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H: The sum of trivial paths for a finite quiver is 1?
let $Q=(E_0, E_1)$ be a quiver and let $P_Q$ be a path algebra of $Q$. Let $p_i$ be the trivial path associated to each vertex $i$ in $E_0$.
Then why is $\sum_{i\in E_0} p_i=1$ for a finite quiver?
Are the $p_i$'s viewed as self-loops (arrows whose source and target is $i$, i.e., $\mathrm{End}(V_i)$ where $V_i$ is a vector space assigned to $i\in E_0$) or are the $p_i$'s viewed as "no arrows" (i.e., $p_i$ is the zero endomorphism from $V_i$ to itself)?
AI: The $p_i$s are self-loops, and special ones, since they're required to compose trivially. That is, if you look at the quiver as a category, they're the identity morphisms. Then in any representation they're mapped to the identity automorphism of each vector space.
As to your question about their sum: take an arbitrary element $\pi=\sum c_j \pi_j,$ where the $\pi_j$ are arbitrary paths in $Q$ and $c_j$ constants from whatever you're taking $P_Q$ over.
Now in considering the product $(\sum p_i) \pi,$ the claim is that each $c_j\pi_j$ is annihilated by all but one of the $p_i,$ and sent to itself by the last. This is because the paths can't concatenate if the end point of $\pi_j$ isn't the start (and end) point of $p_i$, in which case the product is defined as 0, while the $p_i$ that does match the end point of $\pi_j$ is required to composed trivially with it. So each component path in $\pi$ comes back out of the product once, and only once. |
H: What do you call a group that doesn't have a unique identity?
I have a set $M$ and an associative binary relation $+ : M \times M \to M$.
There exists an inversion operator $-$ , such that if $ m \in M$, then $m+(-m) \in Z$ where $Z$ is the set of all zeros ($Z \subset M$).
Additionally, if $z \in Z$ and $m \in M$, then $m+z=m$
If $|Z| = 1,$ then this would form a group, right? But since $|Z|>1$, what would I call it? Is there any theory about categories like this?
Also, what if there was no identity at all? That is, $m_1 + m_2 + (-m_2) = m_1$, but $m_2 + (-m_2)$ is undefined. In this case, associativity would not hold.
AI: Suppose $M$ is a set with an associative binary operator $+$ such that for all $m \in M$ there is a unique $n$ such that $m+n+m = m$ and $n+m+n = n$ (thinking of $z_r=m+n$ we get $z_r + m = m$ and $n+z_r = n$; thinking of $z_l =n+m$ we get $m+z_l = m$ and $z_l+n=n$), then you have a well studied object called an “inverse semigroup”.
The set $Z$ you want is called the set of idempotents, $Z = \{ m \in M : m + m = m \}$. It also consists exactly of all $m + (-m)$ for $m \in M$. I also call them "partial identities", because:
There is a partial order $\leq$ on every inverse semigroup: $x \leq y$ iff $x = x + (-x) + y$. This let's you express how close to an identity each $z \in Z$ is: $z + x = x$ whenever $x \leq z$.
For every partial identity $z=g+(-g)$ there is an honest to goodness group called the $H$-class (or more accurately its Schutzenberger group) associated to $z$ consisting of all invertible $z+m+z$ for $m \in M$.
Inverse semigroups as semigroups of bijections
Every inverse semigroup is isomorphic to a semigroup consisting of bijections between subsets of a set $X$, where multiplication is the best version of composition you can manage: if $f:A\to B$ and $g:C \to D$ then $f\cdot g$ takes $f^{-1}(x)$ to $g(x)$ whenever $x \in B \cap C$.
In terms of bijections, the set $Z$ consists of all identity bijections $f:A \to A :a \mapsto a$ where $A$ is the domain of some $m \in M$. In terms of bijections, $-g$ is the inverse function $g^{-1} : D \to C : g(c) \mapsto c$. The partial identities $g+( -g)$ and $-g+g$ are the identity bijections on the domain and range of $g$.
The natural partial order is actually just “subset” if you consider a bijection to be a set of ordered pairs $(a,f(a))$, so that $f \leq g$ if and only if $f$ is the restriction of $g$ to the domain of $f$.
The H-class of an idempotent $f:A\to A:a\mapsto a$ is all the permutations of $A$, that is, bijections from $A$ to $A$.
Bibliography
I recommend both of these books as very clear and motivated introductions to these structures. Lawson's book is particularly readable and draws connections to topology, symmetry, groupoids, category theory, and computer science.
Lawson, Mark V.
Inverse semigroups: The theory of partial symmetries.
World Scientific Publishing Co., Inc., River Edge, NJ, 1998. xiv+411 pp. ISBN: 981-02-3316-7
MR1694900
DOI10.1142/9789812816689
Petrich, Mario.
Inverse semigroups.
Pure and Applied Mathematics (New York). A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1984. x+674 pp. ISBN: 0-471-87545-7
MR752899 |
H: Finding the all triples
How to find the all positive integer triples such that :
$$ab+c=\gcd (a^2,b^2)+\gcd(a,bc)+\gcd(b,ac)+\gcd(c,ab)=239^2$$
AI: Let $x=\gcd(a,b,c)$. But $\gcd(a^2,b^2)\ge x^2$ and $x$ divides $239^2$, this is impossible due to the equality and 239 is prime. Hence $x=1$.
Let $y=\gcd(c,ab)$. if $y>1$, then once again $y=239$ ($239^2$ would be too big).
Suppose $y$ divides $a$ (the other case is symmetric), $y$ does not divide $b$ ($\gcd(a^2,b^2)$ would be at least $239^2$, too big). Hence $\gcd(c,ab)=y$, $\gcd(a,bc)=y$, and $\gcd(a,bc)=\gcd(a,b)=z$, so
$$ z^2+z+2y=y^2$$ So $y$ divides $z$ (or $z+1$), this is not possible, so $y=1$
So $c$ is prime with $a$ and $b$ and the equation is
$$ab+c=\gcd(a,b)^2+2\gcd(a,b)+1=(\gcd(a,b)+1)^2=239^2$$
$$\gcd(a,b)=238$$
The only small enough possibility is $a=b=238$ and then $c=477$ |
H: A unitary matrix taking a real matrix to another real matrix, is it an orthogonal matrix?
I tried to prove that a real antisymmetric matrix can be taken by an orthogonal tranformation to a form:
where the eigenvalues are $\pm i\lambda_1, \pm i\lambda_2 ... $
which is a statement I saw on wikipedia in http://en.wikipedia.org/wiki/Antisymmetric_matrix
I also know an antisymmetric matrix can be diagonalized by a unitary transformation, and I found a unitary transformation taking the diagonal matrix to the required form.
So by composing the two transformations (diagonalization, then taking the diagonal matrix to the required form), I'll get a unitary transformation taking the real antisymmetric matrix to another real matrix.
My question is if this transformation must be a real matrix? if so I can deduce that the unitary transformation is in fact an orthogonal transformation.
So is this true?
Is a unitary transformation taking a real matrix to another real matrix necessarily an orthogonal transformation?
EDIT: After receiving in the comment here a counterexample, I'm adding:
Alternatively, if it is not necessarily orthogonal, does there necessarily exist an orthogonal transformation taking the two matrices to each other?
AI: Yes. Quoting Halmos's Linear algebra problem book (Solution 160).
“If $A$ and $B$ are real, $U$ is unitary, and $U^*AU = B$, then there exists a real orthogonal $V$ such that $V^*AV = B$.
A surprisingly important tool in the proof is the observation that the unitary equivalence of $A$ and $B$ via $U$ implies the same result for $A^*$ and $B^*$. Indeed, the adjoint of the assumed equation is $U^*A^*U = B^*$.
Write $U$ in terms of its real and imaginary parts $U = E + i F$. It follows from $AU = UB$ that $AE = EB$ and $AF = FB$, and hence that $A(E+\lambda F) = (E+\lambda F)B$ for every scalar $\lambda$. If $\lambda$ is real and different from a finite number of troublesome scalars (the ones for which $\det(E+\lambda F) = 0$), the real matrix $S = E + \lambda F$ is invertible, and, of course, has the property that $AS=SB$.
Proceed in the same way from $U^*A^*U = B^*$: deduce that $A^*(E+\lambda F) = (E+\lambda F)B^*$ for all $\lambda$, and, in particular, for the ones for which $E+\lambda F$ is invertible, and infer that $A^*S = SB^*$ (and hence that $S^*A = BS^*$).
Let $S =VP$ be the polar decomposition of $S$ (that theorem works just as well in the real case as in the complex case, so that $V$ and $P$ are real.) Since $$BP^2 = BS^*S = S^*AS = S^*SB = P^2B,$$ so that $P^2$ commutes with $B$, it follows that $P$ commutes with $B$. Since $$AVP = AS = SB = VPB = VBP$$ and $P$ is invertible, it follows that $AV=VB$, and the proof is complete.”
Needless to say, that isn't the shortest path to prove the reduction of antisymmetric matrices... |
H: Constructing an equation using proportionality
The question is:
If the rate of a certain chemical reaction doubles for every $10$ degree rise in temperature then which is greater:
a) Twice the rate at $10$ degrees
b) Half the rate at $30$ degrees
According to my text they are same. Here is how I am doing it
$$\rm Rate = 2\times Temp$$
For $10$ degrees: $\rm Rate = 2 \times 10 = 20$ so $2r$ (Twice The Rate) $=40.$
For $30$ degrees: we have $R = 2 \times 30 = 60$ so half would be $30.$
Could anyone tell me what I am doing wrong since they are different?
AI: Let $x$ be the rate at $10$ degrees. Then twice the rate at $10$ degrees is $2x$. We know that at $20$ degrees the rate will be $2x$ (since this is a $10$ degree rise). Similarly, the rate at $30$ degrees is $2 \times 2x = 4x$, and half of this is $2x$. |
H: Linear Algebra, eigenvalues and eigenvectors
For any $m \times m$ matrix, I will get a characteristic polynomial of degree $m$
with $m$ eigenvalues.
But for the matrix
$$A = \pmatrix{2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1}$$
I got the characteristic polynomial
$$P(A)=t(t+1)(1-t)^3.$$
This means $5$ eigenvalues: $\{1,1,1,-1,0\}$.
Did I do some thing wrong?
AI: I calculated the same polynomial and I got
$$P(X)= X^2 (X-1)^3 (X-4) \,.$$
Note that $tr(A)=7$ has to be the sum of eigenvalues.
Just to get you started:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 1 & 0 & 0 & 1-t & 0 & 0 \\ 1 & 0 & 0 & 0 & 1-t & 0 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Subtract the 6th row from 4th and 5th:
$$\det(A-tI)= \det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1-t & 0 & t-1 \\ 0 & 0 & 0 & 0 & 1-t & t-1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Now, $(1-t)$ common factor on rows 4 and 5.
$$\det(A-tI)= (t-1)^2\det \pmatrix{2-t & 1 & 1 & 1 & 1 & 1 \\ 1 & 1-t & 0 & 1 & 0 & 1 \\ 1 & 0 & 1-t & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 1-t}$$
Next add Column 4 and column 5 to Column 6, and you can get a smaller $4 \times 4$ determinant.... |
H: Intuition for étale morphisms
Currently working on algebraic surfaces over the complex numbers. I did a course on schemes but at the moment just work in the language of varieties.
Now i encounter the term "étale morphism" every now and then (in the book by Beauville). I know Hartshorne's definition as a smooth morphism of relative dimension zero, and wikipedia states a bunch of equivalent ones. I can work with this, so no problem there. However, some more intuition about the concept would also be nice.
So basically, if you have worked with étale morphisms, could you explain what's your personal intuition for such things, in the case of varieties? If in your answer you could also mention smooth and flat morphisms, that would be really appreciated.
Thanks in advance!
Joachim
AI: Instead of answering your question with general results easily obtainable from the literature, online or traditional, I'll give you a few morphisms.
Deciding whether they are are étale may contribute to developing your intuition.
(Of course I'll gladly help you or anybody else if you had any problem with these morphisms)
a) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^3)): t\mapsto (t^2,t^3)$
b) $\mathbb A^1_\mathbb C\to \operatorname {Spec}(\mathbb C[X,Y]/(Y^2-X^2-X^3)): t\mapsto (t^2-1,t^3-t)$
c) $\mathbb A^2_\mathbb C\to \mathbb A^2_\mathbb C: (x,y)\mapsto (x,xy)$
d) $\operatorname {Spec}\mathbb C[T]\to \operatorname {Spec}\mathbb C[T^2,T^3]$
e) $\operatorname {Spec}\mathbb Q[T]/(T^2-4)\to \operatorname {Spec}\mathbb Q$
f) $\operatorname {Spec}\mathbb Q[T]/(T^2+4)\to \operatorname {Spec}\mathbb Q$
g) $\operatorname {Spec}\mathbb Q[T]/(T^2)\to \operatorname {Spec}\mathbb Q$
h) $\operatorname {Spec}\mathbb F_9\to \operatorname {Spec}\mathbb F_3$
Edit (one day later) : Two useful theorems and how they settle the question of étaleness of the above morphisms
Theorem 1 Given a field $k$ and a $k$-algebra $A$, the morphism $\operatorname {Spec}(A)\to \operatorname {Spec}(k) $ is étale iff $A$ is isomorphic as a $k$-algebra to a finite product $A\cong K_1\times...\times K_n$ of finite separable field extensions $K_i/k$.
Remark In the étale case, $A$ must be reduced ( i.e. $\operatorname {Nil}(A)=0$ )
Example Every finite Galois extension $K/k$ gives rise to an étale morphism $\operatorname {Spec}(K)\to \operatorname {Spec}(k) $. This is the kernel of Grothendieck's famous geometrization of Galois theory
Illustration The morphisms e), f), h) are étale but g) is not because $\operatorname {Spec}\mathbb Q[T]/(T^2)$ is not reduced.
Theorem 2 A morphism of schemes $f:X\to Y$ is étale iff it is flat and unramified.
Illustration The morphisms a) , b), c) and d) are not étale because they are not flat
For the sake of completeness let me mention that a), c) and d) are ramified but that b) is unramified.
Let me also mention that a) and d) are two different presentations of the same morphism. |
H: Can I get a bound like $[u]_\alpha \leq C\lVert u\rVert_{C^0(S)}$?
I badly need a bound like
$$[u]_\alpha = \sup_{S}\frac{|u(x) - u(y)|}{|x-y|^\alpha} \leq C\lVert u \rVert_{C^0(S)} = C\sup_S |u|$$
where $S$ is compact and $C$ is a constant not depending on $u$.
Can it be done? Please let me know. Thank you.
AI: No: If such a $C$ existed then the $C^{\alpha}$ norm would be equivalent to the supremum norm. But $C^{\alpha}$ bounded sets in $C(S)$ are precompact (in the supremum norm). So you can't have such a $C$ working for an infinite dimensional subspace of $C(S)$. |
H: Graded Ring - Finite Sum
I've just read that if $R=R_0\oplus R_1 \oplus \dots$ is a graded ring and $f\in R$ then there's a unique decomposition of $f$ as $f=f_0+\dots+f_n$ with $f_i\in R_i$. I can't see immediately why in general this would have to be a finite sum! Could someone possibly enlighten me? I've got a feeling I'm being stupid, so apologies if it's completely obvious!
AI: It's because the graded ring is a direct sum of the $R_i$'s, not a direct product. Every element of a direct sum has only finitely many non-zero summands by definition.
PS: I always try to keep examples in mind when learning a definition. In the case of graded rings, the first I think of is $k[x_1,\ldots, x_n]$ graded by degree, where $k$ is a field. Here the idea that only finite sums are allowed is exactly what we should expect from experience. |
H: Relation between left and right coset representatives of a subgroup
Let $G$ be a finite group and $H$ a subgroup. Is it true that
a set of right coset representatives of $H$ is also set of left coset
representatives of $H$?
AI: Not every left transversal is also a right transversal. The Group Properties Wiki has a list of subgroup properties that are stronger than "having [at least one] left transversal that is also a right transversal"; among these is normality as William notes.
However the left coset representatives' multiplicative inverses form a right transversal, because
$$\begin{array}{c l}xH=yH & \iff y^{-1}xH=H \\ & \iff y^{-1}x\in H \\ & \iff (y^{-1}x)^{-1}=x^{-1}y\in H \\ & \iff Hx^{-1}y=H \\ & \iff Hx^{-1}=Hy^{-1}. \end{array}$$ |
H: Evaluating an Integral makes it zero
Please point me out some useful notes for evaluating the following integral. I really wonder why it is $0$. $$\int_{-0.5}^{0.5}\cos(x)\ln\frac{1+x}{1-x}dx=0$$ I apply the rules which I know about solving the integral but, they have been useless. Any help will be appreciated.
AI: The reason is this: if $f$ is an odd function then $\displaystyle \int_{-L}^L f(x)\, dx = 0$ for any real number $L$.
Explicitly in this case: we can split the integral as
$$\int_0^{\frac{1}{2}} \cos x \ln \dfrac{1+x}{1-x}\, dx + \int_{-\frac{1}{2}}^0 \cos x \ln \dfrac{1+x}{1-x}\, dx$$
Substituting $u=-x$ into the second integral then gives
$$\int_0^{\frac{1}{2}} \cos x \ln \dfrac{1+x}{1-x}\, dx + \int_{\frac{1}{2}}^0 \cos (-u) \ln \dfrac{1-u}{1+u}\, (-du)$$
Then using symmetry properties, namely $\cos(-u)=\cos u$ and $\ln \dfrac{1-u}{1+u} = -\ln \dfrac{1+u}{1-u}$, we see that the second integral is the negative of the first, so they cancel to give zero. |
H: Given $\forall x \in \mathbb{R} \: h(p^t(x))=th(p(x))$, how to get $h(p(x)) \propto \ln p(x)$?
The whole question is in the title. $p(x)$ is a probability distribution, and $h$ is continuous and monotonic in $p(x)$.
The purpose is to motivate that the "degree of surpise", or the "amount of information" after observing a value of a random variable $x$ having a distribution $p(x)$ is proportional to $\ln p(x)$. The steps leading to the motivation are sketched in Bishop's "Machine Learning and Pattern Recognition", exercise 1.28; this is the last part.
I can't see a why it is so from a constructive point of view, maybe it's obvious? (Of course ensuring that $\ln p$ satisfies is trivial.)
AI: The hypothesis is that $h(u^t)=t\,h(u)$, for every nonnegative $u$ and $t$.
In particular, every solution $h$ is such that $h(2^t)=t\,h(2)$, for every nonnegative $t$. Hence, $h(z)=h(2)\,\log_2(z)$, for every positive $z$. On the other hand, $h_c:z\mapsto c\,\log_2(z)$ solves the equation, for every real number $c$. Hence, the set of solutions is exactly $\{h_c\,;\,c\in\mathbb R\}$. |
H: A group of order $108$ has a proper normal subgroup of order $\geq 6$.
Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$.
My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~|~3^3$.
Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools).
Any ideas are welcome; thanks!
AI: Let $\,P\,$ be a Sylow $3$-subgroup. of $\,G\,$ and let the group act on the left cosets of $\,P\,$ by left (or right) shift. This action determines a homomorphism of $\,G\,$ on $\,S_4\,$ whose kernel has to be non-trivial (why? Compare orders!) and either of order $27$ or of order $9$ (a subgroup of $ \, P \, $, say) , so in any case the claim's proved. |
H: Why is $\frac{y}{x}$ greater than $\frac{q}{p}$ -Figure
From the figure why is $\frac{y}{x}$ greater than $\frac{q}{p}$
AI: Let $\ell_1$ be the line through the origin $O$ and the point $P_1=(x,y)$. Let $\ell_2$ be the line through $O$ and the point $P_2=(p,q)$. Let $B=(0,p)$ and let $C=(p,c_2)$ be the point of intersection of the vertical line through $P_2$ and the line $\ell_1$.
Clearly (cough), $C$ lies above the pictured blue line, so $c_2>q$.
The slope of $\ell_1$ is ${y\over x} = {c_2\over p}$. The slope of $\ell_2$ is $q\over p$. Since $c_2>q$, we have ${c_2\over p}>{q\over p}$; and thus ${y\over x}>{q\over p}$.
(I think you can argue intuitively: the "steeper" the line, the greater its slope.) |
H: Proving that the set of algebraic numbers is countable without AC
A complex number $z$ is said to be algebraic if there is a finite collection of integers $\{a_i\}_{i\in n+1}$, not all zero, such that $a_0z^n + … + a_n = 0$.
