Calculus Quiz

By the Weierstrass M-test $f(x)$ converges uniformly for any $x$ , and since $f(x)$ is also the sum of differentiable functions $f'(x)$ exists with $f'(x) = \displaystyle \sum_{k=1}^{\infty} \dfrac{\cos(kx)}{k}$ . By the Dirichlet test for convergence we can conclude that $f'(x)$ converges for $0 \lt x \lt 2\pi$ .

Now $\cos(y) = \dfrac{e^{iy} + e^{-iy}}{2}$ , so

$f'(x) = \displaystyle\sum_{k=1}^{\infty} \dfrac{\cos(kx)}{k} = \sum_{k=1}^{\infty} \dfrac{e^{ikx} + e^{-ikx}}{2} = \dfrac{-\ln(1 - e^{ikx}) - \ln(1 - e^{-ikx})}{2}$

since in general $\displaystyle\sum_{k=1}^{\infty} \dfrac{z^{k}}{k} = -\ln(1 - z)$ .

(There are radius of convergence issues here to deal with, but for now I will just carry on .....)

So $f'(x) = -\dfrac{1}{2}\ln((1 - e^{ikx})(1 - e^{-ikx})) = \dfrac{1}{2}\ln(2 - 2\cos(x)) = -\ln(2\sin(\frac{x}{2}))$

after making use of the identity $\cos(x) = 1 - 2\sin^{2}(\frac{x}{2})$ . Now $f'(x) = 0$ when

$2\sin(\frac{x}{2}) = 1 \Longrightarrow \sin(\frac{x}{2}) = \dfrac{1}{2} \Longrightarrow \dfrac{x}{2} = \dfrac{\pi}{6} + n*2\pi$ or $\dfrac{5\pi}{6} + n*2\pi$

$\Longrightarrow x = \dfrac{\pi}{3} + n*4\pi$ or $x = \dfrac{5\pi}{3} + n*4\pi$ .

So in the given interval the extrema occur at $x = \dfrac{\pi}{3}$ and $x = \dfrac{5\pi}{3}$ . Plugging these values back into $f(x)$ we see that the first two terms, (which dominate the sum as the denominators are the least in the series), with $x = \dfrac{\pi}{3}$ are positive and with $x = \dfrac{5\pi}{3}$ are negative, so (informally) we can anticipate that the former critical point represents a maximum and the latter a minimum. Thus the maximal value of $f(x)$ on the given interval occurs at $x = \dfrac{\pi}{3}$ , giving us $a + b = 1 + 3 = \boxed{4}$ .