Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
xsin(1/x) Holder on [0,1] I know $x \sin(1/x)$ is not Lipschitz on $[0,1]$, but some experimentation makes me conjecture that it is $1/2$-Holder. What is a good way to prove this?
|
I will sketch the proof that $f(x)=x \sin(x^{-1})$ is 1/2-Holder on $[0,1/2\pi]$.
There are two cases.
*
*If $x,y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$.
Write $x= \frac{1}{2\pi n+ \xi}$ and $y=\frac{1}{2\pi n+\zeta}$ where $0 \le \xi,\zeta \le 2\pi$.
There exists a constant $0<c_1$ such that
$$| f(x)- f(y) | \le c_1 \frac{|\xi-\zeta|}{n} \le c_2 \frac{\sqrt{|\xi-\zeta|}}{n} $$
On the other hand
$$ \left| x-y \right| \ge \frac{|\xi-\zeta|}{\left( 2\pi (n+1) \right)^2} $$
So that
$$| f(x)- f(y) | \le c_3 \sqrt{|x-y|} $$
*Without loss of generality, we can assume $y\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ and that $x<\frac{1}{2\pi (n+1)}$.
Then, by periodicity of sinus, there exists $x'\in [\frac{1}{2\pi (n+1)} , \frac{1}{2\pi n}]$ such that $f(x')=f(x)$ and by 1. we get
\begin{align*}| f(x)- f(y) |=| f(x')- f(y) | &\le c_3 \sqrt{|x'-y|} \\
&\le c_3 \sqrt{|x-y|}
\end{align*}
The exponent $1/2$ is sharp and I guess the best constant is $\sqrt{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
What's wrong with this argument? (Limits) In order to compute $\displaystyle \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x$ we can multiply by the conjugate and eventually arrive at a limit value $1/6$.
But what about the line of reasoning below, what is wrong with the argument and why? I can't think of a simple explanation, I had one involving the limit definition but I believe there should be a less complicated one.
Here's the argument:
Clearly for large $x$ we can say $\sqrt{9x^2 + x} \approx \sqrt{9x^2} = 3x$. Hence $$ \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x = \lim_{x \to \infty} 3x - 3x = 0 \ . $$ So the limit ought to be zero, easy!
What goes wrong and why?
|
The symbol $\approx $ is often misused. It can mean numerical approximation or it can mean asymptotically equal.
For example it is not true that (n+1) 2 $\approx $ n 2 in both senses. This you can see from the difference
${(n + 1)^2} - {n^2} = 1 + 2n$ .
You can observe that the difference is not small. Indeed it tends to infinity as n tends to infinity.
In $\mathop {\lim }\limits_{x \to \infty } \sqrt {9{x^2} + x} - 3x$ you are taking the limit as x tends to infinity. You are looking for the difference of
$\sqrt {9{x^2} + x} $ and $ 3x $ as x gets larger and larger. The limit exists if this difference gets closer and closer to a certain value.
In example in variable $n$ above, the difference tends to infinity and so no limit.
If you now look at the difference
$\sqrt {9{x^2} + x} - 3x = \frac{{\left( {\sqrt {9{x^2} + x} - 3x} \right)}}{{\sqrt {9{x^2} + x} + 3x}}\left( {\sqrt {9{x^2} + x} + 3x} \right) = \frac{x}{{\sqrt {9{x^2} + x} + 3x}}$, you see that
$\sqrt {9{x^2} + x} = 3x + \frac{x}{{\sqrt {9{x^2} + x} + 3x}}$
and
$\frac{x}{{\sqrt {9{x^2} + x} + 3x}} \ge \frac{x}{{\sqrt {16{x^2}} + 3x}} = \frac{1}{7}$ no matter how large x gets and $1/7$ is definitely not a small value. So you cannot say
$\sqrt {9{x^2} + x} \approx 3x$ .
It is precisely this difference that gives you the limiting value. This question is to let you know it is erroneous to think of estimating large value as approximation. It is possible two functions need not have the same value as x gets larger and larger and it is possible that the two functions for large values of x are always a fixed quantity apart, or gets further and further apart or their difference may oscillate about a fixed quantity. For example
x 2 and (x+1)2, x2 and x2+sin(x), x2 and x2+6.
More precislely for this question, you want to show that no matter how small $\epsilon $ is you can find number $K$ so that for all $x > K$ , the difference
$\left| {\sqrt {9{x^2} + x} - 3x - \frac{1}{6}} \right| = \left| {\frac{x}{{\sqrt {9{x^2} + x} + 3x}} - \frac{1}{6}} \right| < \varepsilon $
This is equivalent to
$\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {9{x^2} + x} + 3x}} = \frac{1}{6}$ .
This I think you can prove.
Now you can say $\sqrt {9{x^2} + x} \approx 3x + \frac{1}{6}$
or more precisely $\sqrt {9{x^2} + x} $ is asymptotically equal to $3x + \frac{1}{6}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
}
|
Inconclusive Second derivative test ,Now how shall i proceed While doing maxima and minima questions i have encountered upon question in which i cannot show nature of points
P1 : Given $f(x,y) = 2x^4 - 3x^{2}y + y^{2}$
Doubtful case is as origin .
P2: $f(x,y)$ = $y^{2} + x^{4} +x^{2}y $
Doubtful case is at origin
Thanks for help ..
|
P1: The function is $0$ on the curves $y = x^2$ and $y = 2 x^2$, is negative in between them and positive otherwise. So $(0,0)$ is a saddle point since there are positive and negative values of $f$ arbitrarily close to $(0,0)$. The fact that it is a saddle does not follow simply from the discriminant being $0$, for example the origin is a minimum for $g(x,y,)=x^4+y^4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How do you find the integral, $\int2x\,dx/(1+x)$ Integral: $\displaystyle \int \dfrac{2x}{1+x}dx$. I think I have to use $u$ substitution, but I'm having trouble understanding what to do. Thank you in advance!
|
Here's a trick against long division:
$$
\begin{align}
\int \frac{2x}{1 + x}\mathrm dx &= 2\int \frac{(x + 1) - 1}{(x + 1)}\mathrm dx \\
&= 2\left(\int\mathrm dx -\int\frac{\mathrm d(x+1)}{x+1}\right)\\
&= 2\Big(x - \ln|x+1|\Big) \color{grey}{+ \mathcal C}
\end{align}$$
Note that $2x + \ln\left|\frac{1}{x^2 + 2x + 1}\right| + \color{grey}{\mathcal C}$ is also a valid answer =)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\lim_{n\rightarrow\infty}(1+\frac{1}{n})(1+\frac{2}{n})^{\frac{1}{2}}\cdots(1+\frac{n}{n})^{\frac{1}{n}}$ Evaluate
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots \left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
solve:
$$ \exp\left\{\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^{n} \frac{\ln(\frac{k}{n}+1)}{\frac{k}{n}}\right\}= \exp\left\{\int_0^1 \frac{\ln(1+x)}{x}\,dx\right\}$$
how to evaluate
$$\int_0^1\frac{\ln(1+x)}{x}\,dx$$
|
$$\int_0^1\frac{ln(1+x)}{x}dx$$
$$=Li_2(-1)-Li_2(0)$$
$$=-\sum \frac{0}{n^2}+\sum \frac{(-1)^n}{n^2}$$
$$=\frac{\pi^2}{12}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1023832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
'Easy' question on continuity of integral Here's a problem that I recently stumbled upon. It seems pretty easy, and quite intuitive yet every time I try to solve it, I run into some difficulties. Here it goes :
Let $\phi \in BC(\mathbb{R}^2,\mathbb{R})$ (bounded and continuous) be a function such that
$$\forall_{x \in \mathbb{R}} \ \phi(x, \cdot) \in L_1(\mathbb{R},\mathbb{R})$$
Then the function defined as
$$\psi(x) = \int_{\mathbb{R}} \ |\phi(x,z)| \ dz $$
is continuous.
I got as far as to trying to estimate the expression
$$\int_{\mathbb{R}\backslash [-T,T]} \ |\phi(x,z) - \phi(x_0,z)| \ dz$$
for large $T$ (we want the expression to be small). However, the main problem is that $T$ may depend on $x$, in other words it need not be uniform for $x$ in some neighbourhood of $x_0$. I feel like it would somehow violate the continuity, but maybe I am totally wrong.. I would be grateful for providing a simple counterexample if the statement is false.
|
Here is a counterexample:
$$|\phi(x,z)| = x^2e^{-x}|z|e^{-x|z|}\chi_{[0,\infty)}(x).$$
The function is bounded and continuous, but
$$\psi(0) = 0,$$
and for all $x > 0$,
$$\psi(x) = x^2e^{-x}\int_{-\infty}^{\infty}|z|e^{-x|z|}\,dz= 2x^2e^{-x}\int_{0}^{\infty}ze^{-xz}\,dz=2e^{-x}>0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How do I solve this differential equation $y''^2-2y'y''+3=0$ in parametric form? I do have quite no idea about this one.
The obvious substitution $y'=p, y''=p\frac{dp}{dy}$ doesn't make the situation any better
|
Note that $2y'y'' = ((y')^2)'$
Then let $g = y'$ so the equation reformulates to:
$$(g')^2 - g^2 + 3 =0 $$ which is much simpler an easier to solve. Then find a solution for $g = g(x) = y' $ and integrate to find $y(x) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Factorization of $x^6 - 1$ I started by intuition since I'm familiar with the formula $a^2-b^2 = (a-b)(a+b)$. So in our case $$x^6 - 1 = ({x^3} - 1)({x^3}+1)$$
How should I proceed? I assume there's some sort of algorithm to keep the process till you reach a form of irreducible linear polynomials.
|
${x^3} - 1=(x-1)(x^2+x+1)$
${x^3} + 1=(x+1)(x^2-x+1)$
The last quadratic term is already irreducible in $\mathbb{R}$.
If you mean irreducible in $\mathbb{C}$, you may split it into product of factor $(x-e^{inw})$, where $n=0\cdots 5$,$w=\frac{\pi}{3}$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
How to evaluate $\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$ Im tempted to say that the limit of this sequence is 1 because infinite root of infinite number is close to 1 but maybe Im mising here something? What will be inside the root?
This is the sequence:
$$\lim_{n\to\infty}\sqrt[n]{\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}}$$
|
It is quite easy to show by induction that:
$$\frac{(2n-1)!!}{(2n)!!}\geq\frac{1}{\sqrt{2n}}\tag{1}$$
since the last line is implied by:
$$\frac{2n+1}{2n+2}\geq\sqrt{\frac{2n}{2n+2}}$$
that is equivalent to:
$$(2n+1)^2\geq 2n(2n+2) = (2n+1)^2-1.$$
Using $(1)$ and the trivial bound $\frac{(2n-1)!!}{(2n)!!}\leq 1$, it follows that the limit is $1$ by squeezing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 11,
"answer_id": 2
}
|
Is a projective $R[G]$-module a projective $R[H]$-module if $H$ is a subgroup of $G$? I have a ring $R$ of characteristic $0$ and a finite group $G$. Let $H$ be a subgroup of $G$.
Question: If $M$ is a projective $R[G]$-module where $R[G]$ is the usual group ring then is $M$ also projective as an $R[H]$-module?
This seems easy but I'm having trouble showing that if $M$ is a summand of a free $R[G]$-module then it's also the summand of a free $R[H]$-module.
I would really appreciate some help.
|
The mentioned assumptions are not necessary. If $R$ is any ring and $H$ is a subgroup of any group $G$, then $R[G]$ is a free $R[H]$-module (both from left and from right). This is basically because $G$ is a free $H$-set. We have $G = \coprod_i H g_i$ for some $g_i \in G$, and hence $R[G] = \bigoplus_i R[H] g_i$.
Since $R[G]$ is a free $R[H]$-module, any free $R[G]$-module restricts to a free $R[H]$-module. Hence, the same holds for projective modules.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1$ I have two questions about a prove that I have to do for my mathematic study. I'm now thinking about it the whole day, but can't find the prove.
Let $a,b \in \mathbb Z_{>0}$.
(a) Prove: $\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1$
(b) Is this also true when you replace $2$ with a number $c > 2, c \in \mathbb Z$?
For (b), I think it's true, but I can't explain why.
Thanks in advance!
|
Without loss of generality, let us assume that $a>b$.
You can write $2^a-1 = 2^b2^{a-b} - 2^{a-b} + 2^{a-b} - 1 = 2^{a-b}(2^b-1) + 2^{a-b}-1$.
$\therefore gcd(2^a-1,2^b-1) = gcd(2^b-1,2^{a-b}-1).$
You can now continue the proof on the same lines as Euclid's proof for gcd of two integers.
There is no significance of the number '2' here, so this would be correct also for all integers greater than 2 also.
EDIT : We can continue reducing the larger number, till we get to a point such that $gcd(2^a-1,2^b-1) = gcd(2^k-1,2^k-1)$ for some $k$ after which point we can't further reduce and we get $gcd(2^a-1,2^b-1) = 2^k-1$ . From the equation, it is clear that $k$ will always be a linear combination of $a$ and $b$. The smallest positive number which is a linear combination of $a$ and $b$ is the gcd of $a$ and $b$. Hence, $k=gcd(a,b)$. Hence, $gcd(2^a-1,2^b-1)=2^{gcd(a,b)}-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
By using just the definition of a Cauchy Sequence, how could I show $s_n=n^2$ is not a Cauchy Sequence? Here's what I have,
Let $\epsilon>0$. Then there exists N such that $m,n>N \implies |n^2-m^2|<\epsilon.$
So now we have to find a $N$ that makes the implication true. So, $|n^2-m^2|<\epsilon=n^2<\epsilon+m^2=n<\sqrt{\epsilon+m^2}$. However, this can become arbitrarily large for large $m$. Hence, the sequence $s_n$ is not Cauchy.
I figure this can be written more formally, but I just want to know if what I typed is correct.
|
For $n \neq m$, $|n^{2} - m^{2}| \in \mathbb{Z}$, $|n^{2} - m^{2}| \neq 0$. There isn't an integer between $0$ and $1$, so if you pick $\epsilon = \frac{1}{2}$, then $|m^{2} - n^{2}| > \epsilon$ whenever $m \neq n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Evaluating $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ using complex analysis Again, improper integrals involving $\ln(1+x^2)$
I am trying to get a result for the integral $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ - asked above link- using some complex analysis, however, I couldn't find an appropriate solution. (Of course, for $\alpha \in (0,3)$ as stated above link (o.w. it is divergent))
Any ideas?
|
Differentiation under the integral sign is another possibility. Let:
$$ I(\alpha,\beta) = \int_{0}^{+\infty}\frac{\log(1+\beta x^2)}{x^\alpha}\,dx. \tag{1}$$
Then assuming $1<\alpha<3$ we have:
$$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+\beta x^2}\,dx = \beta^{\frac{\alpha-3}{2}}\int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+x^2}\,dx\tag{2}$$
and by replacing $x$ with $\tan\theta$ we get, through the Euler beta function and the reflection formula for the $\Gamma$ function:
$$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \beta^{\frac{\alpha-3}{2}}\int_{0}^{\pi/2}\left(\tan\theta\right)^{2-\alpha}\,d\theta = -\beta^{\frac{\alpha-3}{2}}\frac{\pi}{2\cos\left(\frac{\pi \alpha}{2}\right)}\tag{3}$$
and by integrating $(3)$ with respect to $\beta$ we have:
$$ \int_{0}^{+\infty}\frac{\log(1+x^2)}{x^\alpha}\,dx = \frac{\pi}{1-\alpha}\,\sec\left(\frac{\pi\alpha}{2}\right).\tag{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Finite state Markov chain Under what conditions a Markov chain can be considered as finite (and not infinite)?
Thank you!
|
The characterization of a Markov Chain as finite or infinite refers to the state space $Ω$ in which the Markov Chain takes it's values. Thus, a finite (infinite) Markov chain is a process which moves among the elements of a finite (infinite) set $Ω$. In other words, if $X_n$ describes the state of the process at time $n$ then $$X_n\inΩ$$ If $Ω$ is finite f.e. $$Ω=\{1, 2, \ldots, n-1, n\}$$ then $(X_n)_n$ is called a finite Markov Chain, otherwise if $Ω$ is infinite. f.e.
$$Ω=\{1, 2, \ldots, n-1, n, \ldots\}$$ then the Markov Chain $(X_n)_n$ is calles infinite.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$\text{arg}(i \: \text{conj}(z))=\text{arg}(i)+\text{arg}(\text{conj}(z))$ This is a Complex Analysis question.
Let z be a complex number, i be the imaginary unit, arg be the argument of z, and $\bar{z}$be the complex conjugate of z.
How do I prove that the following equality holds?
