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H: Permutations: Given $P^4$, how many $P^1$s are possible?
Let $P^0$ be the identity tuple $(1,2,...,N)$
Let $P^{i+1}$ be the tuple after a permutation $P$ is applied to $P^i$.
For example, if $P$ is $(2,1,3,6,4,5)$ than:
$$\begin{align}
P^0 &= (1,2,3,4,5,6) \\
P^1 &= (2,1,3,6,4,5) \\
P^2 &= (1,2,3,5,6,4) \\
P^3 &= (2,1,3,4,5,6) \\
\dots
\end{align}$$
Given the value $P^4$, how many possible values of $P$ are there?
AI: We can restate the question as: how many "fourth roots" does a given permutation have? That is, given a permutation $\sigma$, how many permutations $\tau$ exist such that $\tau^4 = \sigma$? The answer — let's call it $N(\sigma)$ — depends on the structure of $\sigma$.
(The basic idea of what to look at comes from section 4.8 of generatingfunctionology (e.g. here).)
Consider a permutation $\tau$, decomposed into cycles. For a particular cycle in $\tau$ of length $n$, let's look at what happens in $\tau^4$. If $(a_0, a_1, \dots...)$ is a cycle in $\tau$ (that is, $\tau$ takes $a_0$ to $a_1$, takes $a_1$ to $a_2$, etc.), then $\tau^4$ takes $a_0$ to $a_4$, takes $a_1$ to $a_5$, etc. The resulting (sub)permutation may be a single cycle or multiple cycles, depending upon $n$:
If $n$ is odd (relatively prime to $4$), then we have a single cycle of length $n$, which looks like $(a_0, a_4, a_8, \dots)$ with the indices looping around modulo $n$.
If $n$ is even but not a multiple of $4$, then we get two cycles of length $n/2$, one looking like $(a_0, a_4, a_8, \dots)$ and containing all the even indices, and the other looking like $(a_1, a_5, a_9 \dots)$ and containing all the odd ones.
If $n$ is a multiple of $4$, then we get four cycles of length $n/4$ each, namely $(a_0, a_4, \dots)$, $(a_1, a_5, \dots)$, $(a_2, a_6, \dots)$, and $(a_3, a_7, \dots)$.
Turning this around, we can look at a cycle of length $m$ in $\tau^4$ and say the following:
If $m$ is even, then it must have come, along with three others, as the result of a single cycle of length $4m$ splitting into four cycles of length $m$ each.
If $m$ is odd, then either:
One cycle of length $m$ gave rise to one cycle of length $m$,
One cycle of length $2m$ gave rise to two cycles of length $m$ each, or
One cycle of length $4m$ gave rise to four cycles of length $m$ each.
(This immediately gives a condition for $N(\sigma)$ to be positive, i.e. for the permutation $\sigma$ to have at least one fourth root: for every even number $m$, the number of cycles of length $m$ must be a multiple of $4$.)
We can now calculate $N(\sigma)$. For each number $m$, let the number of cycles of length $m$ in $\sigma$ be $c_m$. Then:
If $m$ is even, then ($c_m$ must be a multiple of $4$ and) we can reconstruct a list of $c_m/4$ original cycles of length $4m$ each, by picking $c_m/4$ ordered groups of four, with ordering among them and cyclic order within each group of four being irrelevant, the number of ways of doing which is $$\frac{c_m!}{(c_m/4)!4^{c_m/4}}$$ After picking a particular group of four cycles, we can make a big one out of them (find a fourth root of the product of these four cycles) by choosing where in the cyclic ordering we start in each of the four cycles. WLOG we can fix a point in the first cycle where we start so the choice is only among the other three, so we must multiply the above number by $m^3$.
If $m$ is odd, then for each integer partition of $c_m$ of the form $$c_m = x + 2y + 4z$$ with $x$, $y$ and $z$ being nonnegative integers, we can pick $x$ cycles of length $m$ by themselves, form $y$ pairs of cycles, and $z$ groups of four. The number of ways of doing this is $$\frac{c_m!}{x! (y!2^y) (z!4^z)}.$$ Again in each of the $y$ pairs we have $m$ ways of putting them together, and in each of the $z$ quadruplets we have $m^3$ ways of putting them together.
So using all that, the final expression for $N(\sigma)$ is:
$$N(\sigma) = \left(\prod_{2|m}\frac{c_m!m^{3c_m/4}}{(c_m/4)!4^{c_m/4}}\right) \left(\prod_{2\not|m}\sum_{x,y,z:c_m=x+2y+4z}\frac{c_m! m^y m^{3z}}{x! (y!2^y) (z!4^z)}\right)$$
I've checked this with a computer program for all permutations of size up to 12, so I'm finally convinced the expression is correct.
Examples: Consider your original permutation $\tau = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
2 & 1 & 3 & 6 & 4 & 5\end{pmatrix}$. We can write this in cycle notation as $\tau = (1 2) (3) (4 6 5)$. For this permutation, $\tau^4 = (1)(2)(3)(4 6 5)$. Notice how the cycle of length $2$ has split into two cycles of length $1$ each. If we want to start with this permutation $\sigma = (1)(2)(3)(4 6 5)$ and find fourth roots, we must look at the number of cycles $c_m$ for each $m$:
$m = 1$: $c_m = 3$. We're looking at just the $(1)(2)(3)$ part.
For the partition $3 = x + 2y + 4z = 1 + 2(0) + 4(0)$, we have only one ($\frac{3!}{3!} = 1$) way of picking three individual $1$-cycles: the identity permutation $(1)(2)(3)$ itself.
For the partition $3 = x + 2y + 4z = 1 + 2(1) + 4(0)$, we have three ways ($\frac{3!}{1!2!2^1} = 3$) of picking a single $1$-cycle and a pair, corresponding to the fourth roots $(1)(2 3)$, $(2)(1 3)$ and $(3)(1 2)$ respectively.
$m = 3$: $c_m = 1$. We're looking at the $(4 6 5)$ part.
There's only one partition $1 = 1 + 2(0) + 4(0)$ and only one way of picking a single element out of a single one, and here the fourth root of $(4 6 5)$ happens to be $(4 6 5) itself.
So the four fourth roots of $\sigma = (1)(2)(3)(4 6 5)$ are
$(1)(2)(3)(4 6 5) = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 2 & 3 & 6 & 4 & 5 \end{pmatrix}$,
$(1)(2 3)(4 6 5) = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 3 & 2 & 6 & 4 & 5 \end{pmatrix}$,
$(2)(1 3)(4 6 5) = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 2 & 1 & 6 & 4 & 5 \end{pmatrix}$,
$(3)(1 2)(4 6 5) = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
2 & 1 & 3 & 6 & 4 & 5 \end{pmatrix}$.
Let $\sigma = (1 2 3)(4 5 6) = \begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
2 & 3 & 1 & 5 & 6 & 4\end{pmatrix}$. We can either choose to keep the two $3$-cycles separate ($x = 2$, $y = 0$), giving us the root $(1 2 3)(4 5 6)$ itself, or we can choose to put the pair together ($x = 0, y = 1$), where the choice of the pair is unique but we can interleave the two in three ways — basically the form $(1 ? 3 ? 2 ?)$ with the '?'s filled by $(4 6 5)$ — giving us the roots $(1 4 3 6 2 5)$, $(1 5 3 4 2 6)$ and $(1 6 3 5 2 4)$. So we have four fourth roots.
Let $\sigma = (1 2)(3 4)(5 6)(7 8 9)$. In this case for $m = 2$, $c_m = 3$ is not a multiple of $4$, so $N(\sigma) = 0$.
Let $\sigma = \begin{pmatrix}1 & 7\end{pmatrix} \begin{pmatrix}2\end{pmatrix} \begin{pmatrix}3 & 16 & 9\end{pmatrix} \begin{pmatrix}4 & 10\end{pmatrix} \begin{pmatrix}5\end{pmatrix} \begin{pmatrix}6 & 18\end{pmatrix} \begin{pmatrix}8\end{pmatrix} \begin{pmatrix}11 & 15 & 13\end{pmatrix} \begin{pmatrix}12 & 17\end{pmatrix} \begin{pmatrix}14\end{pmatrix}$. Counting the fourth roots of this permutation by brute-force enumeration is impossible, but we can easily do it by collecting cycles of the same length:
For $m = 1$: the roots of $\begin{pmatrix}2\end{pmatrix} \begin{pmatrix}5\end{pmatrix} \begin{pmatrix}8\end{pmatrix} \begin{pmatrix}14\end{pmatrix}$ are
(all separate) $\begin{pmatrix}2\end{pmatrix} \begin{pmatrix}5\end{pmatrix} \begin{pmatrix}8\end{pmatrix} \begin{pmatrix}14\end{pmatrix}$,
(two pairs) $\begin{pmatrix}2 & 5\end{pmatrix} \begin{pmatrix}8 & 14\end{pmatrix}$, $\begin{pmatrix}2 & 8\end{pmatrix} \begin{pmatrix}5 & 14\end{pmatrix}$, $\begin{pmatrix}2 & 14\end{pmatrix} \begin{pmatrix}5 & 8\end{pmatrix}$,
(all together) $\begin{pmatrix}2 & 5 & 8 & 14\end{pmatrix}$, $\begin{pmatrix}2 & 5 & 14 & 8\end{pmatrix}$, $\begin{pmatrix}2 & 8 & 5 & 14\end{pmatrix}$, $\begin{pmatrix}2 & 8 & 14 & 5\end{pmatrix}$, $\begin{pmatrix}2 & 14 & 5 & 8\end{pmatrix}$, $\begin{pmatrix}2 & 14 & 8 & 5\end{pmatrix}$.
So there are $1 + 3 + 6 = 10$ roots of $\begin{pmatrix}2\end{pmatrix} \begin{pmatrix}5\end{pmatrix} \begin{pmatrix}8\end{pmatrix} \begin{pmatrix}14\end{pmatrix}$.
For $m = 2$, the roots of $\begin{pmatrix}1 & 7\end{pmatrix} \begin{pmatrix}4 & 10\end{pmatrix} \begin{pmatrix}6 & 18\end{pmatrix} \begin{pmatrix}12 & 17\end{pmatrix}$ are $6 \times 8 = 48$ in number (pick one of the $3!$ orderings of the last three, then one of the $2^3$ orderings of each of the three).
For $m = 3$, the roots of $\begin{pmatrix}3 & 16 & 9\end{pmatrix} \begin{pmatrix}11 & 15 & 13\end{pmatrix}$ are
(both separate) $\begin{pmatrix}3 & 16 & 9\end{pmatrix} \begin{pmatrix}11 & 15 & 13\end{pmatrix}$
(both together) $\begin{pmatrix}3 & 11 & 9 & 13 & 16 & 15\end{pmatrix}$, $\begin{pmatrix}3 & 15 & 9 & 11 & 16 & 13\end{pmatrix}$, $\begin{pmatrix}3 & 13 & 9 & 15 & 16 & 11\end{pmatrix}$.
So there are $1 + 3 = 4$ roots of $\begin{pmatrix}3 & 16 & 9\end{pmatrix} \begin{pmatrix}11 & 15 & 13\end{pmatrix}$.
So totally $N(\sigma) = 10 \times 48 \times 4 = 1920$. |
H: Determine convergence of $\sum_{n=1}^\infty\frac{\sin(na)}{n^2} $
I want to detemine the convergence of the next series:
$$\sum_{n=1}^\infty\frac{\sin(na)}{n^2} $$
I've solved the limit:
$$\lim_{n->\infty}\frac{\sin(na)}{n^2}=\frac{[ -1,1]}{\infty}=0$$
The series has the necesary condiction of convergence (limit=0), but I dont go further from here
AI: Use the comparison test. Since $\left|\sin(na)\right| \le 1$, we have:
$$
\left| \frac{\sin(na)}{n^2} \right| \le \left| \frac{1}{n^2} \right|
$$
We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges absolutely, so your series does too. |
H: Can there be a function that's even and odd at the same time?
I woke up this morning and had this question in mind. Just curious if such function can exist.
AI: Others have mentioned that $f(x)=0$ is an example. In fact, we can prove that it is the only example of a function from $\mathbb{R}\to \mathbb{R}$ (i.e a function which takes in real values and outputs real values) that is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Thus $-f(x) = f(x)$, so $f(x)=0.$ |
H: Solve the equation $3x+7=6$ in $\mathbb{Z}_{13}$
The title is exercise 2.2 in The Fundamental Theorem of Algebra.
The hint for the problem is: Find the value of $\frac{1}{3}$ in $\mathbb{Z}_{13}$
(please realize that my knowledge of the subject is what I read in Chapter 2)
I have gotten that $x = -\frac{1}{3}$.
I know that $\mathbb{Z}_{13}$ is the integers modulo 13. Thus $x \equiv n \pmod{13}$? for some integer n, $0\le n < 13$.
Thus for $x – n = km$ for some integer $k$, with $m \neq 0$ and an integer.
How does m divide $(x-n)$, when $(x-n)$ is not an integer, if $x = -\frac{1}{3}$????
AI: The notation can indeed be confusing the first few times one encounters in. When one says $x=-1/3$ in the context of $\mathbb{Z}_{13}$, it is simply short-hand for saying that $x$ is some number from $0$ to $12$ such that $-3x \equiv 1 \pmod{13}.$ So remember that $x$ is not the usual real number $-1/3$, but we use this notation because it behaves like $-1/3$ would, in that $-3x=1$ in $\mathbb{Z}_{13}.$ |
H: Partial Integration - Where did I go wrong?
For a Homework, I need $\int \frac{x}{(x-1)^2} dx$ as an intermediate result. Using partial integration, I derive $x$ and integrate $\frac{1}{(x-1)^2}$, getting: $$ \frac{-x}{x-1} + \int \frac{1}{x-1} dx = \ln(x-1)+\frac{x}{x-1} $$
WolframAlpha tells me this is wrong (it gives me $\frac{1}{1-x}$ where I have $\frac{x}{x-1}$). If I and WA disagree the error is usually somewhere on my side. Unfortunately WA uses partial fractions there instead of partial integration, so I'm not sure which step I screwed up. Supposedly $\int f'g dx = fg - \int fg' dx$ right?
(I leave the constant +C out because it's not relevant for the problem I need this for).
AI: You’ve a sign error: you should have $$\ln(x-1)-\frac{x}{x-1}\;.$$
Now note that
$$\frac{x}{x-1}=1+\frac1{x-1}\;,$$
so your (corrected) answer can be written $$\ln(x-1)-1-\frac1{x-1}=\ln(x-1)+\frac1{1-x}-1\;.$$
Your (corrected) answer differs from WA’s by a constant, which is absorbed in the constant of integration; both are perfectly good antiderivatives of the given function. |
H: Sequence of Lipschitz functions
Let $\{f_{n}\}$ be a sequence of positive continuous functions on $\mathbb R$; $f_{n}:\mathbb R\to \mathbb R$, for all $n\geq 1$, with the folloing properties:
(1) $\{f_{n}\}$ is uniformly bounded by some constant $C>0$,
(2) $\{f_{n}\}$ is uniformly Lipschitz on $\mathbb R$ (so it is uniformly continuous)
(3) $\{f\,'_{n}\}$ is uniformly bounded (sequence of the derivatives), which I think follows from (2).
Does this sequence converges on $\mathbb R$ (or at least contains a subsequence which converges), uniformly on compact sets, or pointwise, to some continuous function $f$ ? If not, what extra (possible) condition must the sequence have to converge to some continuous function?
AI: By Arzela-Ascoli's theorem your sequence has a uniformly convergent subsequence (the limit of which will, of course, be continuous) on each compact subset. You cannot hope for convergence of the sequence itself, though. |
H: Evaluation of $\Xi(z)=\sum_{t=1}^{\infty}\frac{t^z}{e^t}$
I would like to try and evaluate the following gamma function inspired sum.
$$\Xi(z)=\sum_{t=1}^{\infty}\frac{t^z}{e^t}$$
According to my computations, for large $z$,
$$\Xi(z)\approx\Gamma (z+1)$$
and perhaps even
$$\Xi(z) \sim \Gamma (z+1)$$
Does a closed form exist for this sum?
AI: Your sum is in the form of the Polylogarithm. In fact, it is equal to $\operatorname{Li}_{-z}(1/e).$ When $z$ is a negative integer, this sum is easily computed in closed form using the identity $\displaystyle \frac{d}{dx} \operatorname{Li}_n (x) = x\operatorname{Li}_{n-1} (x).$
By applying the Abel-Plana summation to the Polylogarithm series, we get
$$\operatorname{Li}_s(z) = {z\over2} + {\Gamma(1 \!-\! s, -\ln z) \over (-\ln z)^{1-s}} + 2z \int_0^\infty \frac{\sin(s\arctan t \,- \,t\ln z)} {(1+t^2)^{s/2} \,(e^{2\pi t}-1)} \,\mathrm{d}t
$$
and so $$\operatorname{Li}_{-z}(1/e) = \frac{1}{2e} + \Gamma(z+1,1) + \frac{2}{e} \int^{\infty}_0 \frac{ (1+t^2)^{z/2} \sin(-z \tan^{-1}t+t)}{e^{2\pi t} -1} dt .$$
where $\Gamma(s,x)$ is the incomplete gamma function. Your asymptotic would be explained if you could show why the remaining integral is comparatively small. |
H: $ A_0^a B_0^b + A_1 ^a B_1 ^b \leqslant (A_0 + A_1 )^a (B_0 + B_1 )^b$ given $A_0 , B_0 , A_1, B_1 \geqslant 0 $ ; $0 \leqslant a,b <1 $
Let $A_0 , B_0 , A_1, B_1 \geqslant 0 $ and let $0 \leqslant a,b <1 $. Then prove that $$ A_0^a B_0^b + A_1 ^a B_1 ^b \leqslant (A_0 + A_1 )^a (B_0 + B_1 )^b$$
AI: $A_0 = A_1 = B_0 = B_1 = x$ and $a+b < 1$.
Then $$A_0^a B_0^b + A_1^a B_1^b = 2x^{a+b}$$
$$(A_0 + A_1)^a (B_0 + B_1)^b = (2x)^{a+b}$$
Since $a+b < 1$, we have $$(A_0 + A_1)^a (B_0 + B_1)^b = (2x)^{a+b} < 2x^{a+b} = A_0^a B_0^b + A_1^a B_1^b$$
contradicting your claim. |
H: What is the simplest proof that the mutual information $I(X:Y)$ is always non-negative?
What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)\ge0$
AI: By definition,
$$I(X;Y) = -\sum_{x \in X} \sum_{y \in Y} p(x,y) \log\left(\frac{p(x)p(y)}{p(x,y)}\right)$$
Now, negative logarithm is convex and $\sum_{x \in X} \sum_{y \in Y} p(x,y) = 1$, therefore, by applying Jensen Inequality we will get,
$$I(X;Y) \geq -\log\left( \sum_{x \in X} \sum_{y \in Y} p(x,y) \frac{p(x)p(y)}{p(x,y)} \right) = -\log\left( \sum_{x \in X} \sum_{y \in Y} p(x)p(y)\right) = 0$$
Q.E.D |
H: Did Structuralism influence the formulation of Category Theory?
Having only the a very cursory knowledge of Structuralism ( it's a movement generally held to have originated in linguistics, then moving on to philosophy & literature), there does appear to be some points of coincidence:
Structuralism (from wikipedia):
Individual elements of culture must be placed within a System/Structure.
The individual elements of culture must be understood by their inter-relationships within this System, and not by their individual identity, that is their identity is supressed.
and compare this with Category Theory:
Individual objects of mathematical interest must be placed with a Category.
Though these individual objects have their own character, this knowledge is supressed, and only their relationships (morphisms) have import.
There seems to me a clear correspondance here. Of course, it could mean that both paradigms evolved independently from some prior philosophy.
Some more evidence from Structuralism, by Sturrock:
'What is a structure, then, for Husserl, and 'in general'? The broadest definition is that a structure is an abstract model of organisation including a set of elements and the law of their composition...What stands out in a structure is that the relationships between the elements are more important than the intrinsic qualities of each element'.
and the definition of a category can be further elaborated as:
3.Morphisms between objects (i.e. the relationship) follow a law of composition.
Further, if I recall correctly Saunders MaClane remarks in the introduction to Categories for the Working Mathematician, that he purloined the term Category from Kant. I don't think, though, that Kant had any input or influence on Structuralism.
[I've asked this question on Philosophy.StackExchange but have no response from them]
AI: There is a link, but it is not as straightforward as what you describe. Insofar as there has been an influence at all between structuralism in the human and social sciences and the pre-eminence given to the study of structures in mathematics, this influence has predominantly been between the pre-categorical point of view on structures in mathematics and structuralism in the social sciences. Moreover, this influence, if any at all, has been of mathematics on social sciences more than of social sciences on mathematics, as the title of your question might suggest.
To put it more concretely, people like Levi-Strauss, Lacan, Foucault and especially Piaget (to take some emblematic examples of structuralism in anthropology, psycholoy, sociology and epistemology) might have been influenced by their proximity-social and intellectual-with the Bourbaki group and might have thus adapted part of their intellectual framework from the structuralist point of view of Bourbaki, in the sense of the slogan of Dieudonné that "each structures carries with it its own language". I write might because, with the exception of Piaget, it is unclear how much real influence there has actually been and whether this influence amounted to much more than similar choices in some terminological terms. It is very unlikely that there has been any influence in the other direction, and in particular, there is no evidence that I know of that there has been any influence from the social sciences in the foundation of the theory of categories.
