Understanding a Gram-Schmidt example Here's the thing: my textbook has an example of using the Gram Schmidt process with an integral. It is stated thus: 
Let $V = P(R)$ with the inner product $\langle f(x), g(x) \rangle = \int^{1}_{-1}f(t)g(t)dt$.
Consider the subspace $P_2(R)$ with the standard ordered basis $\beta$. We use the Gram Schmidt process to replace $\beta$ by an orthogonal basis {$v_1, v_2, v_3$} for $P_2(R)$ and the use this orthogonal to obtain an orthonormal basis for $P_2(R)$. 
OK, so far so good. But here's where I have trouble following. The text says: 
take $v_1 = 1$. (This is presumably the first part of the basis). 
Then it says: $\rVert v_1\rVert^2= \int^1_{-1}1^2dt = 2$ and $\langle x,v_1 \rangle = \int^1_{-1}t \dot\ 1\ dt = 0$
and that's where I am a little lost. If the standard ordered basis they are talking about here is {$1, x, x^2$} then it makes a bit more sense but I am just trying to follow the steps. Is that what is happening here? Something else? 
 A: You're correct. The standard basis for $P_2(R)$ is $\beta = \{1,x,x^2\}$, as you said, and the author is starting with $\beta$ and following the Gram-Schmidt algorithm to generate an orthogonal basis.
In reply to your comment:
Let me answer your questions about the inner product first. The author has introduced an important type of inner product here. Since the elements of $P_2(R)$ are integrable functions, we can define the inner product of two vectors $f(x),g(x) \in P_2(R)$ as
$$
    \langle f(x),g(x) \rangle = \int_{-1}^1 f(t)g(t) \,dt.
$$
So for example, the inner product of $v = 1-3x + 2x^2$ and $w = 5 + 8x$ is
\begin{align}
    \langle 1-3x + 2x^2, 5 + 8x \rangle &= \int_{-1}^1 (1 - 3t + 2t^2)(5 + 8t) \,dt \\
    &= \int_{-1}^1 5+t+2 t^2+160 t^3 \,dt = \frac{2}{3}
\end{align}
I'll leave it up to you to verify that this is actually an inner product (worth checking, if you haven't already).
(Also, in case you were wondering: the choice of bounds $\pm 1$ doesn't really matter here; if you choose any $a$ and $b$ with $-\infty < a < b < \infty$ you get a new inner product $\langle f(x), g(x)\rangle_{a,b} = \int_a^b f(t)g(t) dt$ on $P_2(R)$!)
So to answer one of your questions, yes, you just substitute $1$ for both $f(x)$ and $g(x)$ when computing $\|v_1\|$ ($=\|1\|$):
$$
    \|v_1\| = \langle v_1,v_1\rangle^{1/2} = \langle1,1\rangle^{1/2}
            = \left(\int_{-1}^1 1 \cdot 1 \,dt\right)^{1/2}.
$$
Summary:
Following the Gram-Schmidt process to generate an orthogonal basis $\{v_1,v_2,v_3\}$ from the ordered basis $\beta = \{1,x,x^2\}$, we obtain the following vectors:
\begin{align}
    v_1 &= 1 \text{ (since $1$ = the first vector in $\beta$)}    \\
    v_2 &= x - \text{proj}_{v_1}(x) = x - \text{proj}_{1}(x) \\
        &= x - \frac{\langle x, 1 \rangle}{\langle 1, 1 \rangle} \cdot 1 = x - \frac{0}{2} \cdot 1 = x \\
    v_3 &= x^2 - \text{proj}_{v_1}(x^2) - \text{proj}_{v_2}(x^2) \\
        &= x^2 - \frac{
                     \langle x^2, 1 \rangle
                 }{
                     \langle 1, 1 \rangle
                 } \cdot 1
               - \frac{
                     \langle x^2, x \rangle
                 }{
                     \langle x, x \rangle
                 } \cdot x \\
        &= x^2 - \frac{
                     \int_{-1}^1 t^2 \cdot 1 \,dt
                 }{
                     \int_{-1}^1 1 \cdot 1 \,dt
                 } \cdot 1
               - \frac{
                     \int_{-1}^1 t^2 \cdot t \,dt
                 }{
                     \int_{-1}^1 t \cdot t \,dt
                 } \cdot x
         = x^2 - \frac{2/3}{2} \cdot 1 - \frac{0}{2/3} \cdot x = x^2 - \frac{1}{3}.