$\text{arg}(i \: \bar{z})=\text{arg}(i)+\text{arg}(\bar{z})$
I haven't been able to prove it using the definition of the argument function together with the trig form of a complex number alone. Please do not prove it using the polar form if possible.
|
For $z=a+bi$, use $\arg(z_1z_2)=\arg(z_1)+\arg(z_2),\,\mod(-\pi,\pi]$:
$$
\arg(i\overline{z})=\arg(i)+\arg(\overline{z}).
$$
To see why the identity is true, note that $z=|z|e^{i\arg z}$, so
$$
\arg(z_1z_2)=\arg(|z_1z_2|e^{i(\arg(z_1)+\arg(z_2))})=\arg(z_1)+\arg(z_2),\,\mod (-\pi,\pi].$$
Throughout I am using $\arg$ as the principal argument, since it seems LaTeX doesn't have $\Arg$ built in (?).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1024853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Verification of an R-Module isomorphism between $R^n$ and its dual With one step at a time, I am getting slightly more used to $R$-Modules.
Let $R$ denote a commutative Ring with $\mathbb{1}$ and $n$ a natural number. For the tuple $a:= (a_i)_{i=1}^n \in R^n$ we have the mapping: $$\varphi_a: R^n \to R ,\ (x_i)_{i=1}^n \mapsto \sum_{i=1}^n a_i x_i $$
It is easy to see that the above mapping is $R$ linear, which means that for scalars $r,s \in R$ and 'vectors' $x,y$ we have $\varphi_a(rx + sy)= r\varphi_a(x) + s\varphi_a(y)$
I now want to show that the following mapping $$\psi : R^n \to \text{dual}(R^n), \ a \mapsto \varphi_a $$
is an $R$-Module isomorphism. Here dual$(R^n)$ just denotes linear mappings from $R^n \to R$
My approach: First I want to show that $\psi$ is a homomorphism. I did that as follows.
Let $r,s \in R$ and $a,b \in R^n$ be arbitrary. Then it follows that $$\psi(ra + sb) = \varphi_{ra + sb} \tag{*} $$
By definition. So lets take $x \in R^n$ arbitrary and continue from there $$\varphi_{ra + sb}(x) =\sum_{i=1}^n (ra_i + sb_i)x_i = r\sum_{i=1}^n a_ix_i + s\sum_{i=1}^nb_i x_i = r \varphi_a(x) + s\varphi_b(x) = (r\varphi_a + s\varphi_b)(x) $$
Since $x \in R^n$ was arbitrary it follows that $\varphi_{ra + sb} = r \varphi_a + s \varphi_b$ which means for the above calculation at (*) that $$\psi(ra + sb) = \varphi_{ra + sb}= r \varphi_a + s \varphi_b = r \psi(a) + s \psi(b) $$
Which means that $\psi$ is R-linear.
Injectivity: I want to show that the kernel of $\psi$ is trivial. Choose $a \in \ker \psi$ arbitrary. Thus I have $\psi(a) = \varphi_a = 0$. I want to show that my 'vector' $a$ is the zero vector.
I thought choosing an arbitrary $x \in R^n$ would be the next thing to do but then I ran into troubles. So instead I choose $x \in R^n \setminus \ker \psi$ and continue my calculations from there.
$\varphi_a(x)=x\varphi_a(1)= x(a_1 + \dots + a_n ) =0 \implies a_1 + \dots + a_n =0, \text{ because } x \neq 0$
And here I am stuck again. I somehow want to argue that it must follow that $a_1 = a_2 = \dots = a_n =0$ but I don't see a rigorous statement for that yet, so I assume my entire approach was flawed.
Surjectivity: I didn't come up with a good idea here yet, I might must give it some more thought or maybe you could provide me a hint here as well.
|
Given $\varphi\colon R^n\to R$ you want to find $a=(a_1,a_2,\dots,a_n)\in R^n$ such that $\varphi=\varphi_a$. In particular, you want that
$$
\varphi(e_1)=\varphi_a(e_1)=a_1
$$
where $e_1=(1,0,\dots,0)$.
Can you go on from here?
A similar idea can be used for injectivity.
The rest is maybe a bit too verbose, but correct.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Compute the limit: $\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)$ Does anyone know how o compute the following limit?
$$\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)$$
I tried to split it up as follows:
$\lim_{x\to \infty}\left[x\ \log(x+c)-x\ \log(x-c)\right]$ but still this is indeterminate form of type $\infty-\infty$. Any thoughts please?
|
Hint; Let $t=\frac1x$${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Do elementary row operations give a similar matrix transformation? So we define two matrices $A,B$ to be similar if there exists an invertible square matrix $P$ such that $AP=PB$. I was wondering if $A,B$ are related via elementary row operations (say, they are connected via some permutation rows for example) then are the necessarily similar?
Obviously swapping rows multiplies the determinant by $-1$ but I was thinking if we permute rows in pairs, would this allow us to construct a similarity transformation?
|
No. For instance, because row permutations do not preserve the trace.
$$
\begin{pmatrix}
1 & 2 \\
4 & 3
\end{pmatrix}\to
\begin{pmatrix}
4 & 3 \\
1 & 2
\end{pmatrix}
$$you can chenge the first matrix to the second via a permutation of the rows, but the trace of the first one is $4$ and the trace of the second one is $6$.
If $AP = PB$ then $\text{tr} B =\text{tr}( P^{-1} AP) =\text{tr}( PP^{-1}AP) =\text{tr} A$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Diffeomorphism: Unit Ball vs. Euclidean Space In my differential geometry class we are being asked to prove that the open unit ball
$B^n$ = { $x$ $\in$ $\mathbb{R}$$^n$ such that |$x$| < $1$}
is diffeomorphic to $\mathbb{R}$$^n$
I am having a hard time with this as I am brand new not only to differential geometry, but also topology.
I know that I need to construct a smooth, differentiable bijection between the two with a differentiable inverse, but beyond that, I am unsure of where to start. Some guidance in the right direction would be greatly appreciated.
|
Let $\phi \colon [0,1) \to [0, \infty)$ a diffeomorphism with inverse $\psi$. Some possible choices: $t \mapsto \frac{t}{1-t}$, $t \mapsto \tan (\frac{\pi}{2}\cdot t)$.
The map
$$x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$$
is a diffeomorphism from $B^n$ to $ \mathbb{R}^n$ with inverse
$$y \mapsto \psi(||y||) \cdot \frac{y}{||y||}$$
$\bf{Added:}$ It turns out that the choice of the diffeomorphism from $[0,1)$ to $[0,\infty)$ matters a lot, since $x \mapsto ||x||$ is not smooth at $0$. This was brought to my attention by @Freeze_S and I thank him a lot! One can check that the map obtained for $\phi(t) = \frac{t}{1-t}$ is only $C^1$ at $0$... However, we can use the map so kindly suggested by @Jesus RS: ( big thanks! ) $\phi(t) = \frac{t}{\sqrt{1-t^2}}$ with inverse
$\psi(s) = \frac{s}{\sqrt{1+s^2}}$ and it will work just fine. The diffeomorphisms are, as written by @Jesus RS:
$$x \mapsto \frac{x}{\sqrt{1-||x||^2}} \\
y \mapsto \frac{y}{\sqrt{1+||y||^2}}$$
In fact, as long as $\phi(t)$ is an odd function of $t$ things will work OK. So, another example is
$$x \mapsto \frac{\tan (\frac{\pi}{2} \cdot ||x|| )}{||x||} \cdot x $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
}
|
Second Degree Polynomial Interpolation, error related We want to create a table of the exponential integral function $$E_{1}(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt, x>0$$
over the interval $x \in [1,10]$ with stepsize $h$. How large can $h$ be if a user of your table expects to obtain values at arbitrary x-locations with an absolute error $\leq 10^{-8}$ when using second degree polynomial interpolation?
I know that I will use the error formula for second degree polynomial interpolation and bound the terms.
I will need to compute $f^{'}(x), f^{''}(x),f^{'''}(x)$
BUT I'm honestly not even sure what I am trying to find here, some stepsize $h$?
How does this relate to my equation?
|
Hints: I will map it out, please fill in the details.
The error formula for second degree polynomial interpolation is given by:
$$\tag 1 |P_2(x) - f(x)| \le \dfrac{|(x-x_0)(x-x_1)(x-x_2)|}{3!}~\mbox{max}_{a \le x \le b} |f^{(3)}(x)|$$
Since we are using three points, we can use equal spacing and take $x_0 = -h, x_1 = 0, x_2 = h$.
Now we need to do three things:
*
*Bound the term $|(x-x_0)(x-x_1)(x-x_2)|$ (in other words, find the max of a cubic in terms of $h$), and
*Find $\mbox{max}_{a \le x \le b} |f^{(3)}(x)| = |E_1^{(3)}(x)|$ (the third derivative under the integral of $E_1(x)$) over $a = 1, b = 10$.
*Using the previous two results in $(1)$ gives us a function in terms of $h$ and we set it $\le 10^{-8}$ and solve for $h$.
Aside: Here are some nice notes by Keith Conrad on differentiation under the integral sign, but it seems like you understand that.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to show a particular set $S$ is a basis for $M_{2\times 2}$? I'm given a set $S$ of four $2\times 2$ matrices with numbers in them and need to show they are a basis for $M_{2\times 2}$. all I know is that the linear combination of these matrices $= [0,0,0,0]$ must only have a solution of a scalar * each of them being 0. All I can think of is making a set of $4$ vectors and using that to test LI somehow. but how exactly am I allowed to make these vectors? Is this even the right method?
also, given another particular $2\times 2$ matrix, how would I figure out how to write it as a linear combination of the $4$ matrices in the set $S$?
|
You need to transform them to coordinate vectors and then your matrix will be converted in form (a,b,c,d) (i,e vector). then procceed with usual method of checking L.I ,by checking determinant or whatever
I highly recommend following for complete understanding of subject
https://www.youtube.com/playlist?list=PLGqzsq0erqU7w7ZrTZ-pWWk4-AOkiGEGp
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Describe all the complex numbers $z$ for which $(iz − 1 )/(z − i)$ is real.
Describe all the complex numbers $z$ for which $(iz − 1 )/(z − i)$ is real.
Your answer should be expressed as a set of the form $S = \{z \in\mathbb C : \text{conditions satisfied by }z\}$.
I started solving for $((iz − 1 )/(z − i)) = \overline{ ((iz − 1 )/(z − i))}$.
I got stuck. Is this the right way to go about this?
|
We need $i\cdot\dfrac{z+i}{z-i}$ purely real
$\iff\dfrac{z+i}{z-i}$ purely imaginary $=iy$(say) where $y$ is real
$$\frac{z+i}{z-i}=iy\implies z+i=iy(z-i)=z(iy)+y\implies z(1-iy)=y-i\implies z=\frac{y-i}{1-iy}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Largest "leap-to-generality" in math history? Grothendieck, who is famous inter alia for his capacity/tendency to look for the most general formulation of a problem, introduced a number of new concepts (with topos maybe the most famous ?) that would generalize existing ones and provide a unified, more elegant and more efficient way to think of a class of objects.
What are your favorite examples of such generalizations and their authors ?
(NB: this is a soft question and largely an excuse to commemorate once more the passing of Alexander Grothendieck this week.)
Grothendieck, who is said to have been both very humble and at times very difficult to cope with, reportedly had (source, the quote comes from L. Schwartz's biography) an argument with Jean Dieudonné who blamed him in his young years for "generalizing for the sole sake of generalizing":
Dieudonné, avec l'agressivité (toujours passagère) dont il était capable, lui passa un savon mémorable, arguant qu'on ne devait pas travailler de cette manière, en généralisant pour le plaisir de généraliser.
|
Perhaps Claude Shannon deserves a mention for his work, such as defining Entropy in the 1948 paper "A Mathematical Theory of Communication" and generally spurring the field of information theory. He created very general models for communication and gave some profound results about them. I don't know much about Shannon's other contributions (besides that they are numerous) so hopefully someone can inform.
Alan Turing's general computing device, the "Turing Machine", could be mentioned. In a world without electronic computers, "this machine can compute anything that you can" seems to me to be a monumental generalization, even to the foundations of mathematics.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 14,
"answer_id": 8
}
|
Why an ideal in a ring is a submodule of a free module over that ring I know an ideal in a ring is a module over this ring, but I don't know why it's a submodule of a free module, what's the free module? Thank you.
|
A fintely generated free $R$-module is isomorphic to $R^n$. Take $n=1$ and you see that $R$ is a free $R$-module. Now it is left to check that an ideal $I\subset R$ defines an $R$-submodule. But this is clear by definition. Is every step and the result clear to you?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How can we be sure of periodicity by testing some terms? A mod $n$ Fibonacci sequence is simply defined as the Fibonacci sequence, except all terms are in mod $n$.
Now to determine periodicity, the worked solutions computed the first 20 or so terms and then observed that it was periodic with period 16.
My question is: Just by testing terms like that, how do we REALLY know for sure that the period is 16?
How do we know that say when we test the first 20 terms, the 17th, 18th, 19th etc match the 2nd, 3rd, 4th etc term respectively, but then everything else goes wrong afterwards?
|
If $F_{16} \equiv F_0 \equiv 0 \text{ modulo }n$ and $F_{17} \equiv F_1 \equiv 1 \text{ modulo }n$ then you can use induction to show $F_{16+k} \equiv F_k \text{ modulo }n$ for all non-negative integer $k$.
This does not guarantee that the period is $16$, but it does ensure the period divides $16$: for example with $n=3$ you have $F_{16+k} \equiv F_k \text{ modulo }3$ but the period is $8$. So to ensure the period is exactly $16$ you also need to check it is not shorter, i.e. that it is not $1,2,4,8$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Verify the identity $\cos^2x-\sin^2x = 2\cos^2x-1$ I am having problems understanding how to verify this identity. I am quite sure that it is to be solved using the Pythagorean identities but, alas, I'm not seeing what might otherwise be obvious.
I need to verify the identity
$$\cos^2x-\sin^2x = 2\cos^2x-1$$
Thank you for your help.
|
Yes, indeed, we can use the Pythagorean Identity:
$$\cos^2 x + \sin^2 x = 1 \iff \color{blue}{\sin^2 x = 1-\cos^2 x}$$
$$\begin{align}\cos^2x-\color{blue}{\sin^2x} & = \cos^2 x - \color{blue}{(1-\cos^2 x)}\\\\
&= \cos^2 x - 1 + \cos^2 x \\ \\
& = 2\cos^2 x - 1\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1025911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
|
Hint:
$$1\cdot 2\cdot 3 + \dots +n(n+1)(n+2) \le n(n+2)^3 \le 8n^4$$
$$(1^2 + 2^2 + \dots + n^2)^2 = \left(\frac{n(n+1)(2n+1)}{6}\right)^2\ge \frac{1}{9}n^6$$
Hence
$$\frac{1\cdot 2\cdot 3 + \dots +n(n+1)(n+2)}{(1^2 + 2^2 + \dots + n^2)^2} \le \frac{8n^4}{1/9\cdot n^6}=\frac{72}{n^2}\to 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
Geometrical Interpretation of Cauchy Riemann equations? Differentiation has an obvious geometric interpretation, and the Cauchy Riemann equations are closely linked with differentiation.
Do the Cauchy Riemann equations have a geometric interpretation?
|
The version of the Cauchy-Riemann equations that I prefer is (assuming we are using $x,y$ as coordinates for the real and imaginary part on the domain)
$$
i\partial_x f = \partial_y f
$$
This has a direct geometrical interpretation: the partial derivative with respect to the real part rotated by $\pi/2$ (which is what multiplying by $i$ does to a complex number) is the partial derivative with respect to the imaginary part.
The geometric interpretation of the complex derivative itself
$$
f(z+h) = f(z) + f'(z)h + o(h)
$$
is that $f$ can be approximated at the first order near $z$ by an affine transformation with a linear part represented by complex multiplication. If $f'(z)=re^{i\theta}$, then $h\mapsto f'(z)h$ is a rotation by $\theta$ followed (or preceded) by a uniform scaling of factor $r$.
This is much more restrictive than real differentiability for functions $\mathbb R^2\to\mathbb R^2$, which allows approximation by arbitrary affine transformations. For example the mapping $c: (x, y) \mapsto (x, -y)$ has itself as a real differential everywhere (because it's a linear transformation). Geometrically $c$ is a reflection, which changes the orientation of the plane (it sends the base $(1, 0),\ (0, 1)$ which is clockwise or anticlockwise oriented depending on how you draw the axes, to $(1, 0),\ (0, -1)$ which has the opposite orientation). When thought of as a transformation of the complex plane, $c$ is just the complex conjugation $z \mapsto \overline z$. From a geometric viewpoint $c$ has no complex derivative anywhere because if such a derivative $c'(z)$ existed the linear transformation $h\mapsto c'(z)h$ would have to be a reflection, which is impossible to obtain by composing uniform scalings with rotations.
