infnorm - Maple Help

numapprox

 infnorm
 compute the L-infinity norm of a function

 Calling Sequence infnorm(f, x=a..b, 'xmax') infnorm(f, a..b, 'xmax')

Parameters

 f - procedure or expression representing the function x - variable name appearing in f, if f is an expression a, b - numerical values specifying the interval [a, b] xmax - (optional) name which will be assigned the point of maximum

Description

 • This procedure computes the ${L}_{\mathrm{\infty }}$ (minimax) norm of a given real function f(x) on the interval [a, b]. Specifically, it computes an estimate for the value

$\mathrm{max}\left(\left|f\left(x\right)\right|,x\in \left[a,b\right]\right)$

 • If the second argument is a range $a..b$ then the first argument is understood to be a Maple operator. If the second argument is an equation $x=a..b$ then the first argument is understood to be an expression in the variable x.
 • If the third argument 'xmax' is present then it must be a name. Upon return, its value will be an estimate for the value of x where the maximum of $\left|f\left(x\right)\right|$ is attained.
 • Various levels of user information will be displayed during the computation if infolevel[infnorm] is assigned values between 1 and 4.
 • The command with(numapprox,infnorm) allows the use of the abbreviated form of this command.

Examples

 > $\mathrm{with}\left(\mathrm{numapprox}\right):$
 > $\mathrm{infnorm}\left(\mathrm{sin},0..2,'\mathrm{xmax}'\right)$
 ${1.0}$ (1)
 > $\mathrm{xmax}$
 ${1.570796332}$ (2)
 > $\mathrm{infnorm}\left(\frac{1}{\mathrm{\Gamma }\left(x\right)},x=1..2\right)$
 ${1.129173885}$ (3)
 > $r\left[2,3\right]≔\mathrm{minimax}\left(\frac{\mathrm{tan}\left(x\right)}{x},x=0..\frac{\mathrm{\pi }}{4},\left[2,3\right]\right)$
 ${{r}}_{{2}{,}{3}}{≔}\frac{{1.130422926}{+}\left({-}{0.07842798254}{-}{0.07066118710}{}{x}\right){}{x}}{{1.130423032}{+}\left({-}{0.07843711579}{+}\left({-}{0.4473405792}{+}{0.02547897687}{}{x}\right){}{x}\right){}{x}}$ (4)
 > $\mathrm{infnorm}\left(\frac{\mathrm{tan}\left(x\right)}{x}-r\left[2,3\right],x=0..\frac{\mathrm{\pi }}{4},'\mathrm{xmax}'\right)$
 ${9.388658362}{×}{{10}}^{{-8}}$ (5)
 > $\mathrm{xmax}$
 ${0.608882102369394}$ (6)