The Construction - thread 1

[Haydon Berrow]

This is an attempt to understand the construction in http://somemath.blogspot.co.uk/2015/11/non-polynomial-factorization-short.html but I've failed to understand the term "integer-like" so I'll avoid it.

Definition

A quadratic polynomal with integer coefficients is primitive if there is no non-trivial integer that divides all the coefficients.

Examples are x²+3x+2, 5x²+x+1, and x²+5x+2

Notation

I'll try to use P() to mean a function, P(x) to mean a polynomial in a variable x, and P[x] to mean P evaluated at the value x but I expect I'll forget and have to backtrack sometimes.

The Construction

Let P(x) be a primitive quadratic polynomial. It's automatically a function from the field of complex numbers to itself.

Suppose we have two further functions from the field of complex numbers to itself with the following properties:

• P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x in the field of complex numbers
• g₁[0] = 0
• g₂[0] = 0 but g₂() is not the zero-function, ie there is a number x such that g₂[x] is not zero

Are there any further constraints on the g_i() before I go further?

[JSH]

No. And I give a simple example of P(x) = (x+1)(x+2), where here you can see that x is a solution for both g's. The conditions simply set up normalized functions, which are NOT limited to being polynomials, though polynomials like x+1 and x+2 are included. Wonder why I don't say that neither of the g's can be zero for all x? My past self considered such things. Must have deemed it unnecessary. Ok.

___JSH 

[Haydon Berrow]

There are no further restrictions of the g_i() so these are all legitimate examples, some of them are quite exotic

1. g₁[x] = 0, g₂[x] = x²+5x
2. if |x| is non-zero and rational then express it as a/b with minimal a,b and b>0;
   
1. now define g₁[x] = (1-a)/a, g₂[x] = a(x²+5x+2)-2 but elsewhere (ie x is zero or irrational) then  g₁[x] = 0, g₂[x] = x²+5x
3. g₂[x] = b(x²+5x+2) if |x| is rational and equal to a/b with a,b coprime positive integers and g₂[x]=x²+5x elsewhere
4. g₁[x] = 0 if |x|<1 and g₁[x] = 1 if |x|≥1, g₂[x] = x²+5x if |x|<1 and g₂[x] = (x²+5x-2)/2 if |x|≥1
5. g₁[x] = π|x|-1, g₂[x] = π^-1(x²+5x+2(1-π))
6. g₁[x] = (x+1 + √(3x²+14x+1))/2, g₂[x] = (x-1 - √(3x²+14x+1))/2

Now let k be an integer greater than 1, and define f₁(), and f₂() by f₁[x] = g₁[x]/k, and f₂[x] = g₂[x]+(k-2). It follows that

k*P(x) =  (f₁(x) + k)(f₂(x) + k) for all x in the complex numbers

Now define h() by h[x] = f₁[x] + f₂[x]; h is a function from the complex numbers to itself but is not generally a polynomial

All right so far? I realise that k could be negative but decided not to complicate things

[JSH]

On Thursday, November 23, 2017 at 11:38:58 AM UTC-5, Haydon Berrow wrote:

This is an attempt to understand the construction in http://somemath.blogspot.co.uk/2015/11/non-polynomial-factorization-short.html but I've failed to understand the term "integer-like" so I'll avoid it.

Definition

A quadratic polynomal with integer coefficients is primitive if there is no non-trivial integer that divides all the coefficients.

Examples are x²+3x+2, 5x²+x+1, and x²+5x+2

Notation

I'll try to use P() to mean a function, P(x) to mean a polynomial in a variable x, and P[x] to mean P evaluated at the value x but I expect I'll forget and have to backtrack sometimes.

The Construction

Let P(x) be a primitive quadratic polynomial. It's automatically a function from the field of complex numbers to itself.

Suppose we have two further functions from the field of complex numbers to itself with the following properties:

• P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x in the field of complex numbers
• g₁[0] = 0
• g₂[0] = 0 but g₂() is not the zero-function, ie there is a number x such that g₂[x] is not zero

There are no further restrictions of the g_i() so these are all legitimate examples, some of them are quite exotic

What is P(x) for each?

