Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
How to calculate modulo of high power of 2 I know there are other such topics but I really can't figure how to calculate the following equation: 2^731 mod 645.
Obviously I can't use the little theorem of Fermat since 645 is not a prime number and I can't also do the step by step rising of powers(multiplying by 2) since the numbers are still really big. Is there any way to do calculate the result in a normal way (without the enormous numbers) ? Thanks in advance!
|
$645 = 15\cdot 43\,$ so we can compute $\,2^{\large 731}\!$ mod $15$ and $43,\,$ then combine them (by CRT or lcm).
${\rm mod}\ 15\!:\,\ 2^{\large\color{#c00} 4}\equiv 1\,\Rightarrow\, 2^{\large{731}}\equiv 2^{\large 3}\,$ by $\,731\equiv 3\pmod{\!\color{#c00}4}$
${\rm mod}\ 43\!:\,\ 2^{\large 7}\equiv -1\,\Rightarrow\,2^{\large\color{#c00}{14}}\equiv 1$ so $\,2^{\large 731}\equiv 2^{\large 3}\,$ by $\,731\equiv 3\pmod{\!\color{#c00}{14}}$
So $2^{\large 731}\!-2^{\large 3}$ is divisible by $15,43\,$ so also by their lcm = product $= 645,\,$ i.e. $\,2^{\large 731}\!\equiv 2^{\large 3}\!\pmod{\!645}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2116939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$
|
Try working mod $3$ and mod $8$. Assuming $x, y>3$, we have $x,y = \pm 1$ mod $3$. Since $x, y$ are odd we have $x^2, y^2=1$ mod $8$, so
$$x^2, y^2 = 1 \text{ mod } 24.$$
Substituting in the equation gives $$x = 24k+11 $$ for some integer $k$.
Rearranging the original equation we get
$$x(5x-7)=(y-1)(y+1), \tag{1}$$
therefore $x |y-1$ or $x|y+1$, since $x$ is a prime number.
Solving for $x$ gives
$$ x = \frac{17}{10} + \frac{1}{10}\sqrt{20y^2+29}>\frac{1}{3}(y+1).$$
Note that $x$ is odd and $y \pm 1$ is even, so $x \ne y\pm1$. This forces $x = \frac{1}{2} (y \pm1)$, or
$$y = 2x \pm 1 = 48k + 22 \pm 1 \Rightarrow y = 48k+23.$$
$48k+21$ is rejected being divisible by 3.
Plugging these in $(1)$ gives the solution $k=0$ or
$$x = 11, \space y = 23.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
}
|
Squaring a Pareto random variable: Is there a proper name for this transformation? Let $Y= X^2$ be a function of a random variable where $X$ has a Pareto Type II (Lomax) distribution with parameters $\alpha = 8$, $\theta = 4000$, $x > 0$. Find the distribution of $Y$.
I basically went through the ropes as typical for this kind of question. I started with $F_Y(y) = P(Y < y) = P(-\sqrt{y} < x < \sqrt{y})$.
Of course, the Lomax distribution is defined for $x > 0$ and so we take $x > 0$ as the lower bound in the integral rather than -$\sqrt{y}$.
This gave me, using the Fundamental Theorem of Calculus,
$[(8*(4000)^8) / ((y^{1/2} + 4000)^9)] * (1/2) * (y^{-1/2})$,
which is just the density of $x$ evaluated at $y^{1/2}$ and using the chain rule where the derivative of $y^{1/2}$ is of course $(1/2)* (y^{-1/2})$.
I am fairly certain this is correct, but I have been asked to name this distribution, in which I am stuck. I can't recognize the distribution of Y, and I've tried manipulating the expression to make it look like something I recognize, with limited success. Are there any slick ways of doing this? Or does this transformation even produce a distribution that is "well known"?
EDIT: For anyone who cares, a Pareto IV where u = 0 is a Burr distribution.
|
The Pareto II (Lomax) CDF is given by $$F_X(x) = 1 - \left(1 + \frac{x}{\sigma}\right)^{-\alpha}, \quad x > 0.$$ Your parametrization is slightly different, but the calculations are comparable.
The Pareto IV CDF is given by $$F_Y(y) = 1 - \left(1 + \left(\frac{y-\mu}{\sigma}\right)^{1/\gamma} \right)^{-\alpha},$$ so it seems natural that if $Y = \mu + (\sigma X)^\gamma$, then we immediately find $$F_Y(y) = \Pr[Y \le y] = \Pr\left[X^\gamma \le \frac{y - \mu}{\sigma} \right] = F_X(((y-\mu)/\sigma)^{1/\gamma}),$$ hence we see that your transformation $Y = X^2$ turns a Pareto II (Lomax) into a Pareto IV with $\alpha$, $\mu = 0$ unchanged, but $\gamma = 2$ and the Pareto IV $\sigma$ parameter is the square of the Pareto II value of $\sigma$.
Okay, so I think you are using the same parametrization, so $\sigma = \theta$, in which case you have the CDF $$F_X(x) = 1 - \left(1 + \frac{x}{\theta}\right)^{-\alpha} = 1 - \frac{\theta^\alpha}{(\theta+x)^\alpha}.$$ With the transformation $Y = X^2$, we obtain $$F_Y(y) = F_X(\sqrt{y}) = \color{red}{\boxed{1 - \left(1 + \frac{y^{1/2}}{\theta}\right)^{-\alpha}}} = 1 - \frac{\theta^\alpha}{(\theta+ y^{1/2})^\alpha},$$ and its derivative is $$f_Y(y) = \alpha \left(1 + \frac{y^{1/2}}{\theta}\right)^{-(\alpha+1)} \frac{1}{2\theta y^{1/2}}.$$ But since we are comparing CDFs, all you need to do is observe that the expression in red corresponds to the Pareto IV distribution I wrote above, but with values $\mu = 0$, $\gamma = 2$, $\theta = \sigma^{1/\gamma}$.
Incidentally, I am using the definitions given by Wikipedia. These may not match the nomenclature or parametrization used by SOA/CAS for the exams.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$m_a(x) = x^n + b_1x^{n−1} + \cdots + b_n$ be the minimal polynomial of $a$ over $E$. Prove that $E = K(b_1, \ldots , b_n)$ Let $L = K(a)$ be an algebraic extension. Let $E \subset L$ be a sub-field containing $K$. Let $m_a(x) = x^n + b_1x^{n−1} + \cdots + b_n$ be the minimal polynomial of $a$ over $E$. Prove that $E = K(b_1, \ldots , b_n)$.
Here $K \subset E \subset L$.
$K(b_1, \ldots , b_n) \subset E$ is trivial; we have to prove the other inclusion.
|
Let $f(x)$ be the minimal polynomial of $a$ over $K(b_1,\ldots,b_n)$. Since $f(x)\in E[x]$ vanishes at $a$, we have that $m_a(x)$ divides $f(x)$. On the other hand, we also have that $m_a(x) \in K(b_1,\ldots,b_n)[x]$ vanishes
at $a$. Thus $f(x)$ divides $m_a(x)$, and then $f(x)=m_a(x)$.
Therefore $$[L: K(b_1,\ldots,b_n)]=\deg f(x)=\deg m_a(x)=[L: E]$$
and
$$K(b_1,\ldots,b_n)\subseteq E \subseteq L$$
gives $E=K(b_1,\ldots,b_n)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Knock out tournament 1 8n players $P_1$, $P_2$, $P_3$, .....$P{_8}{_n}$ play a knock out tournament. It is known that all players are of equal strength. The tournament is held in three rounds where the players are paired at random in each round. If it is given that $P_1$ wins in the third round then what is the conditional probability that $P_2$ loses in the second round.
I tried applying the concept of conditional probability followed by total probability theorem but somehow, there are far too many cases to consider. Any help/ suggestions/ solutions would be highly appreciated.
|
First Round = 8n men. Second Round or Semi-Final = 4n men. 3rd Round or Final = 2n men. The tournament Winner is $P_{_{1}}$.The Probability that he is the winner is given by $\frac{1}{Totalmen -1 } = \frac{1}{8n -1 }$ . But $P_{_{2}}$ lost in the Second Round. Now In Second Round, there are 4n men = 2n Losers + 2n Winners. Therefore, $P_{_{2}}$ is one of the losers. He can be selected in $\binom{2n}{1}$ ways.Hence, P( $P_{_{2}}$ Loses in the Second Round| $P_{_{1}}$ wins The Tournament) = ${\frac{2n}{8n -1 }}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Find the limit $\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}$ Find the following limit:
$$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}$$
My attempt:
$$t:=x-1,\ x \rightarrow 1 \Rightarrow t\rightarrow 0,\ x=t+1$$
$$\lim_{x\to 1}\frac{\sin{(\pi\sqrt x)}}{\sin{(\pi x)}}=\lim_{t\to 0}\frac{\sin{(\pi\sqrt{t+1})}}{\sin{(\pi(t+1))}}=\lim_{t\to 0}\frac{\frac{\sin{(\pi\sqrt{t+1})}}{\pi(\sqrt{t+1})}\cdot \pi \sqrt{(t+1)}}{\frac{\sin{(\pi(t+1))}}{\pi(t+1)}\cdot \pi(t+1)}=\lim_{t\to 0}\frac{1\cdot \pi \sqrt{(t+1)}}{1\cdot\pi(t+1)}=\frac{\pi\sqrt{(0+1)}}{\pi\sqrt{(0+1)}}=1$$
The soulution should be $\frac{1}{2}$. What am I doing wrong?
|
Your mistake is here $$\lim _{ t\to 0 } \frac { \sin { \left( \pi \left( t+1 \right) \right) } }{ \pi \left( t+1 \right) } \neq 1\\ \lim _{ t\to 0 } \frac { \sin { \left( \pi \sqrt { \left( t+1 \right) } \right) } }{ \pi \sqrt { \left( t+1 \right) } } \neq 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Every group of order $56$ has a proper non - trivial normal subgroup? I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.
I got two cases for $n_{2} = 1 , 7 $ and for $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.
Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.
Please help?
|
Count the total number of elements. Suppose neither $n_2$ nor $n_7$ is equal to $1$. Then $n_7=8$, which makes $8\cdot 6$ elements which belong to a $7$-Sylow subgroup, and to no other Sylow subgroup.
On the other hand, since $n_2=7$, each $2$-Sylow subgroup contains at least $4$ elements which belong to no other Sylow subgroup. In all $7\cdot 4$ elements.
Summing up, the group would contain at least $48+28+1$ elements. This is more than $56$, if I'm not mistaken.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Sums of squares are closed under division Surprisingly, we got only one question for our 2-hour exam and I think nobody solved it. Here is the problem:
Assuming that $K$ is a field, show that $S$ is stable under addition, multiplication and division, where $S$ is defined as follow:
$$S=\left\{\sum_{i=1}^{n}{x_i}^2 \mid n\in \Bbb N ,\ x_i\in K\right\}.$$
|
$$\frac{1}{x_1^2+x_2^2+x_3^2}=\frac{x_1^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_2^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_3^2}{(x_1^2+x_2^2+x_3^2)^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
}
|
permutations without any successive digits I need some help with the following problem.
Consider the permutations of the set $ \{1, \dots , n \}$. What is the probability to find a permutation in which the digits $1$, $2$, $3$ are not successive. In other worlds what is the probability to occur a permutation of the form $$(\dots, a_1 , \dots, a_2, \dots, a_3, \dots) \quad, \{ 1,2,3 \} = \{ a_1, a_2, a_3 \} $$ ,while the dots before $a_1$ and after $a_3$ could be omitted. Obviously, every single permutation of the set has the same probability to occur.
The sample space, $\Omega$, of our random experiment is the set of all permutations of $ \{1, \dots , n \}$, so $$N_{\Omega} = n!$$
Now lets name $A$ the event described above. I should count how many permutations, of this specific form, there are. I will count them, in three steps, using the multiplication principle.
step 1
Consider the important digits, $1, 2, 3$, as ones and all the other as zeros. In step one, I will find in how many ways the ones could be arranged.
Let $y_1$ be the number of zeros before the $a_1$, with the same logic I define $y_2$ and $y_3$, finally let $y_4$ be the number of zeros after $a_3$.
It should be $$\sum y_i=n-3$$ with $$ y_1, y_4 \geq 0\, \text{ and }\,y_2, y_3 \geq 1 $$
Setting $$ x_1 = y_1 \quad x_2 = y_2 - 1 \quad x_2 = y_2 - 1 \quad x_4 = y_4 $$ I get the equivalent system $$\sum x_i = n-5\quad, x_i\geq0$$
The number of the integer solutions of the above system is actually the ways the ones could be arranged. In fact there are $$\binom{4 - 1 + n-5}{4-1}=\binom{n-2}{3}=\frac{(n-2)!}{(n-5)! \cdot 3!}$$ ways.
step 2
For every selection of the positions of the ones there are $3!$ ways to arrange the digits $1, 2, 3$
step 3
Finally all the other digits could fill the zeros positions in $(n-3)!$ ways
So, using the the multiplication principle I get $$N_{A} = \binom{n-2}{3}(n-3)! \cdot 3! = \frac{(n-2)!}{(n-5)! \cdot 3!} (n-3)! \cdot 3! $$
and $$P(A)=\frac{N_A}{N_\Omega} = \frac{(n-2)!\cdot(n-3)!\cdot 3!}{n! \cdot(n-5)! \cdot 3!} = \frac{(n-2)!}{n!}\frac{(n-3)!}{(n-5)!} = \frac{(n-4)(n-3)}{(n-1)n}$$
Is it everything alright with my solution? Are their any misconceptions? Is there another way to tackle the problem?
Thank you for your time
|
As you found, the focus set $\{1,2,3\}$ can be arranged $3!$ ways, the remainder $(n-3)!$ ways and the two part-sets can be interleaved by choosing $3$ of the $n{-}2$ gaps between and around the remainder elements in $\binom {n-2}3$ ways, giving the probability: $$ 3!(n-3)!\binom {n-2}3 \frac 1 {n!} = \frac {3!(n-3)!(n-2)!} {n!(n-5)!3!} = \frac{(n-2)!}{n!} \frac{(n-3)!}{(n-5)!} =\frac{(n-3)(n-4)}{n(n-1)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
When is the derivative of an inverse function equal to the reciprocal of the derivative? When is this statement true?
$$\dfrac {\mathrm dx}{\mathrm dy} = \frac 1 {\frac {\mathrm dy}{\mathrm dx}}$$
where $y=y(x)$. I think that $y(x)$ has to be bijective in order to have an inverse and let the expression $\dfrac {\mathrm dx}{\mathrm dy}$ make sense. But is there any other condition?
|
Assume $g(f(x))=x$. Then
$$g'(f(x))f'(x)=1$$ and then
$$g'(f(x))=\frac1{f'(x)}$$
Note that we need also that $f'(x)\neq 0$. All the conditions (the injectivity and the differentability of $f$ and that $f'$ does not vanish) must meet in a neighbourhood of the point where you are differentiating, that is, this works locally.
See the inverse function theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2117928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
}
|
Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
|
$\binom{6}{4}x^2\times(\frac{2}{x})^4$ would give the coefficient for $\frac{1}{x^2}$. What you want instead is $$\binom{6}{2}x^4\times(\frac{2}{x})^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Why is volume of a high-D ball concentrated near its surface? I came across the following sentence while reading a book on applied math:
Volume of a high dimensional unit ball is concentrated near its
surface and is also concentrated at its equator.
This is from book's introduction, and I believe the sentence will be explained at some later point in the book. However, it is hard to me to accept it on intuition level. Can some of you explain this sentence to me, in a sort of laymen-style, if possible?
|
The volume of an n dimensional sphere is proportional to $r^n$. For example the area of a circle (2-sphere) is $\pi r^2$ and the volume of a 3-sphere is $\frac43 \pi r^3$.
This means that the volume grows with a high power of radius, and there is more of that volume near the boundary than near the centre. For a circle, $\frac34$ of the area is closer to the circumference than the centre; for a sphere this becomes $\frac78$ and this increases closer to 1 as the number of dimensions increases.
With enough dimensions, the outer 10% (by radius), or any small proportion you like, will have more volume than the remainder. Go high enough and the outer 1% of the n-sphere will be responsible for over 99% of the volume.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?
|
Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!
A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$
Is to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\tag2$$
Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so$$\begin{align*} & a-b^2=A\tag3\\ & ab=-B\tag4\end{align*}$$
Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.
Examples:
*
*Solve $x^3-6x+4=0$ (your question)
Moving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\-2\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\implies x=2$$
Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.
