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Determining the sampling distribution Let the random variable $X$ represent the number of defective components in
a lot of components. Assume that $X$ can take on four values: $0, 1, 2, 3$. The probability distribution of $X$ is shown in the table below:
X | 0 | 1 | 2 | 3
P(X) | 0.4 | 0.2 | 0.1 | 0.3
1) Randomly pick two lots of components, what is the sampling distribution of average
number of defective components in a lot.
2) Find $Pr(\bar{X} > 2)$.
How do I go about approaching this problem?
|
The possible values of the average number are $0$, $1/2$, and so on up to $3$, a total of seven possible values. If the random variable $Y$ is the average number, we want to compute $\Pr(Y=y)$ for these seven possible values.
For example the probability that $Y=1$ is the probability the sum is $2$, which can happen in three ways, $(0,2)$, $(2,0)$ and $(1,1)$. Thus
$$\Pr(Y=1)=(0.4)(0.1)+(0.1)(0.4)+(0.2)(0.2).$$
Only six more to go.
|
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|
Is a function differentiable in $x$ if $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\infty$? All the definitions of differentiability I found (Wolfram Mathworld for instance) only require this limit to exist, but say nothing about the domain in which that has to happen.
So what if that limit is $\pm\infty$? Wouldn't (in the "normal" interpretation of a function $\mathbb R \rightarrow \mathbb R$, for instance) the slope of the tangent of $f$ in $x$ be undefined?
Edit: a link clearing it up.
|
This is an unfortunate case of terminology getting the better of us. The definition of a limit is
The limit of the function $f(x)$ as $x$ approaches $a$ exists if there is a number $L$ such that for all $\epsilon>0$, there exists a $\delta>0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$. In this case, we say that $$\lim_{x\to a} f(x)=L.$$
So the limit, if it exists is a number. Infinity is not a number, so technically, if $\lim_{x\to a}f(x)=\infty$, the limit doesn't exist. And yet it is infinity. It's confusing at first, but once you get used to it, it's not too bad.
|
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How do I prove that "If prime p does not divide natural number m, then gcd(p,m) = 1" I am having a problem with this.
If prime p does not divide natural number m, then gcd(p,m) = 1
I had to use this for my another proof and because I thought it was quite intuitive, I just assumed this is true and used it for my proof but then I was told that I have to prove it too. But the problem is I can't figure out how I should write a proof for this formally.
Any help would be appreciated! Thank you
|
Your proof might be different depending on your definition of GCD. The definition I prefer is the following: a GCD of two integers $m,p$ is a number $d$ which divides $m$ and $p$, and which has the property that if $e$ is an integer dividing $m$ and $p$, then $e$ divides $d$. A GCD of two nonzero integers exists, is nonzero, and is unique up to sign, so it makes sense to say the GCD (that is, the positive one).
Since $d = \textrm{GCD}(m,p)$ is necessarily a divisor of $p$, your choices for $d$ are $1$ and $p$. But also $d$ must divide $m$, so you cannot have $d = p$. Therefore, the GCD is $1$.
|
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|
Show that $f\big(f(x)\big)$ has at least as many real roots as $f(x)$ $f(x)$ is a real polynomial of odd degree. Show that $f\big(f(x)\big)$ has at least as many real distinct real roots, counted without multiplicity, as $f(x)$.
|
Since $f$ is of odd degree,
$\lim_{x \to \infty} f(x)
=\pm \infty
$
depending on the sign
of the highest order term,
and
$\lim_{x \to -\infty} f(x)
=\mp \infty
$,
with the other sign being used.
Therefore
$f(x)$
takes the reals onto the reals
(with possible multiple occurrances
where $f$ is not monotonic),
so
$f(f(x))$
has at least as many zeros
as $f(x)$.
More explicitly,
for each real root
$x_r$ of $f$,
there is at least one $x_a$
such that
$f(x_a) = x_r$,
so that
$f(f(x_a))
=f(x_r)
=0
$.
|
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|
$G$ is cyclic $iff$ there is an element of order $|G|$. I'm trying to prove that if there is an element of order $|G|=n$ there the group is cyclic. But I have some problem. Let $a\in G\backslash \{1\}$. And consider $\left<a\right>$ which is cyclic. Let $b\in G$. Then, $b^n=1=a^n\in \left<a\right>$. Does the fact that $b^n\in \left<a\right>$ implies that $b\in G$, and if yes, why ?
|
By definition $G$ is cyclic iff $G=\langle a\rangle$ and so order of $a$ should be $|G|$. And if the order of $a$ is $|G|=n$ then $a$ generate $n$ distinct elements, and so $a$ generate whole of $G$.
|
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|
Doubt in the solution of differential equation I know that the solution to the differential equation $\frac{dy}{dt}=y(a-by), a>0, b>0, y(0)=y_0$ can be derived using integration using partial fractions, and the final result is: $$y=a/(b+ke^{-at})$$ where $k$ is $(a/y_0)-b$.
My doubt is: Say $y_0<0$. Then the differential equation itself tells me that $\frac{dy}{dt}<0$ and hence $y$ would just go on falling as $t$ increases, and it would always remain negative. But if you look at the behaviour of the function $y$, there would be a vertical asymptote at some positive value of $t$ (to be precise, where $b+ke^{-at}=0$. Beyond this value of $t$, $y$ would be positive. What am I getting wrong here?
|
It is a classical differential equation called the logistic differential equation http://math.usu.edu/~powell/biomath/mlab3-02/node2.html
|
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|
Prove there are no prime numbers in the sequence $a_n=10017,100117,1001117,10011117, \dots$ Define a sequence as $a_n=10017,100117,1001117,10011117$. (The $nth$ term has $n$ ones after the two zeroes.)
I conjecture that there are no prime numbers in the sequence. I used wolfram to find the first few factorisations:
$10017=3^3 \cdot 7 \cdot 53$
$100117=53\cdot 1889$
$1001117=13 \cdot 53\cdot1453$ and so on.
I've noticed the early terms all have a factor of $53$, so the problem can be restated as showing that all numbers of this form have a factor of $53$. However, I wouldn't know how to prove a statement like this. Nor am I sure that all of the terms do have a factor of $53$.
I began by writing the $nth$ term of the sequence as
$a_n=10^{n+3}+10^n+10^{n-1}+10^{n-2}+10^{n-3}+\cdots+10^3+10^2+10^1+7$ but cannot continue the proof.
|
Another way to find the inductive relationship already cited, from a character manipulation point of view:
Consider any number in the sequence, $a_n$. To create the next number, you must:
*
*Subtract $17$, leaving a number terminating in two zeroes;
*Divide by $10$, dropping one of the terminal zeroes;
*Add $1$, changing the remaining terminal zero to a $1$;
*Multiply by $100$, sticking a terminal double zero back on;
*Add $17$, converting the terminal double zero back to $17$
Expressing this procedure algebraically, and simplifying:
$$a_{n+1}=\left (\frac{a_n-17}{10}+1 \right ) \times 100+17=10a_n-53$$
|
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Let $A$ be a complex matrix and $AA^t=A^2$, is it always right that $A$ is symmetric? It is a well-known result that when $A$ is a real matrix and $AA^t=A^2$ then $A$ is symmetric. I guess the same proposition is right for a complex matrix, but I can't prove it. Could someone give me some hints on this problem? I will be very appreciated for your help!
|
This is not true in $M_n(\mathbb{C})$, $n\geq 3$.
A very simple counterxample is the following:
Let $v^t=(0,0,1,0,...,0)$ and $w^t=(1,i,0,...,0)$.
Notice that $w^tw=1+i^2=0$ and $w^tv=0$. Define $A=vw^t$.
Now, $A^2=(w^tv)A=0$ and $AA^t=(w^tw)vv^t=0$, but $A=vw^t\neq w^tv=A^t$.
|
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|
Find equation for mass in gravity A satellite is moving in circular motion round a planet.
From the physics we know that
$$\Sigma F_r = ma_r = \frac{GMm}{r^2}$$
So I wanted to find the equation for $M$ knowing also that
$$v = \omega r = \frac{2\pi r}{T}$$
and
$$a_r = \frac{v^2}{r}$$
Thus,
$$ma_r = \frac{GMm}{r^2}$$
$$a_r = \frac{GM}{r^2}$$
$$\frac{v^2}{r} = \frac{GM}{r^2}$$
$$\frac{\left(\frac{2\pi r}{T}\right)^2}{r} = \frac{GM}{r^2}$$
$$\frac{\frac{4\pi^2r^2}{T^2}}{r} = \frac{GM}{r^2}$$
$$\frac{4\pi^2r^3}{T^2} = \frac{GM}{r^2}$$
$$\frac{4\pi^2r^5}{T^2} = GM$$
$$\frac{4\pi^2r^5G}{T^2} = M$$
However, this is wrong! It should be:
$$M = \frac{4\pi^2r^3}{GT^2}$$
What was my mistake in Mathematics? Please don't migrate it to physics because my misunderstanding is on math.
Note: I would be very happy if you show my mistake, instead of showing me another way to get to the equation.
|
$\frac{4\pi^2r^2}{T^2}/r=\frac{4\pi r}{T^2}$ and G should go down not up in numerator.
|
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|
Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.
|
AM/GM. $\frac{x}{2}+\frac{1}{x}<x$ iff $x^2>2$. So it is enough to show that if $x^2>2$ then $(\frac{x}{2}+\frac{1}{x})^2>2$ or $\frac{x^2}{4}+\frac{1}{x^2}>1$. But by AM/GM $(\frac{x^2}{4}+\frac{1}{x^2})/2>\sqrt{\frac{1}{4}}$.
|
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|
B-splines locally controlled I have read that in contrast of the thin-plate splines, B-splines are locally controlled, which makes them computationally efficient even for large number of control points.
I didn't understand what does it mean by saying that B-splines are locally controlled. Can anyone help with this?
|
It means that value of B-spline at a point depends just on few control points localized nearby. And vice versa, if you modify a control point, or coefficient of one basisfunction it will affect just some local nighborhood.
In case of cubic B-spline (which is most common e.g. in computer graphics ) in 1D value $f(x)$ at any point depends on value of 4 control points (resp. basisfunctions ) as
$ f(x) = \sum_{i=1..3} \alpha_i \phi ( x - i ) $
It is nice visible here
http://www.brnt.eu/phd/ucbs-basis.png
or here
http://www.cs.berkeley.edu/~sequin/CS284/IMGS/bsplinebasics.gif
|
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what is the relation between $f(x+1)$ and $f(x)$? I searched so much over math sites and google but I didn't find helpful hints and required knowledge or the specific name of this topic in function.
I stuck in relation and operations on $f(x)$ which is really important for solving problems.
for example relation between $f(x)$ and $f(x+1)$
in more complicated form for instance:
1. $$f(x)=3x-5$$
$$f(2x+1)=?$$
true answer is $$2f(x)+8$$
I tried this solution but it doesn't work
$f(2)$*$f(x)$+$f(1)$=?
$f(1)$=$-2$
$f(2)$=$2*3-5=1$
$f(2)$*$f(x)$+$f(1)$=?
$1(3x-5)-2$=$3x-7$ which gives us $f(x)-2$
another instance:
$$f(-x)=2f(x)+6$$
$$f(3)=?$$
true answer is $$-6$$
I tried this solution
$f(x)(-1)$=$2f(x)+6$
$-f(x)$=$2f(x)+6$
$-3f(x)$=$6$
$f(x)$=$-2$
so I will be really appreciative if you tell me the relation between operation on $f(x)$ like $f({1\over x})$ , $f(x-1)$ and ..
|
One can view $f$ as a machine that takes an input (usually called $x$) and outputs something based on that input.
It is usually written in the general case as something like $f(x)=3x-5$ which says whatever our input $x$ happens to be, we output something that is (in this case) $3$ times that input and then subtract five.
In the above example it would be something like $f(\color{fuchsia}{1})=3\cdot \color{fuchsia}{1}-5$, or $f(\color{fuchsia}{2})=3\cdot \color{fuchsia}{2}-5$, or in general $f(\color{fuchsia}{x})=3\cdot \color{fuchsia}{x}-5$
In the case that the input happens to itself be written in a strange way, or even as a different function, that is okay. It is still the input, and we manipulate it in exactly the same way.
$f(\color{fuchsia}{\frac{1}{x}})=3\cdot \color{fuchsia}{\frac{1}{x}}-5$ and $f(\color{fuchsia}{x+1})=3\cdot (\color{fuchsia}{x+1})-5$ for examples.
We may choose to reorganize things afterwards at our convenience.
For example: $f(\color{fuchsia}{2x+1})=3\cdot (\color{fuchsia}{2x+1})-5=6x+3-5=6x-2$
|
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Finding a limit without using L'Hospital's rule I encountered an issue for finding the limit for a function, one small step I just can't get.
I know that this is classic case for L'Hospital's however our calculus course didn't reach it and I can not use it for this one.
$
2.3 .m,n\in\mathbb{N}\quad\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}$
$x\neq1$
\begin{aligned}\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x^{m}-1^{m}}{x^{n}-1^{n}}=\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}=\\
\underset{x\rightarrow1}{\lim}\frac{\left(m-1\right)x^{m-1}}{\left(n-1\right)x^{n-1}}=\underset{x\rightarrow1}{\lim}\frac{mx^{m-1}-x^{m-1}}{nx^{n-1}-x^{n-1}}
\end{aligned}
And here I stuck...
know I need to reach $\frac{m}{n}$ but how can I proceed...?!!? nothing to cancel out here...
|
$\lim_{x\to 1} \dfrac{x^m - 1}{x^n - 1} =
\lim_{x\to 1} \dfrac{(x-1)(x^{m - 1} + x^{m - 2} + x^{m - 3} + ... + 1)}{(x-1)(x^{n - 1} + x^{n - 2} + x^{n - 3} + ... + 1)} =
\lim_{x\to 1} \dfrac{(x-1)\sum_{i=1}^{m} x^{m-i}}{(x-1)\sum_{j=1}^{n} x^{n-j}} =
\lim_{x\to 1} \dfrac{\sum_{i=1}^{m} x^{m-i}}{\sum_{j=1}^{n} x^{n-j}} =
\dfrac{\sum_{i=1}^{m} 1^{m-i}}{\sum_{j=1}^{n} 1^{n-j}} =
\dfrac{\sum_{i=1}^{m} 1}{\sum_{j=1}^{n} 1} =
\dfrac{m}{n}$
|
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|
An exercise on indicator function in a measure space Given $(\Omega, \mathbb F, \mu)$ a measure space, $(A_n)$ a sequence of measurable sets, $f: (\Omega, \mathbb F) \to (\mathbb R, \mathbb B(\mathbb R))$ an integrable function such that: $\displaystyle \lim_{n\to\infty}\int_\Omega |1_{A_n} - f|d\mu = 0$, where $1_{A_n}$ is indicator function.
1) Show that: $|f| \leq 2, \mu$-a.e.
2) Show that: $\exists A \in \mathbb F: f = 1_A, \mu$-a.e.
3) Show that: if $\displaystyle \sum_{n \geq 0}\mu(A_n \Delta A) < \infty$, then $1_{A_n} \to 1_A, \mu$-a.e.
I did the last question as follows:
As $\displaystyle \sum_{n \geq 0}\mu(A_n \Delta A) < \infty$, applying Borel-Cantelli lemma, we obtain: $\mu (\limsup A_n \Delta A) = 0$. Thus,
$$x \in (\limsup A_n \Delta A)^c = \liminf (A_n \Delta A)^c$$
$$\Leftrightarrow \exists n \in \mathbb N: \forall k \geq n, x \in (A_n \Delta A)^c$$
$$\Leftrightarrow \exists n \in \mathbb N: \forall k \geq n, x \in A_n \cap A \text{ or } x \in \Omega \backslash (A_n \cup A)$$
As both cases imply $1_{A_n} = 1_A$, we obtain that $1_{A_n}$ will eventually equal to $1_A$, $\mu$-a.e.
Is my proof correct?
|
Yes, your proof is correct.
