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H: Universal property of sum of ideals
If $A$ is a commutative algebra over the field $K$, and $I,J$ are ideals of $A$ (in particular, they are also algebras over $K$), is it possible to characterize the ideal they generate together $I + J = \langle I,J \rangle$ in terms of a universal property among $K$-algebras?
AI: It is the pushout of the diagram $I \leftarrow I \cap J \rightarrow J$. In other words, given a $K$-algebra $B$ and morphisms $\sigma : I \to B$ and $\eta : J\to B$ such that $\sigma|_{I\cap J} = \tau|_{I \cap J}$, there is a unique morphism $I + J \to B$ inducing $\sigma$ on $I$ and $\tau$ on $B$. |
H: A very simple proof of a divergent sequence
I've already proved that: if $\,0 <x< 1$, then $(x^n)\rightarrow0$. Now I'm wondering about the opposite, when $x>1$ which clearly diverges. I'd like to know if the following approach is correct or maybe I need more sophisticated arguments.
Scratchwork: Let $x>1$ and suppose by contradiction that $(x^n) \rightarrow L$. Since $0<1/x<1$, we can conclude $(1/x)^n\rightarrow 0$ (using the last proposition). We thus see from the limit laws that $x^n(1/x)^n\rightarrow L\cdot0=0$, but $x^n(1/x)^n=1$ and since $(\,1\,)\rightarrow 1$, we thus have $1=0$ by the uniqueness of the limit, which is a contradiction. Thus the sequences diverges.
Thanks in advance.
AI: One way to prove it is this : Suppose $x^n \to L$, then consider the function
$$
f(y) = xy
$$
This is continuous, so if $a_n = x^n \to L$, then $f(a_n) = x^{n+1} \to xL$. But $\{f(a_n)\}$ is the same sequence as $\{a_n\}$, so it must follow that
$$
xL = L
$$
Since $x \neq 0$, $L = 0$, but $a_n > 1$ for all $n \in \mathbb{N}$, so $L \geq 1$ - this is a contradiction. |
H: Let $f(z)=e^x+ie^{2y}$. For what values of $z\in\mathbb{C}$ does $f'(z)$ exist?
Let us define the function $f:\mathbb{C}\longrightarrow\mathbb{C}$ by
$$f(z)=e^x+ie^{2y}$$
where $z=x+iy$ for $x,y\in\mathbb{R}$.
Question: For what values of $z\in\mathbb{C}$ does $f'(z)$ exist?
Attempt: If $f'(z)$ exists at some point $z\in\mathbb{C}$, then $f$ must satisfy the Cauchy-Riemann equations at that point. Namely,
$$u_x=e^x=v_y=2e^y$$
$$u_y=0=-v_x=0$$
Thus, we must have $e^x=2 e^{2y}$ or $x=\ln 2+2y$, so $z$ is on the line
$$z=(\ln 2+2t)+it,\quad t\in\mathbb{R}$$
However, this line is not an open set, so this condition is not sufficient. What can we do then? With this condition we find
\begin{align} \lim_{h\to 0}\frac{f(z+h)-f(z)}{h} &=\lim_{h\to0}\frac{e^{x+h_1}+ie^{2(y+h_2)}-e^x-ie^{2y}}{h_1+ih_2}\\
&=e^{2y}\lim_{h\to0}\frac{2(e^{h_1}-1)+i(e^{2h_2}-1)}{h_1+ih_2}\\
\end{align}
Does this limit exists? How do we know?
AI: The fact the the line is not an open set is irrelevant. For a function $f(z)=u(x,y)+iv(x,y)$, a sufficient condition for the existence of $f'(z)$ at the point $z_0=x_0+iy_0$ is that the first-order partial derivatives $u_x,u_y,v_x,v_y$ exist throughout a neighborhood of $(x_0,y_0)$ and are continuous at $(x_0,y_0)$, and satisfy the Cauchy-Riemann equations at $(x_0,y_0)$.
In your example, inasmuch as the functions $u(x,y)=e^x$ and $v(x,y)=e^{2y}$ are continuously differentiable everywhere, the derivative $f'(z)$ exists at every point where the Cauchy-Riemann equations are satisfied. Of course $f(z)$ is not analytic at any point, because that line has no interior points. (A function $f(z)$ is analytic at a point if it's differentiable throughout some neighborhood of the point.) |
H: Combinatorial Bijection?
I have the following problem, which seems pretty easy, but I'm not sure as to what exactly is meant by a combinatorial bijection. I know what a 'normal' bijection is. The problem and my work follows beneath.
Let $H$ denote the number of ones in a binary string. Give a combinatorial bijection between the set of all binary strings of length $n$
and even $H$ and the set of those that have the same length $n$
and odd $H$.
As for every place in a binary string there are $2$ possibilities, the cardinality of a set of binary strings of length $n$, call this set $S$, is obviously $2^{n}$, corresponding to strings ranging from $H=0$ to $H=(n-1)$. $2^n$ is an even number, so we can partition $S$ into two sets of the same cardinality, one of strings with $H$ even and one with strings of odd $H$. Then just create an arbitrary injective matching and we've got our bijection.
AI: You’re being asked to exhibit some specific bijection between the two sets, thereby showing that they have the same cardinality. Combinatorial here doesn’t really modify bijection: it’s just an ordinary bijection, but it serves a combinatorial purpose.
In this case you could, for example, show that the map that change one fixed bit of the string — the first, the last, whatever — is a bijection between the two sets. |
H: A few questions relating to counting for midterm practise exam?
I'm doing some questions for my midterm practise exam (multiple choice) for discrete structures and would appreciate some help (My answer is bolded):
Using the 26-letter alphabet {a,b,c,...,z}, how many different 20-letter strings are there that start with abxy, end with pq, and contain exactly 4 k’s?
A) $\binom{14}{4}$4!$10^{25}$
B) $\binom{14}{4}$$10^{25}$
C) $\binom{14}{10}$$25^{10}$
D) None of the above
I think it will be C since 6 have a set position, thus 14 and then the 10 is from 4 being k's, thus knowing their value.
-
What is the coefficient of $x^{12}y^{15}$ in the expansion of $(−3x + 7y)^{27}$?
A) $\binom{27}{12}$
B) -$\binom{27}{15} 3^{12} 7^{15}$
C) $\binom{27}{15} 7^{12} 3^{15}$
D) $\binom{27}{15} 3^{12} 7^{15}$
This was a guess just based on the numbers and the negative.
-
Let A be a set of size 6 and let B be a set of size 5. How many different functions are there from A to B that are not one-to-one?
A) $6^5$
B) $5^6$ - 5!
C) $5^6$
D) $6^5$ - $\binom{6}{5}$
I think it's b because the total number of mappings is $5^6$ while 5! removes all the none one-to-one thus receiving the desired answer
-
Any help would be appreciated!
Thank you!
AI: The first one is fine.
For the second you want to look at the binomial theorem:
$$(-3x+7y)^{27}=\sum_{k=0}^{27}\binom{27}k(-3x)^k(7y)^{27-k}\;.$$
Clearly you want the term in which $k=12$, which is $$\binom{27}{12}(-3x)^{12}(7y)^{15}=3^{12}7^{15}\binom{27}{12}x^{12}y^{15}\;,$$ so the correct answer is D. The even power on $x$ kills the minus sign.
For the third one, note that no function from $A$ to $B$ can be one-to-one: there aren’t enough elements in $B$. (This is really just an application of the pigeonhole principle.) Since (as you said) there are $5^6$ functions altogether, the correct answer is C. |
H: Another problem about irreducible polynomials over a (finite) field
I want to know whether it is true that over a finite field $K$ (with characteristic $p$, say), and for any positive integer $m$, does there always exist a prime (or equivalently, irreducible, since the polynomial ring over a field is UFD) polynomial in $K[x]$ with degree $m$. I prefer some elementary proof.
AI: Yes, you can. There is a formula for the number of monic primes of degree $m$ over a finite field of order $q$:
$$\nu_q(m)=\frac{1}{m}\sum_{d\mid m} \mu\left(\frac md\right)q^d$$
Where $\nu_q(m)$ counts the number of prime monic polymomials of degree $m$ over a field of size $q$.
You can easily show this gives $\nu_q(m)\geq \frac{1}{m}\left(q^m - \sum_{i=0}^{m-1} q^i\right)> 0$. |
H: about restricting outer measures
If $\mu_0$ is an outer measure on an algebra, we can extend the premeasure to an outer measure $\mu^*$. By Caratheodory's theorem, the collection of $\mu^*$-measurable sets is a $\sigma$-algebra. Is an outer measure $\mu^*$ restricted to $\mu^*$-measurable sets a measure?
AI: Yes, it is. That's also part of Caratheodory's extension theorem. |
H: Preforming Counting Permutations
Problem: A seven-person committee composed of Alice, Ben, Connie, Dolph, Egbert, Francisco, and Galvin is to select a chairperson, secretary, and treasurer. How many selections are there in which at least one office is held by Dolph or Egbert? Each person may only hold at most one office.
My approach: Because the total number of permutations possible are 210 b/c the first office has 7 options, second office has 6 options, third office has 5 options 7*6*5 = 210
Then I subtract the possible offices with Dolph in an office which is 30 First office has no choice as it is Dolph so it is 1, second office has 6 choices, third office 5 choices gives me 1 * 6 * 5 = 30
Multiply 30 by 2 to account for Egbert resulting in 60.
210 - 60 = 150
So there are 150 permutations possible. Is my methodology correct?
Thanks for your time!
AI: Your "$210-60=150$" is the correct computation, and I see how you got $210$, but I can't follow your explanation of how you got $60$. What you want to subtract from the $210$ total selections is the number of selections in which neither D nor E holds an office, which means all offices are held by the other $5$ people, so the subtrahend is $5\cdot4\cdot3=60$. The final answer is $7\cdot6\cdot5-5\cdot4\cdot3=210-60=150$. |
H: Integral of $1/x$ about the asymptote with Cauchy versus complex
In short, the integral of $1/x$ from $-2$ to $1$ or some other such range is being confusing for me. By means of the Cauchy principal value it receives a value of $-\ln 2$. Whereas with complex integration it would seem I'd get $-\ln 2 + \pi i$. And it can be shown decidedly that $\ln(|x|)$ does not have the derivative of $1/x$, but rather $\operatorname{sgn}(x)/x$. Basically what's the correct integral here?
AI: For a function with a singularity, integrals in the complex plane are in general path-dependent. Depending on what contour you take, you could get $-\ln 2 + k \pi i$ for any odd number $k$.
As for $\ln |x|$, that does not have a derivative at all in the complex sense.
The (complex) antiderivative of $1/x$ is $\ln(x) + C$, but this is multi-valued:
there is no antiderivative that is analytic on all of ${\mathbb C} \backslash \{0\}$. |
H: Intuition Behind The Hyperreals
I know that there are an infinite number of hyperreals. But is it true that there are only two hyperreals with standard part equal to $0$ (the "finite" infinitesimal one and the "infinite" hyperreal)?
Put differently, is it wrong to view the hyperreals as a field "generated" by $\mathbb{R} \cup \{\infty, 1/\infty\}$ whereby every real number $r \in \mathbb{R}$ is associated with its hyperreal shadow $s = r + 1/\infty$ with $s \approx r$ uniquely?
AI: Yes, it’s completely wrong. For example, if $\epsilon$ is any positive infinitesimal (i.e., a positive hyperreal whose standard part is $0$), then so is $\epsilon^2$, and of course $0<\epsilon^2<\epsilon$, so $\epsilon^2\ne\epsilon$, and $\epsilon^2$ is therefore another hyperreal whose standard part is $0$. For that matter, $\epsilon x$ is a positive infinitesimal for each positive standard real number $x$, and no two of these infinitesimals are equal. |
H: Rule for Series
I apologize if this seems really stupid, but I've been stuck in finding the general pattern for the following series:
$$\sum_{n=1}^{\infty}\frac{2\cdot4\cdot6\cdots(2n)}{1\cdot3\cdot 5 \cdots(2n-1)}$$
The numerator is simple enough, it's just $2^nn!$. But what I'm really having trouble is finding the rule for the denominator. I've been racking my brain, but I can't come up with anything. Maybe I'm not thinking correctly? But anyways, if someone could point me in the right direction that'd be great!
AI: If you know what the numerator is, and you know what the numerator times the denominator is, then you know what the denominator is. |
H: inequality of $L^2$ functions
Let $f\in L^2(\mathbb{R})$. Suppose $g(x)=xf(x)\in L^2(\mathbb{R})$. Prove \begin{eqnarray*}
||f||_1\leq \sqrt{2}(||f||_2+||g||_2).
\end{eqnarray*}
How to prove this question?
AI: Hint: Similar to the argument in this post, note that
$$|f(x)|\le\frac{|f(x)|+|g(x)|}{1+|x|}.$$ |
H: Creating one Set from another using Set Builder Notation
I'm a little confused about set builder notation. If I have one set, how do I construct another set from the first set, supposing that I want to alter all the elements?
For example,
Let there be a set $A = \{1,2,3,4,5\}$
and I want to construct, from $A$, a set $B = \{\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\}$.
What is the correct notation to do this? Surely it isn't $B = \{a \in A | \sqrt{a}\}$?
AI: Not quite. You'd want $$\{\sqrt{a} | a \in A\}$$
The elements come first, then the qualification. |
H: Propositional logic-Predicates
I have this problem in my Discrete structures course Show why : ∀x P(x) ∨∀x Q(x) is not logically equivalent to ∀x(P(x)∨Q(x)) . Please help solve this
AI: Let $P(x)$ denote the proposition that $x$ is even ($x \in \mathbb{Z}$) and $Q(x)$ denote the proposition that $x$ is odd. If the domain of discourse is $\mathbb{Z}$, then $$\forall x (P(x) \vee Q(x))$$ is true, but $\forall x P(x) \vee \forall x Q(x)$ is false. |
H: Proof of my conjecture on closed form of $\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}$
Let $a$, $b\in \Bbb R^+$ and $m \in \Bbb N$ then My conjectural closed form is $$\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}\,{\rm d}x
=
\frac{\Gamma(a)}{b^a}\left\lbrack\zeta(a)-\sum^{m-1}_{k=1}\frac{1}{k^a}\right\rbrack$$
Please tell me how to prove it.
AI: If you know what $\zeta(a)$ and $\Gamma(a)$ are, and moreover are able to guess such formula, it is somewhat strange that you are asking for a proof.
