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H: Is there a relationship between $\ln(x\pm y)$ and $\ln(x)\pm\ln(y)$?
I am dealing with points on a 2d space $(x, y)$ where $x$ and $y$ are always positive integers.
In an algorithm, I have pre-computed $\log_2(x)$ and $\log_2(y)$ for given points of interest.
I now need to compute $\log_2(x+y)$ and $\log_2(x-y)$ but don't have the luxury to do so from a computational perspective so I started looking for some sort of correlation.
I wasn't sure if there was a relationship between $\ln(x\pm y)$ and $\ln(x)\pm\ln(y)$ so I charted some numbers in Excel and came across the following:
The ratio of $(Log(X+Y) / (Log(X)+Log(Y)))$ seems to mostly lie between 0.5 and 0.7.
For a large set of numbers the average ratio and median ratio seem to be ALWAYS between 0.56 and 0.58.
Of course, there are corner cases in subtraction where x-y turns out negative. How can I avoid that by the way?
So the question is:
Am I missing some fundamental concepts to have to find this relationship this way?
If the answer to the above is no, how reliable would this correlation be for all integers x and y? The magnitude I am dealing with is around 2 to the power 10,000,000.
SOME EXTRA CONTEXT:
Some have suggested more context may bring about a different approach altogether so here it is. In 2D space, I have a starting point $(x, y)$ and need to move around following some rules. Allowed directions are $\pm (horizontal$, $vertical$, $diagonal$ and $anti-diagonal)$. Some other restrictions include not moving along the diagonal of $(x, y)$ where $(x*y)$ is a power of 2. The target is to get to the top left of the grid where the concentration of power of 2 numbers is high. Lastly, we can only change direction after encountering the diagonal of a power of 2.
So we start looping at the starting point, find neighboring power of 2 coordinates and filter out all diagonal intersections between our current position and power of 2 neighbors (which become potential turning points). Once we have this list, we need to determine optimal direction so we land closest to (1, 1) in euclidean distance. This is where we cannot afford any more multiplication, division, logarithms, etc.
AI: The defining property of the logarithm is that it converts multiplication to addition. Thus we have $\log xy = \log x + \log y$ and $\log \frac{x}{y} = \log x - \log y$. But it does not really convert addition to anything.
More precisely, conditional on Schanuel's conjecture the numbers $\log p$, as $p$ runs over the primes, are algebraically independent; that is, there is no polynomial relationship between them whatsoever (with rational coefficients). In particular, there is no polynomial relationship between the numbers $\log (2 + 3), \log 2, \log 3$ (for example).
However, if you're willing to settle for approximate relationships and if one of $x$ or $y$ is small relative to the other, you can use the Taylor series of the logarithm. |
H: Showing that a linear map $\mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3$ is an isomorphism.
Let $t_1,t_2,t_3$ be distinct real numbers and consider the linear map
$$T : \mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3, \quad \quad p(x) \mapsto \begin{pmatrix}p(t_1)\\p(t_2)\\p(t_3)\end{pmatrix}$$
I want to show that $T$ is an isomorphism. Since it is given that $T$ is linear, I just need to show that $T$ is bijective. My initial approach was to solve the system
$$
\begin{pmatrix}
a + bt_1 + ct_1^2\\
a + bt_2 + ct_2^2\\
a + bt_3 + ct_3^2
\end{pmatrix}
=
\begin{pmatrix}
r_1\\
r_2\\
r_3
\end{pmatrix}
$$
for $a,b,c$ in terms of $r_1,r_2,r_3$, thus determining the inverse map $T^{-1}$ mapping a point in $\mathbb{R}^3$ to a polynomial of degree $\leq 2$.
I solved this system using Maple, and got a solution which was defined when $t_1,t_2,t_3$ where not all equal, which is fine by their definition, but I am wondering if there is a nicer argument, especially since the solution is rather ugly.
(this is not homework)
AI: Hint:
$T$ is onto: Langrange Polynomial.
Injectivity: if $p$ is an polynomial of degree $\le2$, knowing values in 3 distinct points, you know $p$. |
H: Probability without replacement question
I have a bag full of 10 marbles: 2 black, 1 blue, 1 yellow, 3 green, 1 brown, and 2 purple. I draw 5 marbles one at a time without replacement. What is the probability of a black marble to be in the five drawn?
AI: Think of the marbles as having, in addition to colour, an ID number that makes them distinct.
There are two interpretations of "one black:" A: at least one black, and B: exactly one black. The probabilities are of course different. My preferred interpretation of the wording is A. Edit: With the change of wording to "a black" it is clearly A that is meant, but for your interest I will keep the analysis of B.
A: At least one black: It is easier to find first the probability of no black.
There are $\binom{10}{5}$ ways to choose $5$ marbles, all equally likely. Note that there are $\binom{8}{5}$ ways to choose $5$ marbles from the $8$ non-black. So the probability that all the balls are non-black is
$$\frac{\binom{8}{5}}{\binom{10}{5}},$$
and therefore the probability of at least one black is
$$1-\frac{\binom{8}{5}}{\binom{10}{5}}.$$
B: Exactly one black: There are $\binom{2}{1}$ ways of choosing one black from the two available. For each such way, there are $\binom{8}{4}$ ways to choose the non-blacks to go with it. So the total number of ways to pick exactly one black, and the rest non-black, is $\binom{2}{1}\binom{8}{4}$. Thus our probability is
$$\frac{\binom{2}{1}\binom{8}{4}}{\binom{10}{5}}.$$
Remark: We used general techniques. For A, there is a simpler way. The probability that the first marble chosen is non-black is $\frac{8}{10}$. Given the first was non-black, the probability that the second is non-black is $\frac{7}{9}$, since there are $7$ non-blacks left in a total of $9$. So the probability the first two are non-black is $\frac{8}{10}\cdot\frac{7}{9}$. Continue in this way. The probability all five are non-black is
$$\frac{8}{10}\cdot\frac{7}{9}\cdot\frac{6}{8}\cdot\frac{5}{7}\cdot\frac{4}{6}.$$
As in the earlier discussion, subtract the above from $1$. |
H: Estimates of Gaussian Logarithms
I've been implementing logarithmic number system and I came across these functions called Gaussian logarithms:
$f(x) = \log(1 + e^x)$.
$g(x) = \log(e^x - 1)$ for $x > 0$.
$h(x) = \log(1 - e^x)$ for $x < 0$.
Since I am lazy, I assume $\log$ and $\exp$ aren't expensive to compute, but $\exp$ of a very big number causes a problem because I'm using double. So, I found, with some guess work, nice approximations of the above functions:
$\bar f(x) = x + e^{-x}$ for $x$ large.
$\bar g(x) = x - e^{-x}$ for $x$ large.
$\bar h(x) =$ ... Actually this one is not needed because $\exp(x)$ is small for negative $x$. (You see how lazy I am.)
It turns out that $\bar f$ and $\bar g$ are surprisingly good estimates of $f$ and $g$. I can justify why they are decent estimates as follows:
$f(x) = \log(e^x(1 + e^{-x})) \approx \log(e^x(e^{e^{-x}})) = x + e^{-x}$.
Similarly, $g(x) = \log(e^x(1 - e^{-x})) \approx \log(e^x(e^{-e^{-x}})) = x - e^{-x}$.
However, as I try to do a more careful analysis, I get stuck:
$f(x) = \log(e^x(1 + e^{-x})) = \log(e^x(e^{e^{-x}}) + e^xO(e^{-2x}))) = ???$
What is a proper way to do an analysis like this?
Here's my initial attempt: The derivative of $\log(z)$ is bounded by $1$ for $z \ge 1$, so
$f(x) = x + \log(e^{e^{-x}} + O(e^{-2x})) \approx x + e^{-x} + O(e^{-2x})$,
but I have a feeling that this method of analysis is not very good because ... it's so unfamiliar. It requires the knowledge that $e^{e^{-x}} \ge 1$ in order to get the bound, and the bound seems pretty loose in my opinion. Also, I don't see a way to improve the estimate by "adding more terms" like what I could do with Taylor series.
I'd really appreciate if someone can show me a more standard method of analyzing this situation. And it would be really nice if more accurate estimates or an arbitrary precision method can be derived.
AI: $$f(x) = \log(1 + \mathrm e^x)=\log(\mathrm e^x(1+\mathrm e^{-x}))=\log\mathrm e^x+\log(1+\mathrm e^{-x})=x-\sum_{n=1}^\infty\frac{(-1)^n\mathrm e^{-nx}}n\;.$$ |
H: Maximal ideals in $R[x]$
I want to prove the following result:
Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.
I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:
$R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal;
If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification);
The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.
I'll appreciate any comment.
AI: In (2), your argument is valid because of the following more general fact:
Let $f : A \to B$ be a surjective ring homomorphism. If $P$ is a prime ideal of $A$ that contains $\ker f$, then $f(P)$ is a prime ideal in $B$.
Proof: We have $\ker f \subseteq P \subseteq A$. So by the third isomorphism theorem, we have that $P/\ker f$ is an ideal of $A/\ker(f)$. Furthermore, we have that
$$(A/\ker f)/(P/\ker f) \cong A/P.$$
The latter is an integral domain because $P$ is a prime ideal, this proves that $P/\ker f$ is a prime ideal in $A/\ker f$. Furthermore by the first isomorphism theorem you know that because $f$ is surjective,
$$A/\ker f \cong B.$$
It follows that because $P/\ker f$ is a prime ideal in $A/\ker f$ that $f(P)$ is a prime ideal in $B$.
$\hspace{6in} \square$
Once you know this, you just need to apply it with $A = R[x]$, $B = R[x]/M[X]$, $f = \pi$ where $\pi$ is the canonical projection of $R[x]$ onto the quotient $R[X]/M[x]$ and its pre-image $\pi^{-1}$ define bijective correspondences between prime ideals in $R[x]$ that contain $M[x]$ and prime ideals in $B$. The correspondence is as follows:
$$\{P' \in \textrm{Spec}(B) \implies \pi^{-1}(P) \text{ a prime ideal in $A$ that contains} \hspace{2mm} M[x]\}$$
$$\updownarrow$$
$$\{P \hspace{2mm} \text{a prime ideal in $A$ that contains $M[x]$} \implies \pi(P) \in \textrm{Spec}(B) \}$$ |
H: Is the set of irrationals separable as a subspace of the real line?
I am trying to find an example of a separable Hausdorff space which has a non-separable subspace. This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or non-separable?
A space is separable if it contains a countable dense subset. A subset $A$ of a space $X$ is dense in $X$ if $\bar{A}=X$.
It's easy to come up with a dense set and a countable set in $\mathbb{R}\setminus\mathbb{Q}$, since (trivially) $\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}$ in the subspace topology, and as a countable set we can take something like $\{k\pi \mid k \in \mathbb{Z}\}$. But is there a subset that is both dense AND countable? And of course, how do we prove the result?
AI: $\Bbb R$ is second countable (i.e., has a countable base), so it’s hereditarily separable. Specifically, let $\mathscr{B}$ be the set of all open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable. Enumerate $\mathscr{B}=\{B_n:n\in\Bbb N\}$, and for $n\in\Bbb N$ let $x_n$ be any irrational number in $B_n$. Then $\{x_n:n\in\Bbb N\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$. (Clearly the same trick works for any subset of $\Bbb R$, not just the irrationals.)
For an explicit example of such a set, let $\alpha$ be any irrational; then $\{p+\alpha:p\in\Bbb Q\}$ is a countable dense subset of $\Bbb R\setminus\Bbb Q$.
There are many separable Hausdorff spaces with non-separable subspaces. Two are mentioned in this answer to an earlier question. The first is compact; the second is Tikhonov and pseudocompact. Both are therefore quite nice spaces. Both are a bit complicated, however. A simpler example is the Sorgenfrey plane $\Bbb S$: $\Bbb Q\times\Bbb Q$ is a countable dense subset of $\Bbb S$, and $\{\langle x,-x\rangle:x\in\Bbb R\}$ is an uncountable discrete subset of $\Bbb S$ (which is obviously not separable as a subspace of $\Bbb S$). |
H: Block inverse of symmetric matrices
Let us assume we have a symmetric $n \times n$ matrix $A$. We know the inverse of $A$.
Let us say that we now add one column and one row to $A$, in a way that the resulting matrix ($B$) is an $(n+1) \times (n+1)$ matrix that is still symmetric.
For instance,
$A = \begin{pmatrix}a & b \\b & d \\\end{pmatrix}$
and
$B = \begin{pmatrix}a & b & X \\b & d & Y \\X & Y & Z\end{pmatrix}$
Given that I know $A^{-1}$, is there any way of using this information to find $B^{-1}$ without having to compute this latter inverse from scratch? If an exact solution is not possible, approximations would also help.
Thanks,
Bruno
P.S. in case it makes any difference, both $A$ and $B$ are covariance matrices.
AI: Certainly, one can use the bordering method for this (a special case of the usual formula for block inversion):
$$\begin{pmatrix}\mathbf A&\mathbf \delta\\\mathbf \delta^\top&Z\end{pmatrix}^{-1}=\begin{pmatrix}\mathbf A^{-1}+\frac{\mathbf A^{-1}\mathbf \delta\mathbf \delta^\top\mathbf A^{-1}}{\mu}&-\frac{\mathbf A^{-1}\mathbf \delta}{\mu}\\-\frac{\mathbf \delta^\top\mathbf A^{-1}}{\mu}&\frac1{\mu}\end{pmatrix}$$
where $\mathbf \delta^\top=(X\quad Y)$ and $\mu=Z-\mathbf \delta^\top\mathbf A^{-1}\mathbf \delta$. |
H: Limit of a complex number over its conjugate, as z approach the infinity.
this question must be pretty easy but i´m just taking my firs course in complex variables, i need to find the limit of Z over it conjugate as z reach the infinity. I need to know how to do it in a rigorous way and informal way, so i need your help. By formal, i mean with neighborhood and epsilon definition of limit.
I have tried with trig form of Z, but i don´t reach any answer.
Thanks
AI: Hint: The trigonometric form is a good idea. What is $\dfrac{r(\cos\theta+i\sin\theta)}{r(\cos\theta-i\sin\theta)}$ when $r$ is large? Does it depend on $\theta$? |
H: On modules over polynomial rings
Let $\mathbb{A}$ be a polynomial ring in $n$ variables over an algebraically closed field $\mathbb F$. Given a maximal ideal $\mathfrak{m}$ of $\mathbb A$, consider the quotient $\mathbb{A}/\mathfrak{m}$ as a left $\mathbb{A}$-module ($\mathbb{A}/\mathfrak{m}$ is one-dimensional over $\mathbb F$).
Suppose $M$ is a right finitely generated $\mathbb{A}$-module which is infinite dimensional (as $\mathbb F$-vector space) and such that $\dim_{\mathbb F}(M\otimes_{\mathbb{A}} \mathbb{A}/\mathfrak{m})<\infty$ for any maximal ideal $\mathfrak m$.
Question:
Can we say that $\dim_{\mathbb F}(M\otimes_{\mathbb{A}} \mathbb{A}/\mathfrak{m})$ is independent of $\mathfrak m$?
The motivation of the question is to show that $M$ is projective (therefore free, by Quillen's Theorem).
AI: No. If $M$ is f.g. over $A$ then $M\otimes_A (A/\mathfrak m)$ will be finite dimensional over $F$ for every maximal ideal $\mathfrak m$ of $M$ (since $A/\mathfrak m = F$, and change of scalars preserves the property of being finite dimensional).
So your question amounts to asking whether every f.g. module over $A$ that is infinite dim'l over $F$ is free, and the answer is no.
E.g. Suppose that $n = 1$, and take $M = A \oplus F,$ with $x$ and $y$ acting via $0$ on the second summand.
E.g. Suppose that $n = 2$, so $A = F[x,y].$ Let $M = F[x]$, regarded as an $A$-module by having $y$ act via $0$.
E.g. Suppose that $ n =2$, and let $M = (x,y) \subset F[x,y]$. |
H: General questions about equivalence classes and partitions
1) If a set is partitioned into non-overlapping, non-empty subsets, then those subsets are equivalence classes. If each element in a set is unique, how can a set be partitioned into subsets with equivalent elements?
2) What is the criteria for something to be an equivalence class? If the set of natural numbers was partitioned into odds {O} and evens {E}, then if I say E_1 ~ E_2 etc, am I just saying that they are equivalent in terms of how they were partitioned (i.e. divisible by 2)?
Does that mean I could partition the real numbers into subsets with equal numbers of decimal places? If that is a legitimate partition, how do I 'mathematically' formulate it? If it is not, why not?
3) If a set is partitioned into non-overlapping, non-empty subsets at random, how would there be equivalence on the partitioned subsets?
AI: The idea for the first one is the following: Given a partition $\Delta$ of $X$, that partition induces an equivalence relation on $X$. Namely, we say $x$ and $y$ stand in relation if they belong to the same set of the partition. You can check that this is indeed an equivalence relation. Because
$(1)$ $x$ is always in the same set as $x$.
$(2)$ If $x$ is in the same set as $y$, then $y$ is in the same set as $x$.
$(3)$ If $x$ is in the same set as $y$, and $y$ is in the asme set as $z$, then $x$ is in the asme set as $z$.
Given a relation $\sim $ on a set $X$, we say it is an equivalence relation if it has the following three properties:
$(1)$ Reflexivity $x\sim x$. That is, every element stands in relation with itself.
$(2)$ Symmetry If $x\sim y$, then $y\sim x$. That is, if $x$ stands in relation with $y$, then $y$ stands in relation to $x$.
$(3)$ Transitivity If $x\sim y$ and $y\sim z$, then $x\sim z$. That is if $x$ stands in relation with $y$ and $y$ stands in relation with $z$, then $x$ stands in relation with $z$.
You can check that the relation divisibility is transitive and reflexive. You an check the relation of inclusio is transitive and reflexive, too, but not symmetryc. You can check that congruence $\mod{}$ a number $n$ is a equivalence relation, and so is usual equality of numbers.
Given a relation $\sim$ on a set $X$, and an element of $X$, we define the equivalence class $[[x]]$ of $x$ as all elements in $X$ that stand in relation to $x$. That is
$$[[x]]=\{a\in X:a\sim x\}$$
You can check equivalence classes are:
$(1)$ Non-empty: For each $x$, $x\in [[x]]$.
$(2)$ Either disjoint or equal: If there is one element $p$ in $[[x]]\cap[[y]]$ then we have that $p\sim y$ and $p\sim x$. Reflexivity means $y\sim p$ and $p\sim x$ so transitivity means $y\sim x$. Reflexivity once more gives $y\sim x$. Thus whenever $a\in [[x]]$, $a\sim x$ and $x\sim y$, whence $a\sim y$. Similarily whenever $b\in [[y]]$, $b\sim y$ and $y\sim x$, whence $b\sim x$. This means $[[x]]=[[y]]$.
This means that they are a partition of $X$. So any partition induces an equivalence relation, and conversely.
As to the first thing you say: if by unique you mean $x\sim y\iff x=y$, then what you have is the equivalence relation of equality which simply induces the partition of $X$ into singletons of its elements.
On you last question you talk about an equivalence on the partitioned subsets. I presume you mean an equivalence relation , but what do you mean by "on the partitioned subsets"?
Regarding the part on the "partitioning the real numbers into subsets of equal number of decimal places." Recall there are some rational numbers with nonterminating decimal expansion, and any irrational number will have a nonterminating decimal expansion. Thus, maybe you will have to work with rationals whose decimal expansion has the same amount of decimal numbers, and say talk about "number of decimal places" with nonterminating rationals acording to the period of the expansion. For example, $3^{-1}=0.\hat 3$. Else, if the number is irrational, it goesto $\mathbb I$, the set of irrational numbers. |
H: Can anyone explain the intuitive meaning of 'integrating on both sides of the equation' when solving differential equations?
For solving differential equations, especially the ones of the form
$$g(x)dx = h(y)dy$$
we solve the equation by integrating on both sides to reveal the solution.
Understanding this for differentiating the equation on both sides is relatively easy. We know that we can formulate an alternative equation in terms of differentials for the original equation involved and come out with a new differential equation that holds because of the properties of the differentials.