Alternatively, we can notice that the assumption is $\sum_{n\geqslant 0}\lVert \mathbf 1_{A_n}-\mathbf 1_A\rVert_1\lt \infty$. We can use the fact that if $\sum_{n\geqslant 1}\lVert f_n\rVert_1\lt \infty$, then
$\sum_{n\geqslant 1}\lvert f_n\rvert\lt \infty$ almost everywhere hence $f_n\to 0$ almost everywhere.
|
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How do I prove $2^0 + 2^1 + 2^2 + 2^3 +\cdots + 2^{d-1} \le n - 1$ $\space$ if $\space$ $d = \lfloor \log_2 n \rfloor$? I was given this inequality in university, me and my friends can't solve it, hope someone here can explain me:
$2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^{d-1} \le n - 1$ $\space$ if $\space$ $d = \lfloor \log_2 n \rfloor$.
|
Observe that the series $\sum_{k=0}^d 2^k$ is a geometric series which has the closed form $(2^{d}-1)/(2-1) = 2^{d}-1$.
Because $d = \lfloor \log_2n \rfloor$, we have
$$d = \lfloor \log_2n \rfloor \leq \log_2n$$
by definition of the floor function. Raising both sides of the inequality by 2 and subtracting 1 thereafter, we deduce that
$$2^d - 1 \leq 2^{\log_2n} - 1 = n -1 $$
as desired.
|
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Can we always find homotopy of two paths which lies "between" the paths? Let $\gamma_0,\gamma_1:[0,1]\to\mathbb{R}^2$ be paths such that $\gamma_0(0)=\gamma_1(0)$ and $\gamma_0(1)=\gamma_1(1)$. I wish to show that there is a homotopy $\Gamma:[0,1]\times[0,1]\to\mathbb{R}^2$ from $\gamma_0$ to $\gamma_1$ that satisfies the following:
*
*For all $(s,t)\in[0,1]\times[0,1]$, $\Gamma(s,t)$ is not in the unbounded face of $\gamma_0([0,1])\cup\gamma_1([0,1])$.
PS: If the answer is no, are there additional restrictions which we could place on $\gamma_0$ and $\gamma_1$ which would allow this? (Ex. rectifiable, differentiable, etc.)
|
If you allow the contour of the union of those two paths to form a simple closed curve*, this is true, and can be seen as a consequence of the Schoenflies Theorem together with the fact that the disk is contractible.
*I think this is the case, since you are talking about "unbounded face".
|
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Let F be a finite field of characteristic $p$. Show $f(a) = a^p$ is a ring homomorphism, injective, and surjective Let F be a finite field of characteristic $p$. Show that the function $f:F \to F$ defined by $f(a) = a^p$ is a) a ring homomorphism, b) injective and, c) surjective.
I tried to approach this problem by proving $f(a+b) = f(a) + f(b), f(ab) =f(a)f(b) $ for ring homomorphism.
But I don't know how to solve b)and c). Could you please help me to solve it ? Thank you!!!
|
Hint: a field homomorphism is always injective, and an injective map from a finite set to itself is always surjective.
|
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Computing conditional probability with proxy variable Suppose $P(X | Y)$.
If I only know $P(X | Z)$ and $P(Z | Y)$, can I do
$$
P(X|Y) = P(X|Z)P(Z|Y)?
$$
|
We toss a fair coin independently three times. Let $X$ be the event head on the first toss, $Y$ the event head on the second, and $Z$ the event head on the third.
Then $\Pr(X\mid Y)=1/2$ and $\Pr(X\mid Z)\Pr(Z\mid Y)=(1/2)(1/2)$.
|
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|
Find $f'(1)$ if $f$ is continuous and such that $f (f (x))=1+x$ for every $x$
If $ f $ is a continuous function satisfying $f (f (x))=1+x$ find $f'(1)$.
I just guessed $f (x)=x+1/2$.Any formal method for this sum ?
|
The relationship $f(f(x))=x+1$ implies that $f^{-1}(x+1)=f(x)$. This implies that when the inverse function is shifted to the left by one unit it must give back the original function. Indeed, $f(x)=x+1/2$ satisfies this requirement and has $f^{-1}(x)=x-\frac{1}{2}$. Although this doesn't show that this is the only function, there may be more
|
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|
How to prove that every power of 6 ends in 6? Yesterday I had the traditional math matriculation exam, and in it there was a question "In what digit does the number $2016^{2016}$ end in?" After the test The Matriculation Examination Board published a pdf in which they show how to basically solve all the problems in the test, and for the aforementioned question the solution was "Because every power of 6 ends in 6, so does every power of 2016, and thus the last digit is 6". I understand everything else in that problem now, except how to actually show that every power of 6 ends in 6. From there on I know how to solve the last digit if $2016^{2016}$. So,
how to prove that every power of 6 ends in 6?
|
Use induction in order to complete the hint given by @ThePortakal.
First, show that this is true for $n=1$:
$6^1=6$
Second, assume that this is true for $n$:
$6^n=10k+6$
Third, prove that this is true for $n+1$:
$6^{n+1}=$
$6\cdot\color{red}{6^n}=$
$6\cdot(\color{red}{10k+6})=$
$60k+36=$
$60k+30+6=$
$10(6k+3)+6$
Please note that the assumption is used only in the part marked red.
|
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|
Is continuous function finite-valued in $R^n$? Let $f$ be a continuous function defined on $\mathbb{R}^n$. The range of $f$ is in the extended real numbers. Is $f<\infty \ \forall x\in \mathbb{R}^n$ ? And why?
|
In the "usual" topology on the extended real numbers $[-\infty, \infty]$, a neighborhood of $\infty$ is a set containing some interval $(a, \infty] = (a, \infty) \cup \{\infty\}$ with $a$ real, and similarly for neighborhoods of $-\infty$.
If that's true in your setting, $[-\infty, \infty]$ is homeomorphic to a closed, bounded interval of real numbers. For instance, the hyperbolic tangent function $\tanh$ is a homeomorphism from $[-\infty, \infty]$ to $[-1, 1]$. Consequently, asking whether a continuous function can achieve the value $\infty$ is no more mysterious than asking whether a continuous, real-valued function can achieve an absolute maximum.
As MPW and Ethan Bolker note, the constant function with value $\infty$ is continuous. Non-constant continuous functions can achieve the values $\infty$ and/or $-\infty$, as well. For example, if $n = 1$, then
*
*$f(x) = \frac{1}{x^{2}}$ (extended by $f(0) = \infty$) is continuous, since $\lim\limits_{x \to 0} \frac{1}{x^{2}} = \infty$.
*$f(x) = \frac{1}{x}$ (extended by $f(0) = \infty$) is not continuous, since $\lim\limits_{x \to 0^{\pm}} \frac{1}{x} = \pm\infty$ (i.e., the one-sided limits exist as extended real numbers, but are not equal).
And so forth.
|
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|
Simple question about discrete metric and openness. You may think this is silly question, but I'm really confused.
In discrete metric, every singleton is an open set.
And, the proof goes like this
$\forall x \in X$, by choosing $\epsilon < 1$, $N_\epsilon(x) \subset ${$x$}
However, if discrete metric is defined in $X$,
How can we use the radius less than 1 when we only have 0 and 1?
|
A metric on $X$ is a mapping $d:X \times X \to [0,\infty[$, so you can take $d = 1/2$ even if no pairs with distance $1/2$ exist.
|
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|
How to result in moment generating function of Weibull distribution? So I'm trying to get the Weibull distribution moment generating function
$$\sum_{n=0}^\infty \frac{t^n \lambda^n}{n!} \Gamma(1+n/k)$$
(which can be found here https://en.wikipedia.org/wiki/Weibull_distribution)
I'm trying to do it with the definition of moment generating functions, but I can't get to that result. If $X$ has $\mathrm{Weibull}(\lambda, k)$ distribution, then its PDF is
$$f(x) =
\begin{cases}
\frac{k}{\lambda}(\frac{x}{\lambda})^{k-1} e^{(-x/\lambda)^k}, & \text{if $x\ge0$} \\
0, & \text{if $x<0$}
\end{cases} $$
And its moment generating function should be
$$M(t)=E(e^{tX})=\int_0^\infty e^{tx}\frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} dx$$
but I can't get this form from Wikipedia. I have tried many different variable changes, etc., but without good result.
I appreciate any help leading to that form.
|
$$\begin{align}
M(t) &= E(e^{tX}) = \int_0^\infty
e^{tx}\frac{k}{\lambda} \left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k} dx
&\\ &= \int_0^\infty e^{\lambda tu} ku^{k-1} e^{-u^k} du
&\qquad(\text{with }u=x/\lambda\text{, for }\lambda>0)
\\ &= \int_0^\infty e^{\lambda tx^{1/k}} e^{-x} dx
&\qquad(\text{with }x=u^k\text{, for }k>0)
\\ &= \int_0^\infty \sum_{n=0}^\infty
{(\lambda t)^n\over n!} x^{n/k} e^{-x} dx
\\ &= \sum_{n=0}^\infty {(\lambda t)^n\over n!}
\int_0^\infty x^{n/k} e^{-x} dx
\\ &= \sum_{n=0}^\infty {\lambda^n t^n \over n!}
\Gamma\left({n\over k}+1\right)
\end{align}$$
from definitions with two substitutions (as indicated),
the series expansion (for $e^{\lambda tx^{1/k}}$),
and integrating term by term (with the definition of the
Gamma function).
|
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|
Factorisation of a polynomial of degree 5 using limited theorems I have the polynomials $p(X)=X^{5}+P_4X^{4}+P_3X^{3}+P_2X^{2}+P_1X+P_0 \in\mathbb{Z_2[x]}$
I need to determine all $p(x)$ that can be factored into irreducible polynomials of degree three and two.
The catch is that I cannot use most of the techniques usually described to solve this kind of problem. What would be the steps I need to take in order to solve this problem?
For the record I am using Gallian: contemporary abstract algebra as a book and can only use theorems 17.1 and 17.5(plus corollary 1 and 2) from that specific chapter. For reference I wrote them down below:
17.1: Let $F$ be a field. If $f(x) [ F[x]$ and deg $f(x)$ is 2 or 3, then $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in
$F$.
17.5: Let $F$ be a field and let $p(x) \in F[x]$. Then $<p(x)>$ is a maximal ideal in $F[x]$ if and only if $p(x)$ is irreducible over $F$.
Corollary 1: Let $F$ be a field and $p(x)$ be an irreducible
polynomial over $F$. Then $F[x]/<p(x)>$ is a field.
Corollary 2: Let F be a field and let $p(x), a(x), b(x) \in F[x]$. If
$p(x)$ is irreducible over $F$ and $p(x) | a(x)b(x)$, then $p(x) |
> a(x)$ or $p(x) | b(x)$.
|
Use the no roots criterion to determine all irreducible quadratics and cubics. Quadratics are very simple, there is only $x^2+x+1$. Cubics are a little more work. But not much. The cubic has to have shape $x^3+ax^2+bx+1$ where there is an odd number of $1$'s. Once you have your list of cubics, multiply each by $x^2+x+1$.
|
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|
Is it really true that "if a function is discontinuous, automatically, it's not differentiable"? I while back, my calculus teacher said something that I find very bothersome. I didn't have time to clarify, but he said:
If a function is discontinuous, automatically, it's not differentiable.
I find this bothersome because I can think of many discontinuous piecewise functions like this:
$$f(x) =
\begin{cases}
x^2, & \text{$x≤3$} \\
x^2+3, & \text{$x>3$}
\end{cases}$$
Where $f'(x)$ would have two parts of the same function, and give:
$$\begin{align}
f'(x) = &&
\begin{cases}
2x, & \text{$x≤3$} \\
2x, & \text{$x>3$}
\end{cases} \\
= && 2x
\end{align}$$
So I'm wondering, what exactly is wrong with this? Is there something I'm missing about what it means to be "continuous"? Or maybe, are there special rules for how to deal with the derivatives of piecewise functions, that I don't know about.
|
Flagrantly ignoring your specific example: suppose a function $f$ is differentiable at a point $x$. Then by definition of differentiability:
$$\lim_{h\rightarrow0}\frac{f(x+h) - f(x)}{h}$$
must exist (and by this notation I mean the limits exist in both the positive and negative directions and are equal). Since the bottom of that fraction approaches $0$, it's necessary for the top also to approach $0$, or else the fraction diverges. But the top approaching $0$ is just the definition of $f$ being continuous at $x$. So a function that isn't continuous can't be differentiable.
So, your example fails to be differentiable for the same reason that it fails to be continuous, which is that top of that fraction tends to $3$, not $0$, when approached from the positive direction.
|
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|
Subgroup generated by $1 - \sqrt{2}$, $2 - \sqrt{3}$, $\sqrt{3} - \sqrt{2}$ For a number field $K$, Dirichlet's unit theorem says that$$(O_K)^\times = \mathbb{Z}^{r - 1} \oplus (\text{a finite cyclic group}),$$where $r$ is the number of all infinite places of $K$. An infinite place of $K$ is either an embedding of $K$ (as a field) into $\mathbb{R}$ or the complex conjugacy class of an embedding of $K$ (as a field) into $\mathbb{C}$ with dense image.
When $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, how do I see that $1 - \sqrt{2}$, $2 - \sqrt{3}$, and $\sqrt{3} - \sqrt{2}$ generate a subgroup $\Gamma$ of $(O_K)^\times$ such that$$\Gamma \cong \mathbb{Z}^3 = \mathbb{Z}^{r - 1}?$$
|
Expanding what I believe Franz Lemmermeyer meant with his hint. This is closely built into all the other answers as well (+1 to y'all), but I learned a trick from reading all this, so I can't keep the lid on...
Assume contrariwise that there exist a non-trivial triple $(a,b,c)$ of integers such that
$$
|(1-\sqrt2)^a(2-\sqrt3)^b(\sqrt3-\sqrt2)^c|=1.\qquad(i)
$$
I use the absolute value signs so that I don't need to worry about the torsion unit. Let's apply the automorphism $\tau:\sqrt2\mapsto\sqrt2,\sqrt3\mapsto-\sqrt3$ to equation $(i)$. As $\tau(1-\sqrt2)=(1-\sqrt2)$, $\tau(2-\sqrt3)=(2-\sqrt3)^{-1}$ and $\tau(\sqrt3-\sqrt2)=-(\sqrt3-\sqrt2)^{-1}$, we get another equation
$$
|(1-\sqrt2)^a(2-\sqrt3)^{-b}(\sqrt3-\sqrt2)^{-c}|=1.\qquad(ii)
$$
Multiplying $(i)$ and $(ii)$ together gives
$$
|(1-\sqrt2)^{2a}|=1\qquad(iii)
$$
from which we can infer that $a=0$.
There are several ways to continue. By using another automorphism $\sigma:\sqrt2\mapsto-\sqrt2, \sqrt3\mapsto-\sqrt3$ we similarly reach the conclusion $c=0$. Continuing in the same vein, using the automorphism $\sigma\tau$ gives us $b=0$.
Essentially we are taking the relative norms of these units to the three quadratic subfields and checking what that gives us.
Alternatively, after having derived $(iii)$ we can use Mathmo123's idea. The equations $(i)$ and $(iii)$ imply that $(2-\sqrt3)^{2b}$ and $(\sqrt3-\sqrt2)^{2c}$ are inverses of each other. But the former is an element of $\Bbb{Q}(\sqrt3)$ and the latter of $\Bbb{Q}(\sqrt6)$. The intersection of the two fields is $\Bbb{Q}$, so both must be rational. But the only rational units are $\pm1$, so this is absurd.
|
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|
Is there a difference between $\sin^2(x)$ and $\sin(x)^2$? Is there a difference between these? Or are they the same?
If they're the same, then why does $\sin^2(x)$ seem like its used more often?
|
The two possibilities are
$$
P_1 = \big(\sin x\big)^2
\\
P_2 = \sin\big(x^2\big)
$$
The convention (which confuses beginners) is
$$
\sin^2 x = P_1
$$
If you write
$$
\sin x^2
$$
or even
$$
\sin(x)^2
$$
we cannot tell which you mean.
|
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|
Isomorphism of $k$-Algebras Let $k$ be a field an $A$ a finitely generated $k$-algebra (think of the polynomial ring).
