airways1
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I Have Another Maths Problem For You (two In Fact)

Thu Jun 06, 2002 11:23 pm

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!
 
JetService
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RE: I Have Another Maths Problem For You (two In F

Thu Jun 06, 2002 11:44 pm

10001*5000=5,000,5000

Not sure you can have a sum on an infinite equation, can you?
"Shaddap you!"
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Thu Jun 06, 2002 11:57 pm

#2 - the sum will approach 2, but never get there.
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airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Thu Jun 06, 2002 11:57 pm

JetService,

Your answer to the first question is correct.

As for the second, yes you can have a sum of an infinite equation.
 
airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Thu Jun 06, 2002 11:59 pm

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 12:03 am

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.
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airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 12:47 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:04 am

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.
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airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:17 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:29 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:41 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:47 am

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.
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Sunair
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:50 am

Looks good to me Airways1.  Big thumbs up
 
Beefmoney
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 1:59 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 2:06 am

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.
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airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 2:15 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 3:05 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 3:13 am

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.

 
lewis
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 3:16 am

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
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 4:19 am

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...
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D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 4:28 am

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.
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 5:02 am

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.
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JetService
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RE: I Have Another Maths Problem For You (two In F

Fri Jun 07, 2002 5:02 am

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!"
 
D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 7:33 am

"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.
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 7:38 am

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.
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D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 7:38 am

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. So, you are absolutely correct: .999R = 1. (Btw, the key word in this is "after" summing an infinite amount of times.)

(Unless you have a proof that it is not equal, I'm not sure what you're trying to say.)

Math is a funny thing, isn't it?  Smile
 
777236ER
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 7:47 am

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...)

Therefore 9x=9

So x=1.

The shrewd ones among you will notice that this actually is just summing to infinity dressed up, but it's a nice way of showing that 0.9999... = 1 without TOO complex maths.
Your bone's got a little machine
 
D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 8:52 am

777236ER, way to read the earlier posts. Well done...
 
Klaus
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777236ER

Fri Jun 07, 2002 9:16 am

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...)

Therefore 9x=9

So x=1.



That´s not exactly what´s happening, here.

Being precise, lim(0.9999....) = 1. The lim()-operator is basically "built in" to the notation of periodical numbers.

But it´s still there.

SUM(1/2^m) < 2 for any given n.
(m=0...n)

lim(SUM(1/2^m)) is exactly = 2.
(n=0...? ) (m=0...n)

As I said, the lim() is often implied for convenience; But it´s still there.
 
Klaus
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 9:18 am

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 implied for convenience; But it´s still there.
 
D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 10:08 am

McRing, please define infinity. I want to make sure we're on the same page.
 
NWA
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 12:54 pm

#2 will never reach 2, it will get very close though, infnitley close.
23 victor, turn right heading 210, maintain 3000 till established, cleared ILS runwy 24.
 
707CMF
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 6:02 pm

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, right ?
3x(1/3) = 1 right ?
therefore, 0x999999R = 1

another one is "if two numbers, noted X and Y are not equal, then there is *always* one (and actually an infinity of) numbers Z, so that X < Z < Y , or Y < Z < X . If X = 0.9999R and Y = 1, there is no such number. Therefore, 0.999999R = 1"


And the solution of #2 *is* 2.

As for #1, the formula to use is 1+2+3+4+...+n = (n*(n-1))/2

Cheers
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Fri Jun 07, 2002 11:19 pm

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 = x so it is either meaningless or impossible to add to infinity, because the value doesn't change.

By definition, if the 1 + 1/2 + 1/4 + 1/8 +... sequence is EVER equal to exactly 2, the sequence is over. Since it continues to infinity, this is impossible. It will never end. It can't end.

As Klaus points out, the equation determines the limit of the sequence. Not the sum.
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D L X
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 5:49 am

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's not true. I don't get it.

"By definition, if the 1 + 1/2 + 1/4 + 1/8 +... sequence is EVER equal to exactly 2, the sequence is over. "
Correct, and the sequence will be over after summing infinitely many terms. It's only an impossibility in the physical world, not the theoretical universe of math.

My contribution to this thread is complete.
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 6:00 am

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 you don't get that, don't try to blame me.
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airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 8:10 pm

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 wrote: By definition, if the 1 + 1/2 + 1/4 + 1/8 +... sequence is EVER equal to exactly 2, the sequence is over. Since it continues to infinity, this is impossible. It will never end. It can't end.

