problem with cantor diagonal argument

[pzkpfw]

44 minutes ago, studiot said:

I'm sure you actually meant something else.

 

Nope, I meant exactly what I wrote.

 

[phyti]

KJW;

1.

To Cantor, his mathematical views were intrinsically linked to their philosophical and theological implications – he identified the Absolute Infinite with God,[57] and he considered his work on transfinite numbers to have been directly communicated to him by God, who had chosen Cantor to reveal them to the world.[1]

2.

Let M be the totality (n) of all finite numbers n, and M¢ the totality (2n) of all even numbers 2n. Here it is undeniably correct that M is richer in its entity, than M¢; M contains not only the even numbers, of which M¢  consists, but also the odd numbers M¢¢ . On the other hand it is just as unconditionally correct that the same cardinal number belongs to both the sets M and M¢. Both of these are certain, and neither stands in the way of the other if one heeds the distinction between reality and number.[2]

3.

As Leopold Kronecker claimed: "I don't know what predominates in Cantor's theory - philosophy or theology, but I am sure that there is no mathematics there." [3]

"Actual infinity does not exist. What we call infinite is only the endless possibility of creating new objects no matter how many exist already" (Poincaré quoted from Kline 1982). [3]

Wittgenstein denies Hume's principle, arguing that our concept of number depends essentially on counting. "Where the nonsense starts is with our habit of thinking of a large number as closer to infinity than a small one".

reference. [3]

4.

Numbers and counting are reality. Functions plot a y corresponding to each x for y=2x.

His desire/motivation to sell transfinite numbers seems to have affected his interpretation of the 1 to 1 relation as part of the cardinality of the infinite.

A true 1 to 1 correspondence for E=2D.

D 1 3 5 7 9 ...

E 2 4 6 8 10 ...

-----------------------------------------

[1] 'Georg Cantor', Wikipedia

[2] Ewald, W., From Kant to Hilbert, Oxford 1996
[3]en.wikipedia.org/wiki, 'Controversy_over_Cantor's_theory'

 

[phyti]

pzkpfw;

How many elements are in A?
How many elements are in B?
Is there a way to uniquely pair elements from A and B 1-to-1 ?
You are misapplying statistics.

A and B are 'infinite' so there is no value/number/cardinality to associate with them.

The 1 to 1 correspondence/mapping uses N as a reference for counting.

The purpose is to compare a part to the whole. A slice of pie is not equal to the whole pie. In set jargon, It's a subset to the complete set.

Notice in the common 'mapping' of n to 2n, some of the elements are on both sides!

2-4, 4-16, ... redundant information.

A and B each have unique elements forming a unique correspondence, with each containing 1/2 of the set N despite it being infinite. Some ratios remain constant as with H and T for coin tosses.

 

[KJW]

20 hours ago, phyti said:

To Cantor, his mathematical views were intrinsically linked to their philosophical and theological implications – he identified the Absolute Infinite with God,[57] and he considered his work on transfinite numbers to have been directly communicated to him by God, who had chosen Cantor to reveal them to the world.

Earlier in this thread I said that I was "unwilling to debate over a translation of the precise wording of an 1891 paper". I should also have included discussions about the man. The fact of the matter is that the validity of a mathematical theorem is independent of who originated it or how it is expressed in particular writings. Thus, I accept the diagonal argument purely on the basis of its logic, and for you to convince me otherwise, your argument would also have to be purely on the basis of its logic.
 

 

20 hours ago, phyti said:

Let M be the totality (n) of all finite numbers n, and M¢ the totality (2n) of all even numbers 2n. Here it is undeniably correct that M is richer in its entity, than M¢; M contains not only the even numbers, of which M¢  consists, but also the odd numbers M¢¢ .

I don't know about that. It turns out that the mapping from the group of integers [math]\mathbb {Z}[/math] to the group of even integers [math]2\mathbb {Z}[/math] is an endomorphism. This means that properties of the odd numbers map to properties of double of the odd numbers under the endomorphism. And given that this mapping is one-to-one, it is unlikely that the integers [math]\mathbb {Z}[/math] are any richer than the even integers [math]2\mathbb {Z}[/math].
 

 

20 hours ago, phyti said:

As Leopold Kronecker claimed: "I don't know what predominates in Cantor's theory - philosophy or theology, but I am sure that there is no mathematics there."

Leopold Kronecker was a subscriber to the philosophy of finitism and was thus in the minority among mathematicians.
 

 

20 hours ago, phyti said:

Actual infinity does not exist.

