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 /images/graemlins/wink.gif

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 /images/graemlins/smile.gif

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

[kpux]

[ QUOTE ]
By definition actually we know that 0.9999....<1.

[/ QUOTE ]

This is wrong. There is no point on the real number line between these two values.

[GrekeHaus]

[ QUOTE ]
By definition actually we know that 0.9999....<1. This is an expression of the smallest number possible less than 1 in fact.

[/ QUOTE ]

One of the properties of the real numbers (and the rational numbers) is that for any two numbers, there is a number in between them. For example, if our numbers are r and s, then the number (r+s)/2 is between them. As a corrollary to this, for any number, there is no largest number less than it. If a number x < 1, then there exists a number between them (i.e. (1+x)/2).

If you don't believe this, you can use a method similar to the one described above to prove it.

Let x = 0.9999...
then 10x = 9.9999....

The subtract the two lines

10x = 9.9999...
- x = 0.9999...

Gives 9x = 9, so x=1.

[ QUOTE ]
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.

[/ QUOTE ]

After a finite ammount of time you won't be there. The definition of an infinite sum just says that the sum of the terms approaches that number as the number of terms gets arbitrarilly large. The problem is that you're thinking in finite terms.

[K C]

Requiring a number between two numbers is merely a convention as I see it, and it's going to lead us to a reductio ad absurdum anyway, since we'll have to justify the mumber in between as well, and the one in between these, and so on.

As for the equation you put forth, that's certainly the best justification for the two numbers to be identical, yet we end up with two distinct numbers being equal. The main question here is whether it would be correct to assume 10x is 9.999... It may seem so, but perhaps not. In order for this to be possible, we're moving the decimal one place to the left. Can we really do that with an infinite series of decimals though? It appears to me at least that the decimal series of 10x is going to be infinitely smaller than that of x due to all this, at least conceptually, which is all we're dealing with here anyway.

Now as far as the problem goes the way I posed it actually presents it as a finite problem, as looking into at what point in time can it be said that Point B is reached. Given any x where x is a point in the series we do know that it will not.

In a sense it's the failure of this problem which sheds the most light on Zeno's paradox. Instead of the original discussion imagine this taking place instead, taking it from the point where Achilles is conceding defeat to the Tortoise:

Wait a minute Achilles - under these assumptions the tortise will never make it to 12 meters, let alone the finish line.

Achilles: Duh?

You cover ten times as much distance as he does in a given timeframe. So each advance by him represents 1/10 the previous advance he's made. This mumber keeps decreasing by 1/10 each time. Since he covers 1 meter the first time, he only covers 1.1 meters the second, 1.11 the third, and so on. The decimal of 1 keeps increasing but he'll never even get past this, so he cannot possibly win the race under this conception.

Achilles: Hmm I think I see what you mean. (Looking over at the tortoise) Hey, you tricked me, you little *&^%$%

The tortoise meekly retreats into his shell...

KC

[jason_t]

[ 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 ]

The notation

1/2 + 1/4 + 1/8 + ... + 1/(2^n) + ... = 1

stands for

lim n->infinity sum from k = 1 to n of (1/2)^k = 1.

Anyone who tells you anything different is wrong.

[jason_t]

[ QUOTE ]
Requiring a number between two numbers is merely a convention as I see it

[/ QUOTE ]

It's not a convention. It's a consequence of basic properties of real numbers that follow from the axioms governing the natural numbers, the construction leading to the real numbers, and basic principles of logic. Or, if one wants to accept the real numbers as basic objects, it is a consequence of the axioms governing the real numbers that for any two real numbers x < y there is a real number z with x < z < y. Quite simply, one can take z = (x+y)/2.

Again, it's not a convention.

[jason_t]

[ QUOTE ]
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.


[/ QUOTE ]

This is absurdly wrong, as others have already elaborated. But I want to reiterate and elaborate: the statement under discussion here is not a convention. Mathematics doesn't work by making conventions about its facts. Mathematics works by assuming some basic, reasonable principles and deducing facts using the principles of logic.

Convention: 32F is the freezing point of water.
Fact: There is no real number between .999... and 1.

[FNHinVA]

I have nothing to add to this discussion. I proved that in the "Space, Time & Stephen Hawking Jive" thread I initiated (in a momentary lapse in judgement).

I will make an observation, though.

