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Does a statement need to be a biconditional statement to prove by the contrapositive I am trying to write a proof and was wondering if a then b, the converse if b then a might not be true. This leads me to wonder if the statement needs to be an if and only if statement if it can be proven by the contrapositive if not b then not a.
That is do I need a statement like this a if and only if b to prove that if not b then then not a, or is if a then b sufficient to do that.
|
By definition, "$a \implies b$" means "not ($a$ and not $b$)". So then "(not $b$)$\implies$(not $a$)" means "not ((not $b$) and not (not $a$))". This simplifies to "not ($a$ and not $b$)", which is the same thing as "$a \implies b$". So yeah, you only need the one direction of implication.
You should probably think about this definition of $a \implies b$ if you've never seen it. Like, draw a truth table or something.
|
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|
What is the meaning of $ \mathbb{R}^n$ to $\mathbb{R}^{n+1}$? In linear algebra, what does it mean to go from $\mathbb{R}^1$ to $\mathbb{R}^2$ or $\mathbb{R}^2$ to $\mathbb{R}^3$?
|
It means to bring a two dimensional space to a three dimensional space, i.e., mapping vectors of the form $(x,y)$ to vectors of the form $(x,y,z).$ This must be done by applying some kind of linear transformation.
|
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How to express this expression in terms of N? I am trying to express this formula in terms of N:
$$
A=\frac{a^2N} {a\cdot \tan{\frac{180}{N}}}
$$
I really don't know how to do this. I tried and got this:
$$
A \cdot a \cdot \tan{\frac{180}{N}} = a^2N
$$
$$
\tan{\frac{180}{N}}=\frac{a^2N}{Aa}
$$
$$
\frac{180}{N}=\tan^{-1}{\frac{a^2N}{Aa}}
$$
$$
N=\frac{180}{\tan^{-1}{\frac{a^2N}{Aa}}}
$$
But I just can't get the N on the right side. The $\tan$ is so annoying! How do I do this?
|
Equations which mix polynomial and trigonometric terms do not show analytical solutions and numerical methods should be involved.
Let us consider the equation $$A=\frac{a^2N} {a\cdot \tan({\frac{\pi}{N})}}=\frac{aN} {\tan({\frac{\pi}{N})}}$$ So, we can consider that we look for the zero of the function $$f(N)=\frac{aN} {\tan({\frac{\pi}{N})}}-A$$ $$g(N)={aN} -A\,{\tan({\frac{\pi}{N}})}$$ $$h(N)=a N \cos(\frac{\pi}{N})-A \sin(\frac{\pi}{N})$$ Probably, the last equation will be better to consider since discontinuties have been removed.
For illustration purposes, let us consider the case $a=2$, $A=1$. A plot of the function $h(N)$ shows a solution somewhere between $2$ and $3$. So, let us start Newton method using $N_0=2$; this will generates the following iterates $$N_1=2.31830988618379$$ $$N_2=2.31348787275960$$ $$N_3=2.31348827662319$$ which is the solution for fifteen significant figures.
If we consider the case where $N$ could be large (this would correspond to large values of ratio $\frac A a$, Taylor expansion would give $$h(N)=a N-\frac{\frac{\pi ^2 a}{2}+\pi A}{N}+O\left(\frac{1}{N^2}\right)$$ that is to say $$N\approx\sqrt{\frac{\pi }{2}} \sqrt{\frac{\pi a+2 A}{a}}$$ In the case considered above, this would lead to $N\approx \frac{1}{2} \sqrt{\pi (2+2 \pi )}\approx 2.55061$ which is not too bad.
Using $a=2$, $A=35$, the estimate would be $7.74033$ while the solution would be $7.63580$.
Edit
You could be amazed by the fact that, using a $[2,2]$ Pade approximant (do not worry : you will learn about them sooner or later), the solution can be approximated by $$N=\frac 12 \sqrt{\frac \pi {3a}}\sqrt{2 A \left(\frac{2 A}{\pi a+2 A}+5\right)+5 \pi a}$$ which, for the two worked cases gives $\approx 2.34191$ and $\approx 7.63777$.
Still more funny, using a $[2,4]$ Pade approximant, the solution can be approximated by $$N= \frac 15 \sqrt{\frac \pi {6a}}\sqrt{2 A \left(\frac{4 A (37 \pi a+84 A)}{5 \pi ^2 a^2+20 \pi a A+24
A^2}+61\right)+61 \pi a}$$ which, for the two worked cases gives $\approx 2.31681$ and $\approx 7.63584$.
|
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.
Edit: $n$ is a positive integer such that $n>1$.
|
$f(x) = 1/x$ is strictly convex, therefore
$$
\frac{1}{2n} < \frac 12 \left( \frac{1}{n+k} + \frac{1}{3n-k} \right)
$$
for $k = 1, ..., n-1$, or
$$
\frac{1}{n+k} + \frac{1}{3n-k} > \frac {1}{2n} + \frac {1}{2n}
$$
Combining terms pairwise from both ends of the sum shows that
$$
\frac{1}{n+1} + \frac{1}{n+3}+\dots+\frac{1}{3n-3} + \frac{1}{3n-1} >
\underbrace{\frac {1}{2n} + \frac {1}{2n} + \dots +\frac {1}{2n} + \frac {1}{2n}}_{n \text{ terms}}
= \frac 12.
$$
(If $n$ is odd then the middle term $ \frac {1}{2n}$ is not combined with another one.
But since $n> 1$ there is at least one "pair" to combine, which gives
the strict inequality.)
|
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|
Limit of $a_{n+2}=a^2_{n+1}+\frac{1}{6}\cdot a_n+\frac{1}{9}$ Find a limit of sequence:
$$a_{n+2}=a^2_{n+1}+\frac{1}{6}\cdot a_n+\frac{1}{9}$$
$$a_1=0,a_2=0$$
I tried to prove that $a_n$ is bounded and monotonic, but I couldn't prove that $a_n$ is monotonic (by strong induction).
So please give a idea how to show that $a_n$ is monotonic.
|
(this proof takes a little longer, but gives a neat bound on the rate of convergence)
*
*If you want a universal bound of the form $a_n\le M$, the recursion tells you that you're going to need $M\le M^2+\frac{M}{6}+\frac{1}{9}$, which works out to be equivalent to $(6M-1)(3M-2)\le0$. The lowest bound this allows us to take is $M=\frac{1}{6}$, i.e. that $a_n\le\frac{1}{6}$ for all $n$. Feeding this back into the recursion, we see that this inequality does indeed hold.
*If you're hoping for a limit $L$ to exist, the recursion tells you that you're going to need $L=L^2+\frac{L}{6}+\frac{1}{9}$, and similar reasoning to before gives that the only possible limits are $\frac{1}{6}$ and $\frac{2}{3}$. But having established our earlier bounds on $a_n$, we see that if a limit does exist, it will necessarily be $\frac{1}{6}$.
*With the aim of showing that $a_n\to\frac{1}{6}$, we set $a_n=\frac{1}{6}-b_n$ and work through the algebra to obtain the recurrence $b_{n+2}=\frac{b_n}{6}+\frac{b_{n+1}}{3}-b_{n+1}^2$. Our earlier work gives that $0<b_n\le \frac{1}{6}$ for all $n$, and some examination hints that this $b_n$ decays to 0 geometrically, so we seek a bound of the form $b_n\le Cr^n$. Appealing to a strong-induction-style argument, we would need:
$$b_{n+2}=\frac{b_n}{6}+\frac{b_{n+1}}{3}-b_{n+1}^2\le\frac{Cr^n}{6}+\frac{Cr^{n+1}}{3}=Cr^{n+2}\left(\frac{1}{6r^2}+\frac{1}{3r}\right)$$
So, if we can make $\left(\frac{1}{6r^2}+\frac{1}{3r}\right)\le 1$, we can get this inequality to hold (for a suitable $C$) by strong induction. $r=\frac{1+\sqrt7}{6}$ is the smallest $r$ which works for this, and we note that $r\le1$, guaranteeing convergence.
Thus, we see that $0<b_n<\frac{r^{n-2}}{6}$, and hence $b_n\to 0, a_n\to \frac{1}{6}$.
|
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Structure theorem for non connected graded Hopf algebras Let $V^{\bullet}$ be a graded vector space. Assume that $V^{i}=0$ for $i<1$. Then the tensor algebra has the structure of a connected Hopf algebra and in particular its Lie algebra of primitive $P(T(V))$ is naturally ismorphic to the free Lie algebra. This result is a consequence of the structure theorem for connected Hopf algebras: given a connected Hopf algebra $H$ then
$$ H\cong U(P(H)), $$
where $U$ is the universal enveloping algebra. Here my questions
1) Is the connectess of $H$ a necessary requirement for the structure theorem? Do you know some counter examples?
2) Assume $V^{i}=0$ for $i<0$, then the tensor algebra $T(V)$ is again a graded Hopf algebra but it is no longer connected. Let $L(V)$ be the graded free Lie algebra on $V^{\bullet}$, are $P(T(V))$ and $L(V)$ isomorphic?
|
*
*More general hypotheses are possible, but there are counterexamples. Take, for example, the group algebra of a nontrivial finite group, concentrated in degree $0$ (it's not clear to me whether you're talking about graded Hopf algebras here).
*$T(V)$ is, for formal reasons, always the universal enveloping algebra of $L(V)$: this follows from the observation that adjoints compose. The remaining question is whether it's always true that the Lie algebra of primitive elements of a universal enveloping algebra (in this graded setting) recovers the original Lie algebra. I believe this is also true, at least over a field of characteristic zero, but it's less formal; I think you need a PBW-type result to handle the degree $0$ part.
|
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|
Proving the convergnce of a sequence So, I have to prove that the sequence defined as
$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$ converges and then find the limit.
I have few questions; Do i have to assume that $a_n \geq 0$ or $a_n \leq0$. Because if i assume that $a_n \geq 0$, I can prove that
$\left|a_{n+1}-2\right| \leq \frac{4}{7}\left|a_{n}-2\right|$, and by induction, it would be easy to show that $a_n \to 2$.
Any help would be appreciated. Is there a way to prove convergence without knowing the signs of $a_n$?
|
As Varun showed, the fixed points obeying $L=6(1+L)/(7+L)$ are $+2$ and $-3$. For $a_0\gt -3$, one may see that one is getting close to $L=+2$. For $a_0\gt -3$, the subsequent terms are getting more negative, but they inevitably drop below $a_n\lt -7$ at some point, and for $a_0\lt -7$, $a_1$ may already be calculated to be positive i.e. $a_1\gt - 3$, because the graph of $y=6(1+x)/(7+x)$ is a hyperbola, and one starts to converge towards the limit $L=2$, anyway.
As Arpit wrote, Contraction Mapping Theorem guarantees the existence of a limit. The fixed points $L=+2$ and $L=-3$ are the only possible limits. And the sequence diverges away from $L=-3$ for almost all $a_0$. The only exception occurs when $a_0=-3$ when all $a_n=-3$ and $L=-3$ is also the limit. For all other values of $a_0$, the limit has to be $L=+2$.
|
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|
Weak convergence in $L^p$ equivalent to pointwise almost everywhere convergence Can weak convergence of a sequence $f_n\in L^p(\Omega, \mu)$ to some $f\in L^p(\Omega, \mu)$ be characterised as almost everywhere pointwise convergence?
Let us also assume the measure space is $\sigma$-finite.
I'm asking because for continuous functions on the unit interval, weak convergence is equivalent to pointwise convergence.
|
It isn't true.
The standard counterexample is to look at $L^2((0,1))$ with Lebesgue measure and take $f_n(x) = \sqrt{2} \sin(n \pi x)$. The functions $f_n$ are orthonormal in $L^2$, so by Bessel's inequality they converge weakly to 0. But pointwise, the sequence $\{f_n(x)\}$ diverges for every $x \in (0,1)$.
|
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|
Differential Equation $y' = 2y/x - 1$ Can I get help solving this DE?
$$ y' = \frac 2xy - 1$$
Doesn't look too hard but i just can't get to the correct result.
Thank you in advance
|
$$
y' + \left(- \frac 2 x\right) y = -1
$$
$$
My' + \left(-\frac 2 x M\right) y = -M \tag 1
$$
$$
My' + M' y = -M \tag 2
$$
$$
(My)' = -M
$$
$$
My = \int -M\,dx
$$
$$
\text{etc.}
$$
In order that $(1)$ be equal to $(2)$, you need $\displaystyle -\frac 2 x M = M'$. Thus
$$
-\frac 2 x M = \frac {dM}{dx}
$$
$$
\frac{-2\,dx} x = \frac{dM} M
$$
$$
-2\log x + C = \log M
$$
$$
M = x^{-2} e^C
$$
You don't need all solutions to this last differential equation, but only one, so let's go with $M=x^{-2}$.
Then $(1)$ above becomes
$$
x^{-2} y' + \left( -2x^{-3} \right) M = -x^{-2}.
$$
$$
\Big( x^{-2} y \Big)' = -x^{-2}.
$$
$$
x^{-2} y = x^{-1} + C
$$
$$
y = x + Cx^2.
$$
|
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|
Show that the function $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$. Show that $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$.
I know that if $\phi$ is harmonic then it satisfies Laplace's equation but I don't see how considering $w(z)$ will help me.. I've tried looking at the real and imaginary parts of $z$ but they don't seem to match up. I'm guessing the solutions will have something to do with the Cauchy Riemann equations but I've tried differentiating $w(z)$ with respect to $x$ and $y$, where $z=x+iy$, and that hasn't shed any light either.
|
Hint:
$w$ is a fractional linear transformation, hence a conformal mapping (of what domains?). Therefore its inverse is also conformal. Composing a harmonic function with a conformal map gives you another harmonic function. So it remains to see what function you get by composing $\phi$ and $w$.
|
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|
Pointwise convergence to zero, with integrals converging to a nonzero value For $n\in{\mathbb{N}}$ let $$f_n(x)=nx(1-x^2)^n\qquad(0\le x\le 1).$$
Show that $\{f_n\}_{n=1}^\infty$ converges pointwise to $0$ on $[0,1]$.
Show that $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$.
I've already shown both of these statements to be true. What I don't understand is this: how can $f_n$ converge pointwise to $0$, yet the sequence $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$? Isn't that almost like saying $f(x)=0\qquad(a\le x\le b)$, but $\int_a^bf(x)=\frac12$?
Clearly that would be false. I know this has something to do with that this is a sequence of functions but it still baffles my mind. Thanks in advance.
|
Have you tried sketching the first few functions?
As $n$ increases, the area under the function graph "flows" further and further to the right end like a wave while remaining constant.
As $f_n(1)=0$ for all $n$ and every point $x<1$ falls behind the wave sooner or later, the pointwise convergence follows.
The underlying problem is that swapping limits (and integration is a limit) is not always allowed; the basic example for sequences is $\lim_{n\to\infty}\lim_{m\to\infty}\frac{n}{n+m}=0\ne1= \lim_{m\to\infty}\lim_{n\to\infty}\frac{n}{n+m}$.
|
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|
Motorway problem: how to minimize the number of encounters? I was working on the motorway problem:
At time t=0, cars are launched from the same entrance of highway following a Poisson process with parameter $\lambda$, and the speed of cars ($v$) follows a known distribution of $S$.
The observer is on one of the cars with a constant speed of $v_0$, and the amount of observation time is $T$. Since the speed of each car might be different, observer's car can catch up with some cars ahead of it, or be caught up by some cars from behind.
The question is: how to choose $v_0$ such that the expected number of encounters as seen from the observation car is minimized, given only 3 parameters ($\lambda$, $v_0$, observation time $T$) and 2 distributions (Poisson distribution of launch time, and the speed distribution for cars $S$)?
|
EDIT: check Sheldon M. Ross, introduction to probability model, edition 9, example 5.19, or this link.
|
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|
Why don't infinite sums make any sense? Using the infinite sum series, an infinite sum of (1/5)to the nth power, where n goes from zero to infinity, the general summation equation tells us that the answer is 5/4. However, how is this possible, considering that each additional sum is smaller than the previous sum? If you begin to do the math by hand, calculating and adding each sum individually, you get something like .249984...etc... where it appears that the number will extend infinitely to the right, adding decimal places and occasionally revising numbers later to the right of the decimal, but the additional sums should never be large enough to affect the initial answers in the decimal like the .2 part of the answer? It seems very counterintuitive.
There seems to be a basic contradiction between the expected answer and the answer of the equation by a large margin of difference. Can anyone explain this?
|
If you try working in base 5 rather than base 10 then it will become more obvious.
It's kind of like how using 1/9 instead of 1/5 gives you 0.99999... = 1 which I hope you agree with.
|
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|
Determining an upper bound I have a function $$f(\lambda)=n\ln(1-p+pe^{\frac{\lambda}{n}})-\lambda p$$
I need to prove that $$f(\lambda)\leq \frac{\lambda^2}{8n}$$ using Taylor expansion.
I have used the taylor expansion for a function of the form $\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\dots$
Therefore: $f(\lambda)=n\ln[1-p+pe^{\frac{\lambda}{n}}]-\lambda p=n[-p+pe^{\frac{\lambda}{n}}+(p-p^2)e^{\frac{\lambda}{n}}+\frac{1}{2}e^{{2\lambda}{n}}-\dots]-\lambda p$
Not sure how to go on from here.
|
$$f'(\lambda) = \frac{pe^{\frac{\lambda}{n}}}{1-p+pe^{\frac{\lambda}{n}}}$$
$$f''(\lambda) = \frac{\frac{1}{n}(1-p)pe^{\frac{\lambda}{n}}}{(1-p+pe^{\frac{\lambda}{n}})^2}$$
$$f'''(\lambda) = \frac{1}{n}(1-p)p \frac{\frac{1}{n}e^{\frac{\lambda}{n}}(1-p -pe^{\frac{\lambda}{n}})}{(1-p+pe^{\frac{\lambda}{n}})^3}$$
Use GM-AM inequality,
$$ f''(\lambda) \le \frac{1}{n} \frac{1}{4} = \frac{1}{4n}$$
Then $$ f(\lambda) = f(0) + f'(0)\lambda +\frac{1}{2}f''(\eta)(\lambda)^2 - p\lambda$$
$$\le \frac{1}{2}f''(\eta)(\lambda)^2 = \frac {\lambda^2}{8n}$$
Here $0 \le \eta \le \lambda$.
