Mathematics - 2016-05-13

Axoren

01:17

Does anyone know why Munkres avoids starting his definitions with 0? Going out of his way to start with 1 in the set of $\mathbb Z_+$ rather than 0 in the set of $\mathbb N$? Is there a good reason why some mathematicians don't use 0 in their set of counting numbers?

user19405892

01:29

hi

does anyone know how to prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$?

Axoren

@user19405892 Rewrite it as $2^x = 3^{2m+1} + 5$

Does that seem easier to solve?

user19405892

Yes it does, but I don't know how to prove the statement

Eric Stucky

Axo, some people do, some people don't. I don't know much more than that; trying to think of a reason why a topologist would choose not.

I mean, $\Bbb Z_+$ is pretty unambiguously starting at 1. It's $\Bbb N$ that gets you into trouble.

user19405892

@EricStucky Do you know of a way to prove $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$?

PVAL

@Axoren None of the content in that book depends on this.

Axoren

01:44

Assume it was true for some other $x = 2k$, you could factor $3^{2m+1} + 5$ into two factors $(2^k)(2^k)$. $2^k$ is an integer, is $\sqrt{3^{2m+1} + 5}$?

Rather, when is $\sqrt{3^{2m+1} + 5}$ an integer?

Eric Stucky

(u19: no)

Axoren

@EricStucky using the @ symbol notifies people that you've replied to them.

Eric Stucky

You know

as mortifying as it was

Axoren

Just so you don't have to skimp on @user19405892's numbers and at the same time not have to type them all

Eric Stucky

to have the "stop tagging me" comment starred

at least people stopped pinging me for a while...

Axoren

01:48

Not a fan of it? Luckily, they've made a mute button in the top right, now.

Eric Stucky

ooh

Axoren

Mortification: optional.

Eric Stucky

u r hero :D

Axoren

But yeah, it's just good habit to tag everyone else.

user19405892

@Axoren Can you explain how you you get $(2^k)(2^k) \cdot 2^k$?

Eric Stucky

01:49

idk, for me it's like

we're talking, right?

Axoren

@user19405892 That's a period, not a cdot.

EricStucky: Yeah, but if I walk away and come back? Chat scrolls. Your message is lost.

user19405892

$2^k \cdot 2^k = 3^{2m+1}+5$

why did you take the square root?

Axoren

$\sqrt{2^k 2^k} = 2^k$, do you agree?

user19405892

yes

Axoren

So, then square-root both sides.

user19405892

01:52

right, so we need $3^{2m+1}+5$ to be a perfect square

but it turns out that $x$ must be odd

not even

Axoren

Why must $x$ be odd?

Was that part of the original problem?

user19405892

No, but see that $3 \cdot 9^m + 5 \pmod 9 \equiv 5$, which implies $x \equiv 5 \pmod 6$

thus, $x$ is odd

Axoren

Right, either way, we've shown that $x$ can't be even.

user19405892

yes, the tricky part is dealing with $x$ is odd

Axoren

You said it like you knew it before analyzing it.

user19405892

01:56

yes, i had known $x$ is odd from the time i saw this question first

Axoren

Consider now $2 \cdot 2^k 2^k = 3^{2m+1} + 5$

user19405892

ok

Axoren

This is simply $x = 2k +1$, now.

The only other case.

user19405892

how do we deal with it?

Axoren

Actually, this may be kind of silly, but I think it's easier to just show that $2^x = 3^y + 5$ only has 2 integer solutions.

Than to do this with radicals.

Because if you do that, it's a stronger claim.

user19405892

02:04

Actually, that's how i got this question in the first place. I was trying to prove that result

Now how to prove that $2^{2k+1} = 3 \cdot 9^m+5 $ has no solutions?

Double AA

What's the mathjax way to write the alpha-like proportionality relation?

Axoren

You've said above that $x \equiv 5 \mod 6$, so we have instead $2k \equiv 4 \mod 6$, then $k \equiv 2 \mod 6$.

Eric Stucky

02:21

@DoubleAA: \propto

Related, detexify is great

Axoren

Actually, scratch that. I don't like doing it the modulo way. Assume we know it is true for $k=2$. If $k=3$, we have $2^7 = 3 \cdot 9 ^ m + 5$ for some $m$. But we also know the following: $2^5 = 3 \cdot 9^1 + 5 = 32$

Double AA

@EricStucky That does it! Thanks a bunch

Eric Stucky

npnp

Axoren

So, we're really trying to say that $2^2 = (3\cdot 9^m + 5) / 32$

We also know that it held for $2^3 = 3 \cdot 9^0 + 5 = 8$

So we have to also show that $2^4 = (3 \cdot 9^m + 5)/8$

We know that for even $x$, there's no integer solution, so $3 \cdot 9^m + 5 \not \equiv 0 \mod 8$

Since we know that, we know that it can't be divisible by $32$, because it would have to be divisible by $8$.

Eric Stucky

02:58

Hey Mike :)

just wanted to say, I'm interested in what you have to say, and I'll hit you up soon

But I have a final to do tonight, gotta get to it ;)

see you around

Mike Miller

Good luck!

Joshua Lamusga

04:21

I have to find the normal vector of <7cos(t), 7sin(t), 7ln(cos(t))> at point (7, 0, 0). That point is t = 0. The normal is T/T' and T is r/r' where r is the vector I was given. I get r’: <-7sin(t), 7cos(t), -7tan(t)>
|r’|: 7sqrt(sec^2(t))
T(0) : <0, 1, 0>. This is correct.
T: 1/7sqrt(sec^2(t)) * <-7sin(t), 7cos(t), -7tan(t)>
So only the center coordinate can change since sin(0) and tan(0) are 0, but the derivative of sqrt(sec^2(t))cos(t) is 0. How is the normal not <0, 0, 0>?

TheGreatDuck

06:09

does anyone know how to solve this?

4

Q: Explicit formula for floor(x)?

In number theory we have so-called explicit formula's in terms of the Riemann zeta zero's. For instance to count the sum of the logarithms of the primes below some given integer. ( second Chebyshev Function ) Consider the floor function : http://mathworld.wolfram.com/FloorFunction.html Is there...

anon

06:20

the explicit formula for psi(x) in terms of zeta zeros is essentially a fourier series. by asking us not to give a fourier series, OP is essentially telling us not to answer the question. his loss.

and yours too I guess.

TheGreatDuck

well then that sentence needs to be ignored

if it can be answered without using the arctan formula that was given in the other question (simply because people said it was flawed) then it should be and not left to sit and rot

Tobias Kildetoft

06:39

@AlexClark So, is it done or are those last cases taking a long time?

Balarka Sen

06:59

@anon curious; in what sense is Riemann's formula a fourier series?

anon

with $e^{a+bi}=e^ae^{ib}$, the imaginary parts of zeta zeros are frequencies and the real parts contribute to their amplitudes. if there were a real part >1/2, it would mean there are zeta zeros that contribute more than others to the growth of primes.

Balarka Sen

aha, I see.

anon

(of course it's actually $x$ as the base of the exponentials, but that's reaslly $e^{\ln x}$, so things are just a bit logarithmic in $x$)

Balarka Sen

right

interesting aspect

I have never had a particular intuition for Riemann's formula, myself. I "know" how to prove it, but...

Vrouvrou

07:51

hello

user 1618033

@robjohn hey! Did you have a chance to work on my problem?

(the limit)

robjohn