Question

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

21. Compute \(\left\langle {f,g} \right\rangle \), where \(f\left( t \right) = 1 - 3{t^2}\)and \(g\left( t \right) = 1 - {t^3}\).

Short answer

The inner product is \(\left\langle {f,g} \right\rangle = 0\).

Step-by-step solution

Use the given information

It is given that\(f\left( t \right) = 1 - 3{t^2}\),\(g\left( t \right) = 1 - {t^3}\).

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

Find the inner product

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

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

Thus, the inner product is\(\left\langle {f,g} \right\rangle = 0\).