The Cauchy-Riemann equations can be seen as a consequence of the relation between the complex derivative and the real differential. Because of the approximation argument given above and the unicity of the real differential, the complex linear transformation $h\mapsto f'(z)\cdot h$ in $\mathbb C$ corresponds in $\mathbb R^2$ to the real differential
$$
df(z_x, z_y): \begin{bmatrix}h_x\\h_y\end{bmatrix}\mapsto
\begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix}
\begin{bmatrix}h_x\\h_y\end{bmatrix}
$$
where the real partial derivatives $\partial_x f(z_x, z_y)$ and $\partial_y f(z_x, z_y)$ are column vectors and correspond to the complex partial derivatives $\partial_x f(z), \partial_y f(z)$. Taking $h = 1$, i.e. $h_x = 1, h_y = 0$, in $\mathbb R^2$ we have
$$
\begin{bmatrix}\partial_x f(z_x, z_y) & \partial_y f(z_x, z_y)\end{bmatrix}
\begin{bmatrix}1 \\ 0\end{bmatrix} =
\begin{bmatrix}\partial_x f(z_x, z_y)\end{bmatrix}
$$
which translated back to $\mathbb C$ is $f'(z)\cdot 1 = \partial_x f(z)$. Taking $h = i$ (i.e. $h_x = 0, h_y = 1$) yields $f'(z)\cdot i = \partial_y f(z)$. Together these two identities give the Cauchy-Riemann equations $\partial_y f(z) = if'(z) = i\partial_x f(z)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 7,
"answer_id": 5
}
|
Help to determine a basis for eigenspace Please find a basis for the eigenspace corresponding to eigenvalue=3 for the following matrix:
$$
\pmatrix{3&1&0\\0&3&1\\0&0&3}
$$
[3 1 0]
[0 3 1]
[0 0 3]
I have already calculated [A-(lambda)I] and the result is the following augmented matrix:
x1 x2 x3
[0 1 0 0]
[0 0 1 0]
[0 0 0 0]
From here I can see that:
* x1 is a free variable
* x2 y x3 are lead variables.
¿How is the basis determined?
|
Apparently $\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\in\ker (A-3I)$ iff $x_2=x_3=0$. So the eigenspace is spanned by $\begin{pmatrix}1\\0\\0\end{pmatrix}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is $x^2-2$ irreducible over R and Q? I'm not sure if it is valid to say that $x^2 - 2$ can be factorised to $2\cdot\left(\frac 12x^2 - 1\right)$ for it to be reducible in Q.
Though I know $(x + \sqrt{2})(x - \sqrt{2})$ works in the reals.
|
The polynomial is ofcourse reducible over $\mathbb{R}$ as it can be written as a product of polynomials of (strictly) smaller degrees.
Again, by Eisenstein's criteria it is irreducible over $\mathbb{Q}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is there a game theory for doing the dishes in a shared living situation? It occurred to me this morning (when I was intentionally not tidying up my flatmate's dishes) that doing the dishes in a shared living situation, such as at an office, or living with housemates, might be subject to a kind of game theory.
The idea being, that there's a conflict between doing the dishes yourself and immediately producing value for yourself, and the others, but at the sametime enabling the lazy housemate, and decreasing the likelyhood that they'll do the dishes in the future.
Are there any studies or research or theories around this?
|
You could take the approach @Pburg suggests and view it in the repeated games context, but I think it's simpler to view it as a positive externality question. Here's a related question (with solutions) from a game theory class (where improving one's garden, I argue, is comparable to doing the dishes):
Question:
Alice and Bob are neighbors, and each maintains his/her own garden. Each enjoys looking at the other's garden, as well as his/her own. This enjoyment is increasing in the quality of the gardens, but a higher quality garden requires more effort. Alice has Saturday off from work, while Bob has Sunday off, so Alice works on her garden before Bob. That is, Bob observes the quality of Alice's garden in deciding how much effort to put into his own, but Alice does not observe the quality of Bob's garden before making her decision. For all $i \in {A,B}$, $u_i(e_i)=e_i(c+e_{-i}-e_i)$, where $c>0$ is a constant and $e_i$ is $i$'s choice of effort. Find the subgame perfect equilibrium effort levels. Is there a first- or second- mover advantage?
Solution:
When Bob makes his decisions, he will best respond to Alice, whatever her effort choice was. That is Bob solves
$$
\max_{e_B} e_B(c+e_{A}-e_B)
$$
Taking the FOC yields $e_B^* = \frac{c+e_A}{2}$. Alice anticipates this, so she solves
\begin{align*}
&\max_{e_A} e_A(c+e_{B}-e_A) \\
=&\max_{e_A} e_A(c+\frac{c+e_A}{2}-e_A) \\
=&\max_{e_A} e_Ac + e_A\frac{c}{2} - e_A^2\frac{1}{2}
\end{align*}
Taking the FOC yields $e_A^*=\frac{3}{2}c$. That means that $e_B^*=\frac{5}{4}c$. Then $u_A=\frac{3}{2}c\left(c+\frac{5}{4}c-\frac{3}{2}c\right)=\frac{9}{8}c^2$. $u_B=\frac{5}{4}c\left(c+\frac{3}{2}c-\frac{5}{4}c\right)=\frac{25}{16}c^2>u_A$. Bob has a second-mover advantage as his payoffs are higher than they would be in the static game.
Comments:
The related concepts here are dynamic games (i.e. players make decisions sequentially) and subgame perfect equilibrium. The method to solve it was backward induction, in which we first find what the last player to move (Bob) would do, then work out that the first person will do given she knows how the second will respond.
If you really want to capture the notion of punishments and cooperation, however, it is probably best to model this as an infinitely-repeated prisoner's dilemma and look at grim-trigger and limited punishment subgame perfect strategy profiles. The gist of that line of thinking is that your housemate will do the dishes if he believes that failing to do so will yield a punishment (perhaps you not doing your own dishes for a week, or forever in the grim trigger case), and therefore there can be an equilibrium where all parties do their dishes understanding this potential punishment.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
}
|
Probability of selecting three of the same thing from a collection Question: A collection of 6 items is to be randomly drawn from a bin containing 100 good items and 8
defective items. What is the probability that exactly 3 of the items chosen are defective?
My Attempt: Well we know that in total there are 108 items. We know that you have a 8/108 chance of the first item being a defective one. I don't understand however, how we can figure out the other items, because it's dependent upon whether the first one was defective or not.
If anyone could explain this, I would greatly appreciate it.
|
The number $X$ of the defective items in the sample followes the hypergeometric distribution with parameters $N=108$ (population size), $K=8$ ("successes" in the population) and $n=6$ (sample size). Thus $$P(X=3)=\frac{\binom{8}{3}\binom{108-8}{6-3}}{\binom{108}{6}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integration: u substitution problem I'm trying to integrate the following:
$\int_{0}^{\infty} x^2 e^{-2x}$
I know that $\int_{0}^{\infty} x^n e^{-x} dx = n!$
Feels like a $u$ substitution problem but I'm having trouble making use of the above.
Thanks,
Mariogs
|
Look at the more general integral, related to Laplace transforms, of this type:
\begin{align*}
\mathcal{L}\left\{t^n\right\}\left(s\right)&=\int_0^\infty t^n e^{-st}\:dt,
\end{align*}
and you may solve using integration by parts to always get something of the form
\begin{align*}
\int_0^\infty t^n e^{-st}\:dt & = \frac{n!}{s^{n+1}}.
\end{align*}
In terms of the Gamma function you wrote,
\begin{align}
\int_{0}^\infty t^ne^{-st}\:dt&=\frac{\Gamma\left(n+1\right)}{s^{n+1}},
\end{align}
which looks somewhat nicer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
How many even 3 digit numbers contain at least one 7. How many even 3 digit numbers contain at least one 7.
I got 126, but it was not an answer choice for the problem. Can anyone help?
|
Count the number of $3$ digit even numbers that do not contain $7$ and subtract from the total number of $3$ digit even numbers.
Let $xyz$ be the $3$ digit even number. Total number of $3$ digit even numbers is $450$, since $x$ has $9$ options, $y$ has $10$ options and $z$ has $5$ options. To obtain the number of $3$ digit even numbers that do not contain $7$, $x$ has $8$ options, while $y$ has $9$ options and $z$ continues to have $5$ options, i.e., $8 \times 9 \times 5 = 360$. Hence, the number of $3$ digit even numbers that contain at-least one $7$ is $450 - 360 = 90$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
How to find the integer solutions of $\frac{2^m-1}{2^{m+x}-3^x}=2a+1$? Is there a way to find all integer triplets of $(x, m, a)$ for the following equation.
$$\frac{2^m-1}{2^{m+x}-3^x}=2a+1$$
|
@Next did already link to the other Q&A, so it is no more need to discuss this further. But I've looked at this one time with a slightly different focus, and may be the reformulation looks interesting for you for further experimenting.
Let for convenience $2a+1 = k$ and let us express $3^x $ in terms of $2^m$ such that
$ 3^x = n \cdot 2^m + r $ where $0<r<2^m$
Then your formula
$$ { 2^m- 1 \over 2^m2^x - 3^x } = 2a +1 $$
changes to
$${ 2^m- 1 \over 2^m2^x - (n2^m + r) } = k\\
{ 2^m- 1 \over 2^m (2^x - n) - r } = k\\
2^m- 1 = k(2^m (2^x - n) - r)\\
2^m = k 2^m (2^x - n) - (kr -1)\\
1 = k (2^x - n) - {kr -1 \over 2^m}\\
k (2^x - n) = {kr -1 \over 2^m}+1 \qquad \qquad \text{where } {kr -1 \over 2^m}+1\le k\\
$$
The last form of this equation has now an additional interesting property. The rhs can now be at most equal $k$ (because $r$ is smaller than $2^m$) so on the lhs the term $2^x-n$ is not allowed to become greater than 1; thus so we need to have $n=2^x-1$. But if we look now at the decomposition of $3^x$ then we see, that we must have that $3^x = n \cdot 2^m +r = (2^x-1) \cdot 2^m + r = 2^{x+m} - 2^m+r $and the difference between the perfect power of 2 and that of the perfect power of 3 is expected to be $2^{x+m}-3^x = 2^{x+m} - (2^{x+m}-2^m+r) = 2^m-r$ . But this does happen only in the "trivial" small case(s).
The relation of neighboured perfect powers of 2 and 3 have been much studied, and perhaps it is also interesting for you to look at the "Waring's" problem to see some more general relations.
One more tiny remark: we have not only a focus on the difference between perfect powers here, but also some modularity condition: the value $2a+1 = k$ must be the modular inverse of the residual $r = 3^x - n \cdot 2^m$ and is thus restricted by this rule ... and thus one might look at it with even a bit more couriosity...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Need to draw phase portrait near the equilibrium points of differential equation So, this equation
$$\ddot{x}+3\dot{x}-4x+2x^2 = 0.$$
I can write like a system
\begin{equation}
\left\{ \begin{array}{ll}
\dot{x} = v, \\
\dot{v} = 2x^2 - 4x - 3v.
\end{array} \right.
\end{equation}
Finding the equilibrium points:
\begin{equation}\nonumber
\left\{ \begin{array}{ll}
v = 0 \\
2x^2 - 4x - 3v = 0
\end{array} \right. \ \ \Rightarrow 2x^2 - 4x = 0, \ \ \Rightarrow \ \ x_1 = 0, \ x_2 = 2. \ \ \Rightarrow \\ \Rightarrow \boldsymbol{(0,0)}, \
\boldsymbol{(2,0)} - \text{equilibrium points}
\end{equation}
The matrix for the system
\begin{equation}
A(x,y)=\left(\begin{array}{ccc}
0 & 1 \\
4-2х & -3 \\
\end{array}
\right)\end{equation}
Now I need to determine what is happening in each equilibrium point.
For example, first point, $\boldsymbol{(0,0)}$:
$$A(0,0)=\left(\begin{array}{ccc}
0 & 1 \\
4 & -3 \\
\end{array}
\right)$$
$$\lambda^2 + 3\lambda -4 = 0,$$
$$\lambda_1=1, \ \ \lambda_2=-4$$
Here everything looks fine, the different signs of real roots saying to me it's saddle.
But then, let's look at the point $\boldsymbol{(2,0)}$:
$$A(2,0)=\left(\begin{array}{ccc}
0 & 1 \\
0 & -3 \\
\end{array}
\right)$$
$$\lambda^2 - 3\lambda = 0,$$
$$\lambda_1 = 0, \ \ \lambda_2 = -3.$$
It means just one eigen vector. Would it be a sink? On a picture you can see numerical solution:
I think, after I write it here, I get it why the picture should be like this. But I wrote so a lot, so I will post it, maybe someone can explain it better or add something. Thank you for attention.
|
Your first calculation is correct and leads to an unstable saddle.
However, you made a slight error in your Jacobian which affected your second calculation (the first was not affected due to the zero value).
For the Jacobian, you should have:
$$J(x,y)=\begin{bmatrix}
0 & 1 \\
4-4х & -3 \\
\end{bmatrix}$$
When you evaluate the Jacobian at your second critical point, $(2,0)$, you end up with the eigenvalues:
$$\lambda_{1,2} = \dfrac{1}{2}\left(-3 ~ \pm ~ i \sqrt{7}\right)$$
This tells you that you have a stable spiral at $(2,0)$.
Your phase portrait agrees with this. Here it is plotted with a different tool:
I would also like to mention that you are mixing variable names in your reduction and in other places in the calculations and that will surely confuse you and potentially lead to errors, so you should clean that up.
For example, you could have written the system as:
$$x' = y \\ y' = 4 x - 2 x^2 - 3y$$
Notice how that makes everything else jive.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Difference between "limit point" and "points in the closure" Given a topology $(X,T)$, $A\subset X$, $x \in X$ is a limit point of A if $\forall$ open $U$ that contains $x$, $(U\cap A)$\ {$x$} $\neq \emptyset$. $x \in X$ is in $cl(A)$ if $\forall$ open $U$ that contains $x$, $U\cap A$ $\neq \emptyset$. Is there any example that a point in the the closure of $A$ is not a limit point of A? Are the two equivalent in metric space?
|
consider a finite set A of singletons in $\mathbb{R}$, there are no limit points, but they are al in $cl(A)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1026975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Acceleration from Velocity How would you find the acceleration of an automobile at $t$ seconds from the formula giving velocity $v(t)$ in $ft/sec$:
$$v(t) = \frac{85t}{6t+16}$$
$5$ seconds ? _____$ft/sec^2$
$10$ seconds ? _____$ft/sec^2$
$20$ seconds ? _____$ft/sec^2$
Do I use the formula:
$$a = \frac{v_1 - v_0}{t_1-t_0}$$
or
$$a = \frac{340}{(3t+8)^2}$$
Thanks,
-
Update 1—According to comments, I should use the first derivative of the velocity equation. The answers I calculated are $.643, .235$, and $.074$. Would I need to divide by seconds?
Update 2—Thanks for all your help. The aforementioned values were correct! :D
|
I would strongly suggest that instead of simply applying formulas, you strengthen your concepts first.
The first equation you quoted:
$$a = \frac{v_1 - v_0}{t}$$
only applies in the case of a constant acceleration.
The acceleration here is not constant.
To find it you'll need to apply the more general rule:
$$a = \frac{dv}{dt}$$
which means differentiating the expression for velocity that you're given.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Can we make a sequence of real numbers such that polynomial of any degree with co-efficients of the sequence has all its roots real and distinct ? Does there exist a sequence of real numbers $(a_n)$ such that $\forall n \in \mathbb N$ , the polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_o$ has all $n$ real roots ? Can we make a sequence so that all the $n$ real roots are distinct ?
|
Here is an inductive, semi-non-constructive approach: assume I have a polynomial $p(x) = a_0 + \dots + a_n x^n$, with $n$ real roots and which goes to infinity as $x$ goes to infinity. Let M be large enough that all the zeros of $p$ are contained in $[-M,M]$.
Now, pick $a_{n+1}$ positive but so small that $a_{n+1}x^{n+1}$ is essentially zero on $[-2M,2M]$. Then $p(x) + a_{n+1}x^{n+1}$ looks almost the same as $p(x)$ on the interval $[-M,M]$ (in particular it has all the old zeros, slightly perturbed), but it also has a new real root, somewhere to the right of $-2M$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Power series expansion involving non integer exponent I'm working on a real and complex analysis course right now and one power series question has me really stumped:
I'm not sure what to do with the non integer in the exponent, as my initial plan of differentiating the power series of 1/(1-x) won't work.
Any help on this would be great, thanks!
|
I do not see what is the problem after Olivier Oloa's answer (use the generalized binomial theorem).
Doing so, the series expansion you look for is, around $x=a$ $$\frac{1}{(1-x)^{3/2}}=\frac{1}{(1-a)^{3/2}}+\frac{3 (x-a)}{2 (1-a)^{5/2}}+\frac{15 (x-a)^2}{8
(1-a)^{7/2}}+\frac{35 (x-a)^3}{16 (1-a)^{9/2}}+O\left((x-a)^6\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How can I show this equality between inverses of functions? Let $f:X\to Y$ be a function between metric spaces $X$ and $Y$. Show that for any $B\subset Y$, $f^{-1}(B^\complement)=(f^{-1}(B))^\complement$.
