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H: Inline elements denoted by {{}}? Cartesian product of these inline elements?
What does it mean when you have a set like A = {1, {2}, 3, 4}? Is {2} a subset within A?
Say I have the following:
A = {1, {2}, 3, 4}, B = {1, 2, {3}}, C = {1, 2}
How many elements are in $A \cap B \cap C$? 1 or 2?
What if I want to find $A \times B$, is there 6 elements or 12?
AI: The best way to deal with this is to write $a=\{2\}$ and $b=\{3\}$. Now we have: $$A=\{1,a,3,4\}, B=\{1,2,b\}, C=\{1,2\}.$$
In the common assumptions, it's not the case that $X=\{X\}$ or even $X\in X$. So it's safe to assume that $2\neq\{2\}$, and so on. Do note, however, that in some mathematical contexts it might be the case, but when you get to that, you usually know already what is the context and assumptions. |
H: Extended Hamming Code (8,4)
I just finished learning some basic concepts about Hamming code. Now I see this extended Hamming code and they say that you can "detect" up to 3 bits of error. Now, say, my actual codeword was $X=10100011$ and I received $Y=10011011$ (3 bits of error) and my transpose of parity check matrix $H(t)$ is below
1110
0111
1101
1011
1000
0100
0010
0001
Now from the rule I understand in (7,4) Hamming code was to compute syndrome which in this case would be S=1111 which is not present in the parity check matrix so that is probably not correct. So, can someone guide me with an example on how to detect 3 bit error per code in (8,4) Hamming code?
AI: Please recheck your calculations. I got $YH^T=1110$, which is also the first row of $H^T$. This means that your $Y$ is within Hamming distance one of the legal codeword $Z=00011011$.
What they mean when they say that a code of minimum distance four can detect it when $\le 3$ bits are in error is the following (does somebody actually phrase it that way? I wouldn't, but that is besides the point here). If the number or bit errors $t$ is in the range $1\le t\le3$, then the received word is guaranteed not to be a legal codeword. And that you can detect by calculating the syndrome (assuming the code is linear). It does not promise that you could ascertain that exactly one, two or three errors occur. The Hamming code has minimum distance three, but any sequence of seven bits is within Hamming distance one from a valid codeword. In other words, the covering radius of the Hamming code is equal to one. The covering radius of the $(8,4)$ extended Hamming code is two meaning that any sequence of 8 bits is within Hamming distance two of a valide codeword. But the Hamming codes are special in this sense. Usually it is difficult to find the covering radius of a code.
To conclude: with the extended Hamming code the best you can do is the following.
-Ccalculate the syndrome of the received word, if its all zeros, then assume that no errors occurred (or the number of errors was at least four)
- If the syndrome is non-zero, check whether it is a column of the check matrix (or a row, if you use transpose check matrix). If it is you assume that a single bit was in error, and can correct it. If not, you conclude that some two bits (but you don't know which).
In the case of the extended Hamming code you can always be sure whether the number of errorneous bits was even or odd (can you figure out how?). The usual practice (aka Maximum Likelihood decoding) is to assume that the number of errors is as low as possible, but it may, of course have been higher. |
H: Show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ with the box topology is Hausdorff but not metrizable.
Show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ with the box topology is Hausdorff but not metrizable.
$\Bbb{R}$ must be Hausdorff. For $x_1, x_2 \in \Bbb{R}$ (where $x_1 \not= x_2$), if $d$ denotes the distance between these two points, then we can choose an open ball $U_1$ with radius $d/2$ around $x_1$ and an open ball $U_2$ again of radius $d/2$ around $x_2$. Obviously, these two open balls will not intersect at any point.
So let $(x_1, x_2, ...), (x'_1, x'_2, ...) \in \Bbb{R^n}$. Then For each $x_i, x'_i \in \Bbb{R}$ (where $i \in \Bbb{N}$), there exist open balls $U_i$ and $U'_i$ in $\Bbb{R}$ around $x_i$ and $x'_i$, respectively, such that the open balls do not intersect. But then $(U_1 \times U_2 \times ...) \cap (U'_1 \times U'_2 \times ...) = \emptyset$.
We can show that $\displaystyle\prod_{\Bbb{N}} \Bbb{R}$ is not metrizable by proving that it is not first countable.
The first countability condition tells us that $\forall x \in X$ $\exists \{ U_n \}_{n\ \in \Bbb{N}} \subseteq P(X)$ such that
(i) $\forall n \in \Bbb{N}$, $U_n$ is a neighborhood of $x$.
(ii) For every neighborhood $V$ of $x$ $\exists n \in \Bbb{N}$ such that $U_n \subseteq V$.
So let's pick the point $(0,0,0,...) \in \displaystyle\prod_{\Bbb{N}} \Bbb{R}$. We know that, for all $n \in \Bbb{N}$ (in each $\Bbb{R}$) $(-1/n, 1/n)$ is a neighborhood of $0$.
So $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...$ is a neighborhood of $(0,0,0,0,...)$.
But in order to have $U_n \in \{U_n\}_{n \in \Bbb{N}}$ such that $U_n \subseteq (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...$ for all $n$, we know that the sequence $\{(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})...\}_{n \in \Bbb{N}} = \{U_n\}_{n \in \Bbb{N}}$.
But a sequences approaches infinity, and as $n \rightarrow \infty$ $(\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n}) \times (\frac{-1}{n}, \frac{1}{n})... = \{0\} \times \{0\} \times ...$, which is a closed neighborhood of $(0,0,0,...)$, meaning that the 1st condition of first countability fails.
Is my answer correct? If not, can anybody give me a hint?
Thanks.
AI: Your solution to Hausdorffness is not quite correct. Note what happens if your points are $\mathbf{x} = \langle 1 , 2, 3, \ldots \rangle$ and $\mathbf{x^\prime} = \langle 3 , 3 , 3 , \ldots \rangle$. (How do you choose the neighbourhoods $U_3$ and $U^\prime_3$?)
Recall that the usual product topology on $\mathbb{R}^\mathbb{N}$ is Hausdorff. Perhaps we can use this fact (or the proof of this fact.)
Your solution to the non-first-countability of the space is also somewhat off. You appear to be showing that one particular countable family of neighbourhoods of $\mathbf{0} = \langle 0 , 0 , \ldots \rangle$ is not a neighbourhood base. But you instead have to show that every countable family of neighbourhoods of $\mathbf{0}$ is not a neighbourhood base.
What you want to do is start with any old countably family $\{ U_n : n \in \mathbb{N} \}$ of open neighbourhoods of $\mathbf{0}$, and produce an open neighbourhood $V$ of $\mathbf{0}$ for which $U_n \not\subseteq V$ holds for all $n$. As a particularly strong hint, produce $V$ so that for each $n \in \mathbb{N}$ the projection of $V$ onto the $n$th coordinate is a proper subset of the projection of $U_n$ onto the $n$th coordinate. |
H: is it possible to proof that this number is not rational
It is an idea I had when reading the proof that $(0,1)$ is uncountable. There the numbers in $(0,1)$ are written into a list in decimal expansion and then the diagonal is modified and the resulting number is a number not on the list.
Now instead consider $S= \mathbb Q \cap (0,1)$. Like in the proof about $(0,1)$ write the numbers into a list in decimal expansion. Add $1$ to every digit on the diagonal and compute the remainder modulo $10$. I am trying to proof that this new diagonal number is not rational but without using the knowledge that the rationals are countable. Here is the proof:
There are two cases: A rational has either a finite expansion or is periodic. Let the digits in the expansions be called $a_{mn}$. Let the new diagonal number after the modification be $d_n$.
In the first case: If $a_{nn}$ is finite then $d_n$ is finite and $d_n = 0$ for $n>N$ for an $N$. Then, it is possible to find $n+1$ different periodic rational with no $0$ in the expansion. But this is a contradiction. Therefore $a_{n n}$ can not be finite.
In the second case: if $a_{nn}$ is a periodic rational with period $c_1 c_2 \dots c_N$. Then because of a same argument like the finite case this period can not be constant ($N=1$). It is forced to contain all digits in $\{0, ..., 9\}$. But what now? Is it possible to finish this proof without using that the rationals are countable?
AI: You start by explicitly assuming that $S=\mathbb Q\cap (0,1)$ is countable as you write all these numbers into a list.
If your list does not contain all elements of $S$ it is well possible that the antidiagonal number is in fact one of the rationals not in the list.
For example if $S$ consists only of those rational numbers having at least one $2$ in their decimal expansion, it might happen that the diagonal number is simply $0.222\ldots =\frac29$ and your antidiagonal number becomes $0.333\ldots=\frac13$, which is rational but not an element of $S$.
On the other hand, if $S$ really contains all eventually periodic decimal expansions then it is clear that the antidiagonal is not eventually periodic as it differs from each single element of $S$.
By the way, you should have a closer look at how you define your antidiagonal number: You might accidentally end up with $0.000\ldots=0$ or $0.999\ldots = 1$ |
H: Question regarding Lebesgue outer measure.
Given $m\geq1$, $0\leq s<\infty$, $0<\delta\leq\infty$ and $A\subseteq\mathbb{R}^{m}$ define: $$\mathcal{H}_{\delta}^{s}\left(A\right)=\inf\left\{ {\displaystyle \sum_{n=1}^{\infty}d\left(B_{n}\right)^{s}\,|\,\left\{ B_{n}\right\} _{n\in\mathbb{N}}}\:\mbox{is a covering of }A\:\mbox{and }d\left(B_{n}\right)<\delta\,\forall\, n\geq1\right\}$$
For ease of notation this refers to countable coverings but it also includes finite coverings. Now I've shown the following two things:
For all $m\geq1$, $0\leq s<\infty$, $0<\delta\leq\infty$ this defines an external measure on $\mathbb{R}^{m}$.
For all $0\leq s<\infty$ and $A\subseteq\mathbb{R}^{m}$ the limit $ \mathcal{H}^{s}\left(A\right)=\lim\limits _{\delta\downarrow0}\mathcal{H}_{0}^{s}\left(A\right)$ exists.
I'm now trying to show these following two things:
For all $ 0\leq s<\infty\quad$ $\mathcal{H}^{s}$ is an external metric measure on $\mathbb{R}^{m}$.
$\mathcal{H}^{0}$ is the counting measure on $\mathbb{R}^{m}$
Here is what I know so far regarding each of these goals:
I want to show that for each $A,B\subseteq\mathbb{R}^{m}$ if $dist(A,B)>0$ then $$\mathcal{H}^{s}(A\cup B)=\mathcal{H}^{s}(A)+\mathcal{H}^{s}(B)$$I know that $dist(A,B)>0$ is equivalent to $\overline{A}\cap\overline{B}=\emptyset$ so I assume I should use that somehow but I'm not sure how.
I've shown it works for finite sets but I'm having trouble with infinite sets. Obviously for $s=0$ we get that $\mathcal{H}_{\delta}^{0}\left(A\right)$ is a function only of the number of sets in the covering for which the infima is obtained. What I want to show is that given an infinite subset $A\subseteq\mathbb{R}^{m}$ as $\delta\downarrow0$ the number of sets required in order to cover $A$ by sets of diameter less than $\delta$ goes to infinity.
At first this struck me as a bit odd since for a compact infinite set we know that for each $\delta>0$ there is a finite $\delta$-net covering $A$ but then I realized that doesn't mean the number of sets in these nets doesn't go to infinity.
Anyway, I'd really appreciate some thick hints or a full proof of these two claims, I've already spent a couple of hours wrecking my head over it alone :)
AI: For (2) - for any $n$, an infinite set $A$ contains a subset of cardinality $n$. So as $H^0$ is an outer measure, $H^0(A)\geq n$. As this holds for all $n$, we are done. |
H: Is this a hyperbolic PDE?
Is the PDE
$$
(1+x^2)^2\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}+2x(1+x^2)\frac{\partial u}{\partial x}=0\text{ in }\Omega:=\mathbb{R}^2
$$
hyperbolic?
To answer this I set
$a(x,y):=(1+x^2)^2, b(x,y):=0, c(x,y)=-1$, then $b^2-ac=(1+x^2)^2 >0$
and this we had as a criterion for hyperbolicism.
Am I right?
AI: You are right but $\Delta(x, y)=b^2-4ac$. If we have a second-order PDE
$$a(x, y)u_{xx}+b(x ,y)u_{xy}+c(x, y)u_{yy}+ F(x, y, u, u_x, u_y)=0$$ (called semilinear equation), then we can classify the eqn. according to the sign of $\Delta(x, y).$ It is hperbolic on $\Omega$ if $\Delta(x, y)>0$ , parabolic on $\Omega$ if $\Delta(x, y)=0$ and elliptic on $\Omega$ if $\Delta(x, y)<0$ for all $(x, y)\in\Omega$ . |
H: Proving a predicate logic statement to be valid
I've been stuck on this question for the better part of the day, and I've succumbed to asking for help. I'm not sure how to go about it honestly. I've tried to do the contrapositive to prove it, but I get stuck and end up at a dead end.
The problem is as follows:
$$
[(\forall y \forall z \space p(y,z)) \space \vee \space (\exists z \forall y \space q(y,z))] \space \to \space \forall y\exists z(p(y,z) \space \vee \space q(y,z))
$$
I'm supposed to to prove this is valid using interpretations. I can not replace the implies either.
AI: Could you just pull out the quantifiers from the anacedent? Since $\forall y$ is in each anacendent condition, $\forall xp(x) \vee \forall x q(x) \rightarrow \forall x (p(x) \vee q(x))$ Also, the universal quantifier implies the existential quantifier, so $[(\forall y \forall z \space p(y,z))] \rightarrow [(\forall y\exists z \space p(y,z)) ]$ Also,$(\exists z \forall y \space q(y,z))] \to (\forall y \exists z \space q(y,z))]$ Since the $z$ that makes the second form true, is the same as the $z$ from the first form (note the converse is not true). Therefore you get the series of deductions:
$[(\forall y \forall z \space p(y,z)) \space \vee \space ( \exists z\forall y \space q(y,z))] \to [(\forall y \exists z \space p(y,z)) \space \vee \space (\forall y \exists z \space q(y,z))] \to \forall y \exists z \space [p(y,z) \space \vee \space q(y,z)]$ |
H: How do you simplify this big O sum?
I saw someone interpret $\sum_{i=1}^{n}\mathcal{O}\left(i^{k-2}\right)$ as $\mathcal{O}\left(n^{k-1}\right)$. Is this right? If so, can you explain?
AI: This is not quite true in this simplicity or at least depends on interpretation and context.
One might interprete the statement as follows: Suppose that for each $n\in\mathbb N$ we have a function $f_n\colon \mathbb N\to\mathbb R$. Then we can define
$f\colon \mathbb N\to \mathbb R$, $n\mapsto\sum_{i=1}^n f_i(n)$. Now if $f_n(x)=O(x^k)$ as $x\to \infty$ it does not follow that $f(x)=O(x^{k+1})$ as $x\to\infty$.
For example, let $$f_i(n)=\begin{cases}n!&\text{if }n=i\\0&\text{otherwise}.\end{cases}$$ Then $f_i(n)=O(1)$ as $n\to\infty$ and $f(n)=n!$.
On the other hand, assume we have $g\colon\mathbb N\to\mathbb R$ and define $f\colon \mathbb N\to \mathbb R$, $n\mapsto\sum_{i=1}^n g(i)$. Now if $g(x)=O(x^k)$ as $x\to\infty$ we have $f(x)=O(x^{k+1})$ as $x\to\infty$.
Indeed, assume $|g(x)|<cx^k$ for $x>N$. Then with $A=\left|\sum_{i=1}^Ng(i)\right|$ we have $$|f(x)|<A+\sum_{i=N+1}^n |g(i)|<A+c\sum_{i=N+1}^ni^k\le A+c(n-N)n^k<(c+A)n^{k+1}$$
for $n>N$. |
H: Finding possibilities puzzle
The answer: 2
Hi guys, I'm doing the practice papers for an examination, and got stuck on this question. I ended up getting the right answer, but I'm not sure if my thinking is correct. I would love someone else's opinion on it.
My thinking -- >
"Statue to fountain, always pass the bandstand"
The bandstand can only be the * in the middle, because that is the only place where going from one star to another star, the same star must always be crossed.
