interesting problem (3 Viewers)

[hayabusaboston]

1. You say n is a part of 2n which implies it can be a non-integer. This is impossible because you are using n as a limit in product notation ie: it MUST be an integer.

2. I still don't understand what you're trying to convey by saying that n is a subset of 2n and n^2. It's not making much sense to me...

I think this is what you're trying to say:

Yep thanks guys for your help, we're working well here hehe.

o shit yea

I just got my mistake

hmmm


LOL picked it out now.


Well


I meant to imply that all integers n, ie infinity really. So if its 2 times n, it must be even, since its divisible by two.

With the n squared part, it implies for all integers n, but in this case only those that are powers of 2.


Hmmmmmmmmmmmmmmmmmmmmmmmmmm


Well thanks guys for your critique,
now ive picked up my mistake eh.

But wait, what other mistakes do I have?

Im sure you understand what im trying to say right, could you write it the correct way for me plz? I suck at notation lol

 

[Carrotsticks]

Oh my, this thread just sparked an idea for that problem beforehand with the cyclic sums etc.

I hope this turns out well....

Like seanieg89 said, your product will always converge to 0 because you defined an INFINITE sum.

The n'th root of any real number converges to 1 as n approaches infinity (proof of this is required, but it's a fairly intuitive result). So the inside of say... your 1000000000000000000th product term will be something like (1.00000000001 - 1.0000000001), which as you can imagine, is very close to 0.

So your overall product will converge to 0. This is a very basic explanation, but I hope you can understand what I'm saying (moreso than what I'm understanding what you're saying haha).

 

[hayabusaboston]

You mean you missed the condition b,c>0.

In any case, read my previous post. This infinite product will always converge to 0...

Yep I know it converges to zero lol, like I said, its just expressing everything up UNTIL zero, or really close to it anyway. This whole thing was just a fun thing I thought I'd try to do, its actually quite useless but fun to know youve covered everything there is to cover in that particular factorisation

But overall, is my logic Correct, in the sense of expressing all factors (even though they converge to zero)? It might be useless, just wondering if its right haha.

 

[hayabusaboston]

Oh my, this thread just sparked an idea for that problem beforehand with the cyclic sums etc.

I hope this turns out well....

Like seanieg89 said, your product will always converge to 0 because you defined an INFINITE sum.

The n'th root of any real number converges to 1 as n approaches infinity (proof of this is required, but it's a fairly intuitive result). So the inside of say... your 1000000000000000000th product term will be something like (1.00000000001 - 1.0000000001), which as you can imagine, is very close to 0.

So your overall product will converge to 0. This is a very basic explanation, but I hope you can understand what I'm saying (moreso than what I'm understanding what you're saying haha).

WAITUP bro, no one has told me, am I actually right? just wrote it the wrong way???

So ive helped you come to an epiphany Carrotsticks? Do I get rep?

 

[seanieg89]

Yep I know it converges to zero lol, like I said, its just expressing everything up UNTIL zero, or really close to it anyway. This whole thing was just a fun thing I thought I'd try to do, its actually quite useless but fun to know youve covered everything there is to cover in that particular factorisation

But overall, is my logic Correct, in the sense of expressing all factors (even though they converge to zero)? It might be useless, just wondering if its right haha.

Well I haven't understood your explanations of what you are TRYING to achieve, or what you THINK you have achieved, but all I see is an infinite product over some (still ambiguous) set which is equal to zero.

 

[seanieg89]

If "right" means writing down an infinite product that is equal to a/(b-c) for any a,b,c>0, then no you are not...sorry!

 

[hayabusaboston]

Well I haven't understood your explanations of what you are TRYING to achieve, or what you THINK you have achieved, but all I see is an infinite product over some (still ambiguous) set which is equal to zero.

umm yeaa lol I cant really explain more than I did.

Hmm indeed it is a useless little thing that converges to zero.

In essence, it is a proof

Stating that a(b-c) is not fully factorised, nor is a(sqrtb-sqrtc). It is proving that there are an infinite amount of factors to the expression a(b-c).

 

[Carrotsticks]

I understand where your logic is going and what you're trying to say. I think this is what you mean:

Consider a-b. We split this into the difference of two squares:



But the term:



Is also the difference of another two set of squares (square root of a square root gives a 4th root, hence the 2^n that you were trying to say before)

So we break it up as:



And so our new expression is:



And we get another difference of two squares etc etc.