I was able to show that they both map to $B^\complement$, but I know that that's not enough to prove equality. By definition, $(f^{-1}(B))^\complement = X \setminus f^{-1}(B)$. We know that $f^{-1}(B)=\{x\in X: f(x) \in B\}$ so all $\{x\in X : f(x) \notin B\} = f^{-1}(B^\complement)$. So, taking $f(f^{-1}(B^\complement))=B^\complement$. Taking $f[(f^{-1}(B))^\complement] = f(X\setminus f^{-1}(B)) = Y\setminus B = B^\complement$. So, $f^{-1}(B^\complement)\subseteq (f^{-1}(B))^\complement$ since $f^{-1}(B^\complement)\not\subset B$ and $(f^{-1}(B))^\complement\subseteq f^{-1}(B^\complement)$ since they both map to $B^\complement$. I am not sure if this is all correct, so any suggestion would be greatly appreciated. Thanks!
|
Your $f^{-1}(B^\complement)\not\subset B$ is true but probably not what you meant, since the lefthand side is a subset of $X$ and the righthand side a subset of $Y$.
In any case you’re making it much too hard: $x\in f^{-1}[Y\setminus B]$ iff $f(x)\in Y\setminus B$ iff $f(x)\notin B$ iff $x\notin f^{-1}[B]$ iff $x\in X\setminus f^{-1}[B]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Permutations and school timetable If there are 6 periods in each working day of a school. In how many different ways can one arrange 5 subjects such that each subject is allowed at least one period? I tried this way- One of the six periods can be arranged in 5 ways and the remaining 5 periods in 5 factorial ways. Totally 600 ways
|
Lets think how we might pick this time table we can first assign a slot for each of our five subjects then finally pick the subject for the remaining slot.
Subject A has a choice of 6 slots
Subject B has a choice of 5 slots
Subject C has a choice of 4 slots
Subject D has a choice of 3 slots
Subject E has a choice of 2 slots
So far we have created $6! = 720 $ distinct permutations and there is a choice of 5 subjects for the remaining slot creating $5 \cdot 720 = 3600$ possible permutations but we note that these are not all distinct because we could swap the repeated subject with itself in the other slot the total number of distinct
timetables is therefore $\dfrac{3600}{2} = 1800.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Question about derivatives and inequalities Problem: Assume that $f: \mathbb R\rightarrow\mathbb R$ and $g: \mathbb R\rightarrow\mathbb R$ are differentiable and $f(0) = g(0)$ and $f'(0) < g'(0)$. Prove that there exists $h > 0$, so that $f(x) < g(x)$ when $0<x<h$.
I tried to solve this by letting $F(x) = g(x) - f(x)$, where $F(0) = g(0) - f(0) = 0$ and $F'(0) = g'(0) - f'(0) > 0$. Also due to the definition of derivative
$F'(0) = \lim_{x\to0}$ ${F(x) - F(0)\over x-0}$ = $\lim_{x\to0}$ $F(x)\over x$
and as we assumed $F'(0) > 0$, so must be $F(x) > 0$ if $x>0$. This leads to $g(x) - f(x) > 0$ and thus $g(x) > f(x)$ when $x>0$, which doesn't seem right because I can't show that the required $h>0$ exists. Any help would be appreciated!
|
Since $F(x)/x$ tends to a limit $F'(0)>0$ as $x\to0$, there is $h>0$ so that $F(x)/x>0$ for all $x\in(0,h)$.
Thus $F(x)>0$ for all $x\in(0,h)$, and your result follows.
The condition $F(x)/x\to F'(0)$ only means that $F(x)/x$ is close to $F'(0)$ (or positive) near $0$; the limit "does not see" the values for large $x$.
What large means depends on the situation but by the definition of a limit there always is a number $h$ as above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Derivative relation between two equal functions I am stuck with the following problem. Suppose $g: \Bbb R\rightarrow\Bbb R$ is $C^1$. $f(x,y)=g(x^2+y^2)$. I need to show that $xf_y=yf_x$
My attempt was:
$f_x=g_x \cdot 2x$ (1) and $f_y=g_y\cdot 2y$ (2)
Multiplying (1) by y and (2) by x, I get
$yf_x=g_x \cdot2xy$ and $xf_y=g_y\cdot 2xy$
So the only thing now I need to show is that $g_x=g_y$ Does $C^1$ satisfy that condition? If not, how can I prove that? A good hint would be appreciated. I would be very thankful :)
Thanks in advance!
|
Apply the chain rule
$$f_x(x,y)=g'(x^2+y^2)\cdot (2x) \quad\Longrightarrow\quad y\cdot f_x(x,y)=g'(x^2+y^2)\cdot 2xy$$
$$f_y(x,y)=g'(x^2+y^2)\cdot (2y) \quad\Longrightarrow\quad x\cdot f_y(x,y)=g'(x^2+y^2)\cdot 2xy$$
And the equality follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1027916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sum of $\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$ Find the sum of $$\sum_{n=0}^{\infty }\frac{1}{4^{(n/3)+1}}$$
|
prove by induction that for the definite sum holds
$\sum_{n=0}^{m}\frac{1}{4^{\frac{n}{3}+1}}=-1/12\, \left( {2}^{2/3}+2\,\sqrt [3]{2}+4 \right) \left( \left( 1/2
\,\sqrt [3]{2} \right) ^{m+1}-1 \right)
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find all possible combinations of $A, A, A, B, B$ 10 year old daughter has this problem.
She knows that all possible combinations of $A,B,C,D$ are $4! = 24 $
She figured it like this:
If I write down $A$ first, it has $4$ possible places.
If I write down $B$ next, it has $3$ possible places for each of the places of $A (4.3)$
If I write down $C$ next, it has $2$ possible places for each of the above placings $(4.3.2)$
If I write down $D$ last, it has $1$ possible place for each of the above placings $(4.3.2.1)$
How can she extend this method to find out all possible combinations of $A,A,A,B,B$ ?
|
Hint.
First pretend that the $A$'s and $B$'s are actually different - for example, say they're numbered $A_1$, $A_2$, $A_3$, $B_1$, $B_2$. Then this is the same problem you mentioned, and there are $5!$ ways to arrange the letters.
The problem is that you'll obtain many arrangements that differ only because of the numbering. So your problem is to determine how many arrangements there are differing only by the numbering that correspond to a single arrangement where the numbering is ignored.
For example, how many arrangements including numbers correspond to ABAAB? Can you get this by a calculation?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
$2x = 2y \Rightarrow x = y$ with $x,y \in \mathbb{Z}$ How can I show, $2x = 2y \Rightarrow x = y$ with $x,y \in \mathbb{Z}$, when I can only use elements out of $\mathbb{Z}$? It is $\frac{1}{2} \notin \mathbb{Z}$, so I can't multiply both sides with $\frac{1}{2}$. How can I prove it?
|
$$\begin{align}
2x&=2y
\\ \Rightarrow2x-2y&=0
\\ \Rightarrow 2(x-y)&=0
\\ \Rightarrow 2=0&\text{ or }x-y=0
\\ \Rightarrow x&=y.
\end{align}$$
EDIT: This (or rather I) implicitly uses the field axioms of $\mathbb{Q}$. We need this to fix it.
EDIT 2 How about this:
$$\begin{align}
2x&=2y
\\\Rightarrow x+x&=y+y
\\\Rightarrow x-y&=-(x-y)
\end{align}
$$
The only number that is its own inverse is zero. Hence $x=y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Stokes’ Theorem to find integration Use Stokes’ Theorem to evaluate integration $c (xy \,dx+ yz\, dy + zx\, dz)$ where and $C$ is the triangle with vertices $(1,0,0),(0,1,0),(0,0,1)$, oriented counter-clockwise rotation as viewed from above.
can anyone please help me with this?
|
You are planning to evaluate the path integral
$$\oint_C \vec{X} \cdot d\vec{r}$$
where
$$\vec{X} = \left(\begin{array}{cc}xy\\yz\\zx\end{array}\right) \ .$$
Stokes Theorem gives
$$\oint_C \vec{X} \cdot d\vec{r} = \iint_{\text{Inside} \ C} \nabla\times \vec{X} \cdot d\vec{S}$$
so we start by evaluating (google how to compute a curl)
$$\nabla \times \vec{X} = \left(\begin{array}{c}-y\\-z\\-x\end{array}\right) \ .$$
Fortunately, the curl is quite simple. What´s left is evaluating
$$-\iint_{\text{Inside} \ C} \left(\begin{array}{c}y\\z\\x\end{array}\right) \cdot d\vec{S} \ .$$
Now find a two-dimensional parametrisation of the surface that is the inside of triangle C and use the standard procedure to solve the surface integral. Definitely make a 3D-sketch to avoid errors. It´s just a matter of duty but quite a bit of work (as usual in vector analysis), so I will not break it down any further, sorry.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Help with Related Rates problem I'm slowly working my way through a ton of these problems but have come across one that has me stumped.
Here's the problem in full:
Water is leaking out of an inverted conical tank at a rate of
$0.0081$${\frac {m^3}{min}}$. At the same time water is being pumped
into the tank at a constant rate. The tank has height 8 meters and the
diameter at the top is 3 meters. If the water level is rising at a
rate of $0.23$${\frac {m}{min}}$ when the height of the water is 2
meters, find the rate at which water is being pumped into the tank.
So what I've done so far:
Volume of the water $V=\pi r^2\frac {H}{3}$
Height of the water $H = 2$
Radius of the tank at the $r = 0.375$ (via similar triangles when H = 2)
$C$ is the rate at which water is being pumped in and is the rate I am trying to find.
$0.0081$ is the rate at which it is leaking out.
$\frac {dV}{dt} = C - 0.0081$
$C = \frac {dV}{dt} + 0.0081$
So what I need to do is find is $\frac {dV}{dt}$ when H = 2 and then use that to find C.
I set up the problem appropriately, and used implicit differentiating to get:
(using the product an chain rules)
$\frac {dV}{dt} = (2\pi r\frac {dr}{dt})*(\frac {H}{3})+\frac {\pi r^2}{3}\frac {dH}{dt}$
This process is the same as all the other RR problems I've done, and at this point I'd simply sub in the known values and solve for the rate I'm looking for. The issue here is I don't know what $\frac {dr}{dt}$ is. $H=2$, $r=0.375$, and $\frac {dH}{dt} = 0.23$.
Please let me know what I'm missing/what I need to do. Also if I made any mistakes along the way. Thanks in advanced.
|
We can use similar triangles to get rid of a variable. Since we're given $\frac{dH}{dt}$, we want to eliminate $r$ by expressing it as a function of $H$. To this end, notice that at any point in time, we have that:
$$
\frac{r}{H} = \frac{3/2}{8} \iff r = \frac{3}{16}H
$$
Substituting before differentiating, we obtain:
\begin{align*}
V &= \frac{\pi}{3}\left(\frac{3}{16}H\right)^2H =\frac{3\pi}{256}H^3 \\
\frac{dV}{dt} &= \frac{9\pi}{256}H^2 \cdot \frac{dH}{dt} \\
C &= 0.0081 + \frac{9\pi}{256}H^2 \cdot \frac{dH}{dt} \\
\end{align*}
Subsituting the known values at the given snapshot in time, we obtain:
$$
C = 0.0081 + \frac{9\pi}{256}(2)^2 \cdot (0.23) = 0.109711\ldots \frac{\text m^3}{\text {min}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proofs of Sets and Subsets I have these proof problems that I need some help on, any direction would be great. Thanks
Let A, B, and C be subsets of some universal set U
(a) Prove the following:
IF $A \cap B$ $\subseteq$ C, and $'A \cap B$ $\subseteq$ C, THEN $B \subseteq C$
(b) Either prove the following or provide a counterexample:
IF $A \cap B$ = $A \cap C$ and $'A \cap B$ = $'A \cap B$ = $'A \cap C$, THEN B = C
|
Hint. For (a) it is given that
$$A\cap B\subseteq C\ ,\quad A'\cap B\subseteq C$$
and you have to prove $B\subseteq C$. You should know the basic way of proving a subset statement like this: assume $x$ is in the LHS, and use this assumption (and the given facts) to prove that $x$ is in the RHS.
So, let $x\in B$. Consider two cases: either $x\in A$ or $x\in A'$.
*
*Case 1, $x\in A$. Then $x\in A\cap B$, so $x\in C$.
*Case 2, [fill in the details yourself].
In both cases, $x\in C$. Therefore $B\subseteq C$.
You can use (a) to answer (b). We have
$$A\cap B=A\cap C\subseteq C\ ,\quad A'\cap B=A'\cap C\subseteq C\ ,$$
so by (a) we get $B\subseteq C$. See if you can write out a similar argument to show $C\subseteq B$ and thereby prove $B=C$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to decompose a matrix into an antisymmetric matrix plus a multiple of the identity I was given a problem to solve earlier that I couldn't figure out. I don't still have it, but it was basically: Given the invertible matrix $A$, find the invertible matrix $P$, such that $A=P^{-1}CP$. Where $$C=\begin{bmatrix} c & -d \\ d & c\end{bmatrix}$$ and $A$ I don't exactly remember, but it was something like $$A=\begin{bmatrix} 3 & 1 \\ 1 & 0\end{bmatrix}$$
I was completely stumped. I see that $A$ and $C$ are similar so they should have the same determinant and trace. So I should be able to find $C$ from that. But how would I find $P$?
And I can see that this is similar to eigendecomposition, but is this decomposition at all useful, or did some textbook writer just come up with this problem with no underlying use?
|
Hint: If two matrices are diagonalizable with the same eigenvalues, then they are similar.
If an $n \times n$ matrix has $n$ distinct eigenvalues, it is necessarily diagonalizable.
The punchline: you should try to find a matrix $C$ that has the same eigenvalues as $A$.
Note that if we have invertible matrices $S,T$ such that
$$
SCS^{-1} = D = TAT^{-1}
$$
then
$$
A = T^{-1}SCS^{-1}T = (S^{-1}T)C(S^{-1}T)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Range conditions on a linear operator While reading though some engineering literature, I came across some logic that I found a bit strange. Mathematically, the statement might look something like this:
I have a linear operator $A:L^2(\Bbb{R}^3)\rightarrow L^2(\Bbb{R}^4)$, that is a mapping which takes functions of three variables to functions of four variables. Then, "because the range function depends on 4 variables while the domain function depends on only three", there must be redundancy in the operator $A$, that is the range of $A$ is a proper subset of $L^2(\Bbb{R}^4)$, characterized by some range conditions.
Is such a statement always true? For the specific example I am reading about (the X-ray transform), it is definitely true - in fact, the range of the operator is characterized by a certain PDE - but I can't image such a thing is true in general.
For instance, I can cook up an operator $A:L^2(\Bbb{R}^3)\rightarrow L^2(\Bbb{R}^4)$ such that the range of $A$ is dense in $L^2(\Bbb{R}^4)$: simply choose orthonormal bases $(e_j)$ and $(f_j)$ for both, then map $e_j$ to $f_j$.
Any thoughts?
|
I think it always has to be true. It is like mapping points in a flat plane to a 3D space. You get a volume 0 manifold in the 3D space if there is a 1 to 1 mapping.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Trigonometric substitution in the integral $\int x^2 (1-x^2)^{\frac{9}{2}} \ \mathrm dx$ I'm trying to solve
$$\int_{-1}^{1} x^2(1-x^2)^{\frac{9}{2}} \, dx$$
The hint said to use the substitution $x=\sin y$
I got $$\int_{-\pi/2}^{\pi/2} \sin^2y \cos^{\frac{11}{2}} y \,dy$$
|
As said in comments, the antiderivative reduces to the evaluation of $$I=\int \cos^{10}(x)dx-\int \cos^{12}(x)dx=J_{10}-J_{12}$$ with $$J_n=\int \cos^{n}(x)dx$$ The $J_n$ terms can easily be computed since we can easily establish (performing two integrations by parts) the recurrence relation $$J_n=\frac 1n \cos^{n-1}(x)\sin(x)-\frac{n-1}{n}J_{n-2}$$ with $$J_0=x$$ $$J_1=-\sin(x)$$ Integrating between the given bounds $(-\pi/2,\pi/2)$, most terms disappear because of the cosines and we can then establish that $$J_n=\frac{\sqrt{\pi } \Gamma \left(\frac{n+1}{2}\right)}{\Gamma
\left(\frac{n}{2}+1\right)}$$ So $J_{10}=\frac{63 \pi }{256}$ and $J_{12}=\frac{231 \pi }{1024}$ and then the result Adhvaitha gave.
Notice that if $n=2m$ this simplify further $$J_{2m}=\frac{\sqrt{\pi } \Gamma \left(m+\frac{1}{2}\right)}{m!}$$ with $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma \left(m+\frac{1}{2}\right)=\left(m-\frac{1}{2}\right) \Gamma
\left(m-\frac{1}{2}\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Is $\sqrt2$ a tricky notation? When someone asked me how to solve $x^2=9$,I can easily say, $x=3$ or $-3$. But what about $x^2=2$? There is NOT any "ordinary" number to solve this question. It's an irrational number. So we say helplessly, the answer is $\pm\sqrt2$, but what does $\sqrt2$ mean? It's a number, when squared, equals $2$.