QED
|
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Number of elements of order $6$ in $\text{Aut}(\mathbb Z_{720})$ I tried to determine the number of elements of order $6$ in $\text{Aut}(\mathbb Z_{720})$.
Please could someone tell me if this is correct?
$$ \text{Aut}(\mathbb Z_{720}) \cong U(720) \cong U(9) \oplus U(16) \oplus U(5) \cong \mathbb Z_{6} \oplus \mathbb Z_{8} \oplus \mathbb Z_{4}$$
We want elements of orders $i,j,k$ such that $\text{lcm}(i,j,k) = 6$.
In $\mathbb Z_{6}$ there is one element of order 1, two of order three, one of order two and two of order six.
In $\mathbb Z_{8}$ there is one of order one, two of order 4, one of order 2 and 4 of order 8.
In $\mathbb Z_{4}$ there is one of order 1, one of order 2 and 2 of order 4.
Writing $\text{lcm}(i,j,k)$ as $(i,j,k)$ we get the following possibilities for order six:
$$ (6,1,1), (6,1,2), (6,2,1), (6,2,2), (3,2,1), (3,1,2) , (3,2,2)$$
Summing these possibilities using the above observation of numbers of elements of certain orders we get:
$$ 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 + 2*1*1 = 7*2 = 14$$
elements of order 6.
This seems plausible since the number of elements of order $6$ must be a multiple of $2$. But this doesn't mean that it is correct, of course.
What is the number of elements of order $6$ in $\text{Aut}(\mathbb
Z_{720})$?
|
As noted by other you are making mistake in assuming that $U(16) \cong \mathbb{Z}_8$. Instead you have that:
$$\text{Aut}(\mathbb{Z}_{720}) \cong \mathbb{Z}_6 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4$$
Now as we want an element of order $6$ we must include an element whose "order" is divisible by $3$ from the first group and there are four such elements, namely $1,2,4,5$.
If we choose $1$ or $5$ then we can include any element of order divisible by $2$ from the other three subgroups, so as each of them has $2$ of them there are $2^4$ options, namely $(1,1,2,2), (1,1,0,0), (1,1,2,0), (1,1,0,2), (1,0,2,2), (1,0,0,2), (1,0,2,0), (1,0,0,0), (5,1,2,2), (5,1,0,0), (5,1,2,0), (5,1,0,2), (5,0,2,2), (5,0,0,2), (5,0,2,0), (5,0,0,0)$
If we choose $2,4$ then we can include any element of order divisible by $2$ from the other three subgroups, but we must not have all of them identity. So therefore there are $2(2^3 - 1) = 14$ options. Namely: $(2,1,2,2), (2,1,0,0), (2,1,2,0), (2,1,0,2), (2,0,2,2), (2,0,0,2), (2,0,2,0), (4,1,2,2), (4,1,0,0), (4,1,2,0), (4,1,0,2), (4,0,2,2), (4,0,0,2), (4,0,2,0)$
Therefore there are $30$ elements of order $6$ in $\text{Aut}(\mathbb{Z}_{720})$
|
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|
Show that $f_{n}^2(x)$ does not converge in $D^1({\Omega})$ Let
$$ f_n(x) = \left\{ \begin{array}{ll}
n & \mbox{if $0 \lt x \lt \frac{1}{n}$};\\
0 & \mbox{otherwise}.\end{array} \right. \ $$
I have to show that
*
*$\lim_{n \to \infty}f_n(x)$ exists in terms of distributions
*$\lim_{n \to \infty}f_n^2(x)$ doesn't exist in terms of distributions.
*$\lim_{n \to \infty}(f_n^2(x)-n\delta)$ exists in terms of distributions
For $(1)$
$$n\int_{0}^{\frac{1}{n}}\phi(x) dx \to \phi(0) , \text{for}, n \to \infty$$
For $(2)$
$$n^2\int_{0}^{\frac{1}{n}}\phi(x) dx $$ Now there exists a $C^{\infty}$ cut off function such that $\phi(x) =1$ for $x \in (0,\frac{1}{n+1})$ and $\phi(x)=0, x \gt \frac{1}{n}$. For this particular choice of $\phi$ this integral goes to $\infty$.
For $(3)$ This will be
$$\int_{-n}^{\frac{1}{n}} -(n\delta)\phi(x) dx +(n^2-n\delta)\int_{0}^{\frac{1}{n}}\phi(x) dx+\int_{\frac{1}{n}}^{n} -(n\delta)\phi(x) dx $$
I am unable to see where this would converge.
Thanks for the help!!
|
A quick remark on your solution of (2): You need to say that $\phi = 1$ on $[0;1]$ or something like that, as $\phi$ has to be independent of $n$. I know it's pedantic, but this is how you can loose marks on exams.
For (3): you actually want to show that the map $$T(\phi) = \lim_{n \to \infty} \int_{0}^{\frac{1}{n}} n^2f(x)dx - n \delta(\phi)$$ is a distribution.
Because $\delta(\phi) = \phi(0) = \int_{0}^{\frac{1}{n}}n\phi(0)dx$, the above map is actually
$T(\phi) = n^2\int_{0}^{\frac{1}{n}}\phi(x) - \phi(0) dx$
There are many ways to find this limit. I will use Taylor's theorem: $\phi(x) \phi(0) + x \phi'(0) + O(n^{-2})$ where I used that in the integral $|x| \leq \frac{1}{n}$. The remainder term is bounded by $\frac{C}{n^2}$ so when you multiply by $n^2$ and integrate from $0$ to $\frac{1}{n}$, this will converge to $0$.
So, $$ T(\phi) = \lim n^2\int_{0}^{\frac{1}{n}} x \phi'(0) dx = \frac{\phi'(0)}{2}$$
Indeed, this is a distribution: it is a linear functional and it is continuous: if $\phi_k \to \phi$ in the topology of $D$ then $\sup_{\mathbb{R}} |\phi_k' - \phi'|\to 0$, i.e $T(\phi_k) \to T(\phi)$. If the last explanation looks meaningless you might use a different definition of the topology of $\mathbb{D}$: you can look at Distributions for more information.
I hope this helps.
|
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|
What is the range of the derivative of $g(x) = f(x)/(x+1)$, where $f(x)$ is differentiable on $x\in[0,5]$? Let $g(x) = f(x)/(x+1)$, where $f(x)$ is differentiable on $x\in[0,5]$, such that $f(0)=4$ and $f(5)=-1$. What is the range of values $g'(c)$ for a $c$ belonging to $[0,5]$?
Considering values of $f(x_i)$, $f(x)$ must decrease at least once from $0$ to $5$. But that is all the information I can use here. Is there anything I am missing?
|
This question indeed seems to be incomplete. For instance, if there exists a
$c\in(0,5)$ such that $$c=\dfrac{-1+2f'(c)\pm\sqrt{4\big(f'(c)\big)^2-8f''(c)f(c)} }{2f''(c)}$$ then $g'(x)$ attains a local maximum on $(0,5)$. To know if this is the case we need way more information on $f$.
If we were given, that $f$ is twice differentiable on $[0,5]$, then we could use the above to conclude at least that the range of $g'$ is bounded. But as I said, this question does not seem solvable, with the information we're given.
|
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|
Existance of an (in)finite theory having infinite model Please help me to study the following simple cases:
Let $P$ be a binary predicate symbol. I am trying to find out, if there exists a satisfiable $T$ having infinite models only, for the following cases:
1) T is under {P} language with "="
2) T is under {P} language with "=" and T is finite
3) T is under {P} language without "="
4) T is under {P} language without "=" and T is finite
Where I am:
1)
$\varphi_i \equiv \forall x_1 \forall x_2 ... \forall x_i \exists y \land ^i_{j=1} \lnot( x_j=y)$
$T = \{\varphi_i | i>0 \}$ so there exists a T having only infinite models
2) 4) can probably have use of Löwenheim-Skolem theorem, but I don't see how to apply it properly.
3) No adequate idea so far
|
To make Mike's answer just a little more explicit:
When $P$ is the inequality symbol of a partial order, it interprets equality in the sense that the definable set $\{m \in M \operatorname{ | } P(m,m)\}$ is exactly the diagonal relation $\{m \in M \operatorname{ | } m = m\}$.
In particular, we know that such a $P$ must satisfy: $$\forall a \forall b, P(a,b) \land P(b,a) \rightarrow P(a,a),$$
$$\forall a \forall b \forall c, P(a,b) \land P(b,c) \rightarrow P(a,c),$$
$$\forall a, P(a,a).$$
After writing these out, can you think of just one more axiom we can add to $T$ such that any chain in our partial order must be infinite? (Hint: given an element $a$ in a partial order $M$, the set of all elements $b$ less than $a$ is $a$-definable.)
|
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|
There are 6 white balls and 9 black balls. Probability of drawing two white, then two black? From A First Course in Probability (9th Edition):
3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be
randomly selected without replacement, what is the probability that
the first 2 selected are white and the last 2 black?
This method is straightforward and results in the correct answer (according to the book):
$$\frac{6}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} \cdot \frac{8}{12} = \frac{6}{91} $$
(This is just the multiplication principle and probability of drawing the color of that ball at that time)
However, I want to understand this in terms of conditional probability. I don't understand why this doesn't work:
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{6 \choose{2}}{9 \choose 2}}{{15 \choose{2}}{13 \choose 2}}}÷{\frac{{6 \choose{2}}}{{15 \choose{2}}}} = {\frac{{9 \choose 2}}{{13 \choose 2}}} = \frac{6}{13} \ne \frac{6}{91}$$
$\frac{6}{13}$ is exactly 7 times more than the previous answer. Why does this method fail to work? What mistake have I made? I tried to use the exact same method used in question 3.3, where this resulted in the correct answer.
Optional – About 3.3
3.3 Use Equation (2.1) to compute in a hand of bridge the conditional
probability that East has 3 spades given taht North and South have a
combined total of 8 spades.
Here, we see that:
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{13 \choose{8}}{39 \choose 18}{5 \choose 3}{21 \choose 10}} {{52 \choose{26}}{26 \choose 13}}}÷{\frac{{13 \choose{8}}{39 \choose 18}}{{52 \choose{26}}}} = {\frac{{5 \choose 3}{21 \choose 10}}{{26 \choose 13}}} = \frac{29}{115} \approx 0.339$$
Which is the answer in the back of the book.
|
Method 1.
$$
\text{Probability}={\frac{{{6}\choose{2}}{{9}\choose{2}}}{{{15}\choose{2}}{{13}\choose{2}}}}=\frac{6}{91}.
$$
$$
\binom{6}{2}\text{ is the quantity of ways to choose 2 any white balls of 6 white balls},
$$
$$
\binom{9}{2}\text{ is the quantity of ways to choose 2 any black balls of 9 black balls},
$$
$$
\binom{15}{2}\text{ is the quantity of ways to choose 2 any white balls of}
$$
$$
\text{ the total amount of balls (15)},
$$
$$
\binom{13}{2}\text{ is the quantity of ways to choose 2 any black balls of}
$$
$$
\text{ the total amount of balls (13), considering that 2 any white balls are chosen already}.
$$
Method 2.
Also this problem can be performed using the more detailed binoms:
$$
\text{Probability}=\frac{\binom{6}{1}\binom{5}{1}\binom{9}{1}\binom{8}{1}}{\binom{15}{1}\binom{14}{1}\binom{13}{1}\binom{12}{1}}=\frac{6}{91}.
$$
|
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|
Find root of the equation Find maximum root of the equation $$x - \frac{1000}{\log 2} \log x = 0$$
It locates between $13746$ and $13747$, but I want to find right solution not using graphing calculators. Thanks in advance.
|
$$x-\frac{1000\ln(x)}{\ln(2)}=0\Longleftrightarrow$$
$$-\frac{1000\ln(x)}{\ln(2)}=-x\Longleftrightarrow$$
$$\frac{1000\ln(x)}{\ln(2)}=x\Longleftrightarrow$$
$$1000\ln(x)=x\ln(2)\Longleftrightarrow$$
$$e^{1000\ln(x)}=e^{\ln(2)x}\Longleftrightarrow$$
$$x^{1000}=2^x\Longleftrightarrow$$
$$x=\exp\left[-\text{W}\left(-\frac{\ln(2)}{1000}\right)\right]\approx13746.809166647028809$$
|
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|
Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\frac{x}{n}}}$ converges pointwise to $\frac{1}{x^3}$. So if we could apply Lebesgue's Dominated Convergence Theorem, we have:
$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}=\lim_{n \rightarrow \infty} \int_{1}^{\infty} \frac{\mathrm dx}{x^3}=\frac{1}{2}$
I have a problem with finding a majorant. Could someone give me a hint?
|
I think one could do this in a conceptually simpler way:
Since
$$
\frac{y}{1+y}<\log(1+y)<y
$$
your integrand is bounded as
$$
\frac{1}{x^3}<\frac{1}{nx^2\ln(1+x/n)}<\frac{1}{nx^2(x/n)/(1+x/n)}=\frac{1}{x^3}(1+x/n).
$$
By monotonicity, your integral satisfies
$$
\frac{1}{2}=\int_1^{+\infty}\frac{1}{x^3}\,\mathrm dx<\int_1^{+\infty}\frac{1}{nx^2\ln(1+x/n)}\,\mathrm dx<\int_1^{+\infty}\frac{1}{x^3}(1+x/n)\,\mathrm dx=\frac{1}{2}+\frac{1}{n}.
$$
Now squeeze.
|
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|
Show $\forall \delta > 0, \exists n \in \mathbb{N}$ such that $\frac{1}{n} < \delta$ The question is in the title, but I have no idea how to solve it, so a few hints would be appreciated, thanks.
|
Suppose the statement is false. Then we have some $\delta>0$ such that $\delta^{-1}>n$ for all $n$. This contradicts the Archimedean property. Assuming $\mathbb{R}$ is defined axiomatically, this property is easily derived as a consequence of the least upper bound axiom.
|
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|
Combining two results from partial integration I have a set of two PDEs:
$$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$
$$\partial_{\tau}\psi=-\partial_{\eta}\theta+\alpha\partial_{\eta}^{2}\psi$$
These can be combined into a wave equation of the form:
$$\partial_{\tau}^{2}\theta=\partial_{\eta}^{2}\left[\theta+\alpha\partial_{\tau}\theta\right]
$$
which with the ansatz:
$$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right)$$
gives the dispersion relation:
$$\omega^2-\alpha\omega+1=0$$
The solution to the dispersion relation is $\omega=\omega_r+i\omega_i$ with $\omega_r=\frac{1}{2}\alpha$ and $\omega_i=\pm\sqrt{1-\frac{1}{4}\alpha^2}$.
Now i would like to determine $\psi\left(\eta,\tau\right)$, I attempted to do this by using the original equations in a rewritten form:
*
*$$\partial_{\eta}\psi = -\partial_{\tau}\theta \rightarrow \psi\left(\eta,\tau\right) = -\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)+K_1$$
*$$\partial_{\tau}\psi = -\partial_{\eta}\theta-\alpha\partial_{\eta}\partial_{\tau}\theta \rightarrow \psi\left(\eta,\tau\right) = \left(\frac{1}{\omega}-\alpha\right)\cos\left(\eta\right)\exp\left(-\omega\tau\right) + K_2$$
but as far as i can see these two partial solutions cannot be combined to satisfy both original equations simulateously. How do i go about getting a solution for $\psi$?
|
As it turns out the two results for $\psi$ are actually equivalent as by the dispersion relation:
$$\frac{1}{\omega}-\alpha=\frac{1}{\omega}\left(1-\alpha\omega\right)=-\omega$$
Alternative method: First, redefine $\theta$ and $\psi$ using a scalar $\phi$:
$$\theta=\partial_{\eta}\phi\quad\psi=-\partial_{\tau}\phi$$
This ensures that:
$$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$
is exactly satisfied and yields a wave equation for $\phi$ from the other equation:
$$\partial_{\tau}^{2}\phi=\partial_{\eta}^{2}\left[\phi+\alpha\partial_{\tau}\phi\right]$$
The solution for $\phi$ has a similar form as previously found for $\theta$:
$$\phi\left(\eta,\tau\right)=\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right)
$$ where the constants $A$ and $B$ are yet to be determined.
From the definitions for $\theta$ and $\psi$ I find:
$$\theta=\partial_{\eta}\phi=\left[A\cos\left(\eta\right)-B\sin\left(\eta\right)\right]\exp\left(-\omega\tau\right)$$
$$\psi=-\partial_{\tau}\phi=\omega\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right)
$$
Now if I am looking for a solution of $\psi$ which corresponds to a $\theta$ distribution:
$$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right)
$$
I require $A=0$ and $B=-1$ such that:
$$\psi=-\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)
$$
|
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|
Proof based on definition of big-$O$ I want to prove that $n! = O(n^n)$ based on the definition of big-$O$.
