JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 1, posted (12 years 9 months 1 day 6 hours ago) and read 1754 times:

10001*5000=5,000,5000

Not sure you can have a sum on an infinite equation, can you?

Mcringring From , joined Dec 1969, posts, RR:
Reply 5, posted (12 years 9 months 1 day 6 hours ago) and read 1740 times:

No, it will never reach 2. If you keep cutting the numbers in half and adding them, even infinitely, it won't get to 2. Essentially, the number will be so close to 2 that it's meaningless to distingush the two, but it will never be exactly 2.

Mcringring From , joined Dec 1969, posts, RR:
Reply 7, posted (12 years 9 months 1 day 5 hours ago) and read 1722 times:

Basically all you're doing is using math to come back to the original equation.

OK, let's think about this another way. You have a piece of paper. You cut it in half. Again and again and again... an infinite number of times. There will always be something left. Applying to this equation; you keep adding half as much as you previously added. That means there is always half as much "space" available between the number and 2. If you do this infinitely, there will always be the 1/infinity difference between the "end" result and 2.

You should graph this equation and see if the numbers ever reach exactly 2. They won't.

Airways1 From United Kingdom, joined Jul 1999, 561 posts, RR: 0
Reply 8, posted (12 years 9 months 1 day 4 hours ago) and read 1714 times:

Mcringring, your paper example is not very good because you're using a practical example for a mathematical problem. No matter how many times you cut the paper in reality, it will never be an infinite number of times.

If it were cut an infinite number of times, then as you rightly said there will always be the 1/infinity difference between the "end" result and 2.

But what does 1/infinity equal? Zero. And zero difference between the end result and 2 means the end result is 2.

Mcringring From , joined Dec 1969, posts, RR:
Reply 11, posted (12 years 9 months 1 day 4 hours ago) and read 1688 times:

OK, the equation converges toward 2. Converge, defined in math means "To approach a limit," key word being "approach." If you look at a graph of this equation, you will see this convergence.

Beefmoney From United States of America, joined Oct 2000, 1118 posts, RR: 3
Reply 13, posted (12 years 9 months 1 day 4 hours ago) and read 1683 times:

Its kind of like the problem, If you stand 10 feet away from a wall, then walk half that distance to the wall (5 feet) then half that distance to the wall (2.5 feet) then half that distance..... you will get extremely close to the wall, but never actually touch the wall.

Mcringring From , joined Dec 1969, posts, RR:
Reply 14, posted (12 years 9 months 1 day 4 hours ago) and read 1677 times:

Yeah, Beefmoney, but according to some people...

your example is not very good because you're using a practical example for a mathematical problem. No matter how many times you [walk the distance] in reality, it will never be an infinite number of times.

Of course, no example is perfect. In fact, the essence of an example is that it is similar, but different. But some people will try to find excuses for anything.

D L X From United States of America, joined May 1999, 11686 posts, RR: 52
Reply 17, posted (12 years 9 months 1 day 2 hours ago) and read 1654 times:

McRing,

Airways1 is 100% correct in his example where he defines s as the infinite series. That solution is straight out of a calculus textbook. The sum over an infinite number of terms often comes to an exact value, in this case exactly 2. (Not just under 2, but exactly 2.) I think the problem with your thinking is that you are finitizing the sum. If you think of it as taking the first million terms and summing them, then yes, the sum is less than 2. If you take the next million, the sum is still just under 2. If you ever stop summing, the sum is less than 2. But the idea of the infinite series is that you never stop summing. So, after an infinite amount of time (assuming of course that it takes non-zero time to sum a term) the sum of these terms will be *exactly* 2.

Send me a PM at http://www.airliners.net/aviation-forums/sendmessage.main?from_username=NULL

Mcringring From , joined Dec 1969, posts, RR:
Reply 19, posted (12 years 9 months 1 day 1 hour ago) and read 1644 times:

OK, the formula is designed to calculate where the series converges. I don't disagree that this converges to 2. Above I have given the definition of converge in the mathematical sense.

D L X: the idea of the infinite series is that you never stop summing

Exactly. If you never stop summing, you will never get to 2.

