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Conference publicationsAbstractsXVII conferenceShowing Properties of Numeric Method on the Basis of Homographic FunctionsRussia, 119333 Moscow, Vavilov st. 40, CC RAS, Belyankov A.Ya. 2 pp. (accepted)SHOWING PROPERTIES OF NUMERIC METHOD ON THE BASIS OF HOMOGRAPHIC FUNCTIONS Let function $f:R\to R$ be sufficiently smooth, $f(x_{*})=0$, $f'(x_{*})\neq 0$, and let successive approximations $x_0$, $x_1$, $x_2$, $\ldots$ be generated by a numeric method for solution of equation $f(x)=0$. Recall well known asymptotic convergence behavior of certain three methods described in [1], namely Newton method, secant method and secant method with the recurrency $x_{k+1}=F(x_{k},x_{k-1})$: $$ \left(x_{k+1}=\right)\qquad x_{k}-{\frac {f(x_{k})}{f'(x_{k})}} =x_{*}+\left(x_{k}-x_{*}\right)^{2}\left(\kappa+o(1)\right), \eqno (1) $$ $$ \left(x_{k+1}=\right)\qquad {\frac {af(x_{k})-x_{k}f(a)}{f(x_{k})-f(a)}} =x_{*}+\left(x_{k}-x_{*}\right)\left(q+o(1)\right), \eqno (2) $$ $$ \left(x_{k+1}=\right)\qquad {\frac {x_{k-1}f(x_{k})-x_{k}f(x_{k-1})}{f(x_{k})-f(x_{k-1})}} =x_{*}+\left(x_{k-1}-x_{*}\right)\left(x_{k}-x_{*}\right)\left(\kappa+o(1)\right). \eqno (3) $$ For homographic functions $\varphi_A(x)=(a_{11}x+a_{12})/(a_{21}x+a_{22})$, $\mbox{det}(A)\neq 0$, the following is established. If $f=\varphi_A$ then formulae (1--3) hold exactly i.e. $o(1)$ can be thrown off; corresponding values for $\kappa$, $q$ are $\kappa=f''(x_{*})/2f'(x_{*})$, $q=\kappa (a-x_{*})$. These properties are specific for homographic functions: if we throw off $o(1)$ in any formula (1--3) and consider this formula as functional equation over unknown function $f$ with independent variable $x_k$ (formulae (1--2)) or with independent variables $x_{k-1},x_k$ (formula (3)), then we deduce that $f$ is homographic. It should be noted also that solving equation $x=\varphi(x)$ using simple iteration method i.e. creating sequence $x_0$, $x_1=\varphi(x_0)$, $x_2=\varphi(x_1)=\varphi(\varphi(x_0))$, $\ldots$ is quite transparent in the case of homographic function $\varphi=\varphi_A$ because $\varphi_A(\varphi_A(x))=\varphi_{A^2}(x)$, $\varphi_A(\varphi_A(\varphi_A(x)))=\varphi_{A^3}(x)$ and powers of $A$ are easy to calculate: $A^k=V\Lambda^k V^{-1}$ where $A=V\Lambda V^{-1}$ is canonical decomposition of 2 by 2 matrix $A$. References 1. Bakhvalov N.S. Numerical methods. Moscow: Nauka, 1975. 632 p.
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