Funcions ortogonals

Donat l’espai euclidià $C[0, \pi]$ amb el producte escalar $\langle f, g \rangle = \int_0^\pi f(x)g(x) \, dx$, hem de trobar $a \in \mathbb{R}$ tal que les funcions $f(x) = 3\sin x + 2\cos x$ i $g(x) = \sin x + a\cos x$ siguin ortogonals, és a dir, $\langle f, g \rangle = 0$. Després, hem de normalitzar $f(x)$ i $g(x)$.

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1. Condició d’ortogonalitat

Perquè siguin ortogonals, cal que $\langle f, g \rangle = 0$:

$$\langle f, g \rangle = \int_0^\pi f(x)g(x) \, dx = \int_0^\pi (3\sin x + 2\cos x)(\sin x + a\cos x) \, dx = 0$$

Calculem $f(x)g(x)$:

$$(3\sin x + 2\cos x)(\sin x + a\cos x) = 3\sin x \cdot \sin x + 3\sin x \cdot a\cos x + 2\cos x \cdot \sin x + 2\cos x \cdot a\cos x$$

$$= 3\sin^2 x + 3a\sin x \cos x + 2\sin x \cos x + 2a\cos^2 x$$

Ara integrem:

$$\langle f, g \rangle = \int_0^\pi (3\sin^2 x + 3a\sin x \cos x + 2\sin x \cos x + 2a\cos^2 x) \, dx = 0$$

Separem la integral:

$$3 \int_0^\pi \sin^2 x \, dx + 3a \int_0^\pi \sin x \cos x \, dx + 2 \int_0^\pi \sin x \cos x \, dx + 2a \int_0^\pi \cos^2 x \, dx = 0$$

Calculem cada terme:

• $\int_0^\pi \sin^2 x \, dx ): Usem ( \sin^2 x = \frac{1 – \cos 2x}{2}$:

$$\int_0^\pi \sin^2 x \, dx = \int_0^\pi \frac{1 – \cos 2x}{2} \, dx = \frac{1}{2} \left[ x – \frac{\sin 2x}{2} \right]_0^\pi = \frac{1}{2} \left( \pi – \frac{\sin 2\pi}{2} \right) = \frac{\pi}{2}$$

• $\int_0^\pi \cos^2 x \, dx$: Usem $\cos^2 x = \frac{1 + \cos 2x}{2}$:

$$\int_0^\pi \cos^2 x \, dx = \int_0^\pi \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_0^\pi = \frac{1}{2} \left( \pi + \frac{\sin 2\pi}{2} \right) = \frac{\pi}{2}$$

• $\int_0^\pi \sin x \cos x \, dx$: Usem $\sin x \cos x = \frac{1}{2} \sin 2x$:

$$\int_0^\pi \sin x \cos x \, dx = \frac{1}{2} \int_0^\pi \sin 2x \, dx = \frac{1}{2} \left[ -\frac{1}{2} \cos 2x \right]_0^\pi = \frac{1}{2} \left( -\frac{1}{2} \cos 2\pi + \frac{1}{2} \cos 0 \right) = \frac{1}{2} \left( -\frac{1}{2} + \frac{1}{2} \right) = 0$$

Substituint:

$$3 \cdot \frac{\pi}{2} + 3a \cdot 0 + 2 \cdot 0 + 2a \cdot \frac{\pi}{2} = 0$$

$$\frac{3\pi}{2} + a\pi = 0 \implies a\pi = -\frac{3\pi}{2} \implies a = -\frac{3}{2}$$

Per tant, $a = -\frac{3}{2}$.

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2. Normalitzem $f(x)$

La norma és $|f| = \sqrt{\langle f, f \rangle}$:

$$\langle f, f \rangle = \int_0^\pi (3\sin x + 2\cos x)^2 \, dx$$

$$(3\sin x + 2\cos x)^2 = 9\sin^2 x + 12\sin x \cos x + 4\cos^2 x$$

$$\langle f, f \rangle = 9 \int_0^\pi \sin^2 x \, dx + 12 \int_0^\pi \sin x \cos x \, dx + 4 \int_0^\pi \cos^2 x \, dx$$

$$9 \cdot \frac{\pi}{2} + 12 \cdot 0 + 4 \cdot \frac{\pi}{2} = \frac{9\pi}{2} + \frac{4\pi}{2} = \frac{13\pi}{2}$$

$$|f| = \sqrt{\frac{13\pi}{2}}$$

La funció normalitzada és $\frac{f(x)}{|f|}$:

$$f_{\text{norm}}(x) = \frac{3\sin x + 2\cos x}{\sqrt{\frac{13\pi}{2}}} = \sqrt{\frac{2}{13\pi}} (3\sin x + 2\cos x)$$

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3. Normalitzem $g(x)$

Com que $a = -\frac{3}{2}$, tenim $g(x) = \sin x – \frac{3}{2}\cos x$:

$$\langle g, g \rangle = \int_0^\pi \left( \sin x – \frac{3}{2}\cos x \right)^2 \, dx$$

$$\left( \sin x – \frac{3}{2}\cos x \right)^2 = \sin^2 x – 3\sin x \cos x + \frac{9}{4}\cos^2 x$$

$$\langle g, g \rangle = \int_0^\pi \sin^2 x \, dx – 3 \int_0^\pi \sin x \cos x \, dx + \frac{9}{4} \int_0^\pi \cos^2 x \, dx$$

$$\frac{\pi}{2} – 3 \cdot 0 + \frac{9}{4} \cdot \frac{\pi}{2} = \frac{\pi}{2} + \frac{9\pi}{8} = \frac{13\pi}{8}$$

$$|g| = \sqrt{\frac{13\pi}{8}}$$

La funció normalitzada és:

$$g_{\text{norm}}(x) = \frac{\sin x – \frac{3}{2}\cos x}{\sqrt{\frac{13\pi}{8}}} = \sqrt{\frac{8}{13\pi}} \left( \sin x – \frac{3}{2}\cos x \right)$$

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Resposta final

• $a = -\frac{3}{2}$
• Funció normalitzada $f$: $f_{\text{norm}}(x) = \sqrt{\frac{2}{13\pi}} (3\sin x + 2\cos x)$
• Funció normalitzada $g$: $g_{\text{norm}}(x) = \sqrt{\frac{8}{13\pi}} \left( \sin x – \frac{3}{2}\cos x \right)$