Note that $a$ should be actually greater than $1$ for the integral to converge. Under such assumption, we have
\begin{align}
\zeta(a)-\sum_{k=1}^{m-1}\frac{1}{k^a}&=\sum_{k=m}^{\infty}\frac{1}{k^a}=\\
&=\sum_{k=m}^{\infty}\frac{1}{\Gamma(a)}\int_0^{\infty}x^{a-1}e^{-kx}dx=\\
&=\frac{1}{\Gamma(a)}\int_0^{\infty}\frac{x^{a-1}e^{-mx}}{1-e^{-x}}dx=\\
&=\frac{b^a}{\Gamma(a)}\int_0^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}dx.
\end{align}
Explanations:
Making the change of variables $kx=y$ in the second line transforms the integral into the definition of $\Gamma(a)$ multiplied by $k^{-a}$;
To pass from the second to the third line we exchange summation and integration and sum the resulting geometric series;
The fourth line is obtained from the third by the change of variables $x\rightarrow bx$. |
H: Can anyone explain to me this square root? step by step?
$$\begin{align}
v(p_1, p_2, w)
& = \sqrt{\frac w{p_1^2\left(\frac1{p_1}+\frac1{p_2}\right)}}
+ \sqrt{\frac w{p_2^2\left(\frac1{p_1}+\frac1{p_2}\right)}}
\\
& = \sqrt{\frac w{\left(\frac1{p_1}+\frac1{p_2}\right)}}
\left(\sqrt{\frac1{p_1^2}}+\sqrt{\frac1{p_2^2}}\right)
\\
& = \sqrt{\frac w{p_1}+\frac w{p_2}}
\end{align}$$
So as the title says, can anyone explain to me this square root?
(Original scan)
AI: Step by step:
Rewrite the original expression as follows:
$$\sqrt{\frac{w}{p_{1}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} + \sqrt{\frac{w}{p_{2}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Factoring, this becomes
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right)$$
Note that
$$\sqrt{\frac{1}{p_{1}^{2}}} = \frac{1}{p_{1}}$$
and
$$\sqrt{\frac{1}{p_{2}^{2}}} = \frac{1}{p_{2}}$$
So
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)$$
This can be rewritten as
$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right) = \sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}$$
And finally, we can simplify this to
$$\sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}} = \sqrt{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)} = \sqrt{\frac{w}{p_{1}} + \frac{w}{p_{2}}}$$ |
H: Invariance and Orthogonal Projection
Suppose $\mathcal{S}$ is a subspace of $\mathbb{R}^n$ and let $P$ be the orthogonal projection of $\mathbb{R}^n$ onto $\mathcal{S}$.
Show that $\mathcal{S}$ is invariant under a square matrix $A$ if and only if $AP=PAP$.
Show that $\mathcal{S}$ and $\mathcal{S}^{\perp}$ are invariant under $A$ if and only if $A$ and $P$ commute.
How should I start/approach to prove these two assertions? Any recommendations?
AI: Note that $x \in S$ iff $Px = x$.
(1)
If $S$ is $A$ invariant, then $x \in S$ implies $Ax \in S$. Hence $APx \in S$ for all $x$ which means $PAPx = APx$ for all $x$, and so $AP=PAP$. If $AP=PAP$, then suppose $x \in S$ (and so $Px=x$). We have $APx = PAPx$, and using $Px=x$, we get $Ax = PAx$. Since $P(Ax) = (Ax)$ we have $Ax \in S$, and so $S$ is $A$ invariant.
(2)
! From the above, we see that $S^\bot$ is $A$ invariant iff $A(I-P) = (I-P)A(I-P)$, hence to finish we need to show that $A,P$ commute iff
$AP=PAP$ and $A(I-P) = (I-P)A(I-P)$. Suppose $A,P$ commute, then $PAP = APP = AP$, since $P$ is a projection. The other equality follows since $A$ and $I-P$ commute and $I-P$ is a also a projection. Now suppose the two equalities hold. Expanding $A(I-P) = (I-P)A(I-P)$ we get $PAP = PA$, and the other equality gives $PAP = AP$ from which we get that $AP=PA$. |
H: Decomposition of an ideal as a product of two ideals
How to show $$5\mathbb{Z}[\sqrt[3]{2}] = (5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1).$$
Firstly, I think that I can say that $$(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1)= (25,5(\sqrt[3]{2}+2),5((\sqrt[3]{2})^2+3\sqrt[3]{2}-1),5((\sqrt[3]{2})^2+\sqrt[3]{2})).$$ Since 5 divides each term, I have that $(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1) \subseteq (5)$. But now how do I show the other direction? Any help would be appreciated. Thanks in advance.
AI: Working in $\Bbb Z[x]/(x^3-2)$ I get
$$(5,x+2)(5,x^2+3x-1)=\big(25,5(x^2+3x-1),5(x+2),(x+2)(x^2+3x-1)\big).$$
Then $(x+2)(x^2+3x-1)=x^3+5x^2+5x-2=5x(x+1)$. Hence the above ideal is
$$=(5)(5,x^2+3x-1,x+2,x^2+x).$$
Use the Euclidean division algorithm (which applies here) to simplify the right-hand factor. For example $x^2+3x-1\equiv x-1$ mod $x+2$ so we may replace $x^2+3x-1$ with $x-1$. |
H: Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n
Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n
Hi everyone, for the proof to the above question,
Can I assume that since $(a, b) = 1$, then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the $nth$ power, they will still have no prime factors in common, and so $(a^n, b^n) = 1$ for all positive integers n.
I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :)
And also, I do not know how to approach the reverse problem where if $(a^n, b^n) = 1 then (a, b) = 1$, any guidance will be much appreciated!
AI: Here's an approach that doesn't use prime factorization.
Lemma: For all $m,r,k\in \Bbb N$, $\gcd(m,k)=1\implies \gcd(m,k^r)=1$.
Proof: Let $m,r,k\in \Bbb N$ be such that $\gcd(m,k)=1$.
Bézout yields $um+vk=1$, for some $u,v\in \Bbb Z$.
Thus $$1=1^r=(um+vk)^r=\sum \limits_{j=0}^r\left({r\choose j}(um)^{r-j}(vk)^j\right)=\sum \limits_{j=0}^{r-1}\left({r\choose j}(um)^{r-j}(vk)^j\right)+(vk)^r,$$
which implies $$m\underbrace{\color{blue}{\sum \limits_{j=0}^r\left({r\choose j}u^{r-j}m^{r-j-1}(vk)^j\right)}}_{\huge \in \Bbb Z}+k^r\color{blue}{v^r}=1.$$
Again Bézout says that $\gcd(m,k^r)=1$ (due to the blue scalars).$\,\square$
You can now use the lemma twice: $\gcd(a,b)=1\implies \gcd(a,b^n)=1\implies \gcd(a^n,b^n)=1$. |
H: Jacobi fields in polar coordinates.
This is from Sakai's Riemannian Geometry:
Let $(r, \theta)$ be polar coordinates of the plane. We define a Riemannian metric $g$ on the plane by $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial r}) = 1$, $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}) = 0$, and $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}) = f^{2}(r, \theta)$ where $f(r, \theta)$ is of class $C^2$ with $f(r\neq 0, \theta) > 0$, $f(0, \theta)=0$, and $\frac{\partial}{\partial r} f(0, \theta) = 1$. Let $\gamma_{\theta}$ are geodesics (so $\theta$ is constant) and let $V(r)$ is a parallel vector field along $\gamma_{\theta}$ perpendicular to $\gamma_{\theta}$. Show that $Y(r) = f(r, \theta) V(r)$ is a Jacobi field along $\gamma_{\theta}$.
I think the notation is causing a fair bit of confusion for me. I want to show that $Y$ satisfies the Jacobi field equation. So since $V(r)$ is parallel, $\nabla_{\partial/\partial r} V \equiv 0$ and also $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$, so the equation reduces to
$\begin{align*}
\nabla_{\partial/\partial r} \nabla_{\partial/\partial r} Y + R(Y, \dot{\gamma})\dot{\gamma}
&= \nabla_{\partial/\partial r} \nabla_{\partial/\partial r} Y
+ \nabla_{Y} \nabla_{\dot{\gamma}} \dot{\gamma}
- \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma}
- \nabla_{[Y, \dot{\gamma}]} \dot{\gamma} \\
&= \nabla_{\partial/\partial r} \left(\frac{\partial f}{\partial r} V \right)
+ \nabla_{Y} \nabla_{\dot{\gamma}} \dot{\gamma}
- \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma}
- \nabla_{[Y, \dot{\gamma}]} \dot{\gamma} \\
&= \frac{\partial^2 f}{\partial r^2} V
- \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma}
- \nabla_{[Y, \dot{\gamma}]} \dot{\gamma}.
\end{align*}$
But I'm unsure where to go from there to ensure that I get zero. I tried playing around with some properties of Levi-Civita connections, but I couldn't come up with anything. Can anyone help me finish up the proof? Many thanks.
AI: As $V$ is perpendicular to $\gamma'$, we may write $V = g \frac{\partial }{\partial \theta}$. Then (Using the definition of the metric)
$$ 0=\nabla_r V = \nabla_r(g\frac{\partial }{\partial \theta}) = g_r \frac{\partial }{\partial \theta}+ g\nabla_r\frac{\partial }{\partial \theta} = \big( g_r + \frac{f_rg}{f}\big) \frac{\partial }{\partial \theta}$$
Thus $ g_r + \frac{f_rg}{f}=0$, which is the same as $\big(\log(fg)\big)_r=0$. Thus $g = C/f$ for some constant $C$ and $Y = fV = C\frac{\partial }{\partial \theta}$. Then $Y$ is a Jacobi field. You can either check directly, but it is clear as $Y$ generates a family of geodesics (varying $\theta$) starting at the origin. |
H: Anti-homomorphism of Hopf Algebra
I've got a quick question regarding the anti-homomorphism property. Specifically, what does it actually mean??
For a bit of context, I have the following question.
We define $U_q = U_q[o(3)]$ to be the associative algebra, with generators $L_0$, $L_+$ and $L_-$, with the defining relations; $[L_0, L_{\pm}] = \pm L_{\pm}$, $[L_+, L_-] = (q^{2L_0} - q^{-2L_0}) / (q - q^{-1})$.
The antipode has been defined such that $$S(I) = I$$ $$S(L_0) = -L_0$$ $$S(L_{\pm}) = -q^{-L_0} L_{\pm}q^{L_0} = -q^{\mp1}L_{\pm}$$
I'm being asked to show that $S$ constitutes an anti-homomorphism, for the $L_0$ and $L_{\pm}$ generators. However, I'm not really sure how to go about this, because I'm not too sure what the anti-homomorphism actually looks like, or really means.
Any insight into this would be great. Thank you!!
AI: If $A$ is an algebra, then anti-homomorphism $\phi:A\to A$ is a linear map such that $$\phi(x\cdot y)=\phi(y)\cdot\phi(x).$$ In your situation, you are probably being asked to show that there exists an anti-homomorphism $S$ which acts as you were told on the generators of the algebra.
A simple example, different from the one you want, so as not to ruin your problem:
Let $A$ be the algebra generated by two letters $x$ and $y$ subject to the relation $$yx-xy=1.$$ I claim there is a unique anti-homomorphism $\phi:A\to A$ such that $\phi(x)=y$ and $\phi(y)=-x$. |
H: Subspaces of $\Bbb R^3$
Given
$U=\{(1, -1, 3)^t\}$ and $V_a=\{(x, 3x-az,z)^t\}$ for any $a\in\mathbb{R}$, how to determine $a$, such that $U \cap V_a = \{o\} \land U+V_a=\mathbb{R}^3$?
I already found out that $V_a$ always describes a plane in this 3D space, because its base is $\{(1,3,0)^t,(0,-a,1)^t\}$ ($\dim(V)=2$) and that in order to satisfy both conditions, the line determined by $U$ should not lie within this plane and the line should cross the plane in $o$.
However, I forgot much of the algebra of my senior year and don't remember how exactly I am supposed to find out if a line is contained in the plane or not. I tried to set a vector equation:
$$(1,-1,3)^t = l_1 (1,3,0)^t + l_2 (0,-a,1)^t + l_3 (0, 0, 0)$$
but that yields a solution for the $l$'s and not for my $a$.
AI: We have $V_a = \operatorname{sp} \{(1,3,0)^T, (0,-a,1)^T \}$.
Then $(1,-1,3)^T \in V_a$ iff $a = \frac{4}{3}$.
Since $\dim V_a = 2$, then $\dim (V_a + \operatorname{sp} ( U)) = 2 $ iff $(1,-1,3)^T \in V_a$. |
H: $\int^\infty_{-\infty} \frac{1}{\pi(1+x^2)} dx = 1$. How?
$$\int^\infty_{-\infty} \frac{1}{\pi(1+x^2)} dx = 1$$. How?
I can do
$$\int^\infty_{-\infty} \frac{1}{\pi(1+x^2)} dx = \frac{1}{\pi} \int^\infty_{-\infty} \frac{d}{dx} \tan^{-1}{(x)} \; dx$$ But how do I proceed? I remember the TA mentioning something about changing the limits of integration to $\pi/2$ but I don't really understand why ...
AI: $$
\begin{align}
\int_{-\infty}^\infty \frac{1}{1+x^2}\,\mathrm dx&=\lim_{n\to\infty}\int_{-n}^n \frac{1}{1+x^2}\,\mathrm dx=\lim_{n\to\infty}\left[\tan^{-1}(x)\right]_{-n}^n\\
&=\lim_{n\to\infty}[\tan^{-1}(n)-\tan^{-1}(-n)]\\
&=\lim_{n\to\infty} 2\tan^{-1}(n)=\pi
\end{align}
$$ |
H: Show that $\frac{(m+n)!}{m!n!}$ is an integer whenever $m$ and $n$ are positive integers using Legendre's Theorem
Show that $\frac{(m+n)!}{m!n!}$ is an integer whenever $m$ and $n$ are positive integers using Legendre's Theorem.
Hi everyone, I seen similar questions on this forum and none of them really talked about how to apply the Legendre's theorem to questions like the one above.
I get that $\frac{(m+n)!}{m!n!}$ = $\binom{m+n}{m}$, which is an integer. But could someone explain how the Legendre proof works in this case and why it proves the above is true?
AI: For a prime $p$, denote by $v_p(r)$ the exponent of $p$ in the prime factorization of $r$. So for example, $v_2(12) = 2$. Legendre's theorem states that for any prime $p$ and integer $n$, we have
$$v_p(n!) = \sum_{i = 1}^\infty \left\lfloor \frac{n}{p^i} \right\rfloor $$
Note that $v_p(rs) = v_p(r) + v_p(s)$. Also $r$ divides $s$ if and only if $v_p(r) \leq v_p(s)$ for all primes $p$.
