I A limits problem (fractal initiator & generator)....

[Look]

I don't know what you mean by "a content of variable n is greater than any natural number", so I haven't any idea about what it would mean to prove it.

Since n is a variable it can have many different numbers, so what I have asked is to prove that one od these numbers is greater than any natural number.

Can you provide such a proof?

You seem to be arguing that any finite portion of the infinite series 2a + 2b + 2c + ... sums to a total that is less than X. That is correct.

Again dear Stephen Tashi, you can't prove that there are infinitely many terms in that series without proving that there is n value (some possible content of variable n) that is greater than any natural number.

Once again, I am not talking about partial sums, but somehow you translate again and again what I say in terms of partial sums.

I suggest that you express whatever you are trying to say about the picture and the infinite series 2a + 2b + 2c + ... in standard mathematical terminology.

You did not prove that there are infinitely many terms in series 2a + 2b + 2c + ..., so all you currently have is an hypothesis.

My life is rather busy at the moment, so I prefer to spend my time on the forum discussing mathematical questions rather than guessing what somebody's mathematical questions are. There may be other forum members interested in interpreting your words into specific mathematical questions, so I'll leave that task to them.

Dear Stephen Tashi indeed thank you very much for sharing your time with me.

I do not think that my arguments can't be understood mathematically, even if they don't follow after the agreed notions of Real-analysis.

Please look at the following diagram:

Let's assume that n (please see https://www.physicsforums.com/threa...nitiator-generator.881023/page-2#post-5540086) is greater than any natural number.

Since all infinitely many non-straight lines have constant length X>0 and series 2a+2b+2c+2d+... is defined by their projected endpoints on the top straight orange line (which its length is also the constant length X>0), series 2a+2b+2c+2d+... can't be but < X>0.

 

[Look]

Let's call the endpoints of the horizontal line segment at the top of the picture P and Q and let's name the endpoints of the intervals in some manner appropriate to discussing an infinite sequence. Let the lengths "a", "b", "c".. on the left side of PQ be denoted as closed intervals by A0 = [ P, a1], A_1 = [a1, a2] , A3 =[a2, a3], etc. Let the intervals on the right of the line segment that have lengths "a", "b", "c" etc. be denoted as B0 = [ b1, Q], B1 = [b2,b1], B2= [b3, b2], etc.

The only thing I see about the geometry that might be confusing is that the union of all the above intervals:
$( \cup_{i=0}^{\infty} Ai) \cup ( \cup_{i=0}^{\infty} Bi ) = S$ does not contain the midpoint M of PQ.

However this does not contradict the fact that the infinite series given by the sum of the lengths of those intervals sums to the length of PQ.

1. As long as one does not prove that some variable n is greater than any natural number, one has no basis to claim that it is a fact that series 2a+1b+2c+2d... has infinitely many terms.

2. Even if one proves that series 2a+1b+2c+2d... has infinitely many terms, stills one can't conclude that X=2a+1b+2c+2d... , exactly because series 2a+1b+2c+2d... is defined by the projected endpoints of non-straight lines on the top the straight line ,where all of the orange lines (straight or non-straight) have the constant length X>0.

So, since no one of them is reducible into the midpoint of the top straight line (which also has the constant length X>0) series 2a+1b+2c+2d... can't be but < the constant length X>0.

One can use partial sums as much as he\she likes, or claim as much as he\she likes that the absence of the midpoint in $( \cup_{i=0}^{\infty} Ai) \cup ( \cup_{i=0}^{\infty} Bi ) = S$ "does not contradict the fact that the infinite series given by the sum of the lengths of those intervals, sums to the length" X>0.

But (2) proves the impossibility of what is claimed to be a fact.

 

[Look]

Do you mean the following (and there are several similar):
If you walk along the diagonal of a unit square, it is $\sqrt{2}$ long, whereas the path along the outside is $2$ long. Now we divide the square into $4$ equal parts. If we now walk along the new borders of the now smaller squares, it is still $2= (\frac{1}{2}+\frac{1}{2})+(\frac{1}{2}+\frac{1}{2})$ long. If we continue this procedure, it will always be a way of $2$ no matter how often we divide the square. But we get closer and closer towards the diagonal. The limit is pointwise arbitrary close to the diagonal. Thus $\sqrt{2}=2$.
One can do the same with half circles and "prove" $\pi = 1$.

Please look at the following diagram:

If series a+b+c+d+...=X, then we are forced also to conclude that 2X=X√2.

If one does not agree with my conclusion, than please rigorously show that my conclusion is false.

 

[micromass]

Please look at the following diagram:

If series a+b+c+d+...=X, then we are forced also to conclude that X²=√X.

If one does not agree with my conclusion, than please rigorously show that my conclusion is false.

Pointswise or even uniform convergence of a curve does not preserve arc length.

