. To prove the converse inclusion, let $p(x)$ be a monic polynomial of smallest degree such that $p(a)=0$. By doing induction on the degree, we can prove in the usual way that every element $f(x)$ of $R[x]$ can be written as $f(x)=q(x)p(x) + r(x)$, where $r(x)=0$ or $\deg(r)\lt deg(p)$; the reason being that the leading coefficient of $p$ is $1$, so we can perform the long-division algorithm without problems. Thus, $f(a) = q(a)p(a)+r(a) = r(a)$, so every element of $R[a]$ can be expressed as in the right hand side.
In the case where $a=\sqrt{2}$, $R=\mathbb{Z}$, $K=\mathbb{R}$ (or $K=\mathbb{Q}(\sqrt{2})$), we have that $a$ satisfies $x^2-2$< so that is why we get
$$\mathbb{Z}[\sqrt{2}] = \{ p(\sqrt{2})\mid p(x)\in\mathbb{Z}[x]\} = \{r_0+r_1\sqrt{2}\mid r_0,r_1\in\mathbb{Z}\}.$$
| {
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The completion of a Boolean algebra is unique up to isomorphism Jech defines a completion of a Boolean algebra $B$ to be a complete Boolean algebra $C$ such that $B$ is a dense subalgebra of $C$.
I am trying to prove that given two completions $C$ and $D$ of $B$, then the mapping $\pi: C \rightarrow D$ given by $\pi(c) = \sum^D \{ u \in B \ \vert \ u \le c \} $ is an isomorphism. I understand that $c \not= 0 \implies \pi(c) \not= 0$, and that given any $d \in D$, I can write it as $d = \sum^D \{ u \in B \ \vert \ u \le d \} $, but I am unsure how to proceed. Any help would be appreciated.
| Edit: Ignore this answer. See Loronegro's.
Apparently I haven't done enough on Math.SE before, so I have to post this as an answer instead of a comment.
Hopefully this is helpful: first notice that $\pi c \le c$ for every $c\in C$ (the supremum of a bunch of things that are $\le c$ is also $\le c$). From this, you can deduce that, for every $u\in B$, $u\le c$ iff $u\le \pi c$. Therefore the sets $\{u\in B : u\le c\}$ and $\{u\in B : u\le \pi c\}$ are identical, so $\pi^{-1}\pi = 1$. By starting the argument with $\pi^{-1}c \le c$ instead of $\pi c \le c$, you can also conclude that $\pi\pi^{-1} = 1$.
You should feel cheated now, as I have avoided any mention of density. I guess proving that $\sum^C \{u\in B: u\le c\} = c$ requires the density of $B$.
| {
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What does integration do? I know that integrals are used to compute the area under a curve. Let's say I have $y = x^2$. It creates smaller rectangles and then add up the sum (assuming that rectangles are going infinitely in number and is like going to a limit).
But I recently encountered a problem in my mind. Suppose we have a function, $y = x^2$. If we integrated it, we simply get the anti derivative of it which is $x^3/3$, assuming that the area is not of concern. What is the correlation of $x^3/3$ to $x^2$? I mean, it simply likes transforms a function into another function, but I can't get a clearer picture. When we graph $x^2$ and $x^3/3$, there is no connection visually. They are simply different graphs.
Thanks and I hope your comments can clear up my mind.
|
The key word here is instantaneous. Although the 2 graphs are "different". They are linked to each other through the "instantaneous area". What I mean is that if you take a point on your function and calculate the value of the area of the function from, say, 0 to that point. That value of area is the same as g(x). $(g(x) = \int(f(x))$
Consider:
$x =3$, the area of the graph from 0 to 3 is the same as the value of g(x) i.e. $\int(f(x))$
| {
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How does $\exp(x+y) = \exp(x)\exp(y)$ imply $\exp(x) = [\exp(1)]^x$? In Calculus by Spivak (1994), the author states in Chapter 18 p. 341 that
$$\exp(x+y) = \exp(x)\exp(y)$$ implies
$$\exp(x) = [\exp(1)]^x$$
He refers to the discussion in the beginning of the chapter where we define a function $f(x + y) = f(x)f(y)$; with $f(1) = 10$, it follows that $f(x) = [f(1)]^x$. But I don't get this either. Can anyone please explain this? Many thanks!
| I like (as I have done here before) to start with a functional equation
and derive properties of the function.
If $f(x+y) = f(x) f(y)$ and $f$ is differentiable (and non-zero somewhere),
$f(0) = 1$ and
$f(x+h)-f(x) = f(x)f(h)-f(x)
=f(x)(f(h)-1)
=f(x)(f(h)-f(0))
$
so
$$(f(x+h)-f(x))/h = f(x)(f(h)-f(0))/h.$$
Letting $h \to 0$,
$f'(x) = f'(0) f(x)$.
From this, $f(x) = \exp(f'(0)x)$.
This also works for $\ln$, $\arctan$, and $\sin$ and $\cos$.
Functional equations are fun!
| {
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If $A^2 = I$ (Identity Matrix) then $A = \pm I$ So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the questions is this:
If $A^2 = I$ (Identity Matrix), then $A = \pm I$ ?
I'm pretty sure it is true but the answer says it's false. How can this be false (maybe it's a typography error in the book)?
| I know $2·\mathbb C^2$ many counterexamples, namely
$$A=c_1\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}+c_2\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}\pm\sqrt{c_1^2+c_2^2\pm1}\begin{pmatrix}
0&-1\\
1&0
\end{pmatrix},$$
see Pauli Matrices $\sigma_i$.
These are all such matrices and can be written as $A=\vec e· \vec \sigma$, where $\vec e^2=\pm1$.
| {
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Prove that $n! > \sqrt{n^n}, n \geq 3$
Problem Prove that $n! > \sqrt{n^n}, n \geq 3$.
I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.
| $(n!)^2 = (n \times 1) \times ((n-1)\times 2) \times \cdots \times (1 \times n) \gt n^n$
since $(n-1)\times 2 = 2n-2 \gt n$ iff $n \gt 2$.
Then take the square root.
| {
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Finding limit of fraction with square roots: $\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$ I have been looking at this for five minutes, no clue what to do.
$$\lim_{r\to 9} \frac {\sqrt{r}} {(r-9)^4}$$
| The limit is $+\infty$ because the numerator approaches a positive number and the denominator approaches $0$ from above. Sometimes one says the limit "doesn't exist" when one means there is no real number that is the limit, so you could put it that way.
| {
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Trying to figure out a complex equality An answer to a comlex equation I was working on was
$$z = \frac{1}{2} + \frac{i}{2}$$
My teacher further developed it to be
$$e^{\frac{i\pi}{4}-\frac{1}{2}\ln{2}}$$
And here's what I tried:
$$z = \frac{1}{2} + \frac{i}{2} = z = \frac{1}{\sqrt{2}}e^{\frac{i\pi}{4}}
= e^{\frac{1}{2}\ln{2}}e^{\frac{i\pi}{4}} = e^{\frac{1}{2}\ln{2}+\frac{i\pi}{4}}$$
I feel this is stupid, but I can't see why we have different answers. Anyone? Thanks!
| The mistake occurs here:
$$\frac{1}{\sqrt{2}}e^{\frac{i\pi}{4}}
= e^{\frac{1}{2}\ln{2}}e^{\frac{i\pi}{4}}.$$
In fact, we have
$$e^{\frac{1}{2}\ln{2}}=2^{\frac{1}{2}}=\sqrt{2}.$$
Therefore, we should have
$$\frac{1}{\sqrt{2}}=(\sqrt{2})^{-1}
= e^{-\frac{1}{2}\ln{2}}.$$
Mixing this, your answer matches with your teacher's answer.
| {
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Finding the n-th root of a complex number I am trying to solve $z^6 = 1$ where $z\in\mathbb{C}$.