I am a subscriber to the philosophy of formalism, and regard mathematics as not being rigorous unless it can be derived from first principles (axioms) by mechanical symbol manipulation. The meanings of the symbols are abstract, not necessarily connected to reality, and manifest in the process of manipulation. I do not ask if notions represented by the symbols exist in reality.

However, I ask you whether our description of reality is in any way adversely affected by the concept of the transfinite numbers, or the notion that an infinite set can be placed into one-to-one correspondence with proper subsets of that set. In other words, I challenge your notion of existence in mathematics.
 

 

20 hours ago, phyti said:

Where the nonsense starts is with our habit of thinking of a large number as closer to infinity than a small one.

I don't think any of the reasoning I put forward in this thread even remotely suggests a thinking of a large number as closer to infinity than a small one. I would say the opposite. For example, in the one-to-one mapping of [math]n \longleftrightarrow 2^n[/math], for large [math]n[/math], [math]2^n[/math] is a lot larger than [math]n[/math], yet there was no suggestion that [math]2^n[/math] was any closer to the "end of the list" than [math]n[/math].
 

 

20 hours ago, phyti said:

A true 1 to 1 correspondence for E=2D.

D 1 3 5 7 9 ...

E 2 4 6 8 10 ...

How is this E=2D?

The mapping:

[math]1 \longleftrightarrow 2[/math]
[math]2 \longleftrightarrow 4[/math]
[math]3 \longleftrightarrow 6[/math]
[math]4 \longleftrightarrow 8[/math]
[math]...[/math]
[math]n \longleftrightarrow 2n[/math]
[math]...[/math]

is just as true a one-to-one correspondence as the mapping [math]2n-1 \longleftrightarrow 2n[/math].
 

 

19 hours ago, phyti said:

A and B are 'infinite' so there is no value/number/cardinality to associate with them.

The cardinality is [math]\aleph_0[/math].

 

 

19 hours ago, phyti said:

A slice of pie is not equal to the whole pie.

Banach-Tarski paradox

 

 

Edited December 24, 2023 by KJW

[phyti]

KJW;

If there are as many even integers n in E as integers in N

and

if there are as many odd integers n in D as integers in N

then there are 2n integers in the union of E and D.

Does that seem logical?

 

Edited December 24, 2023 by phyti

[KJW]

1 hour ago, phyti said:

Does that seem logical?

That the union of two disjoint sets, both of which have a cardinality of [math]\aleph_0[/math], has a cardinality of [math]\aleph_0[/math] is logical because we are not dealing with finite sets. The logic of finite sets does not apply to infinite sets. Infinite sets have a logic of their own.

If we have two disjoint sets A = {a₁, a₂, ...,a_n, ...} and B = {b₁, b₂, ...,b_n, ...}, both of which can be placed into one-to-one correspondence with [math]\mathbb {N}[/math]:

[math]1 \longleftrightarrow a_1[/math]

[math]2 \longleftrightarrow a_2[/math]

[math]...[/math]

[math]n \longleftrightarrow a_n[/math]

[math]...[/math]

and:

[math]1 \longleftrightarrow b_1[/math]

[math]2 \longleftrightarrow b_2[/math]

[math]...[/math]

[math]n \longleftrightarrow b_n[/math]

[math]...[/math]

then we can interleave these two lists to form a list that is also a one-to-one correspondence with [math]\mathbb {N}[/math]:

[math]1 \longleftrightarrow a_1[/math]

[math]2 \longleftrightarrow b_1[/math]

[math]3 \longleftrightarrow a_2[/math]

[math]4 \longleftrightarrow b_2[/math]

[math]...[/math]

[math]2n-1 \longleftrightarrow a_n[/math]

[math]2n \longleftrightarrow b_n[/math]

[math]...[/math]

 

 

Edited December 24, 2023 by KJW

[phyti]

KJW;

1.

The fact of the matter is that the validity of a mathematical theorem is independent of who originated it or how it is expressed in particular writings.

2.

I don't know about that. It turns out that the mapping from the group of integers Z ... 

3.

Leopold Kronecker was a subscriber to the philosophy of finitism and was thus in the minority among mathematicians.

4.

I am a subscriber to the philosophy of formalism, and regard mathematics as not being rigorous unless it can be derived from first principles (axioms) by mechanical symbol manipulation. 

5.

The cardinality is aleph0.

1. It concerns the motivation of the author. There have been real world cases of scientists who manipulate data to support their theories. In the justice system, motive is important.

2. It's NOT about mapping of any class of numbers. It's Cantor's use of the diagonal argument to prove a list of generic symbols is incomplete.

