> There is no term in the sequence that has a value of exactly 1/3 - the sequence has a value of 1/3. You get hung up looking for a specific term in an infinite sequence; that’s nonsensical.
No, the sequence does not have a value of 1/3, the limit of the sequence is 1/3. There’s a difference. But leaving that aside. You agree with me that the sequence never contains a term with the value 1/3. But it is true to say that it does contain every possible way of writing a decimal point followed by a series of 3s, including the one where you never stop writing the threes. So the limit cannot be written that way. The formula used to calculate a repeating decimal is a/(1-r). But this is drawn from the more general formula a(1-r^n)/(1-r) by taking the limit of the formula as n tends to infinity. If you don’t take the limit you always have that a(-r^n)(1-r) left over, and that will be the difference between the fraction and the repeating decimal. As n increases in size that fraction gets smaller and smaller, but it can only become zero when you take the limit. The given sequence must contain 0.3~, but cannot contain 1/3. So in order to be able to express 1/3 as a decimal you have to define 1/3 as 0.3~
No, “the one where you never stop writing the threes” is not one of the terms of the sequence. (If it were, you could tell me which term it is: the 10th term? the 476th term? the 1,000,000,000th term?)
> It’s “actually equal to -1/12” is shorthand for: Many summation methods are used in mathematics to assign numerical values even to a divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of −1/12, which is expressed by a famous formula
If you are talking about Ramanujan summation, call it that, and don’t mix it with ordinary summation, because the two are not the same thing. You can say that the Ramanujan summation of the natural numbers is -1/12, but you can’t say that adding them up will give that result. The two are different operations. Would you say that 5 = 6 because 2 + 3 = 2 * 3? No, that’s just nonsense. You find the result ‘interesting’ because it is an actual number whereas the sequence actually diverges, so the two operations are clearly producing different results. In the video one of the two asks if typing the numbers into a calculator and adding them (if such a thing were possible) would give you -1/12, and the other said yes!
I’ve not looked at the mathematics behind alternate summation methods closely. I can say I pulled three simple errors out of their version of it. I hope that there is a real proof that more solid than that!
You’re not going to be convinced so I’m going to bow out and let the real mathematicians explain where your reasoning has gone off the rails. Having followed the many threads on this board on the subject I can safely say that your starting from an incorrect assumption and it’s leading you astray.
There is no “the one where you never stop writing the threes”. There’s only the limit, which you’ve accepted is equal to 1/3. Everything else is just you being unhappy that mathematicians aren’t speaking precisely enough according to your non-mathematician standards.
I think this sums up the interminable objections to this rather nicely. Mathematicians generally have an easy time of things since 99% of their realm is so far beyond the ken of the opinionated non-mathematician. Feel for evolutionary biologists and cosmologists, since every ignoramus seems to feel qualified to express an opinion on the veracity or mechanisms of biological evolution and the nature of the universe. Then there’s linguistics, of course…
What do you mean by “ordinary summation”? Because the summation operator on real numbers isn’t the same operation as the one on the integers, either, and infinite summation isn’t the same as either of them. The only reason we think of them as the same is that (for instance) for any numbers for which integer summation is defined, real summation gives the same value. But that’s true of Ramanujan summation, too: For any sequence of numbers for which a convergent sum is defined, the Ramanujan summation gives the same value. If we’re willing to give real summation the same name as integer summation, then, why not do the same for Ramanujan?
A long long long long time ago, we had a thing called “long division”. It was a bitch but it gave good answers. All this mathemagical mumbo-jumbo is just giving me a headache. 3 into 10 is 3 plus a remainder, bring down another zero; 3 into 10 is 3 plus a remainder, bring down another zero; rinse, spin, repeat. Unless we’re filling out tax forms, that’s all there is to it.
Why would you bring that up? That’s exactly where engjs is stuck, thinking of a series as a process and unable to escape from the “plus a remainder” part of that process.