So What I have so far is :
$$z^6 = 1 \rightarrow r^6\operatorname{cis}(6\theta) = 1\operatorname{cis}(0 + \pi k)$$
$$r = 1,\ \theta = \frac{\pi k}{6}$$
$$k=0: z=\operatorname{cis}(0)=1$$
$$k=1: z=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3} + i}{2}$$
$$k=2: z=\operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{1 + \sqrt{3}i}{2}$$
$$k=3: z=\operatorname{cis}\left(\frac{\pi}{2}\right)=i$$
According to my book I have a mistake since non of the roots starts with $\frac{\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots.
Can anyone please tell me where my mistake is? Thanks!
| Hint: Imagine that there is a unit circle on the Argand Plane. Now the roots will be the 6 equidistant points on that circle each having the argument as multiple of $\frac{2\pi}{6}=\frac{\pi}{3}$
| {
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Is it possible to generate a unique real number for each fixed length sequence of real numbers? Let A be the set of all sequences of real numbers of size $n$. Does there exist an injection from A to R?
I know this is possible if we are only considering integers instead of real numbers; But I am not sure if it is possible if we consider real numbers instead.
For integers, we can generate a unique integer using the following method:
Let S be a sequence of integers of size n. $S = s_1,s_2,\ldots,s_n$. Let $P = p_1,p_2,\ldots,p_n$ be the sequence of $n$ primes. Then $f(S) = (p_1^{s_1})(p_2^{s_2})\cdots(p_n^{s_n})$ creates a unique integer for each sequence $S$.
If each $s_i$ was a real number instead, would $f(S)$ still be an injection? If not, is there an alternative invective function from A to R?
edit:
I fixed some of my poor wording.
I am trying to find an injection function from A to R. Such a function does exist and the function I proposed clearly does not work (From the comments).
If possible, I would like to find an injective function that does not involve directly manipulating the decimal expansions.
| You can easily create an injection even for infinite sequences of reals (an injective mapping from sequences to real numbers). So your request is too weak.
See Boas Primer of Real Functions Exercise 3.13 and its solution in the back.
Also see my thread about this exercise.
| {
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what is Prime Gaps relationship with number 6? Out of the 78499 prime number under 1 million. There are 32821 prime gaps (difference between two consecutive prime numbers) of a multiple 6. A bar chart of differences and frequency of occurrence shows a local maximum at each multiple of 6. Why is 6 so special?
| take any integer $n> 3$, and divide it by $6$. That is, write
$n = 6q + r$
where $q$ is a non-negative integer and the remainder $r$ is one of $0$, $1$, $2$, $3$, $4$, or $5$.
If the remainder is $0$, $2$ or $4$, then the number $n$ is divisible by $2$, and can not be prime.
If the remainder is $3$, then the number $n$ is divisible by $3$, and can not be prime.
So if $n$ is prime, then the remainder $r$ is either
*
*$1$ (and $n = 6q + 1$ is one more than a multiple of six), or
*$5$ (and $n = 6q + 5 = 6(q+1) - 1$ is one less than a multiple of six).
| {
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A question concerning maps of $G$-coverings I am having difficulties thinking about how an argument for the following exercise should proceed:
Let $p: Y \rightarrow X$ and $q: Z \rightarrow X$ be $G$-coverings (i.e., covering maps such that $X = Y /G = Z/G$ (quotient spaces)), with $X$ connected and locally path connected. Let $\phi: Y \rightarrow Z$ and $\psi: Y \rightarrow Z$ be maps of $G$-coverings (i.e., a covering homomorphism such that $\phi(g \cdot y) = g \cdot \phi(y)$ for all $g \in G$ and $y \in Y$, same with $\psi$). Assume that $\phi(y) = \psi(y)$ for some $y \in Y$. Show that $\phi(y) = \psi(y)$ for every point $y \in Y$.
If $Y$ is connected, this should be an immediate consequence of the unique lifting property for maps, but in the general case I am lost, and I am curious to know if anyone visiting would know how to proceed.
| (Sorry I can't comment)
Show that each connected component of $Y$ contains a preimage of every $z\in Z$. Fix a connected component $Y_0$ of $Y$. Take a $y\in Y$. Using the transitivity of $G$, translate $y$ to $Y_0$. Then use the fact that you know the statement is true for $Y_0$.
| {
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intermediate step in proving old Ramsey lower bound Let $r(n,n)=r(n)$ be the usual Ramsey number of a graph. It is known that $$\frac{1}{e\sqrt{2}}n2^{n/2}<r(n)$$ as a lower bound for $r(n).$
Now, in the proof given in the book Erdős on Graphs by Graham and Chung, as an intermediate step this is given:
$$2^{\binom{m}{2}}>\binom{m}{n}2^{\binom{m}{2}-\binom{n}{2}+1}\;,\tag{*}$$ and that this implies that $$m\ge\frac{1}{e\sqrt{2}}n2^{n/2}\;.\tag{**}$$
I cannot figure out how $(*)$ implies $(**)$. Can someone please explain this?
| In fact, the inequality $(**)$ should be the other way around.
As Austin Mohr noted, Stirling's formula comes in handy here. The form that I will use is
$$n! \sim \biggl(\dfrac{n}{e}\biggr)^n \sqrt{2\pi n}. \tag{1}$$
Also, I assume that $m \to \infty$ and that $m - n \to \infty$.
We start by observing that the inequality $(*)$ is equivalent to
$$\dbinom{m}{n} < 2^{\binom{n}{2} - 1}. \tag{2}$$
Since
$$\dbinom{m}{n} \geq \dfrac{(m - n)^n}{n!},$$
we have from $(2)$ that
$$(m - n)^n < n! 2^{\binom{n}{2} - 1}.$$
Plugging in Stirling's formula $(1)$ on the right-hand side gives
$$m^n \biggl(1 - \dfrac{n}{m}\biggr)^n < \biggl(\dfrac{n}{e}\biggr)^n \sqrt{\dfrac{\pi n}{2}} 2^{\binom{n}{2}}.$$
Taking $n$th roots, we get
$$m \biggl(1 - \dfrac{n}{m}\biggr) < \dfrac{n}{e} 2^{\frac{n - 1}{2}} \biggl(\dfrac{\pi n}{2}\biggr)^{1/2n}.$$
Finally, observing that $(\frac{\pi n}{2})^{1/2n} / (1 - \frac{n}{m}) \to 1$ as $m$, $n \to \infty$, we end up with
$$m < \dfrac{1}{e\sqrt{2}} n 2^{n/2},$$
as desired.
| {
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How to show a matrix is full rank? I have some discussion with my friend about matrix rank.