*Solving $x^3+16x=455$
A factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\cdot7=455$$
Therefore,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
}
|
Sequence Of Primes Hello I have a basic number theory question.
*
*I want to find a list of primes of the form
a, a + d, a + 2d, ... , a + 5d
*
*So a sequence of at least 6 or greater if I want to select a = 101
then what would I choose as my d to get a sequence of primes (they
don't have to be consecutive).
*I am quite lost. Maybe there isn't an answer and it's not possible? I
don't know how to find d in this case.
|
A well-known sequence $a_n:=a+nd$ producing at least $6$ primes with $3$-digit $a$ (this was your requirement here) is $a_n=199+210n$, i.e.,
$$
199, 409, 619, 829,1039,1249,1459,1669,1879,2089,
$$
so that the first consecutive $10$ sequence members are prime numbers. With $a=101$ you can search for such a $d$ by computer. I do not know of a conceptual way to find such a $d$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
two models of ZFC such that there is a isomorphism between their ordinals if two models of ZFC have their ordinals isomorphic then there is a isomorphism between their constructibles?
|
The answer is no.
It's not hard to show - similarly to the situation with respect to PA - that if $M$ is a countable model of ZFC such that $\omega^M$ is non-well-founded, then $ON^M$ has ordertype $(\omega+\zeta\cdot\eta)\cdot(1+\eta)$, where $\zeta$ and $\eta$ represent the ordertypes of $\mathbb{Z}$ and $\mathbb{Q}$ respectively.
In particular, this means:
If $M, N$ are countable models of ZFC with illfounded natural numbers, then $ON^M$ and $ON^N$ are isomorphic.
So now let $p$ be any sentence independent of ZFC+V=L (which exists by Goedel), and let $M, N$ be countable models of ZFC+V=L with illfounded natural numbers such that $M\models p$ and $N\models\neg p$ (which exist by Compactness + Lowenheim-Skolem). Then $ON^M\cong ON^N$ but $L^M=M\not\cong N=L^N$.
Even if we restrict attention to $\omega$-models the answer is no, as long as we allow ill-founded $\omega$-models. This is a bit more complicated - there are lots of ordertypes of ordinals in countable $\omega$-models of ZFC - but can still be handled similarly to the above.
Namely, any theory extending $ZFC$ has a model whose ordinals have ordertype $$\omega_1^{CK}+\omega_1^{CK}\cdot\eta,$$ where $\omega_1^{CK}$ is the first nonrecursive ordinal; the ordertype above is called the Harrison order and is a standard counterexample in computable structure theory. So, again, pick a sentence $p$ which is true in some $\omega$-models of ZFC+V=L and false in others (e.g. "There is an inaccessible cardinal"), and let $M, N$ be countable models of $ZFC+V=L$ whose ordinals each have ordertype the Harrison order, and $M\models p$ but $N\models\neg p$.
Finally, if we restrict attention to well-founded models, the answer is yes - if $M, N$ are transitive models with $ON^M=ON^N$, then by induction on $\alpha$ we have $L_\alpha^{M}=L_\alpha^N$ for every $\alpha\in ON^M$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
If $p^{2k+1} | m^2$, show $p^{k+1} | m$ Question: If $p^{2k+1} | m^2$, show $p^{k+1} | m$.
Answer that was provided: If $p^k$ is the largest power of $p$ that divides $m$, then $p^{2k}$ is the largest power of $p$ that divides $m^2$. Hence if a power of $p$ larger than $2k$ can divide $m$, then $p^{k+1}$ surely also divides m.
Here is my question about the proof provided to me. If $p^{2k}$ is already considered to be the largest power of $p$ that divides $m^2$, how can we suggest there is an even larger power of $p$ that divides $m^2$? I guess I'm wondering if someone could add more detail to the above proof so that it is clearer to me. I would really appreciate that.
Thank you.
|
If $p^k$ is the largest power of p that divides m, then $p^{2k}$ is the largest power of p that divvies $m^2$.
The confusion here lies with (re)using the same variable name for $k$.
The statement would have been easier to follow if it said: "let $p^a$ be the largest power of $p$ that divides $m$, then $p^{2a}$ is the largest power of $p$ that divides $m^2$".
Hence if a power of p larger than $2k$ can divide m, then $p^{k+1}$ surely also divides m.
Rephrased using the above: we know that $p^{2k+1}$ divides $m^2$, therefore $2k+1 \le 2a$. It follows that $a \ge k + \frac{1}{2}$ and, since both $a,k$ are integers, $a \ge k+1$, so in the end $p^{k+1} \mid p^a$ must divide $m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Finding recursive function Range A function $f: \Bbb{N^+} \rightarrow \Bbb{N^+}$ , defined on the set of positive integers $\Bbb{N^+}$,satisfies the following properties:
$$f(n)=\begin{cases} f(n/2) & \text{if } n \text{ is even}\\
f(n+5) & \text{if } n \text{ is odd} \end{cases}$$
Let $R=\{ i \mid \exists{j} : f(j)=i \}$ be the set of distinct values that $f$ takes. The maximum possible size of $R$ is ___________.
Answer of this question is $2$, and solution goes like this:-
every multiple of $5$ has same value, and every other number has same value.
I want to proof it, by NOT using examples, but some real mathematical proof, that can show us that indeed this is true.
Thanks.
|
Suppose that $f(1) = a$ and $f(5) = b$. It is clear that $$f(5n) = b$$ for all $n$. We'll prove by induction that for all $n \ne 5k$, $f(n) = a$. First note that
$$f(2) = f(\frac{2}{2}) = f(1) = a,$$
$$f(3) = f(3+5) = f(8) = f(4) = f(2) = a,$$
$$f(4) = f(2) = a.$$
Now suppose $n = 5k + r$, where $0 < r < 5$, and for all $m<n$ which are not divisible by $5$, $f(m) = a$.
If $n$ is odd, $f(n) = f(n-5)$, and by induction hypothesis, $f(n-5) = a$, so we get $$f(n) = a.$$
If $n$ is even, $f(n) = f(n/2)$, and by induction hypothesis, $f(n/2) = a$, so we get $$f(n) = a.$$
Note that here $\frac{n}{2}$ isn't divisible by $5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Example of an $\mathbb R$-linear map from $\mathbb C^2$ to $\mathbb C^2$ that is not $\mathbb C$-linear Example of an $\mathbb R$-linear map from $\mathbb C^2$ to $\mathbb C^2$ that is not $\mathbb C$-linear. One class of examples that I can think of is the conjugate linear maps, are there any other importatnt examples?
|
All you have to do is completely forget about the $\mathbb{C}$ structure, and pick your favorite endomorphism of $\mathbb{R}^4$. For example, $(a+bi, c+di)\rightarrow (d+ai, b+ci)$ is not $\mathbb{C}$-linear, like most endomorphisms you pick out.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$ Prove that:$$\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$$.
My Attempt,
$$L.H.S= \frac {\sec (16A)-1}{\sec (8A)-1}$$
$$=\frac {\frac {1}{\cos (16A)} -1}{\frac {1}{\cos (8A)} -1}$$
$$=\frac {(1-\cos (16A)).(\cos (8A)}{(\cos (16A))(1-\cos (8A))}$$.
What should I do next?
|
$\frac{\sec 16A -1}{\sec 8A -1}$
= $\frac{\frac{1}{\cos 16A}-1}{\frac{1}{\cos 8A}-1}$
= $\frac{\frac{1 - \cos 16A}{\cos 16A}}{\frac{1 - \cos 8A}{\cos 8A}}$
= $\frac{2 \sin^2 8A}{\cos 16A} × \frac{\cos 8A}{2 \sin^2 4A}$
= $\frac{2 \sin 8A \cos 8A}{\cos 16A} × \frac{\sin 8A}{2 \sin^2 4A}$
= $\frac{\sin 16A}{\cos 16A} × \frac{ 2 \sin 4A \cos 4A}{2 \sin^2 4A}$
= $\tan 16A × \frac{\cos 4A}{\sin 4A}$
= $\tan 16A × \cot 4A$
= $\frac{\tan 16A}{\tan 4A}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2118974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
How to solve $\frac{dy}{dt}=\alpha y-\beta y^n$ for $n\geq 2,\beta>0$? Is there a way to solve $\frac{dy}{dt}=\alpha y-\beta y^n$ for $n\geq 2,\beta>0$?
I know how to solve it for $n=2$ but I am not sure how to solve for $n\geq 2$?
|
$$\text{y}'\left(t\right)=\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}\space\Longleftrightarrow\space\int\frac{\text{y}'\left(t\right)}{\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}}\space\text{d}t=\int1\space\text{d}t\tag1$$
Now, use:
*
*Substitute $\text{u}=\text{y}\left(t\right)$ and $\text{d}\text{u}=\text{y}'\left(t\right)\space\text{d}t$:
$$\int\frac{\text{y}'\left(t\right)}{\alpha\cdot\text{y}\left(t\right)-\beta\cdot\text{y}\left(t\right)^\text{n}}\space\text{d}t=\int\frac{1}{\alpha\cdot\text{u}-\beta\cdot\text{u}^\text{n}}\space\text{d}\text{u}=\frac{\ln\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|}{\alpha\cdot\left(1-\text{n}\right)}+\text{C}_1\tag2$$
*$$\int1\space\text{d}t=t+\text{C}_2\tag3$$
So:
$$\frac{\ln\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|}{\alpha\cdot\left(1-\text{n}\right)}=t+\text{C}\space\Longleftrightarrow\space\left|\alpha\cdot\text{y}\left(t\right)^{1-\text{n}}-\beta\right|=\text{C}\cdot\exp\left[\alpha\cdot t\cdot\left(1-\text{n}\right)\right]\tag4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.
|
Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) ^{ \frac { 1 }{ x } }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) -\ln { \left( { e }^{ 2 }+x \right) } } \right] } }=\\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+2x \right) -x\ln { \left( { e }^{ 2 }+x \right) } } }{ { x }^{ 2 } } } }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 2 }{ 1+2x } -\ln { \left( { { e }^{ 2 }+x } \right) -\frac { x }{ { e }^{ 2 }+x } } }{ 2{ x } } } }\overset { L'hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 4 }{ { \left( 1+2x \right) }^{ 2 } } -\frac { 1 }{ { e }^{ 2 }+x } -\frac { { e }^{ 2 } }{ { \left( { e }^{ 2 }+x \right) }^{ 2 } } }{ 2 } } }={ e }^{ -\frac { 4{ e }^{ 2 }+2 }{ 2{ e }^{ 2 } } }$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
}
|
Find the length of $EF$.
$AB = 60\text{ cm}$
$CD:DB = 3:5$
$AF:AD = 4:5$
$E$ is the midpoint of $AC$
Find $EF$.
I have this following problem. I tried solving it, but it doesn't make sense alot. Can I have a hint or a guide for solving this question. Thanks!
|
First we notice that $CD=\frac{3CB}{8}$ and $DB=\frac{5CB}{8}$
Draw line $l$ parallel to AB from F.Let G be the point of intersection of $l$ and CB, then we have $DG=\frac{CB}{8}$ But then $CG=CD+DG=\frac{4CB}{8}$ and thus G is the mid point of CB and thus line $l$ passes through E.Now cause of similar triangles we have that $FG=\frac{AB}{5}=12$cm and $EG=\frac{AB}{2}=30$cm ,so $EF=EG-FG=18$cm
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Quadratic equation, find $1/x_1^3+1/x_2^3$ In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.
It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?
|
If $x_{1}$ and $x_{2}$ are the roots then
$x_{1} + x_{2} = -\frac{b}{a}$ and $x_{1} \cdot x_{2}=\frac{c}{a}$, now $\frac{1}{x_{1}}+\frac{1}{x_{2}} = -\frac{b}{c}$ and $$\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}} = \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)^3- 3\cdot\frac{1}{x_{1}.x_{2}}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Is the sum of permutations of disjointed sets larger than the count of permutations in their union? My wife and I are having a bit of a disagreement.
Concerning eight-digit passwords like the kind most secure websites require, she believes you could generate more unique password combinations by using only one of the following sets: one lowercase or one uppercase or one symbol or one numeral. In other words, if you created a password using only lowercase letters with no other conventions. e.g.: jtpdlrkc or JTPDLRKC or 29456014 or #^@%#&(%.
I have asserted that, by using at least one each of the aforementioned sets, a person can generate FAR, FAR more unique password combinations, thereby making the password much more difficult for someone to hack. e.g. gy7HU*j9 or $F6ms38@.
She thinks I'm making this up. I need numbers to prove my point.
Anyone interested in running with this one?
|
There is a context in which the wife is right, which may explain her intuition. Suppose we split the permitted symbols into two sets, $S$ and $T$, of size $s=|S|$ and $t=|T|$. Suppose further that our password must be exactly two letters long.
The wife's strategy yields $s^2+t^2$ passwords, either two from $S$ or two from $T$. The OP's strategy yields $2st$ passwords, either one from $S$ then one from $T$, or the other way.
The wife's strategy will give at least as many passwords as the OP's, because $(s^2+t^2)-2st=(s-t)^2\ge 0$. Indeed, if $s\neq t$, then the wife's strategy gives strictly more passwords.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
}
|
Exercise on Manifolds: Transition Maps
Let $M\;$ be a differentiable manifold of distance $3$, $p\in M\;$ and
two charts $(U,φ=(x_1,x_2,x_3))\;\;,(U,ψ=(y_1,y_2,y_3))\;$ of $M$
near $p\;$ with $φ(p)=(1,1,-2)\;$ such that:
$y_1=x_1\;,\;y_2=x_2-{x_1}^3\;,\;y_3=x_3+3x_1 {x_2}^2 \;$ in $U$
Find transition maps : $ψο{φ}^{-1}\;$ and $φο{ψ}^{-1}$
What I thought to do is:
$ψ(x_1,x_2,x_3)=(x_1,x_2-{x_1}^3,x_3+3x_1 {x_2}^2)\;$ and $φ(y_1,y_2,y_3)=(y_1,y_2+{y_1}^3,y_3-3y_1(y_2+{y_1}^3)^2)\;$
So in order to compute $ψ^{-1}\;$ I write $ψ(x_1,x_2,x_3)=(x_1,x_2-{x_1}^3,x_3+3x_1 {x_2}^2)=(z_1,z_2,z_3)\;\Rightarrow \begin{cases} x_1=z_1\\x_2=z_2+{z_1}^3\\x_3=z_3-3z_1(z_2+{z_1}^3)^2 \end{cases}$
Now $φοψ^{-1}(z_1,z_2,z_3)=φ(z_1,z_2+{z_1}^3,z_3-3z_1(z_2+{z_1}^3)^2)=(z_1,z_2+2{z_1}^3,z_3-3z_1(z_2+{z_1}^3)^2-3z_1(z_2+2{z_1}^3)^2)$
In similar way I can compute $ψοφ^{-1}\;$
My question is if the above thought is correct. I feel I'm missing something... I would appreciate if somebody could help me through this. Hints or solutions other than this are of course welcome!
Thanks in advance..
|
Note that $\varphi, \psi \colon U \rightarrow \mathbb{R}^3$ and you have no idea what $U$ is (it is some open subset of some manifold $M$).
The equations
$$y_1 = x_1, y_2 = x_2 - x_1^3, y_3 = x_3 + 3x_1 x_2^2$$ already give you (practically by definition)
$$(\psi \circ \varphi^{-1})(x_1,x_2,x_3) = (y_1,y_2,y_3) = (x_1, x_2 - x_1^3, x_3 + 3x_1 x_2^2). $$
In more details, if $q \in U$ then the coordinates of $q$ with respect to the coordinate system $\varphi$ are $\varphi(q) = (x_1(q), x_2(q),x_3(q))$. Similarly, the coordinates of $q$ with respect to the coordinate system $\psi$ are $\psi(q) = (y_1(q),y_2(q),y_3(q))$. The transition function $\psi \circ \varphi^{-1}$ eats a triple $(x_1,x_2,x_3)$ and returns the $y_i$ coordinates of the point $q = \varphi^{-1}(x_1,x_2,x_3)$.
The meaning of the equations $y_i = f_i(x_1,x_2,x_3)$ you are given is that $y_i(q) = f_i(x_1(q),x_2(q),x_3(q))$ (the $y_i$ coordinate of $q \in U$ is related to the $x_i$ coordinates of $q$ by $f_i$). Letting $q = \varphi^{-1}(x_1,x_2,x_3)$ we get
$$ y_i(\varphi^{-1}(x_1,x_2,x_3)) = f_i(x_1(\varphi^{-1}(x_1,x_2,x_3)),x_2(\varphi^{-1}(x_1,x_2,x_3)),x_3(\varphi^{-1}(x_1,x_2,x_3))) = f_i(x_1,x_2,x_3) $$
but the left hand side is precisely the $i$-th coordinate of $\psi \circ \varphi^{-1}$.
