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H: True or False problem on sum principle
If A $\cap$ B $\cap$ C = $\emptyset$, then the sum principle applies so |A $\cup$ B $\cup$ C| = |A|+|B|+|C|.
I think it would be true since there is nothing in common among A,B and C, but just wondering if there is any exceptions to this problem so it would be false?
AI: Let A = {1,2}, B={2,3} and C={3,1} so |A| = |B| = |C| = 2. Then A $\cap$ B $\cap$ C = $\emptyset$, but $$A \cup B \cup C = \{1,2,3\} \implies |A \cup B \cup C|=3 \not= |A| + |B| + |C|$$
When you apply inclusion-exclusion in this case, you also need to consider the sizes of |A $\cap$ B|, |B $\cap$ C| and |C $\cap$ A|. I suspect the cause of the confusion is that you have generalized from the two set case, in a way that doesn't work in the three set case. The "sum principle" you refer to, is exactly the inclusion-exclusion principle applied to only two sets:
$$|A \cup B| = |A| + |B| - |A\cap B| $$
If and only if A and B are mutually exclusive, so $A \cap B = \emptyset $, we have:
$$|A \cup B| = |A| + |B| $$
But when three sets are involved, the inclusion-exclusion principle gives:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A| + |A\cap B\cap C| $$
You can check why this works visually. Can you see why each subset of A $\cup$ B $\cup$ C is only counted once? For instance the part of A that does not intersect with B or C (i.e. $A \cap B' \cap C'$) is only counted by the first term. More complicated is $A \cap B \cap C$ (in the middle), which is included by the first three terms so it starts off as triple-counted. The next three terms take it away again, and the final term adds it back on, so by the end it has been counted once only.
Knowing that $A\cap B\cap C = \emptyset $ would only reduce this to:
$$|A \cup B \cup C| = |A| + |B| + |C| - |A\cap B| - |B\cap C| - |C\cap A|$$
In my example $A\cap B=\{2\}$, $B\cap C=\{3\}$ and $C\cap A=\{1\}$ which all have cardinality one. The formula correctly gives $3 = 2 + 2 + 2 - 1 -1 -1$. |
H: How does knowing a ratio help me determine a total?
I am working with this question:
A received 1/3 more votes than B. Which of the following could have been the total number of votes cast for the two candidates? Answer options are 12, 13, 14, 15, 16.
I can see that however many votes B got, A got the same number plus 1/3 more, so the total number of votes cast is $x + \frac{4}{3}x = \frac{7}{3}x$. I can also see that x has to be an integer, and that it should be evenly divisible by three.
I got that far, then couldn't see how to get anywhere with algebra. I solved the problem by testing multiples of three (first $x = 9$, then $x = 6$) to see if $2\frac{1}{3}x$ was in the list of answers. Of course it works for $x = 6$, but I feel like this was a very haphazard way of getting the answer.
Is there some way that knowing the total number of votes is $\frac{7}{3}x$ can help me determine the correct answer directly rather than testing values? I don't feel like this was a hard question, but I do feel like I'm missing some obvious connection between the ratio and the total.
AI: You have figured out that $x$ must be divisible by three. Say $x = 3y$.
Then the total number of votes is $7 y$.
So we see that the total is divisible by $7$. Now all you have to do is look at the answers and see that only one of them is divisible by $7$. |
H: Determine the value for which a sequence is an arithmetic progression.
We have the following sequence $$ -a, -\dfrac{a}{b}, \dfrac{a}{b}, a$$
Determine the value of $b$ for which this is an arithmetic progression ($a \neq 0$)
I don't know how to do this. I've tried adding a variable and making a couple of equations:
$-a + x = -\dfrac{a}{b}$
$ -a + 2x = \dfrac{a}{b}$
$-a + 3x = a$
But this doesn't seem to yield any logical answers. Just from looking at the sequence it seems that $b=2$ should be the answer, but I can't seem to get that result..
AI: Since this is an arithmetic progression, $a - \frac{a}{b} = \frac{a}{b} - \frac{-a}{b}$, i.e. $t_{i+1} - t_{i}$ is constant $\forall\; i$.
Note that $a\cdot(1-\frac{1}{b})=a\cdot(\frac{1}{b} - \frac{-1}{b})$. We write $1-\frac{1}{b} = \frac{2}{b}$. So, $b=3$. We still know nothing about a. |
H: Homework - NFA and its regular expression.
So for (a) I got this:
I THINK that is right, I'm not too sure. But I am also having a lot of trouble doing (b). Any pointers to get me started would be appreciated.
AI: There are four parts, $u,c,v,a$. $u \in \{a,b\}^*$, so write $(a|b)^*$ for that, $c$ is easy, $v$ has length one, and is any character, so is $(a|b|c)$, and $a$ is easy. Combining gives $(a|b)^* c(a|b|c)a$. |
H: Connection on complex vector bundle
Let $M$ be a $m$-dimensional
Riemannian manifold. I will follow the notation of the book "From calculus to
cohomology - Madsen and Tornehave"
If $\xi $ is a $k$-dimensional complex vector bundle, what is the connection ?
If $\{e_i\}$ is a basis of $\Omega^0(\xi)$ then any section has the form $$ s=\sum_{i=1}^k [s_j(x_1, ... , x_m) + i t_j(x_1, ... ,x_m) ]e_j$$
Hence $$\nabla\ s = \{\ [ds_j + i\ dt_j ] + (s_l + it_l ) A_{lj}\ \}\ e_j $$
So $$\xi = \xi_0\otimes_{\mathbb R}\varepsilon^1_{\mathbb C}$$ and $A$ is the connection on $\xi_0$.
Further, $$ s=u+iv,\ F^{\nabla^\xi} (s) = F^{\nabla^{\xi_0}} (u) +i F^{\nabla^{\xi_0}} (v)$$
Am I right ? Thank you in advance.
AI: There's a few issues. Firstly, asking "what is the connection" doesn't usually make sense: an arbitrary vector bundle does not have a canonical connection.
Secondly, to be able to decompose $s$ into it's real and imaginary parts as you did may not always be possible. Indeed, this happens if and only if the complex bundle is the complexification of a real vector bundle (which doesn't always happen and a necessary condition is that the Chern classes vanish in degrees $4k+2$).
Unless are you given that $\xi = \xi_0 \otimes \mathbb C$? |
H: If $f = u + iv$ is a complex function, is $|f| = (u^2 + v^2)^{1/2}$?
If $f = u + iv$ is a complex function, is $|f| = (u^2 + v^2)^{1/2}$? Where, $|f|$ is the modulus or absolute of $f$.
I thought this should be correct because for any complex number $z = a + ib$, $|z| = (a^2 + b^2)^{1/2}$. At any point $w = x + iy$, $f(w) = u(x,y) + iv(x,y)$ is just a complex number and so the modulus should defined the same way.
AI: Yes. If $f(z) = u(z) + iv(z)$ where $u$ and $v$ are real-valued functions, then
$$
|f(z)| = \sqrt{|u(z)^2 + v(z)^2|},
$$
and so (writing $|f|$ for the function $|\cdot|\circ f$ as usual) you have $|f| = \sqrt{u^2 + v^2}$. Accordingly, we can think of $|f|$ as a function $\mathbb{C}\to\mathbb{C}$ (or $\mathbb{C} \to \mathbb{R}$, or $\mathbb{R}^2 \to \mathbb{R}$, etc.).
One suggestion: it might make your life easier to get into the habit of writing $\sqrt{\alpha}$ to denote the positive square root of $\alpha$ and reserve the notation $\alpha^{1/2}$ for the set of both roots of $\alpha$. Of course, notations and conventions vary, but following a convention like this will make discussing the branches of complex functions easier when you get to that topic. |
H: How to differentiate CDF of Gamma Distribution to get back PDF?
CDF of a gamma distribution ($X \sim \mathcal{G}(n, \lambda)$) looks like
$$F(x) = \frac{\Gamma_x(n)}{\Gamma(n)}$$
Where $\Gamma_x(n) = \int_0^x t^{n-1} e^{-t} \, dt$ the incomplete gamma function. Ok so far?
But how do I differentiate such an expression?
$$\frac{d}{dx} \frac{\int_0^x t^{n-1} e^{-t} \, dt}{\int_0^\infty t^{n-1} e^{-t} \, dt}$$
UPDATE
With help from @MhenniBenghorbal, I have gotten:
$$\frac{d}{dx} \frac{\int_0^x t^{n-1} e^{-t} \, dt}{\int_0^\infty t^{n-1} e^{-t} \, dt} = \frac{1}{\Gamma(n)} x^{n-1}e^{-\lambda t}$$
but then its missing the $\lambda^n$ term of the original PDF of a gamma distribution. How do I get that back? I must have overlooked something? But I can see where ...
AI: Here is how you apply it
$$ \frac{d}{dx}\int_{0}^{x}t^{n-1}e^{-t}dt= x^{n-1}e^{-x}. $$
Note: In your case you the fundamental theorem of calculus is enough
Theorem: Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by
$$F(x) = \int_a^x\!f(t)\, dt.$$
Then, $F$ is continuous on $[a, b]$, differentiable on the open interval (a, b), and
$$F'(x) = f(x)\,$$
for all $x$ in $(a, b)$. |
H: Finding derivative of three variables
Consider a box with dimensions x, y, and z. x is changing at a rate of 1 m/s, y at -2 m/s and z at 1 m/s. Find the rate that the volume, surface area and diagonal length ($s = \sqrt{x^2+y^2+z^2}$) are changing at the instant when $x = 4$, $y = 3$ and $z = 2$.
I know I need to use to product rule. But my teacher didn't bother to teach it to us how to do it with three variables (ex. xyz for the volume). So I'm having trouble figuring out how to start this problem.
AI: Write out your equations for volume, surface area, and diagonal length and take the derivative with respect to time. I assume you have gone over this in class, but for the surface area it should look like:
$\frac{d}{dt}s(x,y,z) = \frac{\partial s}{\partial x} \frac{dx}{dt} + \frac{\partial s}{\partial y}\frac{dy}{dt} + \frac{\partial s}{\partial z}\frac{dz}{dt}$. Then, you have equations for $\frac{dx}{dt}$ etc. From there it is just plug and chug. |
H: If $A$ has a positive Lebesgue measure then there exist subsets which are not measurable
I was thinking if there is a solution to this problem without trying to explicitly create Vitali sets in $A$. Does anyone have any ideas?
AI: I don't think the Vitali construction is useful here. Instead, I'd use the other (Bernstein's?) method of constructing a nonmeasurable set. We need two facts:
There are only $2^{\aleph_0}$ closed sets of real numbers.
Every closed set of positive measure (in fact every uncountable Borel set but we don't need that) has cardinality $2^{\aleph_0}$.
Let $A$ be a set of positive measure. Using the above facts, a straightforward transfinite induction will serve to construct two disjoint sets $B,C\subseteq A$, each of which has nonempty intersection with every closed subset of $A$ which has positive measure. (If you want to, you can just as easily construct a pairwise disjoint family of $2^{\aleph_0}$ such sets instead of only two.) It is easy to see that $B$ and $C$ are nonmeasurable. ($B$ can't have positive measure because it doesn't contain a closed set of positive measure; $A\setminus B$ can't have positive measure for the same reason. Since $A$ has positive measure, this means that $B$ is nonmeasurable.) |
H: If $f \ge 0$ is zero a.e., then $\int_{\Bbb R}f \,\mathrm d\lambda= 0$
Suppose $f \geq 0$ is measurable, then $f = 0 $ almost everywhere implies $\int\limits_{\mathbb{R}} f \,\mathrm d\lambda= 0 $.
My try
Pick Simple $\phi \leq f $. Since $f = 0 $ ae, then how can we show $\phi = 0 ?$. IF we can show this, then $\int \phi = 0 \implies \int f = 0 $ by definition.
Can someone help me ? thanks
AI: If $\phi$ were not identically $0$, then there would exist a set of positive measure $E$ and a positive number $\alpha$ such that
$$x \in E \implies \phi(x) = \alpha$$
by definition of a simple function. So....? |
H: Big Oh and Big Omega clarification
Can I get an explanation of:
Can g(n) be Big O of $n^{2}$ and also the Big O of $n^{3}$? (at the same time)
Can g(n) be Big Omega of $\Omega (n)$ and also be the Big O of $n$?
AI: For your first question: if $g(n)=O(n^2)$, then $g(n)=O(n^{2+k})$ for all $k>0$. Indeed, $g(n)=O(n^2)$ means that
$$
\limsup_{n\to\infty}\frac{|g(n)|}{n^2}<\infty,
$$
and we also have
$$
\frac{|g(n)|}{n^{2+k}}\leq\frac{|g(n)|}{n^2}
$$
if $k>0$.
The answer to your second question is yes. After all, it is possible that $g(n)=n$. |
H: Let n∈ℕ. Suppose that p is an odd prime number that divides n^2+1 Show that if p=4k+3 for some integer k≥0, then n^(p−1)≡−1(modp).
Let $n \in \mathbb N$. Suppose that $p$ is an odd prime number that divides $n^2+1$. Show that if $p=4k+3$ for some integer $k \ge 0$, then $n^{p−1}\equiv−1 \pmod p$.
The question says to use "if $p=4k+3$ then $\frac {(p-1)} 2$ is odd".
So we know $n^2 \equiv -1 \pmod p$
Do I manipulate that into $n^{4k+2} \equiv -1 \pmod p$ ?
I tried that and ended up with $n^{4k+2} \equiv -1^k \pmod p$.
How do I use "if $p=4k+3$ then $\frac {p-1} 2$ is odd" ?
AI: Modulo $p$ we get $n^{p-1} \equiv (n^2)^{(p-1)/2} \equiv -1^{(p-1)/2} \equiv -1$, which gives the implication you are looking for. By the way this implication is vacuously true, as the odd prime divisors of $n^2 + 1$ are all of the form $4k + 1$. |
H: Prove that $f$ is bounded if it converges as $x \rightarrow \infty$ and $x \rightarrow -\infty$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(x) \rightarrow 7 $ as $x \rightarrow \infty $ and $f(x) \rightarrow -25 $ as $x \rightarrow -\infty $ and
Prove that $f$ is a bounded function.
AI: Hint: Choose an $N_1$ such that $$x > N_1 \implies |f(x) - 7| < 1$$ and an $N_2$ such that $$x < -N_2 \implies |f(x) + 25| < 1$$
Now use the fact that $[-N_2, N_1]$ is compact. |
H: Inner Product of Real Polynomials
Updated improved question:
Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define
$$
\langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1).
$$
(i) Prove that $\langle\cdot,\cdot\rangle$ defines an inner product.
Could we just do this $f(a)=0$ and $f'(a)=0$ then $f(x)$ is divisible by $(x-a)^2$ ?
If so how would we solve this?
Can someone please help me with this proof for part (i). It is frustrating me.
AI: The function you have defined is clearly bi-linear and symmetric, so the only thing one needs to check is positive-definiteness. For this, note that
$$
\langle p,p\rangle = p(1)^2 + p'(1)^2 + p''(1)^2 + p'''(1)^2 \geq 0
$$
And if $\langle p,p\rangle = 0$, then note that
$$
p(1) = p'(1) = p''(1) = p'''(1) = 0
$$
Now write
$$
p(t) = at^3 + bt^2 + ct + d
$$
and see that $a=b=c=d=0$
and conclude that
$$
p = 0
$$ |
H: Can $\Bbb N$ be topologized to be a compact Hausdorff space?
Can $\Bbb N$ be topologized to be a compact Hausdorff space?
I guess it might be a topology that is strictly finer than cofinite topology and strictly coarser than the topology consists of all infinite subsets of $\Bbb N$. But is there such a topology?
AI: Yes. Consider the ordinal $\omega+1$ (respect to the order topology.) |
H: Rotating and Scaling about centroid.
rotating
$x' = x\cos(\text{angle}) - y\sin(\text{angle})$
$y' = x\sin(\text{angle}) + y\cos(\text{angle})$
Scaling
$x' = x\cdot sx$
$y' = y\cdot sy$
but all formulas will doing about origin point. If i want to do about Centroid point.
(I have $(Cx,Cy)$ ). What the formulas will be.
Sorry, about english, I will practice more.
Thanks.
AI: $$
x' = C_x + (x - C_x) \cos(\theta) - (y - C_x) \sin(\theta)\\
y' = C_y + (x - C_x) \sin(\theta) + (y - C_x) \cos(\theta)
$$ |
H: Calculating residue in exponential fraction
I want to calculate the residue of $$\dfrac{e^{it}}{e^t+e^{-t}}$$ at $t=\pi i/2$.
To calculate the residue at $\pi i/2$, I write
$$\frac{e^{it}}{e^t+e^{-t}}=\frac{e^{it}e^t}{e^{2t}+1}=\frac{e^{it}e^t}{(e^t+i)(e^t-i)}$$so the residue is $$\frac{e^{i\pi i/2}e^{\pi i/2}}{(e^{\pi i/2}+i)}=\frac{e^{-\pi/2}i}{2i}=\dfrac{e^{-\pi/2}}{2}$$ Is this a correct way to calculate the residue? I just want to make sure I understand it correctly.
