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H: proving that there is no solution for $n$ in positive integers except $n = 1$ for $2^n = nk + 1$
If $k$ is a positive integer, how to prove that there is no solution for $n$ in positive integers except $n = 1$ for $2^n = nk + 1$
AI: The reason is that the only solution to the congruence $2^n\equiv 1\mod n$ is $n=1$.
Assume there exist solution $n>1$ of the congruence written above and consider smallest prime $p\;|\;n$. Then $2^n\equiv 1\mod p$, and $\mathrm{ord}_p(2)\;|\;n$. Recall that $\mathrm{ord}_p(2)<p$ and $p$ is the smallest prime divisor of $n$, then $\mathrm{ord}_p(2)=1$. This means then $p\;|\; 2^1-1=1$, which is impossible for prime number $p$. Contradiction. |
H: Getting the recurrence formula with a condition
Get the recurrence formula of
$$U_n=2(-3)^n-5n(-3)^n$$
For $$n \geq 1$$
What am I supposed to do with this condition $n\geq 1$?
AI: $U_{n}=(-3)^{n}(2-5(n))$
$U_{n+1}=(-3)^{n+1}(2-5(n+1)) = 3(-3)^n(2-5n-5) $
$U_{n+1}= -3U_{n}+5(-3)^{n+1}$ |
H: Poincare inequality question
Help me please with this question:
Is it true that for all $N>0$ exists $u \in C^{1}[0,1]$ such that $u(0)=0$ and $\frac{\int_{0}^{1}u^2(x)dx}{\int_{0}^{1}\left [ u'(x) \right ]^2dx}>N$;
what happens without assuming that $u(0)=0$?
Thanks a lot!
AI: If $u\in C^1[0,1]$ satisfies $u(0)=0$ then
$$\int_0^1u(x)^2dx=\int_0^1\left(\int_0^xu'(t)dt\right)^2dx\leq \int_0^1x\int_0^x(u'(t))^2dtdx\leq \frac 12\int_0^1(u'(t))^2dt,$$
hence for $K>1/2$ the inequality you want doesn't hold.
If we don't require $u(0)=0$, then take functions like $u(x)=x+C$ to get want we want. |
H: The relation between formal logic and proof writing
I was reading vellmans how to prove it and he forms a link between formal logic and proof writing. For instance, he decomposes if p then q to not(p and not q) and similarly for other such proof writing statements. However what I don't follow is What's the motivation for writing out truth tables and evaluating conditions for them.
Usually, we write out Boolean statements like p and not q etc. and evaluate the truth tables when we wish to evaluate different conditions of p and q. Here however, we know that p is either true or not and q is determined accordingly. Why truth tables?
My question might seem a bit vague and under thought but I can't seem to grasp the fundamentals itself so all else is shaky too.
AI: You say
we know that p is either true or not and q is determined accordingly.
But that's just not true. If we have "if p then q" and p is false, then q may be true or false, since there's no "only if" in our statement. It may be that p being true makes q true, but that some other circumstance could also make q true. All that we have ruled out is "p and not q": in other words, the statement "if p then q" is exactly the statement "not (p and not q)" which, by standard manipulations (or you can use truth tables to work this out!) is the same as "not p, or else q".
Now, why do we care at all? Well, writing out truth tables helps us to understand complex statements and their interaction. So that, for example, we can discover that $(((p\Rightarrow q)\Rightarrow p)\Rightarrow p)$ is true regardless of $p$ and $q$, and we can use it in proofs accordingly. Similarly, we can establish that $p \Rightarrow p$ is true, and that $(q \Rightarrow r) \Rightarrow ((p \Rightarrow q) \Rightarrow (p \Rightarrow r))$ is true, which together sort of tell us that $\Rightarrow$ up to propositional equivalence is a partial order (but that's beyond the scope of this answer). |
H: Does $k$-th power of $p$ divide ${}_n\!C_r$ if the previous divides $n$?
Does $p^k$ divide ${}_n\!C_r$ for all integer r if $p^k|n$ where $0\leq r \leq n$ and $p$ is prime?
AI: The answer is no. Take $p=3$, $n=6$, $k=1$, $r=3$. Then $p^k\mid n$ , but $p^k\nmid{}_n C_r$ |
H: What is Kosambi-Cartan-Chern (KCC) theory?
I'm reading a paper that's basically about stability analysis of the Lane-Emden differential equation. The authors make use of "Kosambi-Cartan-Chern (KCC) theory". I've been trying to find out what this "theory" is about and haven't really gotten anywhere except for a another paper (PDF) that refers to this KCC theory as an established thing.
Given that the papers discuss the theory as an established thing, I expected there to be a Wikipedia or MathWorld article or something. But no luck. References are given to a series of papers by Kosambi, Cartan and Chern from the 1930's but I haven't hit any textbooks that describe it. For interest, the relevant references appear to be
Cartan, E. Observations sur le mémoir précédent, Math. Zeitschrift 37 (1933),
619-622.
Chern, S.S., Sur la géometrie d'un système d'equations differentialles du second
ordre, Bull. Sci. Math. 63 (1939), 206-212
Kosambi, D.D. Parallelism and path-space, Math. Zeitschrift 37 (1933), 608-618
AI: The name 'KCC-theory' was introduced in a book of Antonelli, Ingarden and Matsumoto entitled The Theory of Sprays and Finsler Spaces with Applications in Physics
and Biology published in 1993. It is mostly used in the physics and biology fields.
The KCC theory concerns the following. Consider a second order differential equation $(d^2x_i/dt^2) + g_i(x,x',t) = 0$, for $i = 1, ... , n$, where $x = (x_1, ..., x_n)$, t is the time parameter, $x'$ denotes $((dx_1/dt), ..., (dx_n/dt))$, and $g_i$'s are smooth functions of $(x, x', t)$ defined on a domain in the $(2n + 1)$-dimensional Euclidean space. The aim is to understand what geometric properties of the system of integral curves - the paths associated with the system of differential equations - remain invariant under nonsingular transformation of the coordinates involved. The theory describes certain invariants, which are specific tensors depending on the $g_i$'s, which characterize the geometry of the system, in the sense that two such systems can be locally transformed into each other if and only if the corresponding invariants are equivalent tensors. In particular, a given system as above can be transformed into one for which the $g_i$'s are identically 0, so that the integral curves are all straight lines, if and only if the associated tensor invariants are all zero.
The problem may be viewed also as that of realising the integral curves of a second order differential equation as geodesics for an associated linear connection on the tangent bundle. Kosambi introduced a method using calculus of variations, which involves realizing the paths as extremals of a variational principle; this is related to finding a 'metric' for the path space.
By associating a non-linear connection and a Berwald type connection to the dynamical system, five geometrical invariants are obtained, with the second invariant giving the Jacobi stability of the system.
These review articles may be of help:
http://emis.icm.edu.pl/journals/AMAPN/vol24_1/amapn24_16.pdf
http://www.desy.de/~mnrw2/publications/2009_pre79_046606.pdf |
H: In set theory, what does the symbol $\mathfrak d$ mean?
What's meaning of this symbol in set theory as following, which seems like $b$?
I know the symbol such as $\omega$, $\omega_1$, and so on, however, what does it denote in the lemma?
Thanks for any help:)
AI: The symbol $\mathfrak d$ is used to denote the dominating number of the continuum.
If $g,f\colon\omega\to\omega$ we say that $g$ dominates $f$ if for all but finitely many $n$, $f(n)\leq g(n)$.
The dominating number is the smallest cardinality of a dominating family, namely the minimal $|F|$ such that $F\subseteq\omega^\omega$ and for every $f\colon\omega\to\omega$ there is some $g\in F$ such that $g$ dominates $f$.
Some observations:
$\aleph_0<\frak d\leq c$: the former is true because if we have a countable family of functions by diagonalization argument we can produce a non-dominated function; the latter is true because it is obvious that $F=\omega^\omega$ is a dominating family and its size is exactly $\frak c$.
If $\aleph_1=\frak c$ then $\frak c=d$, which is a trivial consequence of the above.
It is not provable that there is an equality, because by forcing we can ensure that $\frak d<\frak c$. |
H: Immersed curve in $\mathbb{R}^2$ and regular curve
I read that a curve $\gamma:S^1 \to \mathbb{R}^2$ is an immersion iff it's regular. So it's an immersion iff $\gamma'(t) \neq 0$ for all $t \in S^1$.
An immersed curve is one whose derivative is an injective linear map. When the domain is single-variabled, this is the same as requring $\alpha'(t) \neq 0$ for all $t$.
Is this right? But what about $t=0?$ Shouldn't $0$ make the derivative equal $0$? Why isn't this included in this characterisation (since regular means non-zero everywhere)?
How does one check that such a curve is immersive? If $\gamma(t) = (e^t, 1)$, which I know is regular, how do I check that the derivative is an injective linear map? Clearly the "derivative" in this context isn't referring to $\gamma'(t)$ since that is not linear.
Thanks
AI: In one-variable calculus one introduces the derivative of a function $f:{\mathbb R}\to{\mathbb R}$ as derivative at a point $t_0$ by defining
$$f'(t_0):=\lim_{t\to t_0}{f(t)-f(t_0)\over t-t_0}\ .\qquad(1)$$
This is a number. In a second step one envisages this process performed at all points $t_0\in{\rm dom}(f)$, which leads to the idea of the derivative being a function $$f':\quad {\mathbb R}\to{\mathbb R},\qquad t_0\mapsto f'(t_0)\ .$$
This works as well for vector-valued functions of a single variable $t$, e.g. for the function $\gamma:\ t\mapsto(e^t,1)$. Here one gets $\gamma'(t)=(e^t,0)$.
When one has more than one independent variables the derivative $d{\bf f}({\bf p})$ of a function $${\bf f}:\quad {\mathbb R}^n\to{\mathbb R}^m$$ at a point ${\bf p}$ can no longer be defined as a limit in the form $(1)$. Instead $d{\bf f}({\bf p})$ is a linear map mapping the tangent space at ${\bf p}$ into the tangent space at ${\bf q}={\bf f}({\bf p})$. The matrix of $d{\bf f}({\bf p})$ is nothing else but the Jacobian matrix of ${\bf f}$ at ${\bf p}$. The map ${\bf f}$ is an immersion if this matrix has rank $n$ at all points ${\bf p}$ in the domain of ${\bf f}$.
What we have said in 2. also makes sense in the one-variable case, i.e. for $n=1$:
$$t\mapsto{\bf f}(t)=(f_1(t),\ldots,f_m(t))\ .$$ In this case the Jacobian at a point $p$ in the domain of ${\bf f}$ has just one column given by $[f_1'(p),\ldots,f_m'(p)]^T$, and it has rank $1$ iff ${\bf f}'(p)\ne{\bf 0}$, where ${\bf f}'(p)$ is the derivative of ${\bf f}$ at $p$ considered in 1. In your example $\gamma'(t)\ne(0,0)$ for all $t\in{\mathbb R}$, so $\gamma$ defines an immersion. |
H: Top cohomology detecting compactness
Could someone point me to a standard reference for the fact that the top cohomology $H^n(M,A)$ of an $n$-dimensional manifold $M$ is non-trivial for local coefficients $A$ if and only if the manifold is compact?
EDIT: It seems that there are some issues when $M$ is non-orientable. I would like to include the non-orientable case. I figure the result uses (twisted) Poincaré duality and some kind of pairing between the $n$th cohomology and compactly supported cohomology in degree $0$.
I am not sure of its validity, but I am looking for (a reference for) an isomorphism $H_c^0\cong H^n$ which holds for local systems.
Thread on MathOverflow.
AI: As I pointed out in the comments, you need the hypothesis of orientability. For example, $\mathbb{R}P^2$ is nonorientable and compact, but has $H_2(\mathbb{R}P^2;\mathbb{Z}) = 0$.
For a reference for homology, see theorem 3.26, page 236 of Hatcher's Algebraic topology book. For cohomololgy, try Corollary 3.39 on page 250 of the same book.
His book is freely available from his own website. See here. |
H: Formula to this pattern? $1$, $11$, $21$, $1211$, $111221$, $\ldots$
I have this pattern:
1
11
21
1211
111221
I'm guessing it's a fibo pattern, been at it for hours now. Anyone know?
AI: It's known as the look-and-say sequence. |
H: Stronger condition lead to weaker result?
Suppose there are two theorems $A \Rightarrow B$ and $C \Rightarrow A$. Then we have $C \Rightarrow B$.
Now comparing $A \Rightarrow B$ and $C \Rightarrow B$, we know that $C \Rightarrow A$ means C is a stronger condition than A. Is it to say $A \Rightarrow B$ is stronger or weaker than $C \Rightarrow B$? I personally think $A \Rightarrow B$ is a stronger result/theorem than $C \Rightarrow B$, but I also saw $A \Rightarrow B$ is said to be weaker than $C \Rightarrow B$.
Thanks!
AI: Suppose we later find that $D\Rightarrow A$. Then we can use $A\Rightarrow B$ to show $D\Rightarrow B$, but we can't use $C\Rightarrow B$ to do that.
Conversely, if we find that $E\Rightarrow C$, then we can use either $A\Rightarrow B$ or $C\Rightarrow B$ to prove $E\Rightarrow B$, since the latter can be recovered from the former and $C\Rightarrow A$.
So, in the presence of $C\Rightarrow A$, the statement $A\Rightarrow B$ is stronger than $C\Rightarrow B$. |
H: To show $X$ is a complete vector field on $M$
Well, I have solved myself the problem : every smooth vector field on a compact manifold is complete.
Now I have got this problem which I am not able to progress:
let $X$ is a vector field on $M$, suppose $\exists \epsilon >0\ni (-\epsilon,\epsilon) \subsetneq (a(m),b(m))\forall m\in M$, Then show that $X$ is a complete vector field i.e $(a(m), b(m))=\mathbb{R}$
thank you for any help.
where $(a(m),b(m))$ is the domain of maximal integral curve, for each $m\in M$ we will get this interval $(a(m),b(m))$
here I asked some realted things
A series of Lemmas about $C^{\infty}$ vector fields
AI: Let $\gamma(t)$ be the maximal integral curve with initial condition $\gamma(0) = m$, and let's suppose $b(m) < \infty$.
If we denote $\gamma'(t)$ the maximal integral curve with initial condition $\gamma'(0) = \gamma(b(m) -\epsilon/2)$ then
$$
\gamma''(t) =
\begin{cases}
\gamma(t) & \text{if } t\in (a(m), b(m) - \epsilon/2)\\
\gamma'(t - b(m) + \epsilon/2)& \text{if } t\in [b(m) - \epsilon/2, b(m) + \epsilon/2)
\end{cases}
$$
prolongs $\gamma(t)$. That contradicts the assertion that $\gamma(t)$ is maximal.
A similar reasoning can be used to show $a(m) = -\infty$.
Edit - Some clarifications
$\gamma''$ prolongs $\gamma$ means that $\gamma''$ has an interval of existence, $(a(m), b(m) + \epsilon/2)$, that is a proper superset of the interval of existence, $(a(m), b(m))$, of $\gamma$ and the two curves are equal on the common domain, $(a(m), b(m))$.
The "uniformity" of $\epsilon$ ensures that the integral curve with initial condition
$$
\gamma'(0) = \gamma(b(m) - \epsilon/2)
$$
exists on the interval $[0, \epsilon)$. Without "uniformity" it could happen that the maximal right interval of existence of $\gamma'$ is $[0, \epsilon/2)$. In such a case, a curve $\gamma''$, constructed as above, would coincide with $\gamma$: it would not prolong $\gamma$.
The initial condition of $\gamma'$ is chosen in such a way to allow us to smoothly join $\gamma'$ to $\gamma$ in order to form a new curve, $\gamma''$, that prolongs $\gamma$.
The contradiction arises because we constructed an integral curve that prolongs a maximal integral curve, which, as such, cannot be prolonged by definition. |
H: Functional Inequality question where $\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt $ ,$ 0 \leq x \leq 1$ and $f(0) = 0$
$$\int^x_0\frac1{f'(t)}dt = \int^x_02f(t)dt \tag{1}$$ where $0 \leq x \leq 1$ and $f(0) = 0$
I need to prove that $$f(\frac1{\sqrt{2}})> \frac1{\sqrt{2}}$$
$$f(\tan (x))> \tan(x) > x , x \in (0,\frac{\pi}{4}) $$
$$f(e^{-x^2})\geq e^{-x^2}$$
The problem is that i dont know how to derive the function it self.
All I could do was say that $$\frac{d\left(\int^x_0\frac1{f~'(t)}dt)\right )}{dx} = \frac1{f~'(x)}$$
and $$\frac{d\left(\int^x_02f(t)dt\right )}{dx} =2f(x)$$
AI: You get
$$\frac{1}{f'(x)}=2f(x) $$
Thus
$$2f'(x)f(x)=1 \,.$$
or
$$\left( f(x)^2 \right)' =1 \,.$$
Integrate, find the constant and you are done. |
H: Compute $\sum_0^{n-1}2^i11^{n-i-1}\bmod10^9$ when $n=13^{17}$
Given the following function $f$
$f(1)=1$
$f(n)=11\cdot f(n-1)+2^{n-1}$
I would like to compute $f(13^{17})\mod 10^9$ and ended up using the following :
$f(n)=\sum_{i=0}^{n-1}({11^{n-(i+1)}\cdot 2^i})$
though I am able to quickly compute a single term using modular exponentiation, the loop over $13^{17}-1$ takes ages.
So do you have any suggestion to simplify $f(n)$ to avoid the loop ?
AI: Note that $f(n)$ may be simply rewritten as $\displaystyle f(n)=\frac{11^n-2^n}{11-2}$ |
H: for $n=x^2+3y^2$ , $n=\prod p^{a(p)}$ , $a(p)$ is even for all $p \equiv 2 \pmod 3$ where $p$ is prime
I'd really like your help with the following Number Theory question:
I need to show that if I can write an integer $n=x^2+3y^2$ so in the factorization of $n$ to primes, every $p \equiv 2\pmod 3$ would be with a even power, I mean if $n=\prod p^{a(p)}$ so $a(p)$ is even for all $p \equiv 2 \pmod 3$, where $p$ is prime.
I don't rally know how to start this one.
Thank you.
AI: Hint: If $p\neq 3$ is a prime and $p|x^2+3y^2$ and $p\not\mid x$, $p\not\mid y$, then $-3$ is a square mod $p$. |
H: In what base does "100" equal "4 in base 10"
I was reading this answer to an amusing comic related question: https://math.stackexchange.com/a/166891/35132 and I understand that in the linked answer, the examples of how four may be expressed used base (expressed in decimal!!) is 10, 4, 3 for 4, 10, 11.
What I can't figure out is what base would need to be used for his last example (100) to equal 4 in decimal?
P.S. Are maths questions like this always this hard to put in words?!
