| 1. | Lecture 1 |
| 2. | Lecture 2 |
| 3. | Lecture 3 |
| 4. | Lecture 4 |
| 5. | Lecture 5 |
| 6. | Lecture 6 |
| 7. | Lecture 7 |
| 8. | Lecture 8 |
| 9. | Lecture 9 |
| 10. | Lecture 10 |
| 11. | Lecture 11 |
| 12. | Lecture 12 |
| 13. | Lecture 13 |
Definition. Given \(k \in \mathbb{Z}_+\), define \(G(k)\) to be the least integer having the property that whenever \(s \geqslant G(k)\), then all sufficiently large natural numbers are the sum of \(s\) positive integer \(k\)-th powers.
Thus, when \(k \in \mathbb{Z}_+\) and \(s \geqslant G(k)\), there exists \(N_{0}=N_{0}(s, k)\) such that, whenever \(n \geqslant N_{0}\), then there exist \(x_{1}, \ldots, x_{s} \in \mathbb{Z}_+\) such that \[n=x_{1}^{k}+\ldots+x_{s}^{k}.\]
A relatively easy exercise shows that \(G(k) \geqslant k+1\) whenever \(k \geqslant 2\).
The current state of the art for \(k\in[9]\).
\(G(2)=4\), a consequence of Lagrange’s theorem from 1770;
\(G(3) \leqslant 7\), due to Linnik, 1942;
\(G(4)=16\), due to Davenport, 1939;
\(G(5) \leqslant 17\), due to Vaughan and Wooley, 1995;
\(G(6) \leqslant 24\), due to Vaughan and Wooley, 1994;
\(G(7) \leqslant 31\), due to Wooley, 2016;
\(G(8) \leqslant 39\), due to Wooley, 2016 (and it is known that \(G(8) \geqslant 32\) );
\(G(9) \leqslant 47\), due to Wooley, 2016;
In general, for large values of \(k\), it was shown 30 years ago that \[G(k) \leqslant k(\log k+\log \log k+2+o(1)) \quad(\text{Wooley, } 1992 \text { and 1995)},\] where \(o(1) \rightarrow 0\) as \(k \rightarrow \infty\).
Within the past year, this longstanding upper bound has been improved so that for all natural numbers \(k\) one has \[G(k) \leqslant\lceil k(\log k+4.20032)\rceil \quad \text { (Brüdern and Wooley 2022). }\]
Let us now return to Hardy and Littlewood in 1920, and indeed to Hardy and Ramanujan in 1918. They considered a power series \[g_{k}(z)=\sum_{m=1}^{\infty} z^{m^{k}}\]
Note that this series is absolutely convergent for \(|z|<1\). If one now considers the expression \(g_{k}(z)^{s}\), one sees that \[\begin{aligned} g_{k}(z)^{s} =\left(\sum_{m_{1}=1}^{\infty} z^{m_{1}^{k}}\right)\left(\sum_{m_{2}=1}^{\infty} z^{m_{2}^{k}}\right) \cdots\left(\sum_{m_{s}=1}^{\infty} z^{m_{s}^{k}}\right) =\sum_{m_{1}=1}^{\infty} \ldots \sum_{m_{s}=1}^{\infty} z^{m_{1}^{k}+\ldots+m_{s}^{k}} \end{aligned}\]
Then we can further write \[\begin{aligned} g_{k}(z)^{s} & =\left(\sum_{m_{1}=1}^{\infty} z^{m_{1}^{k}}\right)\left(\sum_{m_{2}=1}^{\infty} z^{m_{2}^{k}}\right) \cdots\left(\sum_{m_{s}=1}^{\infty} z^{m_{s}^{k}}\right) \\ & =\sum_{m_{1}=1}^{\infty} \ldots \sum_{m_{s}=1}^{\infty} z^{m_{1}^{k}+\ldots+m_{s}^{k}} =\sum_{n=1}^{\infty} R_{s, k}(n) z^{n}, \end{aligned}\] where \(R_{s, k}(n)=\#\left\{(m_{1}, \ldots, m_{s}) \in \mathbb{Z}_+^s: m_{1}^{k}+\ldots+m_{s}^{k}=n\right\}\).
We can recover the coefficients \(R_{s, k}(n)\) by employing Cauchy’s integral formula to evaluate a suitable contour integral. Thus \[R_{s, k}(n)=\frac{1}{2 \pi i} \int_{\mathcal{C}} g_{k}(z)^{s} z^{-n-1} \mathrm{~d} z\] where \(\mathcal{C}\) denotes a circular contour, centered at \(0\) with radius \(r\in(0, 1)\).
When \(k=1\), the series in question is \(g_{1}(z)=z /(1-z)\), and Hardy and Ramanujan obtained an asymptotic formula for \(R_{s, k}(n)\) by evaluating their generating functions asymptotically for all values of \(\theta\).
The method also applies even in the more delicate situation with \(k=2\). However, when \(k \geqslant 3\), the situation is much more involved and here the innovative circle method of Hardy and Littlewood becomes essential.
The basis of this method is the elementary orthogonality identity \[\begin{aligned} \int_{0}^{1} e(k \theta) d \theta=\int_{0}^{1} e^{2 \pi i k \theta} d \theta= \begin{cases} 1 & \text { if } k=0,\\ 0 & \text { if } k\neq0. \end{cases} \end{aligned}\]
Fix \(n\in\mathbb Z_+\) and let \(X=n^{1/k}\). Using this identity, in a similar way as in the Vinogradov’s mean value theorem we have \[\begin{aligned} R_{s, k}(n)&=\#\left\{(m_{1}, \ldots, m_{s}) \in \mathbb{Z}_+^s: m_{1}^{k}+\ldots+m_{s}^{k}=n\right\}\\ &=\sum_{(m_{1}, \ldots, m_{s}) \in \mathbb{Z}_+^s}\int_0^1e\big((m_{1}^{k}+\ldots+m_{s}^{k}-n)\alpha\big)d\alpha\\ &=\int_0^1\Big(\sum_{1\le x\le X}e(\alpha x^k)\Big)^se(-n\alpha)d\alpha \end{aligned}\]
Define \[f(\alpha)=\sum_{1\le x\le X}e(\alpha x^k).\]
Whenever \(s\ge 2^k+1\), our goal is to asymptotically evaluate the number \[R_{s, k}(n)=\int_{0}^{1} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\]
We divide the interval of integration according to a Hardy–Littlewood dissection with major arcs \(\mathfrak{M}_{\delta}\) equal to the union of the intervals \[\mathfrak{M}_{\delta}(q, a)=\left\{\alpha \in[0,1):|\alpha-a / q| \leqslant X^{\delta-k}\right\}\] with \(0 \leqslant a < q \leqslant X^{\delta}\) and \((a, q)=1\), and with minor arcs \[\mathfrak{m}_{\delta}=[0,1] \backslash \mathfrak{M}_{\delta}.\]
Subject to the condition \(0<\delta<1/5\), the major arcs \(\mathfrak{M}_{\delta}\) defined in this way are a disjoint union of the arcs \(\mathfrak{M}_{\delta}(q, a)\).