I find it pretty easy to show that $n! = O(n^n)$ by simply showing that $n (n-1) < n \cdot n \ldots$ etc. However I can't figure out how to prove this based on the definition of big-$O$. How can I prove this based upon the definition?
|
We have
$\displaystyle
\lim_{n \to \infty} \frac{n!}{n^n}=0,
$
which shows that $n!=o(n^n)$ and hence $n!=O(n^n)$.
To compute the limit one can make use of Stirling's formula:
$$n!\sim \sqrt{2\pi n} \frac{n^n}{e^n}.$$
|
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|
Integrate the following equation. (exponential function) Integrate $$\frac{e^x -2}{e^{x/2}}$$
This is my calculation:
but it is wrong....
|
It looks like you attempted to use substitution to replace
$$ \int \frac{\frac12 dx}{e^{x/2}}$$
with
$$ \int \frac{du}{u} = \ln(u) + C. $$
The fatal flaw here is that if $u = e^{x/2}$,
then $du = \frac12 e^{x/2} dx \neq \frac12 dx$.
The correct substitution is
$$ \tfrac 12 dx = \frac{1}{e^{x/2}} du = \frac{du}{u}.$$
$$ \int \frac{\frac12 dx}{e^{x/2}}
= \int \frac{du}{u^2} = -\frac{1}{u} + C = -\frac{1}{e^{x/2}} + C
= - e^{-x/2} + C.$$
Of course you can get this much quicker by just writing
$$ \int \frac{\frac12 dx}{e^{x/2}}
= \int \tfrac12 e^{-x/2} dx$$
as in the other answers.
It looks like you are not writing out your substitutions explicitly,
nor are you using the $dx$ notation to keep track of the variable of
integration. The first is a shortcut and the second is questionable
notation. Since you make errors doing things this way, perhaps
it would be better to use more explicit notation instead.
|
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|
Comparing matrix norm with the norm of the inverse matrix I need help understanding and solving this problem.
Prove or give a counterexample: If $A$ is a nonsingular matrix, then $\|A^{-1}\| = \|A\|^{-1}$
Is this just asking me to get the magnitude of the inverse of Matrix $A$, and then compare it with the inverse of the magnitude of Matrix $A$?
|
If $A$ is nonsingular, then $AA^{-1} = I$, so
$$ 1 = ||I|| = ||AA^{-1}|| \leqslant ||A||\cdot||A^{-1}||. $$
In general, then $1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||$.
Equality is thus not necessarily guaranteed for arbitrary nonsingular $A$; however, the inequality above implies that equality may occur. Consider an example.
Example:
$A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 0.5 \end{bmatrix} $
$\implies ||A||_{1,2,\infty} = 2$
$\implies ||A^{-1}||_{1,2,\infty} = 1$
$\implies \frac{1}{2} = ||A||_{1,2,\infty}^{-1} \neq ||A^{-1}||_{1,2,\infty} = 1 \implies ||A||^{-1} \neq ||A^{-1}||.$
|
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|
Calculating first and second moments for random sums? Assume that $N$ and $X_1, X_2, \ldots $ are all independent and identically distributed over $(0,1)$ with the density function:
$f (x) = cx^2 (1 − x)^2$.
An integer–valued random variable, $N$ specifies a random sum of first $(N + 1)$ variables,
$$\sum_{j=1}^{N+1}x_j$$
We assume, for integer values of $k$, that $N$ is distributed as:
$$P [N = n] = (1 − p) p^k$$
I want to find the first and second moments of $Z$.
So, first, I integrated $f(x)$ over $(0,1)$ to solve for $c$ and recovered that $c=30$. So, we have:
$$f(x)=30x^2(1-x)^2.$$
My textbook doesn't really provide a method for doing this for a random sum. I understand the standard approach for just one variable, where
$$E[X^k]=\int_0^1 x^kf(x) \, dx$$
However, I don't understand how this relates to the method for a random sum. I also need to repeat this process for when $N$ is Poisson, so I'd really appreciate a clear step-by-step process for how to calculate this. Very much appreciated.
|
From your previous question, you know how to do this if you know $\mathsf E(Z\mid N)$ and $\mathsf {Var}(Z\mid N)$
So: $\mathsf E(Z\mid N) = \sum\limits_{i=1}^{N+1} \mathsf E(X_i)$ By linearity of Expectation.
Similarly: $\mathsf E(Z^2\mid N) =\mathsf E\Big(\big(\sum\limits_{i=1}^{N+1} X_i\big)^2\Big) = \raise{1.5ex}\mathop{\sum\limits_{i=1}^{N+1}\sum\limits_{j=1}^{N+1}}_{j\neq i}\mathsf E(X_i)\mathsf E(X_j)+\sum\limits_{i=1}^{N+1}\mathsf E(X_i^2)$ by that and the iid distribution of $\{X_i\}$.
Where : $\displaystyle\mathsf E(X_i)=\int_0^1 30x^3(1-x)^2\operatorname d x \\ \displaystyle \mathsf E(X_i^2)=\int_0^1 30 x^4(1-x)^2\operatorname d x$
|
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|
Important numerator and denominators in the evaluation of the integral: $\int_0^\infty x^t \operatorname{csch} x\text{ d}x$ $$\int_0^\infty x^t\operatorname{csch}x\text{ d}x=\frac{a\zeta(t+1)}{b}$$
for $t\in\Bbb{N}$
How might one represent $a,b$ in terms of $t$?
(Note that $a,b\in \Bbb{N}$)
If possible, could one also provide a proof please?
|
Using my newfound knowledge of Mellin Transforms, we can note that if we denote
$$F(s)=\int_0^\infty x^{s-1} \space f(x) \text{ d}x$$
Plugging in $f(x)=\text{csch} x$
One might note that $$\frac{a}{b}=F(s+1)$$
which is known as the result given.
|
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|
Prove that $u\cdot v = 1/4||u+v||^2 - 1/4||u-v||^2$ for all vectors $u$ and $v$ in $\mathbb{R}^n$ I need some help figuring out how to work through this problem.
Prove that $ u \cdot v = 1/4 ||u + v||^2 - 1/4||u - v||^2$ for all vectors $u$ and $v$ in $\mathbb{R}^n$.
Sorry, forgot to include my work so far:
I decided to ignore the 1/4 and deal with it later once I had a better understanding of the question.
$= ||u+v||^2 - ||u-v||^2$
$= (u+v)(u+v) - (u-v)(u-v)$
$= u(u+v) + v(u+v) - u(u-v) + v(u-v)$
$= uu + uv + uv + vv - uu + uv + uv - vv$
$u \cdot v= 3(uv)$
This is as far as I've gotten, not sure if I'm on the right track or where to go next.
|
Remember $\vec{x} \cdot \vec{x} = \left \| x \right \|^2 $. That should be really helpful, in my mind.
And as @T.Bongers said, when working with identities like this, begin with the harder side. Try to "simplify" it (or make it a big more complex before having things cancel!) into the more basic side.
|
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Cauchy like inequality $(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$ Problem: Prove that for real $x, y, \alpha, \beta$,
$(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$.
I am looking for an elegant (non-bashy) solution.
It closely resembles Cauchy inequality but $\alpha y +\beta y$ part is creating a problem.
I also tried to define a suitable inner product but couldn't.
Since, the inequality is homogenous, (for non-zeros) it reduces to,
$(5mn+m+n+3)^2 \leq (5m^2+2m+3)(5n^2+2n+3)$, by putting $\alpha = \beta m$ and $x=n y$. But I couldn't take it furthur from here.
So, any hints, solutions (especially along lines of inner product) would be very welcome.
|
By C-S $(5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)=$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)^2+\frac{14\beta^2}{5}\right)\left(\left(\sqrt5x+\frac{y}{\sqrt5}\right)^2+\frac{14y^2}{5}\right)\geq$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)\left(\sqrt5x+\frac{y}{\sqrt5}\right)+\frac{14\beta y}{5}\right)^2=(5\alpha x+\alpha y+\beta x + 3\beta y)^2$
|
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|
Evaluate $\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$ I am trying to re-learn some basic math and I realize I have forgotten most of it.
Evaluate $$\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$$
Call the terms $S_n$ and the total sum $S$.
$$S_n < \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1} = S < \infty$$
$$S_n = \frac{n}{n^4+n^2+1} = \frac{n}{(n^2+1)^2-1}$$
It has been more than a few years since I did these things.
I would like a hint about what method I should try to look for?
Thanks.
|
HINT:
$$n^4+n^2+1=(n^2-n+1)(n^2+n+1)$$
Write $2n$ as $$n^2+n+1-(n^2-n+1)$$
Observe that if $f(m)=m^2-m+1, f(m+1)=?$
which immediately reminds me of Telescoping Series.
|
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|
Expected Value Problem Using Confusing Conditional Probability I am trying this question:
We have a bag with 10 blue jacks and 20 red jacks. We pick 3 jacks from the bag at random and with replacement. We are told that at least one jack is red.
Compute the expected values of blue jacks and red jacks, again being told that we have at least one red jack
My attempt so far:
I am trying to use Bayes' law to calculate the probabilities first, before I multiply by the number of balls to find the final expected values. However, when I am trying to calculate the conditional probability, I do not know how to include the "at least one red jack" part into the Bayes' formula.
$$P(\text{blue jack} \mid \text{given at least one red jack}) = \frac{P(\text{blue jack}\ \cap\ \text{at least one red jack}) }{\text{P(at least one red jack)}}$$
But wouldn't the numerator of the above fraction be $0$? I am new to probability and I would appreciate some help understanding this. Thank you.
|
Since we draw just three jacks, and each is either red or blue,
the condition $R \geq 1$ is equivalent to $B \neq 3$.
That is, $R \geq 1$ rules out just the possible outcome $B = 3$.
This suggests the following shortcut for calculating $E(B \mid R \geq 1)$.
The unconditional expected value of the number of blue jacks is
\begin{align}
E(B)
& = 3\cdot P(B=3) + 2\cdot P(B=2) + 1
\cdot P(B=1) + 0 \cdot P(B=0) \\
& = 3\cdot P(B=3) + 2\cdot P(B=2) + P(B=1) \\
& = 1.
\end{align}
Note that we can actually calculate $E(B)=1$ by summing the expected
number of blue jacks on the first draw plus
the expected number on the second draw plus the
expected number on the third,
that is, $\frac13 + \frac13 + \frac13 = 1.$
I wrote the whole formula out only as a reminder that that formula
also equals $E(B)$.
The conditional expectation is
$$
E(B \mid R \geq 1)
= 3\cdot P(B=3\mid R \geq 1) + 2\cdot P(B=2\mid R \geq 1) + P(B=1\mid R \geq 1).
$$
But
$$P(B=3\mid R \geq 1) = \frac{P(B=3 \cap R \geq 1)}{P(R \geq 1)} = 0,$$
while
\begin{align}
P(B=2\mid R \geq 1) &= \frac{P(B=2 \cap R \geq 1)}{P(R \geq 1)} =
\frac{P(B=2)}{P(R \geq 1)},\\
P(B=1\mid R \geq 1) &= \frac{P(B=1 \cap R \geq 1)}{P(R \geq 1)} =
\frac{P(B=1)}{P(R \geq 1)},
\end{align}
so
\begin{align}
E(B \mid R \geq 1)
& = 2 \frac{P(B=2 \cap R \geq 1)}{P(R \geq 1)}
+ \frac{P(B=1 \cap R \geq 1)}{P(R \geq 1)} \tag 1\\
& = \frac{1}{P(R \geq 1)}(2 P(B=2) + P(B=1)).
\end{align}
Noticing that if we just drop the first term from the formula for
$E(B)$, we get the quantity $2 P(B=2) + P(B=1)$ in the last equation,
we can write
$$2 P(B=2) + P(B=1) = E(B) - 3 P(B=3),$$
and noticing that $P(R \geq 1) = 1 - P(B = 3)$,
we can write
$$
E(B \mid R \geq 1) = \frac{1}{1 - P(B = 3)}(E(B) - 3 P(B=3)). \tag 2
$$
The right side of Equation $(2)$ should be relatively easy to calculate.
There are a lot of equations above, possibly more than if you just wrote
out the definition of $E(B \mid R \geq 1)$ and evaluated it the "hard" way,
but most of the equations are there just to show the thought process that
came up with Equation $(2)$.
I would also write at least Equation $(1)$ as part of my reasoning,
but not most of the other equations.
This isn't much of a shortcut for this particular problem, since
it's not very much more work to just work out the terms from the
definition of $E(B \mid R \geq 1)$, but if we were drawing $9$ jacks
instead of $3$ then it would make quite a difference in the
effort to get the result. We'd just be dealing with $P(B=9)$ and
$9P(B=9)$ instead of $P(B=3)$ and $3P(B=3)$.
|
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|
Proof that $|x|+|y|\leq\sqrt{2(x^2+y^2)}$ How do I prove that for $x,y\in\mathbb{R}$ we have $|x|+|y|\leq\sqrt{2(x^2+y^2)}$?
I thought that $(|x|+|y|)^2=x^2+y^2+2|x||y|\leq2(x^2+y^2)$, but I'm not sure why that holds.
|
The claim immediately follows from
$$\bigl(|x|+|y|\bigr)^2\leq\bigl(|x|+|y|\bigr)^2+\bigl(|x|-|y|\bigr)^2=2\bigl(|x|^2+|y|^2\bigr)\ .$$
|
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|
Problem with Indefinite Integral $\int\frac {\cos^4x}{\sin^3x} dx$ I'm stuck with this integral
$\int\frac {\cos^4x}{\sin^3x} dx$
which I rewrote as
$\int \csc^3x \cos^4xdx$
then after using the half angle formula twice for $\cos^4x$ I got this
$\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$
then after solving those products I got these integrals
$\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$
I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!
|
Let $t=\cos x$, then
$$\int \frac{\cos^4 x}{\sin^3 x}dx=-\int \frac{t^4}{(1-t^2)^2}dt.$$
Can you proceed?
|
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|
Finding all eigenvectors and eigenvalues of a linear operation on a function Here is the question I am stuck on:
Consider $T \in \mathrm{Hom}(\Bbb{R}[x]_{\le 2} ,\Bbb{R}[x]_{\le 2} )$ given by
$$ (Tf)(x)=\int_{-1}^1(x-y)^2f(y)dy-2f(0)x^2$$
for all $f \in \Bbb{R}[x]_{\le 2}$. Find all eigenvalues and eigenvectors for T.
So far, I'm trying to apply T to the monomial basis of $\Bbb{R}[x]_{\le 2}$.
I got that $T(1)=-2x$, but where I'm stuck is how to evaluate $T(x)$ for the $f(y)$ and $f(0)$ terms. I thought that if I applied $T(x)$ to $f(y)$, it would equal $y$ because the function is $x$, and $f(y)$ means $x=y$. When I evaluate this with the $f(y)$ term equal to $y$, I get back that $T(x)=-(4/3)x-2x^3$. The $2x^3$ term's degree is too high for the bounds, so I'm not sure how to proceed from here.
Is it wrong to say that $x$ evaluated at $y$ equals $y$?
Any help would be greatly appreciated.
|
Note that we should integrate with respect to $y$, not $x$. Write $f(x) = x$. Then
$$ T(f)(x) = \int_{-1}^1 (x - y)^2 f(y) \, dy - 2f(0)x^2 = \int_{-1}^1 (x - y)^2 y \, dy = \int_{-1}^1 (x^2y - 2xy^2 + y^3) \, dy =\left[ \frac{x^2y^2}{2} - \frac{2xy^3}{3} + \frac{y^4}{4} \right]^{y = 1}_{y = -1} = -\frac{4}{3} x. $$
|
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|
showing projection is a linear operator Show that the orthogonal projection is linear.
Let $x_i=y_i+z_i$, where $x_i\in X$, $y_i\in Y$, $z_i\in Y^\perp$, and $\alpha,\beta$ be scalars. Then \begin{align}P(\alpha x_1+\beta x_2)&=P(\alpha(y_1+z_1)+\beta(y_2+z_2))\\&=P(\alpha y_1+\beta y_2+\alpha z_1+\beta z_2)\end{align}
Now I know $\alpha x_1 + \beta y_1 \in Y$ but not sure what to do here.
|
The key here is that $X = Y\oplus Y^\perp$, i.e. for any $x\in X$ there are unique $y\in Y, z\in Y^\perp$ such that $x = y+z$.
This is essential in order for $P(x) = P(y+z) = y$ to be well defined in the first place.
Now, what you showed is that $\alpha x_1 +\beta x_2$ can be uniquely written as $(\alpha y_1 + \beta y_2) + (\alpha z_1 + \beta z_2)$ where $\alpha y_1 + \beta y_2\in Y$, $\alpha z_1 + \beta z_2\in Y^\perp$. So, what is $P(\alpha x_1 + \beta x_2)$?
Edit:
Your definition of orthogonal projection assumes that for each $x\in X$ there is unique $y\in Y$ such that $x-y\in Y^\perp$. This is actually equivalent to stating that $Y\oplus Y^\perp = X$, i.e. there are unique $y\in Y$, $z\in Y^\perp$ such that $x = y+z$ (notice that $z = x-y$ from your definition).
So, assume that there is unique $y\in Y$ such that $x-y\in Y^\perp$. Then, $P(x) = P(y+(x-y)) = y$ is just restating my claim if you substitute $z = x-y$.
The bigger question is why such $y$ exists. In finite-dimensional case this follows immediately from existence of orthonormal basis for $Y$. Then you can define $y = \sum \langle x,e_i\rangle e_i$. In infinite-dimensional case we can use projection theorem for Hilbert spaces when we can find such $y \in Y$ that minimizes length $\|x-y\|$ (think of a point and a line: minimum distance between point and line is given by orthogonal projection).