Suppose I have an isomorphism of $k$-algebras
$$
A \cong A/I
$$
for some ideal $I$. If $A$ were a finitely generated $k$-vector space we can conclude $I=0$, as opposed to the non finitely generated case. So can we still get $I=0$, even though $A$ is not f.g. as a vector space over $k$, only as an algebra?
|
The answer is yes for finitely generated commutative $k$-algebras (Mariano's answer deals with the non-commutative case). More generally, if $A$ is a Noetherian ring and there is an isomorphism $f\colon A/I \to A$, then $I = (0)$.
Why? Let $\pi\colon A\to A/I$ be the quotient map. If $I\neq (0)$, then $f^{-1}[I]$ is a nonzero ideal in $A/I$, so $\pi^{-1}[f^{-1}[I]]$ is an ideal in $A$ which properly contains $I$. Repeating, we can build an infinite ascending chain of ideals.
Edit: Here are a few more details. We start with $I_0 = (0)$ in $A$, and we set $I_{i+1} = \pi^{-1}[f^{-1}[I_i]]$ for all $i$. Let's check by induction that we get a proper increasing sequence $I_0 \subsetneq I_1 \subsetneq \dots$ of ideals in $A$. In the base case, $I_1 = \pi^{-1}[f^{-1}[(0)]] = \pi^{-1}[(0)] = I$, and we've assumed that $(0) \subsetneq I$. And given the inductive hypothesis $I_{i-1} \subsetneq I_i$, we have $I_i = \pi^{-1}[f^{-1}[I_{i-1}]] \subsetneq \pi^{-1}[f^{-1}[I_{i}]] = I_{i+1}$.
|
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|
sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \frac{\sum_{n=1}^{\infty} 2^n}{\sum_{n=1}^{\infty} 3^n} + \frac{\sum_{n=1}^{\infty} 3^n}{\sum_{n=1}^{\infty} 6^n}$, am I on the right track? I really don't know how else to solve it.
Thanks.
|
$$\sum_{i=1}^\infty \frac{2^n + 3^n}{6^n}=\sum_{i=1}^\infty (\frac{2}{6})^n + \sum_{i=1}^\infty (\frac{3}{6})^n=\frac{1}{2}+1=\frac{3}{2}$$
sums from forumla of geometric series $1/3^n$ and $1/2^n$
|
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|
Prove: $\csc a +\cot a = \cot\frac{a}{2}$
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$
All I have right now, from trig identities, is
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
|
Another solution,
$${1\over \sin(a)}= {1\over \tan(a/2)}-{{1-\tan^2(a/2)}\over {2\tan(a/2)}}$$
$${1\over \sin(a)}={1+\tan^2(a/2) \over 2\tan(a/2)}$$
|
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|
Boundary value problem in complex plane, how to find general solution?
Hi
I was trying to find out general solution, can anyone explain how to find it?
|
The domain is conformally equivalent to the half-plane, and hopefully you have discussed in class how to solve the Dirichlet problem in the half-plane using the Poisson kernel. Since the Laplace equation is conformally invariant it suffices to determine this conformal map. This can be done by translating the "corner" of the domain to the origin, then considering a power map: something like $z\mapsto z^{2/3}$.
|
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|
Is "Let $M$ is a connected n-manifold, then: $M$ is $R$-orientable iff $H_n(M;R)\cong R$" true? I heard that every topological $n$-manifold $M$ is $\mathbb{F}_2$-orientable, but then for $M=\mathbb{R}^2$ is must be $H_2(\mathbb{R}^2;\mathbb{F}_2)\neq 0$?
In lecture we had the lemma: Let $M$ is a connected n-manifold, then: $M$ is $R$-orientable iff $H_n(M;R)\cong R$.
Something must be wrong, I always thought that $H_2(\mathbb{R}^2;\mathbb{F}_2)=0$ because $M$ is contractible. Or is the lemma only true for compact manifolds?
|
Yes, this is true only for compact manifolds (I assume that your "manifolds" are not allowed to have boundary). In fact, $H_n(M;R)\cong R$ iff $M$ is $R$-orientable and compact. You should be able to find a proof in any text that covers Poincaré duality. For instance, this follows from Theorem 3.26 and Proposition 3.29 of Hatcher's Algebraic Topology (note the hypothesis that $M$ is closed in Theorem 3.26).
|
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|
$R^2-\{x_1,x_2,\dots,x_n\}$ does not have the structure of topological group Let $n>1$. I need to show that the space $X=\mathbb{R}^2-\{x_1,x_2,\dots,x_n\}$ does not have the structure of topological group.
This is an exercise about the Van Kampen theorem. Certainly, we should prove it by contradiction, but I do not know how to get this contradiction.
|
Fundamental group of a topological group is abelian always. And it's easy to show the given space has non-abelian fundamental group.
Proof of the statement in bold:
Let $a$ and $b$ be two loops in a topological group $(G,\bullet )$ starting at the identity element $e$. We need to show $ a\ast b \simeq b\ast a$, where "$\ast$" is the fundamental group operation.
Now for each $t,s\in [0,1]$, define
$F_t(s)=a(st)\ast(a(t)\bullet b(s))\ast \bar a(st)$
Clearly {$F_t$} gives the homotopy between $b$ and $a\ast b \ast \bar a$.
( We can assume the topological group path connected )
|
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|
Probability Problem - Finding a pdf Below is a problem I did. However, I did not come up with the answer in book.
I am thinking that I might have the wrong limits for the integral. I am hoping
somebody can point out what I did wrong.
Bob
Problem
Let $Y = \sin X$, where $X$ is uniformly distributed over $(0, 2 \pi)$. Find
the pdf of $Y$.
Answer:
\begin{eqnarray*}
P(Y<=y_0) &=& P( \sin x <= y_0 ) = P( x <= \arcsin( y_0) ) \\
P(Y<=y_0) &=& \int_0^{\arcsin(y_0)} \frac{1}{2 \pi} dx =
\frac{x}{2 \pi} \Big|_0^{\arcsin(y_0)} =
\frac{\arcsin(y_0)}{2 \pi} - \frac{0}{2 \pi} \\
P(Y<=y_0) &=& \frac{\arcsin(y_0)}{2 \pi} \\
F(Y) &=& \frac{\arcsin(Y)}{2 \pi} \\
f(y) &=& \frac{1}{2 \pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
f(y) &=& \frac{1}{\pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
I believe that book is right because
\begin{eqnarray*}
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=&
\frac{\arcsin{y}}{\pi} \Big|_{-1}^{1} =
\frac{\frac{\pi}{2}}{\pi} - \frac{-\frac{\pi}{2}}{\pi} \\
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=& 1 \\
\end{eqnarray*}
|
your answer is correct indeed just a little mistake. in the first step:
$$P(Y \le y) = P( \sin x \le y ) \Rightarrow P(x_1 \le x \le x_2 )$$ as shown in figure below (sorry figure is for $y=\sin(x+\theta)$ but still is useful .just draw the figure for $\sin(x)$ in your imagination. also figure is from "Probability, Random Variables and Stochastic Processes" by Papoulis A. chapter 5 example 7. which is very similar to your problem).
since $\sin(x)$ intersects with each horizontal line two times. so we can now write the right answer as below:
$$P(Y \le y)= \frac{1}{2\pi} \int_0^{x_{1}}dx+\frac{1}{2\pi} \int_{x_{2}}^{2\pi}dx$$
and we have in mind that $x_2=\pi-\arcsin(y)=\pi-x_1$ so we get:
$$P(Y \le y)= \frac{1}{2\pi} \int_0^{x_{1}}dx+\frac{1}{2\pi} \int_{x_{2}}^{2\pi}dx = \frac{1}{2\pi}\times (x_{1}+2\pi-x_{2})=\frac{1}{2\pi}\times (x_{1}+\pi+x_{1}) \\ = \frac{1}{2\pi}\times (2x_{1}+\pi)=\frac{1}{2}+\frac{1}{\pi}\arcsin(y)$$
now by taking derivative with respect to $y$ we get:
$$f(y) = \frac{1}{\pi \sqrt{1 - y^2}}$$
EDIT
To answer the question in comments, when $y=a\sin(x+\theta)$ and $x$ is uniform on $(0,2\pi)$ and we want to find $F(y),f(y)$ i.e. CDF and PDF of $y$. We want to find $F(y)=P(Y\le y)$. To find where $Y\le y$ notice that according to the figure above, this happens in two intervals (when the sine figure is less than the $y$ line) which is $x\in (-\pi,x_0)$ and when $x \in (x_1,\pi)$. Therefore
$$P(Y \le y) = P((-\pi,x_0) \cup (x_1,\pi)) = P((-\pi,x_0)) + P((x_1,\pi))$$
The total length of this interval is $(x_0 - (-\pi)) + (\pi-x_1) = 2\pi + x_0 - x_1$ and $x_0 = \text{Arcsin}\left(\frac{y}{a}\right)-\theta , x_1 = \pi - \text{Arcsin}\left(\frac{y}{a}\right) - \theta $ which results from solving $y = a\sin(x+\theta)$. Substituting these in
$$P(Y \le y) = \frac{1}{2\pi}(2\pi + x_0 - x_1) = \frac{1}{2} + \frac{1}{\pi}\text{Arcsin}\left(\frac{y}{a}\right)$$.
Also please notice that since $x_1 = \pi - x_0 - 2\theta$ the length of the interval can also be written as $L = 2\pi + x_0 - x_1 = \pi + 2x_0 + 2\theta$ which results in the same solution.
|
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|
Which convergence test to use I need to prove that the following converges:
$$\sum\limits_{n=0}^{\infty} n^3e^{-n^4}$$
I'm not sure which test to use. I've tried ratio test but it gets messy. I don't think I have any sums to use in the comparison test and the integral test obviously wouldn't work. Thanks
|
Directly $\;n-$th root test:
$$\sqrt[n]{n^3e^{-n^4}}=\frac{\left(\sqrt[n]n\right)^3}{e^{n^3}}\xrightarrow[n\to\infty]{}0<1$$
and thus the series converges
|
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|
Find the volume of the shapes defined as: $n$-dimensional cone
Can this be done without using spherical coordinates?
$$i)\ \frac{x_1^2}{a_1^2}+\cdots+\frac{x_{n-1}^2}{a_{n-1}^2}=\frac{x_n^2}{a_n^2}, \quad 0\leq x_n \leq a_n.$$
$$ii)\ \text{The shape bounded by the following:} \\ \sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}}=1, \quad \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=2, \\ \frac{x}{a} = \frac{y}{b}, \quad 4 \frac{x}{a} = \frac{y}{b} \ \ (a,b > 0)$$
Im thinking that some of these last equations are cones too? Im not sure. What geometric shapes are they to begin with? Does anyone have a link to a site, that explains these shapes?
|
It can be, but it's usually not because the equations are really messy. Almost no problem is only solvable in a given coordinate system, it's just a question of which integrals are less awful
|
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|
a one-to-one and onto function that cannot be continuous I'm reading Torchinsky's Real Analysis, and am having difficulty demonstrating if
$$\varphi : [0,1]^2 \rightarrow [0,1]$$ is one-to-one and onto, then it cannot be continuous. (5.13)
Unable to demonstrate this myself, I did some research and found what seems to a proof:
"Suppose such a function exists $$\varphi : [0,1]^2 \rightarrow [0,1]$$
Now, $[0,1]^2 \setminus\{\varphi^{-1}(1/2)\}$ is connected (since $\varphi$ is a bijection we are just removing one point from the square) and by properties of continuous functions we know that $\varphi$ maps connected sets to connected sets (cf. Rudin 4.22) . But $$\varphi\left([0,1]^2 \setminus\{\varphi^{-1}(1/2)\}\right) = [0, 1/2)\cup(1/2, 1]$$ which is not connected, hence no such map can exist."
What is unclear to me: removing point from the domain introduces a discontinuity in the codomain since the function is onto, but how does this prove the assertion for the domain in question: $$ [0,1]^2 \text{ ?}$$
|
We assumed the existence of such a function $\varphi$, did some work to arrive at a contradiction (in particular, the existence of a continuous map that maps a connected set to a set that is not connected). Therefore the assumption was not consistent.
The author might have meant $\varphi$ instead of $\varphi^{-1}$. That might have caused confusion.
|
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|
How to find the stationary point of $f(x,y)=\sin x \sin y \sin (x+y)$ How to find the stationary point(s) of $f(x,y)=\sin x \sin y \sin (x+y)$
With $x,y\in(0,\pi)$
So far I have found $$\nabla f =(\color{red}{\sin x\cos (x+y)+\cos x\sin (x+y)\sin y},\color{blue}{\sin y\cos (x+y)+\cos y\sin (x+y)\sin x)}$$
So we need
$$(\sin x \cos (x+y)+\cos x \sin (x+y)) \sin y=0 \\ (\sin y \cos (x+y)+\cos y\sin (x+y))\sin x=0$$
Problem is I don't know how to solve this.
|
$\left(\sin x \cos(x+y) + \cos x\sin(x+y)\right) \sin y = 0$ when $\sin x \cos(x+y) + \cos x\sin(x+y) = 0$ or when $\sin y = 0$. I trust that you know when $\sin y = 0$. What about that first one?
\begin{align}
\sin x \cos(x+y) + \cos x\sin(x+y) &= 0 \\
\sin x \cos(x+y) &= -\cos x\sin(x+y)\\
-\tan x &= \tan(x+y)\\
\tan(-x) &= \tan(x+y)
\end{align}
Because tangent is periodic with period $\pi$, then this equation is satisfied when $x+y = -x + n\pi$, where $n$ is any integer (provided there are no domain restriction violations). This is equivalent to $y = -2x + n\pi$.
The second equation you mentioned is nearly identical.
|
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How is it posible that $f + g \in O(f)$? I am confused how to do this question. Intuitively it doesn't even make sense how a function $f$ plus another function is in $O(f)$. How can I approach this question:
$$
n\log(n^7)+n^{\frac{7}{2}} \in O(n^{\frac{7}{2}}).
$$
We know the fact that $\log n < n$ and I tried factoring out the $n$ but I am stuck. Any hints would be appreciated, Thanks!
|
The easiest way is to take the limit: $\frac{n^{\frac{7}{2}}+7 n \log n}{n^{\frac{7}{2}}} \to_n 1$, hence these two functions are of the same order.
|
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|
Problem rearranging a trigonometric expression I've been trying to figure out why this equation is satisfied:
$$\frac{1-(\cos(x))^3}{x^2}=\frac{2\cdot(\sin(\frac{x}{2}))^2}{x^2}\cdot(1+\cos(x)+(\cos(x))^2)$$
but I can't find the proper trigonometric changes in order to change from one to another. I know that the sine comes from the double angle formula, but I obtain slightly different results in other parts and it's never the same as the formula above.
I'd appreciate any hint.
Thanks for the time.
|
One may observe that, in general,
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$ giving
$$
1-(\cos x)^3=(1-\cos x)(1+\cos x+(\cos x)^2)
$$ then, by using $2\sin^2(x/2)=1-\cos x$ as noticed by @André Nicolas, you are Ok with your expression.
|
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Find the Maclaurin series for $\ln(2-x)$ A little unsure if the result I got makes sense, so I want to ask here to be sure I am not doing something very silly.