This is a cycle define, just like "what's grandfather mean?father's father - what's father mean、grandfather‘s son." It does NOT tell anything more. And if we can define notations arbitrarily, we can give any questions answer tricky. For example, what's $123456789\times987654321$? We need not calculate, just define $f(x)=123456789x$, the answer is $f(987654321)$.
|
Yes, $\sqrt{2}$ only tells you that it is a number which, when squared, yields $2$. It's a whole lot more informative than any other thing you might write for the same number. However, the fact that $\sqrt{2}^2=2$ is really important in certain contexts. For instance, in higher mathematics, we are often less concerned with the easily determined fact $\sqrt{2}$ is somewhere between $1.4$ and $1.5$ than we are with other questions about it.
In particular, some branches of mathematics stop thinking about the real numbers altogether and stop thinking about arranging things on the number line, and just want to talk about addition and multiplication. They start off in the rational numbers, $\mathbb Q$, equipped with their ordinary addition and multiplication and move further. Quickly, one can create questions which have no solution, like:
What $x$ satisfies $x^2=2$?
which can't be solved in the rationals. However, a very natural question is, "Well, supposing there were a solution to that, what properties might it have?" So, we define a new number, $\sqrt{2}$ and extend the rationals by it to the field $\mathbb{Q}[\sqrt{2}]$. What's this mean?
Well, now we're considering any number which can be written as a polynomial, with rational coefficients, of $\sqrt{2}$ - or equivalently, the numbers that can be written as a sum or product of rational numbers and $\sqrt{2}$. So, we're now interested in things like $\sqrt{2}+1$ and $\frac{1}2-3\sqrt{2}+\sqrt{2}^3$ and how addition and multiplication might work with them. Provably, every such number is of the form
$$a+b\sqrt{2}$$
for rational $a$ and $b$ and we define addition and multiplication as
$$(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})=(a_1a_2+2b_1b_2)+(a_1b_2+b_1a_2)\sqrt{2}$$
$$(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})=(a_1+a_2)+(b_1+b_2)\sqrt{2}$$
which might not look like much at first, but suddenly, we have a new field in which we can perform addition and multiplication (and we even find results like division if we look harder) - and we find curious things like defining $\overline{a+b\sqrt{2}}=a-b\sqrt{2}$ preserves all the structure of multiplication and addition, which tells us that $\sqrt{2}$ and $-\sqrt{2}$ are somehow interchangeable.
This branch of mathematics is too large to summarize in any adequate way, but essentially, my point is that the algebraic properties of a number - that is, how it responds to addition and multiplication - are very worthwhile in their own right, and hence, though the notation involves "inventing" new numbers that we can't write in any satisfying closed form like we can the rationals, the definition "$\sqrt{2}$ is a number which, when squared, gives $2$" actually has a lot of interest to it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
}
|
sum and product of Lipschitz functions I have the following question from my notes, where $f$ and $g$ are Lipschitz functions on $A ⊂ \Bbb R$.
I'm able to show that the sum $f + g$ is also a Lipschitz function, however I'm stuck on trying to show that if $f,g$ are bounded on $A$, then the product $fg$ is Lipschitz
on $A$.
Also, is there a valid example of a Lipschitz function $f$ on $[0,+∞)$ such that $f^2$ is
not Lipschitz on $[0,+∞)$?
|
Suppose $|f| \leq M$ and $|g| \leq M$.
Then
$$ |(fg)(x) - (fg)(y)| \leq |f(x)g(x) - f(x)g(y)| + |f(x)g(y) - f(y)g(y)| \leq M(|g(x) - g(y)|+|f(x)-f(y)|)$$
Consider $f(x) = x \, \forall x \in [0,\infty)$ which is Lipchitz on $[0, \infty)$ but $f^2$ is not.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1028974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
}
|
Closed- form of $\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$ Is there a possibility to find a closed-form for
$$\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$$
|
Focus on the relation
$$\frac{d}{dx}{\rm Li}_k(x)=\frac{1}{x}{\rm Li}_{k-1}(x)$$
Let's look at the derivative of ${\rm Li}_m {\rm Li}_n$ in general:
$$\frac{d}{dx}({\rm Li}_m(-x){\rm Li}_n(-x))=\frac{1}{x}({\rm Li}_{m-1}(-x){\rm Li}_n(-x)+{\rm Li}_m(-x){\rm Li}_{n-1}(-x))$$
Let's define
$$f(m,n)=\int_0^1 \frac{{\rm Li}_m(-x){\rm Li}_n(-x)}{x^2}dx$$
Integration by parts:
$$f(m,n)=-\frac{1}{x}{\rm Li}_m(-x){\rm Li}_n(-x)|_0^1 +\int_0^1 \frac{{\rm Li}_{m-1}(-x){\rm Li}_n(-x)+{\rm Li}_{m}(-x){\rm Li}_{n-1}(-x)}{x^2}dx$$
$$f(m,n)=-\frac{1}{x}{\rm Li}_m(-x){\rm Li}_n(-x)|_0^1 +f(m-1,n)+f(m,n-1)$$
This is a recursive relation that can express $f(3,3)$ with lower terms that are easier to express analytically. Note that the nonintegral part above has to be taken as a limit at $x=0$. You can imagine having $\epsilon$ for the lower integral bound and taking $\epsilon\to 0$ at the end.
Anyway, you can see from the power series definition that
$$\lim_{\epsilon\to 0}\frac{1}{\epsilon}{\rm Li}_m(\epsilon){\rm Li}_n(\epsilon)=0$$
Additionally, ${\rm Li}_n(-1)=-\eta(n)$ where $\eta$ is the Dirichlet eta function. Simplification:
$$f(m,n)=-\eta(m)\eta(n) +f(m-1,n)+f(m,n-1)$$
First of all, $f$ is symmetric in the arguments. Secondly, ${\rm Li}_0(x)=\frac{x}{1-x}$ so recursion can end at
$$f(n,0)=-\int_0^1 \frac{{\rm Li}_n(-x)}{x(1+x)}dx=-\int_0^1 \frac{{\rm Li}_n(-x)}{x}dx+\int_0^1 \frac{{\rm Li}_n(-x)}{1+x}dx=$$
$$=-{\rm Li}_{n+1}(-1)+\int_0^1 \frac{{\rm Li}_n(-x)}{1+x}dx$$
The first one is again the eta function. But the second one I don't know what to do with. Maybe this wasn't such a good idea. Any suggestions?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
}
|
Integration with respect to a signed measure Let $\mu$ be a singed measure, $f\in C_c(X)$, I want to show
$$\int fd(c\mu) = c \int fd\mu, \forall c \in \mathbb{R}$$
Since $c\mu$ is also a singed measure, I think by definitionm I need to show $$\int fd(c\mu)^+-\int fd(c\mu)^- = c \int fd\mu^+-c \int fd\mu^-$$
But I think here I can't directly say $d(c\mu)^+=cd\mu^+,d(c\mu)^-=cd\mu^-$, since the definition of $\mu^+,\mu^-$ involves $|\mu|$ and $|c\mu|=|c||\mu|$, hence I need to discuss wether $c<0$ or $c>0$ or $c=0$.
For example, $c<0$, then $$(c\mu)^+=\frac{1}{2}(|c\mu|+c\mu)=-c\mu^-$$$$(c\mu)^-=\frac{1}{2}(|c\mu|-c\mu)=-c\mu^+$$
Hence the LHS equals to the RHS since $\int fd(c\lambda) = c \int fd\lambda, \forall c >0$ when $\lambda$ is a measure.
Since I am new to this subject, could you help to confirm whether my understanding is correct?
|
Seems correct, I assume that you used the following line of reasoning: if $c<0$ then
$$
(c\mu)^+ = \frac12(|c\mu|+c\mu) = \frac12|c|(|\mu|-\mu) = |c|\mu^- = -c\mu^-.
$$
Alternatively, you could go via the decomposition of the state space into positive and negative sets.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Why is indefinite integral called so? Two questions that are greatly lingering on my mind:
1.
Integral is all about area(as written in Wolfram).
But what about indefinite integral? What is the integral about it?? Is it measuring area?? Nope. It is the collection of functions the derivative of which give the original function and not measuring area. So, why "integral"?? And what about the indefinite??? It is not measuring an infinite area ; just telling about the original functions. So, what is the logic of this name??
*Famous statement:
Differentiation breaks apart the function infinitesimally to calculate the instantaneous rate of change, while, on the other hand, integration sums up or integrates the infinitesimal changes to measure the whole change or area .
Yes, totally correct but in case of definite integrals,where small changes are summed up to give the area. But how is the statement related with indefinite integral?? Do they sum up small changes??? What is the connection between them???
I am confused. Please help me explaining these two problems.
|
A primitive of a function $f$ is another function $F$ such that $F'=f$. If $F$ is a primitive of $f$, so is $F+C$ for any constant $C$, the so called constant of integration. The indefinite integral of $f$ can be thought of as the set of all primitives of $f$:
$$
\int f=F+C.
$$
Why indefinite? Because is there some indefinition due to the constant $C$.
What is the relation to areas, or definite integrals? The fundamental theorem of calculus. If $F$ is a primitive of $f$ then
$$
\int_a^bf=F(b)-F(a).
$$
Indefinite integrals are a tool for the computation of definite integrals.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Is $\nabla$ a vector? The following passage has been extracted from the book "Mathematical methods for Physicists":
A key idea of the present chapter is that a quantity that is properly called a vector must have the transformation properties that preserve its essential features under coordinate transformation; there exist quantities with direction and magnitude that do not transform appropriately and hence are not vectors.
Cross product: $\nabla \times (Vector)=Vector$
From the above equation of cross product we can say that $\nabla$ is a vector (specifically vector operator). However, a vector generally has magnitude and an associated direction. While in case of $\nabla$, it might satisfy essential features under transformation to be a vector, but I don't see whether it has magnitude or not? Does it has magnitude? If so, what is it? Or otherwise is it that a vector need not have magnitude?
|
With $f$ a scalar function of the coordinates, $\nabla f$ is a vector called the gradient of $f$.
With $f$ a vector function of the coordinates, $\nabla.f$ is a scalar called the divergence of $f$.
With $f$ a vector function of the coordinates, $\nabla\times f$ is a vector called the curl of $f$.
These three symbols ($\nabla,\nabla.,\nabla\times$) are differential operators and represent no quantity by themselves.
If you really want to see $\nabla$ as a vector, then it is $$\nabla=i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
A fair game with triangular numbers?
Definition 1
A ball game is a state where you have $n$ white balls and $m$ black balls.
The rule is that you remove first one ball from the cup. And without returning the first ball, you pick another.
*
*$P(A)$ is defined as the probability of drawing two balls with opposite colors
*$P(B)$ is defined as the probability of drawing two balls with the same color
A game is defined as fair if $P(A)=P(B)$.
Conjecture 1
A game is fair if and only if $n$ and $m$ are consecutive triangular numbers.
$(n,m) = (1,3) \ , \ (3,6) \ , \ (6,10) \ \ldots$
Image a cup with 2 white balls and 2 black balls. If you draw two balls with the same color, you win. If on the other hand you draw two balls with opposite colors, I win. Note that once a ball is drawn the ball is gone.
Is this a fair game? No, ofcourse not. After you pull a ball from the cup there are two of my colors, and only one of yours. Giving me a $P = 2/3$ chance of winning. This can be illustrated in the following diagram
Following a doted line means I win. Following a whole line, you win. There are more doted lines, than whole; hence I win. One can make this game fair by changing the balls
just count the lines, or do the simple math. Now a fun generalization is to find all configurations that allow a fair game. Suprisingly this is always two consecutive triangular numbers. I want to explain this to my class in an intuitive way, perhaps let them explore it.
*
*Is there a way to use the diagrams or else to obtain a intuitive explenation why the solutions are always two consecutive triangular numbers?
|
I think the diagram may not generalise in an obviously triangular way. For example with $3$ and $6$ the diagram has $36$ lines ($72$ if you count them in both directions) and the best I could do was something like
I think the most you can say is that this suggests for a fair game you have $$mn = \tfrac12m(m-1) +\tfrac12n(n-1)$$ i.e. $$2mn = m(m-1)+n(n-1)$$
I do not see how to see directly from the diagram that the pair $\frac12k(k-1),\frac12k(k+1)$ provides a solution for positive $k$, or that there are no other essentially different solutions
The mechanical approach would be
*
*Use the quadratic formula to show $m=n +\frac{1\pm\sqrt{8n+1}}{2}$
*$8n+1$ is an odd integer so its square root is either irrational or another odd integer
*$\sqrt{8n+1}=2k+1$ has the solution $n=\frac12k(k+1)$, i.e. a triangular number
*$n=\frac12k(k+1) \implies m=\frac12k(k-1)$ or $\frac12(k+1)(k+2)$ and you are done
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Book recommendation for ordinary differential equations This question has been posted before, but I need book with specific qualifications. I do not need books for engineers, book that is centered around calculations and stuff. I need to find a book that is theoretical, proves the statements and has good presentation of the theoretical structure. I have had the book by Tenenbaum, I did not like it. I would be very very thankful if someone shared their knowledge with us about this matter.
|
I would like to add "Differential Equations With Applications and Historical Notes by George Simmons" to this list. This is a very well written text on ODEs and very approachable and explains concepts very clearly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 2
}
|
How to solve the integral equation? How to solve the integral equation
$$ \int_{-20}^{x} \left| \left| \left| \left| \left| \left|
\left| \left| t \right| -1 \right| -1 \right| -1 \right| -1 \right|
-1 \right| -1 \right| -1 \right| \,{\rm d}t={\frac {4027}{2}}?$$
|
The main tough part is getting a grasp on the function in the integral. Let
$$
A_0(t) = |t|
$$
and for $n > 0$
$$
A_n(t) = |A_{n-1}(t) - 1|
$$
Then the function integrated is $A_7(t)$ (count the -1's).
Now it is easy to prove (by a two-step induction proof on odd and even $n$) that for all $n \in \Bbb{Z}$
$$
|t| \leq n \Longrightarrow A_{n}(t) = |t| - n
$$
and
$$
\int_{-n}^{n} A_{n}(t)dt = n $$
So, for $x > 7$,
$$
\int_{-20}^{-7}A_7(t)dt = \int_{-20}^{-7} (-t-7) dt + 7 + \int_{7}^{x}(t-7)dt = \frac{169}{2} + 7 + \frac{(x-7)^2}{2}
$$
Then
$$\frac{183}{2} + \frac{(x-7)^2}{2}= \frac{4027}{2}$$
$$(x-7)^2 = 3844 $$
$$x = 62+7 = 69 $$
(The other solution, $x = -55$, clearly does not satisfy the original problem, since the integrand is always non-negative, so an integral from $=20$ to $-55 < -20$ is necessarily negative.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Very basic question about submanifolds. I'm beginning to study differential geometry and I'm a litle confused about the concept of submanifold of a differentiable manifolds.
Can someone provide me an example of how to show that a non-open subset of a manifold is a submanifold and other showing it isn't?
|
Here are some very simple examples.
Let $M = \mathbb R^2$. Let $N$ be a straight line in $M$. Then $N$ is not open in $M$ and a submanifold of $M$. For example, you can choose global coordinates in $M$ such that $N$ is the first coordinate axis.
Let $P \subseteq M$ be the union of two straight lines which meet in one point. Then $P$ is not open in $M$, and $P$ isn't a submanifold of $M$. In fact, $P$ isn't even an manifold, because the meeting point of the two lines doesn't have a one-dimensional local trivialization.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
The behavior of BV functions at a point of approximate continuity Given $u\in BV(\mathbb R^N)$, we say $u$ is approximate continuos at $x$ and the approximation limit is $l\in R$ if
$$ \lim_{r\to 0}\frac{\mathcal{L}^N(B(x,r)\cap \{|u-l|>\epsilon\})}{r^N} =0 $$
for all $\epsilon>0$. (I know the approximate continuity can be defined for even just a barely measurable function. But given I am studying $BV$, let's keep $u\in BV$ and maybe it is useful).
Now fix any $x_0\in \mathbb R^N$ such that $u$ is approximate continuous at this point $x_0$. My first question is: for a fixed $\epsilon_0>0$, would it be possible to have $r_{\epsilon_0}$ defined such that
$$ \mathcal{L}^N(B(x,r)\cap \{|u-l|>\epsilon_0\})=0 $$
for all $r<r_{\epsilon_0}$
My second question is not related with above. Given two conditions: for a fixed $x_0\in R^N$,
$$ \lim_{r\to 0}\frac{\mathcal{L}^N(B(x_0,r)\cap \{u>l+\epsilon\})}{r^N} =0, \,\,\text{ for all }\epsilon>0 \tag{1}$$
|
(I'll answer the first question; please post the second one separately.)
The answer is negative. In $N\ge 2$ dimensions, let $\xi$ be some unit vector and define
$$
u(x) =\sum_{n=1}^\infty \left(1-4^n |x-2^{-n}\xi|\right)^+
$$
This function is in BV, because the BV norm of the $n$th term (i.e., $L^1$ norm of its gradient) is of order $4^n4^{-nN} = 4^{-(N-1)n}$. It is approximately continuous at the origin because the set $\{u\ne 0\}$ is the union of balls $B(2^{-n} \xi, 4^{-n})$ which has Lebesgue density $0$ at the origin. Yet, for every $r>0$ the set $\{u>1/2\}$ intersects $B(0,r)$ in a set of positive measure.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1029991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
An odd function $f$ is differentiable at zero. Prove $f'(0)=0$? I know that $f'$ of an even function is odd function, thus I have $f(x)=f(-x)$.