In order to get $\varphi \circ \psi^{-1} = \left( \psi \circ \varphi^{-1} \right)^{-1}$ you need to invert $\psi \circ \varphi^{-1}$. Namely, you need to solve for the $x_i$'s in terms of the $y_i$'s. In your case,
$$ x_1 = y_1, x_2 = y_2 + x_1^3 = y_2 + y_1^3, \\
x_3 = y_3 - 3x_1x_2^2 = y_3 - 3y_1(y_2 + y_1^3)^2 $$
so
$$ (\varphi \circ \psi^{-1})(y_1,y_2,y_3) = (y_1, y_2 + y_1^3, y_3 - 3y_1(y_2 + y_1^3)^2). $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Stuck on this differential equation I want to solve this differential equation with the power-series method: $$ x^{2}\cdot y''(x) +(1+x^{2}) y'(x) + y(x) =0$$ where $$y(0)=1$$
They want the solution given in elementary functions. I managed to get the recursive formula for $$j\ge 2$$$$a_{j+1} = \frac{-((j(j-1)+1)a_{j} + (j-1)a_{j-1})}{j+1} $$
And also got $$a_{0}=y(0)=1 $$,and then$$ a_{1}=-1, a_{2}=1/2 $$
But Im not sure how to answer in an elementary function, I tried to get some more values and its just irrational values. Appreciate help.
|
I agree with you're recurrence relation $(i+1)a_{i+1}=-((i^2-i+1)a_i+(i-1)a_{i-1})$. & the first few values $a_0=1, a_1=-1, a_2=1/2, a_3=-1/6$ from this I guess the general formulae $a_i=(-1)^i/i!$. You can easily show (by induction) that this is indeed true. Now the solution is obviously $y=e^{-x}$ & this does indeed satisfy the original equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to intuitively explain that if kernel(L) equal to the set containing the zero vector, then L is one-to-one? Whenever I try to learn about a relationship, I try to reason intuitively why a theorem or lemma should make sense. I know often times this is increasingly difficult to achieve. However, I have the following:
Let $ L: \mathbb{V} \to \mathbb{W} $ be a linear mapping. $L$ is one-to-one if and only if Ker($L$) = $\{\vec{0}\}$
I can intuitively explain the forward direction, that if $L$ is one-to-one, then only one value can possibly map to $\vec{0}$ and that must be $\vec0$.
When I consider the other direction, it is hard for me to intuitively explain that the property of a linear mapping being one-to-one follows naturally from the fact that its kernel contains only $\vec0$. I know that the algebraic proof of this is trivial, but I feel like without a more intuitive understanding, I am misunderstanding something about linear mappings and hence not fully appreciating the importance of this relationship.
Thanks in advance.
EDIT:
The question is, can I explain the logic behind this principle without implicitly referring to the proof?
|
We know that the kernel is a subvectorspace. Hence if $x_1, x_2$ are (two different elements, nonzero) in the kernel, we have that $x_1 - x_2$ is in the kernel, so we find that
$$0 = L(x_1 - x_2) = L(x_1) - L(x_2)$$
by linearity. This would mean that $L$ is not one-to-one, since we have two different points with the same image.
I think this is just such a basic concept, that any 'non algebraic argument' can be translated to an algebraic argument en vice versa. I could translate my algebraic explanation into words as follows: If we suppose that the map is not injective, then there are different elements having the same image, hence by linearity their difference is mapped onto zero. Therefore the kernel is not trivial. But this is essentially what I wrote above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2119929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
$A+B=AB$ does it follows that $AB=BA$? If $A$, $B$ are two normal operators such that: $A+B=AB$ does it follow that $AB=BA$?
|
First, note that if $C$ is a normal operator which has a left inverse $D$ (so $DC=I$), then $C$ is invertible (and thus $D$ is its inverse and $CD=I$ as well). This follows from the spectral theorem: you can identify your Hilbert space with $L^2(X)$ for some semifinite measure space $X$ and $C$ with multiplication by $f$ for some $f\in L^\infty(X)$. If $0$ is in the essential range of $f$, then this means that for any $\epsilon>0$ there exists $g\in L^2(X)$ such that $\|fg\|<\epsilon\|g\|$ (choose $g$ to be supported on the set where $|f|<\epsilon$). Since $DC=I$, we must have $D(fg)=g$ so $\|D\|\geq\|g\|/\|fg\|>1/\epsilon$. Since $\epsilon$ is arbitrary, this is a contradiction. Thus $0$ is not in the essential range of $f$, so $1/f\in L^\infty(X)$, and multiplication by $1/f$ is an inverse for $C$.
(Probably there is a more elementary argument that avoids spectral theory, but this is all I can come up with at the moment.)
Now note that $A+B=AB$ implies $(I-A)(I-B)=I-A-B+AB=I$. Since $I-B$ is normal, by the result above this implies $(I-B)(I-A)=I$ as well. That is, $I-A-B+BA=I$, or $BA=A+B=AB$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Find a ring that contain $\mathbb{Q}$ as a group and has solution for $x^2 \equiv 2$ but no solution to $x^2 \equiv 3$ I have the following question :
Find a ring that contain $\mathbb{Q}$ as a group and has solution for $x^2 \equiv 2$ but no solution to $x^2 \equiv 3$
Hint : Start from $\mathbb{Q}[x]$
I really don't know how to approach this, how to force that $x^2 \equiv 3$ has not solution???
This is what I did - I don't know if its right
Maybe if I choose $I=x^2-2$? I think its an ideal since no root since there is no such $x\in \mathbb {Q}$ such that $x^2-2=0$,
Yet how that implies solution for $x^2 \equiv 2$? what about no solutions for $x^2 \equiv 3$?
Any help will be appreciated.
|
You have the right idea. The ideal $(x^2 - 2) \subseteq \mathbb{Q}[x]$ is maximal and so $R = \frac{\mathbb{Q}[x]}{(x^2-2)}$ is a field.
Note that $R \cong \{ a + b \sqrt{2} \ | \ a,b \in \mathbb{Q} \}$ and $\mathbb{Q} \subseteq R$.
Suppose that $\sqrt{3} \in R$. Then there exist $a,b \in \mathbb{Q}$ such that
$$\sqrt{3} = a + b\sqrt{2}.$$
We now seek a contradiction.
Case 1: $a = 0$
I'll leave this to you.
Case 2: $b = 0$
I'll leave this to you.
Case 3: $a,b \neq 0$
If you square both sides of that equation and rearrange, you get:
$$ 3 - a^2 - 2b^2 = 2ab \sqrt{2}.$$
This implies that $$\sqrt{2} = \frac{3-a^2-2b^2}{2ab} \in \mathbb{Q}_.$$
So we arrive at a contradiction.
Thus $\sqrt{3} \notin R$. Obviously neither is $-\sqrt{3}$. So $x^2-3$ has no roots in $R$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite differences to be 6 and degree of the polynomial with the given points to be 3 ($n=3$). But how do I find the polynomial function rule out of this information? Given no factors? Please help with a short solution. Thanks a lot in advance. I also did $y=kx^3$, $-9=k(-3)^3$, $k=1/3$. But of course the rule isn't $y=1/3x^3$.
|
Treat your numbers as a sequence with $g(0)$ being the first term, corresponding to $f(-3)$,
$$f(x-3):=g(x) : \color{green}{-9},3,3,-3,-9,-9,3$$
$$\Delta g=g(x+1)-g(x) : \color{green}{12},0,-6,-6,0,12$$
$$\Delta^2 g : \color{green}{-12},-6,0,6,12$$
$$\Delta^3 g : \color{green}{6},6,6$$
Now assume all else is $0$. To get an umbral Taylor series representation,
$$f(x-3)=\color{green}{-9}+\color{green}{12}{x \choose 1}\color{green}{-12}{x \choose 2}+\color{green}{6}{x \choose 3}$$
So that with $x \mapsto x+3$,
$$f(x)=-9+12{{x+3} \choose 1}-12{{x+3} \choose 2}+6{{x+3} \choose 3}$$
$$=x^3-7x-3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Please help solving this inital-value problem The initial-value problem
$$\begin{cases}(ye^{xy} + cos(x))dx + (xe^{xy})dy = 0\\y(\dfrac{\pi}{2}) = 0\end{cases}$$
From my calculations it seems the integrating factor depends on y and x, but I am unsure how to find the correct integrating factor.
Please help
|
Hint:$$m.dx+n.dy=0 \\$$
$$\begin{cases}(ye^{xy} + cos(x))dx + (xe^{xy})dy = 0\\y(\dfrac{\pi}{2}) = 0\end{cases}\\
\frac{\partial (ye^{xy} + cos(x)) }{\partial y}=1.e^{xy}+xy.e^{xy}+0\\
\frac{\partial (xe^{xy})}{\partial x}=1.e^{xy}+yxe^{xy}\\so \\
\frac{\partial n}{\partial x}=\frac{\partial m }{\partial y}\\exact \space equation$$now
$$u(x,y)=\int m.dx=\int (ye^{xy} + cos(x))dx=\\e^{xy}+sinx +g(y) \to $$
$$u_y=x.e^{xy}+0+g'(y) =xe^{xy} \\\to g'(y)=0 \to g(y)=c\\so \\ u(x,y)=e^{xy}+sinx+c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Top does not satisfy the axiom of choice I am trying to think about the axiom of choice from a categorical point of view.
I found out in some book that one possible formalization of this axiom is obtained by stating: "a category $\mathcal{C}$ satisfies the axiom of choice if all epics in $\mathcal{C}$ are split".
In this way $\mathbf{Set}$ does satisfy the axiom of choice (and that is fine). The category of groups doesn't though! What about $\mathbf{Top}$, i.e. the category of topological spaces? Does a not split epimorphism exist in $\mathbf{Top}$?
|
Top indeed does not satisfy the axiom of choice. To show this it suffices to exhibit a fiber bundle without a section, and for example the unit tangent bundle of $S^2$ has this property by the hairy ball theorem. Various other kinds of examples are possible; see, for example, this blog post.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find a basis for the given subspace: confirmation I would like to have a confirmation if possible!
Let $V$ be the subspace of $\mathbb{R}^3$ given by the solution of the system
\begin{equation}
\begin{cases} x+6y-3z=0\\2x+12y-6z=0 \end{cases}
\end{equation}
Find a basis for $V$.
By solving the homogeneous system I can find $\infty^2$ solutions (as the two row vectors are linearly dependent) of the kind $(3z-6y,y,z)$, which can be rewritten as $z(3,0,1)+y(-6,1,0)$. Hence a basis is $(3,0,1),(-6,1,0)$ and the subspace is two-dimensional.
An example of solution is $(0,1,2)$, obtained by observing that for $y=,z=$ one has $2(3,0,1)+1(-6,1,0)=(0,1,2)$.
Is it all correct?
|
Yes, all what you have done is correct !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Be $f:[0,1]\longrightarrow[0,1]$ a continuous function, prove that exists $x\in[0,1]$ so that $f(x)=x$ . Be $f:[0,1]\longrightarrow[0,1]$ a continuous function, prove that exists $x\in[0,1]$ so that $f(x)=x$ .
I am studying mathematical analysis in functions of one variable, and looking through my notes I can't find any theorem or proposition to help me prove it.
|
Hint.
Consider the function
$$g(x)=f(x)-x$$
and use the intermediate value theorem (this is possible because $f$ is continuous) to prove that there exists $x\in [0,1]$ such that $g(x)=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Fourier Series/ fourier transform What is the Fourier series of the following piece-wise function?
$$
f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
$$
|
Given f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
Its nth Fourier polynomial is $S_n(x)=\sum_{v=-n}^{n}\alpha_ve^{ivx}$, where $\alpha_v=\int_{-\pi}^{\pi}f(x)e^{ivx}dx=\int_{-1/2}^{1/2} \cos(3\pi x)e^{ivx}dx$. Notice that $\dfrac{e^{3\pi ix}+e^{-3\pi ix}}{2}=cos(3 \pi x)$, so $$\alpha_v=\int_{-1/2}^{1/2}\dfrac{e^{(v+3\pi) ix}+e^{(v-3\pi) ix}}{2}dx=\dfrac{e^{(v+3\pi)ix}}{2i(v+3\pi)}+\dfrac{e^{(v-3\pi)ix}}{2i(v-3\pi)}\mid_{-1/2}^{1/2}=\dfrac{e^{(v+3\pi)i/2}}{2i(v+3\pi)}+\dfrac{e^{(v-3\pi)i/2}}{2i(v-3\pi)}-\dfrac{e^{-(v+3\pi)i/2}}{2i(v+3\pi)}-\dfrac{e^{-(v-3\pi)i/2}}{2i(v-3\pi)}=\dfrac{\sin(\dfrac{v+3\pi}{2})}{2(v+3\pi)}+\dfrac{\sin(\dfrac{v-3\pi}{2})}{2(v-3\pi)}=\dfrac{3\pi\cos(\dfrac v2)}{v^2-9\pi^2}$$
Notice that if we regard $\alpha_v$ as a function of $v$, it is an even function. That is, $\alpha_v=\alpha_{-v} \;\forall |v|\le n$. Furthermore, $\alpha_0=-\dfrac1{3\pi}$. Hence $S_n(x)=-\dfrac1{3\pi}+3\pi\sum_{v=1}^{n}\dfrac{e^{ivx/2}+e^{3ivx/2}}{v^2-9\pi^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Lexicographic Product I found out that the lexicographic product of a complete graph of order m and a cycle of order n is singular iff n is divisible by 4. But I am having a hard time on how to prove it using eigenvalues. Can someone help me? Thank you!
|
The eigenvalues of the second graph (if regular) are also the eigenvalues of the product (lexicographic) graph. Further, if the second graph is $C_n$, then its eigenvalues are given by $2\cos\left(\frac{2\pi j}{n}\right),$ for $j=0, 1, \ldots, n-1$. Now if $n=4k$, for some $k$, then choosing $j=k$, we get $0$ as an eigenvalue of $C_n$. Hence the result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2120950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate a limit involving definite integral Evaluate the following limit:
$$\lim_{n \to \infty} \left[n - n^2 \int_{0}^{\pi/4}(\cos x - \sin x)^n dx\right]$$
I've tried to rewrite the expression as follows:
$$\lim_{n \to \infty} \left[n - n^2 \sqrt{2}^n \int_{0}^{\pi/4}\sin^n \left( \frac{\pi}{4} - x \right) dx\right]$$
However, this doesn't seem to help too much. Thank you!
|
Writing
$$ (\cos x - \sin x)^n = \color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}} \cdot \color{red}{ (\cos x + \sin x)(\cos x - \sin x)^{n-1}} $$
and applying integrating by parts, we have
$$ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx
= \frac{1}{n} - \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx. $$
Plugging this back,
$$ n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx = 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx. $$
Now from the observation
$$ n \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x)(\cos x - \sin x)^n \, dx
= \frac{n}{n+1}, $$
we can apply the usual approximation-to-the-identity argument to obtain
\begin{align*}
\lim_{n\to\infty} \left[ n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx \right]
&= \lim_{n\to\infty} 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx \\
&= \lim_{x \to 0^+} \frac{2}{(\sin x + \cos x)^3} \\
&= \color{red}{\boxed{2}}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Homogeneous ODE I am a bit confused with the definition and the solutions ways for homogeneous ODE.
I understand that they are 2 different definitions for homogeneous ODE
*
*The ODE is a function of $y$ and its derivatives such that $F(y,y',y'',...,y^{(n)})=0$
*the ODE has the same order of homogeneous such that $M(\lambda x,\lambda y)=\lambda^n M( x, y)$ and $N(\lambda x,\lambda y)=\lambda^n N( x, y)$ which is the same as saying that the ODE can be written in as the function $f(\frac{y}{x})$
If all of the above is correct so if we take $$y(x-y)dx-x^2dy=0$$
We can tell straight away that it is not exact as $M_{y}=x\neq N_{x}=2x$
So we have to find the order of homogeneous:
$M(\lambda x,\lambda y)=\lambda xy-(\lambda y)^2=\lambda xy-\lambda^2 y\neq \lambda^nM(x,y)$
$N(\lambda x,\lambda y)=(\lambda x)^2=\lambda^2x^2=\lambda^2 N(x,y)$
But on the other hand if we divide the ODE by $x^2$ we get:
$$(\frac{y}{x}-\frac{y^2}{x^2})dx-dy=0$$
So it is a non-homogeneous ODE?
|
setting $$y=xu$$ in your equation then we get with $$y'=u+xu'$$
$$u-u^2-u-xu'=0$$ and you will get $$-\frac{dx}{x}=\frac{du}{u^2}$$ which is easy to solve
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Finding a geometric interpretation I recently solved a question of complex numbers which was this:
$A\left( \frac{2}{\sqrt{3}} e^\frac{i\pi}{2}\right)$, $B\left( \frac{2}{\sqrt{3}} e^\frac{-i\pi}{6}\right)$, $C\left( \frac{2}{\sqrt{3}} e^\frac{-5i\pi}{6}\right)$ are the vertices of an equilateral triangle. If $P$ be a point on the incircle of the triangle, prove that $AP^2 + BP^2 + CP^2 = 5$.
