# What is the use of inner products on the inner product space of polynomials?

In inner product spaces of polynomials, what is the point of finding the angle and distance between two polynomials? How does the distance and angle relate back to the polynomial?

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That would depend on the inner product you choose. – tomasz Sep 8 '12 at 12:14
Moreover: you can also upvote the answer. :-) – a.r. Sep 9 '12 at 19:18

Many spaces of functions are naturally equipped with an inner product such as $\langle f, g\rangle =\int_a^b f\bar g$. When you want to approximate an element of an inner product space by a linear combination of some simpler elements (such as low-degree polynomials) $f_1,\dots,f_n$ it helps to have the angles between $f_i,f_j$ bounded away from $0$ and $\pi$. (Ideally, all angles will be $\pi/2$, which means you have an orthogonal set of functions).
To see why, try working in $\mathbb R^2$ with the basis of $v_1=(1,0)$ and $v_2=(0.999, 0.001)$. For example, the expansion of the vector $(1,1)$ in this basis is $(1,1)=-998v_1+1000v_2$. This is just a recipe for the loss of numerical precision and all kinds of ugliness in error estimates.
Unfortunately, the "obvious" choices for basis elements such as monomials $x^n$ on the interval $[0,1]$ have small angles with respect to the aforementioned inner product. For example, the angle between $x^9$ and $x^{10}$ is about $0.05$ radian. It's better to work with Legendre polynomials instead, which are orthogonal.