10.6, #6. f(x) is an odd function on [−π, π]. So f(x) cos nx and f(x) sin nx are respectively odd and even functions on [−π, π]. In particular,

a_{1} = \frac{1}{π} ∫_{−π}^{π} f(x) cos x dx = 0
a_{2} = \frac{1}{π} ∫_{−π}^{π} f(x) cos 2x dx = 0
a_{3} = \frac{1}{π} ∫_{−π}^{π} f(x) cos 3x dx = 0

and

b_{1} = \frac{2}{π} ∫_{0}^{π} sin x dx = \frac{4}{π}
b_{2} = \frac{2}{π} ∫_{0}^{π} sin 2x dx = 0
b_{3} = \frac{2}{π} ∫_{0}^{π} sin 3x dx = \frac{4}{3π}

Moreover, a_{0} = 0 since a_{0} is the average value of f(x) on [−π, π]. Thus F_{1}(x) = F_{2}(x) = (4/π) sin x, F_{3}(x) = (4/π) sin x + (4/3π) sin 3x.

10.6, #7. f(x) is an even function on [−π, π]. So f(x) sin nx and f(x) cos nx are respectively odd and even functions on [−π, π]. In particular,

b_{1} = \frac{1}{π} ∫_{−π}^{π} f(x) sin x dx = 0
b_{2} = \frac{1}{π} ∫_{−π}^{π} f(x) sin 2x dx = 0
b_{3} = \frac{1}{π} ∫_{−π}^{π} f(x) sin 3x dx = 0

and

a_{1} = \frac{2}{π} ∫_{0}^{π} x cos x dx = \frac{2}{π} (x sin x + cos x)|_{0}^{π} = −\frac{4}{π}
a_{2} = \frac{2}{π} ∫_{0}^{π} x cos 2x dx = \frac{2}{π} ((x sin 2x)/2 + (cos 2x)/4)|_{0}^{π} = 0
a_{3} = \frac{2}{π} ∫_{0}^{π} x cos 3x dx = \frac{2}{π} ((x sin 3x)/3 + (cos 3x)/9)|_{0}^{π} = −\frac{4}{9π}

Finally, a_{0} = π/2 since a_{0} is the average value of f(x) on [−π, π]. Thus F_{1}(x) = F_{2}(x) = (π/2) − (4/π) cos x, F_{3}(x) = (π/2) − (4/π) cos x − (4/9π) cos 3x.

10.6, #9. f(x) = x^{2} is an even function on [−π, π]. So f(x) sin nx and f(x) cos nx are respectively odd and even functions on [−π, π]. In particular,

b_{n} = \frac{1}{π} ∫_{−π}^{π} f(x) sin nx dx = 0

and

a_{n} = \frac{2}{π} ∫_{0}^{π} x^{2} cos nx dx

= \frac{2}{π} (\frac{x^{2} sin nx}{n} + \frac{2x cos nx}{n^{2}} − \frac{2 sin nx}{n^{3}}) |_{0}^{π}

= 4/n^{2} cos nπ = (−1)^{n}4/n^{2}

Finally, a_{0} = π^{2}/3 since a_{0} is the average value of f(x) on [−π, π]. Thus

F_{1}(x) = π^{2}/3 − (4/π) cos x,
F_{2}(x) = π^{2}/3 − (4/π) cos x + (1/π) cos 2x,
F_{3}(x) = π^{2}/3 − (4/π) cos x + (1/π) cos 2x − (4/9π) cos 3x.

10.6, #15. g(x) = 1 for −π/2 ≤ x ≤ π/2 and = 0 elsewhere. Therefore g is even with average value 1/2, so a_{0} = 1/2, b_{k} = 0, and

a_{k} = \frac{2}{π} ∫_{0}^{π/2} cos(kx) dx = \{ \begin{matrix}0, & k even, \\ +\frac{2}{kπ} , & k = 1, 5, 9 ⋯ \\ −\frac{2}{kπ} , & k = 3, 7, 11, ⋯ \end{matrix} .

The series is

\frac{1}{2} + \frac{2}{π} (cos x − \frac{1}{3} cos(3x) + \frac{1}{5} cos(5x) − + ⋯).

This square-wave is shifted by π/2 to the left compared to the square-wave of example #1, g(x) = f(x + π/2). Replacing x by x + π/2 in the Fourier series for f gives a series whose general term is

\frac{2}{kπ} sin(k(x + π/2)) = \{ \begin{matrix}+\frac{2}{kπ} cos(kx), & for k = 1, 5, ⋯ \\ −\frac{2}{kπ} cos(kx) & for k = 32, 7, ⋯ \end{matrix} .

10.6, #19. For the flute the first and second harmonics are equally dominant, while higher harmonics are weak. For the basson the dominant harmonic is the third; the second harmonic is moderate, and all other harmonics including the first harmonic are weak.

10.6, #20.