However I'd no idea how to prove that $f'(0)=0$?
Please answer my question...
|
I assume you mean that $f(x)$ is an even function. If so, note that for an even function $f(x)$, it satisfies $f(x)=f(-x)$. Now if a function is differentiable, it satisfies
$$f'(x) =\lim_{h\to0^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}$$
Now note that
$$f'(0)=\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{f(-h)-f(0)}{-h}$$$$=-\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=-f'(0)$$
So we have $f'(0)=-f'(0)\implies f'(0)=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute
$$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$
What I am doing is take
$$f(z)=\frac{(\log z)^2}{1+z^2}$$
and calculating
$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$
which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$
Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$
Any help will be appreciated. thank you in advanced.
|
For this integral, you want to define you branch cut along the negative x axis and use a contour with a semi circle around the origin.
Also, note we are taking the real Cauchy Principal value.
Then
$$
\int_0^{\infty}\frac{(\ln(z))^2}{1+z^2}dz = \int_{\Gamma}f(z)dz+\int_{\gamma}f(z)dz + \int_{-\infty}^0f(z)dz + \int_0^{\infty}f(z)dz
$$
Let $R$ be the radius of big semi circle, $\Gamma$ and $\delta$ the radius of the small semi circle $\gamma$. When $R\to\infty$, $\int_{\Gamma}\to 0$ and $\int_{\gamma}\to 0$ when $\delta\to 0$ by the estimation lemma.
Then
$$
\int_{-\infty}^0f(z)dz + \int_0^{\infty}f(z)dz = 2\pi i \sum\text{Res}_{\text{UHP}}
$$
where UHP is the upper half plane. There is only one pole in the upper half plane and that is $z = i$
$$
\text{Re PV}\biggl[\int_{-\infty}^0\frac{(\ln|z| + i\pi)^2}{z^2+1}dz + \int_0^{\infty}\frac{(\ln|z| + i\cdot 0)^2}{z^2+1}dz\biggr] = 2\pi i\sum\text{Res}_{\text{UHP}}
$$
Recall that $\ln(z) = \ln|z| + i\arg(z)$.
Can you take it from here?
Mouse over for solution.
\begin{align}\text{Re PV}\biggl[\int_{-\infty}^0\frac{\ln^2|z| + 2\pi i\ln|z| - \pi^2}{z^2 + 1}dz +\int_0^{\infty}\frac{\ln^2|z|}{z^2 + 1}dz\biggr]&= 2\pi i\lim_{z\to i}(z - i)\frac{(\ln|z| + i\pi/2)^2}{z^2+1}\\2\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx - \pi^2\int_0^{\infty}\frac{dx}{x^2+1} &= -\frac{\pi^3}{4}\\2\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx - \frac{\pi^3}{2} &= -\frac{\pi^3}{4}\\\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx &= \frac{\pi^3}{4} - \frac{\pi^3}{8}\\&= \frac{\pi^3}{8}\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
}
|
How do I convert this parametric expression to an implicit one I have:
$$x=5+8 \cos \theta$$
$$y=4+8 \sin \theta$$
With $ -\frac {3\pi}4 \le \theta \le 0$
If I wanted to write that implicitly, how would I do it? I get that it's a circle, and I can easily write the circle implicitly, but I'm not sure how to convert the domain from $ -\frac {3\pi}4 \le \theta \le 0$ to a suitable domain in x and y.
I can think of a way to do it that would involve using the $Min$ function, but is that allowed in an implicit representation?
|
Use that $\exists \theta : (v,w) = (\cos\theta, \sin\theta) \iff v^2 + w^2 = 1$.
You immediately get that the parametric curve is a part of the curve defined by
the implicit equation
$$
\left(\frac{x-5}8\right)^2 + \left(\frac{y-4}8\right)^2 = 1$$
To get only the part that you want, you must also make sure that $-\frac{3\pi}4\le \theta\le 0$, which is equivalent to
$$
\cos \theta \ge -\frac 12, \sin\theta< 0
$$
and then you get the equation
\begin{cases}
\left(\frac{x-5}8\right)^2 + \left(\frac{y-4}8\right)^2 = 1\\
\frac{x-5}8 \ge -\frac 12\\
\frac{y-4}8 \le 0
\end{cases}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Do solutions of $\dot{x} = \frac{x}{t^2} + t$ exist satisfying $x(0) =0$ Suppose we have the 1-dimensional ODE
\begin{equation}
\dot{x} = \frac{x}{t^2} + t
\end{equation}
Do there exist solution curves with initial condition $x(0)=0$? If you proceed in a standard way then you would get as solution formula
\begin{equation}
x= C {\rm e}^{-1/t} + {\rm e}^{-1/t} \int^{t}_0 \tau {\rm e}^{1/\tau} d \tau
\end{equation}
but then the expression in the integral explodes! This does not imply that solution curves with $x(0)=0$ do not exist. How to proceed with such existence results?
|
The problem is that your original problem is not a Cauchy problem
The usual setup for solving differential equations is that you are given a function $f: [a,b] \times \mathbb R \to \mathbb R$ and a point $(x_0, y_0)$ and you want to find a function $y:[a,b]\to\mathbb R$ which satisfies two conditions:
*
*$f(x,y(x)) = y'(x)$ for all values $x\in[a,b]$
*$y(x_0)=y_0$
In your case, you only defined the function $f$ on $(a,b]\times\mathbb R$, since $f(t, x)=\frac{x}{t^2}+t$ in your case. Your equation has many solutions. For example, taking $x(t)=0$ creates a function that satisfies $f(t, x(t))=\dot x(t)$ for every $t\in (0,1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Showing that normal line passes through a point. I need to show that a line passes through a point. How should I go about doing this? The question is:
Let $L$ be the normal line at $(1,1,1)$ to the level surface of $f(x,y,z) = x^2 - z$ that passes through $(1,1,1)$. Show that $L$ passes through the point $(3,1,0)$.
Thanks in advance!
Edit: I tried taking the gradient, where the gradient is (2,0,-1). So L would be 2(x-1)-(z-1) = 0. Is this right? If it is, how do I show that it passes through the point?
Edit 2: How do I show that it passes through (3,1,0)? (1+2(3),1,1-(0))?
Edit 3: So you're saying that t_0 = 1? I'm confused as to how does this show that L passes through the point.
Edit 4: Can you please show an example where the point is NOT in L? Thanks!
Edit 5: So conflicting values of t? Thank you very much for your help!
|
If the line passes through the two given points, its direction vector is $(3,1,0)-(1,1,1)=(2,0,-1)$, indeed parallel to the gradient.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Analytic Combinatorics to asymptotically estimate the number of objects of size at most n? I have read some bits of Flajolet's and Sedgewick's book on Analytic Combinatorics.
I am quiet curious as how to asymptotically estimate the number of objects of size at most n.
Suppose for example that I specified a combinatorial class $A$. I also computed its generating function $A(z)=\sum\limits_n a_n z^n$. Therefore, studying the singularities of $A(z)$ yields an asymptotic estimate of the coefficient $a_n$, that represents the number of objects of size n.
How can I asymptotically estimate the number of objects of size at most $n$ ($\sum\limits_{k=0}^{n} a_n$)?
Ideally, I would seek a combinatorial interpretation that would let me re-use the generating function $A(z)$, but I am drying up!
Any thoughts?
|
Indeed, it does the trick!
B(z) = $A(z) \times \sum\limits_n z^n = \sum\limits_{n} \sum\limits_{k = 0}^n a_k z^n$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Can the underlying set functor corresponding to an algebraic theory always be viewed as a model of that theory? Let $\mathsf{T}$ denote a Lawvere theory, and let $\mathbf{C}$ denote its category of models in $\mathbf{Set}$. Write $U : \mathbf{C} \rightarrow \mathbf{Set}$ for the underlying set functor. I think that $U$ can always be viewed as model of $\mathsf{T}$ in the functor category $[\mathbf{C},\mathbf{Set}].$ For example, suppose $\mathsf{T}$ has a binary operation $f : G \times G \rightarrow G$, where $G$ is the generic object of $\mathsf{T}$. Then there should be a corresponding natural transformation $\nu : U \times U \Rightarrow U$ given by writing $\nu_X = X(f)$ for all objects $X$ of $\mathbf{C}$. Hence $U$ becomes equipped with the operations of $\mathsf{T}$ in a natural way.
Is this right? If so, I'd appreciate some kind of a discussion of how this all works. Its currently very fuzzy to me.
|
In more detail: Recall that a model of a Lawvere theory $\mathcal{T}$ in a category (with finite products) $\mathcal{S}$ is a finite-product-preserving functor $\mathcal{T} \to \mathcal{S}$. Thus, for any category $\mathcal{C}$, a $\mathcal{T}$-model in $[\mathcal{C}, \mathbf{Set}]$ is the same thing as a diagram $\mathcal{C} \to \mathbf{Mod}(\mathcal{T}, \mathbf{Set})$; in symbols:
$$\mathbf{Mod}(\mathcal{T}, [\mathcal{C}, \mathbf{Set}]) \cong [\mathcal{C}, \mathbf{Mod}(\mathcal{T}, \mathbf{Set})]$$
In particular, we may take $\mathcal{C} = \mathbf{Mod} (\mathcal{T}, \mathbf{Set})$. Then,
$$\mathbf{Mod}(\mathcal{T}, [\mathbf{Mod} (\mathcal{T}, \mathbf{Set}), \mathbf{Set}]) \cong [\mathbf{Mod} (\mathcal{T}, \mathbf{Set}), \mathbf{Mod}(\mathcal{T}, \mathbf{Set})]$$
and the $\mathcal{T}$-model structure on the forgetful functor $\mathbf{Mod} (\mathcal{T}, \mathbf{Set}) \to \mathbf{Set}$ is precisely the one corresponding to the identity functor on $\mathbf{Mod} (\mathcal{T}, \mathbf{Set})$.
In fact, all of this goes through for any category $\mathcal{S}$ with finite products instead of $\mathbf{Set}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1030984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Is there any holomorphic function in a unit ball Is there any holomorphic function in a unit ball such that $f(1/n)=n^{-5/2}$ for $n=2,3,\dots$
Natural candidate is $f(z)=z^{5/2}$ But it isn't holomorphic obviously inside that ball. Can you tell me whatI should check?
|
If $f$ is such a function, then $g(z)=f(z)^2$ conincides with $z\mapsto z^5$ on a sequence of points converging to $0$. Hence $g(z)=z^5$ for all $z$. If we write $f(z)=z^kh(z)$ with $k\in\mathbb N_0$, $h$ holomorphic, $h(0)\ne0$, we find that $z^{2k}h(z)^2=z^5$, qea.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Subtraction of functions with BigO When trying to assess the Big $O$ of two functions that are added together, we take the max of the two. What happens if there is subtraction instead of addiiton?
for instance: $$f(n) = O(n^3) $$
$$ \text{and} $$ $$g(n) = O(n^3)$$
then
$$ (f-g)(n)$$
|
Note that the sign of a function doesn't matter in $O $-notation:
If $f(n)\in O (h(n)) $ then $-f(n)\in O (h(n))$ follows directly from the definition of the $O $-Notation.
For two functions $f (n)\in O (h_1 (n)) $ and $g (n)\in O (h_2 (n))$ you know
$$f(n)+g (n)\in O (\max (h_1 (n), h_2 (n))). $$ where in this case
$\max (h_1 (n), h_2 (n))=h_1(n)$ means that $h_2 (n)\in O ( h_1 (n))$ respectively $\max (h_1 (n), h_2 (n))=h_2(n)$ means that $h_1 (n)\in O ( h_2 (n))$ .
Therefore, you can follow
$$f (n)-g (n)=f (n)+ (-g (n))\in O (\max (h_1 (n), h_2 (n)))$$
since $-g (n)\in O (h_2 (n))$, too.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Suppose $f$ is integrable on $\mathbb{R}$, and $g$ is locally integrable and bounded. Then $f*g$ is uniformly continuous and bounded?
Suppose $f$ is integrable on $\mathbb{R}$, and $g$ is locally
integrable and bounded. Then $f*g$ is uniformly continuous and
bounded?
I don't even know where to start proving or disproving, but I feel it is true. Can somebody give me a hint?
|
Assuming $g$ bounded
$$
|f\ast g(x)|\le\int_{\mathbb{R}}|g(y)|\,|f(x-y)|\,dy\le\|g\|_\infty\|f\|_1.
$$
Next, for $h\in\mathbb{R}$
$$
|f\ast g(x+h)-f\ast g(x)|\le\|g\|_\infty\int_{\mathbb{R}}|f(x+h-y)-f(x-y)|\,dy=\int_{\mathbb{R}}|f(z+h)-f(z)|\,dz.
$$
Sincc $f$ is integrable,
$$
\lim_{h\to0}\int_{\mathbb{R}}|f(z+h)-f(z)|\,dz=0,
$$
proving the uniform continuity of $f\ast g$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Certain property of supremum Given real numbers $x_{ni}, n\in \mathbb{N}, i\in I$, does it hold that $$\sup \bigg\{ \sum_{n\in\mathbb{N}}x_{ni}|i\in I\bigg\}=\sum_{n\in \mathbb{N}}\sup\bigg\{x_{ni}|i\in I \bigg\}?$$
Thanks in advance.
|
No. A counterexample is $I = \{ \alpha, \beta \}$,
$$x_{n, \alpha} = \frac{1}{n^2}$$
$$x_{1, \beta} = \frac{5}{4}, \ \ x_{n, \beta} = 0 \ \mbox{for $n \geq 2$}$$
Then
$$\sup_{i \in I} \sum_{n \in \mathbb{N}} x_{n,i}= \max \{ \frac{\pi^2}{6} , \frac{5}{4}\} = \frac{\pi^2}{6}$$
while
$$\sum_{n \in \mathbb{N}} \sup_{i \in I} x_{n,i} = \frac{5}{4} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} + \frac{1}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to prove this statement: $\binom{r}{r}+\binom{r+1}{r}+\cdots+\binom{n}{r}=\binom{n+1}{r+1}$ Let $n$ and $r$ be positive integers with $n \ge r$. Prove that
Still a beginner here. Need to learn formatting.
I am guessing by induction? Not sure what or how to go forward with this.
Need help with the proof.
|
What is the number on the left counting?
The number of subsets of size $r$ of $\{1,2,3\dots r\}$ plus the number of subsets of size $r$ of $\{1,2,3\dots r+1\}$ plus the number of subsets of size $r$ of $\{1,2,3\dots r+1,r+2\}$ and so on up until the number of subsets of size $r$ of the set $\{1,2,3\dots n\}$
What is the element in the right counting? Th number of sunsets of size $r+1$ of the set $\{1,2,3\dots n+1\}$
Lets give a bijection between the elements in the right and those in the left.
How? suppose you have a subset $S$ of size $r+1$ of $\{1,2,3\dots n+1\}$ Let $k$ be it's greatest element, we associate to this subset a subset of size $r$ of the set $\{1,2,3\dots k-1\}$ which one?
$S-k$ Clearly this is a function from the obects counted on the right to those counted on the left. It is injective since different subsets go to different sets, and it is surjective since the subset $R$ of size $r$ of $\{1,2,3\dots m\}$ comes precisely from the set $R\cup \{m+1\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
$\mathbf R^2-\{\mathbf 0\}$ is homeomorphic to $S^1\times \mathbf R$. I am trying to to prove the following:
$\mathbf R^2-\{\mathbf 0\}$ is homeomorphic to $S^1\times \mathbf R$.
Since $\mathbf R^+=\{x\in \mathbf R:x>0\}$ is homeomorphic to $\mathbf R$, it suffices to show that $S^1\times \mathbf R^+$ is homeomorphic to $\mathbf R^2-\{\mathbf 0\}$.
There is a natural thing to try.
Define $f:S^1\times \mathbf R^+\to \mathbf R^2-\{\mathbf 0\}$ as
$$g(\mathbf p,t)=t\mathbf p$$
It is clear that $g$ is continuous and bijective.
So we need to show that $g^{-1}$ is continuous.
It is intuitively clear to me that $g^{-1}$ is continuous but I cannot see how I can prove this in a clean way.
Perhaps there is another approach?
Thanks.
|
$g^{-1}$ sends $re^{i\theta}$ to $(e^{i\theta},r)$. It is enough to show $re^{i\theta}\mapsto e^{i\theta}$ and $re^{i\theta}\mapsto r$ are continuous. The latter is the norm. Hence it is continuous. The first is a quotient of the identity by the norm. It remains only to prove that the quotient (reciprocal + multiplication) are continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
A Pigeonhole Principle Question Show that in a party of $n$ people, there are two people having identical number of friends.
I am a beginner at Pigeonhole Principle problems and have produced a solution to this intermediate level question. I am excited about it and I want my solution to be cross-checked and if anything is going wrong, please point it out.