\end{align}
In other words, our new orthogonal basis is
$$
    B = \{v_1,v_2,v_3\}.
$$
If you haven't before, it's worth thinking through why this gives you an orthogonal basis (and validating your conceptual understanding by confirming that $\langle v_i, v_j \rangle = 0$ for $i \neq j$.
We've gotten through the labor-intensive part of Gram-Schmidt.
Since $B$ is an orthogonal basis, we can arrive at an orthonormal basis $\{u_1, u_2, u_3\}$ by simply dividing each of the vectors $v_i$ by its norm $\|v_i\| = \sqrt{\langle v_i, v_i \rangle}$.
This is easily done after having calculated $B$, since you can reduce calculating $\|v_i\|$ into taking the square root of a sum of mostly known inner product values.
The resulting orthonormal basis $\{u_1, u_2, u_3\}$ is
\begin{align*}
    u_1 &= v_1/\|v_1\| = 1/\|1\| = 1/\sqrt{2} \\
    u_2 &= v_2/\|v_2\| = x/\|x\| = x/\left(\int_{-1}^1 t^2 \,dt\right)^{1/2} = \frac{1}{\sqrt{2/3}}x = \sqrt{\frac{3}{2}}x \\
    u_3 &= v_3/\|v_3\| = \frac{x^2 - 1/3}{\left(\int_{-1}^1 (t^2 - 1/3)^2 \,dt\right)^{1/2}} = \frac{3\sqrt{5}}{2\sqrt{2}} \left(x^2 - \frac{1}{3}\right) = \frac{3\sqrt{5}}{2\sqrt{2}} x^2 - \frac{\sqrt{5}}{2\sqrt{2}}
\end{align*}
Edit: As @lojle and @RayBern both noticed, I accidentally omitted a projection term when calculating $v_3$. I've corrected the error and revised the surrounding commentary accordingly.
A: Previous Dan's answer was very nice and he explained it well, but he missed out a detail which could potentially confuse many future readers, so I decided to do the correction.
When you use Gram-Schmidt algorithm to find the (ortho)normal basis $\left\{e_1, e_2, e_3\dots\right\}$ from the standard basis $\beta = \left\{v_1,v_2,v_3\dots\right\}$, you must take care to take away components of all the vectors $e_i$ that you have already found. So in this example we would have:
$v_1' = 1\quad$  (first vector from $\beta$, not yet normalized)
To normalize it we have to use our inner product definition: $\langle f(x), g(x) \rangle = \int^{1}_{-1}f(t)g(t)dt \quad$ and: $\Vert x\Vert^2 = \langle x, x\rangle$
$$e_1 = \frac{v_1'}{\sqrt{\langle v_1', v_1'\rangle}} = \frac1{\sqrt{\int^1_{-1}1dt}} = \frac1{\sqrt2}$$
$$v_2' = v_2 - \langle e_1, v_2\rangle \ e_1 = t - \frac1{\sqrt2} \ \int^1_{-1}\frac1{\sqrt2}tdt = t$$
$$\Rightarrow e_2 = \frac{t}{\sqrt{\int^1_{-1}t^2dt}} = \sqrt{\frac32}\ t$$
Now comes the important bit. We must take away both $e_1$ and $e_2$ components from $v_3'$.
$$v_3' = v_3 - \langle e_1, v_3\rangle \ e_1 - \langle e_2, v_3\rangle \ e_2$$
$$\Rightarrow v_3' = t^2 -\frac12 \int^1_{-1}t^2dt - \frac32 t\int^1_{-1}t^3dt = t^2 - \frac13$$
Since $$\int^1_{-1}\left(t^2-\frac13\right)^2 = \frac8{45} \Rightarrow\quad e_3 = \sqrt{\frac{45}8} \left(t^2 - \frac13\right)$$
So you get a new basis diferent from $\beta$. You could also use the general formula for finding new orthonormal basis vectors. Suppose you already found $e_1\dots e_i$ vectors. To find $i+1$-st vector you would use:
$$e_{i+1} = v_{i+1} - \sum_{k=1}^i \langle e_k, v_{i+1} \rangle e_k$$