My approach:
The point $P$ is given by $z = \frac{1}{\sqrt{3}}e^{i\theta}$, since the radius of the incircle of an equilateral triangle is half its circumradius. Then,
$$
AP² = |A - z|^2 \\
= (A - z)(A^* - z^*) \\
= AA^* - Az^* - A^*z + zz^* \\
= |A|² + |z|² - Az^* - A^*z \\
= \frac43 + \frac13 - Az^* - A^*z \\
= \frac53 - Az^* - A^*z.
$$
Now,
$$
AP^2 + BP^2 + CP^2 \\
= 3\times\frac53 - z^*(A+B+C) - z(A^*+B^*+C^*) \\
= 5
$$
since $A+B+C = A^*+B^*+C^* = 0$ because the position vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ are coplanar and are mutually separated by $120^\circ$.
This, I guess, is a pretty neat solution. But what I'm looking for is a more intuitive solution, rather a geometric interpretation. Does anyone know such an approach?
|
The sides of your triangle are $2$. E.g. the distance of $A$ and $B$ is the length of $$\frac{2}{\sqrt{3}}(e^{-\frac{i\pi}{6}}-e^{\frac{i\pi}{2}})=\frac{2}{\sqrt{3}}(\frac{\sqrt{3}}{2}-\frac{3}{2}i)=1-\sqrt{3}i.$$
We can calculate $PA^2+PB^2+PC^2$ using elementary methods, if $P$ is a point on the incircle of an equilateral triangle $ABC$ of side $a$.
In this case the radius of the incircle is $r=\frac{a\cdot\sqrt{3}}{6}$. Let $O$ be the incenter. Then $OA=OB=OC=\frac{a\sqrt{3}}{3}$. Let $\angle POC$ be $\varphi$. Then using the cosine theorem for the triangle $POC$ $$PC^2=r^2+OC^2-2r\cdot OC\cdot\cos\varphi = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos\varphi.$$
Since $\angle POA=120^{\circ}-\varphi$ and $\angle POB=120^{\circ}+\varphi$, similarly we get
$$PA^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}-\varphi)$$
and
$$PB^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}+\varphi).$$
Summarizing and using the addition theorem
$$PA^2+PB^2+PC^2=a^2(\frac{1}{4}+1-\frac{1}{3}\cos\varphi-\frac{1}{3}(-\frac{1}{2}\cos\varphi+\frac{\sqrt{3}}{2}\sin\varphi)-\frac{1}{3}(-\frac{1}{2}\cos\varphi-\frac{\sqrt{3}}{2}\sin\varphi))=\frac{5}{4}a^2.$$
Since in our case $a=2$, this gives $PA^2+PB^2+PC^2=5.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Evaluation of $\int_{0}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$
For $0<a,b<1.$
Evaluation of $$\int_{0}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{1-x}dx+\int_{1}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
|
You have to study what happens near $x=1$ and when $x\to \infty$, as I suppose you have notice.
Defining $f(x)=\frac{x^{a-1}-x^{b-1}}{1-x}$, $\lim_{x\to 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $\int_0^1 f(x)\ dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Projecting from volume to surface area of a sphere Does it make sense to, and is there a formula for the sphere whose surface consists of all the points inside another sphere?
I realize that I'm trying to compare cubed units with squared units here but I think the assumption I'm making is that these spheres are more like clumps of sand, so another way to ask this question is: given a ball of sand, how big is the spherical shell I could construct out of grains of sand?
Hopefully this isn't nonsense.. I have a slight fear that it could be.
|
The region between two spheres of radius $R_{1} < R_{2}$ has volume $\frac{4}{3}\pi (R_{2}^{3} - R_{1}^{3})$. If it were filled with sand grains of side length $r \ll R_{1}$, and if these grains were rearranged into a spherical shell of radius $R$ and thickness $r$, the shell would have volume (roughly, to within a factor of $2$ or so due to gaps between grains and other "local" irregularities) $4\pi R^{2}r \approx \frac{4}{3}\pi (R_{2}^{3} - R_{1}^{3})$, or radius
$$
R \approx \sqrt{\frac{R_{2}^{3} - R_{1}^{3}}{3r}}.
$$
For instance, a ball $1$ meter in radius ($R_{1} = 0$, $R_{2} = 1$) filled with $1$ millimeter sand grains ($r = 0.001$) would make a spherical shell about
$$
\sqrt{\frac{1}{0.003}} \approx 18.26\ \text{meters}
$$
in radius.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Can Atlas on S^1 only contain one chart? The smallest atlas for the $S^1=\{(x,y)\in \Bbb R^2|\,x^2+y^2=1\}$ must contain two charts. How to prove it?
My route is first to prove that it can only be homeomorphic to $
\Bbb R^1$ by invariant of domain (dimension). Then it restricts on mapping from $S^1$ into $\Bbb R$.
Secondly, I want to show such a homeomorphism cannot exist. But I got stuck, since I do not know how to partition a circle and I noticed that the topological structure on the circle can be defined by those arcs on the circle with one end be filled one end be empty. Then for the topology structure one should have a homeomophism eventually map$: (a,b]\rightarrow (c,d)$ which is not true. Then I do not know what is wrong.
|
If $U$ is any (nonempty) open subset of $\mathbb{R}$, then $U$ minus any point is disconnected. However, $S^1$ minus a point is always still connected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
How many distinct groups of 5 possible from 10 seniors and 6 freshman? A school is forming a group of 5. There are 10 freshman and 6 seniors, and the group must have at least 2 freshman and at least 1 senior. How many distinct groups are possible?
My approach (that I think is definitely flawed) -
This would be a combination problem because order does not matter. There has to be at least 2 freshmen and 1 senior so there are only 2 open spots left with a pool of 8 freshmen and 5 seniors, or 13 people. Would it just be C(13,2) or 13!/(2*11!)?
|
As you say the order doesn't matter. Then you have to think of all allowed groups, that is 2 freshmen 3 seniors, 3 freshmen 2 seniors, and 4 freshmen 1 senior.
2 freshmen 3 seniors:
$\dfrac{10 \cdot 9}{2!} \cdot \dfrac{6 \cdot 5 \cdot 4}{3!} = \dfrac{10!}{2!(10-2)!}\cdot \dfrac{6!}{3!(6-3)!} = 900 $ possible combinations
3 freshmen 2 seniors:
$\dfrac{10 \cdot 9 \cdot 8}{3!} \cdot \dfrac{6 \cdot 5}{2!} = \dfrac{10!}{3!(10-3)!}\cdot \dfrac{6!}{2!(6-2)!} = 1,800 $ possible combinations
4 freshmen 1 senior:
$\dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4!} \cdot 6 = \dfrac{10!}{(10-4)!4!}\cdot \dfrac{6!}{1!(6-1)!} = 1,260 $ possible combinations
TOTAL:
Therefore, in total, there are $3,960$ possibilities
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Show that a subring of a division ring must be a domain.
Show that a subring of a division ring must be a domain.
Let $S$ be a subring of $R$ and let $R$ be a division ring.
Can we just say that since every element has an inverse in $R$, then every element also has an inverse in $S$, then we can deduce that since for all elements in $S$ we can find a non zero inverse element?
Assume by contradiction that there is a zero divisor $k$ in $S$, such that $ks=0$ or $sk=0$ for all $s$ in $S$ well we can find $k$ inverse such that it is not true. Thus contradicting assumption, blah blah blah no zero divisor means $S$ is a domain and fin, drop the mic and such or am I completely off?
I realize I am wrong now because a subring of a division ring does not imply that there is an inverse, right? So I don't know where to go from here.
|
Hint $\ $ If $\,ax=0\,$ has unique root $\,x=0\,$ in $R\,$ then the same holds true in every subring.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2121863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Is L' context free language? Given context free grammar : G=(V,T,P,S).
and L'={a $\in$ (V $\cup$T)* | S$\Rightarrow$*a} when $\Sigma$=V $\cup$ T
Is L' context free language ?
I think no because for the grammer:
G=(S,{a,b},P,S) ,P={ S$\rightarrow$aSb|ab}
we will get that L' is:
L'={$a^nS^nb^n$}$\cup${$a^nb^n$} and L' is not context free language.
|
Given context-free grammar $G = (V, T, P, S)$, define $G' = (V, T \cup V', P \cup P', S)$, such that $V'$ has one element for each element of $V$,
$$ V' = \{ v' \mid v \in V \} \enspace, $$
and, for each element $v \in V$, $P'$ contains the production $v \rightarrow v'$. The language of $G'$ is context-free, and is isomorphic to $L'$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Dimension of an invariant subspace Let $V$ be a vector space of dimension $k$ and $n$ a positive integer. Let $a: V^n \to V$ be the map sending $(v_1,\ldots,v_n)$ to $v_1 + \cdots + v_n$. The symmetric group $S_n$ acts on $V^n$ by permuting the factors and the subspace $K := ker(a)$ is stable under this action.
What is the dimension of the vector spaces $( K^{\otimes l} )^{S_n}$, $(\wedge^lK)^{S_n}$, and $(Sym^lK)^{S_n}$ in terms of $k$, $n$, and $l$? Is there any reference which does systematically this kind of computations?
|
UPDATE: this answer responded to an earlier version of the question, see comments below.
Any element of $V^{\oplus n}$ (this includes elements of $K$) that is invariant under $S_n$ is of the form $(v, v, \ldots, v)$. Now for such an element to lie in $K$ we must have $v + v + \ldots + v = 0$ ($n$ terms). If the characteristic of the ground field over which you have your vectorspace $V$ is zero (i.e. if the ground field are the complex numbers) this can only happen when $v = 0$ and so $K^{S_n} = \{0\}$.
If the characteristic of the ground field is $p$ then $v + v + \ldots + v = 0$ is possible when $p|n$. However, in that case it is not only possible but also inevitable. It follows that for fields of characteristic $p$ the answer is:
$K^{S_n} = \{0\}$ if $p \not| n$ and
$K^{S_n} = (V^{\oplus n})^{S_n} = \{(v, \ldots, v) : v \in V \}$ if $p|n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Determinant of 5x5 matrix with letters I want to find the determinant of the following $5 \times 5$ matrix
\begin{bmatrix} 2 & 0 & p & 0 & q \\r & 2 & s & 1 & 2 \\0 & 0 & 1 & 0 & 0\\u & 1 & v & 1 & w\\ 0 & 0 & x & 0 & 2 \\\end{bmatrix}
I know I have to do some row/column operations and expansions but I really don't get it. Help is appreciated. Thanks in advance :)
|
Let given matrix is A.
\begin{bmatrix} 2 & 0 & p & 0 & q \\r & 2 & s & 1 & 2 \\0 & 0 & 1 & 0 & 0\\u & 1 & v & 1 & w\\ 0 & 0 & x & 0 & 2 \\\end{bmatrix}
Expand it by $A_{33}$
= 1 × \begin{bmatrix} 2 & 0 & 0 & q \\r & 2 & 1 & 2 \\u & 1 & 1 & w\\ 0 & 0 & 0 & 2 \\\end{bmatrix}
Expand it by $A_{44}$
= 1 × 2 × \begin{bmatrix} 2 & 0 & 0 \\r & 2 & 1 \\u & 1 & 1 \\\end{bmatrix}
Expand it by $A_{11}$
= 1 × 2 × (2 - 1)
= 2
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the value of $x$ that satisfy the equation: $3^{11}+3^{11}+3^{11} = 3^x$ I have this question:
$$3^{11}+3^{11}+3^{11} = 3^x$$
Find the value of $x$
|
More simply:
$$3^{11}+3^{11}+3^{11}=3\times 3^{11}=3^{12}$$
so the answer you are looking for is $x=12$ (because $x\mapsto 3^x$ is injective).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Show that $\mathbb{Q_p} $ is locally compact Suppose $\mathbb{Q_p} $ is the fraction field of $\mathbb{Z_p}$ ($p$-adic integers) i.e.
$$\mathbb{Q_p} = \left\lbrace\frac{x}{y} \space \bigg{|} \space x,y \in \mathbb{Z_p} , y\neq 0 \right\rbrace$$
Now with respect to the topology defined by $d(x,y) = e^{-v_p(x-y)}$ ($v_p$ is the $p$-adic valuation) , we need to show that $\mathbb{Q_p} $ is locally compact.
Any suggestions?
|
Every point has a fundamental system of neighborhoods given by $\{x+p^n\Bbb Z_p\}_{n\in\Bbb N}$ which are compact.
In essence this is just the fact that $p^n\Bbb Z_p$ is a fundamental system of compact neighborhoods of $0$: since we are in a vector space--really a topological group is enough, but not everyone is familiar with structures of that generality--we can translate these sets anywhere to form a compact (and open) neighborhood of any point.
If you have any trouble seeing this recall that open balls generate the topology and all open balls are of the form $x+p^n\Bbb Z_p$ for some $n\in\Bbb Z$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
2 Answers for Derivative of $x^2 \sin 1/x$ Derivative of $x^2\sin(1/x)$ :
if we use the derivative rules to take the derivative of this function we conclude that It is not defined at x =0,
but if we use the derivative definition (lim x -> a), we get this:
$$x\sin(1/x)$$
which has a derivative of 1 at 0.
Which is the answer? I mean obviously the formal defintion is right, but why does this happen?
|
The derivative by product rule and chain rule for $x\ne0$ gives us
$$\frac d{dx}x^2\sin(1/x)=2x\sin(1/x)-\cos(1/x)$$
While it is true that this is undefined at $x=0$, this is simply because we are not able to apply chain rule at $x=0$. At $x=0$, we must apply the limit definition:
$$\frac d{dx}\bigg|_{x=0}x^2\sin(1/x)=\lim_{h\to0}\frac{h^2\sin(1/h)}h=\lim_{h\to0}h\sin(1/h)=0$$
which follows from the squeeze theorem:
$$-h\le h\sin(1/h)\le h$$
One may note that it is not always the case that just because a function is differentiable everywhere, then the derivative will be continuous. Such behavior is called the smoothness of a function, and it depends on how many times you can differentiate a function and still have a continuous function.
Your specific example may be found on Wikipedia as the second example of smoothness
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
why $\int_\Omega S(t) div(w) = \int_\Omega div(w)$ ? where $S(t)$ is a heat semigroup with neumann condition In this work
https://www.dropbox.com/s/sygzebrr87ma99z/cao_2015_publicado.pdf?dl=0
page 1984, The author claims that
Let $w \in (C^{\infty}_0(\Omega))^N$ then $\int_{\Omega}S(t)div(w)=\int_{\Omega}div(w)$ See,
where $div$ is the divergence operator, S(t) is a heat semigroup with Neumann condition (Analytic semigroup) and $\Omega$ is a open set of $R^N$
I understand the right side (use divergence theorem and $w=0$ in boundary) and the continuation of the proof is clear for me. I know that if div (w) = constant the statement is true. I appreciate your help.
Thanks and sorry for my English
|
The divergence term is not important here. What is important is that solutions of the heat equation with homogeneous Neumann boundary conditions preserve total heat, so that
$$\int_\Omega S(t) f \, dx = \int_\Omega f \, dx$$
for all $t$. To show this, simply note that $u(x,t)=S(t)f(x)$ solves the heat equation $u_t = \Delta u$ with $\partial u/\partial \nu=0$ on the boundary and $u(x,0) = f(x)$. Integrating by parts we have
$$\frac{d}{dt} \int_\Omega u \,dx = \int_\Omega u_t \, dx = \int_\Omega \Delta u \, dx = \int_{\partial \Omega} \frac{\partial u}{\partial \nu} \, dS = 0.$$
So
$$\int_\Omega S(t) f \, dx = \int_\Omega u(x,t) \, dx = \int_\Omega u(x,0) \, dx = \int_\Omega f(x) \, dx.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
|
you start with a $2×2$ matrix and see the eigenvalues they are $4,6$.
