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H: Homogeneous riemannian manifolds are complete. Trouble understanding proof.
I came across this proof while looking for hints on my homework, and I think it's only gotten me more confused. This is from Global Lorentzian Geometry.
Lemma 5.4 If $(H,h)$ is a homogeneous Riemannian manifold, then $(H,h)$ is complete.
Proof. By the Hopf-Rinow Theorem, it suffices to show that $(H,h)$ is geodesically complete. Thus suppose that $c:[a,1) \rightarrow H$ is a unit speed geodesic which is not extendible to $t=1$. Choosing any $p \in H$, we may find a constant $\alpha > 0$ such that any unit speed geodesic starting at $p$ has length $\ell \geq \alpha$. Set $\delta = \min\{\alpha/2, (1-a)/2\} > 0$. Since isometries preserve geodesics, it follows from the homogeneity of $(H,h)$ that any unit speed geodesic starting at $c(1-\delta)$ may be extended to a geodesic of length $\ell \geq 2 \delta$. In particular, $c$ may be extended to a geodesic $c:[a, 1+\delta) \rightarrow H$, in contradiction to the inextendibility of $c$ to $t=1$.
How exactly does it "follow from homogeneity" that we are able to extend unit speed geodesics? Isn't that what we're trying to show in the lemma? And how does considering geodesics based at $c(1-\delta)$ lead us to extending $c$ to $c:[a,1+\delta) \rightarrow H$?
AI: Homogeneity implies that all metric balls of the same radius are isometric. Therefore if one can extend a geodesic at a point $p$ in each direction by a distance of $\delta$, then one can extend by the same $\delta$ at every point of the manifold. |
H: MMSE (Minimum min square estimate) problem
I have a problem as follows. As of now, I cannot provide the definition of X and Y but can anyone provide a rough overview of what needs to be done ?
An experiment
consist of rolling a single die and the experimental outcome
determines a value for X and Y. (the random variables X and Y are defined Table)
a) Find a constant a that will give a
mininmum mean-square estimate of X and calculate the resulting
mean-square error.
b) Given that Y=4, find the minimum mean-square
estimate of X and the resulting mean-square error
AI: The mean of a distribution is the estimate that minimizes the mean square error.
For example, imagine that we are going to draw $x$, and it can be $0$ or $1$ with equal (50%) probability. We make an estimate $\hat{x}$ to minimize the mean square error:
$$
\min_\hat{x} \ E\left[(x-\hat{x})^2\right]
$$
where the $E$ means expected value.
Using calculus, the mean square error is minimized by a choice of $\hat{x}$ that satisfies:
$$
0 = E\left[-2\left(x-\hat{x}\right)\right]
$$
Dividing both sides by $-2$ and rearranging, we get
$$
\hat{x} = E\left[x\right]
$$
So we should guess $\hat{x} = 0.5$ in this case.
For your problem, in part a), the answer will be the expected value of $X$, and in part b), the answer will be the expected value of $X$ conditional on $Y=4$. |
H: Probability: Normal Distribution
Each item produced by a certain manufacturer is, independently, of acceptable quality with probability $0.95$. Approximate the
probability (by a normal distribution) that at most $10$ of the next
$150$ items produced are unacceptable.
Hits to a high-volume Web site are assumed to follow a Poisson distribution with a mean of $10,000$ per day.
$\,\,\,\,\,\,\,\,\,\,\text{(a)}$ Approximate (by a normal
distribution) the expected number of days in a year ($365$ $\,\,\,\,\,\,\,\,\,\,\,$years) that
exceed $10,200$ hits.
$\,\,\,\,\,\,\,\,\,\,\text{(b)}$ Approximate (by a normal
distribution) the probability that over a year ($365$ days) $\,\,\,\,\,\,\,\,\,\,\,$more than
$15$ days each have more than $10,200$ hits.
For question 1, is the answer $0.8695$?
For question 2(a), is that I have to use P(X<= 10199.5) to find the probability and multiply the probability by $365$ ?
For question 2(b), would anyone mind telling me how to solve it?
AI: For Question 1)
Model the number of unacceptable items as normal, p=.05.
$\hat\mu = .05(150) = 7.5\space,\hat\sigma = \sqrt{150(.05)(.95)} = 2.67$
Normal Approximation: F = No. of Unacceptable Items, $F \sim N(7.5,2.67)$
$P(F\leq 10) = \Phi(\frac{10-7.5+.5}{0.84}) = 0.87$ Using the continuity correction.
So, your first answer is correct.
Question 2)
a) Your approach is generally correct. You want to calcluate $P(X\geq 10,201)$ using the normal approximation ($\hat \mu = \hat \sigma^2 = 10,000 $and multiply by 365.
b) HINT: You'll need to first approximate $P(X\geq 10,201)$ as before then use the Binomial distribution to get the probability for the number of days. |
H: Hermite polynomials recurrence relation
Hermite polynomials $H_n (x)$ can be obtained using the recurrence relation
$$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x).$$
To prove this, I started by calculating the first derivative of the Hermite's Rodrigues formula $H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2 } $. The process goes like this:
$$
\frac{d}{dx}H_n (x)=(-1)^n 2xe^{x^2}\frac{d^n}{dx^n} e^{-x^2 }+(-1)^n e^{x^2} \frac{d^{n+1}}{dx^{n+1}}e^{-x^2 }
$$
Rearranging the terms in the previous equation produces
$$H_{n+1} (x)=2xH_n (x)-\frac{d}{dx}H_n (x)$$ or
$$H_{n+1} (x)=2xH_n (x)-H_n'(x).$$
That's it. Im stuck here. I don't know how to show that $H_n' (x)=2nH_{n-1} (x)$.
Can you help me?
AI: Let’s write $D = \frac{\operatorname{d}}{\operatorname{d}x}$, and take the derivative of $H_n$
$$
\begin{aligned}D H_n &= D \left( e^{x^2} D^n e^{-x^2} \right) \\
&= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} D^n \left( - 2 x e^{-x^2} \right) \\
&= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^n \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2} \\
& \left(\text{note that $D^k(-2x)=0$ for $k\ge 2$, so $\sum_{k=0}^n=\sum_{k=0}^1$}\right)\\
&= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^1 \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2}\\
&= {{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\underbrace{\binom{n}{0}}_1\underbrace{D^0(-2x)}_{-2x}D^{n}e^{-x^2}+\underbrace{\binom{n}{1}}_n\underbrace{D^1(-2x)}_{-2}D^{n-1}e^{-x^2} \right) \\
&= \color{red}{{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\color{red}{-{2 x D^{n} e^{-x^2} }}- 2 n D^{n-1} e^{-x^2} \right)
\end{aligned}
$$
So we have the rather simple end result
$$\frac{\operatorname{d}}{\operatorname{d}x} H_n(x)=2 n H_{n-1}(x)$$ |
H: Another question of finding eigenvalues with parameters
Let
$$A=\left(\begin{matrix}a&b&c\\a-s&b+s&c\\a-t&b&c+t\end{matrix}\right)$$
Find A's eigenvalues.
So, I did some row operations without changing the value of A's determinant, so that I got
$$A'=\left(\begin{matrix}a&b&c\\-s&0&0\\-t&0&t\end{matrix}\right)$$
But I still can't get a simple way for finding the eigenvalues. Are there more row operations I can do?
Thanks in advance for any assistance!
AI: According to Maple, the characteristic polynomial of $A$ is
$$\begin{align}
x^3 &{}+(-c-a-t-b-s)x^2 \\&{}+ (-a^2+2as+st+at+sc+tc+ab+bt)x \\&{}+ (tac-asc-2ast+scb-abt+a^2t-tsc-tcb),\end{align}$$ and it does not factor. This means the problem is pretty hopeless. The constant term is also clearly not $-\det(A')$, so either you mistyped the matrix $A$ in the first place, or you made errors in the row operations leading to $A'$ (but which would not lead to solving this problem anyway if done correctly, just to get the right constant term of the characteristic polynomial).
Added. With the now modified matrix, the simplest thing is to conjugate by $I_3-E_{1,2}-E_{1,3}$ (with $E_{i,j}$ an elementary matrix; conjugations means left-multiply by this matrix and right-multiply by the inverse $I_3+E_{1,2}+E_{1,3}$); this transforms $A$ into the similar matrix
$$
A'=\begin{pmatrix}a+b+c&b&c\\0&s&0\\0&0&t\end{pmatrix},
$$
which being triangular clearly has eigenvalues $a+b+c$, $s$, and $t$, and so has$~A$. |
H: Why is this an ellipse?
On a textbook, I've arrived at the following function:
$\displaystyle \phi(z)=\log{\frac{|z-\sqrt{(z²-1})|}{2}}$
and it says that the formula has a simple interpretation: the level curves of $\phi(z)$ are the ellipses with foci $-1, 1$. I know the problem is reduced to proving $|z-\sqrt{(z²-1})| = k$ is an ellipse, $k$ constant, but I don't know why this is true. So, my question is: Why is this an ellipse?
AI: Without caring about choosing a branch of the square root etc., when $z$ traverses a level curve of $\phi$, then
$$w = z-\sqrt{z^2-1}$$
traverses a circle. So let's identify the inverse mapping of $z \mapsto z - \sqrt{z^2-1}$:
$$\begin{align}
w = z - \sqrt{z^2-1} &\iff \sqrt{z^2-1} = z-w\\
&\Rightarrow z^2-1 = z^2-2zw+w^2\\
&\iff 2zw = w^2+1\\
&\iff z = \frac12\left(w+\frac1w\right).
\end{align}$$
Now write $w = r e^{i\varphi}$ to see that that maps circles to ellipses. |
H: Probability of rolling a cuboid dice
It's easy to count the probability of events on a regular dice because we know the probabilities ($P(1)$, $P(2)$, $P(3)$, $P(4)$, $P(5)$, $P(6)$) of all the basic outcomes ($P(i)=\frac{1}{6}$).
But... Is there any (simple) way how to determine the probabilities of basic outcomes of a cuboid dice?
Let's suppose for example a cuboid of sizes 1 cm, 1.1 cm, 1.2 cm...
AI: Diaconis, Holmes, and Montgomery have shown that when you look closely at the actual dynamics, taking angular momentum, etc., into account, even a coin toss is rather complicated. |
H: Divide $x=123456$ into three factors $x=uvw$ such that $uv^iw$ is divisible by 3
I have the problem of dividing the string 123456 into three factors uvw that such $uv^iw$ as a number is divisible by three, where $\left|uv\right|\le4$ and $\left| v\right|>0$, i.e. the factors u and v should together only be 4 digits long.
I found the answer but just by simply trying out different values. My knowledge in number theory is very limited, I basically just know that a number is divisible by 3 if it contains three in its prime factorization... And I just cannot explain why it is that if I e.g. divide 123456 such that u=12, v=34, w=56 it's the case that for all small values of i, it's the case that $uv^iw$ is divisible by 3: $uv^2w=121156656, uv^3w=123930456, uv^4w=12133633656$ are all divisible by 3, why is that the case?
AI: Shouldn't $v^i$ be repetition of $v$, not a power of the number $v$? That seems more likely since we are talking about strings. In that case $u=12,v=34,w=56$ is not correct. The classic divisibility rule says that a number is a multiple of $3$ if and only if the sum of the digits is. So as long as the sum of the digits in your string is a multiple of $3$, the whole number is. If the sum of the digits in $v$ is a multiple of $3$, repeating it will not change the status of the number. |
H: Countability of real numbers in $(0, 1)$
I've been reading a basic book on set theory for a while now. And recently ran into an uncountability of $(0, 1)$. I guess like almost everyone who tries to understand what he or she reads I tried to come up with a rule that matches every real number in $(0, 1)$ with a corresponding natural number to see why it is not trivially possible.
Well I came up with something similar to what's described here:
Proof that the real numbers are countable: Help with why this is wrong
(Finitely) decimal expressible real numbers between [0,1] countable?
...
I also recognised that my example doesn't take into account real numbers with infinite decimal representation. But here's the thing. Yes, I cannot write down a natural number that corresponds to $0.3(3)$, but for the same reason I cannot write down the natural number that corresponds to a natural number $33...$. This fact (as far as I know) doesn't make natural numbers uncountable.
If we accept $33...$ as a valid representation of the natural number that has all $3$s in it's decimal representation, will it make $(0, 1)$ countable?
Thanks.
AI: If you allow natural numbers to have infinitely many digits, then this collection is not countable anymore.
Yes, if you map a real number to its string of digits (even if you have two possible representations, e.g. $0.4\bar9=0.5$ you can always choose the finite one) then this is a well-defined injection. But you no longer inject the real numbers into a countable set. So you in fact changed the definition of countable.
So no, the real numbers are not countable. But they are of the same cardinality as the cardinality of all infinite sequences of digits, of natural numbers. |
H: Prove on residue theorem
I have try to use the equation
$$
Res(f;z_0)=\lim_{z\to z_0}\frac1{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]
$$
But very soon I stuck, is that a good way to solve it?
AI: A few hints:
1) Since you have a double pole at $z=z_0$, your $f$ has the form
$$f(z) = \frac{p(z)}{(z-z_0)^2 r(z)}$$
i.e., $q(z) = (z-z_0)^2 r(z)$, where $r$ is analytic at $z=z_0$.
2) Use the given expression for the residue with $m=2$. This means you will be taking the derivative of $p(z)/r(z)$.
3) Finally, you need things in terms of $q$, not $r$, you need to evaluate the second and third derivatives of $q(z)=(z-z_0)^2 r(z)$ at $z=z_0$. This will give you expressions for $r(z_0)$ and $r'(z_0)$ as needed if you did everything right in step 2). |
H: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
AI: since
$$a^2+1\ge 2a$$
so
$$a^2+b^2+c^2+3\ge a+b+c+(a+b+c)\ge a+b+c+3\sqrt[3]{abc}=a+b+c+3$$
so
$$a^2+b^2+c^2\ge a+b+c$$ |
H: Simple module and homomorphisms
Are the following statements equivalent?
i) $M$ is an $R$-simple module.
ii) Every $R$-homomorphism (nonzero) from $M$ to an $R$-module $N$ is a monomorphism.
iii) Every $R$-homomorphism (nonzero) from an $R$-module $N$ to $M$ is an epimorphism.
AI: The answer is yes.
Hint: You can easily see that considering the kernel, image resp. of an $R$-homomorphism. Conversely, if assumptions $ii)$ or $iii)$ hold for such $R$-homomorphisms, then $M$ has to be simple, where you can consider the possible submodules of $M$ to check. |
H: How to prove the compactness of the set of Hermitian positive semidefinite matrices
I am dealing with convex optimization problems. There are some useful theories for optimization problems where real-valued vector parameter, e.g., $x \in \mathbb{R}^n$, is considered. I manage to apply the theories to my semidefinite programming problem. I mean, I verify it by simulations. However, to prove the convergence of the algorithm analytically, I need to verify whether the set of Hermitian positive semidefinite matrices, where $\bf{X}=\bf{X}^{\rm H}$ and $\bf{X} \succeq 0$, is compact given that ${\rm trace}\{\bf{X}\} \leq 1$. Can anyone help me with this or at least point out a direction I should go for? Thank you very much in advance.
AI: Think of it as a subset of $\mathbb R^{n^2}$. You need to show that the set is closed and bounded. Closed is easy. For boundedness, show that each matrix entry is bounded by $1$. You know this is true for the diagonal entries (because they are all non-negative and add up to 1). Now use the fact that each $2\times 2$ submatrix $\left[\begin{matrix} a_{ii} & a_{ij} \\ a_{ij} & a_{jj} \end{matrix}\right]$ is positive semi-definite, and hence its determinant is non-negative.
Let me add one more thing. That the condition $X$ is positive semidefinite is closed under limits is perhaps most easily shown by noting that this is true if and only if all the diagonal minor matrices have non-negative determinant. |
H: how do we prove $p|q\cdot r \rightarrow p=q$ or $p=r$ (all primes)?