The Maclaurin series is given by $f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^n$
First I need a formula for the $n$th derivative of $x$. The derivative of $\ln(2-x)=(x-2)^{-1}$, and the derivative of $(u)^{-n}=-n(u)^{-n-1}u'$, and $u'=1$ for $u=x-2$ so:
$$f^n(x)=\ln(2-x), (x-2)^{-1}, -(x-2)^{-2}, 2(x-2)^{-3}, -6(x-2)^{-4},...$$
For a Maclaurin series, we are interested in $f^n(0)$:
$$f^n(0)=\ln(2), \frac{1}{(-2)^{1}}, \frac{-1}{(-2)^{2}}, \frac{2}{(-2)^{3}}, \frac{-6}{(-2)^{4}}={(-1)\frac{(n-1)!}{2^{n}}, n=1,2,3... \choose \ln(2), n=0}$$
Plugging into the Maclaurin series $f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^n$
$$f(x)=\ln(2)\frac{1}{0!}*x^0+\sum_{n=1}^{\infty}(-1)\frac{(n-1)!}{2^{n}}\frac{1}{n!}x^n=\ln(2)-\sum_{n=1}^{\infty}\frac{x^n}{n2^{n}}$$
Is this answer sensible? A bit confused since there's a term ($\ln(2)$) that doesn't follow the same formula as the rest and I had to put it "outside" the sum (when you write out the sum, that shouldn't matter I suppose, but I haven't encountered this partiuclar scenario before)
EDIT: $\ln(2-x)$, not $\ln(x-2)$, was the initial fuction
|
The given answer is not only sensible, it is correct. The "extra" $\ln{2}$ term is part of the solution. Not every function has all its terms in the Maclaurin series fitting a perfect pattern.
|
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Is the tensor product over $B$ of two flat $A$-modules flat over $A$? Given a morphism of commutative rings $A\to B$ such that $B$ is a flat $A$-module and given $M$, $N$ two $B$-modules flat as $A$-modules, is the tensor product $M\otimes_B N$ flat over $A$??
The tensor product $M\otimes_A N$ is flat over $A$, the proof is not hard:
Given an exact sequence $0\to X\to Y\to Z\to 0$ of $A$-modules since $M$ is
a flat $A$-module the sequence $0\to X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M\to 0$ is exact. Again $N$ is a flat $A$-module, and since the tensor commute we have the statement.
I've tried some similar arguments but without any success, and I can't find a counterexample.
There is a morphism $M\otimes_A N\to M\otimes_B N$ (it is shown in here), but I don't know how to use it.
|
Assume that as $B$-modules, you have that $N= B\otimes_A N$: note that this is generally false. There is an obvious map $B\otimes_A N\to N$ that sends $b\otimes n\to bn$. One might be tempted to use the functional inverse $n\to 1\otimes n$, but this is not necessarily $B$-linear, unless say $A\to B$ is onto. If it were true, then the argument goes through. See this. As an example of how $B$ might definitely not be isomorphic to $B\otimes_A B$, take $B$ to be a $k$-algebra with $k$ a field, $k\to B$ the structure map, and say $B$ is finitely dimensional of dimension $>1$. Then $B\otimes_k B$ has dimension strictly greater than $B$.
Now take $C$ an exact sequence. Then $N\otimes_A C$ is exact, and since $B$ is flat, so is $B\otimes_A N\otimes_A C$. Now write $$M\otimes_B N =(M\otimes_A B)\otimes_B (B\otimes_A N)$$
Then $M\otimes_B N\otimes C$ is obtained from $B\otimes_A N\otimes_A C$ by tensoring first over $B$ with $B$, which preserves exactness, and then with $M$ over $A$, which also does.
|
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How do I find the vector $T\begin{pmatrix} 5 & 0 \\ -10 & -13 \end{pmatrix}$? I defined a function $T: M^R_{2x2} \rightarrow R_4[x]$ and I defined:
$T\begin{pmatrix} 2 & 3 \\ 1 & 0 \end{pmatrix} = x^2$
$T\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = 3x - 4$
$T\begin{pmatrix} 0 & 2 \\ 4 & 5 \end{pmatrix} = 2x^2 - 7$
How can I find the following function if I know that all three vectors above are linearly independent:
$T\begin{pmatrix} 5 & 0 \\ -10 & -13 \end{pmatrix} = ?$
What is the method to handle such questions?
|
Hint: You'll need to determine whether there exist $a,b,c$ for which
$$
\pmatrix{5&0\\-10&-13} =
a\pmatrix{2&3\\1&0} +
b\pmatrix{1&0\\0&2} +
c \pmatrix{0&2\\4&5}
$$
and, if such $a,b,c$ exist, find them.
|
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Determine if the function $f$ is even, odd or neither given the graph of $f$ The following picture is a graph of a function $f$. I am to determine whether the function is even, odd, or neither. I reasoned that $f$ is odd, because if the graph is rotated $\pi$ radians, the graph is reproduced perfectly. The graph is clearly not even, because it is not symmetrical with respect to the vertical axis. However, the solution states that $f$ is neither even or odd, and I do not understand why.
|
You're right that the graph has a symmetry - if you rotate about a certain point on the $y$-axis, you preserve the graph. Algebraically, it has the property:
$$f(x)=2c-f(-x)$$
where $c=f(0)$ is the intersection of the graph and the $y$-axis. The condition of oddness is more strict, however, it demands:
$$f(x)=-f(-x)$$
meaning the function must have this rotational symmetry about the origin, not about some arbitrary point. Note that oddness implies $f(0)=0$.
|
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Showing $u'=v$ a.e. given $u_k \to u$ and $u'_k \to v$ in $L_2(\mathbb{R})$. Suppose $(u_k)$ is a sequence of differentiable functions in $L_2(\mathbb{R})$ satisfying
(1) There is a $u \in L_2(\mathbb{R})$ so that $\| u_k - u\|_2 \to 0$.
(2) There is a $v \in L_2(\mathbb{R})$ so that $\|u'_k - v \|_2 \to 0$.
If $u$ is differentiable, show that $u'=v$ almost everywhere.
This is a problem on a practice qualifying exam that I am having trouble with.
|
Define $w(x) := \int_0^x v(t)\,dt$. Then $w$ is differentiable a.e. and $w' = v$ a.e. Let $\varphi\in C_0^\infty(\mathbb R)$. Then
$$
\langle u'-w',\varphi\rangle = \langle w-u,\varphi'\rangle = \langle w-u_k,\varphi'\rangle + \langle u_k-u,\varphi'\rangle.
$$
Now,
$$
\langle w-u_k,\varphi'\rangle = \langle u_k'-v,\varphi\rangle\to 0
$$
as $k\to\infty$. Therefore, and since $u_k\to u$ in $L^2$, we have $\langle u'-w',\varphi\rangle = 0$ for all $\varphi\in C_0^\infty(\mathbb R)$. Since $C_0^\infty(\mathbb R)$ is dense in $L^2(\mathbb R)$, this shows $u' = w' = v$ a.e., which we had to prove.
|
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Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + \frac{3yx}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 3x + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 2.5x = 3y$$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?
|
Assuming your calculations so far are correct (I didn't check), you are almost there. Divide both sides by $y$, you will get
$$\frac {3x^2} {y^2} + \frac {2.5x} y = 3,$$
a quadratic equation for $\frac xy$.
|
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Algebraic Varieties vs Smooth Manifolds There are many posts I have read on that subject which seem unclear for me. My main question (it might be silly) is:
"Every non-singular algebraic variety over $\mathbb{C}$ is a smooth
manifold."
(see: http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds)
How? For an algebraic variety we have the Zariski topology which is not even Hausdorff? How can they be diffeomorphic then?
|
The statements it that if the variety is endowed with the relative topology as a subset of $\Bbb C^n$ then it is a manifold.
|
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About a Possible Decomposition of a Set $E$ of Infinite Measure Can we decompose $m(E)=\infty$ into $E=\bigcup_{n\in \mathbb{Z}}E_n$ where $E_n=E\bigcap[n-1,n]$ where $n \in \mathbb{Z}$? Then we have $\{E_n\}_{n\in \mathbb{Z}}$, a countable collection of measurable subsets of $E$ and of finite measure since $E_n \subseteq [n-1,n]$ so $m(E_n)\leq 1< \infty$. Also, each pair of $E_n$'s are not necessarily disjoint.
Now, if there is a closed set $F_n \subseteq E_n$ for each $n\in\mathbb{Z}$, this means that the collection $\{F_n\}_{n\in \mathbb{Z}}$ do not have pairwise disjoint sets in it right?
My question might be lacking in motivation, but I was hoping that if my constructions above are true then it might be a remedy in my proof (invalid) on the extension of Lusin's theorem to the case where $m(E)=\infty$ which appears in this post: Lusin's Theorem when $m(E)=+\infty$. Proof Verification.
|
*
*First question: Yes. From $E_n \triangleq E \cap [n-1,n]$, it's trivial that $E=\bigcup_{n\in \Bbb Z} E_n$.
*Second question: No. Consider $E = \Bbb R$, then $m(E) = \infty$. $$E_n \triangleq E \cap [n-1,n] = [n-1,n],$$ so "each pair of $E_n$'s is not necessarily disjoint" (as $E_n \cap E_{n+1} = \{n\}$). $$F_n \triangleq E \bigcap \left [n-\dfrac34,n-\dfrac14 \right] = \left [n-\dfrac34,n-\dfrac14 \right] \subseteq E_n \quad \forall\,n \in \Bbb N,$$ but the collection of closed subsets $\{F_n\}_{n\in \Bbb Z}$ is pairwise disjoint: $$F_i \cap F_j = \varnothing \text{ whenever } i \ne j.$$
|
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Mean value theorem question. Show that if $f(2) = -2$ and $f ' (x) \geq 5$ for $x > 2$, then $f(4) \geq 8$.
I know that the $b$ value is $4$ and the $a$ value is $2$ for the mean value theorem equation.
$$f'(c)=\frac{f(4) - f(2)}{4-2}$$
We know the value of $f(2)$ but for $f(4)$ I'm confused there and with the $f ' (x) \geq 5$ for $x > 2$.
Any help will be appreciated.
|
Your equation can be rewritten as
$$f(4)=f(2)+(4-2)f'(c).\tag{1}$$
We also know that $c$ is between $2$ and $4$, and therefore in particular $c\gt 2$.
So $f'(c)\ge 5$. Also, $f(2)=-2$. It follows from (1) that
$$f(4)\ge -2+(4-2)(5).$$
|
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Computing degrees of projective varieties via Chern classes I know that the degree of a projective hypersurface $H \subset \mathbb{P}^n$ can be computed in terms of the Chern class of the normal (line) bundle of $H$. Is there a similar formula for the degree of a higher codimension projective variety in terms of Chern classes of the normal bundle?
In general, does degree just depend on the normal bundle of the projective variety in projective space? I feel like the answer is no, which would make it impossible to compute the degree in terms of the Chern class of the normal bundle.
|
Here is a general answer in terms of Chow rings.
Let $i:Y\hookrightarrow \mathbb P^n$ be a smooth closed subvariety of codimension $r$ and degree $d$, so that for the corresponding cycle class we have $[Y]=dH^r\in A^r(\mathbb P^n)=\mathbb Z\cdot H^r$ .
We have (Hartshorne, page431): $$c_r(N_{Y/X})=i^\ast [Y]=i^\ast (dH^r)=dh^r\in A^r(Y)$$ where $h=i^\ast H\in A^r(Y)$ and where $c_r(N_{Y/X})\in A^r(Y)$ is the Chow Chern class
[which, in case the base field is $\mathbb C$, is infinitely more precise than its image in singular cohomology $c_r^\mathbb C(N_{Y/X})\in H^{2r}(Y(\mathbb C),\mathbb Z)$ ].
If $c_r(N_{Y/X})$ is known, we may often extract the degree $d$ of $Y$ from the the above formula $c_r(N_{Y/X})=dh^r$.
However if $h^r=0$, as is the case for curves in $\mathbb P^3$ for example, the equality $c_r(N_{Y/X})=dh^r$ reduces to $0=d\cdot0$, which (in conformity with Pooh Bear's great counterexample) doesn't allow us to compute $d$.
|
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How to find the eigenvector when there are multiple instances of the eigenvalues $$\begin{pmatrix}0&1&-1\\2&1&-2\\-1&-1&0\end{pmatrix}$$
The characteristic polynomial is $$\lambda^3-\lambda^2-5\lambda-3=(\lambda+1)(\lambda^2-2\lambda-3)=(\lambda+1)(\lambda+1)(\lambda-3)$$
So I have two eigenvectors with the same eigenvalue $-1$ and I already know one of them is $(1,0,1)^t$.
I found the eigenvector for $3$ but how do I find the other eigenvector for $-1$? If I do it the usual way by placing it in the matrix $(\lambda I -A)$ I get to $$-x-y+z=0$$
|
When there is an eigenvalue of multiplicity $k>1$, there is an eigenspace of dimension at most $k$. If $k=1$, the dimension is always $1$ as there is always at least one eigenvector $\vec u$, thus all $t\vec u$ are also eigenvectors, thus an eigenspace of dimension $1$.
When $k>1$, it can happen that there are not "enough" eigenvectors (that is, the eigenspace has a dimension $<k$). Here it's not the case though.
You have to solve
$$\left(\begin{matrix}1&1&-1\\2&2&-2\\-1&-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\end{matrix}\right)$$
This amount to the single equation $x+y-z=0$, as you found. This means there are two arbitrary parameters, say $x$ and $y$, then $z=x+y$.
To find an eigenspace of dimension $2$, just find two non-collinear eigenvectors, by letting $x=1,y=0$, then $x=0,y=1$, hence the eigenvectors $(1,0,1)^T$ and $(0,1,1)^T$.
Notice there are infinitely many eigenvectors, and any pair of non-collinear vectors lying in the plane defined by the above two would be equally valid.
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Understanding the unit tangent vector The vector $\dot x (s) $ is called the unit tangent vector to the oriented curve $x=x(s)$.
I am told that $x=x(s)$ is a natural representation of a regular curve C.
What does natural representation mean?
The derivative $\dot x(s)=\frac{dx}{ds}$ is defined as the tangent direction to C at the point $x(s)$.
If $x=x(s)$ is a representation of a curve, how can it be a point?
Why does the length of $\dot x(s)=1$ ?
What does it mean to say the quantity $\dot x(s)$ is an oriented
quantity?
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$s$ is the arclength. Namely, (and we just work in $\mathbb R^2$ here):
if $\gamma:[0,1]\to \mathbb R^2$, is a paramterization of your curve $C$, then
$\gamma (t)=(x(t),y(t))\in C$ and $\gamma(0)=(x_0,y_0); \ \gamma(1)=(x_1,y_1)$
then the length of $C$ from $(x_0,y_0)$ to $(x_1,y_1)$ is given by
$l=\int_{0}^{1}\sqrt{\dot x^2+\dot y^{2}}dw$ and this is just $ \int_{0}^{1}\left \| \frac{d\gamma}{dw} \right \|dw$.
This formula is dervied easily from the Riemann Sums that arise from considering polygonal approximations to $C$.
Generalizing this, i.e. we have the function
$$\tag1l(t)=\int_{0}^{t}\left \| \frac{d\gamma}{dw} \right \|dw,\ $$ which measures the length of $\gamma $ from $\gamma (0)=(x_0,y_0)$ to $\gamma (t)= (x,y)$.
Now if we set $s=l(t)$ then $t=l^{-1}(s)$ which means that we can define a new parameterization $\overline \gamma (s)=\gamma (l^{-1}(s))$. That is $$\tag2\overline \gamma =\gamma \circ l^{-1}$$
$\overline \gamma $ is what is meant by the natural representation of a regular curve $C$, and here is why:
if we use $s$ in (1) with $\overline \gamma $ we obtain $l(s)=\int_{0}^{s}\left \| \frac{d\overline \gamma}{dw} \right \|dw$. But $l(s)=s$ because a change in parameterization does not affect the length of the curve (why?), and now using FTC we have $1=\frac{ds}{ds}=\left \| \frac{d\overline \gamma }{ds} \right \|$ and so we see that $\overline \gamma $ is the parameterization that makes the rate at which the curve is traversed $constant$ and equal to $1$. You can show that it is unique with this property.
