Given the inner product of two polynomials $p(X), q(X) \in P(d)$, where $P(d)$ is the vector space of all polynomials of degree less than or equal to d, with real coefficients, and using the inner product $$\langle p(x),q(X) \rangle = \int_{-1}^{1} p(X)q(X)dX$$

How can the angle $$\cos(\alpha)=\frac{\langle p(x),q(X) \rangle}{\|p(X)\|\|q(X)\|}$$

be interpreted geometrically?

  • $\begingroup$ Well, perhaps a more or less "easy" way is: remember that $\;P(d)\cong\Bbb R^{d+1}\;$ (assuming you meant real polynomials), so you can choose a basis (better: an orthonormal one) in your space and denote each vector (polynomial) in it by coordinates as in $\;\Bbb R^{d+1}\;$, and thus these coordinate vectors' angle is what you want ... $\endgroup$ – DonAntonio Jun 6 '14 at 15:03

The angle, the normalized inner-product, can be seen as a measure on the scale from 0 to 1 of how "orthogonal" (independent) or "parallel" (dependent) are the two polynomials.


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