Numerical Linear Agebra

Question

how to Prove the backward stability of the inner product ? http://books.google.com/books?id=-tW8-FUoxWwC&pg=PA55&lpg=PA55&dq=inner+product+backward+stability&source=bl&ots=9VY-kLlRC3&sig=dRdmkf9QJuam22uf2wtY30dL7-o&hl=en&sa=X&ei=eu84U4alHoHIsATfpIIw&ved=0CDEQ6AEwATgK#v=onepage&q=inner%20product%20backward%20stability&f=false

Answer 1

I have given a try, kindly check.

Consider $x,y \in \mathbb{R^{n}}$ with components, $x=(x_1,x_2,x_3,...,x_n)$ and $y=(y_1,y_2,y_3,...,y_n)$. Suppose $P$ represent the floating point calculation then,

Let $P_1 = P(x_1y_1)$ and recursively define $P_n , n \ge 2$ represent the floating point calculation of $P(P_{n-1} + P(x_ny_n))$. Then analyze the given digression,

$P_1= x_1y_1(1+\delta_1)$ where $\delta_1$ is "small". $P_2= P(P_1+ x_2y_2(1+\delta_2)= (x_1y_1(1+\delta_1)+x_2y_2(1+\delta_2))(1+ \beta_2) $ where perturbation $ \beta_2, \delta_2$ are "small".

Compute the $n-1$ terms and substitution in $P_n$ with rearrangements to get,

$P_n= \sum_{i=1}^{n} x_iy_i(1+\gamma_i) $ where, $1+\gamma_i=(1+\delta_i)(1+\beta_i)(1+\beta_{i+1})...(1+\beta_n)$ with $\beta_1=0$ further which $1+\gamma_i$ can be shown to be small as terms like $\beta_j\beta_m$ are small.