1. g₁[x] = 0, g₂[x] = x²+5x
2. if |x| is non-zero and rational then express it as a/b with minimal a,b and b>0;
   
1. now define g₁[x] = (1-a)/a, g₂[x] = a(x²+5x+2)-2 but elsewhere (ie x is zero or irrational) then  g₁[x] = 0, g₂[x] = x²+5x
3. g₂[x] = b(x²+5x+2) if |x| is rational and equal to a/b with a,b coprime positive integers and g₂[x]=x²+5x elsewhere
4. g₁[x] = 0 if |x|<1 and g₁[x] = 1 if |x|≥1, g₂[x] = x²+5x if |x|<1 and g₂[x] = (x²+5x-2)/2 if |x|≥1
5. g₁[x] = π|x|-1, g₂[x] = π^-1(x²+5x+2(1-π))
6. g₁[x] = (x+1 + √(3x²+14x+1))/2, g₂[x] = (x-1 - √(3x²+14x+1))/2

For every single example you presented as if I should consider, give P(x) or admit you were trying to waste MY time.

And why?

 

Now let k be an integer greater than 1, and define f₁(), and f₂() by f₁[x] = g₁[x]/k, and f₂[x] = g₂[x]+(k-2). It follows that

k*P(x) =  (f₁(x) + k)(f₂(x) + k) for all x in the complex numbers

Now define h() by h[x] = f₁[x] + f₂[x]; h is a function from the complex numbers to itself but is not generally a polynomial

All right so far? I realise that k could be negative but decided not to complicate things

For every construction, go back through and show P(x). Remember, you have a factorization.

___JSH

[Haydon Berrow]

This is an attempt to understand the construction in http://somemath.blogspot.co.uk/2015/11/non-polynomial-factorization-short.html but I've failed to understand the term "integer-like" so I'll avoid it.

 

Definition

A quadratic polynomal with integer coefficients is primitive if there is no non-trivial integer that divides all the coefficients.

Examples are x²+3x+2, 5x²+x+1, and x²+5x+2

 

Notation

I'll try to use P() to mean a function, P(x) to mean a polynomial in a variable x, and P[x] to mean P evaluated at the value x but I expect I'll forget and have to backtrack sometimes.

 

The Construction

 

Let P(x) be a primitive quadratic polynomial. It's automatically a function from the field of complex numbers to itself.

   

Suppose we have two further functions, g₁() and g₂(), from the field of complex numbers to itself with the following properties: 

• P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x in the field of complex numbers
• g₁[0] = 0
• g₂[0] = 0 but g₂() is not the zero-function, ie there is a number x such that g₂[x] is not zero

There are no further restrictions of the g_i() so these are all legitimate examples, some of them are quite exotic. In every case the Polynomial is x²+5x+2 and I'll give a proof that P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x

Example 1. g₁[x] = 0, g₂[x] = x²+5x

    Proof. P[x] = (g₁[x] + 1)(g₂[x] + 2) = (1)(x²+5x+2) = x²+5x+2

Example 2. if |x| is non-zero and rational then express it as a/b with minimal a,b and b>0;

now define g₁[x] = (1-a)/a, g₂[x] = a(x²+5x+2)-2 but elsewhere (ie x is zero or irrational) then  g₁[x] = 0, g₂[x] = x²+5x

Proof. If x is irrational then the functions are as i example 1, so look at the case when x is rational

P[x] = (g₁[x] + 1)(g₂[x] + 2)

= ((1-a)/a+1) (a(x²+5x+2)-2+2)                                              

= ((1-a+a)/a) (a(x²+5x+2))

          = (1/a)(a(x²+5x+2)) = x²+5x+2

Example 3. g₁[x] = 0 if |x|<1 and g₁[x] = 1 if |x|≥1, g₂[x] = x²+5x if |x|<1 and g₂[x] = (x²+5x-2)/2 if |x|≥1

Proof. If |x|<1 then this example is as in example 1 so I'll consider the case of |x|≥1

P[x] = (g₁[x] + 1)(g₂[x] + 2)

= (1+1)((x²+5x-2)/2+2)
= (2)((x²+5x-2+4)/2)
= (x²+5x-2+4)
= x²+5x+2

Example 4. g₁[x] = πx-1, g₂[x] = π^-1(x²+5x+2(1-π))

Proof.

P[x] = (g₁[x] + 1)(g₂[x] + 2)

= (πx-1+1)(π^-1(x²+5x+2(1-π))+2)

= (πx)(π^-1(x²+5x+2(1-π)+2π)

= (πx)(π^-1(x²+5x+2))

=x²+5x+2

Example 5. g₁[x] = (x+1 + √(3x²+14x+1))/2, g₂[x] = (x-1 - √(3x²+14x+1))/2

Proof.

