10.5, #10. P_{n}(1) approximates cos 1 to four decimal places if the corresponding error E_{n} satisfies |E_{n}| < 0.00005. By the preceding problem we want 1/(n + 1)! < 0.00005. This happens for n = 7 since 8! = 40320 and 1/8! = 0.000025. For 6 decimal places, we want 1/(n + 1)! < 0.0000005. This happens for n = 9.

10.5, #12. The higher derivatives of f(x) = cos x are ± sin x and ± cos x. So |f^{n+1}(x)| ≤ 1 for all x and we may take M = 1 in the boxed remark on page 631. Then

|E_{n}| ≤ \frac{1}{(n + 1)!}|x|^{n+1}.

But as noted on page 633 of the text, the ratio x^{n+1}/(n + 1)! → 0 as n → ∞. So the Taylor series for cos x converges to cos x.

10.5, #14. For −0.1 ≤ x ≤ 0.1, E_{1} lies between the x-axis and the graph of x^{2} and looks like \frac{1}{2}x^{2}. For −0.1 ≤ x ≤ 0.1, E_{2} lies between the x-axis and the graph of x^{3}. A good range for y is −0.001 ≤ y ≤ 0.001.

10.6, #1. This is a Fourier series.

10.6, #2. This is not a Fourier series.

10.6, #3. This is not a Fourier series. Although sin(x+n) = (cos n) sin x+(sin n) cos x is a combination of sin x and cos x that could appear in a Fourier series, the fact that sin(x+n) occurs for n = 1, 2, 3, … means that such a Fourier series would have

1 + cos 1 + cos 2 + cos 3 + cos 4 + ⋯

as the coefficient of sin x. This sum does not converge.

10.6, #5 The next Fourier approximation would be

sin x + \frac{sin 3x}{3} + \frac{sin 5x}{5} + \frac{sin 7x}{7}.

The square wave function has period 2π and has the form

f(x) = \{ \begin{matrix}−h & −π ≤ x < 0 \\ h & 0 ≤ x < π. \end{matrix}

Since the coefficient of sin x in our series is 1 we must have

1 = b_{1} = \frac{1}{π} ∫_{−π}^{π} f(x) sin x dx = \frac{2}{π} ∫_{0}^{π} h sin x dx = \frac{4h}{π}.

Hence h = π/4. f(x) is not continuous at k ∗ π for any integer k.