Mathematical Analysis Apostol Solutions Chapter 11 «CERTIFIED – 2025»

Next, we set the partial derivatives equal to zero and solve for x and y:

We know uniform continuity of (f) on ([a,b]) (Theorem 4.19). Given (\epsilon > 0), find (\delta > 0) such that (|x-y|<\delta \implies |f(x)-f(y)|<\frac\epsilon\alpha(b)-\alpha(a)+1). Choose partition (P) with (|P|<\delta). Then on each subinterval, (M_k - m_k < \frac\epsilon\alpha(b)-\alpha(a)+1). Multiply by (\Delta \alpha_k) and sum: [ U(P,f,\alpha)-L(P,f,\alpha) < \frac\epsilon\alpha(b)-\alpha(a)+1 \sum \Delta \alpha_k = \frac\epsilon\alpha(b)-\alpha(a)+1 (\alpha(b)-\alpha(a)) < \epsilon. ] Thus the Riemann-Stieltjes condition holds. ✅ Mathematical Analysis Apostol Solutions Chapter 11

Forgetting that the Fourier series converges to the periodic extension, not to (f) at discontinuities. Apostol’s solutions explicitly mention the Gibbs phenomenon near (x=\pm\pi). Next, we set the partial derivatives equal to