Interestingly, apart from the case of Piaget, who really did organize his theory of learning in analogy with the logical organization of mathematics in the Bourbaki style, the deepest and most significant link that I can think of between structuralism in mathematics and in the social science is about Foucault, who was a student of Canguilhem, himself a close friend of Cavaillès, who was very influential (at the time, that it is to say the 1920s/1930s) in mathematics and philosophy at École Normale Supérieure. |
H: How to find every possible scalar product on $V$
I am given the following task:
"For which $a$, $b \in \mathbb{R}$ there exists a scalar product, such that
$$A = \left( \begin{matrix} 0 & 0 & a b \\ 1 & 0 & a \\ 0 & 1 & b\end{matrix}\right) $$
is a self-adjoint matrix".
A hint says, the task breaks down in finding $a$, $b$ such that $A$ is diagonalizable. But I can't figure out why.
As far as I got is, that Self-adjoint means that $\left< A v, w\right> = \left< v, A w\right> \forall v, w \in V$. So we are looking for $a,b$ such that there is a scalar product on $V$ such that this equality holds. I know that for every Bilinearform $f$ there is a Matrix $B$ such that $f(v,w) = <v,Bw>$ in means of the canonical scalar product. Is there a similar way to represent every scalar product in means of the canonical scalar product? And anyone has a hint why this should break down to find $a,b$ such that $A$ is diagonalizable?
Thanks for any help
AI: Every self-adjoint matrix can be diagonalized in an orthonormal basis and has real eigenvalues, that's the spectral theorem for matrices.
On the other hand, the converse is also true (it's easy to check), and for every basis, we can construct an inner product with respect to which it is orthonormal (just by inducing it through a bijection with $\mathbf C^n$ which maps the basis onto the standard basis of $\mathbf C^n$), so in particular, the basis diagonalizing a given matrix can be made orthonormal.
Therefore, a matrix is self-adjoint for some inner product iff it is diagonalizable, and its eigenvalues are all real.
As to your second question, a scalar product is really just a positive-definite sesquilinear form (bilinear in the real case) that is conjugate-symmetric (or just symmetric in the real case), so you really have answered it. |
H: Truncation in Lorentz spaces
I am reading a paper, whose author state the following: if $f \in L^{(q,\infty)}(\mathbb{R}^N)$, then $f_\delta \in L^p(\mathbb{R}^N)$ for every $p \in [1,q)$, where $\delta > 0$ and
$$
f_\delta = f\; \boldsymbol{1}_{\{x\in X: f(x) \geq \delta\}}.
$$
But is this really true?
AI: I will use the following fact
Lemma. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space and $p\in[1,+\infty)$. Then for measurable non-negative function $f$ and measurable set $A\in M$ we have
$$
\int\limits_A f(x)^p d\mu(x)=p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt
$$
where $F_{f,A}(t)=\mu(\{x\in A:f(x)>t\})$
Proof. Using Fubini theorem for positive functions we can say that
$$
\int\limits_A f(x)^p d\mu(x)=
\int\limits_A \int\limits_{(0,f(x))}pt^{p-1}dt d\mu(x)=
\int\limits_A \int\limits_{(0,+\infty)}pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)dt d\mu(x)=
$$
$$
\int\limits_{(0,+\infty)}\int\limits_A pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
\int\limits_{(0,+\infty)}pt^{p-1}\int\limits_A \boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
$$
$$
\int\limits_{(0,+\infty)}pt^{p-1}\mu(\{x\in A:f(x)>t\}) dt =
\int\limits_{(0,+\infty)}pt^{p-1}F_{f,A}(t) dt
$$
Theorem. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space, and $f\in L^{(q,\infty)}(X,M,\mu)$. Then for all $\delta>0$ the function $f_\delta = f\; \boldsymbol{1}_{\{x\in X:f(x)>\delta\}}$ is in $L^p(X,M,\mu)$.
Proof. From definition of quasi-norm in Lorentz space we see that there exist $C>0$ such that
$$
\mu(\{x\in X:|f(x)|>t\})\leq \frac{C}{t^q}
$$
Denote $A=\{x\in X: f(x)>\delta\}$. Since $\delta>0$, then the function $f_\delta$ is non-negative and measurable. Then from previous lemma for $p\in[1,q)$ we have
$$
\Vert f_\delta\Vert_p^p=
\int\limits_{X} |f_\delta(x)|^pd\mu(x)=
\int\limits_A f(x)^pd\mu(x)=
p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt
$$
Note that for $t\in(0,\delta)$ we have $\mu(\{x\in A:f(t)>t\})=0$ so
$$
\Vert f_\delta\Vert_p^p=
p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,A}(t)dt\leq
p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,X}(f)dt\leq
p\int\limits_{[\delta,+\infty)}t^{p-1}\frac{C}{t^q}dt=
Cp\int\limits_{[\delta,+\infty)}\frac{1}{t^{q+1-p}}dt
$$
Since $p<q$ the last integral is convergent, so $\Vert f_\delta\Vert_p <+\infty$ and $f_\delta\in L^p(X,M,\mu)$. |
H: Rectangles Diagonal Calculation
I was having a problem with the following question, and could use some help:
If a rectangle with a perimeter of 48 inches is equal in area to a right triangle with legs of 12 inches and 24 inches, what is the rectangle's diagonal?
The answer to the above question is $12\sqrt{2}$. Frankly I am a bit confused by a part of the question stating
in area to a right triangle with legs of 12 inches and 24 inches
Is the area of rectangle equal to the area of the triangle? Frankly, the phrase "equal to triangle" doesn't say much. And 12 and 24 the lengths of which sides of a triangle?
Am I missing something here or are my concerns valid?
Edit:
After reading the suggestions posted here, here is what I did, and I am getting a square instead of a rectangle:
$$2x + 2y = 48 \qquad (A)$$
$$xy = 144 \qquad (B)$$ so $x=144/y$, inserting in $(A)$ I get $x=12$. So is this actually a square?
AI: There is a specific rectangle that the question is referring to, but we don't know all of its properties. The question tells us that this rectangle has a perimeter of 48 inches. When the question says that this rectangle is
equal in area to a right triangle with legs of 12 inches and 24 inches
it just means that the area of this rectangle is equal to the area of such a triangle.
The area of a shape doesn't depend on how you orient it. If you look at a right triangle that has legs of lengths $A$ and $B$, and draw it like this:
then the "height" is $B$, the "base" is $A$, and the area is $$\frac{1}{2}\times \text{base}\times\text{height}=\frac{1}{2} AB.$$
If you take the same right triangle and draw it like this:
then the "height" is $A$, the "base" is $B$, and the area is $$\frac{1}{2}\times \text{base}\times\text{height}=\frac{1}{2} BA=\frac{1}{2}AB.$$
So the area of a right triangle that has legs of length $A$ and $B$ is always $\frac{1}{2}AB$.
Your calculations are correct; the rectangle is a square. Note that a square is a quadrilateral with four right angles, and therefore is a rectangle (all squares are rectangles, though of course not all rectangles are squares).
Now, it should be fairly straightforward to show that, for a square with side lengths of $12$ inches, the length of the diagonal is $12\sqrt{2}$ inches (hint: use the Pythagorean theorem). |
H: An Inequality involving integrations (Hölder-like).
Let $ p,r \geqslant 1, \; f \in L^r (\mathbb R), g \in L^p (\mathbb R), \; 2/q = 1/p + 1/r$. Here $ 2p / q > 1 $ and $ 2r / q > 1 $ . Also $ \frac{q}{2p} + \frac{q}{2r} = 1$. I want to prove following. $$ \sum_{k=1}^n \left( \int_{a_{k-1}}^{a_k} |g|^p \right)^{\frac{q}{2p}} \left( \int_{a_{k-1}}^{a_k} |f|^r \right)^{\frac{q}{2r}} \leqslant \left( \sum_{k=1}^n \int_{a_{k-1}}^{a_k} | g| ^p \right)^{\frac{q}{2p}}\left( \sum_{k=1}^n \int_{a_{k-1}}^{a_k} | f| ^r \right)^{\frac{q}{2r}} $$
for given real sequence $\{a_k\}$.
AI: Let $b_k:=\left(\int_{a_{k-1}}^{a_k}|g|^p\right)^{\frac q{2p}}$ and $c_k:=\left(\int_{a_{k-1}}^{a_k}|f|^q\right)^{\frac q{2r}}$. We have by Hölder's inequality for sums, applied to the exponents $\frac{2r}q$ and $\frac{2p}r$ that,
$$\sum_{k=1}^nb_kc_k\leq \left(\sum_{k=1}^nb_k^{\frac{2p}q}\right)^{\frac q{2p}} \left(\sum_{j=1}^nc_j^{\frac{2r}q}\right)^{\frac q{2r}}.$$
The LHS corresponds to the LHS in the wanted inequality, and the same for the RHS. |
H: Proving the Möbius formula for cyclotomic polynomials
We want to prove that
$$ \Phi_n(x) = \prod_{d|n} \left( x^{\frac{n}{d}} - 1 \right)^{\mu(d)} $$
where $\Phi_n(x)$ in the n-th cyclotomic polynomial and $\mu(d)$ is the Möbius function defined on the natural numbers.
We were instructed to do it by the following stages:
Using induction we assume that the formula is true for $n$ and we want to prove it for $m = n p^k$ where $p$ is a prime number such that $p\not{|}n$.
a) Prove that $$\prod_{\xi \in C_{p^k}}\xi = (-1)^{\phi(p^k)} $$ where $C_{p^k}$ is the set of all primitive $p^k$-th roots of unity, and $\phi$ is the Euler function. I proved that.
b) Using the induction hypothesis show that
$$ \Phi_m(x) = (-1)^{\phi(p^k)} \prod_{d|n} \left[ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} - 1 \right) \right]^{\mu(d)} $$
c) Show that
$$ \prod_{\xi \in C_{p^k}} \left( (\xi^{-1}x)^{\frac{n}{d}} - 1 \right) = (-1)^{\phi(p^k)} \frac{x^{\frac{m}{d}}-1}{x^{\frac{m}{pd}} - 1} $$
d) Use these results to prove the formula by substituting c) into b).
I am stuck in b) and c).
In b) I tried to use the recursion formula $$ x^m - 1 = \prod_{d|m}\Phi_d(x) $$ and
$$ \Phi_m(x) = \frac{x^m-1}{ \prod_{\stackrel{d|m}{d<m}} \Phi_d(x)} . $$
In c) I tried expanding the product by Newton's binom using $\phi(p^k) = p^k ( 1 - 1/p)$. I also tried replacing the product by $\xi \mapsto [ \exp(i2\pi / p^k) ]^j$ and let $j$ run on numbers that don't divide $p^k$. In both way I got stuck.
I would appreciate help here.
AI: For b), you have to prove this by saying that $\Phi_n(x) = \prod_{\xi \in C_n} (x - \xi)$, so you have to relate $C_m$ with $C_n$ and $C_{p^k}$.
Can you find a relation between $\phi(m)$, $\phi(n)$, and $\phi(p^k)$ ? If so it should give you an idea for a way to describe $C_m$ in terms of $C_n$ and $C_{p^k}$.
For c), maybe you can compute $\Phi_{p^k}$, and use that result to show that both expressions are equal to $(-1)^{\phi(p^k)} \Phi_{p^k}(x^{n/d})$ |
H: Getting the shortest paths for chess pieces on n*m board
I originally posted this question of stackoverflow but I was suggested to post it here.
So:
I am stuck solving a task requiring me to calculate the minimal number of steps required to go from point A to point B with different chess pieces on a n*m board that has obstacles and save the all the shortest paths.
So far I have an algorithm that for all squares calculates the amount of steps to reach them:
Example for a knight starting from "A2" (the square marked with zero). Obstacles are marked with "-2" and free squares with "-1"
Before algorithm:
https://dl.dropbox.com/u/65978956/before.JPG
After algorithm:
https://dl.dropbox.com/u/65978956/after.JPG
The algorithm looks like this:
public int[,] BFS(int[,] chessBoard, Tuple<int, int> startXY, int obstacles, int piece) {
int dimensionX = chessBoard.GetLength(0);
int dimensionY = chessBoard.GetLength(1);
// Look at all the movements a knight can make
int[] knightMovementY = { -2, -2, -1, -1, 1, 1, 2, 2 };
int[] knightMovementX = { -1, 1, -2, 2, -2, 2, -1, 1 };
var allMoves = new Tuple<int[],int[]>(knightMovementX, knightMovementY);
chessBoard[startXY.Item1, startXY.Item2 - 1] = 0;
int cnt = dimensionX * dimensionY - obstacles - 1;
int step = 0;
// loop until the whole board is filled
while (cnt > 0) {
for (int x = 0; x < dimensionX; x++) {
for (int y = 0; y < dimensionY; y++) {
if (chessBoard[x, y] == step) {
for (int k = 0; k < allMoves.Item1.Length; k++) {
int dx = allMoves.Item1[k] + x, dy = allMoves.Item2[k] + y;
if (dx >= 0 && dx < dimensionX && dy >= 0 && dy < dimensionY) {
if (chessBoard[dx, dy] == -1) {
chessBoard[dx, dy] = step + 1;
cnt--;
}
}
}
}
}
}
step++;
}
return chessBoard;
}
The algorithm works also for the king, but for the queen, rook and bishop I need something else, because they can move many squares at a time and cannot jump over obstacles.
I am very thankful for suggestions as to what I am supposed to do to solve this problem for the queen, bishop and rook and how to save the shortest paths from the picture "after algorithm" to an array.
I have been looking for days and know that A* algorithm is often used for finding paths, but I don't know how it is going to work when some chess pieces can move such large distances in one move.
AI: Here's a recursive process to find the number of moves from point $X$ to point $Y$ in a grid $G$ with obstacles $O$.
Search algorithms work by generation the set of squares that can be reached in $n$ moves.
Let $B_0 = \{X\}$ - the border of expansion, $I_0 = \emptyset$ - interior, $E_0 = (G / O) / B_0$ -exterior.
Let $f$ - a function such that for any $g \in G$, $f(g)$ is the set of squares that can be reached from $g$ in one move. This function is the definition of the chess piece you use.
Now, iterate:
$B_n = \bigcup \limits_{x \in B_{n-1}} f(x) \cap E_{n-1}$
$I_n = I_{n-1} \cup B_{n-1}$
$E_n = E_{n-1} / B_n$
Continue until $Y \in B_n$ or $B_n = \emptyset$. In the first case $\text{distance}(X, Y) = n$, in the second, there is no path from $X$ to $Y$.
I hope that explains the general idea. There is plenty of code examples for pathfinding problems. Note that instead of sets of squares, lists of Nodes are used, where a Node is an object containing the position of a square, a reference to the Node that generated it, for the purposes of backtracking and also a priority number if heuristics are used. Heuristic is the simple idea that if $Y$ lies to the left from $X$, the path from $X$ to $Y$ most likely goes to the left (which is not true in case of obstacles). There isn't much point to use heuristics here since $f$ is usually not trivial and the grid is small. |
H: ODE theory. Need someone to jog my memory
I have two questions regarding the same thing.
Let's say I have a homogeneous ODE (is that what they are called?) $ay'' + by' + cy = 0$
The trick in this problem is to multiply both sides by $e^{rt}$ and do some trick to get the homogeneous solution which usually includes strictly exponential or sometimes a mixed with trigonometric functions. This is what most book does, but now that I had forgotten most of this stuff, could someone remind me why we ignore the trivial solution $y = 0$?
Let's say the ODE $ay'' + by' + cy = 0$ has the homogeneous solution $y = Ae^{-r_1t} + Be^{r_2t}$ (Let $r_1, r_2 > 0$ and are roots of the characteristic equation). If it happens that $\lim_{x\to \infty} y = 1$. Why can't there be a solution? Is there a theorem that says this? I can't remember it out for the life of me.
Thank you for reading
AI: Consider the homogeneous ODE of second order,
$$ \tag 1 y''+ay'+by=0$$
Where $a$ and $b$ are constant.
The trivial solution is ignored simply because it is trivial. It's not significant in the study of the ODE. We're interested in non-trivial solutions.
What can be done is assume a solution of the form $y=e^{rx}$, $r$ not necessarily real. Then
$$r^2e^{rx}+ar e^{rx}+b e^{rx}=0 $$
Since $e^{rx}$ is never zero,
$$\tag 2 r^2+ar +b =0 $$
Thus, let $r_1$ and $r_2$ be the roots of this equation, we have two solutions:
$$y_1= e^{r_1 x}$$
$$y_2= e^{r_2 x}$$
If the roots aren't identical, then the solutions are linearly independent, so that the general solution of the hom. ODE is given by
$$y_h = c_1 y_1 +c_2y_2$$
where the $c$s are arbitrary constants.
In genera, we call $(2)$ the characteristic equation of $(1)$, $r_1$, $r_2$ the roots of the equation, and each $y_1$ ,$y_2$ the partical solution.
Depending on the nature of the roots (complex or real), we'll get solutions of the form:
$(1)$ Real roots, $r_1$,r_2$
$$y = c_1 e^{r_1 x} +c_2 e^{r_2 x}$$
$(2)$ Conjugate complex roots, $r_1=a+bi$,$r_2=a-bi$
$$ y = {e^{ax}}\left( {{C_1}\sin bx + {C_2}\cos bx} \right) $$
I guess you can now see it is not possible that the limit of such functions is $1$.
Thus, no solution can exists fullfilling that initial condition.
Suppose now we have the following non-homogeneous ODE,
$$ \tag 3 y''+ay'+by=\rm F$$
where $\mathrm F$ is a function of $x$.
There's a theorem that states that if $y_p$ is a particular solution of $(3)$ and $y_h$ is the solution to the related hom. ODE, then all the solutions are given by
$$ \tag 4 Y=y_h+y_p$$
We can get particular solutions by various methods, and they will depend on $F$.
In particular, one method gives the general solution, and thus the particular solution:
If $y_h$ is $y_h = c_1 y_1 +c_2 y_2$ then the particular solution is
$$ y_p=y_2 \int \frac{{\rm F} \cdot y_1}{{\rm W}(y_1,y_2)}dx-y_1\int \frac{{\rm F} \cdot y_2}{{\rm W}(y_1,y_2)}dx$$
where ${\rm W} (y_1,y_2)$ is the wronskian determinant of $y_1$ and $y_2$, $=y_1y'_2-y_1'y_2$ |
H: How to calculate all the four solutions to $(p+5)(p-1) \equiv 0 \pmod {16}$?
This is a kind of a plain question, but I just can't get something.
For the congruence and a prime number $p$: $(p+5)(p-1) \equiv 0\pmod {16}$.
How come that the in addition to the solutions
$$\begin{align*}
p &\equiv 11\pmod{16}\\
p &\equiv 1\pmod {16}
\end{align*}$$
we also have
$$\begin{align*}
p &\equiv 9\pmod {16}\\
p &\equiv 3\pmod {16}\ ?
\end{align*}$$
Where do the last two come from? It is always 4 solutions? I can see that they are satisfy the equation, but how can I calculate them?
Thanks
AI: First note that $p$ has to be odd. Else, $(p+5)$ and $(p-1)$ are both odd.
Let $p = 2k+1$. Then we need $16 \vert (2k+6)(2k)$ i.e. $4 \vert k(k+3)$.
Since $k$ and $k+3$ are of opposite parity, we need $4|k$ or $4|(k+3)$.
Hence, $k = 4m$ or $k = 4m+1$. This gives us $ p = 2(4m) + 1$ or $p = 2(4m+1)+1$.
Hence, we get that $$p = 8m +1 \text{ or }8m+3$$ which is what your claim is as well.
EDIT
You have obtained the first two solutions i.e. $p = 16m+1$ and $p=16m + 11$ by looking at the cases $16 \vert (p-1)$ (or) $16 \vert (p+5)$ respectively.
However, note that you are leaving out the following possibilities.
$2 \vert (p+5)$ and $8 \vert (p-1)$. This combination also implies $16 \vert (p+5)(p-1)$
$4 \vert (p+5)$ and $4 \vert (p-1)$. This combination also implies $16 \vert (p+5)(p-1)$
$8 \vert (p+5)$ and $2 \vert (p-1)$. This combination also implies $16 \vert (p+5)(p-1)$
Out of the above possibilities, the second one can be ruled out since $4 \vert (p+5)$ and $4 \vert (p-1)$, then $4 \vert ((p+5)-(p-1))$ i.e. $4 \vert 6$ which is not possible.
The first possibility gives us $ p = 8m+1$ while the second possibility gives us $p = 8m +3$.
Combining this with your answer, we get that $$p = 8m +1 \text{ or }8m+3$$
In general, when you want to analyze $a \vert bc$, you need to write $a = d_1 d_2$, where $d_1,d_2 \in \mathbb{Z}$ and then look at the cases $d_1 \vert a$ and $d_2 \vert b$. |
H: Convergence in the upper half-plane
I have a sequence $\{F_{n}(z)\}_{n=1}^{\infty}$ of analytic functions in the open upper half plane $\mathbb H$ and continuous on $\mathbb R$, such that $|F_{n}(z)|\leq 1$ for all $n\geq 1$, and all $z$ in the closed upper half plane $\overline{\mathbb H}=\mathbb H\cup \mathbb R$. Also, restricting to the real line, $\{F_{n}(x)\}$ is continuous and uniformly Lipschitz on $\mathbb R$.
