Some More Infinite Series Jive

[K C]

Zeno's paradox doesn't have much meat on it when it's all said and done, so I've come up with a variation of this which is a bit trickier [img]/images/graemlins/wink.gif[/img]

We know already that an object under a constant rate of speed will traverse A to B in a finite time. Whatever notions that we may have that contradict this will be false simply by virtue of the logical contridiction. Ho hum [img]/images/graemlins/smile.gif[/img]

I was reading a quote from one of our posters who suggests the following as a solution to the original paradox:

"Basically, you can add up an infinite number of positive real numbers and have it come out to a finite sum. For simplicity, let's say the distance A|B = 1. In this example, 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 1."

This isn't true at all actually. You'll never get to 1 that way, and this gave me an idea for a variation of the paradox.

We assume that an object in constant positive motion will eventually get from point A to point B. After all, with each second that passes it will get closer, so it must eventually get there right? Certainly if we allow an infinite time to do it, right?

So here's the scenario. We have an object at Point A which is currently travelling 1 foot per second. After the first foot, it will decellerate at the rate of 50% per second, and we're going to assume that's the average rate. In other words, in the first second it will travel 1 foot, in the second 1/2 foot, in the third 1/4 foot, etc.

We could make the distance several miles but to make this an easy target to hit we're going to set point B at exactly 2 feet away from Point A. Starting at 1 foot per second, only two feet to cover, an infinite amount of time allowed - shouldn't be too difficult right?

Does the object ever reach Point B, ever?

KC

[PairTheBoard]

"Does the object ever reach Point B, ever? "

No.

Neither would it reach point B if it stopped altogether before reaching it.

PairTheBoard

[K C]

The interesting thing about this is that it never does stop moving forward though.

KC

[K C]

Here's a little more to chew on here. The so called infinite sum postulation has its defenders outside of this forum to be sure. Here's the rationale behind it which I've taken from an online site:

"A little reflection will reveal that this isn't so strange after all: if I can divide up a finite distance into an infinite number of small distances, then adding all those distances together should just give me back the finite distance I started with."

A little reflection may lead you here, and with a little more we can discover the error. The space between A and B is infinitely divisible that's true. And if we add up all these infinite parts we will indeed get A-B. However, the sum in the above example does not contain the whole series. In fact it is designed purposely not to.

KC

[PairTheBoard]

KC --
"However, the sum in the above example does not contain the whole series. In fact it is designed purposely not to."

Unclear what "sum" you're talking about now.

PairTheBoard

[gumpzilla]

[ QUOTE ]

"Basically, you can add up an infinite number of positive real numbers and have it come out to a finite sum. For simplicity, let's say the distance A|B = 1. In this example, 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 1."

This isn't true at all actually. You'll never get to 1 that way, and this gave me an idea for a variation of the paradox.

[/ QUOTE ]

Sigh.

People just really seem to hate convergent series. I suppose you are also of the mindset that .999999... != 1.

[gumpzilla]

Because I'm a glutton for punishment, I'll elaborate somewhat on 1/2 + 1/4 + . . . .

Series that are of the form 1 + x + x^2 + x^3 + . . . can be summed formally by a neat little trick. Suppose our series sums to the value S. So,

1 + x + x^2 + . . . = S

Now, notice that we can write the series as follows:

1 + x(1 + x + x^2 + . . .) = S

The term in parentheses is just the original sum, which we are supposing to have value S. So, we find that doing a little algebra, S = 1 / (1 - x). In our case, we are interested in the sum 1/2 + 1/4 + 1/8 . . . . This is a sum of this form with x = 1/2 and the leading 1 subtracted. So, plugging 1/2 into our handy formula, we get S = 1 / (1 - 1/2) = 2, but remember we need to subtract the leading 1. So, we get 1/2 + 1/4 + . . . = 1, as promised, and we haven't used any fancy tricks or limits - yet.

