The first coordinate of $x_n$ is $P_n$, the second is $Q_n$, the third is $R_n$. The fourth is the sum $P_{n-1} + \cdots + P_0$, the fifth is $R_{n-1} + \cdots R_0$, the sixth is always $1$.
No, in position 1 of the $6 \times 1$ column vector $x_n$.
Essentially I just added two more variables representing the cumulative sums of $P$ and $R$ so that I could express each of the five as a combination of the results for $n-1$.
Okay I will look at that. Thankyou. But just for context this is a problem from Niven Zuckerman and he does not give anything about snake oil@Semiclassical
sure. main reason i suggest inclusion-exclusion is that the sum is alternating. (that, and the conclusion of the counting problem would be that "either exactly one exists, or none exists at all" which is cute)
@AlexClark I was acknowleged in a paper recently for reasons unclear to me. I asked the author, and he told me that it was because I talked to him about the project at some point.
So I agree with Tobias that you should strive to be cited, not acknowledged. :)
@Albas The binomial coefficient is be defined as $$\binom{r}{n} = \frac{r^{\underline n}}{n!},\quad \text{integer } n \geqslant 0,$$ and 0 for $n < 0$.
Here $r^{\underline n} = r(r-1)\cdots(r-n+1)$ is the falling factorial.
Yep. By the way, the definition in terms of the falling factorial is more general than the more common one involving factorials, since it defines the binomial coefficient for all real upper indices.
Why is it that if we have ratio as 121212/131313, the simplification is just 12/13? I've found that extends to any ratio where the same set of numbers repeats.
Suppose I am on the ground, and I throw some object perpendicularly (wrt the ground) up with the amount of force required to make it's initial velocity $v$ say. Due to gravity, my object will go up for some time (say till time $t$) and stabilize with velocity $0$. So $0 = v - gt$ where $g$ is acceleration due to gravity, hence $t = v/g$. Plug and chug further to get $s = vt - 1/2gt^2 = v^2/g - 1/2 v^2/g = 1/2 v^2/g$. So I know my object goes distance $1/2 v^2/g$ 'till it starts to fall.
But if there was no gravity, it'd go $d = vt = v^2/g$ distance in time $t = v/g$. So it seems under the influence of gravity the distance the object travels is halved. Is there an intuitive way to see this?
Decompose the distance traveled as a function of two things: initial velocity and the downward velocity induced by gravity over time. You're really asking why the latter is (negative) half the former. That's a double integral, giving 1/2 gt^2. Since we're evaluating precisely at the time for which gt = -v, and the integral coming from v is vt, you get the desired answer.
So I would say the reason is "When you integrate from acceleration to distance, a factor of 1/2 appears that doesn't when you integrate from velocity to distance."
@MikeMiller @Semiclassical Both of you raise interesting points. But I suppose I was expecting for more insightful or "physically obvious" explanations, which may not really exist.
So a good way to remember this is: if you throw a small heavy rock up at a 45 degree angle, it will land four times further away than the maximum height it reaches.
@SemiC Yup, that picture with the lines makes things much more geometrically obvious. I feel like I should have figured that out by myself, seeing the simplicity. Thanks.
@MikeMiller academia obscura has a lot of interesting posts
there's a classmate who's bugged me with solving his math problems yesterday- which i did and now he's bugging me with solving his physics problems tomorrow, and i had to learn physics to prepare for answering his questions (i didn't tell him i didn't study anything)
It's just a cutting down dimensions argument. My manifold by definition sits inside some huge Euclidean space. I choose some appropriate point/vector not in the manifold so that projecting the ambient Euclidean space to the orthogonal complement of that vector and postcomposing that with the embedding also gives me an immersion.
Excuse me @DanielFischer, since this conversation wasn't required from you, but I would like to know if there are counterexamples, or it is possible to prove that $-2N+3S_+(N)+S_-(N)+2S_0(N)=M(N)$, where $M(N)$ is the Mertens function and with the $S_{*}$ we take the count of those integers $n\leq N$ with $\mu(n)=1,-1$ or $0$ respectively the subscript $+,-$ or $0$. Thanks.
and with the details you've given me, by induction, I can embed my manifold in $\Bbb R^0$
The point I'm not successfully making is that if your sketch isn't detailed enough to see why we get the particular dimension you stated as the answer, then it's not a detailed enough sketch.
