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
>>> 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!)
If the fact that the sum is finite is not known, you might not know if
the operation is valid, so you may need to carry the condition around in
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
>> 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.
The only hidden false rule that is assumed is that you can manipulate
infinite sums like normal number, even if they don't converge.