Indeed, if some real number \(\alpha\) lies in two distinct major \(\operatorname{arcs} \mathfrak{M}_{\delta}\left(q_{1}, a_{1}\right)\) and \(\mathfrak{M}_{\delta}\left(q_{2}, a_{2}\right)\) lying in \(\mathfrak{M}_{\delta}\), then by the triangle inequality, one has \[\frac{1}{q_{1} q_{2}} \leqslant\left|\frac{a_{1} q_{2}-a_{2} q_{1}}{q_{1} q_{2}}\right| \leqslant\left|\frac{a_{1}}{q_{1}}-\frac{a_{2}}{q_{2}}\right| \leqslant\left|\alpha-\frac{a_{1}}{q_{1}}\right|+\left|\alpha-\frac{a_{2}}{q_{2}}\right| \leqslant 2 X^{\delta-k} .\]
Thus, one finds that \(1 \leqslant 2 q_{1} q_{2} X^{\delta-k} \leqslant 2 X^{3 \delta-k}\). This is plainly impossible when \(\delta<1 / 3\) and \(X\) is large.
The exponential sum \[f(\alpha)=\sum_{1\le x\le X}e(\alpha x^k)\] can be approximate by the integral on major arcs, whereas on minor arcs one expects that the sequence from the phase is equidistributed due to Weyl’s inequality.
We will write \[R_{s, k}(n)=\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha +\int_{\mathfrak{M}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\] We first handle the integral over minor arcs.
Let \(\alpha \in \mathbb{R}\), and suppose that \(X \geqslant 1\) is a real number. Then there exist \(a \in \mathbb{Z}\) and \(q \in \mathbb{N}\) with \((a, q)=1\) and \(1 \leqslant q \leqslant X\) such that \(|\alpha-a / q| \leqslant 1 /(q X)\).
Proof. Exercise!$$\tag*{$\blacksquare$}$$
Given \(\alpha \in[0,1)\), by Dirichlet’s approximation theorem, there exist \(a \in \mathbb{Z}\) and \(q \in \mathbb{N}\) with \(1 \leqslant q \leqslant X^{k-\delta},(a, q)=1\) and \[|\alpha-a / q| \leqslant 1 /\left(q X^{k-\delta}\right) \leqslant \min\{X^{\delta-k}, q^{-2}\}.\]
If \(q \leqslant X^{\delta}\), then we would have \(\alpha \in \mathfrak{M}_{\delta}\). Thus, when \(\alpha \in \mathfrak{m}_{\delta}\), we may suppose that \(X^{\delta}<q \leqslant X^{k-\delta}\). We thus conclude from Weyl’s inequality that, whenever \(0<\delta<1\), one has \[\begin{aligned} |f(\alpha)| &=O \Big(X^{1+\varepsilon}\left(q^{-1}+X^{-1}+q X^{-k}\right)^{2^{1-k}}\Big)\\ &=O \Big(X^{1+\varepsilon}\left(X^{-\delta}+X^{-1}+X^{k-\delta} / X^{k}\right)^{2^{1-k}}\Big)\\ &= O\big(X^{1-\delta 2^{1-k}+\varepsilon}\big). \end{aligned}\]
Provided that \(s>(k / \delta) 2^{k-1}\), we may conclude that \[\begin{aligned} \left|\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\right| & \leqslant\left(\sup _{\alpha \in \mathfrak{m}_{\delta}}|f(\alpha)|\right)^{s} \int_{\mathfrak{m}_{\delta}} \mathrm{d} \alpha \\ & =O\Big(\big(X^{1-\delta 2^{1-k}+\varepsilon}\big)^{s}\Big)=o\left(X^{s-k}\right) . \end{aligned}\]
But our goal is to asymptotically evaluate \(R_{s, k}(n)\) assuming that \(s\ge 2^k+1\).
When \(s \geqslant 2^{k}+1\), one has \[\Big|\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\Big| =O \big(X^{s-k-\delta 2^{-k}}\big).\]
Proof. By Weyl’s inequality in combination with Hua’s lemma, one obtains \[\begin{aligned} \left|\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\right| & \leqslant\left(\sup _{\alpha \in \mathfrak{m}_{\delta}}|\: f(\alpha)|\right)^{s-2^{k}} \int_{0}^{1}|\:f(\alpha)|^{2^{k}} \mathrm{~d} \alpha \\ & =O\bigg(\left(X^{1-\delta 2^{1-k}+\varepsilon}\right)^{s-2^{k}} X^{2^{k}-k+\varepsilon}\bigg) \\ & =O\Big(X^{s-k-\left(s-2^{k}\right) \delta 2^{1-k}+s \varepsilon}\Big). \end{aligned}\] The conclusion of the corollary follows on recalling that \(s \geqslant 2^{k}+1\).$$\tag*{$\blacksquare$}$$
For \(s \geqslant 2^{k}+1\) we have shown that \[\Big|\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha\Big| =O \big(X^{s-k-\delta 2^{-k}}\big)=o(n^{s/k-1}).\]
Let \(\alpha \in \mathfrak{M}_{\delta}(q, a) \subseteq \mathfrak{M}_{\delta}\). Write \(\beta=\alpha-a / q\), so that \(|\beta| \leqslant X^{\delta-k}\). By breaking the summand into arithmetic progressions modulo \(q\), one has
\[\begin{align*} \sum_{1 \leqslant x \leqslant X} e\left(\alpha x^{k}\right) & =\sum_{r=1}^{q} \sum_{(1-r) / q \leqslant y \leqslant(X-r) / q} e\left((\beta+a / q)(y q+r)^{k}\right) \tag{A}\\ & =\sum_{r=1}^{q} e\left(a r^{k} / q\right) \sum_{(1-r) / q \leqslant y \leqslant(X-r) / q} e\left(\beta(y q+r)^{k}\right) . \end{align*}\]
Since \(\beta\) is small, we can hope to approximate the inner sum here by a smooth function with control of the accompanying error terms.