I know this is one of the most fundamental basis of arithmetic but I can't find the result by myself.
how do we prove $p|q\cdot r\rightarrow p=q$ or $p=r$? ($p, q, r$ being prime numbers)
AI: If $p\mid q$, we are done.
Suppose that $p\not\mid q$. Since $p$ is only divisible by itself and $1$, $\gcd(p,q)=1$. By Bezout's Identity, there are $a$ and $b$ so that
$$
ap+bq=1\tag{1}
$$
Multiply $(1)$ by $r$ to get
$$
(ar)p+b(qr)=r\tag{2}
$$
Since $p\mid qr$, there is a $k$ so that $kp=qr$. Using this in $(2)$ gives
$$
r=p(ar+bk)\tag{3}
$$
which says that $p\mid r$.
QED |
H: Probability question with trees and fruit using probability generating functions
Each year a tree of a particular type flowers once and the probability that it has n flowers is $(1-p)p^n$, $n=0,1,2...,$ where $0<p<1$. Each flower has probability $1/2$ of producing a ripe fruit, independently of all other flowers. Find the probability generating function for the number of ripe fruits produced by the tree. Hence find the probability that in a given year
(a) the tree produces $r$ ripe fruits.
(b) the tree had $n$ flowers if it produces $r$ ripe fruits.
I have called the number of flowers $N$ where it's generating function is $G_N(s)=\frac{1-p}{1-ps}$. If we say that the number of fruits coming from flower $i$ is $X_i$ where $X_i\sim$ Ber(1/2) then $G_X(s)=\frac{1}{2}-\frac{1}{2}s$. This should mean the generating function of $F$ (The number of fruit, i.e. $\sum_{i=1}^{N}X_i$ )is $G_N(G_X(s))=\frac{2(1-p)}{2-p-ps}$. I was wondering how I might get answers to (a) and (b) from this. Thank you for any help.
AI: Given that the tree produced $N=n$ flowers, the number of ripe fruits $R$ follows binomial distribution $\operatorname{Bin}\left(n,\frac{1}{2}\right)$. Hence the probability of having $r$ ripe fruits is
$$
\Pr\left(R=r\right) = \sum_{n=r}^\infty \Pr(N=n)\Pr(R=r\mid N=n) = \sum_{n=r}^\infty (1-p)p^n \binom{n}{r} \frac{1}{2^n}
$$
We are tasked with computing the probability generating function:
$$\begin{eqnarray}
\mathcal{P}_R(z) &=& \sum_{r=0}^\infty \Pr(R=r)z^r = \sum_{r=0}^\infty \sum_{n=r}^\infty\Pr(R=r\mid N=n) \Pr(N=n)z^r = \\ &=& \sum_{n=0}^\infty \Pr(N=n) \sum_{r=0}^n \Pr(R=r\mid N=n) \Pr(N=n)z^r \\
&=& \sum_{n=0}^\infty \Pr(N=n) \mathcal{P}_{R\mid N=n}(z) = \sum_{n=0}^\infty (1-p) p^n \left(\frac{1+z}{2}\right)^n = \frac{1-p}{1-p \frac{1+z}{2}} \\ &=& \frac{2-2p}{2-p - p z} = \sum_{r=0}^\infty \frac{2-2p}{2-p} \left(\frac{p}{2-p}\right)^r z^r
\end{eqnarray}
$$
Hence
$$
\Pr(R=r) = \frac{2-2p}{2-p} \left(\frac{p}{2-p}\right)^r
$$
For the second question we can use Bayes theorem:
$$\begin{eqnarray}
\Pr\left(N=n\mid R=r\right) &=& \frac{\Pr\left(N=n, R=r\right) }{\Pr\left(R=r\right)} = \frac{ \Pr\left(R=r\mid N=n\right) \Pr(N=n) }{\Pr\left(R=r\right)} \\
&=& \frac{ \binom{n}{r} 2^{-n} (1-p) p^n}{\frac{2-2p}{2-p} \left(\frac{p}{2-p}\right)^r }
\end{eqnarray}
$$ |
H: How to solve for $g(x)$ in $f(x)=\int_a^b g(x) dx$
$$f(x)=\int_a^b g(x) dx$$
Do you just take the derivative and evaluate $f(x)$ from $a$ to $b$? Sorry if it seems obvious, but I have never done this before.
AI: $$\text{If}\;\;F(x) = \underbrace{\int g(x) \,dx}_{\text{indefinite integral}}\;\text{ then }\;\underbrace{f = \int_a^b g(x) \,dx = F(a) - F(b)}_{\text{evaluation of definite integral}},$$ so $f$ as defined will be a constant $C$, and not a function of $x$. As a result, taking the derivative of $f = C$ would then be $(C)' = 0$.
You can not recover $g(x)$ when given only the evaluation of its integral between the bounds $a$ and $b$. |
H: Definite Integral questions
Evaluate the definite integral of the function.
$$ \int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx $$
My solution was :
$$ \cos \left(\frac{\pi}{2}\right) - \cos\left(\frac{-\pi}{4}\right)$$
$$ \frac{-\sqrt2}{2} $$
But the answer in the book is $ 2 - \frac{\sqrt2}{2} $. So what's wrong with my solution ?
AI: You seem to have neglected the absolute value signs. The derivative of $\cos x$ is $\sin x$, not $|\sin x|$. So you cannot apply the fundamental theorem of calculus the way you did. $\int_a^b f(x)\;dx = g(b) - g(a)$ only when $f$ is the derivative of $g$, and that is not the case in your proposed solution. You have instead calculated $\int_{-\pi/4}^{\pi/2} \sin x\;dx$, which is different.
The usual technique to deal with this kind of problem is to divide the interval of integration into pieces on which the integrand is simpler. For example, $|x| = x$ when $x$ is positive and $-x$ when $x$ is negative, so it is easy to integrate on intervals where its argument doesn't change sign; on such intervals $|f(x)|$ can be replaced by $f(x)$, if $f(x)$ is positive, and by $-f(x)$, if $f(x)$ is negative.
In this case we divide the interval of integration $\left[-\frac\pi4, \frac\pi2\right]$ into two parts, $L = \left[-\frac\pi4, 0\right]$
and
$R = \left[0, \frac\pi2\right]$. On interval $L$, $\sin x$ is everywhere negative, and $|\sin x| = -\sin x$; on interval $R$, $\sin x$ is everywhere positive, and $|\sin x| = \sin x$. This allows us to get rid of the absolute value signs, as follows:
$$\begin{align}
\int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx &=
\int_{-\pi/4}^0 \; |\sin x| \; dx
+
\int_0^{\pi/2} \; |\sin x| \; dx\\
& =
\int_{-\pi/4}^0 \; -\sin x \; dx
+
\int_0^{\pi/2} \; \sin x \; dx
\end{align}
$$
Now we can apply the fundamental theorem of calculus to the two terms on the right:
$$
\begin{align}
\hphantom{\int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx} &=
\cos x\left.\right|_{-\pi/4}^0 +
-\cos x\left.\right|_0^{\pi/2} \\ &=
\left(1 - \frac{\sqrt2}2\right) - \left(0-1\right) \\&=
2-\frac{\sqrt2}2
\end{align}
$$ |
H: Greatest common divisor of $2 + 3i$ and $1-i$ in $\mathbb{Z}[i]$
Greatest common divisor of $2 + 3i$ and $1-i$ in $\mathbb{Z}[i]$
Here is my attempt at solving this using a generalized Euclid's Algorithm. Does it look alright?
Step 1
$2 + 3i = M(1-i) + N$
$$\frac{2 + 3i}{1-i} = \frac{2 + 3i}{1-i}\frac{1 + i}{1 + i}$$
$$= \frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5i}{2}$$
Now, this point in the complex plane is equidistant from the nearest four $a + bi \in \mathbb{Z}[i]$, so we can pick any of them to be $M$, I will pick $2i$ for the sake of simplcitiy.
$N = 2 + 3i - M(1 - i) = 2 + 3i - (2i)(1 - i) = 2 + i - 2 = i$
So at the conclusion of step 1 we have
$$2 + 3i = 2i(1-i) + i$$
Step 2
$1 - i = M(i) + N$
$\implies M=-1$ and $N = 1$
So at the conclusion of step 1 we have
$$1 - i = (-1)(i) + 1$$
Step 3
$i = i(1) + 0$
A remainder of $0$ means we conclude the algorithm. The last non-vanishing remainder was $1$, so that means the GCD of $2 + 3i$ and $1-i$ is $1$?
AI: I would solve this be looking at the modulus square of each number and taking the common divisor of that. in this case $|2+3i|^2 = 4+9=13$ and $|1-i|^2=1+1=2$ the modulus square of any common divisor must therefore divide 2 and 13 so it must be 1. Therefore the only common divisors are i,-1,i,-i |
H: Orthogonal Projections in Hilbert space
I am stuck with the following exercise about projections in Rudin 12.26.
Let $H$ be a Hilbert space $P,Q\in B(H)$ self-adjoint projections (A projection has the property that $P^2=P$), then the following are equivalent.
(a) $P\geq Q$
(b) $R(P)\supset R(Q)$
(c) $PQ=Q$
(d) $QP=Q$
Theorem 12.4 says the following about a projection : $P$ self-adjoint is equivalent with $P$ is normal, with $R(P) = N(P)^{\perp}$, and with $(Px,x) = \left\| Px\right\|^2$.
If we just assume (a).$ P\geq Q$ then $(Px,x)\geq (Qx,x)$ which is then the same as $\left\|Px\right\|^2\geq \left\|Qx\right\|^2 $. Can we conclude (b)? Implementing my intuition about projections in $\mathbb{R}^n$ I can see why these things would be equivalent, but working straight from the definitions I can't really work it out.
Some ideas or suggestions?
AI: The equivalence $(b)\iff (c)$ is true in general for "projections" $P, Q$ (i.e. idempotency $P^2=P$ and $Q^2=Q$ suffices). Hint: $PQ=Q \iff (I-P)Q=0$ and $N(I-Q)=R(Q)$.
The equivalence $(c)\iff (d)$ is trivial. Hint: adjoint.
It remains to show that $(a)$ is equivalent to, say, $(b)$. You were off to a good start. One way to do that, is to prove the following useful fact. The equivalence follows immediately given what you already observed.
Fact If $P$ is a self-adjoint projection, then $\|Px\|\leq \|x\|$ for every $x\in H$ and $$R(P)=\{x\in H\;;\; \|Px\|=\|x\|\}$$
Hint Pythagoras. |
H: A basic theorem on field isomorphisms
I'm reading A Book Of Abstract Algebra by Charles C. Pinter. On page 314 is the following theorem:
Let $h:F_1\to F_2$ be an isomorphism, and let $p(x)$ be irredicible in $F_1[x]$. Suppose $a$ is a root of $p(x)$, and $b$ is a root of $h(p(x))$. Then $h$ can be extended to an isomorphism $\bar h:F_1(a)\to F_2(b)$, and $\bar h(a)=b$.
By $h(p(x))$ the author is referring to the obvious extension of $h$ to $F_1[x]$ (applying $h$ to each coefficient).
Noting that each element of $F_1(a)$ can be expressed in a unique fashion as $\sum c_i a^i$, the author defines $$\bar h(\sum c_i a^i)=\sum h(c_i) b^i$$
This seems to me to be well-defined, always a homomorphism, and bijective as long as $a$ has the same degree over $F_1$ as $b$ has over $F_2$. The author's requirement that $b$ be a root of $h(p(x))$ seems unnecessarily strong.
Question: Is this theorem true so long as $a$ has the same degree over $F_1$ as $b$ has over $F_2$?
AI: No, the requirement that $b$ be a root of $(h(p))(x)$ cannot be dropped. Indeed,
$$0= h(0) = h(p(a)) = (h(p))(h(a)) = (h(p))(b)$$
shows that $b$ is a root of $h(p)$.
Vaguely speaking, a map between two objects must preserve existing relations. The condition that the relation $p(a)=0$ be preserved by $h$ is precisely that $h(a)$ be a root of $h(p)$. |
H: How to show that $\operatorname{Spec}(A)=\bigcup_{i=1}^n D(g_i)$ implies that $(g_1, \ldots, g_n)=A$?
$\newcommand{\Spec}{\operatorname{Spec}}$
Let $A$ be an algebra and $\Spec(A)$ the scheme consisting of all prime ideals of $A$.
How to show that $\Spec(A)=\bigcup_{i=1}^n D(g_i)$ implies that $(g_1, \ldots, g_n)=A$? By definition, $D(g_i)=\{p \in \Spec(A): g_i \not\in p\}$. Let $a \in A$. Then $a \in p$ for some $p \in \Spec(A)$. Therefore $a \in p \in D(g_i)$ for some $i$. Hence $g_i \not\in p$. But how could we show that $a \in (g_1, \ldots, g_n)$? Thank you very much.
AI: Hint: every proper ideal of $A$ is contained in a prime ideal.
Apply this to the ideal $I=\left<g_1, \dots, g_n\right>$. What does the condition that the $D(g_i)'s$ cover $\text{Spec }A$ tell you? |
H: Notation of a function that maps a random element
Let there be a functions $f$ and $g$ such that,
$$f:A \times B \mapsto \Re$$
$$g: B \mapsto A$$
where $\forall b \in B$, $g(b)$ is some $a$ such that, $\forall a' \in A, f(a,b) \geq f(a',b)$. (This is because there could be many elements in $A$ that suffices $\forall a' \in A, f(a,b) \geq f(a',b)$)
My question is, is there a simple notation to write this function $g$?
e.g. I can write $g$ as a relation as,
$$g(b) = \{a|\forall a' \in A, f(a,b) \geq f(a',b)\}$$
But $g$ is a function that chooses one element from the set $\{a|\forall a' \in A, f(a,b) \geq f(a',b)\}$. The mechanism of choosing an element from the set is random.
--EDIT--
I said random to mean that it could be any of those values. No statistics involved. What I have is a computer program that computes the maximum (using a linear searching method) as described and returns (i.e. outputs) it. I want to explain it mathematically. Pardon me since I am a novice to mathematics.
AI: I think it would be best if you would define $g$ in plain words, e.g. first define $$A_b = \{a \mid \forall a' \in A.\ f(a,b) \geq f(a', b)\}$$ and then say that $$g : B \to A \quad \quad \text{ or } \quad \quad g : B \to \bigcup_{b \in B}A_b$$ is a function that assigns any $b \in B$ a random element from $A_b$ (I guess that your $A_b$ is finite and random means uniform distribution).
However, be aware that this isn't very formal. The problem isn't that its text (there is no problem with text-only definition being formal), but the notion of randomness. To speak about random events and probability you need to define a probability space (usually denoted $\Omega$) and your function would have to depend on it, that is, in math you could have one function
$$g : \Omega \times B \to A$$
or a set
$$g_b : \Omega \to A_b$$
for $b \in B$. To emphasize the difference, functions are deterministic by definition, i.e. for same inputs you get the same outputs. The usual computer-science $\mathtt{rand}$ is not a function, because for same input it returns different outputs (the same goes for things which read from $\mathtt{stdin}$, etc.). Therefore, to simulate the probabilistic behavior, you introduce a dependency on the probability space. That might get messy (I am not saying that it will, only that it might), so I would advise you to avoid formality in this case (unless you really need it), just make the textual description clear enough to get the meaning through.
Some other thoughts, I'm not sure, but it might be easier for you to define a relation $\leq_b$ by $x \leq_b y \iff f(x,b) \leq f(y,b)$ instead of writing $f(x,b)$ everywhere. On the other hand, the construction of $A_b$ suggests that $\leq_b$ is not a partial order, as so consider using some other symbol, as $\leq_b$ might be misleading.