The fraction $\frac{(m+n)!}{m!n!}$ is an integer if and only if $m!n!$ divides $(m+n)!$. So to prove your statement, it suffices to prove $v_p(n!) + v_p(m!) \leq v_p((n+m)!)$. Using Legendre's theorem, this follows from $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x + y \rfloor$. |
H: Generalized Bernoulli's inequality
I was able to prove Bernoulli's inequality, easily by simple induction.
However, I'm not sure how to prove the generalized inequality (generalized = for each sequence of numbers $i=1,\ldots,n$):
$$\prod\limits_{i = 1}^n {(1 + {x_i})} \ge 1 + \sum\limits_{i = 1}^n {{x_i}},\qquad {x_i} \ge 0$$
How do you prove it? Thanks!
AI: Let $$P(n):(1+x_1)(1+x_2)\cdots(1+x_n)\ge1+\sum_{1\le i\le n}x_i$$
Clearly, $P(n)$ holds for $n=1$
Let $P(n)$ holds for $n=m$
$$\implies (1+x_1)(1+x_2)\cdots(1+x_m)\ge1+\sum_{1\le i\le m}x_i$$
For $n=m+1,$
$$ (1+x_1)(1+x_2)\cdots(1+x_m)(1+x_{m+1})\ge(1+\sum_{1\le i\le m}x_i)(1+x_{m+1})$$
$$=1+\sum_{1\le i\le m+1}x_i+x_{m+1}\sum_{1\le i\le m}x_i\ge1+\sum_{1\le i\le m+1}x_i$$ as $x_i>0\forall i\in[1,n]$ |
H: Uniform convergent and lipschitz continuous
I want to prove that if I have a sequence $ f_n\in C[0,1]$ that is uniform convergent to zero and all functions are lipschitz continuous, then the lipschitz constants form a zero sequence. Does anybody here know to show this?
AI: It is not true.
If I take $f_n(x) = \frac{1}{n} \sin (n^2x)$, then $f_n$ converges to zero uniformly, each $f_n$ is Lipschitz continuous, but the Lipschitz constants are $L_n = n$ (since $\sup_{x \in [0,1]} f_n'(x) = n$). |
H: In a finite field $F$, how is $a^{|F|}=a$?
Let $F$ be a finite field with $|F|=q$. If $a\in F$, I have to prove $$a^q=a$$.
We know that if $a^r=1$, then $r|q$ (treating $F$ is a group under multiplication, and using Lagrange's theorem). Hence, $a^q=a^{r\frac{q}{r}}=1^{\frac{q}{r}}=1$. This is because $\frac{q}{r}$ is an integer.
How is it possible then that $a^q=a$?
Thanks in advance!
AI: Hint: If $a=0$ this is obvious. If not, then you can say $a^{|F^\times|}=1$ (why?), and what is $|F^\times|$? |
H: Defining a map based on a group action on left cosets
If $H$ is subgroup of $G$ such that the index of $H$ in $G$ is $n$ and $\pi_H$ is the permutation representation of the action of $G$ on the left cosets of $H$, is $\pi_H$ a map from $H$ to $S_n$? I am a bit confused about how $\pi_H$ relates to the symmetric group.
AI: The map $\pi_H$ is technically a homomorphism from $G$ to $S_X$ where $X = G/H$. Now if $G$ has finite index, then $|X| = n$, and so $S_X \cong S_n$. Under this isomorphism you can think of $\pi_H$ as a homomorphism from $G$ to $S_n$. |
H: Let $ (x_{1}, x_{2}) \sim (y_{1}, y_{2})$ iff $x_{2} =y_{2}$ on $ \mathbb{R} ^{2}$ . Then $\mathbb{R} ^2 /\sim$ is homeomorphic to $\mathbb{R}$
Let $ (x_{1}, x_{2}) \sim (y_{1}, y_{2})$ iff $x_{2} =y_{2}$ on $ \mathbb{R} ^{2}$ . Then $\mathbb{R} ^2/\sim $ is homeomorphic to $\mathbb{R}$
I am using Willard's book and there the "identification space " $X / \sim$ is defined to be the decomposition space $ D$ whose elements are the equivalence classes for $ \sim$.
I think that by following the definition the topology on the decomposition space is the trivial one( if we consider $\mathbb{ R} ^2 $ to have the standard topology). But, if this is true I don't see how the spaces are homeomorphic. Thanks for any hints / suggestions!
AI: Note that $\pi : \mathbb{R}^2 \to \mathbb{R}$ given by
$$
\pi(x,y) = y
$$
is a continuous surjection. Furthermore, $(x_1,x_2)\sim (y_1,y_2)$ iff $\pi((x_1,x_2)) = \pi((y_1,y_2))$. Thus you have an induced map :
$$
\overline{\pi} : \mathbb{R}^2/\sim \to \mathbb{R}
$$
given by
$$
\overline{\pi}([(x,y)]) = \pi(x,y) = y
$$
This is well-defined (by what I said above), surjective, and also injective (why?).
Furthermore, it is continuous (this you need to explain to yourself). Once you do that, it is the required homeomorphism. |
H: If a subset of the real numbers has no cluster points it is countable
Prove that if a subset of the real numbers has no cluster points it is countable.
I'd just like to see how this proof goes; preferably directly, if convenient.
AI: Here is a brief outline of the contrapositive. We show that an uncountable set has a limit point; let $A$ be our set.
At least one of the sets $A_n := [-n, n] \cap A$ must be uncountable (why?)
Suppose that we've fixed $n$ from the above point. Bisect and get $[0, n] \cap A$ and $[-n, 0] \cap A$; one of these has to be uncountable (why?); call the appropriate interval $B_1$.
Bisect $B_1$ and get an interval that's half as large, and contains uncountably many elements of $A$. Call it $B_2$.
Bisect and get $B_3$. Continue.
Now $$\bigcap_{n = 1}^{\infty} B_n$$
contains a unique point (why?) that is an accumulation point of $A$ (why?).
Now a follow-up for you to think about: Why does this not work if $A$ is merely countable? |
H: Probability theory - Dice
Two guys are playing dice with each wagering $50. Player 1 chooses 2 as his lucky number, and Player 2 chooses 6. Every time their lucky number appears as a result, the player gets one point.
The player who gets 3 points first wins $100
Suddenly, the game has to be stopped. Player 1 chalks up 2 points and Player 2 chalks 1 point by then. What is a fair way of splitting $100?
Edit: the dice has 6 sides
AI: Denote the event that (by going on) the first result that belongs
to the lucky numbers $2$ and $6$ is a $2$ by $A$ and that it is a $6$ by $B$. Denote
the event that player 1 wins by $W$. Then $P\left(W\right)=P\left(W\mid A\right)P\left(A\right)+P\left(W\mid B\right)P\left(B\right)=1.\frac{1}{2}+P\left(W\mid B\right).\frac{1}{2}=\frac{1}{2}\left(1+P\left(W|B\right)\right)$.
Here $P\left(W|B\right)$ stand for the probability that player 1 wins by going on if
both players have $2$ points, so it is evident that $P\left(W\mid B\right)=\frac{1}{2}$.
This leads to $P\left(W\right)=\frac{3}{4}$. Player 1 should get
$75$ and 2 should get $25$ |
H: Find all $f: \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(x+y) = f(x)+f(y)$
i have to find all functions $f: \mathbb{Q} \rightarrow \mathbb{R}$, such that $f(x+y)=f(x)+f(y)$.
So functions of the form $f(x) := ax, a \in \mathbb{R}$ satisfy the above condition:
$$
f(x+y)=a(x+y)=ax+ay=f(x)+f(y)
$$
But how do i proove that all functions that satisfy the above condition have the form $f(x) := ax, a \in \mathbb{R}$?
Thanks in advance!
AI: Let $f:\mathbb{Q}\rightarrow\mathbb{R}$ be such that $f(a+b)=f(a)+f(b)$ for all $a,b\in\mathbb{Q}$. It should be easy for you to prove the following:
for every $m\in\mathbb{Z}$, $p\in\mathbb{Q}$, $f(m\cdot p)=m\cdot f(p)$ (use induction on $m$).
From that, conclude that $f(m^{-1})=m^{-1}f(1)$ for every $m\in\mathbb{Z}\setminus\left\{0\right\}$.
Now, given $q\in\mathbb{Q}$, write $q=m/n$, with $m,n\in\mathbb{Z}$, $n\neq 0$. Then,
$$f(q)=f(m/n)=m\cdot f(n^{-1})=m\cdot n^{-1}f(1)=q\cdot f(1).$$
Concluding, $f$ is of the form $f(x)=f(1)\cdot x$. |
H: sum of divisors for given range of numbers from 1 to n
we are given a function F(n) for a number n which is defined as sum of the divisors of n (including 1 and n) ... now given an integer N we have to calculate
G(n) = F(1) + F(2) + F(3) + ..... + F(n)...
is there any formula for it???
i saw "Peter Gustav Lejeune Dirichlet " formula that can be used while surfing the net...can this be used??if yes, please provide an explanation for this formula??if no, then provide another method that can be used
AI: If $f\left(k,n\right)$ denotes the cardinality of $\left\{ i\in\left\{ 1,\ldots,n\right\} \mid k\text{ divides }i\right\} $
then $G\left(n\right)=\sum_{k=1}^{n}kf\left(k,n\right)$. Here $f\left(k,n\right)=\lfloor\frac{n}{k}\rfloor$ so $G\left(n\right)=\sum_{k=1}^{n}k\times\lfloor\frac{n}{k}\rfloor$.
Take a number, $3$ for instance, and wonder in what $F\left(i\right)$
it will turn up as a term. This is the case if $3$ is a divisor of
$i$. Here $f\left(3,n\right)=\lfloor\frac{n}{3}\rfloor$ expresses the number of times that
$3$ turns up as a term of $G\left(n\right)$
Edit:
Let say that $X_{k,i}=1$ if $k|i$ and $X_{k,i}=0$ otherwise.
Then
for $i\leq n$ we have $F\left(i\right)=\sum_{k=1}^{n}k\times X_{k,i}$.
So $G\left(n\right)=\sum_{i=1}^{n}F\left(i\right)=\sum_{i=1}^{n}\sum_{k=1}^{n}k\times X_{k,i}=\sum_{k=1}^{n}\sum_{i=1}^{n}k\times X_{k,i}=\sum_{k=1}^{n}k\sum_{i=1}^{n}X_{k,i}=\sum_{k=1}^{n}k\lfloor\frac{n}{k}\rfloor$ |
H: Are side lengths enough to find the ratio of the diagonals of a quadrilateral?
Is it possible to find the ratio of two diagonals of a quadrilateral when the length of all sides are given??
AI: If you don't know some of the angles, then no. This is not to difficult to see by drawing some pictures (or even better, bending something like a straw into the shape of a quadrilateral, and then deforming it). Keeping all the lengths of the sides fixed, you can vary the angles to produce some fairly extreme shapes - in particular, you can draw lots of quadrilaterals with the same four side lengths, but wildly different diagonal lengths.
Note: The question has significantly changed in an edit, to ask about the ratio of the diagonals, rather than the lengths. However, the same answer applies (imagine taking a square, where the two diagonals have equal length, and squashing two opposite corners together without changing the side lengths, to get a rhombus, where the diagonals are different). |
H: Solve the following Diffrential Equation $(x+y+1)dx+(2x+2y-1)dy=0$
I want to seperate variables in the following equation and need some advice:
$$(x+y+1)dx+(2x+2y-1)dy=0$$
what I tried to do so far is:
$$ydx+(x+1)dx+(2y-1)dy+2xdy=0$$
now how I should I continue? thanks.
AI: Hint:
Setting $$(x+y+1)=u\longrightarrow 2x+2y-1=2u-3,~~~dx+dy=du$$ makes your ODE separable. |
H: Prove if $f(x) = g(x)$ for each rational number x and $f$ and $g$ are continuous, then $f = g$
$f,g: \mathbb{R} \to \mathbb{R}$
I'd like to see a sketch for this proof.
[sorry for posting errors, I am on a cell phone]
AI: Hint. For every real number $r$ there is a rational sequence $(q_n)$ such that $q_n\to r$ as $n\to\infty$. You can check that $f(q_n)=g(q_n)$ for all $n$. Take $n\to\infty$. |
H: What does $\sigma(A_n)$ look like?
Let $\Omega=\mathbb{N}$, $A_n:=\left\{\left\{1\right\},\left\{2\right\},\ldots,\left\{n\right\}\right\}$.
What does then $\sigma(A_n)$ (generated $\sigma$-Algebra) look like?
Surely, $\mathbb{N}\in\sigma(A_n)$ and $A_n\in\sigma(A_n)$. But which sets are in $\sigma(A_n)$, too?
Unions and complements, okay. But how can I determine them?
AI: For each $0 \leq k\leq n$, define
$$
\mathcal{A}_k = \{B \subset \{1,2,\ldots, n\} : |B| = k\}
$$
and set
$$
\mathcal{A} = \bigcup_{k=0}^n \{B : B\in \mathcal{A}_k, \text{ or } B^c \in \mathcal{A}_k\}
$$
Clearly, any $\sigma$-algebra containing $A_n$ must contain $\mathcal{A}$, so it suffices to prove that $\mathcal{A}$ is a $\sigma$-algebra.
The only non-trivial thing to check is that $\mathcal{A}$ is closed under countable unions. So suppose $B_1, B_2 \ldots$ are in $\mathcal{A}$, and
$$
B = \cup B_j
$$
Then you have two choices, either $|B| < \infty$ or $|B| = +\infty$.
If $|B|<\infty$, each $B_j \in \mathcal{A}_k$, and so $B \in \mathcal{A}_{k_0}$ where $k_0$ would be the max of those $k$'s.
If $|B| = +\infty$, then there must be a $B_j$ such that $|B_j| = +\infty$ (why?), but then
$$
|B_j^c| \leq n \Rightarrow |B^c| \leq n \Rightarrow B \in \mathcal{A}
$$
Given all this, you can actually "write down" all the elements of $\mathcal{A}_k$, and thus all the elements of $\mathcal{A}$. |
H: Limit of the sequence: $\frac{k(k+1)}{(k+1)^2-qk^2}$
I am stuck on the proof of this limit:
given the sequence:
$$S(k,q)=\frac{k(k+1)}{(k+1)^2-qk^2}$$
$$\lim_{k\to\infty}S(k,q)=-\frac{1}{q-1}$$
How can I prove this limit?
AI: Hint: Expand the products and squares in the nominator and denominator and multiply both with $1/k^2$. As $k\to\infty$, all the terms that look like $1/k^i$ vanish and you are done. This is the standard trick to calculate limits of quotients of polynomials. |
H: Fermat's Little Theorem and polynomials
We know that in $F_p[y]$, $y^p-y=y(y-1)(y-2)\cdots (y-(p-1))$. Let $g(y)\in F_p[y]$. Why is it valid to set $y=g(y)$ in the above equation to obtain $g(y)^p-g(y)=g(y)(g(y)-1)\cdots (g(y)-(p-1))$. This is done in Theorem 1 of Chapter 22 of A Concrete Introduction to Higher algebra by Lindsay Childs.