So here it goes:
Let us consider an $n$ by $n$ symmetric matrix $A$ such that $a_{ij}=1$ if $i$-th and $j$-th persons are friends, otherwise $0$.
Please understand that I have assumed that friendship is mutual in the ordinary sense and also, $a_{ii}=1$ for all $i$. The matrix visualization is not necessary, but it just helped me in my thinking process.
So all we need to do is count the number of $1$'s in a particular row as this will give the number of friends of that particular person to whom that row corresponds. For example, the total number of $1$'s in row $3$ will give the total number of friends the third person has, after we have arbitrarily chosen a first person and begun counting.
So it boils down to showing that there will be two rows with same number of $1$'s. First consider that no person is friend only with himself, and no person has everyone a friend i.e. no row in the matrix has only one $1$ and no row has all $n$ entries $1$. Then the sum of $1$'s in a row can be maximum $n-1$ and minimum $2$, giving a total of $n-2$ possible sums, but there are $n$ rows. So, two rows must have the same sum of $1$'s, and those two persons have the same number of friends.
Now consider the case where one person, say person $1$ is friend with no one except himself. That means $a_{11}=1$ and $a_{1j}=0$ for $j=2,3,...,n$. We claim that in this case, the sum of $1$'s in any row other than the first row, has a minimum value of $2$ and a maximum of $n-1$, which is easy to check. The minimum is not a problem, but in case questions arise regarding the maximum, it is clear that the maximum is not $n$ because that would imply there exists a person with $n$ friends, contradicting the fact that the first person is friend only with himself. So there are $n-1$ rows and $n-2$ possible sums, for which two rows have the same sum.
The case where one person is friends with all others is similar.
We have exhausted all the three cases and thus we are done.
|
Clearly there is a minimum value of $n$ for which this statement holds, and
equally obviously it's not $n=1.$ The statement should be something of the form,
"in a party of $n$ people, where $n$ is at least ..., ... ."
I would strongly recommend being much clearer about the fact that you are going
to examine three separate cases, prior to starting to describe the first case.
Otherwise it can appear that you're making unjustified assumptions, at least
until we start reading the other cases and see that you weren't actually
supposing that there could not be someone who was friends with nobody else
or with everyone.
Edit: I initially thought this was an inductive proof, but
on closer examination, I do not think it is. I was misled by the fact that two
cases of the proof seem to be eliminating one of the members of the party.
The proof can be simplified (as already explained), which would also help to
avoid that potential confusion.
One could do an inductive proof, but that's another matter.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$ I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$
I think I am stuck on two fronts. First, I don't know how to express the leading terms on the left hand side before the $\dfrac{n}{(n+1)!}$ (or if doing so is even necessary to solve the problem). I am also assuming that the right high side should initially be expressed $1 - \dfrac{1}{(n+2)!}$. But where to go from there.
I'm actually not sure if I'm even thinking about it the right way.
|
Induction.
*
*Base. $n = 1: \frac{1}{2!} = 1 - \frac{1}{2!}$.
*Step. $n = m$ $-$ true. Let's prove for $m + 1$:
$$
\frac{1}{2!} + \dots + \frac{m}{(m+1)!} + \frac{m+1}{(m+2)!} = 1 - \frac{1}{(m+1)!} = 1 - \frac{m+2}{(m+2)!} + \frac{m+1}{(m+2)!} =
$$
$$
= 1 - \frac{1}{(m+2)!}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1031909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
$\lim_{x \to \infty} x\sin(\frac{1}{x}) = 1$ (epsilon-delta like condition) So, it is clear in many ways that this limit holds. However, I am interested in proving this with an "epsilon-delta" like condition. The start of my proof:
Let $f(x) = x\,\sin\left(\frac{1}{x}\right)$. We would like to show that for any $\epsilon > 0$, there exists an $\alpha > 0$ such that $x >\alpha$ implies $|f(x)-1| < \epsilon$. But observe that
$$|f(x)-1| = \left|x\,\sin\left(\frac{1}{x}\right) - 1\right| \leq \left|x\,\sin\left(\frac{1}{x}\right)\right| + 1 \leq |x| + 1 = x + 1\,\,\,.$$
This is where I am not seeing how to proceed. How can we choose $\alpha$ such that $x > \alpha$, but yet $x + 1 < \epsilon$ ? Any guidance would be much appreciated (without the Taylor theorem and relevant series expansions).
|
$\displaystyle \left|x\,\text{sin}\left(\frac{1}{x}\right) - 1\right| \leq \left|x\,\text{sin}\left(\frac{1}{x}\right)\right| + 1$ is true but not helpful, as the left hand side is about $0$ and the right about $2$.
For $x\gt 0$ you have $\dfrac1x \gt \sin\left(\dfrac1x\right) \gt \dfrac1x-\dfrac1{6x^3}$ and so: $$1 \gt x\sin\left(\dfrac1x\right) \gt 1-\dfrac1{6x^2}.$$
Now consider what happens as $x \to \infty$ using the squeeze theorem.
If $x \gt \sqrt{\dfrac1{6\epsilon}}$ then $\left|x\,\text{sin}\left(\frac{1}{x}\right) - 1\right| \lt \epsilon$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Given $a,b,c$ are the sides of a triangle. Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$ Given $a,b,c$ are the sides of a triangle. Prove that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}<2$.
My attempt:
I could solve it by using the semiperimeter concept. I tried to transform this equation since it is a homogeneous equation as $f(a,b,c)=f(ta,tb,tc)$. I considered $a+b+c=1$ and thus the inequality reduces to
$$\dfrac{a}{1-a}+\dfrac{b}{1-b}+\dfrac{c}{1-c}<2$$.
This is getting quite difficult to prove, as I don't have any clue to approach. I want to solve this only by homogeneous equation transformation, so any other transformation is welcome, but not any other idea(I don't mean to be rude but that's my necessity). Please help. Thank you.
|
As these are sides of a triangle, let $a=x+y, b= y+z, c=z+x$, and using homogeneity, set $x+y+z=1$. The inequality is then to show:
$$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z} < 2$$
Note that $x \in (0, 1) \implies 1+x > 1 \implies \dfrac{1-x}{1+x}< 1-x$. Sum that across $x, y, z$ to get the above inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Volume of water in a cone Let, slant height of a cone be 6cm, and radius be 3cm and the cone be uniform. Let a uniform & solid sphere of radius 1cm be put in the cone & fill the cone by water. What is the minimum volume of water such that the sphere is under water?
I find that the height of the cone is $\sqrt(27)$cm & I can not proceed further.
|
You can easily compute the demi-angle $\alpha$ of the cone.
$\cos \alpha=\dfrac {\sqrt{27}}{6}$
Consider that the sphere is in contact with the edge of the cone. Thus the distance between the center of the sphere $O$ with the edge is $1$, radius of the sphere.
Let's call $d$ the distance between the apex of the cone and the center of the sphere.
Then $\sin \alpha=\dfrac 1d$
From there you deduce $d$. In order to cover the sphere, you need to be at the upper edge, whose height id $d+1$.
You just need to compute the volume of the cone of demi-angle $\alpha$, of height $d+1$, and then substract the volume of the sphere.
EDIT: based on a comment, I used the height being $5$, which is wrong... height is $\sqrt{27}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Combinatorics - pick every object in a set Suppose I have a set of N objects and I pick from this set (with replacement) n>N times. How many permutations are there that include all N objects?
I've tried a number of different things, but am kind of going in circles and expect it should be a pretty simple solution if you do it right.
-- EDIT: My confusion of combinations vs. permutations has opened up a slightly bigger (or smaller perhaps) bag of worms on my methodology. The greater problem I am considering is as follows: Suppose you have N objects from which you draw with replacement. How many objects must you draw, n, s.t. the probability of having all N objects in your set is >p%, for some p. I thought I would determine the number of total permutations, N^n, and the set of all permutations that contain all N objects, say P(N). Then I would let q(n) = P(N)/N^n and find n s.t. q(n) >= p.
This started as a thought experiment and is much trickier than I expected.
|
Assuming the answer in my comments above comes back positive, that we are interested simply in the number of times each object is selected and not the order in which it was selected, suppose there are $N$ objects and you want select $n$ times with replacement, tallying how many times each was selected. Let the $N$ objects be labeled $a_1, a_2, \dots a_N$ and let $x_i$ be the number of times $a_i$ was selected.
Note, $x_1 + x_2 + \dots + x_N = n$ and $x_i\geq 1$ for every $i$ since we selected objects a total of $n$ times and we are curious about the situation that we selected each object at least once. The question is then to find the number of integral solutions to this.
To continue, consider a change of variable, $y_i = x_i - 1$. Then $y_i \geq 0$ and $y_1 + y_2 + \dots + y_N = n - N$
By formula, http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29
The number of combinations is then $\binom{N+(n-N)-1}{(n-N)-1} = \binom{ n-1}{n-N-1}$
In the case that it you are curious of the permutations instead of combinations, i.e., the order the elements were chosen in matters instead of simply the number of times each element was selected, you could take each possible combination (found above) and count the number of rearrangements of each and sum them all. At each step, the number of permutations would be $\frac{n!}{x_1!x_2!\cdots x_N!}$. With a computer you might be able to complete this train of thought but this doesn't seem to simplify well, so I'll scratch this method.
Instead thinking of an inclusion-exclusion approach, consider all sequences of length $n$ with entries chosen from $N$ elements $a_i$. The generalized inclusion-exclusion principle essentially states (#none violate condition) = (#all possible no restriction) - (#at least one violation) + (#at least two violations) - (#at least three violations) ... +(#at least 2k violations) - (#at least 2k+1 violations) + ...
There are $N^n$ possible such sequences without restriction. So our running total begins as $N^n$.
We subtract the number of sequences which at least don't include $a_1$ and subtract the number of sequences that at least don't include $a_2$ and $\dots$.
For each $a_i$ there are $(N-1)^n$ violating cases, and there are $N$ choices for which $a_i$ it was that was missing. So, our running total is currently $N^n - N\cdot (N-1)^n$
Adding now the number of sequences which at least don't include $a_i$ and $a_j$ for $i\neq j$, there are $(N-2)^n$ violating cases, and there are $\binom{N}{2}$ number of ways to choose $i$ and $j$. So, our running total is currently $N^n - N\cdot(N-1)^n + \binom{N}{2} (N-2)^n$
Continuing on to the general term, if there are $2k$ terms missing from the sequence, there are $(N-2k)^n$ number of sequences and $\binom{N}{2k}$ number of $2k$-tuples that might have been missing.
So, the answer would be:
$$N^n - N(N-1)^n + \binom{N}{2}(N-2)^n -\cdots + \binom{N}{2k}(N-2k)^n - \binom{N}{2k+1}(N-(2k+1))^n + \cdots \pm \binom{N}{N}(N-N)^n$$
Simplified into a single sum is $\sum\limits_{i=0}^N (-1)^i\binom{N}{i} (N-i)^n$
I personally do not recognize this as being able to be simplified further, but perhaps you might recognize it or someone else might later.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $M$ is a $R$-module with $R=R_1\times \cdots \times R_n$ then $M\simeq M_1\times \cdots \times M_n$ where $M_i$ is a $R_i$-module? Let $R_1, \ldots, R_n$ be rings and consider the ring $R:=R_1\times \cdots\times R_n$. How can I show every $R$-module $M$ is isomorphic to a product $M_1\times \cdots\times M_n$ where each $M_i$ is a $R_i$-module?
Obs: By module I always mean left module.
|
It's easier with just two rings and then you can do induction.
So consider $R=R_1\times R_2$ and the two idempotents $e_1=(1,0)$, $e_2=(0,1)$. If $M$ is an $R$-module, you can consider $M_1=e_1M$ and $M_2=e_2M$.
For all $r\in R$, $e_1r=re_1$, and similarly for $e_2$, so $M_1$ and $M_2$ are $R$-submodules of $M$ and $M=M_1\oplus M_2$ is clear:
$$
x=1x=(e_1+e_2)x=e_1x+e_2x\in M_1+M_2
$$
Note that $M_1=\{x\in M:e_1x=x\}$ and similarly for $M_2$. Thus if $x\in M_1\cap M_2$, then $x=e_1x=e_1e_2x=0$.
That $M_1$ and $M_2$ are in a natural way modules over $R_1$ and $R_2$ is straightforward.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Probability: find the probability of event B given that event A occurs My Problem,
Suppose a family has 2 children,
if one children is randomly selected and it is a girl then what is the probability of the second child to be a girl ?
Please help. Thanks
|
Let's work by definition of conditional probabilities:
Let A = first child is girl, and B = second child is girl.
$$P(B \mid A) = \frac{P(A \cap B) }{P(A)}= \frac{0.5 \times 0.5}{0.5}=0.5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
When can ZFC be said to have been "born"? The "History" section of the Wikipedia article on ZFC isn't particularly helpful. The only thing I have understood from it is that ZFC had appeared after 1922. In what book or paper was ZFC first explicitly formulated and proposed?
|
Here are some relevant quotes from Fraenkel, Bar-Hillel, Levy's "Foundations of set theory":
"Zermelo's vague notion of a definite statement did not live up to the standard of rigor customary in mathematics ... In 1921/22, independently and almost simultaneously, two different methods were offered [by Fraenkel and Skolem] for replacing in the axiom of subsets the vague notion of a definite statement by a well-defined, and therefore much more restricted, notion of a statement ... The second method, proposed by Skolem and, by now, universally accepted because of its universality and generality ... It [The axiom schema of replacement] was suggested first by Fraenkel and independently by Skolem."
An english translation of Skolem's "Some remarks on axiomatized set theory" appears in Heijenoort's "From Frege to Godel - A source book in mathematical logic". The commentary on Skolem's paper says (and I agree):
"These indications do not exhaust the content of a rich and clearly written paper, which when it was published did not receive the attention it deserved, although it heralded important future developments".
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
}
|
One measurable $\lim$ and one theorem How we can prove following theorem?
Let $f_n \ge 0 $ be measurable, $\lim f_n = f $ and $f_n \le f$ for each $n$. Show that $$\int f(x)dx=\lim_n \int f_n(x)dx $$
Any idea would be highly appreciated.
|
Fatou's lemma gives $\int f = \int \liminf_n f_n \le \liminf_n \int f_n$ (the values may be infinite).
If $\liminf_n \int f_n < \infty$, then $f$ is integrable and the result follows from the dominated convergence theorem.
Otherwise, we have $\lim_n \int f_n = \infty$, and since $0 \le f_n \le f$, we have
$\int f_n \le \int f$ and so $\int f = \infty$.
Aside: The dominated convergence theorem states that if $|f_n| \le g$, $f_n(x) \to f(x)$ and $g$ is integrable, then $\int f_n \to \int f$.
To apply dominated convergence theorem, let $g=f$. Note that all the functions are non-negative, so $|f_n| = f_n$ and by assumption $f_n \le f = g$.
Since $\int f \le \liminf_n \int f_n < \infty$
we see that $g=f$ is integrable, hence the conditions of the theorem are satisfied and so $\lim_n \int f_n = \int f$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1032875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
At Most Two Distinct Members of A The quantified predicate logic statement that describes at most two distinct members of A, where A, is some arbitrary set is:
$\forall$xyz( (Px $\land$ Py $\land$ Pz) $\Rightarrow$ (x=y $\lor$ x=z $\lor$ y=z) )
I can parse this quantified statement into three cases:
*
*There are No Members of A
*There is 1 Member of A
*There are 2 Members of A
In case 2, x must equal y, then in turn for any z candidate, z is either equal to x, or to y, but x=y.
(So all three "or" statements are true.)
In case 3, x does not equal y, therefore for any z, z will either be equal to x, or z will be equal to y. (The first "or" fails, but the the other two are true.)
In case 1 I don't see how the quantified predicate statement guarantees that no membership can exist, i.e. there are non such members of A. If we don't choose any x, y, z then there are no such x,y, or z in A and the quantified statement "short circuits"? How do I interpret this or reason about it?
|
In the case 1 (no elements) the antecedent $(Px ∧ Py ∧ Pz)$ is is false and the implication is true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$
|
Both the function that you are integrating as the region over which you are integrating it get unchanged if you exchange $x$ with $y$. Therefore, your integral is equal to$$2\int_0^a\int_0^x\frac1{(a^2+x^2+y^2)^{3/2}}\,\mathrm dy\,\mathrm dx.$$You can compute this integral using polar coordinates: $\theta$ can take values in $\left[0,\frac\pi4\right]$ and, for each $\theta$, $r$ can take values in $\left[0,\frac a{\cos\theta}\right]$. And\begin{align}\int_0^{\pi/4}\int_0^{a/\cos(\theta)}\frac r{(a^2+r^2)^{3/2}}\,\mathrm dr\,\mathrm d\theta&=\int_0^{\pi/4}\frac{1-\frac1{\sqrt{\sec ^2(\theta)+1}}}a\,\mathrm d\theta\\&=\frac1a\left(\frac\pi4-\int_0^{\pi/4}\frac1{\sqrt{\sec ^2(\theta)+1}}\,\mathrm d\theta\right)\\&=\frac\pi{12a}.\end{align}Note that the final equality is equivalent to$$\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta=\frac\pi6.$$This can be justified as follows: you do $\theta=\arccos\left(\sqrt x\right)$ and $\mathrm d\theta=-\frac1{2\sqrt{x-x^2}}\,\mathrm dx$. Doing this, you will get\begin{align}\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta&=\int_1^{1/2}-\frac1{2\sqrt{1-x^2}}\\&=\frac12\int_{1/2}^1\frac1{\sqrt{1-x^2}}\,\mathrm dx\\&=\frac12\left(\arcsin\left(1\right)-\arcsin\left(\frac12\right)\right)\\&=\frac12\left(\frac\pi2-\frac\pi6\right)\\&=\frac\pi6.\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
}
|
Trace of nilpotent matrix over a ring Let $R$ be a commutative ring with unity, and $n$ a positive integer.