"However, there is a proof of this proposition that is much simpler, and which does not depend on considering the irrational numbers".

3. On another forum someone thought the solution to a different subject involved classifying me in a particular belief system. He was eventually banned, and at that point I wished I had told him I was a librarian.

4. I would have to side with the constructivist.

5. What is it's value? Is it the largest number you can think of?

Infinite implies without a boundary/limit. It does not imply a largest number.

Measurement in a world of finite things is based on boundaries, and why the rod with one end. Apply the hook end of your industrial strength tape measure to the near end, extend it until the far end is aligned to a mark on the tape. But you reply, 'there is no end'. The event 'reaching the far end' never happens!

[KJW]

2 hours ago, phyti said:

Quote

The fact of the matter is that the validity of a mathematical theorem is independent of who originated it or how it is expressed in particular writings.

It concerns the motivation of the author. There have been real world cases of scientists who manipulate data to support their theories. In the justice system, motive is important.

There is one important difference that you are overlooking: the mathematical theorem is completely transparent. There is no reliance on trust. I can see from the theorem itself that the motivation of the author is irrelevant, there is no data to manipulate, and this is not the justice system.

 

 

2 hours ago, phyti said:

It's NOT about mapping of any class of numbers. It's Cantor's use of the diagonal argument to prove a list of generic symbols is incomplete.

"However, there is a proof of this proposition that is much simpler, and which does not depend on considering the irrational numbers".

I disagree. Cantor's theorem is about the existence of different transfinite numbers, based on the mappings of sets of numbers.

 

 

2 hours ago, phyti said:

I would have to side with the constructivist.

Have you ever seen the following identity?:

[math]1 + 2 + 3 + 4\> +\> ...\> = -\dfrac{1}{12}[/math]

This is fairly well-known because it is recognised as one the most bizarre identities in mathematics. It emerges as a result of analytic continuation of the Riemann zeta function. But surely such apparent absurdity has no relevance to physical reality? Actually, a related equally bizarre identity manifests itself in the Casimir effect. I find this quite extraordinary, to be honest.

 

 

2 hours ago, phyti said:

Quote

The cardinality is [math]\aleph_0[/math].

What is its value?

The value is [math]\aleph_0[/math]. Why do you think its value should be something else?

 

 

2 hours ago, phyti said:

Infinite implies without a boundary/limit. It does not imply a largest number.

Where have I said that it does? I think I have indicated that "infinite" is a property. But you also have to recognise that there are different properties that lead to different infinities. In particular, the property of being able to be placed into one-to-one correspondence with the natural numbers, and the property of not being able to be placed into one-to-one correspondence with the natural numbers are two different properties leading to two different infinities.

 

 

Edited December 25, 2023 by KJW

[phyti]

KJW;

[quote

Earlier in this thread, you indicated that there is only one infinity. Have you changed your mind on that?

[/quote]

No. 'Infinite' has no superlative state. Comparison of 2 quantities results in equal or not equal. If not equal, one is greater than the other. For a sample size >2, if one is greater than all others in a sample s, then it is the greatest relative to s.

Cantor quote:

"When I conceive the infinite … there follows for me a genuine pleasure … in seeing how the concept of integer [der ganze Zahlbegriff], which in the finite has only the background of number [Anzahl], as it were splits into *two* concepts when we ascend to the infinite – one one of number [Anzahl] which is necessarily bound to a lawlike ordering of the set [Menge] by virtue of which it becomes well-ordered (wohlgeordneten)."

So again I ask, what is the magic n in N which is infinite by definition, where the set becomes aleph0?

Cantor's error is assuming the diagonal D extends to the end of the list forming a geometric square, excluding its negation E0. That's not possible using established math procedures plotting the relation between u and v.

The variable v is a linear function, the variable u is an exponential function. They are never equal as shown in the graph. The (u, v) relation depends on the number of symbols used in the formation of sequences.

--------------------------------------------------------

I can't imagine 'approaching infinity', yet it's a common phrase in most text books.

Do you know how?

[ref]
'Cantor on Set Theory',Source:  Ewald, W., From Kant to Hilbert, Oxford 1996.

 

[KJW]

12 hours ago, phyti said:

So again I ask, what is the magic n in [math]\mathbb {N}[/math] which is infinite by definition, where the set becomes [math]\aleph_0[/math]?

Again?? You previously asked me a similar question which I answered. The answer to this question is that [math]\aleph_0[/math] is not an element of [math]\mathbb {N}[/math]. [math]\mathbb {N}[/math] contains only finite numbers and [math]\aleph_0[/math] is not a finite number.

