If S = 1 + 2 + 4 + 8 + 16...

S = 1 + (2 + 4 + 8 + 16...)

S = 1 + 2*(1 + 2 + 4 + 8...) — (since S = 1 + 2 + 4 + 8... we multiply it by two)

Then

S = 1 + 2S

S - 2S = 1

-S = 1

S = -1

How can that be?

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- Jul 2nd 2006, 11:19 PMcmart022Negative sum
If S = 1 + 2 + 4 + 8 + 16...

S = 1 + (2 + 4 + 8 + 16...)

S = 1 + 2*(1 + 2 + 4 + 8...) — (since S = 1 + 2 + 4 + 8... we multiply it by two)

Then

S = 1 + 2S

S - 2S = 1

-S = 1

S = -1

How can that be? - Jul 2nd 2006, 11:24 PMmalaygoelQuote:

Originally Posted by**cmart022**

S= 1+2S

Now you have subracted 2S from both sides to get

-S=1

but wait.

I think that you can't subtrct 2S from both sides of the equation until you are sure that 2S is a real number.

Keep Smiling

Malay - Jul 2nd 2006, 11:31 PMcmart022
Malay,

Thanks, but if you had to find a solution to the problem by generalizing a formula, how would you go about generalizing such a progression?

Do you agree that the term inside the parentheses is not the same as "S" and therefore cannot be simplified as:

S = 1 + 2S

Or is the above equation "correct"? - Jul 3rd 2006, 01:15 AMCaptainBlackQuote:

Originally Posted by**cmart022**

S=2S=3S=S+1=...

in effect you cannot consistently do arithmetic with it. - Jul 3rd 2006, 02:00 AMcmart022Do you agree with...?
Captain

Do you agree with:

http://www.wordpower.com.br/maths/maths015.png

As a generalized formula?

"Se" stands for "If" and "Ou" stands for "or".

Alternatively:

http://www.wordpower.com.br/maths/maths016.png - Jul 3rd 2006, 02:47 AMCaptainBlackQuote:

Originally Posted by**cmart022**

You have now switched to a finite sum, and when you play your game

the upper limit of summation should change, but you are leaving it unchanged.

This is now a finite geometric progression which has a well known sum:

$\displaystyle

S_n=1+2+2^2+2^3+..+2^n={2^{n+1}-1}

$

this is unproblematic, and $\displaystyle S_n \to \infty, \mbox{ as } x \to \infty$

RonL - Jul 3rd 2006, 03:18 AMcmart022
Captain,

Thanks for the reply. This is a problem I found in another group where, according to the person who posted the original problem, the "correct solution" to the problem is:

S = 1 + 1/2 + 1/4 + 1/8...

S = 1 + (1/2 + 1/4 + 1/8 ...)

S = 1 + 1/2(1 + 1/2 + 1/4...), substituting (1 + 1/2 + 1/4...) = S

S = 1 + S/2

S - S/2 = 1

S/2 = 1

Now, since "S" is an infinite sum, how could someone divide S by 2 and get 1?

To me, this is complete nonsense because if you define S as a value, then 2S is twice the value of S and cannot take any other arbitrary value since it is the same variable.

Sorry, I am just trying to get other views becuase they insist that "S-2S=1", and I insist that can only be the case if the "S" on "2S" is something else other than "S".

What are your views? Anyone else has any views on this? - Jul 3rd 2006, 04:04 AMCaptainBlackQuote:

Originally Posted by**cmart022**

made rigourous. The argument goes:

Let:

$\displaystyle

S_n=1+1/2+1/2^2+..+1/2^n

$

so:

$\displaystyle

2(S_n-1)=1+1/2+..+1/s^{n-1}=S_{n-1}

$

Now if $\displaystyle S_n \to S < \infty \mbox{ as } n \to \infty$, then:

$\displaystyle \lim_{n \to \infty} 2(S_n-1)=\lim_{n \to \infty}S_{n-1}$,

so:

$\displaystyle

2(S-1)=S

$

and so $\displaystyle S=2$. But the GP can be shown to converge so

$\displaystyle S=2$.

RonL - Jul 3rd 2006, 04:13 AMcmart022
Captain

Thanks again.

Just to wrap it up... Your final argument is very clarifying, but I am still in double that your final views represent the same thing as the original one.

After all, we originallythat S = 1 + 2 + 4 + 8 + 16... to infinity. So, it is my view that "S" is NOT 2, but an infinite sum.__defined__

But since on the proposed answer, the problem is actually changed, then, yes, we end up in a situation where S = 2.

So, it is my view that the problem they presented is not the same for the answer they presented. Am I right on this?

