interesting problem (3 Viewers)

[hayabusaboston]

I thought of a problem today, I know thr answer, I wonder if you guys know how to do it. It is within reach of most high schoolers, I would think

Express a(b-c) as a product of an infinite series of factors

I would lile to see your method carrotsticks. Also spirals

 

[Carrotsticks]

I joke =p

 

[seanieg89]

I joke =p

haha. its actually probably the best answer to give. Most other expressions in terms of a,b,c could probably be made to diverge for suitable choices of a.b,c.

 

[seanieg89]

Unless we write something silly, like:

.

Where the infinite product is just a constant.

 

[hayabusaboston]

Mm I split it up into an integer a, sum of two cubes sqrtb+sqrtc and difference of two squares,then wrote the three factors in infinite terms. That is a valid solution right?

 

[Carrotsticks]

I didn't really get what you meant by that.

Regardless, I don't think the idea of making an infinite product works in this case.

Given some constants a, b and c, it is difficult to produce an *general* infinite product that converges to something. Anything dealing with infinitismals tends to be quite fragile.

An example of an infinite product converging to some value is the Wallis Product: http://en.wikipedia.org/wiki/Wallis_product

Which converges to pi/2. And I can assure you, the process of finding the convergent value for the product (or visa versa) was by no means a simple process.

If only there existed some general infinite product for an expression consisting of any 3 constants in the form a(b-c).... life would be wonderful.

 

[seanieg89]

Mm I split it up into an integer a, sum of two cubes sqrtb+sqrtc and difference of two squares,then wrote the three factors in infinite terms. That is a valid solution right?

Be a little more explicit and we might be able to tell you...write it up?

 

[hayabusaboston]

Okay I wont bother with latex lol will just write up verbally
for a(b-c)
I left a as it was

then using the product symbol (You know, the one that looks like a big pi symbol), the 2nd factor was

"the product of all n, such that n is greater than zero, and a member of 2n and n squared, from n=1, for the formula (nth root of b minus the nth root of c)"

Then the third factor was the same as above, but nth root of b PLUS nth root of c

Fourth factor was "the product of all n such that n is greater than 0, a member of n^3 and 3n, from n=3, for the formula nth root of b minus nth root of c."

Then again, fifth factor was like above but nth root b PLUS nth root c

That, shall we say "quantifies" the difference of two cubes lol, though I am working on the sum of two cubes, it is rather more difficult than squares HAHA.

Anyway is my logic valid?

Wait I think that might be a bit fractured, anyway simplifying it all I still had
(3rd root of b)^2-(3rd root b)(3rd root c)-(3rd root c)^2

Thats the final bit im trying to quantify

I thought perhaps to make the middle bit

"The product of all n such that n is greater than 0, a member of 3n and a member of n^3, from n=3, for the formula (nth root b)(nth root c)"

But im not sure if that is sensible? it leaves the other two parts out

 

[hayabusaboston]

If only there existed some general infinite product for an expression consisting of any 3 constants in the form a(b-c).... life would be wonderful.

Yea lol I think I might have found one, plz check for me. As I still dont have the sum of cubes, you could try to quantify it. Teamwork

 

[Carrotsticks]

You said a few times something like:

"the product of all n such that n is greater than 0, a member of n^3 and 3n, from n=3"

I don't quite get what you mean for this part.

Just latex one expression. Surely if you can't be bothered typing out YOUR question in a way that we can understand it, then why should we bother answering it? =p

 

[hayabusaboston]

You said a few times something like:

"the product of all n such that n is greater than 0, a member of n^3 and 3n, from n=3"

I don't quite get what you mean for this part.

Just latex one expression. Surely if you can't be bothered typing out YOUR question in a way that we can understand it, then why should we bother answering it? =p

ohh im sorry man really I just cant use latex. What I mean by that part, you know the symbol used in set notation, the one that looks like a reverse capital letter E, but rounded as well? I mean like that, so let that symbol be E. So

n E n^2,2n

Put this:

\prod_{n=1}^{n\in n^2,2n}(\sqrt[n]{b}-\sqrt[n]{c})


Into this site

http://www.codecogs.com/latex/eqneditor.php


Thats what I mean

 

[seanieg89]

still makes no sense, you are using n to denote both a variable and a set? And the factor inside the product you latexed is independent of n...