"Going from the lake to the bowling green, I never pass the fountain unless I pass the bandstand twice."
Not quite sure how to make that one out.
I ended up somehow getting at 2, but can't even seem to replicate my thinking in writing, which makes me think that it was more of a guess.
Would really appreciate other people's thinking on this!
AI: The star in the southeast has to be either the statue or the fountain, right? So lake and the bowling green must be west of the bandstand. If the lake, the bowling green, and the fountain were all on the circle west of the bandstand, you could certainly go from lake to fountain to bowling green without going past the bandstand more than once. So the fountain must be the star in the southeast.
That's two stars identified. Among the other three, no way of telling which is which. |
H: If a function is differentiable almost everywhere, can it be written as an integral?
Consider a function $f:\mathbb{R}^n \to \mathbb{R}$. If $f$ is differentiable with Lebesgue integrable derivative, we may write
$$
f(x+y) - f(x) = \sum_{i=1}^p \int_0^1 y_i \nabla f_i(x+ty)dt
$$
by the fundamental theorem of calculus. If it only holds that $f$ is differentiable on a set $A\subseteq\mathbb{R}^n$ such that the complement $A^c$ has Lebesgue measure zero (that is, $f$ is differentiable Lebesgue almost everywhere), does the same formula hold? If yes, how can this be proven rigorously?
AI: Consider the function $f:\mathbb{R}\to\mathbb{R}$ defined
$$f(x) = \begin{cases}1 & x\geq 0\\ 0 & x<0\end{cases}$$
Does your premise hold? Does the proposition? |
H: Are two Hilbert spaces with the same algebraic dimension (their Hamel bases have the same cardinality) isomorphic?
We know that two Hilbert spaces that have orthonormal bases of the same cardinality are isomorphic (as an inner product spaces).
My question is: what can we say when we know that their Hamel bases have the same cardinality? It clearly implies they are isomorphic as vector spaces (just send a basis to a basis and extend linearly), but are they isomorphic also as inner product spaces? (i.e. via an isomorphism that also preserves the inner-product).
AI: No, Hilbert spaces of different Hilbert dimensions can have the same algebraic dimension as vector spaces over $\mathbb R$ (or $\mathbb C$, take your pick).
For a cardinal $\kappa$, let $H_{\kappa}$ be the $\kappa$-dimensional real (or complex, whatever) Hilbert space, i.e., it has an orthonormal basis of cardinality $\kappa$. Let $\mathcal H=\{H_{\kappa}:\aleph_0\le\kappa\le2^{\aleph_0}\}$. The set $\mathcal H$ contains at least two nonisomorphic Hilbert spaces ($H_{\aleph_0}$ and $H_{2^{\aleph_0}}$) and maybe as many as $2^{\aleph_0}$ of them depending on your set theory. I claim that they are all algebraically isomorphic because they all have algebraic dimension $2^{\aleph_0}$ as vector spaces over $\mathbb R$.
First, since the number of points in the space $H_{\kappa}\in\mathcal H$ with orthonormal basis $\mathcal B$ is $|H_{\kappa}|\le|\mathcal B|^{\aleph_0}2^{\aleph_0}\le(2^{\aleph_0})^{\aleph_0}2^{\aleph_0}=2^{\aleph_0}$, the algebraic dimension of each $H\in\mathcal H$ is at most $2^{\aleph_0}$.
Next, we show that the infinite-dimensional separable Hilbert space $H_{\aleph_0}$ has algebraic dimension at least $2^{\aleph_0}$, by exhibiting a continuum of (algebraically) linearly independent elements in the space $\ell^2$ of square-summable sequences.
Let $(r_n:n\in\mathbb N)$ be an enumeration of the rational numbers. For $t\in\mathbb R$ and $n\in\mathbb N$ define $\varepsilon(t,n)$ to be $1$ if $r_n\lt t$ and $0$ otherwise. Finally, let $x_t=\left<\dfrac{\varepsilon(t,n)}n:n\in\mathbb N\right>\in\ell^2$. It is easy to see that every finite subset of $\{x_t:t\in\mathbb R\}$ is linearly independent. |
H: How to work out sinh^2(x)
I just did a question where I had $sinh^2(x)$
I know this is simply $(sinh(x))^2$ however couldn't work out where the extra 2 came from when working out.
$sinh(x) = \frac{e^x-e^{-x}}{2}$
so
$sinh^2(x) = (\frac{e^x-e^{-x}}{2})^2$
which I figured
= $\frac{e^xe^x-e^{-x}e^{-x}}{4}$
= $\frac{e^{2x}-e^{-2x}}{4}$
However I am told the answer
= $\frac{e^{2x}-2+e^{-2x}}{4}$
And I don't know where the 2 came from
AI: You are trying to do $(a-b)^2=a^2-b^2$, which is not correct. The correct expansion is $(a-b)^2=a^2-2ab+b^2$ Inserting $a=e^x, b=e^{-x}$, gives the formula you were given. The $2$ comes because $ab=1$ |
H: Why do my professors ignore my work?
I'm a high-school student finishing in December and about to pursue a career in mathematics.
In my free time, I like to ''research'' hard problems and come up with unique proofs or combine already-established proofs to come up with a beautiful solution. As a high-school student, I am obviously not researching anything of importance, but I am solving IMO (Mathematical olympiad) problems in unique ways and such, constructing beautiful reasoning behind them spanning several pages. I take great pride in my work (even though it is trivial to real mathematicians) and I would love some feedback...However, when I send my work to my professors and teachers, I always get ignored completely. I never receive anything back, even if I remind them. This is making me doubt myself and my abilities.
What on earth could be the reason for this? Do teachers loathe it when students put extra work on them like that? If so, why won't they tell me?
AI: I don't know which country you are from, but in the US, I doubt many HS math teachers would be able to solve math olympiad problems or would need to expend a lot of work to do so. Keep in mind their point of view and pressures.
If you feel that your work is truly unique and worthy of consideration, you may want to write up your proofs and post them on the General Math ArXiv site, where mathematicians can review your work and provide comments. Note that if you are not proving anything new, but just solving problems (albeit hard ones), then don't expect a ton of responses. Most researchers are quite focused on their work or work related to theirs, and don't have time to review problem solutions (especailly if they are already swamped with reviewing their PhD candidates work and assisting with class grading).
In general, I wouldn't take it personally. You said "your" professors...so I assume you have person-to-person contact with them..why don't you drop by after class and ask if they would be willing to meet to discuss (as opposed to just sending them something). They may not feel that the problems you work on are within their area of expertise. |
H: Probability, that the computer network will work
So a single computer blows up after turning it on with probability $0.05$. We have an order to create a network of $50$ computers, and we gathered $52$ computers for this purpose. What's the probability, that we can create a network with these computers?
My solution:
So in order to work either $50$ computers are good and $2$ are faulty, $51$ are good and $1$ is faulty or $52$ are good.
Therefore the probability is:
${{52}\choose{50}}(0.95)^{50}*(0.05)^2+{{52}\choose{51}}(0.95)^{50}*(0.05)^1+0.95^{52} \approx 0.5$
Now how many computers do I need, for the probability to be $\approx 0.99$?
AI: Your work is fine. To answer the $0.99$ question, you need to increase $52$ and repeat the calculation until the probability gets that high.
To reduce the calculation, you can reason that if the odds are only $0.01$ that less than $50$ computers are working, the chance of less than $49$ working is negligible. Then you can solve ${50+n \choose 49}0.95^{49}\cdot 0.05^{n+1}=0.01$ You can just plug this into a spreadsheet or use bisection with a calculator. $n$ came out higher than I expected. |
H: Is $f(n) = 2^{\frac{1}{2}(n^2-n)} / n!$ polynomially bounded?
The numerator counts the number of different adjacency matrices. I think Sterlings approximation helps to anwser my question but I fail to derive the answer.
So, is there a polynomial function $g(x)$ such that
$$f(x) \leq g(x)$$
AI: It's not polynomially bounded. For $n \geqslant 3$, we have $\frac12(n^2-n) \geqslant \frac13 n^2$, so
$$\frac{2^{\frac12(n^2-n)}}{n!} > \frac{2^{\frac13 n^2}}{n^n} = \left(\frac{2^{n/3}}{n}\right)^n.$$
$2^{n/3}$ has exponential growth, so altogether we have superexponential growth. |
H: Changing an exponential function to logarithmic
I have a question stating that $P=75e^{-0.005t}$ and they want to get t by itself.
I used the example $y=2^x = x=log_2(y)$
To find that $-0.005t = 75ln(P)$
So $t=\frac{75ln(P)}{-0.005}$
However apparently this isn't correct. Can someone please show me where I went wrong (not so much another way of doing it, but what was wrong in the working I used)
Thanks
AI: $$\begin{align} P=75e^{-0.005t} & \iff \dfrac P{75} = e^{-0.005t} \\ \\ &\iff \ln\left(\frac P{75} \right) = \ln\left(e^{-0.005t}\right) \\ \\ & \iff \ln\left(\frac P{75} \right) = -0.005t \\ \\ & \iff t = -\left(\frac{\ln P - \ln (75)}{0.005}\right)\end{align}$$ |
H: Open set and sequences in $\mathbb{R}$
Let $A \subset \mathbb{R}$. Prove that $A$ is an open set if, and only if, the following condition is satisfied: " if a sequence $(x_n)$ converges for a point $a \in A$, then $x_n \in A$ for all $n$ sufficiently large".
I have doubts in the second implication $( \Leftarrow )$. Let $A \subset \mathbb{R} $ and $x_n$ a sequence of real numbers such thar $x_n \to a \in A$ and for all sufficiently large $n$ we have that $x_n \in A$. But I can't find an $ \varepsilon$ such that $B_{\varepsilon}(a) \subset A$ because I don't see why my $A$ cannot be $[a, +\infty)$, for example. I know that, if this $\varepsilon$ exists, it will come from the definition of this limit.
Thanks for your help!
AI: Hint: Try by contradiction. That is, if $a_n\to a$ as $n\to\infty$, we know from the definition that $|a_n-a|<\varepsilon$ for all sufficiently large $n$. Now, if $A$ is not open, then can you construct a sequence converging to $a$ such that none of the points are in $A$? This provides a contradiction to the hypothesis. |
H: Find intersection points of two functions
I have
$f(x)=\sqrt{3x}+1$
$g(x)=x+1$
My thinking was that at the intersection points both will be equal to each other so
$\sqrt{3x}+1=x+1$
$\sqrt{3x}=x$
However I don't know where to go from here.
AI: $$\sqrt{3x} = x \implies 3x = x^2 \iff x^2 - 3x = x(x - 3) = 0$$
$$\implies x = 0 \quad \text{OR}\quad x = 3$$
Hence, the intersection points will be $(0, f(0)),$ and $(3, f(3))$. |
H: Coded language puzzle!!
Here is a puzzle I can't crack. It goes like this:
In a certain coded language
MANGO=3/5
ORANGE=2/6
APPLE=1/5
Then, POTATO=??
The answer is 5/6. I would like to know to arrive at the answer.
AI: HINT: It appears that the denominator is the number of letters in the word. The numerator also appears to be a count: specifically, it appears to be the number of letters of the word that come from a certain $3$-letter set. That is, there are three letters, $\ell_1,\ell_2$, and $\ell_3$, such that MANGO contains all $3$ of them, ORANGE contains $2$ of them, and APPLE contains just one of them. Once you identity these three letters (which can be arranged to form a common word), see how many of them are in POTATO.
Note that there may be other solutions leading to different answers. |
H: Field and Algebra
What is the difference between "algebra" and "field"? In term of definition in Abstract algebra.
(In probability theory, sigma-algebra is a synonym of sigma-field, does this imply
algebra is the same as field?)
AI: An algebra is a ring that has the added structure of a field of scalars and a coherent (see below) multiplication. Some examples of algebras:
M_n(F), where $F$ is any field.
$C(T)$, continuous real (or complex)-valued functions on a topological space $T$ (here the scalars could be either the real or the complex numbers).
$B(X)$, bounded operators over a Banach space $X$, with complex (or real) scalars.
$F[x]$, polynomials over a field $F$.
A field, on the other hand, is a commutative ring where every nonzero elements is invertible (i.e. a commutative division ring).
Note: "coherent multiplication" means that given $x,y$ in the algebra and $\alpha,\beta$ in the field,
$$
\alpha(x+y)=\alpha x+\alpha y,\ \ (\alpha+\beta)x=\alpha x+\beta x, \ \ (\alpha\beta)x=\alpha(\beta x),\ \ (\alpha x)y=x(\alpha y).
$$ |
H: Percent spread in ratio
Say I have five solutions of various concentrations such as the below:
A = 16%
B = 12%
C = 12.5%
D = 17%
E = 5%
Their quantities do not matter, only their concentrations. Say I want to mix them and the final spread between them should follow a ratio of 1:2:1:2:6. Is this enough information given to determine the final % of each solution in the final mixture? If so, how would I calculate the final %s of each solution?
AI: Riista..based on your description, if you know the dilution ratios, then since each solution contains a different solute, each solution is a relative dilutant to the others. For example, the concentration of A after mixing with your ratios will be: $\frac{1\cdot16\%}{1+2+1+2+6=12}$. In general, the final % will be $\frac{ratio \cdot conc.\%}{12}$ |
H: The limit $\left( \sin x \right)^{x}$ when $x \rightarrow\; 0$
Take the limit $\left( \sin x \right)^{x}$ when $x \rightarrow\; 0$
I tried using $e^{\ln \left( \left( \sin x \right)^{x} \right)}=e^{x\cdot \ln \left( \sin x \right)}$ and then saying sinx → 0 but ln(0) is undefined. Stopped there.
AI: For $x$ near zero, $\sin x\sim x$, so you are looking at $x\log x$ when $x\to0$, which goes to zero. This suggests the limit is $e^0=1$.
We can confirm this by using L'Hopital:
$$
x\log\sin x=\frac{\log\sin x}{1/x},
$$
which is of the form "$\infty/\infty$". So
$$
\lim_{x\to0}x\log\sin x=\lim_{x\to0}\frac{\log\sin x}{1/x}=\lim_{x\to0}\frac{\cos x}{\sin x}\,\frac{1}{\frac{-1}{x^2}}=\lim_{x\to0}\frac{-x\cos x}{\frac{\sin x}x}=0
$$
Edit: here is a way to see that the sine function has no impact in how this limit behaves: it involves the change of variable $t=\sin x$.
$$
\lim_{x\to0}x\log\sin x=\lim_{x\to0} \frac{x}{\sin x}\,\sin x\,\log\sin x=\lim_{x\to0} \frac{x}{\sin x}\,\lim_{x\to0} \sin x\log \sin x\\ =\lim_{x\to0} \sin x\log x=\lim_{t\to0}\,t\log t=0
$$
Edit 2: for the sake of completeness, I will include a proof of $\lim_{x\to0}x\log x=0$ without using L'Hopital. All we need is the Mean Value Theorem and some basic knowledge about the exponential and logarithm functions.
First, note that
$$
\lim_{x\to0}x\log x=\lim_{x\to0^+}x\log x=\lim_{u\to\infty}\frac1u\,\log\frac1u=-\lim_{u\to\infty}\frac1u\,\log u=-\lim_{t\to\infty}\frac{t}{e^t}.
$$
To calculate this last limit, we will use the MVT. Given $s>0$, there exists $\xi\in(0,s)$ with
$$
e^s-1=e^{\xi}(s-0)=se^\xi.
$$
As $e^\xi\geq1$, we get $e^s-1\geq s$, or $e^s\geq1+s$. Integrating from $0$ to $t$, we get (as both sides of the inequality are positive functions, integrating will preserve the inequality)
$$
e^t-1=\int_0^te^s\,ds\geq\int_0^t(1+s)\,ds=t+\frac{t^2}2.
$$
Then $e^t\geq1+t+\frac{t^2}2\geq\frac{t^2}2$ (of course, if you know Taylor's expansion, you can get this right away). Now, for $t>0$,
$$
0\leq\frac{t}{e^t}\leq\frac{2t}{t^2}=\frac2t.
$$
As $2/t\to0$ as $t\to\infty$, we get $\lim_{t\to\infty} t/e^t=0$. |
H: Finding a,b,c,d in a quartic expression
Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far.