 

 

12 hours ago, phyti said:

Cantor's error is assuming the diagonal D extends to the end of the list forming a geometric square, excluding its negation E0. That's not possible using established math procedures plotting the relation between u and v.

The variable v is a linear function, the variable u is an exponential function. They are never equal as shown in the graph. The (u, v) relation depends on the number of symbols used in the formation of sequences.

We've already discussed this, so I suggest you go back to that discussion.

 

 

12 hours ago, phyti said:

I can't imagine 'approaching infinity', yet it's a common phrase in most text books.

Do you know how?

The term "approaching infinity" is often used in connection to limits. Suppose a₁, a₂, …, a_n, … is a sequence of real numbers. When the limit of the sequence exists, the real number L is the limit of this sequence if and only if for every real number ε > 0, there exists a natural number N such that for all n > N, we have |a_n − L| < ε. The common notation:

[math]\displaystyle \lim _{n \to \infty} a_{n}=L[/math]

is read as: "The limit of a_n as n approaches infinity equals L" or "The limit as n approaches infinity of a_n equals L". But the formal definition itself doesn't mention "infinity". Indeed, it seems to me that the whole notion of limits is about dealing with the infinite and infinitesimal in a way that remains firmly in the realm of the finite.

 

 

Edited December 26, 2023 by KJW

[phyti]

KJW;

Have you ever seen the following identity?:

Yes, and not impressed. It's confusing since the initial question is the value for the summation of the elements of N. It's a series of discrete values, not continuous. The equation resulting in -1/12 looks like nonsense, not an equality at face value. This quote from the Wiki page:

"Because the sequence of partial sums fails to converge to a finite limit, the series does not have a sum".

Why would anyone expect a limit for a positively divergent series?

The value is ℵ0. Why do you think its value should be something else?

Already covered this. ℵ0 is a symbol representing a number.

'c' represents the speed of light. It's a convention to avoid writing its value (3*108) each time it is used. Cantor had no value to assign to ℵ0 because there is no way of counting/measuring a set that has no limit.

That the union of two disjoint sets, both of which have a cardinality of ℵ0, has a cardinality of ℵ0 is logical because we are not dealing with finite sets. 

[/quote]

N {1 2 3 4 5...}

E {2 4 6 8 10...}

D {3 5 7 9 11...}

Looks more like a 1 to 2 correspondence, each n to a pair.

Where have I said that it does? I think I have indicated that "infinite" is a property. 

Didn't say you did. I'm referring to Cantor's ideas.

[KJW]

18 hours ago, phyti said:

Have you ever seen the following identity?:

Yes, and not impressed. It's confusing since the initial question is the value for the summation of the elements of N. It's a series of discrete values, not continuous. The equation resulting in -1/12 looks like nonsense, not an equality at face value. This quote from the Wiki page:

"Because the sequence of partial sums fails to converge to a finite limit, the series does not have a sum".

Why would anyone expect a limit for a positively divergent series?

Do you know where the value [math]-\dfrac{1}{12}[/math] comes from? It is the value of [math]\zeta(-1)[/math], where [math]\zeta(z)[/math] is the Riemann zeta function over the complex number field. The Riemann zeta function is defined as:

[math]\zeta(z) \buildrel\rm def\over= \dfrac{1}{1^z} + \dfrac{1}{2^z} + \dfrac{1}{3^z} + \dfrac{1}{4^z} + \dfrac{1}{5^z} \> + \> ...[/math]

Substituting [math]z = -1[/math] gives the series [math]1 + 2 + 3 + 4 + 5 \> + \> ... \>[/math], but that's not how [math]\zeta(-1) = -\dfrac{1}{12}[/math] is calculated. Calculating [math]\zeta(-1) = -\dfrac{1}{12}[/math] involves a process called "analytic continuation" where the domain of a function is extended as the domain of a continuous function beyond where it would otherwise fail to converge. The series converges for real values [math]> 1[/math], but there is a "pole" at [math]z = 1[/math], and analytic continuation to negative real values from real values [math]> 1[/math] involves going around the pole in the complex number domain.

In the case of the Casimir effect, one uses the identity:

[math]\zeta(-3) = 1 + 8 + 27 + 64 + 125 \> + \> ... \> = \dfrac{1}{120}[/math]

A simple example of a function that is continuous over the entire real number domain, yet its infinite series only converges within a limited domain of the real numbers is:

[math]\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + x^8 - x^{10} \> + \> ... [/math]

Although the function is well-defined for [math]|x| \ge 1[/math], the infinite series fails to converge for [math]|x| \ge 1[/math]. But note that for [math]-1 < x < 1[/math], the infinite series converges to the same value as the function. Therefore, one could say that:

[math]1 - 4 + 16 - 64 + 256 - 1024 \> + \> ... \> = \dfrac{1}{5}[/math]

 

 

 

Edited December 27, 2023 by KJW

[phyti]

KJW;

The term "approaching infinity" is often used in connection to limits.