Watching Math Mystics argue among themselves to the point of describing another's view as "absurdly wrong" is infinitely amusing.

That said, I will run for cover.

[jason_t]

[ QUOTE ]
I have nothing to add to this discussion. I proved that in the "Space, Time & Stephen Hawking Jive" thread I initiated (in a momentary lapse in judgement).

I will make an observation, though.

Watching Math Mystics argue among themselves to the point of describing another's view as "absurdly wrong" is infinitely amusing.

That said, I will run for cover.

[/ QUOTE ]

Your attempt to add to this discussion is akin to the following:

Person A: Flushes beat straights because the event (five cards are of the same suit) is less likely than the event (five cards are in sequence).
Person B: It's just a convention that that is true.
Person A: That's absurdly wrong. It's not a convention. It's a fact because the probability of the two events can be calculated exactly and it turns out that flushes are less likely than straights.
You: Person A, you can't call his view "absurdly wrong." It's his view, his opinion, his way of looking at things.

This isn't art, this isn't philosophy. These aren't opinions. They aren't open to debate. They are right or they are wrong. The OP is wrong.

[FNHinVA]

Your off-point response to my observation is also amusing.

[PairTheBoard]

[ QUOTE ]
I have nothing to add to this discussion. I proved that in the "Space, Time & Stephen Hawking Jive" thread I initiated (in a momentary lapse in judgement).

I will make an observation, though.

Watching Math Mystics argue among themselves to the point of describing another's view as "absurdly wrong" is infinitely amusing.

That said, I will run for cover.

[/ QUOTE ]

What I'm watching is someone who doesn't know math trying to argue math with someone who does know math. I think the "Math Mystics" are posting on these Forums:

http://p205.ezboard.com/fembracingthemoonfrm38

PairTheBoard

[ QUOTE ]

As for the equation you put forth, that's certainly the best justification for the two numbers to be identical, yet we end up with two distinct numbers being equal.


[/ QUOTE ]

Here is your misconception. 0.999999... and 1 are not two distinct numbers, but merely two different names for the same number. Just as 1/2 and 2/4 are the same.

[kpux]

this is pretty trivial, and it's not really a proof, but it is mildly entertaining:

1/9 = .11111....
2/9 = .22222....
.
.
.
.
9/9 = .99999....
but 9/9 = 1

[K C]

[ QUOTE ]
Mathematics doesn't work by making conventions about its facts. Mathematics works by assuming some basic, reasonable principles and deducing facts using the principles of logic.

Convention: 32F is the freezing point of water.
Fact: There is no real number between .999... and 1.

[/ QUOTE ]

Why is this a "fact"? Actually convention was too strong a word - it's merely an assumption, then you're using this assumption to refute. That won't work. Show me how this statement must be necessarily true. This doesn't include using some mathematical assumptions about it and claim that if it conflicts with these assumptions it is false.

KC

[K C]

[ QUOTE ]
this is pretty trivial, and it's not really a proof, but it is mildly entertaining:

1/9 = .11111....
2/9 = .22222....
.
.
.
.
9/9 = .99999....
but 9/9 = 1

[/ QUOTE ]

How is 9/9 .99999....? My trusty calculator does not agree /images/graemlins/smile.gif

By definition, .99999... does not reach 1.0. It is fractional. We have a whole number and one that is infinitely fractional. A principle here that we cannot overlook is the fact that if we can round up a number it is by definition distinct from the number it is rounded up to.

This has been fun but I'll let you guys each have one more flame at this and as I'm going to mercifully move on here /images/graemlins/smile.gif

KC

[Neal_Schon]

Trust me, Neal Schon, Journey guitarist, Rock god, that 0.999999999... does, in fact = 1.

[jason_t]

[ QUOTE ]
[ QUOTE ]
Mathematics doesn't work by making conventions about its facts. Mathematics works by assuming some basic, reasonable principles and deducing facts using the principles of logic.

Convention: 32F is the freezing point of water.
Fact: There is no real number between .999... and 1.

[/ QUOTE ]

Why is this a "fact"? Actually convention was too strong a word - it's merely an assumption, then you're using this assumption to refute. That won't work. Show me how this statement must be necessarily true. This doesn't include using some mathematical assumptions about it and claim that if it conflicts with these assumptions it is false.

