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Jan 19, 2022, 4:11:28 AMJan 19

to

Rule3 of Infinite Series is added into NumberView-en.txt

https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

These rules are commonly used, but some issue may be controversial:

Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

numbers, the notion that the whole is greater than the part is conflicted

by this rule.

But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?

An idea that using C-language's for loop expression might solve this

problem (or, at least, better than the traditional Σ notation):

for(n=0;;++n) n; or f(n=0;;n+=2) n;

Any idea?

https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

These rules are commonly used, but some issue may be controversial:

Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

numbers, the notion that the whole is greater than the part is conflicted

by this rule.

But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?

An idea that using C-language's for loop expression might solve this

problem (or, at least, better than the traditional Σ notation):

for(n=0;;++n) n; or f(n=0;;n+=2) n;

Any idea?

Jan 19, 2022, 5:24:42 AMJan 19

to

So if we sum the even numbers, is that a lower infinity, because half

of the series has been removed? Or is it a higher infinity, because each

term is bigger?

It's quite hard to work out from first principles what the most logical

answers are.

Jan 19, 2022, 6:42:18 AMJan 19

to

wij <wyn...@gmail.com> writes:

> Rule3 of Infinite Series is added into NumberView-en.txt

> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

>

> Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

> Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

>

> These rules are commonly used, but some issue may be controversial:

>

> Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

That's not an application of the rule you cite.
> Rule3 of Infinite Series is added into NumberView-en.txt

> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

>

> Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

> Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

>

> These rules are commonly used, but some issue may be controversial:

>

> Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

> If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

> numbers, the notion that the whole is greater than the part is conflicted

> by this rule.

your intuitions. You'd have to start by picking a definition of "greater

than" that matches your intuition and then you would have to follow

through with the consequences of that decision.

You could always just say there are no non-finite sets in your

mathematics. Do you need them? You could leave them just for people

who want them!

> But, how do we express "the sum of even numbers"?

--

Ben.

Jan 20, 2022, 4:02:33 AMJan 20

to

*Spammer Alert*

Julio

Jan 21, 2022, 10:43:54 AMJan 21

to

that the infinite sum converges to a finite value. When you try to think

of a unconverging sum as 'having a value', you can get all sorts of

issues. Even if the sum converges, there can be limits to what sort of

operations you can do on the elements if the sum has positive and

negative elements, and the sum of just the elements of one sign don't

also converge. Such sums can change their value if you reorder the terms.

Basically, 'infinity' doesn't act like a normal number, and you can't

expect it to, and infinite sums that don't converge fall into that category.

Ignoring that can get you into all sorts of problems, for example if we

let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to

infinity.)

X = sum(n)

Y = sum(1)

The sum of the odds can be expressed as either

sum(2n+1) = 2X+Y

or, if we skip the first element and add it in explicitly)

1 + sum(2n+3) = 1 + 2X + 3Y

thus it seems we could say that

2X+Y = 1 + 2X + 3Y

But this says that Y = -1/2 ( 0 = 1 + 2Y)

Jan 22, 2022, 4:04:44 AMJan 22

to

On Friday, 21 January 2022 at 23:43:54 UTC+8, Richard Damon wrote:

> On 1/19/22 4:11 AM, wij wrote:

> > Rule3 of Infinite Series is added into NumberView-en.txt

> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

> >

> > Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

> > Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

> >

> > These rules are commonly used, but some issue may be controversial:

> >

> > Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

> > If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

> > numbers, the notion that the whole is greater than the part is conflicted

> > by this rule.

> > But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?

> > An idea that using C-language's for loop expression might solve this

> > problem (or, at least, better than the traditional Σ notation):

> > for(n=0;;++n) n; or f(n=0;;n+=2) n;

> >

> > Any idea?

> Fundamentally, the issue is that these rules are all under the condition

> that the infinite sum converges to a finite value.

If the result is not known, how does one know the infinite series converges or not?
> On 1/19/22 4:11 AM, wij wrote:

> > Rule3 of Infinite Series is added into NumberView-en.txt

> > https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

> >

> > Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

> > Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

> >

> > These rules are commonly used, but some issue may be controversial:

> >

> > Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

> > If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

> > numbers, the notion that the whole is greater than the part is conflicted

> > by this rule.

> > But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?

> > An idea that using C-language's for loop expression might solve this

> > problem (or, at least, better than the traditional Σ notation):

> > for(n=0;;++n) n; or f(n=0;;n+=2) n;

> >

> > Any idea?