But how does it work for integration on both sides? Am I missing any point here? I have referred to multiple books but none give a satisfactory explanation. Integrating an equation on both sides seems really wrong, if I may dare to use the word.
Please help. I'm stuck with this thing and I can only begin to understand differential equations once this is cleared from my head.
Thank you very much!
AI: The original equation was presumably
$$h(y)\frac{dy}{dx}=g(x),$$
or something equivalent to this.
You are given the mysterious rule about "splitting" $\frac{dy}{dx}$. You probably were told at one time that $\frac{dy}{dx}$ is not a fraction, and now all of a sudden we are treating it as a fraction!
So let us not split it. Suppose that $H(y)$ is an antiderivative of $h(y)$, that is, a function whose derivative wiith respect to $y$ is $h(y)$. Let $G(x)$ be a function whose derivative with respect to $x$ is $g(x)$.
We recognize $h(y)\frac{dy}{dx}$ as the derivative with respect to $x$ of $H(y)$ (Chain Rule). So our equation can be written as
$$\frac{d}{dx} H(y)=\frac{d}{dx}G(x).$$
Thus $H(y)$ and $G(x)$ have the same derivative with respect to $x$. So they differ by a constant, and we find
$$H(y)=G(x)+C.$$
Now the important part: this is exactly what we get when we "split" $\frac{dy}{dx}$ and integrate on both sides. So whether or not the splitting and integrating makes sense, it gives the right answer.
If you wish, splitting and integrating can be treated as a senseless mnemonic that works, a "shortcut" to the real calculation using the Chain Rule. In fact, the individual terms $dy$ and $dx$ can be given meaning, but it is a little complicated. And Applied (and less Applied) people have an essentially correct intuition based on adding up "infinitely small" quantities. Unfortunately, it takes considerable effort to make that intuition rigorous. |
H: A non-square matrix with orthonormal columns
I know these 2 statements to be true:
1) An $n$ x $n$ matrix U has orthonormal columns iff. $U^TU=I=UU^T$.
2) An $m$ x $n$ matrix U has orthonormal columns iff. $U^TU=I$.
But can (2) be generalised to become "An $m$ x $n$ matrix U has orthonormal columns iff. $U^TU=I=UU^T$" ? Why or why not?
Thanks!
AI: The $(i,j)$ entry of $U^T U$ is the dot product of the $i$'th and $j$'th columns of $U$, so
the matrix has orthonormal columns if and only if $U^T U = I$ (the $n \times n$ identity matrix, that is). If $U$ is $m \times n$, this requires $m \ge n$, because the rank of $U^T U$ is at most $\min(m,n)$. On the other hand, $U U^T$ is $m \times m$, and this again has rank at most $\min(m,n)$, so if $m > n$ it can't be the $m \times m$ identity matrix. |
H: Random walk on finite graph
I know that the stationary distribution of a random walk on the graph is given by,
(degree of the node)/($2\times$ total number of links in graph). My question is, how do we get this solution?
AI: I you start out in a distribution where the probability of being at a given node is proportional to the degree of the node, and you take one step (choosing at random a link from the current node), each link has equal probability of being the one taken, and moreover with equal probabilities in both directions. So the probability of the result of that step taking you to a given node is proportional to the number of links going to that node, i.e. the degree of the node. Thus this distribution is stationary. |
H: Trouble with representing power series as polynomials.
I am a math student trying to wrap my head around complex analysis through self-study. I am using Complex Analysis by Serge Lang, but I find myself struggling with some of his power series manipulations and his representations of power series as polynomials. For instance, given the following theorem/proof (taken from Theorem 6.1, p.76):
Theorem
Let $f(T) = a_1 T + higher$ $terms$ be a formal power series with $a_1 \not= 0$. Then there exists a unique power series $g(T)$ such that $f(g(T)) = T$. This power series also satisfies $g(f(T)) = T$.
Proof:
For convenience of notation we write $f(T)$ in the form
$$f(T) = a_1T - \sum_{2}^{\infty} a_nT^n$$
We seek a power series
$$g(T) = \sum_{1}^{\infty} b_nT^n$$
such that
$$f(g(T)) = T$$
The solution to this problem is given by solving the equation in terms of the coefficients of the power series
$$a_1g(T) - a_2g(T)^2 - ... = T$$
These equations are of the form
$$a_1b_n - P_n(a_2, ..., a_n, b_1,..., b_{n - 1}) = 0 \quad \text{and} \quad a_1b_1 = 1 \quad \text{for} \quad n = 1$$
where $P_n$ is a polynomial with positive integer coefficients (generalized binomial coefficients)
...
I can follow this all right up until the polynomial representation is used. What I would like to do is follow this type of argument with greater ease. What readings and/or exercises should I do accomplish this?
AI: It's probably clearest if you look at the first few terms explicitly.
$$\eqalign{f(g(T)) &= a_1 g(T) - a_2 g(T)^2 - a_3 g(T)^3 - a_4 g(T)^4 + \ldots\cr
&= a_1 (b_1 T + b_2 T^2 + b_3 T^3 + b_4 T^4 + \ldots) - a_2 (b_1 T + b_2 T^2 + b_3 T_3 + \ldots)^2 \cr& \ \ \ \ - a_3 (b_1 T + b_2 T^2 + \ldots)^3 - a_4 (b_1 T + \ldots)^4 + \ldots\cr
&= a_1 b_1 T + (a_1 b_2 - a_2 b_1^2) T^2 + (a_1 b_3 - 2 a_2 b_1 b_2 - a_3 b_1^3) T^3\cr& \ \ \ \ +
(a_1 b_4 - a_2 (b_2^2 + 2 b_1 b_3) - 3 a_3 b_1^2 b_2 - a_4 b_1^4) T^4 + \ldots\cr}$$
Do you see the pattern? For the coefficient of $T^m$, $a_1 g(T)$ contributes $a_1 b_m $ and $-a_j g(T)^j$ contributes $-a_j$ times a sum of products of $j$ of the $b_k$ with indices adding up to $m$, for $2 \le j \le m$. |
H: A question on calculating probabilities for the random walk
I am currently working on a high school project revolving around the 'Cliff Hanger Problem' taken from ”Fifty Challenging Problems in Probability with Solutions” by Frederick Mosteller.
The problem is 'From where he stands, one step toward the cliff would send the drunken man over the bridge. He takes random steps, either toward or away from the cliff. At any step his probability of taking a step away is $\frac{2}{3}$ of a step toward the cliff $\frac{1}{3}$
. What is his chance of escaping the cliff?'
Although the book provided a solution for the eventual probability of falling via the use of recursive equations, I decided to see if I can derive an expression to compute the probability of one falling down prior step N.
What I did was to calculate the number of paths that one can take such that he reaches $X_n=1$, where n=2m-1 for some m (this is because at even steps the person would not fall so I'm only considering the case with odd steps) without reaching $X_j=1$ for any $0\le j< 2m-1$.
The following expression is what I have calculated.
$P_n=1-\sum\limits_{i=0}^{m} p_{2i+1}\\
=1-\sum\limits_{i=0}^{m} \left({{2i+1}\choose{i+1}} - \sum\limits_{k=1}^{i}{{2k}\choose{k}}\right)\left(\left(\frac{1}{3}\right)^{i+1}\left(\frac{2}{3}\right)^{i}\right)$
(I actually made a mistake here as I considered 2m+1 rather than 2m-1, which left the summation sign with $\sum\limits_{k=1}^{i}{{2k}\choose{k}}$ undefined when i=0)
Where $p_2i+1=$ the probability of the path touching 1 at $n=2i+1$ without touching $1$ prior to the step.
The first binomial expression corresponds to choosing n+1 steps towards the cliff out of the 2n+1 steps. The second binomial expression is to subtract the paths that stemmed from previous 1's (in order to ensure that the path did not touch 1 prior to $n=2i+1$.)
However as I plotted this into excel I found that the probability does not converge to $\frac{1}{2}$ as n->infinity, which is the answer the book obtained through recursive relations.
I reviewed my argument but I don't know what did I do wrong (whether I've overcounted or undercounted).
Can anyone help?
AI: I'll try a slightly different approach -- for each $n>0$, find the probability $P_n$ that he falls off exactly on step $n$. As you noticed, $P_{2k}=0$ since an even number of steps can only end in him having moved an even number of spots away from his starting position; hence he either is safe, or fell off a turn earlier.
A useful observation here is that if you consider moves to the right as open parentheses, and moves to the left as closed parentheses, and arrange them in order as he takes the respective move, then a sequence of moves which end in him falling on turn $n$ must look like a 'proper' (or correctly matched) arrangement of parentheses, followed by one extra close parens. That is, he must return to his original location eventually without ever having stepped past it, and then finally stepping past it (verify this for yourself if it isn't clear). So, for $n=2k+1$, the number of ways he can fall on turn $n$ is $C_k$, the $k^\text{th}$ Catalan number, which counts the number of ways to correctly match $k$ pairs of parentheses. Each of these paths to fall have equal likelihood, $ \left(\dfrac{1}{3}\right)^{k+1} \left(\dfrac{2}{3}\right)^k $ . Further, as we have that $C_k = \frac{1}{k+1}{2k \choose k}$ , then
$$
P_{2k+1} = \frac{1}{k+1}{2k \choose k} \frac{2^k}{3^{2k+1}} ~~.
$$
So, the probability that after turn $n=2k+1$ , the man has not yet fallen, is
$$
1 - \sum_{\ell=0}^k P_{2\ell+1} ~~.
$$
W|A can't seem to find a closed-form for the summation, but we just want to proceed straight to the limit, and see what the probability $P$ is that he survives indefinitely. So, enlisting this or using the generating function as detailed by Brian's comment,
$$
P = 1 - \sum_{\ell=0}^\infty P_{2\ell+1} = 1 - \frac{1}{2} = \frac{1}{2} ~~.
$$
It is natural that he should survive about half of the time; (I believe that) there is a bijection between those infinite strings of (properly placed) parentheses for which no finite cutoff is balanced, and those which at some point are. |
H: Complex function limit
I asked time ago about the limit of a complex number $z$ over its conjugate, as $z$ goes to infinity.
Now I have a strategy, it is to convert $z$ to trigonometric form, and the limit depends uniquely on the norm of $z$, which can be eliminated from the fraction, so you got a fraction $\cos\theta+i\sin\theta$ over $\cos\theta-i\sin\theta$ as the norm of $z$ goes to infinity.
Since the function doesn't depend on that, does the limit exist?
Thank you!
AI: Let's be explicit: consider $z=x+iy$ . Then we have
$$
\frac{z}{\bar z} = \frac{x+iy}{x-iy} = \frac{x^2-y^2+2ixy}{x^2+y^2} = 1 + \frac{-2y^2+2ixy}{x^2+y^2} = -1 + \frac{2x^2+2ixy}{x^2+y^2} ~~,
$$
the latter two appearing by adding and subtracting in one case $x^2$, in the other $y^2$ from the top of the fraction.
When we say $z\to\infty$, as you mentioned this means that the norm of $z$ must grow without bound, and so $x^2+y^2 \to \infty$. However, there are many, many ways for this quantity to grow to infinity; for instance, take the simple cases of traveling to the right along the $x$-axis (so $y=0$, $x\to\infty$) and traveling up the $y$-axis (so $x=0$, $y \to\infty$) . Clearly both satisfy $x^2+y^2 \to \infty$ .
But notice that
$$
\lim_{y=0,x\to\infty} 1 + \frac{-2y^2+2ixy}{x^2+y^2} = 1 ~~,
$$
while
$$
\lim_{x=0,y\to\infty} -1 + \frac{2x^2+2ixy}{x^2+y^2} = -1 ~~.
$$
Hence, the limit cannot exist. The axes are just the simplest choices -- taking different paths to infinity can give you different limits. |
H: How do you prove the trichotomy law for cardinal numbers?
Law of trichotomy is that for any two cardinals $a$ and $b$, exactly one of the
conditions $a<b$, $a=b$, or $a>b$ holds.
AI: First let us prove this only for $\aleph$ numbers.
Suppose that $\aleph_\alpha$ and $\aleph_\beta$ are two cardinals. We take $\omega_\alpha$ and $\omega_\beta$ to be the corresponding initial ordinals (i.e. transitive sets which are well-ordered by $\in$, and are not in bijection with any of their members).
Since every two ordinals are comparable in $\subseteq$, either $\omega_\alpha\subseteq\omega_\beta$, and then $\aleph_\alpha\leq\aleph_\beta$; either $\omega_\beta\subseteq\omega_\alpha$ and then $\aleph_\beta\leq\aleph_\alpha$.
If both things occur then $\omega_\alpha=\omega_\beta$ by extensionality and therefore $\aleph_\alpha=\aleph_\beta$.
Now comes the point where we use the axiom of choice (or rather the well-ordering theorem), given two infinite sets they can be put in bijection with an ordinal and therefore with an initial ordinal. The above shows that the trichotomy holds.
See that as a direct and nice corollary we have the Cantor-Bernstein theorem. Take two [infinite] sets, use bijections to replace them momentarily with the corresponding initial ordinals, and you will have that the initial ordinals are equal and the bijections you used can form the bijection between your two original sets.
(This, to my knowledge, was Cantor's original proof of the theorem) |
H: An Interesting Question about Pythagorean Triples
I have recently thought about a interesting question about Pythagorean Triples.
Consider such a right-angled trapezium formed by 3 right-angled triangle. Determine does it exist integral solutions for lengths of sides $AB, BC, CD, DE, EA, AC$ and $CE$.
Here's my ideas.
I know that Pythagorean Triple can be generated by substituting integer into $x^2-y^2, 2xy, x^2+y^2.$
So let $AB=m^2-n^2$, $BC=2mn$, $CD=2pq$, $DE=p^2-q^2$, $AC=m^2+n^2=u^2-v^2$, $CE=p^2+q^2=2uv$ and $ AE=u^2+v^2.$ To answer the question, I have to show that if $m^2+n^2=u^2-v^2$ and $p^2+q^2=2uv$ have integral solution(*). But I don't know how to show this.
Can anyone tell me if I'm right about (*)? If I'm right, how to show it?
If I'm wrong, how to solve the question?
Thank you.
Sorry, I'm a poor question-tagger.
AI: Note that triangles ABC and CDE must be similar if BCD is to be a straight line, so both must be built from integer multiples of the same primitive pythagorean triple T= {a,b,c} with hypotenuse c.
Suppose ABC is $pT$ and CDE is $qT$ then $AC=pc$ and $CE=qc$ and $AE^2=p^2c^2+q^2c^2$ so that $AE$ is divisible by $c$ and equals $rc$.
You can build your trapezium from any pair of pythagorean triples (not necessarily primitive)
T={$a,b,c$} and U= {$p,q,r$}
with ABC = {$pa,pb,pc$}, CDE = {$qb,qa,qc$}, ACE = {$pc,qc,rc$}.
And this is essentially the only way of doing it. |
H: Help in understanding integration by changing the variable
I need help better understanding how, and why integration by changing the variable works (I've seen it's related to the derivative of a composite function $f(g(x))$), and generally tips and tricks, an intuitive understanding and exercises.
I sort of understand it, but when it comes to evaluate such an integral, I'm very sluggish and it seems harder than it should be.
LE: My mistake, I originally said integration by parts, but I meant variable change.
AI: Hint: How about you take a look here http://en.wikipedia.org/wiki/Integration_by_parts. |
H: Interpret overall fitness in prediction
I came through this equation and would like to learn more about how to inteprete it.
d is defined as actual value, and $\hat{d}$ is defined as predicted value.
Why does this equation
(a) divides sum of least square error to the sum of $d_{i,j}^2$ ?
(b) takes square root of (a) ?
May I know what's the name of this equation, that I can read further...
$$ \sqrt{\frac{\sum_{i,j} (d_{i,j} - \hat{d}_{i,j})^2}{\sum_{i,j} d_{i,j}^2}} $$
AI: a) it is a normalization factor. Assume that you have perfect prediction. Then you will get $0$. Say your predictor is restricted to a closed interval $[0,1]$ and your estimates are all zeros and actual values are all ones. Then you will get $1$.
b)square root is to make it as a distance in $N$ dimensional Euclidean space, such as $z=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+...}$ |
H: Switching between $h$ and $-h$ in the definition of derivative of a real function.
Suppose that $f$ is a real function and that $f'(a)$ exists:
$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. $$
By replacing all instances of $h$ with $-h$, I can get an equivalent definition
$$ f'(a) = \lim_{-h \to 0} \frac{f(a-h) - f(a)}{-h}. $$
Question:
Is it right to say that I can replace $\lim_{-h \to 0}$ with $\lim_{h \to 0}$ in the second definition because $-h$ tending to $0$ is the same as saying $h$ tends to $0$ from the left/bottom?
Since $f'(a)$ exists, the left limit must be equal to the right limit and hence the switch is valid?
AI: Neither statement involves a one-sided limit. In both statements $h$, whether as $h$ or as $-h$, can be either positive or negative.
Both statements are shorthand. The first says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a+h)-f(a)}h-f'(a)\right|<\epsilon\tag{1}$$ whenever $0<|h|<\delta$. The second says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a-h)-f(a)}{-h}-f'(a)\right|<\epsilon\tag{2}$$ whenever $0<|-h|<\delta$. Now $|-h|=|h|$, so in both cases you're simply saying that a certain inequality involving $f'(a)$ is true whenever $-\delta<h<0$ or $0<h<\delta$. There is no implication that $h$ is positive or that $-h$ is negative. The question is whether saying that $(1)$ is true whenever $-\delta<h<0$ or $0<h<\delta$ is the same thing as saying that $(2)$ is true whenever $-\delta<h<0$ or $0<h<\delta$.
The answer is yes. Let $$q_1(h)=\frac{f(a+h)-f(a)}h$$ and $$q_2(h)=\frac{f(a-h)-f(a)}{-h}\;.$$
A little algebra shows that $q_1(-h)=q_2(h)$ for all non-zero $h$. As $h$ takes on all non-zero values between $-\delta$ and $\delta$, so does $-h$; it just does so in reverse order, so to speak. Thus, saying (as in $(1)$) that $|q_1(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$ is exactly the same as saying that $|q_1(-h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$, which is the same as saying (as in $(2)$) that $|q_2(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$. |
H: Probability of Bridge hand having at most two suits
I am reviewing MIT 18.440 slides. I am wondering why the question that is worded as How many bridge hands have at most two suits represented? has a given answer of ${4\choose 2}{26 \choose 13} - 8$.
I understand that the ${4\choose 2}$ takes in account choosing the two suits that the hand will have, and ${26 \choose 13}$ indicates we're choosing 13 cards out of a limited subset of the 52 total cards.
But I'm not sure where the extra $-8$ comes from. Does it include special cases such as the 4 hands that have all 13 cards from one suit?
Possible question to use as reference: A bridge hand void in one suit
AI: The formula
$$
{4\choose 2}{26\choose 13}
$$
includes the hand with, say 13 diamonds, thrice. Once as an extremal case of hands with clubs and diamonds only, once among the hands with diamonds and hearts only, ditto with diamonds and spades. But that hand should be counted exactly once, so we need to subtract two from the answer. Of course, the same reasoning applies to all single suit hands, so altogether we need to subtract 8. |
H: Evaluating partial derivatives to apply finite differences
I need to solve the following partial differential equation (Cahn-Hilliard) using finite differences:
$$\frac{\partial c}{\partial t} = \nabla^2h + \cdots$$
where $h = c(1-c)(1-2c)$.
The question I want to ask is, which of
$$\nabla^2 h = \frac{h(x+h) + h(x-h) -2h(x)}{\Delta x^2}$$
or
$$\nabla^2 h = \frac{h(x+h) + h(x-h) -2h(x)}{\Delta x^2} \frac{\partial^2h}{\partial c^2},$$
is correct? (Although the second one seems dimensionally wrong.)
Also, is $\nabla^2 c^n = n(n-1)\nabla c^{n-2}$ correct?
The question is simple, but I am not able to find the answer. Any help will be appreciated.
AI: The first option:
$$\nabla^2 h = \frac{h(x+h) + h(x-h) -2h(x)}{\Delta x^2} = \frac{h(x+h) + h(x-h) -2h(x)}{\Delta c^2} \frac{\partial^2c}{\partial x^2},$$
Regarding the power law:
$$\nabla^2 c^n = n(n+1)c^{n-2}$$
For a proof, see here. |
H: Given the width and height of an ellipse find n number of points around the ellipse?