Then can I prove the set of all algebraic numbers is countable without AC? I can only prove this by assuming 'countable union of countable sets is countable'. I guess and hope there's a way to prove this in ZF. Help.
AI: You do not need AC to prove the result because you can explicitly define an enumeration of the algebraic numbers within ZF. To do so, we start with any of the standard enumerations of the set $\bigcup_{n \in \omega} (\omega \times \mathbb{Z}^{n+1})$. This set consists of all sequences of the form $(k, a_0, \ldots, a_n)$ where $k \in \omega$ and $(a_0, \ldots, a_n)$ is a nonempty tuple of integers. We will use this as a starting point to define an enumeration of the algebraic numbers.
First, suppose that $w, z$ are two roots of the same nonzero polynomial $p(x) \in \mathbb{Z}[x]$. Say that $w <_{p(x)} z$ if either $|w| < |z|$ or else if $|w| = |z|$ and $\operatorname{arg}(w) < \operatorname{arg}(z)$. So $<_{p(x)}$ is a well ordering of the finite set of roots of $p(x)$ for every nonzero $p(x) \in \mathbb{Z}[x]$, and this ordering is uniformly definable with the tuple of coefficients of $p(x)$ as a parameter.
We form an enumeration of the algebraic numbers as follows. Say that a finite tuple of integers $(k, a_0, \ldots, a_n)$ represents a complex number $w$ if $w$ is a root of $p(x) = a_0 + a_1 x + \cdots + a_n x^n$, and $p(x)$ is not identically zero, and $k \in \omega$, and there are exactly $k$ other roots $z$ of $p(x)$ with $z <_{p(x)} w$. The relation "$(k, a_0, \ldots, a_{n})$ represents $w$" is definable in ZF, and ZF proves that for every algebraic $z$ there is at least one tuple that represents $z$, and every complex number that is represented is algebraic. Therefore, we can enumerate the algebraic numbers in the same order that they are represented by tuples, using the enumeration of the tuples from above.
Thus ZF proves that there is a surjection from $\omega$ to the set of algebraic numbers. Because $\omega$ is well ordered already, ZF can turn this into a bijection $f$ from $\omega$ to the set of algebraic numbers (namely, $f(n)$ is the $n$th distinct algebraic number to appear in the enumeration). Hence ZF can prove the set of algebraic numbers is countable. |
H: CW complex structure on a disk with 2 smaller disks removed
For some reason, I'm having trouble visualizing how to put a CW structure on a disk with 2 smaller disks removed. What I'd like to do is have three 0-cells, five 1-cells, and a single 2-cell. Three of the one cells would be glued to the three vertices to give three disjoint circles, and the other two 1-cells would be used to connect the three circles together. Then I can't see how to glue the 2-cell in, because when I visualize it, it always ends up covering one of the holes in the disk.
I realize my ability to visualize CW-complexes is not so good. For another example, it seems to me that if you take a wedge of two circles A and B, and glue a 2-cell on by $AB$, you should get a wedge of two disks, which would be contractible. But using cellular homology I see this has a nontrivial first homology group. I'd appreciate help sorting these two examples out to aid my intuition.
AI: Your thoughts on how to arrange the 0- and 1-cells are spot on. If you can draw the structure with just the 0- and 1-cells, you should see a good place to put the 2-cell. If you get stuck, I've drawn a picture:
https://i.stack.imgur.com/9zFDM.jpg
For the second example, recall that you're not actually filling in either of $A$ or $B$. Attaching a 2-cell via $AB$ means creating something like what you get when you pinch one of these:
https://i.stack.imgur.com/RYzAf.jpg
by holding a point on the top and bottom rings in your fist. Alternately, you can picture a piece of macaroni noodle with the smaller arc connecting the ends shrunk to a point. In fact, such a space is homotopy equivalent to a cylinder- simply unscrunch the join point of the circles into an interval. If you're curious about this sort of thing, some treatment is given in chapter 0 of Hatcher. |
H: Euler's theorem for powers of 2
According to Euler's theorem,
$$x^{\varphi({2^k})} \equiv 1 \mod 2^k$$
for each $k>0$ and each odd $x$. Obviously, number of positive integers less than or equal to $2^k$ that are relatively prime to $2^k$ is
$$\varphi({2^k}) = 2^{k-1}$$
so it follows that
$$x^{{2^{k-1}}} \equiv 1 \mod 2^k$$
This is fine, but it seems like even
$$x^{{2^{k-2}}} \equiv 1 \mod 2^k$$
holds, at least my computer didn't find any counterexample.
Can you prove or disprove it?
AI: You have it. For any odd $x$ we get $ x^2 \equiv 1 \pmod 8,$ which is where it starts. Note that we can write any odd $x$ as $4m \pm 1, $ then we get
$$ x^2 = (4m \pm 1)^2 = 16 m^2 \pm 8m + 1 \equiv 1 \pmod 8. $$ After that there should be no trouble applying induction.
Sure, just to save typing, take $A = 2^{k-2}.$ Then we begin with
$$ x^A \equiv 1 \pmod {4A}, $$
or
$$ x^A = 1 + 4 A t. $$ Then
$$ x^{2A} = (x^A)^2 = 1 + 8 A t + 16 A^2 t^2 \equiv 1 \pmod {8A}. $$ |
H: About fractional differentiation under the integral sign
$1.$ Does $\dfrac{d^n}{dx^n}\int_a^bf(x,t)~dt=\int_a^b\dfrac{\partial^n}{\partial x^n}f(x,t)~dt$ correct when $n$ is a positive real number?
$2.$ How about $\dfrac{d^n}{dx^n}\int_{a(x)}^{b(x)}f(x,t)~dt$ when $n$ is a positive real number?
AI: The first rule of Leibniz you mention will still hold for the fractional case, as the definitions of fractional derivatives are equivalent to whole-order differentiation of an integral which is operating orthogonal to the integral in another variable. Symbolically the proof for 1 using the traditional differintegral:
$$
D^n_x = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^xf(X,t)(X-x)^{\lceil n \rceil - n -1}dX
$$
so:
$$
D^n_x\int_a^bf(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_a^b f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX
$$
by Fubini's theorem:
$$
= \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_a^b \int_\alpha^x f(X,t) (X-x)^{\lceil n \rceil - n -1}dXdt
$$
By Leibniz theorem (for integer integrals and derivatives):
$$
= \int_a^b\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x (X-x)^{\lceil n \rceil - n -1} f(X,t)dXdt
$$
$$ = \int_a^bD_x^nf(x,t)dt $$
For Caputo's fractional differential the differential is inside of the integral with respect to x, so the use of Leibniz for integers is used before Fubini's theorem (the steps are just reversed).
For part 2:
$$
D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX
$$
Fubini's theorem applies here but it must be noted that you're operating over an entire measure space simultaneously (the symbolic idea of reversing the integrals just doesn't make sense here):
$$
=\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_{[a(X),b(X)]\times [\alpha,x]} f(X,t)(X-x)^{\lceil n \rceil - n -1}d(t,X)
$$
This is where things get a bit dicey for me, as the integral in this case involves a bit more measure theory than I'm entirely comfortable with (due to the interdependence of the product that forms the integral). My inclination is that this fails to make sense, as you wind up with differentiation of a variable which is caught in the bounds of both directions in the integral.
if, however the Caputo differ-integral were used:
$$
D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}}{dX^{\lceil n \rceil}}\int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX
$$
applying Leibniz here makes more sense since the derivative variable matches the variables in the limits of integration (for 0 < n < 1):
$$
D^n_x\int_{a(x)}^{b(x)}f(x,t)dt =
$$
$$
\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} -\frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} + \int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX
$$
At this point some breakdown can occur into three integrals:
$$
= B(x)- A(x) + C(x)
$$
By substituting the appropriate integral values for f(x,a(x)) and f(x,b(x)) so that they would cancel in the product of derivatives:
$$
A(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} dX
$$
$$
B(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} dX
$$
$$
C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x\int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX
$$
By the product rule for fractional differentiation we can then say:
$$
A(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX)
$$
$$
B(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX)
$$
The term for C(x) then solvable by substitution and the fundamental theorem:
$$
C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x f(X,b(X))(X-x)^{\lceil n \rceil - n -1}dX
$$
$$
- \frac{1}{\Gamma(\lceil n \rceil
- n)}\int_\alpha^x f(X,a(X))(X-x)^{\lceil n \rceil - n -1}dX
$$
Which implies, by substituting the operators back in:
$$
C(x) = D_x^nf(x,b(x)) - D^n_xf(x,a(x))
$$
So the total equation would then be:
$$
D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX)
$$
$$
- \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX)
$$
$$
+ D_x^nf(x,b(x)) - D^n_xf(x,a(x))
$$
So I would be confident in saying that there is a general formula for the fractional derivative, but it doesn't necessarily match the Leibniz rule for normal integration and differentiation. I've already noted my issues trying to find an expression for the normal fractional derivative but after this I'd guess one exists.
It's also a bit late so there might be some typos in my work I can't see at the moment so I'll double check it tomorrow just to be sure. |
H: What exactly is a manifold?
Wikipedia's "Simple English" entry describes a 2D map of the Earth as a manifold of the planet Earth.
Does this mean that in mathematics a manifold is essentially a representation of something that otherwise would difficult to "model in another way" in order to use it for some other purpose?
I have no idea what I'm talking about, but I am curious.
AI: First off, the definition is not totally standardized. I ran across the following comment on the talk page of the WP article, which I thought was very helpful in explaining this possible source of confusion:
Those of us who were introduced to manifolds via point set topology
(as in Munkres) have a gut feeling that this is what manifolds are,
and that the differential structure is an overlay. Those of us who
were introduced to manifolds via the differential structure (as in
Spivak) have a gut feeling that that is what manifolds are.
Both the WP article and this book have helpful lists of things that are and aren't manifolds. I would suggest having these lists handy while going through actual definitions of manifolds, because otherwise it's hard to understand why the different aspects of the definitions make sense.
It's also helpful to prepare yourself with an intuitive, informal idea of what we're trying to encapsulate in a formal definition. The basic idea is that we want to be able to describe a geometry stripped of (1) any notion of measurement, and (2) any notion of what is a straight line. However, we want to preserve distinctions like the distinction between a torus and a sphere.
Informally, here is a definition that I like. An n-dimensional manifold is a space M with the following properties:
M1. Dimension: M’s dimension is n.
M2. Homogeneity: No point has any property that distinguishes it from any other point.
M3. Completeness: M is complete, in the sense that specifying an arbitrarily small neighborhood gives a unique definition of a point.
If you work through some of the examples collected earlier, you'll see that this pretty much does the job. You can verify that the following are manifolds: the real line, a circle, the open half-plane $y>0$, the union of two disjoint planes. And that the following are not: a line glued to a plane, the rational numbers, the closed half-plane $y \ge 0$.
Here is a previous question I asked about formalizing the above definition.
The more typical definition is that a manifold is a space that is locally like $\mathbb{R}^n$. I dislike it philosophically, but it's easier to formalize than M1-M3 above. To formalize it, you can say that a manifold is a space in which any sufficiently small neighborhood is homeomorphic to an open set in $\mathbb{R}^n$. Homeomorphic means that you can find a homeomorphism between them. A homeomorphism is, intuitively, a process of stretching and distorting something without cutting or gluing. More formally, a homeomorphism is a function that is invertible and continuous in both directions. |
H: Why is $\int 1/(t ~\log^2 t) ~dt$ convergent?
$\displaystyle \int \frac{1}{t} dt = \ln t$ diverges.
How do I show that $\displaystyle \int_2^\infty \frac{1}{t ~\log^2 t} dt$ is convergent?
AI: To help yourself figure out the appropriate substitution, reorganize the integral into:
$$ \int \frac{1}{\log^2 t} \frac{dt}{t} $$
Does this ring a bell? $\dfrac{dt}{t}$? If we take $u = \log t,$ then we have $du = \dfrac{dt}{t}$ and $\dfrac{1}{\log^2 t} = \dfrac{1}{u^2}.$
So the integral is:
$$ \int \frac{1}{t\log^2 t} dt = \int \frac{1}{u^2} du = - \frac{1}{u} + \text{const} = -\frac{1}{\log t} + \text{const}.$$
Now $$ \int_2^{\infty} \frac{1}{t\log^2 t} dt = - \lim_{n \to \infty} \frac{1}{\log n} + \frac{1}{\log 2} = 0 + \frac{1}{\log 2}.$$ |
H: Sum of mutliples b/w $2$ and $10$ . What is wrong with this method
I am trying to find the sum of multiples b/w $2$ and $10$ end points non-inclusive using the following mechanism and I don't know why I am getting the wrong answer. (I selected this small range intentionally just to check the result)
First I find the no. of multiples b/w $2$ and $10$ using
No. of Multiples = $\frac{10-2}{2} - 1 = 3$ So there are three multiples of 2 b/w $2$ and $10$
Sum of Multiples now is = Multiple$(1+2+3)$=$2(6)=12$
But The answer is supposes to be $4+6+8 = 18$
Why am I getting the wrong answer.
Edit: I think the problem is with the formula:
Sum of Multiples now is = Multiple$(1+2+3)$=$2(6)=12$
I got this from here 3rd Comment after the question
AI: Maybe things will be clearer if we look at another example. Find the sum of all the multiples of 3 between 10 and 100. In this case, we are being asked to find $$12+15+18+21+\cdots+93+96+99.$$ That's the same as $$3(4+5+6+7+\cdots+31+32+33).$$ Why do the numbers inside the bracket start at 4, not at 1? Because the numbers we're adding start at 12, which is $3\times4$, not $3\times1$. Why do the numbers inside the bracket go up to 33? Because the numbers we're adding go up to 99, and $99=3\times33$.
This still leaves you with the problem of finding $$4+5+6+7+\cdots+31+32+33,$$ of course. |
H: Extending continuous and uniformly continuous functions
What do you say about the following statement:
Let $f\colon(a,b)\rightarrow \mathbb{R}$ be a uniformly continuous function. Then $f$ can be extended to a uniformly continuous function with domain $[a,b]$.
Let $f\colon(a,b)\rightarrow \mathbb{R}$ be a continuous function. Then $f$ can be extended to a continuous function with domain $[a,b]$.
So, I think that 2) is true, will be like add the points $a,b$ to domain used the definition limit, but I'm not sure. And 1) is true, too. But I don't know a good explanation.
AI: If $x_n$ is a Cauchy sequence, then uniform continuity of $f$ allows us to conclude that the sequence $f(x_n)$ is also Cauchy.
The sequences $a_n=a+\frac{1}{n}$ and $b_n =b-\frac{1}{n}$ are Cauchy, hence so are the sequences $f(a_n)$, $f(b_n)$. Since $\mathbb{R}$ is complete, these sequences converge to some numbers $f_a, f_b$ respectively. Define the function $\overline{f}:[a,b] \to \mathbb{R}$ by $\overline{f}(a) = f_a$, $\overline{f}(b) = f_b$ and $\overline{f}(x) = f(x)$ for $x \in (a,b)$. Clearly $\overline{f}$ is continuous in $(a,b)$, it only remains to show continuity at $a,b$.
Suppose $x_n\in [a,b]$, and $x_n \to a$. We have $|\overline{f}(x_n) - \overline{f}_a| \leq |\overline{f}(x_n) - \overline{f}(a_n)| + | \overline{f}(a_n) - \overline{f}_a|$. Let $\epsilon >0$, then by uniform continuity, there exists $\delta>0$ such that if $|x-y|< \delta$, with $x,y \in (a,b)$, then $|f(x)-f(y)| < \epsilon$. Choose $n$ large enough such that $|x_n-a_n| < \delta$, and $| f(a_n) - f_a| < \epsilon$. If $x_n = a$, then $|\overline{f}(x_n) - \overline{f}_a| = 0$, otherwise we have $|\overline{f}(x_n) - \overline{f}_a| \leq |f(x_n) - f(a_n)| + | f(a_n) - f_a| < 2 \epsilon$. Consequently $\overline{f}(x_n) \to \overline{f}_a$, hence $\overline{f}$ is continuous at $a$. Similarly for $b$.
For the second case, take $f(x) = \frac{1}{x-a}$. Then $f$ is continuous on $(a,b)$, but the domain cannot be extended to $[a,b]$ while keeping $f$ continuous, and $\mathbb{R}$ valued. To prove this, take the sequence $a_n$ above, then $f(a_n) = n$, and clearly $\lim_n f(a_n) = \infty$. If $f$ could be continuously extended to $\overline{f}$, then $\overline{f}(a) \in \mathbb{R}$, which would be a contradiction. |
H: Find a bijection from $(A^B)^C$ into $A^{B \times C}$
Possible Duplicate:
How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers?
Notation: Let A and B be sets. The set of all functions $f:A \rightarrow B$ is denoted by $B^A$.
Problem: Let A, B, and C be sets. Show that there exists a bijection from $(A^B)^C$ into $A^{B \times C} $. You should first construct a function and then prove that it is a bijection.
What I have so far:
$f:C \rightarrow A^B$
$ f(c) : B \rightarrow A $
and
$ g : B \times C \rightarrow A $
$ g(b, c) \in A$
How do I find a bijection between these functions?
AI: Let $f \in (A^B)^C, g \in A^{B \times C}$. Define $\Phi: (A^B)^C \to A^{B \times C}$ by setting
$$\Phi(f)(b,c) = f(b)(c)$$
This is a bijection because it has an inverse $\Psi: A^{B \times C} \to (A^B)^C$
$$\Psi(g)(b)(c) = g(b,c)$$ |
H: Show inclusion of a sum of subspaces in another subspace
Let $V$ be a vector space, and let $V',V'',W$ be subspaces of $V$. I want to show that
$$
(V' \cap W) + (V'' \cap W)
\subseteq
(V' + V'') \cap W
$$
So I take an element $a \in (V' \cap W) + (V'' \cap W)$ and show that it is also an element of $(V' + V'') \cap W$.
I know that $(V' \cap W)$ and $(V'' \cap W)$ are also subspaces of $V$, and that the sum of elements in a given subspace is also inside the subspace. I do not really know how to attack this question, since I usually work with regular sets, and not vector spaces, and presume that the solution to this problem relies on the definition of vector spaces.
Actually the problem is not really homework, but merely an exercise for a course, but I figured it would be best to tag it as homework anyways.
AI: Take a vector $a+b\in (V'\cap W)+(V''\cap W)$. So $a\in V'\cap W$ and $b\in V''\cap W$.
Now in particular, $a\in V'$ and $b\in V''$, so $a+b\in\dots$
Also, $a\in W$ and $b\in W$, so $a+b\in W$ because $\dots$ |
H: Calculating area between circle and line - where have I gone wrong here
I think the initial set up is wrong below I should just integrate over the area as a multiple integral here. correct? the 1step below seems wrong.
Problem is asking for the area inside a region of a circle of radius 2 (centered on origin) and with x> 0 and y > 1. Here's my attempt at a solution
$A = \int_0^\sqrt3 \sqrt{4-x^2}dx $
since when x varies from $\sqrt3$ to 0 as y varies from 1 to the circles edge. So then I tried a substitution
$x= 2\sin\theta\qquad$
$ dx = 2\cos\theta d\theta$
$A = \int_0^{\frac{\pi}{3}} \sqrt{4-(2sin\theta)^2}dx$
$A = 2\int_0^{\frac{\pi}{3}} \cos\theta dx $
$A = 2\int_0^{\frac{\pi}{3}} \cos\theta dx = 4\int_0^{\frac{\pi}{3}} \cos^2\theta d\theta = 4\int_0^{\frac{\pi}{3}} \frac{1}{2}\left( 1 + \cos2\theta\right) d\theta$
Here I substituted for $2\theta = v$ so $dv = 2d\theta$; and the limits change by a factor of two on the second integral so it should be
$ A= \frac{\theta}{2}|^\frac{\pi}{3}_0 + \sin{v} |^\frac{2\pi}{3}_{0} = \frac{2\pi + 3\sqrt3}{6} $
whereas the textbook gives the solution as = $\frac{4\pi -3\sqrt3}{6}$
I know the book answers is right since the total area of the quadrant is only pi but I'm not sure where exactly I'm screwing this up any tips appreciated!
AI: If we use formula for the area of circular segment
$$2A = \frac{R^2}{2} \left(\theta - \sin\theta \right)$$
with $R=2$, $\theta = \frac{2\pi}3$ and $\sin\theta=\sin\frac{2\pi}3=\sin\frac\pi3=\frac{\sqrt3}2$, we get
$$2A=2\left(\frac{2\pi}3-\frac{\sqrt3}2\right)\\
A=\frac{2\pi}3-\frac{\sqrt3}2$$
which is the same result as in the book.
