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7:00 PM
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: $367 = 3.67 \times 10^2$
So $\log_{10}(367) = \log_{10}(3.67) + \log_{10}(10^2) = $ something $+ 2$.
If "something" is less than 1, then the floor of the entire expression, will be 2.
Since $1 < 3.67 < 10$, the log will be between 0 and 1.
So the "something" doesn't matter, taking the floor obliterates it.
 
@DavidWheeler You just made a huge rectangle.
 
Now we have to "add 1" because the "third digit place" is hundreds, and we start with "ones" ($10^0$)
@ABeautifulMind Wait, what?
 
@DavidWheeler Just look and you will see.
 
Look...where?
 
Sorry, I am too obscure. The huge chunk of your text in chat.
 
7:12 PM
At the block of writing :-)
 
Henceforth, I should request a chat basis of order 3, in order to minimize content.
Of course, only two dimensions will be visible here.
 
Please don't pass your 3 dimensional objects through our 2 dimensional world :-)
 
Ok... Thank you!! @DavidWheeler
 
I try to much, I might when forget.
Damn. Sorry about that.
I will try not to overly much, but I still might when I forget.
 
Hi everyone!
do you have any idea for:
0
Q: Greatest Commom Divisor

user159870In 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_...

 
7:25 PM
@robjohn Theoretically, one can use the contour integration for compution this one too $$\int_0^{\pi/2}\frac{x^n\log{\sin{(x)}}}{\sin(x)}\,dx$$
 
Greetings
 
@infinitesimal Hi
 
How are you @Chris'ssis?
 
@infinitesimal Not that bad, thanks. You?
 
@Chris'ssis fine thanks.
 
7:28 PM
@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 $\sin(x)$.
@infinitesimal Great. :-)
 
A person says "good morning" and exactly 12 hours later he says "good morning" again. How is that possible?
 
@robjohn The same rectangle contour should work nicely.
 
@Chris'ssis May I hope that one day your book will be entitled "Integrals of rational expressions of transcendental functions"?
 
Both statements being true at the time he says them.
 
@DavidWheeler I cannot promise you anything for the future. :-)
 
7:33 PM
When I learned some functions did not have primitives, I lost interest in the general case. I see you are much braver :)
 
@robjohn at the same time, one can also go the real analysis by using the substitution $-\log(\sin(x))=y$.
@DavidWheeler Well, I simply like what I do. That's all. :-)
@robjohn No, that contour doesn't work ... (actually not in that plane)
Let me check another version ...
 
@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.
 
@DavidWheeler I like engineering pretty much, I attended such a job in the engineering area. I might return to such a place one day.
 
@DavidWheeler I found the following:

$| \mathbb{Z}_n^{\star} |=\phi(n)$

We need $\log n$ digits to represent the elements of the group.

So, does it stand that $\lfloor \log_2 \phi(n) \rfloor +1=\log n$ ??
 
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.
 
7:50 PM
Hi @DavidWheeler
 
@BalarkaSen Hi?
 
I see that you finally hit 2K.
 
@Chris'ssis @robjohn are these integrals known ? $\int_0^{\pi } \cos ^{2 n}(x) dx$
 
@LeGrandDODOM Yes, they are.
2
 
@robjohn I meant even indices
 
8:01 PM
@LeGrandDODOM Well-known ...
 
@LeGrandDODOM $W_{2n}$ more or less ?
 
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...
 
@LeGrandDODOM You should know since from your profile pic I guess that you did Math Spé :P
 
@BalarkaSen I did. Toyed with the idea of losing rep by madly editing things to humorous effect, but meh. Too lazy.
 
@Chris'ssis Waiit. he wrote from $0$ to $\pi$ :O Doesn't that make 0 ?
 
8:03 PM
hah. seems like a good idea for timepass.
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ enough
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ And? For even indices?
 
@Chris'ssis -_- sorry
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ :D
 
@Chris'ssis That is still $2W_{2n}$ then
 
8:05 PM
@robjohn It looks like that an integral here doesn't survive more than a few seconds ... :-)
@ɧɿρρԹʅȝՇԵՐՎԾՌ The idea is that one can reduce it to $0, \pi/2$ interval (and hence the Wallis' result.)
 
Yeah
@Chris'ssis What did he mean by 'enough' though ?
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ Not sure.
 
Have the Dehn functions of most of the groups that you first meet when studying group theory been computed?
 
You're studying geometric group theory, @Alyosha?
As to the answer, almost any group is Gromov hyperbolic [*], and Dehn function of Gromov hyperbolic groups are known...
And of course, that applies to finite groups, which are trivially quasi-isometric to a point.
[*] : Theorem by Gromov - any given group is hyperbolic with probability 1.
 

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