I am looking to find $n$ number of points around an ellipse.
They don't necessarily have to be equidistant.
Similar to what this forum is asking:
I found several answers that are similar but I am having a hard time expressing it in code.
AI: You can take the points $P_k=(x_k,y_k)$, with
$$
x_k=a\cos(2k\pi/n)\\
y_k=b\sin(2k\pi/n)
$$
for $k=0,\ldots,n-1$, and where $a,b$ are the semi-wight and the semi-heigth, respectively. |
H: Is there any example of a discontinuous one real variable integral.
I know that an integral from $0$ to $x$ of a function of $x$ is itself a function, so I want to know if is there any condition of it to be discontinuous.
I haven't restrict $f(x)$ to be continuous or not, so I think that one needs to take both options.
So if the integrand is continuous then the integral is too.
AI: No, at least not with the usual calculus notion of integral (with Lebesgue measure in the background) and $f$ being bounded. Let $(x_n)$ be a nonnegative sequence converging to $x$ and $f$ be a bounded function such that the integral $I_x=\int_0^xf(y)~dy$ is finite and well defined for all $x\geq 0$.
Suppose $I_{x_n}$ does not converge to $I_x$. Then there is an $\epsilon>0$ such that for all $\delta>0$, there is $x'$ such that $|I_x-I_{x'}|>\epsilon$ and $|x-x'|<\delta$. Let $B>0$ be such that $-B< f(x)<B$ for all $x\geq 0$. Then the integral $$\bigg|\int_{x-\delta}^{x+\delta}f(y)~dx\bigg|\leq 2\delta B$$ for all $\delta>0$. In particular for, $\delta<\epsilon/(B2)$, the integral can not vary that much and we get a contradiction.
A continuous function on a closed and bounded interval is of course already bounded. |
H: Proving that $2^{2^n} + 5$ is always composite by working modulo $3$
By working modulo 3, prove that $2^{2^n} + 5$ is always composite for every positive integer n.
No need for a formal proof by induction, just the basic idea will be great.
AI: Obviously $2^2 \equiv 1 \pmod 3$.
If you take the above congruence to the power of $k$ you get
$$(2^2)^k=2^{2k} \equiv 1^k=1 \pmod 3$$
which means that $2$ raised to any even power is congruent to $1$ modulo $3$.
What can you say about $2^{2k}+5$ then modulo 3?
It is good to keep in mind that you can take powers of congruences, multiply them and add them together.
If you have finished the above, you have shown that $3\mid 2^{2k}+5$. Does this imply that $2^{2k}+5$ is composite? |
H: If $x \geq \min\{y,w\}$ then $x \leq y+w$?
If $x \geq \min\{y,w\}$ then $x \leq y+w$.
This is kind intuitive,very trivial. Or not? It's like triangular inequality, isn't it?
AI: This is wrong. Take $y,w=1$ and $x=3$. |
H: Find all integer solutions to $7595x + 1023y=124$
Find all integer solutions to $7595x + 1023y=124$
Using the Euclidean algorithm I have found the $\gcd(7595,1023)=31$ and found the Bezout identity $31=52\cdot1023-7\cdot7595$ but I'm not really sure how to go about finding all solutions to that equation.
I believe you can divide the equation through by the $\gcd$ - which gives $245x+33y=4$ - but I'm not sure what to do next.
AI: From your remark $31 = 52\times 1023-7\times7595$, you get $124=4\times 31 = 7595\times(-4\times7)+1023\times(52\times 4)$, so you have a first solution $x_0=-28$ and $y_0=208$.
So you have $7595(x-x_0)+1023(y-y_0)=0$, or $245(x-x_0) = - 33(y-y_0)$ (*).
Now $245$ and $33$ are coprime, so for this equality to hold, you need to have:
$$
x-x_0=33\lambda\\
y-y_0=245\mu
$$
When you replace this into (*), you get $\lambda = -\mu$, so at the end the solution is given by:
$$
x=33\lambda+x_0\\
y=-245\lambda+y_0
$$
Where $\lambda\in\mathbb{Z}$. |
H: An infinite set having "one more element" than another infinite set
A classic example of homeomorphism is between a sphere missing one point and a plane
To see this, place a sphere on the plane so that the sphere is tangent to the plane. Given any point in the plane, construct a line through that point from the "north" pole (assuming the point of tangency is the south pole); this necessarily intersects the sphere in a unique point other than the north pole. Similarly, given any point on the sphere other than the north pole, the line through the north pole and that point will intersect the plane at a unique point. Thus, the homeomorphism is established.
Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable?
AI: Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable?
It can certainly be given meaning. For example, given any inclusion of subsets $A \subseteq B$, we can define the phrase "$B$ has $\alpha$ more elements than $A$" to mean that $|B \setminus A| = \alpha$.
More generally, given any injection $i: A \to B$, we can define "$B$ has $\alpha$ more elements than the subset defined by $i$" or "$B$ has $\alpha$ more elements than $A$ relative to the inclusion $i$" to mean $|B \setminus i(A)| = \alpha$.
These notions can even be useful. For example, the notion of a cofinite subset often comes up: a subset $S$ of a set $X$ such that $|X \setminus S| < \aleph_0$. |
H: How to simplify the following basic equation
I just wanna know if you can find a way to simplify this following equation :
originalState + ( ( animatedState * ( 100 - ( finishPos - x ) * 100 / (finishPos - startPos) ) / 100 )
Thanks
AI: Being lazy, I'll replace each variable by its first letter, and I get this (with its unbalanced parens):
$o + ( ( a * ( 100 - ( f - x ) * 100 / (f - s) ) / 100 )$.
The "100"s seems to cancel out, and. balancing the parens, I get
$o + a * ( 1 - \frac{ f - x}{f - s} )$.
Simplifying the expression inside the parens I get this:
$o + a * ( \frac{ x-s}{f - s} )$. |
H: Hahn-Banach theorem
Hope this won't turn out to be a stupid question. If I want to apply the Hahn-Banach lemma to prove the Hahn-Banach theorem, then I want to define a function $p$ that provides an upper bound on my functional defined on a linear subspace $Y \subset X$. Say, I want to extend $f \in Y^\ast$. Then the notes define $p(x) = \|f\| \|x\|$. I don't quite believe this works since if $\|y\| \lt 1$ we might not have $f(y) \leq p(y) = \|f\| \|y\| = \sup_{\|x\|=1} |f(x)| \|y\|$. What am I missing? Thanks for your help.
AI: The thing you're probably missing is the fact that the left hand side depends on the norm of $y$. Maybe you will be more convinced if you write it that way (for $y\neq 0$, if $y$ is zero both sides are zero, so the inequality holds):
$$f(y)=f(y/\lVert y\rVert)\cdot\lVert y\rVert\leq \lVert f\rVert\lVert y\rVert$$
(notice that $\lVert (y/\lVert y\rVert) \rVert=1$). |
H: Why do characters on a subgroup extend to the whole group?
As background, I am trying to do exercise 3.10 in Deitmar's "Principles of Harmonic Analysis." I can do most of the problem but I'm stuck on the third part proving surjectivity.
Given a locally compact abelian group $G,$ a closed subgroup $H,$ and a character $\chi: H \rightarrow S^1,$ I need to construct an extension of $\chi$ to all of $G.$ Any extension will do, but it's not clear to me that one should be able to construct an extension. For example, we can't just define an extension to be 1 outside of $H,$ since elements outside of $H$ can multiply to something within $H.$
Can anyone help me out? Thanks!
AI: We can use the following results, from Rudin's book Fourier Analysis on Groups.
Denote
$$A(H):=\{\gamma\in\widehat G,\forall h\in H,\gamma(h)=0\}$$
the annihilator of $H$. It's a closed subgroup of $\widehat G$, which ensures us that $\widehat G/A(H)$ is Hausdorff and locally compact.
Theorem 2.1.2. $A(H)$ is isomorphically homeomorphic to the dual group of $G/H$, and $\widehat G/A(H)$ is isomorphically homeomorphic to the dual group of $H$.
Sketch of proof: Let $\pi$ the projection $\pi\colon G\to G/H$. We define an one-to-one correspondence between $A(H)$ and $\widehat{\left(\frac GH\right)}$ by
$$\gamma(x):=\phi(\pi(x)).$$
Furthermore, it's compatible with the group structure.
By Pontryagin duality theorem, and using the fact that for $x_0\notin H$ we can find $\gamma\in \widehat G$ such that $\gamma(x_0)\neq 1$, we deduce the second assertion. To show it's an homeomorphism, we can use the fact that projection is continuous and open. |
H: Help in evaluating $\sum\limits_{k=1}^\infty \frac1{k}\sin (\frac{a}{k})$
I would like to try to evaluate
$$\sum\limits_{k=1}^\infty \frac{\sin (\frac{a}{k})}{k}$$
However, all of my attempts have been fruitless. Even Wolfram Alpha cannot evaluate this sum. Can someone help me evaluate this interesting sum?
AI: This is the Hardy-Littlewood function (see this and this as well), which is known to be very slowly convergent. Walter Gautschi, in this article, shows that
$$\sum_{k=1}^\infty \frac1{k}\sin\frac{x}{k}=\int_0^\infty \frac{\operatorname{bei}(2\sqrt{xu})}{\exp\,u-1}\mathrm du$$
where $\operatorname{bei}(x)=\operatorname{bei}_0(x)$ is a Kelvin function, through Laplace transform techniques (i.e., $\mathcal{L} \{\operatorname{bei}(2\sqrt{xt})\}=\sin(x/s)/s$), and gives a few methods for efficiently evaluating the integral.
Here's a plot of the Hardy-Littlewood function: |
H: Optimization with non-negativity and norm constraint
I am facing the following optimization problem:
$$\min_x w^tx \\
s.t. ||x|| = 1, \forall i: x_i \geq 0
$$
where $w$ and $x$ are real valued vectors. How would I solve this?
My background is not optimization (only some experience with equality constraints via Lagrange multipliers).
AI: Assume $\mathbf{w}\neq\mathbf{0}$. We have
$$
<\mathbf{w},\mathbf{x}>=||\mathbf{w}||\, ||\mathbf{x}|| \cos(\varphi)=||\mathbf{w}|| \cos(\varphi),
$$
where $0\leq\varphi\leq\pi$ the angle between $\mathbf{w}$ and $\mathbf{x}$.
If $w_i\leq 0$ for all $i$ we obtain the minimal value in the only case $\varphi=\pi$ and $\mathbf{x}=\frac{\mathbf{w}}{||\mathbf{w}||}$.
If there are $i_0, j_0$ such that $w_{i_0}>0$ and $w_{j_0}<0$ then choose $\mathbf{y}$ such that $y_{i_0}:=0$ and $y_{j_0}:=-w_{j_0}$. Then $\mathbf{x}=\frac{\mathbf{y}}{||\mathbf{y}||}$.
If $w_i\geq 0$ for all $i$ then project $\mathbf{w}$ to the coordinate-axes. We obtain the vectors $\mathbf{w}_1,\ldots,\mathbf{w}_n$. The angles between $\mathbf{w}$ and $\mathbf{w}_i$ are $\varphi_i$. Choose $i_0$ such that the angle would be maximal (it is not necessarily unique). Then $\mathbf{x}=\frac{\mathbf{w}_{i_0}}{||\mathbf{w}_{i_0}||}$. |
H: How to find the ratio for a combination of two chemical solutions
Lets say you have a chemical solution $A$ that has a ratio of $4:1$ of water to some chemical substance, and a solution $B$ that has a ratio of $9:1$ of water to some other chemical substance.
If you combine the same amount of each solution, the resulting solution is not in a ratio of $6.5 : 1$ (the average), but is instead $5.6666 : 1$.
Can someone help me understand why one cant just average the two ratios if the same amount of solution is added together? For some reason its just not clicking.
AI: Take 10x litre of each solution, so we are adding 8x & 9x litre of water (not 8x litre each), 2x & x litre of that chemical(not 2x or x litre each).
So, total water = (8x + 9x) litre = 17x litre and total chemical=(2x + x) litre = 3x litre.
So, the resultant ratio is $\frac{17}{3}=5.666...$
Clearly, when we are adding the same amount of the solutions, the same amount of water (or the chemical) are not added.
Trivially, we could take average, only if we had added same amount of water (or the chemical) separately.
Let us assume that the ratio of water and the chemical be a:b in the first solution and c:d in the second.
Let's take (a+b)x litre of the 1st solution and (c+d)y litre of the second.
So, water taken = (ax + cy) litre and the chemical taken = (bx + dy) litre.
If $\frac{ax+cy}{bx+dy}=\frac{\frac{a+c}{2}}{\frac{b+d}{2}}$
$=>\frac{ax+cy}{bx+dy}=\frac{a+c}{b+d}=>(bc-ad)(x-y)=0$
So if bc=ad=>c:d=a:b (i.e., if the ratios of water & the chemical are same), the ratio of water & the chemical will be the ratio of their average for any values of x,y i.e., we can mix the solutions in any ratio.
Here is this question $a=4,b=1,c=9,d=1=>bc-ad=1.9-4.1≠0=>x=y$
So, we need to take (4+1)x and (9+1)x litres of the solutions, respectively to make the ratio of water and the chemical as the ratio of their average. |
H: How would I express this code in math notation?
I'm writing a paper on an application I wrote, and I'm trying to describe mathematically what I did to determine the rotation of an object. What I did is really simple, but my math lags far behind my computer science.
What it does is select a random number between min and max. 0 <= min, max < 1 If min > max, in which case it selects a random number between min and max and subtracts 1 if the number is greater than 1. (i.e. the number under (mod 1))
This is what I'm saying right now:
When determining an orientation, the orientation is selected from the
valid orientations, then a random number x is selected such that
Where min is the minimum offset, max is the maximum offset, and
0≤min,max<1. The radian measure of the orientation would be 2πx, and
the degree measure would be 360x°.
This is the code in question:
if (min > max)
{
double d = Utility.r.NextDouble() * (max + 1 - min) + min;
if (d >= 1)
d -= 1;
return d;
}
else
{
double d = Utility.r.NextDouble() * (max - min) + min;
return d;
}
AI: I think this is much easier to state in words than in mathematical notation. "We are given two real numbers min and max between 0 and 1. The function we coded selects a random real number between 0 and 1 as follows: if min < max, then we select a real number uniformly from the interval [min, max); if min > max, we select a real number uniformly from the interval [max, min+1), and then reduce the resulting number modulo 1." |
H: Best Notation for Introducing Finite Numbers of Elements
What is the best way to quantify over a finite number of elements of a set? For instance, suppose that one wished to quantify over $n$ real numbers $a_1,\ldots,a_n$ in order to state a property of their sum. In such a case, I often see authors quantify over the upper bound of the sum. For instance:
$$ \phi_1 \equiv \forall n \in \mathbb{N}~.~\sum_{i = 1}^n \ln a_i = \ln \left ( \prod_{i = 1}^n a_i \right ) $$
If an author wishes to be explicit, he or she will sometimes quantify over the $a_i$ using ellipses:
$$ \phi_2 \equiv \forall n \in \mathbb{N}\forall a_1,\ldots,a_n \in \mathbb{R}~.~\sum_{i = 1}^n \ln a_i = \ln \left ( \prod_{i = 1}^n a_i \right ) $$
Sometimes, I see authors use index notation with a suitable data type. For instance, if $A^*_n$ denotes the set of multisets of size $n$ of elements of $A$, then one could introduce elements more precisely. Otherwise, one could sum or multiply over the elements of the multiset in the following way:
$$ \phi_3 \equiv \forall n \in \mathbb{N}\forall \alpha \in \mathbb{R}^*_n~.~\sum_{a \in \alpha} \ln a = \ln \left ( \prod_{a \in \alpha} a \right ) $$
Of course, a multiset is usually defined in terms of a family of elements. If we are given a function indexing the elements of a set, then we can write:
$$ \phi_4 \equiv \forall n \in \mathbb{N}\forall f : n + 1 \rightarrow \mathbb{R}~.~\sum_{i = 1}^n \ln f(i) = \ln \left ( \prod_{i = 1}^n f(i) \right ) $$
If none of the $\phi_i$ contain ideal notations for introducing a finite number of elements, what is the most ideal notation? For brevity, I would usually opt for the first, but I feel that it's incomplete. I hope this isn't a completely trivial question.
AI: This is nothing to fret about. There are many different ways to say the same thing.
Use what you consider to be the clearest, and always be considerate of your intended audience. For advanced audiences, you are not going to need to be 100% rigorous about indexing conventions. Even for weaker audiences you might want to avoid overloading them with notation.
You will definitely need exact notation if the validity of the result depends on the clarity, but otherwise you can use your discretion to pick the "best" notation for the application. |
H: Create a new image from an image set's patterns
So, I'm interested in learning how to detect patterns in a set of images and then use those patterns to create a new image of similar style. For example, say there is a group of 20~ish images (hopefully a large enough sample) with some interesting characteristics and one wished to produce a new image of a set size with such characteristics for an application's background. The catch is that the background should be different each time it is loaded, so it must be generated - not produced in an image editing software.
Now, once I've the base concepts down, I should be able to write something genetic to spit out images until I like the results, but I'm not sure how to start researching this one. I can't really use fractal algorithms because I want to avoid the repeating behavior. So, which metrics do you think I should obtain, and do you know of any good resources for using said metrics to create new images that incorporate the patterns/tendencies of an image set?
AI: You can start with reading and implementing these two papers: "Texture Synthesis by Non-parametric Sampling"
Alexei A. Efros and Thomas K. Leung
IEEE International Conference on Computer Vision (ICCV'99), Corfu, Greece, September 1999, and J Portilla and E P Simoncelli. A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients. Int'l Journal of Computer Vision. 40(1):49-71, October, 2000. |
H: Proving that $\mu$ is $\sup S$
Suppose a set $S$ of real numbers is bounded and let $\mu$ be an upper bound for $S$. Show that $\mu$ is the least upper bound of $S$ $\Longleftrightarrow$ for every $\epsilon > 0$ there is an element of $S$ in the interval $[\mu - \epsilon, \mu]$.
My Work
($\Rightarrow$)
If there is no element of $S$ in the interval $[\mu - \epsilon, \mu]$, then $\mu - \epsilon$ could also be an upper bound for $S$, but since $\mu = \sup S$ and $\mu - \epsilon < \mu$ there is a contradiction.
($\Leftarrow$)
Considering the least upper bound of $S$, I need to show that it cannot be smaller than $\mu$. But I am not sure how to do this using the condition I am given.
Edit
By the definition of a supremum, if $\lambda$ is another upper bound of $S$, then if $\mu = \sup S \Rightarrow \mu \le \lambda$. So proof by contradiction, assuming that $\mu \ne \sup S$, does that mean that there is an element $\lambda \in [\mu - \epsilon, \mu], \lambda \notin S, \lambda < \mu$? How does this prove that there is no element of $S$ in $[\mu - \epsilon, \mu]$ for every $\epsilon > 0$?
AI: Recall the definition of supremum
An element $\mu$ is a supremum of a set $S$ if
$(1)$ It is an upper bound. That is, for $x$ in $S$, we have $x \leq \mu$.
$(2)$ If $\eta$ is any other upper bound, $\mu \leq \eta$.
Now the theorem you might want is
THEROEM $\mu$ is supremum of $S$ if and only if for each $\epsilon >0$, there is some $x \in S$ for which $$\mu-\epsilon \color{red}{<} x \leq \mu$$
Note the $<$ and not the $\leq$. You will see why this is so.
$(\Rightarrow)$ Suppose $\mu$ is a supremum. Then clearly for any $x\in S$, $x\leq \mu$. Now argue by contradiction. Suppose there were some $\epsilon \;>0$ such that $$\mu-\epsilon \geq x$$ for each element of $A$.
This means that $x \leq \mu-\epsilon$ for each $x$. But that would mean $\mu-\epsilon$ is upper bound with $\mu-\epsilon \leq \mu$ which is impossible, since $\mu$ is the supremum.