Once again, thank you very much for taking the time in answering my questions. - Jul 3rd 2006, 04:37 AMCaptainBlackQuote:

Originally Posted by**cmart022**

In one case you have:

S=1+2+4+..,

and in the other you are using:

S=1+1/2+1/4+...

These are not the same thing the first is divergent and you have to be

careful what you do with it because it is not a conventional number,

while the second is convergent, and so the sum is a number (in this case 2).

RonL - Jul 3rd 2006, 04:56 AMcmart022
Captain,

I am not changing the question. I apologize for not leaving it clear.

Like I said, I came across this problem in another forum in another language. My argument in that forum was that the guy who posted the problem gave an answer which is different from the original problem.

However, he swears by God that:

S = 1+2+4+... is the same thing as S=1+1/2+1/4... which yields S=2. At one point he even said S = infinity; so I could not believe that S could be infinity and 2 at the same time, unless, of couse, we are dealing with two separete issues, which is what I told him originally that the two "S's" as not the same...

I posted the problem here because I wanted to make sure I am not going nuts when I say that these are different problems.

I told the guy in the other forum that: S = 1 + 2 + 4 + ... was a completely differente matter from S = 1 + 2S (which is the original post that yields -1); but then he posted the answer as being S=1+1/2+1/4, which converges to 2 and yields the S-S/2=1.

I keep repeating to him that he mixing things up, but he insists that S = 1+2+4+... is the same thing as S=1+1/2+1/4...

Your answer above shows me I am not crazy when I say they're not the same stuff and he's changing the goal post...

Thanks for you patience and time in explaining things. It is very much appreciated. - Jul 3rd 2006, 05:52 AMThePerfectHacker
That guy in the forum is wrong.

As, CaptainBlack said.

$\displaystyle S=1+2+3+4+...$ is what? It is (in formal terms) the convergent value of the sequence,

$\displaystyle 1$

$\displaystyle 1+2$

$\displaystyle 1+2+3$

....

Since it has no value, you cannot define what $\displaystyle S$ is.

In the other case,

$\displaystyle S=1+1/2+1/4+1/8+...$ is what? It is the convergent vale of the sequence, which is $\displaystyle 2$.

Question answered.

That guy is probably one of those who never taken math on a high level and loves to get involved in useless philosophical topics on math.

-------

(Maybe, he want to say,)

$\displaystyle S=1+2+3+...$ is the same as, $\displaystyle S=1+1/2+1/3+1/4+...$- he just wrote that otherwise. Then, I can understand how he says that they are the same, (because they both diverge)-though that is still wrong. - Jul 3rd 2006, 07:14 AMcmart022
ThePerfectHacker,

Thanks. I feel much lighter now... :D

For a minute or two I thought I was crazy. :D - Jul 3rd 2006, 10:25 AMSoroban
Hello, cmart022!

Quote:

If $\displaystyle S \:= \:1 + 2 + 4 + 8 + 16 + \hdots$

$\displaystyle S \:= \:1 + (2 + 4 + 8 + 16 + \hdots)$

$\displaystyle S \;= \;1 + 2\underbrace{(1 + 2 + 4 + 8 + \hdots)}$

. . . . . . . . . . This is $\displaystyle S$

We have: .$\displaystyle S \:= \:1 + 2S\quad\Rightarrow\quad-S\,=\,1$

Therefore: .$\displaystyle S = -1$

How can that be?

The same reason that: .$\displaystyle S\;=\;1 - 1 + 1 - 1 + \hdots$ has**three**values.

$\displaystyle [1]\;\;S\;= \;(1 - 1) + (1 - 1) + (1 - 1) + \hdots$

. . .$\displaystyle S\;=\;0 + 0 + 0 + \hdots$

Therefore: .$\displaystyle \boxed{S\:=\:0}$

$\displaystyle [2]\;\;S\;=\;1 - (1 - 1) - (1 - 1) - (1 - 1) - \hdots$

. . .$\displaystyle S\;=\;1 - 0 - 0 - 0 - \hdots$

Therefore: .$\displaystyle \boxed{S\:=\:1}$

$\displaystyle [3]$ We have a geometric series with $\displaystyle a = 1,\;r = -1$

. . .Hence, it's sum is: .$\displaystyle S\;=\;\frac{1}{1 - (-1)}$

Therefore: .$\displaystyle \boxed{S\:=\:\frac{1}{2}}$

How did I do this? . . . Simple,*I broke some rules.*

- Jul 4th 2006, 04:37 AMcmart022
Soroban

I did enjoy reading your post!! :D

And how you "broke the rules" without breaking any mathematical rules... The second one was really great!