Another way to see this is the following:
If we differentiate $\overline \gamma $ wrt to $s$, we get: $\overline \gamma'(s)=\gamma'(l^{-1}(s))(l'^{-1}(s))=(\gamma'(t))t'(s)$ where we have used the fact that $t=l^{-1}(s)$. Or using Leibnitz notation:
$\frac{d\overline \gamma}{ds}=\frac{d\gamma }{dt}\frac{dt}{ds}=\frac{\frac{d\gamma }{dt}}{\frac{ds}{dt}}$. But $\frac{ds}{dt}=\left \| \frac{d\gamma}{dt} \right \|$ from (1), so we have $$\tag 3\left \| \frac{d\overline \gamma}{ds} \right \|=1$$
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Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k}$
Question: The curve in the figure is the parabola $y=kx^2$ where $k>0$.
Several normal lines to this parabola are also shown. Consider the points in the first quadrant from which the normal lines are drawn. Notice that as the $x$ coordinate gets smaller, the $y$-coordinate of the intersection of the normal with the other arm of the parabola also decreases until it reaches a minimum, and then it increases. The normal line with the minimum $y$ coordinate is dotted.
$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$
(b) Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k} $
What I have done:
$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$
$$ f(x) = kx^2 $$
$$ f( x_{0}) = kx_{0}^2 $$
$$ f'(x) = 2kx $$
$$ f'(x_{0}) = 2kx_0 $$
$$ Normal = -1/m $$
$$ m= {-1\over 2kx_0} $$
$$ y-y_1 = m (x-x_1) $$
$$ y-kx_0^2 = {-1\over 2kx_0}(x-x_0) $$
$$ y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k} $$
(b) Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k} $
$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$
$$ {-1\over m} = \frac{-\sqrt{2}}{2} $$
$$ m = \sqrt{2} $$
$$ f'(x) = 2kx $$
$$ 2kx = \sqrt{2} $$
$$ x = {\sqrt{2}\over 2k} $$
$$ f({\sqrt{2}\over 2k}) = {1\over 2k}$$
$$ y-y_1 = m (x-x_1) $$
$$ y - {1\over 2k} = \frac{-\sqrt{2}}{2} (x-{\sqrt{2}\over 2k}) $$
$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$
However I am stuck trying to prove that this is the minimum y coordinate.
I've attempted
$$ y = y$$
$$ kx^2 = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$$
$$ kx^2 + (\frac{1}{2kx_0})x -(kx_0^2 + \frac{1}{2k}) = 0$$
$$ x={-b\pm\sqrt{b^2-4ac} \over 2a}$$
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ (\frac{1}{4k^2x_0^2})-4(k)(-kx_0^2 - \frac{1}{2k})} \over 2k}$$
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ \frac{1}{4k^2x_0^2}-4(-k^2x_0^2 - \frac{k}{2k})} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ \frac{1}{4k^2x_0^2}+4k^2x_0^2 + 2} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ (2kx_0 + \frac{1}{2kx_0})^2} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$
$$ x = x_0 $$
$$ x = \frac{-1}{2k^2x_0} - x_0 $$
Now I am lost..
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You did almost all the work. From
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$
the $x$ you are looking for is the one with the '$-$':
$$x={-\frac{1}{2kx_0}- { (2kx_0 + \frac{1}{2kx_0})} \over 2k}=-\frac{1}{2k^2x_0}-x_0$$
(the other one is simply $x=x_0$).
Now
$$y=k\left(-\frac{1}{2k^2x_0}-x_0\right)^2$$
so you just have to minimize
$$\frac{1}{2k^2x_0}+x_0$$
and it's easy to see that the minimum is at $x_0=\frac{1}{\sqrt2k}$.
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How can I mentally calculate $\cos(x), x∈(0.7, 1.2)$ I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $\cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $\color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $\color{red}{\cos(x)},\ \ \color{blue}{1-x^2/2},\ \ \color{green}{1.3-x/1.3},\ \ \color{blue}{π/2-x}$:
Error:
How can I quickly approximate $\cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?
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Here's a quick approximation. The second-order Taylor series around $x=\pi/3$ is
$$T_2(x) = \frac12 - \frac{\sqrt3}2\left(x-\frac\pi3\right)-\frac14\left(x-\frac\pi3\right)^2.$$
Now, $\frac{\sqrt3}2 = 0.866... \approx \frac{13}{15}$, and $\frac\pi3 \approx 1.05$, so we have
$$T_2(x) \approx \frac12 - \frac{13}{15}(x-1.05) - \frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $\frac{\sqrt3}2 \approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
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$a,b,c \in \mathbf{Z}$ such that $a^7+b^7+c^7=45$ Do there exist integers $a,b,c$ such that $a^7+b^7+c^7=45$?
[I have an ugly argument for a negative answer, is it possible to give a "manual" solution?]
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The seventh powers modulo $49$ are $0,\pm 1,\pm 18,\pm 19.$ There is no way to combine three of these to get $45$ modulo $49$.
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What are the principal (different) mechanisms of infinite descent proof? I’m interested in building a list (including, where possible, links to proofs/papers/examples) which presents all known mechanisms of infinite descent (ID).
I think this list would best be presented in two parts:
*
*Answers, each of which outlines a fundamentally (or at least demonstrably) different mechanism which comprises the critical step(s) in an ID proof.
*Comments to each answer, giving concrete examples of that type of ID proof.
I’ve started by giving several [community wiki] answers myself, in roughly the form I think will be helpful to future readers.
Now… Do any ID proofs use the fact that every odd number is the sum of the squares of four integers which sum to 1? Or Bezout’s identity? etc. What are all of the methods and mechanisms?
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Set theoretic elementary result:
Every subset of a finite set is finite.
(for definition of finite and argument see this math.stackexhange post).
Keep decrementing a positive integer $n$ by $1$.
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Show that $\mathbb{Q}(\sqrt2, \sqrt[3]2)$ is a primitive field extension of $\mathbb{Q}$. I've tried a method similar to showing that $\mathbb{Q}(\sqrt2, \sqrt3)$ is a primitive field extension, but the cube root of 2 just makes it a nightmare.
Thanks in advance
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Try to express both $\sqrt{2}$ and $\sqrt[3]{2}$ as rational functions of $a = \sqrt{2}+\sqrt[3]{2}$. The job is simple and easily done via equation $$(a -\sqrt{2})^{3}=2\tag{1}$$ so that $$a^{3}-3\sqrt{2}a^{2}+6a-2\sqrt{2}=2$$ or $$\sqrt{2}=\frac{a^{3}+6a-2}{3a^{2}+2}\tag{2}$$ and we have $$\sqrt[3]{2}=a-\sqrt{2}$$ and using equation $(2)$ we can replace $\sqrt{2}$ by a rational function of $a$, so that $\sqrt[3]{2}$ is also a rational function if $a$. It thus follows that $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.
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Where am I violating the rules? Being fascinated by the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed, more than 1400 years ago by Mahabhaskariya of Bhaskara I (a seventh-century Indian mathematician) (see here), I considered the function $$\sin \left(\frac{1}{2} \left(\pi -\sqrt{\pi ^2-4 y}\right)\right)$$ which I expanded as a Taylor series around $y=0$. This gives $$\sin \left(\frac{1}{2} \left(\pi -\sqrt{\pi ^2-4 y}\right)\right)=\frac{y}{\pi }+\frac{y^2}{\pi ^3}+\left(\frac{2}{\pi ^5}-\frac{1}{6 \pi ^3}\right)
y^3+O\left(y^4\right)$$ Now, I made (and may be, this is not allowed) $y=(\pi-x)x$. Replacing, I obtain
$$\sin(x)=\frac{(\pi -x) x}{\pi }+\frac{(\pi -x)^2 x^2}{\pi ^3}+\left(\frac{2}{\pi ^5}-\frac{1}{6 \pi ^3}\right) (\pi -x)^3 x^3+\cdots$$ I did not add the $O\left(.\right)$ on purpose since not feeeling very comfortable.
What is really beautiful is that the last expansion matches almost exactly the function $\sin(x)$ for the considered range $(0\leq x\leq\pi)$ and it can be very useful for easy and simple approximate evaluations of definite integrals such as$$I_a(x)=\int_0^x \frac{\sin(t)}{t^a}\,dt$$ under the conditions $(0\leq x\leq \pi)$ and $a<2$.
I could do the same with the simplest Padé approximant and obtain $$\sin(x)\approx \frac{(\pi -x) x}{\pi \left(1-\frac{(\pi -x) x}{\pi ^2}\right)}=\frac{5\pi(\pi -x) x}{5\pi ^2-5(\pi -x) x}$$ which, for sure, is far to be as good as the magnificent approximation given at the beginning of the post but which is not very very bad (except around $x=\frac \pi 2$).
The problem is that I am not sure that I have the right of doing things like that.
I would greatly appreciate if you could tell me what I am doing wrong and/or illegal using such an approach.
Edit
After robjohn's answer and recommendations, I improved the approximation writing as an approximant $$f_n(x)=\sum_{i=1}^n a_i \big(\pi-x)x\big)^i$$ and minimized $$S_n=\int_0^\pi\big(\sin(x)-f_n(x)\big)^2$$ with respect to the $a_i$'s.
What is obtained is $$a_1=\frac{60480 \left(4290-484 \pi ^2+5 \pi ^4\right)}{\pi ^9} \approx 0.31838690$$ $$a_2=-\frac{166320 \left(18720-2104 \pi ^2+21 \pi ^4\right)}{\pi ^{11}}\approx 0.03208100$$ $$a_3=\frac{720720 \left(11880-1332 \pi ^2+13 \pi ^4\right)}{\pi ^{13}}\approx 0.00127113$$ These values are not very far from those given by Taylor ($\approx 0.31830989$), ($\approx 0.03225153$), ($\approx 0.00116027$) but, as shown below, they change very drastically the results.
The errors oscillate above and below the zero line and, for the considered range, are all smaller than $10^{-5}$.
After minimization, $S_3\approx 8.67\times 10^{-11}$ while, for the above Taylor series, it was $\approx 6.36\times 10^{-7}$.
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@Claude Leibivici use the following two point Taylor series in x=-Pi, Pi
$$\frac{z (z-\pi )^3 (z+\pi )^3}{48 \pi ^4}-\frac{5 z (z-\pi )^3 (z+\pi )^3}{16 \pi ^6}+\frac{3 z (z-\pi )^2 (z+\pi )^2}{8 \pi ^4}-\frac{z (z-\pi ) (z+\pi )}{2 \pi ^2}$$ the cuadratic error is superior to any formula above at the same grade
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Is the autoparallel equation same as the geodesic equation? My question may sound silly but i am self learning differential geometry using watching these lectures (Lecture 8) from 2015 by professor Frederic Schuller.
Can somebody please tell that the auto-parallel equation same as the geodesic equation?
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You cannot talk about geodesics until you have a notion of distance, and curvature is insufficient for that. However, in lecture 10 he introduces the metric tensor, and he shows that the geodesic equation for a given metric takes the form of an autoparallel equation for one specific curvature / connection / covariant derivative.
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How is the binomial expansion of the vectors? I'm trying to find out if there is an attempt to define binomial expansion of vectors. i.e
$$(\overrightarrow a + \overrightarrow b)^n = ?? $$
I tried to google around this (e.g : binomial expansion of vectors), but simple searches do not give any useful keywords as how this approach is named.
I'm trying to visualize how the following generalization goes. When two vectors are on the number line (one dimension), their sum is just the magnitude addition and the result is simply lying in the same number line. When the lines lie in general two dimensional space their 'sum' is equivalent to the square of the magnitude of their vector sum (which is essentially the binomial expansion of $(a + b)^2$ ).
So how is the interpretation of $ (a+b)^3 $? Definitely it is not related to the 'normal' vector sum of the two vectors in 3d (that would be simply $(a + b)^2$).
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You can take a binomial type expansion of the vector expression
$$|a-b|^n,$$
where $|\cdot|$ is the norm. For instance, in $\mathbb{R}^2$ consider $f(x) = |x+y|^n$ Taylor expanded about the point $x = (0, 0)$. For $n=2$ we have
\begin{align}
f(x) & = f(0) + x_1f_{x_1}(0) + x_2f_{x_2}(0) + \frac{1}{2!}\bigg(x_1^2f_{x_1x_1}(0) + 2x_1x_2f_{x_1x_2}(0) + x_2^2f_{x_2x_2}(0)\bigg) \\
& = |y|^2 + 2\langle x,y\rangle + |x|^2.
\end{align}
For $n = 4$ you would obtain
$$
|x+y|^4 = |y|^4 + 4 \langle x, y\rangle|y|^2 + (2|x|^2|y|^2 + 4\langle x, y \rangle^2) + 4 \langle x, y\rangle|x|^2 + |x|^4.
$$
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How to show $\text{Im}~ \theta=\Bbb{Q} [\sqrt 2]$ for a homomorphism? How to show $\text{Im} ~\theta=\Bbb{Q} [\sqrt 2]$ for the homomorphism defined as $\theta:\mathbb{Q}[X] \rightarrow \mathbb{R}$ given by $\theta(f(X))=f(\sqrt2)$.
I can show $\Bbb{Q} [\sqrt 2] \subseteq \text{Im}~\theta$ because we can simply take an arbitrary element of $\Bbb{Q} [\sqrt 2]$ and show it is in $\text{Im} ~\theta$ because $a+b\sqrt2=f(\sqrt2)$ where $f(X)=bX+a$.
Now I can't seem to show $\Bbb{Q} [\sqrt 2] \supseteq \text{Im}~\theta$ now it seems obvious that I can write anything in the image as $a+b\sqrt2$ for some $a,b \in \mathbb{Q}$ because we are going to alternate between even and odd powers of $\sqrt 2$ so in the end we will end up with $a+b\sqrt2$ but I can't seem to write it very well even though I get the gist of it.
Any takers?
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Note that for every $n$: $\sqrt{2}^n\in \mathbb{Q}$ for $n$ even and $\sqrt{2}^n\notin \mathbb{Q}$ for $n$ odd. So, if $f(x)=q_nx^n+q_{n-1}x^{n-1}+\ldots+q_1x+q_0\in \mathbb{Q}[x]$, then: $$f(\sqrt{2})=q_n\sqrt{2}^n+q_{n-1}\sqrt{2}^{n-1}+\ldots+q_1 \sqrt{2}+q_0=\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}+\sum_{1\leq i\leq n ~odd}q_i\sqrt{2}^{~i}$$
$$=\bigg(\sum_{0\leq i\leq n ~even}q_i\sqrt{2}^{~i}\bigg)+\bigg(\sum_{0\leq i\leq n ~even}q_i'\sqrt{2}^{~i}\bigg) \sqrt{2}\in \mathbb{Q}[\sqrt{2}].$$
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Homomorphic image of a free group Given a group $G$ with a generating set $S$, such that $|S|=n$, I need to prove that $G$ is a homomorphic image of $F_n$.
Right now I'm just looking for any tips for how to even start this proof or how to start thinking about the question.
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Hint: $F_n$ is a free object in the category of groups i.e $Hom_{Group}(F_n,G)=Hom_{Set}(\{1,...,n\},G)$.
This is equivalent to saying that the forgetful functor from the category of sets to the category of groups has a left adjoint which associates to a set $S$ the free group $F_S$.
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Unable to prove a simple inequality Let the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be strictly increasing. That is: $x_2>x_1 \Rightarrow f(x_2)>f(x_1)$.
In the segment $[0;a] \subset \mathbb{Z}$, we have $f(a)<a$
I want to prove that $\forall i \in [0;a] , f(i)<i$ .
I am getting a bit frustrated because it looks so obvious.
One proof I came up with is using reverse induction:
$\bullet$ Base Case: $k=a$, because it is given that $f(a)<a$
$\bullet$ Inductive Case:
Assume $f(k)<k$. $f$ is increasing so $f(k-1)<f(k)$. (1)
$f(k)<k \Rightarrow f(k) \leq k-1$ . (2)
Combining (1) and (2) we have: $f(k-1)<k-1$.
Thus $\forall k>0 , f(k)<k \Rightarrow f(k-1)<k-1$.
So the conclusion is that starting from $a$ and moving backwards we have $\forall i \leq a$ , $f(i)<i$. Of course we are only interested in the segment $[0;a]$.
