15.9 Function space

$P_2$ = {all quadratic polynomials}. Let $b_0(t)=1$ , $b_1(t)=t$ , $b_2(t)=t^2$ .

$\{b_0(t), b_1(t), b_2(t)\}$ span $P_2$ , i.e. you can write any $f(t)\in P_2$ as $$f(t)={\alpha }_0{b}_0(t)+{\alpha }_1{b}_1(t)+{\alpha }_2{b}_2(t)$$ for some ${\alpha }_i\in \mathbb{R}$ .

$$f(t)=t^2-3t-4$$

Alternate basis $$\{b_0(t), b_1(t), b_2(t)\}=\{1, t, \frac{1}{2}(3t^2-1)\}$$ write $f(t)$ in terms of this new basis $d_0(t)=b_0(t)$ , $d_1(t)=b_1(t)$ , $d_2(t)=\frac{3}{2}{b}_2(t)-\frac{1}{2}{b}_0(t)$ . $$f(t)=t^2-3t-4=4b_0(t)-3b_1(t)+b_2(t)$$ $$f(t)={\beta }_0{d}_0(t)+{\beta }_1{d}_1(t)+{\beta }_2{d}_2(t)={\beta }_0{b}_0(t)+{\beta }_1{b}_1(t)+{\beta }_2(\frac{3}{2}{b}_2(t)-\frac{1}{2}{b}_0(t))$$ $$f(t)={\beta }_0{b}_0(t)+{\beta }_1{b}_1(t)+\frac{3}{2}{\beta }_2{b}_2(t)$$ so $${\beta }_0-\frac{1}{2}=4$$ $${\beta }_1=-3$$ $$\frac{3}{2}{\beta }_2=1$$ then we get $$f(t)=4.5d_0(t)-3d_1(t)+\frac{2}{3}{d}_2(t)$$

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$(n, )$∞ ∞ ω 0 n t is a basis for $L^2()$

0 T

, $T=\frac{2\pi }{{\omega }_0}$ , $f(t)=\sum C_{n}e^{i{\omega }_{0}nt}$ .

We calculate the expansion coefficients with

Infinite-dimensional spaces are hard to visualize. We can get a handle on the intuition by recognizingthey 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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