The floor function just "chops off the decimal" part (essentially the same as "ignoring all but the first digit")
So, if we know $\log_{10}(10^n) = n$, then together with the fact that $\log_{10}$ is increasing, we know that the floor function will return the same integer until $\log_{10}(x)$ reaches $n+1$.
And since $\log_{10}$ is increasing, the FIRST place that happens is at $10^{n+1}$
Which is also the next time we go "up another digit"
Let's look at it another way: $367 = 3.67 \times 10^2$
In my lecture nores there is the following:
$a, b \in \mathbb{N}, gcd(a, b)=d \Rightarrow \exists s, t \in \mathbb{Z} : sa+tb=d \Rightarrow tb \equiv d \pmod a$
If $gcd(a, b)=1$ then $tb \equiv 1 \pmod a \ $ $ \ \ \ [b]_a \in \mathbb{Z}_a^{\star}$ and $[b]_a^{-1}=[t]_a$
$$r_0=a, r_1=b \\ r_...
@robjohn And related to $$\int_0^{\pi/2}\frac{x \log{\sin{(x)}}}{\sin(x)}\,dx$$ it should also be interesting to study the case $$\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$$ where I only swapped $x$ and $\sin(x)$.
@Chris'ssis I think you consider it a "fun game", like being good at chess. You'd probably make a great engineer, but it might not be as enjoyable for you.
For me, I enjoy the change-of-variable theorem, as a theorem, but as for particular substitutions, not so much.
@MaryStar I think you're mixing apples and oranges
The totient function is "unpredictable" (having to do, as it does, with the distribution of prime integers, or more properly, prime factors of integers)
The number of digits required to represent the integers mod n, has only to do with n.
But the number of UNITS can be much smaller. I'm not sure where the base 2 log is coming from, either.
Usually the same number of digits is used for representing the units, as in the integers mod n. This isn't necessarily optimal.
In mathematics, and more precisely in analysis, the Wallis' integrals constitute a family of integrals introduced by John Wallis.
== Definition, basic properties ==
The Wallis' integrals are the terms of the sequence defined by:
or equivalently (through a substitution: ):
In particular, the first few terms of this sequence are:
The sequence is decreasing and has strictly positive terms. In fact, for all :
, because it is an integral of a non-negative continuous function which is not all zero in the integration interval
(by the linearity of integration and because the last integral i...
Inspired by a question I saw these days, I try to calculate in closed form
$$\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$$
So far no fruitful idea that is worth sharing. What way would you propose? Note I prefer ways,
not necessarily solutions, but I have nothing against any of the opti...
@PaulPlummer yeah, you should read up about it. it's an interesting problem (i'm biased, because i know the guy who proved it for pairs $(H, G)$ with $H$ normal in $G$). only recently they found a counterexample to it, i think.
A man is a distance $s$ from his home, and has his loyal dog with him. He walks home with a speed of $v_m$, and his dog runs home at a speed of $v_d$. When the dog returns, he runs back to his man and so forth.
@teadawg1337 I am trying to find the times $t[n]$ when the dog is home. My goal is try to try to write a parametric equation that gives the dogs position at time $t$. any ideas? :p
I did a numerical test if you let $d = 10$, $v_d = 6$ and $v_h = 4$. Then $t[1] = 5/3$, $t[2] = 2$, $t[3] = 7/5$. do you get the same?
user129943
Anyway I need a favour, two people really helped me on a question that didn't get much attention. (especially the guy who didn't get my tick) math.stackexchange.com/questions/1172594/… can you +1 the answers here please - they really helped me and I don't like that the one guy only got +10 for it
@ABeautifulMind to the guy with 277k - yeah you're right, but the other new guy. It's wrong that an easy real-analysis answer will earn ya loads but this - gains only 10.
The "basic set-theoretical definitions" of sets mathematicians commonly employ can get pretty horrendous. Logically airtight, sure, but lacking a certain transparency.
Hello @DanielFischer!!! I am given some recurrence relations $T(n)$ and I have to give asymptotic upper and lower bounds for $T(n)$. We assume that $T(n)$ is continuous for $n \leq 2$. How can we use the fact that $T(n)$ is continuous for $n \leq 2$?
@TedShifrin: My area was downgraded to a winter weather advisory, though I live a couple of miles south of the county line. The county directly north has a winter storm warning, so idk what to expect
The way he told me enough, I took it as if someone was asking me without any bad intent 'haven't you seen this/that at school' and I was answering don't question me !, it feels weird
I will pay for someone to help me with a math assignment. I could not think of a better place to find someone to help me.
I would like to skype or use google hangouts to have someone aid me in an assignment. Want to make some extra money?
I'm so frustrated. I told my students to expect a problem on the exam where they'd have to (as in homework) use the contraction mapping principle to prove that $f(x)=y$ for $y$ in a certain domain has a root. And almost no one is getting it. Very upset.
@Hippa: Consider $B=B(0,4)\subset\Bbb R^n$. Suppose $f:B\to\Bbb R^n$ is $C^1$ and $\|Df(x)-I\|\le 1/3$ for all $x\in\bar B(0,3)$. Suppose, moreover, that $f(0)=a$ and $\|a\|\le 1$. Prove that if $\|y\|\le 1$, then there is a unique $x\in\bar B(0,3)$ with $f(x)=y$.
@David: The students in my class thrive on being challenged and learning, but they just have gone through life with classes and tests being a formality.
i still remember my first math class in college. i had aced every class i took up until then, and got AP credit, so jumped right in to vector calculus. i nearly flunked the first quarter.
@TedShifrin I've never seen the contraction mapping th. before, so just to be sure, is it the following : if $f$ is a contraction mapping ($\exists a<1,\forall x,y,|f(x)-f(y)|\le a|x-y|$) then it has a unique fixed-point on a given metric space ?
Sure, @Hippa. We spent over a week on this stuff and using it to prove the inverse function theorem. So one needs to set up a contraction map $\phi$ so that $\phi(x)=x \iff f(x)=y$.