I hope that helps $\ddot\smile$ |
H: Generator for the ideal $I + J$ where $I = (2 + 3i)$ and $J = (1 - i)$
On a related question I calculated the GCD of $I = (2 + 3i)$ and $J = (1 - i)$ to be $1$.
Now I know that $\mathbb{Z}[i]$ is a principal ideal domain. And I also know that the greatest common divisor of two elements $a, b$ in a P.I.D. generates the ideal $(a) + (b)$, yes?
So does that mean that $1$ in this case generates the ideal $I + J$? And hence that the ideal $I + J$ is actually equal to the ring $\mathbb{Z}[i]$ itself? Have I got all my assumptions correct here?
AI: Yes, in this case $I+J = \mathbb{Z}[i]$ itself for the reasons you specified. You could also see this explicitly by finding $a = a_1 + a_2 i$ and $b = b_1 + b_2i$ in the ring such that $a(2 + 3i) + b(1-i) = u$ for some unit $u$ in the ring. Indeed, for $a = -1$ and $b = 3i$ you can see that
$$
-(2+3i) + 3i(1-i) = (-2-3i) + (3i+3) = 1 \in I+J.
$$ |
H: How to show that h is constant by using **Liouville's theorem**
Let $h$ be an entire function.
$\exists$ some $R \gt 0$ and $z_0\in \Bbb C$ s.t. open ball $B_R(z_0)$ isnt in the range of $h$. $B_R(z_0)\cap h(\Bbb C) \not = \emptyset{}{}$
How to show that h is constant by using Liouville's theorem
I dont have any idea to prove this. Honestly, I would like to write but I cannot. Please help me. In fact, really, this question seems so difficult, so I asked. Thank you.
AI: Define $g(z)=1/(h(z)-z_0)$. $g$ is obviously entire, and your condition for the range of $h$ gives you that for every $z\in\mathbb{C}$ you have
$$|g(z)|=\frac{1}{|h(z)-z_0|}\leq \frac1R,$$
hence $g$ is bounded. Liouville's theorem now tells us that $g$ is constant, and so the same must be true for $h$. |
H: Help on how to read and understand a question on vector space and subspace
V is a vector space, S is its subset. Determine if the following subsets are its subspace:
$$\begin{align} &S_1=\{(x_1,x_2,...,x_n)|x_1+x_2...+x_n=0\};V=\Bbb R^n \\
&S_2=\{(x_1,x_2,...,x_n)|x_1x_2=0\};V=\Bbb R^n \end{align}$$
I'm not sure I understand how to read this. This is how far I got, for $S_1$:
$u=(a_1,...,a_n) \\v=(b_1,...,b_n) $
And to determine whether or not its a subspace we use:
$\alpha u+ \beta v \in S : \forall \alpha,\beta \in S : \forall u, v \in S$
Now I don't know how to use the right side of the given subset here ($|x_1+x_2...+x_n=0$). Same with $S_2$.
Any insight would be appreciated.
AI: For $S_1$: Note $S_1 \neq \varnothing$ because $(0, \dotsc, 0) \in S_1$. Next, choose two elements $x = (x_1, \dotsc, x_n)$ and $y = (y_1, \dotsc, y_n)$ in $S_1$. Then $x + y \in S_1$, because
$$ \sum_i (x_i + y_i) = \sum_i x_i + \sum_i y_i = 0. $$
Similarly, $\alpha x \in S_1$ for any $\alpha \in \mathbb{R}$ because
$$ \sum_i \alpha x_i = \alpha \sum_i x_i = 0. $$
Thus, $S_1$ is a subspace of $V$.
I claim that $S_2$ is not a subspace of $V$ because $e_1 = (1, 0, \dotsc, 0) \in S_2$ and $e_2 = (0, 1, \dotsc, 0) \in S_2$, but $e_1 + e_2 = (1, 1, \dotsc, 0) \notin S_2$. |
H: When should we take direct limit and when should we take inverse limit?
We know that we can take direct limit for a direct system and inverse limit for an inverse system. For example, when can defined the stalk of a presheaf $\mathcal{F}$ on a topological space $X$ at a point $P\in X$ by $$ \lim_{\rightarrow_{p \in U \text{ open}}} \mathcal{F}(U). $$
It seems that we get something small. We can take inverse limit of the inverse system $\{k[x]/(x)^n, n=0, 1, 2, \ldots\}$. Then we obtain $k[[x]]$ which is larger than $k[x]$. It seems that by taking inverse limit we add something to $k[x]$. It seems that taking inverse limit is to add something and taking direct limit is to get something small. Is this correct? If a system is both direct system and inverse system, how do we know that we need to take direct limit or inverse limit? Thank you very much.
AI: This is not a full answer, but some thoughts about direct and inverse limits.
I always think of the direct limit of a system $(A_i,\varphi_i)_{i \in I}$ as a union $A = \bigcup A_i$ but with compatible transition maps, i.e. an element is in this union if and only if it is compatible by those maps and the canonical injections $A_i \longrightarrow A$.
In contrast to inverse limit, where I think of them not as the union but as the cartesion product (or to be exact: a subset of the cartesian product). There you don't have canonical injections, but canonical projections $A \longrightarrow A_i$ and the transition maps of the inverse system have to be compatible with this canonical projections.
In conclusion, for me a direct limit is a really tiny subset of the union of all spaces and the inverse limit is a really big subset of the cartesian product, roughly speaking.
I hope that intuition helped you! |
H: Find the rest of the division when $23^{84292}$ is divided by $7$, is the procedure and the result correct?
I want to know if the procedure I have followed in order to get the result for the next problem is correct. The problem is this:
Find the rest of the division when $23^{84292}$ is divided by $7$.
As $7$ is a prime number and $23$ is not divided by $7$ Small Fermat's Theorem can be used. So, I know,
$23^{6}\equiv1\pmod7$ and $23^{6\cdot 14048} = 23^{84288} \equiv1\pmod7$
Then,
$23^{84288}\cdot 23^{4}=23^{84292}$
And,
$23 \equiv 2\pmod7$
So, finally,
$23^{84288}\cdot 23\cdot 23\cdot 23\cdot 23=23^{84292} \equiv 16\pmod7$
So,
$23^{84292} \equiv 2\pmod7$
Is the procedure and the result all correct?
AI: Yes, that is correct. $\tiny{\text{(Converting to an answer.)}}$
By the way, you could have saved a little bit of work by noting $ 23 \equiv 2 \pmod 7 $ from the start. This way, you could write $2$ for every $23$. |
H: Finding the solutions of $\cos (x) +x = a$
What is the approach to finding the solutions of the following function? I was not able to analytically resolve the solutions - but rather resorted to a graphical approach.
$$\cos (x) + x = 1$$
or in general where $a$ could be any real number,
$$\cos(x) + x = a.$$
EDIT:
Or when more than one solution exists.
$$\cos (\pi x) + x = a.$$
AI: Numerical root finding is a perfectly good way to go. If you want to solve
$\cos (x) + x = a$
you can do a little shuffling to get
$a - \cos(x) = x$
Since $\cos(x)$ will be a number between $-1$ and $1$, you know that that $a - \cos(x)$ will be between $a-1$ and $a + 1$. That means that the $x$ you're looking for must be in the interval $[a-1, a+1]$. Once you know that, a few steps of bisection followed by Newton's method should get you a very accurate answer very fast. |
H: Matrix question regarding symmetric
I need help with the following problem
Express the matrix $$B=\begin{pmatrix} 2 &-2 &-4 \\
-1 &3 &4\\ 1 &-2 &-3 \end{pmatrix}$$ as the and sum of a symmetric and a skew symmetric matrices.
AI: In general,
$$\mathbf{M} = \dfrac{\mathbf{M} + \mathbf{M}^t}{2} + \dfrac{\mathbf{M} - \mathbf{M}^t}{2}.$$
The first term is a symmetric matrix; the second is skew-symmetric.
This is an instance of a much more general theorem. The "transpose" operation, done twice, is the identity. Such operations are called "involutions". And involutions all "split" their domains this way:
If you call that involution (think of "transpose") $T$, we have $T^2 - I = 0$. That means that $+1$ and $-1$ are the only eigenvalues for $T$,. Furthermore, the domain of $T$ splits into a direct sum of the $+1$ and $-1$ eigenspaces. That's the larger theorem I was referring to.
In fact, the solution above (for transpose) generalizes as well: for any vector in the domain of $T$, we have
$$
v = \frac{1}{2} (v + T(v)) + \frac{1}{2} (v - T(v))
$$
which splits $v$ as a sum of a vector that's invariant under $T$ and one that negates when $T$ is applied, i.e., it splits $v$ as a linear combination of a multiple of a $+1$ eigenvector and a multiple of a $-1$ eigenvector.
Notice that the fraction $\frac{1}{2}$ appears here. That means that this trick, as written, only works over fields of characteristic not equal to two. You might ask whether it's still true in characteristic two. You could try some examples to find out.
The theorem can be applied to show that every function on the reals is the sum of an even and an odd function, that every complex number is the sum of a real and a pure imaginary (OK, that's pretty trivial), etc. |
H: Rows of Change of Basis Matrix
Maybe a stupid question, but since I don't find a confirmation to my doubt on the Internet, I'll also ask here.
To change basis from $A$ to $B$ we use a matrix whose columns are the basis vectors of $A$ expressed in the new basis $B$. But we can also say that its rows are the basis vectors of $B$ expressed in the old basis $A$, can't we?
AI: No. That's not in general true. If it were, then the change-of-basis matrix from $B$ to $A$ would be the transpose of the c-o-b matrix from $A$ to $B$. It's not, in general. Instead, it's the inverse of that matrix.
If the change of basis matrix is orthogonal, then the inverse is the transpose, in which case your statement about the rows is correct. |
H: Is an integer a sum of two rational squares iff it is a sum of two integer squares?
Let $a\in \mathbb Z$. Is it true that $a$ is a sum of two squares of rational numbers if and only if it is a sum of two squares of integers?
I came to face this problem while dealing with quaternion algebras.
AI: Yes, if an integer is the sum of two rational squares then it is a sum of two integer squares. For suppose that $n\ge 1$ is the sum of two rational squares. Without loss of generality we may assume that $n=\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2$, where $a,b,c$ are integers with $c\ge 1$.
Then $nc^2=a^2+b^2$.
Any prime of the form $4k+3$ in the prime power factorization of $a^2+b^2$ occurs to an even power.
It follows that the same is true of $n$, and therefore $n$ is a sum of the squares of two integers.
Remark: We have used without proof the fact that if $m$ is a positive integer, then $m$ is a sum of two squares if and only if every prime of the form $4k+3$ in the prime power factorization of $m$ occurs to an even power. This is an elementary but non-trivial result that goes back to Fermat. It is most easily proved by using Gaussian integers.
The idea of the proof can be used to show directly that a sum of two rational squares is the sum of two integer squares. But there is little saving. |
H: What is the difference between Symmetric vs Skew Symmetric?
I want to know the difference between Symmetric Symmetric vs Skew Symmetric?
AI: A symmetric matrix satisfies $A^T = A$
A skew-symmetric matrix satisfies $A^T = -A$.
Additionally, it is a fact that every matrix can be written as the sum of a symmetric matrix and a skew-symmetric matrix:
$$
A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)
$$
then $B = \frac{1}{2}(A + A^T)$ is symmetric since
$$
B^T = \frac{1}{2}(A + A^T)^T = \frac{1}{2}(A^T + A) = B
$$
and $C = \frac{1}{2}(A - A^T)$ is skew-symmetric:
$$
C^T = \frac{1}{2}(A - A^T)^T = \frac{1}{2}(A^T - A) = -C
$$ |
H: Calculate the probability that $x \lt 5$ given Poisson distribution states the mean is $6$
Two grocers agree that the daily demand for a particular item has Poisson distribution. However, grocer $A$ claims that the mean demand is $3$ items per day, while grocer $B$ claims that the mean demand is $6$ items per day.
They agree to resolve the disagreement by observing the demand on one particular day: $B$ agrees to
accept $A$’s claim if the observed demand is $4$ or less, and $A$ agrees to accept $B$’s claim if the observed demand
is $5$ or more.
(a) Calculate the probability that $A$’s claim is accepted when, in fact, $B$’s claim is correct
AI: Let $X$ be the observed demand. The problem states that $X$ is Poisson distributed random variable with mean of $6$ items per day. Denote $X(\omega)$ the actual observed value of the demand, also known as realization of the random variable $X$.
The A's claim is accepted if $X(\omega) \leqslant 4$. The probability of this event is
$$
\Pr\left(X \leqslant 4\right) = \sum_{n=0}^4 \Pr(X=n)
$$
Can you finish this off? |
H: $L^p $ is not uniformly convex for $p=1, \infty$
As the title says: how can we prove that $L^p(\mathbb R^n)$ is not uniformly convex for $p=1$ and $p=\infty$.
Does anyone knows a counter-example for the cases $ p=1$ and $ p =
\infty$ for the space $L^p( \mathbb R^n)$ ?
I am studying N.L. Carothers book "A short course to Banach space theory".
Any help?
AI: For $p=\infty$ and $n=2$, consider $\mathbf{x}=(1,1)$ and $\mathbf{y}=(0,1)$.
You might also make use of the fact that uniformly convex Banach spaces are reflexive.
Edit: As I was typing up a more detailed answer, I found the first comment to this answer, which puts things more succinctly than I was going to do (and saves me a bit of typing). |
H: Extending exponentiation to reals
I've been reading through a course on exponential functions, starting from integer-valued exponents to rational ones as in: $x^r$ from $r\in \Bbb{N}$ to $\Bbb{Z}$, and combining them to rigorously construct for $r\in\ \Bbb{Q}$. Still, this book adresses high-schoolers, and therefore "summons" an extension of the exponential notation for real exponents. Is there any formal basis for this extension? Could it be related to the density of $\Bbb{Q}$ in $\Bbb{R}$ ? Or the limit of a sequence $(x^{(r_n)})$ where $(r_n)_{n=1}$ is a sequence of rationals that converge to $r$ a non-rational exponent?
AI: The answer is: yes, it is related to both those things. Given any real $\alpha,$ the density of $\Bbb Q$ in $\Bbb R$ allows us to find an increasing sequence $(r_n)_{n=1}^\infty$ of rational numbers converging to $\alpha$. Depending on which positive $x$ we are considering, the sequence $(x^{r_n})_{n=1}^\infty$ will be increasing, decreasing, or constant. In the last case, the sequence readily converges, and its limit is defined to be $x^\alpha.$ In the first two cases, we note that the sequence is bounded, so by completeness of $\Bbb R$, has a least upper/greatest lower bound, one of which will be the limit of the sequence, and in either case, we define $x^\alpha$ to be the limit of the sequence.
Showing that our choice of sequence doesn't matter (so that $x^\alpha$ is well-defined in this fashion) is a bit trickier, and likely beyond the scope of your course, but that's the idea. |
H: Functions on finite metric spaces
Is a function f from a finite metric space M to itself always continuous? I have tried proving it, but I have gotten stuck. Any help would be appreciated.
AI: Hint: If $M$ is a finite metric space, then which subsets of $M$ are open? |
H: Trouble with l'Hôpital's rule for $\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$
This is an assignment and I am stuck:
Find the limit, whether finite or infinite, or indicate that the limit does not exist. Use l'Hôpital's Rule if appropriate.
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}.$$
When I try l'Hôpital's rule I get $\frac 8{10}$. Can someone tell me why this isn't right?
AI: Original posted question:
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\sin x} = \lim_{x\to 0} \frac{(4x + 4\sin x)'}{(10x + 10 \sin x)'} = \lim_{x\to 0} \frac{4+4\cos x}{10 + 10\cos x} = \frac {4 + 4}{10 + 10} = \frac {8}{20} = \frac 25$$
Since you meant to post $$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$ note that in this case, l'Hôpital is not applicable (the limit does not at first evaluate to an indeterminate limit). Nor would we want to use it! It is easily solved by evaluating immediately:
$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x} = \frac{ 0 + 0}{0 + 10(1)} = \frac {0}{10} = 0$$
IMPORTANT TO REMEMBER: We apply l'Hôpital's rule if and only if a limit evaluates to an indeterminate form. That bold-face link will take you to Wikipedia's concise list of "what counts" as an indeterminate form, and why. |
H: Does this packing problem even have an optimal solution?
