\documentclass[a4paper,12pt]{article} \begin{document} \parindent=0pt QUESTION The inner product is sometimes written $<\textbf{u,v}>$ rather than \textbf{u.v}. In this notation the basic properties become \begin{tabular}{cll} (a)&$<\textbf{u,v}>=<\textbf{v,u}>$&symmetry,\\ (b)&$<\textbf{u+v,w}>=<\textbf{u,w}>+<\textbf{v,w}>$& additivity,\\ (c)&$<\lambda\textbf{u,v}>=\lambda<\textbf{u,v}>$&homogeneity,\\ (d)&$<\textbf{v,v}>\geq0$ with $<\textbf{v,v}>=0\Leftrightarrow \textbf{v}=\textbf{0}$ &positivity.\\ \end{tabular} More generally if $V$ is a vector space then a function $<>:V\times V\rightarrow\textbf{R}$ which associates a real number with each pair of ordered vectors is called an inner product if the above four properties hold; $V$ itself is called an inner product space. The concepts of length of a vector, distance between vectors, angle between vectors, orthogonal bases etc. can be defined for such spaces and the Gram-Schmidt process still works. \textbf{Example} The vector space $P_2$ of polynomials of degree less than or equal to two can be turned into an inner product space by defining $$=\int_{-1}^1p(x)q(x)\,dx.$$ One basis for $P_2$ is the set $\{1,x,x^2\}$ and since $$|p|=\sqrt{}=\sqrt{\{\int_{-1}^1(p(x))^2\,dx\}},$$ the length of the \lq\lq vector" 1 is $$\sqrt{\{\int_{-1}^11^2\,dx\}}=\sqrt{\{[x]_{-1}^1\}}=\sqrt{2}$$ and the length of the \lq\lq vector" $x$ is $$\sqrt{\left\{\int_{-1}^1x^2\,dx\right\}} =\sqrt{\left\{\left[\frac{x^3}{3}\right]_{-1}^1\right\}}=\sqrt{\frac{2}{3}}.$$ Furthermore since $$<1,x>=\int_{-1}^1(1 \times x)\,dx=[\frac{x^2}{2}]_{-1}^1=0$$ the functions 1 and $x$ are orthogonal with respect to the inner product and the functions $\frac{1}{\sqrt{2}}$ and $\sqrt{\frac{3}{2}}x$ are orthonormal. Since $x^2$ is not orthogonal to 1, however, the basis $\{1,x,x^2\}$ is not an orthogonal basis. \textbf{Exercise} Apply the Gram-Schmidt process to the basis $\{1,x,x^2\}$ to turn it into an orthogonal basis and the normalise the new basis. (The resulting polynomials are the first three normalised Legendre polynomials.) \bigskip ANSWER Since 1 and $x$ are orthogonal one can take \begin{eqnarray*} \textbf{w}_0&=&1,\\ \textbf{w}_1&=&x. \end{eqnarray*} Then \begin{eqnarray*} \textbf{w}_2&=&x^2-\frac{[\int_{-1}^1(x^2\times x)\,dx]x}{\frac{2}{3}}-\frac{[\int_{-1}^1(x^2\times 1)\,dx]1}{2}\\ &=&x^2-\frac{[\frac{x^4}{4}]_{-1}^1x}{\frac{2}{3}}-\frac{[\frac{x^3}{3}]_{-1}^1}{2}\\ &=&x^2-0-\frac{1}{3}\\ &=&\frac{(3x^2-1)}{3} \end{eqnarray*} Now \begin{eqnarray*} \textbf{w}_2.\textbf{w}_2&=&\int_{-1}^1\frac{(3x^2-1)^2}{9}\,dx\\ &=&\int_{-1}^1\frac{(9x64-6x^2+1)}{9}\,dx\\ &=&\frac{[\frac{9x^5}{5}-2x^3+x]_{-1}^1}{9}\\ &=&\frac{2}{9}[\frac{9}{5}-2+1]\\ &=&\frac{8}{45}.\\ |\textbf{w}_2|&=&\frac{2\sqrt{2}}{3\sqrt{5}} \end{eqnarray*} The orthonormal basis giving the first few normalised Legendre polynomials is: \begin{eqnarray*} \hat{\textbf{w}}_0&=&\frac{1}{\sqrt{2}},\\ \hat{\textbf{w}}_1&=&\sqrt{\frac{3}{2}}x\\ \hat{\textbf{w}}_2&=&\frac{\sqrt{5}}{2\sqrt{2}}(3x^2-1) \end{eqnarray*} \end{document}