AI: Let the required base be $b$. Then,
$$1 \cdot b^2 + 0 \cdot b^1 + 0 \cdot b^0 = 4 \implies b = 2.$$ |
H: Isomorphism $KG \cong K[x]/(x^p-1)$
Why is $KG \cong K[x]/(x^p-1)$, for $K=\mathbb{Z}/p \mathbb{Z}$ and $G = \langle g | g^p=1 \rangle$ a ring isomorphism?
If I take $g \mapsto [x]$, I'd have an group homomorphism. And than an algebra(?) homomorphism f: $KG \rightarrow K[x]$, $a_g g \mapsto a_g[x]$, with $a_g \in K$. Now I need $\ker(f)=(x^p-1)$, not?
AI: You should map the other way, namely consider $K[x] \to K[G]$ taking $x\mapsto g.$ Then it should be easy to see it's a homomorphism, and clear that $\operatorname{Ker}(f)=(x^p-1).$ |
H: Change of order of summation.
I feel like an idiot for asking this, so bear my stupidity.
I have the sum $\sum_{n\leq N} \sum_{p | n ; \ p \ prime} 1$, and I want to change the order of summation of these two sums I think it should be:
$$\sum_{n | N} \sum_{p\leq n ; \ p \ prime} 1 = \sum_{n | N} \pi(n)$$
Is this right or wrong, in which case if it's right then how do I simplify the term
$$\sum_{n\mid N}\pi(n)$$
any further? Here $\pi(n)$ is the prime-counting function.
Thanks in advance.
AI: It looks like your answer is wrong.
The correct change is:
$$\sum_{n\leq N} \sum_{\substack p|n \\ p \text{ prime}} 1= \sum_{\substack p\leq N \\ p\text{ prime}} \sum_{\substack n\leq N \\ p|n} 1 =\sum_{\substack p\leq N \\ p\text{ prime}} \left\lfloor \frac N p\right\rfloor$$
The last step is because:
$$\sum_{\substack n\leq N \\ p|n} 1 = \left\lfloor \frac N p\right\rfloor$$
The key to changing the order of summation is to write out the set of pairs that you are summing over.
In this case, you are summing over all pairs, $(n,p)$ with the condition $p|n$, $n\leq N$ and $p$ is prime. The original form is to sum over all $n$ and then find the corresponding set of $p$. The "change" is to list all possible $p$ first, namely, the primes $p\leq N$, and then list all the $n\leq N$ which are multiples of $p$.
Edit
You can actually remove the condition $p\leq N$ since it is redundant:
$$\sum_{n\leq N} \sum_{\substack p|n \\ p \text{ prime}} 1= \sum_{p\text{ prime}} \sum_{\substack n\leq N \\ p|n} 1 =\sum_{p\text{ prime}} \left\lfloor \frac N p\right\rfloor$$
This is the same result because $\left\lfloor\frac N p\right\rfloor = 0$ when $p>N$. |
H: What is the algorithm for solving an equation like this one?
The solutions of the equation : $\sqrt{x+2\sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}} = 2$ are:
A) $x=1$;
B) $x=2$;
C) $x\in [1,2]$;
D) $x\in \begin{bmatrix}
\frac{3}{2},2
\end{bmatrix}$;
E) $x=\frac{3}{2}$;
AI: As you have some answer possibilities you can just check if they solve your equation. Denoting your right hand side by
\[ f(x) = \sqrt{x + 2\sqrt{x-1}} + \sqrt{x - 2\sqrt{x-1}}, \]
we have
$f(1) = 2$,
$f(2) = 2$,
$f(\frac 32) = 2$,
so C) is the only possibility.
If you don't have choices given, you could try to rewrite $f$, noting that the maximal domain of definition of $f$ is $[1,2]$. We have
\begin{align*}
f(x)^2 &= x + 2\sqrt{x-1} + 2\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1})} + x - 2\sqrt{x-1}\\
&= 2x + 2\sqrt{x^2 - 4(x-1)}\\
&= 2x + 2\sqrt{x^2 - 4x + 4}\\
&= 2x + 2\sqrt{(x-2)^2}\\
&= 2x + 2\left|x-2\right|\\
&= 2x + 2(2-x)\\
&= 4
\end{align*}
As $f$ is non-negative, this gives $f(x) = 2$ for all $x \in [1,2]$ (that is, where $f$ is defined). |
H: Written as disjuctions, conjunctions and negations?
With a domain from -2 to 2 I'm trying to write the following using disjunctions conjunctions and nagations. I'm not sure how correct I am and wanted to know if I did them correct? Could someone help with the last one I just cant figure out how to start that expression?
$∀ x\ P(x) = P(-2) ∧ P(-1) ∧ P(0) ∧ P(1) ∧ P(2)$
$\neg ∀x\ P(x) = (P(−2)∨P(−1)∨P(0)∨P(1)∨P(2))∧¬(P(−2)∧P(−1)∧P(0)∧P(1)∧P(2))
In English doesn't that mean: At least one element is true and one element is false?
$∃x\ \neg P(x) = $
Edit: The last two with the correct answer of "At least one is not true".
∃x￢P(x) = ￢P(-2) ∨￢P(-1) ∨￢P(0) ∨￢P(1) ∨￢P(2)
￢∀xP(x) = ￢(P(-2) ∧ P(-1) ∧ P(0) ∧ P(1) ∧ P(2))
AI: The last two should have the same answer, since the second one, $\lnot\forall x.P(x)$ says that it is not true that $P(x)$ holds for every $x$, while the third one, $\exists c.\lnot P(x)$ says that there exists an $x$ for which $P(x)$ does not hold. These mean exactly the same thing. "Not every crow is black" is the same as "There is a crow that is not black."
But your answer for the second one is not correct.
Mouse over for hint:
$\lnot\forall x.P(x)$ says that it is not true that every element of the domain satisfies $P$. But $P(-2)\lor P(-1)\lor\ldots\lor P(2)$ says that $P(x)$ is true for -2, or for -1, or…
EDIT: Now your second one is correct, if you and I have the same idea for the part you indicated by "…", but it could be much simpler.
EDIT: Now your second one is incorrect again. $\lnot\forall x.P(x)$ says that $P(x)$ is not true for every $x$. It does $not$ say that $P(x)$ is true for any $x$; it might be false for all $x$. |
H: interchange sum and integral
suppose I have a family of i.i.d standard normal random variables $Y_{n,k}$ and I define $X^N_t:=\sum_{n=0}^N\sum_{k=1}^{2^n}Y_{n,k}\phi_{n,k}(t)$ for $t\in [0,1]$ where $\phi_{n,k}$ are the Schauder functions.Furthermore, I know that $(X_t^N)$ is a martingale bounded in $L^2$ and therefore converges a.s. and in $L^2$ to a random variable $X$.
Why am I allowed to interchange expectation and the sum in the following expression
$$E[X_sX_t]=\sum\sum E[Y_{n,k}Y_{l,m}]\phi_{n,k}\phi_{l,m}$$
Note: The two sums are running both over two variables. The first over $n,m$ and the second over $k,l$.
AI: By the Cauchy-Schwarz inequality, the mapping
\[
L^2(P)\times L^2(P) \ni (X,Y) \mapsto E(XY)
\]
is continuous, which can be seen as follows: Let $(X_0, Y_0) \in L^2(P)^2$ and $\epsilon > 0$. Let $\delta = \min\{\epsilon, \frac{\epsilon}{\max\{\|X_0\|_2, \|Y_0\|_2+\epsilon\}}\} $, then for $(X,Y) \in L^2(P)^2$ with $\|X-X_0\|_2 + \|Y-Y_0\|_2 < \delta$ it holds
\begin{align*}
\left|E(X_0Y_0) - E(XY)\right| &\le \left|E[X_0(Y_0 - Y)]\right| + \left|E[Y(X_0-X)]\right|\\
&\le \|X_0\|_2 \|Y_0 - Y\|_2 + \|Y\|_2 \|X-X_0\|_2\\
&\le \max\{\|X_0\|_2, \|Y_0\|_2 + \epsilon\} \bigl(\|X-X_0\|_2 + \|Y-Y_0\|_2\bigr)\\
&< \epsilon.
\end{align*}
Now we can continue:
As $X^N_t \to X_t$ and $X^M_s \to X_s$ in $L^2(P)$, we have $E(X_sX_t) = \lim_{N,M} E(X^N_s Y^M_s)$. As expectation is linear, the last term equals $\lim_{N,M} \sum_{n\le N} \sum_{m\le M} \sum_{k=1}^{2^n}\sum_{l=1}^{2^m} E(Y_{n,k}Y_{m,l})\phi_{n,k}(t)\phi_{m,l}(s)$. |
H: Work out the number of edges given conditions of a graph
Given a set of nodes $V$, and a parameter $c \lt |V|$,
How can we show that whether we can derive an un-directed graph $G=(V,E)$ where the degree of each node equals to $c$?
If a graph as above exists, how many un-directed edges are there in $G$?
AI: For 2., you want to use that the sum of the degrees is twice the number of edges, as Matt points out in the comments to the question. For 1., you should just construct a graph --- try putting the points in a circle, and try connecting each vertex to some of them that are near it on the circle. Hint: you'll have to do cases for if $c$ is even or odd. |
H: Tell if $(\mathbb Z_6, \odot)$ is a semigroup and if the identity element belongs to it
Let the operation $\odot$ be defined in $\mathbb Z_6$ as follows:
$$a \odot b = a +4b+2$$
check if $(\mathbb Z_6, \odot)$ is a semigroup and if the identity element belongs to it.
This is the way I have solved this exercise:
Let $x,y,z \in \mathbb Z_6$ then in order for $(\mathbb Z_6, \odot)$ to be a semigroup, the following condition must be met:
$$(x\odot y)\odot z = x\odot (y\odot z)$$
Considering only the first part of the equation:
$$\begin{aligned} (x\odot y)\odot z
&= (x+4y+2)\odot z \\
&= (x+4y+2)+4z+2 \\
&=x+4y+4z+4
\end{aligned}$$
now considering the second part of the equation:
$$\begin{aligned} x\odot (y\odot z)
&= x \odot (y+4z+2) \\
&= x+4(y+4z+2)+2 \\
&= x+4y+16z+10 \\
&= x+4y+4z+4
\end{aligned}$$
So I conclude stating that $(\mathbb Z_6, \odot)$ is a semigroup. When it comes to verifying the presence of the identity element within the semigroup, some confusion arises:
$$x \odot 1_{\mathbb Z_6} = x+4\cdot 1_{\mathbb Z_6} + 2 \neq x $$
and also
$$1_{\mathbb Z_6} \odot x = 1_{\mathbb Z_6} +4x+2 \neq x$$
so the identity element does not belong to $(\mathbb Z_6, \odot)$. Is my solution right or am I wrong?
AI: I suppose addition and multiplication is interpreted modulo 6. (Otherwise it would not be a binary operation on $\mathbb Z_6$.)
I guess you have to find out whether the given semigroup has identity.
This means: Is there an element $e$ such that $a\odot e=e\odot a=e$ for all elements.
$a\odot e=a$ means
$$a+4e+2=a\\4e+2=0.$$
We can easily check that this is fulfilled by $e\in\{1,4\}$. So this semigroup has two right identities.
Since there can be only one identity element, there cannot be left identity. But we can check this anyway.
Existence of left identity $e$ would mean that for each $a$ we have $e\odot a=a$, i.e.
$$e+4a+2=a\\e=4+3a.$$
The expression $4+3a$ has various values for various $a$'s (namely the values $1$ and $4$), so there is no element $e$ fulfilling this for each $a\in\mathbb Z_6$. |
H: Convergence Properties of the Taylor Series for $\frac{1+z}{1-z}$
I just ran into the following exercise:
Find and state the convergence properties of the Taylor series for the following:
$$\frac{1+z}{1-z}$$
around $z_0=i$.
First of all, let
$$f(z)=\frac{1+z}{1-z}.$$
Then we have that
$$\begin{align}
f^{(1)}(z)&=\frac{2}{(1-z)^2},\\
f^{(2)}(z)&=\frac{-4}{(1-z)^3},\\
f^{(3)}(z)&=\frac{12}{(1-z)^4},\cdots
\end{align}$$
leading us to conclude that
$$\frac{d^j}{dz^j}\left[\frac{1+z}{1-z}\right]=2(-1)^{j+1}\left(1-z\right)^{-j-1}j!,$$
for $j\geqslant1$. Now, evaluating these at $z=i$ should give the Taylor series
$$i+\sum_{j=1}^{\infty}\frac{2}{(1-i)^{j+1}}(z-i)^j.$$
I have two questions: I could derive that same series without problems if the $(-1)^{j+1}$ term were not there, but in this case, where does it go? Also, how does the author determine that this holds whenever $|z-i|<\sqrt{2}$? I suppose that this has something to do with the fact that
$$\sum_{j=0}^{\infty}c^j$$
converges whenever $|c|<1$. Thanks in advance!
AI: You shouldn't have that alternating sign term showing up. $$f''(z)=\frac{d}{dz}\bigl[2(1-z)^{-2}\bigr]=2\cdot-2(1-z)^{-3}\cdot\frac{d}{dz}[1-z]=-4(1-z)^{-3}\cdot-1=\frac{4}{(1-z)^3}.$$ Something similar will happen in all other cases, too.
It holds whenever $|z-i|<\sqrt{2}$ because the nearest (in fact, only) singularity is the pole at $z=1$. The distance from $i$ to $1$ is $\sqrt{2}$. The radius of convergence of a Taylor series is the distance from the center to the nearest point of nonanalyticity. You can also, of course, see this by rearranging this as a geometric series. Note that $\frac{2}{1-i}=1+i$, so we can rewrite as $$\frac{1+z}{1-z}=i+(1+i)\sum_{j=1}^\infty\left(\frac{z-i}{1-i}\right)^j=-1+(1+i)\sum_{j=0}^\infty\left(\frac{z-i}{1-i}\right)^j,$$ and the series converges for $$\left|\frac{z-i}{1-i}\right|<1,$$ or equivalently, for $$|z-i|<|1-i|=\sqrt{2}.$$ |
H: Quadric surface graphing applet
Does anybody know of any online tool/applet that can be used to graph quadric surfaces? i.e. If I want an elliptic paraboloid, I can click on "elliptic paraboloid" and enter my own specified values for a, b, and c. I couldn't find one on google, any ideas?
AI: The POV-Ray 3D renderer can render quadric surfaces, as well as plenty of others. It's not an applet, and it may not be quite as easy to use at first as you envision, but the results can certainly be pretty. |
H: Density function of a function of random variable by expectation?
Suppose $Z$ is a continuous random variable on $\mathbb{R}^n$.
$f: \mathbb{R}^n \to \mathbb{R}^+ \cup \{0\}$ is a function, such that $\mathrm{E} Z = \int_{\mathbb{R}^n} z \times f(z)dz$. Then we know $f$ is not necessarily the density function of $Z$.
What else conditions can make $f$ the density function of $Z$?
For example, if for any measurable and bounded function $u: \mathbb{R}^n \to \mathbb{R}^n$, $\mathrm{E} (u(Z)) = \int_{\mathbb{R}^n} u(z) \times f(z)dz$. Then is $f$ the density function of $Z$? This is inspired from did's reply.
Thanks and regards!
AI: This works since you can take indicator functions $u = \mathbb{I}_A$ for any measurable subset $A$ and see that $P(Z \in A) = \int_A f(z) dz$. |
H: Is there an infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$ analogous to the Rogers-Ramanujan identity?
Given
$$
\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\;\; =\;\; \frac{r^5}{1-11r^5-r^{10}},\;\;\;\;\;\text{with}\;\;r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$
where $\eta(\tau)$ is the Dedekind eta function,
$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$
and $q = \exp(2\pi i\tau)$, is there an analogous identity for,
$$
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2}\;\; =\;\; ???$$
AI: Yes. Michael Somos just today found the identity,
$$
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$
where,
$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$
thus completing the family for $N = 2,3,5,7,13$.
Kindly see this MSE post for the other N. Does address Matt E and Loeffler's comments in that post? Is "s" a modular function? |
H: How to explain what it means to say a function is "defined" on an interval?
I am having difficulty in explaining the terminology "defined" to the students I am assisting. Here is the sentence: If a real-valued function $f$ is defined and continuous on the closed interval $[a,b]$ in the real line, then $f$ is bounded on $[a,b]$. Can I have some thoughts on how to explain the word "defined" used in the sentence? Thank you.
AI: What age students? If it's just a precalculus or calculus course, I would just give examples of a nice looking formula that "isn't defined" on all of an interval, e.g. $\log(x)$ on [-.5, 2] or $1/x$ on [-1, 1].
If it's an analysis course, I would interpret the word defined in this sentence as saying, "there's some function $f$, taking values in $\mathbb{R}$, whose domain is a subset of $\mathbb{R}$, and whatever the domain is, definitely it includes the closed interval $[a, b]$."
In general the mathematician's notion of "domain" is not the same as the nebulous notion that's taught in the precalculus/calculus sequence, and this is one of the few cases where I agree with those who wish we had more mathematical precision in those course. (Often "domain" means something like "I wrote down a formula, but my formula doesn't make sense everywhere. Tell me where it does make sense," which I hate, especially because students are so apt to confuse functions with formulas representing functions.) |
H: The role of maps in algebraic structures?
In group theory we have encountered the concepts of homomorphisms as maps from one group to another, and the same thing with ring and field theory and we ave called them respectively homomorphisms of groups (of rings, and of fields). Also in linear algebra we have linear maps.
So my question is this:
What is the role of such maps in this algebraic structures?
AI: One of my professors once said this about ring theory:
To study rings, you have two choices. You can take a particular ring, and stare at it intently for a long period of time until you discover interesting things about that ring. Then you write them down.
Or you can study how the rings "acts" on other structures, by studying the collection of all modules on that ring, and thus deduce interesting things about the ring. It turns out that the latter method is more fruitful, easier to generalize, and allows one to go from one ring to another ring more easily.
Studying the modules of a ring can be seen as a special case of studying homomorphisms (an $R$-module structure on an abelian group $A$ is equivalent to a ring homomorphism from $R$ to the endomorphisms of $A$, which are a ring under pointwise addition and composition. (Which is why I mention it)
The example of vector spaces is very apt. Vector spaces are nice objects, no doubt; but if all you do is stare at a particular vector space, you can say some interesting things (dimensions, subspaces, and so on), but it's not until you bring in linear transformations that the full power of vector spaces really becomes apparent. It is through linear transformations that we can actually do things with vector spaces: use them to solve systems of equations (linear and differential), to study Markov systems, to solve least squares problems (which amount to constructing certain kinds of projections, which are types of linear transformations), etc.
There is a whole philosophy of mathematics that says that the best way to study objects is to study the maps that "respect the structure", and not only for algebraic objects: topology is not just the study of topological spaces, it's the study of topological spaces and continuous maps between them. Analysis is not just the study of the real numbers, it is the study of the real numbers and real valued functions. Differential geometry concerns itself with differentiable maps. Etc. Turns out that maps give you a fruitful way of studying the structure.