I'm not sure about the $(\Leftarrow)$. I've always seen this used in one direction. The theorem is sometimes called the approximation theorem, and it says that given a bounded nonempty subset $S$ of the reals, there is a sequence of elements of $S$, call it $s_n$ such that $s_n \to \sup S$,and another sequence, call it $u_n$, such that $u_n \to \inf S$ |
H: Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition?
This problem comes from baby Rudin, Chapter 1, Number 6(c).
For the problem, given the condition $b > 1$, one is asked to show that for rational $r$, $$b^r = \sup B(r),$$
where $B(x) = \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$.
I have demonstrated this property. However, the problem statement concludes,
Hence, it makes sense to define $$b^x = \sup B(x)$$ for every real $x$.
I am not sure I entirely follow this statement. Why does it make sense to define $b^x$ in this way?
What I have tried to do so far is the following:
For $x$ rational, the statement is given by the above analysis for $b^r$. For $x$ irrational, define $x$ by its Dedekind cut, and let $A = \left\{ s \mid s \in \Bbb Q, s < x \right\}$. Then, for any $a \in A$, $b^a = \sup B(a)$, and since for $a' < a$, $b^{a'} \in B(a)$, then $b^a > b^{a'}$ by the definition of supremum. If $b^x$ is not the supremum, then there exists some $y$ such that $b^a \le b^y <b^x$ for all $a$.
If $y$ is rational, then by the previous argument, $y = a$. But because $A$ has no greatest element, then there is some $y' > y$ that contradicts that $b^y$ is an upper bound. Thus, $y$ must be irrational. However, if $y<x$, then by density of the rationals, there is some $s' \in \Bbb Q$ such that $y < s' < x$.
However, I have not yet shown that $b^{s'} > b^y$, because the current exercise is to define irrational exponents, so I cannot legitimately make that argument.
Am I on the right track? Have I over-complicated it, or am I missing something?
Edit: I should add, this is not homework.
Second edit: $b > 1$.
AI: Probably it is easier to accept the definition of $b^x$ as the unique element of separation of the two sets
$$
B(x)=\{b^t|t\in\mathbb{Q},t\leq x\}\\
C(x)=\{b^t|t\in\mathbb{Q},t\geq x\}
$$
which implies
$$b^{t_1}\leq b^x\leq b^{t_2}\quad\forall\;t_1,t_2\in\mathbb{Q}\;|\;t_1\leq x\leq t_2\quad .$$
As a consequence you can set indifferently
$$b^x=\sup B(x)$$
or
$$b^x=\inf C(x)$$
(some assumption about $b>1$ and $t>0$ should obviously be done). |
H: Linear Transformation
Let $T:V\rightarrow V$ be a linear transformation and dimension of $V$ be infinite. Then show that either the nullity or the rank of $T$ is infinite.
The problem is that I cannot use rank nullity theorem.
Thanks for any help.
AI: Let $\,\{v_1,...\}\,$ be any (obviously infinite) basis of $\,V\,$ , then:
(1) if $\,\{Tv_1,...\}\,$ contains an infinite linearly independent subset then $\,\dim Im\,\,(T)=rank\,\,T=\infty\,$ , other wise:
(2) there only exists a finite linearly independent subset in $\,\{Tv_1,...\}\,$ ,so it must be that an infinite subset of $\,\{v_1,...\}\,$ is mapped to zero by $\,T\,$ and thus $\,\dim\ker T=null\,\,T=\infty\,$
Added: Please do read the comment below by Tunococ as it proves (2) accurately. |
H: Can lines connecting nearest neighbors ever cross?
I don't have any background in graph theory so I am sorry if this is a basic question. I want to know if lines connecting nearest neighbors can cross in a few different cases.
I have convinced myself that if there are no ties for nearest
neighbor it is impossible for lines to cross. Is this correct?
I think it might still be impossible even if ties are allowed, in which case either or both lines can be drawn but I haven't come up with solid reasoning to back that up.
This case is a bit more complicated. Say that points can be connected to their second nearest neighbor if a line originating from them has yet to be drawn and they are already connected to their nearest neighbor. In this condition, two pairs of points very far from one another would be connected into one long line. Each pair would get a line connecting it to its nearest neighbor and then one or two lines would connect between the pairs. Connecting to the third nearest neighbor is not allowed. It seems to me that in this case as well no lines would cross.
I have no idea where to begin looking for formal math to back up these statements. Is there a general theory of lines crossing in a network of nearest neighbor connections? Any help would be appreciated.
AI: If $A$ has nearest neighbor $B$, then any point $x$ on the line segment joining $A$ and $B$ is nearest to either $A$ or $B$, but not to any other point (not even tied):
Assume that there exists a point $C$, such that $d(x,C) \leq d(x,A),d(x,B)$ for some $x$ on $AB$. Then, by the triangle inequality (the inequality is strict because we know $C$ can't lie on $AB$)
\begin{align*}
d(A,C) &< d(A,x) + d(x,C) \\
&< d(A,x) + d(x,B) \\
&< d(A,B)
\end{align*}
contradicting the fact that $B$ is the nearest neighbor of $A$.
Now assume there are 4 points $A,B,C,D$, such that there's an edge between $A,B$ and between $C,D$. Then if the lines $AB$ and $CD$ would cross, the point of intersection $x$ would be strictly nearer to at least one of $A$, $B$ than to any other point, and also strictly nearer to at least one of $C,D$ than to any other point, which is absurd.
Also, for some general theory: Voronoi diagrams and Delaunay triangulations can be very helpful when thinking about such problems. The above is actually equivalent to the fact that in a Voronoi diagram, the line connecting two nearest neighbors only crosses the Voronoi cells of these two points (which must be adjacent).
EDIT: I overlooked something: If $A$ is the point nearest to $B$, then that doesn't necessarily imply that $B$ is also the nearest point to $A$. I assume you want to consider the graph formed by connecting every point to its nearest neighbor (so it can happen that multiple lines are adjacent to a certain point). The argument was easy to fix though. |
H: Complex Integration inside a Domain with a singularity
$f(z)= \dfrac{2z+3i}{z^{2}+1/4}$, $C$ is the unit circle centered at zero.
Though the answer is $12\pi$, but my answer is coming to be $4\pi i$
If we decompose it into partial fractions then $f(z)$ reduces to $\dfrac{4}{z-\frac{i}{2}} - \dfrac{2}{z+ \frac{i}{2}}$
Applying Cauchy Integral formula, the first part integrates to $8\pi i $ and the second part reduces to $4\pi i$. Thus my answer, but the actual answer is otherwise
Help appreciated,
Soham
AI: $$f(z):=\frac{2z+3i}{z^2+\frac{1}{4}}=\frac{2z+3i}{\left(z-\frac{i}{2}\right)\left(z+\frac{i}{2}\right)}\Longrightarrow$$
$$\begin{align*}Res_{z=i/2}(f)=&\lim_{z\to i/2}\left(z-\frac{i}{2}\right)f(z)=\lim_{z\to i/2}\frac{2z+3i}{z+\frac{i}{2}}=4\\Res_{z=-i/2}=&\lim_{z\to -i/2}\left(z+\frac{i}{2}\right)f(z)=\lim_{z\to -i/2}\frac{2z+3i}{z-\frac{i}{2}}=-2\end{align*}
$$
so we finally get
$$\oint_C\frac{2z+3i}{z^2+\frac{1}{4}}dz=2\pi i(4-2)=4\pi i$$ |
H: Use Chebychev's Inequality to find a lower bound.
Let $X$ be the number of heads one would obtain in $140$ flips of a fair coin.
Use Chebychev's Inequality to find a lower bound on the probability $P(60 < X < 80)$.
Okay so Chebychev's Inequality is $P(|X - E(X)| > kσ) \le 1/k^2$ for $ k > 0$, where $σ^2$ is the variance of $X$.
I'm not sure how to fill this in or anything. My probabilty test is tomorrow so help is much appreciated! Descriptive answers would be awesome.
AI: Note that $X$ is binomial with $n=140, p = 0.5$ so $\mathbb{E}[X] = np = 70$, $\sigma^2 = Var(X) = np(1-p) = 35,$ giving $\sigma = \sqrt{35} $ .
Now,
$\begin{split}
\mathbb{P}[60 < X < 80]
&= \mathbb{P}[-10 < X - \mathbb{E}[X] < 10] \\
&= \mathbb{P}[|X - \mathbb{E}[X]| < 10] \\
&= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10}{\sqrt{35}} \sigma]\\
&= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10\sigma}{\sqrt{35}}]
\end{split}
$
By Chebyshev's Inequality,
$\mathbb{P}[X \not \in (60,80)]
= \mathbb{P}[|X - \mathbb{E}[X]| > \frac{10\sigma}{\sqrt{35}}]
\leq (\frac{\sqrt{35}}{10})^2 = 35/100 = 7/20$.
Hence,
$\mathbb{P}[60 < X < 80]
= 1 - \mathbb{P}[X \not \in (60,80)]
\geq 1-7/20 = 13/20.
$ |
H: Intuition behind Snake Lemma
I've been struggling with this for some time. I can prove the Snake Lemma, but I don't really “understand” it. By that I mean if no one told me Snake Lemma existed, I would not even attempt to prove it. I believe I am missing some important intuition that tells me that the lemma “could” be true before actually trying to prove it. Please give me some pointers.
AI: A special case is easy to see: Imagine $A \leq A' \leq B=B'$, and $C=B/A$ and $C'=B'/A'$ with the obvious maps from a module $M$ to $M'$. The kernels are $0$, $0$, and $A'/A$. The cokernels are $A'/A$, $0$, $0$. The last kernel and the first cokernel are linked.
The snake lemma is then just one of the isomorphism theorems. As you deform $B$ and $B'$ more, how do the kernels and cokernels deform? The last kernel and first cokernel no longer need be isomorphic, but the kernel and cokernel of that linking (snakey) map can be described in terms of the kernels and cokernels already there. A specific relation is the snake lemma.
Example deformation
An example deformation might be helpful: distort the $A' \to B' \to C'$ sequence by quotienting out by some $M \leq B$ (imagine $M=IB$ for some ideal $I$, so we are tensoring the second line with $R/I$). How does this change the kernels and cokernels?
The first line is $$0 \to A \to B \to B/A \to 0,$$ and the second line becomes $$ 0 \to (A'+M)/M \to B'/M \to B'/(A'+M) \to 0$$ so the kernels are $A \cap M$, $M$, and $(A'+M)/A$ and the cokernels become $(A'+M)/(A+M)$, $0$, and $0$. The last kernel and the first cokernel are related, but not equal. One clearly has the relation $0 \to (A+M)/A \to (A'+M)/A \to (A'+M)/(A+M) \to 0$ where the last two nonzero terms are the last kernel and the first cokernel. The first term is weird though. We apply another isomorphism theorem to write $(A+M)/A \cong M/(A\cap M)$ and then the solution is clear: We already have $0 \to A \cap M \to M \to M/(M\cap A) \to 0$ so we splice these two together to get the snake lemma: $$ 0 \to A \cap M \to M \to (A'+M)/A \to (A'+M)/(A+M) \to 0 \to 0 \to 0$$ |
H: Fibres in algebraic geometry: multiplicity
Currently I am studying varieties over $\mathbb{C}$ and I know some scheme theory.
My professor mentioned the other day that given a morphism of varieties over an alg. closed field $k$: $f: X \rightarrow Y$, one should count fibres "with multiplicity". I have been trying to make sense of this. It is especially important for me if $Y$ is a curve and the fibres are divisors of $X$.
So I looked in Hartshorne, dusted off my scheme knowledge, and rediscovered the scheme theoretic fibre $X_y := X \times_Y \text{Spec}(k(y))$ for some $y \in Y$. I worked out an example: $f: \mathbb{A}^2 \rightarrow \mathbb{A}^2$, $(a,b) \mapsto (a^2,b)$. A calculation shows that the fibre over a point $(p,q)$ is the spectrum of $k[x,y]/(x^2-p,y-q) \cong k[x]/(x^2-p)$, which is reduced and consists of two different points if $p \neq 0$, and a nonreduced one point scheme if $p=0$.
Really similarly $g: \mathbb{A}^1 \rightarrow \mathbb{A}^1$, $a \mapsto a^2$, then the fibre is the spectrum of $k[x]/(x^2-p)$. I mention these two examples since in one case, the points and fibres are divisors, in the other they are not.
This is all no problem at all, however i am trying to translate this back into varieties, trying to make sense of what "counting with multiplicities" should mean.
Now the questions.
How does one translate this into the language of varieties? It seems obvious from the above that the fibre of $(0,q)$ should be $2(0,q)$, since this "doubling" is hidden in the structure sheaf. This would be especially important if the points were divisors, since then 2 times a point makes a lot of sense. But how to make this mathematically precise? As in, what is the actual mathematical procedure? I guess if we can factor the ideal by which we are modding out into prime ideals, we can count the prime factors (i.e. varieties) with multiplicity, BUT: is this factoring always possible? And is this the right method?
I did a course on Riemann surfaces, there multiplicity was also defined: a holomorphic map is locally at a point always of the form $z \mapsto z^n$, then $n$ is the multiplicity of the map at that point. I'm quite sure the definitions agree, of course assuming $k=\mathbb{C}$ and the varieties to be smooth. Can anyone give an argument why?
An ideal answer could include a reference to the two examples, point out differences if there are any (coming from the fact that fibres are/are not divisors) and a general method for writing down a fibre in the language of varieties. If this is only possible in the case of fibres being divisors, i'd still be really happy.
Thanks a lot!
Edit: Thanks a lot for your answers, Andrew and Froggie! There is a small question left. I should give you a motivation for my question: somewhere in Beauville there is a map $f: S \rightarrow C$ where $C$ is a curve, then the inverse image of a point turns out to be the divisor $nE$, where $E$ is a curve on $S$ and $n>1$. So this is why i expected that there at least would be a notion of "multiplicity" in the language of varieties if the fibres were divisors. But as i realize now, there is also the notion of $f^*: \text{Div}(C) \rightarrow \text{Div}(S)$, which is most probably what Beauville used (i havent checked this). So new question: does this $n$ always agree with the multiplicity that you defined Froggie? Is there any other way to relate the scheme theoretic multiplicity and the multiplicity of divisors?
AI: I'm not really an algebraic geometer, so in my answer I'll stick to the simple situation where $f\colon X\to Y$ is a finite surjective morphism between smooth irreducible varieties over $k$. Both of your examples fall into this category.
If $x\in X$ is a (not necessarily closed) point and $y = f(x)$, then the multiplicity you are probably looking for is the integer I'll denote by $m_f(x)$, which is $$m_f(x):= \dim_{\kappa(y)}\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x} = \dim_{\kappa(y)}\mathcal{O}_{X,x}\otimes_{\mathcal{O}_{Y,y}}\kappa(y),$$ where here you use $f$ to make $\mathcal{O}_{X,x}$ into a $\mathcal{O}_{Y,y}$-module.
Another way of computing this integer is the following. The space $X_y$ you defined previously is a scheme, and its generic points correspond to the preimages of $y$. If $x$ is a generic point of $X_y$, then $\mathcal{O}_{X_y,x}$ is an artinian ring, and one can take its length, which I will denote $v_f(x):= length\,\mathcal{O}_{X_y,x}$. Then one can derive $m_f(x) = v_f(x)\times [\kappa(x):\kappa(y)]$. In both of your examples, $x$ and $y$ were closed points, so $\kappa(x) = \kappa(y) = k$, and the factor $[\kappa(x):\kappa(y)]$ is $1$. Thus $m_f(x) = v_f(x)$.
Let's apply these definitions to your examples.
(1) $f(a,b) = (a^2,b)$. Let $x = (p,q)$ and $y = (p^2,q) = f(x)$. Then $\mathcal{O}_{X,x} = k[u,v]_{(u-p,v-q)}$ and $\mathfrak{m}_y\mathcal{O}_{X,x} = (u^2-p^2,v-q)\subset k[u,v]_{(u-p,v-q)}$. If $p\neq 0$, then $$\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}\cong k[u,v]_{(u-p,v-q)}/(u^2-p^2,v-q) = k,$$ which has dimension 1. If $p = 0$, then $$\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}\cong k[u,v]_{(u,v-q)}/(u^2,v-q) = k[u]/(u^2),$$ which has dimension 2. Thus $m_f(x) = 1$ if $p\neq 0$ and $m_f(x) = 2$ if $p = 0$.
To compute this multiplicity the other way, we start from your observation that $X_y$ has two components so long as $p\neq 0$ and $1$ component (a double point) if $p = 0$. In the first case, $\mathcal{O}_{X_y,x}\cong k[t]/(t\pm p)$ for a preimage $x$ of $y$, which has length $1$, so $v_f(x) = m_f(x) = 1$. If $p = 0$, however, then $\mathcal{O}_{X_y,x}\cong k[t]/(t^2)$, which has length $2$. Thus $v_f(x) = m_f(x) = 2$.
Your second example is worked out similarly. The fact that $x$ and $y$ are or are not divisors doesn't make any real difference.
One can show that in the specific situation I'm considering (finite morphism between smooth varieties), every (not necessarily closed) point $y\in Y$ has the same number of preimages when counted with the multiplicity $m_f$, and that integer is the degree $d$ of the map $f$. A sketch of the proof of this fact is the following: Finite morphisms between smooth irreducible varieties over any algebraically closed field are flat, and hence $f_*\mathcal{O}_X$ is a locally free $\mathcal{O}_Y$-module of rank $d$. The fiber of $f_*\mathcal{O}_X$ at $y\in Y$ is exactly $\bigoplus_{f(x) = y} \mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$, so that $$d = \sum_{f(x) = y} \dim\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x} = \sum_{f(x) = y} m_f(x).$$
Edit: Multiplicities when pulling back divisors. Let $f\colon X\to Y$ be a (surjective) morphism between smooth irreducible varieties. Let $D$ be a prime divisor on $Y$, and let $f^{-1}(D)$ have irreducible decomposition $D_1\cup\cdots\cup D_n$. Let $x\in X$ be a general point of $D_i$ and $y = f(x)$. Since $Y$ is smooth, $D$ has a local defining equation at $y$, that is, $D = \{\varphi = 0\}$ for some irreducible function $\varphi\in \mathcal{O}_{Y,y}$. This $\varphi$ pulls back to a function $\varphi\circ f\in \mathcal{O}_{X,x}$ which will define $D_i$ near $x$. However, $\varphi\circ f$ will not necessarily be irreducible, it might be $\varphi\circ f = \psi^{n_i}$, where $\psi$ is an irreducible local defining equation for $D_i$ at $x$. This integer $n_i$ should be the multiplicity of $D_i$ when pulling back divisors --- it doesn't depend on the choice of $x$, so long at $x$ is chosen generic. I think $f^*D = \sum_i n_iD_i$ is how you would pull back $D$. Of course, you can extend $f^*$ linearly to all divisors $D$, not just prime divisors.
When $f\colon X\to Y$ is finite surjective like we had considered before, the integers $n_i$ won't be the multiplicities $m_f$, but should be the multiplicities $v_f$. That is, if $D$ is a prime divisor of $Y$ with generic point $y$, then $f^{-1}(y) = \{x_1,\ldots, x_n\}$ consists of the generic points of the components $D_1,\ldots, D_n$ of $f^{-1}(D)$, and $n_i = v_f(x_i)$. Of course, if both $X$ and $Y$ are curves, then divisors are closed points, and $m_f = v_f$ on closed points, so in the special case of curves the $n_i$ should be given by $m_f$. |
H: Parametrization of a conic and rational solutions
How can we parametrize the conic $C$: $x^2+y^2 = 5$, by considering a variable line through $(2,1)$ and hence all rational solutions of $x^2 + y^2 = 5$?
I'm thinking let $x = \sqrt{5}\cos t$, and $y = \sqrt{5} \sin t$, and somehow I need to get the coordinates $(2,1)$ in also.
AI: What you need to do is this:
Find the equation of a line through $(2,1)$ whose gradient $t$ is a rational number. You should get $y-1 = t(x-2)$.
Take a moment to visualise what is happening. As $t$ increases from large and negative, to large and positive, this line rotates about $(2, 1)$, through almost a half turn. We need to find the other intersection of the line and the conic.
