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Some geometry: Cauchy-Schwarz inequality.

$\overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^3$ $f,g \in C[a,b]$
$\cos \angle \overrightarrow{x} ,\overrightarrow{y} =\frac {x_1y_1+x_2y_2+x_3y_3...
...+x_2^2+x_3^2} \cdot \sqrt{y_1^2+y_2^2+y_3^2}}
\quad \overset{?}{\leq} 1 \qquad$ $\qquad \cos \angle f,g =
\frac{\int_a^b f(x)g(x)dx}
{\sqrt{\int_a^b f(x)^2dx} \cdot \sqrt{\int_a^b g(x)^2dx}}
\quad \overset{?}{\leq} 1$

Question: Why is it possible to define $\cos \phi$ by the previous formulae? Answer: If the conditions for a function to be an inner product are fulfilled, then we can prove the Cauchy-Schwarz inequality:

Theorem 13.0.1   \fbox{$\forall \overrightarrow{x},\overrightarrow{y} \in V, \quad
\langle \over...
...htarrow{x}\end{Vmatrix} \cdot
\begin{Vmatrix}\overrightarrow{y}\end{Vmatrix}$ }.


\begin{proof}.
\par\begin{center}
$\forall \overrightarrow{x} ,\overrightarrow{y...
...tarrow{x} \vert \cdot \vert\overrightarrow{y} \vert$ }.
\end{center}\end{proof}

Corollary 13.0.2   $\forall \overrightarrow{x} ,\overrightarrow{y}\in V, \quad
\left\vert \frac{\l...
...trix}
\cdot \begin{Vmatrix}\overrightarrow{y}\end{Vmatrix}} \right\vert \leq 1$.



Noah Dana-Picard
2001-02-26