AI: You're off to a good start! From your work, you can see that $\pi i/2$ is a simple pole, so you can calculate the residue as
$$
\lim_{t \to \pi i/2} (t - \pi i/2) \left( \frac{e^{it}e^t}{(e^t+i)(e^t-i)} \right).
$$
From here it might help to notice that
$$
\frac{t-\pi i/2}{e^t - i} = \frac{t-\pi i/2}{e^t - e^{\pi i/2}}
$$
and think derivatives.
In general, if you have a simple pole at $t_0$ of a map $f$, then the residue of $f$ at $t_0$ can be calculated by
$$
\operatorname{Res}(f,t_0) = \lim_{t\to t_0} (t-t_0)f(t).
$$ |
H: Are there non commutative rings with no zero divisors?
If there are,
Are there unity (but not division) rings of this kind?
Are there non-unity rings of this kind?
Sorry, I forgot writting the non division condition.
AI: Take the "polynomials" with integer coefficients in two non-commuting variables $x$ and $y$. If you don't want a unit, use even integers only.
A related example replaces integer coefficients by coefficients in $\mathbb{Z}_2$. |
H: What is wrong with my reasoning about this conditional probability problem?
There is a box containing 6 red balls and 5 white balls. If the first drawn ball is red and not returned to the box, find the probability that a white ball is taken in the second drawn.
I have 3 solutions.The first two solutions produce the same but not for the last one. What is wrong with my reasoning?
Solution 1 (simple logic)
After the first drawn there are 5 red balls and 5 white balls. The probability to take a white ball is $5/10 = 1/2$.
Solution 2 (conditional probability)
Let
\begin{align*}
A&=\{(x,y)\mid x \text{ is a red ball and } y \text{ is any ball} \}\\
A\cap B &=\{(x,y)\mid x \text{ is a red ball and } y \text{ is white ball} \}\\
S&=\{(x,y)\mid x \text{ is any ball and } y \text{ is any ball} \}
\end{align*}
\begin{align*}
n(A)&=6\times 10\\
n(A\cap B)&=6\times 5\\
n(S)&=11\times 10
\end{align*}
\begin{align*}
P(A)&=\frac{6\times 10}{11\times 10}\\
P(A\cap B)&=\frac{6\times 5}{11\times 10}
\end{align*}
\begin{align*}
P(B\mid A)&=\frac{P(A\cap B)}{P(A)}\\
&= \frac{\frac{6\times 5}{11\times 10}}{\frac{6\times 10}{11\times 10}}\\
&=\frac12
\end{align*}
Solution 3 (permutation point of view)
All balls have numbers to uniquely identify each other. The first and second drawn balls are placed separately in a different place each.
The number of elements in the sample space is $11\times 10$ (there are 11 ways to occupy the first place and 10 ways to occupy the second place). The number elements of the space of event in question is $6\times 5$ (there are 6 ways for red balls to occupy the first place and 5 ways for the white balls to the second place). The probability is $\frac{6\times 5}{11\times 10} = \frac{3}{11}$.
AI: In your solution $3$ the sample space conists of only $6 \times 10$ possibilities, as the first ball must be red. |
H: Adjoining two primitive n-th roots
Let $\omega_n$ denote a primitive $n^{th}$ root of unity. If $m$ and $n$ are positive integers with $lcm(m,n)=k$, show that $\mathbb{Q}(\omega_n,\omega_m)=\mathbb{Q}(\omega_k)$.
To start, I am aware that $(\omega_n\omega_m)^k=1$, and so $o(\omega_n\omega_m)|k$, I am working towards showing that $o(\omega_n\omega_m)$ is in fact equal to $l$ (although I don't know this to be true at this point). This would show that $\mathbb{Q}(\omega_n,\omega_m)$ contains a primitive $k-th$ root of unity and thus all of $\mathbb{Q}(\omega_k)$. Any help would be appreciated, I think I am over thinking things (especially regarding the other direction of the inclusion).
AI: This may become obvious if you write everything additively instead of multiplicatively.
That is, you know the $k$-th roots of unity are a cyclic group under multiplication of order $k$, and so it is isomorphic to the integers modulo $k$ under addition. Use the isomorphism to translate the original problem into a question about integers modulo $k$. |
H: Is there an axiomatic definition of the concept "field equipped with a conjugation operator"?
In some sense, $\mathbb{C}$ is more than just a field, since aside from the usual field operations, it is also equipped with a conjugation operator $\mathbb{C} \rightarrow \mathbb{C}$.
This means a couple of things.
We can formulate the notion of an inner product space $V$ over $\mathbb{C}$, which includes the axiom
$$\langle x,y\rangle = \overline{\langle y,x\rangle}$$
If $U$ and $V$ are vector spaces over $\mathbb{C},$ and if $f : U \rightarrow V$ is a function, then not only can we make sense of the question "Is $f$ linear?" but also, we can make sense of the question: "Is $f$ conjugate-linear?"
So, I think the notion of "field equipped with a conjugation operator" is very important for linear algebra.
Question. Is there a structuralist/axiomatic definition of this concept? And what about the more general concept of a ring equipped with a conjugation operator? (so that may consider modules over such a ring, and conjugate-homomorphisms between them.)
AI: One can do things rather straightforwardly. A fancier statement might be to talk about a field along with an automorphism of the field which is an involution.
This generalizes easily to rings, and in fact the resulting theory is a variety of universal algebra, so the basics are rather well-understood.
This even has a name: a *-ring |
H: Sum of Powers of Coprime Ideals
Let $R$ be a commutative unital ring and let $I$, $J$ be ideals in $R$ such that $I+J=R$.
Show that $II+JJ=R$.
What I've tried:
Clearly $II+JJ \subseteq I+J =R$ as $I$ and $J$ are closed under addition and multiplication. So it suffices to show that $R \subseteq II+JJ$. Since $1\in R$, there exist $i_1 \in I, j_1 \in J$ such that $i_1+j_1 = 1$. Similarly, for each $r\in R$, there exist $i_r, j_r$ such that $i_r + j_r = r$. Now $r = r1= (i_r+j_r)(i_1+j_1)= i_ri_1+i_rj_1+j_ri_1+j_rj_1$. From here I would like to show $ i_rj_1+j_ri_1 \in II+JJ$.
Any suggestions?
AI: If $I^2+J^2$ is not equal to $R$ then it is contained in a maximal ideal, call it $M$. Then we have that $I^2\in M$ and $J^2\in M$.
Claim: If $P$ is a prime ideal and $AB\subset P$ where $A$ and $B$ are ideals, then we must have either $A\subset P$ or $B\subset P$.
Proof: You try it :)
Use the claim ($M$ is prime, why?) to show that since $I^2\subset M$, then we have $I\subset M$. Similarly $J\subset M$, how does this create a contradiction? |
H: Help with determining trigonometric limit
Use the relation $\lim_{\theta \to 0}\frac{\textrm{sin}\theta}{\theta}=1$ to determine the limit of $f(x)=\frac{\textrm{tan}(2x)}{x}$
I understand the identity $\textrm{tan}(2x)=\frac{\textrm{sin}(2x)}{\textrm{cos}(2x)}$.
So, $$\frac{\textrm{tan}(2x)}{x} = \frac{\frac{\textrm{sin}(2x)}{\textrm{cos}(2x)}}{x},$$
But then I don't understand this manipulation: $$(\frac{2}{2}) \frac{\frac{\textrm{sin}(2x)}{\textrm{cos}(2x)}}{x} = 2 (\frac{\textrm{sin}(2x)}{2x})(\frac{1}{\textrm{cos}(2x)})$$
Can you please show some in-between steps on how they arrived at this manipulation? I'm really not seeing how it works. Thank you.
AI: ${\tan 2x\over x}={1\over x}\times{\sin 2x\over \cos 2x}={\sin 2x\over 2x}\times{2\over \cos 2x}$
Now take $y=2x$ so $y\to 0$ as $x\to 0$ right?
so puting $y=2x$ in above expression and taking limit $y\to 0$ we get
$$\lim_{y\to 0}{\sin y\over y}\times {2\over \cos y}=1\times {2\over 1}=2$$ |
H: $\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$
Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$$ where $\left( \frac{a}{p}\right)$ is the Legendre symbol.
This seems to be a tricky one! I've tried using the property $\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)$ and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like $\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right]$ but that didn't bring anything either. And I've been looking for pairwise cancellation with $1$ term leftover, but that does not seem to work.
Can you help?
AI: Let $a^\ast$ be the inverse of $a$ modulo $p$. Then
$$\left(\frac{a(a+1)}{p}\right)=\left(\frac{a(a+aa^\ast)}{p}\right)=\left(\frac{a^2(1+a^\ast)}{p}\right)=\left(\frac{1+a^\ast}{p}\right).$$
As $a$ ranges from $1$ to $p-2$, the number $1+a^\ast$ ranges, modulo $p$, through the integers from $2$ to $p-1$. But the sum from $1$ to $p-1$ of the Legendre symbols is $0$, so our sum is $-1$. |
H: Show that $\lim_{x\to a^{+}} g(x) =g(a)$
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a bounded function and suppose that $g(x)=\sup_{t>x}f(t)$. Show that $\lim_{x\to a^{+}}g(x) = g(a)$ for all real $a$.
I have a hard time with these kind of proofs because in high school, I would simply replace $x$ by $a$ in $g(x)$ to get that $g(x) \rightarrow g(a)$. Obviously, Real Analysis requires a more formal proof.
I can always apply the definition of a limit of a function :
Given $\varepsilon>0$, there exists $\delta>0$ such that if $x\in A$ and $0<|x-c|<\delta$, then $|f(x)-L| < \varepsilon$.
In particular, we have that if
$$ |x-c| = |x-a^{+}|<\delta$$
then
$$|g(x)-g(a)| < \varepsilon$$
How can I find a relation between $f$ and $g$, the sup of $f(t)$ and between $x$ and $a^{+}$?
AI: It should be clear that $g$ is non-increasing. If $x<y$, then since $(x,\infty) \supset (y,\infty)$ we have $g(x) \ge g(y)$.
Now suppose $x_n \downarrow a$. We have $g(x_n) \uparrow \gamma$, and $g(x_n) \le \gamma \le g(a)$ (since $x_n > a$).
Suppose $\gamma < g(a)$, and let $\epsilon= \frac{1}{2}(g(a)-\gamma)>0$. Then there is some $x>a$ such that $g(x) > g(a)-\epsilon$ (by definition of $\sup$), and so for some $n$ we have $a \le x_n \le x$. Since $g$ is non-increasing we have $g(x_n) \ge g(x) > g(a)-\epsilon = \frac{1}{2}(\gamma+g(a)) \ge \gamma$, which is a contradiction. Hence $\gamma = g(a)$. |
H: Is this equation to prove that $aRb \iff a^2 - b^2 = 1$ is antisymmetric correct?
Over $\mathbb{R}$, $aRb \iff a^2 - b^2 = 1$.
I tried determining if it was antisymmetric. I seem to have done it, but while doing the equation, I stumbled upon a scenario that always made me doubt my decisions:
Have
$$aRb \land bRa$$
$$a^2 - b^2 = 1 \land b^2 - a^2 = 1$$
$$-a^2 + b^2 = -1 \land b^2 - a^2 = 1$$
$$-2a^2 + 2b^2 = 0$$
$$\color{#C00}{2(-a^2 + b^2) = 0}$$
$$\color{#C00}{-a^2 + b^2 = 0}$$
$$b^2 = a^2$$
$$b = a$$
I passed the $2$ to divide to the other side. The other side has a $0$, so the $2$ is essentially killed off with no apparent consequence.
Is this equation correct?
AI: The deduction you have does not work, as $a^2 = b^2$ does not imply $a = b$. It is actually much simpler. Assuming $aRb$ and $bRa$, you get (third line of equations) $-a^2 + b^2 = -1$ and $b^2 - a^2 = 1$. So $b^2 - a^2$ must be equal to $1$ and $-1$ at the same time, which is not possible. |
H: Residue of $f(z)=\frac{\cot(z)\coth(z)}{z^3}$
How can I find the residue of the following function at the point $z=0$
$$f(z)=\frac{\cot(z)\coth(z)}{z^3}$$
AI: Keep in mind that each of $\cot{z}$ and $\text{coth}{z}$ has a simple pole at $z=0$. Thus $z=0$ is a pole of order $5$, and the residue is equal to
$$\frac{1}{4!} \left [ \frac{d^4}{dz^4} \left ( z^2 \cot{(z)} \, \text{coth}(z) \right) \right ]_{z=0}$$
ADDENDUM
The above expression, while correct, is not really practical without access to a computer algebra software package. (I'm sure in the olden days, someone had to work out something like this by hand. But...why?) Better to work directly with the Taylor expansions of $z \cot{z}$, etc. about $z=0$ directly as follows:
$$z \cot{z} = 1-\frac{z^2}{3} - \frac{z^4}{45}+ O\left (z^6\right)$$
$$z \coth{z} = 1+\frac{z^2}{3} - \frac{z^4}{45}+ O\left (z^6\right)$$
So we want the coefficient of $1/z$ in the product of the above divided by $z^5$. This coefficient is straightforward as the sum of three terms:
$$-\frac{1}{45} - \frac{1}{45} - \frac13 \cdot \frac13 = -\frac{7}{45}$$
This of course agrees with the result of taking the 4th derivative above, but is much easier to see. |
H: limit of sequence of functions
Suppose
$$ f_n = \frac{1}{(1 + \frac{x}{n})^n x^{\frac{1}{n}}} $$
What is $\lim_{n \to \infty } f_n $ ??
I am having hard time with this sequence which seems like it is going to be something like the exponential, but I cannot see how to simplify it and make it look like the exponential. Any help would be appreciated.
AI: HINT:
Recall that if $\lim_{n\rightarrow\infty}a_n$ and $\lim_{n\rightarrow\infty}b_n$ both exist then $\lim_{n\rightarrow\infty}a_nb_n$=($\lim_{n\rightarrow\infty}a_n$)($\lim_{n\rightarrow\infty}b_n$) |
H: Given $f_{X,Y}(x,y)$, what is the pdf of $Z=XY$?
I approached the problem as following:
$f_{Z}(z) = \int_{0}^{\infty} f_{X,Y}(X=\frac{z}{y},Y=y)dy$
However, according to the textbook, the problem should be solved as below:
$f_{Z}(z) = \int_{0}^{\infty} f_{X}(X=\frac{z}{y}) f_{Y}(Y=y) dy$
Why do we use the marginal pdfs multiplied, instead of the joint pdf? They are equivalent when $X$ & $Y$ are independent, but when they are not doesn't it make more sense to use the joint distribution of $X$ & $Y$ in the integral, since that would account for any dependence?
I am looking to understand this conceptually. Any insights would be appreciated!
AI: Why do we use the marginal pdfs multiplied, instead of the joint pdf?
We should not (hence either your book is wrong or you took a formula valid in the discrete setting for a formula valid in the continuous setting).
In full generality, if $Z=XY$, then
$$
f_Z(z)=\int f_{X,Y}(y^{-1}z,y)y^{-1}\mathrm dy.
$$
In your case,
$$
f_Z(z)=\int_0^\infty y^{-2}z\mathrm e^{-z-z/y}\mathrm dy\,\mathbf 1_{z\gt0}=\left[\mathrm e^{-z-z/y}\,\mathbf 1_{z\gt0}\right]_{y=0}^{y=+\infty}=\mathrm e^{-z}\mathbf 1_{z\gt0}.
$$ |
H: Extending an embedding $:S^1\rightarrow \mathbb R^{n}$
Assume we have an embedding $f:S^1\rightarrow \mathbb R^n$. I want to extend $f$ to an embedding $\tilde{f}:B\rightarrow \mathbb R^n$, where $B$ is the closed unit ball of $\mathbb R^2$. In fact, I want to see $f(S^1)$ as the boundary of a smooth manifold. I would appreciate some ideas on the construction of $\tilde{f}$.
Edit: I am thinking of contracting $f$ to a point $w_0\in \mathbb R^n$, let's say through the homotopy $F:S^1\times [0,1]\rightarrow \mathbb R^n$, with $F(x,1)=f(x)$ and $F(x,0)=w_0$. Now I define $\tilde f(x)=F(\frac{x}{|x|},|x|)$ for $x\in B$. However, how can I make sure that $\tilde{f} (B)$ is a manifold?
AI: If $n=3$ what you want to prove is not true. The only knot which bounds a disk is the unknot. The page about Seifert surfaces at Wikipedia has details.
When $n>3$, you can use the general position theorem stated here to conclude that all $S^1$s in $\mathbb R^n$ are isotopic (smoothly if the knot is smooth) and reduce the problem to the standardly embedded unknot, which is easy.