We can do that by substituting for $y$, in your original equation; that gives us a quadratic equation in $x$, for any given value of $t$. We already know one of the roots of this quadratic, since it must be satisfied by $x=2$, since the line and your conic both go through $(2,1)$.
The quadratic you get is $$x^2(1+t^2) + x(-4t^2+2t) + 4t^2-4t-4 = 0$$
Using algebraic long division, we can therefore factorise this to get
$$\big[x-2\big]\big[x(1+t^2) -2t^2+2t+2\big] = 0$$
So the other root is given by $$x = \frac{2(t^2-t-1)}{1+t^2}$$
and then, using the fact that $y = tx-2t+1$ with a bit of algebra, we get
$$y = \frac{-t^2-4t+1}{1+t^2}$$
You can double-check that $x^2+y^2=5$, and thus we have a parametrisation giving the rational points on your curve, by taking rational values for $t$. |
H: Cancellation of addition on convex sets
I recently found a question about a property of the Minkowski sums. However the question was not properly answered (it used a projection argument which might not be true in a general Banach space).
I was wondering whether the following (weaker) statement holds:
Let $X$ be a Banach space and suppose $A,B,C_0\subset X$ are bounded, closed, convex and non-empty subset. Do we then have $$A+C_0=B+C_0\implies A=B?$$
AI: Modulo result presented in this question the solution is extremely simple
$$
A+C_0=B+C_0\Longleftrightarrow (A+C_0\subset B+C_0)\wedge(B+C_0\subset A+C_0)\Longrightarrow
$$
$$
(A\subset B)\wedge (B\subset A)\Longleftrightarrow A=B
$$ |
H: How long does it take to distribute a file among n computers
So i am at a LAN party where there are $n$ computers of which one $c_0$ computer has a file he wants everybody to have. Let's assume that transferring the file always takes $t$ time between any two computers.
Initially only the one $c_0$ computer has the file. He then transfers the file to a computer $c_1$ which takes $t$ time. Now there are 2 computers in the LAN that have the file. They then transfer the file to one new computer each, $c_2$ and $c_3$, which takes $t$ time. So after $2t$ time 4 computers have the file. And so in $xt$ time $2^x$ computers have the file.
How long does it take for a file that takes $t$ time to be transferred, to be distributed among all $n$ computers?
edit: What if only one-to-one transfers are allowed? E.g. two computers cannot 'work together' to transfer the file to one computer in $t/2$.
AI: $t*log_2(n)$ time. Each transfer between two computers takes the same $t$ time, so we can assume that all transfers at a "stage" of the process happen concurrently. After each stage, double the number of computers as the last stage have the file (base-2 exponential growth), so the number of stages of transfers that must occur is logarithmically bound, base-2, to the number of computers in total that must have the file. As $log(1) = 0$ regardless of base, the operation is bound to the entire number of computers, even though one of them already has the file (and thus it would take 0 time for the trivial case of a group consisting of the one computer that already has the file). In computer-science lingo, this operation is logarithmic-time, or O(logN).
EDIT: From my comment, if $n$ is not an exact power of two, the logarithm will produce a decimal result. If the time $t$ required to transfer one file to one computer is a minimum bound, then the correct answer is the next higher integer value for the log, $t*ceil(log_2(n))$. However, if $t$ is the time it takes for one computer to transfer one file to one recipient, and it is possible in this system for $n$ computers to each transfer $1/n$ of the file to 1 recipient in $t/n$ time, then the decimal log value is correct, because for $n = 2^x+y < 2^{n+1}$, each of the $y$ computers can receive a fraction $y/2^x$ of the file from $2^x/y$ other machines, taking $ty/2^x$ additional time after $t*log(2^x)$, approaching the decimal value of $t*log_2(n)$. This is the ideal behavior of P2P systems like BitTorrent where every computer that has the file (or any portion of it) can re-distribute portions of it to any computer requesting the file, and thus the file's data can come from as many sources as are available, making the bound on transfer time the much larger download bandwidth of the average consumer internet connection and a fraction of $t$. |
H: Notation and naming for two operations with $p$-form valued $n$-forms
While trying to answer my other question I found I never heard about vector-valued differential forms. I've been searching for them in various mathematical physics books, but didn't get too much.
I'm interested in notion and names of two operations with $p$-form valued $n$-forms.
I treat p-form valued n-form as
$$ (TM)^n \to \Omega^p(M) $$
or
$$ (TM)^n \to (TM)^p \to \mathbb R $$
I think ususal contraction with a vector field is denoted as $\rfloor$:
$$\rfloor : TM \to \left(TM^n \to TM^p \to \mathbb R \right) \to \left(TM^{n-1} \to TM^{p} \to \mathbb R \right) $$
How do I denote the similar operation, but when contracting with the resulting $p$-form? :
$$\lfloor : TM \to \left(TM^n \to TM^p \to \mathbb R \right) \to \left(TM^n \to TM^{p-1} \to \mathbb R \right) $$
Or maybe there is a notation for switching position, that is $p$-form valued $n$-forms is turned into $n$-form valued $p$-forms.
Given an $n$-form and a $p$-form I can produce $p$-form valued $n$-form. How do I denote and name this operation?
$$(TM^n \to \mathbb R) \to (TM^p \to \mathbb R) \to \left( TM^n \to TM^p \to \mathbb R \right)$$
$$(\omega,\nu) \mapsto \left(\left(v^n,u^p\right) \mapsto \omega(v^n) \nu(u^p)\right)$$
Maybe these operations are not specific only to this setting and have some universal nature and notation.
AI: If $E \to M$ is a vector bundle then $E$-valued $n$-forms are sections of the tensor product bundle $E \otimes \Lambda^n(T^*M)$. From basic properties of the tensor product of bundles, sections are spanned by $\sigma \otimes \mu$ where $\sigma \in \Gamma(E), \mu \in \Omega^n(M)$. In your situation, you're looking at sections of the bundle $\Lambda^p(T^*M) \otimes \Lambda^n(T^*M)$ so the last construction you mention is just tensoring a section of $\Lambda^p(T^*M)$ with one of $\Lambda^n(T^*M)$ to get a section of the tensor product of the bundles.
As for contraction, this is from the general fact that any bundle map $E \to F$ extends to a bundle map $E \otimes \Lambda^n(T^*M) \to F \otimes \Lambda^n(T^*M)$ by acting as the identity on the second factor of the tensor product. In your situation $E = \Lambda^p(T^* M)$ and $F = \Lambda^{p-1}(T^* M)$. |
H: $2$ question on matrix with some condition
we need to tell whether $a,b$ true or false,
I know that there does not exist $n\times n$ $A,B$ such that $(AB-BA)=I$,as if we take trace of both side they are not equal, so $a$ is false?
$b$ I have no idea, could any one give me hint?
AI: (a) Let $C = AB - BA$. Then we have that the eigenvalues of $(I-C)^n$ are all $0$, and hence the eigenvalues of $I - C$ are also all zero. (Here we use the fact that if $\lambda$ is an eigenvalue of $A$ then $\lambda^n$ is an eigenvalue of $A^n$).
Hence, the eigenvalues of $C$ are all $1$, which implies that the trace of $C$ is $n$. On the other hand, $tr(AB) = tr(BA)$ implies that $tr(C) = tr(AB - BA) = 0$, a contradiction.
So, no such matrices exist.
(b) Take $a_1,a_2, \cdots, a_n$ to be the eigenvalues of $A$, and note that $\displaystyle tr(A) = \sum_{i=1}^n a_i, \ \ \ \det(A) = \prod_{i=1}^n a_i$.
As positive definiteness of $A$ gives us that $a_1, a_2, \cdots, a_n$ are all positive, therefore we apply the AM -GM Inequality which immediately gives us what is required. |
H: Probability distributions - Exam paper question - $\mathrm{Cov}(X,Y)$, PDF
Two people have decided to meet at a certain point in a forest sometime between noon and 2pm.
Their respective independent arrival times are $X$ and $Y$ such that $X \sim \mathrm{Unif}(0,2)$ and $Y \sim \mathrm{Unif}(0,2)$.
Hence the joint density of $X$ and $Y$ is
$$f_{X,Y} {(x,y)} = \begin{cases} 1/4, & 0< x <2 , 0< y <2 \\ 0, & \text{otherwise.} \end{cases} $$
They have agreed that whoever arrives first will wait for at most $20$ minutes for the arrival of the other.
a) Sketch the region in the $xy$ plane of times values for which they will meet and specify precisely the appropriate bounds (in terms of $x$ and $y$) for this region; then find the probabilty that they will meet by integrating the joint PDF $f_{X,Y} {(x,y)}$ over this region.
b) Since $X$ and $Y$ are independent, what value must $\mathrm{Cov}(X,Y)$ have?
c) Calculate explicitly $\mathrm{Cov}(X,Y)$ starting from its definition. Recall that
$$\mathrm{Cov}(X,Y) = E[(X - E(X))(Y - E(Y))].$$
I know this is quite a long question but I didn't know how to break it down into smaller parts without just having to type it into three different questions to ask.
If you could give me an idea about the sketch great! I'm not sure how to integrate the PDF as it's only $1/4$? Would it not just be $x/4 + C$?
Also for $\mathrm{Cov}(X,Y)$, I don't seem to have any notes on this, so detail would be good too.
Test is in the morning and you guys have been a big help so far!
AI: I forgot part a)
a- When you shade the square in $\{[0,2],[0,2]\}$ then $20 min=1/3$ and they can not meet only if you start waiting from $2-1/3=5/3$ and it follows that you have the probability that thay cannot meet
$$\int_0^{5/3}\int_0^{5/3}\frac{1}{4} \, dx \, dy=\frac{25}{36}$$
and they meet with probability
$$1-\frac{25}{36}=\frac{11}{36}$$
b- If $X$ and $Y$ are independent then their covariance should be zero as follows
$$E[XY]=E[X]E[Y]$$
$$Cov(X,Y)=E[(X-E(X))(Y-E(Y))]=E[XY]-E[X]E[Y]=0$$
c- $$E[XY]=\int_0^2\int_0^2 \frac{1}{4}xy \, dx \, dy=1$$
and you have
$$f_X(x)=\int_{-\infty}^\infty\frac{1}{4}f_{XY}(x,y)\,dy=\int_0^2 \frac{1}{4}1 \, dy=\frac{1}{2} \in [{0,2}]$$
you get
$$E[X]=E[Y]=\int_0^2f_X(x)xdx=\int_0^2\frac{1}{2}x \, dx=1$$ |
H: Integer coefficient polynomial $p(x)$ has no integer roots of $\,p(0)$ and $p(1)$ are odd [Parity Root Test]
This came up an a training piece for the Putnam Competition and also in Ireland and Rosen.
The question posed was basically:
Let $p(x)$ be a polynomial with integer coefficients satisfying that $p(0)$ and $p(1)$ are odd. Show that $p$ has no integer zeros.
I&R give an example:
$p(x) = x^2 - 117x + 31$ and show (no problem) that for any $n$ whether even or odd, $p(n)$ will be odd. And claim that this shows $p(n)$ will never be $0$.
I can see, e.g., that $x^2 + 2x + 1$ will be odd substituting an even $n$ and even for an odd $n$.
But would appreciate help in understanding the underlying math and what is happening here.
Also, as a second part, can a general statement about the existence of an integer solution be made if $n$, even and odd, generates an even and an odd as in the last example.
I can see that if you look at these equations (mod $2$), you can distinguish whether there is an integer solution. And I would guess this is intimately connected with the question.
Thanks as always.
AI: Hint $ $ If an integer coefficient polynomial has an integer root $\rm\,n,\,$ i.e.$\rm\ p(n) = 0,\ $ then $\rm\,n\,$ remains a root modulo $\:\!2,\,$ i.e. $\rm\ p(n)\equiv 0\pmod{\!2}\,$ by the Polynomial Congruence Rule. So, contrapositively, if a polynomial has no roots $\!\bmod 2\,$ then it has no integer roots. This leads to the following simple
Parity Root Test $\ $ A polynomial $\rm\,p(x)\,$ with integer coefficients
has no integer roots if its constant coefficient and coefficient sum are both odd.
Proof $\ $ The test verifies that $\rm\, p(0) \equiv 1\equiv p(1)\pmod{\!2},\,$ i.e.
that $\rm\,p(x)\,$ has no roots $\!\bmod 2,\,$ hence no integer roots. $\ \bf\small QED$
E.g. $\rm\, a\,x^2 + b\,x + c\ $ has no integer roots
if $\rm\:c\:$ is odd and $\rm\,a,b\,$ have equal parity $\rm\:a\equiv b\pmod{\!2}$.
The Parity Root Test generalizes to any ring with a sense of parity, e.g. the Gaussian integers $\rm\,a + b\,{\it i}\ $ for integers $\rm\,a,b.\,$ For much further discussion see this post and also these related posts. |
H: Is there a relationship between products and integrals? Is there a way to convert a product into an integral?
I know that the Euler-Maclaurin formula establishes a relationship between sums and integrals, but is there some sort of formula that establishes a relationship between products and integrals? I don't mean the product rule, for products of functions. I mean a way to convert the sort of product you use the Π notation for into an integral. Specifically, a product of all the values of a function, from f(1) to f(n), into an integral.
Sorry if this is a silly question; I'd still like to know the answer.
Thanks very much.
AI: Assuming $f(i) > 0, \forall i$, use can use the identity
$$\prod_{i=1}^n f(i) = \exp \left[\sum_{i=1}^n \log f(i)\right]$$
and then employ Euler-Maclaurin for the sum on the right hand side. |
H: Convergent Subsequence in $\mathbb R$
Let $\langle a_n \rangle = \dfrac{(-1)^n}{1+n}$ be a sequence in $\mathbb R$.
Considering the limit point(s) of this sequence and the subsequences that converge to this point, I have two subsequences: $$ a_{2k} = \frac{1}{1+2k} \to 0 \text{ and } a_{2k+1} = \frac{-1}{1+2k+1} \to 0 $$
If the question asks for the limit points of the sequence, and a subsequence that converges to this limit point, do I leave out the second subsequence? I don't think I've missed a limit point, but it doesn't hurt to check.
AI: If you want to find the limit points, you have to think about the subsequential limits for every subsequence, not just those two. But luckily, the original sequence converges (to 0) and in a convergent sequence, every subsequence converges to the same limit as the whole sequence does. |
H: Counterexample Check for Sum of Limit Points of Subsequences
Let $c$ be a limit point of a sequence of real numbers $\langle a_n \rangle$ and $d$ a limit point of $\langle b_n \rangle$. Is $c+d$ necessarily a limit point of $\langle a_n + b_n \rangle$?
My Question:
When considering this question, do I have to sum over the same index or can the indices for the different subsequences differ? My intuition is that the subsequences must be summed over identical indices, in which case I believe that the following example serves as a counterexample:
Let $\langle a_n \rangle = (-1)^n, \ \langle b_n \rangle = (-1)^{n+1}$. Then summing over even and odd indices, I get $0 \ne 2$ or $-2$, which is the sum of their limit points. Have I done this correctly?
AI: Your example is fine. Both $\langle(-1)^n:n\in\Bbb N\rangle$ and $\langle(-1)^{n+1}:n\in\Bbb N\rangle$ have $1$ as a limit point, but $\langle(-1)^n+(-1)^{n+1}:n\in\Bbb N\rangle=\langle 0:n\in\Bbb N\rangle$ converges to $0$ and so does not have $1+1=2$ as a limit point.
It doesn’t matter what subsequence of $\langle a_n:n\in\Bbb N\rangle$ has $c$ as limit or what subsequence of $\langle b_n:n\in\Bbb N\rangle$ has $d$ as a limit; all that matters is whether some subsequence of $\langle a_n+b_n:n\in\Bbb N\rangle$ has $c+d$ as a limit. In your example that’s not the case, so yours is a genuine counterexample to the conjecture. |
H: $F$ closed and connected then...
Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon>0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)<\epsilon$ for $1\leq k\leq n$.
Is the hypothesis that $F$ is closed needed?
AI: $F$ can be any connected subset of $X$. Given $\epsilon>0$, let $\mathscr{U}=\{F\cap B(x,\epsilon/2):x\in F\}$, where $B(x,r)$ is the open $d$-ball of radius $r$ centred at $x$. Now apply the chain characterization of connectedness proved in this answer (mentioned by David Mitra in the comments) to conclude that for any $a,b\in F$ there is a finite sequence $x_0=a,x_1,\dots,x_n=b\in F$ such that $B(x_k,\epsilon/2)\cap B(x_{k+1},\epsilon/2)\ne\varnothing$ for $k=0,\dots,n-1$. Clearly $d(x_k,x_{k+1})<\epsilon/2+\epsilon/2=\epsilon$ for $k=0,\dots,n-1$. |
H: Question about transitivity convergence
Let us define $E^*=\{f:E\rightarrow \mathbb{R} \mid f $ is linear and continuous$\}$, where $E$ is separable space equipped with a norm. Let $A = \{x_1,x_2,\ldots\}$ be a countable dense subset of $E$. Suppose that $(f_n)_{n\in\mathbb{N}}$ is a sequence of functions from $E^*$ such that $$\lim_{n\to\infty}f_n(x_i)=f(x_i)$$for all $i\in\{1,2,\ldots\},$ i.e. $(f_n)$ converges pointwise to $f$ on $A$. My teacher writes then:
Consequently, $f_n(x) \rightarrow f(x) $ $\forall x \in E$.
My question is: why is it true?
I am trying to demonstrate this as follows:
Choose any $x_i$ close to $x$. Then $f_n(x - x_i) = 0 = f_n(x) -f_n(x_i)$. We have $$\lim_{n\rightarrow \infty}|f_n(x) - f(x_i)| = 0 \iff f_n(x) \rightarrow f(x_i)$$ as $n\rightarrow \infty$. By hypothesis $f_n(x_i) \rightarrow f(x_i)$ whilst by continuity $f(x_i) \rightarrow f(x)$ and then I will be able to conclude that $f_n(x) \rightarrow f(x) $ $\forall x \in E$. Is it correct?
AI: Your proof is quite strange because it starts from the assumption that $f_n(x-x_i)=0$ for all $n\in\mathbb{N}$. This is not true in general.
I'll additionally assume that $\{f_n:n\in\mathbb{N}\}\subset \mathrm{Ball}_{E^*}(0,C)$ for some $C>0$, i.e. norms of functionals $\{f_n:n\in\mathbb{N}\}$ is uniformly bounded by some constant $C$.
Take arbitrary $x\in E$. Fix $\varepsilon>0$, and find $i\in\mathbb{N}$ such that
$$
\Vert x- x_i\Vert\leq\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}
$$
Since $\lim\limits_{n\to\infty} f_n(x_i)=f(x_i)$, then there exist $N\in\mathbb{N}$ such that $n>N$ implies $|f_n(x_i)-f(x_i)|\leq\varepsilon/3$. Now we use $\varepsilon/3$-trick, i.e. for all $n>N$ we have
$$\eqalign{
|f_n(x)-f(x)|&\leq
|f_n(x)-f_n(x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i)-f(x)|\phantom{A\over B}\cr
&=
|f_n(x-x_i)|+|f_n(x_i)-f(x_i)|+|f(x_i-x)|\phantom{A\over B}\cr
&\leq
\Vert f_n\Vert\Vert x-x_i\Vert+\varepsilon/3+\Vert f\Vert\Vert x-x_i\Vert\phantom{A\over B}\cr
&\leq
C\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}+\varepsilon/3+\Vert f\Vert\frac{\varepsilon}{3\max\{ C,\Vert f\Vert\}}\cr
&\leq
\varepsilon/3+\varepsilon/3+\varepsilon/3\phantom{A\over B}\cr
&=\varepsilon}
$$
Thus for all $x\in E$ and $\varepsilon>0$ we found $N\in\mathbb{N}$ such that $n>N$ implies
$|f_n(x)-f(x)|<\varepsilon$. This means that for all $x\in E$ we have $\lim\limits_{n\to\infty} f_n(x)=f(x)$.
P.S.
If we won't require uniform boundedness of norms of $\{f_n:n\in\mathbb{N}\}$ we can construct a counterexample. See Davide's comment. |
H: How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$?
How to construct a $2\times 2$ real matrix $A$ not equal to Identity such
that $A^3=I$?