How I can get the following result:
(*) Given the sequence $\{F_{n}\}$ above,we can find a subsequence $\{F_{n'}\}$ which converges uniformly on compact subsets of $\overline{\mathbb H}$ to a function $F$, and $F$ will be analytic on $\overline{\mathbb H}$.
I tried the following:
Suppose that $\{F_{n_{k}}(x)\}$ is a subsequence of $\{F_{n}(x)\}$ which converges uniformly on compact subsets of $\mathbb R$ to some continuous function, say $F_{R}(x)$ (this subsequence exists because of the uniformly Lipschitz property).
Now, consider the subsequence $\{F_{n_{k}}(z)\}$, $z\in \mathbb H$: By Montel's theorem, we can find a subsequence of $F_{n_{k}}(z)$, say $\{F_{n_{k_{j}}}\}$, which converges uniformly on compact subsets of $\mathbb H$ to an analytic function, say $F_U$. (As far as I know, the theorem doesn't say anything about convergence on the real line).
So, in this case, the new subsequence $\{F_{n_{k_{j}}}(x)\}$ will converge uniformly on compact subsets of $\mathbb R$ to $F_{R}$.
(**) Is this correct?
Now to answer my question in (*), can we get such $F$ using $F_{U}$ and $F_{R}$ above?
Edit: I changed the statement from analytic on $\overline{\mathbb H}$ to analytic on $\mathbb H$ and continuous on $\mathbb R$ to avoid confusion.
AI: I think the statement is false. Let $\mathbb D$ be the unit disk. Take a function $f\colon \mathbb D\to\mathbb D$ that is Lipschitz in $\overline{\mathbb D}$ but is not analytic in the closed disk (i.e., does not have an analytic extension to a larger disk). For example, $\displaystyle f(z)=\frac{1}{10}\sum_{n=1}^{\infty}\frac{z^n}{n^3}$. Transplant this function to the upper half-plane via substitution $F(z)=f((i-z)/(i+z))$. The composition $F$ is also Lipschitz, but does not extend analytically to a neighborhood of $0$, since $f$ does not extend to a neighborhood of $1$. Finally, let $F_n(z)=F(z+i/n)$: this sequence has all the properties stated, and converges to $F$.
[added] Let $\mathbb H=\{z : \mathrm{Im}\ z>0\}$. Here is a proof of the following statement: "Given the sequence $\{F_n\}$ above, we can find a subsequence $\{F_{n_k}\}$ which converges uniformly on compact subsets of $\overline{\mathbb H}$ to a function $F$, where $F$ is analytic on $\mathbb H$ and continuous on $\overline{\mathbb H}$."
Step 1: the functions $F_n$ are actually uniformly Lipschitz on $\overline{\mathbb H}$. Indeed, let $L$ be the Lipschitz constant of ${F_n}\big|_{\mathbb R}$. For any $h\in \mathbb R$ the difference $g(z)=F_n(z+h)-F_n(z)$ is bounded by $2$ in $\mathbb H$ and bounded by $Lh$ on $\mathbb R$. By the maximum principle, $|g(z)|\le Lh$ on $\mathbb H$. Passing to the limit $h\to 0$, we obtain $|F_n'|\le L$ on $\mathbb H$. It follows that $F_n$ is $L$-Lipschitz on $\overline{\mathbb H}$.
Step 2: for each integer $R$ there is a subsequence that converges uniformly on $\{z\in \overline{\mathbb H} : |z|\le R\}$. This follows from 1 and the Arzela-Ascoli theorem.
Step 3: The usual diagonal argument gives a subsequences that converges uniformly on $\{z\in \overline{\mathbb H} : |z|\le R\}$ for every $R$. Let $F$ be its limit.
Step 4, conclusion: The function $F$ is continuous on $\overline{\mathbb H}$, and is analytic on $\mathbb H$. |
H: Linear transformation for projection of a point on a line
This is what my textbook wants me to do:
The matrix of the linear transformation $P_L$ that projects $\mathbb{R}^2$ on de straight line $l \leftrightarrow y = mx$ is:
\begin{pmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{pmatrix}
And I get this picture, which probably is their to inspire me:
Now, I tried to tackle this with the perpendicular line between A and C, but I got nowhere:
\begin{align*}
A &= (x_A, y_A) \\
&\Downarrow \\
AC \leftrightarrow y - y_A &= \frac{-1}{m}(x - x_A) \\
&\Downarrow \\
C &\leftrightarrow \begin{cases}
y = \frac{-x}{m} + \frac{x_A}{m} + y_A \\
y = mx \\
\end{cases}
\end{align*}
But that's were I am stuck, I can't get it to the form
\begin{align*}
\begin{bmatrix}
x_C \\
y_C
\end{bmatrix} &= \begin{bmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{bmatrix}
\begin{bmatrix}
x_A \\
y_A
\end{bmatrix}
\end{align*}
I am probably not doing it right. All tips are greatly appreciated.
AI: You are off to a great start!
If you substitute your second equation into your first, you find $$mx=-\frac{x}{m}+\frac{x_A}{m}+y_A,$$ so $$\frac{1+m^2}{m}x=\frac{x_A}{m}+y_A,$$ and so $$x=\frac{1}{1+m^2}x_A+\frac{m}{1+m^2}y_A,\tag{$\star$}$$ which corresponds to what your first row should be.
Once you've found that, use $(\star)$ to substitute into your second equation, and you readily see that $$y=\frac{m}{1+m^2}x_A+\frac{m^2}{1+m^2}y_A,$$ which corresponds to the second row. |
H: Optimization of Unconstrained Quadratic form
So I'm learning about optimization of quadratic forms and this textbook goes through definiteness of matrices and principle minors etc. and then goes straight onto optimizing with constraints but never mentions how to solve the general problem of finding stationary points on $f(x)$ where
$$f(x)=x^TAx + \mathbf b^Tx+c$$
$A$ being an n x n matrix and b an n x 1
Any help is appreciated
AI: Let $\vec\nabla f$ be a $n \times 1$ vector, we will call this $\dfrac{\partial f}{\partial x}$. Then we get that $$\dfrac{\partial f}{\partial x} = Ax + A^Tx + b.$$
The differentiation with respect to a vector is obtained by recalling the following vector differentiation identities.
If $y \in \mathbb{R}^{m \times 1}$, $x \in \mathbb{R}^{n \times 1}$, then $\dfrac{\partial y^T}{\partial x}$ is a $n \times m$ matrix $A$ with $A(i,j) = \dfrac{\partial y_j}{\partial x_i}$.
Also, the chain rule goes as follows.
$$\dfrac{\partial (\cdot)}{\partial x} = \sum_k \dfrac{\partial y_k^T}{\partial x}\dfrac{\partial (\cdot)}{\partial y_k}$$
Hence, we have the following relations
$$\dfrac{\partial x^T}{\partial x }= I_{n \times n}$$
$$\dfrac{\partial (c^Tx)}{\partial x} = \dfrac{\partial (x^Tc)}{\partial x} = c$$
Hence, $$\dfrac{\partial b^Tx}{\partial x} = b$$ and
\begin{align}
\dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial x} & = \dfrac{\partial y_1^T}{\partial x}\dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x}\dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_2}\\
& = \dfrac{\partial y_1^T}{\partial x} \dfrac{\partial \left(y_2(x)^TA^Ty_1(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x} \dfrac{\partial \left(y_1(x)^TAy_2(x) \right)}{\partial y_2}\\
& = \dfrac{\partial y_1^T}{\partial x} \dfrac{\partial \left((Ay_2(x))^Ty_1(x) \right)}{\partial y_1} + \dfrac{\partial y_2^T}{\partial x} \dfrac{\partial \left((A^Ty_1(x))^Ty_2(x) \right)}{\partial y_2}\\
& = \dfrac{\partial y_1^T}{\partial x} Ay_2(x) + \dfrac{\partial y_2^T}{\partial x} A^Ty_1(x)
\end{align}
In our case, $y_1(x) = y_2(x) = x$ and hence $\dfrac{\partial y_1^T}{\partial x} = \dfrac{\partial y_2^T}{\partial x} = I_{n \times n}$. Hence, $$\dfrac{\partial (x^TAx)}{\partial x} = I A x + I A^T x = \left(A+A^T \right)x$$
Setting the gradient to zero, we get the equation $$(A+A^T)x = -b$$ Solve this linear system to get the stationary points. |
H: $ \text{if} \;\;a^q \leqslant b^q + c^q \;\;\text{then}\;\; a \leqslant b+c. $
Let $a,b,c >0$ and $q >1$. Then $$ \text{if} \;\;a^q \leqslant b^q + c^q \;\;\text{then}\;\; a \leqslant b+c. $$ How can I prove this?
AI: Suppose to the contrary that $a \gt b+c$. Then $a^q \gt (b+c)^q \gt b^q+c^q$.
To prove that $(b+c)^q \gt b^q+c^q$, consider the function $f(x)=(b+x)^q-b^q-x^q$. At $x=0$, $f(x)=0$. Now calculate $f'(x)$. Because $q\gt 1$, one easily shows that $f'(x)\ge 0$ if $x \gt 0$. So $f$ is an increasing function of $x$, and therefore $f(c)\gt 0$. |
H: Understanding Algebraic Multiplicity
Can you help me understand this statement:
An eigenvalue c has algebraic multiplicity $k$ if $(t-c)^k$ is the highest power of $(t-c)$ that divides the characteristic polynomial.
I am not sure, what does $t$ stand for. I have lifted this statement from the first statement under Algebraic Multiplicity heading from this link
AI: The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial.
Then a root $c$ of $P(t)$ has multiplicity $\mu$ if $\mu$ is the highest integer such as $(t-c)^\mu$ divides $P(t)$. |
H: Sum of Sines Interval
Possible Duplicate:
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
How is it possible to show for integer $m$:
$$\frac{1}{M}\sum_{k=1}^{M}\sin(m\cdot y_{k})=0$$
Thank you very much
Interval $[-\pi,\pi]$ split into $M$ equal intervals, with the mid point of interval is $y_{k}$
AI: The sine function is odd function, meaning that in general $\sin(-t)=-\sin(t)$.
The interval $[-\pi,\pi]$ is symmetrical about $0$. So if $x$ is one of the $x_k$, then $-x$ also is one of the $x_k$, and the values of $\sin(mx)$ and $\sin(-mx)$ add up to $0$. (If $N$ is odd, there isn't the perfect twinning, but the untwinned point is $0$, and $\sin(0)=0$.)
Exactly the same argument works for any odd function and any interval symmetric about the origin. |
H: How you prove that $p^2 \mid m$ and $p^2 \mid n \Rightarrow p^2 \mid mn$?
We know that (Characterization of prime number) $p \mid ab \Rightarrow p \mid a$ or $p \mid b$, where $p$ is prime number. How you prove that $p^2 \mid m$ and $p^2 \mid n \Rightarrow p^2 \mid mn$?
AI: You surely do not mean to ask this, since if $p^2$ divides $m$, then automatically $p^2$ divides $mn$, whether or not $p$ is prime. |
H: Finding the velocity of a rock given its height as a function of time
I'm trying to learn calculus here, but I know I have to set the $h$ equal to 0 and find the time at when it's equal to 0, but I have no idea what to do after. Here is the question. How do I find out the velocity at that time?
If a rock is thrown upward on the planet Mars with a velocity of $10\;m/s$, its height (in meters) after $t$ seconds is given by $H = 10t − 1.86t^2$. Find the velocity of the rock when it hits the ground.
AI: First we find the time(s) when the rock is at ground level. So set $10t-1.86t^2=0$ and solve for $t$. We get $t=0$ and $t=\frac{10}{1.86}$.
The velocity at time $t$ is the derivative of the displacement function $H(t)$. So the velocity at time $t$ is $10-(2)(1.86)t$. Substitute the value of $t$ we found above.
Remark: We can solve the problem instantly without calculus. The initial velocity is $10$. So by symmetry the velocity when it hits the ground on its return trip must be $-10$. |
H: Difference between $\left< x\right> \cap \left< x,y\right>^2$ and $\left< x,y\right>^3$
Consider the ideals $I = \left< x\right> \cap \left< x,y\right>^2 = \left<x^3,x^2y, xy^2\right>$ and $J=\left< x,y\right>^3=\left< x^3, x^2y, xy^2, y^3
\right>$ in $k[x,y]$.
What is the geometric difference between $I$ and $J$?
I know that the zero set for $J$ is a triple point at the origin on $k^2$ while the zero set for $I$ is the $y$-axis together with a double point at the origin on $k^2$.
But aren't they both colength $3$ ideals in $k[x,y]$?
AI: The ideals are hugely different since $\sqrt I=\left< x\right> $ whereas $\sqrt J=\left< x,y\right> $, so that the schemes $V(I)$ and $V(J)$ have dimensions $1$ and $0$: a big difference even at the gross level of their underlying topological spaces!
As to your actual question : the length of of $k[x,y]/I$ is infinite whereas that of $k[x,y]/J$ is $6$.
I think that this is what people mean by colength and I don't understand what you mean when you say that both ideals have colength $3$. |
H: Intermediate Value Theorem Confusion
I have a question on this online website I'm trying to learn calculus on. What I am really confused about the Intermediate Value Theorem is: it says I should set it to 0 but I'm totally lost at the steps used to approach and take this? Perhaps someone can guide me in the right direction.
Do I plug in 1 and 0 for $x$ and then compare it?
Use the Intermediate Value Theorem to show that there is a solution to
the equation $e^{-x^2} = x$ in the interval (0, 1).
AI: The Intermediate Value Theorem states:
Theorem. Let $f(x)$ be a function that is continuous on $[a,b]$. If $k$ is a number between $f(a)$ and $f(b)$, then there exists (at least) one $c$, $a\lt c \lt b$, such that $f(c) = k$.
I often tell my Calculus students to think of the Intermediate Value Theorem as the "Goldilocks Theorem". You are trying to find a point where $f$ takes the value $k$. If one of $f(a)$ and $f(b)$ is "too much", and
the other is "too little", then somewhere in between is a point where $f$ is "just right."
The most common (but by no means only) use of the Intermediate Value Theorem is to find points where $f(x)$ is equal to $0$. If you can find a point where $f(x)$ is positive and one where $f(x)$ is negative, then somewhere in between there will be a point where $f(x)$ is exactly $0$ (provided $f$ is continuous).
In the case at hand, you want to find a solution to $e^{-x^2}=x$. So we can take $f(x) = e^{-x^2}-x$; we are looking for a point where $f(x)=0$. This function is continuous everywhere. Since $f(0) = e^{0}-0 = 1$ and $f(1) = e^{-1}-1 \lt 0$, we have:
$f(x)$ is continuous on $[0,1]$, it is positive at $0$ (value is "too much"), and it is negative at $1$ (value is "too little"). Therefore, by the Intermediate Value Theorem, there is at least one point $c$ somewhere in $(0,1)$ where $f(c)=0$ ("just right").
But if $f(c)=0$, then that means that $0 = e^{-c^2}-c$, so $e^{-c^2}=c$. That is, the equation $e^{-x^2} = x$ has at least one solution in $(0,1)$. |
H: Sum Cosine Mod?
interval $-\pi:\pi$ split into M equal intervals.
midpoints are $y_K$
but i dont understand how to show
$$
\frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})=\begin{cases} 1, & \ m \equiv 0\pmod{M}\\ 0, & \text{else} \end{cases}$$
thank you very much
AI: It is convenient to push the interval forward by $\pi$. Since we are working over a full period of the cosine function, the sum does not change.
Now use the fact that
$$\cos x=\frac{e^{ix}+e^{-ix}}{2}.$$
Then our sum turns out to be the sum of two finite geometric series. The case $m\equiv 0\pmod{M}$ corresponds to the trivial geometric series of all $1$'s.
Some detail: After the shift, the midpoints $x'_k$ at which we evaluate $\cos(mx)$ are
$x'_k=\frac{2\pi(2k+1)}{2M}$, where $k$ ranges from $0$ to $M-1$. So one of the geometric series that we calculate the sum of is
$$\sum_{k=0}^{M-1}\exp\left(\frac{2\pi i(2k+1)m}{2M}\right).\tag{$1$}$$
The other is the conjugate, obtained by replacing $i$ by $-i$.
For the sum $(1)$, by taking out the obvious common factor, we can see that all we need is $\sum_{k=0}^{M-1} \exp(2\pi i m k/M)$. This reduces to finding $\sum_{k=0}^{M-1} t^k$, where $t=2\pi i m/M$. If $1-t\ne 0$, then by the usual formula for the sum of a finite geometric series, the sum is
$$\frac{1-t^M}{1-t}.$$
This sum (if $1-t\ne 0$) is $0$, since $t^M=1$. The only situation in which $1-t=0$ is when $m/M$ is an integer, meaning that $m \equiv 0\pmod M$.
Remark: If we do not push forward by $\pi$, again we obtain a geometric series, with somewhat nicer symmetries. We went to the interval $[0,2\pi]$ mainly because it is more familiar.
The relevant facts can be also proved without appealing to complex numbers. What we need then is a formula for the sum of the cosines of numbers in arithmetic progression. Please see Wikipedia for the
relevant formula. The formula can be proved by a telescoping series argument. |
H: Implication Laurent series to polynomial
Let $f$ be holomorphic on $\mathbb{C}$. We have $f(z)=\sum_{n=0}^\infty a_nz^n$. Let $g$ be defined on $\mathbb C\setminus\{0\}$ by the Laurent series $g(z)=\sum_{n=0}^\infty \frac{a_n}{z^n}$.
If $0$ is an essential singularity of $g$, then the coefficients satisfy $a_n \neq 0$ for infinitely many $n$, so we have an infinite number of terms in the Laurent series. Why does this imply that $f$ is not a polynomial?
AI: This comes down to the uniqueness of Taylor coefficients. By assumption, $f(z)=a_0+a_1z+a_2z^2+\cdots$, where the series converges on all of $\mathbb C$. If $f$ were a polynomial, then there would exist an $n\in\mathbb N$ and complex numbers $b_0,b_1,\ldots,b_n$ such that $f(z)=b_0+b_1z+b_2z^2+\cdots+b_nz^n$ for all $z\in\mathbb C$.
But then for $k\in\{0,1,2,\ldots,n\}$, $b_k=a_k=\dfrac{1}{k!}f^{(k)}(0)$, and for $k>n$, $a_k=\dfrac{1}{k!}f^{(k)}(0)=0$. Therefore, if $f$ is a polynomial, then $a_k=0$ for all but finitely many $k$. |
H: Implication injective holomorphic function on the zeroes of derivative
We know the following things:
$f(z)$ is holomorphic on $\mathbb{C}$, $f(z)$ is injective and $f(z)$ is a polynomial, so $f'(z)$ is a polynomial too.
(1) Why is $f'(z) \neq 0$? Or why does $f$ have only one zero?
(2) Why is $f'(z)$ constant if (1) holds?
AI: Injective means that, in particular, $f(z_1) = f(z_2) =0$ implies $z_1=z_2.$ This means $f(z)=a(z-z_1)^n$, with $a\ne 0,$ $n= \deg f.$ Since $f(z_1 + \omega )=1$ where $\omega $ is an $n$th root of unity, injectivity gives once again that $n=1.$ Hence $f'(z) = a.$
What's more interesting is that you don't need to require that $f$ is a polynomial - here you actually use the fact that $f$ is entire.
EDIT: Here's a sketch.
$f$ is injective and analytic, so the open mapping theorem says that $f( B_{\epsilon }(1) )$ contains a neighborhood of $f(1).$ Define $g(z) = f(\frac{1}{z}).$ If $f$ does not have a power series expressible as a polynomial, $g$ has an essential singularity at $0.$ Since $g$ is holomorphic on $\mathbb{C} / \{ 0 \} ,$ the Casorati-Weierstrass Theorem lets us construct a sequence $\{ \zeta_n \}$ converging to $0$ such that $g( \zeta_n ) $ converges to $f(1).$ For $n$ sufficiently large, $z_n \notin B_{\epsilon }(1),$ contradicting injectivity. |
H: Natural deduction proof of $\forall x (\exists y (P(x) \vee Q(y))) \vdash \exists y (\forall x (P(x) \vee Q(y)))$
I'm trying to do a Fitch proof of
$$
\forall x (\exists y (P(x) \vee Q(y))) \vdash \exists y (\forall x (P(x) \vee Q(y)))
$$
Edit: using only the axioms on http://www.proofwiki.org/wiki/Category:Natural_Deduction_Axioms, along with universal/existential generalisation/instantiation
The following is my first attempt.
$$
\begin{array}{lll}
1 & \begin{array}{l}\forall x (\exists y (P(x) \vee Q(y))) \\ \hline \end{array} & \text{assumption} \\
2 & \exists y (P(v) \vee Q(y)) & \text{$\forall$E, 1} \\
3 & \begin{array}{ll} & \begin{array}{l} P(v) \vee Q(w) \\ \hline\end{array} \end{array} & \text{assumption} \\
4 & \begin{array}{ll} & \forall x (P(x) \vee Q(w)) \end{array} & \text{$\forall$I, 3} \\
5 & \forall x (P(x) \vee Q(w)) & \text{$\exists$E, 2, 3, 4} \\
6 & \exists y (\forall x (P(x) \vee Q(y))) & \text{$\exists$I, 5} \\
\end{array}
$$
Edit: I know that this proof is incorrect, since replacing $P(x) \vee Q(y)$ with $R(x, y)$ would yield the result
$$
\forall x (\exists y (R(x, y)) \vdash \exists y (\forall x (R(x, y)))
,$$
which is clearly not true in general.