I see it as a contradiction of terms. You can't approach something that is out of reach.

Here is an analogy.

 

[KJW]

17 hours ago, phyti said:

Quote

The term "approaching infinity" is often used in connection to limits.

I see it as a contradiction of terms. You can't approach something that is out of reach.

Here is an analogy.

 

Did you continue to read what I said after "The term "approaching infinity" is often used in connection to limits."? You seem to have put in a lot of effort to argue against a phrase that is intended to be intuitive. Why can't you approach something that is out of reach? The notion of limits is about approaching something but not actually reaching it.

 

 

[phyti]

35 minutes ago, KJW said:

Did you continue to read what I said after "The term "approaching infinity" is often used in connection to limits."? You seem to have put in a lot of effort to argue against a phrase that is intended to be intuitive. Why can't you approach something that is out of reach? The notion of limits is about approaching something but not actually reaching it.

 

 

You answered your question in the last line!

The anaut can't get any closer to the horizon.

The horse keeps moving forward to reach the carrot on a stick but never reaches it. Infinity can appear in cases other than math.

KJW;

Ref. '1+2+3+4+', Wiki.

The partial sums are discrete integers. There are no bands of areas forming a step function as shown in the Wiki article. The green curve appears to be enclosing an equivalent area of the bands. No reason for y to be <0 when x=0.

Ref. png graphic:

Using y=x(x+1)/2 the green curve contains the partial sums for all n.

The y increment remains increasing without limit, thus no last term.

alternating series s=(1-1+1-1+1...)

1. For self referential, s-1= -s, thus s=1/2 avg .

2. If summed sequentially, s=0 or 1, with an avg. of 1/2.

3. If sorted and summed, s=(n-n) = 0, or

s=[(1-1)+ (1-1)+ (1-1)+ (1-1)...] = 0.

Depends on the method of manipulation.

alternating series s=(1-2+3-4+5-6+7-8...)

s=(1+3+5+7...)-(2+4+6+8...)=A-B

s=Σ(2x-1)-2Σ(x), x varies 1 to n.

s= 2Σ(x)-n-2Σ(x)=-n.

Ref. png graphic

The measurements are alternating which skews the differences. Using the envelope measurements which are linear, and B' is the mirror image of B, the difference is .5.

 

Edited December 28, 2023 by phyti

[KJW]

Quote

[math]1 − 4 + 16 − 64 + 256 − 1024\>+\>...\>= \dfrac{1}{5}[/math]

Proof:

Let [math]X = 1 − 4 + 16 − 64 + 256 − 1024\>+\>...[/math]

[math]X - 1 = −4 + 16 − 64 + 256 − 1024 + 4096\>-\>...[/math]

[math]= −4\>(1 − 4 + 16 − 64 + 256 − 1024\>+\>...)[/math]

[math]= −4\,X[/math]

[math]5\,X - 1 = 0[/math]

[math]X = \dfrac{1}{5}[/math]

Q.E.D.

 

 

In general:

Quote

[math]1 − x^2 + x^4 − x^6 + x^8 − x^{10}\>+\>...\>= \dfrac{1}{1 + x^2}[/math]

Proof:

Let [math]y = 1 − x^2 + x^4 − x^6 + x^8 − x^{10}\>+\>...[/math]

[math]y - 1 = −x^2 + x^4 − x^6 + x^8 − x^{10} + x^{12}\>-\>...[/math]

[math]= −x^2\>(1 − x^2 + x^4 − x^6 + x^8 − x^{10}\>+\>...)[/math]

[math]= −x^2\>y[/math]

[math](1 + x^2)\>y - 1 = 0[/math]

[math]y = \dfrac{1}{1 + x^2}[/math]

Q.E.D.

