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H: Non-square tensors?
I learnt tensor algebra for physics and I never saw a non-square (or non-cubic...) tensor. But, from a mathematical point of view, do non-square tensors exist? And if so, are they used in some area in physics?
AI: I don't know much about the physical uses of a non-square tensor, but there is no mathematical reason for them not to exist.
If we have two vector spaces $V$ and $W$, with dimensions $n$ and $m$ and bases $v_1,v_2,\dots , v_n$ and $w_1, w_2, \dots , w_m$, then the tensor product $V \otimes W$ has basis $v_1 \otimes w_1, v_1 \otimes w_2, \dots, v_1\otimes w_m, \dots, v_n\otimes w_m$, and so is of dimension $nm$. Usually, when writing the components of tensor, we write them in an $m\times n$ grid. So, if $V$ and $W$ are of different dimensions, we will have non-square tensors.
The reason that you see square tensors most often, however, is that we most often care about tensor powers of a given vector space, rather than more general tensor products. |
H: A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0\le k \le n$
This is problem 3 from Hungerford's section about the Sylow theorems.
I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. This is what I have so far:
Suppose $|G| = p^n$. For $k = 0$, $\{e\}$ is a normal subgroup of order $p^0$. Suppose $N_1, ..., N_k$ are normal subgroups of $G$ with orders as described (ie. $|N_i| = p^i$) and $k < n$.
$G/N_k$ is a $p$-group, so it has a non-trivial centre $C(G/N_k)$. $\pi^{-1}(C(G/N_k))$ is a subgroup of $G$, where $\pi : G \to G/N_k$ is the quotient map. It is normal in $G$ because $\pi$ is a homomorphism and $C(G/N_k)$ is a normal subgroup of $G/N_k$.
I know it contains $N_k$ and some stuff not in $N_k$, so it has to have order at least $p^{k+1}$, but I do not know how to argue it must be equal, or indeed if it even is equal.
AI: Hint: Apply Cauchy's theorem to $C(G/N_k)$. This will give you a normal subgroup of $G$ of order $p^{k+1}$. |
H: Bijection between column space and row space
Suppose that $A_{mn}$ is a matrix over some field, and that $C, R$ is its column space and row space, without using the fact that $rank(C) = rank(R)$, can we show that, there exists a bijection between $C$ and $R$?
AI: Hint: For $x\in R$, consider $Ax\in C$. Show this map is a bijection. (It might be useful to note that $\Bbb R^n=R\oplus N$, where $N=\{x: Ax=0\}$.) |
H: Function that converge uniformly
Does the sequence $x, ..., x^n $ of functions converge uniformly on the interval $[0,k]$ for $k\in(0,1)$? If it does, prove it. How about on the interval $[0,1]?$
Can you help me complete the proof?
For the interval $[0,k]$ I got it does and for $[0,1]$ it doesn't. So let $\{f_n\} = \{x^n\}$, and suppose $f^n \to f$. Then I must show that for $\epsilon>0$, there exists an $N$ such that $d(f,f^n)<\epsilon$ whenever $n>N$ for all $x.$
For $k\in (0,1),$ it is clear to see that $x^n\to 0$ as $n$ approaches infinity. Then I must show $|x^n|<\epsilon$ whenever $n$ is greater than some $N$.
On $[0,k]$, $x^n$ attains its max at $x=k$ so $x^n<k^n$. Then note that $k^n$ decreases with increasing $n$, so we choose $N$ such that $k^N<\epsilon$.
$\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.
AI: Let $x\in[0,1)$ and $t\in[0,x]$. Suppose $n\in\mathbb{N}$. Then
$$|t^n| \le x^n. $$
Since $x^n\to 0$ as $n \to\infty$, the series converges uniformly to $0$ on $[0,x]$.
Furthermore, note that the $x^n$ converge pointwise to $0$ on $(0,1)$ and to $1$ at $x = 1$. Since the target functions is not continuous, the convergence on $[0,1]$ cannot be uniform. |
H: Showing that the sequence converges knowing that three other sequences converge
I have a question in Analysis. Knowing that $x_{2n}$, $x_{2n-1}$, $x_{3n}$ converge, how can I show that $x_{n}$ converges?
AI: Hint: 1) $x_{2n}$ and $x_{3n}$ should converge to the same limit (by looking at a profitable subsequence of each). 2) A similar argument for another profitable choice of two of $x_{2n}, x_{2n-1}, x_{3n}$. 3) All three given subsequences actually converge to the same limit. We can now conclude via $x_{2n},x_{2n-1}$. |
H: Proving Vector Space with Scalar Multiplication
I'm trying to prove a vector space with the following definitions:
$$[x,y]+[a,b]=[x+a+1,y+b]$$
$$r[x,y]=[rx+r-1,ry]$$
I'm working on the distributive law but I'm running into a problem. It's not coming out correctly for reason. Here's my work:
$$r([x,y]+[a,b])=r[x+a+1,y+b]=[r(x+a+1)+r-1,ry+rb]=[(rx+r)+(ra+r-1),ry+rb]$$
Shouldn't I be getting another $-1$ in there? (After the $rx+r$ term).
AI: Preservation of addition is given in terms of the vector space. So in fact,
$$r[x,y] + r[a,b] = [rx+r-1, ry] + [ra+r-1,rb] = [(rx+r-1) + (ra+r-1) + 1,ry+rb] = [(rx+r) + (ra+r-1), ry+rb]$$
as desired.
(I've posted this as an answer to avoid the odd typesetting issues I was seeing in the comments. ) |
H: How to find the points of tangency of a parabola using Calculus?
How can someone find the points of tangency of a parabola in this situation?
I need to find two points of tangency so that the triangle formed by the two tangent lines at those points and the x axis is an equilateral triangle.
What approach should I take?
I know that I can find the slope of a tangent line passing through the point (x,y) by just taking the derivative of that function and by using the point-slope formula I can get the equation of that tangent line. However, I would need to know the point...
AI: Hint: In order to form an equilateral triangle with the $x$ axis, one of the tangents must slope upwards by 60° and the other must slope downwards by 60°. But as long as they do this (and don't happen to have the same $x$ intercept) they will form an equilateral triangle somewhere. |
H: Connected Set and connected subset
Question: Let $Y \subset X$; let $X$ and $Y$ be connected. Show that if $A$ and $B$ form a separation of $X-Y$, then $Y \cup A$ and $Y \cup B$ are connected.
Attempt at an answer: since $A,B$ form a separation of $X-Y$ then $X-Y = A \cup B$ and $A \cap B = \emptyset$. Then $X = A \cup B \cup Y = (A\cup Y) \cup (B \cup Y)$, but $X$ is connected hence it must be that $(A \cup Y) \cap ( B \cup Y) \neq \emptyset$, but this must imply that $A \cup Y$ and $B \cup Y$ are connected otherwise $X$ could possibly be disconnected if they are not connected. Is this enough, I feel like I a missing some details ( $Y$ is not necessarily an open set in $X$).
Thank you.
AI: Your first question:
$x\in U$ if and only if $x\in U\vee x\in U$.
Your second question:
It's the exercise 12 in section 23 Munkres.
I think your proof isn't quite right.
$Proof:$
Suppose $Y\cup A$ disconnected (The other one is similar, just swap $A$ and $B$ through out), let $U,V$ be a separation. Then $\overline{U}\cap V=\emptyset$ and $U\cap\overline{V}=\emptyset$ and $Y\cup A=U\cup V$. Similarly, since $A,B$ separates $X\setminus Y$, we have $\overline{A}\cap B=\emptyset$ and $A\cap\overline{B}=\emptyset$ and $X\setminus Y=A\cup B$.
Since $Y$ is connected, either $Y\subseteq U$ or $Y\subseteq V$. For simplicity, assume $Y\subseteq U$. Since $Y\cup A=U\cup V$ and both of which are disjoint unions, we have $V\subseteq A$, $\overline{V}\subseteq\overline{A}$ as well.
Now we can infer $\overline{V}\cap B=\emptyset$ and $V\cap\overline{B}=\emptyset$.
Since $(U\cup V)\cup B=(A\cup B)\cup Y=(X\setminus Y)\cup Y=X$. We will show $U\cup B$ and $V$ form a separation for $X$. Note $(\overline{U\cup B})\cap V=(\overline{U}\cup\overline{B})\cap V=(\overline{U}\cap V)\cup(\overline{B}\cap V)=\emptyset$. And $(U\cup B)\cap\overline{V}=(U\cap\overline{V})\cup(B\cap\overline{V})=\emptyset$. Hence they are indeed a separation for $X$, which is a contradiction.
$\mathrm{QED}$ |
H: Classifying complex $2\times 2$ matrices up to similarity
I would like to prove the following proposition, which is given as an exercise in Hoffman and Kunze:
If $A$ is a $2\times 2$ matrix with coefficients in $\mathbb{C}$, then $A$ is similar either to a matrix of the form $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ or to a matrix of the form $\begin{pmatrix} a & 0 \\ 1 & a \end{pmatrix}$.
A hint directs the reader to prove that if $N$ is an nilpotent matrix (also in $M_{2}(\mathbb{C})$), then either $N=0$ or $N$ is similar to $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. I have proven this claim (by supposing that $N \neq 0$ and showing that the transformation induced by multiplication by $N$ has matrix representation $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ under a particular basis).
However, I'm not really sure how to use this knowledge to prove the proposition in question. Any steps in the right direction would be appreciated. Also, I'd love to see any other proofs of the proposition.
Thanks in advance!
AI: Hint:
If a matrix has two distinct eigenvalues, it is similar to $\begin{pmatrix} a&0 \\ 0&b\end{pmatrix}$ with $a\neq b$. Now, suppose that $A$ has the eigenvalue $a$ of algebraic multiplicity $2$. We note that $A - aI$ must be nilpotent. From there, note that for any invertible $S$,
$$
S(A - aI)S^{-1} = SAS^{-1}-SaIS^{-1} = SAS^{-1}-aI
$$ |
H: Symmetric System of Equations
I'm new on studying Systems of equations. I just want to know the number of real solutions of this system of equations:
$$x^2-y^2=z$$
$$y^2-z^2=x$$
$$z^2-x^2=y$$
I also want to know how was your solution and explanation on how did you find your answer.
AI: Add $x^2 - y^2 = z$ and $y^2 - z^2 = x$ to get $x^2 - z^2 = x + z$. This means either $x + z = 0$, or $x - z = 1$. Similarly, either $y + x = 0$ or $y - x = 1$, and either $z + y = 0$ or $z - y = 1$.
Now there are a total of $8$ different combinations of equations to deal with.
$x + z = 0$, $y + x = 0$, $z + y = 0$
$x + z = 0$, $y + x = 0$, $z - y = 1$
$x + z = 0$, $y - x = 1$, $z + y = 0$
$x + z = 0$, $y - x = 1$, $z - y = 1$
$x - z = 1$, $y + x = 0$, $z + y = 0$
$x - z = 1$, $y + x = 0$, $z - y = 1$
$x - z = 1$, $y - x = 1$, $z + y = 0$
$x - z = 1$, $y - x = 1$, $z - y = 1$
And then you just proceed to solve for $x$, $y$, and $z$, and find which cases make sense and which ones don't. |
H: Is it possible to solve this nonlinear equation analytically?
Is it possible to solve the following equation analytically?
$B_1\exp(\beta_1 x) + B_2\exp(\beta_2 x) = C_1\exp(\alpha_1 x) + C_2\exp(\alpha_2 x)$
where, $B_1$, $B_2$, $C_1$, $C_2$, $\beta_1$, $\beta_2$, $\alpha_1$ and $\alpha_2$ are constants. And $x$ is the independent variable.
Many many thanks in advance.
AI: No. For example, $1+e^x=e^{2x}+e^{17x}$ becomes $y^{17}+y^2-y-1=0$ on substituting $y=e^x$, a polynomial equation of degree 17. Polynomials of degree exceeding 4 are, in general, not solvable analytically, only numerically. |
H: Prove that the polynomial $x^6+x^4-5x^2+1$ has at least four real roots.
Prove that the polynomial $x^6+x^4-5x^2+1$ has at least four real roots.
Talking analysis here, using the definition of continuity, intermediate value theorem, and extreme value theorem.
AI: Giving your polynomial the name $f$,
$f(-2)$ is positive.
$f(-1)$ is negative.
$f(0)$ is positive.
$f(1)$ is negative.
$f(2)$ is positive.
So the intermediate value theorem says there must be a root in each of $(-2,-1)$, $(-1,0)$, $(0,1)$, and $(1,2)$. |
H: Using residue to compute real fractional integral
Compute the integral $$\int_{-\infty}^\infty \dfrac{t-1}{t^5-1}dt$$ The hint is to use residues. I tried taking a look at the residue theorem, but I don't know which curve in the complex plane I should be integrating over, and what the complex function should be (it should match $\dfrac{t-1}{t^5-1}$ on the real line?)
AI: Note this function has a removable singularity at $t=1$. Take a large semicircle in the upper half-plane with its diameter $[-R,R]$. Let $f(z)=\dfrac1{z^4+z^3+z^2+z+1}$. |
H: Proof for the property of a fixed point of $f$.
Suppose that $f \colon [a, b] \to [a, b]$ is continuous. (Note that the range of $f$ is a subset of $[a, b]$)
Prove that there exists at least one point $x \in [a, b]$ such that $f(x) = x$. A point with this property is known as a fixed point of $f$.
AI: If $f(a)=a$ or $f(b)=b$ then we are done, suppose $f(a)\ne a, f(b) \ne b$
Now what can you say about $g$ below? does it vanish somewhere in $(a,b)$?
$g(x)=f(x)-x$ |
H: Zeros of polynomials are continuous
For two sets $A,B$, let $d(A,B)=\sup_{x\in A}\inf_{y\in B}|x-y|+\sup_{y\in B}\inf_{x\in A}|x-y|$.
Let $p(z)=a_nz^n+\ldots+a_0$, and let $\epsilon>0$. Show that there exists $\delta>0$ such that for any $q(z)=b_nz^n+\ldots+b_0$ such that $\max_k|a_k-b_k|<\delta$, we have $d(Z_p,Z_q)<\epsilon$, where $Z_p,Z_q$ denote the sets of zeros of $p$ and $q$.
I think it must be somehow related to Rouche's theorem, but I can't see how.
AI: Hint: Write $q=p+(q-p)$ and choose $\delta$ small enough so that $|q-p|<|p|$ on the circles of radius $\epsilon$ centered at the roots of $p$. |
H: What's the exact meaning of this sentence from George Peacock?
I am reading the book "A history of abstract algebra" by Israel Kleiner. The following sentence is said by George Peacock. I am not a native English speaker. So could someone translate it into plain English?
In symbolical algebra, the rules determine the meaning of the
operations... we might call them arbitrary assumptions, inasmuch as
they are arbitrarily imposed upon a science of symbols and their
combinations, which might be adapted to any other assumed system of
consistent rule.
AI: The context makes the meaning clear. In 1830, George Peacock wrote a book called "A Treatise on Algebra". In it, he attempted to justify things like $(-x)(-y) = xy$ (note that negative numbers were still far from universally accepted then, much less the meaning of multiplying them), by arguing as follows: that $(a-b)(c-d) = ac + bd - ad - bc$ is a law of "arithmetical algebra" whenever $a > b$ and $c > d$, so it is a law of symbolical algebra (always holding, whatever the values of $a, b, c, d$). With $a = 0$ and $d = 0$, you get $(-b)(-d) = bd$ for all values of $b$ and $d$.