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?
AI: You have $p(x) = x^4 + ax^3 + bx^2 + cx + d$, and you're given that $a + b + c + d = 9$, $8a + 4b + 2c + d = 4$, and $27a + 9b + 3c + d = -51$
Now, $p(12) + p(-8) = 12^4 + 8^4 + (12^3 - 8^3) a + (12^2 + 8^2) b + (12 - 8) c + 2d = 24832 + 1216 a + 208 b + 4c + 2d = 24832 + 1216 a + 208 b + 2 (2c+d)$. You note that $2c + d = 4 - 8a - 4b$, and substitute that into the equation to get $p(12) + p(-8) = 24832 + 1216 a + 208 b + 8 - 16a - 8b = 24840 + 1200 a + 200b = 24840 + 200(6a + b).$
Plug in the $6a + b = -25$ and you get $p(12) + p(-8) = 24840 - 5000 = 19840$. Divide it by $10$ and you get $\displaystyle \frac{p(12) + p(-8)}{10} = 1984$.
Remember, $d + d \ne d$. |
H: geometric progression calculation
In a given geometric progression, a1 equals 30, q in absolute value is smaller than one, and the sum of all arguments in even positions is 11.25. what is the value if q?
attempt at a solution: for the even arguments, a1 is 30q, q is 2q and we know the sum in infinity is 11.25. plugging that to the summation formula for this kind of progression yields 3/14. official answer is 1/3 . any suggestions?
AI: It appears that the terms are numbered starting with $a_1$, so that $a_2=a_1q$, $a_3=a_1q^2$, and in general $a_n=a_1q^{n-1}$. Thus, the subseries of even-numbered terms is
$$11.25=a_1q+a_1q^3+a_1q^5+\ldots=a_1q\sum_{n\ge 0}q^{2n}=a_1q\sum_{n\ge 0}(q^2)^n=\frac{10q}{1-q^2}\;,$$
and $11.25(1-q^2)=10q$. This is just a quadratic in $q$, so you can solve it. (It even factors nicely once you simplify it to a form with integer coefficients, though of course you can also just use the quadratic formula.) You’ll find that one solution is the answer that you were given, and the other is unusable, because it’s too big. |
H: Find integers $a$, $b$ and $c$ with $55a + 65b + 143c= 1$.
I'm not sure if this is a diophantine equation with three variables or not, but I can't find any resources for it. I am thinking there must be some sort of solving for two and then substituting the answer into another two variables. Unfortunately my notes are not sufficient.
How would I go about solving? Any hints?
AI: Hint: Use standard two-variable theory to solve $55s+65t=5$. (Find $x$ and $y$ such that $11x+13y=1$.)
Then use standard theory to solve $5u+143v=1$. Then put things together. |
H: Calculating $\operatorname{Res} \left(\frac{f(z)}{g(z)}, z=a\right)$ with $a$ a double zero of $g$.
I have to show that for $f,g$ analytic on some domain and $a$ a double zero of $g$, we have:
$$\operatorname{Res} \left(\frac{f(z)}{g(z)}, z=a\right) = \frac{6f'(a)g''(a)-2f(a)g'''(a)}{3[g''(a)]^2}.$$
The problem is that direct calculation using the formula (for pole of order $2$):
$$\operatorname{Res}(h(z),z=a)=\lim_{z \to a} \frac{d}{dz}\left( (z-a)^2h(z) \right)$$
is extremely ugly, given that we're dealing with a quotient. Is there some sort of trick to make the calculation more manageable?
AI: A little Taylor expansion takes you a long way. Say
$$g(z) = (z-a)^2\left(a_2 + a_3(z-a) + (z-a)^2\cdot \tilde{g}(z)\right)$$
with $a_2 \neq 0$, and
$$f(z) = b_0 + b_1(z-a) + (z-a)^2\cdot \tilde{f}(z).$$
Then
$$\begin{align}
\frac{f(z)}{g(z)} &= \frac{b_0 + b_1(z-a) +(z-a)^2\tilde{f}(z)}{(z-a)^2\left(a_2 + a_3(z-a) + (z-a)^2\tilde{g}(z)\right)}\\
&= \frac{1}{a_2(z-a)^2}\frac{b_0 + b_1(z-a) + (z-a)^2\tilde{f}(z)}{1 + \frac{a_3}{a_2}(z-a) + (z-a)^2h(z)}\\
&= \frac{1}{a_2(z-a)^2}\left(b_0 + b_1(z-a)\right)\left(1-\frac{a_3}{a_2}(z-a)\right) + \tilde{h}(z)\\
&= \frac{c}{(z-a)^2} + \frac{b_1 - (b_0a_3)/a_2}{a_2(z-a)} + k(z),
\end{align}$$
so the residue is
$$\frac{b_1a_2 - b_0a_3}{a_2^2} = \frac{f'(a)\frac12g''(a) - f(a)\frac16g'''(a)}{\left(\frac12g''(a)\right)^2} = \frac{6f'(a)g''(a) - 2f(a)g''(a)}{3g''(a)^2}.$$
That didn't hurt, doctor. |
H: Describe the kernel and the range, and any vector such that T(p)=y.
Let $T: \mathbb{P_2} \to \mathbb{R^2}$ be a linear transformation given by
$T(p)=[ p(0) ; p'(0)]$
Describe the kernel and range. That is to say, for what polynomials of the form
$p(t)= at^3 + bt^2 + ct + d$ does T map to zero?
Also, for what vectors of the form $y$ does $T(p) = y$ (The question also asked something
about $\mathbb{R^2}$ here but I just left the test and cannot remember completely. Any explanation is graciously accepted.
AI: You made one error in your post. If $p\in\Bbb P_2,$ then $p(t)=at^2+bt+c$ for some constants $a,b,c$.
Now, $p'(t)=2at+b.$ Consequently, $p(0)=c$ and $p'(0)=b.$ What do we need to know about $b$ and $c,$ then, to know that $p$ is in the kernel of $T$? How, then, can you describe the kernel of $T$?
Also, noting that $$T(p)=\begin{bmatrix}c\\b\end{bmatrix},$$ and that $b,c$ can be any real constants we like, what can you say about the range of $T$?
As for the last question, my suspicion is that it says something like: "Given $y\in\Bbb R^2,$ find some $p\in\Bbb P_2$ such that $T(p)=y,$ if possible." (That's what it would say, if I wrote the exam.) Once you've worked your way through my hint about the range, you should be able to answer this question readily. |
H: Proof that one large number is larger than another large number
Let $a = (10^n - 1)^{(10^n)}$ and $b=(10^n)^{(10^n - 1)}$
Which of these numbers is greater as n gets large?
I believe it is $a$ after looking at some smaller special cases, but I'm not sure how to prove it.
AI: Divide:
$$\frac{a}{b} = \frac{(10^n-1)^{10^n}}{(10^n)^{10^n-1}} = (10^n-1)\left(1- \frac{1}{10^{n}}\right)^{10^n-1} \approx (10^n-1)e^{-1} > 1.$$ |
H: Equation of the tangent to the curve
I want to find the equation of the tangent to the curve:
$F(x) = \sqrt x$ at $x=4$. (Write answer in slope intercept form)
I am very confused on this and would need step by step directions. Thanks
AI: Given $f(x) = \sqrt x$.
"Step-by-Step Directions" (with "spoilers" so you can check your answers: just hover over the gray with your cursor).
Find $f'(x)$
$f'(x) = \frac{1}{2\sqrt{x}}$
Evaluate $f'(x)$ at $x = 4$. The slope of your line is equal to $m = f'(4)$.
$m = \frac 14$
Then, you have a point on the line (the point at which the desired
line is tangent. What is $y = f(x) = \sqrt x\;$ when $\;x = 4$?
$y = \sqrt 4 = 2$
Now you have slope, a point on the line $(x_0, y_0) = (4, f(4))$, and
you can use the point-slope form of an equation: $$y = m(x - x_0) +
y_0$$ to obtain the equation,
$y = \frac 14(x - 4) + 2$
Finally, manipulate this equation to obtain slope-intercept form.
$ y = \frac 14 x + 1$ |
H: Step in finding $\sin^{-1}z = w$ for a fixed complex $w$ and unknown complex $z$
This is in the section of the book preceding a general formula but I don't know how the author arrives to the second equation in the picture. The closes I have gotten to it is
$$2iz = e^{iw}(1-e^{-1})$$
but I don't see how I can get an $(e^{iw})^2$ term without also having a $z^2$ term.
AI: Note that $$z=\frac{e^{iw}-e^{-iw}}{2i}$$ can be rewritten (through some arithmetic) as $$e^{iw}-2iz-e^{-iw}=0.$$ Now, multiply by the non-zero quantity $e^{iw}$ to get $$(e^{iw})^2-2iz(e^{iw})-1=0,$$ as desired. |
H: Limit of f(x). What does it tell about Limit of f(x)^n?
Suppose $\lim_{x\to\infty}f(x) = L$. Is it telling us something about $\lim_{x\to\infty}f(x)^n$?
AI: The function $\phi(x) = x^n$ is continuous, and $\lim_{x \to \infty} f(x) = L$, hence $\lim_{x \to \infty} \phi(f(x)) = \phi(\lim_{x \to \infty} f(x)) = \phi(L) = L^n$. |
H: How to find all perfect squares in a given range of numbers?
I need to write a program that finds all perfect squares between two given numbers a and b such that the range can also be a = 1 and b = 10^15 what is the best way I can do this, how do I list down all such square numbers, is there some abstract math hidden underneath this problem?
AI: One thing that makes this pretty straight forward is this: $(n+1)^2-n^2=2n+1$ Start with one, and keep on adding that. |
H: Prove Multinomial Coefficient (Probability Theory)
Prove that the multinomial coefficient given by:
$$
\binom{n}{n_1}\binom{n-n_1}{n_2}\binom{n-n_1-n_2}{n_3}\cdots\binom{n-n_1-n_2-\dots-n_{k-1}}{n_k}
$$
equals the following expression
$$
\frac{n!}{n_1!n_2!n_3!\cdots n_k!}
$$
Thank you for any help you can give me
AI: Just write it out and see that most terms cancel out.
$$
{n \choose n_1}{n-n_1\choose n_2}\cdots{n-n_1-n_2-\cdots n_{k-2} \choose n_{k-1}}{n-n_1-n_2-\cdots n_{k-2}-n_{k-1} \choose n_{k}}
$$
Let's look at the first part:
$$
{n \choose n_1}{n-n_1\choose n_2}=\frac{n!}{n_1!(n-n_1)!}\frac{(n-n_1)!}{n_2!(n-n_1-n_2)!}=\frac{n!}{n_1!n_2!(n-n_1-n_2)!}
$$
and then with the next term:
$$
{n \choose n_1}{n-n_1\choose n_2}{n-n_1-n_2\choose n_3}=\frac{n!}{n_1!n_2!(n-n_1-n_2)!}\frac{(n-n_1-n_2)!}{n_3!(n-n_1-n_2-n_3)!}=\frac{n!}{n_1!n_2!n_3!(n-n_1-n_2-n_3)!}
$$
and that way it continues. In the end you will have
$$
{n \choose n_1}\cdots{n-n_1-n_2-\cdots n_{k-2}-n_{k-1} \choose n_{k}}=\frac{n!}{n_1!n_2!\cdots n_{k-1}!(n-n_1-n_2-\cdots-n_{k-1})!}\frac{(n-n_1-n_2-\cdots-n_{k-1})!}{n_k!(n-n_1-n_2-\cdots-n_{k})!}=\frac{n!}{n_1!n_2!\cdots n_k!}\frac{1}{(n-n_1-n_2-\cdots-n_k)!}
$$
However, since
$$
n=\sum_{i=1}^kn_i
$$
it follows that
$$
(n-n_1-n_2-\cdots-n_k)!=1
$$
and the result is that
$$
{n \choose n_1}\cdots{n-n_1-n_2-\cdots n_{k-2}-n_{k-1} \choose n_{k}}=\frac{n!}{n_1!n_2!\cdots n_k!}.
$$ |
H: Path of an ellipse
A path is described by the position vector $\mathbf{r}$:
$$\mathbf{r}=a\cos(\omega t)\mathbf{\hat{i}}+b\sin{\omega t}\mathbf{\hat{j}}$$
I am asked to show that the path is the ellipse in the form of:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Is this not converting from polar to cartesian using $x=\mathbf{r}\cos(\omega t)$ and $y=\mathbf{r}\sin(\omega t)$? When I tried this I got as far as $x^2+y^2-ax-by=0$ and I am not sure what to do next, thanks.
AI: Hint: Consider above $\bf{r}$ as $$\mathbf{r}=x(t)\mathbf{\hat{i}}+y(t)\mathbf{\hat{j}}$$ and then satisfy $x(t),~y(t)$ in the given Cartesian equation. |
H: Checking step by step proof of $\lim_{n\to\infty}(\sqrt2-\sqrt[n]2)^n=0$
$$\mathop {\lim }\limits_{n \to \infty } {(\sqrt 2 - \root n \of 2 )^n}$$
Is it right to say:
the limit of $\sqrt2$ is $\sqrt2$
the limit of $\root n \of 2$ is 1
then, $\sqrt2 - 1$ is between 0 to 1.
so, the limit of $(\sqrt2 - 1)^n $ must be 0.
in conclusion,
$$\mathop {\lim }\limits_{n \to \infty } {(\sqrt 2 - \root n \of 2 )^n} = 0$$
is it right?
AI: Yes, that works just fine and is perfectly rigorous, once you take ABC's comment into consideration. |
H: GRE counting problem
I have approached this question the following way.
For Quantity $A$: The number of ways to pick $3$ cards including $1$ is : $1\cdot 4\cdot 3=12$ ways. [$1$ is fixed]
For Quantity $B$: The number of ways to pick $3$ cards excluding $1$ is : $4\cdot 3\cdot 2=24$ ways resulting $24$ which means $B$ is a greater quantity, but I got it wrong.
Can someone explain to me where I went wrong? I did not find Manhattan's explanation to this problem sufficient, they use some weird shortcuts. Can someone break this down if possible?
AI: They’re counting $3$-card sets, not sequences of $3$ cards. There are $\binom42=6$ pairs of cards that could be combined with the $1$, so Quantity A is $6$. There are $\binom43=4$ ways to choose $3$ of the non-$1$ cards, so Quantity B is only $4$.
Since the numbers are so small, here are the $3$-card hands that include the $1$:
$$\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,3,5\},\{1,4,5\}$$
And here are the $3$-card hands that don’t include it:
$$\{2,3,4\},\{2,3,5\},\{2,4,5\},\{3,4,5\}$$ |
H: How to isolate y?
I've got an equation: $$-4y=12-3x$$
I want to simply isolate the y variable so i could get rid of the -4 coefficient of the y variable.
What can I do to isolate the y variable? I've done the following but i'm not sure:
$$ y = \frac{12}{-4}-\frac{3x}{-4} $$
Is that the right way for solving this kind of equation? if not, what can i do?
Thanks in advance, really appreciate that.
AI: What you've done is correct. Basically, you want the $y$-term to be on one side and then divide: $$ -4y = 12-3x \implies 4y = 3x - 12 \implies y = \boxed {\dfrac {3x-12}{4} = \dfrac {3}{4}x - 3}. $$ |
H: field extension-notation problem $\mathbb{Q}(\sqrt{2})=\mathbb{Q}[\sqrt{2}]$
let $K\subseteq L$ be a field extension and let $K=\mathbb{Q}$ and $L=\mathbb{C}$ also let $\alpha =\sqrt{2}$.
Then $\mathbb{Q}(\sqrt{2})=\mathbb{Q}[\sqrt{2}]$
What is the difference in the above notations? and why do we get them?
$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\}$
what is $\mathbb{Q}[\sqrt{2}]$?
AI: For fields $K \subset L$, and an $\alpha\in L$,
$K[\alpha]$ denotes the smallest subring of $L$ that contains both $K$ and $\alpha$, and
$K(\alpha)$ denotes the smallest subfield of $L$ that contains both $K$ and $\alpha$.
