Sponsor Message:
Non Aviation Forum
My Starred Topics | Profile | New Topic | Forum Index | Help | Search 
I Have Another Maths Problem For You (two In Fact)  
User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Posted (11 years 10 months 2 weeks 6 days 10 hours ago) and read 1544 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:
 
User currently offlineJetService From United States of America, joined Feb 2000, 4798 posts, RR: 12
Reply 1, posted (11 years 10 months 2 weeks 6 days 10 hours ago) and read 1519 times:

10001*5000=5,000,5000

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



"Shaddap you!"
User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 2, posted (11 years 10 months 2 weeks 6 days 9 hours ago) and read 1515 times:

#2 - the sum will approach 2, but never get there.

User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 3, posted (11 years 10 months 2 weeks 6 days 9 hours ago) and read 1512 times:

JetService,

Your answer to the first question is correct.

As for the second, yes you can have a sum of an infinite equation.


User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 4, posted (11 years 10 months 2 weeks 6 days 9 hours ago) and read 1509 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....


User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 5, posted (11 years 10 months 2 weeks 6 days 9 hours ago) and read 1505 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.

User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 6, posted (11 years 10 months 2 weeks 6 days 8 hours ago) and read 1494 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




User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 7, posted (11 years 10 months 2 weeks 6 days 8 hours ago) and read 1487 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.


User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 8, posted (11 years 10 months 2 weeks 6 days 8 hours ago) and read 1479 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.



User currently offlineSunAir From India, joined Jun 2009, 0 posts, RR: 1
Reply 9, posted (11 years 10 months 2 weeks 6 days 8 hours ago) and read 1474 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.





User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 10, posted (11 years 10 months 2 weeks 6 days 8 hours ago) and read 1459 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)



User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 11, posted (11 years 10 months 2 weeks 6 days 7 hours ago) and read 1453 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.

User currently offlineSunAir From India, joined Jun 2009, 0 posts, RR: 1
Reply 12, posted (11 years 10 months 2 weeks 6 days 7 hours ago) and read 1453 times:

Looks good to me Airways1.  Big thumbs up


User currently offlineBeefmoney From United States of America, joined Oct 2000, 1111 posts, RR: 4
Reply 13, posted (11 years 10 months 2 weeks 6 days 7 hours ago) and read 1448 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.

User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 14, posted (11 years 10 months 2 weeks 6 days 7 hours ago) and read 1442 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.


User currently offlineAirways1 From United Kingdom, joined Jul 1999, 559 posts, RR: 0
Reply 15, posted (11 years 10 months 2 weeks 6 days 7 hours ago) and read 1436 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.


User currently offlineKlaus From Germany, joined Jul 2001, 21353 posts, RR: 54
Reply 16, posted (11 years 10 months 2 weeks 6 days 6 hours ago) and read 1423 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.


User currently offlineD L X From United States of America, joined May 1999, 10998 posts, RR: 52
Reply 17, posted (11 years 10 months 2 weeks 6 days 6 hours ago) and read 1419 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
User currently offlineLewis From Greece, joined Jul 1999, 3592 posts, RR: 5
Reply 18, posted (11 years 10 months 2 weeks 6 days 6 hours ago) and read 1418 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.



User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 19, posted (11 years 10 months 2 weeks 6 days 5 hours ago) and read 1409 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...


User currently offlineD L X From United States of America, joined May 1999, 10998 posts, RR: 52
Reply 20, posted (11 years 10 months 2 weeks 6 days 5 hours ago) and read 1408 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
User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 21, posted (11 years 10 months 2 weeks 6 days 4 hours ago) and read 1400 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.


User currently offlineJetService From United States of America, joined Feb 2000, 4798 posts, RR: 12
Reply 22, posted (11 years 10 months 2 weeks 6 days 4 hours ago) and read 1400 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



"Shaddap you!"
User currently offlineD L X From United States of America, joined May 1999, 10998 posts, RR: 52
Reply 23, posted (11 years 10 months 2 weeks 6 days 2 hours ago) and read 1385 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
User currently offlineMcringring From , joined Dec 1969, posts, RR:
Reply 24, posted (11 years 10 months 2 weeks 6 days 2 hours ago) and read 1381 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 Post contains images 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 Post contains images Mcringring : Thank you Klaus. Maybe they will listen to you. They sure won't listen to me.
44 Post contains images Killjoy : Oops, I just noticed the topic of Klaus' post was "Killjoy" so I guess he *was* answering me .
45 Post contains images 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 Post contains images 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 Post contains images 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
Top Of Page
Forum Index

This topic is archived and can not be replied to any more.

Printer friendly format

Similar topics:More similar topics...
I Have A Math Problem For You! posted Thu Jun 6 2002 12:12:18 by BA777
777236ER! I Have Found The Car For You! posted Tue Sep 16 2003 12:31:49 by JAL777
For Whom Will You Vote In Next Us Pres. Election? posted Fri Apr 7 2006 03:19:22 by DL021
For All Of You People In Cymru! posted Sat Mar 11 2006 15:05:06 by RootsAir
Have You Been In Combat? posted Fri Oct 21 2005 16:22:05 by F4N
I Have A Question For You... Official Ending posted Tue Oct 11 2005 06:54:23 by CanadianNorth
Another - How Much Do You Pay For Gas - Thread posted Thu Oct 6 2005 05:55:59 by Searpqx
Sorry Mate, No Virgins In Heaven For You.... posted Fri Jul 22 2005 11:51:46 by WhiteHatter
Do You Have A Storm Named For You? posted Mon May 16 2005 20:33:37 by FlyingTexan
Thank You For Your Help In English posted Mon Jan 10 2005 21:51:00 by Kilavoud