Then see for $3×3$ matrix the eigenvalues are $9,6,-6$
for $4×4$ they are $14,6,-6,6.$.
Hence whenever the order is even the eigenvalues 6 exceeds the eigenvalue $-6$ by 1 in multiplicity.
and hence for even $n$ the snswer is
$det= (5(n-1)-1).6^{n/2}.(-6)^{\frac{n}{2}-1}$
In your case the answer is $494.(6)^{50}.(-6)^{49}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Non identity character of an Abelian group How does one show that for any non-trivial abelian group, there exists some non-trivial character?
After looking up for a while, the best solution I could find was Pontrayagin duality, which seems like too heavy a machinery for this problem.
|
For arbitrary abelian groups something non-trivial has to be used, it seems. For finite abelian groups, though, we have a very elementary Lemma:
Lemma: Let $G$ be a finite abelian group and $a,b$ distinct elements in $G$. Then there exists a character $\chi$ of $G$ such that $\chi(a)\neq \chi(b)$.
For a proof see here, Lemma $1.3.33$. So for $n=|G|\ge 2$ we see that there exists a non-trivial character of $G$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2122982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $\sum g_n$ converges uniformly , then does $(g_n)$ converge uniformly to $0$? If $\sum g_n$ converges uniformly , then does $(g_n)$ converge uniformly to $0$?
I think that it does basically from the fact that the convergence of numerical series implies that the numerical sequence of terms goes to $0$. But I feel I may be missing something.
Is the claim true?
UPDATE: Would $\sum x^n$ be a counterexample on some compact subset of $(0,1)$?
|
Note that
$$g_n(x) = \left(\sum_{k=1}^{n} g_k(x) - g(x)\right) - \left(\sum_{k=1}^{n-1} g_k(x) - g(x) \right)$$
where both series converge uniformly to the same function $g$ on some set $D$.
Using the triangle inequality we have
$$|g_n(x)| \leqslant \left|\sum_{k=1}^{n} g_k(x) - g(x)\right| + \left|\sum_{k=1}^{n-1} g_k(x) - g(x) \right|.$$
Given uniform convergence of the series, for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n > N$and for all $x \in D$, each term on the right-hand side is smaller than $\epsilon/2$.
Therefore, for all $n > N$ and for all $x \in D$ we have $|g_n(x) - 0| < \epsilon$ and $g_n \to 0$ uniformly.
The case of $\sum x^n$ for $x$ in some compact subset of $(0,1)$ is not a counterexample. If $D \subset (0,1)$ is compact, then there exists $b < 1$ such that $|x| \leqslant b$ and $|x^n| \leqslant b^n$ for all $x \in D$. Consequently $\sum b^n$ is a convergent geometric series and $\sum x^n$ converges uniformly by the Weirstrass test. Furthermore, $b^n \to 0$ which implies $x^n \to 0$ uniformly as $n \to \infty.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Beth number - justify existence of Beth omega I read through the answers to previous questions regarding Beth numbers and was unable to find the answer to my question, so I hope this isn't a duplicate.
I am studying the definition of Beth numbers, specifically:
$\beth_0:=\aleph_0$
$\beth_{\alpha+1}:=2^{\beth_\alpha}$
$\beth_\lambda:=\displaystyle\sup_{\alpha<\lambda}\beth_\alpha$ for limit ordinals $\lambda$
How is the third line of the definition justified? How do we know that the power set operation can be applied an infinite number of times? Is there a way to show rigorously that $\beth_\omega$ for example, exists? Would I need a version of the Axiom of Choice?
|
To expand on my comment, recall the following version of the Recursion Theorem reads, where $\phi$ is a formula in the language of set theory:
Suppose that $\forall x \exists! y\phi(s,y)$, and define $G(s)$ to be the unique $y$ such that $\phi(s,y)$ (note the use Replacement). Then we can define a formula $\psi$ for which the following are provable.
*
*$\forall x \exists! y \psi(x,y)$, so $\psi$ defines a functions $F$, where $F(x)$ is the $y$ such that $\psi(x,y)$.
*$\forall\xi\in ON [F(\xi)=G(F\upharpoonright\xi)]$.
where $F\upharpoonright\xi$ means $F$ restricted to $\xi$. For a proof and some comments on the statement of the theorem, see Kunen's Foundations of Mathematics, page 45.
To apply this theorem, we need only specify $G$. We let:
$$
G(s)=
\begin{cases}
\aleph_0 & \text{if $s=0$}\\
2^{s(\eta)} & \text{if $s$ is a function with dom$(s)=\eta+1$ a successor ordinal}\\
\sup_{\alpha<\lambda} s(\alpha) & \text{if $s$ is a function with dom$(s)=\lambda$ a limit ordinal}\\
0 & \text{otherwise}
\end{cases}
$$
You can now apply (2) in the theorem to this $G$ to see that $F$ is the desired $\beth$ function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Methods for choosing $u$ and $dv$ when integrating by parts? When doing integration by parts, how do you know which part should be $u$ ?
For example,
For the following:
$$\int x^2e^xdx$$
$u = x^2$?
However for:
$$\int \sqrt{x}\ln xdx$$
$u = \ln x$?
Is there a rule for which part should be $u$ ? As this is confusing.
|
My general principle is "Which bit gets nicer to work with when differentiated than when integrated?", which roughly lines up with the LIATE approach:
*
*The derivative of $\ln x$ is $\frac{1}{x}$, which will interact nicely with polynomial terms, whereas the integral is some weird rubbish on the order of $x \ln x$ which looks terrible.
*The derivatives of the inverse trigonometric functions tend to be rational functions or square roots of rational functions, which again tend to interact with polynomial terms decently, whereas the integrals are horrible things with logarithms in them.
*Algebraic stuff (polynomials) differentiates into other algebraic stuff, and if you're lucky it will get simpler as it goes (but not always).
*Trig and exponential functions tend to sit at about the same complexity no matter how much you differentiate or integrate them. So if you're left picking them as your $u$ it probably means you're going to be setting up a recurrence relation of some kind.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
}
|
$A=\emptyset $ if and only if $B = A \bigtriangleup B$ Is this true?
If $A$ and $B$ are sets, then $A=\emptyset $ if and only if $B = A \bigtriangleup B$.
If $A=\emptyset$ then $B=\emptyset$ too?
Could someone help me please?
|
$A \Delta B$ is the set of things that belong to exactly one of $A$ and $B$.
So, if $A = \emptyset$, then $A\Delta B = (A\backslash B) \cup (B\backslash A) = B$.
If $A\Delta B = B$, then $A\backslash B = \emptyset$, or $A\subset B$, but $B\backslash A = B$ then $A = \emptyset$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Basic Question on the real and imaginary part of a complex number I was introduced to the use of Re and Im to denote the real and imaginary parts of complex numbers today, but I'm still feeling unclear on how exactly they work.
I think understand them for a basic example: $$Re(a+ib)=a$$
But we were given this equality, which I'm unsure about:
$$Re\left(\frac{1}{z}\right)=Re\left(\frac{z}{{\lvert z\rvert}^2}\right)$$
If someone could help me out, maybe with a simple proof of this, it would be helpful!
|
Recall $|z|^2=z\overline{z}$ where $\overline{z}$ is the complex conjugate of $z$ (that is, if $z=a+ib$. then $\overline{z}=a-ib$).
Then use $$ \frac{1}{a+ib}=\frac{1}{a+ib}\frac{a-ib}{a-ib}=\frac{a-ib}{a^2+b^2}=\frac{a}{a^2+b^2}-i\frac{b}{a^2+b^2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
find the least value of $n$ such $2017^{{2017}^{2017}}~~|~n!$
Let $n$ be positive integer. Find the minimum of the $n$ such
$$2017^{{2017}^{2017}}~~|~n!$$
[Note that 2017 is a prime]
use this formula:
$$v_{2017}(n!)=\dfrac{n-S_{2017}(n)}{2016}$$so find least $n$ such $$ n-S_{2017}(n)\ge 2017^{2017}$$
$S_{p}(n)$ denotes the sum of the standard base-p digits of n, then How find this least $n?$
|
idea
If $n=2017^{k}$, then
$$\nu_{2017}(n!)=\sum_{j=1}^{k}\left\lfloor\frac{n}{2017^j}\right\rfloor=1+\dotsb +2017^{k-1}=\frac{2017^k-1}{2016}$$
Now perhaps we want $k$ such that
$$\frac{2017^k-1}{2016} \geq 2017^{2017}.$$
Note: this is just an idea which may require a bit more massaging to get the least value of $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\frac {x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}$$
$$=\frac {x+y+z-xz^2-xy^2+xy^2z^2-yz^2-yx^2+x^2yz^2-zy^2-zx^2+zx^2y^2}{(1-x^2)(1-y^2)(1-z^2)}$$.
I could not move on from here.Please help.
Thanks
|
Continuing from where you left, expressing terms of the numerator as:
$$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$
$$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$
$$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$
$$x+y+z=xyz $$
Now add everything and the result follows. Hope it helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2123892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos x) = 1-\cos x$
some help me., thanks
|
$$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^2}=\frac12$$
thus $1-\cos x\to\dfrac12x^2$ with substantiation
$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (\dfrac12x^2))))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots \dfrac12(\dfrac12x^2)^2)))}{x^{2^n}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{\left(\dfrac12\right)^{2^n-1}x^{2^n}}{x^{2^n}}$$
$$=\left(\dfrac12\right)^{2^n-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Why is $\int_0^1 \frac{t^x - 1}{\log(t)} dt$ finite for $x \in [0,1]$? I want to show that the following integral
$$\int_0^1 \frac{t^x - 1}{\log(t)} dt$$
is finite for all $x \in [0,1]$. But I struggle to come up with a good argument for that. Thanks in advance :)
|
IF we differentiate under the integral sign we have
\begin{array}
$ I(x)&=&\displaystyle\int_0^1\frac{t^x-1}{\log t}dt\\
I'(x)&=&\displaystyle\int_0^1\frac{t^x\log t}{\log t}dt\\
&=&\displaystyle\int_0^1 t^x dt\\
&=&\displaystyle\frac{1}{1+x}, \quad x\neq -1
\end{array}
Next integrating we have $I(x)=\log(1+x)+c$. Since $I(0)=0$, $c=0$, we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
why $x^{4} + y^{4} - 4 x y + 1$ graph differently in my textbook and mathematica? why this function $x^{4} + y^{4} - 4 x y + 1$ graph differently between mathematica and my textbook? did I make some error with my mathematica syntax?
|
If $f(x,y)=x^4+y^4-4xy+1$, then
$$
D_1f(x,y)=4x^3-4y\\
D_2f(x,y)=4y^3-4x
$$
which has critical points at $(0,0)$, $(-1,-1)$ and $(1,1)$.
The Hessian is $H(x,y)=16(9x^2y^2-1)$; since $H(0,0)<0$ and $H(-1,-1)=H(1,2)>0$, we have that $(0,0)$ is a local maximum, while $(-1,-1)$ and $(1,1)$ are local minimum.
Now $f(0,0)=0$ and $f(-1,-1)=f(1,1)=-1$. With your $z$-axis scaled to fit the interval $(0,300\,000)$, the difference between $0$ and $-1$ cannot be appreciated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Middle school problem - Percentage of students passing an exam This is a middle school problem that my nephew got, the teacher asked to solve it without using proportions:
The $\frac23$ of boys and the $\frac34$ of girls have passed an exam. Knowing that the
number of boys enrolled in the exam is three times the number of
girls, do we have enough information to calculate the percentage of
the group that have passed the exam? [$68.75\%$]
That's how I solved it:
$x =$ number of boys
$y =$ number of girls
We know that $x = 3y$, so the total of students that enrolled in the exam is $4y$.
To calculate the number of students that have passed the exam:$$\left(\frac23\right)\times3y + \left(\frac34\right)\times y = \left(\frac{11}4\right)\times y$$ This is the number of students that have passed the exam related to the number of girls. To calculate the percentage related to the number of boys $+$ girls: $$\frac{\left(\frac{11}4\right)\times y}{4y} = 0.6875 = 68.75\%$$ But to calculate it I have used proportions. Is it even possible to get $68.75\%$ without using proportions?
|
Obviously you can't solve the problem without using facts about proportions, since all the input data as well as the answer are themselves proportions.
But here's an attempt to do it without performing any division operation,
which is the usual way of getting a proportion.
Instead, we use a probabilistic argument.
Let us select a student at random from all students who took the test,
with uniform probability of choosing each student.
The percentage probability that the randomly selected student has passed the exam is the same as the percentage of students who passed.
Let $A$ be the probability that the student passed, $B$ the probability that the student was a boy, $G$ the probability that the student was a girl.
Then by the law of total probability,
$$
P(A) = P(A\mid B)P(B) + P(A\mid G)P(G).
$$
From the problem statement we can "guess" that the proportion of girls in the population is $0.25$ (and confirm that this is correct, because
$3\times 0.25 + 0.25 = 1$),
and we assume $P(B) = 1 - P(G),$ so we can deduce the probability values
\begin{align}
P(B) &= 0.75, \\
P(G) &= 0.25, \\
P(A\mid B) &= \frac23, \\
P(A\mid G) &= \frac34. \\
\end{align}
Therefore
$$
P(A) = \frac23 \cdot 0.75 + \frac34 \cdot 0.25
= 0.5 + 0.75 \cdot 0.25 = 0.6875.
$$
To get a percentage, we multiply by $100\%.$
But this has used a lot of knowledge about proportions (for example,
how we know that the solution to $3P(G) + P(G) = 1$ is unique),
and it has sneakily performed some division to conclude that $\frac34=0.75$
(and also to "guess" the proportion of girls, we presumably mentally
divide $1$ by $4$).
So I think the teacher's secret almost surely involves some sneaky way
of hiding the operations that produce proportions.
On the other hand, to deduce the direct answer to the question
(which is, "Yes, there is enough information"),
we only need to know how proportions work;
we do not actually need to evaluate any specific proportions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Identifying random variables from moment generating functions and using characteristic functions Suppose that $X$ and $Y$ have moment generating functions $M_X(t),M_Y(t)$ respectively and $U$ has uniform distribution on $[0,1]$.
What is a random variable that is a function of X,Y and U such that it's moment generating function is 1) $\int_0^1 M_X(tu) \, du$ and 2) $\frac{M_X(t)+M_Y(t)}2$?
I know the linear properties of moment generating functions, such as $M_U(t) = \int_0^1 e^{ut}f(u)\,du$ and the properties $M_{X+Y}(t) = M_X(t)M_Y(t)$ as well as $M_{aX+b}(t) = e^{bt} M_X(at)$ but don't know how other properties would apply. Would the product of UV yield a moment generating function corresponding to $\int_0^1 M_X(tu) \, du$.
Any help would be really appreciated!
|
Suppose $W=\begin{cases} 1 & \text{with probability } 1/2, \\ 0 & \text{with probability } 1/2, \end{cases}$
$\vphantom{\dfrac11}$and $Z=WX + (1-W)Y,$ so that $Z=X$ if $W=1$ and $Z=Y$ if $W=0.$ Then
$$
M_Z(t) = \operatorname{E}(e^{tZ}) = \operatorname{E}(\operatorname{E}( e^{tZ}\mid W)).
$$
Oberserve that
$$
M_{Z\,\mid\, W=1}(t) = \operatorname{E}(e^{tZ}\mid W=1) = \operatorname{E}(e^{tX}) \\ M_{X\,\mid\,W=0}(t) = \operatorname{E}(e^{tZ}\mid W=0) = \operatorname{E}(e^{tY})
$$
so
$$
\operatorname{E}(e^{tZ}\mid W) = \begin{cases} \operatorname{E}(e^{tX}) & \text{with probability } 1/2, \\ \operatorname{E}(e^{tY}) & \text{with probability } 1/2. \end{cases}
$$
Therefore
$$
\operatorname{E}(\operatorname{E}(e^{tZ}\mid W)) = \frac 1 2 \operatorname{E}(e^{tX}) + \frac 1 2 \operatorname{E}(e^{tY}) = \frac{M_X(t) + M_Y(t)} 2,
$$
so
$$
M_Z(t) = \frac{M_X(t) + M_Y(t)} 2.
$$
The question involving $\displaystyle \int_0^1 M_X(tu)\,du$ is a bit subtler.