> Fundamentally, the issue is that these rules are all under the condition

> that the infinite sum converges to a finite value.

(converge OK, not converge not OK? what is the rule!)

> When you try to think

> of a unconverging sum as 'having a value', you can get all sorts of

> issues. Even if the sum converges, there can be limits to what sort of

> operations you can do on the elements if the sum has positive and

> negative elements, and the sum of just the elements of one sign don't

> also converge. Such sums can change their value if you reorder the terms.

>

> Basically, 'infinity' doesn't act like a normal number, and you can't

> expect it to, and infinite sums that don't converge fall into that category.

>

> Ignoring that can get you into all sorts of problems, for example if we

> let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to

> infinity.)

>

> X = sum(n)

> Y = sum(1)

>

> The sum of the odds can be expressed as either

>

> sum(2n+1) = 2X+Y

> or, if we skip the first element and add it in explicitly)

> 1 + sum(2n+3) = 1 + 2X + 3Y

> thus it seems we could say that

>

> 2X+Y = 1 + 2X + 3Y

>

> But this says that Y = -1/2 ( 0 = 1 + 2Y)

No valid rule for the derivation ∑ 2n*1= 1+ 2*(∑2*n+3) (trick of magic show)

A hidden false rule is assumed.

Jan 22, 2022, 5:59:03 AMJan 22

to

On 1/22/22 4:04 AM, wij wrote:

> On Friday, 21 January 2022 at 23:43:54 UTC+8, Richard Damon wrote:

>> On 1/19/22 4:11 AM, wij wrote:

>>> Rule3 of Infinite Series is added into NumberView-en.txt

>>> https://sourceforge.net/projects/cscall/files/MisFiles/NumberView-en.txt/download

>>>

>>> Rule3: Σ(n=0,∞) f(n) ± Σ(n=0,∞) g(n) = Σ(n=0,∞) f(n)±g(n)

>>> Σ(n=0,∞) c*f(n)= c*(Σ(n=0,∞) f(n))

>>>

>>> These rules are commonly used, but some issue may be controversial:

>>>

>>> Ex: Σ 2*n =Σ (n+n) =Σ n + Σ n

>>> If Σ 2*n is said the sum of all even numbers, Σ n the sum of all natural

>>> numbers, the notion that the whole is greater than the part is conflicted

>>> by this rule.

>>> But, how do we express "the sum of even numbers"? Or Σ(n=0,∞/2) 2*n ?

>>> An idea that using C-language's for loop expression might solve this

>>> problem (or, at least, better than the traditional Σ notation):

>>> for(n=0;;++n) n; or f(n=0;;n+=2) n;

>>>

>>> Any idea?

>> Fundamentally, the issue is that these rules are all under the condition

>> that the infinite sum converges to a finite value.

>

> If the result is not known, how does one know the infinite series converges or not?

> (converge OK, not converge not OK? what is the rule!)

the operation is valid, so you may need to carry the condition around in

the argument.

Just like if you have that X * Y = X * Z, you can't just divide by X and

say that Y = Z, you have to say something like if X != 0, then Y = Z.

>

>> When you try to think

>> of a unconverging sum as 'having a value', you can get all sorts of

>> issues. Even if the sum converges, there can be limits to what sort of

>> operations you can do on the elements if the sum has positive and

>> negative elements, and the sum of just the elements of one sign don't

>> also converge. Such sums can change their value if you reorder the terms.

>>

>> Basically, 'infinity' doesn't act like a normal number, and you can't

>> expect it to, and infinite sums that don't converge fall into that category.

>>

>> Ignoring that can get you into all sorts of problems, for example if we

>> let: (let sum(f(n)) be the sum of f(n) for all values of n from 0 to

>> infinity.)

>>

>> X = sum(n)

>> Y = sum(1)

>>

>> The sum of the odds can be expressed as either

>>

>> sum(2n+1) = 2X+Y

>> or, if we skip the first element and add it in explicitly)

>> 1 + sum(2n+3) = 1 + 2X + 3Y

>> thus it seems we could say that

>>

>> 2X+Y = 1 + 2X + 3Y

>>

>> But this says that Y = -1/2 ( 0 = 1 + 2Y)

>

> Lots of inconsistency bypassed for comment. I was asking for opinion, I won't say too much.

>

> No valid rule for the derivation ∑ 2n*1= 1+ 2*(∑2*n+3) (trick of magic show)

> A hidden false rule is assumed.

infinite sums like normal number, even if they don't converge.

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