exercise:D8cd0664e8: Difference between revisions

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Show that <math>f(x)L</math> satisfies [[guide:7d4d2d242c#eq11.3.1 |equations]] [[guide:7d4d2d242c#eq11.3.2 |and]] and hence
Show that <math>f(x)L</math> satisfies [[guide:7d4d2d242c#eq11.3.1 |equation]] and  [[guide:7d4d2d242c#eq11.3.2 |equation]] and hence
is a linear operator.
is a linear operator.

Latest revision as of 01:21, 26 November 2024

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

Let [math]f(x)[/math] be a given function and [math]L[/math] a linear operator. Define [math]f(x)L[/math] by the equation

[[math]] (f(x)L)y =f(x)(L_y) . [[/math]]

Show that [math]f(x)L[/math] satisfies equation and equation and hence is a linear operator.