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

• I
• Look
In summary: Since the exponent value n of number 4n is no more than an ever larger natural number, there are no more than finitely many non-straight orange lines of the same finite length (X>0) that their projected endpoints upon the straight orange line, define 2(a+b+c+d+...) .Since there are no more than finitely many projected endpoints that define 2(a+b+c+d+...) , the limit as done by real-analysis is undefined.In summary, Stephen Tashi is confused by the following diagram and
Look
Hi,

I am confused by the following diagram when I try to understand it in terms of limit as done by Real-analysis:

What I currently understand is as follows:

Let the finite length of the straight orange line be X>0.

The rest of the non-straight orange lines (in this particular case, the non-straight orange lines have forms of different degrees of Koch fractal) are actually the same line with finite length X>0, such that its end points are projected upon itself, and as a result we get the convergent series 2*(a+b+c+d+...) .

Each one of the non-straight lines is constructed by 4n straight parts, where n is some natural number and the length of each part is X/4n, so given any arbitrary non-straight line, it has a finite amount of parts ( X=(X/4n)*4n ).

By using limit as done by Real-analysis X-2*(a+b+c+d+...)=0, but 2*(a+b+c+d+...) is the projected result of finite amount of non-straight lines, where each one of them has the same finite length, which is X>0 (as observed above).

2*(a+b+c+d+...) (which is the result of the projection of a finite length X>0 upon itself) is equal to X only if a projected non-straight line somehow collapsed into length 0.

This is definitely not the case in the diagram above (there are finitely many non-straight lines, where each one of them has the same finite length X>0).

So, I still do not understand how X-2*(a+b+c+d+...)=0 by Real-analysis.

Can I get some help?

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Look said:
2*(a+b+c+d+...) is the projected result of finite amount of non-straight lines, where each one of them has the same finite length, which is X>0 (as observed above).
I do not understand what you mean by this. Can you explain?

But if you do not understand why X=2(a+b+c...) I can explain that to you.

andrewkirk said:
I do not understand what you mean by this. Can you explain?
Since the exponent value n of number 4n is no more than an ever larger natural number, there are no more than finitely many non-straight orange lines of the same finite length (X>0) that their projected endpoints upon the straight orange line, define 2(a+b+c+d+...) .

In that case I do not see how X=2(a+b+c+d+...), since:

1. All orange lines (whether they are straight or non-straight) have the same constant finite length (X>0).

2. There are no more than finitely many projected endpoints that define 2(a+b+c+d+...) .

andrewkirk said:
But if you do not understand why X=2(a+b+c...) I can explain that to you.
By using what was explained above, please explain how X=2(a+b+c+d+...) .

Thank you.

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The summation of a series of the form ##\sum_{i = 0}^{\infty} a_i ## is to defined to be a limit of partial sums of the form ##\sum_{i=0}^{n} a_i##.

##\ \sum_{i = 0}^{\infty} a_i = L = \lim_{n \rightarrow \infty} ( \sum_{i=0}^{n} a_i) ##. There are only finitely many terms in each partial sum. Nothing in the definition of the limit requires that any of the partial sums will be equal to ##L##.

Stephen Tashi said:
The summation of a series of the form ##\sum_{i = 0}^{\infty} a_i ## is to defined to be a limit of partial sums of the form ##\sum_{i=0}^{n} a_i##.

##\ \sum_{i = 0}^{\infty} a_i = L = \lim_{n \rightarrow \infty} ( \sum_{i=0}^{n} a_i) ##. There are only finitely many terms in each partial sum. Nothing in the definition of the limit requires that any of the partial sums will be equal to ##L##.
Are there infinitely many partial sums by this definition?

Look said:
Are there infinitely many partial sums by this definition?

The definition relevant to the cardinality of the set of partial sums is the definition for partial sum. A partial sum exists for each positive integer, so the set of all possible partial sums is infinite. However, nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together.

Stephen Tashi said:
The definition relevant to the cardinality of the set of partial sums is the definition for partial sum. A partial sum exists for each positive integer, so the set of all possible partial sums is infinite. However, nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together.
Dear Stephen Tashi, if nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together (since the sequence of partial sums of natural numbers fails to converge to a finite limit, the series does not have a sum (it is a divergent series)).

