Question

Exercises 21–24 refer to \(V = C\left( {0,1} \right)\) with the inner product given by an integral, as in Example 7.

24. Compute \(\left\| g \right\|\) for \(g\) in Exercise 22.

Short answer

The required value is \(\left\| g \right\| = \frac{1}{{\sqrt {105} }}\).

Step-by-step solution

Use the given information

For the pair of vectors\(\left\langle {f,g} \right\rangle \), the inner product is given by \(\left\langle {f,g} \right\rangle = \int_0^1 {f\left( t \right)g\left( t \right)dt} \).

It is given that \(g\left( t \right) = {t^3} - {t^2}\), then \(\left\langle {g,g} \right\rangle = \int_0^1 {g\left( t \right)g\left( t \right)dt} \)and\(\left\| g \right\| = \sqrt {\left\langle {g,g} \right\rangle } \).

Find the inner product

Plug the expression for\(g\left( t \right)\)into inner product\(\left\langle {g,g} \right\rangle = \int_0^1 {g\left( t \right)g\left( t \right)dt} \), as follows:

\(\begin{aligned}{}\left\langle {g,g} \right\rangle &= \int_0^1 {g\left( t \right)g\left( t \right)dt} \\ &= \int_0^1 {\left( {{t^3} - {t^2}} \right)\left( {{t^3} - {t^2}} \right)dt} \\ &= \int_0^1 {\left( {{t^6} - 2{t^5} + {t^4}} \right)\,dt} \\ &= \left( {\frac{{{t^7}}}{7} - \frac{{2{t^6}}}{6} + \frac{{{t^5}}}{5}} \right)_0^1\\ &= \left( {\frac{1}{7} - \frac{1}{3} + \frac{1}{5} - 0} \right)\\ &= \frac{1}{{105}}\end{aligned}\)

Compute \(\left\| g \right\|\) 

Plugthe expression for\(\left\langle {g,g} \right\rangle \)into\(\left\| g \right\| = \sqrt {\left\langle {g,g} \right\rangle } \)and simplify as follows:

\(\begin{aligned}{}\left\| g \right\| &= \sqrt {\left\langle {g,g} \right\rangle } \\ &= \sqrt {\frac{1}{{105}}} \\ &= \frac{1}{{\sqrt {105} }}\end{aligned}\)

Hence, the required value is, \(\left\| g \right\| = \frac{1}{{\sqrt {105} }}\).