## Linear Algebra and Its Applications, Exercise 3.4.20

Exercise 3.4.20. Given the vector $v = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{4}}, \frac{1}{\sqrt{8}}, \ldots)$ what is the length $\|v\|$? Given the function $f(x) = e^x$ for $0 \le x \le 1$ what is the length of the function over the interval? Given the function $g(x) = e^{-x}$ for $0 \le x \le 1$ what is the inner product of $f(x)$ and $g(x)$? $\|v\|^2 = v^Tv = (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{4}})^2 + (\frac{1}{\sqrt{8}})^2 + \ldots$ $= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = 1$

so that $\|v\| = \sqrt{1} = 1$.

We have $\|f\|^2 = \int_0^1 f(x)^2 \mathrm{d}x = \int_0^1 (e^x)^2 \mathrm{d}x = \int_0^1 e^{2x} \mathrm{d}x$ $= \frac{e^{2x}}{2} \big|_0^1 = \frac{e^2}{2} - \frac{e^0}{2} = \frac{e^2 - 1}{2}$

so that $\|f\| = \sqrt{\frac{e^2 - 1}{2}}$.

Finally, the inner product of $f(x)$ and $g(x)$ is $\int_0^1 f(x)g(x) \mathrm{d}x = \int_0^1 e^x e^{-x} \mathrm{d}x = \int_0^1 e^0 \mathrm{d}x = \int_0^1 1 \mathrm{d}x = 1$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books .

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