Showing functions form a basis

[gpax42]

Homework Statement

Show that the functions p_o(t)=1, p₁(t)=t, p₂(t)=1/2(3t²-1), and p₃(t)=(3/2)*[(5/3)t³-t) also form a basis for the vector space P³(R) ... "R" meaning all real numbers

Homework Equations

I know these polynomials are the first four Legendre polynomials

The Attempt at a Solution

I know that proving functions form a basis involves proving that each funciton has a unique representation as a linear combination... I'm not certain on what this means exactly but I'm trying my best to figure it out... here's my work so far on the problem

a(1) + b(t) + c/2(3t²-1) + d/2(5t³-3t) = a + bt + ct² + dt³

at t=0 ; -c/2 = 0 ; so c=0

a + b(t) + d/2(5t³-3t)=0
at t=0 ; a=0
at t=1 ; b=-d
at t=2 ; -2d + d/2(34) = 15d = 0 ; d=0 ; b=0

so the legendre polynomials are a linearly independent set at; 1/2 + t - 3/2t + 3/2t² +5/3t³ = 1/2-1/2t+3/2t²+5/3t³
; a polynomial in the span of P(R) that forms a basis

thanks in advance for any advice you have to offer me

 

[vela]

I know that proving functions form a basis involves proving that each funciton has a unique representation as a linear combination... I'm not certain on what this means exactly but I'm trying my best to figure it out... here's my work so far on the problem

a(1) + b(t) + c/2(3t²-1) + d/2(5t³-3t) = a + bt + ct² + dt³

To show linear independence, you want to set the righthand side to zero and show that requires a=b=c=d=0.

You also need to show the four polynomials span P₃(R). This time, you want the righthand side to be an arbitrary 3rd-degree polynomial -- so don't use a, b, c, and d that appear on the lefthand side on the righthand side -- and show you can solve for a, b, c, and d.

 

[Dick]

Ok, so you want to show a+b*t+c*t^2+d*t^3 has a unique representation of the form A*p0(t)+B*p1(t)+C*p2(t)+D*p3(t). That means you just want to find A, B, C, and D in terms of a, b, c and d, right? You really don't have to put t values in. You know a+b*t+c*t^2+d*t^3=a'+b'*t+c'*t^2+d'*t^3 only if a=a', b=b', c=c' and d=d', yes? So just start at the top. What is the relation between d and D?

 

[gpax42]

is it D = dt²/[(5/2)t²-(3/2)]

 

[Dick]

is it D = dt²/[(5/2)t²-(3/2)]

Nooo. D is supposed to be constant. Expand the Legendre side and equate the cubic terms.

 

[HallsofIvy]

Since $\{ 1, x, x^2\}$ form a basis for P3, you know that it has dimension 3. That, in turn, tells you that any independent set of three functions will be a basis. After you have shown that the given functions are independent, you don't have to show that they span the space.

 

[gpax42]

ooo alright, thanks for all the helpful explanations =D