There is a correspondence between the ring of complex numbers and the
ring of $2\times2$ matrices (0 matrix
is included!) i.e.,$$a+ib\leftrightarrow\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$
Can I apply this result and construct such matrix?
AI: Find the three cube roots of $1$. Let $a+bi$ be one of those. They are solutions of $x^3=1$. Hence $x^3-1=0$.
Since $1$ is one of the solutions, $x-1$ must be one of the factors, thus:
$$
x^3-1 = (x-1)(\cdots\cdots\cdots).
$$
Fill in the blanks by doing long division. You should get
$$
(x-1)(x^2+x+1).
$$
So the equation is
$$
(x-1)(x^2+x+1) = 0.
$$
That implies
$$
x-1=0\quad\text{or}\quad x^2+x+1 = 0.
$$
Solve the quadratic equation. |
H: Classify all alphabets into homeomorphism classes $\{ M N B H\}$,
classify all alphabets into homeomorphism classes $\{ M N B H\}$,
what does it mean by homeomorphism classes?
They're looking upon letters as drawings of topological spaces, perhaps!?
connectedness distinguishes $B$ from other $3$ as if we remove mid point of vertical line of $B$, it is still one single component, but if we remove one point from other three they will be into two component. but if we remove 2 point suitably from $H$ it will be broken into $5$ component, but the case is not with $M$ and $N$ they will be broken into 3 component, so the homeomorphism classes?
$\{H\},\{M,N\},\{B\}$, am I right? is there any other approach?
AI: I'll describe a general approach here, using for illustration purposes the serif font that appears in the question. We are given compact connected Hausdorff spaces, also known as continua. A point $x$ of continuum $C$ is a cut point if $C\setminus\{x\}$ is not connected. The number of components of $C\setminus\{x\}$ is the order of cut point $x$. The cardinality of the set of cut points of order $n$ is a topological invariant of the space.
All of our continua have infinitely many cut points of order 2, so that does not help much (although we could look at the structure of the set of all such cut points, e.g., the number of its components). Let's count the cut points of order 3: $M$ has 4, $N$ has $3$, $B$ has none and $H$ has $6$. Hence, none of them are homeomorphic in the serif version.
In addition, $B$ is distinguished from the rest by having infinitely many non-cut points. It's a very neat theorem of general topology that every nondegenerate continuum has at least two non-cut points. |
H: Primes Not Dividing $\binom{2n}{n}$
Let $n \geq 3$, show ${2n \choose n}$ is not divisible by $p$ for all primes $\frac{2n}{3} <p\leq n$
Note: This fact along with other facts about ${2n \choose n}$ are used in a proof of Bertrand's postulate.
AI: I think that the idea is rather simple (not verifying Brian's hint...) :
We need to divide $(2n)!$ by $n!^2$ so let's write the primes of the decomposition of $(2n)!$ and $n!$.
Let's suppose that $\frac n2<p \le n$ is such a prime larger than $2$ then :
$$
\binom{2n}{n} =
\frac{(2n)!}{(n!)^2}=
\frac{
2n\cdot (2n-1)\cdots(2p)\cdots (n)\cdots (p)\cdots 2\cdot 1}
{(n)\cdots (p)\cdots 2\cdot 1\cdot (n)\cdots (p)\cdots 2\cdot 1}
$$
When $3p\le 2n$ we will have at least $3$ $p$'s at the top and the fraction will be divisible by $p$ else the two $p$ will cancel ! |
H: Inverse of symmetric matrix $M = A A^\top$
I have a matrix, generated by the product of a non-square matrix with its own transpose:
$$M = A A^\top.$$
I need the inverse of $M$, assuming $\det(M) \neq 0$.
Given the nature of the matrix $M$, are there any specialised computational methods to find its inverse, prioritising precision over speed? Gauss-Jordan is prone to error, and I hope to find something nicer than and with comparable precision to the adj/det method.
$M$ will either have be around $70 \times 70$, or $1000 \times 1000$ in size.
I've had a quick read of the Matrix Cookbook and of this page, but (at the present time of 1 am) I'm struggling to see how it could help me.
In case it helps, I'm actually trying to calculate:
$$B = (A A^\top)^{-1} A.$$
AI: Cholesky decomposition! It's a very stable method and it works on sparse matrices too.
But if you are just trying to solve the normal equation, I may suggest conjugate gradient or SOR. |
H: Matrix-product integrals
Suppose you multiply the values of $f$ between $a$ and $b$ at intervals of $\Delta x = (b-a)/n$ and then raise the product to the power $\Delta x$, and take the limit as $n\to\infty$. What you get would bear the same relation to products that integrals bear to sums, and there's a trivial reduction to ordinary integrals, in that what you get is $\displaystyle\exp\int_a^b\log f$, provided the addition and multiplication are reasonably like the ordinary ones on real numbers.
But what if the things getting multiplied are matrices, so the multiplication is non-commutative? I think I've seen it asserted somewhere ("somewhere" is a horrible word sometimes, isn't it?) that that is when such "product integrals" are not trivially reducible to more familiar things.
Last time I checked, Wikipedia's article titled product integral didn't go into matrix products, but only products of real numbers. If what I think I heard somewhere is right, then that makes it not all that good an article.
So:
Is what I think I heard correct?
If so, where's the literature on this?
And what are such matrix-product integrals used for? Or, what is done with them?
AI: I think you're talking about "time-ordered exponentials". These are used a lot in physics.
EDIT: Mathematically, the finite-dimensional case is just the fundamental matrix for a first-order linear ODE system: $\Psi(t)$ is the solution of $\Psi'(t) = A(t) \Psi(t)$ with $\Psi(0) = I$, where $A$ is a continuous function of time $t$ with values in the $n \times n$ matrices.
One way to obtain it is as
$$ \Psi(t) = \lim_{n \to \infty} e^{A((n-1)t/n)/n} e^{A((n-2)t/n)/n} \ldots e^{A(t/n)/n} e^{A(0)/n} $$
Of course physicists are prone to use it in infinite-dimensional settings with unbounded operators $A(t)$, where things can be trickier.
You might look at H. Araki, "Expansional in banach algebras", Annales scientifiques de l'École Normale Supérieure, Sér. 4, 6 no. 1 (1973), p. 67-84 http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1973_4_6_1/ASENS_1973_4_6_1_67_0/ASENS_1973_4_6_1_67_0.pdf |
H: $L^p$ Norm continuity boundedness
I'm studying for a Qualifying exam and can't figure out this problem. I see that the limit must be $f(x)$ and can get the boundedness but had trouble with continuity. Any suggestions?
Let $f\in L^{\infty}(\mathbb{R}^{d})$ and let $\phi:\mathbb{R}^{d} \times (0,\infty)\rightarrow \mathbb{R}$ be the following map:
$$\phi(x,r)=\frac{1}{r}\int_{B_{r}(x)}f(y)dy.$$
Prove that $\phi$ is continuous in $x$, and in $r$, and is uniformly bounded. What can you say about $\lim_{r\rightarrow 0}\phi(x,r)$?
AI: I assume $\mathfrak R$ means reals (first time I see someone write them using fraktur, I think :) ).
Notice that if $f\in L^{\infty}$, then for any $A,B$, $\lvert \int_A f-\int_B f\rvert\leq \lVert f\rVert\cdot \mu(A\triangle B)$.
That should make it easy to show that it is continuous in $x$.
For $r$ it is much the same: write the map as a composition of $r\mapsto (\int_{B_r(x)} f,r)$ and $(a,r)\mapsto a/r$. The former map is continuous -- you can show it the same way as you've shown that $\phi$ is continuous in $x$, and the second is obviously continuous.
On the other hand, I don't see why would it be bounded. It also seems to me that the limit is actually $0$ if $d>1$, for example if $f\equiv 1$, then, up to a multiplicative constant, $\phi(x,r)\cong r^{d-1}$, which approaches $0$ as $r\to 0$... |
H: Absolute Continuity, Lipschitz
I'm studying for a qualifying exam and can't seem to solve this problem. Any suggestions would be appreciated!
Let $f:[a,b] \rightarrow \mathbb R$ be absolutely continuous. Show, for each $\epsilon>0$, that there is a uniformly Lipschitz function(global) $g:[a,b] \rightarrow \mathbb R$ such that $|f(x)-g(x)|<\epsilon$ for all $x\in [a,b]$.
AI: To expand on my comment:
One approach is just to invoke the Weierstrass approximation theorem. This works even if $f$ is merely continuous, and it gives a $g$ which is a polynomial, which is drastically stronger than just being Lipschitz or even $C^\infty$.
You could also give a more direct proof. An absolutely continuous function has a derivative which is $L^1$; a Lipschitz function has a derivative which is bounded. You can approximate $L^1$ functions by bounded functions. Now, to get from the derivative back to the function, what could you do...?
Indeed, integrate. So find a bounded measurable function $h$ which is close to $f'$ in $L^1$ norm. What can you say about the difference between the integrals (from $a$ to $x$) of $f'$ and $h$?
So we can get $\int_a^x f'(t) dt - \int_a^x h(t)dt$ to be small, right? Or in other words, we can get $f(x) - f(a) - \int_a^x h(t)dt$ to be small. So what if we set $g(x) = f(a) + \int_a^x h(t)dt$? |
H: Fundamental Theorem of Calculus problem
I don't understand the intuition behind this. Why can we just plug in $x$ for $t$ here and that gives us the result? I thought I was understanding the Fundamental Theorem of Calculus, but I don't see how it applies here. I thought the Theorem mainly stated that the area under a function can be found by taking the the value of the anti derivative over the specified interval. It doesn't make sense to me why we just plug in $x$ and voila that's our answer.
$$\frac {d}{dx} \int_{a}^{x} (t^3 + 1) \ dt = x^3 + 1$$
AI: Since the function $\,t^3+1\,$ is continuous (and derivable) everywhere, it has a primitive function $\,G(t)\,$in any finite interval. Using the FTC , write
$$\int_a^x(t^3+1)dt=G(x)-G(a)\\\Longrightarrow \frac{d}{dx}{\left(\int_a^x(t^3+1)dt\right)}=\frac{d}{dx}{(G(x)-G(a))}=G'(x)=x^3+1$$ |
H: About the sequence satisfying $a_n=a_{n-1}a_{n+1}-1$
"Consider sequences of positive real numbers of the form x,2000,y,..., in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence?
(A) 1 (B) 2 (C) 3 (D) 4 (E) More than 4"
Can anyone suggest a systematic way to solve this problem? Thanks!
AI: Perhaps a bit of a tedious solution but this is the most straight forward approach. By direct substitution it is not difficult to show that the sequence necessarily repeats itself after the $6$th term. We have
$$a_1 = x$$
$$a_2 = 2000$$
$$a_3 = \frac{2001}{x}$$
$$a_4 = \frac{2001 + x}{2000x}$$
$$a_5 = \frac{1 + x}{2000}$$
$$a_6 = x$$
$$a_7 = 2000$$
and so on....
Clearly we can take $x=2001$ for a trivial appearance. For $a_3$ to be $2001$ we require $x=1$. For $a_4$ we require $x=0.0005$. For $a_5$ to be $2001$ we require $x = 4001999$. This gives a total of $4$ solutions for $x$. |
H: A question on iterated sums
We are to prove that the set
$$\Bigg\{ \sum^m_{i=1}\sum^n_{j=1}|a_ib_j| :m,n\in \mathbb N \Bigg\}$$
is bounded. Also, We are to use this to show that the iterated sum $\sum^\infty_{i=1} \sum^\infty_{j=1}|a_ib_j|$ converges.
I could prove that the set is bounded, however, I can't solve the second part.
I become very confused when I have to deal with iterated sums. Could anyone please help me with this problem.
AI: Although by first proving that the sum is bounded, it can be proved that the sum converges. But, there is a way to directly prove that it converges.
For that, let
$$\sum^\infty_{i=1}|a_i|=L~~~~~\text{and}~~~~~\sum^\infty_{j=1}|b_j|=M$$
For each fixed $i \in \mathbb N$, the Algebraic Limit Theorem will allow us to write $\sum^\infty_{j=1}|a_ib_j|$=$|a_i|\sum^\infty_{j=1}|b_j|$.
If one continues the process, we will see that:
$$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|=\sum^\infty_{i=1}|a_i|\sum^\infty_{j=1}|b_j|=ML$$
Hence,$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|$ converges (to $ML$). |
H: Prove that $ \frac{n}{\phi(n)} = \sum\limits_{d \mid n} \frac{\mu^2(d)}{\phi(d)} $
I am trying to show that:
\begin{equation}
\frac{n}{\phi(n)} = \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)}
\end{equation}
where $\phi(n)$ is Euler's totient function and $\mu$ is the Möbius function.
The identity clearly holds for $n=1$, so if we write $n=p_1^{a_1} \cdots p_k^{a_k}$ for the prime decomposition of $n$, the left-hand side becomes:
\begin{eqnarray*}
\frac{n}{\phi(n)} & = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{\phi(p_1^{a_1}) \cdots \phi(p_k^{a_k})} \\
& = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{(p_1^{a_1-1})(p_1-1) \cdots (p_k^{a_k})(p_k-1)} \\
& = & \frac{p_1 \cdots p_k}{(p_1-1)\cdots (p_k-1)}
\end{eqnarray*}
Whereas the right-hand side becomes:
\begin{eqnarray*}
\sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} & = & \sum_{\substack{d \mid n \\ p^2 \nmid d}} \frac{1}{\phi(d)} \\
& = & 1 + \frac{1}{\phi(p_1)} + \dots + \frac{1}{\phi(p_k)} + \frac{1}{\phi(p_1 p_2)} + \dots + \frac{1}{\phi(p_{k-1}p_k)} \\
& + & \dots + \frac{1}{\phi(p_1 \cdots p_k)} \\
& = & 1 + \frac{1}{p_1 - 1} + \dots + \frac{1}{p_k - 1} + \frac{1}{(p_1 - 1)(p_2 - 1)} + \dots + \frac{1}{(p_{k-1}-1)(p_k-1)} \\
& + & \dots + \frac{1}{(p_1 - 1) \cdots (p_k - 1)}
\end{eqnarray*}
I am unsure of how to proceed from this point however. Any help would be much appreciated.
AI: Prove both sides (of the original equation) are multiplicative, then prove they're equal for prime powers. |
H: Example of a quasinilpotent operator
Can anybody please give me an example of a quasinilpotent operator $T$, i.e. an operator such that $\sigma(T)=\{0\}$ on $l_2$ such that it has finite dimensional but non-trivial kernel and is not compact?
This is probably easy and well known but I just can't figure it out it and I am getting frustrated.
Thanks!
AI: Take a quasinilpotent operator with trivial kernel and a finite Jordan Block and glue them together. |
H: Compact operators and uniform convergence
Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in operator norm to $AT$.
I'm attempting to prove this by contradiction: That is, there exists some $\epsilon > 0$ such that $ \|A_nT-AT\| = \sup_{\|f\|=1}\|(A_nT-AT)f\| > \epsilon$ for all $n$. How can I exploit the compactness of $T$?
AI: You can use the property of the compact operators: that they map bounded sequences in X to sequences in Y with convergent subsequences. So you can choose your sequence to be an orthonormal basis in your Hilbert space and then you can advance with your proof. |
H: Windowed Fourier transform of Gaussian distributed random time series
If I have a discrete time series $x(t_i)$, and each of the $x(t_{i})$ are normally distributed, i.e., come from a Gaussian distribution with mean zero and variance one, would a windowed finite Fourier transform of $x(t_0)$ through $x(t_{N-1})$ also be Gaussian distributed? In other words, would the real and imaginary parts of:
$$y(f) = \sum_{t=0}^{N-1}exp(-i 2 \pi f t) x(t) a(t)$$
also have Gaussian distributions? a(t) is a window function that decays to 0 at $t=0$ and $t=N-1$.
AI: Yes, provided one assumes that the family $(x(t))_{0\leqslant t\leqslant N-1}$ is gaussian (recall that, as soon as $N\geqslant2$, this asks strictly more than each $x(t)$ being normally distributed). Then $y(f)$, the real part of $y(f)$ and the imaginary part of $y(f)$ are all linear combinations of a gaussian family hence they are normally distributed (complex-valued for $y(f)$ and real-valued for the others).
Edit: A standard way to ensure that a family is gaussian is to assume that each random variable is normally distributed and that the family is independent. |
H: Residue Rings of Polynomial Rings over $\mathbb F_p$
Let $\mathbb Z$ be the ring of integers, $p$ a prime and $\mathbb F_p = \mathbb Z/p\mathbb Z$ the field with $p$ elements. Let $x$ be an indeterminate. Set $R_1 = \mathbb F_p[x]/(x^2-2)$, $R_2 = \mathbb F_p[x]/(x^2-3)$. Determine whether the rings $R_1$ and $R_2$ are isomorphic in each of the following cases:
(a) $p = 2$
(b) $p = 5$
(c) $p = 11$
I'm pretty sure about my answer to (c), but not very sure about (a) & (b). Any comments would be greatly appreciated for PhD Quals prep. Thank you.
Attempt at Solution:
(c) When $p = 11$, $x^2 - 2$ is irreducible but $x^2 - 3$ is reducible [$2$ is a quadratic nonresidue $\bmod 11$; $3$ is a quadratic residue]. So $(x^2 - 2)$ is a maximal ideal and hence $R_1$ is a field, whereas $R_2$ is not. So they are not isomorphic.
(b) When $p = 5$, both $x^2 - 2$ and $x^2 - 3$ are irreducible, so both $R_1$ and $R_2$ are fields. As any polynomial in $R_1$ or $R_2$ of degree $\ge2$ is equal to some polynomial of degree $0$ or $1$, effectively, the elements of $R_1$ and $R_2$ can be represented by $a_0 + a_1x$, where $a_0, a_1 \in \{0,1,2,3,4\}$. So $R_1 = R_2 =$ finite field with $5^2$ elements, i.e. they are isomorphic.
(a) When $p=2$, both $x^2 - 2 = x^2$ and $x^2 - 3 = x^2-1$ are reducible. So although $R_1$ and $R_2$ can each be represented by $a_0 + a_1x$, where $a_0, a_1 \in \{0,1\}$. I am not sure whether each of them is isomorphic to $\mathbb Z/4\mathbb Z$ or the Klein $4$-group.
AI: (a) $R_1 = F_p[t]$, where $t$ is the image of $x$.
$R_2 = F_p[s]$, where $s$ is the image of $x + 1$.
Since $t^2 = 0$ and $s^2 = 0$, $R_1$ and $R_2$ are isomorphic.
(b) and (c) were solved by you. |
H: A question about convolution of two distributions
Generally,when taking convolution of two distributions,at least one of which is supposed to be of compact support.
But when u,$v\in S'(\mathbb{R})$ ( temperate distributions) have suports on the positive half axis,then $u\ast v \in S'(\mathbb{R})$
how to prove this and generalize to high dimensions?
AI: Since this is homework, I probably shouldn't write down a complete solution. But let's at least write down a definition for the convolution general enough for the situation described above (taken from my lecture notes of the course "Distribution et équations aux derivées partiélles" by André Cérezo):
Théorême Soient $S,T \in \mathcal{D}'(\mathbb{R}^n)$, $F=(\operatorname{supp} S_x)\times(\operatorname{supp} T_y)\subset \mathbb{R}^{2n}$, et $\Delta=\{ (x,-x)|x\in \mathbb{R}^n\}\subset \mathbb{R}^{2n}$. Supposons que, pour tout $K\Subset\mathbb{R}^n$, le fermé $(K\times\{0\}+\Delta)\cap F$ soit un compact de $\mathbb{R}^{2n}$. Alors la formule
$$(*)\qquad\forall \varphi\in \mathcal{D}(\mathbb{R}^n)\qquad <S*T,\varphi>=<S_x\otimes T_y,\varphi(x+y)>$$
définit une distribution sur $\mathbb{R}^n$, appelée "produit de convolution" de $S$ et $T$.
Here $K\Subset\mathbb{R}^n$ means that $K$ is compact. We have $\mathcal{S}'(\mathbb R)\subset\mathcal{D}'(\mathbb R)$, so the first step is to verify the additional condition. This gives us $u*v\in\mathcal{D}'(\mathbb R)$. Now all that is left to show is $u*v\in\mathcal{S}'(\mathbb R)$.