In algebra, an important role is played by "congruences", which are equivalence relations on the underlying set that are compatible with the operations. For example, in the ring of integers, the usual modular congruences, $a\equiv b\pmod{n}$, give an equivalence relation that "respects" the ring structure on $\mathbb{Z}$, in the sense that if $a\equiv b\pmod{n}$ and $x\equiv y\pmod{n}$, then $a+x\equiv b+y\pmod{n}$ and $ax\equiv by\pmod{n}$ (the sum of classes is the same as the class of the sum; the product of classes is the same as the class of the product). Congruences correspnd, via the isomorphism theorems, to surjective morphisms.
Morphisms let you take one particular structure and study it from other points of view, in other contexts; it allows you (via "quotients") to focus on particular details while ignoring other part of the structure that may be "in the way" (e.g., the study of "parity" is made simpler by considering the quotient $\mathbb{Z}/2\mathbb{Z}$, rather than $\mathbb{Z}$ itself).
Morphisms let you understand inherent symmetry in an object; Klein proposed studying geometry almost exclusively in terms of certain kinds of symmetries and certain kinds of morphisms (projective transformations). Galois theory is founded on the idea of understanding the "symmetries" that may exist between different roots of the same polynomial (in the form of automorphisms).
All in all, morphisms are just a very rich way of studying objects. You can think of them as a way of letting you poke, examine, cut open, x-ray, fold, and generally manipulate your objects of interest so that you can get a better sense, not only of what they are, but how they interact with other objects and other contexts. |
H: entire function is constant
Let $ f $ a complex entire function such that:
$$ |f(z)| \leq \sqrt{2|z|} + \frac{1}{\sqrt{2|z|}} \quad \forall z \neq 0 $$
Prove that $ f$ is constant.
Thank's in advance!
AI: Let $g(z)=\frac{f(z)-f(0)}{z}$. Then $g(z)$ is entire and for all $z$ with $|z| \geq 1$ you have
$$ \left| g(z) \right| \leq \left|\frac{f(z)}{z} \right| + \left|\frac{f(0)}{z} \right|
\leq \left|\frac{\sqrt{2|z|} + \frac{1}{\sqrt{2|z|}}}{z} \right| + \left|f(0) \right| $$
Now from here it is easily to prove that $g$ is bounded (note that by continuity $g$ is bounded on $|z| \leq 1$), thus, since entire is constant.
This proves that $f(z)=az+b $ for some $a,b$. Plug this in your equation and see that $a =0$. |
H: Jacobian physical meaning
Possible Duplicate:
What is Jacobian Matrix?
Is there any physical intuition for the Jacobian?
I understand that it is the matrix of partial derivatives and how to construct it. What I want to know is
what's the use of it? Application wise
is there a nice intuitive explanation for it? I mean regarding its significance and otherwise
AI: The Jacobian is the matrix that represents the linear transformation that takes a small change in the input of a function to the corresponding small change in output:
For $f:\mathbb R^n\to\mathbb R^m$, a fixed $x \in \mathbb R^n$, we have
$$f(x+h) = f(x) + J(x)h + o(|h|) \qquad\text{for }h\in\mathbb R^m, h\to 0$$
provided that the various partial derivatives exist and behave sufficiently nicely.
In this way the Jacobian is the direct analogue of the derivative in ordinary real analysis. |
H: Determine largest area of house using geometry
The question is as follows. Given a house as shown in the figure below
$AB = 3a\,,\,BC = 10\,,\,CD = a\,,\,DE= a\,,\,EF=2a,$ and $FA=10-a$
Where a is some arbitrary constant such that $a \in [0,10]$
Now one could show that the area of the figure is given by the function
$$T(a) = 30a -2a^2$$
Whereas completing the square, or solving $T'(a)=0$, gives that the largest
area of the house is $15^2/2$ for $a=15/2$.
Is there any other way of finding the largest area of said figure?
(Hopefully using geometric reasoning.)
AI: Here is an approach for the answer, that relies on a small (correct assumption) that would likely be possible to make more formal.
Step 1. We transform the problem into something more manageable. I would like to cut off the square on the top, together with the corresponding part of the rectangle, to get one rectangle with dimensions $2a \times (10-a)$ and that leaves another rectangle with dimensions $a \times 10$. Since we are into computing area, we can transform the second rectangle into an equivalent one with one side $2a$ and another side of $5$ (equivalent because halving one side and doubling another leaves the same area). Now connect the two rectangles together to get one rectangle of dimensions $2a \times (15-a)$.
Now we are asking how to geometrically prove that the maximum area of this rectangle happens at $a=15/2$.
Step 2. We know that if we need to maximize the area of a rectangle with a fixed perimeter of $30$, this results in finding the best $a$ such that rectangle with dimensions $a \times (15-a)$ contains the largest area. Geometrically, the optimal solution is a square with sides $a = 15-a = 15/2$.
This is exactly our problem, except we have one of the sides a linear scale larger (i.e. $2a$ instead of $a$). I would like to assume that such an extension (or any other simple multiplication of any of the sides by a constant) would not alter the correct solution.
Note: This assumption is correct, it is easy to verify using Calculus that since sucha transformation alters the objective function by a constant factor, it will have no effect on the location of the extrema, but proving this geometrically, as the author of the question requested, is non-trivial to me.
Hence the correct solution does indeed occur at $a=15/2$ and the maximum area will be $2a \times (15-a) = 15*15/2$ as desired. |
H: Interpretation of the Fourier coefficients?
Suppose I have a discrete function $f( x_i ) = y_i$.
I can use these pairs $(x_i, y_i)$ as complex number $z_i = x_i + j \, y_i$.
Now, having this set $z_i$, I can apply discrete Fourier transform, as show in Wikipedia.
Now, suppose the calculated Fourier coefficients are ${X_i}$, where each ${X_i}$ is, of course, a complex number.
So, what is the interpretation of these numbers? For example:
what does the real part of these number means (if anything at all)?
what does the imaginary part of these number means (if anything at all)?
what does the module $|{X_i}|$ means? Does these values give the spectrum of the function? If so, what it's used for?
AI: This makes little sense to me in the original setting, i.e., if we regard the points as $(x,y)$ values of a function. For one thing, a mere permutation of the points (which is not significant) gives different Fourier transforms.
But if you regard your input $\{ (x_1,y_1), (x_2,y_2) \cdots\}$, not only as a set of points belonging to the graph of the function $f(x)$, but rather as true sequence (a list, where order is significant) which travels along that graph (like succesive points along a parametric curve), then it makes sense and can be quite useful: see for example Fourier descriptors.
Of course, in this case we don't need to restrict to true functions, we just deal with general parametric curves in the $(x,y)$ plane. |
H: How many Fourier coefficients are enough ( in discrete fourier transform)
Probably there's a similar question, but I could't find it through all questions about FT
As the title says, how many Fourier coefficients are enough, to be able to "resume" the original function, using inverse discrete Fourier transform?
For example, in the definition from Wikipedia, it looks like we need N coefficients, where N is the number of given points from the original discrete function. I also noticed, that for FFT (fast Fourier transform), the number of calculated coefficients is the same as the number of given points.
Is this always like this? Or we may have fewer coefficients? Or more? And is there a way to estimate this number of necessary coefficients?
I ran some test with randomly generated "control points" of a discrete function and applied DFT and IDFT (in this order) and all control points were recreated.
AI: The discrete Fourier transform of a signal $\{x_j\}$ is given by a linear combination of the $x_j$'s with some factors of the form $e^{j \pi i /N}$ or something similar. This can be shortly written as
$$x_k=A_{ki} x_i$$
where $A_{ki}$ is the transformation matrix. It is invertible (this is why you also have the inverse transform).
Having understood that, you see that the Fourier transform is nothing more than a change of basis in the space $\mathbb{C}^N$ where $N$ is the signal length. Since any basis will be of size $N$, you see that in order to fully describe your signal you always need exactly $N$ complex numbers (or $2N$ real ones).
Therefore, if your signal is complex you will always need all the coefficients of the DFT. If your signal is purely real then the coefficients in the DFT are related by $x_k=x_{N-k}^*$ and you need only half of them. |
H: relating flatness, equidimensional, and complete intersection
I am a bit confused and am trying to clarify some notions. First consider the following well-known statement.
A dominant map $f:X\rightarrow Y$ between regular varieties is flat if and only if it is equidimensional.
Question 1. Doesn't a regular variety mean that it is nonsingular?
Question 2. I am not sure what equidimensional means in this context. Does it mean that the fibers $f^{-1}(c)$ for each $c\in Y$ are equidimensional?
Question 3. What are some examples of a variety that is equidimensional but not a complete intersection?
Thank you.
AI: Googling, I find this article, which seems worth a look: www.emis.de/journals/UIAM/actamath/PDF/35-243-246.pdf. Particularly, saying that a morphism is equidimensional at a point of the domain means that nearby fibres of the morphism have the same dimension. This is different from equidimensionality for a scheme, which means that all irreducible components of the scheme should have the same dimension.
An example for question 3 is the twisted cubic, which is not a (global) complete intersection, but is an equidimensional variety of dimension $1$ (i.e. a curve). |
H: Why does $L^2$ convergence not imply almost sure convergence
What's wrong with this argument?
Let $f_n$ be a sequence of functions such that $f_n \to f$ in $L^2(\Omega)$. This means $$\lVert f_n - f \rVert_{L^2(\Omega)} \to 0,$$ i.e.,
$$\int_\Omega(f_n - f)^2 \to 0.$$
Since the integrand is positive, this must mean that $f_n \to f$ a.e.
Why is this not true? Apparently this only true for a subsequence $f_n$ (and in all $L^p$ spaces).
AI: Consider the following sequence of characteristic functions $f_n \colon [0,1] \to R$ defined as follows:
$f_1 = \chi[0, 1/2]$
$f_2 = \chi[1/2, 1]$
$f_3 = \chi[0, 1/3]$
$f_4 = \chi[1/3, 2/3]$
$f_5 = \chi[2/3, 1]$
$f_6 = \chi[0, 1/4]$
$f_7 = \chi[1/4, 2/4]$
and so on.
Then $f_n \to 0$ in $L^2$, but $f_n$ does not converge pointwise. |
H: Solving Fibonaccis Term Using Golden Ratio ConvergEnce
While solving this problem, I discovered that there is a relationship between the Fibonacci sequence and the golden ratio. After I got the correct answer via brute force, I discovered this relationship. One of the posters said this:
The nth Fibonacci number is $[\phi^n / \sqrt{5}\;]$, where the brackets
denote "nearest integer".
So we need $\phi^n/\sqrt{5} > 10^{999}$
$n \log(\phi) - \log{5}/2 > 999 \log(10)$
$n \log(\phi) > 999 \log(10) + \log(5)/2$
$n > (999 \log(10) + \log(5) / 2) / \log(\phi)$
A handheld calculator shows the right hand side to be 4781.8593, so
4782 is the first integer n with the desired property.
I can't quite make sense of this. How does the poster know that the $n$-th term is $\phi^n / \sqrt{5}$? Can you also walk me through the operations?
AI: Using standard techniques for finding closed-form solutions to recurrences, one can show that $$F_n=\frac{\varphi^n-\hat\varphi^n}{\sqrt5}=\frac{\varphi^n}{\sqrt5}-\frac{\hat\varphi^n}{\sqrt5}\;,\tag{1}$$ where $\varphi=\frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$. Now $$\left|\frac{\hat\varphi}{\sqrt5}\right|\approx -0.2764<\frac12\;,$$ and $|\hat\varphi|\approx 0.618<1$, so $$\left|\frac{\hat\varphi^n}{\sqrt5}\right|<\frac12$$ for all $n\ge 0$.
Thus, $$\left|F_n-\frac{\varphi^n}{\sqrt5}\right|=\left|\frac{\hat\varphi^n}{\sqrt5}\right|<\frac12$$ for all $n\ge 0$. But we know that $F_n$ is an integer, so $F_n$ is an integer less than half a unit away from $\dfrac{\varphi^n}{\sqrt5}$. That means that $F_n$ is the unique integer closest to $\dfrac{\varphi^n}{\sqrt5}$, so that rounding $\dfrac{\varphi^n}{\sqrt5}$ to the nearest integer gives you $F_n$. In particular,
$$F_n=\left\lfloor\frac{\varphi^n}{\sqrt5}+\frac12\right\rfloor\;,$$ where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$. However, I’m not actually going to use that result; the argument that you posted is slightly defective, and it’s easier to do it right using $(1)$.
A number $m$ has $1000$ digits if and only if $10^{999}\le m<10^{1000}$: $10^{999}$ is the first integer with $1000$ digits, and $10^{1000}-1$ is the last. Thus, we need the smallest $n$ such that $F_n\ge 10^{999}$, or, using $(1)$, the smallest $n$ such that
$$\frac{\varphi^n}{\sqrt5}-\frac{\hat\varphi^n}{\sqrt5}\ge 10^{999}\;.$$
Clearly $n$ is going to be quite large, and we saw earlier that the second term on the left is going to be very small, so to a good first approximation we want the smallest $n$ such that $$\frac{\varphi^n}{\sqrt5}\ge 10^{999}\;;\tag{2}$$ we’ll find that and then come back to make sure that the second term is too small to matter.
$(2)$ is equivalent to $\varphi^n\ge 10^{999}\sqrt5$ and hence, after taking logarithms, to $$n\ln\varphi\ge 999\ln 10+\ln\sqrt5=999\ln 10+\frac12\ln 5\;.\tag{3}$$
Dividing both sides of $(3)$ by $\ln\varphi$, we get
$$n\ge\frac{999\ln 10+\frac12\ln 5}{\ln\varphi}\approx 4781.86\;,$$
where the numerical approximation is taken from a calculator. Clearly the smallest integer $n\ge 4781.86$ is $4782$. A little more work with a calculator shows that $$\log_{10}\frac{\varphi^{4782}}{\sqrt5}\approx 999.0294106732$$ and hence that $$\frac{\varphi^{4782}}{\sqrt5}\approx 1.07\times 10^{999}\;.$$
Since $\dfrac{\hat\varphi^{4782}}{\sqrt5}$ is clearly much less than $7\times 10^{997}$, $F_{4782}\ge 10^{999}$, and our approximation in $(2)$ did no harm.
As an extra check, note that $$\frac{\varphi^{4781}}{\sqrt5}<7\times 10^{998}\;,$$ and $\dfrac{\hat\varphi^{4781}}{\sqrt5}$ is clearly much less than $3\times 10^{998}$, so $F_{4781}<10^{999}$ and has only $999$ digits. |
H: Error estimation in $H^1(\sigma)$
I have a question. If $u$ and $u_h$ are the solutions of the continuous and discrete variational equations, respectively, then for $u\in H^1_0$, how does one prove that $\lim_{h\rightarrow 0} \|u - u_h\| = 0$ where the norm is taken in the $H^1(\sigma)$. Sigma is the domain.
Thank you.
AI: This is a broadly posed question, so I'll paint a picture in similarly broad strokes.
Let $E$ denote the functional you are minimizing on some space $V$. Let $E_*=\inf_V E$, and assume it's attained at $u_*$. You need two ingredients:
(*) $\qquad \forall \epsilon>0$ $\exists \delta>0$ such that $E(u)<E_*+\delta\implies \|u-u_*\|<\epsilon$
(**)$\qquad$The union of all finite-dimensional subspaces $V_h\subset V$ (which correspond to discretization of the problem) is dense in $V$.
Given $\epsilon>0$, pick $\delta>0$ from (*). Since the set $\Omega_\delta=\{u\in V\colon E(v)<E_*+\delta\}$ is open, there exists $h>0$ such that $V_h\cap\Omega_\delta$ is nonempty. The minimum of $E$ on $V_h$ will be attained by some $u_h\in\Omega_\delta$, which guarantees $\|u_h-u_*\|<\epsilon$. |
H: Finding arithmetic mean, standard deviation, mode and median
On the market quality of the fruit was measured and following results came out:
Quality of Fruit( in measuring units ) 65 70 75 80 85 90 95 100
Number 2 3 2 5 8 7 5 3
Define:
a) arithmetic mean and standard deviation
b) mode and median
How to find what is wanted here?
AI: There are $35$ items that have been assessed.
a) To find the mean, you need to calculate
$$\tfrac{(2)(65)+(3)(70)+(2)(75)+(5)(80)+(8)(85)+(7)(90)+(5)(95)+(3)(100)}{35}.\tag{$1$}$$
The standard deviation would could be defined in a couple of different ways. I will use the one I guess is the one more likely for your course.
For the sample variance, calculate first
$$\tfrac{(2)(65^2)+(3)(70^2)+(2)(75^2)+(5)(80^2)+(8)(85^2)+(7)(90^2)+(5)(95^2)+(3)(100^2)}{35}.\tag{$2$}$$
Subtract the square of the sample mean calculated in $(1)$. That gives you the sample variance $s^2$. For the sample standard deviation, take the square root.
But perhaps in your course, the formula for the sample variance and standard deviation involves an $n-1$ instead of an $n$. In that case, you should multiply the $s^2$ that I described by $\frac{35}{34}$. Then for the sample standard deviation, take the square root as usual.
b) Since there are $35$ items, and the median is the "middle" number, count $18$ from the bottom, or $18$ from the top. We end up in the "$85$" slot, so the median is $85$.
The mode is the value that occurs most often. A quick scan shows that the value $85$ is the one. |
H: Visualization of Cantor set by Mathematica or Maple!
We all know what it is the Cantor set. The Cantor set is created by repeatedly deleting the open middle thirds of a set of line segments. One starts by deleting the open middle third $(\frac{1}{3}, \frac{2}{3})$ from the interval $[0, 1]$, leaving two line segments: $[0, \frac{1}{3}] ∪ [\frac{2}{3}, 1]$. Next, the open middle third of each of these remaining segments is deleted, leaving four line segments: $[0, \frac{1}{9}] ∪ [\frac{2}{9}, \frac{1}{3}] ∪ [\frac{2}{3}, \frac{7}{9}] ∪ [\frac{8}{9}, 1]$. This process is continued ad infinitum, where the nth set. The Cantor ternary set contains all points in the interval $[0, 1]$ that are not deleted at any step in this infinite process. An explicit formula for the Cantor set is $$C=[0,1]\bigcup_{m=1}^{\infty}\bigcup_{k=0}^{3^{m-1}-1}(\frac{3k+1}{3^m},\frac{3k+2}{3^m})$$
I know Maple just for doing some calculus and some basic modeling. I am asking if someone can note me a program in which we visualize the Cantor set. As I don't know if this job can be done in Maple, I added Mathematica in the title. Maybe its environment is more powerful than Maple in this question. Thanks.