But we find that even we know how to compute rank, we don't know how to show the matrix is full rank.
How to show this?
| If you are talking about square matrices, just compute the determinant.
If that is non-zero, the matrix is of full rank.
If the matrix $A$ is $n$ by $m$, assume wlog that $m\leq n$ and compute all determinants of $m$ by $m$ submatrices. If one of them is non-zero, the matrix has full rank.
Also, you can solve the linear equation $Ax=0$ and figure out what dimension the space of solutions has. If the dimension of that space is $n-m$, then the matrix is of full rank.
Note that a matrix has the same rank as its transpose.
| {
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Two ants on a triangle puzzle Last Saturday's Guardian newspaper contained the following puzzle:
Two soldier ants start on different vertices of an equilateral triangle. With each move, each ant moves independently and randomly to one of the other two vertices. If they meet, they eliminate each other. Prove that mutual annihilation is eventually assured. What are the chances they survive... exactly N moves.
I understand that the probability of a collision on any one move is 1/4, but I don't understand the quoted proof of eventual annihilation:
If the chances of eventual mutual annihilation are are P, then P = 1/4 + 3/4 P, so P = 1.
I scratched my head for a while but I still couldn't follow it. Do they mean the probability in the limit of an infinite number of moves? Or is there something crucial in that calculation of P that I'm not getting?
| Note that the fact that it's a triangle is irrelevant. The ants could move to random vertices on any $n$-gon and the result would be the same.
Put another way, if two people repeatedly choose random integers from $1$ to $n$, they will eventually choose the same number.
| {
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Does $\int_{1}^{\infty}\sin(x\log x)dx $ converge? I'm trying to find out whether $\int_{1}^{\infty}\sin(x\log x)dx $ converges, I know that $\int_{1}^{\infty}\sin(x)dx $ diverges but $\int_{1}^{\infty}\sin(x^2)dx $ converges, more than that, $\int_{1}^{\infty}\sin(x^p)dx $ converges for every $p>0$, so it should be converges in infinity. I'd really love your help with this.
Thanks!
| Since $x\log(x)$ is monotonic on $[1,\infty)$, let $f(x)$ be its inverse. That is, for $x\in[0,\infty)$
$$
f(x)\log(f(x))=x\tag{1}
$$
Differentiating implicitly, we get
$$
f'(x)=\frac{1}{\log(f(x))+1}\tag{2}
$$
Then
$$
\begin{align}
\int_1^\infty\sin(x\log(x))\;\mathrm{d}x
&=\int_0^\infty\sin(x)\;\mathrm{d}f(x)\\
&=\int_0^\infty\frac{\sin(x)}{\log(f(x))+1}\mathrm{d}x\tag{3}
\end{align}
$$
Since $\left|\int_0^M\sin(x)\;\mathrm{d}x\right|\le2$ and $\frac{1}{\log(f(x))+1}$ monotonically decreases to $0$, Dirichlet's test (Theorem 17.5) says that $(3)$ converges.
| {
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An epimorphism from $S_{4}$ to $S_{3}$ having the kernel isomorphic to Klein four-group Exercise $7$, page 51 from Hungerford's book Algebra.
Show that $N=\{(1),(12)(34), (13)(24),(14)(23)\}$ is a normal subgroup
of $S_{4}$ contained in $A_{4}$ such that $S_{4}/N\cong S_{3}$ and
$A_{4}/N\cong \mathbb{Z}_{3}$.
I solved the question after many calculations. I would like to know if is possible to define an epimorphism $\varphi$ from $S_{4}$ to $S_{3}$ such that $N=\ker(\varphi)$.
Thanks for your kindly help.
| Here is an approach:
Proof Idea: $S_4/N$ is a group with 6 elements. There are only two such groups, one is cyclic and the other is $S_3$, and $S_4/N$ cannot have elements of order $6$ thus must be $S_3$.
First it is easy to show that $N$ is normal in $S_4$. It follows that $S_4/N$ is a group with $6$ elements. Let us call this factor $G$.
Now, $G$ is a group of order $6$. As no element of $S_4$ has order $6$, it follows that $G$ has no element of order $6$.
Pick two elements $x,y \in G$ such that $\operatorname{ord}(x)=2$ and $\operatorname{ord}(y)=3$. Then, $e, y, y^2, x, xy, xy^2$ must be 6 distinct elements of $G$, and hence
$$G= \{ e, y, y^2, x, xy, xy^2 \}$$
Now, let us look at $yx$. This cannot be $xy$, as in this situation we would have $\operatorname{ord}(xy)=6$. This cannot be $e, x, y, y^2$ either. This means that
$$yx=xy^2$$
Now it is trivial to construct an isomorphism from $G$ to $S_3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eigenvalues of a infinitesimal generator matrix Consider a Markov process on a finite state space $S$, whose dynamic is determined by a certain infinitesimal generator $Q$ (that is a matrix in this case) and an initial distribution $m$.
1) Is there anything general that can be said on the spectrum of the matrix $-Q$?
2) Is there any bounds on the eigenvalues of $-Q$? And on their ratios?
I am interested in linear algebra results as well as more probabilistic ones, maybe using information about the structure of the state space or on the process itself.
If can be of any help, for the problem I am addressing, I usually have a quite big state space and the Q-matrix is quite sparse, but unfortunately without any symmetry or nice structure.
For discrete-time Markov chains there is an huge literature linking properties of the spectrum of the transition matrix $P$ to various mixing times, geometric properties of the state space, etc. Of course one can move from the continuous setting to the discrete one via uniformization and thus "translate" most of these results, but I was wondering the following
3) Is there any bounds on the eigenvalues developed specifically for the continuous time setting? And generally speaking, does anyone know any references for such continuous-time theory?
| The Gershgorin Circle Theorem can be used to construct a set of closed balls (i.e., circles) in the complex plane that are guaranteed to contain the eigenvalues of the matrix.
This theorem is not specific to generator matrices, so there may be some other related result that takes advantage of these matrices' specific structure... I'm looking into this question myself. I'll post an update if I find something better.
| {
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How to directly prove that $M$ is maximal ideal of $A$ iff $A/M$ is a field? An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field.
This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove it directly? Take $x + M \in A/M$. How can you construct $y + M \in A/M$ such that $xy - 1 \in M$? All I can deduce from the maximality of $M$ is that $(M,x) = A$.
| From $(M,x)=A$ you can infer that there are $m\in M, y\in A$ so that $m+xy=1$. Thus, $xy+M=1+M$.
| {
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Why does $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$? I don't understand the identity $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$, where $k$ is fixed.