 

[Look]

Dear micromass, here is a corrected version of my argumet:

Please look at the following diagram:

If series a+b+c+d+...=X, then we are forced also to conclude that 2X=X√2.

If you do not agree with my conclusion, then please rigorously show that my conclusion is false.

By following this corrected version, please explain (by using more details) why your reply "Pointswise or even uniform convergence of a curve does not preserve arc length." rigorously proves that my conclusion is false?

Thank you.

 

[micromass]

If series a+b+c+d+...=X, then we are forced also to conclude that 2X=X√2.

Why don't you prove rigorously why $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$. Showing a picture is not a proof.

 

[Look]

Why don't you prove rigorously why $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$. Showing a picture is not a proof.

Dear micromass, since $a+b+c+d+...$ is defined by "zig-zag" lines that all of them have the constant length $2X$, then $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$.

If you disagree with this implementation then please show that $a+b+c+d+...=X$ doe not imply $2X = X\sqrt{2}$.

Without loss of generality you can use the case $X=1$ (in that case 1/2+1/4+1/8+1/16...=1 which implies $2 = 1\sqrt{2}$).

If you agree that 1/2+1/4+1/8+1/16...=1, then the diagram is a proof without words that 1/2+1/4+1/8+1/16...=1 implies $2 = 1\sqrt{2}$.

 

[fresh_42]

Dear micromass, since $a+b+c+d+...$ is defined by "zig-zag" lines that all of them have the constant length $2X$, then $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$.

If you disagree with this implementation then please show that $a+b+c+d+...=X$ doe not imply $2X = X\sqrt{2}$.

Without loss of generality you can use the case $X=1$ (in that case 1/2+1/4+1/8+...=1 which implies $2 = 1\sqrt{2}$).

The point is: You seemingly want to establish deduction principles that differ from the framework, that mathematics provides. Even in mathematics there are frameworks that differ from the mainstream. However, in these cases it is carefully and rigorously said to what extend and which are the distinct principles. You haven't done this at all. What you are trying to do is, to engage people arguing within the standard mathematical framework where you simultaneously use an obscure and undefined system of logic without telling it. This cannot match by construction. Drawn pictures and off-label usage of standard mathematical terminology are not appropriate means to overcome this lack of fundamental agreement - it only tries to disguise it.
[And I refuse to participate in this kind of debate.]

 

[micromass]

Dear micromass, since $a+b+c+d+...$ is defined by "zig-zag" lines that all of them have the constant length $2X$, then $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$.

That is not a rigorous proof, you just made an assertion.

If you disagree with this implementation then please show that $a+b+c+d+...=X$ doe not imply $2X = X\sqrt{2}$.

I can't prove it. It has been proven by Gödel that we can never proof that there is no proof. So while it's true that I cannot prove that an even number is odd, nobody can prove that it cannot be proven. This is coined in usual language "one can never prove a negative".

So because you made the claim that $a+b+c+d+...=X$ implies $2X = X\sqrt{2}$, I would like to see a rigorous proof of that.

Without loss of generality you can use the case $X=1$ (in that case 1/2+1/4+1/8+...=1 which implies $2 = 1\sqrt{2}$).

Definition:
In $\mathbb{R}$, we define $\sum_{n=1}^{+\infty} a_n = a$ if for all $\varepsilon>0$, there exists an $N\in \mathbb{N}$ such that for all $N'\geq N$ holds that $\left|a - \sum_{n=1}^{N'}\right|<\varepsilon$.

Part 1: If $k$ is a natural number, then we define $x_k = \sum_{n=1}^k \frac{1}{2^n}$. This is a real number since the sum is finite. Note that
2x_k = 2\sum_{n=1}^k \frac{1}{2^n} = \sum_{n=1}^k \frac{1}{2^{n-1}} = 1 + \sum_{n=0}^{k-1} \frac{1}{2^n} = 1 + x_k - \frac{1}{2^k}
Hence by rearranging, we have
x_k = 1 - \frac{1}{2^k}

Part 2: Choose $n\in \mathbb{N}$ arbitrarily with $n\geq 1$. We show that $n<2^n$ by induction.
Induction step: $1<2^1 = 2$.
Assume that $n<2^n$. Since $1\leq n$, we have in particular that $1<2^n$. So then $n+1< 2^n + 2^n = 2\cdot 2^n = 2^{n+1}$.

Final proof:
Choose $\varepsilon>0$ arbitrary. By the Archimedes axiom, we can choose an $N\mathbb{N}$ such that $N\geq 1$ and $N\geq \frac{1}{\varepsilon}$. If $N'\geq N$, then by part $2$, we have
\frac{1}{\varepsilon} \leq N\leq N&#039; &lt; 2^{N&#039;} = |2^{N&#039;}|
Thus
\left|\frac{1}{2^{N&#039;}}\right|&lt;\varepsilon
Hence
\left|1 - \left(1 - \frac{1}{2^{N&#039;}}\right)\right|&lt;\varepsilon
By using part $1$, we have
\left|1 - \sum_{n=1}^{N&#039;}\frac{1}{2^n}\right|&lt;\varepsilon
This is what we had to show by the definition.