You need $A = \int_0^\sqrt3 (\sqrt{4-x^2} -1) \,dx = I-\sqrt 3$, where $I$ denotes the integral you have computed.
In general $\int_a^b (g(x)-f(x)) \,dx$ is area between two functions $f$ and $g$.
You also missed factor 2 in the last step in $2\int_0^{\pi/3} 1\, d\theta$, see Rahul Narain's comment. |
H: integration by parts question
$\frac {2}{L}(-i\hbar)\int_{0}^{n\pi}\sin(u)\cos(u) \, du$
I tried to solve this using integration by parts
and I got
$\frac {2}{L}(-i\hbar)(\sin^2 u +\cos u)|_{0}^{n\pi}$.
But the answer is zero, and according to the above equation, it is zero or some number.
What did I do wrong here?
($n$ is constant positive integer, but not defined.)
AI: Just make a substitution $\sin u \cos u = \frac12 \sin 2u$ so that your integral is
$$
\frac1L(-\mathrm i\hbar)\int_0^{2n\pi}\sin t\mathrm dt
$$
which is clearly zero due to the $2\pi$-periodicity of $\sin t$. |
H: Are Cumulative Distribution Functions measurable?
It is well-known that CDFs (Cumulative Distribution Functions) of one-dimensional random variables are Borel measurable. But does the same apply to CDFs of multi-dimensional random variables (rvecs)? It suffices, for my purposes, to consider finite dimensional rvecs.
Relevant definitions
Let $n$ be an integer $\geq 2$.
Call a probability measure over the Borel field on $\mathbb R^n$ an $n$-dimensional probability measure.
To every $n$-dimensional probability measure, $m$, define the following function $F:\mathbb R^n\rightarrow\mathbb R$, called $m$'s CDF: $F(x)=m\left((-\infty, x]\right)$ where $(-\infty, x]$ is the set of all $y$ in $\mathbb R^n$ such that $y\leq x$ component-wise.
An $n$-dimensional CDF is a function $F:\mathbb R^n\rightarrow\mathbb R$ that can be obtained as some $n$-dimensional probability measure's CDF.
$n$-dimensional CDFs are known to be characterized by certain properties.
AI: For any $x \in \mathbb R^n$, consider the measurable box-like function $M_x(y) = m((-\infty,x]) \cdot \mathbb1_{[x,\infty)}(y)$. Here $\mathbb1_{[x,\infty)}(y)$ is the indicator function of "$y \ge x$ component-wise".
It should be easy to show that $F(y) = \sup_{x \in \mathbb Q^n} M_x(y)$ for any $y$ (take an increasing sequence of $x$ converging to $y$), and this makes $F$ the countable supremum of measurable functions. |
H: Polynomial coefficients in exponential-series: how can I convert this into a composite of $\exp(x)$?
Assume we have the exponential-series $ \small \exp(x) = 1+ {x \over 1!} + {x^2 \over 2! } + \cdots = \sum\limits_{k=0 }^\infty {x^k \over k!} $ modified with a polynomial in the coefficients
say $ \qquad \displaystyle f_1(x)= \sum_{k=0}^\infty { k^2 + k \over 2} {x^k \over k!} $
or $ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( 15 k^3 + 15 k^2 - 10k - 8) {x^k \over k!} $
or in general
$ \qquad \displaystyle f_3(x)= \sum_{k=0}^\infty ( d k^3 + c k^2 + b k + a) {x^k \over k!} $
-: is there a good formula/algorithm/scheme how this has to be expressed as composition of the $\exp(x)$-function? (I know this can be solved using the derivatives and cancelling of k's in the polynomial with the factorials in the denominator - I'm asking for a handy/memorizable translation-formula )
Because I can factor my examples under study: is there possibly a special handy scheme, if the polynomials are given in a form like this
$ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( k-1)(k-2) {x^k \over k!} $ ?
AI: \begin{eqnarray}
\sum_{k=0}^\infty k\frac{x^k}{k!}&=&\sum_{k=1}^\infty \frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty\frac{x^k}{k!}=xe^x;\cr
\sum_{k=0}^\infty k^2\frac{x^k}{k!}&=&x\sum_{k=1}^\infty k\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x(x+1)e^x\cr
\sum_{k=0}^\infty k^3\frac{x^k}{k!}&=&\sum_{k=1}^\infty k^2\frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty(k+1)^2\frac{x^k}{k!}\cr
&=&x\sum_{k=0}^\infty(k^2+2k+1)\frac{x^k}{k!}=x[x(x+1)+2x+1]e^x\cr
&=&x(x^2+3x+1)e^x.
\end{eqnarray}
Hence
\begin{eqnarray}
\sum_{k=0}^\infty (dk^3+ck^2+bk+a)\frac{x^k}{k!}&=&d\sum_{k=0}^\infty k^3\frac{x^k}{k!}+c\sum_{k=0}^\infty k^2\frac{x^k}{k!}+b\sum_{k=0}^\infty k\frac{x^k}{k!}+a\sum_{k=0}^\infty \frac{x^k}{k!}\cr
&=&dx(x^2+3x+1)e^x+c(x^2+x)e^x+bxe^x+ae^x\cr
&=&[a+bx+c(x^2+x)+d(x^3+3x^2+x)]e^x\cr
&=&[a+(b+c+d)x+(c+3d)x^2+dx^3]e^x
\end{eqnarray} |
H: Geometric intuition for the inequality $(f(y) - c) ( y - d ) \geq (f(d) - c) ( f^{-1}(c) - d )$
Good day to everyone. I am interested in the geometric intuition for the following statement:
Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a monotonically increasing, invertible function and $c,d \in \mathbb{R}$. Then for any $y \in \mathbb{R}$ we have
\begin{equation}
\Big(f(y) - c\Big) \Big( y - d \Big) \geq \Big(f(d) - c\Big) \Big( f^{-1}(c) - d \Big).
\end{equation}
It is a lemma which I have encountered in the article (Lemma 3.1):
T. Kerkhoven and J. W. Jerome.
$L_\infty$ stability of finite element approximations of elliptic gradient equations. Numerische Mathematik, 57:561, 1990.
The lemma is proven in the paper. The proof is analytic and it is rather elementary. However, it seems to me that this statement is releated to some geometric property of graphs of real increasing functions. And I am interested in this geometric intuition. Perhaps it is a standard argument, but I cannot see it. Thus my question is:
What are the geometrical reasons for the statement cited above?
Some thoughts:
It is maybe relevant to multiply the inequality by (-1), reversing its direction. It is so, because if we take $f(x):=x$, then also $f^{-1}(x) = x$ and we obtain $LHS:=(y-c)(y-d) \geq -(d-c)^2=:RHS$, so RHS is negative for all $c \neq d$. Thus if LHS is positive, then the signs differ and probably no geometrical argument can be given. So the interesting case would be when LHS is also negative.
AI: I get very confused thinking geometrically with $y$ on the horizontal axis, so let us call the variable $x$ instead, and let $y = f(x)$. Also, let's say $x_1 = f^{-1}(c)$, $y_1 = c$, and $x_2 = d$, $y_2 = f(d)$, so that the points $(x_1,y_1)$ and $(x_2,y_2)$ lie on the monotonically increasing curve $y=f(x)$. The inequality becomes
$$(y-y_1)(x-x_2)\ge(y_2-y_1)(x_1-x_2).$$
As $f$ is increasing, $y_2-y_1$ has the same sign as $x_2-x_1$, so the right-hand side is always negative. If $x$ is outside the interval between $x_1$ and $x_2$, the left-hand side is positive, so the inequality holds. The interesting part is when $x$ lies between $x_1$ and $x_2$, and here we will look at it geometrically.
Negate the inequality to obtain
$$(y-y_1)(x_2-x)\le(y_2-y_1)(x_2-x_1).$$
Assuming $x_1<x_2$, the right-hand side is the area of the rectangle $[x_1,x_2]\times[y_1,y_2]$, while the left-hand side is the area of $[x,x_2]\times[y_1,y]$. The latter rectangle is a subset of the former, so the inequality holds. If $x_1>x_2$, negate all four terms and you again get the positive areas of two rectangles, and the same reasoning applies. |
H: A Time Calculation Problem: 3yrs or 3yrs and 1 day?
Suppose, a person A has his birthday on 30 Sept 1994 and another person B has his birthday on 30 Sept 1997, then which of the following statement would be correct and why?
The difference between the ages of the two persons is
Three years correct to the number of Days
Three years and 1 day correct to the number of Days (because of 29th Feb 1996 that comes in between)?
AI: You need to be precise about what you mean by a year, but in practise all alternatives will give you 1 as the correct answer.
Either you define a year so that it has constant length, for example 365.24 days, and then you observe that the difference between the two dates differs from 3 years with less than a day. This is the approach astronomers would take in their work.
Or you define a year so that it always has a whole number of days, like in the Julian or the Gregorian calendar. Then a year will not have constant length, which is messy when you work with astronomy, but is OK for most everyday purposes. Also in that case you get the same answer of 3 years, or Alternative 1. |
H: Best constant in an integral inequality
Which is the smallest constant $B_d$ such that the following inequality
$$\left|\int_{0}^{1}t^d(1-t)\psi(t) dt\right|^2\le
B_d\int_{0}^{1}t^d|\psi(t)|^2dt$$ holds, provided that $\psi(t)$ is a polynomial?
AI: A hint: Consider the scalar product
$$\langle u,v\rangle:=\int_0^1 u(t) v(t)\ t^d\ dt\ .$$ |
H: $\sigma$-algebra of $\theta$-invariant sets in ergodicity theorem for stationary processes
Applying Birkhoff's ergodic theorem to a stationary process (a stochastic process with invariant transformation $\theta$, the shift-operator - $\theta(x_1, x_2, x_3,\dotsc) = (x_2, x_3, \dotsc)$) one has a result of the form
$ \frac{X_0 + \dotsb + X_{n-1}}{n} \to \mathbb{E}[ X_0 \mid J_{\theta}]$ a.s.
where the right hand side is the conditional expectation of $X_0$ concerning the sub-$\sigma$-algebra of $\theta$-invariant sets... How do these sets in $J_{\theta}$ look like? (I knew that $\mathbb{P}(A) \in \{0,1\}$ in the ergodic case, but I don't want to demand ergodicity for now).
AI: One asks that $A$ is such that $(x_n)_{n\geqslant1}\in A$ if and only if $(x_{n+1})_{n\geqslant1}\in A$. The surprising fact is that such events $A$ do exist, whose definition is not trivial, and in fact a lot of them. For example, $A$ is invariant as soon as the fact that $(x_n)_{n\geqslant1}\in A$ depends only on:
the liminf and/or the limsup of $\frac1{b_n}\sum\limits_{k=1}^na_ku(x_k)$ when $n\to\infty$, for some function $u$ and some sequences $(a_n)_{n\geqslant1}$ and $(b_n)_{n\geqslant1}$ such that $b_n\to\infty$.
the liminf and/or the limsup of $\frac1{b_n}\sum\limits_{k=1}^na_ku(x_k,\ldots,x_{k+N})$ when $n\to\infty$, for some fixed $N$, some function $u$ and some sequences $(a_n)_{n\geqslant1}$ and $(b_n)_{n\geqslant1}$ such that $b_n\to\infty$.
the liminf and/or the limsup of $\frac1{b_n}\sum\limits_{k=1}^na_ku((x_{k+i})_{i\geqslant0}))$ when $n\to\infty$, for some function $u$ and some sequences $(a_n)_{n\geqslant1}$ and $(b_n)_{n\geqslant1}$ such that $b_n\to\infty$.
the fact that $\{n\geqslant1\,;\,x_n\in B\}$ is finite or infinite, for some $B$.
the fact that $\{n\geqslant1\,;\,(x_n,\ldots,x_{n+N})\in B\}$ is finite or infinite, for some fixed $N$ and some $B$.
the fact that $\{n\geqslant1\,;\,(x_{n+i})_{i\geqslant0}\in B\}$ is finite or infinite, for some $B$. |
H: Number Theory and combinatorial
Today, I took this observation from my note book. I am looking the strategy to deal this statement. The difference between $$\binom{n}{p}$$ and $$\left\lfloor\frac{n}{p}\right\rfloor\,$$ is divisible by p for a positive integer n and p is prime with >1. Here $$\binom{n}{p}$$ is the number of ways one can choose p out of n elements and $$\left\lfloor{x}\right\rfloor\,$$ is the greatest integers not exceeding the real number x.
The above one is I found from the following problem.
5 divides the difference between $$\binom{n}{5}$$and$$\left\lfloor\frac{n}{5}\right\rfloor\,$$
Numerically we can solve. I would like to learn how to solve or prove the above cited statement mathematically?
Thank you.
I got good reply from one of the MATH STACK USER. I studied as per his guidance about the LUCAS Theorem, I encounter the following facts with doubts and difficulties.
If we express the p (not prime) in terms of $q^x$ k where q and k are relatively primes with q is prime,. Then my example given above fails. Of course x and k are not equal to 1 simultaneously.
With reference to the above fact, how we generalize the above fact mathematically?
Now, my second doubt/question is, why to solve my statement by Lucas Theorem? If we can do the same by Wilson’s theorem? This is I am just guessing. I am not sure how far I am correct. Kindly discuss, if I am wrong/correct?
If Lucas Theorem only will solve my statement, how to encounter the fgollowing fact from Lucas theorem?
For a and q are positive integers and greater than 1, such that
$$\binom{na}{ma}$$ $\equiv 3\ $$\binom{n}{m}$ (mod p)
For every pair of integers n greater than equal to m greater than equal to 0 with a & q are powers of the same prime p ?
I am so exited to encounter the above facts during my study on Lucas theorem to complete my statement given above. Kindly discus and thank you so much for every replier.
AI: This is a very pretty instance of Lucas's Theorem: http://en.wikipedia.org/wiki/Lucas%27_theorem
If you write out $n$ and $p$ in base $p$ you'll see that only the "tens" digit of $n$ contributes to the product, and this digit is equivalent to $\lfloor n/p \rfloor$ modulo $p$.
Here is a more elementary argument by induction: notice that $\lfloor n/p \rfloor - \lfloor (n-1)/p \rfloor$ is either $0$ or $1$ depending on whether $n$ is a multiple of $p$.
Therefore we want to show that the remainder of $\binom{n}{p}$ is exactly the same as $\binom{n-1}{p}$ if $n$ is not divisible by $p$, and that it is exactly $1$ higher in the case that $n$ is divisible by $p$. Expand out:
$$\binom{n}{p}-\binom{n-1}{p} = \binom{n-1}{p-1} = \frac{(n-1)(n-2)\cdots (n-p+1)}{(p-1)(p-2)\cdots1}.$$
If $n$ is not divisible by $p$, then the numerator has one term divisible by $p$, but the denominator doesn't, so $\binom{n}{p}-\binom{n-1}{p}$ is a multiple of $p$. This proves half of what we wanted to show.
If $n$ is divisible by $p$, then both the numerator and denominator are congruent to $(p-1)!$ mod $p$, so they cancel out to exactly $1$. This proves the other half. |
H: divergence of $(2^n-n)$
Can anyone give me a satisfactory proof that the real sequence $(x_n)$ defined by $x_n = 2^n - n$ diverges to $+\infty$?
The heuristic reason is that
$$
\lim_{n\to\infty} \frac{n}{2^n} = 0,
$$
but I can't seem to turn this into a rigorous proof.
More generally is there a theorem which says that $(z_n-y_n)$ diverges to $+\infty$ if $(y_n)$ and $(z_n)$ both diverge to $+\infty$ and $\lim_{n\to\infty} y_n/z_n = 0$?
AI: As you pointed out $$\frac{n}{2^n} \underset{n \rightarrow \infty}{\longrightarrow} 0$$
Thus you can find $n_0 \geq 0$ such that $\forall n \geq n_0$
$$\frac{n}{2^n} \leq \frac{1}{2}$$
Thus $\forall n \geq n_0$,
$$ x_n = 2^n - n = 2^n \left( 1 - \frac{n}{2^n}\right) \geq 2^n \left( 1 - \frac{1}{2} \right) \geq 2^{n-1} \underset{n \rightarrow \infty}{\longrightarrow}+\infty $$
Thus $x_n \underset{n \rightarrow \infty}{\longrightarrow}+\infty $. The same exact proof can be applied to the generalized case you mentionned. |
H: Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
Show that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
AI: I enjoyed working this out-cool question man. Here's a straightforward inductive proof:
I'm writing $c$ instead of $n$.
Inductively assume $(c + \sqrt{c^2 - 1})^k = a + b \sqrt{c^2 - 1}$, where $b^2(c^2 - 1) = a^2 - 1$ (the base case is clear). That is, not only can you write it as $t + \sqrt{t^2 - 1}$, but also $t^2 - 1$ is a square times $c^2 - 1$ (I guessed this by working out the first $k$'s for $c =2,3$).
Now to see $k \Rightarrow k+1$, we have $$(c + \sqrt{c^2 - 1})^{k+1} = (a + b \sqrt{c^2 - 1})(c + \sqrt{c^2 - 1}) $$$$= (ac + b(c^2 - 1)) + (a + cb)\sqrt{c^2 - 1}$$It suffices to check that $$(ac + b(c^2 - 1))^2 - 1 = (a + cb)^2(c^2 - 1)$$Expanding both sides gives $$c^2 a^2 + 2c(c^2 - 1)ab + (c^2 - 1)^2 b^2 - 1 = (c^2 - 1)a^2 + 2c(c^2 - 1)ab + c^2(c^2 - 1)b^2$$Cancelling the $ab$ terms, pulling the $a^2$ term to the LHS, and the $b^2$ terms to the RHS, we get $$a^2 - 1=b^2(c^2 - 1) $$our inductive hypothesis. |
H: 'Linux' math program with interactive terminal?
Are there any open source math programs out there that have an interactive terminal and that work on linux?
So for example you could enter two matrices and specify an operation such as multiply and it would then return the answer or a error message specifying why an answer can't be computed? I am just looking for something that can perform basic matrix operations and modular arithmetic.
AI: Sage is basically a Python program/interpreter that aims to be an open-source mathematical suite (ala Mathematica and Magma etc.). There are many algorithms implemented as a direct part of Sage, as well as wrapping many other open-source mathematics packages, all into a single interface (the user never has to tell which package or algorithm to use for a given computation: it makes the decisions itself). It includes GP/Pari and maxima, and so does symbolic manipulations and number theory at least as well as them.
It has a command-line mode, as well as a web notebook interface (as an example, a public server run by the main developers).
And, although this might not be relevant since your use-case sounds very simple, the syntax is just Python with a small preprocessing step to facilitate some technical details and allow some extra notation (like [1..4] which expands to [1,2,3,4]), so many people already know it and, if not, learning it is very easy.
As a slight tangent, Sage is actually the origin of the increasingly popular Cython language for writing fast "Python", and even offers an easy and transparent method of using Cython for sections of code in the notebook. |
H: Checking efficiency of randomness with entropy
I was going through random numbers and found that the randomness of certain observations is measured by the entropy as given in here. Here, $p(x_i)$ is the probability that $x_i$ will take place. But if I have fair dice then $p(x_i)$ is $\frac{1}{6}$. So, I am assuming that less is the entropy, better is the randomness. Is my conclusion correct? Also, entropy is calculated in terms of - so while determining the efficiency should I consider this negative sign also or only the positive number? I hope my question is clear. Any help will be appreciated!
AI: Your conclusion is wrong. Entropy in a measure of randomness it is correct but it increases if the randomness increases. There are many different characterizations to maximize the entropy. For example from all densities with equal variance Gaussian density is the one which maximizes the entropy.
On the other hand given a number of events, uniform distribution maximizes the entropy. This means if you have a fair dice it has the maximum entropy and also maximum randomness. If some events are correlated to the others, then entropy decreases as well as the randomness because you can estimate it using the correlation patterns! Read:
http://en.wikipedia.org/wiki/Maximum_entropy_probability_distribution |
H: Atiyah-Macdonald, Exercise 2.17 (direct limit)
I have solved the following exercise, can you tell me if this is correct? Thanks.
