Ecmandu
(Ecmandu)
January 26, 2020, 4:44pm
1743
(\sum_{i=1}^{n} \frac{9}{10^i} = 1 - \frac{1}{10^n})
In order for (\sum_{i=1}^{n} \frac{9}{10^i}) to be equal to (1), there must be (n) such that (1 - \frac{1}{10^n} = 1).
This means that there must be (n) such that (\frac{1}{10^n} = 0).
Suppose that (x) is such a number: (1 \div 10^x = 0).
This means that (0 \times 10^x = 1).
But zero times any number is zero, right?
Magnus! Honestly! sigh
This is what I was trying to explain to you before.
If you halve 1, it’s 1/2 + 1/2
If you halve THAT!
It’s 1/4+1/4+1/4+1/4. Etc…
If you keeps doing this:
At infinity (convergence)
EVERY!!! Whole number solves as zero.
I still think you are confusing all your arguments. You switch back and forth when it suits you without understanding (or caring) about the implications.
Well, you didn’t say much. What you said is basically:
(1 = \frac{1}{2} + \frac{1}{2} = 2 \times \frac{1}{2})
(1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 4 \times \frac{1}{4})
(\dotso)
(1 = 0 + 0 + 0 + \cdots = \infty \times 0)
How can that be the case when zero times anything is zero?
Properly speaking, (1 = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + \cdots = \infty \times \frac{1}{\infty}).
Ecmandu
(Ecmandu)
January 26, 2020, 5:15pm
1745
Well, you didn’t say much. What you said is basically:
(1 = \frac{1}{2} + \frac{1}{2} = 2 \times \frac{1}{2})
(1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 4 \times \frac{1}{4})
(\dotso)
(1 = 0 + 0 + 0 + \cdots = \infty \times 0)
How can that be the case when zero times anything is zero?
Properly speaking, (1 = \frac{1}{\infty} + \frac{1}{\infty} + \frac{1}{\infty} + \cdots = \infty \times \frac{1}{\infty}).
I said a lot! I said that it’s impossible for convergence theory to be true (standard calculus and other mathematics)
Ultimately it equals (according to convergence theory)
0+0+0+0 (etc…) at infinity! (Convergence)
That means that every whole number equals zero. (Contradiction)
How’s that relevant to anything of what I’m saying?
Ecmandu
(Ecmandu)
January 26, 2020, 6:01pm
1747
How’s what you said relevant to what I said?!?!
1/infinity is a meaningless mathematical phrase.
You quoted me. I guess it’s somehow relevant to what I said in that quote.
(\frac{1}{\infty}) is a number greater than (0) but smaller than every number of the form (\frac{1}{n}, n \in N). That’s what it means (so it’s not really meaningless.)
(0 < \frac{1}{\infty} < \cdots < \frac{1}{4} < \frac{1}{3} < \frac{1}{2} < \frac{1}{1})
Ecmandu
(Ecmandu)
January 26, 2020, 6:57pm
1749
You quoted me. I guess it’s somehow relevant to what I said in that quote.
(\frac{1}{\infty}) is a number greater than (0) but smaller than every number of the form (\frac{1}{n}, n \in N). That’s what it means (so it’s not really meaningless.)
(0 < \frac{1}{\infty} < \cdots < \frac{1}{4} < \frac{1}{3} < \frac{1}{2} < \frac{1}{1})
1/infinity is not a number. It’s hypothetically, a procedure. It is most definitely not a number.
It’s not a procedure.
(\infty) represents a number greater than every number of the form (n, n \in N).
Similarly, (\frac{1}{\infty}) represents a number greater than (0) but less than every number of the form (\frac{1}{n}, n \in N).
(0 < \frac{1}{\infty} < \cdots < \frac{1}{3} < \frac{1}{2} < \frac{1}{1} < 2 < 3 < \cdots < \infty)
Also:
(0.\dot9 + 0.\dot01 = 1)
Note that (0.\dot01) or (\frac{1}{10^\infty}) is actually smaller than (\frac{1}{\infty}).
Ecmandu
(Ecmandu)
January 26, 2020, 7:09pm
1751
It’s not a procedure.
(\infty) represents a number greater than every number of the form (n, n \in N).
Similarly, (\frac{1}{\infty}) represents a number greater than (0) but less than every number of the form (\frac{1}{n}, n \in N).
(0 < \frac{1}{\infty} < \cdots < \frac{1}{3} < \frac{1}{2} < \frac{1}{1} < 2 < 3 < \cdots < \infty)
Also:
(0.\dot9 + 0.\dot01 = 1)
Note that (0.\dot01) or (\frac{1}{10^\infty}) is actually smaller than (\frac{1}{\infty}).
Ok, fine. I don’t buy this, but let’s say that what you’re saying is absolutely true.