One should be able to prove this «by hand»: if the knot $K$ is smooth, there is a point $p$ such that the lines through $p$ pass through at most one point of $K$. Indeed, the set $S$ of points on the lines which pass through two points of $K$ is the image of a smooth map $U\to\mathbb R^n$ with $U\subseteq S^1\times S^1\times\mathbb R$ an open set, so $S$ is not all of $\mathbb R^n$: it suffices to pick a $p$ in $\mathbb R^n\setminus S$. Now you can construct a "cone" with vertex at $p$ and base the knot. This is not smooth, but it becomes smooth if you smooth it at the apex. |
H: How to find all surjective functions $f:M_n(\Bbb R)\to\{0,1,2,\cdots,n\}$ satisfying $f(XY)\le\min{(f(X),f(Y))}$
Let $M_n(\Bbb R)$ be the set of all real $n\times n$ matrices. Find all surjective functions $f:M_n(\Bbb R)\to\{0,1,2,\cdots,n\}$ such that
$$f(XY)\le\min{(f(X),f(Y))}$$
for all $X,Y\in M_n(\Bbb R)$.
My try: since
$$f(XY)\le\min{(f(X),f(Y))}$$
so we have
$f(XY)\le f(X)$ and $f(XY)\le f(Y)$, but I can't proceed.Thank you.
AI: Claim: $f(X)={\rm rank} X$, $~\forall X\in M_n(\Bbb R)$( $\Bbb R$ can be replaced with any field).
Hint:
Show that $f(I)=n$ by letting $Y=I$.
Show that if $X$ is invertible($~{\rm rank} X=n$), then $f(X)=n$ by letting $Y=X^{-1}$ and using 1.
Show that if $X$ and $Z$ are invertible, then $f(XYZ)=f(Y)$ by using 2.
Show that if ${\rm rank} X={\rm rank} Y$, then $f(X)=f(Y)$ by using 3.
Show that if ${\rm rank X}\le {\rm rank Y}$, then $f(X)\le f(Y)$ by using 4.
Make the conclusion from 4, 5 and the surjectivity of $f$. |
H: Help with numerical analysis proof
Let $u$ be a nonzero vector in $\mathbb{R}^n$, and define $\gamma=\frac{2}{||u||_2^2}$ and $Q=I-\gamma uu^T$.
Prove Q is a reflector satisfying
A) $ Qu=-u$
B) $Qv=v$ if $<u,v>=0$
My approach: I'm letting some $$\hat u=\frac{u}{||u||_2}$$ so that $||\hat u||=1$.
I don't know where this is getting me though I'm missing a few details to prove A and B. Any help would be appreciated. Thanks guys.
AI: $Qu = u - \frac{2}{||u||_2^2}uu^Tu = u - \frac{2}{||u||_2^2}u||u||_2^2 = u-2u = -u$
$Qv = v-\gamma uu^Tv = v-\gamma u \langle u,v \rangle = v$, if $\langle u,v \rangle = 0$
and because by definition $||u||_2^2 = \langle u,u \rangle = u^Tu$. |
H: Prove $h(x)=\sqrt{x^2-1}$ continuous by $\epsilon,\delta$
Proof:
Let $h\colon (1, \infty)\to \Bbb R$ be a function. Let $h(x)= \sqrt{x^2-1}$.
Let $\epsilon>0$ be arbitrary.
Let $x_0\geq 1$ be arbitrary.
Suppose $x_0 > 1$.
Let $$\delta=\min\left\{1, \frac{\epsilon\sqrt{x_0^2-1}}{2|x_0|+1}\right\}$$
Let $x\geq1$ and $|x-x₀|<\delta$.
Thus $|h(x)-h(x_0)|=\left|\sqrt{\vphantom{x_0^2}x^2-1}-\sqrt{x_0^2-1}\right|={?}$ That's where I get stuck.
Suppose $x_0=1$.
I don't know what $\delta$ should equal.
Let $x\geq1$ and $|x-x_0|<\delta$.
Thus $|h(x)-h(1)|=\left|\sqrt{x^2-1}-\sqrt 0\right|=\left|\sqrt{x^2-1}\right|=\sqrt{|x^2-1|}={?}$ And that's where I get stuck again.
It would end with: Since $\epsilon$ is arbitrary, $h$ is continuous at $x₀$. And since $x_0$ is arbitrary, $h$ is continuous for all $x\in (1, \infty)$
AI: rationalize by $|\sqrt{x^2-1}+\sqrt{x_0^2-1}|$
so,
$$ |\sqrt{x^2-1}-\sqrt{x_0^2-1}| = |\sqrt{x^2-1}-\sqrt{x_0^2-1}|\frac{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$
$$ = \frac{|(x^2-1)-(x_0^2-1)|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|} $$
$$\leq \frac{|x^2 - x_0^2|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$
$$\leq |x-x_0|\frac{|x|+|x_0|}{|\sqrt{x^2-1}+\sqrt{x_0^2-1}|}$$
$$\leq |x-x_0|$$
Where the last inequality is becuase on $(1,\infty)$ the denominator is always positive. So given $\epsilon>0$, let $\delta = \epsilon$, so that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$ |
H: Finding a vector that is in two subspaces
The following question has been posed to me:
You are given two subspaces $U$, $V$, both in $\mathbb{R}^n$, each with a basis of column vectors forming the columns of the respective matrices $U$, $V$. Find a vector $\vec{x}$ in both subspaces $U$, $V$. That is, find a vector $\vec{x}$ that can be written as a linear combination of the basis vectors for $U$ and as a linear combination of the basis vectors for $V$. That means $\vec{x} = U\vec{a} = V\vec{b}$ for some coefficient vectors $\vec{a}$, $\vec{b}$. Do this for the following two examples:
$$U = \begin{bmatrix}1&0\\-2&1\\-3&2\\-1&2\end{bmatrix}, V = \begin{bmatrix}0&4\\2&-8\\7&-6\\5&-2\end{bmatrix}$$
I have tried setting up a system of equations by distributing out the $a$s from $\vec{a}$, and the $b$s from $\vec{b}$, and ended up with the system that went
$$a_1 = 4b_2$$
$$-2a_1 + x_2 = 2y_1 - 8y_2$$
$$...$$
and so on, but this lead me to a dead end; they seem to me to be inconsistent unless the vector $\vec{x}$ is the zero vector.
The zero vector seems like it technically should work - It's a vector that's by definition in both subspaces and so can be written as a linear combination of those two, but it seems too trivial of an answer. Am I missing something, or is it really that simple?
AI: Note that $U (-4,4)^T + V (-2,1)^T = 0$.
In particular, $U (-4,4)^T = (-4,12,20,12)^T = V(2,-1)$. |
H: Prove $\left\{\frac{n}{2n+3}\right\}$ and $\left\{\frac{n}{2n-3}\right\}$ converge?
Question: prove that the sequences $\left\{\frac{n}{2n+3}\right\}$ and $\left\{\frac{n}{2n-3}\right\}$ converge using the definition.
What I have: I know both of them have limit $1/2$.
The definition says: a sequence $\{x_n\}$ is said to converge to a number $x \in \mathbb R$, if for every $\epsilon >0$, there exists an $M \in \mathbb N$ such that $|x_n-x|<\epsilon$ for all $n\geq M$.
Start: let $\epsilon >0$ such that.... Where do I go from here?
AI: Let $\varepsilon >0$. We want do show the existence of a number $M\in \mathbb{N}$ such that $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$.
$$\left|\frac{n}{2n+3}-\frac{1}{2}\right|=\frac{3}{2(2n+3)}<\varepsilon\iff \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right)<n$$
so if we choose $M> \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right),$ we get $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$. |
H: Find $x$ for inequality of $1+x+x^{2}+x^{3}+...+x^{99}\le0$
Finding the range of $x$ for inequality of $1+x+x^{2}+x^{3}+...+x^{99}\le0$
AI: We observe that $x=1$ is not a solution. Now note that $1+x+...+x^{99}=\frac{1-x^{100}}{1-x}$ (show this!), so that the inequality becomes $\frac{1-x^{100}}{1-x}\ge 0$. From here we observe if $x>1$, the bottom part is negative. We also have the case $x<1$. These you should be able to do.
Good luck! |
H: Calculating integral with standard normal distribution.
I have a problem to solving this,
Because I think that for solving this problem, I need to calculate cdf of standard normal distribution and plug Y value and calculate.
However, at the bottom I found that Integral from zero to infinity of 1 goes to infinity and I cannot derive the answer. Can you tell me what's the problem and what can I do?
I appreciate for your help in advance:)
AI: You have managed to state a closed form for $F_x(x)$.
This is not in fact possible for a normal distribution, so you have an error in your integral about a third of the way down: the integral of $\displaystyle e^{-\frac12 t^2}$ is not $\displaystyle -\frac{1}{t} e^{-\frac12 t^2}$
There is an easier solution, but it uses a shortcut which your teacher might not accept here:
For a non-negative random variable, you have $E[Y] = \displaystyle\int_{t=0}^\infty \Pr(Y \gt t) dt$.
$X^2$ is indeed non-negative, so you want $E[X^2]$.
For $X$ with a standard $N(0,1)$ Gaussian distribution, $E[X^2]=1$. |
H: Proof that two norms $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ are equivalent
Two norms $\def\norm#1{\lVert#1\rVert}\norm\cdot_1$ and $\norm\cdot_2$ are equivalent iff $\;\exists\;c_1,c_2>0$ such that $c_1\norm x_1\le \norm x_2\le c_2\norm x_1$
Show that $\norm x_1=\sum_{i=1}^n \lvert x_i \rvert$ and $\norm x_2=\left ( \sum_{i=1}^n x_i^2 \right )^{1/2}$ are equivalent.
It looks like $c_2$ is $1$, and that this can be proven with induction. But what could $c_1$ be?
edit
I mean what I don't understand is the following: If I square both terms, and expand $(\sum_{i=1}^n \lvert x_i \rvert)^2 = \left ( \sum_{i=1}^n x_i^2 \right ) + \sum_{i=1}^n x_i(x_k+\dotsb+x_n)_{x_k\neq x_i}$. However, the second term grows with $n$, so how can $\frac{\norm x_1}{\norm x_2} \leq C$ at all?
AI: You want to show that for a fixed $n$ the norms $\|\|_1,\|\|_2$ are equivalent in $\mathbb R^n$. The constant $c_1$ will depend on $n$.
Hint:
Cauchy Schwarz inequality. |
H: Is $A$ s.t $A_{i, j} = x^T_i x_j$ semi-positive definite?
Let $x_1, x_2, \ldots, x_k \in \mathbb{R}^n$ and set define a $k$ by $k$ matrix $A$ by setting $A_{i, j} = x^T_i x_j$. Is $A$ semi-positive definite? If so, how can I show it?
AI: Let $t:=(t_1,\dots,t_k)\in\mathbb R^k$. Then
$$t^TAt=\sum_{i,j=1}^kx_i^Tx_jt_it_j=\sum_{i,j=1}^k((t_ix_i)^T)\cdot (t_jx_j)=\left(\sum_{i=1}^k(t_ix_i)^T\right)\cdot\left(\sum_{j=1}^kt_jx_j\right),$$
which is the square of the Euclidian norm of $\sum_{j=1}^kt_jx_j$. |
H: How many different ways can a number be expressed as a sum of any number of integers when order matters?
How many different ways can a number $n \in \mathbb{N} $ be expressed as a sum of any number of positive numbers when order matters?
My solution:
Since I know, that $n$ can be represented as a sum of $k$ positive integers ${{n-1}\choose{k-1}}$ different ways, and $n$ can be represented as a sum of at most $n$ positive integers (the case where they are all $1$), then we can present $n$ as a sum of of any number of integers $\sum\limits_{k=1}^{n} {{n-1}\choose{k-1}}$ different ways.
AI: Your solution is correct. $n \in \mathbb{N}$ can be expressed as the sum of a sequence of natural numbers in $$\sum\limits_{k=1}^{n}{n-1 \choose k-1} = 2^{n-1}$$ different ways. This is called a composition of an integer $n$. See Wikipedia Article for more information. |
H: Existence of $x \in E$ such that $\sup(E)-\epsilon
If $\alpha=\sup\left(E\right)$ exists, I often encounter the argument
that there exists a point $x\in E$ such that
$\sup\left(E\right)-\epsilon<x\leq\sup\left(E\right)$
I realize that if $\sup\left(E\right)\in E$, then I can just chose
$x=\sup\left(E\right)$ and the argument would be valid.
But what if $\sup\left(E\right)\notin E$? Is it then true that, there exists
a point $x\in E$ such that
$\sup\left(E\right)-\epsilon<x<\sup\left(E\right)$
?
Related to the question: Are the following two statements correct?
$\alpha=\sup\left(E\right)\in E\implies\alpha\text{ can be either an isolated point or a limit point of }E$
$\alpha=\sup\left(E\right)\notin E\implies\alpha\text{ can only be a limit point of }E$
AI: Suppose no such $x$ existed. Would this be consistent with the definition of $\sup(E)$ as the least upper bound?
And indeed, your assessment of the limit point/isolated point properties of $\sup(E)$ is correct. |
H: Evaluating $\lim\limits_{n\to \infty}\left\{(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})\right\}^{\frac{1}{n}}$
Question is to evaluate :
$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$
I tried to do something like this but it is not getting better...
$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})=(\frac{n+1}{n})(\frac{n+2}{n})\cdots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$
please give some hint to proceed further.
AI: Consider the log of $[(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}$:
$$\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right).$$
The limit of this sum is equal to $\int_0^1\ln(1+x)dx$, so
$$
\begin{aligned}
\lim_{n\to\infty} [(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}
&= \lim_{n\to\infty} \exp\left(\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp\left(\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp \left( \int_0^1\ln(1+x)dx\right)
\end{aligned}$$ |
H: $\lambda_k \to 0$ implies $T$ is compact?
I am doing an exercise which asks to show that if $\{\varphi_k\}$ is an orthonormal basis in a Hilbert space with $T$ a bounded operator satisfying $T\varphi_k = \lambda_k \varphi_k$, then $\lambda_k \to 0$ implies that $T$ is compact. Now I will be done if I can show that $TP_n \to T$ in norm for $P_n$ is a compact operator. Here $P_n$ is the orthogonal projection onto the linear span $\{\varphi_1\ldots,\varphi_n\}$.
However the hint given is that $\lambda_k\to 0$ already implies this. Why is this so? I guess if I can understand why $||T|| = \sup_k |\lambda_k|$ then I would be done. For I believe $||P_nT - T|| = \sup_{k\geq n} |\lambda_k|$ and then $\lim_{n\to \infty} ||P_nT - T|| = \limsup_{n\to \infty} |\lambda_n| = 0.$
My question is: Why is $||T|| = \sup_{k} |\lambda_k|$?
AI: Any vector $x$ can be expanded in the orthonormal basis as $x = \sum_k x_k \varphi_k$, and any vector expressed in such a way has norm given by $\Vert x \Vert^2 = \sum_k |x_k|^2$. Thus $$ \Vert T(x) \Vert^2 = \Vert \sum_k x_k T(\varphi_k) \Vert^2 = \Vert \sum_k x_k\lambda_k \varphi_k\Vert^2 \\= \sum_k |x_k \lambda_k|^2 \le \sum_k\sup_j|\lambda_j|^2x_k^2 = \sup_k|\lambda_k|^2 \lVert x \rVert^2.$$
This shows $\Vert T \Vert \le \sup_k |\lambda_k|$. For the other direction, note that $\Vert T(\varphi_k)\Vert= \Vert \lambda_k \varphi_k \Vert = |\lambda_k|$ implies $\Vert T\Vert\ge |\lambda_k|$ for every $k$; i.e. $\Vert T\Vert \ge \sup_k |\lambda_k|$. |
H: Solving $\int_{-\infty}^{\infty}\frac{x^2e^x}{(1+e^x)^2}dx$
I am attempting to use residues to solve $\int_{-\infty}^{\infty}\frac{x^2e^x}{(1+e^x)^2}dx$; the answer is $\frac{\pi^2}{3}$. I have tried to split $\frac{x^2e^x}{(1+e^x)^2}$ into two parts
$$\frac{x^2}{e^x+1}-\frac{x^2}{\left(e^x+1\right)^2},$$
however no progress was to be made. I could not solve
$$
\int_{-\infty}^{\infty}\frac{z^2}{e^z+1}-\frac{z^2}{\left(e^z+1\right)^2}dz.
$$
any easier than the problem at hand. As a matter of fact neither of these two integrals converge individually. The way I approached the problem was by finding the singularities, namely when $e^z=-1$. So I got, when we let $z=x+iy$, that $z_k=i\pi+2\pi k$ for $k\in\mathbb{Z}$. This is my attempt,
$$
\int_{C}\frac{z^2e^{z}}{(e^z+1)^2}=2i\pi\sum_{\forall n}(Res(f,z_n)).
$$
Since $\{-1\}$ is a second order pole, our residue is given by
$$
\lim_{z\to z_k}\frac{d}{dz}[(z-z_k)^2\frac{z^2e^{z}}{(e^z+1)^2}].
$$
I found these with the aid of Mathematica. However if we were to calculate all the residues we would get an alternating sequence, if arranged properly. It becomes more apparent when we solve the equation that is split. For instance if we take $z_{-4}$ we get
$$
2 \pi i \left(\lim_{z\to z_{-4}} \, \frac{z^2 (z-z_{-4})}{e^z+1}\right)\ \text{and}\ 2 \pi i \left(\lim_{z\to z_{-4}} \, \frac{\partial }{\partial z}\frac{\left(z-z_{-4}\right){}^2 \left(-z^2\right)}{\left(e^z+1\right)^2}\right)
$$
or,
$$
98i\pi^3 \text{ and } 14 i (-7 \pi +2 i) \pi ^2.
$$
So the total contribution would be $-28 \pi ^2$. Were as if we take $z_4$ we would get
$$
162 i \pi ^3 \text{ and } -18 i \pi ^2 (9 \pi +2 i)
$$
and the contribution would be $36 \pi ^2$.
