Question

1 point) use the inner product ⟨f,g⟩=∫10f(x)g(x)dx ⟨f,g⟩=∫01f(x)g(x)dx in the vector space c0[0,1]c0[0,1] of continuous functions on the domain [0,1][0,1] to find ⟨f,g⟩⟨f,g⟩, ‖f‖‖f‖, ‖g‖‖g‖, and the angle αf,gαf,g between f(x)f(x) and g(x)g(x) for f(x)=10x2−3 and g(x)=6x−9.

Answer 1

$f(x)=10x^2-3$
$g(x)=6x-9$

$\langle f,g\rangle=\displaystyle\int_0^1(10x^2-3)(6x-9)\,\mathrm dx=3$

$\|f\|=\sqrt{\langle f,f\rangle}=\sqrt9=3$

$\|g\|=\sqrt{\langle g,g\rangle}=\sqrt{39}$

$\cos\alpha=\dfrac{\langle f,g\rangle}{\|f\|\|g\|}=\dfrac3{9\sqrt39}=\dfrac1{3\sqrt{39}}\implies \alpha=\cos^{-1}\dfrac1{3\sqrt{39}}$