I'm not exactly sure on which line the flaw is. I would appreciate it if someone could point that out, and explain why it's wrong.
I suspect that I'm supposed to use the distributivity of $\exists$ over $\vee$ for this one. But I don't know how to formally justify this distributivity, and I can't find a natural deduction proof for it.
Edit: I think I have successfully proved this, based on @ZevChonoles's answer. Here is a screencap:
AI: You can't go from step 3 to step 4. For the particular $v$ you chose, you know that you can find $w$ such that $P(v)\vee Q(w)$. If $Q(w)$ is true, then certainly $P(x)\vee Q(w)$ is true for all $x$, but if $Q(w)$ is false there's no reason why it should be the case that $P(x)\vee Q(w)$ for all $x$.
I don't know how to write this in the Fitch format, but the essential idea is just that either $Q(y)$ is true for some $y$, or $Q(y)$ is false for all $y$.
If $Q(v)$ is true for some particular $v$, then $P(x)\vee Q(v)$ is true for all $x$, and so we certainly have that $\exists y(\forall x(P(x)\vee Q(y)))$; the $y$ that exists is just $y=v$.
If $Q(y)$ is false for all $y$, then $P(x)\vee Q(y)\iff P(x)$, so $\forall x(\exists y(P(x)\vee Q(y)))$ implies that $P(x)$ is true for all $x$, and hence $\exists y(\forall x(P(x)\vee Q(y)))$; any $y$ at all will work. |
H: Limit of $x^2\cos(1/x^2)$ when $x\to0$ by squeeze theorem
How can I argue that $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$$
I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.
AI: Use $-1 \le \cos(\frac{1}{x^2}) \le 1$ and multiply through by $x^2$. Since $x^2 \ge 0$, the inequalities remain valid. |
H: Finding Area of a Triangle without Trignometric ratios.
Hi I need to figure out the area of the following triangle, without using Trigonometric ratios. Any suggestions on the best approach.
The answer is 12 square units
Edit:
I also think that the above triangle can't qualify for a $30-60-90$ triangle since it fails the $x,x.\sqrt3,2x$ rule/
AI: If you use a 30-60-90 triangle with hypotenuse 6, then the height is 3. So the height of this triangle is 3. Thus $\frac{b\cdot h}{2}=\frac{8\cdot3}{2}=12$
triangle |
H: Limit Find Value Question
If $f$ and $g$ are continuous functions, with $f(3) = 5$ and
$$\lim_{x \to 3} (2f(x) − g(x)) = 4$$
find $g(3)$.
I am confused at how to tackle this question, I understand I have to find $g(3)$ but do I plus in $3$ for $x$? How do I go about getting the solution because apparently someone got $g(3) = 6$ as the solution?
AI: Define $h(x)$ as:
$$
h(x) = 2f(x) - g(x)
$$
We can write $g(x)$ in terms of $f(x)$ and $h(x)$ as:
$$
g(x) = 2f(x) - h(x)
$$
Since $f(x)$ is continuous:
$$
\lim_{x \to 3} f(x) = f(3) = 5
$$
From the problem statement, we have:
$$
\lim_{x \to 3} h(x) = \lim_{x \to 3} \left(2f(x) - g(x)\right)= 4
$$
Therefore:
$$
\lim_{x \to 3} g(x) = \lim_{x \to 3} \left(2f(x) - h(x)\right) = 2 \times 5 - 4 = 6
$$ |
H: Tangent Line to $\sin x+\cos x$
When is the tangent line to
$y = \sin x + \cos x$
horizontal?
I have no idea how to solve this problem. Would I use the equation of a tangent line here? Because if so i have no idea how to apply that.
AI: Hint 1. A line is horizontal when its slope is $0$.
Hint 2. The slope of the tangent line to $y=f(x)$ at $a$ is $f'(a)$. |
H: Indices in differential geometry
Often times in differential geometry it is convenient to use Einstein summation notation, and there it is presented to beginning graduates and advanced undergraduates alike that if you see two indices that are the same letter with one upper and the other lower, written next to each other, then there is an implied summation symbol.
No seriously mathematically rigorous notion can depend on the particular way in which we choose to communicate, and so even though you could rip apart the Einstein summation convention, asking all sorts of fringe case questions of the type "does Einstein summation notation apply in this case" it is generally understood enough that it is just accepted as a way to communicate, no more or less treacherous than the fact that we write symbols on paper. (Or that in general we do not write out all the quantifiers or use logical symbols in prose.) It is in the same light that I ask the following questions, but I hope that any answerers will also alert me if there is rigorous content in any of this:
Sometimes people seem to derive more meaning than just a bookkeeping device for Einstein Summation out of whether an index is written in a particular position. I don't understand what people are trying to communicate, but I can reconstruct enough of it that you can hopefully help me:
Sometimes writing upper indices indicates something of a "dual" nature. For instance, Einstein summation is often used when we contract a dual vector applied to a vector.
Sometimes writing something upper indicates inversion. For instance, $g^{ij}$ is the $ij_{th}$ entry of the inverse matrix $(g_{ij})$ which is the matrix of the Riemannian metric from a particular coordinate basis to itself.
Sometimes the left to right order of the indices on a complicated tensor matters.
I believe I've seen indices written to the left of a mathematical symbol.
My questions are:
a. Is there any system to what the location of an index means? Or is it just special case by special case? For instance, is 2. an instance of a more general rule, or is it that inversion is only meant in the case of $g_{ij}$. 2. and 1. seem to be in conflict? Or maybe neither 1. nor 2. hold, there is no meaning, and it just happens that for both the case of inversion and transpose, it is convenient to simply write upper indices so that Einstein summation is in effect?
b. Which of these imply more significance than simply a bookkeeping device for Einstein summation notation, among points 1 and 2. above?
c. In points 3. and 4. I am not even aware of what the "face-value" meaning is, as in I would not know how to perform a computation with these things. For instance, is a left-side index supposed to mean transpose or something?
Feel free to simply connect me to a resource, but the most pedestrian searches on my part haven't revealed anything authoritative. I was unable to rigorously understand introductory differential geometry for most of my undergrad because few people stopped to explain their notation, so I hope that in addition to helping me right now, this might become useful to someone else struggling in the same way in the future.
AI: 1. Sometimes writing upper indices indicates something of a "dual" nature. For instance, Einstein summation is often used when we contract a dual vector applied to a vector.
This is the only meaning of upper indices that I am aware of. That is, subscripts denote covariant components of a tensor, while superscripts denote contravariant components. See this Wikipedia section.
2. Sometimes writing something upper indicates inversion. For instance, $g^{ij}$ is the $ij$th entry of the inverse matrix $(g_{ij})$ which is the matrix of the Riemannian metric from a particular coordinate basis to itself.
This is a special case of #1. In particular, $g^{ij}$ represents the $ij$th component of the contravariant metric tensor. As a matrix, the contravariant metric tensor happens to be the inverse matrix of the covariant metric tensor $(g_{ij})$. See the Wikipedia article on raising and lowering indices.
3. Sometimes the left to right order of the indices on a complicated tensor matters.
In principle, the left to right order of the indices always matters. Of course, if a tensor happens to be symmetric, then the indices can be swapped without affecting the value.
4. I believe I've seen indices written to the left of a mathematical symbol.
I am not aware of this notation.
Which of these imply more significance than simply a bookkeeping device for Einstein summation notation, among points 1 and 2. above?
Perhaps the abtract index notation article will shed some light on this for you. Index notation is really fundamental to the nature of tensors.
Edit: Personally, I also feel like I gained a lot of insight into Einstein index notation when I learned about Penrose graphical notation. See the Wikipedia article, or The Road to Reality by Roger Penrose. |
H: Directional derivative (muiltivariable calculus)
I have an encountered an example in my text book which I don't fully see the intuition of. I will write out the part of the example I'm struggling with:
A hiker is standing beside a stream on the side of a mountain examining her map of the region. The height of the land (in meters) at any point $(x,y)$ is given by the function
$$h(x,y) = \frac{20000}{3 + x^2 + 2y^2}$$
where $x$ and $y$ (in kilometers) denote the coordinates of the point on the hiker's map. The biker is at the point $(3,2)$
At what angle to the path of the stream (on the map) should the hiker set out if she wishes to climb at a $15^o$ inclination to the horizontal?
ANSWER (from textbook)
Suppose the hiker moves away from $(3,2)$ in the direction of the unit vector $\vec{u}$. She will be ascending at an inclination of $15^o$ if the directional derivative of $h$ in the direction of $\vec{u}$ is $1000 tan(15^{o}) \approx 268$ (The 1000 compensates for the fact that the vertical units are meters while the horizontal units are kilometers.). If $\theta$ is the angle between $\vec{u}$ and the upstream direction, then
$$500cos(\theta) = |\bigtriangledown h(3,2)|cos(\theta) \approx 268$$
OK, so what I have some problems seeing here is is the part where it is stated that we know that the ascending will be at an inclination of $15^{o}$ if the directional derivative of $h$ in the direction of $\vec{u}$ is $1000 tan(15^{o}) \approx 268$
Am I correct when I interpret the $15^{o}$ angle here as the angle between $\bigtriangleup y$ and $\bigtriangleup x$ in the $(x,y)$ plane? If so, why would this angle then give us the desired inclination to the horizontal? Or am I totally off here?
I would truly appreciate it if anyone could make me see this more intuitively :).
AI: Am I correct when I interpret the $15^{o}$ angle here as the angle between
$\Delta y$ and $\Delta x$ in the $(x,y)$ plane?
I don't think so. The $\theta=15^{o}$ angle refers to the climbing inclination, so $\tan(\theta) = \frac{\Delta z}{|u|}$ where $u$ is the displacement in the $(x,y)$ plane (so $|u| = \sqrt{\Delta x^2+\Delta y^2}$) and $\Delta z = f[(x,y)+u]-f[(x,y)]$ is the difference in "altitude". And $\frac{\Delta z}{|u|}$ is, when $|u|\to 0$, the directional derivative. |
H: Is there any geometric way to characterize $e$?
Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting $\sqrt2$ is really easy: constructing a $45^\circ$-$90^\circ$-$45^\circ$ triangle with unitary sides will make me have an idea of what $\sqrt2$ is. Extending this to $\sqrt5$, $\sqrt{13}$, and other algebraic numbers is easy using Trigonometry; however, it turned difficult working with some transcendental constants. Constructing $\pi$ is easy using circumferences; but I couldn't figure out how I should work with $e$. Looking at
made me realize that $e$ is the point $\omega$ such that $\displaystyle\int_1^{\omega}\frac{1}{x}dx = 1$. However, I don't have any other ideas. And I keep asking myself:
Is there any way to "see" $e$ geometrically? And more: is it true that one can build any real number geometrically? Any help will be appreciated. Thanks.
AI: For a certain definition of "geometrically," the answer is that this is an open problem. You can construct $\pi$ geometrically in terms of the circumference of the unit circle. This is a certain integral of a "nice" function over a "nice" domain; formalizing this idea leads to the notion of a period in algebraic geometry. $\pi$, as well as any algebraic number, is a period.
It is an open problem whether $e$ is a period. According to Wikipedia, the answer is expected to be no.
In general, for a reasonable definition of "geometrically" you should only be able to construct computable numbers, of which there are countably many. Since the reals are uncountable, most real numbers cannot be constructed "geometrically." |
H: Solve $y' = x + y$
I am suppose to use the substitution of $u = x + y$
$y' = x + y$
$u(x) = x + y(x)$
I actually forget the trick to this and it doesn't really make much sense to me. I know that I need to get everything in a variable with x I think but I am not sure how to manipulate the problem according to mathematical rules that will make sense. Also I know that at some point I will get an integral or something and that I have no idea how to do that with multiple variables.
AI: $$y'=x+y$$
Then we let $u=x+y$
This gives $u'=1+y'$, so that the equation becomes
$$u'-1=u$$
$$u'-u=1$$
Can you solve that for $u$?
Hint $(e^x-1)'=e^x$
Moving on with the solution:
$$\frac{du}{dx}-u=1$$
$$\frac{du}{dx}=1+u$$
And the classic abuse in DE's
$$\frac{du}{u+1}=dx$$
Now
$$\int\frac{du}{u+1}=\int dx$$
$$\log(u+1)=x+C$$
We take logarithms
$$u+1=e^{x+C}$$
We use the property of the exponential function $f(x+y)=f(x)f(y)$
$$u+1=e^C e^x$$
Here $K=e^C$
$$y+x+1=Ke^x$$
$$y=K e^x-x-1$$ |
H: How to solve this recurrence $T(n) = 2T(n/2) + n\log n$
How can I solve the recurrence relation $T(n) = 2T(n/2) + n\log n$? It almost matches the Master Theorem except for the $n\log n$ part.
AI: Let us take $n = 2^m$. Then we have the recurrence $$T(2^m) = 2T(2^{m-1}) + 2^m \log_2(2^m) = 2T(2^{m-1}) + m 2^m$$
Calling $T(2^m)$ as $f(m)$, we get that
\begin{align}
f(m) & = 2 f(m-1) + m 2^m\\
& = 2(2f(m-2) + (m-1)2^{m-1}) + m2^m\\
& = 4f(m-2) + (m-1)2^m + m2^m\\
& = 4(2f(m-3) +(m-2)2^{m-2}) + (m-1)2^m + m2^m\\
& = 8f(m-3) +(m-2)2^m + (m-1)2^m + m2^m\\
\end{align}
Proceeding on these lines, we get that
\begin{align}
f(m) &= 2^m f(0) + 2^m (1+2+3+\cdots+m) = 2^m f(0) + \dfrac{m(m+1)}{2}2^m\\
& = 2^m f(0) + m(m+1)2^{m-1}
\end{align}
Hence, $T(n) = n T(1) + n \left(\dfrac{\log_2(n) (1+\log_2(n))}{2} \right) = \mathcal{\Theta}(n \log^2 n)$. |
H: Find the vertical and horizontal asymptotes of the function.
I am asked to find the vertical and horizontal asymptotes of the equation:
$$f(x)=(a^{-1}+x^{-1})^{-1}$$
I simplify this to
$$f(x)=\frac{1}{a^{-1}+x^{-1}}$$
$$f(x)=a^1+x^1$$$$f(x)=a+x$$Which is some constant, graphed as horizontal line - that will not have a vertical or horizontal asymptote. Is my algebra terribly off?
AI: $$
(a^{-1}+x^{-1})^{-1}=\frac{1}{a^{-1}+x^{-1}}=\frac{1}{\frac{1}{a}+\frac{1}{x}} =
\frac{1}{\frac{x}{ax}+\frac{a}{ax}}=
\frac{1}{\frac{x+a}{ax}}=
\frac{ax}{x+a}
$$
The horizontal asymptotes:
$$\lim_{x \to \infty} \frac{ax}{x+a}=\lim_{x \to -\infty} \frac{ax}{x+a}=a$$
i.e. at $y=a$
The vertical asymptotes:
$$\lim_{x \to -a^+} \frac{ax}{x+a} = \infty$$
$$\lim_{x \to -a^-} \frac{ax}{x+a} = -\infty$$
i.e. at $x=-a$ |
H: Rewriting a power series as a geometric series?
For this series, find the radius of convergence and write it as a geometric series and give a formula if $x>3$
$$\sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n$$
Now finding the radius of convergence wasn't too difficult and I'll save you guys the trouble of doing it because the interval of convergence is $ -1 = 3 - 4< x < 4 + 3 = 7$ and so the radius of convergence is 2
The second part confuses me because I don't understand (if it is even possible) to convert a power series to a geometric series.
AI: For this particular function, shouldn't one note that
$$ \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}(x-3)^n = \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n}}(x-3)^n = \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x-3}{2}\right)^{n}$$
which is a geometric series! |
H: $K$ finite extension of $F$ s.t. for every 2 subextensions $M_1, M_2$, $M_1\subset M_2$ or $M_2\subset M_1$. Then there's $a\in K$ such that $K=F(a)$
Let K be a finite extension of a field F such that for every two intermediate field $M_1$, $M_2$ we have $M_1\subset M_2$ or $M_2\subset M_1$. I need to show that there is an element $a\in K$ such that $K=F(a)$.
I have an idea that goes like this: If I show that there are finite intermediate fields, then I could use the Primitive Element Theorem (which demonstration wasn't given during my algebra course, so I would have to include this in my solution so it can be whole).
There's also some doubts here: through the internet we can see that some people enunciate the Primitive Element theorem excluding the hypothesis of the extension (finite) having finite intermediate subfields. Which one is the correct?
AI: I don't think you need the primitive element theorem. Pick an element $\alpha \in K \setminus F$. If $F(\alpha) = K$ then great; otherwise, choose a $\beta \in K \setminus F(\alpha)$. Using the hypothesis, what's the relationship between $F(\beta)$ and $F(\alpha)$? Try to continue this process. To make the resulting argument formal, you could use induction on the degree of $K$ over $F$.
You probably saw a version of the primitive element theorem that assumes separability, which is a stronger condition: the extension $\mathbf F_p(t^{1/p})$ of the function field $\mathbf F_p(t)$ is inseparable, for example. |
H: To understand some terminology of proof.
Let $|\sigma_{n}(x)|\leq K$
$\displaystyle \Rightarrow \frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx\leq 2K^{2}.$
Now, $$\sigma_{n}(x)=\sum_{k=0}^{n}\left(1-\frac{k}{n+1}\right)(a_{k}\cos kx+b_{k}\sin kx).$$
By Parseval's identity,
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx=\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2}).$$
From this it follows that if $m$ is any integer, $m\leq n$ then
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}\tag{1}$$
Taking $n\to \infty$ and keeping $m$ constant, we get
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}.\tag{2}$$
I read this thing in some proof of theorem from real analysis book. In this proof i can't understand to derive equation $(1)$ why are they taking that integer $m\leq n$? and from equation $(1)$ how can they get equation $(2)$? Please explain this terminology me! Thanks in advance!
AI: If $m \leq n$
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{m}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
is not greater than
$$\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
thus since
$$\frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx=\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left(1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})$$
the inequality follows.
For the second one, note that as $n \to \infty$,
$$\frac{k}{n+1} \to 0$$ |
H: Diameter of wheel
If a wheel travels 1 mile in 1 minute at a rate of 600 revolutions per minute. What is the diameter of the wheel in feet ? The answer to this question is 2.8 feet.
Could someone please explain how to solve this problem ?
AI: We are told that every minute, the wheel made $600$ revolutions. When a wheel makes one revolution, it travels forward by one circumference-length - here is an illustrative animation from Wikipedia (in this animation, the circle has radius 1):
Thus, in one minute, the wheel travels $600C$, where $C$ is the circumference of the wheel. Recall that $C=2\pi R=\pi D$ where $R$ and $D$ are the radius and diameter, respectively.
Therefore, we want to solve
$$600\cdot \pi D\approx 1885D =5280\text{ft}.$$ |
H: A complex map with "bounded" derivative is injective
The exercise I try to solve states: "Let $\,f\,$ be analytic in $\,D:=\{z\in\mathbb{C}\;|\;|z|<1\}\,$ , and such that $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$
Prove that $\,f\,$ is $\,1-1\,$ in $\,D\,$.
My thoughts: The condition $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$
means the range of the analytic function $\,f'\,$ misses lots of points on the complex plane, so applying Picard's Theorem (or some extension of Liouville's) we get that $\,f'(z)=w=\,$ constant, from which it follows that $\,f\,$ is linear on $\,D\,$ and thus $\,1-1\,$ there.
Doubts: $\,\,(i)\,\,$ This exercise is meant to be from an introductory first course in complex functions, so Picard's theorem seems overkill here...yet I can't see how to avoid it.
$\,\,(ii)\,\,$ Even assuming we must use Picard's Theorem, the versions of it I know always talk of "entire functions", yet our function $\,f\,$ above is analytic only in the open unit disk. Is this a problem? Perhaps it is and thus something else must be used...?
Any help will be much appreciated.
AI: There is a very similar answer here (feel free to ignore the question if it looks intimidating). The basic idea is to use the triangle inequality applied to $f(z)-z$ to show that $|f(x)-f(y)| > 0$ for any distinct $x,y$. More precisely your assumption allows us to show $|f(x)-f(y)| \ge \frac12 |x-y|$. |
H: Given a cubic function, and its quadratic derivative- can I recover the cubic from quadratic?
Background: I'm trying to learn how to work with cubic and quadratic bezier splines for various drawing libraries, and working through how to approximate a cubic spline with a quadratic spline. It's occured to me that it should be possible to approximate any continuous parametric function with an arbitrary series of interconnected quadratic splines, as long as you have a suitable fitness function for each section.
It has further occured to me that the derivative of any cubic spline is expressible as a quadratic spline. So now I am curious.
if I have an abitrary function f(t);
its derivative f'(t);
and I can approximate f'(t) with a series of quadratic splines...
can I have some way of assuming those quadratics as derivatives of cubics, and directly compute a series of cubics that then approximates f(t) from the quadratics?
is there something wrong with my reasoning here?