D L X From United States of America, joined May 1999, 11686 posts, RR: 52
Reply 20, posted (12 years 9 months 1 day 1 hour ago) and read 1643 times:

McRingRing,

try this infinite series:

1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + ... = ?

Using a method like Airways1's textbook solution, the answer is clear. But, doing it in your head, or trying to find a physical model will definitely lead you to the wrong answer.

Send me a PM at http://www.airliners.net/aviation-forums/sendmessage.main?from_username=NULL

JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 22, posted (12 years 9 months 1 day 1 hour ago) and read 1635 times:

Airways1, you were using the same method that proves .999R = 1

Mcringring From , joined Dec 1969, posts, RR:
Reply 24, posted (12 years 9 months 22 hours ago) and read 1616 times:

DLX, if in fact you do reach 2/3, you've reached the end. Unless you're working with another definition of infinity, this is impossible.

25 D L X
: Well, I assume that .999R means 9/10 + 9/100 + 9/1000 + ... right? In that case, after summing an infinite amount of times, the sum will in fact be 1.

26 777236ER
: You can prove that 0.9999... is equal to 1 without summing. Let x = 0.9999..... So 10x = 9.9999.... Take them away, 9x=(9.99999...)-(0.99999...) There

27 D L X
: 777236ER, way to read the earlier posts. Well done...

28 Klaus
: 777236ER: You can prove that 0.9999... is equal to 1 without summing. Let x = 0.9999..... So 10x = 9.9999.... Take them away, 9x=(9.99999...)-(0.99999

29 Klaus
: Dang, that evil filter did it again. The last part should read: lim(SUM(1/2^m)) is exactly = 2. (n=0...oo) (m=0...n) As I said, the lim() is often imp

30 D L X
: McRing, please define infinity. I want to make sure we're on the same page.

31 NWA
: #2 will never reach 2, it will get very close though, infnitley close.

32 707cmf
: I've always loved the 0.99999R thing. It is indeed equal to 1, and there several ways to prove it. 0.99999R = 3x0.333333R right ? 0.3333333R = 1/3, ri

33 Mcringring
: Well, definition mathematically: the limit of 1/x as x approaches zero. more user friendly definition: a boundless number. so if x = infinity, x + 1 =

34 D L X
: Oy. Instead of finding anything wrong with the mathmatical proofs that Airways1 has provided, you continue to try to manipulate words to show that it'

35 Mcringring
: There is nothing wrong with the proofs. It's just that you and Airways1 are trying to use them to prove something they were not designed to prove. If

36 Airways1
: Mcringring, on the contrary, the equation was 'designed' exactly for this purpose. See my post above where I derived the equation. And as for what you

37 Klaus
: Airways1: But as a mathematical concept, you can sum an infinite number of terms at which point the sum becomes exactly 2. That´s a popular misconcep

38 Airways1
: OK, Klaus, I didn't word that very well, I shouldn't have written 'at which point', because as you rightly said, there is no such point. But you say '

39 Klaus
: Airways1: But you say 'the infinite sum implies a limit transition'. Well I did specify initially that this is an infinite sum, so with 'limit transit

40 Killjoy
: My math book is a bit unclear on the subject. It says: "The series sum[n=1 to inf] 1/(2^n) has the following partial sums. -listing of partial sums le

41 Klaus
: For most practical purposes, the individual partial sums aren´t really interesting. So the lim() operator takes care of it and gives you the limit in

42 Killjoy
: Klaus, I'm not exactly sure who you're answering, but if your last comment was in response to my question... I understand the concept of getting infin

43 Mcringring
: Thank you Klaus. Maybe they will listen to you. They sure won't listen to me.

44 Killjoy
: Oops, I just noticed the topic of Klaus' post was "Killjoy" so I guess he *was* answering me .

45 Klaus
: Yes, I was talking to you. The limit transition is usually implied with an infinite sum. There are many such "shorthands" in math. It would often be b

46 Killjoy
: Ah, now I get it, thanks Klaus. I'd never actually heard the term "limit transition" before, damn that book for not explaining the thing properly... N

47 Klaus
: Yeah, I think you´ve got it right. The "0.9999..." is another place where the "..." signifies infinity and therefore implies yet another limit transi