However I am sure there is a simpler proof.
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I don't think there is much of a simpler proof. Certainly every proof needs some kind of method equivalent to induction, be it applying the well-ordering principle or infinite descent.
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Probability of choosing 5 out of 60 in ascending order. The title may be a little misleading. Let's say we choose 5 out of 60 balls. We write down the result which are in a form as $k_1,k_2,k_3,k_4,k_5$.
I have to calculate the probability of this happening :
\begin{aligned}k_1<k_2<k_3<k_4<k_5\end{aligned}
Also, the probability of this happening:
\begin{aligned}k_1>\max\{k_2,k_3,k_4,k_5\}\end{aligned}
We do care for the order so the number of the elements in the sample space is : $$\frac{60!}{(60-5)!}$$
I am stuck there. I can't think of anything to do to calculate those two probabilities. I would appreciate it if someone could help me. Thanks in advance!
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Each choice in which $k_1 < \cdots k_5$ corresponds one-to-one to a way to pick five balls from the set of 60. Prove this statement. Then think about how many ways there are to pick 5 balls from 60.
This should be easy. The second one is a little harder, but having thought about the first one in these terms should help. Please write back if you get stuck again.
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|
Generating Numbers Proof We can do the following actions: multiply by $10$ (add $0$ at the end of number) , multiply by $10$ and add $4$ (add $4$ at the end of number) and divide by $2$. I need to prove that we can get every natural number from $4$ using these actions.
Is it easier to prove that from any number I can get $4$ by multiplying it $2$ , dividing by $10$ and extract it's last $4$, or it is the same problem?
|
This is a trickier problem than I expected, so this is only a partial answer for now.
Working backwards, it's clear that you can get from any number that ends in a $0$ or a $4$ to a smaller number in a single step, simply by deleting the final digit (i.e., divide by $10$, in the first case, or subtract $4$ then divide by $10$, in the second case). Similarly, if $N$ ends in a $2$, $5$, or $7$, you get to a smaller number in two steps: first multiply by $2$ and then remove the resulting $0$ or $4$ in the ones digit. Likewise, if $N$ ends in a $1$ or a $6$, three steps will suffice to arrive at a smaller number, while if it ends in a $3$ or $8$, four steps produce a smaller number -- that is, three doublings followed by a reduction by at least a factor of $10$.
The sticking point is if $N$ ends in a $9$. In that case it takes four doublings before you can remove a $4$, but at that point you're left with a number that's larger than what you started with. For example, let's look at $N=49$:
$$49\to98\to196\to392\to784\to78$$
Even continuing for another round leaves us something bigger than $49$:
$$78\to156\to312\to624\to62$$
It's only by going one more round that we wind up with something smaller than $49$:
$$62\to124\to12$$
It feels like it shouldn't ever take more than three rounds to get from $N$ to a number smaller than $N$ (at which point one can say "strong induction" and call it a day), but it'll take a closer look than I've given it here to be sure. The problem has enough of the flavor of the classic $3x+1$ problem for me to be sure there isn't some lingering loophole. On the other hand, maybe there's some elementary argument that I'm overlooking. If I have a chance I'll give the problem some more thought, but I'd be happy if someone else posted a complete answer.
Update (April 7, 2016): I was wrong about it never needing more than three rounds to get to a smaller number (where each "round" ends in the removal of trailing $4$'s and $0$'s). Here's an example that takes four rounds:
$$749\to1498\to2996\to5992\to11984\to1198\\
1198\to2396\to4792\to9584\to958\\
958\to1916\to3832\to7664\to766\\
766\to1532\to3064\to306$$
|
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Proving that a countable set in an arbitrary set $ X $ has outer measure zero . It is well known that in $\mathbb{R} $ (well, also in $\mathbb{R}^n$) a countable set $A $ has outer measure zero. It is not really hard to prove and it is a common exercise among the books which I've already done.
However, I'm somewhat stuck at trying to prove it for an arbitrary set because I can't figure out how to let the $\varepsilon$-thing play into the game.
Is it still true that for every countable set $A\subseteq X$ one has that it's outer measure is zero ? And if so, which conditions one must have over $X$ ?
Thanks!
|
It depends on the (outer) measure and set $X$ you look at. For example let $X = \mathbb{Z}$ and define an outer measure that is not zero on each subset of $\mathbb{Z} $. Or on $\mathbb{R}$ define $\nu(\emptyset) := 0, \nu(A) := 1 \, (\emptyset \neq A \in \mathcal{P}(X))$.
|
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Let $f : E \rightarrow \mathbb{R}$. Show that if $|f|$ is measurable on $E$ and the set $\{f > 0\}$ is measurable, then $f$ is measurable on $E$. I'm learning about Measure Theory (specifically measurable functions) and need help with the following problem:
Let $f : E \rightarrow \mathbb{R}$. Show that if $|f|$ is measurable on $E$ and the set $\{f > 0\}$ is measurable, then $f$ is measurable on $E$.
What I don't understand is that the implication $$|f| \ \text{measurable} \implies f \ \text{measurable} \ \ (1)$$ is clearly false.
For example, let $A$ be a non-measurable subset of $\mathbb{R}$, and let $f$ be the function $$f(x) = \begin{cases} 1\text{ if }x\in A,\\ -1\text{ if }x\notin A.\end{cases}$$
Then $f$ is not measurable ($f^{-1}[\{1\}] = A$) but $|f|$ is measurable (it's the constant function $1$).
How does the additional condition that $\{f > 0\}$ is measurable makes the implication $(1)$ true and how do I prove it?
|
You have that
if $b \geq 0, f^{-1}(]b,a[) = (|f|)^{-1}(]b,a[) \cap \{f > 0 \}$, so it's measurable
if $b < 0, f^{-1}(]b,0]) = (|f|)^{-1}([0,-b[ ) \cap \{f > 0 \}^c$, so it's also measurable
This imply that if $a>0, b<0$, $f^{-1}(]b,a[) = f^{-1}(]0,a[) \cup f^{-1}(]b,0])$ is measurable
if $a < 0, b<a, f^{-1}(]b,a[) = (|f|)^{-1}(]-a,-b[) \cap \{f > 0 \}^c$, so it's also measurable
hence $f^{-1}(]b,a[)$ is measurable for all $a>b$
|
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Prove that every set formed by polynomials of different degrees is linearly independent How to prove that every set formed by polynomials of different degrees is linearly independent.
My problem is in how to make a general set of polynomials of different degrees.
|
I would order them in increasing degree. Then, if $\langle S \rangle$ denotes the subspace generated by a subset $S$, I would use the following fact:
FACT: If $u_1, \ldots , u_r$ are linearly independent and $u_{r+1}$ does not belong to $\langle u_1, \ldots, u_r\rangle$, then $u_1, \ldots , u_{r+1}$ are linearly independent.
This follows from considering a linear combination which yields $0$, i.e. $\sum_{i=1}^{r+1} \lambda_i u_i=0,$ and studying the cases $\lambda_{r+1}\neq 0$ and $\lambda_{r+1}=0$ separately.
Applying the above to our collection of polynomials with increasing degrees, the result follows.
|
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|
Roll two dice. What is the probability that one die shows exactly two more than the other die?
Two fair six-sided dice are rolled. What is the probability that one die shows exactly two more than the other die (for example, rolling a $1$ and $3$, or rolling a $6$ and a $4$)?
I know how to calculate the probabilities of each event by itself, but I do not know how to proceed with this problem.
|
The probability of rolling a 1 and 3 is 1/18. Same for the probability of 2&4, 3&5, and 4&6.
So the overall probability of the dice being two apart equals 4/18 = 2/9.
|
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Balls and boxes pigeonhole problem One has $60$ boxes and many (one colored) balls with $8$ different colors. In every box one puts $3$ balls with different colors. Must there exist (at least) two boxes with the same three colored balls?
If one puts $3$ balls with different colors each in a box, we must find first how many combinations we can have from $8$ different colors choosing 3 each time. That is $(\frac{8}{3}) = 56$ different combinations of colors, which means that there are exactly $4$ boxes with the same color (according to the pigeonhole principle).
I just want to check if my way of thinking is right?
|
Your explanation is partially correct.
There are $\binom{8}{3} = 56$ ways of selecting three of the eight colors. Since there are $60$ boxes, this means that there must be at least two boxes containing the same combination of three colors.
However, we cannot conclude that there are four boxes containing the same combination of colors, as you seem to be stating. Crostul has provided you with an example in the comments. One extreme case is all $60$ boxes contain the same three colors. The other extreme is that all $56$ color combinations are used, with four of them being used twice. Is that what you meant?
To guarantee that there are at least three boxes containing balls with the same combination of three colors, we would require that there be at least $2 \cdot 56 + 1 = 112 + 1 = 113$ boxes. If we had only $112$ boxes, we could use every color combination exactly twice.
The strong form of the pigeonhole principle states that if $k$ objects are to be placed in $n$ boxes, then the number of boxes that must contain the same number of objects is
$$\left\lceil \frac{k}{n} \right\rceil$$
where $\lceil x \rceil$ is the least integer greater than or equal to $x$. In our case, $k = 60$ and $n = \binom{8}{3} = 56$, so we are guaranteed at most
$$\left\lceil \frac{60}{56} \right\rceil = 2$$
boxes that contain the same combination of three colors.
|
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Solve for the values of $x$ in $|x+k|=|x|+k$ where $k$ is a positive real number The question asks me for which values of the real number $x$ is $|x+k|=|x|+k$ where $k$ is a positive real number.
How do I go about this? Can I square both sides to get rid of the absolute value signs? When I do it this way, I get a single $x$ sandwiched in the middle of a quadratic equation which I don't know what to do with, after I square and expand $|x|+k$. Is there another way of solving this problem?
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If $x\ge0$ and $x+k\ge0$, then the identity holds always. You get $x\ge0$ and $k\ge-x$.
If $x\ge0$ and $x+k<0$, then $-x-k=x+k$, and $x+k=0$ (impossible).
If $x<0$ and $x+k<0$, then $-x-k=-x+k$, and $k=0$ (impossible).
If $x<0$ and $x+k\ge0$, then $x+k=-x+k$, and $x=0$ (impossible).
|
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Show that $17$ divides $p-1$ I'm given that $s=2^{17}-1$ and that $p$ is a prime factor of $s$.
First I'm asked to show that $2^{17}\equiv 1(\mod p).$
For this I have simply said that since $p$ divides $s$, this means that $2^{17}-1=pk$ and so $2^{17}\equiv 1(\mod p).$
Next I'm asked to show that $17$ divides $p-1.$
I'm think I'm expected to be using Fermat's Little Theorem here, I'm not getting anywhere with it at the moment though. Is this the correct approach and how would one do it?
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$2^{17}\equiv 1\mod p$ says the order of $2$ modulo $p$ is a divisor of $17$. However $17$ is prime, and the order of $2$ is not $1$. Hence this order is $17$. From Little Fermat there results $17$ divides $p-1$.
|
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Does the Series $\sum_{n=1}^{\infty} (1-\cos\frac{\pi}{n})$ Converge?
Does the serie $\displaystyle\sum_{n=1}^{\infty}\Bigl(1-\cos\frac{\pi}{n}\Bigr)$ converge?
Limit test of $1-\cos\frac{\pi}{n}$
$$
\lim_{x\to\infty} 1-\cos\frac{\pi}{n} =
1 -\lim_{n\to\infty}\cos\frac{\pi}{n}= 1-1 = 0
$$
I've checked the necessary condition that $\lim_{n\to\infty} 1-\cos\frac{\pi}{n} = 0 $
But how do I check if $\sum_{n=1}^{\infty}(1-\cos\frac{\pi}{n})$ converges?
|
You can use L'hopital's rule (twice) to show that
$$\lim_{x \rightarrow 0} {1 - \cos x \over x^2} = {1 \over 2}$$
So you can use the limit comparison test with the series ${\displaystyle \sum_{n = 1}^{\infty} {1 \over n^2}}$ to show your series converges.
|
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How to determine if a sphere is locally isometric to a plane?
Is a sphere of radius one locally isometric to a plane? (Briefly justify your answer.)
How do I go about answering this question. Could someone provide an explanation to what exactly locally isometric means? Also is the radius value of 1 an irrelevant piece of information so basically what they are asking is whether sphere , regardless of the radius locally isometric to a plane?
Any help would be much appreciated.
|
Hint:
1) We say that a smooth map $F : S_1 →\rightarrow S_2$ between the two surfaces $S_1$,$S_2$ is a local isometry if it preserves distances between two points close to each other.
2) From the Teorema egregium of Gauss we know that the gaussian curvature of a surface is invariant under isometries. And the gaussian curvature of a sphere of radius $R$ is $1/R^2$ but the gaussian curvature of a plane is $0$.
|
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Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.
|
This seems like an exercise in Lagrange Multipliers. You need to minimize the distance function $f(x,y,z) = (x-1)^2 + (y-4)^2 + z^2$ (which is the square distance from $(x,y,z)$ to $(1,4,0)$) subject to the constraint that $g(x,y,z) = x^2 + y^2 - z^2 = 0$. Any point which minimizes such $f$ subject to the constraint will satisfy $$\nabla f = \lambda \nabla g$$ for some $\lambda \in \mathbb R$. This gives us 4 equations: \begin{align*} 2(x-1) &= 2\lambda x \\ 2(y-4) &= 2\lambda y \\ 2z&= -2\lambda z \\ x^2 + y^2 - z^2 &= 0.\end{align*} The third equation gives $z = 0$ or $\lambda = -1$. If $z=0$, then the fourth equation gives $x=0, y= 0$. But then the first two equations couldn't be satisfied, so this is impossible. We conclude that $\lambda = -1$. In this case $x = 1/2$, $y=2$ so by the fourth equation $z = \pm\sqrt{17/4}$. By the symmetry in $z$, both sides of the $\pm$ work, so the points which minimize the distance are $$\left(\tfrac 1 2, 2, \sqrt{17/4}\right) \,\,\,\, \text{ and } \,\,\,\, \left(\tfrac 1 2, 2, -\sqrt{17/4}\right).$$ The distance is then $$d_{\text{min}} = \sqrt{\left(\tfrac 1 2 - 1 \right)^2 + (2-4)^2 + \left( \pm\sqrt{17/4}\right)^2}= \sqrt{\tfrac 1 4 + 4 + \tfrac{17}4} = \sqrt{\tfrac{34}{4}} = \tfrac{\sqrt{34}}{2}.$$
|
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Geometric intuition of graph Laplacian matrices I am reading about Laplacian matrices for the first time and struggling to gain intuition as to why they are so useful. Could anyone provide insight as to the geometric significance of the Laplacian of a graph? For example, why are the eigenvectors of a Laplacian matrix helpful in interpreting the corresponding graph?
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Perhaps the most natural definition is the (oriented) incidence transformation.
The incidence transformation of an edge $(u,v)$ is defined by $E(u,v) = v - u$.
Its adjoint $E^*$ has a dual definition, taking vertices to the sum of incoming edges minus the sum of outgoing ones.
The Laplacian is just $L=EE^*$. Thus, for instance, its eigenvalues and eigenvectors are related to the singular values and vectors of the incidence transformation.
|
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Find all positive integers n such that $2^2 + 2^5+ 2^n$ is a perfect square. Find all positive integers n such that $2^2 + 2^5 + 2^n$ is a perfect square. Explain your answer.
|
We have:
$$2^n+36=m^2$$
$$2^n=(m-6)(m+6)$$
Both $m-6$ and $m+6$ must be powers of $2$ (and note that they differ by $12$). Can you continue?
|
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What's "$\mathbb Z = (1)$ is cyclic"? As I understand it, $(1)$ is a cyclic group with $1$ being its generator. So, $1^n$ with $n \in \mathbb N$ generates $\mathbb Z^+$ , but what about $\mathbb Z^{-1}$ ? Do we say $(1)$ is a cyclic group with two operations $(+, -)$ defined on it so that $\mathbb Z = (1)$?
|
In algebra, when you say that a set $S$ generates some structure $A$, you mean that any element of $A$ can be represented as repeated applications of operations of the structure in question to elements of $S$.