So we are left with no more than ever larger natural numbers that grow without bounds (yet each natural number is a finite (or bounded) mathematical object).

In that case exponent n of number 4n is no more than an ever larger natural number (we have at most what is known as potential infinity (as explained, for example, in https://en.wikipedia.org/wiki/Actual_infinity)).

So I still do not see how the used definition actually determines the equality X=2*(a+b+c+d+...) .

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For simplicity, and without loss of generality, let's assume X=1, and rename ##a,b,c,...## as ##a_1,a_2,a_3,...##
Given that, here is a proof that ##\sum_{j=1}^\infty a_j=0.5##, which is I think what you are asking for @Look .

Call the difference between two x (horizontal) coordinates of the two ends of a non-straight line its 'width'. Then the width of each line is ##\frac34## times that of the line above it. Hence ##a_{j+1} = \frac34 a_j##. We also observe that ##a_1=\frac18##.

Define ##s_n=\sum_{j=1}^n a_n##. This is equal to
$$\frac18\times \sum_{j=0}^{n-1}\left(\frac34\right)^j =\frac18\times \frac{1-\left(\frac34\right)^n}{1-\frac34} =\frac12\times \left(1-\left(\frac34\right)^n\right)$$

Since ##\lim_{n\to\infty}\left(\frac34\right)^n=0## we have
$$\lim_{n\to\infty}s_n=\frac12$$
And ##\sum_{j=1}^\infty a_n## is defined to mean ##\lim_{n\to\infty}s_n##.

Hence we have ##\sum_{j=1}^\infty a_n=\frac12## as required.

That covers the question that has been asked.

However, I have an elusive memory that there is a puzzle that uses this shape, involving an apparent contradiction, and that resolving it requires noticing a false, hidden assumption that one did not realize one was making. Unfortunately, I cannot remember the puzzle, and googling things like 'Koch triangle' did not turn up anything that looked familiar.

andrewkirk said:
However, I have an elusive memory that there is a puzzle that uses this shape, involving an apparent contradiction, and that resolving it requires noticing a false, hidden assumption that one did not realize one was making.
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##.

It wasn't that one fresh, but I really like that. I'm going to remember it and use it in future.

The one that I'm unable to remember (assuming it's not a phantom memory) uses exactly the diagram above, and the marked items a,b,c... and then calculates some sum involving those two different ways and gets different answers.

Look said:
Dear Stephen Tashi, if nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together (since the sequence of partial sums of natural numbers fails to converge to a finite limit, the series does not have a sum (it is a divergent series)).

The notation indicates partial sums add sequences of real numbers, not necessarily the sequence of natural numbers.
The notation ##a_k## doesn't denote the natural number ##k##. The notation ##\sum_{i=0}^n a_n ## doesn't indicate the same sum as ##\sum_{i=0}^n i ##.
So I still do not see how the used definition actually determines the equality X=2*(a+b+c+d+...) .

You'd have to write an explicit description of the series before you can apply a mathematical definition for its sum. You've only drawn a picture.

Stephen Tashi said:
The notation indicates partial sums add sequences of real numbers, not necessarily the sequence of natural numbers.
The notation ##a_k## doesn't denote the natural number ##k##. The notation ##\sum_{i=0}^n a_n ## doesn't indicate the same sum as ##\sum_{i=0}^n i ##.You'd have to write an explicit description of the series before you can apply a mathematical definition for its sum. You've only drawn a picture.
The numbers of 2a+2b+2c+2d+... are not natural numbers, but the mechanism that defines them is derived form 4n where n is no more than an ever larger natural number.

Therefore I still do not see how X=2(a+b+c+d+...) .

andrewkirk said:
For simplicity, and without loss of generality, let's assume X=1, and rename ##a,b,c,...## as ##a_1,a_2,a_3,...##
Given that, here is a proof that ##\sum_{j=1}^\infty a_j=0.5##, which is I think what you are asking for @Look .

Call the difference between two x (horizontal) coordinates of the two ends of a non-straight line its 'width'. Then the width of each line is ##\frac34## times that of the line above it. Hence ##a_{j+1} = \frac34 a_j##. We also observe that ##a_1=\frac18##.