AI: For any field $F$, and element $a\in A$ of an (associative) $F$-algebra $A$, the substitution $X:=a$ defines a ring morphism $F[X]\to A: P\mapsto P[a]$. This works in particular when $A=F[X]$, which is what happens in the question (with $F=\Bbb F_p$, and $X=y$). The morphism property implies that the substitution $y:=g(y)$ can be performed separately in the factors of the product $y^p-y=\prod_{r\in\Bbb F_p}(y-r)$, giving as result the identity $g(y)^p-g(y)=\prod_{r\in\Bbb F_p}(g(y)-r)$. |
H: Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative?
Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative? (Since $\sqrt{a}$ by itself is undefined in $\mathbb{R}$ when $a$ is negative)
I was wondering about how we can handle the above problem in $\mathbb{R}$ when I had to do some exercise that involved squaring the squareroot in $\mathbb{C}$
AI: Surely you could define a square-root in $\mathbb{C}$ that is either nonnegative real, or has positive imaginary part. Then, I think, $(\sqrt{a})^2$ would equal $a$ because the square-roots would be consistent.
Unfortunately, you lose $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if you do that. |
H: What is the maximum amount of eigenvectors that a square matrix n x n might have?
having the following arbitrary matrix:
$A=\begin{pmatrix}
a_{1,1} & \cdots & a_{1,n}\\
\vdots & \ddots & \vdots \\
a_{n,1} & \cdots & a_{n,n}
\end{pmatrix}$
What is the maximum amount of (linearly independent) eigenvectors and eigenvalues of $A$ ? How to calculate the actual number of eigenvectors and eigenvalues of $A$?
thanks!
AI: Think about $A=I$, how many (linearly independent) eigenvectors does $A$ have?
With regards to computing the eigenvectors/eigenvalues of a general $A$ you have to find pairs $(\lambda,v)$ such that
$$Av=\lambda v.\quad\quad (*)$$
One way to go around this is first compute the possible $\lambda$s by solving for the roots of
$$\det(A-I\lambda).$$
Then compute the corresponding $v$s by solving the linear equations defined by $(*)$.
For more info, I'd recommend you have a look at these camcasts. |
H: Hilbert Spaces are Reflexive
I want to show that all Hilbert spaces are reflexive. I have found the following proof on StackExchange:
Hilbert Space is reflexive
However, I do not understand it. Essentially, we want to show that for all $g \in X^{**}$, (X is some Hilbert space) there exists a unique $x \in X$ such that $g(h) = h(x)$ for all $h \in X^*$. Following the OP's logic, we should apply the Riesz-Fréchet Representation Theorem (RRT) twice:
Pick any $g \in X^{**}$. Then, since $g$ is bounded on $X^*$, by RRT there exists a unique $f \in X^*$ such that $||g|| = ||f||$, and $g(h) = \langle h,f \rangle$ for all $h \in X^*$. Now apply RRT to $f$ to get a unique $x \in X$ such that $||f|| = ||x||$ and $f(y) = \langle y,x \rangle$ for all $y \in X$. It follows that:
$f(x) = \langle x,x\rangle = ||x||^2 = ||f||^2 = \langle f,f \rangle = g(f)$.
We have shown that for any $g \in X^{**}$ that there exist unique $x \in X$, $f \in X^*$ such that $f(x) = g(f)$. This is not quite what we want. We want this to hold for a general $h \in X^*$.
According to icurays1, we have basically defined a bijective mapping $T:X^{**} \to X$. Why is T bijective, and why does this give us that $X$ is reflexive?
AI: Was this same question posted earlier? I have reposted the answer here in any case.
For any normed linear space, there is a natural map $T:X \to X^{\ast \ast}$ given by
$$
T(x)(f) = f(x) \quad\forall f\in X^{\ast}
$$
One can show that this is a linear isometry.
Reflexivity means that $T$ is surjective.
Now, in the case of a Hilbert space, there is a natural anti-linear isometry
$$
S_1 : H\to H^{\ast}
$$
given by $S_1(y)(x) = \langle x,y\rangle$. The Riesz Representation theorem says that this map is surjective, and hence an isomorphism.
In particular, $H^{\ast}$ is a Hilbert space, and so there is an anti-linear isomorphism
$$
S_2 : H^{\ast} \to H^{\ast\ast}
$$
Now what you need to check is that $T = S_2\circ S_1$ |
H: Banach space with cardinality bigger than $\mathfrak{c}$.
By using the information contained in this post, we have that the cardinality of every Banach space is equal to its dimension, which in turn, is bigger or equal to $\mathfrak{c}$.
In my area of study, I have always beem studying spaces like $W^{1,p}$ for $p\in (1,\infty)$ which are separable. Moreover, the space that I know which are not separable is $L^\infty$, but I don't know how to calculate its cardinality.
My question is: Is there a example of a Banach space with cardinality bigger than $\mathfrak{c}$?
Remark: I'm not used to study those things, hence, if I have posted something stupid here, please neglect it.
AI: Let $S$ be any set, and $\ell^\infty(S)$ the space of bounded maps $S \to \mathbb R$, with the norm
$$ \|x\|_{\ell^\infty(S)} := \sup_{s\in S}\left|x(s)\right| $$
It is complete and $\left|\ell^{\infty}(S)\right| \ge \left|S\right|$. Now choose $S$ with $\left|S\right|>\mathfrak c$. |
H: Probability of having lots of unique elements
If you sample $n$ integers from the range $1$ to $n$ inclusive it seems intuitive that you are likely to get a lot of numbers exactly once. Call $X_n$ the number of integers you get that occur exactly once in your sample. Is there a nice simple way of showing the following for all $n\geq 2$.
$$\exists a,b >0\text{ such that } P(X_n \leq an) \leq 2^{-bn} $$
AI: A good model for large $n$ is a Poisson distribution with mean $\lambda=1$. You would expect about $\frac ne \approx 0.3688n$ numbers not to be selected at all and the same number to be selected once. I don't have an easy way to calculate the distribution of the number that are selected exactly once. |
H: Why do I see i and k as the indices of summation?
I'm working on linear algebra and just wanted to clear up an uncertainty regarding whether there is a difference in the use of i and k as the dummy variables for the index of summation?
$\sum\limits_{i=1}^\infty {i^2} = \sum\limits_{k=1}^{\infty} {k^2}$ ?
I got confused at first since I was working with vectors [i, j] with a summation indicating k=1 , although k was indicating the z dimension (i and j indicating x and y respectively).
Just to clarify, the choice of i and k as dummy variables is completely arbitrary - right?
I found my way to this page at Wolfram MathWorld and it actually switches from i to k in the course of a short piece of text, is this normal and nothing to concern me or should I take note of differences like this?
AI: Summation indices are dummy variables that are completely arbitrary.
If $i$, $j$, and $k$ are already being used for vector notation, it would be good to use a different index for summation. The letter $m$ would be one sensible choice, if you are writing things like $\sum_{m = 0}^n$, and $l$ is another possibility (just because it is close in the alphabet to $i,j,k,m,$ and $n$). Of
course you are free to use any variable that hasn't already been given a meaning, but it is good to use letters that will have the psychological connotation of being an index (so letters like $x$, $y$, and $z$ are fairly uncommon as summation indices). |
H: Suppose $f$ is twice differentiable function such that ...
I am stuck with the following problem :
I did integration by parts which gives the result $\,\,f'(1)$.
Can someone explain? Thanks in advance for your time.
AI: Can you think of a reason that $f\,'(1)=f\,'(0)$? |
H: Changing variable
I've problem with formulating the following problem. I guess I need to express $v(d)$ in $v(t)$ but since $d=v*t$ I can't just replace $d$ with $v*t$ since I would get $v(t) = v...$, a recursive function.
A particle moves in a straight line. The velocity of the particle ($v$) depends on the traveled distance ($d) according to:
$$ v = \frac{3d+4}{2d+1} $$
where $d$ is the distance from the particle to its starting point. Calculate the particle acceleration when d = 2. (The derivate of speed depending on time gives the acceleration).
AI: $d$ is a function $d(t)$ of time, and by extension $v(t) = d'(t)$ is as well. You have an expression for the first derivative $d'(t) = v(t)$, the text asks you to find the second derivative at the time when $d(t) = 2$. So, we need to differentiate with respect to $t$:
$$
a(t) = v'(t) = \frac{ 3(2d(t) + 1)-2(3d(t) + 4)}{(2d(t) + 1)^2}\cdot d'(t)\\
= \frac{-5}{(2d(t) + 1)^2}\cdot v(t) = \frac{-5(3d(t) + 4)}{(2d(t) + 1)^3}
$$
entering $d(t) = 2$, we get $a(t) = -2/5$ |
H: Evaluate $\binom{12}0+\binom{12}2+\ldots+\binom{12}{12}$ using binomial theorem
Solve the sum:
$$
{12 \choose 0}+
{12 \choose 2}+
{12 \choose 4}+
{12 \choose 6}+
{12 \choose 8}+
{12 \choose 10}+
{12 \choose 12}
$$
using the binomial theorem.
I know the binomial theorem:
$$ \left(a+b\right)^2 = \sum_{k=0}^{n} {n \choose k} a^nb^{n-k}$$
However I fail to translate the sum into the theorem. I probably need to choose a variable ($b$?) to be negative to get half of the terms to disappear.
AI: With $a = 1, b = -1$ you get that
$$
\sum_{i= 0}^{12}(-1)^i\binom{12}{i} = (1-1)^{12} = 0
$$
and $a = 1, b = 1$ gives you
$$
\sum_{i = 0}^{12}\binom{12}{i} = (1 + 1)^{12} = 2^{12}
$$
Add these two together and see what terms are left.
Edit
Expanding, another way to write the $b = -1$ equation over is
$$
0 = \binom{12}{0} -\binom{12}{1} +\binom{12}{2} -\binom{12}{3}+ \binom{12}{4}- \binom{12}{5}+ \binom{12}{6}\\- \binom{12}{7} +\binom{12}{8} -\binom{12}{9} +\binom{12}{10} -\binom{12}{11} +\binom{12}{12}
$$
The $b = 1$ equation ca be written
$$
2^{12}= \binom{12}{0} +\binom{12}{1} +\binom{12}{2} +\binom{12}{3}+ \binom{12}{4}+ \binom{12}{5}+ \binom{12}{6}\\+ \binom{12}{7} +\binom{12}{8} +\binom{12}{9} +\binom{12}{10}+\binom{12}{11} +\binom{12}{12}
$$
If we add these two together, we get
$$
2^{12} = 2\binom{12}{0} +2\binom{12}{2} +2\binom{12}{4} +2\binom{12}{6}+ 2\binom{12}{8}+ 2\binom{12}{10}+ 2\binom{12}{12}
$$
Dividing by $2$ gets you the answer you're after. |
H: Prove that $f$ is a convex function if $f=d(x,C)$ and $C$ is convex.
Question: Suppose $V$ is a normed space and $f:V \rightarrow \mathbb R$ defined as $f(x)=d(x,C)=\inf_{c \in C}||x-c||_V$ where $C$ is a convex subset of $V$. Prove $f$ is a convex function.
Attempt at a solution: (I would like to know if it is correct to say the first line equals the second line.)
For some $\lambda \in [0,1]$ we have $$f(\lambda x+(1-\lambda )y)=d(\lambda x+(1-\lambda )y,C)=\inf_{c \in C} ||\lambda x+(1-\lambda )y-c||_V=$$
$$\inf_{a, b \in C}||\lambda x+(1-\lambda )y-\lambda a-(1-\lambda )b||_V =$$
$$\inf_{a, b \in C}||\lambda x-\lambda a+(1-\lambda )y-(1-\lambda )b||_V $$
$$\leq \lambda \inf_{a\in C}||x-a||_V+(1-\lambda)\inf_{b \in C}||y-b||_V=\lambda f(x)+(1-\lambda )f(y)$$
So we can conclude $f$ is convex.
AI: The two expressions are equal, but this begs a justification of the type:
Since $C = \{\lambda a + (1-\lambda)b : a,b \in C\}$ ($C$ is convex), it follows that: $$\inf_{c \in C} \|\lambda x+(1-\lambda)y-c\|_V = \inf_{a,b\in C}\|\lambda x +(1-\lambda)y-(\lambda a + (1-\lambda)b)\|_V$$
However, one can argue that your approach is needlessly complicated: just take $a = b = c$ -- since $c = \lambda c+(1-\lambda)c$, the conclusion isn't altered.
The stricken-through argument doesn't work. For, we would have to assert that:
$$\inf_{c \in C} g(c)+h(c) = \inf_{c \in C} g(c)+\inf_{c\in C} h(c)$$
which is definitely false in general. (I was brought to this by considering non-convex $C$, where $x,y \in C,\lambda x+(1-\lambda)y \notin C$ clearly invalidates the result.) |
H: Roots of a Quadratic Problem
I'm struggling with this problem and was hoping I could get some advice. Here is the problem:
Let a and b be the roots of the quadratic equation $x^2−x−1/27=0$.
Without calculating the a and b show that $a^{1/3}+b^{1/3}$ is a root of the equation $x^3+x−1=0$.
Any help would be much appreciated!
Thanks
AI: Using Vieta's formula, $a+b=1$ and $\displaystyle ab=\frac{-27}1$
Let $y=a^{\frac13}+b^{\frac13}$
$$\implies y^3=a+b+3(ab)^{\frac13}(a+b)=1+3(-27)^{-\frac13}y$$
Now, one of three values of $(-27)^{-\frac13}$ is $\{(-3)^3\}^{-\frac13}=-\frac13$
$$\implies y^3=1-y$$ |
H: Complete ordered field is an Archimedean field that cannot be extended to an Archimedean field
As a bonus problem, our professor of real analysis asked us to prove that the real numbers (a complete ordered field) cannot be extended into an Archimedean field, with no definition of what he meant by extending.
I have tried using proof by contradiction to show that if we have some set, $\mathbb{R}^{*}$ such that $\mathbb{R}$ is a proper subset of $\mathbb{R}^{*}$, and that $\mathbb{R}^{*}$ forms an Archimedean field, then because for there to be new elements in $\mathbb{R}^{*}$ as opposed to $\mathbb{R}$, they would have to be larger (or smaller) than all the elements of R. But then we would have reached a contradiction with the Archimedean property.
Professor returned this solution and said it's not the right solution (without any further comment). Can anyone offer some enlightenment on what I have done wrong or what I could try now?
