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Originally posted by shavixmirAfter 1 step we have the first partial sum S(1) = 1/2, after 2 steps 1/2 + 1/4 = 3/4 = S(2) = 1 - 1/4, after 3 steps we have 1/2 + 1/4 + 1/8 = 7/8 = 1 - 1/8. After n steps S(N) = 1/2 + 1/4 + ... + 1/2^N = 1 - (1/2)^N.
So, like, yeah... that got a lot more simple...
As for the 2S - S = 1S
Yes. But if you conclude that 1S isn't exactly 1S, but actually 1S - an ever decreasing portion, then surely this doesn't prove that S = 1.
?
Choose arbitrary ŋ in the range 0 < ŋ < 1. Then we can always find some n so that the nth partial sum S(n) is bigger than 1 - ŋ.
The nth partial sum is: S(n) = 1 - (1/2)^n < 1, so all the partial sums are less than 1.
We are looking for N such that 1 - ŋ < S(N), which is true if ŋ > (1/2)^N, and so if N > -log_2(ŋ) the partial sum S(N) will be larger than 1 - ŋ.
This means that there is no number smaller than 1 for which all the partial sums are smaller than it. So 1 is the limit point of the sequence of partial sums. The sum of the series is the limit point of the sequence.
In the case of your falling glass for any point above the floor there is a time (corresponding to a step in the supertask) when it has passed that point. There is a time at which the glass has passed all points above the floor, which is when it smashes.
Originally posted by shavixmirThere are a sequence of partial sums S(n), the S(n) change with n. The limit of the sequence of partial sums is the sum of the series S. It doesn't change as it includes all the terms.
Because S is not S. It's a part of S. That part just keeps getting bigger and bigger... so it's constantly changing.
Originally posted by KazetNagorraAlright, it still sounds incredibly silly to me, and looks like some clever use of mathematics and the difficulty of understanding infinity, but I can't poke a hole in this other than using common sense and gut feeling which is often not helpful in these cases.
We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ...
Note how the terms after 1 + are again equal to S itself, so that:
2S = 1 + 1/2 + 1/4 + 1/8 + ... = 1 + S
Now we subtract S on both sides:
2S - S = 1
Therefore S = 1. (this is only an informal proof at best, though)
You can "prove" in a similar fashion that 0.999... (infinite sequence of 9s) = 1.
What if you use the same line of reasoning, but instead of "solving" it using 2S you use 3S?
So 3S = 1.5 + 3/4 + 3/8 +...
Does it still hold up, or does it only work when using even numbers (2S, 4S...)?
I like maths, but not when it spits in the face of logic.
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Originally posted by Great King RatIt wasn't really offered as a definitive proof, so I don't mean to criticize it, but you are correct that this isn't a proof. All it is is an algebra equation (2S=1+S) which asks you to solve for S.
Alright, it still sounds incredibly silly to me, and looks like some clever use of mathematics and the difficulty of understanding infinity, but I can't poke a hole in this other than using common sense and gut feeling which is often not helpful in these cases.
What if you use the same line of reasoning, but instead of "solving" it using 2S you use ...[text shortened]... hen using even numbers (2S, 4S...)?
I like maths, but not when it spits in the face of logic.
Look at the following statement:
"We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ... "
The second line assumes that S equals one (The second line could be rewritten as 2S=1+S, which defines S as equal to one). S must equal one, or the second statement cannot be true. Therefore, the second statement cannot be used to prove that S=1 in the first line, because it already assumes that S=1. This is a begging the question fallacy.
Again, not a criticism of the proof, because it wasn't offered as one, but just to help show you that your gut feeling is correct in this case.
Originally posted by PatNovakNo, it doesn't. That is the conclusion. The second line is obtained by multiplying S by two. Now it may be the case that multiplying an infinite series by two is a bad idea and that one cannot assume the total sum of an infinite series also multiplies by two when each term is multiplied by two.
The second line assumes that S equals one ...
But at no point is the assumption made that S equals one.
Originally posted by twhiteheadThis is not true. The second line is obtained by multiplying S by two on one side of the equation, while adding one to S on the other side of the equation. This is not a conclusion (or at least a justified conclusion), but is merely a definition of S.
The second line is obtained by multiplying S by two.
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Originally posted by PatNovakNo, it is not. It is obtained by multiplying both sides by two.
This is not true. The second line is obtained by multiplying S by two on one side of the equation, while adding one to S on the other side of the equation.
Maybe Wikipedia will make it clearer:
https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%E2%8B%AF
Originally posted by twhiteheadAs I think you know, I am not arguing with the conclusion (which as I have clearly stated earlier in the thread that I agree with), but arguing with the proof.
No, it is not. It is obtained by multiplying both sides by two.
Maybe Wikipedia will make it clearer:
https://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%E2%8B%AF
No, it is not. It is obtained by multiplying both sides by two.
