## Elementary Linear Algebra 7th Edition

(a) $\langle f, g\rangle= -\frac{2}{e}$ (b) $\langle f, f\rangle= \sqrt{\langle f, f\rangle}=\sqrt \frac{2}{3}$ (c) $\| g \| =\sqrt {\frac{1}{2}(e^2-e^{-2})}$ (d) $d(f,g) = \sqrt{\frac{2}{3}+\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}}$
$f(x)=x, \quad q(x)=e^{-x}.$ (a) $\langle f, g\rangle=\int_{-1}^{1}f(x)g(x) d x=\int_{-1}^{1} xe^{-x} d x=\left[-xe^{-x}-e^{-x}\right]_{-1}^{1}=-\frac{2}{e}$ (b) $\langle f, f\rangle=\int_{-1}^{1} x^2 d x=\left.\frac{1}{3}x^3\right|_{-1} ^{1}=\frac{2}{3}\Longrightarrow\| f \| =\sqrt{\langle f, f\rangle}=\sqrt \frac{2}{3}$ (c) \begin{aligned} \langle g, g\rangle &=\int_{-1}^{1}e^{-2x} d x \\ &=-\frac{1}{2}\left.e^{-2x} \right|_{-1} ^{1} \\ &=\frac{1}{2}(e^2-e^{-2}). \end{aligned} Hence, we have $\| g \| =\sqrt{\langle g, g\rangle}=\sqrt {\frac{1}{2}(e^2-e^{-2})}$ (d) \begin{aligned}\langle f-g, f-g\rangle &=\int_{-1}^{1}\left(x-e^{-x}\right)^{2} d x \\ &=\int_{-1}^{1}\left(x^{2}+e^{-2x}-2xe^{-x}\right) d x \\ &=\left.\frac{1}{3} x^{3}-\frac{1}{2}e^{-2x}+2xe^{-x}+2e^{-x}\right|_{-1} ^{1} \\ &=\frac{2}{3}+\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}\end{aligned}. Hence, we have $d(f,g)=\| f-g \|= \sqrt{\langle f-g, f-g\rangle}=\sqrt{\frac{2}{3}+\frac{4}{e}+\frac{e^2}{2}-\frac{1}{2e^2}}$