(edited for brevity)
(clarification)
I mentioned I was working through how to approximate a cubic spline with a quadratic spline. Let us assume for the purposes of this question that I have already solved this problem, and I have a procedure P which takes any arbitrary function f(t) (which could be a cubic spline, or anything else) with the constraint that we will only approximate over intervals where f'(t) is continuous.
The question therefore is not how do I approximate a cubic with a quadric but, given that I have procedure P, if I apply procedure P to f'(t) to compute a series of quadratic splines, can I use that information to recover a cubic spline approximation of f(t).
From the answers, if I am understanding corrrectly, I think that I can so long as I can still use f(t) to recover this "integral constant". This is an acceptable solution since what I want is something like a procedure P2 which can take a f(t) and give me a sequence of cubic splines which approximate it.
edit: further, it seems that if we have f(t) and its f'(t) and its f''(t) we could even do a straight forward linear approximation of f''(t) and recover a cubic spline approximation of f(t) by solving for the integral constants in f'(t) and f(t).
neat.
AI: In general, we can only recover $f$ from $f'$ up to a constant summand, the famous constant of integration. For example, $x^3 + 2x + 4$ and $x^3 + 2x - 5$ have the same derivative, $3x^2 + 2$. Suppose we have a derivative of second degree $f' = ax^2 + bx + c$. Then, $f$ could be any of $\frac{1}{3}ax^3 + \frac{1}{2}bx^2 + cx + d$, with $d \in \mathbb{R}$. So, in order to uniquely recover $f$, we would have to find $d$. We need more information; the derivative is never enough to determine the original function.
If, for example, we know the value of $f$ at any point $x_0$, we can get the original function back, because we could then pose the equation $f(x_0) = \frac{1}{3}ax_0^3 + \frac{1}{2}bx_0^2 + cx_0 + d$ and solve for $d$.
It seems that I have answered the question in the title, which doesn't correspond to the question in the post. I agree with Gerry's comment, anyway, that we can't simply differentiate a cubic spline for a set of points to get to a quadratic spline for the same set of points. This is somewhat related to what I wrote above: a derivative of a function doesn't say much about the exact values of the function at each point, precisely because it "forgets" a constant. So, once we differentiate the cubic spline, we don't know anything about the exact images of any value, so in particular, the derivative doesn't have to pass by the original set of points at all.
Edit:
Ok, so I'm a bit confused about what you're asking, to be honest. I don't know if my answer is relevant, but I'd like to add a third comment regarding the same issue: if we have a spline for $f'$, we won't be able to get a spline for $f$ if we don't have any further information; because the spline for $f'$ is also a spline for $(f+C)'$, with $C \in \mathbb{R}$ any constant, but it's certainly not true that there's one unique spline that works for all such $f+C$. In any case, you need more information about $f$, something that goes beyond what you can deduce from $f'$. |
H: Piecewise Function Calc Confusion
Help me understand this question.
Consider The Function
$$f(x)= \begin {cases} \dfrac{2x-2}{x-1}&\text{if }x \leq2\\
\dfrac{8}{x}\
&\text{if }x \in_\ (2,4)\\
\sqrt x&\text{if }x \geq 4 \end {cases}$$
Where is the function continuous? If there are any removeable
discontinuities, list them and then redefine the function so that is
now continuous in those places.
What exactly is the question asking for and how do i solve this? Or explain it on a test?
AI: You are given a function, which is defined in different ways depending on where the input is.
You will want to first consider each part of the function separately; on $(-\infty,2]$, is it always defined? Is it continuous there? When you are looking at the part of the function, all you need to worry about is the formula
$$f(x) = \frac{2x-2}{x-1}.$$
For instance, you'll notice that the function is not defined at $x=1$. So it cannot be continuous there; everywhere else it is continuous (quotient of two continuous functions, denominator not equal to $0$). What is happening at $x=1$? Is it a removable discontinuity, or not?
Then consider what happens on $(2,4)$. Then what happens on $[4,\infty)$.
Once you are done considering each piece separately, you need to think about how the pieces fit together. What happens at $x=2$? The function is defined one way on the left, and a different way on the right. Do the two piece fit together properly (that is, are the limits from the right and from the left the same?) If so, what does that tell you about continuity? If not, what does that tell you?
You'll want to then do the same thing at $x=4$. |
H: How to calculate a linear transformation given its effect on some vectors
Im not sure if my question is worded very well, but I'm having trouble understanding how to tackle this problem.
Let $T\colon\mathbb{R}^3\to\mathbb{R}^2$ be the linear transformation such that $T(1,-1,2)=(-3,1)$ and $T(3,-1,1) = (-1,2)$. Find $T(9,-1,10)$.
Thanks
AI: Recall that a linear transformation is, well, linear. If you know the value at vectors $\mathbf{v}_1$ and $\mathbf{v}_2$, then you can compute the value at any linear combination of those two vectors, by using linearlity:
$$T(\alpha\mathbf{v}_1+\beta\mathbf{v}_2) = \alpha T(\mathbf{v}_1) + \beta T(\mathbf{v}_2).$$
Here, for example, you know the value of $T$ at $(1,-1,2)$ and at $(3,-1,1)$. So, for instance, you can easily calculate the value of $T$ at $(5,-3,5) = 2(1,-1,2)+(3,-1,1)$:
$$\begin{align*}
T(5,-3,5) &= T\Bigl( 2(1,-1,2) + (3,-1,1)\Bigr)\\
&= 2T(1,-1,2) + T(3,-1,1)\\
&= 2(-3,1) + (-1,2)\\
&= (-7,4).\end{align*}$$
So... if you can find a way of writing $(9,-1,10)$ as a linear combination of $(1,-1,2)$ and $(3,-1,1)$, then you'll be set.
Added. Unfortunately, this cannot be done here. Namely, there are no scalars $a$ and $b$ such that $(9,-1,10) = a(1,-1,2) + b(3,-1,1)$. Note that you would need $a+3b=9$ and $-a-b=-1$. Adding these two you get $2b = 8$, so $b=4$; then $a=-3$, but then the last coordinate does not work out:
$$-3(1,-1,2) + 4(3,-1,1) = (-3+12, 3-4, -6+4) = (9,-1,-2).$$
So the information you have does not determine the value of $T$ at $(9,-1,10)$. Given any vector $(r,s)$ in $\mathbb{R}^2$, you can find a linear transformation $U$ that agrees with $T$ on $(1,-1,2)$, on $(3,-1,1)$, and that sends $(9,-1,10)$ to $(r,s)$.
That suggests to me that you have either miscopied the problem, or else that whoever assigned the problem made a mistake. One possibility is that the first vector should have been $(1,-1,-2)$, as then you do get $(9,-1,10)$. |
H: Question about a step in a proof (that nonempty open sets are unions of open sets)
Thm: Every nonempty open set $G$ of real numbers is the union of a finite or countable family of pairwise disjoint open intervals.
Proof: Let $x$ be a point in a nonempty open set $G$. There is an open interval $(y, z)$ such that $x\in (y, z)\subset G$. Then, $(y, x)\subset G$ and $(x, z)\subset G$. We define (possibly extended) numbers $a_x$ and $b_x$ by
$a_x = inf (y : (y, x)\subset G)$ and $b_x = sup (z : (x, z)\subset G)$...
My question. Since $(y,x)$ and $(x,z)$ are intervals in G (hence, connected) wouldn't $a_x = b_x =x$?
AI: No. Suppose we are on the real line, and $G = (0,1)$, and $x = 1/2$ for a moment (I assume this ($\mathbb{R}$) is where the whole question takes place - I don't understand it otherwise). Then $a_x = 0$, as $0$ is the smallest element (usual ordering) such that $(a_x, 1/2)$ is in $(0,1)$. Similarly, $b_x = 1$. |
H: $f$ closed iff $y\in N$ and open $V\supset f^{-1}\left(\{y\}\right)$ exists $U$ open such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$
Prove that $f\colon M\to N$ (topological spaces) is closed if and only if for all $y\in N$ and all open sets $V\supset f^{-1}\left(\{y\}\right)$ in $M$ there exists an open set $U$ in $N$ containing $y$ such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$.
I can't prove this in any way. I tried for 3 days.
AI: I was able to obtain one direction:
Fix some $y\in N$. Note that $V\supseteq f^{-1}(U)\supseteq f^{-1}(\{y\})$ if and only if $$V^c\subseteq f^{-1}(U)^c= f^{-1}(U^c)\subseteq f^{-1}(\{y\})^c.$$
If $f$ is closed, then for any closed set $C\subseteq M$ disjoint from $f^{-1}(\{y\})$, we have that $D=f(C)$ is a closed set disjoint from $\{y\}$, and $$C\subseteq f^{-1}(f(C))=f^{-1}(D)\subseteq f^{-1}(\{y\})^c.$$ Then $V=C^c$ and $U=D^c$ are open subsets of $M$ and $N$ respectively satisfying the necessary property, and every open subset $V\supseteq f^{-1}(\{y\})$ of $M$ can of course be obtained as the complement of a closed subset $C$ of $M$ that is disjoint from $f^{-1}(\{y\})$. |
H: Showing that an algebraic number is not a root of a real
While answering this question on mathoverflow, I stumbled across a question that I expect may be easily answered by someone knowing a bit more algebra than me.
Let's make it really specific.
Consider the polynomial equation $X^4-X^3-X^2-X-1=0$. It has two real roots and a pair of complex roots.
How can one show that the complex roots are not roots of a real number?
AI: Let $p(x)$ be a fixed monic polynomial with integer coefficients and degree $d$. If $r$ is one of its roots and $r^n = b$ is real and non-negative (and $n$ is minimal with this property), then $b$ is a product of algebraic integers, hence an algebraic integer. We therefore have
$$r = \zeta_n \sqrt[n]{b}$$
for some primitive $n^{th}$ root of unity $\zeta_n$. Since $\sqrt[n]{b}$ is real, $\bar{r} = \zeta_n^{-1} \sqrt[n]{b}$ is also a root of $p$, and so it follows that
$$\frac{r}{\bar{r}} = \zeta_n^2$$
lies in the splitting field $F$ of $p$. But $\zeta_n^2$ generates a cyclotomic field of degree $\varphi \left( \frac{n}{\gcd(2, n)} \right)$, and so this must divide the degree of $F$, which divides $d!$. It is known that there are only finitely many numbers with a given totient, so there are only finitely many possibilities for $n$, and so for fixed $p$ this indeed reduces to a finite problem as Gerry says.
When $p(x) = x^4 - x^3 - x^2 - x - 1$, we compute that $\bmod 2$ we have $p(x) \equiv \frac{x^5 - 1}{x - 1}$, which is irreducible (the smallest finite field over $\mathbb{F}_2$ which has elements of order $5$ is $\mathbb{F}_{2^4}$), so $p$ is irreducible and its splitting field has degree dividing $4! = 24$. So $\zeta_n^2$ lies in its splitting field only if
$$\varphi \left( \frac{n}{\gcd(2, n)} \right) | 24.$$
If $q$ is a prime dividing $n$, then $q - 1 | 24$, so we can only have $q = 2, 3, 5, 7, 13$. Of these, the only odd prime which also divides $24$ is $3$ and it only does so once, so $3$ can occur with multiplicity at most $2$ and the other odd primes occur with multiplicity at most $1$. Since $2^3 | 24$, the prime $2$ occurs with multiplicity at most $5$.
Summarizing, to prove that $p(x)$ does not have a root which is the root of a real number, it suffices to prove that $r^{2^5 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13}$ is not real for any of the roots $r$, and this can be done by a finite calculation. Of course this is a large exponent, and the actual size of the possible values of $n$ is smaller, but the possible values of $n$ are somewhat tedious to list out.
Here's another idea for ruling out values of $n$. Recall that if $K \subset L$ is an inclusion of number fields, then the discriminant $\Delta_K$ of $K$ divides $\Delta_L$. The discriminants of the cyclotomic fields $\mathbb{Q}(\zeta_n)$ are known (see Wikipedia, although the general formula is somewhat complicated), and in particular every odd prime divisor of $n$ divides them, so by computing the discriminant of $p$ we can rule out some prime factors.
WolframAlpha tells me that the discriminant of $x^4 - x^3 - x^2 - x - 1$ is $-563$. This is prime so it must be the discriminant of the splitting field, and this already rules out all of the possible values of $n$ above. |
H: Compute Limit Question Confused
$$\lim_{r \to 0^+} \frac{\sqrt r}{(r-9)^4}\
$$
How do i compute this limit? I was told to see what 1/x is approaching and rewrite it but can someone guide me in the right direction?
How can i find which infinity it is approaching?
Also
What does it approach if the limit approach 9 instead of 0 from the right?
AI: We need to look at what happens to top and to bottom as $r$ approaches $0$ from the right. The bottom behaves very nicely: as $r$ approaches $0$, $(r-9)^4$ approaches $(-9)^4$.
The top also behaves nicely: as $r$ approaches $0$ from the right, $\sqrt{r}$ approaches $0$.
So the quotient approaches $0/(-9)^4$, which is $0$.
Things get substantially more difficult when top and bottom both approach $0$. In that sort of situation, the analysis can be quite a bit harder to do.
Remark: You might wish to confirm this with some calculator experimentation. Pick a very small positive $r$, like $r=10^{-6}$. Calculate $\frac{\sqrt{r}}{(r-9)^4}$. You will find it is close to $0$. |
H: Compute Limit Question with tangent
$$\lim_{x \to 0^+} \tan^{-1}\ \left(\frac{1}{x}\right)\
$$
I am not sure how to solve this limit either, it says i should first see what 1/x is approaching but im confused as how to do that, and how to solve it. How would i rewrite this or what rules would i use?
AI: This problem can be rewritten by changing $\lim_{x \to 0^+}$ to $\lim_{x \to \infty}$ and $\frac 1x$ to $(x)$. |
H: compute area under two curve
suppose we are give task to calculate area of figure,which is bounded by two curve
$y=[x]$ and $y=(2-x^2)$, here $[x]$ denotes modulus,not ceiling or rounding of x.
i use wolframalpha to figure out what kind of location,intersection points has this two figure,here is link of this
http://www.wolframalpha.com/input/?i=abs%28x%29%3D2-x%5E2
i see that points of intersection are $-1$ and $1$,also i know that area under two curve
$y=f_1(x)$ and $y=f_2(x)$ and intersection points are $x_1$ and $x_2$ is
$$\int_{x_1}^{x_2}(f_2(x)-f_1(x))dx$$
but i am confused if i should take $y=[x]$ directly or consider two interval $[-1..0]$ and $[0...1]$ and use $-x$ and $x$ for each interval? please give me some hint
AI: Do it separately, it is safer. Anyway, you only need to deal with the first quadrant part. By symmetry for the full area you double the result.
Because the curve $y=2-x^2$ is above the curve $y=x$ in the region of interest, the first quadrant part has area $\int_0^1((2-x^2)-x)\,dx$.
I would somewhat prefer to draw the vertical line from the point of intersection $(1,1)$ to the $x$-axis. Then the desired area is $\int_0^1 (2-x^2)\,dx -\int_0^1 x\,dx$. It feels more concrete: area under $y=2-x^2$, above $x$-axis, from $0$ to $1$, minus the area of the "hole." |
H: Characterizing all ring homomorphisms $C[0,1]\to\mathbb{R}$.
This is something I've been trying to work out this evening.
Let $R$ be the ring of continuous real-valued functions on $[0,1]$ with pointwise addition and multiplication. For $t\in [0,1]$, the map $\phi_t\colon f\to f(t)$ is a ring homomorphism of $R$ to $\mathbb{R}$. I'm trying to show that every ring homomorphism of $R\to\mathbb{R}$ has this form.
Suppose otherwise, that there is some $\phi\neq\phi_t$, and thus there is some $f_t\in R$ such that $\phi(f_t)\neq \phi_t(f_t)=f_t(t)$. Define $g_t=f_t-\phi(f_t)1\in R$. Here $\phi(f_t)1$ is the constant function sending $[0,1]$ to $\phi(f_t)$. Then $g_t(t)\neq 0$. My first small question is why does $\phi(g_t)=0$? It seems only that $\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)$.
I would like to conclude that there are only finitely many $t_i$ such that $g(x)=\sum g_{t_i}^2(x)\neq 0$ for all $x$. Then $g^{-1}=1/g(x)\in R$, but $\phi(g)=0$, contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many $g_{t_i}$ such that the sum of their squares is never $0$? Thanks.
AI: Use compactness. For each $t$ the set $\{x : g_t(x) \ne 0\}$ is open and contains $t$, so the union of all these sets is $[0,1]$, meaning they form an open cover. |
H: Numbers between real numbers
I wonder if there can be numbers (in some extended theory) for which all reals are either smaller or larger than this number, but no real number is equal to that number?!
Is there some extension of number which allows that? Under what conditions (axiom etc.) there is no such number.
AI: Under the axioms of the real numbers this cannot occur. You must add new elements to the real numbers, note that if $\varepsilon$ is smaller than all $\frac1n$ but still positive then $\frac1\varepsilon$ is larger than any real number.
Such $\varepsilon$ is called infinitesimal and their existence is incompatible with the real numbers per se. There is a branch, however, called non-standard analysis in which these numbers play an important role.
One example to such field is called Hyperreal numbers. |
H: $\gcd(n!+1,(n+1)!)$
The recent post didn't really provide sufficient help. It was too vague, most of it went over my head.
Anyway, I'm trying to find the $\gcd(n!+1,(n+1)!)$.
First I let $d=ab\mid(n!+1)$ and $d=ab\mid(n+1)n!$ where $d=ab$ is the GCD.
From $ab\mid(n+1)n!$ I get $a\mid(n+1)$ and $b|n!$.
Because $b\mid n!$ and $ab\mid(n!+1)$, $b$ must be 1.
Consequently, $a\mid(n!+1)$ and $a\mid(n+1)$.
So narrowing down options for $a$ should get me an answer. At this point I've tried to somehow bring it around and relate it to Wilson's theorem as this problem is from that section of my textbook, but I seem to be missing something. This is part of independent study, though help of any kind is appreciated.
AI: The previous posts have I think carefully explained why the gcd is $1$ if $n+1$ is composite. It comes down to this: if $q$ is a prime that divides $(n+1)!$, and $n+1$ is composite, then $q \lt n+1$, and therefore $q \le n$. But then $q$ divides $n!$, and therefore $q$ cannot divide $n!+1$.
You have shown that any common divisor of $n!+1$ and $(n+1)!$ must divide $n+1$.
Suppose now that $n+1$ is prime, say $n+1=p$. Then by Wilson's Theorem, $(p-1)!\equiv -1 \pmod p$. This says that $p$ divides $(p-1)!+1$, meaning that $n+1$ divides $n!+1$.
It follows that if $n+1$ is prime, then $n+1$ is a common divisor of $n!+1$ and $(n+1)!$. It is the greatest common divisor, since all common divisors must divide $n+1$, and nothing bigger than $n+1$ can divide $n+1$.
We conclude that $\gcd(n!+1,(n+1)!)=1$ if $n+1$ is composite, and $\gcd(n!+1,(n+1)!)=n+1$ if $n+1$ is prime. |
H: $G$ is a finite group. $ H,K \leq G $ and $ K \lhd G $. $G:H$ and $|K|$ are coprime. Show that $K \leq H $
Let $G$ be a finite group. $ H,K \leq G $ and $ K \lhd G $.
$G:H$ and $|K|$ are coprime. Show that $K \leq H $
I started like this:
$G:H = (G:KH)(KH:H)$
Therefore, both $(G:KH)$ and $(KH:H)$ are coprime to $|K|$, but have no idea how to continue. Any clues?
AI: $\eqalign{
& [G:KH]=\frac{\vert G\vert}{\vert KH \vert} \cr
& =\frac{\vert G \vert}{\vert H \vert}\frac{\vert H\cap K\vert}{\vert K \vert} \cr
& =[G:H]\frac{\vert H\cap K\vert}{\vert K \vert}
}$
$[G:H]$ and $\vert K\vert$ are coprime,hence $\vert K \vert$ divides $\vert H\cap K\vert$.
But $\vert H\cap K\vert \leq \vert K \vert$,so,... |
H: Chromatic Number Identity Involving Edges
I'm trying the prove the following:
Let $G$ be a simple graph with $m$ edges. Show that $\chi(G)\leq \frac{1}{2}+\sqrt{2m+\frac{1}{4}}.$
A very minute bit of algebraic manipulation shows that this is equivalent to proving $$\chi(G)(\chi(G)-1)\leq 2m.$$ From here I am a bit stuck. Could someone suggest a direction to head in?
Please no full solutions, just hints.
AI: HINT: A little more manipulation turns it into $$m\ge\binom{\chi(G)}2\;,$$ which can be understood as saying that there must be at least as many edges as there are pairs of colors in a minimal coloring. |
H: Fourier-Series of a part-wise defined function?
