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15.10: Espacio de función

  • Page ID
    86590
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    También podemos encontrar vectores base (Sección 15.9) para espacios vectoriales (Sección 15.2) distintos a\(\mathbb{R}^n\).

    Dejar\(P_n\) ser el espacio vectorial de polinomios de orden\(n\) -ésimo en (-1, 1) con coeficientes reales (verificar\(P_2\) es un v.s. en casa).

    Ejemplo\(\PageIndex{1}\)

    \(P_{2}=\{\text { all quadratic polynomials }\}\). Vamos\(b_0(t)=1\),\(b_1(t)=t\),\(b_2(t)=t^2\).

    \(\left\{b_{0}(t), b_{1}(t), b_{2}(t)\right\}\)span\(P_2\), es decir, puede escribir cualquiera\(f(t) \in P_2\) como

    \[f(t)=\alpha_{0} b_{0}(t)+\alpha_{1} b_{1}(t)+\alpha_{2} b_{2}(t) \nonumber \]

    para algunos\(\alpha_{i} \in \mathbb{R}\).

    Nota

    \(P_2\)es de 3 dimensiones.

    \[f(t)=t^2−3t−4 \nonumber \]

    Bases alternas

    \[\left\{b_{0}(t), b_{1}(t), b_{2}(t)\right\}=\left\{1, t, \frac{1}{2}\left(3 t^{2}-1\right)\right\} \nonumber \]

    escribir\(f(t)\) en términos de esta nueva base\(d_0(t)=b_0(t)\),\(d_1(t)=b_1(t)\),\(d_{2}(t)=\frac{3}{2} b_{2}(t)-\frac{1}{2} b_{0}(t)\).

    \ [\ begin {array} {c}
    f (t) =t^ {2} -3 t-4=4 b_ {0} (t) -3 b_ {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}\ izquierda (\ frac {3} {2} b_ {2} (t) -\ frac {1} {2} b_ {0} (t)\ derecha)\
    f (t) =\ beta_ {0} b_ {0} (t) +\ beta_ {1 } b_ {1} (t) +\ frac {3} {2}\ beta_ {2} b_ {2} (t)
    \ end {array}\ nonumber\]

    por lo

    \ [\ begin {array} {c}
    \ beta_ {0} -\ frac {1} {2} =4\\
    \ beta_ {1} =-3\\
    \ frac {3} {2}\ beta_ {2} =1
    \ end {array}\ nonumber\]

    entonces conseguimos

    \[f(t)=4.5 d_{0}(t)-3 d_{1}(t)+\frac{2}{3} d_{2}(t) \nonumber \]

    Ejemplo\(\PageIndex{2}\)

    \(\left.e^{j \omega_{0} n t}\right|_{n=-\infty} ^{\infty}\)es una base para\(L^2([0,T])\),\(T=\frac{2 \pi}{\omega_0}\),\(f(t)=\sum_{n} C_{n} e^{j \omega_{0} n t}\).

    Calculamos los coeficientes de expansión con

    Fórmula de “cambio de base”:

    \[C_{n}=\frac{1}{T} \int_{0}^{T}\left(f(t) e^{-\left(j \omega_{0} n t\right)}\right) d t \nonumber \]

    Nota

    Hay un número infinito de elementos en el conjunto de bases, eso significa que\(L^2([0,T])\) es infinito dimensional (¡da miedo!).

    Los espacios infinito-dimensionales son difíciles de visualizar. Podemos manejar la intuición reconociendo que comparten muchas de las mismas propiedades matemáticas con espacios dimensionales finitos. Muchos conceptos se aplican a ambos (como “expansión de base”). Algunos no (el cambio de base no es una buena fórmula matricial).


    This page titled 15.10: Espacio de función is shared under a CC BY license and was authored, remixed, and/or curated by Richard Baraniuk et al..