$K[\alpha]$ is the set of all polynomial expressions in $\alpha$,
$$K[\alpha] = \left\lbrace f(\alpha) : f \in K[X]\right\rbrace,$$
and $K(\alpha)$ is the set of all rational functions evaluated in $\alpha$ subject to the restriction that the denominator doesn't vanish in $\alpha$,
$$K(\alpha) = \left\lbrace \frac{f(\alpha)}{g(\alpha)} : f,g\in K[X],\, g(\alpha)\neq 0\right\rbrace.$$
It turns out that for algebraic $\alpha$, the two coincide, $K[\alpha] = K(\alpha)$. In fact, that is a necessary and sufficient condition, $\alpha$ is algebraic over $K$ if and only if $K[\alpha]$ is a field. |
H: Stuck with Taylor expansion of $f(x+x')$
I know that the Taylor series of $f(x)$ around $a$ is given by:
$$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2}+\dots=\sum_{n=0}^\infty \frac{f^{(n)}(a) }{n!} (x-a)^n$$
In my textbook I see the following formula for
$f(x+x')$ which I however don't understand:
$$f(x+x')=f(x)+f'(x)x' +\frac{f''(x)}{2}(x')^2+\dots=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!} (x')^n $$
I don't understand how they obtained this formula. Can someone explain me that?
AI: Rewrite your original formula, using $t$ instead of $x$:
$$
f(t)=f(a)+f'(a)(t-a)+f''(a)\frac{(t-a)^2}{2}+\ldots=\sum_{n=0}^{\infty} \frac{f^{(n)}(a) }{n!} (t-a)^n.
$$
Now, let $t = x + x'$:
$$
f(x + x')=f(a)+f'(a)(x + x'-a)+f''(a)\frac{(x + x'-a)^2}{2}+\ldots=\sum_{n=0}^{\infty} \frac{f^{(n)}(a) }{n!} (x + x'-a)^n
$$
Last, take $a = x$:
\begin{align*}
f(x + x')&=f(x)+f'(x)(x + x'-x)+f''(x)\frac{(x + x'-x)^2}{2}+\ldots\\
&= f(x)+f'(x)x'+f''(x)\frac{x'^2}{2}+\ldots\\
&= \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} (x')^n.
\end{align*} |
H: roots of polynomial equation
How to find the roots of $x^5-2^5$ by hand. I see that we get a root of $x=2$ and 4 complex roots (should come in pairs). Not sure how to work out the complex roots. Do we need to convert to polar? Would that make it easier to see the other roots?
AI: Write as $x^5 - 2^5e^{2\pi ik} = 0$
So your solutions are $x = 2{e^{2\pi ik/5}}$ for $k=-2,-1,0,1,2$ |
H: integrating to clear differential
I'm unsure if that's the correct term.
I have an equation $v^2 = \frac{2ILB}{m}x$ and I need to find distance with respect to time (yes, physics.. but the silly math is what trips me up so I'm posting here)
Here's my attempt
$v^2 = [\frac{dx}{dt}]^2 = ...$
$\frac{dx}{dt} = \sqrt{\frac{2ILB}{m}x}$
$\frac{dx}{\sqrt{x}} = dt \sqrt{\frac{2ILB}{m}}$
$2\sqrt{x} = t \sqrt{\frac{2ILB}{m}}$
$x(t) = t^2 \frac{2ILB}{4m}$
My rule of thumb when I don't know if I'm doing something is "well, if I'm not lying at any point how could I be wrong?". I've finished calc3 and honestly never recall dealing with these outside of physics, so I can never know for sure if I'm lying at some point !
Anyways, I'm just wondering if I approached this right and if not, what exactly am I supposed to do?
AI: Your answer is correct.
Assuming that $v$ is velocity rather than speed, your second line should be
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$
If it's speed, your choice to ignore the negative term is ok.
You should add a constant when you integrate, giving
$$2\sqrt{x} = \pm t \sqrt{\frac{2ILB}{m}}+C$$
Also, sometimes (but rarely), squaring both sides of an equation can introduce erroneous solutions. See http://www.jimloy.com/algebra/square.htm for examples. To be certain, you really need to plug solutions back into your original equation to make sure they work, unless you can show analytically that they're ok.
If $x(0) = 0$, this gives $C=0$.
Your result is
$$x(t) = t^2 \frac{2ILB}{4m}$$
If we check that this satisfies
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$
we find the answer is yes in the positive case and no in the negative.
If $x(0) = 0$, squaring both sides of
$$2\sqrt{x} = \pm t \sqrt{\frac{2ILB}{m}}+C$$
gives us your result for $x$.
We already know that this only satisfies
$$\frac{dx}{dt} = \pm \sqrt{\frac{2ILB}{m}x}$$
for the positive case and so
$$\frac{dx}{dt} = \sqrt{\frac{2ILB}{m}x}$$
As such, your answer
$$x(t) = t^2 \frac{2ILB}{4m}$$
is correct. |
H: Stochastics with induction
prove that for all $n \in \mathbb{N}$:
$\sum_{r=0}^n \binom{n}{r}(-1)^{r} = 0$.
The base step is easy, i only get lots of problems when i try to mess with the sum boundries....
so far i've tried:
$\sum_{r=0}^{n+1} \binom{n+1}{r}(-1)^{r} = \sum_{r=0}^{n+1}(\binom{n}{r}+\binom{n}{r-1})(-1)^{r} = \sum_{r=0}^{n+1}\binom{n}{r}(-1)^{r}+\sum_{r=0}^{n+1}\binom{n}{r-1}(-1)^{r}$
Don't know how to proceed... Please help?
AI: Perhaps this way:
$$
\begin{split}
\sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^r
&= \binom{n+1}{0} + \binom{n+1}{n+1} (-1)^{n+1}
+ \sum_{r=1}^n \binom{n+1}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^{r+1} \\
&= 1 + (-1)^{n+1}
- \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^r \\
&= 1 + (-1)^{n+1}
- \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^r
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^{r+1}
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=1}^n \binom{n}{r} (-1)^r
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
& \text{cancel terms in the sum indexed $1...n-1$} \\
&= 1 + (-1)^{n+1} + \binom{n}{n}(-1)^n- \binom{n}{0} \\
&= (-1)^{n+1} + (-1)^n \\
&= 0
\end{split}
$$
as desired. |
H: Expansion of $ \frac{1}{|\vec r -\vec r'|} $
I would like to show that:
$$ \frac{1}{|\vec r -\vec r'|} =\frac{1}{r} + \frac{\vec r'\cdot r'}{r^3}+\frac{3 ((\vec r \cdot \vec r)^2 -\vec r^2 \vec r'^2 )}{2r^5} +\dots$$
What I derived so far is:
$$\frac{1}{|\vec r -\vec r'|}= \sum_{n=0}^{\infty} \frac{1}{n!} (-\vec r' \cdot \nabla_r )^n \frac{1}{r} $$
The last part is however unclear to me, since I don't know how to rewrite, e.g. $(-\vec r' \cdot \nabla_r )$
AI: The notation $(-\vec r'\cdot \nabla_r)$ means "first take the gradient with respect to $r$, then do the dot product with $-\vec r'$. For example
$$ (-\vec r'\cdot \nabla_r)^2 f(r) = -\vec r' \cdot \left[ \nabla_r(-\vec r'\cdot \nabla_r f(r)) \right].$$
So let's have a look at the first few summands. Starting with $n=0$, we get just $\frac 1 r$, since $0!$ is $1$ and the operator is applied $0$ times. Now $n=1$:
$$
(-\vec r'\cdot\nabla_r)\frac 1 r = -\vec r' \cdot \left(\nabla_r \frac{1}{r}\right) = -\vec r' \cdot \left(\frac{-\vec r}{r^3}\right) = \frac{\vec r'\cdot\vec r}{r^3}.
$$
Here we used the fact that the gradient of $\frac 1 r=\frac{1}{|\vec r|}$ is $\frac{-\vec r}{r^3}$. For $n=2$ we can use what we already calculated:
$$
\frac 1 {2!} (-\vec r'\cdot \nabla_r)^2 \frac 1 r
= \frac 1 {2!} (-\vec r'\cdot \nabla_r) \frac{\vec r'\cdot\vec r}{r^3}
= \frac {-\vec r'} {2} \cdot \nabla_r\left[
\frac{\vec r'\cdot\vec r}{r^3} \right] = \dots
$$
Here you need to involve the product rule for gradients of dot products, it involves the vector gradient which is in this case the same as the Jacobian matrix. Have you done calculations like that before? |
H: Find all positive/negative integers N for which $N^2+20N+11$ is a perfect square?
I know that there might be a duplicate of this. But I don't know where.
I tried equatin this to $X^2$ and and then bringing it to the other side and completing the square. What next? Is there a way to solve these types of questions?
AI: Your proposed procedure is right. We get $(N+10)^2-89=y^2$, that is, $x^2-y^2=89$.
So we want to solve $(x-y)(x+y)=89$. Note that the possibilities for $x-y$ are $1$, $-1$, $89$, and $-89$. Write down the corresponding value of $x+y$, and solve for $x$. |
H: Solutions to equation involving trigonometry and geometric series
$$\cos^2 x + \cos ^3 x +\dots = 1+ \cos x$$
I want to find values of $x$ between $0$ and $ 180$ degrees for which the above equation holds true.
Attempt at a solution: left side is a converging geometric progression, for which $a_1$ is $\cos^2 x$ and $q = \cos x$. Plugging into the known formula for such series yields $\cos^2 x = 1$, which yields $45 $ and $135$ degrees as solutions.
Is this okay?
AI: As mentioned, the LHS is a geometric series, with first term $ \cos^2 x $ and common ratio $ \cos x $, so it evaluates to $ \dfrac {\cos x}{1 - \cos x} $ and our equation becomes $$ \dfrac {\cos^2 x}{1-\cos x} = 1 + \cos x \implies \cos^2 x = 1 - \cos^2 x \implies \cos^2 x = \dfrac {1}{2}. $$Finally, take the square root of both sides and we have $ \cos x = \pm \dfrac {\sqrt{2}}{2} $ and our answers are $ \boxed {\dfrac{\pi}{4}, \dfrac{3\pi}{4}} $.
Seems I've been beaten by a couple of minutes by ncmathsadist and a few seconds by Oliver! |
H: Are there such things as non-extensional set theories?
I have always assumed that extensionality is a paradigmatic example of a property of mathematical objects (sets) which is essential to those objects--- if your set theory doesn't obey extensionality, it isn't set theory.
Given the existence of alternative set theories, such as non-well-founded set theories, though, it occurred to me that maybe I shouldn't be so certain of this.
Which brings me to my question: have any set theories been developed which don't include the axiom of extensionality, or at least which include set-like objects which are not extensional?
This question is certainly related to my own, but doesn't consider theories which have been developed, only ways that extensionality might fail by manipulating models of ZFC.
AI: Various kinds of type theories are intensional by nature, and in those theories one can usually carry out the constructions of typical set theory, and therefore deal with sets in an intensional way.
I suggest that you look at the book Intensional Mathematics, which is part of the series Studies in Logic and the Foundations of Mathematics published by Elsevier. In it, you will find two articles, one by J. Myhill called Intensional Set Theory, and one by N. D. Goodman called A Genuinely Intensional Set Theory (which is kind of a follow-up to Myhill's article). They develop intensional modal set theories, and Goodman also gives an interpretation of ZFC in ZFM (which is his proposal of a modal Zermelo-Fraenkel). |
H: Show that $\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)}$
Show that $$\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)},$$ where $0\leq r <1$.
Using this, prove that $\sum_{n=0}^\infty r^n \cos(n\theta)$ and $\sum_{n=0}^\infty r^n \sin(n\theta)$ are convergent.
AI: The series $\sum_{n=0}^\infty z^n$ is convergent provided $|z|<1$, and we have
$$
\sum_{n=0}^\infty z^n=\frac{1}{1-z} \quad |z|<1.
$$
So if we set $z=re^{i\theta}$, we have $r=|z|$, and therefore for every $0 \le r <1$ we get
$$
\sum_{n=0}^\infty r^ne^{in\theta}=\frac{1}{1-re^{i\theta}}=\frac{1-re^{-i\theta}}{(1-re^{i\theta})(1-re^{-i\theta})}=\frac{1-r\cos\theta+ir\sin\theta}{1-2r\cos\theta+r^2}.
$$
I'm guessing that what want to deduce is that the series $\sum_{n=0}^\infty r^n\cos(n\theta)$ and $\sum_{n=0}^\infty r^n\sin(n\theta)$ are convergent provided $0 \le r <1$.
This follows from the fact
$$
\sum_{n=0}^\infty r^ne^{in\theta}=\sum_{n=0}^\infty r^n\cos(n\theta)+i\sum_{n=0}^\infty r^n\sin(n\theta)
$$ is convergent and so we have
$$
\sum_{n=0}^\infty r^n\cos(n\theta)=\frac{1-r\cos\theta}{1-2r\cos\theta+r^2},\quad
\sum_{n=0}^\infty r^n\sin(n\theta)=\frac{r\sin\theta}{1-2r\cos\theta+r^2}.
$$ |
H: Draw lattice for a join-semillatice
I am working with a set that I believe is a join-semilattice (all elements $x,y\in S$ have a least upper bound). I am a little bit confused as to how I would draw a lattice for this set $S=\{1,2,3,12,18,36\}$ when it is partially ordered by divisibility.
My confusion stems from how I would draw the $12$ and $18$ in the lattice. Initially, I would draw a $1$ as the base. To the NW of $1$, I would draw a $2$ with a line going from $1$ to $2$. I would do the same for $1$ and $3$, except I would put the $3$ to the NE of $1$. This is just where I get a tiny bit confused. $2$ and $3$ divide both $12$ and $18$, but $12$ doesn't divide $18$. How would I make this work?
AI: Like this rough sketch:
Note that $2$ and $3$ do not have a least upper bound, since $12$ and $18$ are unrelated. |
H: Cardinality of sets discrete math
Given $A = \{\text{Vine}, \text{Tree}, \text{Shrub}\}$; $B = \{ \text{Tree}\}$; $C = \{ \text{Vine}, \text{Moss}\}$; $D = \{\text{Red}, \text{Green}\}$; $E = \{ \text{Red} \}$
what is $A \cup B \cup C$?
AI: $$A\cup B\cup C=\{\text{Vine, Tree, Shrub, Tree, Vine, Moss}\}$$ How many distinct elements are in this set? |
H: $x^6+x^3+1$ is irreducible over $\mathbb{Q}$
I have been trying to prove that $x^6+x^3+1$ is irreducible over $\mathbb{Q}$ (or $\mathbb{Z}$ since by Gauss' Lemma is the same), but I can't. Any idea of how to do so?
AI: HINT: Let $y=x-1$, and apply Eisenstein's criterion for $p=3$. |
H: Vector Calculus
I want to show that given $ax+by=c$ the vector $(a,b)$ is perpendicular to the line determined by the equation.
I was thinking using dot product.
AI: The line $\;ax+by=c\;$ can be given in parametric form as (assuming $\;b\neq 0\;$ to avoid trivialities)
$$\left\{\left(x,-\frac ab x+\frac cb\right)\;;\;x\in\Bbb R\right\}=\left\{\left(0,\frac cb\right)+x\left(1,-\frac ab\right)\right\}$$
The direction vector is, of course, $\;\left(1,-\frac ab\right)\;$ , so the dot product is
$$(a,b)\cdot\left(1,-\frac ab\right)=a-a=0\iff (a,b)\perp\left(1,-\frac ab\right)$$ |
H: set identities discrete mathematics help?
$A-C\subseteq A-(B-C)$
how would you do this question?
and also b - c(with a line through b-c) $\subseteq C$
AI: I prefer the modern notation $A\setminus C$ to $A-C$, so I’ll use it. One way to attack the first problem is to try to prove that the statement is always true. If you succeed, you’ve solved the problem, and if you fail, the attempt is likely to have given you some insight into how to construct a specific example in which the statement is false. I’ll illustrate that approach in detail.
The most straightforward way to show that $A\setminus C\subseteq A\setminus(B\setminus C)$ is to let $x$ be a completely arbitrary element of $A\setminus C$ and prove that it must therefore be an element of $A\setminus(B\setminus C)$. Suppose, then, that $x\in A\setminus C$. At this point you must turn to the definition of set difference to see just what this tells you about $x$: $x\in A$, and $x\notin C$.
