Find the first-order and third-order Fourier approximations to [latex]f(t)=3-2 \sin t+5 \sin 2 t-6 \cos 2 t[/latex].

The third-order Fourier approximation to f is the best approximation in C[0, 2π] to f by functions (vectors) in the subspace spanned by 1, [latex]\cos t, \cos 2 t, \cos 3 t[/latex] ,[latex]\sin t, \sin 2 t, \text { and } \sin 3 t[/latex] . But f is obviously in this subspace, so f is its own best approximation: [latex]f(t)=3-2 \sin t+5 \sin 2 t-6 \cos 2 t[/latex]. For the first-order approximation, the closest function to f in the subspace W = [latex]\operatorname{Span}\{1, \cos t, \sin t\} \text { is } 3-2 \sin t[/latex] . The other two terms in the formula for f (t ) are orthogonal to the functions in W , so they contribute nothing to the integrals that give the Fourier coefficients for a first-order approximation.

Question 6.8.P.2: Find the first-order and third-order Fourier approximations ...

Linear Algebra and Its Applications

Find the first-order and third-order Fourier approximations to

$f(t)=3-2 \sin t+5 \sin 2 t-6 \cos 2 t$ .

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The third-order Fourier approximation to f is the best approximation in C[0, 2π] to f by functions (vectors) in the subspace spanned by 1, $\cos t, \cos 2 t, \cos 3 t$ , $\sin t, \sin 2 t, \text { and } \sin 3 t$ . But f is obviously in this subspace, so f is its own best approximation:

$f(t)=3-2 \sin t+5 \sin 2 t-6 \cos 2 t$ .

For the first-order approximation, the closest function to f in the subspace W = $\operatorname{Span}\{1, \cos t, \sin t\} \text { is } 3-2 \sin t$ . The other two terms in the formula for f (t ) are orthogonal to the functions in W , so they contribute nothing to the integrals that give the Fourier coefficients for a first-order approximation.