AI: The Cantor $\frac13$ set is kind of hard to visualize, since it has infinitely many points but zero Lebesgue measure. The best visualization I could think of was to plot an approximation of its indicator function:
> f := a -> [seq(x/3, x in a), seq((2+x)/3, x in a)]:
> cantor := n -> (f@@n)([0,1]):
> delta2plot := points -> [seq(seq([x,y], y in [0,1,0]), x in points))]:
> plots:-listplot(delta2plot(cantor(5)), color=gray, view=[0..1,0..1]);
$\hspace{110px}$ |
H: Conformal mappings:general tips
Has anyone any general tips for finding conformal tranformations from a domain to another?
For example from $D=\{ z=x+iy | x^2+y^2<1, x^2-x+y^2>0 \}$ to the unit disc. This set is a disc
minus a smaller disc, so what should i imagine to try? If someone could provide his reasoning
step by step telling what ideas make him use certain transformations (also providing some variants,
for example: 'i'd do this because, while if there had been this i would have done that') i
would be very grateful, because i don't seem to get a general idea of what is happening. Thanks
AI: Looks like the disks are internally tangent at $z=1$. This looks like a special point for this domain. It's a good idea to map the special point to $\infty$ by a fractional linear transformation. This makes both circles a special kind of circles: namely, lines. And the domain becomes a strip. |
H: Is it unlikely to get the same number of heads/tails?
A question in probability by a non-mathematician:
A fair coin is tossed $2N$ times. Is it unlikely that we get exactly $N$ heads and $N$ tails?
From one side, this must be the most likely result! But intuitively, if someone reports to me that they threw 2,000,000 coins and got exactly 1,000,000 heads and 1,000,000 tails I would be a little suspicious.
If this really is suspicious, how does it compare to biased results? Obviously, 2,000,000 heads is a much more suspicious result. But, for what value of $K$ is $N+K$ heads and $N-K$ tails as suspicious as $N$ heads and $N$ tails?
Equations are very welcome, but please try to provide some aid in understanding them.
ADDED: Henning Makholm's answer helped me understand better what I want to ask: I always assume a fair coin and would like to compare the probability of the split N/N with the probability of getting either more than more than N+K or less than N-K heads. By ncmathsadist's answer , the probability for the N/N split goes like 1 over the square root of N. What about the probability to get more than either more than N+K or less than N-K heads? For what value of K (approximately) does it equal to 1/sqrt(n pi)?
CLARIFICATION: If f(N) is the probability to get an N/N split in 2N tosses of a fair coin, and g(N,K) is the probability to get more that N+K heads or less than N-K heads in 2N tosses of a fair coin, then I am asking for a formula which, given N, approximates the values of K for each f(N) is closest to g(N,K). This is a formula that takes N and returns K so we can call it h(N)=K. Which kind of function is h? My guess from a little experementing with the online software in Brian M. Scott's answer is that the function h is apporximately the SQRT function, multiplied by some constant, or something like that.
AI: It is fairly unlikely if $n$ is large. The probability of $n$ heads and $n$ tails is
$${2n\choose n} {1\over 4^n}.$$
We have $${2n\choose n} = {(2n)!\over n!n!}.$$
Stirling's formula states that as $n$ becomes large,
$$n! \sim {n^ne^{-n}\sqrt{2\pi n}},$$
where we say $a_n\sim b_n$ if $a_n/b_n \to 1$ as $n\to\infty$.
Using this formula,
$${2n\choose n}{1\over 4^n}\sim {1\over 4^n}\left((2n)^{2n}e^{-2n} \sqrt{4\pi n}\right)
\left(1 \over\sqrt{2\pi n} n^ne^{-n}\right)^2
$$
Cancel the exponentials and the $n^n$s and we get
$${2n\choose n}{1\over 4^n}\sim {2^{2n}\over 4^n}{2\sqrt{\pi n}\over 2\pi n }
= {1\over \sqrt{n\pi}}.$$
So you can see that this probability decays like $1/\sqrt{n\pi}$ as $n$ gets large.
So if $n$ gets 100 times larger, this probability diminishes by a factor of 10. |
H: Circular Rotation
I'm trying to emulate planetary rotation. I have a 'planet' rotating around the 'sun', but when trying to rotate the 'moon' around the 'planet', the motion is skewed.
If I stop the motion of the 'planet', the rotation is fine. I think this is just outside of my understanding.
I have demo at http://jsfiddle.net/btW7j/
AI: Wow, that's a pretty neat effect! What's happening is this:
First, you rotate the moon around the planet by some angle, say, $\alpha$.
Then you rotate the planet around the sun by some angle, say, $\beta$.
Now, you try to rotate the moon around the planet by angle $\alpha$ again... but the planet has moved in the mean time!
Thus, you get the effect seen in your demo: the absolute velocity of the moon is proportional to its distance from the planet, so when the moon is close, it's moving slower than the planet and so lags behind. But that puts it further away from the planet, and so makes it move faster, so it catches up again — but then its orbit takes it in front of the planet, so that their trajectories converge and they move closer again. Thus, you end up with these spirograph-like trajectories.
Varying the ratio $\alpha/\beta$ changes the trajectory; $\alpha/\beta > 1$ gives effects similar to what you observed, while if $\alpha/\beta = 1$, the moon escapes to infinity and never comes back. For $0 < \alpha/\beta < 1$, the planet actually catches up to the moon after completing one extra rotation around the sun, while $\alpha/\beta < 0$ causes the moon to dip inwards closer to the sun (and, for $-1 < \alpha/\beta < 0$, to trace neat quasi-polygonal patterns as it does so).
Anyway, the fix is pretty simple: after rotating the planet around the sun, also rotate the moon around the sun by the same angle (jsFiddle):
function do_rotation() {
moon.attr(rotate(get_point(moon), get_point(earth), 5));
earth.attr(rotate(get_point(earth), get_point(sun), 1));
moon.attr(rotate(get_point(moon), get_point(sun), 1)); // <- NEW
setTimeout(do_rotation, 1000 / 60);
}
That way, the planet and the moon stay at the same distance from each other. As a side effect, the rotation speed of the moon around the planet, from a non-rotating viewpoint, becomes $\alpha + \beta$ instead of just $\alpha$. If you don't want that, just subtract the rotation angle of the planet from that of the moon. |
H: Showing that an entire function is a polynomial
Let $f(z)$ be an entire function, $R_n$ a sequence of positive real numbers tending to $\infty$ such that $f(z) \neq 0$ on $|z|=R_n$ and there exists $M>0$ such that
$$\int_{|z|=R_n} \left|\frac{f'(z)}{f(z)}\right| ~dz<M$$
for all $n$. Show that $f$ is a polynomial.
What came to my mind is to consider that $f(z)=a_0+a_1z+\cdots \;\;\forall z\in\mathbb{C}$, and to try proving that $a_k=0$ from a certain $k$, maybe using the Cauchy formula for these coefficients, but I can't use the hypotesis on that bounded integral. Is observing that there is a logarithmic derivative of any use?
AI: As $\,\displaystyle{\frac{1}{2\pi i}\oint \frac{f'}{f}\,dz}\,$ is the number of roots inside the circle, the total number of roots is finite
(bounded by $\,\frac{M}{2\pi}\,$). Let
$$g(z)=\frac{f(z)}{(z-a_1)\cdot\ldots\cdot(z-a_k)}$$, where $a_1,\dots,a_k$ are the roots of $f$. Then $g$ satisfies the same condition as $f$ (with a different $\tilde M$). As $g$ has no root, it is of the form $g(z)=\exp(h(z))$ for some entire function $h$. We thus have $\,\displaystyle{\oint |h'(z)|\,|dz|<\tilde M}\,$ for circles of radii $R_n\to\infty$. This implies $h'=0$, hence $f$ is a polynomial.
edit: why $h'=0$: if
$$h'(z)=c_1 z^m+c_2 z^{m+1}+\dots$$ ($c_1\neq0$) then
$$\oint \frac{h'(z)}{z^{m+1}}\,dz=2\pi i c_1$$, which certainly implies $\oint |h'(z)|\,|dz|\to\infty$. |
H: Why is decomposing rational functions by assigning selected numerical values to x mathematically consistent?
Possible Duplicate:
How does partial fraction decomposition avoid division by zero?
Say you have the rational function:
$\frac{x^2 + 1}{(x-1)(x-2)(x-3)}$
This means that the function is undefined when x is equal to 1, 2, or 3.
Then to decompose it, you can equate that function to:
$\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$
If you clear the fraction then you will get:
x^2 + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
Then if you let x = 1,2,3 you can find A,B, and C but why are you even allowed to do that? From the beginning don't we define the domain of the function to be all real numbers besides 1,2, and 3? So why can we can go against the domain of the function to solve for the coefficients ?
AI: Once we are looking at the equation $x^2 + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ , we are no longer solving the initial problem -- rather, we are ascertaining that the polynomials on the left hand and right hand sides are equal for all values of $x$, and as such we can figure out what the values of $A$, $B$, and $C$ are.
Traditionally, to the best of my knowledge, this is done by simply expanding out the polynomial on the right hand side, then getting three linear equations in three unknowns between the coefficients of $x^2$, $x$, and the constants $(x^0)$. However, because polynomials are perfectly nice and continuous, etc., and because a degree $n$ polynomial can always be precisely specified by $n+1$ points with different $x$ values, then the values you find by constraining the above quadratic with 3 points will be the same as if you used the 'traditional' method, in every case. Is this sufficient? |
H: Is there more than one infinitessimal among the hyperreal numbers
Take $\mathbb{H}=\mathbb{R}^\mathbb{N}/\mathcal{U}$, where $\mathcal{U}$ is some ultrafilter. Questions:
Are there more than one independent infinitessimal in this field. This means $\epsilon_1 > 0$ and $\epsilon_2 > 0$ such that $0 < \epsilon_1 < \epsilon_2 < r$ for all $0 < r \in \mathbb{R}$, and there is no relation (lineair or algebraic) between them ?
If there are, are there examples ?
Notes:
This is like asking for an "hyperlarge number comparable to the ordinal $\omega_n$ with $n>1$", I only wish to know if they can be constructed in $\mathbb{H}$.
I suspect you can proof the existence of more infinitessimals in $\mathbb{H}$ by roughly the same diagonal argument which is used to proof the uncountability of $\mathbb{R}$, but I can't lay my finger on it, and it doesn't lead to a construction (I am not one of the constructionalist. but in this case I wish to know if you can do more than an existence result).
Edit:
I thought of introducing "a second infinitessimal" to $\mathbb{H}=\mathbb{R}^\mathbb{N}/\mathcal{U}$ by looking at $(\mathbb{R}^\mathbb{N}/\mathcal{U})^\mathbb{N}/\mathcal{V}$, with a second ultrafilter, but (aside from the seond ultrafilter), I don't know in wich swamp I am moving then. Any thoughts on this would be appreciated.
AI: Let $$\epsilon_2=\left\langle\frac1{n+1}:n\in\omega\right\rangle^{\mathscr{U}}$$ and $$\epsilon_1=\left\langle\frac1{n!}:n\in\omega\right\rangle^{\mathscr{U}}\;;$$ clearly $a\epsilon_1^b<\epsilon_2$ for any $a,b\in\Bbb R^+$. Is that sufficient independence? |
H: Inequality between chromatic number and number of edges of a graph
I have not been able to find a proof to the statement that if a graph $G$ has $\chi(G)=k$, then it must have at least $\binom{k}{2}$ edges. Would you be able to show me a simple proof?
AI: Suppose we have a $k$-coloring of our graph, with $\chi(G)=k$. Then for any two colors, call them red and blue, there must be some edge that connects them; if there weren't, we could paint every red vertex blue, and we would have a $(k-1)$-coloring of our graph. There are $\binom{k}{2}$ pairs of colors, and any given edge cannot connect more than $2$ colors, so there must be at least $\binom{k}{2}$ edges in our graph. |
H: invariance of cross product under coordinates rotation
Question goes as
If $\vec A$ and $\vec B$ are invariant under rotation, the prove that $ \vec A \times \vec B $ is also invariant.
However solution of on the other page is not given. Says that if you replace A and B with A' and B' and i,j,k with i', j', k' you will get the desired result.
I tried to change $ (A_2 B_3 - A_3 B_2 ) \hat i$ into $ A' $ and $ B'$ but I am getting zero.I would like to have a hint on this.
AI: I'm not exactly sure what you are asking, however the following may be useful as a less verbose way of obtaining the same result.
The key fact is that the cross product of $A,B$ is the unique element $A\times B$ such that
$\langle x, A\times B \rangle = \det \begin{bmatrix} A & B & x\end{bmatrix}$, $\forall x$.
Let $Q$ be a rotation (ie, $Q^TQ = I$), then using the properties of $\det$ we have
$$\det \begin{bmatrix} A & B & x\end{bmatrix} =\det Q^T Q \det \begin{bmatrix} A & B & x\end{bmatrix} = \det Q \det \begin{bmatrix} QA & QB & Qx\end{bmatrix},$$
from which we obtain $\langle x, A\times B \rangle = \det Q \langle Qx, QA\times QB \rangle = \langle x, (\det Q) Q^T(QA\times QB) \rangle$.
Since this is true for all $x$, we have $A\times B = (\det Q) Q^T(QA\times QB)$,
or
$$Q(A\times B) = (\det Q) QA\times QB.$$
Thus, if $Q$ is a proper rotation ($\det Q = +1$), you have $Q(A\times B) = QA\times QB$ (ie, the cross product is invariant under proper rotations).
Your question posits that both $A,B$ are invariant under rotation (which seems like a fairly restrictive condition), in which case $A = QA$, $B= QB$, from which it follows that $Q(A\times B) = A\times B$, ie, $A\times B$ is invariant under rotation too (assuming a proper rotation, of course). |
H: Value of $n$ for which an improper integral is convergent.
A question from the Calculus book that I'm self-studying is asking me to determine the value of $n$ for which the improper integral below is convergent:
$$\int_1^{+\infty}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$
My attempt is below:
Using the definition of improper integral:
$$\lim_{b\to+\infty} \int_1^{b}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$
I will first solve the indefinite integral:
$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = \int \frac{n}{x+1}dx-\int \frac{3x}{2x^2+n}dx$$
For the first integral, $\int \frac{n}{x+1}dx=n\ln (x+1) + \text{constant}$. For the second integral, substituting $u = 2x^2+n$ (and therefore $dx=\frac{du}{4x}$), $\int \frac{3x}{2x^2+n}dx=\frac{3}{4}\int \frac{1}{u}du=\frac{3}{4}\ln(2x^2+n) + \text{constant}$. So, the result for the whole integral is:
$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = n\ln(x+1) - \frac{3}{4}\ln(2x^2+n) + \text{constant} = \ln\left(\frac{(x+1)^n}{(2x^2+n)^{3/4}}\right) + \text{constant}$$
Thus, the value of the definite integral from $x = 1$ to $x = b$ is:
$$\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{(2)^n}{(2+n)^{3/4}}\right)$$
Thus, the value of the improper integral is:
$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)\right]$$
For this limit to exist, it only depends on the limit of the term $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$, because the term $\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)$ is a constant. To calculate the limit of $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$ as $b\to +\infty$, I suppose we can neglect the terms $1$ in $(b+1)^n$ and $n$ in $(2b^2+n)^{3/4}$, because they become negligible as $b$ gets larger. Thus we get, for the limit of this term:
$$\lim_{b\to+\infty} \ln\left(\frac{(b)^n}{(2b^2)^{3/4}}\right)$$
For the above limit to exist, it seems that $n = 3/2$ is the only possible value, because $\lim_{b\to+\infty} \ln\left(\dfrac{(b)^{3/2}}{(2b^2)^{3/4}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{b^{3/2}}{2^{3/4}b^{3/2}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{1}{2^{3/4}}\right)$. If any different value for $n$ were chosen, the logarithm would approach $-\infty$ or $+\infty$, and therefore the limit would not exist.
For $n = 3/2$, the value of the limit is:
$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b)^{3/2}}{(2b^2)^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(2+3/2)^{3/4}}\right)\right]=\lim_{b\to+\infty} \left[\ln\left(\frac{1}{2^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(7/2)^{3/4}}\right)\right]=\frac{3}{4}\ln\frac{7}{16}$$
This is equal to the answer given by the book. So, this informal argument of neglecting the terms $1$ and $n$ worked.
But I would like to ask if the reasoning above is correct, and if this informal way of finding the value of $n$ is good. Is there a more formal way?
Edit: A more formal way of showing that $n$ should be $3/2$ is suggested by David Mitra in the comments to this question below:
$$\lim_{b\to+\infty} \ln \left( {(b+1)^n\over (2b^2+n)^{3/4} } \right) = \lim_{b\to+\infty} \ln \left( {b^n\over b^{3/2}} \cdot {\bigl(1+{1\over b}\bigr)^n\over \bigl(2 +{n\over b^2}\bigr)^{3/4}} \right)$$
It is clear from above that $n$ should equal $3/2$. If it were a different value, the limit would not exist, because the expression inside the logarithm would approach either $0$ (causing the logarithm to approach $-\infty$) or $+\infty$ (causing the logarithm to tend to $+\infty$).
AI: Your method is correct and it is easy to formalize your neglecting of minor terms.
We can see that $$\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right) - \ln\left( \frac{b^n}{(2b^2)^{3/4}}\right)= \ln \left( (1+1/b)^n (1+n/(2b^2))^{-3/4} \right) \to \ln 1=0$$
as $b\to \infty.$ So the limit of the first term equals the limit of the second term if either limit exists.
However, this is probably not the best way to do this problem. The point of these types of exercises is to get you used to estimating the value of an integral, by doing something to get a feel for the growth of the integrand. Often we can't explicitly integrate the term.
Here, if you put the terms on a common denominator you get $$ \frac{n^2 + 2nx^2 - 3x(x+1)}{(x+1)(2x^2+n)}.$$
The critical fact used often is: $\displaystyle \int^{\infty}_1 \frac{1}{x^a} dx $ converges only for $a>1.$
So the $\frac{n^2}{(x+1)(2x^2+n)}$ part clearly converges. So we only need to worry about $$\frac{2nx^2 - 3x(x+1) }{(x+1)(2x^2+n)}.$$ If $n=3/2$ then $2nx^2=3x^2$ is able to cancel with the $-3x^2$ term in the next term, so we have a linear term in the top, cubic denominator, so behaviour like $1/x^2$ which is convergent. But if not, then we always get some quadratic term in the numerator, so the behaviour is like some constant term times $1/x$, which diverges. It is instructive to formalize these ideas with some inequalities, which I will leave you to try. |
H: r-colorable simple graph
What $r$ colorable simple graph on $n$ vertices has the most edges? Is there a unique such graph? I am told this has something to do with Turan's theorem...
(This is the progress I have made- deleting a vertex $u$ and replacing it with a copy of a vertex $v$ preserves the number of vertices and $r$ colorability. When does, however, increase the number of edges? I also think that the same graph should work for the graph with the most edges and no $K_{r+1}$ subgraph.
