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Function Space

Module by: Justin Romberg. E-mail the author

Summary: This module gives an example on function space.

We can also find basis vectors for vector spaces other than Rn n .

Let PnPn be the vector space of n-th order polynomials on (-1, 1) with real coefficients (verify P2P2 is a v.s. at home).

Example 1

P2P2 = {all quadratic polynomials}. Let b0 t=1 b0 t 1 , b1 t=t b1 t t , b2 t=t2 b2 t t 2 .

b0 t b1 t b2 t b0 t b1 t b2 t span P2P2, i.e. you can write any ft P2 f t P2 as ft= α0 b0 t+ α1 b1 t+ α2 b2 t f t α0 b0 t α1 b1 t α2 b2 t for some αi R αi .

Note:

P2P2 is 3 dimensional.
ft=t23t4 f t t 2 3 t 4

Alternate basis b0 t b1 t b2 t=1t12(3t21) b0 t b1 t b2 t 1 t 1 2 3 t 2 1 write ft f t in terms of this new basis d0 t= b0 t d0 t b0 t , d1 t= b1 t d1 t b1 t , d2 t=32 b2 t12 b0 t d2 t 3 2 b2 t 1 2 b0 t . ft=t23t4=4 b0 t3 b1 t+ b2 t f t t 2 3 t 4 4 b0 t 3 b1 t b2 t ft= β0 d0 t+ β1 d1 t+ β2 d2 t= β0 b0 t+ β1 b1 t+ β2 (32 b2 t12 b0 t) f t β0 d0 t β1 d1 t β2 d2 t β0 b0 t β1 b1 t β2 3 2 b2 t 1 2 b0 t ft= β0 b0 t+ β1 b1 t+32 β2 b2 t f t β0 1 2 b0 t β1 b1 t 3 2 β2 b2 t so β0 12=4 β0 1 2 4 β1 =-3 β1 -3 32 β2 =1 3 2 β2 1 then we get ft=4.5 d0 t3 d1 t+23 d2 t f t 4.5 d0 t 3 d1 t 2 3 d2 t

Example 2

ej ω0 nt|n= n ω0 n t is a basis for L2 0T L2 0 T , T=2π ω0 T 2 ω0 , ft=n Cn ej ω0 nt f t n Cn ω0 n t .

We calculate the expansion coefficients with

"change of basis" formula

Cn =1T0T(fte(j ω0 nt))dt Cn 1 T t 0 T f t ω0 n t
(1)

Note:

There are an infinite number of elements in the basis set, that means L2 0T L2 0 T is infinite dimensional (scary!).
Infinite-dimensional spaces are hard to visualize. We can get a handle on the intuition by recognizing they share many of the same mathematical properties with finite dimensional spaces. Many concepts apply to both (like "basis expansion"). Some don't (change of basis isn't a nice matrix formula).

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