MATH 414 Lecture 5

« previous | Friday, January 24, 2014 | next »

Hyperbolic Trigonometric Functions

$\sinh x=\frac{{\mathrm{e}}^x-{\mathrm{e}}^{-x}}{2}=i\sin (-ix)$

Gram-Schmidt Process

INPUT: basis vectors $U=\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_n\}$ for a vector space $V=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_n\}$.

OUTPUT: orthonormal basis $O=\{{\hat{e}}_1,{\hat{e}}_2,\dots ,{\hat{e}}_n\}$ for $V$

${\hat{e}}_1:=\frac{{\overrightarrow{v}}_1}{\Vert {\overrightarrow{v}}_1\Vert }$

$O:=\left\{{\hat{e}}_1\right\}$

for $i$ from $2$ to $n$

${\overrightarrow{p}}_i:=\sum _{k=1}^{i-1}⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩{\hat{e}}_k$

${\hat{e}}_i:=\frac{{\overrightarrow{v}}_i-{\overrightarrow{p}}_i}{\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert }$

$O:=O\cup \left\{{\hat{e}}_i\right\}$

Python Implementation

def gram_schmidt(basis):
    on_basis = [ normalize(basis[0]) ]
    for i in xrange(1, len(basis)):
        p = sum(inner_product(basis[i], e) * e for e in on_basis)
        next_e = normalize(basis[i] - p)
        on_basis.append(next_e)
    return on_basis

Trick

Calculating $\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert =⟨{\overrightarrow{v}}_i-{\overrightarrow{p}}_i,{\overrightarrow{v}}_i-{\overrightarrow{p}}_i⟩$; in $L^2[a,b]$ space, $\underset{a}{\overset{b}{\int }}{(v_i\left(x\right)-p_i\left(x\right))}^2\mathrm{d}x$; can be a long and cumbersome process, but we can compute this length using values we already know:

Theorem. $\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert =\sqrt{{\Vert \overrightarrow{v}\Vert }^2-\sum _{k=1}^{n-1}{⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩}^2}$

Proof. Observe that ${\overrightarrow{p}}_i$ and ${\overrightarrow{v}}_i-{\overrightarrow{p}}_i$ are orthogonal by construction, so ${\overrightarrow{v}}_i$, ${\overrightarrow{p}}_i$, and ${\overrightarrow{v}}_i-{\overrightarrow{p}}_i$ form a right triangle.

Therefore ${\Vert {\overrightarrow{p}}_i\Vert }^2+{\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert }^2={\Vert {\overrightarrow{v}}_i\Vert }^2$, in particular, ${\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert }^2={\Vert {\overrightarrow{v}}_i\Vert }^2-{\Vert {\overrightarrow{p}}_i\Vert }^2$, by the Pythagorean theorem.

By definition, ${\overrightarrow{p}}_i=\sum _{k=1}^{n-1}⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩{\hat{e}}_k$. By generalizing the pythagorean theorem into multiple dimensions, we have ${\Vert {\overrightarrow{p}}_i\Vert }^2={\Vert \sum _{k=1}^{n-1}⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩{\hat{e}}_k\Vert }^2=\sum _{k=1}^{n-1}{\Vert ⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩{\hat{e}}_k\Vert }^2$ because all $⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩{\hat{e}}_k$ are orthogonal.

Factoring out the scalar value $⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩$ from the length of each component in the sum leaves just ${⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩}^2{\Vert {\hat{e}}_k\Vert }^2$, and of course the length of ${\hat{e}}_k$ is $1$. Therefore,

${\Vert {\overrightarrow{p}}_i\Vert }^2=\sum _{k=1}^{n-1}{⟨{\overrightarrow{v}}_i,{\hat{e}}_k⟩}^2$

Plugging this definition into the pythagorean identity above yields the desired equation for $\Vert {\overrightarrow{v}}_i-{\overrightarrow{p}}_i\Vert$.

quod erat demonstrandum