Here, we apply the mean value theorem to the inner sum.
By the mean value theorem, when \(F(z)\) is a differentiable function on \([a, b]\) with \(a<b\), one sees that \(F(a)-F(b)=(a-b) F^{\prime}(\xi)\) for some \(\xi \in(a, b)\). Also, trivially, one has \[e(F(z))=\int_{-1 / 2}^{1 / 2} e(F(z)) \mathrm{d} \eta\]
Hence \[\begin{aligned} \left|e(F(z))-\int_{-1 / 2}^{1 / 2} e(F(z+\eta)) \mathrm{d} \eta\right| & \leqslant \sup _{|\eta| \leqslant 1 / 2}|e(F(z+\eta))-e(F(z))| \\ & =O\big(\sup _{|\eta| \leqslant 1 / 2}\left|F^{\prime}(z+\eta)\right|\big). \end{aligned}\]
Using this approximation, we obtain \[\begin{aligned} \sum_{(1-r) / q \leqslant y \leqslant(X-r) / q} &e\left(\beta(y q+r)^{k}\right)- \int_{-r / q}^{(X-r) / q} e\left(\beta(z q+r)^{k}\right) \mathrm{d} z \\ & =O \Big(1+(X / q) \sup _{0 \leqslant z \leqslant X / q}\left|k \beta q(q z+r)^{k-1}\right| \Big)\\ & =O \big(1+X^{k}|\beta|\big). \end{aligned}\]
By substituting the last relation into (A), we deduce that \[f(\alpha)=\sum_{r=1}^{q} e\left(a r^{k} / q\right)\left(\int_{-r / q}^{(X-r) / q} e\left(\beta(z q+r)^{k}\right) \mathrm{d} z+O\left(1+X^{k}|\beta|\right)\right)\] so that \[f(\alpha)-\sum_{r=1}^{q} e\left(a r^{k} / q\right) \int_{-r / q}^{(X-r) / q} e\left(\beta(z q+r)^{k}\right) \mathrm{d} z = O\big(q+X^{k}|q \beta|\big) . \tag{B}\]
By the change of variable \(\gamma=z q+r\), moreover, we have \[\int_{-r / q}^{(X-r) / q} e\left(\beta(z q+r)^{k}\right) \mathrm{d} z=q^{-1} \int_{0}^{X} e\left(\beta \gamma^{k}\right) \mathrm{d} \gamma. \tag{C}\]
Introducing, for \(a \in \mathbb{Z}\), \(q \in \mathbb{Z}_+\) and \(\beta \in \mathbb{R}\) the following objects \[S(q, a)=\sum_{r=1}^{q} e\left(a r^{k} / q\right), \quad \text{ and } \quad v(\beta)=\int_{0}^{X} e\left(\beta \gamma^{k}\right) \mathrm{d} \gamma,\] we can summarize our discussion in the form of a lemma.
Suppose that \(\alpha \in \mathbb{R}, a \in \mathbb{Z}\) and \(q \in \mathbb{Z}_+\). Then one has \[|f(\alpha)-q^{-1} S(q, a) v(\alpha-a / q)| =O \big(q+X^{k}|q \alpha-a|\big) .\]
Proof. The desired conclusion follows by substituting (C) into (B).$$\tag*{$\blacksquare$}$$
When \(\alpha \in \mathfrak{M}_{\delta}(q, a) \subseteq \mathfrak{M}_{\delta}\), one has \[|f(\alpha)-q^{-1} S(q, a) v(\alpha-a / q)| =O (X^{2 \delta}) .\]
Proof. When \(\alpha \in \mathfrak{M}_{\delta}(q, a) \subseteq \mathfrak{M}_{\delta}\), one has \(|q \alpha-a|=q|\alpha-a / q| \leqslant X^{\delta} \cdot X^{\delta-k},\) whence \(q+X^{k}|q \alpha-a| =O (X^{2 \delta})\). The claimed bound now follows from the previous lemma.$$\tag*{$\blacksquare$}$$
Let us now substitute the conclusion of the previous lemma into the formula for the major arc contribution. Since \[\mathfrak{M}_{\delta}=\bigcup_{\substack{0 \leqslant a < q \leqslant X^{\delta} \\(a, q)=1}} \mathfrak{M}_{\delta}(q, a),\] then \[\int_{\mathfrak{M}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha=\sum_{1 \leqslant q \leqslant X^{\delta}} \sum_{\substack{a=1 \\(a, q)=1}}^{q} \int_{-X^{\delta-k}}^{X^{\delta-k}} f(\beta+a / q)^{s} e(-n(\beta+a / q)) \mathrm{d} \beta .\]
Assuming that \(\alpha\in \mathfrak{M}_{\delta}(q, a) \subseteq \mathfrak{M}_{\delta}\), we set \[f^{*}(\alpha)=q^{-1} S(q, a) v(\alpha-a / q),\] and write \[E(\alpha)=f(\alpha)-f^{*}(\alpha)\]
It follows from the previous lemma that \(E(\alpha) =O (X^{2 \delta})\).