All I have found on Google are mentions in textbooks that go along the lines of "every Archimedean field is isomorphic to a subfield of real numbers". That would imply that we cannot extend reals into an Archimedean field, but how can one go about proving that?
I can only use basic definition of an ordered (Archimedean) field and other "basics", we have not yet covered sequences, etc.
AI: Outline: Let $F$ be a proper Archimedean extension field of $\mathbb{R}$. Let $\eta$ be an element of $F$ which is not in $\mathbb{R}$.
By taking the additive inverse if necessary, we can assume that $\eta$ is positive. By the assumed Archimedean property of $F$, there is an integer greater than $\eta$.
Thus the set of $x$ in $\mathbb{R}$ which are $\lt \eta$ is bounded above, and hence has a least upper bound $b$.
Now comes the key fact, that is being left for you to prove: There is no rational number between $0$ and $|\eta-b|$.
From this you can conclude that $\frac{1}{|\eta-b|}$ is greater than any integer, contradicting the assumption that $F$ is Archimedean.
Remark: In a sense, this fleshes out your notion that any "new" element is bigger or smaller than every element of $\mathbb{R}$. That part was not correct. However, from a new element $\eta$, assumed positive and smaller than some integer, we produce an element $\frac{1}{|\eta-b|}$ which is larger than any integer. |
H: How do you call functions integrable over any compact subset of their domain?
I'm quite sure there is a name for such class of functions but can not remember or figure out what terms to search for. A simple and very practical example would be "periodic Lebesgue" functions: functions which are periodic—and thus, except for null cases, can not in principle have a Lebesgue integral over $\mathbb{R}$–but are integrable in their restriction to the fundamental interval. Or any finite interval whatsoever, for that matter.
AI: Such a function is called locally integrable, and the space of such functions is generally notated as $L^1_{\text{loc}}$ for this reason. An alternative, and equivalent, definition, is that for all infinitely differentiable $\varphi$ with compact support, we have
$$\int |f \varphi| dx < \infty$$ |
H: Showing that $f$ and $g$ are analytic continuations of themselves.
Let $$f(z)=\sum_{n=0}^{\infty} a^n z^n$$
$$g(z)=\sum_{n=0}^{\infty} (-1)^n \frac{(1-a)^n z^n}{(1-z)^{n+1}}$$
From this, I have:
$$f(z)=\frac{1}{1-az}, |z|<\frac{1}{|a|}$$
$$g(z)=\frac{1}{1-az}, |z||a-1|<|1-z|$$
I want to show that the two regions have some sort of overlap where an open set could be stuck, so that would prove they are analytic continuations, but I don't know how to do that. Would appreciate any hints. From what I've looked at Wolfram for some particular values of $a$, the region looks pretty complicated.
AI: Hint: The intersection of two open sets is open and $0$ belongs to both regions. |
H: simple math: finding what percentage B is performing when compared to A
Boy A's performance is $500$ and boy B's performance is $525$ for a particular task.
How can we calculate how much percentage B is performing better than A?
AI: The percentage at which B is performing better than A is given by:
$$\dfrac{(\text{B's performance}) - (\text{A's performance})}{\text{A's performance}}\times 100\%$$
In this case, that means that B performs
$\quad \dfrac{525 - 500}{500}\times 100\%\quad$
better than A. |
H: bilinear form decomposition
Let $V$ be a vector space over a field of $\text{char} \neq 2$. Then if $g: V \times V \to F$ is a bilinear form, and $U \subseteq V$ is a subspace, do we have $V = U + U^\perp$, where $U^\perp = \{x \in V | g(x,U) = 0\}$? Of course, this is the case when $g$ is an inner product. Is it the case for an arbitrary bilinear form?
I don't recall seeing this is any of my linear algebra courses.
EDIT: Ok, so it's not true. Does it hold when $g$ is nondegenerate?
AI: This is not true in general. Take $g$ to be skew-symmetric, say, on $\mathbb{R}^2$. Say, it's matrix in some basis $u,v$ is $\pmatrix{0 & 1\\ -1 & 0}$, and take $U=Span(u)$. Then $U^\perp=U$.
Such decomposition is true if $g$ is symmetric. |
H: Explaining and using the $N$-term Taylor series for $\sin x$
So I'm given the Taylor Series expansion of the sine function and I've been asked to prove it (Done) and then construct the following by my lecturer:
Explain why the Taylor series containing $N$ terms is:
$$\sin x = \sum_{k=0}^{N-1} \frac{(-1)^k}{(2k+1)!}x^{2k+1}+r_{2N-1}(x)$$
with a remainder $r_{2N-1}(x)$ that satisfies:
$$|r_{2N-1}(x)| \leqslant \frac{|x|^{2N}}{(2N)!}$$
How many terms of the series do you need to include if you want to compute $\sin x$ with an error of at most $10^{-3}$ for all $x \in [-\pi/2, \pi/2]$?
Compute $\sin(\pi/2)$ from the Taylor Series. How large is the actual error?'
I'm fairly certain I can do the third part, but the first and second have completely thrown me. Can anybody help, or at least point me in the right direction?
Cheers.
AI: The first part is a special case of a standard error estimate for alternating series (see e.g. here) and is proven in any introductory book on analysis. Here is an online proof for this and for other error estimates of truncated series: http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx (see the section on the alternating series test).
For the second part, you just need to notice that the upper bound for the error term is maximised at the end points of the interval. So as long as you arrange for $r_{2N-1}(\pi/2)$ to be smaller than $10^{-3}$, the same will hold for all the other $x$ in the interval. |
H: How to solve this natural logarithms problem?
How do I take natural logarithm of the following?
$A - (Be^{-xy})$
AI: If you have the equation shown below on the left hand side of $\iff$, we can do the following: $$A - Be^{-xy} = 0 \iff Be^{-xy} = A$$ Now we can take the natural logarithm of each side:
$$\begin{align} \ln\left(Be^{-xy}\right) & = \ln A \\ \\ \iff \ln B + \ln\left(e^{-xy}\right) & = \ln A \\ \\ \iff -xy & = \ln A - \ln B = \ln \left(\dfrac AB\right)\end{align}$$
Otherwise, if you are simply trying to take the logarithm of an expression, there's nothing you can really do to simplify the expression, so you'd have nothing simpler than $$\ln\left(A - B^{-xy}\right)$$ |
H: Laurent series for $\frac{e^z}{1 - z}$ for $|z| > 1$
I do this like $\dfrac{e^z}{1-z}$ = $-e \dfrac{e^{z-1}}{z-1}$ = $-e \sum_{k=0}^{\infty} \dfrac{(z-1)^k}{k!(z-1)}$ = $-e \sum_{k=0}^{\infty} \dfrac{(z-1)^{k-1}}{k!}$
However doesn't this give me the Laurent series around $|z-1| > 0$? Should I do a taylor expansion of the first term $\dfrac{-e}{z-1}$ to a geometric series to get the expansion for $|z|>1$?
Maybe I have just messed up some definitions, sorry if that's the case.
AI: $|z|>1$ is an annulus with center $0$, thus all powers of the relevant Laurent series should be of the form $(z-0)^k$.
On the one hand $$e^z=1+z+\frac{z^2}{2}+ \dots=\sum_{k=0}^\infty \frac{z^k}{k!}$$
On the other $$\frac{1}{1-z}= -\frac{1}{z} \frac{1}{1-1/z}=-\frac{1}{z} \sum_{k=0}^\infty \frac{1}{z^k}=\sum_{k=0}^\infty- \frac{1}{z^{k+1}} $$
Try multiplying both series and collecting the terms of the same degree. |
H: What is an "incongruent" solution?
For example, "Solve the congruence (if possible), listing all the incongruent solutions:"
$$561x\equiv 3575\mod{1562}$$
I found $x\equiv 37+142t,\ 0\leq t\leq 10,\ t\in\mathbb{Z}$... There are 11 "incongruent solutions" because $(561,1562)=11$ and $11\mid 3575$... but what does "incongruent" mean?
AI: Incongruent (in this case) means distinct modulo $1562$. For example, $1$ and $1561$ are incongruent modulo $1562$, but $1$ and $1563$ are not (rather, they are congruent modulo $1562$). |
H: Geometrical interpretation of $P(x) + Q(y) = 0 $ when P,Q are polynomials of degree 2?
Special cases are circles ( $ (x-x_0)^2 + (y-y_0)^2 = R $ ) and ellipses.
Is there a geometric interpretation in the general case $ ( ax^2 + bx + c ) + ( dy^2 + ey + f ) = 0$?
AI: By completing the square, you find that
$$
(ax^2 + bx + c) + (dy^2 + ey + f) = (a(x- \alpha)^2 + s) + (d(y - \beta)^2 + t))
$$
so that your equation has the form:
$$
a (x - \alpha)^2 + d (y - \beta)^2 = \kappa.
$$
Assume first that $a, d > 0$ (or both are negative).
If $\kappa < 0$, then there are no solutions, while if $\kappa = 0$, then the only solution is the point $(\alpha,\beta)$.
The interesting case is when $\kappa > 0$. Then you get a circle if $a = d$ and an ellipse otherwise, centered at the point $(\alpha, \beta)$.
Similar analysis applies to other cases. If $a$ and $d$ are both non-zero, but have opposite signs, then you get a hyperbola. If either $a$ or $d$ is zero but the other is not, you get a parabola (or two lines or a double line, in degenerate cases). If both $a$ and $d$ are zero, then you get a line (or no solutions in a degenerate case). |
H: Uniform convergence of $f'_n/f_n$
The following is the proof of Hurwit'z theorm in wiki
Let f be an analytic function on an open subset of the complex plane
with a zero of order m at z0, and suppose that {fn} is a sequence of
functions converging uniformly on compact subsets to f. Fix some ρ > 0
such that f(z) ≠ 0 in 0 < |z−z0| < ρ. Choose δ such that |f(z)| > δ
for z on the circle |z−z0| = ρ. Since fk(z) converges uniformly on the
disc we have chosen, we can find N such that |fk(z)| ≥ δ/2 for every k
≥ N, ensuring that the quotient fk′(z)/fk(z) is well defined for all z
on the circle |z−z0| = ρ. By Morera's theorem we have a uniform
convergence:
$$\frac{f_{k}'(z)}{f_{k}(z)} \to \frac{f'(z)}{f(z)}.$$ ..........
But I don't know why it is a uniform convergence. (I think that Morera's theorem is not relevant.) Why does it hold?
AI: You are right, Morera's theorem is not relevant here. What is relevant is Weierstraß' theorem that locally uniform convergence of a sequence of holomorphic functions implies locally uniform convergence of the sequences of derivatives,
$$f_k \xrightarrow{\text{loc. unif.}} f \Rightarrow f_k^{(n)} \xrightarrow{\text{loc. unif.}} f^{(n)}$$
for all $n\in\mathbb{N}$. Since the $\lvert f_k\rvert$ are bounded below by $\delta/2 > 0$ on $\{\lvert z-z_0\rvert = \rho\}$, that implies uniform convergence
$$\frac{f_k'}{f_k} \to \frac{f'}{f}$$
on $\{\lvert z-z_0\rvert = \rho\}$. |
H: Show: $\limsup$ does not change when changing finite many sets
Let $(A_n)_{n\in\mathbb{N}}$ be a series of sets. Define
$$
A^+:=\limsup\limits_{n\to\infty}A_n:=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k,~~~~~A^-:=\liminf\limits_{n\to\infty}A_n:=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k
$$
Show that $A^+$ and $A^-$ are not changing, when one changes finite many sets $A_i$.
Hello, I tell you my idea concerning $A^+$ and would be very thankful to get a feedback from you if my idea is right or wrong, thanks!
Consider the changed series $(\tilde{A}_n)_{n\in\mathbb{N}}$ with $\tilde{A}_n\neq A_n$ for a finite set $I=\left\{i_1,i_2,\ldots,i_m\right\}, m\in\mathbb{N}$ and $\tilde{A}_n=A_n$ when $n\notin I$. What is to show is
$$
A^+=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\tilde{A}_k=:\tilde{A}^+.
$$
To show the conclusion "$\subseteq$", suppose that $x\in A^+$.
$$
\Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}A_k\Rightarrow\forall~n\in\mathbb{N}~\exists~j\in\left\{n,n+1,n+2,\ldots\right\}: x\in A_j\\\Rightarrow\forall~n\in\mathbb{N}~\exists~s>\max\limits_{i_l\in I, 1\leq l\leq m}\left\{i_l\right\}: x\in \tilde{A}_s (=A_s)\\\Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}\tilde{A}_k\Rightarrow x\in\tilde{A}^+
$$
To show the other conclusion, assume that $x\in\tilde{A}^+$.
$$
\Rightarrow\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}\tilde{A}_k\Rightarrow\forall~n\in\mathbb{N}~\exists~j\in\left\{n,n+1,n+2,\ldots\right\}: x\in\tilde{A}_j\\\Rightarrow\forall~n\in\mathbb{N}~\exists s>\max\limits_{i_l\in I, 1\leq l\leq m}\left\{i_l\right\}: x\in\tilde{A}_s=A_s\\\Rightarrow~\forall~n\in\mathbb{N}: x\in\bigcup_{k=n}^{\infty}A_k\Rightarrow x\in A^+
$$
With kind regards,
math12
AI: An easier thing to do is to note that $x \in \limsup A_n$ if and only if $x$ is in infinitely many of the $A_n$, and $x \in \liminf A_n$ if and only if $x$ is in all but finitely many of the $A_n$. From these descriptions, it is trivial to see that changing finitely many of the $A_n$ does not affect either the limsup or liminf. |
H: Ordinary generating function question
Let $a_1\le a_2\le\cdots\le a_6$ and $b_1\le b_2\le\cdots\le b_6$ be positive integers, not necessarily distinct, such that $a_1+a_2+\cdots+a_6<b_1+b_2+\cdots+b_6$. When the $36$ sums $a_i+b_j$ are computed, it turns out that $k$ appears $k-1$ times for $k=2,3,\cdots,7$ and $13-k$ times for $k=8,9,\cdots,12$. Find the possible values of $a_1,\cdots,a_6,b_1,\cdots,b_6$.
I'm totally clueless on how to approach this question and how to come up with the generating function. Any help is appreciated.
AI: To get you started with the generating function approach, construct generating functions for your sequences, which will be polynomials since your sequences have only finitely many terms:
$$
f(z) = a_1 z + a_2 z^2 + \cdots + a_6 z^6 \\
g(z) = b_1 z + b_2 z^2 + \cdots + b_6 z^6
$$
Now you must figure out how to translate the condition on the sums $a_i + b_j$ into some condition on $f(z)g(z)$. You should be able to explicitly write down $f(z)g(z)$ as a specific polynomial with positive integer coefficients.