I hope this clarifies things a bit.
|
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|
subsets in the cartesian product Let $A,B,C,D$ be sets. Consider $A\times B$ and $X\subseteq A\times B$. Is it true that $X$ has the form $A'\times B'$ where $A'\subseteq A$ and $B'\subseteq B$ ?
At the same time is it true that $(A\times B)\cup (C\times D)=(A\cup C)\times (B\cup D)$?
For me are both false but I don't know how to prove them.
thank you
|
For the first, let $A=\{1,2\}, B=\{a,b\}, X=\{(1,a),(2,b)\}$ For the second, note that the pairs on the left have a first element that might come from $B$ while the pairs on the right have a first element that might come from $C$. Maybe, as in my first example, they are different things.
|
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Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from here?
|
One has $$\int \frac{u+\frac{3}{2}}{u^2+\frac{59}{4}} du = \int \frac{u}{u^2+\frac{59}{4}}du + \int \frac{\frac{3}{2}}{u^2+\frac{59}{4}}du.$$ The first term can be computed be setting $v = u^2+\frac{59}{4}$ and the second thanks to the $\arctan$ function.
|
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|
Wronskian of two independent solutions equaling zero at a specific point only? Given $y_1(x)=\sin(x^2)$ and $y_2(x)=\cos(x^2)$, I constructed a linear, homogenic ODE of order 2 by solving:
$$
\begin{vmatrix}
y & y_1 & y_2 \\
y' & y_1' & y_2' \\
y'' & y_1'' & y_2'' \\
\end{vmatrix}=0
$$
Now, I noticed that the Wronskian of $y_1$ and $y_2$ at $x=0$ equals $0$. But the Wronskian of independent solutions is never $0$. And if it $0$ at one point, it is zero everywhere, which I don't see happening here, as
$$
W(y_1,y_2)=\begin{vmatrix}
y_1 & y_2 \\
y_1' & y_2' \\
\end{vmatrix}=-2x
$$
How does this make sense?
|
make sure that the coefficient of the highest order derivative doesn't vanish anywhere in the domain and the coefficient of the rest of the derivatives are continuous.specially check at x=0
|
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Limit of derivative does not exist, while limit of difference quotient is infinite Can anyone show an example of a function $f$ of a real variabile such that
*
*$f$ is differentiable on a neighborhood of a point $x_0 \in \mathbb{R}$, except at $x_0$ itself;
*$f$ is continuous at $x_0$;
*$\displaystyle \lim_{x \to x_0} f'(x)$ does not exist;
*$\displaystyle \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = +\infty$ or $\displaystyle \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = -\infty$ ?
|
What about
$$f(x)=\sqrt{x}+x\sin\frac{1}{x}$$
defined on $(0,+\infty)$, that can be defined by continuity at $0$ with $f(0)=0$? (I let you check that it works, if I am not wrong)
If you don't like it because it is only $\mathbb{R}_+$, you can replace with
$$f(x)=\sqrt[3]{x}+x\sin\frac{1}{x},$$
graph
that can be defined on $\mathbb{R}$.
Using google, you get a fairly obvious drawing explaining what is happening (personally before having an explicit example, I had a fairly good drawing in my head to convince myself that such an example was possible, and then help me to produce a formula)
|
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|
Three fair dice are rolled one time. What is the probability of at least one $6$? I think the answer is $10/216$, but I am not sure. I did it by brute force though, and would like to know the background, as well as knowing if $10/216$ is correct.
Edit: now i think via brute force, the answer is 93/216. if you roll a 1 on any dice there are 3x6 = 18 ways of getting a 6 using any number of dice. similarly for the other numbers.. giving 5 x 18 = 90. now add in the three 6's which adds three more ways... giving 93 total. is the answer 93/216 correct? and if so is there a more elegant was of doing it than manually writing down all the combinations? :)
|
Roll the three dice. Record the result as $(a,b,c)$ where $a$ is the number on the blue die, $b$ the number on the white, and $c$ the number on the red. All $6^3$ such sequences are equally likely.
There are $5^3$ sequences consisting of non-$6$, and therefore $216-125$ sequences with at least one $6$. Thus the probability of at least one $6$ is $\frac{216-125}{216}$.
Note that $216-125$ is $91$, not your $93$. You counted everything correctly, except there is only one sequence $(6,6,6)$, not three.
For this problem, I prefer to work directly with probabilities. The probability of a non-$6$ is $5/6$, so the probability of three of them in a row is $(5/6)^3$, so the probability of at least one $6$ is $1-(5/6)^3$.
|
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Is There a Problem with This Professor's Proof Concerning Interior and Boundary Points? Here is a professor's solution to the exercise which states, " Prove that if $x$ is an isolated point of a set $S \subseteq \mathbb{R}$, then $x$ is a boundary point of $S$."
The professor derived a contradiction upon assuming that for an arbitrary neighborhood $N$ centered at an isolated point $x$ of $S$, we have $N \subseteq S$, so he then proceeds in saying that every $N$ must have a noenmpty intersection with $S^c$. But I feel that there is something wrong with his argument: Using mathematical logic, we can represent his assumption that every $N$ is a subset of $S$ as $\forall \, N : p(N)$ where $p(N)$ means, "$N$ is contained in $S$." He then showed that this assumption is false, so $\sim \forall \, N : p(N) \equiv \exists \, N : \, \sim p(N)$ is true, which means there are some $N$'s that are not contained in $S$. But this information is not sufficient in concluding as to whether or not there is at least one $N$ contained in $S$; it just means not every $N$ is contained in $S$. But if I am right, how can he proceed with his proof since whatever result he deduces about $N$ after the contradiction does not necessarily apply to all $N$?
I hope what I am trying to convey is clear, and I thank you for your help
|
The quantifiers you have do not reflect the proof. Here, the professor wants to prove $$\forall N, \lnot(N\subseteq S)$$ and proceeds by contradiction. After taking an arbitrary $N$, (s)he shows that $N\subseteq S$ would lead to a contradiction. I.e., what is showed is $$\forall N, ((N\subseteq S)\Rightarrow \bot)$$ which is logically equivalent to
$$\forall N, (\bot \lor \lnot(N\subseteq S))$$
i.e.
$$\forall N, \lnot(N\subseteq S).$$
|
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Group orderable iff all its finitely-generated subgroups are orderable I want to proof this specifically using the Compactness Theorem from propositional logic (this is an exercise from Model Theory, Hodges).
$G$ orderable means there is a total ordering s. t. for all $g,h$, $g\leq h$ implies $gk\leq hk$ and $kg\leq kh$ for all $k$.
My idea is to take the following set of formulae:
$$\{p_{gh}\lor p_{hg}|g,h\in G\}\cup \{(p_{gh}\land p_{hi})\implies p_{gi}|g,h,i\in G\}\cup \{p_{gh}\implies\neg p_{hg}|g\neq h\in G\}\cup \{p_{gh}\implies(p_{ab}\land p_{cd})|a=kg, b=kh, c=gk, d=hk; g,h,k\in G\}$$
Models of this set should precisely be total orderings that meet the orderable requirement if I'm not mistaken? But I struggle to use compactness from here. All finitely-generated subgroups being orderable doesn't correspond to all subsets of above formulae-set having a model, or does it?
|
To apply compactness, you don't need all subsets of your set to have models; you just need all finite subsets to have models. A finite subset $S$ of your set involves only finitely many of the propositional variables $p_{gh}$ which in turn involve only finitely many elements of the group $G$. The subgroup $H$ of $G$ generated by these elements is finitely generated, so by assumption $H$ is orderable. Picking an order on $H$ and using it to determine the truth of $p_{gh}$ whenever $g,h\in H$ will then make all the formulas in $S$ true: that is, it will give a model of $S$. Thus every finite subset of your set has a model, so by compactness the whole set has a model.
|
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How to find integral of the form $e^xf(x)$? I always face trouble with these type of integrals.
I need to find $$\int{e^x \frac{x(\cos x -\sin x)-\sin x}{x^2}}dx$$
My problem would be solved if can express $f(x)$ like $g(x)+g'(x)$ but identifying $g(x)$ by trial and error method is sometimes tedious.Is there any easier approach?
|
As already said in comments, I do not think that there is a solution in terms of elementary functions.
However, the problem can be approached using $$\cos(x)=\frac{1}{2} \left(e^{i x}+e^{-i x}\right)\qquad \sin(x)=-\frac{1}{2} i \left(e^{i x}-e^{-i x}\right)$$ So, the numerator of the integrand becomes$$\left(\frac{1}{2}-\frac{i}{2}\right) e^{(1-i) x}
x+\left(\frac{1}{2}+\frac{i}{2}\right) e^{(1+i) x} x-\frac{1}{2} i e^{(1-i)
x}+\frac{1}{2} i e^{(1+i) x}$$ and we are then left with integrals looking like$$I=\int \frac{e^{\alpha x}} x \, dx\qquad J=\int \frac{e^{\alpha x}}{x^2} \, dx$$ Changing variable $\alpha x=y$ then makes $$I=\int \frac{e^y}{y}\,dy=\text{Ei}(y)$$ $$J= \alpha\int\frac{ e^y}{y^2}\,dy=\alpha \left(\text{Ei}(y)-\frac{e^y}{y}\right)$$ where appears the exponential integral function (for the computation of $J$, one integration by parts is needed).
Using these last results and back to $x$ $$\int{e^x \frac{x(\cos x -\sin x)-\sin x}{x^2}}\, dx=i \Big(\text{Ei}((1+i) x)- \text{Ei}((1-i) x)\Big)+\frac{e^x \sin (x)}{x}$$
Edit
Looking at $$\int e^x\frac{ x (\cos (x)\pm \sin (x))\pm \sin (x)}{x^2}\,dx$$ the only combination which leads to elementary functions is the combination $(+,-)$ and then $$\int e^x\frac{ x (\cos (x)+ \sin (x))- \sin (x)}{x^2}\,dx=\frac{ \sin (x)}{x}e^x$$ So, one more typo in a textbook.
|
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|
How to define gradient of an affine connection I heard somewhere (and just read on a physics forum) that the gradient of a smooth function $f$ on a manifold $M$ can be defined when $M$ is equipped with an affine connection on its tangent bundle, i.e. the Riemannian metric is not strictly necessary.
Is this true? If so, how can this be done?
|
No, or at least not if you want to coincide with the usual gradient in the case where the connection is derived from a metric. To show this, just note that whenever $\nabla$ is the metric connection of $g$, it is also that of $2g$ (since $\nabla(2g) = 2\nabla g = 0$); but the latter metric will produce gradients with half the magnitude of the former.
It seems to me that any sensible definition of the gradient of a function will involve a linear 1-1 relationship to the differential; i.e. will have the structure of a vector bundle isomorphism $$\xi: TM \to {TM}^*.$$ This is almost exactly the same data as a metric: if we impose some positivity conditions (which boil down to "if you move for some short time in the direction $\nabla f=\xi^{-1}(df)$ then $f$ increases unless $df=0$") then given either of the two we can get to the other using the equation $g(u,v) = (\xi(u))(v)$.
|
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|
How limiting/ heavy is the "triangle inequality" assumption? Suppose a theorem proves something about a family of distance measures, with this the triangle inequality assumption. How limiting this assumption is in reality?
What are some real-world examples of distance measures which don't obey the triangle inequality?
Edit:
The triangle inequality is the following requirement upon the distance function on $M$:
$$
\operatorname{dist}(x,z) \leq \operatorname{dist}(x,y) + \operatorname{dist}(y,z), \qquad \forall x,y,z \in M.
$$
|
That pretty much depends on your definition of distance "in the real world". In my opinion, the triangle inequality is extremely important for the definition of any notion of distance.
The triangle inequality tells you that going from $A$ to $B$ and taking a detour over $C$ will at most increase the distance that needs to be covered to get from $A$ to $B$.
Assuming that the triangle inequality does not hold would mean that the distance from $A$ to $C$ plus the distance from $C$ to $B$ is less than the distance from $A$ to $B$. But then nothing would stop you from declaring that smaller distance as the "real" distance from $A$ to $B$...
One thing to maybe keep in mind is that the mathematical notion of distance refers to the minimal effort one needs to make to go from one point to another.
|
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|
Solve equation of inverse functions I have two different functions $y_1=f_1(x)$ and $y_2=f_2(x)$, both invertible but quite complex. I am able to find their inverse functions numerically, i.e. $f^{-1}_1(x)$ and $f^{-1}_2(x)$, by solving for $x_1$ and $x_2$ two equations
$$
f_1(y_1) - x_1 = 0, \\
f_2(y_2) - x_2 = 0,
$$
for specific realizations of $y_1$ and $y_2$. Since $f_1 \neq f_2$, I get two different solutions for both equations $x_1 \neq x_2$. The problem is that I would like to solve equation on those inverse functions. The goal is to find a single point $x_0$ for
$$
f^{-1}_1(x_0) - f^{-1}_2(x_0) =0
$$
Is this possible? If I plug in the above inverse equations, assuming that $x_1=x_2$, into the main equation I get $f_1(y_1)=f_2(y_2)$ which is a contradiction. Is there a method to solve this problem, even numerically? If not possible, maybe finding $\min |f^{-1}_1(x) - f^{-1}_2(x)|$ would be an option?
|
If I well understand (which is not sure), you have two given functions $f_1(x)$ and $f_2(x)$
The inverse function of $y=f_1(x)$ is $x=f_1^{-1}(y)$
The inverse function of $y=f_2(x)$ is $x=f_2^{-1}(y)$
You want to find a root $y_0$ of the equation $f_1^{-1}(y)=f_2^{-1}(y)$
So, you first have to
$$\text{solve}\quad f_1(x)=f_2(x)\quad \text{for } x\quad \text{which leads to}\quad x_0$$
Then, knowning $x_0$, compute :
$$y_0=f_1(x_0)\quad \text{or}\quad =f_2(x_0)$$
$y_0$ is a root of $f_1^{-1}(y)=f_2^{-1}(y)$
Note that, in order to avoid confusion, I changed the symbol $x_0$ in the wording of the question to the symbol $y_0$ but anyways this doesn't matter : any symbol used doesn't change the solution.
For intuitive understanding only, draw $y=f_1(x)$ and $y=f_2(x)$ on a graph. On the same graph draw the symetric curves relatively to the line $y=x$. Those two last curves represent the inverse functions. The intersection represents the root of the inverse equation. This point is the symetrical point representing the root of the direct equation. That is why we solve first the direct equation $f_1(x)=f_2(x)$ : The intersection point is $(x_0\:,\:y_0)$ while it is $(y_0\:,\:x_0)$ on the inverse curves.
|
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|
Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.
|
Yes, for example $\tan(\frac{\pi}{4})=1$
Write $\tan$ as an infinite Taylor expansion in powers of $\pi$.
Other trigonometric functions of irrational arguments can also have rational outputs, eg. $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
|
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|
Integrating triangle in a 2D plane I am interested in integrating $(x^2y+y^2x)$ on the following loop: $(x=1,y=2)\rightarrow(x=2,y=1)\rightarrow(x=3,y=3)\rightarrow(x=1,y=2)$.
I know this loop forms a triangle with all three sides changing with respect to the $x$ and $y$ axis. I understand how to do the integration part, but unfortunately I am really stuck on how to set up the integration itself. Should I do a double integral for each side of the triangle and then add them?
If anyone can help me set up the integration for this problem it would be greatly appreciated. Thank you!
|
You can always parameterize a triangle with $u,v$, where $u,v>0$ and $u+v<1$. This always leads to an integral
$$\int_0^1 \int_0^u\ldots dv\,du$$
Convex interpolation inside the triangle with vertices A,B,C is then
$$T=Au+Bv+C(1-u-v)$$
Imagine that "u" is the variable that tells you how much you move from C to A, and "v" the same from C to B. (u=0,v=0) is now C, (u=1,v=0) is now A, and (u=0,v=1) is B. That defines your variable change for both coordinates.
For illustration, see sketches for Barycentric coordinates (we are integrating over two of barycentric coordinates, and the third one is expressed with the other two $1-u-v$).
From this you must also compute the Jacobian (which is the ratio how much your original triangle is bigger than the unit triangle (0,0)→(1,0)→(0,1) in (u,v) space where you are doing the integral).
In your case:
$$(x,y)=(u+2v+3(1-u-v),2u+v+3(1-u-v))=(3-2u-v,3-u-2v)$$
and the jacobian:
$$J=\det\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} =(-2)(-2)-(-1)(-1)=3$$
This leaves you with
$$\int_0^1 \int_0^u \left((3-2u-v)^2(3-u-2v)+(3-u-2v)^2(3-2u-v)\right) 3 dv\,du$$
|
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Number of Labels used in reduction of Isomorphism of Labelled Graph to Graph Isomorphism From "Lecture Notes in Computer Science" by Christoph M. Hoffmann ,
Assume that both $X$ and $X'$ have $n$ vertices. We plan to code the
graph labels as suitable subgraphs which we attach to the vertices of
$X$ and of $X'$. In time polynomial in the length of the input we can rename
the labels and may assume, therefore, that $L = \{1 ..... k\}$ is
the set of labels assigned by $\lambda$ and $\mu$. Note that $k\leq2n$ .We describe
how to construct $Z$ from $(X,\lambda)$. The construction of
$Z'$ is done in the same way. Let $X = (V,E), V = \{v_1..... v_n\}$.
Intuitively, we obtain $Z$ from $X$ by attaching to each vertex $v_i$ a
complete graph with $n + r$ vertices, where $\lambda (v_i) = r$. Note that $r\leq2n$.
The vertices of the attached complete graph will be, for $v_i$, $\{v_{(i,1)},..... v_{(i,n+r)} \}$
The subgraph is atached to $v_i$ by an edge $(v_i, v_{i,1})$.