The significance was looking at operations like $(a-b)(c-d)$ in an axiomatic sort of way, in terms of rules (like that the distributive law should hold). Rather than looking at such expressions only as place-holders for what happens when dealing with positive integers (which is what he meant by "arithmetical algebra"), he was looking at them more abstractly, letting $a, b, c, d$ be truly and purely just symbols, not necessarily positive integers or even integers.
Under certain circumstances (certain values of $a$, $b$, $c$, etc.), it may be possible to interpret a result of symbolical algebra, or give meaning to it. Under other circumstances, it may not be, but still the result holds in symbolical algebra. For instance, he says that a product of three symbols, like $abc$, can be interpreted as volume, in the case when they are all positive numbers. Something like $abcd$ has no interpretation (he says), because there are only three dimensions, but still we can work with expressions like that in symbolical algebra, as symbolical algebra is just a system of rules for operating on symbols. We don't have to worry at every step about the meaning of the operations; the meaning is just whatever the rules say.
Thus, going over (the quoted part of) his quote line by line:
In symbolical algebra, the rules determine the meaning of the operations...
In symbolical algebra, the meaning of operations (like $(-b)(-d)$ or $(a-b)(c-d)$) is determined by the rules (like the distributive law).
we might call them arbitrary assumptions,
You can take these rules to be arbitrary assumptions.
inasmuch as they are arbitrarily imposed upon a science of symbols and their combinations,
The symbolical algebra can be thought of as "a science of symbols and their combinations", with rules arbitrarily attached to it. For instance, Boolean algebra was another "science of symbols and their combinations", with different arbitrary rules attached: it had rules like $x + x = x$, for instance.
which might be adapted to any other assumed system of consistent rules.
You could assume different rules, and as long as they were consistent, you would still have a symbolical algebra. The Boolean algebra above is an example. |
H: Linear homogeneous recursive sequence of constant sign
Let recursive sequence be defined by the formula
$$
s_{j+1}=as_j-s_{j-1},
$$
where $a>1$ is some integer number. Is it true that $s_0<0$, $s_1<0$ implies $s_j<0$ for $j \geq 0$?
Edit: No, its wrong. Under what conditions on $s_0$ and $s_1$ property $s_j<0$ holds?
AI: First assume $a > 2$, the characteristic equation of the recurrence equation
$$s_{j+1} = a s_{j} - s_{j-1}$$
is given by
$$\lambda^2 - a \lambda + 1 = 0$$
It has roots at $\mu$ and $\mu^{-1}$ where $\displaystyle \mu = \frac{a + \sqrt{a^2-4}}{2} > 1$.
The general solution of the recurrence equation have the form
$$s_j = A \mu^j + B \mu^{-j}$$
Solving $A$ and $B$ from $s_0$ and $s_1$, we get:
$$s_j = \frac{\mu s_1 - s_0}{\mu^2 - 1}\mu^j + \frac{\mu^2 s_0 - \mu s_1}{\mu^2 - 1} \mu^{-j}$$
Since $|\mu| > 1 > |\mu^{-1}|$, $s_j$ will be dominated by the term $A \mu^j$ for large $j$. If $A > 0$,
then $s_j$ will be positive for sufficient large $j$. This means $A \le 0$ is an necessary
condition for all $s_j < 0$. This condition is also sufficient. If $A \le 0$, then for $j > 1$, we have
$$s_j = A \mu^j + B \mu^{-j} = A \mu ( \mu^{j-1} - \mu^{1-j} ) + \mu^{1-j} (A \mu + B \mu^{-1})\\ = A \mu ( \mu^{j-1} - \mu^{1-j} ) + \mu^{1-j} s_1 \le \mu^{1-j} s_1 < 0$$
Conclusion, the necessary and sufficient condition for all $s_j < 0$ is $A \le 0 \iff \mu s_1 - s_0 \le 0$.
For $a = 2$, the general solution has the form
$$s_j = A j + B$$
It is easy to see the necessary and sufficient condition is again $A \le 0 \iff s_1 - s_0 \le 0$. |
H: Calculating the expected profit with Probability A level maths CIE
Company sets up display of 20 fireworks! for each firework, the probability that it fails is 0.05,independently of other fireworks
the probability that more than 1 firework fails is 0.264
the 20 firework cost company 24 dollars each .450 pay the company 10 dollars each to watch the display.if more than 1 firework fails to work, they get their money back.
Calculate the expected Profit of the company.?
Answer given to me is 2830 dollars [2832]
Hint: -480+4500(1-0264)
Please help me with why they multiplied 4500 with (1-0.264)
I've exams tomorrow and i will be so thankful to you
AI: Hint: If the probability (call it $P(X \ge 1)$) that more than $1$ firework fails, what is $1-P(X \ge 1)$ going to give you? Keeping in mind that $1$ is the probability of the sample space. |
H: Evaluating a Real Improper Integral by Residues
I am having trouble evaluating this improper integral due to its integrand and the singularities that are present. The question reads as
Show that $\int_{-\infty}^{\infty}\frac{dx}{x^4+1}=\frac{\pi}{\sqrt{2}}$.
The contour is assumed to be the boundary of the half disc $|z|\leq R, Im(z)\geq 0$ taken once anticlockwise. You may assume that the integrals converge.
The questions that I have been solving have had "nice" singularities; in the form of $x+iy,x,y\in \mathbb{R}$. I understand the process that takes place in this case. However, this particular integrand has singularities (inside the boundary) of $z=i^{1/2},i^{3/2}$. This is already an intuitive problem. How does one situate these singularities inside the boundary and thus progress with calculations? Furthermore, if I continue with the methodic process I will not get the answer.
Thank you in advanced for you help.
AI: Factorization is $(x+0.5+0.5i) \sqrt2)(x-0.5+0.5i) \sqrt2)(x+0.5-0.5i)\sqrt2)(x-0.5-0.5i) \sqrt2)$ The third and fourth factor provide the root in the upper half plane.
To find the residues: Knock out the third factor and substitute its roots in the fraction. That's one residue. And then, knock out the fourth factor (put the third back of course) and then put in its root in the fraction. Now you have two residues. Add them up and multiply by $2 \pi i$. That should do it. It's time for me to sleep now. Good luck |
H: Is there an example where the maximal and maximum elements are different?
I know by heart the definitions of both maximal and maximum elements but cannot grasp examples when these dont coincide. Usually with numbers it is easy to see that maximum and maximal coincide. But what about when they dont coincide?
I recently read an example where we have three presidential (a,b and c) candidates and two groups that have preferences over them. So group one prefers a to b and b to c. Group 2 prefers b to a and a to c. As result they both rank c at the bottom. Thus a and b are maximum elements in society as a whole, which makes them both maximal and maximum elements.
Can you offer an example where we can have maximal elements and maximum elements in a set and that they dont coincide?
Finally, is it possible to have more than one element in the maximal set? My intuition says yes, but doesnt this make it coincide with maximal elements, as in the example of presidential candidates where both a and b are preferred to c?
Thanks
AI: Consider the partially ordered set $\mathcal{P}(\Bbb{N})-\{\Bbb{N}\}$ (ordered by inclusion relation $\subset$). It has maximal element. In fact, $\Bbb{N}-\{n\}$ is maximal elements of $\mathcal{P}(\Bbb{N})-\{\Bbb{N}\}$ so it has maximal elements infinitely many. But it does not have maximum.
You can easily check that every maximum is maximal element. If $(P,\le)$ is totally ordered then maximum and maximal element are coincide. (That is, every maximal element in totally ordered set is maximum in this set.) |
H: Prove a function is not uniformly continuous.
Use the definition of uniform continuity to prove the function G(x) = x^3 is not uniformly on [0, infinity).
AI: Equivalent sequential Criterion for Uniform continuity is violated by taking
$x_n=n+{1\over n},y_n=n, \text{Then} (x_n-y_n)\to 0\text {but } f(x_n)-f(y_n)\nrightarrow 0$ |
H: Orthogonal projection on vector space convergence in $L^p$
Let $R_N$ be the set of $2^N$ intervals $$\left\{\left[0,\frac{1}{2^N}\right), \left[\frac{1}{2^N}, \frac{2}{2^N}\right),\ldots,\left[\frac{2^N-1}{2^N}, 1\right)\right\}.$$ Let $$V_N=\operatorname{span}\{1_I\mid I\in R^N\}$$ Let $P_N:L^2([0,1])\rightarrow V_N$ be the orthogonal projection onto $V_N$, which is defined by $$P_Nf=\sum_{I\in R_N}2^N\langle f,1_I\rangle 1_I$$
Prove that if $f\in L^p([0,1])$, then $P_Nf\rightarrow f$ in $L^p([0,1])$ as $N\rightarrow \infty$.
We must show that $\int_0^1|f(x)-P_Nf(x)|^pdx\rightarrow 0$ as $N\rightarrow \infty$.
Suppose the point $x$ belongs to the interval $I_x$ in $R_N$, we have $$|f(x)-P_Nf(x)|^p=|2^N\langle f,1_I\rangle -f(x)|^p=\left|2^N\int_{I_x}(f(y)-f(x))dy\right|^p$$
But how would this yield the desired convergence $$\int_0^1\left|2^N\int_{I_x}(f(y)-f(x))dy\right|^pdx\rightarrow 0$$ as $N\rightarrow \infty$?
EDIT: Prahlad Vaidyanathan provided an answer below, but I think the proof is incorrect (or at least, incomplete), as I pointed out in the comment. How can we fix it?
AI: By this earlier problem, you know that
$$
P_N(g) \to g\quad\forall g\in C[0,1]
$$
Now if $f\in L^p[0,1]$ (with $1\leq p < \infty$), and $\epsilon > 0$, there is $g\in C[0,1]$ such that
$$
\|f-g\| < \epsilon/3 \Rightarrow \|P_N(f) - P_N(g)\| < \epsilon/3
$$
and hence
$$
\|P_N(f) - f\| \leq \epsilon/3 + \|P_N(g) - g\| + \epsilon/3
$$
so for $N$ large enough, you will get $\|P_N(f) - f\| < \epsilon$ |
H: How to find 'k' from this equation
I have a problem to calculate 'k' from this equation :
$$X = \frac{\left(\rho-\rho^{k+1}\right)\left(1-\frac{\gamma}{2}\rho\right)^{2}-k\rho^{k}\left(1-\frac{\gamma}{2}\rho\right)}{\lambda\left(1-\rho\right)\left(1-\rho^{k}\right)}$$
When I expand this equation, it will be :
$\rho^{k}$(some equations) - $k\rho^{k}$(some equations) = value containing X and $\lambda$
because there is 'k' in $k\rho^{k}$, I can not calculate k from given X and $\lambda$.
Could anyone know how to find k from this equation?
Thank you
AI: By multiplying both sides of equation by some constant, you will get the equation like
$$(A+k)\rho^k=B(X,\lambda)$$
Or,
$$(A+k)\rho^{A+k}=(A+k)e^{(A+k)\ln\rho}=B\rho^A\\
\left((A+k)\ln\rho\right)e^{(A+k)\ln\rho}=B\rho^A\ln\rho$$
Then you will refer to Lambert W function and obtain
$$(A+k)\ln\rho=W_e(B\rho^A\ln\rho)$$
and then the solution is
$$k=\frac{W_e(B\rho^A\ln\rho)}{\ln\rho}-A$$ |
H: Borel sets on the plane
Let's say we have two sigma algebras $D_1$ and $D_2$ both of which contain open intervals. We know that the Borel sigma algebra $B(R)\subset D_{1}\cap D_{2}$. I'm having difficulty proving that $B(R)\otimes B(R)\subset D_{1}\otimes D_{2}$. Any help will be greatly appreciated.
Thanks.
AI: Every measurable rectangle, $A \times B$ ($A, B \in B(R)$) is in $D_1 \otimes D_2$, and since $D_1 \otimes D_2$ is a sigma algebra and $B(R) \otimes B(R)$ is the smallest sigma algebra containing such rectangles, the containment follows. |
H: Logarithm of singular matrix
How do we define logarithm of a singular matrix(Say it is real square symmetric and has distinct eigen values). I tried searching online but could not find much information(Something that someone as dumb as me could understand).
MATLAB logm help says something about principal and non principal logarithm and MATLAB does give log of singular matrix. Can someone explain?
So, if I am to calculate it by hand using either decoposition into eigen value matrix and eigen vector matrices or jordan form, how do I do it?
AI: The online documentation from MathWorks says that
If $A$ is singular or has any eigenvalues on the negative real axis, the principal logarithm is undefined. In this case, logm computes a nonprincipal logarithm and returns a warning message.
This is poorly written. The first part of the second sentence (about nonprincipal logarithm) actually refers to the case where $X$ is nonsingular and it has a negative real eigenvalue, but not the case where $X$ is singular. Singular matrices have no matrix logarithms.
A matrix logarithm of a complex square matrix $X$ is some matrix $Y$ such that $e^Y=X$. The eigenvalues of $e^Y$, however, are the exponentials of the eigenvalues of $Y$. Since a singular matrix always has a zero eigenvalue but $e^\lambda=0$ is not solvable, singular matrices have no matrix logarithms.
The documentation for an earlier version of Matlab is clearer, although there is still room of improvement:
... Complex results are produced if A has negative eigenvalues. A warning message is printed if the computed expm(L) is not close to A.
... Some matrices, like A = [0 1; 0 0], do not have any logarithms, real or complex, and LOGM cannot be expected to produce one. |
H: Hackenbush game strategy for stalk
There are some piles of numbers. Numbers are divided in 2 groups, A and B. Player x plays with group A and player y plays with group B. x makes the move first. On each step a player chooses a pile and chooses a number of his group and remove all numbers smaller and equal to the number (number of both group A and B) from that piles only.
The person who can't make the move which means he finds no number of his group, loses.
Is that a nim game? What should be the strategy? Who will win?
AI: It's not a nim game, it's an Hackenbush game (Blue-Red version if $A\cap B=\emptyset$)
Replace each number by a line segment of color red for number from $A$, and blue for $B$. To each pile of number, build a pile of line segment, by decreasing order of number.
For example, if $A$ are even numbers, and $B$ are odd numbers, the pile $[3,4,7,8,10]$ will be replace by ground-red(for 10)-red(for 8)-blue(for 7)-red(4)-blue(3).
Hackenbush games are not nim games. To each finite position can be assigned a dyadic number whose sign will give the player who can win.
EDIT : strategy explained for player x :
To each number (from $A$) in each pile, $x$ will assign a value that will be :
$1$ if this the largest number of the pile
$1$ if the next larger number was assigned 1 and is from the same set ($A$ or $B$)
$2^{-n}$ where $n$ is the distance to the closest number with value $1$.
Example : in pile [10,8,7,4,3,1], 10 will have value 1 (by rule 1), 8 value 1 (by rule 2), and 4 value $\frac{1}{4}$ (by rule 3).
For player y, 7 will be assigned value $\frac{1}{2}$ (rule 2 does not apply here because the next larger number 8 is not from the same set, so rule 3), 3 value $\frac{1}{8}$, and 1 value $\frac{1}{16}$.
The best strategy is to remove the number of lowest value. (This is not the quickest strategy !)
If you add all values for player x and subtract all values for player y, a positive sum means player x has a winning strategy, negative sum is for player y while a zero sum means second player. |
H: Evaluating $\int_0^\infty \frac{\log t}{1+t^2}\,\mathrm dt$ using residues
I want to integrate $$\int_0^\infty \dfrac{\log t}{1+t^2}\,\mathrm dt$$ using the residue theorem. The poles are at $i,-i$. If the integral were from $-\infty$ to $\infty$, I would consider integrating along the semicircle with large radius $R$ and centered at the origin. But here it is from $0$ to $\infty$. What contour would be good to integrate along, if any?