 

[Carrotsticks]

I am still very confused about the intentions of your notation.

1. N is used as the upper limit for product notation, so it must be an integer. But you said n is an element of n^2, 2n (I presume you mean the interval?) but this order can only hold if n is between 0 and 1, which contradicts it being used for a limit.

2. You also used N as a dummy variable for the summation and had it as a limit as well, which doesn't work.

3. Your expression for the product has no N in it, so this defeats the purpose of using product notation.

 

[seanieg89]

The factor contains n-th roots now but my other complaints remain...

 

[hayabusaboston]

still makes no sense, you are using n to denote both a variable and a set? And the factor inside the product you latexed is independent of n...

It does doesnt it? For (nth root b minus nth root c), the rule is implying that you start at n=1, then plug in all other integers of n when n is part of the set 2n and part of the set n squared.
The 2n is to make the roots even, as any number times 2 is even, and the n squared makes the roots literally one square root after the next, which is what you do in the difference of two squares. You keep sqrt'ing the function. So the root of root 2 is
4th root 2, the root of that is 8th root, root of that is 16th etc etc.
So I just restricted the numbers n can be.


That makes sense I thinK? Did I write it wrongly?

Wait am I meant to use k?


Sorry about the notation, i think words express it better. The above big paragraph should make sense. And yea in the formula I had 3rd root accidentally lol, I meant nth root.

 

[hayabusaboston]

Wait so where is my error here:

on the bottom of the product symbol, start at n=1.

Then on top of the symbol, put all integers n where n is an element of 2n and n^2

So it goes on to infinity, picking out all n that are elements of 2n and n^2.

Then the formula for that is nth root b minus nth root c

So dont you just put in n=1

then all other n, as specified by the condition on top?


Isnt that the right way? Sorrry guys

So you would be multiplying

(\sqrt[]{b}-\sqrt[]{c})(\sqrt[4]{b}-\sqrt[4]{c})(\sqrt[8]{b}-\sqrt[8]{c})

etc etc until infinity for 2n and n^2


Put into latex for my meaning

So I wrote the formula wrong then im guessing?


Do my words make sense at least..? Of my intentions i mean?

 

[seanieg89]

So you are taking the product of b^(1/n)-c^(1/n) over all even squares n in addition to n=1? For starters, that doesn't make sense when either of b,c is negative. Secondly for positive b,c this should converge to 0...since b^1/n and c^1/n both approach 1.

Again, being a member of n^2 doesnt make sense, so I don't know what your notation means there. My guess above cannot be right if 8 is included...it is not a square.

REGARDLESS, this product will converge to zero if it is taken over ANY infinite subset of the natural numbers.

 

[Carrotsticks]

It does doesnt it? For (nth root b minus nth root c), the rule is implying that you start at n=1, then plug in all other integers of n when n is part of the set 2n and part of the set n squared.
The 2n is to make the roots even, as any number times 2 is even, and the n squared makes the roots literally one square root after the next, which is what you do in the difference of two squares. You keep sqrt'ing the function. So the root of root 2 is
4th root 2, the root of that is 8th root, root of that is 16th etc etc.
So I just restricted the numbers n can be.


That makes sense I thinK? Did I write it wrongly?

Wait am I meant to use k?

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:

 

[hayabusaboston]

So you are taking the product of b^(1/n)-c^(1/n) over all even squares n in addition to n=1? For starters, that doesn't make sense when either of b,c is negative. Secondly for positive b,c this should converge to 0...since b^1/n and c^1/n both approach 1.

Um I missed the condition n>0

hmm

I could swear that makes sense


Doesnt it? Even the words?

 

[seanieg89]

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

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