Annoying maths puzzle.

[Gruthar]

And my calculator has run out of battery. I'll take that as a sign to give up. I'll defer to Chief's answer, since I haven't a clue about complex numbers. D:

 

[BaronAsh]

Are you bad at geometry?

All your doing is making a spiral infinitely.

You do realize that because you're halving the distance traveled at each turn, the spiral does eventually approximate a coordinate? You know, a limit.

No, I do not understand.

If you simply halve a value you end up with half that value.

2/2=1 or 2/2=2/2 and (2/2)/2=1/2 and finally (1/2)/2=1/4

As you can plainly see it's infinite.

 

[Gruthar]

Are you bad at geometry?

All your doing is making a spiral infinitely.

You do realize that because you're halving the distance traveled at each turn, the spiral does eventually approximate a coordinate? You know, a limit.

No, I do not understand.

If you simply halve a value you end up with half that value.

2/2=1 or 2/2=2/2 and (2/2)/2=1/2 and finally (1/2)/2=1/4

As you can plainly see it's infinite.

And if you keep dividing by two, the number gets smaller and smaller. If you write your series as 1/(2^x), then as x grows infinitely large, you should be able to see that the result starts getting closer and closer to 0. In other words, the limit of 1/(2^x) as x approaches infinity is 0. This means that the sum of the series is finite (convergent). Meaning that at some point, all you're doing is adding infinitely insignificant numbers. This is best illustrated if you sum up all the terms in your series (do it by hand), you will eventually start getting infinitely close to a value of 2, but you'll never cross it.

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 and so forth.

In math terms, you say that the infinite summation of 1/(2^x) when the series starts at 0 is 2. If a geometric series is convergent (has an 'end'), you can figure out what the sum of all the terms is by using the equation a/(1-r), where a is the first term in the series, and r is the rate of change. So, to use the 1/(2^x) example, I can write that as (1/2)^x. The first term is 1, and the rate of change is 1/2 (each term is the previous term * 1/2.) So:

1/(1-1/2) = 1/(1/2) = 2

If I were to do the same thing with a series of (1/4)^x beginning at x = 0, then again the first term is 1, and the rate this time is 1/4. So:

1/(1-1/4) = 1/(3/4) = 4/3

The infinite summation of (1/4)^x thus gets infinitely close to 4/3.

Again, you can verify this by doing the first couple of sums manually.
1 + 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

I hope that clears things up a bit. I'm a terrible teacher.

 

[Lukeje]

50% the puppy is male.

I see what you did there. Of course 0.(9) = 1.

 

[Ancalagon]

50% the puppy is male.

I still hear those beagles late at night. Don't bring back the madness.

 

[hazuka3377]

I don't see how you can define your location by the degrees of rotation n.
Also, given the progression of distance (halving each instance) you won't actualy stop.

You can only give an approximation of where the sequence will no longer be worth measuring due to being that small.

 

[Daveman]

EDIT: Fuck it I don't know. Argh. Lame.

 

[BaronAsh]

Are you bad at geometry?

All your doing is making a spiral infinitely.

You do realize that because you're halving the distance traveled at each turn, the spiral does eventually approximate a coordinate? You know, a limit.

No, I do not understand.

If you simply halve a value you end up with half that value.

2/2=1 or 2/2=2/2 and (2/2)/2=1/2 and finally (1/2)/2=1/4

As you can plainly see it's infinite.

And if you keep dividing by two, the number gets smaller and smaller. If you write your series as 1/(2^x), then as x grows infinitely large, you should be able to see that the result starts getting closer and closer to 0. In other words, the limit of 1/(2^x) as x approaches infinity is 0. This means that the sum of the series is finite (convergent). Meaning that at some point, all you're doing is adding infinitely insignificant numbers. This is best illustrated if you sum up all the terms in your series (do it by hand), you will eventually start getting infinitely close to a value of 2, but you'll never cross it.

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 and so forth.

In math terms, you say that the infinite summation of 1/(2^x) when the series starts at 0 is 2. If a geometric series is convergent (has an 'end'), you can figure out what the sum of all the terms is by using the equation a/(1-r), where a is the first term in the series, and r is the rate of change. So, to use the 1/(2^x) example, I can write that as (1/2)^x. The first term is 1, and the rate of change is 1/2 (each term is the previous term * 1/2.) So:

1/(1-1/2) = 1/(1/2) = 2

If I were to do the same thing with a series of (1/4)^x beginning at x = 0, then again the first term is 1, and the rate this time is 1/4. So:

1/(1-1/4) = 1/(3/4) = 4/3

The infinite summation of (1/4)^x thus gets infinitely close to 4/3.

Again, you can verify this by doing the first couple of sums manually.
1 + 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

I hope that clears things up a bit. I'm a terrible teacher.

Your math is wrong, 1/(1-1/4) = 1/(3/4) = 1/0.75 = 1.333(repeating) = 1+1/3

Understand? Also, we're infinitely dividing by 2 not 1/2 so what was the point of your speech?

 

[Lukeje]

Your math is wrong, 1/(1-1/4) = 1/(3/4) = 1/0.75 = 1.333(repeating) = 1+1/3 = 4/3

 

[BaronAsh]

Your math is wrong, 1/(1-1/4) = 1/(3/4) = 1/0.75 = 1.333(repeating) = 1+1/3 = 4/3

Thanks for pointing that out, I just realized that. XD

This is a formal apology.

 

[ThreeWords]

*maths lesson*

I hope that clears things up a bit. I'm a terrible teacher.

Not at all. I'm doing GCSE maths, and you educated me a bit. You're a fine teacher =D