Now factor this polynomial into irreducibles. Any solution to your problem must correspond to a factorization into a product of two degree 6 polynomials. Use your irreducible decomposition of $f(z) g(z)$ to determine how many such factorizations are possible. Then for each one, check whether the coefficients are positive integers, (weakly) increasing, and have different total sum of all coefficients. For each such factorization, you'll be able to read off a solution to your problem. (Hint: there's no guarantee that even one such factorization is possible.) |
H: Integral of product of two measurable functions
I have to show that for $\psi$ and $\phi$ two positive and measurable functions:
$$\int_X \phi\psi \, d\mu \le \sqrt{\int_X \phi^2 \, d\mu} \sqrt{\int_X \psi^2 \, d\mu}$$
I know that for $f,g$ two measurable functions it holds that:
$$\int_X fg \, d\mu \le \frac{1}{2} \left(\int_X f^2 \, d\mu + \int_X g^2 \, d\mu\right)$$
I have set $f=\frac{ \phi}{\sqrt{\int \phi^2}}$ and $g=\frac{\psi }{ {\sqrt{\int \psi^2 d\mu}}}$ and then tried to plug them into the equation $2fg \le f^2 +g^2$
which lead to the following expression:
$$\frac{\phi\psi}{\sqrt{\int \phi^2}\sqrt{\int \psi^2}} \le \frac{\phi^2}{\int \phi^2} + \frac{\psi^2}{\int \psi^2}$$
$$\phi\psi \le \sqrt{\int \phi^2} \sqrt{\int \psi^2} \left(\frac{\phi^2 \int \psi^2 + \psi^2 \int \phi^2}{\int \phi^2 \int \psi^2}\right) = \sqrt{\int \phi^2}\sqrt{\int \psi^2}(\phi^2+\psi^2)$$
I'm not sure if I have taken the wrong approach or if there is a way I can continue from this point? Thanks for any help! :)
AI: The $f$ and $g$ that you defined have the property that $\int f^2=\int g^2=1$. So you have
$$
\int fg\,d\mu\leq\frac12(1+1)=1.
$$
So
$$
\int_X\frac{\phi}{\sqrt{\int_X\phi^2\,d\mu}}\,\frac{\psi}{\sqrt{\int_X\psi^2\,d\mu}}\,d\mu\leq1,
$$
which is what you want to prove. |
H: How to prove that a series is equal to a recursive algorithm
I have the following sequence:
$$
y_n = \int_0^1 \frac{x^n}{x+5}\,dx, n = 0,1,\dots
$$
Now I have the following recursive algorithm:
$$
y_0 = \log{6} - \log{5}
$$
$$
y_n = \frac{1}{n} - 5y_{n-1}, n = 1,...
$$
I tried to prove that this algorithm is equal to the sequence.
I thought I would have to do complete induction:
Induction start:
$$
y_n = 1 - 5\log6 + 5\log5 = \left.(x- 5\log(x+5))\right|_0^1 = \int_0^1 \frac{x}{x+5}\,dx
$$
Induction step $(n \longrightarrow n+1)$:
$$
\frac{1}{n+1} - 5 y_n = \frac{1}{n+1} -5\left(\frac{1}{n}-5y_{n-1}\right)= \frac{1}{n+1} - \frac{5}{n} + 25 y_{n-1} = -5\left(\frac{1}{-5(n+1)}+\frac{1}{n}-5y_{n-1}\right) = \frac{1}{n+1} -5\left(\int_0^1 \frac{x^n}{x+5}\,dx\right) = ...?
$$
Can you help me?
Thanks.
AI: $$y_n=\int_0^1\frac{x^{n-1}(x+5-5)}{x+5}dx=\int_0^1x^{n-1}\left[\frac{x+5}{x+5}-\frac{5}{x+5}\right]=$$ $$=\int_0^1x^{n-1}dx-5\int_0^1\frac{x^{n-1}}{x+5}dx=\frac{1}{n}-5y_{n-1}$$ |
H: Grade calculation
My teacher said that $\frac{1}{3}$rd of the final grade will be based on Exam 1, $\frac{1}{12}$th each based on Exam 2-5, and $\frac{1}{3}$rd again based on Exam 6, how will my teacher calculate my final grade?
AI: If $x_1, x_2, \dots, x_6$ are your scores on Exams $1,2,\dots,6$, then your final grade will be
$$
\frac{x_1}{3} + \frac{x_2}{12} + \frac{x_3}{12} + \frac{x_4}{12} + \frac{x_5}{12} + \frac{x_6}{3}.
$$ |
H: Prove that $G/N$ is Abelian
Below is a homework problem I have gotten stuck on. I would really appreciate a hint, but please do not just give the answer away.
Let $G \subset \mathcal{M}_2(\mathbb{R})$ such that each $m \in G$ is
upper triangular with a nonzero determinant. Let $N = \left\{
\bigl(\begin{smallmatrix} 1&b\\ 0&1 \end{smallmatrix} \bigr) : b \in
\mathbb{R}\right\}$. Prove that:
• $N$ is a normal subgroup of $G$.
• $G/N$ is abelian.
Solution: Let $m \in G$; then $\exists \alpha, \beta, \delta \in \mathbb{R} : m = \bigl(\begin{smallmatrix}
\alpha & \beta \\ 0&\delta
\end{smallmatrix} \bigr)$; $\alpha\delta \neq 0$. $m$ is invertible, hence $m^{-1} = \frac{1}{\alpha \delta}\bigl(\begin{smallmatrix}
\delta & -\beta \\ 0&\alpha
\end{smallmatrix} \bigr)$. Let's look at $m N m^{-1}$.
\begin{align*}
m N m^{-1} &= \left\{ \left(\begin{matrix}
\alpha & \beta \\ 0&\delta
\end{matrix} \right) \left(\begin{matrix}
1&b\\ 0&1
\end{matrix} \right) \frac{1}{\alpha \delta}\left(\begin{matrix}
\delta & -\beta \\ 0&\alpha
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{\left(\begin{matrix}
\alpha& \alpha b + \beta \\ 0&\delta
\end{matrix} \right) \frac{1}{\alpha \delta}\left(\begin{matrix}
\delta & -\beta \\ 0&\alpha
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{ \frac{1}{\alpha\delta}\left( \begin{matrix}
\alpha\delta& -\beta\alpha + \alpha^2 b + \beta\alpha \\ 0 &\alpha\delta
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
&= \left\{\left( \begin{matrix}
1& \frac{\alpha b}{\delta} \\ 0 & 1
\end{matrix} \right) : \alpha, \beta, \delta, b \in \mathbb{R} \right\} \\
\end{align*}
Since $\alpha \neq 0 $ and $ \delta \neq 0$, it follows that $\frac{\alpha b}{\delta}$ is a valid real number. Therefore $mNm^{-1} \in G$, hence $N$ is normal.
Now consider $G/N$: the group of right cosets of $N$ in $G$. Let $Nm_1$ and $Nm_2$ be two right cosets of $N$ in $G$. Then
\begin{align*}
Nm_1Nm_2 &= Nm_1m_2\\
&= N \left(\begin{matrix}
a & b \\ 0& d
\end{matrix} \right) \left(\begin{matrix}
\alpha & \beta \\ 0&\delta
\end{matrix} \right)\\
&= N \left(\begin{matrix}
a\alpha & a\beta+ b \delta \\ 0& d\delta
\end{matrix} \right) \\
....
\end{align*}
This is where I get stuck. I figure there must be a matrix in $n \in N$ such that, for any two upper triangular matrices $m_1$ and $m_2, m_1 n m_2 = m_2 n m_1$, but I can't seem to find it.
AI: I never know whether to call cosets left or right, but $G/N$ stands for cosets of the form $gN$ with $g \in G$. I guess people call these left cosets, even though you are multiplying by $N$ on the right. Anyway, it does not materially affect your calculation, but I thought I would mention it.
To show that $G/N$ is abelian, you need
$$
(m_1 N) (m_2 N) = (m_2 N) (m_1 N),
$$
which is equivalent to
$$
m_1 m_2 N = m_2 m_1 N,
$$
which in turn is equivalent to $m_1^{-1} m_2^{-1} m_1 m_2 \in N$. This expression is called the commutator of $m_1^{-1}$ and $m_2^{-1}$ (the inverses just appear because of how we wrote the initial calculation). So you just need to consider $m_1^{-1} m_2^{-1} m_1 m_2$ and argue that it must have $1$'s on the diagonal.
Further hint:
Instead of writing out the entire matrix multiplication, it is more transparent if you observe that the diagonal entries in the product of two upper triangular matrices are the products of the diagonal entries in those two upper triangular matrices. |
H: Finding Integers to Satisfy a Condition
Below is the question I have to answer. I'm not sure how to attack it. Any tips?
For how many integers
$n$ between $1$ and $6 \cdot 10^6$
does there exist at least
one pair of integers $(x, y)$ such that
$xn
+ 60y
= 1$?
AI: Hint: The equation
$$xn + 60 y = 1$$ has a solution $(x, y)$ if and only if the greatest common divisor of $n$ and $60$ is $1$. Now try counting how many integers between $1$ and $6 \cdot 10^6 = (60) \cdot 10^5$ are relatively prime to $60$. |
H: Orthonormal bases for Hilbert spaces
In Reed and Simon (Functional Analysis) Theorem II.6 states that, given an orthonormal basis $\{ x_\alpha \}_{\alpha \in A}$ (not necessarily countable)for a Hilbert space $H$, every $y \in H$ can be written as a sum
$$
\sum_{\alpha \in A} (x_{\alpha}, y) x_\alpha
$$
where $(\cdot,\cdot)$ denotes the inner product).
The proof uses a conclusion that I cannot quite follow:
By Bessel's inequality we know that for any finite subset $A' \subset A$, $\sum_{\alpha \in A'} |(x_\alpha ,y )|^2 \le \|y\|^2$. Thus $(x_\alpha,y) \ne 0$ for at most a countable number of $\alpha's$ in $A$.
Why can we deduce the last statement ?
AI: For each $n \in \mathbb{N}$, consider the set
$$
\{x_{\alpha} : |(x_{\alpha}, y)| \geq 1/n\}
$$
Show that this set is finite, so the union over all the $n$'s is countable, which is the set you are talking about. |
H: How you'd show that $f$ is not continuous?
How do you show that
$$f(x)=\begin{cases}2,&x < c\\1,&x\geq c\end{cases}$$ is not continuous at $c$ by using $\epsilon$-$\delta$ - formalism like here?
AI: Suppose $f$ were continuous. Let $\epsilon=0.1$, and $\delta$ be chosen so that $|f(x)-f(c)|<\epsilon$ for all $|x-c|<\delta$. But $|(c-\delta/2)-c|<\delta$ and $|f(c-\delta/2)-f(c)|=|2-1|=1>\epsilon$. Contradiction. |
H: Is it ok to prove a subset of a group is an abelian group this way?
I'll admit from the start I'm being lazy, but all the same it makes thing's neater in my opinion - if it's valid.
Now it's known that if we have a group $G$ such that $g^2=e,\ \ \forall g \in G$ then $G$ is abelian. But what if we don't know G is a group yet? We know the mentioned condition, but thats it. Is it ok to just check, say $x,y \in G \implies xy \in G$ and take the other order for granted? Since.. well, the group, if it is one, will be abelian!
In particular, I'd like to show that $G=\{e,(12)(34),(13)(24),(14)(23)\}\subset S_4$ is an abelian group.
AI: As every element has order two you do not need to check both directions of every multiplication, i.e., you have $3$ non-identity elements and you only need to check $\binom32 = 3$ multiplications instead of the $6$ that a brute force approach would normally require.
The reason is because if $xy = z$ and $z$ is in your subset, then because $z$ has order two we know that $z^{-1} = z$ is in your subset, and $z^{-1} = (xy)^{-1} = y^{-1}x^{-1} = yx$ (because $x$ and $y$ also have order $2$).
That trick I just used is exactly how you prove that if every element of a group is order $2$ it is abelian, but I don't want to say that it's because of that theorem that you don't have to check the other multiplications. Really it's because of the proof of that theorem; because the method of proof still applies in this circumstance. I make this distinction because if you have a theorem that says "If a group satisfies condition $P$ then that group is a $Q$" and then you check that a subset satisfies condition $P$, then in general it is NOT true that this implies that your subset is in fact a subgroup of type $Q$. |
H: How to solve this probability exercise?
We have a box and we have on it 6 balls with numbers from 0 to 5. We push out 3 balls in the way that after pushing out a ball we turn it back again in the box. What is the probability that the sum of the numbers will be equal to 5. Thanks in advance.
AI: Hint: how many choices are there for the first ball? Given what the first is, how many choices are there for the second ball? The third? How many total choices does that make? That is your denominator. For he number of successes, you can do it as a stars and bars problem, or (about as easily) just count by hand. |
H: Number of points on $Y^2 = X^3 + A$ over $\mathbb{F}_p$
Let
$p\equiv 2\pmod{3}$ be prime and let $A\in\mathbb{F}^{∗}_p$
. Show that the number of points
(including the point at infinity) on the curve
$Y^2
=
X^
3
+
A$
over
$\mathbb{F}_
p$
is exactly
$p
+ 1$
I'm having trouble bring in the fact that $p
\equiv 2
\pmod 3$
Any help greatly appreciated!
AI: The multiplicative group of $\Bbb{F}_p$ is cyclic of order $p-1$. As $p\equiv2\pmod3$, this is not a multiple of three. Therefore cubing is a bijection from $\Bbb{F}_p$ to itself. Hence for each $Y$ ($p$ choices) there is a unique solution $X\in\Bbb{F}_p$, namely the cube root of $Y^2-A$. Including the point at infinity gives a total of $p+1$ solutions. |
H: Dense curve on torus not an embedded submanifold
In reference to Showing a subset of the torus is dense, the responders helped show the poster that the image set $f(\mathbb{R})$ is dense in the torus. But, it's not immediately clear to me why the image set is not an embedded submanifold. If $f(\mathbb{R})$ is an embedded submanifold, we must have that it's
a smooth manifold in the subspace topology and
that the inclusion map from $f(\mathbb{R})$ to $T^2$ is a smooth embedding.
It's visually clear to me that under the subspace topology, $f(\mathbb{R})$ is not locally Euclidean (as it's not locally path connected). But, I can't seem to formalize this or any argument, using that $f(\mathbb{R})$ is dense in $T^2$, which says that $f(\mathbb{R})$ fails 1) or 2). Thanks for any help!
AI: For convenience, let's call $f(\mathbb{R}) = A$. It's clear that $A$ is not locally path connected. As you point out, if $A$ were an embedded submanifold, it would be a smooth manifold in the subspace topology. If it were a smooth manifold in the subspace topology, every $a\in A$ would have a neighborhood homeomorphic to $\mathbb{R}$. In particular, $A$ would be locally path-connected. However, it's not, so it can't be a submanifold.