AI: You can use a keyhole contour about the positive real axis, but you would consider the contour integral
$$\oint_C dz \frac{\log^2{z}}{1+z^2}$$
where $C$ is the above described keyhole contour. The integrals over the large and small arcs about the origin vanish as the respective radii go to infinity and zero. Thus, we have that the above contour integral is equal to
$$\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{1+x^2} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^2}+4 \pi^2 \int_0^{\infty} dx \frac{1}{1+x^2}$$
This contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$:
$$\frac{i 2 \pi}{i 2} \left ( -\frac{\pi^2}{4} +\frac{9 \pi^2}{4}\right ) = 2 \pi^3$$
Setting the above two expressions equal to each other, you can show that the desired integral is zero. |
H: How is it possible to have a transitive yet not complete relation?
How can we say we cannot compare between pairs (no completeness) and yet we can have transitivity?
Let's assume the relation is a preference relation.
For instance if my set has: $(a, b, c)$ How can I say that a is preferred to b, b is preferred to c and hence a is preferred to c (transitivity) if the relation is not complete and thus I cannot compare a to b and b to c?
AI: EDIT: This question is about microeconomic theory, not set theory. Directly following is an answer related to microeconomics. Further below is a set-theoretic answer, which I'm going to leave there in case someone searches for similar keywords.
Microeconomic Answer:
In economic preference, we say that a relation is complete if and only if for all $A,B$, we have $A \succsim B$ or $B \succsim A$ or both. So consider $A,B,C$ such that $A\succsim B$ and $B \succsim C$. Then the relation is not complete because we do not have an explicit relationship $A \succsim C$. However, we can infer one by transitivity: if $A\succsim B$ and $B \succsim C$, then by transitivity $A \succsim C$, though technically the relation is not `complete'.
Set-Theoretic Answer:
I don't understand your terms (what do you mean with completeness?). So I am somewhat guessing at what your question actually asks. Please comment with clarifications so I can provide a better answer.
A relation $R$ on a set $X$ is a subset of $X \times X$.
An equivalence relation on $X$ is a relation $R$ on $X$ such that:
Reflexivity: For all $a \in X, (a,a) \in R$
Symmetry: For all $a,b \in X$, if $(a,b) \in R$, then $(b,a) \in R$.
Transitivity: For all $a,b,c\in X$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.
Your notion of completeness seems to be a restatement of trichotomy: in an ordered set, one of the three is true: $a=b$, or $a>b$, or $a<b$.
Notably, we can have transitivity in terms of an equivalence relation on a set, while the set itself is unordered and not transitive. Is this what you mean? |
H: How find the $\sum_{n=1}^{\infty}\frac{1}{n^p}\left(1-\frac{x\ln{n}}{n}\right)^n$ convergent
Question:
Study the series
$$\sum_{n=1}^{\infty}\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n$$
convergence,when $p$ and $x$ such what conditions?
My try:
since
$$\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n=\dfrac{1}{n^p}e^{n\ln{\left(1-\dfrac{x\ln{n}}{n}\right)}}=\dfrac{1}{n^p}e^{n\left(-\dfrac{xln{n}}{n}+o(-\dfrac{xln{n}}{n})\right)}=\dfrac{1}{n^p}e^{-x\ln{n}+o(-x\ln{n})}$$
and my try is usefull? and How solve this problem, Thank you.
AI: We can find easily that
$$\dfrac{1}{n^p}\left(1-\dfrac{x\ln{n}}{n}\right)^n\sim_\infty\dfrac{1}{n^p}\dfrac{1}{n^x}=\dfrac{1}{n^{p+x}}$$
so the necessary and sufficient condition of the convergence is
$$p+x>1$$
Added Following the Sundaycat's request
Recall that $\ln(1- x)\sim_0 -x$ so
$$\left(1-\dfrac{x\ln{n}}{n}\right)^n=\exp\left(n\ln\left(1-\dfrac{x\ln{n}}{n}\right)\right)\sim_\infty\exp(-x\ln n)=\frac{1}{n^x}$$ |
H: Linear System of ODE
Would you mind telling me how do we use Matrix Algebra to get a general solution of the system:
$$x'=-{\delta}^{2}x+y+\delta z$$
$$y'=-x-{\delta}^{2}y+\delta z$$
$$z'=-\delta z$$
where $\delta$ is a parameter.
AI: We rewrite this as:
$$x' = A x = \begin{bmatrix} -\delta^2 & 1 & \delta \\ -1 & -\delta^2 & \delta \\ 0 & 0 & -\delta \\ \end{bmatrix}x$$
We find the eigenvalues of this matrix using $|A - \lambda I| = 0$, which gives us:
$$-\delta - \delta^5 - \lambda - 2 \delta^3 \lambda - \delta^4 \lambda -
\delta \lambda^2 - 2 \delta^2 \lambda^2 - \lambda^3 = -(\delta+\lambda) ((\delta^2+\lambda)^2+1) = 0$$
This gives us the three eigenvalues:
$$ \lambda_1 = -\delta,~ \lambda_2 = -\delta^2 + i, ~\lambda_3 = -\delta^2 - i$$
Now, you can find the eigenvectors and then write the solution for $x(t)$.
The three eigenvectors are:
$v_1 = \left(\dfrac{\delta (\delta^2- \delta +1 )}{\delta^4-2 \delta^3 + \delta^2+1}, -\dfrac{-\delta^3 +\delta^2+\delta}{\delta^4-2 \delta^3 + \delta^2+1},1\right)$
$v_2 = (-i, 1, 0)$
$v_3 = (i, 1, 0)$
Also note, that $z$ is a decoupled equation and you can solve for it straight off, substitute back into the $x'$ and $y'$ and reduce this to a $2 x 2$ system, which is much easier to work with when dealing with parameters. |
H: Sum $\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+...+\binom{n+m}{k}$
Evaluate the following series sum which n, m, k are nonnegative integers.
$$\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+...+\binom{n+m}{k}$$
I have no idea about it@@
AI: I don't know if this is the kind of thing you are looking for, but here goes : Consider
$$
(1+x)^n + (1+x)^{n+1} + \ldots + (1+x)^{n+m}
$$
The expression you want to evaluate is the coefficient of $x^k$ of this polynomial. So re-write it as
$$
(1+x)^n\{1 + (1+x) + \ldots + (1+x)^m\} = (1+x)^n \left \lbrace \frac{1-(1+x)^{m+1}}{1-(1+x)} \right\rbrace = \frac{(1+x)^n}{x}\left((1+x)^{m+1} - 1\right )
$$
Now expand both sides out using the Binomial theorem, and, keeping in mind the $1/x$ factor there, the coefficient of $x^k$ turns out to be
$$
\sum_{i+j=k} {n\choose i+1}{m+1\choose j}
$$ |
H: How to bound away integral over complex rectangle?
I am integrating the following integral
$$\int_{-\infty}^\infty\frac{\cos t}{e^t+e^{-t}}dt$$
by computing residues inside the rectangle with vertices $-R,R,-R+\pi i,R+\pi i$. On the left and right side of the rectangles, $\cos z$ is bounded, while $e^z+e^{-z}$ gets large, so the integral goes to zero. But what about the top side near the imaginary axis? How can we bound $\frac{\cos t}{e^t+e^{-t}}$ from above?
AI: First note you can change the integral to
$$\int_{-\infty}^\infty \frac{e^{it}}{e^t+e^{-t}}dt$$
because $\frac{i\sin(t)}{e^t+e^{-t}}$ is odd.
Then if you proceed the same way you suggested, integrating along the top edge of the rectangle comes down to integrating
$$\frac{e^{i(t+i\pi)}}{e^{t+i\pi}+e^{-t-i\pi}}= (-e^{\pi}) \frac{e^{it}}{e^{t}+e^{-t}}$$
which is just a multiple of the bottom edge. That means when you add the four integrals you get the two side edges plus $(1+e^\pi)$ times the bottom edge. Note the change in sign because the top integral will be in the opposite direction of the bottom one. Then your final answer will be $(1+e^\pi)^{-1}$ times whatever the residue calculation is. |
H: Is multiplication by zero in an equation allowed?
If we have equal quantities, we cannot divide with zero. But, we can multiply both sides with zero. But, my friend said, even multiplication with zero also wrong it seems. Unfortunately, he is not explaining the why wrong, if multiplication with zero on both sides?
Ex: $7 = 7$
If we multiply with zero on both sides, $0\times7 = 0\times7 \Rightarrow 0 = 0.$
But, multiplication with zero on both sides also illegal it seems. Why?
Otherwise, in what cases, the multiplication with zero on both sides are not applicable. Please explain?
Thank you.
AI: Multiplication of both sides of an equation by zero will always result in another true statement, it just won't be very useful. Multiplication by zero destroys information. The reason you can't divide both sides by zero is that division by zero just isn't defined, and there are plenty of threads here explaining why. |
H: How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$
let $f:\mathbb R\to \mathbb R$,and such
$$f(f(x))=\dfrac{x+1}{x+2}$$
Find the $f(x)$
My try
I found $f(x)=\dfrac{1}{x+1}$
because when $f(x)=\dfrac{1}{x+1}$,then
$$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$
so $f(x)=\dfrac{1}{x+1}$ such this condition,But $f(x)$ Have other form? Thank you
AI: You can easily check that if we define
$$f_{A} = \frac{ax+b}{cx+d} \quad \text{for} \quad A = \begin{pmatrix}a & b \\ c & d \end{pmatrix}, $$
then $f_{A} \circ f_{B} = f_{AB}$. Thus any matrix $A$ satisfying
$$ A^{2} = k \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad \text{for some} \ k \neq 0 $$
gives rise to a solution. Mathematica yields two different solutions
$$ f(x) = \frac{1}{x+1} \quad \text{and} \quad f(x) = \frac{2x+1}{x+3}, $$
but I'm not sure if other solutions exist. |
H: Polynomial functions of odd degree are surjective
Prove if the function $f: \mathbb{R} \to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}.$
We know a polynomial, $f(x)=a_nx^n +a_{n−1}x^{n−1} ...a_1x+a_0$ with real coefficients is continuous. Also, $\mathbb{R}$ is connected now since $\mathbb{R}$ is connected then $f(\mathbb{R})$ is connected, thus we can apply the intermediate value theorem. Now any polynomial of odd degree has at least one real root since every polynomial, with real coefficients, has as many roots as it has degrees. Also, any complex roots are paired with their complex conjugates thus they take up an even number of roots. So, an odd degree polynomial has only an even number of roots that can be complex therefore at least one root must be real. Therefore, there exists $p,q\in \mathbb{R}$, $p<q$ such that $f(p)<0$ and $f(q)>0$. Thus we can then choose $p\in \mathbb{R}$ such that $f(p) \ge 0,$ $f(p)\le 0$ and obtain any $f(\mathbb{R})\in \mathbb{R}$.
Similarly, suppose that the leading coefficient of $f$ is positive then, $\lim_{x\to \infty} f(x) = \infty$, and $\lim_{x\to -\infty} f(x) = -\infty$. So, any arbitrarily large or small value can be attained. Due to the intermediate value theorem and since $f$ is continuous, any value between such a large value $M$ and a small value $m$ is attained. Hence, $f(\mathbb{R}) = \mathbb{R}$. And a similar argument follows if the leading coefficient of $f$ is negative. Therefore, $f(\mathbb{R}) = \mathbb{R}$.
Is this correct?
AI: That is correct, but all you really need is the last part : Assume without loss of generality that the leading coefficient is positive (else apply this result to $-f$), and note that
$$
\lim_{x\to +\infty} f(x) = +\infty, \text{ and } \lim_{x\to -\infty} f(x) = -\infty
$$
Then, for any $y\in \mathbb{R}$, there is $M > 0$ such that
$$
x > M \Rightarrow f(x) > y, \text{ and } x < -M \Rightarrow f(x) < y
$$
Hence, by intermediate value theorem, there is $x_0 \in [-M,M]$ such that $f(x_0) = y$.
This is true for all $y \in \mathbb{R}$, and hence $f(\mathbb{R}) = \mathbb{R}$ |
H: Show $ n + \operatorname{lcm}(a,b)\mathbb{Z} \subseteq (n + a \mathbb{Z}) \cap (n + b \mathbb{Z}) $
let $n \in \mathbb{Z}$. I want to show
$$ n + \operatorname{lcm}(a,b)\mathbb{Z} \subseteq (n + a \mathbb{Z}) \cap (n + b \mathbb{Z}) $$
for integers $a,b$
TRY:
Since $lcm(a,b)$ is multiple of $a,b$, then $lcm(a,b) = ka = k'b$. Therefore $n + lcm(a,b) = n + ka \mathbb{Z} = n + bk' \mathbb{Z} $. and hence $ n + \operatorname{lcm}(a,b)\mathbb{Z} \subseteq (n + a \mathbb{Z}) \cap (n + b \mathbb{Z}) $
AI: HINT: Show that if $r$ is a multiple of $s$, then $n+r\Bbb Z\subseteq n+s\Bbb Z$. |
H: Finding value(s) of a for which f is continuous
A function $f$ is defined as follows: $$f(x)=\begin{cases}\sin x&\text{if }\;x\leq c,\\ax+b&\text{if }\;x>c,\end{cases}$$ where $a,b,c$ are constants. If $b$ and $c$ are given, find all values of $a$ (if any exist) for which $f$ is continuous at the point $x=c$.
So, I'm pretty new to thinking about continuity. I know that I need to show that $$f(c)=\lim_{x\to c}f(x)$$ (and that both exist). However, I'm a little lost. I see that $f(c)=\sin c$, and I think that I've got to find the point where the two cases meet, which should be the limit. $$\sin c=ac+b\implies a=\frac{\sin c-b}{c}.$$ At this point, though, I'm not sure I'm on the right track, and I'd appreciate your help.
AI: A function is said to be continous if LHL=RHL=value of the function at that point.
Therefore
$$\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)$$
is the condition of for continuity.
In your problem
$$\lim_{x\to c^-}f(x)=\lim_{x\to c^-}\sin x=\sin c$$
$$\lim_{x\to c^+}f(x)=\lim_{x\to c^+}ax+b=ac+b$$
and the value of the fucntion at x=c is $\sin c$
.
Hence for the function to be continuous you must have
$$ac+b=\sin c$$ |
H: Representing functions as power series: Homework questions
Completed all my homework exept for this problem. Our teacher likes to give us a problem at the end that is for a future lesson. That is this question above. So it is kind of tricky...
I don't have any written solution so far. I believe there is something to do with the power rule in the problem?
Thanks!
AI: njguliyev’s answer points you at the slick way, which I recommend, but the problem can be done by brute force as well. If you differentiate $f$ a few times, it’s not too hard to pick up the pattern:
$$\begin{align*}
f(x)&=\frac3{4-x}=3(4-x)^{-1}\\
f\,'(x)&=1\cdot3(4-x)^{-2}\\
f''(x)&=2\cdot1\cdot3(4-x)^{-3}\\
f'''(x)&=3\cdot2\cdot1\cdot3(4-x)^{-4}\\
&\;\vdots\\
f^{(n)}(x)&=n!\cdot3(4-x)^{-n-1}=\frac{3n!}{(4-x)^{n+1}}\;.
\end{align*}$$
Now just substitute into the usual formula for the coefficient of $x^n$ in the Taylor series expansion about $1$. |
H: Existance of Hamiltonian cycle in the connected graph.