I have a function f given as
$$
f(x) =
\begin{cases}
ax&\text{ if }\quad-\pi \leq x \leq 0\\
bx&\text{ if }\quad 0<x\leq\pi
\end{cases}
$$
I am supposed to develop the fourier series of this function, using the scalar product
$$
\varphi(f,g) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)g(x)dx
$$
and
$$a_n=\varphi(f,\cos(n x))\\
b_n=\varphi(f,\sin(n x))\\
f(x) = \frac{a_0}{2} +\sum\limits_{n=1}^{\infty} a_n\cos(nx)+b_n\sin(nx)
$$
I came up with $$f(x)=\sum\limits_{n=1}^{\infty} \frac{-2a \cos(n\pi)}{n}\sin(nx)$$ for the first case and $$f(x) = \sum\limits_{n=1}^{\infty} \frac{-2b \cos(n\pi)}{n}\sin(nx)$$ for the other one - but that's two fourier series, not one, so I'm sure this is wrong. That being said, I have no idea how else I would go about this. Especially troubling is that the task demands that I calculate $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+2)^2}$$ somehow by setting $x$ to $0$ - but my series is $0$ at $x=0$ (and at least that seems to be right, because the function is too).
So if anyone could clear me up on how to solve this tasks and what misunderstandings I'm sure I've made, that would be great.
AI: while calculating the Fourier coefficients you have to use the definition of the function (not only one of its parts). You calculated the series for $f_1(x) = ax$ and $f_2(x) = bx$ but not for your $f$. So for example
\begin{align*}
a_0 &= \frac 1\pi\int_{-\pi}^\pi f(x)\, dx\\\
&= \frac 1\pi\left(\int_{-\pi}^0 ax\, dx + \int_0^\pi bx\, dx\right)\\\
&= \frac 1\pi\left( -\frac{a\pi^2}2 + \frac{b\pi^2}2 \right)\\\
&= \pi \cdot \frac{b-a}2
\end{align*}
I think yoou can do the other coefficients now for yourself. |
H: Differentiability of $(x,y)\mapsto|x|\cdot y$
Check the differentiability of the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ of two variables given by the formula $$f(x)=|x_1|\cdot x_2$$
I still have problems with this. I started by trying to count partial derivatives:
$\displaystyle\frac{\partial f}{\partial x_1}(x)=\lim_{h\to 0}\frac{f(x_1+h,x_2)-f(x_1,x_2)}{h}=\lim_{h\to 0}\frac{|x_1+h|x_2-|x_1|x_2}{h}$, so I think the problems can be at the points: $(0,x_2)$, where $x_2\neq 0$, because then we have that this limit is equal to $\displaystyle\lim_{h\to 0}\frac{|h|x_2}{h}$ which doesn't exist (left and right limits are not equal).
$\displaystyle\frac{\partial f}{\partial x_2}(x)=\lim_{h\to 0}\frac{f(x_1,x_2+h)-f(x_1,x_2)}{h}=\lim_{h \to 0}\frac{|x_1|(x_2+h)-|x_1|x_2}{h}=|x_1|$, so I think we haven't any problems here, this partial derivative always exists.
But what exactly can we deduce from these speculations about partial derivatives?
I've also tried to proudly find the differential of this function. I was taught that the function is differentiable at the point $x$ iff there exists (if there exists, there is only one) a linear mapping $L$ such that $(*)\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)-L(h)}{\|h\|}=0$. Then we say that $Df(x)=L$ is a differential of function $f$ at the point $x$. For example consider function $g:\mathbb{R}^2\rightarrow \mathbb{R}, \ g(x)=x_1\cdot x_2$. We can find differential of this function by looking at the increment of this function: $g(x+h)-g(x)=(x_1+h_1)(x_2+h_2)-x_1x_2=x_1h_2+x_2h_1+h_1h_2$ . Then the candidate for $Df(x)$ is linear part of this increment: $L(h)=x_1h_2+x_2h_1$. When we check $(*)$ it appears that indeed it is a desired differential.
But in my example: $f(x+h)-f(x)=|x_1+h_1|(x_2+h_2)-|x_1|x_2$ I'm confused, it seems hard. Do I have to consider a few cases depending on a sign of $x_1, \ x_2$ ?
Can anybody make it clear for me? It is really important to me to finally understand this topic.
AI: A necessary condition for a function $f\colon \mathbb{R}^2 \to \mathbb{R}$ to be differentiable at $x_0$ is that both partial derivatives at $x_0$ exist. In this case, $$Df(x_0) \colon (h_1,h_2) \mapsto \frac{\partial f}{\partial x_1}(x_0)h_1 +\frac{\partial f}{\partial x_2}(x_0)h_2.$$
In other words, if the differential exists, then it must have this precise form.
If one partial derivative does not exist at $x_0$, then you can stop. Otherwise, you need to check that the linear mapping $$(h_1,h_2) \mapsto \frac{\partial f}{\partial x_1}(x_0)h_1 +\frac{\partial f}{\partial x_2}(x_0)h_2$$ satisfies the definition of derivative. A sufficient condition for differentiability is that both partial derivatives exist and are continuous around $x_0$. |
H: find point with equidistant from two points
Suppose there is given two point $A=(-4;-2)$ and $B=(3,2)$ we have to find such $C$ point on
$OY$ axis, such that
a) $C$ is equidistant from $A$ and $B$
b) $ACB$ spline must be minimum.
As I know for solving part (a), we should write equation of line, which is perpendicular of $AB$ segment and goes through it's midpoint as well, in a) case coordinates of C is $(0,y)$
midpoint of $AB=(-0.5,0)$,so equation would be
$y=k(x+0.5)$ where $k$ is equal to $-7/4$,because the slope of segment $AB$ is $4/7$ and we know that for perpendicular lines,$\text{slope}_1*\text{slope}_2=-1$, so we get that $y=-7/4*(x+0.5)$.
If $x=0$ then $y=-0.5*7/4=-7/8$, so I have got that for a $C$ coordinates are $(0,-7/8)$.
Am i correct for part (a)?
As for (b) I think that I can take symmetry point of $B'$ related to $B$ along
so that $B'=(-3,2)$, equation of $AB'$ would be
$y+2=4(x+4)$ or $y=4*x+14$ coordinated of $C$ is if we put $x=0$ we get $(0,14)$.
Am i right? please help me to check my work
AI: I thing your method is corretc.A short method to solve a) is : if $I$ is the middle of $[AB]$ then $C$ is determined by : $\overrightarrow{CI}.\overrightarrow{AB}=0$ and $x_C=0$ |
H: $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral
Compute the following integral:
$$\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$$
Any hint, suggestion is welcome.
AI: Yet a different approach: parametric integration. Let
$$
F(\lambda)=\int_{0}^{\infty} \frac{e^{-\lambda x} \sin(x)}{x}\,dx,\qquad\lambda>0.
$$
Then
$$
F'(\lambda)=-\int_{0}^{\infty} e^{-\lambda x} \sin(x)\,dx=-\frac{1}{1+\lambda^2}.
$$
Integrating and taking into account that $\lim_{\lambda\to\infty}F(\lambda)=0$ we have
$$
F(\lambda)=\frac\pi2-\arctan\lambda
$$
and
$$
\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x}\,dx=F(1)=\frac\pi4.
$$ |
H: property to be exact a 1-form on $\mathbb R^2 -\{(0,0)\}$
(a) Let $\omega$ a $1-$form defined on the open set $ U \subset \mathbb R ^n$ and $ c:[a,b] \to U$ a $ C^1 -$differentiable curve such that $ |\omega (c(t))| \leq M \quad \forall t \in [a,b]$
Prove that
$$ \displaystyle{\Bigg| \int_c \omega \Bigg| \leq ML}$$
where $L$ is the length of the curve $c$.
(b) Let $\omega=a_1dx +a_2dy$ a closed $1-$form defined on $\mathbb R^2 -\{(0,0)\}$. If $\omega$ is bounded ( which means that $a_1 ,a_2$ are bounded) on a disk with center the origin $O(0,0)$ prove that:
$$ \omega \text{ is exact on } \mathbb R^2 -\{(0,0)\}$$
(c) If $\omega$ is a closed $1-$form defined on $\mathbb R^2 -\{(0,0)\}$ such that $\displaystyle{ \lim_{x^2 + y^2 \to 0} \left( \sqrt{x^2 + y^2} \omega \right) =0 }$ prove that :
$$ \omega \text{ is exact on } \mathbb R^2 -\{(0,0)\}$$
I have done (a) but I am stuck in (b) and (c). I think I have to use (a) and Poincare's lemma for $1-$ forms but I don't know how.
AI: I don't know what your proof of ii) looks like, but I'd expect that you estimate $$| \int_{\gamma} \omega | \le C M r $$
for closed curves $\gamma$ surrounding the origin, e.g. $\gamma(t)= r(\cos t , \sin t)$ and then let $r\rightarrow 0$. That approach should do the trick, for iii), as well. (in ii) you know the integrals approach $0$ as fast as $r$, but you don't need that rate of decay, convergence to $0$ is sufficient).
Edit (added explanation): if you write down the integral with the curves I defined earlier, you get
$$\left| \int_\gamma \omega\,\right| =\left| \int_0^{2\pi}r(\omega_1\circ\gamma\sin(t) + \omega_2\circ\gamma \cos(t)) dt \right|\le \int_0^{2\pi}|\cdots |\le C r|\sup_{x=r} \omega |$$
for some constant $C$ not depending on $r$ or $\omega$ which swallows the integral over $\sin, \cos$. Note $r=\sqrt{x^2+y^2}$, so the expression on the rhs tends to $0$ when $r$ does. Since $\omega$ is closed this implies that $\int_\gamma \omega$ vanishes for all these $\gamma$, since they are homotopic to each other in the domain of definition of $\omega$, so the integral is the same for all these curves. You should be in the posession of a theorem which allows you to conclude from this that $\omega$ is exact. If this is not true you should explain how you solved ii)
Another edit: In simply connected domains every closed form is exact, this you seem to know. Recall the proof of this. In general, this is not true. What remains true is that a closed differential form $\omega$ is exact if the integral along any closed curve vanishes. This applies here, as I outlined earlier.
(You should have seen a theorem claiming something like that if you are given this kind of exercise. If not, it is not too hard to see, though a bit technical if you want to make it rigorous. You pick and fix any point $p$ in the domain you are looking at, $U$ say, (assume it's open and connected, then it's path connected), and for $q$ in $U$ choose any smooth curve $\gamma_{pq}$ joining $p$ and $q$. Then define
$$F(p) = 0; \,\, F(q) :=\int_{\gamma_{pq}} \omega $$
You need to verify that this is well defined, i.e. does not depend on the particular choice of $\gamma_{pq}$, but this is what follows from the fact that the integral over any closed curve vanishes. The reasoning above does show this only for curves winding once around the origin, but is easily generalizd accordingly. I don't go into the details of this particular claim since I don't know which tools you have available.
Once you know $F$ is well defined you show $dF=\omega$, this is similar to the case of simply connected domains). |
H: Lebesgue integral vs area under a curve
Possible Duplicate:
Lebesgue measure on Riemann integrable function in $\mathbb{R}^2$
Is the Lebesgue integral of a positive real function of a real variable equivalent to the Lebesgue measure of the set (in $\mathbb{R}^2$) of all the points between the interval of integration and the graph of the function?
I'm asking this because all the different definitions of "length", "area", "volume", "measure" I was exposed to (Euclidean geometry, path length, measure of a set, integral, scalar product, ...) seem to be different from one another and I would like to see what are the points in common
AI: Let $f \ge 0$ be measurable. Then (where $\chi_A$ denotes the characteristic function of a set $A$)
\begin{align*}
\int_{\mathbb R} f(x)\, dx
&= \int_{\mathbb R} \int_{\mathbb R} \chi_{[0, f(x))}(y)\, dy\, dx\\\
\text{and by the Fubini-Tonelli theorem,} &\\
&= \int_{\mathbb R^2} \chi_{[0, f(x))}(y)\, d(x,y)\\
&= \int_{\mathbb R^2} \chi_{\{(a,b) \mid 0 \le b < f(a)\}}(x,y)\, d(x,y)\\\
&= \lambda\bigl(\{(x,y) \mid 0 \le y < f(a)\}\bigr).
\end{align*} |
H: Continuous images of open sets are Borel?
Consider a Polish space $(X,d)$ and any metric space $(Y,e)$. If we have a continuous surjection $f:X\to Y$ then is the image $f(U)$ of any open subset $U\subset X$ a Borel set in $Y$?
I know that this is true if we allow $X$ to be compact, since every open subset of a metric space is $F_{\sigma}$ and closed subsets of a compact metric space are compact, and moreover continuous images of compact sets are compact. So in this case, the image of an open set would be $F_{\sigma}$ in $Y$.
If this is not the case for the given $f$, would it do any difference if we allowed $Y$ to be in addition Polish? Or compact? Or should $f$ have more properties? How about $X$? Or is this all just a triviality that I have overlooked?
Thanks for all the input in advance.
AI: The answer is no. Endow both $\mathbb{R}^2$ and $\mathbb{R}$ with the usual Borel $\sigma$-algebra and let $\pi_1:\mathbb{R}^2\to\mathbb{R}$ be the projection onto the first coordinate. Now it is a well known fact that there exists Borel sets $B\subseteq\mathbb{R}^2$ such that $\pi_1(B)$ is not Borel. Now $B$ will be in general not open, but we can make it so. We just apply the following theorem (Lemma 4.58 in Alipranits & Border 2006):
Let $\mathcal{C}$ be a countable family of Borel subsets of a Polish space
$(X, \tau)$. Then there is a Polish topology $\tau'\supseteq\tau$ on $X$ with the same Borel $\sigma$-algebra for which each set in $\mathcal{C}$ is both open and closed.
In particular, we can make $B$ open. Then $B$ is open and $\pi_1$ is a continuous surjection, but $\pi_1(B)$ not Borel. Since one can take $B$ in the original space to be bounded, the example works even when $Y$ is compact. |
H: Euler's product formula for $\sin(\pi z)$ and the gamma function
I want to derive Euler's infinite product formula
$$\displaystyle \sin(\pi z) = \pi z \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right)$$
by using Euler's reflection equation $\Gamma(z)\Gamma(1-z) \sin(\pi z) = \pi$ and the definition of $\Gamma(z)$ as an infinite product, namely
$$\displaystyle \Gamma(z) := \frac{1}{z} \prod_{k=1}^\infty \frac{(1+\frac{1}{k})^z}{1+\frac{z}{k}}.$$
To be precise, I obtain that
$$\sin(\pi z) = \pi z(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right)$$
hence I wish to prove
$$(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right) = \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right).$$
I multiplied things out and got it to the form
$$(1-z) \prod_{k=1}^\infty \frac{1 + \frac{1}{k} + \frac{z(1-z)}{k^2}}{1 + \frac{1}{k}} = (1-z) \prod_{k=1}^\infty \left( 1 + \frac{\frac{z(1-z)}{k}}{1+\frac{1}{k}}\right)$$ however the $(1-z)$ factor out front is giving me some trouble; I'm not sure how to proceed.
AI: Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:
$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$ |
H: Automorphisms of the field of complex numbers
Using AC one may prove that there are $2^{\mathfrak{c}}$ field automorphisms of the field $\mathbb{C}$. Certainly, only the identity map is $\mathbb{C}$-linear ($\mathbb{C}$-homogenous) among them but are all these automorphisms $\mathbb{R}$-linear?
AI: An automorphism of $\mathbb C$ must take $i$ into $i$ or $-i$. Thus an automorphism that is $\mathbb R$-linear must be the identity or conjugation. |
H: Symbol for finite
I understand there is a symbol for infinite. Is there one for finite?
I searched and found there is none. How is finite represented symbolically?
AI: I have never seen a notation for 'finite,' but what I do very often see is denoting something finite as simply being less than infinity. For example, $|A| < \infty$, or $[G:H] < \infty$.
Small thing I'd like to add: Of course something like $[G:H] < \infty$ isn't technically meaningful, but it certainly gets the point across and in my experience at least seems to be pretty standard. |
H: Homomorphisms into complex numbers
Let $I$ be a set of huge cardinality, that is, let $|I|>\mathfrak{c}$. Consider the real product algebra $\mathbb{R}^I$ of all real functions defined on $I$. Can we determine:
1) all algebra homomorphisms $\varphi\colon \mathbb{R}^I\to \mathbb{R}$?
2) all algebra homomorphisms $\varphi\colon \mathbb{R}^I\to \mathbb{C}$, where $\mathbb{C}$ is considered as a real algebra over $\mathbb{R}$?
At least when $|I|\leqslant \mathfrak{c}$, for (1) we can describe the homomorphisms as point evaluations.
Apologies if my questions are trivial. I am not an algebraist and I am just trying to build a general picture of this world.
AI: Edit: Jason DeVito's comment below correctly indicates that this argument is incomplete. It remains to show that the ultraproduct with respect to a non-principal ultrafilter is always nonstandard (in that it contains elements not equivalent to the constant sequences). Equivalently, it remains to show that no ultrafilter $F$ on $I$ can have the property that in any partition of $I$ into at most $|\mathbb{R}|$ disjoint subsets, exactly one subset lies in $F$. (Being an ultrafilter is equivalent to having this property for finite partitions.) I do not actually know if this is true.
You only get evaluation maps.
Let $\varphi : \mathbb{R}^I \to \mathbb{R}$ be an algebra homomorphism. Let $S$ be any subset of $I$. The indicator function $1_S \in \mathbb{R}^I$ is idempotent (it is its own square) and so must be sent to an idempotent of $\mathbb{R}$. The only such idempotents are $0$ or $1$, so this determines a distinguished collection $F$ of subsets of $S$, namely those subsets whose indicator functions map to $1$ under $\varphi$.
$F$ has the following properties. The empty set is never in $F$. Since $\varphi$ is a homomorphism we must have
$$\varphi(1_S) = 1, \varphi(1_R) = 1 \Rightarrow \varphi(1_S 1_R) = \varphi(1_{S \cap R}) = 1$$
so $F$ is closed under intersection. Moreover, if $\varphi(1_S) = 1$ and $S \subset R$, then $1_S 1_R = 1_S$, hence $\varphi(1_S 1_R) = \varphi(1_S) = 1$, hence $\varphi(1_R) = 1$, so $F$ is also upward closed. Finally, since $\varphi(1) = 1$, we have
$$\varphi(1 - 1_S) = 1 - \varphi(1_S) = \varphi(1_{S^c})$$
hence $S \in F$ if and only if $S^c \not \in F$.
These are precisely the axioms defining an ultrafilter on $I$. Now suppose $f, g \in \mathbb{R}^I$ are two functions such that the set $S$ of $x \in I$ for which $f(x) = g(x)$ lies in $F$. Then $(f - g) = (f - g) 1_{S^c}$ and $\varphi(1_{S^c}) = 0$, hence $\varphi(f) = \varphi(g)$. It follows that $\varphi$ factors through the ultraproduct of $I$ copies of $\mathbb{R}$ defined by the ultrafilter $F$.
This ultraproduct is already a field. If $F$ is a principal ultrafilter, $F$ consists of all sets containing a fixed $x \in I$, and the ultraproduct reduces to $\mathbb{R}$; in this case $\varphi$ is just the evaluation homomorphism at $x$. If $F$ is non-principal, then the resulting ultraproduct is a nonstandard version of $\mathbb{R}$ (it satisfies the same first-order properties but has oddities like infinitesimal and infinitely large elements). The standard version of $\mathbb{R}$ sits properly inside it, and since it is a field it admits no nontrivial quotients, so $\varphi$ cannot exist in this case. |
H: Lattices of Subgroups and Graph
In Dummit and Foote´s Abstract Algebra, when talking about the lattice of subgroups of $A_4$, the authors make the statement that, unlike virtaully all groups, $A_4$ has a planar lattice? My question is
What do they mean when they say virtually all groups? Is there a reference for this statement?
On a somewhat different note, are there classes of graphs which can be realized as the subgroup lattice of some group?
AI: Starr–Turner (2004) classifies abelian groups with planar subgroup lattices, and Schmidt (2006) and Bohanon–Reid (2006) complete the classification. Dummit–Foote made this claim prior to this work, so it may have just been the observation that fairly small groups have non-planar lattices, so that it would intuitively be pretty hard for a large group to have a planar lattice.
Starr, Colin L.; Turner, Galen E., III.
"Planar groups."
J. Algebraic Combin. 19 (2004), no. 3, 283–295.
MR2071475
DOI:10.1023/B:JACO.0000030704.77583.7b
Schmidt, Roland.
"Planar subgroup lattices."
Algebra Universalis 55 (2006), no. 1, 3–12.
MR2217274
DOI:10.1007/s00012-006-1964-2
Bohanon, Joseph P.; Reid, Les.
"Finite groups with planar subgroup lattices."
J. Algebraic Combin. 23 (2006), no. 3, 207–223.
MR2228925
DOI:10.1007/s10801-006-7392-8
James Wilson's catalog of groups by lattice (from 2004-2005) is mostly in tact at his new webpage. |
H: Derivative of a function defined by the divided difference of another function.
Given a function $f$ of class $C$ $^{n+2}$ in an interval $[a,b]$ and $x_{0}=a<x_1<x_2 ... <x_n = b$ a subdivision of $[a,b]$ into $n+1$ points. Given another function $g$ defined in the same interval $[a,b]$ by the divided difference such that $g(x) = f[x_0, x_1, ... , x_n, x]$. Prove that $g'(x)=f[x_0, x_1, ... , x_n, x, x]$.
AI: The divided difference is the same whatever the arrangement of the inputs i.e.