I am trying to find out how to say what's the lower bound for dimension cutting $(2k + 1)$ is without getting too detailed. I want to say 'something something secant variety'. You're essentially looking at $M \times M \times \Bbb R$ and immersing that in $\Bbb R^m$ via $(x, y, r) \mapsto (f(x) -f(y))r$, and you choose by Sard's some $a$ which does not belong to the image of neither this nor my embedding $f : M \to \Bbb R^m$.
So $m > 2k + 1$ has to happen. What this accomplishes is that if $f$ composed with the projection is not injective, then image of $M \times M \times \Bbb R \to \Bbb R^m$ contains $a$.
Which is a contradiction. I guess geometrically one should think about something secant variey like construction happening but I can't quite see it precisely yet.
OK, suppose $p$ is the projection of $\Bbb R^m$ into the complement of $(a)$. Then I have to prove $p \circ f$ is an immersion.
So $D(p \circ f)(x) = Dp(f(x)) \circ Df(x)$ which is $p(f(x)) \circ Df(x)$. All I need to prove is that this has nonzero kernel.
Oh, yikes, that's not obvious. Hmm. I need to show $Df(x)$ is not in the span of $(a)$.
Ah, I remember now.
$T(M) \to \Bbb R^m$ given by $(x, v) \mapsto Df(x)(v)$ is an immersion. So I need to use Sard's theorem again to make sure $a$ is not in the image of this either ($m > 2k = \dim T(M)$ so I can do that).
if $e(n)$ is the minimal dimension in which all compact manifolds of dimension $n$ embed, we've provided the bound $e(n) \leq 2n$. this is the best possible polynomial bound.
Yes, that is what I was trying to say. I cannot improve that uniformly for all manifolds of dimension $n$. But I suppose you were trying to say something else?
Thanks @DanielFischer, the right one is $M(N) = -2N+3S_{+}(N) +S_{-}(N)+2S_0(N)$, I get it from substituion in the relationship between the zeta and Dirichlet eta functions, for $\Re>1$ (if there a no mistake in this claims) with the substitution $s_n=3+\mu(n)$, then taking logarithms and using particular values of the zeta and eta you get it.
I think it's pretty clear what I said above, but it's not very helpful, because obviously a polynomial bound cannot be better than a linear bound. So let me say instead that no algebraic function $g$ is a better bound than 2n; so there is no algebraic function with $e(n) \leq g(n) \leq 2n$ excerpt for $2n$ itself.
Excuse me, if there are no mistakes I say that you need to show $$2^{M(N)}=\prod_{n=1}^N\frac{1}{4}\cdot\frac{4\cdot2^{\mu(n)}\zeta(3+\mu(n))-\zeta(3+\mu(n))}{\eta(3+\mu(n))}.$$ Now I hope tomorrow, if you want answer if there are any counterexample, and don't disturb to you more, now. Very thanks much @DanielFischer and sorry me for this puntual interference.
Let $n$ be an integer that is not divisible by any square greater than $1$. Denote by $x_m$ the last digit of the number $x^m$ in the number system with base $n$. For which integers $x$ is it possible for $x_m$ to be $0$?
Well, there's something you should know about square-free integers. A positive integer is square-free if and only if it is the product of distinct primes.
Not it's not. We have $2^2 \equiv 0 \pmod 4$ but $2 \not \equiv 0 \pmod 4$.
Let's take the simple case where $n$ is prime, so we'll denote $n$ by $p$ now. We're looking for all integers $x$ for which there exists some integer $m \ge 0$ such that $x^m$ is divisible by $p$.
So you're saying that the answer is "exactly those $x$ which are divisible by $n$". You're right. But I don't think we really proved it. Can you prove it?
yes: It is easy to see that for any such squarefree $n$, $x \equiv 0 \pmod{n}$ since any integer less than $n$ doesn't share at least one prime in common with $n$. Thus, $x = nk$ for some $k \in \mathbb{Z}$.
I'm not sure exactly what it means for the period of the sequence $x_m$ to be independent of $x$, but whatever.
But back to the first part, if you like you can consider this to be an application of this theorem: If $n$ has prime factorization $\prod_p p^{n_p}$, then $$a \equiv b \pmod m \iff a \equiv b \pmod {p^{n_p}} \text{ for all }p.$$
And then let $a = x^m$ and $b=0$. So the square-free case reduces to the prime case.
Anyway, I gotta go. Bye.
^^^ that $\mod m$ on the left side should be a $\mod n$.