I know the fact, that if a graph is connected and each of its vertices has a degree of $2$, then graph is a cycle graph and it has a Hamiltonian path. From that I easily conclude, that, if graph with n vertices is connected and each of its vertices has a degree at least $2$, then there must be a Hamiltonian cycle in this graph. I dont have a strict mathematical proof of this, but, I think it's obvious from the first fact I mentioned. I can conclude even more : if graph with $n$ vertices is connected and each of its vertices has a degree at least $2$, then this graph must have a subgraph, that is a cycle graph.
The question is, am I thinking correctly?
AI: What about this graph:
x x
|\ /|
| x---x |
|/ \|
x x
It’s connected, and every vertex has degree $2$ or $3$, but there is no Hamilton cycle, thanks to the bridge in the middle. |
H: Finding the inverse function
The question is to find the inverse function of
$$f(x)=x-(2\sqrt{x})+1$$
I first found that the domain of definition is $\,x\ge 0$
Then studied the variation of the function and it is decreasing between $0$ and $1$ and increasing otherwise. Thus there are $2$ inverse functions to be found.
How can I find them. Any hints?
AI: So $y = x-2\sqrt{x}+1 = (\sqrt{x}-1)^2$
$x = (1\pm\sqrt{y})^2=1 \pm 2\sqrt{y} + y$
So $f^{-1}(x)=x\pm2\sqrt{x}+1$
Specifically $f^{-1}(x) = x+2\sqrt{x}+1$ when $x\ge1$ and $f^{-1}(x) = x-2\sqrt{x}+1$ when $0\le x\le1$. |
H: Example of smooth function without compact support on open real interval
What is an explicit example of a smooth function on a real open interval that does not have compact support, i.e. for given $a,b \in \mathbb{R}$, a function (in common notation)
$$f \in C^{\infty}((a,b))\setminus C_0^{\infty}((a,b))$$
More precisely, I have the following technical issue: the support of a function is defined as a closure, i.e.
$$\text{supp } f = \overline{\{x\in(a,b) \vert f(x)\neq 0\}}$$
Intuitively I would think that a function $g\in C_0^{\infty}((a,b))$ has to vanish at the boundaries $a,b$ (e.g. to use for integration by parts), but because of this formal closure, it can be non-zero on $(a,b)$ and $\overline{(a,b)}=[a,b]$ is compact. Thus a non-zero function on the interval $(a,b)$ is in $C_0^{\infty}((a,b))$.
Where is the mistake in this reasoning?
AI: Well the closure $\overline{\{x\in(a,b) \vert f(x)\neq 0\}}$ is meant in the space you are working in and that is $(a,b)$. So $\overline{\{x\in(a,b) \vert f(x)\neq 0\}} \subset (a,b)$ has to hold. And if you have $g$ nonzero in whole $(a,b)$ than its support is $(a,b)$ but $(a,b)$ is not compact. |
H: does the max function holds the triangle inequality?
I need to prove if the following is a norm:
$$||f||:=\max_{-1<x<0}|f| + \max_{0<x<1}|f|$$
when $f$ is a continuous on $[-1,1]$.
The only problem I have is with showing it holds the triangle inequality.
My question is, what I can say about the max function that can help me with the last part?
AI: Let $N_{-}(f)={\sf max}_{-1<x<0}|f(x)|$ and $N_{+}(f)={\sf max}_{0<x<1}|f(x)|$
It suffices to show that $N_{-}$ and $N_{+}$ each satisfy the triangle inequality
individually.
Let $x\in (0,1)$. Then $|(f+g)(x)| \leq |f(x)|+|g(x)| \leq N_{-}(f)+N_{-}(g)$.
Since this is true for all $x$, we deduce $N_{-}(f+g) \leq N_{-}(f)+N_{-}(g)$.
Similarly for $N_{+}$. |
H: Countable disjoint union of non-measurable sets
Can a countable union of non-measurable sets of reals be measurable?
For instance, can we partition $\mathbb{C}$ into countably many disjoint non-measurable sets?
AI: Sure. Just take any non-measurable set and its complement. For example, let $V$ be a Vitali set, which is known to be non-measurable. We have
$$
\mathbb R = V \cup V^\complement.
$$ |
H: special parameter integral
Does anyone know a proof for the following formula ?
$$\int_{0}^{\infty} \frac {1}{x^y+1} dx=\frac{\frac{\pi}{y}}{\sin(\frac{\pi}{y})}$$
for $y>1$?
If $y$ is an even positive integer than the integral can be calculated using the
residue theorem. But even the case, that $y$ is an odd positive integer makes it
more difficult because the integral cannot be taken from $-\infty$ to $\infty$. For arbitrary $y$, it should be even more difficult.
AI: Consider the following contour integral:
$$\oint_C \frac{dz}{z^y+1}$$
where $C$ is a wedge which has four sections: 1) along the positive real axis from $[\epsilon,R]$; 2) along an arc of radius $R$ from the positive real axis, counterclockwise to the point $z=R e^{i 2 \pi/y}$; 3) along the line segment $z=e^{i 2 \pi/y} t$, $t \in [R,\epsilon]$; 4) a small arc of radius $\epsilon$ about the origin of angle $2 \pi/y$. Thus the contour integral is equal to
$$\int_{\epsilon}^R \frac{dx}{x^y+1} + i R \int_0^{2 \pi/y} d\theta \, \frac{e^{i \theta}}{R^y e^{i y \theta}+1} + e^{i 2 \pi/y} \int_R^{\epsilon} \frac{dt}{t^y+1} + i \epsilon \int_{2 \pi/y}^0 d\phi \, \frac{ e^{i \phi}}{\epsilon^y e^{i y \phi}+1}$$
As $R \to \infty$, the second integral vanishes because $y \gt 1$; as $\epsilon \to 0$, the fourth integral vanishes. By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at $z=e^{i \pi/y}$. Thus,
$$\left (1-e^{i 2 \pi/y}\right ) \int_0^{\infty} \frac{dx}{x^y+1} = \frac{i 2 \pi}{y \,e^{i \pi (y-1)/y}} = -\frac{i 2 \pi}{y} e^{i \pi/y}$$
Thus,
$$\int_0^{\infty} \frac{dx}{x^y+1} = -\frac{i 2 \pi}{y} \frac{e^{i \pi/y}}{1-e^{i 2 \pi/y}} = \frac{\pi/y}{\sin{(\pi/y)}}$$
ADDENDUM
I added the small arc over which the integral vanishes so as to avoid a branch point at $z=0$ for nonintegral $y$. Although it has no impact on the result, the derivation was not quite correct without it. It should be a habit to define integration contours so as to avoid branch points if Cauchy's theorem is to be invoked. |
H: A mean square derivative
I'm doing an exercise where I have to check some properties about these two stochastical processes:
$X(t)=At+B\;\;$ and $\;\;Y(t)=\frac{1}{t}\displaystyle\int_{0}^{t}X(\tau)\;d\tau$, $t>0$.
Where $A$ and $B$ are uncorrelated 2-random variables with $\mathbb{E}[A]=m_A$, $\mathbb{E}[B]=m_B$, $\mathbb{V}[A]=\sigma_A^2$ and $\mathbb{V}[B]=\sigma_B^2$.
I have checked that both verify that are continuous, differentiable and integrable, and I have their correlation functions $\Gamma_X$ and $\Gamma_Y$.
I have checked too using the definition that that $X'(t)=A$.
Now, I need to check that $Y'(t)=A\frac{t}{2}+B$. I have tried using the definition ( that is, checking if $\mathbb{E}[(\frac{Y(t+\epsilon)-Y(t)}{\epsilon}-(A\frac{t}{2}+B))^2]$ goes to zero when $\epsilon$ approaches zero), but i don't know how to manage this...
Thanks for any help.
AI: By definition and the linearity of the integral,
$$\begin{align*} Y(t)(\omega) &= \frac{1}{t} \int_0^t X(\tau)(\omega) \, d\tau = \frac{1}{t} \int_0^t A(\omega) \cdot \tau \, d\tau +\frac{1}{t} \int_0^t B(\omega) \, d\tau \\ &= A(\omega) \cdot \frac{t}{2} + B(\omega) \end{align*}$$
i.e. $Y$ is of the same form as $X$. You already proved $X'(t)=A$; exactly the same calculation shows $Y'(t) = A/2$.
On the other hand, if we consider
$$Y(t) = \int_0^t X(\tau) \, d\tau$$
it's not difficult to show that
$$\frac{Y(t+\varepsilon)-Y(t)}{\varepsilon} = A \frac{2t +\varepsilon}{2} + B$$
using a similar reasoning as above. This implies
$$\mathbb{E} \left[ \left( \frac{Y(t+\varepsilon)-Y(t)}{\varepsilon}- (A \cdot t + B) \right)^2 \right] \stackrel{\varepsilon \to 0}{\to} 0$$
i.e. $Y'(t) = A \cdot t+B=X(t)$.
This shows that in both cases the mean-square derivative coincides with the pointwise derivative. Nevertheless, you can not simply apply the fundamental theorem of calculus; you have to check the $L^2$-convergence. |
H: Difference $\Delta P_t$ approaches 0, then its relative difference $\Delta P_t / P_(t-1) \approx ln(P_t / P_(t-1))$.
When difference $\Delta P_t$ approaches 0, its relative difference $\dfrac{\Delta P_t}{P_{t-1}} \approx \ln(\dfrac{P_t}{P_{t-1}})$.
I know that it can be shown somehow with Taylor series: $\ln(1+x)=x-\frac12x^2+\frac13x^3...$ I just can't figure how.
AI: Note that $\Delta P_t=P_t-P_{t-1}$,
the right hand side
$$\ln\frac{P_t}{P_{t-1}}=\ln\frac{P_{t-1}+\Delta P_t}{P_{t-1}}=\ln\left(1+\frac{\Delta P_t}{P_{t-1}}\right)\approx \frac{\Delta P_t}{P_{t-1}}$$ |
H: How to prove properties of the family of closed sets in a metric space
I know that is true:
Let $(X, d)$ a metric space. The family $\mathcal {U}$ of all open subsets of $X$ has these properties:
$1)$ $\phi, X\in \mathcal {U}$;
$2)$ $U_1, U_2 \in \mathcal {U}\Rightarrow U_1\cap U_2 \in \mathcal {U}$;
$3)$ $\{U_i\}_{i\in I}\subset\mathcal{U}\Rightarrow \bigcup_{i\in I} U_i\in\mathcal{U}$
Prove: $1)$ it is clear (because $\phi$, is a subset for every set $\Rightarrow\phi \in\mathcal {U}$, every set is a subset of its own $\Rightarrow X \in\mathcal {U};$ If I'm wrong please correct)
$2)$ Let $x\in U_1\cap U_2.$ Then exist $r_1, r_2>0$ such that $T(x,r_i)\subset U_i, i=1,2.$ Let $r=\min\{r_1,r_2\}.$ Then $r>0$ and $T(x,r)\subset U_1\cap U_2.$
$3)$ Let $\{U_i\}_{i\in I}\subset\mathcal{U} \text{and} x\in\bigcup_{i\in I} U_i,$ then $x\in U_{i_0}$ for some $i_0\in I$. For opennes of the $U_{i_0}$ (why is $U_{i_0}$ open, please tell me), exist $r>0$ such that $T(x,r)\subset U_{i_0}\subset \bigcup_{i\in I} U_i$
But I didn`t now how to prove this theorem:
Let $(X, d)$ a metric space. The family $\mathcal {F}$ of all closed subsets of $X$ has these properties:
$1)$ $\phi, X\in \mathcal {F}$;
$2)$ $F_1, F_2 \in \mathcal {F}\Rightarrow F_1\cup F_2 \in \mathcal {F}$;
$3)$ $\{F_i\}_{i\in I}\subset\mathcal{F}\Rightarrow \bigcap_{i\in I} F_i\in\mathcal{F}$
I now that using a formula of DeMorgan but I didn`t now how to should do. Help me plase to prove. Thanks for your help and your attention.
AI: Atahualpa has answered the main part of your question; here are some comments on the first part.
I’m assuming that your definition of open set is that $U$ is open if and only if for each $x\in U$ there is an $r_x>0$ such that $T(x,r_x)\subseteq U$, where $T(x,r_x)$ is the open ball of radius $r_x$ centred at $x$.
$X$ is open because for each $x\in X$, $T(x,1)\subseteq X$. $\varnothing$ is open because it’s vacuously true that $T(x,1)\subseteq\varnothing$ for each $x\in\varnothing$, since there is no $x\in\varnothing$.
Your argument is fine.
What you’re proving here is that if $\{U_i:i\in I\}$ is a family of open sets, then $\bigcup_{i\in I}U_i$ is also open. Thus, $U_{i_0}$ is open by hypothesis. |
H: Two complex integrals of the function $1/z$
Let $a,b \in \mathbb{R}^{*}$ and $\alpha, \beta : [0,1]\rightarrow\mathbb{C}$ be defined by:
$$\alpha(t):=a\cos(2\pi t)+ia\sin(2\pi t)$$
$$\beta(t):=a\cos(2\pi t)+ib\sin(2\pi t)$$
Show that $\int_\alpha 1/z \,dz = \int_\beta 1/z \,dz$.
Progress
I thought that the first integral can be found as
$$ \int_{0}^{1} \frac{2\pi a(i\cos(2\pi t)-\sin(2\pi t))}{a(\cos(2\pi t)+i\sin(2\pi t))}\,dt$$
and similarly for the second integral. But I don't know how to compute these.
AI: Direct computation
The curves $\alpha$ and $\beta$ given in the OP can be written as
$$\alpha(t)=ae^{2\pi i t}, $$
and similarly for $\beta$. Then
$$\int_{\alpha} f:=\int_0^1 f(\alpha(t))\frac{d}{dt}\alpha(t)dt= \int_0^1 \frac{1}{ae^{2\pi i t}}2\pi ia e^{2\pi i t}dt=2\pi i,$$
and similarly for $\beta$. I used $\frac{d}{dt}\alpha(t)= 2\pi ia e^{2\pi i t}$.
The thesis follows.
A second approach
$f(z)=\frac{1}{z}$ is meromorphic with a pole in $z=0$. The closed curves $\alpha$ and $\beta$ describe circles in the complex plane with center $0$. Now use the Cauchy residue theorem. |
H: Find the Laurent series for $f(z) = (z^2 - 4)/(z-1)^2 $ for $z=1$
What I understand is that we have to expand $f(z$) in the positive and negative powers of $(z-1)$.
Hence I tried factorizing the numerator $(z^2-4)=(z+2)(z-2)$ , which can then be written in terms of $(z-1)$ as:
$(z-1-1)(z-1+3)/(z-1)^2$ . however i cannot expand it using the geometric series expansion:
$$1/(1-z) = 1 + z + z^2 + \cdots$$
Help.
AI: Use
$$z^2-4 = (z-1+1)^2 -4 = (z-1)^2+ 2 (z-1)-3$$
so that
$$f(z) = -\frac{3}{(z-1)^2} + \frac{2}{z-1}+1$$ |
H: Proving that $R$ is antisymmetric if and only if $G \cap G^{-1} \subseteq D$
I found this relations exercise that I couldn't finish. It involves a set's diagonal. I got it almost done - had trouble wording some parts related to the diagonal.
Is the procedure correct?
How can I better word the indicated parts?
Did it make sense to prove both implications separately?
Any kind of feedback is welcome. Still new to this stuff.
If $R = (G,A,A)$ is a relation over $A$, prove that $R$ is
antisymmetric if and only if $G \cap G^{-1} \subseteq D$ where $D$ is
the diagonal of $A$.
We have to prove that
$$Antisymmetric \iff (G \cap G^{-1} \subseteq D)$$
We have two implications. Let's prove each of them:
Proving
$$Antisymmetric \implies (G \cap G^{-1} \subseteq D)$$
Our premise is the fact that the relation is antisymmetric.