Note that
$$
M_X(tu)=\operatorname{E}(e^{tXu}) = \operatorname{E}(e^{tXU} \mid U=u),
$$
so
$$
M_X(tU) = \operatorname{E}(e^{tXU} \mid U).
$$
It follows that
\begin{align}
\int_0^1 M_X(tu) \,du = \operatorname{E}(M_X(tU)) = \operatorname{E}(\operatorname{E}(e^{tXU} \mid U)) = \operatorname{E}(e^{tXU}) = M_{XU}(t).
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$. Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$.
Since $S(999)=27 $ it is clear that the number of digits $n>3.$ Let $x_i$ be digits then we have two equations
\begin{cases}
x_1+x_2+\cdots+x_n=27=5 \mod 11,\\
x_1+x_3+x_5+\cdots=x_2+x_4+\cdots \mod 11.
\end{cases}
It follows that
$2(x_2+x_4+\cdots+)=5 \mod 11$
or
$$
x_2+x_4+\cdots = 8 \mod 11.
$$
For the case $n=5$ it reduced to $x_2+x_4=8$ and I manage to pick up the solution $10989.$
Is there better solution?
|
Since $\sum x_i=27$, $\sum (-1)^{i-1}x_i=27-2(x_2+x_4+\cdots)$ must be odd and divisible by $11$, and hence either $11$ or $-11$ (it must be between $-27$ and $27$.)
If $\sum (-1)^{i-1}x_i=11$ then $x_1+x_3+\cdots = 19, x_2+x_4+\cdots = 8$. The only way to get $19$ is with three digits, $(x_1,x_3,x_5)=(1,9,9)$ yielding the smallest values in lexicographical order. Then $x_2+x_4=8$ has $(0,8)$ for the smallest lexicographical order.
So you get $10989$.
If $\sum (-1)^{i-1}x_i=-11$, then $x_2+x_4+\cdots = 19$ and thus that there must be at least $6$ digits.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Show that $I$ is an interval iff any function $f:I\rightarrow \{0,1\}$ is continuous. Let $I\in R$ be a nonemplty set. Show that $I$ is an interval iff any function $f:I\rightarrow \{0,1\}$ is continuous.
I've no idea how to handle this problem. Thanks in advvance to anyone who comes up with a step - solution.
|
Can't be true.
Let $I$ be any set, interval or not, then $f(x) = 0$ is a continuous function whether or not $I$ is an interval. And $I$ = $A\cup B$ where $A$ and $B$ are separated then if $f(A) = 0$ and $f(B) = 1$, $f$ is continuous.
Likewise on an interval one can easily find an non-continuous function to $\{0,1\}$. Simply map any arbitrary points to $0$ and all the rest to $1$. Say, $I = [0,1]; f(1) = 1;f(x) = 0$ if $x<1$ .
As egreg puts in the comments, a true statement would be $I$ is an interval if and only if every continuous function $f:I\rightarrow \{0,1\}$ is constant.
Proof. (I'm going to use notation $<a,b>$ to be an interval without distinguishing whether is open, closed or clopen.)
If $I= <a,b>$ is an interval and $f:I \rightarrow \{0,1\}$ be a continuous function. Note: for any $x,y \in R$ that $|f(x) - f(y)|$ is either $0$ or $1$. There is no other option.
Let $m = \frac {a+b}2 \in I$. $x$ is the mid point of $I$. Let $f(m) = k$. Let $K = \{x \in [<a,m]| f(x) \ne f(m)\}$. If $K$ is non-empty, let $d = \sup K$. Then for every neighbor of $d$ you will have a $y \le d: y \in K$ and a $w > d: w\not \in K$ so $f(y)\ne f(w)$ and $|f(y) - f(w)| = 1$ which contradicts that $f$ is continuous. (Continuity says for $\epsilon > 0$ there is a neighborhood, $N$, of $d$ where $x,y \in N$ then $|f(x) - f(y)| <\epsilon$ which is not the case for $\epsilon \le 1$.)
So $K$ is empty. So $f$ is constant when restricted to $<a,m]$.
A similar argument shows $f$ is constant when restricted to $[m, b>$. So $f$ is constant on $<a,b> = I$.
If $I$ is not an interval and $I$ has two or more points, then there exists an $m \not \in I$ so that there is an $a < m$ with $a \in I$ and a $b > m$ with $b \in I$. Let $f(x) = 0$ for all $x \in I$ where $x < m$ and let $f(x) = 1$ for all $x \in I$ where $x > m$. This function is continuous and non-constant.
... because ... for any $x \in I$ and any $\epsilon > 0$. Let $\delta = |x-m|$. If $y \in I$ and $|x - y| < \delta$ then either $x,y$ are both less than $m$ or $x,y$ are both greater then $m$. Either way, then $f(x) = f(y)$ and $|f(x)-f(y)| = 0 < \epsilon$.
Hmmm.... if $I$ has one point or $I$ is empty, then all functions from $I$ to $\{0,1\}$ are continuous and constant. But single pointed sets and empty sets are trivial intervals.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determining function is onto a. Consider the set $ℝ^+ = \{x∈ℝ|x>0\}$ together. Let $f:ℝ^+→ℝ^+$ be the function given by $f(x) = x^2.$ Is $f$ onto?
b. Consider the set $ℚ^+ = \{x∈ℚ|x>0\}$ together. Let $f:ℚ^+→ℚ^+$ be the function given by $f(x) = x^2.$ Is $f$ onto?
Workings:
a. Let $y \in \mathbb{R}^+$. Let $x = \sqrt{y}$.
Then we have
$f(x)=x^2 = (\sqrt y)^2 = y$
Therefore $f$ is onto.
What I am wondering is. If the same would follow b. Any help will be appreciated.
|
Your complete proof for (a) should be as follows (the red is the bit you left out).
Let $y\in{\Bbb R}^+$. Let $x=\sqrt y$. Then $\color{red}{x\in{\Bbb R}^+}$ and we have
$$f(x)=x^2=(\sqrt y)^2=y\ .$$
Therefore f is onto.
The corresponding proof for (b) would be:
Let $y\in{\Bbb Q}^+$. Let $x=\sqrt y$. Then $\color{red}{x\in{\Bbb Q}^+}$ and we have
$$f(x)=x^2=(\sqrt y)^2=y\ .$$
Therefore f is onto.
Can you decide whether or not this is correct?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2124955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determining Limits
The Question above states determine the limits, for problem a) I evaluated for when lim x->2 instead of 4. This was marked wrong im simply asking for a clarification on what the question is asking me to do.
|
[2, 3, 6] might be the number of points given for each correct solution. In any case, it doesn't mean that you should calculate $\lim_{x\to 2}$ instead of $\lim_{x\to 4}$.
So what you should do for a) is
$$\lim_{x\to 4}\sqrt{4x+\sqrt x}=\sqrt{4\cdot 4+\sqrt 4}=\sqrt{18}=3\sqrt 2$$
The first answer is 2 points because you don't have any indeterminate form, just plug in $x=4$. The other answers value more points because there is more work involved to handle the indeterminate forms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
On a characterization of the material conditional $\to$ I understand the usual motivation behind the truth table for the logical connective $\to$.
However, I would like to know if there is a more fundamental reason for that truth table. Something that would have to do with arguments and validity.
A.G.Hamilton writes in Logic for Mathematicians that "the significance of the conditional statement $A\to B$ is that its truth enables the truth of $B$ to be inferred from the truth of $A$, and nothing in particular to be inferred from the falsity of $A$".
Question: What does Hamilton mean precisely? Something like $\to$ is the only binary truth function such that
*
*The argument form $(p\to q),p;\therefore q$ is valid
*The argument form $(p\to q),\sim p;\mathcal{A}$ is invalid unless $\mathcal{A}$ is a statement form logically equivalent to $\sim p$ ?
|
From the statements of $\{A{\to}B, A\}$ we can infer something new; that $B$ is true.
However, from the statements of $\{A{\to}B, \neg A\}$ we cannot infer anything new.
That is all.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Triple integral bounds
*
*Let $W$ be the region bounded by the planes
$x = 0$, $y = 0$, $z = 0$, $x + y = 1$, and $z = x + y$.
*$(x^2 + y^2 + z^2)\, \mathrm dx\, \mathrm dy\, \mathrm dz$; $W$ is the region bounded by $x + y + z = a$ (where $a > 0$), $x = 0$, $y = 0$, and $z = 0$.
$x,y,z$ being $0$ is throwing me off because I'm not sure how to graph it and get a bounded area; it seems like it would be infinite to me. What would the bounds for the triple integrals be?
|
1) give you nothing to integrate. But we can still find the limits
$z = 0$ lower limit for $z$
$z = x+y$ upper limit for $z$
$y = 0$ lower limit for $y$
$y = 1-x$ upper limit for $y$
$x = 0$ lower limit for $x$
$x = 1$ upper limit for $x$
For the last one, because that is where $y = 0$ intersects $x+y = 1$
2)
$\int_0^a\int_0^{a-x}\int_0^{a-x-y} x^2 + y^2 + z^2 \;dz\;dy\;dx$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
First order logic to English statement? Assume
$A(x) = x$ is an American.
$D(y) = y$ is a dream.
$H(x,y) = x$ has $y$.
Then, Convert below first order logic to English statements :
*
*$∀x ∃y \left ( A(x)\rightarrow D(y) ∧ H(x,y) \right )$
I tried to translate this as "Every American has his own set of dreams".
*
*$∀x ∃y \left ( A(x) ∧ D(y) \rightarrow H(x,y) \right )$
Not getting how is this pronounced ?
For this, I guess it is like "For all x if x is an American and there exists some y, such that y is a dream then x has y".
How much both of them are correct ?
|
The first one is pretty close. I would say "Every American has a dream," because we only know that there exists one $y$.
The second one is weird. If there exists any y that is not a dream, then it is vacuously true. Literally it's something like "For every American there is a thing y such that if y is a dream, then the American has that dream." But y could just not be a dream.
Edit: I think the second one is logically equivalent to "Either there exists something that is not a dream, or every American has a dream."
Formally, ($A(x) \wedge D(Y) \implies H(x,y)$) is equivalent to $(\neg(A(x) \wedge D(y)) \vee H(x,y))$, and subsequently $(\neg A(x) \vee \neg D(y) \vee H(x,y))$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$. Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$.
Using Euclidean algorithm:
$$2^{2004}-1=4(2^{2002}-1)+3$$
$$2^{2002}-1=x\cdot 3+y$$
The solution manual says that $2^{2002}-1$has the remainder $0$ when divided by $3$, that is $y=0$ so GCD is $3$. But how do I find that remainder?
|
We have,
$2^{2002} - 1$
$= (2^4)^{500}.2^2 - 1$
Now $2^4 \equiv 1 (\mod 3)$
From above,
$= (1)^{500} .2^2 - 1$
$= 1.2^2 - 1$
$= 4 - 1 = 3$
Divisible by 3.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solution of a equation .
Solve the equation:
$\frac{a}{ax-1}$+$\frac{b}{bx-1}$=$a+b$
It is my problem.I simply evaluate the equation and I found $ab(a+b)x^2-[(a+b)^2+2ab]x+2(a+b)=0$.I use Sridhar Achharya's theorem but it became complicate . Somebody please help me to solve the equation.
|
$ab(a+b)x^2-[(a+b)^2+2ab]x+2(a+b)=0$
$(abx-a-b)(ax+bx-2)=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Understanding an example of equivalence relation On $\mathbb{N}\times\mathbb{N}$, Let $\left( a,b\right) \equiv \left( c,d\right) \Leftrightarrow a+d=c+b$.
Show that it is equivalence relation. Find equivalence of class of this.
My answer.
For all $a,b\in\mathbb{N}$, we have $a+b=a+b$. Clear. So, The relation is reflexivity.
For all $a,b,c,d\in\mathbb{N}$, we assume $a+b=c+d$. Hence, also we know $c+d=a+b$. Thus, the relation is symetric.
For all $a,b,c,d,e,f\in\mathbb{N}$, we assume $a+d=c+b$ and $c+f=e+d$. We will show that $a+f=e+b$. Sum of both of sides $a+b=c+d$ and $c+f=e+d$, then we obtain $(a+f)+(c+d)=(c+d)+(e+b)$. Hence, by the cancelation, we obtain $(a+f)=(e+b)$. Thus, The relation $R$ is transitivity.
Therefore, the relation $R$ is equivalence relation. Can you check my answer,so, what is the equivalence classes of this?
|
Note that $a+d=c+b$ is true iff $a-b=c-d$. Let us interpret a pair such as $(a,b)$ as: $a$ is the amount of income this month, $b$ as the expenditure this month for a single person. We say two persons are equivalent if the money they saved this month (that is income minus the expenses) is the same for these two persons.
Now have fun solving your problem without using plus and minus signs or brackets.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Generalise this. I'm reading R. Courant & H. Robbins' "What is Mathematics: An Elementary Approach to Ideas and Methods" for fun. I'm on page $60$ and $61$ of the second edition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.
Exercise $3$: Prove that $\phi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Try to make up similar and more general examples.
My Attempt:
Lemma: The number $\sqrt{2}+\sqrt{3}$ is irrational. (This is part of Exercise 2.)
Proof: Suppose $\sqrt{2}+\sqrt{3}=r$ is rational. Then
$$\begin{align}
2&=(r-\sqrt{3})^2 \\
&=r^2-2r\sqrt{3}+3
\end{align}$$ is rational, so that
$$\sqrt{3}=\frac{r^2+1}{2r}$$ is rational, a contradiction. $\square$
Let $\psi=\sqrt{2}+\sqrt{3}$. Then, considering $\phi$,
$$\begin{align}
5&=(\phi-\psi)^2 \\
&=\phi^2-2\psi\phi+5+2\sqrt{6}.
\end{align}$$
I don't know what else to do from here.
My plan is/was to use the Lemma above as the focus for a contradiction, showing $\psi$ is rational somehow.
Please help :)
Thoughts:
The "try to make up similar and more general examples" bit is a little vague.
The question is not answered here as far as I can tell.
|
Using the straightforward fact that the sum of two (or more) algebraic integers is an algebraic integer (see here), one has the following very short argument: since $\sqrt{2},\sqrt{3},\sqrt{5}$ are algebraic integers, their sum $A=\sqrt{2}+\sqrt{3}+\sqrt{5}$ must be also. Hence if $A$ is rational, it must be an integer. However, as a rudimentary estimate of the three square roots shows, we have $5 < A < 6$: contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
}
|
Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt,
$a,b,c$ are real and positive numbers, then
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$$a^2+b^2+ c^2-2(a+b+c)+3\ge 0$$.
I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method.
|
By AM-GM $$6(a^2+b^2+c^2)=\sum_{cyc}(4a^2+b^2+c^2)\geq6\sum_{cyc}\sqrt[6]{a^8b^2c^2}=6(a+b+c)$$
and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2125952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.
|
Yes, this holds in general. Recall that an extension $F/\Bbb Q$ is algebraic iff for each $\alpha\in F$ we have
$$[\Bbb Q(\alpha):\Bbb Q]=\dim_{\Bbb Q} \Bbb Q(\alpha)< \infty$$
that is, if every element in it generates a finite extension. Similarly any finite extension is algebraic since if $\beta\in F$ then $\Bbb Q(\beta)\subseteq F\implies [\Bbb Q(\beta):\Bbb Q]\le [F:\Bbb Q]$. For a nested radical expression, this is easy then, take the most nested radical, adjoin the needed roots and proceed by induction.
In your first example $\alpha=\sqrt{11+2\sqrt 7}$, then we see that the field $\Bbb Q(\sqrt 7)$ has a polynomial for this which is ${1\over 2}(x^2-11)-\sqrt 7 = 0$ so the splitting field of this polynomial, which has degree at most $2$ over $\Bbb Q(\sqrt 7)$, and therefore degree at most $4$ over $\Bbb Q$, is algebraic, hence $\alpha$ is algebraic.
For your other example and more generally any example, you do similarly, since there are only finitely many nested radicals each time, you get splitting fields of finite degree, implying all such expressions are algebraic.
Just to work out the more complicated example, $\alpha = \sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$ first adjoin $\sqrt 6$, and let $F_1=\Bbb Q(\sqrt 6)$. Then look at the splitting field, $F_2$, of $(x/3)^7-5-3\sqrt 6$ over $\Bbb Q(\sqrt 6)$. Then the splitting field, $F_3$ of $x^3-2-3\sqrt[7]{5+3\sqrt 6}$ over $F_2$, finally let $F_4=F_3(\sqrt[9]{2})$. Because each step had finite degree, $[F_4:\Bbb Q]<\infty$ showing $\alpha$ is algebraic.
