Let's discuss Mathematics

[sanabas]

Nobody has proved this? I did not read all the posts here, but I did read the Wikipedia link provided by Paradigm Shifter.

EDIT: The Wikipedia link does specify a positive integer. My mathematical notation is rusty, and typing out a proof in mathematical notation on a forum like this is a bit awkward.

Second EDIT:
It looks like in fact we eventually do get to a number divisible by 4.
If we express an odd number as A*2^N-1, (where A is an odd number and N is an integer greater than zero) then:
a. After (2N-1) iterations, we will arrive at a number divisible by 4.
b. This number divisible by 4 will be 2*(A*3^N-1).

Nobody has proved you'll always reach 1.

Proving you always reach a number divisible by 4 is easier, and you have come up with the proof I was hinting at with:

I think it's merely something interesting that I came up with the first time I met this series. But I couldn't see how it helped me to prove the series. I'll drop an even bigger hint.

Any odd number can be written as (p.2^n) - 1, where p is odd. Correct? What happens when you iterate from that?

 

[Atticus]

What do you mean by p.2^n ?

 

[sanabas]

p * 2^n

Was using . for multiply

 

[Atticus]

Ok. I thought for a while it doesn't work, but some more thought made it obvious that it does.

 

[Harv]

Nobody has proved you'll always reach 1.

Proving you always reach a number divisible by 4 is easier, and you have come up with the proof I was hinting at with:

Any odd number can be written as (p.2^n) - 1, where p is odd. Correct? What happens when you iterate from that?

I found where you said that. Actually, I did not give the proof above. So here it is.

First we define the iterative function f, so that:
f[n+1] = f[n]/2 if f[n] is even
f[n+1] = 3*f[n]+1 if f[n] is odd

It follows that if f[n] is odd, then
f[n+2] = 1.5*f[n]+0.5 and
f[n+2] = 1.5*(f[n]+1)-1

If f[n+2] is odd, then
f[n+4] = 1.5*(f[n+2]+1)-1 = 1.5^2*(f[n]+1)-1

If we express an odd number f[n] = A*2^N-1 where A is an odd number, then
f[n+2N] = 1.5^N*(A*2^N) - 1 = A*3^N - 1 which is an even number.

It follows that
f[n+2N-1] = 2*(A*3^N-1) which is divisible by 4.

I am sorry about the notations which may be difficult to follow. The square brackets following the f as in f[n] is intended as a subscript.

 

[nc-1701]

<snip>
It follows that if f[n] is odd, then
f[n+2] = 1.5*f[n]+0.5 and
f[n+2] = 1.5*(f[n]+1)-1

Not true...
Take n=7
f[9]=28
But 1.5f[7] +0.5 = 1.5*22+0.5=33.5

 

[sanabas]

Not true...
Take n=7
f[9]=28
But 1.5f[7] +0.5 = 1.5*22+0.5=33.5

You've misinterpreted something in there.

If f[n] = 7

then f[n+1] = 3 * 7 + 1 = 22

then f[n+2] = 22/2 = 11 = 1.5 * 7 + 0.5

 

[nc-1701]

You've misinterpreted something in there.

If f[n] = 7

then f[n+1] = 3 * 7 + 1 = 22

then f[n+2] = 22/2 = 11 = 1.5 * 7 + 0.5

<facepalm>

 

[Mise]

This is cool: Snooker break simulation

 

[dutchfire]

Nice.
Someone in my class in uni did a similar project once, though a bit simpler IIRC.

 

[Mise]

Stats question: if I have a model that fits data points going back to 2010, and another that fits the same data but only to 2013, is there any statistic that will say "model A is better than model B because, even though the fit is worse over the respective periods, it fits more data" or alternatively "model B is better than model A because, even though the fit is worse over the whole period, model B fits recent data better than model A"? Basically I can either fit 2013 data really really well, or 2010-present data "kind of okay". I just want a number that will tell me which is better.

I mean so far I've just weighted the square differences by recency before calculating the R^2, which I'm hoping is good enough if I just want to compare the two models, but I'd rather use some sort of standard statistical measure instead.

 

[azzaman333]

I can't think of any statistical measure that would compare something like that. I can't remember even looking at a statistic that might rate a lot of ok errors better than a small set of really low errors.

 

[nerdfighter13]

a=b
a+a=a+b
2a=a+b
2a-2b=a+b - 2b
2(a-b)=a+b -2b
2(a-b)=a-b
2=1

Saw this in a youtube comment. Is this sound reasoning or not?

 

[Harv]

a=b
a+a=a+b
2a=a+b
2a-2b=a+b - 2b
2(a-b)=a+b -2b
2(a-b)=a-b
2=1

Saw this in a youtube comment. Is this sound reasoning or not?

No. It is not.

2(a-b)=a-b

Given a=b, a-b=0 so you really have 2x0=0. In the next line you have a divide 0 by 0, and when you just used 0/0 = 1, you got a bizarre result like 2 = 1.

 

[azzaman333]

No. It is not.



Given a=b, a-b=0 so you really have 2x0=0. In the next line you have a divide 0 by 0, and when you just used 0/0 = 1, you got a bizarre result like 2 = 1.

And this is why you don't listen to youtube comments for anything remotely intellectual.

 

[Mise]

Even if you didn't see any of the preceding lines of workings, there is only 1 possible solution to the equation "2(a-b)=a-b", and that's that a=b=0.

Thanks for your help azzaman!

 

[sanabas]

Even if you didn't see any of the preceding lines of workings, there is only 1 possible solution to the equation "2(a-b)=a-b", and that's that a=b=0.

Thanks for your help azzaman!

Not sure if you've made a typo or have it wrong. If you meant to say 'and that's a-b=0', then it's a typo. If you did mean to say a=b=0, then you're wrong. a=b=1 is a solution, so is a=b=2, so is a=b=anything else.

 

[Mise]

Yeah, I meant to say a-b=0, not a=b=0 What I meant was that, when you see that formula, you can either believe that 2=1, or you can believe that a-b=0. Since 2=1 isn't true, the only solution is that a-b=0.

 

[nerdfighter13]

Thanks, I thought it wasn't right, but i didn't know why.

 

[Harv]

You know it is not right when you get a bizarre result like 2=1. Once you know the answer is not right then you can go through everything and find the error.

It was fun!