Define ##s_n=\sum_{j=1}^n a_n##. This is equal to
$$\frac18\times \sum_{j=0}^{n-1}\left(\frac34\right)^j =\frac18\times \frac{1-\left(\frac34\right)^n}{1-\frac34} =\frac12\times \left(1-\left(\frac34\right)^n\right)$$

Since ##\lim_{n\to\infty}\left(\frac34\right)^n=0## we have
$$\lim_{n\to\infty}s_n=\frac12$$
And ##\sum_{j=1}^\infty a_n## is defined to mean ##\lim_{n\to\infty}s_n##.

Hence we have ##\sum_{j=1}^\infty a_n=\frac12## as required.

That covers the question that has been asked.

However, I have an elusive memory that there is a puzzle that uses this shape, involving an apparent contradiction, and that resolving it requires noticing a false, hidden assumption that one did not realize one was making. Unfortunately, I cannot remember the puzzle, and googling things like 'Koch triangle' did not turn up anything that looked familiar.
Please read carefully https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5538138 and https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5538399 .

Since exponent n of 4n is no more than an ever larger natural number, I still do not see how X=2(a+b+c+d+...) .

Look said:
Please read carefully https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5538138 and https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5538399 .

Since exponent n of 4n is no more than an ever larger natural number, I still do not see how X=2(a+b+c+d+...) .
I'm afraid I don't understand what you are saying in either of those posts. Your terminology is non-standard. For instance I do not know what is meant by
Look said:
the mechanism that defines them is derived form 4n

Did you understand the proof that I wrote for you?

andrewkirk said:
Did you understand the proof that I wrote for you?
Yes, and it does not cover the following fact (as clearly obsereved in https://www.physicsforums.com/threads/a-limits-problem-fractal-initiator-generator.881023/):
Look said:
Each one of the non-straight lines is constructed by 4n straight parts, where n is some natural number and the length of each part is X/4n, so given any arbitrary non-straight line, it has a finite amount of parts ( X=(X/4n)*4n ).

The numbers of 2a+2b+2c+2d+... are not natural numbers, but the mechanism that defines them is derived form 4n ( as clealy seen in https://www.physicsforums.com/threads/a-limits-problem-fractal-initiator-generator.881023/) where n is no more than an ever larger natural number (the definition of partial sums in the issue at hand is not about 2a+2b+2c+2d+..., but it is about the ever larger n's (which is a divergent series that actually stands at the basis of the convergent serias 2a+2b+2c+2d+...) .

I don't see how Real-analysis can ignore the following:
Look said:
So we are left with no more than ever larger natural numbers that grow without bounds (yet each natural number is a finite (or bounded) mathematical object).

In that case exponent n of number 4n is no more than an ever larger natural number (we have at most what is known as potential infinity (as explained, for example, in https://en.wikipedia.org/wiki/Actual_infinity)).

So I still do not see how your proof ( which is based on ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## ) actually determines the equality X=2*(a+b+c+d+...), exactly because no non-striehgt orange line (where each arbitratry non-striehgt orange line has the same constant length X>0) is actually reducible into length 0.

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Look said:
(the definition of partial sums in the issue at hand is not about 2a+2b+2c+2d+..., but it is about the ever larger n's (which is a divergent series that actually stands at the basis of the convergent serias 2a+2b+2c+2d+...) .

There is a distinction between a "sequence" and a "series". Unless you think your diagram somehow implies evaluation of a "series" like 1+2+3+4+..., I think you are trying to say something about a "sequence" of numbers like: 1,2,3,4,...

When you say "which is a divergent series that actually stands at the basis of the convergent serias 2a+2b+2c+2d+...", it isn't clear what it would mean for a divergent series to "stand at the basis" of another series. That isn't standard mathematical terminology.

You aren't making it clear whether you understand the summation of series as it is done "in real analysis", even though you mention real analysis. The objections about n being "ever larger" could be made about any convergent series.

One cannot tell just from the picture whether the series 2a+2b+2c+... converges or not. Evaluating convergence from the point of view of real analysis would require dealing with the numerical values involved.

