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

1. Aug 4, 2016

### 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?

Last edited: Aug 4, 2016
2. Aug 5, 2016

### andrewkirk

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.

3. Aug 7, 2016

### Look

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+...) .

By using what was explained above, please explain how X=2(a+b+c+d+...) .

Thank you.

Last edited: Aug 7, 2016
4. Aug 7, 2016

### Stephen Tashi

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 $L$.

5. Aug 7, 2016

### Look

Are there infinitely many partial sums by this definition?

6. Aug 7, 2016

### Stephen Tashi

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.

7. Aug 7, 2016

### Look

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+...) .

Last edited: Aug 7, 2016
8. Aug 7, 2016

### andrewkirk

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 realise one was making. Unfortunately, I cannot remember the puzzle, and googling things like 'Koch triangle' did not turn up anything that looked familiar.

9. Aug 7, 2016

### Staff: Mentor

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" $\pi = 1$.

10. Aug 7, 2016

### andrewkirk

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.

11. Aug 7, 2016

### Stephen Tashi

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.

12. Aug 7, 2016

### Look

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+...) .

13. Aug 7, 2016

### Look

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+...) .

14. Aug 7, 2016

### andrewkirk

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
Did you understand the proof that I wrote for you?

15. Aug 8, 2016

### Look

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/):
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:
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.

Last edited: Aug 8, 2016
16. Aug 8, 2016

### Stephen Tashi

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.

Last edited: Aug 8, 2016
17. Aug 8, 2016

### Look

Dear Stephen Tashi, you are right, I wrongly wrote:
The right one is:
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+...

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+...

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

Last edited: Aug 8, 2016
18. Aug 8, 2016

### Stephen Tashi

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

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

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.

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 ?

19. Aug 9, 2016

### Look

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.

Last edited: Aug 9, 2016
20. Aug 9, 2016

### Stephen Tashi

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