Per the argument I leveled. That means every whole number is EXACTLY equal to the lowest possible “number” (your argument, not mine) that’s not zero.
My argument still stands. It’s absurd.
How did you arrive at the conclusion that every whole number is exactly equal to the lowest number?
That’s not even true for (\frac{1}{\infty}) let alone for whole numbers.
Ecmandu
(Ecmandu)
January 26, 2020, 7:51pm
1753
Really Magnus ?!
I’ll use your own post for it!
viewtopic.php?p=2758485#p2758485
You’re right. 1=0 is a constradiction.
My argument proves that when numbers converge at infinity (and in saying this, infinity is NOT A NUMBER!)
That 1=0.
Thus, infinities do not converge.
All you did was change infinity to “lowest possible ‘number’ that’s not equal to zero, which by my argument, makes every whole number equal to “the lowest possible number not equal to zero” which is still a contradiction.
That means that 1=2!! Contradiction !
What argument, Ecmandu? Where is it?
Ecmandu
(Ecmandu)
January 26, 2020, 8:00pm
1755
You replied to the post yourself ! Honestly! This is getting absurd!
viewtopic.php?p=2758484#p2758484
What exactly does that prove?
Ecmandu
(Ecmandu)
January 26, 2020, 9:36pm
1757
Just what I said it does. Infinite series don’t converge.
gib
(gib)
January 27, 2020, 2:40am
1758
To answer your question.
This started out with you asking:
“You’d have to explain why you’re limiting yourself to integers.”
To which I said: “Try it without integers. It doesn’t work.”
I was showing you it doesn’t work. Of course, what you really meant was the non-integer (\infty).
In the case of (\sum_{i=1}^{n}), (i) starts with (1), increases by (1) and ends with (n). The number of terms is (n), so the sum stops (is complete) after (n) number of terms.
Yes.
In the case of (\sum_{i=1}^{\infty}), (i) starts with (1), increases by (1) and does not end. The fact that (i) does not end tells us that the number of terms is (\infty). This means the sum stops (is complete) after an infinite number of terms. Which is to say it doesn’t stop. (I assume you’re one of the people in this thread who have no problem with the concept of “actual or completed infinity”.)
I have no problem with saying things like “suppose there is an infinite number of items.” But I don’t think you can “build up” to infinity.
(\sum_{i=1}^{\infty}) does not mean that (i) ends with (\infty). Agreed. In other words, (\sum_{i=1}^{\infty} i) is not equal to something like (1 + 2 + 3 + \cdots + \infty). (Note that such a sum would have more than (\infty) terms.)
The value of (i) is always a natural number. So if we are asking a question such as “What’s the value of the sum of terms whose index is (i, 1 \leq i \leq x)?” then (x) cannot be anything other than a natural number since the range of (x), in such a case, must be the range of (i) Well, x must be the end of the range of i-- and this means that (x) can’t be (\infty). But if we’re asking a question such as “What’s the value of the sum after (x) number of terms?” then the range of (x) goes from (1) to the number of terms of the sum. If the number of terms is infinite, then (x) can be (\infty). And it is precisely this question that we’re asking.
I don’t get the distinction between these two cases. Sounds like the exact same case just worded differently. In the one case you’re saying x can equal (\infty), in the other that it can’t.
What is the value of the sum (\sum_{i=1}^{\infty} \frac{9}{10^i}) after an infinite number of terms?
You’re saying it’s (1), I am saying it’s less than (1).
My argument (which is basically James’s argument) is that the pattern of this sum prohibits its value after (x) number of terms to be equal to (1) for any (x > 0). (You can limit the value of (x) to numbers that have no fractional component, if you want.) Since (\infty) is greater than (0), it applies to (\infty) as well.
I don’t see anything new in this argument that I haven’t already addressed.
By what logic does the value of this sum become (1) after an infinite number of terms?
This logic:
(x = 0.\dot9)
(10x = 9.\dot9)
(10x = 9 + 0.\dot9)
(10x = 9 + x)
(9x = 9)
(x = 1)
A sum cannot stop after an infinite number of terms because if it could it would be finite so the concept of completed infinity is entirely fallacious
And so your first sentence and third sentence contradict each other because if (1) increases by (1) and does not end then logically the sum cannot stop
I0x = 9.999…
I0x = 9 + .999…
I0x = 9 + x
I0x = 9 + I
I0x = I0
x = I
I addressed this “proof” around 20 pages ago and I can restate what’s wrong with it but I think it’s pointless since you don’t agree that we can do arithmetic with infinite quantities.
Basically, you don’t agree that adding a green apple to an infinite line of red apples increases the number of apples in the line. Instead, you prefer to contradict yourself by saying that the number of apples remains the same.