 

 

Edited December 28, 2023 by KJW

[KJW]

The above proof is incomplete:

Quote

[math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1+x^{n}}[/math]

Proof:

Let [math]y = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math]

[math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + \displaystyle \sum_{k=m}^{\infty} (-1)^{k} x^{nk}[/math]

[math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + (-1)^{m} x^{nm} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk}[/math]

[math]y = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + (-1)^{m} x^{nm}\>y[/math]

[math]y\>(1 - (-1)^{m} x^{nm}) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math]

[math]y \left(1 + \displaystyle \sum_{k=1}^{m-1} (-1)^{k} x^{nk} - \displaystyle \sum_{k=1}^{m-1} (-1)^{k} x^{nk} - (-1)^{m} x^{nm}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math]

[math]y\left(\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} - \displaystyle \sum_{k=1}^{m} (-1)^{k} x^{nk}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math]

[math]y\left(\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} + x^{n} \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}\right) = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math]

[math]y\>(1 + x^{n}) \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk} = \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}[/math]

[math]y = \dfrac{\displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}}{(1 + x^{n}) \displaystyle \sum_{k=0}^{m-1} (-1)^{k} x^{nk}}[/math]

[math]y = \dfrac{1}{1 + x^{n}}[/math]

 

Therefore:

 

[math]\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{nk} = \dfrac{1}{1 + x^{n}}[/math]

 

Q.E.D.
 

Two things to note about this proof is that it generalises the previous result to other powers of [math]x[/math], and that it splits the series at an arbitrary location, thus establishing the consistency of the result with respect to the location of the split (the result is independent of [math]m[/math]).

 

 

Edited December 29, 2023 by KJW

[KJW]

20 hours ago, phyti said:

Depends on the method of manipulation.

But the correct method leads to the correct result. The correct method is the method I have shown above. In the case of [math]1 - 1 + 1 - 1 + 1 - 1\>+\>...\>[/math], the correct method is the method you labelled as "self referential", leading to a value of [math]\dfrac{1}{2}[/math] (in the above formula, [math]x = 1[/math], [math]n =[/math] whatever). It's important to note that no attempt was made to sum the series. For [math]|x| \ge 1[/math], the series does not converge. Instead, the series was assumed to have a value, which was then determined algebraically as shown above. The correctness of the value is the result of being the only value that is consistent with the definition of the series. Whether or not a non-convergent series can be regarded as having a value is a matter of debate, but if it does have a value, then that value must be the value obtained by the method I have shown above. In the case of [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math], the above formula with [math]x = -1[/math], [math]n = 1[/math] does not produce a finite result. That is, it is not just the series that fails to converge, but the corresponding function is also unbounded.

 

 

Edited December 29, 2023 by KJW

[KJW]

Given that the above infinite series are alternating, one might assume that their values are some sort of average of the terms. But the non-alternating series:

[math]\displaystyle \sum_{k=0}^{\infty} x^k[/math] where [math]x > 1[/math]

dispels that notion:

[math]\displaystyle \sum_{k=0}^{\infty} x^k = \displaystyle \sum_{k=0}^{m-1} x^k + \displaystyle \sum_{k=m}^{\infty} x^k =\dfrac{x^m - 1}{x - 1} + x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math]

[math]\displaystyle \sum_{k=0}^{\infty} x^k - \dfrac{x^m - 1}{x - 1} = x^m \displaystyle \sum_{k=0}^{\infty} x^k[/math]

Note that removing initial terms from the infinite series yields the same infinite series multiplied by some value, and that removing more initial terms yields an even larger multiple of the infinite series. This behaviour of the infinite series is clearly different to the behaviour of any corresponding finite partial series. Also note that if [math]x = 10[/math], then the removed partial sum has a decimal value of the form [math]11111 \cdots 11111[/math] which is [math]\dfrac{1}{9}[/math] of [math]99999 \cdots 99999[/math], hence the general form for base-[math]x[/math] numbers. Continuing:

[math](x^m - 1) \displaystyle \sum_{k=0}^{\infty} x^k = -\dfrac{x^m - 1}{x - 1}[/math]

[math]\displaystyle \sum_{k=0}^{\infty} x^k = \dfrac{-1}{x - 1} < 0[/math] for [math]x > 1[/math]

All the partial sums of the infinite series are positive-valued, yet the infinite series itself is negative-valued. In the case of [math]x = 2[/math], the value of the infinite series is [math]-1[/math]. But treated as an infinite sum, the value of the infinite series can be expressed in binary as:

[math]\cdots 1111111111[/math]

This is the same value that was excluded by Cantor's diagonal argument in an earlier post for the mapping of the set of all finite subsets of the natural numbers to the natural numbers. In other words, the sum of the infinite series goes beyond [math]\aleph_0[/math]. However, the binary expression for the infinite sum also corresponds to the two's complement representation of [math]-1[/math] used in computing, albeit extended leftward to infinite bits.