P[x] = (g₁[x] + 1)(g₂[x] + 2)

= ( (x+1 + √(3x²+14x+1))/2 + 1 )( (x-1 - √(3x²+14x+1))/2 +2 )

= ( (x+3 + √(3x²+14x+1))/2  )( (x+3 - √(3x²+14x+1))/2 )

= (x+3 + √(3x²+14x+1)) (x+3 - √(3x²+14x+1)) / 4

= ( (x+3)² - (3x²+14x+1) ) ) / 4

= ( (x²+6x+9+(3x²+14x+1) ) / 4

= (4x²+20x+10)/4

= x²+5x+2

 

Now let k be an integer greater than 1, and define f₁(), and f₂() by f₁[x] = g₁[x]/k, and f₂[x] = g₂[x]+(k-2). It follows that

k*P(x) =  (f₁(x) + k)(f₂(x) + k) for all x in the complex numbers

and now define h() by h[x] = f₁[x] + f₂[x]; h is a function from the complex numbers to itself but is not generally a polynomial

All right so far? I realise that k could be negative but decided not to complicate things

---------

You asked why I constructed these strange examples. The answer is that it's second nature for mathematicians to look at the boundary conditions of constructions and proofs as sanity-checks, otherwise it's too easy to build castles on sand. It's not all that dissimilar from competent computer programmers' practice of checking edge-cases.

[JSH]

You don't get to set the g's though, now do you? You only know they are normalized. Correct?

So why would you try to set them?

They are further defined by H(x) ,which you DO get to set.

Was just kind of curious why you talked as if you can just set them. Hey if you wish to waste mental energy, go right ahead.

There are an INFINITE number of possible solutions for the g's, now how do you narrow down that infinity?

I show you how. H(x) is a handle.

___JSH 

[Haydon Berrow]

Oh! I understood from this text on the web page

In the complex plane given:

P(x) = (g₁(x) + 1)(g₂(x) + 2)

where P(x) is a primitive quadratic with integer coefficients, g₁(0) = g₂(0) = 0, but g₁(x) does not equal 0 for all x.

that the construction would be for a fixed P(), g₁() and g₂(), and we would be varying the value at which they were being evaluated.

Should I have instead written

Let P(x) be a primitive quadratic polynomial. It's automatically a function from the field of complex numbers to itself   

and consider all g₁() and g₂(), from the field of complex numbers to itself with the following properties: 

• P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x in the field of complex numbers
• g₁[0] = 0
• g₂[0] = 0 but g₂() is not the zero-function, ie there is a number x such that g₂[x] is not zero

these are not fixed in the following construction but possible amongst the many possible values for g₁() and g₂() are ...   ?

[JSH]

Here is the given example: P(x) = (x+1)(x+2)

Its simplicity does not remove reality it tells you all you need to know to get abstract concepts. The g's are functions which equal 0 at x = 0. P(x) is a quadratic, which is primitive. While with the example have linear functions that is not given as a requirement. They just also work with the requirements.

The g's are simply defined, and then later there is a method for determining them, with primary control by a single function called H(x) through f's which are also other functions. But the control is with H(x). And the H Is for handle.

Not complicated really.

___JSH

[Haydon Berrow]

This is an attempt to understand the construction in http://somemath.blogspot.co.uk/2015/11/non-polynomial-factorization-short.html but I've failed to understand the term "integer-like" so I'll avoid it.

 

Definition

A quadratic polynomal with integer coefficients is primitive if there is no non-trivial integer that divides all the coefficients.

Examples are x²+3x+2, 5x²+x+1, and x²+5x+2

 

Notation

I'll try to use P() to mean a function, P(x) to mean a polynomial in a variable x, and P[x] to mean P evaluated at the value x but I expect I'll forget and have to backtrack sometimes.

 

The Construction

 

Let P(x) be a primitive quadratic polynomial. It's automatically a function from the field of complex numbers to itself.

   

Suppose we have two further functions, g₁() and g₂(), from the field of complex numbers to itself with the following properties: 

• P[x] = (g₁[x] + 1)(g₂[x] + 2) for all x in the field of complex numbers
• g₁[0] = 0
• g₂[0] = 0 but g₂() is not the zero-function, ie there is a number x such that g₂[x] is not zero

There are no further restrictions of the g_i() so these are all legitimate examples, some of them are quite exotic. In every case the Polynomial is x²+5x+2 and a previous text showed that they satisfied the requirements

Example 1. g₁[x] = 0, g₂[x] = x²+5x

Example 2. if |x| is non-zero and rational then express it as a/b with minimal a,b and b>0; now define g₁[x] = (1-a)/a, g₂[x] = a(x²+5x+2)-2 but elsewhere (ie x is zero or irrational) then  g₁[x] = 0, g₂[x] = x²+5x

