Which of the following sets are orthonormal? 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.
 A: 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?
A: In polynomial space, the standard orthonormal basis (with respect to the inner-product you gave above) is $\{1, x, x^2, x^3, ...\}$ and so, for example, a polynomial such as $2 + 3x^2 - 4x^4$ can be represented by the vector $(2, 0, 3, 0, -4, 0, ...)$ since
$$2\cdot 1 + 0 \cdot x + 3 \cdot x^2 + 0 \cdot x^3 + (-4) \cdot x^4 = 2 + 3x^2 - 4x^4.$$
In your case, for (a), you then have the identifications:
$$
\left\{1,\frac{x}{\sqrt2}+\frac{x^3}{\sqrt2},\frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}\right\} \longrightarrow \left\{ \begin{pmatrix}1 \\0\\0\\0\\0\\ 0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\1/\sqrt{2}\\0\\1/\sqrt{2}\\0\\0\\\vdots\end{pmatrix}, \begin{pmatrix} 0 \\0\\1/\sqrt{2}\\0\\-1/\sqrt{2}\\ 0\\\vdots\end{pmatrix} \right\}
$$
Now that you have these understood componentwise, you can use the inner-product you have given.