Under this answer, user Bruno Joyal asks:
This might be a naive question, but... how do we know there is a best possible solution?
I (but that's just me) assume that he might be thinking of a logical possibility that there is a sequence of ever better solutions, but that the limit itself is not a solution. But, whether he is or not doesn't quite matter. So, here goes...
Can it be proven that there exists an optimal solution?
How big a square do you need to hold eleven little squares?
We don't even know if this is the best possible [solution.]
YouTube
AI: The configuration space can be represented as a compact set, specifying for each square the position of one corner and the angle of rotation, as well as the size of the big square (which can be assumed to be bounded by some ridiculously large constant). The constraints are all of the form $f(\omega) \ge c$ where $\omega$ is the configuration and $f$ is a continuous function, and the objective (the size of the big square) is a continuous function. So yes, the minimum does exist. |
H: Find the Second Point Of a 3D Line Segment Given Some Information
Given this information about a 3D line segment:
XYZ of the first point
Length of the line
Pitch and yaw angles
How do I determine the XYZ of the second point?
I know how to do this with a 2D segment described by one angle:
point2x = point1x + sin(angle) * length
point2y = point1y + cos(angle) * length
But given a 3D segment described by two angles, I am stuck.
AI: With pitch angle $p$ and yaw angle $y$ (in radians) and distance $r$, your displacement will be
$\mathbf{R} = \begin{bmatrix}
\cos(p) & 0 & \sin(p) \\
0 & 1 & 0 \\
-\sin(p) & 0 & \cos(p)
\end{bmatrix}
\begin{bmatrix}
\cos(y) & -\sin(y) & 0 \\
\sin(y) & \cos(y) & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
r \\
0 \\
0
\end{bmatrix}
$.
Then your new position from $\mathbf{X} = \begin{bmatrix}
x \\
y \\
z\end{bmatrix}
$ is $\mathbf{X+R}$. |
H: For fractional ideal, why $AB=R$ implies $B=A^{-1}$?
Let $A,B$ be two fractional ideals of $R$ (an integral domain). Could anyone tell me why $AB=R$ implies $B=A^{-1}$?
AI: Also for the other readers: $A^{-1}$ is the largest fractional ideal with the property that $A A^{-1} \subseteq R$. Explicitly, we have $A^{-1} = \{x \in K : xA \subseteq R\}$, where $K$ is the field of fractions of $A$.
Thus, $AB=R$ clearly implies $B \subseteq A^{-1}$. Conversely, we have
$$A^{-1} = A^{-1} (A B) = (A^{-1} A) B \subseteq R B = B.$$
This proves $A^{-1} = B$. All this works for any integral domain $R$. |
H: Calculating limit of function
To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
AI: Using Prosthaphaeresis Formulas,
$$\cos(\sin x)-\cos x=2\sin\frac{x-\sin x}2\sin\frac{x+\sin x}2$$
So, $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin\frac{x-\sin x}2}{\frac{x-\sin x}2}\frac{\sin\frac{x+\sin x}2}{\frac{x+\sin x}2}\cdot\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}x\cdot\frac14$$
We know, $\lim_{h\to0}\frac{\sin h}h=1$
$$\text{Apply L'Hospital's Rule on }\lim_{x\to0}\frac{x-\sin x}{x^3}$$
$$\text{and we get }\lim_{x\to0}\frac{x+\sin x}x=1+\lim_{x\to0}\frac{\sin x}x$$ |
H: Is matrix multiplication really a group operator?
A group has an operation that can be performed over ANY two elements in a set. Given that an $n \times m$ matrix can only be multiplied by an $m \times o$ matrix, doesn't that mean that matrix multiplication can't be a group operator except for sets of commonly sized square matrices?
I ask this because it's called linear algebra yet this aspect seems inconsistent with groups.
AI: Usually, when we talk about groups of matrices, we talk about square matrices, as only then can we be guaranteed that $AB$ and $BA$ are valid operations. In fact, if we refer to matrices as a group, we talk about invertible matrices, and we usually write it as $GL_n(\mathbb{F})$, or the general linear group of $n \times n$ matrices over some field $\mathbb{F}$.
The set of all matrices does not form a group under matrix multiplication or addition. |
H: Why is $ \hbox{Ext}_R^* (M,M) = H^*(\hbox{Hom}_R^*(P^*,P^*))$?
Let me first fix some notation and conventions.
Let $ R$ be a ring and $ M$ a left $R$-module. Given chain complexes $P^*$ and $Q^*$ in $R$-mod, define $ \hbox{Hom}^*_R(P^*,Q^*)$ to be the graded space such that $ \hbox{Hom}_R ^i (P^*, Q^*) = \bigoplus_n \hbox{Hom}_R(P^n, Q^{n-i})$. $\hbox{Hom}^*_R (P^*,Q^*)$ has a differential with degree $+1$. Now let $ P^*$ be a projective resolution of $M$.
Why is $ \hbox{Ext}^*_R (M,M)= H^* ( \hbox{Hom}_R^* (P^*, P^*))$?
I've seen this result stated in a few places but I dont see why it's true. I think I could believe that $ \hbox{Ext}^*_R (M,M) = H^* ( \hbox{Hom}_R^*(P^*, M))$. But why is there a projective resolution in the second slot of the Total Hom and not an injective resolution?
AI: They key here is that the differential of $\hom^\bullet(P_\bullet, Q_\bullet)$ is designed so that
$$H^i(\hom^\bullet(P_\bullet, Q_\bullet)) = \hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i])$$
where $\mathsf{K}(R)$ is the homotopy category (you might want $i$ instead of $-i$, it depends on your indexing conventions and I can't remember what you should get in this case).
If $P_\bullet$ is a projective resolution then it is colocal with respect to acyclic complexes. This is what Daniel Murfet calls a hoprojective resolution (google that term to find his notes on derived categories) and it has the property that
$$\hom_{\mathsf{K}(R)}(P_\bullet, Q_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, Q_\bullet[-i])$$
where $\mathsf{D}(R)$ is the derived category. Now if $Q_\bullet = P_\bullet$ then it is quasi-isomorphic to $M$ as a complex ($M$ as a complex has an $M$ in the zeroth position and has $0$'s elsewhere). Quasi-isomorphisms in $\mathsf{D}(R)$ are actually isomorphisms so
$$H^i(\hom^\bullet(P_\bullet, P_\bullet)) = \hom_{\mathsf{D}(R)}(P_\bullet, P_\bullet[-i]) = \hom_{\mathsf{D}(R)}(P_\bullet, M[-i]) = H^i(\hom^\bullet(P_\bullet, M))$$
Now just check that $\hom^\bullet(P_\bullet, M)$ when $M$ is considered a complex is the same as $\hom(P_\bullet, M)$ when $M$ is merely a module, and the homology of that last thing becomes $\mathrm{Ext}^i(M, M)$.
Edit: Just fyi, my experience is with dg-algebras in characteristic $2$, so forgive me if I've left out any necessary $\pm1$'s, because they don't matter to me! |
H: Proof related with mathematical induction
I tried to prove this claim using mathematical induction.
$ a^2 + 15a + 5 ≤ 21 a^2 $ $\;\; ∀a∈\mathbb Z^+$
The way is as the following:
Basis: for a = 1 is true since 21 = 21
Inductive step: If $P(t)$ then $P(t+1)$ should be proved , so it goes
$ t^2 + 15t + 5≤21t^2 $
add $2t + 16$ to both sides to obtain $t^2 + 17t+21≤21t^2+2t+16$
From here it follows that $2t+16≤21t^2 $ $ ∀t∈Z^+$ So I can replace it with $42t^2+21$
i.e. $(t+1)^2+15(t+1)+5≤21(t+1)^2$
Is this way correct? If not can anyone suggest any other alternative?
AI: You've got the base case and the inductive hypothesis down pat.
What we assume to be true is called the inductive hypothesis: $$P(t): t^2 + 15t + 5≤21t^2\tag{P(t)}$$
Then, we need to prove that given $P(t)$, $P(t+1)$ is true, as you note:
$$P(t+1) = (t+1)^2 + 15(t+1) + 5 \leq 21(t+1)^2\tag{P(t+1)}$$
The objective is to manipulate the left hand side of $P(t)$, and to use the inductive hypothesis $P(t)$ to validate the inequality by showing the left-hand side is necessarily less than or equal to right-hand side of $P(t+1)$.
We can expand the left hand side of $P(t+1)$:
$$\begin{align}(t+1)^2 + 15(t+1) + 5 & = t^2 + 17t + 21 \\ \\ & = \color{blue}{\bf (t^2 + 15t + 5)} + (2t + 16) \\ \\ & \overset{\text{I.H.}}{\leq} \color{blue}{\bf 21t^2} + 2t + 16 \\ \\ &\leq 21t^2 + 42t + 21 \\ \\ &= 21(t^2 + 2 t + 1) \\ \\ &= 21(t+1)^2 \end{align}$$
...and end with the desired right-hand side of $P(t+1)$. Hence, we've proven $P(t) \implies P(t+1)$, and together with your base case, we're done! |
H: Catalan number basic question - combinatorics
I have a question regarding catalan numbers:
1) Find the number of sequences $a_1 \leq ... \leq a_n$ where $a_i \in \mathbb N$ and $0 \leq a_i \leq i-1$ for all $i \in \{1,2,...,n\}$
For example:
When $n=1$ the only available sequence is: $0$, so only 1 sequence, which is equal to $C_1$.
When $n=2$ the only available sequences are $0,0$ and $0,1$, so 2 sequences, which is $C_2$
When $n=3$ the options are: $0,0,0$ and $0,0,1$ and $0,0,2$ and $0,1,1,$ and $0,1,2$ so 5 options, which is equal to $C_3$ etc...
I know the answer. The answer is $C_n$ (catalan number of index $n$) but I don't understand why. What do catalan numbers have to do with this question?
AI: Catalan numbers count the number of monotonous paths from $(0,0)$ to $(n,n)$ not crossing the diagonal. Interpret each $a_i$ as the max height of the path at $x={i-1}$. Then each of your sequences bijects to a monotonous path through the points $(i-1,\ a_i)$. |
H: Help with a proof involving normality.
Prove that $A_n$ is normal in $S_n$.
We know, in order to be normal, the centers of each group need to commute with everybody. We also know, the groups are normal if every left coset is equal to the right coset.
How can we do this using these groups?
AI: Hint: $|S_n:A_n|=2$ is all I want to say |
H: multiplicative character evaluated at -1 (from Ireland and Rosen's number theory book)
I'm self studying from Ireland and Rosen's "A Classical Introduction to Number Theory" Second edition. Near the beginning (page 153 in my book) of Chapter 11 the authors discuss the number of solutions of a hypersurface in a finite field.
At one point they involve the character of order 2 on a finite field of $q^s$ elements. Let this character be denoted by X$q^s$.
They then mention that if $-1$ is not a square in the prime subfield of q elements, then then X$q^s$(-1)=-1 for s odd and X$q^s$(-1)=1 for s even.
I don't understand why this is.
Thanks for the help.
AI: I don't have the book with me, but I think your problem lies in the sentence
They then mention that if -1 is not a square in the field...
which should really read
... if -1 is not a square in $\mathbf F_q$...
(in other words, in this sentence "the field" is $\mathbf F_q$, not $\mathbf F_{q^s}$.)
On a finite field with $q^s$ elements (where $q$ is odd), there is precisely one quadratic character, the Legendre symbol:
$$\mathbf F_{q^s}^* \to \pm 1$$
$$ a \mapsto \left(\frac{a}{q^s}\right) = a^{(q^s-1)/2}.$$
Now $(-1/q^s) = (-1)^{(q^s-1)/2}$ is $+1$ if and only if $q^s \equiv 1 \mod 4$. This happens either if $q \equiv 1 \mod 4$ and $s$ is arbitrary, or $q \equiv 3 \mod 4$ and $s$ is even. In other words, if $-1$ is not a square in $\mathbf F_{q^s}$, then $s$ is odd and $-1$ is not a square in $\mathbf F_{q}$ (and conversely). |
H: Are these equivalent?
$\forall x \in D, (P(x) \Rightarrow Q(x))$ is equivalent to $(\forall x \in D \cap P,Q(x))$.
However, is this also equivalent to $(\forall x\in D)( P(x)\land Q(x))$?
If not, what's the difference? Thank you.
AI: No.
$\forall x \in D, (P(x)\rightarrow Q(x)) \not\equiv \forall x\in D(P(x) \land Q(x))$.
$P(x) = x$ is a pink unicorn and
$Q(x) = x$ can fly.
Then it's certainly true that $\forall x\in D(P(x) \rightarrow Q(x))$, but not $\forall x\in D(P(x) \land Q(x))$.
In the first case, $P(x) \rightarrow Q(x)$ holds for all $x$ because for every x, $P(x)$ is false, and therefore the implication is true, vacuously.
In the second case, we have a proposition asserting that "Every x in the domain is both a pink unicorn and can fly." That's a different statement altogether, which happens not to be true: it never holds. |
H: Best approximation for $\displaystyle \sum_{k=2}^n\ln\ln k$?
I need the best approximation for
$\displaystyle{\sum_{k = 2}^{n}\ln\left(\ln\left(k\right)\right)}$. Any suggestion or hint is welcomed.
I derived $n\ln\left(\ln\left(n!\right) \over n\right)$ so is there any better one ?
AI: Euler-Maclaurin series:
$$ \sum_{k=2}^n \ln(\ln(k)) = C + \int_2^n \ln(\ln(t))\ dt + \dfrac{1}{2} \ln(\ln(n)) + \dfrac{1}{12 n \ln(n)} - \dfrac{1}{360 n^3 \ln(n)} - \dfrac{1}{240 n^3 \ln(n)^2} - \dfrac{1}{360 n^3 \ln(n)^3} + O(1/n^5) $$
for some constant $C$,
where $$\eqalign{\int_2^n \ln(\ln(t))\ dt &= n \ln(\ln(n)) - 2 \ln(\ln(2)) - \int_2^n \dfrac{dt}{\ln(t)}\cr &= n \ln(\ln(n)) - Li(n) - 2 \ln(\ln(2)) + Li(2)\cr}$$
Numerically it appears $C \approx -.2412388$. |
H: Quadrature Rule Error
Question: Suppose that $S(f,h)$ is a quadrature rule for the integral $I= \int^b_a f(x) dx $ and that the error series is $c_4h^4+c_6h^6+...$ Combine $S(f,h)$ with $S(f,\frac{h}{3})$ to find a more accurate approximation of I.
So I got the easy part, that
$I-S(f,h)=c_4h^4+c_6h^6+...$
$I-S(f,\frac{h}{3})=c_4(\frac{h}{3})^4+c_6(\frac{h}{3})^6+...$
And now I need to combine them somehow. There is always adding the two together and solving for $I$, but I don't think that is what they want. As a side note if this was supposed to represent Simpson's Rule it would be much more obvious, but I don't think that $S(f,h)$ is referring to Simpson's but an arbitrary quadrature rule. If there were some way of making both right sides of the equations the same I would know where to go, but don't see how.
I don't want you to solve the problem for me, but if I could use some assistance on the next step to take (assuming this one is even right) that would be helpful.
AI: The idea is to combine the two and cancel out the terms proportional to $h^4$. Can you find a linear combination of $S(f,h)$ and $S(h,\frac h3)$ that does that? |
H: Residue Theorem for Denominator with $e^z$
$$ f(z)=\frac{z^3}{e^z-1} $$
Is this a simple pole at $z=0$ or some other types of pole?