A full fledged solution would be fantastic.
AI: Hint: The maximal $K_{r+1}$-free graph happens to be $r$-colorable.
More on the hint: Turán's theorem bounds the number of edges that a $K_{r+1}$-free graph can have. The tight example is $r$-colorable. Since $K_{r+1}$ isn't $r$-colorable, Turán's theorem answers your question.
Another way to solve this uses David's method. Let $n_1,\ldots,n_r$ be the sizes of the color classes. We can assume that all edges connecting different color classes are there. The only edges not there are the within-class edges, which number
$$ \binom{n_1}{2} + \cdots + \binom{n_r}{2} = \frac{n_1^2 + \cdots + n_r^2}{2} - \frac{n}{2}. $$
Since $n$ is constant, we might as well minimize $n_1^2 + \cdots + n_r^2$. Consider any two $n_1,n_2$ satisfying $n_2 - n_1 \geq 2$. If we increase $n_1$ by one and decrease $n_2$ by one then the new sum of squares is
$$ (n_1+1)^2 + (n_2-1)^2 = n_1^2 + n_2^2 + 2(n_1 - n_2 + 1) < n_1^2 + n_2^2. $$
This shows that the optimum is achieved for a setting in which $|n_i - n_j| \leq 1$ for all $i,j$. Thus for some $m$ and $k$, wlog
$$n_1 = \cdots = n_k = m+1, \, n_{k+1} = \cdots = n_r = m.$$
Since the total number of vertices is $n$, we must have
$$ n = k(m+1) + (r-k)m = rm + k. $$
We can assume that $0 \leq k < r$ (i.e. $n_r = m$), hence $k = n \mod{r}$ and $m = \lfloor n/r \rfloor$. |
H: Multiple variables for a logical expression?
I wanted to know if what I did is even on the correct path for how this question is worded. How can you have two variables when it's dealing with a single unhappy person? I'm guessing the third way will just be using De Morgan’s Laws for quantifiers?
Translate each of these statements into logical expressions in three different ways by varying the domain and by using predicates with one and with two variables.
There is a person in your school who is not happy.
My first solution:
S(x) statement “x is in my school” and H(x) statement “x is happy”
∃x(S(x) ∧ ￢H(x))
AI: You might want to say that $\exists x\in \hbox{Human Race}$ s.t. $S(x)\wedge \sim H(x)$. What you have is basically correct. Just bound the $x$ in some set to give context. |
H: On $T_2$, first countable, countably compact space
As we know,
For every $T_2$, first countable, compact space, its cardinality is not more than $2^\omega$. (See chapter 3 of Engelking's book.)
However, I want to know whether the result is same for the $T_2$, first countable, countably compact space, i.e., for every $T_2$, first countable, countably compact space, its cardinality is also not more than $2^\omega$?
Thanks for any help:)
AI: For any uncountable regular cardinal $\kappa$ consider the subspace $S=\{\alpha\in \kappa: cof(\alpha)=\omega\}$ with the order topology. $S$ has cardinality $\kappa$ and it is easy to see that satisfies all your requirements. |
H: Am I too young to learn more advanced math and get a teacher?
I am still 15 years old, but I am very interested in pure math. I have been teaching myself though books, from the internet and from others for the past year or so. I haven't mastered all the topics that are covered in university, just the ones that happen to interest me (elements of differential and integral calculus, complex analysis, etc. You can see what I am interested in by looking at the questions I've asked and answered).
Now, a few months ago, back in school, I asked my math teacher for help on a differential calculus question whose solution I did not understand. I was told by this teacher that I should not be doing calculus and I should wait until I learn it in school. Other math teachers either did not understand what I was asking or shared the same view as my math teacher. For awhile this had distressed me very much, because some of my own math teachers were telling me to stop learning math and to wait three or four years to continue! Should I stop learning math by myself? I decided that I would keep going, because this is a hobby and interest of mine and I didn't think teachers should have the right to stop me from learning.
I find it more and more difficult to proceed learning on my own without a mentor who can and will help me, and I don't know what to do. I went to my school's math club, but alas, no one there was that interested in doing math for fun like me, and no one was interested in answering or helping me with my questions. This website has proven very helpful to me, however, it is not like talking ans asking a person face-to-face.
What should I do? Am I learning math too early? Should I wait until university to continue learning calculus? If not, how should I get a teacher or continue to learn on my own?
AI: I was recently in a similar situation. After finishing precalculus at my high school, when I was 15 I started taking calculus at my local university and studying higher mathematics on my own (out of the book "Modern Algebra: An Introduction" by John Durbin, which in retrospect seems laughably basic but at the time blew my mind). Three years later, I can say without a doubt that it is the best decision I've ever made. I ended up learning mathematics through a combination of taking classes at university, talking with students/professors, reading textbooks, and using this site. I did have one major advantage over you though, as my parents are both professors (although neither of them math professors) which made it easier for me to get into classes. However, I know of other people doing the same thing without any connection to the university. Here are some things I would recommend based on my experience:
Get an introductory textbook for some relatively advanced subject, such as Calculus, Linear Algebra, or Abstract Algebra. Read reviews online before choosing one to find one that is both rigorous and easy enough for beginners. I'd recommend Spivak for Calculus (take this with a grain of salt though, as I never read it but have heard good things about it) or Durbin for Abstract Algebra. Make sure it comes with plenty of exercises, and DO THEM. If you don't know how to do a problem, or if you've done it correctly, ask here!
If you have a university nearby, take advantage of it. Email a professor teaching an upcoming introductory course and explain your situation to him/her, and ask if you can sit in on the class. You might even be able to enroll in classes as a non-degree-seeking student, if the university allows this (most do) and you can afford it (if it's a state school, the tuition for a single course might not be too bad). Don't be afraid to talk about math with professors. It can be intimidating, but remember, these people have dedicated their lives to math. Almost uniformly, they LOVE it. Half of the time I had to find a way to break off a conversation with a professor because they were so engrossed in the math at hand.
Find something specific you don't understand. It may be a theorem, a proof, a concept, or even an unsolved problem, so long as it fascinates you. Figure out what you need to know in order to understand it, and start down the rabbit hole. The experience of coming to understand something like this can be very rewarding in addition to teaching you a great deal of mathematics. I've had several of these over the past few years, most recently an unsolved problem known as the Triangular Billiards Conjecture which I'm studying right now.
If you have any questions about my experience, feel free to ask. Good luck! |
H: When standard deviation is unknown?
I am reading a book about Statistics and I have encountered a text line:
...except in the case where standard deviation of the basic set is unknown...
I am not really sure what it means, could you please help me. For what types of data can't we define the standard deviation?
AI: What they are talking about is the value of the standard deviation. When the value is assumed to be known you can divide the sample mean by it and if the samples have a normal distribution the sample mean minus the population mean divided by the "known" standard deviation divided by the square root of the sample size n has a standard normal distribution (a normal distribution with mean 0 and variance 1). This fact is then used to construct confidence intervals for the population mean based on teh standard normal distribution.
However in most prectical situations the value of the standard deviation is unknown and we use a sample estimate to scale since we can't use a known value. Under the same assumptions about the sample distribution the sample mean- the population mean divided by its estimated standard error has a t distribution with n-1 degrees of freedom. So when the population variance is unknown the t distribution is used to construct confidence intervals for the population mean. The standard error of the mean is the sample standard deviation divided by the square root of n.
I hope this makes it clear. Statistics text categorize the estimate of the mean into two case (1) standard deviation known where the standard normal is used and the case where the standard deviation is unknown and the t distribution is used. |
H: Determining the minimum number of pixels on the boundary of a circle drawn in discrete space
I am trying to draw a circle in discrete space (actual image pixel space). I have the center (x,y) and radius r of a circle that I am supposed to draw. The manner in which I draw this circle is the following:
Starting from the center position (x,y), I have a for loop over angles $\theta \in \{0,2\pi\}$. Lets say the angle is incremented by $\Delta\theta$ in every iteration. In each iteration, I calculate an x-deviation and a y-deviation based on,
$$\Delta x = r cos(\theta)\\ \Delta y = r sin(\theta).$$
The point on the circumference of the circle is then calculated as
$$x' = \text{round}(x + \Delta x)\\ y'= \text{round}(y + \Delta y).$$
This gives a location $(x', y')$ in discrete space at which I can color a pixel. How do I determine for a given radius, what is the minimum number of discrete "pixels" I will have along the circumference.
In other words, lets say if I have a radius of 10, then how many unique discrete points would I have along the boundary of the circle? Is this problem well defined? I know there is a pitfall here of what consists of a discrete circumference. I consider every connected pixel to be a circumference point.
AI: You should probably be using Bresenham's Circle Algorithm?
The "distance" between adjacent pixels is either 1 or $\sqrt{2}$, i.e., two adjacent pixels either share a row or column or they are diagonal. This limits how much of the circumference consecutive pixels can consume. So the number of pixels needed, $x$, to complete the circumference of a circles with a radius of $r$ pixels is bounded $\sqrt{2} \pi r \leq x \leq 2 \pi r$, or roughly $4.443 \times r \leq x \leq 6.283\times r$. This gives a first approximation, by which to check the later estimate.
To get more accuracy we see that, by symmetry, only the pixels drawn in the first octant (i.e., $\pi/4$) need to be computed; the remaining can be found by using the same offsets but in different directions. With each consecutive pixel in the first octant progressing upward, there are at least $\lfloor r \sin(\pi/4)\rfloor$ pixels. Multiply this by 8, and you get roughly $5.657\times r$, which is within the earlier bounds. |
H: Unit distance graph in $\mathbb{R}^2$
Suppose $G$ is the simple graph with vertex set $\mathbb{R}^2$ created by connecting two points iff they are distance one from each other in the plane. How can we prove that $4\le \chi(G)\le7$? I think I have seen this problem somewhere before but can't pinpoint it. No hints please.
AI: The fact that there is a $4$-chromatic unit distance graph implies that $\chi \geq 4$, while the appropriate colouring of a hexagonal tiling shows that $\chi \leq 7$. For details, here's the Wikipedia article. |
H: Difference between permutation and combination?
Permutation:
$$P(n,r) = \frac{n!}{(n-r)!}$$
Combination:
$$C(n,r) = \frac{n!}{(n-r)!r!}$$
Apparently, you use combination when the order doesn't matter. Great. I see how a combination will give you all the possible well, combinations. However, I don't see what exactly does a permutation do then.
AI: If you see how combinations work then you're most of the way there. Say I want to pick 3 letters out of ABCDE. There are $C(5,3)$ ways of doing this. But if order matters, then several things that I counted as the same are now different. Picking $ABC$ now generates $ABC,ACB,BAC,BCA,CBA,CAB$ as different choices, when they weren't before. How many different choices are there? Well that's the number of ways I can rearrange the $r$ chosen letters, which is $r!$. So if permutations matter:
$$P(n,r)=r!\cdot C(n,r)=\frac{r!n!}{(n-r)!r!}=\frac{n!}{(n-r)!}$$
Permutations are the number of different ordered selections of $r$ elements from a set of $n$. |
H: Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$
Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals.
Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$, how does one calculate the Krull dimension of $R$? This variety is well-known as the twisted cubic in $\mathbb{P}^3$.
Question 2. In general for any ring $R$, how are the Krull dimension of $R$ and the dimension of Spec$(R)$ related?
Thank you.
AI: Question 1: It is a theorem that $\mathrm{dim}\ A[x]=\mathrm{dim}\ A+1$ for any Noetherian ring $A$, where $\mathrm{dim}$ denotes Krull dimension. Thus $\mathrm{dim}\ \mathbb C[x_1,x_2,x_3,x_4]=4$, as $\mathrm{dim}\ \mathbb C=0$ trivially. The easiest way to compute the dimension of $R$ is to verify that
$$P=\langle x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\rangle$$
is a prime ideal, i.e. by the methods I employ here, and computing its height. As a caution, I have not proved that $P$ is prime myself but I believe that it is. To determine its height, notice that $\mathbb C[x_1,x_2,x_3,x_4]$ is a UFD so all prime ideals of height $1$ are principal, while $P$ is easily shown not to be principal. By the Generalized Principal Ideal Theorem, $P$ has height at most it's number of generators, i.e. $3$. Hence we need only determine whether $P$ has height $2$ or $3$. In fact, it has height $3$, which can be shown by verifying that
$$0\subset \langle x_1x_3-x_2^2\rangle\subset \langle x_1x_3-x_2^2,x_2 x_4-x_3^2\rangle\subset P$$
is a strict chain of prime ideals, e.g. by the methods in the post I linked to.
Question 2: Yes, the dimensions are equal. See for example Wikipedia.
Edit: Not doing calculations can get you in trouble. As Mariano's answer shows, $P$ has height $2$ rather than $3$ (so apparently the second ideal in the chain I wrote is not prime), hence the dimension of $R$ is $4-2=2$. |
H: Counting men and women around a circular table such that no 2 men are seated next to each other
Possible Duplicate:
Circular Permutation
I was looking through old homework to work on my counting skills, and I had this problem that I didn't get right.
Question: In how many ways can 4 men and 5 women be seated around a circular table so that no 2 men are seated next to one another?
My Solution: First we sit the women, and leave an open spot in between each to assure a man doesn’t sit next
to a man. There are 5! ways to do this. The there are 5 open spots, we only want to choose 4
of those for men to stay in, so $5 \choose 4$. Then there are 4! possible different ways to arrange the men 4 within those spaces. Finally, since the table is round, we want to make sure that rotating it doesn’t make a difference. We can rotate the table 9 times that will produce the same arrangement, thus divide by 9. With the product rule, we find that the total number of arrangements to be:
$$\frac{5!{5 \choose 4}4!}{9} = 1600$$
I believe I undercounted the total arrangements for this problem, but I am not sure where I went wrong.
AI: To begin with: I dont' think it's right to divide by 9. That would be right if you had counted as distinct configurations that differ only in rotations, but that's not the case. When you said "there are 5! ways to sit women..." you implied that you were counting only relative ordering of women, regardless of their "absolute" positions.
Furthermore: if you are counting different relative configuration of women (permutations), you must take into account that you are interested in ciclic permutations (the permutation 13245 is actually the same as 32451), so the number 5! is actually wrong. |
H: $f, g$ entire functions with $f^2 + g^2 \equiv 1 \implies \exists h $ entire with $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$
I am studying for a qualifier exam in complex analysis and right now I'm solving questions from old exams. I am trying to prove the following:
Prove that if $f$ and $g$ are entire functions such that $f(z)^2 + g(z)^2 = 1$ for all $z \in \mathbb{C}$, then there exists an entire function $h$ such that $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$.
My Attempt
The approach that occurred to me is the following. Since $f(z)^2 + g(z)^2 = 1$ then we have $(f(z) + ig(z))(f(z) - ig(z)) = 1$. Then each factor is nonvanishing everywhere in $\mathbb{C}$ and thus by the "holomorphic logarithm theorem" we know that since $\mathbb{C}$ is simply connected, there exists a holomorphic function $H:\mathbb{C} \to \mathbb{C}$ such that
$$e^{H(z)} = f(z) + ig(z)$$
and then we can write $\exp(H(z)) = \exp\left(i\dfrac{H(z)}{i} \right) = \exp(ih(z))$,
where $h(z) := \dfrac{H(z)}{i}$.
Thus so far we have an entire function $h(z)$ that satisfies
$$e^{ih(z)} = f(z) + ig(z)$$
On the other hand, we also know that $e^{iz} = \cos{z} + i \sin{z}$ for any $z \in \mathbb{C}$, thus we see that
$$e^{ih(z)} = \cos{(h(z))} + i \sin{(h(z))} = f(z) + ig(z)$$
Thus at this point I would like to conclude somehow that we must have $f(z) = \cos(h(z))$ and $g(z) = \sin(h(z))$, but I can't see how and if this is possible.
My questions
Is the approach I have outlined a correct way to proceed, and if so how can I finish my argument?
If my argument does not work, how can this be proved?
Thanks for any help.
AI: You approach appears to be correct, and it can be finished with the following thought: not only do complex exponentials split into combinations of trigonometric functions, but trig functions also split into combinations of complex exponentials. Indeed:
$$\cos\alpha=\frac{e^{i\alpha}+e^{-i\alpha}}{2},\quad \sin\alpha=\frac{e^{i\alpha}-e^{-i\alpha}}{2i}.$$
This is applicable for not just real $\alpha$, but complex as well. You've deduced $e^{ih(z)}=f(z)+ig(z)$ for some entire function $h$, and taking inverses gives $e^{-ih(z)}=f(z)-ig(z)$, so averaging these two will give you $\cos h(z)=f(z)$ (and similarly, $\sin h(z)=g(z)$). |
H: Multiple variable quantifications in english?
Problem 1 is really simple but 2 is confusing me a little. I'm going on the assumption that $x$ and $y$ are both students and they are saying $z$ is the class? Is it really as simple as what I wrote or am I missing something?
Let $C(x, y)$ mean that student $x$ is enrolled in class $y$, where the domain for $x$ consists of all students in your school and the domain for $y$ consists of all classes being given at your school. Express each of these statements by a simple English sentence.
1) C(Randy Goldberg, CS 252)
2) $\exists x\exists y\forall z((x\neq y) \land (C(x, z)\to C(y, z)))$
1) The student Randy Goldberg is enrolled in the class CS252.
2) All classes have multiple students.
Or should 2 be: Each class has two different students enrolled in it.
AI: Statement 2 means that, there is some distinct pair of students $x,y$ such that for any class $z$ if student $x$ is enrolled in class $z$ then student $y$ is enrolled in class $z$.
Edit: I read the quantifiers backwards. Fixed. |
H: Implicit differentiation question
Differentiate given
$$\frac{y}{x-y}=x^2+1$$
Initially I wanted to use the quotient rule to solve this, but then I tried differentiating it as it is:
$$\frac {y_\frac{dy}{dx}}{1-y_\frac{dy}{dx}}=2x$$
$$\frac{dy}{dx}(y y^{-1})=2x$$
$$\frac{dy}{dx}=\frac{2x}{yy^{-1}}$$
$$\frac{dy}{dx}=\frac{2xy}{y}$$
$$\frac{dy}{dx}=2x$$
I am wondering how I can check to see if this is a valid answer?
AI: Your equation is
$$\frac{y}{x-y}=x^2+1 $$
You claim that
$$y'=2x$$ so that
$y=x^2+C$
This means
$$\frac{x^2+C}{x-x^2-C}=x^2+1 $$
This is absurd, since the quotient of two second degree polynomials can't be a second degree polynomial. In fact you get two non vanishing terms $x^3$ and $x^4$ which are off.