2.17. Let $(M_i)_{i \in I}$ be a family of submodules of an $A$-module such that for each pair $i,j$ in $I$ there exists $k$ in $I$ such that $M_i + M_j \subset M_k$. Define $i \leq j$ to mean $M_i \subset M_j$ and let $\mu_{ij}: M_i \to M_j$ be the embedding of $M_i$ in $M_j$. Show that
$$ \varinjlim_n M_n = \bigcup M_n = \sum M_n$$
$ \varinjlim_n M_n = \bigcup M_n$:
We want to show that the union together with inclusions $i_i : M_i \hookrightarrow \bigcup M_n$ satisfies the universal property of the direct limit of the direct system $M_i, \mu_{ij}$. To this end, let $Y$ be a module and $i_{i}^\prime$ be inclusions $i_{i}^\prime : M_i \hookrightarrow Y$ such that for $x_k \in M_k$ we have $i_i^\prime (\mu_{ki}(x_k)) = i_k^\prime(x_k)$. We want to show that there exists a unique homomorphism $\varphi: \bigcup M_n \to Y$ such that for all $i,j$: $\varphi \circ i_i = i_j^\prime \circ \mu_{ij}$. Define $\varphi: \bigcup M_i \to Y$ as follows: for $m$ in $\bigcup M_i $ there is $M_i$ such that $m \in M_i$. Set $\varphi(m) = i^\prime_j (\mu_{ij} (i_i^{-1}(m)))$. Now we have existence, since this (as a concatenation of homomorphisms) is clearly a homomorphism. (well-define since inclusions are injective). To verify uniqueness, let $\varphi^\prime$ be a second homomorphism making the diagram commute. Let $m \in M$ and for a random $k$, let $m_k = i_k^{-1}(m)$. Then $\varphi (m) = \varphi (i_k (m_k)) = i^\prime_j (\mu_{kj} (m_k)) = \varphi^\prime (i_k (m_k) ) = \varphi^\prime (m)$, hence showing uniqueness.
Similarly in the case $\sum M_n$.
Edit
Dear BenjaLim, or anyone else, I'm sorry but I don't understand why I cannot do $i_i^{-1}(m)$. Would someone explain it to me? Thank you.
AI: Edit: We first check that $\bigcup M_i$ is an $A$ - module: We will check the only non-trivial part here which is closure under addition. Suppose that we have $y,z \in \bigcup M_i$. Then there exist $j,k$ such that $y \in M_j$ and $z \in M_k$. Now by assumption there exists $l \in I$ such that $M_j + M_k \subseteq M_l$. Hence it is clear that $y+z \in M_l \subset \bigcup M_i$ showing that $\bigcup M_i$ is closed under addition.
I don't think your proof is quite right because you can't just say "concatenation of homomorphisms": how do you define $i^{-1}(m)$? I think your proof breaks down here because you cannot just do that. The way I would do it is as follows:
We want to prove that whenever we have maps $f_i : M_i \rightarrow N$ where $N$ is some $A$ - module and $f_i = f_j \circ \mu_{ij}$ whenever $i\leq j$, there is a unique $A$ - module homomorphism $L : \bigcup M_i \to N$. Take an element $x \in \bigcup M_i$. Then $x$ is in at least one $M_j$, so we can define $L(x)$ to be equal to $f_j(x)$. We need to check that this is well defined. Suppose $x$ is also in some other $M_k$. Then we know that we can choose an $l \in I$ such that $l \geq j,k$. By definition of $f_l$ we have that $f_j(x) = f_l\circ \mu_{jl}(x)$. But the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. Similarly writing down $f_k(x) = f_l \circ \mu_{kl}(x)$, the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. It follows that $f_j(x) = f_k(x)$ so that $L$ is well defined.
We now check that $L$ is an $A$ - module homomorphism; the only non-trivial part to check is that $L$ is additive. So suppose we have $x,y \in \bigcup M_i$. Then there is $j,k$ such that $x \in M_j$ and $y \in M_k$. Then we know that there exists an $l \geq j,k$ such that $x+y \in M_l$ and so $L(x+y) = f_l (x+y)$. Now $L(x) = f_j(x)$, $L(y) = f_k(y)$. Since
$$f_j(x) = f_l\circ \mu_{jl}(x), \hspace{4mm} f_k(y) = f_l\circ \mu_{kl}(y)$$
we can say that $f_j(x) + f_k(y) = f_l(x) +f_l(y)$. This is because $\mu_{jl}(x)$ and $\mu_{kl}(y)$ just views $x$ and $y$ respectively as elements of $M_l$. It follows that $L$ is additive. It now remains to check that $L$ is uniquely determined by the $f_i$. Suppose we have another map $\varphi : \bigcup M_i \longrightarrow N$ such that $\varphi \circ \tau_i = f_i$. Then it follows that
$$\varphi \circ \tau_i(x) = L \circ \tau_i(x)$$
given any $i \in I$ and $x \in M_i$. But then since the $\tau_i$ are inclusion maps this is effectively saying that $L(x) = \varphi(x)$ for all $x \in M_i$ for any $M_i$. Since $x$ is arbitrary we conclude that $L = \varphi$ showing uniqueness. This completes the proof that
$$\varinjlim M_i \cong \bigcup M_i.$$
Edit: I think you have your universal properties wrong: The universal property of the direct limit is this: If you have $N$ an $A$ - module and for each $i \in I$ let $\alpha_i : M_i \to N$ be an $A$ - module homomorphism such that $\alpha_i = \alpha_j \circ \mu_{ij}$ whenver $i \leq j$. The there exists a unique homomorphism $\alpha : M \to N$ such that $\alpha_i = \alpha \circ \mu_i$ for all $i \in I$.
Edit: Let me show you why it suffices to show that $\bigcup M_i$ satisfies the universal property of $\varinjlim M_i$ to prove that $\bigcup M_i \cong \varinjlim M_i$. Now we have inclusion maps $\tau_i : M_i \to \bigcup M_i$ such that clearly $\tau_i = \tau_j \circ \mu_{ij}$ for any $i \leq j$ because the $\mu_{ij}$ are just inclusion maps. So now we just need to show that $\bigcup M_i$ has the property that given compatible $A$ - module homomorphisms $f_i : M_i \to N$ for some $A$ - module $N$ such that $f_i = f_j \circ \mu_{ij}$ whenever $i \leq j$, there is a unique $A$ - linear map $L : \bigcup M_i \to N$ such that
$$f_i = L \circ \tau_i \hspace{3mm} \forall i\in I.$$
To see why we only need to verify this, suppose the result above about $\bigcup M_i$ holds. Then putting in place of $f_i$ the usual maps $\phi_i$ from $M_i$ to $\varinjlim M_i$ and $N = \varinjlim M_i$, we have a unique linear map $L : \bigcup M_i \longrightarrow \varinjlim M_i$ such that
$$\phi_i = L \circ \tau_i \hspace{3mm} \forall i \in I.$$
Recall how the maps $\phi_i$ are defined. We have $M_i \stackrel{\lambda_i}{\longrightarrow} \bigoplus_{i\in I} M_i \stackrel{ \mu}{\longrightarrow} \varinjlim M_i$ so
$$\phi_i \stackrel{\text{def}}{\equiv} \mu \circ \lambda_i$$
where $\mu$ is the canonical projection from the direct sum onto the direct limit, and the $\lambda_i$ the canonical injections from $M_i$ into the direct sum. Now because we have maps $\tau_i$ out of $M_i$ to $\bigcup M_i$ such that $\tau_i = \tau_j \circ \mu_{ij}$ whenever $i\leq j$, then by the universal property of the direct limit, there is a unique $A$ - module homomorphism $L': \varinjlim M_i \rightarrow \bigcup M_i$ such that
$$\tau_i = \phi_i \circ L' \hspace{3mm} \forall i \in I.$$
From here it is not hard to see that $L$ and $L' $ are mutual inverses so that $\bigcup M_i \cong \varinjlim M_i$.
$\hspace{6in} \square$ |
H: why do we need the fourth derivative of the function to check the error bound in the simspon's rule?
I understand that trapezoidal or midpoint rule's error bound needs the second derivative,
but I just don't get why the fourth derivative in simpson's rule
please help me :)
AI: Simpson's Rule is exact for polynomials of degree 3 or less, so the error term can't depend on any derivative less than the 4th. Put another way: the 3rd derivative of a 3rd degree polynomial is nonzero, so it can't be involved in the error term for Simpson, since the error for 3rd degree polynomials is zero. |
H: convergence of a particular series
Let be $ \Lambda\subseteq \mathbb C$ a lattice, I don't understand why the series $$\sum_{\lambda\in\Lambda\setminus\{0\}} \frac{1}{|\lambda|^s}$$ converges for $s>2$. Can someone help me?
AI: Start by looking initially, the the first 8 terms (with $\lambda$ nearest to $0$), in terms of the smallest distance of these 8 values of $\lambda$ from $0$ (call it $r$). You should get
$$\frac{1}{|\lambda^s|} \leq \frac{1}{r^s}.$$
Then for the next 16 terms (slightly further away from $0$), you should be able to get:
$$\frac{1}{|\lambda^s|} \leq \frac{1}{(2r)^s}$$
and so on for values of $\lambda$ further out from $0$.
If you then add the first $8(1+2+\dots+n)$ values of $\lambda$, you find that the partial sum is bounded above by $$\frac{8}{r^s}\sum_{k=1}^n \frac{1}{k^{s-1}},$$ which should give the convergence for $s>2$. You could also look at lower bounds in the same way if you wanted to show that the series converges if and only if $s>2$. |
H: Usage of Addition principle or Combination.
The question is, " How many ways are there to draw a heart or a club from an ordinary deck of card ? "
The correct method is to use Addition Principle to solve it .
But can I use Combination to solve it as well ? Technically, it is selecting 13 cards out of the 52 cards without any order.
I tried 52C13 + 52C13 .. the results is not logical at all.
AI: I tried 52C13 + 52C13 .. the results is not logical at all.
The reason $\binom{52}{13}$ is not logical is because what you're saying is, from the entire deck of cards--i.e. from all 52 cards--be it a heart or a club or a spade or a diamond, pick any combination of 13 cards.
So, one such possibility out of the $\binom{52}{13}$ = 635,013,559,600 ways is the entire spade suit, which does not come near what your problem asks.
There is a way to find it using the choose function. Find out how many cards fit the criteria (in this case 13 hearts and 13 diamonds) and specify how many cards (among these) you want to choose (1 card). So that's $\binom{13 + 13}{1} = \binom{26}{1} = 26$. |
H: Randomly selecting a natural number
In the answer to these questions:
Probability of picking a random natural number,
Given two randomly chosen natural numbers, what is the probability that the second is greater than the first?
it is stated that one cannot pick a natural number randomly.
However, in this question:
What is the probability of randomly selecting $ n $ natural numbers, all pairwise coprime?
it is assumed that we can pick $n$ natural numbers randomly.
A description is given in the last question as to how these numbers are randomly selected, to which there seems to be no objection (although the accepted answer is given by one of the people explaining that one cannot pick a random number in the first question).
I know one can't pick a natural number randomly, so how come there doesn't seem to be a problem with randomly picking a number in the last question?
NB: I am happy with some sort of measure-theoretic answer, hence the probability-theory tag, but I think for accessibility to other people a more basic description would be preferable.
AI: It really depends on what you mean by the "probability of randomly selecting n natural numbers with property $P$". While you cannot pick random natural number, you can speak of uniform distribution.
For the last problem, the probability is calculated, and is to be understood as the limit when $N \to \infty$ from the "probability of randomly selecting n natural numbers from $1$ to $N$, all pairwise coprime".
Note that in this sense, the second problem also has an answer. And some of this type of probabilities can be connected via dynamical systems to an ergodic measure and an ergodic theorem.
Added The example provided by James Fennell is good to understand the last paragraph above.
Consider ${\mathbb Z}_2 = {\mathbb Z}/2{\mathbb Z}$, and the action of ${\mathbb Z}$ on ${\mathbb Z}_2$ defined by
$$m+ ( n \mod 2)=(n+m) \mod 2$$
Then, there exists an unique ergodic measure on ${\mathbb Z}_2$, namely $P(0 \mod 2)= P(1 \mod 2)= \frac{1}{2}$.
This is really what we intuitively understand by "half of the integers are even".
Now, the ergodic theory yields (and is something which can be easily proven directly in this case)
$$\lim_{N} \frac{\text{amount of even natural numbers} \leq N}{N} = P( 0 \mod 2) =\frac{1}{2} \,.$$ |
H: Modulus Distributing Over Multiplication?
Given positive integers a,b,c and k:
Define a function $M: \mathbb{Z^2} \rightarrow \mathbb{Z}$ as
$$M(x,y) = (x \bmod y)$$
i.e. the remainder of integer division
The following is always true:
$$a+b=c \implies M(M(a,k) + M(b,k), k) = M(c,k)$$
Under which values of k is the following true:
$$ab=c \implies M(M(a,k)M(b,k), k) = M(c,k)$$
That is when does mod distribute over multiplication?
The answer is always:
Proof:
Let $a = q_ak + r_a$ and $b = q_bk + r_b$ where $ 0 \le r_a, r_b < k$
$$\begin{align*}
c &= ab \\
&= (q_ak + r_a)(q_bk + r_b) \\
&= q_aq_bk^2 + q_ar_bk + q_br_ak + r_ar_b \\
&= (q_aq_bk + q_ar_b + q_br_a)k + r_ar_b \\
\end{align*}$$
$$\begin{align*}
M(c,k) &= M((q_aq_bk + q_ar_b + q_br_a)k + r_ar_b,k) \\
&= M(r_ar_b,k)
\end{align*}$$
$$\begin{align*}
M(M(a,k)M(b,k), k) &=(M(q_ak + r_a,k)M(q_bk + r_b,k)) \\
&= M(r_ar_b, k)
\end{align*}$$
QED
AI: Hint $\rm\ mod\ k\!:\ A\equiv a,\, B\equiv b\:\Rightarrow\: AB\equiv ab,\ $ so $\rm\ AB\, mod\, k\, =\, ab\, mod\, k$
Yours is the special case $\rm\ A = (a\,mod\,k),\,\ B = (b\,mod\,k)$ |
H: Proving the subset $C$ is a left coset of a certain subgroup iff $x,y,z \in C \Rightarrow xy^{-1}z\in C$
Revising for Group Theory at the moment and I'm sort of stumped on this exercise:
Given a subset $C$ of $G$, prove that $C$ is a left coset of a certain subgroup of $G$ iff $$xy^{-1}z \in C $$ when $$ x,y,z \in C.$$
I'm not sure how to start solving this, so any hints are welcome.
AI: If $C=cH$, for some subgroup $H$, then you can determine $H$ by taking the set of all values $y^{-1}z$ where $y,z\in C$.
So let's try that. Given a $C$ with the property above, define:
$$H=\{y^{-1}z: y,z\in C\}$$
Step: Prove $H$ is a subgroup - that it is non-empty, and it is closed under multiplication and inverses. (You need to add the condition that $C$ is non-empty to prove this.)
Step: Given any element $c\in C$, all elements can be written as $ch$ for some $h\in H$
So if $C$ has this property and is non-empty then $C$ is a left coset.
If $C$ is a left coset, it is pretty clear it is non-empty, and it's direct to prove that $C$ has this property. Specifically, if $C=cH$, then $x=ch_1$, $y=ch_2$ and $z=ch_3$, for $h_1,h_2,h_3\in H$, and we get $$xy^{-1}z = ch_1h_2^{-1}c^{-1}ch_3 = c(h_1h_2^{-1}h_3)\in C$$ |
H: diagonalizable matrix
$A$ is $n\times n$ matrix over $\mathbb C$.
must exist that:
$A^*A$ is diagonalizable over C
$AA^*$ is unitary matrix
if $A$ is not diagonalizable over $\mathbb C$ so $AA^*$ is not diagonalizable over $\mathbb C$
$i+1$ is not eigenvalue of $A$
I know the answer is 1+4 but I really dont understand why!
sorry for bad english
AI: It's true, as any Hermitian matrix is diagonalizable over $\Bbb C$.
It's not necessarily true, for example if $A=2I$, $AA^*=4I$ is not unitary.
$AA^*$ is always diagonalizable over $\Bbb C$ (as a Hermitian matrix), independently of the fact that $A$ is diagonalizable or not.
May happen ($A=I$ for example) or not (if $A=(1+i)I$). |
H: How do I calculated probabilities for cards?
I am trying to gain basic understanding on how to calculate probabilities. Below are a few examples of what I am trying to calculate. I would prefer (if possible) for formulas to be given on how to solve these using Microsoft Excel. Also, educating me on Probability Terminology involved would be helpful.
Situation 1:
I am going to draw "X" cards (5, for 5-card stun poker) from a "Y" card deck (52), what are the odds of getting a flush in hearts (Variable I have labeled as "Z" = 13... in my current ignorance, I refer to this as the "Targets" variable)?
My algebraic solution:
= (13/52) * (12/51) * (11/50) * (10/49) * (9/48)
Solution to chance of getting any flush would be the above times 4 to represent a flush in any suit.
The problem with the above solution is that it becomes impractical when the situational probabilities I am trying to calculate become more complicated.
QUESTION 1: What is a more statistical way of calculating the above? (using excel)
Situation 2:
I am going to draw "X=5" 5 cards from a "Y=10" 10 card deck. In this deck there are only two different draw possibilities, Card 1 or Card 2... There are 7 copies of Card 1 which are all identical and 3 copies of Card 2 which are all identical.
What is the probability of getting 0-3 copies of Card 2?
X = 5, Y = 10, Z = 3
QUESTION 2: What is the statistical way of calculating the above? (using excel)
Situation 3:
X = 7, Y = 60, Z = 20
QUESTION 3: How do I calculate, using excel, what the odds are of getting 0-7 of the target cards in a draw of 7?
AI: While computing the probabilities directly using algebra are possible, it's much easier to take an approach based on combinatorics.
We can do this using binomial coefficients, which are sometimes read as "Choose $k$ from $n$ objects," or "n Choose k."
Basically, for the chances of any flush of clubs, you need to compute the probability of choosing 5 out of 13 cards out of the 52 card deck. Probability is, of course, represented by a number $0 \le p \le 1$, so what we want to compute is the number of possible flushes of clubs, and divide it by the total number of hands.
The total number of hands can be written as $\begin{pmatrix} 52 \\ 5\end{pmatrix} = \frac{52!}{5!(52-5)!}$. The total number of club flushes (including straight and royal flushes) is $\begin{pmatrix} 13 \\ 5\end{pmatrix}$, or "choose any 5 from 13 total cards." So your probability is $$P(\mathrm{club flush}) = \begin{pmatrix}13 \\ 5 \end{pmatrix}/\begin{pmatrix}52 \\ 5\end{pmatrix}.$$
You can extend this arbitrarily. Say you want the probability of any flush. Well, obviously it's better, because there are four possible suits. So you want to multiply the previous result by the number given by "choosing one suit for the flush out of four possible suits," or $\begin{pmatrix}4 \\ 1\end{pmatrix}$, resulting in
$$P(\mathrm{any flush}) = \begin{pmatrix}4 \\ 1\end{pmatrix}\begin{pmatrix}13 \\ 5 \end{pmatrix}/\begin{pmatrix}52 \\ 5\end{pmatrix}.$$
That is the general approach to answer your first question. To answer your second question, we can do the same thing: You want to know the probability that if you choose 5 cards, you will get between 0 and 3 copies of a 2.
To answer questions 2 and 3, which can be handled quite similarly, we have 10 cards, and draw 5. So the total possible number of hands is $\begin{pmatrix}10 \\ 5\end{pmatrix}$. Right now, it doesn't matter that all of the Card 1s/Card 2s are the same -- we're going to formulate the problem in such a way that it doesn't matter.
To choose 0 Card 2s in 5 cards is the same thing as saying "choose 0 out of 3 card 2s, and choose 5 out of 7 card 1s." Using the combinatorical technique described above, that's $\begin{pmatrix}3 \\ 0\end{pmatrix}\begin{pmatrix}7 \\ 5\end{pmatrix}$. Moving on, choosing 1 Card 2 is like saying "choose 1 out of 3 Card 2s, and 4 out of 7 Card 1s," or $\begin{pmatrix}3 \\ 1\end{pmatrix}\begin{pmatrix}7 \\ 4\end{pmatrix}$. Continue this for each value 0,1,2,3, and add up the results. This gives you the total number of events; to get the total probability, divide that by the total number of possible events, $\begin{pmatrix}10 \\ 5\end{pmatrix}$.