This is demonstrably false. Read this again (bolding mine):
"We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ... "
The left side of the equation has clearly been multiplied by 2, while the right side of the equation has clearly had 1 added to it. This is not a legitimate math operation. The same action must be made to both sides of the equation.
If I changed the statement to a variable so it read:
We have S = x
Therefore 2S = 1 + x
This would be an incorrect conclusion. It would only be a correct conclusion if we have already decided x = 1 before we even make it to the conclusion.
Originally posted by PatNovakNot true Pat, the right side has also been multiplied by 2. Two times 1/2 is 1, two times 1/4 is 1/2 etc. etc.
"We have S = 1/2 + 1/4 + 1/8 + ...
Therefore [b]2S = 1 + 1/2 + 1/4 + 1/8 + ... "
The left side of the equation has clearly been multiplied by 2, while the right side of the equation has clearly had 1 added to it. This is not a legitimate math operation. The same action must be made to both sides of the equation.[/b]
Doesn't make the end result any less fishy though.
Originally posted by Great King RatAll it really is in the end is arguing that S=1 because 2S=2. So it isn't much of a proof.
Not true Pat, the right side has also been multiplied by 2. Two times 1/2 is 1, two times 1/4 is 1/2 etc. etc.
Doesn't make the end result any less fishy though.
Your version of the statement would read:
We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 2(1/2 + 1/4 + 1/8 + ...)
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ...
The original statement skips the middle step, and it is, as you say, a fishy conclusion to state that we are proving the value of 1/2 + 1/4 + 1/8 + ... by saying that we know the value of 1 + 1/2 + 1/4 + 1/8 + ...
Originally posted by PatNovakIf 2S=2, then S=1 is most definitely true and it would be a rock solid proof.
All it really is in the end is arguing that S=1 because 2S=2. So it isn't much of a proof.
The original statement skips the middle step,
Yes, it kind of assumes you know what is going on, so maybe the words "multiply both sides by two" was in order. Nevertheless, it remains the case that both sides were multiplied by two.
and it is, as you say, a fishy conclusion to state that we are proving the value of 1/2 + 1/4 + 1/8 + ... by saying that we know the value of 1 + 1/2 + 1/4 + 1/8 + ...
No, that is not how the conclusion is obtained.
The conclusion comes from the claim that:
1/2+1/3+1/4+.... times two is 1+1/2+1/3+1/4+.... and that the portion after the 1+ in the second sequence has the same total value as the first sequence. It sure looks identical - but is it?
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Originally posted by PatNovakthe way I am thinking of this, instead of arguing it:
All it really is in the end is arguing that S=1 because 2S=2. So it isn't much of a proof.
Your version of the statement would read:
We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 2(1/2 + 1/4 + 1/8 + ...)
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ...
The original statement skips the middle step, and it is, as you say, a fishy conclusion to state that ...[text shortened]... the value of 1/2 + 1/4 + 1/8 + ... by saying that we know the value of 1 + 1/2 + 1/4 + 1/8 + ...
“...
We have S = 1/2 + 1/4 + 1/8 + ...
Therefore 2S = 2(1/2 + 1/4 + 1/8 + ...)
Therefore 2S = 1 + 1/2 + 1/4 + 1/8 + ...
...”
you could argue:
“...
We have S = 1/2 + 1/4 + 1/8 + ...
Therefore if we multiplied by 2 each of the terms on the right hand side, regardless of what the value of S is, we get;
2S = 2(1/2) + 2(1/4) + 2(1/8) + ...
Therefore,
2S = 1 + 1/2 + 1/4 + 1/8 + ...
...”
That way you can argue we are not implying we already know the value of 1 + 1/2 + 1/4 + 1/8 + ...
Anyone: Is there anything wrong with this above argument of mine?
Originally posted by humyThe question is whether or not multiplying every element in an infinite series by two results in the sum of the infinite series doubling. I don't think it is a valid assumption to make without more argument. Infinity is a tricky animal.
Anyone: Is there anything wrong with this above argument of mine?
You will notice that Wikipedia carefully steered away from the technique being discussed here and insisted on talking about the first n terms.
Originally posted by twhiteheadThe series 1 + 1/2 + 1/4 + ... + (1/2)^N + ... is absolutely convergent because you can write the nth partial sum as 2 - (1/2)^(N+1), so the sequence of partial sums is bounded from above by 2. This means you can reorder the sum to your hearts content and it will not change the result of the sum. It also means the sum has the property that doubling each element will increase the sum by two.
The question is whether or not multiplying every element in an infinite series by two results in the sum of the infinite series doubling. I don't think it is a valid assumption to make without more argument. Infinity is a tricky animal.
You will notice that Wikipedia carefully steered away from the technique being discussed here and insisted on talking about the first n terms.