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 I Have Another Maths Problem For You (two In Fact)
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0Posted Thu Jun 6 2002 16:23:40 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1799 times:

 1) What is the sum of all the integers (whole numbers) from 1 to 10000. In other words: 1 + 2 + 3 + 4 + .... + 9999 +10000 = ? 2) What is the sum if we take 1, then half it and add it on, then half that and add it on, infinitely. In other word: 1 + 1/2 + 1/4 + 1/8 + 1/16 +1/32 + .... (infinitely)... = ? And by the way, this isn't my homework!
 47 replies: All unread, showing first 25:
 JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11 Reply 1, posted Thu Jun 6 2002 16:44:17 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1774 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 2, posted Thu Jun 6 2002 16:57:18 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1770 times:

 #2 - the sum will approach 2, but never get there.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 3, posted Thu Jun 6 2002 16:57:38 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1767 times:

 JetService, Your answer to the first question is correct. As for the second, yes you can have a sum of an infinite equation.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 4, posted Thu Jun 6 2002 16:59:40 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1764 times:

 Mcringring, You are kind of correct. The answer is in fact 2 exactly. If you take the sum over an infinite number of terms, it will actually reach 2. I guess it's time to lock this thread....
 Mcringring From , joined Dec 1969, posts, RR: Reply 5, posted Thu Jun 6 2002 17:03:33 UTC (12 years 9 months 3 weeks 6 days 16 hours ago) and read 1760 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.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 6, posted Thu Jun 6 2002 17:47:19 UTC (12 years 9 months 3 weeks 6 days 15 hours ago) and read 1749 times:

 Let s = 1 + 1/2 + 1/4 +1/8 + 1/16 + 1/32 + ... then double everything 2s = 2 ( 1 + 1/2 + 1/4 +/18 + 1/16 + 1/32 + ... ) = 2 + 1 + 1/2 +1/4 +1/8 + 1/16 + 1/32 + ... then subtract the first equation from the second 2s - s = 2 + 1 + 1/2 + 1/4 + ... - ( 1 + 1/2 + 1/4 + ... ) = 2 so s = 2
 Mcringring From , joined Dec 1969, posts, RR: Reply 7, posted Thu Jun 6 2002 18:04:18 UTC (12 years 9 months 3 weeks 6 days 15 hours ago) and read 1742 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, 562 posts, RR: 0 Reply 8, posted Thu Jun 6 2002 18:17:51 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1734 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.
 SunAir From India, joined Jun 2009, 0 posts, RR: 1 Reply 9, posted Thu Jun 6 2002 18:29:46 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1729 times:

 Are any of you aware there is a formula to work out the sum of an infinate Geometric progression? I used the infinate sum formula and... I would agree with Airways1 that the answer is 2. I have worked on paper...it's a bit tough to do it on computer. Apologies for the handwriting...it's a bit rough.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 10, posted Thu Jun 6 2002 18:41:11 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1714 times:

 SunAir, Yes, I am aware of this formula. In fact, I nearly derived it in my previous post. Here I'll derive it properly. Let's say the starting value of the progression is A and that the ratio of terms is R (where R<1). So the sum, S, is given by S = A + AR +AR^2 +AR^3 + AR^4 + AR^5 + ... Now multiply both sides by R, giving SR = AR + AR^2 + AR^3 +AR^4 +AR^5 + ... Now subtract this from the previous equation to give S - SR = A S(1-R) = A so S = A/(1-R)
 Mcringring From , joined Dec 1969, posts, RR: Reply 11, posted Thu Jun 6 2002 18:47:40 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1708 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.
 SunAir From India, joined Jun 2009, 0 posts, RR: 1 Reply 12, posted Thu Jun 6 2002 18:50:11 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1708 times:

 Looks good to me Airways1.
 Beefmoney From United States of America, joined Oct 2000, 1122 posts, RR: 3 Reply 13, posted Thu Jun 6 2002 18:59:48 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1703 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 Thu Jun 6 2002 19:06:50 UTC (12 years 9 months 3 weeks 6 days 14 hours ago) and read 1697 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.
 Airways1 From United Kingdom, joined Jul 1999, 562 posts, RR: 0 Reply 15, posted Thu Jun 6 2002 19:15:27 UTC (12 years 9 months 3 weeks 6 days 13 hours ago) and read 1691 times:

 Well, Mcringring, I showed it mathematically. If you won't accept that, then I don't think there's anything I can say to convince you. Perhaps you should ask a mathematician.
 Klaus From Germany, joined Jul 2001, 21578 posts, RR: 53 Reply 16, posted Thu Jun 6 2002 20:05:53 UTC (12 years 9 months 3 weeks 6 days 13 hours ago) and read 1678 times:

 You´d have to make the explicit transition from the numerical sequence to it´s limit (which can´t ever be reached). There is a difference between the two. It´s just a mathematical convention to usually imply the transition. But it´s still there.
 D L X From United States of America, joined May 1999, 11736 posts, RR: 52 Reply 17, posted Thu Jun 6 2002 20:13:25 UTC (12 years 9 months 3 weeks 6 days 12 hours ago) and read 1674 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.
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 Lewis From Greece, joined Jul 1999, 3711 posts, RR: 4 Reply 18, posted Thu Jun 6 2002 20:16:06 UTC (12 years 9 months 3 weeks 6 days 12 hours ago) and read 1673 times:

 It is just a matter of series and sequences. The first example is : let a be the first term, meaning a=1 let n be the total number of the numbers, ie n=10,000 Their sum is defined by the formula {sum of n=(n/2)*(a+un)} where un is the nth number of the sequence (un=10,000) The second example is already explained, the Sinfinite is a standard formula.
 Mcringring From , joined Dec 1969, posts, RR: Reply 19, posted Thu Jun 6 2002 21:19:50 UTC (12 years 9 months 3 weeks 6 days 11 hours ago) and read 1664 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. Later...
 D L X From United States of America, joined May 1999, 11736 posts, RR: 52 Reply 20, posted Thu Jun 6 2002 21:28:24 UTC (12 years 9 months 3 weeks 6 days 11 hours ago) and read 1663 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.
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 Mcringring From , joined Dec 1969, posts, RR: Reply 21, posted Thu Jun 6 2002 22:02:31 UTC (12 years 9 months 3 weeks 6 days 11 hours ago) and read 1655 times:

 DLX, I don't see your point. The answer for the convergence may be clear. But if you keep summing infinitely, you will never reach 2/3. Of course I'm not saying the formula is wrong or useless. For finding where the formula converges, it is helpful.
 JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11 Reply 22, posted Thu Jun 6 2002 22:02:37 UTC (12 years 9 months 3 weeks 6 days 11 hours ago) and read 1655 times:

 Airways1, you were using the same method that proves .999R = 1 x = .999R 10x = 9.999R <--multiply both sides by 10 9x = 9 <--subtract x (.999R) from both sides x = 1 <--simplify Result: x equals both .999R and 1, thus .999R = 1
 D L X From United States of America, joined May 1999, 11736 posts, RR: 52 Reply 23, posted Fri Jun 7 2002 00:33:11 UTC (12 years 9 months 3 weeks 6 days 8 hours ago) and read 1640 times:

 "But if you keep summing infinitely, you will never reach 2/3." No, you will reach 2/3 after summing an infinite amount of times.
 Send me a PM at http://www.airliners.net/aviation-forums/sendmessage.main?from_username=NULL
 Mcringring From , joined Dec 1969, posts, RR: Reply 24, posted Fri Jun 7 2002 00:38:27 UTC (12 years 9 months 3 weeks 6 days 8 hours ago) and read 1636 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
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