I don't understand what your procedure is, also. I would proceed as follows:
$$\displaylines{
\frac{y}{{x - y}} = {x^2} + 1 \cr
\frac{d}{{dx}}\left( {\frac{y}{{x - y}}} \right) = \frac{d}{{dx}}\left( {{x^2} + 1} \right) \cr
\frac{{y'\left( {x - y} \right) - \left( {1 - y'} \right)y}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr
\frac{{y'x - yy' - y + yy'}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr
\frac{{y'x - y}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr
y'x = 2x{\left( {x - y} \right)^2} + y \cr
y' = 2{\left( {x - y} \right)^2} + \frac{y}{x} \cr} $$ |
H: Limit of a sequence of real numbers
If $(a_n), (b_n)$ are two sequences of real numbers so that $(a_n)\rightarrow a,\,\,(b_n)\rightarrow b$ with $a, b\in \mathbb{R}^+$. How to prove that $a_n^{b_n}\rightarrow a^b$ ?
AI: Since $a_n\to a$ and $a>0$ by assumption, we have $a_n>0$ for $n\geq N$ for some sufficiently large positive integer $N$. So we can just consider $\log a_n$ for $n\geq N$. Note that $\log$ is a continuous function, we have
$$\lim_{n\to\infty}(\log a_n)=\log(\lim_{n\to\infty} a_n)=\log a.$$
Therefore, we have
$$\log\Big(\lim_{n\to\infty} a_n^{b_n}\Big)=\lim_{n\to\infty}(\log a_n^{b_n})=\lim_{n\to\infty}(b_n\log a_n)=(\lim_{n\to\infty}b_n)(\lim_{n\to\infty}\log a_n)=b\log a=\log a^b,$$
which implies that (by taking exponential on both sides)
$$\lim_{n\to\infty} a_n^{b_n}=a^b$$
as required. |
H: proof for members of ideals
If $I$ is an ideal, could you show that if $ x\in I$ and $y\notin$ I, then $x+y \notin I$? It seems like an intuitively obvious statement and yet my rigor is failing me. So if you could show me all the steps of the proof that would be much appreciated.
AI: Assume for the sake of contradiction that $x\in I$, $y\not\in I$ and $x+y\in I$. Since $I$ is a subgroup of the additive group of the ring, we have that $y = (x+y) - x \in I$ which is a contradiction. As mentioned by Dylan Moreland, this is a general fact about subgroups of any group. |
H: Rules of Inference argument has multiple steps?
I need to write the rules of inference and explain which rules of inference are used for each step. This looks like it's just Modus Ponens from what I wrote the equations as, am I missing some steps or is this really that simple?
“There is someone in this class who has been to France. Everyone who goes to France visits the Louvre. Therefore, someone in this class has visited the Louvre.”
C(x) “x is in this class”, F(x) “x has been to France”, L(x) “x visited the Louvre”
∃x(C(x) ∧ F(x))
∀x(F(x)→L(x))
∃x(C(x) ∧ L(x))
AI: I ended up figuring it out and I think this is 100% correct for the answer. My logic could be incorrect thought since I'm still learning this stuff.
C(x) “x is in this class”, F(x) “x has been to France”, L(x) “x visited the Louvre”
1. ∃x(C(x) ∧ F(x))
2. C(x) ∧ F(x) existential instantiation
3. F(x) simplification from 2
4. ∀x(F(x)→L(x))
5. F(x)→L(x) universal instantiation
6. L(x) modus ponens from 3 and 5
7. C(x) simplification from 2
8. C(x) ∧ L(x) conjunction from 6 and 7
9. ∃x(C(x) ∧ L(x)) existential generalization |
H: Ring map-integers proof
Prove that there is a ring map from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$ iff $m|n$.
I am not able to prove this with the appropriate rigour, could someone show the steps for a proof?
AI: $\def\Z{\mathbb Z}$
First suppose that there is a ring morphism $\phi\colon\Z/n\Z \to \Z/m\Z$. Then $0= \phi(0) = \phi(n \cdot 1) = n \cdot \phi(1) = n$, hence $n + m\Z = 0$, which gives $m \mid n$.
If on the other hand $m \mid n$, define $\phi\colon \Z/n\Z \to \Z/m\Z$ by $\phi(a + n\Z) = a + m\Z$, this is well defined: If $a + n\Z = b+n\Z$, then $n \mid a-b$, hence $m \mid a-b$, so $a + m\Z = b+m\Z$. Obviously $\phi$ is a homomorphism. |
H: Exponentiation of reals
Let $1<b\in \mathbb{R}$ and $x\in \mathbb{R}$.
I want to prove that $\sup${$b^t\in \mathbb{R}$|$x≧t\in \mathbb{Q}$} = $\inf${$b^t\in \mathbb{R}$|$x≦t\in \mathbb{Q}$}.
I have proved that $\sup$≦$\inf$, but dont know how to show that $\inf$≦$\sup$..
AI: We know that $f(t) = b^t$ is continuous and monotonic. Take two sequences in $\mathbb{Q}$ that converge to $x$, say $(t_n)$ and $(s_n)$. Due to continuity, $\lim b^{t_n} = \lim b^{s_n} = b^x$. Then you only need to show that these limits are really $\inf$ and $\sup$. Use monotony for that. |
H: Rings, ideals and units-proof
Let $R$ be a ring and let $J$ be an ideal of $R$. Assume $J$ contains
a unit of $R$. Prove $J=R$, rigorously.
Again I feel here as if the statement is intuitive so I am not able to make it rigorous. If you could show all steps in the proof that would be nice.
AI: If $J$ contains a unit of $R$, say $a$ then we must have $a^{-1} \cdot a = 1$ being in $J$. However if $J$ contains $1$, recall that an ideal must "absorb" multiplication from the outside. In our case, this means in particular that for all $r\in R$, we have
$$r \cdot 1 = r \in J$$
so that $R \subseteq J$. Since a priori $J \subseteq R$ we have $J = R$. |
H: How to confirm if my explicit formula is right?
I have to determine an explicit formula for
$$a_n=5a_{n-1}+6a_{n-2}$$
Initial values are $$a_0=2\\a_1=-1\\n>=2$$
My answer is
$$a_n = \frac{1}{7}\cdot 6^n+\frac{13}{7}\cdot (-1)^n$$
Which I suspect is wrong. But, how to "test" it?
AI: You are right. All solutions of the recurrence have the shape $(A)6^n+(B)(-1)^n$, where $A$ and $B$ are constants. To determine the values of $A$ and $B$, you need to know the initial conditions.
You were told that $a_0=2$ and $a_1=-1$. That determines $A$ and $B$, which are correct. So your answer is completely right.
To verify that your answer is correct, you can (i) substitute your solution in the recurrence and (ii) check by substitution that the values you get for $n=0$ and $n=1$ match the given ones.
At the practical "informal check" level, you can calculate the first $4$ or $5$ terms from the recurrence, and check them against the values given by any formula you arrive at. If there is a match, your answer is very unlikely to be wrong. |
H: Rigorous definition of what it means to be the same group?
Consider the following groups: $(\mathbb{Z}_4,+)$, $(U_5,.)$, $(U_8,.)$ and the set of symmetries for a rhombus if I am not mistaken the first and last are equivalent. What other justifiable equivalencies and nonequivalencies are there and what does it mean rigorously to be in the same group in general?
AI: Two groups $G$ and $H$ are "the same" if there is an isomorphism between them, ie a map $f: G\to H$ which is bijective ans such that for all $g,h\in G$ it satisfies $f(gh) = f(g)f(h)$ (so here the multiplication on the left is inside $G$ and on the right it is inside $H$).
The reason that this is the condition we want is that in that case, anything we can say about $G$ which only uses that it is a group, we can transfer to $H$ via $f$ and vice versa.
The first two groups you mention are isomorphic, while the thirds is not isomorphic to the first two. What the symmetries of a rhombus are depends on whether you allow a square to be a rhombus. If you do not, then the group of symmetries of a rhombus is isomorphic to $U(8,\cdot)$ |
H: Ring map one-to one kernel-proof
How can we show that a ring map $f: R\rightarrow S$ is one-to-one iff $\ker(f)=\{0\}$? I have seen this for a while axiomatically so I am unsure of my rigor.
AI: $(\Rightarrow)$ If $f$ is injective, then $\text{ker}(f) = \{0\}$ since $f(0) = 0$ for all homomorphisms and injectivity.
$(\Leftarrow)$ Suppose $f$ is not injective. Then there exists $x \neq y$ in $R$ such that $f(x) = f(y)$. Since $x \neq y$, $x - y \neq 0$. Then $f(x - y) = f(x) - f(y) = 0$. Hence $x - y \in \text{ker}(f)$. So $\text{ker}(f) \neq \{0\}$. |
H: Prove that (X,Y) is bivariate normal if X is normal and Y conditionally on X is normal
$X$ has a normal distribution. The conditional distribution of another random variable $Y$ given that $X=x$ is a normal distribution with mean $ax+b$ and variance $t^2$, where $a$, $b$, and $t^2$ are constants. How can I prove that the joint distribution of $X$ and $Y$ is bivariate normal distribution?
AI: By definition,
$$f_{X,Y}(x,y)=f_X(x)\cdot f_{Y\mid X}(y\mid x)=\frac1{\sqrt{2\pi\sigma^2}}\mathrm e^{-(x-\mu)^2/(2\sigma^2)}\cdot\frac1{\sqrt{2\pi t^2}}\mathrm e^{-(y-ax-b)^2/(2t^2)}
$$
Hence the task is to solve for $M$ in $\mathbb R^2$ and $C$ a definite positive $2\times2$ matrix, the fact that for every $z=(x,y)^t$, one has the identity
$$
f_{X,Y}(z)=\frac1{2\pi\sqrt{\det C}}\mathrm e^{-\frac12(z-M)^tC^{-1}(z-M)}.
$$
Hints: note that the diagonal of $C^{-1}$ is $(1/\sigma^2+a^2/t^2,1/t^2)$ and that $M=(\mu,a\mu+b)^t$. The rest of the proof relies on some simple algebraic manipulations of a degree $2$ polynomial in $(x,y)$. |
H: Evaluating $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x$
I have to evaluate:
$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$
I can't get the right answer! So please help me out!
AI: Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi }{2}.$ Hence $I = \frac{\pi }{4}.$
In general, $ \displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) $ $dx$ whenever $f$ is integrable, and $\displaystyle\int_0^{\frac{\pi }{2}} \frac{\cos^a x}{\cos^a x + \sin^a x } dx = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sin^a x}{\cos^a x + \sin^a x } dx = \frac{\pi }{4}$ for $a>0$ (same trick.) |
H: Find the maximum of a linear function, given other linear equations with infinite solutions
$a_{1,1}x_1 + a_{1,2}x_2 + \dots + a_{1,20}x_{20}\leq b _1$
$a_{2,1}x_1 + a_{2,2}x_2 + \dots + a_{2,20}x_{20} \leq b_2$
$x_1 \geq 0, x_2 \geq 0, \dots, x_{20} \geq 0$
$f(x) = a_{3,1}x_1 + a_{3,2}x_2+ \dots + a_{3,20}x_{20} $
All $a$'s and $b$'s are known. There are infinite solutions. I want to find values for $x$'s that solve the linear equations that maximizes $f(x)$. Not sure how to go about this. Not sure if my title is a very good way to summarize the question.
AI: This is called Linear Programming problem, one famous method is Simplex Algorithm. For reference see here : http://en.wikipedia.org/wiki/Linear_programming |
H: Help deriving a vorticity equation
I am reading Majda & Bertozzi (Vorticity and Incompressible Flow). In page 12 the following equation appears:
$$\frac{D \Omega}{Dt} + \Omega \mathcal{D} + \mathcal{D} \Omega = \nu \Delta \Omega$$
where $\frac{D}{Dt}$ is the convective/lagrangian/material derivative. $\Omega$ and $\mathcal{D}$ are $3$ by $3$ matrices, the first antisymmetric and the second symmetric, and $\nu$ is a scalar. Using that $\Omega$ is defined by $\Omega h = \frac{1}{2} \omega \times h $, where $\omega$ is a vector function representing vorticity, I should be able to get the following vorticity equation (which apparently plays a crucial role in the rest of the book):
$$ \frac{D \omega}{Dt} = \mathcal{D} \omega + \nu \Delta \omega. $$
Any idea how?
Here is a link to the book
AI: Re-write the main given equation in index notation (following the Einstein summation convention)
$$ D_t \Omega_{ij} + \Omega_{ik}\mathcal{D}_{kj} + \mathcal{D}_{ik}\Omega_{kj} = \nu\triangle \Omega_{ij} \tag{1}$$
Small $\omega$ is defined by
$$ \Omega_{ik}h^k = \frac12 \epsilon_{ijk}\omega_j h^k \tag{2}$$
which is the cross product definition. The $\epsilon_{ijk}$ is the Levi-Civita symbol (or fully antisymmetric tensor with $\epsilon_{123} = 1$).
Plugging in (2) (which implies that $\Omega_{ij} = \frac12 \epsilon_{ikj}\omega_k$) into (1) we have that
$$ \epsilon_{ilj} D_t\omega_l + \epsilon_{ilk}\mathcal{D}_{kj}\omega_l + \mathcal{D}_{ik}\epsilon_{klj}\omega_l = \nu \epsilon_{ilj}\triangle \omega_l \tag{3}$$
Next we use the property of the Levi-Civita tensor,
$$ \epsilon_{jik}\epsilon_{jlk} = 2 \delta_{jl} \tag{4}$$
which means that multiplying (3) by $\epsilon_{imj}$ gives
$$ 2D_t\omega_m + \left(\epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} + \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik}\right) \omega_l = \nu \triangle \omega_m \tag{5}$$
The antisymmetry properties of the Levi-Civita tensor, as well as the symmetry of the tensor $\mathcal{D}$ can be used to show that
$$ \epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} = \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik} $$
So by another property of the Levi-Civita tensor,
$$ \epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} \tag{6}$$
we conclude that (5) is equivalent to
$$ D_t\omega_m + \omega_m (\delta_{kj}\mathcal{D}_{kj} - \mathcal{D}_{jm}) = \nu \triangle \Omega_m ~.$$
Which shows that you in fact omitted one necessary condition for your equation to hold, which is that $\mathcal{D}$, in addition to being symmetric, is also trace-free.
If you have learned about differential forms, one should treat $\Omega$ as a differential two form on $\mathbb{R}^3$ and $\omega$ as a differential one form on $\mathbb{R}^3$ related by the Hodge star operator $\Omega = *\omega$. From this point of view the equation you want (the one for $\omega$) is merely obtained by taking the Hodge dual of the equation you are given (the one for $\Omega$) plus a little bit of multilinear algebra. |
H: Construction of $\Bbb R$ from $\Bbb Q$
As it is true that we can construct all rational numbers $\Bbb Q$ from the set of integers $\Bbb Z$, is it possible to construct the set of real numbers $\Bbb R$ from $\Bbb Q$? If yes, how? Is there any procedure? And if no, is there any proof that we can't ? Thanks!
AI: Consider the following set of rational numbers:
$S =\{x \in \Bbb{Q}:x < 0, \text{ or } x^2 < 2\}$
This set is clearly bounded above, for example, $2$ is an upper bound. The thing is, the set has no rational least upper bound. In fact, it seems like we can get quite a good idea of "how big" such a least upper bound would have to be: if we call it "$x$", you can see that: $1.4 < x < 1.5$, for example.
So, the general idea is to capture all the "magnitudes" that rational numbers can approximate, but never hit "exactly". There are a few different ways to do this: Dedekind cuts, Cauchy sequences and "infinite decimals" being the most popular. All of these are designed to do one thing: ensure that the least upper bounds of sets like $S$ always exist (equivalently: ensure that (rational) Cauchy sequences converge "to something").
This process is called "completion", and can be thought of as "filling the gaps" in the rationals (so we get a continuum). It is, at the heart of it, a topological construction, and the need for it doesn't become clear until one starts to investigate continuity. For algebra (that is, solving polynomial equations with rational coefficients), algebraic numbers would suffice (things involving square, cube and n-th roots, and the like), but analysis of functions often requires we get an estimate of "the size of something". Intuitively, we want such "sizes" to be bona-fide numbers, that we can manipulate according to the familiar rules we learn early in life (that is, we want an ordered field, at the very least).
So, yes, there are several such constructions...what is fascinating is that each construction gives us "the same object" (only the names are changed, to protect the innocent). This justifies the real numbers being called "THE" complete ordered field (although to be 100% correct, one should say "complete archimedean ordered field", because there are complete ordered fields which are not (isomorphic to) the real numbers, such as the hyperreals). |
H: Ramification of primes
Let $K \subset L$ be two fields with ring of integers $\mathcal O_K$ and $\mathcal O_L$.
If a prime $p$ is totally ramified in $\mathcal O_K$, is it true that $p$ is also ramified in $\mathcal O_L$?
AI: If $p$ is (not necessarily totally) ramified in $K$ then there is a prime ideal $\mathfrak p$ of $\mathcal O_K$ s.t. $p \in \mathfrak p^2$. Let $\mathfrak P$ be any prime ideal of $\mathcal O_L$ lying over $\mathfrak p$. Then $p \in \mathfrak p^2 \subseteq \mathfrak P^2$, so $p$ is also ramified in $\mathcal O_L$.
In general, $(p) = p\mathcal O_K$ can be uniquely written as a product of prime ideals of $\mathcal O_K$:
$$(p) = \mathfrak p_1^{e_1} \cdots \mathfrak p_r^{e_r}.$$
The number $e(\mathfrak p_i | p) := e_i$ is called the ramification index of $\mathfrak p_i$ over $p$. Similarly, each $\mathfrak p_i$ can be written as a unique product of prime ideals of $\mathcal O_L$. Substitute this into the product of $\mathfrak p_i$'s to find
$$e(\mathfrak P | p) = e(\mathfrak P | \mathfrak p) \cdot e(\mathfrak p | p)$$
whenever $\mathfrak P \subset\mathcal O_L$ lies over $\mathfrak p$ and $\mathfrak p \subset \mathcal O_K$ lies over $p$. Now $p$ is ramified in $K$ iff there is a prime ideal $\mathfrak p $ of $K$ s.t. $e(\mathfrak p | p) > 1$. By the multiplicativity of the ramification index, this implies $e(\mathfrak P|p) > 1$ for a prime ideal $\mathfrak P$ of $L$ lying over $\mathfrak p$, so $p$ is also ramified in $L$. |
H: trivial Picard group
let $S=\operatorname{Spec}(A)$ be an affine scheme. For which ring $A$, not field is it known that $H^1(S,\mathcal{O}_S^{*})$ is trivial?
If $X\to S$ is a finite map and $H^1(S,\mathcal{O}_S^{*})$ is trivial, is it true that also $H^1(X,\mathcal{O}_X^{*})$ is trivial?
Thanks
AI: Sample answers to your first question: If $S$ is Spec of a local ring, or of a UFD, then $H^1(S, \mathcal O_S^{\times})$ is trivial.
