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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*}

Problem:Let \(B_1(x) = x - 1/2\), as in Appendix B.

(a) Show that \[ \sum_{n \le x} \frac{1}{n} = \log x + C_0 - {B_1(\{x\})}/{x} + O(1/x^2). \] (b) Write \(\sum_{n \le x} d(n) = x \log x + (2C_0 - 1)x + \Delta(x)\). Show that \[ \Delta(x) = -2 \sum_{n \le \sqrt{x}} B_1(\{x/n\}) + O(1). \] (c) Show that \(\int_0^X \Delta(x)\, dx \ll X\).

(d) Deduce that \begin{align*} \sum_{n \le X} d(n)(X-n) &= \int_0^X \left( \sum_{n \le x} d(n) \right)dx \\ &= \frac{1}{2} X^2 \log X + \left( C_0 - \frac{3}{4} \right) X^2 + O(X). \end{align*}

Solution: (a) \begin{align*} \sum_{n \le x} \frac{1}{n} &= \int_{1-}^{x} \frac{1}{u} d[u] = \int_{1-}^{x} \frac{1}{u} du - \int_{1-}^{x} \frac{1}{u} d\{u\} \\ &=\log x - \int_{1-}^{x} \frac{1}{u} d\left(\{u\}- 1/2 \right)=\log x - \int_{1-}^{x} \frac{1}{u} dB_{1}(\{u\})\\ &=\log x - \frac{1}{u} B_{1}(\{u\})\Bigg|_{1}^{x} - \int_{1-}^{x} \frac{1}{u^2} B_{1}(\{u\}) du\\ &= \log x - {B_1(\{x\})}/{x} + \frac{1}{2} - \int_{1}^{x} \frac{1}{u^2}\left(\{u\}- 1/2 \right) du\\\ &= \log x - {B_1(\{x\})}/{x} + 1 - \int_{1}^{\infty} \frac{\{u\}}{u^2} du + \int_{x}^{\infty}\frac{1}{u^2}\left(\{u\}- 1/2 \right) du \\ &= \log x + C_0 - {B_1(\{x\})}/{x} + O(1/x^2). \end{align*}

(b) Note that \begin{align*} \sum_{n \le \sqrt{x}}[x/n] &= \sum_{n \le \sqrt{x}}\left(x/n- 1/2 - B_1(\{x/n\}) \right) \\ &= x\left(\frac{1}{2}\log x + C_0 - {B_1(\{\sqrt{x}\})}/{\sqrt{x}} + O(1/x) \right) - [\sqrt{x}]/2 -\sum_{n \le \sqrt{x}} B_1(\{x/n\})\\ &= \frac{1}{2}x\log x+ C_0 x - \sqrt{x} \{\sqrt{x}\} + \sqrt{x}/2 - [\sqrt{x}]/2 -\sum_{n \le \sqrt{x}} B_1(\{x/n\})+ O(1) . \end{align*} Hence, \begin{align*} \sum_{n \le x} d(n) &= 2 \sum_{n \le \sqrt{x}}[x/n] - [\sqrt{x}]^2 \\ &= x\log x+2 C_0 x - 2\sqrt{x} \{\sqrt{x}\} + \sqrt{x} - [\sqrt{x}]- [\sqrt{x}]^2 -2\sum_{n \le \sqrt{x}} B_1(\{x/n\})+ O(1)\\ &= x\log x+2 C_0 x - 2\sqrt{x} \{\sqrt{x}\} + \{\sqrt{x}\}- x + 2\sqrt{x}\{\sqrt{x}\} + \{\sqrt{x}\}^2 + \Delta (x)\\ &=x \log x + (2C_0 - 1)x + \Delta(x). \end{align*}

(c) Note that \[\left|\int_{a}^{b} B_{1}\left( \{x\} \right)dx \right| \le 1/4\] for every \(a,b\). Hence, \begin{align*} \left| \int_{0}^{X} \sum_{n \le \sqrt{x}} B_{1} (\{x/n\}) dx\right| &= \left|\sum_{0 \le k \le \sqrt{X}-1} \int_{k^2}^{(k+1)^2} \sum_{n \le k} B_{1} (\{x/n\})dx \right|\\ &\le \sum_{0 \le k \le \sqrt{X}-1} \sum_{n \le k}\left| \int_{k^2}^{(k+1)^2} B_{1} (\{x/n\})dx \right|\\ &\le \sum_{0 \le k \le \sqrt{X}-1} \frac{1}{4}k \ll X. \end{align*}

(d) \begin{align*} \int_0^X \left( \sum_{n \le x} d(n) \right)dx &= \int_0^X ( x \log x + (2C_0 - 1)x + \Delta(x)) dx\\ &= \frac{1}{2} X^2 \log X + \left( C_0 - \frac{3}{4} \right) X^2 + O(X). \end{align*}