Mathematics

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.

A Beautiful Mind

@DavidWheeler You just made a huge rectangle.

David Wheeler

Now we have to "add 1" because the "third digit place" is hundreds, and we start with "ones" ($10^0$)

@ABeautifulMind Wait, what?

A Beautiful Mind

@DavidWheeler Just look and you will see.

David Wheeler

Look...where?

A Beautiful Mind

Sorry, I am too obscure. The huge chunk of your text in chat.

infinitesimal

7:12 PM

At the block of writing :-)

David Wheeler

Henceforth, I should request a chat basis of order 3, in order to minimize content.

Of course, only two dimensions will be visible here.

infinitesimal

Please don't pass your 3 dimensional objects through our 2 dimensional world :-)

Mary Star

Ok... Thank you!! @DavidWheeler

David Wheeler

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.

user159870

Hi everyone!

do you have any idea for:

0

Q: Greatest Commom Divisor

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_...

abstract-algebra

Chris's sis

7:25 PM

@robjohn Theoretically, one can use the contour integration for compution this one too $$\int_0^{\pi/2}\frac{x^n\log{\sin{(x)}}}{\sin(x)}\,dx$$

infinitesimal

Greetings

Chris's sis

@infinitesimal Hi

infinitesimal

How are you @Chris'ssis?

Chris's sis

@infinitesimal Not that bad, thanks. You?

infinitesimal

@Chris'ssis fine thanks.

Chris's sis

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. :-)

infinitesimal

A person says "good morning" and exactly 12 hours later he says "good morning" again. How is that possible?

Chris's sis

@robjohn The same rectangle contour should work nicely.

David Wheeler

@Chris'ssis May I hope that one day your book will be entitled "Integrals of rational expressions of transcendental functions"?

infinitesimal

Both statements being true at the time he says them.

Chris's sis

@DavidWheeler I cannot promise you anything for the future. :-)

David Wheeler

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 :)

Chris's sis

@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 ...

David Wheeler

@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.

Chris's sis

@DavidWheeler I like engineering pretty much, I attended such a job in the engineering area. I might return to such a place one day.

Mary Star

@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 $\lfloor \log_2 \phi(n) \rfloor +1=\log n$ ??

David Wheeler

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.