It is easy to see that $(X,\lambda)$ is
isomorphic to $(X,\mu)$ iff $Z$ is isomorphic to $Z'$. Since there are at most
$2n$ distinct labels, the graphs $Z$ and $Z'$ have no more than
$2n^2+n$ vortices each and can thus be constructed in
polynomial time.
the definition of graph label is given as-
Let $X = (V,E)$ be a graph, $\lambda$ is a mapping from $V$ onto a set $L = \{ l_1,...l_k\}$.
Then the pair $(X,\lambda)$ is a labelled graph.
Question 1: Why $k\leq 2n$ which implies $r\leq2n$? It seems that it should be $k\leq n$ since the passage of wikipedia here tells
When used without qualification, the term labeled graph generally
refers to a vertex-labeled graph with all labels distinct. Such a
graph may equivalently be labeled by the consecutive integers $\{1, …,
> |E |\}$, where $|E |$ is the number of vertices in the graph.
Question 2: What is the explanation of -
Since there are at most
$2n$ distinct labels, the graphs $Z$ and $Z'$ have no more than
$2n^2+n$ vortices each and can thus be constructed in
Thanks in advance.
|
My interpretation:
The Confusion arises, here-
Assume that both $X$ and $X'$ have $n$ vertices. We plan to code the
graph labels as suitable subgraphs which we attach to the vertices of
$X$ and of $X'$. In time polynomial in the length of the input we can rename
the labels and may assume, therefore, that $L = \{1 ..... k\}$ is
the set of labels assigned by $\lambda$ and $\mu$. Note that $k\leq2n$ .
One needs to note that-
we can rename the labels and may assume, therefore, that $L = \{1 ..... k\}$ is the set of labels
assigned by $\lambda$ and $\mu$.
So, if you have $k\leq n$, then there is a guarantee or certainty,that there exists an Isomorphism from $X$ to $X'$ already, since, if $k\leq n$ then it is certain that $\exists i,j$ such that $ \lambda(v_i)=r=\mu(v_j)$ where $ 1\leq i,j \leq n$.
But if $k\leq 2n$, then $\lambda(v_i)$ may take $n$ different label, and $\mu(v_j)$ may take $n$ different label.
For example, consider Cyclic Graph $C_4$ and Complete Graph $K_4$. Add 1 vertex to 1st,2 vertices to 2nd,3 vertices to 3rd, 4 vertices to 4th vertex of $C_4$ and $K_4$ and rename them as $C'_4$ and $K'_4$. Now you have to use 14 different labels in $C'_4$, and none of them can be used in $K'_4$. Note that the number of vertices is same in both $C'_4$ and $K'_4$. Here, $|L|=2n=28$;
So, $k\leq2n$.
|
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A sequence of continuous functions which is pointwise convergent to zero and not uniformly convergent on any interval. The exercise is to construct a sequence of continuous functions $f_n:\mathbb{R}\rightarrow \mathbb{R}, n\in \mathbb{N}$ , which is pointwise convergent to $f(x)=0 , x\in \mathbb{R}$ and not uniformly convergent on any interval $(a,b)$.
I posted this question once and I got the answer, but I had forgotten to mention that $f_n$ must be continuous. The example of sequence of discontinuous functions looks like this:
Let $\{r_n\}_{n\geq 1}$ be an enumeration of the rationals, and define $f_n(x)$ as follows:
$$
f_n(x)=\begin{cases}
1,& x\in \{r_k\}_{k\geq n}\\
0,& else
\end{cases}
$$
The person, who posted this, made the remark:
If you want to make the $\{f_n\}$ continuous, you can modify this example by using bump functions supported on $(r_n-2^{-n},r_n+2^{-n})$ in place of the spikes to $1$ at $r_n$.
I'm not realy sure what he meant. I was thinking of defining $g_k$ as a broken line or maybe more smooth function going through $(r_k-2^{-k},0), (r_k,1), (r_k+2^{-k},0)$ and zero on $\mathbb{R}\setminus [r_k-2^{-k},r_k+2^{-k}]$ and then defining $f_n(x)=\sup\{g_k(x):k\ge n\}$. But I don't know whether it converges to zero and whether $f_n$ are even continuous.
On the other hand I have found this topic
A sequence of continuous functions on $[0,1]$ which converge pointwise a.e. but does not converge uniformly on any interval
and there is only an example of a sequence converging to zero almost everywhere. That's why I begin to doubt that there exists a sequence converging to zero everywhere (that would mean there is a mistake in my book because the exercise says "construct such a sequence").
|
According to the hint in "Problems in Mathematical Analysis", Biler and Witkowski, problem 4.112.
Define $g_n$ as the function which is $0$ on the intervals $(-\infty, 0]$ and $[2/n,\infty)$, $1$ at $1/n$, and linear otherwise. Then $g_n$ converges pointwise to $0$ but is not uniform in any interval containing $0$.
Now define $$f_n(x) = \sum_{k=0}^{\infty} 2^{-k} g_n(x - r_k).$$
Then the $f_n$ are continuous, converge pointwise to $0$, but do not converge uniformly on any open interval.
|
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support of an operator on a Hilbert space Let $x\colon\mathcal{H}\to\mathcal{H}$ be a self-adjoint operator,
the support $s(x)$ of $x$ is defined as the smallest projection $e\in B(\mathcal{H})$ such that $ex=xe=x$.
Let $x=\int\lambda \, dE_x$ be the spectral decomposition.
Then $s(x)=E_x(\mathbb{R}\setminus\{0\})$ und $E_x(\{0\})=\ker x$.
How can these properties be proven, or where can I find such elementary things with proofs?
|
Because $x$ commutes with $e$, then $e$ commutes with $E$ as well, including with $f=E_x(\mathbb{R}\setminus\{0\})$. Because $fx=xf=x$ and $fe=ef$, then $ef$ is also a projection such that $(ef)x=x(ef)=x$. Therefore $ef=fe=e$ follows from the minimality of $e$. Furthermore $x(f-e)=0$, which puts the range of $f-e$ in $\mathcal{N}(x)=\mathcal{R}(E_x\{0\})$; hence $0=f(f-e)=f^2-fe=f-e$.
|
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|
Vector norm and relationship with euclidean distance If $y\in E_n$ (n dimensional euclidean space) show that $||\textbf{y}||\leq|\textbf{y}|\leq \sqrt{n}||\textbf{y}||$
Where $||\textbf{y}||$ is the euclidean length of the vector $\textbf{y}$
and $|\textbf{y}|$ is the norm of $\textbf{y}$
The hint: Use the inequality $2|uv|\leq |u|^2+|v|^2$
to show
$||\textbf{y}||^2\leq|\textbf{y}|^2\leq n||\textbf{y}||^2$
By letting $u=|y|$ and $v=y$ I get
$2||y|y|\leq||y||^2+|y|^2$ I don't really see how this helps.
|
Hint: for any nonnegative number $a$ and $b$ you have
$$
(a+b)^2=a^2+b^2+2ab\ge a^2+b^2
$$
and
$$
2(a^2+b^2)-(a+b)^2=a^2+b^2-2ab=(a-b)^2\ge0.
$$
Hence,
$$
a+b\ge\sqrt{a^2+b^2}\quad\text{and}\quad \sqrt2\sqrt{a^2+b^2}\ge a+b.
$$
|
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|
$\mathbb C$-dimension of vector space $\mathbb C\otimes_{\mathbb R}\mathbb C$
Let $\mathbb R$ be the field of real numbers, $\mathbb C$ be the field of complex numbers. Consider $\mathbb C\otimes_{\mathbb R}\mathbb C$ as a $\mathbb C$-vector space via $a(b\otimes c) :=
ab \otimes c,$ for $a, b, c \in \mathbb C$. Compute the $\mathbb C$-dimension of this vector space.
I guess the dimension is 1 and a basis is $\{1\otimes 1\}$. Is it correct? Thank you in advance!
|
Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on.
1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$.
2) If $\{b_1,\dotsc,b_n\}$ is a $K$-basis of $V$, then $\{1 \otimes b_1,\dotsc,1 \otimes b_n\}$ is an $L$-basis of $L \otimes_K V$.
3) We have $\dim_L(L \otimes_K V) = \dim_K(V)$.
In 1) you have to check that this scalar multiplication exists at all. Use the universal property of the tensor product ( = definition of the tensor product) for this.
For the proof of 2) there are several methods. I suggest that you prove more generally that the tensor product commutes with direct sums in each variable (since this is useful anyway); then the claim follows immediately.
Of course, 3) follows from 2).
In particular, we see that $\{1 \otimes 1, 1 \otimes i\}$ is a $\mathbb{C}$-basis of $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$.
|
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Closed form of $\sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$
Let
$$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$$
1. Find the radius of convergence.
2. Calculate $S(x)$.
3. Find $S^{(n)}(x)$ without computing the derivatives of $S(x)$.
From the root test I find $R = 1/4$. It's the second point that troubles me. This is my attempt:
$$\begin{align}
S(x) &= \sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n =\\
&= \sum\limits_{n=1}^\infty 2^{2n+1}\int_{-4}^x (t + 4)^{2n-1}\mathrm dt =\\
&= \int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n+1}(t + 4)^{2n-1}\mathrm dt =\\
&= 4\int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n-1}(t + 4)^{2n-1}\mathrm dt =\\
&=\ ???
\end{align}$$
I don't know how to continue from there. I know that I should transform the inner sum into a known Taylor expansion or a geometric series, but I don't see how I could do that.
As for the last point, we have that
$$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n = \sum_{n = 0}^\infty \frac{S^{(n)}(x)}{n!}(x + 4)^n,$$
as per the Taylor series definition. However, I don't know how to reconcile the indices and the two powers $2n$ and $n$.
|
Let $t=4(x+4)^2$ and the series becomes
$$\sum_{k=1}^\infty\frac{t^n}n.$$
You can recognize the Taylor development of $-\ln(1-t)$, or derive the series to get
$$\sum_{k=1}^\infty t^{n-1}=\frac1{1-t}.$$
|
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Ms. A selects a number $X$ randomly from the uniform distribution on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers. Ms. A selects a number $X$ randomly from the uniform distribution
on $[0, 1]$. Then Mr. B repeatedly, and independently, draws numbers
$Y_1, Y_2, ...$ from the uniform distribution on $[0, 1]$, until he gets a number
larger than $\frac{X}{2}$, then stops. The expected number of draws that Mr. B
makes equals?
The answer to this is $2\ln(2)$. Could someone post the explanation? I got the result as $1.333...$ which is a bit off.
Edit1: I assumed that since it is a uniform distribution, the expected value of the number drawn by A will be o.5. The probability of Mr. B drawing a number bigger than 0.5/2 is then 0.75, which gives expectation of number of draws by Mr. B to be 1.333..
|
Given $X=x$, B draws a large enough number if they draw something between $x/2$ and $1$, which occurs with probability $1-x/2$. Using the expectation of the geometric distribution, you find that then the expected number of trials is $\frac{1}{1-x/2}$.
Now we need to deal with the fact that $X$ is itself random. The density of $X$ is $1$ on $[0,1]$ and zero elsewhere. Using the total expectation formula, the expected number of trials is
$$\int_0^1 \frac{1}{1-x/2} dx.$$
I think you can probably work from here.
|
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Solving $yy''=(y')^2-y'$ I want to solve the ODE $yy''=(y')^2-y'$ with the initial conditions $y(0)=1, y'(0)=2$.
My attempt:
$$yy''=(y')^2-y'$$
$$(\frac {y'}y)'=(\frac 1y)'$$
$$\frac {y'}y=\frac 1y+c$$
This holds for all $x$. Plugging the initial conditions for $x=0$, we get $c=1$.
$$y'=1+y$$
Solving this got me to $y=-1$. It seems like a singular solution, but I used all methods I know and didn't get anything else.
Your ideas, please?
edit: I'm sorry about the silly question. I put my effort on getting through the hard part and lost my ability to perform simple calculations. Thank you for your answers.
|
The full solution with the homogeneous part is
$$
y=-1+C·e^x
$$
and from the initial conditions $C=2$.
|
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Find the fixed field of the following subgroup? I am trying to understand some concepts via random exercises I found from past papers but this particular one, I am not sure even where to start.
There aren't any solutions for the paper so would someone be kind enough to provide me with a standard solution, so I can study the flow and try figuring out the ideas myself? I might comment and ask if simply I cannot digest a part of it
Let $\omega = e^{\frac{2 \pi i}{5}}$. So then the $\mathbb{Q}(\omega)$ consists of all elements of the form $p+q \omega+r\omega^2+s\omega^3+t\omega^4 $. Let the $\mathbb{Q}$-automorphisms be
$$\alpha_1: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+q \omega+r\omega^2+s\omega^3+t\omega^4 $$
$$\alpha_2: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+s \omega+q\omega^2+t\omega^3+r\omega^4 $$
$$\alpha_3: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+r \omega+t\omega^2+q\omega^3+s\omega^4 $$
$$\alpha_4: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+t \omega+s\omega^2+r\omega^3+q\omega^4 $$
Take the subgroup $\{\alpha_1, \alpha_4\}$ and find its fixed field.
So I am starting with the Galois group $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$, and considering a particular subgroup. I don't have a method to proceed, so it would be very much appreciated if I can find a way to answer it, thank you in advance!
|
Here’s the method I use for this very special case, and other similar ones. You have $\omega^4+\omega^3+\omega^2+\omega+1=0$, and you can rewrite this as
$$
0=\omega^2+\omega+1+\omega^{-1}+\omega^{-2}\,.
$$
Now, your nontrivial automorphism interchanges $\omega$ and $\omega^{-1}$, so has to have $\Bbb Q(\omega+\omega^{-1},\omega\omega^{-1})$ as its fixed field. Of course the second quantity is automatically fixed, so let’s see what kind of equation $\xi=\omega+\omega^{-1}$ might satisfy.
\begin{align}
\xi^2&=\omega^2+2+\omega^{-2}\\
&=-\omega+1-\omega^{-1}\qquad\text{(subtracting zero)}\\
&=-\xi+1\,,
\end{align}
which tells you your equation.
|
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|
Probability: Finding the Number of Pears Given Two Scenarios
You have a bag containing 20 apples, 10 oranges, and an unknown number
of pears. If the probability that you select 2 apples and 2 oranges is
equal to the probability that you select 1 apple, 1 orange, and 2
pears, then what is the number of pears originally in the bag?
Using $n$ as the number of pears, I found the probability of selecting 2 apples and 2 oranges to be: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}$$
seeing as the number of ways to choose two apples and 2 oranges (order shouldn't matter) would be given by $\dbinom{20}{2}*\dbinom{10}{2}$. I put this over the total number of possibilities, which was found by choosing $4$ fruits from a total of $30+n$ fruits.
As the problem stated, this equaled the probability of choosing 1 apple, 1 orange, and 2 pears, which would be:
$$\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
And thus: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}=\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
$$\Rightarrow\dbinom{20}{2}*\dbinom{10}{2}=\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}$$
$$\Rightarrow\dbinom{n}{2}=42.75$$
which doesn't seem to be correct. Can anyone tell me where I went wrong? I believe it is most likely a conceptual mistake; is this not how you calculate the respective probabilities?
|
It is good that you were alert. I believe everything you are doing is correct. If we continue, then
$$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$
I think it is ok to have a decimal number here.
This gives
\begin{align*}
\frac{n!}{2!(n-2)!} &= \frac{171}{4}\\
\implies \frac{n!}{(n-2)!} &= \frac{171}{2}\\
n(n-1) &= \frac{171}{2}\\
n^2-n+\frac{1}{4} &= \frac{171}{2}+\frac{1}{4}\\
\left(n-\frac{1}{2}\right)^2 &= \frac{343}{4}\\
\implies n &= \sqrt{\frac{343}{4}}+\frac{1}{2}\\
n&=9.76013
\end{align*}
which I feel is an appropriate approach. The numbers are usually chosen so that when you complete the square, you will end up with a whole number $n$, but I think there was an error.
|
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|
How do I prove that if $2\nmid n$ then $2|(n+1)$? I'd like to prove a very simple fact, but it's stumping me: namely, that if $2 \nmid n$ then $2\mid(n+1)$.
How would this usually be done?
|
If $n \not\equiv 0 \pmod2$, then $n+1 \not\equiv 1 \pmod2$, so we must have $n+1 \equiv 0 \pmod2$
|
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|
Prove that for any integer $n$, if $b^2$ divides $n$, then $b$ divides $n$.
Prove that for any integer $n$, if $b^2$ divides $n$, then $b$ divides $n$.
Trying to figure out this proof. The proof I'm looking at is written as
$n$ = any integer, if $25|n \implies 5|n$.
I've been trying to figure this for days and have been running around in circles. Would appreciate a general proof for this.
$n$ = any integer, if $(b^2)|n \implies b|n$.
|
Since an answer has been given ...
The $\mid$ relation is transitive; if $x \mid y$ and $y\mid z$, then $x\mid z$. Let $x=b$, $y=b^2$, and $z=n$. Done.
|
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|
Pairwise independence vs independence
Two fair dice are thrown. We have three events:
*
*A: The first die shows an odd number
*B: The second die shows an even number
*C: Both are odd or both are ven
Show that $A,B,C$ are piecewise independent but not independent.
My answer:
$P(A) = P(B) = P(C) = \frac{1}{2}$.
$P( A \cap B) = P( A \cap C) = P( B \cap C) = \frac{1}{4}.$
This means that all the events are pairwise independent. However:
$P(A \cap B \cap C) = 0$ while $P(A)P(B)P(C) = \frac{1}{8}$, so the events are not independent.
