Λύση

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Έχουμε ότι

$\begin{displaymath}(f,g)=\int_0^1(2t^3-t^2)dt=\left[\frac{t^4}{2}-\frac{t^3}{3}\right]_0^1=\frac{1}{6}\end{displaymath}$

$\begin{displaymath}\Vert f\Vert^2=(f,f)=\int_0^1(4t^2-4t+1)dt=\frac{1}{3}\end{displaymath}$

$\begin{displaymath}\Vert g\Vert^2=(g,g)=\int_0^1t^4dt=\frac{1}{5}.\end{displaymath}$

Άρα

$\begin{displaymath}\cos\theta =\frac{\frac{1}{6}}{\frac{1}{\sqrt{3}}\frac{1}{\sqrt{5}}}=\frac{\sqrt{15}}{6}.\end{displaymath}$

Vassilis Metaftsis
1999-09-15