Polynomials of the 4th order, basis

G

God

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Hi.

Let P4 represent all the polynomials of the 4th order : a+bx+cx²+dx3+ex4.
We have E={P∈P4, P(1)=0 and P(-1)=0}

-Show that P(t) = a+bx+cx²+dx3+ex4 ∈ E if and only if (a,b,c,d,e) is the solution of a linear system :

Ok so we have a+b+c+d+e=0 and a-b+c-d+e=0, which gives a+c+e=0 and b+d=0

-Solve the system and give a basis of the E space :

So the system has too many unknown values, I'm not sure how the solution can be shown... either as a+c+e=0 and b+d=0, or a=-c-e and b=-d... something along the lines I guess

My problem lies in finding the basis

I thought I'd do it that way : (a,b,c,d,e) = (-c-e,-d,c,d,e) = c*(-1,0,1,0,0) + d*(0,-1,0,1,0) + e*(-1,0,0,0,1)

So the basis would be ((-1,0,1,0,0),(0,-1,0,1,0),(-1,0,0,0,1)) (these are three vectors)
but I feel like something's missing... since these vectors don't make polynomials !
I don't know where to place the x's, x²'s etc...
Actually I have a hard time transforming my vectors into hybrid polynomial-vectors... because they should be able to form E, that's to say all polynomials of the 4th order with P(1)=0 and P(-1)=0

Any help is very appreciated, thanks for your attention

 
God said:
Hi.

Let P4 represent all the polynomials of the 4th order : a+bx+cx²+dx3+ex4.
We have E={P∈P4, P(1)=0 and P(-1)=0}

-Show that P(t) = a+bx+cx²+dx3+ex4 ∈ E if and only if (a,b,c,d,e) is the solution of a linear system :

Ok so we have a+b+c+d+e=0 and a-b+c-d+e=0, which gives a+c+e=0 and b+d=0

-Solve the system and give a basis of the E space :

So the system has too many unknown values, I'm not sure how the solution can be shown... either as a+c+e=0 and b+d=0, or a=-c-e and b=-d... something along the lines I guess
Click to expand...
If P(1) = 0, then (x - 1) must be a factor of the polynomial,
and if P(-1) = 0 then (x + 1) is also a factor, so the polynomial must be divisible by (x^2 - 1)

This is not new information .. it is totally equivalent to setting \(\displaystyle d = -b\) and \(\displaystyle e = -(a + c)\). That substitution in the original equation leads to something like this: [CHECK THIS CAREFULLY! IT TAKES 5 OR 6 STEPS TO GET HERE.]

\(\displaystyle a + bx + cx^2 + dx^3 + ex^4 = (1 - x^2)[a + bx + (a+c)x^2]\)

The "solution" of a set of 2 equations in 5 unknowns leaves 3 unknowns. Hence you need a set of three linearly independent basis vectors to describe the set E. Does the factored form suggest a set of 3 vectors?
 
By the way, you mean, of course, that P4 is the set of all polynomials of degree 4 or less.
 
HallsofIvy said:
By the way, you mean, of course, that P4 is the set of all polynomials of degree 4 or less.
Click to expand...

Indeed. 4 or less sorry.

DrPhil said:
If P(1) = 0, then (x - 1) must be a factor of the polynomial,
and if P(-1) = 0 then (x + 1) is also a factor, so the polynomial must be divisible by (x^2 - 1)

This is not new information .. it is totally equivalent to setting \(\displaystyle d = -b\) and \(\displaystyle e = -(a + c)\). That substitution in the original equation leads to something like this: [CHECK THIS CAREFULLY! IT TAKES 5 OR 6 STEPS TO GET HERE.]

\(\displaystyle a + bx + cx^2 + dx^3 + ex^4 = (1 - x^2)[a + bx + (a+c)x^2]\)

The "solution" of a set of 2 equations in 5 unknowns leaves 3 unknowns. Hence you need a set of three linearly independent basis vectors to describe the set E. Does the factored form suggest a set of 3 vectors?
Click to expand...

Okay, I understand how you got the equation : \(\displaystyle a + bx + cx^2 + dx^3 + ex^4 = (1 - x^2)[a + bx + (a+c)x^2]\)
Thanks a lot.

I am, however, still missing the point as to how it will help me to find the vectors.

The only way I was taught to find them was replacing that way : (a,b,c,d,e) = (-c-e,-d,c,d,e) = c*(-1,0,1,0,0) + d*(0,-1,0,1,0) + e*(-1,0,0,0,1)
perhaps what's missing is just the "x's", but I don't know where I should place them
 
Well, you have, at this point, that any polynomial in this set is of the form \(\displaystyle (1- x^2)(ax^2+ bx+ c)\) (I have replaced "a+ c" with just "c". Since a and c are arbitrary constants, so is a+ c.) That tells us that any polynoial in this set can be written as \(\displaystyle a(1-x^2)x^2+ b(1- x^2)x+ c(1- x^2)\). Now, what do you think basis "vectors" are?
 