In your case, the structure is a group. A group has not only the binary operation, but also the inverse (and, technically, the identity element, considered as a nullary operation). So, $\{1\}$ indeed generates $\mathbb{Z}$, since the inverse of it (which is $-1$) and powers of inverse are also allowed.
|
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Proving a trigonometric equation Knowing that : $$ \sin t - \cos t = x $$
Prove that : $$ \cos^3t - \sin^3t = \frac{x}{2}(3-x^2) $$
I tried to solve it by the important corresponding $$ a^3 - b^3 = (a-b)(a²+ab+b²) $$
But I got stuck in the middle and I don't even know if it's correct what I did
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Note that $$a^2=(\sin(t)-\cos(t))^2=\sin^2(t)+\cos^2(t)-2\sin(t)\cos(t)=1-2\sin(t)\cos(t).$$ Hence $$\cos^3(t)-\sin^3(t) = a(\cos^2(t)+\sin^2(t)+\cos(t)\sin(t))=a(1+\frac{1-a^2}{2})=\frac{a}{2}(3-a^2).$$
|
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Complex analysis: Calculating $\int_{-\infty}^{\infty} \frac{\sin x}{x} dx$ by using $f(z) = \frac{e^{iz} - 1}{iz}$ So I integrate the holomorphic function $\frac{e^{iz} -1}{iz}$ over the half disk in the upper half plane, let me name it $\Gamma$. By using Cauchy's theorem, it equals 0.
$$0 = \int_{\Gamma} \frac{e^{iz} -1}{iz} = \int_{0}^{\pi}(e^{iRe^{i\theta}}-1)d\theta + \int_{-R}^{R} \frac{e^{it}-1}{it}dt$$
($Re^{[0, \pi]}$ and $t: [-R, R]$ being the respective parametrizations.)
By splitting the left integral in two, I get $\int_{0}^{\pi} e^{iRe^{i\theta}}d\theta \rightarrow 0$ by dominated convergence, and $\int_{0}^{\pi} -1d\theta = -\pi$. By making $R \rightarrow \infty$, I'm left with
$$\pi = \int_{-\infty}^{\infty} \frac{e^{it} - 1}{it}dt = \int_{-\infty}^{\infty}\frac{\sin t}{t}dt + \int_{-\infty}^{\infty}\frac{\cos t -1}{it}dt.$$
How do I get rid of the integral on the right? I welcome all feedback.
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$\frac{\cos(t)-1}{it}$ is an odd function so its integral over any $[-R,R]$ is zero.
|
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Prove that $\mathbb Q[\sqrt[3]2]$ is a field We define the set:
$$\mathbb{Q}[\sqrt[3]2]=\{a_{0}+a_{1}\sqrt[3]{2}+a_{2}\sqrt[3]{2^{2}}:a_{0}, a_1,a_2\in\mathbb{Q}\}$$
It's easy to prove all the properties of fields, except for the unit elements.
So, how can we prove that
$$\forall x\in\mathbb{Q^{*}}[\sqrt[3]2],\exists x^{-1} \in\mathbb{Q}[\sqrt[3]2]:xx^{-1}=1$$
And how can we prove this in general for the set
$$\mathbb{Q}[\sqrt[n]2]=\{a_{0}+a_{1}\sqrt[n]{2}+a_{2}\sqrt[n]{2^{2}}+...+a_{n-1}\sqrt[n]{2^{n-1}}:a_{0}, a_1,a_2,...,a_{n-1}\in\mathbb{Q}\}$$
|
Doing this for $\sqrt[3]{2}$ is a waste of time. ;-) Exactly because doing it for $\sqrt[n]{2}$ would lead to gigantic computations.
Suppose $r\in\mathbb{C}$ is algebraic over $\mathbb{Q}$. We want to see that the set $\mathbb{Q}[r]$ consisting of all the expressions of the form $a_0+a_1r+\dots+a_nr^n$ is a field.
Let $r$ be an algebraic element over the field $F$. Then $F[r]$ is precisely the image of the ring homomorphism $\varphi\colon F[X]\to F[r]$ which is the identity on $F$ and $\varphi(X)=r$.
By the homomorphism theorem,
$$
F[r]\cong F[X]/\ker\varphi
$$
Now, if $f(X)$ is the minimal polynomial for $r$ over $F$, we can easily see that $\ker\varphi=(f(X))$, the principal ideal generated by $f(X)$.
As $f(X)$ is irreducible, $(f(X))$ is a maximal ideal, so $F[X]/(f(X))$ is a field.
For the particular case, consider the map $\mathbb{Q}[\sqrt[3]{2}]\to\mathbb{Q}[\sqrt[3]{2}]$ given by $t\mapsto t(a+b\sqrt[3]{2}+c\sqrt[3]{4}\,)$. It is a $\mathbb{Q}$-linear map and, with respect to the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, its matrix is
$$
\begin{bmatrix}
a & 2c & 2b \\
b & a & 2c \\
c & b & a
\end{bmatrix}
$$
whose inverse (one of the few cases where using the adjugate is simpler than other methods) is
$$
\frac{1}{a^3+2b^3+4c^3-6abc}
\begin{bmatrix}
a^2 - 2bc & -2ac + 2b^2 & -2ab + 4c^2 \\
-ab + 2c^2 & a^2 - 2bc & -2ac + 2b^2 \\
-ac + b^2 & -ab + 2c^2 & a^2 - 2bc
\end{bmatrix}
$$
which, by the way, proves that $a^3+2b^3+4c^3-6abc\ne0$ as soon as one among $a$, $b$ and $c$ is nonzero.
The inverse of $a+b\sqrt[3]{2}+c\sqrt[3]{4}\ne0$ is thus
$$
\frac{(a^2-bc)+(-ab+2c^2)\sqrt[3]{2}+(-ac+b^2)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}
$$
|
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Dimension of Basis of Subspace I was just wondering if there is any "rule" for what the dimension of a basis of a given subspace will be.
For example, a problem I just did involved a vector $v = (1, 2, 3, 4)$ in $\mathbb{R}^4$, and I had to find a basis for the subspace in $\mathbb{R}^4$ consisting of all vectors perpendicular to $v$.
My intuition for this was to note that the subspace of vectors perpendicular to v is the plane with v as its normal vector. Thus, any two vectors in the plane which are linearly independent would be a basis, and the dimension of the basis would be two.
However, the answer the book gave had a dimension of three. They solved it in a way that makes sense to me as well but I'm still confused as to where I went wrong with my logic. I'm sensing the answer lies in the fact that R4 planes don't behave the same way as R3 planes.
Another problem involved finding the basis for the orthogonal complement of a subspace in R4 spanned by two vectors: $W = Span\{(1, 2, 3, 4),(5, 6, 7, 8)\}$, and the answer had a dimension of two (not three).
I guess I'm just confused on where the dimension is coming from. Any help would be great.
Thanks!
|
My intuition for this was to note that the subspace of vectors perpendicular to v is the plane with v as its normal vector. Thus, any two vectors in the plane which are linearly independent would be a basis, and the dimension of the basis would be two.
Unfortunately planes are not always two-dimensional. The correct intuition is that the orthogonal complement of a subspace $U$ has dimension that "complements" the dimension of $U$ (assuming everything is finite-dimensional). So in a 5-dimensional space, the orthogonal complement of a one-dimensional subspace is 4-dimensional, the orthogonal complement of a two-dimensional subspace is 3-dimensional, and so forth.
Until you build a strong geometric intuition for vectors in dimensions higher than 3, I would suggest going with the direct approach of forming a basis and counting the number of elements.
|
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|
What are the limit points of $\sin(2n \pi/3)$? I want to find the limit points of $a_n = \sin(2n \pi/3)$. So far I've identified the subsequences $a_{3n} = 0$ and $a_{1} = \frac{\sqrt{3}}{2}$ and $a_2 = -\frac{\sqrt{3}}{2}$ so are the limit points $\left\{0, \pm \frac{\sqrt{3}}{2}\right\}$? How do I show there aren't more?
|
A limit point is a point (number) for which there is a subsequence that converges to that point. Now any subsequence must have either infinitely many $0$'s or infinitely many $\dfrac {\sqrt 3} 2$'s or infinitely many $-\dfrac {\sqrt 3} 2$'s, so the limit can ONLY be one of these three numbers.
|
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|
Find $x$ in the equation $x^x = n$ for a given $n$ Simply: How do I solve this equation for a given $n \in \mathbb Z$?
$x^x = n$
I mean, of course $2^2=4$ and $3^3=27$ and so on. But I don't understand how to calculate the reverse of this, to get from a given $n$ to $x$.
|
Simpler:
If
$x^x = n$,
then $x\ln(x) = \ln(n)
=y$.
Let
$f(x) = x\ln(x)-y
$.
$f'(x)
=\ln(x)+1
$.
Applying Newton's iteration,
starting with $x = \frac{y}{\ln y}$,
$x_{new}
=x-\frac{f(x)}{f'(x)}
=x-\frac{x\ln(x)-y}{\ln(x)+1}
=\frac{x\ln(x)+x-x\ln(x)+y}{\ln(x)+1}
=\frac{x+y}{\ln(x)+1}
$.
Iterate until cooked.
|
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|
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
Find the remainder of $9^2\cdot 13\cdot 21^2$ when divided by $4$
How should I approach this type of questions?
Without calculator of course
I did this:
$9^2\cdot 13\cdot 21^2=81\cdot 13\cdot 441=81\cdot 5733=464,373=33\bmod 4=1 \bmod 4$
|
Hint: What is $9\bmod 4$? What is $13\bmod 4$?
How would you proceed from here?
Also, you wrote that $464\,373=1\bmod 4$. Always keep in mind that the "$=$" sign means that two things are exactly the same, in every way. In stead one should write
*
*$\overline{464\,373}=1\bmod 4$,
*$464\,373\bmod 4=1\bmod 4$,
*$\overline{464\,373}=\overline 1$ (when it is understood that we are working in $\mathbb Z/4\mathbb Z$)
or
*
*$464\,373\equiv 1\bmod 4$.
|
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|
Expected value of two successive heads or tails ( I do not understand the answer) Problem: This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis Introduction to Probability 2nd:
We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What is the expected value of the number of tosses?
Solution:
,
I don't understand the answer.
Why $$E[X\mid H1,T2] = 1+E[X\mid T1]$$ Is there any relationship?
|
The equation $$E[X\mid H_1,T_2]=1+E[X\mid T_1]$$ reads as follows:
*
*LHS: The expected number of trials after you have tossed a head in the first toss and tails in the second is equal to
*RHS: 1 plus the expected number of tosses given that in the last toss you tossed tails.
The $1+$ in the RHS stands for the first toss that missed to bring a result (the first was Heads and the second Tails, so the first toss in know irrelevant, it is a failure. But keep the Tails in the second toss, you might toss again Tails in the third so this is not yet a failure).
So, to make it clearer: In the first toss you tossed a Head and in the second Tails. Where are you standing now, just before the third toss? You have tossed Tails and you are starting over. So, the Heads in the first toss are irrelevant now and the only thing that matters is the Tails that you tossed in the second toss. Name this toss $1$ and start over. Do you see it now?
|
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|
A simple binomial identity Is there a simple way of showing that a prime $p$ must divide the binomial coefficient $p^n\choose{k}$ for all $n\geq 1$ and $1\leq k\leq p^n-1$?
|
Just a quick remark after the fact: If you accept that $$ (a +b )^{p} \equiv a^p +b^p\pmod p ,$$ for $a$ and $b$ indeterminants,
then
$$(a+b)^{p^n} = \left(\ (a + b )^p\ \right)^{p^{n-1}}\equiv \left(\ a^p + b^p\ \right)^{p^{n-1}}\equiv a^{p^n}+ b^{p^n}\pmod p,$$
which also gives the result.
|
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|
Show there in no non constant analytic functions in disk unit s.t $f(z)=f(2z)$ I'm trying to solve the following
Show that there is no non constant analytic function in the unit disc such that $f(z)=f(2z)$.
My try: let $f$ be an analytic function in the unit disk such that $f(z)=f(2z)$.
Now, we can write $\displaystyle f\left(\frac{z}{2}\right)=f(z)$. The function is analytic, hence we can write $f(z)=\sum_{n=0}^{\infty}a_n z^n$. We have $\displaystyle \left|\frac{z}{2}\right|\le \left|z\right|<R$ where $R$ is the radius of convergence, thus $\displaystyle f\left(\frac{z}{2}\right)=\sum_{n=0}^{\infty}\frac{a_n}{2^n}z^n$.
Equating coefficients we get $\displaystyle \forall n: \frac{a_n}{2^n}=a_n$, thus we have $\forall n>0:a_n=0$, i.e $f(z)=a_0$, i.e $f$ is constant.
Is my reasoning correct? Why should we notice the unit disk?
Thank you!
|
[1]. Another way to show that $a_n=0$ for $n\geq 1$ is that for $0<r<1/2$ and $n\geq 1$ we have $$2 \pi i a_n=(2 \pi i/n!)(d^nf/dz^n)(0)=\int_{|z|=r}f(z)z^{-n-1} \;dz=$$ $$=\int_{|z|=r}f(2 z)z^{-n-1}\;dz=\int_{|z|=r}f(2 z)(2 z)^{-n-1}2^n\;d(2 z)=$$ $$=2^n\int_{|y|=2 r} f(y)y^{-n-1}\;dy= 2^n(2 \pi i/n!)(d^nf/dz^n)(0) =2^n(2 \pi i a_n).$$
[2]. Different proof. (i) Prove that if $g:[0,1)\to \mathcal C$ is continuous and $g(x)=g(2 x)$ for $x\in [0,1)$ then $g$ is constant. (ii) For $r\in [0,1)$ and $t\in \mathcal R,$ we have $$f'(r e^{i t})=\lim_{s\to r^+} (f(s e^{i t})-f(r e^{i t})) /((s-r)e^{i t}).$$ Observe that $g_t(x)=f(x e^{i t})$, satisfies $g_t(x)=g_t(2 x)$ for $x\in [0,1).$ So $f'=0$.
For a proof of (i), if $x\in (0,1)$ then $g(0)=\lim_{n\to \infty}g(x 2^{-n})=\lim_{n\to \infty}g(x)=g(x).$
|
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|
What's the condition for a matrix $A$ ($2N\times 2N$ dimension) to have eigenvalues in pairs $\pm\lambda$?
For a given even dimension square complex matrix $A$ ($2N\times 2N$ dimension),
what's the sufficient and necessary condition for the matrix $A$ such that:
if $\lambda_{1}$ is an eigenvalue, then $\lambda_{2}=-\lambda_{1}$ is also an eigenvalue?
I have found one sufficient condition: skew-symmetric matrix, but
obviously there is a lot of matrices having eigenvalues in pairs are
not skew-symmetric.
So I wonder is there a sufficient and necessary condition?
Or is there a name for this family of matrices (skew-symmetric matrix
is a member)?
|
Take any Hermitian matrix of the form $\mathfrak{B}=\left[\begin{array}{c|c}\mathbf{0}_n & A\\\hline A^* & \mathbf{0}_m\end{array}\right],$ where $A$ is $n\times m$. Then if $\lambda$ is an eigenvalue of $\mathfrak{B}$ with multiplicity $k$, then $-\lambda$ is also an eigenvalue with the same multiplicity. This is another sufficient condition.
|
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|
Is there a way to study "how good" is the Newton's Raphson method applied to a function? My question is as simple as that.
When we're applying the fixed point algorithm, we can see if it's going to converge or diverge finding the derivative and checking if the absolute value of that function is less or greater than $1$ on a given interval, and we can estimate if at any given point if the convergence will be fast or slow.
Is there a way to study the "contractivity" of the Newton's Raphson method?