Since \[\begin{aligned} f(\alpha)^{s}-f^{*}(\alpha)^{s} & =\left(\: f(\alpha)-f^{*}(\alpha)\right)\left(\: f(\alpha)^{s-1}+\ldots+f^{*}(\alpha)^{s-1}\right) \\ & =O (X^{s-1}|E(\alpha)|) =O (X^{s-1+2 \delta}), \end{aligned}\] we obtain the asymptotic relation \[\begin{aligned} \int_{\mathfrak{M}_{\delta}} f(\alpha)^{s}& e(-n \alpha) \mathrm{d} \alpha\\ = & \sum_{1 \leqslant q \leqslant X^{\delta}} \sum_{\substack{a=1 \\ (a, q)=1}}^{q} \int_{-X^{\delta-k}}^{X^{\delta-k}}\left(q^{-1} S(q, a) v(\beta)\right)^{s} e(-n(\beta+a / q)) \mathrm{d} \beta \\ & +\sum_{1 \leqslant q \leqslant X^{\delta}} \sum_{\substack{a=1 \\ (a, q)=1}}^{q} \int_{-X^{\delta-k}}^{X^{\delta-k}} X^{s-1+2 \delta} \mathrm{~d} \alpha. \end{aligned}\]
The second sum is \[O\Big(X^{s-1+2 \delta} \sum_{1 \leqslant q \leqslant X^{\delta}} q \cdot X^{\delta-k}\Big) =O(X^{s-k-1+3 \delta} \cdot X^{2 \delta}) =O (X^{s-k+(5 \delta-1)}).\]
This is \(o\left(X^{s-k}\right)\) whenever \(\delta<1 / 5\).
Turning to the first sum, we find that it factorises in the shape \[\sum_{1 \leqslant q \leqslant X^{\delta}} \sum_{\substack{a=1 \\(a, q)=1}}^{q}\left(q^{-1} S(q, a)\right)^{s} e(-n a / q) \int_{-X^{\delta-k}}^{X^{\delta-k}} v(\beta)^{s} e(-\beta n) \mathrm{d} \beta.\]
When \(Q\in\mathbb R_+\), we define the truncated singular series \[\mathfrak{S}_{s, k}(n ; Q)=\sum_{1 \leqslant q \leqslant Q} \sum_{\substack{a=1 \\(a, q)=1}}^{q}\left(q^{-1} S(q, a)\right)^{s} e(-n a / q),\] and the truncated singular integral \[J_{s, k}(n ; Q)=\int_{-Q X^{-k}}^{Q X^{-k}} v(\beta)^{s} e(-\beta n) \mathrm{d} \beta.\]
Now we can summarize our discussion in the form of a lemma.
When \(0<\delta<1\), one has \[\int_{\mathfrak{M}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha=J_{s, k}\left(n ; X^{\delta}\right) \mathfrak{S}_{s, k}\left(n ; X^{\delta}\right)+O\left(X^{s-k+(5 \delta-1)}\right) .\]
When \(s \geqslant 2^{k}+1\) and \(0<\delta<1 / 5\), one has \[R_{s, k}(n)=J_{s, k}\left(n ; X^{\delta}\right) \mathfrak{S}_{s, k}\left(n ; X^{\delta}\right)+o\left(X^{s-k}\right)\] in which \(X=n^{1 / k}\).
Proof. Since \([0,1)\) is the disjoint union of \(\mathfrak{m}_{\delta}\) and \(\mathfrak{M}_{\delta}\), one has \[R_{s, k}(n)=\int_{\mathfrak{M}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha+\int_{\mathfrak{m}_{\delta}} f(\alpha)^{s} e(-n \alpha) \mathrm{d} \alpha.\] The conclusion follows from the previous results.$$\tag*{$\blacksquare$}$$
Our objective is now to analyse the truncated singular series \(\mathfrak{S}_{s, k}\left(n ; Q\right)\) and singular integral \(J_{s, k}\left(n ; Q\right)\).
We first consider the truncated singular integral \(J_{s, k}(n ; Q)\), our first step being to complete this integral to obtain the (complete) singular integral \[J_{s, k}(n)=\int_{-\infty}^{\infty} v(\beta)^{s} e(-n \beta) \mathrm{d} \beta .\]
Whenever \(\beta \in \mathbb{R}\), one has \[v(\beta) =O \Big(X\left(1+X^{k}|\beta|\right)^{-1 / k}\Big).\]
Proof.
Recall that \[v(\beta)=\int_{0}^{X} e\left(\beta \gamma^{k}\right) \mathrm{d} \gamma.\]
The estimate \(|v(\beta)| \leqslant X\) is trivial. Also, since \(|v(\beta)|=|v(-\beta)|\), we may assume henceforth that \(\beta>X^{-k}\).
Changing the variable \(u=\beta \gamma^{k}\), we find that when \(\beta>0\), one has \[v(\beta)=k^{-1} \beta^{-1 / k} \int_{0}^{\beta X^{k}} u^{-1+1 / k} e(u) \mathrm{d} u,\] whence \[|v(\beta)| \leqslant k^{-1} \beta^{-1 / k}\bigg|\int_{0}^{\beta X^{k}} u^{-1+1 / k} e(u) \mathrm{d} u\bigg|.\]
Notice that \(u^{-1+1 / k}\) decreases monotonically to 0 as \(u \rightarrow \infty\). By Dirichlet’s test for convergence of an infinite integral the last integral is uniformly bounded, and indeed \[\bigg|\int_{0}^{\beta X^{k}} u^{-1+1 / k} e(u) \mathrm{d} u\bigg| \leqslant \sup _{Y \geqslant 0}\left|\int_{0}^{Y} u^{-1+1 / k} e(u) \mathrm{d} u\right|<\infty\]
When \(0<Y<1\), we are also making use of the inequality \[\left|\int_{0}^{Y} u^{-1+1 / k} e(u) \mathrm{d} u\right| \leqslant \int_{0}^{Y} u^{-1+1 / k} \mathrm{~d} u =O(1).\]
Hence we deduce that when \(|\beta|>X^{-k}\), one has \[|v(\beta)| =O\big(|\beta|^{-1 / k}\big) =O \Big(X\left(1+X^{k}|\beta|\right)^{-1 / k}\Big).\]
The desired conclusion follows on combining this estimate with our earlier bound \(|v(\beta)| \leqslant\) \(X\), applied in circumstances wherein \(|\beta| \leqslant X^{-k}\).
The proof is completed. $$\tag*{$\blacksquare$}$$
Suppose that \(s \geqslant k+1\). Then the singular series \(J_{s, k}(n)\) converges absolutely, and moreover, \[|J_{s, k}(n ; Q)-J_{s, k}(n)| =O (X^{s-k} Q^{-1 / k}).\]
Proof.