(Note that in example 2, there are only three Card 2s, so you should have a 100% probability of picking between 0 and 3 (inclusive) Card 2s in the draw.)
The solution is exactly the same for Question 3.
To compute these numbers in Excel, $\begin{pmatrix}10 \\ 5\end{pmatrix}$ can be computed using
combin(10,5) |
H: How do I calculate the probability distribution of the percentage of a binary random variable?
I have an urn containing balls that are all either black or red. I'm interested in discovering the percentage of balls that are red. But I can only sample from the urn (without replacement), so the best I can do is calculate a probability distribution over possible percentages.
Obviously, if I've drawn no balls, I have no information, so the probability distribution is uniform from 0 to 1. But what is it once I start drawing balls?
AI: Theory: if $reds$ is the number of red balls you've seen and $blacks$ is the number of black balls you've seen, then the distribution is:
$Beta(reds+1, blacks+1)$
This starts out as $Beta(1,1)$ which is the uniform distribution we want. As we see blacks, it shifts toward zero; as we see reds, it shifts toward one.
Does anyone know if this is right? |
H: Technical question on integral ring extensions
Let $A$ be an integral domain, integrally closed in its field of quotients $K$ and
let $L$ be a finite Galois extension of $K$ with group $G$. Let $B$ be the integral closure of $A$ in $L$. Let $p$ be a maximal
ideal of $A$ and let $\beta$ be a maximal ideal of $B$ such that
$A \cap \beta=p$. Let $G_{\beta}$ be the subgroup of $G$ consisting of those automorphisms
$\sigma$ such that $\sigma \beta = \beta$. Let $L^{dec}$ be the fixed field of $G_{\beta}$ in $L$ and let $B^{dec}$ be the integral closure of $A$ in $L^{dec}$.
Let $\sigma, \tau \in G$ be such that $(\sigma \beta) \cap B^{dec} = (\tau \beta) \cap B^{dec}$. Does this then imply that $\sigma|_{L^{dec}} = \tau|_{L^{dec}}$?
AI: Let me suppose that $A$ is in fact a Dedekind domain, so that I don't have to think through various foundational questions.
Then $G$ acts transitively on the set of $\beta$ lying over $p$, and $G_{\beta}$ is the stabilizer of $\beta$. So we see that $\sigma \beta = \tau \beta$ if and only if $\sigma G_{\beta} = \tau G_{\beta}$, which holds if and only if $\sigma_{| L^{dec}} = \tau_{| L^{dec}}$.
So your question amounts to asking if $\sigma\beta \cap B^{dec} = \tau\beta\cap B^{dec}$ implies that $\sigma \beta = \tau\beta$. So you are asking if a prime above $p$ is determined by its restriction to $B^{dec}$, or equivalently, you are asking if for every prime $\beta'$ above $p$ in $B^{dec}$, there is a unique prime above $\beta'$ in $B$.
Continuing to translate, since the decomposition group at $\sigma\beta$ for the extension $L/L^{dec}$ is $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta}$, you are asking if $\sigma G_{\beta} \sigma^{-1} \cap G_{\beta} = G_{\beta}$ for each
$\sigma \in G$, or equivalently, if $G_{\beta}$ is normal in $G$.
So the answer to your question will be yes if $G_{\beta}$ is normal in $G$,
and no otherwise. |
H: Is an automorphism of a normal extension determined by its image of the maximal separable sub extension?
Let $L / K$ be a normal, algebraic field extension. Suppose that the maximal separable sub- extension $M/K$ is finite, $K \subseteq M \subseteq L$. By the primitive element theorem, $M=K(x)$ for some $x \in L$. Is it true that any automorphism of $L$ over $K$ is determined by its image on $x$? If yes, why?
AI: Let $L$ and $F$ be algebraic extensions of a field $K$ with $L_s$ and $F_s$ the $K$-separable parts. Then the restriction map
$\mathrm{Hom}_K(L,F)\rightarrow\mathrm{Hom}_K(L_s,F_s)$
is injective (these are $K$-algebra homomorphisms). Note that if $\sigma:L\rightarrow F$ is an element of the source, then for any $\alpha\in L_s$, $\sigma(\alpha)$ has the same minimal polynomial, and therefore is also separable over $K$, i.e., $\sigma(\alpha)\in F_s$.
The assertion is tautological if $K$ has characteristic zero, so I'll assume the characteristic is $p>0$. Suppose $\sigma\vert_{L_s}=\tau\vert_{F_s}$. Given $\alpha\in L$, there exists $n\geq 0$ such that $\alpha^{p^n}\in L_s$. Then
$\sigma(\alpha)^{p^n}=\sigma(\alpha^{p^n})=\tau(\alpha^{p^n})=\tau(\alpha)^{p^n}$,
so $\sigma(\alpha)=\tau(\alpha)$ because the map $\beta\mapsto\beta^{p^n}$ is injective on any field of characteristic $p$ (being a ring map).
In your setup, $L=F$, and $L_s=M=F_s$. Since a $K$-algebra endomorphism of $M$ is determined entirely by where $x$ goes (it can go to any other root of its minimal polynomial in $M$), the above shows that a $K$-algebra endomorphism of $L$ is determined by the image of $x$. |
H: Expectation and proofs on $(\Omega,\mathcal{B},P)$ involving moments and MGF
Ok, suppose we have a random variable, X, on $(\Omega,\mathcal{B},P)$ and $r>0$.
I am trying to prove the following 4 things:
1- If $E(|X|^r)<\infty$ then $E(|X|^s)<\infty \;\;\;\forall s\in(0,r]$
2- If $E(e^{t|X|})<\infty$ for some $t>0$, then $E(|X|^s)<\infty \;\;\;\forall s\in(0,\infty)$
side note for #2...is this just saying that if the moment-generating function is finite somewhere, then all of the moments are finite?
3- If $\sup_{x\geq 0}(x^{r+\delta}P(|X|>x))<\infty$ for some $\delta>0$ then $E(|X|^r)<\infty$
4- $E(|X|^r)<\infty\Rightarrow \lim_{x\to\infty}x^rP(|X|>x)=0$
Here is what I have for the first 2:
1: For $0<s<r, |x|^s\leq\max(|x|^r,1)<|x|^r+1$
$E(|X|^s)\leq E(|X|^r)+1<\infty$
2: This follows from the above and the relation:
$|x|^n\leq n!e^{t|x|}t^{-n}$ for $n\geq 1$
Could anyone please offer some detail into how to complete 3 and 4? For some reason limits and sup/inf often elude me. Also, if I have made any mistakes in the first 2, please note those as well. This is all being learned on my own, so the more detail and critique, the better for me!
Thanks.
AI: For 3., a consequence of Fubini's theorem is that
$$E|X|^r=\int_0^{+\infty}rt^{r-1}P(|X|\geq t)dt.$$
(the equality has to be understood as follows: the LHS is finite if and only if the RHS is finite, and if this is so, they are equal).
Call $M$ the supremum; then
$$E|X|^r\leq M\int_1^{+\infty}rt^{-\delta-1}dt+\int_0^1rt^{r-1}P(|X|\geq t)dt\leq M\int_1^{+\infty}rt^{-\delta-1}dt+r\int_0^1\frac 1{t^{1-r}}dt,$$
and the last integrals are convergent.
For 4., we have
$$x^rP(|X|\geq x)=x^rP(|X|^r\geq x^r)\leq \int_{\Omega}|X|^r\chi_{\{|X|\geq x\}}dP,$$
and use monotone convergence theorem. |
H: Complex variety with Zariski dense set of algebraic points
Let $V$ be an irreducible algebraic variety in $\mathbb{C}^n$ containing a Zariski dense set of points such that every coordinate is algebraic. Then is $V$ a product of one dimensional components?
AI: No. If you take any (say projective, irreducible ) variety $X$ defined over $\overline{\mathbb{Q}}$, then its $\overline{\mathbb{Q}}$-points -- i.e., points in which every coordinate in a suitable projective embedding is $\overline{\mathbb{Q}}$-rational -- are Zariski dense in its $\mathbb{C}$-points. You can see this e.g. by noting that the dimension of the closure in each case is the transcendence degree of the function field, and the transcendence degree of a field extension is unchanged by base extension. There are many other ways as well...
So it comes down to showing that there are varieties over $\overline{\mathbb{Q}}$ which are not products of one-dimensional varieties. The easiest such example seems to be the projective plane $\mathbb{P}^2$: the fact that $H^2(\mathbb{P}^2(\mathbb{C}),\mathbb{C}) = 1$ means, by the Kunneth formula, that it cannot be a product of curves. |
H: Calculating the average speed using only velocities
For the following question:
Pedro travels by bus to school at an average speed of $40$ km/hr. He is driven home by the same route by a friend's car at an average speed of $50$ km/hr. Which of the following is greatest:
(a) Average speed of both legs of the journey.
(b) $45$.
According to the book, the answer is (b). How did they calculate average speed here? I know that the formula for average speed is $A_v=\frac{Total~ Distance~ Covered}{Total~ Time ~Taken }$. Here we only know the velocity.
AI: Let $d$ be the distance from Pedro's house to school. The time it took him to travel to school was $d/50$, and the time it took him to return home was $d/40$. Thus the average speed is $$\frac{2d}{d/50+d/40}=\frac{400d}{9d}=44.\overline{4}<45$$ |
H: Taylor polynomial of $f(x) = 1/(1+\cos x)$
I'm trying to solve a problem from a previous exam. Unfortunately there is no solution for this problem.
So, the problem is:
Calculate the Taylor polynommial (degree $4$) in $x_0 = 0$ of the function:
$$f(x) = \frac{1}{1+\cos(x)}$$
What I tried so far:
calculate all $4$ derivatives
$1+\cos(x) = 2\cos^2(\frac{x}{2})$ and work with this formula
$\int\frac{1}{1+\cos(x)}dx = \tan(\frac{x}{2})$ and then use the Taylor series of $\tan(\frac{x}{2})$
$\frac{1}{1 + \cos(x)} = \frac{1}{1 + \left(1 + \frac{x^2}{2!} + \cdots\right)}$
What do you think, is there a good way to calculate the Taylor polynomial of this function or is there just the hard way (derivatves)?
AI: Let's do this in a couple of different ways. We want to expand $\dfrac{1}{1 + \cos x}$.
Method 1
If it's not obvious what the best method is, we might just choose one and go. A fourth degree Taylor Expansion isn't so bad - it's only a few derivatives. So you might just go, suffer through a few derivatives, and save yourself the trouble of deciding the best way.
Alternately, if you happen to know the series for $\tan x$, then that's a great way to proceed (referring to your idea of using the series expansion for $\tan (x/2)$
Method 2
If we are certain it has a Taylor expansion, and we are comfortable then we know it will look like $a_0 + a_1x + a_2x^2/2 + \ldots$ We know that $\cos x = 1 - x^2/2 + x^4/4! + \ldots$, so that $\dfrac{1}{1 + \cos x} = \dfrac{1}{2 - x^2/2 + x^4/4! + \dots}$
So we consider $\dfrac{1}{2 - x^2/2 + x^4/4! + \dots} = a_0 + a_1x + a_2x^2/2 + \ldots$, or equivalently $$(a_0 + a_1x + a_2x^2/2 + \ldots)(2 - x^2/2 + x^4/4! + \dots) = 1$$
By equating the coefficients of powers of $x$ on the left and on the right (which are all $0$ except for $x^0$, which has coefficient $1$), we get that $2a_0 = 1$, $a_1 = 0$, $a_0(-x^2/2) + (a_2x^2/2)(2) = 0$, etc. This isn't too bad, and is just a set of linear equations. |
H: Why does Strassen's algorithm work for $2\times 2$ matrices only when the number of multiplications is $7$?
I have been reading Introduction to Algorithms by Cormen. Before explaining Strassen algorithm the book says this:
Strassen’s algorithm is not at all obvious. (This might be the biggest understatement in this book.)
The book just states the algorithm but don't explain why it works. So if we take the case of multiplication of two $2 \times 2$ matrices, it reduces the number of multiplications from $8$ to $7$ thereby reducing the complexity from $n^{3}$ to $n^{\lg 7}$ which is roughly $n^{2.8}$.
My question is why $7$ and say not $5$ or even lesser number? I mean they could have made the recursion tree even less "bushier".
AI: Allow me to shamelessly copy an answer of mine to this question first.
The idea behind Strassen multiplication is the same as the idea behind Karatsuba multiplication.
In this, if we have two numbers and a base $B$ (maybe $10^6$, say), and the numbers are written $a = x_0B + x_1$, $b = y_0B + y_1$, and we want to calculate $ab$, then naively we might say that it's
$ab = x_0 y_0 B^2 + (x_0 y_1 + x_1 y_0)B + x_1 y_1 \qquad$ (requiring 4 multiplications)
But we can be wittier, as $x_0y_1 + x_1y_0 = (x_0 + x_1)(y_0 + y_1) - x_0y_0 - x_1y_1 $
So we can calculate $ab$ by knowing $x_0y_0, x_1y_1$ and $(x_0 + x_1)(y_0 + y_1)$, which only uses 3 multiplications. This is just a slick idea, akin to the many other slick arithmetic tricks out there that might be found by inspection, and it became popular in the 1960s. (There's a good story behind it - Karatsuba responded to a challenge of Kolmogorov, ultimately culminating with Kolmogorov writing a paper in Karatsuba's name).
For matrices, a direct port of Karatsuba doesn't quite work. But it's easy to verify that Strassen is true (just write it out). But it's very unclear how Strassen came across them. But it should be mentioned that Strassen was working on many numerical-methods style techniques to improve matrix calculations. It should also be mentioned that Strassen has an improved algorithm, better than this technique, using fast fourier transforms as well.
One might ask: well, we can do it with 7 multiplications. Can we do it with 6? Strassen multiplication has been around since 1969, so surely it's known? This was an open problem until just a few years ago, when Landsberg showed that 7 is optimal. It's nontrivial and a bit far afield for me, but his (corrected after it was published) article can be found on the arXiv.
One might also wonder about $3 \times 3$ matrices. How few multiplications are necessary? Naively, it takes 27 multiplications. But it can be done in 23. (See Julian D. Laderman, A noncommutative algorithm for multiplying 3×3 matrices using 23 muliplications, Bull. Amer. Math. Soc. 82 (1976) 126–128, MR0395320 (52 #16117).) Is that optimal? That's unknown - all that's known is that it takes at least 19. (But if you're just interested in computational savings, even 19 gives less savings when iterated than the $2 \times 2$ case).
So in short, $7$ is optimal. It's not obvious that it's optimal. It was a very clever idea, even, to find the $7$. |
H: Solving for unknown, trouble with $\ln$ and $\exp$
Having some trouble understanding $\ln$ and $\exp$ rules and what to do in this situation. Perhaps it has just been a very long day...
$$\hat{Y} = \exp \left[\left(\hat{\beta_0} + \sum_i \hat{\beta_i}{x_i}\right)\space A' \right] $$
Solving for $A'$.
AI: Hint:
$\ln$ and $\exp$ are inverse of each others. So if
$$ y = \exp(x) $$
this means
$$ \ln(y) = \color{red}{\ln(\exp}(x)) = x. $$
In your case take $\ln$ of both sides. Can you take it from here? |
H: Characterization of uniform continuity via sequence
For $f(x)$, defined on the interval $X$, $f(x)$ is uniformly continuous in X if and only if for every sequences $x_{n}$, $y_{n}\in X$, when we have $\lim_{n\rightarrow\infty}(x_{n}-y_{n})=0$, then $\lim_{n\rightarrow\infty}[f(x_{n})-f(y_{n})]=0$.
For this characterization, I am confused with three point.
It is shown on my textbook that, for $f(x)=x^2$, defined in $[0,+\infty)$, $x_{n}=\sqrt {n+1}$, $y_{n}=\sqrt{n}$ were chosen. I wonder why not to choose some apparent sequence such as $x_{n}=1/n$, $y_{n}=1/2n$.
In my opinion, I think that $1/n$ and $1/2n$ are not defined on zero point.
For function $f(x)=1/x$, defined in $(\xi,1)$, $0\lt\xi\lt1$), I wonder why it's unwarranted to use $x_{n}=1/n$, $y_{n}=1/2n$.
As far as I am concerned, we need to confirm $\lim_{n\rightarrow\infty}[f(x_{n})-f(y_{n})]$ and if $x_{n}$ is defined in $(\xi,1)$ then $n$ should be in $(1,1/\xi)$. Finally, $n$ can't tend to $\infty$.
I noticed that this characterization mention that for all $x_{n}$, $y_{n}$. What if we want to prove uniform continuity for certain function? How can we choose all suited sequences?
thanks.
AI: For 1., your textbook is presumably showing that $f$ is not uniformly continuous on $[0,\infty)$. To prove this, they would show that the property in your characterization of uniform continuity does not hold. Towards this end, sequences $(x_n)$ and $(y_n)$ have to be found such that $x_n-y_n$ tends to $0$ but $f(x_n)-f(y_n)$ does not tend to $0$. The sequences you wish to use do satisfy $f(x_n)-f(y_n)\rightarrow0$ and would not serve to demonstrate that $f$ is not uniformly continuous. The sequences in the text, however, do.
For 2., it's unwarranted to use your sequences simply because not all of its terms are in the given interval $(\zeta,1)$ (for $n$ sufficiently large, $x_n=1/n<\zeta$; so, for such $n$, $x_n\notin(\zeta,1)$). Further, if, as you seem to suggest, you just look at the $x_n$ and $y_n$ that are in $(\zeta,1)$, you no longer have $x_n-y_n\rightarrow0$. In fact, you would no longer have infinite sequences, as needed.
In fact, the function $f(x)=1/x$ is uniformly continuous on $(\zeta,1)$, $1>\zeta>0$. To prove this (and to address your third question), you have to show that whenever $(x_n)$ and $(y_n)$ are sequences in $(\zeta,1)$ with $x_n-y_n\rightarrow0$, we necessarily have $f(x_n)-f(y_n)\rightarrow 0$. You "choose all suited sequences" by fixing (arbitray) sequences $(x_n)$ and $(y_n)$ in $(\zeta,1)$ with $x_n-y_n\rightarrow0$ and then demonstrating that you must have $f(x_n)-f(y_n)\rightarrow0$. So, let $(x_n)$ and $(y_n)$ have this property. Now you have to somehow prove that $f(x_n)-f(y_n)\rightarrow0$. If you can do this, you will have shown what $f$ is indeed uniformly continuous on $(\zeta,1)$.
I presume your text does show that the function $f$ of the preceding paragraph is uniformly continuous on the given interval. If not, a hint here is that the quantity $1/(x_ny_n)$ is bounded away from $0$ on $(\zeta,1)$ and ${1\over x_n}-{1\over y_n}={y_n-x_n\over x_n y_n}$. From this it will follow that $f(x_n)-f(y_n)$ tends to $0$, since $x_n-y_n$ does. |
H: Unsolved arithmetic exercise for printing numbers in book
Here is the exercise from the "Arithmetic" book (ISBN: 5-02-013764-2).
There are 1989 numbers had printed to enumerate book's pages. How many
pages in book?
Honestly I decided to refresh my arithmetic knowledge at all but stuck with that problem.
Here's my decision:
there are 9 numbers to print for pages 1-9 (9 * 1);
there are 200 numbers to print for pages 10-99 (100 * 2);
1989 - 209 = 1780;
1780 /3 = 593,3(3) - that is the problem
I clearly understand how simple it must be and maybe something is wrong with my brain. But please, tell me where I made mistake.
AI: There are not 100 numbers between 10 and 99. There are 10 between 10 and 19, 20 between 10 and 29, ..., 90 between 10 and 99. Each of those numbers has two digits. $2 \times 90 = 180$.
The rest of you're reasoning is perfectly sound. You just over-zealously counted the total number of two-digit numbers. |
H: Closed set on Euclidean space that is not compact
I have read that a subset of Euclidean space may be called compact if it is both closed and bounded. I was wondering what a good example of a closed but unbounded set would be?
Would a closed ball inside a sphere with an infinite radius do the trick? If that example works are there any other examples people could think of?
AI: Being closed means nothing but being the complement of an open set. So take any bounded open subset $S \subset \mathbb R^n$, then $\mathbb R^n \setminus S$ is closed but not bounded. What you are looking for.
I.e:, Any complement of any open ball! |
H: Question about Lie superalgebra.
What are the generators and relations for the Lie superalgebra $\mathfrak{psu}(2, 2 | 4)$? Thank you very much.