I first approached it by considering the identity
$$
\sum_{n,k\geq 0} \binom{n}{k} x^n y^k = \sum_{n=0}^\infty x^n \sum_{k=0}^n \binom{n}{k} y^k = \sum_{n=0}^\infty x^n (1+y)^n = \frac{1}{1-x(1+y)}.
$$
So setting $y=1$, shows $\sum_{n,k\geq 0}\binom{n}{k}x^n=\frac{1}{1-2x}$. What happens if I fix some $k$ and let the sum range over just $n$? Thank you.
| You can work directly with properties of the binomial coefficient. For $k\ge 0$ let $$f_k(x)=\sum_{n\ge 0}\binom{n}kx^n\;.$$ Then
$$\begin{align*}
f_k(x)&=\sum_{n\ge 0}\binom{n}{k}x^n\\
&=\sum_{n\ge 0}\left[\binom{n-1}{k-1}+\binom{n-1}{k}\right]x^n\\
&=\sum_{n\ge 0}\left[\binom{n}{k-1}+\binom{n}k\right]x^{n+1}\\
&=x\sum_{n\ge 0}\binom{n}{k-1}x^n+x\sum_{n\ge 0}\binom{n}kx^n\\
&=xf_{k-1}(x)+xf_k(x)\;,
\end{align*}$$
so $(1-x)f_k(x)=xf_{k-1}(x)$, and $$f_k(x)=\frac{x}{1-x}f_{k-1}(x)\;.\tag{1}$$
Since $$f_0(x)=\sum_{n\ge 0}\binom{n}0x^n=\sum_{n\ge 0}x^n=\frac1{1-x}\;,$$ an easy induction yields the desired result.
| {
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Rittner equation I would like to know if the Rittner equation :
$$\partial_{t}{\varPhi(x,t)=k\partial_{xx}{\varPhi(x,t)}}-\alpha{\partial_{x}{\varPhi(x,t)}-\beta{\varPhi(x,t)}+g(x,t)}$$
can be solved using the Lax pair method or the Fokas method.
Thanks
| Setting $\psi(x,t) = e^{(\alpha^2/4k + \beta)t - \alpha x/2k} \Phi(x,t)$ and plugging in I get
$$
\begin{align*} \psi_t(x,t) - k \psi_{xx}(x,t) &= e^{(\alpha^2/4k + \beta)t - \alpha x/2k}g(x,t) \\
&\equiv f(x,t).
\end{align*}
$$
If the forcing $f(x,t)$ is in $L^1(\mathbf{R}\times \mathbf{R})$ then we can use the heat kernel
$$
K(x,t)=\begin{cases} (4\pi k t)^{-1/2} e^{-x^2/2kt},& t>0 \\ 0,& t\leq 0 \end{cases}
$$
and a solution to the problem on the domain $\mathbf{R}\times\mathbf{R}$ is $\psi= K*f$. Unfortunately this puts very harsh restrictions on the original forcing term $g=g(x,t)$ because of the exponential growth of the factor $e^{(\alpha^2/4k + \beta)t - \alpha x/2k}$.
Similarly, if one is interested in the Cauchy problem with $\Phi(x,0)=\Phi_0(x)$, then the new problem has initial data $\psi_0(x) = e^{-\alpha x/2k} \Phi_0(x)$. Using heat kernel methods this can again lead to severe restrictions. Recall the solution to the homogeneous Cauchy problem would be
$$
\psi(x,t)=K_t * \psi_0(x)
$$
which is well defined in a classical sense if, for instance, $\psi_0 \in L^p(\mathbf{R})$ for $1\leq p\leq \infty$. This means the associated initial data $\Phi_0(x)$ would need exponential decay, which is a little artificial.
For the Cauchy problem for the Rittner equation on the full line (for simplicity consider the homogeneous problem $g=0$, the more general case is similar) one can use the Fourier transform. Setting $\psi = e^{\beta t} \Phi$ we have the Cauchy problem
$$
\begin{align*}\psi_t - k \psi_{xx} + \alpha \psi_x &=0, \\
\psi(x,0) &= \Phi_0(x), \end{align*}
$$
the solution to which is
$$
\psi(x,t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{\mathrm{i} \lambda x -\lambda (k\lambda + \mathrm{i}\alpha) t} \hat{\Phi}_0(\lambda)\, \mathrm{d}\lambda.
$$
The problem on the half-line is more subtle. The Fokas method provides a solution in this case (for given boundary data at $x=0$, $\Phi(0,t)=\Phi_1(t)$ say) in terms of an integral along a smooth contour in the complex $\lambda$-plane. Now if the initial data $\Phi_0(x)$ has sufficient exponential decay (which is a big restriction), then one can show that this contour can be deformed onto the real $\lambda$-axis, giving a Fourier type integral as in the case of the Cauchy problem on the full line. However, in general, this contour cannot be shifted.
This problem is treated in Fokas' book, A unified approach to boundary value problems. He emphasizes that this inability to shift the contour back to the real axis (in the half-line problem) reflects the fact there is no classical transform for the problem.
| {
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Reflect a complex number about an arbitrary axis This should be really obvious, but I can't quite get my head round it:
I have a complex number $z$. I want to reflect it about an axis specified by an angle $\theta$.
I thought, this should simply be, rotate $z$ by $\theta$, flip it (conj), then rotate by $-\theta$.
But this just gives $z^* (e^{-i\theta})^* e^{i\theta} $...
but this can't be right - as it's just $z^*$ rotated by angle $2\theta$, surely?
| Indeed, it's just $z^*$ rotated by $2\theta$... And it's almost the right answer! You know that a symmetry composed with a rotation is still a symmetry, and you know that an (orthogonal) symmetry is characterized by its fixed points. So you want to get a symmetry that fixes the axis spanned by $e^{i\theta}$ (as an $\mathbb{R}$ vector space); an easy way to do that is, as you've noticed, to rotate by $-\theta$ (and not $\theta$ actually), flip over the real line (conjugation) and then rotate by $\theta$. So you get $(ze^{-i\theta})^* e^{i\theta} = z^* e^{2i\theta}$. It is easily checked that this fixes $e^{i\theta}$, and it's an orthogonal symmetry, therefore it's the one you're looking for.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finite dimensional vector space with subspaces
Possible Duplicate:
Could intersection of a subspace with its complement be non empty.
Is it possible for a finite dimensional vector space to have 2 disjoint subspaces of the same dimension ?
Any help would be much appreciated.
| As David Giraudo points out, any subspace $U\subseteq V$ is going to contain the zero vector (it has to: the subspace is a vector space so is closed under multiplication by elements of the underlying scalar field, and zero is in the scalar field, so multiplying by it tells us the zero vector is in the subspace). In this way, no two vector subspaces are disjoint as sets. In fact, the smallest subspace is trivial, $\{0\}$.
There is a linear notion of "disjoint" here: orthogonal. For example, both a line and a plane through the origin are subspaces, and if they intersect at a right angle they are orthogonal. More generally, two vector subspaces $U$ and $W$ are orthogonal if for every $u\in U, w\in W$, the vectors $u,w$ are perpendicular. The orthogonal complement of a subspace is the maximal orthogonal subspace to it (see Wikipedia). So saying $U,W$ are orthogonal is equivalent both $U\subseteq W^\perp$ and $W\subseteq U^\perp$.
| {
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Essay about the art and applications of differential equations? I teach a high school calculus class. We've worked through the standard derivatives material, and I incorporated a discussion of antiderivatives throughout. I've introduced solving "area under a curve" problems as solving differential equations. Since it's easy to see the rate at which area is accumulating (the height of the function), we can write down a differential equation, take an antiderivative to find the area function, and solve for the constant.