By applying the last lemma, one sees that \[|J_{s, k}(n)| =O \bigg(\int_{-\infty}^{\infty} \frac{X^{s}}{\left(1+X^{k}|\beta|\right)^{s / k}} \mathrm{~d} \beta\bigg) =O (X^{s-k}).\]
Thus, the integral defining \(J_{s, k}(n)\) is indeed absolutely convergent, and the singular integral exists. Moreover, and similarly, \[|J_{s, k}(n ; Q)-J_{s, k}(n)| =O\bigg(\int_{Q X^{-k}}^{\infty} \frac{X^{s}}{\left(1+X^{k} \beta\right)^{1+1 / k}} \bigg)= O(X^{s-k} Q^{-1 / k})\]
This completes the proof.$$\tag*{$\blacksquare$}$$
When \(s \geqslant k+1\), one has \[J_{s, k}(n)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1}\] in which \[\Gamma(z)=\int_{0}^{\infty} t^{z-1} e^{-t} \mathrm{~d} t \quad \text{ for } \quad \operatorname{Re}z>0.\]
Proof.
We begin by observing that \[\begin{aligned} J_{s, k}(n) & =\lim _{B \rightarrow \infty} \int_{-B}^{B} v(\beta)^{s} e(-\beta n) \mathrm{d} \beta \\ & =\lim _{B \rightarrow \infty} \int_{-B}^{B} \int_{[0, X]^{s}} e\left(\beta\left(\gamma_{1}^{k}+\ldots+\gamma_{s}^{k}-n\right)\right) \mathrm{d} \boldsymbol{\gamma} \mathrm{~d} \beta \\ & =\lim _{B \rightarrow \infty} \int_{[0, X]^{s}} \int_{-B}^{B} e\left(\beta\left(\gamma_{1}^{k}+\ldots+\gamma_{s}^{k}-n\right)\right) \mathrm{d} \beta \mathrm{~d} \boldsymbol{\gamma}. \end{aligned}\]
We make use of the observation that when \(\phi \neq 0\), one has \[\int_{-B}^{B} e(\beta \phi) \mathrm{d} \beta=\frac{\sin (2 \pi B \phi)}{\pi \phi} .\]
For \(\phi=0\), we interpret the right hand side of this formula to be \(2 B\). Thus we obtain the relation \[J_{s, k}(n)=\lim _{B \rightarrow \infty} \int_{[0, X]^{s}} \frac{\sin \left(2 \pi B\left(\gamma_{1}^{k}+\ldots+\gamma_{s}^{k}-n\right)\right)}{\pi\left(\gamma_{1}^{k}+\ldots+\gamma_{s}^{k}-n\right)} \mathrm{d} \boldsymbol{\gamma}\]
We substitute \(u_{i}=\gamma_{i}^{k}\) for \(i\in [s]\), and recall that \(n=X^{k}\). Thus \[J_{s, k}(n)=k^{-s} \lim _{B \rightarrow \infty} I(B),\] where we write \[I(B)=\int_{[0, n]^{s}} \frac{\sin \left(2 \pi B\left(u_{1}+\ldots+u_{s}-n\right)\right)}{\pi\left(u_{1}+\ldots+u_{s}-n\right)}\left(u_{1} \ldots u_{s}\right)^{-1+1 / k} \mathrm{~d} \mathbf{u}.\]
A further substitution reduces our task to one of evaluating an integral in just one variable. We put \(v=u_{1}+\ldots+u_{s}\) and make the change of variable \(\left(u_{1}, \ldots, u_{s}\right) \mapsto\) \(\left(u_{1}, \ldots, u_{s-1}, v\right)\), obtaining the relation \[I(B)=\int_{0}^{s n} \Psi(v) \frac{\sin (2 \pi B(v-n))}{\pi(v-n)} \mathrm{d} v,\] in which \[\Psi(v)=\int_{\mathfrak{B}(v)}\left(u_{1} \ldots u_{s-1}\right)^{\frac{1}{k}-1}\left(v-u_{1}-\ldots-u_{s-1}\right)^{\frac{1}{k}-1} \mathrm{~d} u_{1} \ldots \mathrm{~d} u_{s-1},\] and \[\mathfrak{B}(v)=\left\{\left(u_{1}, \ldots, u_{s-1}\right) \in[0, n]^{s-1}: 0 \leqslant v-u_{1}-\ldots-u_{s-1} \leqslant n\right\}.\]
Notice that the condition on \(u_{1}, \ldots, u_{s-1}\) in the definition of \(\mathfrak{B}(v)\) may be rephrased as \(v-n \leqslant u_{1}+\ldots+u_{s-1} \leqslant v\).