One dilemma is finding the summation of all these residues. Is my work flawed? Is there a better contour to make this problem work? Thank you for the assistance.
AI: $\newcommand{\+}{^{\dagger}}%
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\newcommand{\yy}{\Longleftrightarrow}$
This integral is an usual one in Statistical Mechanics $\pars{~\mbox{the integrand is an}\ {\large\tt\mbox{even}}\ \mbox{function}~}$:
\begin{align}\color{#0000ff}{\large%
\int_{-\infty}^{\infty}\!\!{x^{2}\expo{x}\,\dd x \over \pars{\expo{x} + 1}^{2}}}
&=
-2\int_{0}^{\infty}x^{2}\,\totald{}{x}\pars{1 \over \expo{x} + 1}\,\dd x
=
4\int_{0}^{\infty}{x \over \expo{x} + 1}\,\dd x
\\[3mm]&=
4\int_{0}^{\infty}x\pars{{1 \over \expo{x} + 1} - {1 \over \expo{x} - 1}}\,\dd x
+
4\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
\\[3mm]&=
-8\int_{0}^{\infty}{x \over \expo{2x} - 1}\,\dd x
+
4\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
\\[3mm]&=
-2\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
+
4\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
=
2\int_{0}^{\infty}x\expo{-x}\,{1 \over 1 - \expo{-x}}\,\dd x
\\[3mm]&=
2\int_{0}^{\infty}x\expo{-x}\sum_{\ell = 0}^{\infty}\expo{-\ell x}\,\dd x
=
2\sum_{\ell = 0}^{\infty}\int_{0}^{\infty}x\expo{-\pars{\ell + 1}x}\,\dd x
\\[3mm]&=
2\sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1}^{2}}\
\overbrace{\int_{0}^{\infty}x\expo{-x}\,\dd x}^{=\ 1}
=
2\ \overbrace{\sum_{\ell = 1}^{\infty}{1 \over \ell^{2}}}^{\ds{=\,\pi^{2}/6}}
=
\color{#0000ff}{\large{\pi^{2} \over 3}}
\end{align} |
H: Bijection map from a set of subgroup to another set of subgroup under some condition.
Let $G$ be a finite group. Let $N \trianglelefteq G$ and $U \leq G$ such that $G = NU$. Then there exists a bijection, preserving inclusion, from the set of subgroups $X$ satisfying $U ≤ X ≤ G$ to the set of $U-$invariant subgroups $Y$ satisfying $U \cap N ≤ Y ≤ N$.
This is a problem from Kurzweil-Stellmacher book. I dont know how to do it. Thanks for any help.
AI: Hint: you should use Dedekind's Lemma:Let $U$ and $V$ be subgroups of a group $G$ and let $U \subseteq X \subseteq G$, with $X$ also a subgroup. Then $U(V \cap X)=UV \cap X$.Note that $UV$ here is just a subset of $G$, not necessarily a subgroup! Anyway, in your case define a map $\phi$ from the set {$X$: $X$ is a subgroup of $G$ with $U \subseteq X \subseteq G$} to the set {$Y$: $Y$ is a $U$-invariant subgroup of $N$ with $N \cap U \subseteq Y \subseteq N$} by $\phi:X \mapsto X \cap N$. Observe that $\phi$ is well-defined by the normality of $N$. The injectivity of this map can be proved with Dedekind's Lemma. For the surjectivity look at an $X=YU$, of which you have to show it is a subgroup (remember $Y$ was $U$-invariant!) and again apply the lemma. |
H: prove if $(A_n)$ limit is $L$ then $(A_n)^2$ limit is $L^2$
Hello,
What I need to prove is: if $(A_n)$ limit is $L$ then $(A_n)^2$ limit is $L^2$.
I've added my attempt to prove it. I got stuck so I'm guessing I'm missing something here.
help will be appreciated :)
AI: $f(x)=x^2$ is continuous, which by definition in equivalent to $$\forall \varepsilon>0 \exists\delta>0 \hbox{ s.t.} : |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon.$$
By the assumption that $(A_n)$ converges to $L$ you got that $\forall \varepsilon_1>0 \exists k\in \mathbb{N} $ s.t. : $\forall n\geq k \implies |A_n -L|<\varepsilon_1.$
Now you are so use these two statements in order to show that
$$
\forall \varepsilon >0 \exists k \in \mathbb{N} \hbox{ s.t} : \forall n\geq k \implies |f(A_n)-f(L)|<\varepsilon
$$
If you cant use that $f(x)=x^2$ is continuous, it goes as follows:
fix $\varepsilon>0$.
$$
|A_n^2-L^2|=|(A_n-L)(A_n+L)|=|A_n-L||A_n+L|=|A_n-L||A_n-L+2L|$$
$$
\leq |A_n-L|(|A_n-L|+2|L|)
$$
Now fix $k_1$ s.t. $\forall n\geq k_1 $ : $|A_{n}-L|< \varepsilon_1$, and $k_2$ s.t. $\forall n\geq k_2$ : $|A_{n}-L|< 1.$ (note that this can be done because of the assumption, that $A_n$ converges to $L$)
Now for all $n\geq \max\{k_1,k_2\}=h$ we got that
$$
|A_n^2-L^2| \leq |A_n-L|(|A_n-L|+2|L|) < \varepsilon_1|(1+2|L|)
$$
If we started to set $\varepsilon_1=\varepsilon / (1+2|L|)$ we would get that $|A_n^2-L^2|<\varepsilon$.
This was done for an arbitrary $\varepsilon>0$, thus it must hold that
$$\forall \varepsilon>0 \exists h\in\mathbb{N} \text{ s.t. } \forall n\geq h : |A_n^2-L^2| < \varepsilon
$$ |
H: Injective $\alpha:A \to B$ has surjective $\beta: B \to A$ such that $\alpha\beta = {\rm id}_A$
Let $\alpha : A \to B$ be an injective function. Show that there is a surjective function
$\beta: B\to A$ such that $\alpha; \beta = {\rm id}_A$.
AI: For every $y$ that belongs to the image of $\alpha$ there is exactly
one $x\in A$ with $\alpha\left(x\right)=y$. For such $y$ define
$\beta\left(y\right)=x$. Pick some $x_{0}\in A$ and define $\beta\left(y\right)=x_{0}$
for every $y$ that is not in the image of $\alpha$. It is evident
that $\beta\circ\alpha=1_{A}$ (not $\alpha\circ\beta$ as you mention
in your question).
Caution: If $A=\emptyset$ then there is a unique function $\alpha:\emptyset\rightarrow B$
that is vacuously injective. However, if $B\neq\emptyset$ then there
are no functions from $B$ to $\emptyset$. So this only works under
the extra (and mild) condition that $A\neq\emptyset\vee B=\emptyset$. |
H: Calculate a total percentage based on individual percentages but without original values.
(I hope the question/title made sense.)
Let's say I have the following list:
+----------+--------------+------------+-----------+
| Item | Expected | Actual | % |
+----------+--------------+------------+-----------+
| Item A | 141068 | 45714 | 32.41% |
| Item B | 44942 | 20844 | 46.38% |
| Item C | 193052 | 50597 | 26.21% |
| Item D | 67152 | 27084 | 40.33% |
+----------+--------------+------------+-----------+
| Total | 746214 | 144239 | 19.33% |
+----------+--------------+------------+-----------+
The way for calculating $total\ \%$ is pretty straight forward.
But is it possible to figure out the $total\ \%$ without knowing the $Expected$ And $Actual$ values?
ie. is there a formula for "summing" up the percentages, without actually adding them together?
Visual illustration:
+----------+--------------+------------+-----------+
| Item | Expected | Actual | % |
+----------+--------------+------------+-----------+
| Item A | ?? | ?? | 32.41% |
| Item B | ?? | ?? | 46.38% |
| Item C | ?? | ?? | 26.21% |
| Item D | ?? | ?? | 40.33% |
+----------+--------------+------------+-----------+
| Total | ?? | ?? | 19.33% | <-- Can I somehow find this number?
+----------+--------------+------------+-----------+
AI: No, there is not. To see this, consider a simpler example. Suppose that there are two items, each at $50\%$. The following three arrays show clearly that the total percentage depends heavily on the amounts: it’s a weighted average of the individual percentages, so it depends both on those percentages and on their weights, which in turn depend on the absolute quantities involved.
$$\begin{array}{rrr}
100&10&10.00\%\\
100&90&90.00\%\\ \hline
200&100&50.00\%
\end{array}$$
$$\begin{array}{rrr}
100&10&10.00\%\\
500&450&90.00\%\\ \hline
600&460&76.67\%
\end{array}$$
$$\begin{array}{rrr}
100&10&10.00\%\\
1000&900&90.00\%\\ \hline
1100&910&82.73\%
\end{array}$$
There is one situation in which you can infer the total percentage: if the individual percentages are all the same, the total percentage will agree with them. It’s also the case that the overall percentage must lie between the smallest and the largest of the individual percentages, so if that range is narrow, you do have a good estimate of the overall percentage. |
H: About vector bundles on algebraic varieties
If $X$ is an irreducible algebraic variety (over $\mathbb C$), an algebraic vector bundle of rank $r$ over $X$ is a couple $(E,\pi)$ where $E$ is an algebraic variety and $\pi: E\longrightarrow X$ is a surjective morphism, with the following properties:
1) $\pi^{-1}(x)$ is a vector space isomorphic to $\mathbb C^r$ for every $x\in X$
2) For every $x\in X$ there are an open neighborhood $U$ and an isomorphism of varieties $\phi_U:\pi^{-1}(U)\longrightarrow U\times\mathbb C^r$ such that:
2a) $ \pi_1\circ\phi_U=\pi$ where $\pi_1:U\times\mathbb C^r\rightarrow U$ is the canonical projection on $U$.
2b) $\phi_U|_{\pi^{-1}(x)}:\pi^{-1}(x)\longrightarrow \{x\}\times\mathbb C^r$ is an isomorphism of vector spaces.
One can also construct an algebraic vector bundle by an open cover $\{U_\alpha\}$ of $X$ and some functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\longrightarrow GL_r(\mathbb C)$ which satisfy some cocicle conditions. Now my question is the following:
Are the functions $g_{\alpha\beta}$ morphisms between varieties? (remember that $GL_r(\mathbb C)$ is an open affine subvariety of $\mathbb C^{r^2}$)
For example Cox in his book on Toric varieties says only that $g_{\alpha\beta}$ is a function. But for smooth manifolds, $g_{\alpha\beta}$ must be a smooth function, so by analogy I think that in the algebraic case $g_{\alpha\beta}$ should be a morphism.
AI: Yes.
First of all, this is not quite the correct defintion. The vector space structure of the fibers (more generally, on the sections) is not a property. It is an additional structure which belongs to the vector bundle. The modern definition of a vector bundle (which works for manifolds, analytic spaces and schemes at once, in fact arbitrary ringed spaces) is that of a locally free sheaf of modules. Actually this definition simplifies lots of constructions for vector bundles. And it also makes clear what is part of the structure (the sheaf of modules) and what is a property (local triviality). $(\star)$
If $X$ is a ringed space and $F$ is a vector bundle of rank $n$ on $X$, then the definition tells us that there is an open covering $X = \cup_i X_i$ such that $F|_{X_i}$ is free of rank $n$, i.e. isomorphic to $\mathcal{O}_{X_i}^n$. On the intersections we get an automorphism of the sheaf of modules $\mathcal {O}_{X_i \cap X_j}^n$. This can be desribed as an element of $\mathrm{GL}_n(\mathcal{O}(X_i \cap X_j))$. Just for the sake of completeness (skip this if you don't know cohomology yet), this cocycle description may be refined and summarized as $\mathrm{Vect}_n(X) \cong \check{H}^1(X,\mathrm{GL}_n)$.
If $X$ is a scheme (or just a variety), then there is an isomorphism
$$\mathrm{GL}_n(\mathcal{O}(X)) \cong \mathrm{Hom}_{\mathrm{Sch}}(X,\mathrm{GL}_n).$$
Actually this could serve as a functorial defintition of the (group) scheme $\mathrm{GL}_n$. More concretely, $\mathrm{GL}_n$ is the spectrum of $k[\{x_{ij}\}][\mathrm{det}^{-1}]$ if $k$ is our base ring which I have omitted so far from the notation.
Of course, $\mathrm{Hom}_{\mathrm{Sch}}(X,\mathrm{GL}_n)$ differs a lot from the (huge and uninteresting) set of all maps of the underlying sets $|X| \to |\mathrm{GL}_n|$. Hence, cocycles $X_i \cap X_j \to \mathrm{GL}_n$ should be morphisms of schemes (or varieties if $X$ is a variety).
You can also specialize to topological vector bundles: here $X$ is a topological space equipped with the sheaf of continuous functions, say with values in $\mathbb{C}$. The cocycles should be continuous maps $X_i \cap X_j \to \mathrm{GL}_n(\mathbb{C})$, not just arbitrary maps. For holomorphic vector bundles, take holomorphic cocycles, etc.
$(\star)$ The connection to the classical / geometric vector bundles is given as follows: If $F$ is a sheaf of modules on a scheme $X$, then $V(F) := \mathrm{Spec}(\mathrm{Sym}(F^*))$ is a scheme which is affine over $X$. For $F=\mathcal{O}_X^n$ we have $V(F)=\mathbb{A}^n_X$. Thus, if $F$ is locally free, we have that $V(F)$ is locally trivial in the usual sense. |
H: Find all positive integers m, n, p such that $(m+n)(mn+1)=2^p$
Find all positive integers m, n, p such that
$$(m+n)(mn+1)=2^p$$
Please give me some hints
Thanks
AI: We have $m+n=2^a$ and $mn+1=2^b$ with $a+b=p$. First note that both $m$ and $n$ need to be odd.
First case: Suppose $m=1$ or $n=1$, then the other is equal to $2^{p/2}-1$ and this is a solution for even $p$.
Now suppose $m>1$ so $b>a>1$. Adding the two equations above we get
$(m+1)(n+1)=mn+1+m+n=2^a(2^{b-a}+1)$. Let $m+1=2^xw$ and $n+1=2^zy$ with $w,y$ odd, $x,z>0$ and $x+z=a$. Then $2^a=2^xw+2^zy-2$, so one of $x$ or $z$ needs to be $1$ (if both were $>1$ the RHS would be $\equiv 2\mod 4$.
Case $x=1$: We have $x+z=a$, so $z=a-1$. The equation for $2^a$ becomes
$$
2^a=2w+2^{a-1}y-2\geq 2^{a-1}y.
$$
This implies $y=1$, since it is odd. We get $m-1=2w-2=2^{a-1}$ and $n+1=2^z=2^{a-1}$ and indeed this gives a solution.
Case $z=1$: The same just with $m$ and $n$ swapped.
So the only solutions are $m=1$ and $y=2^{p/2}-1$ and
$(m,n)=(2^k\pm 1,2^k\mp 1)$ |
H: Contraction mapping- Help needed
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a)<0$ and $F(b)>0$ and
$$
0<K_1\leq F'(x)\leq K_2\quad (a\leq x\leq b)
$$
Find the unique root of equation $F(x)=0$.
The given hint is to use the contraction mapping theorem i.e., if $f(.)$ is a contraction mapping it has a fixed point. Define $f(x)=x-\lambda F(x)$
$$
|f(x)-f(y)|=\Bigg|(x-y)\Big(1-\lambda\frac{F(x)-F(y)}{x-y}\Big)\Bigg|=|x-y|\Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big|
$$
So if I can show $ \Big|1-\lambda\frac{F(x)-F(y)}{x-y}\Big|$ is smaller than $1$, I can say $f(x)=x$. However, I can't proceed from thereon. Many thanks for any help!
AI: There are two things you have to impose on $\lambda$ in order to apply the Banach fixed point theorem (and the unique fixed point of $f$ that the theorem guarantees is the unique zero of $F$),
you must choose $\lambda$ so that there is a $c < 1$ with $$\left\lvert1 - \lambda\frac{F(x)-F(y)}{x-y} \right\rvert \leqslant c$$ for all $x\neq y\in[a,b]$, and
you must choose $\lambda$ so that $f([a,b]) \subset [a,b]$.