Is this what you were trying to say?

 

[hayabusaboston]

If "right" means writing down an infinite product that is equal to a/(b-c) for any a,b,c>0, then no you are not...sorry!

nope, by "right" I mean proving that a(b-c) has an infinite amount of factors.

 

[hayabusaboston]

I understand where your logic is going and what you're trying to say. I think this is what you mean:

Consider a-b. We split this into the difference of two squares:



But the term:



Is also the difference of another two set of squares (square root of a square root gives a 4th root, hence the 2^n that you were trying to say before)

So we break it up as:



And so our new expression is:



And we get another difference of two squares etc etc.

Is this what you were trying to say?

YES!!!

THen the root a + root b can be expressed as the sum of two cubes

And that sum of two cubes can also be continuously increased

YAY CARROTSTICKS!!


you can explain to everybody what i mean perhaps

 

[seanieg89]

"Stating that a(b-c) is not fully factorised, nor is a(sqrtb-sqrtc). It is proving that there are an infinite amount of factors to the expression a(b-c). "

Having an 'infinite amount of factors' is a vague property unless you specify what these factors must be. Eg we factorise polynomials into a product of polynomials, we factorise an integer into a product of integers. You have made no such specification...

It is fully factorised in the sense that a, and (b-c) have no factors that are polynomials in a,b,c. So a and b-c are called irreducible elements of the ring of polynomials in a,b,c.

If we are working over something like the real numbers then nothing will ever be "fully factorised"...eg 1=1*2*1/2*69*(1/69)... I can make this list as long as I want.

Allowing the taking of n-th roots makes it unsurprising that we can write (b-c) as a finite product of arbitrarily large length. But note that arbitrarily large is NOT the same as infinite. Taking an actual infinite product here makes no sense.

 

[hayabusaboston]

"Stating that a(b-c) is not fully factorised, nor is a(sqrtb-sqrtc). It is proving that there are an infinite amount of factors to the expression a(b-c). "

Having an 'infinite amount of factors' is a vague property unless you specify what these factors must be. Eg we factorise polynomials into a product of polynomials, we factorise an integer into a product of integers. You have made no such specification...

It is fully factorised in the sense that a, and (b-c) have no factors that are polynomials in a,b,c. So a and b-c are called irreducible elements of the ring of polynomials in a,b,c.

If we are working over something like the real numbers then nothing will ever be "fully factorised"...eg 1=1*2*1/2*69*(1/69)... I can make this list as long as I want.

Allowing the taking of n-th roots makes it unsurprising that we can write (b-c) as a finite product of arbitrarily large length. But note that arbitrarily large is NOT the same as infinite. Taking an actual infinite product here makes no sense.

But the roots can go onto infinity..... even if converge to zero.......as long as the roots obey the condition of being powers of 2 and divisible by 2, you could go onto infinity....


And reiterating it again

Repeated taking of nth roots in this case is simply for amusement, like multiplying 2x2x2x2x2x2..... as many times as you can. I do random mathematical things for fun, even though they might be pointless, it is fun getting to the very core of the concept.

Using the conditions and projecting the formula onto infinity quantifies the infinite factors into one expression, which is just amusing in itself lol. Thats just the kind of thing I do. Random eh? haha. My favourite is doing 2x2x2x2x2x2..... loads and loads of times. That is fun

 

[seanieg89]

The infinite expression makes sense if you are letting a,b,c be real numbers. But then this is saying nothing other than that a real number can be written as a product of arbitrary length. Nothing amazing...

Anyway, I'm done here. I think enough has been said.

 

[hayabusaboston]

The infinite expression makes sense if you are letting a,b,c be real numbers. But then this is saying nothing other than that a real number can be written as a product of arbitrary length. Nothing amazing...

Anyway, I'm done here. I think enough has been said.

lol pick out every flaw haha thats good though, you are very meticulate. indeed i did not talk about real vs unreal, i may incorporate that now, thanks for the reminder

 

[seanieg89]

I didn't mean real vs nonreal I meant real vs indeterminate. Eg the role x plays in polynomials.

 

[hayabusaboston]

Oh dear I did indeed miss out a lot. Thanks for the input :-