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Chapter 6: Q-6.8-7E (page 331)

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

7. Show that \({\left\| {\cos kt} \right\|^2} = \pi \) and \({\left\| {\sin kt} \right\|^2} = \pi \) for \(k > 0\).

Short Answer

Expert verified

It is proved that\({\left\| {\cos kt} \right\|^2} = \pi {\rm{ and }}{\left\| {\sin kt} \right\|^2} = \pi ,\,\,\,\forall k > 0\).

Step by step solution

01

Inner Product

The Inner Productfor any two arbitrary functions is given by:

\(\left\langle {f,g} \right\rangle = \int_0^{2\pi } {f\left( t \right)g\left( t \right)dt} \)

02

Proving the statement

It is given that,\(k \in + {\rm I}\).

Using theinner productrule, we have:

\(\begin{array}{c}{\left\| {\cos kt} \right\|^2} = \int_0^{2\pi } {{{\cos }^2}ktdt} \\ = \frac{1}{2}\int_0^{2\pi } {\left\{ {1 + \cos 2kt} \right\}dt} \\ = \pi \end{array}\)

Similarly,

\(\begin{array}{c}{\left\| {\sin kt} \right\|^2} = \int_0^{2\pi } {{{\sin }^2}ktdt} \\ = \frac{1}{2}\int_0^{2\pi } {\left\{ {1 - \cos 2kt} \right\}dt} \\ = \pi \end{array}\)

Hence, \({\left\| {\cos kt} \right\|^2} = \pi {\rm{ and }}{\left\| {\sin kt} \right\|^2} = \pi ,\,\,\,\forall k > 0\).

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Most popular questions from this chapter

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)—the sum of the squares of the “regression term.” Denote this number by \(SS\left( R \right)\).

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)—the sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)—the “total” sum of the squares of the -values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of \(y\)-values.

20. Show that \({\left\| {X\hat \beta } \right\|^2} = {\hat \beta ^T}{X^T}{\bf{y}}\). (Hint: Rewrite the left side and use the fact that \(\hat \beta \) satisfies the normal equations.) This formula for is used in statistics. From this and from Exercise 19, obtain the standard formula for \(SS\left( E \right)\):

\(SS\left( E \right) = {y^T}y - \hat \beta {X^T}y\)

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

7. \(\left\| {\mathop{\rm w}\nolimits} \right\|\)

A healthy child’s systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)
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