We have to prove that $G \cap G^{-1} \subseteq D$.
Have an arbitrary pair of elements $(a,b)$ in $G \cap G^{-1}$:
$$(a,b) \in G \cap G^{-1}$$
$$(a,b) \in G \land (a,b) \in G^{-1}$$
$$aRb \land bRa$$
Since the relation $R$ is antisymmetric:
$$a = b$$
I'm pretty sure that I'm almost done here (because, err... clearly (a,b) belongs to the diagonal...), but I don't know how to word it. Can you advice me here?
Proving
$$(G \cap G^{-1} \subseteq D) \implies Antisymmetric$$
Our premise is that $(G \cap G^{-1} \subseteq D)$.
We have to prove that the relation $R$ is antisymmetric.
Basically, we have to prove that
$$(aRb \land bRa) \implies a = b$$
Have arbitrary elements $a$ and $b$:
$$aRb \land bRa$$
Since $aRb$, the pair $(a,b)$ must be in $G$. Similarly, given that $bRa$, I know that $aR^{-1}b$, which means that $(a,b)$ is also in $G^{-1}$:
$$(a,b) \in G \land (a,b) \in G^{-1}$$
$$(a,b) \in G \cap G^{-1}$$
From our premise, we conclude that
$$(a,b) \in D$$
If a pair of elements belongs to a set's diagonal, both elements are the same (again I have problems wording this), so:
$$a = b$$
Effectively proving $(G \cap G^{-1} \subseteq D) \implies Antisymmetric$.
The exercise concludes as both implications were proven.
AI: It’s wordier than necessary, but I can live with that. :-)
There’s really almost nothing to do, which is perhaps why you’re having trouble. To finish up the first part, after you get $a=b$, just conclude:
Thus, $\langle a,b\rangle=\langle a,a\rangle\in D$.
There’s nothing needed to finish up the second part: it’s immediate from the fact that $\langle a,b\rangle\in D$ that $a=b$, just by the definition of $D$. |
H: Does inclusion of an affine open into an affine scheme correspond to restriction?
Let $X$ be an affine scheme and $U\subset X$ be an affine open. Let $i:U\to X$ be the inclusion, and let $\phi:\mathcal{O}_X(X) \to \mathcal{O}_X(U)$ be the induced morphism of rings.
Is $\phi$ the restriction map?
AI: Yes. But this comes from the definition of the morphism $U \to X$.
If $X$ is any ringed space and $U$ is an open subset of $X$ (i.e. the underlying topological space), then there is a morphism of ringed spaces $U \to X$ (with $\mathcal{O}_U := \mathcal{O}_X|_U$). The map on spaces is just the inclusion map. The map on sheaves is induced by the restriction maps of $\mathcal{O}_X$ (write it down!). |
H: How find this $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}$
Suppose
$$n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}$$
for every postive integers $n\ge 1$,Give that $a_{0}=1,a_{1}=2$
find the $a_{n}$
My try:
$$a_{2}=\dfrac{1}{2}=\dfrac{1}{2!},a_{3}=\dfrac{1}{6}=\dfrac{1}{3!},\cdots$$
so I guess
$$a_{n}=\dfrac{1}{n!}(n\ge 2)$$
and we easy use the Mathematical induction prove it.
My question:
This problem have without mathematical induction to solve it?
and following is my idea:
let $n(n+1)a_{n+1}=b_{n+1}$
so
$$b_{n+1}=b_{n}-\dfrac{b_{n-1}}{n-1},n\ge 2$$
then I can't,Thank you for your help!
This problem from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1290603&sid=5c95d4057706d4a61097b007952afb70#p1290603
AI: Let $c_n = na_n -a_{n-1}$. Then, rearranging:
$$n[(n+1)a_{n+1}-(n-1)a_n] = -(n-2)a_{n-1}$$
$$n[(n+1)a_{n+1}-a_n - (n-2)a_n] = -(n-2)a_{n-1}$$
$$nc_{n+1} - n(n-2)a_n = -(n-2)a_{n-1}$$
$$nc_{n+1} = (n-2)(na_n - a_{n-1})$$
$$c_{n+1} = (1-2/n)\cdot c_n$$
Plugging in $n=2$, we have $c_3=0$, and it follows that $c_4=0$, $c_5=0$, etc. So, for $n\geq 3$, $c_n=0$ and $na_n = a_{n-1}$.
We have to calculate one value "by hand": plugging in $n=1$ into the original equation gives $a_2=1/2$.
Then, for $n\geq 2$, we calculate: $a_n = 1/n \cdot a_{n-1} = 1/n \cdot 1/(n-1) \cdot \ldots \cdot a_2 = 1/n!$. |
H: Does this reduce to finding PDF of a function of a random variable?
In the below question in image, there is a deterministic non-service period ($\tau$) between serving customers, and then time to serve a customer is given to be $t \sim \varepsilon(\lambda)$. I need to find the PDF of time between consecutive customers.
My initial attempt is this: since $\tau$ is constant, it affects the total time by a constant $\tau$ only. So it becomes $(t+\tau) \sim \varepsilon(\lambda)$ then, $f(T) = f(t+\tau) = \lambda e^{-\lambda(t+\tau)}$. But is this right?
Doing the 2nd part of the question, which I also have posted a related question about. I found that if $Y = g(X)$, then $\text{PDF}_y = f(g^{-1}(y))$. So in this case, it looks like $T = t + \tau$, then $g^{-1}(T) = T - \tau$ then $f(T) = \lambda e^{-\lambda(T - \tau)}$.
But which is correct? Not a math major, my intuition feels that the 1st attempt makes more sense but the 2nd seems to be correct answer ... can someone help me make sense of this? If the 2nd is correct, why might my initial attempt be wrong?
AI: If the time between serving customers is $T$ with $T=t+\tau$ with $t$ having an exponential distribution with a PDF of $\lambda e^{-\lambda t}$ for $x\ge 0$ and $\tau$ being constant,
then $\Pr(t \le x)=1- e^{-\lambda x}$ for $x\ge 0$
so $\Pr(t + \tau \le x+\tau)=1- e^{-\lambda x}$ for $x\ge 0$
so $\Pr(T \le x+\tau)=1- e^{-\lambda x}$ for $x\ge 0$
so $\Pr(T \le y)=1- e^{-\lambda (y-\tau)}$ for $y\ge \tau$
so the PDF for $T$ is $\lambda e^{-\lambda (T-\tau)}$ for $T\ge \tau$
so your second formulation is correct |
H: Cantor's Theorem for $\Bbb N$.
Hi I would like to prove this statement.
Show that there is no one-to-one correspondence from
the set of positive integers to the power set of the set of
positive integers.
[Hint: Assume that there is such a one-
to-one correspondence. Represent a subset of the set of
positive integers as an infinite bit string with ith bit 1 if i
belongs to the subset and 0 otherwise. Suppose that you
can list these infinite strings in a sequence indexed by the
positive integers. Construct a new bit string with its ith
bit equal to the complement of the ith bit of the ith string
in the list. Show that this new bit string cannot appear in
the list.]
AI: Let $S$ be the set of positive integers. Suppose there is a bijection $f : S \to P(S)$. Then every subset of $S$ is equal to $f(s)$ for some $s \in S$. For any $s\in S$, $f(s)$ is a subset of $S$, and it is certainly the case that either $s\in f(s)$ or $s\notin f(s)$. [For example, there exists $s_1$ such that $f(s_1)=
S$, and then $s_1 \in f(s_1)$; likewise, there exists $s_2$ such that $f(s_2) = \emptyset$, and then $s_2 \notin f(s_2)$.] Define $A$ to be the set of all elements $s$ of $S$ such that $s \notin f(s)$; symbolically, $$A = \{s\in S\,|\,s \notin f(s)\}.$$
(In the above notation, $s_1 \notin A$ but $s_2 \in A$.)
Certainly $A$ is a subset of $S$; that is, $A\in P(S)$. Therefore, as $f$ is a bijection, $A = f(a)$ for some $a \in S$. We now ask the question: does $a$ belong to $A$? If $a \notin A$, then $a \notin f(a)$, so by definition of $A$, we have $a \in A$. This is a contradiction. And if $a \in A$ then $a \in f(a)$, so by definition of $A$ we have $a \notin A$, again a contradiction.Thus we have reached a contradiction in any case. So we conclude that there cannot be a bijection from $S$ to $P(S)$. |
H: Let $A$ be a dense subset of $X$ and suppose $A$ is connected in the induced subspace topology, then $X$ is also connected
PROBLEM
Let $A$ be a dense subset of $X$ and suppose $A$ is connected in the
induced subspace topology, then $X$ is also connected
ATTEMPT
Suppose $X = U \cup V $ for open, nonempty, disjoint $U,V$. By Denseness of $A$, $Cl(A) = U \cup V $. We know that $A \subseteq Cl(A)$. Hence $A \subseteq U \cup V $. Well, here it is pretty obvious that $A $ must be $U \cup V$ and hence having a contradiction. But I think probably I need to prove this obvious thing. Can someone help me? Is this approach correct?
Thanks,
AI: HINT: No, $A$ need not be $U\cup V$. Consider the sets $U\cap A$ and $V\cap A$. |
H: Euler Totient clarification
I'm asked to determine what $\varphi{(p^k)}$ is for an arbitrary prime $p$. By definition, $\varphi{(p^k)}=p^k\left(1-\frac1{p}\right)=p^k\left(\frac{p-1}{p}\right)=p^{k-1}(p-1)$. But I thought that since the Totient function was multiplicative that $\varphi{(p^k)}=\varphi{(p)}^k=(p-1)^k$. I've checked small values and the first formula is correct. But why if the Totient function is a multiplicative function is the second formula not correct?
AI: The term multiplicative means (in this context) that when we have $\rm{gcd}(m,n) = 1$ we get $\varphi(mn) = \varphi(m)\varphi(n)$. |
H: Why does an algebraic curve over an algebraic closed field have smooth points?
Is there an easy way to see this fact? I could try to show that the differential of the defining polynomial cannot vanish at all the zeros. However, I don't see how this could be done. Also there should be a more sophisticated way.
Thanks for any answers.
AI: The basic idea is as follows. We can assume without loss of generality that $f$ is non-zero and irreducible.
If $f$ was not smooth anywhere, then $f_{x_i}$ vanishes everywhere on $f$ and so each $f_{x_i}$ would be divisible by $f$. By degree considerations this implies that each $f_{x_i}$ is zero. Clearly this is impossible in characteristic zero (why?), and so we may assume that our field has characteristic $p$. Then, each $f_{x_i}$ being zero implies that $f$ is actually of the for $g(x_1^p,\ldots,x_n^p)$, and since the coefficients are also $p^{\text{th}}$-powers this implies that $f=g^p$, which contradicts that $f$ is irreducible.
EDIT: I assumed you meant plane curve, for some reason, in the above. You can make the above work for any variety though (after this edit) by first assuming that we're working with a hypersurface in some projective space. This is ok since any variety is birational to a hyperurface in some projective space. |
H: Proving that $A \cap (A \cup B) = A$ . Please check solution
For homework I need to prove the folloving:
$$
A \cap (A \cup B) = A
$$
I did that in the following manner:
$$
A \cap (A \cup B)\\
x \in A \land (x \in A \lor x \in B)\\
(x \in A\ \land x \in A) \lor (x \in A \land x \in B)\\
x \in A \lor (x \in A \land x \in B)\\
\text{now I have concluded that x belongs to A,}\\
\text{because of the boolean expression, because I get something like this:}\\
x \in A \lor (x \in A \land x \in B) = x \in A\\
\text{when $x \in A$ is true the whole expression will be true}\\
$$
Did I do it right???
thanks
AI: Yes, you are correct. The given equality can be expressed by the two implications: $x \in A\cap (A \cup B) \implies x \in A$, and the implication $x \in A \implies x \in A \cap(A\cup B)$. Since both implications are satisfied, so is the set equality.
Just be careful in your "equations" (i.e., your use of the "equal sign here"). You wrote:
$$x \in A \lor (x \in A \land x \in B) \underbrace{=}_{\uparrow ?} x \in A$$
It would be more correct to conclude: $$x \in A \lor (x \in A \land x \in B)\;\underbrace{\iff}_{\text{if and only if}}\; x \in A$$
from which it follows $$A \cap (A \cup B) = A$$ |
H: How to determine Depedent and Span of matrices?
$ \displaystyle s= (2,4,6)^T ,(0,0,0)^T ,(0,1,1)^T \in R^3 $
Does S are dependent linear?
Does S are Span of $R^3$ ?
AI: A set of vectors $\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n$ is said to be linearly independent if
$$\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2+\ldots+\alpha_n\mathbf{v}_n=\mathbf{0} \iff \alpha_1=\alpha_2=\ldots=\alpha_n=0;$$
otherwise, it is linearly dependent.
In your case, we have to find the solutions to
$$\alpha_1(2,4,6)+\alpha_2(0,0,0)+\alpha_3(0,1,1)=(0,0,0).$$
Clearly, the equation is satisfied by $\alpha_1=\alpha_3=0$, $\alpha_2\in\mathbb{R}$, so the condition is false and the set is linearly dependent.
It also follows that the set cannot span all of $\mathbb{R}^3$, since there are at most two linearly independent vectors. |
H: chain of compact subspaces must be nonempty
Suppose $X$ is a compact space and suppose $\{ F_i \} $ is a collection of closed subsets of $X$ such that $F_{i+1} \subseteq F_i $ for all $i$. , then we must have $\bigcap_i F_i \neq \varnothing $.
MY try: Suppose $\bigcap_i F_i = \varnothing $. We know the $\{ F_i \} $ must all be compact since they are closed sets living inside a compact space. Notice the collection $\mathcal{O} = \{ X \setminus F_i \} $ is definitely an open cover for $X$. Since $X$ is compact, $\mathcal{O} $ must have an finite open subcollection that covers $X$. How can I use my hypothesis that $F_{i+1} \subseteq F_i$ to get a contradiction. thanks.
AI: You went astray when you said that $\mathcal{O}$ is definitely an open cover of $X$. It’s definitely a family of open subsets of $X$, but it doesn’t necessarily cover $X$.
Show that $\mathcal{O}$ covers $X$ if and only if $\bigcap_{n\in\Bbb N}F_n=\varnothing$.
Now use that to get your contradiction: if $\bigcap_{n\in\Bbb N}F_n=\varnothing$, then something, and hence contradiction. |
H: How to take this limit? $\lim_{x \to \infty}(x-ln^3(x))$
Need to take the limit:
$$\lim_{x \to \infty}(x-ln^3(x))$$
My idea is to come to indeterminate form $(\frac{0}{0})$, then use L'Hospital's rule:
$$\lim_{x \to \infty}(x-ln^3(x))=ln\lim_{x \to \infty}\frac{e^x}{e^{ln^3(x)}}=\left(\frac{0}{0}\right) = ln\left( \lim_{x \to \infty}\left(\frac{e^x}{e^{ln^3(x)}\cdot 3ln^2(x)\cdot \frac{1}{x}}\right)\right)=ln\lim_{x \to \infty}\left(\frac{e^x\cdot x}{e^{ln^3(x)} \cdot 3ln^2(x)}\right)$$
But I'm confused now. What I have to do next?