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Stephen Tashi said:
There is a distinction between a "sequence" and a "series". Unless you think your diagram somehow implies evaluation of a "series" like 1+2+3+4+..., I think you are trying to say something about a "sequence" of numbers like: 1,2,3,4,...
Dear Stephen Tashi, you are right, I wrongly wrote:
Look said:
Dear Stephen Tashi, if nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together (since the sequence of partial sums of natural numbers fails to converge to a finite limit, the series does not have a sum (it is a divergent series)).
The right one is:
Look said:
Dear Stephen Tashi, nothing in the definition of ##\lim_{n\rightarrow \infty} (\sum_{i=0}^n a_i) ## asserts that an "infinite number" of things are being added together, because the sequence of partial sums of the associated series of natural numbers (I am talking here about exponents n of 4n) fails to converge to a finite limit (the series of exponents n do not have a sum (it is a divergent series)).

Stephen Tashi said:
When you say "which is a divergent series that actually stands at the basis of the convergent serias 2a+2b+2c+2d+...", it isn't clear what it would mean for a divergent series to "stand at the basis" of another series. That isn't standard mathematical terminology.
Since series 2a+2b+2c+2d+... is the result of the projection of each non-straight orange line on the straight orange line, where the constant length X>0 of each non-straight orange line is the sum of 4n straight parts, then it is mathematically clear that the divergent series of n's (which is series 1+2+3+4+... (that is the associated series of sequence 1,2,3,4,...)) indeed stands in the basis of series 2a+2b+2c+2d+...

Stephen Tashi said:
You aren't making it clear whether you understand the summation of series as it is done "in real analysis", even though you mention real analysis. The objections about n being "ever larger" could be made about any convergent series.
Please look at the diagram. Each orange line (straight or non-straight) has the same constant length X>0, and since there are no more than ever larger exponents n (that by 4n define the sum of the straight parts of each non-straight line with constant length X>0) there are always non-straight lines with constant length X>0, which prevent the sum of series 2a+2b+2c+2d+...

In other words, I still do not see how the definitions of Real-analysis are sufficient in order to determine that X=2a+2b+2c+2d+...

Stephen Tashi said:
One cannot tell just from the picture whether the series 2a+2b+2c+... converges or not. Evaluating convergence from the point of view of real analysis would require dealing with the numerical values involved.
Please see https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5538276 .

Yet I don't see its ability to determine that X=2a+2b+2c+2d+... because of the (corrected) reasons mentioned above.

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Look said:
Since series 2a+2b+2c+2d+... is the result of the projection of each non-straight orange line on the straight orange line,

I don't see any projection done in the picture. It seems to me that 2a represents a difference in two distances.

where the constant length X>0 of each non-straight orange line is the sum of 4n straight parts, then it is mathematically clear that the divergent series of n's (which is series 1+2+3+4+... (that is the associated series of sequence 1,2,3,4,...)) indeed stands in the basis of series 2a+2b+2c+2d+...

What do you mean by "the basis" of a series ?

Please look at the diagram. Each orange line (straight or non-straight) has the same constant length X>0, and since there are no more than ever larger exponents n (that by 4n define the sum of the straight parts of each non-straight line with constant length X>0) there are always non-straight lines with constant length X>0, which prevent the sum of series 2a+2b+2c+2d+...

An infinite series can converge to a limit X even if no finite number of terms of it ever sum to X. That's a consequence of the definition of the limit of an infinite series.

In other words, I still do not see how the definitions of Real-analysis are sufficient in order to determine that X=2a+2b+2c+2d+...

Then the place to start is to understand the implications (and non-implications) of the definition of the sum of an infinite series as used in real analysis.

Start with a simpler picture. Draw horizontal line segment PQ with midpoint M1. Let distance a = PM1. Draw the midpoint M2 of segment M1Q. Let b be the length of M1M2. Draw M3 the midpoint of M2Q. Let c be the distance M3Q, etc. Now are you going to argue that the infinite series a + b + c + ... doesn't sum to the length PQ ?

Stephen Tashi said:
I don't see any projection done in the picture. It seems to me that 2a represents a difference in two distances.
Dear Stephen Tashi,

Let's do it step by step.

Here is the considered diagram:

The straight top orange line has some finite length X>0.

The all non-straight orange lines have the same finite length X>0 as the straight top orange line.

Series 2*(a+b+c+d+...) is the result of the projected endpoints of all non-straight orange lines (which have the constant finite length X>0) on the top orange straight line.