Since \(\Psi(v)\) is a function of bounded variation, it follows from Fourier’s integral theorem that since \(n \in(0, s n)\), one has \[\lim _{B \rightarrow \infty} I(B)=\Psi(n)=\int_{\mathfrak{B}(n)}\left(u_{1} \ldots u_{s-1}\right)^{\frac{1}{k}-1}\left(n-u_{1}-\ldots-u_{s-1}\right)^{\frac{1}{k}-1} \mathrm{~d} \mathbf{u} .\]
Note that \[\mathfrak{B}(n) =\left\{\left(u_{1}, \ldots, u_{s-1}\right) \in[0, n]^{s-1}: 0 \leqslant u_{1}+\ldots+u_{s-1} \leqslant n\right\}.\]
Thus \[J_{s, k}(n)=k^{-s} \Psi(n)=k^{-s} \int_{\mathfrak{B}(n)}\left(u_{1} \ldots u_{s-1}\right)^{\frac{1}{k}-1}\left(n-u_{1}-\ldots-u_{s-1}\right)^{\frac{1}{k}-1} \mathrm{~d} \mathbf{u} .\]
We now apply induction to show that \[J_{s, k}(n)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1}.\]
First, when \(s=2\), we have \[\begin{aligned} J_{2, k}(n) & =k^{-2} \int_{0}^{n} u_{1}^{\frac{1}{k}-1}\left(n-u_{1}\right)^{\frac{1}{k}-1} \mathrm{~d} u_{1} \\ & =k^{-2} n^{\frac{2}{k}-1} \int_{0}^{1} v^{\frac{1}{k}-1}(1-v)^{\frac{1}{k}-1} \mathrm{~d} v . \end{aligned}\]
Thus, on recalling the classical Beta function, we obtain the formula \[J_{2, k}(n)=k^{-2} n^{\frac{2}{k}-1} \mathrm{~B}(1 / k, 1 / k)=k^{-2} n^{\frac{2}{k}-1} \frac{\Gamma(1 / k)^{2}}{\Gamma(2 / k)}=\frac{\Gamma(1+1 / k)^{2}}{\Gamma(2 / k)} n^{\frac{2}{k}-1}.\]
Thus, the inductive hypothesis holds for \(s=2\). Suppose now that the inductive hypothesis holds for \(s=t\). Then we have \[\begin{aligned} J_{t+1, k}(n) & =k^{-1} \int_{0}^{n} u_{t}^{\frac{1}{k}-1} J_{t, k}\left(n-u_{t}\right) \mathrm{d} u_{t} \\ & =k^{-1} \frac{\Gamma(1+1 / k)^{t}}{\Gamma(t / k)} \int_{0}^{n} u_{t}^{\frac{1}{k}-1}\left(n-u_{t}\right)^{\frac{t}{k}-1} \mathrm{~d} u_{t}. \end{aligned}\]
Recalling once again the classical Beta function, we see that \[\begin{aligned} J_{t+1, k}(n) & =k^{-1} \frac{\Gamma(1+1 / k)^{t}}{\Gamma(t / k)} n^{\frac{t+1}{k}-1} \mathrm{~B}(1 / k, t / k) \\ & =k^{-1} \frac{\Gamma(1+1 / k)^{t}}{\Gamma(t / k)} n^{\frac{t+1}{k}-1} \frac{\Gamma(1 / k) \Gamma(t / k)}{\Gamma((t+1) / k)} \\ & =\frac{\Gamma(1+1 / k)^{t+1}}{\Gamma((t+1) / k)} n^{\frac{t+1}{k}-1}. \end{aligned}\]
This yields the inductive hypothesis with \(t\) replaced by \(t+1\). We have therefore shown that whenever \(s \geqslant k+1\), one has \[J_{s, k}(n)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1}.\]
Suppose that \(s \geqslant k+1\). Then one has \[J_{s, k}(n ; Q)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1}+O\left(n^{s / k-1} Q^{-1 / k}\right),\] as \(Q \rightarrow \infty\).
Proof. The conclusion follows by the previous two results, since \(X=\) \(n^{1 / k}\).$$\tag*{$\blacksquare$}$$
We next consider the truncated singular series \(\mathfrak{S}_{s, k}(n ; Q)\). Our first step is to complete this series to obtain the (complete) singular series \[\mathfrak{S}_{s, k}(n)=\sum_{q=1}^{\infty} \sum_{\substack{a=1 \\(a, q)=1}}^{q}\left(q^{-1} S(q, a)\right)^{s} e(-n a / q).\]
Again, we must consider the tail of the infinite sum.
Whenever \(a \in \mathbb{Z}\) and \(q \in \mathbb{N}\) satisfy \((a, q)=1\), one has \[|S(q, a)| =O (q^{1-2^{1-k}+\varepsilon}).\]
Proof. We apply Weyl’s inequality with \(\alpha_{k}=a / q\) and \(X=q\) to obtain \[\Big|\sum_{r=1}^{q} e\left(a r^{k} / q\right)\Big| =O\Big (q^{1+\varepsilon}\left(q^{-1}+q^{-1}+q^{1-k}\right)^{2^{1-k}}\Big).\]$$\tag*{$\blacksquare$}$$
Suppose that \(s \geqslant 2^{k}+1\). Then \(\mathfrak{S}_{s, k}(n)\) converges absolutely, and \[|\mathfrak{S}_{s, k}(n)-\mathfrak{S}_{s, k}(n ; Q)| =O \big(Q^{-2^{-k}}\big)\] uniformly in \(n\in\mathbb Z_+\).
By the previous lemma we estimate the tail of the truncated singular series as follows \[\sum_{q>Q} \sum_{\substack{a=1 \\(a, q)=1}}^{q}\left|\left(q^{-1} S(q, a)\right)^{s} e(-n a / q)\right| =O \Big(\sum_{q>Q} \phi(q)\left(q^{\varepsilon-2^{1-k}}\right)^{s}\Big).\]
Thus, when \(s \geqslant 2^{k}+1\), we deduce that \[\sum_{\substack{q>Q}} \sum_{\substack{a=1 \\(a, q)=1}}^{q}\left|\left(q^{-1} S(q, a)\right)^{s} e(-n a / q)\right| =O \Big(\sum_{q>Q} q^{\varepsilon-1-2^{1-k}}\Big) = O(Q^{-2^{-k}}).\]
It follows that the infinite series \(\mathfrak{S}_{s, k}(n)\) converges absolutely under these conditions, and moreover that \[|\mathfrak{S}_{s, k}(n)-\mathfrak{S}_{s, k}(n ; Q)| =O (Q^{-2^{-k}}).\]
Notice that this estimate is uniform in \(n\).
We shall see shortly that there is a close connection between the singular series \(\mathfrak{S}_{s, k}(n)\) and the number of solutions of the congruence \[x_{1}^{k}+\ldots+x_{s}^{k} \equiv n\pmod q,\] as \(q\) varies. This suggests a multiplicative theme.
Suppose that \((a, q)=(b, r)=(q, r)=1\). Then one has the quasimultiplicative relation \[S(q r, a r+b q)=S(q, a) S(r, b).\]
Proof.
Each residue \(m\) modulo \(q r\) with \(m \in [q r]\) is in bijective correspondence with a pair \((t, u)\) with \(t \in [q]\) and \(u \in [r]\), with \(m \equiv t r+u q\pmod {q r}\).
Indeed, if we write \(\bar{q}\) for any integer congruent to the multiplicative inverse of \(q\pmod r\), and \(\bar{r}\) for any integer congruent to the multiplicative inverse of \(r\pmod q\), then claimed bijection is as follows \(m \equiv(m \bar{r}) r+(m \bar{q}) q\pmod {q r}\), which follows from the Chinese remainder theorem.
