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Function Space

Module by: Justin Romberg. E-mail the author

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Summary: This module gives an example on function space.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

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 b0t=1 b0 t 1 , b1t=t b1 t t , b2t=t2 b2 t t 2 .

b0tb1tb2t b0 t b1 t b2 t span P2P2, i.e. you can write any ft∈P2 f t P2 as ft=α0b0t+α1b1t+α2b2t f t α0 b0 t α1 b1 t α2 b2 t for some αi∈ℝ αi .

Note:

P2P2 is 3 dimensional.
ft=t2−3t−4 f t t 2 3 t 4

Alternate basis b0tb1tb2t=1t123t2−1 b0 t b1 t b2 t 1 t 1 2 3 t 2 1 write ft f t in terms of this new basis d0t=b0t d0 t b0 t , d1t=b1t d1 t b1 t , d2t=32b2t−12b0t d2 t 3 2 b2 t 1 2 b0 t . ft=t2−3t−4=4b0t−3b1t+b2t f t t 2 3 t 4 4 b0 t 3 b1 t b2 t ft=β0d0t+β1d1t+β2d2t=β0b0t+β1b1t+β232b2t−12b0t 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=β0−12b0t+β1b1t+32β2b2t 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 32β2=1 3 2 β2 1 then we get ft=4.5d0t−3d1t+23d2t f t 4.5 d0 t 3 d1 t 2 3 d2 t

Example 2

ⅇⅈω0nt|n=-∞∞ n ω0 n t is a basis for L20T L2 0 T , T=2πω0 T 2 ω0 , ft=∑nCnⅇⅈω0nt f t n Cn ω0 n t .

We calculate the expansion coefficients with

"change of basis" formula

Cn=1T∫0Tftⅇ-ⅈω0ntdt 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 L20T 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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