Let $A\in \mathfrak{M}_n(R)$ such that there exists $m\in \mathbb N$, for which $A^m=0$.
Is it true that there exists $\ell\in \mathbb N$, such that $\bigl(\text{tr}(A)\bigr)^\ell=0$ ?
Remark : It is true for $n=1$ and $n=2$.
|
If $\mathfrak{p}$ is any prime ideal of $R$, then $A/\mathfrak{p}$ is a domain. We can embed it in its fraction field, then in an algebraic closure. Then your matrix mod $\mathfrak{p}$ is triangularisable, with of course eigenvalues $0$.
Hence, the reduction mod $\mathfrak{p}$ of your matrix has vanishing trace. Since this is true for every prime ideal, this shows that the trace of $A$ is in the intersection of all prime ideals of $R$, that is in the nilradical of $R$, which is the ideal of nilpotent elements of $R$.
This concludes the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Holder continuity using Sobolev imbedding We assume for any $V\subset \subset U$ and $1<p<\infty$
$||u||_{W^{2,p}(V)}\le C(||\Delta u||_{L^p(U)}+||u||_{L^1(U)})$ for some $C=C(V,U,p)$.
Given, $B=\{x∈R^3,|x|<1/2\}$ and we suppose $u∈H^1(B)$ is a weak solution of $-\Delta u+cu=f$ for some $c(x){\in L^q(B)}, 3/2<q<2$ and $f\in C^\infty (B)$. How would I show that $u$ is Hölder continuous inside $B$?
|
First notice that, by Hölder's inequality, we have $cu\in L^p$ for $1/p=1/q+1/6$ (since $u\in L^6$ by the Sobolev embedding). Next your inequality gives $u\in W^{2,p}$ for this same $p$. Applying Sobolev embedding again, we get $\nabla u \in L^{p*}$ where as usual $1/p^* =1/p-1/n$. If we substitute we get
$$
\frac{1}{p^*} = \frac{1}{q} +\frac{1}{6} -\frac{1}{3} = \frac{1}{q}-\frac{1}{3} < \frac{1}{3},
$$
where the first equality is the expression we got for $p$ at the start, and the last inequality is because $q>3/2$. This means that $\nabla u \in L^{p^*}$ with $p^*>3=n$, so that Morrey's inequality gives $u$ Hölder continuous (in the interior of $B$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Why base of a logrithim function must be greater than one? I'm in domain section of my textbook. It says that for logarithm functions, the base must be greater than one.
I can understand why base shouldn't be one but what is problem with negative numbers?
for example, What's wrong with $\log_{-2} x$ ?
|
Because when working only with real numbers, we do not know how to make sense of the expression $(-2)^\sqrt{2}$, for example.
In general the expression $a^x$, with $x \in \mathbb R$, makes sense only if $a > 0$.
Since $a$ is the basis of the logarithm, you can see why we limit ourselves with positive base.
With complex number the situation is different; we can always define $a^x = e^{x \ln^\mathbb C a}$, where $\ln^\mathbb C$ indicates the complex logarithm, which is a multi-values function, formally given by
$$\ln^\mathbb C z = \ln^\mathbb R |z| + (2k\pi + \arg z)i$$
(note though that this way $a^x$ is not a well-defined complex number, it is an infinite collection of complex numbers )
In principle then one could define the complex logarithm for bases different than $e$ (also negative ones!), but it is not useful to do so.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
On Galois closure I'm working on this problem in Hungerford: For $\sigma \in Aut_F \bar{F}$, show that every finite extension of $K$, the fixed field of $\sigma$, is cyclic.
For a finite extension $L$ of $K$, let $M$ be a Galois closure of $L$ then I can show that $M/K$ is cyclic so $L/K$ is cyclic.. but there is one problem; is $M/K$ is finite? When $F$ has characteristic zero, $L$ is separable over $K$ so simple by primitive element theorem and so $M$ is finite. However, if $F$ has nonzero characteristic, this proof is not applicable. How can I show that $M/K$ is finite?
|
In this setting $L/K$ will always be separable.
To see that let $z\in L\setminus K$ be arbitrary. Because $L/K$ is finite, $z$ is algebraic over $K$. Therefore $z$ has a minimal polynomial $m(x)\in K[x]$. Over $\overline{F}$ $m(x)$ splits into linear factors
$$
m(x)=(x-z_1)(x-z_2)\cdots (x-z_n)
$$
with $z_1=z$. The elements $\sigma^i(z)$, $i=0,1,2,\ldots,$ are all roots of $m(x)$, so they are among the $z_j$:s. In particular there are only finitely many $\sigma^i(z)$:s. Let $k$ be the smallest positive integer such that $\sigma^k(z)$ is among the $\sigma^j(z), 0\le j<k$. Because $\sigma$ is bijective this implies that $\sigma^k(z)=z$.
Next consider the polynomial
$$
f(x)=(x-z)(x-\sigma(z))(x-\sigma^2(z))\cdots (x-\sigma^{k-1}(z)).
$$
Its zeros are all distinct. And we see that the coefficients of $f$ are fixed under $\sigma$. Therefore $f(x)\in K[x]$. This implies that $m(x)$, as the minimal polynomial, is a factor of $f(x)$. That shows that $m(x)$ is separable proving the claim.
We also see that we must have $m(x)=f(x)$. I guess that we could turn that into a solution by-passing your separability concern altogether, but I haven't thought it through.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to draw or plot illustrative figures? stackexchange users
I would like to plot or draw some illustrative figures for my research paper. I've tried GeoGebra already. But couldn't draw them as I wanted.
So my question is How can I draw them?
Or would you tell me what are their functions so that I can plot them in Matlab?
I post the figures in here, so that you can understand what am I looking for.
Thanks for your reply.
P.S. If I am in a wrong forum please help me to find the appropriate forum for asking this question.
|
I like the program Asymptote. See a tutorial and a gallery of examples. Also, you can see the drawing in Calculating line integrals via Stokes theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
A question about the properties of the pseudospectrum Assume that $A\in \mathbb{C}^{n\times n}$. The $\epsilon-$pseudospectrum of $A$ is defined by
$$\sigma_{\epsilon}(A)=\{z\in C \quad | \quad \Arrowvert (zI-A)^{-1} \Arrowvert>\frac{1}{\epsilon}\}.$$
Why $\sigma_{\epsilon}(A)$ is a bounded open subset of $\mathbb{C}$?
|
Take $x\in \mathbb C^n$, $z\in \mathbb C$. Then it holds
$$
|z|\cdot \|x\| = \|zx\| \le \|(zI-A)x\| + \|Ax\| \le \|(zI-A)x\| + \|A\|\cdot \|x\|,
$$
which implies
$$
\|(zI-A)x\| \ge (|z|-\|A\|)\|x\|.
$$
If $|z| \ge \|A\| + \epsilon$, then this implies
$$
\|(zI-A)^{-1}\| \le \frac1{|z|-\|A\|} \le \frac1\epsilon.
$$
Hence $\sigma_\epsilon(A)$ is bounded.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Equality condition for convolution's $L^p$ norm. Suppose that $1< p< \infty$, $f\in L^1(R)$, and $g\in L^p(R)$ and that $\|f*g\|_p=\|f\|_1\|g\|_p$. Show that then either $f=0$ a.e. or $g=0$ a.e.
I have solved for $g=0$ a.e. if $\|f\|_1>0$ using the equality condition for Hölder ; $\alpha f^p=\beta g^q$ for some $\alpha, \beta \ge 0$. Is this still useful for supposing $\|g\|_p>0$ and deriving $f=0$ a.e?
|
In general, if you want to prove a statement of the form $A\implies (B\ \text{or}\ C)$, it is logically equivalent to show $(A\ \text{and}\ (\text{not}\ B))\implies C$. The other answer essentially shows how this works in this particular case, but also you could convince yourself of this by making a truth table with columns
*
*$A$,
*$B$,
*$C$,
*$A\implies (B\ \text{or}\ C)$,
*$(A\ \text{and}\ (\text{not}\ B))\implies C$,
and then checking that no matter what values of $0$ or $1$ ($\text{False}$ or $\text{True}$) you assign to the first three statements $A, B, C$, both of the last two statements will agree.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
''Differential equation'' with known solution $\sin$ and $\cos$ I am given the following two two equations
$f,g : \mathbb{R} \to \mathbb{R}$ are differentiable on $\mathbb{R}$ and they satisfy $\forall x,y \in \mathbb{R}$ $$f(x+y) = f(x)g(y)+f(y)g(x)\\g(x+y)=g(x)g(y)-f(x)f(y) $$ with $f(-x)=-f(x), \forall x \in \mathbb{R}$ and $f'(0)=1$.
I want to show that $f= \sin$ and $g = \cos$
My approach: My general idea was to show that $$f'=g, \ g'=-f$$ and $$f(0)=0, \ g(0)=1$$
If I'd manage to do that, then the solution follows because the above differential equation has unique solution $f= \sin, g = \cos$.
Since $f$ is an 'uneven' function it follows that $f(0)=-f(0) \implies f(0)=0$ so that's one check out of the above list $\checkmark$
Differentiating the first given expression I'd obtain $$f'(x+y)=f'(x)g(y)+f(y)g'(x) \implies f'(0)=f'(0)g(0)+f(0)g'(0)=g(0)=1 $$
So that takes care of $g(0)=1 \ \checkmark$
Now to the tricky part. I will differentiate both equations and obtain: $$f'(x+y)=f'(x)g(y) + f(y) g'(x) \\ g'(x+y)=g'(x)g(y)-f'(x)f(y) $$
It seems like it is crucial that $g'(0)=0$ in order to show the remaining two properties out of my list above. Which so far I didn't manage to do because all my steps are circular: $$f'(y)= f'(0)g(y)+f(y)g'(0)=g(y)+f(y)g'(0) \tag{*} $$
which would only help me if $g'(0)=0$, similarly $$g'(y)=g'(0)g(y)-f'(0)f(y)=g'(0)g(y)-f(y) $$
Which doesn't help me either, maybe there is a substitution I have to perform but I yet fail to see it.
|
Well, you can note that $$0=f(x-x)=f(x)g(-x)+f(-x)g(x)=f(x)g(-x)-f(x)g(x)=f(x)\bigl[g(x)-g(-x)\bigr].$$ This holds for all $x\in\Bbb R,$ and since $f'(0)\ne 0,$ then $f$ isn't the constant zero function. It follows that $g(x)=g(-x)$ whenever $f(x)\ne 0.$ In particular $f$ is non-zero in some punctured neighborhood about $x=0$ (why?), and clearly $g(0)=g(-0),$ so for all $x$ sufficiently close to $0,$ we have $g(x)=g(-x).$ From this, you can show that $g'(0)=0$ directly, using a symmetric derivative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1033831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Get the numbers from (0-30) by using the number $2$ four times How can I get the numbers from (0-30) by using the number $2$ four times.Use any common mathematical function and the (+,-,*,/,^)
I tried to solve this puzzle, but I couldn't solve it completely. Some of my results were:
$$2/2-2/2=0$$ $$(2*2)/(2*2)=1$$ $$2/2+2/2=2$$ $$2^2-2/2=3$$ $$\frac{2*2}{2}+2=4$$ $$2^2+2/2=5$$ $$2^2*2-2=6$$ $$\frac{2^{2*2}}{2}=8$$ $$(2+2/2)^2=9$$ $$2*2*2+2=10$$ $$2*2*2*2=16$$ $$22+2/2=23$$ $$(2+2)!+2/2=25$$ $$(2+2)!+2+2=28$$
|
$\left\lfloor \exp\left(2^2\right)\right\rfloor+2+2=11$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1034122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 11,
"answer_id": 4
}
|
Semi-infinite plate problem when 2 edges are equal temperature and one edge is a function of position Here is the problem posted:
Now here is my solution for a)
I knew to discard the $e^{ky}$ and $\sin(kx)$ due to $T\to 20$ when $y\to\infty$ and $T= 20$ at $x = 0$.
Which leaves me with $\cos(kx)*e^{-ky} = T$.
Now I have two main problems.
1) I don't know how to solve for $k$.
If I plug in $(0,0)$ I get $cos(0*k) = 20$
2) I don't know if I'm understanding the problem correctly, because if I plug in $x = 0$ to the bottom edge equation, I get $T = 20$ which equals the given left edge temperature.
But if I plug in $x = 30$ I get $T = 320$, not 20, which should be the right edge.
I really need help on this.
|
I suggest you start making $t=T-20$ your new temperature function; recall that the Laplace operator is linear, and if $T$ is a solution so is $t$ and vice versa. The advantage of this new temperature is that it makes 3 out of 4 boundary conditions homogeneous, making the problem easier to handle.
More precisely, if you set $t=T-20$ then $t(x=0)=t(x=30)=t(x,y=\infty)=0$.
Now let $t=f(x)g(y)$, separation of variables yields:
$$f=-k^2f_{xx} \text{ and } g=k^2g_{yy}.$$
Making $f(x)=A\sin{kx}+B\cos{kx}$ and using the boundary conditions at $x=0$ and $x=30$ we get $B=0$ and $k=\frac{n\pi}{30}, n\in\{1,2,3,\dots\}$.
The solutions for $f$ have the form:
$$f_n=A_n \sin{\left(\frac{n\pi}{30}x\right)}$$
Proceeding in an analogous way for $g$ we get
$g(y)=Ce^{ky}+De^{-ky}$ and from $t\rightarrow 0$ as $y\rightarrow \infty$ we deduce $C=0$, so, taking into account the previously calculated values for $k$:
$$g_n=D_n e^{-\frac{n\pi}{30}y}.$$
Now note that each $t_n=f_n g_n$ is a solution, and so is their sum; the solution can be therefore written as:
$$t(x,y)=\sum_{n=1}^\infty A_n \sin{\left(\frac{n\pi}{30}x\right)}\exp{\left(-\frac{n\pi}{30}y\right)}.$$
Where I have renamed the constants. We have only to determine the coefficients $A_n$, we use the condition at $y=0$ for this. From $t(x,y=0)=10x$ we obtain
$$10x=\sum_{n=1}^\infty A_n \sin{\left(\frac{n\pi}{30}x\right)}.$$
Multiplying by $\sin{\left( \frac{m\pi}{30}x\right)}$, changing variables, integrating and using the orthogonality relations between sines we finally arrive at
$$A_n=10\frac{2}{\pi}\frac{\pi}{30}\int_0^{30}x \sin{\left( \frac{n\pi}{30}x \right)} \operatorname d\!x .$$
This is the Fourier series of your lower boundary conditions (I suggest you check this calculation as I might have made mistakes).
Once you calculate the coefficients you can go back to $T$ by summing the baseline 20 to $t$, the constant term of the Fourier series for the solution..
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1034209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Secant line and diameter of a circle A secant line incident to a circle at points $A$ and $C$ intersects the circle's diameter at point $B$ with a $45^\circ$ angle. If the length of $AB$ is $1$ and the length of $BC$ is $7$, then what is the circle's radius?
|
We can set up a system of equations for $BE$ and $BD$, then that will give us the radius, which is
$$r=\frac{BE+BD}{2}$$
First, by the chord-chord theorem (AKA the power chord theorem), we can write
$$AB\times BC=BE\times BD$$
$$\Rightarrow BE\times BD=7\quad\quad\quad(1)$$
Next, we can use the cosine law in $\triangle BCO$ (note that $CO$ is also a radius):
$$CO^2=BC^2+BO^2-2\times BC\times BO\times \cos \angle CBO$$
$$\Rightarrow r^2=49+(BE-r)^2-14(BE-r)\cos(45)$$
$$\Rightarrow r^2=49+(BE-r)^2-7\sqrt{2}(BE-r)$$
$$\Rightarrow \left(\frac{BE+BD}{2}\right)^2=49+\left(\frac{BE-BD}{2}\right)^2-7\sqrt2\left(\frac{BE-BD}{2}\right)$$
$$\dots$$
$$\Longrightarrow BE\times BD+\frac{7\sqrt2 BE}{2}-\frac{7\sqrt2 BD}{2}=49$$
$$\Rightarrow \left(BE-\frac{7\sqrt2}{2}\right)\left(BD+\frac{7\sqrt2}{2}\right)=\frac{49}{2}\quad\quad\quad(2)$$
Can you continue the algebra?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1034313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.