Real Polynomials in inner product space projections?

Question

Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define $$ \langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1). $$

(i) Prove that $\langle\cdot,\cdot\rangle$ defines an inner product.

(ii) Find the angle between the polynomials given by $p(t)=-4+9t-6t^2+t^3$ and $q(t)=3-6t+3t^2$.

Out of curious from looking at this question:

Curiosity Question: (iii) If we define $\langle\cdot,\cdot\rangle$ on the space of all continuous functions I can see that it isn’t an inner product. right? Can someone explain to me why not though? I know we can probably still use the projection formula to project a continuous function onto the subspace V.

Can you show me how to project $\sqrt{t}$ onto $V$ to obtain a cubic polynomial that is “near” $\sqrt{t}$.

I saw this in a numerical analysis book which explained a little about these things a couple days ago. I think it was by Isaacson or Keller. This question came up. From reading, I know some things about this problem as you can probably tell from my curiousity question. I have come to the point where I am frustrated trying to solve this problem though. I am trying to self teach myself. Can someone please help me?

Answer 1

Assume

$$ \sqrt{t}=a+bt+ct^2+dt^3 \longrightarrow(*) $$

We need to find the coefficients $a,b,c,d$. This can be achieved by taking the inner product of both sides of $(*)$ with the functions $1,t,t^2,t^3$ will result in a system of four equations which solving them gives the desired coefficients. For instance,

$$ <\sqrt{t},1>=< a+bt+ct^2+dt^3, 1 >. $$

Now, just use the definition of the inner product you have given and the linearity property of the inner product to evaluate the above.