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(M) Let V be the space \(C\left( {0,2\pi } \right)\)with the inner product of Example 7. Use the Gram–Schmidt process to create an orthogonal basis for the subspace spanned by \(\left\{ {1,\cos t,{{\cos }^2}t,{{\cos }^3}t} \right\}\). Use a matrix program or computational program to compute the appropriate definite integrals.

Short Answer

Expert verified

The orthogonal basis are \({f_0}\left( t \right) = 1\), \({f_1}\left( t \right) = \cos t\), \({f_2}\left( t \right) = \frac{1}{2}\cos 2t\) and \({f_3}\left( t \right) = \frac{1}{4}\cos 3t\) .

Step by step solution

01

Find the basis

It is given that \(V\) be the space \(C\left( {0,2\pi } \right)\) and an orthogonal basis for the subspace is spanned by \(\left\{ {1,\cos t,{{\cos }^2}t,{{\cos }^3}t} \right\}\).

Let the matrix be defined as \(A = \left( {\begin{array}{*{20}{c}}1\\{\cos t}\\{{{\cos }^2}t}\\{{{\cos }^3}t}\end{array}} \right)\).Use the following steps to find the associated values for the obtained data in MATLAB.

Formulate the matrix A using the commands as:

>>A=(1,cos(t),(cos(t))^2,(cos(t))^3);
>>(m,n) = size(A);

And to find\(QR\)by using the following commands:

>>fori = 1:n
>>v = A(:,i);
>>for j=1:i-1
>>R(j,i) = Q(:,j)'*A(:,i);
>>v = v - R(i,j)*Q(:,j);
>>end
>>R(i,i) = norm(v);
>>Q(:,i) = v/R(i,i);
>>end

By using the matrix program, the new orthogonal polynomials, are obtained as \({f_0}\left( t \right) = 1\), \({f_1}\left( t \right) = \cos t\), \({f_2}\left( t \right) = \frac{1}{2} - {\cos ^2}t\) and \({f_3}\left( t \right) = \frac{3}{4}\cos t - {\cos ^3}t\) .

02

Simplify the polynomials

By using the trigonometric identities, the simplified orthogonal basis is \({f_0}\left( t \right) = 1\), \({f_1}\left( t \right) = \cos t\), \({f_2}\left( t \right) = \frac{1}{2}\cos 2t\) and \({f_3}\left( t \right) = \frac{1}{4}\cos 3t\).

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Most popular questions from this chapter

Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).

Question: In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)

Compute the quantities in Exercises 1-8 using the vectors

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

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Compute the quantities in Exercises 1-8 using the vectors

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

4. \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

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