Is this correct (disregarding that I didn't explain how I got those probabilities)?
|
Yes, this is correct. The definitions are as follow (see here).
A finite set of events $\{A_i\}$ is pairwise independent if and only if every pair of events is independent — that is, if and only if for all distinct pairs of indices $m, k$,
$$
\mathrm{P}(A_m \cap A_k) = \mathrm{P}(A_m)\mathrm{P}(A_k).
$$
A finite set of events is mutually independent if and only if every event is independent of any intersection of the other events — that is, if and only if for every $n$-element subset ${A_i}$,
$$
\mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i).
$$
You have shown that the events $A,B$ and $C$ are pairwise independent,but not mutually independent.
|
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|
How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$ To prove: $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
My attempt at the solution:
\begin{gather}\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(x)\cos(2x)\sin(3x)-\cos(2x)\cos(3x)\sin(x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{ \cos(2x)[ \cos(x)\sin(3x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(2x)[\sin(4x)\sin(2x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(2x)[2\sin(4x)\sin(2x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(4x)\cos(2x)}{2\sin(3x)\sin(x)}\end{gather}
The problem is, I don't know where to go from here (and due to so many calculations involved, I'm also not sure of the above result).
Also, if you see a more elegant way to solve this, please provide a hint (not the complete solution).
|
Go slowly:
\begin{align}
\cot x\cot2x-\cot2x\cot3x
&=
\cot2x\left(\frac{\cos x}{\sin x}-\frac{\cos3x}{\sin3x}\right)\\[6px]
&=\frac{\cos2x}{\sin2x}\frac{\sin3x\cos x-\cos3x\sin x}{\sin x\sin 3x}
\\[6px]
&=\frac{\cos2x}{\sin2x}\frac{\sin2x}{\sin x\sin 3x}\\[6px]
&=\frac{\cos2x}{\sin x\sin 3x}
\end{align}
So you want to compute
$$
\frac{\cos2x}{\sin x\sin 3x}-\frac{\cos3x}{\sin3x}\frac{\cos x}{\sin x}
=\frac{\cos2x-\cos3x\cos x}{\sin x\sin3x}
$$
Now
$$
\cos2x=\cos(3x-x)=\cos3x\cos x+\sin3x\sin x
$$
and you are done.
|
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|
'Obtain' the Fourier transform If $g(t) = e^{-a|t|}$ and a is a real positive constant, obtain the fourier transform.
I'm a bit unsure what this is asking. I can write out the expression for the fourier transform. Should I stop there? Or do I do the integration as well? If I do the integration, how do I deal with the absolute value symbol? And the fact that the limits are positive and negative infinity?
|
We note that $\displaystyle \mathcal{F}\left(f(t)\right)(x)=\int_{-\infty}^{+\infty}f(t)e^{-2i\pi x t}\text{ d}t.$
\begin{eqnarray*}
\mathcal{F}\left(e^{-a |t|}\right)(x) &= & \int_{-\infty}^0 e^{ at}e^{-2i\pi x t} dt+\int_{0}^{+\infty} e^{-at}e^{-2i\pi x t} dt \\
&= &\left[\frac{e^{ at-2i\pi x t}}{ a-2i\pi x}\right]_{-\infty}^0+\left[\frac{e^{-at-2i\pi x t}}{- a-2i\pi x}\right]_{0}^{+\infty}\\
&= &\frac{1}{ a-2i\pi x}- \frac{1}{- a-2i\pi x}\\
&= & \frac{2 a}{a^2+(2\pi x)^2}\\
&= &\frac{2 a}{ a^2+ 4 \pi^2 x^2}
\end{eqnarray*}
If we assume the $\mathcal{F}$ourier transform can be writen as:
$$\displaystyle \mathcal{F}\left(f(t)\right)(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}f(t)e^{-i x t}\text{ d}t.$$
We can find $$
\mathcal{F}\left(e^{-a |t|}\right)(x) =\frac{1}{\pi}\frac{ a}{ a^2+ x^2}
\qquad \forall a \in \mathbb{R}_+^*. $$
|
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|
Prime counting function; when is it true that $\pi(n) > \pi(2n) -\pi(n)$? Let $\pi$ be the prime counting function.
Under what conditions is it proven true that $\pi(n) > \pi(2n) -\pi(n)$, if at all?
|
I've written the program Akiva Weinberger suggested above. This is just a straightforward interpretation of the sieve of Eratosthenes, in R.
n = 30092
top = 2*n
isPrime = rep(TRUE, top)
isPrime[1] = FALSE
nextprime = 2
while (nextprime < sqrt(top)){
isPrime[seq(2*nextprime, floor(top/nextprime)*nextprime, nextprime)] = FALSE
nextprime = min(which(isPrime[(nextprime+1):top])) + nextprime
}
#isPrime[n] is now TRUE if n is prime and FALSE otherwise
primePi = cumsum(isPrime) #prime counting function, denoted as pi above
f = primePi[seq(2, 2*n, 2)] - 2*primePi[1:n]
which(f>0)
The output is the list [1]. That is, $\pi(2k) > 2\pi(k)$ for $k = 1$ and no other $k <= 30092$. As Barry Cipra showed above, we can prove the desired inequality for larger values from of $k$ from known bounds.
If we want to consider the possibility that $\pi(2k) = 2\pi(k)$, we can replace the last line with which(f>=0). The output here is [1, 2, 4, 10]. And in fact we have $2 = \pi(4) = 2 \pi(2), 4 = \pi(8) = 2 \pi(4), 8 = \pi(20) = 2\pi(10)$.
|
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|
Classifying groups of order $6$ using semidirect products Let G be a group of order 6. I am able to do the exercise without semidirect products($G \cong Z_6 $ or $S_3$) but I don't know how to use semidirect products to do this.
By Sylow's theorem, there is only subgroup of order 3, say P. As it is unique, P is normal in G. Also, let Q be a 2-Sylow subgroup. Note that $P\cap Q=\{e\}$. That means G is a semidirect product $P \rtimes_{\phi} Q$ for $\phi:Q \to Aut(P)$. So there are two possible automorphisms on P, names $\alpha$ which is the identity on $P$ and $\beta$ which takes $x\in P$ to $x^{-1}$. That gives us two homomorphisms $\phi:Q \to Aut(P)$. How do I proceed after this? (I suspect I don't understand semidirect products well enough and I really don't know how to explicitly show that one gets two nonisomorphic groups)
|
By Cauchy's theorem, there must be an element $s$ of order $2$ in $G$, and an element $r$ of order $3$, because $2$ and $3$ divide $6$. Clearly $r$ generates a cyclic group $C_3$ which has index $2$ in $G$, hence is a normal subgroup. Then $G$ is a semidirect product of $C_2$ and $C_3$. In case it is abelian, this product is direct. One can say this more precisely. We have
$$
G=\{1,r,r^2,s,sr,sr^2\}
$$
and since $C_3$ is a normal subgroup, we must have $srs^{-1}=r^k$ for some $k$. So we have
$$
r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
$$
This means $k^2\equiv 1 \bmod 3$, which has exactly $2$ solutions, namely $k\equiv \pm 1 \bmod 3$. In the first case, $G$ is commutative and we have $G=C_2\times C_3=C_6$, and in the second case we have $G=D_3=S_3$.
|
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|
The line $2x-y=5$ turns about a point.... The line $2x-y=5$ turns about a point on it, whose ordinate and abscissae are equal, through an angle of $45°$, in anti clockwise direction. Find the equation of line in the new position.
My attempt to solve:
Let AB be the line with the equation $2x-y=5$ and $B(a,a)$ be a point on it whose ordinate and abscissae are equal. Then
$$2x-y=5$$
$$2a-a=5$$
$$a=5$$
Hence, $B(a,a)=B(5,5)$.
My solution ends up here. What should I do next?
|
The line turns by an angle of $45°$ anti-clockwise about the point $(5,5)$. Thus, you need to find the equation of a line whose angle with the $x-axis$ is $45°$ more than this line and which passes through the point $(5,5)$.
The line $2x-y=5$ has a slope of $2$, i.e., it makes an angle of $\arctan2$ with the $x-axis$.
Slope of the required line=$\tan(\frac\pi4+\arctan2)=\frac{1+2}{1-2}=-3$
Its equation, therefore, is
$$y-5=-3(x-5)$$
or
$$3x+y=20$$
|
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|
The complex version of the chain rule I want to prove the following equality:
\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = (\frac{\partial g}{\partial z} \frac{\partial f}{\partial z}) + (\frac{\partial g}{\partial \bar{z}} \frac{\partial \bar{f}}{\partial z})
\end{eqnarray}
So I decide to do the following:
\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = \frac{1}{2}[(\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x}) + \frac{1}{i}(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y})]
\end{eqnarray}
but the thing is that I am doing something wrong here since I don't get any conjugate function and any derivative with respect to $\bar{z}$ so Can someone help me to see where I am wrong and fix it please?
In fact I don't see what to do next, so I appreciate your help.
Thanks a lot in advance.
Edition:
What I've got so far is the following:
$$\frac{1}{2}[(\frac{\partial g}{\partial x} \circ f + \frac{\partial g}{\partial y} \circ f)\frac{\partial f}{\partial z} ]$$
but I'm still stuck.
|
The question is taken in the context of Wirtinger Derivatives.
To that end, we let $g$ and $f$ be functions of both $z$ and $\bar z$. Then, the composite function $g\circ f$ can be expressed as
$$g\circ f=g(f(z,\bar z),\bar f(z,\bar z))$$
The partial derivative of $g\circ f$ with respect to $z$ is then given by
$$\begin{align}
\frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\
&=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\
&=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z}
\end{align}$$
|
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|
Integrate $\int_{-\infty}^\infty xe^{-\alpha x^2+\beta x}dx$ I am familiar with the gauusian integral
$$\int_{-\infty}^\infty e^{-\alpha x^2+\beta x}dx=\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/(4\alpha)}$$
Could anyone help me to find out the value of the following?
$$\int_{-\infty}^\infty xe^{-\alpha x^2+\beta x}dx$$
Integration by parts turns out to contain error function which I am not familiar with.
|
$$\int_{-\infty}^{\infty}xe^{-\alpha x^2+\beta x}\ dx=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta-\beta)e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}\ dx$$
$$=-\frac{1}{2\alpha}\int_{-\infty}^{\infty}(-2\alpha x+\beta)e^{-\alpha x^2+\beta x}\ dx+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
since first integral diminishes,
$$=0+\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
$$=\frac{\beta}{2\alpha}\sqrt{\frac{\pi}{\alpha}}e^{\beta^2/4\alpha}$$
|
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|
Multiplication of transpositions? I can't seem to understand how the multiplication of two transpositions yield the results below:
$(x b)(x a) = (x a)(a b) \\
(c a)(x a) = (x c)(c a)$
I can't figure it out for the life of me. I'm trying to learn permutation multiplication on my own, and I feel confident in my abilities, but here I feel as though I'm missing something essential. Could someone please enlighten me as to the process (in depth) of multiplying these transpositions and getting the desired results?
|
Without loss of generality, suppose the only elements in the set are in fact $a,b,x$ for the first case. Try writing each as a single permutation.
$(x~b)(x~a)$ read from right to left says:
$\begin{array}{c}a\mapsto x\mapsto b\\ x\mapsto a\mapsto a\\ b\mapsto b\mapsto x\end{array}$
I.e. $(x~b)(x~a) = \begin{pmatrix}a&b&x\\b&x&a\end{pmatrix}$
On the other hand, $(x~a)(a~b)$ read from right to left says:
$\begin{array}{c}a\mapsto b\mapsto b\\ b\mapsto a\mapsto x\\ x\mapsto x\mapsto a\end{array}$
So we have $(x~a)(a~b)=\begin{pmatrix} a&b&x\\b&x&a\end{pmatrix}$
These are indeed equal. We have then $(x~b)(x~a)= (x~a)(a~b)$
Try reading through the second example to see if it is true.
|
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|
How to solve the differential equation $y' + \sec(x)*y = \tan(x)$ I am really struggling to solve the differential equation: $y' + \sec(x)y = \tan(x)$. If someone could point me in the right direction or give me a step by step plan it would be much appreciated!
So far I have tried taking the common factor to be $\exp(\int(\sec(x))$ (which simplifies to $\tan(x)+\sec(x)+c$ if I am not mistaken) however I end up with an equation which is beyond my integration abilities because it has both $X$ and $Y$ in it. I'm not sure if it is a case of trig identities or imaginary numbers which are letting me down.
Thanks in advance!!
|
This is in the form of a Linear Differential Equation
and here the integrating factor would be $e^{\int(\sec(x))}$ so that you will get $$\begin{align}
(\tan(x)+\sec(x))\frac{dy}{dx}+(\tan(x)+\sec(x))y\sec(x) = &\, \tan^2(x)+\sec(x)\tan(x) \\
((\tan(x)+\sec(x))y)' = & \\
\end{align}$$
So integrating on both sides you have, $$(\tan(x)+\sec(x))y=\tan(x)+\sec(x)-x+C$$ $$y=\frac{\tan(x)+\sec(x)-x+C}{\tan(x)+\sec(x)}.$$
I have just outlined what you need to do here, you can go to the hyperlink given to know more about linear differential equations. Hope it helps.
EXTRA EDITS:
Links to the integrals: integrating factor, and the integral of the right side. For both links, press the Go!button next to the formula.
|
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|
What is the the integral of $\sqrt{x^a + b}$? How do you evaluate $\displaystyle\int\sqrt{x^a + b}\,\,\text{dx}$, where $a \neq 0$ and $a \neq 1$?
For example, how do you evaluate $\displaystyle\int\sqrt{x^2 + 1}\,\text{dx}$? If we let $u=x^2+1$, then $du=2x\,\text{dx}$. We cannot do this because there is no $2x$ in the original function. Of course you cannot let $u=\sqrt{x^2 + 1}$, because $du=\displaystyle\frac{x}{\sqrt{x^2+1}}\,\text{dx}$. There is no $\displaystyle\frac{x}{\sqrt{x^2+1}}$ in the original function. So how do we solve this?
How about $\displaystyle\int\sqrt{x^3 + 1}\,\text{dx}$?
|
Kim Peek's "funny" hypergeometric solution is really the series solution near $x=0$. We have, for $|x^a/b| < 1$,
$$ \sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} =
\sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$
so integrating term-by-term
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x^{ak+1}}{(ak+1) \; b^{k-1/2}}$$
where $${1/2 \choose k} = \dfrac{\Gamma(3/2)}{\Gamma(k+1)\; \Gamma(3/2-k)}
= \dfrac{(2k)!}{(-4)^k (k!)^2 (1-2k)}$$
If you take the definition of the hypergeometric function as a power series, you get exactly this series.
Alternatively, for $|x^a/b| > 1$ you get a different series involving negative powers of $x$:
$$\sqrt{x^a + b} = x^{a/2} \sqrt{1 + b/x^a} = \sum_{k=0}^\infty {1/2 \choose k} b^k x^{(1/2 - k)a}$$
so that
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{b^k x^{(1/2-k)a+1}}{(1/2-k)a + 1}$$
which also has a hypergeometric representation.
|
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Analyze the Complex Function by using the Principal log Branch I am trying to analyze the function $\sqrt{1-z^2}$, where the square root function is defined by the principal branch of the log function. I want to locate the the discontinuities.
I know the discontinuities will lie on the negative real axis but I cannot figure out for which values of $z$ this will occur. I first tried rewriting the function as $(1-e^{2i\theta})$. But this seems to be getting me nowhere. Any thoughts?
|
Firstly we write $$f\left ( z \right )=\sqrt{1-z^2}=e^{\frac{1}{2}\ln\left ( 1-z^2 \right )}$$
We have branch points for all $z$ such that the argument of the log vanishes. We see that $$1-z^2=0\Leftrightarrow z_{1,2}=\pm1$$
We now need to investigate the point $z=\infty$. In that manner we need to see if if there is a branch point az $z=0$ for the function $$g\left ( z \right )=f\left ( \frac{1}{z} \right )=e^{\frac{1}{2}\ln\left ( z^2-1 \right )}e^{-\frac{1}{2}\ln z^2}$$
We can now conclude that the original function has two finite branch points, and one branch point at infinity. We chose a branch cut such that it connects $-1$ and $\infty$ and also $1$ and $\infty$. Note that this is actually one branch cut(in terms of the Riemann sphere)
|
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Checking whether a given polynomial is reducible or irreducible.
We're given the polynomial $x^{2}-2$ , and we need to prove that it's irreducible in $\mathbb Q$ but reducible in $\mathbb R$.
Writing the polynomial as $(x^{2}-2) = 1.(x^{2}-2)$ ,
$(x^{2}-2)$ can't be a unit in $Q[x]$ whereas $1$ is a unit in $Q[x]$ , hence the given polynomial is irreducible in $Q$.
Can't we apply the same argument for $R$ also ? Since rationals are contained in reals , won't that imply that it's irreducible in $R$ too ?
But we have to prove that it's reducible in $R$ , am I missing something ?
|
Like the comments said, $x^2-1 = (x-1)(x+1)$, which is reducible in $\mathbb{Q}[x]$, clearly. So, I am going to assume that you mean $x^2+1$, which is not reducible in $\mathbb{R}$, since $\text{discriminate}(x^2+1) = 0^2 - 4(1)(1) = -4 < 0$.
But, to show that it is not reducible in $\mathbb{Q}[x]$, suppose that it is, then it is the product of two degree one polynomials in $\mathbb{Q}[x]$; I.e.
$$x^2+ 0x +1 = (x - a)(x -b) =x^2 -(a+b)x + ab $$
Implying $a+b = 0$, and $ab = 1$. Further, $a = -b$ implying $ab = (-b)(b) = 1$, a contradiction.