Thanks for your time.
|
As a fixed point iteration
$$
N(x)=x-\frac{f(x)}{f'(x)}
$$
has derivative
$$
N'(x)=\frac{f(x)f''(x)}{f'(x)^2}
$$
and the usual estimates apply. Since at the solution $f(x_*)=0$, you can get intervals around $x_*$ with arbitrarily small contraction constants, compatible with the quadratic convergence speed.
|
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|
How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$? How to prove the convergence of $\int_0^\infty \frac{\ln{x}}{1+x^2}dx$.Since it's unbounded on both sides, we need to prove the convergence of both $\int_0^1 \frac{\ln{x}}{1+x^2}dx$ and $\int_1^\infty \frac{\ln{x}}{1+x^2}dx$.
$\frac{\ln{x}}{1+x^2}\le \frac{x}{1+x^2}$ on $[1,\infty)$ and both of them are nonnegative, and $\int_1^\infty \frac{x}{1+x^2}dx=\frac12[\ln{|1+x^2|}]^\infty_1=\infty$. Apparently basic comparison does not work in this case.
For $\int_0^1 \frac{\ln{x}}{1+x^2}dx$, the function$\frac{\ln{x}}{1+x^2}$ is not even nonnegative, we can't apply any comparision test. How are we supposed to know its convergence?
|
You say that the function is not even non-negative. Have you studied its sign properly on $[0,1]$? What is the matter about its sign?
The function is actually negative on $(0,1]$, so minus the function is positive, and now all comparison tests work just fine.
Since $1+x^2$ lies between $1$ and $2$, we have for $\epsilon\gt 0$:
$$0\leq \int_{\epsilon}^1 \frac{-\log x}{1+x^2} dx\leq \int_{\epsilon}^1 (-\log x) dx,$$
so the function $|f(x)|$ (where $f$ is your integrand) is integrable, and hence so is $f$.
|
{
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|
Geometric reasons finite fields have prime power orders? All variations of proofs that finite fields have prime power orders have a very algebraic feel to them. I was wondering - is there a more geometric way to see why this is true?
|
A finite field is a vector space over some $\mathbb{F}_p$, so it has a basis and its elements are in bijection with some $n$-uples of coordinates in $\mathbb{F}_p$, which tells you it has cardinal $p^n$. I don't know if you consider that a geometric or algebraic argument though.
|
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|
Showing that every polynomial over the Algebraic Numbers has a $0$ in the Algebraic Numbers. Let $\mathbb{A}$ denote the field of Algebraic Numbers: the field of all complex numbers that are algebraic over $\mathbb{Q}$. Assuming that every polynomial over $\mathbb{C}$ has a $0$ in $\mathbb{C}$ how would you go about proving that every polynomial over $\mathbb{A}$ has a $0$ in $\mathbb{A}$? I've tried contradiction but all that has lead me to is concluding that the $0$ of some $p(x)\in \mathbb{A}[x]$, say $\alpha$, is transcendental over $\mathbb{Q}$, which doesn't help much. I'd appreciate a hint as where to start the proof.
|
Let $P\in\mathbb A[X]$. Then there is $a\in\mathbb C$ such that $P(a)=0$. This shows that $a$ is algebraic over $\mathbb A$, so over $\mathbb Q$ (why?). Thus we get $a\in\mathbb A$.
|
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|
Gradient of Function I am trying to find the $\nabla F$ with respect to $x_i$ where $F$ is as follows:
$$F(x_0,...,x_n) = c_1\sum_{i=0}^{n}\sum_{j=1}^{k}\frac{1}{||x_i - r_j||_2^2} + c_2\sum_{i=0}^{n-1}||x_{i+1}-x_i||_2^2$$
For clarity I will show my attempts at derivation for each seperate part of the sum.
First part of the sum:
$$c_1\sum_{i=0}^{n}\sum_{j=1}^{k}\frac{1}{||x_i - r_j||_2^2}$$
Since it is with respect to $x_i$ there is only a single contribution from the outer summation (fix the value of i):
$$c_1\sum_{j=1}^k\frac{-2}{(x_i - r_j)^3}$$
As for the second part of the sum:
$$c_2\sum_{i=0}^{n-1}||x_{i+1}-x_i||_2^2$$
If you were to expand the summation there would be 2 terms that contain $x_i$, namely $c_2(||x_{i+1}-x_{i}||_2^2 + ||x_{i}-x_{i-1}||_2^2)$. Because of this all other values are considered constants and don't add anything to the final sum.
Thus the partial derivative with respect to $x_i$ would be:
$$c_2(4x_i - 2x_{i+1} - 2x_{i-1})$$
All together I end up with
$$\frac{\partial F}{\partial x_i} = c_1(\sum_{j=1}^k\frac{-2}{(x_i - r_j)^3}) + c_2(4x_i - 2x_{i+1} - 2x_{i-1})$$
I am not very confident about my derivation of the 2-norm squared $|| \cdot ||_2^2$, especially when it is the denominator. Does this derivation look correct? If not can somebody please show me where my errors stem from?
|
The partial derivative $\partial_\alpha = \partial/\partial x_\alpha$ of $F$ is:
\begin{align}
\partial_\alpha F(x_0,\dotsc,x_n)
&=
c_1\sum_{i=0}^{n}\sum_{j=1}^{k}\partial_\alpha \frac{1}{\lVert x_i - r_j\rVert_2^2} +
c_2\sum_{i=0}^{n-1} \partial_\alpha \lVert x_{i+1}-x_i\rVert_2^2 \\
&=
c_1\sum_{i=0}^{n}\sum_{j=1}^{k}(-1) \frac{1}{\lVert x_i - r_j\rVert_2^4} \partial_\alpha \lVert x_i - r_j\rVert_2^2 +
c_2\sum_{i=0}^{n-1} \partial_\alpha \lVert x_{i+1}-x_i\rVert_2^2 \\
&=
c_1\sum_{i=0}^{n}\sum_{j=1}^{k}(-1) \frac{1}{\lVert x_i - r_j\rVert_2^4}
2(x_\alpha - r_j) +
c_2\sum_{i=0}^{n-1} 2 ((x_\alpha - x_{\alpha - 1}) - (x_{\alpha+1} - x_\alpha)) \\
&=
\begin{cases}
-2\left(
c_1\sum_{i=0}^{n}\sum_{j=1}^{k} \frac{x_\alpha - r_j}{\lVert x_i - r_j\rVert_2^4} +
c_2\sum_{i=0}^{n-1} (x_{\alpha+1} - x_\alpha) \right)
&; \alpha = 0 \\
-2\left(
c_1\sum_{i=0}^{n}\sum_{j=1}^{k} \frac{x_\alpha - r_j}{\lVert x_i - r_j\rVert_2^4} +
c_2\sum_{i=0}^{n-1} (x_{\alpha+1} - 2x_\alpha + x_{\alpha-1}) \right)
&; 0 < \alpha < n \\
-2\left(
c_1\sum_{i=0}^{n}\sum_{j=1}^{k} \frac{x_\alpha - r_j}{\lVert x_i - r_j\rVert_2^4} +
c_2\sum_{i=0}^{n-1} (-x_\alpha + x_{\alpha-1}) \right)
&; \alpha = n \\
\end{cases}
\end{align}
so we get
$$
\DeclareMathOperator{grad}{grad}
\grad F(x_0,\dotsc,x_n) =
-2\left(
c_1\sum_{i=0}^{n}\sum_{j=1}^{k} \frac{x - r_j}{\lVert x_i - r_j\rVert_2^4} +
c_2\sum_{i=0}^{n-1} \sum_{\alpha=0}^n (
\dotsb ) e_\alpha \right)
$$
|
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|
Formula for finding the nth number in this sequence: $[0, 1, 3, 11, 50, 274...]$? The sequence here is what I have discovered to be this: https://oeis.org/A000254, which references Stirling numbers of the First Kind. The formulas provided in that link are very ambiguous with their notation, and I have no idea what to take from them. Attempting to come up with a formula that will give me the nth number in the sequence, I've had to delve into research of Stirling numbers of the first kind and its definitions, which have proved largely ineffective in helping me derive this formula. I'm sure a formula for this already exists and is on the internet (or in someone's memory), I just haven't found it, and if anyone could provide this formula, or lead me toward deriving this formula, that would be fantastic. Thanks!
|
The link you provide has the following formula:
$$a(n)=n!\cdot\sum_{k=0}^{n-1}\frac{1}{n-k}$$
where $a(n)$ is the $n$th number in the series and $n\ge 1$. Thus,
$a(0)=0$
$a(1)=1!\cdot\frac{1}{1-0}=1$
$a(2)=2!\cdot\left(\frac{1}{2-0}+\frac{1}{2-1}\right)=3$
$a(3)=3!\cdot\left(\frac{1}{3-0}+\frac{1}{3-1}+\frac{1}{3-2}\right)=11$
and so on.
There are a lot of other formulas on the link you provide, for example, $a(n)=a(n−1)\cdot(2n−1)−a(n−2)\cdot(n−1)^2$, if $n>1$.
$a(0)=0$
$a(1)=1$
$a(2)=a(2-1)\cdot(2\cdot 2-1)-a(2-2)\cdot(2-1)^2=a(1)\cdot 3-a(0)\cdot 1=1\cdot 3-0\cdot 1=3$
$a(3)=a(3-1)\cdot(2\cdot 3-1)-a(3-2)\cdot(3-1)^2=a(2)\cdot 5-a(1)\cdot 4=3\cdot 5-1\cdot 4=11$
and so on.
Another formula is $a(n+1)=(n+1)*a(n)+n!$, where $n\ge 0$
$a(0)=0$
$a(1)=1\cdot a(0)+0!=1\cdot 0+1=1$
$a(2)=2\cdot a(1)+1!=2\cdot 1+1=3$
$a(3)=3\cdot a(2)+2!=3\cdot 3+2\cdot 1=11$
and so on.
|
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Proof by induction that $1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$ I need to know if I am doing this right. I have to prove that
$1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$
So first I did the base case which would be $1$.
$1^2 = (1(2(1)-1)(2(1)+1)) / 3
1 = 3/3
1 = 1$ Which is right.
Then I assumed true for k so
$1^2 + 3^2 + 5^2 + ... + ((2k-1)^2 = k(2k-1)(2k+1)) / 3$
This is where I get lost. I know that the next step is to prove its true for k+1, but I am lost on what to do here. Do I actually simplify down what I assumed true for k? Am I right so far?
|
Let your statment be $A(n)$. You want to show it holds for all $n \in \mathbb{N}$. You use the principle of induction to establish a chain of implications starting at $A(1)$ (you did that one).
What is left to show is
$$
A(n) \Rightarrow A(n+1)
$$
This means you consider $n$ fixed and try to proof $A(n+1)$. For this you can assume that $A(n)$ is true, or in another variant of the principle of induction: $A(1), \dotsc, A(n)$ are true.
You have:
\begin{align}
A(n) &: \sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3} \\
A(n+1) &: \sum_{k=1}^{n+1} (2k-1)^2 = \frac{(n+1)(2(n+1)-1)(2(n+1)+1)}{3} \\
\end{align}
One way is to split
$$
\sum_{k=1}^{n+1} (2k-1)^2 = \left(\sum_{k=1}^{n} (2k-1)^2 \right) + (2(n+1)-1)^2
$$
and to use $A(n)$.
|
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Is there a math function to find an element in a vector? I would like to write mathematically, if possible, the following statement:
Given a vector $x=[1,4,5,3]$ and an integer $j=3$, find the position of $j$ in $x$?
How to write this mathematically?
If I am looking for the position of the minimum value in $x$, I would achieve this by $\arg\min x$.
I guess $j^*=\operatorname{arg\,find} (x=j)$ but $\LaTeX$ does not recognize this.
|
If you consider the vector $x$ as a function from $[1, n]$ to $\mathbb{N}$, you can use the inverse $x^{-1}$. See https://en.wikipedia.org/wiki/Inverse_function#Preimages.
In your example, if $x=(1,4,5,3)$ then $x^{-1} (\lbrace 3 \rbrace) = \lbrace 4 \rbrace$.
As Jonathan Gafar remarked, the inverse set might contain more than one element. Then you can use a minimum to get the first one.
For example, if $x=(1,2,5,7,2)$ then $x^{-1} (\lbrace 2\rbrace)= \lbrace 2, 5 \rbrace$ and $\min x^{-1} (\lbrace 2\rbrace) = 2$.
However, I don't recommend to use this notation without a proper introduction or definition, since the inverse symbol might be ambiguous.
|
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|
The textbook's way of deriving a natural deduction proof of $\vdash((\phi\leftrightarrow\psi)\leftrightarrow\phi)$ feels wrong. The problem is
"Show that if we have a derivation $D$ of $\psi$ with no undischarged assumptions, then we can use it to construct, for any statement $\phi$, a derivation of $((\phi\leftrightarrow\psi)\leftrightarrow\phi)$ with no undischarged assumptions."
Here is my solution.
Below is the textbook's solution.
As you can see on the textbook solution, $\phi$ and $\psi$ cross over to another branch to introduce $\rightarrow$. The textbook didn't mention it was possible. Is the textbook solution a correct derivation of the proof?
Update 1 : I want to solve exercises with what the author explicitly allowed in the textbook. The author wrote:
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There is no error; an allowed way to use $\to$-intro is to discharge a "non-existent" assumption, i.e.:
$\psi \vdash \phi \to \psi$,
as in the two top-right branches.
This corresponds to the fact that $\psi \to (\phi \to \psi)$ is a tautology; thus, assuming $\psi$, by modus ponens $(\phi \to \psi)$ follows, for $\phi$ whatever.
See the textbbok's comment on the rule, page 17:
The rule is still correctly applied if we do not discharge all of them; in fact the rule is correctly applied even if $\phi$ is not an assumption of $D$ at all, so that there is nothing to discharge. Example 2.4.4 (page 19) will illustrate these points.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1719906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
The "obvious" symmetry group $C_3 \times S_4$ related to the hexacode I am studying the large Mathieu groups and more specifically the hexacode from Robert Wilson's book "The Finite Simple Groups". The following paragraph is from page 184:The hexacode
My question is related to what is underlined: How should I interpret this? What does the group "$C_3 \times S_4$ generated by scalar multiplications and those three coordinate permutations" look like?
I figured that the "$C_3$-part" of the group is there because of the (cyclic) multiplicative group {$1, \omega, \overline{\omega}$} of the field $\mathbb{F}_4$, but how come $<(1,2)(3,4); (1,3,5)(2,4,6); (1,3)(2,4)> \cong S_4$? I tried constructing $<(1,2)(3,4); (1,3,5)(2,4,6); (1,3)(2,4)>$ element by element and became only 16 different elements to be in this group. Can anyone help me to get some more insight in how this group looks like and perhaps how this is important in relation to the hexacode and the large Mathieu groups?
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Write $a:=(12)(34)$, $b:=(135)(246)$, and $c:=(13)(24)$. Let $G=\langle a,b,c\rangle$. Note that $a^c=a$ while $(a^b)^c=a^{b^2}$. It follows that $N:=\{1,a,a^b,a^{b^2}\}$ is a normal subgroup of $G$ of order $4$. (Isomorphic to a Klein group.) Moreover, $b^c=b^{-1}$ so $H:=\langle b,c\rangle$ is isomorphic to $S_3$. Moreover, $H\cap N=1$, so $G=N\rtimes H$ and $G$ has order $24$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Proving that if $\sum_0^\infty f_n$ converges uniformly on $D$ to f and $f_n$ is bounded, then f is bounded. "Prove that if each $f_n$ is a bounded function and $\sum_0^\infty f_n$ converges uniformly on $D$ to $f$, then
$f$ is a bounded function"
I don't know how to do this at all. Any help appreciated.
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By Cauchy's Convergence Principle of uniformly convergence, $\forall\varepsilon>0\exists N>0$ s.t.$\forall n,m>N$,$|f_n-f_m|<\varepsilon,\forall x\in D$. In particular, $|f_n-f_{N+1}|<\varepsilon,\forall x\in D$. Thus $|f_n|<sup|f_{N+1}|+\varepsilon$ i.e. $f_n$ is uniformly bounded. Then you can use the definition of uniformly convergence on $f$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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|
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