Example 3. g₁[x] = 0 if |x|<1 and g₁[x] = 1 if |x|≥1, g₂[x] = x²+5x if |x|<1 and g₂[x] = (x²+5x-2)/2 if |x|≥1

Example 4. g₁[x] = πx-1, g₂[x] = π^-1(x²+5x+2(1-π))

Example 5. g₁[x] = (x+1 + √(3x²+14x+1))/2, g₂[x] = (x-1 - √(3x²+14x+1))/2

 Let k be an integer greater than 1, and define f₁(), and f₂() by f₁[x] = g₁[x]/k, and f₂[x] = g₂[x]+(k-2). It follows that

k P[x] =  (f₁[x] + k) (f₂[x] + k) for all x in the complex numbers

and now define h() by h[x] = f₁[x] + f₂[x]; h is a function from the complex numbers to itself but is not generally a polynomial

It follows that f₂[x] = h[x] - f₁[x] and substituting this into k P[x] =  (f₁[x] + k) (f₂[x] + k) gives

k P[x] =  (f₁[x] + k) (h[x] - f₁[x] + k) and then expand to get

k P[x] =  f₁[x] h[x] - f₁[x]² + k f₁[x] + k h[x] - k f₁[x] + k² and rearrange terms to give

f₁[x]² - h[x] f₁[x] - k h[x] - k² + k*P[x] = 0 which is a quadratic in f₁[x]

Solving the quadratic gives

f₁[x] = ( h[x] +/- sqrt( h[x]² - 4( -k h[x] - k² + k P[x] )  )/2

= ( h[x] +/- sqrt( h[x]² + 4k h[x] + 4k² - 4k P[x] )  )/2

= ( h[x] +/- sqrt( (h[x] + 2k)² - 4k P[x] )  )/2

So far, so good, but I don't understand this next bit

And sqrt[(H(x) + 2k)² - 4k*P(x)] will only resolve if it is to a linear function

what does resolve mean?

[JSH]

Give explicit solutions. For example: sqrt(4) resolves to 2 or -2. While sqrt(2) does not so resolve.

___JSH 

[Haydon Berrow]

I don't understand. Why isn't +/- sqrt(2) an explicit solution ?

[JSH]

Sqrt(2) is not singular. Sqrt(4) equals 2 or -2. There are two singular solutions. So sqrt(4) resolves to two explicit solutions, while sqrt(2) does not.

___JSH 

[Haydon Berrow]

x²=2 has two solutions,  x²=4 has two solutions. why aren't the two solutions to the first equation explicit?

[JSH]

Because sqrt(2) is not a single number any more than sqrt(4) is a single number. However, with sqrt(4) can explicitly give 2 and -2. But sqrt(2) is sqrt(2) and its solutions cannot be so displayed. Am surprised you did not suggest approximation. However 2 is exact in case you try, and -2 is exact as explicit solutions for sqrt(4).

___JSH 

[Haydon Berrow]

If I type "define:explicit" into google I get "stated clearly and in detail, leaving no room for confusion or doubt."

If I ask anyone with a science degree what is sqrt(2) there will never be any confusion or doubt as to what is meant. You are using some other private definition of explicit. What is it and do you mean a "rational number"?

[JSH]

I have a B.Sc. in Physics. 

Forget about the word explicit. Yeah, rational number solution. Like from when I say: sqrt(4) has solutions 2 or -2 (dropping that word which befuddled you so much). While sqrt(2) does not so resolve.

___JSH 

[Haydon Berrow]

If I type "define:explicit" into google I get "stated clearly and in detail, leaving no room for confusion or doubt."

If I ask anyone with a science degree what is sqrt(2) there will never be any confusion or doubt as to what is meant. You are using some other private definition of explicit. What is it and do you mean a "rational number"?

I have a B.Sc. in Physics. 

Forget about the word explicit. Yeah, rational number solution. Like from when I say: sqrt(4) has solutions 2 or -2 (dropping that word which befuddled you so much). While sqrt(2) does not so resolve.

I know, were you in any doubt what is meant by sqrt(2)

OK,

And sqrt[(H(x) + 2k)² - 4k*P(x)] will only resolve if it is to a linear function

becomes

And sqrt[(H(x) + 2k)² - 4k*P(x)] will only give a rational solution if it is to a linear function

but I still don't understand what that means

[JSH]

Ok.

___JSH