The answer to your second question is no: $X = $Spec $\mathbb C[x,y]/(y^2 - x^3 +x) \to S =$ Spec $\mathbb C[x]$ gives a counterexample of a geometric nature, and $X =$Spec $\mathbb Z[\sqrt{-5}] \to S =$ Spec $\mathbb Z$ gives a counterexample of an arithmetic nature. |
H: $f$ is irreducible in $\Bbb F[x]$
Let $\Bbb F$ be a field of characteristic $p\gt 0$ and $f(x)=x^{p^n}-c \in\Bbb F[x]$ where $n$ is a positive integer. If $c \notin \{a^p:a\in \Bbb F \}$, show that $f$ is irreducible in $\Bbb F[x]$.
I recently started studying Field theory, so I don't know How to approach this problem? Any help would be appreciated.
AI: Let $\gamma\in \mathbb F^{alg}$ be a root of $f(x)=x^{p^n}-c \in\Bbb F[x]$ in an algebraic closure of $\mathbb F$.
Thus we have $f(x)=x^{p^n}-c=(x-\gamma)^{p^n}$.
On the other hand let $m(x)\in \Bbb F[x]$ be the minimal polynomial of $\gamma$ over $\Bbb F$.
If $n(x)$ is an irreducible monic factor of $f(x)$ over $\Bbb F$, its only root in $\mathbb F^{alg}$ is $\gamma$, so that $n(x)$ must also be the minimal polynomial of $\gamma$ in $\Bbb F[x]$. Hence $n(x)=m(x)$.
This means that the decomposition of $f(x)$ into irreducibles in $\mathbb F[x]$ is $f(x)=m(x)^{p^e}$ for some integer $e\geq 0$ [the exponent is a power of $p$ because it divides $p^n=deg (f(x))$].
More explicitly we have $f(x)=x^{p^n}-c =m(x)^{p^e}=(x^s+...+m_0)^{p^e}$.
But then $c=-m_0^{p^e}=(-m_0)^{p^e}$ and since $c$ is supposed not to be a $p$th-power in $\Bbb F$, this forces $e=0$ and so $f(x)=m(x)^{p^e}=m(x)$ is irreducible in $\Bbb F[x]$ |
H: The Betti numbers of a triangulated ball relative to a disk in its boundary
Fix $n \geq 1$ and let $B$ denote a triangulated closed $n$-ball. Let $D$ be a subset of the boundary of $B^n$ that is homeomorphic to the closed $(n-1)$-ball and such that is properly triangulated by the same triangulation as well.
I would like to compute the Betti numbers of $B$ relative to $D$. I have tried to regard the long exact sequence on homology of the chain complexes, and that the $0$-th homology is trivial as it is isomorphic to the $0$-th homology group of the de-Rham complex. Is it correct that all Betti number of $B$ relative to $D$ vanish?
I am not used to algebraic topology, and need the result only in a secondary (if even that) context. I could use a reference which clarifies under which circumstances to use the long exact sequence on homology in the context of simplicial complexes.
AI: Yes, you can use the long exact sequence of relative homology. In fact, both the ball and the disk are contractible so they only have nonzero homology in degree 0, and there the inclusion is an isomorphism on homology. Examining the long exact sequence, this is enough to conclude that the relative homology is always zero. (So the Betti numbers all vanish.)
You can also see this by noting that there is a homotopy equivalence between the pair (Ball,disk) and (Disk, Disk) essentially by crushing the non-disk parts of the ball down onto the boundary disk. Then your statement holds by the homotopy invariance of homology. |
H: Proof of $R/I$ integral over $S/(S \cap I)$
Can you tell me if my reasoning is correct?
I want to prove if $S \subset R$ are rings and $R$ is integral over $S$ and $I$ is an ideal of $R$ then $R/I$ is integral over $S/ (S\cap I)$.
Let $R$ be integral over $S$. $(S \cap I) \subset I$ is an ideal of $S$ and hence of $R$.
Since $R$ is integral over $S$ we have that $R/(S \cap I)$ is integral over $S/(S \cap I)$ and since $(S \cap I) \subset I$ we have $R/I \subset R/(S\cap I)$ and hence $R/I$ is integral over $S/(S\cap I)$.
AI: Let $r+I\in R/I$. Find a monic polynomial $f(x)\in S[x]$ that $r$ satisfies.
It's natural then to look at the image of $f(x)$ in $S/(S\cap I)[x]$ to see if it works for $r+I$!
Here is the viewpoint from homomorphisms' perspective:
If $R$ is integral over $S$ and $\phi:R\rightarrow T$ is a ring homomorphism, then $\mathrm{Im}(R)$ is integral over $\mathrm{Im}(S)$. |
H: How do I calculate a change of coordinates given two lines as new axis?
Suppose we have an old coordinate system using the variables $x$ and $y$. We are given two equations for lines to form the axis for new coordinates. e.g. The line $z=0$ is given by the equation $y = m_1 x + b_1$, and the line $w=0$ is given by the equation $y = m_2 x + b_2$. Now suppose we are given a point $(x_3, y_3)$. What are the $w$ and $z$ coordinates of this point?
AI: Let $\bf p$ be the intersection point of the lines, and let $\bf u$ and $\bf v$ be unit direction vectors associated to the two lines. The first part of the transformation should translate $\bf p$ to $\bf 0$; so wlog assume $\bf p=0$.
We want to write $\mathbf{x}=a\mathbf{u}+b\mathbf{v}$; to do this, form the matrix $\mathrm{U}=[\bf u~v]$ (we assume vectors are column vectors), so we can write $\mathbf{x}=\mathrm{U}[a~b]^T$ and hence we solve for $a,b$ via $[a~b]^T=\mathrm{U}^{-1}\bf x$. Then $\bf x$ in the old coordinates corresponds to $[a~b]^T$ in the new coordinates.
Two changes-of-coordinates of the sort we want are related by a third coordinate change taking place in the second coordinates. This new coordinate change preserves the origin and two axes, and therefore is precisely a rescaling of the $a$- and $b$-axes individually. |
H: Irreducible factors of $X^p-1$ in $(\mathbb{Z}/q \mathbb{Z})[X]$
Is it possible to determine how many irreducible factors has $X^p-1$ in the polynomial ring $(\mathbb{Z}/q \mathbb{Z})[X]$ has and maybe even the degrees of the irreducible factors? (Here $p,q$ are primes with $\gcd(p,q)=1$.)
AI: It has one factor of degree $1$, namely $x-1$. All the remaining factors have the same degree, namely the order of $q$ in the multiplicative group $(\mathbb{Z}/p \mathbb{Z})^*$. To see it: this is the length of every orbit of the action of the Frobenius $a\mapsto a^q$ on the set of the roots of $(x^p-1)/(x-1)$ in the algebraic closure. |
H: Decreasing functions
Let $u:[a,b]\to \mathbb{R}$ be a continuous and a. e. differentiable function (with respect to the Lebesgue measure).
Is it true that $u' < 0$ a. e. in $[a,b]$ implies $u$ strictly decreasing everywhere in $[a,b]$?
(New question added on 12/21/2012)
I know the answer is negative (thanks to Jonas and Cameron).
But, what appens if $u$ is absolutely continuous in $[a,b]$?
In other words, is it true that $u^\prime \leq 0$ implies $u$ decreasing in $[a,b]$ when $u$ lies in the Sobolev space $W^{1,1}(a,b)$?
AI: Hint: Cantor-Lebesgue function minus $x$. |
H: How can I disprove the following statements?
1) $ a \neq b , a^0 = b^0, a = b$
Usually we do $ a^n = b^n \implies a= b, $ can't we do it when $n$ is zero?
2) $ i = \sqrt{ -1 \over 1} = \sqrt{1 \over -1 } = {1 \over i} \implies -1 = 1$
What's wrong with the above statements?
AI: What you're doing is basically this.
$$(-2)^2=4$$
$$2^2=4$$
$$\therefore -2=2$$
both $i$ and $-i$ are square roots of $-1$. Since squaring isn't a one to one function, this doesn't imply that the numbers inside the square roots are equal. This also applies to your first question. The general pattern is
$$f(a)=f(b)\implies a=b$$
only if $f(a)$ is one to one. $f(a)=a^2$ isn't one to one. $f(a)=a^0$ isn't one to one either, since everything gets turned into $1$ (except $0$, sort of, but that's more complicated). Basically you're assuming that there's a nice "undo" operation when one doesn't exist. So what you should ask yourself first is this. "I have the answer. Can I figure out where I came from with this information?" If you can't, then this idea of just getting rid of the function doesn't work. Let's take $f(a)=a^0$.
$$a^0=b^0$$
So far so good. But what does it mean to isolate $a$ and $b$ here? You have to apply an inverse function to both sides. But such a function doesn't exist (i.e. you can't work backwards), so this doesn't work.
As for explaining this to someone in $9^{th}$ grade, I would use the idea of "going backwards". That is, you can't always go backwards, you have to show that you can. Say I have identical balls in both my left and right hands. I put them into different bags. I then point to a bag and say, "which hand did that ball come from?" You can answer. Next I take the balls out again, one in each hand, and put them into the same bag. Now if I pick a ball, you can't say which hand it came from, because they ended up in the same place. You can only undo something if different things always end up in different places, which doesn't happen in either of these. |
H: Is this polynomial positive?
Let $p\geq 2$, and $p$ is not a half odd integer. $t\in R$.
Is the following polynomial positive:
$$
T_k(t)=\left(\frac t2\right)^p\sum_{j=0}^k\frac{\left(-\frac{t^2}{4}\right)^j\Gamma(p+1)}{j!\Gamma(p+j+1)}.
$$
Thank you for your help
AI: Consider series
$$
T(t)=\left(\frac{t}{2}\right)^p\sum\limits_{j=0}^\infty\frac{\left(-\frac{t^2}{4}\right)^j\Gamma(p+1)}{j!\Gamma(j+p+1)}
$$
This is related to the series representation of the Bessel function of order $p$ of the first kind. Indeed
$$
T(t)=\Gamma(p+1)\sum\limits_{j=0}^\infty\frac{(-1)^j}{j!\Gamma(j+p+1)}\left(\frac{t}{2}\right)^{2j+p}=\Gamma(p+1)J_p(t)
$$
Since this series converges, then
$$
\lim\limits_{k\to\infty} T_k(t)=\Gamma(p+1)J_p(t)\tag{1}
$$
It is known that Bessel functions of the first kind take negative and positive values infinitely many times on $(0,+\infty)$. Hence we may consider $t_0$ such that
$\Gamma(p+1)J_p(t_0)<0$. From $(1)$ it follows that for some $k_0$ we would have
$$
T_k(t_0)<0\quad\text{ for all}\quad k>k_0.
$$ |
H: Derivative of Multi Variable Equation for Diablo 3 Damage Equation
It has been years since I took calculus and I am trying to figure out if it is possible to calculate the derivative of a mutli variable equation and what the result would be. This equation is from the video game Diablo 3 that represents your damage based on your critical chance and critical damage. I am trying to determine where are the optimal points (or best bang for buck stat) by using derivatives.
x = critical chance, y = critical damage, z = damage output
Equation: $z = 1 + xy$
AI: There are two variables in $z$, $x$ and $y$. Taking the partial derivative in respect to $x$ (treating $y$ like a constant), we get:
$$\frac{\partial z}{\partial x}=y$$
similarly, the partial derivative in terms of $y$ is
$$\frac{\partial z}{\partial y}=x$$ |
H: Heat equation with bounded or compactly supported initial data
Let $u$ denote to the solution of the heat equation
$$\begin{cases} u_t(x,t)-\Delta u(x,t) & = & 0 & t>0 \\ u(x,0) & = & g(x) \end{cases}$$
where $x\in\mathbb{R}^n$.
I want to show that
if $||g||_\infty<\infty$ then $u$ tends to some constant as $t\to\infty$
if $\text{supp}(g) \Subset \mathbb{R}$ (that is compact) then this constant is $0$.
I started from $$u(x,t) = (g*K_t)(x) \ \ \ \ \ \ \ \ \ (\diamond)$$ where $K_t$ is the heat kernel and tried to prove 1. by showing $||u'(x,t)||_{\infty}\to 0$ as $t\to\infty$. Unfortunately this might be the wrong way since formula $(\diamond)$ solves the heat equation (with initial data) only if $g \in L^p$ (what is not clear as we only claim $g$ to be bounded).
Who can help?
AI: Without additional assumptions on $u$, I believe this is not true.
There exist nontrivial solutions to the heat equation with initial condition 0. Indeed, one can find nontrivial $v(x,t)$ such that $v_t - \Delta v = 0$ for all $(x,t) \in \mathbb{R}^{n} \times \mathbb{R}$ and $v(x,t) = 0$ for all $x$ and all $t \le 0$. See for example
P. C. Rosenbloom and D. V. Widder. A temperature function which vanishes initially. Amer. Math. Monthly, 65:607–609, 1958.
The relevant example is at the very end.
I did not check whether this $v$ converges as $t \to \infty$. If it does not, we are done. If it does, we could take something like $u(x,t) = \sum_{n=1}^\infty v(x, t-n)$ so that disturbances keep appearing and prevent convergence. |
H: Probability that 2 sequences of k trials (success/failure) are identical.
A single trial has probability p of success and (1-p) of failure. For simplicity, we assume p = 1/2 = (1-p). An experiment is defined as a sequences of trials until 4 consecutive success or failures.
Suppose two such experiments were conducted. What is the probability that they will be identical?
Attempt
Given an outcome of experiment 1 that is made up of k trials. Then the probability of experiment 2 being exactly the same is, $$ P(k)(1/2^k) $$
where P(k) is the probability of an experiment being k trials in length. Since experiment 1 can be of length 4 to infinity, the final probability is, $$\sum_{k=4}^{\infty} P(k)^2(1/2^k)$$
Question
How do I find P(k)?
Is this the right approach? Or is there a simpler way?
AI: Call $T$ the number of trials needed to complete a full experiment. For $1\leqslant k\leqslant3$, call $T_k$ the number of supplementary trials needed to complete an experiment, starting from $k$ identical trials. Thus, $T\stackrel{\text{dist}}{=}1+T_1$, $T_1\stackrel{\text{dist}}{=}1+T_2$ or $1+T_1$ with equal probabilities, $T_2\stackrel{\text{dist}}{=}1+T_3$ or $1+T_1$ with equal probabilities, and $T_3\stackrel{\text{dist}}{=}1$ or $1+T_1$ with equal probabilities.
Fix $|s|\leqslant1$ and call $u_k=\mathrm E(s^{T_k})$. The identities between distributions stated above are translated into the relations $\mathrm E(s^T)=su_1$, $u_1=\frac12s(u_1+u_2)$, $u_2=\frac12s(u_3+u_1)$ and $u_3=\frac12s(1+u_1)$. Solving for $\mathrm E(s^T)$ this system yields
$$
\mathrm E(s^T)=\frac{s^4}{8-4s-2s^2-s^3}.
$$
The probability that the first experiment has length $k$ is $\mathrm P(T=k)$. Conditionally on $[T=k]$, the second experiment is identical to the first one with probability $\frac1{2^k}$. Hence, the probability $Q$ that the two experiments are identical is
$$
Q=\sum\limits_k\mathrm P(T=k)\frac1{2^k}=\left.\mathrm E(s^T)\right|_{s=\frac12}=\frac1{86}.$$ |
H: Finding the equation of a curve that has a perpendicular distance of $d$ from another curve
Let's say we have an equation of a curve as $y = f(x)$.
I want to find the curve $y = g(x)$ where $(x_1, f(x_1))$ has a perpendicular distance of $d$ from that curve.
Doing this with straight lines is pretty easy. For example:
If $f(x) = mx + b$, then $g(x) = mx + b \pm \frac{d}{sin(atan(1 / m))}$
But how would I be able to do this with curves?
Sorry if it's a bit hard to understand since I'm not too sure how to word this question.
AI: If you insist that the second curve is given in the form $\gamma_d:\ y=g(x)$ it can be done only in very special cases, e.g., when $\gamma$ is a line or a circle. But it is easy to give a parametric representation for $\gamma_d\ $:
Assume the curve $\gamma$ is given by a parametric representation of the form
$$\gamma:\quad t\mapsto z(t)=\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$
(The form $y=f(x)$ can easily be recoded as $t\mapsto\bigl(t,f(t)\bigr)$.) Then the tangent vector at $z(t)$ is given by $\dot z(t)=\bigl(\dot x(t),\dot y(t)\bigr)$, and the unit normal vector pointing to the left by
$$n(t)=\Bigl({-\dot y(t)\over|\dot z(t)|},{\dot x(t)\over|\dot z(t)|}\Bigr)\ ,$$
where $|\dot z(t)|=\sqrt{\dot x^2(t)+\dot y^2(t)}$. Therefore a parametric representation of the curve $\gamma_d$ at distance $d$ to the left of $\gamma$ reads as follows:
$$\gamma_d:\quad t\mapsto z(t)+d\ n(t)=\Bigl(x(t)-d{\dot y(t)\over|\dot z(t)|}, \ y(t)+ d {\dot x(t)\over|\dot z(t)|}\Bigr)\ .$$ |
H: Maximum likelihood estimate for pdf
I am attempting a problem from Larsen and Marx, 4th edition that asks to find the maximum likelihood estimate for $\theta$ in the pdf: $$ f(y; \theta) = \dfrac{2y}{1-\theta^{2}}, \theta \leq y \leq 1$$
It also states that a random sample of size 6 yielded measurements 0.70, 0.63, 0.92, 0.86, 0.43, and 0.21. I used the definition of the likelihood function to get: $$L(\theta) = \prod_{i=1}^n \dfrac{2y_{i}}{1-\theta^{2}} = 2^{n}(1-\theta^{2})^{-n} * \prod_{i=1}^n y_{i}$$
Because it is easier to deal with $\ln L(\theta)$ for the purpose of deriving and finding a $\theta$ that maximizes $L(\theta)$: $$\ln L(\theta) = -n \ln (1-\theta^{2}) + n \ln 2 + \ln \prod_{i=1}^n y_{i}$$ $$\dfrac{d \ln L(\theta)}{d\theta} = \dfrac{2n\theta}{(1-\theta^{2})}$$
However, setting this derivative to zero would mean $\theta_{e} = 0$ regardless of sample size and leaves me with a pdf of $f_{Y} = 2y$. Did I go wrong somewhere in my solution?
AI: Did I go wrong somewhere in my solution?
Apart for the typo (see did's comment), you made the most common error student make when computing ML: to just compute the derivative and set it to zero, before even asking about the domain of your function. Remember that your goal is to find the maximum of the function; now, you're supposed to know (Calculus I) that the global maximum of a function does not necessary happen on a critical point (null derivative), one should also look for the domain boundaries and non differentiable points (if they exist).
Always, when doing MLE, before computing the derivative, look at the function, ask which is the domain, if it's differentiable, perhaps try to graph it, see if it's easy to deduce where it's increasing/decreasing, etc. |
H: What does "increases in proportion to" mean?