AI: The article arXiv:0505234, "Non-linear Realization of $\operatorname{PSU}(2,2|4)$ on the Light-Cone" by Pierre Ramond et al. has details and further references.
The algebra has 30 bosonic generators forming direct product of conformal algebra $\mathfrak{su}(2,2) \sim \mathfrak{so}(4,2)$ (15 generators) and internal $\mathfrak{su}(4) \sim \mathfrak{so}(6)$ (also 15 generators), as well as 32 fermionic generators.
Fermionic generators form a representation of $\mathfrak{su}(2,2) \otimes \mathfrak{su}(4)$ under adjoint action of bosonic sub-algebra. |
H: How can I construct a $2^{63}$-gon with a straightedge and compass?
I entered $2^{63}$ as a stand alone value at WolframAlpha. Among
the responses was a factoid that 'A regular 9223372036854775808-gon is constructible with a straightedge and compass.'
What is such a shape and how can I construct one?
AI: Start with a circle and a diameter. Bisect the diameter and extend the bisector to the circle to make a square. Bisect each right angle at the center of the circle $61$ times, extending the bisector to the circle. You have $2^{63}$ points, equally spaced, around the circle. As Dario says, you won't be able to tell it from the original circle.
The regular polygons constructible with straightedge and compass have the number of sides of the form $2^n$ times a product of zero or more of the Fermat primes: $3, 5, 17, 257, 65537$ (each to only the first power) |
H: What is the difference between following approaches to line integrals?
What is the difference between following approaches concerning line integrals:
first approach is for complex function with parametrization $\gamma(t) = \cos t+i\sin t$ (Line_integral :: Example from Wikipedia).
Second approach is for $f(x,y)$ with parametrization $x=\cos t, y=\sin t$ (Line_integral :: Example from khanacademy.org).
Yes complex function $f(z)=1/z$ is $f: C \rightarrow C$ while the other is $f:R^2 \rightarrow R$, but what other differences there are or are they same thing said in different method?
AI: Let me add one more kind of line integration to make the difference clearer:
Line integral of a scalar function. This is your second approach. There is not much going on here. An intuitive description of the integral would be to sum the function values encountered along the curve.
Line integral of a vector function. Take a vector function with components $(f(x,y),g(x,y))$. Then the integral is defined by
$$
\int \left( f(x(t),y(t))x'(t)+g(x(t),y(t))y'(t) \right) {\mathrm{d}}t.
$$
A rough description would be to sum the tangential component of the vector field $(f,g)$ along the curve.
Line integral of a complex function. Take a complex function $f(x,y)+ig(x,y)$. Then the integral is
$$
\int \left( f(x(t),y(t))x'(t)-g(x(t),y(t))y'(t) \right) {\mathrm{d}}t + i\int \left( f(x(t),y(t))y'(t)+g(x(t),y(t))x'(t) \right) {\mathrm{d}}t.
$$
As you see, this is somewhat similar to the vector integral, in the sense that the integral depends on the interaction between the components of the function and the curve in a nontrivial way. However, this is "more scalar" than the vector integral because of the complex multiplication, in the same way complex numbers are "more scalar" than just two dimensional vectors. |
H: Distributing an item equally.
For the following question
A $10$ foot plank of wood is cut to give three equal lengths with a shorter length left over. Which is more
a)The length of one of equal pieces
b) $3$ feet
Now here is how I am solving it --> Each piece gets =$\frac{10}{3}$ and remaining is $1$ foot.
So I think both are equal. However the text states that "Not enough information is given to solve the problem" Why is that.
AI: You have $10=3x+y$, where $x$ is the number of feet of each of the lengths cut off and $y$ is the length left over. And you have $y<x$, and so $10 = 3x+y < 4x$. This is all the information you have. So why does this mean that you can't infer the answer? |
H: How do I show that $f(z) \equiv \sum_{n \in \mathbb{Z}} \frac{1}{(z-n)^2}$ is a meromorphic function?
Let
$$f(z)=\sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}.$$
Show $f$ is meromorphic on $\mathbb{C}$ with double poles at each integer.
I think I got it to be meromorphic. I fixed an integer $m$ and considered the series
$$\sum_{n=m+1}^\infty \frac{1}{(z-n)^2}\qquad\text{and}\qquad\sum_{n=-\infty}^{-m+1} \frac{1}{(z-n)^2}.$$
I showed these were analytic which implies the whole series is meromorphic. How do I show $f$ has poles at each integer?
AI: To show your series defines a meromorphic function you can do the following:
Show that the series converges uniformly in every compact subset of $\mathbb C\setminus\mathbb Z$. This implies that the series defines on that set a holomorphic function.
Now let us show that it is meromorphic with poles at the integers. To do this, it is enough to show that for each $n\in\mathbb Z$ we can write $f(z)=g(z)+\frac{1}{(z-n)^2}$ with $g(z)$ a function which is holomorphic in a neighborhood of $n$. Can you do this?
Let us do the first one.
Let $N\in\mathbb N$ and let $\Omega_N=\{z\in\mathbb C\setminus\mathbb Z:|z|<N\}$. If $z\in\Omega_N$ and $n\in\mathbb Z$ we have $|z-n|\geq |n|-|z|\geq |n|-N$, so that the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}$ is majorated, term by term, by the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(n-N)^2}$, which converges. Weierstrass' criterion tells us then that the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}$ in fact converges absolutely and uniformly on $\Omega_N$. It follows that the last series defines an holomorphic function on $\Omega_N$, and then so does $$\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}+\sum\limits_{\substack{n\in\mathbb Z\\|n|\leq N}}\frac{1}{(z-n)^2}=\sum\limits_{n\in\mathbb Z}\frac{1}{(z-n)^2}.$$
Doing this for all $N\in\mathbb Z$ shows that the series, in fact, defines a function which is holomorphic in $\mathbb C\setminus\mathbb Z$. |
H: Riemann Sum question
limit as $n$ goes to infinity of $$\sum_{k=1}^n \left(\frac k {n^2}-\frac{k^2}{n^3}\right)$$ I know I need to make this an integral but I cannot figure out how to acquire the limits of integration. Any help would be great.
AI: We actually don't need to perform an integral here, we can simply take out constant factors:
$$\lim_{n\to\infty}\left(\sum_{k=1}^{n}\left[\frac{k}{n^{2}}-\frac{k^{2}}{n^{3}}\right]\right)=\lim_{n\to\infty}{\left(\frac{1}{n^{2}}\sum_{k=1}^{n}{k}-\frac{1}{n^{3}}\sum_{k=1}^{n}{k^{2}}\right)}$$
Using our known formulae, we have $\sum_{k=1}^{n}{k}=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n}{k^{2}}=\frac{n(n+1)(2n+1)}{6}$, so we can rewrite our above expression as:
$$\lim_{n\to\infty}{\left(\frac{n^{2}+n}{2n^{2}}-\frac{n+3n^{2}+2n^{3}}{6n^{3}}\right)}=\lim_{n\to\infty}{\left(\frac{1}{6}-\frac{1}{6n^{2}}\right)}=\frac{1}{6}$$
So we managed to do this particular case without the Riemann Integral. Although I'm sure the other answers will help you to do it using the integral if you are required to do so for homework.
Hope this gives a slightly different and helpful angle on the summation. |
H: What does $\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}$ equal?
Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?
I encountered the following sum in Boros & Moll's "Irresistible Integrals" question 5.2.11 (pg. 78):
$$\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}$$
How do I evaluate this sum to get a closed form?
AI: If $H_x=1+\frac{1}{2}+\cdots+\frac{1}{x}$ (i.e. the $x^{th}$ Harmonic Number), we see
$$
\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}=
\lim_{n \to \infty} \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}=
\lim_{n \to \infty}\left(1+\frac{1}{2}+\cdots + \frac{1}{2n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n-1}\right)=
\lim_{n \to \infty} H_{2n}-H_{n-1}
$$
You may recall that
$$\log x +\gamma+O(1)= H_x \quad (\text{as } x\to\infty)$$
Now, because $m \ge n \to \infty$ we may use $\log x$ instead of $H_x$. Then we have
$$
\lim_{n \to \infty} \sum_{m=n}^{2n} \frac{1}{m}=
\lim_{n \to \infty} H_{2n}-H_{n-1}=
\lim_{n \to \infty} (\log (2n)+\gamma)-(\log (n-1)+\gamma)=
\lim_{n \to \infty} \log (2)+\log (n)-\log (n-1)=
\log (2)
$$
A quick numerical check seems to confirm this: If $n=1000$, the sum approximately equals $0.6938972430$ - three correct decimal places of $\log 2$. |
H: Example that $u\in W^{1,2}$, but $u \notin W^{1,3}$
I'm doing the calculations about the following assertion
Let $\Omega$ be $\{(x,y):0<y<x^2, 0<x<1\}$. The function $u(x,y)=\log (x^2+y^2)$ belongs to $W^{1,2}(\Omega)$, which you can check by integrating $|\nabla u|^2\approx 1/x^2$ within $\Omega$. We have $\Delta u=0$, which is the nicest equation one could ask for. However, $u$ does not belong to $W^{1,3}(\Omega)$, which you also can check by integration.
My efforts:
$| \nabla u |^2 = 4/(x^2+y^2)$ and
\begin{equation}
\int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{x^2+y^2}dy dx = \int_{0}^{1} \dfrac{\arctan}{x}dx < \infty
\end{equation}
because $\lim_{x\rightarrow 0}\dfrac{\arctan}{x} = 1$ and $\dfrac{\arctan}{x}$ is bounded in(0,1). Hence $\nabla u \in L^{2}(\Omega)$.
Am I right here?
I don't know how $\int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{(x^2+y^2)^{^3/2}}dy dx$ diverges.
Thank you.
3.
\begin{eqnarray}
\int_{0}^{1}\int_{0}^{x^2} \ln(x^2+y^2)dydx &=& \int_{0}^{1}\int_{0}^{x^4} u\ln udu dx\\
& = &\int_{0}^{1} \left \{\dfrac{1}{2} u^2 \ln + \dfrac{1}{4}u^2 \right \}_{0}^{x^4}dx\\
&=& \int_{0}^{1} (x^8 \ln(x^4) -\dfrac{1}{4}x^8) dx.
\end{eqnarray}
In the second equality used that $\lim_{x\rightarrow 0}x^2 \ln x = 0$. Moreover, we obatain that $x^2 \ln(x^2)$ is finite in $(0,\infty)$ and the integral above is finite.
Am I right here too?
AI: It's correct. Maybe you could justify why is $u$ is $L^2$.
Let $I(x):=\int_0^{x^2}\frac{dy}{(x^2+y^2)^{3/2}}$. Put $tx=y$, ($xdt=dy$) to get
$$I(x)=\int_0^x\frac{x dt}{(x^2+x^2t^2)^{3/2}}=\frac x{x^3}\int_0^x\frac{dt}{(1+t^2)^{3/2}}\geq \frac 1{x^2}\frac 1{2^{3/2}}x=\frac 1{2^{3/2}}\frac 1x,$$
and the integral over $(0,1)$ of the last term is infinite. |
H: OEIS A000255 recursion.
I encountered the sequence A000255. $a(n)$ counts permutations of $[1,...,n+1]$ having no substring $[k,k+1]$
I am finding difficulty in proving it.
Can you please give any clues or hints on how to attack the problem?
AI: The recurrence is $a(n)=na(n-1)+(n-1)a(n-2)$ The first term is the number of ways to insert $n+1$ into a proper permutation of $[1,\ldots,n]$-you can put it anywhere except after $n$. The second counts the ways to have a permutation of $[1,\ldots,n]$ with exactly one pair $[k,k+1]$, which you then break up by putting $n+1$ between them. You choose which number will be $k$ (to be followed by $k+1$), but it can't be $n$. Then link $k$ and $k+1$ as a pair and make a proper permutation (of $n-1$ items). |
H: Is every monomial over the UNIT OPEN BALL bounded by its L^{2} norm?
Let $m\geq 2$ and $B^{m}\subset \mathbb{R}^{m}$ be the unit OPEN ball . For any fixed multi-index $\alpha\in\mathbb{N}^{m}$ with $|\alpha|=n$ large and $x\in B^{m}$
$$|x^{\alpha}|^{2}\leq \int_{B^{m}}|y^{\alpha}|^{2}dy\,??$$
AI: No. For a counterexample, take $\alpha=(n,0,\ldots,0)$. Obviously, $\max_{S^m}|x^\alpha|=1$, but an easy calculation shows
$$
\int_{S^m}|y^\alpha|^2{\mathrm{d}}\sigma(y) \to 0,
$$
as $n\to\infty$.
For the updated question, that involves the open unit ball, the answer is the same. With the same counterexample, we have
$$
\int_{B^m}|y^\alpha|^2{\mathrm{d}}y \to 0,
$$
as $n\to\infty$. |
H: Relationship between real inverses of analytic functions
Take some analytic function, $f(x)$, that goes from $-\infty$ to $\infty$, with a finite number of points such that $\frac{df}{dx}=0$. You can divide the y axis into intervals, where the boundary between each interval is the y value at a critical points (see graph).
Within each interval, there are an odd number of real inverse functions, $g_n(y)$. Number them in ascending order. Then, for every function I've tried, $\sum_n (-1)^n g_n(y)=ag_m(y)+b$ for some a,b and m that is constant across the interval. Does anyone know why this would be the case?
AI: It's not true in general. What is true is this. If $f(x) = a_n x^n + \ldots + a_0$ is a polynomial with $n$ odd, then in any interval where there are $n$ real inverse functions, the sum of those is constant, namely $-a_{n-1}/a_n$. So if you happened to take $n=3$, you would get $-g_1(y) + g_2(y) - g_3(y) = a_2/a_3 + 2 g_2(y)$. |
H: Number of bit strings with 3 consecutive zeros or 4 consecutive 1s
I am trying to count the number of bit-strings of length 8 with 3 consecutive zeros or 4 consecutive ones. I was able to calculate it, but I am overcounting. The correct answer is $147$, I got $148$.
I calculated it as follows:
Number of strings with 3 consecutive zeros = $2^5+5\times2^4 = 112$, because the 3 zeros can start at bit number 1, 2, 3, .., 6
Number of strings with 4 consecutive ones = $2^4+4\times2^3 = 48$, I used the same reasoning.
Now I am trying to count the number of bit-strings that contain both 3 consecutive zeros and 4 consecutive 1s. I reasoned as follows:
the strings can be of the following forms: 0001111x, 000x1111, x0001111..thus there are $2+2+2 = 6$ possibilities for bit-strings where the 3 consecutive zeros come first. Symmetrically there are $6$ bit-strings where the 4 consecutive ones come first.
Thus the answer should be = $112+48-12 = 148$.
clearly there's something wrong with my reasoning, if someone could point it out, that would be awesome. Thanks
AI: You've left out accounting for strings that have two triple zeroes. So $00010000,00010001,00001000,00011000,10001000$ were added to your total twice. This didn't cause any problems in your count of strings with four $1$s, however, since we can't put four $1$s in two separated places in an $8$-bit string. So the union now has $155$ elements, and cutting out the two duplicates from each symmetry of your intersection calculation turns that to $8$, for a total $107+48-8=147$. |
H: is division by zero automatically irrational?
I know that division by zero is undefined and is also not rational, but I am not sure whether this means it's irrational because it is undefined.
Can anyone clarify?
AI: An irrational number is a real number that is not rational. Dividing by zero doesn't give you a number at all. Calling irrationals "not rationals" oversimplifies what's going on. Is $i$ rational, for example? Cardinal numbers? You have to be very careful with how you apply definitions.
Irrationals aren't merely "not rational", they're real numbers that aren't rational. Since $\frac{1}{0}$ doesn't evaluate to a real number (or any kind of number at all, if you're working in $\mathbb{R}$), it's neither rational nor irrational. It's non-existent.
See this interfaith description for more information. There are things that aren't in the rationals or irrationals (all reals are one or the other though, so that diagram is somewhat misleading). |
H: Can't follow a proof involving Prime-Power Fields
Theorem: Let $p$ be a prime and let $n\in\mathbb{Z}^{+}$. If $E$ and $E'$ are fields of order $p^{n}$, then $E\cong E'$.
Proof: Both $E$ And $E'$ have $\mathbb{Z}_{p}$ as prime fields (up to isomorphism). By Corollary 33.6, $E$ is a simple extension of $\mathbb{Z}_{p}$ of degree $n$, so there exists an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Z}_{p}[x]$ such that $E\cong \mathbb{Z}_{p}[x] / \langle f(x)\rangle$.
I'm hoping someone can get me unstuck! Thanks!
Corollary 33.6 that is referenced states: A finite extension $E$ of a finite field $F$ is a simple extension of $F$.
Edit: The original question below has been answered by DonAntonio.
I understand how Corollary 33.6 is being applied, but I don't see where the polynomial comes from. This looks like the construction of $\mathbb{Z}_{p}(\alpha)$, if $\alpha$ were a root of the polynomial $f(x)$, so I'm guessing that is related. But what if $E = \mathbb{Z}_{p}(\alpha)$ for some transcendental $\alpha$ over $\mathbb{Z}_{p}$? This isn't outlawed by the Corollary.
AI: It can't be $\,E=\Bbb F_p(\alpha)=\Bbb Z_p(\alpha)\,\,,\,\,\alpha$ transcendental over $\,\Bbb F_p\,$ , as then $\,E\,$ wouldn't be a finite extension of $\,\Bbb F_p\,$ and thus not even a finite field, of course. |
H: How to prove that the language of a DFA is some $L$
Consider the following DFA:
It is quite clear that the language of this FDA is all the words that don't have the word $aa$ as a subword.
My question is: How can I formally prove that this is the language of this FDA ?
My efforts: I tried to determine $L(q_0)$ and $L(q_1)$ (that we denote as $L_0$ and $L_1$ accordingly) and prove that these are indeed what I determined using induction (this is the type of method used in the book I am studing from), I had some problems determining $L(q_0)$ and $L(q_1)$ and I am not sure if I should show equality 'straight out' or should I do two proofs showing $L(q_i)\subset L_i$ and $L_i\subset L(q_i)$.
Help is very much appreciated!
AI: Show by induction on the length of $w$ that the state of the machine after reading $w$ is
$q_2$ if and only if $w$ does contain $\mathtt{aa}$
$q_1$ if and only if $w$ does not contain $\mathtt{aa}$, but ends with an $\mathtt{a}$.
$q_0$ if and only if $w$ does not contain $\mathtt{aa}$ and does not end with an $\mathtt{a}$.
The base case is trivial -- the empty word neither contains $\mathbb{aa}$ nor ends with an $\mathtt{a}$, which matches the initial state $q_0$.
In the induction case we either have $w=v\mathtt{a}$ or $w=v\mathtt{b}$, and in each of these cases simply consider the three subcases that tell what happens with $v$.
For example, in the subcase where $w=v\mathtt{a}$ and the machine ends in $q_1$ after reading $v$, the induction hypothesis says that $v$ must end with an $\mathtt{a}$. But then $v\mathtt{a}$ ends with two $\mathtt{a}$s, so $w=v\mathtt{a}$ certainly contains $\mathtt{aa}$. And that matches the fact that reading $\mathtt{a}$ in state $q_1$ lands us in $q_2$. |
H: Perfect set in $\mathbb{R}$ which contains no rational number
Possible Duplicate:
Perfect set without rationals
Does there exist a nonempty perfect set in $\mathbb{R}$ which contains no rational number?
This problem is on p.44 PMA - Rudin
I found a proof of this on google but the proof is not 'suitable' for me, since the proof uses the concept of 'measure', which is in chapter 11, while this problem is on chapter 2.
Can one show this by direct construction?
AI: Since the rational are countable, let $a_n$ enumerate all the rationals. Let $B_{2^{-n}}(a_n)$ be the open interval centered at $a_n$ of radius $2^{-n}$. Let $A = \bigcup_{n \in \mathbb{N}} B_{2^{-n}}(a_n)$. $A$ is open since it is it is a union of open sets. Note that $\mathbb{Q} \subset A$. Clearly $A$ is not all of $\mathbb{R}$ because "length" of $A$ is less than or equal to $2\sum_{n = 1}^\infty 2^{-n} < \infty$. (This is essentially the measure idea.) Because of this and fact that $\mathbb{R}$ is uncountable, you have that $C:= \mathbb{R} - A$ is an uncountable set. $C$ closed since it is the complement of an open set. Also $C \cap \mathbb{Q} = \emptyset$ since $\mathbb{Q} \subset A$.