Thus, we see that \[\begin{aligned} S(q r, a r+b q) & =\sum_{m=1}^{q r} e\left(\frac{a r+b q}{q r} m^{k}\right) \\ & =\sum_{t=1}^{q} \sum_{u=1}^{r} e\left(\frac{(a r+b q)(t r+u q)^{k}}{q r}\right) \\ & =\sum_{t=1}^{q} \sum_{u=1}^{r} e\left(\frac{a}{q}(t r)^{k}+\frac{b}{r}(u q)^{k}\right) . \end{aligned}\]
By the change of variable \(t r \mapsto t^{\prime}(\bmod q)\) and \(u q \mapsto u^{\prime}(\bmod r)\), bijective owing to the coprimality of \(q\) and \(r\), we obtain the relation \[S(q r, a r+b q)=\left(\sum_{v=1}^{q} e\left(a v^{k} / q\right)\right)\left(\sum_{w=1}^{r} e\left(b w^{k} / r\right)\right)=S(q, a) S(r, b) .\]
This completes the proof of the lemma. $$\tag*{$\blacksquare$}$$
Now define the quantity \[A(q, n)=\sum_{\substack{a=1 \\(a, q)=1}}^{q}\left(q^{-1} S(q, a)\right)^{s} e(-n a / q)\]
The quantity \(A(q, n)\) is a multiplicative function of \(q\).
Proof.
Suppose that \((q, r)=1\). Then by the Chinese remainder theorem, there is a bijection between the residue classes \(a\) modulo \(q r\) with \((a, q r)=1\), and the ordered pairs \((b, c)\) with \(b\pmod q\) and \(c\pmod r\) satisfying \((b, q)=(c, r)=1\), via the relation \(a \equiv b r+c q\pmod {q r}\).
Thus, we obtain \[\begin{aligned} A(q r, n) & =\sum_{\substack{a=1 \\ (a, q r)=1}}^{q r}\left((q r)^{-1} S(q r, a)\right)^{s} e(-n a / q r) \\ & =\sum_{\substack{b=1 \\ (b, q)=1}}^{q} \sum_{\substack{c=1 \\ (c, r)=1}}^{r}\left((q r)^{-1} S(q r, b r+c q)\right)^{s} e\left(-\frac{b r+c q}{q r} n\right) . \end{aligned}\]
By applying the previous lemma, we infer that \[\begin{aligned} A(q r, n)= & \sum_{\substack{b=1 \\ (b, q)=1}}^{q} \sum_{\substack{c=1 \\ (c, r)=1}}^{r}\left(q^{-1} S(q, b)\right)^{s}\left(r^{-1} S(r, c)\right)^{s} e(-b n / q) e(-c n / r) \\ & =A(q, n) A(r, n). \end{aligned}\]
Since \(A(1, n)=1\), this confirms the multiplicative property for \(A(q, n)\) and completes the proof of the lemma.$$\tag*{$\blacksquare$}$$
Observe that \[\mathfrak{S}_{s, k}(n)=\sum_{q=1}^{\infty} A(q, n).\]
The multiplicativity of \(A(q, n)\) therefore suggests that \(\mathfrak{S}_{s, k}(n)\) should factor as a product over prime numbers \(p\) of the \(p\)-adic densities \[\sigma(p)=\sum_{h=0}^{\infty} A\left(p^{h}, n\right).\]
Suppose that \(s \geqslant 2^{k}+1\). Then the following hold:
The series \(\sigma(p)\) converges absolutely, and one has \[|\sigma(p)-1| = O(p^{-1-2^{-k}}).\]
The infinite product \[\prod_{p \in\mathbb P} \sigma(p)\] converges absolutely.
One has \(\mathfrak{S}_{s, k}(n)=\prod_{p\in\mathbb P} \sigma(p)\).
There exists a natural number \(C=C(k)\) with the property that \[1 / 2<\prod_{p \in\mathbb P_{\geqslant C(k)}} \sigma(p)<3 / 2 .\]
We begin by establishing (i). We recall from estimates of complete exponential sums that whenever \((a, p)=1\), one has \[|S\left(p^{h}, a\right)| =O\big(p^{h(1-2^{1-k}+\varepsilon)} \big).\]
Then, whenever \(s \geqslant 2^{k}+1\), one finds that \[\begin{aligned} A\left(p^{h}, n\right)&=\sum_{\substack{a=1 \\(a, p)=1}}^{p^{h}}\left(p^{-h} S\left(p^{h}, a\right)\right)^{s} e\left(-n a / p^{h}\right)\\ &=O\big(p^{h\left(1-s 2^{1-k}\right)+\varepsilon}\big) =O \big(p^{-h\left(1+2^{-k}\right)}\big). \end{aligned}\]
Hence \[\begin{aligned} \sigma(p)-1=\sum_{h=1}^{\infty} A\left(p^{h}, n\right) =O \Big(\sum_{h=1}^{\infty} p^{-h\left(1+2^{-k}\right)}\Big) =O (p^{-1-2^{-k}}). \end{aligned}\]
Thus \(\sigma(p)\) converges absolutely, and one has \(|\sigma(p)-1|=O (p^{-1-2^{-k}})\).
We next turn to the proof of (ii). By part (i), there is a positive number \(B=B(k)\) with the property that \(|\sigma(p)-1| \leqslant B p^{-1-2^{-k}}\).
Hence, whenever \(p\) is sufficiently large, one sees that \[\log (1+|\sigma(p)-1|) \leqslant \log \left(1+B p^{-1-2^{-k}}\right) \leqslant B p^{-1-2^{-k-1}},\] whence \[\sum_{p\in\mathbb P} \log (1+|\sigma(p)-1|) =O\Big (B \sum_{p\in\mathbb P} p^{-1-2^{-k-1}}\Big) = O(1).\]
Thus we deduce that the infinite product \(\prod_{p} \sigma(p)\) converges absolutely.