If it is a simple pole, what is its residue?
Is it using this formula or other else?
$$ \lim_{z\to 0}=zf(z) $$
AI: $$e^z=\sum_{k=0}^\infty\frac{x^k}{k!}\implies\frac{z^3}{e^z-1}=\frac{z^3}{z+\frac{z^2}2+\frac{z^3}6+\mathcal O(z^4)}=$$
$$=\frac{z^2}{1+\frac z2+\frac{z^2}6+\mathcal O(z^3)}$$
The above shows at once that $\;z=0\;$ is a removable singularity...though the last step requires a little justification (factor out $\;z\;$, and then you can cancel. Why?) |
H: Fixed roots and quotient rings
I'm reading a paper by Smart and Vercauteren on homomorphic encryption (http://eprint.iacr.org/2011/133). I don't understand a specific statement around quotient rings. The authors state that for a polynomial $F(X)$ in indeterminate $X$, elements of quotient rings such as $\mathbb F_2 (X)/F(X)$ and $\mathbb Q(X)/F(X)$ can be represented as polynomials in some fixed root of $F(X)$ in the algebraic closure of the base field. I don't understand this statement, could someone please explain?. Would some of the more comprehensive texts on algebra such as Dummit and Foote be helpful in explaining such constructs? If not, any other recommendations for texts?
AI: The statement is only true if $F$ is irreducible over the base field, which is to say that it cannot be factored into two polynomials of lower degree over that field.
Let $k$ be a field and $f(x)\in k[x]$ an irreducible polynomial over $k$. Let $\alpha$ be a root of $f$ in $\bar k$, the algebraic closure of $k$.
Consider the homomorphism $$\phi_\alpha: k[x]\to \bar k$$ given by evaluation at $\alpha$, i.e. $\phi_\alpha(g) = g(\alpha)$. Since $f$ is irreducible, we have that $g(\alpha)=0$ if and only if $g(x) = q(x)f(x)$ for some $q\in k[x]$. In other words, the kernel of $\phi_\alpha$ is precisely $(f)$, the principal ideal generated by $f$. Note also that the image of $\phi_\alpha$ is $k[\alpha]$, the subring of $\bar k$ generated by $k$ and $\alpha$. Then by the first isomorphism theorem, $\phi_\alpha$ descends to an isomorphism
$$k[x]/(f(x))\stackrel{\sim}{\to} k[\alpha].$$
This is the correspondence between elements of the quotient ring and polynomials in the fixed root $\alpha$. |
H: Is the following set connected?
Let $R \subset \mathbb{R}^2$ denote the unit square $R = [0,1] \times [0,1]$. If $F \subset R$ is finite, is $R \backslash F$ connected?
AI: You can even prove that $R\setminus F$ is connected if $F$ is countable. In fact, it’s path-connected.
HINT: Given two points $p$ and $q$ in $R\setminus F$, draw their perpendicular bisector; note that this has uncountably many points within $R\setminus F$. Show that there must be a point $r$ on this bisector such that the segments $\overline{pr}$ and $\overline{rq}$ lie entirely within $R\setminus F$. |
H: Team winning probability
If a team has a 2/3rd chance of winning any game, what is the probability that it wins at most 4 out of 5 games? The answer I got was
The chances that they win no games = 1/243
The chances that they win one game = 10/243
The chances that they win two games = 40/243
The chances that they win three games = 80/243
The chances that they win four games = 80/243
For a total probability of 211/243 that they win at least 4 out of 5 games.
AI: That is correct, but you could just use complementary counting. The probability that the team will win all five games is $ \left( \dfrac {2}{3} \right)^5 $. Thus, the probability that the team will not win all $5$ games is $$ 1 - \left( \dfrac {2}{3} \right)^5 = 1 - \dfrac {32}{243} = \boxed {\dfrac {211}{243}}, $$which is what you got. |
H: The automorphism $\varphi(k)$ in the semidirect product is $h \mapsto kh k^{-1}$ What do they mean?
Let $H \rtimes K$ be the semidirect product of groups $H$ and $K$, not necessarily subgroups of anything.
In Theorem 10, part (5) of Dummit & Foote, Algebra, there's
Identifying $H$ and $K$ with their isomorphic copies in $H\rtimes K$, namely $\{(h,1)\}$ and $\{(1,k)\}$, we have
E) for all $h \in H$, $k \in K$, $k h k^{-1} = k \cdot h = \varphi(k)(h)$.
In other words, the hom into $Aut(H)$ associated with this semidirect product, $\varphi$, sends $k$ to conjugation by $k$ in $H \rtimes K$. Or something like that...
Their proof is:
$$
(1, k) (h, 1) (1, k)^{-1} = \\
(1 k\cdot h, k) (k^{-1} \cdot 1, k^{-1}) = \\
(k\cdot h, k)(1, k^{-1}) = \\
(k\cdot h, 1)
$$
$(k\cdot h, 1) = (\varphi(k)(h), 1) = \tilde{\varphi}(k)(h)$
This is all very confusing. Is there a way to clear things up?
I'm not getting how you can mix the two isomorphisms and $\varphi$
AI: By definition, we have that for $\;a,c\in H\,,\,\,b,d\in K\;$
$$(a,b)(c,d):=(a\cdot c^b,bd)$$
Note that we usually denote $\;\phi(c)(b)=:b^c\;$ and then clearly we get conjugation in the semidirect product.
Now, first
$$(h,k)\in H\rtimes K\implies (h,k)^{-1}=\left(h^{-k^{-1}}\,,\,k^{-1}\right)\;,\;\;\text{since}$$
$$(h,k)\left(h^{-k^{-1}},k^{-1}\right)=\left(h\cdot\left(h^{-k^{-1}}\right)^k\,,\,kk^{-1}\right)=(h\cdot h^{-k^{-1}k},1)=(1,1)$$
So that now
$$(1,k)(h,1)(1,k)^{-1}=(1\cdot h^k,k)(1,k^{-1})=(h^k,1)\in H\times 1$$ |
H: If the diagonals of an isosceles trapezoid are perpendicular to each other, prove that the area is $S=H^2$.
Where H is the altitude of the whole trapezoid (the distance between the bases).
Thanks.
AI: Here's another way to look at it.
Let $ABCD$ be the given isosceles trapezoid with $BC=a$ the top line segment and $AD=b$ the bottom line segment (see figure). Then the area of the trapezoid is given by $$S=bH - \left( \frac{b-a}{2}\right) H = H \frac{a+b}{2}.$$ It therefore suffices to show that $H=(a+b)/2$. To do this, extend $a$ and draw the line perpendicular to $b$ through $D$ such that it intersects $a$ at $X$. Similarly, draw the line perpendicular to $b$ through $B$ intersecting $b$ at $Y$. To show that $H=(a+b)/2$ it suffices to show that the quadrilateral $BXDY$ is a square. But this is immediate, as can be seen by a translation of the diagonal $AC$ by $AY$ along $b$ (since the diagonals are of equal length, are perpendicular and span a quadrilateral, the quadrilateral must be a square). Hence $H=(a+b)/2$ and we are done. |
H: Cosets & Groups.
I am a newbie to group theory. I am confused by something that I would like clarified.
Say $G$ is the integers, a group under addition, and $H$ is the even integers, a subgroup of $G$.
It is usually stated that left cosets of $G$ are a partition of $G$.
Now for all $g \in G$, the left coset of $G$ by $g$ is defined as $gH$ in multiplicative notation, which for this example is really $g+H$.
So, here a left coset of $H$ by $g$ should be the set $\left\{g+h_1, g+h_2, g+h_3,\ldots\right\}$, where $h_1, h_2,h_3,\ldots$ are the elements of $H$.
How can these sets be disjoint?
For example, the two
cosets $2+H$ and $4+H$ both contain the element 8.
I don't understand what am I getting it wrong.
Thanks in advance.
AI: The distinct cosets of $H$ partition $G$. If $g_1+H\cap g_2+H \neq \emptyset$, then $g_1 + 2m_1 = g_2 + 2m_2$ for integers $m_1$ and $m_2$, and therefore $g_2 = g_1 + 2n$ for $n = m_2-m_1$. It follows that $g_1 + H = g_2 + H$.
The more precise statement is that if we choose a complete family of representatives $g_1,\ldots,g_n$ (or more generally $g_i$ for $i \in I$ with $I$ a (possibly infinite) index set), then $G$ is the disjoint union of $g_i + H$ for $i = 1,\ldots,n$.
Here a complete family of coset representatives is a set of distinct elements $g_1, \ldots, g_n$ of $G$ such that the cosets of $H$ in $G$ are exactly $g_1+H, \ldots, g_n+H$ and $g_i + H \neq g_j + H$ for $i \neq j$.
Edit: For your particular example, notice that $2+H = 4+H$. Indeed $g \in 2+H$ if and only if $g = 2+2n$ for some integer $n$, which in turn is true if and only if $g = 4 + 2m$ for some integer $m$. (Here $m = n-1$ since $g = 2+2n = 4 + 2(n-1)$.) Consequently $g \in 2+H$ if and only if $g \in 4+H$, i.e., $2+H = 4+H$. |
H: Exam question, complicated chain rule
I have this old exam question that I don't understand.
Calculate $\frac{d}{dt}ln(x^2+y^2)$ where $x=e^t+e^{-t}$ and $y=e^t-e^{-t}$. Write the answer in x and y's.
The solution given does the following:
$$\frac{dx}{dt}=e^t-e^{-t}=y$$
$$\frac{dy}{dt}=e^t+e^{-t}=x$$
then use the chain rule to:
$$\frac{d}{dt}ln(x^2+y^2)=\frac{\partial}{\partial x}ln(x^2+y^2)\frac{dx}{dt}+\frac{\partial}{\partial y}ln(x^2+y^2)\frac{dy}{dt}=$$
$$=\frac{1}{x^2+y^2}*2x*y+\frac{1}{x^2+y^2}*2y*x=\frac{4xy}{x^2+y^2}$$
Question:
How does the teacher do that chain rule step? I would really need a step by step explanation how it's made. I guess I don't understand how dt relates to dx and dy.
Thank you.
EDIT: Updated with last $\frac{dy}{dt}$term that the teacher had left out of the solution sheet.
AI: I think you are having trouble with the concept of a total derivative of a function of several variables. Consider $f=f(x,y)$ where $x=x(t)$ and $y=y(t)$. Then $${d\over dt}f(x,y)={\partial f\over \partial x}\cdot {dx\over dt}+{\partial f\over \partial y}\cdot {dy\over dt}.$$
Here, $f(x,y)=\ln(x^2+y^2)$ with $x=e^t-e^{-t}$ and $y=e^t+e^{-t}$. Thus,
\begin{align}
{d\over dt}(\ln(x^2+y^2))
&={\partial f\over \partial x}\cdot {dx\over dt}+{\partial f\over \partial y}\cdot {dy\over dt}\\
&={1\over x^2+y^2}\cdot 2x\cdot (e^t+e^{-t})+{1\over x^2+y^2}\cdot 2y\cdot (e^t-e^{-t})\\
&={1\over x^2+y^2}\cdot 2x\cdot y+{1\over x^2+y^2}\cdot 2y\cdot x\\
&={4xy\over x^2+y^2}.
\end{align} |
H: Is this element-of_{ij} - looking symbol the Levi-Civita symbol?
I'm reading this formula:
from a page
Is the symbol that looks like an element-of symbol with two indices i and j the Levi-Civita symbol?
Mathematics is my weak-side so I'm not sure.
Actually I guessed on Levi-Civita not based on my expertise but by tried to look up the symbol from the wikipedia table of mathematical symbols. As it didn't help, so I thought I look up Einstein-notation and Kronecker delta although I knew it's not them... but the wikipedia footer lists Kronecker as "notable tensors" and by checking the whole list I've got Levi-Civita which seem to make sense in the context.
AI: The line following the equation in your link to the page says
The $\epsilon_{ij}$ is a constant known factor |
H: Proof inequality with log
How can I prove that there exist $n_0$, $c$ such that for all $n>n_0$:
$$n^{\log_2{n}}\le c2^{n}$$
(So I mean the log of n with base 2). Can anybody help me?
AI: First, let's show that
$$
\lim_{n\to\infty}\frac{\log(n)^2/\log(2)}{\log(c)+n\log(2)}=0
$$
Using L'Hospital twice gives
$$
\begin{align}
\lim_{n\to\infty}\frac{\log(n)^2/\log(2)}{\log(c)+n\log(2)}
&=\lim_{n\to\infty}\frac{2\log(n)/n}{\log(2)^2}\\
&=\lim_{n\to\infty}\frac{2\log(n)}{\log(2)^2n}\\
&=\lim_{n\to\infty}\frac{2/n}{\log(2)^2}\\[6pt]
&=0
\end{align}
$$
Thus, there is an $n_0$ so that for $n\ge n_0$
$$
\frac{\log(n)^2/\log(2)}{\log(c)+n\log(2)}\le1
$$
Thus,
$$
\log(n)^2/\log(2)\le\log(c)+n\log(2)
$$
and therefore,
$$
n^{\log_2(n)}\le c\,2^n
$$ |
H: Show that a $5 \times 4$ Matrix of row rank $3$ has nontrivial solutions for $Ax=0$
Statement: Let $A$ be a $5 \times 4$-Matrix with row rank 3. Show that $Ax=0$ has nontrivial solutions.
This is a homework problem, as a hint I am given one formula (which we didn't discuss in class yet) but intuition wise I don't understand how it should help me at all to verify this statement. My tutor went about this as follows, consider: $$ F: \begin{cases} V & \longrightarrow W \\ v & \longmapsto A \circ v \end{cases} $$
know that: Row Rank of A = Column Rank of A = $\dim(\text{Im}(A))=\dim(\text{Im}(F))$ and make use of
$\dim(V)=\dim(\text{Kernel}(F))+\dim(\text{Im}(F)) \tag{given}$
I do understand the given Formula, but I don't understand how to apply it, since the Row rank of $A$ is equal to $3$, so is the Column Rank and therefore the dimension of the Image of $F$ is also equal to $3$. Can I make a statement about the dimension of $V$? I know that $v \in \mathbb{R}^4$ but that doesn't help me much with the dimension of $V$.
Some hints or insights, especial intuition wise insights, would be very helpful and appreciated.
AI: The nullity-rank theorem is stated for the general case and you can apply it to your situation:
Let $V=\mathbb{R}^4$, $W=\mathbb{R}^5$ and define $F(v)=Av$, for $v\in\mathbb{R}^4$. Then the image of $F$ is the column space, which has dimension $3$ because $A$ has rank $3$. Therefore $N(A)$, which is the solution set of $Ax=0$ and the kernel of $F$, has dimension $4-3=1$ and so contains nonzero vectors. |
H: Strong induction proof with polygon
How can we show that if a simple polygon with at least four sides is triangu-lated, then at least two of the triangles in the triangulation have two sides that border the exterior of the polygon using strong induction
AI: Proposition. If $n\ge 3$ and $AB$ is an edge of a triangulated $n$gon then there exists a triangle with two sides bordering the exterior of the polygon and $AB$ is not among these two sides.
Proof:
For $n=3$ this is clear.
For $n>3$ note that removing the triangle $ABP$ having $AB$ as edge produces two smallerpolygons, a $k$gon and an $(n+1-k)$gon. At most one of these is degenerate (i.e. either $k\ge 3$ or $n+1-k\ge3$ because $k+(n+1-k)=n+1\ge 5$). By assumption, this smaller polygon has a triangle with two edges on the boundary of the smaller polygon and the edge $BP$ resp. $PA$ is not among these two edges. Then this triangle also works for the original $n$gon. $_\square$
Now the original problem is solved as follows: Pick any edge $AB$ to find a first triangle $XYZ$ having two boundary edges $XY$ and $YZ$. Now pick $XY$ instead of $AB$ to find a second triangle $UVW$ having two boundary edges different from $XY$. Since $n>3$, the triangles cannot be the same (as not all three edges can be boundary edges). |
H: Help with 'If, then'- and 'Only if'-sentences in Predicate Logic
So, I have to ask now, because I've spent so much time on these two translations.