Alternatively, if $A$ were a submanifold, every $a\in A$ would have a neighborhood $U$ in $T^2$ with a coordinate chart taking $a$ to zero, $U$ to $\mathbb{R}^2$, and $U\cap A$ to the $x$-axis. This cannot happen, for every open ball $U_\epsilon$ of $a$ in $T^2$ (for $\epsilon$ sufficiently small) has that $U\cap A$ is disconnected (with countably many components). |
H: $\sup_n \inf_{y \in X} (F(y) + n d(x,y))= F(x)$
Let $(X,d)$ be a metric space and $F : X \rightarrow [0, +\infty)$ a lower semicontinuous function. Then
$$ \sup_n \inf_{y \in X} (F(y) + n d(x,y))= F(x). $$
Is this true?
Intuitively it works since the increasing distance increases the sum of the two functions everywhere except in $x$, but can't figure out how to prove it formally.
AI: Since $\inf_{y \in \mathbb{R}} (f(y) + n d(x,y) ) \le f(x)$, we have $\sup_n \inf_{y \in \mathbb{R}} (f(y) + n d(x,y) ) \le f(x) $.
Let $\epsilon>0$. Since $f$ is lsc. at $x$, there is some $\delta>0$ such that if $d(x,y) < \delta$, then $f(y) \ge f(x) -\epsilon$. If we choose $n \ge \frac{f(x)}{\delta}$, then if $d(x,y) \ge \delta$, we have $f(y) + n d(x,y) \ge \frac{f(x)}{\delta} d(x,y) \ge f(x)$. Hence
$\inf_{y \in \mathbb{R}} (f(y) + n d(x,y) ) \ge f(x) -\epsilon$ and hence
$\sup_n \inf_{y \in \mathbb{R}} (f(y) + n d(x,y) ) \ge f(x) -\epsilon$. Since
$\epsilon>0$ was arbitrary, we have the desired result. |
H: Is $\{(x,y)\in \Bbb R^2: x^2+4y^2+6x+4y+12=0\}$ the empty set?
I am attempting the following problem and can someone verify it ?
We see that $\,x^2+4y^2+6x+4y+12=0 \implies (x+3)^2+4(y+\frac 12)^2=-2$ and hence option (d) is the correct choice. Am I right?
AI: Yes, that is correct and a correct technique. |
H: Irreducibility of Polynomial in $\mathbb{Q}$
How can we show that, if $a>1$ is the product of distinct primes, then $x^n-a$ is irreducible in $\mathbb{Q}$ for all $n \geq 2$ and that it has no repeated roots in any extension of $\mathbb{Q}$?
Thoughts:
Perhaps we can do an inductive proof on $x^{k+2}-a$ as the $k=0$ and $k=1$ cases I know how to deal with.
Any help would be appreciated.
AI: Hint: Eisenstein's Criterion. |
H: Question regarding polygons
Can you prove, that if a equilateral lattice n-gon is constructible, then there will be such a polygon for which the sides have minimal length?
AI: The set of possible side lengths is exactly the set $S$ of all positive numbers of the form $\sqrt{a^2+b^2}$ for integers $a$ and $b$. In increasing order, its elements are $$\{1, \sqrt2, 2, \sqrt 5, \sqrt8, 3, \sqrt{10}, \ldots\}.$$
This set $S$ is well-ordered, which means that every nonempty subset of $S$ contains a minimal element. (It is well-ordered because it is order-isomorphic to $\Bbb N$, the positive integers, which is also well-ordered.)
Since any constructible lattice $n$-gon must have a side length from $S$, the set of constructible side lengths $C$ is a subset of $S$. Since $S$ is well-ordered, $C$ is either empty or has a minimum element, say $m$. So either there is no constructible lattice $n$-gon at all, or else there is at least one with the minimum side length $m$.
I hope this is something like what you were looking for. |
H: $A \oplus B = A \oplus C$ imply $B = C$?
I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both.
But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.
For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$ implies $B = C$ is true.
I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:
I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.
On the other hand, I could be completely wrong about my reasoning.
Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.
AI: Think of $\oplus$ as $\neq$. That is $A \oplus B$ iff $A \neq B$.
Note that $A \oplus A$ is always false, and $\text{False}\oplus A = A$.
Then
$A \oplus (A \oplus B) = (A \oplus A) \oplus B = \text{False} \oplus B = B $.
Similarly, $A \oplus (A \oplus C) = C$, hence $B=C$.
Aside: A 'cute' (as in amusing but not of any practical significance) use of $\oplus$ is to swap the values of two bit variables in a programming language without using an intermediate variable:
\begin{eqnarray}
x = y \oplus x \\
y = y \oplus x \\
x = y \oplus x \\
\end{eqnarray}
Show that the values of $x,y$ are swapped! |
H: Why the binary division of the two has the same quotent
I am trying to divide 110000 with 1101 in binary (long division). I use XOR each time. The divisions above should have quotent 100. However I might must misunderstood something. Could please anyone help me with that?
basically I am doing the following
---011
1101/110000
0000
11000
11010
1101
----
0111
AI: 110000 : 1101 = 11 rem 1001
1101
-----
10110
1101
-----
1001 |
H: partioning of $X$ such that $f_n$ converges uniformly to $f$
Let a measure be $\sigma$-finite and suppose $f_n \rightarrow f$ a.e.. Show that there exists $E_k$ and a null-set $F$ partioning $X$ such that $f_n$ converges uniformly to $f$ on each $E_k$.
I was trying to deduce it from Egorov's Theorem http://mathworld.wolfram.com/EgorovsTheorem.html, but I can't work it out, and it looks like it contradicts with the statement that Egorov's Theorem doesn't hold for $\epsilon=0$.
Thanks!
AI: Let $X=F_1\cup F_2\cup F_3\cup\cdots$ where $F_1\subseteq F_2\subseteq F_3\subseteq\cdots$ and $F_k$ has finite measure. Pick $E_k\subseteq F_k$ so that $f_n\to f$ uniformly on $E_k$, and $\mu(F_k\setminus E_k)<2^{-n}$. I think you can build from there. |
H: Formula for the length of line that connects two sides of a triangle.
For the triangle in the picture, coordinates of $A$, $B$ and $C$ are known. Is there an explicit formula for length $XY$, as a function of height $h$? It's a function of other variables as well, but it's important that it depends explicitly on $h$.
$XY$ is parallel to side $c$, but I'm not good at drawing.
AI: $Cxy$ is similar to $CAB$, so the sides are in the same ratio as the altitudes. |
H: show that torus is compact
I am having difficuties in showing a torus is compact. Initially I wanted to use Heine-Borel theorem, but after that I realise we are not working in $\mathbb{R}^n$ space. So a simple way to show torus is compact is by definition. But after defining an open cover for torus, I don't know how to proceed. Can anyone guide me?
AI: If you are thinking of the torus as $S^1 \times S^1$:
Products of compact sets are compact (you only need the finite case)
$S^1$ compact (as it is a closed and bounded subset of $R^2$).
If you are thinking of the torus as $R^2 / Z^2$:
This quotient map makes the same identifications as the exponential map $(e^{2 \pi i x}, e^{2 \pi i y})$, so since both are quotients $R^2 / Z^2$ is homeomorphic to $S^1 \times S^1$. (You will have to prove that if $q_1: X \to Y_1$ and $q_2 : X \to Y_2$ are quotient maps, and $q_1(x) = q_1(x')$ iff $q_2(x) = q_2(x')$, then $Y_1$ and $Y_2$ are (naturally) homeomorphic.) |
H: How to go about calculating this finite summation?
I have the summation
$$
\sum_{n = 0}^{10} \frac {1 + (-1)^n} {2^n}
$$
I have looked up how to work out sequences without manually finding each term and adding it up, but I have only found out how to work out problems like $n^2$, $n^3$, etc.
How would I solve this? Is there some way to simplify it to be able to solve it? Please be detailed.
Thanks
AI: $$\sum_{0\le n\le10}\frac{1+(-1)^n}{2^n}=\sum_{0\le m\le5}\frac2{2^{2m}}$$ which is a Geometric Series |
H: Proving a function is constant, under certain conditions?
The problem:
Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t) - f(x)| \leq |t - x|^2$ for all $t, x$. Prove $f$ is constant.
I believe I have some intuition about why this is the case; i.e. if $t$ and $x$ are very close, ($|t - x| = \epsilon$), then $\sqrt{\epsilon}$ will converge to 0, so $|f(t) - f(x)|$ will also converge to zero if you keep taking $t$ and $x$ closer and closer to each other. However, how do I formalize this argument? Thanks.
AI: For all $x,t\in \Bbb R$ such that $x\neq t$, the equivalence below holds:
$$|f(t) - f(x)| \leq |t - x|^2\iff \left\vert \dfrac{f(t)-f(x)}{t-x}\right\vert\leq |t-x|,$$
taking the limit as $t$ approaches $x$ yields
$$\lim \limits_{t\to x}\left(\left\vert \dfrac{f(t)-f(x)}{t-x}\right\vert\right)\leq \lim \limits_{t\to x}|t-x|=0.$$
This proves that $f$ is $\bbox[5px,border:2px solid #FFFFFF]{\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$ and $\forall x\in \Bbb R(f'(x)=\bbox[5px,border:2px solid #FFFFFF]{\_})$, thus $f$ is $\bbox[5px,border:2px solid #FFFFFF]{\_\_\_\_\_\_\_\_\_\_\_}$. |
H: Proof of orthogonal matrix property: $A^{-1} = A^t$
I have prooved this orthogonal property. Please correct it or show your version of the proof if I am wrong:
$A^{-1} = A^t$
$A^{-1} \times A = A^t \times A$
$I = I$
I appreciate your answer.
AI: There are two main definitions of orthogonality. Accepting one you can prove another. Since you need to prove $Q^T = Q^{-1}$, you should define orthogonality as follows:
An orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors.
So, let's say you have a matrix $Q = [q_1, q_2, \ldots, q_n]$, where $q_i$ is a unit column vector and $q_i^T q_j = \delta_{ij}$ due to the orthogonality. Now, find its transpose
$$
Q^T = \left [ \begin{array}{c}
q_1^T \\ q_2^T \\ \vdots \\ q_n^T
\end{array} \right ]
$$
where $q_i^T$ is a row vector. So
$$
Q^T \cdot Q = \left [ \begin{array}{c}
q_1^T \\ q_2^T \\ \vdots \\ q_n^T
\end{array} \right ] \cdot [q_1, q_2, \ldots, q_n] = \left [ \begin{array}{ccccc}
q_1^T q_1 & q_1^T q_2 & \cdots & q_1^T q_n \\
q_2^T q_1 & q_2^T q_2 & \cdots & q_2^T q_n \\
\vdots & \vdots & \ddots & \vdots \\
q_n^T q_1 & q_n^T q_2 & \cdots & q_n^T q_n
\end{array}\right ] = \left [ \begin{array}{ccccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{array}\right ] = I
$$
which means $Q^T = Q^{-1}$. |
H: Trigonometric inequality solving
How to solve this inequality $\left|\dfrac{\cos 2x + 3}{\cos x}\right|\geq 4$ ?
I tried to consider 2 cases:
1) When $\cos 2x \geq 0$ and $0<\cos x<1$
2) $\cos 2x\leq 0$ and $-1 < \cos x < 0$.
But I think that's wrong.
AI: As $\cos2x=2\cos^2x-1$
$$\frac{\cos2x+3}{\cos x}=\frac{2\cos^2x-1+3}{\cos x}=2\left(\cos x+\frac1{\cos x}\right)$$
If $\displaystyle \cos x>0, \cos x+\frac1{\cos x} \ge 2\sqrt{\cos x\cdot\frac1{\cos x}}=2$ (using A.M.$\ge$ G.M.)
I leave the case $\cos x<0$ for you |
H: Why does knowing where two adjacent vertices of regular $n$-gon move under rigid motion determines the motion?
I am reading the book Abstract Algebra by Dummit and Foote.
In the section about the group $D_{2n}$ (of order $2n$) the authors
claim that knowing where two adjacent vertices move to, completely
determine the entire motion.
Based on this, the book gets the conclusion that $|D_{2n}|\leq2n$
and by showing existence of $2n$ different motions it concludes $|D_{2n}|=2n$.
My question is about the claim maid: Why does knowing where two adjacent
vertices of a regular $n$-gon move to, completely determine the entire
motion ?
I am looking for a convincing argument, not neccaseraly a formal proof,
I just want to get the idea.
AI: An Euclidean movement is determined by the effect on three (noncollinear) points. If $A\mapsto A'$, $B\mapsto B'$, $C\mapsto C'$, we can consider the movement composed of
the translation along $\vec{A'A}$
the rotation about $A$ that maps the translated image of $B'$ to $B$
the reflection at $AB$ if necesary for the third point
Performing these movements after the given movement leaves $A,B,C$ fixed in the end and the only movement that leaves a (nondegenerate) triangle fixed is the identity.
In the case of a regular $n$-gon being mapped to itself, the center must remain fixed(!) Together with two vertices we thus know th eeffect of the movement on three noncollinera points as required.
Alternatively, if you feel uncomfortable about the center being ficed: If we know where a vertex $A$ and a neighbour(!) vertex $B$ go, we also know that "the other" neighbour $C$ (assuming the polygon has more than two vertices) of $A$ must be mapped to "the other" neighbour of the image of $A$ |
H: Norm space, linear operator exercise, help please!
$f \in L_2[a,b]$
$Uf(s):=\int_a^bk(s,t)f(t)dt$
$k(s,t):[a,b]^2\to R $ continuous.
show
1) $U:L_2[a,b]\to L_2[a,b]$, in other words, $Uf(s)\in L_2[a,b] \quad \forall f$
2) $U$ is linear and continuous
3) $||U||\leq ||k(s,t)||_2$
AI: Here are some hints:
Let $g_s(t) = k(s,t)$. Since $k$ is continuous, it is bounded on $[a,b]^2$, say $|k(s,t)| \le B$. We can consider $g_s $ as an element of $L_2[a,b]$.
Then $(Uf)(s) = \langle g_s, f \rangle$. Use this to find a bound on $|(Uf)(s)|^2$. This shows that $U$ is well defined and $Uf \in L_2[a,b]$.
Linearity is straightforward using the definition. The above shows that $U$ is bounded, hence continuous.
The above will also show the bound on $\|U\|$.
Addendum:
As I wrote above, let $g_s(t) = k(s,t)$. Then $(Uf)(s) = \langle g_s, f \rangle$, from which we have $|(Uf)(s)|^2 \le \|g_s\|^2 \|f\|^2 = \int_a^b k(s,t)^2 dt \|f\|^2$. If we now integrate over $s$, we get $\|Uf\|^2 = \int_a^b |(Uf)(s)| ds \le \int_a^b\int_a^b k(s,t)^2 ds dt \|f\|^2$. Taking square roots gives the desired answer. |
H: Category for measure spaces?