Now back up and look at what you’re trying to prove: you want to show that $x\in A\setminus(B\setminus C)$, which means showing that $x\in A$ and $x\notin B\setminus C$. You do know that $x\in A$, so it only remains to show that $x\notin B\setminus C$. Now $x\notin B\setminus C$ if and only if $x\notin B$, or $x\in C$ (why?). You know that in fact $x\notin C$, so the only hope of showing that $x\notin B\setminus C$ is to show that $x\notin B$. Unfortunately, the hypothesis that $x\in A\setminus C$ says nothing at all about $B$, so it seems very unlikely that we can prove that $x\notin B$ from the available hypotheses. This suggests that perhaps it isn’t necessarily true that $A\setminus C\subseteq A\setminus(B\setminus C)$, and that we should try to find an example in which it fails.
Our attempt at a proof suggests that a failure can occur if there’s an $x\in A\setminus C$ that is also in $B$. That’s easy to arrange: let $A=B=\{1\}$ and $C=\varnothing$. Then $$A\setminus C=\{1\}\setminus\varnothing=\{1\}\;,$$ and $$A\setminus(B\setminus C)=\{1\}\setminus(\{1\}\setminus\varnothing)=\{1\}\setminus\{1\}=\varnothing\;.$$
$\{1\}\nsubseteq\varnothing$, so in this case $A\setminus C\nsubseteq A\setminus(B\setminus C)$. |
H: Prove that the following element is an invertible element
Question in Abstract Algebra:
How can I prove that: $r+s \sqrt{2}$ , when $r$ and $s$ are rational, is an invertible element in $\mathbb{Q}(\sqrt{2})$?
(In fact I need to prove that's a number field but I only have problem with this.)
AI: Note that if $r+s\sqrt{2}\ne 0$, then
$$\frac{1}{r+s\sqrt{2}}=\frac{1}{r+s\sqrt{2}}\cdot\frac{r-s\sqrt{2}}{r-s\sqrt{2}}=\frac{r}{r^2-2s^2}+\frac{-s}{r^2-2s^2}\sqrt{2}.$$
You will have to verify that $r-s\sqrt{2}\ne 0$. |
H: Infinite Dyadic Rationals in any open interval $(a,b)$ where $a
Given that there exists at least one dyadic rational of the form $2^{-n}m$ between any two distinct real numbers $a<b$, show that there infinite such rationals between $a$ and $b$.
My Attempt
Given $a<b\in\mathbb{R}$ we have $$a<\frac{m_1}{2^{n_1}}<b,$$where $n$ is a natural number. But then $a<(b+a)/2<b\in\mathbb{R}$ so that $$a<\frac{m_{21}}{2^{n_{21}}}<\frac{b+a}{2}\text{ and }\frac{b+a}{2}<\frac{m_{22}}{2^{n_{22}}}<b.$$So there are at least two dyadic rationals between the reals $a$ and $b$. However, $$a<\frac{a+(b+a)/2}{2}<\frac{b+a}{2}\text{ and } \frac{b+a}{2}<\frac{(b+a)/2+b}{2}<b.$$By applying the given result again, we have $$a<\frac{m_{31}}{2^{n_{31}}}<\frac{a+(b+a)/2}{2}\text{ , }\frac{a+(b+a)/2}{2}<\frac{m_{32}}{2^{n_{32}}}<\frac{b+a}{2}$$and $$\frac{b+a}{2}<\frac{m_{33}}{2^{n_{33}}}<\frac{(b+a)+b/2}{2}\text{ , }\frac{(b+a)/2+b}{2}<\frac{m_{34}}{2^{n_{34}}}<b.$$ So, there at least four dyadic rationals in $(a,b)$.
After the $k$-th iteration we will have partitioned the interval $(a,b)$ exactly $2^{k-1}$ times. Thus, we will have reached the conclusion that there exist at least $2^{k-1}$ dyadic rationals in $(a,b)$. Repeating the process indefinitely shows that there exist $$\lim_{k\to\infty}2^{k-1}=\infty$$dyadic rationals in $(a,b)$.
Note: the first number in the indices stands for the number of intervals.
A PhD student in a support class we have for Analysis told me that my proof is a bit flawed because as $k$ grows large then the intervals in which I have partitioned $(a,b)$ become smaller and smaller and tend to zero. If at infinity they become zero, then there cannot be a rational dyadic (or any number for that matter) inside those intervals, and hence my proof is invalid. He didn't seem to be sure about this though but he suggested that I try a different proof altogether so that I avoid this issue.
In my opinion, his assertion is not so sound, because this would imply that because intervals get smaller and smaller then all the intervals are countable, which is not the case. However, I haven't yet been taught anything on countability of sets, so I hope you can enlighten me on the matter.
Thanks in advance!
AI: Your proof is fine except for some informal "and so on" in it.
Even if the interval lengths tend to zero, they never are zero. Instead of invoking the $\lim$, however, it may be better to state that $2^{k-1}$ exceeds any natural number $N$ (then again, this is exactly what $\lim 2^{k-1}=\infty$ says), hence the assumption that there are only $N$ such numbers leads to a contradiction, no matter how big $N$ is.
A variation of the theme might be this:
Let $S$ be the sets of dyadic numbers in $(a,b)$.
Assume $|S|$ is finite.
As there is at least one dyadic in $(a,\frac{a+b}2)$ and one in $(\frac{a+b}2,b)$, we conclude $|S|\ge 2$ so that we can
let $x=\min S$ and $y=\min(S\setminus\{x\})$. [This uses that every finite nonempty set of reals has a minimal element.] Then $S\cap (x,y)=\emptyset$, but as $x<y$ there exists a dyadic $z\in(x,y)\subseteq (a,b)$, hence $z\in S$ - contradiction. |
H: Permutations and combinations ! 9 different fruit pies divided between three
Different fruit pies are divided between 3 people so that each person gets and odd number of pies. Find the number of ways this can be done??
hint- so many combinations are added to get this answer ..
AI: We can do it by splitting into cases: $7$-$1$-$1$, $5$-$3$-$1$, $3$-$3$-$3$.
First case: The lucky person who gets $7$ pies can be chosen in $\binom{3}{1}$ ways. Her pies can be chosen in $\binom{9}{7}$ ways. For each choice, the older of the unlucky people can be assigned her pie in $\binom{2}{1}$ ways, for a total of $\binom{3}{1}\binom{9}{7}\binom{2}{1}$.
Next two cases: it's your turn. Then add up. |
H: Cardinality of set-Discrete math
Take $A=\{1,2,3\}$ and $B=\{1,2,5\}$.
If we unionized them together it would be $A\cup B=\{1,2,3,5\}$ and if we intersected them it would be $A\cap B=\{1,2\}$.
However, if we change $B$ to $B=\{\{1,2\},2,5\}$ would a set within a set change the previous answers and why?
AI: Yes, it changes the union and intersection. We now have
$$A\cup B=\{1,2,3\}\cup\big\{\{1,2\},2,5\big\}=\big\{1,2,3,\{1,2\},5\big\}$$
and
$$A\cap B=\{1,2,3\}\cap\big\{\{1,2\},2,5\big\}=\{2\}\;.$$
The set $\{1,2\}$ is an object in its own right, distinct from the objects $1$ and $2$; $1$ and $2$ are elements of $A$, and $\{1,2\}$ is now an element of $B$, so these three distinct objects are all elements of the union $A\cup B$, along with the other elements of $A$ and $B$.
$1$ is now an element of an element of $B$, but it is not itself an element of $B$. Thus, the only object that $A$ and $B$ now have in common is $2$, and the intersection is therefore $\{2\}$. |
H: A graph with a degree sequence 0,1,2,3,4
Please help me to prove whether or not a there exists a graph with the degree sequence 0,1,2,3,4. I do not know the formal way of writing the proof hence any advises and proofs will be much appreciated. Thank you.
AI: Hint There are 5 vertices. If one has degree $4$, it is connected with all four other vertices. Then which one can have degree 0?
Of course, the above argument only holds if the graph is simple, if you allow multiedges/loops, the answer is different. |
H: KKT maximization problem
$x^2y \rightarrow$ max,
such that $x^2 + 4xy \leq 1, x \geq 0$ and $y \geq 0$.
I think I need to use the KKT conditions here. I did however not yet succeed in solving it, so could someone please give me an example of how this should be done? And should I include the constraints $x \geq 0$ and $y \geq 0$ into the Lagrangian function?
AI: The KKT conditions in this case for $x,y$ that are feasible ($x\geq0,y\geq0,x^2+4xy\leq1$) is that $\exists \alpha,\beta,\gamma\geq0$ (feasible dual variables for the constraints in the last parens in respective order) such that
Stationary: $2xy=-\alpha+\gamma(2x+4y)$, $x^2=-\beta+\gamma x^2$, and
Complementary slackness: $\alpha x=\beta y=\gamma(x^2+4xy-1)=0$
However, the problem is not convex therefore KKT conditions are not sufficient. In particular, the eigenvalues of the hessian of your objective are $y\pm\sqrt{4x^2+y^2}$ so that one of them is always positive (so not concave for a maximization problem) and also one is negative whenever $x>0$ which is certainly feasible (so it's not even definite in either direction).
But they are necessary. So you can guess which constraints are active or not and go through all permutations of this to find all KKT-satisfying points. For example, guess that the nonnegativity constraints are slack ($\alpha=\beta=0$) and that the other constraint is tight ($x^2+4xy=1$). Then plugging this into the KKT conditions we get three equations:
$$2xy=\gamma(2x+4y)$$
$$(1-\gamma)x^2=0$$
$$x^2+4xy=1$$
Which solve for $x=1/\sqrt{3}$ and $y=1/\sqrt{12}$ and give objective value $1/\sqrt{108}$. (Hint: this is optimal.) |
H: Integral curves in the plane
Maybe this is a stupid question but i can not solve this mechanical problem...
How can I find the integral curves of the vector field
$$X_{(x,y)} = x \dfrac{\partial}{\partial x} − y\dfrac{\partial}{\partial y} = \begin{bmatrix}x \\ -y\end{bmatrix}\,.$$
AI: Well, the integral curves satisfy
$\begin{bmatrix} x \\ y \end{bmatrix}' = \begin{bmatrix} x \\ -y \end{bmatrix}, \tag{1}$
so we've got
$x' = x \tag{2}$
and
$y' = -y; \tag{3}$
the solutions of (2) and (3) are
$x(t) = x_0 e^t \tag{4}$
and
$y(t) = y_0e^{-t}, \tag{5}$
where $x_0$ and $y_0$ are the values of $x$ and $y$ at $t = 0$. And that just about does it, doesn't it?!???!?
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: Show why given set is not a frame
I am rather new to this material and an explanation of what is happening would be greatly appreciated. At first glance, it seems like the sum of squares is bounded at both ends but I guess I'm looking at it completely wrong. The question is as follows:
Consider the set $\Phi = \{\varphi_1, \varphi_2, ...\} \subset \ell^2 (\Bbb Z_+)$ with
$\varphi_1 = (1, 0, 0, 0, ...)$
$\varphi_2 = (0, 1/2, 0, 0)$
$\varphi_3 = (0, 0, 1/4, 0, ...)$
...
Why is $\Phi$ not a frame for $\ell^2 (\Bbb Z_+)$?
AI: For your set to be a frame, you need
$$
\alpha\,\|v\|^2\leq\sum_k|\langle \varphi_k,v\rangle|^2\leq\beta\|v\|^2
$$
for every vector $v$ and certain fixed constants $\alpha,\beta>0$.
Now if you take $v=\varphi_j$ for some integer $j$,
then
$$
\sum_k|\langle \varphi_k,\varphi_j\rangle|^2=|\langle\varphi_j,\varphi_j\rangle|^2=(2^{-j+1})^4=2^{-4j+4}.
$$
If the set was a frame, we would have some constant $\alpha>0$ such that
$$
2^{-4j+4}=\sum_k|\langle \varphi_k,\varphi_j\rangle|^2\geq\alpha\|\varphi_j\|^2=\alpha\,2^{-2j+2}.
$$
The inequality above is $2^{-2j}\geq\alpha/4$. This fails for $j$ big enough. |
H: Limit of nth power of operator norm
I am given a compact operator $A$ which lives in a Banach algebra and whose spectral radius obeys $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{\frac{1}{n}}<1$. Now I want to prove that this implies that $||A^n||\rightarrow 0$ as $n\rightarrow \infty$ (if true).
I tried the following. We know that for $m$ big enough, $||A^k||^{\frac{1}{k}}<1$ for all $k\geq m$, so (choosing such a $k$) also $||A^k||<1$. Therefore $\lim_{n\rightarrow\infty} ||A^k||^n=0$. But since $||A^{nk}||\leq||A^n||^k$, we have $\lim_{n\rightarrow\infty} ||A^{kn}||=0$. But then I have only proven that a subsequence of $||A^n||$ converges to 0!
Is my approach completely futile? Or is limit I want to prove incorrect, and if so, what's a counterexample?
AI: Since $\rho(A) = \lim\limits_{n\to\infty} \lVert A^n\rVert^{1/n} < 1$, you know that for all large enough $n$ you have
$$\lVert A^n\rVert^{1/n} < q := \frac{1+\rho(A)}{2} < 1,$$
whence $\lVert A^n\rVert < q^n \to 0$. |
H: At which parameter value $c>0$ do the number of solutions of $\log(1+x^2)=x^c$ change?
I'm looking at the functions $x\mapsto \log(1+x^2)$ and $x\mapsto x^c,\ c>0$ on the interval $\mathbb R^+_0$.
I'm interested in the properties of
$$\log(1+x^2)=x^c.$$
Graphically, for small $c$, the function $x^c$ is concave and intersects $\log(1+x^2)$ once or twice. Then at some value $c=c_1$, they don't seem to intersect anymore and then at a value $c=c_2$ they start touching again.
What are $c_1, c_2$, and, generally, what is the enclosed area between 0 and the first intersection point of the two functions, as a function of $c$?
AI: Consider the function $f_c(x)=\log(1+x^2)-x^c$.
Its derivative with respect to $x$ is $f_c'(x)=\frac{2x}{1+x^2}-cx^{c-1}$ We want to find $x$ and $c$ such that $f_c(x)=f_c'(x)=0$.
We can eliminate $c$ from the equations. $$c=\frac{2x^2}{(1+x^2)(-\log(1+x^2))}.$$
Replacing in $f_c(x)=0$ we get $$\log(1+x^2)-e^{\frac{2x^2\log(x)}{(1+x^2)(-\log(1+x^2))}}=0$$
which can be solved numerically. From the values of $x$ we can then get those of $c$.
Since the solution(s) will be obtained by numerical methods, it would be convenient to think better what equation(s) to use. |
H: Recursion problem.
$$A_{n+1}=A_{n}+\frac{G(n+1)-A(n)}{n+1}$$ $$A(n)=G(1)=80$$ $$G(2)=70$$
$$G(3)=60$$
$$G(4)=70$$
$$G(5)=100$$
My question is how did following equation is formed.
$$A(2)=A(1)+\frac{G(2)-A(1)}{2}=80-10/2=75$$
I want to know how it is done.
AI: It appears you are equating $A_n$ and $A(n)$, which is not normal. Then we substitute $n=1$ in the equation, giving $$A_{n+1}=A_{n}+\frac{G(n+1)-A(n)}{nn+1}\\
A_2=A_1+\frac{G(2)-A(1)}{1\cdot 1+1}\\A_2=80+\frac{70-80}2=75$$ |
H: Is $2^n \mod m \equiv (2^{n/2} \pmod m ) ^ 2 \pmod m$?
I'm trying to write a procedure that solves (2^n - 1) mod 1000000007 for a given n.
n can be very large (~10^9).
Say m = 1000000007
So this is what I have so far:
func(n):
if n = 1 return 2
if n is even return ((func(n/2) mod m)^2)mod m
else return ((func(n/2) mod m)^2 * 2)mod m
I'll call func and subtract 1 from the result to get the final answer.
Is it right to use that recursion?
AI: If you are asking whether $(2^{n/2})^2 \equiv 2^n \pmod{m}$ then the answer is yes, since they are already equal as integers, so they will be equal after you pass to the residue classes. There is no need to drag the symbol "mod" along everywhere in your title. Just put it at the end of the congruence.
