Finding W perp of inner product space

[trap101]

Consider P₅(R) together with innner product < p ,q > = ∫p(x)q(x) dx. Find W-perp if W = {p(x) $\in$ P₅(R) : p(0) = p'(0) = p''(0) = 0}

Attempt: I am having trouble with the condition. I always have trouble with these conditions. SO as of now I am going to let q(x) be the standard basis of P₅(R). Now I don't know how to apply the condition to p(x).

After I do apply the condition I would take the inner product and have it set equal to 0. I should have a set of equations that I can solve in matrix form. This should produce some free variables from where I can obtain vectors for W-perp. So the concept is understood...I just can't seem to use the conditions appropriately...

thanks.

 

[Dick]

Write p(x) out as a polynomial. Your conditions are giving you information about the coefficients of the polynomial. What does p(0)=0 tell you?

 

[LCKurtz]

Consider P₅(R) together with innner product < p ,q > = ∫p(x)q(x) dx. Find W-perp if W = {p(x) $\in$ P₅(R) : p(0) = 0,p'(0) =0, p''(0) = 0}

Attempt: I am having trouble with the condition. I always have trouble with these conditions. SO as of now I am going to let q(x) be the standard basis of P₅(R). Now I don't know how to apply the condition to p(x).

After I do apply the condition I would take the inner product and have it set equal to 0. I should have a set of equations that I can solve in matrix form. This should produce some free variables from where I can obtain vectors for W-perp. So the concept is understood...I just can't seem to use the conditions appropriately...

thanks.

If $p(x) = a + bx + cx^2 + dx^3 + ex^4 +fx^5$, set $p(0)=0,p'(0)=0,p''(0)=0$. What does that tell you?

 

[trap101]

If $p(x) = a + bx + cx^2 + dx^3 + ex^4 +fx^5$, set $p(0)=0,p'(0)=0,p''(0)=0$. What does that tell you?

Well I set it up the other way so my "a" was with ax⁵. Well after doing that it turns out that d=e=f= 0 in order for the polynomial to satisfy the condition.

Doing that it turns into p(x) = ax⁵+bx⁴+cx³

now I have to multiply this by q(x) and then integrate. Is there any substitution I could possibly do or am I just going to have to literally multiply it out and integrate each component...if that's the case wouldn't that just be mean?

 

[Dick]

Well I set it up the other way so my "a" was with ax⁵. Well after doing that it turns out that d=e=f= 0 in order for the polynomial to satisfy the condition.

Doing that it turns into p(x) = ax⁵+bx⁴+cx³

now I have to multiply this by q(x) and then integrate. Is there any substitution I could possibly do or am I just going to have to literally multiply it out and integrate each component...if that's the case wouldn't that just be mean?

It might seem a little more manageable if you notice a basis for W is {x^5,x^4,x^3}, so q(x) is in W-perp if it's perpendicular to each vector in the basis.