Edit (the requested translation of the cited theorem)
Theorem Let $S,T \in \mathcal{D}'(\mathbb{R}^n)$, $F=(\operatorname{supp} S_x)\times(\operatorname{supp} T_y)\subset \mathbb{R}^{2n}$, and $\Delta=\{ (x,-x)|x\in \mathbb{R}^n\}\subset \mathbb{R}^{2n}$. Assume that for all $K\Subset\mathbb{R}^n$, the closed set $(K\times\{0\}+\Delta)\cap F$ is always compact. Then the formula
$$(*)\qquad\forall \varphi\in \mathcal{D}(\mathbb{R}^n)\qquad <S*T,\varphi>=<S_x\otimes T_y,\varphi(x+y)>$$
defines a distribution on $\mathbb{R}^n$. It is called the "convolution" of $S$ and $T$. |
H: Calculus: volumes
My tutor gave me this question and I don't know how to tackle this type of question.
Consider the solid of revolution obtained when the region surrounded by $x=2$, $x=3$, $y=1$ and $y=\frac{x}{x+3}$ is rotated about the line $y=1$. Find an expression for the cross-sectional area of this solid at a point $x$.
AI: Here’s a rough sketch of the region, which I’ve labelled $R$.
When you revolve the red line about the line $y=1$, it traces out a disk; this disk is one of the cross-sections of the resulting solid of revolution. The radius of the disk is the length of the red line, which is the $y$-coordinate at the top of the line minus the $y$-coordinate at the bottom. The $y$-coordinate at the top is $1$; the one at the bottom is $\frac{x}{x+3}$. Thus, the radius of the disk is
$$1-\frac{x}{x+3}=\frac3{x+3}\;,$$
and the area of the disk is $$\pi\left(\frac3{x+3}\right)^2=\frac{9\pi}{(x+3)^2}\;.$$ |
H: Convention of digit grouping after decimal point
I read that different cultures have different ways of grouping digits before the decimal point for readability e.g. 1234567890 can be grouped as 1 234 567 890 (English), 12 3456 7890 (Chinese) or 1 23 45 67 890 (Indian), but is there an established convention for grouping digits of a real number after the decimal point for readability e.g. should 3.14159265358979 be grouped as 3.141 592 653 589 79, 3.14159 26535 8979 or something else?
AI: The only convention I have come across is that many publications of mathematical vales tend to group the digits after the decimal point in blocks of 5, so that Pi would appear as 3.14159 26535 89793 ...
You can see examples such as this on Wikipedia, but I have never heard of any official or widespread convention about this. |
H: Probability Density Function: Small question
Let $X \sim N(\mu, \sigma^2)$ and $f_{X} {(x)} = \frac{1}{\sigma\sqrt{2\pi}} \exp \left( -\frac{(x-\mu)^2}{2\sigma^2} \right)$, where $-\infty < x < \infty.$
Find the PDF of $X^3$.
Do you literally just sub $x^3$ into the function given?
AI: Let $Y=X^3$. Find first an expression for the probability that $Y\le y$, that is, for the cumulative distribution function $F_Y(y)$ of $Y$. Since $Y \le y$ iff $X^3\le y$ iff $X \le y^{1/3}$, we want the probability that $X\le y^{1/3}$. Thus
$$F_Y(y)=\Pr(X\le y^{1/3})=\int_{-\infty}^{y^{1/3}} f_X(x)\,dx.$$
To find the density function $f_Y(y)$ of $Y$, differentiate $F_Y(y)$. To do this, use the Fundamental Theorem of Calculus. By the Chain Rule, you will get
$$f_Y(y)=\frac{1}{3}y^{-2/3}f_X(y^{1/3}).$$
If differentiating under the integral sign is no longer a familiar process to you, see the Wikipedia article.
Remark: Note that substituting worked, sort of. Except that it was $y^{1/3}$ that we substituted, and there is an extra term needed for the derivative of $y^{1/3}$. What made the calculation relatively straightforward is that $t^3$ is an increasing function of $t$. Things are less nice if you want the density function of $g(X)$, where $g(t)$ is not a monotone function. A similar strategy works, but the details get more complicated. |
H: Do these two constructions lead to the same set?
Be $A=\bigcup_{n=0}^\infty A_n$ where $A_0=\emptyset$, $A_{n+1}=P(A_n)$.
Be $B=\bigcup_{n=0}^\infty B_n$ where $B_0=\{\emptyset\}$, $B_{n+1}=\{P(X):X\in B_n\}\cup\{X\setminus Y:X,Y\in B_n\}$.
Question: Is $A=B$?
Note: Here $P(X)$ denotes the power set of $X$.
AI: Clearly, $B\subseteq A$, because $A=V_\omega$, the set of all hereditarily finite well-founded sets, and all members of $B$ are hereditarily finite and well-founded. Thus it is enough to show that $A\subseteq B$.
For this, it is enough to show that for each $n$, there exists $m$ such that $A_n\subseteq B_m$. We will prove it by induction with respect to $n$.
for $n=0$, $A_0\subseteq B_0$.
Choose arbitrary $n\geq 0$, and suppose $A_n\subseteq B_{m_n}$ for some $m_n$.
Notice that $B_m$ is nondecreasing.
Notice that $A_n\in B_{n}$ (so $A_n\in B_m$ for all $m\geq n$), so to show that every subset of $A_n$ is a member of some $B_{m}$, it is enough to show that every singleton subset is (because then we can subtract successive singletons from $A_n$ to eventually obtain each subset, so if some $B_m$ has as a member $A_n$ as well as all its singleton subsets, $B_{m+\lvert A_n\rvert}$ will have all subsets of $A_n$).
Choose arbitrary $x\in A_n$. We need to find $m$ such that $\{x\}\in B_m$.
Notice that every subset of $x$ is also a member of $A_n$ (so $B_{m_n}$ too), and that $\{x\}=P(x)\setminus(\bigcup_{y\subsetneq x} P(y))$ (it would be enough to choose $y$ whose complement in $x$ is a singleton, but that does not matter).
Since for every $y\subsetneq x$ we have $P(y)\in B_{m_n+1}$, we also have that $\{x\}\in B_{m_n+1+\lvert P(x)\rvert}$
Therefore, all singletons of elements of $A_n$ are in $B_{m_n+1+\lvert P(A_n)\rvert}$, and all subsets of $A_n$ are in $B_{m_n+1+\lvert P(A_n)\rvert+\lvert A_n\rvert}$.
These bounds are by no means optimal, but that's not what we needed. |
H: (Continued:) finiteness of étale morphisms
I was writing a question, it became too long, and i decided to split it into two parts. I hope posting two questions at the same time is not a problem.
First question: Checking flat- and smoothness: enough to check on closed points?
Now let $f: X \rightarrow Y$ be a morphism of varieties.
If $f$ is smooth of relative dimension 0, i.e. étale, its preimage of a point should be a 0-dimensional regular scheme, i.e. a collection of points. But a zero dimensional union of varieties (i.e. a zero-dimensional scheme that is a variety except that integral is replaced by reduced) always has a finite set as space, right? So
Am i correct to say that fibers of étale morphisms are always finite, i.e. étale implies quasi-finite?
Moreover, i understood that étale morphisms are not always finite, so to finish the picture could you give an example of a non-finite étale morphism?
Thanks a lot!
PS tag "complex-geometry" is included since i'm happy to assume $k=\mathbb{C}$.
AI: For an exemple of a non-finite étale morphism, simply consider an open immersion like $\mathbb{A}^1 \setminus \{0\} \to \mathbb{A}^1$. You can even give an example of a surjective étale non-finite morphism by considering an open covering like $(\mathbb{A}^1 \setminus \{0\}) \amalg (\mathbb{A}^1 \setminus \{1\}) \to \mathbb{A}^1$. |
H: Show that $\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x\, f(x)^2\,dx $
Let $f$ be a nondecreasing, integrable, function defined on $[0, 1]$. Show that $$\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x \,f(x)^2\,dx $$
AI: Fact: Let $a$ and $b$ denote two nondecreasing functions on $[0,1]$, then $\int\limits_0^1ab\geqslant\int\limits_0^1a\cdot\int\limits_0^1b$.
To solve your question, apply this to the functions $a:x\mapsto2x$ and $b=f^2$, then apply Cauchy-Schwarz inequality to get $\int\limits_0^1b\geqslant\left(\int\limits_0^1f\right)^2$.
To prove the Fact recalled above, consider the function $c:(x,y)\mapsto(a(x)-a(y))(b(x)-b(y))$ defined on $[0,1]^2$ and note that $\iint\limits_{[0,1]^2} c=2\int\limits\limits_0^1 ab-2\int\limits_0^1 a\cdot\int\limits_0^1 b$ and that $c\geqslant0$ on $[0,1]^2$. |
H: Inverse of symmetric matrix M = A*At
Possible Duplicate:
Inverse of symmetric matrix $M = A A^\top$
I have a matrix, generated by the product of a non-square matrix with its own transpose:
M = A * A^T
I need the inverse of M, assuming det(M) != 0.
Given the nature of the matrix M, are there any specialised computational methods to find its inverse, prioritising precision over speed? Gauss-Jordan is prone to error, and I hope to find something nicer than and with comparable precision to adj(M^T)/det(M).
I've had a quick read of the Matrix Cookbook and of this page, but (at the present time of 1am) I'm struggling to see how it could help me.
In case it helps, I'm actually trying to calculate:
B = (A * A^T)^-1 * A
AI: Actually, you don't need to calculate (A * A^T)^-1 to compute B. What you are trying to calculate is the left inverse of A^T, and it is give by
$B = \left(A A^T\right)^{-1} A = \left(A^T\right)^{-1_{L}} = \left(A^{TT} A^T \right)^{-1} * A^{TT}$
Since in your case the left inverse exists, it's equivalent to the pseudo-inverse of A^T, which can be computed by SVD. And since the left (right) singular vectors of a matrix are right (left) singular vectors of it's transpose or left/pseudo inverse, and a matrix and its transpose have the same singular values, but the pseudo-inverse has inversed singular values, the pseudo-inverse of A^T has the same singular vectors with A, and has the inversed singular values of A.
So, all you have to do is calculating the SVD of A, and inversing it's singular values, then you will get B. |
H: Checking flat- and smoothness: enough to check on closed points?
I am currently studying varieties over $\mathbb{C}$, i know some scheme theory.
Let $f: X \rightarrow Y$ be a morphism of varieties. If we want to show flatness, is it enough to check the condition only at the closed points of $Y$? If yes, could you give an argument, if not, could you give an example where $f$ is flat over all closed points of $Y$, but not over some non-closed point?
Same question with $f$ smooth of relative dimension $n$, using the following characterization of smooth (Hartshorne III.10.5):
$f$ is called smooth if it is flat and for all $y \in Y$ the "algebraic closure of the fibre" $X_{\overline{y}} := X_y \times_{k(y)} \overline{k(y)}$ is regular (and equidimensional of dimension $n$). Now it must be clear what is meant by smooth "at a point" $y \in Y$.
The question originally continued after this, but i decided to split it since it became too long. Second part: (Continued:) finiteness of étale morphisms
Thanks a lot!
PS tag "complex-geometry" is included since i'm happy to assume $k=\mathbb{C}$.
AI: The answer is yes.
For the flatness: a module $M$ over a ring $A$ is flat if and only if for all maximal ideal $\mathfrak m$ of $A$, $M_\mathfrak m$ is flat over $A_\mathfrak m$.
The smooth locus is open in $X$. If it contains all closed points of $X$, then it is equal to $X$. |
H: Why does this text insist on changing the variable name here?
In What is mathematics? by Courant, Robbins, and Stewart, "5. An important inequality", the authors change $n$ in this example:
$$(1+p)^n\geq1+np$$
to $r$ in this example:
$$(1+p)^r\geq1+rp$$
In other examples given by the book's author, he also switches the variable. I also recall seeing something similar in some other book. Why would one do that?
AI: Courant & Robbins' style in that book was to reserve the letter $n$ in an inductive argument for the general case, like this:
We want to prove that some statement $P(n)$ is true for all natural numbers $n$.
First we prove that it is true for $P(0)$.
Now we assume it is true for some natural number $r$, and we will prove it must also be true for $r+1$. That is, we will show $P(r)\implies P(r+1)$.
Here the authors are trying to make explicit the idea that $n$ represents any natural number, and in their induction step they are choosing one particular natural number $r$. This is simply the way that induction is introduced in the book at the top of page 11.
The essential idea in the preceding arguments is to establish a general theorem $A$ for all values of $n$ by successively proving a sequence of special cases, $A_1, A_2, \dots$ . The possibility of doing this depends on two things: a) There is a general method for showing that if any statement $A_r$ is true then the next statement, $A_{r+1}$, will also be true. b) The first statement $A_1$ is known to be true.
Once you are comfortable with this idea, many people simply use $n$ in the second step as well, remembering that in that second step $n$ represents a particular natural number. |
H: Intuition on proof of Cauchy Schwarz inequality
To prove Cauchy Schwarz inequality for two vectors $x$ and $y$ we take the inner product of $w$ and $w$ where $w=y-kx$ where $k=\frac{(x,y)}{|x|^2}$ ($(x,y)$ is the inner product of $x$ and $y$) and use the fact that $(w,w) \ge0$ . I want to know the intuition behind this selection. I know that if we assume this we will be able to prove the theorem, but the intuition is not clear to me.
AI: Pick $k$ so as to minimize the distance from $kx$ to $y$, or rather its square, noting that
$$\def\Re{\operatorname{Re}} 0\le(y-kx,y-kx)=(x,x)k^2-2\Re((x,y)k)+(y,y)$$
for all $k$. Now pick the $k$ that minimizes the right hand side (in order to get the most out of the inequality), and find that it is the very same $k$ used in the standard proof.
You can determine this $k$ by standard calculus methods, or by completing the square. But first, pick the phase of $k$ to make $(x,y)k$ positive, thus avoiding the difficulty of working with the real part. |
H: Normal subgroup of a group of automorphisms
let $A$ be a non empty set with a given structure, and let $G$ be the group of automorphisms of $A$, that is, the set of bijective maps from $A$ to itself that preserve its structure, with composition as group law.
Is it true that for any $a\in A$, the subgroup $G(a)$ of $G$ consisting of all elements of $G$ that preserve $a$ is a normal subgroup of $G$?
Thanks in advance.
AI: No. Let $A = \{1,2,3\}$ with the trivial structure, so that $G=\operatorname{Aut}(A) = \operatorname{Sym}(A)$ is the full symmetric group. For $a=3$, $G(a)$ is the symmetric group on $\{1,2\}$, which is not normal in $G$. |
H: $f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer.
$f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. find the divisors of $f(n)$ for odd and even values of $n$. Is there a general solution for the divisors.
$$f(1)=7^6-6^6=(7^3)^{2}-(6^3)^{2}$$
$$f(1)=(7-6)(7^2+(7)(6)+6^2)(7^3+6^3)$$
$$f(1)=(1)(127)(7^3+6^3)$$
AI: As you have said, you always have $7^{6n}-6^{6n}=(7^{3n})^2-(6^{3n})^2=(7^{3n}-6^{3n})(7^{3n}+6^{3n})$
As Mark Bennet has said $7^{3n}+6^{3n}=(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
Also, $7^{3n}-6^{3n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})$
So we have $7^{6n}-6^{6n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$
Alpha confirms this is the best we can do for $n=1$, but there are more for $2$ through $5$. |
H: $(x-1)^{p-1} = 1+x+\dots+x^{p−1}$ mod p?
It is well known that in characteristic p we have the "freshman dream" $x^p -1 = (x - 1)^p$. Some experimentation seems to suggest that the heuristic computation $(x-1)(x-1)^{p-1} = (x-1)^p = x^p - 1 = (x-1)(1 + x + \dots + x^{p-1}) \implies (x-1)^{p-1} = 1 + x + \dots + x^{p-1}$ actually yields a correct result.
Playing around with binomial coefficients leads me to believe that $ p-1 \choose k$ = $(-1)^k$ mod p, which would imply the result, but I do not see how to show this. It seems like there should be an easy way, I just do not see it.
AI: $\binom{p-1}{0} \equiv 1$ and $\binom{p-1}{k+1} \equiv \binom{p-1}{k} \frac{p-1-k}{k+1} = \binom{p-1}{k} (-1)$ which proves your claim. |
H: Totally ordered set with greater cardinality than the continuum
Does there exist a totally ordered set $S$ with cardinality greater than that of the real numbers? Sequences are continuous functions with domain $\mathbb{N}$ and paths are continuous functions with domain $\mathbb{R}$; both of them are very important "traversals" of the points inside a space. Would there be another interesting "traversal", a continuous function with domain $S$? In a sense, $S$ would have to be "denser" than $\mathbb{R}$.
AI: As mentioned in the comments, (assuming axiom of choice) there are many examples of linearly ordered sets of cardinality greater than $\mathfrak c$, for example the cardinal $\mathfrak c^+$ (successor of $\mathfrak c$).
Perhaps more interestingly, you can indeed find linear orders that are "denser" than $\mathbf R$ in the intuitive sense. If you take the real interval $[0,1]$, then you can extend it to a dense linear order of arbitrary cardinality while preserving endpoints: you can choose some arbitrary cardinal $\kappa\geq \mathfrak c$ and find $T\supseteq [0,1]$ with linear ordering $\leq$ which agrees with the ordering on $[0,1]$, and has the property that $0$ is the least element of $T$, $1$ is the greatest, and between each pair of distinct elements of $T$ (including any pair of real numbers) there are at least $\kappa$ other elements (e.g. if we choose $T$ to be a $\kappa$-saturated elementary extension of $[0,1]$ in the language $\{0,1,\leq\}$, if you're into model theory).
Of course, finding interesting "traversals" in your sense from such a $T$ for $\kappa>\mathfrak c$ will be quite troublesome, since most commonly used (outside set-theoretical topology) spaces have cardinality at most $\mathfrak c$. |
H: Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy
Suppose the terms of the sequence of real numbers $\{a_n\}$ satisfy $|a_{n+1} - a_n| \le 2^{-n}$ for all $n$. Prove that $\{a_n\}$ is Cauchy.
My Work
So by the definition of a Cauchy sequence, for all $\varepsilon > 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. However, questions like this one make me understand that the $2^{-n}$ condition is necessary for this to be a true statement.
So I am wondering how to appeal to the Cauchy definition for this proof. Do I prove that every convergent sequence is therefore Cauchy, and then try to prove convergence?
AI: As you said, you want to show that for any $\epsilon>0$ there is some $n_0\in\Bbb N$ such that $|a_m - a_n|<\epsilon$ whenever $m, n \ge n_0$. The trick is to figure out how big an $n_0$ you’re going to need to make sure that $|a_m-a_m|<\epsilon$ no matter how far apart $m$ and $n$ are, as long as they’re both at least $n_0$. Okay, suppose that we look at $|a_m-a_n|$ when $m$ and $n$ are not necessarily consecutive. There’s no harm in assuming that $m\le n$; then $k=n-m\ge 0$, and we’re looking at $|a_m-a_{m+k}|$. We only have a handle on the size of this number when $k=1$: if $k=1$, $|a_m-a_{m+k}|\le 2^{-m}$. But we also have the triangle inequality:
$$\begin{align*}
|a_m-a_{m+k}|&=|(a_m-a_{m+1})+(a_{m+1}-a_{m+2})+\ldots+(a_{m+k-1}-a_{m+k})|\\
&\le|a_m-a_{m+1}|+|a_{m+1}-a_{m+2}|+\ldots+|a_{m+k-1}-a_{m+k}|\\
&<2^{-m}+2^{-(m+1)}+\ldots+2^{-(m+k-1)}\\
&<\sum_{k\ge m}\frac1{2^k}\\
&=\frac{\frac1{2^m}}{1-\frac12}\\
&=\frac1{2^{m-1}}\;.