 

 

Edited December 31, 2023 by KJW

[phyti]

KJW;

 

fig.1 shows the alternating series s=(1-2+3-4+5-6+7-8...) as a discontinuous function.

fig.2 shows continuous functions A and B.

For A y=x.

For B y=-x.

If A and B are applied to fig.1, the difference A-B=0 for all x.

 

The B function has been shifted by -1 in fig.2.

The equation for B is transformed from y=-x to y=-x-1.

Now the difference A-B=x-(x+1)=(-1) for all x.

The shift results in subtracting the B value for x from the A value for x-1.

 

[KJW]

14 hours ago, phyti said:

s=(1-2+3-4+5-6+7-8...)

[math]\zeta(z) = \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math]

[math]\eta(z) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k^{-z} = \displaystyle \sum_{k=1}^{\infty} (2k-1)^{-z} - \displaystyle \sum_{k=1}^{\infty} (2k)^{-z} = \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2 \displaystyle \sum_{k=1}^{\infty} (2k)^{-z}[/math]

[math]= \displaystyle \sum_{k=1}^{\infty} k^{-z} - 2^{1-z} \displaystyle \sum_{k=1}^{\infty} k^{-z} = (1 - 2^{1-z}) \displaystyle \sum_{k=1}^{\infty} k^{-z}[/math]

[math]= (1 - 2^{1-z})\>\zeta(z)[/math]

For [math]z = -1[/math], [math]\eta(-1) = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = 1 - 2 + 3 - 4 + 5 - 6\>+\>...\>= -3\>\zeta(-1) = \dfrac{1}{4}[/math]

 

 

On 12/30/2023 at 2:33 AM, KJW said:

In the case of [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math], the above formula with [math]x = −1[/math], [math]n = 1[/math] does not produce a finite result. That is, it is not just the series that fails to converge, but the corresponding function is also unbounded.

Although the series [math]1 + 1 + 1 + 1 + 1 + 1\>+\>...\>[/math] can't be determined as the geometric series, it can be determined from the Riemann zeta function:

[math]\zeta(0) = \displaystyle \sum_{k=1}^{\infty} k^0 = 1 + 1 + 1 + 1 + 1 + 1\>+\>...\>= -\dfrac{1}{2}[/math]

Note that this could also have been obtained from:

[math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math]

For [math]z = 0[/math], [math]\zeta(0) = -\eta(0) = -\dfrac{1}{2}[/math], where [math]\eta(0) = 1 - 1 + 1 - 1 + 1 - 1\>+\>...\>= \dfrac{1}{2}[/math] had already been obtained in a previous post.
 

 

[joigus]

 

ℵ₀ is obviously a symbol used in a definition. The definition involves a bijection. A bijection to the natural numbers. Any set for which a bijection can be constructed to the natural numbers is said to have the cardinality of the natural numbers. We call this cardinality ℵ₀. Mind you, we call this abstract concept ℵ₀.

The question,

On 12/26/2023 at 12:07 AM, phyti said:

what is the magic n in N which is infinite by definition, where the set becomes aleph0?

Proves that you do not understand the definition of ℵ₀. Repeat: You do not understand the definition of aleph naught. Nothing becomes anything. It is what it is. Your question is as meaningless as, eg,

What is the magic in the natural numbers that makes n(n-1)...2  become n! ? 

Cardinalities aren't numbers, although sometimes they can be. They are what they are, and what they are is what they are defined to be. They are defined via bijection, therefore no numbers necessarily, but abstract properties of relations between sets that are equivalence relations, and only sometimes happen to "become something" in the sense that you suggest.

 

[phyti]

KJW;

You still don't understand, my argument with Cantor's argument doesn't involve properties of numbers. As he states in the 1891 paper "However, there is a proof of this proposition that is much simpler, and which does not depend on considering the irrational numbers."

It involves two alphabetical symbols used in forming sequences/patterns as members of a set M. The elements are sequences, NOT numbers. The integers in the set N are used for the purpose of forming an array within a coordinate system (v, u). L His goal is to convince the reader that the list will be incomplete. Thus the cardinality (number of elements in a set) L is greater than N. This allows him to sell his transfinite numbers.

I will leave you with your math diversions.

 

[KJW]

On 1/3/2024 at 10:28 AM, phyti said:

You still don't understand, my argument with Cantor's argument doesn't involve properties of numbers.

What I do understand is that Cantor's theorem is about sets, their elements, and their subsets, as well as things that are represented by sets, their elements, and their subsets, and also things that represent sets, their elements, and their subsets. My vision is not limited to an 1891 paper.
 