Anyhow, I find myself wanting to share with my students what differential equations are all about. I don't know so much about them, but I have a sense that they are both a beautiful pure mathematics subject and a subject that has many, many applications.
I'm hoping to hear suggestions for an essay or a chapter from a book--something like 3 to 10 pages--that would discuss differential equations as a subject, give some interesting examples, and point to some applications. Any ideas?
| Differential equations is a rather immense subject. In spite of the risk of overwhelming you with the amount of information, I recommend looking in the Princeton Companion to Mathematics, from which the relevant sections are (page numbers are within parts)
*
*Section I.3.5.4 for an introductory overview
*Section I.4.1.5
*Section III.23 on differential equations describing fluids (including the Navier-Stokes equation which is the subject of one of the Millennium problems)
*Section III.36 especially on the heat equation and its relation to various topics in mathematical physics and finance
*Section III.51 on wave phenomenon
*Section IV.12 on partial differential equations as a branch of mathematics
*Section IV.14 on dynamical systems and ordinary differential equations
*Section V.36 on the three body problem
*Section VII.2 on mathematical biology
Some of these material may be too advanced or too detailed for your purposes. But they may on the other hand provide keywords and phrases for you to improve your search.
| {
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"source": "stackexchange",
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Symmetries of the singular vectors of the line graph Consider the matrix $$ A = \left( \begin{matrix} 1/2 & 1/2 & 0 & 0 & 0 & 0 \\ 1/3 & 1/3 & 1/3 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1/2 & 0 & 0 \\
\vdots & \ & \ddots & & \ddots & \\
0 & \cdots & 0 & 1/3 & 1/3 & 1/3 \\
0 & \cdots & 0 & 0 & 1/2 & 1/2 \end{matrix} \right) $$
This question is about the singular vectors of $A$, i.e., the eigenvectors of $A^T A$. Ill denote them by $v_1, \ldots, v_n$. Suppose $n$ is even, and let $P$ be the permutation matrix which flips the vector around the midpoint, i.e., $P$ flips entries $1$ and $n$, entries $2$ and $n-1$, and finally entries $n/2$ and $n/2+1$.
Observation: A few numerical experiments in MATLAB suggest that exactly half of the $v_i$ satisfy $Pv=-v$ and the remainder satisfy $Pv=v$.
My question: Can someone provide an explanation for why this is the case?
Update: I suppose I should note that it isn't surprising that we can find eigenvectors which satisfy either $Pv=v$ or $Pv=-v$. Indeed, we can go through the list of $v_1, \ldots, v_n$ and if $Pv_i \neq v_i$, then because $Pv_i$ is also an eigenvector of $A^T A$ with the same eigenvalue, we can replace $v_i$ with $v_i'=v_i+Pv_i, v_i''=v_i-Pv_i$; then $Pv_i' = v_i', Pv_i'' = -v_i''$; finally we can throw out the redundant vectors. What is surprising to me is that the orthogonal basis returned by the MATLAB eig command has exactly half of the vectors which satisfy $Pv=v$ and exactly half which satisfy $Pv=-v$.
| The vectors (general vectors, not eigenvectors) with either kind of parity form $n/2$-dimensional subspaces. For instance, a basis for the even vectors is $(1,0,0,\dotsc,0,0,1)$, $(0,1,0,\dots,0,1,0)$, $\dotsc$
You've already explained that we can choose an eigenbasis for $A$ in which all vectors have definite parity. Since there can be at most $n/2$ of either kind, there must be exactly $n/2$ of either kind.
| {
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Does $\sum_{n=0}^{\infty}e^{-|x-n|}$ converge and uniformly converge? For $ 0 \leq x < \infty$, I'd like your help with deciding whether the following series converges and uniformly converges: $\sum_{n=0}^{\infty}e^{-|x-n|}$.
$$\sum_{n=0}^{\infty}\frac{1}{e^{|x-n|}}=\sum_{n=0}^{N}\frac{1}{e^{|x-n|}}+\sum_{i=0}^{\infty}\frac{1}{e^{i}}.$$
I divided the sum into two sums where $N$ is the place where $x=n$, so the first sum is finite and the second one converges, so the total sum converges. Am I allowed to say that the sum is uniformly converges, since the second series, where the tail converges, does not depend on $x$?
Thanks.
| Looking at the given series we see that uniform convergence is endangered by the fact that for arbitrary large $n$ we can find an $x$ where the $n$-th term is large, namely $\ =e^0=1$. Therefore we try to prove that the convergence is not uniform by exploiting this fact.
Put $s_n(x):=\sum_{k=0}^n e^{-|x-n|}$. By Cauchy's criterion uniform convergence means that for any $\epsilon>0$ there is an $n_0$, depending on $\epsilon$, such that
$$\bigl|s_m(x)-s_n(x)\bigr|<\epsilon\qquad\forall m>n_0,\ \forall n>n_0,\ \forall x\in{\mathbb R}\ .$$
Assume now there is such an $n_0$ for $\epsilon:={1\over2}$ and put
$$n:=n_0+1, \quad m:=n_0+2, \quad x:=m\ .$$
Then $s_n$ and $s_m$ differ by just one term (the $m$th), and one has
$$\bigl|s_m(x)-s_n(x)\bigr|=e^{-|x-m|}=e^0>{1\over2}=\epsilon\ ,$$
which is a contradiction.
| {
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badly approximated numbers on the real line form a meagre set Let $S$ be the set of real numbers $x$ such that there exist infinitely many (reduced) rational numbers $p/q$ such that $$\left\vert x-\frac{p}{q}\right\vert <\frac{1}{q^8}.$$
I would like to prove that $\mathbb{R}\setminus S$ is a meagre set (i.e. union of countably many nowhere dense sets).
I have no idea about how to prove this, as I barely visualise the problem in my mind. I guess that the exponent $8$ is not just a random number, as it seems to me that with lower exponents (perhaps $2$?) the inequality holds for infinitely many rationals for every $x\in\mathbb{R}$.
Could you help me with that?
Thanks.
| The idea is to transform the quantifiers into unions/intersections.
For example, let $T$ be the same as $S$ but dropping the infinitely many assumption.
Consider $A_{\frac p q}=(-\frac 1 {q^8}+\frac p q, \frac p q +\frac 1{q^8})$ then $T=\bigcup_{\frac p q\in\mathbb Q}A_{\frac p q}$. Thus, $T$ is a countable union of open sets ($\mathbb Q$ is countable). The same idea applies to $S$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalization of Pythagorean triples Is it known whether for any natural number $n$, I can find (infinitely many?) nontrivial integer tuples $$(x_0,\ldots,x_n)$$ such that $$x_0^n + \cdots + x_{n-1}^n = x_n^n?$$
Obviously this is true for $n = 2$.
Thanks.
| These Pythagorean triples can appear in the most unexpected place.