By the mean value theorem, the difference quotient is a value of the derivative, so the given bounds of $F'$ yield
$$1 - \lambda K_2 \leqslant 1 - \lambda \frac{F(x)-F(y)}{x-y} \leqslant 1 - \lambda K_1$$
for $\lambda \geqslant 0$. Thus choosing $0 < \lambda < \frac{2}{K_2}$ will satisfy the first requirement. It remains to show that for all sufficiently small $\lambda > 0$ you have
$$a \leqslant x - \lambda F(x) \leqslant b$$
for all $x\in [a,b]$. |
H: Prove that $\mathbb{Z}\left[\sqrt{-3}\right]$ is not a Dedekind domain.
Prove that $\mathbb{Z}\left[\sqrt{-3}\right]$
is not a Dedekind domain.
AI: Consider $(1+\sqrt{-3})(1-\sqrt{-3})=2\cdot2$ |
H: Does $A \cap (E_1\cup E_2) = [A\cap E_1]\cup [A\cap E_2\cap E_1^c]$?
$$A \cap (E_1\cup E_2) = [A\cap E_1]\cup [A\cap E_2\cap E_1^c]$$
Is this a correct relationship?
I tried not returned substantiated.
Thanks for all.
AI: Yes it is.
Use:
$E_1 \cup E_2 = E_1 \cup (E_2\cap E_1^c)$;
$A \cap (B \cup C) = (A \cap B)\cup (A\cap C)$.
I leave the justification to you. |
H: Show that: $\sum_{n=1}^{\infty} n(n-1)s^{n-2} = \frac{2}{(1-s)^3}$
How can I show that:
$\sum_{n=1}^{\infty} n(n-1)s^{n-2} = \frac{2}{(1-s)^3}$
I'm struggling to figure out how to start on this question. Should I sum the series and then differentiate it and also differentiate the sum term by term and then equate the two together? Would that be the right method of going about it?
Thanks.
AI: Hint:
$$
\sum_{n=2}^{\infty} n (n-1) s^{n-2} = \sum_{n=0}^{\infty} \frac{d^2}{ds^2} s^n
$$ |
H: Avoid evaluation of a very large matrix in non-negative matrix factorization
This is somewhere in between a math and a programming question, so please
send me back to SO if you think it's off-topic.
I'm implementing non-negative sparse coding, a regularized variant of
non-negative matrix factorization. This entails finding $W \in \mathbb{R}^{n\times k}$ and $H \in \mathbb{R}^{k\times m}$ that minimize
$\frac{1}{2} ||X - WH||_2^2 + \lambda ||W||_1$.
$k$ is a free parameter, but $n,m$ depend on the data and can be very large, so that evaluating $WH$ is prohibitively expensive. I can avoid it by evaluating the first term, the squared Frobenius norm, as
$||X||_2^2 + \operatorname{tr}(((W^T W)H)H^T) - 2\operatorname{tr}((X H^T)W^T) $,
which can be evaluated cheaply by noticing that
$\operatorname{tr}(A B^T) = \operatorname{vec}(A)^T\operatorname{vec}(B)$.
Now comes the problem: as part of the algorithm, I have to evaluate a gradient update
$W - \mu (WH - X)H^T$,
which again involves $WH$. $\mu$ is a learning rate, and the subtraction is element-wise. However, I don't see how I can avoid evaluating $WH$ this time; is there a way to rewrite this so I don't have to?
(I could evaluate it piece by piece, but I'm really looking for an expression that lends itself to computation using a Matlab-like linear algebra library.)
AI: Forming each entry of $WH$ requires $k$ multiplications (and $k-1$ adds) using a conventional matrix multiplication, for a total of $kmn$ multiplications.
Alternatively one can form $HH^T$ with $k^2 m$ multiplications, and then $W(HH^T)$ requires a further $k^2 n$ multiplications.
So if you have to evaluate it, I'd try:
$$ (WH - X)H^T = W(HH^T) - XH^T $$
since $m+n$ is much less than $mn$.
Still $XH^T$ looks expensive to evaluate. Perhaps this piece should be left unevaluated/accessed implicitly (by storing $X$ and $H^T$ stepwise).
Comparing the notation to the article linked in the Question, presumably $H$
(and therefore $H^T$) are sparse but in a random fashion. It's possible
that this sparsity can be exploit to perform $XH^T$ quickly, but this is not
likely the case with a high-level matrix library/interface similar to MATLAB. |
H: $\operatorname{Ran} \lambda I - T$ is closed for compact operator $T$ and $\lambda \neq 0$
Let $T$ be a compact operator on a Hilbert space $\mathcal{H}$ and $\lambda \in \Bbb{C} - \{0\}$. I want to show that $\operatorname{ran} \lambda I - T$ is closed. So suppose we have $g_j = (\lambda I - T)f_j \in \operatorname{ran} \lambda I - T$ converging to some $f$. We need to show $f \in \operatorname{ran} \lambda I - T$.
First because $\lambda I - T$ is a bounded operator its kernel $V_\lambda$ is closed, hence $\mathcal{H} = V_\lambda^\perp \oplus V_\lambda$. With this we may assume wlog that $f_j \in V_\lambda^\perp$. Now the next hint in the problem is to show that $\{f_j\}$ is bounded under the assumption that $\{f_j\} \subseteq V_\lambda^\perp$. I have tried for about 1 hour and I can't show this.
How can I go around showing this? Here's a partial proof for this fact that I have. We may write $$\begin{eqnarray*}||\lambda f_j||& =& \|\lambda f_j - Tf_j + Tf_j\| \\
&\leq& \| (\lambda I - T)f_j\| + \|Tf_j\|\\
&\leq& \|g_j\| + \|T\|\|f_j\|\\
&\leq& N + \|T\|\|f_j\|\end{eqnarray*}$$
where we have used that $g_j \to g$ impies $\|g_j\| \leq N$. Now we may rearrange to get $\|f_j\| (|\lambda - \|T\|) \leq N$ and so I will be done if I know $\|\lambda \| - \|T\|) \neq 0$. However this may not be true in general. So how can I show $\{f_j\}$ is bounded using the hint in Stein and Shakarchi? Please do not give full answers.
Next, assuming that $\{f_j\}$ is bounded and that $T$ is compact we know $Tf_{n_j} \to f$ for some $f \in \mathcal{H}$. However I don't know how to show this implies $g$ that we started out with is in $\operatorname{ran} \lambda I - T$.
What's strange about all of this is they claim in Stein and Shakarchi that the result is not true if $\lambda = 0$. My guess is that we somehow have to use the result that if $\lambda \neq 0$ that $V_\lambda $ is finite dimensional. All this is very puzzling at the moment and I would appreciate some help.
AI: The key is to show that $S = (\lambda I - T)\lvert_{V_\lambda^\perp}$ has the property
$$\bigl(\forall x \in V_\lambda^\perp\bigr)\bigl(\lVert Sx\rVert \geqslant \delta \lVert x\rVert\bigr)\tag{1}$$
for some $\delta > 0$. Everything follows quite easily from that.
To show the existence of such a $\delta > 0$, suppose the contrary. Then there is a sequence $(x_n)$ in $V_\lambda^\perp$ with $\lVert x_n\rVert = 1$ for all $n$ and $Sx_n \to 0$. Use the compactness of $T$ and then $\lambda \neq 0$ to obtain a contradiction.
I assume the OP has meanwhile filled in the details, so let me add them here too:
If we had a sequence $x_n \in V_\lambda^\perp$ with $\lVert x_n\rVert = 1$, and $z_n = Sx_n \to 0$, by the compactness of $T$ there is a subsequence such that $Tx_{n_k}$ converges. Suppose without loss of generality that the subsequence is the entire sequence, so $y_n = Tx_n \to y$. Then
$$\lambda x_n = Sx_n + Tx_n = z_n + y_n \to 0 + y = y.$$
Since $\lambda\neq 0$, we have $\lVert y\rVert = \lvert\lambda\rvert > 0$, it follows that $x_n \to \lambda^{-1}y \in V_\lambda^\perp\setminus\{0\}$ and
$$S(\lambda^{-1}y) = \lim_{n\to\infty} Sx_n = \lim_{n\to\infty} z_n = 0,$$
contradicting $V_\lambda^\perp \cap V_\lambda = \{0\}$.
So the existence of a $\delta > 0$ with $\lVert Sx\rVert \geqslant \delta \lVert x\rVert$ for all $x\in V_\lambda^\perp$ is established.
Since $\mathcal{H} = V_\lambda \oplus V_\lambda^\perp$, we have the (linear) bijection $S\colon V_\lambda^\perp \to \mathcal{R}(\lambda I - T)$. Now if $(y_n)$ is a Cauchy sequence in $\mathcal{R}(\lambda I-T)$, then $(S^{-1}y_n)$ is a Cauchy sequence in $V_\lambda^\perp$ by $(1)$, and we have
$$\lim_{n\to\infty} y_n = \lim_{n\to\infty} S(S^{-1}y_n) = S(\lim_{n\to\infty} S^{-1}y_n) \in \mathcal{R}(S) = \mathcal{R}(\lambda I - T).$$
If we had not assumed $\lambda\neq 0$, we couldn't have deduced the convergence of $x_n$ from the convergence of $Sx_n\to 0$ and $Tx_n \to y$, and in fact the range of a compact operator is in general not closed.
It should be noted that the result does not depend on $\mathcal{H}$ being a Hilbert space, that only makes the proof a little bit more easy. When working with an arbitrary Banach space, the proof is exactly the same if one can establish a closed complement of $V_\lambda$. Although generally closed subspaces need not be complemented in Banach spaces, the compactness of $T$ implies the finite-dimensionality of $V_\lambda$, and finite-dimensional subspaces (as well as subspaces of finite codimension) are always complemented in Banach spaces. A slight variation of the proof considers the quotient $\mathcal{H}/V_\lambda$ instead of a complementing subspace. |
H: Question about polynomials of odd degree with no zeros in formally real fields which are maximal to the property of being ordered
I have encountered this argument while reading Tent and Ziegler's "Course in model theory", and I don't know why it is justified.
It arises during the proof that every ordered field has a real closure.
Let $R$ be an ordered field, and let $f\in R[X]$ be a polynomial of odd degree without a zero. Now let $\alpha$ be a zero for $f$ in the algebraic closure of $R$ and put $L=R(\alpha)$. It follows from prior arguments that $L$ cannot be ordered (as $R$ was constructed to be maximal with regard to this property), and so $-1$ is a sum of squares in $L$. Why does it follow that there are polynomials $g_i \in R[X]$ of degree less than $n$ such that $f$ divides $1+\sum g_i^2$?
Thanks
AI: The elements of $L$ are of the form $g(\alpha)$ with $g(x)\in R[x]$, $\deg g<n$. If
$$
-1=\sum_{i=1}^kg_i(\alpha)^2
$$
for some $g_i(x)\in R[x], i=1,2,\ldots,k,$ then $\alpha$ is a zero of the polynomial
$$
p(x)=1+\sum_i g_i(x)^2\in R[x].
$$
Because $R$ is ordered, $p(x)$ cannot be the constant polynomial zero (think about the coefficient of degree $2\max\{\deg g_i\mid i=1,2,\ldots,k\}$). Therefore $p(x)$ must be divisible by the minimal polynomial of $\alpha$, i.e. $f(x)$ |
H: Does there exist a commutative magma such that $\mid$ is transitive, which is not a semigroup?
Let $M$ denote a commutative magma, and write $x \mid y$ iff $xa=y$ for some $a \in M$. If $M$ is a semigroup, then $\mid$ is transitive. Does there exist a commutative magma such that $\mid$ is transitive, but which is not a semigroup?
I'm guessing "yes." (But can't think of any examples).
AI: $M=\{a,b,c\},aa=bc=cb=a,bb=ac=ca=b,cc=ab=ba=c$.
Moreover, any commutative magma can be embedded in a commutative magma in which $\mid$ is transitive. |
H: Euclidean algorithm to find integers $s$ and $t$ such that $sa+tb=1$
$a=19845$, $b=218$
I got this far but am now stuck and don't know what to do.
19845 % 218 = 7
218 % 7 = 1
1 = 218 - (7*31)
7 = 19845 - (218*91)
1 = 218 - ((19845-218*91)*31)
Then from there I do not know how to simplify it to get the values for $s$ and $t$.
AI: Use the distributive law:
\begin{align}
1 &= 218 - (19845-218\cdot 91)\cdot 31 \\
&= 218 - 19845\cdot 31 + 218\cdot 91\cdot 31 \\
&= 218 \cdot (1 + 91\cdot 31) + 19845\cdot (-31) \\
&= 218\cdot 2822 + 19845\cdot (-31)
\end{align}
Thus, $s=-31$ and $t=2822$. |
H: Is there a name for sequences like these?
Starting from an integer value (say $0$ in these cases), I need a sequence of integers to add in a cycle that progress through the integers visiting each exactly once.
For example, the most obvious and simplest sequence would be $(+1)$ which would obviously generate the sequence $1,2,3,4,5,...$.
Getting slightly more complicated there would be $(+2,-1,+2)$ which would generate the sequence $2,1,3,5,4,6,...$.
Sequences like $(+3,-1,-1,+3)$, $(+3,-2,+1,+2)$, $(+4,-1,-1,-1,+4)$, $(+4,-2,-1,+2,+2)$ and $(+2,+1,+1,-3+4)$ also fit the bill.
Is there a name for these? Can they be generated automatically?
AI: I don't know any name for this kind of sequence, though it might well exist.
If your cycle has length $n$ the you can see that its sum must be $\pm n$. Since if the sum is $s$, then every repetition of the cycle will advance by$~s$, and every integer must be obtained in exactly one way by adding an integer multiple of$~s$ to a number encountered during the first cycle; this means that every congruence class for$~s$ must be present once in the numbers encountered during the first cycle, and there are $|s|$ distinct such classes.
Now if you fix the sum to be $+n$ (the other case being symmetric), you can get any sequence in the following way: choose a permutation of $\{1,2,\ldots,n-1\}$ and extend it to a permutation$~\pi$ of $\{0,1,\ldots,n-1\}$ by requiring $\pi(0)=0$, and for every congruence class (modulo$~n$) choose some representative $r_i$, so $r_i\equiv i\pmod n$ of $i=0,1,
\ldots,n-1$. Then you sequence will be $(a_1,\ldots,a_n)$ where $a_i=r_{\pi(i)}-r_{\pi(i-1)}$ for $0<i<n$ and $a_n=r_0+n-r_{\pi(n-1)}$.
This is closely related to the group of so-called affine permutations, the (affine) Coxeter group of type $\tilde A_{n-1}$. It is by definition the set of permutations of the set$~\Bbb Z$ of all integers with the additional property that restricted to any congruence class modulo$~n$ they act by a translation (addition of a constant). The sequence of the partial sums obtained from your periodic sequence is the image of an affine permutation acting on the integers.
In fact affine permutation have the additional requirement that they avoid any global "drift": the image of $S=\{0,1,\ldots,n-1\}$, or of any other set of representatives of the congruence classes, must have the same sum as the set $S$ itself. You partial sums have the extra condition that the sum at$~0$ must be$~0$. There is a unique translation of the image sequence that will transform one restriction into the other, giving a natural bijection between the two sets.
The affine permutations are generated by the "elementary transpositions" that interchange two neighbouring congruence classes (adding$~1$ to all numbers congruent to some given$~i$, while subtracting$~1$ from all numbers congruent to$~i+1$) which is indeed a set of Coxeter generators. |
H: What is the maximum value of $4(\sin x)^2 + 3(\cos x)^2$
The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must be $4$. However my book says that the answer is $5$. Where did I go wrong?
AI: Your argument is OK. You can check it taking the derivative of the function:
$$f(x)=4\sin(x)^2+3\cos(x)^2$$
$$y'(x)=2\sin(x)\cos(x)$$
which is null for $x=0,x=\frac{\pi}{2}$
The first value gives $y(x)=3$, with the second one you get $y(x)=4$. So, $4$ is the maximum value of $f(x)$ |
H: Spectral theorem for $n$-tuples of selfadjoint operators
I need a 'good' reference to the following version of the Spectral Theorem:
Given $n$ commuting selfadjoint operators on an infinite-dimensional Hilbert space, there exist a Borel measure $\mu$ on $\mathbb R^n$ and auxiliary Hilbert spaces $h(x)$ such that the construction is unitarily equivalent to $\int\oplus h(x)\,d\mu(x)$ with selfadjoint operators of multiplication by $x_k$, $k=1, \dots, n$.
It is important that the construction be based on a measure in $\mathbb R^n$. Thanks in advance.
AI: I don't think you can find such theorem precisely in that form. Instead you can find the following
Spectral theorem. For $n$ bounded commuting self-adjoint operators $X_1,\dots,X_n$ on a Hilbert space $H$ there exists a unique projection valued measure $E$ on $\mathbb R^n$ such that $X_k=\int x_k dE(x_1,\dots,x_n)$ for all $k=1,\dots,n.$
Then using the following two (easy) Lemmas you get your version of the spectral theorem.