I don’t know much about math but the Dope has worked this one to death. 0.333… is 1/3. It’s just a different notation for the same value. You have to imagine that the … means a continuing sequence of 3s that goes on forever, to infinity and beyond, it never stops, never ends. So in order to ever finish writing down the number we abbreviate to 0.333…, or use 1/3. And we know they are the same value because there are no other numbers between 1/3 and 0.333… If you add or subtract anything but 0 to 0.333… you will get a value less or greater than 1/3. The same applies to 0.999… = 1. In fact any integer can be expressed in the same way, it’s a characteristic of the notation we use, not some tricky mathematical principle. If I can understand this, I who has to count my fingers three times to be sure there are 10, then anyone can by forgetting about some possible exception that does not exist. It’s really no different than 1 = 1.000… There’s nothing magical about the infinite sequence of 0s there, if you add or subtract anything but 0 to 1.000… you will get a value less or greater than 1. It comes down to that old bugaboo that has thrown me in the past, and I feel better that it has thrown others as well, the graphic representation of a number is not the number itself, the number is a concept, the notation is a representation of that concept as agreed on by people. Changing the representation doesn’t change the concept.
I disagree … engjs is stuck when he stops the process … and I’m guessing throwing away the remainder and saying “see, it’s not 1/3”. Of course at some point it is okay to stop when we’re close enough to 1/3 to make no difference, three or four iterations is usually good enough. To say 1.00 / 3.00 = 0.333 is close enough is fine generally.
There is no "the one where you never stop writing the threes". to quote a previous poster.
Spot on correct TriPolar … “it’s a characteristic of the notation we use, not some tricky mathematical principle” … 1/3 = 10sup[/sup] is no better. Where … means irrational.
FTR, Autocad and the CAM systems it talks to cares deeply about where the decimal place stops. I’ve had drawings crash the system because my computer couldn’t handle trilionths.
“The fact that it is infinite means it can never reach its limit.”
There’s your problem. The fact that it is infinite means no finite subsequence (e.g. the first N terms for some integer N) can reach its limit.
But plenty of infinite sequences reach their limit.
Take the sequence of partial sums of (1/2) minute + (1/4) minute + (1/8) minute + … + (1/2^n) minute + … that started a minute or so ago. The fact that we’re in another minute says that sequence reached its limit. Time did not come shuddering to a nearly full stop approaching the end of that minute, just because I defined a converging infinite sequence on it. That sequence, that minute, finished up, and we went on to the next minute.
Again, you’re confusing the sequence itself with…what? A passenger who is riding the infinite length of the sequence?
For the sequence is the whole thing, the whole countably infinite set of terms, in a specific order. Maybe someone who’s stopping at the first term in the sequence today, then at the second term tomorrow, then the term after that the day after that, forever after, will never get to 1/3. But the sequence as a whole is already there. Because ‘being there’ means ‘given any epsilon > 0, there’s an N such that every term after the Nth term is within epsilon of ‘there’.’
You say the sequence never gets all the way to 1/3? Good - tell me how far away from 1/3, at what infinitesimal distance short of 1/3, the infinite sequence (.3, .33, .333, …) stalls out. I’ll prove you wrong.
I always like to think of calculus, and epsilon-delta proofs, as a kind of legal challenge, as in a courtroom.
Go ahead and specify a delta (greater than zero) that is as small as you want. Go ahead, pick one. I can construct an epsilon (also greater than zero) that’s closer to the limit than that delta.
Same with .999~ Go ahead and name a tolerance. Suppose the other guy says “one part in ten million.” Easy: .9999999. Suppose the other guy says “one part in 100 million.” Easy: .99999999. No matter what he says, I can provide the answer. The really important thing is: he can never win. It is always possible for me to provide an answer within any tolerance he demands. In the end, he has to go home, stymied.
RTFirefly makes the same point. “You say the sequence never gets all the way to 1/3? Good - tell me how far away from 1/3, at what infinitesimal distance short of 1/3, the infinite sequence (.3, .33, .333, …) stalls out. I’ll prove you wrong.”