The proof of (iii) employs the multiplicative property of \(A(q, n)\) established in the previous lemma. One finds that \[\mathfrak{S}_{s, k}(n)=\sum_{q=1}^{\infty} A(q, n)=\sum_{q=1}^{\infty} \prod_{p^{h} \| q} A\left(p^{h}, n\right)\]
Then since \(\prod_{p\in\mathbb P} \sigma(p)\) converges absolutely as a product, and \(\sum_{q=1}^{\infty} A(q, n)\) converges absolutely as a sum, we may rearrange summands to deduce that \[\mathfrak{S}_{s, k}(n)=\prod_{p\in\mathbb P} \sum_{h=0}^{\infty} A\left(p^{h}, n\right)=\prod_{p\in\mathbb P} \sigma(p).\]
Finally, we establish (iv). We begin by observing that from part (i), it follows that whenever \(p\) is sufficiently large in terms of \(k\), one has \[1-p^{-1-2^{-k}} \leqslant \sigma(p) \leqslant 1+p^{-1-2^{-k}}\]
Hence, provided that \(C=C(k)\) is sufficiently large, one finds that \[\bigg|\prod_{p \in\mathbb P_{\geqslant C(k)}} \sigma(p)-1\bigg| \leqslant \sum_{n \geqslant C(k)} n^{-1-2^{-k}} =O \big(C(k)^{-2^{-k}}\big).\]
Then, if \(C(k)\) is chosen sufficiently large in terms of \(k\), we have that \[\bigg|\prod_{p \in\mathbb P_{\geqslant C(k)}} \sigma(p)-1\bigg|<1 / 2,\] and we conclude that \[1 / 2<\prod_{p \in\mathbb P_{\geqslant C(k)}}\sigma(p)<3 / 2.\]
The final conclusion of the theorem therefore follows, and the proof of the theorem is complete.$$\tag*{$\blacksquare$}$$
Our plan is to show that there exists a constant \(c_0>0\) such that \(\mathfrak{S}_{s, k}(n) \ge c_0\) uniformly in \(n\in\mathbb Z_+\).
In view of item (iv) of the previous theorem it suffices to prove that \(\sigma(p)>0\) for \(p \leqslant C(k)\) with sufficient uniformity in \(n\).
When \(q \in \mathbb{Z}_+\), we put \[M_{n}(q)=\#\left\{\mathbf{m} \in(\mathbb{Z} / q \mathbb{Z})^{s}: m_{1}^{k}+\ldots+m_{s}^{k}=n\right\}.\]
For each natural number \(q\in\mathbb Z_+\), one has \[\sum_{d \mid q} A(d, n)=q^{1-s} M_{n}(q).\]
Proof. We make use of the orthogonality relation \[q^{-1} \sum_{r=1}^{q} e(h r / q)= \begin{cases}1, & \text { when } q \mid h, \\ 0, & \text { when } q \nmid h .\end{cases}\]
Then \[M_{n}(q)=q^{-1} \sum_{r=1}^{q}\left(\sum_{m_{1}=1}^{q} \cdots \sum_{m_{s}=1}^{q} e\left(r\left(m_{1}^{k}+\ldots+m_{s}^{k}-n\right) / q\right)\right) .\]
Classifying the values of \(r\) according to their common factors \(q / d\) with \(q\), we obtain the relation \[\begin{aligned} M_{n}(q) & =q^{-1} \sum_{d \mid q} \sum_{\substack{a=1 \\ (a, d)=1}}^{d}(q / d)^{s} \sum_{m_{1}=1}^{d} \cdots \sum_{m_{s}=1}^{d} e\left(a\left(m_{1}^{k}+\ldots+m_{s}^{k}-n\right) / d\right) \\ & =q^{-1} \sum_{d \mid q} q^{s} \sum_{\substack{a=1 \\ (a, d)=1}}^{d}\left(d^{-1} S(d, a)\right)^{s} e(-n a / d) \\ & =q^{s-1} \sum_{d \mid q} A(d, n) \end{aligned}\]
Hence \[\sum_{d \mid q} A(d, n)=q^{1-s} M_{n}(q),\] and the proof of the lemma is complete. $$\tag*{$\blacksquare$}$$
For each prime number \(p\in\mathbb P\), one has \[\begin{align*} \sigma(p)=\lim _{h \rightarrow \infty} p^{h(1-s)} M_{n}\left(p^{h}\right). \tag{*} \end{align*}\]
Proof. Take \(q=p^{h}\) in the previous lemma to obtain the relation \[\sum_{l=0}^{h} A\left(p^{l}, n\right)=\left(p^{h}\right)^{1-s} M_{n}\left(p^{h}\right).\] Taking the limit as \(h \rightarrow \infty\), we obtain (*), since \(\sigma(p)=\sum_{l=0}^{\infty} A\left(p^{l}, n\right)\).$$\tag*{$\blacksquare$}$$
Exercise. Show that for the small primes \(p\) with \(p<C(k)\), and for all large enough values of \(h\), one has \(M_{n}\left(p^{h}\right) \ge c_0 p^{h(s-1)}\) for some \(c_0>0\). From this we deduce that \(\sigma(p)>0\), and the desired conclusion follows from item (iv) of the previous theorem.
We have shown that when \(s \geqslant 2^{k}+1\) and \(0<\delta<1 / 5\), one has \[R_{s, k}(n)=J_{s, k}\left(n ; X^{\delta}\right) \mathfrak{S}_{s, k}\left(n ; X^{\delta}\right)+o\left(X^{s-k}\right).\]
We also know that \[J_{s, k}\left(n ; X^{\delta}\right)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1}+O\left(n^{s / k-1-\delta / k^{2}}\right),\] and \[\mathfrak{S}_{s, k}\left(n ; X^{\delta}\right)=\mathfrak{S}_{s, k}(n)+O\left(n^{-\delta 2^{-k} / k}\right),\] where \(c < \mathfrak{S}_{s, k}(n) < C\) for some \(C>c>0\).
Thus we conclude that when \(s \geqslant 2^{k}+1\), one has \[R_{s, k}(n)=\frac{\Gamma(1+1 / k)^{s}}{\Gamma(s / k)} n^{s / k-1} \mathfrak{S}_{s, k}(n)+o\left(n^{s / k-1}\right).\]
Then \(R_{s, k}(n) \rightarrow \infty\) as \(n \rightarrow \infty\), whence \(G(k) \leqslant 2^{k}+1\).