By the way, the multiplicative order of $2$ in $\mathbb{Z}/1000000007\mathbb{Z}$ is $500000003$. So let $r$ be the remainder after division of $n$ by $500000003$, and we'll have $2^n \equiv 2^r \pmod{1000000007}$. That is probably the most efficient way to it. |
H: Did I take the derivative correctly? $x^y=y^x$
Need to differentiate following equation:
$$x^y=y^x$$
My attempt:
$$x^y\log(x) \cdot y' = y^x\log(y)\cdot1; $$
$$y'=\frac{y^x\log(y)}{x^y\log(x)}$$
Please tell me if I've made a mistake.
AI: Not quite. The idea is to take the $\log$ of both sides and implicitly differentiate. $$ \begin {eqnarray*} x^y &=& y^x \\ y \cdot \log (x) &=& x \cdot \log (y) \\ y' \cdot \log(x) + \dfrac {y}{x} &=& \dfrac {x}{y} \cdot y' + \log (y) \\ y' \cdot \left( \log(x) - \dfrac {x}{y} \right) &=& \log(y) - \dfrac {y}{x} \\ y' &=& \dfrac {\log (y) - \dfrac {y}{x}}{\log (x) - \dfrac {x}{y}}, \end {eqnarray*} $$so our answer is $$ y' = \boxed {\dfrac {x \cdot \log (y) - y}{y \cdot \log (x) - x} \cdot \dfrac {y}{x}}. $$ |
H: Showing a topology is not metrizable
Show $\prod_{N} \mathbb{R}$ with the box topology is not metrizable.
The Box Topology on $\prod_{j \in J} X_j$ ($X_j$ topological spaces) is generated by the basis $\left\{\prod_{j \in J} U_j \; \Big| \; U_j \text { is open in } X_j \right\}$.
I made an attempt, and found the following:
Consider the contrary. Suppose that there is a countable basis at $0$ say $ (U_i ) $ where $U_i =V_i^1 \times V_i^2 \times \cdots \times V_i^j \times \cdots$. The set $W =\left(\frac{1}{2} V_1^1 \right)\times \left(\frac{1}{2} V_2^2 \right)\times \cdot \times \left(\frac{1}{2} V_i^i\right) \times \cdots$ is open but there exists no $n$ such $U_n\subset W$, so we are done. $ \blacksquare $
StackExchange is a great place for comments on this solution and alternate solutions!
AI: Be careful: Simply putting a factor $\frac12$ before the neighborhood does not necessarily make it strictly smaller. If $V_i^i=X_i$, then $\frac12V_i^i=X_i=V_i^i$. But you can assume that the first one of the local base elements, $U_1$, is bounded, because there is a bounded neighborhood of $0$ ("bounded" in the sense that each projection is bounded). Furthermore, we can assume that $U_1\supset U_2\supset...$ and that all of them are open and, as you correctly noted, are of the box-form.
Now it remains to show that indeed $V_i^i\not\subseteq\frac12 V_i^i$. But since $V=V_i^i$ is bounded, the connected component $C$ of $V$ containing the $0$ is just an interval $(a,b)$ with $-\infty<a<0<b<∞$. Then $\frac12b\in V\setminus\frac12V$
This way, $W=\frac12V^1_1×\frac12V^2_2×...$ is the neighborhood not containing any $U_i$. |
H: Ruler and compass construction of the unit-distance petersen graph embedding
The Petersen graph is a unit distance graph, and this embedding is shown below, where each edge of the graph is one unit in length.
Is there a ruler and compass construction for this embedding? If so, what is it?
It seems like the kind of thing that you can construct with a ruler and compass, but my brief attempt and google search failed.
I tried by constructing the outer pentagon, and then drawing unit radius arcs inside it, centered at each vertex. The inner star should have it's vertices on those arcs.
I did come fairly close by eyeballing the "radius" of the star and drawing the smaller circle, but I suspect this was cheating :p
Image credit: http://en.wikipedia.org/wiki/File:Petersen_graph,_unit_distance.svg
AI: If the inner star's side are $1$ the radius of its circumflexing circle is $\frac1{2r}=\sin\frac25\pi=\sqrt{\frac{5+\sqrt5}8}$ which leads that $$r=\sqrt{\frac{5-\sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem. |
H: Does this general solution look right?
Here is the starting system:
\begin{cases}
-x+y+4z= -1 \\
3x-y+2z=2\\
2x-2y-8z=2
\end{cases}
I started by performing $2E1+E3\to E3$. This showed that they cancelled each other out and I was left with $0=0$ for $E3$ (this left me assigning $z$ to $t$). Then I performed $E1+E2\to E2$ and got $2x+6z=1$ and solved for $x$, getting $x=-3t+1/2$. After this, I simply back solved for $y$ and got $y=-3t-1/2$. Does all this look right?
AI: I get:
$$y = \dfrac{-1}{2} - 7z$$
$$x = \dfrac{1}{2} - 3z$$
Of course $z$ is a free variable.
It looks like you may have just copied your $y$ incorrectly. |
H: Proving something with Wilson's Theorem [continued.]
At first I asked this: Proving something with Wilson theorem.
Now I have to prove that if $p=4n+3$ it's impossible to represent $-1$ in the form $x^2$ modulo $p$. How can I prove it?
Thank you!
AI: It is generally true that $a$ is a quadratic remainder (i.e. is of the form $x^2$ for some $x$) if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. Once you know this, just note that for $p \equiv 3 \pmod{4}$ you have $(-1)^{\frac{p-1}{2}} = -1$ and you are done.
Because you only need to show that $-1$ is not a quadratic residue, it will suffice to show one implication (the easy one). So, suppose that $a = x^2$ and $a^{\frac{p-1}{2}} \equiv -1 \pmod{p}$. We get by substituting one equation to another that $x^{p-1} \equiv -1 \pmod{p}$. But this is in contradiction with Fermat's theorem, which finishes the proof. |
H: Determine run-time of an algorithm
Probably a stupid question but I don't get it right now. I have an algorithm with an input n. It needs n + (n-1) + (n-2) + ... + 1 steps to finish. Is it possible to give a runtime estimation in Big-O notation?
AI: Indeed, it is. Since we have $$n+(n-1)+(n-2)+\cdots+1=\frac{n(n+1)}2,$$ then you should be able to show rather readily that the runtime will be $\mathcal{O}(n^2)$. |
H: radical of I is the entire ring R implies I=R?
Can anyone prove that:
radical of I (ideal) is the entire ring R implies I=R?
The ring has a unit and commutative.
Thanks...
AI: If $rad(I)=R$ then $1 \in rad(I)$. Since $1$ is idempotent, this implies $1 \in I$. Therefore $I=R$. |
H: Question About a Classical Root Appending Extension Field Proof
Let $F$ be a field and $f(x) \in F[x]$ s.t. $f(x)$ is irreducible and of degree $n \ge 1$. I've seen it proved that there exists an extension field $E$ of $F$ s.t. that there is a root $\alpha \in E$ of $f(x)$ and the degree of $E$ over $F$ is $n$. In particular, the construction works like this.
Consider $(f(x)) \subset F[x]$ and let $V = (f(x))$.
Then $F[x]/V$ is a field with basis $\{1 + V, x + V, \ldots , x^{n-1} + V\}$.
The polynomial $f(x) + V$ has root $x + V$.
Now this seems to be skirting the issue somewhat to me. We originally wanted to extend $F$, but instead we have moved from the setting of $F$ and $F[x]$ to some other setting.
It's been pointed out that if $\overline{F} = \{f_i + V : f_i \in F\}$, then $F \cong \overline{F}$. In particular, this is done via the isomorphic mapping $\psi(f_i) = f_i + V$. But I'm still not clear on how this guarantees us that there exists some $E$ over $F$ of degree $n$ with a root of $f(x)$. Presumably that $E$ is isomorphic to $F[x]/V$. But what is that $E$?
So to sum up my question: we have moved from the setting of $F$ and $f(x) \in F[x]$ to the new setting of $\overline{F}$ and $f(x) + V \in F[x] / V = \overline{F}[x]$. We've shown some analogous result in this new setting, but how exactly does this net us the desired result in our original setting?
AI: Consider $(F[x]/V\,\setminus \{\,a\cdot x^0+V\mid a\in F\,\})\cup F$.
Define addition as
$$ a+b=\begin{cases}a+ b&\text{if } a,b\in F\\
(a\cdot x^0+V)+b&\text{if }a \in F, b\notin F\\
a+(b\cdot x^0+V)&\text{if }a \notin F, b\in F\\
a+b&\text{if }a,b\notin F\end{cases}$$
and similarly for multiplication. Now go through the hell of showing that this givs us a field. Or please don't, work with $E$ and note that there is a canonical embedding $F\to E$.
This is essentially thge same "trick" we use when saying that $\mathbb Z\subseteq \mathbb Q$, although $\mathbb Q$ is constructed as a set of equivalence classes of pairs of integers and does not contain a single integer.
Edit: In fact, those canonical maps are so natual that I forgot to explicitly mention $F\to F[x]$, $a\mapsto a\cdot x^0$. If you don't trust that one may view $F$ as a subset of $F[x]/V$, you shuld not believe that $F\subseteq F[x]$ either: $F[x]$ is a ring together with a ring morphism $i\colon F\to F[x]$ and a map $\iota\colon\{x\}\to F[x]$ with the universal property that for each ring morhism $f\colon F\to Y$ and map $\phi\colon\{x\}\to Y$ there exists a unique ring morphism $h\colon F[x]\to Y$ with $h\circ i=f$ and $h\circ\iota=\phi$. A priori, $F$ need not be a subring of $F[x]$; and yet we identify $F$ with $i(F)$ and $x$ with $\iota(x)$. - And even in this excursion I ought to add a handful of forgetful functors :) |
H: Quicker way to solve 10! congruent to x (mod 11)
I am new to modular arithmetic and solving congruences and the way I went about this was to write out $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot2$, then multiply numbers until I get a number greater than $11$, replace it with a smaller number in its congruence class and repeat. Is there a quicker way to go about this? The next question I must solve is $100!\equiv x \pmod {101}$.
AI: Wilson's Theorem states that for any prime number $n$, you get $$(n-1)! \equiv -1 (mod\ n).$$
$10!$ makes $n = 11$, which is prime, so $10! \equiv -1 (mod\ 11) \equiv 10 (mod\ 11)$
Likewise, $100!$ makes $n = 101$, which is prime, so $100! \equiv -1 (mod\ 101) \equiv 100 (mod\ 101)$. |
H: How do I calculate a weighted average purchase date?
A farmer bought 100 kg of seed on the following dates:
20 kg on 13-may-2007 (20%)
30 kg on 4-oct-2007 (30%)
50 kg on 31-jul-2008 (50%)
How do I calculate the weighted average purchase date? I don't want weighted average age because that changes daily. I'm thinking of a specific point in time.
The total days between first and last purchase is 445 days.
AI: Hint: If you take 13-may-2007 as day zero, he bought 30 kg on day 144 and 50 kg on day 445. How many kg-days is that? Then divide by 100 and go that many days after 13-may. |
H: How to find the height of a 2D coordinate on a four-sided 3D polygon plane?
How do I find the height of a given 2D coordinate on a four-sided 3D polygon plane? The polygon has no volume. I'm trying to match 3D terrain vectors to a 3D polygon. I'll always know that the 2D version of the 3D poly contains the 2D coordinate, but I need to get the height at that 2D coordinate on the polygon surface.
How can I figure out the height of point F1 in the image example?
AI: You can get a normal vector to the plane.
$n = (P_4 -P_1) \times (P_2 - P_1)$
From there you can find the point-normal form of the plane.
$(p-P_1) \cdot n = 0$
You want the intersection of this plane and the line $l =<\!312,0,190\!> + <\!0,1,0\!>\,t$ , so solve
$(<\!312,0,190\!> + <\!0,1,0\!>\,t - P_1) \cdot n = 0$ for $t$:
$ t = \Large{(P_1 - <312,0,190>) \cdot n \over <0,1,0> \cdot n}$
Then plug that value into your line equation to find your point. |
H: Number Theory: Solutions of $ax^2+by^2\equiv1 \pmod p$
Assume $p$ is a prime number and $\gcd(ab, p)=1$. Show that the number of integer solutions $(x, y)$ of $ax^2+by^2 \equiv 1 \pmod p$ is
$$p - \left(\dfrac{-ab}{p}\right)$$ where $\left(\dfrac{x}{y}\right)$ is the Legendre symbol.
Ok, so here's my partial solution to the above question.
Suppose that $-ab$ is a quadratic residue of $p$, then $-ab\equiv c^2 \pmod p$ for some $c$.
Then $(ax)^2+aby^2\equiv(ax)^2-(cy)^2\equiv a \pmod p$ by multiplying both sides by $a$.
Then $(ax+cy)(ax-cy)\equiv a \pmod p$. Considering every possibility, it is not hard to see that there are $p-1$ solutions $(x, y)$ for the equation.
The problem is that I couldn't figure out how to prove this when $-ab$ is a quadratic nonresidue.
And even the solution I had shown to you is not truly mine; I got some help from my peers.
So I was wondering if there is a simpler solution to this question that includes the case when $-ab$ is a quadratic nonresidue. Nevertheless, I would also be really glad if someone could show me how to treat the nonresidue case separately.
Thanks!
AI: Note: A caveat: It is to be noted that the analysis below only works for $p>2$ an odd prime. In any case, the result to be proved does not hold for $p=2$, and we shall only concern ourselves with $p>2$.
We have $(ax)^2 \equiv -aby^2+a \pmod{p}$, so we get $2$ solutions for $x$ when $(\frac{-aby^2+a}{p})=1$, $1$ solution when $(\frac{-aby^2+a}{p})=0$, and $0$ solutions when $(\frac{-aby^2+a}{p})=-1$. Thus the number of solutions is $$\sum_{y=0}^{p-1}{\left(1+\left(\frac{-aby^2+a}{p}\right)\right)}=p+\sum_{y=0}^{p-1}{\left(\frac{-aby^2+a}{p}\right)}=p+\left(\frac{-ab}{p}\right)\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}$$
It is fairly well known that $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=-1$, so this will give the number of solutions as $p-(\frac{-ab}{p})$ indeed. I shall prove that $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=-1$ below:
Proof of $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=-1$: By Euler's criterion, $$\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)} \equiv \sum_{y=0}^{p-1}{(y^2-b^{-1})^{\frac{p-1}{2}}} \equiv \sum_{y=0}^{p-1}{(y^{p-1}+Q(y))} \pmod{p}$$
where $Q(y)$ is a polynomial in $y$ with degree $\leq p-2$, over $\mathbb{Z}_p$.
It is easy to prove that $\sum_{y=0}^{p-1}{y^n} \equiv \begin{cases} 0 \pmod{p} & n=0 \, \text{or} \, p-1 \nmid n \\ -1 \pmod{p} & n>0, p-1 \mid n \end{cases}$.
Indeed, we have $\sum_{y=0}^{p-1}{y^0}=\sum_{y=0}^{p-1}{1} \equiv 0 \pmod{p}$, and for $n>0$, consider a primitive root $g \pmod{p}$, then $$\sum_{y=0}^{p-1}{y^n}=\sum_{y=1}^{p-1}{y^n}\equiv \sum_{i=0}^{p-2}{(g^i)^n} \equiv \begin{cases} \frac{1-(g^n)^{p-1}}{1-g^n} \equiv 0 \pmod{p}& p-1 \nmid n \\ \sum_{y=1}^{p-1}{1} \equiv -1 \pmod{p} & p-1 \mid n \end{cases}$$
Thus we now have
$$\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)} \equiv \sum_{y=0}^{p-1}{(y^{p-1}+Q(y))} \equiv -1\pmod{p}$$
Note that $$\left|\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}\right| \leq \sum_{y=0}^{p-1}{\left|\left(\frac{y^2-b^{-1}}{p}\right)\right|} \leq \sum_{y=0}^{p-1}{1}=p$$
Therefore we must have $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=-1 \, \text{or} \, p-1$. If $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=p-1$, then we must have $p-1$ terms equal to $1$ and exactly $1$ term $\left(\frac{c^2-b^{-1}}{p}\right)$ which is $0$. However $\left(\frac{(-c)^2-b^{-1}}{p}\right)=0$ as well, which is only possible if $c \equiv -c \pmod{p}$, i.e. $c=0$. This gives $\left(\frac{-b^{-1}}{p}\right)=0$, a contradiction. Therefore $\sum_{y=0}^{p-1}{\left(\frac{y^2-b^{-1}}{p}\right)}=-1$, as desired.
