Producte escalar i normes

Calcular el producte escalar i les normes de les funcions $f(x) = x + 1$ i $g(x) = x^2 + x$ en l’espai lineal $C[0,1]$, amb el producte escalar definit com $\langle f, g \rangle = \int_0^1 f(x)g(x) \, dx$ i la norma com $|f| = \sqrt{\langle f, f \rangle}$.

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1. Producte escalar $\langle f, g \rangle$

$$\langle f, g \rangle = \int_0^1 f(x)g(x) \, dx = \int_0^1 (x + 1)(x^2 + x) \, dx$$

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

$$(x + 1)(x^2 + x) = x \cdot x^2 + x \cdot x + 1 \cdot x^2 + 1 \cdot x = x^3 + x^2 + x^2 + x = x^3 + 2x^2 + x$$

Ara integrem:

$$\int_0^1 (x^3 + 2x^2 + x) \, dx = \int_0^1 x^3 \, dx + \int_0^1 2x^2 \, dx + \int_0^1 x \, dx$$

Calculem cada terme:

• $\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} – 0 = \frac{1}{4}$
• $\int_0^1 2x^2 \, dx = 2 \int_0^1 x^2 \, dx = 2 \cdot \left[ \frac{x^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3}$
• $\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} – 0 = \frac{1}{2}$

Sumem:

$$\frac{1}{4} + \frac{2}{3} + \frac{1}{2}$$

Amb denominador comú 12:

• $\frac{1}{4} = \frac{3}{12}$
• $\frac{2}{3} = \frac{8}{12}$
• $\frac{1}{2} = \frac{6}{12}$

$$\frac{3}{12} + \frac{8}{12} + \frac{6}{12} = \frac{3 + 8 + 6}{12} = \frac{17}{12}$$

Per tant, $\langle f, g \rangle = \frac{17}{12}$.

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2. Norma de $f$, $|f|$

$$|f| = \sqrt{\langle f, f \rangle} = \sqrt{\int_0^1 [f(x)]^2 \, dx} = \sqrt{\int_0^1 (x + 1)^2 \, dx}$$

Calculem $(x + 1)^2$:

$$(x + 1)^2 = x^2 + 2x + 1$$

Integrem:

$$\int_0^1 (x^2 + 2x + 1) \, dx = \int_0^1 x^2 \, dx + \int_0^1 2x \, dx + \int_0^1 1 \, dx$$

• $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$
• $\int_0^1 2x \, dx = 2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = 2 \cdot \frac{1}{2} = 1$
• $\int_0^1 1 \, dx = \left[ x \right]_0^1 = 1 – 0 = 1$

Sumem:

$$\frac{1}{3} + 1 + 1 = \frac{1}{3} + \frac{3}{3} + \frac{3}{3} = \frac{7}{3}$$

La norma és:

$$|f| = \sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} = \frac{\sqrt{21}}{3}$$

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3. Norma de $g$, $|g|$

$$|g| = \sqrt{\langle g, g \rangle} = \sqrt{\int_0^1 [g(x)]^2 \, dx} = \sqrt{\int_0^1 (x^2 + x)^2 \, dx}$$

Calculem $(x^2 + x)^2$:

$$(x^2 + x)^2 = x^4 + 2x^3 + x^2$$

Integrem:

$$\int_0^1 (x^4 + 2x^3 + x^2) \, dx = \int_0^1 x^4 \, dx + \int_0^1 2x^3 \, dx + \int_0^1 x^2 \, dx$$

• $\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}$
• $\int_0^1 2x^3 \, dx = 2 \cdot \left[ \frac{x^4}{4} \right]_0^1 = 2 \cdot \frac{1}{4} = \frac{1}{2}$
• $\int_0^1 x^2 \, dx = \frac{1}{3}$

Sumem:

$$\frac{1}{5} + \frac{1}{2} + \frac{1}{3}$$

Amb denominador comú $30$:

• $\frac{1}{5} = \frac{6}{30}$
• $\frac{1}{2} = \frac{15}{30}$
• $\frac{1}{3} = \frac{10}{30}$

$$\frac{6}{30} + \frac{15}{30} + \frac{10}{30} = \frac{31}{30}$$

La norma és:

$$|g| = \sqrt{\frac{31}{30}} = \frac{\sqrt{31}}{\sqrt{30}}$$

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Resposta final

• Producte escalar: $\langle f, g \rangle = \frac{17}{12}$
• Norma de $f$: $|f| = \frac{\sqrt{21}}{3}$
• Norma de $g$: $|g| = \frac{\sqrt{31}}{\sqrt{30}}$