 

[Look]

The point is: You seemingly want to establish deduction principles that differ from the framework.

Dear fresh, since you agree that 1/2+1/4+1/8+1/16...=1, then the diagram is a proof without words that 1/2+1/4+1/8+1/16...=1 implies $2 = 1\sqrt{2}$.

 

[micromass]

Dear fresh, since you agree that 1/2+1/4+1/8+1/16...=1, then the diagram is a proof without words that 1/2+1/4+1/8+1/16...=1 implies $2 = 1\sqrt{2}$.

A proof without words doesn't exist. You need to show rigorously why $1/2 + 1/4 + 1/8 + ... = 1$ implies $2 = \sqrt{2}$.

 

[Look]

That is not a rigorous proof, you just made an assertion.

Dear micromass, since you agree that 1/2+1/4+1/8+1/16...=1, then the diagram is a proof without words that 1/2+1/4+1/8+1/16...=1 implies $2 = 1\sqrt{2}$ (which is a rigorous implementation of implication, without loss of generality).

 

[micromass]

Dear micromass, since you agree that 1/2+1/4+1/8+1/16...=1, then the diagram is a proof without words that 1/2+1/4+1/8+1/16...=1 implies $2 = 1\sqrt{2}$ (which is a rigorous implementation without loss of generality).

You demand from us a rigorous proof from real analysis? But when we ask you the same thing you just call it a "proof without words"? Very dishonest from you. I provided a rigorous real analysis proof. Now it's your turn to prove rigorously that $2 = \sqrt{2}$.

 

[Look]

A proof without words doesn't exist.

Please see https://en.wikipedia.org/wiki/Proof_without_words:

Such proofs can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature.

 

[micromass]

Please see https://en.wikipedia.org/wiki/Proof_without_words .

I know they exist. But they're not rigorous. Mathematicians will never accept them as actual proofs.

 

[micromass]

But sure, I can provide a proof without words that $X = a+b+c+d+...$ too, you know.

 

[Look]

I know they exist. But they're not rigorous.

They are mathematically rigorous.

 

[micromass]

They are mathematically rigorous.

Not according to 100% of the mathematicians. Sorry.

 

[Look]

They are mathematically rigorous.

Are going to ignore the following? :

Such proofs can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature.

 

[Look]

Not according to 100% of the mathematicians. Sorry.

Realy?, please support this claim.

 

[micromass]

Are going to ignore the following? :

Your quote just confirms that I'm right by contrasting such proofs with "more formal and mathematically rigorous proofs". Meaning that proofs without words are not rigorous.

 

[micromass]

Realy?, please support this claim.

Do you know the definition of a proof and the definition of a theorem?

 

[Look]

Not according to 100% of the mathematicians. Sorry.

Again, you simply ignore the following quote, taken from Wikipedia ( https://en.wikipedia.org/wiki/Proof_without_words ):

Such proofs can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature.

 

[micromass]

Again, you simply ignore the following quote, taken from Wikipedia ( https://en.wikipedia.org/wiki/Proof_without_words ):

I just addressed it: it implies that actual proofs are more mathematically rigorous than proof without words.

Also, relying on wikipedia to make your argument, really?? Refer to an actual math book if you want me to take you seriously.

 

[Look]

Do you know the definition of a proof and the definition of a theorem?

Do you know that proofs without words "can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature" ?

 

[micromass]

Do you know that proofs without words "can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature" ?

Yes, they are more elegant. They are also less formal and less mathematically rigorous which is stated by the very quote you give!

 

[Look]

I just addressed it: it implies that actual proofs are more mathematically rigorous than proof without words.

Wrong, by Wikipedia proofs without words "can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature"

Also, relying on wikipedia to make your argument, really?? Refer to an actual math book if you want me to take you seriously.

Wikipedia is a reliable source for mathematics, which is not less reliable than math books.

 

[micromass]

Wrong, by Wikipedia proofs without words "can be considered more elegant than more formal and mathematically rigorous proofs due to their self-evident nature"

I am not disagreeing that they can be considered more elegant. I'm saying that they're not mathematically rigorous. Your quote doesn't disagree with that assertion.

 

[Look]

Yes, they are more elegant. They are also less formal and less mathematically rigorous which is stated by the very quote you give!

A proof without words is a self-evident true that uses symbols without the need of further verbal-symbolic definitions.

 

[micromass]

A proof without words is a self-evident true that uses symbols without the need of further verbal-symbolic definitions.

For centuries, people thought it was self-evident that the Earth is the center of the universe. Self-evident doesn't mean true. There are many self-evident things in math that turn out to be false. This is why we need mathematical rigor.