If: $a^2+b^2=c^2$
Then alignment: $N_1^3+N_2^3+N_3^3+N_4^3+N_5^3=N_6^3$
$N_1=cp^2-3(a+b)ps+3cs^2$
$N_2=bp^2+3bps-3bs^2$
$N_3=ap^2+3aps-3as^2$
$N_4=-bp^2+3(2c-b)ps+3(3c-3a-2b)s^2$
$N_5=-ap^2+3(2c-a)ps+3(3c-2a-3b)s^2$
$N_6=cp^2+3(2c-a-b)ps+3(4c-3a-3b)s^2$
And more:
$N_1=cp^2-3(a+b)ps+3cs^2$
$N_2=bp^2+3bps-3bs^2$
$N_3=ap^2+3aps-3as^2$
$N_4=(3c+3a+2b)p^2-3(2c+b)ps+3bs^2$
$N_5=(3c+2a+3b)p^2-3(2c+a)ps+3as^2$
$N_6=(4c+3a+3b)p^2-3(2c+a+b)ps+3cs^2$
$a,b,c$ - can be any sign what we want.
And I would like to tell you about this equation:
$X^5+Y^5+Z^5=R^5$
It turns out the solution of integral complex numbers there. where: $j=\sqrt{-1}$
We make the change:
$a=p^2-2ps-s^2$
$b=p^2+2ps-s^2$
$c=p^2+s^2$
Then the solutions are of the form:
$X=b+jc$
$Y=-b+jc$
$Z=a-jc$
$R=a+jc$
$p,s$ - what some integers.
| {
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$k$-out-of-$n$ system probabilities
An engineering system consisting of $n$ components is said to be a $k$-out-of-$n$ system ($k \le n$) when the system functions if and only if at least $k$ out of the $n$ components function. Suppose that all components function independently of each other.
If the $i^{th}$ component functions with probability $p_i$, $i = 1, 2, 3, 4$, compute the probability that a 2-out-of-4 system functions.
This problem in itself does not seem very difficult to solve, but I suspect I am not doing it the way it was intended to be done, because the formulas that come out are very ugly. I calculated the probability by conditioning on whether or not the $1^{st}$ and $2^{nd}$ components worked, and it came out to be
$$
p_3 p_4 + p_2 (p_3 + p_4 - 2 p_3 p_4) + p_1 (p_3 + p_4 - 2 p_3 p_4 + p_2 (1 - 2 p_3 - 2 p_4 + 3 p_3 p_4))
$$
Even if this is right, there's no way it's what the answer is supposed to look like. Can someone give me a push in the right direction?
| If you want something "prettier" you could take all the possible cases and write them using products and sums, such as $$\prod_{i=1}^4 p_i \left(1+\sum_{j=1}^4\frac{1-p_j}{p_j} + \sum_{k=1}^3 \sum_{l=k+1}^4 \frac{(1-p_k)(1-p_l)}{p_k \; p_l} \right)$$
or you could work out the probability that one or none work and subtract that from $1$ to get $$1 - \prod_{i=1}^4(1-p_i)\left(1+\sum_{j=1}^4\frac{p_j}{1-p_j}\right)$$
| {
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Prove that a finite union of closed sets is also closed Let $X$ be a metric space. If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.
I'm looking for a direct proof of this theorem. (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.") Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.
Here are the definitions I'm using:
Let $X$ be a metric space with distance function $d(p, q)$. For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) < r\}$. Any $p \in X$ is a limit point of $E$ if $\forall r > 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$. Any subset $E$ of $X$ is closed if it contains all of its limit points.
| It is sufficient to prove this for a pair of closed sets $F_1$ and $F_2$. Suppose $F_1 \cup F_2$ is not closed, even though $F_1$ and $F_2$ are closed. This means that some limit point $p$ of $F_1 \cup F_2$ is missing. So there is a sequence $\{ p_i\} \subset F_1 \cup F_2$ converging to $p$. By pigeonhole principle, at least one of $F_1$ or $F_2$, say $F_1$, contains infinitely many points of $\{p_i\}$, hence contains a subsequence of $\{p_i\}$. But this subsequence must converge to the same limit, so $p \in F_1$, because $F_1$ is closed. Thus, $p \in F_1 \subset F_1 \cup F_2$.
Alternatively, if you do not wish to use sequences, then something like this should work. Again, it is sufficient to prove it for a pair of closed sets $F_1$ and $F_2$. Suppose $F_1 \cup F_2$ is not closed. That means that there is some points $p \notin F_1 \cup F_2$ every neighbourhood of which contains infinitely many points of $F_1 \cup F_2$. By pigeonhole principle again, every such neighbourhood contains infinitely many points of at least one of $F_1$ or $F_2$, say $F_1$. Then $p$ must be a limit point of $F_1$; so $p \in F_1 \subset F_1 \cup F_2$.
| {
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Matrix multiplication, equivalent to numeric multiplication, or just shares the name? Is matrix multiplication equivalent to numeric multiplication, or do they just share the same name?
While there are similarities between how they work, and one can be thought of being derived from the other, I ask because they have different properties such as not being commutative, a × b ≠ b × a, and sometimes multiplication between matrices is referred to by the alternative name apply instead of multiply. For example applying a transformation matrix, where this is the same as multiplying by it.
Additionally sometimes in programming operations can be defined between new types of things, allowing the language to expand with new concepts, however the link between the name and rules such as commutative are supposed to continue to hold true.
| "Multiplication" is often used to describe binary operations that share only some of the properties of ordinary multiplication.
The case of matrix multiplication is special. There, multiplication is in some sense a generalization of ordinary multiplication. Let $M_n(a)$ be the $n\times n$ matrix whose diagonal entries are all equal to $a$, and whose off-diagonal entries are $0$.
It is easy to verify that
$$M_n(x)M_n(y)=M_n(xy).$$
So we can think of the real numbers as, for example, special $7\times 7$ matrices. Then the multiplication of real numbers can be viewed as a special case of matrix multiplication.
More interestingly, define the $2\times 2$ matrix $M(a,b)$ by
$$M(a,b)=\pmatrix{x & -y\\ y & x}.$$
It is not hard to verify that
$$\pmatrix{a & -b\\ b & a}\pmatrix{c & -d\\ d & c}=\pmatrix{ac-bd & -(ad+bc)\\ ad+bc & ac-bd}.$$
Note that the product of the two complex numbers $a+ib$ and $c+id$ is equal to $ad+bc +i(ad+bc)$. So the special matrices $M(x,y)$ multiply like the complex numbers. They also add like the complex numbers, and can be identified with the complex numbers.
So the multiplication of $2\times 2$ matrices can be thought of as a generalization of the multiplication of complex numbers. This is of practical usefulness: rotations about the origin can be thought of as either multiplying by a special kind of complex number, or as multiplying by a special type of matrix.
| {
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Prove or refute: If $f$ is Riemann integrable on $[0,1]$, then so is $\sin(f)$ I'd love your help with the following question: I need to prove or refute the claim that for a Riemann integrable function $f$ in $[0,1]$ also $\sin(f)$ is integrable on $[0,1]$.
My translation for this claim: If $\int_{0}^{1} f(x) dx < \infty$, so does $\int_{0}^{1} \sin(f(x))dx < \infty$, Am I right?