Lemma 1. $H$ can be decomposed into direct sum of common invariant subspaces $H_i,\ i\in I$ such that every $H_i$ is cyclic for the set $\{X_1,\dots,X_n\}$ with a cyclic vector $\xi_i.$
Lemma 2. Let $H_i,\xi_i$ be as in Lemma 1. Define $\mu_i(\cdot)=\langle E(\cdot)\xi_i,\xi_i\rangle.$ Then every $H_i$ is isomorphic to $L^2(\mathbb R^n,\mu_i)$ and $X_k$ are unitarily equivalent (under that isomorphism) to multiplication operators by $x_k.$
That spectral theorem you can find in the books:
Birman, Solomyak "Spectral theory of self-adjoint operators in Hilbert space,"
Yu. Samoilenko "Spectral theory of families of self-adjoint operators",
Yu. Berezanskii "Self-adjoint operators in spaces of functions of infinitely many variables" (in russian) |
H: Proof of an elementary property of Projection Operators
I'm asked to show the following:
Let $X$ be a linear space, and let $P : X \rightarrow X$ be a projection operator. Restricted to the linear space $range(P)$, the projection $P$ is the identity operator, that is, $Px = x$ for all
$x \in range(P)$.
If anyone could offer a hint for how to show this. I'm rusty with linear algebra so don't really know what I'm supposed to bring in outside of what's specified in the lemma/definition of projection. What is so special about the range of P that would make this so?
AI: Hint: projection operators satisfy the property $P^2=P$ (by definition).
If $x\in\operatorname{range}(P)$, then $x=P(y)$, so $P(x)=P(P(y))=\ldots$ |
H: Anti symmetrical relation
Currently learning about symmetrical and anti symmetrical relations. Been working on assignments but I cant seem to bend my head around this one, especially because I can't understand the solution.
What I thought an anti symmetrical relation is: a pair like $(x,y)$ that also exists as $(y,x)$ with $x = y.$
The question is: describe the relation of the following collection.
$$R = \{(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)\}$$
Answer: anti symmetrical.
Why is this anti symmetrical?
AI: Suppose that $R$ is a relation on a set $A$. When we say that $R$ is symmetric, we mean that if one of the pairs $\langle a,b\rangle$ and $\langle b,a\rangle$ is in $R$, then so is the other. When we say that $R$ is antisymmetric, we mean that if $a\ne b$, and one of the pairs $\langle a,b\rangle$ and $\langle b,a\rangle$ is in $R$, then the other one is not in $R$.
This is the case with your relation
$$R=\{\langle 1,2\rangle,\langle 1,3\rangle,\langle 1,4\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,4\rangle\}\;:$$
$\langle 1,2\rangle$ is in it, but $\langle 2,1\rangle$ is not; $\langle 2,3\rangle$ is in it, but $\langle 3,2\rangle$ is not; and so on for every possible pair of distinct elements of $\{1,2,3,4\}$.
Note that for both symmetry and antisymmetry it’s perfectly fine to have neither pair in the relation. The relation
$$\{\langle 1,2\rangle,\langle 2,1\rangle,\langle 1,3\rangle,\langle 3,1\rangle\}$$
is symmetric even though it has neither of the pairs $\langle 2,3\rangle$ and $\langle 3,2\rangle$, and the relation
$$\{\langle 1,2\rangle,\langle 3,1\rangle\}$$
is antisymmetric even though it has neither of the pairs $\langle 2,3\rangle$ and $\langle 3,2\rangle$. |
H: Combinations' Problem
A person has six friends and during a certain vacation, he met them during several dinners. He found that he dined:-
with all the six exactly on one day,
with every five of them on $2$ days,
with every four of them on $3$ days,
with every three of them on $4$ days, and,
with every two of them on $5$ days.
Further every friend was present at $7$ dinners and every friend was absent at seven dinners.
How many dinners did he have alone?
I got puzzled with so much of information and couldn't get the answer. Please help.
AI: He had one mean alone and one meal with each friend alone. One meal had all 6 friends present. All other meals involved 5 attendees and one absent friend. The easiest way to solve this is to make a table for all 14 dinners and all 6 friends.
Below is a solution where 1 means present (though it would be identical practically if 0 ment present).
[ \begin{array}{lcr}
\mbox{Friend} & Adam & Bob&Charlie&Dave&Evgeni&Fred \\
\mbox{Dinner 1} & 1 & 1&1&1&1&1 \\
\mbox{Dinner 2} & 0 & 1&1&1&1&1 \\
\mbox{Dinner 3} & 1 & 0&1&1&1&1 \\
\mbox{Dinner 4} & 1 & 1&0&1&1&1 \\
\mbox{Dinner 5} & 1 & 1&1&0&1&1 \\
\mbox{Dinner 6} & 1 & 1&1&1&0&1 \\
\mbox{Dinner 7} & 1 & 1&1&1&1&0 \\
\mbox{Dinner 8} & 0 & 0&0&0&0&0 \\
\mbox{Dinner 9} & 1 & 0&0&0&0&0 \\
\mbox{Dinner 10} & 0 & 1&0&0&0&0 \\
\mbox{Dinner 11} & 0 & 0&1&0&0&0 \\
\mbox{Dinner 12} & 0 & 0&0&1&0&0 \\
\mbox{Dinner 13} & 0 & 0&0&0&1&0 \\
\mbox{Dinner 14} & 0 & 0&0&0&0&1 \end{array}] |
H: Prove that when $f(A) \subseteq f(B)$ doesn't always mean that $A \subseteq B$
How to prove, when $f(A) \subseteq f(B)$ doesn't "always" mean that $$A \subseteq B$$
when $ f\colon X \to Y $ is total function (not partial)
AI: Consider $f: \mathbb R \to \mathbb R^+, x\mapsto x^2$
And then see
$$f([0, 2]) = f([-2, 0])$$ |
H: Finding $x+y+z$
If $x+1/x = y$, $y+1/y=z$, $z+1/z=x$, then find $x+y+z$. Is there any way to do so without taking out the values of $x$, $y$, $z$?
Please help,
AI: Let us first write down the equations:
$$x+\frac{1}{x}=y, \; y + \frac{1}{y}=z,\; z+\frac{1}{z}=x$$
We now add these equations together:
$$x+\frac{1}{x}+ y + \frac{1}{y}+z+\frac{1}{z}=x +y+z$$
So we can see cancelling out $x+y+z$ gives us something, which was our motivation in adding them up:
$$\frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$$
Expanding, we get $xy+yz+xz=0$.
Here, we can note the identity $(x+y+z)^2 = x^2 +y^2+z^2 + 2(xy+yz+zx)$. Hence, if we can find $x^2+y^2+z^2$, we are done.
This can be done by realizing that by multiplying $x$ on both sides of $x+\frac{1}{x} = y$, gives us $x^2+1=xy$. Doing this with other equations and adding them all, we have:
$$x^2+1+y^2+1+z^2+1=xy+yz+xz$$
Substituing $xy+yz+xz=0$, we have $x^2 +y^2+z^2 = -3$. So,
$$x+y+z=\pm\sqrt{x^2 +y^2+z^2 + 2(xy+yz+zx)} = \pm\sqrt{-3+2(0)} = \pm\sqrt{-3} = \pm\sqrt{3}i$$ |
H: Finding real part of fourier series
I have encountered the following problem in one of my textbooks but I'm not really getting anywhere:
Let $f$ be complex-valued and piecewise continuous on the interval $[-\pi,\pi]$. Find the complex fourier series of $Re(f)$ on the basis of the complex Fourier series of $f$.
i.e we're given that $f$ has the complex Fourier coefficient $c_n$ and we want to express $Re(f)$'s fourier coefficients, $C_n$, in terms of $c_n$. Did i interpret this correct?
I've tried splitting $f=g+ih$ but i'm not managing to express $C_n$ in terms of known objects.
Thanks in advance.
AI: $\Re(f) = \frac{1}{2}(f + \overline{f})$, so $C_n = \hat{\Re(f)} = \frac{1}{2}(\hat{f}+\hat{\overline{f}}) = \frac{1}{2}(c_n + \overline{c_n}).$ Is that what you'd like? |
H: Reverse of number (numerical)
You can find reverse number $R$ using $x_{n+1} = x_n(2 - x_n \cdot R) \ \ $ where $ n = 0,1,..$
Prove it using Newton method for finding $0's$ of some function $f$
Anyone have idea what that function $f$ might be?
AI: The Newton iteration is
$$
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
$$
and to find the inverse you should root-find $f(x) = 1/x - R$ which has a root at $1/R$. Note that $f'(x) = -1/x^2$ so $f(x_n)/f'(x_n)$ will give you exactly the iteration you are looking for. |
H: Graph Theory : Job Assignment Problem
Problem is to assign 5 jobs to five people. How many non-planar graphs can be drawn such that no vertex is isolated?
AI: The mothers of all non-planar graphs are $K_5$ and $K_{3,3}$ as you know. The second of those is the one you want to avoid. That restricts how you can partition jobs into groups, since you cannot have 3 and 3 from both sides.
Can you think of what the valid partitions now would be? Can you take it from here?
EDIT Some further hints. Consider the bipartite graph $G$ with $V = J \cup P$ where $J = \{j_1, \ldots, j_5\}$ and similarly $P = \{p_i\}_{i=1}^5$. If a job $j_k$ is assigned to the person $p_n$, add the edge $e = \{j_k, p_n\}$ to the edge set $E$. Clearly, $G$ is bipartite by construction.
Each different assignment of jobs to workers corresponds to a different edge set of $G$. How many possible assignments are there?
You cannot have isolated vertices, so this means at least one edge to everything on both sides - i.e. must assign every job and every worker. How many ways to do that?
Can you now enforce the planarity by yourself? |
H: How to factorize the quadratic $a(b-c)x^2 + b(c-a)x + c(b-a) = 0$?
How do i factorize this equation:
$a(b-c)x^2 + b(c-a)x + c(b-a) = 0$
I tried the quadratic formula, but the discriminant is not factorising into a perfect square.
Please help!
AI: HINT:
Observe that $$a(b-c)-c(b-a)=-b(c-a)$$
Put the value of $b(c-a),$ and take out common from the first two terms & the last two terms and see what happens? |
H: Question about criteria of connectedness
I have to prove the following :
A) No proper non empty subset of $X$ is both open and closed in $X$ $implies$ that $X$ is not the union of two disjoint open subsets of itself.
Attempt at the proof:
I assumed that, No proper non empty subset of $X$ is both open and closed in $X$.Suppose on the contrary that $X$ = $U union U^c$ where both $U$ and $U^c$ are open.
Then their complements $U^c$ and $U$ respectively, are closed.
Hence $U$ and $U^c$ are both open and closed, a contradiction.
B) $X$ is not the union of two disjoint open subsets of itself $implies$ that $X$ is not the union of two disjoint closed subsets of itself.
Attempt at the proof:
Assume the first criterion holds true.
And suppose that $X$ is the disjoint union of two closed subsets $A$ and $B$ of $X$.
Then $A^c$ = $B$ and $B^c$ = $A$ and so both $A$ and $B$ should be open, and $X$ is their union, a contradiction.
Am i correct ?
AI: Hint: For (A) and (B), try proving the contrapositive of the statements. |
H: $a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$ Does there exist $k\in\mathbb N$ such that $a_n,n=1,2,3,\cdots$ are all composite numbers?
Let $a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$
If $k=1$ then $a_n=1,3,7,15,31,63,\cdots$ here $3,7,31$ are prime numbers. I'm interested in this problem:
Does there exist $k\in\mathbb N$ such that $a_n,n=1,2,3,\cdots$ are all composite numbers?
If $k=147$ then $a_n,n=1,2,\cdots 2551$ are all composite, but $a_{2552}$ is prime. So I doubt the existence of such number.
AI: The numbers you mention are Riesel numbers http://en.wikipedia.org/wiki/Riesel_number
and there is the similar Sierpinski numbers where it is $2a_{n-1}-1$ instead.
http://en.wikipedia.org/wiki/Sierpinski_number |
H: Example to $\lim f(x)g(x)$ may not exist
Let $A\subset\mathbb{R}$, $c$ a cluster point of $A$ and $f,g:A\rightarrow \mathbb{R}$. Suppose that $f$ is bounded on some neighbourhood of $c$ show by example that if $\lim_{x\rightarrow c}g(x)$ exists, then $\lim_{x\rightarrow c}g(x)f(x)$ may not exist.
By hypothesis, $\lim_{x\rightarrow c}f(x)$. I had some ideas:
$\lim_{x\rightarrow c}1/x$ does not exist
define $g(x)=1$, and $f(x)$ by part, then $\lim fg$ does not exist
But I could not find $fg=1/x$ and both $\lim f$, $\lim g$ exist.
And I don't think we can multiply functions defined by part...
Is there a classic example of that in Analysis?
AI: Consider $$f(x)=\begin{cases}\sin\frac1x & \text{if }x\ne 0\\5 &\text{otherwise}.\end{cases}$$ and $g$ any non-$0$ constant function with $c=0$. |
H: Problem book,typing error?
I have the following problem :
Let $f:(0,\infty)\rightarrow \mathbb R$ be an arbitrary function satisfying the hypothesis, $\lim_{x \rightarrow 0} x(f(x)-1)=0$. Show that $\lim_{x \rightarrow 0 } f(x)$ exists.
The problem has also other parts and 2 other hypothses on $f$. But the solution starts like this: by hypothesis one it follows $\lim_{x \rightarrow 0} xf(x)=1$... So it doesn't use the other two hypothesis ( why would he explicitly state that the result follows from hyp. one if the others are also used?).
But than $ \ \sin(\frac{1}{x}):=f(x) \ $ would be a conter example.
Can somone say if my counterexample is right?I can't see the error. And I'm pretty sure he doesn't use the other hypotheses since he doesn't state them in the problem.
here are the other two conditions on $f$ (wich i didn't use in the counterexample by the reasons explained above)
two : $f(1)=2$
three: $(x+2)f(x+2)-2(x+1)f(x+1)+xf(x)=0$
AI: Your counterexample is correct. (Converting to an answer.) |
H: What is the probability of this event?
The question is: A card is drawn from an ordinary pack(52 cards) and a gambler bets that either a spade or an ace is going to appear. The probability of his winning are?
I think the answer is $\frac{16}{52} = \frac{4}{13}$. Did I go "probably" go wrong somewhere?
AI: There are 16 `good' cards (the 13 spades and the three other aces), out of 52 total, so you're correct. |
H: Probability that $j$ persons to get off on the same floor, and $k-j$ persons to get off on separate floors
There are $k$ persons and $n$ floors. Assuming that the probability of any person to get off on any floor is $\frac{1}{n}$, and the decisions taken by the persons are independent, what is the probability that $j$ persons get off on the same floor, and the rest $k-j$ persons to get off on different floors?
I know that the answer is $\dfrac{C_k^j \cdot A_n^{k-j+1}}{n^k}$ but I don't know how to get there.
Thanks!
AI: The thing to keep in mind is that it doesn't matter which $j$ people get off on a floor, nor does it matter which floor this happens to be. So think about the number of ways a `good' outcome can happen.
First, by definition of the choose functions, there are $C_k^j$ ways to choose $j$ people which get off together, and $n$ different floors for them to get off on. Then, once these people have gotten off, the next person has $(n-1)$ floors to choose from, the next has $(n-2)$, and so on.
So there are $n C_k^j A_{n-1}^{k-j+1}$ good outcomes, in which $j$ people get off at the same floor, and everyone else gets off on a separate one. There are $n^k$ total ways of assigning floors, so you get
$$ \frac{n C_k^j A_{n-1}^{k-j+1}}{n^k} = \frac{C_k^j A_n^{k-j+1}}{n^k}$$ |
H: Understanding proofs from paper on Game Theory (Price of Anarchy)
I'm trying to distill the arguments in the paper "Worst-Case Equilibria" (http://cgi.di.uoa.gr/~elias/publications/paper-kp09.pdf). But there are some things I do not understand and would appreciate some help.
Theorem 1: The price of anarchy for $m$ links is $\Omega(\log(n)/\log(\log(n)))$. For $m = 2$ links, it is at least $\frac{3}{2}$. (note: number of agents $n$ = $m$)
I understand that the social cost (expected maximum amount of traffic) is equal to the problem of throwing $m$ balls into $m$ bins. The paper states $\Theta(\log m/\log\log m)$.
Now here is the part I don't understand: (a) "Given that the optimal solution has cost 1, the lower bound follows." I just don't understand what this sentence means. Maybe someone could reword it. (b) "For $m = 2$, this gives a lower bound of $\frac{3}{2}$." How do you show this? I've tried plugging in $m =2$ into $(\log m/\log \log m)$ but you don't get $\frac{3}{2}$. How do you show how to arrive at lower bound of $\frac{3}{2}$?
One more question (for now): There is a proof that says the expected cost of $c_i$ for agent $i$ is at most $2 - \frac{1}{m}$. But it seems what they prove is: $c_i \leq \frac{\sum_kw_k}{m} + \frac{m-1}{m}w_i$ (Aside: can't get the summation symbol to work.)