I came across a multiple choice problem where a function $f(x) = \frac{x^2 - 1}{x+1} - x$ is given. One has to choose the statement that is correct about the function. The different statements about the function included:
(1) the function increases in proportion to $x^2$.
(2) the function increases in proportion to $x$
(3) the function is constant.
Obviously the function is constant, but my questions is what the "increases in proportion to" means. I haven't come across it before, and was wondering if it is standard for something. I would like so see an example of a function that can be said to satisfy this statement.
AI: To say "$f$ increases in proportion to $g$" is just another way of saying "$f$ is proportional to $g$", i.e., $f$ is equal to $g$ times a nonzero constant.
(In other contexts this may only be meant approximately, but the "literal" usage is that it is equal to $g$ times a constant.) |
H: Question about polynomials and congruences
Let be $m(X), n(X), a(X), b(X), m(X), n(X), g(X) \in F[X]$ a polynomials, where $F$ is a finite field with characteristic 2. Let be a relations $a \equiv b \mod g(X)$ and $m \equiv n \mod g(X)$. Are there any relation, property, etc, between $a(X), b(X), m(X), n(X)$?.
AI: Let $F=\mathbb{F}_2$. The congruence
$$
p(X)\equiv q(X)\pmod{g(X)}
$$
means simply that there exists a polynomial $r(X)\in F[X]$ such that
$$
p(X)-q(X)=r(X) g(X).
$$
For example we can say that
$$
X^8\equiv X \pmod{X^3+X+1},
$$
because
$$
X^8-X=X(X^7-1)=X(X-1)(X^3+X^2+1)(X^3+X+1),
$$
(as an exercise check that the product on the right hand side is what I claim it to be)
so $r(X)=X(X-1)(X^3+X^2+1)$ works. Note that this is exactly analogous to the definition of congruence of integers: $a\equiv b \pmod{n}$ means that $a-b=rn$ for some integer $r$.
The ring $F[X]$ is a PID meaning that in their concepts like "greatest common divisor", "factorization into products of primes irreducible polynomials" and "Bezout's identity" make sense. As an example I claim here that the two polynomials that appeared, $a(X)=X^3+X^2+1$ and $b(X)=X^3+X+1$, have no common factors. IOW $\gcd(a(X),b(X))=1$. This follows because they are both irreducible: an eventual factorization would include a linear factor, but such a fact cannot exist, because then $a$ or $b$ would have a zero in $F$. Bezout the commands me to produce polynomials $u,v\in F[X]$ such that the identity
$$
ua+bv=1
$$
is true. As is always the case in Bezout's identity, the left hand side will be divisible by any common factor of $a$ and $b$, so that common factor will also be a factor of the right hand side - here 1. A little bit of tinkering (we can run the generalized Euclidean algorithm here, if we can't do without) shows us that
$$
(X+1)a+(X)b=(X^4+X^2+X+1)+(X^4+X^2+X)=1,
$$
so the choices $u=X+1$ and $v=X$ work.
So this hopefully gives you an idea what congruence and Bezout's identity mean in the ring $F[X]$. Is that what you really wanted to ask? Also, why did you include the coding-theory -tag? What did you mean by lattices here? |
H: Division and number scaling
I'm trying to implement an interactive (secure) protocol which operates only on integers. Here's what I have:
$$ f(x) = \sum_i{a_i K_i} + b $$
$$ K_i = \dfrac{1}{1 + \gamma \|x - s_i\|^2}$$
where $ 0 < i \le M $ and $$ \|x - s_i \|^2 = \sum_{j=1}^{N}{(x_j - s_{ij})^2}.$$
Because the server has to work only with encrypted values, it needs to engage in an interactive protocol with the client in order to perform the division. So, basically, the server computes every $ 1 + \gamma \|x - s\|^2 $ value, blinds each of them multiplicatively with a random factor $ r_i $ ($ 0 < r_i < 2^{100} $) and sends the value vector to the client to perform the division. After the client divides 1 by each blinded value, it sends the results back to the server, which multiplies each received value by $ r_i $, to remove the blinding, and then it computes $ f(x) $.
Now, the above protocol works nice, if the numbers are floating point, but, unfortunately, the server can only work with encrypted integers (it performs homomorphic additions on them). In order to overcome this issue, I need to scale all the values accordingly:
$$ K_{ri} = \dfrac{s_{\gamma} s_f^2 s_k s_r}{r_i (s_{\gamma} s_f^2 + \gamma s_{\gamma} \|x s_f - s_i s_f\|^2)}$$
where:
$ s_{\gamma} $ is the scaling applied to $ \gamma $
$ s_f $ is the scaling applied to $ x $ and $ s_i $
$ s_r $ is the scaling needed to compensate for the blinding factor $ r_i $
$ s_k $ specifies how many decimals should be preserved after the division, because the value of $ K $ itself also needs to be scaled
After the client sends back the $ K_{ri} $ values, the server computes $ f(x) $:
$$ f(x_i) = \sum_i{a_i s_a K_{ri} r_i} + b s_b $$
where:
$ s_a $ is the scaling applied to $ a_i $
$ s_b $ is the scaling which needs to be applied to $ b $
Since $ b $ is usually large enough, it's sufficient to scale it with the scaling that is implied from the other factors.
Now, my problem is that I am having a really hard time figuring out how to construct $ s_b $ in order to compensate for $ s_r $. I think it should be a trivial thing, but I've been staring at the formulas for many hours and I can't figure it out. For starters, it needs to contain $ s_a s_k s_{\gamma} s_f^2 $, but what do I do about $ s_r $, given that $ 0 < r_i < 2^{100} $? There is no way to communicate the size of each $ r_i $ to the client, and the multiplication $ K_{ri} r_i $ seems to mess everything up...
AI: To avoid confusion, let me use the superscript $\,^*$ to denote the scaled variables. So the unscaled return value $$K_{ri} = K_i / r = \frac{1}{r_i(1 + \gamma\|x - s_i\|^2)}$$ corresponds to the scaled return value $$K^*_{ri} = \frac{s_{\gamma} s_f^2 s_k s_r}{r_i (s_{\gamma} s_f^2 + \gamma s_{\gamma} \|x s_f - s_i s_f\|^2)} = \frac{s_{\gamma} s_f^2 s_k s_r}{s_{\gamma} s_f^2 r_i (1 + \gamma \|x - s_i\|^2)} = s_k s_r K_{ri}.$$
Now, you want to calculate the scaled version of $$f(x) = \sum_i{a_i K_i} + b =\sum_i{a_i K_{ri} r_i} + b,$$ namely $$f^*(x) = \sum_i{a_i s_a K^*_{ri} r_i} + b s_b = \sum_i{a_i s_a s_k s_r K_{ri} r_i} + b s_b.$$
If $b = 0$, we have $f^*(x) = s_a s_k s_r f(x)$. To make this equation apply even when $b \ne 0$, we need to choose $$s_b = s_a s_k s_r.$$ |
H: Chronicles of Discoveries
First I'd like to bring an example to make myself more clear.
I know what the Jacobian matrix is and where, how and why it is used (some examples, at least) . But still I can't get it's geometrical interpretation. And I don't understand how and why it works.
I wonder how Jacobian matrix was discovered. What theoretical issues laid behind this discovery? What mathematical problems of the time did it solve?
Generally, is are there any books that describe the history of various mathematical discoveries, not [auto]biographies of the author, but rather chronicles of the discovery itself, but the demands of the time, the difficulties, approaches that didn't work, and eventually the discovery?
AI: At Alain Lascoux's web page you can find .pdf files of Thomas Muir's encyclopedic history of determinants. Look at the citations for Jacobian on p. 391 of the file "History, Index (pdf)" and track down those citations in Muir's work.
I'm sure there are some historical survey papers on what you're asking about, but off-hand I don't know of any. However, Muir does something that may be unique in mathematical historical writing by providing detailed summaries of virtually every paper written in a certain area of mathematics up to a certain time. |
H: Could someone remind me why is incorrect to switch an infinite sum and an integral?
Could someone jog my memory on this?
The order of operation between an $\int$ and $\sum_{n\in \mathbb{N}}$ is not always interchangable? Note that the sum is an INFINITE sum
Why is it that $\int \sum_{n \in \mathbb{N}} \neq \sum_{n \in \mathbb{N}} \int$
Is the reason because the integral itself is a sum and the order of "summing" actually matters? (I think it's Multivariable calculus related stuff now)
AI: It is easier and more instructive to give a counterexample using sequences: For example $$\lim_{n\to\infty}\int_0^1 n^2x^n(1-x)\,dx=1$$ even though the integrand goes to zero everywhere in [0,1].
To understand what is going on here, note that the function in the integrand has a graph which is a tall, thin peak getting taller and taller and thinner and thinner as $n\to\infty$, while approaching $x=1$ from the left.
Taking differences, you can easily realize the sequence as partial sums of a series, thus providing the counterexample you seek. To be precise, consider $$\lim_{n\to\infty}\int_0^1\sum_{k=0}^{n-1} \bigl((k+1)^2x^{k+1}(1-x)-k^2x^k(1-x)\bigr)\,dx,$$ which is just a difficult way to write the limit above. |
H: Limit of the sequence of rational numbers above a given real
I need to prove that for any $a \in\mathbb{R}^+$ the sequence $S[a]_n=a+b_n$, where $b_n = \min\{|{x \over n}-a|\;\colon\;x \in \mathbb{N}\}$, converges to $a$. The only way I know how to prove convergence is with an epsilon-delta argument but I don't think that would work here. Any ideas?
AI: The statement that $S[a]_n\to a$ is, by definition, that for any $\epsilon>0$, there is some $N\in\mathbb{N}$ such that
$$|S[a]_n-a|=\big|\min\{|\tfrac{x}{n}-a|: x\in\mathbb{N}\}\big|<\epsilon\text{ for all }n>N.$$
Note that, for any given $n\in\mathbb{N}$, $|\frac{x}{n}-a|=\frac{1}{n}\cdot|x-an|$, and thus
$$b_n=\min\{|\tfrac{x}{n}-a|:x\in\mathbb{N}\}=\tfrac{1}{n}\cdot\min\{|x-na|:x\in\mathbb{N}\}.$$
Because $a>0$, we have that $an>0$ for any $n\in\mathbb{N}$. Note that $\lceil s\rceil\in\mathbb{N}$ for any $s>0$, and that $$\big|\lceil an\rceil -an\big|<1.$$
We've found one $x\in\mathbb{N}$ such that $|x-na|<1$ (namely, $x=\lceil an\rceil$), so that the smallest of the quantities $|x-na|$ as $x$ ranges over all natural numbers can't be any bigger than $1$. Thus $$\min\{|x-na|:x\in\mathbb{N}\}< 1$$
for any $n$, so that
$$b_n=\tfrac{1}{n}\cdot\min\{|x-na|:x\in\mathbb{N}\}<\tfrac{1}{n}$$
for any $n$. Because $b_n$ is non-negative, we have that $$b_n=|b_n|=\big|\min\{|\tfrac{x}{n}-a|: x\in\mathbb{N}\}\big|<\tfrac{1}{n}$$ for any $n$. Thus, for any $\epsilon>0$, we have that
$$|S[a]_n-a|<\epsilon\text{ for all }n>\lceil\tfrac{1}{\epsilon}\rceil.$$ |
H: Find $\lim \ a_n$ if $a_n = \frac{1}{\sqrt[3]{n^3+1}} + \frac{1}{\sqrt[3]{n^3+2}}+\cdots+\frac{1}{\sqrt[3]{n^3+n}}.$
I would greatly appreciate some help in finding $\lim \ a_n$ if
$$a_n = \frac{1}{\sqrt[3]{n^3+1}} + \frac{1}{\sqrt[3]{n^3+2}}+\cdots+\frac{1}{\sqrt[3]{n^3+n}}.$$
AI: HINT:
Note that $$\dfrac1{\sqrt[3]{n^3+n}} \leq \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac1{\sqrt[3]{n^3+1}}$$
for all $k \in \{1,2,3,\ldots,n\}$.
Hence,
$$\dfrac1{\sqrt[3]{1+1/n^2}} = \dfrac{n}{\sqrt[3]{n^3+n}} \leq \sum_{k=1}^{n} \dfrac1{\sqrt[3]{n^3+k}} \leq \dfrac{n}{\sqrt[3]{n^3+1}} = \dfrac1{\sqrt[3]{1+1/n^3}}$$
Now apply the squeeze/sandwich theorem. |
H: Number of $1$s in a binary grid
Consider a binary grid of size $4\times 4$, each of cell can either have $0$ or $1$. Among all possible $2^{16}$ arrangement how many arrangement of such grid exist in which each row and column contains even number of $1$s.
Solution which I thought
There will be $2$ possibilities for the answer of this question $1$st all ones in the $4\times 4$ grid that will count up to $1$ possible arrangement and $2$nd possibility will be $2$ ones in each row and column so how can i find the possible arrangement which will have $2$ ones in each row and column.
Am I right?
AI: Let the positions be $a_{i,j}$, where $1\le i,j\le 4$. You can fill the $3\times 3$ square in the upper lefthand corner, i.e., positions $a_{i,j}$ with $1\le i,j\le 3$, any way you like. Once those $9$ positions are filled, there is exactly one way to fill the remaining $7$ positions to get an even number of $1$’s in each row and column. Can you see why? (HINT: Fill $a_{4,4}$ last.) |
H: Explain Carmichael's Function To A Novice
I understand that the Carmichael Function (I'm going to call λ) is essentially the smallest positive integer m, where $a^m$ is congruent $1 \pmod n$ for all $a$ co-prime to $n$ and less than $n$.
6 makes sense to me. The only co-prime is 5 and $5^2 = 25\equiv 1 \pmod 6$. So $λ(6) = 2$.
I also know that the answer to $λ(49)$ is $42$. I'm not clear on what the "shortcut" is though. I realized I could brute force the answer, but that's not feasible for any significant values of n. I understand there is some sort of mathematical "shortcut" to derive this answer... something to do with LCM of prime powers. I would appreciate an explanation on this shortcut in terms an avg 14 year old could understand. No Fermat or Euler.
AI: The first thing we want is this:
If $\gcd(a,b) = 1$ and $a|bc$, then $a|c$.
This is fairly easy to establish. For example, you can use the Bezout identity: $a$ and $b$ are coprime if and only if there exist $x$ and $y$ such that $ax+by=1$. Multiplying through by $c$ we get $c = acx + bcy$. Since $a$ divides both $acx$ and $bcy$ (the latter by virtue of dividing $bc$), then $a$ divides $c$.
From this we have:
Suppose that $a$ and $n$ are coprime integers, and $x$ and $y$ are integers such that $ax\equiv ay\pmod{n}$. Then $x\equiv y\pmod{n}$.
Indeed, if $ax\equiv ay\pmod{n}$, the $n$ divides $ax-ay = a(x-y)$. Since $\gcd(n,a)=1$ by assumption, it follows that $n|x-y$, so $x\equiv y\pmod{n}$.
Hence, for each $a$ that is coprime to $n$, there is at least one positive integer $k$ such that $a^k\equiv 1\pmod{n}$:
If $\gcd(a,n)=1$, then there exists a positive integer $k$ such that $a^k\equiv 1\pmod{n}$.
There are only finitely many positive integers smaller than $n$ that are coprime to $n$. Let's say there are $t$ of them. Then the numbers $a\bmod n$, $a^2\bmod n$, $a^3\bmod n,\ldots, a^{t+1}\bmod n$ cannot all be distinct, because each of them is coprime to $n$, and there are $t+1$ of them, all smaller than $n$. So there must exist exponents $i$ and $j$, $i\lt j$, such that $a^i\equiv a^j\pmod{n}$. Write $j=i+\ell$, and we have
$$a^i(1) \equiv a^ia^{\ell}\pmod{n}.$$
Since $\gcd(a^i,n)=1$, then we can cancel and we get $1\equiv a^{\ell}\pmod{n}$, which shows there must be some exponent that works.
If $a^k\equiv 1\pmod{n}$, and $r$ is a multiple of $k$, then $a^r\equiv 1\pmod{n}$.
This is fairly straightforward: we can write $r=kt$ for some $t$. Then using the laws of exponents, we have:
$$a^r = a^{kt} = (a^k)^t \equiv 1^t \equiv 1\pmod{n}.$$
So, given a positive integer $n$, why does there exist a smallest $m$ such that $a^m\equiv 1\pmod{n}$ for every positive $a$ that is smaller than $n$ and coprime to $m$? First, note that there has to exist some $m$ that works: there are only finitely many positive integers coprime to $m$; call them $a_1,\ldots,a_f$. For each of them, we know there exists some positive integer $k_1$, $k_2,\ldots,k_f$, with the property that $a_i^{k_i}\equiv 1\pmod{n}$. Now, if $k=k_1k_2\cdots k_f$, then $k$ is a multiple of each $k_i$, so $a_i^k\equiv 1\pmod{n}$.
That is, there certainly are positive integers that "work" for all $a$ coprime to $n$, positive, and smaller than $n$. Since there are such integers, there must be a smallest one. That smallest one is the value of the Carmichael function.
Using the division algorithm, it is easy to check that in fact any number that "works" is a multiple of the smallest one.
Now, as to the "short-cut": it relies on two facts:
We know what the value is for prime powers. And
The Carmichael function satisfies the following: if $m$ and $n$ are relatively prime, then $C(mn) = \mathrm{lcm}(C(m),C(n))$.
The latter assertion follows from the Chinese Remainder Theorem: the positive integers smaller than $mn$ that are coprime to $mn$ are in one-to-one correspondence to pairs of integers $(a,b)$ with $0\lt a\lt m$, $0\lt b\lt n$, and $\gcd(a,m)=\gcd(b,n)=1$ (namely, given $x$, $0\lt x\lt mn$ coprime to $mn$, set $a=x\bmod m$ and $b=x\bmod n$ (the Chinese Remainder Theorem guarantees that this is a bijection). And $x^r\equiv 1\pmod{mn}$ if and only if $a^r\equiv 1\pmod{m}$ and $b^r\equiv 1\pmod{n}$. Thus, any number that is a multiple of both $C(m)$ and $C(n)$ will "work" for $mn$, and any number that "works" for $mn$ must be a multiple of $C(m)$ and of $C(n)$. Hence, the set of integers that "work" are the common multiples of $C(m)$ and $C(n)$, and so the smallest integer that works is the least common multiple.
As to the value in the prime factors, for odd primes, this is just Euler's Theorem; it takes a bit of work, but one can show that if $p$ is an odd prime, then $C(p^r) = (p-1)p^{r-1}$.
For $p=2$, we have that $C(2)=1$, $C(4) = 2$, and $C(2^{r}) = 2^{r-2}$ if $r\geq 3$.
These formulas give you a way to compute $C$ for any integer that you can factor into primes. |
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