Welcome to Doyeob Lee's Blog!

Problem: (a) Let \(d_n = [1,2, \dots, n]\). Show that \(d_n = e^{\psi(n)}\).

(b) Let \( P \in \mathbb{Z}[x] \), \(\deg P \le n\). Put \(I = I(P) = \int_{0}^{1}P(x)dx\). Show that \(Id_{n+1}\in \mathbb{Z}\), and hence that \(d_{n+1} \ge 1/|I|\) if \(I \neq 0\).

(c) Show that there is a polynomial \(P\) as above so that \(Id_{n+1} = 1\).

(d) Verify that \(\max_{0\le x\le 1} | x^2 (1-x)^2 (2x-1)| = 5^{-5/2}\).

(e) For \(P(x) = \left(x^2(1-x)^2 (2x-1) \right)^{2n} \), verify that \(0 \lt I \lt 5^{-5n}\).

(f) Show that \(\psi (10n+1) \ge (\tfrac{1}{2}\log 5)\cdot 10n\).

Solution:

(d) Put \(f(x) = x^2 (1-x)^2 (2x-1)\), then \(f(x)=0\) for \(0\lt x\lt 1\) implies \(x = \frac{1}{2}\pm\frac{\sqrt{5}}{10}\). Since \(|f\left(\frac{1}{2}\pm\frac{\sqrt{5}}{10}\right)| = 5^{-5/2}\) and \(f(0) = f(1)= 0\), \(\max_{0\le x\le 1} | x^2 (1-x)^2 (2x-1)| = 5^{-5/2}\).

(e) \[I = I(P) = \int_{0}^{1}P(x) dx \le \int_{0}^{1} \max_{0\le t\le 1}P(t) dx =\int_{0}^{1} 5^{-5n} dx = 5^{-5n}\]

Problem: Let \(R(x)\) be as in Exercise 24(c). Show that \(R(x)\ll x^{1/3}\log x\).

Solution: Put \(f(t) = \sqrt{x-t^{2}}\), then \(x^{-1/2} \le -f''(t) \le 2\sqrt{2}x^{-1/2}\) for \(0\le t \le \sqrt{x/2}\). By Exercise 25, \[\sum_{0 \le n \le \sqrt{x/2}} B_{1} (\{\sqrt{x-n^2}\}) \ll x^{1/3}\log x.\] Hence, \[R(x) = -8\sum_{1 \le n \le \sqrt{x/2}} B_{1} (\{\sqrt{x-n^2}\})+O(1) \ll x^{1/3}\log x.\]

Problem: Show that if \(U \le \sqrt{x}\), then \[\sum_{U \lt n \le 2U} B_1(\{x/n\}) \ll x^{1/3}\log x.\] Let \(\Delta(x)\) be as in Exercise 23(b). Show that \(\Delta(x)\ll x^{1/3}(\log x)^2\).

Solution: Put \(f(t) = x/t\), then \(1/(2U/(2x^{1/3}))^{3}\le f''(t)\le 8/(2U/(2x^{1/3}))^{3}\) for \(U \lt t \le 2U\). By Exercise 25, \[\sum_{U \lt n \le 2U} B_1(\{x/n\}) \ll U (2U/(2x^{1/3}))^{-1} \log 2(2U/(2x^{1/3})) + (2U/(2x^{1/3}))^2\ll x^{1/3}\log x.\] By applying this \(O(\log x)\) times, \begin{align*} \Delta (x) = -2 \sum_{n \le \sqrt{x}}B_1 (\{x/n\}) + O(1)\ll x^{1/3}(\log x)^2. \end{align*}

Problem: (a) Show that if \((a,q)=1\), and \(\beta\) is real, then \[ \sum_{n=1}^{q} B_1\!\left( \left\{ \frac{a}{q}n + \beta \right\} \right) = B_1(\{ q\beta \}). \] (b) Show that if \(A \geq 1\), \(\; |f'(x) - a/q| \leq A/q^2\) for \(1 \leq x \leq q\), and \((a,q)=1\), then \[ \sum_{n=1}^{q} B_1(\{ f(n) \}) \;\ll\; A. \] (c) Suppose that \(Q \geq 1\) is an integer, \(B \geq 1\), and that \({1}/{Q^3} \leq \pm f''(x) \leq {B}/{Q^3}\) for \(0 \leq x \leq N\) where the choice of sign is independent of \(x\). Show that numbers \(a_r, q_r, N_r\) can be determined, \(0 \leq r \leq R\) for some \(R\), so that (i) \((a_r, q_r) = 1\), (ii) \(q_r \leq Q\), (iii) \(\lvert f'(N_r) - a_r/q_r \rvert \leq 1/(q_r Q)\), (iv) \(N_0 = 0\), \(N_r = N_{r-1} + q_{r-1}\) for \(1 \leq r \leq R\), and \(N - Q \leq N_R \leq N\).