I tried to think of an elementary function that will fit the conditions, one that will blow up in $0$ or $1$ or both, but I didn't find any. Can I just use the fact that $\int_{0}^{1} \sin(f(x))dx \leq \int_{0}^{1} 1dx < \infty$ and that's it or Am I missing something?
Thanks!
| I do not think your translation is correct (unless you meant Lebesgue, and not Riemann integrable). The concept of Riemann integrable and Lebesgue integrable are not the same.
Riemann integrable: $f\colon[a,b]\to\mathbb{R}$, $-\infty<a<b<+\infty$, $f$ bounded and the upper integral equal to the lower integral.
Lebesgue integrable: $f\colon E\subset\mathbb{R}\to\mathbb{R}$, $E$ Lebesgue measurable set, $f$ Lebesgue measurable and $\int_E|f|<\infty$.
There are several ways of showing that if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable so is $\sin(f)$. The easiest way is to use the following fact: a bounded function $g\colon[a,b]\to\mathbb{R}$ is Riemann integrable if and only if the set of points where $g$ is discontinuous has measure $0$.
First of all it is cleat that $\sin(f)$ is bounded. Since $\sin$ is a continuous function,
$$
\{x\in[0,1]:\sin(f)\text{ is discontinuous at }x\}\subset\{x\in[0,1]:f\text{ is discontinuous at }x\}.
$$
Since $f$ is Riemann integrable, the set on the right hand side is of measure $0$, and so are all its subsets.
| {
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$f \geq 0$ continuous, $\lim_{x \to \infty} f(x)$ exists, $\int_{0}^{\infty}f(x)dx< \infty$, Prove: $\int_{0}^{\infty}f^2(x)dx< \infty$ Something that bothers me with the following question: $f: [0, \infty] \to \mathbb{R}$,$f \geq 0$, $\lim_{x \to \infty} f(x)$ exists and finite, and $\int_{0}^{\infty}f(x)dx$ converges, I need to show that $\int_{0}^{\infty}f^2(x)dx$
I separated the integral in the following way: $\int_{0}^{1}f^2(x)dx$+$\int_{1}^{\infty}f^2(x)dx$, while for the second one we know that $\lim_{x \to \infty}\frac{f^2(x)}{f(x)}$ so they converges together, but what happens in the first range, does my $\lim_{x \to 0}f(x)$ has to be finite? Do you have an example for an integral between $0$ and $\infty$ which converges and the $\lim_{x \to 0}$ is not finite?
What should I do in my case?
Thanks!
| To answer your questions regarding the behaviour at $0$ first. Note that $f:[0;\infty[ \ \to \mathbb{R}$, hence $f$ is well-defined and continuous at $0$ - according to the assumptions you state. In particular this means that $\lim_{x\to 0}f(x) = f(0)$, so there is really no issue at $x=0$; your function is well-defined at this end of the interval. Had this not been the case then you would have had to consider what might go wrong here (the function $f(x) = x^{-1/2}$ could be an instructional case to consider when integrating).
I am not sure I follow your reasoning in the case when $x\to \infty$, but I think it might be a good idea to try and deduce what the possible values of $\lim_{x\to \infty}f(x)$ could be, if the integral $\int_0^{\infty}f(x)dx$ is suppose to converge, and then maybe use $0<f(x)^2\leq f(x)$ under the right circumstances...
| {
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If a series converges, then the sequence of terms converges to $0$. Following the guidelines suggested in this meta discussion, I am going to post a proposed proof as an answer to the theorem below. I believe the proof works, but would appreciate any needed corrections.
Theorem If a series $\sum_{n=1}^{\infty}a_n$ of real numbers converges then $\lim_{n \to \infty}a_n = 0$
| Another view of this may be useful.
First, recall a basic fact that if $a_n$ is a convergent sequence of numbers, then the sequence $b_n = a_{n+1} - a_n$ converges to $0$. This is easy to prove and does not require the notion of a Cauchy sequence.
Therefore, if the partial sums $s_n$ are convergent, then $b_n = s_{n+1} - s_n$ converges to $0$. But the terms of this sequence are easily seen to be $b_n = a_{n+1}$. Hence $a_n \rightarrow 0$.
| {
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Generalized "Duality" of Classical Propositional Logical Operations Duality in propositional logic between conjunction and disjunction, $K$ and $A$ means that for any "identity", such as $KpNp = 0$ (ignoring the detail of how to define this notion in propositional logic), if we replace all instances of $K$ by $A$, all instances of $A$ by $K$, all instances of 1 by 0, and all instances of 0 by 1, the resulting equation will also consists of an "identity", $ApNp = 1$. Suppose that instead of conjunction "$K$" and disjunction "$A$", we consider any pair of "dual" operations $\{Y, Z\}$ of the 16 logical operations such that they qualify as isomorphic via the negation operation $N$, where $Y$ does not equal $Z$. By "isomorphic" I mean that the sub-systems $(Y, \{0, 1\})$, $(Z, \{0, 1\})$ are isomorphic in the usual way via the negation operation $N$, for example $K$ and $A$ qualify as "isomorphic" in the sense I've used it here.
If we have an identity involving operations $A_1, \dots, A_x$, and replace each instance of each operation by its "dual" $A'_1$, ..., $A'_x$, replace each instance of 1 by 0, and each instance of 0 by 1, is the resulting equation also an identity? If so, how does one prove this? How does one show that the equation obtained via the "duality" transformation here is also an identity?
| The resulting equation is also an identity. This is because any of the $16$ operations can be put in canonical disjunctive normal form using only $\land$, $\lor$, and $\lnot$. Then the replacement procedure described in the post becomes the standard one, and we are dealing with ordinary duality.
| {
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Combinatorial interpretation of this identity of Gauss? Gauss came up with some bizarre identities, namely
$$
\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}.
$$
How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.
| The typical analytic proof is not difficult and is an easy consequence of Jacobi's triple product $$\sum_{n=-\infty} ^{\infty} z^{n} q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$ for all $z, q$ with $z\neq 0,|q|<1$. Let's put $z=-1$ to get the sum in question. The corresponding product is equal to $$\prod(1-q^{2n})(1-q^{2n-1})^{2}=\prod(1-q^{n})(1-q^{2n-1})=\prod \frac{(1-q^{2n})(1-q^{2n-1})} {1+q^{n}}=\prod\frac{1-q^{n}} {1+q^{n}}$$ which completes the proof.
The proof for Jacobi's triple product is non-trivial / non-obvious and you may have a look at the original proof by Jacobi in this blog post.
On the other hand Franklin obtained a nice and easy combinatorial proof of the Euler's Pentagonal theorem which is equivalent to Jacobi Triple Product.
| {
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For $\sum_{0}^{\infty}a_nx^n$ with an infinite radius of convergence, $a_n \neq0$ , the series does not converge uniformly in $(-\infty , \infty)$. I'd like your help with the following question:
Let $\sum_{0}^{\infty}a_nx^n$ be a power series with an infinite
radius of convergence, such that $a_n \neq 0 $ , for infinitely many values of
$n$. I need to show that the series does not converge uniformly in
$(-\infty , \infty)$.
I quite don't understand the question, since I know that within the radius of convergence for a power series, there's uniform convergence, and since I know that the radius is "infinite", it says that the uniform convergence is infinite, no?