AI: write $x-ln^3x=x(1-\frac {ln^3x}{x})$ and use Hospital at $\frac {ln^3x}{x}$ |
H: Show that $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ is not a cyclic group
Show that $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ is not a cyclic group. This question is from the book 'Of Abstract Algebra' by Pinter. Now $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ containt 8 elements. I found them to be as follows:
$$(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)$$
Now if $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ were cyclic, one of these elements should be a generator of the entire group. However, for every element listed we find that repeated repetitions of the group operation yields the identity element already before having to apply the group operation eight times. Therefore, using any of these elements as a single element would result in a subgroup of order less than 8, in this case it will have either order 4, order 2 or order 1. Therefore the group is not cyclic. Now I am quite sure this constitutes a correct proof but I am wondering if a more elegant way exist to show this result. This might be especially useful when investigating similar questions for much larger groups. Thanks in advance
P.S. I have been posting more questions from this book because I find them very interesting, but I am reading it on my own so I am sometimes unsure if the methods I use are the most elegant and quick. So far I have received great help and I am grateful to the stackexchange community for this :)
AI: I think your approach of studying the order of the elements is the right one. If you want to do that in a more systematic way, you may study order of elements in a direct product. Namely, say we have $G_1, \ldots, G_r$ some groups, some element $g = (g_1, \ldots g_r) \in G_1 \times \ldots \times G_r$ in the direct product, and we want to know the order of $g$ inside $G_1 \times \ldots \times G_r$. It turns out that if $n_i$ denotes the order $g_i \in G_i$, then the order of $g$ is $\operatorname{lcm}(n_1,\ldots,n_r)$. In particular, since the order $g_i$ divides $|G_i|$ (the number of elements of the group), then order of $g$ divides $\operatorname{lcm}(|G_1|,\ldots,|G_r|)$.
In your example, take $G_1 = \mathbb{Z}_2$ and $G_2 = \mathbb{Z}_4$. We know the order of any element inside $G_1 \times G_2$ divides $\operatorname{lcm}(|G_1|,|G_2|) = \operatorname{lcm}(2,4) = 4$. So there is no element of order $8$.
Using the same principle we may see for example that $ \mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic. Indeed, we know $(1,1)$ has order $\operatorname{lcm}(2,3)= 6$. |
H: Some basic questions about vectors
I've got two quite basic questions about vectors. I'm sorry if it isn't right to put two questions at the same thread. I'm quite confused about the technique of solving such problems.
Let $\vec v=(3,-4)$, $\vec u=(1,2)$. Find two vectors $\vec w_1, \vec w_2$ so that:
(i) $\vec u=\vec w_1+\vec w_2$.
(ii) $\vec v || \vec w_1$
(iii) $\vec v \bot \vec w_2$
And the second question:
Show that for all four points $A,B,C,D \in \mathbb R^n$ from some $n$.:
$\vec {AB} \cdot \vec {CD} + \vec {AC} \cdot \vec {DB} + \vec {AD} \cdot \vec {BC}=0$
Thanks in advance for any help!
AI: For your first, $\vec w_1$ will be of the form $(3a,-4a)$ and $\vec w_2$ will be of the form $(4b,3b)$ for real numbers $a$ and $b$, so setting their sum equal to $\vec u$ you need to solve: $$3a+4b=1$$ $$-4a+3b=2$$
Added: For your second, you could start by showing that $$\sum_i (b_i-a_i)(d_i-c_i) + \sum_i (c_i-a_i)(b_i-d_i) + \sum_i (d_i-a_i)(c_i-b_i)=0$$ by looking at each $i$ individually |
H: Denesting Phi, Denesting Cube Roots
I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial equations. $y= x^2-x-1$ and $y=x^6-4x^3-1$. I do not know how to solve a degree 6 polynomial by hand like that. My objective is to denest the original $\sqrt[3]{2+\sqrt{5}}$ without knowledge of the simplified phi.
AI: Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have
$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(3a^2b+5b^3)}_1\sqrt5$$ $$\iff a^3+15ab^2=2(3a^2b+5b^3)\iff a^3+15ab^2-6a^2b-10b^3=0\quad|:b^3\iff$$ $$\iff\left(\frac ab\right)^3-6\left(\frac ab\right)^2+15\left(\frac ab\right)-10=0\iff x^3-6x^2+15x-10=0\iff x=1$$ $$\iff\frac ab=1\iff a=b\iff a^3+15a^3=2\iff16a^3=2\iff a=\sqrt[3]\frac18=\frac12$$ or $3b^3+5b^3=1\iff b=\sqrt[3]\frac18=\frac12$ . Similarly for when the second term is $-1$. |
H: Integrate $\frac{x}{1+x^4}$
How do I integrate something like this? $$\int \frac{x}{1+x^4}\mathrm{d}x$$
I've tried trig substitutions but none have worked.
AI: Let $u = x^2$. Then $\,du = 2x \,dx $.
$$\int \dfrac x{1 + x^4} \,dx = \dfrac 12 \int \dfrac{2x\,dx}{1 + (x^2)^2} = \dfrac 12 \int \dfrac{\,du}{1 + u^2}$$
Then use the trigonometric substitution $u = \tan \theta$. |
H: $f$ uniformly continuous , $a_n$ Cauchy $\Rightarrow f(a_n)$ is Cauchy
Let $I$ be an interval and let $f: I\to \mathbb{ R}$ be uniformly continuous on I. Suppose that $\{a_n\}$ is a Cauchy sequence in $I$. Prove that $\{f(a_n)\} $is a Cauchy sequence.
AI: Fix $\epsilon>0$, Since $f$ is uniformly continuous,$\exists \delta>0 \ni |f(x)-f(y)|<\epsilon$ Whenever $|x-y|<\delta$.
$a_n$ is Cauchy, so $\exists N\in\mathbb{N}\ni |a_n-a_m|<\delta \forall m,n>N$
So $|f(a_n)-f(a_m)|<\epsilon \forall m,n>N$. So $\{f(a_n)\}$ is Cauchy |
H: evaluation of $\int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$
$\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx$
By Using Substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
I have Tried it without using the given substution.
$\bf{My\; Try}::$ $\displaystyle \int_{0}^{1}\frac{1-x^2}{\left(1+x^2\right)\sqrt{1+x^4}}dx = \int_{0}^{1}\frac{1-x^2}{x\cdot \left(x+\frac{1}{x}\right)\cdot x \cdot \sqrt{\left(x^2+\frac{1}{x^2}\right)}}dx$
$\displaystyle = -\displaystyle \int_{0}^{1}\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)\cdot \sqrt{\left(x+\frac{1}{x}\right)^2-\left(\sqrt{2}\right)^2}}dx$
Now Let $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
$\displaystyle = -\int_{0}^{1}\frac{1}{t\sqrt{t^2-\left(\sqrt{2}\right)^2}}dt = -\frac{1}{\sqrt{2}}\left[\sec^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)\right]_{0}^{1} = -\frac{1}{\sqrt{2}}\left(\sec^{-1}\left(\sqrt{2}\right)-\sec^{-1}\left(\infty \right)\right)$
$\displaystyle = \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}$
Would anyone explain me how can i solve using the substution $\sqrt{1+x^4} = (1+x^2)\cdot \cos \theta$
Help Required
Thanks
AI: Nicely done. So they suggest setting
$$
\cos \theta =\frac{\sqrt{1+x^4}}{1+x^2} \quad\Rightarrow\quad \frac{2x(1-x^2)}{\sqrt{1+x^4}(1+x^2)^2}dx=\sin \theta \,d\theta
$$
and, since $\sin\theta\geq 0$ here:
$$
\sin\theta =\sqrt{1-\cos^2\theta}=\frac{\sqrt{2} \;x}{1+x^2} \quad\Rightarrow \quad \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\frac{d\theta}{\sqrt{2}}
$$
This is indeed a valid change of variable $\theta =\arccos\left( \frac{\sqrt{1+x^4}}{1+x^2}\right)$. The bounds become $\arccos(1)=0$ and $\arccos(\sqrt{2}/2)=\pi/4$. This yields
$$
\int_0^1 \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} dx=\int_0^{\pi/4}\frac{d\theta}{\sqrt{2}}=\frac{\pi}{4\sqrt{2}}
$$ |
H: How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$
Find this limit
$$\lim_{n\to\infty}n^2\left(\dfrac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\dfrac{1}{k+1}-\dfrac{1}{2n}\right)\tag{1}$$
I can only solve this limit
$$I=\lim_{n\to\infty}\left(\dfrac{1^k+2^k+\cdots+n^k}{n^k}-\dfrac{n}{k+1}\right)=\dfrac{1}{2}$$
proof:
use Stolz therom:let
$$a_{n}=(k+1)(1^k+2^k+\cdots+n^k)-n^{k+1},b_{n}=(k+1)n^k$$
then
$$I=\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}}=\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\dfrac{1}{k+1}\dfrac{(k+1)(n+1)^k-(n+1)^{k+1}+n^{k+1}}{(n+1)^k-n^k}$$
so
$$I=\lim_{n\to\infty}\dfrac{1}{k+1}\dfrac{[k(k+1)/2]n^{k-1}+\cdots+k}{kn^{k-1}+\cdots+1}=\dfrac{1}{2}$$
But for (1),I can't,Thanks.
AI: For every $k\geqslant0$ and $n\geqslant0$, let
$$
s_n^k=\sum\limits_{i=1}^ni^k.
$$
We first recall how to get a simple equivalent of $s_n^k$ when $n\to\infty$. One starts from the fact that, for every $i\geqslant1$, $(i-1)^k\leqslant t^k\leqslant i^k$ for every $t$ in $(i-1,i)$. Summing this from $i=1$ to $i=n$ yields
$$
(k+1)s_{n-1}^k\leqslant (k+1)\int_0^nt^k\mathrm dt\leqslant(k+1)s_n^k,
$$
that is,
$$
n^{k+1}\leqslant (k+1)s_n^k=(k+1)s_{n-1}^k+(k+1)n^k\leqslant n^{k+1}+(k+1)n^k,
$$
which is enough to get
$$
s_n^k=\frac{n^{k+1}}{k+1}+O(n^k)\tag{$\dagger$}.
$$
To be more precise than $(\dagger)$, one needs to refine the inequalities $(i-1)^k\leqslant t^k\leqslant i^k$. For every $i\geqslant1$ and every $t$ in $(i-1,i)$, Taylor formula reads
$$
t^k=i^k+ki^{k-1}(t-i)+\frac12k(k-1)r^{k-2}(t-i)^2,
$$
for some $r$ in $(t,i)$. If $k\geqslant2$, $r^{k-2}\leqslant i^{k-2}$. Furthermore, for every $i\geqslant1$, $\int\limits_{i-1}^i(t-i)\mathrm dt=-\frac12$ and $\int\limits_{i-1}^i(t-i)^2\mathrm dt=\frac13$ hence, integrating both sides from $t=i-1$ to $t=i$ and summing over $i$ from $i=1$ to $i=n$ yields
$$
n^{k+1}=(k+1)s_n^k-\frac12(k+1)ks_n^{k-1}+\frac16(k+1)k(k-1)s_n^{k-2}\theta_n^k,
$$
where $\theta_n^k$ is in $(0,1)$. Since $ks_n^{k-1}=n^k+O(n^{k-1})$ and $s_n^{k-2}=O(n^{k-1})$, this guarantees that
$$
s_n^k=\frac{n^{k+1}}{k+1}+\frac12n^k+O(n^{k-1}).\tag{$\ddagger$}
$$
The expansion $(\ddagger)$ refines $(\dagger)$ but we need one more term. To get it, use $(\ddagger)$ for $k-1$. This reads
$$
ks_n^{k-1}=n^{k}+\frac12kn^{k-1}+O(n^{k-2}),
$$
Additionally, note that $s_n^{k-1}\leqslant s_n^{k-2}\theta_n^k\leqslant s_n^{k-2}$ hence $\theta_n^k\to1$ when $n\to\infty$.; Since $(k-1)s_n^{k-2}n^{k-1}+O(n^{k-2})$, putting everything together yields
$$
n^{k+1}=(k+1)s_n^k-\frac12(k+1)\left(n^{k}+\frac12kn^{k-1}\right)+\frac16(k+1)kn^{k-1}+O(n^{k-2}),
$$
that is,
$$
\frac{n^{k+1}}{k+1}=s_n^k-\frac12n^{k}-\frac1{12}kn^{k-1}+O(n^{k-2}),
$$
which is equivalent to
$$
n^2\left(\frac{s_n^k}{n^{k+1}}-\frac1{k+1}-\frac1{2n}\right)=\frac1{12}k+O\left(\frac1n\right).
$$ |
H: Understanding issue of basic sets
I have some answers here, that I can barely understand.
1) $\{a,b,\{a,b\}\} - \{a,b\} = \{a,b\}$. The answer indicates it is wrong. I think it is correct, what is it that I cant see? (very confused here).
2) $\{a\} \subseteq \{a,b,\{\{\{a\},b\}\}\}$. Correct. Mr. obvious (?) ...
3) $\{a\} \in \{a,b,\{\{\{a\},b\}\}\}$. Wrong. Why is it wrong??? I thought it was correct.
Anyone here could perhaps explain these to me?
AI: If $A$ and $B$ are sets, the elements of $A\setminus B$ are the objects that are in $A$ but not in $B$. The elements of $\big\{a,b,\{a,b\}\big\}$ are $a,b$, and $\{a,b\}$, and the elements of $\{a,b\}$ are $a$ and $b$. Thus, the only element of $\big\{a,b,\{a,b\}\big\}$ that is not an element of $\{a,b\}$ is the object $\{a,b\}$, and the set difference $\big\{a,b,\{a,b\}\big\}\setminus\{a,b\}$ is the set whose only member is that object; that set is $\big\{\{a,b\}\big\}$.
You’re okay here.
The elements of the set $\big\{a,b,\{\{\{a\},b\}\}\big\}$ are $a$, $b$, and the set $\big\{\{\{a\},b\}\big\}$. None of these is $\{a\}$, so $\{a\}\notin\big\{a,b,\{\{\{a\},b\}\}\big\}$. Remember, $a$ and $\{a\}$ are not the same thing.
For the record, the set $\big\{\{\{a\},b\}\big\}$ has one element, the set $\big\{\{a\},b\big\}$. This set in turn has two members, the set $\{a\}$ and the object $b$. |
H: Let $G$ be a group, then let $f :G\to G$ via $f(x) = x^2$. Now, is $f$ injective and/or surjective?
Let $G$ be a group, then let $f :G\to G$ via $f(x) = x^2$. Now, is $f$ injective and/or surjective? To this end let $f(x)=f(y)$. Then
$$x^2 =y^2 $$
If we take a look at the group $( \Bbb Q \setminus\{0\} , \cdot )$, the non-zero rationals with multiplication we find that $f(1)=f(-1)$ where obviously $-1=1$. This means that this $f$ is not injective for this group.
Now we will try to determine if $f$ is surjective. Let $b \in G$ be arbitrary. Then if $G$ is a group of order 2, i.e. the elements of $G$ are $\{e,b\}$ there is no element satisfying $f(x)=b$ because $b^2 = e^2=e$ in this case. So $f$ is not surjective on this group
My questions: Are these arguments correct? Is there a way to show these results without having to find specific examples where $f$ does not constitute an injective or surjective function? Thanks for your time!
AI: Yes, the examples are correct. Even more simply, you could have used the cyclic group of order $4$ for both examples. Since the map can be surjective/injective, counterexamples are the natural way to go. However, you could prove instead that every group of even order has an element of order $2$ and conclude that if $G$ has even order, the map is neither injective nor surjective: it’s clearly not injective, since if $g$ is an element of order $2$, then $f(g)=f(1_G)$, and it’s not surjective because a map from a finite set to itself is surjective if and only if it’s injective. |
H: Prove that every manifold is regular.
Prove that every manifold is regular and hence metrizable.