I think the problem is that you're seeing infinity as some 'end-point' where the series terminates. Infinitity is not a number, and there is nothing beyond it. It is a MATHEMATICAL CONCEPT. Sure in reality no matter how many terms of the series you add together, you will never reach 2. But as a mathematical concept, you can sum an infinite number of terms at which point the sum becomes exactly 2.
 
Klaus
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Airways1

Sat Jun 08, 2002 8:37 pm

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 misconception. There is no "such point". In fact, the infinite sum implies a limit transition and is therefore a slightly different concept from a finite sum.

As I said, it´s implied for convenience; But the distinction stays there.
 
airways1
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 8:43 pm

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 'the infinite sum implies a limit transition'. Well I did specify initially that this is an infinite sum, so with 'limit transition implied', as you put it, the sum is 2, just as I said.
 
Klaus
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Airways1

Sat Jun 08, 2002 9:14 pm

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 transition implied', as you put it, the sum is 2, just as I said.

Which doesn´t change the fact that there is no iteration n which would ever reach the limit.
 
killjoy
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 10:13 pm

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 left out from this quote-

Since

lim[n->inf] ((2^n)-1)/(2^n) = 1

we conclude that the series converges and its sum is 1."

In essence, my math book just converted a limit into a normal value! Wtf?? Is it not telling me everything about the definition of a series? Could someone please tell me why it's ok to do this?

If it weren't for that peculiarity, I'd side with Mcringring and Klaus without hesitation, but now I'm just confused...
 
Klaus
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Killjoy

Sat Jun 08, 2002 10:33 pm

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 instead.

After all, the limit is approached to any desirable proximity beyond a sufficiently large iteration index. So under practical circumstances, you can assume it as "the" sum of the series, although technically it isn´t for any given n. It just comes infinitely close without ever actually getting there. Good enough for most purposes. Just not exactly the same thing as a finite sum.
 
killjoy
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RE: I Have Another Maths Problem For You (two In Fact)

Sat Jun 08, 2002 11:40 pm

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 infinitely close to a value, and that saying =1 is "Good enough for most purposes", but the text I quoted never made that distinction. It simply removed the limit and said "sum(1, inf)=1" instead of "sum(1, inf)=1 for most purposes" or even better, "sum(1, inf)->1".

If I hadn't read it myself, I'd've had a hard time believing a math book would make such a generalization, but apparently it has. I certainly can't find another explanation, unless there's something it's leaving unsaid (which is what I wanted to find out).
 
McRingRing
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RE: I Have Another Maths Problem For You (two In Fact)

Sun Jun 09, 2002 12:43 am

Thank you Klaus. Maybe they will listen to you. They sure won't listen to me.  Insane
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killjoy
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RE: I Have Another Maths Problem For You (two In Fact)

Sun Jun 09, 2002 12:46 am

Oops, I just noticed the topic of Klaus' post was "Killjoy" so I guess he *was* answering me  Smile/happy/getting dizzy .
 
Klaus
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Killjoy

Sun Jun 09, 2002 5:12 am

Yes, I was talking to you.  Wink/being sarcastic

The limit transition is usually implied with an infinite sum.

There are many such "shorthands" in math. It would often be better for understanding (or explanations) to explicitly specify what´s going on. But for "real world" applications, it´s often omitted.
 
killjoy
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RE: I Have Another Maths Problem For You (two In Fact)

Sun Jun 09, 2002 6:50 am

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... Neglecting to mention that the authors didn't in fact consider the two things to be exactly the same made me doubt my whole understanding of the issue.

Now I don't, so I'll join the discussion. The series will not reach 2, it will only approach it. Likewise, 1 != 0,9999...

Let's do it with some finite numbers first:

x=0,99 //*10
10x=9,90 //-x
9x=8,91

x=0,9999 //*10
10x=9,9990 //-x
9x=8,9991

Now, if you continue inserting a finite number of nines, the difference of 9-8,9...1 will still remain there preventing x from equaling 1, it'll just be smaller. Problem is, at infinity it'll be infinitely small, and you won't be able to manipulate it normally, which is why you can manage to prove 0,99...=1.

However, if you write it properly as lim(x->1) = 1, you won't. Limit transitions aside, that is  Wink/being sarcastic .

I hope that made any sense, it's just my interpretation based on normal schoolwork.
 
Klaus
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Killjoy

Sun Jun 09, 2002 7:32 am

Yeah, I think you´ve got it right.  Smile

The "0.9999..." is another place where the "..." signifies infinity and therefore implies yet another limit transition. (Which I´m not sure if it´s the proper english term; It´s the literal translation of the german expression.)

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