EDIT: Note, similar logic holds in showing that $x^2-2$ is irreducible in $\mathbb{Q}$. To show that it is reducible in $\mathbb{R}$ it is enough to show that $x^2-2 =(x-\sqrt{2})(x +\sqrt{2})$, noting that $\sqrt{2} \in \mathbb{R}$ and that $0 \neq \text{deg}(x \pm \sqrt{2}) = 1 < \text{deg}(x^2-2) = 2.$
|
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subobject classifier for partial orders Does the category of partial orders have a subobject classifier? (Edit: No, see Eric's answer.)
If not, what is a category which is "close" to the category of partial orders (e.g. it should consists of special order-theoretic constructs) and has a subobject classifier? Bonus question: Is there also such an elementary topos? Notice that the category of partial orders has all limits, colimits and it is cartesian closed.
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The fact that all internal co-categories in a coherent category are necessarily co-equivalence relations [see Peter Lumsdaine's TAC article A small observation on co-categories] provides a telltale sign that the category of posets fails to be a topos.
For the inclusion functor $\textbf{Poset} \to \textbf{Cat}$ is represented by the internal co-category whose (co-?)nerve is the inclusion of the non-empty finite ordinals $\Delta \to \textbf{Poset}$, which is evidently not a co-equivalence relation (since not all posets are equivalence relations).
|
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On proving that $\sin x > x \cos x$ in a right neighbourhood of zero. I was recently asked to prove rigorously that there exists a $\delta$ s.t. $\sin x > x \cos x$ in $]0,\delta[$ .
The best I could come up with was that Taylor expanding $\sin x - x \cos x $ I obtain $\frac{ x^3}{3} + o(x^3)$ and noticing that the dominant term goes to zero for positive values I know that there must exist such a $\delta$.
But I was told this is not rigorous enough, how could I formalize it better? Any other paths I could take?
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$${\sin x\over x}=\int_0^1\cos(t\,x)\>dt>\cos x\qquad(0<|x|\leq\pi)\ .$$
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What is the sum of all the natural numbers between $500$ and $1000$. What is the sum of all the natural numbers between $500$ and $1000$ (extremes included) that are multiples of $2$ but not of $7$?
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In general, the sum of an arithmetic progression is $$S_n = \sum_{i=1}^{n} a_i=\frac{n}{2}(a_1+a_n)$$
So, the sum of all even numbers in your interval
$$ = \frac{251}{2}(500+1000)$$
And the sum of all multiples of $14$ in your interval
$$ = \frac{36}{2}(504+994)$$
Subtracting these two answers will give you the result you are looking for.
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Group of exponent $2$. When I have a group $G$ of exponent $2$ and I know that all elements in $G-\{e\}$ are conjugated, is it right that $G$ is of order $2$?
My try: For $g,h \in G - \{e\}$ the conjugation assumption implies $$ga=ah$$ but since $\exp G=2$ then $G$ is abelian and therefore $$ga=ha$$
It follows that $g=h$.
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Note that if $x, y \in G$. Then $yx = y^{-1}x^{-1}= (xy)^{-1} = xy$ by the hypothesis. Hence $G$ having an exponent 2 implies that $G$ is abelian.
Choose $a, b \ G\backslash\{1 \}$.
Then there exists a $g$ such that $b = gag^{-1}$.
Then $h = ba^{-1} = gag^{-1}a^{-1} = 1$ since $G$ is abelian.
It follows that $a = b$ and we have the result as it was promised.
|
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Linear vs. bilinear I'm tripping over something elementary:
Suppose $f:\mathbb{R^2}\rightarrow X$ is linear, then $f(x+y)=f(x)+f(y)$ for all vectors $x$ and $y$. Now suppose that $f$ is also bilinear and in particular linear in the first argument, then
$$f\Big((a+b,c)\Big)= f\Big((a,c)\Big)+f\Big((b,c)\Big)$$
But now setting $x=(a,0)$ and $y=(b,c)$ contradicts linearity of f.
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A (non-trivial) linear function from $\mathbb{R}^n$ for $n > 1$ will never be a multilinear function (bilinear when $n=2$) if it is a linear function. Instead, a multilinear function is a linear function from a tensor product. So a bilinear function on $\mathbb{R}$ is a linear function $\mathbb{R}\otimes\mathbb{R} \to X$, but $\mathbb{R}$ is the unit for the tensor product, so $\mathbb{R}\otimes\mathbb{R} \cong \mathbb{R}$ which is very much not $\mathbb{R}^2$.
|
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Distance between two circles on a cube I found this problem in a book on undergraduate maths in the Soviet Union (http://www.ftpi.umn.edu/shifman/ComradeEinstein.pdf):
A circle is inscribed in a face of a cube of side a. Another circle is circumscribed about a neighboring face of the cube. Find the least distance between points of the circles.
The solution to the problem is in the book (page 61), but I am wondering how to find the maximum distance between points of the circles and I cannot see how the method used there can be used to find this.
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I placed the inscribed circle on the top face (+y direction) and the circumscribed circle on the front face (+z direction). Their locus of points is
$$ \begin{align}
\vec{r}_1 & = \begin{bmatrix} r_1 \cos \theta_1 & \frac{a}{2} & r_1 \sin \theta_1 \end{bmatrix} \\
\vec{r}_2 & = \begin{bmatrix} r_2 \cos \theta_2 & r_2 \sin \theta_2 & \frac{a}{2} \end{bmatrix}
\end{align} $$
where $r_1 = \frac{a}{2}$, $r_2 = \frac{a}{\sqrt{2}}$ and $a$ is the side of the cube.
The distance (squared) is $$d^2 = \| \vec{r}_1 -\vec{r}_2 \|^2 $$ which is a function of $\theta_1$ and $\theta_2$.
It comes out as
$$ \frac{d^2}{a^2} = \frac{5}{4} - \frac{\sqrt{2} \cos\theta_1 \cos \theta_2 + \sin \theta_1 + \sqrt{2} \sin \theta_2}{2} $$
To minimize this you have to set $$\frac{\partial}{\partial \theta_1} \frac{d^2}{a^2} = 0$$ and at the same time $$\frac{\partial}{\partial \theta_2} \frac{d^2}{a^2} = 0$$
The system to solve is $$\begin{align} \frac{\sin \theta_1 \cos \theta_2}{\sqrt{2}} - \frac{\cos\theta_1}{2} & = 0 \\ \frac{\cos \theta_1 \sin \theta_2}{\sqrt{2}} - \frac{\cos\theta_2}{\sqrt{2}} & = 0 \end{align} $$
From the first equation $\cos \theta_2 = \frac{\cot \theta_1}{\sqrt{2}}$. When used in the second equation I get $\theta_1 = \arctan \sqrt{2}$.
The result I get was $$\frac{d^2}{a^2} = \frac{5}{4} - \frac{\sqrt{6}}{2} $$
|
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Proof of the Stratonovich integral? Computing the integral $\int \Phi(x_t,t)dx_t$
writing the equation in the form we can write the integral as the mean square limit
$$\int \Phi(x_t,t)dx_t=\lim_{\Delta \to 0} \sum^{j=1}_{N-1} [\Phi((\frac {x(t_j)+x(t_{j+1})}{2},t_j)][x(t_j+1)-x(t_j)]\ \ \ \ \ \ .\ (3)$$
and in Ito's form as $$\int \Phi(x_t,t)dx_t=\lim_{\Delta \to 0} \sum^{j=1}_{N-1} [\Phi x(t_j),t_j][x(t_j+1)-x(t_j)] \ \ \ \ \ \ \ \ \ \ .\ (4)$$
Let us prove the existence of the limit in $(3)$ and find the formula relating
the two indicated integrals. To do this we select the $\Delta$ partitioning and consider
the difference between the limit expressions on the right-hand sides of
$(3)$ and $(4)$. Making use of the differentiability with respect to $x$ of the
function $\Phi(x_t, t)$ we get
$$D_{\Delta}=\sum^{j=1}_{N-1} [\Phi((\frac {x(t_j)+x(t_{j+1})}{2}),t_j-\Phi(x(t_j),t_j][x(t_{j+1})-x(t_j)] \ \ \ \ \ \ .(5)$$ $$= \frac{1}{2} \sum ^{N-1}_{j=1} \frac {\partial \Phi}{\partial x}[(1-\theta)x(t_j)+\theta x(t_{j+1}),t_j)][x(t_{j+1})-x(t_j)]^2 ,0\le \theta \le 1/2,t_j=t_j^{\Delta}. $$ my question is how to proceed after $(5)$, how this final equation comes .
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If $x \mapsto \Phi(x,t)$ is differentiable, it follows from Taylor's formula that
$$\Phi(x,t) = \Phi(y,t) + (x-y) \frac{\partial}{\partial x} \Phi(\zeta,t)$$
for some intermediate point $\zeta$ between $x$ and $y$ (i.e we can find $\lambda \in (0,1)$ such that $\zeta = \lambda x+ (1-\lambda) y$). Using this identity for
$$x :=\frac{x(t_j)+x(t_{j+1})}{2} \qquad y := x(t_j) \qquad t = t_j$$
we find
$$\Phi \left( \frac{x(t_j)+x(t_{j+1})}{2}, t_j \right) = \Phi(x(t_j),t_j)+ \frac{x(t_{j+1})-x(t_j)}{2} \frac{\partial}{\partial x} \Phi(\zeta,t_j) \tag{1}$$
with
$$\zeta = \lambda x(t_j)+ (1-\lambda) \frac{x(t_j)+x(t_{j-1})}{2} \tag{2} $$
for some $\lambda \in (0,1)$. Note that $(2)$ is equivalent to
$$\begin{align*} \zeta &= x(t_j) \left[ \frac{2\lambda}{2} + \frac{(1-\lambda)}{2} \right] + \underbrace{\frac{1-\lambda}{2}}_{=:\theta} x(t_{j+1}) \\ &= x(t_j)(1-\theta) + \theta x(t_{j+1}) \end{align*}$$
for some $\theta \in (0,1/2)$. Hence, by $(1)$,
$$\Phi \left( \frac{x(t_j)+x(t_{j+1})}{2}, t_j \right) -\Phi(x(t_j),t_j)= \frac{x(t_{j+1})-x(t_j)}{2} \frac{\partial}{\partial x} \Phi ( x(t_j)(1-\theta) + \theta x(t_{j+1}), t_j).$$
Multiplying this expression with $x(t_{j+1})-x(t_j)$ and summing over $j=1,\ldots,N-1$ yields the identity you are looking for. (Mind that $\theta = \theta(j)$; we cannot expect to find one $\theta$ which works for all $j=1,\ldots,N-1$).
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Let $F : X → X$ be continuous. Prove that the set $\{x ∈ X : F(x) = x\}$ of fixed points of F is closed in X Here X is a Hausdorff Space.
I know that singleton sets, {x}, are closed in a Hausdorff space. Although Im not sure if thats how to use the Hausdorff property.
Should I investigate $h=F(x)-x$?
Can anyone give a hint
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I have this idea that I'd love somebody with more knowledge than I will check: define a function
$$\;\phi:X\to X\times X\;,\;\;\text{by}\;\;\;\phi(x):=(x,F(x))\;$$
Since each coordinate function is continous also $\;\phi\;$ is, and if $\;\Delta:=\{(x,x)\in X\times X\}\;$ is the diagonal in the cartesian product, then
$$\phi^{-1}(\Delta\cap\phi(X))=\left\{x\in X\;;\;F(x)=x\;\right\}=:K$$
Since $\;\Delta\cap\phi(X)\;$ is closed (see comments by Ayman Hourieh and Thomas Andres below the question) and $\;\phi\;$ is continous, then $\;K\;$ is closed.
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Compute a natural number $n\geq 2$ s.t. $p\mid n \Longrightarrow p^2\nmid n$ AND $p-1\mid n \Longleftrightarrow p\mid n$ for all prime divisor p of n. Question: Compute a natural number $n\geq 2$ that satisfies:
*
*For each prime divisor $p$ of $n$, $p^2$ does not divide $n$.
*For each prime number $p$, $p-1$ divides $n$ if and only if $p$ divides $n$.
I'm stuck with this one like forever.
My first intuition was to find the canonical form of such $n$, which is $\prod_{i} p_i^{\alpha_i}$, where $\alpha_i$ must be equal to $1$ for all $i\in [1,r]$ for some $r\in\mathbb{N}$.
But it got me nowhere.
Then, I started trying to find such $n$ from the second condition. Yet again, I have no clue whatsoever.
Hence, I'm here asking for help. Any hints?
Many thanks,
D.
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Here's a partial answer: Suppose $n$ is a solution. Then $n(n+1)$ is a solution if $n+1$ is prime, since $n+1\mid n(n+1)$ and $n\mid n(n+1)$, and the condition is clearly satisfied for all primes less than $n$. So in addition to $2$, $3$, $6$, and $42$, we also get
\begin{align*}
42(43) = 1806 = 2\cdot 3\cdot 7\cdot 43 .
\end{align*}
This fails at the next step since $1807=13\cdot 139$. This doesn't prove that there aren't other solutions, however.
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Prove that if G contains an odd vertex then every vertex of G is odd Let G be a graph with degree of each vertex either $m$ or $n$, where there are $m$ vertices of degree $m$ and $n$ vertices of degree $n$. Prove that if G contains an odd vertex then every vertex of G is odd.
how do I go about this? What does it imply if G contains an odd vertex? Also, how am I to use the first line of the given problem? Please help.
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A very useful elementary result in graph theory is the degree sum formula:
$$\sum_{v\in V}\deg(v) = 2\lvert E \rvert$$
for a graph $G=(V,E)$.
In particular the sum of the degrees of the vertices is even.
Now what can you say if some vertex has odd degree?
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Distribtution of the maximum of three uniform random variables. How do I get the cumulative density function of $Y$? $X$ is a continuous random variable with pdf
$$f(x) = 1,\quad 0 < x < 1. $$
Three independent observations of $X$ are made. Find the pdf of $Y$ where $$Y = \max\{X_1,X_2,X_3\}.$$
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The pdf or cdf you are looking for is part of what is called the "order statistics": https://en.wikipedia.org/wiki/Order_statistic
The easiest path is through the cdf, using the following transformations:
$F_Y(y) \ = \ P(Y\leq y) \ = P(\max{\{X_1,X_2,X_3\}} \leq y)$
$F_Y(y) \ = \ P(X_1 \leq y \ \& \ X_1 \leq y \ \& \ X_1 \leq y).$
Using independence of the $X_k$:
$F_Y(y) \ = \ P(X_1 \leq y) P(X_2 \leq y) P(X_3 \leq y)=(y)^3$
It suffices then to derive wrt $y$ to obtain the pdf:
$$f_Y(y)=3~y^2$$
Edit : Very sorry : I made a mistake that I have corrected now (confusion between $\min$ and $\max$)
|
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Check if this is the example of $x$ as a limit point of $C$ but ($x_t$) does not converge to $x$. Let ($x_t$) be a sequence in a metric space, and let $C$ be the range of ($x_t$). I want to give an example in which $x$ is a limit point of $C$ but ($x_t$) does not converge to $x$.
Here's my example: let $C$ = (0,1], where 1 is a limit point of $C$, let ($x_t$) = ($1/t$)=($1,1/2,1/3,...$), but ($x_t$) converges to zero instead of 1.
Is my example correct? If not, could someone give another example please? Thanks.
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$1$ is not a limit point of $C$.
The definition for limit point is,($x$ of $C$)
$$ \forall r > 0, \exists y \in C\mid y \neq x, \operatorname{d}(x,y) < r
$$
so taking $r = 1/3$...
You can choose $x_j = 1/j$ for all even $j$ and $x_j = 1-1/j$ for all odd $j$. Then $0$ and $1$ are limit points of $C$, but the sequence does not converge.
|
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Example of inverse semigroup with at least two idempotent elements We say that the semigroup $S$ is inverse semigroup if for any $s\in S$ there is a unique $t\in S$ such that $sts=s,\ tst=t$. Suppose that $E(S)=\{e:\ e\in S,\ e^2=e\}$ and define $$s\sim t\Leftrightarrow \exists\ e\in E(S):\ se=te,\qquad (s,t\in S); $$
and suppose that $[s]=\{t\in S:\ s\sim t\}$.
I have two request:
*
*Could we say that $[s]=\{se: e\in E(S)\}$ for any $s\in S$?
*I need an inverse semigroup $S$ such that $S\neq E(S)$ and $E(S)$ has at least two elements.
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We know that the set of natural number with multiplication defined by
$m\cdot n=\min\{m,n\}$ is an inverse semigroup. Now suppose that $G$ is a group such that there is an element $h\in G$ such that $h^2\neq h$ and suppose that $e$ is its identity. Obviously $S:=N\times G$ is an inverse semigroup with mutiplication
$$(n,g)(m,k)=(\min\{n,m\}, gk).$$
Now because $(1,h)(1,h)=(1,h^2)\neq (1,h)$ we get $S\neq E(S)$. Also for any $n\in N$ we have
$$(n,e)(n,e)=(\min\{n,n\},e^2)=(n,e)\in E(S)$$
which shows that $|E(S)|\geq 2$.
From the previous part
$E(S)=\{(n,0): n\in N\}$ and $[(m,q)]=\{(n,q):n\in N\}$. Also
$$\{(2,g)E: E\in E(S)\}=\{(2,g)(n,0): n\in N\}=\{(1,g),(2,g)\}$$
but $(2,g)\sim (3,g)$.
|
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A sequence in a Hausdorff space and in a space that is not Hausdorff.
