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# 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 $f⁢t∈ P2 f t P2$ as $f⁢t= α0 ⁢ b0 ⁢t+ α1 ⁢ b1 ⁢t+ α2 ⁢ b2 ⁢t f t α0 b0 t α1 b1 t α2 b2 t$ for some $αi ∈R αi$.

Note:
$P2P2$ is 3 dimensional.
$f⁢t=t2−3⁢t−4 f t t 2 3 t 4$

Alternate basis $b0 ⁢t b1 ⁢t b2 ⁢t=1t12⁢(3⁢t2−1) b0 t b1 t b2 t 1 t 1 2 3 t 2 1$ write $f⁢t f t$ in terms of this new basis $d0 ⁢t= b0 ⁢t d0 t b0 t$, $d1 ⁢t= b1 ⁢t d1 t b1 t$, $d2 ⁢t=32⁢ b2 ⁢t−12⁢ b0 ⁢t d2 t 3 2 b2 t 1 2 b0 t$. $f⁢t=t2−3⁢t−4=4⁢ b0 ⁢t−3⁢ b1 ⁢t+ b2 ⁢t f t t 2 3 t 4 4 b0 t 3 b1 t b2 t$ $f⁢t= β0 ⁢ d0 ⁢t+ β1 ⁢ d1 ⁢t+ β2 ⁢ d2 ⁢t= β0 ⁢ b0 ⁢t+ β1 ⁢ b1 ⁢t+ β2 ⁢(32⁢ b2 ⁢t−12⁢ b0 ⁢t) f t β0 d0 t β1 d1 t β2 d2 t β0 b0 t β1 b1 t β2 3 2 b2 t 1 2 b0 t$ $f⁢t= β0 ⁢ b0 ⁢t+ β1 ⁢ b1 ⁢t+32⁢ β2 ⁢ b2 ⁢t f t β0 1 2 b0 t β1 b1 t 3 2 β2 b2 t$ so $β0 −12=4 β0 1 2 4$ $β1 =-3 β1 -3$ $32⁢ β2 =1 3 2 β2 1$ then we get $f⁢t=4.5⁢ d0 ⁢t−3⁢ d1 ⁢t+23⁢ d2 ⁢t f t 4.5 d0 t 3 d1 t 2 3 d2 t$

Example 2

$ei⁢ ω0 ⁢n⁢t|n=−∞∞ n ω0 n t$ is a basis for $L2 ⁢0T L2 0 T$, $T=2⁢π ω0 T 2 ω0$, $f⁢t=∑n Cn ⁢ei⁢ ω0 ⁢n⁢t f t n Cn ω0 n t$.

We calculate the expansion coefficients with

"change of basis" formula
$Cn =1T⁢∫0T(f⁢t⁢e−(i⁢ ω0 ⁢n⁢t))dt Cn 1 T t 0 T f t ω0 n t$
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).