My key is
Domain: living things
Px: x is a Pokerplayer
Cx: x is a Chessplayer
Yx: x is Professional
Rx: x is Rich
and the sentences are:
1) "Pokerplayers and Chessplayers are rich if they are professional"
2) "Pokerplayers and Chessplayers are rich only if they are professional"
Now after messing around with using combinations of material conditionals and conjunctions, I ended on only using material conditionals for the first one, like this:
1) ∀x((Px v Cx) --> (Yx --> Rx))
for it seems to be saying that for all living things, if it is a pokerplayer/chessplayer then, if it is professional, then it is rich ...
My question is then, for the second, if I can only "revert"/"switch" the predicates in the main consequent to get the 'only if'-version, like this:
2) ∀x((Px v Cx) --> (Rx --> Yx))
for this seems to be saying that for all living things, if it is a chessplayer/pokerplayer then, if it is rich, then it is profession; which seems to exclude the possibility of any non-professional rich pokerplayers/chessplayers ... But ... doesn't this also allow there to be chessplayer/pokerplayer who are not rich but still professional - since the last material conditional is true even when 'Rx' is false and 'Yx' is true ... Or have I gotten this wrong?
Thank you!
(I now notice that I have forgot to use your fancy symbolization tools, but I will try to change it ... I was just in a bit of a hurry to post this question)
AI: Both your translations are correct.
The bare appearance of "if" (stripped of the word "only") usually precedes the antecedent of a conditional.
When "if" is preceded by the qualifier "only", as in "only if", then we need to know that "only if" precedes the consequent of a conditional.
In general: $A\rightarrow B$ says "if
$A$ then $B$", but it is also equivalent to "$A$ only if $B$."
In your first statement, the bare "if" correctly precedes the antecedent $P(x)$: we have, in effect,
"$R(x)$, if $P(x)$" $\iff \;P(x)\rightarrow R(x) \iff\;$ "if x is a professional, then x is rich".
In your the second statement, indeed, you have correctly translated the stattement, noting that the consequent of the implication "x is rich only if x is professional" is equivalent to $\;R(x) \rightarrow P(x)$ |
H: Sequence with a contraction mapping of the sum
Consider a continuous function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with the following property. There exists $c \in (0,1)$ such that for all $x,y \in \mathbb{R}^n$ it holds that $\left\| f(x) - f(y) \right\| \leq c \left\| x-y \right\|$, where $\left\| \cdot \right\|$ is the Euclidean norm.
From Banach's fixed-point theorem, we know that, for any $x^{(0)} = x_0 \in \mathbb{R}^n$, the sequence
$$ x^{ (k+1) } := f \left( x^{ (k) } \right) $$
converges to the unique fixed-point $\bar{x}$ of $f$, i.e. $\lim_{ k \rightarrow \infty } f( x^{(k)} ) = \bar{x} = f( \bar{x} )$.
Now consider $y^{ (0) } = x_0$ are the sequence
$$ y^{ (k) } := f \left( \frac{1}{k} \sum_{i=0}^{k-1} y^{(i)} \right). $$
For instance, $y^{(1)} = x^{(1)} = f(x_0)$, $y^{(2)} = f\left( \frac{1}{2}\left(x_0 + f(x_0) \right) \right)$, $y^{(3)} = f \left( \frac{1}{3}\left( x_0 + f(x_0) + f\left( \frac{1}{2}\left(x_0 + f(x_0) \right) \right) \right) \right)$.
What happens to $\lim_{k \rightarrow \infty} y^{(k)} $? Does it converge to some point? In particular to $\bar{x}$?
AI: Let $z_k =\displaystyle \dfrac{1}{k} \sum_{i=0}^{k-1} y^{(i)}$, Thus
$$z_k = \dfrac{1}{k} (y^{(k-1)} + (k-1) z_{k-1}) = \dfrac{1}{k} \left(f(z_{k-1}) + (k-1) z_{k-1} \right)$$
while
$$\overline{x} = \dfrac{1}{k} (f(\overline{x}) + (k-1) \overline{x})$$
and so
$$ \eqalign{ \|z_k - \overline{x}\| &\le \dfrac{1}{k} \left\|f(z_{k-1}) - f(\overline{x})\right\| + \dfrac{k-1}{k} \left\| z_{k-1} - \overline{x}\right\|\cr
&\le \dfrac{c + k-1}{k} \|z_{k-1} - \overline{x}\|}$$
Since $\sum_{k=1}^\infty 1/k = \infty$, $\prod_{k=1}^\infty (1 - (1-c)/k) = 0$, so $\|z_k - \overline{x}\| \to 0$. |
H: $\varepsilon$-$\delta$ proof of $\sqrt{x+1}$
need to prove that $ \lim_{x\rightarrow 0 } \sqrt{1+x} = 1 $
proof of that is:
need to find a delta such that $ 0 < |x-1| < \delta \Rightarrow 1-\epsilon < \sqrt{x+1} < \epsilon + 1 $ if we choose $ \delta = (\epsilon + 1)^2 -2 $ and consider $ |x-1| < \delta = (\epsilon + 1)^2 - 2 $
$ 4 - (\epsilon + 1)^2 < x +1 < (\epsilon + 1)^2 $
$ \sqrt{4-(\epsilon + 1)^2} -1 < \sqrt{x+1} -1 < \epsilon$ but I need to show that $\sqrt{4-(\epsilon + 1)^2} -1 > -\epsilon $ before this proof is complete...
any help on how to finish the proof?
AI: You need to show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon.$$
It's helpful in this case to multiply by the conjugate of $\sqrt{x+1}-1$, which is $\sqrt{x+1}+1$, so that the rightmost inequality becomes $$|\sqrt{x+1}-1|\cdot|\sqrt{x+1}+1|<\varepsilon(\sqrt{x+1} + 1)$$ $$|x|<\varepsilon(\sqrt{x+1}+1). \tag{1}$$
Now, if you require that $0 < |x| < 1$, then we know that $0<\sqrt{x+1}<\sqrt{2}$, so $1<\sqrt{x+1} + 1<\sqrt{2}+1.$
So we now know that if we require $0 < |x| < 1$, then $(1)$ tells us that $|x| < \varepsilon(\sqrt{2}+1)$, which we have also shown can be rearranged into the form $|\sqrt{x+1}-1| < \varepsilon.$
So, we take $\delta = \min\left(1,\ \varepsilon(\sqrt{2}+1)\ \right).$ Then, $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon$$
which is exactly what we wanted to show. |
H: Conditional probability distribution question
You are given that $N$ is a Poisson random variable with mean $4$. Define a new random variable
$M=(N - 4 \mid\ N \ge 4)$. Find the mean of $M$. Lost myself trying to solve this, help would be appreciated.
AI: Hint: $E[N-4] = P(N=0) E[N-4|N=0] + \ldots + P(N=3) E[N-4|N=3] + P(N \ge 4) E[N-4|N \ge 4]$. |
H: What happens if we overlap two points in Pascal's theorem?
In projective geometry, we know that:
For any inscribed hexagon in a conic, the intersection of the segments formed by the union of opposite vertices, are collinear.
But now, what happens if two vertices overlap, ie, if one side of the hexagon becomes tangent to the conic.
I have tried to make multiple drawings, and I see that the intersections are no longer collinear, but... do they? I know that I draw terrible, and I think that they are really colinear because the theorem never needed the points to be different.
AI: The result still holds. Yay to projective geom.
In fact, there is an interesting 'degenerate case': $ABCD$ is a cyclic quad (or quadrilateral on a conic), then $AB \cap CD$, $BC \cap DA$, the point intersection of tangents at $A$ and $C$, and the point of intersection of tangents at $B$ and $D$ are collinear. |
H: Integrating $e^x$
I'm wondering if there are any rules to integrate $e^x$? For example $e^{-0.05x}$. Can I use the reversed chain rule, u-sub or something to easily integrate most examples of $e^x$?
AI: Hint: Given $\lambda \in \Bbb R$, differentiate $x\mapsto \dfrac{e^{\lambda x}}{\lambda} $.
More generally, given a function $f$ such that there exists $F$ with the property that $F'=f$, to obtain an antiderivative of $(f\circ u)u'$ where $u$ is some differentiable function such that the composition makes sense, differentiate $F\circ u$: $(F\circ u)'=(F'\circ u)u'=(f\circ u)u'$. |
H: Function iteration and intervals of attraction for fixed points
I am currently studying iteration sequences and I am a bit hung up on one specific bit which involves determining intervals of attraction of fixed points.
I've been given a graphical method to determine the intervals of attraction, and a more formal method called the "gradient criterion" which states that if $f$ is a smooth function with an attracting fixed point $a$, then if an interval $I$ with midpoint $a$, exists for which $|f'(x)| < 1$, for $x \in I$ then $I$ is a interval of attraction.
Right, so far so good. Now, what I have a harder time understanding are is that if I solve the inequality $-1 < f'(x) < 1$ and then use the result to identify a interval $I$ with midpoint $a$ (where a is an attracting fixed point), then isn't the complete interval, its simply a interval where I know that a start value from this interval will tend to $a$. However, larger intervals exits, which is mentioned in my text book, where its simply pointed out that other start values might end up producing terms which are in $I$, and they will then tend to $a$ from there. I get that, but its rather informal, and I feel like I'm missing how I can formally attack the problem of finding the full interval.
By looking various graphs of functions being iterated, it seems to me like I need to look at the boundries of the open interval $I$, if for example, the gradient is increasing though the lower boundry, I need to extend the interval with the largest possible interval where f is increasing and no other fixed points are present.
Am I going in the right direction here?
AI: If a maximal interval of attraction exists, then by continuity it has to be open. If that interval is $(b,c)$, and if the function is still defined in the closed interval $[b,c]$, then $b$ and $c$ can not be attracted to $a$, and since $f$ maps $(b,c)$ into itself, it has to map the endpoints into the closure $[b,c]$. This shows that $f(\{b,c\}) \subseteq \{b,c\}$, so that either $b$ and $c$ are both fixed points, or one of them is fixed and the other one its preimage, or they form a two-cycle.
These possibilities are realized with $f_1(x) = x^3$, $f_2(x) = x^2$, $f_3(x) = -x^3$, where in each case $[b,c]=[-1,1]$. So you can find the maximal interval of attraction by finding all fixed points, their preimages, and all two-cycles. |
H: Must the intersection of connected sets be connected?
Must the intersection of two connected sets be connected?
I believe the answer is no, but I am not entirely sure. I think a counter example would be a set that intersects another set in more than one area, yet those intersections are disjoint. (Think of a cylinder that extends through one end of a donut, the hole in the middle, and then the other end of the donut).
AI: No. Consider the intersection of the line segment and the circle in $\varnothing$. |
H: Drawing the subgroup lattice of D10
I've been tasked with drawing a subgroup lattice of the dihedral group of order 10. I know from Lagrange's theorem that non-trivial subgroups must have order 2 or 5. Finding the subgroups of order 2 is straightforward, since there can be only one element besides the identity, but what about the groups of order 5? I could feasibly brute-force my way through this problem by going through the elements of $D_{10}$ one by one and trying to construct groups of order 5, but there must be a better way to solve this.
AI: As $5$ is a prime number, groups of order $5$ are cyclic. So you should find all the elements of order $5$ in $D_{10}$ and then determine what subgroups they generate.
Note that any non-identity element of a cyclic group of order $5$ generates that group. So if an element of order $5$ is contained in a subgroup that you have already found, then it generates that subgroup and you do not need to check it again. |
H: How do we prove this fact about cyclic groups?
Prove that an Abelian group of order 33 is cyclic.
Can we take an element a of order 3 and an element b of order 11 and say, |ab|=33?
AI: Yes, that's definitely one way to do it. But you'll have to be explicit about how you know that you can choose $a$ and $b$ to have those orders and once they're chosen you'll have to be explicit about how you know that the order of $ab$ is $33$. As long as you can fill in those details your proof will be correct. |
H: Limit of $f(x)=x^2\left(1+2+\cdots+\left\lfloor \frac1{|x|} \right\rfloor\right) $
Suppose $f$ a function defined on $\left[\dfrac{-1}{2};\dfrac{1}{2}\right]$ as $f(x)=x^2\left(1+2+\cdots+\left\lfloor \dfrac{1}{|x|} \right\rfloor\right) $.
How can I prove that $f$ has a finite limite on $0$?
Calculate for $k \in \mathbb{N}^*$ : $\displaystyle \lim_{x\to0+}\left(x\left(\left\lfloor \dfrac{1}{x} \right\rfloor+\left\lfloor \dfrac{2}{x} \right\rfloor+\cdots+\left\lfloor \dfrac{k}{x} \right\rfloor\right)\right)$.
AI: For the second question, note that if $x=\frac1n$, then
$$x\left(\left\lfloor\frac1x\right\rfloor+\left\lfloor\frac2x\right\rfloor+\ldots+\left\lfloor\frac{k}x\right\rfloor\right)=\frac1n(n+2n+\ldots+kn)=\frac12k(k+1)\;,$$
which tells you what the limit has to be. Moreover, if $\frac1{n+1}<x<\frac1n$, then $n<\frac1x<n+1$, so
$$\frac1{n+1}\sum_{m=1}^kmn\le x\left(\left\lfloor\frac1x\right\rfloor+\left\lfloor\frac2x\right\rfloor+\ldots+\left\lfloor\frac{k}x\right\rfloor\right)<\frac1n\sum_{m=1}^km(n+1)\;,$$
and a little more work shows that it really is the limit. |
H: Prove $13|19^n-6^n$ by congruences
I am trying to prove $13|19^n-6^n$. With induction its not so bad but by congruences its quite difficult to know how to get started.
Any hints?
AI: HINT: Just use the congruence $19\equiv 6\pmod{13}$. |
H: A Double Limit Question
Maybe it's easy, but: Is it true that
$$\lim_{(x,y) \rightarrow (0,0)} \frac{x}{\sqrt{x^2+y^2}}=0$$
If it is, could you help me prove it?
Thanks
AI: For the function $\frac{x^2}{\sqrt{x^2+y^2}}$, the limit exists. To see why, find a bound that trivially tends toward 0:
$$\frac{x^2}{\sqrt{x^2+y^2}} \le \frac{x^2}{\sqrt{x^2}} = |x| \to 0$$
For the function $\frac{x}{\sqrt{x^2+y^2}}$, the limit does not exist.
To see why, find two paths that tend toward (0,0) for which the limit is different (if the limit exists, it must be the same whatever the path).
Let $y=\lambda x$ and $x > 0$, then
$$ \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{\sqrt{x^2+\lambda^2x^2}} = \frac{x}{\sqrt{x^2 (1+\lambda^2)}} = \frac{1}{\sqrt{1+\lambda^2}}$$
Thus the function is constant and obviously "tends" toward $\frac{1}{\sqrt{1+\lambda^2}}$, which will vary with $\lambda$. Hence the limit does not exist.
Note: of course, the function of $x$ alone (that is, with $y=\lambda x$) is only constant for $x > 0$. For $x<0$, the constant is negative, $- \frac{1}{\sqrt{1+\lambda^2}}$, since in this case $\frac{x}{|x|} = -1$. |
H: Joukowski Conformal Mapping
I'm having trouble understanding how to map the streamlines from one plane to another using the Joukowski transform.
In the $\zeta$ plane, I'm considering flow around a cylinder, with the complex potential given by $$w(\zeta)=U(\zeta e^{-i \alpha} + \frac {a^2}{\zeta}e^{i\alpha})$$
where $\zeta = re^{it}$.