$f[x_0, x_1, ... , x_n, x, x]=f[x,x_0, x_1, ... , x_n, x]$
and $f[x_0, x_1, ... , x_n, x]=f[x,x_0, x_1, ... , x_n]=g(x)$
$g[u,v]=\cfrac{g(u)-g(v)}{u-v}$
Also $\displaystyle \lim_{(u,v)->(x,x)}$$g[u,v]=g'(x)$
we have $g(u)=f[u,x_0, x_1, ... , x_n]$ and $g(v)=f[x_0, x_1, ... , x_n, v]$, then it follows that $g'(x)=\displaystyle\lim_{(u,v)->(x,x)}g[u,v]=\lim_{(u,v)->(x,x)}\cfrac{f[u,x_0, x_1, ... , x_n]-f[x_0, x_1, ... , x_n, v]}{u-v}$
since $\cfrac{f[u,x_0, x_1, ... , x_n]-f[x_0, x_1, ... , x_n, v]}{u-v}$$=f[u,x_0, x_1, ... , x_n, v ]=f[x_0, x_1, ... , x_n, u,v ]$
$\displaystyle g'(x)=\lim_{(u,v)->(x,x)}f[x_0, x_1, ... , x_n, u,v ]=f[x_0, x_1, ... , x_n, x, x]$ |
H: automorphisms of varieties with respect to a cover
Let $X$ and $Y$ be (smooth projective connected) varieties over $\mathbf{C}$.
Let $\pi:X\to Y$ be a finite surjective flat morphism.
Does this induce (by base change) a map $\mathrm{Aut}(Y) \to \mathrm{Aut}(X)$?
I think it does. Given an automorphism $\sigma:Y\to Y$, the base change via $\pi:X\to Y$ gives an automorphism of $X$.
My real question is as follows:
Is $\mathrm{Aut}(Y)\to \mathrm{Aut}(X)$ injective?
If not, under which hypotheses is $\mathrm{Aut}(Y)\to \mathrm{Aut}(X)$ injective? Does $\pi$ etale do the trick?
What if $\dim X=\dim Y =1$?
AI: I don't think that there is any (canonical) map $Aut(Y) \to Aut(X)$ of the kind you presume exists.
E.g. if $X$ is a curve of genus $g \geq 2$ and $Y$ is $\mathbb P^1$, then $Aut(X)$ is finite (often trivial), while $Aut(Y)$ equals $PGL_2(\mathbb C)$, which is simple. What is the map that you have in mind? (In any case, whatever it it is, it won't be injective.) |
H: Prove that $fg\in L^r(\Omega)$ if $f\in L^p(\Omega),g\in L^q(\Omega)$, and $\frac1 p+\frac1 q=\frac1 r$
Can anyone give me a hint for proving the following:
Let $\Omega$ be a measure space. Assume $f \in L^p(\Omega)$ and $g \in L^q(\Omega)$ with $1 \leq p, q \leq \infty$ and $\frac1p + \frac1q \leq 1$. Prove that $fg \in L^r(\Omega)$ with $\frac1r = \frac1p + \frac1q$.
Note: One should be able to use (the standard) Hölder inequality. Notice that if you have $\frac1p + \frac1q = 1$ you recover the former result.
AI: Hint:
$$
1 = \frac{r}{p}+\frac{r}{q} = \frac{1}{\frac{p}{r}}+\frac{1}{\frac{q}{r}}
$$ |
H: How to handle the following percentage scenarios?
I have the following scenarios, but I am unclear on how to handle them correctly. I start off with a value like 200 and the following scenarios are:
Remove 10% from 200, and then remove a compound 20% from that value.
Remove 10% from 200, and then remove 20% from 200. I guess this is just like removing 30%, correct?
Add 10% to 200, and then remove a compound 20% from that value.
Add 10% to 200, and then remove 20% from 200.
I am also unclear on on a certain on something else because I have been given 2 different answers. If I take 100 and want to remove 10%, is it:
100 * 0.9 = 90
or
100 / 1.1 = 90.9090....
What is the difference above?
Lastly, is there a formula than can be converted into an algorithm that will handle more than removing/adding between 1 and 2 percentages (≥3)? If so, what is it?
Initial Value = 252
Remove 10% and then a 20% Compound
Based on your answer, I am doing:
252 - (252 * (10/100)) = 226.8
226.8 - (226.8 * (20/100)) = 181.44
However, they are doing:
252 * 0.9 = 226.8
226.8 / 1.2 = 189
AI: First, I will answer your second question. To reduce a number by 10%, you always multiply by 0.9. I don't know the context in which you think you were told that you sometimes divide by 1.1, but I am sure there is something here you have misunderstood. (I have a guess; since you asked about it, I have added it below.)
1: To remove 10% from 200, and then remove a compound 20% from that value:
Remove 10% from 200, giving 200 - 200×10% = 200 - 20 = 180.
Remove 20% from 180, giving 180 - 180×20% = 180 - 36 = 144.
The answer is 144.
2: Your idea here is correct.
3: To add 10% to 200, and then remove a compound 20% from that value:
Add 10% to 200, giving 200 + 200×10% = 200 + 20 = 220.
Subtract 20% from 220, giving 220 - 220×20% = 220 - 44 = 176.
4: You can use your idea from #2 here.
To handle multiple increases or reductions, just do them one step at a time.
I hope this is some help.
Suppose you see a price $p$ and you know that this price was marked up by 10% sometime in the past. You want to know what the original price was before the markup. This is $p ÷ 1.1$.
Notice that this is not the same as if you reduce $p$ by 10%! That would be $p × 0.9$, which is different.
I guess that the reason you thought you might reduce a number by 10% by dividing by 1.1 is that you were confused about this case. It seems as though reducing a number by 10% should be the same as undoing an increase of 10%, but it isn't.
For example, if you see a price of \$110 and you know it was marked up by 10%, then the original price was \$110 ÷ 1.1 = \$100. But to reduce \$110 by 10% you calculate 110 × 0.9 = \$99. Undoing a markup of 10% is not the same as reducing by 10%. |
H: Prove that the function is constant
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be a differentiable function such that for all $x\in\mathbb{R}^2$,
$$
\frac{\partial f}{\partial x_2}(x)=2\cdot\frac{\partial f}{\partial x_1}(x).
$$
Prove that for every $c\in\mathbb{R}$ function $f$ is constant on $$M_c=\left\{x\in\mathbb{R}^2: 2x_2+x_1=c\right\}.$$
I don't know how to approach.
AI: Consider the map $\phi \colon \mathbb{R} \to \mathbb{R}^2$ defined by
$$
\phi(t) = \begin{pmatrix}
c-2t \\
t \end{pmatrix}.
$$
Now try to compute
$$
\frac{d}{dt}\left( f\circ \phi \right) = \frac{\partial f}{\partial x_1} \cdot (-2) + \frac{\partial f}{\partial x_2}
$$
and use the assumption. What can you conclude from this? |
H: What is the length of a maximal deranged sequence of permutations
We were playing a home-made scribblish and were trying to figure out how to exchange papers. During each round, you'll trade k times and each time you need to give your current paper to someone who has never had it, and you need to receive a paper that you've never had. There are n papers. For example, if everyone passes their paper to the left, then you can trade $n-1$ times, and on the $n$th trade everyone gets their papers back. Clearly $k < n$ no matter how you trade.
It is suboptimal for one player to always trade with the same player, so we want to use different permutations each time. When writing a webpage to choose permutations randomly, I ran into a theoretical problem: if we don't know k, can we still generate a good sequence of shuffles? As long as any good sequence of shuffles can be extended to a maximal sequence, we are ok. For n ≤ 6 this is true. Is it true in general?
For n a positive integer, call a sequence of permutations $g_i \in S_n$ deranged if $\prod_{i=a}^b g_i$ has no fixed points on $\{1,\dots,n\}$ for any $1 \leq a\leq b \leq k$, where $k$ is the length of the sequence. Must every maximal deranged sequence have $k=n-1$?
A deranged sequence of length 1 is just called a derangement.
We partial order deranged sequences by $a \leq b$ if $a$ is an initial segment of $b$, so that $(1,2,3,4,5) \leq (1,2,3,4,5), (1,2,3,4,5) \leq (1,2,3,4,5), (1,2,3,4,5), (1,3,5,2,4)$. Hence a deranged sequence $g_1, \dots, g_k\in S_n$ is maximal iff for every $g_{k+1} \in S_n$, the sequence $g_1, \dots, g_k, g_{k+1}$ is not deranged.
Examples:
In order to make the problem more symmetrical, it can be helpful to append $g_{k+1} = (g_1 \cdots g_k)^{-1}$ to the sequence. This corresponds to a final step of "handing everyone back their original paper." Then every consecutive $k-1$ subsequence of every cyclic permutation has the property that it is deranged. Thus these deranged "cycles" of length $k+1$ are acted on both by $S_n$ (relabeling the people) and by $C_{k+1}$ (cyclic permutations). This can help reduce the number of truly distinct examples.
For two players, obviously you just pass it to each other and it is over.
(1,2) [ add (1,2) to complete the cycle ]
For three players, you can either pass clockwise twice or counterclockwise twice.
(1,2,3), (1,2,3) [ add (1,2,3) to complete the cycle ]
(1,3,2), (1,3,2) [ this is the previous one with players 2 and 3 swapped ]
For four players, there are 24 deranged sequences, but after completing them to deranged cycles, there are only 3 distinct orbits under $S_n$ and $C_{k+1}$. Notice that $k+1=n$ in each case:
(1,2)(3,4), (1,3)(2,4), (1,2)(3,4), (1,3)(2,4) -- trade within, across, within, across
(1,2)(3,4), (1,3,2,4), (1,2)(3,4), (1,4,2,3) -- across, left, across, right
(1,2,3,4), (1,2,3,4), (1,2,3,4), (1,2,3,4) -- four lefts
For five players there are 1344 deranged sequences, and once completed they fall into 4 orbits. In each case $n=k+1$.
(1,2)(3,4,5), (1,3)(2,4,5), (1,2,3,4,5), (1,2,4,5,3), (1,4,5,2,3)
(1,2)(3,4,5), (1,3)(2,4,5), (1,4,3,5,2), (1,3,5,4,2), (1,3,4,2,5)
(1,2,3,4,5), (1,2,3,4,5), (1,2,3,4,5), (1,2,3,4,5), (1,2,3,4,5) -- five left
(1,2,3,4,5), (1,2,3,4,5), (1,3,5,2,4), (1,5,4,3,2), (1,3,5,2,4) -- left, left, double-left, right, double-left
For six players, the number of possibilities seems to explode (1128960 deranged sequences, 362 orbits of deranged cycles), but in each case $n=k+1$.
The sequence OEIS:A000479 may be relevant.
AI: Form a $k\times n$ matrix
$$A=\left[\matrix{a_{11}&a_{12}&\dots&a_{1n}\\
a_{21}&a_{22}&\dots&a_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{k1}&a_{k2}&\dots&a_{kn}}\right]$$
as follows: $a_{ij}$ is the number of the person holding $i$’s paper after $j-1$ rounds. For example, your first deranged sequence for four players produces the matrix
$$\left[\matrix{1&2&3&4\\
2&1&4&3\\
4&3&2&1\\
3&4&1&2}\right]\;.$$
Each row of $A$ must be a permutation of $1,\dots,n$, and no column of $A$ may contain any number more than once, so $A$ is a $k\times n$ Latin rectangle, which can always be extended to a Latin square. Thus, every maximal deranged sequence has length $n-1$.
Added: Since OEIS A000479 counts the number of such Latin squares with first row $[\matrix{1&\dots&n}]$, it is indeed relevant: it counts the number of unreduced maximal deranged sequences. Dividing by $(n-1)!$ gives the number of reduced Latin squares of order $n$. According to Wikipedia, this number is known only for $n\le 11$, so you’re unlikely to get any nice expression for it. |
H: Does this multivariate function have only one maximum?
Let $X_1$ and $X_2$ be random variables (not of the same distribution and not independent). Both have a zero probability of being below $-1$. Their joint density is $\rho(x_1,x_2)$. Also, they both have finite expectations.
Now, define the region $A = \{ (t_1,t_2)\in\mathbb{R}^2 \mid 0\le t_1,t_2<1 \text{ and }t_1+t_2<1 \}$, and define the function $f:A\to\mathbb{R}$ with
$$f(t_1,t_2) = \int_{-1}^\infty\int_{-1}^\infty \log(1+t_1x_1+t_2x_2)\rho(x_1,x_2)\,dx_1dx_2.$$
Can we say something interesting about $f$? For example, does $f$ have at most only one local maximum?
AI: It does, because $f$ is strictly concave. Indeed, consider its restriction to a line through $(t_1,t_2)$ in the direction $(v_1,v_2)$. In terms of parameter $s$ this restriction is
$$f(t_1+sv_1,t_2+sv_2)=\int_{-1}^\infty\int_{-1}^\infty \log(1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))\rho(x_1,x_2)\,dx_1dx_2$$
The one-variable function $s\mapsto \log(1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))$ is strictly concave unless $v_1x_1+v_2x_2=0$. Assuming the joint density is indeed a density and not some singular measure, we conclude that for $\rho$-a.e. $(x_1,x_2)$ strict concavity holds. Thus, integration against $\rho$ gives a strictly concave function.
If you want a more explicit argument, take the second derivative in $s$:
$$\frac{d^2}{ds^2}f(t_1+sv_1,t_2+sv_2)\\=-\int_{-1}^\infty\int_{-1}^\infty (v_1x_1+v_2x_2)^2 (1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))^{-2}\rho(x_1,x_2)\,dx_1dx_2<0$$ |
H: Help with complex number phasor notation
I am having trouble understanding how $10jy$ is converted to $10 e^{j\pi/2}$. Here $x$ and $y$ are unit vectors:
(original image)
$$\large=\operatorname{Re}\left[(10\hat{x}-10j\hat{y})e^{-j10\pi z}e^{jwt}\right]$$
$$\large=\operatorname{Re}\left[(10\hat{x}-10e^{j\pi/2}\hat{y})e^{-j10\pi z}e^{jwt}\right]$$
$$\large=\underbrace{10\hat{x}\cdot\cos(\omega t-10\pi z)}_{Ex}+\underbrace{10\hat{y}\cos(\omega t-10\pi z-\tfrac{\pi}{2})}_{Ey}$$
Thank you.
AI: Euler's formula says that for any $\theta$,
$$e^{j\theta}=\cos(\theta)+j\sin(\theta).$$
Therefore,
$$e^{j\pi/2}=\cos(\tfrac{\pi}{2})+j\sin(\tfrac{\pi}{2})=0+j\cdot 1=j$$
and thus $10j=10e^{j\pi/2}$ (or, if you want to talk about vectors, $10j\hat{y}=10e^{j\pi/2}\hat{y}$; but note that it is incorrect to say that $10j\hat{y}=10e^{j\pi/2}$, because the left side is a vector, and the right side is a scalar). |
H: Calculation of atan2
I am familiar with the basics of atan2. The doubt I have in the computation of atan2 came across from an image processing sofware.
This is a portion of the code segment when x>y. x and y are absolute values.
const_1 = 57.2836266;
const_2 = -18.6674461;
const_3 = 8.91400051;
const_4 = -2.53972459;
c = sqrt(y/x);
c2 = c*c;
angleInTheta = (((const_4*c2 + const_3)*c2 + const_2)*c2 + const_1)*c;
What confuses me is the formula of angleInTheta. The results are perfectly correct when applied. I would like to know get a brief explanation of this formula.
Thanks,
AI: This produces a not-too-bad approximation to $\dfrac{180}{\pi} \arctan(\sqrt{y/x})$ for $0 < y < x$. I suppose the $x$ and $y$ are squares of the actual coordinate values. If you call your function $f(\sqrt{y/x})$, it appears that $f(t)/t$ is close to a best uniform approximation of $\dfrac{180}{\pi} \dfrac{\arctan(t)}{t}$ on the interval $[-1,1]$ by polynomials of degree $6$. That best approximation would have coefficients 57.28492047,
-18.69087134, 8.976572844, 2.581481000, according to Maple. |
H: Why do mathematicians care so much about zeta functions?
Why is it that so many people care so much about zeta functions? Why do people write books and books specifically about the theory of Riemann Zeta functions?
What is its purpose? Is it just to develop small areas of pure mathematics?
AI: For one thing, the Riemann Zeta function has many interesting properties. No one knew of a closed form of $\zeta (2)$ until Euler famously found it, along with all the even positive integers:
$$\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$
However, to this day, no nice closed form is known for values in the form $\zeta(2n+1)$.
Another major need of the Zeta function is relating to the Riemann hypothesis. This conjecture if fairly simple to understand. It essentially hypothesizes that the nontrivial zeros of the zeta function have a real part of 1/2. This hypothesis, if proven true, has major implications in number theory and the distribution of primes.
The Riemann zeta function also occurs in many fields and appears occasionally when evaluating different equations, just as many other functions do.
Lastly, the sum
$$\sum_{n=1}^{\infty} \frac{1}{n^s}$$
is a very natural one to try and study and evaluate and is especially interesting because of the above-mentioned properties and more. |
H: Probability of vertices in a complete bipartite graph being disconnected such that no path of length 2 remains between them?
My problem is the following. I have a set of vertices $N$ and a set of vertices $H$. Each vertex $n \in N$ is connected by means of an edge to each vertex $h \in H$. So the two sets of vertices and the set of edges form a complete bipartite graph. The graph is also undirected. Let us say that two vertices $n_1,n_2 \in N$ can communicate with each other if there exists a path $n_1h_1n_2$ of length 2, where $h_1 \in H$. Since we start out with a complete bipartite graph, initially all vertices in $N$ can communicate with each other. However, now there is a probability $p$ assigned to each edge. This probability is the same for all edges and it models the probability that a vertex $n \in N$ is disconnected from a vertex $h \in H$ after some fixed amount of time $t$. Moreover, the probability of two vertices being disconnected is independent of the probability of two other vertices being disconnected. Thus, given that vertices can be disconnected, what is the probability that there does not exist a path $n_1h_1n_2$, for some $n_1, n_2 \in N$ and some $h_1 \in H$ when the time $t$ has elapsed? In other words, what is the probability that two vertices from the set $N$ cannot communicate after $t$ has elapsed?
Edit: To give some background, what I am trying to model is a computer network comprised of computing nodes, represented by the set of vertices $N$, and hubs, represented by the set of vertices $H$. Computing nodes cannot communicate directly with each other, but only by means of an intermediate hub. The edges basically represent cables connecting nodes and hubs. The probability $p$ represents the probability that a cable suffers a failure, e.g., breaks, in an interval of time $[0, t]$.
AI: so let's denote by $D_{n,h}$ the event that $n$ and $h$ are disconnected at time $t$. $n_1$ and $n_2$ can't communicate if the event $A_{n_1,n_2} := \bigcap_{h \in H} (D_{n_1,h} \cup D_{n_2,h})$ occurred (that is, each path of length 2 is destroyed). We have using indenpence and denoting complementation by ${}^c$:
\begin{align*}
P(A_{n_1, n_2}) &= \prod_{h\in H} P(D_{n_1, h} \cup D_{n_2,h})\\\
&= \prod_{h\in H} \bigl(1 - P(D_{n_1, h}^c \cap D_{n_2, h}^c)\bigr)\\\
&= \prod_{h\in H} \bigl(1 - P(D_{n_1,h}^c)P(D_{n_2,h}^c)\bigr)\\\
&= \prod_{h\in H} \bigl(1 - (1-p)^2\bigr)\\
&= \bigl(1 - (1-p)^2\bigr)^{|H|}
\end{align*} |
H: sum of even-valued and odd-valued Fibonacci numbers
I was solving the Project Euler problem 2
*By starting with 1 and 2, the first 10 terms of Fibonacci Series will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
Find the sum of all the even-valued terms in the sequence which do not exceed 4 million.*
here is my code in Python
a=0
b=1
nums = []
while True:
x=a+b
a=b
b=x
if(b>4000000):
break
nums.append(x)
sum=0
for x in nums:
if(x%2==0):
sum+=x
print sum
I noticed that the answer comes out to be 4613732
However I initially did a mistake by doing x%2!=0 and the answer turned out to be 4613731 (4613732-1)
Is this some property or just luck??
AI: Each even valued term (which are the 2nd, 5th, and every 3rd one after) is the sum of the preceding two odd terms. There is a startup transient that 2 doesn't have two 1's before it (you would if you started with 1,1). So the sum of the even valued terms up to some value will be 1 more than the sum of the preceding odd terms. |
H: Chain rule and inverse in matrix calculus
I am having trouble understanding the derivation of some seemingly simple matrix derivatives and am wondering if there is an intuitive (perhaps geometric) explanation. I am reasonably well-versed in multivariate calculus and linear algebra, but am not comfortable with tensor math.
The function I am interested in is $f(t)=\mathbf{B}^T(\mathbf{X}+t\mathbf{Y})^{-1}\mathbf{A}$, where $t$ is a scalar, and $\mathbf{A},\mathbf{B},\mathbf{X},\mathbf{Y}$ are matrices with conformant dimensions.
On the page 24 of the pdf of the appendix on matrix calculus in the book by Jon Dattorro (page 600 of the book), I find the formula for the first derivative of $f(t)$:
$$\frac{df}{dt}=-\mathbf{B}^T(\mathbf{X}+t\mathbf{Y})^{-1}\mathbf{Y}(\mathbf{X}+t\mathbf{Y})^{-1}\mathbf{A}$$
This sort of makes sense to me from my knowledge of calculus of functions of single variable: if you have $g(t)=a(x+ty)^{-1}b=ab(x+ty)^{-1}$, then $\frac{dg}{dt}=-ab(x+ty)^{-2}y=-a(x+ty)^{-1}y(x+ty)^{-1}b$ (from the chain rule and the power rule). That is, there is a clear similarity in the form.