(d) Show that under the above hypotheses \[ \sum_{n=0}^{N} B_1(\{ f(n) \}) \;\ll\; B(R+1) + Q. \] (e) Show that the number of \(s\) for which \(a_s/q_s = a_r/q_r\) is \(\ll Q^2/q^2\). Let \(1 \leq q \leq Q\). Show that the number of \(r\) for which \(q_r = q\) is \[ \ll \; (Q/q)^2 \left( BNq / Q^3 + 1 \right). \] (f) Conclude that under the hypotheses of (c), \[ \sum_{n=0}^{N} B_1(\{ f(n) \}) \;\ll\; B^2 N Q^{-1} \log 2Q + B Q^2. \]

Solution: (a) \[\exp\left(2\pi i \sum_{n=1}^{q} B_1\!\left( \left\{ \frac{a}{q}n + \beta \right\} \right) \right) = \exp(2\pi i B_1(\{ q\beta \})).\] Since \[\sum_{n=1}^{q} B_1\!\left( \left\{ \frac{a}{q}n + \beta \right\} \right) = \sum_{n=1}^{q} B_1\!\left( \left\{ \frac{n}{q} + \beta \right\} \right),\] \[-\frac{1}{2}\le \sum_{n=1}^{q} B_1\!\left( \left\{ \frac{a}{q}n + \beta \right\} \right) \le \frac{1}{2}.\]

(b) \begin{align*} \sum_{n=1}^{q} B_1(\{ f(n) \}) &= \sum_{n=1}^{q} \left( \{ f(n) \} - \left\{ \frac{a}{q}n + f(1) \right\} \right) + \sum_{n=1}^{q} B_1\!\left( \left\{ \frac{a}{q}n + f(1)\right\} \right)\ll A. \end{align*}

(c) By Dirichlet's approximation theorem, it's done.

(d) With (b), it's done.

Problem: Let \(r(n)\) be the number of ordered pairs \((a,b)\) of integers for which \(a^2+b^2=n\).

(a) Show that \[ \sum_{n \le x} r(n) = 1 + 4\left[ \sqrt{x} \right] + 8 \sum_{1 \le n \le \sqrt{x/2}} \left[ \sqrt{x - n^2} \right] - 4 \left[ \sqrt{x/2} \right]^2. \] (b) Show that \[ \sum_{1 \le n \le \sqrt{x/2}} \sqrt{x - n^2} = \left( \frac{\pi}{8} + \frac{1}{2} \right) x - B_1\!\left(\left\{\sqrt{x/2}\right\}\right) - \frac{1}{2}\sqrt{x} + O(1). \] (c) Write \(\sum_{0 \le n \le x} r(n) = \pi x + R(x)\). Show that \[ R(x) = -8 \sum_{1 \le n \le \sqrt{x/2}} B_1\!\left(\left\{\sqrt{x - n^2}\right\}\right) + O(1). \]

Solution: (a) \begin{align*} \sum_{n \le x} r(n) &= \#\{(a,b)\in \mathbb{Z}^2: a^2+b^2 \le x\}\\ &= \#\{(0,0)\} +4\times \#\left\{(a,0):a\le\sqrt{x}\right\}\\ &\qquad\qquad+ 8\times \#\left\{(a,b): a^2+b^2 \le x,\ a\le b\right\} \\ &\qquad\qquad\qquad- 4\times \#\left\{(a,b): a,b\le \sqrt{x/2}\right\}\\ &= 1 + 4\left[ \sqrt{x} \right] + 8 \sum_{1 \le n \le \sqrt{x/2}} \left[ \sqrt{x - n^2} \right] - 4 \left[ \sqrt{x/2} \right]^2. \end{align*}

(b) \begin{align*} \sum_{1 \le n \le \sqrt{x/2}} \sqrt{x - n^2} &= \int_{0+}^{\sqrt{x/2}}\sqrt{x - u^2}\ d[u] \\ &= \int_{0+}^{\sqrt{x/2}}\sqrt{x - u^2}\ d\left(u-B_1(\{u\}) \right)\\ &= \left( \frac{\pi}{8} + \frac{1}{2} \right) x - \left(\left. \sqrt{x-u^2}B_1(\{u\})\right|_{0}^{\sqrt{x/2}} + \int_{0+}^{\sqrt{x/2}}\frac{u}{\sqrt{x - u^2}}B_1(\{u\})du\right)\\ &=\left( \frac{\pi}{8} + \frac{1}{2} \right) x - \sqrt{x/2}\ B_1\!\left(\left\{\sqrt{x/2}\right\}\right) - \frac{1}{2}\sqrt{x} + O(1). \end{align*}

(c) \begin{align*} \sum_{n \le x} r(n) &= 1 + 4\left[ \sqrt{x} \right] + 8 \sum_{1 \le n \le \sqrt{x/2}} \left[ \sqrt{x - n^2} \right]- 4 \left[ \sqrt{x/2} \right]^2\\ &= 1 + 4\left[ \sqrt{x} \right] + 8 \sum_{1 \le n \le \sqrt{x/2}} \left( \sqrt{x - n^2}- 1/2 -B_1\!\left(\left\{\sqrt{x - n^2}\right\}\right) \right) - 4 \left[ \sqrt{x/2} \right]^2\\ &= 4\left[ \sqrt{x} \right] - 4 \left[ \sqrt{x/2} \right]^2 + {\pi}x + 4 x - 8\sqrt{x/2}\ B_1\!\left(\left\{\sqrt{x/2}\right\}\right)\\ &\qquad\qquad- 4\sqrt{x} -8 \sum_{1 \le n \le \sqrt{x/2}} B_1\!\left(\left\{\sqrt{x - n^2}\right\}\right)-4\left[\sqrt{x/2}\right] + O(1)\\ &=\pi x -8 \sum_{1 \le n \le \sqrt{x/2}} B_1\!\left(\left\{\sqrt{x - n^2}\right\}\right) + O(1). \end{align*}