I don't see how to go from $c_i \leq \frac{\sum_kw_k}{m} + \frac{m-1}{m}w_i$ to $2 - \frac{1}{m}$.
AI: a)
In this setting, the price of anarchy is
$$\frac{\max_{s \in \mathrm{NE}}\mathrm{cost}(s)}{\mathrm{cost}(\mathrm{OPT})},$$
and to show it is $\Omega(\mathrm{something})$ you need both the nominator and the denominator of appropriate order. The nominator bound $\Omega\left(\frac{\log n}{\log \log n}\right)$ is not enough by itself, you need also that the denominator is $O(1)$. So the bound (also) follows from the fact, that $\mathrm{cost}(\mathrm{OPT}) = 1$, and this is what that sentence means.
b)
The symbols $\Omega$ and $\Theta$ and $O$ notations describe only the order (see e.g. Big $O$ notation), so plugging numbers there isn't very meaningful (at least if you have only one variable). The $\frac{3}{2}$ comes from direct calculation, namely throwing $2$ balls into $2$ bins you have maximum of $2$ and $1$ both with probability $\frac{1}{2}$, hence the expected value is $\frac{2}{2} + \frac{1}{2} = \frac{3}{2}$.
I hope this helps $\ddot\smile$ |
H: "Comfort" function with tunable parameter
I'm trying to create a "comfort" function with the following characteristics:
its domain is $(-\infty, +\infty)$;
its range is $[0,1]$;
it is at or near its maximum value ($1$) in some interval $[x_{c}-\delta, x_{c}+\delta]$;
it is at or near its minimum value ($0$) in the intervals $(-\infty, x_{c}-\delta-\gamma]$ and $[x_{c}+\delta+\gamma, +\infty)$;
in the interval $(x_{c}-\delta-\gamma, x_{c}-\delta)$ it has behavior similar to the sigmoid function $\dfrac{1}{1 + e^{-x}}$, while in the interval $(x_{c}+\delta, x_{c}+\delta+\gamma)$ similar to the sigmoid function $\dfrac{1}{1 + e^{x}}$;
it should have tunable parameters to control $\delta$, $\gamma$, and the transition from minimum to maximum value and from maximum to minimum (in the same way that a sigmoid function can be controlled).
Can anybody help? Thank you.
AI: Here is my function. Through dumb luck tweaking another construct of mine this appears to fit the bill:
$$f(x)=\displaystyle\frac{e^{\delta^2r}}{e^{\delta^2r}+e^{rx^2}}.$$
Parameter descriptions:
The parameter $\delta$ corresponds to your description. The jump/dip occurs at $\pm \delta$.
The parameter $r$ controls the shape of the sigmoid turn at $\pm \delta$. If positive it jumps from $0$ to $1$ then back. If negative, it does the reverse.
An additional parameter improvement:
Here I just added an additional parameter $c$ which corresponds to your $x_c$ allowing for you to modify the center as well. :))
$$f(x)=\displaystyle\frac{e^{\delta^2r}}{e^{\delta^2r}+e^{r(x-c)^2}}.$$
That was alot of fun. Keep them coming! |
H: Show a set is in Borel sigma algebra
First of all, sorry for asking again a question.
For all functions, $f:[a,b] \to \mathbb{R}_{+} $ define
$S(f)=\{(x,y)\in \mathbb{R}^{2}: 0\leq y \leq f(x)\}$.
Show if $f$ is measurable, then $S(f) \in \mathbb{B} (\mathbb {R}^{2})$. Compute that two-dimensional Lebesgue measure of $S(f)$.
For the first part, I thought maybe if I define a function $g$ such that $g(x,y)=f(x)-x$ and if I can show this is continuous , it may work.. will it ?
For the second part, can I approach with integrable simple functions ?..
Since I don't have good knowledge of this subject , I don't know what to do anymore. Maybe it will be good to push me a little just by giving hints not the answer. Thanks in advance.
AI: For the first part, let $\phi(x,y) = f(x)-y$. Then $\phi$ is measurable and $S(f) = (\mathbb{R} \times [0,\infty) ) \cap \phi^{-1} [0,\infty)$ and both $\mathbb{R} \times [0,\infty)$ and $\phi^{-1} [0,\infty)$ are measurable, hence $S(f)$ is measurable.
For the second, use Fubini-Tonelli (I used $m$ to denote the product measure), and note that $1_{S(f)}(x,y) = 1_{[0,f(x)]}(y)$.
Then
$\int 1_{S(f)} = \int 1_{S(f)}(x,y) dm = \int \left( \int 1_{S(f)}(x,y) dy \right) dx = \int \left( \int 1_{[0,f(x)]}(y) dy \right) dx = \int f(x) dx = \int f.$ |
H: Find $f$ using equation involving real and imaginary part
Suppose $f=u+iv$ is differentiable in the entire complex plane. The real and imaginary parts of $f$ are related by $au(x,y)+bv(x,y)+c=0$ where $a,b,c \in \mathbb{R}$ and $a$ and $b$ are not simultaneously zero. Find $f$.
My attempt: Differentiate the equation with respect to $x$ and $y$. Then we obtain two equations $$au_x+bv_x=0$$ $$au_y+bv_y=0$$
Since $f$ is differentiable, by Cauchy Riemann equation, $u_x=v_y$ and $u_y=-v_x$. By substituting the CR-equations into the two equations above, we have $$av_y+bv_x=0$$ $$au_y+bu_x=0$$
Then I stuck here. Can anyone guide me?
AI: Write down everyting wrt $\;u\;$ , say:
$$\begin{align*}au_x-bu_y&=0\\bu_x+au_y&=0\end{align*}$$
Now multiply the first equation by $\;a\;$ , the second one by $\;b\;$ and add them:
$$(a^2+b^2)u_x=0\implies u_x=0\;\text{(as we're given}\;\;a^2+b^2\neq0)$$
So $\;u\;$ is constant wrt $\;x\;$ ...etc. Go on, now. |
H: Convex hull of the union of two nonempty sets
I was reading about convex hulls on Wikipedia (Convex hull) and I read :
$ Conv(A \cup B)= Conv(Conv(A) \cup Conv (B))$ where $A$ and $B$ are nonempty sets.
I can see intuitively that this equality is true, but I do not know how to write it formally down.
AI: For simplicity, let us denote $T$ as the operation of taking the convex hull of a set.
We observe following basic principles.
$A \subset T(A)$.
If $A \supset B$, then $T(A) \supset T(B)$.
$T(T(A))=T(A)$.
Now we prove the proposition.
From 2, $T(A \cup B) \supset T(A)$ and also $T(A \cup B ) \supset T(B)$, so $T(A \cup B) \supset T(A) \cup T(B)$. Hence $T(A \cup B) \supset T(T(A) \cup T(B))$.
On the other hand, we have $ A \cup B \subset T(A) \cup T(B)$. Therefore, $T(A \cup B) \subset T(T(A) \cup T(B))$. |
H: How find this limit $\lim_{n\to\infty}\frac{(2n+1)!}{(n!)^2}\int_{0}^{1}(x(1-x))^nf(x)dx$
Let $f:[0,1]\longrightarrow R$ be a continuous function,Calculate the limit
$$\lim_{n\to\infty}\dfrac{(2n+1)!}{(n!)^2}\int_{0}^{1}(x(1-x))^nf(x)dx$$
My try:use this
$$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$
so
$$\dfrac{(2n+1)!}{(n!)^2}\approx\dfrac{\left(\dfrac{2n+1}{e}\right)^{2n+1}\sqrt{2(2n+1)\pi}}{\left(\dfrac{n}{e}\right)^{2n}\cdot 2n\pi}=\dfrac{\sqrt{2n+1}}{e\sqrt{2\pi}}\left(2+\dfrac{1}{n}\right)^{2n+1}$$
Then following I can't it,and I guess this problem answer is $f(\dfrac{1}{2})$,But I can't prove it.
Thank you for your help.
AI: Hint. If we put
$$ K_{n}(x) = \frac{(2n+1)!}{(n!)^{2}} x^{n}(1-x)^{n}, $$
then the total mass $\int_{0}^{1} K_{n}(x) \, dx$ equals $1$ for any $n$. (Just apply integration by parts $n$ times!) Show that this sequence of functions is an approximation to the identity by checking that for any $\delta > 0$ we have
$$\lim_{n\to\infty} \int\limits_{\delta \leq |x - 1/2| \leq 1/2} K_{n}(x) \, dx = 0.$$
You may find the Stirling's formula useful. Once this is proved, you can check that
$$ \int_{0}^{1} f(x) K_{n}(x) \, dx - f\left(\tfrac{1}{2}\right)
= \int_{0}^{1} \left\{ f(x) - f\left( \tfrac{1}{2}\right) \right\} K_{n}(x) \, dx $$
goes to zero as $n \to \infty$. |
H: Limit of metric of sequences
I'm not sure if I'm overcomplicating this, but I'm trying to prove that if $x_n \to x$ and $y_n \to y$, then $\lim_{n\to \infty} \rho(x_n, y_n) = \rho(x,y)$.
So far I have that I want to show that $\rho(\rho(x_n,y_n), \rho(x,y)) \to 0$, and I have tried a tricky triangle inequality:
$$
\rho(\rho(x_n,y_n), \rho(x,y)) \leq \rho(\rho(x_n,y_n), \rho(x,y_n)) + \rho(\rho(x,y_n), \rho(x,y))
$$
But I'm pretty stuck here. A hint would be great. Thanks!
AI: $$\rho(x_n,y_n)\leq\rho(x_n,x)+\rho(x,y_n)\leq\rho(x_n,x)+\rho(x,y)+\rho(y,y_n)$$
$$\rho(x,y)\leq\rho(x,x_n)+\rho(x_n,y)\leq\rho(x,x_n)+\rho(x_n,y_n)+\rho(y_n,y)$$
Can you continue from here? |
H: Proving Riemann integral does not change when finite values of a function is changed.
I know how to prove that the Riemann integral of a function does not change if one point of the function is changed. However, extending that result to a finite set by use of induction is something I have struggled to prove. I just need a hint as to how I should start off the proof. Looking forward to an exchange of ideas. An answer I found for a single point is given below.
AI: Now, let $f_1$ be equal to $f_0$, except at $x_1$; you know that the integral of $f_1$ is the same as that of $f_0$ by the above argument (so then it is also the same as $f$!).
Then, let $f_2$ be equal to $f_1$, except at $x_2$, etc., etc. |
H: Combinatorics - Find the coefficient of $x^{12}$ in...
Would someone be able to help me figure out these two binomial coefficient problems using generating functions? Its a rough concept for me to understand, so a good explanation would be very much appreciated!
$a$) $(1-x)^8$
$b$) $(1-4x)^{-5}$
Thank you in advanced for your help!!!
AI: Consider $$(1-x)^8=1-8 x+28 x^2-56 x^3+70 x^4-56 x^5+28 x^6-8 x^7+x^8.$$ The highest power in this polynomial is $x^8$. Thus the coefficient of $x^{12}$ is $0$.
Consider $${1\over (1-4x)^5}=\sum_{n=0}^\infty{n+5-1\choose n}4^nx^n.$$ Let $n=12$. The coefficient of $x^{12}$ is ${16\choose 12}4^{12}=30534533120$.
We know that $${1\over (1-x)^k}=\sum_{n=0}^\infty{n+k-1\choose n}x^n.$$
We can use this fact to calculate the coefficients of $x^n$ as we please. |
H: Example of 3-regular graph with chromatic index > 3
folks.
For a homework assignment I've been asked to prove that a 3-regular Hamiltonian graph has a chromatic index of 3. I really would like to work through the proof myself, but am having trouble thinking of a 3-regular graph that's NOT Hamiltonian as a negative example.
Any help would be appreciated! Thanks!
AI: Hint: how many vertices has a 3-regular graph (parity). So how many colors do you need to edge-color the Hamiltonian cycle. Now finish the proof. |
H: Riemann Integral (Rudin)
I was reading Rudin's, "Principles of Mathematical Analysis", specifically the section about the Riemann Integral and I've ran into some "shaky" notation. Can someone just explain to me geometrically what is going on here (by here I mean what is def $6.2$ saying)?
$6.2$. Definition. Let $\alpha$ be a monotonically increasing function on $[a,b]$ .....he goes on to say, "Corresponding to each partition $P$ of $[a,b]$, we write $\Delta \alpha_{i} = \alpha(x_{i})-\alpha(x_{i-1})$." It is clear that $\Delta\alpha_{i}\geq0$. For any real function $f$ which is bounded on $[a,b]$ we put, $U(P,f,\alpha)=\sum_{i=1}^{n}M_{i}\Delta\alpha_{i}$.
Definitons which were given before this:
$M_i$ is defined as, $M_i=lub\ f(x)$ with $(x_{i-1}\leq x\leq x_i)$. Also, the partition $P$ is given on $[a,b]$ a finite set of points $x_0\leq x_1\leq ...\leq x_{n-1}\leq x_n$ , where $a= x_0\leq x_1\leq ...\leq x_{n-1}\leq x_n=b$
I hope it is clear what I'm asking, and if anyone has the book it is pages $104-105$.
AI: You may think of the following. In Riemann integral we chop the interval $[a, b]$ into smaller intervals $[x_i x_{i+1}]$ and calculate the area of the two rectangles which is
$$M_i(x_{i+1} - x_i) \text{ and }m_i(x_{i+1} - x_i)$$
and using these two area to approximate the "area under graph" of $f$.
But when $\alpha$ is given, we are giving the intervals $[a, b]$ another measurement: we define the "length" of $[x_i, x_{i+1}]$ to be $\alpha(x_{i+1} - \alpha(x_i)$ (This is nonnegative as $\alpha$ is increasing). So we can still estimate the "area under graph" with respect to this measurement. By taking limit we come up to the integral
$$\int_a^b f d\alpha\ .$$
We may also think (as Tyler stated) that $\alpha$ is a way to give a weight to each element in the interval. To the extreme, if $\alpha$ is the function
$$\alpha (x) = 0\ \text{ when } x<c \ \text{ and } \alpha(x) = 1 \text{ when }x\geq c\ ,$$
when this is given weight solely to the point $c$ (This is sometimes called the point measure). In this case, one would have
$$\int_a^b f d\alpha = f(c)$$
if $f$ is continuous at $c$. |
H: Prove that $P_{S_N}(t) = P_N(P_X(t))$ for $S_N = X_1 + \cdots + X_N$.
Let $N$ and $(X_i)_{(i \ge 1)}$ be independent random variables ($X_i$ have the same density). Let $S_N = X_1 + \cdots + X_N$. Prove that $P_{S_N}(t) = P_N(P_X(t))$ where $$P_X(t) = E(t^X) = \sum_{k=0}^\infty t^x \Pr(X=k) $$
is probability-generating function. I have this proof at my university, but I can't understand it.
$$P_{S_N}(t) = E(t^{S_N}) \stackrel{(1)}{=} \\ \sum_{n=0}^\infty E(t^{S_N}\mid N=n) \Pr(N=n) \stackrel{(2)}{=} \\ \sum_{n=0}^{ \infty } E(t^{S_n}) \Pr(N=n) \\ \sum_{n=0}^\infty P_{X_1 + \cdots + X_n}(t) \Pr(N=n)= \\ \sum_{n=0}^\infty [P_X(t)]^n \Pr(N=n) = \\P_N(P_X(t)) $$
I don't understand equal $(1)$ and $(2)$. Could explain me that? I will grateful for your help. Thanks in advance!
AI: $\newcommand{\E}{\mathbb E}$
The identity labeled $(1)$ in your question is an instance of the identity
$$
\E(Y) = \sum_x \E(Y \mid X=x) \Pr(X=x).
$$
This is often written as
$$
\E(Y) = \E(\E(Y\mid X))
$$
and is sometimes called the "law of total expectation" (except that the law of total expectation applies not only in the discrete case but more generally).
$$
\begin{align}
\E(Y) & = \sum_y y\Pr(Y=y) \\
& = \sum_y \left(y \sum_x \Pr(Y=y\ \&\ X=x) \right) \\
& = \sum_y \left( y \sum_x \Pr(Y=y\mid X=x)\Pr(X=x) \right) \\
& = \sum_x \sum_y \left( y \Pr(Y=y\mid X=x)\Pr(X=x) \right) \\
& = \sum_x \left(\Pr(X=x)\sum_y y\Pr(Y=y\mid X=x)\right) \tag{$\star$} \\
& = \sum_x \left(\Pr(X=x)\cdot\E(Y\mid X=x)\right).
\end{align}
$$
The "$=$" in $(\star)$ is true because the factor $\Pr(X=x)$ does not change as $y$ changes, going through the list of all possible values of the random variable $Y$.
The expression following your step $(2)$ would be correct if the superscript were $S_n$ rather than $S_N$, since $\E(t^{S_N}\mid N=n)=\E(t^{S_n})$.