AI: HINT: A space $X$ is regular if and only for each point $x\in X$ and each open nbhd $U$ of $x$ there is an open set $V$ such that $x\in V\subseteq\operatorname{cl}V\subseteq U$. Use the fact that a manifold is locally Euclidean to prove that it has this property. |
H: Solving partial fraction expansion with all variables
Okay so I have an equation in my book which is as follows..
$$
\frac {a}{s(s+a)}
$$
it says "using partial fractions this can be expanded to
$$
\frac {1}{s} + \frac {-1}{s+a}
$$
My usual method would be to cross multiply and do something like this
$$
\frac {a}{s(s+a)} = \frac {A(s+a)}{s(s+a)} + \frac {B(s)}{s(s+a)}
$$
Then cancel off the denominators and solve..
$$
a = A(s+a) + B(s)
$$
usually though the a would be some constant but here I have no values to play around with.. how has he done it in the book?
AI: We have:
$$\dfrac{a}{s(s+a)} = \dfrac{A}{s}+\dfrac{B}{s+a}$$
So,
$$a = A(s+a) + Bs = (A+B)s + A a$$
we have $A = 1, B = -1$
Final result:
$$\dfrac{a}{s(s+1)} = \dfrac{1}{s}-\dfrac{1}{s+a}$$ |
H: subsequence of unitary matrices has two limit
Consider the sequence of unitary matrices $U_k=\left(\begin{array}{cc} 0 & 1 \\
1 & 0 \end{array}
\right)^k,k=1,2,\dots$
I am not able to show that there are two possible limits of subsequences
Could anyone help me how?
AI: HINT: Calculate a few powers of $U$ to see what’s happening:
$$\begin{align*}
U^1&=\pmatrix{0&1\\1&0}\\
U^2&=\pmatrix{0&1\\1&0}\pmatrix{0&1\\1&0}=\pmatrix{1&0\\0&1}\\
U^3&=\pmatrix{0&1\\1&0}U^2=\pmatrix{0&1\\1&0}\pmatrix{1&0\\0&1}=\pmatrix{0&1\\1&0}=U\\
U^4&=\pmatrix{0&1\\1&0}U^3=\pmatrix{0&1\\1&0}\pmatrix{0&1\\1&0}=U^2=\pmatrix{1&0\\0&1}
\end{align*}$$
This looks a lot like what happens with the real sequence $\langle(-1)^n:n\in\Bbb Z^+\rangle$, which has a subsequence converging to $1$ and another converging to $-1$. |
H: How to find the biggest m that a function is in O($h^m$)
i try to find the biggest m that
$$f(h) = h^{-2}(\sin(1+h)-2\sin(1)+\sin(1-h))+\sin(1)$$
is $\in O(h^m)$ ($h \to 0, h > 0$)
I tried this:
$$ f(h) = h^{-2} (((1+h)-\frac{(1+h)^3}{3!}+\frac{(1+h)^5}{5!} \mp ...) -2(1-\frac{1}{3!}+\frac{1}{5!} \mp...) + ((1-h)-\frac{(1-h)^3}{3!}+\frac{(1-h)^5}{5!} \mp ...))+(1-\frac{1}{3!}+\frac{1}{5!} \mp...) =h^{-2}(2(1-h)-\frac{(1+h)^3-(1-h)^3+2}{3!} + \frac{(1+h)^5-(1-h)^5-2}{5!} \mp ...)+\sin(1)$$
But what now?
Can you help me?
Thanks!
AI: Hint: Use the Taylor approximation around $1$,
$$f(1+h) = f(1) + f'(1)\cdot h + \frac{f''(1)}{2}h^2 + \frac{f'''(1)}{3!}h^3 + \dotsb$$
Use the similar approximation of $f(1-h)$ to obtain
$$f(1+h) + f(1-h) = 2\cdot f(1) + f''(1)\cdot h^2 + \dotsb$$
Use approximations of high enough order to achieve your goal. |
H: If 2 vectors form a basis for $\mathbb{R}^2$, must these 2 vectors always be orthogonal to each other?
If 2 vectors form a basis for $\mathbb{R}^2$, must these 2 vectors always be orthogonal to each other?
For instance, the standard bases in $\mathbb{R}^2$ are definitely orthogonal (easily drawn). How about other bases?
AI: No, they definitely need not be orthogonal, just non-parallel. |
H: Not injective given cardinalities of sets
How do I prove that a function $f:G \rightarrow H$ is not one-to-one if $|G|=20$ and $|H|=24$?
AI: You can't prove that $f$ is not one-to-one. It may very well be.
What you can prove is that $f$ is not onto (surjective).
Aside:
If you are working with a group homomorphism $f: G\to H$, with groups $G, H$, then the homomorphism $f$ cannot be an isomorphism because it is neither injective (use Lagrange's Theorem) nor surjective. |
H: Degree of continuous maps from S1 to S1 - Two equivalent properties
I understand what is meant by the degree of a continuous map $f$ from $S^1$ to $S^1$. If we let $[S^1, S^1]$ denote the set of homotopy classes of continuous maps from $S^1$ to $S^1$, it turns out that the degree map gives a bijection from $[S^1, S^1]$ to the integers. I am also cool with this.
My problem is I heard that this bijection fact is equivalent to the following:
The degree map from $C(S^1,S^1)$ to the integers is a continuous map, and whenever $deg(f_0) = deg(f_1)$, there exists a path in $C(S^1,S^1)$ from $f_0$ to $f_1$.
How are the two notions equivalent? I don't have a very good grasp (or intuition) for continuous maps from $C(S^1,S^1)$ to the integers, and how that is related to modding functions out by homotopy.
(You may assume that I have background knowledge equivalent to Munkres chapter 9)
AI: If you lift the map $f\colon S^1\to S^1$ to a map $\tilde f\colon \Bbb R\to\Bbb R$, then the degree is given by $\dfrac{\tilde f(2\pi)-\tilde f(0)}{2\pi}$, and you can easily construct a path joining two maps of the same degree by taking the straight-line homotopy between their respective lifts. |
H: CF grammar on this language
I'm trying to write a context-free grammar for this language:
$L = \{a^n b a^m (bb)^n : m \ge 1, n \ge 0\}$
I was getting lost with maintaining $n$ number of $a$'s and $(bb)$'s and I'm not sure how to fix it. The terminals don't seem to be working out because I think splitting it up between lines is just asking for the $n$ to not be the same between the 2 terminals that require it (as I stated above). The $m$ seems ok though:
$S \rightarrow a S b b \mid A$
$A \rightarrow b B$
$B \rightarrow a B \mid a$
Would be very helpful if someone can get me back on track.
AI: It looks fine: there’s nothing there to be fixed. Any derivation must look like this:
$$S\Rightarrow^n a^nSb^{2n}\Rightarrow a^nb^{2n}\Rightarrow a^nbBb^{2n}\Rightarrow^m a^nba^mBb^{2n}\Rightarrow a^nba^{m+1}b^{2n}\;,$$
where $n$ is the number of times you use $S\to aSbb$, and $m$ is the number of times you use $B\to aB$. Since $n,m\ge 0$, this gives you exactly what you want. |
H: Is the empty set always part of the result of an intersection?
From very basic set theory we have that:
"The empty set is inevitably an element of every set."
Then, is it correct to assume that the intersection of
$A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$ is actually $\{\emptyset, 3\}$, and not just $\{3\}$?
Thank you
AI: While it is true that for all sets $A$, we have $\emptyset \subseteq A$, it is not true that for all sets $A$, $\emptyset \in A$.
The intersection of $A,B$ is defined by $A \cap B = \{x | x \in A$ and $x \in B\}$.
What is important here is the difference between being a subset and being an element of a set. Do you now see why the empty set is not contained in $A\cap B$, with reference to your example? |
H: Infinite sum $\sum_{n=1}^\infty{\frac{1}{n2^n}}$
How do I evaluate this sum:
$$\sum_{n=1}^\infty{\frac{1}{n2^n}}$$
AI: Another way is
$$\sum_{n=1}^\infty{\frac{1}{n2^n}}=\sum_{n=1}^\infty{\int_0^{1/2}x^{n-1}\,\mathrm{d}x=\int_0^{1/2}\sum_{n=1}^\infty x^{n-1}\,\mathrm{d}x=\int_0^{1/2}{\frac{1}{1-x}}}\,\mathrm{d}x=\ln{2}$$ |
H: How to determine basis by Reducing a sets
How to do Reducing set of $(x_1,x_2,x_3,x_4)$
So that form basis for $\mathbb{R}^3$
for vectors:
$\displaystyle x_1 = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}x_2=\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}x_3=\begin{bmatrix} 3\\ 2\\ -1 \end{bmatrix}
x_4=\begin{bmatrix} -2\\ 1 \\ 1 \end{bmatrix}$
AI: Observing that
$$\det(\mathbf{x}_1\,\mathbf{x}_2\,\mathbf{x}_3)=-16\neq0,$$
$$\det(\mathbf{x}_1\,\mathbf{x}_2\,\mathbf{x}_4)=4\neq0,$$
$$\det(\mathbf{x}_1\,\mathbf{x}_3\,\mathbf{x}_4)=8\neq0,$$
$$\det(\mathbf{x}_2\,\mathbf{x}_3\,\mathbf{x}_4)=-4\neq0,$$
we see that any set of three vectors will form a basis for $\mathbb{R}^3$. |
H: Using limit of sinx/x as x approaches 0 to simplify the equation
So, I know that I'm supposed to use the rule that the limit of $\sin(x)/x$ as x approaches 0 is equal to 1 to simplify the following:
$$\lim_{x\to0}\frac{\sin(x)\cos(4x)}{x+x\cos(5x)} $$
However, I'm not sure where to simplify. Which $x$ on the denominator do I use, and what do I do with the remaining cosines afterwards? According to the review for my test, the final answer is 1/2, but I'm not sure how to reach that after simplification.
AI: Factor $x$ from $x+x\cos 5x$ then you have
\begin{align*}
\lim_{x\to 0} \frac{\sin x \cdot \cos 4x}{x(1+\cos 5x)} &= \left(\lim_{x\to 0} \frac{\sin x}{x}\right)\cdot\left(\lim_{x\to 0} \frac{\cos 4x}{1+\cos 5x}\right) \\
&= 1\cdot\frac{1}{1+1} = \frac 1 2.
\end{align*} |
H: $H_0 ( X )$ is free abelian on the path components of $X$ .
I do not quite understand this statement. Does this mean if there is one path-components, then $H_0 ( X )$ is generated by one generator, or cyclic. And if there is two path-components, then $H_0 ( X )$ is generated by two generator, but of the form $a^i b^j$?
And why this is true?
AI: It is customary to write homology groups additively rather than multiplicatively. So if you mean $a$ and $b$ to be the generators, then an arbirary element will be a linear combination of $a$ and $b$ with arbitrary coefficients (presumably you mean integer coefficients). Other than that your guess is correct: the number of generators is the number of path-connected components. |
H: find a generator for the group $G =\{ f(x) = x+n\mid n\in \Bbb Z \}$ with the group operation being composition.
Another question from 'A book of Abstract Algebra' by Pinter.
For each $n\in \Bbb Z$ define $f_n = x+n$. Then $f_n\in S_{\Bbb R}$, the symmetric set on $\Bbb R$. The group operation being composition. Now $f_n \circ f_m =f_{m+n}$ and $f^{-1}_n=f_{-n}$
Let $G=\{ f_n \mid n\in \Bbb Z \}$ such that $G$ is a subgroup of $S_{\Bbb R}$. Show that $G$ is cyclic by finding a generator.
I cannot seem to find the right generator. Using $f_1$ as a generator produces all $f_n$ for $n = 1,2,3,\ldots$ but then the identity $f_0$ is not included and the inverses are missing. Similarly for $f_{-1}$. What am I missing? I just need to find the generator but I cannot seem to find it. Thanks in advance.
AI: Hint: Prove $G \cong \mathbb Z$. |
H: divisibility of $n^{2}+n+1$ by $6k-1$ when $n,k$ are integers.
I wanted to prove that for any $n\in\mathbb{Z}$, the integer $n^{2}+n+1$ does not have
any divisors of the form $6k-1$. So far, I proved $n^{2}+n+1\neq-1 \pmod 6$ and an integer which is $-1\pmod6$ should have a prime factor which is also $-1 \pmod6$. And I was trying to assume $n^{2}+n+1\equiv0\mathrm\,{(mod \;6k-1)}\Rightarrow n^{2}\equiv-n-1(mod\;6k-1)\,$, where $6k-1$ is prime here, and use the quadratic reciprocity law to find some contradiction (since the question was an exercise in the chapter of quadratic reciprocity from some lecture note) but after a while, I couldn't find anything useful yet. Any hint please? I only know some elementary number theory.
AI: Promoting the hint to an answer:
If a prime $p$ divides $n^2+n+1$, then it also divides $(n-1)(n^2+n+1) = n^3-1$.
Now, if a prime $p$ divides $n^3-1$, then either $p$ divides $n-1$, or the multiplicative order of $n$ modulo $p$ is $3$, which implies $3 \mid (p-1)$, and thus $p = 6k+1$ for some $k > 0$. If $p$ divides $n-1$ and $n^2+n+1$, it also divides their greatest common divisor, which is a divisor of $(n^2+n+1) - (n-1)(n+2) = 3$. So a prime dividing $n^2+n+1$ can only be $3$ or of the form $6k+1$. |
H: Negation of statement and determining truth
a,b $\in$ $\mathbb{R}$
Original statement: $\exists a$ such that $\forall b$, $a+b>0$
My negation: $\forall a$, $\exists b$ such that $a+b \leq 0$
Is my negation correct? If it is, is the negation true whereas the original statement is false?
I drew this conclusion because my interpretation of the original statement was that there was one $a$ that would satisfy the inequality regardless of what value of $b$ was chosen (obviously not true). My understanding of the negation statement was that you could choose different values of $b$ to satisfy the inequality based on the value of $a$ you are dealing with (so $b$ is not fixed unlike $a$ in the original statement). Or am I wrong and in fact $b$ is fixed like $a$ was before? If I am wrong then I am stuck on trying figure out which statement is true. Any help is appreciated.
AI: A basic principle worth remembering is this, in headline terms
When you push a negation sign past a quantifier, the quantifier "flips" into its dual.
So $\neg\forall x \varphi \Leftrightarrow \exists x\neg \varphi$, and $\neg\exists x \varphi \Leftrightarrow \forall x\neg \varphi$. [Before reading on make you understand why that has to be right!] And moreover, you can apply this equivalence inside a wff. [Why?]
Applied to this case, the negation of
$\exists a \forall b\, a+b > 0$
is, of course
$\neg\exists a \forall b\, a+b > 0$
Which applying the principle is equivalent to
$\forall a \neg\forall b\, a+b > 0$
which is equivalent to
$\forall a \exists b\neg\, a+b > 0$
which is equivalent to
$\forall a \exists b \, a+b \leq 0$.
As you rightly said! |
H: Proving using vectors, that if a median is also a height, then the triangle is isosceles.
Proving using vectors, that if a median is also a height, then the triangle is isosceles.
*Better wording would be very helpful.
Thanks in advance for any help.
AI: Let the vertices be the origin $O$, and the points $A$ and $B$. Let $u$ be the vector $OA$, and let $v$ be the vector $OB$. Then if $M$ is the midpoint of $AB$, the vector $OM$ is $\frac{u+v}{2}$.
Because the median is an altitude, we have $\left(\frac{u+v}{2}\right)\cdot (u-v)=0$. Thus
$(u+v)\cdot(u-v)=0$.