Finally, by the Cantor Bendixson Theorem (Exercise 28 on page 45) which states that every uncountable closed set is (uniquely) the union of a perfec set and a countable. Hence there exists a perfect set $C'$ and a countable set $F$ such that $C = C' \cup F$. Since $C \cap \mathbb{Q} = \emptyset$, you also have $C' \cap \mathbb{Q} = \emptyset$. Hence $C'$ is a nonempty perfect set that does not contain any rational numbers.
By the way $C'$ can be constructed, look at the proof of the Cantor Bendixson Theorem to see exactly how it is made. |
H: Question about algebraic field extensions
If I have a subfield $F$ of a field $E$, and an algebraic (over $F$) $\alpha\in E$, I can form $F(\alpha)$ which is isomorphic to $F[x]/\langle f(x)\rangle$ for $f(x) = irr(\alpha, F)$. That is, $f(x)$ is the minimal degree and monic element of $F[x]$ such that $f(\alpha) = 0$.
The book I'm using defines $F(\alpha)$ officially as the image of $F[x]$ under $\phi_{\alpha}$, the map $f(x)\mapsto f(\alpha)$, and the isomorphism mentioned above comes from the fact that its kernel is $\langle f(x)\rangle$.
My question is: if there are other roots of the polynomial $f(x)$ in $E$, then are they necessarily contained in $F(\alpha)$? My intuitive guess is yes, since the field $F[x]/\langle f(x)\rangle$ doesn't know the difference between distinct roots of $f(x)$. But if this is correct then why?
AI: No. Take $F = \mathbb{Q}, E = \mathbb{C}$ and $\alpha = \sqrt[3]{2}$. Then $F(\alpha)$ lies in $\mathbb{R}$ but $f(x) = x^3 - 2$ has two complex roots which do not lie in $\mathbb{R}$, hence cannot lie in $F(\alpha)$.
The fact that this can be false motivates the definition of a normal extension. |
H: Is There a Continuous Analogue of the Hypergeometric Distribution?
As the title states, is there a continuous analogue of a Hypergeometric distribution?
If $ X \sim H(m,n,N)$ is a common Hypergeometric distribution, where $N$ is the population size, $n$ is the number of draws, and $m$ is the number of success. In my case $N,n,m$ are not integer, therefore I could have a not integer number of success.
AI: Short version, yes.
Long version, it's complicated. There is an interesting link between probability distributions and orthogonal polynomials. For instance, consider the Hermite polynomials, $H_0, H_1, \ldots$. These are orthogonal with respect to the weight function $e^{-x^2/2}$ on the support $(-\infty,\infty)$. In other words, $\int_{-\infty}^{\infty}H_i(x)H_j(x)e^{-x^2/2}dx = \sqrt{2\pi}i!\delta_{ij}$.
The weighting function and the coefficient having a factor $\sqrt{2\pi}$ should ring a bell; specifically, you find these represented in a standard normal distribution.
Norbert Wiener recognized that the two are intricately linked and developed what is called the Gauss-Hermite Polynomial Chaos. In short, we can write a random variable $k$ of any distribution as an infinite series about a random variable $\zeta$ of a normal (Gaussian) distribution by using the Hermite polynomials as basis functions: $$ k = \sum_{i=0}^{\infty}k_i H_i(\zeta).$$
It turns out, you can generalize this to other distributions, and to discrete distributions.
There is something called the Askey Scheme, which is essentially a family tree relating hypergeometric orthogonal polynomials. Many of the weighting functions for these polynomials are distribution functions for probability distributions, meaning we can perform a Wiener-Askey Polynomial Chaos expansion for any random variable about another random variable of almost any distribution of our choosing. See also: http://www.dam.brown.edu/scicomp/media/report_files/BrownSC-2003-07.pdf
The hypergeometric family linked to the discrete hypergeometric distribution is the Hahn family of polynomials. The continuous analogue is uncreatively called the "Continuous Hahn" family. This leads to the following answer:
The weighting function of the Continuous Hahn polynomials will give you the continuous analogue of the discrete hypergeometric distribution. In fact, it is most likely identical to within a scaling parameter. These functions are quite complicated, so it is not intuitive.
The polynomial chaos distribution is quite interesting and powerful. If you look at the series expansion, it is analogous to a Taylor expansion of a random variable. Importantly, it allows you to write a random variable in a possibly unknown or complicated distribution in terms of any distribution of your choosing; as with any series expansion, the challenge is to then compute the deterministic coefficients. This becomes quite potent when you have, say, a dynamical system parameterized about a random parameter -- instead of using complicated stochastic models, or time-consuming Monte-Carlo analysis, you can frame the problem as one of computing a handful of deterministic coefficients, and generate statistical moments and/or Monte Carlo samples from them easily.
Interestingly, the first coefficient, $k_0$ always represents the distribution mean. |
H: Determining limits on variable change
Ok, I've seen some questions similar to mine but it didn't really get me what I want so I figured I'd ask. I was given the following problem to solve by making the change of variables $ u = x-y, v = x+y$ in the following integral
$ I = \int_0^1dy\int_0^{1-y}e^\frac{x-y}{x+y}dx$
This is an iterated integral over an region of the x-y plane. The Jacobian of the transformation is 1/2 so the integral becomes
$I = \frac{1}{2}\int\int_Ae^\frac{u}{v}dudv $
At this point I had some difficulties. I eventually realized that if I played with the numbers, that u would have upper and lower bounds of 1 and -1 while v would have upper and lower bounds of 1 and 0. So my first attempt was this:
$ I = \frac{1}{2} \int_0^1dv\int_{-1}^1e^\frac{u}{v}du$
however evaluating this integral was practically impossible (if you don't believe me, try finding an antiderivitive for $xsinh(1/x)$ with elementary techniques!), eventually I realized the integral would simplify if changed the limits as follows:
$ I = \frac{1}{2} \int_0^1dv\int_{-v}^ve^\frac{u}{v}du$
this yielded the correct answer [$I=0.5sinh(1)$] however I would like to know how to determine how to find regions of integration on change of variables in ways other than guessing at the answer! Is there an algorithmic procedure? Is there an easy graphical procedure? I know I could graph in the u-v plane but that seems like a lot more work than necessary.
edit: the way I am thinking about this now is that the first set of boundaries are incorrect because they correspond to a region that is a box (rectangle) in the uv plane. Cleary, the region is triangular in the x-y plane, and I'm assuming it is also triangular in the uv plane. How do I determine the dependence between u and v in of this region?
AI: This is an attempt to describe a picture by using a thousand words. The original integral is over the triangle with corners $(0,0)$, $(1,0)$, and $(0,1)$.
Because of the shape of the integrand, we let $u=x-y$ and $v=x+y$. To see what this means for the geometry, note that $u=c$ is the equation of a line $x-y=c$, that is, of a line with slope $1$. Similarly, $v=c$ is the equation of a line $x+y=c$ with slope $-1$.
The $u$-axis (that is, the line $v=0$) is the line with slope $-1$ through the origin. Similarly, the $v$-axis is the line with slope $1$ through the origin.
Now turn our diagram through $45^\circ$, or twist one's neck a bit. Our original triangle is symmetrical about the $v$-axis. The $x$-axis is the line $u=v$, and the $y$-axis is the line $u=-v$. The other boundary of our triangle of integration is just the line $x+y=1$, or more simply $v=1$. In terms of $u$ and $v$, our triangle has coordinates $(0,0)$, $(1,1)$, and $(-1,1)$.
A natural way to integrate over our triangle is to integrate first from $u=-v$ to $u=v$, and then to integrate from $v=0$ to $v=1$. That is exactly what you arrived at.
Remark: Can't argue with success, you got to the right integral. But I would be unable to get at the answer without using the above-described geometry. The point is that we use a geometric transformation, in this case a simple rotation, to make the problem collapse. |
H: Consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find $x(1-t)$
consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find the value of $x(1-t)$
AI: Note that $(2+\sqrt{3})^6+(2-\sqrt{3})^6$ is an integer, indeed an even integer. For imagine expanding each term, using the Binomial Theorem. The terms involving odd powers of $\sqrt{3}$ cancel.
We have $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. So $(2-\sqrt{3})^6=\frac{1}{x}$, and
$$(2+\sqrt{3})^6+(2-\sqrt{3})^6=x+\frac{1}{x}.$$
Since $2-\sqrt{3}$ is between $0$ and $0.3$, the number $(2-\sqrt{3})^6$ is positive but close to $0$. So $x$ is close to but below the integer $x+\frac{1}{x}$, and therefore
$$\lfloor x\rfloor=x+\frac{1}{x}-1.$$
Also, $t=x-(x+\frac{1}{x}-1)=1-\frac{1}{x}$, so $1-t=\frac{1}{x}$. It follows that $x(1-t)=(x)(1/x)=1$.
Remark: The result is obviously structural. In particular, the exponent $6$ is irrelevant. The same argument works with, for example, $(3+2\sqrt{2})^{99}$, and in many analogous situations.
A crucial role was played by the conjugate $2-\sqrt{3}$ of the number $2+\sqrt{3}$. This sort of thing happens very frequently.
The fact that medium sized powers of $2+\sqrt{3}$ are almost integers (but just a little bit smaller than an integer) is at first a little startling. It may be enlightening to use the calculator to compute a few powers. For similar but not identical behaviour, compute some powers of $2+\sqrt{5}$. |
H: Integral dependence and rings of fractions
I have a question on a Proposition in Atiyah and MacDonald's text. It concerns Proposition 5.12 ($A$ and $B$ are commutative rings with an identity) pictured here:
Here's my concern: After multiplying the equation of integral dependence in the ring of fractions through by $(st)^n$ we'll obtain something of the form
$$\frac{ (bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'}{1}=0$$
in $S^{-1}B$ where $a_i'\in A$ ($1\leq i \leq n$). Thus we conclude there exists $u\in S$ such that
$$u\left[(bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'\right]=0.$$
If $B$ were an integral domain (and $0 \notin S$), then we arrive at the desired equation of integral dependence for $bt$ over $A$. But $B$ is assumed arbitrary---we might have zero-divisors. Am I missing something? How do we obtain their conclusion without the assumption that $B$ is an integral domain?
You can see the title of the section, so maybe this was a missing hypothesis...?
AI: $u\left[(bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'\right]=0.$
Multiplying the both sides by $u^{n-1}$, we get
$(ubt)^n + ua_1'(ubt)^{n-1} + \cdots + u^na_n'=0.$
Hence $ubt \in C$.
Therefore $b/s = ubt/stu \in S^{-1}C$. |
H: $\sum\limits_{i=1}^n z_i=0\Longleftrightarrow z_1,\ldots ,z_n$ are the vertices of ...
Suppose for $n\geq 3$ we have, $z_1,\ldots ,z_n\in\mathbb{C}$ and $|z_1|=|z_2|=\cdots=|z_n|=1$. Now I need to determine a property $P$ such that the following is true :
$$\sum_{i=1}^n z_i=0\Longleftrightarrow z_1,\ldots ,z_n\mbox{ are the vertices of a polygon satisfying } P$$
I have solved the cases for $n=3$ and $n=4$, in the first case, $P$ is equilateral triangle and for the second, $P$ is rectangle. But my methods does not generalize for general case. Can we solve it for general $n$ ? At least I would like to know what happens for $n=5$.
AI: You can get all the solutions for $n=5$, but it isn't pretty.
There must be a sector of size $4\pi/5$ containing at least 3 of the points. By rotating and reflecting, if necessary, we may assume $z_j=e^{2\pi i\theta_j}$ with $\theta_1=0$, $0\le\theta_1\le1/5$, $\theta_1\le\theta_2\le2/5$. Now provided only that $$|z_1+z_2+z_3|\le2$$ we can find unique $\theta_4\le\theta_5$ to make $\sum^5z_j=0$.
Thinking of $\theta_1$ and $\theta_2$ as parameters, this gives a 2-parameter family of solutions to $$z_1+z_2+z_3+z_4+z_5=0,\qquad|z_j|=1$$ |
H: What does String mean in context of counting theory?
What does "String" mean in context of counting theory ? For example :
There are 8 binary strings of length 3 .
What does binary and length signify here ?
Edit : I want to specifically why it is called a String ?
AI: binary string means $\textit{a sequence of 0's and 1's}$ and the length defines the number of elements in the sequence.
Like $10011$ is a binary string of length $5$, $110011100$ is a binary string of length $9$ etc. |
H: Maximum Profit at which combination
A dealer deals only in colour TVs and VCRs . He wants to spend up to Rs.12 lakhs to buy 100 pieces. He can purchase colour TV at Rs. 10,000 and a VCR at Rs. 15,000. He can sell a colour TV at Rs. 12,000 and a VCR at Rs. 17,500. His objective is to maximiz profits
Fpr the maximum profit number of colour TVs and VCRs that he should stock is
I have found answer using hit and trial
But is there a mathmatical way to guarantee that maximum profit happens for 60 colour TV and 40 VCR
AI: Yes. You must consider the boundary conditions. If we let $t$ denote the number of TVA sold and $v$mdenote the number of vcrs sold, the we know that $t+v=100$, and $10000t+15000v \leq 1200000$.
Graphing these lines on the $t,v$ axes, along with the lines $t =0, v=0$, we get an enclosed region representing all the possible sales combinations. If you think about it, you can determine that the optimal point will always be on the boundary. And because all the relationships are linear, the optimal point will be at the intersection of two boundary conditions. (otherwise you shift along one of the lines in whichever direction increasesrofits)
Plugging in $t = 100 - v$ into $10t+15v=1200$, we get $1000 +5v = 200$, so that their intersection is at $(40,60)$. Comparing the profits from only selling TVA, only selling vcrs, selling nothing, and selling $(40,60)$, we see that it's the latter that has the best profit. |
H: Analysis of how-many-squares and rectangles are are there on a chess board?
I know the formula $$f(n) = \frac{n \cdot (n + 1) \cdot (2n + 1) } 6$$
Since chess board consists $8\times 8$, hence here $n=8$.
But I want to know how it has been concluded? Also how to tackle the number of rectangles?
AI: Here is an easy way to count the number of rectangles. There are $9$ horizontal lines on the chessboard, and $9$ vertical lines. Choose two distinct horizontal lines, and two distinct vertical lines. These determine a unique rectangle. And any rectangle determines a pair of horizontal lines and a pair of vertical lines.
So the number of rectangles is $\binom{9}{2}^2$. That is $1296$.
Exactly the same idea can be used to count the number of rectangles on an $m$ by $n$ chessboard. It is
$$\binom{m+1}{2}\binom{n+1}{2}.$$
The number of squares is a bit less nice. It is easy to see that there are $8^2$ small $1\times 1$ squares, $7^2$ $2\times 2$ squares, and so on down to $1^2$ $1\times 1$ squares, for a total of
$$1^2+2^2+3^2+\cdots+8^2.$$
Now we can add up, but there is also a simple formula for the sum of the first $n$ squares. The same idea works for counting the squares on an $n \times n$ chessboard. |
H: Can $(\Bbb{R}^2,+)$ be given the structure of a matrix Lie group?
I have an assignment problem that is coming from Brian Hall's book Lie Groups, Lie Algebras and Representations: An Elementary Introduction.
Suppose $G \subseteq GL(n_1;\Bbb{C})$ and $H \subset GL(n_2;\Bbb{C})$ are matrix Lie group and that $\Phi:G \to H$ is a Lie group homomorphism. Is the image of $G$ under $\Phi$ a matrix Lie group?
The definition of a matrix lie group that I have is a closed subgroup of $GL(n;\Bbb{C})$ for some $n$. I do not know about things like differentiable manifolds and the like. By a Lie group homomorphism, we mean a continuous group homomorphism from $G$ to $H$ (continuity is with respect to the usual topology coming from the euclidean metric).
Now an example I have in mind is the Heisenberg group $G$ that consists of all matrices of the form
$$\left\{\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) : a,b,c\in \Bbb{R} \right\}.$$
I can get a surjective group homomorphism (which I believe is also continuous) from $G$ to $\Bbb{R}^2$ under addition simply by sending
$$\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) \to (a,c) \in \Bbb{R}^2.$$
This would be a counterexample to the problem above if only I were to know how to show that $\Bbb{R}^2$ does not embed as a matrix lie group in $GL(n;\Bbb{C})$ for any $n \in \Bbb{N}$. The difficulty in showing this is it is not obvious if such an embedding exists or not.
For example $\Bbb{R}$ under addition surprisingly is a matrix lie group. I have a copy of $\Bbb{R}$ sitting inside of $GL(2;\Bbb{C})$, namely as the set of all matrices
$$\left\{ \left(\begin{array}{CC} 1 & a \\ 0 & 1 \end{array}\right) : a\in \Bbb{R} \right\}.$$
How do I know if $(\Bbb{R}^2,+)$ can or cannot be given the structure of a matrix Lie group? If only I were to know representation theory, I believe this is asking if there exists a finite dimensional complex faithful representation of $\Bbb{R}^2$.
Thanks.
AI: In fact $(\mathbb R^n,+)$ is a matrix Lie group for all $n$. Note that the map
$$\begin{pmatrix}
x_1\\
\vdots\\
x_n
\end{pmatrix}\mapsto \begin{pmatrix}
1 & 0 & \cdots & x_1\\
& \ddots & & \vdots\\
0 & \cdots & 1 & x_n\\
0 & \cdots & 0 & 1
\end{pmatrix}$$
which sends a vector $x$ to the identity matrix with the zeroes of the last column replaced by coordinates of $x$, is an injective homomorphism from $(\mathbb R^n,+)$ to $GL(n+1,\mathbb C)$. |
H: Find the probability of the the sum
A new drug has been released and produces some minor side effects. 8% of users suffer
only a loss of sleep and 12% of users suffer only bouts of nausea. 75% of users will have
no side effects at all. What percentage of users will suffer from both loss of sleep and
nausea?
What I learn in school is: if there is or in question then sum ($+$) , if there is and then multiply ($\times$)
Ans: $12*8=96$
and I have try many ways, I can't get the right answer 5%
I have no idea how to find the probability, please explain me.
Highly appreciate it. thanks
AI: You just need to add these numbers so out of 100 - (8 + 12 + 75) = 100 - 95 = 5%.
Essentially we are accounting for percentage of people out of a group having a particular property, whatever is left are the bunch who have both since every one else who has been counted has only one specific property. |
H: What's the proof of correctness for Robert Floyd's algorithm for selecting a single, random combination of values?
I read about it in a SO answer:
Algorithm to select a single, random combination of values?
initialize set S to empty
for J := N-M + 1 to N do
T := RandInt(1, J)
if T is not in S then
insert T in S
else
insert J in S
What is the proof of correctness for this? I've spent quite a bit of time on it and can prove to myself the $N^\text{th}$ value and the $1^\text{st}$ - $(N-M+1)^\text{th}$ values have $P$ (chosen) $= M/N$, but not the ones that remain.
e.g.
For $N$ choose $M$, each item enumerated from $1$ to $N$, I can prove to myself that using this algorithm, items $1$, $2$, $\ldots$, $N-M$, $N-M+1$, and $N$ each have the probability of $M/N$ to be chosen. I don't know for the other remaining items though.
e.g. 2
For $8$ choose $4$, each item enumerated from $1$ to $8$, I can prove to myself that using this algorithm, items $1$, $2$, $3$, $4$, $5$, and $8$ each have the probability of $4/8$ to be chosen.
AI: Consider a fixed number $k\in [n]$ and denote by $T_j$ $\,(n-m\leq j\leq n)$ the time after the value $j$ has been processed; whence $T_{n-m}$ is the starting time. We argue about the probability $P_j$ that $k$ is not yet chosen at time $T_j$.
When $k\leq n-m$ then $P_{n-m}=1$, and $$P_j=\biggl(1-{1\over j}\biggr)P_{j-1}\qquad(n-m+1\leq j\leq n)\ .$$
It follows that
$$P_n=\prod_{j=n-m+1}^n\biggl({j-1\over j}\biggr)={n-m\over n}\ ,$$
as required.
When $k>n-m$ then $P_{k-1}=1$, because up to time $T_{k-1}$ only numbers $\leq k-1$ have been selected, in fact exactly $k-1-(n-m)$ of them. Therefore, when processing the value $j:=k$ there are $n-m$ numbers amongst the $\leq k-1$ left, and if one of these is chosen in step $j:=k$, the number $k$ is not jet chosen at time $T_k$. This implies that we have
$$P_k={n-m\over k} P_{k-1}={n-m\over k}\ .$$
For the remaining steps we get as before
$$P_n={n-m\over k}\prod_{j=k+1}^n\biggl({j-1\over j}\biggr)={n-m\over n}\ .$$ |
H: Question about proof that $C(X)$ is separable
In my notes we prove Stone-Weierstrass which tells us that if we have
a subalgebra $A$ of $C(X)$ such that it separates points and contains
the constants then its closure (w.r.t. $\|\cdot\|_\infty$) is $C(X)$.
