Minimize integral

Question

Find numbers A and B such that the integral is minimal

$$ \int_{0}^{\infty}\left\vert% \,\vphantom{\Large A}{\rm e}^{-x} - A{\rm e}^{-2x} - B{\rm e}^{-3x}\, \right\vert^{2}\,{\rm d}x $$

I have tried to find an orthonormal basis so I can compute the projection between the functions without success. All help is very appreciated, and what is the best way of finding an orthonormal basis of similar problems ?.

Answer 1

Instead of computing an ONB it is easier to work from basic principles: the vector connecting $e^{-x}$ to the nearest point in the plane is orthogonal to the plane. Thus, $e^{-x}-Ae^{-2x} -Be^{-3x}$ must be orthogonal to both $e^{-2x}$ and $e^{-3x}$. This requirement amounts to $$\frac13 - \frac{A}{4}-\frac{B}{5}=0\quad \text{and}\quad \frac{1}{4}-\frac{A}{5}-\frac{B}{6}=0$$ Hence $A=10/3$ and $B=-5/2$.

^{My original answer was wrong, I corrected it after seeing Felix Martin's answer, thinking there is still some merit to this approach.}

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal I} \equiv \int_{0}^{\infty}\verts{\sum_{n = 1}^{3}a_{n}\expo{-nx}}^{2} \,\dd x\,,\quad a_{1} = 1,\quad a_{2} = -A\,,\quad a_{3} = -B}$. $$ {\cal I}= \sum_{m = 1}^{3}\sum_{n = 1}^{3}a_{m}^{*}a_{n}\int_{0}^{\infty} \expo{-\pars{m + n}x}\,\dd x = \sum_{m = 1}^{3}\sum_{n = 1}^{3}{a_{m}^{*}a_{n} \over m + n} $$ With $m = 2, 3$: $$ 0 = \partiald{{\cal I}}{{a_{m}^{*}}} = \sum_{n = 1}^{3}{a_{n} \over m + n} ={1 \over m + 1} + \sum_{n = 2}^{3}{a_{n} \over m + n}\quad\imp\quad \sum_{n = 2}^{3}{a_{n} \over m + n} = -\,{1 \over m + 1} $$ $$ \left\lbrace% \begin{array}{rcrcl} -\,{1 \over 4}\,A & - & {1 \over 5}\,B & = & -\,{1 \over 3} \\ -\,{1 \over 5}\,A & - & {1 \over 6}\,B & = & -\,{1 \over 4} \end{array}\right. \quad\imp\quad \left\lbrace% \begin{array}{rcrcl} 15A & + & 12B & = & 20 \\ 12A & + & 10B & = & 15 \end{array}\right. $$ $$\color{#0000ff}{\large% A = {10 \over 3}\,,\qquad B = -\,{5 \over 2}} $$