The trickier among you may now point out that we can use this same trick to come up with some ridiculous conclusions. Namely, let's say I pick x = 2. So I get that the infinite series 1 + 2 + 4 + . . . = 1 / (1 - 2) = -1. Huh. So by adding a bunch of positive numbers, I get -1? Something must be wrong here. And there is something wrong. The formula derived above assumes that a value S exists for the series, in other words that the series converges. But this isn't so for the series 1 + 2 + 4 . . . . If you tell me some number n that you think this sum converges to, I can always find some number of terms to sum such that I can get a higher value than n, so there can't be any n that this sums to. These difficulties will occur whenever x is greater than or equal to 1, which you can sort of see in the formula for S: when x = 1, we get S = 1 / 0, which is clearly a sign of trouble. To use technical terminology, the radius of convergence of the series 1 + x + x^2 + . . . is 1, and you can't talk meaningfully about the series if x is 1 or larger.

So how do we know that our original series is convergent and it's okay for me to do the S trick? I'm not going to give a rigorous proof, but here's a basic idea. Let's consider S_n = 1/2 + 1/4 + . . . 1/2^n. S_1 = 1/2, S_2 = 3/4, S_3 = 7/8, . . . . Notice a pattern? We could prove by induction, if we so desired, that S_n = (2^n - 1) / 2^n. Notice that no matter how big we make n, this number is still less than 1. So unlike before, I can bound the series. And since every S_n is bigger than the one before it, they don't have any place to go other than 1, the bound. This is the kind of argument that I think people who are uncomfortable with infinite series rebel against, but remember, all we need this for is to justify the original trick we used to sum the series. If you say that the series has a sum, you can sum it exactly, without these limiting processes.

This also knocks down the .999999... = 1 question. How? What the decimal system means is that when I write .999...., what I'm really doing is concisely expressing the notion of a number that is 9*(1/10) + 9*(1/100) + . . . = 9*(1/10 + 1/100 + ...) The thing in parentheses is a sum of the same form that we had before. Using our trick, we see that S = 1 / (1 - 1/10) = 10/9. Subtracting the missing leading 1, we see that the sum in parentheses is 1/9, and so 9*(1/9) = 1.

[kpux]

[ QUOTE ]
"Basically, you can add up an infinite number of positive real numbers and have it come out to a finite sum. For simplicity, let's say the distance A|B = 1. In this example, 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 1."

This isn't true at all actually. You'll never get to 1 that way

[/ QUOTE ]

No, no. It is true. This series is equal to 1. Just because there are an infinite amount of terms doesn't mean that someone who is traversing the distance is required to sum up an infinite amount of positive numbers. This infinite series is simply another way to express the number 1. I think Zeno intuitively realized this, and asserted his paradox in a confusing, convoluted way just to, well, be a jerk.

[ QUOTE ]
We could make the distance several miles but to make this an easy target to hit we're going to set point B at exactly 2 feet away from Point A. Starting at 1 foot per second, only two feet to cover, an infinite amount of time allowed - shouldn't be too difficult right?

Does the object ever reach Point B, ever?

[/ QUOTE ]

No.

[kpux]

[ QUOTE ]
This also knocks down the .999999... = 1 question. How? What the decimal system means is that when I write .999...., what I'm really doing is concisely expressing the notion of a number that is 9*(1/10) + 9*(1/100) + . . . = 9*(1/10 + 1/100 + ...) The thing in parentheses is a sum of the same form that we had before. Using our trick, we see that S = 1 / (1 - 1/10) = 10/9. Subtracting the missing leading 1, we see that the sum in parentheses is 1/9, and so 9*(1/9) = 1.

[/ QUOTE ]

I was gonna post exactly this but you beat me to it.

[K C]

By definition actually we know that 0.9999....<1. This is an expression of the smallest number possible less than 1 in fact. If we do come up with some equations that seem to conflict with this, it is the equation that has been disproven as correct since this cannot possibly be true under this most basic mathematical premise that it is competing with.

The only premise that we need to justify this in fact is 1>0.x, where x can represent any decimal. This is a convention in fact not a postulation. It doesn't really matter how many decimal places we go here, including an infinite amount of them.

Similarly, divisions such as 2/3 don't work out to 0.67, even though it does give you a result that is as close to this as possible. To get to 0.67 or 0.667 or whatever we need to round the number up. We're not entitled to round up in these examples though.

Getting back to the original problem, we also know for a fact that the object cannot possibly reach Point B. As the series progresses, we'll get closer and closer, but at no point can it be shown that we will reach it. Where x is a given point in the series, as the distance traversed during the last second, we will always be x distance away.

KC