God said:
Indeed. 4 or less sorry.



Okay, I understand how you got the equation : \(\displaystyle a + bx + cx^2 + dx^3 + ex^4 = (1 - x^2)[a + bx + (a+c)x^2]\)
Thanks a lot.

I am, however, still missing the point as to how it will help me to find the vectors.

The only way I was taught to find them was replacing that way : (a,b,c,d,e) = (-c-e,-d,c,d,e) = c*(-1,0,1,0,0) + d*(0,-1,0,1,0) + e*(-1,0,0,0,1)
perhaps what's missing is just the "x's", but I don't know where I should place them
Click to expand...
The powers of x are represented by components of the vectors, that is, they define the vector space. The polynomial P4 can be written

\(\displaystyle P4 = \begin{pmatrix} a & b & c & d & e \end{pmatrix}
× \begin{pmatrix}1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{pmatrix} \)

You arrived at set \(\displaystyle E\) by eliminating coefficients \(\displaystyle a\) and \(\displaystyle b\):

\(\displaystyle E = \begin{pmatrix} c & d & e \end {pmatrix} ×
\begin{pmatrix} -1 & 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \end{pmatrix}
× \begin{pmatrix}1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{pmatrix} \)

The difference in what I did was to eliminate the highest-power coefficients, \(\displaystyle e\) and \(\displaystyle f\):

\(\displaystyle E = \begin{pmatrix} a & b & c \end {pmatrix} ×
\begin{pmatrix} 1 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & -1 \end{pmatrix}
× \begin{pmatrix}1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{pmatrix} \)

Either should work. AGAIN, you have to check my work! It is too easy to make typos.
 
Okay, I think I got it now.
The "vectors" (can we really talk about vectors here ? or just polynomials forming a basis ?) are (x2-x4),(x-x3),(1-x²).

It tallies with what DrPhil posted to explain.
So if I understood correctly, (x2-1),(x3-x),(x4-1) is also another basis.

Thanks to both of you !

And yes, I did check your work DrPhil. Thanks again !
 
God said:
IThe only way I was taught to find them was replacing that way : (a,b,c,d,e) = (-c-e,-d,c,d,e) = c*(-1,0,1,0,0) + d*(0,-1,0,1,0) + e*(-1,0,0,0,1)
Click to expand...
This is essentially correct in your originally post. The subset E exists in the 5-dimensional space P4. Since there are two constraints, it requires only 3 basis vectors to span E, but those basis vectors have 5 elements representing coefficients of 5 powers of x. Thus I would consider the three 5-vectors to be an adequate representation.

When you write the basis set as three polynomials, such as either
{(x2-x4), (x-x3), (1-x²)} or {(x2-1), (x3-x), (x4-1)},
note that all basis polynomials are divisible by (1-x2), so any linear combination is also divisible by (1-x2). That statement embodies the two constraints.

Enjoy :p !
 
DrPhil said:
This is essentially correct in your originally post. The subset E exists in the 5-dimensional space P4. Since there are two constraints, it requires only 3 basis vectors to span E, but those basis vectors have 5 elements representing coefficients of 5 powers of x. Thus I would consider the three 5-vectors to be an adequate representation.

When you write the basis set as three polynomials, such as either
{(x2-x4), (x-x3), (1-x²)} or {(x2-1), (x3-x), (x4-1)},
note that all basis polynomials are divisible by (1-x2), so any linear combination is also divisible by (1-x2). That statement embodies the two constraints.

Enjoy :p !
Click to expand...

Actually I'm still not convinced
Let's take the derivative of a+bt+ct²+dt^3+et^4
it gives b+2ct+3dt²+4et^3
now let F be polynomials such as P4'(1)=0 and P4'(-1)=0
we have b+2c+3d+4e=0, b-2c+3d-4e=0, <=> b+3d=0 and c+2e=0 so that's (b,c,d,e)=(-3d, -2e, d, e)=d*(-3,0,1,0) + e*(0,-2,0,1)
so (x²-3),(x^3-2x) is a basis.
too bad (x²-3) is not divisible by (x²-1) !
 
Last edited:
God said:
Actually I'm still not convinced
Let's take the derivative of a+bt+ct²+dt^3+et^4
it gives b+2ct+3dt²+4et^3
now let F be polynomials such as P4'(1)=0 and P4'(-1)=0
we have b+2c+3d+4e=0, b-2c+3d-4e=0, <=> b+3d=0 and c+2e=0 so that's (b,c,d,e)=(-3d, -2e, d, e)=d*(-3,0,1,0) + e*(0,-2,0,1)
so (x²-3),(x^3-2x) is a basis.
too bad (x²-3) is not divisible by (x²-1) !
Click to expand...
Is this the same problem?
Taking the derivative does nothing to help find basis vectors for the set.
You do not know the values of the derivative at 1 or at -1.
 
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