 

On 1/3/2024 at 10:28 AM, phyti said:

It involves two alphabetical symbols used in forming sequences/patterns as members of a set M. The elements are sequences, NOT numbers.

Infinite sequences of binary symbols can be placed into one-to-one correspondence with subsets of the natural numbers. As such, it doesn't really matter if one considers them as sequences or subsets. I actually avoided considering the sequences as real numbers due to a complication that I did not wish to discuss.
 

 

On 1/3/2024 at 10:28 AM, phyti said:

This allows him to sell his transfinite numbers.

Cantor was a mathematician, not a snake oil salesman. What Cantor demonstrated was a mathematical truth concerning infinite sets. He wasn't "selling" anything.
 

 

On 1/3/2024 at 10:28 AM, phyti said:

I will leave you with your math diversions.

It is disappointing that you didn't see any value in the mathematics I presented about non-convergent infinite series. You started a topic that was ultimately about the infinite, but you failed to accept that the infinite is more complicated than the common simplistic understanding of it.

I do wish to conclude with the following:

 

Consider the negative of the derivative:

[math]-\dfrac{d}{dx} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} = -\displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k}}{dx} = \displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k\,x^{k-1}[/math]

[math]= -\dfrac{d}{dx} \dfrac{1}{1 + x} = \dfrac{1}{(1 + x)^2}[/math]

 

For [math]x = 1[/math]:

[math]\displaystyle \sum_{k=1}^{\infty} (-1)^{k-1} k = \eta(-1) = \dfrac{1}{4}[/math]

[math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math]

[math]\zeta(-1) = -\dfrac{1}{3}\>\eta(-1) = -\dfrac{1}{12}[/math]

 

Note that I have evaluated [math]\zeta(-1)[/math], not simply obtained its value from elsewhere.

 

[math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} \dfrac{dx^{k+1}}{dx} = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)x^{k}[/math]

[math]= \dfrac{d}{dx} \dfrac{x}{1 + x} = \dfrac{1}{1 + x} - \dfrac{x}{(1 + x)^2} = \dfrac{1 + x}{(1 + x)^2} - \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2}[/math]

 

This is the same as above. However, by multiplying by [math]x[/math] before differentiation, the exponents of [math]x[/math] in the series remains unchanged, leading to coefficients that are powers of [math]k + 1[/math] (equivalent to powers of [math]k[/math]) rather than factorials when the operation is iterated.

 

[math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1) x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}[/math]

[math]= \dfrac{d}{dx} \dfrac{x}{(1 + x)^2} = \dfrac{1}{(1 + x)^2} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 + x}{(1 + x)^3} - \dfrac{2x}{(1 + x)^3} = \dfrac{1 - x}{(1 + x)^3}[/math]

[math]\dfrac{1}{1 - 2^{3}} \dfrac{1 - x}{(1 + x)^3} = \zeta(-2) = 0[/math] for [math]x = 1[/math]

 

[math]\dfrac{d}{dx} \left(x\displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^2 x^{k}\right) = \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^3 x^{k}[/math]

[math]= \dfrac{d}{dx} \dfrac{x - x^2}{(1 + x)^3} = \dfrac{1 - 2x}{(1 + x)^3} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - x - 2x^2}{(1 + x)^4} - \dfrac{3x - 3x^2}{(1 + x)^4} = \dfrac{1 - 4x + x^2}{(1 + x)^4}[/math]

[math]\dfrac{1}{1 - 2^{4}} \dfrac{1 - 4x + x^2}{(1 + x)^4} = \zeta(-3) = \dfrac{1}{120}[/math] for [math]x = 1[/math]

 

[math]\dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n}\displaystyle \sum_{k=0}^{\infty} (-1)^{k} x^{k} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \dfrac{1}{1 - 2^{n+1}} \displaystyle \sum_{k=0}^{\infty} (-1)^{k} (k+1)^n x^{k}[/math]

[math]= \dfrac{1}{1 - 2^{n+1}} \overbrace{\dfrac{d}{dx} \Biggl( \Biggr. x \cdots \dfrac{d}{dx} \Biggl( \Biggr. x}^{n} \>\>\dfrac{1}{1 + x} \Biggl. \Biggr) \cdots \Biggl. \Biggr) = \zeta(-n)[/math] for [math]x = 1[/math]

 

Edited January 5 by KJW

[phyti]

In case there is anyone else interested in the subject of Cantor's diagonal argument,

the latest paper is here:

https://drive.google.com/file/d/17Zz39YzAEEvD5pWB0f4NihEucQfZ3RHU/view?usp=sharing