Expand. We get $u\cdot u-v\cdot v=0$, so $u\cdot u=v\cdot v$.
But $u\cdot u$ is the square of the length of $OA$, and $v\cdot v$ is the square of the length of $OB$. It follows that the length of $OA$ is equal to the length of $OB$, and therefore the triangle is isosceles. |
H: $5^a - 5^b$ is divisible by $n$ (prove)
Prove that for every n natural number exist natural numbers $a,b \leq 4n, a\not= b $, which accomplish, that number $ 5^a - 5^b $ is divisible by n. How many of these pairs exist?
Help please, I'm stuck with this problem. Any help is appreciated.
AI: Let $n=5^km$ with $gcd(5,m)=1$.
Now, the pair $(a,b)$ is a solution if and only if $(b,a)$ is a solution. So it suffices to look for the solutions where $a>b$. Don't forget at the end to double the number of solutions.
Then
$$n|5^a-5^b =5^b(5^{a-b}-1)) \Leftrightarrow k \leq b \mbox{ and } m | 5^{a-b}-1$$
The existence is easy then, pick $b =k $ and $a-b =\phi(m) \leq m \leq n$.
For the number of solutions, you need
$$k \leq b \leq 4n$$
Thus you have $4n-k$ choices for $b$.
Morever, for each $b$, if $j$ is the order of $5$ modulo $m$, you must have
$$j |a-b \,.$$
Thus, $a-b$ has to be a multiple of $j$ between $1$ and $4n-b$. From here you figure out how many choices of $a$ you have for each $b$. Add them, double and you are done. |
H: Show that interval $(a, b)$ is not open in $\mathbb{R}^2$
I know that interval $(a, b)$ is open in $\mathbb{R}$. To show that interval $(a,b)$ is open in $\mathbb{R}$, I have done so: Let it be $x\in (a,b)$. Enough to find an open ball containing the point $x$, and that is included in the interval $(a,b)$. Suffice to get $0<\epsilon\leq \min \{\vert b-x\vert, \vert a-x \vert\}$. In this case $D(x,\epsilon)$ containing the point $x$ and $D(x,\epsilon)\subseteq (a,b)$. I do not know a good act. If it is that I did very well then correct, but I do not know how to prove this fact: To show that interval $(a,b)$ is not open in $\mathbb{R^2}.$
Please if anyone has the opportunity to help, to make verification of keti example, thank you preliminarily
AI: As kahen pointed out, what you want to say is that $$(a,b)\times\{0\}$$ is not open in $\Bbb R^2$. Now, pick any point ${\bf x}=(x,0)$ with $a<x<b$. Then $B({\bf x},\varepsilon)$ contains elements of the form $(x_1,x_2)$ with $x_2\neq 0$ (prove this!), so $B({\bf x},\varepsilon)$ cannot be contained in $(a,b)\times \{0\}$. This means $(a,b)$ is not open, since we've found a point (actually all of them) that are not interior points. In fact, ${\rm int}_{\Bbb R^2}\;(a,b)=\varnothing$. |
H: Computing $\int_{0}^\infty\frac{t^a}{1+t^2}dt$ for $-1
I am integrating the following integral
$$\int_{0}^\infty\frac{t^a}{1+t^2}dt$$ for $-1<a<0$.
by computing residues inside some contour. But I'm not sure what contour to use here, since $z^{a}$ gets large when $z$ is small, so I have to be careful about the part of the contour that gets near the origin.
AI: Use the keyhole contour suggested, with outer radius $R$ and inner radius $\epsilon$. The contour integral becomes
$$\int_{\epsilon}^R dx \frac{x^a}{1+x^2} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{R^a e^{i a \theta}}{1+R^2 e^{i 2 \theta}}\\ + e^{i 2 \pi a} \int_R^{\epsilon} dx \frac{x^a}{1+x^2} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^a e^{i a \phi}}{1+\epsilon^2 e^{i 2 \phi}}$$
Because $a \in (-1,0)$, the second and fourth integrals vanish as $R \to \infty$ and $\epsilon \to 0$, respectively.
On the other hand, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$. Thus we have
$$\left ( 1-e^{i 2 \pi a}\right ) \int_0^{\infty} dx \frac{x^a}{1+x^2} = \frac{i 2 \pi}{2 i} \left (e^{i \pi a/2} - e^{i 3 \pi a/2} \right )$$
or
$$\int_0^{\infty} dx \frac{x^a}{1+x^2} = \frac{\pi}{2 \cos{(\pi a/2)}}$$ |
H: $Y$ is a function of $X$: making an inference based on the markovity of $ X$
In the information theory book by Cover and Thomas it is written:
if $X$ is markov and $Y$ is a function of $X$ then:
$H(Y_n|Y_{n-1},Y_{n-2},...,Y_1,X_1)=H(Y_n|Y_{n-1},Y_{n-2},...,Y_1,X_1,X_0,X_{-1},...,X_{-k})$ because $X$ is markov.
Can someone please explain how they made this inference? It's not very clear to me.
Thank you!!
AI: The expression should follow from the property of Markov process. For a Markov process X(t) (with t being a time-index) the future states don't depend on the past states when the present state is given. In other words, knowledge of history of the random process doesn't add any new information.
So, for X is Markov and Y is a function of X, adding historical information (conditionally) to the function H, that is, having X(t) only for t=1 conditionally in the function H is the same as having X(t) for t = 1, 0, -1, -2, ..., -k conditionally. What matters here is the X1 in the condition. Hope this makes sense. |
H: Does $\text{Aut}(A)\cong \text{Aut}(B)\cong \text{Aut}(A\bigcap B)$ imply $A=B$?
The question is in the title: given two mathematical structures $A$ and $B$ that we conjecture to be equal, is it sufficient to prove that $\text{Aut}(A)\cong \text{Aut}(B)\cong \text{Aut}(A\bigcap B)$, where $\text{Aut}(X)$ is the automorphism group of $X$, to establish the desired equality $A=B$?
Thanks in advance.
EDIT December 27th 2013: Stated the way it is, the question can't be answered affirmatively. What if we add as an hypothesis that there exists a measure $\mu$ such that $\mu(A\Delta B)=0$?
AI: Presumably $A$ and $B$ are subsets of a bigger set, because otherwise you cannot speak of their intersection. Assuming this so that the question makes sense, it is false. For instance, consider the abelian groups $A=2\mathbf Z$ and $B=3\mathbf Z$ sitting inside $\mathbf Z$. The intersection is $6 \mathbf Z$. The automorphism groups of all three objects are isomorphic to $\pm 1$, but $A \neq B$.
Edit: Even if you want to relax the conclusion from an equality $A=B$ to an isomorphism $A\cong B$, it is still false. For instance, the automorphism group of the one-point set is isomorphic to the automorphism group of its empty subset (they are both trivial), but the empty set and the one-point set are not isomorphic. |
H: What does this huge X mean that is written like the sigma notation?
I hope that you will not mind if I do not explain the background of this formula. The problem which I encounter is probably a simple one: What does the huge $\large \times$ mean and why is it written like a sigma? I had no clue how to search for this.
My understanding is: $(l_1, ..., l_{|M|})$ is a set that is defined by counting from $j$ up to $|M|$ and that projects on $\mathbb{R}$ while using $r(m_i, m_j)$ as a function for $l_j$.
Is that correct?
Thank you very much!
$$\large{L_i=(l_1, \ldots l_{|M|})\in \mathop{\huge \times}^{|M|}_{j=1}\Bbb R, \,l_j=r(m_i, m_j)}$$
AI: It's probably an $|M|$-fold Cartesian product. This link to the TeX stackexchange supports this interpretation. |
H: Order of a homomorphism of groups
Let $f: G \to H$ be a homomorphism of groups. Assume that $a\in G$ and $\operatorname{ord}(a)=n$. Prove that the order of $f(a)$ is a divisor of $n$.
I know that if $H$ is a subgroup of $G$, then $$|G|=|H|\cdot|\text{distinct cosets of $H$}|$$ so $|H|$ must be divisor of $|G|$, but I don't understand how to apply that logic to homomorphisms (or if that will even work).
Any help is appreciated!
AI: Hint
$$ \left( f(a) \right)^n=f(a^n)=f(e)=e \,.$$
You don
't need to calculate the order of $f$, you need to calculate the order of $f(a)$ which is an element in $H$. |
H: Locus in complex z-plane given eqn
I have the question:
$
\text{Find the locus in the complex }z\text{-plane that satisfies the equation: }
z-c=\rho\dfrac{1+it}{1-it}, \text{where }c\text{ is complex, }\rho\text{ is real, and }t\text{ is a real parameter that varies in the range }-\infty<t<\infty.
$
[1]
But I am unsure how to proceed - I haven't seen such a question with as many variables before; I don't really understand what should be the format of a solution. The only similar questions I can find I understand, but I can't translate that to what I should be doing here.
I realise that $c$ is essentially a transformation, so I assume that could be ignored to begin with, similarly I assume it could be assumed momentarily that $\rho=1$. But what more of an answer am I trying to find? The question already gives a condition for $z$, and wants a solution for $z$ too? Or have I totally misunderstood?
1 - Mathematical Methods for Physics and Engineering - Ex3.4a
AI: Let $\alpha=1+it$. Then
$$z-c=\rho\frac{\alpha}{\bar{\alpha}}.$$
It follows that
$$|z-c|^2=(z-c)(\bar{z}-\bar{c})=\rho\frac{\alpha}{\bar{\alpha}}\rho\frac{\bar{\alpha}}{\alpha}=\rho^2,$$
as $|\alpha|^2\neq 0$.
The desired locus is then
$$\mathcal L=\{z\in\mathbb C ~|~~ |z-c|=|\rho|\}. $$ |
H: Let $A$ be an infinite set, $B\subset A$, let $G \subset S_A$ where $f$ in $G$ implies $x\in B\implies f(x)\in B$. find example where $G$ not a group
Let $A$ be a finite set and $B$ a subset. Let $G$ be the subset of $S_A$ (the symmetric group on $A$) consisting of all the permutations $f$ on $A$ such that $f(x)\in B$ for all $x\in B$. Show that $G$ is a group. The next question asked is now, find an example of an infinite set $A$ where the above conclusion does not hold. I will first present the solution to the first part which i think I solved well, but want to add for completeness
SOLUTION FIRST PART
Solution: Because $G \subset S_A$ we need only show that $G$ is closed under multiplication (in this case this operation is composition) and closed under inverses. Let $f,g\in G$. Then $g(x) \in B$ for all $x\in B$, say $g(x) = x'\in B$. Now this implies that $f(x') \in B$ because $f\in G$ and $x' \in B$. Thus
$$[f \circ g] (x) \in B$$
And we find that $G$ is closed under the group operation of composition.
Now let $f\in G$ again. Certainly $f \in S_A$ and thus $f^{-1} \in S_A$. Now if $x\in B$ we have $f(x)\in B$. Now $f^{-1}(f(x))=x \in B$. Because for every $x \in B$ we have a corresponding unique $x' \in B$ such that $f(x)=x'$, and $f^{-1}(x') = x$ we find that for all $x\in B$ it follows that $f^{-1}(x)\in B$.
Thus $G$ is a group.
SECOND PART
Now what is asked is to find an example of an infinite set $A$ with subset $B$, where $G \subset S_A$ such that $G$ consiststs of permutations $f$ such that $f(x)\in B\quad \forall x\in B$, but where $G$ does not constitute a subset. That is, $G$ is either not closed under composition, or not closed under inverses. I was thinking of using $A=\Bbb R$ and $B=\Bbb R _{\geq0}$ and then considering $f(x)=x^2$ but then I realized that this function is not a permutation because it is not injective or surjective on $\Bbb R$. Can anyone find an example where the claim is shown? Thanks a lot in advance!
AI: For the second part, the key is that $f(B) \subset B$ doesn't imply $f(B)=B$.
Take $A= \mathbb Z$ and $B=2\mathbb Z$. Now pick a bijection so that $f(x)=2x$ for all $x \in B$, anything else outside.
Prove that $f^{-1}$ is not in your group, by showing that $f^{-1}(2) \notin B$. |
H: Decide with proof if the groups are isomorphic.
Decide with proof if the groups are isomorphic.
$(a)\quad (\mathbb Z, +)$
$(b)\quad (3\mathbb Z, +)$
$(c) \quad$ The additive group of the rational numbers.
I've just done questions of the form groups "$C_2\times C_2$" etc but not really sure how to go about this.
Thank you
AI: $$\mathbb Z \cong 3\mathbb Z$$ under the isomorphism $\phi: \mathbb Z \to 3\mathbb Z$. Note, it is good to know that both $3\mathbb Z = \langle 3\rangle $ and $\mathbb Z = \langle 1 \rangle$ are cyclic groups of infinite order. And every infinite cyclic group is isomorphic to $\mathbb Z$.
However, no single element generates all of $(\mathbb Q, +)$ (this is a good exercise to prove) and hence the additive group of the rationals is not cyclic, unlike the first two groups.
Any isomorphism between groups must preserve the groups' structural properties. Being cyclic is a structural property of a group.
Hence, $(\mathbb Q, +)$ is not isomorphic to either of the other groups. |
H: Why $\arctan x \to \pi/2$ when $x \to \infty$
The limit of $\arctan x$ as $x \to +\infty$ is $\pi/2$. Maybe I don't understand the arctan enough, because by looking at the graph I see that indeed it approaches pi. But: why? How can I understand that?
AI: $\tan{x}$ is the ratio of the opposite side to the adjacent side in a right angled triangle.
As angle $A\to\ \pi/2$, the opposite side tends to $\infty$, while the adjacent side is fixed, so since $\tan{A}=\frac{\text{opposite}}{\text{adjacent}}$, $\tan{A} \to\infty$.
Obviously then as $\arctan(x)\to\pi/2$, $x\to\infty$. |
H: The intervals at which $f(x) = \cos x + \sin x$ is concave up and down and $f$'s inflection point(s)
Describe the intervals at which $f(x) = \cos x + \sin x$ is concave up and down and $f$'s inflection point(s).
I totally went blank on this once I hit the analysis of second derivative, $f''(x)=-\cos x-\sin x$. I found the solution as $n\pi-\frac{\pi}{4}$ but I can't figure out how to analyze the concave up and down.
I would really appreciate an explanation not only the solution.
AI: Write the function as
$$f(x) = \sqrt2 \sin\left(x+\dfrac{\pi}4\right)$$ and the rest should follow. |
H: Probability of Combinations which have different probabilties.
I cannot figure this out. The problem goes
A boy has a bag filled with balls:
1 red
2 green
4 Blue
7 White
A child plays a game in which he withdraws one ball, writes down its color, and then replaces the ball. In order to win the game he must write down at least one of each color of ball. What is the probability that this will happen on, or before the N-th trial???
I initially assumed this problem involved a geometric probability distribution function, based on the 4 probabilities multiplied together, for N trials. But I dont know. Please help!
AI: This is $1$ minus the probability that after $N$ trials, one or more of the colours is missing.
So let us find the probability that some colour is missing. We will use the Method of Inclusion/Exclusion.
The probability red is missing is $\left(\frac{14}{15}\right)^N$.
We can write down similar expressions for the probability green is missing, blue is missing, white is missing.
If we add up these $4$ probabilities, we will have double-counted the situations in which $2$ of the colours are missing. So we must subtract the probability that red and green is missing, together with $5$ other similar expressions. The probability that red and green is missing is, for example, $\left(\frac{12}{15}\right)^N$.
But we have subtracted too much, for we have subtracted once too many times the situations in which $3$ colours are missing. So we must add back $4$ terms that represent these probabilities. |
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