A few chapters later there is a lemma that if $X$ is a compact metric
space then $C(X)$ is separable. The proof constructs a subalgebra that
separates points by taking a dense countable subset of $X$, $\{x_n\}$,
and defining $f_n (x) = d(x,x_n)$.
Question: could we treat this as a corollary of Stone-Weierstrass and
say that polynomials with rational coefficients are a subalgebra
containing $1$ and separating points? Thank you.
AI: In the hope I did understand the question correctly by now: It is a corollary of Stone-Weierstrass. Add the constant functions to the algebra you constructed and you are done. |
H: Evaluating $I(x)=\int_{0}^{\infty}\frac{e^{-xy}}{y^2+a^2}dy$
I examine currently this integral:
$$I(x)=\int_{0}^{\infty}\frac{e^{-xy}}{y^2+a^2}dy;x\geqslant0$$ where $x$ and $a$ are real.
It seems that the integral has no closed form in terms of elementary functions. But perhaps it has a closed form in terms of special functions?
The end result should be series expansion of $I(x)$ at $x=0$
Thanks!
AI: Note that $$I''(x)=\int_0^\infty \frac{y^2 e^{-xy}}{y^2+a^2}dy$$
so that $$\tag{1} I''(x)+a^2 I(x)= \int_0^\infty e^{-xy} dy=\frac 1x$$
Solutions of the homogenous equation are $I(x)=\sin(ax)$ and $I(x)=\cos(ax)$ but specific solutions will require (as noticed by Norbert) sine and cosine integrals as we will show using variation of constants :
$$\tag{2} I(x)=c\;\sin(ax)$$
$$I'(x)=c'\sin(ax)+c\;a\cos(ax)$$
$$I''(x)=c''\sin(ax)+2\,c'a\cos(ax)-c\;a^2\sin(ax)$$
$$I''(x)+a^2 I(x)=\frac 1x=c''\sin(ax)+2\,c'a\cos(ax)$$
Let's set $\;b:=c'$ then we want $$b'\sin(ax)+2\,b\,a\cos(ax)=\frac 1x$$
multiply this by $\,\sin(ax)$ to get :
$$b'\sin(ax)^2+b\,2\,a\sin(ax)\cos(ax)=\frac {\sin(ax)}x$$
$$b'\sin(ax)^2+b \frac d{dx} \sin(ax)^2=\frac {\sin(ax)}x$$
$$\frac d{dx} (b\;\sin(ax)^2)=\frac {\sin(ax)}x$$
or $$c'=b=\frac 1{\sin(ax)^2} \left(C_0+\int \frac {\sin(ax)}{ax} d(ax)\right)=\frac {C_0+\rm{Si(ax)}}{\sin(ax)^2}$$
and $c$ will be :
$$c=C_1+\int \frac {C_0+\rm{Si(ax)}}{\sin(ax)^2} dx$$
but (int. by parts and since $\cot(x)'=-\frac 1{\sin(x)^2}$, $\rm{Si}'(x)=\frac {\sin(x)}x$ and $\rm{Ci}'(x)=\frac {\cos(x)}x$) :
$$\int \frac {\rm{Si}(ax)}{\sin(ax)^2} dx=\frac 1a\left[-\rm{Si}(ax)\cot(ax)\right]+\int \frac {\sin(ax)}{x} \cot(ax)dx$$
$$\int \frac {\rm{Si}(ax)}{\sin(ax)^2} dx=-\frac 1a\left[\rm{Si}(ax)\cot(ax)\right]+\int \frac {\cos(ax)}{x}dx$$
so that (for $C_0=C_1=0$) $c$ will simply be :
$$\tag{3} c=\frac{\rm{Ci}(ax)-\rm{Si}(ax)\cot(ax)}a$$
multiplying $c$ by $\,\sin(ax)$ in $(2)$ we get the specific solution :
$$\tag{4} I(x)=\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)$$
with the general solution of the O.D.E. given by :
$$\tag{5} I(x)=C\,\sin(ax)+D\,\cos(ax)+\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)$$
You may use the specific case $I(0)=\left[\frac {\arctan(ax)}a\right]_0^\infty=\frac {\pi}
{2a}$ and the derivative $I'(0)=-\log(a)$ to find Norbert's expression :
$$\tag{6}\boxed{\displaystyle I(x)=\frac {\pi\cos(ax)}{2a}+\frac 1a\left(\rm{Ci}(ax)\sin(ax)-\rm{Si}(ax)\cos(ax)\right)}$$
For numerical evaluation perhaps that the DLMF link or A&S' book will be helpful.
I got following series expansion of $\displaystyle (a\;I(x))$ (as explained by J.M. there is a logarithmic term coming from the $\rm{Ci}$ function that can't be expanded at $0$) :
$$\left[(\gamma+\ln(ax))\sin(ax)+\frac {\pi}2-(ax)-\frac{\pi}4 (ax)^2+\frac {11}{36}(ax)^3+\frac{\pi}{48}(ax)^4-\frac{137}{7200}(ax)^5+\rm{O}\left((ax)^6\right)\right]$$
(with $\gamma\approx 0.5772156649$ the Euler constant)
To get more terms you may use following expansions :
$$\rm{Ci}(z)=\gamma+ \ln(z) +\sum_{n=1}^\infty \frac{(-1)^nz^{2n}}{(2n)(2n)!}$$
$$\rm{Si}(z)=\sum_{n=0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)(2n+1)!}$$
P.S. It is quite interesting to notice that there is a reference to nearly the same integral in this other recent S.E. thread, the paper is Coffey 2012 'Certain logarithmic integrals, including solution of Monthly problem #tbd, zeta values, and expressions for the Stieltjes constants' where equation $(4.4)$ reads $\ \displaystyle I(k)\equiv \int_0^\infty \frac{e^{-(k-1)v}}{v^2+\pi^2}\,dv$. |
H: Number Expected of Tests
A medical clinic tests blood for certain disease from which approximately one person in a hundred suffers. People come to the clinic in group of 50. The operator of the clinic wonders whether he can increase the efficiency of the testing procedure by conducting pooled tests. In the pooled tests, the operator would pool the 50 blood samples and test them altogether. If the pooled test was negative, he could pronounce the whole group healthy. If not, he could then test each person‘s blood individually. The expected number of tests the operator will have to perform if he pools the blood samples are:
AI: If $p=1/100$ is the probability of an individual suffering from this disease, then the probability of none of the $n=50$ people suffering from it is $(1-p)^n$. Thus, the expected number of tests in the pooled approach is $1+(1-(1-p)^n) n=n+(1-n(1-p)^n)$, compared to $n$ in the direct approach, so the pooled approach is worthwhile if $n(1-p)^n\gt1$, or $p\lt1-n^{-1/n}$. For $n=50$, we have $50^{-1/50}\approx0.075$, so the pooled approach works for $p=1/100$.
[Edit in respone to a comment under the question:]
As you rightly pointed out, this misses the special case where the batch test is positive and the first $n-1$ individual tests are negative. The probability for that to happen is $(1-p)^{n-1}p$, so the expected number of tests for the batch approach is actually
$$
\begin{align}
1+\left(1-(1-p)^n\right)n-(1-p)^{n-1}p
&=
n+\left(1-n(1-p)^n-(1-p)^{n-1}p\right)
\\
&=n+\left(1-(n(1-p)+p)(1-p)^{n-1}\right)\;.
\end{align}
$$
The condition for the added term to be negative can no longer be solved for $p$ in closed form. With your numbers, the expected number is
$$50+\left(1-\left(50\left(1-\frac1{100}\right)+\frac1{100}\right)\left(1-\frac1{100}\right)^{50-1}\right)\approx21\;.$$
The substracted term $(1-p)^{n-1}p$ is just
$$\frac1{100}\left(1-\frac1{100}\right)^{50-1}\approx0.006\;,$$
so it doesn't make an appreciable difference. The ratio between the two subtracted terms is $n(1-p)/p=50\cdot99=4950$. |
H: All possibilities of seven numbers in ascending order?
I had posted a question about a pretty old permutation cipher here on Stack Overflow.
One @Mark Adler has commented that all possibilities of seven numbers with range $0$-$255$ in ascending order is $9,503,812,464$.
I tried to speculate by saying "Ok you mean each byte can be having value $0$-$255$ if first byte is $255$ then all others are also $255$. This is one way. If first byte is $254$ then there can be seven ways for remaining bytes ie $255,255,255,255,255,255$ or $254,255,255,255,255,255$ or $254,254,255,255,255,255$ or .... $254,254,254,254,254,254$ yes i seem to get it"
But I am not sure how to count all possibilities. Can anyone please help?
AI: If $$0\le a_1\le a_2\le\cdots\le a_7\le255$$ then let $$b_0=a_1, b_1=a_2-a_1,b_2=a_3-a_2,\dots,b_6=a_7-a_6,b_7=255-a_7$$ and you have $$b_0+b_1+\cdots+b_7=255,\quad b_j\ge0.\tag1$$ Conversely, any solution of (1) gives you numbers $a_j$. So, do you know how to count the number of solutions of (1)? |
H: Is a countable product of compact intervals in $\mathbf R$ compact (without using the AC)?
Let $\{I_n=[a_n,b_n]\}_{n\in\mathbf N}$ be a countable collection of closed, bounded intervals in $\mathbf R$. Is the infinite Cartesian product $$\prod_{n=1}^\infty I_n$$ compact without using the Axiom of Choice?
AI: See Herrlich's book The Axiom of Choice [1], Theorem 3.13:
(ZF) $[0,1]^\mathbb N$ is compact.
Of course your question is about a general product of closed intervals, however note that the functions $f_n\colon[a_n,b_n]\to[0,1]$ defined by $$f_n(x)=\frac{x-a_n}{b_n-a_n}$$ are homeomorphisms, and therefore we can transfer the product of closed intervals to a product of $[0,1]$.
It should also be noted that this is no longer true for general uncountable products, indeed even $[0,1]^\mathbb R$ may fail to be compact (see [1, Section 4.8: theorem 4.70 and E 13]).
Bibliography:
Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006. |
H: What's the supremum of this?
I would like to know how to show that the supremum
$$\sup_S\frac{(x-y)^2 + (t-s)^2}{(|x-y|^2 + |t-s|)^{\epsilon}}$$
is less than infinity, where $0<\epsilon \leq 1$ and
$$S = \{ (x,t), (y,s) \in [0,2\pi] \times[0,T] \mid (x,t) \neq (y,s)\}$$
where $0 < T \leq 1$.
What can I use?
AI: We have
\begin{align}
\frac{(x-y)^2+(t-s)^2}{((x-y)^2+|t-s|)^{\varepsilon}}&\leq ((x-y)^2+|t-s|)^{1-\varepsilon}+\frac{|t-s|(1+|t-s|)}{((x-y)^2+|t-s|)^{\varepsilon}}\\
&\leq ((x-y)^2+|t-s|)^{1-\varepsilon}+|t-s|^{1-\varepsilon}(1+T),
\end{align}
and using the fact that $s\mapsto x^s$ is increasing for $s\in (0,1)$, we get
$$\sup_S\frac{(x-y)^2+(t-s)^2}{((x-y)^2+|t-s|)^{\varepsilon}}\leq (4\pi^2+T^2)^{1-\varepsilon}+(1+T)T^{1-\varepsilon}.$$ |
H: Prove that the solution tends to $0$ as $t$ goes to infinity
I have to prove that, if $A(t)$ is a real symmetric $N\times N$ matrix whose
eigenvalues are all less than $-1$ for all $t$, then if we consider $u$ to be the solution of
$$\dot u(t)=A(t)u(t),$$
Then
$$\lim_{t\to\infty}|u(t)|^2=0.$$
The only thing I was able to do was to write down the solution, but nothing more.. can you help
me? Many thanks..
AI: Note that
$$\frac{\mathrm d}{\mathrm dt}|u(t)|^2=2\langle u(t),u'(t)\rangle=2\langle u(t),A(t)u(t)\rangle\leqslant-2\langle u(t),u(t)\rangle=-2|u(t)|^2,
$$
hence
$$
|u(t)|^2\leqslant|u(0)|^2\,\mathrm e^{-2t}.
$$ |
H: characteristic polynomial of the adjacency matrix of a tree
I have read that if $A$ is the adjacency matrix of a tree $T$, then we have that
$$\det(\lambda I - A) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k N_k(T) \lambda^{n-2k} $$
where $N_k(T)$ is the number of matchings of size $k$ of $T$.
However, I couldn't find a proof. Does anyone know how to do it? Thanks!
AI: This is just a direct application of the definition of the determinant as the signed sum of all permuted products. Every term in that sum corresponds to a permutation of the vertices, which you can write as a product of cycles, and because the tree contains no cycles, for the term to be non-zero the permutation can only contain $1$- and $2$-cycles, with the $2$-cycles corresponding to edges. Each $2$-cycle yields a factor $-1$, and each $1$-cycle yields a factor $\lambda$, and the result follows. |
H: upper bound for derivatives of analytic functions on upper half-plane
Let $f$ be analytic mapping of the upper half-plane into the unit disc. Given $f(i)=\alpha$, how does one rigorously obtain an upper bound for $|f'(i)|$?
AI: We can assume that $f(i)=i$ without loss of generality (scale and rotate). Then we use Cayley's transform $h(z):=\frac{z-i}{z+i}$, from the upper half-plane to the unit disc, and its inverse $h^{—1}(z)=-i\frac{z+1}{z-1}$. Let $g(z):=h\circ f\circ h^{-1}$ from the unit disk to itself. We have $g(0)=0$ and by Schwarz lemma, $|g'(0)|\leq 1$. Hence we have, after having used the chain rule,
$$|f'(i)|\leq\frac 1{|h'(i)|\cdot |(h^{-1})'(0)|}.$$ |
H: What is the image of $|z-4i|+|z+4i|=10$?
What is the image of $|z-4i|+|z+4i|=10$?
I tried to simplify this equation but it is too difficult for me. I tried squaring both side but it was too long and I can't get equation like any curve.
Can anyone give me any simple simplification? Or is there any result of complex number used in this example? Please help me. Thank in advance.
AI: If we represent $Z$ by $x + iy$, then the equation above given by you after simplification becomes that of an ellipse: $25x^2 + 9y^2 = 225.$
Steps for simplification are:
Insert the value of $Z$ as $x + iy$ and apply the magnitude formula of the complex numbers: $\sqrt{x^2 + y^2}$
Take the part obtained from $|z+4i|$ to the RHS and then square both the sides; you will get on simplification
$\sqrt{x^2 + (y-4)^2} + \sqrt{x^2 + (y+4)^2} = 10$
$\sqrt{x^2 + (y-4)^2} = 10 - \sqrt{x^2 + (y+4)^2}$
(square both sides)
$x^2 + y^2 +16 -8y = 100 +x^2 + y^2 +16 +8y - 20\sqrt{x^2 + (y+4)^2}$
(removing common terms and common factors)
$4y + 25 = 5 \sqrt{x^2 + (y+4)^2}$
Again square both sides and then simplify to obtain the equation of ellipse
$16x^2 + 625 +200y = 25x^2 + 25 y^2 +200y + 400$
$25x^2 + 9y^2 = 225$
Process the output as you want to take the image. |
H: Find the probability of three tosses of a fair coin
Find the probability that, in three tosses of a fair coin, there are three heads, given that
there is at least one head.
I manage to get $\frac{3}{6}$ or $\frac{1}{6}$ but the right answer is
$\frac{1}{7}$
I have no idea, Can you please explain?
thanks! Appraciate it!
AI: A fair coin has etqual probabilitiers for heads and tails. Therefore the probability can simply be found by counting.
There are eight different possibilities for outcomes of three tosses:
Head, Head, Head
Head, Head, Tail
Head, Tail, Head
Head, Tail, Tail
Tail, Head, Head
Tail, Head, Tail
Tail, Tail, Head
Tail, Tail, Tail
Possibilities 1 to 7 have at least 1 head, while possibility 8 doesn't. Therefore remove that last one.
From the 7 remaining possibilities, 1 has all three heads, so the probability is one out of seven, or $\frac17$.
OK, but what to do if there are too many cases to explicitly list them? Well, it is easy to see that you cannot have three heads without having at least one head. So your probability is
$$p_{\text{3 heads if at least 1 head}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of cases with at least 1 head}}$$
However you know that
$$p_{\text{3 heads}} = \frac{\text{Number of cases with 3 heads}}{\text{Number of all cases}}$$
and
$$p_{\text{at least 1 head}} = \frac{\text{Number of cases with at least 1 head}}{\text{Number of all cases}}$$
From this, it is not hard to see that
$$p_{\text{3 heads if at least 1 head}} =\frac{p_{\text{3 heads}}}{p_{\text{at least 1 head}}}$$
Now you surely know that $p_{\text{3 heads}}=\left(\frac12\right)^3=\frac18$ and $p_{\text{at least 1 head}} = 1-p_{\text{3 tails}}=\frac78$. Inserting then again gives $p_{\text{3 heads if at least 1 head}}=\frac18/\frac78=\frac17$.
Indeed, that latter formula works even for arbitrary probabilities (i.e. not only fair coins). Note however that it still depends on the fact that having 3 heads implies having at least 1 head. |
H: Why this proof $0=1$ is wrong?(breakfast joke)
We have $$e^{2\pi i n}=1$$
So we have $$e^{2\pi in+1}=e$$
which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$
Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$
This implies $$e^{-4\pi^{2}n^{2}}=1$$
Taking the limit when $n\rightarrow \infty$ gives $0=1$.
AI: Your error is (as in most of those fake-proofs) in the step where you use the power law $(a^b)^c=a^{bc}$ without the conditions of that power law being fulfilled. |
H: Additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity
I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.
Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} < 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$ where $\frac{p}{q}$ is the mentioned representative.
Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.
Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?
AI: To prove it is a bijection, one can use rather "primitive" methods. suppose that:
$f\left(\frac{p}{q} + \Bbb Z\right) = f\left(\frac{p'}{q'} + \Bbb Z\right)$,
then: $e^{2\pi ip/q} = e^{2\pi ip'/q'}$, so $e^{2\pi i(p/q - p'/q')} = 1$.
This, in turn, means that $\frac{p}{q} - \frac{p'}{q'} \in \Bbb Z$, so the cosets are equal. Hence $f$ is injective.
On the other hand, if $e^{2\pi i p/q}$ is any $q$-th root of unity, it clearly has the pre-image $\frac{p}{q} + \Bbb Z$ in $\Bbb Q/\Bbb Z$ (so $f$ is surjective).
One caveat, however. You haven't actually demonstrated $f$ is a function (i.e., that it is well-defined, although if you stare hard at the preceding, I'm sure it will come to you). |
H: Why there may be no single "maximum" element in a partially ordered set?
From Appendix B.2 (relations) of Introduction to Algorithms by Cormen et al:
In a partially ordered set $A$, there may be no single "maximum" element $a$ such that $b R a$ for all $b ∈ A$. Instead, there may several maximal elements a such that for no $b ∈ A$, where $b ≠ a$, is it the case that $a R b$. For example, in a collection of different-sized boxes there may be several maximal boxes that don't fit inside any other box, yet no single "maximum" box into which any other box will fit.
I know the definition of a partial ordered set , but do not get why the author/authors say that "there may be no single "maximum" element" in a partially ordered set ?
AI: Remember that we're dealing with a partial order! This is fundamentally different from a total order like $\leq$ is on numbers.
Because: For all numbers, we have either $a\leq b$ or $b\leq a$, i.e. everything is comparable.
However a partial order is partial because elements need not be comparable. Solely if elements should happen to be comparable, then the partial order imposes some laws like transitivity.
Now take as an example the proper subsets of $\{1,2,3\}$ with the partial order of $\subseteq$. E.g. $\{1\} \subseteq \{1,2\}$, but the sets $\{1,2\}$ and $\{2,3\}$ simply are not comparable because neither includes the other. However there are no "bigger" proper subsets to each of them, so both are maximal.
Now if in turn all elements were comparable, the maximum was unique (hence in total orders it is). But in partial orders, this isn't usually the case. |
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