In that case the series 2*(a+b+c+d+...) can't be equal to the constant finite X>0 even if (by Real-analysis) its limit is the middle point of the top straight orange line.

If you disagree with my conclusion, then please explicitly demonstrate (by using Real-analysis) that X=2*(a+b+c+d+...) .

Thank you.

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Look said:
If you disagree with my conclusion, then please explicitly demonstrate (by using Real-analysis) that X=2*(a+b+c+d+...) .

Are you familiar with the theorem that a bounded increasing sequence converges to its least upper bound ?

Stephen Tashi said:
Are you familiar with the theorem that a bounded increasing sequence converges to its least upper bound ?
Dear Stephen Tashi, I am talking about the series 2a+2b+2c+2d+... as observed in the diagram, so please use Real-analysis in order to prove that X=2a+2b+2c+2d+...

I do not see how X=2a+2b+2c+2d+... (because X>0 is a constant among all the non-straight orange lines (which their projected endpoints on the top straight orange finite line, actually define series X=2a+2b+2c+2d+...) .

Please explain exactly how "the theorem that a bounded increasing sequence converges to its least upper bound" is relevant to the issue at hand?Thank you.

Look said:
Please explain exactly how "the theorem that a bounded increasing sequence converges to its least upper bound" is relevant to the issue at hand?

The sequence 2a + 2b + 2c + ... is increasing. It is bounded above by the total length of the top line segment. If you think some shorter segment is an upper bound for the sequence, please draw that segment.

Stephen Tashi said:
The sequence 2a + 2b + 2c + ... is increasing. It is bounded above by the total length of the top line segment. If you think some shorter segment is an upper bound for the sequence, please draw that segment.
2a + 2b + 2c + ... is a convergent series (not a sequence) that approaches to the middle point of the top straight line with constant length X>0.

Since series 2a + 2b + 2c + ... is defined by the projected endpoints of all non-straight lines with the same constant length X>0, the series 2a + 2b + 2c + ... is ever smaller than X>0, exactly as very clearly seen in https://www.physicsforums.com/threa...ctal-initiator-generator.881023/#post-5539303 .

So I still do not see how exactly Real-analysis determines the equality X=2a + 2b + 2c + ... exactly because X>0 is a constant among all the non-straight orange lines (which their projected endpoints on the top straight orange finite line, actually define series 2a + 2b + 2c + ...) .

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Look said:
2a + 2b + 2c + ... is a convergent series (not a sequence) that approaches to the middle point of the top straight line with constant length X>0.

I think you mean that the right endpoints of a, a+b, a+b+c, ... can be visualized a sequence of points that approaches the midpoint of top straight line. So 2a, 2a + 2b, 2a + 2b + 2c is a sequence of numbers that approaches the total length of the top straight line.

Since series 2a + 2b + 2c + ... is defined by the projected endpoints of all non-straight lines with the same constant length X>0, the series 2a + 2b + 2c + ... is ever smaller than X>0,

As I said before, the fact that any finite number of terms in a series add to less than L doesn't prevent L from being the sum of the series. For example, consider ##\sum_{k=0}^\infty \frac{1}{2^k} = 2 ##.

Stephen Tashi said:
I think you mean that the right endpoints of a, a+b, a+b+c, ... can be visualized a sequence of points that approaches the midpoint of top straight line. So 2a, 2a + 2b, 2a + 2b + 2c is a sequence of numbers that approaches the total length of the top straight line
I am talking exactly about the series 2a + 2b + 2c + ... and not about partial sums of it.

As I said before, the fact that any finite number of terms in a series add to less than L doesn't prevent L from being the sum of the series. For example, consider ##\sum_{k=0}^\infty \frac{1}{2^k} = 2 ##.
I am talking exactly about the series 2a + 2b + 2c + ... and not about partial sums of it.

n is no more than an ever larger natural number, which is the exponent of 4n.

Therefore 4n is also no more than an ever larger natural number, that defines the number of straight parts for all non-straight lines (that their projected end points on the top straight line, actually define the series 2a + 2b + 2c + ... , and not just partial sums of it).

In order to prove that there are infinitely many terms in 2a + 2b + 2c + ... , you first must prove that n is larger than any natural number.

Mathematical induction (whether it is used as an axiom schema in Peano, or as an axiom in ZF) can't help you to prove that n is more than an ever larger natural number.