Let $X$ be a topological space and $\{x_n\}_{n=1}^{\infty}$ a sequence in $X$.
Show that if $X$ is Hausdorff, $x_n \rightarrow x \:$, $x_n \rightarrow y \:$ implies $x=y$.
Give an example that shows that if $X$ is not Hausdorff then this is not necessarily true.
A sequence is convergent, $x_n \rightarrow x \:$, if there is an element $x \in X$ such that for every open neighborhood $U$ of $x$ there exists a $n_0$ such that $n > n_0$ implies $x_n \in U$.
If $X$ is Hausdorff and $x$ and $y$ are different there exists open disjoint sets $x \in U$ and $y \in V$.
Because all but a finite number of $x_n$ lies in $U$ according to the definition of that $x_n \rightarrow x \:$ then only a finite number of $x_n$ can be in $V$, which is a contradiction to that $x_n \rightarrow y \:$.
But i have a hard time to come up with an example that shows that this not necessarily is true when $X$ isnt Hausdorff.
Can anyone give me a hint?
Thanks.
|
Take $\mathbb{N}$ in the cofinite topology (the only closed sets are the finite ones (including the empty set) and $\mathbb{N}$ itself). Take $a_n$ to be any sequence where all values are different, like $a_n = n$ or $a_n = 2n$ etc.. Then $(a_n)$ converges to every point $m$ of $\mathbb{N}$, because the only open sets that contain $m$ are of the form $O = \mathbb{N} \setminus F$, where $F \subset \mathbb{N}$ is finite. But after some initial segment the values of $a_n$ are never in $F$ (as $F$ is only finite), so all values from some value onwards are in $O$. As this holds for all $O$, $(a_n) \rightarrow m$ for all $m$.
So any two of these sequences have all points of the space as their limit as well and so the limits are very non-unique. The cofinite topology is $T_1$, so one separation axiom below Hausdorff, making it a "sharper" example than the indiscrete topology.
|
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|
Probability Type I error I have these values:
A sample of size $100$ is taken from the population.
Standard Deviation is $5$.
Average is $125$.
$H_0$: $\mu =125$ Against $H_a$: $\mu <125$
Accept $H_0$ if the sample mean is $\bar{\mathbb{x}} ≥ 124$
Reject $H_0$ if the sample mean $\bar{\mathbb{x}} < 124 $
Find the probability of making a Type I error.
I did:
Variance$\; =25$, because S.D. is $5$ then,
$$Z= \frac{\bar{\mathbb{x}} -\mu}{\text{S.D.}/\sqrt{100}}$$
which is : $ Z=\frac{124-125}{5/10}=-2 $
$1- \Phi (-2)=1-0.97725=0.023$
The probability of making a type I error is $2.3%$.
Am I correct? Can anyone let me know if this is correct?
thanks
|
You work seems fine, except for the final step as Augustin pointed out.
In terms of probability, recall that we are seeking the probability that $\bar x < 124$ given that the $\mu = 125$. This means
$$P(\bar x <124|\mu = 125)$$
By standardizing, we get
\begin{align*}
P(\bar x <124|\mu = 125) &= P\left(Z < \frac{124-125}{5/\sqrt{100}}\bigg|\mu
= 125\right) \\
&= P(Z<-2|\mu = 125) \\
&= \Phi(-2) \\
&= 0.02275013
\end{align*}
In simpler terms (if you don't know probability), we are looking for the "area to the left" of $-2$ under the normal curve.
|
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|
Evaluating the limit $\lim_{x \to 0}\left(x+e^{\frac{x}{3}}\right){}^{\!\frac{3}{x}}$
$$y=\left(x+e^{\frac{x}{3}}\right)^{\frac{3}{x}}$$
I'm looking at this equation, and need to solve for the limit as $ \to 0$. I've graphed it, and know the solution is $e^4$, but am clueless as to the steps to actually solve this.
(Note, I am an adult working as a math aide in a high school. I help students at Algebra, Trig, Geometry, and Intro to Calculus. 35 years out of HS myself, there are clearly some things I need to brush up on. i.e. I know my limits. Pun intended.)
Yes, L'Hopital is fine. The student is in a BC calc class)
|
Intuition:
A way to see what is going on is to see the affine approximation of $e^x$ around $0$: $$e^u \simeq e^0 + (e^\prime)(0) x = 1 + x$$ (this can be made formal by Taylor approximations to order $1$, for instance). This implies that your quantity is roughly $\left(x+ 1+ \frac{x}{3}\right)^{3/x} = \left(1+ \frac{4x}{3}\right)^{3/x}$, where you recognize, setting $t = \frac{3}{x}\to \infty$, the limit $$\left(1+\frac{4}{t}\right)^t \xrightarrow[t\to\infty]{} e^4.$$ The only key is to make this first approximation $\simeq$ rigorous, which is done below.
An approach based on Taylor expansions: (but which requires no knowledge of them besides the Landau notation $o(\cdot)$ — justifying what is needed as we go)
Start (as often when you have both a base and an exponent depending on $x$) by rewriting it in exponential form:
$$
\left(x+e^{\frac{x}{3}}\right)^\frac{3}{x} = e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)}
$$
Now, when $u\to 0$, we have $\frac{e^u-1}{u}\to \exp^\prime 0 = e^0 = 1$, so that $e^u = 1+u + o(u)$; which gives $$x+e^{\frac{x}{3}} = x+1+ \frac{x}{3} + o(x) = 1+\frac{4}{3}x.$$
Similarly, since $\frac{\ln(1+u)}{u}\xrightarrow[u\to 0]{} 1$, we have $\ln(1+u) = u + o(u)$. Combining the two, we get
$$\ln\left(x+e^{\frac{x}{3}}\right) = \ln\left(1+\frac{4}{3}x\right) = \frac{4}{3}x + o(x).$$
Putting it together,
$$
\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right) = \frac{3}{x}\left(\frac{4}{3}x + o(x)\right) = 4 + o(1) \xrightarrow[x\to 0]{} 4
$$
and, by continuity of $\exp$,
$$e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)} \xrightarrow[x\to 0]{} e^4.
$$
|
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|
Average of cosine waves What looks like the graph of the function $A=E[cos(\omega t)]$, where $E$ means "average" and $\omega$ is a random variable?
|
Assuming that $\omega$ is uniformly distributed in range $[a,b]$, we have
$$E[\cos(\omega t)]=\frac1{b-a}\int_a^b\cos(\omega t)d\omega=\frac{\sin(bt)-\sin(at)}{(b-a)t}.$$
For example, with $a=1,b=3.14$:
In the case of a Gaussian, another Gaussian.
|
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|
Express reverse inversion, major index, descents in terms of the forward direction. Given $w=a_1a_2...a_n \in S_n $, then the reverse of $w$ is $w^r=a_n....a_2a_1$.
Express inv($w^r$), des($w^r$) and maj($w^r$) in terms inv($w$), des($w$), maj($w$), respectively.
I know the definition of an inversion is a partition(i, j) s.t. i $<$ j and part(i) > part(j), descent is where the number goes down, and major index is the sum of the positions of the descents of the permutation.
Can someone point me in the right direction as to how to go about solving these?
Perhaps some formulas, or theorems.
|
HINT: For the inversions, show that $\langle i,j\rangle$ is an inversion of $w^r$ if and only if it is not an inversion of $w$. How many pairs $\langle i,j\rangle$ with $i<j$ are there?
For the descents, show that each descent of $w^r$ is an ascent of $w$ and vice versa, so $w^r$ has exactly as many descents as $w$ has ascents. If $w$ has $k$ descents, how many ascents does it have?
For the major index, show that if $i$ is an ascent of $w$, then the corresponding descent of $w^r$ is $n-i$. Thus, if $s$ is the sum of the ascents of $w$, and $w$ has $a$ ascents, then $\operatorname{maj}(w^r)=an-s$. You know $a$ from the second part, so it only remains to find $s$ in terms of $n$ and $\operatorname{maj}(w)$. Further HINT: Express $\operatorname{maj}(w)+s$ in terms of $n$.
|
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|
Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$
I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.
Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$:
$$\overline{f}^G = \frac{29}{3}$$
How can this remainder be of use to me?
Thanks. (Note: I am using Macaualay2)
|
Using just Macaulay2, you can do the following
Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
ReesAlgebra, TangentCone
i1 : R=QQ[a,b,c]
o1 = R
o1 : PolynomialRing
i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7)
2 2 2 3 3 3
o2 = ideal (a + b + c - 3, a + b + c - 5, a + b + c - 7)
o2 : Ideal of R
i3 : S=R/i1
o3 = S
o3 : QuotientRing
i4 : phi=map(S,R)
o4 = map(S,R,{- b - c + 3, b, c})
o4 : RingMap S <--- R
i6 : use R
o6 = R
o6 : PolynomialRing
i7 : phi(a^5+b^5+c^5)
29
o7 = --
3
o7 : S
(I deleted i5 and o5 as I made a typo in the input there)
|
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|
Convergent sequence of irrational numbers that has a rational limit. Is it possible to have a convergent sequence whose terms are all irrational but whose limit is rational?
|
If $q$ is any rational number at all and $n$ is a positive integer then $q+\frac 1 n \sqrt 2$ is irrational (it's a simple algebra exercise to prove that), and $\lim\limits_{n\to\infty}\left(q + \frac 1 n \sqrt 2\right) = q$.
|
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|
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?
|
We have $\sqrt[6]{3} \approx 1.2009$ and $\sqrt[6]{4} \approx 1.2599$. Let $f(x)=x^3+x^2-x-2$. Then $f(1.2) \approx -0.032$ and $f(1.25) \approx 0.2656$. So, $a$ must be between $1.2$ and $1.25$.
EDIT: As I said in the comments below, I see no way of showing that there is only one root using only precalculus. But, for completeness of my answer: If there were two roots, then the Mean Value Theorem would imply that the derivative is $0$ between the roots. But, $f'(x) = 3x^2+2x-1$. The roots of this are $x=-1$ and $x=\frac{2}{3}$. Using whichever method one likes, you can see that there is a local maximum at $x=-1$ and a local minimum at $x=\frac{2}{3}$. And, both $f(-1)$ and $f\left(\frac{2}{3}\right)$ are negative. So, there cannot be another root.
|
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|
Prove $\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$ is always divisible by $6$ when $n$ is an integer.
Prove $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!}$$ is always divisible by $6$ when $n$ is an integer.
I have done a similar proof that $\binom{2n}{n}$ is divisible by $2$ by showing that $$\binom{2n}{n}=\binom{2n-1}{n-1}+\binom{2n-1}{n}=2\binom{2n-1}{n-1}$$ but I am at a loss for how to translate this to divisible by $6$. Another way to do this proof would be to show that when you shoot an $n$-element subset from $2n$ you can always match it with another subset (namely the $n$-elements that were not chosen). Again, no idea how to translate this to $6!$.
|
Notice that $$\binom{3n}{n,n,n}=\frac{(3n)!}{n!n!n!} = \binom{3n}{n}\binom{2n}{n}.$$ We have shown that $\binom{2n}{n}$ is divisible by 2. Now, all we must do is show that $\binom{3n}{n}$ is divisible by 3. You know that $\binom{3n}{n}$ is the number of n-element subsets of a 3n-element set (or, the number of ways to choose n objects among 3n distinct objects), which is always an integer. Note that $$\binom{3n}{n} = \frac{(3n)!}{(n!)(2n)!} = \frac{(3)(n)(3n - 1)!}{(n)(n-1)!(2n)!} = \frac{(3)(3n - 1)!}{(n-1)!(2n)!} = (3)\frac{(3n - 1)!}{(n-1)!(2n)!} = (3)\binom{3n-1}{n-1},$$ which is an integer and is divisible by 3.
By Peter's suggestion, you can generalize to say that for all integers k, $$\binom{kn}{n_{1},n_{2}, ... , n_{k}} = \binom{kn}{n}\binom{(k-1)n}{n}...\binom{(k-k+1)n}{n}$$ and $$\binom{kn}{n} = \frac{(kn!)}{n!(kn-n)!} = \frac{(kn)(kn-1)!}{(n)(n-1)!(kn-n)!} = k\binom{kn-1}{n-1}, $$ and use induction to prove that $$\binom{kn}{n_{1},n_{2}, ... , n_{k}}$$ is divisible by $k!.$
|
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|
Show that $a_n = 1 + \frac{1}{2} + \frac{1}{3} +\dotsb+ \frac{1}{n}$ is not a Cauchy sequence
Let $$
a_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}
\quad (n \in \mathbb{N}).
$$
Show that $a_n$ is not a Cauchy sequence even though
$$
\lim_{n \to \infty} a_{n+1} - a_n = 0
$$
(Therefore $a_n$ does not have a limit).
|
Hint: Showing it doesn't converge (specifically that it goes to infinity) would help.
|
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$G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate Let $G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate and any proper subgroup is contained in a maximal subgroup . Then is $G$ cyclic ? I know that if $G$ is a finite group such that any two maximal subgroups are conjugate then $G$ is cyclic . But I cannot handle the infinite case . Please help . Thanks in advance
|
Here is a reduction to the finite case. Suppose $G$ is a group satisfying your requirements, and $M$ a maximal subgroup. Then all maximal subgroups are conjugate to $M$. This implies that the Frattini subgroup $\Phi(G)$ of $G$ (which is defined to be the intersection of all maximal subgroups of $G$) is the intersection of the finitely many conjugates of $M$, all of which have finite index, so $\Phi(G)$ has finite index. Therefore $G/\Phi(G)$ is finite.
Now it is easy to see that $G/\Phi(G)$ still has all maximal subgroups conjugate, so by the finite case, $G/\Phi(G)$ is cyclic. Let $\overline a$ be a generator of this group, where $a \in G$. Then we get that $G = \langle a, \Phi(G)\rangle$.
We now claim that $\langle a\rangle = G$. If not, then there is some maximal subgroup $N$ of $G$ containing $\langle a\rangle$. But by definition of the Frattini subgroup, $N$ also contains $\Phi(G)$, so then $N$ would contain $\langle a, \Phi(G)\rangle = G$. Contradiction, so there is no maximal subgroup containing $\langle a\rangle$, and therefore we conclude that $G = \langle a \rangle$.
|
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|
Classifying Covering Spaces using First Cohomology I am familiar with the classification of covering spaces of a space $X$ in terms of subgroups of $\pi_1(X)$ (up to conjugation). However, if $X$ is a manifold, I know that $H^1(X; G)$ classifies G-bundles over $X$ (using Cech cohomology here). I think finite regular covering spaces are $\mathbb{Z}/k \mathbb{Z}$-bundles; regular means that the deck transformations act transitively on the fiber (and regular covers correspond to normal subgroups of $\pi_1(X)$).
Does this mean that $H^1(X; \mathbb{Z}/k\mathbb{Z})$ is in bijection with k-sheeted regular covering spaces over $X$. I could not find such a statement anywhere and so am a bit suspicious.
Also, if this is correct, what does $H^1(X; \mathbb{Z})$ classify? I'm not sure what a $\mathbb{Z}$-bundle is - what has automorphism group equal to $\mathbb{Z}$?
Also, $H^1(X; \mathbb{Z}) = [X, S^1]$ so if $H^1(X; \mathbb{Z})$ classifies some kind of bundles, there should be universal bundle over $S^1$ which pulls back to these bundles. What is this bundle?
|
Regular cyclic $q$-coverings are classified by normal subgroups $$N\hookrightarrow \pi_1X \to \pi_1X/N\cong \mathbb Z/q,$$
hence there is a natural isomorphism to $Hom(\pi_1X,\mathbb Z/q)$ mod automorphism of $\mathbb Z/q$.
Note that $K_q=K(\mathbb Z/q,1)$ is universal in the sense that the map $$[X,K_q] \to Hom(\pi_1X,\mathbb Z/q) = H^1(X,\mathbb Z/q),$$ $$ [f]\mapsto f_*$$ is an isomorphism. In particular you see that every cyclic $q$-covering is obtained by pulling back the universal cover of $K_q$.
Corollary If $X$ is a smooth orientable manifold, then all infinite cyclic covers (or cyclic which factor through $\mathbb Z$) are obtained by cutting and gluing along a hypersurface.
|
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|
$S_1 \subset S_2$. To show, $Span(S_1) \subset Span(S_2)$ Prove that if $S_{1} \subset S_{2}$, then $Span(S_{1}) \subset Span(S_{2})$
Approach: Suppose $S_{1} \subset S_{2}$
Let $x \in S_{1}$, then by definition of a subset, $x \in S_{2}$
All possible linear combinations of $x$ are $cx$ with $c \in \mathbb{R}$
So $cx \in Span(S_{1})$
But $x$ is also in $S_{2}$, therefore $Span(S_{2})$ contains all linear combinations of $x$ as well.
So $cx \in Span(S_{2})$
We have $cx \in Span(S_{1})$, then $cx \in Span({S2})$ $\forall c \in \mathbb{R}$
We have shown: if $S_{1} \subset S_{2} \Rightarrow Span(S_{1}) \subset Span(S_{2})$
Is my approach to the question correct? If so, is this enough to answer the question or have I missed something?
|
No, that is not a valid approach.
In order to show that $\mathit{Span}(S_1)\subseteq \mathit{Span}(S_2)$ you need to start with an arbitrary element of $\mathit{Span}(S_1)$ and show that it is also an element of $\mathit{Span}(S_2)$.
You have shown this only for elements of $\mathit{Span}(S_1)$ that happen to have the form $cx$ for some $c$ and and $x\in S_1$, but there will generally be elements of $\mathit{Span}(S_1)$ that do not have this simple form.
|
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