Thanks!
| Hint: put $s_N(x):=\sum_{n=0}^Na_nx^n$. If the sequence $\{s_N\}$ is uniformly convergent on $\mathbb R$ then the sequence $\{s_{N+1}-s_N\}$ converges uniformly on $\mathbb R$ to $0$ so $\{a_{N+1}x^{N+1}\}$ converges uniformly to $0$. Do you see the contradiction?
| {
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Where should the exponent be written in numbers with units of measurement? If you are to calculate the hypotenuse of a triangle, the formula is:
$h = \sqrt{x^2 + y^2}$
If you don't have any units for the numbers, replacing x and y is pretty straightforward:
$h = \sqrt{4^2 + 6^2}$
But what if the numbers are in meters?
$h = \sqrt{4^2m + 6^2m}$ (wrong, would become $\sqrt{52m}$)
$h = \sqrt{4m^2 + 6m^2}$ (wrong, would become $\sqrt{10m^2}$)
$h = \sqrt{(4m)^2 + (6m)^2}$ (correct, would become $\sqrt{52m^2}$)
Or should I just ignore the unit of measurement in these cases?
| Suppose you have been given $x$ and $y$ in metres, and you'd like to know the quantity, $z=\sqrt{x^2+y^2}$. Then, as you have predicted this quatity will be in metres.
Two things have been involved:
Homogeneity of Dimension
Two quantities of different dimensions cannot be added. This is one of the axioms of numerical physics.
Example: It is clear that adding $5$ metres to $3$ seconds does not give a physically meaningful quantity that can be interpreted in real life.
Certain functions only take values in dimensionless quantities
For instance, $\sin (\sqrt{x^2+y^2})$ would not make sense even if $x$ and $y$ have same dimensions. This is a bit subtler, but this is what it is!
*
*Coming to your question, the first quantity you tell us in dimension of $m^{1/2}$ which is against your guess!
*The second quantity is dimensionally fine while numerically this is not what you want.
*The third quantity is fine in all ways.
My suggestion:
First manipulate the numbers and then the units separately. This is a good practice in Numerical Physics. The other answers have done it all at one go. But, I don't prefer it that way!
| {
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What is the equation of an ellipse that is not aligned with the axis? I have the an ellipse with its semi-minor axis length $x$, and semi major axis $4x$. However, it is oriented $45$ degrees from the axis (but is still centred at the origin). I want to do some work with such a shape, but don't know how to express it algebraically. What is the equation for this ellipse?
| Let the center of the ellipse be at $C = (x_c, y_c)$. Let the major axis be the line that passes through $C$ with a slope of $s$; points on that line are given by the zeros of $L(x,y) = y - y_c - s(x - x_c)$. Let the minor axis be the line perpendicular to $L$ (and also passing through $C$); points on that line are given by the zeros of $l(x,y) = s(y-y_c)+(x-x_c)$.
The ellipse is then defined by the zeros of
$$E(x,y) = L(x,y)^2/a + l(x,y)^2/b - 1$$
Requiring that the
distance between the intersections of $E$ and $L$ be $2M$ identifies $$b=M^2(1+s^2)$$
and similarly, requiring that the intersections between $E$ and $l$ be separated by $2m$ identifies
$$a=m^2(1 + s^2)$$
This is demonstrated in the following SymPy session:
>>> from sympy import *
>>> a, b, x, y, m, M, x_c, y_c, s = var('a,b,x,y,m,M,x_c,y_c,s')
>>> L = (y - y_c) - s*(x - x_c)
>>> l = s*(y - y_c) + (x - x_c)
>>> idiff(L, y, x) == -1/idiff(l, y, x) # confirm they are perpendicular
True
>>> E = L**2/a + l**2/b - 1
>>> xy = (x, y)
>>> sol = solve((E, L), *xy)
>>> pts = [Point(x, y).subs(zip(xy, p)) for p in sol]
>>> solve(pts[0].distance(pts[1]) - 2*M, b)
[M**2*(s**2 + 1)]
>>> sol = solve((E, l), *xy)
>>> pts = [Point(x,y).subs(zip(xy, p)) for p in sol]
>>> solve(pts[0].distance(pts[1]) - 2*m, a)
[m**2*(s**2 + 1)]
So the general equation of the ellipse centered at $(x_c, y_c)$ whose major axis (with radius of $M$) is on a line with slope $s$, and whose minor axis has radius of $m$, is given by the solutions of:
$$\frac{((y - y_c) - s(x - x_c))^2}{m^2(1 + s^2)} + \frac{(s(y - y_c) + (x - x_c))^2}{M^2(1 + s^2)} = 1$$
| {
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An application of the General Lebesgue Dominated convergence theorem I came across the following problem in my self-study:
If $f_n, f$ are integrable and $f_n \rightarrow f$ a.e. on $E$, then $\int_E |f_n - f| \rightarrow 0$ iff $\int_E |f_n| \rightarrow \int_E |f|$.
I am trying to prove (1) and the book I am using suggests that it follows from the Generalized Lebesgue Dominated Convergence Theorem:
Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $E$ that converge pointwise a.e. on $E$ to $f$. Suppose there is a sequence $\{g_n\}$ of integrable functions on $E$ that converge pointwise a.e. on $E$ to $g$ such that $|f_n| \leq g_n$ for all $n \in \mathbb{N}$. If $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $g_n$ = $\int_E$ $g$, then $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $f_n$ = $\int_E$ $f$.
I suspect that I need the right inequalities to help satisfy the hypothesis of the GLDCT, but I am not certain about what these inequalities should be.
| Take $g_n = |f_n| + |f|$ and use the triangle inequality to get the bound.
| {
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continuous map of metric spaces and compactness
Let $f:X\rightarrow Y$ be a continuous map of metric spaces. Show that if $A\subseteq X$ is compact, then $f(A)\subseteq Y$ is compact.
I am using this theorem: If $A\subseteq X$ is sequentially compact, it is compact.
Also this definition: A set $A\subseteq X$ is sequentially compact if every sequence in $A$ has a convergent subsequence in $A$.
Attempt at a proof:
Let $\{y_n\}\subseteq f(A)$. Since $f$ is continuous, $\{y_n\}=f(x_n)$ for some $\{x_n\}\subseteq A$. If $A\subseteq X$ is compact, every sequence $\{x_n\}\subseteq A$ has a subsequence that converges to a point in $A$, say $\{x_{n_k}\}\rightarrow a\in A$. Since $f$ is continuous, $f(x_{n_k})\rightarrow f(a)\in f(A)$. Then $f(x_{n_k})\subseteq f(A)$ is a convergent sequence in $f(A)\implies f(A)$ is compact since $\{y_n\}\subseteq f(A)$ was arbitrary.
| Yet another formulation for topological spaces: If $f:X \to Y$ continuous and $f(x_\iota)$ is a net in $f(X)$, then $x_\iota$ has a converging subnet, say $x_\tau \to x$. Then $f(x_\tau) \to f(x)$, hence $f(x_\iota)$ has a converging subnet, so $f(X)$ is compact.
| {
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Subsets and Splits