Postscript on 11/6 responding to a comment below:
\begin{align}
\E(t^{S_N}\mid N=n) & = \sum_s t^s \Pr(S_N=s\mid N=n) \\[8pt]
& = \sum_s t^s \frac{\Pr(S_N=s\ \&\ N=n)}{\Pr(N=n)} \\[8pt]
& = \sum_s t^s \frac{\Pr(S_n=s)\cdot\Pr(N=n)}{\Pr(N=n)} \\[8pt]
& = \sum_s t^s \Pr(S_n=s) \\[8pt]
& = \E(s^{S_n}).
\end{align} |
H: find $\int _\gamma \frac{1}{z+\frac {1}{2}}dz$
I'm asked to find $$\int _\gamma \frac{1}{z+\frac {1}{2}}$$ where $\gamma (t)=e^{it}, 0\leq t\leq 2\pi$.
To do this I deal with two different logarithms, one without the negative imaginary axis which I call $\log _1$ and another without the positive imaginary axis, $\log _2$.
Then $$\int _\gamma \frac{1}{z+\frac {1}{2}}=\log _1(z+\frac {1}{2})\bigg|^{z=-1}_{z=1}+\log _2(z+\frac {1}{2})\bigg|^{z=1}_{z=-1}=\\
(\log _1(-\frac {1}{2})-\log _1(\frac{3}{2}))+(\log _2(\frac {3}{2})-\log _2(-\frac{1}{2}))$$
now
$\log _1(-\frac {1}{2})=\log (\frac{1}{2})+i\pi$
$\log _1(\frac {3}{2})=\log (\frac{3}{2})+0$
$\log _2(\frac {3}{2})=\log (\frac{3}{2})+\color{green}{2i\pi}$, maybe $0$ instead of green??
$\log _2(-\frac {1}{2})=\log (\frac{1}{2})+i\pi $
so if green is right the integral is $2\pi i$ if instead of green it is $0$ the integral is $0$.
I think green is right.
can you please tell me the right answer?
AI: Your $\;\log_2(3/2)\;$ is wrong, imo:
$$\log_1\left(-\frac12\right)-\log_1\frac32+\log_2\left(\frac32\right)-\log_2\left(-\frac12\right)=$$
$$=\log\frac12+\pi i-\log\frac32+\log\frac32+2\pi i-\log\frac12-\pi i=2\pi i$$ |
H: Matrix norm inequality involving max and stacked matrices
In a paper I found the following inequality for matrices $A$ and $B$:
$\max\left\{||A||, ||B||\right\} \le \left\| \begin{align}A \\ B\end{align} \right\|_2 $
I suspect that this is a well-known inequality, which I just did not happen to find. Can somebody point me to the proof for this one?
AI: $\| \begin{bmatrix} A \\ B \end{bmatrix} \| = \sqrt{\| \begin{bmatrix} A^* & B^* \end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix}\|} = \sqrt{ \| A^*A + B^*B\|}$.
Since $A^*A, B^*B$ are positive semi-definite Hermitian matrices, we have $A^*A + B^*B \ge A^*A$ and similarly for $B^*B$. In particular, we have $\| A^*A + B^*B\| \ge \|A^* A \|$ and similarly for $B^*B$. Since $\sqrt{\|A^*A\|} = \|A\|$, we have the desired result.
Addendum: Note that $\|A\|^2 = \lambda_\max (A^*A)$. Furthermore,
$\lambda_\max (A^*A+ B^*B) \|x\|^2 \ge \langle x, (A^*A + B^*B) x \rangle \ge \langle x, A^*A x \rangle$. If we choose $x$ to be a unit eigenvector of $A^*A$ corresponding to $\lambda_\max (A^*A)$, then we obtain
$\lambda_\max (A^*A+ B^*B) \ge \lambda_\max (A^*A)$. And so we obtain
$\| A^*A + B^*B\| \ge \|A^* A \|$ |
H: Is $g(x,y) = f(\frac{x}{2},\frac{y}{2})$ correct notation?
I was a bit confused when I saw this statement $g(x,y) = 2f(\frac{x}{2},\frac{y}{2})$, and seeing it used in a double integral $\int \int g(x,y) = 2 \int \int f(\frac{x}{2},\frac{x}{2}) \, dx dy$.
I understand that idea is for us to use the change of variables $u = \frac{x}{2}$ and $v = \frac{y}{2}$. However, is $f(\frac{x}{2},\frac{y}{2})$ proper notation?
AI: Given any function $g \colon {\mathbb R}^2 \to {\mathbb R}$, you can define another function $f \colon {\mathbb R}^2 \to {\mathbb R}$ such that $g(x,y) = 2 f(\frac{x}{2}, \frac{y}{2})$ for all $x, y \in {\mathbb R}$. Just define $f(x, y) := \frac{1}{2} g(2x, 2y)$ for all $x, y \in {\mathbb R}$.
With that definition $\int\!\!\int g(x,y) \, dx dy = \int\!\!\int 2 f(\frac{x}{2}, \frac{y}{2}) \, dx dy = 2 \int\!\!\int f(\frac{x}{2}, \frac{y}{2})\, dx dy$. Changing variables as you indicated ($x = 2u$, $y = 2v$) then gives that this is equal to $2 \int\!\!\int f(u,v) \,2 du \,2 dv = 8 \int\!\!\int f(u,v) \,du dv$. |
H: Prove that $\left( \frac{p-1}{2} \right)! \equiv (-1)^n \mod p$, $n$ is quad. nonres. of $p$ $< p/2$.
Let $p$be a prime number with $p \equiv 3 \mod 4$. Prove that $\left( \frac{p-1}{2} \right)! \equiv (-1)^n \mod p$ where $n$ is the number of positive integers less than $p/2$ that are quadratic nonresidues of $p$.
I have tried solving this problem using the fact that $\left( \frac{p-1}{2}\right)! \equiv \pm 1 \mod p$ if and only if $p \equiv 3 \mod 4$. But I can't seem to bring the logic with the $n$ into the picture. Can you help?
AI: You can write
$$1 \equiv \left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv \prod_{k=1}^{\frac{p-1}{2}}k^2.$$
Now write the squares $> \frac{p-1}{2}$ as $k^2 = p-r$ with $r \leqslant \frac{p-1}{2}$. Count the $-1$s. |
H: Compact but not Hausdorff space
I think this space might be a space which is compact and non-Hausdorff, but I don't how to prove this.
Let $x,y\in\mathbb{R}^n$, define $x\sim y\leftrightarrow\exists t\neq 0(x=ty)$. Then $\sim$ is an equivalence relation and $\mathbb{R}^n/\sim$ is a compact and non-Hausdorff space.
The strategy might be constructing a "natural" compact non-Hausdorff space and proving they are homeomorphic. But how to come up with such spaces?
AI: Let $X = \mathbb{R}^n / \sim$, and consider the projection map $\pi : \mathbb{R}^n \to X$ which maps each point to its equivalence class. By definition of the quotient topology, $\pi$ is continuous. If we let $E = S^{n-1} \cup \{0\}$ consist of the unit sphere in $\mathbb{R}^n$ together with the origin, then $E$ is compact in $\mathbb{R}^n$, and $\pi(E) = X$ (as $E$ contains at least one point from every equivalence class). Hence $X$ is the continuous image of a compact set.
To see it is not Hausdorff, notice that the only open set in $X$ which contains the equivalence class $[0]$ is $X$ itself. Indeed, if $U$ is open in $X$ and contains $[0]$, then $\pi^{-1}(U)$ contains a neighborhood of $0$ in $\mathbb{R}^n$; in particular it contains a point of every equivalence class. So in fact $U$ contains every equivalence class in $X$. Now this prevents $X$ from being Hausdorff, since we cannot find disjoint open neighborhoods of $[0]$ and any other point. |
H: About norm and duality in $\mathcal S(\mathbb R^n)$
Reading the book "Classical and multilinear harmonic analysis, Vol. 1" by Muscalu, Schlag, 2013; I have a problem understanding the first step of the proof of Lemma 11.3.
The relevant parts are:
Let $\mu$ be a finite measure on $\mathbb R^n$ with $n\geq 2$ and $g\in L^2(\mu) \cap \mathcal S(\mathbb R^n)$. Then
$$\Vert g\Vert_{L^2(\mu)} = \sup_{f\in \mathcal S(\mathbb R^n), \Vert f\Vert_{L^2(\mu)} = 1} \left| \int_{\mathbb R^n} \hat g(\xi) f(\xi) d\mu(\xi) \right|$$
I understand that $(L^2(\mu))' = L^2(\mu)$, but why is it justified to say that
$$\Vert f \Vert_{L^2(\mu)} = 1 \Leftrightarrow \Vert \overline{\hat f} \Vert_{L^2(\mu)} = 1$$
Because this is what I'd expect for the usual notation of
$$\Vert x \Vert_E = \sup_{y\in E', \Vert y \Vert_{E'}=1} |y[x]|$$
With
$$y[x] = \int_{\mathbb R^n} x(\xi) \overline{y(\xi)} d\xi$$
AI: Okay, a bit of thought put into it and using Parseval's identity I found the solution myself:
$$\Vert f \Vert_{L^2(\mu)}^2 = \int f\bar{f} d\mu = \int \hat f \check{\bar f} d\mu = \int\hat f \overline{\hat f} d\mu = \Vert \hat f \Vert_{L^2(\mu)}^2$$ |
H: Show that prime $p=4n+1$ is a divisor of $n^{n}-1$
Show that the prime number $p=4n+1$ is a divisor of $n^{n}-1$
Ok, the question itself is simple as hell, but I couldn't think of a simple way to solve this question. I tried to solve the question by using $p\equiv 1 \pmod n$ but only to fail miserably... I couldn't use quadratic residue concept neither since n could be odd.
It would be really glad if some could figure it out using methods of elementary number theory such as primitive root, Jacobian symbols and so on (since I'm just a beginner for number theory stuff).
Thanks!
AI: We know that $n$ is a quadratic residue modulo $p$, since $1 = \left( \frac{-1}{p}\right) = \left(\frac{4n}{p}\right) = \left(\frac{n}{p}\right)$. Let $k^2 \equiv n \pmod{p}$. Then
$$(k+2n)^2 = k^2 + 4nk + 4n^2 \equiv n -k - n \equiv -k \pmod{p},$$
so $n$ is in fact a fourth power, $n \equiv (k+2n)^4 \pmod{p}$. Hence
$$n^n \equiv (k+2n)^{4n} = (k+2n)^{p-1} \equiv 1 \pmod{p}.$$ |
H: Problem with change of variables
I have this integral: $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}xyz\,dx\,dy\,dz=\frac{1}{8}$$
But when I make this change of variables:$$x=t$$$$y=t$$$$z=t$$ I have $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}t^3\,dt\,dt\,dt=\frac{1}{4} ?!$$
What am I doing wrong?
AI: If $D$ is the region you are integrating over (the cube in the first octant with sides parallel to the $x$,$y$, $z$ axes on side length $1$). Then notice that the variables $x$, $y$, and $z$ give you a coordinate system in $D$ where you can describe the location of any point $P$ in $D$. Now, let's say we adjust to your "coordinate system" with $x=y=z=t$. Then we suddenly run into a problem, since now we can no longer describe every point in $D$! Indeed, if $P = (1/2, 1/2, 1/3)$, say, in regular cartesian coordinates, then how would you describe $P$ in your coordinate system? You would need $t = x = 1/2, t = y = 1/2$, and $t = z = 1/3$, meaning that you'd have to deduce that $1/2=1/3$, which is clearly a problem. So, the real problem with your change of coordinates is that it is not a change of coordinates. That is, your variable change no longer lets you give coordinates to describe all points in $D$, so you can't integrate over $D$ with this change of variables. |
H: estimate for $\int |f|d\mu$
Let $(\Omega,\mathcal A,\mu)$ be a measure-space with $\mu$ a finite measure (i.e. $\mu(A)<\infty$ for all $A\in\mathcal A$) and $f:\Omega\to\mathbb R$ a measurable function. Prove that:
1) $$\sum_{i=1}^\infty \mu(\{|f|\geq n\})\leq\int |f|d\mu\leq\mu(\Omega)+\sum_{n=1}^\infty \mu(\{|f|\geq n\})$$
2) $f$ is $\mu$-integrable if and only if $\sum_{n=1}^\infty \mu(\{|f|\geq n\})$ converges.
I don't really know how to start this. I think the $\Leftarrow$-part of 2) follows from 1) because if the sum converges the integral $\int |f|d\mu$ is finite, too.
But how can I estimate the sums by the integral?
AI: Let $S_n = |f|^{-1} ([n,n+1))$. We have $n \le |f(x)| < n+1$ for all $x \in S_n$.
Hence $\sum_n n 1_{S_n} \le |f| < \sum_n (n+1) 1_{S_n} $.
Now find a relationship between the $S_n$ and $\mu \{ x | |f(x)| \ge n \}$.
Addendum:
Note that $\Omega = \cup_{n=0}^\infty S_n$.
Note that $U_n = \{ x | |f(x)| \ge n \} = \cup_{k=n}^\infty\{ x | |f(x)| \in [k,k+1) \} = \cup_{k=n}^\infty S_k$. Since the $S_k$ are disjoint, we can write
$1_{U_n} = \sum_{k=n}^\infty 1_{S_k}$. Then
\begin{eqnarray}
\sum_{n=1}^\infty 1_{U_n} &=& \sum_{n=1}^\infty \sum_{k=n}^\infty 1_{S_k} \\
&=& \sum_{k=1}^\infty \sum_{n=1}^k 1_{S_k} \\
&=& \sum_{k=1}^\infty k 1_{S_k} \\
&=& \sum_{n=1}^\infty n 1_{S_n}
\end{eqnarray}
We have
\begin{eqnarray}
\sum_{n=1}^\infty \mu \{ x | |f(x)| \ge n \} &=& \sum_{n=1}^\infty \int 1_{U_n} \\
&=& \int \sum_{n=1}^\infty 1_{U_n} \\
&=& \int \sum_{n=1}^\infty n 1_{S_n} \\
&=& \sum_{n=1}^\infty \int n 1_{S_n}
\end{eqnarray}
Finally,
$$
\sum_{n=1}^\infty n 1_{S_n} = \sum_{n=0}^\infty n 1_{S_n} \le |f| < \sum_{n=0}^\infty (n+1) 1_{S_n} = \sum_{n=0}^\infty n 1_{S_n} + \sum_{n=0}^\infty 1_{S_n} = 1+ \sum_{n=0}^\infty n 1_{S_n} = 1+ \sum_{n=1}^\infty n 1_{S_n}
$$
and integrating gives
$\sum_{n=1}^\infty \mu \{ x | |f(x)| \ge n \} \le \int |f| \le \mu \Omega + \sum_{n=1}^\infty \mu \{ x | |f(x)| \ge n \} $. |
H: Homework - set theory infinite union
A question from my homework I'm having trouble understanding.
We are given:
$A(1) = \{\varnothing\}$, $A(n+1) = A(n)\cup (A(n)\times A(n))$
$A=A(1)\cup A(2)\cup A(3)\cup \cdots \cup A(n)\cup A(n+1) \cup \cdots$ to infinity
The questions are:
1) show that $A\times A \subseteq A$
2) Is $A \times A = A$?
Thank you for your help.
I've tried writing $A(2)$ but it gets really complicated and I'm having trouble understanding what the sets are. Let alone solve the question.
AI: HINT: For (2), note that not all the elements of $A$ are ordered pairs.
Also, let's write $A(2)$, but to make it simpler let's call $A(1)=X$. Then $A(2)=X\cup(X\times X)=\{\varnothing\}\cup\{\langle\varnothing,\varnothing\rangle\}=\{\varnothing,\langle\varnothing,\varnothing\rangle\}$.
Not very difficult, $A(3)$ will have six elements. |
H: Finding The Order of Elements
This is a homework problem from my Group Theory class.
What is the order of $6$ in $\mathbb Z_{16}$?
I know $\mathbb Z_{16} = \{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$.
I know that in order to find the $\operatorname{ord}(6)$ I need to find $n,$ such that $6^n=e,$ where $e$ is the identity.
My main problems/questions here are:
Is the operation on $\mathbb Z_{16}$ addition [modular to be exact]?
If so, $e=0$.
$$6+6=12+6=2+6=8+6=14+6=4+6=10+6=0$$
Therefore $6^8=0.$
I am not sure if my work is correct. And if it is, is there a different way of going about it? For instance if I am trying to find the $\operatorname{ord}(6) \in \mathbb Z_{100}$ instead...or any larger group.
Thank you!
AI: You've got it!
You've found the order of $6 \in \mathbb Z_{16}$: the additive group of integers, modulo $16$. $$|\langle 6\rangle| = 8$$
Once you understand that the order of an element $g$ in a finite group is given by the least positive integer $n \neq 0$ such that $g^n = e$, or in an additive group, the least positive integer $n \neq 0$ such that $ng = 0$, and compute a few cycles like you did above:
Then you can more quickly use the fact that for the additive group of integers modulo $n$, the order of an element $g$ in $\mathbb Z_n$ is given by $$\operatorname{ord}(g) = \dfrac{n}{\gcd(g,n)}$$ |
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