From here I understand it is more useful to consider the streamlines first in the $\zeta$ plane and then use the Joukowski map $z=\zeta+\frac{b^2}{\zeta}$ to transform them to the $z$ plane. It's not a problem finding the streamlines in the $\zeta$ plane, as I can use the stream function given by $$\psi(r,t)=Usin(t-\alpha)(r-\frac{a^2}{r})$$ and equate it to constant values. From here though, the notes I'm looking at say it's easy to then use the transform to get the streamlines in the $z$ plane, but I'm not very sure how. I'm aware it's supposed to look like flow around an ellipse, but my ideas on Maple all draw a blank.
Thanks!
AI: I would find the parametric equation for the streamlines in the $\zeta$ plane; then the transformation to the $z$-plane is indeed easy. As a parameter, take $s=t-\alpha$. The streamline equation is $U\sin s (r-a^2/r) = C$. Solving this for $r$ yields
$$r=r(s)=\frac12\left(\frac{C}{U\sin s}+\sqrt{\left(\frac{C}{U\sin s}\right)^2+4a^2}\right)$$
So, the parametric equation of streamline is $\zeta=r(s)e^{i(s+\alpha)}$. Applying the Жуковский transformation (I hate that transliteration) yields
$$z= r(s)e^{i(s+\alpha)} +\frac{b^2}{r(s)} e^{-i(s+\alpha)}$$
which is something that Maple's complexplot can handle. |
H: Elementary differentiation question on derivation of p.d.f. of function of random variable
Let $G(y) = \Pr(Y \le y) = 1 - F(\frac{1}{y})$.
Then apply the chain rule (assuming $y \ne 0$ and $F(x)$ is differentiable at $x = 1/y$) and we have $$g(y) = \frac {d\ G(y)}{dy} = \frac{-d\ F(x)}{dx} \Large \mid_{\normalsize x = 1/y} \normalsize \left(-\frac{1}{y^2} \right) = f\left( \frac{1}{y}\right)\frac{1}{y^2} $$
This is just a really elementary question on differentiation applied to probability density functions which are one-to-one and differentiable. Still, I'm having trouble seeing the steps which lead to even the first equality after $\frac {d\ G(y)}{dy}$. Is there any way to show the steps to the calculation above more explicitly, particularly for those not so skilled at calculus?
AI: I am using the notation $Df = \frac{\partial f}{\partial x}$.
The basic rules are: If $g=g_1+g_2$, then $Dg = Dg_1+Dg_2$. If $h= a \circ b$, then $D h(x) = Da(b(x)) Db(x)$.
Here $g_1(x) = 1$$, g_2(x) = -F(\frac{1}{y})$.
This gives $Dg_1(x) = 0$.
To compute $Dg_2$, write $g_2=a \circ b$ where $a(x) = -F(x)$, $b(x) = \frac{1}{y}$. Then we have $Da(x) = -F'(x)$, and $Db(x) = - \frac{1}{x^2}$.
Putting them altogether gives: $Dg(x) = 0 - (F'(\frac{1}{x}) (- \frac{1}{x^2}) = \frac{1}{x^2} F'(\frac{1}{x})$. |
H: Does 1 distinct eigenvalue guarantee 1 eigenvector?
I am trying to figure out when 2x2 matrices are not diagonalizable. Right now, my conditions are:
the matrix has only 1 distinct eigenvalue
the matrix yields only 1 linearly independent eigenvector
But when I know that there are 2 eigenvalues, can I safely assume that each eigenvalue will have at least 1 linearly independent eigenvector?
AI: By definition, an eigenvalue always has an eigenvector. And it's easy to prove that eigenvectors for distinct eigenvalues are linearly independent. If a $2 \times 2$ matrix has only one eigenvalue, there may be either $1$ or $2$ linearly independent eigenvectors. |
H: Prove $(a + b) \bmod n = (a \bmod n + b \bmod n) \bmod n$
I find I am in trouble to prove:
$$(a + b) \bmod n = (a \bmod n + b \bmod n) \bmod n ?$$
Can anyone help?
AI: Let $a = hn + (a \bmod n)$, $b = kn + (b \bmod n)$, $h,k\in \mathbb Z$. Then the left hand side
$$\begin{align*}
(a+b)\bmod n =& [a+b-(h+k)n]\bmod n\\
=& [(hn +a\bmod n) + (kn+b\bmod n) - hn - kn]\bmod n\\
=& (a\bmod n + b\bmod n)\bmod n
\end{align*}$$ |
H: The Tangent Disc Topology
Let $X$ be the tangent disc topology, $X=P\cup L$ where $P=\{(x,y):x,y\in \mathbb{R}, y>0\}$ and $L$ is the real line.Then,
$X$ is completely regular but not normal,
$X$ is separable,
$X$ is countably metacompact.
$X$ is not compact
I see all of them. But I wonder $X$ can be metrizable?
AI: No: a metrizable space is paracompact and therefore normal. Also, the tangent disk space is separable but not second countable, which is impossible for a metrizable space. |
H: Application of fixed point theorem in $R^n$
Let $A=(a_{ij}) \in \mathbb R^{n \times n}$ a matrix such that $|a_{ij}|<\frac{1}{n}$ for every $i,j$. Prove that $I-A$ is invertible.
My attempt at a solution:
$I-A$ is invertible $\iff$ $(I-A)v=0$ has the trivial solution $v=0$. But $(I-A)v=0 \iff Iv-Av=0 \iff Av=v$. Let $T:\mathbb R^n \to \mathbb R^n$ be the linear transformation with $|T|_e=A$, as T is a linear transformation, $v=0$ satisfies $T(v)=v$. In order for $I-A$ to be invertible, $T$ must have a unique fixed point (as $v=0$ is a fixed point, we've already proved existence). The space $\mathbb R^{n \times n}$ is complete, so if we prove $T$ is a contraction, then, by the fixed point theorem, we can assure there is a unique point $v$ such that $T(v)=v$, and as $T$ is linear, this point has to be $0$.
I don't know how to prove $T$ is a contraction. I must prove that for $x,y \in \mathbb R^n$, $\|T(x)-T(y)\|<\alpha \|x-y\|$, with $0<\alpha <1$. I have to use the fact that $|a_{ij}|<\frac{1}{n}$, can anyone guide me on how to use this information properly?
AI: Since $T$ is linear, it is sufficient to prove $\|Tx\| \le \alpha \|x\|$ for some $0 \le \alpha < 1$.
Also, it is sufficient to prove it is a contraction for any norm on $\mathbb R^n$, and the problem is much easier if you use the taxicab norm $\|x\| = \sum \limits_{i=1}^n |x_i|$ or the max norm $\|x\| = \max \limits_{1 \le i \le n} |x_i|$.
But if you want to use the Euclidean norm, start with
$$ \|Tx\|^2 = \sum_{i=1}^n \left|\sum_{j=1}^n a_{ij}x_j\right|^2 \le n\left(\max_{1 \le i,j \le n} |a_{ij}|^2\right) \left(\sum_{j=1}^n |x_j|\right)^2 $$
Then use Cauchy-Schwarz to show
$$ \sum_{j=1}^n |x_j| \le \sqrt n \|x\| .$$
You will get the result with $\alpha = n\left(\max \limits_{1 \le i,j \le n} |a_{ij}|\right)$. |
H: Permutation of natural numbers
Find the number of permutation of {1,2,3,4,5,6} such that the patterns 13 and 246 do not appear. Show the steps .
AI: It is easier to first find the number of permutations in which the pattern $13$ or $246$ does appear.
For the pattern $13$, tie $1$ and $3$ together to make a supersymbol. Together with $2,4,5,6$, and the supersymbol. we have $5$ "symbols" that can be arranged in $5!$ ways.
Similarly, there are $4!$ ways in which the pattern $246$ appears.
There are $3!$ ways in which both patterns appear. These are counted twice in the sum $5!+4!$.
Thus there are $5!+4!-3!$ ways thast one or both of our patterns can appear. It follows that there are $6!-(5!+4!-3!)$ ways to arrange our symbols so that neither pattern appears. |
H: Why use the biconditional in the Axiom of Extensionality
I'm studying the Axiom of Extensionality in the following form:
$$
\forall a \forall b[\forall x(x\in a\leftrightarrow x\in b)\rightarrow a=b]
$$
(where quantification of a,b is restricted to sets and quantification of x can range over domain objects as well as sets)
What is the advantage of the preceding formulation over the following, where the biconditional is replaced with conjunction? E.g.
$$
\forall a \forall b[\forall x(x\in a\wedge x\in b)\rightarrow a=b]
$$
I am aware that conjunction and the biconditional are not logically equivalent since their truth tables differ in the case where both argument propositions are false.
So it seems that these two forms should differ in the case where an element of the domain is neither in a nor b.
Using the conjunction form, this case makes the main conditional of the axiom vacuously true ($False\rightarrow True$). Using the biconditional form, this case makes the main conditional of the form $True\rightarrow True$. Either way, the main conditional is true.
So, what is the advantage of the biconditional form?
AI: Recall that the inference rule for implication states that if we know $\phi$ is true, and if we know that $\phi\rightarrow\psi$ is true, then we can deduce that $\psi$ is true. Notice that it does not also say that if we know $\phi$ is false, and $\phi\rightarrow\psi$ is true, that we can deduce $\psi$ is true. Indeed, it better not because then (because of LEM) whenever $\phi\rightarrow\psi$ is true, $\psi$ would be true, which is absurd).
Now, there is a slight with your analysis of the two axioms, a small circularity: you consider only how the axioms instantiates when you already have $a=b$, and you do case analysis on whether there is an $x$ in $a$ or not. You need to also consider the case when $a\neq b$. This error doesn't make a difference though, it only obscures what happens because you should consider if $x\in a$ or not independently of whether $a=b$.
You see, we want the axiom to allow us to conclude $a=b$ under appropriate conditions. With the biconditional form, it is clear that we can derive $a=b$ if we have a way of showing that for every $x$, $x\in a\rightarrow x\in b$ and $x\in b\rightarrow x\in a$, which is exactly what we want.
With the conjunction form, we can derive $a=b$ if for every $x$, $x\in a$ and $x\in b$. In particular, we can only derive $a=b$ if $a$ and $b$ both contained every possible $x$. If there were an $x\not\in a$ (or an $x\not\in b$), then the axiom becomes False${}\rightarrow a=b$, so we cannot derive $a=b$. To think what would happen to Set theory when this is the case, imagine that you have different colored sets that have the same elements (unless they contain everything, which i think Russel's paradox might still go through so such sets don't exist). Then applying comprehension to craft an empty set, will give you many differently colored empty sets, which you wouldn't be able to prove are equal. |
H: A question about multivariable concave functions
Consider a concave function $f: \mathbb{R}^N \rightarrow \mathbb{R}$. Is it possible for it to be convex in a single argument when I fix the remaining $N-1$ variables or does concavity of $f$ in $N$ variables imply that it should be concave in each individual argument?
AI: It will be concave in each individual argument. (Concavity is a condition on how the function behaves on lines, so if it behaves a certain way along all lines, it will behave a certain way along a specific line.)
There is a chance it might be linear in an argument. Then it would be both concave and convex at the same time. |
H: Can any metric space be completed?
Completion defined in Real Analysis, Carothers, 1ed has been captured below.
Can any metric space be completed?
AI: Yes. Rudin outlines a proof in Principles of Mathematical Analysis, Chapter 3, Problem 24.
Given a metric space $(X,d)$, say two Cauchy sequences $(p_n)$ and $(q_n)$ are equivalent if $\lim_{n\to\infty} d(p_n,q_n) = 0$. This is an equivalence relation on the set of all Cauchy sequences, as you can verify.
The set of all equivalence classes, say $X^*$, can be given a metric: if $P$ and $Q$ are equivalence classes in $X^*$ and $(p_n)\in P$, $(q_n)\in Q$ are representatives for the equivalence classes, set $\Delta(P,Q) = \lim_{n\to\infty} d(p_n,q_n)$. It can be shown that this is a metric on $X^*$. Moreover, $(X^*, \Delta)$ is a complete metric space, and there is an isometry of $X$ into $X^*$, so $(X^*, \Delta)$ is a completion of $(X,d)$ by Carothers's definition of the term.
This is an excellent exercise, and I highly encourage you to find a copy of Rudin and try it out. He gives more details and hints in his outline, so with substantial effort you should be able to give it a good go. |
H: Prove that D is normal
Let G be a group, $T = G \times G$ and let $D = \{(g,g)\in G \times G | g\in G\}$. Prove that D is normal in T if and only if G is abelian.
I assume that D is normal in T, then for any $x,y \in T$ we have $(xgx^{-1}, ygy^{-1})$ for $g\in D.$ How can I show that $G$ is abelian?
AI: If $D$ is normal, then $(xgx^{-1}, ygy^{-1}) \in D$ for every $x, y \in G$, that is, $xgx^{-1} = ygy^{-1}$. Take $y$ to be the identity element of $G$ and $x$ to be an arbitrary element of $G$. |
H: The vectors a and b are non-collinear. For what value(s) of m is it true that (m^2 + 2m - 3)a + (m^2 + m -6)b = 0?
Since they are both non-collinear they can never be equal to zero. So what do i do now?
Any help is appreciated. Thank you very much!
AI: Hint: Two vectors are non-collinear if and only if they are linearly independent (prove this!). Linear independence in this context means
$$c_1 a + c_2 b = 0 \iff c_1 = 0 = c_2$$ |
H: Resolvent recurrence relation
Let the resolvent matrix of $\mathbf{X}$, a symmetric matrix with real entries, be defined as
\begin{align}
R_{\mathbf{X}}(\lambda):=\bigl(\mathbf{X}-\lambda\mathbf{I}\bigr)^{-1}, \qquad \lambda \in \mathbb{C}\backslash\mathbb{R}.
\end{align}
with $\mathbf{I}$ the identity of matrix (same dimensions as $\mathbf{X}$).
I must show that
\begin{align}
R_{\mathbf{X}}(\lambda)= - \frac{1}{\lambda} \mathbf{I} - \frac{1}{\lambda}\mathbf{X}R_{\mathbf{X}}(\lambda).
\end{align}
My naive approach is:
By definition
\begin{align}
\bigl(\mathbf{X}-\lambda\mathbf{I}\bigr) R_{\mathbf{X}}(\lambda)= \mathbf{I}
\end{align}
Thus, direct inversion yields
\begin{align}
\mathbf{X}R_{\mathbf{X}}(\lambda)-\lambda R_{\mathbf{X}}(\lambda) = \mathbf{I}\\
R_{\mathbf{X}}(\lambda) = -\frac{1}{\lambda}\mathbf{I}+\frac{1}{\lambda}\mathbf{X}R_{\mathbf{X}}(\lambda)
\end{align}
which seems to contradictory the stated result... what did I missed?
AI: When there is only one operator present, you can treat them as if they are numbers. So what you were asked to prove was
$$ \frac1{x-\lambda} = - \frac1\lambda - \frac x{x-\lambda} .$$
Now it is really obvious that the statement you were asked to prove is wrong. |
H: Calculating $\int_{0}^{\infty} x^{a-1} \cos(x) \ \mathrm dx = \Gamma(a) \cos (\pi a/2)$
My goal is to calculate the integral
$\int_{0}^{\infty} x^{a-1} \cos(x) dx = \Gamma(a) \cos (\pi a/2)$,
where $0<a<1$,
and my textbook provides the hint: integrate $z^{a-1} e^{iz}$ around the boundary of a quarter disk.
However, I couldn't figure out how to control the integral over the quarter arc. Any hints?
AI: Consider the contour integral
$$\oint_C dz \, z^{a-1} \, e^{i z}$$
where $C$ is a quarter circle of radius $R$ in the 1st quadrant (real and imaginary $> 0$), with a small quarter circle of radius $\epsilon$ about the origin cut out (to avoid the branch point at the origin).
