I have three sets to determine whether they're orthonormal or not. These are;
(a) $\{1,\frac{x}{\sqrt2}+\frac{x^3}{\sqrt2},\frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}$}
(b) $\{\frac{x}{2}-\frac{x^2}{2}-\frac{x^3}{2}, -\frac{x}{2}-\frac{x^2}{2}+\frac{x^4}{2}, \frac{x^2}{2}-\frac{x^3}{2}$}
(c) $\{\frac{x^3}{3}-\frac{2x^4}{3}+\frac{2x^6}{3}, \frac{x^2}{3}+\frac{2x^5}{3}-\frac{2x^7}{3}$}
The inner product for two arbitrary polynomials is
$$a.b= \sum_{k=0}^n a_kb_k$$
I know that to prove that they are I need to do the dot product of each pair and for them to all equal 0 and that the magnitude of each should equal 1 for the sets to be determined as orthonormal but I'm having trouble starting off with this exercise, the x's are throwing me off and I'm wondering if there's something else I have to do before working out the magnitude and scalar product. Thanks in advance.
Hint: just convert the polynomial into vectors. For the first set, the highest degree of the polynomials is 4 so you need at least 5 coordinates to take into account all the non-zero coefficients: $$1\rightarrow (1,0,0,0,0), \frac{x}{\sqrt2}+\frac{x^3}{\sqrt2}\rightarrow (0,1/\sqrt2,0,1/\sqrt2,0), \frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}\rightarrow (0,0,1/\sqrt2,0,1/\sqrt2)$$ then take the usual scalar product in $\mathbb{R}^5$. Now $$\begin{align*} &(1,0,0,0,0)\cdot(1,0,0,0,0)=1 \rightarrow\mbox{it is a unit vector}\\ &(1,0,0,0,0)\cdot(0,1/\sqrt2,0,1/\sqrt2,0)=0 \rightarrow\mbox{they are orthogonal} \end{align*}$$ and so on. Now are you able to start your exercise?