# solutions to 2.14

Exercise:
Let $$V$$ be the space of real polynomial functions of degree at most $$3$$. Consider the quadratic form $$q_1:V \to \mathbb{R}$$ given by $$q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2.$$ Is this form positive definite?
Consider another form $$q:V\to \mathbb{R}$$ given by $$q(f):=\int_{0}^\infty e^{-x} |f(x)|^2 dx$$ Diagonalize the form using Gram-Schmidt procedure starting with the standard monomial basis $$\{1,x,x^2,x^3\}$$.
Solution:
In order for $$q_1$$ be positive definite, it should vanish only for the zero polynomial, namly $$q_1(f)=0$$ if and only if $$f(x)=0$$. However, the polynomial $$g(x) = x(x-1)(x+1) \in V$$ is nonzero and yet has $$q_1(g) = |g(-1)|^2+|g(0)|^2+|g(1)|2 = 0^2+0^2+0^2 = 0,$$ hence $$q_1$$ is not positive definite on $$V$$.
For the second part, note the bi-linear form corresponding to the quadratic form: $$Q(f,g) = \int_{0}^\infty e^{-x} f(x)g(x) dx$$ and we will utilize the identity $$\int_0^\infty e^{-x} x^n = n!$$ for any integer $$n\geq 1$$.
Starting with the standard basis is $$f_k(x) = x^k$$, the first Gram-Schmidt vector is simply be \begin{align} e_0(x) & = f_0(x) = \boxed{1} \\ \Rightarrow ~~Q(e_0,f_k) &= \int_0^\infty e^{-1} x^k dx = k! \\ Q(e_0,e_0) &= \int_0^\infty e^{-1} 1 dx = 1 \end{align} The second Gram-Schmidt vector follows from that: \begin{align} e_1(x) &= f_1(x) – \frac{Q(e_0,f_1)}{Q(e_0,e_0)}e_0(x) = \boxed{x – 1}, \\ \Rightarrow ~~Q(e_1,e_1) &= \int_0^\infty e^{-1} (x-1)^2 dx = \int_0^\infty e^{-x}(x^2 – 2x +1) dx = 2!-2+1 = 1\\ Q(e_1,f_k) &= \int_0^\infty e^{-1} (x-1)x^k dx = (k+1)!-k! = k\cdot k! \end{align} The third vector: \begin{align} e_2(x) &= f_2(x) – \frac{Q(e_1,f_2)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_2)}{Q(e_0,e_0)}e_0(x) = x^2 – 4 (x-1) -2 = \boxed{x^2 -4x + 2}. \\ \Rightarrow ~~Q(e_2,e_2) &= \int_0^\infty e^{-1} (x^2-4x+2)^2 dx = \int_0^\infty e^{-1}(x^4-8x^3 +20x^2 -16x + 4)dx = 4 \\ Q(e_2,f_k) &= \int_0^\infty e^{-1} (x^2-4x+2)x^k dx = (k+2)!-4(k+1)! +2k! \end{align} And the last is: \begin{align} e_3(x) &= f_3(x) – \frac{Q(f_3,e_2)}{Q(e_2,e_2)}e_2(x)- \frac{Q(e_1,f_3)}{Q(e_1,e_1)}e_1(x)- \frac{Q(e_0,f_3)} {Q(e_0,e_0)}e_0(x) \\ &= x^3 – \frac{36}{4}(x^2-4x+2) – \frac{3\cdot 3!}{1}(x-1) – 3! = \boxed{x^3-9x^2 +18x -6} \end{align}
The diagonalized basis $$\{e_0=1,~e_1=x-1,~e_2= x^2-4x+2,~e_3=x^3-9x^2 +18x -6\}$$ is part of the renowned Laguerre polynomials, which are (among other things) orthogonal with respect to the weighted inner product defined by $$Q(f,g)$$ above. That is, for every polynomial $$p(x) = \sum_{k=0}^3 a_k e_k(x)$$ we have $$q(p) = \sum_{k=0}^3 \frac{1}{q(e_k)} a_k^2$$
Exercise: Find the signature of the quadratic form $$q_1(f) =|f(-1)|^2+|f(0)|^2+|f(1)|^2.$$ above.
Solution:
First note that being a sum of non-negative numbers, the form $$q_1(f)\geq 0$$ (i.e no vectors have negative form value) is non-negative definite (sometimes called positive semidefinite). Its signature has $$n_{-} = 0$$ Our next step is to find $$n_0$$ – the dimension of the nullspace of the quadratic form $$q_1$$. If $$q_1(f)=0$$ then it must have three roots at \ (x=\pm 1,0\), hence $$f(x) = h(x) x(x-1)(x+1).$$ However, as we require that $$f\in V$$, or $$\text{deg}(f) \leq 3$$, the factor $$h(x)$$ must be some real constant $$h(x) = \alpha$$, hence all nullvectors of $$q_1(f)=0$$ have the form $$f(x) = \alpha x(x-1)(x+1),$$ forming a one dimensional span, hence $$n_0 = 1$$. Lastly, we compute $$n_{+}$$: since $$\text{dim}(V) = 4 = n_{+} + n_0 + n_{-}$$ we have $$n_{+} = 3$$.
Note: the form $$q_1$$ is positive definite on the space of polynomials of degree $$\leq 2$$