What I don't understand is why the matrix equation for $\frac{df}{dt}$ looks the way it does. Is it due to non-commutativity of matrix multiplication? But how does that come in to this problem exactly? I've found the chain rule for matrix-valued function in the same pdf on page 8 (eq 1749) but I am not sure how to apply it here. Maybe I don't understand something about the calculus of the single-variable functions.
I guess I am asking if there is a way to derive the equation for $\frac{df}{dt}$ "from first principles" without using tensors.
AI: I think this follows more quickly from the product rule. The derivative of $t \mapsto X + tY$ is $Y$. You have $$0 = \frac{d}{dt}I = \frac{d}{dt} [(X+tY)^{-1}(X+tY)] = \frac{d}{dt}(X+tY)^{-1} * (X+tY) + (X+tY)^{-1}Y$$ and so $$\frac{d}{dt} (X+tY)^{-1} = -(X+tY)^{-1} Y (X+tY)^{-1}.$$ Multiplying on the left and right by $B^T$ and $A$ won't change much. |
H: Sign of an inequality
I have the following:$$\log (0.46)^{k+1}<\log 0.018$$
I solved this by writing$$k<\frac{\log 0.018}{\log 0.46}-1,$$
so $k < 4.17$. The result should be $k > 4.17$. Why is that? Where am I getting wrong?
AI: Since $\log(0.46)$ is negative (because $0.46 \leq e$, or $0.46 \leq 10$, depending on what base we're using for the $\log$), when we divide through we must change the direction of the inequality. The rest looks fine.
The root of this matter is that $x \geq 0$ implies $-x \leq 0$. We can invent any example with concrete numbers: consider $-2 \leq 6$, and divide through by $-2$ to get $1 \geq -3$, not $1 \leq -3$. However, we could divide by $6$ and get $-\frac{1}{3} \leq 1$, still valid, because we're dividing by a positive number.
To answer your question in the comments:
When we have something like $\displaystyle \frac{2x + 3}{5x - 1} \geq 2$, we might be tempted to multiply by $5x-1$, but since you don't know its sign, it's better to avoid multiplying or dividing by a number which, for all we know, might be negative, and for example substract $2$ from both sides: $\displaystyle \frac{2x + 3}{5x - 1} - 2 = \frac{2x + 3 - 2(5x - 1)}{5x - 1} =\frac{-8x+5}{5x-1} \geq 0$, and keep the division as a last resource, when we'll have to use the next tip. To multiply or divide by an unknown quantity we have to consider every possible sign of what we're multiplying or dividing by, for example:
If $5x-1 \gt 0$, then we multiply without changing the direction: $\displaystyle 2x + 3 \geq 2(5x - 1)$, and if $5x-1 \lt 0$, then $\displaystyle 2x + 3 \leq 2(5x - 1)$; finally, we continue manipulating the expression to solve for $x$. In each case we must remember that we imposed a condition on $x$ (like $5x-1 \gt 0$ or $5x-1 \lt 0$), and take it into account at the end, even though it's not written on the equation.
The basic idea is:
try to do simpler steps whenever possible; and
we definitely have to change the direction if we're multiplying or dividing by a negative number, so if we don't know the sign, we must separate in cases. |
H: How to prove that $\operatorname{lcm}\{1,\ldots,n\}\geq (\sqrt{n})^{\pi(n)}$?
Let $\operatorname{LCM}[n]:=\operatorname{lcm}\{1,\ldots,n\}$. It is easy to verify $\operatorname{LCM}[n]\geq 2^{\pi(n)}$, where $\pi(n)$ counts the number of distinct primes up to $n$.
But how can I prove the bound $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$?
For the bound $2^{\pi(n)}$, I used the identity:
$$\operatorname{LCM}[n]=\!\!\!\prod_{p\in\mathbb{P},p\leq n} p ^{\lfloor \log_p(n) \rfloor}$$
But this doesn't work with the lower bound of $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$.
Can someone give me a hint?
AI: The exact same identity works. You can do much better than $p^{\lfloor \log_p(n) \rfloor} \ge 2$, and show that $p^{\lfloor \log_p(n) \rfloor} > \sqrt{n}$. Here's the idea: $p^{\lfloor \log_p(n) \rfloor}$ is just the largest power of $p$ that fits into $[1,n]$.
If $p$ is in the range $(n^{1/2},n]$ there is nothing to prove. But if $p$ is in $(n^{1/3},n^{1/2}]$ then $p^2 \le n$ and $p^2 > \sqrt{n}$. If you think about it, no matter what prime you start with, there is some power of $p$ that is larger than $\sqrt{n}$. |
H: Main differences between analytic number theory and algebraic number theory
What are some of the big differences between analytic number theory and algebraic number theory?
Well, maybe I saw too much of the similarities between those two subjects, while I don't see too much of analysis in analytic number theory.
AI: The obvious answer is that Algebraic Number Theory uses techniques from algebra to answer number theory questions, while Analytic Number Theory uses techniques from analysis.
But that's a bit pedantic.
There is a certain degree to which analytic number theory is "about" the order properties of the natural numbers. For example, a lot of analytic number theory is trying to answer "how many examples of (some property) exists between $1$ and $N$?" The seminal example of this is "How many primes are between $1$ and $N$?" The answers to these are often estimations, and, in computing bounds on the errors, we are always using analysis.
Algebraic number theory is relatively unconcerned with this sort of question, and is often ill-equipped to answer it. Algebraic number theory can often be used to answer "Are there infinitely many?" but it often has a hard time dealing with bounds. It tends to be about bigger structural properties of the integers (and related number rings.)
As with any such generalizations, there are always exceptions. |
H: Orthogonal Trajectory of $x^2 + 2y^2 = k^2$
$x^2 + 2y^2 = k^2$
I first take the derivative like the instructions say.
$2x + 4y \frac{dy}{dx} = 0$
I am not entirely sure why a dy and dx appears but it does in the instructions so I go with it.
Now I need to solve for $y'$
$ + 4y \frac{dy}{dx} = -2x$
$ \frac{dy}{dx} = \frac{-x}{2y}$
$ \frac{dy}{dx} = \frac{-x}{2y}$
So now I need to find the inverse negative
$ \frac{dy}{dx} = \frac{2y}{x}$
And that should be my slope at each line.
And now I need to solve that.
$ \frac{1}{2}ydy = xdx$
Take the integratal and I get
$\frac{1}{4}y^2 = \frac{1}{2}x^2 + C$
$y^2 = 2x^2 + C$
$y = \sqrt{2x^2 + C}$
This is wrong and I do not know why.
AI: From the line David noted you; You would have $\frac{dy}{2y}=\frac{dx}{x}$. So by integrating from both sides, you get $\frac{1}{2}$Ln$|y|$=Ln$|x|+c$:
$$\int\frac{dy}{2y}=\int\frac{dx}{x}$$ $$\frac{1}{2}\int\frac{dy}{y}=\int\frac{dx}{x}$$ $$\frac{1}{2}Ln|y|=Ln|x|+c$$ wherein $c$ is a constant. You can simplify the result as $y=cx^2$. |
H: A simple quadratic inequality
For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
AI: To avoid fractions, we multiply the left-hand side by $4$, obtaining
$$(n-c)^2+8(n-c)+4c.$$
Complete the square. We get
$$(n-c+4)^2 +4c -16.$$
Now calculate $[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]$.
The difference of squares factors as $(c-5)(2n-c+3)$, so
$$\begin{align}[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]&=(c-5)(2n-c+3)-4(c-5)\\
&=(c-5)(2n-c-1).\end{align}$$
The condition $c \ge 5$ ensures that $c-5\ge 0$. And since $n \ge c$, the term $2n-c-1$ is positive. |
H: norm for estimating the error of the numerical method
In most of the books on numerical methods and finite difference methods the error is measured in discrete $L^2$ norm. I was wondering if people do the in Sobolev norm. I have never see that done and I want to know why no one uses that.
To be more specific look at the $$Au=f,$$ where assume $A_h$ is some approximation for $A$ and $U$ is the numerical solution for the system. Then if we plug the actual function $u$ into $A_hU=f$ and substruct we have $$A_h(u-U)=\tau$$ for $\tau$ being a local error. Thus I have an error equation $$e=A_h^{-1}\tau$$ What are the problems I am facing If I use discrete Sobolev norm?
AI: For one thing, it's a question of what norm measures how "accurate" the solution is. Which of the two error terms would you rather have: $0.1\sin(x)$ or $0.0001\sin(10000x)$? The first is smaller in the Sobolev norm, the second is smaller in the $L^2$ norm. |
H: Context-free grammar for words of a CF grammar starting with a certain symbol
Let $L$ be a context-free language on the alphabet $\Sigma$.
I need to show that for each $a \in \Sigma$ the language $L_a = \{ x \in \Sigma^* \; | \; a.x \in L\}$ is context-free as well.
I wanted to prove it by generating a context-free grammar for $L_a$ by the generation of a new nonterminal for each existing one, to check if an 'a' has been read, but this doesn't work as what I need are all words from $L$ with a leading 'a'.
Could you please help me to find a solution?
Thanks in advance!
AI: Let $G$ be a context-free grammar the produces $L$. You can assume that $G$ is in Chomsky normal form, i.e. all productions are of the form $X \to YZ$ or $X\to c$ or $S \to \varepsilon$ (in this case $S$ must not appear on the right side of any production).
In order to get a context-free grammar for $L_a$ you have to modify $G$ in such a way that it produces the words in $L$ that start with an $a$ but produces them without it.
You can achieve this by adding for every non-terminal $X$ of $G$ a new non-terminal $\tilde X$ together with the production rule $\tilde X\to \tilde Y Z$ whenever $X\to YZ$ is a production in $G$. This way the left-most symbol in the intermediate steps of a production is always one of the added non-terminals $\tilde X$. Furthermore, for every rule $X\to a$ add the new rule $\tilde X \to \varepsilon$. This way, if $G$ would produce an $a$ at the first position this $a$ is now deleted. Finally, declare $\tilde S$ as the new start symbol. This should result in a context-free grammar that produces $L_a$. |
H: Proving that $A_n$ is the only proper nontrivial normal subgroup of $S_n$, $n\geq 5$
There is a famous Theorem telling that:
For $n≥5$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.
For the proof, we firstly start with assuming a subgroup of $S_n$ which $1≠N⊲S_n$. We proceed until at the last part of proof's body, we assume $N∩A_n=\{1\}$. This assumption should be meet a contradiction with normality of $N$ in $S_n$. There; we get $N=\{1,\pi $} in which $\pi$ is an odd permutation of order $2$. Now for meeting desire inconsistency, I have two approaches:
(a) Since every normal subgroup, having two elements, lies in the center of $G$ so, our $N⊆ Z(S_n)=\{1\}$ for $n≥5$ and then $N=\{1\}$.
(b) Clearly, $1≠N$ acting on set $\Omega=\{1,2,...,n\}$ is intransitive wherein $|\Omega|≥5$ and according to the following Proposition $S_n$ would be imprimitive.
Proposition 7.1: If the transitive group $G$ contains an intransitive normal subgroup different from $1$, then $G$ is imprimitive
(Finite Permutation Groups by H.Wielandt).
May I ask if the second approach is valid? I am fond of knowing new approach if exists. Thanks.
AI: You are almost there. Try to prove that $Z(S_n)= 1$ for all $n \geq 3$. Then if $N$ is non-trivial and normal, you assume $N \cap A_n = 1$. This implies $N \subseteq Z(S_n)$. Why? Because in general, if $N \unlhd G$ and $N \cap [G,G] = 1$ then $N \subseteq Z(G)$.
We conclude that the normal subgroup $N \cap A_n \neq 1$. At this point I assume that you know that $A_n$ is a simple group for $n \geq 5$. Hence $N \cap A_n = A_n$, so $A_n \subseteq N \subseteq S_n$. Since $index[S_n:A_n] = 2$, it follows that $N=A_n$ or $N=S_n$. |
H: Approximation of $\log(x)$ as a linear combination of $\log(2)$ and $\log(3)$
I wonder if it's possible to approximate $\log(n)$, n integer, by using a linear combination of $\log(2)$ and $\log(3)$.
More formally, given integer $n$ and and real $\epsilon>0$, is it always possible to find integer $x,a,b$ where:
$$\left|n^x-2^a 3^b\right|<\epsilon$$
For example, I can approximate $11$ by $$2^{-33} 3^{23}=10.959708460955880582332611083984375 \approx 10.96.$$
AI: Yes.
Let $a=\frac{\log(2)}{\log(3)}$. Then, $a$ is irrational, thus by Dirichclet Theorem, the set $\{ ma+n | m,n \in Z \}$ is dense. Thus, there exists some $m,n \in Z$ so that
$$\left| \frac{\log(n)}{\log(3)} - ma -k \right| < \frac{\epsilon}{\log(3)}$$
Multiply by $\log(3)$ and you are done.
P.S. It is irrelevant that $n$ is integer. Also, the proof works if you replace $2$ and $3$ by any numbers $x,y$ so that $\log_x(y)$ is irrational.
P.P.S. I think that for $n$ positive integer, it is enough to use one $\log(2)$. Indeed, if $n$ is a power of 2, you are done, otherwise, $\frac{\log(n)}{\log(2)}$ is irrational, and then the set $m\frac{\log(n)}{\log(2)} - k$ is dense. Thus, you can find some integers so that
$$\left|m\frac{\log(n)}{\log(2)} - k \right| < \frac{\epsilon}{\log(2)}$$
Of course, you get rational coefficients in this case. |
H: Matrix Representation of the Tensor Product of Linear Maps
I'm trying to work out some examples of applying the tensor product in some concrete cases to get
a better understanding of it. Within this context, let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$
be a linear map with matrix $A$ and let $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear map with matrix $B$.
It follows almost immediately from the universal property of the tensor product that there exists a unique linear map
$$
f \otimes g: \mathbb{R}^2 \otimes \mathbb{R}^2 \rightarrow \mathbb{R}^2 \otimes \mathbb{R}^2
$$
such that
$$
(f \otimes g)(u \otimes v) = f(u) \otimes g(v).
$$
Suppose we take a basis for $\mathbb{R}^2$, say the canonical one, $(e_1, e_2)$, and let
$u = a^ie_i$ and $v = b^je_j$. Then, using the linearity of $f$ and $g$ and multilinearity of
$\otimes$ we arrive at the general expression
$$
(f \otimes g)(u \otimes v) = a^ib^j f(e_i) \otimes g(e_j)
$$
where the summation convention is in force. Since $f \otimes g$ is a linear map
we ought to be able to represent it by a matrix $A \otimes B$ whose columns represent the action
of $f \otimes g$ a basis of $\mathbb{R}^2 \otimes \mathbb{R}^2$.
Now, since $\{e_i \otimes e_j| i,j = 1, 2 \}$ is a basis for $\mathbb{R}^2 \otimes \mathbb{R}^2$ the resulting matrix
with respect to this basis should have columns that are given by $(f \otimes g)(e_i \otimes e_j)$ where $i,j=1, 2$.
The first column of this matrix, for example, is the vector obtained by
$$
(f \otimes g)(e_1 \otimes e_1) = f(e_1) \otimes g(e_1)
$$
which is the tensor product of the first column of $A$ and the first column of $B$ since $A$ and $B$
respectively represent $f$ and $g$.
This is where I am stuck. I understand $f(e_1) \otimes g(e_1)$
is essentially an equivalence class that is obtained in the existence proof of the tensor
product but I'm not so sure how one works this out concretely based on the definition of the tensor product
and its universal construction.
So my question is, how does one concretely compute $f(e_1) \otimes g(e_1)$?
AI: $f$ has matrix $A$ with respect to $\{e_1, e_2\}$. Hence $f(e_1) = a^i_1e_i$, analogously $g(e_1) = b^j_1e_j$. By bilinearity of $\otimes$, therefore
\[ f(e_1) \otimes g(e_1) = a^i_1b^j_1 (e_i \otimes e_j) \]
So if say, you decide that $(e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2)$ is your ordered basis of $\mathbb R^2 \otimes \mathbb R^2$, then the first column of your matrix is $(a_1^1b_1^1, a_1^1b_1^2, a_1^2b_1^1, a_1^2b_1^2)^t$. |
H: Notation question. Piecewise function.
I have observed the below statement in a report.
$
f(x,y) = \left\{
\begin{array}{lr}
f<0 & : (x,y) \in A\\
f = 0 & : (x,y) \in B \\
f>0 & : (x,y) \in C
\end{array}
\right.$
I understand the meaning behind it, but is the notation correct? And in this case how do I write the statement then?
AI: It’s at best confusing: I’m not entirely certain what is intended. If the intended meaning is that $f(x,y)$ is negative on $A$, $0$ on $B$, and positive on $C$, you could use the signum function and write
$$\operatorname{sgn}\big(f(x,y)\big)=\begin{cases}-1,&\text{if }(x,y)\in A\\
0,&\text{if }(x,y)\in B\\
1,&\text{if }(x,y)\in C\;.
\end{cases}$$
And if that’s the intended meaning, I don’t consider what’s in the report correct, though others may be more generous than I. |
H: Are There Any Symbols for Contradictions?
Perhaps, this question has been answered already but I am not aware of any existing answer. Is there any international icon or symbol for showing Contradiction or reaching a contradiction in Mathematical contexts? The same story can be seen for showing that someone reached to the end of the proof of a theorem (i.e. as shown the tombstone symbol ∎, Halmos).
AI: Different sources use different symbols (if they use symbols at all). I've seen $\Rightarrow\Leftarrow$ most often. For some others, see "Symbolic Representation" here. |
H: What are operators that map a number to a probability distribution function called?
I'm a scientist who has stumbled upon an idea that I think might be helpful in my field. I'm looking for information about whether it has been treated in mathematics before – and I would be surprised if it hasn't – and if so, a pointer to some of the results/interesting applications/ideas/etc that have been written about it.
The idea is simple and as follows: higher-order functions that map (ie. take as input) a number to (ie. produce as output) a probability distribution function. The probability distribution function might be parameterised by a function of the number, or it might not be.
In what branches of mathematics are such operators used, and what are they called?
AI: I can't think of a more narrow branch to look in than "statistics". I imagine someone who knows more about the topic will quickly post a better answer than this, but since it is currently unanswered I'll take a stab at the problem.
I had luck googling "assigns a probability density function" in quotes. The idea of a "field" of probability distributions assigned to points of a space sounds pretty natural to me. I also saw "kernel density estimator" appearing several times, so you might look into that.
Lots of the important probability density functions are purely functions of their parameters, for example:
The normal distribution $\frac{1}{\sigma\sqrt{2\pi}}\exp({-\frac{1}{2}(\frac{x-\mu}{\sigma})^2})$
the Chi-squared distribution: $\frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})}x^{(k/2)-1}\exp(-x/2)$
the t-distribution
It would be interesting to know more about any special properties you want the mapping to have. |
H: What is a good book to study linear algebra?
I'm looking for a book to learn Algebra. The programme is the following. The units marked with a $\star$ are the ones I'm most interested in (in the sense I know nothing about) and those with a $\circ$ are those which I'm mildly comfortable with. The ones that aren't marked shouldn't be of importance. Any important topic inside a unite will be boldfaced.
U1: Vector Algebra.
Points in the $n$-dimensional space. Vectors. Scalar product. Norm. Lines and planes. Vectorial product.
$\circ$ U2: Vector Spaces.
Definition. Subspaces. Linear independence. Linear combination. Generating systems. Basis. Dimesion. Sum and intersection of subspaces. Direct sum. Spaces with inner products.
$\circ$ U3: Matrices and determinants.
Matrix Spaces. Sum and product of matrices. Linear ecuations. Gauss-Jordan elimination. Range. Roché Frobenius Theorem. Determinants. Properties. Determinant of a product. Determinants and inverses.
$\star$ U4: Linear transformations.
Definition. Nucleus and image. Monomorphisms, epimorphisms and isomorphisms. Composition of linear transformations. Inverse linear tranforms.
U5: Complex numbers and polynomials.
Complex numbers. Operations. Binomial and trigonometric form. De Möivre's Theorem.
Solving equations. Polynomials. Degree. Operations. Roots. Remainder theorem. Factorial decomposition. FTA. Lagrange interpolation.
$\star$ U6: Linear transformations and matrices.
Matrix of a linear transformation. Matrix of the composition. Matrix of the inverse. Base changes.
$\star$ U7: Eigen values and eigen vectors
Eigen values and eigen vectors. Characteristc polynomial. Aplications. Invariant subspaces. Diagonalization.
To let you know, I own a copy of Apostol's Calculus $\mathrm I $ which has some of those topics, precisely:
Linear Spaces
Linear Transformations and Matrices.
I also have a copy of Apostol's second book of Calc $\mathrm II$which continues with
Determinants
Eigenvalues and eigenvectors
Eigenvalues of operators in Euclidean spaces.
I was reccommended Linear Algebra by Armando Rojo and have Linear Algebra by Carlos Ivorra, which seems quite a good text.
What do you reccomend?
AI: "Linear Algebra Done Right" by Sheldon Axler is an excellent book. |
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