KC

[/ QUOTE ]

A few posters have already given you two very nice explanations as to why it's true. It's a fact because its truth can be derived from basic mathematical principles and the rules of logic. It is not merely an assumption that .999... = 1. It follows from definitions and basic mathematical reasoning.

Quite frankly, you have no idea how mathematics works and your understanding of it seems incredibly limited. Either that, or you're intentionally being difficult. If it's the former then there is no point discussing this topic further. If it's the latter, stop it. Whatever the case may be, either think about what others have already posted or give up.

[jason_t]

[ QUOTE ]
[ QUOTE ]
this is pretty trivial, and it's not really a proof, but it is mildly entertaining:

1/9 = .11111....
2/9 = .22222....
.
.
.
.
9/9 = .99999....
but 9/9 = 1

[/ QUOTE ]

How is 9/9 .99999....? My trusty calculator does not agree /images/graemlins/smile.gif

[/ QUOTE ]


1/9 = .111...

Now multiply both sides by 9 to obtain

9 * (1/9) = .111...

and thus

1 = .999....

[ QUOTE ]
By definition, .99999... does not reach 1.0. It is fractional. We have a whole number and one that is infinitely fractional. A principle here that we cannot overlook is the fact that if we can round up a number it is by definition distinct from the number it is rounded up to.

[/ QUOTE ]

You don't understand the definitions. In my last post I gave you credit for possibly being intelligent but just being difficult. Now it's clear: you have no understanding of mathematics at any level. For example, among other things, you use the meaningless phrase "infinitely fractional."

[ QUOTE ]

This has been fun but I'll let you guys each have one more flame at this and as I'm going to mercifully move on here /images/graemlins/smile.gif


[/ QUOTE ]

You should "mercifully" move away so you don't continue to look retarded. At the least, stop trying to argue in an area where you don't know anything.

[JoshuaD]

[ QUOTE ]

By definition, .99999... does not reach 1.0. It is fractional.

[/ QUOTE ]

That doesn't mean anything.

[ QUOTE ]
infinitely fractional.

[/ QUOTE ]

huh.

[ QUOTE ]
A principle here that we cannot overlook is the fact that if we can round up a number it is by definition distinct from the number it is rounded up to.

[/ QUOTE ]

We're not rounding. .999... = 1. The end. .9999999999999999999999999999999999999999999999999 999999999999999999 doesn't equal 1. If you put a bazillion 9's after that . it still doesn't equal 1. Only when there are an infinte number of 9's do we have the case that the two numbers are the same.

As posters have already said in this thread, the way we know two numbers are different is by showing that there's a number between them (coincidentally, if there's a number, there's an infinite number of numbers between them). 1 and 2 are different because 1.5 sits between them.

Either demonstrate there exists a number between .9999.. and 1, or realize they're the same.

P.S. The number N = (1 + .9999...) / 2 doesn't work because it begs the question. It assumes the two numbers aren't equal, if they are equal, N = 1.

[Dov]

[ QUOTE ]
.S. The number N = (1 + .9999...) / 2 doesn't work because it begs the question. It assumes the two numbers aren't equal, if they are equal, N = 1.

[/ QUOTE ]

Maybe I missed a post somewhere, but why does this need to be said?

What is wrong with N=(1+1)/2=1 ?

What 2 numbers are you assuming to be unequal? (I think you mean the 1 and .999... but I just want to be sure.)

Maybe I just missed the context and you were trying to get a useful N value for something and it can't be 1.

Sorry for rambling.

[JoshuaD]

[ QUOTE ]
[ QUOTE ]
.S. The number N = (1 + .9999...) / 2 doesn't work because it begs the question. It assumes the two numbers aren't equal, if they are equal, N = 1.

[/ QUOTE ]

Maybe I missed a post somewhere, but why does this need to be said?

What is wrong with N=(1+1)/2=1 ?

What 2 numbers are you assuming to be unequal? (I think you mean the 1 and .999... but I just want to be sure.)

Maybe I just missed the context and you were trying to get a useful N value for something and it can't be 1.

Sorry for rambling.

[/ QUOTE ]

Hey. It's usually pretty easy to find a number (N) between two numbers (A and B) by using this formula: N = (A + B) / 2.

In the case of A = 4, B =2, we get N = 3, and we see that 4 and 2 are in fact different numbers.

It doesn't work for this thing because there's nothing about (1 + .999.. ) / 2 in anyway that suggests it's a different number than .999... or 1.