One more since we're on a roll, look at $\sqrt{\sqrt{23}+\sqrt 5}$ this gives you a slight problem at first it seems since $\sqrt{23}$ and $\sqrt{5}$ are both the most nested radical, but as usual $[F(\alpha,\beta): E]\le [F(\alpha):E][F(\beta):E]$ and we can reduce to the simpler case by the classical result that sums and products of algebraic numbers are algebraic (proved in the same way with finite degrees).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 5
}
|
Given $K \subset L^2(\mathbb{R})$, how to determine $K^{\perp}$? Problem: Define $K \subset L^2(\mathbb{R})$ given by $$ K = \left\{f \in L^2(\mathbb{R}) \mid f(-x) = 2f(x) \ \text{for almost all} \ x \geq 0 \right\}. $$
I need to give an explicit expression for $K^{\perp}$.
My attempt: By definition $$K^{\perp} = \left\{f \in L^2(\mathbb{R}) \mid \langle f, g \rangle = 0 \ \text{for all} \ g \in K\right\}.$$ So I take $f \in L^2(\mathbb{R})$ and ask when we have $\langle f, g \rangle = 0$. Now, $$\langle f, g \rangle = \int_{\mathbb{R}} f(x) \overline{g(x)} dx = \int_{-\infty}^{0} f(x) \overline{g(x)} dx + \int_0^{\infty} f(x) \overline{g(x)} dx. $$ In the first integral, I do the substitution $x \mapsto -y$. The first integral is then $$ - \int_{\infty}^0 f(-y) \overline{g(-y)} dy = \int_0^{\infty} f(-x) \overline{g(-x)} dx. $$ But $g \in K$, so $g(-x) = 2g(x)$ for almost all $x \geq 0$. So together with the other integral I have $$\langle f, g \rangle = 2 \int_0^{\infty} f(-x) \overline{g(x)} dx + \int_0^{\infty} f(x) \overline{g(x)} dx = \int_{0}^{\infty} \big[ 2f(-x) + f(x) \big] \overline{g(x)} dx = 0 $$ if $f(x) = -2 f(-x) $ for almost all $x \geq 0$. So I would say $$ K^{\perp} = \left\{f \in L^2(\mathbb{R}) \mid f(x) = -2 f(-x) \ \text{for almost all} \ x \geq 0 \right\}. $$ Is this reasoning correct, and the answer? Thank you in advance.
|
$g \in K^\perp$ iff
$$
0= \int_{0}^{\infty}g(t)f(t)dt+\int_{-\infty}^{0}2g(t)f(-t)dt \\
=\int_{0}^{\infty}\{g(t)+2g(-t)\}f(t)dt.
$$
This must hold for all $f \in K$. However $f\in K$ is an arbitrary $L^2$ function on $(0,\infty)$, which forces $g(-t)=-\frac{1}{2}g(t)$ for $t > 0$. This checks because
$$
(fg)(-t)=-(fg)(t) \mbox{ for } t > 0, \\
\implies \int_{-\infty}^{\infty}(fg)(t)dt = 0.
$$
Working in the complex space doesn't change the basic argument.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope there'a an exploitable link between them. They fall easily under Muirhead, yet I cannot figure out how to prove them using the Rearrangement Inequality.
Any hints greatly appreciated.
|
by AM-GM we get $$\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2}{6}\geq \sqrt[6]{a^{12}b^{12}c^{12}}=a^2b^2c^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
}
|
Calculate $\int \frac{1}{\sqrt{4-x^2}}dx$ Calculate $$\int \dfrac{1}{\sqrt{4-x^2}}dx$$
Suppose that I only know regular substitution, not trig.
I tried to get help from an integral calculator, and what they did was:
$$\text{Let u = $\frac{x}{2}$} \to\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{2}$$
Then the integral became:
$$={\displaystyle\int}\dfrac{1}{\sqrt{1-u^2}}\,\mathrm{d}u = \arcsin(u) = \arcsin(\frac{x}{2})$$
And I'm not sure how they accomplished this, where did the 4 go? I understand the arcsin part but not sure how they got rid of the 4? Also how did they know to substitute $\frac{x}{2}$? It doesn't seem very obvious to me.
|
$$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4-(2u)^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4(1-u^2)}}=\int \frac{2 \ \text{d}u}{2\sqrt{1-u^2}}=\int \frac{ \text{d}u}{\sqrt{1-u^2}} $$
Why especially this substitution: Notice that
$$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{\text{d}x}{\sqrt{4\left(1-\frac14x^2\right)}}=\int \frac{\text{d}x}{2\sqrt{1-\left(\frac{x}{2} \right)^2}}$$
so you can see that it is quite nice to substitute $u=\frac{x}{2}$; we get a function $\frac{1}{\sqrt{1-u^2}}$ and we already know the integral to this one.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Decomposition Theorem for Posets There is in module theory the following (krull-shmidt) theorem:
If $(M_i)_{i\in I}$ and $(N_j)_{j\in J}$ are two families of simple module such that $\bigoplus\limits_{i\in I} M_i \simeq \bigoplus\limits_{j\in J} M_j$, then there exists a bijection $\sigma:I\rightarrow J$ such that $M_i \simeq N_{\sigma(i)}$.
Is there a similar kind of theorem for partially ordered sets? More precisely, is there a class of 'simple' partial orders such that if $(P_i)_{i\in I}$ and $(Q_j)_{j\in J}$ are 'simple' orders such that
$\prod\limits_i P_i \simeq \prod\limits_j Q_j$, then there exists a bijection $\sigma:I \rightarrow J$ such that $P_i \simeq Q_{\sigma(i)}$.
|
The question concerns identifying a class of partial orders whose products have unique factorizations. I will give positive and negative partial answers, both (co)authored by Hashimoto.
The positive answer may be found in
Hashimoto, Junji
On direct product decomposition of partially ordered sets.
Ann. of Math. (2) 54, (1951). 315-318.
Here it is proved that
any two direct product decompositions of a connected poset have a common refinement. Arbitrary products of connected orders are not always connected, but they are sometimes connected. Thus, one can almost take the word simple in the original question to mean connected and directly indecomposable. For example, you can take simple to mean connected and directly indecomposable if you are only concerned about finite products of posets.
The negative result (from an earlier paper of Hashimoto and Nakayama) concerns the nonconnected case. Hashimoto and Nakayama point out that $(1+x)(1+x^2+x^4)=(1+x^3)(1+x+x^2)$, and translate this fact into a statement about posets by letting $1={\bf 1}$ stand for the $1$-element chain, $x={\bf 2}$ stand for the $2$-element chain, multiplication stand for cartesian product, and sum stand for disjoint (parallel) sum.
In other words, there is an isomorphism $A\times B\cong C\times D$ where $A = {\bf 1}+{\bf 2}$ is the disjoint sum of a $1$-element chain and a $2$-element chain, $B={\bf 1}+{\bf 2}^2+{\bf 2}^4$,
$C = {\bf 1}+{\bf 2}^3$, and $D = {\bf 1}+{\bf 2}+{\bf 2}^2$. Hashimoto and Nakayama argue that these factorizations have no common refinement.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
How to evaluate an infinite sum involving remainders I've been trying to evaluate the sum
$$\sum_{k=0}^\infty \frac{m^k\bmod n}{m^k}$$
where $m$ and $n$ are positive integers greater than $1$ and $a\bmod b$ is the remainder when $a$ is divided by $b$. This came up in a combinatorics problem I was doing, and I know how to evaluate it given $m$ and $n$ (the numerators repeat, so it ends up just being geometric), but I'm not sure how to evaluate it generally.
Any ideas?
|
The numerators must repeat because only finitely many possible remainders exist. Suppose the repeating part starts after the first $K$ terms, so you have
\begin{align}
& \sum_{k=1}^K \frac{m^k\bmod n}{m^k} + \sum_{k=K+1}^\infty \frac{m^k\bmod n}{m^k} \\[10pt]
= {} & \sum_{k=1}^K + \sum_{k=K+1}^{K+R} + \sum_{k=K+R+1}^{K+2R} + \sum_{k=K+2R+1}^{K+3R} + \cdots \\
& \text{where $R$ is the length of the repeating part} \\[10pt]
= {} & \sum_{k=1}^K + \left(\sum_{k=K+1}^{K+R}\right)\left( 1 + \frac 1 {m^R} + \frac 1 {m^{2R}} + \frac 1 {m^{3R}} + \cdots \right) \\[10pt]
= {} & \sum_{k=1}^K + \left(\sum_{k=K+1}^{K+R}\right)\left( \frac 1 {1- \dfrac 1 {m^R}} \right)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
In a regular 12-sided polygon, how many triangles can be formed using vertices of the polygon that do not share a side with the polygon? I think is is 12 choose 3 minus 12, because that is the number of triangles that share a side with the polygon. Is that correct?
|
You're not quite right. For each side of the 12-gon, there are 12-4=8 non-adjacent vertices which you can use to form a triangle, so there are 12*8=96 such triangles. There are also those triangles which share two sides with the 12-gon, of which there are 12 (one for each vertex of the 12-gon--choose the two sides adjacent to the vertex). So there are 108 total excluded triangles, and thus the answer is $\binom{12}{3}-108 = 220-108 = 112$.
This figure shows two of the 8 possible triangles sharing the right edge of the 12-gon:
Here's another way to get the answer:
For a triangle to not share an edge with the 12-gon, between each pair of triangle vertices there must be another vertex of the 12-gon. Number the vertices of the 12-gon 1 through 12. Consider two cases for a valid triangle:
*
*Vertex 1 is not on the triangle. In this case, consider getting the 12-gon by starting with a 9-gon and putting 3 triangle vertices between the existing 9 vertices. There are $\binom{9}{3} = 84$ ways to do this.
*Vertex 1 is on the triangle. In this case, consider getting the 12-gon by starting with a 10-gon consisting of vertex 1 and the 9 other vertices which are not triangle vertices. Then the remaining 2 triangle vertices can go between vertices 2 & 3, 3 & 4, ..., 9 & 10, so there are 8 places to put them. There are $\binom{8}{2} = 28$ ways to do this.
Any valid triangle falls into case 1 or 2, so there are $84+28 = 112$ possible triangles.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Geometry Derivative Show that the normals to the curve $y=4x^2$ from points the same distance on either side of the y-axis intersect on the y-axis.
My attempt, I differentiated so it becomes $8x$, but I don't understand the question. Can anyone explain it to me ? Thanks in advance:
|
That parabola is symmetric with respect to $y$-axis, hence the normal lines issued from two symmetric points (as in your case) are corresponding one another through a reflection across $y$-axis. Unless they are both parallel (which is not possible if they are issued from different points) they must therefore meet on the $y$-axis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.
How do I prove this directly?
|
First, the partial fraction of the summand can be written
$$\begin{align}
\frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\
&=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right)
\end{align}$$
Second, we note that
$$\begin{align}
\sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\
&=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\
&=\frac{\pi^2}{4}
\end{align}$$
Third, it is easy to show that
$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$
Putting it all together we have
$$\sum_{n,\,\,\text{odd}}\frac{n^2}{(4-n^2)^2}=\frac{\pi^2}{8}$$
If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
find maximum and minimum value of $|z|$ If $a,b,c$ are complex number of equal magnitude and satisfy $az^2+bz+c=0,$
then finding maximum and minimum value of $|z|$
with the help of triangle inequality $|az^2+bz+c|\leq |az^2|+|bz|+|c|=|a||z|^2+|b||z|+|c|$
now let $|a|=|b| = |c| = k>0$
so $|az^2+bz+c|\leq k(|z|^2+|z|+1)$
so $|z|^2+|z|+1\geq 0$
wan,t be able to go after that, help me
|
We can divide across by $a$ and get $z^2+bz+c= 0$ with $|b|=|c| = 1$. Solving
the quadratic gives
$z = {1 \over 2} (-b \pm \sqrt{b^2 -4c} ) $. From this we get
$\sqrt{5}-1 \le 2 |z| \le \sqrt{5}+1$, and by choosing $b=1,c=-1$ we see
that these bounds are attained.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2126974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Proving det(B) = -det(A) theorem - am I on the right track? Theorem
Let $A$ be some $n \times n$ matrix, and $B$ be an $n \times n$ matrix
which is the result of performing a row swap operation on $A$. Column swaps won't be touched on for brevity's sake.
Let $b_{ij}$ be an element of $B$ at row $i$ column $j$.
Let $C_{ij} = (-1)^{i+j}M_{ij}$ be the cofactor of $b_{ij}$, and $M_{ij}$ be its corresponding minor. Substitute $a_{ij}$ and $A$ for the same definitions of minor/cofactor etc.
Then
$$\det B = \sum^{n}_{k=1} b_{ik}C_{ik}=-\det A= -\sum^n_{k=1}a_{ik}C_{ik}$$
Given that using row expansion off of any row of $A$ always produce $\det A$. Ditto for $\det B$ and any row of $B$.
Problem
I've been in the process of trying to prove this. Here are my thoughts, so far.
If $i$ is the row that's been interchanged with some other row $l \ne i$, then we can make the connection that:
$$a_{ij} = b_{lj}$$
and
$$a_{lj} = b_{ij}$$
Since $i \ne l$, then $(-1)^{l+j}$ doesn't necessarily equal $(-1)^{i+j}$.
But, if both $i$ and $l$ are even or both $i$ and $l$ are odd, then the sign itself of the cofactor won't change, which means that we can't necessarily factor -1 from either row via the outer sign of the cofactor at least.
Question
I'm not looking for someone to give me the proof; I'd like to solve it myself, but what I'm trying to figure out is whether or not I'm on the right track for finding the relationship; if not, what is a good direction to point me in?
|
To swap the i and j'th row of A, multiply A on the right side such that $M_{i,j} = 1, M_{j,i} = 1, M_{k,k} = 1$ if $k\ne i$, or $k\ne j$ and $0's$ elsewehere.
$MA = B\\
\det MA = \det M \det A\\
\det M = -1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to decide whether an equation in index notation is valid. I am given the following equation in index notation: $k_{ijkl} = a_{i}b_{kl}c_{njm}d_{mn} + e_{ik}e_{jn}f_{n}$. I am told that this is a valid equation, but can anyone explain why? It doesn't violate the summation convention, and there's no obvious illegal characters in there, but how does one decide whether an equation as complicated as this is valid?
|
Hint: Check if all non-free indices are on both sides.
Left side we have $i,j,k,l$. On the right-hand side, we have for the first term $i,k,l,j$ ($m$ and $l$ are repeated indices, which by itself is not valid as far as I remember) and for the second term we have $i,k,j$ ($n$ is a repeated index).
Hence, this expression is not valid.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$
I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$
So it's enough to find: $$\lim_{x\to 0^+}\frac{\ln(\frac {\sin x}{x})}{x}$$
I tried to use L'Hôpital's rule but it got me nowhere. Any ideas?
|
There is an interesting massive overkill: we may prove that in a right neighbourhood of the origin we have
$$1-\frac{x^2}{6}\leq\frac{\sin x}{x}\leq \exp\left(-\frac{x^2}{6}\right) \tag{1}$$
hence the wanted limit is trivially $\exp(0)=1$. By the Weierstrass product for the sine function:
$$ \log\left(\frac{\sin x}{x}\right) = \sum_{n\geq 1}\log\left(1-\frac{x^2}{\pi^2 n^2}\right)\leq\sum_{n\geq 1}-\frac{x^2}{\pi^2 n^2}=-\frac{\zeta(2)}{\pi^2}x^2=-\frac{x^2}{6}\tag{2} $$
and we are done, since the inequality on the left of $(1)$ just follows from the Taylor series of $\text{sinc}(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
|
If $(9+x)^\frac 13 + (9-x)^\frac 13 = 3$,
then there is a $y$ such that $(9+x)^\frac 13 = (\frac 32+y)$ and $(9-x)^\frac 13 = (\frac 32-y)$.
Taking cubes of both equations you get $9 \pm x = (\frac {27} 8+\frac 92y^2) \pm (\frac {27} 4 +y^2)y$,
and so $9 = \frac {27} 8+\frac 92y^2$ and $x = (\frac {27} 4 +y^2)y$.
This gives you $y^2 = \frac 54$, then $x = \frac {32} 4 y = 8y$, and so $x^2 = 64y^2 = 16 \times 5 = 80$.
So $(9 + \sqrt {80})^\frac 13 + (9 - \sqrt {80})^\frac 13 = (\frac {3 + \sqrt 5}2) + (\frac {3 - \sqrt 5}2) = 3$ .