Note: The above method can be modified to help evaluate $\sum_{y=0}^{p-1}{\left(\frac{f(y)}{p}\right)}$ for any polynomial $f(y)$. |
H: Logic Puzzle, Two Dollar Bills
I read the following online, and I can't seem to figure it out: A bank robber is planning a heist of a high-tech bank vault. In order to exploit a weakness in the bank's security system, he needs to set off his dynamite exactly 45 seconds after triggering the banks silent alarm - If he sets it off earlier, a failsafe will kick in and electrocute him. If he sets it off later, the cops will be notified. The bank's security system will prevent him from bringing in any electronics or clocks, but he knows that it takes 1 minute for a 100 dollar bill to burn completely. He also knows that a 100 dollar bill burns at a non-uniform rate (there's more ink in the portraits). If he's able to take 2 100 dollar bills and a box of matches in to the bank with him, how can the robber time his dynamite?
Does anyone know a solution?
AI: He starts by lighting one of them by both sides and the other by only one side. When the first dollar is done burning 30 seconds will have passed and the other dollar will have burned to one half. It is at this time that he starts to burn the other dollar by the other side. It will take an additional 15 seconds for half of the dollar to burn at this rate: giving you the 45 seconds.
Edit: to burn it on both sides at the same time you can fold it first to light both and the unfold it. |
H: How to interpret this formula from dsp?
Can you describe me what this formula does and what result in the end?
$$\delta(x)=\begin{cases}
0, & \text{if $x \neq 0$} \\
1, & \text{if $x = 0$} \\
\end{cases}$$
$$x(t)=\sum_{k=-\infty}^{\infty}x(t_k)\delta(t-t_k)$$
AI: $\delta(x)$ is also known as the Kronecker Delta Function. Looking at this page will help. I'm posting as an answer because it does help with achieving the desired sum. |
H: Why must complex linear maps be of the form $h \mapsto ah$, $a \in \mathbb{C}$?
On page 4 of David Ullrich's "Complex Made Simple", he says that complex linear maps from $\mathbb{C}$ to $\mathbb{C}$ are precisely of the form $h \mapsto ah$ for some complex number $a$.
How do I prove that all complex linear maps must be of this form (multiplication by a complex number)?
AI: If $\varphi$ is a complex linear map from $\mathbb{C}$ to $\mathbb{C}$, and $z$ is a given complex number, we see that
$$\varphi(z) = \varphi(z \cdot 1) = z \varphi(1)$$
by linearity. Set $a = \varphi(1)$. |
H: Subgroup of the centralizer
On pg.124 of Abstract Algebra by Dummit and Foote the observation is made that In any group $G$, $<\hspace{0.2mm}g \hspace{0.2mm}> \hspace{1mm} \le \hspace{1mm} C_G(g)$
I am having a difficult time proving this observation to myself. The book gives a particular example but does not provide any proof or general explanation.
AI: To make it trivial write down with words:
$$\langle g\rangle \le C_g(g)\iff \text{any power of $\,g\,$ commutes with $\;g\;$}$$ |
H: Is $\varphi$ a homomorphism?
Define $\varphi:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}_6$ by $\varphi(a,b)=[a+b]_6$.
To show $\varphi$ preserves addition,
$\hspace{100pt} \varphi((a,b)+(c,d))=\varphi((a,b))+\varphi((c,d))$.
$\varphi((a,b)+(c,d))=\varphi((a+c),(b+d))=[a+c]_6+[b+d]_6$
$\hspace{73pt}$$=[a+b]_6+[c+d]_6=\varphi((a,b))+\varphi((c,d))$
$\\$
To show $\varphi$ preserves multiplication,
$\hspace{100pt} \varphi((a,b)\cdot(c,d))=\varphi((a,b))\cdot\varphi((c,d))$.
$\varphi((a,b)\cdot(c,d))=\varphi(ac,bd)=[ac]_6+[bd]_6$
$\varphi((a,b))\cdot \varphi((c,d))=[a+b]_6\cdot[c+d]_6=[(a+b)\cdot(c+d)]_6=[ac+ad+bc+bd]_6$
$\\$
From what I can tell, multiplication is not preserved. However, my book says that $\varphi$ is a ring homomorphism, so either multiplication must be preserved or the book is wrong. Am I missing something?
AI: You are right, this $\phi$ is not a ring homomorphism, only a homomorphism between the underlying ableian groups. Consider for example $(1,0)$ and $(0,1)$, then $(1,0)\cdot(0,1)=(0,0)$ so
$$\phi((1,0)\cdot(0,1))=[0]_6\ \ne \ [1]_6=\phi(1,0)\cdot\phi(0,1)\,.$$ |
H: Show that $\frac{|F(z)-F(a)|}{|F(z)-\bar{F(a)}|}\le\frac{|z-a|}{|z-\bar{a}|}$ if $z\in\Pi^{+}=\{z\in\mathbb{C}:Im(z)>0\}$
Consider $\Pi^{+}=\{z\in\mathbb{C}:Im(z)>0\}$ and let $a\in\Pi^{+}$. Suppose that $F:\Pi^{+} \rightarrow \Pi^{+}$ is holomorphic. Prove that for all $z\in\Pi^{+}$ we have:
$$\frac{|F(z)-F(a)|}{|F(z)-\bar{F(a)}|}\le\frac{|z-a|}{|z-\bar{a}|}$$
Also, show that:
$$|F´(z)|\le\frac{Im(F(a))}{Im(a)}$$
AI: By Pick's Lemma, we know that $\frac{|f(z_1) - f(z_2)|}{|1-\overline{f(z_1)}f(z_2)|} \leq \frac{z_1-z_2}{1-\overline{z_1}z_2}$ for any holomorphic function that maps the unit disk $D$ to itself. Now, note that the map $H(z) =\frac{z-i}{z+i}$ maps the upper half plane $\Pi^{+}$ to the unit disk. Then, we see that $H \circ F \circ H^{-1}$ is a map from $D$ to $D$. Now use Pick's Lemma above and simplify to get your result. |
H: Harmonic function takes both positive and negative values
I am a little confused on the following question:
Suppose that $u$ is harmonic nonconstant on a $D(z_0,R)$ and $u(z_0)=0$. Is it true that on each circle $C(z_0,r)$, with $0<r<R$, the function $u$ will take both positive and negative values?
I think the following should be a counterexample, but I am not so sure:
$f(z)=0$ if $|z|\le 1$, $f(z)=\ln|z|$ for $|z|>1$, $R=2$.
AI: If a function is harmonic then it satisfies the mean value property on any circle inside the region in which it is harmonic:
$$u(z_0) = \frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta$$
If $u(z_0)=0$, then if the function takes any positive values (strictly greater than 0) on the circle $|z-z_0|=r$, then it must clearly also take negative values in order for the mean to equal zero.
It is easy to see that your example $f(z)$ cannot be harmonic, since if we take a small circle centered at $z=1$, then $f$ takes values equal to zero on the part of the circle inside the unit disk and strictly positive values on the part of the circle outside the unit disc, which would make it impossible for the mean to be zero. |
H: Probability of not picking a particular ball, $W1$, out of an urn on the 1st pick, and not picking $W2$ on the 2nd pick?
3 balls are chosen from an urn (without replacement) containing 5 white and 8 red balls. The white balls are numbered (that is, they are distinct), and the red balls are not.
Find $P\{Y_{1}=0,Y_{2}=0\}$.
Here is what I did:
$P\{Y_{1}=0,Y_{2}=0\}=P(Y_{1}=0)P(Y_{2}=0|Y_{1}=0)$
It is clear that,
$P(Y_{1}=0)=12/13$
Using the law of total probability,
$P(Y_{2}=0|Y_{1}=0) = P(Y_{2}=0|Y_{1}=0\, \bigcap \, X_{1}=W_{2} )P(X_{1}=W_{2}) + P(Y_{2}=0|Y_{1}=0\, \bigcap \, X_{1} \neq W_{2} )P( X_{1} \neq W_{2} )$
Where, $X_{1}=W_{2}$ is the event where the 1st ball picked was the White ball #2.
And, $X_{1} \neq W_{2}$ is the event where the 1st ball picked was not White ball #2.
So,
$P(Y_{2}=0|Y_{1}=0\, \bigcap \, X_{1}=W_{2} )=1$
$P(X_{1}=W_{2})= 1/13$
$P(Y_{2}=0|Y_{1}=0\, \bigcap \, X_{1} \neq W_{2} )=11/12$
$P( X_{1} \neq W_{2} ) = 12/13$
And,
$P(Y_{2}=0|Y_{1}=0)=1(1/13)+(11/12)(12/13)=12/13$
Leading to the answer,
$P\{Y_{1}=0,Y_{2}=0\}=P(Y_{1}=0)P(Y_{2}=0|Y_{1}=0)=(12/13)(12/13)=144/169 \approx 0.85$
However, the correct answer apparently is, $15/26$. I am not sure where I messed up. Maybe I need to account for the fact that three balls are chosen from the urn? (The 3rd chosen ball seems to be irrelvant for the problem at hand though). Any insight would be appreciated!
AI: The probability that white ball 1 is chosen is $\frac{3}{13}$, so $\Pr(Y_1=0)=\frac{10}{13}$.
Given that white ball 1 is not chosen, the probability that white ball 2 is chosen is $\frac{3}{12}$, so $\Pr(Y_2=0|Y_1=0)=\frac{9}{12}$.
Multiply. |
H: When is $1\vec{u} \neq \vec{u}$?
In a linear algebra textbook they define a vector space to be a nonempty set $V$ of objects that satisfy certain properties.
One of these properties is that $\forall\vec{u}\in V(1\vec{u}=\vec{u})$
The only way I can think of that property NOT holding would be if scalar multiplication were not defined for some object (not necessarily a vector).
Are there any other cases where multiplication of $n$ by 1 (the scalar multiplicative identity) would not equal $n$?
(not really sure what to tag this with)
AI: Think of it this way. If you did not explicitly make $1\vec{u}=\vec{u}$ into one of the vector space axioms, you would have no justifiable way to conclude that $1\vec{u}=\vec{u}$; this identity does not follow from the other vector space properties. (You can try to derive it, but you'll never get there using only the other axioms.)
Your intuition is good: you want $1\vec{u}=\vec{u}$ to be true in order to have an abstract vector space be in good correspondence to your perception of physical vector spaces and scaling. So you just need to put this relation $1\vec{u}=\vec{u}$ into the list of axioms.
To answer your question more directly, any mathematical structure that has $1\vec{u}\neq\vec{u}$ is, in my opinion, just being needlessly confusing, using the symbol "$1$" to mean something different from all common meanings of that symbol, or using the multiplication operation to mean something different from all common meanings. (Added later: as MarkS. points out in the comments, concatenation is indeed a common meaning for two items written adjacent to each other that I had overlooked.) |
H: One to one and bijection in $\mathbb{Z}^2$
I have the following:
$f(m,n) = (3m+7n, 2m+5n)$ and I want to know if it is a bijection and if so, fine the inverse as well.
Here's my approach:
Suppose $f(m_1,n_1)=f(m_2,n_2)$
then:
$$ (3m_1+7n_1,2m_1+5n_1)=(3m_2+7n_2,2m_2+5n_2)$$
$$3m_1+7n_1=3m_2+7n_2 $$ and $$2m_1+5n_1=2m_2+5n_2$$
Now my issue begins here and don't know what the next step should be... Perhaps adding both equations? I'm not sure.
AI: You can solve them simultaneously : Rewrite them as
$$
3(m_1-m_2) + 7(n_1-n_2) = 0, \text{ and } 2(m_1-m_2) + 5(n_1-n_2) = 0
$$
So set $m=m_1-m_2, n=n_1-n_2$ and see that
$$
6m + 14n = 0, \text{ and } 6m + 15n = 0
$$
Subtracting, gives $n=0$, and hence $m=0$, which is what you want.
As for the inverse, solve
$$
3m+7n=x, \text{ and }2m+5n = y
$$
Multiplying gives
$$
6m+14n = 2x, \text{ and } 6m+15n = 3y
$$
and so
$$
n = 3y-2x
$$
Can you solve for $m$ similarly? This will give you a formula
$$
g(x,y) = (\ldots, 3y-2x)
$$
which is $f^{-1}$ |
H: $ 7^{50} \cdot 4^{102} ≡ x \pmod {110} $
The way I would solve this would be: $$ (7^3)^{15} \cdot 7^5 \cdot (4^4)^{25} \cdot 4^2 $$ and take it from there, but I know that this is most likely in an inefficient way. Does anyone have more efficient methods?
AI: Do prime factorization on $110$. It's $110 = 5 \cdot 2 \cdot 11$.
Now work with the prime factors as moduli.
$$7^{50} \cdot 4^{102} \equiv 0 \pmod 2 \text{ trivial, as 2 \mid 4}$$
Then apply Fermat's Little Theorem so we have:
$$7^4 \equiv 1 \pmod 5 \implies 7^{50} \equiv 49 \equiv 4 \pmod 5$$
$$4^4 \equiv 1 \pmod 5 \implies 4^{102} \equiv 4^2 \equiv 1 \pmod 5$$
Now multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 4 \cdot 1 \equiv 4 \pmod 5$$
Now repeat the method with the last factor:
$$7^{10} \equiv 1 \pmod {11} \implies 7^{50} \equiv 1 \pmod {11}$$
$$4^{10} \equiv 1 \pmod {11} \implies 4^{102} \equiv 4^2 \equiv 16 \equiv 5 \pmod {11}$$
Multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 1 \cdot 5 \equiv 5 \pmod {11}$$
Now just apply CRT to the three congruence relation to get the final answer. |
H: Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$
Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$
I use software to complete the series is $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$
I have no idea about it. :|
AI: Consider the function
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
$f(x)$ has a Maclurin expansion as follows:
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$
Differentiating, we get
$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$
Evaluate at $x=1/2$
$$\sum_{n=0}^{\infty} \frac{1}{\displaystyle \binom{2 n}{n}} = \frac{\frac12 \arcsin{\frac12}}{3 \sqrt{3}/8} + \frac{4}{3} = \frac{2\sqrt{3} \pi+36}{27}$$
ADDENDUM
There are many derivations here of the above result for the Maclurin series for $f(x)$; I refer you to this one. |
H: Neumann BV problem on disk (weak vs classical solution)
I am tring to solve
$\bigtriangleup u =-1$ such that the normal derivative vanishes at the boundary where the domain is the unit disc.
In polar coordinates I got
I got $u(r)=-1/4 r^{2} +1/2 \ln(r)$
as a solution. Does this qualify as a weak solution (since it has a singularity).
Are there any smooth solutions?
AI: There are no solutions, weak or smooth. The homogeneous Neumann condition is incompatible with the Poisson equation $\Delta u = f$ unless $\int_\Omega f =0$.
For your $u$, the weak Laplacian has a delta function component at the origin, which violates the equation. |
H: Clarification on a proof involving cluster point
Definition of cluster point- Let $A \subseteq \mathbb{R}$. A point $c\in\mathbb{R}$ is a cluster point of $A$ if for evert $\delta>0$ there exists at least one point $x\in A$, $x\neq c$ such that $|x-c|<\delta$.
Theorem- A number $c\in\mathbb{R}$ is a cluster point of a subset $A$ of $\mathbb{R}$ iff there exists a sequence $(a_n)$ in A such that lim$(a_n)=c$ and $a_n \neq c$ for all $n\in\mathbb{N}$. I'm stuck on understanding the foward direction.
Proof: $(\rightarrow)$"If $c$ is a cluster point of $A$ then for any $n\in\mathbb{N}$ the $\frac{1}{n}$-neighborhood $V_{\frac{1}{n}}(c)$ contains atleast one point $a_n\in A$ distinct from $c$. Then $a_n\in A$, $a_n \neq c$, and $|a_n-c|<\frac{1}{n}$ implies lim$(a_n)=c$."