\end{align*}$$
Thus, if $m,n\ge n_0$, we automatically have $|a_m-a_n|<\dfrac1{2^{m-1}}\le\dfrac1{2^{n_0-1}}$. If we choose $n_0$ big enough so that $\dfrac1{2^{n_0-1}}\le\epsilon$, we’ll be in business. Is this always possible? Sure: just make sure that $2^{n_0-1}\ge\dfrac1\epsilon$, i.e., that $n_0\ge\log_2\dfrac2\epsilon$; this is certainly always possible. |
H: Value of functions when approximating area using Riemann sums
When approximating the area beneath the curve $f(x)=x^2+1$ on the interval $[0,3]$ using a left Riemann sum with 4 rectangles, I calculated that the height of each rectangle, in order from left to right, would be
$f(0)$, $f(\frac{3}{4})$, $f(\frac{3}{2})$, $f(\frac{9}{4})$
and therefore, the area would be
$A=\frac{3}{4}\cdot f(0)+\frac{3}{4}\cdot f(\frac{3}{4})+\frac{3}{4}\cdot f(\frac{3}{2})+\frac{3}{4}\cdot f(\frac{9}{4})$ (I'll call this equation 1)
Now I know the next step is
$A=\frac{3}{4}\cdot1+\frac{3}{4}\cdot \frac{25}{16}+\frac{3}{4}\cdot \frac{13}{4}+\frac{3}{4}\cdot\frac{97}{16}=\frac{285}{32}\approx8.906$ (I'll call this equation 2)
But it's this step that confuses me.
How does one know for example that $f(\frac{3}{4})$ from equation 1 goes to $\frac{25}{16}$ in equation 2 and $f(\frac{9}{4})$ from equation 1 goes to $\frac{97}{16}$ in equation 2? (The same goes for the other variables).
I apologise if this is written in a confusing way.
AI: Well $f(x) = x^2+1$
$$f\left(\frac 3 4\right) = \left(\frac 3 4\right)^2+1 = \frac 9 {16} + \frac {16}{16} = \frac {25} {16}$$
and
$$f\left(\frac 9 4\right) = \left(\frac 9 4\right)^2+1 = \frac {81} {16} + \frac {16}{16} = \frac {97} {16}$$ |
H: A problem for Matlab solver
I can numercally solve a system of two ODEs for x and p:
dy1 = -y1 + y2
dy2 = -1 + y2
with the following Matlab code:
%main.m
[t y]=rk4sys(20,0.05);
[t' y]
%rk4sys.m
function [t,y] = rk4sys(n,h)
C = [0;-1];
M = [-1,1;0,1];
y(1, :) = [0, 0];
f =@(t,y)(C+M*y')';
t(1) = 0;
for j = 1 : n
k1 = f(t(j), y(j,:));
k2 = f(t(j)+h/2, y(j,:)+h*k1/2 );
k3 = f(t(j)+h/2, y(j,:)+h*k2/2 );
k4 = f(t(j)+h, y(j,:)+h*k3);
y(j+1,:) = y(j,:)+h*(k1+2*(k2+k3)+k4)/6;
t(j+1)=t(j)+h;
end
end
The main program produces:
ans =
0 0 0
0.0500 -0.0013 -0.0513
0.1000 -0.0050 -0.1052
0.1500 -0.0113 -0.1618
0.2000 -0.0201 -0.2214
0.2500 -0.0314 -0.2840
0.3000 -0.0453 -0.3499
0.3500 -0.0619 -0.4191
0.4000 -0.0811 -0.4918
0.4500 -0.1030 -0.5683
0.5000 -0.1276 -0.6487
0.5500 -0.1551 -0.7333
0.6000 -0.1855 -0.8221
0.6500 -0.2188 -0.9155
0.7000 -0.2552 -1.0138
0.7500 -0.2947 -1.1170
0.8000 -0.3374 -1.2255
0.8500 -0.3835 -1.3396
0.9000 -0.4331 -1.4596
0.9500 -0.4862 -1.5857
1.0000 -0.5431 -1.7183
Now, I have to change the element of (1,3) in the matrix above (that is 0(zero) for now) in order for the last number of the third column (that is -1.7183) to be equal to 0(zero).
In other words, the element (1,3), which is 0(zero) for now, is just my initial guess. I have to change this number to meet that y2 (at t = 21), that is -1.7183 for now, be 0(zero).
I can do this in Excel, but I have failed finding the way to do it in Matlab. Any help would be greatly appreciated.
AI: You are looking for a shooting method.
In other words, you have the following problem description: $$\mathbf{y}'(t) = f(\mathbf{y},t),$$
$$y_1(0) = 0,$$
$$y_2(\tau) = 0.$$
This is a boundary value problem, and you need to transform it into an initial value problem.
To do so, instead of solving the equations above, you are really solving
$$\mathbf{y}'(t) = f(\mathbf{y},t),$$
$$y_1(0) = 0,$$
$$y_2(0) = b.$$
for some estimate of $b$. Your problem is to find $b$ such that the original condition is met.
One technique is known as shooting. You make an initial guess for $b$, compute the error at the right boundary for $y_2(\tau)$, and adjust $b$ based on your error. This is essentially a root-finding problem, and now you can use root-finding techniques to find the "root" $b$.
Essentially, the solution to an ODE initial value problem can be written as $y(t;y_0)$, or the solution $y(t)$ parameterized by the initial conditions $y_0$. This is because, broadly speaking, solutions to ODEs are not unique until you specify initial conditions. So the actual solution trajectory depends greatly on your choice of initial conditions $y_0$.
In your case, you have a second-order system, so you can compute $\mathbf{y}(t;\mathbf{y}_0)$, where $\mathbf{y}_0 = \begin{pmatrix} 0 \\ b\end{pmatrix}$. Since the trajectory that leads you to $y_2(\tau) = 0$ is unique, you can iterate over different values of $b$ to find the right result, and indeed, formulate the problem as solving $F(b) = 0$, where $F$ is your "function" that computes the solution to your ODE. In other words, treat the MATLAB code that you've written above as a function of the parameter $b$ that returns the value $y_2(\tau)$. Then, compute the root.
There are a number of ways to do this; the simplest method, although by far not the fastest, is through bisection. Newton's method may also work, but that is more difficult to set up. Broyden's method (the secant method) will also work.
EDIT: Below is a solution. If this is homework, you must attribute this code properly. This code is presented to show you how shooting works, not to solve the problem for you
The following code can be pasted in a single file called "odetest.m". This is not the highest-quality code; I banged it together to illustrate a point, and I tried to design it in such a way so as to be mathematical in how it was structured. Certainly it can be optimized. That's not the point.
First, we define an RK4 ODE solver. Then, we define a function $F(b)$ such that $F$ returns the value of $y_2(t=1)$. This value depends entirely on the parameter $b$; as such, we wish to find $b$ such that $F(b) = 0$.
We then use Broyden's method, which is a secant method, to compute the estimate for $b$ from an initial guess of both the value and the derivative. Finally, we call this, and then plot some things.
The resulting plot is here:
Broyden's method is not the only way to solve this problem; however, it converges quite quickly and as you can see, it is easy to implement.
function odetest
[i,lambda,lh] = broyden(@F,0,3);
figure;
hold on
colors = 'brgky';
legend_string = {};
for j=1:length(lh)
F(lh(j),1,colors(mod(j-1,length(colors))+1));
legend_string{j} = ['j = ' num2str(j) ', b = ' num2str(lh(j))];
end
legend(gca,legend_string);
plot([0 1],[0 0],'k'); % Plot the x-axis
set(gcf,'PaperPositionMode','auto')
print('-f1','-r360','-dpng','broydenshooting.png');
answer = lambda
end
function x=F(b,ploton,color)
% Extra arguments for plotting mode; when the function is called as a
% function, just use a single argument
ode = @(t,y)[-y(1)+y(2);-1+y(2)];
y0 = [0; b];
T = [0 1];
h = 0.05;
t = T(1):h:T(2);
y = RK4(ode,.05,t,y0);
if nargin > 1
plot(t,y(2,:),color,'LineWidth',2)
end
x = y(2,end);
end
function y=RK4(f,h,t,y0)
n = length(t);
y = zeros(length(y0),n);
y(:,1) = y0;
for i=1:n-1
k1 = f(t(i),y(:,i));
k2 = f(t(i)+h/2, y(:,i)+h/2*k1);
k3 = f(t(i)+h/2, y(:,i)+h/2*k2);
k4 = f(t(i)+h, y(:,i)+h*k3);
y(:,i+1) = y(:,i)+h/6*(k1+2*k2+2*k3+k4);
end
end
function [i,lambda,lh] = broyden(func,IX,IJ)
% lambda = broyden(func,IX,IJ)
% IX -- initial guess for X
% IJ -- initial guess for J = dF/dX
%
% i -- number of steps needed to converge
% lambda -- solution
% lh -- lambda history, the computed lambda at each step
%
% Attempt to compute the solution to a system of nonlinear equations
% without explicitly requiring the computation of the derivatives of the
% function.
lh = [];
lambda = IX;
B = IJ;
for i=1:30
lh = [lh; lambda];
f = func(lambda);
s = B\(-f);
x = s + lambda;
y = func(x)-f;
B = B + ((y-B*s)*s')/(s'*s);
if norm(x-lambda)/norm(lambda) < 1e-10
lambda = x;
break
end
lambda = x;
end
end |
H: Question about Sobolev embedding theorem
The Sobolev embedding theorem as stated in my notes says that if we have $k > l + d/2$ then we can continuously extend the inclusion $C^\infty(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$ to $H^k(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$.
We define $H^k$ to be the closure of $C^\infty$ with respect to the Sobolev norm, see my previous question for the definition.
What I'm confused about is, why we need the condition $k > l + d/2$. What exactly does it give us? If $H^k$ is the closure of $C^\infty$ we already get that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. What am I missing? Thanks for your help.
AI: You said that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. This is true, provided that the topology on $C^\infty$ is inherited from that of $H^k$. In other words, one has to use $H^k$-norms in the domain to define continuity of $T:C^\infty\to C^l$. The condition $k>\frac{n}2+l$ guarantees this continuity when $T$ is the canonical inclusion. |
H: (ZF) Equivalent statements to Baire Category Theorem
So far, I have proved following two for a polish space $X$;
1.If $\{F_n\}$ is a family of closed subset of $X$, where $X=\bigcup_{n\in \omega} F_n$, then at least one $F_n$ has a nonempty inteior.
2.If $G_n$ is a dense open subset of $X$, then $\bigcap_{n\in \omega}G_n ≠ \emptyset$.
I have proved these two respectively, but can't prove the equivalence in ZF. ( I can prove the equivalence in ZF+AC$_\omega$ though)
AI: First we observe that $G_n$ is open dense if and only if $F_n=X\setminus G_n$ is closed and has an empty interior. If $G_n$ is dense it intersects every open sets; so its complement does not contain any open set; and vice versa.
By De-Morgan laws we have that $\bigcap G_n=X\setminus\bigcup F_n$.
If the intersection of open dense is non-empty then the union of closed with empty interior is not everything.
If the union of closed sets with empty interior is not everything, the intersection of open dense is non-empty. |
H: What is the number of triangles with integer sides, given the length of the longest side?
Suppose $a,b,c \in\mathbb N$, and the value of $c$ is known and fixed, while $a$ and $b$ are unknown and are both smaller than $c$. What is the total number of unique triangles possible with $a, b$ and $c$ as its sides?
AI: What is the total number of unique triangles possible with $a$, $b$ and $c$ as its sides?
The formula you need is $$\left\lfloor\frac{(c-1)^2}{4}\right\rfloor$$
As noted here, it counts "the number of noncongruent integer-sided triangles with largest side $c$". See this article as well. |
H: What is the outer measure of Vitali set?
Possible Duplicate:
Vitali-type set with given outer measure
Given that the construction of Vitali set is based on the axiom of choice. How can the outer measure of this set be calculated?
AI: On Vital's request:
There isn't just one Vitali set: each choice of representatives of the equivalence relation on $\mathbb{R}$ given by $x \sim y$ if and only if $y - x \in \mathbb{Q}$ yields what one calls a Vitali set. You can arrange them to have any given positive outer measure you want.
There are many threads on this site where Vitali sets were discussed, among which:
Vitali-type set with given outer measure
Are Vitali sets dense in [0,1)?
Vitali set of outer-measure exactly $1$.
You can find a few more by Googling for "Vitali set" site:math.stackexchange.com |
H: An inequality involving the $\sup$ and $\inf$ of a bounded function and its absolute value
Let $f:[a,b]\to\mathbb{R}$ be a bounded function and set $$M=\sup_{[a,b]}f(x)\,,\; m=\inf_{[a,b]}f(x)\,,\;M^*=\sup_{[a,b]}|f(x)|\,,\;m^*=\inf_{[a,b]}|f(x)|\,.$$ Prove that $M^*-m^*\le M-m$.
First, since $f$ is bounded, all of $M$, $M^*$, $m$, $m^*$ are finite.
Next, we have $M^*-m^*\le M-m$ if and only if $$0\le M-m+m^*-M^*=(M-M^*)+(m^*-m)\,,$$ so it suffices to show that $M-M^*\ge 0$ and $m^*-m\ge 0$.
We obviously have $f\le |f|$, so it follows that $\inf f\le \inf|f|$; that is, $m\le m^*$. So $0\le m^*-m$.
But I can't figure out the other inequality. Thanks!
AI: For any real $x$ and $y$, there holds $||x|-|y|| \leq |x-y|$. If $|f(x_n)| \to M^*$ and $|f(y_n)| \to m^*$, then $||f(x_n)|-|f(y_n)|| \leq |f(x_n)-f(y_n)|$. But $|f(x_n)-f(y_n)| \leq M-m$, and therefore, letting $n \to +\infty$, $M^*-m^* \leq M-m$. |
H: $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y).
$\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
AI: There is a neat trick for solving such equations in positive integers, which is to note that $x$ must be larger than 12 and $y$ larger than 48, so we can put $x=12+s$ and $y=48+t$ with $s$ and $t$ positive integers and $t$ odd. It looks complicated, but it simplifies through.
More generally, for integer solutions of $$\frac a x + \frac by=\frac 1 z$$ where everything is an integer set $x=az+s$ and $y=bz+t$. |
H: Inequality. $3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq (x+y+z)^2(xy+yz+zx)^2$
Let $x,y,z$ be positive real numbers. We have to prove that:
$$3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq (x+y+z)^2(xy+yz+zx)^2.$$
Thanks :)
AI: Note that $x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\ge \frac{3}{4}(x+y)^2 $
Repeating the argument for y and z and then for x and z,we get $$3(x^2+y^2+z^2)(y^2+yz+z^2)(x^2+xz+z^2)\ge \frac{81}{64}(x+y)^2(y+z)^2(z+x)^2$$
or $$8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)\dots (1)$$
iff $\sum x(y-z)^2\ge 0$ which is true! where the sum is taken cyclically over x,y and z.
P.S. Alternatively,you can proceed from (1) as follows:
$(x+y)(y+z)(z+x)=(x+y+z)(xy+xz+yz)-xyz$ which reduces the problem to proving $(x+y)(y+z)(z+x)\ge 8xyz$ which is true by AM-GM. |
H: Trigonometry problem involving oblique triangle
How would I solve the following problem?
A ship sails $15$ miles on a course $S40^\circ10'W$(south 40 degrees 10 minutes west) and then $21$ miles on a course $N28^\circ20'W$(north 28 degrees 20 minutes west). Find the distance and direction of the last position from the first.
I think that to solve this problem I must make an oblique triangle. One side of the triangle would be $15$ and have an angle of $40^\circ10'$ and the other side would be $21$ and have a angle of $28^\circ20'$.
I think what I have to find is the side which connect these two sides.
AI: The distances sailed are small enough that the curvature of the Earth makes no significant difference and we can use plane trigonometry.
Let $P$ be our start point, $Q$ where we change course, and $R$ the end point. First choose $P$. Then draw a North-South thin line through $P$ as a guide. To reach $Q$ we turned $40^\circ 10'$ clockwise from due South, and sailed $15$ km. Draw $Q$. Draw a thin North-South line through $Q$ as a guide. We sail $21$ km in a direction $28^\circ 20'$ counterclockwise from due North. Draw the point $R$ that we reach.
By properties of transversals to parallel lines (our two guide lines),
$$\angle Q=40^\circ 10'+28^\circ 20'=68^\circ30'.$$
We can now find the required distance $PR$ we are from our start position by using the Cosine Law. For
$$(PR)^2=15^2+21^2-2(15)(21)\cos 68^\circ30'.$$
(I get $PR\approx 20.86$, but my calculations are not to be trusted.)
Now for the direction. We will know everything once we know $\angle A$. For this, we could use the Cosine Law, but the Sine Law is easier. We have
$$\frac{\sin A}{21}=\frac{\sin 68.5^\circ}{PR}.$$
I get (but again don't trust me) that $A\approx 69.5^\circ$.
If we use the North-South guideline, our position at $R$, as viewed from $P$ is obtained by facing due South and turning clockwise through about $40^\circ10'+69^\circ 30'$. To express this in the notation that your problem was put, subtract from $180^\circ$. We get $70^\circ 20'$. So $R$ is North, $70^\circ 20'$ West from $P$. |
H: Understanding algebraic method of successive approximations to solve quadratic equations
I have straight away copied and pasted a worked example from john birds higher engineering mathematics {page 80 problem 4} which reads as follows.
Use an algebraic method of successive
approximations to determine the value
of the negative root of the quadratic equation:
$4x^2 −6x −7=0$ correct to 3 significant figures.
Check the value of the root by using the quadratic
formula.
The solution goes like this.
A first estimate of the values of the roots is made by
using the functional notation method
$f(x) = 4x2 − 6x − 7$,
$f(0) = 4(0)2 − 6(0) − 7 = −7$
$f(−1) = 4(−1)2 − 6(−1) − 7 = 3$
These results showthat the negative root lies between
0 and −1, since the value of $f(x)$ changes sign
between $f(0)$ and $f(−1)$ (see Section 9.1). The procedure
given above for the root lying between 0 and
−1 is followed.
What i'am not able to understand is this.
**First approximation
(a) Let a first approximation be such that it divides
the interval 0 to −1 in the ratio of −7 to 3, i.e.
let $x_{1}=−0.7$.**
What do they mean when they say "Let the first approx be such that it divides the interval 0 to -1 in the ratio of -7 to 3"?
AI: Answer in the initial spirit (supposing proportionality) :
$$\begin{array} {c|ccccc}
x&-1&&x_0&&0\\
\hline\\
f(x)&3&&0&&-7\\
\end{array}$$
$x_0$ must be at a place in $(-1,0)$ such that the ratio $−7$ to $3$ (on both sides of $f(x)=0$) is respected. Since the increase in $x$ is $1$ while the increase in $f(x)$ is $-10$ this means that $3$ and $-7$ must be replaced by an offset around $x_0$ of $-0.3$ and $0.7$ respectively getting $x_0-0.3=-1$.
(I started from $x=-1$ while they started from $0$ but the principle is the same!)
Visual answer (we want $x_0$ starting from $-1$) :
$$\begin{array} {c|lccccc}
&|\rightarrow&&+1&&\rightarrow|\\
&&\Delta_x=0.3&&&\\
x&-1&&x_0&&0\\
\hline\\
f(x)&3&&0&&-7\\
&&\Delta_l=-3&&\Delta_r=-7&\\
&|\rightarrow&&-10&&\rightarrow|\\
\end{array}$$
Observe that $\ \frac {\Delta_l}{\Delta_l+\Delta_r}=\frac {-3}{-10}=0.3\ $ so that $\ \frac {\Delta_x}{1}=0.3\ $ and $\ x_0=-1+\Delta_x=-0.7$
Text : Suppose that we move in a straight line from $f(-1)=3$ to $f(0)=-7$ (this is an interpolation!) then $x$ will progress of $\ 1=0-(-1)\ $ while $f(x)$ progresses of $\ -10=-7-(3)$. If you want $f(x_0)=0$ then $f(x)$ must progress of $-3=0-3$ while $x$ will progress of $1\cdot \frac {-3}{-10}=0.3$ to get to $-1+0.3=-0.7$.
Algebraic :
Suppose that $f$ is linear : $f(x)=ax+b$ with $f(-1)=3$ and $f(0)=-7$ then :
$-a+b=3$ and $b=-7$ so that (getting $a$ and $b$) we obtain $f(x)=-10x-7$.
So that $f(x_0)=0$ for $x_0=?$ |
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