A collection of all natural numbers is no more than a collection ever larger values, so a request like "please show me a natural number that is not in the collection of all natural numbers" is irrelevant to the issue at hand.

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Look said:
n is no more than an ever larger natural number, which is the exponent of 4n.

I have no idea what you mean by that. What "n" are you talking about ?
In order to prove that there are infinitely many terms in 2a + 2b + 2c + ... , you first must prove that n is larger than any natural number.

Do you think the series 2a + 2b + 2c + ... had a finite number of terms?

Mathematical induction (whether it is used as an axiom schema in Peano, or as an axiom in ZF) can't help you to prove that n is more than an ever larger natural number.

I don't know what "n" you're talking about.

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.

Stephen Tashi said:
I have no idea what you mean by that. What "n" are you talking about ?
Dear Stephe Tashi, please look at the diagram.

1. All orange lines (straight or non-straight) have the same constant length X>0.

2. For all non-straight lines with constant length X>0 (which their projected endpoints actually define the convergent series 2a+2b+2c+2d+... along the top orange straight line with finite constant length X>0) there are no more than ever larger exponents n, which by 4n define the number of the straight parts for these non-straight lines.
Stephen Tashi said:
Do you think the series 2a + 2b + 2c + ... had a finite number of terms?
Yes, and it is inevetable by (2), unless you prove that n is larger than any natural number.

Look said:
Mathematical induction (whether it is used as an axiom schema in Peano, or as an axiom in ZF) can't help you to prove that n is more than an ever larger natural number.

A collection of all natural numbers is no more than a collection of ever larger values, so a request like "please show me a natural number that is not in the collection of all natural numbers" is irrelevant to the issue at hand.

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Look said:
2. For all non-straight lines with constant length X>0 (which their projected endpoints actually define the convergent series 2a+2b+2c+2d+... along the top orange straight line with finite constant length X>0) there are no more than ever larger exponents n, which by 4n define the number of the straight parts for these non-straight lines.

You appear to be using "n" to denote a variable. I think you are observing that the n-th set of orange line segments has ##4^n## sides.
Yes, and it is inevetable by (2), unless you prove that n is larger than any natural number.

What you say in 2) has nothing to do with the convergence of the series and it certainly doesn't limit the number of terms in the series to a finite number.

You seem to think that the fact that the total length of the n-th set of orange line segments is computed from adding the length of ##4^n## line segments somehow puts an upper bound on how many sets of orange line segments can exist. It doesn't.

Stephen Tashi said:
You appear to be using "n" to denote a variable. I think you are observing that the n-th set of orange line segments has ##4^n## sides.
Exactly.

Stephen Tashi said:
What you say in 2) has nothing to do with the convergence of the series and it certainly doesn't limit the number of terms in the series to a finite number.

You seem to think that the fact that the total length of the n-th set of orange line segments is computed from adding the length of ##4^n## line segments somehow puts an upper bound on how many sets of orange line segments can exist. It doesn't.
Dear Stephen Tashi, I agree with you that (the series 2a+2b+2c+2d+... have infinite terms) AND (the number of non-straight orange lines are not bounded to finite amount) only if you prove that a content of variable n is greater than any natural number.

Without this proof your two claims in your quote above, are no more than hypotheses.

Moreover, let's assume that n is greater that 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, series 2a+2b+2c+2d+... can't be but < X>0.

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Look said:
Dear Stephen Tashi, I agree with you that (the series 2a+2b+2c+2d+... have infinite terms) AND (the number of non-straight orange lines are not bounded to finite amount) only if you prove that a content of variable n is greater than any natural number.

Without this proof your two claims in your quote above, are no more than hypotheses.

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 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, series 2a+2b+2c+2d+... can't be but < X>0.

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. However, it does not prevent the sum of the infinite series from being X. If you wish to prove that the infinite series sums to a number less than X then demonstrate that there exists a number X1 < X such that any finite portion of the infinite series sums to a total less than or equal to X1.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. 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.

fresh_42
Stephen Tashi said:
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?

Stephen Tashi said:
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.

Stephen Tashi said:
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.
Stephen Tashi said:
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.

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Stephen Tashi said:
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.

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fresh_42 said:
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.

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Look said:
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.

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.

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