This integral is equal to
$$\int_{\epsilon}^R dx \, x^{a-1} \, e^{i x} + i R^a \int_0^{\pi/2} d\theta \, e^{i a \theta} e^{i R \cos{\theta}} \, e^{-R \sin{\theta}}\\+ i \int_R^{\epsilon} dy \, e^{i \pi (a-1)/2} y^{a-1} e^{-y} + i \epsilon^a \int_{\pi/2}^0 d\phi \, e^{i a \phi} \, e^{i \epsilon e^{i \phi}}$$
We note that the second integral vanishes as $R\to\infty$ because $\sin{\theta} \gt 2 \theta/\pi$, so that the magnitude of that integral is bounded by
$$R^a \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le R^a \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{2}{\pi R^{1-a}}$$
We also note that the fourth integral vanishes as $\epsilon^a$ as $\epsilon \to 0$. In the third integral, we write $i=e^{i \pi/2}$ to make a simplification.
The contour integral is zero by Cauchy's Theorem (no poles in the interior of $C$). Ths we have (+)
$$\int_{0}^{\infty} dx \, x^{a-1} \, e^{i x} - e^{i \pi a/2} \int_0^{\infty} dy \, y^{a-1} \, e^{-y}=0$$
We use the definition of the gamma function:
$$\Gamma(a) = \int_0^{\infty} dy \, y^{a-1} \, e^{-y}$$
and take real parts of (+) to obtain the sought-after result. |
H: Calderón-Zygmund operators with positive kernel
Let $T$ be a Calderón-Zygmund operator. That is, $T$ maps $L^2(\mathbb{R}^d)$ to itself and satisfies the representation formula
$$
Tf(x) = \int_{\mathbb{R}^d}K(x,y)f(y)\, dy
$$
for all $f \in L^2$ with compact support and $x \notin \operatorname{supp} f$, for some kernel $K$ defined outside the diagonal and satisfying
$$
|K(x,y)| \leq C|x-y|^{-d},
$$
$$
\int_{|x-y|>2|y-z|} |K(x,y)-K(x,z)|\, dx \leq C
$$
and
$$
\int_{|x-y|>2|x-w|} |K(x,y)-K(w,y)|\, dy \leq C.
$$
If $\psi$ is any nonnegative $C^\infty_c$ function then $f \mapsto f \ast \psi$ is cleary a Calderón-Zygmund operator with positive kernel.
My question is, do there exist Calderón-Zygmund operators with positive kernels which are not so well-behaved. For example: does there exist a Calderón-Zygmund operator with a positive kernel which is not locally $y$-integrable near the diagonal $\{x=y\}$?
AI: No, the fact that some non-integrable kernels yield bounded operators is due to cancellation. Suppose $K$ is a nonnegative kernel which is nonintegrable with respect to $y$ along the diagonal. Let $f$ be the characteristic function of an open ball $B$. Then for every $x\in B$ we have
$$Tf(x)=\int_B K(x,y)\,dy = \infty$$
which is rather far from having $Tf\in L^2$. |
H: Example of a continuous bijective function on and to the closure of the complex numbers, with an inverse that is not continuous?
Note that by the closure of the complex number, I mean the union of the complex numbers and infinity.
I have been stumbling over this questions for a wile now, and I understand many examples of this for subsets of the closure of the complex numbers, but I wish to demonstrate it for the closure of the complex numbers.
I only know the very basics of topology, but is appear the Invariance of Domain Theorem implies that no such function would exist if this were for the complex numbers (rather than their closure). So my questions are as follows:
Am I mistaken in the claim I made about the Invariance of Domain Theorem; if not, can this idea be extended is some way to the closure of the complex numbers?
Does such a function exist; and if so, what is an example?
(Please comment if my question is unclear, I will monitor regularly, and clarify anything if requested.)
(I apologize for not knowing how to typeset this.)
AI: Such a function can't exist, and this is an easier result than the invariance of domain.
The extended complex numbers form a compact Hausdorff space $\hat{\mathbb{C}}$.
In general, every closed subspace of a compact space is compact, every continuous image of a compact set is compact, and every compact subset of a Hausdorff space is closed. Therefore if $f: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is continuous, then it maps closed sets to closed sets. If it is a bijection, then this means it also maps open sets to open sets, so $f^{-1}$ is continuous. |
H: Local solutions over $\mathbb{Q}_p$ but no solutions over $\mathbb{Q}$
I was looking at a set of notes that states the equation $x^4-17=2y^2$ is solvable locally over $\mathbb{Q_p}$ for every $p$ , but is not solvable over $\mathbb{Q}$. Now, this is not a homework problem and is just for my own reading. Is there a simple way to see why this is the case? Is either direction "trivial"?
AI: This one is hard. There is a complete discussion in Local Fields by Cassels. The following, Exercise 121 in Gouvea (yours is exercise 122) is easier:
$$ (x^2-2)(x^2 - 17)(x^2-34)=0. $$
For one thing, there is just one variable.
Let's see, the Hasse-Minkowski principle says that this kind of example cannot happen for a quadratic form (form means homogeneous). But the same stuff does happen for Selmer's example:
$$ 3 x^3 + 4 y^3 + 5 z^3 = 0, $$ where we require that $x,y,z$ not all be $0.$
I had been convinced for a long time that Fernando Q. Gouvea and Mariano Suárez-Alvarez were the same person. I met Gouvea in January. I've never met Mariano, but I now believe they are separate persons. |
H: Help debunk a proof that zero equals one (no division)?
Unlike the more common variant of proof that 0=1, this does not use division.
So, the reasoning goes like this:
\begin{align}
0 &= 0 + 0 + 0 + \ldots && \text{not too controversial} \\
&= (1-1) + (1-1) + (1-1) + \ldots && \text{by algebra}\\
&= 1 + (-1 + 1) + (-1 + 1) \ldots && \text{by associative property}\\
&= 1\\
\\
&\therefore 0 =1
\end{align}
I can't help but feel that something went wrong here, specifically with the use of the associative property. However, I can't come up with a mathematically compelling reason.
Where's the error?
AI: The error is that the "..." denotes an infinite sum, and such a thing does not exist in the algebraic sense. The usual way to make sense of adding infinitely many numbers is to use the notion of an infinite series: We define the sum of an infinite series to be the limit of the partial sums. (So the notion of convergence from analysis is involved in addition to algebra.)
Not all algebraic rules generalize to infinite series in the way that one might hope. When they fail, it is because something fails to converge. In this case, what fails to converge is the series that should appear between the two lines in the middle of the "proof":
$$1-1+1-1+1 \cdots.$$
Indeed, this series fails to converge because the
sequence of partial sums $\{1, 1-1, 1-1+1,\ldots\}$ oscillates between $1$ and $0$ and does not converge to any value. |
H: Why does $( \operatorname e^x)' = \operatorname e^x?$
It's known the the derivative of exponential function $a^x$ is $xa^{x-1}$.
If I play $e$ as $a$, we'll get $(a^x = \operatorname e^x)' = x \operatorname e^{x-1}$.
Why does $(\operatorname e^x)' = \operatorname e^x$?
AI: As T. Bongers has pointed out, you don't have the correct rule for exponential functions. The real rule for functions like these is
$$
\frac d{dx} a^x = \ln(a)\cdot a^x
$$
Once you have this rule, it's clear what should happen for $e^x$. Whether or not this derivation is inherently circular depends on your definition of $e$. |
H: Let $G$ be a group of order 6. Suppose that $a,b\in G$ with $a$ of order 3 and $b$ of order 2. Show that either $G$ is cyclic or $ab\not=ba$.
Let $G$ be a group of order 6. Suppose that $a,b\in G$ with $a$ of order 3 and $b$ of order 2. Show that either $G$ is cyclic or $ab\not=ba$.
Attempt at Answer:
Suppose $ab=ba$. Then $o(ab)=o(a)o(b)$ since $o(a)$ and $o(b)$ are relatively prime (o being shorthand for 'order'). From here, I'd guess that I want to show that $G$ is NOT cyclic for $ab=ba$. Then I'd show the converse.
The problem is that I don't get how to get from here to there.
Any help/hints would be greatly appreciated.
AI: You've essentially finished the proof already. If $ab = ba$ then $o(ab) = o(a) o(b) = 2 \cdot 3 = 6$ and $G$ is cyclic. If $ab \ne ba$ then $G$ can't be cyclic, since it's not even abelian. |
H: Iran Math Olympiad 2012 (perfect power)
Prove that if $t$ is a natural number then there exists a natural number $n > 1$ such that $(n, t) = 1$ and none of the numbers $n + t, n^2 + t, n^3 + t…$ are perfect powers.
There is a solution posted at AOPS, by considering $n = t(t+1)^2 + 1$. Two cases are discussed: $t + 1$ is not a perfect power and $t + 1$ is a perfect power.
I have no problem understand the first case: if $t + 1$ is not a perfect power, then $n^k + t$ can be expressed in the form of $(t + 1)(b(t + 1) + 1)$, since $(t + 1)\nmid(b(t + 1) + 1)$, $(t + 1)$ has to be perfect power if $n^k + t$ is a perfect power. But this is contradiction.
But I got lost when trying to understand the second case, especially from "Consider $n^k+t=c^d$ so by proof...", can anyone help?
AI: The problem comes from a not so exact translation.
The claim is that that $n$ works, and to prove it, let assume the contrary, that is, there exist $k,c,d$ s.t. $n^k+t=c^d$ then it continues the same.
I hope it makes it clear. |
H: If dim(V) + dim(W) > dim(R^n), show that some nonzero vector is in V & W. [GStrang P183, 3.5.45]
Inside $\mathbb{R^n}$, suppose dimension$(\mathbf{V})$ + dimension$(\mathbf{W}) > n$. Show that some nonzero vector is in both $\mathbf{V}$ and $\mathbf{W}$.
Answer : Since $\dim\mathbf{V} + \dim\mathbf{W} = \dim\mathbf{(V \cap W)} + \dim(\mathbf{V + W}) \quad> n$,
$\Large{\color{red}{[}}$ thus $\dim\mathbf{(V \cap W)} > 0$ $\Large{{\color{red}{]}}}$. So $V \cap W$ contains nonzero vectors. $\quad \square$
$\Large{1.}$ This question doesn't specify what are $\mathbf{V}, \mathbf{W}$. Why not? Should they be subspaces?
$\Large{2.}$ How and why does the red bracket follow? $\dim(\mathbf{V + W})$ is unknown?
AI: The "inside $\mathbb{R}^n$" implies that $V$ and $W$ are subsets of $\mathbb{R}^n$; in particular, since we're referring to dimension, it must be assumed that $V$ and $W$ are vector subspaces of $\mathbb{R}^n$.
Given this assumption, we see that $V + W \subseteq \mathbb{R}^n$ so that $\dim(V + W) \le n$. But since
$$n < \dim(V \cap W) + \dim(V + W) \le \dim(V \cap W) + n$$
In particular, $\dim(V \cap W)$ can't be zero, or the right side is too small. |
H: Real Analysis: Example related to irrational number and rational limit and Cauchy
A) a sequence $(x_n)$ of irrational numbers that converges to a rational number
Are the following answer correct?
$x_n = e^1$
or
$\sqrt{2}/n$
Give a little explanation if you could
B) a sequence $(x_n)$ that is not Cauchy, but for which $|x_{n+1} - x_n|$ converges to $0$
please give an example
AI: Your answer $a_n = \sqrt{2}/n$ is valid for the first part. I'm not sure what you meant to type when you put $a_n = e^1$, but as written it is not valid.
This answer to B) is one of those "classic" counterexamples that are helpful, in general, to remember.
$$
x_n = \sum_{k=1}^n\frac 1k
$$
Can you show that $|x_{n+1}-x_n|\to 0$? Can you show that $x_n$ is not Cauchy? |
H: Confused about the definition of a group as a groupoid with one object.
A groupoid is defined to be a category where every morphism is an isomorphism. So sometimes a group is said to just be a groupoid with one object.
When I try to make sense of this, I denote the single object as $G$. I view the morphisms as the analogue of "elements." The identity $1_G$ is the analogue of the usual identity $e$, we can compose any two morphisms since they are all arrows on $G$, and for any arrow $f$, we have some $f^{-1}$ such that $f\circ f^{-1}=f^{-1}\circ f=1_G$, so the idea of inverses is still there.
So I informally associate the elements of the group in the usual definition to be the arrows in the groupoid. But what does the sole object $G$ in the groupoid "correspond to" if I were to try to informally make sense of a group in the usual sense? Does it even correspond to anything?
AI: The single object in the groupoid corresponding to a group $G$ does not really correspond to anything in the group - but you can think of it as a thing which has as its group of symmetries the group $G$.
As an example, the category whose only object is the set $[n]=\{1,\dotsc,n\}$, and morphisms are bijections from this object onto itself. Then the symmetries of $[n]$ are given by the group $S_n$.
For a more subtle example, consider a topological space $X$, and fix a point $x\in X$. Consider the category with one object, namely the point $x$. A morphism from $x\to x$ is a homotopy-class of paths from $x$ to $x$. Composition of morphisms comes from concatenation of paths. The automorphism group of $x$ is the fundamental group $\pi_1(X,x)$, which is a group of symmetries of the based topological space $(X,x)$.
However, if you consider groupoids with many objects all of which are isomorphic to each other, then the different objects correspond to different realizations of the same group. Moreover, each isomorphism between different objects gives rise to an isomorphism between the corresponding realizations of the group.
To make this precise, suppose $\mathcal G$ is a groupoid where all objects are isomorphic. For each object $x$ of $\mathcal G$, let $G_x$ denote the group $\mathrm{Mor}_\mathcal G(x, x)$. A morphism $\phi:x\to y$ defines an isomorphism $G_y\to G_x$ by taking $g\mapsto \phi\circ g\circ \phi^{-1}$.
A nice example of this is the category $\mathrm{FB}_n$ whose objects are finite sets of cardinality $n$ and morphisms are bijections. Then the automorphism group of each object is isomorphic to $S_n$. Isomorphisms between objects determine isomorphisms of their automorphism groups.
Another example is the Fundamental Groupoid of a path connected topological space $X$. Its objects are the points of $X$. The set $\mathrm{Mor}(x,y)$ is the set of homotopy classes of paths from $x$ to $y$. In this groupoid, $G_x$ is the fundamental group $\pi_1(X,x)$ of $X$ based at the point $x$. Different base points result in isomorphic fundamental groups, and isomorphisms are determined by homotopy classes of paths between these points. |
H: Show that the multiplicative group $\mathbb{Z}_{10}^{\times}$ is isomorphic to the additive group $\mathbb{Z}_4$.
Show that the multiplicative group $\mathbb{Z}_{10}^{\times}$ is isomorphic to the additive group $\mathbb{Z}_4$.
I'm completely lost with this one.
AI: First off, we have $\mathbb{Z}_4^+=\{0,1,2,3\}$, and $\mathbb{Z}_{10}^\times = \{1,3,7,9\}$.
Now, I'm going to tell you that $f:\mathbb{Z}_4^+ \to \mathbb{Z}_{10}^\times$ is a homomorphism for which $f(1) = 3$. What you have to figure out is: what are the values of $f(2),f(3),$ and $f(4)$, and why is this an isomorphism?
At this point, you should have the following:
$$
\begin{align}
f(1) &= 3\\
f(2) &= f(1+1) = f(1)f(1) = f(1)^2 = 9\\
f(3) &= \cdots = f(1)^3=7\\
f(4) &= \cdots = f(1)^4 = 1
\end{align}
$$
In other words, we've defined $f$ by $f(n) = 3^n \pmod{10}$ for $n = 0,1,2,3$. Now, to prove $f$ is a homomorphism, it suffices to state that
$$
f(n+m) = 3^{n+m}=3^n\cdot3^m \pmod{10} = f(n)f(m)
$$ |
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