What I understand is that if we let $\delta>0$ then it follows by the Archimdean property that there exists a $k\in\mathbb{N}$ such that $\frac{1}{k}<\delta$. It follows if $n\geq k$ then $\frac{1}{n}<\frac{1}{k}<\delta$. Since $c$ is a cluster point there exists a $x_{n}\in A$ where $x_{n}\neq c$ such that $|x_{n}-c|<\frac{1}{n}<\delta$ since $\frac{1}{n}>0$. Thus by the definition of convergence we have $lim (a_n)=c$. Would this be a correct way of understanding the proof to this theorem?
AI: Maybe you're thinking about it too hard. For any positive number $\delta$, you're promised one point $x =x _\delta$ different from $c$. In particular, for any $n\in\Bbb N$, $n^{-1}$ is a positive number, so you're promised a point $x=x_n$ different from $c$. And since $x_n\in(c-n^{-1},c+n^{-1})$, we have $$|x_n-c|<n^{-1}$$ so that $x_n\to c$. You are right that to show that $x_n\to c$ we use the Archiemedean property of the real numbers. Given $\varepsilon >0$ all we need to do is take $n$ large enough so that $n^{-1}<\varepsilon$. |
H: Changing operator to polar coordinates
Let $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$ be the Laplace operator on the $(x,y)$-plane. Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. Show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$
I don't know how to change the coordinates for an operator like this one. What is the method that should be used?
AI: Write
$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}$$
where $r=\sqrt{x^2+y^2}$ and $\tan{\theta} = y/x$. Then $\partial r/\partial x = x/r$ and $\partial \theta/\partial x = -y/r^2$, and
$$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$
Similarly, you may show that
$$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$
The Laplacian is then
$$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$
This should be plenty to go on. |
H: Orthogonal projection on vector space of intervals
Let $R_N$ be the set of $2^N$ intervals $$\left\{\left[0,\frac{1}{2^N}\right), \left[\frac{1}{2^N}, \frac{2}{2^N}\right),\ldots,\left[\frac{2^N-1}{2^N}, 1\right)\right\}.$$ Let $$V_N=\operatorname{span}\{1_I\mid I\in R^N\}$$ Let $P_N:L^2([0,1])\rightarrow V_N$ be the orthogonal projection onto $V_N$.
(a) Find the formula for $a_I$ such that $$P_N(f)=\sum_{I\in R_N}a_I1_I$$ for $f\in L^2([0,1])$.
(b) Suppose $f\in C([0,1])$. Show that $P_N(f)$ converges uniformly on $[0,1)$ to $f$.
For part (a) I think we should have $a_I=2^N\langle f,1_I\rangle$. Then we will have $P_N(1_I)=2^N\langle 1_I,1_I\rangle 1_I=1_I$ , so $P_N(f)=f$ for all $f\in V_N$. Now let $f\in L^2([0,1])$. How can I show that $P_N(f)\perp (f-P_N(f))$?
Also, what should I use for part (b) to show uniform convergence?
AI: Your answer to (a) appears to be correct.
To check $P_N(f)\perp f-P_N(f)$, just compute their inner product. You will end up with something that looks like
$$
\sum a_I\langle 1_I, f\rangle - \sum |a_I|^2\langle 1_I, 1_I\rangle
$$
now use the fact that $a_I = 2^n\langle f,1_I\rangle$ to get this to be zero.
For any $f\in C[0,1]$, and $\epsilon > 0$, by uniform continuity of $f$, there is a $N \in \mathbb{N}$ such that
$$
x \in I_j := \left [ \frac{j-1}{2^N}, \frac{j}{2^N} \right ] \Rightarrow |f(x) - f(j/2^N)|<\epsilon/2
$$
Now
$$
|a_{I_j} - f(j/2^N)| = \left |2^N \int_{I_j} f(x)dx - 2^N \int_{I_j}f(j/2^N)dx \right |
$$
$$
\leq 2^N \int_{I_j} |f(x) - f(j/2^N)|dx < \epsilon/2
$$
Hence,
$$
\|P_N(f) - f\| \leq \|\sum a_{I_j} 1_{I_j} - \sum f(j/2^N)1_{I_j}\| + \|\sum f(j/2^n)1_{I_j} - f\|
$$
and each term is $<\epsilon/2$ |
H: Elimination of complex variable in integral
I have the equation: $$\frac{1}{\tau}\intop_{0}^{\tau}A\sin\left(\Omega t\right)\cdot A\sin\left(\Omega\left(t-\lambda\right)\right)\mathrm{d}t$$ for which the attempted solution is to convert the sine terms into complex natural exponents (engineering notation using j as imaginary unit) as $$\frac{A^{2}}{\tau}\intop_{0}^{\tau}\frac{e^{j\Omega t}-e^{-j\Omega t}}{2j}\cdot\frac{e^{j\Omega\left(t-\lambda\right)}-e^{-j\Omega\left(t-\lambda\right)}}{2j}\mathrm{d}t$$ the next step in the solution moves the $\frac{1}{2j}$ term outside of the integral to form $$\frac{-A^{2}}{4\tau}\intop_{0}^{\tau}\left(e^{j\Omega t}-e^{-j\Omega t}\right)\cdot\left(e^{j\Omega\left(t-\lambda\right)}-e^{-j\Omega\left(t-\lambda\right)}\right)\mathrm{d}t$$ I'm struggling to understand how $\frac{1}{2j}\rightarrow\frac{-1}{4}$ when being moved out of the integral.
AI: You're moving out $\dfrac1{2j}\cdot\dfrac1{2j}=-\dfrac14,$ not just $\dfrac1{2j}.$ |
H: Prove that the mapping $U(16)$ to itself by $x \rightarrow x^3$ is an automorphism
Prove that the mapping $U(16) = \{{1,3,5,7,9,11,13,15}\}$ to itself by $x \rightarrow x^3$ is an automorphism. What about $x \rightarrow x^5$ and $x \rightarrow x^7$? any generalization?
So far i have prove the first part.
Let $\psi : x \rightarrow x^3$,
Since $U(16)$ is a group under multiplication modulo 16, hence $U(16)$ is closed under mult.modulo 16. Take any $x,y \in U(16)$ and let $\psi(x) = \psi(y)$, then $x^3 = y^3mod(16)$. This implies $16 | (x^3 - y^3)$ and since $x^3$ and $y^3$ always less than 16, then $x = ymod( 16)$. So that $\psi$ is one-to-one. By this, it clear that $\psi$ is also onto.
Now since $\psi(xy)=x^{3}y^{3}mod(16) = x^{3}mod(16)y^{3}mod(16) = \psi(x)\psi(y)$ and $\psi$ is both one-to-one and onto, thus $\psi$ is an automorphism.
For proving $x \rightarrow x^5$ and $x \rightarrow x^7$, just use the same step above. But how to generalize?
My conclusion so far is that $\psi : x \rightarrow x^n$ where $n$ is odd positive integers are an automorphism. Because if i take $n=2$, then $\psi(3) = 9 = \psi(5)$ and $\psi$ is not one-to-one.
AI: Note that if $x$ is odd, then by Euler's Theorem we have $x^8\equiv 1\pmod{16}$. Indeed by direct calculation $x^4\equiv 1\pmod{16}$. So once you have done $x^1$ and $x^3$, the fact that $x^k$ is an automorphism for $k$ odd is automatic. |
H: $f:U(\mathbb{Z}/(n)) \to Aut(G)$, defined by $f([s]_n) = \phi_s$ is surjective
Let $G = \left \langle {g} \right \rangle$ be a cyclic group of order $n \geq 2.$ Define a map $\phi_s : G \to G (x \to x^s)$ and $Aut(G): = \{f:G \to G | f\ is\ isomorphism\}. $ Then, $\phi_s$ is an isomorphism iff $gcd(s,n)=1. $
Could anyone advise me on how to show the map $f:U(\mathbb{Z}/(n)) \to Aut(G)$, defined by $f([s]_n) = \phi_s$ is surjective? i.e. given $f \in Aut(G),$ how do I find $[s]_n \in U(\mathbb{Z}/(n))$ such that $f(g^i) = g^j = (gi)^s?$ Thank you.
AI: Let $f\in Aut(G)$, then $f(g) = g^s$ for some $s\in \mathbb{N}$. Then for any $x\in G$, there exists $k \in \mathbb{N}$ such that $x = g^k$, whence
$$
f(x) = f(g)^k = g^{sk} = x^s
$$
Hence, $f = \phi_s$ |
H: Can the Poisson Distribution be used to find the expected value of time of arrival given an expected arrivals per unit time?
My understanding of the Poisson Distribution is that its PMF $P(x=k) = \dfrac {\lambda^k e^{-\lambda}} {k!}$ refers to the probability of finding k events given an expected arrival expectancy $\lambda$. This gives me, rather trivially, that the expected value for the number of arrivals is equal to the average number of arrivals $\lambda$. However, suppose I know $\lambda$ is 3 events per day. How can I calculate the expected number of days before $n$ events happen? Can I just invert my $\lambda$, so that my units are now days/event, and use the same distribution?
A supplemental question: Currently, the units in my exponent appears to be events/time. Shouldn't I have to multiply by some time $t$, so that the distribution looks like $P(x=k) = \dfrac {(\lambda t)^k e^{-\lambda t}} {k!}$? (I'm taking "events" to be unitless...) If so, I would expect my new distribution to be $P(t=k) = \dfrac {(\frac n {\lambda})^{k} e^{ \frac {-n} {\lambda}}} {k!}$, where $n$ is the number of events, $k$ is the amount of time, and $\lambda$ is still in events/time. Thus if, in the above example, I want to know the probability that it would take 1 day for 5 arrivals, I would set $k$ = 1, $n$ = 5, and $\lambda$ = 3. Is there anything wrong with this formulation?
AI: If the number of events per unit time has Poisson distribution with parameter $\lambda$, then the waiting time for the first event has exponential distribution with parameter $\lambda$, and therefore expectation $\frac{1}{\lambda}$.
The waiting time until the $n$-th event is the sum of $n$ exponentials with parameter $\lambda$. It therefore has mean $\frac{n}{\lambda}$.
Added: The following may deal with your second question. Let the number of events per unit time have Poisson distribution with parameter $\lambda$. Then the number $Y$ of events in time $t$ has Poisson distribution with parameter $\lambda t$. This is a special property of the Poisson. |
H: Finding the Hopf Algebra Coproduct coming from an Affine Group Scheme
I was wondering if anyone could help with how to, strictly from Yoneda's Lemma, obtain the coproduct map on the Hopf Algebra for an Affine Group Scheme. Particularly for something like $\text{SL}_2$
So if $G=\text{SL}_2$, let $m:G \times G \to G$ be the multiplication map. Yoneda'e lemma tells us that this induces a map $\Delta:A \to A \otimes A$ and there is a specific construction:
$m: G\times G(A \otimes A) \to G(A \otimes A)$ which (since $G$ is representable is just a map $\text{Hom}(A \otimes A, A \otimes A) \to \text{Hom}(A, A \otimes A)$ and element-wise $id$ gets sent to $\Delta$.
This is hypothetically how you should "find" $\Delta$. In the case of $\text{SL}_2$ I guess the part I am having trouble with is finding the corresponding element in $G\times G(A \otimes A)$ that goes with $id$. That way I can just do the multiplication map and see what the entries are to get the $\Delta$ map.
AI: How about we do $\mathbb G_m$ (or $GL_1$, if you prefer to denote it that way),so
there is less notation.
The ring $A$ is equal to $k[x,x^{-1}]$, so $A \otimes A $ equals $k[x,x^{-1},y,y^{-1}]$ (thinking of $x$ as $x\otimes 1$ and $y$ as $1 \otimes x$, if you see what I mean).
Now the identity map from $A$ to $A$ corresponds to the element $x \in \mathbb G_m(A)$ (since the identification of $\mathrm{Hom}(A,A')$ with $\mathbb G_m$ is given by sending a homomorphism $\varphi$ to the element $\varphi(x) \in A'$), and so the identity map of $A\otimes A$ corresponds to the pair $(x,y)
\in (\mathbb G_m\times \mathbb G_m)(A\otimes A) = \mathbb G_m(A\otimes A) \times
\mathbb G_m(A\otimes A)$.
When we multiply these two elements, we get $xy \in \mathbb G_m(A\otimes A)$,
and thus $\Delta$ is given by $x \mapsto xy$.
In general, $\Delta$ just expresses the formula for $m$ in coordinates.
So for $SL_2$, if we label the generators of $A$ as $a,b,c,d$ (with
the relation $ad - bc = 1$), and label the generators of $A \otimes A$
as $a,b,c,d,a',b',c',d'$ (rather than $a\otimes 1$, etc., $1\otimes a$, etc.),
then $\Delta$ will have the formula
$$a \mapsto a a' + bc', \text{ etc.},$$
just coming from the formula
$$\pmatrix{a & b \\ c & d}\cdot \pmatrix{a' & b'\\ c' & d'} =
\pmatrix{a a' + b c' & a b' + b d' \\ c a' + d c' & c b' + d d'}.$$
If you want to derive this via your Yoneda strategy, it will work exactly
as with the $\mathbb G_m$ example (just with more notation, because
there are more generators for $A$). |
H: very basic calculus doubt
Suppose $a,b \in \mathbb{R}^{\geq0}$. Let $\epsilon \in (0,1)$
$$ a \geq b \epsilon \implies a \geq b $$
My try: If $a < b $, then can find $n$ such that $b - a > \frac{1}{n} $. But I am stuck here. Maybe the result in not true??
if it is not true, why then is this book author assumes this in the green part??
AI: Your statement should read for all $\epsilon \in (0,1)$.
I think contradiction is easiest here.
First, if $b=0$, then it is clear there is nothing to prove.
Suppose $b>0$, and $b>a$.
Then $b= a + (b-a)$, and since $b>a$, we have $b-a >0$, and so $b-a > \frac{1}{2}(b-a) >0$.
Then $b > a + \frac{1}{2}(b-a)$, and so $b(1-\frac{1}{2}\frac{b-a}{b}) > a$.
Since $\frac{1}{2}\frac{b-a}{b} \in (0,1)$, we have $1-\frac{1}{2}\frac{b-a}{b} \in (0,1)$. Hence letting $\epsilon =1-\frac{1}{2}\frac{b-a}{b} \in (0,1)$, we have $b \epsilon > a$, a contradiction. |
H: Monic and epic implies isomorphism in an abelian category?
Is it true that monic and epic implies isomorphism in an abelian category?
AI: The answer is yes, as a consequence of two facts. First, in any category (abelian or not), if the equalizer of a pair of maps is epic, then it is an isomorphism. (Proof: By definition of epic, the two maps that it equalizes must be equal, and then an isomorphism clearly serves as the equalizer.) Second, in an abelian category, every monomorphism is the kernel of some map (this is axiom A3 in the definition of abelian category, on page 35 of Freyd's book "Abelian Categories), which means it's the equalizer of that map and a zero map. |
H: Role of differentiation in a polynomial
I have learnt that, to find maximum or minimum value of a polynomial $p(x)$, we take its derivative, equate it to zero, solve for x, find the maxima and minima, and then put the value of x in the original equation. But when we take the derivative of $p(x)$, what exactly are we doing? Also, what do we do if we are unable to solve for x even after taking the derivative?
AI: When I take the derivative of a function $p(x)$, at least in the back of my mind, I am finding another function $p'(x)$ with the property that
$$p'(x)=\lim_{h\rightarrow0}\frac{p(x+h)-p(x)}{h}.$$
Essentially, I am finding another function that contains the instantaneous rate of change of the function $p(x)$ at various values for the independent variable $x$.
When I think of the zeroes of the derivative corresponding to the extrema (max/min) of the function, I think about how the extrema correspond to the points at which the instantaneous rate of change (or the derivative) on the original function $p(x)$ is identically zero, which is to say that the slope of the tangent line at that point is $0$.
Some derivatives are not defined for all $x$. This just means the derivative does not exist at those values. Other derivatives are defined at various values of $x$, but the task of actually finding them by pure algebraic methods is quite beyond any human's reach. We use approximation methods to find these when necessary. |
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