I know some things about measures/probabilities and I know some things about categories. Shortly I realized that uptil now I have never encountered something as a category of measure spaces. It seems quite likely to me that something like that can be constructed. I am an amateur however and my scope is small. I have two questions:
1 Is there indeed material of this sort and can you tell me about it? The whereabouts for instance.
2 Is there a reason for the fact that uptil now I did not find anything of the sort? Is it indeed rare for some reason?
AI: I'm just going to record some references that you might want to take a look at. They might be a little too heavy on the category theory for your taste, but they also might provide further places to look.
$n$-Category Café post: Category Theoretic Probability Theory
Related in content, the nLab article on probability theory; see especially the section "Probability theory from the nPOV"
The nLab article on measurable spaces. See especially the connection with von Neumann algebras, where it is localizable measurable spaces which are the pertinent concept for the appropriate Gelfand-Naimark duality, and see also references to posts by Dmitri Pavlov at MathOverflow. |
H: Limit of $x_n/n$ for sequences of the form $x_{n+1}=x_n+1/x_n^p$
Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{x_n} (n\ge1)$, Prove whether the limit as follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$
Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{\sqrt{x_n}} (n\ge1)$, Prove whether the limit as follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$
In both cases, $x_n$ is increasing, so I tried to get an upper bound on $x_n$ (possibly depending on $n$) to apply a squeeze theorem; but failed.
AI: Combining the ideas in the answers given by kedrigern and N.S., we can obtain a little more general conclusion as below.
Proposition: Let $f:(0,+\infty)\to (0,+\infty)$ be continuously differentiable with $f'>0$. In addition, assume that there exists $\delta>0$, such that (i) $f'(x)\ge\frac{1}{\delta}$ when $x$ is large, and (ii) for any $\theta:(0,+\infty)\to[0,\delta]$,
$$\lim_{x\to\infty}\frac{f'(x+\theta(x))}{f'(x)}=1.\tag{1}$$
Then for any sequence $(x_n)$ satisfying
$$x_1>0\quad\text{and}\quad x_{n+1}=x_n+\frac{1}{f'(x_n)},\ \forall n\ge 1, \tag{2}$$
we have:
$$\lim_{n\to\infty}\frac{f(x_n)}{n}=1.\tag{3}$$
Proof: Form $(2)$ and $f'>0$ we know that $(x_n)$ is positive and increasing; in particular, $L:=\lim\limits_{n\to\infty}x_n$ eixsts(either finite or $+\infty$). If $L<+\infty$, then by $(2)$ and continuity,
$$L=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n+\frac{1}{\lim\limits_{n\to\infty}f'(x_n)}=L+\frac{1}{f'(L)}>L,$$
a contradiction. Therefore, $\lim\limits_{n\to\infty}x_n=+\infty$.
Since $f'>0$, $f$ is increasing. Then from $(x_n)$ being increasing we know that $\big(f(x_n)\big)$ is also increasing, so by Stolz–Cesàro theorem,
$$\lim_{n\to\infty}\frac{f(x_n)}{n}=\lim_{n\to\infty}\big(f(x_n)-f(x_{n-1})\big).\tag{4}$$
Denote $\delta_n=\frac{1}{f'(x_n)}$. By $(2)$ and mean value theorem, there exists $\theta_n\in (0,\delta_n)$, such that
$$f(x_{n+1})-f(x_n)=f(x_n+\delta_n)-f(x_n)=f'(x_n+\theta_n)\delta_n.\tag{5}$$
Since $\lim\limits_{n\to\infty}x_n=+\infty$, by assumption (i), when $n$ is large, $\theta_n<\delta_n\le \delta$. Letting $n\to\infty$ in $(5)$, by assumption (ii), the limit exists and is $1$, so $(3)$ follows from $(4)$, which completes the proof.
Exampes: It is easy to check that for every $c>0$ and every $p\ge 1$, $f(x)=cx^p$ satisfies all the assumptions in the proposition. In particular, for your original question, we have:
For $f(x)=\frac{1}{2}x^2$, $f'(x)=x$ and $x_{n+1}=x_n+\frac{1}{x_n}$, so if $x_1>0$,
$$\lim_{n\to\infty}\frac{f(x_n)}{n}=1\Longleftrightarrow \lim_{n\to\infty}\frac{x_n}{\sqrt{n}}=\sqrt{2}\Longrightarrow \lim_{n\to\infty}\frac{x_n}{n}=0.$$
For $f(x)=\frac{2}{3}x^{\frac{3}{2}}$, $f'(x)=\sqrt{x}$ and $x_{n+1}=x_n+\frac{1}{\sqrt{x_n}}$, so if $x_1>0$,
$$\lim_{n\to\infty}\frac{f(x_n)}{n}=1\Longleftrightarrow \lim_{n\to\infty}\frac{x_n}{n^{\frac{2}{3}}}=(\frac{3}{2})^{\frac{2}{3}}\Longrightarrow\lim_{n\to\infty}\frac{x_n}{n}=0.$$ |
H: Show that $f(X)$ has a dense subset.
Let $X$ be a topological space which has a dense countable subset $D$, and suppose $f\colon X \to Y$ a continuous function. Show that $f(D)$ is dense countable in $f(X)$.
AI: I see that you've decided that $f(D)$ is a good candidate. (Right on target with that one.) Since $f$ maps $D$ onto $f(D)$ and $D$ is countable, then so is $f(D)$. To show that it is dense, consider any non-empty relatively open subset $U$ of $f(X),$ meaning that there is some non-empty open subset $V$ of $Y$ such that $U=V\cap f(X).$ What can you say about $f^{-1}(U)$ by continuity? Since $D$ is dense and $f^{-1}(U)$ is non-empty (why?), what can you conclude from this? |
H: $n\times n$ matrix determinant of rook configuration
We place $n$ rooks on an $n\times n$ chessboard in such a way that they don't threaten each other. To each such placement corresponds an $n\times n$ matrix in which there is a $1$ at the position of the rooks and $0$ at the other places. What is the determinant of the matrices obtained this way? Thanks!
AI: It's going to be $(-1)^k$ where $k$ is the number of row and column swaps you need to make from the identity matrix to obtain the matrix that you get from the rook configuration. |
H: Is there a definition of cylinder that these equations satisfy
Our teacher is claiming that (in $\mathbb{R}^3$) the following surfaces are "cylinders":
$3x+y+\frac{7}{2}=0$
$y=x^2$
$z^2 = y$
$\frac{x^2}{4} + \frac{y^2}{4} = 1$
Is there any definition of cylinder that can justify this statement, and if so, which definition?
The last one is the only one that fits the definition that I know (which he calls an "elliptic cylinder")
AI: When most people heat the word "cylinder" they are thinking of a "right circular cylinder" -- that is -- a circle which is copied up and down along a line perpendicular to the plane that circle lies in.
The term "cylinder" can more generally refer to a curve copied along any line. So for example,
$y=x^2$ is a parabola in the plane, but since there is no reference to $z$, the parabola $y=x^2$ and $z=5$ also satisfies this equation (same for $z=$any constant). So in 3-space, the graph of $y=x^2$ is a parabola copied up and down the $z$-axis. Instead of a circular (regular old) cylinder, you have a "parabolic cylinder".
If you notice, all of the equations you have listed above only list 2 out of the 3 variables (x,y,z). So the graphs of these surfaces are graphs of a curve copied along the remain variable's axis.
For example: Here is (part of) the graph of $z^2=y$ (another parabolic cylinder)...
Notice that the parabola $z^2=y$ (in the $yz$-plane) has been copied along the $x$-axis.
Wikipedia Page on Cylinders -- see "other types of cylinders" |
H: Mapping behavior of imaginary axis via $v=\frac{z-a}{z+a}$
I would like to know what the bilinear transform $v=\frac{z-a}{z+a}$ does to the imaginary axis, where $a$ is a real number.
I substituted $z=yi$ and calculated $|v|$ giving me $|v| =1$.
Is this enough proof to say that $v$ maps the imaginary axis to on the unit circle?
AI: It's enough to show that it maps the imaginary axis to the unit circle, but not quite enough to show that it maps the imaginary axis onto the unit circle. To draw that conclusion, we need to know that linear fractional transforms map circles to circles (in the Riemann sphere).
Alternately, you must show explicitly that it maps onto the unit circle (with one point deleted, if you don't count the "point at $\infty$" as being on the unit circle). |
H: laws of probability
Suppose that there is a $60\%$ probability that the product will be a success on the market (that means, the probability of failure is $40\%$). If the product is a success, you will get a profit of $\$200,000$, and if it is a failure, you will incur a loss of $\$100,000$. Should you develop this product? How do you make a decision in this situation? Also, how can one come up with the probability of success (or failure)?
AI: There really is no right or wrong answer to whether or not you should develop the product. We can at least come up with the expected profit (with the convention that a loss is negative profit). This is just:
$$\mathbb{E}(P)=\$200,000\times60\%+(-\$100,000\times40\%)=\$80,000$$
Since this is positive it might be decided to develop the product.
To come up with success and failure probabilities you would really need to define what you mean by "success" and "failure". It could be defined in terms of total number of units sold or perhaps total generated revenue or perhaps something else quantifiable and essential to the company's bottom line. Failure would then be defined as the complementary event or the even where say a minimum number of units is NOT sold or if you fall short of a certain revenue threshold. Again, the risk factors impacting "success" here are fairly open ended and depend on the uniqueness of the product. |
H: How to find non-isomorphic trees?
"Draw all non-isomorphic trees with 5 vertices."
I have searched the web and found many examples of the non-isomorphic trees with 5 vertices, but I can't figure out how they have come to their answer. How exactly do you find how many non-isomorphic trees there are and what they look like? Thanks for your time
AI: One systematic approach is to go by the maximum degree of a vertex. Clearly the maximum degree of a vertex in a tree with $5$ vertices must be $2,3$, or $4$. If there is a vertex of degree $4$, the tree must be this one:
*
|
*--*--*
|
*
At the other extreme, if the maximum degree of any vertex is $2$, the tree must be the chain of $5$ vertices:
*--*--*--*--*
That leaves the case in which there is a vertex of degree $3$. In this case the fifth vertex must be attached to one of the leaves of this tree:
*
\
*--*
/
*
No matter to which leaf you attach it, you get a tree isomorphic to this one:
*
\
*--*--*
/
*
Thus, there are just three non-isomorphic trees with $5$ vertices. |
H: Laurent expansion of $f (z) = \frac1{z(z − 1)(z − 2)},$ (in powers of $z$) for $0 < |z| < 1,$ $1 < |z| < 2,$ and $|z| > 2.$
Find the Laurent expansion of $$f (z) = \frac{1}{z(z-1)(z-2)}$$
(in powers of $z$) for
a. $0 < |z| < 1$
b. $1 < |z| < 2$
c. $|z| > 2$
AI: The first thing you'll want to do, here, is what is known as a partial fraction decomposition. Write $$\frac{1}{z(z-1)(z-2)}=\frac{A}{z}+\frac{B}{z-1}+\frac{C}{z-2},$$ and solve for $A,B,C.$
Note that for $z\neq0,2,$ we can write $$\frac1{z-2}=-\frac1{2-z}=-\frac12\cdot\cfrac1{1-\left(\frac{z}{2}\right)}$$ and $$\frac1{z-2}=\frac1{z}\cdot\cfrac1{1-\left(\frac{2}{z}\right)}.$$
Now, one of these can be expanded as a multiple of a geometric series in the disk $|z|<2,$ and the other can be expanded as a multiple of a geometric series in the annulus $|z|>2$. That is, we will use the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $|w|<1$. You should figure out which region works for which rewritten version, and find the respective expansions in both cases.
Likewise, we can rewrite $\frac1{z-1}$ in two similar forms, one of which is expandable in $|z|<1$ and one of which is expandable in $|z|>1.$
Using the partial fraction decomposition with the expansions you found above will give you three different Laurent expansions of $f(z),$ one for each of the regions $0<|z|<1,$ $1<|z|<2,$ and $|z|>2$. |
H: New ways to light the fire again
Recently I've been studying a lot of analytic geometry and this subject made my motivation drop. The thing is, the courses aren't stopping and I'm beginning to lose the passion I had before. I need ideas on how to "light the fire" again. Help would be much appreciated.
AI: "Shake it up" a bit! Supplement your text with another text that approaches the topic differently. Also, make some changes in your approach to math: see how what you're learning connects to what you already know, and how each topic you encounter connects to the earlier topics. Try to anticipate what you'll soon encounter by making conjectures using what you have already covered, and/or are presently studying.
While homework may not feel like fun, you can find ways to keep math fun:
Take breaks when necessary.
Try to be proactive when you study (e.g., skim through and read sections of the text prior to lectures on the same topics) and don't fall into the unpleasant trap of barely keeping up, or being "one step behind the game."
As you study, focus on the "key ideas" in the material you're reading and the problem types that are assigned. Focus on understanding the concepts and ideas as opposed to just trying to memorize formulas and just churn out answers, mechanically.
Reward yourself for even small successes.
Also, take heart. It is completely normal to "hit brick walls", for our level of passion for anything to perk up at times, and diminish at other times. Math is a HUGE field, and it is common to encounter phases and topics that just don't interest us as much as other topics have, and as other topics refuel our passion. So be patient and persistent. Chances are your enthusiasm will peak again. In the meantime, think of what you're going through as a temporary challenge to acquire the discipline to persist. If you plan to pursue math over the long hall - indeed, if you plan to succeed in life, in general - you'll need to learn how to get through the tough times and the boring times, in addition to making the most of those times in which you're aided by the fuel of passion. |
H: Congruences with prime number and factorial
Prove that if $p\equiv 1 \pmod{4}$ is a prime number and $$x\equiv \pm \left(\frac{p-1}{2}\right)! \pmod{p}$$
then $x^2\equiv -1 \pmod{p}$
I think Wilson's theorem will come in handy here, used in some clever way, but I can't see it. I would be very grateful for any help.
AI: Using Wilson's theorem and $p \equiv 1 \pmod{4}$ gives
$$-1 \equiv (p-1)! \equiv \prod_{i=1}^{\frac{p-1}{2}}{i(p-i)} \equiv \prod_{i=1}^{\frac{p-1}{2}}{(-i^2)} \equiv (-1)^{\frac{p-1}{2}}\left[\left(\frac{p-1}{2}\right)!\right]^2 \equiv \left[\pm\left(\frac{p-1}{2}\right)!\right]^2 \pmod{p}$$ |
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