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

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
Posts: 4611
Joined: Mon Feb 21, 2000 1:12 pm

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

10001*5000=5,000,5000

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

McRingRing
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Joined: Mon Dec 10, 2001 2:59 am

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

#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)

JetService,

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

airways1
Topic Author
Posts: 536
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### RE: I Have Another Maths Problem For You (two In Fact)

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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
Posts: 346
Joined: Thu Jun 11, 2009 6:59 am

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

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
Topic Author
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Joined: Mon Jul 19, 1999 3:05 am

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

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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Joined: Mon Dec 10, 2001 2:59 am

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

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)

Looks good to me Airways1.

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

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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Joined: Mon Dec 10, 2001 2:59 am

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

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)

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)

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)

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
Posts: 3581
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### RE: I Have Another Maths Problem For You (two In Fact)

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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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Joined: Mon Feb 21, 2000 1:12 pm

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

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
Posts: 12001
Joined: Thu May 27, 1999 3:30 am

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

"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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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?

777236ER
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### RE: I Have Another Maths Problem For You (two In Fact)

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)

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

Klaus
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### 777236ER

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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Joined: Wed Jul 11, 2001 7:41 am

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

Dang, that evil filter did it again.

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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Joined: Thu May 27, 1999 3:30 am

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

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)

#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
Posts: 4698
Joined: Thu Mar 14, 2002 5:39 pm

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

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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
Posts: 1028
Joined: Mon Dec 10, 2001 2:59 am

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

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)

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

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

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

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
Posts: 601
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### RE: I Have Another Maths Problem For You (two In Fact)

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

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
Posts: 601
Joined: Thu Dec 02, 1999 6:00 am

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

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)

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

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

Klaus
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### Killjoy

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 better for understanding (or explanations) to explicitly specify what´s going on. But for "real world" applications, it´s often omitted.

killjoy
Posts: 601
Joined: Thu Dec 02, 1999 6:00 am

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

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   .

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

Klaus
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Joined: Wed Jul 11, 2001 7:41 am

### Killjoy

Yeah, I think you´ve got it right.

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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