As a bonus, you also know the values of the individual cube roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Algorithm for Generating Well-Formed Formulas in Polish Notation I am trying to write an algorithm that constructs only well-formed formulas in PN. I have some list of symbols for binary connectives, unary connectives, and propositional variables (trying to make program robust for any length).
Here Polish Notation mentions an algorithm to test whether or not it is well-formed. It is much easier to make it so it is well-formed and not to have to test it later for computational purposes. It isn't difficult to make all possible permutations and then test. When looking EKKCprCqrKCpsCqsCApqKrs on the provide link, it seemed that I could "smoosh" the words EKKCCKCCCAK and prqrpsqspqrs at all possible places by creating a bunch of insertion sights, but not sure if this is always correct or can get every possible combination. I am teaching myself these things and having a difficult time thinking how to approach this. I am judging the answer based on explanation and also efficiency to generating wff's.
|
Here's one solution in Python that generates formulae up to a certain depth. To keep clutter to a minimum, no arguments are passed to the script, but adding such parameters and passing in signature and depth should be easy. This version seems to work with both Python 2.7 and 3.5.
""" Enumerates wffs in Polish notation up to prescribed depth."""
from __future__ import print_function
from itertools import product
def generate(depth):
if depth == 1:
return variables
else:
wfflist = []
for letter, arity in symbols:
extensions = [[letter]]
if arity > 0:
extensions.extend([generate(depth-1)] * arity)
wfflist.extend([list(w) for w in product(*extensions)])
else:
wfflist.extend(extensions)
return wfflist
functions = ['C', 'K', 'A', 'E', 'D', 'N']
arities = [ 2, 2, 2, 2, 2, 1 ]
variables = ['p', 'q', 'r'];
symbols = list(zip(functions + variables, arities + [0] * len(variables)));
for wff in generate(3): print(wff)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
How to find length of a part of a curve? How can I find the length of a curve, for example $f(x) = x^3$, between two limits on $x$, for example $1$ and $8$?
I was bored in a maths lesson at school and posed myself the question:
What's the perimeter of the region bounded by the $x$-axis, the lines $x=1$ and $x=8$ and the curve $y=x^3$?
Of course, the only "difficult" part of this is finding the length of the part of the curve between $x=1$ and $x=8$.
Maybe there's an established method of doing this, but I as a 16 year-old calculus student don't know it yet.
So my attempt at an approach was to superimpose many triangles onto the curve so that I could sum all of their hypotenuses.
Just use many triangles like the above,
$$
\lim_{\delta x\to 0}\frac{\sqrt{\left(1+\delta x-1\right)^2+\left(\left(1+\delta x\right)^3-1^3\right)^2}+\sqrt{\left(1+2\delta x-\left(1+\delta x\right)\right)^2+\left(\left(1+2\delta x\right)^3-\left(1+\delta x\right)^3\right)^2}+\cdots}{\frac7{\delta x}}
$$
I'm not entirely sure if this approach is correct though, or how to go on from the stage I've already got to.
|
Your idea is along the right lines.
To find the arc length from $x=1$ to $x=8$ we first split the interval $[1,8]$ into sub-intervals $[x_0,x_1], \ldots [x_{n-1},x_n]$ where $$1=x_0 < x_1 < x_2 < \ldots < x_n = 8$$
This is called a partition of $[1, 8]$. Like you suggested in your question, we can then superimpose triangles on the curve where the width of each triangle is the width of a sub-interval in our partition.
Summing up the hypotensuse of each of these triangles gives an approximation of the arc length
$$\sum_{i=1}^{n}\sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}$$
Now if $f$ is differentiable (which in your case it is) we can apply the mean value theorem to $f$ on each sub-interval $[x_{i-1}, x_i]$ to get that there is $c_i \in [x_{i-1},x_i]$ such that
$$ f'(c_i)(x_i - x_{i-1}) = f(x_i) - f(x_{i-1}) $$
Substituting this into our approximation for the arc length and simplifying a bit we get
$$ \sum_{i=1}^{n}(x_i - x_{i-1})\sqrt{1 + (f'(c_i))^2} $$
If you've seen the definition of the Reimann integral before you will recognise that this is precisely a Reimann sum for the function $g(x)=\sqrt{1 + f'(x)}$. This means that our approximation of the arc length of $f$ is the same as an approximation for the area under $g$.
Going from this Reimann sum to the integral is exactly the idea of taking '${\delta}x \rightarrow 0$' as you wrote.
Therefore we can say that the arc length is given by
$$ \int_{1}^{8}{g(x)dx} = \int_{1}^{8}{\sqrt{1+(f'(x))^2}dx} $$
(provided that $g$ is integrable)
As pointed out in the other answers, in your case the integral is $\int_{1}^{8}{\sqrt{1+9x^4}dx}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
}
|
Alternative to Maplesoft I want to use MapleSoft for small project (few days) and want don't to buy it. Since, it doesn't provide any free trial period, I am looking for some alternative. I am planning to use Maple to solve some sequence and series that it seems solves seamlessly.
|
*
*Mathematica has a free trial that you could take a look at (and in my opinion is the most powerful CAS)
*Sage is also pretty good, but may be more difficult to get work on all platforms
*Personally, I think the easiest one to use for quick stuff is SymPy; it has virtually no learning curve if you already know Python.
Also check out this question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating supremum and infinum I have no verified solution for this question.
I had a few questions regarding this:
$X = [1,3]$
$Y = (1,3]$
$X-Y = \{ x-y| x\in X, y\in Y\}$
The two questions that I have are that:
$a)$ Find the value of: $X-Y$.
$b)$ Are $\sup(X-Y)$ and $\inf(X-Y)$ elements of $X-Y$?
Firstly is the answer to $a)$ Simply; $1-1=0, 3-3=0\Rightarrow X-Y= (0,0]$
And for $b)$ I have: $\sup(X)=1, \inf(X)=3$. And $\sup(Y)=3, \inf(Y)$ not possible.
$\sup(X-Y) = 1-3 = -2$
$\inf(X-Y)$ not possible.
If anyone has any feedback it would be greatly appreciated.
|
The solution is [-2,2) let me know if you need further clarification.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution. Find all conditions for $x$ so that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm 1395=x$ has a solution.
My attempt: $x$ cannot be odd because the left hand side is always even then we have $x=2k(k \in \mathbb{N})$ also It has a maximum and minimum
$1-2-3-4-\dots-1395\le x \le 1+2+3+4+\dots +1395$
But I can't show if these conditons are enough or make some other conditions.
|
Given that the $1$ in the series is not $\pm1$, I can see one other condition, namely that $x$ cannot have the values $Max-2$ or $Min + 2$, where Max and Min are the maximum and minimum values of the series.
The Max occurs when all signs are positive. The only way to get a sum of $Max-2$ is by changing the sign of $1$ from positive to negative, thus decreasing the sum by $2$. Changing the sign of $2$ from positive to negative would decrease the sum by $4$. Changing the sign of $3$ from positive to negative would decrease the sum by $6$, etc. It is thus not possible to achieve the number $Max - 2$.
The Min occurs when all signs are negative, except for the $1$ which must be positive. The smallest positive increase it is possible to make is by changing the sign of the $2$ from negative to positive. This increases the sum by $4$. It is thus not possible to achieve the number $Min + 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 3
}
|
What is the Mathematical Property that justifies equating coefficients while solving partial fractions? The McGraw Hill PreCaculus Textbook gives several good examples of solving partial fractions, and they justify all but one step with established mathematical properties.
In the 4th step of Example 1, when going from:
$$1x + 13 = (A+B)x+(4A-5B)$$
they say to "equate the coefficients", writing the linear system
$$A+B = 1$$
$$4A-5B=13$$
It is a simple step, color coded in the textbook for easy understanding, but McGraw Hill does not justify it with any mathematical property, postulate or theorem. Addition and/or multiplication properties of equality don't seem to apply directly.
Can someone help me justify this step?!
|
Lemma: If $px+q=0$ for all values of $x$, then $p=q=0$.
Proof: In particular, $p(0)+q=0$, which means that $q=0$. So $px=0$ for all $x$, which means that $p(1)=0$, and so $p=0$.
Theorem: If $px+q$ and $rx+s$ are equal for all values of $x$, then $p=r$ and $q=s$.
Proof: If $px+q$ and $rx+s$ are equal for all values of $x$, then
$$
px+q-(rx+s)=0
$$
for all values of $x$. But this expression can be rewritten as
$$
(p-r)x+(q-s)
$$
and so, by the lemma, $p-r=0$ and $q-s=0$. That is, $p=r$ and $q=s$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
}
|
Prob. 18, Sec. 2.3, in I.N. Herstein's TOPICS IN ALGEBRA, 2nd ed: For any $n > 2$ construct a non-abelian group of order $2n$ Here is Prob. 18, Sec. 2.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:
For any $n > 2$ construct a non-abelian group of order $2n$.
Herstein gives the following hint to this question: "imitate the relations in $S_3$ (permutation group of order $3$)".
I've already seen an answer to this problem using dihedral groups, but I still couldn't solve the question based on the hint.
Despite the fact I'm asking for a particular solution, please feel free to share different ways of doing it. It will be very interesting.
Thanks in advance.
|
I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 18 on p.36.
I solved this problem as follows:
Let $\phi:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$.
Let $\psi:=\begin{pmatrix}\cos\frac{2\pi}{n}&-\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}$.
Let $D:=\{I_2,\psi,\psi^2,\dots,\psi^{n-1},\phi,\phi\cdot\psi,\phi\cdot\psi^2,\dots,\phi\cdot\psi^{n-1}\}$.
Obviously, $\psi^i\neq\psi^j$ for $0\leq i<j\leq n$.
Obviously, $\phi\psi^i\neq\phi\psi^j$ for $0\leq i<j\leq n$.
Since $\det\psi^i=1$ and $\det\phi\psi^j=-1$, $\psi^i\neq\phi\psi^j$ for $0\leq i<j\leq n$.
So, $\#D=2n$.
$\phi^2=I_2$.
$\psi^n=I_2$.
$\phi\cdot\psi=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\cdot\begin{pmatrix}\cos\frac{2\pi}{n}&-\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}=\begin{pmatrix}-\cos\frac{2\pi}{n}&\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}$.
$\psi^{-1}\cdot\phi=\begin{pmatrix}\cos\frac{-2\pi}{n}&-\sin\frac{-2\pi}{n}\\\sin\frac{-2\pi}{n}&\cos\frac{-2\pi}{n}\end{pmatrix}\cdot\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}-\cos\frac{-2\pi}{n}&-\sin\frac{-2\pi}{n}\\-\sin\frac{-2\pi}{n}&\cos\frac{-2\pi}{n}\end{pmatrix}$.
So, $\phi\cdot\psi=\psi^{-1}\cdot\phi$.
$\phi\cdot\psi^{-1}=\phi\cdot\psi^{n-1}=(\phi\cdot\psi)\cdot\psi^{n-2}=\psi^{-1}\cdot(\phi\cdot\psi^{n-2})=\cdots=(\psi^{-1})^{n-1}\cdot\phi=\psi^n\cdot(\psi^{-1})^{n-1}\cdot\phi=\psi\cdot\phi.$
$\psi^i\cdot\phi\psi^j=\phi\psi^{-i}\psi^j=\phi\psi^{j-i}$.
Since $\psi^n=I_2$, we can write $\phi\psi^{j-i}=\phi\psi^k$, where $0\leq k<n$.
So, $\psi^i\cdot\phi\psi^j\in D$.
$\phi\psi^i\cdot\psi^j=\phi\psi^{i+j}$.
Since $\psi^n=I_2$, we can write $\phi\psi^{i+j}=\phi\psi^k$, where $0\leq k<n$.
So, $\phi\psi^i\cdot\psi^j\in D$.
$\psi^i\cdot\psi^j=\psi^{i+j}$.
Since $\psi^n=I_2$, we can write $\psi^i\cdot\psi^j=\psi^k$, where $0\leq k<n$.
$\phi\psi^i\cdot\phi\psi^j=\phi\phi\psi^{j-i}=\psi^{j-i}$.
Since $\psi^n=I_2$, we can write $\phi\psi^i\cdot\phi\psi^j=\psi^k$, where $0\leq k<n$.
So, $D$ is closed under the matrix multiplication.
$\psi^i\cdot\psi^{n-i}=I_2$.
$\phi\psi^i\cdot\phi\psi^i=I_2$.
So, any element of $D$ has its inverse.
Therefore, $D$ is a subgroup of $GL(2,\mathbb{R})$.
$\psi\cdot\phi=\phi\psi^{-1}=\phi\psi^{n-1}\neq\phi\psi$.
So, $D$ is non-abelian.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.
My Attempt
This can be rewritten as
$$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$
Now consider
$$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$
Taking the derivative
$$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$
By symmetry we have
$$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$
Using W|A I got
$$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$
By integeration we have
$$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$
Let $x = 2/a$
$$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
Question
I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
|
Another approach to break down the last integral might be to consider the integral of $\displaystyle \frac{\log^3 (1-iz)}{z^2}$ along a positively oriented semi-circular contour $\gamma_R = [-R,R]\cup Re^{i[0,\pi]}$ in the upper half-plane. (We choose the branch of logarithm $\log (1-iz)$ in the lower half-plane along $[-i,-i\infty)$).
The integral along the arc is $\displaystyle \mathcal{O}\left(\frac{\log^3 R}{R}\right)$, which vanishes as $R \to +\infty$ we have,
\begin{align*}0 = \lim\limits_{R \to \infty} \int\limits_{\gamma_R} \frac{\log^3 (1-iz)}{z^2}\,dz = \int_{-\infty}^{\infty} \frac{\log^3\left((1+x^2)^{1/2} - i\arctan x\right)}{x^2}\,dx\end{align*}
Comparing the real parts on both sides,
\begin{align*} \int_{-\infty}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx &= \frac{1}{12}\int_{-\infty}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx \\&= \frac{1}{6}\int_{0}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx\\&\underset{\text{(IBP)}}{=} \int_0^{\infty} \frac{\log^2 (1+x^2)}{1+x^2}\,dx \\&= \int_0^{\pi/2} \log^2 (\cos \theta)\,d\theta \\&= \lim\limits_{b \to \frac{1}{2}}\frac{1}{2}\frac{\partial^2}{\partial b^2} B\left(\frac{1}{2},b\right)\\&= \frac{\pi}{2}\left(4\log^2 2 + \frac{\pi^2}{3}\right)\end{align*}
Hence, $$\displaystyle \int_{0}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx = \frac{\pi^3}{12} + \pi\log^2 2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
}
|
How to find the point on the sphere that is closest to a plane? Consider the plane $x+2y+2z=4$, how to find the point on the sphere $x^2+y^2+z^2=1$ that is closest to the plane?
I could find the distance from the plane to the origin using the formula $D=\frac{|1\cdot 0+2\cdot 0+2\cdot 0-4|}{\sqrt{1^2+2^2+2^2}}=\frac43$, and then I can find the distance between the plane and sphere by subtracting the radius of sphere from plane-origin distance:$\frac43-1=\frac13$. But then I am stuck here because I don't know how to convert this distance into a direction vector, so I could subtract it from the plane to find the sphere point. Any help would be appreciated.
|
A little different:
Plane: $f(x,y,z) = x + 2y + 2z - 4 = 0 $; Circle $ g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 $.
Problem:
Point on a sphere with minimum distance to the plane.
Normal vector to the plane, $n_p$:
$\nabla f = (\frac {\partial f}{\partial x},\frac {\partial f}{\partial y},\frac {\partial f}{\partial z})$.
We get $n_p = (1,2,2)$.
Normal vector to the circle, $ n_c$:
$\nabla g = (\frac {\partial g}{\partial x},\frac {\partial g}{\partial y},\frac {\partial g}{\partial z})$.
We get $n_c = (2x,2y,2z)$.
At the desired point on the circle the two normals are parallel or anti parallel.
$(2x,2y,2z) = \alpha (1,2,2)$.
Hence:
$\, 2x = \alpha, 2y = 2\alpha , 2z = 2\alpha $.
Combining the above with the equation of the circle:
$ \alpha ^2 + ( 2\alpha)^2 + (2\alpha)^2 = 4$,
$ \alpha ^2 + 4 \alpha ^2 + 4 \alpha ^2 = 4$,
$ 9 \alpha ^2 = 4$,
$ \alpha_1 = 2/3$ and $\alpha_2 = - 2/3$.
The closest point:
1) $P_1 (1/3,2/3,2/3)$ for $\alpha_1 = 2/3$,
the farthest:
2) $P_2 (-1/3,-2/3,-2/3)$ for $ \alpha_2 = - 2/3.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.